APPLIED NUMERICAL

METHODS USING

MATLAB

Won Young Yang

Chung-Ang University, Korea

Wenwu Cao

Pennsylvania State University

Tae-Sang Chung

Chung-Ang University, Korea

John Morris

The University of Auckland, New Zealand

A JOHN WILEY & SONS, INC., PUBLICATION

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Library of Congress Cataloging-in-Publication Data

Yang, Won-young, 1953–

Applied numerical methods using MATLAB / Won Y. Yang, Wenwu Cao, Tae S.

Chung, John Morris.

p. cm.

Includes bibliographical references and index.

ISBN 0-471-69833-4 (cloth)

1. Numerical analysis–Data processing. 2. MATLAB. I. Cao, Wenwu. II.

Chung, Tae-sang, 1952– III. Title.

QA297.Y36 2005

518–dc22

2004013108

Printed in the United States of America.

10 9 8 7 6 5 4 3 2 1

To our parents and families

who love and support us

and

to our teachers and students

who enriched our knowledge

CONTENTS

Preface xiii

1 MATLAB Usage and Computational Errors 1

1.1 Basic Operations of MATLAB / 1

1.1.1 Input/Output of Data from MATLAB Command

Window / 2

1.1.2 Input/Output of Data Through Files / 2

1.1.3 Input/Output of Data Using Keyboard / 4

1.1.4 2-D Graphic Input/Output / 5

1.1.5 3-D Graphic Output / 10

1.1.6 Mathematical Functions / 10

1.1.7 Operations on Vectors and Matrices / 15

1.1.8 Random Number Generators / 22

1.1.9 Flow Control / 24

1.2 Computer Errors Versus Human Mistakes / 27

1.2.1 IEEE 64-bit Floating-Point Number Representation / 28

1.2.2 Various Kinds of Computing Errors / 31

1.2.3 Absolute/Relative Computing Errors / 33

1.2.4 Error Propagation / 33

1.2.5 Tips for Avoiding Large Errors / 34

1.3 Toward Good Program / 37

1.3.1 Nested Computing for Computational Efficiency / 37

1.3.2 Vector Operation Versus Loop Iteration / 39

1.3.3 Iterative Routine Versus Nested Routine / 40

1.3.4 To Avoid Runtime Error / 40

1.3.5 Parameter Sharing via Global Variables / 44

1.3.6 Parameter Passing Through Varargin / 45

1.3.7 Adaptive Input Argument List / 46

Problems / 46

vii

viii CONTENTS

2 System of Linear Equations 71

2.1 Solution for a System of Linear Equations / 72

2.1.1 The Nonsingular Case (M = N) / 72

2.1.2 The Underdetermined Case (M <N): Minimum-Norm

Solution / 72

2.1.3 The Overdetermined Case (M >N): Least-Squares Error

Solution / 75

2.1.4 RLSE (Recursive Least-Squares Estimation) / 76

2.2 Solving a System of Linear Equations / 79

2.2.1 Gauss Elimination / 79

2.2.2 Partial Pivoting / 81

2.2.3 Gauss–Jordan Elimination / 89

2.3 Inverse Matrix / 92

2.4 Decomposition (Factorization) / 92

2.4.1 LU Decomposition (Factorization):

Triangularization / 92

2.4.2 Other Decomposition (Factorization): Cholesky, QR,

and SVD / 97

2.5 Iterative Methods to Solve Equations / 98

2.5.1 Jacobi Iteration / 98

2.5.2 Gauss–Seidel Iteration / 100

2.5.3 The Convergence of Jacobi and Gauss–Seidel

Iterations / 103

Problems / 104

3 Interpolation and Curve Fitting 117

3.1 Interpolation by Lagrange Polynomial / 117

3.2 Interpolation by Newton Polynomial / 119

3.3 Approximation by Chebyshev Polynomial / 124

3.4 Pade Approximation by Rational Function / 129

3.5 Interpolation by Cubic Spline / 133

3.6 Hermite Interpolating Polynomial / 139

3.7 Two-dimensional Interpolation / 141

3.8 Curve Fitting / 143

3.8.1 Straight Line Fit: A Polynomial Function of First

Degree / 144

3.8.2 Polynomial Curve Fit: A Polynomial Function of Higher

Degree / 145

3.8.3 Exponential Curve Fit and Other Functions / 149

3.9 Fourier Transform / 150

3.9.1 FFT Versus DFT / 151

3.9.2 Physical Meaning of DFT / 152

3.9.3 Interpolation by Using DFS / 155

Problems / 157

CONTENTS ix

4 Nonlinear Equations 179

4.1 Iterative Method Toward Fixed Point / 179

4.2 Bisection Method / 183

4.3 False Position or Regula Falsi Method / 185

4.4 Newton(–Raphson) Method / 186

4.5 Secant Method / 189

4.6 Newton Method for a System of Nonlinear Equations / 191

4.7 Symbolic Solution for Equations / 193

4.8 A Real-World Problem / 194

Problems / 197

5 Numerical Differentiation/Integration 209

5.1 Difference Approximation for First Derivative / 209

5.2 Approximation Error of First Derivative / 211

5.3 Difference Approximation for Second and Higher

Derivative / 216

5.4 Interpolating Polynomial and Numerical Differential / 220

5.5 Numerical Integration and Quadrature / 222

5.6 Trapezoidal Method and Simpson Method / 226

5.7 Recursive Rule and Romberg Integration / 228

5.8 Adaptive Quadrature / 231

5.9 Gauss Quadrature / 234

5.9.1 Gauss–Legendre Integration / 235

5.9.2 Gauss–Hermite Integration / 238

5.9.3 Gauss–Laguerre Integration / 239

5.9.4 Gauss–Chebyshev Integration / 240

5.10 Double Integral / 241

Problems / 244

6 Ordinary Differential Equations 263

6.1 Euler’s Method / 263

6.2 Heun’s Method: Trapezoidal Method / 266

6.3 Runge–Kutta Method / 267

6.4 Predictor–Corrector Method / 269

6.4.1 Adams–Bashforth–Moulton Method / 269

6.4.2 Hamming Method / 273

6.4.3 Comparison of Methods / 274

6.5 Vector Differential Equations / 277

6.5.1 State Equation / 277

6.5.2 Discretization of LTI State Equation / 281

6.5.3 High-Order Differential Equation to State Equation / 283

6.5.4 Stiff Equation / 284

x CONTENTS

6.6 Boundary Value Problem (BVP) / 287

6.6.1 Shooting Method / 287

6.6.2 Finite Difference Method / 290

Problems / 293

7 Optimization 321

7.1 Unconstrained Optimization [L-2, Chapter 7] / 321

7.1.1 Golden Search Method / 321

7.1.2 Quadratic Approximation Method / 323

7.1.3 Nelder–Mead Method [W-8] / 325

7.1.4 Steepest Descent Method / 328

7.1.5 Newton Method / 330

7.1.6 Conjugate Gradient Method / 332

7.1.7 Simulated Annealing Method [W-7] / 334

7.1.8 Genetic Algorithm [W-7] / 338

7.2 Constrained Optimization [L-2, Chapter 10] / 343

7.2.1 Lagrange Multiplier Method / 343

7.2.2 Penalty Function Method / 346

7.3 MATLAB Built-In Routines for Optimization / 350

7.3.1 Unconstrained Optimization / 350

7.3.2 Constrained Optimization / 352

7.3.3 Linear Programming (LP) / 355

Problems / 357

8 Matrices and Eigenvalues 371

8.1 Eigenvalues and Eigenvectors / 371

8.2 Similarity Transformation and Diagonalization / 373

8.3 Power Method / 378

8.3.1 Scaled Power Method / 378

8.3.2 Inverse Power Method / 380

8.3.3 Shifted Inverse Power Method / 380

8.4 Jacobi Method / 381

8.5 Physical Meaning of Eigenvalues/Eigenvectors / 385

8.6 Eigenvalue Equations / 389

Problems / 390

9 Partial Differential Equations 401

9.1 Elliptic PDE / 402

9.2 Parabolic PDE / 406

9.2.1 The Explicit Forward Euler Method / 406

9.2.2 The Implicit Backward Euler Method / 407

CONTENTS xi

9.2.3 The Crank–Nicholson Method / 409

9.2.4 Two-Dimensional Parabolic PDE / 412

9.3 Hyperbolic PDE / 414

9.3.1 The Explicit Central Difference Method / 415

9.3.2 Two-Dimensional Hyperbolic PDE / 417

9.4 Finite Element Method (FEM) for solving PDE / 420

9.5 GUI of MATLAB for Solving PDEs: PDETOOL / 429

9.5.1 Basic PDEs Solvable by PDETOOL / 430

9.5.2 The Usage of PDETOOL / 431

9.5.3 Examples of Using PDETOOL to Solve PDEs / 435

Problems / 444

Appendix A. Mean Value Theorem 461

Appendix B. Matrix Operations/Properties 463

Appendix C. Differentiation with Respect to a Vector 471

Appendix D. Laplace Transform 473

Appendix E. Fourier Transform 475

Appendix F. Useful Formulas 477

Appendix G. Symbolic Computation 481

Appendix H. Sparse Matrices 489

Appendix I. MATLAB 491

References 497

Subject Index 499

Index for MATLAB Routines 503

Index for Tables 509

PREFACE

This book introduces applied numerical methods for engineering and science

students in sophomore to senior levels; it targets the students of today who do

not like or do not have time to derive and prove mathematical results. It can

also serve as a reference to MATLAB applications for professional engineers

and scientists, since many of the MATLAB codes presented after introducing

each algorithm’s basic ideas can easily be modified to solve similar problems

even by those who do not know what is going on inside the MATLAB routines

and the algorithms they use. Just as most drivers only have to know where to

go and how to drive a car to get to their destinations, most users only have to

know how to define the problems they want to solve using MATLAB and how

to use the corresponding routines to solve their problems. We never deny that

detailed knowledge about the algorithm (engine) of the program (car) is helpful

for getting safely to the solution (destination); we only imply that one-time users

of any MATLAB program or routine may use this book as well as the students

who want to understand the underlying principle of each algorithm.

In this book, we focus on understanding the fundamental mathematical concepts

and mastering problem-solving skills using numerical methods with the

help of MATLAB and skip some tedious derivations. Obviously, basic concepts

must be taught so that students can properly formulate the mathematics

problems. Afterwards, students can directly use the MATLAB codes to solve

practical problems. Almost every algorithm introduced in this book is followed

by example MATLAB code with a friendly interface so that students can easily

modify the code to solve real life problems. The selection of exercises follows

the some philosophy of making the learning easy and practical. Students

should be able to solve similar problems immediately after taking the class using

the MATLAB codes we provide. For most students—and particularly nonmath

majors—understanding how to use numerical tools correctly in solving their

problems of interest is more important than studying lengthy proofs and derivations.

MATLAB is one of the most developed software packages available today.

It provides many numerical methods and it is very easy to use, even for people

without prior programming experience. We have supplemented MATLAB’s builtin

functions with more than 100 small MATLAB routines. Readers should find

xiii

xiv PREFACE

these routines handy and useful. Some of these routines give better results for

some problems than the built-in functions. Students are encouraged to develop

their own routines following the examples.

The knowledge in this book is derived from the work of many eminent scientists,

scholars, researchers, and MATLAB developers, all of whom we thank.

We thank our colleagues, students, relatives, and friends for their support and

encouragement. We thank the reviewers, whose comments were so helpful in

tuning this book. We especially thank Senior Researcher Yong-Suk Park for his

invaluable help in correction. We thank the editorial and production staff of John

Wiley & Sons, Inc. including Editor Val Moliere and Production Editor Lisa

VanHorn for their kind, efficient, and encouraging guide.

WON YOUNG YANG

WENWU CAO

TAE-SANG CHUNG

JOHN MORRIS

October 2004

1

MATLAB USAGE AND

COMPUTATIONAL ERRORS

1.1 BASIC OPERATIONS OF MATLAB

MATLAB is a high-level software package with many built-in functions that

make the learning of numerical methods much easier and more interesting. In

this section we will introduce some basic operations that will enable you to

learn the software and build your own programs for problem solving. In the

workstation environment, you type “matlab” to start the program, while in the

PC environment, you simply double-click the MATLAB icon.

Once you start the MATLAB program, a Command window will open with the

MATLAB prompt >>. On the command line, you can type MATLAB commands,

functions together with their input/output arguments, and the names of script files

containing a block of statements to be executed at a time or functions defined

by users. The MATLAB program files must have the extension name ***.m to

be executed in the MATLAB environment. If you want to create a new M-file

or edit an existing file, you click File/New/M-file or File/Open in the top left

corner of the main menu, find/select/load the file by double-clicking it, and then

begin editing it in the Editor window. If the path of the file you want to run

is not listed in the MATLAB search path, the file name will not be recognized

by MATLAB. In such cases, you need to add the path to the MATLAB-path

list by clicking the menu ‘File/Set Path’ in the Command window, clicking the

‘Add Folder’ button, browsing/clicking the folder name, and finally clicking the

SAVE button and the Close button. The lookfor command is available to help

you find the MATLAB commands/functions which are related with a job you

Applied Numerical Methods Using MATLAB, by Yang, Cao, Chung, and Morris

Copyright 2005 John Wiley & Sons, Inc., ISBN 0-471-69833-4

1

2 MATLAB USAGE AND COMPUTATIONAL ERRORS

want to be done. The help command helps you know the usage of a particular

command/function. You may type directly in the Command window

>>lookfor repeat or >>help for

to find the MATLAB commands in connection with ‘repeat’ or to obtain information

about the “for loop”.

1.1.1 Input/Output of Data from MATLAB Command Window

MATLAB remembers all input data in a session (anything entered through direct

keyboard input or running a script file) until the command ‘clear()’ is given or

you exit MATLAB.

One of the many features of MATLAB is that it enables us to deal with the

vectors/matrices in the same way as scalars. For instance, to input the matrices/

vectors,

A = 1 2 3

4 5 6, B=

3

−2

1

, C= 1 −2 3 −4

type in the MATLAB Command window as below:

>>A = [1 2 3;4 5 6]

A=1 2 3

4 5 6

>>B = [3;-2;1]; %put the semicolon at the end of the statement to suppress

the result printout onto the screen

>>C = [1 -2 3 -4]

At the end of the statement, press <Enter> if you want to check the result

of executing the statement immediately. Otherwise, type a semicolon “;” before

pressing <Enter> so that your window will not be overloaded by a long display

of results.

1.1.2 Input/Output of Data Through Files

MATLAB can handle two types of data files. One is the binary format matfiles

named ***.mat. This kind of file can preserve the values of more than one

variable, but will be handled only in the MATLAB environment and cannot be

shared with other programming environments. The other is the ASCII dat-files

named ***.dat, which can be shared with other programming environments, but

preserve the values of only one variable.

Below are a few sample statements for storing some data into a mat-file in

the current directory and reading the data back from the mat-file:

>>save ABC A B C %store the values of A,B,C into the file ’ABC.mat’

>>clear A C %clear the memory of MATLAB about A,C

BASIC OPERATIONS OF MATLAB 3

>>A %what is the value of A?

??? Undefined function or variable ’A’

>>load ABC A C %read the values of A,C from the file ’ABC.mat’

>>A %the value of A

A=1 2 3

4 5 6

If you want to store the data into an ASCII dat-file (in the current directory),

make the filename the same as the name of the data and type ‘/ascii’ at the

end of the save statement.

>>save B.dat B /ascii

However, with the save/load commands into/from a dat-file, the value of only

one variable having the lowercase name can be saved/loaded, a scalar or a vector/

matrix. Besides, non-numeric data cannot be handled by using a dat-file. If

you save a string data into a dat-file, its ASCII code will be saved. If a dat-file

is constructed to have a data matrix in other environments than MATLAB, every

line (row) of the file must have the same number of columns. If you want to read

the data from the dat-file in MATLAB, just type the (lowercase) filename ***.dat

after ‘load’, which will also be recognized as the name of the data contained in

the dat-file.

>>load b.dat %read the value of variable b from the ascii file ’b.dat’

On the MATLAB command line, you can type ‘nm112’ to run the following

M-file ‘nm112.m’ consisting of several file input(save)/output(load) statements.

Then you will see the effects of the individual statements from the running

results appearing on the screen.

%nm112.m

clear

A = [1 2 3;4 5 6]

B = [3;-2;1];

C(2) = 2; C(4) = 4

disp(’Press any key to see the input/output through Files’)

save ABC A B C %save A,B & C as a MAT-file named ’ABC.mat’

clear(’A’,’C’) %remove the memory about A and C

load ABC A C %read MAT-file to recollect the memory about A and C

save B.dat B /ascii %save B as an ASCII-file named ’b.dat’

clear B

load b.dat %read ASCII-file to recollect the memory about b

b

x = input(’Enter x:’)

format short e

x

format rat, x

format long, x

format short, x

4 MATLAB USAGE AND COMPUTATIONAL ERRORS

1.1.3 Input/Output of Data Using Keyboard

The command ‘input’ enables the user to input some data via the keyboard.

For example,

>>x = input(’Enter x: ’)

Enter x: 1/3

x = 0.3333

Note that the fraction 1/3 is a nonterminating decimal number, but only four

digits after the decimal point are displayed as the result of executing the above

command. This is a choice of formatting in MATLAB. One may choose to

display more decimal places by using the command ‘format’, which can make

a fraction show up as a fraction, as a decimal number with more digits, or even

in an exponential form of a normalized number times 10 to the power of some

integer. For instance:

>>format rat %as a rational number

>>x

x = 1/3

>>format long %as a decimal number with 14 digits

>>x

x = 0.33333333333333

>>format long e %as a long exponential form

>>x

x = 3.333333333333333e-001

>>format hex %as a hexadecimal form as represented/stored in memory

>>x

x = 3fd5555555555555

>>format short e %as a short exponential form

>>x

x = 3.3333e-001

>>format short %back to a short form (default)

>>x

x = 0.3333

Note that the number of displayed digits is not the actual number of significant

digits of the value stored in computer memory. This point will be made clear in

Section 1.2.1.

There are other ways of displaying the value of a variable and a string on the

screen than typing the name of the variable. Two useful commands are ‘disp()’

and ‘fprintf()’. The former displays the value of a variable or a string without

‘x = ’ or ‘ans = ’; the latter displays the values of several variables in a specified

format and with explanatory/cosmetic strings. For example:

>>disp(’The value of x = ’),disp(x)

%disp(’string_to_display’ or variable_name)

The value of x = 0.3333

Table 1.1 summarizes the type specifiers and special characters that are used in

‘fprintf()’ statements.

Below is a program that uses the command ‘input’ so that the user could

input some data via the keyboard. If we run the program, it gets a value of the

BASIC OPERATIONS OF MATLAB 5

Table 1.1 Type Specifiers and Special Characters Used in fprintf() Statements

Type

Specifier

Printing Form:

fprintf(‘**format string**’, variables to be printed,..)

Special

Character Meaning

%c Character type \n New line

%s String type \t Tab

%d Decimal integer number type \b Backspace

%f Floating point number type \r CR return

%e Decimal exponential type \f Form feed

%x Hexadecimal integer number %% %

%bx Floating number in 16 hexadecimal digits(64 bits) ’’ ’

temperature in Fahrenheit [◦F] via the keyboard from the user, converts it into

the temperature in Centigrade [◦C] and then prints the results with some remarks

both onto the screen and into a data file named ‘nm113.dat’.

%nm113.m

f = input(’Input the temperature in Fahrenheit[F]:’);

c = 5/9*(f-32);

fprintf(’%5.2f(in Fahrenheit) is %5.2f(in Centigrade).\n’,f,c)

fid=fopen(’nm113.dat’, ’w’);

fprintf(fid, ’%5.2f(Fahrenheit) is %5.2f(Centigrade).\n’,f,c);

fclose(fid);

In case you want the keyboard input to be recognized as a string, you should

add the character ’s’ as the second input argument.

>>ans = input(’Answer <yes> or <no>: ’,’s’)

1.1.4 2-D Graphic Input/Output

How do we plot the value(s) of a vector or an array? Suppose that data reflecting

the highest/lowest temperatures for 5 days are stored as a 5 × 2 array in an ASCII

file named ‘temp.dat’.

The job of the MATLAB program “nm114_1.m” is to plot these data. Running

the program yields the graph shown in Fig. 1.1a. Note that the first line is a

comment about the name and the functional objective of the program(file), and

the fourth and fifth lines are auxiliary statements that designate the graph title

and units of the vertical/horizontal axis; only the second & third lines are indispensable

in drawing the colored graph. We need only a few MATLAB statements

for this artwork, which shows the power of MATLAB.

%nm114_1: plot the data of a 5×2 array stored in “temp.dat”

load temp.dat

clf, plot(temp) %clear any existent figure and plot

title(’the highest/lowest temperature of these days’)

ylabel(’degrees[C]’), xlabel(’day’)

6 MATLAB USAGE AND COMPUTATIONAL ERRORS

20

25

15

10

5

11 12 14

The highest/lowest temperature of days

day day

degrees [°C]

16 17

20

25

15

10

5

1 2 3

(b) Domain of the horizontal

variable specified

(a) Domain of the horizontal

variable unspecified

The highest/lowest temperature of days

degrees [°C]

4 5

Figure 1.1 Plot of a 5 × 2 matrix data representing the highest/lowest temperature.

Here are several things to keep in mind.

ž The command plot() reads along the columns of the 5 × 2 array data given

as its input argument and recognizes each column as the value of a vector.

ž MATLAB assumes the domain of the horizontal variable to be [1 2 .. 5] by

default, where 5 equals the length of the vector to be plotted (see Fig. 1.1a).

ž The graph is constructed by connecting the data points with the straight lines

and is piecewise-linear, while it looks like a curve as the data points are

densely collected. Note that the graph can be plotted as points in various

forms according to the optional input argument described in Table 1.2.

(Q1) Suppose the data in the array named ‘temp’ are the highest/lowest temperatures

measured on the 11th,12th,14th,16th, and 17th days, respectively. How should we

modify the above program to have the actual days shown on the horizontal axis?

(A1) Just make the day vector [11 12 14 16 17] and use it as the first input argument

of the plot() command.

>>days = [11 12 14 16 17]

>>plot(days,temp)

Executing these statements, we obtain the graph in Fig. 1.1b.

(Q2) What statements should be added to change the ranges of the horizontal/vertical

axes into 10–20 and 0–30, respectively, and draw the grid on the graph?

Table 1.2 Graphic Line Specifications Used in the plot() Command

Line Type Point Type (Marker Symbol) Color

– solid line . (dot) + (plus) * (asterisk) r : red m : magenta

: dotted line ^ : > : > o (circle) g : green y : yellow

— dashed line p : v : x : x-mark b : blue c : cyan (sky blue)

-. dash-dot d : ♦ < : < s : k : black

BASIC OPERATIONS OF MATLAB 7

(A2) >>axis([10 20 0 30]), grid on

>>plot(days,temp)

(Q3) How do we make the scales of the horizontal/vertical axes equal so that a circle

appears round, not like an ellipse?

(A3) >>axis(’equal’)

(Q4) How do we have another graph overlapped onto an existing graph?

(A4) If you use the ‘hold on’ command after plotting the first graph, any following

graphs in the same section will be overlapped onto the existing one(s) rather

than plotted newly. For example:

>>hold on, plot(days,temp(:,1),’b*’, days,temp(:,2),’ro’)

This will be good until you issue the command ‘hold off’ or clear all the graphs

in the graphic window by using the ‘clf’ command.

Sometimes we need to see the interrelationship between two variables. Suppose

we want to plot the lowest/highest temperature, respectively, along the

horizontal/vertical axis in order to grasp the relationship between them. Let us

try using the following command:

>>plot(temp(:,1),temp(:,2),’kx’) % temp(:,2) vs. temp(:,1) in black ’x’

This will produce a pointwise graph, which is fine. But, if you replace the third

input argument by ‘b:’ or just omit it to draw a piecewise-linear graph connecting

the data points as Fig. 1.2a, the graphic result looks clumsy, because the data on

the horizontal axis are not arranged in ascending or descending order. The graph

will look better if you sort the data on the horizontal axis and also the data on

the vertical axis accordingly and then plot the relationship in the piecewise-linear

style by typing the MATLAB commands as follows:

>>[temp1,I] = sort(temp(:,1)); temp2 = temp(I,2);

>>plot(temp1,temp2)

The graph obtained by using these commands is shown in Fig.1.2b, which looks

more informative than Fig.1.2a.

(a) Data not arranged

4 6 8 10

10

15

20

25

10

15

20

25

12 4 6 8 10 12

(b) Data arranged along the horizontal axis.

Figure 1.2 Examples of graphs obtained using the plot() command.

8 MATLAB USAGE AND COMPUTATIONAL ERRORS

We can also use the plot() command to draw a circle.

>>r = 1; th = [0:0.01:2]*pi; % [0:0.01:2] makes [0 0.01 0.02 .. 2]

>>plot(r*cos(th),r*sin(th))

>>plot(r*exp(j*th)) %alternatively,

Note that the plot() command with a sequence of complex numbers as its first

input argument plots the real/imaginary parts along the horizontal/vertical axis.

The polar() command plots the phase (in radians)/magnitude given as its

first/second input argument, respectively (see Fig.1.3a).

>>polar(th,exp(-th)) %polar plot of a spiral

Several other plotting commands, such as semilogx(), semilogy(), loglog(),

stairs(), stem(), bar()/barh(), and hist(), may be used to draw various

graphs (shown in Figs.1.3 and 1.4). Readers may use the ‘help’ command to get

the detailed usage of each one and try running the following MATLAB program

‘nm114 2.m’.

%nm114_2: plot several types of graph

th = [0: .02:1]*pi;

subplot(221), polar(th,exp(-th))

subplot(222), semilogx(exp(th))

subplot(223), semilogy(exp(th))

subplot(224), loglog(exp(th))

pause, clf

subplot(221), stairs([1 3 2 0])

subplot(222), stem([1 3 2 0])

subplot(223), bar([2 3; 4 5])

subplot(224), barh([2 3; 4 5])

pause, clf

y = [0.3 0.9 1.6 2.7 3 2.4];

subplot(221), hist(y,3)

subplot(222), hist(y,0.5 + [0 1 2])

Moreover, the commands sprintf(), text(), and gtext() are used for combining

supplementary statements with the value(s) of one or more variables to

construct a string and printing it at a certain location on the existing graph.

For instance, let us try the following statements in the MATLAB Command

window:

>>f = 1./[1:10]; plot(f)

>>n = 3; [s,errmsg] = sprintf(’f(%1d) = %5.2f’,n,f(n))

>>text(3,f(3),s) %writes the text string at the point (3,f(3))

>>gtext(’f(x) = 1/x’) %writes the input string at point clicked by mouse

The command ginput() allows you to obtain the coordinates of a point

by clicking the mouse button on the existent graph. Let us try the following

BASIC OPERATIONS OF MATLAB 9

(b) semilogx (x, y)

0

100 101 102

(d) loglog (x, y)

100 101 102

10

20

30

(c) semilogy (x, y)

100

0 20 40 60

101

102

100

101

102

60

30

330

270 300

90

(a) polar (th, r)

0.5

240

210

180 0

150

120

Figure 1.3 Graphs drawn by various graphic commands.

(a) stairs ([1 3 2 0])

0

1

2

1 2 3 4

3

(b) stem ([1 3 2 0])

0

1

2

1 2 3 4

3

(e) hist ([0.3 .. 2.4], 3)

0

1

2

0 1 2 3

3

(f) hist ([..], [0.5 1.5 2.5])

0

1

2

0 1 2 3

3

(c) bar ([2 3; 4 5])

0

2

4

1 2

6

(d) barh ([2 3; 4 5])

1

2

0 2 4 6

Figure 1.4 Graphs drawn by various graphic commands.

10 MATLAB USAGE AND COMPUTATIONAL ERRORS

commands:

>>[x,y,butkey] = ginput %get the x,y coordinates & # of the mouse button

or ascii code of the key pressed till pressing the ENTER key

>>[x,y,butkey] = ginput(n) %repeat the same job for up to n points clicked

1.1.5 3-D Graphic Output

MATLAB has several 3-D graphic plotting commands such as plot3(), mesh(),

and contour(). plot3() plots a 2-D valued-function of a scalar-valued variable;

mesh()/contour() plots a scalar valued-function of a 2-D variable in a

mesh/contour-like style, respectively.

Readers are recommended to use the help command for detailed usage of each

command. Try running the MATLAB program ‘nm115.m’ to see what figures

will appear (Figs.1.5 and 1.6).

%nm115: to plot 3D graphs

t = 0:pi/50:6*pi;

expt = exp(-0.1*t);

xt = expt.*cos(t); yt = expt.*sin(t);

%dividing the screen into 2 x 2 sections

subplot(221), plot3(xt, yt, t), grid on %helix

subplot(222), plot3(xt, yt, t), grid on, view([0 0 1])

subplot(223), plot3(t, xt, yt), grid on, view([1 -3 1])

subplot(224), plot3(t, yt, xt), grid on, view([0 -3 0])

pause, clf

x = -2:.1:2; y = -2:.1:2;

[X,Y] = meshgrid(x,y); Z = X.^2 + Y.^2;

subplot(221), mesh(X,Y,Z), grid on %[azimuth,elevation] = [-37.5,30]

subplot(222), mesh(X,Y,Z), view([0,20]), grid on

pause, view([30,30])

subplot(223), contour(X,Y,Z)

subplot(224), contour(X,Y,Z,[.5,2,4.5])

1.1.6 Mathematical Functions

Mathematical functions and special reserved constants/variables defined in MATLAB

are listed in Table 1.3.

MATLAB also allows us to define our own function and store it in a file

named after the function name so that it can be used as if it were a built-in

function. For instance, we can define a scalar-valued function:

f1(x) = 1/(1 + 8×2)

and a vector-valued function

f49(x) = f1(x1, x2)

f2(x1, x2) = x2

1 + 4×2

2 − 5

2×2

1 − 2×1 − 3×2 − 2.5

BASIC OPERATIONS OF MATLAB 11

20

10

0

1

0 0

−1

1

−1

(a) plot3(cos(t), sin(t), t)

1

0

−1

0 10

0

−1

1

20

(c) plot3( ), view [(1 −3 1)]

(b) plot3( ), view([0 0 1])

−1

0

−1 −0.5 0 0.5 1

1

(d) plot3( ), view ([0 −3 0])

−1

0

0 5 10 15 20

1

Figure 1.5 Graphs drawn by the plot3() command with different views.

10

5

0

2

2

0

−2

−2 −1 0 1 2

2

0

−2

−2 −1 0 1 2

−2 −2

2

0 0

(a) mesh( ), view (−37.5, 30) (b) mesh( ), view (30, 20)

(c) contour(X,Y,Z) (d) contour(X,Y,Z, [0.5, 2, 4.5])

10

5

0

−2

2 −2

2

0

0

Figure 1.6 Graphs drawn by the mesh() and contour() commands.

as follows.

function y = f1(x)

y = 1./(1+8*x.^2);

function y = f49(x)

y(1) = x(1)*x(1)+4*x(2)*x(2) -5;

y(2) = 2*x(1)*x(1)-2*x(1)-3*x(2) -2.5;

12 MATLAB USAGE AND COMPUTATIONAL ERRORS

Table 1.3 Functions and Variables Inside MATLAB

Function Remark Function Remark

cos(x) exp(x) Exponential function

sin(x) log(x) Natural logarithm

tan(x) log10(x) Common logarithm

acos(x) cos−1(x) abs(x) Absolute value

asin(x) sin−1(x) angle(x) Phase of a complex

number [rad]

atan(x) −π/2 ≤ tan−1(x) ≤ π/2 sqrt(x) Square root

atan2(y,x) −π ≤ tan−1(y, x) ≤ π real(x) Real part

cosh(x) (ex + e−x)/2 imag(x) Imaginary part

sinh(x) (ex − e−x)/2 conj(x) Complex conjugate

tanh(x) (ex − e−x)/(ex + e−x ) round(x) The nearest integer

(round-off)

acosh(x) cosh−1(x) fix(x) The nearest integer

toward 0

asinh(x) sinh−1(x) floor(x) The greatest integer

≤ x

atanh(x) tanh−1(x) ceil(x) The smallest integer

≥ x

max Maximum and its index sign(x) 1(positive)/0/-

1(negative)

min Minimum and its index mod(y,x) Remainder of y/x

sum Sum rem(y,x) Remainder of y/x

prod Product eval(f) Evaluate an expression

norm Norm feval(f,a) Function evaluation

sort Sort in the ascending

order

polyval Value of a polynomial

function

clock Present time poly Polynomial with given

roots

BASIC OPERATIONS OF MATLAB 13

Table 1.3 (continued)

find Index of element(s) roots Roots of polynomial

flops(0) Reset the flops count to

zero

tic Start a stopwatch timer

flops Cumulative # of floating

point operations

(unavailable in

MATLAB 6.x)

toc Read the stopwatch

timer (elapsed time

from tic)

date Present date magic Magic square

Reserved Variables with Special Meaning

i,j √−1 pi π

eps Machine epsilon floating

point relative accuracy

realmax realmin Largest/smallest

positive number

break Exit while/for loop Inf, inf Largest number (∞)

end The end of for-loop or

if, while, case statement

or an array index

NaN Not a Number

(undetermined)

nargin Number of input

arguments

nargout Number of output

arguments

varargin Variable input argument

list

varargout Variable output

argument list

Once we store these functions into the files named ‘f1.m’ and ‘f49.m’ after the

function names, respectively, we can call and use them as needed inside another

M-file or in the MATLAB Command window.

>>f1([0 1]) %several values of a scalar function of a scalar variable

ans = 1.0000 0.1111

>>f49([0 1]) %a value of a 2-D vector function of a vector variable

ans = -1.0000 -5.5000

>>feval(’f1’,[0 1]), feval(’f49’,[0 1]) %equivalently, yields the same

ans = 1.0000 0.1111

ans = -1.0000 -5.5000

(Q5) With the function f1(x) defined as a scalar function of a scalar variable, we enter

a vector as its input argument to obtain a seemingly vector-valued output. What’s

going on?

14 MATLAB USAGE AND COMPUTATIONAL ERRORS

(A5) It is just a set of function values [f1(x1) f1(x2) . . .] obtained at a time for several

values [x1 x2. . .] of x. In expectation of one-shot multi-operation, it is a good

practice to put a dot(.) just before the arithmetic operators *(multiplication),

/(division), and ^ (power) in the function definition so that the term-by-term

(termwise) operation can be done any time.

Note that we can define a simple function not only in an independent M-file,

but also inside a program by using the inline() command or just in a form of

literal expression that can be evaluated by the command eval().

>>f1 = inline(’1./(1+8*x.^2)’,’x’);

>>f1([0 1]), feval(f1,[0 1])

ans = 1.0000 0.1111

ans = 1.0000 0.1111

>>f1 = ’1./(1+8*x.^2)’; x = [0 1]; eval(f1)

ans = 1.0000 0.1111

As far as a polynomial function is concerned, it can simply be defined as its

coefficient vector arranged in descending order. It may be called to yield its

value for certain value(s) of its independent variable by using the command

polyval().

>>p = [1 0 -3 2]; %polynomial function p(x) = x3 − 3x + 2

>>polyval(p,[0 1])

ans = 2.0000 0.0000

The multiplication of two polynomials can be performed by taking the convolution

of their coefficient vectors representing the polynomials in MATLAB,

since

(aNxN + ·· ·+a1x + a0)(bNxN + ·· ·+b1x + b0) = c2Nx2N + ·· ·+c1x + c0

where

ck =

min(k,N)

m=max(0,k−N)

ak−mbm for k = 2N, 2N − 1, . . . , 1, 0

This operation can be done by using the MATLAB built-in command conv() as

illustrated below.

>>a = [1 -1]; b=[1 1 1]; c = conv(a,b)

c = 1 0 0 -1 %meaning that (x − 1)(x2 + x + 1) = x3 + 0 · x2 + 0 · x − 1

But, in case you want to multiply a polynomial by only xn, you can simply

append n zeros to the right end of the polynomial coefficient vector to extend

its dimension.

>>a = [1 2 3]; c = [a 0 0] %equivalently, c = conv(a,[1 0 0])

c = 1 2 3 0 0 %meaning that (x2 + 2x + 3)x2 = x4 + 2×3 + 3×2 + 0 · x + 0

BASIC OPERATIONS OF MATLAB 15

1.1.7 Operations on Vectors and Matrices

We can define a new scalar/vector/matrix or redefine any existing ones in terms

of the existent ones or irrespective of them. In the MATLAB Command window,

let us defineA and B as

A = 1 2 3

4 5 6, B=

3

−2

1

by typing

>>A = [1 2 3;4 5 6], B = [3;-2;1]

We can modify them or take a portion of them. For example:

>>A = [A;7 8 9]

A = 1 2 3

4 5 6

7 8 9

>>B = [B [1 0 -1]’]

B = 3 1

-2 0

1 -1

Here, the apostrophe (prime) operator (’) takes the complex conjugate transpose

and functions virtually as a transpose operator for real-valued matrices. If you

want to take just the transpose of a complex-valued matrix, you should put a

dot(.) before ’, that is, ‘.’’.

When extending an existing matrix or defining another one based on it, the

compatibility of dimensions should be observed. For instance, if you try to annex

a 4 ×1 matrix into the 3 × 1 matrix B, MATLAB will reject it squarely, giving

you an error message.

>>B = [B ones(4,1)]

???All matrices on a row in the bracketed expression must have

the same number of rows

We can modify or refer to a portion of a given matrix.

>>A(3,3) = 0

A=1 2 3

4 5 6

7 8 0

>>A(2:3,1:2) %from 2nd row to 3rd row, from 1st column to 2nd column

ans = 4 5

7 8

>>A(2,:) %2nd row, all columns

ans = 4 5 6

The colon (:) is used for defining an arithmetic (equal difference) sequence

without the bracket [] as

>>t = 0:0.1:2

16 MATLAB USAGE AND COMPUTATIONAL ERRORS

which makes

t = [0.0 0.1 0.2 … 1.9 2.0]

(Q6) What if we omit the increment between the left/right boundary numbers?

(A6) By default, the increment is 1.

>>t = 0:2

t = 0 1 2

(Q7) What if the right boundary number is smaller/greater than the left boundary

number with a positive/negative increment?

(A7) It yields an empty matrix, which is useless.

>>t = 0:-2

t = Empty matrix: 1-by-0

(Q8) If we define just some elements of a vector not fully, but sporadically, will we

have a row vector or a column vector and how will it be filled in between?

(A8) We will have a row vector filled with zeros between the defined elements.

>>D(2) = 2; D(4) = 3

D = 0 2 0 3

(Q9) How do we make a column vector in the same style?

(A9) We must initialize it as a (zero-filled) row vector, prior to giving it a value.

>>D = zeros(4,1); D(2) = 2; D(4) = 3

D = 0

2

0

3

(Q10) What happens if the specified element index of an array exceeds the defined

range?

(A10) It is rejected. MATLAB does not accept nonpositive or noninteger indices.

>>D(5)

??? Index exceeds matrix dimensions.

>>D(0) = 1;

??? Index into matrix is negative or zero.

>>D(1.2)

??? Subscript indices must either be real positive

integers ..

(Q11) How do we know the size (the numbers of rows/columns) of an alreadydefined

array?

BASIC OPERATIONS OF MATLAB 17

(A11) Use the length() and size() commands as indicated below.

>>length(D)

ans = 4

>>[M,N] = size(A)

M = 3

N = 3

MATLAB enables us to handle vector/matrix operations in almost the same

way as scalar operations. However, we must make sure of the dimensional compatibility

between vectors/matrices, and we must put a dot (.) in front of the

operator for termwise (element-by-element) operations. The addition of a matrix

and a scalar adds the scalar to every element of the matrix. The multiplication

of a matrix by a scalar multiplies every element of the matrix by the scalar.

There are several things to know about the matrix division and inversion.

Remark 1.1. Rules of Vector/Matrix Operation

1. For a matrix to be invertible, it must be square and nonsingular; that is, the

numbers of its rows and columns must be equal and its determinant must

not be zero.

2. The MATLAB command pinv(A) provides us with a matrix X of the same

dimension as AT such that AXA = A and XAX = X. We can use this

command to get the right/left pseudo- (generalized) inverse AT [AAT ]−1/

[ATA]−1AT for a matrix A given as its input argument, depending on

whether the number (M) of rows is smaller or greater than the number

(N) of columns, so long as the matrix is of full rank; that is, rank(A) = min(M,N)[K-1, Section 6.4]. Note that AT [AAT ]−1/[ATA]−1AT is called

the right/left inverse because it is multiplied onto the right/left side of A

to yield an identity matrix.

3. You should be careful when using the pinv(A) command for a rankdeficient

matrix, because its output is no longer the right/left inverse, which

does not even exist for rank-deficient matrices.

4. The value of a scalar function having an array value as its argument is also

an array with the same dimension.

Supposewe have defined vectors a1, a2, b1, b2 andmatricesA1,A2,B as follows:

>>a1 = [-1 2 3]; a2 = [4 5 2]; b1 = [1 -3]’; b2 = [-2 0];

a1 = [−1 2 3], a2 = [4 5 2], b1 = 1

−3 , b2 = [−1 2 3]

>>A1 = [a1;a2], A2 = [a1;[b2 1]], B = [b1 b2’]

A1 = −1 2 3

4 5 2, A2 = −1 2 3

−2 0 1, B= 1 −2

−3 0

18 MATLAB USAGE AND COMPUTATIONAL ERRORS

The results of various operations on these vectors/matrices are as follows (pay

attention to the error message):

>>A3 = A1 + A2, A4 = A1 – A2, 1 + A1 %matrix/scalar addition/subtraction

A3 = -2 4 6 A4 = 0 0 0 ans = 0 3 4

2 5 3 6 5 1 5 6 3

>>AB = A1*B % AB(m, n) =

k

A1(m, k )B(k , n) matrix multiplication?

??? Error using ==> *

Inner matrix dimensions must agree.

>>BA1 = B*A1 % regular matrix multiplication

BA1 = -9 -8 -1

3 -6 -9

>>AA = A1.*A2 %termwise multiplication

AA = 1 4 9

-8 0 2

>>AB=A1.*B % AB(m, n) = A1(m, n)B(m, n) termwise multiplication

??? Error using ==> .*

Matrix dimensions must agree.

>>A1 1 = pinv(A1),A1’*(A1*A1’)^-1,eye(size(A1,2))/A1 % AT1

[A1AT1

]−1

A1 1 = -0.1914 0.1399 %right inverse of a 2 x 3 matrix A1

0.0617 0.0947

0.2284 -0.0165

>>A1*A1 1 %A1/A1 = I implies the validity of A1 1 as the right inverse

ans = 1.0000 0.0000

0.0000 1.0000

>>A5 = A1’; % a 3 x 2 matrix

>>A5 1 = pinv(A5),(A5’*A5)^-1*A5’,A5\eye(size(A5,1)) % [AT5A5]−1AT5

A5 1 = -0.1914 0.0617 0.2284 %left inverse of a 3×2 matrix A5

0.1399 0.0947 -0.0165

>>A5 1*A5 % = I implies the validity of A5 1 as the left inverse

ans = 1.0000 -0.0000

-0.0000 1.0000

>>A1 li = (A1’*A1)^-1*A1’ %the left inverse of matrix A1 with M < N?

Warning: Matrix is close to singular or badly scaled.

Results may be inaccurate. RCOND = 9.804831e-018.

A1 li = -0.2500 0.2500

0.2500 0

0.5000 0.5000

(Q12) Does the left inverse of a matrix having rows fewer than columns exist?

(A12) No. There is no N ×M matrix that is premultiplied on the left of an M × N

matrix with M <N to yield a nonsingular matrix, far from an identity matrix.

In this context, MATLAB should have rejected the above case on the ground

that [AT1

A1] is singular and so its inverse does not exist. But, because the roundoff

errors make a very small number appear to be a zero or make a real zero

appear to be a very small number (as will be mentioned in Remark 2.3), it is

not easy for MATLAB to tell a near-singularity from a real singularity. That is

why MATLAB dares not to declare the singularity case and instead issues just a

warning message to remind you to check the validity of the result so that it will

not be blamed for a delusion. Therefore, you must be alert for the condition

BASIC OPERATIONS OF MATLAB 19

mentioned in item 2 of Remark 1.1, which says that, in order for the left inverse

to exist, the number of rows must not be less than the number of columns.

>>A1_li*A1 %No identity matrix, since A1_li isn’t the left inverse

ans = 1.2500 0.7500 -0.2500

-0.2500 0.5000 0.7500

1.5000 3.5000 2.5000

>>det(A1’*A1) %A1 is not left-invertible for A1’*A1 is singular

ans = 0

(cf) Let us be nice to MATLAB as it is to us. From the standpoint of promoting mutual

understanding between us and MATLAB, we acknowledge that MATLAB tries to

show us apparently good results to please us like always, sometimes even pretending

not to be obsessed by the demon of ‘ill-condition’ in order not to make us feel uneasy.

How kind MATLAB is! But, we should be always careful not to be spoiled by its

benevolence and not to accept the computing results every inch as it is. In this case,

even though the matrix [A1’*A1] is singular and so not invertible, MATLAB tried

to invert it and that’s all. MATLAB must have felt something abnormal as can be

seen from the ominous warning message prior to the computing result. Who would

blame MATLAB for being so thoughtful and loyal to us? We might well be rather

touched by its sincerity and smartness.

In the above statements, we see the slash(/)/backslash(\) operators. These operators

are used for right/left division, respectively; B/A is the same as B*inv(A) and

A\B is the same as inv(A)*B when A is invertible and the dimensions of A and B

are compatible. Noting that B/A is equivalent to (A’\B’)’, let us take a close look

at the function of the backslash(\) operator.

>>X = A1\A1 % an identity matrix?

X = 1.0000 0 -0.8462

0 1.0000 1.0769

0 0 0

(Q13) It seems that A1\A1 should have been an identity matrix, but it is not, contrary

to our expectation. Why?

(A13) We should know more about the various functions of the backslash(\), which

can be seen by typing ‘help slash’ into the MATLAB Command window. Let

Remark 1.2 answer this question in cooperation with the next case.

>>A1*X – A1 %zero if X is the solution to A1*X = A1?

ans = 1.0e-015 * 0 0 0

0 0 -0.4441

Remark 1.2. The Function of Backslash (\) Operator. Overall, for the command

‘A\B’, MATLAB finds a solution to the equation A*X = B. Let us denote the

row/column dimension of the matrix A by M and N.

1. If matrix A is square and upper/lower-triangular in the sense that all of

its elements below/above the diagonal are zero, then MATLAB finds the

solution by applying backward/forward substitution method (Section 2.2.1).

20 MATLAB USAGE AND COMPUTATIONAL ERRORS

2. If matrix A is square, symmetric (Hermitian), and positive definite, then

MATLAB finds the solution by using Cholesky factorization (Section 2.4.2).

3. If matrix A is square and has no special feature, then MATLAB finds the

solution by using LU decomposition (Section 2.4.1).

4. If matrix A is rectangular, then MATLAB finds a solution by using QR

factorization (Section 2.4.2). In case A is rectangular and of full rank with

rank(A) = min(M,N), it will be the LS (least-squares) solution [Eq. (2.1.10)]

for M > N (overdetermined case) and one of the many solutions that is not

always the same as the minimum-norm solution [Eq. (2.1.7)] for M < N

(underdetermined case). But for the case when A is rectangular and has

rank deficiency, what MATLAB gives us may be useless. Therefore, you

must pay attention to the warning message about rank deficiency, which

might tell you not to count on the dead-end solution made by the backslash

(\) operator. To find an alternative in the case of rank deficiency, you

had better resort to singular value decomposition (SVD). See Problem 2.8

for details.

For the moment, let us continue to try more operations on matrices.

>>A1./A2 %termwise right division

ans = 1 1 1

-2 Inf 2

>>A1.\A2 %termwise left division

ans = 1 1 1

-0.5 0 0.5

>>format rat, B^-1 %represent the numbers (of B−1) in fractional form

ans = 0 -1/3

-1/2 -1/6

>>inv(B) %inverse matrix, equivalently

ans = 0 -1/3

-1/2 -1/6

>>B.^-1 %termwise inversion(reciprocal of each element)

ans = 1 -1/2

-1/3 Inf

>>B^2 %square of B, i.e., B2 = B ∗ B

ans = 7 -2

-3 6

>>B.^2 %termwise square(square of each element)

ans = 1(b2

11) 4(b2

12)

9(b2

21) 0(b2

22)

>>2.^B %2 to the power of each number in B

ans = 2 (2b11) 1/4(2b12)

1/8(2b21) 1 (2b22)

>>A1.^A2 %element of A1 to the power of each element in A2

ans = -1 (A1(1, 1)A2(1,1)) 4(A1(1, 2)A2(1,2)) 27(A1(1, 3)A2(1,3))

1/16(A1(2, 1)A2(2,1)) 1(A1(2, 2)A2(2,2)) 2(A1(2, 3)A2(2,3))

>>format short, exp(B) %elements of eB with 4 digits below the dp

ans = 2.7183(eb11) 0.1353(eb12)

0.0498(eb21) 1.0000(eb22)

There are more useful MATLAB commands worthwhile to learn by heart.

BASIC OPERATIONS OF MATLAB 21

Remark 1.3. More Useful Commands for Vector/Matrix Operations

1. We can use the commands zeros(), ones(), and eye() to construct a

matrix of specified size or the same size as an existing matrix which has

only zeros, only ones, or only ones/zeros on/off its diagonal.

>>Z = zeros(2,3) %or zeros(size(A1)) yielding a 2 x 3 zero matrix

Z=0 0 0

0 0 0

>>E = ones(size(B)) %or ones(3,2) yielding a 3 x 2 one matrix

E = 1 1

1 1

1 1

>>I = eye(2) %yielding a 2 x 2 identity matrix

I = 1 0

0 1

2. We can use the diag() command to make a column vector composed

of the diagonal elements of a matrix or to make a diagonal matrix with

on-diagonal elements taken from a vector given as the input argument.

>>A1, diag(A1) %column vector consisting of diagonal elements

A1= -1 2 3

4 5 2

ans = -1

5

3. We can use the commands sum()/prod() to get the sum/product of elements

in a vector or a matrix, columnwisely first (along the first nonsingleton

dimension).

>>sa1 = sum(a1) %sum of all the elements in vector a1

sa1 = 4 %

a1(n) = −1 + 2 + 3 = 4

>>sA1 = sum(A1) %sum of all the elements in each column of matrix A1

sA1 = 3 7 5 %sA1(n) =

Mm

= 1 A1(m, n) = [− 1 + 4 2 + 5 3 + 2]

>>SA1 = sum(sum(A1)) %sum of all elements in matrix A1

SA1 = 15 %SA1 =

Nn

= 1

Mm

= 1 A1(m, n) = 3 + 7 + 5 = 15

>>pa1 = prod(a1) %product of all the elements in vector a1

pa1 = 4 % a1(n) = ( − 1) × 2 × 3 = −6

>>pA1=product(A1) %product of all the elements in each column of matrix A1

pA1 = -4 10 6 %pA1(n) = Mm

= 1 A1(m, n) = [−1 × 4 2× 5 3× 2]

>>PA1 = product(product(A1)) %product of all the elements of matrix A1

PA1 = -240 %PA1 = Nn = 1 Mm

= 1 A1(m, n) = ( − 4) × 10 × 6 = −240

4. We can use the commands max()/min() to find the first maximum/minimum

number and its index in a vector or in a matrix given as the input argument.

>>[aM,iM] = max(a2)

aM = 5, iM = 2 %means that the max. element of vector a2 is a2(2) = 5

>>[AM,IM] = max(A1)

AM = 4 5 3

IM = 2 2 1

%means that the max. elements of each column of A1 are

A1(2,1) = 4, A1(2,2) = 5, A1(1,3) = 3

22 MATLAB USAGE AND COMPUTATIONAL ERRORS

>>[AMx,J] = max(AM)

AMx = 5, J = 2

%implies that the max. element of A1 is A1(IM(J),J) = A1(2,2) = 5

5. We can use the commands rot90()/fliplr()/flipud() to rotate a matrix

by an integer multiple of 90◦ and to flip it left-right/up-down.

>>A1, A3 = rot90(A1), A4 = rot90(A1,-2)

A1 = -1 2 3

4 5 2

A3 = 3 2 %90◦ rotation

2 5

-1 4

A4 = 2 5 4 %90◦x(-2) rotation

3 2 -1

>>A5 = fliplr(A1) %flip left-right

A5 = 3 2 -1

2 5 4

>>A6 = flipud(A1) %flip up-down

A6= 4 5 2

-1 2 3

6. We can use the reshape() command to change the row-column size of a

matrix with its elements preserved (columnwisely first).

>>A7 = reshape(A1,3,2)

A7 = -1 5

4 3

2 2

>>A8 = reshape(A1,6,1), A8 = A1(:) %makes supercolumn vector

A8 = -1

4

2

5

3

2

1.1.8 Random Number Generators

MATLAB has the built-in functions, rand()/randn(), to generate random

numbers having uniform/normal (Gaussian) distributions, respectively ([K-1],

Chapter 22).

rand(M,N): generates an M x N matrix consisting of uniformly distributed

random numbers

randn(M,N): generates an M x N matrix consisting of normally distributed

random numbers

BASIC OPERATIONS OF MATLAB 23

1. Random Number Having Uniform Distribution

The numbers in a matrix generated by the MATLAB function rand(M,N) have

uniform probability distribution over the interval [0,1], as described by U(0,1).

The random number x generated by rand() has the probability density function

fX(x) = us(x) − us(x − 1) (us

(x) = 1 ∀x ≥ 0

0 ∀x < 0

: the unit step function)

(1.1.1)

whose value is 1 over [0,1] and 0 elsewhere. The average of this standard uniform

number x is

mX =

∞

−∞

xfX(x)dx =

1

0

x dx =

x2

2

1

0 =

1

2

(1.1.2)

and its variance or deviation is

σ2

X =

∞

−∞

(x − mX)2fX(x)dx =

1

0

(x −

1

2

)2dx =

1

3

(x −

1

2

)3

1

0 =

1

12

(1.1.3)

If you want another random number y with uniform distribution U(a, b), transform

the standard uniform number x as follows:

y = (b − a)x +a (1.1.4)

For practice, we make a vector consisting of 1000 standard uniform numbers,

transform it to make a vector of numbers with uniform distribution U(−1, +1),

and then draw the histograms showing the shape of the distribution for the two

uniform number vectors (Fig. 1.7a,b).

>>u_noise = rand(1000,1) %a 1000×1 noise vector with U(0,1)

>>subplot(221), hist(u_noise,20) %histogram having 20 divisions

0

(a) Uniform noise U[0, 1]

0

20

40

60

−1 −0.5 0.5 1

0

(c) Gaussian noise N(0, 1)

0

50

100

150

−5 5 0

(d) Gaussian noise N(0, 1/22)

0

50

100

150

−5 5

0

(b) Uniform noise U[−1, 1]

0

20

40

60

−1 −0.5 0.5 1

Figure 1.7 Distribution (histogram) of noise generated by the rand()/randn() command.

24 MATLAB USAGE AND COMPUTATIONAL ERRORS

>>u_noise1 = 2*u_noise-1 %a 1000×1 noise vector with U(-1,1)

>>subplot(222), hist(u_noise1,20) %histogram

2. Random Number with Normal (Gaussian) Distribution

The numbers in a matrix generated by the MATLAB function randn(M,N) have

normal (Gaussian) distribution with average m = 0 and variance σ2 = 1, as

described by N(0,1). The random number x generated by rand() has the probability

density function

fX(x) =

1

√2π

e−x2/2 (1.1.5)

If you want another Gaussian number y with a general normal distribution

N(m, σ2), transform the standard Gaussian number x as follows:

y = σ x +m (1.1.6)

The probability density function of the new Gaussian number generated by this

transformation is obtained by substituting x = (y − m)/σ into Eq. (1.1.5) and

dividing the result by the scale factor σ (which can be seen in dx = dy/σ)

so that the integral of the density function over the whole interval (−∞, +∞)

amounts to 1.

fY (y) =

1

√2πσ

e−(y−m)2/2σ2

(1.1.7)

For practice, we make a vector consisting of 1000 standard Gaussian numbers,

transform it to make a vector of numbers having normal distribution N(1,1/4),

with mean m = 1 and variance σ2 = 1/4, and then draw the histograms for the

two Gaussian number vectors (Fig. 1.7c,d).

>>g_noise = randn(1000,1) %a 1000×1 noise vector with N(0,1)

>>subplot(223), hist(g_noise,20) %histogram having 20 divisions

>>g_noise1 = g_noise/2+1 %a 1000×1 noise vector with N(1,1/4)

>>subplot(224), hist(g_noise1,20) %histogram

1.1.9 Flow Control

1. if-end and switch-case-end Statements

An if-end block basically consists of an if statement, a sequel part, and an end

statement categorizing the block. An if statement, having a condition usually

based on the relational/logical operator (Table 1.4), is used to control the program

flow—that is, to adjust the order in which statements are executed according to

whether or not the condition is met, mostly depending on unpredictable situations.

The sequel part consisting of one or more statements may contain else or

elseif statements, possibly in a nested structure containing another if statement

inside it.

The switch-case-end block might replace a multiple if-elseif-..-end

statement in a neat manner.

BASIC OPERATIONS OF MATLAB 25

Table 1.4 Relational Operators and Logical Operators

Relational

operator

Remark Relational

operator

Remark Logical

operator

Remark

< less than > greater than & and

<= less than or equal to >= greater than or equal to | or

== equal ~= not equal( =) ~ not

Let us see the following examples:

Example 1. A Simple if-else-end Block

%nm119_1: example of if-end block

t = 0;

if t > 0

sgnt = 1;

else

sgnt = -1;

end

Example 2. A Simple if-elseif-end Block

%nm119_2: example of if-elseif-end block

if t > 0

sgnt = 1

elseif t < 0

sgnt = -1

end

Example 3. An if-elseif-else-end Block

%nm119_3: example of if-elseif-else-end block

if t > 0, sgnt = 1

elseif t<0, sgnt = -1

else sgnt = 0

end

Example 4. An if-elseif-elseif-..-else-end Block

%nm119_4: example of if-elseif-elseif-else-end block

point = 85;

if point >= 90, grade = ’A’

elseif point >= 80, grade = ’B’

elseif point >= 70, grade = ’C’

elseif point >= 60, grade = ’D’

else grade = ’F’

end

26 MATLAB USAGE AND COMPUTATIONAL ERRORS

Example 5. A switch-case-end Block

%nm119_5: example of switch-case-end block

point = 85;

switch floor(point/10) %floor(x): integer less than or equal to x

case 9, grade = ’A’

case 8, grade = ’B’

case 7, grade = ’C’

case 6, grade = ’D’

otherwise grade = ’F’

end

2. for index = i 0:increment:i last-end Loop

A for loop makes a block of statements executed repeatedly for a specified

number of times, with its loop index increasing from i_0 to a number not

greater than i_last by a specified step (increment) or by 1 if not specified.

The loop iteration normally ends when the loop index reaches i_last, but it

can be stopped by a break statement inside the for loop. The for loop with a

positive/negative increment will never be iterated if the last value (i_last) of

the index is smaller/greater than the starting value (i_0).

Example 6. A for Loop

%nm119_6: example of for loop

point = [76 85 91 65 87];

for n = 1:length(point)

if point(n) >= 80, pf(n,:) = ’pass’;

elseif point(n) >= 0, pf(n,:) = ’fail’;

else %if point(n)< 0

pf(n,:) = ’????’;

fprintf(’\n\a Something wrong with the data??\n’);

break;

end

end

pf

3. while Loop

A while loop will be iterated as long as its predefined condition is satisfied and

a break statement is not encountered inside the loop.

Example 7. A while Loop

%nm119_7: example of while loop

r = 1;

while r < 10

r = input(’\nType radius (or nonpositive number to stop):’);

if r <= 0, break, end %isempty(r)| r <= 0, break, end

v = 4/3*pi*r*r*r;

fprintf(’The volume of a sphere with radius %3.1f = %8.2f\n’,r,v);

end

COMPUTER ERRORS VERSUS HUMAN MISTAKES 27

Example 8. while Loops to Find the Minimum/Maximum Positive Numbers

The following program “nm119 8.m” contains three while loops. In the first

one, x = 1 continues to be divided by 2 until just before reaching zero, and it

will hopefully end up with the smallest positive number that can be represented

in MATLAB. In the second one, x = 1 continues to be multiplied by 2 until just

before reaching inf (the infinity defined in MATLAB), and seemingly it will get

the largest positive number (x_max0) that can be represented in MATLAB. But,

while this number reaches or may exceed inf if multiplied by 2 once more, it still

is not the largest number in MATLAB (slightly less than inf) that we want to

find. How about multiplying x_max0 by (2 − 1/2n)? In the third while loop, the

temporary variable tmp starting with the initial value of 1 continues to be divided

by 2 until just before x_max0*(2-tmp) reaches inf, and apparently it will end

up with the largest positive number (x_max) that can be represented in MATLAB.

%nm119_8: example of while loops

x = 1; k1 = 0;

while x/2 > 0

x = x/2; k1 = k1 + 1;

end

k1, x_min = x;

fprintf(’x_min is %20.18e\n’,x_min)

x = 1; k2 = 0;

while 2*x < inf

x = x*2; k2 = k2+1;

end

k2, x_max0 = x;

tmp = 1; k3 = 0;

while x_max0*(2-tmp/2) < inf

tmp = tmp/2; k3 = k3+1;

end

k3, x_max = x_max0*(2-tmp);

fprintf(’x_max is %20.18e\n’,x_max)

format long e

x_min,-x_min,x_max,-x_max

format hex

x_min,-x_min,x_max,-x_max

format short

1.2 COMPUTER ERRORS VERSUS HUMAN MISTAKES

Digital systems like calculators and computers hardly make a mistake, since they

follow the programmed order faithfully. Nonetheless, we often encounter some

numerical errors in the computing results made by digital systems, mostly coming

from representing the numbers in finite bits, which is an intrinsic limitation of digital

world. If you let the computer compute something without considering what

is called the finite-word-length effect, you might come across a weird answer. In

28 MATLAB USAGE AND COMPUTATIONAL ERRORS

that case, it is not the computer, but yourself as the user or the programmer, who

is to blame for the wrong result. In this context, we should always be careful not

to let the computer produce a farfetched output. In this section we will see how

the computer represents and stores the numbers. Then we think about the cause

and the propagation effect of computational error in order not to be deceived by

unintentional mistakes of the computer and, it is hoped, to be able to take some

measures against them.

1.2.1 IEEE 64-bit Floating-Point Number Representation

MATLAB uses the IEEE 64-bit floating-point number system to represent all

numbers. It has a word structure consisting of the sign bit, the exponent field,

and the mantissa field as follows:

63 62 52 51 0

S Exponent Mantissa

Each of these fields expresses S,E, and M of a number f in the way described

below.

ž Sign bit

S = b63 = 0 for positive numbers

1 for negative numbers

ž Exponent field (b62b61b60 · · · b52): adopting the excess 1023 code

E = Exp − 1023 = {0, 1, . . . , 211 − 1 = 2047} − 1023

= {−1023,−1022, . . . ,+1023,+1024}

=

−1023 +1 for |f | < 2−1022(Exp = 00000000000)

−1022 ∼ +1023 for 2−1022 ≤ |f | < 21024(normalized ranges)

+1024 for ±∞

ž Mantissa field (b51b50 . . . b1b0):

In the un-normalized range where the numbers are so small that they can be

represented only with the value of hidden bit 0, the number represented by the

mantissa is

M = 0.b51b50 · · · b1b0 = [b51b50 · · · b1b0] × 2−52 (1.2.1)

You might think that the value of the hidden bit is added to the exponent, instead

of to the mantissa.

In the normalized range, the number represented by the mantissa together with

the value of hidden bit bh = 1 is

M = 1.b51b50 · · · b1b0 = 1 + [b51b50 · · · b1b0] × 2−52

= 1 + b51 × 2−1 + b50 × 2−2 + ·· ·+b1 × 2−51 + b0 × 2−52

COMPUTER ERRORS VERSUS HUMAN MISTAKES 29

= {1, 1 + 2−52, 1 + 2 × 2−52, . . . , 1 + (252 − 1) × 2−52}

= {1, 1 + 2−52, 1 + 2 × 2−52, . . . , (2 − 2−52)}

= {1, 1 + , 1 + 2, . . . , 1 + (252 − 1) = 2 − } ( = 2−52) (1.2.2)

The set of numbers S,E, and M, each represented by the sign bit S, the

exponent field Exp and the mantissa field M, represents a number as a whole

f = ±M · 2E (1.2.3)

We classify the range of numbers depending on the value (E) of the exponent

and denote it as

RE = [2E, 2E+1) with − 1022 ≤ E ≤ +1023 (1.2.4)

In each range, the least unit—that is, the value of LSB (least significant bit) or

the difference between two consecutive numbers represented by the mantissa of

52 bits—is

E = × 2E = 2−52 × 2E = 2E−52 (1.2.5)

Let us take a closer look at the bitwise representation of numbers belonging

to each range.

0. 0(zero)

63

S 000 . . 0000 0000 0000 . . . 0000 0000

62 52 51 0

1. Un-normalized Range (with the value of hidden bit bh = 0)

R−1023 = [2−1074, 2−1022) with Exp = 0,E = Exp − 1023 + 1 = −1022

( )

( ( ) )

( )

( )

Value of LSB: −1023 = −1022 = 2−1022−52 = 2−1074

2. The Smallest Normalized Range (with the value of hidden bit bh = 1)

R−1022 = [2−1022, 2−1021) with Exp = 1,E = Exp − 1023 = −1022

S 000 0000 0000 0000 0000 (1 + 0) × 2E = (1 + 0) × 2−1022

(1 + 2−52) × 2−1022

{(1 + (252 − 1) 2−52) = (2 − 2−52)} × 2−1022

. . . 0001 . . . .

S 000 . . . 0001 1111 1111 . . . . 1111 1111

S 000 . . . 0001 0000 0000 0000 0001

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . . .

Value of LSB: −1022 = 2−1022−52 = 2−1074

3. Basic Normalized Range (with the value of hidden bit bh = 1)

30 MATLAB USAGE AND COMPUTATIONAL ERRORS

R0 = [20, 21) with Exp = 210 − 1 = 1023, E = Exp − 1023 = 0

S 011 0000 0000 0000 0000 (1 + 0) × 2E = (1 + 0) × 20 = 1

(1 + 2−52) × 20

{(1 + (252 − 1) 2−52) = (2 − 2−52)} × 20

. . . 1111 . . . .

S 011 . . . 1111 1111 1111 . . . . 1111 1111

S 011 . . . 1111 0000 0000 0000 0001

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . . .

Value of LSB: 0 = 2−52

4. The Largest Normalized Range (with the value of hidden bit bh = 1)

R1024 = [21023, 21024) with Exp = 211−2 = 2046,E = Exp−1023 = 1023

S 111 0000 0000 0000 0000 (1 + 0) × 2E = (1 + 0) × 21023

(1 + 2−52) × 21023

{(1 + (252 − 1) 2−52) = (2 − 2−52)} × 21023

. . . 1110 . . . .

S 111 . . . 1110 1111 1111 . . . . 1111 1111

S 111 . . . 1110 0000 0000 0000 0001

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . . .

Value of LSB: −1022 = 2−1022−52 = 2−1074

5. ±∞(inf) Exp = 211 − 1 = 2047, E = Exp − 1023 = 1024 (meaningless)

0 111 0000 0000 0000 0000 +∞ ≠ (1 + 0) × 2E = (1 + 0) × 21024

−∞ ≠ −(1 + 0) × 2E = −(1 + 0) × 21024

invalid (not used)

. . . 1111 . . . .

S 111 . . . 111 1111 1111 . . . . 1111 1111

1 111 . . . 1111 0000 0000 0000 0000

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . . .

S 111 . . . 1111 0000 0000 . . . . 0000 0001 invalid (not used)

From what has been mentioned earlier, we know that the minimum and maximum

positive numbers are, respectively,

fmin = (0 + 2−52) × 2−1022 = 2−1074 = 4.9406564584124654 × 10−324

fmax = (2 − 2−52) × 21023 = 1.7976931348623157 × 10308

This can be checked by running the program “nm119_8.m” in Section 1.1.9.

Now, in order to gain some idea about the arithmetic computational mechanism,

let’s see how the addition of two numbers, 3 and 14, represented in the

IEEE 64-bit floating number system, is performed.

)

)

) )))

1

COMPUTER ERRORS VERSUS HUMAN MISTAKES 31

In the process of adding the two numbers, an alignment is made so that the

two exponents in their 64-bit representations equal each other; and it will kick

out the part smaller by more than 52 bits, causing some numerical error. For

example, adding 2−23 to 230 does not make any difference, while adding 2−22 to

230 does, as we can see by typing the following statements into the MATLAB

Command window.

>>x = 2^30; x + 2^-22 == x, x + 2^-23 == x

ans = 0(false) ans = 1(true)

(cf) Each range has a different minimum unit (LSB value) described by Eq. (1.2.5). It

implies that the numbers are uniformly distributed within each range. The closer the

range is to 0, the denser the numbers in the range are. Such a number representation

makes the absolute quantization error large/small for large/small numbers, decreasing

the possibility of large relative quantization error.

1.2.2 Various Kinds of Computing Errors

There are various kinds of errors that we encounter when using a computer for

computation.

ž Truncation Error: Caused by adding up to a finite number of terms, while

we should add infinitely many terms to get the exact answer in theory.

ž Round-off Error: Caused by representing/storing numeric data in finite bits.

ž Overflow/Underflow: Caused by too large or too small numbers to be represented/

stored properly in finite bits—more specifically, the numbers having

absolute values larger/smaller than the maximum (fmax)/minimum(fmin)

number that can be represented in MATLAB.

ž Negligible Addition: Caused by adding two numbers of magnitudes differing

by over 52 bits, as can be seen in the last section.

ž Loss of Significance: Caused by a “bad subtraction,” which means a subtraction

of a number from another one that is almost equal in value.

ž Error Magnification: Caused and magnified/propagated by multiplying/dividing

a number containing a small error by a large/small number.

ž Errors depending on the numerical algorithms, step size, and so on.

Although we cannot be free from these kinds of inevitable errors in some degree,

it is not computers, but instead human beings, who must be responsible for

the computing errors. While our computer may insist on its innocence for an

unintended lie, we programmers and users cannot escape from the responsibility

of taking measures against the errors and would have to pay for being careless

enough to be deceived by a machine. We should, therefore, try to decrease the

magnitudes of errors and to minimize their impact on the final results. In order

to do so, we must know the sources of computing errors and also grasp the

computational properties of numerical algorithms.

32 MATLAB USAGE AND COMPUTATIONAL ERRORS

For instance, consider the following two formulas:

f1(x) = √x(√x + 1 − √x), f2(x) =

√x

√x + 1 + √x

(1.2.6)

These are theoretically equivalent, hence we expect them to give exactly the

same value. However, running the MATLAB program “nm122.m” to compute

the values of the two formulas, we see a surprising result that, as x increases,

the step of f1(x) incoherently moves hither and thither, while f2(x) approaches

1/2 at a steady pace. We might feel betrayed by the computer and have a doubt

about its reliability. Why does such a flustering thing happen with f1(x)? It is

because the number of significant bits abruptly decreases when the subtraction

(√x + 1 − √x) is performed for large values of x, which is called ‘loss of

significance’. In order to take a close look at this phenomenon, let x = 1015.

Then we have

√x + 1 = 3.162277660168381 × 107 = 31622776.60168381

√x = 3.162277660168379 × 107 = 31622776.60168379

These two numbers have 52 significant bits, or equivalently 16 significant digits

(252 ≈ 1052×3/10 ≈ 1015) so that their significant digits range from 108 to 10−8.

Accordingly, the least significant digit of their sum and difference is also the

eighth digit after the decimal point (10−8).

√x + 1 + √x = 63245553.20336761

√x + 1 − √x = 0.00000001862645149230957 ≈ 0.00000002

Note that the number of significant digits of the difference decreased to 1 from

16. Could you imagine that a single subtraction may kill most of the significant

digits? This is the very ‘loss of significance’, which is often called ‘catastrophic

cancellation’.

%nm122

clear

f1 = inline(’sqrt(x)*(sqrt(x + 1) – sqrt(x))’,’x’);

f2 = inline(’sqrt(x)./(sqrt(x + 1) + sqrt(x))’,’x’);

x = 1;

format long e

for k = 1:15

fprintf(’At x=%15.0f, f1(x)=%20.18f, f2(x) = %20.18f’, x,f1(x),f2(x));

x = 10*x;

end

sx1 = sqrt(x+1); sx = sqrt(x); d = sx1 – sx; s = sx1 + sx;

fprintf(’sqrt(x+1) = %25.13f, sqrt(x) = %25.13f ’,sx1,sx);

fprintf(’ diff = %25.23f, sum = %25.23f ’,d,s);

COMPUTER ERRORS VERSUS HUMAN MISTAKES 33

>> nm122

At x= 1, f1(x)=0.414213562373095150, f2(x)=0.414213562373095090

At x= 10, f1(x)=0.488088481701514750, f2(x)=0.488088481701515480

At x= 100, f1(x)=0.498756211208899460, f2(x)=0.498756211208902730

At x= 1000, f1(x)=0.499875062461021870, f2(x)=0.499875062460964860

At x= 10000, f1(x)=0.499987500624854420, f2(x)=0.499987500624960890

At x= 100000, f1(x)=0.499998750005928860, f2(x)=0.499998750006249940

At x= 1000000, f1(x)=0.499999875046341910, f2(x)=0.499999875000062490

At x= 10000000, f1(x)=0.499999987401150920, f2(x)=0.499999987500000580

At x= 100000000, f1(x)=0.500000005558831620, f2(x)=0.499999998749999950

At x= 1000000000, f1(x)=0.500000077997506340, f2(x)=0.499999999874999990

At x= 10000000000, f1(x)=0.499999441672116520, f2(x)=0.499999999987500050

At x= 100000000000, f1(x)=0.500004449631168080, f2(x)=0.499999999998750000

At x= 1000000000000, f1(x)=0.500003807246685030, f2(x)=0.499999999999874990

At x= 10000000000000, f1(x)=0.499194546973835970, f2(x)=0.499999999999987510

At x= 100000000000000, f1(x)=0.502914190292358400, f2(x)=0.499999999999998720

sqrt(x+1) = 31622776.6016838100000, sqrt(x) = 31622776.6016837920000

diff=0.00000001862645149230957, sum=63245553.20336760600000000000000

1.2.3 Absolute/Relative Computing Errors

The absolute/relative error of an approximate value x to the true value X of a

real-valued variable is defined as follows:

εx = X(true value) − x(approximate value) (1.2.7)

ρx =

εx

X =

X − x

X

(1.2.8)

If the least significant digit (LSD) is the dth digit after the decimal point, then

the magnitude of the absolute error is not greater than half the value of LSD.

|εx| = |X − x| ≤ 12

10−d (1.2.9)

If the number of significant digits is s, then the magnitude of the relative error

is not greater than half the relative value of LSD over MSD (most significant

digit).

|ρx| = |εx |

|X| = |X − x|

|X| ≤

1

2

10−s (1.2.10)

1.2.4 Error Propagation

In this section we will see how the errors of two numbers, x and y, are propagated

with the four arithmetic operations. Error propagation means that the errors in the

input numbers of a process or an operation cause the errors in the output numbers.

Let their absolute errors be εx and εy , respectively. Then the magnitudes of

the absolute/relative errors in the sum and difference are

εx±y = (X ± Y) − (x ± y) = (X − x) ± (Y − y) = εx ± εy

|εx±y| ≤ |εx| + |εy| (1.2.11)

|ρx±y| = |εx±y|

|X ± Y | ≤ |X||εx/X| + |Y ||εy/Y|

|X ± Y | = |X||ρx| + |Y ||ρy|

|X ± Y |

(1.2.12)

34 MATLAB USAGE AND COMPUTATIONAL ERRORS

From this, we can see why the relative error is magnified to cause the “loss

of significance” in the case of subtraction when the two numbers X and Y are

almost equal so that |X − Y| ≈ 0.

The magnitudes of the absolute and relative errors in the multiplication/division

are

|εxy| = |XY − xy| = |XY − (X + εx)(Y + εy)| ≈ |Xεy ± Yεx |

|εxy| ≤ |X||εy| + |Y ||εx| (1.2.13)

|ρxy| = |εxy|

|XY| ≤ |εy|

|Y | + |εx |

|X| = |ρx| + |ρy | (1.2.14)

|εx/y| =

X

Y −

x

y

=

X

Y −

X + εx

Y + εy

≈ |Xεy − Yεx |

Y 2

|εx/y| ≤ |X||εy| + |Y ||εx|

Y 2 (1.2.15)

|ρx/y| = |εx/y|

|X/Y | ≤ |εx|

|X| + |εy |

|Y | = |ρx| + |ρy | (1.2.16)

This implies that, in the worst case, the relative error in multiplication/division

may be as large as the sum of the relative errors of the two numbers.

1.2.5 Tips for Avoiding Large Errors

In this section we will look over several tips to reduce the chance of large errors

occurring in calculations.

First, in order to decrease the magnitude of round-off errors and to lower the

possibility of overflow/underflow errors, make the intermediate result as close to

1 as possible in consecutive multiplication/division processes. According to this

rule, when computing xy/z, we program the formula as

ž (xy)/z when x and y in the multiplication are very different in magnitude,

ž x(y/z) when y and z in the division are close in magnitude, and

ž (x/z)y when x and z in the division are close in magnitude.

For instance, when computing yn/enx with x 1 and y 1, we would program

it as (y/ex )n rather than as yn/enx, so that overflow/underflow can be avoided. You

may verify this by running the following MATLAB program “nm125_1.m”.

%nm125_1:

x = 36; y = 1e16;

for n = [-20 -19 19 20]

fprintf(’y^%2d/e^%2dx = %25.15e\n’,n,n,y^n/exp(n*x));

fprintf(’(y/e^x)^%2d = %25.15e\n’,n,(y/exp(x))^n);

end

COMPUTER ERRORS VERSUS HUMAN MISTAKES 35

>>nm125_1

y^-20/e^-20x = 0.000000000000000e+000

(y/e^x)^-20 = 4.920700930263814e-008

y^-19/e^-19x = 1.141367814854768e-007

(y/e^x)^-19 = 1.141367814854769e-007

y^19/e^19x = 8.761417546430845e+006

(y/e^x)^19 = 8.761417546430843e+006

y^20/e^20x = NaN

(y/e^x)^20 = 2.032230802424294e+007

Second, in order to prevent ‘loss of significance’, it is important to avoid a

‘bad subtraction’ (Section 1.2.2)—that is, a subtraction of a number from another

number having almost equal value. Let us consider a simple problem of finding

the roots of a second-order equation ax2 + bx + c = 0 by using the quadratic

formula

x1 = −b + √b2 − 4ac

2a

, x2 = −b − √b2 − 4ac

2a

(1.2.17)

Let |4ac| ≺ b2. Then, depending on the sign of b, a “bad subtraction” may be

encountered when we try to find x1 or x2, which is the smaller one of the two

roots. This implies that it is safe from the “loss of significance” to compute the

root having the larger absolute value first and then obtain the other root by using

the relation (between the roots and the coefficients) x1x2 = c/a.

For another instance, we consider the following two formulas, which are analytically

the same, but numerically different:

f1(x) =

1 − cos x

x2 , f2(x) =

sin2 x

x2(1 + cos x)

(1.2.18)

It is safe to use f1(x) for x ≈ π since the term (1 + cos x) in f2(x) is a ‘bad subtraction’,

while it is safe to use f2(x) for x ≈ 0 since the term (1 − cos x) in f1(x)

is a ‘bad subtraction’. Let’s run the following MATLAB program “nm125_2.m”

to confirm this. Below is the running result. This implies that we might use some

formulas to avoid a ‘bad subtraction’.

%nm125_2: round-off error test

f1 = inline(’(1 – cos(x))/x/x’,’x’);

f2 = inline(’sin(x)*sin(x)/x/x/(1 + cos(x))’,’x’);

for k = 0:1

x = k*pi; tmp = 1;

for k1 = 1:8

tmp = tmp*0.1; x1 = x + tmp;

fprintf(’At x = %10.8f, ’, x1)

fprintf(’f1(x) = %18.12e; f2(x) = %18.12e’, f1(x1),f2(x1));

end

end

36 MATLAB USAGE AND COMPUTATIONAL ERRORS

>> nm125_2

At x = 0.10000000, f1(x) = 4.995834721974e-001; f2(x) = 4.995834721974e-001

At x = 0.01000000, f1(x) = 4.999958333474e-001; f2(x) = 4.999958333472e-001

At x = 0.00100000, f1(x) = 4.999999583255e-001; f2(x) = 4.999999583333e-001

At x = 0.00010000, f1(x) = 4.999999969613e-001; f2(x) = 4.999999995833e-001

At x = 0.00001000, f1(x) = 5.000000413702e-001; f2(x) = 4.999999999958e-001

At x = 0.00000100, f1(x) = 5.000444502912e-001; f2(x) = 5.000000000000e-001

At x = 0.00000010, f1(x) = 4.996003610813e-001; f2(x) = 5.000000000000e-001

At x = 0.00000001, f1(x) = 0.000000000000e+000; f2(x) = 5.000000000000e-001

At x = 3.24159265, f1(x) = 1.898571371550e-001; f2(x) = 1.898571371550e-001

At x = 3.15159265, f1(x) = 2.013534055392e-001; f2(x) = 2.013534055391e-001

At x = 3.14259265, f1(x) = 2.025133720884e-001; f2(x) = 2.025133720914e-001

At x = 3.14169265, f1(x) = 2.026294667803e-001; f2(x) = 2.026294678432e-001

At x = 3.14160265, f1(x) = 2.026410772244e-001; f2(x) = 2.026410604538e-001

At x = 3.14159365, f1(x) = 2.026422382785e-001; f2(x) = 2.026242248740e-001

At x = 3.14159275, f1(x) = 2.026423543841e-001; f2(x) = 2.028044503269e-001

At x = 3.14159266, f1(x) = 2.026423659946e-001; f2(x) = Inf

It may be helpful for avoiding a ‘bad subtraction’ to use the Taylor series

expansion ([W-1]) rather than using the exponential function directly for the

computation of ex . For example, suppose we want to find

f3(x) =

ex − 1

x

at x = 0 (1.2.19)

We can use the Taylor series expansion up to just the fourth-order of ex about x = 0

g(x) = ex ≈ g(0) + g(0)x +

g(0)

2!

x2 +

g(3)(0)

3!

x3 +

g(4)(0)

4!

x4

= 1 + x +

1

2!

x2 +

1

3!

x3 +

1

4!

x4

to approximate the above function (1.2.19) as

f3(x) =

ex − 1

x ≈ 1 +

1

2!

x +

1

3!

x2 +

1

4!

x3 = f4(x) (1.2.20)

Noting that the true value of (1.2.9) is computed to be 1 by using the L’Hˆopital’s

rule ([W-1]), we run the MATLAB program “nm125_3.m” to find which one of

the two formulas f3(x) and f4(x) is better for finding the value of the expression

(1.2.9) at x = 0. Would you compare them based on the running result shown

below? How can the approximate formula f4(x) outrun the true one f3(x) for

the numerical purpose, though not usual? It is because the zero factors in the

numerator/denominator of f3(x) are canceled to set f4(x) free from the terror of

a “bad subtraction.”

TOWARD GOOD PROGRAM 37

%nm125_3: reduce the round-off error using Taylor series

f3 = inline(’(exp(x)-1)/x’,’x’);

f4 = inline(’((x/4+1)*x/3) + x/2+1’,’x’);

x = 0; tmp = 1;

for k1 = 1:12

tmp = tmp*0.1; x1 = x + tmp;

fprintf(’At x = %14.12f, ’, x1)

fprintf(’f3(x) = %18.12e; f4(x) = %18.12e’, f3(x1),f4(x1));

end

>> nm125_3

At x=0.100000000000, f3(x)=1.051709180756e+000; f4(x)=1.084166666667e+000

At x=0.010000000000, f3(x)=1.005016708417e+000; f4(x)=1.008341666667e+000

At x=0.001000000000, f3(x)=1.000500166708e+000; f4(x)=1.000833416667e+000

At x=0.000100000000, f3(x)=1.000050001667e+000; f4(x)=1.000083334167e+000

At x=0.000010000000, f3(x)=1.000005000007e+000; f4(x)=1.000008333342e+000

At x=0.000001000000, f3(x)=1.000000499962e+000; f4(x)=1.000000833333e+000

At x=0.000000100000, f3(x)=1.000000049434e+000; f4(x)=1.000000083333e+000

At x=0.000000010000, f3(x)=9.999999939225e-001; f4(x)=1.000000008333e+000

At x=0.000000001000, f3(x)=1.000000082740e+000; f4(x)=1.000000000833e+000

At x=0.000000000100, f3(x)=1.000000082740e+000; f4(x)=1.000000000083e+000

At x=0.000000000010, f3(x)=1.000000082740e+000; f4(x)=1.000000000008e+000

At x=0.000000000001, f3(x)=1.000088900582e+000; f4(x)=1.000000000001e+000

1.3 TOWARD GOOD PROGRAM

Among the various criteria about the quality of a general program, the most

important one is how robust its performance is against the change of the problem

properties and the initial values. A good program guides the program users who

don’t know much about the program and at least give them a warning message

without runtime error for their minor mistake. There are many other features

that need to be considered, such as user friendliness, compactness and elegance,

readability, and so on. But, as far as the numerical methods are concerned, the

accuracy of solution, execution speed (time efficiency), and memory utilization

(space efficiency) are of utmost concern. Since some tips to achieve the accuracy

or at least to avoid large errors (including overflow/underflow) are given in the

previous section, we will look over the issues of execution speed and memory

utilization.

1.3.1 Nested Computing for Computational Efficiency

The execution speed of a program for a numerical solution depends mostly on

the number of function (subroutine) calls and arithmetic operations performed in

the program. Therefore, we like the algorithm requiring fewer function calls and

arithmetic operations. For instance, suppose we want to evaluate the value of a

38 MATLAB USAGE AND COMPUTATIONAL ERRORS

polynomial

p4(x) = a1x4 + a2x3 + a3x2 + a4x + a5 (1.3.1)

It is better to use the nested structure (as below) than to use the above form as

it is.

p4n(x) = (((a1x + a2)x + a3)x + a4)x + a5 (1.3.2)

Note that the numbers of multiplications needed in Eqs. (1.3.2) and (1.3.1) are

4 and (4 + 3 + 2 + 1 = 9), respectively. This point is illustrated by the program

“nm131_1.m”, where a polynomial N−1

i=0 aixi of degree N = 106 for a certain

value of x is computed by using the three methods—that is, Eq. (1.3.1), Eq.

(1.3.2), and the MATLAB built-in function ‘polyval()’. Interested readers could

run this program to see that Eq. (1.3.2)—that is, the nested multiplication—is

the fastest, while ‘polyval()’ is the slowest because of some overhead time for

being called, though it is also fabricated in a nested structure.

%nm131_1: nested multiplication vs. plain multiple multiplication

N = 1000000+1; a = [1:N]; x = 1;

tic % initialize the timer

p = sum(a.*x.^[N-1:-1:0]); %plain multiplication

p, toc % measure the time passed from the time of executing ’tic’

tic, pn=a(1);

for i = 2:N %nested multiplication

pn = pn*x + a(i);

end

pn, toc

tic, polyval(a,x), toc

Programming in a nested structure is not only recommended for time-efficient

computation, but also may be critical to the solution. For instance, consider a

problem of finding the value

S(K) =

K

k=0

λk

k!

e−λ for λ = 100 and K = 155 (1.3.3)

%nm131_2_1: nested structure

lam = 100; K = 155;

p = exp(-lam);

S = 0;

for k = 1:K

p=p*lam/k; S=S+p;

end

S

%nm131_2_2: not nested structure

lam = 100; K = 155;

S = 0;

for k = 1:K

p = lam^k/factorial(k);

S = S + p;

end

S*exp(-lam)

The above two programs are made for this computational purpose. Noting that

this sum of Poisson probability distribution is close to 1 for such a large K, we

TOWARD GOOD PROGRAM 39

can run them to find that one works fine, while the other gives a quite wrong

result. Could you tell which one is better?

1.3.2 Vector Operation Versus Loop Iteration

It is time-efficient to use vector operations rather than loop iterations to perform a

repetitive job for an array of data. The following program “nm132_1.m” compares

a vector operation versus a loop iteration in terms of the execution speed. Could

you tell which one is faster?

%nm132_1: vector operation vs. loop iteration

N = 100000; th = [0:N-1]/50000*pi;

tic

ss=sin(th(1));

for i = 2:N, ss = ss + sin(th(i)); end % loop iteration

toc, ss

tic

ss = sum(sin(th)); % vector operation

toc, ss

As a more practical example, let us consider a problem of finding the DtFT

(discrete-time Fourier transform) ([W-3]) of a given sequence x[n].

X() =

N−1

n=0

x[n]e−jn for = [−100 : 100]π/100 (1.3.4)

The following program “nm132_2.m” compares a vector operation versus a loop

iteration for computing the DtFT in terms of the execution speed. Could you tell

which one is faster?

%nm132_2: nested structure

N = 1000; x = rand(1,N); % a random sequence x[n] for n = 0:N-1

W = [-100:100]*pi/100; % frequency range

tic

for k = 1:length(W)

X1(k) = 0; %for for loop

for n = 1:N, X1(k) = X1(k) + x(n)*exp(-j*W(k)*(n-1)); end

end

toc

tic

X2 = 0;

for n = 1:N %for vector loop

X2 = X2 +x(n)*exp(-j*W*(n-1));

end

toc

discrepancy = norm(X1-X2) %transpose for dimension compatibility

40 MATLAB USAGE AND COMPUTATIONAL ERRORS

1.3.3 Iterative Routine Versus Nested Routine

In this section we compare an iterative routine and a nested routine performing the

same job. Consider the following two programs fctrl1(n)/fctrl2(n), whose

common objectives is to get the factorial of a given nonnegative integer k.

k! = k(k − 1) · · · 2 · 1 (1.3.5)

They differ in their structure.While fctrl1() uses a for loop structure, fctrl2()

uses the nested (recursive) calling structure that a program uses itself as a subroutine

to perform a sub-job. Compared with fctrl1(), fctrl2() is easier to program as

well as to read, but is subject to runtime error that is caused by the excessive use

of stack memory as the number of recursive calls increases with large n. Another

disadvantage of fctrl2() is that it is time-inefficient for the number of function

calls, which increases with the input argument (n). In this case, a professional

programmer would consider the standpoint of users to determine the programming

style. Some algorithms like the adaptive integration (Section 5.8), however, may

fit the nested structure perfectly.

function m = fctrl1(n)

m = 1;

for k = 2:n, m = m*k; end

function m = fctrl2(n)

if n <= 1, m = 1;

else m = n*fctrl2(n-1);

end

1.3.4 To Avoid Runtime Error

A good program guides the program users who don’t know much about the

program and at least gives them a warning message without runtime error for

their minor mistake. If you don’t know what runtime error is, you can experience

one by taking the following steps:

1. Make and save the above routine fctrl1() in an M-file named ‘fctrl.m’

in a directory listed in the MATLAB search path.

2. Type fctrl(-1) into the MATLAB Command window. Then you will see

>>fctrl(-1)

ans = 1

This seems to imply that (−1)! = 1, which is not true. It is caused by the mistake

of the user who tries to find (−1)! without knowing that it is not defined. This

kind of runtime error seems to be minor because it does not halt the process.

But it needs special attention because it may not be easy to detect. If you are a

good programmer, you will insert some error handling statements in the program

fctrl() as below. Then, when someone happens to execute fctrl(-1) in the

Command window or through an M-file, the execution stops and he will see the

error message in the Command window as

??? Error using ==> fctrl

The factorial of negative number ??

TOWARD GOOD PROGRAM 41

function m = fctrl(n)

if n < 0, error(’The factorial of negative number ??’);

else m = 1; for k = 2:n, m = m*k; end

end

This shows the error message (given as the input argument of the error()

routine) together with the name of the routine in which the accidental “error”

happens, which is helpful for the user to avoid the error.

Most common runtime errors are caused by an “out of domain” index of array

and the violation of matrix dimension compatibility, as illustrated in Section 1.1.7.

For example, consider the gauss(A,B) routine in Section 2.2.2, whose job is to

solve a system of linear equations Ax = b for x. To appreciate the role of the fifth

line handling the dimension compatibility error in the routine, remove the line

(by putting the comment mark % before the line in the M-file defining gauss())

and type the following statements in the Command window:

>>A = rand(3,3); B = rand(2,1); x = gauss(A,B)

?? Index exceeds matrix dimensions.

Error in ==> C:\MATLAB6p5\nma\gauss.m

On line 10 ==> AB = [A(1:NA,1:NA) B(1:NA,1:NB)];

Then MATLAB gives you an error message together with the suspicious statement

line and the routine name. But it is hard to figure out what causes the

runtime error, and you may get nervous lest the routine should have some bug.

Now, restore the fifth line in the routine and type the same statements in the

Command window:

>>x = gauss(A,B)

?? Error using ==> gauss

A and B must have compatible dimension

This error message (provided by the programmer of the routine) helps you to

realize that the source of the runtime error is the incompatible matrices/vectors A

and B given as the input arguments to the gauss() routine. Very like this, a good

program has a scenario for possible user mistakes and fires the error routine for

each abnormal condition to show the user the corresponding error message.

Many users often give more/fewer input arguments than supposed to be given

to the MATLAB functions/routines and sometimes give wrong types/formats of

data to them. To experience this type of error, let us try using the MATLAB

function sinc1(t,D) (Section 1.3.5) to plot the graph of a sinc function

sin c(t/D) =

sin(πt/D)

πt/D

with D = 0.5 and t = −2, 2 (1.3.6)

With this purpose, type the following statements in the Command window.

42 MATLAB USAGE AND COMPUTATIONAL ERRORS

0.5

1

0

−2 −1 0

(a) sinc1() with division-by-zero handling

1 2

0.5

1

0

−2 −1 0 1 2

(b) sinc1() without division-by-zero handling

Figure 1.8 The graphs of a sinc function defined by sinc1().

>>D = 0.5; b1 = -2; b2 = 2; t = b1+[0:200]/200*(b2 – b1);

>>plot(t,sinc1(t,D)), axis([b1 b2 -0.4 1.2])

>>hold on, plot(t,sinc1(t),’k:’)

The two plotting commands coupled with sinc1(t,D) and sinc1(t) yield the

two beautiful graphs, respectively, as depicted in Fig. 1.8a. It is important to

note that sinc1() doesn’t bother us and works fine without the second input

argument D. We owe the second line in the function sinc1() for the nice errorhandling

service:

if nargin < 2, D = 1; end

This line takes care of the case where the number of input arguments (nargin) is

less than 2, by assuming that the second input argument is D = 1 by default. This

programming technique is the key to making the MATLAB functions adaptive

to different number/type of input arguments, which is very useful for breathing

the user-convenience into the MATLAB functions. To appreciate its role, we

remove the second line from the M-file defining sinc1() and then type the same

statement in the Command window, trying to use sinc1() without the second

input argument.

>>plot(t,sinc1(t),’k:’)

??? Input argument ’D’ is undefined.

Error in ==> C:\MATLAB6p5\nma\sinc1.m

On line 4 ==> x = sin(pi*t/D)./(pi*t/D);

This time we get a serious (red) error message with no graphic result. It is implied

that the MATLAB function without the appropriate error-handling parts no longer

allows the user’s default or carelessness.

Now, consider the third line in sinc1(), which is another error-handling statement.

t(find(t==0))=eps;

TOWARD GOOD PROGRAM 43

or, equivalently

for i = 1:length(t), if t(i) == 0, t(i) = eps; end, end

This statement changes every zero element in the t vector into eps (2.2204e-

016). What is the real purpose of this statement? It is actually to remove the

possibility of division-by-zero in the next statement, which is a mathematical

expression having t in the denominator.

x = sin(pi*t/D)./(pi*t/D);

To appreciate the role of the third line in sinc1(), we remove it from the M-file

defining sinc1(), and type the following statement in the Command window.

>>plot(t,sinc1(t,D),’r’)

Warning: Divide by zero.

(Type “warning off MATLAB:divideByZero” to suppress this warning.)

In C:\MATLAB6p5\nma\sinc1.m at line 4)

This time we get just a warning (black) error message with a similar graphic

result as depicted in Fig. 1.8b. Does it imply that the third line is dispensable?

No, because the graph has a (weird) hole at t = 0, about which most engineers/

mathematicians would feel uncomfortable. That’s why authors strongly

recommend you not to omit such an error-handling part as the third line as

well as the second line in the MATLAB function sinc1().

(cf) What is the value of sinc1(t,D) for t = 0 in this case? Aren’t you curious? If so,

let’s go for it.

>>sinc1(0,D), sin(pi*0/D)/(pi*0/D), 0/0

ans = NaN (Not-a-Number: undetermined)

Last, consider of the fourth line in sinc1(), which is only one essential

statement performing the main job.

x = sin(pi*t/D)./(pi*t/D);

What is the .(dot) before /(division operator) for? In reference to this, authors

gave you a piece of advice that you had better put a .(dot) just before the

arithmetic operators *(multiplication), /(division), and ^(power) in the function

definition so that the term-by-term (termwise) operation can be done any time

(Section 1.1.6, (A5)). To appreciate the existence of the .(dot), we remove it from

the M-file defining sinc1(), and type the following statements in the Command

window.

>>clf, plot(t,sinc1(t,D)), sinc1(t,D), sin(pi*t/D)/(pi*t/D)

ans = -0.0187

44 MATLAB USAGE AND COMPUTATIONAL ERRORS

What do you see in the graphic window on the screen? Surprise, a (horizontal)

straight line running parallel with the t-axis far from any sinc function graph!

What is more surprising, the value of sinc1(t,D) or sin(pi*t/D)/(pi*t/D)

shows up as a scalar. Authors hope that this accident will help you realize how

important it is for right term-by-term operations to put .(dot) before the arithmetic

operators *, / and ^ . By the way, aren’t you curious about how MATLAB deals

with a vector division without .(dot)? If so, let’s try with the following statements:

>>A = [1:10]; B = 2*A; A/B, A*B’*(B*B’)^-1, A*pinv(B)

ans = 0.5

To understand this response of MATLAB, you can see Section 1.1.7 or Section

2.1.2.

In this section we looked at several sources of runtime error, hoping that it

aroused the reader’s attention to the danger of runtime error.

1.3.5 Parameter Sharing via Global Variables

When we discuss the runtime error that may be caused by user’s default in passing

some parameter as input argument to the corresponding function, you might feel

that the parameter passing job is troublesome. Okay, it is understandable as a

beginner in MATLAB. How about declaring the parameters as global so that

they can be accessed/shared from anywhere in the MATLAB world as far as the

declaration is valid? If you want to, you can declare any varable(s) by inserting

the following statement in both the main program and all the functions using

the variables.

global Gravity_Constant Dielectric_Constant

%plot_sinc

clear, clf

global D

D = 1; b1 = -2; b2 = 2;

t = b1 +[0:100]/100*(b2 – b1);

%passing the parameter(s) through arguments of the function

subplot(221), plot(t, sinc1(t,D))

axis([b1 b2 -0.4 1.2])

%passing the parameter(s) through global variables

subplot(222), plot(t, sinc2(t))

axis([b1 b2 -0.4 1.2])

function x = sinc1(t,D)

if nargin<2, D = 1; end

t(find(t == 0)) = eps;

x = sin(pi*t/D)./(pi*t/D);

function x = sinc2(t)

global D

t(find(t == 0)) = eps;

x = sin(pi*t/D)./(pi*t/D);

Then, how convenient it would be, since you don’t have to bother about passing

the parameters. But, as you get proficient in programming and handle many

TOWARD GOOD PROGRAM 45

functions/routines that are involved with various sets of parameters, you might

find that the global variable is not always convenient, because of the following

reasons.

ž Once a variable is declared as global, its value can be changed in any of the

MATLAB functions having declared it as global, without being noitced by

other related functions. Therefore it is usual to declare only the constants as

global and use long names (with all capital letters) as their names for easy

identification.

ž If some variables are declared as global and modified by several functions/

routines, it is not easy to see the relationship and the interaction among

the related functions in terms of the global variable. In other words, the program

readability gets worse as the number of global variables and related

functions increases.

For example, let us look over the above program “plot sinc.m” and the function

“sinc2()”. They both have a declaration of D as global; consequently,

sinc2() does not need the second input argument for getting the parameter

D. If you run the program, you will see that the two plotting statements adopting

sinc1() and sinc2() produce the same graphic result as depicted in Fig. 1.8a.

1.3.6 Parameter Passing Through Varargin

In this section we see two kinds of routines that get a function name (string)

with its parameters as its input argument and play with the function.

First, let us look over the routine “ez_plot1()”, which gets a function name

(ftn) with its parameters (p) and the lower/upper bounds (bounds = [b1 b2])

as its first, third, and second input argument, respectively, and plots the graph of

the given function over the interval set by the bounds. Since the given function

may or may not have its parameter, the two cases are determined and processed

by the number of input arguments (nargin) in the if-else-end block.

%plot_sinc1

clear, clf

D = 1; b1 = -2; b2 = 2;

t = b1+[0:100]/100*(b2 – b1);

bounds = [b1 b2];

subplot(223), ez_plot1(’sinc1’,bounds,D)

axis([b1 b2 -0.4 1.2])

subplot(224), ez_plot(’sinc1’,bounds,D)

axis([b1 b2 -0.4 1.2])

function ez_plot1(ftn,bounds,p)

if nargin < 2, bounds = [-1 1]; end

b1 = bounds(1); b2 = bounds(2);

t = b1+[0:100]/100*(b2 – b1);

if nargin <= 2, x = feval(ftn,t);

else x = feval(ftn,t,p);

end

plot(t,x)

function

ez_plot(ftn,bounds,varargin)

if nargin < 2, bounds = [-1 1]; end

b1 = bounds(1); b2 = bounds(2);

t = b1 + [0:100]/100*(b2 – b1);

x = feval(ftn,t,varargin{:});

plot(t,x)

46 MATLAB USAGE AND COMPUTATIONAL ERRORS

Now, let us see the routine “ez_plot()”, which does the same plotting job

as “ez_plot1()”. Note that it has a MATLAB keyword varargin (variable

length argument list) as its last input argument and passes it into the MATLAB

built-in function feval() as its last input argument. Since varargin can represent

comma-separated multiple parameters including expression/strings, it paves

the highway for passing the parameters in relays. As the number of parameters

increases, it becomes much more convenient to use varargin for passing

the parameters than to deal with the parameters one-by-one as in ez_plot1().

This technique will be widely used later in Chapter 4 (on nonlinear equations),

Chapter 5 (on numerical integration), Chapter 6 (on ordinary differential equations),

and Chapter 7 (on optimization).

(cf) Note that MATLAB has a built-in graphic function ezplot(), which is much more

powerful and convenient to use than ez_plot(). You can type ‘help ezplot’ to see

its function and usage.

1.3.7 Adaptive Input Argument List

A MATLAB function/routine is said to be “adaptive” to users in terms of input

arguments if it accepts different number/type of input arguments and makes a

reasonable interpretation. For example, let us see the nonlinear equation solver

routine ‘newton()’ in Section 4.4. Its input argument list is

(f,df,x0,tol,kmax)

where f, df, x0, tol and kmax denote the filename (string) of function (to

be solved), the filename (string) of its derivative function, the initial guess (for

solution), the error tolerance and the maximum number of iterations, respectively.

Suppose the user, not knowing the derivative, tries to use the routine with just

four input arguments as follows.

>>newton(f,x0,tol,kmax)

At first, these four input arguments will be accepted as f,df,x0, and tol,

respectively. But, when the second line of the program body is executed, the

routine will notice something wrong from that df is not any filename but a

number and then interprets the input arguments as f,x0,tol, and kmax to the

idea of the user. This allows the user to use the routine in two ways, depending

on whether he is going to supply the routine with the derivative function or not.

This scheme is conceptually quite similar to function overloading of C++, but

C++ requires us to have several functions having the same name, with different

argument list.

PROBLEMS

1.1 Creating a Data File and Retrieving/Plotting Data Saved in a Data File

(a) Using the MATLAB editor, make a program “nm1p01a”, which lets its

user input data pairs of heights [ft] and weights [lb] of as many persons

PROBLEMS 47

as he wants until he presses <Enter> and save the whole data in the

form of an N ×2 matrix into an ASCII data file (***.dat) named by

the user. If you have no idea how to compose such a program, you

can permutate the statements in the box below to make your program.

Store the program in the file named “nm1p01a.m” and run it to save

the following data into the data file named “hw.dat”:

5.5162

6.1185

5.7170

6.5195

6.2191

%nm1p01a: input data pairs and save them into an ASCII data file

clear

k = 0;

while 1

end

k = k + 1;

x(k,1) = h;

h = input(’Enter height:’)

x(k,2) = input(’Enter weight:’)

if isempty(h), break; end

cd(’c:\matlab6p5\work’) %change current working directory

filename = input(’Enter filename(.dat):’,’s’);

filename = [filename ’.dat’]; %string concatenation

save(filename,’x’,’/ascii’)

(b) Make a MATLAB program “nm1p01b”, which reads (loads) the data

file “hw.dat” made in (a), plots the data as in Fig. 1.1a in the upperleft

region of the screen divided into four regions like Fig. 1.3, and

plots the data in the form of piecewise-linear (PWL) graph describing

the relationship between the height and the weight in the upper-right

region of the screen. Let each data pair be denoted by the symbol ‘+’

on the graph. Also let the ranges of height and weight be [5, 7] and

[160, 200], respectively. If you have no idea, you can permutate the

statements in the below box. Additionally, run the program to check if

it works fine.

%nm1p01b: to read the data file and plot the data

cd(’c:\matlab6p5\work’) %change current working directory

weight = hw(I,2);

load hw.dat

clf, subplot(221)

plot(hw)

subplot(222)

axis([5 7 160 200])

plot(height,weight,height,weight,’+’)

[height,I] = sort(hw(:,1));

48 MATLAB USAGE AND COMPUTATIONAL ERRORS

1.2 Text Printout of Alphanumeric Data

Make a routine max_array(A), which uses the max() command to find one

of the maximum elements of a matrix A given as its input argument and

uses the fprintf() command to print it onto the screen together with its

row/column indices in the following format.

’\n Max(A) is A(%2d,%2d) = %5.2f\n’,row_index,col_index,maxA

Additionally, try it to have the maximum element of an arbitrary matrix

(generated by the following two consecutive commands) printed in this

format onto the screen.

>>rand(’state’,sum(100*clock)), rand(3)

1.3 Plotting the Mesh Graph of a Two-Dimensional Function

Consider the MATLAB program “nm1p03a”, whose objective is to draw

a cone.

(a) The statement on the sixth line seems to be dispensable. Run the program

with and without this line and see what happens.

(b) If you want to plot the function fcone(x,y) defined in another M-file

‘fcone.m’, how will you modify this program?

(c) If you replace the fifth line by ‘Z = 1-abs(X)-abs(Y);’, what difference

does it make?

%nm1p03a: to plot a cone

clear, clf

x = -1:0.02:1; y = -1:0.02:1;

[X,Y] = meshgrid(x,y);

Z = 1-sqrt(X.^2+Y.^2);

Z = max(Z,zeros(size(Z)));

mesh(X,Y,Z)

function z = fcone(x,y)

z = 1-sqrt(x.^2 + y.^2);

1.4 Plotting The Mesh Graph of Stratigraphic Structure

Consider the incomplete MATLAB program “nm1p04”, whose objective is

to draw a stratigraphic structure of the area around Pennsylvania State

University from the several perspective point of view. The data about

the depth of the rock layer at 5 × 5 sites are listed in Table P1.4. Supplement

the incomplete parts of the program so that it serves the purpose

and run the program to answer the following questions. If you complete

it properly and run it, MATLAB will show you the four similar

graphs at the four corners of the screen and be waiting for you to press

any key.

PROBLEMS 49

(a) At what value of k doesMATLABshow you themesh/surface-type graphs

that are the most similar to the first graphs? From this result, what do you

guess are the default values of the azimuth or horizontal rotation angle and

the vertical elevation angle (in degrees) of the perspective view point?

(b) As the first input argument Az of the command view(Az,E1) decreases,

in which direction does the perspective viewpoint revolve round the

z-axis, clockwise or counterclockwise (seen from the above)?

(c) As the second input argument El of the command view(Az,E1) increases,

does the perspective viewpoint move up or down along the z-axis?

(d) What is the difference between the plotting commands mesh() and

meshc()?

(e) What is the difference between the usages of the command view()

with two input arguments Az,El and with a three-dimensional vector

argument [x,y,z]?

Table P1.4 The Depth of the Rock Layer

x Coordinate

y Coordinate 0.1 1.2 2.5 3.6 4.8

0.5 410 390 380 420 450

1.4 395 375 410 435 455

2.2 365 405 430 455 470

3.5 370 400 420 445 435

4.6 385 395 410 395 410

%nm1p04: to plot a stratigraphic structure

clear, clf

x = [0.1 .. .. . ];

y = [0.5 .. .. . ];

Z = [410 390 .. .. .. .. ];

[X,Y] = meshgrid(x,y);

subplot(221), mesh(X,Y,500 – Z)

subplot(222), surf(X,Y,500 – Z)

subplot(223), meshc(X,Y,500 – Z)

subplot(224), meshz(X,Y,500 – Z)

pause

for k = 0:7

Az = -12.5*k; El = 10*k; Azr = Az*pi/180; Elr = El*pi/180;

subplot(221), view(Az,El)

subplot(222),

k, view([sin(Azr),-cos(Azr),tan(Elr)]), pause %pause(1)

end

1.5 Plotting a Function over an Interval Containing Its Singular Point Noting

that the tangent function f (x) = tan(x) is singular at x = π/2, 3π/2, let us

plot its graph over [0, 2π] as follows.

50 MATLAB USAGE AND COMPUTATIONAL ERRORS

(a) Define the domain vector x consisting of sufficiently many intermediate

point xi’s along the x-axis and the corresponding vector y consisting

of the function values at xi ’s and plot the vector y over the vector x.

You may use the following statements.

>>x = [0:0.01:2*pi]; y = tan(x);

>>subplot(221), plot(x,y)

Which one is the most similar to what you have got, among the graphs

depicted in Fig. P1.5? Is it far from your expectation?

(b) Expecting to get the better graph, we scale it up along the y-axis by

using the following command.

>>axis([0 6.3 -10 10])

Which one is the most similar to what you have got, among the graphs

depicted in Fig. P1.5? Is it closer to your expectation than what you

got in (a)?

(c) Most probably, you must be nervous about the straight lines at the

singular points x = π/2 and x = 3π/2. The more disturbed you become

by the lines that must not be there, the better you are at the numerical

stuffs. As an alternative to avoid such a singular happening, you can

try dividing the interval into three sections excluding the two singular

points as follows.

0 2 4

(a) (b)

(c)

6

−500

500

1000

1500

0

0 2 4 6

−10

0

5

10

−5

0 2 4 6

−10

0

5

10

−5

Figure P1.5 Plotting the graph of f(x) = tan x.

PROBLEMS 51

>>x1 = [0:0.01:pi/2-0.01]; x2 = [pi/2+0.01:0.01:3*pi/2-0.01];

>>x3 = [3*pi/2+0.01:0.01:2*pi];

>>y1 = tan(x1); y2 = tan(x2); y3 = tan(x3);

>>subplot(222), plot(x1,y1,x2,y2,x3,y3), axis([0 6.3 -10 10])

(d) Try adjusting the number of intermediate points within the plotting

interval as follows.

>>x1 = [0:200]*pi/100; y1 = tan(x1);

>>x2 = [0:400]*pi/200; y2 = tan(x2);

>>subplot(223), plot(x1,y1), axis([0 6.3 -10 10])

>>subplot(224), plot(x2,y2), axis([0 6.3 -10 10])

From the difference between the two graphs you got, you might have

guessed that it would be helpful to increase the number of intermediate

points. Do you still have the same idea even after you adjust the range

of the y-axis to [−50, +50] by using the following command?

>>axis([0 6.3 -50 50])

(e) How about trying the easy plotting command ezplot()? Does it answer

your desire?

>>ezplot(’tan(x)’,0,2*pi)

1.6 Plotting the Graph of a Sinc Function

The sinc function is defined as

f (x) =

sin x

x

(P1.6.1)

whose value at x = 0 is

f (0) = lim

x→0

sin x

x =

(sin x)

x x=0 =

cos x

1 x=0 = 1 (P1.6.2)

We are going to plot the graph of this function over [−4π,+4π].

(a) Casually, you may try as follows.

>>x = [-100:100]*pi/25; y = sin(x)./x;

>>plot(x,y), axis([-15 15 -0.4 1.2])

In spite of the warning message about ‘division-by-zero’, you may

somehow get a graph. But, is there anything odd about the graph?

(b) How about trying with a different domain vector?

>>x = [-4*pi:0.1:+4*pi]; y = sin(x)./x;

>>plot(x,y), axis([-15 15 -0.4 1.2])

52 MATLAB USAGE AND COMPUTATIONAL ERRORS

Surprisingly, MATLAB gives us the function values without any complaint

and presents a nice graph of the sinc function. What is the

difference between (a) and (b)?

(cf) Actually, we would have no problem if we used the MATLAB built-in function

sinc().

1.7 Termwise (Element-by-Element) Operation in In-Line Functions

(a) Let the function f1(x) be defined without one or both of the dot(.)

operators in Section 1.1.6. Could we still get the output vector consisting

of the function values for the several values in the input vector?

You can type the following statements into the MATLAB command

window and see the results.

>>f1 = inline(’1./(1+8*x^2)’,’x’); f1([0 1])

>>f1 = inline(’1/(1+8*x.^2)’,’x’); f1([0 1])

(b) Let the function f1(x) be defined with both of the dot(.) operators as in

Section 1.1.6. What would we get by typing the following statements

into the MATLAB command window?

>>f1 = inline(’1./(1+8*x.^2)’,’x’); f1([0 1]’)

1.8 In-Line Function and M-file Function with the Integral Routine ‘quad()’

As will be seen in Section 5.8, one of the MATLAB built-in functions for

computing the integral is ‘quad()’, the usual usage of which is

quad(f,a,b,tol,trace,p1,p2, ..) for b

a

f (x,p1, p2, . . .)dx

(P1.8.1)

where

f is the name of the integrand function (M-file name should be categorized

by ’ ’)

a,b are the lower/upper bound of the integration interval

tol is the error tolerance (10−6 by default [])

trace set to 1(on)/0(off) (0 by default []) for subintervals

p1,p2,.. are additional parameters to be passed directly to function f

Let’s use this quad() routine with an in-line function and an M-file function

to obtain

m+10

m−10

(x − x0)f (x)dx (P1.8.2a)

and

m+10

m−10

(x − x0)2f (x)dx (P1.8.2b)

PROBLEMS 53

where

x0 = 1, f(x)=

1

√2πσ

e−(x−m)2/2σ2 with m = 1, σ = 2 (P1.8.3)

Below are an incomplete main program ‘nm1p08’ and an M-file function

defining the integrand of (P1.8.2a). Make another M-file defining the integrand

of (P1.8.2b) and complete the main program to compute the two

integrals (P1.8.2a) and (P1.8.2b) by using the in-line/M-file functions.

function xfx = xGaussian_pdf(x,m,sigma,x0)

xfx = (x – x0).*exp(-(x – m).^2/2/sigma^2)/sqrt(2*pi)/sigma;

%nm1p08: to try using quad() with in-line/M-file functions

clear

m = 1; sigma = 2;

int_xGausspdf = quad(’xGaussian_pdf’,m – 10,m + 10,[],0,m,sigma,1)

Gpdf = ’exp(-(x-m).^2/2/sigma^2)/sqrt(2*pi)/sigma’;

xGpdf = inline([’(x – x0).*’ Gpdf],’x’,’m’,’sigma’,’x0’);

int_xGpdf = quad(xGpdf,m – 10,m+10,[],0,m,sigma,1)

1.9 μ-Law Function Defined in an M-File

The so-called μ-law function and μ−1-law function used for non-uniform

quantization is defined as

y = gμ(x) = |y|max

ln(1 + μ|x|/|x|max)

ln(1 + μ)

sign(x) (P1.9a)

x = g−1

μ (y) = |x|max

(1 + μ)|y|/|y|max − 1

μ

sign(y) (P1.9b)

Below are the μ-law function mulaw() defined in an M-file and a main

program nm1p09, which performs the following jobs:

ž Finds the values y of the μ-law function for x = [-1:0.01:1], plots the

graph of y versus x.

ž Finds the values x0 of the μ−1-law function for y.

ž Computes the discrepancy between x and x0.

Complete the μ−1-law function mulaw_inv() and store it together with

mulaw() and nm1p09 in the M-files named “mulaw inv.m”, “mulaw.m”,

and “nm1p09.m”, respectively. Then run the main program nm1p09 to plot

the graphs of the μ-law function with μ = 10, 50 and 255 and find the

discrepancy between x and x0.

54 MATLAB USAGE AND COMPUTATIONAL ERRORS

function [y,xmax] = mulaw(x,mu,ymax)

xmax = max(abs(x));

y = ymax*log(1+mu*abs(x/xmax))./log(1+mu).*sign(x); % Eq.(P1.9a)

function x = mulaw_inv(y,mu,xmax)

%nm1p09: to plot the mulaw curve

clear, clf

x = [-1:.005:1];

mu = [10 50 255];

for i = 1:3

[y,xmax] = mulaw(x,mu(i),1);

plot(x,y,’b-’, x,x0,’r-’), hold on

x0 = mulaw_inv(y,mu(i),xmax);

discrepancy = norm(x-x0)

end

1.10 Analog-to-Digital Converter (ADC)

Below are two ADC routines adc1(a,b,c) and adc2(a,b,c), which assign

the corresponding digital value c(i) to each one of the analog data belonging

to the quantization interval [b(i), b(i+1)]. Let the boundary vector

and the centroid vector be, respectively,

b = [-3 -2 -1 0 1 2 3]; c = [-2.5 -1.5 -0.5 0.5 1.5 2.5];

(a) Make a program that uses two ADC routines to find the output d for

the analog input data a = [-300:300]/100 and plots d versus a to see

the input-output relationship of the ADC, which is supposed to be like

Fig. P1.10a.

function d = adc1(a,b,c)

%Analog-to-Digital Converter

%Input a = analog signal, b(1:N + 1) = boundary vector

c(1:N)=centroid vector

%Output: d = digital samples

N = length(c);

for n = 1:length(a)

I = find(a(n) < b(2:N));

if ~isempty(I), d(n) = c(I(1));

else d(n) = c(N);

end

end

function d=adc2(a,b,c)

N = length(c);

d(find(a < b(2))) = c(1);

for i = 2:N-1

index = find(b(i) <= a & a <= b(i+1)); d(index) = c(i);

end

d(find(b(N) <= a)) = c(N);

PROBLEMS 55

input

output

4

2

0

−2

−4

−4 −2 0

(a) The input-output relationship of an ADC

2 4

time

analog input

digital

output

0 2 4

(b) The output of an ADC to a sinusoidal input

6 8

Figure P1.10 The characteristic of an ADC (analog-to-digital converter).

(b) Make a program that uses two ADC routines to find the output d for

the analog input data a = 3*sin(t) with t = [0:200]/100*pi and

plots a and d versus t to see how the analog input is converted into the

digital output by the ADC. The graphic result is supposed to be like

Fig. P1.10b.

1.11 Playing with Polynomials

(a) Polynomial Evaluation: polyval()

Write a MATLAB statement to compute

p(x) = x8 −1 forx = 1 (P1.11.1)

(b) Polynomial Addition/Subtraction by Using Compatible Vector Addition/

Subtraction

Write a MATLAB statement to add the following two polynomials:

p1(x) = x4 + 1, p2(x) = x3 − 2×2 + 1 (P1.11.2)

(c) Polynomial Multiplication: conv()

Write aMATLAB statement to get the following product of polynomials:

p(x) = (x4 + 1)(x2 + 1)(x + 1)(x − 1) (P1.11.3)

(d) Polynomial Division: deconv()

Write a MATLAB statement to get the quotient and the remainder of

the following polynomial division:

p(x) = x8/(x2 − 1) (P1.11.4)

(e) Routine for Differentiation/Integration of a Polynomial

What you see in the below box is the routine “poly_der(p)”, which

gets a polynomial coefficient vector p (in the descending order) and

outputs the coefficient vector pd of its derivative polynomial. Likewise,

you can make a routine “poly_int(p)”, which outputs the coefficient

56 MATLAB USAGE AND COMPUTATIONAL ERRORS

vector of the integral polynomial for a given polynomial coefficient

vector.

(cf) MATLAB has the built-in routines polyder()/polyint() for finding the

derivative/integral of a polynomial.

function pd = poly_der(p)

%p: the vector of polynomial coefficients in descending order

N = length(p);

if N <= 1, pd = 0; % constant

else

for i = 1: N – 1, pd(i) = p(i)*(N – i); end

end

(f) Roots of A Polynomial Equation: roots()

Write a MATLAB statement to get the roots of the following polynomial

equation

p(x) = x8 − 1 = 0 (P1.11.5)

You can check if the result is right, by using the MATLAB command

poly(), which generates a polynomial having a given set of roots.

(g) Partial Fraction Expansion of a Ratio of Two Polynomials: residue()/

residuez()

(i) The MATLAB routine [r,p,k] = residue(B,A) finds the partial

fraction expansion for a ratio of given polynomials B(s)/A(s) as

B(s)

A(s) =

b1sM−1 + b2sM−2 + ·· ·+bM

a1sN−1 + a2sN−2 +· · ·+aN = k(s) +i

r(i)

s − p(i)

(P1.11.6a)

which is good for taking the inverse Laplace transform. Use this

routine to find the partial fraction expansion for

X(s) =

4s + 2

s3 + 6s2 + 11s + 6 = s + + s + + s + (P1.11.7a)

(ii) The MATLAB routine [r,p,k] = residuez(B,A) finds the partial

fraction expansion for a ratio of given polynomials B(z)/A(z)

as

B(z)

A(z) =

b1 + b2z−1 + ·· ·+bMz−(M−1)

a1 + a2z−1 + ·· ·+aNz−(N−1) = k(z−1) +i

r(i)z

z − p(i)

(P1.11.6b)

which is good for taking the inverse z-transform. Use this routine

to find the partial fraction expansion for

X(z) =

4 + 2z−1

1 + 6z−1 + 11z−2 + 6z−3 =

z

z + +

z

z + +

z

z + (P1.11.7b)

PROBLEMS 57

(h) Piecewise Polynomial: mkpp()/ppval()

Suppose we have an M × N matrix P, the rows of which denote

M (piecewise) polynomials of degree (N − 1) for different (nonoverlapping)

intervals with (M + 1) boundary points bb = [b(1)

.. b(M + 1)], where the polynomial coefficients in each row are

supposed to be generated with the interval starting from x = 0. Then

we can use the MATLAB command pp = mkpp(bb,P) to construct a

structure of piecewise polynomials, which can be evaluated by using

ppval(pp).

Figure P1.11(h) shows a set of piecewise polynomials {p1(x + 3),

p2(x + 1), p3(x − 2)} for the intervals [−3, −1],[−1, 2] and [2, 4],

respectively, where

p1(x) = x2, p2(x) = −(x − 1)2, and p3(x) = x2 − 2 (P1.11.8)

Make a MATLAB program which uses mkpp()/ppval() to plot this

graph.

0

p1(x + 3) = (x + 3)2

p3(x − 2) = (x − 2)2 −2

p2(x + 1) = −x 2

−3

−5

0

5

−2 −1 1 2 3 x 4

Figure P1.11(h) The graph of piecewise polynomial functions.

(cf) You can type ‘help mkpp’ to see a couple of examples showing the usage

of mkpp.

1.12 Routine for Matrix Multiplication

Assuming that MATLAB cannot perform direct multiplication on vectors/

matrices, supplement the following incomplete routine “multiply_matrix

(A,B)” so that it can multiply two matrices given as its input arguments only

if their dimensions are compatible, but displays an error message if their

dimensions are not compatible. Try it to get the product of two arbitrary

3 × 3 matrices generated by the command rand(3) and compare the result

with that obtained by using the direct multiplicative operator *. Note that

58 MATLAB USAGE AND COMPUTATIONAL ERRORS

the matrix multiplication can be described as

C(m, n) =

K

k=1

A(m, k)B(k, n) (P1.12.1)

function C = multiply_matrix(A,B)

[M,K] = size(A); [K1,N] = size(B);

if K1 ~= K

error(’The # of columns of A is not equal to the # of rows of B’)

else

for m = 1:

for n = 1:

C(m,n) = A(m,1)*B(1,n);

for k = 2:

C(m,n) = C(m,n) + A(m,k)*B(k,n);

end

end

end

end

1.13 Function for Finding Vector Norm

Assuming that MATLAB does not have the norm() command finding us the

norm of a given vector/matrix, make a routine norm_vector(v,p), which

computes the norm of a given vector as

||v||p = p

N

n=1

|vn|p (P1.13.1)

for any positive integer p, finds the maximum absolute value of the elements

for p = inf and computes the norm as if p = 2, even if the second input

argument p is not given. If you have no idea, permutate the statements in the

below box and save it in the file named “norm_vector.m”. Additionally, try

it to get the norm with p = 1,2,∞ (inf) and of an arbitrary vector generated

by the command rand(2,1). Compare the result with that obtained by using

the norm() command.

function nv = norm_vector(v,p)

if nargin < 2, p = 2; end

nv = sum(abs(v).^p)^(1/p);

nv = max(abs(v));

if p > 0 & p ~= inf

elseif p == inf

end

PROBLEMS 59

1.14 Backslash(\) Operator

Let’s play with the backslash(\) operator.

(a) Use the backslash(\) command, the minimum-norm solution (2.1.7) and

the pinv() command to solve the following equations, find the residual

error ||Aix − bi ||’s and the rank of the coefficient matrix Ai, and fill in

Table P1.14 with the results.

(i) A1x =

1 2 3

4 5 6

x1

x2

x3

=

6

15 = b1 (P1.14.1)

(ii) A2x =

1 2 3

2 4 6

x1

x2

x3

=

6

8 = b2 (P1.14.2)

(iii) A3x =

1 2 3

2 4 6

x1

x2

x3

=

6

12 = b3 (P1.14.3)

Table P1.14 Results of Operations with backslash (\) Operator and pinv( ) Command

backslash(\) Minimum-Norm or

LS Solution

pinv() Remark

on rank(Ai )

x ||Ai redundant/ x − bi || x ||Aix − bi || x ||Aix − bi || inconsistent

1.5000 4.4409e-15

A1x = b1 0 (1.9860e-15)

1.5000

0.3143 1.7889

A2x = b2 0.6286

0.9429

A3x = b3

A4x = b4 2.5000 1.2247

0.0000

A5x = b5

A6x = b6

(cf) When the mismatching error ||Aix − bi ||’s obtained from MATLAB 5.x/6.x version are slightly different, the

former one is in the parentheses ().

60 MATLAB USAGE AND COMPUTATIONAL ERRORS

(b) Use the backslash (\) command, the LS (least-squares) solution (2.1.10)

and the pinv() command to solve the following equations and find the

residual error ||Aix − bi ||’s and the rank of the coefficient matrix Ai ,

and fill in Table P1.14 with the results.

(i) A4x =

1 2

2 3

3 4

x1

x2 =

2

6

7

= b4 (P1.14.4)

(ii) A5x =

1 2

2 4

3 6

x1

x2 =

1

5

8

= b5 (P1.14.5)

(iii) A6x =

1 2

2 4

3 6

x1

x2 =

3

6

9

= b6 (P1.14.6)

(cf) If some or all of the rows of the coefficient matrix A in a set of linear equations

can be expressed as a linear combination of other row(s), the corresponding

equations are dependent, which can be revealed by the rank deficiency, that is,

rank(A) < min(M,N) where M and N are the row dimension and the column

dimension, respectively. If some equations are dependent, they may have either

inconsistency (no exact solution) or redundancy (infinitely many solutions),

which can be distinguished by checking if augmenting the RHS vector b to the

coefficient matrix A increases the rank or not—that is, rank([A b]) > rank(A)

or not [M-2].

(c) Based on the results obtained in (a) and (b) and listed in Table P1.14,

answer the following questions.

(i) Based on the results obtained in (a)(i), which one yielded the

non-minimum-norm solution among the three methods, that is,

the backslash(\) operator, the minimum-norm solution (2.1.7) and

the pinv() command? Note that the minimum-norm solution

means the solution whose norm (||x||) is the minimum over the

many solutions.

(ii) Based on the results obtained in (a), which one is most reliable

as a means of finding the minimum-norm solution among the

three methods?

(iii) Based on the results obtained in (b), choose two reliable methods

as a means of finding the LS (least-squares) solution among the

three methods, that is, the backslash (\) operator, the LS solution

(2.1.10) and the pinv() command. Note that the LS solution

PROBLEMS 61

means the solution for which the residual error (||Ax − b||) is the

minimum over the many solutions.

1.15 Operations on Vectors

(a) Find the mathematical expression for the computation to be done by

the following MATLAB statements.

>>n = 0:100; S = sum(2.^-n)

(b) Write a MATLAB statement that performs the following computation.

10000

n=0

1

(2n + 1)2 −

π2

8

(c) Write a MATLAB statement which uses the commands prod() and

sum() to compute the product of the sums of each row of a 3 × 3

random matrix.

(d) How does the following MATLAB routine “repetition(x,M,m)” convert

a given row vector sequence x to make a new sequence y ?

function y = repetition(x,M,m)

if m == 1

MNx = ones(M,1)*x; y = MNx(:)’;

else

Nx = length(x); N = ceil(Nx/m);

x = [x zeros(1,N*m – Nx)];

MNx = ones(M,1)*x;

y = [];

for n = 1:N

tmp = MNx(:,(n – 1)*m + [1:m]).’;

y = [y tmp(:).’];

end

end

(e) Make a MATLAB routine “zero_insertion(x,M,m)”, which inserts

m zeros just after every Mth element of a given row vector sequence

x to make a new sequence. Write a MATLAB statement to apply the

routine for inserting two zeros just after every third element of x = [1 3 7 2 4 9] to get

y = [1 3 7 0 0 2 4 9 0 0]

(f) How does the following MATLAB routine “zeroing(x,M,m)” convert

a given row vector sequence x to make a new sequence y?

62 MATLAB USAGE AND COMPUTATIONAL ERRORS

function y = zeroing(x,M,m)

%zero out every (kM – m)th element

if nargin < 3, m = 0; end

if M<=0, M = 1; end

m = mod(m,M);

Nx = length(x); N = floor(Nx/M);

y = x; y(M*[1:N] – m) = 0;

(g) Make a MATLAB routine “sampling(x,M,m)”, which samples every

(kM – m)th element of a given row vector sequence x to make a new

sequence. Write a MATLAB statement to apply the routine for sampling

every (3k − 2)th element of x = [1 3 7 2 4 9] to get

y = [1 2]

(h) Make a MATLAB routine ‘rotation_r(x,M)”, which rotates a given

row vector sequence x right by M samples, say, making rotate_r([1

2 3 4 5],3) = [3 4 5 1 2].

1.16 Distribution of a Random Variable: Histogram

Make a routine randu(N,a,b), which uses the MATLAB function rand()

to generate an N-dimensional random vector having the uniform distribution

over [a, b] and depicts the graph for the distribution of the elements of

the generated vector in the form of histogram divided into 20 sections as

Fig.1.7. Then, see what you get by typing the following statement into the

MATLAB command window.

>>randu(1000,-2,2)

What is the height of the histogram on the average?

1.17 Number Representation

In Section 1.2.1, we looked over how a number is represented in 64 bits.

For example, the IEEE 64-bit floating-point number system represents the

number 3(21 ≤ 3 < 22) belonging to the range R1 = [21, 22) with E = 1 as

0 100 0000 0000 0000 0000 . . . . . . . . . . . . 0000 0000 0000 0000 0000

4 0 0 8 0 0 . . . . . . . . 0 0 0 0 0

1000

where the exponent and the mantissa are

Exp = E + 1023 = 1 + 1023 = 1024 = 210 = 100 0000 0000

M = (3 × 2−E − 1) × 252 = 251

= 1000 0000 0000 . . . . 0000 0000 0000 0000 0000

PROBLEMS 63

This can be confirmed by typing the following statement into MATLAB

command window.

>>fprintf(’3 = %bx\n’,3) or >>format hex, 3, format short

which will print out onto the screen

0000000000000840 4008000000000000

Noting that more significant byte (8[bits] = 2[hexadecimal digits]) of a

number is stored in the memory of higher address number in the INTEL

system, we can reverse the order of the bytes in this number to see the

number having the most/least significant byte on the left/right side as we

can see in the daily life.

00 00 00 00 00 00 08 40 → 40 08 00 00 00 00 00 00

This is exactly the hexadecimal representation of the number 3 as we

expected. You can find the IEEE 64-bit floating-point number representation

of the number 14 and use the command fprintf() or format hex to

check if the result is right.

<procedure of adding 2−1 to 23> <procedure of subtracting 2−1 from 23>

alignment

right result

truncation of guard bit

1 .0000 × 23

+ 1 .0000 × 2−1

.00000 × 23

.00010 × 23 alignment

2’s

complement

1 .0000 × 23

− 1 .0000 × 2−1

1 .00000 × 23

− 0 .00010 × 23

1 .00000 × 23

+ 1 .11110 × 23

1 .00010 × 23

1 .0001 × 23

= (1 + 2−4) × 23

right result

normalization

truncation of guard bit

0 .11110 × 23

1 .1110 × 22

= (1 + 1 − 2−3) × 22

<procedure of adding 2−2 to 23>

alignment

.0000 × 23

+ 1 .0000 × 2−2

1 .00000 × 23

+ 0 .00001 × 23

no difference

truncation of guard bit

(cf) : hidden bit,

1 .00001 × 23

1 .0000 × 23

= (1 + 0) × 23

<procedure of subtracting 2−2 from 23>

alignment

1 .0000 × 23

− 1 .0000 × 2−2

1 .00000 × 23

− 0 .00001 × 23

1 .00000 × 23

+ 1 .11111 × 23

2’s

complement

right result

normalization

truncation of guard bit

0 .11111 × 23

1 .1111 × 22

= (1 + 1 − 2−4) × 22

1

+ 0

1

: guard bit

Figure P1.18 Procedure of addition/subtraction with four mantissa bits.

1.18 Resolution of Number Representation and Quantization Error

In Section 1.2.1, we have seen that adding 2−22 to 230 makes some difference,

while adding 2−23 to 230 makes no difference due to the bit shift

by over 52 bits for alignment before addition. How about subtracting 2−23

from 230? In contrast with the addition of 2−23 to 230, it makes a difference

as you can see by typing the following statement into the MATLAB

64 MATLAB USAGE AND COMPUTATIONAL ERRORS

command window.

>>x = 2^30; x + 2^ – 23 == x, x – 2^ – 23 == x

which will give you the logical answer 1 (true) and 0 (false). Justify this

result based on the difference of resolution of two ranges [230, 231) and [229,

230) to which the true values of computational results (230 + 2−23) and (230 − 2−23) belong, respectively. Note from Eq. (1.2.5) that the resolutions—that is,

the maximum quantization errors—are E = 2E−52 = 2−52+30 = 2−22 and

2−52+29 = 2−23, respectively. For details, refer to Fig. P1.18, which illustrates

the procedure of addition/subtraction with four mantissa bits, one hidden bit,

and one guard bit.

1.19 Resolution of Number Representation and Quantization Error

(a) What is the result of typing the following statements into the MATLAB

command window?

>>7/100*100 – 7

How do you compare the absolute value of this answer with the resolution

of the range to which 7 belongs?

(b) Find how many numbers are susceptible to this kind of quantization

error caused by division/multiplication by 100, among the numbers

from 1 to 31.

(c) What will be the result of running the following program? Why?

%nm1p19: Quantization Error

x = 2-2^-50;

for n = 1:2^3

x = x+2^-52; fprintf(’%20.18E\n’,x)

end

1.20 Avoiding Large Errors/Overflow/Underflow

(a) For x = 9.8201 and y = 10.2199, evaluate the following two expressions

that are mathematically equivalent and tell which is better in terms of

the power of resisting the overflow.

(i) z = x2 + y2 (P1.20.1a)

(ii) z = y(x/y)2 + 1 (P1.20.1b)

Also for x = 9.8−201 and y = 10.2−199, evaluate the above two expressions

and tell which is better in terms of the power of resisting the

underflow.

(b) With a = c = 1 and for 100 values of b over the interval [107.4, 108.5]

generated by the MATLAB command ‘logspace(7.4,8.5,100)’,

PROBLEMS 65

evaluate the following two formulas (for the roots of a quadratic

equation) that are mathematically equivalent and plot the values of the

second root of each pair. Noting that the true values are not available

and so the shape of solution graph is only one practical basis on which

we can assess the quality of numerical solutions, tell which is better in

terms of resisting the loss of significance.

(i)

x1, x2 =

1

2a

(−b ∓ sign(b)√b2 − 4ac) (P1.20.2a)

(ii)

x1 =

1

2a

(−b − sign(b)√b2 − 4ac), x2 =

c/a

x1 (P1.20.2b)

(c) For 100 values of x over the interval [1014, 1016], evaluate the following

two expressions that are mathematically equivalent, plot them, and

based on the graphs, tell which is better in terms of resisting the loss

of significance.

(i) y = √2×2 + 1 − 1 (P1.20.3a)

(ii) y =

2×2

√2×2 + 1 + 1

(P1.20.3b)

(d) For 100 values of x over the interval [10−9, 10−7.4], evaluate the following

two expressions that are mathematically equivalent, plot them,

and based on the graphs, tell which is better in terms of resisting the

loss of significance.

(i) y = √x + 4 − √x + 3 (P1.20.4a)

(ii) y =

1

√x + 4 + √x + 3

(P1.20.4b)

(e) On purpose to find the value of (300125/125!)e−300, type the following

statement into the MATLAB command window.

>>300^125/prod([1:125])*exp(-300)

What is the result? Is it of any help to change the order of multiplication/

division? As an alternative, make a routine which evaluates the

expression

p(k) =

λk

k!

e−λ for λ = 300 and an integerk (P1.20.5)

in a recursive way, say, like p(k + 1) = p(k) ∗ λ/k and then, use the

routine to find the value of (300125/125!)e−300.

66 MATLAB USAGE AND COMPUTATIONAL ERRORS

(f) Make a routine which computes the sum

S(K) =

K

k=0

λk

k!

e−λ for λ = 100 and an integerK (P1.20.6)

and then, use the routine to find the value of S(155).

1.21 Recursive Routines for Efficient Computation

(a) The Hermite Polynomial [K-1]

Consider the Hermite polynomial defined as

H0(x) = 1, HN(x) = (−1)Nex2 dN

dxN

e−x2

(P1.21.1)

(i) Show that the derivative of this polynomial function can be written

as

HN (x) = (−1)N2xex2 dN

dxN

e−x2 + (−1)Nex2 dN+1

dxN+1 e−x2

= 2xHN(x) − HN+1(x) (P1.21.2)

and so the (N + 1)th-degree Hermite polynomial can be obtained

recursively from the Nth-degree Hermite polynomial as

HN+1(x) = 2xHN(x) − HN (x) (P1.21.3)

(ii) Make a MATLAB routine “Hermitp(N)” which uses Eq. (P1.21.3)

to generate the Nth-degree Hermite polynomial HN(x).

(b) The Bessel Function of the First Kind [K-1]

Consider the Bessel function of the first kind of order k defined as

Jk(β) =

1

π π

0

cos(kδ − β sin δ)dδ (P1.21.4a)

= β

2 k ∞

m=0

(−1)mβ2m

4mm!(m + k)! ≡ (−1)kJ−k(β) (P1.21.4b)

(i) Define the integrand of (P1.21.4a) in the name of ‘Bessel_integrand(

x,beta,k)’ and store it in an M-file named “Bessel_

integrand.m”.

(ii) Complete the following routine “Jkb(K,beta)”, which uses

(P1.21.4b) in a recursive way to compute Jk(β) of order k = 1:K for given K and β (beta).

(iii) Run the following program nm1p21b which uses Eqs. (P1.21.4a)

and (P1.21.4b) to get J15(β) for β = 0:0.05:15. What is the norm

PROBLEMS 67

of the difference between the two results? How do you compare

the running times of the two methods?

(cf) Note that Jkb(K,beta) computes Jk(β) of order k = 1:K, while the integration

does for only k = K.

function [J,JJ] = Jkb(K,beta) %the 1st kind of kth-order Bessel ftn

tmpk = ones(size(beta));

for k = 0:K

tmp = tmpk; JJ(k + 1,:) = tmp;

for m = 1:100

tmp = ?????????????????????;

JJ(k + 1,:) = JJ(k + 1,:)+ tmp;

if norm(tmp)<.001, break; end

end

tmpk = tmpk.*beta/2/(k + 1);

end

J = JJ(K+1,:);

%nm1p21b: Bessel_ftn

clear, clf

beta = 0:.05:15; K = 15;

tic

for i = 1:length(beta) %Integration

J151(i) = quad(’Bessel_integrand’,0,pi,[],0,beta(i),K)/pi;

end

toc

tic, J152 = Jkb(K,beta); toc %Recursive Computation

discrepancy = norm(J151-J152)

1.22 Find the four routines in Chapter 5 and 7, which are fabricated in a nested

(recursive calling) structure.

(cf) Don’t those algorithms, which are the souls of the routines, seem to have been

born to be in a nested structure?

1.23 Avoiding Runtime Error in Case of Deficient/Nonadmissible Input Arguments

(a) Consider the MATLAB routine “rotation_r(x,M)”, which you made

in Problem 1.15(h). Does it work somehow when the user gives a

negative integer as the second input argument M ? If not, add a statement

so that it performs the rotation left by −M samples for M < 0, say,

making

rotate_r([1 2 3 4 5],-2) = [3 4 5 1 2]

(b) Consider the routine ‘trpzds(f,a,b,N)’ in Section 5.6, which computes

the integral of function f over [a, b] by dividing the integration

interval into N sections and applying the trapezoidal rule. If the user

tries to use it without the fourth input argument N, will it work? If not,

make it work with N = 1000 by default even without the fourth input

argument N.

68 MATLAB USAGE AND COMPUTATIONAL ERRORS

function INTf = trpzds(f,a,b,N)

%integral of f(x) over [a,b] by trapezoidal rule with N segments

if abs(b – a) < eps | N <= 0, INTf = 0; return; end

h = (b – a)/N; x = a+[0:N]*h;

fx = feval(f,x); %values of f for all nodes

INTf = h*((fx(1)+ fx(N + 1))/2 + sum(fx(2:N))); %Eq.(5.6.1)

1.24 Parameter Passing through varargin

Consider the integration routine ‘trpzds(f,a,b,N)’ in Section 5.6. Can

you apply it to compute the integral of a function with some parameter(

s), like the ‘Bessel_integrand(x,beta,k)’ that you defined in Problem

1.21? If not, modify it so that it works for a function with some parameter(

s) (see Section 1.3.6) and save it in the M-file named ‘trpzds_par.m’.

Then replace the ‘quad()’ statement in the program ‘nm1p21b’ (introduced

in P1.21) by an appropriate ‘trpzds_par()’ statement (with N = 1000) and

run the program. What is the discrepancy between the integration results

obtained by this routine and the recursive computation based on Problem

1.21.4(b)? Is it comparable with that obtained with ‘quad()’? How do you

compare the running time of this routine with that of ‘quad()’? Why do

you think it takes so much time to execute the ‘quad()’ routine?

1.25 Adaptive Input Argument to Avoid Runtime Error in the Case of Different

Input Arguments

Consider the integration routine ‘trpzds(f,a,b,N)’ in Section 5.6. If some

user tries to use this routine with the following statement, will it

work?

trpzds(f,[a b],N) or trpzds(f,[a b])

If not, modify it so that it works for such a usage (with a bound vector as

the second input argument) as well as for the standard usage and save it in

the M-file named ‘trpzds_bnd.m’. Then try it to find the intergal of e−t

for [0,100] by typing the following statements in the MATLAB command

window. What did you get?

>>ftn=inline(’exp(-t)’,’t’);

>>trpzds_bnd(ftn,[0 100],1000)

>>trpzds_bnd(ftn,[0 100])

1.26 CtFT(Continuous-Time Fourier Transform) of an Arbitrary Signal

Consider the following definitions of CtFT and ICtFT(Inverse CtFT) [W-4]:

X(ω) = F{x(t)} = ∞

−∞

x(t)e−jωt dt: CtFT (P1.26.1a)

x(t) = F−1{X(ω)} =

1

2π ∞

−∞

X(ω)ejωtdω: ICtFT (P1.26.1b)

PROBLEMS 69

(a) Similarly to the MATLAB routine “CtFT1(x,Dt,w)” computing the

CtFT (P1.26.1a) of x(t) over [-Dt,Dt ] for w, make a MATLAB routine

“ICtFT1(X,Bw,t)” computing the ICtFT (P1.26.1b) of X(w) over

[-Bw, Bw] for t. You can choose whatever integral routine including

‘trpzds_par()’ (Problem 1.24) and ‘quad()’, considering the running

time.

(b) The following program ‘nm1p26’ finds the CtFT of a rectangular pulse

(with duration [−1,1]) defined by ‘rDt()’ for ω = [−6π,+6π] and the

ICtFT of a sinc spectrum (with bandwidth 2π) defined by ‘sincBw()’

for t = [−5,+5]. After having saved the routines into M-files with the

appropriate names, run the program to see the rectangular pulse, its

CtFT spectrum, a sinc spectrum, and its ICtFT. If it doesen’t work,

modify/supplement the routines so that you can rerun it to see the

signals and their spectra.

function Xw = CtFT1(x,Dt,w)

x_ejkwt = inline([x ’(t).*exp(-j*w*t)’],’t’,’w’);

Xw = trpzds_par(x_ejkwt,-Dt,Dt,1000,w);

%Xw = quad(x_ejkwt,-Dt,Dt,[],0,w);

function xt = ICtFT1(X,Bw,t)

function x = rDt(t)

x = (-D/2 <= t & t <= D/2);

function X = sincBw(w)

X = 2*pi/B*sinc(w/B);

%nm1p26: CtFT and ICtFT

clear, clf

global B D

%CtFT of a Rectangular Pulse Function

t = [-50:50]/10; %time vector

w = [-60:60]/10*pi; %frequency vector

D = 1; %Duration of a rectangular pulse rD(t)

for k = 1:length(w), Xw(k) = CtFT1(’rDt’,D*5,w(k)); end

subplot(221), plot(t,rDt(t))

subplot(222), plot(w,abs(Xw))

%ICtFT of a Sinc Spectrum

B = 2*pi; %Bandwidth of a sinc spectrum sncB(w)

for n = 1:length(t), xt(n) = ICtFT1(’sincBw’,B*5,t(n));

end

subplot(223), plot(t,real(xt))

subplot(224), plot(w,sincBw(w))

1:45 am, 6/4/05

2

SYSTEM OF LINEAR

EQUATIONS

In this chapter, we deal with several numerical schemes for solving a system of

equations

a11x1 + a12x2 + ·· ·+a1NxN = b1

a21x1 + a22x2 + ·· ·+a2NxN = b2

. . . . . . . . . = .

aM1x1 + aM2x2 + ·· ·+aMNxN = bM

(2.0.1a)

which can be written in a compact form by using a matrix–vector notation as

AM×Nx = b (2.0.1b)

where

AM×N =

a11 a12 · · a1N

a21 a22 · · a2N

· · · · · aM1 aM2 · · aMN

, x =

x1

x2

· xN

, b =

b1

b2

· bM

We will deal with the three cases:

(i) The case where the number (M) of equations and the number (N) of

unknowns are equal (M = N) so that the coefficient matrix AM×N is

square.

Applied Numerical Methods Using MATLAB, by Yang, Cao, Chung, and Morris

Copyright 2005 John Wiley & Sons, Inc., ISBN 0-471-69833-4

71

72 SYSTEM OF LINEAR EQUATIONS

(ii) The case where the number (M) of equations is smaller than the number

(N) of unknowns (M <N) so that we might have to find the minimumnorm

solution among the numerous solutions.

(iii) The case where the number of equations is greater than the number of

unknowns (M >N) so that there might exist no exact solution and we

must find a solution based on global error minimization, like the “LSE

(Least-squares error) solution.”

2.1 SOLUTION FOR A SYSTEM OF LINEAR EQUATIONS

2.1.1 The Nonsingular Case (M = N)

If the number (M) of equations and the number (N) of unknowns are equal

(M = N), then the coefficient matrix A is square so that the solution can be

written as

x = A−1 b (2.1.1)

so long as the matrix A is not singular. There are MATLAB commands for

this job.

>>A = [1 2;3 4]; b = [-1;-1];

>>x = A^-1*b %or, x = inv(A)*b

x = 1.0000

-1.0000

What if A is square, but singular?

>>A = [1 2;2 4]; b = [-1;-1];

>>x = A^-1*b

Warning: Matrix is singular to working precision.

x = -Inf

-Inf

This is the case where some or all of the rows of the coefficient matrix A are

dependent on other rows and so the rank of A is deficient, which implies that

there are some equations equivalent to or inconsistent with other equations. If

we remove the dependent rows until all the (remaining) rows are independent of

each other so that A has full rank (equal to M), it leads to the case of M <N,

which will be dealt with in the next section.

2.1.2 The Underdetermined Case (M < N): Minimum-Norm Solution

If the number (M) of equations is less than the number (N) of unknowns, the

solution is not unique, but numerous. Suppose the M rows of the coefficient

matrix A are independent. Then, any N-dimensional vector can be decomposed

into two components

x = x+ + x− (2.1.2)

SOLUTION FOR A SYSTEM OF LINEAR EQUATIONS 73

where the one is in the row space R(A) of A that can be expressed as a linear

combination of the M row vectors

x+ = AT α (2.1.3)

and the other is in the null space N(A) orthogonal(perpendicular) to the row

space1 so that

Ax− = 0 (2.1.4)

Substituting the arbitrary N-dimensional vector representation (2.1.2) into

Eq. (2.0.1) yields

A(x+ + x−) = AAT α + Ax− (2.1.4) = AAT α = b (2.1.5)

Since AAT is supposedly a nonsingular M × M matrix resulting from multiplying

an M × N matrix by an N ×M matrix, we can solve this equation for α to get

αo = [AAT ]−1b (2.1.6)

Then, substituting Eq. (2.1.6) into Eq. (2.1.3) yields

xo+ (2.1.3) = AT αo (2.1.6) = AT [AAT ]−1b (2.1.7)

This satisfies Eq. (2.0.1) and thus qualifies as its solution. However, it is far

from being a unique solution because the addition of any vector x− (in the

null space) satisfying Eq. (2.1.4) to xo+ still satisfies Eq. (2.0.1) [as seen from

Eq. (2.1.5)], yielding infinitely many solutions.

Based on the principle that any one of the two perpendicular legs is shorter

than the hypotenuse in a right-angled triangle, Eq. (2.1.7) is believed to represent

the minimum-norm solution. Note that the matrix AT [AAT ]−1 is called the right

pseudo- (generalized) inverse of A (see item 2 in Remark 1.1).

MATLAB has the pinv() command for obtaining the pseudo-inverse. We

can use this command or the slash(/) operator to find the minimum-norm solution

(2.1.7) to the system of linear equations (2.0.1).

>>A = [1 2]; b = 3;

>>x = pinv(A)*b %x = A’*(A*A’)^ – 1*b or eye(size(A,2))/A*b, equivalently

x = 0.6000

1.2000

Remark 2.1. Projection Operator and Minimum-Norm Solution

1. The solution (2.1.7) can be viewed as the projection of an arbitrary solution

xo onto the row space R(A) of the coefficient matrix A spanned by the

1 See the website @http://www.psc.edu/∼burkardt/papers/linear glossary.html

74 SYSTEM OF LINEAR EQUATIONS

row vectors. The remaining component of the solution xo

xo− = xo − xo+ = xo − AT [AAT ]−1b = xo − AT [AAT ]−1Axo

= [I − AT [AAT ]−1A]xo

is in the null space N(A), since it satisfies Eq. (2.1.4). Note that

PA = [I − AT [AAT ]−1A]

is called the projection operator.

2. The solution (2.1.7) can be obtained by applying the Lagrange multiplier

method (Section 7.2.1) to the constrained optimization problem in which

we must find a vector x minimizing the (squared) norm ||x||2 subject to the

equality constraint Ax = b.

Min l(x, λ)

Eq.(7.2.2) = 12

||x||2 − λT (Ax − b) = 12

xT x − λT (Ax − b)

By using Eq. (7.2.3), we get

∂

∂x

J = x − AT λ = 0; x = AT λ = AT [AAT ]−1b

∂

∂λ

J = Ax − b = 0; AAT λ = b; λ = [AAT ]−1b

Example 2.1. Minimum-Norm Solution. Consider the problem of solving the

equation

[1 2] x1

x2 = 3; Ax = b, where A = [1 2], b = 3 (E2.1.1)

This has infinitely many solutions and any x = [ x1 x2 ]T satisfying this

equation, or, equivalently,

x1 + 2×2 = 3; x2 = −

1

2

x1 +

3

2

(E2.1.2)

is a qualified solution. Equation (E2.1.2) describes the solution space as depicted

in Fig. 2.1.

On the other hand, any vector in the row space of the coefficient matrix A

can be expressed by Eq. (2.1.3) as

x+ = AT α = 1

2 α (αis a scalar, since M = 1) (E2.1.3)

SOLUTION FOR A SYSTEM OF LINEAR EQUATIONS 75

1 2 3

(3, 0)

solution

space

row

space

(A)

x+

(3/5, 6/5) = xo+

xo−

x−

x1

x2

(1, 1) = xo

1

2

1.5

null

space

0

(A)

Figure 2.1 A minimum-norm solution.

and any vector in the null space of A can be expressed by Eq. (2.1.4) as

Ax− = [1 2] x−1

x−2 = 0; x−2 = −

1

2

x−1 (E2.1.4)

We use Eq. (2.1.7) to obtain the minimum-norm solution

xo+ = AT [AAT ]−1b = 1

2 [1 2] 1

2

−1

3 =

3

5 1

2 = 0.6

1.2 (E2.1.5)

Note from Fig. 2.1 that the minimum-norm solution xo+ is the intersection of

the solution space and the row space and is the closest to the origin among the

vectors in the solution space.

2.1.3 The Overdetermined Case (M > N): LSE Solution

If the number (M) of (independent) equations is greater than the number (N)

of unknowns, there exists no solution satisfying all the equations strictly. Thus

we try to find the LSE (least-squares error) solution minimizing the norm of the

(inevitable) error vector

e = Ax − b (2.1.8)

Then, our problem is to minimize the objective function

J = 12

||e||2 = 12

||Ax − b||2 = 12

[Ax − b]T [Ax − b] (2.1.9)

76 SYSTEM OF LINEAR EQUATIONS

whose solution can be obtained by setting the derivative of this function (2.1.9)

with respect to x to zero.

∂

∂x

J = AT [Ax − b] = 0; xo = [AT A]−1AT b (2.1.10)

Note that the matrix A having the number of rows greater than the number of

columns (M >N) does not have its inverse, but has its left pseudo (generalized)

inverse [ATA]−1AT as long as A is not rank-deficient—that is, all of its columns

are independent of each other (see item 2 in Remark 1.1). The left pseudo-inverse

matrix can be computed by using the MATLAB command pinv().

The LSE solution (2.1.10) can be obtained by using the pinv() command or

the backslash (\) operator.

>>A = [1; 2]; b = [2.1; 3.9];

>>x = pinv(A)*b %A\b or x = (A’*A)^-1*A’*b, equivalently

x = 1.9800

function x = lin_eq(A,B)

%This function finds the solution to Ax = B

[M,N] = size(A);

if size(B,1) ~= M

error(’Incompatible dimension of A and B in lin_eq()!’)

end

if M == N, x = A^-1*B; %x = inv(A)*B or gaussj(A,B); %Eq.(2.1.1)

elseif M < N %Minimum-norm solution (2.1.7)

x = pinv(A)*B; %A’*(A*A’)^-1*B; or eye(size(A,2))/A*B

else %LSE solution (2.1.10) for M > N

x = pinv(A)*B; %(A’*A)^-1*A’*B or x = A\B

end

The above MATLAB routine lin_eq() is designed to solve a given set of

equations, covering all of the three cases in Sections 2.1.1, 2.1.2, and 2.1.3.

(cf) The power of the pinv() command is beyond our imagination as you might have

felt in Problem 1.14. Even in the case of M <N, it finds us a LS solution if the

equations are inconsistent. Even in the case of M >N, it finds us a minimum-norm

solution if the equations are redundant. Actually, the three cases can be dealt with

by a single pinv() command in the above routine.

2.1.4 RLSE (Recursive Least-Squares Estimation)

In this section we will see the so-called RLSE (Recursive Least-Squares Estimation)

algorithm, which is a recursive method to compute the LSE solution.

Suppose we know the theoretical relationship between the temperature t [◦] and

SOLUTION FOR A SYSTEM OF LINEAR EQUATIONS 77

the resistance R[] of a resistor as

c1t + c2 = R

and we have lots of experimental data {(t1,R1), (t2,R2), . . . , (tk,Rk)} collected

up to time k. Since the above equation cannot be satisfied for all the data with any

value of the parameters c1 and c2, we should try to get the parameter estimates

that are optimal in some sense. This corresponds to the overdetermined case dealt

with in the previous section and can be formulated as an LSE problem that we

must solve a set of linear equations

Akxk ≈ bk, where Ak =

t1 1

t2 1

· · tk 1

, xk = c1,k

c2,k , and bk =

R1

R2

·R

k

for which we can apply Eq. (2.1.10) to get the solution as

xk = [AT

k Ak]−1AT

k bk (2.1.11)

Now, we are given a new experimental data (tk+1, Rk+1) and must find the

new parameter estimate

xk+1 = [AT

k+1Ak+1]−1AT

k+1bk+1 (2.1.12)

with

Ak+1 =

t1 1

· · tk 1

tk+1 1

, xk+1 = c1,k+1

c2,k+1 , and bk+1 =

R1

·R

k

Rk+1

How do we compute this? If we discard the previous estimate xk and make direct

use of Eq. (2.1.12) to compute the next estimate xk+1 every time a new data pair

is available, the size of matrix A will get bigger and bigger as the data pile up,

eventually defying any powerful computer in this world.

How about updating the previous estimate by just adding the correction term

based on the new data to get the new estimate? This is the basic idea of the

RLSE algorithm, which we are going to trace and try to understand. In order to

do so, let us define the notations

Ak+1 = Ak

aT

k+1 , ak+1 = tk+1

1 , bk+1 = bk

Rk+1 , and Pk = [AT

k Ak]−1

(2.1.13)

78 SYSTEM OF LINEAR EQUATIONS

and see how the inverse matrix Pk is to be updated on arrival of the new data

(tk+1, Rk+1).

Pk+1 = [AT

k+1Ak+1]−1 = [AT

k ak+1 ] Ak

aT

k+1 −1

= [AT

k Ak + ak+1aT

k+1]−1 = [P−1

k + ak+1aT

k+1]−1 (2.1.14)

(Matrix Inversion Lemma in Appendix B)

Pk+1 = Pk − Pkak+1[aT

k+1Pkak+1 + 1]−1aT

k+1Pk (2.1.15)

It is interesting that [aT

k+1Pkak+1 + 1] is nothing but a scalar and so we do

not need to compute the matrix inverse thanks to the Matrix Inversion Lemma

(Appendix B). It is much better in the computational aspect to use the recursive

formula (2.1.15) than to compute [AT

k+1Ak+1]−1 directly. We can also write Eq.

(2.1.12) in a recursive form as

xk+1

(2.1.12, 14) = Pk+1AT

k+1bk+1

(2.1.13) = Pk+1[AT

k ak+1] bk

Rk+1

= Pk+1[AT

k bk + ak+1Rk+1] (2.1.11) = Pk+1[AT

k Akxk + ak+1Rk+1]

(2.1.13) = Pk+1[(AT

k+1Ak+1 − ak+1aT

k+1)xk + ak+1Rk+1]

(2.1.13) = Pk+1[P−1

k+1xk − ak+1aT

k+1xk + ak+1Rk+1]

xk+1 = xk + Pk+1ak+1(Rk+1 − aT

k+1xk) (2.1.16)

We can use Eq. (2.1.15) to rewrite the gain matrix Pk+1ak+1 premultiplied by

the ‘error’ to make the correction term on the right-hand side of Eq. (2.1.16) as

Kk+1 = Pk+1ak+1

(2.1.15) = [Pk − Pkak+1[aT

k+1Pkak+1 + 1]−1aT

k+1Pk]ak+1

= Pkak+1[I − [aT

k+1Pkak+1 + 1]−1aT

k+1Pkak+1]

= Pkak+1[aT

k+1Pkak+1 + 1]−1{[aT

k+1Pkak+1 + 1] − aT

k+1Pkak+1}

Kk+1 = Pkak+1[aT

k+1Pkak+1 + 1]−1 (2.1.17)

and substitute this back into Eq. (2.1.15) to write it as

Pk+1 = Pk − Kk+1aT

k+1Pk (2.1.18)

The following MATLAB routine “rlse_online()” implements this RLSE

(Recursive Least-Squares Estimation) algorithm that updates the parameter

estimates by using Eqs. (2.1.17), (2.1.16), and (2.1.18). The MATLAB program

SOLVING A SYSTEM OF LINEAR EQUATIONS 79

“do_rlse.m” updates the parameter estimates every time new data arrive and

compares the results of the on-line processing with those obtained by the off-line

(batch job) processing—that is, by using Eq.(2.1.12) directly. Noting that

ž the matrix [AT

k Ak] as well as bk consists of information and is a kind of

squared matrix that is nonnegative, and

ž [AT

k Ak] will get larger, or, equivalently, Pk = [AT

k Ak]−1 will get smaller and,

consequently, the gain matrix Kk will get smaller as valuable information

data accumulate,

one could understand that Pk is initialized to a very large identity matrix, since

no information is available in the beginning. Since a large/small Pk makes the

correction term on the right-hand side of Eq. (2.1.16) large/small, the RLSE

algorithm becomes more conservative and reluctant to learn from the new data

as the data pile up, while it is willing to make use of the new data for updating

the estimates when it is hungry for information in the beginning.

function [x,K,P] = rlse_online(aT_k1,b_k1,x,P)

K = P*aT_k1’/(aT_k1*P*aT_k1’+1); %Eq.(2.1.17)

x = x +K*(b_k1-aT_k1*x); %Eq.(2.1.16)

P = P-K*aT_k1*P; %Eq.(2.1.18)

%do_rlse

clear

xo = [2 1]’; %The true value of unknown coefficient vector

NA = length(xo);

x = zeros(NA,1); P = 100*eye(NA,NA);

for k = 1:100

A(k,:) = [k*0.01 1];

b(k,:) = A(k,:)*xo +0.2*rand;

[x,K,P] = rlse_online(A(k,:),b(k,:),x,P);

end

x % the final parameter estimate

A\b % for comparison with the off-line processing (batch job)

2.2 SOLVING A SYSTEM OF LINEAR EQUATIONS

2.2.16 Gauss Elimination

For simplicity, we assume that the coefficient matrix A in Eq. (2.0.1) is a nonsingular

3 ×3 matrix with M = N = 3. Then we can write the equation as

a11x1 + a12x2 + a13x3 = b1 (2.2.0a)

a21x1 + a22x2 + a23x3 = b2 (2.2.0b)

a31x1 + a32x2 + a33x3 = b3 (2.2.0c)

80 SYSTEM OF LINEAR EQUATIONS

First, to remove the x1 terms from equations (2.2.0.m) other than (2.2.0.a), we

subtract (2.2.0a)×am1/a11 from each of them to get

a(0)

11 x1 + a(0)

12 x2 + a(0)

13 x3 = b(0)

1 (2.2.1a)

a(1)

22 x2 + a(1)

23 x3 = b(1)

2 (2.2.1b)

a(1)

32 x2 + a(1)

33 x3 = b(1)

3 (2.2.1c)

with

a(0)

mn = amn, b(0)

m = bm for m, n = 1, 2, 3 (2.2.2a)

a(1)

mn = a(0)

mn − (a(0)

m1/a(0)

11 )a(0)

1n, b(1)

m = b(0)

m − (a(0)

m1/a(0)

11 )b(0)

1 for m, n = 2, 3

(2.2.2b)

We call this work ‘pivoting at a11’ and call the center element a11 a ‘pivot’.

Next, to remove the x2 term from Eq. (2.2.1c) other than (2.2.1a,b), we subtract

(2.2.1b)×a(1)

m2/a(1)

22 (m = 3) from it to get

a(0)

11 x1 + a(0)

12 x2 + a(0)

13 x3 = b(0)

1 (2.2.3a)

a(1)

22 x2 + a(1)

23 x3 = b(1)

2 (2.2.3b)

a(2)

33 x3 = b(2)

3 (2.2.3c)

with

a(2)

mn = a(1)

mn − (a(1)

m2/a(1)

22 )a(1)

2n, b(2)

m = b(1)

m − (a(1)

m2/a(1)

22 )b(1)

2 for m, n = 3

(2.2.4)

We call this procedure ‘Gauss forward elimination’ and can generalize the updating

formula (2.2.2)/(2.2.4) as

a(k)

mn = a(k−1)

mn − (a(k−1)

mk /a(k−1)

kk )a(k−1)

kn for m, n = k + 1, k + 2, . . .,M (2.2.5a)

b(k)

m = b(k−1)

m − (a(k−1)

mk /a(k−1)

kk )b(k−1)

k for m = k + 1, k + 2, . . .,M (2.2.5b)

After having the triangular matrix–vector equation as Eq. (2.2.3), we can solve

Eq. (2.2.3c) first to get

x3 = b(2)

3 /a(2)

33 (2.2.6a)

and then substitute this result into Eq. (2.2.3b) to get

x2 = (b(1)

2 − a(1)

23 x3)/a(1)

22 (2.2.6b)

Successively, we substitute Eqs. (2.2.6a,b) into Eq.(2.2.3a) to get

x1 = b(0)

1 −

3

n=2

a(0)

1n xn

/a(0)

11 (2.2.6c)

SOLVING A SYSTEM OF LINEAR EQUATIONS 81

We call this procedure ‘backward substitution’ and can generalize the solution

formula (2.2.6) as

xm = b(m−1)

m −

M

n=m+1

a(m−1)

mn xn

/a(m−1)

mm for m = M,M − 1, . . . , 1

(2.2.7)

In this way, the Gauss elimination procedure consists of two steps, namely,

forward elimination and backward substitution. Noting that

ž this procedure has nothing to do with the specific values of the unknown

variable xm’s and involves only the coefficients, and

ž the formulas (2.2.5a) on the coefficient matrix A and (2.2.5b) on the RHS

(right-hand side) vector b conform with each other,

we will augment A with b and put the formulas (2.2.5a,b) together into one

framework when programming the Gauss forward elimination procedure.

2.2.2 Partial Pivoting

The core formula (2.2.5) used for Gauss elimination requires division by a(k−1)

kk

at the kth stage, where a(k−1)

kk is the diagonal element in the kth row. What if

a(k−1)

kk = 0? In such a case, it is customary to switch the kth row and another row

below it having the element of the largest absolute value in the kth column. This

procedure, called ‘partial pivoting’, is recommended for reducing the round-off

error even in the case where the kth pivot a(k−1)

kk is not zero.

Let us consider the following example:

0 1 1

2 −1 −1

1 1 −1

x1

x2

x3

=

b1 = 2

b2 = 0

b3 = 1

(2.2.8)

We construct the augmented matrix by combining the coefficient matrix and the

RHS vector to write

a11 a12 a13 b1

a21 a22 a23 b2

a31 a32 a33 b3

=

0 1 1 2

2 −1 −1 0

1 1 −1 1

: r1

: r2

: r3

(2.2.9)

and apply the Gauss elimination procedure.

In the stage of forward elimination, we want to do pivoting at a11, but a11

cannot be used as the pivoting element because it is zero. So we switch the first

row and the second row having the element of the largest absolute value in the

first column.

82 SYSTEM OF LINEAR EQUATIONS

a(1)

11 a(1)

12 a(1)

13 b(1)

1

a(1)

21 a(1)

22 a(1)

23 b(1)

2

a(1)

31 a(1)

32 a(1)

33 b(1)

3

=

2 −1 −1 0

0 1 1 2

1 1 −1 1

: r(1)

1

: r(1)

2

: r(1)

3

(2.2.10a)

Then we do pivoting at a(1)

11 by applying Eq. (2.2.2) to get

r(1)

1 →

r(1)

2 − a(1)

21 /a(1)

11 × r(1)

1 →

r(1)

3 − a(1)

31 /a(1)

11 × r(1)

1 →

a(2)

11 a(2)

12 a(2)

13 b(2)

1

a(2)

21 a(2)

22 a(2)

23 b(2)

2

a(2)

31 a(2)

32 a(2)

33 b(2)

3

=

2 −1 −1 0

0 1 1 2

0 3/2 −1/2 1

: r(2)

1

: r(2)

2

: r(2)

3

(2.2.10b)

Here, instead of pivoting at a(2)

22 , we switch the second row and the third row

having the element of the largest absolute value among the elements not above

a(2)

22 in the second column.

a(3)

11 a(3)

12 a(3)

13 b(3)

1

a(3)

21 a(3)

22 a(3)

23 b(3)

2

a(3)

31 a(3)

32 a(3)

33 b(3)

3

=

2 −1 −1 0

0 3/2 −1/2 1

0 1 1 2

: r(3)

1

: r(3)

2

: r(3)

3

(2.2.10c)

And we do pivoting at a(3)

22 by applying Eq. (2.2.4)—more generally, Eq.

(2.2.5)—to get the upper-triangularized form:

r(3)

1 →

r(3)

2 →

r(3)

3 − a(3)

31 /a(3)

11 × r(3)

2 →

a(4)

11 a(4)

12 a(4)

13 b(4)

1

a(4)

21 a(4)

22 a(4)

23 b(4)

2

a(4)

31 a(4)

32 a(4)

33 b(4)

3

=

2 −1 −1 0

0 3/2 −1/2 1

0 0 4/3 4/3

: r(4)

1

: r(4)

2

: r(4)

3

(2.2.10d)

Now, in the stage of backward substitution, we apply Eq. (2.2.6), more generally,

Eq. (2.2.7) to get the final solution as

x3 = b(4)

3 /a(4)

33 = (4/3)/(4/3) = 1

x2 = (b(4)

2 − a(4)

23 x3)/a(4)

22 = (1 − (−1/2) × 1)/(3/2) = 1 (2.2.11)

x1 = b(4)

1 −

3

n=2

a(4)

1n xn

/a(4)

11 = (0 − (−1) × 1 − (−1) × 1)/2 = 1

[x1 x2 x3] = [1 1 1] (2.2.12)

SOLVING A SYSTEM OF LINEAR EQUATIONS 83

Let us consider another system of equations.

1 0 1

1 1 1

1 −1 1

x1

x2

x3

=

b1 = 2

b2 = 3

b3 = 1

(2.2.13)

We construct the augmented matrix by combining the coefficient matrix and the

RHS vector to write

a11 a12 a13 b1

a21 a22 a23 b2

a31 a32 a33 b3

=

1 0 1 2

1 1 1 3

1 −1 1 1

: r1

: r2

: r3

(2.2.14)

and apply the Gauss elimination procedure.

First, noting that all the elements in the first column have the same absolute

value and so we don’t need to switch the rows, we do pivoting at a11.

a(1)

11 a(1)

12 a(1)

13 b(1)

1

a(1)

21 a(1)

22 a(1)

23 b(1)

2

a(1)

31 a(1)

32 a(1)

33 b(1)

3

=

1 0 1 2

0 1 0 1

0 −1 0 −1

: r(1)

1

: r(1)

2

: r(1)

3

(2.2.15a)

Second, without having to switch the rows, we perform pivoting at a(1)

22 .

r(1)

1 →

r(1)

2 →

r(1)

3 − a(1)

32 /a(1)

22 × r(1)

2 →

a(2)

11 a(2)

12 a(2)

13 b(2)

1

a(2)

21 a(2)

22 a(2)

23 b(2)

2

a(2)

31 a(2)

32 a(2)

33 b(2)

3

=

1 0 1 2

0 1 0 1

0 0 0 0

:

r

(

2)

1

: r(2)

2

: r(2)

3

(2.2.15b)

Now, we are at the stage of backward substitution, but a(2)

33 , which is supposed

to be the denominator in Eq. (2.2.7), is zero. We may face such a weird situation

of zero division even during the forward elimination process where the pivot is

zero; besides, we cannot find any (nonzero) element below it in the same column

and on its right in the same row except the RHS element. In this case, we

cannot go further. This implies that some or all rows of coefficient matrix A are

dependent on others, corresponding to the case of redundancy (infinitely many

solutions) or inconsistency (no exact solution). Noting that the RHS element

of the zero row in Eq. (2.2.15.2) is also zero, we should declare the case of

redundancy and may have to be satisfied with one of the infinitely many solutions

being the RHS vector as

[x1 x2 x3] = [b(2)

1 b(2)

2 b(2)

3 ] = [2 1 0] (2.2.16)

Furthermore, if we remove the all-zero row(s), the problem can be treated as an

underdetermined case handled in Section 2.1.2. Note that, if the RHS element

were not zero, we would have to declare the case of inconsistency, as will be

illustrated.

Suppose that b1 = 1 in Eq. (2.2.14). Then, the Gauss elimination would have

proceeded as follows:

84 SYSTEM OF LINEAR EQUATIONS

1 0 1 1

1 1 1 3

1 −1 1 1

→

1 0 1 1

0 1 0 2

0 −1 0 0

→

1 0 1 1

0 1 0 2

0 0 0 2

(2.2.17)

This ended up with an all-zero row except the nonzero RHS element, corresponding

to the case of inconsistency. So we must declare the case of ‘no exact

solution’ for this problem.

The following MATLAB routine “gauss()” implements the Gauss elimination

algorithm, and the program “do_gauss” is designed to solve Eq. (2.2.8) by using

“gauss()”. Note that at every pivoting operation in the routine “gauss()”, the

pivot row is divided by the pivot element so that every diagonal element becomes

one and that we don’t need to perform any computation for the kth column at

the kth stage, since the column is supposed to be all zeros but the kth element

a(k)

kk = 1.

function x = gauss(A,B)

%The sizes of matrices A,B are supposed to be NA x NA and NA x NB.

%This function solves Ax = B by Gauss elimination algorithm.

NA = size(A,2); [NB1,NB] = size(B);

if NB1 ~= NA, error(’A and B must have compatible dimensions’); end

N = NA + NB; AB = [A(1:NA,1:NA) B(1:NA,1:NB)]; % Augmented matrix

epss = eps*ones(NA,1);

for k = 1:NA

%Scaled Partial Pivoting at AB(k,k) by Eq.(2.2.20)

[akx,kx] = max(abs(AB(k:NA,k))./ …

max(abs([AB(k:NA,k + 1:NA) epss(1:NA – k + 1)]’))’);

if akx < eps, error(’Singular matrix and No unique solution’); end

mx = k + kx – 1;

if kx > 1 % Row change if necessary

tmp_row = AB(k,k:N);

AB(k,k:N) = AB(mx,k:N);

AB(mx,k:N) = tmp_row;

end

% Gauss forward elimination

AB(k,k + 1:N) = AB(k,k+1:N)/AB(k,k);

AB(k,k) = 1; %make each diagonal element one

for m = k + 1: NA

AB(m,k+1:N) = AB(m,k+1:N) – AB(m,k)*AB(k,k+1:N); %Eq.(2.2.5)

AB(m,k) = 0;

end

end

%backward substitution for a upper-triangular matrix eqation

% having all the diagonal elements equal to one

x(NA,:) = AB(NA,NA+1:N);

for m = NA-1: -1:1

x(m,:) = AB(m,NA + 1:N)-AB(m,m + 1:NA)*x(m + 1:NA,:); %Eq.(2.2.7)

end

%do_gauss

A = [0 1 1;2 -1 -1;1 1 -1]; b = [2 0 1]’; %Eq.(2.2.8)

x = gauss(A,b)

x1 = A\b %for comparison with the result of backslash operation

SOLVING A SYSTEM OF LINEAR EQUATIONS 85

(cf) The number of floating-point multiplications required in this routine ‘gauss()’ is

NA

k=1

{(NA − k + 1)(NA + NB − k) + NA − k + 1} +

NA−1

k=1

(NA − k)NB

=

NA

k=1

k(k + NB − 1) − NB

NA

k=1

k +

NA

k=1

NA · NB

=

1

6

(NA + 1)NA(2NA + 1) −

1

2

NA(NA + 1) + NA2NB

=

1

3

NA(NA + 1)(NA − 1) + NA2NB

≈

1

3

NA3 for NA NB (2.2.18)

where NA is the size of the matrix A, and NB is the column dimension of the RHS

matrix B.

Here are several things to note.

Remark 2.2. Partial Pivoting and Undetermined/Inconsistent Case

1. In Gauss or Gauss–Jordan elimination, some row switching is performed

to avoid the zero division. Even without that purpose, it may be helpful

for reducing the round-off error to fix

Max{|amk|, k ≤ m ≤ M} (2.2.19)

as the pivot element in the kth iteration through some row switching, which

is called ‘partial pivoting.’ Actually, it might be better off to fix

Max |amk|

Max{|amn|, k ≤ n ≤ M}

, k ≤ m ≤ M (2.2.20)

as the pivot element in the kth iteration, which is called ‘scaled partial

pivoting’ or to do column switching as well as row switching for choosing

the best (largest) pivot element, which is called ‘full pivoting.’ Note that

if the columns are switched, the order of the unknown variables should be

interchanged accordingly.

2. What if some diagonal element akk and all the elements below it in the

same column are zero and, besides, all the elements in the row including

akk are also zero except the RHS element? It implies that some or all

rows of the coefficient matrix A are dependent on others, corresponding

to the case of redundancy (infinitely many solutions) or inconsistency (no

86 SYSTEM OF LINEAR EQUATIONS

exact solution). If even the RHS element is zero, it should be declared

to be the case of redundancy. In this case, we can get rid of the all-zero

row(s) and then treat the problem as the underdetermined case handled in

Section 2.1.2. If the RHS element is only one nonzero in the row, it should

be declared to be the case of inconsistency.

Example 2.2. Delicacy of Partial Pivoting. To get an actual feeling about the

delicacy of partial pivoting, consider the following systems of linear equations,

which apparently have xo = [1 1]T as their solutions.

(a) A1x = b1 with A1 = 10−15 1

1 1011 , b1 = 1 + 10−15

1011 + 1 (E2.2.1)

Without any row switching, the Gauss elimination procedure will find us

the true solution only if there is no quantization error.

[A1 b1] = 10−15 1 1+ 10−15

1 1011 1011 + 1

forward

elimination

−−−−−−−→ 1 1015 1015 + 1

0 1011 − 1015 1011 − 1015

backward

substitution

−−−−−−−−→x = 1

1

But, because of the round-off error, it will deviate from the true solution.

forward

elimination

−−−−−−−→

1 1015 = 9.999999999999999e+014 1015 + 1 = 1.000000000000001e+015

0 1011 − 1015 1011 + 1 − (1015 − 1)

= −9.998999999999999e+014 = −9.999000000000000e+014

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

backward

substitution

−−−−−−−−→ x = 8.750000000000000e-001

1.000000000000000e+000

If we enforce the strategy of partial pivoting or scaled partial pivoting, the

Gauss elimination procedure will give us much better result as follows:

[A1 b1]

row swap

−−−−−−→ 1 1011 1011 + 1

10−15 1 1+ 10−15

forward

elimination

−−−−−−−→ 1 1011 = 1.000e+011 1011 + 1 = 1.000000000010000e+011

0 1− 10−4 = 9.999e-001 9.999000000000001e-001

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

backward

substitution

−−−−−−−−→x = 9.999847412109375e-001

1.000000000000000e+000

(b) A2 x = b2 with A2 = 10−14.6 1

1 1015 , b2 = 1 + 10−14.6

1015 + 1 (E2.2.2)

SOLVING A SYSTEM OF LINEAR EQUATIONS 87

Without partial pivoting, the Gauss elimination procedure will give us a

quite good result.

[A1 b1] = 1 1014.6 = 3.981071705534969e+014 1014.6 + 1 = 3.981071705534979e+014

0 6.018928294465030e+014 6.018928294465030e+014

→ 1 3.981071705534969e+014 3.981071705534979e+014

0 1 1

backward

substitution

−−−−−−−−→x = 1

1

But, if we exchange the first row with the second row having the larger

element in the first column according to the strategy of partial pivoting, the

Gauss elimination procedure will give us a rather surprisingly bad result

as follows:

row

swapping

−−−−−−−→ forward

elimination

1 1015 = 1.000000000000000e+015 1015 + 1 = 1.000000000000001e+015

0 1− 1015 · 10−14.6 1 + 10−14.6 − (1 + 1015) · 10−14.6

= −1.5118864315095819 = −1.5118864315095821

backward

substitution

−−−−−−−−→ x = 0.7500000000000000

1.0000000000000002

One might be happy to have the scaled partial pivoting scheme

[Eq. (2.2.20)], which does not switch the rows in this case, since the

relative magnitude (dominancy) of a11 in the first row is greater than that

of a21 in the second row, that is, 10−14.6/1 > 1/1015.

(c) A3x = b3 with A3 = 1015 1

1 10−14.6 , b3 = 1015 + 1

1 + 10−14.6 (E2.2.3)

With any pivoting scheme, we don’t need to switch the rows, since the

relative magnitude as well as the absolute magnitude of a11 in the first row

is greater than those of a21 in the second row. Thus, the Gauss elimination

procedure will go as follows:

forward

elimination

−−−−−−−→ 1 1.000000000000000e-015 1.000000000000001e+000

0 1.511886431509582e-015 1.332267629550188e-015

backward

substitution

−−−−−−−−→ x = 1.000000000000000

0.811955724875121

(cf) Note that the coefficient matrix, A3 is the same as would be obtained by applying

the full pivoting scheme for A2 to have the largest pivot element. This example

implies that the Gauss elimination with full pivoting scheme may produce a worse

result than would be obtained with scaled partial pivoting scheme. As a matter of

88 SYSTEM OF LINEAR EQUATIONS

factor, we cannot say that some pivoting scheme always yields better solution than

other pivoting schemes, because the result depends on the random round-off error as

well as the pivoting scheme (see Problem 2.2). But, in most cases, the scaled partial

pivoting shows a reasonably good performance and that is why we adopt it in our

routine “gauss()”.

Remark 2.3. Computing Error, Singularity, and Ill-Condition

1. As the size of the matrix grows, the round-off errors are apt to accumulate

and propagated in matrix operations to such a degree that zero may

appear to be an absolutely small number, or a nonzero number very close

to zero may appear to be zero. Therefore, it is not so simple a task to

determine whether a zero or a number very close to zero is a real zero or

not.

2. It is desirable, but not so easy, for us to discern the case of singularity

from the case of ill-condition and to distinguish the case of redundancy

from the case of inconsistency. In order to be able to give such a qualitative

judgment in the right way based on some quantitative analysis,

we should be equipped with theoretical knowledge as well as practical

experience.

3. There are several criteria by which we judge the degree of ill-condition,

such as how discrepant AA−1 is with the identity matrix, how far

det{A}det{A−1} stays away from one(1), and so on:

AA−1 ?=

I, [A−1]−1 ?=

A, det(A)det(A−1)

?=

1 (2.2.21)

The MATLAB command cond() tells us the degree of ill-condition for a

given matrix by the size of the condition number, which is defined as

cond(A) = ||A||||A−1|| with ||A|| = largest eigenvalue of AT A,

i.e., largest singular value of A

Example 2.3. The Hilbert matrix defined by

A = [amn] = 1

m + n − 1 (E2.3)

is notorious for its ill-condition.

We increase the dimension of the Hilbert matrix from N = 7 to 12 and make

use of the MATLAB commands cond() and det() to compute the condition

number and det(A)det(A−1) in the MATLAB program “do_condition”. Especially

for N = 10, we will see the degree of discrepancy between AA−1 and

SOLVING A SYSTEM OF LINEAR EQUATIONS 89

the identity matrix. Note that the number RCOND following the warning message

about near-singularity or ill-condition given by MATLAB is a reciprocal condition

number, which can be computed by the rcond() command and is supposed

to get close to 1/0 for a well-/badly conditioned matrix.

%do_condition.m

clear

for m = 1:6

for n = 1:6

A(m,n) = 1/(m+n-1); %A = hilb(6), Eq.(E2.3)

end

end

for N = 7:12

for m = 1:N, A(m,N) = 1/(m + N – 1); end

for n = 1:N – 1, A(N,n) = 1/(N + n – 1); end

c = cond(A); d = det(A)*det(A^- 1);

fprintf(’N = %2d: cond(A) = %e, det(A)det(A^ – 1) = %8.6f\n’, N, c, d);

if N == 10, AAI = A*A^ – 1, end

end

>>do_condition

N = 7: cond(A) = 4.753674e+008, det(A)det(A^-1) = 1.000000

N = 8: cond(A) = 1.525758e+010, det(A)det(A^-1) = 1.000000

N = 9: cond(A) = 4.931532e+011, det(A)det(A^-1) = 1.000001

N = 10: cond(A) = 1.602534e+013, det(A)det(A^-1) = 0.999981

AAI =

1.0000 0.0000 -0.0001 -0.0000 0.0002 -0.0005 0.0010 -0.0010 0.0004 -0.0001

0.0000 1.0000 -0.0001 -0.0000 0.0002 -0.0004 0.0007 -0.0007 0.0003 -0.0001

0.0000 0.0000 1.0000 -0.0000 0.0002 -0.0004 0.0006 -0.0006 0.0003 -0.0000

0.0000 0.0000 -0.0000 1.0000 0.0001 -0.0003 0.0005 -0.0006 0.0003 -0.0000

0.0000 0.0000 -0.0000 -0.0000 1.0001 -0.0003 0.0005 -0.0005 0.0002 -0.0000

0.0000 0.0000 -0.0000 -0.0000 0.0001 0.9998 0.0004 -0.0004 0.0002 -0.0000

0.0000 0.0000 -0.0000 -0.0000 0.0001 -0.0002 1.0003 -0.0004 0.0002 -0.0000

0.0000 0.0000 -0.0000 -0.0000 0.0001 -0.0002 0.0003 0.9997 0.0002 -0.0000

0.0000 0.0000 -0.0000 -0.0000 0.0001 -0.0001 0.0003 -0.0003 1.0001 -0.0000

0.0000 0.0000 -0.0000 -0.0000 0.0001 -0.0002 0.0003 -0.0003 0.0001 1.0000

N = 11: cond(A) =5.218389e+014, det(A)det(A^-1) = 1.000119

Warning: Matrix is close to singular or badly scaled.

Results may be inaccurate. RCOND = 3.659249e-017.

> In C:\MATLAB\nma\do_condition.m at line 12

N = 12: cond(A) =1.768065e+016, det(A)det(A^-1) = 1.015201

2.2.3 Gauss–Jordan Elimination

While Gauss elimination consists of forward elimination and backward substitution

as explained in Section 2.2.1, Gauss–Jordan elimination consists of

forward/backward elimination, which makes the coefficient matrix A an identity

matrix so that the resulting RHS vector will appear as the solution.

90 SYSTEM OF LINEAR EQUATIONS

For simplicity, we start from the triangular matrix–vector equation (2.2.3)

obtained by applying the forward elimination:

a(0)

11 a(0)

12 a(0)

13 b(0)

1

0 a(1)

22 a(1)

23 b(1)

2

0 0 a(2)

33 b(2)

3

(2.2.22)

First, we divide the last row by a(2)

33

a(0)

11 a(0)

12 a(0)

13 b(0)

1

0 a(1)

22 a(1)

23 b(1)

2

0 0 a

[1]

33 = 1 b

[1]

3 = b(2)

3 /a(2)

33

(2.2.23)

and subtract (the third row ×a(m−1)

m3 (m = 1, 2)) from the above two rows to get

a(0)

11 a(0)

12 a

[1]

13 = 0 b

[1]

1 = b(0)

1 − a(0)

13 b

[1]

3

0 a(1)

22 a

[1]

23 = 0 b

[1]

2 = b(1)

2 − a(1)

23 b

[1]

3

0 0 a

[1]

33 = 1 b

[1]

3

(2.2.24)

Now, we divide the second row by a(1)

22 :

a(0)

11 a(0)

12 0 b

[1]

1

0 a

[2]

22 =1 0 b

[2]

2 = b

[1]

2 /a

[1]

22

0 0 a

[1]

33 = 1 b

[1]

3

(2.2.25)

and subtract (the second row ×a(m−1)

m2 (m = 1)) from the above first row to get

a(0)

11 0 0 b

[2]

1 = b

[1]

1 − a(0)

12 b

[2]

2

0 1 0 b

[2]

2

0 0 1 b

[1]

3

(2.2.26)

Lastly, we divide the first row by a(0)

11 to get

1 0 0 b

[3]

1 = b

[2]

1 /a(0)

11

0 1 0 b

[2]

2

0 0 1 b

[1]

3

(2.2.27)

SOLVING A SYSTEM OF LINEAR EQUATIONS 91

which denotes a system of linear equations having an identity matrix as the

coefficient matrix

I x = b[] = [ b

[3]

1 b

[2]

2 b

[1]

3 ]T

and, consequently, take the RHS vector b[] as the final solution.

Note that we don’t have to distinguish the two steps, the forward/backward

elimination. In other words, during the forward elimination, we do the pivoting

operations in such a way that the pivot becomes one and other elements

above/below the pivot in the same column become zeros.

Consider the following system of linear equations:

−1 −2 2

1 1 −1

1 2 −1

x1

x2

x3

=

−1

1

2

(2.2.28)

We construct the augmented matrix by combining the coefficient matrix and the

RHS vector to write

a11 a12 a13 b1

a21 a22 a23 b2

a31 a32 a33 b3

=

−1 −2 2 −1

1 1 −1 1

1 2 −1 2

: r1

: r2

: r3

(2.2.29)

and apply the Gauss–Jordan elimination procedure.

First, we divide the first row r1 by a11 = −1 to make the new first row r(1)

1

have the pivot a(1)

11 = 1 and subtract am1 × r(1)

1 (m = 2, 3) from the second and

third row r2 and r3 to get

r1 ÷ (−1) →

r2 − 1 × r(1)

1 →

r3 − 1 × r(1)

1 →

a(1)

11 a(1)

12 a(1)

13 b(1)

1

a(1)

21 a(1)

22 a(1)

23 b(1)

2

a(1)

31 a(1)

32 a(1)

33 b(1)

3

=

1

2

−

2

1

0 −1 1 0

0 0 1 1

: r(1)

1

: r(1)

2

: r(1)

3

(2.2.30a)

Then, we divide the second row r(1)

2 by a(1)

22 = −1 to make the new second row

r(2)

2 have the pivot a(2)

22 = 1 and subtract a(1)

m2 × r(2)

2 (m = 1, 3) from the first and

third row r(1)

1 and r(1)

3 to get

r(1)

1 − 2 × r(2)

2 →

r(1)

2 ÷ (−1) →

r(1)

3 − 0 × r(2)

2 →

a

(

2)

11 a(2)

12 a(2)

13 b(2)

1

a(2)

21 a(2)

22 a(2)

23 b(2)

2

a(2)

31 a(2)

32 a(2)

33 b(2)

3

=

1 0 0 1

0 1 −1 0

0 0 1 1

: r(2)

1

: r(2)

2

: r(2)

3

(2.2.30b)

Lastly, we divide the third row r(2)

3 by a(2)

33 = 1 to make the new third row

r(3)

3 have the pivot a(3)

33 = 1 and subtract a(2)

m3 × r(3)

3 (m = 1, 2) from the first and

92 SYSTEM OF LINEAR EQUATIONS

second row r(2)

1 and r(2)

2 to get

r(2)

1 − 0 × r(3)

3 →

r(2)

2 − (−1) × r(3)

3 →

r(2)

3 →

a(3)

11 a(3)

12 a(3)

13 b(3)

1

a(3)

21 a(3)

22 a(3)

23 b(3)

2

a(3)

31 a(3)

32 a(3)

33 b(3)

3

=

1 0 0 1= x1

0 1 0 1= x2

0 0 1 1= x3

: r(3)

1

: r(3)

2

: r(3)

3

(2.2.30c)

After having the identity matrix–vector form like this, we take the RHS vector

as the solution.

The general formula applicable for Gauss–Jordan elimination is the same as

Eq. (2.2.5), except that the index set is m = k—that is, all the numbers from

m = 1 to m = M except m = k. Interested readers are recommended to make

their own routines to implement this algorithm (see Problem 2.3).

2.3 INVERSE MATRIX

In the previous section, we looked over some algorithms to solve a system of

linear equations. We can use such algorithms to solve several systems of linear

equations having the same coefficient matrix

Ax1 = b1,Ax2 = b2, . . . , AxNB = bNB

by putting different RHS vectors into one RHS matrix as

A[ x1 x2 · · · xNB ] = [ b1 b2 · · · bNB ], AX= B

X = A−1B

(2.3.1)

If we substitute an identity matrix I for B into this equation, we will get the matrix

inverse X = A−1I = A−1. We, however, usually use the MATLAB command

inv(A) or A^-1 to compute the inverse of a matrix A.

2.4 DECOMPOSITION (FACTORIZATION)

2.4.1 LU Decomposition (Factorization): Triangularization

LU decomposition (factorization) of a nonsingular (square) matrix A means

expressing the matrix as the multiplication of a lower triangular matrix L and

an upper triangular matrix U, where a lower/upper triangular matrix is a matrix

having no nonzero elements above/below the diagonal. For the case where some

row switching operation is needed like in the Gauss elimination, we include a

permutation matrix P representing the necessary row switching operation(s) to

write the LU decomposition as

P A= L U (2.4.1)

DECOMPOSITION (FACTORIZATION) 93

The usage of a permutation matrix is exemplified by

PA =

0 0 1

1 0 0

0 1 0

a11 a12 a13

a21 a22 a23

a31 a32 a33

=

a31 a32 a33

a11 a12 a13

a21 a22 a23

(2.4.2)

which denotes switching the first and third rows followed by switching the second

and third rows. An interesting and useful property of the permutation matrix is

that its transpose agrees with its inverse.

PT P = I, PT = P−1 (2.4.3)

To take a close look at the LU decomposition, we consider a 3 × 3 nonsingular

matrix:

a11 a12 a13

a21 a22 a23

a31 a32 a33

=

1 0 0

l21 1 0

l31 l32 1

u11 u12 u13

0 u22 u23

0 0 u33

a11 a12 a13

a21 a22 a23

a31 a32 a33

=

u11 u12 u13

l21u11 l21u12 + u22 l21u13 + u23

l31u11 l31u12 + l32u22 l31u13 + l32u23 + u33

(2.4.4)

First, equating the first rows of both sides yields

u1n = a1n, n= 1, 2, 3 (2.4.5a)

Then, equating the second rows of both sides yields

a21 = l21u11, a22 = l21u12 + u22, a23 = l21u13 + u23

from which we can get

l21 = a21/u11, u22 = a21 − l21u12, u23 = a23 − l21u13 (2.4.5b)

Now, equating the third rows of both sides yields

a31 = l31u11, a32 = l31u12 + l32u22, a33 = l31u13 + u32u23 + u33

from which we can get

l31 = a31/u11, l32 = (a32 − l31u12)/u22, u33 = (a33 − l31u13) − l32u23

(2.4.5c)

In order to put these formulas in one framework to generalize them for matrices

having dimension greater than 3, we split this procedure into two steps

and write the intermediate lower/upper triangular matrices into one matrix for

compactness as

94 SYSTEM OF LINEAR EQUATIONS

step 1:

a11 a12 a13

a21 a22 a23

a31 a32 a33

→

u11 = a11 u12 = a12 u13 = a13

l21 = a21/u11 a(1)

22 = a22 − l21u12 a(1)

23 = a23 − l21u13

l31 = a31/u11 a(1)

32 = a32 − l31u12 a(1)

33 = a33 − l31u13

(2.4.6a)

step 2: →

u11 u12 u13

l21 u22 = a(1)

22 u23 = a(1)

23

l31 l32 = a(1)

32 /u22 a(2)

33 = a(1)

33 − l32u23

(2.4.6b)

This leads to an LU decomposition algorithm generalized for an NA × NA

nonsingular matrix as described in the following box. The MATLAB routine

“lu_dcmp()” implements this algorithm to find not only the lower/upper

triangular matrix L and U, but also the permutation matrix P. We run it for

a 3 ×3 matrix to get L, U, and P and then reconstruct the matrix P−1LU = A

from L, U, and P to ascertain whether the result is right.

function [L,U,P] = lu_dcmp(A)

%This gives LU decomposition of A with the permutation matrix P

% denoting the row switch(exchange) during factorization

NA = size(A,1);

AP = [A eye(NA)]; %augment with the permutation matrix.

for k = 1:NA – 1

%Partial Pivoting at AP(k,k)

[akx, kx] = max(abs(AP(k:NA,k)));

if akx < eps

error(’Singular matrix and No LU decomposition’)

end

mx = k+kx-1;

if kx > 1 % Row change if necessary

tmp_row = AP(k,:);

AP(k,:) = AP(mx,:);

AP(mx,:) = tmp_row;

end

% LU decomposition

for m = k + 1: NA

AP(m,k) = AP(m,k)/AP(k,k); %Eq.(2.4.8.2)

AP(m,k+1:NA) = AP(m,k + 1:NA)-AP(m,k)*AP(k,k + 1:NA); %Eq.(2.4.9)

end

end

P = AP(1:NA, NA + 1:NA + NA); %Permutation matrix

for m = 1:NA

for n = 1:NA

if m == n, L(m,m) = 1.; U(m,m) = AP(m,m);

elseif m > n, L(m,n) = AP(m,n); U(m,n) = 0.;

else L(m,n) = 0.; U(m,n) = AP(m,n);

end

end

end

if nargout == 0, disp(’L*U = P*A with’); L,U,P, end

%You can check if P’*L*U = A?

DECOMPOSITION (FACTORIZATION) 95

(cf) The number of floating-point multiplications required in this routine lu_dcmp() is

NA−1

k=1

(NA − k)(NA − k + 1) =

NA−1

k=1

{NA(NA + 1) − (2NA + 1)k + k2}

= (NA − 1)NA(NA + 1) −

1

2

(2NA + 1)(NA − 1)NA +

1

6

(NA − 1)NA(2NA − 1)

=

1

3

(NA − 1)NA(NA + 1) ≈

1

3

NA3 (2.4.7)

with NA: the size of matrix A

0. Initialize A(0) = A, or equivalently, a(0)

mn = amn for m, n = 1 : NA.

1. Let k = 1.

2. If a(k−1)

kk = 0, do an appropriate row switching operation so that

a(k−1)

kk = 0.

When it is not possible, then declare the case of singularity and stop.

3. a(k)

kn = a(k−1)

kn = ukn for n = k : NA (Just leave the kth row as it is.)

(2.4.8a)

a(k)

mk = a(k−1)

mk /a(k−1)

kk = lmk for m = k + 1 : NA (2.4.8b)

4. a(k)

mn = a(k−1)

mn − a(k)

mka(k)

kn for m, n = k + 1 : NA (2.4.9)

5. Increment k by 1 and if k < NA − 1, go to step 1; otherwise, go to step 6.

6. Set the part of the matrix A(NA−1) below the diagonal to L (lower triangular

matrix with the diagonal of 1’s) and the part on and above the

diagonal to U (upper triangular matrix).

>>A = [1 2 5;0.2 1.6 7.4; 0.5 4 8.5];

>>[L,U,P] = lu_dcmp(A) %LU decomposition

L = 1.0 0 0 U = 1 2 5 P = 1 0 0

0.5 1.0 0 0 3 6 0 0 1

0.2 0.4 1.0 0 0 4 0 1 0

>>P’*L*U – A %check the validity of the result (P’ = P^-1)

ans = 0 0 0

0 0 0

0 0 0

>>[L,U,P] = lu(A) %for comparison with the MATLAB built-in function

What is the LU decomposition for? It can be used for solving a system of

linear equations as

Ax = b (2.4.10)

Once we have the LU decomposition of the coefficient matrix A = PT LU, it is

more efficient to use the lower/upper triangular matrices for solving Eq. (2.4.10)

96 SYSTEM OF LINEAR EQUATIONS

than to apply the Gauss elimination method. The procedure is as follows:

PT LU x = b, LU x = P b, U x = L−1 P b, x = U−1 L−1 P b

(2.4.11)

Note that the premultiplication of L−1 and U−1 by a vector can be performed

by the forward and backward substitution, respectively. The following

program “do_lu_dcmp.m” applies the LU decomposition method, the Gauss

elimination algorithm, and the MATLAB operators ‘\’ and ‘inv’ or ‘^-1’ to

solve Eq. (2.4.10), where A is the five-dimensional Hilbert matrix (introduced

in Example 2.3) and b = Axo with xo = [1 1 1 1 1]T . The residual error

||Axi − b|| of the solutions obtained by the four methods and the numbers of

floating-point operations required for carrying out them are listed in Table 2.1.

The table shows that, once the inverse matrix A−1 is available, the inverse matrix

method requiring only N2 multiplications/additions (N is the dimension of the

coefficient matrix or the number of unknown variables) is the most efficient in

computation, but the worst in accuracy. Therefore, if we need to continually

solve the system of linear equations with the same coefficient matrix A for different

RHS vectors, it is a reasonable choice in terms of computation time and

accuracy to save the LU decomposition of the coefficient matrix A and apply the

forward/backward substitution process.

%do_lu_dcmp

% Use LU decomposition, Gauss elimination to solve Ax = b

A = hilb(5);

[L,U,P] = lu_dcmp(A); %LU decomposition

x = [1 -2 3 -4 5 -6 7 -8 9 -10]’;

b = A*x(1:size(A,1));

flops(0), x_lu = backsubst(U,forsubst(L,P*b)); %Eq.(2.4.11)

flps(1) = flops; % assuming that we have already got L\U decomposition

flops(0), x_gs = gauss(A,b); flps(3) = flops;

flops(0), x_bs = A\b; flps(4) = flops;

AI = A^-1; flops(0), x_iv = AI*b; flps(5) = flops;

% assuming that we have already got the inverse matrix

disp(’ x_lu x_gs x_bs x_iv’)

format short e

solutions = [x_lu x_gs x_bs x_iv]

errs = [norm(A*x_lu – b) norm(A*x_gs – b) norm(A*x_bs – b) norm(A*x_iv – b)]

format short, flps

function x = forsubst(L,B)

%forward substitution for a lower-triangular matrix equation Lx = B

N = size(L,1);

x(1,:) = B(1,:)/L(1,1);

for m = 2:N

x(m,:) = (B(m,:)-L(m,1:m – 1)*x(1:m-1,:))/L(m,m);

end

function x = backsubst(U,B)

%backward substitution for a upper-triangular matrix equation Ux = B

N = size(U,2);

x(N,:) = B(N,:)/U(N,N);

for m = N-1: -1:1

x(m,:) = (B(m,:) – U(m,m + 1:N)*x(m + 1:N,:))/U(m,m);

end

DECOMPOSITION (FACTORIZATION) 97

Table 2.1 Residual Error and the Number of Floating-Point Operations of Various

Solutions

tmp = forsubst(L,P*b)

backsubst(U,tmp) gauss(A,b) A\b A^-1*b

||Axi − b|| 1.3597e-016 5.5511e-017 1.7554e-016 3.0935e-012

# of flops 123 224 155 50

(cf) The numbers of flops for the LU decomposition and the inverse of the matrix A are not counted.

(cf) Note that the command ‘flops’ to count the number of floating-point operations is no longer

available in MATLAB 6.x and higher versions.

2.4.2 Other Decomposition (Factorization): Cholesky, QR, and SVD

There are several other matrix decompositions such as Cholesky decomposition,

QR decomposition, and singular value decomposition (SVD). Instead of looking

into the details of these algorithms, we will simply survey the MATLAB built-in

functions implementing these decompositions.

Cholesky decomposition factors a positive definite symmetric/Hermitian matrix

into an upper triangular matrix premultiplied by its transpose as

A = UTU (U: an upper triangular matrix) (2.4.12)

and is implemented by the MATLAB built-in function chol().

(cf) If a (complex-valued) matrix A satisfies A∗T = A—that is, the conjugate transpose

of a matrix equals itself—it is said to be Hermitian. It is said to be just symmetric

in the case of a real-valued matrix with AT = A.

(cf) If a square matrix A satisfies x∗T A x > 0 ∀ x = 0, the matrix is said to be positive

definite (see Appendix B).

>>A = [2 3 4;3 5 6;4 6 9]; %a positive definite symmetric matrix

>>U = chol(A) %Cholesky decomposition

U = 1.4142 2.1213 2.8284

0 0.7071 0.0000

0 0 1.0000

>>U’*U – A %to check if the result is right

QR decomposition is to express a square or rectangular matrix as the product

of an orthogonal (unitary) matrix Q and an upper triangular matrix R as

A = QR (2.4.13)

where QTQ = I (Q∗TQ = I). This is implemented by the MATLAB built-in

function qr().

98 SYSTEM OF LINEAR EQUATIONS

(cf) If all the columns of a (complex-valued) matrix A are orthonormal to each other—that

is, A∗T A = I , or, equivalently, A∗T = A−1—it is said to be unitary. It is said to be

orthogonal in the case of real-valued matrix with AT = A−1.

SVD (singular value decomposition) is to express an M × N matrix A in the

following form

A = USV T (2.4.14)

where U is an orthogonal (unitary) M ×M matrix, V is an orthogonal (unitary)

N × N matrix, and S is a real diagonal M × N matrix having the singular

values of A (the square roots of the eigenvalues of AT A) in decreasing

order on its diagonal. This is implemented by the MATLAB built-in function

svd().

>>A = [1 2;2 3;3 5]; %a rectangular matrix

>>[U,S,V] = svd(A) %Singular Value Decomposition

U = 0.3092 0.7557 -0.5774 S = 7.2071 0 V = 0.5184 -0.8552

0.4998 -0.6456 -0.5774 0 0.2403 0.8552 0.5184

0.8090 0.1100 0.5774 0 0

>>err = U*S*V’-A %to check if the result is right

err = 1.0e-015* -0.2220 -0.2220

0 0

0.4441 0

2.5 ITERATIVE METHODS TO SOLVE EQUATIONS

2.5.1 Jacobi Iteration

Let us consider the equation

3x + 1 = 0

which can be cast into an iterative scheme as

2x = −x − 1; x = −

x + 1

2 → xk+1 = −

1

2

xk −

1

2

Starting from some initial value x0 for k = 0, we can incrementally change k

by 1 each time to proceed as follows:

x1 = −2−1 − 2−1×0

x2 = −2−1 − 2−1×1 = −2−1 + 2−2 + 2−2×0

x3 = −2−1 − 2−1×2 = −2−1 + 2−2 − 2−3 − 2−3×0

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Whatever the initial value x0 is, this process will converge to the sum of a

geometric series with the ratio of (−1/2) as

ITERATIVE METHODS TO SOLVE EQUATIONS 99

xk =

a0

1 − r = −1/2

1 − (−1/2) = −

1

3 = x0 as k→∞

and what is better, the limit is the very true solution to the given equation.

We are happy with this, but might feel uneasy, because we are afraid that this

convergence to the true solution is just a coincidence. Will it always converge,

no matter how we modify the equation so that only x remains on the LHS?

To answer this question, let us try another iterative scheme.

x = −2x − 1 → xk+1 = −2xk − 1

x1 = −1 − 2×0

x2 = −1 − 2×1 = −1 − 2(−1 − 2×0) = −1 + 2 + 22×0

x3 = −1 − 2×2 = −1 + 2 − 22 − 23×0

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

This iteration will diverge regardless of the initial value x0. But, we are never

disappointed, since we know that no one can be always lucky.

To understand the essential difference between these two cases, we should

know the fixed-point theorem (Section 4.1). Apart from this, let’s go into a system

of equations.

3 2

1 2x1

x2 = 1

−1 , Ax = b

Dividing the first equation by 3 and transposing all term(s) other than x1 to the

RHS and dividing the second equation by 2 and transposing all term(s) other

than x2 to the RHS, we have

x1,k+1

x2,k+1 = 0 −2/3

−1/2 0 x1,k

x2,k + 1/3

−1/2

xk+1 = A xk +

b (2.5.1)

Assuming that this scheme works well, we set the initial value to zero (x0 = 0)

and proceed as

xk → [I + A + A2 +· · ·]

b

= [I − A]−1

b

= 1 2/3

1/2 1 −1 1/3

−1/2

=

1

1 − 1/3 1 −2/3

−1/2 1 1/3

−1/2 =

1

2/3 2/3

−2/3 = 1

−1 = xo

(2.5.2)

which will converge to the true solution xo = [1 − 1]T . This suggests another

method of solving a system of equations, which is called Jacobi iteration. It can

be generalized for an N × N matrix–vector equation as follows:

100 SYSTEM OF LINEAR EQUATIONS

am1x1 + am2x2 + ·· ·+ammxm +· · ·+amNxN = bm

x(k+1)

m = −

N

n=m

amn

amm

x(k)

n +

bm

amm

for m = 1, 2, . . . , N

xk+1 = A xk +

b for each time stage k (2.5.3)

where

AN×N =

0 −a12/a11 · · · −a1N/a11

−a21/a22 0 · · · −a2N/a22

· · ··· ·

−aN1/aNN −aN2/aNN · · · 0

,

b =

b1/a11

b2/a22

· bN/aNN

This scheme is implemented by the following MATLAB routine “jacobi()”.

We run it to solve the above equation.

function X = jacobi(A,B,X0,kmax)

%This function finds a soltuion to Ax = B by Jacobi iteration.

if nargin < 4, tol = 1e-6; kmax = 100; %called by jacobi(A,B,X0)

elseif kmax < 1, tol = max(kmax,1e-16); kmax = 100; %jacobi(A,B,X0,tol)

else tol = 1e-6; %jacobi(A,B,X0,kmax)

end

if nargin < 3, X0 = zeros(size(B)); end

NA = size(A,1);

X = X0; At = zeros(NA,NA);

for m = 1:NA

for n = 1:NA

if n ~= m, At(m,n) = -A(m,n)/A(m,m); end

end

Bt(m,:) = B(m,:)/A(m,m);

end

for k = 1: kmax

X = At*X + Bt; %Eq. (2.5.3)

if nargout == 0, X, end %To see the intermediate results

if norm(X – X0)/(norm(X0) + eps) < tol, break; end

X0 = X;

end

>>A = [3 2;1 2]; b = [1 -1]’; %the coefficient matrix and RHS vector

>>x0 = [0 0]’; %the initial value

>>x = jacobi(A,b,x0,20) %to repeat 20 iterations starting from x0

x = 1.0000

-1.0000

>>jacobi(A,b,x0,20) %omit output argument to see intermediate results

X = 0.3333 0.6667 0.7778 0.8889 0.9259 ……

-0.5000 -0.6667 -0.8333 -0.8889 -0.9444 ……

2.5.2 Gauss–Seidel Iteration

Let us take a close look at Eq. (2.5.1). Each iteration of Jacobi method updates

the whole set of N variables at a time. However, so long as we do not use a

ITERATIVE METHODS TO SOLVE EQUATIONS 101

multiprocessor computer capable of parallel processing, each one of N variables

is updated sequentially one by one. Therefore, it is no wonder that we could

speed up the convergence by using all the most recent values of variables for

updating each variable even in the same iteration as follows:

x1,k+1 = −

2

3

x2,k +

1

3

x2,k+1 = −

1

2

x1,k+1 −

1

2

This scheme is called Gauss–Seidel iteration, which can be generalized for an

N × N matrix–vector equation as follows:

x(k+1)

m =

bm −m−1

n=1 amnx(k+1)

n −N

n=m+1 amnx(k)

n

amm

for m = 1, . . . , N and for each time stage k (2.5.4)

This is implemented in the following MATLAB routine “gauseid()”, which

we will use to solve the above equation.

function X = gauseid(A,B,X0,kmax)

%This function finds x = A^-1 B by Gauss–Seidel iteration.

if nargin < 4, tol = 1e-6; kmax = 100;

elseif kmax < 1, tol = max(kmax,1e-16); kmax = 1000;

else tol = 1e-6;

end if nargin < 4, tol = 1e-6; kmax = 100; end

if nargin < 3, X0 = zeros(size(B)); end

NA = size(A,1); X = X0;

for k = 1: kmax

X(1,:) = (B(1,:)-A(1,2:NA)*X(2:NA,:))/A(1,1);

for m = 2:NA-1

tmp = B(m,:)-A(m,1:m-1)*X(1:m – 1,:)-A(m,m + 1:NA)*X(m + 1:NA,:);

X(m,:) = tmp/A(m,m); %Eq.(2.5.4)

end

X(NA,:) = (B(NA,:)-A(NA,1:NA – 1)*X(1:NA – 1,:))/A(NA,NA);

if nargout == 0, X, end %To see the intermediate results

if norm(X – X0)/(norm(X0) + eps)<tol, break; end

X0 = X;

end

>>A = [3 2;1 2]; b = [1 -1]’; %the coefficient matrix and RHS vector

>>x0 = [0 0]’; %the initial value

>>gauseid(A,b,x0,10) %omit output argument to see intermediate results

X = 0.3333 0.7778 0.9259 0.9753 0.9918 ……

-0.6667 -0.8889 -0.9630 -0.9877 -0.9959 ……

As with the Jacobi iteration in the previous section, we can see this Gauss–Seidel

iteration converging to the true solution xo = [1 − 1]T and that with fewer iterations.

But, if we use a multiprocessor computer capable of parallel processing,

102 SYSTEM OF LINEAR EQUATIONS

the Jacobi iteration may be better in speed even with more iterations, since it can

exploit the advantage of simultaneous parallel computation.

Note that the Jacobi/Gauss–Seidel iterative scheme seems unattractive and

even unreasonable if we are given a standard form of linear equations as

Ax = b

because the computational overhead for converting it into the form of Eq. (2.5.3)

may be excessive. But, it is not always the case, especially when the equations

are given in the form of Eq. (2.5.3)/(2.5.4). In such a case, we simply repeat

the iterations without having to use such ready-made routines as “jacobi()” or

“gauseid()”. Let us see the following example.

Example 2.4. Jacobi or Gauss–Seidel Iterative Scheme. Suppose the temperature

of a metal rod of length 10 m has been measured to be 0◦C and 10◦C at

each end, respectively. Find the temperatures x1, x2, x3, and x4 at the four points

equally spaced with the interval of 2 m, assuming that the temperature at each

point is the average of the temperatures of both neighboring points.

We can formulate this problem into a system of equations as

x1 =

x0 + x2

2

, x2 =

x1 + x3

2

, x3 =

x2 + x4

2

,

x4 =

x3 + x5

2

with x0 = 0 and x5 = 10 (E2.4)

This can easily be cast into Eq. (2.5.3) or Eq. (2.5.4) as programmed in the

following program “nm2e04.m”:

%nm2e04

N = 4; %the number of unknown variables/equations

kmax = 20; tol = 1e-6;

At = [0 1 0 0; 1 0 1 0; 0 1 0 1; 0 0 1 0]/2;

x0 = 0; x5 = 10; %boundary values

b = [x0/2 0 0 x5/2]’; %RHS vector

%initialize all the values to the average of boundary values

xp=ones(N,1)*(x0 + x5)/2;

%Jacobi iteration

for k = 1:kmax

x = At*xp +b; %Eq.(E2.4)

if norm(x – xp)/(norm(xp)+eps) < tol, break; end

xp = x;

end

k, xj = x

%Gauss–Seidel iteration

xp = ones(N,1)*(x0 + x5)/2; x = xp; %initial value

for k = 1:kmax

for n = 1:N, x(n) = At(n,:)*x + b(n); end %Eq.(E2.4)

if norm(x – xp)/(norm(xp) + eps) < tol, break; end

xp = x;

end

k, xg = x

ITERATIVE METHODS TO SOLVE EQUATIONS 103

The following example illustrates that the Jacobi iteration and the Gauss–Seidel

iteration can also be used for solving a system of nonlinear equations, although there

is no guarantee that it will work for every nonlinear equation.

Example 2.5. Gauss–Seidel Iteration for Solving a Set of Nonlinear Equations.

We are going to use the Gauss–Seidel iteration to solve a system of nonlinear

equations as

x2

1 + 10×1 + 2×2

2 − 13 = 0

2×3

1 − x2

2 + 5×2 − 6 = 0

(E2.5.1)

In order to do so, we convert these equations into the following form, which

suits the Gauss–Seidel scheme.

x1

x2 = (13 − x2

1 − 2×2

2 )/10

(6 − 2×3

1 + x2

2 )/5 (E2.5.2)

We make the MATLAB program “nm2e05.m”, which uses the Gauss–Seidel

iteration to solve these equations. Interested readers are recommended to run

this program to see that this simple iteration yields the solution within the given

tolerance of error in just six steps. How marvelous it is to solve the system of

nonlinear equations without any special algorithm!

(cf) Due to its remarkable capability to deal with a system of nonlinear equations, the

Gauss–Seidel iterative method plays an important role in solving partial differential

equations (see Chapter 9).

%nm2e05.m

% use Gauss–Seidel iteration to solve a set of nonlinear equations

clear

kmax = 100; tol = 1e-6;

x = zeros(2,1); %initial value

for k = 1:kmax

xp = x; % to remember the previous solution

x(1) = (13 – x(1)^2 – 2*x(2)^2)/10; % (E2.5.2)

x(2) = (6 – x(1)^3)/5;

if norm(x – xp)/(norm(xp) + eps)<tol, break; end

end

k, x

2.5.3 The Convergence of Jacobi and Gauss–Seidel Iterations

Jacobi and Gauss–Seidel iterations have a very simple computational structure

because they do not need any matrix inversion. So, it may be of practical use, if

only the convergence is guaranteed. However, everything cannot always be fine,

104 SYSTEM OF LINEAR EQUATIONS

as illustrated in Section 2.5.1. Then, what is the convergence condition? It is the

diagonal dominancy of coefficient matrix A, which is stated as follows:

|amm| >

N

n=m

|amn| for m = 1, 2, . . . , N (2.5.5)

This implies that the convergence of the iterative schemes is ensured if, in

each row of coefficient matrix A, the absolute value of the diagonal element

is greater than the sum of the absolute values of the other elements. It should

be noted, however, that this is a sufficient, not a necessary, condition. In other

words, the iterative scheme may work even if the above condition is not strictly

satisfied.

One thing to note is the relaxation technique, which may be helpful in accelerating

the convergence of Gauss–Seidel iteration. It is a slight modification of

Eq. (2.5.4) as

x(k+1)

m = (1 − ω)x(k)

m + ω

bm −m−1

n=1 amnx(k+1)

n −N

n=m+1 amnx(k)

n

amm

with 0 < ω < 2 (2.5.6)

and is called SOR (successive overrelaxation) for the relaxation factor 1 < ω <

2 and successive underrelaxation for 0 < ω < 1. But regrettably, there is no

general rule for selecting the optimal value of the relaxation factor ω.

PROBLEMS

2.1 Recursive Least-Squares Estimation (RLSE)

(a) Run the program ‘do_rlse.m’ (in Section 2.1.4) with another value of

the true parameter

xo = [1 2]’

What is the parameter estimate obtained from the RLS solution?

(b) Run the program “do_rlse” with a small matrix P like

P = 0.01*eye(NA);

What is the parameter estimate obtained from the RLS solution? Is it

still close to the value of the true parameter?

(c) Insert the statements in the following box at appropriate places in the

MATLAB code “do_rlse.m” appeared in Section 2.1.4. Remove the

last two statements and run it to compare the times required for using

the RLS solution and the standard LS solution to get the parameter

estimates on-line.

PROBLEMS 105

%nm2p01.m

.. .. .. .. .. .. .. ..

time_on = 0; time_off = 0;

.. .. .. .. .. .. .. ..

tic

.. .. .. .. .. .. .. ..

time_on = time_on + toc;

tic

xk_off = A\b; %standard LS solution

time_off = time_off + toc;

.. .. .. .. .. .. .. ..

solutions = [x xk_off]

discrepancy = norm(x – xk_off)

times = [time_on time_off]

2.2 Delicacy of Scaled Partial Pivoting

As a complement to Example 2.2, we want to compare no pivoting, partial

pivoting, scaled partial pivoting, and full pivoting in order to taste the

delicacy of row switching strategy. To do it in a systematic way, add the

third input argument (pivoting) to the Gauss elimination routine ‘gauss()’

and modify its contents by inserting the following statements into appropriate

places so that the new routine “gauss(A,b,pivoting)” implements the

partial pivoting procedure optionally depending on the value of ‘pivoting’.

You can also remove any unnecessary parts.

– if nargin < 3, pivoting = 2; end %scaled partial pivoting by default

– switch pivoting

case 2, [akx,kx] = max(abs(AB(k:NA,k))./…

max(abs([AB(k:NA,k + 1:NA) eps*ones(NA – k + 1,1)]’))’);

otherwise, [akx,kx] = max(abs(AB(k:NA,k))); %partial pivoting

end

– &pivoting > 0 %partial pivoting not to be done for pivot = 1

(a) Use this routine with pivoting = 0/1/2, the ‘\

’ operator and the

‘inv()’ command to solve the systems of linear equations with the

coefficient matrices and the RHS vectors shown below and fill in

Table P2.2 with the residual error ||Aix − bi || to compare the results

in terms of how well the solutions satisfy the equation, that is,

||Aix − bi || ≈ 0.

(1) A1 = 10−15 1

1 1011 , b1 = 1 + 10−15

1011 + 1

(2) A2 = 10−14.6 1

1 1015 , b2 = 1 + 10−14.6

1015 + 1

(3) A3 = 1011 1

1 10−15 , b3 = 1011 + 1

1 + 10−15

106 SYSTEM OF LINEAR EQUATIONS

Table P2.2 Comparison of gauss() with Different Pivoting Methods in Terms of

||Axi − b||

A1x = b1 A2x = b2 A3x = b3 A4x = b4

gauss(A,b,0) (no pivoting) 1.25e-01

gauss(A,b,1) (partial pivoting) 4.44e-16

gauss(A,b,2) (scaled partial pivoting) 0

A\b 6.25e-02

A^-1*b

(4) A4 = 1014.6 1

1 10−15 , b4 = 1014.6 + 1

1 + 10−15

(b) Which pivoting strategy yields the worst result for problem (1) in (a)?

Has the row swapping been done during the process of partial pivoting

and scaled partial pivoting? If yes, did it work to our advantage? Did

the ‘\’ operator or the ‘inv()’ command give you any better result?

(c) Which pivoting strategy yields the worst result for problem (2) in (a)?

Has the row swapping been done during the process of partial pivoting

and scaled partial pivoting? If yes, did it produce a positive effect for

this case? Did the ‘\’ operator or the ‘inv()’ command give you any

better result?

(d) Which pivoting strategy yields the best result for problem (3) in (a)? Has

the row swapping been done during the process of partial pivoting and

scaled partial pivoting? If yes, did it produce a positive effect for this

case?

(e) The coefficient matrix A3 is the same as would be obtained by applying

the full pivoting scheme for A1 to have the largest pivot element. Does

the full pivoting give better result than no pivoting or the (scaled) partial

pivoting?

(f) Which pivoting strategy yields the best result for problem (4) in (a)? Has

the row swapping been done during the process of partial pivoting and

scaled partial pivoting? If yes, did it produce a positive effect for this

case? Did the ‘\’ operator or the ‘inv()’ command give you any better

result?

2.3 Gauss–Jordan Elimination Algorithm Versus Gauss Elimination Algorithm

Gauss–Jordan elimination algorithm mentioned in Section 2.2.3 is trimming

the coefficient matrix A into an identity matrix and then takes the RHS

vector/matrix as the solution, while Gauss elimination algorithm introduced

with the corresponding routine “gauss()” in Section 2.2.1 makes the matrix

an upper-triangular one and performs backward substitution to get the solution.

Since Gauss–Jordan elimination algorithm does not need backward

substitution, it seems to be simpler than Gauss elimination algorithm.

PROBLEMS 107

Table P2.3 Comparison of Several Methods for Solving a Set of Linear Equations

gauss(A,b) gaussj(A,b) A\b Aˆ-1*b

||Axi − b|| 3.1402e-016 8.7419e-016

# of flops 1124 1744 785 7670

(a) Modify the routine “gauss()” into a routine “gaussj()” which implements

Gauss–Jordan elimination algorithm and count the number of

multiplications consumed by the routine, excluding those required for

partial pivoting. Compare it with the number of multiplications consumed

by “gauss()” [Eq. (2.2.18)]. Does it support or betray our expectation

that Gauss–Jordan elimination would take fewer computations than

Gauss elimination?

(b) Use both of the routines, the ‘\’ operator and the ‘inv()’ command or

‘^-1’ to solve the system of linear equations

Ax = b (P2.3.1)

where A is the 10-dimensional Hilbert matrix (see Example 2.3) and

b = Axo with xo = [1 1 1 1 1 1 1 1 1 1]T . Fill in Table P2.3 with the

residual errors

||Axi − b|| ≈ 0 (P2.3.2)

as a way of describing how well each solution satisfies the equation.

(cf) The numbers of floating-point operations required for carrying out the

computations are listed in Table P2.3 so that readers can compare the computational

loads of different approaches. Those data were obtained by using

the MATLAB command flops(), which is available only in MATLAB of

version below 6.0.

2.4 Tridiagonal System of Linear Equations

Consider the following system of linear equations:

a11x1 + a12x2 = b1

a21x1 + a22x2 + a23x3 = b2

· · · · · · · · · · · · · · · · · · · · · · · · (P2.4.1)

aN−1,N−2xN−2 + aN−1,N−1xN−1 + aN−1,NxN = bN−1

aN,N−1xN−1 + aN,NxN = bN

which can be written in a compact form by using a matrix–vector notation as

AN×Nx = b (P2.4.2)

108 SYSTEM OF LINEAR EQUATIONS

Table P2.4 The Computational Load of the Methods to Solve a Tri-diagonal

System of Equations

gauss(A,b) trid(A,b) gauseid() gauseid1() A\b

# of flops 141 50 2615 2082 94

where

AN×N =

a11 a12 0 0 0

a21 a22 a23 0 0

0 · · · · · · · · · 0

0 0 aN−1,N−2 aN−1,N−1 aN−1,N

0 0 0 aN,N−1 aNN

,

x =

x1

x2

.

xN−1

xN

, b =

b1

b2

.

bN−1

bN

This is called a tridiagonal system of equations on account of that the

coefficient matrix A has nonzero elements only on its main diagonal and

super-/subdiagonals.

(a) Modify the Gauss elimination routine “gauss()” (Section 2.2.1) in such

a way that this special structure can be exploited for reducing the computational

burden. Give the name ‘trid()’ to the modified routine and

save it in an m-file named “trid.m” for future use.

(b) Modify the Gauss–Seidel iteration routine “gauseid()” (Section 2.5.2)

in such a way that this special structure can be exploited for reducing

the computational burden. Let the name of the modified routine be

“Gauseid1()”.

(c) Noting that Eq. (E2.4) in Example 2.4 can be trimmed into a tridiagonal

structure as (P2.4.2), use the routines “gauss()”, “trid()”, “gauseid()”,

“gauseid1()”, and the backslash (\) operator to solve the

problem.

(cf) The numbers of floating-point operations required for carrying out the

computations are listed in Table P2.4 so that readers can compare the computational

loads of the different approaches.

2.5 LU Decomposition of a Tridiagonal Matrix

Modify the LU decomposition routine “lu_dcmp()” (Section 2.4.1) in such a

way that the tridiagonal structure can be exploited for reducing the

PROBLEMS 109

computational burden. Give the name “lu_trid()” to the modified routine

and use it to get the LU decomposition of the tridiagonal matrix

A =

2 −1 0 0

−1 2 −1 0

0 −1 2 −1

0 0 −1 2

(P2.5.1)

You may type the following statements into the MATLAB command window:

>>A = [2 -1 0 0; -1 2 -1 0; 0 -1 2 -1; 0 0 -1 2];

>>[L,U] = lu_trid(A)

>>L*U – A % = 0 (No error)?

2.6 LS Solution by Backslash Operator and QR Decomposition

The backslash (‘A\b’) operator and the matrix left division

(‘mldivide(A,b)’) function turn out to be the most efficient means for solving

a system of linear equations as Eq. (P2.3.1). They are also capable of

dealing with the under/over-determined cases. Let’s see how they handle the

under/over-determined cases.

(a) For an underdetermined system of linear equations

A1x = b1, 1 2 3

4 5 6

x1

x2

x3

= 14

32 (P2.6.1)

find the minimum-norm solution (2.1.7) and the solutions that can be

obtained by typing the following statements in the MATLAB command

window:

>>A1 = [1 2 3; 4 5 6]; b1 = [14 32]’;

>>x_mn = A1’*(A1*A1’)^-1*b1, x_pi = pinv(A1)*b1, x_bs = A1\b1

Are the three solutions the same?

(b) For another underdetermined system of linear equations

A2x = b2, 1 2 3

2 4 6

x1

x2

x3

= 14

28 (P2.6.2)

find the solutions by using Eq. (2.1.7), the commands pinv(), and backslash

(\). If you are not pleased with the result obtained from Eq. (2.1.7),

you can remove one of the two rows from the coefficient matrix A2 and

try again. Identify the minimum solution(s). Are the equations redundant

or inconsistent?

110 SYSTEM OF LINEAR EQUATIONS

Table P2.6.1 Comparison of Several Methods for Computing the LS Solution

QR LS: Eq. (2.1.10) pinv(A)*b A\b

||Axi − b|| 2.8788e-016 2.8788e-016

# of flops 25 89 196 92

(c) For another underdetermined system of linear equations

A2x = b3, 1 2 3

2 4 6

x1

x2

x3

= 21

21 (P2.6.3)

find the solutions by using Eq. (2.1.7), the commands pinv(), and backslash

(\). Does any of them satisfy Eq. (P2.6.3) closely? Are the equations

redundant or inconsistent?

(d) For an overdetermined system of linear equations

A4x = b4,

1 2

2 3

4 −1

x1

x2 =

5.2

7.8

2.2

(P2.6.4)

find the LS (least-squares) solution (2.1.10), that can be obtained from

the following statements. Fill in the corresponding blanks of Table P2.6.1

with the results.

>>A4 = [1 2; 2 3; 4 -1]; b4 = [5.2 7.8 2.2]’;

>> x_ls = (A4’*A4)\A4’*b4, x_pi = pinv(A4)*b4, x_bs = A4\b4

(e) We can use QR decomposition to solve a system of linear equations as

Eq. (P2.3.1), where the coefficient matrix A is square and nonsingular or

rectangular with the row dimension greater than the column dimension.

The procedure is explained as follows:

Ax = QRx = b, Rx = Q−1b = Qb, x = R−1Qb (P2.6.5)

Note that QQ = I ; Q = Q−1 (orthogonality) and the premultiplication

of R−1 can be performed by backward substitution, because R is

an upper-triangular matrix. You are supposed not to count the number

of floating-point operations needed for obtaining the LU and QR

decompositions, assuming that they are available.

(i) Apply the QR decomposition, the LU decomposition, Gauss elimination,

and the backslash (\) operator to solve the system of linear

PROBLEMS 111

Table P2.6.2 Comparison of Several Methods for Solving a System of Linear

Equations

LU QR gauss(A,b) A\b

||Axi − b|| 7.8505e-016 8.7419e-016

# of flops 453 327 1124 785

equations whose coefficient matrix is the 10-dimensional Hilbert

matrix (see Example 2.3) and fill in the corresponding blanks of

Table P2.6.2 with the results.

(ii) Apply the QR decomposition to solve the system of linear equations

given by Eq. (P2.6.4) and fill in the corresponding blanks of

Table P2.6.2 with the results.

(cf) This problem illustrates that QR decomposition is quite useful for solving

a system of linear equations, where the coefficient matrix A is square and

nonsingular or rectangular with the row dimension greater than the column

dimension and no rank deficiency.

2.7 Cholesky Factorization of a Symmetric Positive Definite Matrix:

If a matrix A is symmetric and positive definite, we can find its LU

decomposition such that the upper triangular matrix U is the transpose of

the lower triangular matrix L, which is called Cholesky factorization.

Consider the Cholesky factorization procedure for a 4 ×4 matrix

a11 a12 a13 a14

a12 a22 a23 a24

a13 a23 a33 a34

a14 a24 a34 a44

=

u11 0 0 0

u12 u22 0 0

u13 u23 u33 0

u14 u24 u34 u44

u11 u12 u13 u14

0 u22 u23 u24

0 0 u33 u34

0 0 0 u44

=

u2

11 u11u12 u11u13 u11u14

u12u11 u2

12 + u2

22 u12u13 + u22u23 u12u14 + u22u24

u13u11 u13u12 + u23u22 u2

13 + u2

23 + u2

33 u13u14 + u23u24 + u33u34

u14u11 u14u12 + u24u22 u14u13 + u24u23 + u34u33 u2

14 + u2

24 + u2

34 + u2

44

(P2.7.1)

Equating every row of the matrices on both sides yields

u11 = √a11, u12 = a12/u11, u13 = a13/u11, u14 = a14/u11 (P2.7.2.1)

u22 = a22 − u2

12, u23 = (a23 − u13u12)/u22, u24 = (a24 − u14u12)/u22

(P2.7.2.2)

u33 = a33 − u2

23 − u2

13, u34 = (a43 − u24u23 − u14u13)/u33 (P2.7.2.3)

u44 = a44 − u2

34 − u2

24 − u2

14 (P2.7.2.4)

112 SYSTEM OF LINEAR EQUATIONS

which can be combined into two formulas as

ukk = akk −k−1

i=1

u2

ik for k = 1 : N (P2.7.3a)

ukm = akm −k−1

i=1

uimuik

ukk for m = k + 1 : N and k = 1 : N

(P2.7.3b)

(a) Make a MATLAB routine “cholesky()”, which implements these formulas

to perform Cholesky factorization.

(b) Try your routine “cholesky()” for the following matrix and check if

UT U − A ≈ O (U: the upper triangular matrix). Compare the result with

that obtained by using the MATLAB built-in routine “chol()”.

A =

1 2 4 7

2 13 23 38

4 23 77 122

7 38 122 294

(P2.7.4)

(c) Use the routine “lu_dcmp()” and the MATLAB built-in routine “lu()”

to get the LU decomposition for the above matrix (P2.7.4) and check if

PT LU − A ≈ O, where L and U are the lower/upper triangular matrix,

respectively. Compare the result with that obtained by using the MATLAB

built-in routine “lu()”.

2.8 Usage of SVD (Singular Value Decomposition)

What is SVD good for? Suppose we have the singular value decomposition

of an M × N real-valued matrix A as

A = USV T (P2.8.1)

where U is an orthogonal M × M matrix, V an orthogonal N × N matrix,

and S a real diagonal M × N matrix having the singular value σi’s of A (the

square roots of the eigenvalues of AT A) in decreasing order on its diagonal.

Then, it is possible to improvise the pseudo-inverse even in the case of

rank-deficient matrices (with rank(A) < min(M,N)) for which the left/right

pseudo-inverse can’t be found. The virtual pseudo-inverse can be written as

Aˆ−1 = Vˆ Sˆ−1 Uˆ T (P2.8.2)

where Sˆ−1 is the diagonal matrix having 1/σi on its diagonal that is reconstructed

by removing all-zero(-like) rows/columns of the matrix S and substituting

1/σi for σi = 0 into the resulting matrix; Vˆ and Uˆ are reconstructed

by removing the columns of V and U corresponding to the zero singular

value(s). Consequently, SVD has a specialty in dealing with the singular

cases. Let us take a closer look at this through the following problems.

PROBLEMS 113

(a) Consider the problem of solving

A1x = 1 2 3

2 4 6

x1

x2

x3

= 6

12 = b1 (P2.8.3)

Since this belongs to the underdetermined case (M = 2 < 3 = N), it

seems that we can use Eq. (2.1.7) to find the minimum-norm solution.

(i) Type the following statements into the MATLAB command window.

>>A1 = [1 2 3; 2 4 6]; b1 = [6;12]; x = A1’*(A1*A1’)^-1*b1 %Eq. (2.1.7)

What is the result? Explain why it is so and support your answer by

typing

>>r = rank(A1)

(ii) Type the following statements into the MATLAB command window

to see the SVD-based minimum-norm solution. What is the value of

x = Aˆ−1

1 b1 = Vˆ Sˆ−1 Uˆ T b1 and ||A1x − b1||?

[U,S,V] = svd(A1); %(P2.8.1)

u = U(:,1:r); v = V(:,1:r); s = S(1:r,1:r);

AIp = v*diag(1./diag(s))*u’; %faked pseudo-inverse (P2.8.2)

x = AIp*b1 %minimum-norm solution for singular underdetermined

err = norm(A1*x – b1) %residual error

(iii) To see that the norm of this solution is less than that of any other

solution which can be obtained by adding any vector in the null space

of the coefficient matrix A1, type the following statements into the

MATLAB command window. What is implied by the result?

nullA = null(A1); normx = norm(x);

for n = 1:1000

if norm(x + nullA*(rand(size(nullA,2),1)-0.5)) < normx

disp(’What the hell smaller-norm sol – not minimum norm’);

end

end

(b) For the problem

A2x = 1 2 3

2 3 4

x1

x2

x3

= 6

9 = b2 (P2.8.4)

compare the minimum-norm solution based on SVD and that obtained

by Eq. (2.1.7).

114 SYSTEM OF LINEAR EQUATIONS

(c) Consider the problem of solving

A3x =

1 2 3

4 5 9

7 11 18

−2 3 1

x1

x2

x3

=

1

2

3

4

= b3 (P2.8.5)

Since this belongs to the overdetermined case (M = 4 > 3 = N), it

seems that we can use Eq. (2.1.10) to find the LS (least-squares) solution.

(i) Type the following statements into the MATLAB command window:

>>A3=[1 2 3; 4 5 9;7 11 18;-2 3 1];

>>b3=[1;2;3;4]; x=(A3’*A3)^-1*A3’*b3 %Eq. (2.1.10)

What is the result? Explain why it is so in connection with the rank

of A3.

(ii) Similarly to (a)(ii), find the SVD-based least-squares solution.

[U,S,V] = svd(A3);

u=U(:,1:r); v = V(:,1:r); s = S(1:r,1:r);

AIp = v*diag(1./diag(s))*u’; x = AIp*b

(iii) To see that the residual error of this solution is less than that of

any other vector around it, type the following statements into the

MATLAB command window. What is implied by the result?

err = norm(A3*x-b3)

for n = 1:1000

if norm(A3*(x+rand(size(x))-0.5)-b)<err

disp(’What the hell smaller error sol – not LSE?’);

end

end

(d) For the problem

A4x =

1 2 3

4 5 9

7 11 −1

−2 3 1

x1

x2

x3

=

1

2

3

4

= b4 (P2.8.6)

compare the LS solution based on SVD and that obtained by Eq. (2.1.10).

(cf) This problem illustrates that SVD can be used for fabricating a universal

solution of a set of linear equations, minimum-norm or least-squares, for

all the possible rank deficiency of the coefficient matrix A.

PROBLEMS 115

2.9 Gauss–Seidel Iterative Method with Relaxation Technique

(a) Try the relaxation technique (introduced in Section 2.5.3) with several

values of the relaxation factor ω = 0.2, 0.4, . . . , 1.8 for the following

problems. Find the best one among these values of the relaxation factor

for each problem, together with the number of iterations required for

satisfying the termination criterion ||xk+1 − xk||/||xk|| < 10−6.

(i) A1x = 5 −4

−9 10x1

x2 = 1

1 = b1 (P2.9.1)

(ii) A2x = 2 −1

−1 4x1

x2 = 1

3 = b2 (P2.9.2)

(iii) The nonlinear equations (E2.5.1) given in Example 2.5.

(b) Which of the two matrices A1 and A2 has stronger diagonal dominancy

in the above equations? For which equation does Gauss–Seidel iteration

converge faster, Eq. (P2.9.1) or Eq. (P2.9.2)? What would you conjecture

about the relationship between the convergence speed of Gauss–Seidel

iteration for a set of linear equations and the diagonal dominancy of the

coefficient matrix A?

(c) Is the relaxation technique always helpful for improving the convergence

speed of the Gauss–Seidel iterative method regardless of the value of

the relaxation factor ω?

3

INTERPOLATION

AND CURVE FITTING

There are two topics to be dealt with in this chapter, namely, interpolation1 and

curve fitting. Interpolation is to connect discrete data points in a plausible way

so that one can get reasonable estimates of data points between the given points.

The interpolation curve goes through all data points. Curve fitting, on the other

hand, is to find a curve that could best indicate the trend of a given set of data.

The curve does not have to go through the data points. In some cases, the data

may have different accuracy/reliability/uncertainty and we need the weighted

least-squares curve fitting to process such data.

3.1 INTERPOLATION BY LAGRANGE POLYNOMIAL

For a given set of N + 1 data points {(x0, y0), (x1, y1), . . . , (xN, yN)}, we want

to find the coefficients of an Nth-degree polynomial function to match them:

pN(x) = a0 + a1x + a2x2 + ·· ·+aNxN (3.1.1)

The coefficients can be obtained by solving the following system of linear

equations.

a0 + x0a1 + x2

0a2 + ·· ·+xN

0 aN = y0

a0 + x1a1 + x2

1a2 + ·· ·+xN

1 aN = y1

· · · · · · · · · · · · · · · · · · · · · · · · ·

a0 + xNa1 + x2

Na2 + ·· ·+xN

N aN = yN

(3.1.2)

1 If we estimate the values of the unknown function at the points that are inside/outside the range

of collected data points, we call it the interpolation/extrapolation.

Applied Numerical Methods Using MATLAB, by Yang, Cao, Chung, and Morris

Copyr ight 2005 John Wiley & Sons, I nc., ISBN 0-471-69833-4

117

118 INTERPOLATION AND CURVE FITTING

But, as the number of data points increases, so does the number of unknown

variables and equations, consequently, it may be not so easy to solve. That is

why we look for alternatives to get the coefficients {a0, a1, . . . , aN}.

One of the alternatives is to make use of the Lagrange polynomials

lN(x)=y0

(x − x1)(x − x2) · · · (x − xN)

(x0 − x1)(x0 − x2) · · · (x0 − xN) + y1

(x − x0)(x − x2) · · · (x − xN)

(x1 − x0)(x1 − x2) · · · (x1 − xN)

+· · ·+yN

(x − x0)(x − x1) · · · (x − xN−1)

(xN − x0)(xN − x1) · · · (xN − xN−1)

lN(x)=

N

m=0

ymLN,m(x) with LN,m(x)= N

k=m(x − xk)

N

k=m(xm − xk) =

N

k=m

x − xk

xm − xk

(3.1.3)

It can easily be shown that the graph of this function matches every data point

lN(xm) = ym ∀ m = 0, 1, . . . , N (3.1.4)

since the Lagrange coefficient polynomial LN,m(x) is 1 only for x = xm and zero

for all other data points x = xk (k = m). Note that the Nth-degree polynomial

function matching the given N + 1 points is unique and so Eq. (3.1.1) having

the coefficients obtained from Eq. (3.1.2) must be the same as the Lagrange

polynomial (3.1.3).

Now, we have the MATLAB routine “lagranp()” which finds us the coefficients

of Lagrange polynomial (3.1.3) together with each Lagrange coefficient

polynomial LN,m(x). In order to understand this routine, you should know that

MATLAB deals with polynomials as their coefficient vectors arranged in descending

order and the multiplication of two polynomials corresponds to the convolution

of the coefficient vectors as mentioned in Section 1.1.6.

function [l,L] = lagranp(x,y)

%Input : x = [x0 x1 … xN], y = [y0 y1 … yN]

%Output: l = Lagrange polynomial coefficients of degree N

% L = Lagrange coefficient polynomial

N = length(x)-1; %the degree of polynomial

l = 0;

for m = 1:N + 1

P = 1;

for k = 1:N + 1

if k ~= m, P = conv(P,[1 -x(k)])/(x(m)-x(k)); end

end

L(m,:) = P; %Lagrange coefficient polynomial

l = l + y(m)*P; %Lagrange polynomial (3.1.3)

end

%do_lagranp.m

x = [-2 -1 1 2]; y = [-6 0 0 6]; % given data points

l = lagranp(x,y) % find the Lagrange polynomial

xx = [-2: 0.02 : 2]; yy = polyval(l,xx); %interpolate for [-2,2]

clf, plot(xx,yy,’b’, x,y,’*’) %plot the graph

INTERPOLATION BY NEWTON POLYNOMIAL 119

0

(−1, 0) l3(x) = x 3 − x (1, 0)

(−2, −6)

(2, 6)

−2

−2 −1 1 2

−4

−6

6

4

2

Figure 3.1 The graph of a third-degree Lagrange polynomial.

We make the MATLAB program “do_lagranp.m” to use the routine

“lagranp()” for finding the third-degree polynomial l3(x) which matches the

four given points

{(−2,−6), (−1, 0), (1, 0), (2, 6)}

and to check if the graph of l3(x) really passes the four points. The results from

running this program are depicted in Fig. 3.1.

>>do lagranp

l = 1 0 -1 0 % meaning l3(x) = 1 · x3 + 0 · x2 − 1 · x + 0

3.2 INTERPOLATION BY NEWTON POLYNOMIAL

Although the Lagrange polynomial works pretty well for interpolation irrespective

of the interval widths between the data points along the x-axis, it requires

restarting the whole computation with heavier burden as data points are appended.

Differently from this, the Nth-degree Newton polynomial matching the N + 1

data points {(x0, y0), (x1, y1), . . . , (xN, yN)} can be recursively obtained as the

sum of the (N − 1)th-degree Newton polynomial matching the N data points

{(x0, y0), (x1, y1), . . . , (xN−1, yN−1)} and one additional term.

nN(x) = a0 + a1(x − x0) + a2(x − x0)(x − x1) + ·· ·

= nN−1(x) + aN(x − x0)(x − x1) · · · (x − xN−1) with n0(x) = a0

(3.2.1)

In order to derive a formula to find the successive coefficients {a0, a1, . . . , aN} that make this equation accommodate the data points, we will determine a0 and

a1 so that

n1(x) = n0(x) + a1(x − x0) (3.2.2)

120 INTERPOLATION AND CURVE FITTING

matches the first two data points (x0, y0) and (x1, y1). We need to solve the two

equations

n1(x0) = a0 + a1(x0 − x0) = y0

n1(x1) = a0 + a1(x1 − x0) = y1

to get

a0 = y0, a1 =

y1 − a0

x1 − x0 =

y1 − y0

x1 − x0 ≡ Df0 (3.2.3)

Starting from this first-degree Newton polynomial, we can proceed to the seconddegree

Newton polynomial

n2(x) = n1(x) + a2(x − x0)(x − x1) = a0 + a1(x − x0) + a2(x − x0)(x − x1)

(3.2.4)

which, with the same coefficients a0 and a1 as (3.2.3), still matches the first

two data points (x0, y0) and (x1, y1), since the additional (third) term is zero

at (x0, y0) and (x1, y1). This is to say that the additional polynomial term does

not disturb the matching of previous existing data. Therefore, given the additional

matching condition for the third data point (x2, y2), we only have to

solve

n2(x2) = a0 + a1(x2 − x0) + a2(x2 − x0)(x2 − x1) ≡ y2

for only one more coefficient a2 to get

a2 =

y2 − a0 − a1(x2 − x0)

(x2 − x0)(x2 − x1) =

y2 − y0 −

y1 − y0

x1 − x0

(x2 − x0)

(x2 − x0)(x2 − x1)

=

y2 − y1 + y1 − y0 −

y1 − y0

x1 − x0

(x2 − x1 + x1 − x0)

(x2 − x0)(x2 − x1)

=

y2 − y1

x2 − x1 −

y1 − y0

x1 − x0

x2 − x0 =

Df1 − Df0

x2 − x0 ≡ D2f0 (3.2.5)

Generalizing these results (3.2.3) and (3.2.5) yields the formula to get the Nth

coefficient aN of the Newton polynomial function (3.2.1) as

aN =

DN−1f1 − DN−1f0

xN − x0 ≡ DNf0 (3.2.6)

This is the divided difference, which can be obtained successively from the

second row of Table 3.1.

INTERPOLATION BY NEWTON POLYNOMIAL 121

Table 3.1 Divided Difference Table

xk yk Dfk D2fk D3fk —

x0 y0 Df0 =

y1 − y0

x1 − x0

D2f0 =

Df1 − Df0

x2 − x0

D3f0 =

D2f1 − D2f0

x3 − x0

—

x1 y1 Df1 =

y2 − y1

x2 − x1

D2f1 =

Df2 − Df1

x3 − x1

—

x2 y2 Df2 =

y3 − y2

x3 − x2

—

x3 y3 —

function [n,DD] = newtonp(x,y)

%Input : x = [x0 x1 … xN]

% y = [y0 y1 … yN]

%Output: n = Newton polynomial coefficients of degree N

N = length(x)-1;

DD = zeros(N + 1,N + 1);

DD(1:N + 1,1) = y’;

for k = 2:N + 1

for m = 1: N + 2 – k %Divided Difference Table

DD(m,k) = (DD(m + 1,k – 1) – DD(m,k – 1))/(x(m + k – 1)- x(m));

end

end

a = DD(1,:); %Eq.(3.2.6)

n = a(N+1); %Begin with Eq.(3.2.7)

for k = N:-1:1 %Eq.(3.2.7)

n = [n a(k)] – [0 n*x(k)]; %n(x)*(x – x(k – 1))+a_k – 1

end

Note that, as mentioned in Section 1.3, it is of better computational efficiency to

write the Newton polynomial (3.2.1) in the nested multiplication form as

nN(x) = ((· · · (aN(x − xN−1) + aN−1)(x − xN−2) + ·· ·) + a1)(x − x0) + a0

(3.2.7)

and that the multiplication of two polynomials corresponds to the convolution

of the coefficient vectors as mentioned in Section 1.1.6. We make the MATLAB

routine “newtonp()” to compose the divided difference table like Table 3.1 and

construct the Newton polynomial for a set of data points.

For example, suppose we are to find a Newton polynomial matching the following

data points

{(−2,−6), (−1, 0), (1, 0), (2, 6), (4, 60)}

From these data points, we construct the divided difference table as Table 3.2

and then use this table together with Eq. (3.2.1) to get the Newton polynomial

122 INTERPOLATION AND CURVE FITTING

Table 3.2 Divided differences

xk yk Dfk D2fk D3fk D4fk

−2 −6

0 − (−6)

−1 − (−2) = 6

0 − 6

1 − (−2) = −2

2 − (−2)

2 − (−2) = 1

1 − 1

4 − (−2) = 0

−1 0

0 − 0

1 − (−1) = 0

6 − 0

2 − (−1) = 2

7 − 2

4 − (−1) = 1

1 0

6 − 0

2 − 1 = 6

27 − 6

4 − 1 = 7

2 6

60 − 6

4 − 2 = 27

4 60

as follows:

n(x) = y0 + Df0(x − x0) + D2f0(x − x0)(x − x1)

+ D3f0(x − x0)(x − x1)(x − x2) + 0

= −6 + 6(x − (−2)) − 2(x − (−2))(x − (−1))

+ 1(x − (−2))(x − (−1))(x − 1)

= −6 + 6(x + 2) − 2(x + 2)(x + 1) + (x + 2)(x2 − 1)

= x3 + (−2 + 2)x2 + (6 − 6 − 1)x − 6 + 12 − 4 − 2 = x3 − x

We might begin with not necessarily the first data point, but, say, the third one

(1,0), and proceed as follows to end up with the same result.

n(x) = y2 + Df2(x − x2) + D2f2(x − x2)(x − x3)

+ D3f2(x − x2)(x − x3)(x − x4) + 0

= 0 + 6(x − 1) + 7(x − 1)(x − 2) + 1(x − 1)(x − 2)(x − 4)

= 6(x − 1) + 7(x2 − 3x + 2) + (x2 − 3x + 2)(x − 4)

= x3 + (7 − 7)x2 + (6 − 21 + 14)x − 6 + 14 − 8 = x3 − x

This process is cast into the MATLAB program “do_newtonp.m”, which illustrates

that the Newton polynomial (3.2.1) does not depend on the order of the

data points; that is, changing the order of the data points does not make any

difference.

INTERPOLATION BY NEWTON POLYNOMIAL 123

%do_newtonp.m

x = [-2 -1 1 2 4]; y = [-6 0 0 6 60];

n = newtonp(x,y) %l = lagranp(x,y) for comparison

x = [-1 -2 1 2 4]; y = [ 0 -6 0 6 60];

n1 = newtonp(x,y) %with the order of data changed for comparison

xx = [-2:0.02: 2]; yy = polyval(n,xx);

clf, plot(xx,yy,’b-’,x,y,’*’)

Now, let us see the interpolation problem from the viewpoint of approximation.

For this purpose, suppose we are to approximate some function, say,

f (x) =

1

1 + 8×2

by a polynomial. We first pick up some sample points on the graph of this

function, such as listed below, and look for the polynomial functions n4(x), n8(x),

and n10(x) to match each of the three sets of points, respectively.

xk −1.0 −0.5 0 0.5 1.0

yk 1/9 1/3 1 1/3 1/9

xk −1.0 −0.75 −0.5 −0.25 0 0.25 0.5 0.75 1.0

yk 1/9 2/11 1/3 2/3 1 2/3 1/3 2/11 1/9

xk −1.0 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 1.0

yk 1/9 25/153 25/97 25/57 25/33 1 25/33 25/57 25/97 25/153 1/9

We made the MATLAB program “do_newtonp1.m” to do this job and plot the

graphs of the polynomial functions together with the graph of the true function

f (x) and their error functions separately for comparison as depicted in Fig. 3.2,

where the parts for n8(x) and n10(x) are omitted to provide the readers with

some room for practice.

%do_newtonp1.m – plot Fig.3.2

x = [-1 -0.5 0 0.5 1.0]; y = f31(x);

n = newtonp(x,y)

xx = [-1:0.02: 1]; %the interval to look over

yy = f31(xx); %graph of the true function

yy1 = polyval(n,xx); %graph of the approximate polynomial function

subplot(221), plot(xx,yy,’k-’, x,y,’o’, xx,yy1,’b’)

subplot(222), plot(xx,yy1-yy,’r’) %graph of the error function

function y = f31(x)

y=1./(1+8*x.^2);

124 INTERPOLATION AND CURVE FITTING

(a) 4/8/10th -degree polynomial

approximation

0

−0.5 0 0.5

−0.5 0 0.5

0.5

1

−0.5

0

f (x) 0.5

n4(x): O

n10(x):

n8(x):

n8(x) − f (x)

n4(x) − f (x)

n10(x) − f (x)

(b) The error between the approximating

polynomial and the true function

Figure 3.2 Interpolation from the viewpoint of approximation.

Remark 3.1. Polynomial Wiggle and Runge Phenomenon. Here is one thing to

note. Strangely, increasing the degree of polynomial contributes little to reducing

the approximation error. Rather contrary to our usual expectation, it tends to make

the oscillation strikingly large, which is called the polynomial wiggle and the error

gets bigger in the parts close to both ends as can be seen in Fig. 3.2, which is

called the Runge phenomenon. That is why polynomials of degree 5 or above are

seldom used for the purpose of interpolation, unless they are sure to fit the data.

3.3 APPROXIMATION BY CHEBYSHEV POLYNOMIAL

At the end of the previous section, we considered a polynomial approximation

problem of finding a polynomial close to a given (true) function f (x) and have

the freedom to pick up the target points {x0, x1, . . . , xN} in our own way. Once

the target points have been fixed, it is nothing but an interpolation problem that

can be solved by the Lagrange or Newton polynomial.

In this section, we will think about how to choose the target points for better

approximation, rather than taking equidistant points along the x axis. Noting that

the error tends to get bigger in the parts close to both ends of the interval when

we chose the equidistant target points, it may be helpful to set the target points

denser in the parts close to both ends than in the middle part. In this context, a

possible choice is the projection (onto the x axis) of the equidistant points on the

circle centered at the middle point of the interval along the x axis (see Fig. 3.3).

That is, we can choose in the normalized interval [−1, +1]

x

k = cos

2N + 1 − 2k

2(N + 1)

π for k = 0, 1, . . . ,N (3.3.1a)

and for an arbitrary interval [a, b],

xk =

b − a

2

x

k +

a + b

2 =

b − a

2

cos

2N + 1 − 2k

2(N + 1)

π +

a + b

2

for k = 0, 1, . . . ,N

(3.3.1b)

APPROXIMATION BY CHEBYSHEV POLYNOMIAL 125

−1 1

x0′ = 0

cos 9p/10

x1′ =

cos 7p/10 x2′ = cos 5p/10

3p/10

5p/10

p/10

x3′ =

cos 3p/10

x4′ =

cos p/10

Figure 3.3 Chebyshev nodes (N = 4).

which are referred to as the Chebyshev nodes. The approximating polynomial

obtained on the basis of these Chebyshev nodes is called the Chebyshev polynomial.

Let us try the Chebyshev nodes on approximating the function

f (x) =

1

1 + 8×2

We can set the 5/9/11 Chebyshev nodes by Eq. (3.3.1) and get the Lagrange

or Newton polynomials c4(x), c8(x), and c10(x) matching these target points,

which are called the Chebyshev polynomial. We make the MATLAB program

“do_lagnewch.m” to do this job and plot the graphs of the polynomial functions

together with the graph of the true function f (x) and their error functions separately

for comparison as depicted in Fig. 3.4. The parts for c8(x) and c10(x)

are omitted to give the readers a chance to practice what they have learned in

this section.

%do_lagnewch.m – plot Fig.3.4

N = 4; k = [0:N];

x=cos((2*N + 1 – 2*k)*pi/2/(N + 1)); %Chebyshev nodes(Eq.(3.3.1))

y=f31(x);

c=newtonp(x,y) %Chebyshev polynomial

xx = [-1:0.02: 1]; %the interval to look over

yy = f31(xx); %graph of the true function

yy1 = polyval(c,xx); %graph of the approximate polynomial function

subplot(221), plot(xx,yy,’k-’, x,y,’o’, xx,yy1,’b’)

subplot(222), plot(xx,yy1-yy,’r’) %graph of the error function

Comparing Fig. 3.4 with Fig. 3.2, we see that the maximum deviation of the

Chebyshev polynomial from the true function is considerably less than that of

126 INTERPOLATION AND CURVE FITTING

(a) 4/8/10th -degree polynomial approximation (b) The error between the Chebyshev

approximating polynomial and the true function

0.3

0.2

0.1

0

−0.1

−0.2

−0.3

1.25

1

0.5

0

0

c8(x):

c10(x):

c4(x) − f (x)

f (x)

c8(x) − f (x)

c10(x) − f (x)

c4(x):

−0.5 0.5

−0.5 0 0.5

−0.25

Figure 3.4 Approximation using the Chebyshev polynomial.

Lagrange/Newton polynomial with equidistant nodes. It can also be seen that

increasing the number of the Chebyshev nodes—or, equivalently, increasing

the degree of Chebyshev polynomial—makes a substantial contribution towards

reducing the approximation error.

There are several things to note about the Chebyshev polynomial.

Remark 3.2. Chebyshev Nodes and Chebyshev Coefficient Polynomials Tm(x)

1. The Chebyshev coefficient polynomial is defined as

TN+1(x ) = cos((N + 1) cos−1 x ) for − 1 ≤ x ≤ +1 (3.3.2)

and the Chebyshev nodes defined by Eq. (3.3.1a) are actually zeros of this

function:

TN+1(x ) = cos((N + 1) cos−1 x ) = 0, (N+ 1) cos−1 x = (2k + 1)π/2

2. Equation (3.3.2) can be written via the trigonometric formula in a recursive

form as

TN+1(x ) = cos(cos−1 x + N cos−1 x )

= cos(cos−1 x ) cos(N cos−1 x ) − sin(cos−1 x ) sin(N cos−1 x )

= xTN(x ) +

1

2{cos((N + 1) cos−1 x ) − cos((N − 1) cos−1 x )}

= xTN(x ) +

1

2

TN+1(x ) −

1

2

TN−1(x )

TN+1(x ) = 2xTN(x ) − TN−1(x ) for N ≥ 1 (3.3.3a)

T0(x ) = cos 0 = 1, T1(x ) = cos(cos−1 x ) = x (3.3.3b)

3. At the Chebyshev nodes x

k defined by Eq. (3.3.1a), the set of Chebyshev

coefficient polynomials

{T0(x ), T1(x), . . . , TN(x )}

APPROXIMATION BY CHEBYSHEV POLYNOMIAL 127

are orthogonal in the sense that

N

k=0

Tm(x

k )Tn(x

k) =0 form = n (3.3.4a)

N

k=0

T 2

m(x

k ) =

N + 1

2

for m = 0 (3.3.4b)

N

k=0

T 2

0 (x

k ) = N +1 form = 0 (3.3.4c)

4. The Chebyshev coefficient polynomials TN+1(x ) for up to N = 6 are collected

in Table 3.3, and their graphs are depicted in Fig. 3.5. As can be

seen from the table or the graph, the Chebyshev coefficient polynomials of

even/odd degree (N + 1) are even/odd functions and have an equi-ripple

characteristic with the range of [−1,+1], and the number of rising/falling

(intervals) within the domain of [−1,+1] is N + 1.

We can make use of the orthogonality [Eq. (3.3.4)] of Chebyshev coefficient

polynomials to derive the Chebyshev polynomial approximation formula.

f (x)∼=

cN(x) =

N

m=0

dmTm(x )

x=

2

b−a x−a+b

2 (3.3.5)

−1 0 1

−1

−1

0

1

−1 0 1

−1

0

1

−1 0 1

0

1

−1 0 1

−1

0

1

−1 0 1

0

1

2

0

(c) T2 (x′) (d) T3 (x′)

(e) T4 (x′) (f) T5 (x′)

(a) T0 (x′) = 1 (b) T1 (x′) = x′

−1 1

−1

0

1

Figure 3.5 Chebyshev polynomial functions.

128 INTERPOLATION AND CURVE FITTING

Table 3.3 Chebyshev Coefficient Polynomials

T0(x ) = 1

T1(x ) = x (x : a variable normalized onto [−1, 1])

T2(x ) = 2x2 − 1

T3(x ) = 4x3 − 3x

T4(x ) = 8x4 − 8x2 + 1

T5(x ) = 16x5 − 20x3 + 5x

T6(x ) = 32x6 − 48x4 + 18x2 − 1

T7(x ) = 64x7 − 112x5 + 56x3 − 7x

where

d0 =

1

N + 1

N

k=0

f (xk)T0(x

k ) =

1

N + 1

N

k=0

f (xk) (3.3.6a)

dm =

2

N + 1

N

k=0

f (xk)Tm(x

k )

=

2

N + 1

N

k=0

f (xk) cos

m(2N + 1 − 2k)

2(N + 1)

π for m = 1, 2, . . . ,N

(3.3.6b)

function [c,x,y] = cheby(f,N,a,b)

%Input : f = function name on [a,b]

%Output: c = Newton polynomial coefficients of degree N

% (x,y) = Chebyshev nodes

if nargin == 2, a = -1; b = 1; end

k = [0: N];

theta = (2*N + 1 – 2*k)*pi/(2*N + 2);

xn = cos(theta); %Eq.(3.3.1a)

x = (b – a)/2*xn +(a + b)/2; %Eq.(3.3.1b)

y = feval(f,x);

d(1) = y*ones(N + 1,1)/(N+1);

for m = 2: N + 1

cos_mth = cos((m-1)*theta);

d(m) = y*cos_mth’*2/(N + 1); %Eq.(3.3.6b)

end

xn = [2 -(a + b)]/(b – a); %the inverse of (3.3.1b)

T_0 = 1; T_1 = xn; %Eq.(3.3.3b)

c = d(1)*[0 T_0] +d(2)*T_1; %Eq.(3.3.5)

for m = 3: N + 1

tmp = T_1;

T_1 = 2*conv(xn,T_1) -[0 0 T_0]; %Eq.(3.3.3a)

T_0 = tmp;

c = [0 c] + d(m)*T_1; %Eq.(3.3.5)

end

PADE APPROXIMATION BY RATIONAL FUNCTION 129

We can apply this formula to get the polynomial approximation directly for

a given function f (x), without having to resort to the Lagrange or Newton

polynomial. Given a function, the degree of the approximate polynomial, and the

left/right boundary points of the interval, the above MATLAB routine “cheby()”

uses this formula to make the Chebyshev polynomial approximation.

The following example illustrates that this formula gives the same approximate

polynomial function as could be obtained by applying the Newton polynomial

with the Chebyshev nodes.

Example 3.1. Approximation by Chebyshev Polynomial. Consider the problem

of finding the second-degree (N = 2) polynomial to approximate the function

f (x) = 1/(1 + 8×2). We make the following program “do_cheby.m”, which uses

the MATLAB routine “cheby()” for this job and uses Lagrange/Newton polynomial

with the Chebyshev nodes to do the same job. Readers can run this program

to check if the results are the same.

%do_cheby.m

N = 2; a = -2; b = 2;

[c,x1,y1] = cheby(’f31’,N,a,b) %Chebyshev polynomial ftn

%for comparison with Lagrange/Newton polynomial ftn

k = [0:N]; xn = cos((2*N + 1 – 2*k)*pi/2/(N + 1));%Eq.(3.3.1a):Chebyshev nodes

x = ((b-a)*xn +a + b)/2; %Eq.(3.3.1b)

y = f31(x); n = newtonp(x,y), l = lagranp(x,y)

>>do_cheby

c = -0.3200 -0.0000 1.0000

3.4 PADE APPROXIMATION BY RATIONAL FUNCTION

Pade approximation tries to approximate a function f (x) around a point xo by a

rational function

pM,N(x − xo) =

QM(x − xo)

DN(x − xo)

with M = N or M = N + 1

=

q0 + q1(x − xo) + q2(x − xo)2 + ·· ·+qM(x − xo)M

1 + d1(x − xo) + d2(x − xo)2 + ·· ·+dN(x − xo)N

(3.4.1)

where f (xo), f (xo), f (2)(xo), . . . , f (M+N)(xo) are known.

How do we find such a rational function? We write the Taylor series expansion

of f (x) up to degree M + N at x = xo as

130 INTERPOLATION AND CURVE FITTING

f (x) ≈ TM+N(x − xo) = f (xo) + f (xo)(x − xo)

+

f (2)(xo)

2

(x − xo)2 + ·· ·+

f (M+N)(xo)

(M + N)!

(x − xo)M+N

= a0 + a1(x − xo) + a2(x − xo)2 + ·· ·+aM+N(x − xo)M+N (3.4.2)

Assuming xo = 0 for simplicity, we get the coefficients of DN(x) and QM(x)

such that

TM+N(x) −

QM(x)

DN(x) = 0

(a0 + a1x + ·· ·+aM+NxM+N)(1 + d1x + ·· ·+dNxN)

−(q0 + q1x + ·· ·+qMxM)

1 + d1x + d2x2 + ·· ·+dNxN = 0

(a0 + a1x + ·· ·+aM+NxM+N)(1 + d1x + ·· ·+dNxN)

= q0 + q1x + ·· ·+qMxM (3.4.3)

by solving the following equations:

a0 = q0

a1 + a0d1 = q1

a2 + a1d1 + a0d2 = q2

· · · · · · · · · · · · · · · aM + aM−1d1 + aM−2d2 · · · + aM−NdN = qM

(3.4.4a)

aM+1 + aMd1 + aM−1d2 · · · + aM−N+1dN = 0

aM+2 + aM+1d1 + aMd2 · · · + aM−N+2dN = 0

· · · · · · · · · · · · · · · · · · aM+N + aM+N−1d1 + aM+N−2d2 · · · + aMdN = 0

(3.4.4b)

Here, we must first solve Eq. (3.4.4b) for d1, d2, . . . , dN and then substitute di’s

into Eq. (3.4.4a) to obtain q0, q1, . . . , qM.

The MATLAB routine “padeap()” implements this scheme to find the coefficient

vectors of the numerator/denominator polynomial QM(x)/DN(x) of the

Pade approximation for a given function f (x). Note the following things:

ž The derivatives f (xo), f (2)(xo), . . . , f (M+N)(xo) up to order (M + N) are

computed numerically by using the routine “difapx()”, that will be introduced

in Section 5.3.

ž In order to compute the values of the Pade approximate function, we substitute

(x − xo) for x in pM,N(x) which has been obtained with the assumption

that xo = 0.

PADE APPROXIMATION BY RATIONAL FUNCTION 131

function [num,den] = padeap(f,xo,M,N,x0,xf)

%Input : f = function to be approximated around in [xo, xf]

%Output: num = numerator coeffs of Pade approximation of degree M

% den = denominator coeffs of Pade approximation of degree N

a(1) = feval(f,xo);

h = .01; tmp = 1;

for i = 1:M + N

tmp = tmp*i*h; %i!h^i

dix = difapx(i,[-i i])*feval(f,xo+[-i:i]*h)’; %derivative(Section 5.3)

a(i + 1) = dix/tmp; %Taylor series coefficient

end

for m = 1:N

n = 1:N; A(m,n) = a(M + 1 + m – n);

b(m) = -a(M + 1 + m);

end

d = A\b’; %Eq.(3.4.4b)

for m = 1: M + 1

mm = min(m – 1,N);

q(m) = a(m:-1:m – mm)*[1; d(1:mm)]; %Eq.(3.4.4a)

end

num = q(M + 1:-1:1)/d(N); den = [d(N:-1:1)’ 1]/d(N); %descending order

if nargout == 0 % plot the true ftn, Pade ftn and Taylor expansion

if nargin < 6, x0 = xo – 1; xf = xo + 1; end

x = x0+[xf-x0]/100*[0:100]; yt = feval(f,x);

x1 = x-xo; yp = polyval(num,x1)./polyval(den,x1);

yT = polyval(a(M + N + 1:-1:1),x1);

clf, plot(x,yt,’k’, x,yp,’r’, x,yT,’b’)

end

Example 3.2. Pade Approximation for f (x) = ex . Let’s find the Pade approximation

p3,2(x) = Q3(x)/D2(x) for f (x) = ex around xo = 0. We make the

MATLAB program “do_pade.m”, which uses the routine “padeap()” for this

job and uses it again with no output argument to see the graphic results as

depicted in Fig. 3.6.

>>do_pade %Pade approximation

n = 0.3333 2.9996 11.9994 19.9988

d = 1.0000 -7.9997 19.9988

%do_pade.m to get the Pade approximation for f(x) = e^x

f1 = inline(’exp(x)’,’x’);

M = 3; N = 2; %the degrees of Numerator Q(x) and Denominator D(x)

xo = 0; %the center of Taylor series expansion

[n,d] = padeap(f1,xo,M,N) %to get the coefficients of Q(x)/P(x)

x0 = -3.5; xf = 0.5; %left/right boundary of the interval

padeap(f1,xo,M,N,x0,xf) %to see the graphic results

To confirm and support this result from the analytical point of view and to help

the readers understand the internal mechanism, we perform the hand-calculation

132 INTERPOLATION AND CURVE FITTING

0

−0.5

−1

−3.5 −3 −2.5

f (x )

p3,2(x )

Ty (x )

−2 −1.5 −1 −0.5 0 0.5

0.5

1

Figure 3.6 Pade approximation and Taylor series expansion for f(x) = ex(Example 3.2.).

procedure. First, we write the Taylor series expansion at x = 0 up to degree

M + N = 5 for the given function f (x) = ex as

Ty(x) =

M+N

k=0

f (k)(x)

k!

xk = 1 + x +

1

2

x2 +

1

3!

x3 +

1

4!

x4 +

1

5!

x5 + ·· ·

(E3.2.1)

whose coefficients are

a0 = 1, a1 = 1, a2 =

1

2

, a3 =

1

6

, a4 =

1

24

, a5 =

1

120

, . . . (E3.2.2)

We put this into Eq. (3.4.4b) with M = 3,N = 2 and solve it for di’s to get

D2(x) = 1 + d1x + d2x2.

a4 + a3d1 + a2d2 = 0

a3 + a2d1 + a1d2 = 0 , 1/6 1/2

1/24 1/6d1

d2 = −1/24

−1/120 , d1

d2 = −2/5

a1/20

(E3.2.3)

Substituting this to Eq. (3.4.4a) yields

q0 = a0 = 1

q1 = a1 + a0d1 = 1 + 1 × (−2/5) = 3/5

q2 = a2 + a1d1 + a0d2 = 1/2 + 1 × (−2/5) + 1 × (1/20) = 3/20

q3 = a3 + a2d1 + a1d2 = 1/6 + (1/2) × (−2/5) + 1 × (1/20) = 1/60

(E3.2.4)

INTERPOLATION BY CUBIC SPLINE 133

With these coefficients, we write the Pade approximate function as

p3,2(x) =

Q3(x)

D2(x) =

1 + (3/5)x + (3/20)x2 + (1/60)x3

1 + (−2/5)x + (1/20)x2

=

(1/3)x3 + 3×2 + 12x + 20

x2 − 8x + 20

(E3.2.5)

3.5 INTERPOLATION BY CUBIC SPLINE

If we use the Lagrange/Newton polynomial to interpolate a given set of N + 1

data points, the polynomial is usually of degree N and so has N − 1 local extrema

(maxima/minima). Thus, it will show a wild swing/oscillation (called ‘polynomial

wiggle’), particularly near the ends of the whole interval as the number of data

points increases and so the degree of the polynomial gets higher, as illustrated

in Fig. 3.2. Then, how about a piecewise-linear approach, like assigning the

individual approximate polynomial to every subinterval between data points?

How about just a linear interpolation—that is, connecting the data points by

a straight line? It is so simple, but too short of smoothness. Even with the

second-degree polynomial, the piecewise-quadratic curve is not smooth enough

to please our eyes, since the second-order derivatives of quadratic polynomials

for adjacent subintervals can’t be made to conform with each other. In real

life, there are many cases where the continuity of second-order derivatives is

desirable. For example, it is very important to ensure the smoothness up to order 2

for interpolation needed in CAD (computer-aided design)/CAM (computer-aided

manufacturing), computer graphic, and robot path/trajectory planning. That’s why

we often resort to the piecewise-cubic curve constructed by the individual thirddegree

polynomials assigned to each subinterval, which is called the cubic spline

interpolation. (A spline is a kind of template that architects use to draw a smooth

curve between two points.)

For a given set of data points {(xk, yk), k = 0 : N}, the cubic spline s(x)

consists of N cubic polynomial sk(x)’s assigned to each subinterval satisfying

the following constraints (S0)–(S4).

(S0) s(x) = sk(x) = Sk,3(x − xk)3 + Sk,2(x − xk)2 + Sk,1(x − xk) + Sk,0

for x ∈ [xk, xk+1], k = 0 : N

(S1) sk(xk) = Sk,0 = yk for k = 0 : N

(S2) sk−1(xk) ≡ sk(xk) = Sk,0 = yk for k = 1 : N − 1

(S3) sk −1(xk) ≡ sk(xk) = Sk,1 for k = 1 : N − 1

(S4) s k−1(xk) ≡ s k (xk) = 2Sk,2 for k = 1 : N − 1

These constraints (S1)–(S4) amount to a set of N + 1 + 3(N − 1) = 4N − 2

linear equations having 4N coefficients of the N cubic polynomials

{Sk,0, Sk,1, Sk,2, Sk,3, k = 0 : N − 1}

134 INTERPOLATION AND CURVE FITTING

Table 3.4 Boundary Conditions for a Cubic Spline

(i) First-order derivatives

specified

s0(x0) = S0,1, sN (xN) = SN,1

(ii) Second-order

derivatives specified

(end-curvature adjusted)

s 0 (x0) = 2S0,2, s N(xN) = 2SN,2

(iii) Second-order derivatives s 0 (x0) ≡ s 1 (x1) +

h0

h1

(s 1 (x1) − s 2 (x2))

extrapolated

s N(xN) ≡ s N−1(xN−1) +

hN−1

hN−2

(s N−1(xN−1) − s N−2(xN−2))

as their unknowns. Two additional equations necessary for the equations to be

solvable are supposed to come from the boundary conditions for the first/secondorder

derivatives at the end points (x0, y0) and (xN, yN) as listed in Table 3.4.

Now, noting from (S1) that Sk,0 = yk, we will arrange the constraints (S2)–(S4)

and eliminate Sk,1, Sk,3’s to set up a set of equations with respect to the N + 1

unknowns {Sk,2, k = 0 : N}. In order to do so, we denote each interval width by

hk = xk+1 − xk and substitute (S0) into (S4) to write

s k (xk+1) = 6Sk,3hk + 2Sk,2 ≡ s k+1(xk+1) = 2Sk+1,2

Sk,3hk =

1

3

(Sk+1,2 − Sk,2) (3.5.1a)

Sk−1,3hk−1 =

1

3

(Sk,2 − Sk−1,2) (3.5.1b)

We substitute these equations into (S2) with k + 1 in place of k

sk(xk+1) = Sk,3(xk+1 − xk)3 + Sk,2(xk+1 − xk)2 + Sk,1(xk+1 − xk) + Sk,0 ≡ yk+1

Sk,3h3

k + Sk,2h2

k + Sk,1hk + yk ≡ yk+1

to eliminate Sk,3’s and rewrite it as

hk

3

(Sk+1,2 − Sk,2) + Sk,2hk + Sk,1 =

yk+1 − yk

hk = dyk

hk(Sk+1,2 + 2Sk,2) + 3Sk,1 = 3 dyk (3.5.2a)

hk−1(Sk,2 + 2Sk−1,2) + 3Sk−1,1 = 3 dyk−1 (3.5.2b)

We also substitute Eq. (3.5.1b) into (S3)

sk −1(xk) = 3Sk−1,3h2

k−1 + 2Sk−1,2hk−1 + Sk−1,1 ≡ sk (xk) = Sk,1

INTERPOLATION BY CUBIC SPLINE 135

to write

Sk,1 − Sk−1,1 = hk−1(Sk,2 − Sk−1,2) + 2hk−1Sk−1,2 = hk−1(Sk,2 + Sk−1,2)

(3.5.3)

In order to use this for eliminating Sk,1 from Eq. (3.5.2), we subtract (3.5.2b)

from (3.5.2a) to write

hk(Sk+1,2 + 2Sk,2) − hk−1(Sk,2 + 2Sk−1,2) + 3(Sk,1 − Sk−1,1) = 3(dyk − dyk−1)

and then substitute Eq. (3.5.3) into this to write

hk(Sk+1,2 + 2Sk,2) − hk−1(Sk,2 + 2Sk−1,2) + 3hk−1(Sk,2 + Sk−1,2)

= 3(dyk − dyk−1)

hk−1Sk−1,2 + 2(hk−1 + hk)Sk,2 + hkSk+1,2 = 3(dyk − dyk−1) (3.5.4)

for k = 1 : N − 1

Since these are N − 1 equations with respect to N + 1 unknowns {Sk,2, k = 0 :

N}, we need two more equations from the boundary conditions to be given as

listed in Table 3.4.

How do we convert the boundary condition into equations? In the case where

the first-order derivatives on the two boundary points are given as (i) in Table 3.4,

we write Eq. (3.5.2a) for k = 0 as

h0(S1,2 + 2S0,2) + 3S0,1 = 3 dy0, 2h0S0,2 + h0S1,2 = 3(dy0 − S0,1)

(3.5.5a)

We also write Eq. (3.5.2b) for k = N as

hN−1(SN,2 + 2SN−1,2) + 3SN−1,1 = 3 dyN−1

and substitute (3.5.3)(k = N) into this to write

hN−1(SN,2 + 2SN−1,2) + 3SN,1 − 3hN−1(SN,2 + SN−1,2) = 3 dyN−1

hN−1SN−1,2 + 2hN−1SN,2 = 3(SN,1 − dyN−1) (3.5.5b)

Equations (3.5.5a) and (3.5.5b) are two additional equations that we need to solve

Eq. (3.5.4) and that’s it. In the case where the second-order derivatives on the

two boundary points are given as (ii) in Table 3.4, S0,2 and SN,2 are directly

known from the boundary conditions as

S0,2 = s 0 (x0)/2, SN,2 = s N(xN)/2 (3.5.6)

136 INTERPOLATION AND CURVE FITTING

and, subsequently, we have just N − 1 unknowns. In the case where the secondorder

derivatives on the two boundary points are given as (iii) in Table 3.4

s 0 (x0) ≡ s 1 (x1) +

h0

h1

(s 1 (x1) − s 2 (x2))

s N(xN) ≡ s N−1(xN−1) +

hN−1

hN−2

(s N−1(xN−1) − s N−2(xN−2))

we can instantly convert these into two equations with respect to S0,2 and SN,2 as

h1S0,2 − (h0 + h1)S1,2 + h0S2,2 = 0 (3.5.7a)

hN−2SN,2 − (hN−1 + hN−2)SN−1,2 + hN−1SN−2,2 = 0 (3.5.7b)

Finally, we combine the two equations (3.5.5a) and (3.5.5b) with Eq. (3.5.4)

to write it in the matrix–vector form as

2h0 h0 0 ž ž

h0 2(h0 + h1) h1 ž ž

0 ž ž ž 0

ž ž hN−2 2(hN−2 + hN−1) hN−1

ž ž 0 hN−1 2hN−1

S0,2

S1,2

ž

SN−1,2

SN,2

=

3(dy0 − S0,1)

3(dy1 − dy0)

ž

3(dyN−1 − dyN−2)

3(SN,1 − dyN−1)

(3.5.8)

After solving this system of equation for {Sk,2, k = 0 : N}, we substitute them

into (S1), (3.5.2), and (3.5.1) to get the other coefficients of the cubic spline as

Sk,0

(S1) = yk, Sk,1

(3.5.2) = dyk −

hk

3

(Sk+1,2 + 2Sk,2), Sk,3

(3.5.1) =

Sk+1,2 − Sk,2

3hk

(3.5.9)

The MATLAB routine “cspline()” constructs Eq.(3.5.8), solves it to get the

cubic spline coefficients for given x, y coordinates of the data points and the

boundary conditions, uses the mkpp() routine to get the piecewise polynomial

expression, and then uses the ppval() routine to obtain the value(s) of the piecewise

polynomial function for xi—that is, the interpolation over xi. The type of

the boundary condition is supposed to be specified by the third input argument

KC. In the case where the boundary condition is given as (i)/(ii) in Table 3.4,

the input argument KC should be set to 1/2 and the fourth and fifth input arguments

must be the first/second derivatives at the end points. In the case where

the boundary condition is given as extrapolated like (iii) in Table 3.4, the input

argument KC should be set to 3 and the fourth and fifth input arguments do not

need to be fed.

INTERPOLATION BY CUBIC SPLINE 137

function [yi,S] = cspline(x,y,xi,KC,dy0,dyN)

%This function finds the cubic splines for the input data points (x,y)

%Input: x = [x0 x1 … xN], y = [y0 y1 … yN], xi=interpolation points

% KC = 1/2 for 1st/2nd derivatives on boundary specified

% KC = 3 for 2nd derivative on boundary extrapolated

% dy0 = S’(x0) = S01: initial derivative

% dyN = S’(xN) = SN1: final derivative

%Output: S(n,k); n = 1:N, k = 1,4 in descending order

if nargin < 6, dyN = 0; end, if nargin < 5, dy0 = 0; end

if nargin < 4, KC = 0; end

N = length(x) – 1;

% constructs a set of equations w.r.t. {S(n,2), n = 1:N + 1}

A = zeros(N + 1,N + 1); b = zeros(N + 1,1);

S = zeros(N + 1,4); % Cubic spline coefficient matrix

k = 1:N; h(k) = x(k + 1) – x(k); dy(k) = (y(k + 1) – y(k))/h(k);

% Boundary condition

if KC <= 1 %1st derivatives specified

A(1,1:2) = [2*h(1) h(1)]; b(1) = 3*(dy(1) – dy0); %Eq.(3.5.5a)

A(N + 1,N:N + 1) = [h(N) 2*h(N)]; b(N + 1) = 3*(dyN – dy(N));%Eq.(3.5.5b)

elseif KC == 2 %2nd derivatives specified

A(1,1) = 2; b(1) = dy0; A(N + 1,N+1) = 2; b(N + 1) = dyN; %Eq.(3.5.6)

else %2nd derivatives extrapolated

A(1,1:3) = [h(2) – h(1) – h(2) h(1)]; %Eq.(3.5.7)

A(N + 1,N-1:N + 1) = [h(N) – h(N)-h(N – 1) h(N – 1)];

end

for m = 2:N %Eq.(3.5.8)

A(m,m – 1:m + 1) = [h(m – 1) 2*(h(m – 1) + h(m)) h(m)];

b(m) = 3*(dy(m) – dy(m – 1));

end

S(:,3) = A\b;

% Cubic spline coefficients

for m = 1: N

S(m,4) = (S(m+1,3)-S(m,3))/3/h(m); %Eq.(3.5.9)

S(m,2) = dy(m) -h(m)/3*(S(m + 1,3)+2*S(m,3));

S(m,1) = y(m);

end

S = S(1:N, 4:-1:1); %descending order

pp = mkpp(x,S); %make piecewise polynomial

yi = ppval(pp,xi); %values of piecewise polynomial ftn

(cf) See Problem 1.11 for the usages of the MATLAB routines mkpp() and ppval().

Example 3.3. Cubic Spline. Consider the problem of finding the cubic spline

interpolation for the N + 1 = 4 data points

{(0, 0), (1, 1), (2, 4), (3, 5)} (E3.3.1)

subject to the boundary condition

s0 (x0) = s0(0) = S0,1 = 2, sN (xN) = h3(3) = h3,1 = 2 (E3.3.2)

With the subinterval widths on the x-axis and the first divided differences as

h0 = h1 = h2 = h3 = 1

138 INTERPOLATION AND CURVE FITTING

dy0 =

y1 − y0

h0 = 1, dy1

=

y2 − y1

h1 = 3, dy2

=

y3 − y2

h2 = 1 (E3.3.3)

we write Eq. (3.5.8) as

2 1 0 0

1 4 1 0

0 1 4 1

0 0 1 2

S0,2

S1,2

S2,2

S3,2

=

3(dy0 − S0,1)

3(dy1 − dy0)

3(dy2 − dy1)

3(S3,1 − dy1)

=

−3

6

−6

3

(E3.3.4)

Then we solve this equation to get

S0,2 = −3, S1,2 = 3, S2,2 = −3, S3,2 = 3 (E3.3.5)

and substitute this into Eq. (3.5.9) to obtain

S0,0 = 0, S1,0 = 1, S2,0 = 4 (E3.3.6)

S0,1 = dy0 −

h0

3

(S1,2 + 2S0,2) = 1 −

1

3

(3 + 2 × (−3)) = 2 (E3.3.7a)

S1,1 = dy1 −

h1

3

(S2,2 + 2S1,2) = 3 −

1

3

(−3 + 2 × 3) = 2 (E3.3.7b)

S2,1 = dy2 −

h2

3

(S3,2 + 2S2,2) = 1 −

1

3

(3 + 2 × (−3)) = 2 (E3.3.7c)

S0,3 =

S1,2 − S0,2

3h0 =

3 − (−3)

3 = 2 (E3.3.8a)

S1,3 =

S2,2 − S1,2

3h1 = −3 − 3

3 = −2 (E3.3.8b)

S2,3 =

S3,2 − S2,2

3h2 =

3 − (−3)

3 = 2 (E3.3.8c)

%do_csplines.m

KC = 1; dy0 = 2; dyN = 2; % with specified 1st derivatives on boundary

x = [0 1 2 3]; y = [0 1 4 5];

xi = x(1)+[0:200]*(x(end)-x(1))/200; %intermediate points

[yi,S] = cspline(x,y,xi,KC,dy0,dyN); S %cubic spline interpolation

clf, plot(x,y,’ko’,xi,yi,’k:’)

yi = spline(x,[dy0 y dyN],xi); %for comparison with MATLAB built-in ftn

hold on, pause, plot(x,y,’ro’,xi,yi,’r:’)

yi = spline(x,y,xi); %for comparison with MATLAB built-in ftn

pause, plot(x,y,’bo’,xi,yi,’b’)

KC = 3; [yi,S] = cspline(x,y,xi,KC);%with the 2nd derivatives extrapolated

pause, plot(x,y,’ko’,xi,yi,’k’)

HERMITE INTERPOLATING POLYNOMIAL 139

0

−1

0

1

2

3

4

5

0.5

: 1st derivatives on the end points specified

: 2nd derivatives on the end points extrapolated

1 1.5 2 2.5 3

Figure 3.7 Cubic splines for Example 3.3.

Finally, we can write the cubic spline equations collectively from (S0) as

s0(x) = S0,3(x − x0)3 + S0,2(x − x0)2 + S0,1(x − x0) + S0,0

= 2×3 − 3×2 + 2x + 0

s1(x) = S1,3(x − x1)3 + S1,2(x − x1)2 + S1,1(x − x1) + S1,0

= −2(x − 1)3 + 3(x − 1)2 + 2(x − 1) + 1

s2(x) = S2,3(x − x2)3 + S2,2(x − x2)2 + S2,1(x − x2) + S2,0

= 2(x − 2)3 − 3(x − 2)2 + 2(x − 1) + 4

We make and run the program “do_csplines.m”, which uses the routine

“cspline()” to compute the cubic spline coefficients {Sk,3, Sk,2, Sk,1, Sk,0, k = 0 : N − 1} and obtain the value(s) of the cubic spline function for xi (i.e., the

interpolation over xi) and then plots the result as depicted in Fig. 3.7. We also

compare this result with that obtained by using the MATLAB built-in function

“spline(x,y,xi)”, which works with the boundary condition of type (i) for the

second input argument given as [dy0 y dyN], and with the boundary condition

of type (iii) for the same lengths of x and y.

>>do_csplines %cubic spline

S = 2.0000 -3.0000 2.0000 0

-2.0000 3.0000 2.0000 1.0000

2.0000 -3.0000 2.0000 4.0000

3.6 HERMITE INTERPOLATING POLYNOMIAL

In some cases, we need to find the polynomial function that not only passes

through the given points, but also has the specified derivatives at every data

point. We call such a polynomial the Hermite interpolating polynomial or the

osculating polynomial.

140 INTERPOLATION AND CURVE FITTING

For simplicity, we consider a third-order polynomial

h(x) = H3x3 + H2x2 + H1x + H0 (3.6.1)

matching just two points (x0, y0), (x1, y1) and having the specified first derivatives

y0, y1 at the points. We can obtain the four coefficients H3,H2,H1,H0 by solving

h(x0) = H3x3

0 + H2x2

0 + H1x0 + H0 = y0

h(x1) = H3x3

1 + H2x2

1 + H1x1 + H0 = y1

h(x0) = 3H3x2

0 + 2H2x0 + H1 = y0

h(x1) = 3H3x2

1 + 2H2x1 + H1 = y1

(3.6.2)

As an alternative, we approximate the specified derivatives at the data points by

their differences

y0 =

h(x0 + ε) − h(x0)

ε =

y2 − y0

ε

, y1 =

h(x1) − h(x1 − ε)

ε =

y1 − y3

ε

(3.6.3)

and find the Lagrange/Newton polynomial matching the four points

(x0, y0), (x2 = x0 + ε, y2 = y0 + y0ε), (x3 = x1 − ε, y3 = y1 − y1 ε), (x1, y1)

(3.6.4)

The MATLAB routine “hermit()” constructs Eq. (3.6.2) and solves it to get

the Hermite interpolating polynomial coefficients for a single interval given the

two end points and the derivatives at them as the input arguments. The next

routine “hermits()” uses “hermit()” to get the Hermite coefficients for a set

of multiple subintervals.

function H = hermit(x0,y0,dy0,x1,y1,dy1)

A = [x0^3 x0^2 x0 1; x1^3 x1^2 x1 1;

3*x0^2 2*x0 1 0; 3*x1^2 2*x1 1 0];

b = [y0 y1 dy0 dy1]’; %Eq.(3.6-2)

H = (A\b)’;

function H = hermits(x,y,dy)

% finds Hermite interpolating polynomials for multiple subintervals

%Input : [x,y],dy – points and derivatives at the points

%Output: H = coefficients of cubic Hermite interpolating polynomials

for n = 1:length(x)-1

H(n,:) = hermit(0,y(n),dy(n),x(n + 1)-x(n),y(n + 1),dy(n + 1));

end

Example 3.4. Hermite Interpolating Polynomial. Consider the problem of finding

the polynomial interpolation for the N + 1 = 4 data points

{(0, 0), (1, 1), (2, 4), (3, 5)} (E3.4.1)

TWO-DIMENSIONAL INTERPOLATION 141

subject to the conditions

h0(x0) = h0(0) = 2, h1(1) = 0, h2(2) = 0, hN(xN) = h3(3) = 2

(E3.4.2)

For this problem, we only have to type the following statements in the MATLAB

command window.

>>x = [0 1 2 3]; y = [0 1 4 5]; dy = [2 0 0 2]; xi = [0:0.01:3];

>>H = hermits(x,y,dy); yi = ppval(mkpp(x,H), xi);

3.7 TWO-DIMENSIONAL INTERPOLATION

In this section we deal with only the simplest way of two-dimensional

interpolation—that is, a generalization of piecewise linear interpolation called

x1

x1

x1

x2

x2

x2

x3

x3

x3

y1

y2

y3

y1

y2

y3

y1 (x1, y1)

(x1, y2)

(x1, y3)

(x2, y1)

(x2, y2)

(x2, y3)

(x3, y1)

(x3, y2)

(x3, y3)

z1, 1

Z

Zi

Xi Yi (Xi, Yi)

[Xi, Yi] = meshgrid (xi, yi)

Zi = interp2 (X, Y, Z, Xi, Yi)

or interp2 (x, y, Z, Xi, Yi)

xi = [x1 x2 x3 x4 x5]

z2, 1

z3, 1

z1, 2

z2, 2

z3, 2

z1, 3

z2, 3

z3, 3

y2

y3

z1, 1

i

z2, 1

i

z3, 1

i

z4, 1

i

z5, 1

i

x1

i

x1

i

x1

i

x1

i

x1

i

x2

i

x2

i

x2

i

x2

i

x2

i

x3

i

x3

i

x3

i

x3

i

x3

i

x4

i

x4

i

x4

i

x4

i

x4

i

x5

i

i i i i i yi = [y1 y2 y3 y4 y5] i i i i i

x5

i

x5

i

x5

i

x5

i

y1

i (x1, y1) i i

(x1, y2) i i

(x1, y3) i i

(x1, y4) i i

(x1, y5) i i

(x2, y1) i i

(x2, y2) i i

(x2, y3) i i

(x2, y4) i i

(x2, y5) i i

(x3, y1) i i

(x3, y2) i i

(x3, y3) i i

(x3, y4) i i

(x3, y5) i i

(x4, y1) i i

(x4, y2) i i

(x4, y3) i i

(x4, y4) i i

(x4, y5) i i

(x5, y1) i i

(x5, y2) i i

(x5, y3) i i

(x5, y4) i i

(x5, y5) i i

y2

i

y3

i

y4

i

y5

i

y1

i

y2

i

y3

i

y4

i

y5

i

y1

i

y2

i

y3

i

y4

i

y5

i

y1

i

y2

i

y3

i

y4

i

y5

i

y1

i

y2

i

y3

i

y4

i

y5

i

z1, 2

i

z2, 2

i

z3, 2

i

z4, 2

i

z5, 2

i

z1, 3

i

z2, 3

i

z3, 3

i

z4, 3

i

z5, 3

i

z1, 4

i

z2, 4

i

z3, 4

i

z4, 4

i

z5, 4

i

z1, 5

i

z2, 5

i

z3, 5

i

z4, 5

i

z5, 5

i

x = [x1 x2 x3]

X

y = [y1 y2 y3]

Y

[X, Y] = meshgrid (x, y)

(X, Y)

Figure 3.8 A two-dimensional interpolation using Zi = interp2() on the grid points

[Xi,Yi] generated by the meshgrid() command.

142 INTERPOLATION AND CURVE FITTING

the bilinear interpolation. The bilinear interpolation for a point (x, y) on the rectangular

sub-region having (xm−1, yn−1) and (xm, yn) as its left-upper/right-lower

corner points is described by the following formula.

z(x, yn−1) =

xm − x

xm − xm−1

zm−1,n−1 +

x − xm−1

xm − xm−1

zm,n−1 (3.7.1a)

z(x, yn) =

xm − x

xm − xm−1

zm−1,n +

x − xm−1

xm − xm−1

zm,n (3.7.1b)

z(x, y) =

yn − y

yn − yn−1

z(x, yn−1) +

y − yn−1

yn − yn−1

z(x, yn)

=

1

(xm − xm−1)(yn − yn−1) {(xm − x)(yn − y)zm−1,n−1

+ (x − xm−1)(yn − y)zm,n−1 + (xm − x)(y − yn−1)zm−1,n

+ (x − xm−1)(y − yn−1)zm,n} for xm−1 ≤ x ≤ xm, yn−1 ≤ y ≤ yn

(3.7.2)

function Zi = intrp2(x,y,Z,xi,yi)

%To interpolate Z(x,y) on (xi,yi)

M = length(x); N = length(y);

Mi = length(xi); Ni = length(yi);

for mi = 1:Mi

for ni = 1:Ni

for m = 2:M

for n = 2:N

break1 = 0;

if xi(mi) <= x(m) & yi(ni) <= y(n)

tmp = (x(m)-xi(mi))*(y(n)-yi(ni))*Z(n – 1,m – 1)…

+(xi(mi) – x(m-1))*(y(n) – yi(ni))*Z(n – 1,m)…

+(x(m) – xi(mi))*(yi(ni) – y(n – 1))*Z(n,m – 1)…

+(xi(m) – x(m-1))*(yi(ni) – y(n-1))*Z(n,m);

Zi(ni,mi) = tmp/(x(m) – x(m-1))/(y(n) – y(n-1)); %Eq.(3.7.2)

break1 = 1;

end

if break1 > 0 break, end

end

if break1 > 0 break, end

end

end

end

This formula is cast into the MATLAB routine “intrp2()”, which is so named

in order to distinguish it from the MATLAB built-in routine “interp2()”. Note

that in reference to Fig. 3.8, the given values of data at grid points (x(m),y(n))

and the interpolated values for intermediate points (xi(m),yi(n)) are stored in

Z(n,m) and Zi(n,m), respectively.

CURVE FITTING 143

%do_interp2.m

% 2-dimensional interpolation for Ex 3.5

xi = -2:0.1:2; yi = -2:0.1:2;

[Xi,Yi] = meshgrid(xi,yi);

Z0 = Xi.^2 + Yi.^2; %(E3.5.1)

subplot(131), mesh(Xi,Yi,Z0)

x = -2:0.5:2; y = -2:0.5:2;

[X,Y] = meshgrid(x,y);

Z = X.^2 + Y.^2;

subplot(132), mesh(X,Y,Z)

Zi = interp2(x,y,Z,Xi,Yi); %built-in routine

subplot(133), mesh(xi,yi,Zi)

Zi = intrp2(x,y,Z,xi,yi); %our own routine

pause, mesh(xi,yi,Zi)

norm(Z0 – Zi)/norm(Z0)

Example 3.5. Two-Dimensional Bilinear Interpolation. We consider interpolating

the sample values of a function

f (x, y) = x2 + y2 (E3.5.1)

for the 5 × 5 grid over the 21 × 21 grid on the domain D = {(x, y)| − 2 ≤ x ≤ 2,−2 ≤ y ≤ 2}.

We make the MATLAB program “do_interp2.m”, which uses the routine

“intrp2()” to do this job, compares its function with that of the MATLAB

built-in routine “interp2()”, and computes a kind of relative error to estimate

how close the interpolated values are to the original values. The graphic results

of running this program are depicted in Fig. 3.9, which shows that we obtained

a reasonable approximation with the error of 2.6% from less than 1/16 of the

original data. It is implied that the sampling may be a simple data compression

method, as long as the interpolated data are little impaired.

3.8 CURVE FITTING

When many sample data pairs {(xk, yk), k = 0 : M} are available, we often need

to grasp the relationship between the two variables or to describe the trend of the

5

02

−2 −2

2

0 0

(a) True function (b) The function over (c) Bilinear interpolation

sample grid

10

5

02

−2 −2

2

0 0

10

5

02

−2 −2

2

0 0

10

Figure 3.9 Two-dimensional interpolation (Example 3.5).

144 INTERPOLATION AND CURVE FITTING

data, hopefully in a form of function y = f (x). But, as mentioned in Remark 3.1,

the polynomial approach meets with the polynomial wiggle and/or Runge phenomenon,

which makes it not attractive for approximation purpose. Although the

cubic spline approach may be a roundabout toward the smoothness as explained

in Section 3.5, it has too many parameters and so does not seem to be an efficient

way of describing the relationship or the trend, since every subinterval

needs four coefficients. What other choices do we have? Noting that many data

are susceptible to some error, we don’t have to try to find a function passing

exactly through every point. Instead of pursuing the exact matching at every data

point, we look for an approximate function (not necessarily a polynomial) that

describes the data points as a whole with the smallest error in some sense, which

is called the curve fitting.

As a reasonable means, we consider the least-squares (LS) approach to minimizing

the sum of squared errors, where the error is described by the vertical

distance to the curve from the data points. We will look over various types of

fitting functions in this section.

3.8.1 Straight Line Fit: A Polynomial Function of First Degree

If there is some theoretical basis on which we believe the relationship between

the two variables to be

θ1x + θ0 =y (3.8.1)

we should set up the following system of equations from the collection of many

experimental data:

θ1×1 + θ0 = y1

θ1×2 + θ0 = y2

· · · · · · · · ·

θ1xM + θ0 = yM

Aθ = y with A =

x1 1

x2 1

· · xM 1

, θ = θ1

θ0 , y =

y1

y2

· yM

(3.8.2)

Noting that this apparently corresponds to the overdetermined case mentioned

in Section 2.1.3, we resort to the least-squares (LS) solution (2.1.10)

θo = θo

1

θo

0 = [AT A]−1AT y (3.8.3)

which minimizes the objective function

J = ||e||2 = ||Aθ − y||2 = [Aθ − y]T [Aθ − y] (3.8.4)

CURVE FITTING 145

Sometimes we have the information about the error bounds of the data, and it is

reasonable to differentiate the data by weighing more/less each one according to

its accuracy/reliability. This policy can be implemented by the weighted leastsquares

(WLS) solution

θo

W = θo

W1

θo

W0 = [ATWA]−1ATW y (3.8.5)

which minimizes the weighted objective function

JW = [Aθ − y]TW[Aθ − y] (3.8.6)

If the weighting matrix is W = V −1 = R−T R−1, then we can write the WLS

solution (3.8.5) as

θo

W = θo

W1

θo

W0 = [(R−1A)T (R−1A)]−1(R−1A)T R−1y = [AT

RAR]−1AT

RyR

(3.8.7)

where

AR = R−1A, yR = R−1y, W= V −1 = R−T R−1 (3.8.8)

One may use the MATLAB built-in routine “lscov(A,y,V)” to obtain this

WLS solution.

3.8.2 Polynomial Curve Fit: A Polynomial Function of Higher Degree

If there is no reason to limit the degree of fitting polynomial to one, then we may

increase the degree of fitting polynomial to, say, N in expectation of decreasing

the error. Still, we can use Eq. (3.8.4) or (3.8.6), but with different definitions of

A and θ as

A =

xN

1 · x1 1

xN

2 · x2 1

· · · ·

xN

M · xM 1

, θ =

θN

· θ1

θ0

(3.8.9)

The MATLAB routine “polyfits()” performs the WLS or LS scheme to

find the coefficients of a polynomial fitting a given set of data points, depending

on whether or not a vector (r) having the diagonal elements of the weighting

matrix W is given as the fourth or fifth input argument. Note that in the case of

a diagonal weighting matrix W, the WLS solution conforms to the LS solution

with each row of the information matrix A and the data vector y multiplied by

the corresponding element of the weighting matrix W. Let us see the following

examples for its usage:

146 INTERPOLATION AND CURVE FITTING

function [th,err,yi] = polyfits(x,y,N,xi,r)

%x,y : the row vectors of data pairs

%N : the order of polynomial(>=0)

%r : reverse weighting factor array of the same dimension as y

M = length(x); x = x(:); y = y(:); %Make all column vectors

if nargin == 4

if length(xi) == M, r = xi; xi = x; %With input argument (x,y,N,r)

else r = 1; %With input argument (x,y,N,xi)

end

elseif nargin == 3, xi = x; r = 1; % With input argument (x,y,N)

end

A(:,N + 1) = ones(M,1);

for n = N:-1:1, A(:,n) = A(:,n+1).*x; end %Eq.(3.8.9)

if length(r) == M

for m = 1:M, A(m,:) = A(m,:)/r(m); y(m) = y(m)/r(m); end %Eq.(3.8.8)

end

th = (A\y)’ %Eq.(3.8.3) or (3.8.7)

ye = polyval(th,x); err = norm(y – ye)/norm(y); %estimated y values, error

yi = polyval(th,xi);

%do_polyfit

load xy1.dat

x = xy1(:,1); y = xy1(:,2);

[x,i] = sort(x); y = y(i); %sort the data for plotting

xi = min(x)+[0:100]/100*(max(x) – min(x)); %intermediate points

for i = 1:4

[th,err,yi] = polyfits(x,y,2*i – 1,xi); err %LS

subplot(220+i)

plot(x,y,’k*’,xi,yi,’b:’)

end

%xy1.dat

-3.0 -0.2774

-2.0 0.8958

-1.0 -1.5651

0.0 3.4565

1.0 3.0601

2.0 4.8568

3.0 3.8982

Example 3.6. Polynomial Curve Fit by LS (Least Squares). Suppose we have

an ASCII data file “xy1.dat” containing a set of data pairs {(xk, yk), k = 0:6} in

two columns and we must fit these data into polynomials of degree 1, 3, 5, and 7.

x −3 −2 −1 0 1 2 3

y −0.2774 0.8958 −1.5651 3.4565 3.0601 4.8568 3.8982

We make the MATLAB program “do_polyfit.m”, which uses the routine

“polyfits()” to do this job and plot the results together with the given data

CURVE FITTING 147

−4

−4 −2 0 2 4

−2

0

2

4

6

8

(a) Polynomial of degree 1

−4

−4 −2 0 2 4

−2

0

2

4

6

8

(c) Polynomial of degree 5

−4

−4 −2 0 2 4

−2

0

2

4

6

8

(d) Polynomial of degree 7

−4

−4 −2 0 2 4

−2

0

2

4

6

8

(b) Polynomial of degree 3

Figure 3.10 Polynomial curve fitting by the LS (Least-Squares) method.

points as depicted in Fig. 3.10. We can observe the polynomial wiggle that the

oscillation of the fitting curve between the data points becomes more pronounced

with higher degree.

Example 3.7. Curve Fitting byWLS (Weighted Least Squares).Most experimental

data have some absolute and/or relative error bounds that are not uniform for

all data. If we know the error bounds for each data, we may give each data a

weight inversely proportional to the size of its error bound when extracting valuable

information from the data. The WLS solution (3.8.7) enables us to reflect such

a weighting strategy on estimating data trends. Consider the following two cases.

(a) Suppose there are two gauges A and B with the same function, but different

absolute error bounds ±0.2 and ±1.0, respectively. We used them

to get the input-output data pair (xm,ym) as

{(1, 0.0831), (3, 0.9290), (5, 2.4932), (7, 4.9292), (9, 7.9605)} from gauge A

{(2, 0.9536), (4, 2.4836), (6, 3.4173), (8, 6.3903), (10, 10.2443)} from gauge B

Let the fitting function be a second-degree polynomial function

y = a2x2 + a1x + a0 (E3.7.1)

148 INTERPOLATION AND CURVE FITTING

0

0

2

4

6

8

10

2 4 6 8 10 0

0

2

4

6

8

10

5 10 15 20

linearly

interpolated

data

linearly

interpolated

data

LS

LS

WLS

(a) Fitting to a polynomial y = a 2x 2 + a1x + a0 (b) Fitting to y = ax b

WLS

Figure 3.11 LS curve fitting and WLS curve fitting for Example 3.7.

To find the parameters a2, a1, and a0, we write the MATLAB program

“do_wlse1.m”, which uses the routine “polyfits()” twice, once without

weighting coefficients and once with weighting coefficients. The results

are depicted in Fig. 3.11a, which shows that the WLS curve fitting tries

to be closer to the data points with smaller error bound, while the LS

curve fitting weights all data points equally, which may result in larger

deviations from data points with small error bounds.

(b) Suppose we use one gauge that has relative error bound ±40[%] for

measuring the output y for the input values x = [1, 3, 5, . . . , 19] and so

the size of error bound of each output data is proportional to the magnitude

of the output. We used it to get the input–output data pair (xm,ym) as

{(1, 4.7334), (3, 2.1873), (5, 3.0067), (7, 1.4273), (9, 1.7787)

(11, 1.2301), (13, 1.6052), (15, 1.5353), (17, 1.3985), (19, 2.0211)}

Let the fitting function be an exponential function

y = axb (E3.7.2)

To find the parameters a and b, we make the MATLAB program

“do_wlse2.m”, which uses the routine “curve_fit()” without the weighting

coefficients one time and with the weighting coefficients another time.

The results depicted in Fig. 3.11b shows that the WLS curve fitting tries to

get closer to the data points with smaller |y|, while the LS curve fitting pays

equal respect to all data points, which may result in larger deviation from

data points with small |y|. Note that the MATLAB routine “curve_fit()”

appears in Problem 3.11, which implements all of the schemes listed in

Table 3.5 with the LS/WLS solution.

CURVE FITTING 149

(cf) Note that the objective of the WLS scheme is to put greater emphasis on more

reliable data.

%do_wlse1 for Ex.3.7

clear, clf

x = [1 3 5 7 9 2 4 6 8 10]; %input data

y = [0.0831 0.9290 2.4932 4.9292 7.9605 …

0.9536 2.4836 3.4173 6.3903 10.2443]; %output data

eb = [0.2*ones(5,1); ones(5,1)]; %error bound for each y

[x,i] = sort(x); y = y(i); eb = eb(i); %sort the data for plotting

errorbar(x,y,eb,’:’), hold on

N = 2; %the degree of the approximate polynomial

xi = [0:100]/10; %interpolation points

[thl,errl,yl] = polyfits(x,y,N,xi);

[thwl,errwl,ywl] = polyfits(x,y,N,xi,eb);

plot(xi,yl,’b’, xi,ywl,’r’)

%KC = 0; thlc = curve_fit(x,y,KC,N,xi); %for cross-check

%thwlc = curve_fit(x,y,KC,N,xi,eb);

%do_wlse2

clear, clf

x = [1:2:20]; Nx = length(x); %changing input

xi = [1:200]/10; %interpolation points

eb = 0.4*ones(size(x)); %error bound for each y

y = [4.7334 2.1873 3.0067 1.4273 1.7787 1.2301 1.6052 1.5353 …

1.3985 2.0211];

[x,i] = sort(x); y = y(i); eb = eb(i); %sort the data for plotting

eby = y.*eb; %our estimation of error bounds

KC = 6; [thlc,err,yl] = curve_fit(x,y,KC,0,xi);

[thwlc,err,ywl] = curve_fit(x,y,KC,0,xi,eby);

errorbar(x,y,eby), hold on

plot(xi,yl,’b’, xi,ywl,’r’)

3.8.3 Exponential Curve Fit and Other Functions

Why don’t we use functions other than the polynomial function as a candidate

for fitting functions? There is no reason why we have to stick to the polynomial

function, as illustrated in Example 3.7(b). In this section, we consider the case

in which the data distribution or the theoretical background behind the data tells

us that it is appropriate to fit the data into some nonpolynomial function.

Suppose it is desired to fit the data into the following exponential function.

c eax =y (3.8.10)

Taking the natural logarithm of both sides, we linearize this as

a x + ln c = lny (3.8.11)

150 INTERPOLATION AND CURVE FITTING

Table 3.5 Linearization of Nonlinear Functions by Parameter/Data Transformation

Function to Fit Linearized Function

Variable Substitution/

Parameter Restoration

(1) y =

a

x +b y= a

1

x + b →y = ax +b x =

1

x

(2) y =

b

x + a

1

y =

1

b

x +

a

b → y = ax + b y =

1

y

, a =

b

a

, b =

1

a

(3) y = a bx ln y = (ln b)x + lna y = ln y, a = eb, b = ea

→y = ax + b

(4) y = b eax ln y = ax + ln b → y = ax + b y = ln y, b = eb

(5) y = C − b e−ax ln(C − y) = −ax + lnb y = ln(C − y)

→y = ax + b a = −a, b = eb

(6) y = a xb ln y = b(ln x) + lna y = ln y, x = ln x

→y = ax + b a = eb, b = a

(7) y = ax ebx ln y − ln x = bx + lna y = ln(y/x)

→y = ax + b a = eb, b = a

(8) y =

C

1 + b eax

lnC

y − 1 = ax + lnb y = ln C

y − 1, b = eb

(a0, b0,C = y(∞)) →y = ax + b

(9) y = a ln x + b → y = ax +b x = ln x

so that the LS algorithm (3.8.3) can be applied to estimate the parameters a and

ln c based on the data pairs {(xk, ln yk), k = 0 : M}.

Like this, there are many other nonlinear relations that can be linearized to fit

the LS algorithm, as listed in Table 3.5. This makes us believe in the extensive

applicability of the LS algorithm. If you are interested in making a MATLAB

routine that implements what are listed in this table, see Problem 3.11, which lets

you try the MATLAB built-in function “lsqcurvefit(f,th0,x,y)” that enables

one to use any type of function (f) for curve fitting.

3.9 FOURIER TRANSFORM

Most signals existent in this world contain various frequency components, where

rapidly/slowly changing one contains high/low-frequency components. Fourier

series/transform is a mathematical tool that can be used to analyze the frequency

characteristic of periodic/aperiodic signals. There are four similar definitions

of Fourier series/transform, namely, continuous-time Fourier series (CtFS),

continuous-time Fourier transform (CtFT), discrete-time Fourier transform

(DtFT), and discrete Fourier series/transform (DFS/DFT). Among these tools,

DFT can easily and efficiently be programmed in computer languages and that’s

why we deal with just DFT in this section.

FOURIER TRANSFORM 151

Suppose a sequence of data {x[n] = x(nT ), n = 0 : M − 1}(T : the sampling

period) is obtained by sampling a continuous-time/space signal once every T

seconds. The N(≥ M)-point DFT/IDFT (inverse DFT) pair is defined as

DFT: X(k) =

N−1

n=0

x[n]e−j2πnk/N, k= 0 : N − 1 (3.9.1a)

IDFT: x[n] =

1

N

N−1

k=0

X(k)ej2πnk/N, n= 0 : N − 1 (3.9.1b)

Remark 3.3. DFS/DFT (Discrete Fourier Series/Transform)

0. Note that the indices of the DFT/IDFT sequences appearing in MATLAB

range from 1 to N.

1. Generally, the DFT coefficient X(k) is complex-valued and denotes the

magnitude and phase of the signal component having the digital frequency

k = k0 = 2πk/N[rad], which corresponds to the analog frequency ωk = kω0 = k0/T = 2πk/NT [rad/s]. We call 0 = 2π/N and ω0 = 2π/NT

(N represents the size of DFT) the digital/analog fundamental or resolution

frequency, since it is the minimum digital/analog frequency difference that

can be distinguished by the N-point DFT.

2. The DFS and the DFT are essentially the same, but different in the range

of time/frequency interval. More specifically, a signal x[n] and its DFT

X(k) are of finite duration over the time/frequency range {0 ≤ n ≤ N − 1} and {0 ≤ k ≤ N − 1}, respectively, while a signal ˜x[n] (to be analyzed by

DFS) and its DFS X˜ (k) are periodic with the period N over the whole set

of integers.

3. FFT (fast Fourier transform) means the computationally efficient algorithm

developed by exploiting the periodicity and symmetry in the multiplying

factor ei2πnk/N to reduce the number of complex number multiplications

from N2 to (N/2) log2 N (N represents the size of DFT). The MATLAB

built-in functions “fft()”/“ifft()” implement the FFT/IFFT algorithm for

the data of length N = 2l (l represents a nonnegative integer). If the length

Mof the original data sequence is not a power of 2, it can be extended by

padding the tail part of the sequencewith zeros,which is called zero-padding.

3.9.1 FFT Versus DFT

As mentioned in item 3 of Remark 3.3, FFT/IFFT (inverse FFT) is the computationally

efficient algorithm for computing the DFT/IDFT and is fabricated into

the MATLAB functions “fft()”/“ifft()”. In order to practice the use of the

MATLAB functions and realize the computational advantage of FFT/IFFT over

DFT/IDFT, we make the MATLAB program “compare_dft_fft.m”. Readers are

recommended to run this program and compare the execution times consumed by

the 1024-point DFT/IDFT computation and its FFT/IFFT scheme, seeing that the

152 INTERPOLATION AND CURVE FITTING

resulting spectra are exactly the same and thus are overlapped onto each other

as depicted in Fig. 3.12.

%compare_DFT_FFT

clear, clf

N = 2^10; n = [0:N – 1];

x = cos(2*pi*200/N*n)+ 0.5*sin(2*pi*300/N*n);

tic

for k = 0:N – 1, X(k+1) = x*exp(-j*2*pi*k*n/N).’; end %DFT

k = [0:N – 1];

for n = 0:N – 1, xr(n + 1) = X*exp(j*2*pi*k*n/N).’; end %IDFT

time_dft = toc %number of floating-point operations

plot(k,abs(X)), pause, hold on

tic

X1 = fft(x); %FFT

xr1 = ifft(X1); %IFFT

time_fft = toc %number of floating-point operations

clf, plot(k,abs(X1),’r’) %magnitude spectrum in Fig. 3.12

3.9.2 Physical Meaning of DFT

In order to understand the physical meaning of FFT, we make the MATLAB

program “do_fft” and run it to get Fig. 3.13, which shows the magnitude spectra

of the sampled data taken every T seconds from a two-tone analog signal

x(t) = sin(1.5πt) + 0.5 cos(3πt) (3.9.2)

Readers are recommended to complete the part of this program to get Fig. 3.13c,d

and run the program to see the plotting results (see Problem 3.16).

What information do the four spectra for the same analog signal x(t) carry?

The magnitude of Xa(k) (Fig. 3.13a) is large at k = 2 and 5, each corresponding

to kω0 = 2πk/NT = 2πk/3.2 = 1.25π ≈ 1.5π and 3.125π ≈ 3π. The magnitude

of Xb(k) (Fig. 3.13b) is also large at k = 2 and 5, each corresponding to

kω0 = 1.25π ≈ 1.5π and 3.125π ≈ 3π. The magnitude of Xc(k) (Fig. 3.13c) is

0

0 100 200 300 400 500 600 724 824 900 1023

200

400

600

k

digital frequency

Ω200 = 2p × 200/N [rad]

Ω300 = 2p × 300/N [rad]

X(k)

Figure 3.12 The DFT(FFT) {X(k), k = 0 : N − 1} of x[N] = cos(2π × 200n/N) + 0.5sin

(2π × 300n/N) for n = 0 : N − 1(N = 210 = 1024).

FOURIER TRANSFORM 153

%do_fft (to get Fig. 3.13)

clear, clf

w1 = 1.5*pi; w2=3*pi; %two tones

N = 32; n = [0:N – 1]; T = 0.1; %sampling period

t = n*T; xan = sin(w1*t) + 0.5*sin(w2*t);

subplot(421), stem(t,xan,’.’)

k = 0:N – 1; Xa = fft(xan);

dscrp=norm(xan-real(ifft(Xa))) %x[n] reconstructible from IFFT{X(k)}?

subplot(423), stem(k,abs(Xa),’.’)

%upsampling

N = 64; n = [0:N – 1]; T = 0.05; %sampling period

t = n*T; xbn = sin(w1*t)+ 0.5*sin(w2*t);

subplot(422), stem(t,xbn,’.’)

k = 0:N – 1; Xb = fft(xbn);

subplot(424), stem(k,abs(Xb),’.’)

%zero-padding

N = 64; n = [0:N-1]; T = 0.1; %sampling period

………………….

0

0 2 5 16 27 30

0

10

20

30

0 5 10 32 54 59

0

10

20

30

0 5 10 32 54 59

0

10

20

30

0 2 5 32 59 62

0

10

20

30

1 2 3 0 1 2 3

−2 −2

0

2

0

2

xa [n]

xc [n] xd [n]

xb [n]

t = nT

0 2 4 6

−2

0

2

t = nT

t = nT

(a) N = 32, T = 0.1

(c) N = 64, T = 0.1

0 2 4 6

−2

0

2

t = nT

(d) N = 64, T = 0.1

(b) N = 64, T = 0.05

w0 = 2p

NT

: resolution frequency

(fundamental frequency)

|Xa(k )| |Xb(k )|

|Xc (k )| |Xd (k )|

zero-padding

Figure 3.13 DFT spectra of a two-tone signal.

154 INTERPOLATION AND CURVE FITTING

large at k = 4,5 and 9,10, and they can be alleged to represent two tones of kω0 = 2πk/NT = 2πk/6.4 ≈ 1.25π ∼ 1.5625π and 2.8125π ∼ 3.125π. The magnitude

of Xd(k) (Fig. 3.13d) is also large at k = 5 and 10, each corresponding to

kω0 = 1.5625 π ≈ 1.5 π and 3.125 π ≈ 3π.

It is strange and interesting thatwe havemany different DFT spectra for the same

analog signal, depending on the DFT size, the sampling period, the whole interval,

and zero-padding. Compared with spectrum (a), spectrum (b) obtained by decreasing

the sampling period T from 0.1s to 0.05s has wider analog frequency range

[0,2π/Tb], but the same analog resolution frequency is ω0 = 0/Tb = 2π/NbTb = π/1.6 ≡ 2π/NaTa; consequently, it does not present us with any new information

over (a) for all increased number of data points. The shorter sampling periodmay be

helpful in case the analog signal has some spectral contents of frequency higher than

π/Ta. The spectrum (c) obtained by zero-padding has a better-looking, smoother

shape, but the vividness is not much improved compared with (a) or (b), since the

zeros essentially have no valuable information in the time domain. In contrast with

(b) and (c), spectrum (d) obtained by extending the whole time interval shows us

the spectral information more distinctly.

Note the following things:

ž Zero-padding in the time domain yields the interpolation (smoothing) effect

in the frequency domain and vice versa, which will be made use of for data

smoothing in the next section (see Problem 3.19).

ž If a signal is of finite duration and has the value of zeros outside its domain

on the time axis, its spectrum is not discrete, but continuous along the

frequency axis, while the spectrum of a periodic signal is discrete as can be

seen in Fig. 3.12 or 3.13.

ž The DFT values X(0) and X(N/2) represent the spectra of the dc component

(0 = 0) and the virtually highest digital frequency components (N/2 = N/2 × 2π/N = π [rad]), respectively.

Here, we have something questionable. The DFT spectrum depicted in Fig. 3.12

shows clearly the digital frequency components 200 = 2π × 200/N and 300 = 2π × 300/N[rad](N = 210 = 1024) contained in the discrete-time signal

x[n] = cos(2π × 200n/N) + 0.5 sin(2π × 300n/N), N = 210 = 1024

(3.9.3)

and so we can find the analog frequency components ωk = k/T as long as

the sampling period T is known, while the DFT spectra depicted in Fig. 3.13

are so unclear that we cannot discern even the prominent frequency contents.

What’s wrong with these spectra? It is never a ‘right-or-wrong’ problem. The

only difference is that the digital frequencies contained in the discrete-time signal

described by Eq. (3.9.3) are multiples of the fundamental frequency 0 = 2π/N,

but the analog frequencies contained in the continuous-time signal described by

Eq. (3.9.2) are not multiples of the fundamental frequency ω0 = 2π/NT; in

other words, the whole time interval [0,NT ) is not a multiple of the period of

each frequency to be detected. The phenomenon whereby the spectrum becomes

FOURIER TRANSFORM 155

blurred like this is said to be the ‘leakage problem’. The leakage problem occurs

in most cases because we cannot determine the length of the whole time interval

in such a way that it is a multiple of the period of the signal as long as we don’t

know in advance the frequency contents of the signal. If we knew the frequency

contents of a signal, why do we bother to find its spectrum that is already known?

As a measure to alleviate the leakage problem, there is a windowing technique

[O-1, Section 11.2]. Interested readers can see Problem 3.18.

Also note that the periodicity with period N(the DFT size) of the DFT

sequence X(k) as well as x[n], as can be manifested by substituting k + mN

(m represents any integer) for k in Eq. (3.9.1a) and also substituting n + mN

for n in Eq. (3.9.1b). A real-world example reminding us of the periodicity of

DFT spectrum is the so-called stroboscopic effect whereby the wheel of a carriage

driven by a horse in the scene of a western movie looks like spinning at

lower speed than its real speed or even in the reverse direction. The periodicity

of x[n] is surprising, because we cannot imagine that every discrete-time signal

is periodic with the period of N, which is the variable size of the DFT to be

determined by us. As a matter of fact, the ‘weird’ periodicity of x[n] can be

regarded as a kind of cost that we have to pay for computing the sampled DFT

spectrum instead of the continuous spectrum X(ω) for a continuous-time signal

x(t), which is originally defined as

X(ω) = ∞

−∞

x(t)e−jωt dt (3.9.4)

Actually, this is to blame for the blurred spectra of the two-tone signal depicted

in Fig. 3.13.

3.9.3 Interpolation by Using DFS

function [xi,Xi] = interpolation_by_DFS(T,x,Ws,ti)

%T : sampling interval (sample period)

%x : discrete-time sequence

%Ws: normalized stop frequency (1.0=pi[rad])

%ti: interpolation time range or # of divisions for T

if nargin < 4, ti = 5; end

if nargin < 3 | Ws > 1, Ws = 1; end

N = length(x);

if length(ti) == 1

ti = 0:T/ti:(N-1)*T; %subinterval divided by ti

end

ks = ceil(Ws*N/2);

Xi = fft(x);

Xi(ks + 2:N – ks) = zeros(1,N – 2*ks – 1); %filtered spectrum

xi = zeros(1,length(ti));

for k = 2:N/2

xi = xi+Xi(k)*exp(j*2*pi*(k – 1)*ti/N/T);

end

xi = real(2*xi+Xi(1)+Xi(N/2+1)*cos(pi*ti/T))/N; %Eq.(3.9.5)

156 INTERPOLATION AND CURVE FITTING

%interpolate_by_DFS

clear, clf

w1 = pi; w2 = .5*pi; %two tones

N = 32; n = [0:N – 1]; T = 0.1; t = n*T;

x = sin(w1*t)+0.5*sin(w2*t)+(rand(1,N) – 0.5); %0.2*sin(20*t);

ti = [0:T/5:(N – 1)*T];

subplot(411), plot(t,x,’k.’) %original data sequence

title(’original sequence and interpolated signal’)

[xi,Xi] = interpolation_by_DFS(T,x,1,ti);

hold on, plot(ti,xi,’r’) %reconstructed signal

k = [0:N – 1];

subplot(412), stem(k,abs(Xi),’k.’) %original spectrum

title(’original spectrum’)

[xi,Xi] = interpolation_by_DFS(T,x,1/2,ti);

subplot(413), stem(k,abs(Xi),’r.’) %filtered spectrum

title(’filtered spectrum’)

subplot(414), plot(t,x,’k.’, ti,xi,’r’) %filtered signal

title(’filtered/smoothed signal’)

We can use the DFS/DFT to interpolate a given sequence x[n] that is supposed

to have been obtained by sampling some signal at equidistant points (instants).

The procedure consists of two steps; to take the N-point FFT X(k) of x[n] and

to use the formula

ˆx(t) =

1

N |k|<N/2 X(k)ej2πkt/NT

=

1

N {X(0) + 2

N/2−1

k=1

Real{X(k)ej2πkt/NT} + X(N/2) cos(πt/T )} (3.9.5)

This formula is cast into the routine “interpolation_by_dfs”, which makes

it possible to filter out the high-frequency portion over (Ws·π,(2-Ws)π) with

Ws given as the third input argument. The horizontal (time) range over which

you want to interpolate the sequence can be given as the fourth input argument

ti. We make the MATLAB program “interpolate_by_dfs”, which applies the

routine to interpolate a set of data obtained by sampling at equidistant points

along the spatial or temporal axis and run it to get Fig. 3.14. Figure 3.14a shows

a data sequence x[n] of length N = 32 and its interpolation (reconstruction)

x(t) from the 32-point DFS/DFT X(k) (Fig. 3.14b), while Figs. 3.14c and 3.14d

show the (zero-padded) DFT spectrum X(k) with the digital frequency contents

higher than π/2[rad](N/4 < k < 3N/4) removed and a smoothed interpolation

(fitting curve) x(t) obtained from X(k), respectively. This can be viewed as the

smoothing effect in the time domain by zero-padding in the frequency domain,

in duality with the smoothing effect in the frequency domain by zero-padding in

the time domain, which was observed in Fig. 3.13c.

PROBLEMS 157

0

–1

0

1

2

0.5 1 1.5 2 2.5 3 3.5

(a) A given data sequence x [n ] and its interpolation x (t ) by using DFS

t = nT

0

0

5

10

15

8 16(p) 24 31

(b) The original DFS/DFT spectrum X(k )

k

|X(k )|

: x [n ]

: x (t )

x (t )

0

–1

0

1

2

0.5 1 1.5 2 2.5 3 3.5

(d) The filtered signal x ′(t )

t = nT

: x [n ]

: x ′(t )

x ′(t )

digital

frequency

0

0

5

10

15

8 16(p) 24 31

(c) The spectrum X′(k ) of the filtered signal x ′(t )

k

|X′(k )|

digital

frequency

Ws · p

zero-padding(zero-out)

(2 – Ws)p

Figure 3.14 Interpolation/smoothing by using DFS/DFT.

PROBLEMS

3.1 Quadratic Interpolation: Lagrange Polynomial and Newton Polynomial

(a) The second-degree Lagrange polynomial matching the three points

(x0, f0), (x1, f1), and (x2, f2) can be written by substituting N = 2

158 INTERPOLATION AND CURVE FITTING

into Eq. (3.1.3) as

l2(x) =

2

m=0

fmL2,m(x) =

2

m=0

fm

N

k=m

x − xk

xm − xk

(P3.1.1)

Check if the zero of the derivative of this polynomial—that is, the root

of the equation l

2 (x) = 0—is found as

l

2 (x) = f0

(x − x1) + (x − x2)

(x0 − x1)(x0 − x2) + f1

(x − x2) + (x − x0)

(x1 − x2)(x1 − x0)

+ f2

(x − x0) + (x − x1)

(x2 − x0)(x2 − x1)

f0(2x − x1 − x2)(x2 − x1) + f1(2x − x2 − x0)(x0 − x2)

+ f2(2x − x0 − x1)(x1 − x0) = 0

x = x3 =

f0(x2

1 − x2

2 ) + f1(x2

2 − x2

0 ) + f2(x2

0 − x2

1 )

2{f0(x1 − x2) + f1(x2 − x0) + f2(x0 − x1)}

(P3.1.2)

You can use the symbolic computation capability of MATLAB by

typing the following statements into the MATLAB command window:

>>syms x x1 x2 x3 f0 f1 f2

>>L2 = f0*(x – x1)*(x – x2)/(x0 – x1)/(x0 – x2)+…

f1*(x – x2)*(x – x0)/(x1 – x2)/(x1 – x0)+…

f2*(x – x0)*(x – x1)/(x2 – x0)/(x2 – x1)

>>pretty(solve(diff(L2)))

(b) The second-degree Newton polynomial matching the three points

(x0, f0), (x1, f1), and (x2, f2) is Eq. (3.2.4).

n2(x) = a0 + a1(x − x0) + a2(x − x0)(x − x1) (P3.1.3)

where

a0 = f0, a1 = Df0 =

f1 − f0

x1 − x0

a2 = D2f0 =

Df1 − Df0

x2 − x0 =

f2 − f1

x2 − x1 −

f1 − f0

x1 − x0

x2 − x0

(P3.1.4)

Find the zero of the derivative of this polynomial.

(c) From Eq. (P3.1.1) with x0 = −1, x1 = 0, and x2 = 1, find the coefficients

of Lagrange coefficient polynomials L2,0(x), L2,1(x), and L2,2(x).

You had better make use of the routine “lagranp()” for this job.

PROBLEMS 159

(d) From the third-degree Lagrange polynomial matching the four points

(x0, f0), (x1, f1), (x2, f2), and (x3, f3) with x0 = −3, x1 = −2, x2 =

−1, and x3 = 0, find the coefficients of Lagrange coefficient polynomials

L3,0(x), L3,1(x), L3,2(x), and L3,3(x). You had better make use

of the routine “lagranp()” for this job.

3.2 Error Analysis of Interpolation Polynomial

Consider the error between a true (unknown) function f (x) and the interpolation

polynomial PN(x) of degree N for some (N + 1) points of y = f (x),

that is,

{(x0, y0), (x1, y1), . . . , (xN, yN)}

where f (x) is up to (N + 1)th-order differentiable. Noting that the error is

also a function of x and becomes zero at the (N + 1) points, we can write

it as

e(x) = f (x) − PN(x) = (x − x0)(x − x1) · · · (x − xN)g(x) (P3.2.1)

Technically, we define an auxiliary function w(t) with respect to t as

w(t) = f (t) − PN(t) − (t − x0)(t − x1) · · · (t − xN)g(x) (P3.2.2)

Then, this function has the value of zero at the (N + 2) points t = x0, x1, . . . ,

xN, x and the 1/2/ · · · /(N + 1)th-order derivative has (N + 1)/N/ · · · /1

zeros, respectively. For t = t0 such that w(N+1)(t0) = 0, we have

w(N+1)(t0) = f (N+1)(t0) − 0 − (N + 1)!g(x) = 0;

g(x) =

1

(N + 1)!

f (N+1)(t0) (P3.2.3)

Based on this, show that the error function can be rewritten as

e(x) = f (x) − PN(x) = (x − x0)(x − x1) · · · (x − xN)

1

(N + 1)!

f (N+1)(t0)

(P3.2.4)

3.3 The Approximation of a Cosine Function

In the way suggested below, find an approximate polynomial of degree 4

for

y = f (x) = cosx (P3.3.1)

(a) Find the Lagrange/Newton polynomial of degree 4 matching the following

five points and plot the resulting polynomial together with the

true function cos x over [−π,+π].

160 INTERPOLATION AND CURVE FITTING

k 0 1 2 3 4

xk −π −π/2 0 +π/2 +π

f (xk) −1 0 1 0 −1

(b) Find the Lagrange/Newton polynomial of degree 4 matching the following

five points and plot the resulting polynomial on the same graph

that has the result of (a).

k 0 1 2 3 4

xk π cos(9π/10) πcos(7π/10) 0 π cos(3π/10) πcos(π/10)

f (xk) −0.9882 −0.2723 1 −0.2723 −0.9882

(c) Find the Chebyshev polynomial of degree 4 for cos x over [−π,+π]

and plot the resulting polynomial on the same graph that has the result

of (a) and (b).

3.4 Chebyshev Nodes

The current speed/pressure of the liquid flowing in the pipe, which has irregular

radius, will be different from place to place. If you are to install seven

speed/pressure gauges through the pipe of length 4 m as depicted in Fig.

P3.4, how would you determine the positions of the gauges so that the maximum

error of estimating the speed/pressure over the interval [0, 4] can

be minimized?

0 1 2 3 4

x

Figure P3.4 Chebyshev nodes.

3.5 Pade Approximation

For the Laplace transform

F(s) = e−sT (P3.5.1)

representing the delay of T [seconds], we can write its Maclaurin series

expansion up to fifth order as

Mc(s)∼=1 − sT +

(sT )2

2! −

(sT )3

3! +

(sT )4

4! −

(sT )5

5!

(P3.5.2)

(a) Show that we can solve Eq. (3.4.4) and use Eq. (3.4.1) to get the Pade

approximation as

F(s)∼=

p1,1(s) =

q0 + q1s

1 + d1s =

1 − (T /2)s

1 + (T /2)s∼=

e−T s (P3.5.3)

PROBLEMS 161

(b) Compose a MATLAB program “nm3p05.m” that uses the routine

“padeap()” to generate the Pade approximation of (P3.5.1) with T = 0.2 and plots it together with the second-order Maclaurin series expansion

and the true function (P3.5.1) for s = [−5, 10]. You also run it to

see the result as

p1,1(s) =

1 − (T /2)s

1 + (T /2)s = −s + 10

s + 10

(P3.5.4)

3.6 Rational Function Interpolation: Bulirsch–Stoer Method [S-3]

Table P3.6 shows the Bulirsch–Stoer method, where its element in the mth

row and the (i + 1)th column is computed by the following formula:

Ri+1

m = Ri

m+1 +

(x − xm+i)(Ri

m+1 − Ri−1

m+1)(Ri

m+1 − Ri

m)

(x − xm)(Rim − Ri−1

m+1) − (x − xm+i)(Ri

m+1 − Ri−1

m+1)

with Ro

m = 0 and R1

m = ym for i = 1 : N and m = 1 : N − i

(P3.6.1)

function yi = rational_interpolation(x,y,xi)

N = length(x); Ni = length(xi);

R(:,1) = y(:);

for n = 1:Ni

xn = xi(n);

for i = 1:N – 1

for m = 1:N – i

RR1 = R(m + 1,i); RR2 = R(m,i);

if i > 1,

RR1 = RR1 – R(m + 1,???); RR2 = RR2 – R(???,i – 1);

end

tmp1 = (xn-x(???))*RR1;

num = tmp1*(R(???,i) – R(m,?));

den = (xn – x(?))*RR2 -tmp1;

R(m,i + 1) = R(m + 1,i) ????????;

end

end

yi(n) = R(1,N);

end

Table P3.6 Bulirsch–Stoer Method for Rational Function Interpolation

Data i = 1 i = 2 i = 3 i = 4

(x1, y1) R1

1 = y1 R2

1 R3

1 R4

1

(x2, y2) R1

2 = y2 R2

2 R3

2 .

(x3, y3) R1

3 = y3 R2

3 .

. . .

(xm, ym) . .

162 INTERPOLATION AND CURVE FITTING

(a) The above routine “rational_interpolation(x,y,xi)” uses the

Bulirsch–Stoer method to interpolate the set of data pairs (x,y) given

as its first/second input arguments over a set of intermediate points

xi given as its third input argument. Complete the routine and apply

it to interpolate the four data points {(−1, f (−1)), (−0.2, f (−0.2)),

(0.1, f (0.1)), (0.8, f (0.8))} on the graph of f (x) = 1/(1 + 8×2) for

xi = [-100:100]/100 and plot the interpolated curve together with the

graph of the true function f (x). Does it work well? How about doing

the same job with another routine “rat_interp()” listed in Section 8.3

of [F-1]? What are the values of yi([95:97]) obtained from the two

routines? If you come across anything odd in the graphic results and/or

the output numbers, what is your explanation?

(cf) MATLAB expresses the in-determinant 0/0 (zero-divided-by-zero) as NaN

(Not-a-Number) and skips the value when plotting it on a graph. It may,

therefore, be better off for the plotting purpose if we take no special

consideration into the case of in-determinant.

(b) Apply the Pade approximation routine “padeap()” (with M = 2 & N =

2) to generate the rational function approximating f (x) = 1/(1 + 8×2)

and compare the result with the true function f (x).

(c) To compare the rational interpolation method with the Pade approximation

scheme, apply the routines rational_interpolation() and

padeap() (with M = 3 & N = 2) to interpolate the four data points

{(−2, f (−2)), (−1, f (−1)), (1, f (1)), (2, f (2))} on the graph of

f (x) = sin(x) for xi = [-100:100]*pi/100 and plot the interpolated

curve together with the graph of the true function. How do you compare

the approximation/interpolation results?

3.7 Smoothness of a Cubic Spline Function

We claim that the cubic spline interpolation function s(x) has the smoothness

property of

xk+1

xk

(s(x))2 dx ≤ xk+1

xk

(f (x))2 dx (P3.7.1)

for any second-order differentiable function f (x) matching the given grid

points and having the same first-order derivatives as s(x) at the grid points.

This implies that the cubic spline functions are not so rugged. Prove it by

doing the following.

(a) Check the validity of the equality

xk+1

xk

f (x)s(x) dx = xk+1

xk

(s(x))2 dx (P3.7.2)

PROBLEMS 163

where the left-hand and right-hand sides of this equation are

LHS: xk+1

xk

f (x)s(x) dx

= f (x)s(x)|xk+1

xk − xk+1

xk

f (x)s(x) dx

= f (xk+1)s(xk+1) − f (xk)s(xk) − C(f (xk+1) − f (xk)) (P3.7.3a)

RHS: xk+1

xk

s(x)s(x) dx

= s(xk+1)s(xk+1) − s(xk)s(xk) − C(s(xk+1) − s(xk)) (P3.7.3b)

(b) Check the validity of the following inequality:

0 ≤ xk+1

xk

(f (x) − s(x))2 dx

= xk+1

xk

(f (x))2 dx − 2 xk+1

xk

f (x)s(x) dx + xk+1

xk

(s(x))2 dx

(P3.7.2) = xk+1

xk

(f (x))2 dx − xk+1

xk

(s(x))2 dx

xk+1

xk

(f (x))2 dx ≤ xk+1

xk

(s(x))2 dx (P3.7.4)

3.8 MATLAB Built-in Routine for Cubic Spline

There are two MATLAB built-in routines:

>>yi = spline(x,y,xi);

>>yi = interp1(x,y,xi,’spline’);

Both receive a set of data points (x,y) and return the values of the cubic

spline interpolating function s(x) for the (intermediate) points xi given as

the third input argument. Write a program that uses these MATLAB routines

to get the interpolation for the set of data points

{(0, 0), (0.5, 2), (2,−2), (3.5, 2), (4, 0)}

and plots the results for [0, 4]. In this program, append the statements that

do the same job by using the routine “cspline(x,y,KC)” (Section 3.5) with

KC = 1, 2, and 3. Which one yields the same result as the MATLAB builtin

routine? What kind of boundary condition does the MATLAB built-in

routine assume?

164 INTERPOLATION AND CURVE FITTING

3.9 Robot Path Planning Using Cubic Spline

Every object having a mass is subject to the law of inertia and so its

speed described by the first derivative of its displacement with respect to

time must be continuous in any direction. In this context, the cubic spline

having the continuous derivatives up to second order presents a good basis

for planning the robot path/trajectory. We will determine the path of a robot

in such a way that the following conditions are satisfied:

ž At time t = 0 s, the robot starts from its home position (0, 0) with zero

initial velocity, passing through the intermediate point (1, 1) at t = 1 s

and arriving at the final point (2, 4) at t = 2 s.

ž On arriving at (2, 4), it starts the point at t = 2 s, stopping by the

intermediate point (3, 3) at t = 3 s and arriving at the point (4, 2) at

t = 4 s.

ž On arriving at (4, 2), it starts the point, passing through the intermediate

point (2,1) at t = 5 s and then returning to the home position (0, 0) at

t = 6 s.

More specifically, what we need is

ž the spline interpolation matching the three points (0, 0),(1, 1),(2, 2) and

having zero velocity at both boundary points (0, 0) and (2, 2),

ž the spline interpolation matching the three points (2, 2),(3, 3),(4, 4) and

having zero velocity at both boundary points (2, 2) and (4, 4), and

ž the spline interpolation matching the three points (4, 4), (5, 2), (6, 0) and

having zero velocity at both boundary points (4, 4) and (6, 0) on the tx

plane.

On the ty plane, we need

ž the spline interpolation matching the three points (0, 0),(1, 1),(2, 4) and

having zero velocity at both boundary points (0, 0) and (2, 4),

ž the spline interpolation matching the three points (2, 4),(3, 3),(4, 2) and

having zero velocity at both boundary points (2, 4) and (4, 2), and

ž the spline interpolation matching the three points (4, 2),(5, 1),(6, 0) and

having zero velocity at both boundary points (4, 2) and (6, 0).

Supplement the following incomplete program “robot_path”, whose objective

is to make the required spline interpolations and plot the whole robot

path obtained through the interpolations on the xy plane. Run it to get the

graph as depicted in Fig. P3.9c.

%robot_path

x1 = [0 1 2]; y1 = [0 1 4]; t1 = [0 1 2]; ti1 = [0: 0.05: 2];

xi1 = cspline(t1,x1,ti1); yi1 = cspline(t1,y1,ti1);

…………………………………

plot(xi1,yi1,’k’, xi2,yi2,’b’, xi3,yi3, ’k’), hold on

plot([x1(1) x2(1) x3(1) x3(end)],[y1(1) y2(1) y3(1) y3(end)],’o’)

plot([x1 x2 x3],[y1 y2 y3],’k+’), axis([0 5 0 5])

PROBLEMS 165

+

+

+

0 2 4 6

0

1

2

3

4

5

x

t

tx plane

(a) x coordinate varying along t

0 2 4 6

0

1

2

3

4

5

y

t

ty plane

(b) y coordinate varying along t

+

+

+

+

+

+ +

0 1 2 3 4 5

0

1

2

3

4

5

y

x

xy plane

(c) Robot path on the xy plane

+

+

+

+

+

+

+

+

+

+

1

2

3

4

5

0

Figure P3.9 Robot path planning using the cubic spline interpolation.

3.10 One-Dimensional Interpolation

What do you have to give as the fourth input argument of the MATLAB

built-in routine “interp1()” in order to get the same result as that would

be obtained by using the following one-dimensional interpolation routine

“intrp1()”? What letter would you see if you apply this routine to interpolate

the data points {(0,3), (1,0), (2,3), (3,0), (4,3)} for [0,4]?

function yi = intrp1(x,y,xi)

M = length(x); Mi = length(xi);

for mi = 1: Mi

if xi(mi) < x(1), yi(mi) = y(1)-(y(2) – y(1))/(x(2) – x(1))*(x(1) – xi(mi));

elseif xi(mi)>x(M)

yi(mi) = y(M)+(y(M) – y(M – 1))/(x(M) – x(M-1))*(xi(mi) – x(M));

else

for m = 2:M

if xi(mi) <= x(m)

yi(mi) = y(m – 1)+(y(m) – y(m – 1))/(x(m) – x(m – 1))*(xi(mi) – x(m – 1));

break;

end

end

end

end

3.11 Least-Squares Curve Fitting

(a) There are several nonlinear relations listed in Table 3.5, which

can be linearized to fit the LS algorithm. The MATLAB routine

“curve_fit()” implements all the schemes that use the LS method

to find the parameters for the template relations, but the parts for the

relations (1), (2), (7), (8), and (9) are missing. Supplement the missing

parts to complete the routine.

(b) The program “nm3p11.m” generates the 12 sets of data pairs according to

various types of relations (functions), applies the routines

“curve_fit()”/“lsqcurvefit()” to find the parameters of the template

relations, and plots the data pairs on the fitting curves obtained from the

template functions with the estimated parameters. Complete and run it

to get the graphs like Fig. P3.11. Answer the following questions.

166 INTERPOLATION AND CURVE FITTING

(i) If any, find the case(s) where the results of using the two routines

make a great difference. For the case(s), try with another initial

guess th0 = [1 1] of parameters, instead of th0 = [0 0].

(ii) If the MATLAB built-in routine “lsqcurvefit()” yields a bad

result, does it always give you a warning message? How do you

compare the two routines?

function [th,err,yi] = curve_fit(x,y,KC,C,xi,sig)

% implements the various LS curve-fitting schemes in Table 3.5

% KC = the # of scheme in Table 3.5

% C = optional constant (final value) for KC! = 0 (nonlinear LS)

% degree of approximate polynomial for KC = 0 (standard LS)

% sig = the inverse of weighting factor for WLS

Nx = length(x); x = x(:); y = y(:);

if nargin == 6, sig = sig(:);

elseif length(xi) == Nx, sig = xi(:); xi = x;

else sig = ones(Nx,1);

end

if nargin < 5, xi = x; end; if nargin < 4 | C < 1, C = 1; end

switch KC

case 1

………………………..

case 2

………………………..

case {3,4}

A(1:Nx,:) = [x./sig ones(Nx,1)./sig];

RHS = log(y)./sig; th = A\RHS;

yi = exp(th(1)*xi + th(2)); y2 = exp(th(1)*x + th(2));

if KC == 3, th = exp([th(2) th(1)]);

else th(2) = exp(th(2));

end

case 5

if nargin < 5, C = max(y) + 1; end %final value

A(1:Nx,:) = [x./sig ones(Nx,1)./sig];

y1 = y; y1(find(y > C – 0.01)) = C – 0.01;

RHS = log(C-y1)./sig; th = A\RHS;

yi = C – exp(th(1)*xi + th(2)); y2 = C – exp(th(1)*x + th(2));

th = [-th(1) exp(th(2))];

case 6

A(1:Nx,:) = [log(x)./sig ones(Nx,1)./sig];

y1 = y; y1(find(y < 0.01)) = 0.01;

RHS = log(y1)./sig; th = A\RHS;

yi = exp(th(1)*log(xi) + th(2)); y2 = exp(th(1)*log(x) + th(2));

th = [exp(th(2)) th(1)];

case 7 ………………………..

case 8 ………………………..

case 9 ………………………..

otherwise %standard LS with degree C

A(1:Nx,C + 1) = ones(Nx,1)./sig;

for n = C:-1:1, A(1:Nx,n) = A(1:Nx,n + 1).*x; end

RHS = y./sig; th = A\RHS;

yi = th(C+1); tmp = ones(size(xi));

y2 = th(C+1); tmp2 = ones(size(x));

for n = C:-1:1,

tmp = tmp.*xi; yi = yi + th(n)*tmp;

tmp2 = tmp2.*x; y2 = y2 + th(n)*tmp2;

end

end

th = th(:)’; err = norm(y – y2);

if nargout == 0, plot(x,y,’*’, xi,yi,’k-’); end

PROBLEMS 167

%nm3p11 to plot Fig.P3.11 by curve fitting

clear

x = [1: 20]*2 – 0.1; Nx = length(x);

noise = rand(1,Nx) – 0.5; % 1xNx random noise generator

xi = [1:40]-0.5; %interpolation points

figure(1), clf

a = 0.1; b = -1; c = -50; %Table 3.5(0)

y = a*x.^2 + b*x + c + 10*noise(1:Nx);

[th,err,yi] = curve_fit(x,y,0,2,xi); [a b c],th

[a b c],th %if you want parameters

f = inline(’th(1)*x.^2 + th(2)*x+th(3)’,’th’,’x’);

[th,err] = lsqcurvefit(f,[0 0 0],x,y), yi1 = f(th,xi);

subplot(321), plot(x,y,’*’, xi,yi,’k’, xi,yi1,’r’)

a = 2; b = 1; y = a./x + b + 0.1*noise(1:Nx); %Table 3.5(1)

[th,err,yi] = curve_fit(x,y,1,0,xi); [a b],th

f = inline(’th(1)./x + th(2)’,’th’,’x’);

th0 = [0 0]; [th,err] = lsqcurvefit(f,th0,x,y), yi1 = f(th,xi);

subplot(322), plot(x,y,’*’, xi,yi,’k’, xi,yi1,’r’)

a = -20; b = -9; y = b./(x+a) + 0.4*noise(1:Nx); %Table 3.5(2)

[th,err,yi] = curve_fit(x,y,2,0,xi); [a b],th

f = inline(’th(2)./(x+th(1))’,’th’,’x’);

th0 = [0 0]; [th,err] = lsqcurvefit(f,th0,x,y), yi1 = f(th,xi);

subplot(323), plot(x,y,’*’, xi,yi,’k’, xi,yi1,’r’)

a = 2.; b = 0.95; y = a*b.^x + 0.5*noise(1:Nx); %Table 3.5(3)

[th,err,yi] = curve_fit(x,y,3,0,xi); [a b],th

f = inline(’th(1)*th(2).^x’,’th’,’x’);

th0 = [0 0]; [th,err] = lsqcurvefit(f,th0,x,y), yi1 = f(th,xi);

subplot(324), plot(x,y,’*’, xi,yi,’k’, xi,yi1,’r’)

a = 0.1; b = 1; y = b*exp(a*x) +2*noise(1:Nx); %Table 3.5(4)

[th,err,yi] = curve_fit(x,y,4,0,xi); [a b],th

f = inline(’th(2)*exp(th(1)*x)’,’th’,’x’);

th0 = [0 0]; [th,err] = lsqcurvefit(f,th0,x,y), yi1 = f(th,xi);

subplot(325), plot(x,y,’*’, xi,yi,’k’, xi,yi1,’r’)

a = 0.1; b = 1; %Table 3.5(5)

y = -b*exp(-a*x); C = -min(y)+1; y = C + y + 0.1*noise(1:Nx);

[th,err,yi] = curve_fit(x,y,5,C,xi); [a b],th

f = inline(’1-th(2)*exp(-th(1)*x)’,’th’,’x’);

th0 = [0 0]; [th,err] = lsqcurvefit(f,th0,x,y), yi1 = f(th,xi);

subplot(326), plot(x,y,’*’, xi,yi,’k’, xi,yi1,’r’)

figure(2), clf

a = 0.5; b = 0.5; y = a*x.^b +0.2*noise(1:Nx); %Table 3.5(6a)

[th,err,yi] = curve_fit(x,y,0,2,xi); [a b],th

f = inline(’th(1)*x.^th(2)’,’th’,’x’);

th0 = [0 0]; [th,err] = lsqcurvefit(f,th0,x,y), yi1 = f(th,xi);

subplot(321), plot(x,y,’*’, xi,yi,’k’, xi,yi1,’r’)

a = 0.5; b = -0.5; %Table 3.5(6b)

y = a*x.^b + 0.05*noise(1:Nx);

[th,err,yi] = curve_fit(x,y,6,0,xi); [a b],th

f = inline(’th(1)*x.^th(2)’,’th’,’x’);

th0 = [0 0]; [th,err] = lsqcurvefit(f,th0,x,y), yi1 = f(th,xi);

subplot(322), plot(x,y,’*’, xi,yi,’k’, xi,yi1,’r’)

(cf) If there is no theoretical basis on which we can infer the physical relation

between the variables, how do we determine the candidate function suitable

for fitting the data pairs? We can plot the graph of data pairs and choose one

of the graphs in Fig. P3.11 which is closest to it and choose the corresponding

template function as the candidate fitting function.

168 INTERPOLATION AND CURVE FITTING

0

–50

0

50

10 20 30 40

(0) y = ax 2 + bx + c (a = 0.1, b = –1, c = –50)

(3) y = a bx (a = 2, b = 0.95)

0

0

20

40

60

10 20 30 40

(4) y = b eax (a = 0.1, b = 1)

0

1

1.5

2

10 20 30 40

(5) y = C – b e–ax (a = 0.1, b = 1, C = 1)

0

1.0

1.5

10 20 30 40

(1) y = ax

+b (a = 2, b = 1)

0

–5

5

0

10 20 30 40

(2) y = b

x + a (a = –20, b = –9)

(8) y =

C = y (∞)

1 + b eax (a = –0.2, b = 20, C = 5)

0

0

2

1

10 20 30 40

0

0

2

4

10 20 30 40

(6a) y = a x b (a = 0.5, b = 0.5)

0

0

0.5

10 20 30 40

(6b) y = a x b (a = 0.5, b = –0.5)

0

0

1

2

10 20 30 40

(7) y = ax ebx (a = 0.5, b = −0.1)

0

0

5

10 20 30 40

0

0

5

10

10 20 30 40

(9a) y = a ln x + b (a = 2, b = 1)

0

–15

–10

–5

0

10 20 30 40

(9b) y = a ln x + b (a = –4, b = 1)

Figure P3.11 LS fitting curves for data pairs with various relations.

3.12 Two-Dimensional Interpolation

Compose a routine “z = find_depth(xi,yi)” that finds the depth z of a

geological stratum at a point (xi,yi) given as the input arguments, based

on the data in Problem 1.4.

PROBLEMS 169

(cf) If you have no idea, insert just one statement involving ‘interp2()’ into

the program ‘nm1p04.m’ (Problem 1.4) and fit it into the format of a MATLAB

function.

3.13 Polynomial Curve Fitting by Least Squares and Persistent Excitation

Suppose the theoretical (true) relationship between the input x and the

output y is known as

y = x + 2 (P3.13.1)

Charley measured the output data y 10 times for the same input value

x = 1 by using a gauge whose measurement errors has a uniform distribution

U[−0.5,+0.5]. He made the following MATLAB program “nm3p13”,

which uses the routine “polyfits()” to find a straight line fitting the data.

(a) Check the following program and modify it if needed. Then, run the

program and see the result. Isn’t it beyond your imagination? If you use

the MATLAB built-in function “polyfit()”, does it get any better?

%nm3p13.m

tho = [1 2]; %true parameter

x = ones(1,10); %the unchanged input

y = tho(1)*x + tho(2)+(rand(size(x)) – 0.5);

th_ls = polyfits(x,y,1); %uses the MATLAB routine in Sec.3.8.2

polyfit(x,y,1) %uses MATLAB built-in function

(b) Note that substituting Eq. (3.8.2) into Eq.(3.8.3) yields

θo = ao

bo = [ATA]−1AT y

= M

n=0 x2

n M

n=0 xn

M

n=0 xn M

n=0 1

−1

M

n=0 xnyn

M

n=0 yn

(P3.13.2)

If xn = c(constant) ∀ n = 0 : M, is the matrix AT A invertible?

(c) What conclusion can you derive based on (a) and (b), with reference to

the identifiability condition that the input must be rich in some sense

or persistently exciting?

(cf) This problem implies that the performance of the identification/estimation

scheme including the curve fitting depends on the characteristic of input

as well as the choice of algorithm.

3.14 Scaled Curve Fitting for an Ill-Conditioned Problem [M-2]

Consider Eq. (P3.13.2), which is a typical least-squares (LS) solution. The

matrix AT A, which must be inverted for the solution to be obtained, may

become ill-conditioned by the widely different orders of magnitude of its

elements, if the magnitudes of all xn’s are too large or too small, being far

170 INTERPOLATION AND CURVE FITTING

from 1 (see Remark 2.3). You will realize something about this issue after

solving this problem.

(a) Find a polynomial of degree 2 which fits four data points (106, 1), (1.1 × 106, 2), (1.2 × 106, 5), and (1.3 × 106, 10) and plot the polynomial

function (together with the data points) over the interval [106, 1.3 × 106] to check whether it fits the data points well. How big is the relative

mismatch error? Does the polynomial do the fitting job well?

(b) Find a polynomial of degree 2 which fits four data points (107, 1),(1.1 × 107, 2), (1.2 × 107, 5), and (1.3 × 107, 10) and plot the polynomial

function (together with the data points) over the interval [107, 1.3 × 107] to check whether it fits the data points well. How big is the relative

mismatch error? Does the polynomial do the fitting job well? Did you

get any warning message on the MATLAB command window? What

do you think about it?

(c) If you are not satisfied with the result obtained in (b), why don’t you

try the scaled curve fitting scheme described below?

1. Transform the xn’s of the data point (xn, yn)’s into the region

[−2, 2] by the following relation.

xn ←−2 +

4

xmax − xmin

(xn − xmin) (P3.14.1)

2. Find the LS polynomial p(x) fitting the data point (xn, yn)’s.

3. Substitute

x ←−2 +

4

xmax − xmin

(x − xmin) (P3.14.2)

for x into p(x).

(cf) You can complete the following program “nm3p14” and run it to get the

numeric answers.

%nm3p14.m

clear, clf

format long e

x = 1e6*[1 1.1 1.2 1.3]; y = [1 2 5 10];

xi = x(1)+[0:1000]/1000*(x(end) – x(1));

[p,err,yi] = curve_fit(x,y,0,2,xi); p, err

plot(x,y,’o’,xi,yi), hold on

xmin = min(x); xmax = max(x);

x1 = -2 + 4*(x-xmin)/(xmax – xmin);

x1i = ??????????????????????????;

[p1,err,yi] = ?????????????????????????; p1, err

plot(x,y,’o’,xi,yi)

%To get the coefficients of the original fitting polynomial

ps1 = poly2sym(p1);

syms x; ps0 = subs(ps1,x, – 2 + 4/(xmax – xmin)*(x – xmin));

p0 = sym2poly(ps0)

format short

PROBLEMS 171

3.15 Weighted Least-Squares Curve Fitting

As in Example 3.7, we want to compare the results of applying the LS

approach and the WLS approach for finding a function that we can believe

will describe the relation between the input x and the output y as

y = a x ebx (P3.15)

where the data pair (xm, ym)’s are given as

{(1, 3.2908), (5, 3.3264), (9, 1.1640), (13, 0.3515), (17, 0.1140)} from gauge A with error range ± 0.1

{(3, 4.7323), (7, 2.4149), (11, 0.3814), (15,−0.2396), (19,−0.2615)} from gauge B with error range ± 0.5

Noting that this corresponds to the case of Table 3.5(7), use the MATLAB

routine “curve_fit()” for this job and get the result as depicted in Fig.

P3.15. Identify which one of the two lines a and b is the WLS fitting curve.

How do you compare the results?

0

0

2

4

6

10 20

a

b

Figure P3.15 The LS and WLS fitting curves to y = axebx.

3.16 DFT (Discrete Fourier Transform) Spectrum

Supplement the part of the MATLAB program “do_fft” (Section 3.9.2),

which computes the DFT spectra of the two-tone analog signal described by

Eq. (3.9.2) for the cases of zero-padding and whole interval extension and

plots them as in Figs. 3.13c and 3.13d. Which is the clearest one among

the four spectra depicted in Fig. 3.13? If you can generalize this, which

would you choose among up-sampling, zero-padding, and whole interval

extension to get a clear spectrum?

172 INTERPOLATION AND CURVE FITTING

3.17 Effect of Sampling Period, Zero-Padding, and Whole Time Interval on

DFT Spectrum

In Section 3.9.2, we experienced the effect of zero-padding, sampling period

reduction, and whole interval extension on the DFT spectrum of a two-tone

signal that has two distinct frequency components. Here, we are going

to investigate the effect of zero-padding, sampling period reduction, and

whole interval extension on the DFT spectrum of a triangular pulse depicted

in Fig. P3.17.1c. Additionally, we will compare the DFT with the CtFT

(continuous-time Fourier transform) and the DtFT (discrete-time Fourier

transform) [O-1].

(a) The definition of CtFT that is used for getting the spectrum of a

continuous-time finite-duration signal x(t) is

X(ω) = ∞

−∞

x(t)e−jωt dt (P3.17.1)

–1 0 1 t

–1 0 1 t

r (t )

r (t )

–2 0 2 –4 –2 0 2 4

t t

convolution

Λ(t ) Λ(t + 2)

−Λ(t − 2)

time

shifting

time shifting

(a) Two rectangular pulses (b) r (t) ∗ r (t) = Λ(t ) (c) x (t) = Λ(t + 2) − Λ(t − 2)

Figure P3.17.1 A triangular pulse as the convolution of two rectangular pulses.

The CtFT has several useful properties including the convolution

property and the time-shifting property described as

x(t) ∗ y(t)

(CtFT)

−−−→ X(ω)Y(ω) (P3.17.2)

x(t − t1)

(CtFT)

−−−→ X(ω)e−jωt1 (P3.17.3)

Noting that the triangular pulse is the convolution of the two rectangular

pulse r(t)’s whose CtFTs are

R(ω) = CtFT{r(t)} = 1

−1

ejωt dt = 2

sin ω

ω

we can use the convolution property (P3.17.2) to get the CtFT of the

triangular pulse as

CtFT{(t)} = CtFT{r(t) ∗ r(t)} (P3.17.2) = R(ω)R(ω)

= 4

sin2 ω

ω2 = 4 sin c2 ω

π (P3.17.4)

PROBLEMS 173

•

•

•

• •

• •

•

• • • •

xa [n]

2

−2

8

4

0

10

5

0

0

0 digital frequency p

p

p

p

p

p

Ω

w

2π

0 analog frequency 2p

2 4 6 7

0

0 4 t = nT 7

(a) xa[n] and its spectrum

•

•

•

•

•

•

•

• •

•

• •

• •

•

•

xb [n]

2

−2

0

0 4 t = nT 7.5

(b) Reduction of sampling period

•

• •

•

• •

• • • • • • • • • •

xc [n]

2

−2

0

0 8 t = nT 15

(c) Zero-padding/interval extension

• •

•

•

•

•

•

•

•

•

•

• • •

•

•

xd [n]

2

−2

0

0 8 t = nT 15

(d) Double repetition

• •

• •

• •

• • • • • • • • • • • •

• •

• •

•

8

0

4

0

0 digital frequency

0 analog frequency

8 k15

Ω

w

2p

2p

0 digital frequency

0 analog frequency

Ω

k

w

2p

2p

10

0

5

0 8 15

• •

• •

• •

CtFT Xa

DtFT Xa (Ω)

• DFT Xa (k )

CtFT Xc

DtFT Xc (Ω)

DFT Xc (k )

CtFT Xd

DtFT Xd (Ω)

DFT Xd (k )

CtFT spectrum Xb

DtFT spectrum Xb (Ω)

DFT spectrum Xb (k )

• •

•

• • •

0

0 digital frequency π

2p

Ω

w

2π

0 analog frequency 4p

k 8 k 15

•

• • • •

•

• • • •

• •

• • •

• •

• •

Figure P3.17.2 Effects of sampling period, zero-padding, and whole interval on DFT spectrum.

174 INTERPOLATION AND CURVE FITTING

Successively, use the time shifting property (P3.17.3) to get the CtFT of

x(t) = (t + 2) − (t − 2) (P3.17.5)

as

X(ω)

(P3.17.3, 4) = T (ω)ej2ω − T (ω)e−j2ω = j8 sin(2ω) sin c2 ω

π

(P3.17.6)

Get the CtFT Y(ω) of the triangular wave that is generated by repeating

x(t) two times and described as below.

y(t) = x(t + 4) + x(t − 4) (P3.17.7)

Plot the spectrum X(ω) for 0 ≤ ω ≤ 2π and check if the result is the

same as depicted in a solid line in Fig. P3.17.2a or P3.17.2c. You can

also plot the spectrum X(ω) for 0 ≤ ω ≤ 4π and check if the result

is the same as the solid line in Fig. P3.17.2b. Additionally, plot the

spectrum Y(ω) for 0 ≤ ω ≤ 2π and check if the result is the same as

the solid line in Fig. P3.17.2d.

(b) The definition of DtFT, which is used for getting the spectrum of a

discrete-time signal x[n], is

X() =

∞

n=−∞

x[n]e−jn (P3.17.8)

Use this formula to compute the DtFTs of the discrete-time signals

xa[n], xb[n], xc[n], xd [n] and plot them to see if the results are the

same as the dotted lines in Fig. P3.17.2a–d. What is the valid analog

frequency range over which each DtFT spectrum is similar to the

corresponding CtFT spectrum, respectively? Note that the valid analog

frequency range is [−π/T,+π/T ] for the sampling period T .

(c) Use the definition (3.9.1a) of DFT to get the spectra of the discrete-time

signals xa[n], xb[n], xc[n], and xd [n] and plot them to see if the results

are the same as the dots in Fig. P3.17.2a–d. Do they match the samples

of the corresponding DtFTs at k = 2kπ/N? Among the DFT spectra

(a), (b), (c), and (d), which one describes the corresponding CtFT or

DtFT spectra for the widest range of analog frequency?

3.18 Windowing Techniques Against the Leakage of DFT Spectrum

There are several window functions ready to be used for alleviating the

spectrum leakage problem or for other purposes. We have made a MATLAB

routine “windowing()” for easy application of the various windows.

PROBLEMS 175

Applying the Hamming window function to the discrete-time signal xd [n]

in Fig. 3.13d, get the new DFT spectrum, plot its magnitude together with

the windowed signal, check if they are the same as depicted in Fig. P3.18b,

and compare it with the old DFT spectrum in Fig. 3.13d or Fig. P3.18a.

You can start with the incomplete MATLAB program “nm3p18.m” below.

What is the effect of windowing on the spectrum?

20

10

0

0 5 10 k 54 59

Xd1 (k )

(b) Bartlett/triangular windowing

2

1

0

0 2 4 t = nT 6

−1

−2

xd1 [n ]

(a) Rectangular window

2

1

0

0 2 4 t = nT 6

−1

−2

xd [n ]

20

30

10

0

0 5 10 k 54 59

Xd (k )

Figure P3.18 The effect of windowing on DFT spectrum.

function xw = windowing(x,w)

N = length(x);

if nargin < 2 | w == ’rt’ | isempty(w), xw = x;

elseif w == ’bt’, xw = x.*bartlett(N)’;

elseif w == ’bk’, xw = x.*blackman(N)’;

elseif w == ’hm’, xw = x.*hamming(N)’;

end

%nm3p18: windowing effect on DFT spectrum

w1 = 1.5*pi; w2 = 3*pi; %two tones

N = 64; n = 1:N; T = 0.1; t = (n – 1)*T;

k = 1:N; w0 = 2*pi/T; w = (k – 1)*w0;

xbn = sin(w1*t) + 0.5*sin(w2*t);

xbwn = windowing(xbn,’bt’);

Xb = fft(xbn); Xbw = fft(xbwn);

subplot(421), stem(t,xbn,’.’)

subplot(423), stem(k,abs(Xb),’.’)

…………..

176 INTERPOLATION AND CURVE FITTING

3.19 Interpolation by Using DFS: Zero-Padding on the Frequency Domain

The fitting curve in Fig. 3.14d has been obtained by zeroing out all the

digital frequency components higher than π/2 [rad](N/4 < k <3N/4) of the

sequence x[n] in Fig. 3.14a. Plot another fitting curve obtained by removing

all the frequency components higher than π/4 [rad](N/8 < k < 7N/8) and

compare it with Fig. 3.14d.

3.20 On-Line Recursive Computation of DFT

For the case where you need to compute the DFT of a block of data every

time a new sampled data replaces the oldest one in the block, we derive

the following recursive algorithm for DFT computation.

Defining the first data block and the mth data block as

{x0[0], x0[1], . . . , x0[N − 1]} = {0, 0, . . . , 0} (P3.20.1)

{xm[0], xm[1], . . . , xm[N − 1]} = {x[m], x[m + 1], . . . , x[m + N − 1]} (P3.20.2)

the DFT for the (m + 1)th data block

{xm+1[0], xm+1[1], . . . , xm+1[N − 1]} = {x[m + 1], x[m + 2], . . . , x[m + N]} (P3.20.3)

can be expressed in terms of the DFT for the mth data block

Xm(k) =

N−1

n=0

xm[n]e−j2πnk/N, k= 0 : N − 1 (P3.20.4)

as follows:

Xm+1(k) =N−1

n=0

xm+1[n]e−j2πnk/N =N−1

n=0

xm[n + 1]e−j2πnk/N

=N−1

n=0

xm[n + 1]e−j2π(n+1)k/N ej2πk/N

=N

n=1

xm[n]e−j2πnk/Nej2πk/N

= N−1

n=0

xm[n]e−j2πnk/N + x[N] − x[0]ej2πk/N

= {Xm(k) + x[N] − x[0]}ej2πk/N (P3.20.5)

You can compute the 128-point DFT for a block composed of 128 random

numbers by using this RDFT algorithm and compare it with that obtained

PROBLEMS 177

by using the MATLAB built-in routine “fft()”. You can start with the

incomplete MATLAB program “do_RDFT.m” below.

%do_RDFT

clear, clf

N = 128; k = [0:N – 1];

x = zeros(1,N); %initialize the data block

Xr = zeros(1,N); % and its DFT

for m = 0:N

xN = rand; %new data

Xr = (Xr + xN – x(1)).*???????????????? %RDFT formula (P3.20.5)

x = [x(2:N) xN];

end

dif = norm(Xr-fft(x)) %difference between RDFT and FFT

4

NONLINEAR EQUATIONS

4.1 ITERATIVE METHOD TOWARD FIXED POINT

Let’s see the following theorem.

Fixed-Point Theorem: Contraction Theorem[K-2, Section 5.1]. Suppose a function

g(x) is defined and its first derivative g(x) exists continuously on some interval

I = [xo − r, xo + r] around the fixed point xo of g(x) such that

g(xo) = xo (4.1.1)

Then, if the absolute value of g(x) is less than or equal to a positive number α

that is strictly less than one, that is,

|g(x)| ≤ α < 1 (4.1.2)

the iteration starting from any point x0 ∈ I

xk+1 = g(xk) with x0 ∈I (4.1.3)

converges to the (unique) fixed point xo of g(x).

Applied Numerical Methods Using MATLAB, by Yang, Cao, Chung, and Morris

Copyright 2005 John Wiley & Sons, Inc.

179

180 NONLINEAR EQUATIONS

Proof. The Mean Value Theorem (MVT) (Appendix A) says that for any two

points x0 and xo, there exists a point x between the two points such that

g(x0) − g(xo) = g(x)(x0 − xo); x1 − xo (4.1.3),(4.1.1) = g(x)(x0 − xo) (1)

Taking the absolute value of both sides of (1) and using the precondition

(4.1.2) yields

|x1 − xo| ≤ α|x0 − xo| < |x0 − xo| (2)

which implies that x1 is closer to xo than x0 and thus still stays inside the interval

I . Applying this successively, we can get

|xk − xo| ≤ α|xk−1 − xo| ≤ α2|xk−2 − xo| ≤ · ·· ≤ αk|x0 − xo| → 0 as k→∞ (3)

which implies that the iterative sequence {xk} generated by (4.1.3) converges to xo.

(Q) Is there any possibility that the fixed point is not unique—that is, more than one

point satisfy Eq. (4.1.1) and so the iterative scheme may get confused among the

several fixed points?

(A) It can never happen, because the points xo1 and xo2 satisfying Eq. (4.1.1) must

be the same:

|xo1 − xo2| = |g(xo1) − g(xo2)| ≤ α|xo1 − xo2| (α < 1); |xo1 − xo2| = 0; xo1 ≡ xo2

In order to solve a nonlinear equation f (x) = 0 using the iterative method based on

this fixed-point theorem, we must somehow arrange the equation into the form

x = g(x) (4.1.4)

and start the iteration (4.1.3) with an initial value x0, then continue until some stopping

criterion is satisfied; for example, the difference |xk+1 − xk| between the successive

iteration values becomes smaller than some predefined number (TolX) or the iteration

number exceeds some predetermined number (MaxIter). This scheme is cast into the

MATLAB routine “fixpt()”. Note that the second output argument (err) is never the

real error—that is, the distance to the true solution—but just the last value of |xk+1 − xk| as an error estimate. See the following remark and examples.

Remark 4.1. Fixed-Point Iteration. Noting that Eq. (4.1.4) is not unique for a

given f (x) = 0, it would be good to have g(x) such that |g(x)| < 1 inside

the interval I containing its fixed point xo which is the solution we are looking

for. It may not be so easy, however, to determine whether |g(x)| < 1 is

ITERATIVE METHOD TOWARD FIXED POINT 181

satisfied around the solution point if we don’t have any rough estimate of the

solution.

function [x,err,xx] = fixpt(g,x0,TolX,MaxIter)

% solve x = g(x) starting from x0 by fixed-point iteration.

%input : g,x0 = the function and the initial guess

% TolX = upperbound of incremental difference |x(n + 1) – x(n)|

% MaxIter = maximum # of iterations

%output: x = point which the algorithm has reached

% err = last value |x(k) – x(k – 1)| achieved

% xx = history of x

if nargin < 4, MaxIter = 100; end

if nargin < 3, TolX = 1e-6; end

xx(1) = x0;

for k = 2:MaxIter

xx(k) = feval(g,xx(k – 1)); %Eq.(4.1.3)

err = abs(xx(k) – xx(k – 1)); if err < TolX, break; end

end

x = xx(k);

if k == MaxIter

fprintf(’Do not rely on me, though best in %d iterations\n’,MaxIter)

end

Example 4.1. Fixed-Point Iteration. Consider the problem of solving the nonlinear

equation

f41(x) = x2 − 2 = 0 (E4.1.1)

In order to apply the fixed-point iteration for solving this equation, we need

to convert it into a form like (4.1.4). Let’s try with the following three forms and

guess that the solution is in the interval I = (1, 1.5).

(a) How about x2 − 2 = 0 → x2 = 2 → x = 2/x = ga(x)? (E4.1.2)

Let’s see if the absolute value of the first derivative of ga(x) is less than

one for the solution interval, that is, |ga(x)| = 2/x2 < 1 ∀ x ∈ I. This

condition does not seem to be satisfied and so we must be pessimistic

about the possibility of reaching the solution with (E4.1.2). We don’t need

many iterations to confirm this.

x0 = 1; x1 =

2

x0 = 2; x2 =

2

x1 = 1; x3 =

2

x2 = 2; x4 =

2

x3 = 1; · · ·

(E4.1.3)

The iteration turned out to be swaying between 1 and 2, never approaching

the solution.

(b) How about x2 − 2 = 0 → (x − 1)2 + 2x − 3 = 0 → x = −1

2 {(x − 1)2 − 3} = gb(x)? (E4.1.4)

This form seems to satisfy the convergence condition

|gb(x)| = |x − 1| ≤ 0.5 < 1 ∀ x ∈I (E4.1.5)

182 NONLINEAR EQUATIONS

and so we may be optimistic about the possibility of reaching the solution

with (E4.1.4). To confirm this, we need just a few iterations, which can

be performed by using the routine “fixpt()”.

>>gb=inline(’-((x-1).^2-3)/2’,’x’);

>>[x,err,xx]=fixpt(gb,1,1e-4,50);

>>xx

1.0000 1.5000 1.3750 1.4297 1.4077 …

The iteration is obviously converging to the true solution√2 = 1.414 . . . ,

which we already know in this case. This process is depicted in Fig. 4.1a.

(c) How about x2 = 2 → x = 2

x → x + x = 2

x + x → x = 12

x + 2

x = gc(x)? (E4.1.6)

This form seems to satisfy the convergence condition

|gc(x)| =

1

2

1 −

2

x2

≤ 0.5 < 1 ∀x ∈I (E4.1.7)

which guarantees that the iteration will reach the solution. Moreover, since

this derivative becomes zero at the solution of x2 = 2, we may expect fast

convergence, which is confirmed by using the routine “fixpt()”. The

process is depicted in Fig. 4.1b.

>>gc = inline(’(x+2./x)/2’,’x’);

>>[x,err,xx] = fixpt(gc,1,1e-4,50);

>>xx

1.0000 1.5000 1.4167 1.4142 1.4142 …

(cf) In fact, if the nonlinear equation that we must solve is a polynomial equation,

then it is convenient to use the MATLAB built-in command “roots()”.

1.55

1.5

1.45

1.0000

1.5000

1.3750

1.4297

1.4077

1.4

1.35

1 1.2 1.4

x2

y = x y = x

y = gb (x)

1.0000

1.5000

1.4167

1.4142

1.4142

y = gc (x)

x0 x3 x1

1.6

(a) xk + 1 = gb (xk) = − {(xk − 1)2 1 −3}

2

1.55

1.5

1.45

1.4

1.35

1 1.2 1.4

x0 x2 x1

1.6

(b) xk + 1 = gc (xk) = 1 xk +

2

2

xk

Figure 4.1 Iterative method to solve nonlinear equations based on the fixed-point theorem.

BISECTION METHOD 183

(Q) How do we make the iteration converge to another solution x = −√2 of

x2 − 2 = 0?

4.2 BISECTION METHOD

The bisection method can be applied for solving nonlinear equations like f (x) = 0, only in the case where we know some interval [a, b] on which f (x) is continuous

and the solution uniquely exists and, most importantly, f (a) and f (b) have

the opposite signs. The procedure toward the solution of f (x) = 0 is described

as follows and is cast into the MATLAB routine “bisct()”.

Step 0. Initialize the iteration number k = 0.

Step 1. Letm = 12

(a + b). Iff (m) ≈ 0or 12

(b − a) ≈ 0, then stop the iteration.

Step 2. If f (a)f(m) > 0, then let a ← m; otherwise, let b ← m. Go back to

step 1.

function [x,err,xx] = bisct(f,a,b,TolX,MaxIter)

%bisct.m to solve f(x) = 0 by using the bisection method.

%input : f = ftn to be given as a string ’f’ if defined in an M-file

% a/b = initial left/right point of the solution interval

% TolX = upperbound of error |x(k) – xo|

% MaxIter = maximum # of iterations

%output: x = point which the algorithm has reached

% err = (b – a)/2(half the last interval width)

% xx = history of x

TolFun=eps; fa = feval(f,a); fb = feval(f,b);

if fa*fb > 0, error(’We must have f(a)f(b)<0!’); end

for k = 1: MaxIter

xx(k) = (a + b)/2;

fx = feval(f,xx(k)); err = (b-a)/2;

if abs(fx) < TolFun | abs(err)<TolX, break;

elseif fx*fa > 0, a = xx(k); fa = fx;

else b = xx(k);

end

end

x = xx(k);

if k == MaxIter, fprintf(’The best in %d iterations\n’,MaxIter), end

Remark 4.2. Bisection Method Versus Fixed-Point Iteration

1. Only if the solution exists on some interval [a, b], the distance from the

midpoint (a + b)/2 of the interval as an approximate solution to the true

solution is at most one-half of the interval width—that is, (b − a)/2, which

we take as a measure of error. Therefore, for every iteration of the bisection

method, the upper bound of the error in the approximate solution decreases

by half.

184 NONLINEAR EQUATIONS

2. The bisection method and the false position method appearing in the next

section will definitely give us the solution, only if the solution exists

uniquely in some known interval. But the convergence of the fixedpoint

iteration depends on the derivative of g(x) as well as the initial

value x0.

3. The MATLAB built-in routine fzero(f,x) finds a zero of the function

given as the first input argument, based on the interpolation and the bisection

method with the initial solution interval vector x = [a b] given as

the second input argument. The routine is supposed to work even with an

initial guess x = x0 of the (scalar) solution, but it sometimes gives us a

wrong result as illustrated in the following example. Therefore, it is safe

to use the routine fzero() with the initial solution interval vector [a b] as

the second input argument.

Example 4.2. Bisection Method. Consider the problem of solving the nonlinear

equation

f42(x) = tan(π − x) − x = 0 (E4.2.1)

Noting that f42(x) has the value of infinity at x = π/2 = 1.57 . . ., we set

the initial solution interval to [1.6, 3] excluding the singular point and use the

MATLAB routine “bisct()” as follows. The iteration seems to be converging

to the solution as we expect (see Fig. 4.2b).

>>f42 = inline(’tan(pi – x)-x’,’x’);

>>[x,err,xx] = bisct(f42,1.6,3,1e-4,50);

>>xx

2.3000 1.9500 2.1250 2.0375 1.9937 2.0156 … 2.0287

But, if we start with the initial solution interval [a, b] such that f (a) and f (b)

have the same sign, we will face the error message.

>>[x,err,xx] = bisct(f42,1.5,3,1e-4,50);

??? Error using ==> bisct

We must have f(a)f(b)<0!

Now, let’s see how the MATLAB built-in routine fzero(f,x) works.

>> fzero(f42,[1.6 3])

ans = 2.0287 %good job!

>> fzero(f42,[1.5 3])

??? Error using ==> fzero

The function values at interval endpoints must differ in sign.

>> fzero(f42,1.8) %with an initial guess as 2nd input argument

ans = 1.5708 %wrong result with no warning message

(cf) Not all the solutions given by computers are good, especially when we are careless.

FALSE POSITION OR REGULA FALSI METHOD 185

k ak xk bk f (xk)

0 1.6 3.0

5

2.0375

x4

x3

x2

x1

0

1.6 1.8 2 2.2 2.4 2.6 2.8

1.95

2.125

2.3

32.6, −2.86

1 1.6 2.3 3.0 −1.1808

2 1.6 1.95 2.3 0.5595

3 1.95 2.125 2.3 −0.5092

4 1.95 2.0375 2.125

(a) Process of the bisection method (b) The graph of f (x) = tan(p − x) − x

−0.5027

. . . . .

. . . . .

. . . . .

Figure 4.2 Bisection method for Example 4.2.

4.3 FALSE POSITION OR REGULA FALSI METHOD

Similarly to the bisection method, the false position or regula falsi method starts

with the initial solution interval [a, b] that is believed to contain the solution of

f (x) = 0. Approximating the curve of f (x) on [a, b] by a straight line connecting

the two points (a, f (a)) and (b, f (b)), it guesses that the solution may be the

point at which the straight line crosses the x axis:

x = a −

f (a)

f (a) − f (b)

(b − a) = b −

f (b)

f (b) − f (a)

(b − a) =

af (b) − bf (a)

f (a) − f (b)

(4.3.1)

function [x,err,xx] = falsp(f,a,b,TolX,MaxIter)

%bisct.m to solve f(x)=0 by using the false position method.

%input : f = ftn to be given as a string ’f’ if defined in an M-file

% a/b = initial left/right point of the solution interval

% TolX = upperbound of error(max(|x(k)-a|,|b-x(k)|))

% MaxIter = maximum # of iterations

%output: x = point which the algorithm has reached

% err = max(x(last)-a|,|b-x(last)|)

% xx = history of x

TolFun = eps; fa = feval(f,a); fb=feval(f,b);

if fa*fb > 0, error(’We must have f(a)f(b)<0!’); end

for k = 1: MaxIter

xx(k) = (a*fb-b*fa)/(fb-fa); %Eq.(4.3.1)

fx = feval(f,xx(k));

err = max(abs(xx(k) – a),abs(b – xx(k)));

if abs(fx) < TolFun | err<TolX, break;

elseif fx*fa > 0, a = xx(k); fa = fx;

else b = xx(k); fb = fx;

end

end

x = xx(k);

if k == MaxIter, fprintf(’The best in %d iterations\n’,MaxIter), end

186 NONLINEAR EQUATIONS

6

4

2

0

1.7 2 2.5 3

(a) Bisection method

x3 x1 x4 x3 x2 x1

x4

x2

−2

−4

6

4

2

0

1.7 2 2.5 3

(b) False position method

−2

−4

Figure 4.3 Solving the nonlinear equation f(x) = tan(π − x) − x = 0.

For this method, we take the larger of |x − a| and |b − x| as the measure of error.

This procedure to search for the solution of f (x) = 0 is cast into the MATLAB

routine “falsp()”.

Note that although the false position method aims to improve the convergence

speed over the bisection method, it cannot always achieve the goal, especially

when the curve of f (x) on [a, b] is not well approximated by a straight line as

depicted in Fig. 4.3. Figure 4.3b shows how the false position method approaches

the solution, started by typing the following MATLAB statements, while Fig. 4.3a

shows the footprints of the bisection method.

>>[x,err,xx] = falsp(f42,1.7,3,1e-4,50) %with initial interval [1.7,3]

4.4 NEWTON(–RAPHSON) METHOD

Consider the problem of finding numerically one of the solutions, xo, for a

nonlinear equation

f (x) = (x − xo)mg(x) = 0

where f (x) has (x − xo)m(m is an even number) as a factor and so its curve

is tangential to the x-axis without crossing it at x = xo. In this case, the signs

of f (xo − ε) and f (xo + ε) are the same and we cannot find any interval [a, b]

containing only xo as a solution such that f (a)f(b) < 0. Consequently, bracketing

methods such as the bisection or false position ones are not applicable to

this problem. Neither can the MATLAB built-in routine fzero() be applied to

solve as simple an equation as x2 = 0, which you would not believe until you try

it for yourself. Then, how do we solve it? The Newton(–Raphson) method can

NEWTON(–RAPHSON) METHOD 187

be used for this kind of problem as well as general nonlinear equation problems,

only if the first derivative of f (x) exists and is continuous around the solution.

The strategy behind the Newton(–Raphson) method is to approximate the

curve of f (x) by its tangential line at some estimate xk

y − f (xk) = f (xk)(x − xk) (4.4.1)

and set the zero (crossing the x-axis) of the tangent line to the next estimate

xk+1.

0 − f (xk) = f (xk)(xk+1 − xk)

xk+1 = xk −

f (xk)

f (xk)

(4.4.2)

This Newton iterative formula is cast into the MATLAB routine “newton()”,

which is designed to generate the numerical derivative (Chapter 5) in the case

where the derivative function is not given as the second input argument.

Here, for the error analysis of the Newton method, we consider the seconddegree

Taylor polynomial (Appendix A) of f (x) about x = xk:

f (x) ≈ f (xk) + f (xk)(x − xk) +

f (xk)

2

(x − xk)2

function [x,fx,xx] = newton(f,df,x0,TolX,MaxIter)

%newton.m to solve f(x) = 0 by using Newton method.

%input: f = ftn to be given as a string ’f’ if defined in an M-file

% df = df(x)/dx (If not given, numerical derivative is used.)

% x0 = the initial guess of the solution

% TolX = the upper limit of |x(k) – x(k-1)|

% MaxIter = the maximum # of iteration

%output: x = the point which the algorithm has reached

% fx = f(x(last)), xx = the history of x

h = 1e-4; h2 = 2*h; TolFun=eps;

if nargin == 4 & isnumeric(df), MaxIter = TolX; TolX = x0; x0 = df; end

xx(1) = x0; fx = feval(f,x0);

for k = 1: MaxIter

if ~isnumeric(df), dfdx = feval(df,xx(k)); %derivative function

else dfdx = (feval(f,xx(k) + h)-feval(f,xx(k) – h))/h2; %numerical drv

end

dx = -fx/dfdx;

xx(k+1) = xx(k)+dx; %Eq.(4.4.2)

fx = feval(f,xx(k + 1));

if abs(fx)<TolFun | abs(dx) < TolX, break; end

end

x = xx(k + 1);

if k == MaxIter, fprintf(’The best in %d iterations\n’,MaxIter), end

188 NONLINEAR EQUATIONS

We substitute x = xo (the solution) into this and use f (xo) = 0 to write

0 = f (xo) ≈ f (xk) + f (xk)(xo − xk) +

f (xk)

2

(xo − xk)2

and

−f (xk) ≈ f (xk)(xo − xk) +

f (xk)

2

(xo − xk)2

Substituting this into Eq. (4.4.2) and defining the error of the estimate xk as

ek = xk − xo, we can get

xk+1 ≈ xk + (xo − xk) +

f (xk)

2f (xk)

(xo − xk)2,

|ek+1| ≈

f (xk)

2f (xk)

e2

k = Ake2

k = |Akek||ek| (4.4.3)

This implies that once the magnitude of initial estimation error |e0| is small

enough to make |Ae0| < 1, the magnitudes of successive estimation errors get

smaller very quickly so long as Ak does not become large. The Newton method

is said to be ‘quadratically convergent’ on account of the fact that the magnitude

of the estimation error is proportional to the square of the previous estimation

error.

Now, it is time to practice using the MATLAB routine “newton()” for solving

a nonlinear equation like that dealt with in Example 4.2. We have to type the

following statements into the MATLAB command window.

>>x0 = 1.8; TolX = 1e-5; MaxIter = 50; %with initial guess 1.8,…

>>[x,err,xx] = newton(f42,x0,1e-5,50) %1st order derivative

>>df42 = inline(’-(sec(pi-x)).^2-1’,’x’); %1st order derivative

>>[x,err,xx1] = newton(f42,df42,1.8,1e-5,50)

Remark 4.3. Newton(–Raphson) Method

1. While bracketing methods such as the bisection method and the false position

method converge in all cases, the Newton method is guaranteed to

converge only in case where the initial value x0 is sufficiently close to the

solution xo and A(x) = |f (x)/2f (x)| is sufficiently small for x ≈ xo.

Apparently, it is good for fast convergence if we have small A(x)—that is,

the relative magnitude of the second-order derivative |f (x)| over |f (x)| is

small. In other words, the convergence of the Newton method is endangered

if the slope of f (x) is too flat or fluctuates too sharply.

2. Note two drawbacks of the Newton(–Raphson) method. One is the effort

and time required to compute the derivative f (xk) at each iteration; the

SECANT METHOD 189

2

1

0

−1

−2

1.8 2 2.2

(a) f42 (x) = tan (p − x) − x

10

0

20

−10

0 5 10 15 20

(b) f44b (x) = (x 2 − 25)(x − 10) − 5

2.4 2.6

0

−10

−20

−30

−40

−15 −10 −5 0 5 10 15

x1

x0 x1 x2 x3

x2 x3

x0

x0

x0

x3

x3

x2

x2

x1

x1

1

125

1

0

−1

2

−2

−5 0 5 10

(c) f44b (x) = (x 2 − 25)(x − 10) − 5 (d) f44d (x) = tan−1(x − 2) 1

125

Figure 4.4 Solving nonlinear equations f(x) = 0 by using the Newton method.

other is the possibility of going astray, especially when f (x) has an abruptly

changing slope around the solution (e.g., Fig. 4.4c or 4.4d), whereas it converges

to the solution quickly when f (x) has a steady slope as illustrated

in Figs. 4.4a and 4.4b.

4.5 SECANT METHOD

The secant method can be regarded as a modification of the Newton method in

the sense that the derivative is replaced by a difference approximation based on

the successive estimates

f (xk) ≈

f (xk) − f (xk−1)

xk − xk−1

(4.5.1)

which is expected to take less time than computing the analytical or numerical

derivative. By this approximation, the iterative formula (4.4.2) becomes

xk+1 = xk −

f (xk)

dfdx k

with dfdx k =

f (xk) − f (xk−1)

xk − xk−1

(4.5.2)

190 NONLINEAR EQUATIONS

function [x,fx,xx] = secant(f,x0,TolX,MaxIter,varargin)

% solve f(x) = 0 by using the secant method.

%input : f = ftn to be given as a string ’f’ if defined in an M-file

% x0 = the initial guess of the solution

% TolX = the upper limit of |x(k) – x(k – 1)|

% MaxIter = the maximum # of iteration

%output: x = the point which the algorithm has reached

% fx = f(x(last)), xx = the history of x

h = 1e-4; h2 = 2*h; TolFun=eps;

xx(1) = x0; fx = feval(f,x0,varargin{:});

for k = 1: MaxIter

if k <= 1, dfdx = (feval(f,xx(k) + h,varargin{:})-…

feval(f,xx(k) – h,varargin{:}))/h2;

else dfdx = (fx – fx0)/dx;

end

dx = -fx/dfdx;

xx(k + 1) = xx(k) + dx; %Eq.(4.5.2)

fx0 = fx;

fx = feval(f,xx(k+1));

if abs(fx) < TolFun | abs(dx) < TolX, break; end

end

x = xx(k + 1);

if k == MaxIter, fprintf(’The best in %d iterations\n’,MaxIter), end

This secant iterative formula is cast into the MATLAB routine “secant()”,

which never needs anything like the derivative as an input argument. We can

use this routine “secant()” to solve a nonlinear equation like that dealt with

in Example 4.2, by typing the following statement into the MATLAB command

window. The process is depicted in Fig. 4.5.

>>[x,err,xx] = secant(f42,2.5,1e-5,50) %with initial guess 1.8

2.5

2

1.5

1

0.5

0

1.8 1.9

x1 x4

x5 x3 x2 x0

2 2.1 2.2 2.3 2.4 2.5 2.6

−0.5

−1

−1.5

−2

Figure 4.5 Solving a nonlinear equation by the secant method.

NEWTON METHOD FOR A SYSTEM OF NONLINEAR EQUATIONS 191

4.6 NEWTON METHOD FOR A SYSTEM OF NONLINEAR EQUATIONS

Note that the methods and the corresponding MATLAB routines mentioned so

far can handle only one scalar equation with respect to one scalar variable. In

order to see how a system of equations can be solved numerically, we rewrite

the two equations

f1(x1, x2) = 0

f2(x1, x2) = 0

(4.6.1)

by taking the Taylor series expansion up to first-order about some estimate point

(x1k, x2k) as

f1(x1, x2) ∼=

f1(x1k, x2k) +

∂f1

∂x1(x1k,x2k)

(x1 − x1k) +

∂f1

∂x2(x1k,x2k )

(x2 − x2k) = 0

f2(x1, x2) ∼=

f2(x1k, x2k) +

∂f2

∂x1(x1k,x2k)

(x1 − x1k) +

∂f2

∂x2(x1k,x2k )

(x2 − x2k) = 0

(4.6.2)

This can be arranged into a matrix–vector form as

f1(x1, x2)

f2(x1, x2) ∼= f1(x1k, x2k)

f2(x1k, x2k) + ∂f1/∂x1 ∂f1/∂x2

∂f2/∂x1 ∂f2/∂x2 (x1k,x2k ) x1 − x1k

x2 − x2k

= 0

0 (4.6.3)

which we solve for (x1, x2) to get the updated vector estimate

x1,k+1

x2,k+1 = x1k

x2k − ∂f1/∂x1 ∂f1/∂x2

∂f2/∂x1 ∂f2/∂x2

−1

(x1k,x2k ) f1(x1k, x2k)

f2(x1k, x2k)

(4.6.4)

xk+1 = xk − J −1

k f(xk) with the Jacobian Jk(m, n) = [∂fm/∂xn]|xk

This is not much different from the Newton iterative formula (4.4.2) and is cast

into the MATLAB routine “newtons()”. See Eq. (C.9) in Appendix C for the

definition of the Jacobian.

Now, let’s use this routine to solve the following system of nonlinear equations

x2

1 + 4×2

2 = 5

2×2

1 − 2×1 − 3×2 = 2.5

(4.6.5)

In order to do so, we should first rewrite these equations into a form like

Eq. (4.6.1) as

f1(x1, x2) = x2

1 + 4×2

2 − 5 = 0

f2(x1, x2) = 2×2

1 − 2×1 − 3×2 − 2.5 = 0

(4.6.6)

192 NONLINEAR EQUATIONS

function [x,fx,xx] = newtons(f,x0,TolX,MaxIter,varargin)

%newtons.m to solve a set of nonlinear eqs f1(x)=0, f2(x)=0,..

%input: f = 1^st-order vector ftn equivalent to a set of equations

% x0 = the initial guess of the solution

% TolX = the upper limit of |x(k) – x(k – 1)| % MaxIter = the maximum # of iteration

%output: x = the point which the algorithm has reached

% fx = f(x(last))

% xx = the history of x

h = 1e-4; TolFun = eps; EPS = 1e-6;

fx = feval(f,x0,varargin{:});

Nf = length(fx); Nx = length(x0);

if Nf ~= Nx, error(’Incompatible dimensions of f and x0!’); end

if nargin < 4, MaxIter = 100; end

if nargin < 3, TolX = EPS; end

xx(1,:) = x0(:).’; %Initialize the solution as the initial row vector

%fx0 = norm(fx); %(1)

for k = 1: MaxIter

dx = -jacob(f,xx(k,:),h,varargin{:})\fx(:);/;%-[dfdx]ˆ-1*fx

%for l = 1: 3 %damping to avoid divergence %(2)

%dx = dx/2; %(3)

xx(k + 1,:) = xx(k,:) + dx.’;

fx = feval(f,xx(k + 1,:),varargin{:}); fxn = norm(fx);

% if fxn < fx0, break; end %(4)

%end %(5)

if fxn < TolFun | norm(dx) < TolX, break; end

%fx0 = fxn; %(6)

end

x = xx(k + 1,:);

if k == MaxIter, fprintf(’The best in %d iterations\n’,MaxIter), end

function g = jacob(f,x,h,varargin) %Jacobian of f(x)

if nargin < 3, h = 1e-4; end

h2 = 2*h; N = length(x); x = x(:).’; I = eye(N);

for n = 1:N

g(:,n) = (feval(f,x + I(n,:)*h,varargin{:}) …

-feval(f,x – I(n,:)*h,varargin{:}))’/h2;

end

and convert it into a MATLAB function defined in an M-file, say, “f46.m”

as follows.

function y = f46(x)

y(1) = x(1)*x(1) + 4*x(2)*x(2) – 5;

y(2) = 2*x(1)*x(1)-2*x(1)-3*x(2) – 2.5;

Then, we type the following statements into the MATLAB command window:

>>x0 = [0.8 0.2]; x = newtons(’f46’,x0) %initial guess [.8 .2]

x = 2.0000 0.5000

SYMBOLIC SOLUTION FOR EQUATIONS 193

1

1

0 2

2 1

0

654 3

1

0

0

2×1

2 − 2×1 − 3×2 = 2.5

x1

2 + 4x 2

2 = 5

1

0

−1

−3 −2 −1 0

(a) Newton method with (x10, x20) = (0.8, 0.2)

1 2 3

1

0

−1

−3 −2 −1 0

(b) Newton method with (x10, x20) = (−1.0, 0.5)

1 2 3

4

0

2

−2

−4

−5 0

(c) Newton method with (x10, x20) = (0.5, 0.2)

5 −3 −2 −1 0

(d) Damped Newton method with

(x10, x20) = (0.5, 0.2)

1 2 3

Figure 4.6 Solving the set (4.6.6) of nonlinear equations by vector Newton method.

Figure 4.6 shows how the vector Newton iteration may proceed depending on

the initial guess (x10, x20). With (x10, x20) = (0.8, 0.2), it converges to (2, 0.5),

which is one of the two roots (Fig. 4.6a) and with (x10, x20) = (−1, 0.5), it converges

to (−1.2065, 0.9413), which is another root (Fig. 4.6b). However, with

(x10, x20) = (0.5, 0.2), it wanders around as depicted in Fig. 4.6c. From this figure,

we can see that the iteration is jumping too far in the beginning and then going

astray around the place where the curves of the two functions f1(x) and f2(x)

are close, but not crossing. One idea for alleviating this problem is to modify the

Newton algorithm in such a way that the step size can be adjusted (decreased) to

keep the norm of f(xk) from increasing at each iteration. The so-called damped

Newton method based on this idea will be implemented in the MATLAB routine

“newtons()” if you activate the six statements numbered from 1 to 6 by deleting

the comment mark(%) from the beginning of each line. With the same initial guess

(x10, x20) = (0.5, 0.2) as in Fig. 4.6c, the damped Newton method successfully

leads to the point (2, 0.5), which is one of the two roots (Fig. 4.6d).

MATLAB has the built-in function “fsolve(f,x0)”, which can give us a

solution for a system of nonlinear equations. Let us try it for Eq. (4.6.5) or (4.6.6),

which was already defined in the M-file named ‘f46.m’.

>>x = fsolve(’f46’,x0,optimset(’fsolve’)) %with default parameters

x = 2.0000 0.5000

4.7 SYMBOLIC SOLUTION FOR EQUATIONS

MATLAB has many commands and functions that can be very helpful in dealing

with complex analytic (symbolic) expressions and equations as well as in getting

194 NONLINEAR EQUATIONS

numerical solutions. One of them is “solve()”, which can be used for obtaining the

symbolic or numeric roots of equations. According to what we could see by typing

‘help solve’ into the MATLAB command window, its usages are as follows:

>>solve(’p*sin(x) = r’) %regarding x as an unknown variable and p as a parameter

ans = asin(r/p) %sin−1(r/p)

>>[x1,x2] = solve(’x1^2 + 4*x2^2 – 5 = 0’,’2* x 1^2 – 2*x1 – 3*x2-2.5 = 0’)

x1 = [ 2.] x2 = [ 0.500000]

[ -1.206459] [ 0.941336]

[0.603229 -0.392630*i] [-1.095668 -0.540415e-1*i]

[0.603229 +0.392630*i] [-1.095668 +0.540415e-1*i]

>>S = solve(’x^3 – y^3 = 2’,’x = y’) %returns the solution in a structure.

S = x: [3×1 sym]

y: [3×1 sym]

>>S.x

ans = [ 1]

[ -1/2+1/2*i*3^(1/2)]

[ -1/2-1/2*i*3^(1/2)]

>>S.y

ans = [ -1]

[ 1/2 – 1/2*i*3^(1/2)]

[ 1/2 + 1/2*i*3^(1/2)]

>>[u,v] = solve(’a*u^2 + v^2 = 0’,’u – v = 1’)%regarding u,v as unknowns and a as a parameter

u = [1/2/(a + 1)*(-2*a + 2*(-a)^(1/2)) + 1] v = [1/2/(a + 1)*(-2*a + 2*(-a)^(1/2))]

[1/2/(a + 1)*(-2*a – 2*(-a)^(1/2)) + 1] [1/2/(a + 1)*(-2*a – 2*(-a)^(1/2))]

>>[a,u] = solve(’a*u^2 + v^2’,’u-v = 1’,’a,u’) %regards only v as a parameter

a = -v^2/(v^2 + 2*v + 1) u = v + 1

Note that in the case where the routine “solve()” finds the symbols more

than the equations in its input arguments—say, M symbols and N equations with

M >N—it regards the N symbols closest alphabetically to ‘x’ as variables and

the other M − N symbols as constants, giving the priority of being a variable to

the symbol after ‘x’ than to one before ‘x’ for two symbols that are at the same

distance from ‘x’. Consequently, the priority order of being treated as a symbolic

variable is as follows:

x > y > w > z > v > u > t > s > r > q > · · ·

Actually, we can use the MATLAB built-in function “findsym()” to see the

priority order.

>>syms x y z q r s t u v w %declare 10 symbols to consider

>>findsym(x + y + z*q*r + s + t*u – v – w,10) %symbolic variables?

ans = x,y,w,z,v,u,t,s,r,q

4.8 A REAL-WORLD PROBLEM

Let’s see the following example.

Example 4.3. The Orbit of NASA’s “Wind” Satellite. One of the previous NASA

plans is to launch a satellite, called Wind, which is to stay at a fixed position

along a line from the earth to the sun as depicted in Fig. 4.7 so that the solar

wind passes around the satellite on its way to earth. In order to find the distance

A REAL-WORLD PROBLEM 195

G = 6.67 × 10−11

Ms = 1.98 × 1030[kg]

Me = 5.98 × 1024[kg]

m = the mass of satellite [kg]

R = 1.49 × 1011[m]

g = the distance of satellite from

sun [m]

T = 3.15576 × 107[sec]

w = 2p/T

E

E s

s

Sun E

s: satellite

E: earth

s

E

s

R

g

Figure 4.7 The orbit of a satellite.

of the satellite from earth, we set up the following equation based on the related

physical laws as

G

Msm

r2 = G

Mem

(R − r)2 + mrω2 → GMS

r2 −

Me

(R − r)2 − rω2 = 0 (E4.3.1)

(a) This might be solved for r by using the (nonlinear) equation solvers like

the routine ‘newtons()’ (Section 4.6) or the MATLAB built-in routine

‘fsolve()’. We define this residual error function (whose zero is to be

found) in the M-file named “phys.m” and run the statements in the following

program “nm4e03.m” as

x0 = 1e6; %the initial (starting) guess

rn = newtons(’phys’,x0,1e-4,100) % newtons()

rfs = fsolve(’phys’,x0,optimset(’fsolve’)) % fsolve()

rfs1 = fsolve(’phys’,x0,optimset(’MaxFunEvals’,1000)) %more iterations

x01 = 1e10 %with another starting guess closer to the solution

rfs2 = fsolve(’phys’,x01,optimset(’MaxFunEvals’,1000))

residual_errs = phys([rn rfs rfs1 rfs2])

which yields

rn = 1.4762e+011 <with residual error of -1.8908e-016>

rfs = 5.6811e+007 <with residual error of 4.0919e+004>

rfs1 = 2.1610e+009 <with residual error of 2.8280e+001>

rfs2 = 1.0000e+010 <with residual error of 1.3203e+000>

It seems that, even with the increased number of function evaluations and

another initial guess as suggested in the warning message, ‘fsolve()’ is

not so successful as ‘newtons()’ in this case.

196 NONLINEAR EQUATIONS

(b) Noting that Eq. (E4.3.1) may cause ‘division-by-zero’, we multiply both

sides of the equation by r2(R − r)2 to rewrite it as

r3(R − r)2ω2 − GMS(R − r)2 + GMer2 = 0 (E4.3.2)

We define this residual error function in the M-file named “physb.m” and

run the following statements in the program “nm4e03.m”:

rnb = newtons(’physb’,x0)

rfsb = fsolve(’physb’,x0,optimset(’fsolve’))

residual_errs = phys([rnb rfsb])

which yields

rnb = 1.4762e+011 <with residual error of 4.3368e-018>

rfsb = 1.4762e+011 <with residual error of 4.3368e-018>

Both of the two routines ‘newtons()’ and ‘fsolve()’ benefited from the

function conversion and succeeded in finding the solution.

(c) The results obtained in (a) and (b) imply that the performance of the nonlinear

equation solvers may depend on the shape of the (residual error)

function whose zero they aim to find. Here, we try applying them with

scaling. On the assumption that the solution is known to be on the order

of 1011, we divide the unknown variable r by 1011 to scale it down into

the order of one. This can be done by substituting r = r/1011 into the

equations and multiplying the resulting solution by 1011. We can run the

following statements in the program “nm4e03.m”:

scale = 1e11;

rns = newtons(’phys’,x0/scale,1e-6,100,scale)*scale

rfss = fsolve(’phys’,x0/scale,optimset(’fsolve’),scale)*scale

residual_errs = phys([rns rfss])

which yields

rns = 1.4762e+011 <with residual error of -6.4185e-016>

rfss = 1.4763e+011 <with residual error of -3.3365e-006>

Compared with the results with no scaling obtained in (a), the routine

‘fsolve()’ benefited from scaling and succeeded in finding the solution.

(cf) This example implies the following tips for solving nonlinear equations.

ž If you have some preliminary knowledge about the approximate value of

the true solution, scale the unknown variable up/down to around one and

then scale the resulting solution back down/up to get the solution to the

original equation.

ž It might be better for you to apply at least two methods to solve the

equations as a cross-check. It is suggested to use ‘newtons()’ together with

‘fsolve()’ for confirming the solution of a system of nonlinear equations.

PROBLEMS 197

%nm4e03 – astrophysics

clear, clf

global G Ms Me R T

G = 6.67e11; Ms = 1.98e30; Me = 5.98e24;

R = 1.49e11; T = 3.15576e7; w = 2*pi/T;

x0 = 1e6 %initial guess

format short e

disp(’(a)’)

rn = newtons(’phys’,x0)

rfs = fsolve(’phys’,x0 ,optimset(’fsolve’))

%fsolve(’phys’,x0)/fsolve(’phys’,x0,foptions) in MATLAB 5.x version

rfs1=fsolve(’phys’,x0,optimset(’MaxFunEvals’,1000)) %more iterations

%options([2 3 14])=[1e-4 1e-4 1000];

%fsolve(’phys’,x0,options) in MATLAB 5.x

x01 = 1e10; %with another starting guess closer to the solution

rfs2 = fsolve(’phys’,x01,optimset(’MaxFunEvals’,1000))

residual_errs = phys([rn rfs rfs1 rfs2])

disp(’(b)’)

rnb = newtons(’physb’,x0)

rfsb = fsolve(’physb’,x0,optimset(’fsolve’))

residual_errs = phys([rnb rfsb])

disp(’(c)’)

scale = 1e11;

rns = newtons(’phys’,x0/scale,1e-6,100,scale)*scale;

rfss = fsolve(’phys’,x0/scale,optimset(’fsolve’),scale)*scale

residual_errs = phys([rns rfss])

function f = phys(x,scale);

if nargin < 2, scale = 1; end

global G Ms Me R T

w = 2*pi/T; x = x*scale; f = G*(Ms/(x.^2 + eps) – Me./((R – x).^2 + eps))-x*w^2;

function f = physb(x,scale);

if nargin < 2, scale = 1; end

global G Ms Me R T

w = 2*pi/T; x = x*scale; f = (R-x).^2.*(w^2*x.^3 – G*Ms) + G*Me*x.^2;

PROBLEMS

4.1 Fixed-Point Iterative Method

Consider the simple nonlinear equation

f (x) = x2 − 3x + 1 = 0 (P4.1.1)

Knowing that this equation has two roots

xo = 1.5 ± √1.25 ≈ 2.6180 or 0.382; xo1 ≈ 0.382, xo2 ≈ 2.6180

(P4.1.2)

investigate the practicability of the fixed-point iteration.

(a) First consider the following iterative formula:

xk+1 = ga(xk) =

1

3

(x2

k + 1) (P4.1.3)

198 NONLINEAR EQUATIONS

6

5

4

3

2

1

0

0 2

y = x y = x

y = ga (x ) = (x 2 + 1)

x o1

x o1

x o2

x o2

4 6

−1

1

3 y = gb (x ) = 3 − 1x

(a) xk + 1 = ga (xk) = (xk

2 1 + 1)

3

3

2.5

2

1.5

1

0.5

0

0 1 2 3

−0.5

(b) xk + 1 = gb (xk) = 3 − 1

xk

Figure P4.1 Iterative method based on the fixed-point theorem.

Noting that the first derivative of this iterative function ga(x) is

ga (x) =

2

3

x (P4.1.4)

determine which solution attracts this iteration and certify it in Fig.

P4.1a. In addition, run the MATLAB routine “fixpt()” to perform

the iteration (P4.1.3) with the initial points x0 = 0, x0 = 2, and x0 = 3.

What does the routine yield for each initial point?

(b) Now consider the following iterative formula:

xk+1 = gb(xk) = 3 −

1

xk

(P4.1.5)

Noting that the first derivative of this iterative function gb(x) is

gb (x) = −

1

x2 (P4.1.6)

determinewhich solution attracts this iteration and certify it in Fig. P4.1b.

In addition, run the MATLAB routine “fixpt()” to carry out the iteration

(P4.1.5) with the initial points x0 = 0.2, x0 = 1, and x0 = 3. What

does the routine yield for each initial point?

(cf) This illustrates that the outcome of an algorithm may depend on the starting

point.

PROBLEMS 199

4.2 Bisection Method and Fixed-Point Iteration

Consider the nonlinear equation treated in Example 4.2.

f (x) = tan(π − x) − x = 0 (P4.2.1)

Two graphical solutions of this equation are depicted in Fig. P4.2, which

can be obtained by typing the following statements into the MATLAB

command window:

>>ezplot(’tan(pi-x)’,-pi/2,3*pi/2)

>>hold on, ezplot(’x+0’,-pi/2,3*pi/2)

(a) In order to use the bisection method for finding the solution between

1.5 and 3, Charley typed the statements shown below. Could he get the

right solution? If not, explain him why he failed and suggest him how

to make it.

>>fp42 = inline(’tan(pi-x)-x’,’x’);

>>TolX = 1e-4; MaxIter = 50;

>>x = bisct(fp42,1.5,3,TolX,MaxIter)

(b) In order to find some interval to which the bisection method is applicable,

Jessica used the MATLAB command “find()” as shown below.

>>x = [0: 0.5: pi]; y = tan(pi-x) – x;

>>k = find(y(1:end-1).*y(2:end) < 0);

>>[x(k) x(k + 1); y(k) y(k + 1)]

ans = 1.5000 2.0000 2.0000 2.5000

-15.6014 0.1850 0.1850 -1.7530

This shows that the sign of f (x) changes between x = 1.5 and 2.0

and also between x = 2.0 and 2.5. Noting this, Jessica thought that she

might use the bisection method to find a solution between 1.5 and 2.0

by typing the following command.

>>x=bisct(fp42,1.5,2,TolX,MaxIter)

Check the validity of the solution—that is, check if f (x) = 0 or not—by

typing

>>fp42(x)

If her solution is not good, explain the reason. If you are not sure about

it, you can try plotting the graph in Fig. P4.2 by typing the following

statements into the MATLAB command window.

>>x = [-pi/2+0.05:0.05:3*pi/2 – 0.05];

>>plot(x,tan(pi – x),x,x)

200 NONLINEAR EQUATIONS

−1 0

−5

0

5

1 2 3 4

y = x

y = tan (p − x )

Figure P4.2 The graphical solutions of tan(π − x) − x = 0 or tan(π − x) = x.

(cf) This helps us understand why fzero(fp42,1.8) leads to the wrong solution

even without any warning message as mentioned in Example 4.2.

(c) In order to find the solution around x = 2.0 by using the fixed-point

iteration with the initial point x0 = 2.0, Vania defined the iterative function

as

>>gp421 = inline(’tan(pi – x)’,’x’); % x = g1(x ) = tan (π − x )

and typed the following statement into the MATLAB command window.

>>x = fixpt(gp421,2,TolX,MaxIter)

Could she reach the solution near 2? Will it be better if you start the

routine with any different initial point? What is wrong?

(d) Itha, seeing what Vania did, decided to try with another iterative formula

tan−1 x = π, x = g2(x) = π − tan−1x (P4.2.2)

So she defined the iterative function as

>>gp422 = inline(’pi-atan(x)’, ’x’); % x = g(x ) = π − tan−1(x )

and typed the following statement into the MATLAB command window:

>>x = fixpt(gp422,2,TolX,MaxIter)

What could she get? Is it the right solution? Does this command work

with different initial value, like 0 or 6, which are far from the solution

we want to find? Describe the difference between Vania’s approach and

Itha’s.

PROBLEMS 201

4.3 Recursive (Self-Calling) Routine for Bisection Method

As stated in Section 1.3, MATLAB allows us to make nested (recursive) routines

which call itself. Modify the MATLAB routine “bisct()” (in Section

4.2) into a nested routine “bisct_r()” and run it to solve Eq. (P4.2.1).

4.4 Newton Method and Secant Method

As can be seen in Fig. 4.5, the secant method introduced in Section 4.5

was devised to remove the necessity of the derivative/gradient and improve

the convergence. But, it sometimes turns out to be worse than the Newton

method. Apply the routines “newton()” and “secant()” to solve

fp44(x) = x3 − x2 − x + 1 = 0 (P4.4)

starting with the initial point x0 = −0.2 one time and x0 = −0.3 for another

shot.

4.5 Acceleration of Aitken–Steffensen Method

A sequence converging to a limit xo can be described as

xo − xk+1 = ek+1 ≈ Aek = A(xo − xk)

with lim

k→∞

xo − xk+1

xo − xk = A(|A| < 1) (P4.5.1)

In order to think about how to improve the convergence speed of this

sequence, we define a new sequence pk as

xo − xk+1

xo − xk ≈ A ≈

xo − xk

xo − xk−1

; (xo − xk+1)(xo − xk−1) ≈ (xo − xk)2

(xo)2 − xk+1xo − xk−1xo + xk+1xk−1 ≈ (xo)2 − 2xoxk + x2

k

xo ≈

xk+1xk−1 − x2

k

xk+1 − 2xk + xk−1 = pk (P4.5.2)

(a) Check that the error of this sequence pk is as follows.

xo − pk = xo −

xk+1xk−1 − x2

k

xk+1 − 2xk + xk−1

= xo −

xk−1(xk+1 − 2xk + xk−1) − x2

k−1 + 2xk−1xk − x2

k

xk+1 − 2xk + xk−1

= xo − xk−1 +

(xk − xk−1)2

xk+1 − 2xk + xk−1

= xo − xk−1 +

(−(xo − xk) + (xo − xk−1))2

−(xo − xk+1) + 2(xo − xk) − (xo − xk−1)

= xo − xk−1 +

(−A + 1)2(xo − xk−1)2

(−A2 + 2A − 1)(xo − xk−1) = 0 (P4.5.3)

202 NONLINEAR EQUATIONS

Table P4.5 Comparison of Various Methods Applied for Solving Nonlinear

Equations

Newton Secant Steffensen Schroder fzero() fsolve()

x0 = 1.6 x 2.0288

f42 f (x) 1.19e-8 1.72e-9

Flops 158 112 273 167 986 1454

x0 = 0 x 1.0000

fp44 f (x)

Flops 53 30 63 31 391 364

x0 = 0 x 5.0000 NaN

fp45 f (x) NaN

Flops 536 434 42 19 3683 1978

(cf) Since the flops() command is no longer available in MATLAB 6.x version, the numbers of

floating-point operations are obtained from MATLAB 5.x version so that the readers can compare

the various algorithms in terms of their computational loads.

(b) Modify the routine “newton()” into a routine “stfns()” that generates

the sequence (P4.5.2) and run it to solve

f42(x) = tan(π − x) − x = 0 (with x0 = 1.6) (P4.5.4)

fp44(x) = x3 − x2 − x + 1 = 0 (with x0 = 0) (P4.5.5)

fp45(x) = (x − 5)4 = 0 (with x0 = 0) (P4.5.6)

Fill in Table P4.5 with the results and those obtained by using the

routines “newton()”, “secant()” (with the error tolerance TolX = 10−5), “fzero()”, and “fsolve()”.

4.6 Acceleration of Newton Method for Multiple Roots: Schroder Method

In order to improve the convergence speed, Schroder modifies the Newton

iterative algorithm (4.4.2) as

xk+1 = xk −M

f (xk)

f (xk)

(P4.6.1)

with M : the order of multiplicity of the root we want to find

Based on this idea, modify the routine “newton()” into a routine

“schroder()” and run it to solve Eqs. (P4.5.4.6). Fill in the corresponding

blanks of Table P4.5 with the results.

PROBLEMS 203

4.7 Newton Method for Systems of Nonlinear Equations

Apply the routine “newtons()” (Section 4.6) and the MATLAB built-in

routine “fsolve()” (with [x0 y0] = [1 0.5]) to solve the following systems

of equations. Fill in Table P4.7 with the results.

(a) x2 + y2 = 1

x2 − y = 0

(P4.7.1)

(b) 5cosθ1 + 6cos(θ1 + θ2) = 10

5sinθ1 + 6sin(θ1 + θ2) = 4 (P4.7.2)

(c) 3×2 + 4y2 = 3

x2 + y2 = √3/2

(P4.7.3)

(d) x3

1 + 10×1 − x2 = 5

x1 + x3

2 − 10×2 = −1

(P4.7.4)

(e) x2 − √3xy + 2y2 = 10

4×2 + 3√3xy + y = 22

(P4.7.5)

(f) x3y − y − 2×3 = −16

x − y2 = −1

(P4.7.6)

(g) x2 + 4y2 = 16

xy2 = 4

(P4.7.7)

(h) xey − x5 + y = 3

x + y + tan x − sin y = 0 (P4.7.8)

(i) 2 log y − x = 0

xy − y = 1 (P4.7.9)

(j) 12xy − 6x = −1

60×2 − 180x2y − 30xy = 1

(P4.7.10)

4.8 Newton Method for Systems of Nonlinear Equations

Apply the routine “newtons()” (Section 4.6) and the MATLAB built-in

routine “fsolve()” (with [x0 y0 z0] = [1 1 1]) to solve the following

systems of equations. Fill in Table P4.8 with the results.

(a) xyz = −1

x2 + 2y2 + 4z2 = 7

2×2 + y3 + 6z = 7

(P4.8.1)

(b) xyz = 1

x2 + 2y3 + z2 = 4

x + 2y2 − z3 = 2

(P4.8.2)

(c) x2 + 4y2 + 9z2 = 34

x2 + 9y2 − 5z = 40

x2z − y = 7

(P4.8.3)

(d) x2 + 2 sin(yπ/2) + z2 = 0

−2xy + z = 3

ex+y − z2 = 0

(P4.8.4)

204 NONLINEAR EQUATIONS

Table P4.7 Applying newtons()/fsolve() for Systems of Nonlinear Equations

newtons() fsolve()

x0 = [1 0.5] x

(P4.7.1) ||f (x)||

Flops 1043 1393

x0 = [1 0.5] x [0.1560 0.4111]

(P4.7.2) ||f (x)|| 3.97e-15 (3.66e-15)

Flops 2489 3028

x0 = [1 0.5] x

(P4.7.3) ||f (x)||

Flops 1476 3821

x0 = [1 0.5] x [0.5024 0.1506]

(P4.7.4) ||f (x)|| 8.88e-16 (1.18e-6)

Flops 1127 1932

x0 = [1 0.5] x

(P4.7.5) ||f (x)||

Flops 2884 3153

x0 = [1 0.5] x [1.6922 -1.6408]

(P4.7.6) ||f (x)|| 1.83e-15

Flops 9234 12896

x0 = [1 0.5] x

(P4.7.7) ||f (x)||

Flops 2125 2378

x0 = [1 0.5] x [0.2321 1.5067]

(P4.7.8) ||f (x)|| 1.07 (1.07)

Flops 6516 6492

x0 = [1 0.5] x

(P4.7.9) ||f (x)||

Flops 1521 1680

x0 = [1 0.5] x [0.2236 0.1273]

(P4.7.10) ||f (x)|| 0 (1.11e-16)

Flops 1278 2566

(cf) The numbers of floating-point operations and the residual (mismatching) errors in the parentheses are obtained

from MATLAB 5.x version.

PROBLEMS 205

Table P4.8 Applying newtons()fsolve() for Systems of Nonlinear Equations

newtons() fsolve()

x0 = [1 1 1] x [1.0000 -1.0000 1.0000]

(P4.8.1) ||f (x)|| 1.1102e-16 (1.1102e-16)

Flops 8158 12964

x0 = [1 1 1] x [1 1 1]

(P4.8.2) ||f (x)|| 0

Flops 990 854

x0 = [1 1 1] x

(P4.8.3) ||f (x)||

Flops 6611 4735

x0 = [1 1 1] x [1.0000 -1.0000 1.0000]

(P4.8.4) ||f (x)|| 4.5506e-15 (4.6576e-15)

Flops 18,273 21,935

x0 = [1 1 1] x

(P4.8.5) ||f (x)||

Flops 6811 5525

x0 = [1 1 1] x [2.0000 1.0000 3.0000]

(P4.8.6) ||f (x)|| 3.4659e-8 (2.6130e-8)

Flops 6191 4884

x0 = [1 1 1] x [1.0000 3.0000 2.0000]

(P4.8.7) ||f (x)|| 1.0022e-13 (1.0437e-13)

Flops 8055 6102

(e) x2 + y2 + z2 = 14

x2 + 2y2 − z = 6

x − 3y2 + z2 = −2

(P4.8.5)

(f) x3 − 12y + z2 = 5

3×2 + y3 − 2z = 7

x + 24y2 − 2 sin(πz/18) = 25

(P4.8.6)

206 NONLINEAR EQUATIONS

(g) x2 + y2 − 2z = 6

x2 − 2y + z3 = 3

2xz − 3y2 − z2 = −27

(P4.8.7)

4.9 Newton Method for a System of Nonlinear Equations with Varying Parameter(

s)

In order to find the average modulation order xi for each user of an OFDM

(orthogonal frequency division multiplex) system that has N(128) subchannels

to assign to each of the four users in the environment of noise power

N0 and the bit error rate (probability of bit error) Pe, a communication

system expert, Mi-hyun, formulated the problem into the system of five

nonlinear equations as follows:

fi(x) = (2xi (xi ln 2 − 1) + 1)

N0

3

2(erfc−1(Pe/2))2 − λ = 0 (P4.9.1)

for i = 1, 2, 3, 4

f5(x) =

4

i=1

ai

xi − N = 0 (P4.9.2)

where N = 128 and ai is the data rate of each user

where erfc−1(x) is the inverse function of the complementary error function

erfc(x) =

2

√π ∞

x

e−t 2

dt = 1 −

2

√π x

0

e−t 2

dt = 1 − erf(x) (P4.9.3)

and defined as the MATLAB built-in function ‘erfcinv()’. She defined

the mismatching error (vector) function as below and save it in the M-file

named “fp_bits.m”.

function y = fp_bits(x,a,Pe)

%x(i),i = 1:4 correspond to the modulation order of each user

%x(5) corresponds to the Lagrange multiplier (Lambda)

if nargin < 3, Pe = 1e-4;

if nargin < 2, a = [64 64 64 64]; end

end

N = 128; N0 = 1;

x14 = x(1:4);

y = (2.^x14.*(log(2)*x14 – 1)+1)*N0/3*2*erfcinv(Pe/2).^2 – x(5);

y(5) = sum(a./x14) – N;

Compose a program which solves the above system of nonlinear equations

(with N0 = 1 and Pe = 10−4) to get the modulation order xi of each user

PROBLEMS 207

for five different sets of data rates

a = [32 32 32 32],[64 32 32 32], [128 32 32 32], [256 32 32 32], and [512 32 32 32]

and plots a1/x1 (the number of subchannels assigned to user 1) versus a1

(the data rate of user 1).

4.10 Temperature Rising from Heat Flux in a Semi-infinite Slab

Consider a semi-infinite slab whose temperature rises as a function of position

x > 0 and time t > 0 as

T (x, t) =

Qx

k

e−s2

√πs − erfc(s) with s2 = x2/4at (P4.10.1)

where the function erfc() is defined by Eq. (P4.9.3) and

Q (heat flux) = 200 J/m2s, k(conductivity) = 0.015 J/m/s/◦C,

a (diffusivity) = 2.5 × 10−5 m2/s

In order to find the heat transfer speed, a heating system expert, Kyungwon,

wants to solve the above equation to get the positions x(t) with a

temperature rise of T = 30 ◦C at t = 10:10:200 s. Compose the program

which does this job and plots x(t) versus t .

4.11 Damped Newton Method for a Set of Nonlinear Equations

Consider the routine “newtons()”, which is made for solving a system of

equations and introduced in Section 4.6.

(a) Run the routine with the initial point (x10, x20) = (0.5, 0.2) to solve

Eq. (4.6.5) and certify that it does not yield the right solution as depicted

in Fig. 4.6c.

(b) In order to keep the step size adjusted in the case where the norm of the

vector function f(xk+1) at iteration k + 1 is larger than that of f(xk) at

iteration k, insert (activate) the statements numbered from 1 to 6 of the

routine “newtons()” (Section 4.6) by deleting the comment mark (%) at

the beginning of each line to make a modified routine “newtonds()”,

which implements the damped Newton method. Run it with the initial

point (x10, x20) = (0.5, 0.2) to solve Eq. (4.6.5) and certify that it yields

the right solution as depicted in Fig. 4.6d.

(c) Run the MATLAB built-in routine “fsolve()” with the initial point

(x10, x20) = (0.5, 0.2) to solve Eq. (4.6.5). Does it present you a right

solution?

5

NUMERICAL

DIFFERENTIATION/

INTEGRATION

5.1 DIFFERENCE APPROXIMATION FOR FIRST DERIVATIVE

For a function f (x) of a variable x, its first derivative is defined as

f (x) = lim

h→0

f (x + h) − f (x)

h

(5.1.1)

However, this gives our computers a headache, since they do not know how

to take a limit. Any input number given to computers must be a definite number

and can be neither too small nor too large to be understood by the computer.

The ‘theoretically’ infinitesimal number h involved in this equation is a

problem.

A simple approximation that computers might be happy with is the forward

difference approximation

Df 1(x, h) =

f (x + h) − f (x)

h

(h is step size) (5.1.2)

How far away is this approximation from the true value of (5.1.1)? In order to do

the error analysis, we take the Taylor series expansion of f (x + h) about x as

f (x + h) = f (x) + hf (x) +

h2

2

f (2)(x) +

h3

3!

f (3)(x) + ·· · (5.1.3)

Applied Numerical Methods Using MATLAB, by Yang, Cao, Chung, and Morris

Copyr ight 2005 John Wiley & Sons, I nc., ISBN 0-471-69833-4

209

210 NUMERICAL DIFFERENTIATION/ INTEGRATION

Subtracting f (x) from both sides and dividing both sides by the step size h yields

Df 1(x, h) =

f (x + h) − f (x)

h = f (x) +

h

2

f (2)(x) +

h2

3!

f (3)(x) + ·· ·

= f (x) + O(h) (5.1.4)

where O(g(h)), called ‘big Oh of g(h)’, denotes a truncation error term proportional

to g(h) for |h| ≺ 1. This means that the error of the forward difference

approximation (5.1.2) of the first derivative is proportional to the step size h, or,

equivalently, in the order of h.

Now, in order to derive another approximation formula for the first derivative

having a smaller error, let’s remove the first-order term with respect to h from

Eq. (5.1.4) by substituting 2h for h in the equation

Df 1(x, 2h) =

f (x + 2h) − f (x)

2h = f (x) +

2h

2

f (2)(x) +

4h2

3!

f (3)(x) + ·· ·

and subtracting this result from two times the equation. Then, we get

2Df 1(x, h) − Df 1(x, 2h) = 2

f (x + h) − f (x)

h −

f (x + 2h) − f (x)

2h

= f (x) −

2h2

3!

f (3)(x) + ·· ·

Df 2(x, h) =

2Df 1(x, h) − Df 1(x, 2h)

2 − 1

= −f (x + 2h) + 4f (x + h) − 3f (x)

2h

= f (x) + O(h2) (5.1.5)

which can be regarded as an improvement over Eq. (5.1.4), since it has the

truncation error of O(h2) for |h| ≺ 1.

How about the backward difference approximation?

Db1(x, h) =

f (x) − f (x − h)

h ≡ Df 1(x,−h) (h is step size) (5.1.6)

This also has an error of O(h) and can be processed to yield an improved version

having a truncation error of O(h2).

Db2(x, h) =

2Db1(x, h) − Db1(x, 2h)

2 − 1 =

3f (x) − 4f (x − h) + f (x − 2h)

2h

= f (x) + O(h2) (5.1.7)

In order to derive another approximation formula for the first derivative, we

take the Taylor series expansion of f (x + h) and f (x − h) up to the fifth order

APPROXIMATION ERROR OF FIRST DERIVATIVE 211

to write

f (x + h) = f (x) + hf (x) +

h2

2

f (2)(x) +

h3

3!

f (3)(x) +

h4

4!

f (4)(x) +

h5

5!

f (5)(x)+· · ·

f (x − h) = f (x) − hf (x) +

h2

2

f (2)(x) −

h3

3!

f (3)(x) +

h4

4!

f (4)(x) −

h5

5!

f (5)(x)+· · ·

and divide the difference between these two equations by 2h to get the central

difference approximation for the first derivative as

Dc2(x, h) =

f (x + h) − f (x − h)

2h = f (x) +

h2

3!

f (3)(x) +

h4

5!

f (5)(x) + ·· ·

= f (x) + O(h2) (5.1.8)

which has an error of O(h2) similarly to Eqs. (5.1.5) and (5.1.7). This can also be

processed to yield an improved version having a truncation error of O(h4).

22Dc2(x, h) − Dc2(x, 2h) = 4

f (x + h) − f (x − h)

2h −

f (x + 2h) − f (x − 2h)

2 · 2h

= 3f (x) −

12h4

5!

f (5)(x) − ·· ·

Dc4(x, h) =

22Dc1(x, h) − Dc1(x, 2h)

22 − 1

=

8f (x + h) − 8f (x − h) − f (x + 2h) + f (x−2h)

12h

= f (x) + O(h4) (5.1.9)

Furthermore, this procedure can be formularized into a general formula, called

‘Richardson’s extrapolation’, for improving the difference approximation of the

derivatives as follows:

<Richardson’s extrapolation>

Df,n+1(x, h) =

2nDf,n(x, h) − Df,n(x, 2h)

2n − 1

(n: the order of error) (5.1.10a)

Db,n+1(x, h) =

2nDb,n(x, h) − Db,n(x, 2h)

2n − 1

(5.1.10b)

Dc,2(n+1)(x, h) =

22nDc,2n(x, h) − Dc,2n(x, 2h)

22n − 1

(5.1.10c)

5.2 APPROXIMATION ERROR OF FIRST DERIVATIVE

In the previous section, we derived some difference approximation formulas

for the first derivative. Since their errors are proportional to some power of

212 NUMERICAL DIFFERENTIATION/ INTEGRATION

the step-size h, it seems that the errors continue to decrease as h gets smaller.

However, this is only half of the story since we considered only the truncation

error caused by truncating the high-order terms in the Taylor series expansion

and did not take account of the round-off error caused by quantization.

In this section, we will discuss the round-off error as well as the truncation

error so as to gain a better understanding of how the computer really works. For

this purpose, suppose that the function values

f (x + 2h), f (x + h), f (x), f (x − h), f (x − 2h)

are quantized (rounded-off) to

y2 = f (x + 2h) + e2, y1 = f (x + h) + e1

y0 = f (x) + e0 (5.2.1)

y−1 = f (x − h) + e−1, y−2 = f (x − 2h) + e−2

where the magnitudes of the round-off (quantization) errors e2, e1, e0, e−1, and

e−2 are all smaller than some positive number ε, that is, |ei| ≤ ε. Then, the total

error of the forward difference approximation (5.1.4) can be derived as

Df 1(x, h) =

y1 −y0

h =

f (x +h)+e1 −f (x)−e0

h

(5.1-4) = f (x)+

e1 −e0

h +

K1

2

h

|Df 1(x, h)−f (x)| ≤

e1 −e0

h

+ |K1|

2

h ≤

2ε

h + |K1|

2

h with K1 = f (2)(x)

Look at the right-hand side of this inequality—that is, the upper bound of error.

It consists of two parts; the first one is due to the round-off error and in inverse

proportion to the step-size h, while the second one is due to the truncation error

and in direct proportion to h. Therefore, the upper bound of the total error can

be minimized with respect to the step-size h to give the optimum step-size ho as

d

dh 2ε

h + |K1|

2

h = −

2ε

h2 + |K1|

2 = 0, ho = 2 ε

|K1|

(5.2.2)

Thetotal error of the central difference approximation (5.1.8) can also be derived

as follows:

Dc2(x, h) =

y1 − y−1

2h =

f (x + h) + e1 − f (x − h) − e−1

2h

(5.1.8) = f (x) +

e1 − e−1

2h +

K2

6

h2

|Dc2(x, h) − f (x)| ≤

e1 − e−1

2h

+ |K1|

6

h2 ≤

2ε

2h + |K2|

6

h2 with K2 = f (3)(x)

APPROXIMATION ERROR OF FIRST DERIVATIVE 213

The right-hand side of this inequality is minimized to yield the optimum step

size ho as

d

dh ε

h + |K2|

6

h2 = −

ε

h2 + |K2|

3

h = 0, ho = 3 3ε

|K2|

(5.2.3)

Similarly, we can derive the total error of the central difference approximation

(5.1.9) as

|Dc4(x, h) − f (x)| ≤

8e1 − 8e−1 − e2 + e−2

12h

+ |K4|

30

h4

≤

18ε

12h + |K4|

30

h4 with K4 = f (5)(x)

and find out the optimum step size ho as

d

dh 3ε

2h + |K4|

30

h4 = −

3ε

2h2 +

2|K4|

15

h3 = 0, ho = 5 45ε

4|K4|

(5.2.4)

From what we have seen so far, we can tell that, as we make the step size h

smaller, the round-off error may increase, while the truncation error decreases.

This is called ‘step-size dilemma’. Therefore, there must be some optimal step

size ho for the difference approximation formulas, as derived analytically in

Eqs. (5.2.2), (5.2.3), and (5.2.4). However, these equations are only of theoretical

value and cannot be used practically to determine ho because we usually don’t

have any information about the high-order derivatives and, consequently, we

cannot estimate K1,K2, . . . . Besides, noting that ho minimizes not the real error,

but its upper bound, we can never expect the true optimal step size to be uniform

for all x even with the same approximation formula.

Now, we can verify the step-size dilemma and the existence of some optimal

step size ho by computing the numerical derivative of a function, say, f (x) = sin x, whose analytical derivatives are well known. To see how the errors of the

difference approximation formulas (5.1.4) and (5.1.8) depend on the step size h,

we computed their values for x = π/4 together with their errors as summarized

in Tables 5.1 and 5.2. From these results, it appears that the errors of (5.1.4) and

(5.1.8) are minimized with h ≈ 10−8 and h ≈ 10−5, respectively. This may be

justified by the following facts:

ž Noting that the number of significant bits is 52, which is the number of mantissa

bits (Section 1.2.1), or, equivalently, the number of significant digits

is about 52 × 3/10 ≈ 16 (since 210 ≈ 103), and the value of f (x) = sin x is

less than or equal to one, the round-off error is roughly

ε ≈ 10−16/2

214 NUMERICAL DIFFERENTIATION/ INTEGRATION

Table 5.1 The Forward Difference Approximation (5.1.4) for the First Derivative of f(x) = sin x and Its Error from the True Value (cos π/4 = 0.7071067812) Depending on the Step

Size h

hk = 10−k D1k|x=π/4 D1k − D1(k−1) D1k|x=π/4 − cos(π/4)

h1 = 0.1000000000 0.6706029729 −0.03650380828

h2 = 0.0100000000 0.7035594917 0.0329565188 −0.00354728950

h3 = 0.0010000000 0.7067531100 0.0031936183 −0.00035367121

h4 = 0.0001000000 0.7070714247 0.0003183147 −0.00003535652

h5 = 0.0000100000 0.7071032456 0.0000318210 −0.00000353554

h6 = 0.0000010000 0.7071064277 0.0000031821 −0.00000035344

h7 = 0.0000001000 0.7071067454 0.0000003176 −0.00000003581

h8 = 0.0000000100∗ 0.7071067842 0.0000000389 0.00000000305∗

h9 = 0.0000000010 0.7071068175 0.0000000333∗ 0.00000003636

h10 = 0.0000000001 0.7071077057 0.0000008882 0.00000092454

ho = 0.0000000168 (the optimal value of h obtained from Eq. (5.2.2))

Table 5.2 The Forward Difference Approximation (5.1.8) for the First Derivative of f(x) = sin x and Its Error from the True Value (cos π/4 = 0.7071067812) Depending on the Step

Size h

hk = 10−k D2k|x=π/4 D2k − D2(k−1) D2k|x=π/4 − cos(π/4)

h1 = 0.1000000000 0.7059288590 −0.00117792219

h2 = 0.0100000000 0.7070949961 0.0011661371 −0.00001178505

h3 = 0.0010000000 0.7071066633 0.0000116672 −0.00000011785

h4 = 0.0001000000 0.7071067800 0.0000001167 −0.00000000118

h5 = 0.0000100000∗ 0.7071067812 0.0000000012 −0.00000000001∗

h6 = 0.0000010000 0.7071067812 0.0000000001∗ 0.00000000005

h7 = 0.0000001000 0.7071067804 −0.0000000009 −0.00000000084

h8 = 0.0000000100 0.7071067842 0.0000000039 0.00000000305

h9 = 0.0000000010 0.7071067620 −0.0000000222 −0.00000001915

h10 = 0.0000000001 0.7071071506 0.0000003886 0.00000036942

ho = 0.0000059640 (the optimal value of h obtained from Eq. (5.2.3))

ž Accordingly, Eqs. (5.2.2) and (5.2.3) give the theoretical optimal values of

step size h as

ho = 2 ε

|K1| = 2 ε

|f (π/4)| = 2 10−16/2

| − sin(π/4)| = 1.68 × 10−8

ho = 3 3ε

|K2| = 3 3ε

|f (3)(π/4)| = 3 3 × 10−16/2

| − cos(π/4)| = 0.5964 × 10−5

APPROXIMATION ERROR OF FIRST DERIVATIVE 215

10−8

10−6

10−4

10−2

100

10−10

10−5

100

ho: optimal value ho: optimal value

Df 1(x, h) − f ′(x ) ≤ +

2

h

K1

h

2e Dc 2(x, h) − f ′(x ) ≤ +

6

K2 h 2

2h

2e

10−10 ho h 100

(a) Error bound of Eq. (5.1.4) vs. step size h

10−10 ho h 100

(b) Error bound of Eq. (5.1.8) vs. step size h

Figure 5.1 Forward/central difference approximation error of first derivative versus step size h.

Figure 5.1a/b shows how the error bounds of the difference approximations

(5.1.4)/(5.1.8) for the first derivative vary with the step-size h, implying that there

is some optimal value of step-size h with which the error bound of the numerical

derivative is minimized. It seems that we might be able to get the optimal stepsize

ho by using this kind of graph or directly using Eq. (5.2.2),(5.2.3) or (5.2.4).

But, as mentioned before, it is not possible, as long as the high-order derivatives

are unknown (as is usually the case). Very fortunately, Tables 5.1 and 5.2 suggest

that we might be able to guess the good value of h by watching how

small |Dik − Di(k−1)| is for a given problem. On the other hand, Fig. 5.2a/b

shows the tangential lines based on the forward/central difference approximations

(5.1.4)/(5.1.8) of the first derivative at x = π/4 with the three values of stepsize

h. They imply that there is some optimal step-size ho and the numerical

approximation error becomes larger if we make the step-size h larger or smaller

than the value.

0.2

0.4

0.6

0.8

1 h = 10−16 h = 10−8

h = 0.5

f (x ) = sin x

0 0.5 1 1.5 x 2

(a) Forward difference approximation by Eq. (5.1.4) (b) Central difference approximation by Eq. (5.1.8)

0.2

0.4

0.6

0.8

1

h = 10−16

h = 10−5 h = 1

f (x ) = sin x

0 0.5 1 1.5 x 2

Figure 5.2 Forward/central difference approximation of first derivative of f(x) = sin x.

216 NUMERICAL DIFFERENTIATION/ INTEGRATION

5.3 DIFFERENCE APPROXIMATION FOR SECOND

AND HIGHER DERIVATIVE

In order to obtain an approximation formula for the second derivative, we take

the Taylor series expansion of f (x + h) and f (x − h) up to the fifth order to

write

f (x + h) = f (x) + hf (x) +

h2

2

f (2)(x) +

h3

3!

f (3)(x) +

h4

4!

f (4)(x) +

h5

5!

f (5)(x)+· · ·

f (x − h) = f (x) − hf (x) +

h2

2

f (2)(x) −

h3

3!

f (3)(x) +

h4

4!

f (4)(x) −

h5

5!

f (5)(x)+· · ·

Adding these two equations (to remove the f (x) terms) and then subtracting

2f (x) from both sides and dividing both sides by h2 yields the central difference

approximation for the second derivative as

D(2)

c2 (x, h) =

f (x + h) − 2f (x) + f (x − h)

h2

= f (2)(x) +

h2

12

f (4)(x) +

2h4

6!

f (6)(x) + ·· · (5.3.1)

which has a truncation error of O(h2).

Richardson’s extrapolation can be used for manipulating this equation to

remove the h2 term, which yields an improved version

22D(2)

c2 (x, h) − D(2)

c2 (x, 2h)

22 − 1 = −f (x + 2h) + 16f (x + h) − 30f (x) + 16f (x − h) − f (x − 2h)

12h2

= f (2)(x) −

h4

90

f (5)(x)+· · ·

D(2)

c4 (x, h) = −f (x + 2h) + 16f (x + h) − 30f (x) + 16f (x − h) − f (x − 2h)

12h2

= f (2)(x) + O(h4) (5.3.2)

which has a truncation error of O(h4).

The difference approximation formulas for the first and second derivatives

derived so far are summarized in Table 5.3, where the following notations are

used:

D(N)

f i /D(N)

bi /D(N)

ci is the forward/backward/central difference approximation for

the Nth derivative having an error of O(hi)(h is the step size)

fk = f (x + kh)

DIFFERENCE APPROXIMATION FOR SECOND AND HIGHER DERIVATIVE 217

Now, we turn our attention to the high-order derivatives. But, instead of deriving

the specific formulas, let’s make an algorithm to generate whatever difference

approximation formula we want. For instance, if we want to get the approximation

formula of the second derivative based on the function values f2, f1, f0, f−1,

and f−2, we write

D(2)

c4 (x, h) =

c2f2 + c1f1 + c0f0 + c−1f−1 + c−2f−2

h2 (5.3.3)

and take the Taylor series expansion of f2, f1, f−1, and f−2 excluding f0 on the

right-hand side of this equation to rewrite it as

D(2)

c4 (x, h)

=

1

h2

c2 f0 + 2hf 0 +

(2h)2

2

f (2)

0 +

(2h)3

3!

f (3)

0 +

(2h)4

4!

f (4)

0 + ·· ·

+c1 f0 + hf 0 +

h2

2

f (2)

0 +

h3

3!

f (3)

0 +

h4

4!

f (4)

0 + ·· · + c0f0

+c−1 f0 − hf 0 +

h2

2

f (2)

0 −

h3

3!

f (3)

0 +

h4

4!

f (4)

0 − ·· ·

+c−2 f0 − 2hf 0 +

(2h)2

2

f (2)

0 −

(2h)3

3!

f (3)

0 +

(2h)4

4!

f (4)

0 −· · ·

=

1

h2

(c2 + c1 + c0 + c−1 + c−2)f0 + h(2c2 + c1 − c−1 − 2c−2)f 0

+h2 22

2

c2 +

1

2

c1 +

1

2

c−1 +

22

2

c−2f (2)

0

+h3 23

3!

c2 +

1

3!

c1 −

1

3!

c−1 −

23

3!

c−2f (3)

0

+h4 24

4!

c2 +

1

4!

c1 +

1

4!

c−1 +

24

4!

c−2f (4)

0 + ·· ·

(5.3.4)

We should solve the following set of equations to determine the coefficients

c2, c1, c0, c−1, and c−2 so as to make the expression conform to the second

derivative f (2)

0 at x + 0h = x.

1 1 1 1 1

2 1 0 −1 −2

22/2! 1/2! 0 1/2! 22/2!

23/3! 1/3! 0 −1/3! −23/3!

24/4! 1/4! 0 1/4! 24/4!

c2

c1

c0

c−1

c−2

=

0

0

1

0

0

(5.3.5)

218 NUMERICAL DIFFERENTIATION/ INTEGRATION

Table 5.3 The Difference Approximation Formulas for the First and Second Derivatives

O(h) forward difference approximation for the first derivative:

Df 1(x, h) =

f1 − f0

h

(5.1.4)

O(h2) forward difference approximation for the first derivative:

Df 2(x, h) =

2Df 1(x, h) − Df 1(x, 2h)

2 − 1 = −f2 + 4f1 − 3f0

2h

(5.1.5)

O(h) backward difference approximation for the first derivative:

Db1(x, h) =

f0 − f−1

h

(5.1.6)

O(h2) backward difference approximation for the first derivative:

Db2(x, h) =

2Db1(x, h) − Db1(x, 2h)

2 − 1 =

3f0 − 4f−1 + f−2

2h

(5.1.7)

O(h2) central difference approximation for the first derivative:

Dc2(x, h) =

f1 − f−1

2h

(5.1.8)

O(h4) forward difference approximation for the first derivative:

Dc4(x, h) =

22Dc2(x, h) − Dc2(x, 2h)

22 − 1 = −f2 + 8f1 − 8f−1 + f−2

12h

(5.1.9)

O(h2) central difference approximation for the second derivative:

D(2)

c2 (x, h) =

f1 − 2f0 + f−1

h2 (5.3.1)

O(h4) forward difference approximation for the second derivative:

D(2)

c4 (x, h) =

22D(2)

c2 (x, h) − D(2)

c2 (x, 2h)

22 − 1 = −f2 + 16f1 − 30f0 + 16f−1 − f−2

12h2

(5.3.2)

O(h2) central difference approximation for the fourth derivative:

D(4)

c2 (x, h) =

f−2 − 4f−1 + 6f0 − 4f1 + f2

h4 (from difapx(4,[-2 2]) (5.3.6)

DIFFERENCE APPROXIMATION FOR SECOND AND HIGHER DERIVATIVE 219

function [c,err,eoh,A,b] = difapx(N,points)

%difapx.m to get the difference approximation for the Nth derivative

l = max(points);

L = abs(points(1)-points(2))+ 1;

if L < N + 1, error(’More points are needed!’); end

for n = 1: L

A(1,n) = 1;

for m = 2:L + 2, A(m,n) = A(m – 1,n)*l/(m – 1); end %Eq.(5.3.5)

l = l-1;

end

b = zeros(L,1); b(N + 1) = 1;

c =(A(1:L,:)\b)’; %coefficients of difference approximation formula

err = A(L + 1,:)*c’; eoh = L-N; %coefficient & order of error term

if abs(err) < eps, err = A(L + 2,:)*c’; eoh = L – N + 1; end

if points(1) < points(2), c = fliplr(c); end

The procedure of setting up this equation and solving it is cast into the

MATLAB routine “difapx()”, which can be used to generate the coefficients

of, say, the approximation formulas (5.1.7), (5.1.9), and (5.3.2) just for practice/

verification/fun, whatever your purpose is.

>>format rat %to make all numbers represented in rational form

>>difapx(1,[0 -2]) %1st derivative based on {f0, f−1, f−2}

ans = 3/2 -2 1/2 %Eq.(5.1-7)

>>difapx(1,[-2 2]) %1st derivative based on {f−2, f−1, f0, f1, f2}

ans = 1/12 -2/3 0 2/3 -1/12 %Eq.(5.1.9)

>>difapx(2,[2 -2]) %2nd derivative based on {f2, f1, f0, f−1, f−2}

ans = -1/12 4/3 -5/2 4/3 -1/12 %Eq.(5.3.2)

Example 5.1. Numerical/Symbolic Differentiation for Taylor Series Expansion.

Consider how to use MATLAB to get the Taylor series expansion of a function—

say, e−x about x = 0—which we already know is

e−x = 1 − x +

1

2

x2 −

1

3!

x3 +

1

4!

x4 −

1

5!

x5 + ·· · (E5.1.1)

As a numerical method, we can use the MATLAB routine “difapx()”. On

the other hand, we can also use the MATLAB command “taylor()”, which

is a symbolic approach. Readers may put ‘help taylor’ into the MATLAB

command window to see its usage, which is restated below.

ž taylor(f) gives the fifth-order Maclaurin series expansion of f.

ž taylor(f,n + 1) with an integer n > 0 gives the nth-order Maclaurin

series expansion of f.

ž taylor(f,a) with a real number(a) gives the fifth-order Taylor series expansion

of f about a.

220 NUMERICAL DIFFERENTIATION/ INTEGRATION

ž taylor(f,n + 1,a) gives the n th-order Taylor series expansion of f about

default variable = a.

ž taylor(f,n + 1,a,y) gives the nth-order Taylor series expansion of f(y)

about y = a.

(cf) The target function f must be a legitimate expression given directly as the first

input argument.

(cf) Before using the command “taylor()”, one should declare the arguments of the

function as symbols by putting the statement like “syms x t”.

(cf) In the case where the function has several arguments, it is a good practice to put the

independent variable as the last input argument of “taylor()”, though taylor()

takes one closest (alphabetically) to ‘x’ as the independent variable by default only

if it has been declared as a symbolic variable and is contained as an input argument

of the function f.

(cf) One should use the MATLAB command “sym2poly()” if he wants to extract the

coefficients from the Taylor series expansion obtained as a symbolic expression.

The following MATLAB program “nm5e01” finds us the coefficients of fifthorder

Taylor series expansion of e−x about x = 0 by using the two methods.

%nm5e01:Nth-order Taylor series expansion for e^-x about xo in Ex 5.1

f=inline(’exp(-x)’,’x’);

N = 5; xo = 0;

%Numerical computation method

T(1) = feval(f,xo);

h = 0.005 %.01 or 0.001 make it worse

tmp = 1;

for i = 1:N

tmp = tmp*i*h; %i!(factorial i)*h^i

c = difapx(i,[-i i]); %coefficient of numerical derivative

dix = c*feval(f,xo + [-i:i]*h)’; %/h^i; %derivative

T(i+1) = dix/tmp; %Taylor series coefficient

end

format rat, Tn = fliplr(T) %descending order

%Symbolic computation method

syms x; Ts = sym2poly(taylor(exp(-x),N + 1,xo))

%discrepancy

format short, discrepancy=norm(Tn – Ts)

5.4 INTERPOLATING POLYNOMIAL AND NUMERICAL

DIFFERENTIAL

The difference approximation formulas derived in the previous sections are applicable

only when the target function f (x) to differentiate is somehow given. In

this section, we think about how to get the numerical derivatives when we are

INTERPOLATING POLYNOMIAL AND NUMERICAL DIFFERENTIAL 221

given only the data file containing several data points. A possible measure is

to make the interpolating function by using one of the methods explained in

Chapter 3 and get the derivative of the interpolating function.

For simplicity, let’s reconsider the problem of finding the derivative of f (x) = sin x at x = π/4, where the function is given as one of the following data

point sets:

π

8

, sin

π

8 , π

4

, sin

π

4 , 3π

8

, sin

3π

8

(0, sin 0), π

8

, sin

π

8 , π

4

, sin

π

4 , 3π

8

, sin

3π

8 , 4π

8

, sin

4π

8

2π

16

, sin

2π

16 , 3π

16

, sin

3π

16 , 4π

16

, sin

4π

16 , 5π

16

, sin

5π

16 , 6π

16

, sin

6π

16

We make the MATLAB program “nm540”, which uses the routine “lagranp()”

to find the interpolating polynomial, uses the routine “polyder()” to differentiate

the polynomial, and computes the error of the resulting derivative from the true

value. Let’s run it with x defined appropriately according to the given set of data

points and see the results.

>>nm540

dfx( 0.78540) = 0.689072 (error: -0.018035) %with x = [1:3]*pi/8

dfx( 0.78540) = 0.706556 (error: -0.000550) %with x = [0:4]*pi/8

dfx( 0.78540) = 0.707072 (error: -0.000035) %with x = [2:6]*pi/16

This illustrates that if we have more points that are distributed closer to the target

point, we may get better result.

%nm540

% to interpolate by Lagrange polynomial and get the derivative

clear, clf

x0 = pi/4;

df0 = cos(x0); % True value of derivative of sin(x) at x0 = pi/4

for m = 1:3

if m == 1, x = [1:3]*pi/8;

elseif m == 2, x = [0:4]*pi/8;

else x = [2:6]*pi/16;

end

y = sin(x);

px = lagranp(x,y); % Lagrange polynomial interpolating (x,y)

dpx = polyder(px); % derivative of polynomial px

dfx = polyval(dpx, x0);

fprintf(’ dfx(%6.4f) = %10.6f (error: %10.6f)\n’, x0,dfx,dfx – df0);

end

One more thing to mention before closing this section is that we have the

MATLAB built-in routine “diff()”, which finds us the difference vector for a

given vector. When the data points {(xk, f (xk)), k = 1, 2, . . .} are given as an

222 NUMERICAL DIFFERENTIATION/ INTEGRATION

ASCII data file named “xy.dat”, we can use the routine “diff()” to get the

divided difference, which is similar to the derivative of a continuous function.

>>load xy.dat %input the contents of ’xy.dat’ as a matrix named xy

>>dydx = diff(xy(:,2))./diff(xy(:,1)); dydx’ %divided difference

dydx = 2.0000 0.50000 2.0000

k

xk

xy(:,1)

f (xk)

xy(:,2)

xk+1 − xk

diff(xy(:,1))

f (xk+1) − f (xk)

diff(xy(:,2))

Dk =

f (xk+1) − f (xk)

xk+1 − xk

1 −1 2 1 2 2

2 0 4 2 1 1/2

3 2 5 −1 −2 2

4 1 3

5.5 NUMERICAL INTEGRATION AND QUADRATURE

The general form of numerical integration of a function f (x) over some interval

[a, b] is a weighted sum of the function values at a finite number (N + 1) of

sample points (nodes), referred to as ‘quadrature’:

b

a

f (x) dx∼=

N

k=0

wkf (xk) with a = x0 < x1 < · · · < xN =b (5.5.1)

Here, the sample points are equally spaced for the midpoint rule, the trapezoidal

rule, and Simpson’s rule, while they are chosen to be zeros of certain polynomials

for Gaussian quadrature.

Figure 5.3 shows the integrations over two segments by the midpoint rule,

the trapezoidal rule, and Simpson’s rule, which are referred to as Newton–Cotes

formulas for being based on the approximate polynomial and are implemented

by the following formulas.

midpoint rule

xk+1

xk

f (x) dx∼=

hfmk (5.5.2)

with h = xk+1 − xk, fmk = f (xmk), xmk =

xk + xk+1

2

trapezoidal rule

xk+1

xk

f (x) dx∼=

h

2

(fk + fk+1) (5.5.3)

with h = xk+1 − xk, fk = f (xk)

Simpson’s rule

xk+1

xk−1

f (x) dx∼=

h

3

(fk−1 + 4fk + fk+1) (5.5.4)

with h =

xk+1 − xk−1

2

NUMERICAL INTEGRATION AND QUADRATURE 223

xk − 1

h xk h xk + 1 xk − 1 h xk h xk + 1

(a) The midpoint rule (b) The trapezoidal rule

xk − 1 h xk h xk + 1

(c) Simpson’s rule

Figure 5.3 Various methods of numerical integration.

These three integration rules are based on approximating the target function

(integrand) to the zeroth-, first- and second-degree polynomial, respectively. Since

the first two integrations are obvious, we are going to derive just Simpson’s rule

(5.5.4). For simplicity, we shift the graph of f (x) by −xk along the x axis,

or, equivalently, make the variable substitution t = x − xk so that the abscissas

of the three points on the curve of f (x) change from x = {xk − h, xk, xk + h} to t = {−h, 0,+h}. Then, in order to find the coefficients of the second-degree

polynomial

p2(t) = c1t2 + c2t + c3 (5.5.5)

matching the points (−h, fk−1), (0, fk), (+h, fk+1), we should solve the following

set of equations:

p2(−h) = c1(−h)2 + c2(−h) + c3 = fk−1

p2(0) = c102 + c20 + c3 = fk

p2(+h) = c1(+h)2 + c2(+h) + c3 = fk+1

to determine the coefficients c1, c2, and c3 as

c3 = fk, c2 =

fk+1 − fk−1

2h

, c1 =

1

h2 fk+1 + fk−1

2 − fk

Integrating the second-degree polynomial (5.5.5) with these coefficients from

t = −h to t = h yields

224 NUMERICAL DIFFERENTIATION/ INTEGRATION

h

−h

p2(t) dt =

1

3

c1t3 +

1

2

c2t2 + c3t

h

−h =

2

3

c1h3 + 2c3h

=

2h

3 fk+1 + fk−1

2 − fk + 3fk =

h

3

(fk−1 + 4fk + fk+1)

This is the Simpson integration formula (5.5.4).

Now, as a preliminary work toward diagnosing the errors of the above integration

formulas, we take the Taylor series expansion of the integral function

g(x) = x

xk

f (t) dt with g(x) = f (x), g(2)(x) = f (x), g(3)(x) = f (2)(x)

(5.5.6)

about the lower bound xk of the integration interval to write

g(x) = g(xk)+g(xk)(x −xk)+

1

2

g(2)(xk)(x −xk)2 +

1

3!

g(3)(xk)(x −xk)3 + ·· ·

Substituting Eq. (5.5.6) together with x = xk+1 and xk+1 − xk = h into this yields

xk+1

xk

f (x)dx = 0 + hf (xk) +

h2

2

f (xk) +

h3

3!

f (2)(xk) +

h4

4!

f (3)(xk) +

h5

5!

f (4)(xk) + ·· ·

(5.5.7)

First, for the error analysis of the midpoint rule, we substitute xk−1 and −h = xk−1 − xk in place of xk+1 and h in this equation to write

xk−1

xk

f (x)dx = 0 − hf (xk) +

h2

2

f (xk) −

h3

3!

f (2)(xk) +

h4

4!

f (3)(xk) −

h5

5!

f (4)(xk) + ·· ·

and subtract this equation from Eq. (5.5.7) to write

xk+1

xk

f (x)dx − xk−1

xk

f (x) dx = xk+1

xk

f (x)dx + xk

xk−1

f (x) dx

= xk+1

xk−1

f (x) dx = 2hf (xk) +

2h3

3!

f (2)(xk) +

2h5

5!

f (4)(xk) + ·· · (5.5.8)

Substituting xk and xmk = (xk + xk+1)/2 in place of xk−1 and xk in this equation

and noting that xk+1 − xmk = xmk − xk = h/2, we obtain

xk+1

xk

f (x) dx = hf (xmk) +

h3

3 × 23 f (2)(xmk)

+

h5

5 × 4 × 3 × 25 f (4)(xmk) + ·· ·

xk+1

xk

f (x) dx − hf (xmk) =

h3

24

f (2)(xmk) +

h5

1920

f (4)(xmk) + ·· · = O(h3)

(5.5.9)

NUMERICAL INTEGRATION AND QUADRATURE 225

This, together with Eq. (5.5.2), implies that the error of integration over one

segment by the midpoint rule is proportional to h3.

Second, for the error analysis of the trapezoidal rule, we subtract Eq. (5.5.3)

from Eq. (5.5.7) to write

xk+1

xk

f (x) dx −

h

2

(f (xk) + f (xk+1))

= hf (xk) +

h2

2

f (xk) +

h3

3!

f (2)(xk) +

h4

4!

f (3)(xk) +

h5

5!

f (4)(xk) + ·· ·

−

h

2 f (xk) + f (xk) + hf (xk) +

h2

2

f (2)(xk) +

h3

3!

f (3)(xk)

+

h4

4!

f (4)(xk) + ·· ·

= −

h3

12

f (2)(xk) −

h4

24

f (3)(xk) −

h5

80

f (4)(xk) + O(h6) = O(h3) (5.5.10)

This implies that the error of integration over one segment by the trapezoidal

rule is proportional to h3.

Third, for the error analysis of Simpson’s rule, we subtract the Taylor series

expansion of Eq. (5.5.4)

h

3

(f (xk−1) + 4f (xk) + f (xk+1))

=

h

3 f (xk) + 4f (xk) + f (xk) +

2h2

2

f (2)(xk) +

2h4

4!

f (4)(xk) + ·· ·

= 2hf (xk) +

h3

3

f (2)(xk) +

h5

36

f (4)(xk) + ·· ·

from Eq. (5.5.8) to write

xk+1

xk−1

f (x) dx −

h

3

(f (xk−1) + 4f (xk) + f (xk+1)) = −

h5

90

f (4)(xk) + O(h7)

= O(h5) (5.5.11)

This implies that the error of integration over two segments by Simpson’s rule

is proportional to h5.

Before closing this section, let’s make use of these error equations to find

a way of estimating the error of the numerical integral from the true integral

without knowing the derivatives of the target (integrand) function f (x). For

this purpose, we investigate how the error of numerical integration by Simpson’s

rule

IS(xk−1, xk+1, h) =

h

3

(f (xk−1) + 4f (xk) + f (xk+1))

226 NUMERICAL DIFFERENTIATION/ INTEGRATION

will change if the segment width h is halved to h/2. Noting that, from Eq. (5.5.11),

ES(h) = xk+1

xk−1

f (x) dx − IS(xk−1, xk+1, h) ≈ −

h5

90

f (4)(c)(c ∈ [xk−1, xk+1])

ES(

h

2

) = xk+1

xk−1

f (x) dx − IS xk−1, xk+1,

h

2

= xk

xk−1

f (x) dx − IS xk−1, xk,

h

2 + xk+1

xk

f (x) dx

− IS xk, xk+1,

h

2(c ∈ [xk−1, xk+1])

≈ −2

(h/2)5

90

f (4)(c) =

1

16

ES(h)

we can express the change of the error caused by halving the segment width

as

ES(h) − ES h

2

=

IS(xk−1, xk+1, h) − IS xk−1, xk+1,

h

2

≈

15

16|ES(h)| ≈ 15

ES h

2

(5.5.12)

This suggests the error estimate of numerical integration by Simpson’s rule as

ES h

2

≈

1

24 − 1

IS(xk−1, xk+1, h) − IS xk−1, xk+1,

h

2

(5.5.13)

Also for the trapezoidal rule, similar result can be derived:

ET h

2

≈

1

22 − 1

IT (xk−1, xk+1, h) − IT xk−1, xk+1,

h

2

(5.5.14)

5.6 TRAPEZOIDAL METHOD AND SIMPSON METHOD

In order to get the formulas for numerical integration of a function f (x) over

some interval [a, b], we divide the interval into N segments of equal length

h = (b − a)/N so that the nodes (sample points) can be expressed as {x = a + kh, k = 0, 1, 2, . . . , N}. Then we have the numerical integration of f (x) over

[a, b] by the trapezoidal rule (5.5.3) as

b

a

f (x) dx =

N−1

k=0 xk+1

xk

f (x) dx

∼=

h

2 {(f0 + f1) + (f1 + f2) + ·· ·+(fN−2 + fN−1) + (fN−1 + fN)}

TRAPEZOIDAL METHOD AND SIMPSON METHOD 227

IT 2(a, b, h) = hf (a) + f (b)

2 +

N−1

k=1

f (xk) (5.6.1)

whose error is proportional to h2 asN times the error for one segment [Eq. (5.5.10)],

that is,

NO(h3) = (b − a)/h × O(h3) = O(h2)

On the other hand, we have the numerical integration of f (x) over [a, b] by

Simpson’s rule (5.5.4) with an even number of segments N as

b

a

f (x) dx =

N/2−1

m=0 x2m+2

x2m

f (x)dx

∼=

h

3 {(f0 + 4f1 + f2) + (f2 + 4f3 + f4)+· · ·+(fN−2 + 4fN−1 + fN)}

IS4(a, b, h) =

h

3 f (a) + f (b) + 4

N/2−1

m=0

f (x2m+1) + 2

N/2−1

m=1

f (x2m) (5.6.2)

=

h

3 f (a) + f (b) + 2N/2−1

m=0

f (x2m+1) +

N−1

k=1

f (xk)

whose error is proportional to h4 asN times the error for one segment [Eq. (5.5.11)],

that is,

(N/2)O(h5) = (b − a)/2h × O(h5) = O(h4)

These two integration formulas by the trapezoidal rule and Simpson’s rule are

cast into the MATLAB routines “trpzds()” and “smpsns()”, respectively.

function INTf = trpzds(f,a,b,N)

%integral of f(x) over [a,b] by trapezoidal rule with N segments

if abs(b – a) < eps | N <= 0, INTf = 0; return; end

h = (b – a)/N; x = a +[0:N]*h; fx = feval(f,x); values of f for all nodes

INTf = h*((fx(1) + fx(N + 1))/2 + sum(fx(2:N))); %Eq.(5.6.1)

function INTf = smpsns(f,a,b,N,varargin)

%integral of f(x) over [a,b] by Simpson’s rule with N segments

if nargin < 4, N = 100; end

if abs(b – a)<1e-12 | N <= 0, INTf = 0; return; end

if mod(N,2) ~= 0, N = N + 1; end %make N even

h = (b – a)/N; x = a + [0:N]*h; %the boundary nodes for N segments

fx = fevel(f,x,varargin{:}); %values of f for all nodes

fx(find(fx == inf)) = realmax; fx(find(fx == -inf)) = -realmax;

kodd = 2:2:N; keven = 3:2:N – 1; %the set of odd/even indices

INTf = h/3*(fx(1) + fx(N + 1)+4*sum(fx(kodd)) + 2*sum(fx(keven)));%Eq.(5.6.2)

228 NUMERICAL DIFFERENTIATION/ INTEGRATION

5.7 RECURSIVE RULE AND ROMBERG INTEGRATION

In this section, we are going to look for a recursive formula which enables us

to use some numerical integration with the segment width h to produce another

(hopefully better) numerical integration with half the segment width (h/2). Additionally,

we use Richardson extrapolation (Section 5.1) together with the two

successive numerical integrations to make a Romberg table that can be used to

improve the accuracy of the numerical integral step by step.

Let’s start with halving the segment width h to h/2 for the trapezoidal method.

Then, the numerical integration formula (5.6.1) can be written in the recursive

form as

IT 2 a, b,

h

2 =

h

2 f (a) + f (b)

2 +

2N−1

k=1

f (xk/2)

=

h

2 f (a) + f (b)

2 +

N−1

m=1

f (x2m/2) +

N−1

m=0

f (x(2m+1)/2)

=

1

2 IT 2(a, b, h) +

N−1

m=0

f (x(2m+1)/2)(terms for inserted nodes)

(5.7.1)

Noting that the error of this formula is proportional to h2 (O(h2)), we apply a

Richardson extrapolation [Eq. (5.1.10)] to write a higher-level integration formula

having an error of O(h4) as

IT 4(a, b, h) =

22IT 2(a, b, h) − IT 2(a, b, 2h)

22 − 1

(5.6.1) =

1

3 4

h

2 f (a) + f (b) + 2

N−1

k=1

f (xk)

−

2h

2 f (a) + f (b) + 2

N/2−1

m=1

f (x2m)

=

h

3 f (a) + f (b) + 4

N/2

m=1

f (x2m−1) + 2

N/2−1

m=1

f (x2m)

(5.6.2) ≡ IS4(a, b, h) (5.7.2)

which coincides with the Simpson’s integration formula. This implies that we

don’t have to distinguish the trapezoidal rule from Simpson’s rule. Anyway,

RECURSIVE RULE AND ROMBERG INTEGRATION 229

replacing h by h/2 in this equation yields

IT 4 a, b,

h

2 =

22IT 2(a, b, h/2) − IT 2(a, b, h)

22 − 1

which can be generalized to the following formula:

IT,2(n+1)(a, b, 2−(k+1)h) =

22nIT,2n(a, b, 2−(k+1)h) − IT,2n(a, b, 2−kh)

22n − 1

for n ≥ 1, k ≥ 0 (5.7.3)

Now, it is time to introduce a systematic way, called Romberg integration, of

improving the accuracy of the integral step by step and estimating the (truncation)

error at each step to determine when to stop. It is implemented by

a Romberg Table (Table 5.4), that is, a lower-triangular matrix that we construct

one row per iteration by applying Eq. (5.7.1) in halving the segment width

h to get the next-row element (downward in the first column), and applying

Eq. (5.7.3) in upgrading the order of error to get the next-column elements

(rightward in the row) based on the up–left (north–west) one and the left

(west) one. At each iteration k, we use Eq. (5.5.14) to estimate the truncation

error as

ET,2(k+1)(2−kh) ≈

1

22k − 1IT,2k(2−kh) − IT,2k(2−(k−1)h) (5.7.4)

and stop the iteration when the estimated error becomes less than some prescribed

tolerance. Then, the last diagonal element is taken to be ‘supposedly’ the best

function [x,R,err,N] = rmbrg(f,a,b,tol,K)

%construct Romberg table to find definite integral of f over [a,b]

h = b – a; N = 1;

if nargin < 5, K = 10; end

R(1,1) = h/2*(feval(f,a)+ feval(f,b));

for k = 2:K

h = h/2; N = N*2;

R(k,1) = R(k – 1,1)/2 + h*sum(feval(f,a +[1:2:N – 1]*h)); %Eq.(5.7.1)

tmp = 1;

for n = 2:k

tmp = tmp*4;

R(k,n) = (tmp*R(k,n – 1)-R(k – 1,n – 1))/(tmp – 1); %Eq.(5.7.3)

end

err = abs(R(k,k – 1)- R(k – 1,k – 1))/(tmp – 1); %Eq.(5.7.4)

if err < tol, break; end

end

x = R(k,k);

230 NUMERICAL DIFFERENTIATION/ INTEGRATION

estimate of the integral. This sequential procedure of Romberg integration is cast

into the MATLAB routine “rmbrg()”.

Before closing this section, we test and compare the trapezoidal method

(“trpzds()”), Simpson method (“smpsns()”), and Romberg integration

(“rmbrg()”) by trying them on the following integral

4

0

400x(1 − x)e−2x dx = 100 −2e−2xx(1 − x)

4

0 + 4

0

2e−2x(1 − 2x) dx

= 100 −2e−2xx(1 − x)

4

0 − e−2x(1 − 2x)

4

0 − 2 4

0

e−2x dx

= 200x2e−2x

4

0 = 3200e−8 = 1.07348040929 (5.7.5)

Here are the MATLAB statements for this job listed together with the running

results.

>>f = inline(’400*x.*(1 – x).*exp(-2*x)’,’x’);

>>a = 0; b = 4; N = 80;

>>format short e

>>true_I = 3200*exp(-8)

>>It = trpzds(f,a,b,N), errt = It-true_I %trapezoidal

It = 9.9071e-001, errt = -8.2775e-002

>>Is = smpsns(f,a,b,N), errs = Is-true I %Simpson

INTfs = 1.0731e+000, error = -3.3223e-004

>>[IR,R,err,N1] = rmbrg(f,a,b,.0005), errR = IR – true I %Romberg

INTfr = 1.0734e+000, N1 = 32

error = -3.4943e-005

As expected from the fact that the errors of numerical integration by the trapezoidal

method and Simpson method are O(h2) and O(h4), respectively, the

Simpson method presents better results (with smaller error) than the trapezoidal

ADAPTIVE QUADRATURE 231

one with the same number of segments N = 80. Moreover, Romberg integration

with N = 32 shows a better result than both of them.

5.8 ADAPTIVE QUADRATURE

The numerical integration methods in the previous sections divide the integration

interval uniformly into the segments of equal width, making the error

nonuniform over the interval—that is, small/large for smooth/swaying portion

of the curve of integrand f (x). In contrast, the strategy of the adaptive quadrature

is to divide the integration interval nonuniformly into segments of (generally)

unequal lengths—that is, short/long segments for swaying/smooth portion

of the curve of integrand f (x), aiming at having smaller error with fewer

segments.

The algorithm of adaptive quadrature scheme starts with a numerical integral

(INTf) for the whole interval and the sum of numerical integrals (INTf12 =

INTf1 + INTf2) for the two segments of equal width. Based on the difference

between the two successive estimates INTf and INTf12, it estimates the error of

INTf12 by using Eq. (5.5.13)/(5.5.14) depending on the basic integration rule.

Then, if the error estimate is within a given tolerance (tol), it terminates with

INTf12. Otherwise, it digs into each segment by repeating the same procedure

with half of the tolerance (tol/2) assigned to both segments, until the deepest

level satisfies the error condition. This is how the adaptive scheme forms sections

of nonuniform width, as illustrated in Fig. 5.4. In fact, this algorithm really fits

the nested (recursive) calling structure introduced in Section 1.3 and is cast into

0 0.5 1 1.5 2 2.5 3 3.5 4

40

30

20

10

0

−10

−20

the curve of target function

to be integrated

sub-sub interval sub-sub interval

whole interval

sub interval sub interval

f (x) = 400x (1 − x )e−2x

Figure 5.4 The subintervals (segments) and their boundary points (nodes) determined by the

adaptive Simpson method.

232 NUMERICAL DIFFERENTIATION/ INTEGRATION

the routine “adap_smpsn()”, which needs the calling routine “adapt_smpsn()”

for start-up.

function [INTf,nodes,err] = adap smpsn(f,a,b,INTf,tol,varargin)

%adaptive recursive Simpson method

c = (a+b)/2;

INTf1 = smpsns(f,a,c,1,varargin{:});

INTf2 = smpsns(f,c,b,1,varargin{:});

INTf12 = INTf1 + INTf2;

err = abs(INTf12 – INTf)/15; % Error estimate by Eq.(5.5.13)

if isnan(err) | err < tol | tol<eps % NaN? Satisfying error? Too deep level?

INTf = INTf12;

points = [a c b];

else

[INTf1,nodes1,err1] = adap smpsn(f,a,c,INTf1,tol/2,varargin{:});

[INTf2,nodes2,err2] = adap smpsn(f,c,b,INTf2,tol/2,varargin{:});

INTf = INTf1 + INTf2;

nodes = [nodes1 nodes2(2:length(nodes2))];

err = err1 + err2;

end

function [INTf,nodes,err] = adapt smpsn(f,a,b,tol,varargin)

%apply adaptive recursive Simpson method

INTf = smpsns(f,a,b,1,varargin{:});

[INTf,nodes,err] = adap smpsn(f,a,b,INTf,tol,varargin{:});

We can apply these routines to get the approximate value of integration

(5.7.5) by putting the following MATLAB statements into the MATLAB command

window.

>>f = inline(’400*x.*(1 – x).*exp(-2*x)’,’x’);

>>a=0; b = 4; tol = 0.001;

>>format short e

>>true I = 3200*exp(-8);

>>Ias = adapt smpsn(f,a,b,tol), erras=Ias-true I

Ias = 1.0735e+000, erras = -8.9983e-006

Figure 5.4 shows the curve of the integrand f (x) = 400x(1 − x)e−2x together

with the 25 nodes determined by the routine “adapt_smpsn()”, which yields

better results (having smaller error) with fewer segments than other methods

discussed so far. From this figure, we see that the nodes are dense/sparse in the

swaying/smooth portion of the curve of the integrand.

Here, we introduce the MATLAB built-in routines adopting the adaptive recursive

integration scheme together with the illustrative example of their usage.

“quad(f,a,b,tol,trace,p1,p2,..)” / “quadl(f,a,b,tol,trace,p1,p2,..)”

>>Iq = quad(f,a,b,tol), errq = Iq – true I

Iq = 1.0735e+000, errq = 4.0107e-005

>>Iql = quadl(f,a,b,tol), errql = Iql – true I

Iql = 1.0735e+000, errq1 = -1.2168e-008

ADAPTIVE QUADRATURE 233

(cf) These routines are capable of passing the parameters (p1,p2,..) to the integrand

(target) function and can be asked to show a list of intermediate subintervals with

the fifth input argument trace=1.

(cf) quadl() is introduced in MATLAB 6.x version to replace another adaptive integration

routine quad8() which is available in MATLAB 5.x version.

Additionally, note that MATLAB has a symbolic integration routine

“int(f,a,b)”. Readers may type “help int” into the MATLAB command window

to see its usage, which is restated below.

ž int(f) gives the indefinite integral of f with respect to its independent

variable (closest to ‘x’).

ž int(f,v) gives the indefinite integral of f(v) with respect to v given as

the second input argument.

ž int(f,a,b) gives the definite integral of f over [a,b] with respect to its

independent variable.

ž int(f,v,a,b) gives the definite integral of f(v) with respect to v over

[a,b].

(cf) The target function f must be a legitimate expression given directly as the first

input argument and the upper/lower bound a,b of the integration interval can be

a symbolic scalar or a numeric.

Example 5.2. Numerical/Symbolic Integration using quad()/quadl()/int().

Consider how to make use of MATLAB for obtaining the continuous-time

Fourier series (CtFS) coefficient

Xk = P/2

−P/2

x(t)e−jkω0t dt = P/2

−P/2

x(t)e−j2πkt/P dt (E5.2.1)

For simplicity, let’s try to get just the 16th CtFS coefficient of a rectangular

wave

x(t) = 1 for − 1 ≤ t < 1

0 for − 2 ≤ t < 1 or 1 ≤ t < 2 (E5.2.2)

which is periodic in t with period P = 4. We can compute it analytically as

X16 = 2

−2

x(t)e−j2π16t/4 dt = 1

−1

e−j8πt dt =

1

−j8π

e−j8πt

1

−1

=

1

8π

sin(8πt)

1

−1 = 0 (E5.2.3)

234 NUMERICAL DIFFERENTIATION/ INTEGRATION

%nm5e02

%use quad()/quad8() and int() to get CtFS coefficient X16 in Ex 5.2

ftn = ’exp(-j*k*w0*t)’; fcos = inline(ftn,’t’,’k’,’w0’);

P = 4; k = 16; w0 = 2*pi/P;

a = -1; b = 1; tol = 0.001; trace = 0;

X16_quad = quad(fcos,a,b,tol,trace,k,w0)

X16_quadl = quadl(fcos,a,b,tol,trace,k,w0)

syms t; % declare symbolic variable

Iexp = int(exp(-j*k*w0*t),t) % symbolic indefinite integral

Icos = int(cos(k*w0*t),t) % symbolic indefinite integral

X16_sym = int(cos(k*w0*t),t,-1,1) % symbolic definite integral

As a numerical approach, we can use the MATLAB routine “quad()”/

“quadl()”. On the other hand, we can also use the MATLAB routine “int()”,

which is a symbolic approach. We put all the statements together to make the

MATLAB program “nm5e02”, in which the fifth input argument (trace) of

“quad()”/“quadl()” is set to 1 so that we can see their nodes and tell how

different they are. Let’s run it and see the results.

>>nm5e02

X16_quad = 0.8150 + 0.0000i %betrayal of MATLAB?

X16_quadl = 7.4771e-008 %almost zero, OK!

Iexp = 1/8*i/pi*exp(-8*i*pi*t) %(E5.2.3) by symbolic computation

Icos = 1/8/pi*sin(8*pi*t) %(E5.2.3) by symbolic computation

X16_sym = 0 %exact answer by symbolic computation

What a surprise! It is totally unexpected that the MATLAB routine “quad()”

gives us a quite eccentric value (0.8150), even without any warning message. The

routine “quad()” must be branded as a betrayer for a piecewise-linear function

multiplied by a periodic function. This seems to imply that “quadl()” is better

than “quad()” and that “int()” is the best of the three commands. It should,

however, be noted that “int()” can directly accept and handle only the functions

composed of basic mathematical functions, rejecting the functions defined in the

form of string or by the “inline()” command or through an m-file and besides,

it takes a long time to execute.

(cf) What about our lovely routine “adapt_smpsn()”? Regrettably, you had better not

count on it, since it will give the wrong answer for this problem. Actually, “quadl()”

is much more reliable than “quad()” and “adapt_smpsn()”.

5.9 GAUSS QUADRATURE

In this section, we cover several kinds of Gauss quadrature methods—that is,

Gauss–Legendre integration, Gauss–Hermite integration, Gauss–Laguerre integration

and Gauss–Chebyshev I,II integration. Each tries to approximate one of

GAUSS QUADRATURE 235

the following integrations, respectively:

b

a

f (t) dt, +∞

−∞

e−t 2

f (t) dt, +∞

0

e−tf (t) dt,

1

−1

1

√1 − t2

f (t) dt, 1

−1 1 − t2f (t)dt ≈

N

i=1

wif (ti )

The problem is how to fix the weight wi’s and the (Gauss) grid points ti ’s.

5.9.1 Gauss–Legendre Integration

If the integrand f (t) is a polynomial of degree ≤ 3(= 2N − 1), then its integration

I (−1, 1) = +1

−1

f (t) dt (5.9.1)

can exactly be obtained from just 2(N) points by using the following formula

I [t1, t2] = w1f (t1) + w2f (t2) (5.9.2)

How marvelous it is! It is almost a magic. Do you doubt it? Then, let’s find the

weights w1, w2 and the grid points t1, t2 such that the approximating formula

(5.9.2) equals the integration (5.9.1) for f (t) = 1(of degree 0), t(of degree 1),

t2(of degree 2), and t3(of degree 3). In order to do so, we should solve the

following system of equations:

f (t) = 1 : w1f (t1) + w2f (t2) = w1 + w2 ≡ 1

−1

1 dt = 2 (5.9.3a)

f (t) = t : w1f (t1) + w2f (t2) = w1t1 + w2t2 ≡ 1

−1

t dt = 0 (5.9.3b)

f (t) = t2 : w1f (t1) + w2f (t2) = w1t2

1 + w2t2

2 ≡ 1

−1

t2 dt =

2

3

(5.9.3c)

f (t) = t3 : w1f (t1) + w2f (t2) = w1t3

1 + w2t3

2 ≡ 1

−1

t3 dt = 0 (5.9.3d)

Multiplying (5.9.3b) by t2

1 and subtracting the result from (5.9.3d) yields

w2(t3

2 −t2

1 t2) = w2t2(t2 +t1)(t2 −t1) = 0 → t2= −t1, t2 = t1(meaningless)

t2 = −t1 → (5.9.3b), (w1 − w2)t1 = 0,

w1 = w2 → (5.9.3a), w1 + w1 = 2

w1 = w2 = 1 → (5.9.3c), t2

1 + (−t1)2 =

2

3

, t1 = −t2 = −

1

√3

236 NUMERICAL DIFFERENTIATION/ INTEGRATION

so that Eq. (5.9.2) becomes

I [t1, t2] = f −

1

√3 + f 1

√3 (5.9.4)

We can expect this approximating formula to give us the exact value of the

integral (5.9.1) when the integrand f (t) is a polynomial of degree ≤ 3.

Now, you are concerned about how to generalize this two-point Gauss–Legendre

integration formula to an N-point case, since a system of nonlinear equation like

Eq. (5.9.3) can be very difficult to solve as the dimension increases. But, don’t

worry about it. The N grid points (ti ’s) of Gauss–Legendre integration formula

IGL[t1, t2, . . . , tN] =

N

i=1

wN,if (ti) (5.9.5)

giving us the exact integral of an integrand polynomial of degree ≤ (2N − 1)

can be obtained as the zeros of the Nth-degree Legendre polynomial [K-1,

Section 4.3]

LN(t) =

N/2

i=0

(−1)i (2N − 2i)!

2Ni!(N − i)!(N − 2i)!

tN−2i (5.9.6a)

LN(t) =

1

N

((2N − 1)tLN−1(t) − (N − 1)LN−2(t)) (5.9.6b)

Given the N grid point ti ’s, we can get the corresponding weight wN,i ’s of the

N-point Gauss–Legendre integration formula by solving the system of linear

equations

1 1 1 ž 1

t1 t2 tn ž tN

tn−1

1 tn−1

2 tn−1

n ž tn−1

N

ž ž ž ž ž

tN−1

1 tN−1

2 tN−1

n ž tN−1

N

wN,1

wN,2

wN,n

ž

wN,N

=

2

0

(1 − (−1)n)/n

ž

(1 − (−1)N)/N

(5.9.7)

where the nth element of the right-hand side (RHS) vector is

RHS(n) = 1

−1

tn−1 dt =

1

n

tn

1

−1 =

1 − (−1)n

n

(5.9.8)

This procedure of finding the N grid point ti’s and the weight wN,i ’s of the

N-point Gauss–Legendre integration formula is cast into the MATLAB routine

“Gausslp()”. We can get the two grid point ti ’s and the weight wN,i ’s of the twopoint

Gauss–Legendre integration formula by just putting the following statement

into the MATLAB command window.

GAUSS QUADRATURE 237

function [t,w] = Gausslp(N)

if N < 0, fprintf(’\nGauss-Legendre polynomial of negative order??\n’);

else

t = roots(Lgndrp(N))’; %make it a row vector

A(1,:) = ones(1,N); b(1) = 2;

for n = 2:N % Eq.(5.9.7)

A(n,:) = A(n – 1,:).*t;

if mod(n,2) == 0, b(n) = 0;

else b(n) = 2/n; % Eq.(5.9.8)

end

end

w = b/A’;

end

function p = Lgndrp(N) %Legendre polynomial

if N <= 0, p = 1; %n*Ln(t) = (2n – 1)t Ln – 1(t)-(n – 1)Ln-2(t) Eq.(5.9.6b)

elseif N == 1, p = [1 0];

else p = ((2*N – 1)*[Lgndrp(N – 1) 0]-(N – 1)*[0 0 Lgndrp(N – 2)])/N;

end

function I = Gauss_Legendre(f,a,b,N,varargin)

%Gauss_Legendre integration of f over [a,b] with N grid points

% Never try N larger than 25

[t,w] = Gausslp(N);

x = ((b – a)*t + a + b)/2; %Eq.(5.9.9)

fx = feval(f,x,varargin{:});

I = w*fx’*(b – a)/2; %Eq.(5.9.10)

>>[t,w] = Gausslp(2)

t = 0.5774 -0.5774 w = 1 1

Even though we are happy with the N-point Gauss–Legendre integration

formula (5.9.1) giving the exact integral of polynomials of degree ≤ (2N − 1),

we do not feel comfortable with the fixed integration interval [−1,+1]. But,

we can be relieved from the stress because any arbitrary finite interval [a, b]

can be transformed into [−1,+1] by the variable substitution known as the

Gauss–Legendre translation

x =

(b − a)t + a + b

2

, dx=

b − a

2

dt (5.9.9)

Then, we can write the N-point Gauss–Legendre integration formula for the

integration interval [a, b] as

I [a, b] = b

a

f (x) dx =

b − a

2 1

−1

f (x(t)) dt

I [x1, x2, . . . , xN] =

b − a

2

N

i=1

wN,if (xi ) with xi =

(b − a)ti + a + b

2

(5.9.10)

The scheme of integrating f (x) over the interval [a, b] by the N-point Gauss–

Legendre formula is cast into the MATLAB routine “Gauss_Legendre()”. We

238 NUMERICAL DIFFERENTIATION/ INTEGRATION

can get the integral (5.7.5) by simply putting the following statements into the

MATLAB command window. The result shows that the 10-point Gauss–Legendre

formula yields better accuracy (smaller error), even with fewer nodes/segments

than other methods discussed so far.

>>f = inline(’400*x.*(1 – x).*exp(-2*x)’,’x’); %Eq.(5.7.5)

>>format short e

>>true_I = 3200*exp(-8);

>>a = 0; b = 4; N = 10; %integration interval & number of nodes(grid points)

>>IGL = gauss_legendre(f,a,b,N), errGL = IGL-true_I

IGL = 1.0735e+000, errGL = 1.6289e-009

5.9.2 Gauss–Hermite Integration

The Gauss–Hermite integration formula is expressed by Eq. (5.9.5) as

IGH[t1, t2, . . . , tN] =

N

i=1

wN,if (ti) (5.9.11)

and is supposed to give us the exact integral of the exponential e−t 2 multiplied

by a polynomial f (t) of degree ≤ (2N − 1) over (−∞,+∞)

I = +∞

−∞

e−t 2

f (t) dt (5.9.12)

The N grid point ti’s can be obtained as the zeros of the N-point Hermite

polynomial [K-1, Section 4.8]

HN(t) =

N/2

i=0

(−1)i

i!

N(N − 1) · · · (N − 2i + 1)(2t)N−2i (5.9.13a)

HN(t) = 2tHN−1(t) − H(t) (5.9.13b)

function [t,w] = Gausshp(N)

if N < 0

error(’Gauss-Hermite polynomial of negative degree??’);

end

t = roots(Hermitp(N))’;

A(1,:) = ones(1,N); b(1) = sqrt(pi);

for n = 2:N

A(n,:) = A(n – 1,:).*t; %Eq.(5.9.7)

if mod(n,2) == 1, b(n) = (n – 2)/2*b(n – 2); %Eq.(5.9.14)

else b(n) = 0;

end

end

w = b/A’;

function p = Hermitp(N)

%Hn + 1(x) = 2xHn(x)-Hn’(x) from ’Advanced Engineering Math’ by Kreyszig

if N <= 0, p = 1;

else p = [2 0];

for n = 2:N, p = 2*[p 0]-[0 0 polyder(p)]; end %Eq.(5.9.13b)

end

GAUSS QUADRATURE 239

Given the N grid point ti ’s, we can get the weight wN,i ’s of the N-point

Gauss–Hermite integration formula by solving the system of linear equations

like Eq. (5.9.7), but with the right-hand side (RHS) vector as

RHS(1) = ∞

−∞

e−t 2

dt = ∞

−∞

e−x2 dx ∞

−∞

e−y2 dy

= ∞

−∞ ∞

−∞

e−(x2+y2) dx dy = ∞

−∞

e−r22π r dr

= −πe−r2

∞

0 = √π (5.9.14a)

RHS(n) = ∞

−∞

e−t 2

tn−1 dt = ∞

−∞

(−2t)e−t 2 1

−2

tn−2 dt (= 0 if n is even)

= −

1

2

e−t 2

tn−2

∞

−∞ +

1

2

(n − 2) ∞

−∞

e−t 2

tn−3 dt =

1

2

(n − 2)RHS(n − 2)

(5.9.14b)

The procedure for finding the N grid point ti’s and the corresponding weight

wN,i ’s of the N-point Gauss–Hermite integration formula is cast into the MATLAB

routine “Gausshp()”. Note that, even though the integrand function (g(t))

doesn’t have e−t 2 as a multiplying factor, we can multiply it by e−t 2

et 2 = 1 to

fabricate it as if it were like in Eq. (5.9.12):

I = ∞

−∞

g(t) dt = ∞

−∞

e−t 2

(et 2

g(t)) dt = ∞

−∞

e−t 2

f (t) dt (5.9.15)

5.9.3 Gauss–Laguerre Integration

The Gauss–Laguerre integration formula is also expressed by Eq. (5.9.5) as

IGLa[t1, t2, . . . , tN] =

N

i=1

wN,if (ti) (5.9.16)

and is supposed to give us the exact integral of the exponential e−t multiplied

by a polynomial f (t) of degree ≤ (2N − 1) over [0,∞)

I = ∞

0

e−tf (t) dt (5.9.17)

The N grid point ti ’s can be obtained as the zeros of the Nth-degree Laguerre

polynomial [K-1, Section 4.7]

LN(t) =

N

i=0

(−1)i

i!

Ni

(N − i)! i!

t i (5.9.18)

240 NUMERICAL DIFFERENTIATION/ INTEGRATION

Given the N grid point ti ’s, we can get the corresponding weight wN,i ’s of the

N-point Gauss–Laguerre integration formula by solving the system of linear

equations like Eq. (5.9.7), but with the right-hand side (RHS) vector as

RHS(1) = ∞

0

e−t dt = −e−t

∞

0 = 1 (5.9.19a)

RHS(n) = ∞

0

e−t tn−1 dt = −e−t tn−1

∞

0 + (n − 1) ∞

0

e−t tn−2 dt

= (n − 1)RHS(n − 1) (5.9.19b)

5.9.4 Gauss–Chebyshev Integration

The Gauss–Chebyshev I integration formula is also expressed by Eq. (5.9.5) as

IGC1[t1, t2, . . . , tN] =

N

i=1

wN,if (ti) (5.9.20)

and is supposed to give us the exact integral of 1/√1 − t2 multiplied by a

polynomial f (t) of degree ≤ (2N − 1) over [−1,+1]

I = +1

−1

1

√1 − t2

f (t) dt (5.9.21)

The N grid point ti ’s are the zeros of the Nth-degree Chebyshev polynomial

(Section 3.3)

ti = cos

(2i − 1)π

2N

for i = 1, 2, . . .,N (5.9.22)

and the corresponding weight wN,i ’s are uniformly selected as

wN,i = π/N, ∀ i = 1, . . . , N (5.9.23)

The Gauss–Chebyshev II integration formula is also expressed by Eq. (5.9.5) as

IGC2[t1, t2, . . . , tN] =

N

i=1

wN,if (ti) (5.9.24)

and is supposed to give us the exact integral of √1 − t2 multiplied by a polynomial

f (t) of degree ≤ (2N − 1) over [−1,+1]

I = +1

−1 1 − t2f (t)dt (5.9.25)

DOUBLE INTEGRAL 241

The N grid point ti’s and the corresponding weight wN,i ’s are

ti = cos iπ

N + 1, wN,i =

π

N + 1

sin2 iπ

N + 1 for i = 1, 2, . . . , N

(5.9.26)

5.10 DOUBLE INTEGRAL

In this section, we consider the numerical integration of a function f (x, y) with

respect to two variables x and y over the integration region R = {(x, y)|a ≤ x ≤ b, c(x) ≤ y ≤ d(x)} as depicted in Fig. 5.5.

I = R

f (x, y) dx dy = b

a d(x)

c(x)

f (x, y)dydx (5.10.1)

The numerical formula for this double integration over a two-dimensional region

takes the form

I (a, b, c(x), d(x)) =

M

m=1

wm

N

n=1

vnf (xm, ym,n) (5.10.2)

where the weights wm, vn depend on the method of one-dimensional integration

we choose.

hx 0, y 2

hx 1, y 2

hx 1, y 1 hx 0, y 1

d (x )

c (x )

x 0 x1 x2

hx 1 hx 2 a hx 3 hxM b

x

xM

y

Figure 5.5 A region for a double integral.

242 NUMERICAL DIFFERENTIATION/ INTEGRATION

(cf) The MATLAB built-in routine dblquad() can accept the boundaries of integration

region only given as numbers. Therefore, if we want to use the routine in computing

a double integral for a nonrectangular region D, we should define the integrand

function f (x, y) for a rectangular region R ⊇ D (containing the actual integration

region D) in such a way that f (x, y) = 0 for (x, y) /∈ D; that is, the value of the

function becomes zero outside the integration region D, which may result in more

computations.

function INTfxy = int2s(f,a,b,c,d,M,N)

%double integral of f(x,y) over R = {(x,y)|a <= x <= b, c(x) <= y <= d(x)}

% using Simpson’s rule

if ceil(M) ~= floor(M) %fixed width of segments on x

hx = M; M = ceil((b – a)/hx);

end

if mod(M,2) ~= 0, M = M + 1; end

hx = (b – a)/M; m = 1:M+1; x = a + (m – 1)*hx;

if isnumeric(c), cx(m) = c; %if c is given as a constant number

else cx(m) = feval(c,x(m)); %in case c is given as a function of x

end

if isnumeric(d), dx(m) = d; %if c is given as a constant number

else dx(m) = feval(d,x(m)); %in case d is given as a function of x

end

if ceil(N) ~= floor(N) %fixed width of segments on y

hy = N; Nx(m) = ceil((dx(m)- cx(m))/hy);

ind = find(mod(Nx(m),2) ~= 0); Nx(ind) = Nx(ind) + 1;

else %fixed number of subintervals

if mod(N,2) ~= 0, N = N +1; end

Nx(m) = N;

end

for m = 1:M + 1

sx(m) = smpsns fxy(f,x(m),cx(m),dx(m),Nx(m));

end

kodd = 2:2:M; keven = 3:2:M – 1; %the set of odd/even indices

INTfxy = hx/3*(sx(1) + sx(M + 1) + 4*sum(sx(kodd)) + 2*sum(sx(keven)));

function INTf = smpsns fxy(f, x, c, d, N)

%1-dimensional integration of f(x,y) for Ry = {c <= y <= d}

if nargin < 5, N = 100; end

if abs(d – c)< eps | N <= 0, INTf = 0; return; end

if mod(N,2) ~= 0, N = N + 1; end

h = (d – c)/N; y = c+[0:N]*h; fxy = feval(f,x,y);

fxy(find(fxy == inf)) = realmax; fxy(find(fxy == -inf)) = -realmax;

kodd = 2:2:N; keven = 3:2:N – 1; %the set of odd/even indices

INTf = h/3*(fxy(1) + fxy(N + 1) + 4*sum(fxy(kodd)) + 2*sum(fxy(keven)));

%nm510: the volume of a sphere

x = [-1:0.05:1]; y = [0:0.05:1]; [X,Y] = meshgrid(x,y);

f510 = inline(’sqrt(max(1 – x.*x – y.*y,0))’,’x’,’y’);

Z = f510(X,Y); mesh(x,y,Z);

a = -1; b = 1; c = 0; d=inline(’sqrt(max(1 – x.*x,0))’,’x’);

Vs1 = int2s(f510,a,b,c,d,100,100) %with fixed number of segments

error1 = Vs1 – pi/3

Vs2 = int2s(f510,a,b,c,d,0.01,0.01) %with fixed segment width

error2 = Vs2 – pi/3

DOUBLE INTEGRAL 243

0

0

−1

0

x

1

0.5

0.5

y

1

1

(x, y )

z = 1 − x 2 − y 2

y = 1 − x 2

Figure 5.6 One-fourth (1/4) of a sphere with the radius r = 1.

Although the integration rules along the x axis and along the y axis do not need

to be the same, we make a double integration routine “int2s(f,a,b,c,d,M,N)”

which uses the Simpson method in common for both integrations and calls another

routine “smpsns_fxy()” for one-dimensional integration along the y axis. The

left/right boundary a/b of integration region given as the second/third input argument

must be a number, while the lower/upper boundary c/d of integration region

given as the fourth/fifth input argument may be either a number or a function

of x. If the sixth/seventh input argument M/N is given as a positive integer, it

will be accepted as the number of segments; otherwise, it will be interpreted as

the segment width hx/hy. We also constructed a MATLAB program “nm510”

in order to use the routine “int2s()” for finding one-fourth of the volume of a

sphere with the radius r = 1 depicted in Fig. 5.6.

I = 1

−1

√1−x2

0 1 − x2 − y2 dy dx =

π

3 = 1.04719755 . . . (5.10.3)

Interested readers are recommended to work with these routines and run the

program “nm510.m” to see the result.

>>nm510

Vs1 = 1.0470, error1 = -1.5315e-004

Vs2 = 1.0470, error2 = -1.9685e-004

244 NUMERICAL DIFFERENTIATION/ INTEGRATION

PROBLEMS

5.1 Numerical Differentiation of Basic Functions

If we want to find the derivative of a polynomial/trigonometric/exponential

function, it would be more convenient and accurate to use an analytical

computation (by hand) than to use a numerical computation (by computer).

But, in order to test the accuracy of the numerical derivative formulas,

consider the three basic functions as

f1(x) = x3 − 2x, f2(x) = sin x, f3(x) = ex (P5.1.1)

(a) To find the first derivatives of these functions by using the formulas

(5.1.8) and (5.1.9) listed in Table 5.3 (Section 5.3), modify the program

“nm5p01.m”, which uses the MATLAB routine “difapx()” (Section

5.3) for generating the coefficients of the numerical derivative formulas.

Fill in the following table with the error results obtained from running

the program.

First Derivatives h

f1 − f−1

2h

−f2 + 8f1 − 8f−1 + f−2

12h

(x3 − 2x)|x=1 0.1 1.0000e-02

= 1.00000000

0.01 9.1038e-15

(sin x)|x=π/3 0.1 8.3292e-04

= 0.50000000

0.01 8.3333e-06

(ex )|x=0 0.1 3.3373e-06

= 1.00000000

0.01 1.6667e-05

%nm5p01

f = inline(’x.*(x.*x-2)’, ’x’);

n = [1 -1]; x0 = 1; h = 0.1; DT = 1;

c = difapx(1,n); i = 1:length(c);

num = c*feval(f,x0 + (n(1) + 1 – i)*h)’; drv = num/h;

fprintf(’with h = %6.4f, %12.6f %12.4e\n’, h,drv,drv – DT);

(b) Likewise in (a), modify the program “nm5p01.m” in such a way that

the formulas (5.3.1) and (5.3.2) in Table 5.3 are generated and used to

find the second numerical derivatives. Fill in the following table with

the error results obtained from running the program.

PROBLEMS 245

Second Derivatives h

f1 − 2f0 + f−1

h2 −f2 + 16f1 − 30f0 + 16f−1 − f−2

12h2

(x3 − 2x)(2)|x=1 0.1 2.6654e-14

= 6.0000000000

0.01 2.9470e-12

(sin x)(2)|x=π/3 0.1 9.6139e-07

= −0.8660254037

0.01 7.2169e-06

(ex )(2)|x=0 0.1 8.3361e-04

= 1.0000000000

0.01 1.1183e-10

5.2 Numerical Differentiation of a Function Given as a Set of Data Pairs

Consider the three (numerical) functions each given as a set of five data

pairs in Table P5.2.

Table P5.2 Three Functions Each Given as a Set of Five Data Pairs

x f1(x) x f2(x) x f3(x)

0.8000 −1.0880 0.8472 0.7494 −0.2000 1.2214

0.9000 −1.0710 0.9472 0.8118 −0.1000 1.1052

1.0000 −1.0000 1.0472 0.8660 0 1.0000

1.1000 −0.8690 1.1472 0.9116 0.1000 0.9048

1.2000 −0.6720 1.2472 0.9481 0.2000 0.8187

(a) Use the formulas (5.1.8) and (5.1.9) to find the first derivatives of the

three numerical functions (at x = 1, 1.0472 and 0, respectively) and fill

in the following table with the results. Also use the formulas (5.3.1)

and (5.3.2) to find the second derivatives of the three functions (at

x = 1, 1.0472 and 0, respectively) and fill in the following table with

the results.

f

1(x)|x=1 f

2 (x)|x=1.0472 f

3 (x)|x=0

First derivative by Eq. (5.1.8) 1.0000e-02 2.0000e-03

First derivative by Eq. (5.1.9) 2.5000e-04

f (2)

1 (x)|x=1 f (2)

2 (x)|x=1.0472 f (2)

3 (x)|x=0

Second derivative by Eq. (5.3.1) 6.0254e-03

Second derivative by Eq. (5.3.2) 2.4869e-14 8.3333e-04

246 NUMERICAL DIFFERENTIATION/ INTEGRATION

(b) Based on the Lagrange/Newton polynomial matching the three/five

points around the target point, find the first/second derivatives of the

three functions (at x = 1, 1.0472 and 0, respectively) and fill in the

following table with the results.

f

1(x)|x=1 f

2(x)|x=1.0472 f

3 (x)|x=0

First derivative on l2(x) 1.0000e-03

First derivative on l4(x) 4.3201e-12 4.1667e-04

f (2)

1 (x)|x=1 f (2)

2 (x)|x=1.0472 f (2)

3 (x)|x=0

Second derivative on l2(x) 1.4211e-14 0.0000e+00

Second derivative on l4(x) 6.8587e-03

5.3 First Derivative and Step-size

Consider the routine “jacob()” in Section 4.6, which is used for computing

the Jacobian—that is, the first derivative of a vector function with respect

to a vector variable.

(a) Which one is used for computing the Jacobian in the routine “jacob()”

among the first derivative formulas in Section 5.1?

(b) Expecting that smaller step-size h would yield a better solution to the

problem given in Example 4.3, Bush changed h = 1e-4 to h = 1e-5 in

the routine “newtons()” and then typed the following statement into the

MATLAB command window. What solution could he get?

>>rn1 = newtons(’phys’,1e6,1e-4,100)

(c) What baffled him out of his expectation? Jessica diagnosed the trouble

as caused by a singular Jacobian matrix and modified the statement ‘dx

= -jacob()\fx(:)’ in the routine “newtons()” as follows. What solution

(to the problem in Example 4.3) do you get by using the modified

routine, that is, by typing the same statement as in (b)?

>>rn2 = newtons(’phys’,1e6,1e-4,100), phys(rn2)

J = jacob(f,xx(k,:),h,varargin{:});

if rank(J) < Nx

k = k – 1;

fprintf(’Jacobian singular! det(J) = %12.6e\n’,det(J)); break;

else

dx = -J\fx(:); %-[dfdx]^-1*fx;

end

PROBLEMS 247

(d) To investigate how the accident of Jacobian singularity happened, add

h = 1e-5 to the (tentative) solution (rn2) obtained in (c). Does the

result differ from rn2? If not, why? (See Section 1.2.2 and Problem

1.19.)

>>rn2 + 1e-5 ~= rn2

(e) Charley thought that Jessica just circumvented the Jacobian singularity

problem. To remove the source of singularity, he modified the formula

(5.1.8) into

D

c2(x, h) =

f ((1 + h)x) − f ((1 − h)x)

2hx

(P5.5.3)

and implemented it in another routine “jacob1()” as follows.

function g = jacob1(f,x,h,varargin) %Jacobian of f(x)

if nargin<3, h =.0001; end

h2 = 2*h; N = length(x); I = eye(N);

for n = 1:N

if abs(x(n))<.0001, x(n) =.0001; end

delta = h*x(n);

tmp = I(n,:)*delta;

f1 = feval(f,x + tmp,varargin{:});

f2 = feval(f,x – tmp,varargin{:});

f12 = (f1 – f2)/2/delta; g(:,n) = f12(:);

end

With h = 1e-5 or h = 1e-6 and jacob() replaced by jacob1() in

the routine “newtons()”, type the same statement as in (c) to get a

solution to the problem in Example 4.3 together with its residual error

and check if his scheme works fine.

>>rn3 = newtons(’phys’,1e6,1e-4,100), phys(rn3)

5.4 Numerical Integration of Basic Functions

Compute the following integrals by using the trapezoidal rule, the Simpson’s

rule, and Romberg method and fill in the following table with the

resulting errors.

(i) 2

0

(x3 − 2x) dx (ii) π/2

0

sinx dx (iii) 1

0

e−x dx

248 NUMERICAL DIFFERENTIATION/ INTEGRATION

N

Trapezoidal

Rule

Simpson

Rule

Romberg

(tol = 0.0005)

4 0.0000e+0 2

0

(x3 − 2x) dx = 0

8 6.2500e-1

4 1.2884e-2 8.4345e-6 π/2

0

sinx dx = 1

8 8.2955e-6

4 1.3616e-5 1

0

e−x dx = 0.63212055883

8 8.2286e-4

5.5 Adaptive Quadrature and Gaussian Quadrature for Improper Integral

Consider the following two integrals.

(i) 1

0

1

√x

dx = 2×1/2

1

0 = 2 (P5.5.1)

(ii) 1

−1

1

√x

dx = 0

−1

1

√x

dx + 1

0

1

√x

dx = 2 − 2i (P5.5.2)

(a) Type the following statements into the MATLAB command window to

use the integration routines for the above integral. What did you get?

If something is wrong, what do you think caused it?

>>f = inline(’1./sqrt(x)’,’x’); % define the integrand function

>>smpsns(f,0,1,100) % integral over [0,1] with 100 segments

>>rmbrg(f,0,1,1e-4) % with error tolerance = 0.0001

>>adapt_smpsn(f,0,1,1e-4) % with error tolerance = 0.0001

>>gauss_legendre(f,0,1,20) %Gauss-Legendre with N = 20 grid points

>>quad(f,0,1) % MATLAB built-in routine

>>quad8(f,0,1) % MATLAB 5.x built-in routine

>>adapt_smpsn(f,-1,1,1e-4) %integral over [-1,1]

>>quad(f,-1,1) % MATLAB built-in routine

>>quadl(f,-1,1) % MATLAB built-in routine

(b) Itha decided to retry the routine “smpsns()”, but with the singular point

excluded from the integration interval. In order to do that, she replaced

the singular point (0) which is the lower bound of the integration interval

[0,1] by 10−4 or 10−5, and typed the following statements into the

MATLAB command window.

>>smpsns(f,1e-4,1,100)

>>smpsns(f,1e-5,1,100)

>>smpsns(f,1e-5,1,1e4)

>>smpsns(f,1e-4,1,1e3)

>>smpsns(f,1e-4,1,1e4)

PROBLEMS 249

What are the results? Will it be better if you make the lower-bound of

the integration interval closer to zero (0), without increasing the number

of segments or (equivalently) decreasing the segment width? How about

increasing the number of segments without making the lower bound

of the integration interval closer to the original lower-bound which is

zero (0)?

(c) For the purpose of improving the performance of “adap_smpsn()”,

Vania would put the following statements into both of the routines

“smpsns()” and “adap_smpsn()”. Supplement the routines and check

whether her idea works or not.

EPS = 1e-12; fa = feval(f,a,varargin{:});

if isnan(fa)|abs(fa) == inf, a = a + max(abs(a)*EPS,EPS); end

fb = feval(f,b,varargin{:});

?? ??????????????? ?? ????? ? ?? ? ? ???????????????????? ???

5.6 Various Numerical Integration Methods and Improper Integral

Consider the following integrals.

∞

0

sin x

x

dx =

π

2 ∼=

100

0

sin x

x

dx (P5.6.1)

∞

0

e−x2

dx =

1

2

√π (P5.6.2)

Note that the true values of these integrals can be obtained by using the

symbolic computation command “int()” as below.

>>syms x, int(sin(x)/x,0,inf)

>>int(exp(-x^2),0,inf)

(cf ) Don’t you believe it without seeing it? Blessed are those who have not seen

and yet believe.

(a) To apply the routines like “smpsns()”, “adapt_smpsn()”, “Gauss_

Legendre()” and “quadl()” for evaluating the integral (P5.6.1), do

the following.

(i) Note that the integration interval [0, ∞) can be changed into a

finite interval as below.

∞

0

sin x

x

dx = 1

0

sin x

x

dx + ∞

1

sin x

x

dx

= 1

0

sin x

x

dx + 0

1

sin(1/y)

1/y −

1

y2dy

= 1

0

sin x

x

dx + 1

0

sin(1/y)

y

dy (P5.6.3)

250 NUMERICAL DIFFERENTIATION/ INTEGRATION

(ii) Add the block of statements in P5.5(c) into the routines “smpsns()”

and “adap_smpsn()” to make them cope with the cases

of NaN (Not-a-Number) and Inf (Infinity).

(iii) Supplement the program “nm5p06a.m” so that the various routines

are applied for computing the integrals (P5.6.1) and (P5.6.3), where

the parameters like the number of segments (N = 200), the error

tolerance (tol = 1e-4), and the number of grid points (MGL = 20)

are supposed to be used as they are in the program. Noting that

the second integrand function in (P5.6.3) oscillates like crazy with

higher frequency and larger amplitude as y gets closer to zero (0),

set the lower bound of the integration interval to a2 = 0.001.

(iv) Run the supplemented program and fill in Table P5.6 with the

absolute errors of the results.

%nm5p06a

warning off MATLAB:divideByZero

fp56a = inline(’sin(x)./x’,’x’); fp56a2 = inline(’sin(1./y)./y’,’y’);

IT = pi/2; % True value of the integral

a = 0; b = 100; N = 200; tol = 1e-4; MGL = 20; a1 = 0; b1 = 1; a2 = 0.001; b2 = 1;

format short e

e_s = smpsns(fp56a,a,b,N)-IT

e_as = adapt_smpsn(fp56a,a,b,tol)-IT

e_ql = quadl(fp56a,a,b,tol)-IT

e_GL = Gauss_Legendre(fp56a,a,b,MGL)-IT

e_ss = smpsns(fp56a,a1,b1,N) + smpsns(fp56a2,a2,b2,N)-IT

e_Iasas = adapt_smpsn(fp56a,a1,b1,tol)+ …

???????????????????????????? -IT

e_Iqq = quad(fp56a,a1,b1,tol)+??????????????????????????? -IT

warning on MATLAB:divideByZero

%nm5p06b

warning off MATLAB:divideByZero

fp56b = inline(’exp(-x.*x)’,’x’);

fp56b1 = inline(’ones(size(x))’,’x’);

fp56b2 = inline(’exp(-1./y./y)./y./y’,’y’);

a = 0; b = 200; N = 200; tol = 1e-4; IT = sqrt(pi)/2;

a1 = 0; b1 = 1; a2 = 0; b2 = 1; MGH = 2;

e_s = smpsns(fp56b,a,b,N)-IT

e_as = adapt_smpsn(fp56b,a,b,tol)-IT

e_q = quad(fp56b,a,b,tol)-IT

e_GH = Gauss_Hermite(fp56b1,MGH)/2-IT

e_ss = smpsns(fp56b,a1,b1,N) + smpsns(fp56b2,a2,b2,N)-IT

Iasas = adapt_smpsn(fp56b,a1,b1,tol)+ …

+????????????????????????????? -IT

e_qq = quad(fp56b,a1,b1,tol)+????????????????????????? -IT

warning off MATLAB:divideByZero

PROBLEMS 251

Table P5.6 Results of Applying Various Numerical Integration Methods for

Improper Integrals

Simpson adaptive quad Gauss S&S a&a q&q

(P5.6.1) 8.5740e-3 1.9135e-1 1.1969e+0 2.4830e-1

(P5.6.2) 6.6730e-6 0.0000e+0 3.3546e-5

(b) To apply the routines like “smpsns()”, “adapt_smpsn()”, “quad()”, and

“Gauss_Hermite()” for evaluating the integral (P5.6.2), do the following.

(i) Note that the integration interval [0,∞) can be changed into a

finite interval as below.

∞

0

e−x2

dx = 1

0

e−x2

dx + ∞

1

e−x2

dx

= 1

0

e−x2

dx + 0

1

e−1/y2 −

1

y2dy

= 1

0

e−x2

dx + 1

0

e−1/y2

y2 dy (P5.6.4)

(ii) Compose the incomplete routine “Gauss_Hermite” like “Gauss_

Legendre”, which performs the Gauss–Hermite integration introduced

in Section 5.9.2.

(iii) Supplement the program “nm5p06b.m” so that the various routines

are applied for computing the integrals (P5.6.2) and (P5.6.4), where

the parameters like the number of segments (N = 200), the error

tolerance (tol = 1e-4) and the number of grid points (MGH = 2)

are supposed to be used as they are in the program. Note that the

integration interval is not (−∞,∞) like that of Eq. (5.9.12), but

[0,∞) and so you should cut the result of “Gauss_Hermite()” by

half to get the right answer for the integral (P5.6.2).

(iv) Run the supplemented program and fill in Table P5.6 with the

absolute errors of the results.

(c) Based on the results listed in Table P5.6, answer the following questions:

(i) Among the routines “smpsns()”, “adapt_smpsn()”, “quad()”,

and “Gauss()”, choose the best two ones for (P5.6.1) and (P5.6.2),

respectively.

(ii) The routine “Gauss–Legendre()” works (badly, perfectly) even

with as many as 20 grid points for (P5.6.1), while the routine

252 NUMERICAL DIFFERENTIATION/ INTEGRATION

“Gauss_Hermite()” works (perfectly, badly) just with two grid

points for (P5.6.2). It is because the integrand function of (P5.6.1)

is (far from, just like) a polynomial, while (P5.6.2) matches

Eq. (5.9.11) and the part of it excluding e−x2 is (just like, far

from) a polynomial.

function I = Gauss_Hermite(f,N,varargin)

[t,w]=???????(N);

ft = feval(f,t,varargin{:});

I = w*ft’;

(iii) Run the following program “nm5p06c.m” to see the shapes of the

integrand functions of (P5.6.1) and (P5.6.2) and the second integral

of (P5.6.3). You can zoom in/out the graphs by clicking the

Tools/Zoom in menu and then clicking any point on the graphs

with the left/right mouse button in the MATLAB graphic window.

Which one is oscillating furiously? Which one is oscillating

moderately? Which one is just changing abruptly?

%nm5p06c

clf

fp56a = inline(’sin(x)./x’,’x’);

fp56a2 = inline(’sin(1./y)./y’,’y’);

fp56b = inline(’exp(-x.*x)’,’x’);

x0 = [eps:2000]/20; x = [eps:100]/100;

subplot(221), plot(x0,fp56a(x0))

subplot(223), plot(x0,fp56b(x0))

subplot(222), y = logspace(-3,0,2000); loglog(y,abs(fp56a2(y)))

subplot(224), y = logspace(-6,-3,2000); loglog(y,abs(fp56a2(y)))

(iv) The adaptive integration routines like “adapt smpsn()” and

“quad()” work (badly, fine) for (P5.6.1), but (fine, badly) for

(P5.6.2). From this fact, we might conjecture that the adaptive

integration routines may be (ineffective, effective) for the integrand

functions which have many oscillations, while they may be

(effective, ineffective) for the integrand functions which have

abruptly changing slope. To support this conjecture, run the

following program “nm5p06d”, which uses the “quad()” routine

for the integrals

b

1

sin x

x

dx with b = 100, 1000, 10000 . . . . (P5.6.5a)

1

a

sin(1/y)

y

dy with a = 0.001, 0.0001, 0.00001, . . .(P5.6.5b)

PROBLEMS 253

%nm5p06d

fp56a = inline(’sin(x)./x’,’x’);

fp56a2 = inline(’sin(1./y)./y’,’y’);

syms x

IT2 = pi/2 – double(int(sin(x)/x,0,1)) %true value of the integral

disp(’Change of upper limit of the integration interval’)

a = 1; b = [100 1e3 1e4 1e7]; tol = 1e-4;

for i = 1:length(b)

Iq2 = quad(fp56a,a,b(i),tol);

fprintf(’With b = %12.4e, err_Iq = %12.4e\n’, b(i),Iq2-IT2);

end

disp(’Change of lower limit of the integration interval’)

a2 = [1e-3 1e-4 1e-5 1e-6 0]; b2 = 1; tol = 1e-4;

for i = 1:5

Iq2 = quad(fp56a2,a2(i),b2,tol);

fprintf(’With a2=%12.4e, err_Iq=%12.4e\n’, a2(i),Iq2-IT2);

end

Does the “quad()” routine work stably for (P5.6.5a) with the

changing value of the upper-bound of the integration interval?

Does it work stably for (P5.6.5b) with the changing value of the

lower-bound of the integration interval? Do the results support or

defy the conjecture?

(cf) This problem warns us that it may be not good to use only one routine

for a computational work and suggests us to use more than one method

for cross check.

5.7 Gauss–Hermite Integration Method

Consider the following integral:

∞

0

e−x2 cosx dx =

√π

2

e−1/4 (P5.7.1)

Select a Gauss quadrature suitable for this integral and apply it with

the number of grid points N = 4 as well as the routines “smpsns()”,

“adapt_smpsn()”, “quad()”, and “quadl()” to evaluate the integral. In

order to compare the number of floating-point operations required to achieve

almost the same level of accuracy, set the number of segments for Simpson

method to N = 700 and the error tolerance for all other routines to tol = 10−5. Fill in Table P5.7 with the error results.

Table P5.7 The Results of Applying Various Numerical Integration Methods

Simpson

(N = 700)

adaptive

(tol = 10−5) Gauss

quad

(tol = 10−5)

quadl

(tol = 10−5)

(P5.7.1) |error| 1.0001e-3 1.0000e-3

flops 4930 5457 1484 11837 52590 (with quad8)

(P5.8.1) |error| 1.3771e-2 0 4.9967e-7

flops 5024 7757 131 28369 75822

254 NUMERICAL DIFFERENTIATION/ INTEGRATION

5.8 Gauss–Laguerre Integration Method

(a) As in Section 5.9.1, Section 5.9.2, and Problem 5.6(b), compose the

MATLAB routines: “Laguerp()”, which generates the Laguerre polynomial

(5.9.18); “Gausslgp()”, which finds the grid point ti’s and the

coefficient wN,i ’s for Gauss–Laguerre integration formula (5.9.16); and

“Gauss_Laguerre(f,N)”, which uses these two routines to carry out

the Gauss–Laguerre integration method.

(b) Consider the following integral:

∞

0

e−tt dt = −e−t t

∞

0 + ∞

0

e−t dt = −e−t

∞

0 = 1 (P5.8.1)

Noting that, since this integral matches Eq. (5.9.17) with f (t) = t , Gauss–Laguerre method is the right choice, apply the routine

“Gauss_Laguerre(f,N)” (manufactured in (a)) with N = 2 as well as

the routines “smpsns()”, “adapt_smpsn()”, “quad()”, and “quadl()”

for evaluating the integral and fill in Table P5.7 with the error results.

Which turns out to be the best? Is the performance of “quad()”

improved by lowering the error tolerance?

(cf) This illustrates that the routine “adapt_smpsn()” sometimes outperforms the

MATLAB built-in routine “quad()” with fewer computations. On the other

hand, Table P5.7 shows that it is most desirable to apply the Gauss quadrature

schemes only if one of them is applicable to the integration problem.

5.9 Numerical Integrals

Consider the following integrals.

(1) π/2

0 x sinx dx = 1 (2) 1

0 x ln(sin x) dx = −

1

2

π2 ln 2

(3) 1

0

1

x(1 − ln x)2 dx = 1 (4) ∞ 1

1

x(1 + ln x)2 dx = 1

(5) 1

0

1

√x(1 + x)

dx =

π

2

(6) ∞ 1

1

√x(1 + x)

dx =

π

2

(7) 1

0 ln

1

x

dx =

√π

2

(8) ∞ 0 √xe−x dx =

√π

2

(9) ∞ 0 x2e−x cosx dx = −

1

2

(a) Apply the integration routines “smpsns()” (with N = 104), “adapt_

smpsn()”, “quad()”, “quadl()” (tol= 10−6) and “Gauss_legendre()”

(Section 5.9.1) or “Gauss_Laguerre()” (Problem 5.8) (with

N = 15) to compute the above integrals and fill in Table P5.9 with the

relative errors. Use the upper/lower bounds of the integration interval in

Table P5.9 if they are specified in the table.

(b) Based on the results listed in Table P5.9, answer the following questions

or circle the right answer.

PROBLEMS 255

(i) From the fact that the Gauss–Legendre integration scheme worked

best only for (1), it is implied that the scheme is (recommendable,

not recommendable) for the case where the integrand function is

far from being approximated by a polynomial.

(ii) From the fact that the Gauss–Laguerre integration scheme worked

best only for (9), it is implied that the scheme is (recommendable,

not recommendable) for the case where the integrand function

excluding the multiplying term e−x is far from being approximated

by a polynomial.

(iii) Note the following:

ž The integrals (3) and (4) can be converted into each other by a

variable substitution of x = u−1, dx = −u−2 du. The integrals

(5) and (6) have the same relationship.

ž The integrals (7) and (8) can be converted into each other by a

variable substitution of u = e−x , dx = −u−1du.

From the results for (3)–(8), it can be conjectured that the numerical integration

may work (better, worse) if the integration interval is changed from [1,∞)

into (0,1] through the substitution of variable like

x = u−n, dx = −nu−(n+1)du or u = e−nx, dx = −(nu)−1 du (P5.9.1)

Table P5.9 The Relative Error Results of Applying Various Numerical

Integration Methods

Simpson Adaptive Gauss quad quadl

(N = 104) (tol = 10−6) (N = 10) (tol = 10−6) (tol = 10−6)

(1) 1.9984e-15 0.0000e+00 7.5719e-11

(2) 2.8955e-08 1.5343e-06

(3) 9.7850e-02 (a = 10−4) 1.2713e-01 2.2352e-02

(4), b = 104 9.7940e-02 9.7939e-02

(5) 1.2702e-02 (a = 10−4) 3.5782e-02 2.6443e-07

(6), b = 103 4.0250e-02 4.0250e-02

(7) 6.8678e-05 5.1077e-04 3.1781e-07

(8), b = 10 1.6951e-04 1.7392e-04

(9), b = 10 7.8276e-04 2.9237e-07 7.8276e-04

5.10 The BER (Bit Error Rate) Curve of Communication with Multidimensional

Signaling

For a communication system with multidimensional (orthogonal) signaling,

the BER—that is, the probability of bit error—is derived as

Pe,b =

2b−1

2b − 1 1 −

1

√π ∞

−∞

(QM−1(−√2y − √bSNR))e−y2

dy

(P5.10.1)

256 NUMERICAL DIFFERENTIATION/ INTEGRATION

where b is the number of bits, M = 2b is the number of orthogonal waveforms,

SNR is the signal-to-noise-ratio, and Q(·) is the error function

defined by

Q(x) =

1

√2π ∞

x

e−y2/2 dy (P5.10.2)

We want to plot the BER curves for SNR = 0:10[dB] and b = 1:4.

(a) Consider the following program “nm5p10.m”, whose objective is to

compute the values of Pe,b(SNR,b) for SNR = 0:10[dB] and b = 1:4

by using the routine “Gauss_Hermite()” (Problem 5.6) and also by

using the MATLAB built-in routine “quad()” and to plot them versus

SNR[dB] = 10 log10 SNR. Complete the incomplete part which computes

the integral in (P5.10.1) over [−1000, 1000] and run the program

to obtain the BER curves like Fig. P5.10.

(b) Of the two routines, which one is faster and which one presents us with

more reliable values of the integral in (P5.10.1)?

0 1

100

10−2

Pe, b(SNR, b)

10−4

2 3 4 5 6 7 SNR [dB] 9 10

b = 1

b = 2

b = 3

b = 4

Figure P5.10 The BER (bit error rate) curves formultidimensional (orthogonal) signaling.

%nm5p10.m: plots the probability of bit error versus SNRbdB

fs =’Q(-sqrt(2)*x – sqrt(b*SNR)).^(2^b – 1)’;

Q = inline(’erfc(x/sqrt(2))/2’,’x’);

f = inline(fs,’x’,’SNR’,’b’);

fex2 = inline([fs ’.*exp(-x.*x)’],’x’,’SNR’,’b’);

SNRdB = 0:10; tol = 1e-4; % SNR[dB] and tolerance used for ’quad’

for b = 1:4

tmp = 2^(b – 1)/(2^b – 1); spi = sqrt(pi);

for i = 1:length(SNRdB),

SNR = 10^(SNRdB(i)/10);

Pe(i) = tmp*(1-Gauss_Hermite(f,10,SNR,b)/spi);

Pe1(i) = tmp*(1-quad(fex2,-10,10,tol,[],SNR,b)/spi);

Pe2(i) = tmp*(1-?????????????????????????????????)/spi);

end

semilogy(SNRdB,Pe,’ko’,SNRdB,Pe1,’b+:’,SNRdB,Pe2,’r.-’), hold on

end

PROBLEMS 257

5.11 Length of Curve/Arc: Superb Harmony of Numerical Derivative/Integral.

The graph of a function y = f (x) of a variable x is generally a curve and

its length over the interval [a, b] on the x-axis can be described by a line

integral as

I = b

a

dl = b

a dx2 + dy2 = b

a 1 + (dy/dx)2 dx

= b

a 1 + (f (x))2 dx (P5.11.1)

For example, the length of the half-circumference of a circle with the radius

of unit length can be obtained from this line integral with

y = f (x) = 1 − x2, a= −1, b= 1 (P5.11.2)

Starting from the program “nm5p11.m”, make a program that uses the

numerical integration routines “smpsns()”, “adapt_smpsn()”, “quad()”,

“quadl()”, and “Gauss_Legendre()” to evaluate the integral (P5.11.1,2)

with the first derivative approximated by Eq. (5.1.8), where the parameters

like the number of segments (N), the error tolerance (tol), and the

number of grid points (M) are supposed to be as they are in the program.

Run the program with the step size h = 0.001, 0.0001, and 0.00001

in the numerical derivative and fill in Table P5.11 with the errors of the

results, noting that the true value of the half-circumference of a unit circle

is π.

%nm5p11

a = -1; b = 1; % the lower/upper bounds of the integration interval

N = 1000 % the number of segments for the Simpson method

tol = 1e-6 % the error tolerance

M = 20 % the number of grid points for Gauss–Legendre integration

IT = pi; h = 1e-3 % true integral and step size for numerical derivative

flength = inline(’sqrt(1 + dfp511(x,h).^2)’,’x’,’h’);%integrand P5.11.1)

Is = smpsns(flength,a,b,N,h);

[Ias,points,err] = adapt_smpsn(flength,a,b,tol,h);

Iq = quad(flength,a,b,tol,[],h);

Iql = quadl(flength,a,b,tol,[],h);

IGL = Gauss_Legendre(flength,a,b,M,h);

function df = dfp511(x,h) % numerical derivative of (P5.11.2)

if nargin < 2, h = 0.001; end

df = (fp511(x + h)-fp511(x – h))/2/h; %Eq.(5.1.8)

function y = fp511(x)

y = sqrt(max(1-x.*x,0)); % the function (P5.11.2)

258 NUMERICAL DIFFERENTIATION/ INTEGRATION

Table P5.11 Results of Applying Various Numerical Integration Methods for

(P5.11.1,2)/(P5.12.1,2)

Step-size h Simpson Adaptive quad quadl Gauss

(P5.11.1,2)

0.001 4.6212e-2 2.9822e-2 8.4103e-2

0.0001 9.4278e-3 9.4277e-3

0.00001 2.1853e-1 2.9858e-3 8.4937e-2

(P5.12.1,2)

0.001 1.2393e-5 1.3545e-5

0.0001 8.3626e-3 5.0315e-6 6.4849e-6

0.00001 1.3846e-9 8.8255e-7

(P5.13.1) N/A 8.8818e-16 0 8.8818e-16

5.12 Surface Area of Revolutionary 3-D (Cubic) Object

The upper/lower surface area of a 3-D structure formed by one revolution of

a graph (curve) of a function y = f (x) around the x-axis over the interval

[a, b] can be described by the following integral:

I = 2π b

a

y dl = 2π b

a

f (x)1 + (f (x))2 dx (P5.12.1)

For example, the surface area of a sphere with the radius of unit length can

be obtained from this equation with

y = f (x) = 1 − x2, a= −1, b= 1 (P5.12.2)

Starting from the program “nm5p11.m”, make a program “nm5p12.m” that

uses the numerical integration routines “smpsns()” (with the number of

segments N = 1000), “adapt_smpsn()”, “quad()”, “quadl()” (with the

error tolerance tol = 10−6) and “Gauss_Legendre()” (with the number

of grid points M = 20) to evaluate the integral (P5.12.1,2) with the first

derivative approximated by Eq. (5.1.8), where the parameters like the number

of segments (N), the error tolerance (tol), and the number of grid points

(M) are supposed to be as they are in the program. Run the program with

the step size h = 0.001, 0.0001, and 0.00001 in the numerical derivative

and fill in Table P5.11 with the errors of the results, noting that the true

value of the surface area of a unit sphere is 4π .

5.13 Volume of Revolutionary 3-D (Cubic) Object

The volume of a 3-D structure formed by one revolution of a graph (curve)

of a function y = f (x) around the x-axis over the interval [a, b] can be

described by the following integral:

I = π b

a

f 2(x) dx (P5.13.1)

PROBLEMS 259

For example, the volume of a sphere with the radius of unit length (Fig.

P5.13) can be obtained from this equation with Eq. (P5.12.2). Starting from

the program “nm5p11.m”, make a program “nm5p13.m” that uses the numerical

integration routines “smpsns()” (with the number of segments N =

100), “adapt_smpsn()”, “quad()”, “quadl()” (with the error tolerance

tol = 10−6), and “Gauss_Legendre()” (with the number of grid points

M = 2) to evaluate the integral (P5.13.1). Run the program and fill in

Table P5.11 with the errors of the results, noting that the volume of a

unit sphere is 4π/3.

−1 −0.5

−1

0

1

0.5 1

Figure P5.13 The surface and the volume of a unit sphere.

5.14 Double Integral

(a) Consider the following double integral

I = 2

0 π

0

y sinx dx dy = 2

0 −y cos x

π

0 dy = 2

0

2y dy = y2

2

0 = 4

(P5.14.1)

Use the routine “int2s()” (Section 5.10) with M = N = 20, M = N =

50 and M = N = 100 and the MATLAB built-in routine “dblquad()”

to compute this double integral. Fill in Table P5.14.1 with the results

and the times measured by using the commands tic/toc to be taken

for carrying out each computation. Based on the results listed in

Table P5.14.1, can we say that the numerical error becomes smaller

as we increase the numbers (M,N) of segments along the x-axis and

y-axis for the routine “int2s()”?

260 NUMERICAL DIFFERENTIATION/ INTEGRATION

(b) Consider the following double integral:

I = 1

0 1

0

1

1 − xy

dx dy =

π2

6

(P5.14.2)

Noting that the integrand function is singular at (x, y) = (1, 1), use

the routine “int2s()” and the MATLAB built-in routine “dblquad()”

with the upper limit (d) of the integration interval along the y-axis d

= 0.999, d = 0.9999, d = 0.99999 and d = 0.999999 to compute this

double integral. Fill in Tables P5.14.2 and P5.14.3 with the results and

the times measured by using the commands tic/toc to be taken for

carrying out each computation.

Table P5.14.1 Results of Running ‘‘int2s()’’ and ‘‘dblquad()’’ for (P5.14.1)

int2s(), int2s(), int2s(),

M = N = 20 M = N = 100 M = N = 200 dblquad()

|error| 2.1649 × 10−8 1.3250 × 10−8

time

Table P5.14.2 Results of Running ‘‘int2s()’’ and ‘‘dblquad()’’ for (P5.14.2)

a = 0, b = 1 a = 0, b = 1 a = 0, b = 1 a = 0, b = 1

c = 0, c = 0, c = 0, c = 0,

d = 1-10−3 d = 1-10−4 d =1-10−5 d = 1-10−6

int2s() |error| 0.0079 0.0024

M = 2000

N = 2000 time

dblquad |error| 0.0004 0.0006

time

Table P5.14.3 Results of Running the Double Integral Routine ‘‘int2s()’’ for

(P5.14.2)

M = 1000, M = 2000, M = 5000,

N = 1000 N = 2000 N = 5000

int2s() |error| 0.0003

a = 0, b = 1

c = 0, d = 1-10−4 time

Based on the results listed in Tables P5.14.2 and P5.14.3, answer the

following questions.

(i) Can we say that the numerical error becomes smaller as we set the

upper limit (d) of the integration interval along the y-axis closer to

the true limit 1?

PROBLEMS 261

(ii) Can we say that the numerical error becomes smaller as we increase

the numbers (M,N) of segments along the x-axis and y-axis for the

routine “int2s()”? If this is contrary to the case of (a), can you

blame the weird shape of the integrand function in Eq. (P5.14.2)

for such a mess-up?

(cf ) Note that the computation times to be listed in Tables P5.14.1 to P5.14.3

may vary with the speed of CPU as well as the computational jobs which

are concurrently processed by the CPU. Therefore, the time measured by the

‘tic/toc’ commands cannot be an exact estimate of the computational load

taken by each routine.

5.15 Area of a Triangle

Consider how to find the area between the graph (curve) of a function f (x)

and the x-axis. For example, let f (x) = x for 0 ≤ x ≤ 1 in order to find

the area of a right-angled triangle with two equal sides of unit length. We

might use either the 1-D integration or the 2-D integration—that is, the

double integral for this job.

(a) Use any integration method that you like best to evaluate the integral

I1 = 1

0

x dx =

1

2

(P5.15.1)

(b) Use any double integration routine that you like best to evaluate the

integral

I2 = 1

0 f (x)

0

1 dy dx = 1

0 x

0

1 dy dx (P5.15.2)

You may get puzzled with some problem when applying the routine

“int2s()” if you define the integrand function as

>>fp515b = inline(’1’,’x’,’y’);

It is because this function, being called inside the routine

“smpsns_fxy()”, yields just a scalar output even for the vector-valued

input argument. There are two remedies for this problem. One is to

define the integrand function in such a way that it can generate the

output of the same dimension as the input.

>>fp515b = inline(’1+0*(x+y)’,’x’,’y’);

But, this will cause a waste of computation time due to the dead multiplication

for each element of the input arguments x and y. The other is

to modify the routine “smpsns_fxy()” in such a way that it can avoid

the vector operation. More specifically, you can replace some part of

the routine with the following. But, this remedy also increases the computation

time due to the abandonment of vector operation taking less

time than scalar operation (see Section 1.3).

262 NUMERICAL DIFFERENTIATION/ INTEGRATION

function INTf = smpsns_fxy(f,x,c,d,N)

.. .. .. .. .. .. .. .. .. .. ..

sum_odd = f(x,y(2)); sum_even = 0;

for n = 4:2:N

sum_odd = sum_odd + f(x,y(n)); sum_even = sum_even + f(x,y(n – 1));

end

INTf = (f(x,y(1)) + f(x,y(N + 1)) + 4*sum_odd + 2*sum_even)*h/3;

.. .. .. .. .. .. .. .. .. .. ..

(cf ) This problem illustrates that we must be provident to use the vector operation,

especially in defining a MATLAB function.

5.16 Volume of a Cone

Likewise in Section 5.10, modify the program “nm510.m” so that it uses

the routines “int2s()” and “dblquad()” to compute the volume of a cone

that has a unit circle as its base side and a unit height, and run it to obtain

the values of the volume up to four digits below the decimal point.)

6

ORDINARY DIFFERENTIAL

EQUATIONS

Differential equations are mathematical descriptions of how the variables and

their derivatives (rates of change) with respect to one or more independent

variable affect each other in a dynamical way. Their solutions show us how

the dependent variable(s) will change with the independent variable(s). Many

problems in natural sciences and engineering fields are formulated into a scalar

differential equation or a vector differential equation—that is, a system of differential

equations.

In this chapter, we look into several methods of obtaining the numerical solutions

to ordinary differential equations (ODEs) in which all dependent variables

(x) depend on a single independent variable (t ). First, the initial value problems

(IVPs) will be handled with several methods including Runge–Kutta method and

predictor–corrector methods in Sections 6.1 to 6.5. The final section (Section 6.6)

will introduce the shooting method and the finite difference method for solving

the two-point boundary value problem (BVP). ODEs are called an IVP if the

values x(t0) of dependent variables are given at the initial point t0 of the independent

variable, while they are called a BVP if the values x(t0)/ x(tf ) are given

at the initial/final points t0 and tf .

6.1 EULER’S METHOD

When talking about the numerical solutions to ODEs, everyone starts with the

Euler’s method, since it is easy to understand and simple to program. Even though

its low accuracy keeps it from being widely used for solving ODEs, it gives us a

Applied Numerical Methods Using MATLAB, by Yang, Cao, Chung, and Morris

Copyr ight 2005 John Wiley & Sons, I nc., ISBN 0-471-69833-4

263

264 ORDINARY DIFFERENTIAL EQUATIONS

clue to the basic concept of numerical solution for a differential equation simply

and clearly. Let’s consider a first-order differential equation:

y(t) + a y(t) = r with y(0) = y0 (6.1.1)

It has the following form of analytical solution:

y(t) = y0 −

r

a e−at +

r

a

(6.1.2)

which can be obtained by using a conventional method or the Laplace transform

technique [K-1, Chapter 5]. However, such a nice analytical solution does

not exist for every differential equation; even if it exists, it is not easy to

find even by using a computer equipped with the capability of symbolic computation.

That is why we should study the numerical solutions to differential

equations.

Then, how do we translate the differential equation into a form that can easily

be handled by computer? First of all, we have to replace the derivative

y(t) = dy/dt in the differential equation by a numerical derivative (introduced in

Chapter 5), where the step-size h is determined based on the accuracy requirements

and the computation time constraints. Euler’s method approximates the

derivative in Eq. (6.1.1) with Eq. (5.1.2) as

y(t + h) − y(t)

h + a y(t) = r

y(t + h) = (1 − ah)y(t) + hr with y(0) = y0 (6.1.3)

and solves this difference equation step-by-step with increasing t by h each time

from t = 0.

y(h) = (1 − ah)y(0) + hr = (1 − ah)y0 + hr

y(2h) = (1 − ah)y(h) + hr = (1 − ah)2y0 + (1 − ah)hr + hr

y(3h) = (1 − ah)y(2h) + hr = (1 − ah)3y0 + 2

m=0(1 − ah)mhr

(6.1.4)

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

This is a numeric sequence {y(kh)}, which we call a numerical solution of

Eq. (6.1.1).

To be specific, let the parameters and the initial value of Eq. (6.1.1) be a = 1,

r = 1, and y0 = 0. Then, the analytical solution (6.1.2) becomes

y(t) = 1 − e−at (6.1.5)

EULER’S METHOD 265

%nm610: Euler method to solve a 1st-order differential equation

clear, clf

a = 1; r = 1; y0 = 0; tf = 2;

t = [0:0.01:tf]; yt = 1 – exp(-a*t); %Eq.(6.1.5): true analytical solution

plot(t,yt,’k’), hold on

klasts = [8 4 2]; hs = tf./klasts;

y(1) = y0;

for itr = 1:3 %with various step size h = 1/8,1/4,1/2

klast = klasts(itr); h = hs(itr); y(1)=y0;

for k = 1:klast

y(k + 1) = (1 – a*h)*y(k) +h*r; %Eq.(6.1.3):

plot([k – 1 k]*h,[y(k) y(k+1)],’b’, k*h,y(k+1),’ro’)

if k < 4, pause; end

end

end

and the numerical solution (6.1.4) with the step-size h = 0.5 and h = 0.25 are

as listed in Table 6.1 and depicted in Fig. 6.1. We make a MATLAB program

“nm610.m”, which uses Euler’s method for the differential equation (6.1.1), actually

solving the difference equation (6.1.3) and plots the graphs of the numerical

solutions in Fig. 6.1. The graphs seem to tell us that a small step-size helps

reduce the error so as to make the numerical solution closer to the (true) analytical

solution. But, as will be investigated thoroughly in Section 6.2, it is only

partially true. In fact, a too small step-size not only makes the computation time

longer (proportional as 1/h), but also results in rather larger errors due to the

accumulated round-off effect. This is why we should look for other methods to

decrease the errors rather than simply reduce the step-size.

Euler’s method can also be applied for solving a first-order vector differential

equation

y(t) = f(t, y) with y(t0) = y0 (6.1.6)

which is equivalent to a high-order scalar differential equation. The algorithm

can be described by

yk+1 = yk + hf(tk, yk) with y(t0) = y0 (6.1.7)

Table 6.1 A Numerical Solution of the Differential Equation (6.1.1) Obtained by the

Euler’s Method

t h = 0.5 h = 0.25

0.25 y(0.25) = (1 − ah)y0 + hr = 1/4 = 0.25

0.50 y(0.50) = (1 − ah)y0 + hr = 1/2 = 0.5 y(0.50) = (3/4)y(0.25) + 1/4 = 0.4375

0.75 y(0.75) = (3/4)y(0.50) + 1/4 = 0.5781

1.00 y(1.00) = (1/2)y(0.5) + 1/2 = 3/4 = 0.75 y(1.00) = (3/4)y(0.75) + 1/4 = 0.6836

1.25 y(1.25) = (3/4)y(1.00) + 1/4 = 0.7627

1.50 y(1.50) = (1/2)y(1.0) + 1/2 = 7/8 = 0.875 y(1.50) = (3/4)y(1.25) + 1/4 = 0.8220

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

266 ORDINARY DIFFERENTIAL EQUATIONS

0

0

0.5 1

1

1.5 t 2

0.2

0.4

0.6

0.8

the (true) analytical solution

y (t ) = 1 – e –at

h = 0.25

h = 0.5

Figure 6.1 Examples of numerical solution obtained by using the Euler’s method.

and is cast into the MATLAB routine “ode_Euler()”.

function [t,y] = ode_Euler(f,tspan,y0,N)

%Euler’s method to solve vector differential equation y’(t) = f(t,y(t))

% for tspan = [t0,tf] and with the initial value y0 and N time steps

if nargin<4 | N <= 0, N = 100; end

if nargin<3, y0 = 0; end

h = (tspan(2) – tspan(1))/N; %stepsize

t = tspan(1)+[0:N]’*h; %time vector

y(1,:) = y0(:)’; %always make the initial value a row vector

for k = 1:N

y(k + 1,:) = y(k,:) +h*feval(f,t(k),y(k,:)); %Eq.(6.1.7)

end

6.2 HEUN’S METHOD: TRAPEZOIDAL METHOD

Another method of solving a first-order vector differential equation like Eq. (6.1.6)

comes from integrating both sides of the equation.

y(t) = f(t, y), y(t)|tk+1

tk = y(tk+1) − y(tk) = tk+1

tk

f(t, y) dt

y(tk+1) = y(tk) + tk+1

tk

f(t, y) dt with y(t0) = y0 (6.2.1)

If we assume that the value of the (derivative) function f(t ,y) is constant

as f(tk,y(tk)) within one time step [tk,tk+1), this becomes Eq. (6.1.7) (with h = tk+1 − tk), amounting to Euler’s method. If we use the trapezoidal rule (5.5.3), it

becomes

yk+1 = yk +

h

2 {f(tk, yk) + f(tk+1, yk+1)} (6.2.2)

RUNGE–KUTTA METHOD 267

function [t,y] = ode_Heun(f,tspan,y0,N)

%Heun method to solve vector differential equation y’(t) = f(t,y(t))

% for tspan = [t0,tf] and with the initial value y0 and N time steps

if nargin<4 | N <= 0, N = 100; end

if nargin<3, y0 = 0; end

h = (tspan(2) – tspan(1))/N; %stepsize

t = tspan(1)+[0:N]’*h; %time vector

y(1,:) = y0(:)’; %always make the initial value a row vector

for k = 1:N

fk = feval(f,t(k),y(k,:)); y(k+1,:) = y(k,:)+h*fk; %Eq.(6.2.3)

y(k+1,:) = y(k,:) +h/2*(fk +feval(f,t(k+1),y(k+1,:))); %Eq.(6.2.4)

end

But, the right-hand side (RHS) of this equation has yk+1, which is unknown at

tk. To resolve this problem, we replace the yk+1 on the RHS by the following

approximation:

yk+1∼=

yk + hf(tk, yk) (6.2.3)

so that it becomes

yk+1 = yk +

h

2 {f(tk, yk) + f(tk+1, yk + hf(tk, yk))} (6.2.4)

This is Heun’s method, which is implemented in the MATLAB routine

“ode_Heun()”. It is a kind of predictor-and-corrector method in that it predicts

the value of yk+1 by Eq. (6.2.3) at tk and then corrects the predicted value by

Eq. (6.2.4) at tk+1. The truncation error of Heun’s method is O(h2) (proportional

to h2) as shown in Eq. (5.6.1), while the error of Euler’s method is O(h).

6.3 RUNGE–KUTTA METHOD

Although Heun’s method is a little better than the Euler’s method, it is still not

accurate enough for most real-world problems. The fourth-order Runge–Kutta

(RK4) method having a truncation error of O(h4) is one of the most widely used

methods for solving differential equations. Its algorithm is described below.

yk+1 = yk +

h

6

(fk1 + 2fk2 + 2fk3 + fk4) (6.3.1)

where

fk1 = f(tk, yk) (6.3.2a)

fk2 = f(tk + h/2, yk + fk1h/2) (6.3.2b)

fk3 = f(tk + h/2, yk + fk2h/2) (6.3.2c)

fk4 = f(tk + h, yk + fk3h) (6.3.2d)

268 ORDINARY DIFFERENTIAL EQUATIONS

function [t,y] = ode_RK4(f,tspan,y0,N,varargin)

%Runge-Kutta method to solve vector differential eqn y’(t) = f(t,y(t))

% for tspan = [t0,tf] and with the initial value y0 and N time steps

if nargin < 4 | N <= 0, N = 100; end

if nargin < 3, y0 = 0; end

y(1,:) = y0(:)’; %make it a row vector

h = (tspan(2) – tspan(1))/N; t = tspan(1)+[0:N]’*h;

for k = 1:N

f1 = h*feval(f,t(k),y(k,:),varargin{:}); f1 = f1(:)’; %(6.3.2a)

f2 = h*feval(f,t(k) + h/2,y(k,:) + f1/2,varargin{:}); f2 = f2(:)’;%(6.3.2b)

f3 = h*feval(f,t(k) + h/2,y(k,:) + f2/2,varargin{:}); f3 = f3(:)’;%(6.3.2c)

f4 = h*feval(f,t(k) + h,y(k,:) + f3,varargin{:}); f4 = f4(:)’; %(6.3.2d)

y(k + 1,:) = y(k,:) + (f1 + 2*(f2 + f3) + f4)/6; %Eq.(6.3.1)

end

%nm630: Heun/Euer/RK4 method to solve a differential equation (d.e.)

clear, clf

tspan = [0 2];

t = tspan(1)+[0:100]*(tspan(2) – tspan(1))/100;

a = 1; yt = 1 – exp(-a*t); %Eq.(6.1.5): true analytical solution

plot(t,yt,’k’), hold on

df61 = inline(’-y + 1’,’t’,’y’); %Eq.(6.1.1): d.e. to be solved

y0 = 0; N = 4;

[t1,ye] = oed_Euler(df61,tspan,y0,N);

[t1,yh] = ode_Heun(df61,tspan,y0,N);

[t1,yr] = ode_RK4(df61,tspan,y0,N);

plot(t,yt,’k’, t1,ye,’b:’, t1,yh,’b:’, t1,yr,’r:’)

plot(t1,ye,’bo’, t1,yh,’b+’, t1,yr,’r*’)

N = 1e3; %to estimate the time for N iterations

tic, [t1,ye] = ode_Euler(df61,tspan,y0,N); time_Euler = toc

tic, [t1,yh] = ode_Heun(df61,tspan,y0,N); time_Heun = toc

tic, [t1,yr] = ode_RK4(df61,tspan,y0,N); time_RK4 = toc

Equation (6.3.1) is the core of RK4 method, which may be obtained by substituting

Simpson’s rule (5.5.4)

tk+1

tk

f (x) dx∼=

h

3

(fk + 4fk+1/2 + fk+1) with h =

xk+1 − xk

2 =

h

2

(6.3.3)

into the integral form (6.2.1) of differential equation and replacing fk+1/2 with

the average of the successive function values (fk2 + fk3)/2. Accordingly, the

RK4 method has a truncation error of O(h4) as Eq. (5.6.2) and thus is expected

to work better than the previous two methods.

The fourth-order Runge–Kutta (RK4) method is cast into the MATLAB routine

“ode_RK4()”. The program “nm630.m” uses this routine to solve Eq. (6.1.1)

with the step size h = (tf − t0)/N = 2/4 = 0.5 and plots the numerical result

together with the (true) analytical solution. Comparison of this result with those of

Euler’s method (“ode_Euler()”) and Heun’s method (“ode_Heun()”) is given in

Fig. 6.2, which shows that the RK4 method is better than Heun’s method, while

Euler’s method is the worst in terms of accuracy with the same step-size. But,

PREDICTOR–CORRECTOR METHOD 269

+

+

+

0

0

0.5 1 1.5 t 2

0.2

0.4

0.6

0.8

1

Euler solution

Runge-Kutta solution

Heun solution +

y (t ) = 1 – e–at

the (true) analytical solution

h = 0.5

Figure 6.2 Numerical solutions for a first-order differential equation.

in terms of computational load, the order is reversed, because Euler’s method,

Heun’s method, and the RK4 method need 1, 2, and 4 function evaluations (calls)

per iteration, respectively.

(cf) Note that a function call takes much more time than a multiplication and thus the

number of function calls should be a criterion in estimating and comparing computational

time.

The MATLAB built-in routines “ode23()” and “ode45()” implement the

Runge–Kutta method with an adaptive step-size adjustment, which uses a

large/small step-size depending on whether f (t) is smooth or rough. In

Section 6.4.3, we will try applying these routines together with our routines to

solve a differential equation for practice rather than for comparison.

6.4 PREDICTOR–CORRECTOR METHOD

6.4.1 Adams–Bashforth–Moulton Method

The Adams–Bashforth–Moulton (ABM) method consists of two steps. The first

step is to approximate f(t ,y) by the (Lagrange) polynomial of degree 4 matching

the four points

{(tk−3, fk−3), (tk−2, fk−2), (tk−1, fk−1), (tk , fk)}

and substitute the polynomial into the integral form (6.2.1) of differential equation

to get a predicted estimate of yk+1.

pk+1 = yk + h

0

l3(t) dt = yk +

h

24

(−9fk−3 + 37fk−2 − 59fk−1 + 55fk)

(6.4.1a)

270 ORDINARY DIFFERENTIAL EQUATIONS

The second step is to repeat the same work with the updated four points

{(tk−2, fk−2), (tk−1, fk−1), (tk, fk), (tk+1, fk+1)} (fk+1 = f(tk+1, pk+1))

to get a corrected estimate of yk+1.

ck+1 = yk + h

0

l3(t) dt = yk +

h

24

(fk−2 − 5fk−1 + 19fk + 9fk+1) (6.4.1b)

The coefficients of Eqs. (6.4.1a) and (6.4.1b) can be obtained by using the

MATLAB routines “lagranp()” and “polyint()”, each of which generates

Lagrange (coefficient) polynomials and integrates a polynomial, respectively.

Let’s try running the program “ABMc.m”.

>>abmc

cAP = -3/8 37/24 -59/24 55/24

cAC = 1/24 -5/24 19/24 3/8

%ABMc.m

% Predictor/Corrector coefficients in Adams–Bashforth–Moulton method

clear

format rat

[l,L] = lagranp([-3 -2 -1 0],[0 0 0 0]); %only coefficient polynomial L

for m = 1:4

iL = polyint(L(m,:)); %indefinite integral of polynomial

cAP(m) = polyval(iL,1)-polyval(iL,0); %definite integral over [0,1]

end

cAP %Predictor coefficients

[l,L] = lagranp([-2 -1 0 1],[0 0 0 0]); %only coefficient polynomial L

for m = 1:4

iL = polyint(L(m,:)); %indefinite integral of polynomial

cAC(m) = polyval(iL,1) – polyval(iL,0); %definite integral over [0,1]

end

cAC %Corrector coefficients

format short

Alternatively, we write the Taylor series expansion of yk+1 about tk and that

of yk about tk+1 as

yk+1 = yk + hfk +

h2

2

fk +

h3

3!

f (2)

k +

h4

4!

f (3)

k +

h5

5!

f (4)

k + ·· · (6.4.2a)

yk = yk+1 − hfk+1 +

h2

2

fk+1 −

h3

3!

f (2)

k+1 +

h4

4!

f (3)

k+1 −

h5

5!

f (4)

k+1 + ·· ·

yk+1 = yk + hfk+1 −

h2

2

fk+1 +

h3

3!

f (2)

k+1 −

h4

4!

f (3)

k+1 +

h5

5!

f (4)

k+1 − ·· · (6.4.2b)

and replace the first, second, and third derivatives by their difference approximations.

PREDICTOR–CORRECTOR METHOD 271

yk+1 = yk + hfk +

h2

2 −13

fk−3 + 32

fk−2 − 3fk−1 + 11

6 fk

h +

1

4

h3f (4)

k + ·· ·

+

h3

3! −fk−3 + 4fk−2 − 5fk−1 + 2fk

h2 +

11

12

h2f (4)

k + ·· ·

+

h4

4! −fk−3 + 3fk−2 − 3fk−1 + fk

h3 +

3

2

hf (4)

k +· · · +

h5

120

f (4)

k + ·· ·

= yk +

h

24

(−9fk−3 + 37fk−2 − 59fk−1 + 55fk) +

251

720

h5f (4)

k + ·· ·

(6.4.1a)

≈ pk+1 +

251

720

h5f (4)

k (6.4.3a)

yk+1 = yk + hfk+1 −

h2

2 −13

fk−2 + 32

fk−1 − 3fk + 11

6 fk+1

h +

1

4

h3f (4)

k+1 + ·· ·

+

h3

3! −fk−2 + 4fk−1 − 5fk + 2fk+1

h2 +

11

12

h2f (4)

k+1 +· · ·

−

h4

4! −fk−2 + 3fk−1 − 3fk + fk+1

h3 +

3

2

hf (4)

k+1 + ·· · +

h5

120

f (4)

k+1 + ·· ·

= yk +

h

24

(fk−2 − 5fk−1 + 19fk + 9fk+1) −

19

720

h5f (4)

k+1 + ·· ·

(6.4.1b)

≈ ck+1 −

19

720

h5f (4)

k+1 (6.4.3b)

These derivations are supported by running the MATLAB program “ABMc1.m”.

%ABMc1.m

%another way to get the ABM coefficients together with the error term

clear, format rat

for i = 1:3, [ci,erri] = difapx(i,[-3 0]); c(i,:) = ci; err(i) = erri;

end

cAP = [0 0 0 1]+[1/2 1/6 1/24]*c, errp = -[1/2 1/6 1/24]*err’ + 1/120

cAC = [0 0 0 1]+[-1/2 1/6 -1/24]*c, errc = -[-1/2 1/6 -1/24]*err’ + 1/120

format short

From these equations and under the assumption that f (4)

k+1∼=

f (4)

k ∼=

K, we can

write the predictor/corrector errors as

EP,k+1 = yk+1 − pk+1 ≈

251

720

h5f (4)

k ∼=

251

720

Kh5 (6.4.4a)

EC,k+1 = yk+1 − ck+1 ≈ −

19

720

h5f (4)

k+1∼=

−

19

720

Kh5 (6.4.4b)

272 ORDINARY DIFFERENTIAL EQUATIONS

We still cannot use these formulas to estimate the predictor/corrector errors, since

K is unknown. But, from the difference between these two formulas

EP,k+1 − EC,k+1 = ck+1 − pk+1∼=

270

720

Kh5 ≡

270

251

EP,k+1 ≡ −

270

19

EC,k+1

(6.4.5)

we can get the practical formulas for estimating the errors as

EP,k+1 = yk+1 − pk+1∼=

251

270

(ck+1 − pk+1) (6.4.6a)

EC,k+1 = yk+1 − ck+1∼=

−

19

270

(ck+1 − pk+1) (6.4.6b)

These formulas give us rough estimates of how close the predicted/corrected

values are to the true value and so can be used to improve them as well as to

adjust the step-size.

pk+1 → pk+1 +

251

270

(ck − pk) ⇒ mk+1 (6.4.7a)

ck+1 → ck+1 −

19

270

(ck+1 − pk+1) ⇒ yk+1 (6.4.7b)

These modification formulas are expected to reward our efforts that we have

made to derive them.

The Adams–Bashforth–Moulton (ABM) method with the modification formulas

can be described by Eqs. (6.4.1a), (6.4.1b), and (6.4.7a), (6.4.7b) summarized

below and is cast into the MATLAB routine “ode_ABM()”. This scheme needs

only two function evaluations (calls) per iteration, while having a truncation

error of O(h5) and thus is expected to work better than the methods discussed so

far. It is implemented by the MATLAB built-in routine “ode113()” with many

additional sophisticated techniques.

Adams–Bashforth–Moulton method with modification formulas

Predictor: pk+1 = yk +

h

24

(−9fk−3 + 37fk−2 − 59fk−1 + 55fk) (6.4.8a)

Modifier: mk+1 = pk+1 +

251

270

(ck − pk) (6.4.8b)

Corrector: ck+1 = yk +

h

24

(fk−2 − 5fk−1 + 19fk + 9f(tk+1,mk+1)) (6.4.8c)

yk+1 = ck+1 −

19

270

(ck+1 − pk+1) (6.4.8d)

PREDICTOR–CORRECTOR METHOD 273

function [t,y] = ode_ABM(f,tspan,y0,N,KC,varargin)

%Adams-Bashforth-Moulton method to solve vector d.e. y’(t) = f(t,y(t))

% for tspan = [t0,tf] and with the initial value y0 and N time steps

% using the modifier based on the error estimate depending on KC = 1/0

if nargin < 5, KC = 1; end %with modifier by default

if nargin < 4 | N < = 0, N = 100; end %default maximum number of iterations

y0 = y0(:)’; %make it a row vector

h = (tspan(2) – tspan(1))/N; %step size

tspan0 = tspan(1)+[0 3]*h;

[t,y] = rk4(f,tspan0,y0,3,varargin{:}); %initialize by Runge-Kutta

t = [t(1:3)’ t(4):h:tspan(2)]’;

for k = 1:4, F(k,:) = feval(f,t(k),y(k,:),varargin{:}); end

p = y(4,:); c = y(4,:); KC22 = KC*251/270; KC12 = KC*19/270;

h24 = h/24; h241 = h24*[1 -5 19 9]; h249 = h24*[-9 37 -59 55];

for k = 4:N

p1 = y(k,:) +h249*F; %Eq.(6.4.8a)

m1 = pk1 + KC22*(c-p); %Eq.(6.4.8b)

c1 = y(k,:)+ …

h241*[F(2:4,:); feval(f,t(k + 1),m1,varargin{:})]; %Eq.(6.4.8c)

y(k + 1,:) = c1 – KC12*(c1 – p1); %Eq.(6.4.8d)

p = p1; c = c1; %update the predicted/corrected values

F = [F(2:4,:); feval(f,t(k + 1),y(k + 1,:),varargin{:})];

End

6.4.2 Hamming Method

function [t,y] = ode_Ham(f,tspan,y0,N,KC,varargin)

% Hamming method to solve vector d.e. y’(t) = f(t,y(t))

% for tspan = [t0,tf] and with the initial value y0 and N time steps

% using the modifier based on the error estimate depending on KC = 1/0

if nargin < 5, KC = 1; end %with modifier by default

if nargin < 4 | N <= 0, N = 100; end %default maximum number of iterations

if nargin < 3, y0 = 0; end %default initial value

y0 = y0(:)’; end %make it a row vector

h = (tspan(2)-tspan(1))/N; %step size

tspan0 = tspan(1)+[0 3]*h;

[t,y] = ode_RK4(f,tspan0,y0,3,varargin{:}); %Initialize by Runge-Kutta

t = [t(1:3)’ t(4):h:tspan(2)]’;

for k = 2:4, F(k – 1,:) = feval(f,t(k),y(k,:),varargin{:}); end

p = y(4,:); c = y(4,:); h34 = h/3*4; KC11 = KC*112/121; KC91 = KC*9/121;

h312 = 3*h*[-1 2 1];

for k = 4:N

p1 = y(k – 3,:) + h34*(2*(F(1,:) + F(3,:)) – F(2,:)); %Eq.(6.4.9a)

m1 = p1 + KC11*(c – p); %Eq.(6.4.9b)

c1 = (-y(k – 2,:) + 9*y(k,:) +…

h312*[F(2:3,:); feval(f,t(k + 1),m1,varargin{:})])/8; %Eq.(6.4.9c)

y(k+1,:) = c1 – KC91*(c1 – p1); %Eq.(6.4.9d)

p = p1; c = c1; %update the predicted/corrected values

F = [F(2:3,:); feval(f,t(k + 1),y(k + 1,:),varargin{:})];

end

274 ORDINARY DIFFERENTIAL EQUATIONS

Hamming method with modification formulas

Predictor: pk+1 = yk−3 +

4h

3

(2fk−2 − fk−1 + 2fk) (6.4.9a)

Modifier: mk+1 = pk+1 +

112

121

(ck − pk) (6.4.9b)

Corrector: ck+1 =

1

8{9yk − yk−2 + 3h(−fk−1 + 2fk + f(tk+1,mk+1))}(6.4.9c)

yk+1 = ck+1 −

9

121

(ck+1 − pk+1) (6.4.9d)

In this section, we introduce just the algorithm of the Hamming method [H-1]

summarized in the box above and the corresponding routine “ode_Ham()”, which

is another multistep predictor–corrector method like the Adams–Bashforth–

Moulton (ABM) method.

This scheme also needs only two function evaluations (calls) per iteration,

while having the error of O(h5) and so is comparable with the ABM method

discussed in the previous section.

6.4.3 Comparison of Methods

The major factors to be considered in evaluating/comparing different numerical

methods are the accuracy of the numerical solution and its computation

time. In this section, we will compare the routines “ode_RK4()”, “ode_ABM()”,

“ode_Ham()”, “ode23()”, “ode45()”, and “ode113()” by trying them out on

the same differential equations, hopefully to make some conjectures about their

performances. It is important to note that the evaluation/comparison of numerical

methods is not so simple because their performances may depend on the

characteristic of the problem at hand. It should also be noted that there are other

factors to be considered, such as stability, versatility, proof against run-time

error, and so on. These points are being considered in most of the MATLAB

built-in routines.

The first thing we are going to do is to validate the effectiveness of the modifiers

(Eqs. (6.4.8b,d) and (6.4.9b,d)) in the ABM (Adams–Bashforth–Moulton)

method and the Hamming method. For this job, we write and run the program

“nm643_1.m” to get the results depicted in Fig. 6.3 for the differential equation

y(t) = −y(t) + 1 with y(0) = 0 (6.4.10)

which was given at the beginning of this chapter. Fig. 6.3 shows us an interesting

fact that, although the ABM method and the Hamming method, even without

modifiers, are theoretically expected to have better accuracy than the RK4 (fourthorder

Runge–Kutta) method, they turn out to work better than RK4 only with

modifiers. Of course, it is not always the case, as illustrated in Fig. 6.4, which

PREDICTOR–CORRECTOR METHOD 275

1

0.5

0

0 2 4 6 8 10

(a1) Numerical solutions without modifiers

true analytical solution y (t ) = 1 − e−t

and numerical solutions

1

0.5

0

0 2 4 6 8 10

(a2) Numerical solutions with modifiers

true analytical solution y (t ) = 1 − e−t

and numerical solutions

0 2 4 6 8 10

6

4

2

0

(b1) Relative errors without modifiers

× 10−5

1

0.5

0

0 2 4 6 8 10

(b2) Relative errors with modifiers

RK4

ABM

Hamming

1.5

× 10−5

Figure 6.3 Numerical solutions and their errors for the differential equation y(t) = −y(t) + 1.

true analytical solution y (t ) = et − 1

and numerical solutions

0 2 4 6 8 10

0

1

2

3

× 104

(a3) Numerical solutions by ode23,

ode45, ode113

0 2 4 6 8 10

× 10–3

1

0.5

0

ode23 ( )

ode45 ( )

ode113 ( )

(b3) Their relative errors

true analytical solution y (t ) = et − 1

and numerical solutions

2

0

0

2 4 6 8 10 0 2 4 6 8 10

3

× 104 × 10–4

(a1) Numerical solutions without modifiers

1

1.5

1

0.5

0

(b1) Relative errors without modifiers

true analytical solution y (t ) = et − 1

and numerical solutions

2

0 2 4 6 8 10

0

1

3

× 104

(a2) Numerical solutions with modifiers

0 2 4 6 8 10

× 10–4

1.5

1

0.5

0

RK4

ABM

Hamming

(b2) Relative errors with modifiers

Figure 6.4 Numerical solutions and their errors for the differential equation y(t) = y(t) + 1.

276 ORDINARY DIFFERENTIAL EQUATIONS

we obtained by applying the same routines to solve another differential equation

y(t) = y(t) + 1 with y(0) = 0 (6.4.11)

where the true analytical solution is

y(t) = et − 1 (6.4.12)

%nm643_1: RK4/Adams/Hamming method to solve a differential eq

clear, clf

t0 = 0; tf = 10; y0 = 0; %starting/final time, initial value

N = 50; %number of segments

df643 = inline(’-y+1’,’t’,’y’); %differential equation to solve

f643 = inline(’1-exp(-t)’,’t’); %true analytical solution

for KC = 0:1

tic, [t1,yR] = ode_RK4(df643,[t0 tf],y0,N); tR = toc

tic, [t1,yA] = ode_ABM(df643,[t0 tf],y0,N,KC); tA = toc

tic, [t1,yH] = ode_Ham(df643,[t0 tf],y0,N,KC); tH = toc

yt1 = f643(t1); %true analytical solution to plot

subplot(221 + KC*2) %plot analytical/numerical solutions

plot(t1,yt1,’k’, t1,yR,’k’, t1,yA,’k–’, t1,yH,’k:’)

tmp = abs(yt1)+eps; l_t1 = length(t1);

eR = abs(yR – yt1)./tmp; e_R=norm(eR)/lt1

eA = abs(yA – yt1)./tmp; e_A=norm(eA)/lt1

eH = abs(yH – yt1)./tmp; e_H=norm(eH)/lt1

subplot(222 + KC*2) %plot relative errors

plot(t1,eR,’k’, t1,eA,’k–’, t1, eH,’k:’)

end

%nm643_2: ode23()/ode45()/ode113() to solve a differential eq

clear, clf

t0 = 0; tf = 10; y0 = 0; N = 50; %starting/final time, initial value

df643 = inline(’y + 1’,’t’,’y’); %differential equation to solve

f643 = inline(’exp(t) – 1’,’t’); %true analytical solution

tic, [t1,yR] = ode_RK4(df643,[t0 tf],y0,N); time(1) = toc;

tic, [t1,yA] = ode_ABM(df643,[t0 tf],y0,N); time(2) = toc;

yt1 = f643(t1);

tmp = abs(yt1)+ eps; l_t1 = length(t1);

eR = abs(yR-yt1)./tmp; err(1) = norm(eR)/l_t1;

eA = abs(yA-yt1)./tmp; err(2) = norm(eA)/l_t1;

options = odeset(’RelTol’,1e-4); %set the tolerance of relative error

tic, [t23,yode23] = ode23(df643,[t0 tf],y0,options); time(3) = toc;

tic, [t45,yode45] = ode45(df643,[t0 tf],y0,options); time(4) = toc;

tic, [t113,yode113] = ode113(df643,[t0 tf],y0,options); time(5) = toc;

yt23 = f643(t23); tmp = abs(yt23) + eps;

eode23 = abs(yode23-yt23)./tmp; err(3) = norm(eode23)/length(t23);

yt45 = f643(t45); tmp = abs(yt45) + eps;

eode45 = abs(yode45 – yt45)./tmp; err(4) = norm(eode45)/length(t45);

yt113 = f643(t113); tmp = abs(yt113) + eps;

eode113 = abs(yode113 – yt113)./tmp; err(5) = norm(eode113)/length(t113);

subplot(221), plot(t23,yode23,’k’, t45,yode45,’b’, t113,yode113,’r’)

subplot(222), plot(t23,eode23,’k’, t45,eode45,’b–’, t113,eode113,’r:’)

err, time

VECTOR DIFFERENTIAL EQUATIONS 277

Table 6.2 Results of Applying Several Routines to solve a Simple Differential Equation

ode RK4() ode ABM() ode Ham() ode23() ode45() ode113()

Relative error 0.0925 × 10−4 0.0203 × 10−4 0.0179 × 10−4 0.4770 × 10−4 0.0422 × 10−4 0.1249 × 10−4

Computing time 0.05 sec 0.03 sec 0.03 sec 0.07 sec 0.05 sec 0.05 sec

Readers are invited to supplement the program “nm643_2.m” in such a way

that “ode_Ham()” is also used to solve Eq. (6.4.11). Running the program yields

the results depicted in Fig. 6.4 and listed in Table 6.2. From Fig. 6.4, it is noteworthy

that, without the modifiers, the ABM method seems to be better than the

Hamming method; however, with the modifiers, it is the other way around or at

least they run a neck-and-neck race. Anyone will see that the predictor–corrector

methods such as the ABM method (ode_ABM()) and the Hamming method

(ode_Ham()) give us a better numerical solution with less error and shorter computation

time than the MATLAB built-in routines “ode23()”, “ode45()”, and

“ode113()” as well as the RK4 method (ode_RK4()), as listed in Table 6.2. But,

a general conclusion should not be deduced just from one example.

6.5 VECTOR DIFFERENTIAL EQUATIONS

6.5.1 State Equation

Although we have tried using the MATLAB routines only for scalar differential

equations, all the routines made by us or built inside MATLAB are ready to

entertain first-order vector differential equations, called state equations, as below.

x1(t) = f1(t, x1(t), x2(t), . . .) with x1(t0) = x10

x2(t) = f2(t, x1(t), x2(t), . . .) with x2(t0) = x20

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

x(t) = f(t, x(t)) with x(t0) = x0 (6.5.1)

For example, we can define the system of first-order differential equations

x1(t) = x2(t) with x1(0) = 1

x2(t) = −x2(t) + 1 with x2(0) = −1

(6.5.2)

in a file named “df651.m” and solve it by running the MATLAB program

“nm651_1.m”, which uses the routines “ode_Ham()”/“ode45()” to get the

numerical solutions and plots the results as depicted in Fig. 6.5. Note that the

function given as the first input argument of “ode45()” must be fabricated to

generate its value in a column vector or at least, in the same form of vector as

the input argument ‘x’ so long as it is a vector-valued function.

278 ORDINARY DIFFERENTIAL EQUATIONS

%nm651_1 to solve a system of differential eqs., i.e., state equation

df = ’df651’;

t0 = 0; tf = 2; x0 = [1 -1]; %start/final time and initial value

N = 45; [tH,xH] = ode_Ham(df,[t0 tf],x0,N); %with N = number of segments

[t45,x45] = ode45(df,[t0 tf],x0);

plot(tH,xH), hold on, pause, plot(t45,x45)

function dx = df651(t,x)

dx = zeros(size(x)); %row/column vector depending on the shape of x

dx(1) = x(2); dx(2) = -x(2) + 1;

Especially for the state equations having only constant coefficients like Eq.

(6.5.2), we can change it into a matrix–vector form as

x1(t)

x2(t) = 0 1

0 −1x1(t)

x2(t) + 0

1 us (t) (6.5.3)

with x1(0)

x2(0) = 1

−1 and us (t) = 1 ∀ t ≥ 0

x(t) = Ax(t) + Bu(t) with the initial state x(0) and the input u(t) (6.5.4)

which is called a linear time-invariant (LTI) state equation, and then try to find

the analytical solution. For this purpose, we take the Laplace transform of both

sides to write

sX(s) − x(0) = AX(s) + BU(s) with X(s) = L{x(t)}, U(s) = L{u(t)}

[sI − A]X(s) = x(0) + BU(s), X(s) = [sI − A]−1x(0) + [sI − A]−1BU(s)

(6.5.5)

where L{x(t)} and L−1{X(s)} denote the Laplace transform of x(t) and the

inverse Laplace transform of X(s), respectively. Note that

[sI − A]−1 = s−1[I − As−1]−1 = s−1 I + As−1 + A2s−2 + ·· ·

φ(t) = L−1{[sI − A]−1} (6.5.6)

= I + At +

A2

2

t2 +

A3

3!

t3 + ·· · = eAt with φ(0) = I

By applying the convolution property of Laplace transform (Table D.2(4) in

Appendix D)

L−1{[sI − A]−1BU(s)} = L−1{[sI − A]−1} ∗ L−1{BU(s)} = φ(t) ∗ Bu(t)

= ∞

−∞

φ(t − τ)Bu(τ) dτ

u(τ )=0 for τ<0 or τ>t = t

0

φ(t − τ)Bu(τ) dτ (6.5.7)

VECTOR DIFFERENTIAL EQUATIONS 279

we can take the inverse Laplace transform of Eq. (6.5.5) to write

x(t) = φ(t)x(0) + φ(t) ∗ Bu(t) = φ(t)x(0) + t

0

φ(t − τ)Bu(τ) dτ (6.5.8)

For Eq. (6.5.3), we use Eq. (6.5.6) to find

φ(t) = L−1{[sI − A]−1}

= L−1 s 0

0 s − 0 1

0 −1 −1 = L−1 s −1

0 s + 1 −1

= L−1

1

s(s + 1) s +1 1

0 s

= L−1

1/s 1/s − 1/(s + 1)

0 1/(s + 1) = 1 1− e−t

0 e−t (6.5.9)

and use Eqs. (6.5.8), (6.5.9), and u(t) = us (t) = 1 ∀ t ≥ 0 to obtain

x(t) = 1 1− e−t

0 e−t 1

−1 + t

0 1 1− e−(t−τ)

0 e−(t−τ) 0

1 1 dτ

= e−t

−e−t + τ − e−(t−τ)

e−(t−τ)

t

0 = t − 1 + 2e−t

1 − 2e−t (6.5.10)

Alternatively, we can directly take the inverse transform of Eq. (6.5.5) to get

X(s) = [sI − A]−1{x(0) + [sI − A]−1BU(s)}

=

1

s(s + 1) s +1 1

0 s

1

−1 + 0

1 1

s

=

1

s2(s + 1) s +1 1

0 s s

−s + 1 =

1

s2(s + 1) s2 + 1

s(1 − s) (6.5.11)

X1(s) =

s2 + 1

s2(s + 1) =

1

s2 −

1

s +

2

s + 1

, x1(t) = t − 1 + 2e−t (6.5.12a)

X2(s) =

1 − s

s(s + 1) =

1

s −

2

s + 1

, x2(t) = 1 − 2e−t (6.5.12b)

which conforms with Eq. (6.5.10).

The MATLAB program “nm651_2.m” uses a symbolic computation routine

“ilaplace()” to get the inverse Laplace transform, uses “eval()” to evaluate

280 ORDINARY DIFFERENTIAL EQUATIONS

0

−1

1.5

1

0.5

−0.5

0

0.5

x1(t )

1 1.5 2

x2(t )

Figure 6.5 Numerical/analytical solutions of the continuous-time state equation (6.5.2)/(6.5.3).

it, and plots the result as depicted in Fig. 6.5, which supports this derivation procedure.

Additionally, it uses another symbolic computation routine “dsolve()”

to get the analytical solution directly.

>>nm651_2

Solution of Differential Equation based on Laplace transform

Xs = [ 1/s + 1/s/(s + 1)*(-1 + 1/s) ]

[ 1/(s + 1)*(-1 + 1/s) ]

xt = [ -1 + t + 2*exp(-t) ]

[ -2*exp(-t) + 1 ]

Analytical solution

xt1 = -1 + t + 2*exp(-t)

xt2 = -2*exp(-t) + 1

%nm651_2: Analytical solution for state eq. x’(t) = Ax(t) + Bu(t)(6.5.3)

clear

syms s t %declare s,t as symbolic variables

A = [0 1;0 -1]; B = [0 1]’; %Eq.(6.5.3)

x0 = [1 -1]’; %initial value

disp(’Solution of Differential Eq based on Laplace transform’)

disp(’Laplace transformed solution X(s)’)

Xs = (s*eye(size(A)) – A)^-1*(x0 + B/s) %Eq.(6.5.5)

disp(’Inverse Laplace transformed solution x(t)’)

xt = ilaplace(Xs) %inverse Laplace transform %Eq.(6.5.12)

t0 = 0; tf = 2; N = 45; %initial/final time

t = t0 + [0:N]’*(tf – t0)/N; %time vector

xtt = eval(xt:); %evaluate the inverse Laplace transform

plot(t,xtt)

disp(’Analytical solution’)

xt = dsolve(’Dx1 = x2, Dx2 = -x2 + 1’, ’x1(0) = 1, x2(0) = -1’);

xt1 = xt.x1, xt2 = xt.x2 %Eq.(6.5.10)

VECTOR DIFFERENTIAL EQUATIONS 281

6.5.2 Discretization of LTI State Equation

In this section, we consider a discretization method of converting a continuoustime

LTI (linear time-invariant) state equation

x(t) = Ax(t) + Bu(t) with the initial state x(0) and the input u(t) (6.5.13)

into an equivalent discrete-time LTI state equation with the sampling period T

x[n + 1] = Adx[n] + Bdu[n] (6.5.14)

with the initial state x[0] and the input u[n] = u(nT ) for nT ≤ t < (n+ 1)T

which can be solved easily by an iterative scheme mobilizing just simple multiplications

and additions.

For this purpose, we rewrite the solution (6.5.8) of the continuous-time LTI

state equation with the initial time t0 as

x(t) = φ(t − t0)x(t0) + t

t0

φ(t − τ)Bu(τ) dτ (6.5.15)

Under the assumption that the input is constant as the initial value within each

sampling interval—that is, u[n] = u(nT ) for nT ≤ t < (n+ 1)T —we substitute

t0 = nT and t = (n + 1)T into this equation to write the discrete-time LTI state

equation as

x((n + 1)T ) = φ(T )x(nT ) + (n+1)T

nT

φ((n + 1)T − τ)Bu(nT ) dτ

x[n + 1] = φ(T )x[n] + (n+1)T

nT

φ(nT + T − τ) dτBu[n]

x[n + 1] = Adx[n] + Bdu[n] (6.5.16)

where the discretized system matrices are

Ad = φ(T ) = eAT (6.5.17a)

Bd = (n+1)T

nT

φ(nT + T − τ) dτB

σ=nT+T−τ = − 0

T

φ(σ) dσB = T

0

φ(τ) dτB

(6.5.17b)

Here, let us consider another way of computing these system matrices, which

is to the taste of digital computers. It comes from making use of the definition

of a matrix exponential function in Eq. (6.5.6) to rewrite Eq. (6.5.17) as

Ad = eAT =

∞

m=0

AmT m

m! = I + AT

∞

m=0

AmT m

(m + 1)! = I + AT (6.5.18a)

Bd = T

0

φ(τ) dτB = T

0

∞

m=0

Amτm

m!

dτB =

∞

m=0

AmT m+1

(m + 1)!

B = TB (6.5.18b)

282 ORDINARY DIFFERENTIAL EQUATIONS

where

=

∞

m=0

AmT m

(m + 1)!

∼=

I +

AT

2

I +

AT

3

I + ·· ·+

AT

N − 1 I +

AT

N · · · for N 1

(6.5.19)

Now, we apply these discretization formulas for the continuous-time state

equation (6.5.3)

x1 (t)

x2 (t) = 0 1

0 −1x1(t)

x2(t) + 0

1 us (t)

with x1(0)

x2(0) = 1

−1 and us (t) = 1 ∀ t ≥ 0

to get the discretized system matrices and the discretized state equation as

φ(t) = L−1{[sI − A]−1} = L−1 s −1

0 s + 1 −1 (6.5.9) = 1 1− e−t

0 e−t

(6.5.20a)

Ad

(6.5.17a) = φ(T )

(6.5.20a) = 1 1− e−T

0 e−T (6.5.20b)

Bd

(6.5.17b) = T

0

φ(τ) dτB

(6.5.20a) = T

0 1 1− e−τ

0 e−τ dτ 0

1 = T − 1 + e−T

1 − e−T (6.5.20c)

x[n + 1] (6.5.16) = Adx[n] + Bdu[n]

x1[n + 1]

x2[n + 1] = 1 1− e−T

0 e−T x1[n]

x2[n] + T − 1 + e−T

1 − e−T u[n] (6.5.21)

We don’t need any special algorithm other than an iterative scheme to solve

this discrete-time state equation. The formulas (6.5.18a,b) for computing the

discretized system matrices are cast into the routine “c2d_steq()”. The program

“nm652.m” discretizes the continuous-time state equation (6.5.3) by using

the routine and alternatively, the MATLAB built-in routine “c2d()”. It solves

the discretized state equation and plots the results as in Fig. 6.6. As long as

the assumption that u[n] = u(nT ) for nT ≤ t < (n+ 1)T is valid, the solution

(x[n]) of the discretized state equation is expected to match that (x(t)) of

VECTOR DIFFERENTIAL EQUATIONS 283

0

1.5

0.5

1

0

−0.5

−1

0.5 1.5

x 2[n]

x2(t )

x1(t )

x 1[n]

continuous-time

T = 0.05

T = 0.2

1 2

Figure 6.6 The solution of the discretized state equation (6.5.21).

the continuous-time state equation at every sampling instant t = nT and also

becomes closer to x(t) ∀ t as the sampling interval T gets shorter (see Fig. 6.6).

%nm652.m

% discretize a state eqn x’(t) = Ax(t) + Bu(t) to x[n + 1] = Ad*x[n] + Bd*u[n]

clear, clf

A = [0 1;0 -1]; B = [0;1]; %Eq.(6.5.3)

x0 = [1 -1]; t0 = 0; tf = 2; %initial value and time span

T = 0.2; %sampling interval(period)

eT = exp(-T);

AD = [1 1 – eT; 0 eT]%discretized system matrices obtained analytically

BD = [T + eT – 1; 1 – eT] %Eq.(6.5.21)

[Ad,Bd] = c2d_steq(A,B,T,100) %continuous-to-discrete conversion

[Ad1,Bd1] = c2d(A,B,T) %by the built-in routine

t(1) = 0; xd(1,:) = x0; %initial time and initial value

for k = 1:(tf – t0)/T %solve the discretized state equation

t(k + 1) = k*T; xd(k + 1,:) = xd(k,:)*Ad’ + Bd’;

end

stairs([0; t’],[x0; xd]), hold on %stairstep graph

N = 100; t = t0 + [0:N]’*(tf – t0)/N; %time (column) vector

x(:,1) = t-1 + 2*exp(-t); %analytical solution

x(:,2) = 1-2*exp(-t); %Eq.(6.5-12)

plot(t,x)

function [Ad,Bd] = c2d_steq(A,B,T,N)

if nargin < 4, N = 100; end

I = eye(size(A,2)); PSI = I;

for m = N:-1:1, PSI = I + A*PSI*T/(m + 1); end %Eq.(6.5.19)

Ad = I + A*PSI*T; Bd = PSI*T*B; %Eq.(6.5.18)

6.5.3 High-Order Differential Equation to State Equation

Suppose we are given an Nth-order scalar differential equation together with the

initial values of the variable and its derivatives of up to order N − 1, which is

284 ORDINARY DIFFERENTIAL EQUATIONS

called an IVP (Initial Value Problem):

[IVP]N : x(N)(t) = f (t,x(t), x(t), x(2)(t), . . . , x(N−1)(t))

(6.5.22)

with the initial values x(t0) = x10, x(t0) = x20, . . . ,x(N−1)(t0) = xN0

Defining the state vector and the initial state as

x(t) =

x1 = x

x2 = x

x3 = x(2)

…

xN = x(N−1)

, x(t0) =

x10

x20

x30

…

xN0

(6.5.23)

we can rewrite Eq. (6.5.22) in the form of a first-order vector differential

equation—that is, a state equation—as

x1(t)

x2(t)

x3(t)

…

xN (t)

=

x2(t)

x3(t)

x4(t)

…

f (t, x(t), x(t), x(2)(t), . . . , x(N−1)(t))

x(t) = f(t, x(t)) with x(t0) = x0 (6.5.24)

For example, we can convert a third-order scalar differential equation

x(3)(t) + a2x(2)(t) + a1x(t) + a0x(t) = u(t)

into a state equation of the form

x1(t)

x2(t)

x3(t)

=

0 1 0

0 0 1

−a0 −a1 −a2

x1(t)

x2(t)

x3(t)

+

0

0

1

u(t) (6.5.25a)

x(t) = 1 0 0

x1(t)

x2(t)

x3(t)

(6.5.25b)

6.5.4 Stiff Equation

Suppose that we are given a vector differential equation involving more than one

dependent variable with respect to the independent variable t . If the magnitudes

of the derivatives of the dependent variables with respect to t (corresponding

VECTOR DIFFERENTIAL EQUATIONS 285

to their changing rates) are significantly different, such a differential equation is

said to be stiff because it is difficult to be solved numerically. For such a stiff

differential equation, we should be very careful in choosing the step-size in order

to avoid numerical instability problem and get a reasonably accurate solution

within a reasonable computation time. Why? Because we should use a small

step-size to grasp rapidly changing variables, and it requires a lot of computation

to cover slowly changing variables for such a long time as it lasts.

Actually, there is no clear distinction between stiff and non-stiff differential

equations, since stiffness of a differential equation is a matter of degree. Then, is

there any way to estimate the degree of stiffness for a given differential equation?

The answer is yes, if the differential equation can be arranged into an LTI state

equation like Eq. (6.5.4), the solution of which consists of components having

the time constants (modes) equal to the eigenvalues of the system matrix A. For

example, the system matrix of Eq. (6.5.3) has the eigenvalues

|sI − A| = 0, det

s −1

0 s + 1 = s(s + 1) = 0, s= 0 and s = −1

which can be observed as the time constants of two terms 1 = e0t and e−t in

the solution (6.5.12). In this context, a measure of stiffness is the ratio of the

maximum over the minimum among the absolute values of (negative) real parts

of the eigenvalues of the system matrix A:

η(A) =

Max{|Re(λi )|}

Min{|Re(λi )| = 0}

(6.5.26)

This can be thought of as the degree of unbalance between the fast mode and

the slow mode.

Now, what we must know is how to handle stiff differential equations. Fortunately,

MATLAB has several built-in routines like “ode15s()”, “ode23s()”,

“ode23t()”, and “ode23tb()”, which are fabricated to deal with stiff differential

equations efficiently. One may use the help command to see their detailed

usages. Let’s apply them for a Van der Pol equation

d2y(t)

dt2 − μ(1 − y2(t))

dy(t)

dt + y(t) = 0 with y(0) = 2,

dy(t)

dt = 0

(6.5.27a)

which can be written in the form of a state equation as

x1 (t)

x2 (t) = x2(t)

μ(1 − x2

1 (t))x2(t) − x1(t) with x1(0)

x2(0) = 2

0 (6.5.27b)

For this job, we defined this equation in an M-file named “df_van.m” and made

the MATLAB program “nm654.m”, where we declared the parameter μ (mu) as

286 ORDINARY DIFFERENTIAL EQUATIONS

0

3

2

1

0

−1

50

(a) ode_Ham() with N = 8700

100

m = 25

× 10297

diverged because of

numerical instability

0

40

20

0

−20

−40

50

(b) ode_Ham() with N = 9000

100

m = 25

x1 (t)

x2 (t)

0

40

20

0

−20

−40

50

(c) Result obtained by ode45( )

100

m = 25

x1 (t)

x2 (t)

0

100

0

−100

−200

−300

50

(d) Results obtained by ode23( ), ode23s( ),

ode23t( ), ode23tb( ) and ode15s( )

100 150 200

Since the range of x1(t) is

much smaller than that of

x2(t), x1(t) is invisibly

dwarfed by x2(t).

m = 200

Figure 6.7 Numerical solutions of Van der Pol equation obtained by various routines.

a global variable so that it could be passed on to any related routines/functions

as well as “df_van.m”. In the beginning of the program, we set the global

parameter μ to 25 and applied “ode_Ham()” with the number of segments

N = 8700 and 9000. The results are depicted in Figs. 6.7a and 6.7b, which

show how crucial the choice of step-size is for a stiff equation. Next, we

applied “ode45()” to obtain the solution depicted in Fig. 6.7c, which is almost

the same as Fig. 6.7b, but with the computation time less than one fourth

of that taken by “ode_Ham()”. This reveals the merit of the MATLAB builtin

routines that may save the computation time as well as spare our trouble

to choose the step-size, because the step-size is adaptively determined inside

the routines. Then, setting μ = 200, we applied the MATLAB built-in routines

“ode45()”/“ode23()”/“ode15s()”/“ode23s()”/“ode23t()”/“ode23tb()” to get

the results that are little different as depicted in Fig. 6.7d, each taking the

computation time as

time = 24.9530 14.9690 0.1880 0.2650 0.2500 0.2820

The computation time-efficiency of “ode15s()”/“ode23s()”/“ode23t()”/

“ode23tb()” (designed deliberately for handling stiff differential equations) over

“ode45()”/“ode23()” becomes prominent as the value of parameter μ (mu) gets

large, reflecting high stiffness.

BOUNDARY VALUE PROBLEM (BVP) 287

%nm654.m

% to solve a stiff differential eqn called Van der Pol equation

global mu

mu=25, t0=0; tf = 100; tspan = [t0 tf]; xo = [2 0];

[tH1,xH1] = ode_Ham(’df_van’,tspan,x0,8700);

subplot(221), plot(tH1,xH1)

tic,[tH2,xH2] = ode_Ham(’df_van’,tspan,x0,9000); time_Ham = toc

tic,[t45,x45] = ode45(’df_van’,tspan,x0); time_o45 = toc

subplot(222), plot(tH2,xH2), subplot(223), plot(t45,x45)

mu = 200; tf = 200; tspan = [t0 tf];

tic,[t45,x45] = ode45(’df_van’,tspan,x0); time(1) = toc;

tic,[t23,x23] = ode23(’df_van’,tspan,x0); time(2) = toc;

tic,[t15s,x15s] = ode15s(’df_van’,tspan,x0); time(3) = toc;

tic,[t23s,x23s] = ode23s(’df_van’,tspan,x0); time(4) = toc;

tic,[t23t,x23t] = ode23t(’df_van’,tspan,x0); time(5) = toc;

tic,[t23tb,x23tb] = ode23tb(’df_van’,tspan,x0); time(6) = toc;

plot(t45,x45, t23,x23, t15s,x15s, t23s,x23s, t23t,x23t, t23tb,x23tb)

disp(’ ode23 ode15s ode23s ode23t ode23tb’)

time

function dx = df_van(t,x)

%Van der Pol differential equation (6.5.27)

global mu

dx=zeros(size(x));

dx(1) = x(2); dx(2) = mu*(1-x(1).^2).*x(2) – x(1);

6.6 BOUNDARY VALUE PROBLEM (BVP)

A boundary value problem (BVP) is an Nth-order differential equation with some

of the values of dependent variable x(t) and its derivative specified at the initial

time t0 and others specified at the final time tf .

[BVP]N : x(N)(t) = f (t,x(t), x(t), x(2)(t), . . . , x(N−1)(t))

with the boundary values x(t1) = x10, x(t2) = x21, . . . ,x(N−1)(tN) = xN,N−1

(6.6.1)

In some cases, some relations between the initial values and the final values may

be given as a mixed-boundary condition instead of the initial/final values specified.

This section covers the shooting method and the finite difference method

that can be used to solve a second-order BVP as

[BVP]2 : x(t) = f (t, x(t), x(t)) with x(t0) = x0, x(tf ) = xf (6.6.2)

6.6.1 Shooting Method

The idea of this method is to assume the value of x(t0), then solve the differential

equation (IVP) with the initial condition [x(t0) x(t0)] and keep adjusting the value

288 ORDINARY DIFFERENTIAL EQUATIONS

of x(t0) and solving the IVP repetitively until the final value x(tf ) of the solution

matches the given boundary value xf with enough accuracy. It is similar to

adjusting the angle of firing a cannon so that the shell will eventually hit the target

and that’s why this method is named the shooting method. This can be viewed

as a nonlinear equation problem, if we regard x(t0) as an independent variable

and the difference between the resulting final value x(tf ) and the desired one xf

as a (mismatching) function of x(t0). So the solution scheme can be systemized

by using the secant method (Section 4.5) and is cast into the MATLAB routine

“bvp2_shoot()”.

(cf) We might have to adjust the shooting position with the angle fixed, instead of adjusting

the shooting angle with the position fixed or deal with the mixed-boundary

conditions. See Problems 6.6, 6.7, and 6.8.

For example, let’s consider a BVP consisting of the second-order differential

equation

x(t) = 2×2(t) + 4t x(t)x(t) with x(0) =

1

4

, x(1) =

1

3

(6.6.3)

function [t,x] = bvp2_shoot(f,t0,tf,x0,xf,N,tol,kmax)

%To solve BVP2: [x1,x2]’ = f(t,x1,x2) with x1(t0) = x0, x1(tf) = xf

if nargin < 8, kmax = 10; end

if nargin < 7, tol = 1e-8; end

if nargin < 6, N = 100; end

dx0(1) = (xf – x0)/(tf-t0); % the initial guess of x’(t0)

[t,x] = ode_RK4(f,[t0 tf],[x0 dx0(1)],N); % start up with RK4

plot(t,x(:,1)), hold on

e(1) = x(end,1) – xf; % x(tf) – xf: the 1st mismatching (deviation)

dx0(2) = dx0(1) – 0.1*sign(e(1));

for k = 2: kmax-1

[t,x] = ode_RK4(f,[t0 tf],[x0 dx0(k)],N);

plot(t,x(:,1))

%difference between the resulting final value and the target one

e(k) = x(end,1) – xf; % x(tf)- xf

ddx = dx0(k) – dx0(k – 1); % difference between successive derivatives

if abs(e(k))< tol | abs(ddx)< tol, break; end

deddx = (e(k) – e(k – 1))/ddx; % the gradient of mismatching error

dx0(k + 1) = dx0(k) – e(k)/deddx; %move by secant method

end

%do_shoot to solve BVP2 by the shooting method

t0 = 0; tf = 1; x0 = 1/4; xf = 1/3; %initial/final times and positions

N = 100; tol = 1e-8; kmax = 10;

[t,x] = bvp2_shoot(’df661’,t0,tf,x0,xf,N,tol,kmax);

xo = 1./(4 – t.*t); err = norm(x(:,1) – xo)/(N + 1)

plot(t,x(:,1),’b’, t,xo,’r’) %compare with true solution (6.6.4)

function dx = df661(t,x) %Eq.(6.6.5)

dx(1) = x(2); dx(2) = (2*x(1) + 4*t*x(2))*x(1);

BOUNDARY VALUE PROBLEM (BVP) 289

The solution x(t) and its derivative x(t) are known as

x(t) =

1

4 − t2 and x(t) =

2t

(4 − t2)2 = 2t x2(t) (6.6.4)

Note that this second-order differential equation can be written in the form of

state equation as

x1(t)

x2(t) = x2(t)

2×2

1 (t) + 4t x1(t)x2(t) with x1(0)

x2(1) = x0 = 1/4

xf = 1/3

(6.6.5)

In order to apply the shooting method, we set the initial guess of x2(0) = x(0) to

dx0[1] = x2(0) =

xf − x0

tf − t0

(6.6.6)

and solve the state equation with the initial condition [x1(0) x2(0) = dx0[1]].

Then, depending on the sign of the difference e(1) between the final value x1(1)

of the solution and the target final value xf , we make the next guess dx0[2]

larger/smaller than the initial guess dx0[1] and solve the state equation again

with the initial condition [x1(0) dx0[2]]. We can start up the secant method

with the two initial values dx0[1] and dx0[2] and repeat the iteration until the

difference (error) e(k) becomes sufficiently small. For this job, we compose

the MATLAB program “do_shoot.m”, which uses the routine “bvp2_shoot()”

to get the numerical solution and compares it with the true analytical solution.

Figure 6.8 shows that the numerical solution gets closer to the true analytical

solution after each round of adjustment.

(Q) Why don’t we use the Newton method (Section 4.4)?

(A) Because, in order to use the Newton method in the shooting method, we need IVP

solutions instead of function evaluations to find the numerical Jacobian at every

iteration, which will require much longer computation time.

0 0.2 0.4 0.6 0.8 t 1

0.45

0.4

0.35

0.3

0.25

0.2

1/3

x [n] for k = 3

true analytical

solution

x (t)

x [n] for k = 1

1/4 x [n] for k = 2

Figure 6.8 The solution of a BVP obtained by using the shooting method.

290 ORDINARY DIFFERENTIAL EQUATIONS

6.6.2 Finite Difference Method

The idea of this method is to divide the whole interval [t0, tf ] into N segments

of width h = (tf − t0)/N and approximate the first & second derivatives in the

differential equations for each grid point by the central difference formulas. This

leads to a tridiagonal system of equations with respect to (N − 1) variables {xi = x(t0 + ih), i = 1, . . . , N − 1}. However, in order for this system of equations to

be solved easily, it should be linear, implying that its coefficients may not contain

any term of x.

For example, let’s consider a BVP consisting of the second-order linear differential

equation

x(t) + a1(t)x(t) + a0(t)x(t) = u(t) with x(t0) = x0, x(tf ) = xf (6.6.7)

According to the finite difference method, we divide the solution interval

[t0, tf ] into N segments and convert the differential equation for each grid point

ti = t0 + ih into a difference equation as

xi+1 − 2xi + xi−1

h2 + a1i

xi+1 − xi−1

2h + a0ixi = ui

(2 − ha1i)xi−1 + (−4 + 2h2a0i)xi + (2 + ha1i)xi+1 = 2h2ui (6.6.8)

Then, taking account of the boundary condition that x0 = x(t0) and xN = x(tf ),

we collect all of the (N − 1) equations to construct a tridiagonal system of

equations as

−4 + 2h2a01 2 + ha11 0 ž 0 0 0

2 − ha12 −4 + 2h2a02 2 + ha12 ž 0 0 0

0 2− ha13 −4 + 2h2a03 ž 0 0 0

ž ž ž ž ž ž ž

0 0 0 ž −4 + 2h2a0,N−3 2 + ha1,N−3 0

0 0 0 ž 2 − ha1,N−2 −4 + 2h2a0,N−2 2 + ha1,N−2

0 0 0 ž 0 2− ha1,N−1 −4 + 2h2a0,N−1

×

x1

x2

x2

ž

xN−3

xN−2

xN−1

=

2h2u1 − (2 − ha11)x0

2h2u2

2h2u3

ž

2h2uN−3

2h2uN−2

2h2uN−1 − (2 − ha1,N−1)xN

(6.6.9)

This can be solved efficiently by using the MATLAB routine “trid()”, which

is dedicated to a tridiagonal system of linear equations.

BOUNDARY VALUE PROBLEM (BVP) 291

The whole procedure of the finite difference method for solving a second-order

linear differential equation with boundary conditions is cast into the MATLAB

routine “bvp2_fdf()”. This routine is designed to accept the two coefficients

a1 and a0 and the right-hand-side input u of Eq. (6.6.7) as its first three input

arguments, where any of those three input arguments can be given as the function

name in case the corresponding term is not a numeric value, but a function

of time t . We make the program “do_fdf” to use this routine for solving the

second-order BVP

x(t) +

2

t

x(t) −

2

t2 x(t) = 0 with x(1) = 5, x(2) = 3 (6.6.10)

function [t,x] = bvp2_fdf(a1,a0,u,t0,tf,x0,xf,N)

% solve BVP2: x” + a1*x’ + a0*x = u with x(t0) = x0, x(tf) = xf

% by the finite difference method

h = (tf – t0)/N; h2 = 2*h*h;

t = t0+[0:N]’*h;

if ~isnumeric(a1), a1 = a1(t(2:N)); %if a1 = name of a function of t

elseif length(a1) == 1, a1 = a1*ones(N – 1,1);

end

if ~isnumeric(a0), a0 = a0(t(2:N)); %if a0 = name of a function of t

elseif length(a0) == 1, a0 = a0*ones(N – 1,1);

end

if ~isnumeric(u), u = u(t(2:N)); %if u = name of a function of t

elseif length(u) == 1, u = u*ones(N-1,1);

else u = u(:);

end

A = zeros(N – 1,N – 1); b = h2*u;

ha = h*a1(1); A(1,1:2) = [-4 + h2*a0(1) 2 + ha];

b(1) = b(1)+(ha – 2)*x0;

for m = 2:N – 2 %Eq.(6.6.9)

ha = h*a1(m); A(m,m – 1:m + 1) = [2-ha -4 + h2*a0(m) 2 + ha];

end

ha = h*a1(N – 1); A(N – 1,N – 2:N – 1) = [2 – ha -4 + h2*a0(N – 1)];

b(N – 1) = b(N-1)-(ha+2)*xf;

x = [x0 trid(A,b)’ xf]’;

function x = trid(A,b)

% solve tridiagonal system of equations

N = size(A,2);

for m = 2:N % Upper Triangularization

tmp = A(m,m – 1)/A(m – 1,m – 1);

A(m,m) = A(m,m) -A(m – 1,m)*tmp; A(m,m – 1) = 0;

b(m,:) = b(m,:) -b(m – 1,:)*tmp;

end

x(N,:) = b(N,:)/A(N,N);

for m = N – 1: -1: 1 % Back Substitution

x(m,:) = (b(m,:) -A(m,m + 1)*x(m + 1))/A(m,m);

end

292 ORDINARY DIFFERENTIAL EQUATIONS

%do_fdf to solve BVP2 by the finite difference method

clear, clf

t0 = 1; x0 = 5; tf = 2; xf = 3; N = 100;

a1 = inline(’2./t’,’t’); a0 = inline(’-2./t./t’,’t’); u = 0; %Eq.(6.6.10)

[tt,x] = bvp2_fdf(a1,a0,u,t0,tf,x0,xf,N);

%use the MATLAB built-in command ’bvp4c()’

df = inline(’[x(2); 2./t.*(x(1)./t – x(2))]’,’t’,’x’);

fbc = inline(’[x0(1) – 5; xf(1) – 3]’,’x0’,’xf’);

solinit = bvpinit(linspace(t0,tf,5),[1 10]); %initial solution interval

sol = bvp4c(df,fbc,solinit,bvpset(’RelTol’,1e-4));

x_bvp = deval(sol,tt); xbv = x_bvp(1,:)’;

%use the symbolic computation command ’dsolve()’

xo = dsolve(’D2x + 2*(Dx – x/t)/t=0’,’x(1) = 5, x(2) = 3’)

xot = subs(xo,’t’,tt); %xot=4./tt./tt +tt; %true analytical solution

err_fd = norm(x – xot)/(N+1) %error between numerical/analytical solution

err_bvp = norm(xbv – xot)/(N + 1)

plot(tt,x,’b’,tt,xbv,’r’,tt,xot,’k’) %compare with analytical solution

We run it to get the result depicted in Fig. 6.9 and, additionally, use the

symbolic computation command “dsolve()” and “subs()” to get the analytical

solution

x(t) = t +

4

t2 (6.6.11)

and substitute the time vector into the analytical solution to obtain its numeric

values for check.

Note the following things about the shooting method and the finite difference

method:

ž While the shooting method is applicable to linear/nonlinear BVPs, the finite

difference method is suitable for linear BVPs. However, we can also apply

the finite difference method in an iterative manner to solve nonlinear BVPs

(see Problem 6.10).

1 1.2 1.4 1.6 1.8 t 2

3

3.5

4

4.5

5

x (t)

Figure 6.9 A solution of a BVP obtained by using the finite difference method.

PROBLEMS 293

ž Both methods can be modified to solve BVPs with mixed-boundary conditions

(see Problems 6.7 and 6.8).

ž In MATLAB 6.x, the “bvp4c()” command is available for solving

linear/nonlinear BVPs with mixed-boundary conditions (see Problems

6.7–6.10).

ž The symbolic computation command “dsolve()” introduced in Section

6.5.1 can be used to solve a BVP so long as the differential equation is linear,

that is, its coefficients may depend on time t , but not on the (unknown)

dependent variable x(t).

ž The usages of “bvp4c()” and “dsolve()” are illustrated in the program

“do_fdf”, where another symbolic computation command “subs()” is used

to evaluate a symbolic expression at certain value(s) of the variable.

PROBLEMS

6.0 MATLAB Commands quiver() and quiver3() and Differential Equation

(a) Usage of quiver()

Type ‘help quiver’ into the MATLAB command window, and then you

will see the following program showing you how to use the quiver()

command for plotting gradient vectors. You can also get Fig. P6.0.1 by

running the block of statements in the box below. Try it and note that

the size of the gradient vector at each point is proportional to the slope at

the point.

%do_quiver

[x,y] = meshgrid(-2:.5:2,-1:.25:1);

z = x.*exp(-x.^2 – y.^2);

[px,py] = gradient(z,.5,.25);

contour(x,y,z), hold on, quiver(x,y,px,py)

axis image %the same as AXIS EQUAL except that

%the plot box fits tightly around the data

−2 −1.5 −1 −0.5 0 0.5 1 1.5 2

1

0.5

0

−0.5

−1

Figure P6.0.1 Graphs obtained by using gradient(), contour(), quiver().

294 ORDINARY DIFFERENTIAL EQUATIONS

0.5

0

−0.5

2

1

0

−1 −2

−1

0

1

2

3

Figure P6.0.2 Graphs obtained by using surfnorm(), quiver3(), surf().

(b) Usage of quiver3()

You can obtain Fig. P6.0.2 by running the block of statements that you

see after typing ‘help quiver3’ into the MATLAB command window.

Note that the “surfnorm()” command generates normal vectors

at points specified by (x, y, z) on the surface drawn by “surf()” and

the “quiver3()” command plots the normal vectors.

%do_quiver3

clear, clf

[x,y] = meshgrid(-2:.5:2,-1:.25:1);

z = x.*exp(-x.^2 – y.^2);

surf(x,y,z), hold on

[u,v,w] = surfnorm(x,y,z);

quiver3(x,y,z,u,v,w);

(c) Gradient Vectors and One-Variable Differential Equation

We might get the meaning of the solution of a differential equation

by using the “quiver()” command, which is used in the following

program “do_ode.m” for drawing the time derivatives at grid points as

defined by the differential equation

dy(t)

dt = −y(t) + 1 with the initial condition y(0) = 0 (P6.0.1)

The slope/direction field together with the numerical solution in

Fig. P6.0.3a is obtained by running the program and it can be regarded

as a set of possible solution curve segments. Starting from the initial

point and moving along the slope vectors, you can get the solution

curve. Modify the program and run it to plot the slope/direction

PROBLEMS 295

1

0.8

0.6

0.4

0.2

0 0.5 1 1.5 2

0

×

××××××××××××××××××××××××××××××××××××××××

(a) The graph of dy vs. dt for

and its solution for y(0) = 0

y ′(t) = −y(t) + 1

1

0.5

0

−0.5

−1

0.5 1 1.5 2

× × × × × × × × × × ××××××

××

×××××××××××××××××××××××

+ +++++++++++++++++++++++++++++++++++++++++

and its solution for the initial condition

[x1(0) x2(0)] = [1 − 1] or [1.5 − 0.5]

(b) The graph of dx2 vs. dx1 for

x1′(t) = x2(t )

x2′(t) = −x2(t) + 1

Figure P6.0.3 Possible solutions of differential equation and slope/direction field.

field (x2(t) versus x1(t)) and the numerical solution for the following

differential equation as depicted in Fig. P6.0.3b.

x1(t) = x2(t)

x2(t) = −x2(t) + 1

with x1(0)

x2(0) = 1

−1 or 1.5

−0.5 (P6.0.2)

%do_ode.m

% This uses quiver() to plot possible solution curve segments

% called the slope/directional field for y’(t) + y = 1

clear, clf

t0 = 0; tf = 2; tspan = [t0 tf]; x0 = 0;

[t,y] = meshgrid(t0:(tf – t0)/10:tf,0:.1:1);

pt = ones(size(t)); py = (1 – y).*pt; %dy = (1 – y)dt

quiver(t,y,pt,py) %y(displacement) vs. t(time)

axis([t0 tf + .2 0 1.05]), hold on

dy=inline(’-y + 1’,’t’,’y’);

[tR,yR] = ode_RK4(dy,tspan,x0,40);

for k = 1:length(tR), plot(tR(k),yR(k),’rx’), pause(0.001); end

6.1 A System of Linear Time-Invariant Differential Equations: An LTI State

Equation

Consider the following state equation:

x1(t)

x2(t) = 0 1

−2 −3x1(t)

x2(t) + 0

1 us (t) with x1(0)

x2(0) = 1

0

(P6.1.1)

(a) Check the procedure and the result of obtaining the analytical solution

by using the Laplace transform technique.

296 ORDINARY DIFFERENTIAL EQUATIONS

X(s) = [sI − A]−1{x(0) + BU(s)}

=

1

s(s + 3) + 2 s +3 1

−2 s

1

0 + 1

0 1

s

=

1

(s + 1)(s + 2) s + 3 + 1/s

−2 + 1

= (s2 + 3s + 1)/s(s + 1)(s + 2)

−1/(s + 1)(s + 2)

X1(s) =

1/2

s +

1

s + 1 −

1/2

s + 2

, x1(t) =

1

2 + e−t −

1

2

e−2t (P6.1.2a)

X2(s) = −1

s + 1 +

1

s + 2

, x2(t) = −e−t + e−2t (P6.1.2b)

(b) Find the numerical solution of the above state equation by using the

routine “ode_RK4()” (with the number of segments N = 50) and the

MATLAB built-in routine “ode45()”. Compare their execution time

(by using tic and toc) and closeness to the analytical solution.

6.2 A Second-Order Linear Time-Invariant Differential Equation

Consider the following second-order differential equation

x(t) + 3x(t) + 2x(t) = 1 with x(0) = 1, x(0) = 0 (P6.2.1)

(a) Check the procedure and the result of obtaining the analytical solution

by using the Laplace transform technique.

s2X(s) − x(0) − sx(0) + 3(sX(s) − x(0)) + 2X(s) =

1

s

X(s) =

s2 + 3s + 1

s(s + 1)(s + 2)

, x(t)=

1

2 + e−t −

1

2

e−2t (P6.2.2)

(b) Define the differential equation (P6.2.1) in an M-file so that it can be

passed to the MATLAB routines like “ode_RK4()” or “ode45()” as

their input argument (see Section 6.5.1).

6.3 Ordinary Differential Equation and State Equation

(a) Van der Pol Equation

Consider a nonlinear differential equation

d2

dt2 y(t) − μ(1 − y2(t))

d

dt

y(t) + y(t) = 0 with μ = 2 (P6.3.1)

Compose a program to solve this equation with the initial condition

[y(0) y(0)] = [0.5 0] and [−1 2] for the time interval [0, 20] and plot

y(t) versus y(t) as well as y(t) and y(t) along the t-axis.

PROBLEMS 297

(b) Lorenz Equation: Turbulent Flow and Chaos

Consider a nonlinear state equation.

x1 (t) = σ(x2(t) − x1(t)) σ = 10

x2 (t) = (1 + λ − x3(t))x1(t) − x2(t) with λ = 20 ∼ 100 (P6.3.2)

x3 (t) = x1(t)x2(t) − γx3(t) γ = 2

Compose a program to solve this equation with λ = 20 and 100 for the

time interval [0,10] and plot x3(t) versus x1(t). Let the initial condition

be [x1(0) x2(0) x3(0)] = [−8 −16 80].

(c) Chemical Reactor

Consider a nonlinear state equation describing the concentrations of

two reactants and one product in the chemical process.

x1(t) = a(u1 − x1(t)) − bx1(t)x2(t) a = 5

x2(t) = a(u2 − x2(t)) − bx1(t)x2(t) with b = 2

x3(t) = −ax3(t) + bx1(t)x2(t) u1 = 3, u2 = 5

(P6.3.3)

Compose a program to solve this equation for the time interval

[0, 1] and plot x1(t), x2(t), and x3(t). Let the initial condition be

[x1(0) x2(0) x3(0)] = [1 2 3].

(d) Cantilever Beam: A Differential Equation w.r.t a Spatial Variable

Consider a nonlinear state equation describing the vertical deflection of

a beam due to its own weight

JE

d2y

dx2 = ρg 1 +

dy

dx 2

x x −

L

2 +

L2

2 (P6.3.4)

where JE = 2000 kg · m3/s2, ρ = 10 kg/m, g = 9.8 m/s2, L = 2 m.

Write a program to solve this equation for the interval [0,L] and plot

y(t). Let the initial condition be [y(0) y(0)] = [0 0]. Note that the

physical meaning of the independent variable for which we usually use

the symbol ‘t’ in writing the differential function is not a time, but

the x-coordinate of the cantilever beam along the horizontal axis in

this problem.

(e) Phase-Locked Loop (PLL)

Consider a nonlinear state equation describing the behavior of a PLL

circuit depicted in Fig. P6.3.1.

x1 (t) =

au(t) cos(x2(t)) − x1(t)

τ

with

a = 1500

τ = 0.002

u(t) = sin(ωot)

(P6.3.5a)

x2 (t) = x1(t) + ωc

y(t) = x1(t) + ωc (P6.3.5b)

298 ORDINARY DIFFERENTIAL EQUATIONS

u(t ) = sin(wot)

cos(x2 (t ))

oscillator

x2(t ) 1s

1 + τs

a

loop filter

x1(t )

wc

y (t )

Figure P6.3.1 The block diagram of PLL circuit.

where ωo = 2100π [rad/s] and ωc = 2000π [rad/s]. Compose a program

to solve this equation for the time interval [0,0.03] and plot y(t)

and ωo. Let the initial condition be [x1(0) x2(0)] = [0 0]. Is the output

y(t) tracking the frequency ωo of the input u(t)?

(f) DC Motor

Consider a linear differential equation describing the behavior of a DC

motor system (Fig. P6.3.2)

J

d2θ(t)

dt2 + B

dθ(t)

dt = T (t) = KT i(t)

L

di(t)

dt + Ri(t) + Kb

dθ(t)

dt = v(t)

(P6.3.6)

Convert this system of equations into a first-order vector differential

equation—that is, a state equation with respect to the state vector

[θ(t) θ(t) i(t)].

(g) RC Circuit: A Stiff System

Consider a two-mesh RC circuit depicted in Fig. P6.3.3. We can write

the mesh equation with respect to the two mesh currents i1(t) and

i2(t) as

vb (t )

back e.m.f. vb (t ) = Kb w(t) = Kb q'(t)

torque T (t ) = KT i (t )

v (t )

T(t )

q(t )

+

+

−

−

R L

B

i (t ) J

angular

displacement

Figure P6.3.2 A DC motor system.

PROBLEMS 299

v (t) = t [V ]

+

−

R1 = 100[Ω] C1 = 10[mF ]

C2 =

10[mF]

i1(t) R2 =

1[kΩ]

i2(t )

Figure P6.3.3 A two-mesh RC circuit.

R1i1(t) +

1

C1 t

−∞

i1(τ ) dτ + R2(i1(t) − i2(t)) = v(t) = t

R2(i2(t) − i1(t)) +

1

C2 t

−∞

i2(τ ) dτ = 0 (P6.3.7a)

In order to convert this system of differential equations into a state

equation, we differentiate both sides and rearrange them to get

(R1 + R2)

di1(t)

dt − R2

di2(t)

dt +

1

C1

i1(t) =

dv(t)

dt = 1

−R2

di1(t)

dt + R2

di2(t)

dt +

1

C2

i2(t) = 0 (P6.3.7b)

R1 + R2 −R2

−R2 R2 i1(t)

i2(t) = 1 − i1(t)/C1

−i2(t)/C2 (P6.3.7c)

i1 (t)

i2 (t) = R1 + R2 −R2

−R2 R2 −1 1 − i1(t)/C1

−i2(t)/C2 with Gi = 1/Ri

= −G1/C1 −G1/C2

−G1/C1 −(G1 + G2)/C2 i1(t)

i2(t) + G1

G1 us (t)

(P6.3.7d)

where us (t) denotes the unit step function whose value is 1 (one) ∀ t ≥ 0.

(i) After constructing an M-file function “df6p03g.m” which defines

Eq. (P6.3.7d) with R1 = 100[], C1 = 10[μF], R2 = 1[k], C2 = 10[μF], use the MATLAB built-in routines “ode45()” and

“ode23s()” to solve the state equation with the zero initial condition

i1(0) = i2(0) = 0 and plot the numerical solution i2(t) for

0 ≤ t ≤ 0.05 s. For possible change of parameters, you may declare

R1, C1, R2, C2 as global variables both in the function and in the

main program named, say, “nm6p03g.m”. Do you see any symptom

of stiffness from the results?

300 ORDINARY DIFFERENTIAL EQUATIONS

(ii) If we apply the Laplace transform technique to solve this equation

with zero initial condition i(0) = 0, we can get

I1(s)

I2(s) (6.5.5) = [sI − A]−1Bu(s)

= s + G1/C1 G1/C2

G1/C2 s + (G1 + G2)/C2 −1 G1

G1 1

s

I2(s) =

G1

s2 + (G1/C1 + (G1 + G2)/C2)s + G1G2/C1C2

=

1/100

s2 + 2100s + 100000

∼=

1/100

(s + 2051.25)(s + 48.75)

∼=

1

200250 1

s + 48.75 −

1

s + 2051.25

i2(t)∼=

1

200250

(e−48.75t − e−2051.25t ) (P6.3.7e)

where λ1 = −2051.25 and λ2 = −48.75 are actually the eigenvalues

of the system matrix A in Eq. (P6.3.7d). Find the measure of

stiffness defined by Eq. (6.5.26).

(iii) Using the MATLAB symbolic computation command “dsolve()”,

find the analytical solution of the differential equation (P6.3.7b) and

plot i2(t) together with (P6.3.7e) for 0 ≤ t ≤ 0.05 s. Which of the

two numerical solutions obtained in (i) is better? You may refer to

the following code:

syms R1 R2 C1 C2

i = dsolve(’(R1+R2)*Di1 – R2*Di2 + i1/C1 = 1’,…

’-R2*Di1 + R2*Di2 + i2/C2’,’i1(0) = 0’,’i2(0) = 0’); %(P6.3.7b)

R1 = 100; R2 = 1000; C1 = 1e-5; C2 = 1e-5;

t0 = 0; tf = 0.05; t = t0+(tf-t0)/100*[0:100];

i2t = eval(i.i2); plot(t,i2t,’m’)

6.4 Physical Meaning of a Solution for Differential Equation and Its Animation

Suppose we are going to simulate how a vehicle vibrates when it moves

with a constant speed on a rugged way, as depicted in Fig. P6.4a. Based on

Newton’s second law, the situation is modeled by the differential equation

(P6.4.1).

M

d2

dt2 y(t) + B

d

dt

(y(t) − u(t)) + K(y(t) − u(t)) = 0 (P6.4.1)

with y(0) = 0, y(0) = 0

PROBLEMS 301

%do_MBK

clf

t0 = 0; tf = 10; x0 = [0 0];

[t1,x] = ode_Ham(’f_MBK’,[t0 tf],x0);

dt = t1(2) – t1(1);

for n = 1:length(t1)

u(n) = udu_MBK(t1(n));

end

figure(1), clf

animation = 1;

if animation

figure(2), clf

draw_MBK(5,1,x(1,2),u(1))

axis([-2 2 -1 14]), axis(’equal’)

pause

for n = 1:length(t1)

clf, draw_MBK(5,1,x(n,2),u(n),’b’)

axis([-2 2 -1 14]), axis(’equal’)

pause(dt)

figure(1)

plot(t1(n),u(n),’r.’, t1(n),x(n,2),’b.’)

axis([0 tf -0.2 1.2]), hold on

figure(2)

end

draw_MBK(5,1,x(n,2),u(n))

axis([-2 2 -1 14]), axis(’equal’)

end

function [u,du] = udu_MBK(t)

i = fix(t);

if mod(i,2) == 0, u = t-i; du = 1;

else u = 1 – t + i; du = -1;

end

function draw_MBK(n,w,y,u,color)

%n: the # of spring windings

%w: the width of each object

%y: displacement of the top of MBK

%u: displacement of the bottom of MBK

if nargin < 5, color = ’k’; end

p1 = [-w u + 4]; p2 = [-w 9 + y];

xm = 0; ym = (p1(2) + p2(2))/2;

xM = xm + w*1.2*[-1 -1 1 1 -1];

yM = p2(2) + w*[1 3 3 1 1];

plot(xM,yM,color), hold on %Mass

spring(n,p1,p2,w,color) %Spring

damper(xm + w,p1(2),p2(2),w,color) %Damper

wheel_my(xm,p1(2)- 3*w,w,color) %Wheel

function dx = f_MBK(t,x)

M = 1; B = 0.1; K = 0.1;

[u,du] = udu_MBK(t);

dx = x*[0 1; -B/M – K/M]’+[0 (K*u + B*du)/M];

302 ORDINARY DIFFERENTIAL EQUATIONS

function spring(n,p1,p2,w,color)

%draw a spring of n windings, width w from p1 to p2

if nargin < 5, color = ’k’; end

c = (p2(1) – p1(1))/2; d = (p2(2) – p1(2))/2;

f = (p2(1) + p1(1))/2; g = (p2(2) + p1(2))/2;

y = -1:0.01:1; t = (y+1)*pi*(n + 0.5);

x = -0.5*w*sin(t); y = y+0.15*(1 – cos(t));

a = y(1); b=y(length(x));

y = 2*(y – a)/(b – a)-1;

yyS = d*y – c*x + g; xxS = x+f; xxS1 = [f f];

yyS1 = yyS(length(yyS))+[0 w]; yyS2 = yyS(1)-[0 w];

plot(xxS,yyS,color, xxS1,yyS1,color, xxS1,yyS2,color)

function damper(xm,y1,y2,w,color)

%draws a damper in (xm-0.5 xm + 0.5 y1 y2)

if nargin < 5, color = ’k’; end

ym = (y1 + y2)/2;

xD1 = xm + w*[0.3*[0 0 -1 1]]; yD1 = [y2 + w ym ym ym];

xD2 = xm + w*[0.5*[-1 -1 1 1]]; yD2 = ym + w*[1 -1 -1

1];

xD3 = xm + [0 0]; yD3 = [y1 ym] – w;

plot(xD1,yD1,color, xD2,yD2,color, xD3,yD3,color)

function wheel_my(xm,ym,w,color)

%draws a wheel of size w at center (xm,ym)

if nargin < 5, color = ’k’; end

xW1 = xm + w*1.2*[-1 1]; yW1 = ym + w*[2 2];

xW2 = xm*[1 1]; yW2 = ym + w*[2 0];

plot(xW1,yW1,color, xW2,yW2,color)

th = [0:100]/50*pi; plot(xm + j*ym+w*exp(j*th),color)

y(t)

y(t) u(t)

u(t)

1.2

1

0.8

0.6

0.4

0.2

−0.2

0

0 2 4 6 8 10

M

K

(a) The block diagram (b) The graphs of the input u(t) and the output y(t)

B

Figure P6.4 A mass–spring–damper system.

PROBLEMS 303

where the values of the mass, the viscous friction coefficient, and the spring

constant are given as M = 1 kg, B = 0.1 N s/m, and K = 0.1 N/m, respectively.

The input to this system is the movement u(t) of the wheel part

causing the movement y(t) of the body as the output of the system and is

approximated to a triangular wave of height 1 m, duration 1 s, and period

2 s as depicted in Fig. P6.4b. After converting this equation into a state

equation as

x1(t)

x2(t) = 0 1

−K/M −B/Mx1(t)

x2(t) + 0

(B/M)u(t) + (K/M)u(t)

(P6.4.2)

with x1(0)

x2(0) = 0

0

we can use such routines as ode_Ham(), ode45(), . . . to solve this state

equation and use some graphic functions to draw not only the graphs of

y(t) and u(t), but also the animated simulation diagram. You can run

the above MATLAB program “do_MBK.m” to see the results. Does the

suspension system made of a spring and a damper as depicted in Fig.

P6.4a absorb effectively the shock caused by the rolling wheel so that the

amplitude of vehicle body oscillation is less than 1/5 times that of wheel

oscillation?

(cf) If one is interested in graphic visualization with MATLAB, he/she can refer to

[N-1].

6.5 A Nonlinear Differential Equation for an Orbit of a Satellite

Consider the problem of an orbit of a satellite, whose position and velocity

are obtained as the solution of the following state equation:

x1 (t) = x3(t)

x2 (t) = x4(t)

x3 (t) = −GMEx1(t)/(x2

1 (t) + x2

2 (t))3/2 (P6.5.1)

x4 (t) = −GMEx2(t)/(x2

1 (t) + x2

2 (t))3/2

where G = 6.672 × 10−11 N m2/kg2 is the gravitational constant, and

ME = 5.97 × 1024 kg is the mass of the earth. Note that (x1, x2) and (x3, x4)

denote the position and velocity, respectively, of the satellite on the plane

having the earth at its origin. This state equation is defined in the M-file

‘df_sat.m’ below.

(a) Supplement the following program “nm6p05.m” which uses the three

routines ode_RK4(), ode45(), and ode23() to find the paths of the

satellite with the following initial positions/velocities for one day.

304 ORDINARY DIFFERENTIAL EQUATIONS

function dx = df_sat(t,x)

global G Me Re

dx = zeros(size(x));

r = sqrt(sum(x(1:2).^2));

if r <= Re, return; end % when colliding against the earth surface

GMr3 = G*Me/r^3;

dx(1) = x(3); dx(2) = x(4); dx(3) = -GMr3*x(1); dx(4) = -GMr3*x(2);

%nm6p05.m to solve a nonlinear d.e. on the orbit of a satellite

clear, clf

global G Me Re

G = 6.67e-11; Me = 5.97e24; Re = 64e5;

f = ’df_sat’; ;

t0 = 0; T = 24*60*60; tf = T; N = 2000;

R = 4.223e7;

v20s = [3071 3500 2000];

for iter = 1:length(v20s)

x10 = R; x20 = 0; v10 = 0; v20 = v20s(iter);

x0 = [x10 x20 v10 v20]; tol = 1e-6;

[tR,xR] = ode_RK4(f,[t0 tf],x0,N);

[t45,x45] = ode45(????????????);

[t23s,x23s] = ode23s(f,[t0 tf],x0);

plot(xR(:,1),xR(:,2),’b’, x45(:,1),x45(:,2),’k.’, ????????????)

[t45,x45] = ode45(f,[t0 tf],x0,odeset(’RelTol’,tol));

[t23s,x23s] = ode23s(?????????????????????????????????);

plot(xR(:,1),xR(:,2),’b’, x45(:,1),x45(:,2),’k.’, ????????????)

end

(i) (x10, x20) = (4.223 × 107, 0)[m] and (x30, x40) = (v10, v20) = (0, 3071)[m/s].

(ii) (x10, x20) = (4.223 × 107, 0)[m] and (x30, x40) = (v10, v20) = (0, 3500)[m/s].

(iii) (x10, x20) = (4.223 × 107, 0)[m] and (x30, x40) = (v10, v20) = (0, 2000)[m/s].

Run the program and check if the plotting results are as depicted in

Fig. P6.5.

(b) In Fig. P6.5, we see that the “ode23s()” solution path differs from

the others for case (ii) and the “ode45()” and “ode23s()” paths differ

from the “ode_RK4()” path for case (iii). But, we do not know which

one is more accurate. In order to find which one is the closest to the

true solution, apply the two routines “ode45()” and “ode23s()” with

smaller relative error tolerance of tol = 1e-6 to find the paths for the

three cases. Which one do you think is the closest to the true solution

among the paths obtained in (a)?

(cf) The purpose of this problem is not to compare the several MATLAB routines,

but to warn the users of the danger of abusing them. With smaller number

of steps (N) (i.e., larger step size), the routine “ode_RK4()” will also deviate

much from the true solution. The MATLAB built-in routines have too many

good features to be mentioned here. Note that setting the parameters such as

PROBLEMS 305

×

× × × ×

× × × × × × ×

×

×

×

×× ×× ×× ×× ×× ×× ××

××

××

××

××

××

××

×

×

×××××××××××××× × ×× × × × × ×× × ×× ××

××

××

××

×

×

×

××

××

××

××

××××

×× ×× ×× ×× ×× ××××××

××

××

××

××

××××××

××××

×

×

×

×

×××

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

× × × × × × × × × ×

×

×

×

×

×

×

××

× ×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

× × × × × × × ×

×

×

×

×

×

×

×

×

6

x2

4

2

−2

0

−4

ode_RK4( )

ode45( )

ode23s( )

−8 −6 −4 −2 0 2 x1 4

× 107

× 107

Earth

Figure P6.5 The paths of a satellite with the same initial position and different initial velocities.

the relative error tolerance (RelTol) is sometimes very important for obtaining

a reasonably accurate solution.

6.6 Shooting Method for BVP with Adjustable Position and Fixed Angle

Suppose the boundary condition for a second-order BVP is given as

x(t0) = x20, x(tf ) = x1f (P6.6.1)

Consider how to modify the MATLAB routines “bvp2_shoot()” and

“bvp2_fdf()” so that they can accommodate this kind of problem.

(a) As for “bvp2_shootp()” that you should make, the variable quantity

to adjust for improving the approximate solution is not the derivative

x(t0), but the position x(t0) and what should be made close to zero is

still f (x(t0)) = x(tf ) − xf . Modify the routine in such a way that x(t0)

is adjusted to make this quantity close to zero and make its declaration

part have the initial derivative (dx0) instead of the initial position (x0)

as the fourth input argument as follows.

function [t,x] = bvp2_shootp(f,t0,tf,dx0,xf,N,tol,kmax)

306 ORDINARY DIFFERENTIAL EQUATIONS

Noting that the initial derivative of the true solution for Eq. (6.6.3) is

zero, apply this routine to solve the BVP by inserting the following

statement into the program “do_shoot.m”.

[t,x1] = bvp2_shootp(’df661’,t0,tf,0,xf,N,tol,kmax);

and plot the result to check if it conforms with that (Fig. 6.8) obtained

by “bvp2_shoot()”.

(b) As for “bvp2_fdfp()” implementing the finite difference method, you

have to approximate the boundary condition as

x(t0) = x20 →

x1 − x−1

2h = x20, x−1 = x1 − 2hx20, (P6.6.2)

substitute this into the finite difference equation corresponding to the

initial time as

x1 − 2×0 + x−1

h2 + a10

x1 − x−1

2h + a00x0 = u0 (P6.6.3)

x1 − 2×0 + x1 − 2hx20

h2 + a10x20 + a00x0 = u0

(a00h2 − 2)x0 + 2×1 = h2u0 + h(2 − ha10)x20 (P6.6.4)

and augment the matrix–vector equation with this equation. Also, make

its declaration part have the initial derivative (dx0) instead of the initial

position (x0) as the sixth input argument as follows:

function [t,x] = bvp2_fdfp(a1,a0,u,t0,tf,dx0,xf,N)

Noting that the initial derivative of the true solution for Eq. (6.6.10)

is −7, apply this routine to solve the BVP by inserting the following

statement into the program “do_fdf.m”.

[t,x1] = bvp2_fdfp(a1,a0,u,t0,tf,-7,xf,N);

and plot the result to check if it conforms with that obtained by using

“bvp2_fdf()” and depicted in Fig. 6.9.

6.7 BVP with Mixed-Boundary Conditions I

Suppose the boundary condition for a second-order BVP is given as

x(t0) = x10, c1x(tf ) + c2x(tf ) = c3 (P6.7.1)

Consider how to modify the MATLAB routines “bvp2_shoot()” and

“bvp2_fdf()” so that they can accommodate this kind of problem.

PROBLEMS 307

(a) As for “bvp2_shoot()” that you should modify, the variable quantity

to adjust for improving the approximate solution is still the derivative

x(t0), but what should be made close to zero is

f (x(t0)) = c1x(tf ) + c2x(tf ) − c3 (P6.7.2)

If you don’t know where to begin, modify the routine

“bvp2_shoot()” in such a way that x(t0) is adjusted to make this

quantity close to zero. Regarding the quantity (P6.7.2) as a function of

x(t0), you may feel as if you were going to solve a nonlinear equation

f (x(t0)) = 0. Here are a few hints for this job:

ž Make the declaration part have the boundary coefficient vector cf

= [c1 c2 c3] instead of the final position (xf) as the fifth input

argument as follows.

function [t,x] = bvp2m_shoot(f,t0,tf,x0,cf,N,tol,kmax)

ž Pick up the first two guesses of x(t0) arbitrarily.

ž You may need to replace a couple of statements in “bvp2_shoot()” by

e(1) = cf*[x(end,:)’;-1];

e(k) = cf*[x(end,:)’;-1];

Now that you have the routine “bvp2m_shoot()” of your own making,

don’t hesitate to try using the weapon to attack the following

problem:

x(t) − 4t x(t)x(t) + 2×2(t) = 0 with x(0) =

1

4

, 2x(1) − 3x(1) = 0

(P6.7.3)

For this job, you only have to modify one statement of the program

“do_shoot” (Section 6.6.1) into

[t,x] = bvp2m_shoot(’df661’,t0,tf,x0,[2 -3 0],N,tol,kmax);

If you run it to obtain the same solution as depicted in Fig. 6.8, you

deserve to be proud of yourself having this book as well as MATLAB;

otherwise, just keep trying until you succeed.

(b) As for “bvp2_fdf()” that you should modify, you have only to augment

the matrix–vector equation with one row corresponding to the

approximate version of the boundary condition c1x(tf ) + c2x(tf ) = c3,

that is,

c1xN + c2

xN − xN−1

h = c3; −c2xN−1 + (c1h + c2)xN = c3h (P6.7.4)

308 ORDINARY DIFFERENTIAL EQUATIONS

Needless to say, you should increase the dimension of the matrix A

to N and move the xN term on the right-hand side of the (N − 1)th row

back to the left-hand side by incorporating the corresponding statement

into the for loop. What you have to do with “bvp2m_fdf()” for this

job is as follows:

ž Make the declaration part have the boundary coefficient vector cf =

[c1 c2 c3] instead of the final position (xf) as the seventh input

argument.

function [t,x] = bvp2m_fdf(a1,a0,u,t0,tf,x0,cf,N)

ž Replace some statement by A = zeros(N,N).

ž Increase the last index of the for loop to N-1.

ž Replace the statements corresponding to the (N − 1)th row

equation by

A(N,N-1:N) = [-cf(2) cf(1)*h + cf(2)]; b(N) = cf(3)*h;

which implements Eq. (P6.7.4).

ž Modify the last statement arranging the solution as

x = [x0 trid(A,b)’]’;

Now that you have the routine “bvp2m_fdf()” of your own making,

don’t hesitate to try it on the following problem:

x(t) +

2

t

x(t) −

2

t2 x(t) = 0 with x(1) = 5, x(2) + x(2) = 3

(P6.7.5)

For this job, you only have to modify one statement of the program

“do_fdf.m” (Section 6.6.2) into

[t,x] = bvp2m_fdf(a1,a0,u,t0,tf,x0,[1 1 3],N);

You might need to increase the number of segments N to improve the

accuracy of the numerical solution. If you run it to obtain the same

solution as depicted in Fig. 6.9, be happy with it.

6.8 BVP with Mixed-Boundary Conditions II

Suppose the boundary condition for a second-order BVP is given as

c01x(t0) + c02x(t0) = c03 (P6.8.1a)

cf 1x(tf ) + cf 2x(tf ) = cf 3 (P6.8.1b)

Consider how to modify the MATLAB routines “bvp2m_shoot()” and

“bvp2m_fdf()” so that they can accommodate this kind of problems.

PROBLEMS 309

(a) As for “bvp2mm_shoot()” that you should make, the variable quantity

to be adjusted for improving the approximate solution is x(t0) or x(t0)

depending on whether or not c01 = 0, while the quantity to be made

close to zero is still

f (x(t0), x(t0)) = cf 1x(tf ) + cf 2x(tf ) − cf 3 (P6.8.2)

If you don’t have your own idea, modify the routine “bvp2m_shoot()”

in such a way that x(t0) or x(t0) is adjusted to make this quantity

close to zero and x(t0) or x(t0) is set by (P6.8.1a), making its declaration

as

function [t,x] = bvp2mm_shoot(f,t0,tf,c0,cf,N,tol,kmax)

where the boundary coefficient vectors c0 = [c01 c02 c03] and cf =

[cf1 cf2 cf3] are supposed to be given as the fourth and fifth input

arguments, respectively.

Now that you get the routine “bvp2mm_shoot()” of your own making,

try it on the following problem:

x(t) −

2t

t2 + 1

x(t) +

2

t2 + 1

x(t) = t2 + 1 (P6.8.3)

with x(0) + 6x(0) = 0, x(1) + x(1) = 0

(b) As for “bvp2_fdf()” implementing the finite difference method, you

only have to augment the matrix–vector equation with two rows

corresponding to the approximate versions of the boundary conditions

c01x(t0) + c02x(t0) = c03 and cf 1x(tf ) + cf 2x(tf ) = cf 3, that is,

c01x0 + c02

x1 − x0

h = c03, (c01h − c02)x0 + c02x1 = c03h

(P6.8.4a)

cf 1xN + cf 2

xN − xN−1

h = cf 3; −cf 2xN−1 + (cf 1h + cf 2)xN = cf 3h

(P6.8.4b)

Now that you have the routine “bvp2mm_fdf()” of your own making,

try it on the problem described by Eq. (P6.8.3).

(c) Overall, you will need to make the main programs like “nm6p08a.m”

and “nm6p08b.m” that apply the routines “bvp2mm_shoot()” and

“bvp2mm_fdf()” to get the numerical solutions of Eq. (P6.8.3) and

plot them. Additionally, use the MATLAB routine “bvp4c()” to get

another solution and plot it together for cross-check.

6.9 Shooting Method and Finite Difference Method for Linear BVPs

Apply the routines “bvp2_shoot()”, “bvp2_fdf()”, and “bvp4c()” to

solve the following BVPs.

310 ORDINARY DIFFERENTIAL EQUATIONS

%nm6p08a.m: to solve BVP2 with mixed boundary conditions

%x” = (2t/t^2 + 1)*x’ -2/(t^2+1)*x +t^2+1

% with x(0)+6x’(0) = 0, x’(1) + x(1) = 0

%shooting method

f = inline(’[x(2); 2*(t*x(2) – x(1))./(t.^2 + 1)+(t.^2 + 1)]’,’t’,’x’);

t0 = 0; tf = 1; N = 100; tol = 1e-8; kmax = 10;

c0 = [1 6 0]; cf = [1 1 0]; %coefficient vectors of boundary condition

[tt,x_sh] = bvp2mm_shoot(f,t0,tf,c0,cf,N,tol,kmax);

plot(tt,x_sh(:,1),’b’)

%nm6p08b.m: finite difference method

a1 = inline(’-2*t./(t.^2+1)’,’t’); a0 = inline(’2./(t.^2+1)’,’t’);

u = inline(’t.^2+1’,’t’);

t0 = 0; tf = 1; N = 500;

c0 = [1 6 0]; cf = [1 1 0]; %coefficient vectors of boundary condition

[tt,x_fd] = bvp2mm_fdf(a1,a0,u,t0,tf,c0,cf,N);

plot(tt,x_fd,’r’)

y(x) = f (y(x), y(x), u(x)) with y(x0) = y0, y(xf ) = yf (P6.9.0a)

Plot the solutions and fill in Table P6.9 with the mismatching errors (of the

numerical solutions) that are defined as

function err = err_of_sol_de(df,t,x,varargin)

% evaluate the error of solutions of differential equation

[Nt,Nx] = size(x); if Nt < Nx, x = x.’; [Nt,Nx] = size(x); end

n1 = 2:Nt – 1; t=t(:); h2s = t(n1 + 1)-t(n1-1);

dx = (x(n1 + 1,:) – x(n1 – 1,:))./(h2s*ones(1,Nx));

num = x(n1 + 1,:)-2*x(n1,:) + x(n1 – 1,:); den = (h2s/2).^2*ones(1,Nx);

d2x = num./den;

for m = 1:Nx

for n = n1(1):n1(end)

dfx = feval(df,t(n),[x(n,m) dx(n – 1,m)],varargin{:});

errm(n – 1,m) = d2x(n – 1,m) – dfx(end);

end

end

err=sum(errm.^2)/(Nt – 2);

%nm6p09_1.m

%y”-y’+y = 3*e^2t-2sin(t) with y(0) = 5 & y(2)=-10

t0 = 0; tf = 2; y0 = 5; yf = -10; N = 100; tol = 1e-6; kmax = 10;

df = inline(’[y(2); y(2) – y(1)+3*exp(2*t)-2*sin(t)]’,’t’,’y’);

a1 = -1; a0 = 1; u = inline(’3*exp(2*t) – 2*sin(t)’,’t’);

solinit = bvpinit(linspace(t0,tf,5),[-10 5]); %[1 9]

fbc = inline(’[y0(1) – 5; yf(1) + 10]’,’y0’,’yf’);

% Shooting method

tic, [tt,y_sh] = bvp2_shoot(df,t0,tf,y0,yf,N,tol,kmax); times(1) = toc;

% Finite difference method

tic, [tt,y_fd] = bvp2_fdf(a1,a0,u,t0,tf,y0,yf,N); times(2) = toc;

% MATLAB built-in function bvp4c

sol = bvp4c(df,fbc,solinit,bvpset(’RelTol’,1e-6));

tic, y_bvp = deval(sol,tt); times(3) = toc

% Eror evaluation

ys=[y_sh(:,1) y_fd y_bvp(1,:)’]; plot(tt,ys)

err=err_of_sol_de(df,tt,ys)

PROBLEMS 311

Table P6.9 Comparison of the BVP Solver Routines bvp2 shoot()/bvp2 fdf()

BVP Routine

Mismatching Error

(P6.9.0b) Times

(P6.9.1)

bvp2 shoot() 1.5 × 10−6

N = 100, tol = 1e-6, bvp2 fdf()

kmax = 10

bvp4c() 2.9 × 10−6

(P6.9.2)

bvp2 shoot()

N = 100, tol = 1e-6, bvp2 fdf() 1.6 × 10−23

kmax = 10

bvp4c()

(P6.9.3)

bvp2 shoot() 1.7 × 10−17

N = 100, tol = 1e-6, bvp2 fdf()

kmax = 10

bvp4c() 7.8 × 10−14

(P6.9.4)

bvp2 shoot()

N = 100, tol = 1e-6, bvp2 fdf() 4.4 × 10−27

kmax = 10

bvp4c()

(P6.9.5)

bvp2 shoot() 8.9 × 10−9

N = 100, tol = 1e-6, bvp2 fdf()

kmax = 10

bvp4c() 8.9 × 10−7

(P6.9.6)

bvp2 shoot()

N = 100, tol =1e-6, bvp2 fdf() 4.4 × 10−25

kmax =10

bvp4c()

err =

1

N − 1

N−1

i=1

{D(2)y(xi ) − f (Dy(xi ), y(xi ), u(xi ))}2 (P6.9.0b)

with

D(2)y(xi ) =

y(xi+1) − 2y(xi ) + y(xi−1)

h2 , Dy(xi ) =

y(xi+1) − y(xi−1)

2h

(P6.9.0c)

xi = x0 + ih, h =

xf − x0

N

(P6.9.0d)

and can be computed by using the following routine “err_of_sol_de()”.

312 ORDINARY DIFFERENTIAL EQUATIONS

Overall, which routine works the best for linear BVPs among the three

routines?

(a) y(x) = y(x) − y(x) + 3e2x − 2 sin x with y(0) = 5, y(2) = −10

(P6.9.1)

(b) y(x) = −4y(x) with y(0) = 5, y(1) = −5 (P6.9.2)

(c) y(t) = 10−6y(t) + 10−7(t2 − 50t) with y(0) = 0, y(50) = 0 (P6.9.3)

(d) y(t) = −2y(t) + sin t with y(0) = 0, y(1) = 0 (P6.9.4)

(e) y(x) = y(x) + y(x) + ex(1 − 2x) with y(0) = 1, y(1) = 3e (P6.9.5)

(f) d2y(r)

dr2 +

1

r

dy(r)

dr = 0 with y(1) = ln 1, y(2) = ln 2 (P6.9.6)

6.10 Shooting Method and Finite Difference Method for Nonlinear BVPs

(a) Consider a nonlinear boundary value problem of solving

d2T

dx2 = 1.9 × 10−9(T 4 − T 4

a ), Ta = 400 (P6.10.1)

with the boundary condition T (x0) = T0, T(xf

) = Tf

to find the temperature distribution T (x) [◦K] in a rod 4 m long, where

[x0, xf ] = [0, 4].

Apply the routines “bvp2_shoot()”, “bvp2_fdf()”, and “bvp4c()”

to solve this differential equation for the two sets of boundary conditions

{T (0) = 500, T(4) = 300} and {T (0) = 550, T(4) = 300} as listed in

Table P6.10. Fill in the table with the mismatching errors defined by

Eq. (P6.9.0b) for the three numerical solutions

%nm6p10a

clear, clf

K = 1.9e-9; Ta = 400; Ta4 = Ta^4;

df = inline(’[T(2); 1.9e-9*(T(1).^4-256e8)]’,’t’,’T’);

x0 = 0; xf = 4; T0 = 500; Tf = 300; N = 500; tol = 1e-5; kmax = 10;

% Shooting method

[xx,T_sh] = bvp2_shoot(df,x0,xf,T0,Tf,N,tol,kmax);

% Iterative finite difference method

a1 = 0; a0 = 0; u = T0 + [1:N – 1]*(Tf – T0)/N;

for i = 1:100

[xx,T_fd] = bvp2_fdf(a1,a0,u,x0,xf,T0,Tf,N);

u = K*(T_fd(2:N).^4 – Ta4); %RHS of (P6.10.1)

if i > 1 & norm(T_fd – T_fd0)/norm(T_fd0) < tol, i, break; end

T_fd0 = T_fd;

end

% MATLAB built-in function bvp4c

solinit = bvpinit(linspace(x0,xf,5),[Tf T0]);

fbc = inline(’[Ta(1)-500; Tb(1)-300]’,’Ta’,’Tb’);

tic, sol = bvp4c(df,fbc,solinit,bvpset(’RelTol’,1e-6));

T_bvp = deval(sol,xx); time_bvp = toc;

% The set of three solutions

Ts = [T_sh(:,1) T_fd T_bvp(1,:)’];

% Evaluates the errors and plot the graphs of the solutions

err = err_of_sol_de(df,xx,ys)

subplot(321), plot(xx,Ts)

PROBLEMS 313

Table P6.10 Comparison of the BVP routines bvp2 shoot()/bvp2 fdf()

Boundary Condition Routine

Mismatching

Error (P6.9.0b)

Time

(seconds)

(P6.10.1) with Ta = 400

bvp2 shoot()

T (0) = 500, T (4) = 300

bvp2 fdf() 3.6 × 10−6

bvp4c()

(P6.10.1) with Ta = 400

bvp2 shoot() NaN (divergent) N/A

T (0) = 550, T (4) = 300

bvp2 fdf()

bvp4c() 30 × 10−5

(P6.10.2) with

bvp2 shoot()

y(0) = 0, y(1) = 0

bvp2 fdf() 3.2 × 10−13

bvp4c()

(P6.10.3) with

bvp2 shoot() NaN (divergent) N/A

y(1) = 4, y(2) = 8

bvp2 fdf()

bvp4c() 3.5 × 10−6

(P6.10.4) with

bvp2 shoot()

y(1) = 1/3, y(4) = 20/3

bvp4c() 3.4 × 10−10

bvp2 fdf(c)

(P6.10.5) with

bvp2 shoot() 3.7 × 10−14

y(0) = π/2, y(2) = π/4

bvp2 fdf()

bvp4c() 2.2 × 10−9

(P6.10-6) with

bvp2 shoot()

y(2) = 2,y(8) = 1/4

bvp2 fdf() 5.0 × 10−14

bvp4c()

{T (xi ), i = 0 : N} (xi = x0 + ih = x0 + i

xf − x0

N

with N = 500

Note that the routine “bvp2_fdf()” should be applied in an iterative

way to solve a nonlinear BVP, because it has been fabricated to accommodate

only linear BVPs. You may start with the following program

“nm6p10a.m”. Which routine works the best for the first case and the

second case, respectively?

314 ORDINARY DIFFERENTIAL EQUATIONS

(b) Apply the routines “bvp2_shoot()”, “bvp2_fdf()”, and “bvp4c()” to

solve the following BVPs. Fill in Table P6.10 with the mismatching

errors defined by Eq. (P6.9.0b) for the three numerical solutions and

plot the solution graphs if they are reasonable solutions.

(i) y − ey = 0 with y(0) = 0, y(1) = 0 (P6.10.2)

(ii) y −

1

t

y −

2

y

(y)2 = 0 with y(1) = 4, y(2) = 8 (P6.10.3)

(iii) y −

2

y + 1 = 0 with y(1) =

1

3

, y(4) =

20

3

(P6.10.4)

(iv) y = t (y)2 with y(0) = π/2, y(2) = π/4 (P6.10.5)

(v) y +

1

y2 y = 0 with y(2) = 2, y(8) = 1/4 (P6.10.6)

Especially for the BVP (P6.10.6), the routine “bvp2m_shoot()”

or “bvp2mm_shoot()” developed in Problems 6.7 and 6.8 should be

used instead of “bvp2_shoot()”, since it has a mixed-boundary condition

I.

(cf) Originally, the shooting method was developed for solving nonlinear BVPs,

while the finite difference method is designed as a one-shot method for solving

linear BVPs. But the finite difference method can also be applied in an

iterative way to handle nonlinear BVPs, producing more accurate solutions in

less computation time.

6.11 Eigenvalue BVPs

(a) A Homogeneous Second-Order BVP to an Eigenvalue Problem

Consider an eigenvalue boundary value problem of solving

y(x) + ω2y = 0 (P6.11.1)

with c01y(x0) + c02y(x0) = 0, cf 1y(xf ) + cf 2y(xf ) = 0

to find y(x) for x ∈ [x0, xf ] with the (possible) angular frequency ω.

In order to use the finite difference method, we divide the solution

interval [x0, xf ] into N subintervals to have the grid points xi = x0 + ih = x0 + i(xf − x0)/N and then, replace the derivatives in the

differential equation and the boundary conditions by their finite difference

approximations (5.3.1) and (5.1.8) to write

yi−1 − 2yi + yi+1

h2 + ω2yi = 0

yi−1 − (2 − λ)yi + yi+1 = 0 with λ = h2 ω2 (P6.11.2)

PROBLEMS 315

with

c01y0 + c02

y1 − y−1

2h = 0 → y−1 = 2h

c01

c02

y0 + y1 (P6.11.3a)

cf 1yN + cf 2

yN+1 − yN−1

2h = 0 → yN+1 = yN−1 − 2h

cf 1

cf 2

yN

(P6.11.3b)

Substituting the discretized boundary condition (P6.11.3) into (P6.11.2)

yields

y−1 − 2y0 + y1 = −λy0

(P6.11.3a)

−−−−−→

2 − 2h

c01

c02y0 − 2y1 = λy0 (P6.11.4a)

yi−1 − 2yi + yi+1 = −λyi →−yi−1 + 2yi − yi+1 = λyi

for i = 1 : N − 1 (P6.11.4b)

yN−1 − 2yN + yN+1 = −λyN

(P6.11.3b)

−−−−−→

− 2yN−1 + 2 + 2h

cf 1

cf 2yN = λyN (P6.11.4c)

which can be formulated in a compact form as

2 − 2hc01/c02 −2 0 0 0

−1 2 −1 0 0

0 −1 2 −1 0

0 0 −1 2 −1

0 0 0 −2 2+ 2hcf 1/cf 2

y0

y1

·

yN−1

yN

= λ

y0

y1

·

yN−1

yN

Ay = λy; [A − λI]y = 0 (P6.11.5)

For this equation to have a nontrivial solution y = 0, λ must be one of

the eigenvalues of the matrix A and the corresponding eigenvectors are

possible solutions.

316 ORDINARY DIFFERENTIAL EQUATIONS

function [x,Y,ws,eigvals] = bvp2_eig(x0,xf,c0,cf,N)

% use the finite difference method to solve an eigenvalue BVP4:

% y”+w^2*y = 0 with c01y(x0) + c02y’(x0) = 0, cf1y(xf) + cf2y’(xf) = 0

%input: x0/xf = the initial/final boundaries

% c0/cf = the initial/final boundary condition coefficients

% N – 1 = the number of internal grid points.

%output: x = the vector of grid points

% Y = the matrix composed of the eigenvector solutions

% ws = angular frequencies corresponding to eigenvalues

% eigvals = the eigenvalues

if nargin < 5|N < 3, N = 3; end

h = (xf – x0)/N; h2 = h*h; x = x0+[0:N]*h;

N1 = N + 1;

if abs(c0(2)) < eps, N1 = N1 – 1; A(1,1:2) = [2 -1];

else A(1,1:2) = [2*(1-c0(1)/c0(2)*h) -2]; %(P6.11.4a)

end

if abs(cf(2)) < eps, N1 = N1 – 1; A(N1,N1 – 1:N1) = [-1 2];

else A(N1,N1 – 1:N1) = [-2 2*(1 + cf(1)/cf(2)*h)]; %(P6.11.4c)

end

if N1 > 2

for m = 2:ceil(N1/2), A(m,m – 1:m + 1) = [-1 2 -1]; end %(P6.11.4b)

end

for m=ceil(N1/2) + 1:N1 – 1, A(m,:) = fliplr(A(N1 + 1 – m,:)); end

[V,LAMBDA] = eig(A); eigvals = diag(LAMBDA)’;

[eigvals,I] = sort(eigvals); % sorting in the ascending order

V = V(:,I);

ws = sqrt(eigvals)/h;

if abs(c0(2)) < eps, Y = zeros(1,N1); else Y = []; end

Y = [Y; V];

if abs(cf(2)) < eps, Y = [Y; zeros(1,N1)]; end

Note the following things:

ž The angular frequency corresponding to the eigenvalue λ can be

obtained as

ω = λ/a0/h (P6.11.6)

ž The eigenvalues and the eigenvectors of a matrix A can be obtained

by using the MATLAB command ‘[V,D] = eig(A)’.

ž The above routine “bvp2_eig()” implements the above-mentioned

scheme to solve the second-order eigenvalue problem (P6.11.1).

ž In particular, a second-order eigenvalue BVP

y(x) + ω2y = 0 with y(x0) = 0, y(xf ) = 0 (P6.11.7)

corresponds to (P6.11.1) with c0 = [c01 c02] = [1 0] and cf = [cf 1 cf 2] = [1 0] and has the following analytical solutions:

y(x) = a sin ωx with ω =

kπ

xf − x0

, k = 1, 2, . . . (P6.11.8)

PROBLEMS 317

0.1

0.05

−0.05

0

−0.1

0 0.5 1 1.5 2

0.05

−0.05

0

−0.1

0 0.5 1 1.5 2

0.1

(a) Eigenvector solutions for BVP2 (b) Eigenvector solutions for BVP4

Figure P6.11 The eigenvector solutions of homogeneous second-order and fourth-order BVPs.

Now, use the routine “bvp2_eig()” with the number of grid points

N = 256 to solve the BVP2 (P6.11.7) with x0 = 0 and xf = 2, find

the lowest three angular frequencies (ωi ’s) and plot the corresponding

eigenvector solutions as depicted in Fig. P6.11a.

(b) A Homogeneous Fourth-Order BVP to an Eigenvalue Problem

Consider an eigenvalue boundary value problem of solving

d4y

dx4 − ω4y = 0 (P6.11.9)

with y(x0) = 0,

d2y

dx2 (x0) = 0, y(xf ) = 0,

d2y

dx2 (xf ) = 0

to find y(x) for x ∈ [x0, xf ] with the (possible) angular frequency ω.

In order to use the finite difference method, we divide the solution

interval [x0, xf ] into N subintervals to have the grid points xi = x0 + ih = x0 + i(xf − x0)/N and then, replace the derivatives in the

differential equation and the boundary conditions by their finite difference

approximations to write

yi−2 − 4yi−1 + 6yi − 4yi+1 + yi+2

h4 − ω4yi = 0

yi−2 − 4yi−1 + 6yi − 4yi+1 + yi+2 = λyi(λ = h4ω4) (P6.11.10)

with

y0 = 0,

y−1 − 2y0 + y1

h2 = 0 → y−1 = −y1 (P6.11.11a)

yN = 0,

yN−1 − 2yN + yN+1

h2 = 0 → yN+1 = −yN−1

(P6.11.11b)

318 ORDINARY DIFFERENTIAL EQUATIONS

Substituting the discretized boundary condition (P6.11.11) into (P6.11.10)

yields

y−1 − 4y0 + 6y1 − 4y2 + y3 = λy1

(P6.11.11a)

−−−−−→

5y1 − 4y2 + y3 = λy1

y0 − 4y1 + 6y2 − 4y3 + y4 = λy2

(P6.11.11a)

−−−−−→

− 4y1 + 6y2 − 4y3 + y4 = λy2

yi − 4yi+1 + 6yi+2 − 4yi+3 + yi+4 = λyi+2

for i = 1 : N − 5 (P6.11.12)

yN−4 − 4yN−3 + 6yN−2 − 4yN−1 + yN = λyN−2

(P6.11.11b)

−−−−−−→

yN−4 − 4yN−3 + 6yN−2 − 4yN−1 = λyN−2

yN−3 − 4yN−2 + 6yN−1 − 4yN + yN+1 = λyN−1

(P6.11.11b)

−−−−−−→

yN−3 − 4yN−2 + 5yN−1 = λyN−1

which can be formulated in a compact form as

5 −4 1 0 0 0 0

−4 6 −4 1 0 0 0

1 −4 6 −4 1 0 0

0 · · · · · 0

0 0 1 −4 6 −4 1

0 0 0 1 −4 6 −4

0 0 0 0 1 −4 5

y1

y2

y3

· yN−3

yN−2

yN−1

= λ

y1

y2

y3

· yN−3

yN−2

yN−1

Ay = λy, [A − λI]y = 0 (P6.11.13)

For this equation to have a nontrivial solution y = 0, λ must be one

of the eigenvalues of the matrix A and the corresponding eigenvectors

are possible solutions. Note that the angular frequency corresponding

to the eigenvalue λ can be obtained as

ω = √4 λ/h (P6.11.14)

(i) Compose a routine “bvp4_eig()” which implements the abovementioned

scheme to solve the fourth-order eigenvalue problem

(P6.11.9).

function [x,Y,ws,eigvals] = bvp4_eig(x0,xf,N)

PROBLEMS 319

(ii) Use the routine “bvp4_eig()” with the number of grid points N = 256 to solve the BVP4 (P6.11.9) with x0 = 0 and xf = 2, find the

lowest three angular frequencies (ωi ’s) and plot the corresponding

eigenvector solutions as depicted in Fig. P6.11b.

(c) The Sturm–Liouville Equation

Consider an eigenvalue boundary value problem of solving

d

dx

(f (x)y) + r(x)y = λq(x)y with y(x0) = 0, y(xf ) = 0

(P6.11.15)

to find y(x) for x ∈ [x0, xf ] with the (possible) angular frequency ω.

In order to use the finite difference method, we divide the solution

interval [x0, xf ] into N subintervals to have the grid points xi = x0 + ih = x0 + i(xf − x0)/N, and then we replace the derivatives in

the differential equation and the boundary conditions by their finite

difference approximations (with the step size h/2) to write

f (xi + h/2)y(xi + h/2) − f (xi − h/2)y(xi − h/2)

2(h/2) + r(xi)yi = λq(xi)y(xi )

1

h

f xi +

h

2 yi+1 − yi

h − f xi −

h

2 yi − yi−1

h + r(xi )yi = λq(xi)y(xi )

aiyi−1 + biyi + ciyi+1 = λyi for i = 1, 2, . . . , N − 1 (P6.11.16)

with

ai =

f (xi − h/2)

h2q(xi )

, ci =

f (xi + h/2)

h2q(xi )

, and bi =

r(xi )

q(xi) − ai − ci

(P6.11.17)

(i) Compose a routine “sturm()” which implements the abovementioned

scheme to solve the Sturm–Liouville BVP (P6.11.15).

function [x,Y,ws,eigvals] = sturm(f,r,q,x0,xf,N)

(ii) Use the routine “sturm()” with the number of grid points N = 256

to solve the following BVP2:

d

dx (1 + x2)y = −2λy with y(x0) = 0, y(xf ) = 0

(P6.11.18)

Plot the eigenvector solutions corresponding to the lowest three

angular frequencies (ωi ’s).

7

OPTIMIZATION

Optimization involves finding the minimum/maximum of an objective function

f (x) subject to some constraint x ∈ S. If there is no constraint for x to satisfy—

or, equivalently, S is the universe—then it is called an unconstrained

optimization; otherwise, it is a constrained optimization. In this chapter, we

will cover several unconstrained optimization techniques such as the golden

search method, the quadratic approximation method, the Nelder–Mead method,

the steepest descent method, the Newton method, the simulated-annealing (SA)

method, and the genetic algorithm (GA). As for constrained optimization, we

will only introduce the MATLAB built-in routines together with the routines for

unconstrained optimization. Note that we don’t have to distinguish maximization

and minimization because maximizing f (x) is equivalent to minimizing −f (x)

and so, without loss of generality, we deal only with the minimization problems.

7.1 UNCONSTRAINED OPTIMIZATION [L-2, CHAPTER 7]

7.1.1 Golden Search Method

This method is applicable to an unconstrained minimization problem such that

the solution interval [a, b] is known and the objective function f (x) is unimodal

within the interval; that is, the sign of its derivative f (x) changes at most once in

[a, b] so that f (x) decreases/increases monotonically for [a, xo]/[xo, b], where

xo is the solution that we are looking for. The so-called golden search procedure is

summarized below and is cast into the routine “opt_gs()”.Wemade a MATLAB

Applied Numerical Methods Using MATLAB, by Yang, Cao, Chung, and Morris

Copyr ight 2005 John Wiley & Sons, I nc., ISBN 0-471-69833-4

321

322 OPTIMIZATION

program “nm711.m”, which uses this routine to find the minimum point of the

objective function

f (x) = (x2 − 4)2/8 − 1 (7.1.1)

GOLDEN SEARCH PROCEDURE

Step 1. Pick up the two points c = a + (1 − r)h and d = a + rh inside the

interval [a, b], where r = (√5 − 1)/2 and h = b − a.

Step 2. If the values of f (x) at the two points are almost equal [i.e., f (a) ≈ f (b)] and the width of the interval is sufficiently small (i.e., h ≈ 0),

then stop the iteration to exit the loop and declare xo = c or xo = d

depending on whether f (c) < f(d) or not. Otherwise, go to Step 3.

Step 3. If f (c) < f(d), let the new upper bound of the interval b ← d; otherwise,

let the new lower bound of the interval a ← c. Then, go to

Step 1.

function [xo,fo] = opt_gs(f,a,b,r,TolX,TolFun,k)

h = b – a; rh = r*h; c = b – rh; d = a + rh;

fc = feval(f,c); fd = feval(f,d);

if k <= 0 | (abs(h) < TolX & abs(fc – fd) < TolFun)

if fc <= fd, xo = c; fo = fc;

else xo = d; fo = fd;

end

if k == 0, fprintf(’Just the best in given # of iterations’), end

else

if fc < fd, [xo,fo] = opt_gs(f,a,d,r,TolX,TolFun,k – 1);

else [xo,fo] = opt_gs(f,c,b,r,TolX,TolFun,k – 1);

end

end

%nm711.m to perform the golden search method

f711 = inline(’(x.*x-4).^2/8-1’,’x’);

a = 0; b = 3; r =(sqrt(5)-1)/2; TolX = 1e-4; TolFun = 1e-4; MaxIter = 100;

[xo,fo] = opt_gs(f711,a,b,r,TolX,TolFun,MaxIter)

Figure 7.1 shows how the routine “opt_gs()” proceeds toward the minimum

point step by step.

Note the following points about the golden search procedure.

ž At every iteration, the new interval width is

b − c = b − (a + (1 − r)(b − a)) = rh or d − a = a + rh − a = rh

(7.1.2)

so that it becomes r times the old interval width (b − a = h).

UNCONSTRAINED OPTIMIZATION 323

2

1

0

−1

0 0.5 1

(1 − r )h = rh1 = r 2h

c3

c1

c2

c

d3

d2

d1

d

b3

b2

a3

a1

1.5 2 2.5 3

b1 h1 = b1 − a1 = rh

b h = b − a

a2

a

Figure 7.1 Process for the golden search method.

ž The golden ratio r is fixed so that a point c1 = b1 − rh1 = b − r2h in the

new interval [c, b] conforms with d = a + rh = b − (1 − r)h, that is,

r2 = 1 − r, r2 + r − 1 = 0, r= −1 + √1 + 4

2 = −1 + √5

2

(7.1.3)

7.1.2 Quadratic Approximation Method

The idea of this method is to (a) approximate the objective function f (x) by a

quadratic function p2(x) matching the previous three (estimated solution) points

and (b) keep updating the three points by replacing one of them with the minimum

point of p2(x). More specifically, for the three points

{(x0, f0), (x1, f1), (x2, f2)} with x0 < x1 < x2

we find the interpolation polynomial p2(x) of degree 2 to fit them and replace

one of them with the zero of the derivative—that is, the root of p2 (x) = 0 [see

Eq. (P3.1.2) in Problem 3.1]:

x = x3 =

f0(x2

1 − x2

2 ) + f1(x2

2 − x2

0 ) + f2(x2

0 − x2

1 )

2{f0(x1 − x2) + f1(x2 − x0) + f2(x0 − x1)}

(7.1.4)

In particular, if the previous estimated solution points are equidistant with an

equal distance h (i.e., x2 − x1 = x1 − x0 = h), then this formula becomes

x3 =

f0(x2

1 − x2

2 ) + f1(x2

2 − x2

0 ) + f2(x2

0 − x2

1 )

2{f0(x1 − x2) + f1(x2 − x0) + f2(x0 − x1)} x1=x +h

x2=x1+h

= x0 + h

3f0 − 4f1 + f2

2(−f0 + 2f1 − f2)

(7.1.5)

324 OPTIMIZATION

We keep updating the three points this way until |x2 − x0| ≈ 0 and/or |f (x2) − f (x0)| ≈ 0, when we stop the iteration and declare x3 as the minimum point.

The rule for updating the three points is as follows.

1. In case x0 < x3 < x1, we take {x0, x3, x1} or {x3, x1, x2} as the new set of

three points depending on whether f (x3) < f(x1) or not.

2. In case x1 < x3 < x2, we take {x1, x3, x2} or {x0, x1, x3} as the new set of

three points depending on whether f (x3) ≤ f (x1) or not.

This procedure, called the quadratic approximation method, is cast into the

MATLAB routine “opt_quad()”, which has the nested (recursive call) structure.

We made the MATLAB program “nm712.m”, which uses this routine to find the

minimum point of the objective function (7.1.1) and also uses the MATLAB

built-in routine “fminbnd()” to find it for cross-check. Figure 7.2 shows how

the routine “opt_quad()” proceeds toward the minimum point step by step.

(cf) The MATLAB built-in routine “fminbnd()” corresponds to “fmin()” in the MATLAB

of version.5.x.

function [xo,fo] = opt_quad(f,x0,TolX,TolFun,MaxIter)

%search for the minimum of f(x) by quadratic approximation method

if length(x0) > 2, x012 = x0(1:3);

else

if length(x0) == 2, a = x0(1); b = x0(2);

else a = x0 – 10; b = x0 + 10;

end

x012 = [a (a + b)/2 b];

end

f012 = f(x012);

[xo,fo] = opt_quad0(f,x012,f012,TolX,TolFun,MaxIter);

function [xo,fo] = opt_quad0(f,x012,f012,TolX,TolFun,k)

x0 = x012(1); x1 = x012(2); x2 = x012(3);

f0 = f012(1); f1 = f012(2); f2 = f012(3);

nd = [f0 – f2 f1 – f0 f2 – f1]*[x1*x1 x2*x2 x0*x0; x1 x2 x0]’;

x3 = nd(1)/2/nd(2); f3 = feval(f,x3); %Eq.(7.1.4)

if k <= 0 | abs(x3 – x1) < TolX | abs(f3 – f1) < TolFun

xo = x3; fo = f3;

if k == 0, fprintf(’Just the best in given # of iterations’), end

else

if x3 < x1

if f3 < f1, x012 = [x0 x3 x1]; f012 = [f0 f3 f1];

else x012 = [x3 x1 x2]; f012 = [f3 f1 f2];

end

else

if f3 <= f1, x012 = [x1 x3 x2]; f012 = [f1 f3 f2];

else x012 = [x0 x1 x3]; f012 = [f0 f1 f3];

end

end

[xo,fo] = opt_quad0(f,x012,f012,TolX,TolFun,k – 1);

end

UNCONSTRAINED OPTIMIZATION 325

2

1.5

1

0.5

0

−0.5

−1

0 0.5 1 1.5 2 2.5 3

x2

x5

x4 x6 x0 x3 x1

Figure 7.2 Process of searching for the minimum by the quadratic approximation method.

%nm712.m to perform the quadratic approximation method

clear, clf

f711 = inline(’(x.*x – 4).^2/8-1’, ’x’);

a = 0; b = 3; TolX = 1e-5; TolFun = 1e-8; MaxIter = 100;

[xoq,foq] = opt_quad(f711,[a b],TolX,TolFun,MaxIter)

%minimum point and its function value

[xob,fob] = fminbnd(f711,a,b) %MATLAB built-in function

7.1.3 Nelder–Mead Method [W-8]

The Nelder–Mead method is applicable to the minimization of a multivariable

objective function, for which neither the golden search method nor the quadratic

approximation method can be applied. The algorithm of the Nelder–Mead method

summarized in the box below is cast into the MATLAB routine “Nelder0()”.

Note that in the N-dimensional case (N >2), this algorithm should be repeated

for each subplane as implemented in the outer routine “opt_Nelder()”.

We made the MATLAB program “nm713.m” to minimize a two-variable objective

function

f (x1, x2) = x2

1 − x1x2 − 4×1 + x2

2 − x2 (7.1.6)

whose minimum can be found in an analytical way—that is, by setting the partial

derivatives of f (x1, x2) with respect to x1 and x2 to zero as

∂

∂x1

f (x1, x2) = 2×1 − x2 − 4 = 0

∂

∂x2

f (x1, x2) = 2×2 − x1 − 1 = 0

xo = (x10, x2o) = (3, 2)

326 OPTIMIZATION

NELDER–MEAD ALGORITHM

Step 1. Let the initial three estimated solution points be a, b and c, where

f (a) < f(b) < f (c).

Step 2. If the three points or their function values are sufficiently close to each

other, then declare a to be the minimum and terminate the procedure.

Step 3. Otherwise, expecting that the minimum we are looking for may be at

the opposite side of the worst point c over the line ab (see Fig. 7.3),

take

e = m + 2(m − c), where m = (a + b)/2

and if f (e) < f(b), take e as the new c; otherwise, take

r = (m + e)/2 = 2m − c

and if f (r) < f(c), take r as the new c; if f (r) ≥ f (b), take

s = (c + m)/2

and if f (s) < f(c), take s as the new c; otherwise, give up the two points

b,c and take m and c1 = (a + c)/2 as the new b and c, reflecting our

expectation that the minimum would be around a.

Step 4. Go back to Step 1.

a

m

b

r

e

c1

c2

m = (a + b)/2

r = m + (m − c)

e = m + 2(m − c)

s1 = (c + m)/2

s2 = (m + r )/2

c1 = (c + a)/2

c2 = (r + a) /2

c s1

s2

Figure 7.3 Notation used in the Nelder–Mead method.

UNCONSTRAINED OPTIMIZATION 327

function [xo,fo] = Nelder0(f,abc,fabc,TolX,TolFun,k)

[fabc,I] = sort(fabc); a = abc(I(1),:); b = abc(I(2),:); c = abc(I(3),:);

fa = fabc(1); fb = fabc(2); fc = fabc(3); fba = fb – fa; fcb = fc – fb;

if k <= 0 | abs(fba) + abs(fcb) < TolFun | abs(b – a) + abs(c – b) < TolX

xo = a; fo = fa;

if k == 0, fprintf(’Just best in given # of iterations’), end

else

m = (a + b)/2; e = 3*m – 2*c; fe = feval(f,e);

if fe < fb, c = e; fc = fe;

else

r = (m+e)/2; fr = feval(f,r);

if fr < fc, c = r; fc = fr; end

if fr >= fb

s = (c + m)/2; fs = feval(f,s);

if fs < fc, c = s; fc = fs;

else b = m; c = (a + c)/2; fb = feval(f,b); fc = feval(f,c);

end

end

end

[xo,fo] = Nelder0(f,[a;b;c],[fa fb fc],TolX,TolFun,k – 1);

end

function [xo,fo] = opt_Nelder(f,x0,TolX,TolFun,MaxIter)

N = length(x0);

if N == 1 %for 1-dimensional case

[xo,fo] = opt_quad(f,x0,TolX,TolFun); return

end

S = eye(N);

for i = 1:N %repeat the procedure for each subplane

i1 = i + 1; if i1 > N, i1 = 1; end

abc = [x0; x0 + S(i,:); x0 + S(i1,:)]; %each directional subplane

fabc = [feval(f,abc(1,:)); feval(f,abc(2,:)); feval(f,abc(3,:))];

[x0,fo] = Nelder0(f,abc,fabc,TolX,TolFun,MaxIter);

if N < 3, break; end %No repetition needed for a 2-dimensional case

end

xo = x0;

%nm713.m: do_Nelder

f713 = inline(’x(1)*(x(1)-4-x(2)) +x(2)*(x(2)-1)’,’x’);

x0 = [0 0], TolX = 1e-4; TolFun = 1e-9; MaxIter = 100;

[xon,fon] = opt_Nelder(f713,x0,TolX,TolFun,MaxIter)

%minimum point and its function value

[xos,fos] = fminsearch(f713,x0) %use the MATLAB built-in function

This program also applies the MATLAB built-in routine “fminsearch()” to minimize

the same objective function for practice and confirmation. The minimization

process is illustrated in Fig. 7.4.

(cf) The MATLAB built-in routine “fminsearch()” uses the Nelder–Mead algorithm

to minimize a multivariable objective function. It corresponds to “fmins()” in the

MATLAB of version.5.x.

328 OPTIMIZATION

2.5

2

1.5

1

0.5

0

0 0.5 1 1.5 2 2.5 3 3.5

−4.5(r)

−5.3(r)

−6.9(e)

−5.6(e)

−3.0(a)

0.0(b)

0.0(c)

−6.4(s1) −6.9(s2)

−1.3(s1)

Figure 7.4 Process for the Nelder–Mead method (nm713.m-opt Nelder()).

7.1.4 Steepest Descent Method

This method searches for the minimum of an N-dimensional objective function

in the direction of a negative gradient

−g(x) = −∇f (x) = −∂f (x)

∂x1

∂f (x)

∂x2 · · ·

∂f (x)

∂xN T

(7.1.7)

with the step-size αk (at iteration k) adjusted so that the function value is

minimized along the direction by a (one-dimensional) line search technique

like the quadratic approximation method. The algorithm of the steepest descent

method is summarized in the following box and cast into the MATLAB routine

“opt_steep()”.

We made the MATLAB program “nm714.m” to minimize the objective function

(7.1.6) by using the steepest descent method. The minimization process is

illustrated in Fig. 7.5.

STEEPEST DESCENT ALGORITHM

Step 0. With the iteration number k = 0, find the function value f0 = f (x0)

for the initial point x0.

Step 1. Increment the iteration number k by one, find the step-size αk−1 along

the direction of the negative gradient −gk−1 by a (one-dimensional) line

UNCONSTRAINED OPTIMIZATION 329

search like the quadratic approximation method.

αk−1 = ArgMinαf (xk−1 − αgk−1/||gk−1||) (7.1.8)

Step 2. Move the approximate minimum by the step-size αk−1 along the direction

of the negative gradient −gk−1 to get the next point

xk = xk−1 − αk−1gk−1/||gk−1|| (7.1.9)

Step 3. If xk ≈ xk−1 and f (xk) ≈ f (xk−1), then declare xk to be the minimum

and terminate the procedure. Otherwise, go back to step 1.

function [xo,fo] = opt_steep(f,x0,TolX,TolFun,alpha0,MaxIter)

% minimize the ftn f by the steepest descent method.

%input: f = ftn to be given as a string ’f’

% x0 = the initial guess of the solution

%output: x0 = the minimum point reached

% f0 = f(x(0))

if nargin < 6, MaxIter = 100; end %maximum # of iteration

if nargin < 5, alpha0 = 10; end %initial step size

if nargin < 4, TolFun = 1e-8; end %|f(x)| < TolFun wanted

if nargin < 3, TolX = 1e-6; end %|x(k)- x(k – 1)|<TolX wanted

x = x0; fx0 = feval(f,x0); fx = fx0;

alpha = alpha0; kmax1 = 25;

warning = 0; %the # of vain wanderings to find the optimum step size

for k = 1: MaxIter

g = grad(f,x); g = g/norm(g); %gradient as a row vector

alpha = alpha*2; %for trial move in negative gradient direction

fx1 = feval(f,x – alpha*2*g);

for k1 = 1:kmax1 %find the optimum step size(alpha) by line search

fx2 = fx1; fx1 = feval(f,x-alpha*g);

if fx0 > fx1+TolFun & fx1 < fx2 – TolFun %fx0 > fx1 < fx2

den = 4*fx1 – 2*fx0 – 2*fx2; num = den – fx0 + fx2; %Eq.(7.1.5)

alpha = alpha*num/den;

x = x – alpha*g; fx = feval(f,x); %Eq.(7.1.9)

break;

else alpha = alpha/2;

end

end

if k1 >= kmax1, warning = warning + 1; %failed to find optimum step size

else warning = 0;

end

if warning >= 2|(norm(x – x0) < TolX&abs(fx – fx0) < TolFun), break; end

x0 = x; fx0 = fx;

end

xo = x; fo = fx;

if k == MaxIter, fprintf(’Just best in %d iterations’,MaxIter), end

%nm714

f713 = inline(’x(1)*(x(1) – 4 – x(2)) + x(2)*(x(2)- 1)’,’x’);

x0 = [0 0], TolX = 1e-4; TolFun = 1e-9; alpha0 = 1; MaxIter = 100;

[xo,fo] = opt_steep(f713,x0,TolX,TolFun,alpha0,MaxIter)

330 OPTIMIZATION

7.1.5 Newton Method

Like the steepest descent method, this method also uses the gradient to search for

the minimum point of an objective function. Such gradient-based optimization

methods are supposed to reach a point at which the gradient is (close to) zero.

In this context, the optimization of an objective function f (x) is equivalent to

finding a zero of its gradient g(x), which in general is a vector-valued function

of a vector-valued independent variable x. Therefore, if we have the gradient

function g(x) of the objective function f (x), we can solve the system of nonlinear

equations g(x) = 0 to get the minimum of f (x) by using the Newton method

explained in Section 4.4.

The backgrounds of this method as well as the steepest descent method can

be shown by taking the Taylor series of, say, a two-variable objective function

f (x1, x2):

f (x1, x2) ∼=

f (x1k, x2k) + ∂f

∂x1

∂f

∂x2 (x1k,x2k ) x1 − x1k

x2 − x2k

+

1

2 x1 − x1k x2 − x2k ∂2f/∂x2

1 ∂2f/∂x1∂x2

∂2f/∂x2∂x1 ∂2f/∂x2

2 (x1k,x2k ) x1 − x1k

x2 − x2k

f (x)∼=

f (xk)+∇f (x)T |xk [x − xk] +

1

2

[x − xk]T ∇2f (x)|xk [x − xk]

f (x)∼=

f (xk) + gT

k [x − xk] +

1

2

[x − xk]THk[x − xk] (7.1.10)

with the gradient vector gk = ∇f (x)|xk and the Hessian matrix Hk = ∇2f (x)|xk .

In the light of this equation, we can see that the value of the objective function at

point xk+1 updated by the steepest descent algorithm described by

Eq. (7.1.9)

xk+1

(7.1.9) = xk − αkgk/||gk||

is most likely smaller than that at the old point xk, with the third term in

Eq. (7.1.10) neglected.

f (xk+1)∼=

f (xk) + gT

k [xk+1 − xk] = f (xk) − αkgT

k gk/||gk||

f (xk+1) − f (xk)∼=

−αkgT

k gk/||gk|| ≤ 0 ⇒ f (xk+1) ≤ f (xk)

(7.1.11)

Slightly different from this strategy of the steepest descent algorithm, the Newton

method tries to go straight to the zero of the gradient of the approximate objective

function (7.1.10)

gk + Hk[x − xk] = 0, x = xk − H−1

k gk (7.1.12)

by the updating rule

xk+1 = xk − H−1

k gk (7.1.13)

with the gradient vector gk = ∇f (x)|xk and the Hessian matrix Hk = ∇2f (x)|xk

(Appendix C).

UNCONSTRAINED OPTIMIZATION 331

This algorithm is essentially to find the zero of the gradient function g(x) of the

objective function and consequently, it can be implemented by using any vector

nonlinear equation solver. What we have to do is just to define the gradient

function g(x) and put the function name as an input argument of any routine

like “newtons()” or “fsolve()” for solving a system of nonlinear equations

(see Section 4.6).

Now, we make a MATLAB program “nm715.m”, which actually solves

g(x) = 0 for the gradient function

g(x) = ∇f (x) = ∂f

∂x1

∂f

∂x2 T

= 2×1 − x2 −4 2×2 − x1 − 1 (7.1.14)

of the objective function (7.1.6)

f (x) = f (x1, x2) = x2

1 − x1x2 − 4×1 + x2

2 − x2

Figure 7.5 illustrates the process of searching for the minimum point by the Newton

algorithm (7.1.13) as well as the steepest descent algorithm (7.1.9), where the

steepest descent algorithm proceeds in the negative gradient direction until the

minimum point in the line is reached, while the Newton algorithm approaches

the minimum point almost straightly and reaches it in a few iterations.

>>nm715

xo = [3.0000 2.0000], ans = -7

%nm715 to minimize an objective ftn f(x) by the Newton method.

clear, clf

f713 = inline(’x(1).^2 – 4*x(1) – x(1).*x(2) + x(2).^2 – x(2)’,’x’);

g713 = inline(’[2*x(1) – x(2) – 4 2*x(2) – x(1) – 1]’,’x’);

x0 = [0 0], TolX = 1e-4; TolFun = 1e-6; MaxIter = 50;

[xo,go,xx] = newtons(g713,x0,TolX,MaxIter);

xo, f713(xo) %an extremum point reached and its function value

0

0

1

1

2

2

3

3

4

4

5 6

Newton

steepest descent

Figure 7.5 Process for the steepest descent method and Newton method (‘‘nm714.m’’ and

‘‘nm715.m’’).

332 OPTIMIZATION

Remark 7.1. Weak Point of Newton Method.

The Newton method is usually more efficient than the steepest descent method

if only it works as illustrated above, but it is not guaranteed to reach the minimum

point. The decisive weak point of the Newton method is that it may approach one

of the extrema having zero gradient, which is not necessarily a (local) minimum,

but possibly a maximum or a saddle point (see Fig. 7.13).

7.1.6 Conjugate Gradient Method

Like the steepest descent method or Newton method, this method also uses the

gradient to search for the minimum point of an objective function, but in a

different way. It has two versions—the Polak–Ribiere (PR) method and the

Fletcher–Reeves (FR) method—that are slightly different only in the search

direction vector. This algorithm, summarized in the following box, is cast into

the MATLAB routine “opt_conjg()”, which implements PR or FR depending

on the last input argument KC = 1 or 2. The quasi-Newton algorithm used in

the MATLAB built-in routine “fminunc()” is similar to the conjugate gradient

method.

This method borrows the framework of the steepest descent method and needs

a bit more effort for computing the search direction vector s(n). It takes at most N

iterations to reach the minimum point in case the objective function is quadratic

with a positive-definite Hessian matrix H as

f (x) =

1

2

xTHx + bT x + c where x: an N-dimensional vector (7.1.15)

CONJUGATE GRADIENT ALGORITHM

Step 0. With the iteration number k = 0, find the objective function value

f0 = f (x0) for the initial point x0.

Step 1. Initialize the inside loop index, the temporary solution and the search

direction vector to n = 0, x(n) = xk and s(n) = −gk = −g(xk), respectively,

where g(x) is the gradient of the objective function f (x).

Step 2. For n = 0 to N − 1, repeat the following things:

Find the (optimal) step-size

αn = ArgMinαf (x(n) + αs(n)) (7.1.16)

and update the temporary solution point to

x(n + 1) = x(n) + αns(n) (7.1.17)

and the search direction vector to

s(n + 1) = −gn+1 + βns(n) (7.1.18)

UNCONSTRAINED OPTIMIZATION 333

with

βn =

[gn+1 − gn]T gn+1

gT

n gn

(FR) or

gT

n+1gn+1

gT

n gn

(PR) (7.1.19)

Step 3. Update the approximate solution point to xk+1 = x(N), which is the

last temporary one.

Step 4. If xk ≈ xk−1 and f (xk) ≈ f (xk−1), then declare xk to be the minimum

and terminate the procedure. Otherwise, increment k by one and go back

to Step 1.

function [xo,fo] = opt_conjg(f,x0,TolX,TolFun,alpha0,MaxIter,KC)

%KC = 1: Polak–Ribiere Conjugate Gradient method

%KC = 2: Fletcher–Reeves Conjugate Gradient method

if nargin < 7, KC = 0; end

if nargin < 6, MaxIter = 100; end

if nargin < 5, alpha0 = 10; end

if nargin < 4, TolFun = 1e-8; end

if nargin < 3, TolX = 1e-6; end

N = length(x0); nmax1 = 20; warning = 0; h = 1e-4; %dimension of variable

x = x0; fx = feval(f,x0); fx0 = fx;

for k = 1: MaxIter

xk0 = x; fk0 = fx; alpha = alpha0;

g = grad(f,x,h); s = -g;

for n = 1:N

alpha = alpha0;

fx1 = feval(f,x + alpha*2*s); %trial move in search direction

for n1 = 1:nmax1 %To find the optimum step size by line search

fx2 = fx1; fx1 = feval(f,x+alpha*s);

if fx0 > fx1 + TolFun & fx1 < fx2 – TolFun %fx0 > fx1 < fx2

den = 4*fx1 – 2*fx0 – 2*fx2; num = den-fx0 + fx2; %Eq.(7.1.5)

alpha = alpha*num/den;

x = x+alpha*s; fx = feval(f,x);

break;

elseif n1 == nmax1/2

alpha = -alpha0; fx1 = feval(f,x + alpha*2*s);

else

alpha = alpha/2;

end

end

x0 = x; fx0 = fx;

if n < N

g1 = grad(f,x,h);

if KC <= 1, s = – g1 +(g1 – g)*g1’/(g*g’+ 1e-5)*s; %(7.1.19a)

else s = -g1 + g1*g1’/(g*g’+ 1e-5)*s; %(7.1.19b)

end

g = g1;

end

if n1 >= nmax1, warning = warning+1; %can’t find optimum step size

else warning = 0;

end

end

if warning >= 2|(norm(x – xk0)<TolX&abs(fx – fk0)< TolFun), break; end

end

xo = x; fo = fx;

if k == MaxIter, fprintf(’Just best in %d iterations’,MaxIter), end

%nm716 to minimize f(x) by the conjugate gradient method.

f713 = inline(’x(1).^2 – 4*x(1) – x(1).*x(2) + x(2).^2 – x(2)’,’x’);

x0 =[0 0], TolX = 1e-4; TolFun = 1e-4; alpha0 = 10; MaxIter = 100;

[xo,fo] = opt_conjg(f713,x0,TolX,TolFun,alpha0,MaxIter,1)

[xo,fo] = opt_conjg(f713,x0,TolX,TolFun,alpha0,MaxIter,2)

334 OPTIMIZATION

Based on the fact that minimizing this quadratic objective function is equivalent

to solving the linear equation

g(x) = ∇f (x) = Hx + b = 0 (7.1.20)

MATLAB has several built-in routines such as “cgs()”,“pcg()”, and “bicg()”,

which use the conjugate gradient method to solve a set of linear equations.

We make the MATLAB program “nm716.m” to minimize the objective function

(7.1.6) by the conjugate gradient method and the minimization process is

illustrated in Fig. 7.6.

7.1.7 Simulated Annealing Method [W-7]

All of the optimization methods discussed so far may be more or less efficient

in finding the minimum point if only they start from the initial point sufficiently

close to it. But, the point they reach may be one of several local minima and we

often cannot be sure that it is the global minimum. How about repeating the procedure

to search for all local minima starting from many different initial guesses

and taking the best one as the global minimum? This would be a computationally

formidable task, since there is no systematic way to determine a suitable

sequence of initial guesses, each of which leads to its own (local) minimum so

that all the local minima can be exhaustively found to compete with each other

for the global minimum.

An interesting alternative is based on the analogy between annealing and minimization.

Annealing is the physical process of heating up a solid metal above its

melting point and then cooling it down so slowly that the highly excited atoms

can settle into a (global) minimum energy state, yielding a single crystal with

a regular structure. Fast cooling by rapid quenching may result in widespread

0

0

1

1

2

2

3

3

4

4

5 6

Figure 7.6 Process for the conjugate gradient method (‘‘nm716.m’’).

UNCONSTRAINED OPTIMIZATION 335

irregularities and defects in the crystal structure, analogous to being too hasty

to find the global minimum. The simulated annealing process can be implemented

using the Boltzmann probability distribution of an energy level E(≥0)

at temperature T described by

p(E) = α exp(−E/KT ) with the Boltzmann constant K and α = 1/KT

(7.1.21)

Note that at high temperature the probability distribution curve is almost flat over

a wide range of E, implying that the system can be in a high energy state as

equally well as in a low energy state, while at low temperature the probability

distribution curve gets higher/lower for lower/higher E, implying that the system

will most probably be in a low energy state, but still have a slim chance to be

in a high energy state so that it can escape from a local minimum energy state.

The idea of simulated annealing is summarized in the box below and cast

into the MATLAB routine “sim_anl()”. This routine has two parts that vary

with the iteration number as the temperature falls down. One is the size of step

x from the previous guess to the next guess, which is made by generating a

random vector y having uniform distribution U[−1,+1] and the same dimension

as the variable x and multiplying μ−1(y) (in a termwise manner) by the difference

vector (u − l) between the upper bound u and the lower bound l of the domain

of x. The μ−1-law

g−1

μ (y) =

(1 + μ)|y| − 1

μ

sign (y) for |y| ≤ 1 (7.1.22)

implemented in the routine “mu_inv()” has the parameter μ that is increased

according to a rule

μ = 10100 (k/kmax)q with q > 0: the quenching factor (7.1.23)

as the iteration number k increases, reaching μ = 10100 at the last iteration k = kmax. Note the following:

ž The quenching factor q > 0 is made small/large for slow/fast quenching.

ž The value of μ−1-law function becomes small for |y| < 1 as μ increases

(see Fig. 7.7a).

The other is the probability of taking a step x that would result in change

f > 0 of the objective function value f (x). Similarly to Eq. (7.1.21), this is

determined by

p(taking the step x) = exp

−

k

kmaxq f

|f (x)|εf for f > 0 (7.1.24)

336 OPTIMIZATION

(a) The mu-law inverse function gm

−1(y ) (b) The exponential function for

randomness control

0

0

1

0.2

0.4

0.6

0.8

as k (iteration number)

increases

Δf /|f (x)|/ef

Δf /|f (x)|/ef )

p (Δx) =

exp (−(k /kmax)q

0

0

0.5 y 1

1

0.2

0.4

0.6

0.8

Δx = gm

−1(y) (u − l )

with

m = 10100 (k /kmax)q

m = 0.01

m = 102

as k (iteration number)

increases

m = 1050

Figure 7.7 Illustrative functions used for controlling the randomness–temperature in SA.

SIMULATED ANNEALING

Step 0. Pick the initial guess x0, the lower bound l, the upper bound u, the

maximum number of iterations kmax > 0, the quenching factor q > 0 (to

be made small/large for slow/fast quenching), and the relative tolerance

εf of function value fluctuation.

Step 1. Let x = x0, xo = x, f o = f (x).

Step 2. For k = 1 to kmax, do

{Generate an N × 1 uniform random vector of U[−1,+1] and transform

it by the inverse μ law (with μ = 10100 (k/kmax)q) to make x and then

take x1 ← x + x, confining the next guess inside the admissible region

{x|l ≤ x ≤ u} as needed.

If f = f (x1) − f (x) < 0,

{set x ← x1 and if f (x) < fo, set xo ← x and f o ← f (xo).}

Otherwise,

{generate a uniform random number z of U[0,1] and set x ← x1 only in case

z < p(taking the step x)

(7.1.24) = exp(−(k/kmax)qf/|f (x)|/εf )

}

}

Step 3. Regarding xo as close to the minimum point that we are looking for,

we may set xo as the initial value and apply any (local) optimization

algorithm to search for the minimum point of f (x).

UNCONSTRAINED OPTIMIZATION 337

function [xo,fo] = sim_anl(f,x0,l,u,kmax,q,TolFun)

% simulated annealing method to minimize f(x) s.t. l <= x <= u

N = length(x0);

x = x0; fx = feval(f,x);

xo = x; fo = fx;

if nargin < 7, TolFun = 1e-8; end

if nargin < 6, q = 1; end %quenching factor

if nargin < 5, kmax = 100; end %maximum iteration number

for k = 0:kmax

Ti = (k/kmax)^q; %inverse of temperature from 0 to 1

mu = 10^(Ti*100); % Eq.(7.1.23)

dx = mu_inv(2*rand(size(x))- 1,mu).*(u – l);

x1 = x + dx; %next guess

x1 = (x1 < l).*l +(l <= x1).*(x1 <= u).*x1 +(u < x1).*u;

%confine it inside the admissible region bounded by l and u.

fx1 = feval(f,x1); df = fx1- fx;

if df < 0|rand < exp(-Ti*df/(abs(fx) + eps)/TolFun) Eq.(7.1.24)

x = x1; fx = fx1;

end

if fx < fo, xo = x; fo = fx1; end

end

function x = mu_inv(y,mu) % inverse of mu-law Eq.(7.1.22)

x = (((1+mu).^abs(y)- 1)/mu).*sign(y);

%nm717 to minimize an objective function f(x) by various methods.

clear, clf

f = inline(’x(1)^4 – 16*x(1)^2 – 5*x(1) + x(2)^4 – 16*x(2)^2 – 5*x(2)’,’x’);

l = [-5 -5]; u = [5 5]; %lower/upperbound

x0 = [0 0]

[xo_nd,fo] = opt_Nelder(f,x0)

[xos,fos] = fminsearch(f,x0) %cross-check by MATLAB built-in routines

[xou,fou] = fminunc(f,x0)

kmax = 500; q = 1; TolFun = 1e-9;

[xo_sa,fo_sa] = sim_anl(f,x0,l,u,kmax,q,TolFun)

which remains as big as e−1 for |f/f (x)| = εf at the last iteration k = kmax,

meaning that the probability of taking a step hopefully to escape from a local

minimum and find the global minimum at the risk of increasing the value of

objective function by the amount f = |f (x)|εf is still that high. The shapes of

the two functions related to the temperature are depicted in Fig. 7.7.

We make the MATLAB program “nm717.m”, which uses the routine “sim_

anl()” to minimize a function

f (x) = x4

1 − 16×2

1 − 5×1 + x4

2 − 16×2

2 − 5×2 (7.1.25)

and tries other routines such as “opt_Nelder()”, “fminsearch()”, and “fminunc()”

for cross-checking. The results of running the program are summarized

in Table 7.1, which shows that the routine “sim_anl()” may give us the

global minimum even when some other routines fail to find it. But, even this

routine based on the idea of simulated annealing cannot always succeed and its

success/failure depends partially on the initial guess and partially on luck, while

the success/failure of the other routines depends solely on the initial guess.

338 OPTIMIZATION

Table 7.1 Results of Running Several Optimization Routines with Various Initial Values

x0 opt Nelder() fminsearch() fminunc() sim anl()

[0, 0] [2.9035, 2.9035] [2.9035, 2.9036] [2.9036, 2.9036] [2.8966, 2.9036]

(f o = −156.66) (f o = −156.66) (f o = −156.66) (f o = −156.66)

[−0.5,−1.0] [2.9035, −2.7468] [−2.7468, −2.7468] [−2.7468, −2.7468] [2.9029, 2.9028]

(f o = −128.39) (f o = −100.12) (f o = −100.12) (f o = −156.66)

7.1.8 Genetic Algorithm [W-7]

Genetic algorithm (GA) is a directed random search technique that is modeled

on the natural evolution/selection process toward the survival of the fittest.

The genetic operators deal with the individuals in a population over several

generations to improve their fitness gradually. Individuals standing for possible

solutions are often compared to chromosomes and represented by strings of

binary numbers. Like the simulated annealing method, GA is also expected to

find the global minimum solution even in the case where the objective function

has several extrema, including local maxima, saddle points as well as local

minima.

A so-called hybrid genetic algorithm [P-2] consists of initialization, evaluation,

reproduction (selection), crossover, and mutation as depicted in Fig. 7.8

Evaluation

Mutation

Crossover

Reproduction

No

Termination

Initialize

the population

if the function values Yes

for all the chromosomes

are almost equal?

Figure 7.8 Flowchart for a genetic algorithm.

UNCONSTRAINED OPTIMIZATION 339

Np = 8, N = 2, Nb = [8 8]

pool P

01100110 01100110

01001111

11110110

01110111

10101101

11011011

11011000

10011100

10101011

01101000

11101111

10110011

11110110

00000001

00011110

decode

population X

−1.0000

−1.9020 −1.7059

−0.9216

−3.8235

−0.1209 −0.0466

−1.5527

−54.22

−12.54

−4.9608

−0.3333

4.6471

4.3725

1.7843 2.0196

3.5882

3.4706

1.1176

4.6471

−1.0000 −10.00

−40.95

44.67

18.84

evaluate

fx

19.71

85.84

reproduction

random pairing

a1 01111100

a1 01111100

b1 01010111

01111110 a2

c1 11000010

d1 10010011

e1 10101101

f1 11000010

g1 10101101

h1 10100001

10101011 b2

10011100 c2

11001111 d2

10110011 e2

11010010 f2

10110011 g2

01001010 h2

crossover/mutuation

b1′ 01000010

c1′ 11001101

d1 10010011

e1 10101101

f1′ 11010111

g1′ 10100010

h1 10100001

01111110 a2

10101010 b2′

10011111 c2′

11001111 d2

10110011 e2

11010011 f2′

10110000 g2′

01001010 h2

decode

1.7356

2.6259 1.1550

0.7713 3.1452

2.0196

3.2601

2.0196

−2.0846

1.7843

2.6360

1.7843

1.3160

−0.1373

−2.4118

3.0392

0.7647

1.7843

3.4314

1.3529

1.3137

0.28

−36.40

− 50.62

− 44.58

− 54.22

−74.73

− 54.22

− 35.76

− 0.0588

− 2.0980

− 46.12

− 52.96

− 44.73

− 54.22

− 69.94

− 43.50

− 35.88

1.6667

1.2353

3.1176

2.0196

3.2745

1.9020

0.31

encode

Figure 7.9 Reproduction/crossover mutation in one iteration of genetic algorithm.

and is summarized in the box below. The reproduction/crossover process is illustrated

in Fig 7.9. This algorithm is cast into the routine “genetic()” and we

append the following statements to the MATLAB program “nm717.m” in order

to apply the routine for minimizing the function defined by Eq. (7.1.25). Interested

readers are welcome to run the program with these statements appended

and compare the result with those of using other routines. Note that like the

simulated annealing, the routine based on the idea of GA cannot always succeed

and its success/failure depends partially on the initial guess and partially

on luck.

340 OPTIMIZATION

Np = 30; %population size

Nb = [12 12]; %the numbers of bits for representing each variable

Pc = 0.5; Pm = 0.01; %Probability of crossover/mutation

eta = 1; kmax = 100; %learning rate and the maximum # of iterations

[xo_gen,fo_gen] = genetic(f,x0,l,u,Np,Nb,Pc,Pm,eta,kmax)

HYBRID GENETIC ALGORITHM

Step 0. Pick the initial guess x0 = [x01 . . . x0N](N: the dimension of the variable),

the lower bound l = [l1 . . . lN], the upper bound u = [u1 . . .uN],

the population size Np, the vector Nb = [Nb1 . . .NbN ] consisting of the

numbers of bits assigned for the representation of each variable xi, the

probability of crossover Pc, the probability of mutation Pm, the learning

rate η(0 < η ≤ 1, to be made small/large for slow/fast learning),

and the maximum number of iterations kmax > 0. Note that the dimensions

of x0, u, and l are all the same as N, which is the dimension

of the variable x to be found and the population size Np can not be

greater than 2Nb in order to avoid duplicated chromosomes and should

be an even integer for constituting the mating pool in the crossover

stage.

Step 1. Random Generation of Initial Population

Set xo = x0, f o = f (xo) and construct in a random way the initial population

array X1 that consists of Np states (in the admissible region

bounded by u and l) including the initial state x0, by setting

X1(1) = x0 and X1(k) = l + rand.∗(u − l) for k = 2 : Np (7.1.26)

where rand is a random vector of the same dimension N as x0, u,

and l. Then, encode each number of this population array into a binary

string by

P1(n, 1 +m−1

i=1

Nbi :m

i=1

Nbi)

= binary representation of X1(n,m) with Nbm bits

= (2Nbm − 1)

X1(n,m) − l(m)

u(m) − l(m)

for n = 1 : Np and m = 1 : N (7.1.27)

so that the whole population array becomes a pool array, each row of

which is a chromosome represented by a binary string of

N

i=1 Nbi bits.

UNCONSTRAINED OPTIMIZATION 341

Step 2. For k = 1 to kmax, do the following:

1. Decode each number in the pool into a (decimal) number by

Xk(n,m) = decimal representation of

Pk

n, 1 +m−1

i=1

Nbi :m

i=1

Nbi with Nbm bits

= Pk(n, ·)

u(m) − l(m)

2Nbm − 1 + l(m)

for n = 1 : Np and m = 1 : N (7.1.28)

and evaluate the value f (n) of function for every row Xk(n,🙂 = x(n)

corresponding to each chromosome and find the minimum fmin = f (nb)

corresponding to Xk(nb,🙂 = x(nb).

2. If fmin = f (nb) < fo, then set f o = f (nb) and xo = x(nb).

3. Convert the function values into the values of fitness by

f1(n) = MaxNp

n=1{f (n)} − f (n) (7.1.29)

which is nonnegative ∀ n = 1 : Np and is large for a good chromosome.

4. If MaxNp

n=1{f1(n)} ≈ 0, then terminate this procedure, declaring xo as

the best.

Otherwise, in order to make more chromosomes around the best point

x(nb) in the next generation, use the reproduction rule

x(n) ← x(n) + η

f1(nb) − f1(n)

f1(nb)

(x(nb) − x(n)) (7.1.30)

to get a new population Xk+1 with Xk+1(n,🙂 = x(n) and encode it to

reconstruct a new pool array Pk+1 by Eq. (7.1.27).

5. Shuffle the row indices of the pool array for random mating of the chromosomes.

6. With the crossover probability Pc, exchange the tail part starting from

some random bit of the numbers in two randomly paired chromosomes

(rows of Pk+1) with each other’s to get a new pool array P k+1.

7. With the mutation probability Pm, reverse a random bit of each number

represented by chromosomes (rows of P k+1) to make a new pool array

Pk+1.

342 OPTIMIZATION

function [xo,fo] = genetic(f,x0,l,u,Np,Nb,Pc,Pm,eta,kmax)

% Genetic Algorithm to minimize f(x) s.t. l <= x <= u

N = length(x0);

if nargin < 10, kmax = 100; end %# of iterations(generations)

if nargin < 9|eta > 1|eta <= 0, eta = 1; end %learning rate(0 < eta < 1)

if nargin < 8, Pm = 0.01; end %probability of mutation

if nargin < 7, Pc = 0.5; end %probability of crossover

if nargin < 6, Nb = 8*ones(1,N); end %# of genes(bits) for each variable

if nargin < 5, Np = 10; end %population size(number of chromosomes)

%Initialize the population pool

NNb = sum(Nb);

xo = x0(:)’; l = l(:)’; u = u(:)’;

fo = feval(f,xo);

X(1,:) = xo;

for n = 2:Np, X(n,:) = l + rand(size(x0)).*(u – l); end %Eq.(7.1.26)

P = gen_encode(X,Nb,l,u); %Eq.(7.1.27)

for k = 1:kmax

X = gen_decode(P,Nb,l,u); %Eq.(7.1.28)

for n = 1:Np, fX(n) = feval(f,X(n,:)); end

[fxb,nb] = min(fX); %Selection of the fittest

if fxb < fo, fo = fxb; xo = X(nb,:); end

fX1 = max(fxs) – fX; %make the nonnegative fitness vector by Eq.(7.1.29)

fXm = fX1(nb);

if fXm < eps, return; end %terminate if all the chromosomes are equal

%Reproduction of next generation

for n = 1:Np

X(n,:) = X(n,:) + eta*(fXm – fX1(n))/fXm*(X(nb,:) – X(n,:)); %Eq.(7.1.30)

end

P = gen_encode(X,Nb,l,u);

%Mating/Crossover

is = shuffle([1:Np]);

for n = 1:2:Np – 1

if rand < Pc, P(is(n:n + 1),:) = crossover(P(is(n:n + 1),:),Nb); end

end

%Mutation

P = mutation(P,Nb,Pm);

end

function P = gen_encode(X,Nb,l,u)

% encode a population(X) of state into an array(P) of binary strings

Np=size(X,1); %population size

N = length(Nb); %dimension of the variable(state)

for n = 1:Np

b2 = 0;

for m = 1:N

b1 = b2+1; b2 = b2 + Nb(m);

Xnm =(2^Nb(m)- 1)*(X(n,m) – l(m))/(u(m) – l(m)); %Eq.(7.1.27)

P(n,b1:b2) = dec2bin(Xnm,Nb(m)); %encoding to binary strings

end

end

function X = gen_decode(P,Nb,l,u)

% decode an array of binary strings(P) into a population(X) of state

Np = size(P,1); %population size

N = length(Nb); %dimension of the variable(state)

for n = 1:Np

b2 = 0;

for m = 1:N

b1 = b2 + 1; b2 = b1 + Nb(m) – 1; %Eq.(7.1.28)

X(n,m) = bin2dec(P(n,b1:b2))*(u(m) – l(m))/(2^Nb(m) – 1) + l(m);

end

end

CONSTRAINED OPTIMIZATION 343

function chrms2 = crossover(chrms2,Nb)

% crossover between two chromosomes

Nbb = length(Nb);

b2 = 0;

for m = 1:Nbb

b1 = b2 + 1; bi = b1 + mod(floor(rand*Nb(m)),Nb(m)); b2 = b2 + Nb(m);

tmp = chrms2(1,bi:b2);

chrms2(1,bi:b2) = chrms2(2,bi:b2);

chrms2(2,bi:b2) = tmp;

end

function P = mutation(P,Nb,Pm) % mutation

Nbb = length(Nb);

for n = 1:size(P,1)

b2 = 0;

for m = 1:Nbb

if rand < Pm

b1 = b2 + 1; bi = b1 + mod(floor(rand*Nb(m)),Nb(m)); b2 = b2 + Nb(m);

P(n,bi) = ~P(n,bi);

end

end

end

function is = shuffle(is) % shuffle

N = length(is);

for n = N:-1:2

in = ceil(rand*(n – 1)); tmp = is(in);

is(in) = is(n); is(n) = tmp; %swap the n-th element with the in-th one

end

7.2 CONSTRAINED OPTIMIZATION [L-2, CHAPTER 10]

In this section, only the concept of constrained optimization is introduced. The

explanation for the usage of the corresponding MATLAB routines is postponed

until the next section.

7.2.1 Lagrange Multiplier Method

A class of common optimization problems subject to equality constraints may

be nicely handled by the Lagrange multiplier method. Consider an optimization

problem with M equality constraints.

Min f (x) (7.2.1a)

s.t. h(x) =

h1(x)

h2(x)

:

hM(x)

= 0 (7.2.1b)

According to the Lagrange multiplier method, this problem can be converted

to the following unconstrained optimization problem:

Min l(x, λ) = f (x) + λT h(x) = f (x) +

M

m=1

λmhm(x) (7.2.2)

344 OPTIMIZATION

The solution of this problem, if it exists, can be obtained by setting the derivatives

of this new objective function l(x, λ) with respect to x and λ to zero:

∂

∂x

l(x, λ) =

∂

∂x

f (x) + λT ∂

∂x

h(x) = ∇f (x) +

M

m=1

λm∇hm(x) = 0 (7.2.3a)

∂

∂λ

l(x, λ) = h(x) = 0 (7.2.3b)

Note that the solutions for this system of equations are the extrema of the objective

function. We may know if they are minima/maxima, from the positive/negative-

definiteness of the second derivative (Hessian matrix) of l(x, λ) with respect

to x. Let us see the following examples.

Remark 7.2. Inequality Constraints with the Lagrange Multiplier Method.

Even though the optimization problem involves inequality constraints like

gj (x) ≤ 0, we can convert them to equality constraints by introducing the (nonnegative)

slack variables y2

j as

gj (x) + y2

j = 0 (7.2.4)

Then, we can use the Lagrange multiplier method to handle it like an equalityconstrained

problem.

Example 7.1. Minimization by the Lagrange Multiplier Method.

Consider the following minimization problem subject to a single equality constraint:

Min f (x) = x2

1 + x2

2 (E7.1.1a)

s.t. h(x) = x1 + x2 − 2 = 0 (E7.1.1b)

We can substitute the equality constraint x2 = 2 − x1 into the objective function

(E7.1.1a) so that this problem becomes an unconstrained optimization problem

as

Min f (x1) = x2

1 + (2 − x1)2 = 2×2

1 − 4×1 + 4 (E7.1.2)

which can be easily solved by setting the derivative of this new objective function

with respect to x1 to zero.

∂

∂x1

f (x1) = 4×1 − 4 = 0, x1 = 1, x2

(E7.1.1b) = 2 − x1 = 1 (E7.1.3)

Alternatively, we can apply the Lagrange multiplier method as follows:

Min l(x, λ)

(7.2.2) = x2

1 + x2

2 + λ(x1 + x2 − 2) (E7.1.4)

CONSTRAINED OPTIMIZATION 345

f (x) = x1 + x2 = 5 5 2 2 8

50

0

0

0

(a) A mesh-shaped graph

f (x) = x1 + x2

−5 −5

2 2

h(x) = x1 + x2 − 2 = 0

5

0

0 5

(b) A contour-shaped graph

f (x) = 2

−5

−5

h(x) = x1 + x2 − 2 = 0

Figure 7.10 The objective function with constraint for Example 7.1.

∂

∂x1

l(x, λ)

(7.2.3a) = 2×1 + λ = 0, x1 = −λ/2 (E7.1.5a)

∂

∂x2

l(x, λ)

(7.2.3a) = 2×2 + λ = 0, x2 = −λ/2 (E7.1.5b)

∂

∂λ

l(x, λ)

(7.2.3b) = x1 + x2 − 2 = 0 (E7.1.5c)

x1 + x2

(E7.1.5c) = 2 (E7.1.5a,b) → −λ/2 − λ/2 = −λ = 2, λ= −2 (E7.1.6)

x1

(E7.1.5a) = −λ/2 = 1, x2

(E7.1.5b) = −λ/2 = 1 (Fig. 7.10) (E7.1.7)

In this example, the substitution of (linear) equality constraints is more convenient

than the Lagrange multiplier method. However, it is not always the case,

as illustrated by the next example.

Example 7.2. Minimization by the Lagrange Multiplier Method.

Consider the following minimization problem subject to a single nonlinear

equality constraint:

Min f (x) = x1 + x2 (E7.2.1a)

s.t. h(x) = x2

1 + x2

2 − 2 = 0 (E7.2.1b)

Noting that it is absurd to substitute the equality constraint (E7.2.1b) into

the objective function (E7.2.1a), we apply the Lagrange multiplier method as

below.

Min l(x, λ)

(7.2.2) = x1 + x2 + λ(x2

1 + x2

2) (E7.2.3)

346 OPTIMIZATION

f (x) = x1 + x2 = 2

f (x) = x1 + x2 = −2

0

0

−2

−2

h(x) = x1 + x2 −2 2 2 = 0

2

2

f (x) = x1 + x2

h(x) = x1 + x2 −2 = 0

5

0

0 0

−2

−2 −2

2

2 2

−5

Figure 7.11 The objective function with constraint for Example 7.2.

∂

∂x1

l(x, λ)

(7.2.3a) = 1 + 2λx1 = 0, x1 = −1/2λ (E7.2.4a)

∂

∂x2

l(x, λ)

(7.2.3a) = 1 + 2λx2 = 0, x2 = −1/2λ (E7.2.4b)

∂

∂λ

l(x, λ)

(7.2.3b) = x2

1 + x2

2 − 2 = 0 (E7.2.4c)

x2

1 + x2

2

(E7.2.4c) = 2 (E7.2.4a,b) → (−1/2λ)2 + (−1/2λ)2 = 2, λ= ±1/2 (E7.2.5)

x1

(E7.2.4a) = −1/2λ = ∓1, x2

(E7.2.4b) = −1/2λ = ∓1 (E7.2.6)

Now, in order to tell whether each of these is a minimum or a maximum,

we should determine the positive/negative-definiteness of the second derivative

(Hessian matrix) of l(x, λ) with respect to x.

H =

∂2

∂x2 l(x, λ) = ∂2l/∂x2

1 ∂2l/∂x1∂x2

∂2l/∂x2∂x1 ∂2l/∂x2

2 = 2λ 0

0 2λ (E7.2.7)

This matrix is positive/negative-definite if the sign of λ is positive/negative.

Therefore, the solution (x1, x2) = (−1,−1) corresponding to λ = 1/2 is a (local)

minimum that we want to get, while the solution (x1, x2) = (1, 1) corresponding

to λ = −1/2 is a (local) maximum (see Fig. 7.11).

7.2.2 Penalty Function Method

This method is practically very useful for dealing with the general constrained

optimization problems involving equality/inequality constraints. It is really

CONSTRAINED OPTIMIZATION 347

attractive for optimization problems with fuzzy or loose constraints that are not

so strict with zero tolerance.

Consider the following problem.

Min f (x) (7.2.5a)

s.t. h(x) =

h1(x)

:

hM(x)

= 0, g(x) =

g1(x)

:

gL(x)

≤ 0 (7.2.5b)

The penalty function method consists of two steps. The first step is to construct

a new objective function

Min l(x) = f (x) +

M

m=1

wmh2

m(x) +

L

m=1

vmψ(gm(x)) (7.2.6)

by including the constraint terms in such a way that violating the constraints

would be penalized through the large value of the constraint terms in the objective

function, while satisfying the constraints would not affect the objective function.

The second step is to minimize the new objective function with no constraints

by using the method that is applicable to unconstrained optimization problems,

but a non-gradient-based approach like the Nelder method. Why don’t we use

a gradient-based optimization method? Because the inequality constraint terms

vmψm(gm(x)) attached to the objective function are often determined to be zero as

long as x stays inside the (permissible) region satisfying the corresponding constraint

(gm(x) ≤ 0) and to increase very steeply (like ψm(gm(x)) = exp(emgm(x))

as x goes out of the region; consequently, the gradient of the new objective function

may not carry useful information about the direction along which the value

of the objective function decreases.

From an application point of view, it might be a good feature of this method

that we can make the weighting coefficient (wm,vm, and em) on each penalizing

constraint term either large or small depending on how strictly it should be

satisfied.

Let us see the following example.

Example 7.3. Minimization by the Penalty Function Method.

Consider the following minimization problem subject to several nonlinear

inequality constraints:

Min f (x) = {(x1 + 1.5)2 + 5(x2 − 1.7)2}{(x1 − 1.4)2 + 0.6(x2 − 0.5)2}

(E7.3.1a)

348 OPTIMIZATION

s.t. g(x) =

−x1

−x2

3×1 − x1x2 + 4×2 − 7

2×1 + x2 − 3

3×1 − 4×2

2 − 4×2

≤

0

0

0

0

0

(E7.3.1b)

According to the penalty function method, we construct a new objective function

(7.2.6) as

Min l(x) = {(x1 + 1.5)2 + 5(x2 − 1.7)2}{(x1 − 1.4)2 + 0.6(x2 − 0.5)2}

+

5

m=1

vmψm(gm(x)) (E7.3.2a)

where

vm = 1, ψm(gm(x)) = 0 if gm(x) ≤ 0 (constraint satisfied)

exp(emgm(x)) if gm(x) > 0 (constraint violated)

,

em = 1 ∀ m = 1, . . . , 5 (E7.3.2b)

%nm722 for Ex.7.3

% to solve a constrained optimization problem by penalty ftn method.

clear, clf

f =’f722p’;

x0=[0.4 0.5]

TolX = 1e-4; TolFun = 1e-9; alpha0 = 1;

[xo_Nelder,fo_Nelder] = opt_Nelder(f,x0) %Nelder method

[fc_Nelder,fo_Nelder,co_Nelder] = f722p(xo_Nelder) %its results

[xo_s,fo_s] = fminsearch(f,x0) %MATLAB built-in fminsearch()

[fc_s,fo_s,co_s] = f722p(xo_s) %its results

% including how the constraints are satisfied or violated

xo_steep = opt_steep(f,x0,TolX,TolFun,alpha0) %steepest descent method

[fc_steep,fo_steep,co_steep] = f722p(xo_steep) %its results

[xo_u,fo_u] = fminunc(f,x0); % MATLAB built-in fminunc()

[fc_u,fo_u,co_u] = f722p(xo_u) %its results

function [fc,f,c] = f722p(x)

f=((x(1)+ 1.5)^2 + 5*(x(2)- 1.7)^2)*((x(1)- 1.4)^2 + .6*(x(2)-.5)^2);

c=[-x(1); -x(2); 3*x(1) – x(1)*x(2) + 4*x(2) – 7;

2*x(1)+ x(2) – 3; 3*x(1) – 4*x(2)^2 – 4*x(2)]; %constraint vector

v=[1 1 1 1 1]; e = [1 1 1 1 1]’; %weighting coefficient vector

fc = f +v*((c > 0).*exp(e.*c)); %new objective function

CONSTRAINED OPTIMIZATION 349

>> nm722

xo_Nelder = 1.2118 0.5765

fo_Nelder = 0.5322 %min value

co_Nelder = -1.2118

-0.5765

-1.7573 %high margin

-0.0000 %no margin

-0.0000 %no margin

xo_s = 1.2118 0.5765

fo_s = 0.5322 %min value

xo_steep = 1.2768 0.5989

fo_steep = 0.2899 %not a minimum

co_steep = -1.2768

-0.5989

-1.5386

0.1525 %violating

-0.0001

Warning: .. Gradient must be provided

……..

Maximum # of function evaluations

exceeded;

xo_u = 1.2843 0.6015

fo_u = 0.2696 %not a minium

Note that the shape of the penalty function as well as the values of the

weighting coefficients is set by the users to cope with their own problems. Then,

we apply an unconstrained optimization technique like the Nelder–Mead method,

which is not a gradient-based approach. Here, we make the program “nm722.m”,

which applies not only the routine “opt_Nelder()” and the MATLAB built-in

routine “fminsearch()” for cross-check, but also the routine “opt_steep()” and

the MATLAB built-in routine “fminunc()” in order to show that the gradientbased

methods do not work well. To our expectation, the running results listed

above and depicted in Fig. 7.12 show that, for the objective function (E7.3.2a)

augmented with the penalized constraint terms, the gradient-based routines

“opt_steep()” and “fminunc()” are not so effective as the non-gradientbased

routines “opt_Nelder()” and “fminsearch()” in finding the constrained

0.6 0.8 1 1.2 1.4 1.6 1.8 2

0

0.5

1

1.5

: opt_Nelder()

+ : fminsearch()

: opt_steep()

3×1 − x1x2 + 4×2 ≤ 7 × : fminunc()

2×1 + x2 ≤ 3

3×1 − 4×22

− 4×2 ≤ 0

Figure 7.12 The contours for the objective function (E7.3.1a) and the admissible region

satisfying the inequality constraints.

350 OPTIMIZATION

minimum, which is on the intersection of the two boundary curves corresponding

to the fourth and fifth constraints of (E7.3.1b).

7.3 MATLAB BUILT-IN ROUTINES FOR OPTIMIZATION

In this section, we apply severalMATLAB built-in unconstrained optimization routines

including “fminsearch()” and “fminunc()” to the same problem, expecting

that their nuances will be clarified. Our intention is not to compare or evaluate the

performances of these sophisticated routines, but rather to give the readers some

feelings for their functional differences.We also introduce the routine “linprog()”

implementing Linear Programming (LP) scheme and “fmincon()” designed for

attacking the (most challenging) constrained optimization problems. Interested

readers are encouraged to run the tutorial routines “optdemo” or “tutdemo”, which

demonstrate the usages and performances of the representative built-in optimization

routines such as “fminunc()” and “fmincon()”.

7.3.1 Unconstrained Optimization

In order to try applying the unconstrained optimization routines introduced

in Section 7.1 and see how they work, we made the MATLAB program

“nm731_1.m”, which uses those routines for solving the problem

Min f (x) = (x1 − 0.5)2(x1 + 1)2 + (x2 + 1)2(x2 − 1)2 (7.3.1)

where the contours and the (local) maximum/minimum/saddle points of this

objective function are depicted in Fig. 7.13.

−1.5

−1.5

−1

−1

−0.5

−0.5

0

0

0.5

0.5

1

1 minimum

maximum

saddle

steepest descent

Newton

1.5

Figure 7.13 The contours, minima,maxima, and saddle points of the objective function (7.3.1).

MATLAB BUILT-IN ROUTINES FOR OPTIMIZATION 351

%nm731_1

% to minimize an objective function f(x) by various methods.

clear, clf

% An objective function and its gradient function

f = inline(’(x(1) – 0.5).^2.*(x(1) + 1).^2 + (x(2)+1).^2.*(x(2) – 1).^2’,’x’);

g0 = ’[2*(x(1)- 0.5)*(x(1)+ 1)*(2*x(1)+ 0.5) 4*(x(2)^2 – 1).*x(2)]’;

g = inline(g0,’x’);

x0 = [0 0.5] %initial guess

[xon,fon] = opt_Nelder(f,x0) %min point, its ftn value by opt_Nelder

[xos,fos] = fminsearch(f,x0) %min point, its ftn value by fminsearch()

[xost,fost] = opt_steep(f,x0) %min point, its ftn value by opt_steep()

TolX = 1e-4; MaxIter = 100;

xont = Newtons(g,x0,TolX,MaxIter);

xont,f(xont) %minimum point and its function value by Newtons()

[xocg,focg] = opt_conjg(f,x0) %min point, its ftn value by opt_conjg()

[xou,fou] = fminunc(f,x0) %min point, its ftn value by fminunc()

Noting that it depends mainly on the initial value x0 whether each routine

succeeds in finding a minimum point, we summarize the results of running those

routines with various initial values in Table 7.2. It can be seen from this table

that the gradient-based optimization routines like “opt_steep()”, “Newtons()”,

“opt_conj()”, and “fminunc()” sometimes get to a saddle point or even a

maximum point (Remark 7.1) and that the routines do not always approach the

extremum that is closest to the initial point. It is interesting to note that even

the non-gradient-based MATLAB built-in routine “fminsearch()” may get lost,

while our routine “opt_Nelder()” works well for this case. We cannot, however,

conclude that this routine is better than that one based on only one trial,

because there may be some problems for which the MATLAB built-in routine

works well, but our routine does not. What we can state over this happening is

that no human work is free from defect.

Now, we will see a MATLAB built-in routine “lsqnonlin(f,x0,l,u,

options,p1,..)”, which presents a nonlinear least-squares (NLLS) solution to

Table 7.2 Results of Running Several Unconstrained Optimization Routines with

Various Initial Values

x0 opt−Nelder fminsearch opt−steep Newtons opt−conjg fminunc

[0, 0] [−1, 1] [0.5, 1] [0.5, 0] [−0.25, 0] [0.5, 0] [0.5, 0]

(minimum) (minimum) (saddle) (maximum) (saddle) (saddle)

[0, 0.5] [0.5, 1] [0.02, 1] [0.5, 1] [−0.25, −1] [0.5, 1] [0.5, 1]

(minimum) (lost) (minimum) (saddle) (minimum) (minimum)

[0.4, 0.5] [0.5, 1] [0.5, 1] [0.5, 1] [0.5, −1] [0.5, 1] [0.5, 1]

(minimum) (minimum) (minimum) (minimum) (minimum) (minimum)

[−0.5, 0.5] [0.5, 1] [−1, 1] [−1, 1] [−0.25, −1] [−1, 1] [−1, 1]

(minimum) (minimum) (minimum) (saddle) (minimum) (minimum)

[−0.8, 0.5] [−1, 1] [−1, 1] [−1, 1] [−1, −1] [−1, 1] [−1, 1]

(minimum) (minimum) (minimum) (minimum) (minimum) (minimum)

352 OPTIMIZATION

the minimization problem

Min

N

n=1

f 2

n (x) (7.3.2)

The routine needs at least the vector or matrix function f(x) and the initial guess

x0 as its first and second input arguments, where the components of f(x) = [f1(x) · · · fN(x)]T are squared, summed, and then minimized over x. In order to

learn the usage and function of this routine, we made the MATLAB program

“nm731_2.m”, which uses it to find a second-degree polynomial approximating

the function

y = f (x) =

1

1 + 8×2 (7.3.3)

For verification, the result of using the NLLS routine “lsqnonlin()” is compared

with that obtained from directly applying the routine “polyfits()” introduced

in Section 3.8.2.

>> nm731_2

ao_lsq = [-0.1631 -0.0000 0.4653], ao_fit = [-0.1631 -0.0000 0.4653]

%nm731_2 try using lsqnonlin() for a vector-valued objective ftn F(x)

clear, clf

N = 3; a0 = zeros(1,N); %the initial guess of polynomial coefficient vector

ao_lsq = lsqnonlin(’f731_2’,a0) %parameter estimate by lsqnonlin()

xx = -2+[0:400]/50; fx = 1./(1+8*xx.*xx);

ao_fit = polyfits(xx,fx,N – 1) %parameter estimate by polyfits()

function F = f731_2(a)

%error between the polynomial a(x) and f(x) = 1/(1+8x^2)

xx = -2 +[0:200]/50; F = polyval(a,xx) – 1./(1+8*xx.*xx);

7.3.2 Constrained Optimization

Generally, constrained optimization is very complicated and difficult to deal with.

So we will not cover the topic in details here and instead, will just introduce the

powerful MATLAB built-in routine “fmincon()”, which makes us relieved from

a big headache.

This routine is well-designed for attacking the optimization problems subject

to some constraints:

function [c,ceq] = f722c(x)

c = [-x(1); -x(2); 3*x(1) – x(1)*x(2) + 4*x(2)- 7;

2*x(1)+ x(2)- 3; 3*x(1)- 4*x(2)^2 – 4*x(2)]; %inequality constraints

ceq = []; %equality constraints

MATLAB BUILT-IN ROUTINES FOR OPTIMIZATION 353

Usage of the MATLAB 6.x built-in function “fmincon()”

[xo,fo,.] = fmincon(’ftn’,x0,A,b,Aeq,beq,l,u,’nlcon’,options,p1,p2,.)

ž Input Arguments (at least four input arguments ’ftn’,x0,A and b required)

’ftn’ : an objective function f (x) to be minimized, usually defined in an

M-file, but can be defined as an inline function, which will

remove the necessity of quotes(’’).

x0 : an initial guess x0 of the solution

A,b : a linear inequality constraints Ax ≤ b; to be given as [] if not

applied.

Aeq,beq: a linear equality constraints Aeqx = beq ; to be given as [] if not

applied.

l,u : lower/upper bound vectors such that l ≤ x ≤ u; to be given as []

if not applied, set l(i) = -inf/u(i) = inf if x(i) is not

bounded below/above.

’nlcon’: a nonlinear constraint function defined in an M-file, supposed to

return the two output arguments for a given x; the first one being

the LHS (vector) of inequality constraints c(x) ≤ 0 and the

second one being the LHS (vector) of equality constraints

ceq (x) = 0; to be given as [] if not applied.

options: used for setting the display parameter, the tolerances for xo and

f (xo), and so on; to be given as [] if not applied. For details,

type ‘help optimset’ into the MATLAB command window.

p1,p2,.: the problem-dependent parameters to be passed to the objective

function f (x) and the nonlinear constraint functions c(x), ceq (x).

ž Output Arguments

xo : the minimum point (xo) reached in the permissible region

satisfying the constraints

fo : the minimized function value f (xo)

%nm732_1 to solve a constrained optimization problem by fmincon()

clear, clf

ftn=’((x(1) + 1.5)^2 + 5*(x(2) – 1.7)^2)*((x(1)-1.4)^2 + .6*(x(2)-.5)^2)’;

f722o = inline(ftn,’x’);

x0 = [0 0.5] %initial guess

A = []; B = []; Aeq = []; Beq = []; %no linear constraints

l = -inf*ones(size(x0)); u = inf*ones(size(x0)); % no lower/upperbound

options = optimset(’LargeScale’,’off’); %just [] is OK.

[xo_con,fo_con] = fmincon(f722o,x0,A,B,Aeq,Beq,l,u,’f722c’,options)

[co,ceqo] = f722c(xo_con) % to see how constraints are.

354 OPTIMIZATION

Min f (x) (7.3.4)

s.t. Ax ≤ b, Aeqx = beq , c(x) ≤ 0, ceq (x) = 0 and l ≤ x ≤ u (7.3.5)

A part of its usage can be seen by typing ‘help fmincon’ into the MATLAB

command window as summarized in the above box. We make the MATLAB

program “nm732_1.m”, which uses the routine “fmincon()” to solve the problem

presented in Example 7.3. Interested readers are welcomed to run it and observe

the result to check if it agrees with that of Example 7.3.

There are two more MATLAB built-in routines to be introduced in this section.

One is

“fminimax(’ftn’,x0,A,b,Aeq,beq,l,u,’nlcon’,options,p1,..)”,

which is focused on minimizing the maximum among several components of

the vector/matrix-valued objective function f(x) = [f1(x) · · · fN(x)]T subject to

some constraints as described below. Its usage is almost the same as that of

“fmincon()”.

Min

x {Max

n {fn(x)}} (7.3.6)

s.t. Ax ≤ b, Aeqx = beq , c(x) ≤ 0, ceq (x) = 0, and l ≤ x ≤ u (7.3.7)

The other is the constrained linear least-squares (LLS) routine

“lsqlin(C,d,A,b,Aeq,beq,l,u,x0,options,p1,..)”,

whose job is to solve the problem

Min

x ||Cx − d||2 (7.3.8)

s.t. Ax ≤ b, Aeqx = beq and l ≤ x ≤ u (7.3.9)

In order to learn the usage and function of this routine, we make the MATLAB

program “nm732_2.m”, which uses both “fminimax()” and “lsqlin()” to find

a second-degree polynomial approximating the function (7.3.3) and compares

the results with that of applying the routine “lsqnonlin()” introduced in the

previous section for verification. From the plotting result depicted in Fig. 7.14,

note the following.

ž We attached no constraints to the “fminimax()” routine, so it yielded the

approximate polynomial curve minimizing the maximum deviation from

f (x).

ž We attached no constraints to the constrained linear least-squares routine

“lsqlin()” either, so it yielded the approximate polynomial curve

minimizing the sum (integral) of squared deviation from f (x), which is

MATLAB BUILT-IN ROUTINES FOR OPTIMIZATION 355

−2 −1.5 −1 −0.5 0 0.5 1 1.5 2

−0.2

0.2

0.4

0.6

0.8

0

1

least squares

fminimax

Chebyshev

f (x) = 1

1 + 8x 2

Figure 7.14 Approximation of a curve by a second-degree polynomial function based on the

minimax, least-squares, and Chebyshev methods.

the same as the (unconstrained) least squares solution obtained by using the

routine “lsqnonlin()”.

ž Another MATLAB built-in routine “lsqnonneg()” gives us a nonnegative

LS (NLS) solution to the problem (7.3.8).

%nm732_2: uses fminimax() for a vector-valued objective ftn f(x)

clear, clf

f = inline(’1./(1+8*x.*x)’,’x’);

f73221 = inline(’abs(polyval(a,x) – fx)’,’a’,’x’,’fx’);

f73222 = inline(’polyval(a,x) – fx’,’a’,’x’,’fx’);

N = 2; % the degree of approximating polynomial

a0 = zeros(1,N + 1); %initial guess of polynomial coefficients

xx = -2+[0:200]’/50; %intermediate points

fx = feval(f,xx); % and their function values f(xx)

ao_m = fminimax(f73221,a0,[],[],[],[],[],[],[],[],xx,fx) %fminimax sol

for n = 1:N+1, C(:,n) = xx.^(N + 1 – n); end

ao_ll = lsqlin(C,fx) %linear LS to minimize (Ca – fx)^2 with no constraint

ao_ln = lsqnonlin(f73222,a0,[],[],[],xx,fx) %nonlinear LS

c2 = cheby(f,N,-2,2) %Chebyshev polynomial over [-2,2]

plot(xx,fx,’:’, xx,polyval(ao_m,xx),’m’, xx,polyval(ao_ll,xx),’r’)

hold on, plot(xx,polyval(ao_ln,xx),’b’, xx,polyval(c2,xx),’–’)

axis([-2 2 -0.4 1.1])

7.3.3 Linear Programming (LP)

The linear programming (LP) scheme implemented by the MATLAB built-in

routine

“[xo,fo] = linprog(f,A,b,Aeq,Beq,l,u,x0,options)”

is designed to solve an LP problem, which is a constrained minimization problem

as follows.

Min f (x) = fT x (7.3.10a)

subject to Ax ≤ b, Aeqx = beq , and l ≤ x ≤ u (7.3.10b)

356 OPTIMIZATION

%nm733 to solve a Linear Programming problem.

% Min f*x=-3*x(1)-2*x(2) s.t. Ax <= b, Aeq = beq and l <= x <= u

x0 = [0 0]; %initial point

f = [-3 -2]; %the coefficient vector of the objective function

A = [3 4; 2 1]; b = [7; 3]; %the inequality constraint Ax <= b

Aeq = [-3 2]; beq = 2; %the equality constraint Aeq*x = beq

l = [0 0]; u = [10 10]; %lower/upper bound l <= x <= u

[xo_lp,fo_lp] = linprog(f,A,b,Aeq,beq,l,u)

cons_satisfied = [A; Aeq]*xo_lp-[b; beq] %how constraints are satisfied

f733o=inline(’-3*x(1)-2*x(2)’, ’x’);

[xo_con,fo_con] = fmincon(f733o,x0,A,b,Aeq,beq,l,u)

It produces the solution (column) vector xo and the minimized value of the

objective function f (xo) as its first and second output arguments xo and fo,

where the objective function and the constraints excluding the constant term are

linear in terms of the independent (decision) variables. It works for such linear

optimization problems as (7.3.10) more efficiently than the general constrained

optimization routine “fmincon()”.

The usage of the routine “linprog()” is exemplified by the MATLAB program

“nm733.m”, which uses the routine for solving an LP problem described as

Min f (x) = fT x = [−3 − 2][x1 x2]T = −3×1 − 2×2 (7.3.11a)

s.t.

Ax =

−3 2

3 4

2 1

x1

x2 =

≤≤

2

7

3

= b and

l = 0

0 ≤ x = x1

x2 ≤ 10

10 = u (7.3.11b)

−0.5 0 0.5

0.5

1

1

1.5

1.5

2

2

2.5

2.5

0 x1 = 0

x2 = 0

fTx = −3×1 − 2×2 = −5

fTx = −3×1 − 2×2 = −4

−3×1 + 2×2 = 2

3×1 + 4×2 = 7

−2×1 + x2 = 3

x2

x1

Figure 7.15 The objective function, constraints, and solutions of an LP problem.

PROBLEMS 357

Table 7.3 The Names of MATLAB Built-In Minimization Routines in MATLAB 5.x/6.x

Unconstrained Minimization Constrained Minimization

Minimization

Methods Bracketing

Non-Gradient-

Based

Gradient-

Based Linear Nonlinear

Linear

LS

Nonlinear

LS Minimax

MATLAB 5.x fmin fmins fminu lp constr leastsq conls minimax

MATLAB 6.x fminbnd fminsearch fminunc linprog fmincon lsqnonlin lsqlin fminimax

The program also applies the general constrained minimization routine “fmincon()”

to solve the same problem for cross-check. Readers are welcome to run

the program and see the results.

>> nm733

xo_lp = [0.3333 1.5000], fo_lp = -4.0000

cons_satisfied = -0.0000 % <= 0(inequality)

-0.8333 % <= 0(inequality)

-0.0000 % = 0(equality)

xo_con = [0.3333 1.5000], fo_con = -4.0000

In this result, the solutions obtained by using the two routines “linprog()” and

“fmincon()” agree with each other, satisfying the inequality/equality constraints

and it can be assured by Fig. 7.15.

In Table 7.3, the names of MATLAB built-in minimization routines in MATLAB

version 5.x and 6.x are listed.

PROBLEMS

7.1 Modification of Golden Search Method

In fact, the golden search method explained in Section 7.1 requires only

one function evaluation per iteration, since one point of a new interval

coincides with a point of the previous interval so that only one trial point

is updated. In spite of this fact, the MATLAB routine “opt_gs()” implementing

the method performs the function evaluations twice per iteration.

An improvement may be initiated by modifying the declaration type as

[xo,fo] = opt_gs1(f,a,e,fe,r1,b,r,TolX,TolFun,k)

so that anyone could use the new routine as in the following program,

where its input argument list contains another point (e) as well as the new

end point (b) of the next interval, its function value (fe), and a parameter

(r1) specifying if the point is the left one or the right one. Based on this

idea, how do you revise the routine “opt_gs()” to cut down the number

of function evaluations?

358 OPTIMIZATION

%nm7p01.m to perform the revised golden search method

f701 = inline(’x.*(x-2)’, ’x’);

a = 0; b = 3; r = (sqrt(5)-1)/2;

TolX = 1e-4; TolFun = 1e-4; MaxIter=100;

h = b – a; rh = r*h;

c = b – rh; d = a + rh;

fc = f701(c); fd = f701(d);

if fc < fd, [xo,fo] = opt_gs1(f701,a,c,fc,1 – r,d,r,TolX,TolFun,MaxIter)

else [xo,fo] = opt_gs1(f701,c,d,fd,r,b,r, TolX,TolFun,MaxIter)

end

7.2 Nelder–Mead, Steepest Descent, Newton, SA, GA and fminunc(), fminsearch()

Consider a two-variable objective function

f (x) = x4

1 − 12×2

1 − 4×1 + x4

2 − 16×2

2 − 5×2 (P7.2.1)

− 20 cos(x1 − 2.5) cos(x2 − 2.9)

whose gradient vector function is

g(x) = ∇f (x) = 4×3

1 − 24×1 − 4 + 20 sin(x1 − 2.5) cos(x2 − 2.9)

4×3

2 − 32×2 − 5 + 20 cos(x1 − 2.5) sin(x2 − 2.9)

(P7.2.2)

You have the MATLAB functions f7p02(), g7p02() defining the objective

function f (x) and its gradient function g(x). You also have a part of the

MATLAB program which plots a mesh/contour-type graphs for f (x). Note

that this gradient function has nine zeros as listed in Table P7.2.1.

Table P7.2.1 Extrema (Maxima/Minima) and Saddle Points of the Function (P7.2.1)

Points Signs of ∂2f/∂x2

i Points Signs of ∂2f/∂x2

i

(1) [0.6965 −0.1423] −, − M (6) [−1.6926 −0.1183]

(2) [2.5463 −0.1896] (7) [−2.6573 −2.8219] +, + m

(3) [2.5209 2.9027] +, + G (8) [−0.3227 −2.4257]

(4) [−0.3865 2.9049] (9) [2.5216 −2.8946] +, + m

(5) [−2.6964 2.9031]

(a) From the graphs (including Fig. P7.2) which you get by running the

(unfinished) program, determine the characteristic of each of the nine

points, that is, whether it is a local maximum(M)/minimum(m), the

global minimum(G) or a saddle point(S) which is a minimum with

respect to one variable and a maximum with respect to another variable.

Support your judgment by telling the signs of the second derivatives of

f (x) with respect to x1 and x2.

PROBLEMS 359

4

3

2

1

0

−1

−2

−3

−4

−4 −3 −2 −1 0 1 2 3 4

5 4 3

2

9

1

8

7

6

Figure P7.2 The contour, extrema and saddle points of the objective function (P7.2.1).

%nm7p02 to minimize an objective ftn f(x) by the Newton method

f = ’f7p02’; g = ’g7p02’;

l = [-4 -4]; u = [4 4];

x1 = l(1):.25:u(1); x2 = l(2):.25:u(2); [X1,X2] = meshgrid(x1,x2);

for m = 1:length(x1)

for n = 1:length(x2), F(n,m) = feval(f,[x1(m) x2(n)]); end

end

figure(1), clf, mesh(X1,X2,F)

figure(2), clf,

contour(x1,x2,F,[-125 -100 -75 -50 -40 -30 -25 -20 0 50])

… … … …

function y = f7p02(x)

y = x(1)^4 – 12*x(1)^2 – 4*x(1) + x(2)^4 – 16*x(2)^2 – 5*x(2)…

-20*cos(x(1) – 2.5)*cos(x(2) – 2.9);

function [df,d2f] = g7p02(x) % the 1st/2nd derivatives

df(1) = 4*x(1)^3 – 24*x(1) – 4 + 20*sin(x(1) – 2.5)*cos(x(2) – 2.9);%(P7.2.2)

df(2) = 4*x(2)^3 – 32*x(2)-5 + 20*cos(x(1) – 2.5)*sin(x(2) – 2.9);%(P7.2.2)

d2f(1) = 12*x(1)^2 – 24 + 20*cos(x(1) – 2.5)*cos(x(2) – 2.9); %(P7.2.3)

d2f(2) = 12*x(2)^2 – 32 + 20*cos(x(1) – 2.5)*cos(x(2) – 2.9); %(P7.2.3)

∂2f/∂x2

1 = 12×2

1 − 24 + 20 cos(x1 − 2.5) cos(x2 − 2.9)

∂2f/∂x2

2 = 12×2

1 − 32 + 20 cos(x1 − 2.5) cos(x2 − 2.9)

(P7.2.3)

(b) Apply the Nelder–Mead method, the steepest descent method, the Newton

method, the simulated annealing (SA), genetic algorithm (GA), and

the MATLAB built-in routines fminunc(), fminsearch() to minimize

the objective function (P7.2.1) and fill in Table P7.2.2 with the number

and character of the point reached by each method.

360 OPTIMIZATION

Table P7.2.2 Points Reached by Several Optimization Routines

Initial Point Reached Point

x0 Nelder Steepest Newton fminunc fminsearch SA GA

(0, 0) (5)/m

(1, 0) (3)/G

(1, 1) (9)/m

(0, 1) (3)/G

(−1, 1) (5)/m

(−1, 0) ≈(3)/G

(−1, −1) (3)/G

(0, −1) (9)/m

(1, −1) (9)/m

(2, 2) (3)/G

(−2, −2) (7)/m

(c) Overall, the point reached by each minimization algorithm depends on

the starting point—that is, the initial value of the independent variable

as well as the characteristic of the algorithm. Fill in the blanks in

the following sentences. Most algorithms succeed to find the global

minimum if only they start from the initial point ( , ), ( , ), ( , ), or ( , ).

An algorithm most possibly goes to the closest local minimum (5) if

launched from ( , ) or ( , ), and it may go to the closest local minimum

(7) if launched from ( , ) or ( , ). If launched from ( , ), it may go to

one of the two closest local minima (7) and (9) and if launched from

( , ), it most possibly goes to the closest local minimum (9). But, the

global optimization techniques SA and GA seem to work fine almost

regardless of the starting point, although not always.

7.3 Minimization of an Objective Function Having Many Local Minima/

Maxima

Consider the problem of minimizing the following objective function

Min f (x) = sin(1/x)/((x − 0.2)2 + 0.1) (P7.3.1)

which is depicted in Fig. P7.3. The graph shows that this function has

infinitely many local minima/maxima around x = 0 and the global minimum

about x = 0.2.

(a) Find the solution by using the MATLAB built-in routine “fminbnd()”.

Is it plausible?

(b) With nine different values of the initial guess x0 = 0.1, 0.2, . . . , 0.9, use

the four MATLAB routines “opt_Nelder()”, “opt_steep()”, “fminunc()”,

and “fminsearch()” to solve the problem. Among those 36

tryouts, how many times have you got the right solution?

PROBLEMS 361

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

10

−10

5

−5

0

Figure P7.3 The graph of f(x) = sin(1/x)/((x − 0.2)2 + 0.1) havingmanylocal minima/maxima.

(c) With the values of the parameters set to l = 0, u = 1, q = 1, εf = 10−9,

kmax = 1000 and the initial guess x0 = 0.1, 0.2, . . . , 0.9, use the SA

(simulated annealing) routine “sim_anl()” to solve the problem. You

can test the performance of the routine and your luck by running the

routine four times for the same problem and finding the probability of

getting the right solution.

(d) With the values of the parameters set to l = 0, u = 1, Np = 30, Nb = 12, Pc = 0.5, Pm = 0.01, η = 1, kmax = 1000 and the initial guess x0 = 0.1, 0.2, . . . , 0.9, use the GA (genetic algorithm) routine “genetic()”

to solve the problem. As in (c), you can run the routine four times for

the same problem and find the probability of getting the right solution

in order to test the performance of the routine and your luck.

7.4 Linear Programming Method

Consider the problem of maximizing a linear objective function

Max f (x) = fT x = [3 2 −1 ][ x1 x2 x3 ]T (P7.4.1a)

subject to the constraints

Ax =

3 −2 0

−3 −4 0

−2 −1 0

x1

x2

x3

=≥≥

−2

−7

−3

= b and

l =

0

0

0

≤ x =

x1

x2

x3

≤

10

10

10

= u

(P7.4.1b)

Jessica is puzzled with this problem, which is not a minimization but a

maximization. How do you suggest her to solve it? Make the program that

uses the MATLAB built-in routines “linprog()” and “fmincon()” to solve

this problem and run it to get the solutions.

362 OPTIMIZATION

7.5 Constrained Optimization and Penalty Method

Consider the problem of minimizing a nonlinear objective function

Minxf (x) = −3×1 − 2×2 + M(3×1 − 2×2 + 2)2 (P7.5.1a)

(M : a large positive number)

subject to the constraints

3 4

−2 −1x1

x2 ≤ 7

≥ −3

and l = 0

0 ≤ x = x1

x2 ≤ 10

10 = u

(P7.5.1b)

(a) With the two values of the weighting factor M = 20 and 10,000 in

the objective function (P7.5.1a), apply the MATLAB built-in routine

“fmincon()” to find the solutions to the above constrained minimization

problem. In order to do this job, you might have to make the variable

parameter M passed to the objective function (defined in an M-file)

either through “fmincon()” or directly by declaring the parameter as

global both in the main program and in the M-file defining (P7.5.1a). In

case you are going to have the parameter passed through “fmincon()”

to the objective function, you should have the parameter included in

the input argument list of the objective function as

function f=f7p05M(x,M)

f = -3*x(1)-2*x(2)+M*(3*x(1)-2*x(2)+2).^2;

Additionally, you should give empty matrices ([]) as the ninth input

argument (for a nonlinear inequality/equality constraint function ‘nonlcon’)

as well as the 10th one (for ‘options’) and the value of M as

the 11th one of the routine “fmincon()”.

xo = fmincon(’f7p05M’,x0,A,b,[],[],l,u,[],[],M)

For reference, type ‘help fmincon’ into the MATLAB command

window.

(b) Noting that the third (squared) term of the objective function (P7.5.1a)

has its minimum value of zero for 3×1 − 2×2 + 2 = 0 and, thus, it actually

represents the penalty (Section 7.2.2) imposed for not satisfying the

equality constraint

3×1 − 2×2 + 2 = 0 (P7.5.2)

tell which of the solutions obtained in (a) is more likely to satisfy this

constraint and support your answer by comparing the values of the

left-hand side of this equality for the two solutions.

PROBLEMS 363

(c) Removing the third term from the objective function and splitting the

equality constraint into two reversed inequality constraints, we can

modify the problem as follows:

Minxf (x) = −3×1 − 2×2 (P7.5.3a)

subject to the constraints

3 4

−2 −1

3 −2

3 −2

x1

x2

≤ 7

≥ −3

≤ −2

≥ −2

and (P7.5.3b)

l = 0

0 ≤ x = x1

x2 ≤ 10

10 = u

Noting that this fits the linear programming, apply the routine “linprog()”

to solve this problem.

(d) Treating the equality constraint separately from the inequality constraints,

we can modify the problem as follows:

Minxf (x) = −3×1 − 2×2 (P7.5.4a)

subject to the constraints

3 −2

3 4

−2 −1

x1

x2 =

≤≥

−2

7

−3

and l = 0

0 ≤ x = x1

x2 ≤ 10

10 = u

(P7.5.4b)

Apply the two routines “linprog()” and “fmincon()” to solve this

problem and see if the solutions agree with the solution obtained in (c).

(cf) Note that, in comparison with the routine “fmincon()”, which can solve a general

nonlinear optimization problem, the routine “linprog()” is made solely

for dealing with a class of optimization problems having a linear objective

function with linear constraints.

7.6 Nonnegative Constrained LS and Constrained Optimization

Consider the problem of minimizing a nonlinear objective function

Minx ||Cx − d||2 = [Cx − d]T [Cx − d] (P7.6.1a)

subject to the constraints

x = x1

x2 ≥ 0

0 = l (P7.6.1b)

where

C =

1 2

3 4

5 1

, d =

5.1

10.8

6.8

(P7.6.1c)

364 OPTIMIZATION

(a) Noting that this problem has no other constraints than the lower bound,

apply the constrained linear least-squares routine “lsqlin()” to find

the solution.

(b) Noting that the lower bounds for all the variables are zeros, apply the

MATLAB built-in routine “lsqnonneg()” to find the solution.

(c) Apply the general-purpose constrained optimization routine “fmincon()”

to find the solution.

7.7 Constrained Optimization Problems

Solve the following constrained optimization problems by using the MATLAB

built-in routine “fmincon()”.

(a) Minx x3

1 − 5×2

1 + 6×1 + x2

2 − 2×2 + x3 (P7.7.1a)

subject to the constraints

x2

1 + x2

2 − x3 ≤ 0

x2

1 + x2

2 + x2

3 ≥ 6

x3 ≤ 5

and x =

x1

x2

x3

≥

0

0

0

= l (P7.7.1b)

Try the routine “fmincon()” with the initial guesses listed in

Table P7.7.

Table P7.7 The Results of Applying ‘‘fmincon()’’ with Different Initial Guess

Initial Guess

x0

Lower

Bound xo f (xo) Remark (warning ?)

[0 0 0] 0 No feasible solution

[1 1 5] 0 Not a minimum

(a)

[0 0 5] 0 Minimum

[1 0 2] 0 [1.29 0.57 2] 2.74

[0 0 0] 0 [0 0 0] 0

(b1)

[10 10 10] 0 Maximum (good)

[0 0 0] 0 No feasible solution

(b2) [10 10 10] 0 Not a minimum, but the max

[0.1 0.1 3] 0 One of many minima

[0 0 0] 0

(c1) [0.1 0.1 0.1] 0 [1 1 1] 3 Maximum (good)

[0 1 2] 0

[0 0 0] 0 [1 1 1] 3 Not a minimum, but the max

(c2) [0.1 0.1 0.1] 0

[0 1 2] 0 One of many minima

[1.0 0.5] 0 Weird (warning)

[0.2 0.3] 0 (d) [10.25 0] ∞

[2 5] 0 [5.77 8.17] 25.98

[100 10] 0 Minimum

PROBLEMS 365

(b1) Maxx x1x2x3 (P7.7.2a)

subject to the constraints

x1x2 + x2x3 + x3x1 =3 and x =

x1

x2

x3

≥

0

0

0

(P7.7.2b)

Try the routine “fmincon()” with the initial guesses listed in

Table P7.7.

(b2) Minx x1x2x3 (P7.7.3a)

subject to the constraints (P7.7.2b).

Try the routine “fmincon()” with the initial guesses listed in

Table P7.7.

(c1) Maxx x1x2 + x2x3 + x3x1 (P7.7.4a)

subject to the constraints

x1 + x2 + x3 =3 and x =

x1

x2

x3

≥

0

0

0

(P7.7.4b)

Try the routine “fmincon()” with the initial guesses listed in

Table P7.7.

(c2) Minx x1x2 + x2x3 + x3x1 (P7.7.5a)

subject to the constraints (P7.7.4b).

Try the routine “fmincon()” with the initial guesses listed in

Table P7.7.

(d) Minx

10000

x1x2

2

(P7.7.6a)

subject to the constraints

x2

1 + x2

2 = 100 and x = x1

x2 > 0

0 (P7.7.6b)

Try the routine “fmincon()” with the initial guesses listed in

Table P7.7.

(e) Does the routine work well with all the initial guesses? If not, does it

matter whether the starting point is inside the admissible region?

(cf) Note that, in order to solve the maximization problem by “fmincon()”, we

have to reverse the sign of the objective function. Note also that the objective

functions (P7.7.3a) and (P7.7.5a) have infinitely many minima having the value

f (x) = 0 in the admissible region satisfying the constraints.

366 OPTIMIZATION

(cf) One might be disappointed with the reliability of the MATLAB optimization

routines to see that they may fail to find the optimal solution depending on the

initial guess. But, how can a human work be perfect in this world? It implies

the difficulty of nonlinear constrained optimization problems and can never

impair the celebrity and reliability of MATLAB. Actually, it demonstrates the

importance of studying some numerical stuff in addition to just getting used

to the various MATLAB commands and routines.

Here is a tip for the usage of “fmincon()”: it might be better to use with

an initial guess that is not at the origin, but in the admissible region satisfying

the constraints, even though it does not guarantee the success of the routine.

It might also be helpful to apply the routine with several values of the initial

guess and then choose the best result.

7.8 Constrained Optimization and Penalty Method

Consider again the constrained minimization problem having the objective

function (E7.3.1a) and the constraints (E7.3.1b).

Min f (x) = {(x1 + 1.5)2 + 5(x2 − 1.7)2}{(x1 − 1.4)2 + 0.6(x2 − 0.5)2} (P7.8.1a)

s.t. g(x) =

−x1

−x2

3×1 − x1x2 + 4×2 − 7

2×1 + x2 − 3

3×1 − 4×2

2 − 4×2

≤

0

0

0

0

0

(P7.8.1b)

In Example 7.3, we made the MATLAB program “nm722.m” to solve the

problem and defined the objective function (E7.3.2a) having the penalized

constraint terms in the file named “f722p.m”.

Min l(x) = {(x1 + 1.5)2 + 5(x2 − 1.7)2}{(x1 − 1.4)2 + 0.6(x2 − 0.5)2}

+

5

m=1

vmψm(gm(x)) (P7.8.2a)

where

ψm(gm(x)) = 0 if gm(x) ≤ 0 (constraint satisfied)

exp(emgm(x)) if gm(x) > 0 (constraint viloated)

with em = 1 ∀m = 1, . . . , 5 (P7.8.2b)

(a) Whatis theweighting coefficient vector v in thefilenamed“f722p.m”?Do

the points reached by the routines “fminsearch()”/“opt_

steep()”/“fminunc()” satisfy all the constraints so that they are in the

admissible region? If not, specify the constraint(s) violated by the points.

(b) Suppose the fourth constraint was violated by the point in (a). Then,

how would you modify the weighting coefficient vector v so that the

violated constraint can be paid more respect? Choose one of the following

two weighting coefficient vectors:

PROBLEMS 367

(i) v = [1 1 1 1/3 1]

(ii) v = [1 1 1 3 1]

and modify the file “f722p.m” with this coefficient vector. Then, run

the program “nm722.m”, fill in the 22 blanks of Table P7.8 with the

results and see if the fourth constraint is still violated by the points

reached by the optimization routines?

(c) Instead of the penalty method, apply the intrinsically constrained optimization

routine “fmincon()” with the initial guesses x0 = [0.4 0.5]

and [0.2 4] to solve the problem described by Eq. (E7.3.1) or (P7.8.1)

and fill in Table P7.8 with the results concerning the reached point and

the corresponding values of the objective/constraint functions.

(d) Based on the results listed in Table P7.8, circle the right word in each

of the parentheses in the following sentences:

ž For penalty methods, the non-gradient-based minimization routines like

“Nelder()”/“fminsearch()” may work (better, worse) than the gradientbased

minimization routines like “opt_steep()”/“fminunc()”.

ž If some constraint is violated, you had better (increase, decrease) the

corresponding weight coefficient.

(cf) Besides, unconstrained optimization with the penalized constraints in the

objective function sometimes works better than the constrained optimization

routine “fmincon()”.

Table P7.8 The Results of Penalty Methods Depending on the Initial Guess and

Weighting Factor

The Starting Point x0 = [0.4 0.5] x0 = [0.2 4]

v Nelder fminsearch steep fminunc fmincon Nelder fminsearch steep fminunc fmincon

1.21 1.34 1.34 1.34

xo 1 0.58 0.62 0.62 0.62

1 fo 0.53 0.17 0.17 0.17

1 −1.21 −1.34 −1.34 −1.38 −1.21 −1.34 −1.34 −1.34 1.26 0.00

−0.58 −0.62 −0.62 −0.63 −0.58 −0.62 −0.62 −0.62 −1.70 −1.59

1/3

co −1.76 −1.34 −1.34 −1.19 −1.76 −1.34 −1.34 −1.33 −1.84 −0.65

1 −0.00 0.00 0.29 −3.82 −1.41

−0.00 −0.00 −0.00 −0.00 0.00 −0.00 −0.00 −0.00 −22.1 −16.4

1.21 1.21 1.12 1.18 — 1.21 1.21 1.15 −1.26 —

xo

1 0.58 0.58 0.76 0.64 — 0.58 0.58 0.71 1.70 —

f o 0.53 0.53 1.36 0.79 — 0.53 0.53 1.08 0.46 —

1

−1.21 −1.21 −1.12 −1.18 −1.21 −1.21 −1.15 1.26

1 −0.58 −0.58 −0.76 −0.64 −0.58 −0.58 −0.71 −1.70

co

3 −1.76 −1.76 −1.44 −1.65 — −1.76 −1.76 −1.54 −1.84 —

−0.00 −0.00 −3.82

1 −0.00 −0.00 −2.04 −0.70 −0.00 −0.00 −1.39 −22.1

368 OPTIMIZATION

7.9 A Constrained Optimization on Location

A company has three factories that are located at the points (−16,4), (6,5),

and (3,−9), respectively, in the x1x2-plane, and the numbers of deliveries

to those factories are 5, 6, and 10 per month, respectively (Fig. P7.9). The

company has a plan to build a new warehouse in its site bounded by

|x1 − 1| + |x2 − 1| ≤ 2 (P7.9.1)

and is trying to minimize the monthly mileage of delivery trucks in determining

the location of a new warehouse on the assumption that the distance

between two points represents the driving distance.

(a) What is the objective function that must be defined in the program

“nm7p09.m”?

(b) What is the statement defining the inequality constraint (P7.9.1)?

(c) Complete and run the program “nm7p09.m” to get the optimum location

of the new warehouse.

function [C,Ceq] = fp_warehouse_c(x)

C = sum(abs(x – [1 1])) – 2;

Ceq = []; % No equality constraint

%nm7p09.m to solve the warehouse location problem

f = ’sqrt([sum((x – [-16 4]).^2) sum((x – [6 5]).^2) sum((????????).^2)])’;

fp_warehouse = inline([f ’*[?;?;?]’],’x’);

x0 = [1 1]; A = []; b = []; Aeq = []; beq = []; l = []; u = [];

xo = fmincon(fp_warehouse,x0,A,b,Aeq,beq,l,u,’fp_warehouse_c’)

+

factory B

factory C

site

5

0

−5

−10

−20 −10 0 10

factory A

Figure P7.9 The site of a new warehouse and the locations of the factories.

7.10 A Constrained Optimization on Ray Refraction

A light ray follows the path that takes the shortest time when it travels in

the space. We want to find the three angles θ1,θ2, and θ3 (measured between

the array and the normal to the material surface) of a ray traveling from

P = (0, 0) to Q = (L,−(d1 + d2 + d3)) through a transparent material of

thickness d2 and index of refraction n as depicted in Fig. P7.10. Note the

following things.

PROBLEMS 369

ž Since the speed of light in the transparent material is v = c/n (c is the

speed of light in the free space), the traveling time to be minimized

can be expressed as

Min f (θ, d, n,L) =

d1

c cos θ1 +

nd2

c cos θ2 +

d3

c cos θ3

(P7.10.1)

ž The sum of the three horizontal distances traveled by the light ray must

be L:

g(θ, d, n,L) =

3

i=1

di tan θi − L = 0 (P7.10.2)

ž The horizontal distance L and the index of refraction n are additionally

included in the input argument lists of both the objective function

f (θ, d, n,L) and the constraint function g(θ, d, n,L) regardless of

whether or not they are used in each function. It is because the objective

function and the constraint function of the MATLAB routine “fmincon()”

must have the same input arguments.

(a) Compose a program “nm7p10a.m” that solves the above constrained

minimization problem to find the three angles θ1, θ2, and θ3 for n = 1.52, d1 = d2 = d3 = 1[cm], and different values of L = 0.6:0.3:6 and

plots sin(θ1)/sin(θ2) and sin(θ3)/sin(θ2) versus L.

(b) Compose a program “nm7p10b.m” that finds the three angles θ1,θ2,

and θ3 for L = 3 cm, d1 = d2 = d3 = 1 cm, and different values of

n = 1:0.01:1.6 and plots sin(θ1)/sin(θ2) and sin(θ3)/sin(θ2) versus n.

P

Q

L

d3

d2

d1

q3

q2

q1

a light ray

air

air

speed of light = c

speed of light = c /n

transparent material

with refraction index n

Figure P7.10 Refraction of a light ray at an air–glass interface.

7.11 A Constrained Optimization on OFDM System

In order to find the average modulation order xi for each user of an OFDM

(orthogonal frequency division multiplex) system that has N(128) subchannels

to assign to each of the four users in the environment of noise power

370 OPTIMIZATION

N0 and the bit error rate (probability of bit error) Pe, Seung-hee, a communication

system expert, formulated the following constrained minimization

problem:

Min f (x) =

4

i=1

(2xi − 1)

N0

3

2(erfc−1(Pe/2))2 ai

xi

(P7.11.1)

subject to

g(x) =

4

i=1

ai

xi − N = 0 (P7.11.2)

with N = 128, and ai : the data rate of each user

where erfc−1(x) is the inverse function of the complementary error function

defined by Eq. (P4.9.3) and is installed as the MATLAB built-in function

‘erfcinv()’. He defined the objective function and the constraint function

as below and save them in the M-files named “fp_bits1.m” and

“fp_bits_c.m”.

function y = fp_bits1(x,a,N,Pe)

N0 = 1; y = sum((2.^x-1)*N0/3*2*erfcinv(Pe/2).^2.*a./x);

function [C,Ceq] = fp_bits_c(x,a,N,Pe)

C = []; Ceq = sum(a./x) – N;

Compose a program that solves the above constrained minimization problem

(with N0 = 1 and Pe = 10−4) to get the modulation order xi of each user

for five different sets of data rates

a = [32 32 32 32],[64 32 32 32], [128 32 32 32], [256 32 32 32], and [512 32 32 32]

and plots a1/x1(the number of subchannels assigned to user 1) versus a1

(the data rate of user 1). If you feel uneasy about the results obtained with

your initial guesses, try with the initial guesses as follows for each set of

data rates, respectively:

x0 = [0.5 0.5 0.5 0.5], [1 1 1 1],[1 1 1 1], [2 2 2 2], and [4 4 4 4]

8

MATRICES AND

EIGENVALUES

In this chapter, we will look at the eigenvalue or characteristic value λ and its

corresponding eigenvector or characteristic vector v of a matrix.

8.1 EIGENVALUES AND EIGENVECTORS

The eigenvalue or characteristic value and its corresponding eigenvector or characteristic

vector of an N × N matrix A are defined as a scalar λ and a nonzero

vector v satisfying

Av = λv ⇔ (A − λI) v = 0 (v = 0) (8.1.1)

where (λ, v) is called an eigenpair and there are N eigenpairs for the N × N

matrix A.

How do we get them? Noting that

ž in order for the above equation to hold for any nonzero vector v, the matrix

[A − λI] should be singular—that is, its determinant should be zero (|A − λI| = 0)— and

ž the determinant of the matrix [A − λI] is a polynomial of degree N in terms

of λ,

we first must find the eigenvalue λi ’s by solving the so-called characteristic

equation

|A − λI| = λN + aN−1λN−1 + ·· ·+a1λ + a0 = 0 (8.1.2)

Applied Numerical Methods Using MATLAB, by Yang, Cao, Chung, and Morris

Copyr ight 2005 John Wiley & Sons, I nc., ISBN 0-471-69833-4

371

372 MATRICES AND EIGENVALUES

and then substitute the λi ’s, one by one, into Eq. (8.1.1) to solve it for the

eigenvector vi ’s. This is, however, not always so simple, especially if some root

(eigenvalue) of Eq. (8.1.2) has multiplicity k > 1, since we have to generate k

independent eigenvectors satisfying Eq. (8.1.1) for such an eigenvalue. Still, we

do not have to worry about this, thanks to the MATLAB built-in routine “eig()”,

which finds us all the eigenvalues and their corresponding eigenvectors for a given

matrix. How do we use it? All we need to do is to define the matrix, say A, and

type a single statement into the MATLAB command window as follows.

>>[V,Lambda] = eig(A) %e = eig(A) just for eigenvalues

Let us take a look at the following example.

Example 8.1. Eigenvalues/Eigenvectors of a Matrix.

Let us find the eigenvalues/eigenvectors of the matrix

A = 0 1

0 −1 (E8.1.1)

First, we find its eigenvalues as

|A − λI| =

−λ 1

0 −1 − λ

= λ2 + λ = 0

λ(λ + 1) = 0, λ1 = 0, λ2 = −1 (E8.1.2)

and then, get the corresponding eigenvectors as

[A − λ1I ]v1 = 0 1

0 −1 v11

v21 = v21

−v21 = 0

0 ,

v21 = 0, v1 = v11

v21 = 1

0 (E8.1.3a)

[A − λ2I ]v2 = 1 1

0 0v12

v22 = v12 + v22

0 = 0

0 ,

v12 = −v22, v2 = v12

v22 = 1/√2

−1/√2 (E8.1.3b)

where we have chosen v11, v12, and v22 so that the norms of the eigenvectors

become one.

Alternatively, we can use the MATLAB command “eig(A)” for finding eigenvalues/

eigenvectors or “roots(poly(A))” just for finding eigenvalues as the

roots of the characteristic equation as illustrated by the program “nm811.m”.

SIMILARITY TRANSFORMATION AND DIAGONALIZATION 373

%nm811 to get the eigenvalues & eigenvectors of a matrix A.

clear

A = [0 1;0 -1];

[V,L] = eig(A) %V = modal matrix composed of eigenvectors

% L = diagonal matrix with eigenvalues on its diagonal

e = eig(A), roots(poly(A)) %just for eigenvalues

L = V^ – 1*A*V %diagonalize through similarity transformation

% into a diagonal matrix having the eigenvalues on diagonal.

8.2 SIMILARITY TRANSFORMATION AND DIAGONALIZATION

Premultiplying a matrix A by P−1 and post-multiplying it by P makes a similarity

transformation

A → P−1AP (8.2.1)

Remark 8.1 tells us how a similarity transformation affects the eigenvalues/

eigenvectors.

Remark 8.1. Effect of Similarity Transformation on Eigenvalues/Eigenvectors

1. The eigenvalues are not changed by a similarity transformation.

|P−1AP − λI| = |P−1AP − P−1λIP| = |P−1||A − λI||P| = |A − λI| (8.2.2)

2. Substituting v = Pw into Eq. (8.1.1) yields

Av = λv, APw = λPw = Pλw, [P−1AP]w = λw

This implies that the matrix P−1AP obtained by a similarity transformation

has w = P−1v as its eigenvector if v is an eigenvector of the matrix A.

In order to understand the diagonalization of a matrix into a diagonal matrix

(having its eigenvalues on the main diagonal) through a similarity transformation,

we have to know the following theorem:

Theorem 8.1. Distinct Eigenvalues and Independent Eigenvectors.

If the eigenvalues of a matrix A are all distinct—that is, different from each

other—then the corresponding eigenvectors are independent of each other and,

consequently, the modal matrix composed of the eigenvectors as columns is

nonsingular.

Now, for an N × N matrix A whose eigenvalues are all distinct, let us put all

of the equations (8.1.1) for each eigenvalue-eigenvector pair together to write

A[v1 v2 · · · vN] = [v1 v2 · · · vN]

λ1 0 · 0

0 λ2 · 0

· · · · 0 0 · λN

, AV= V

(8.2.3)

374 MATRICES AND EIGENVALUES

Then, noting that the modal matrix V is nonsingular and invertible by Theorem

8.1, we can premultiply the above equation by the inverse modal matrix

V −1 to get

V −1AV = V −1V ≡ (8.2.4)

This implies that the modal matrix composed of the eigenvectors of a matrix A

is the similarity transformation matrix that can be used for converting the matrix

A into a diagonal matrix having its eigenvalues on the main diagonal. Here is an

example to illustrate the diagonalization.

Example 8.2. Diagonalization Using the Modal Matrix.

Consider the matrix given in the previous example.

A = 0 1

0 −1 (E8.2.1)

We can use the eigenvectors (E8.1.3) (obtained in Example 8.1) to construct

the modal matrix as

V = [v1 v2] = 1 1/√2

0 −1/√2 (E8.2.2)

and use this matrix to make a similarity transformation of the matrix A as

V −1AV = 1 1/√2

0 −1/√2 −1

0 1

0 −11 1/√2

0 −1/√2

= 1 1

0 −√20 −1/√2

0 1/√2 = 0 0

0 −1 (E8.2.3)

which is a diagonal matrix having the eigenvalues on its main diagonal.

This job can be performed by the last statement of the MATLAB program

“nm811.m”.

This diagonalization technique can be used to decouple an N-dimensional

vector differential equation so that it can be as easy to solve as N independent

scalar differential equations. Here is an illustration.

Example 8.3. Decoupling of a Vector Equation Through Diagonalization

(a) For the linear time-invariant (LTI) state equation (6.5.3)

x1(t)

x2(t) = 0 1

0 −1x1(t)

x2(t) + 0

1 us (t) (E8.3.1)

with x1(0)

x2(0) = 1

−1 and us (t) = 1 ∀ t ≥ 0

x(t) = Ax(t) + Bu(t) with the initial state x(0) and the input u(t)

SIMILARITY TRANSFORMATION AND DIAGONALIZATION 375

we use the modal matrix obtained as (E8.2.2) in Example 8.2 to make a substitution

of variable

x(t) = Vw(t), x1(t)

x2(t) = 1 1/√2

0 −1/√2w1(t)

w2(t) (E8.3.2)

which converts Eq. (E8.3.1) into

Vw(t) = AVw(t) + Bus (t) (E8.3.3)

We premultiply (E8.3.3) by V −1 to write it in a decoupled form as

w(t)=V −1AVw(t) + V −1Bus (t)=w(t) + V −1Bus (t) with w(0) = V −1x(0);

w1(t)

w2(t) = 0 0

0 −1w1(t)

w2(t) + 1 1

0 −√20

1 us (t) = us (t)

−w2(t) − √2us (t)

(E8.3.4)

with w1(0)

w2(0) = 1 1

0 −√2 1

−1 = 0 √2

where there is no correlation between the variables w1(t) and w2(t). Then we

can solve these two equations separately to have

w1(t) = us (t) with w1(0) = 0;

sW1(s) − w1(0) =

1

s

; W1(s) =

1

s2 ; w1(t) = t us (t) (E8.3.5a)

w2(t) = −w2(t) − √2us (t) with w2(0) = √2;

sW2(s) − w2(0) = −W2(s) −

√2

s

;

W2(s) =

w2(0)

s + 1 −

√2

s(s + 1) = −

√2

s +

2√2

s + 1

,

w2(t) = √2(−1 + 2e−t)us (t) (E8.3.5b)

and substitute this into Eq. (E8.3.2) to get

x1(t)

x2(t) = 1 1/√2

0 −1/√2w1(t)

w2(t) = 1 1/√2

0 −1/√2 t √2(−1 + 2e−t ) us (t)

= t − 1 + 2e−t

1 − 2e−t us (t) (E8.3.6)

This is the same result as Eq. (6.5.10) obtained in Section 6.5.1.

376 MATRICES AND EIGENVALUES

(b) Suppose Eq. (E8.3.1) has no input term and so we can expect only the

natural response resulting from the initial state, but no forced response

caused by the input.

x1(t)

x2(t) = 0 1

0 −1x1(t)

x2(t) with x1(0)

x2(0) = 1

1 (E8.3.7)

We apply the diagonalization/decoupling method for this equation to get

w1(t)

w2(t) = λ1 0

0 λ2 w1(t)

w2(t) = 0 0

0 −1w1(t)

w2(t)

with w(0) = V −1x(0), w1(0)

w2(0) = 1 1

0 −√21

1 = 2

−√2

w1(t)

w2(t) = w1(0)eλ1t

w2(0)eλ2t = 2

−√2e−t (E8.3.8)

x(t)

(E8.3.2) = Vw(t) = [v1 v2] w1(0)eλ1t

w2(0)eλ2t = w1(0)eλ1tv1 + w2(0)eλ2tv2

= 1 1/√2

0 −1/√2 2

−√2e−t = 2 − e−t

e−t (E8.3.9)

As time goes by, this solution converges and so the continuous-time system

turns out to be stable, thanks to the fact that all of the eigenvalues

(0, −1) are distinct and not positive.

Example 8.4. Decoupling of a Vector Equation Through Diagonalization.

Consider a discrete-time LTI state equation

x1[n + 1]

x2[n + 1] = 0 1

0.2 0.1x1[n]

x2[n] + 0

2.2361 us [n]

with x1[0]

x2[0] = 1

−1 and us [n] = 1 ∀ n ≥ 0 (E8.4.1)

In order to diagonalize this equation into a form similar to Eq. (E8.3.4), we use

MATLAB to find the eigenvalues/eigenvectors and the modal matrix composed

of the eigenvectors and finally, do the similarity transformation.

A = [0 1;0.2 0.1]; B = [0; 2.2361]; % Eq.(E8.4.1)

[V,L] = eig(A) % V = modal matrix composed of eigenvectors (E8.4.2)

% L = diagonal matrix with eigenvalues on its diagonal

Ap = V^-1*A*V %diagonalize through similarity transformation (E8.4.3)

% into a diagonal matrix having the eigenvalues on the diagonal

Bp = V^-1*B % (E8.4.3)

SIMILARITY TRANSFORMATION AND DIAGONALIZATION 377

Then, we get

L = λ1 0

0 λ2 = −0.4 0

0 0.5 , V = [v1 v2] = −0.9285 −0.8944

0.3714 −0.4472

(E8.4.2)

Ap = V −1AV = −0.4 0

0 0.5 and Bp = V −1B = 2.6759

−2.7778 (E8.4.3)

so that we can write the diagonalized state equation as

w1[n + 1]

w2[n + 1] = −0.4 0

0 0.5w1[n]

w2[n] + 2.6759

−2.7778 us [n]

= −0.4w1[n] + 2.6759

0.5w2[n] − 2.7778 (E8.4.4)

Without the input term on the right-hand side of Eq. (E8.4.1), we would have

obtained

w1[n + 1]

w2[n + 1] = λ1 0

0 λ2 w1[n]

w2[n] = λn+1

1 w1[0]

λn+1

2 w2[0] with w[0] = V −1x[0]

(E8.4.5)

x[n] = Vw[n] = [v1 v2] w1[0]λn1

w2[0]λn2

= w1[0]λn1

v1 + w2[0]λn2

v2 (E8.4.6)

As time goes by (i.e., as n increases), this solution converges and so the discretetime

system turns out to be stable, thanks to the fact that the magnitude of every

eigenvalue (−0.4, 0.5) is less than one.

Remark 8.2. Physical Meaning of Eigenvalues and Eigenvectors

1. As illustrated by the above examples, we can use the modal matrix to

decouple a set of differential equations so that they can be solved one

by one as a scalar differential equation in terms of a single variable and

then put together to make the solution for the original vector differential

equation.

2. Through the above examples, we can feel the physical significance of the

eigenvalues/eigenvectors of the system matrix A in the state equation on its

solution. That is, the state of a linear time-invariant (LTI) system described

by an N-dimensional continuous-time (differential) state equation has N

modes {eλi t ; i = 1, . . . , N}, each of which converges/diverges if the sign of

the corresponding eigenvalue is negative/positive and proceeds slowly as

378 MATRICES AND EIGENVALUES

the magnitude of the eigenvalue is close to zero. In the case of a discretetime

LTI system described by an N-dimensional difference state equation,

its state has N modes {λn

i ; i = 1, . . . ,N}, each of which converges/diverges

if the magnitude of the corresponding eigenvalue is less/greater than one

and proceeds slowly as the magnitude of the eigenvalue is close to one.

To summarize, the convergence property of a state x or the stability of a

linear-time invariant (LTI) system is determined by the eigenvalues of the

system matrix A. As illustrated by (E8.3.9) and (E8.4.6), the corresponding

eigenvector determines the direction in which each mode proceeds in the

N-dimensional state space.

8.3 POWER METHOD

In this section, we will introduce the scaled power method, the inverse power

method and the shifted inverse power method, to find the eigenvalues of a

given matrix.

8.3.1 Scaled Power Method

This method is used to find the eigenvalue of largest magnitude and is summarized

in the following box.

SCALED POWER METHOD

Suppose all of the eigenvalues of an N × N matrix A are distinct with the

magnitudes

|λ1| > |λ2| ≥ |λ3| ≥ ··· ≥ |λN|

Then, the dominant eigenvalue λ1 with the largest magnitude and its corresponding

eigenvector v1 can be obtained by starting with an initial vector x0

that has some nonzero component in the direction of v1 and by repeating the

following procedure:

Divide the previous vector xk by its largest component (in absolute value)

for normalization (scaling) and premultiply the normalized vector by the

matrix A.

xk+1 = A

xk

||xk||∞ → λ1v1 with ||x||∞ = Max {|xn|} (8.3.1)

POWER METHOD 379

Proof. According to Theorem 8.1, the eigenvectors {vn; n = 1 : N} of an N × N

matrix A whose eigenvalues are distinct are independent and thus can constitute

a basis for an N-dimensional linear space. Consequently, any initial vector x0

can be expressed as a linear combination of the eigenvectors:

x0 = α1v1 + α2v2 + ·· ·+αNvN (8.3.2)

Noting that Avn = λnvn, we premultiply both sides of this equation by A

to get

Ax0 = α1λ1v1 + α2λ2v2 + ·· ·+αNλNvN

= λ1

α1v1 + α2

λ2

λ1

v2 + ·· ·+αN

λN

λ1

vN

and repeat this multiplication over and over again to obtain

xk = Akx0

= λk

1 α1v1 + α2

λ2

λ1k

v2 + ·· ·+αN

λN

λ1 k

vN

→ λk

1α1v1 (8.3.3)

which will converge to an eigenvector v1 as long as α1 = 0. Since we keep

scaling before multiplying at every iteration, the largest component of the limit

vector of the sequence generated by Eq. (8.3.1) must be λ1.

xk+1 = A

xk

||xk||∞ → A

v1

||v1||∞

(8.1.1) = λ1

v1

||v1||∞

(8.3.4)

Note that the scaling prevents the overflow or underflow that would result from

|λ1| > 1 or |λ1| < 1.

Remark 8.3. Convergence of Power Method

1. In the light of Eq. (8.3.3), the convergence speed of the power method

depends on how small the magnitude ratio (|λ2|/|λ1|) of the second largest

eigenvalue λ2 over the largest eigenvalue λ1 is.

2. We often use x0 = [1 1 · · · 1 ] as the initial vector. Note that

if it has no component in the direction of the eigenvector (v1)

corresponding to the dominant eigenvalue λ1—that is, α1 = x0žv1/||v1||2 = 0 in Eq. (8.3.2)—the iteration of the scaled power method leads to the limit

showing the second largest magnitude eigenvalue λ2 and its corresponding

eigenvector v2. But, if there is more than one largest (dominant) eigenvalue

of equal magnitude, it does not converge to either of them.

380 MATRICES AND EIGENVALUES

8.3.2 Inverse Power Method

The objective of this method is to find the (uniquely) smallest (magnitude) eigenvalue

λN by applying the scaled power method to the inverse matrix A−1 and

taking the inverse of the largest component of the limit. It works only in cases

where the matrix A is nonsingular and thus has no zero eigenvalue. Its idea is

based on the equation

Av = λv → A−1v = λ−1v (8.3.5)

obtained from multiplying both sides of Eq. (8.1.1) by λ−1A−1. This implies

that the inverse matrix A−1 has the eigenvalues that are the reciprocals of the

eigenvalues of the original matrix A, still having the same eigenvectors.

λN =

1

the largest eigenvalue of A−1 (8.3.6)

8.3.3 Shifted Inverse Power Method

In order to develop a method for finding the eigenvalue that is not necessarily

of the largest or smallest magnitude, we subtract s v (s: a number that does not

happen to equal any eigenvalue) from both sides of Eq. (8.1.1) to write

Av = λv → [A − sI ]v = (λ − s)v (8.3.7)

Since this implies that (λ − s) is the eigenvalue of [A − sI ], we apply the inverse

power method for [A − sI ] to get its smallest magnitude eigenvalue (λk − s) with

min{|λi − s|, i = 1 : N} and add s to it to obtain the eigenvalue of the original

matrix A which is closest to the number s.

λs =

1

the largest eigenvalue of [A − sI ]−1 + s (8.3.8)

The prospect of this method is supported by Gerschgorin’s disk theorem,

which is summarized in the box below. But, this method is not applicable to the

matrix that has more than one eigenvalue of the same magnitude.

Theorem 8.2. Gerschgorin’s Disk Theorem.

Every eigenvalue of a square matrix A belongs to at least one of the disks

(in the complex plane) with center amm (one of the diagonal elements of A) and

radius

rm =n=m

|amn|(the sum of all the elements in the row except the diagonal element)

JACOBI METHOD 381

Moreover, each of the disks contains at least one eigenvalue of the

matrix A.

The power method introduced in Section 8.3.1 is cast into the routine

“eig_power()”. The MATLAB program “nm831.m” uses it to perform the power

method, the inverse power method and the shifted inverse power method for

finding the eigenvalues of a matrix and compares the results with that of the

MATLAB built-in routine “eig()” for cross-check.

function [lambda,v] = eig_power(A,x,EPS,MaxIter)

% The power method to find the largest eigenvalue (lambda) and

% the corresponding eigenvector (v) of a matrix A.

if nargin < 4, MaxIter = 100; end % maximum number of iterations

if nargin < 3, EPS = 1e-8; end % difference between successive values

N = size(A,2);

if nargin < 2, x = [1:N]; end % the initial vector

x = x(:);

lambda = 0;

for k = 1:MaxIter

x1 = x; lambda1 = lambda;

x = A*x/norm(x,inf); %Eq.(8.3.4)

[xm,m] = max(abs(x));

lambda = x(m); % the component with largest magnitude(absolute value)

if norm(x1 – x) < EPS & abs(lambda1-lambda) < EPS, break; end

end

if k == MaxIter, disp(’Warning: you may have to increase MaxIter’); end

v = x/norm(x);

%nm831

%Apply the power method to find the largest/smallest/medium eigenvalue

clear

A = [2 0 1;0 -2 0;1 0 2];

x = [1 2 3]’; %x = [1 1 1]’; % with different initial vector

EPS = 1e-8; MaxIter = 100;

%the largest eigenvalue and its corresponding eigenvector

[lambda_max,v] = eig_power(A,x,EPS,MaxIter)

%the smallest eigenvalue and its corresponding eigenvector

[lambda,v] = eig_power(A^ – 1,x,EPS,MaxIter);

lambda_min = 1/lambda, v %Eq.(8.3.6)

%eigenvalue nearest to a number and its corresponding eigenvector

s = -3; AsI = (A – s*eye(size(A)))^ – 1;

[lambda,v] = eig_power(AsI,x,EPS,MaxIter);

lambda = 1/lambda+s %Eq.(8.3.8)

fprintf(’Eigenvalue closest to %4.2f = %8.4f\nwith eigenvector’,s,lambda)

v

[V,LAMBDA] = eig(A) %modal matrix composed of eigenvectors

8.4 JACOBI METHOD

This method finds us all the eigenvalues of a real symmetric matrix. Its idea is

based on the following theorem.

382 MATRICES AND EIGENVALUES

Theorem 8.3. Symmetric Diagonalization Theorem.

All of the eigenvalues of an N × N symmetric matrix A are of real value and

its eigenvectors form an orthonormal basis of an N-dimensional linear space.

Consequently, we can make an orthonormal modal matrix V composed of the

eigenvectors such that V T V = I ; V −1 = V T and use the modal matrix to make

the similarity transformation of A, which yields a diagonal matrix having the

eigenvalues on its main diagonal:

V T AV = V −1AV = (8.4.1)

Now, in order to understand the Jacobi method, we define the pq-rotation

matrix as

Rpq(θ) =

pth column qth column

1 0 · 0 · 0 · 0

0 1 · 0 · 0 · 0

· · · · · · · · 0 0 · cos θ · −sin θ · 0

· · · · · · · · 0 0 · sin θ · cos θ · 0

· · · · · · · · 0 0 · 0 · 0 · 1

pth row

qth row

(8.4.2)

Since this is an orthonormal matrix whose row/column vectors are orthogonal

and normalized

RT

pqRpq = I, RT

pq = R−1

pq (8.4.3)

premultiplying/postmultiplying a matrix A by RT

pq/Rpq makes a similarity transformation

A(1) = RT

pqA Rpq (8.4.4)

Noting that the similarity transformation does not change the eigenvalues (Remark

8.1), any matrix resulting from repeating the same operations successively

A(k+1) = RT

(k)A(k)R(k) = RT

(k)RT

(k−1) · · ·RT AR · · ·R(k−1)R(k) (8.4.5)

has the same eigenvalues. Moreover, if it is a diagonal matrix, it will have all

the eigenvalues on its main diagonal, and the matrix multiplied on the right of

the matrix A is the modal matrix V

V = R · · ·R(k−1)R(k) (8.4.6)

as manifested by matching this equation with Eq. (8.4.1).

JACOBI METHOD 383

function [LAMBDA,V,ermsg] = eig_Jacobi(A,EPS,MaxIter)

%Jacobi method finds the eigenvalues/eigenvectors of symmetric matrix A

if nargin < 3, MaxIter = 100; end

if nargin < 2, EPS = 1e-8; end

N = size(A,2);

LAMBDA =[]; V = [];

for m = 1:N

if norm(A(m:N,m) – A(m,m:N)’) > EPS

error(’asymmetric matrix!’);

end

end

V = eye(N);

for k = 1:MaxIter

for m = 1:N – 1

[Am(m),Q(m)] = max(abs(A(m,m + 1:N)));

end

[Amm,p] = max(Am); q = p + Q(p);

if Amm < EPS*sum(abs(diag(LAMBDA))), break; end

if abs(A(p,p)-A(q,q))<EPS

s2 = 1; s = 1/sqrt(2); c = s;

cc = c*c; ss = s*s;

else

t2 = 2*A(p,q)/(A(p,p)- A(q,q)); %Eq.(8.4.9a)

c2 = 1/sqrt(1 + t2*t2); s2 = t2*c2; %Eq.(8.4.9b,c)

c = sqrt((1 + c2)/2); s = s2/2/c; %Eq.(8.4.9d,e)

cc = c*c; ss = s*s;

end

LAMBDA = A;

LAMBDA(p,:) = A(p,:)*c + A(q,:)*s; %Eq.(8.4.7b)

LAMBDA(:,p) = LAMBDA(p,:)’;

LAMBDA(q,:) = -A(p,:)*s + A(q,:)*c; %Eq.(8.4.7c)

LAMBDA(:,q) = LAMBDA(q,:)’;

LAMBDA(p,q) = 0; LAMBDA(q,p) = 0; %Eq.(8.4.7a)

LAMBDA(p,p) = A(p,p)*cc +A(q,q)*ss + A(p,q)*s2; %Eq.(8.4.7d)

LAMBDA(q,q) = A(p,p)*ss +A(q,q)*cc – A(p,q)*s2; %Eq.(8.4.7e)

A = LAMBDA;

V(:,[p q]) = V(:,[p q])*[c -s;s c];

end

LAMBDA = diag(diag(LAMBDA)); %for purification

%nm841 applies the Jacobi method

% to find all the eigenvalues/eigenvectors of a symmetric matrix A.

clear

A = [2 0 1;0 -2 0;1 0 2];

EPS = 1e-8; MaxIter =100;

[L,V] = eig_Jacobi(A,EPS,MaxIter)

disp(’Using eig()’)

[V,LAMBDA] = eig(A) %modal matrix composed of eigenvectors

What is left for us to think about is how to make this matrix (8.4.5) diagonal.

Noting that the similarity transformation (8.4.4) changes only the pth

rows/columns and the qth rows/columns as

vpq = vqp = aqp(c2 − s2) + (aqq − app)sc

= aqp cos 2θ +

1

2

(aqq − app) sin 2θ (8.4.7a)

384 MATRICES AND EIGENVALUES

vpn = vnp = apnc + aqns for the pth row/column with n = p, q (8.4.7b)

vqn = vnp = −apns + aqnc for the qth row/column with n = p, q (8.4.7c)

vpp = appc2 + aqqs2 + 2apqsc = appc2 + aqq s2 + apq sin 2θ (8.4.7d)

vqq = apps2 + aqqc2 − 2apqsc = apps2 + aqqc2 − apq sin 2θ (8.4.7e)

(c = cos θ, s = sin θ)

we make the (p, q) element vpq and the (q, p) element vqp zero

vpq = vqp = 0 (8.4.8)

by choosing the angle θ of the rotation matrix Rpq(θ) in such a way that

tan 2θ =

sin 2θ

cos 2θ =

2apq

app − aqq

, cos 2θ =

1

sec 2θ =

1

√1 + tan2 2θ

,

sin 2θ = tan 2θ cos 2θ

cos θ =

√cos2 θ = (1 + cos 2θ)/2, sin θ =

sin 2θ

2 cos θ

(8.4.9)

and computing the other associated elements according to Eqs. (8.4.7b–e).

There are a couple of things to note. First, in order to make the matrix closer

to a diagonal one at each iteration, we should identify the row number and the

column number of the largest off-diagonal element as p and q, respectively, and

zero-out the (p, q) element. Second, we can hope that the magnitudes of the

other elements in the pth,qth row/column affected by this transformation process

don’t get larger, since Eqs. (8.4.7b) and (8.4.7c) implies

v2

pn + v2

qn = (apnc + aqns)2 + (−apns + aqnc)2 = a2

pn + a2

qn (8.4.10)

This so-called Jacobi method is cast into the routine “eig_Jacobi()”. The

MATLAB program “nm841.m” uses it to find the eigenvalues/eigenvectors of a

matrix and compares the result with that of using the MATLAB built-in routine

“eig()” for cross-check. The result we may expect is as follows. Interested

readers are welcome to run the program “nm841.m”.

A =

2 0 1

0 −2 0

1 0 2

→ RT

13AR13 =

3 0 0

0 −2 0

0 0 1

=

with R13 =

1/√2 0 −1/√2

0 1 0

1/√2 0 1/√2

= V

PHYSICAL MEANING OF EIGENVALUES/EIGENVECTORS 385

8.5 PHYSICAL MEANING OF EIGENVALUES/EIGENVECTORS

According to Theorem 8.3 (Symmetric Diagonalization Theorem), introduced in

the previous section, the eigenvectors {vn, n = 1 : N} of an N × N symmetric

matrix A constitute an orthonormal basis for an N-dimensional linear space.

V T V = I, vT

mvn = δmn = 1 form = n

0 form = n

(8.5.1)

Consequently, any N-dimensional vector x can be expressed as a linear combination

of these eigenvectors.

x = α1v1 + α2v2 +· · ·+αNvN =

N

n=1

αnvn (8.5.2)

Thus, the eigenvectors are called the principal axes of matrix A, and the squared

norm of a vector is the sum of the squares of the components (αn’s) along the

principal axis.

||x||2 = xT x = N

m=1

αmvmT N

n=1

αnvn =

N

m=1

N

n=1

αmαnvT

mvn =

N

n=1

α2

n

(8.5.3)

Premultiplying Eq. (8.5.2) by the matrix A and using Eq. (8.1.1) yields

Ax = λ1α1v1 + λ2α2v2 + ·· ·+λNαNvN =

N

n=1

λnαnvn (8.5.4)

This shows that premultiplying a vector x by matrix A has the same effect as

multiplying each principal component αn of x along the direction of eigenvector

vn by the associated eigenvalue λn. Therefore, the solution of a homogeneous

discrete-time state equation

x(k + 1) = Ax(k) with x(0) =

N

n=1

αnvn (8.5.5)

can be written as

x(k) =

N

n=1

λk

nαnvn (8.5.6)

which was illustrated by Eq. (E8.4.6) in Example 8.4. On the other hand, as illustrated

by (E8.3.9) in Example 8.3(b), the solution of a homogeneous continuoustime

state equation

x(t) = Ax(t) with x(0) =

N

n=1

αnvn (8.5.7)

386 MATRICES AND EIGENVALUES

can be written as

x(t) =

N

n=1

eλntαnvn (8.5.8)

Equations (8.5.6) and (8.5.8) imply that the eigenvalues of the system matrix

characterize the principal modes of the system described by the state equations.

That is, the eigenvalues determine not only whether the system is stable or

not—that is, whether the system state converges to an equilibrium state or

diverges—but also how fast the system state proceeds along the direction of

each eigenvector. More specifically, in the case of a discrete-time system, the

absolute values of all the eigenvalues must be less than one for stability and

the smaller the absolute value of an eigenvalue (less than one) is, the faster the

corresponding mode converges. In the case of a continuous-time system, the real

parts of all the eigenvalues must be negative for stability and the smaller a negative

eigenvalue is, the faster the corresponding mode converges. The difference

among the eigenvalues determines how stiff the system is (see Section 6.5.4).

This meaning of eigenvalues/eigenvectors is very important in dynamic systems.

Now, in order to figure out the meaning of eigenvalues/eigenvectors in static

systems, we define the mean vector and the covariance matrix of the vectors

{x(1), x(2), . . . , x(K)} representing K points in a two-dimensional space called the

x1x2 plane as

mx =

1

K

K

k=1

x(k), Cx =

1

K

K

k=1

[x(k) − mx][x(k) − mx ]T (8.5.9)

where the mean vector represents the center of the points and the covariance

matrix describes how dispersedly the points are distributed. Let us think about

the geometrical meaning of diagonalizing the covariance matrix Cx. As a simple

example, suppose we have four points

x(1) = 0

−1 , x(2) = −1

0 , x(3) = 2

3 , x(4) = 3

2 (8.5.10)

for which the mean vector mx , the covariance matrix Cx , and its modal matrix

are

mx = 1

1 , Cx = 2.5 2

2 2.5 , V = [ v1 v2 ] =

1

√2 1 1

−1 1

(8.5.11)

Then, we can diagonalize the covariance matrix as

V T CxV =

1

√2 1 −1

1 12.5 2

2 2.5 1

√2 1 1

−1 1

= 0.5 0

0 4.5 = λ1 0

0 λ2 = (8.5.12)

PHYSICAL MEANING OF EIGENVALUES/EIGENVECTORS 387

which has the eigenvalues on its main diagonal. On the other hand, if we transform

the four point vectors by using the modal matrix as

y = V T (x − mx) (8.5.13)

then the new four point vectors are

y(1) = 1/√2

−3/√2 , y(2) = −1/√2

−3/√2 , y(3) = −1/√2

3/√2 , y(4) = 1/√2

3/√2

(8.5.14)

for which the mean vector mx and the covariance matrix Cx are

my = V T (mx − mx) = 0

0 , Cy = V T CxV = 0.5 0

0 4.5 =

(8.5.15)

The original four points and the new points corresponding to them are depicted

in Fig. 8.1, which shows that the eigenvectors of the covariance matrix for a set of

point vectors represents the principal axes of the distribution and its eigenvalues

are related with the lengths of the distribution along the principal axes. The

difference among the eigenvalues determines how oblong the overall shape of

the distribution is.

Before closing this section, we may think about the meaning of the determinant

of a matrix composed of two two-dimensional vectors and three threedimensional

vectors.

3

2

2 4

1

0

0

y(2) y(1)

y(3) y(4)

x(1)

x2

x1

x(2)

x(3)

x(4)

v1

v2

−1

−2

−3

−4 −2

Figure 8.1 Eigenvalues/eigenvectors of a covariance matrix.

388 MATRICES AND EIGENVALUES

First, let us consider a 2 × 2 matrix composed of two two-dimensional vectors

x(1) and x(2).

X = [ x(1) x(2) ] = x11 x12

x21 x22 (8.5.16)

Conclusively, the absolute value of the determinant of this matrix

det(X) = |X| = x11x22 − x12x21 (8.5.17)

equals the area of the parallelogram having the two vectors as its two neighboring

sides. In order to certify this fact, let us make a clockwise rotation of the two

vectors by the phase angle of x(1)

−θ1 = −tan−1

x21

x11 (8.5.18)

so that the new vector y(1) corresponding to x(1) becomes aligned with the x1-axis

(see Fig. 8.2). For this purpose, we multiply our matrix X by the rotation matrix

defined by Eq. (8.4.2)

R(−θ1) = cos θ1 −sin(−θ1)

sin(−θ1) cos θ1 =

1

x2

11 + x2

21

x11 x21

−x21 x11 (8.5.19)

to get

Y = R(−θ1)X =

1

x2

11 + x2

21

x11 x21

−x21 x11 x11 x12

x21 x22 (8.5.20a)

[ y(1) y(2) ] =

1

x2

11 + x2

21

x2

11 + x2

21 x11x12 + x21x22

0 −x12x21 + x11x22 (8.5.20b)

The parallelograms having the original vectors and the new vectors as their two

neighboring sides are depicted in Fig. 8.2, where the areas of the parallelograms

turn out to be equal to the absolute values of the determinants of the matrices X

and Y as follows:

Area of the parallelograms

= Length of the bottom side × Height of the parallelogram

= (x1 component of y(1)) × (x2 component of y(2)) = y11y22 = det(Y )

=

x2

11 + x2

21

x2

11 + x2

21

× −x12x21 + x11x22

x2

11 + x2

21

≡ det(X) (8.5.21)

EIGENVALUE EQUATIONS 389

x (1) = (x11, x21)

x (2) = (x12, x22)

x21

q2 x11

q2 − q1

q1 < 0

x22

x2 x2

y (2)

y (1)

x1

R(−q1)

x1

x12

x11 + x21

2 2

(a) A parallelogram (b) The rotated parallelogram

Figure 8.2 Geometrical meaning of a determinant.

On extension of this result into a three-dimensional situation, the absolute

value of the determinant of a 3 × 3 matrix composed of three three-dimensional

vectors x(1), x(2), and x(3) equals the volume of the parallelepiped having the

three vectors as its three edges.

det(X) = |X| = |x(1) x(2) x(3) | =

x11 x12 x13

x21 x22 x23

x31 x32 x33

≡ x(1) × x(2) · x(3)

(8.5.22)

8.6 EIGENVALUE EQUATIONS

In this section, we consider a system of ordinary differential equations that can

be formulated as an eigenvalue problem.

For the undamped mass–spring system depicted in Fig. 8.3, the displacements

x1(t) and x2(t) of the two masses m1 and m2 are described by the following

system of differential equations:

x

1(t)

x

2(t) = −(k1 + k2)/m1 −k2/m1

−k2/m2 k2/m2 x1(t)

x2(t)

with x1(0)

x2(0) and x1(0)

x2(0)

x(t) = −Ax(t) with x(0) and x(0) (8.6.1)

Let the eigenpairs (eigenvalue–eigenvectors) of thematrixAbe (λn = ω2

n, vn) with

Avn = ω2

nvn (8.6.2)

390 MATRICES AND EIGENVALUES

spring

constant

k1

spring

constant

k2

m1 m2

x1(t ) x2(t )

Figure 8.3 An undamped mass–spring system.

Noting that the solution of Eq. (8.5.7) can be written as Eq. (8.5.8) in terms of

the eigenvectors of the system matrix, we write the solution of Eq. (8.6.1) as

x(t) =

2

n=1

wn(t)vn = [ v1 v2 ] w1(t)

w2(t) = Vw(t) (8.6.3)

and substitute this into Eq. (8.6.1) to have

2

n=1

w

n(t)vn = −A

2

n=1

wn(t)vn

(8.6.2) = −

2

n=1

wn(t)ω2

nvn (8.6.4)

w

n(t) = −ω2

nwn(t) for n = 1, 2 (8.6.5)

The solution of this equation is

wn(t) = wn(0) cos(ωnt) +

w

n (0)

ωn

sin(ωnt) with ωn = λn for n = 1, 2

(8.6.6)

where the initial value of w(t) = [w1(t) w2(t ]T can be obtained via Eq. (8.6.3)

from that of x(t) as

w(0)

(8.6.3) = V −1x(0)

(8.4.1) = V T x(0), w(0) = V T x(0) (8.6.7)

Finally, we substitute Eq. (8.6.6) into Eq. (8.6.3) to obtain the solution of

Eq. (8.6.1).

PROBLEMS

8.1 Symmetric Tridiagonal Toeplitz Matrix

Consider the following N × N symmetric tridiagonal Toeplitz matrix as

a b 0 ·· 0 0

b a b ·· 0 0

0 b a ·· 0 0

··

··

··

··

··

··

··

0 0 0 ·· a b

0 0 0 ·· b a

(P8.1.1)

PROBLEMS 391

(a) Verify that the eigenvalues and eigenvectors of this matrix are as follows,

with N = 3 for convenience.

λn = a + 2b cos

nπ

N + 1 for n = 1 to N (P8.1.2)

vn = 2

N + 1 sin

nπ

N + 1 sin

2nπ

N + 1· · · sin

Nnπ

N + 1T

(P8.1.3)

(b) Letting N = 3, a = 2, and b = 1, find the eigenvalues/eigenvectors of

the above matrix by using (P8.1.2,3) and by using the MATLAB routine

“eig_Jacobi()” or “eig()” for cross-check.

8.2 Circulant Matrix

Consider the following N × N circulant matrix as

h(0) h(N− 1) h(N− 2) ·· h(1)

h(1) h(0) h(N− 1) ·· h(2)

h(2) h(1) h(0) ·· h(3)

· · · ·· ·

· · · ·· · h(N − 1) h(N− 2) h(N− 3) ·· h(0)

(P8.2.1)

(a) Vertify that the eigenvalues and eigenvectors of this matrix are as follows,

with N = 4 for convenience.

λn = h(0) + h(N − 1)ej2πn/N + h(N − 2)ej2π2n/N (P8.2.2)

+ ·· ·+h(1)ej2π(N−1)n/N

vn = [1 ej2πn/N ej2π2n/N · · · ej2π(N−1)n/N ]T (P8.2.3)

for n = 0 to N − 1

(b) Letting N = 4, h(0) = 2, h(3) = h(1) = 1, and h(2) = 0, find the eigenvalues/

eigenvectors of the above matrix by using (P8.2.2,3) and by using

the MATLAB routine “eig_Jacobi()” or “eig()”. Do they agree? Do

they satisfy Eq. (8.1.1)?

8.3 Solving a Vector Differential Equation by Decoupling: Diagonalization.

Consider the following two-dimensional vector differential equation (state

equation) as

x1(t)

x2(t) = 0 1

−2 −3x1(t)

x2(t) + 0

1 us (t) (P8.3.1)

with x1(0)

x2(0) = 1

0 and us (t) = 1 ∀ t ≥ 0

392 MATRICES AND EIGENVALUES

which was solved by using Laplace transform in Problem P6.1. In this problem,

we solve it again by the decoupling method through diagonalization of

the system matrix.

(a) Show that the eigenvalues and eigenvectors of the system matrix are as

follows.

λ1 = −1, λ2 = −2; v1 = 1

−1 , v2 = 1

−2 (P8.3.2)

(b) Show that the diagonalization of the above vector differential equation

using the modal matrix V = [ v1 v2 ] yields the following equation:

w1(t)

w2(t) = −1 0

0 −2w1(t)

w2(t) + 1

−1 us (t) (P8.3.3)

= −w1(t) + us (t)

−2w2(t) − us (t) with w1(0)

w2(0) = 2

−1

(c) Show that these equations can be solved individually by using Laplace

transform technique to yield the following solution, which is the same

as Eq. (P6.1.2) obtained in Problem P6.1(a).

W1(s) =

1

s +

1

s + 1

, w1(t) = (1 + e−t)us (t) (P8.3.4a)

W2(s) = −1/2

s −

1/2

s + 2

, w2(t) = −

1

2

(1 + e−2t)us (t) (P8.3.4b)

x1(t)

x2(t) = 1/2 + e−t − (1/2)e−2t

−e−t + e−2t us (t) (P8.3.5)

8.4 Householder Method and QR Factorization

This method can zero-out several elements in a column vector at each iteration

and make any N × N matrix a (lower) triangular matrix in (N − 1)

iterations.

(a) Householder Reflection (Fig. P8.4)

Show that the transformation matrix by which we can multiply a vector

x to generate another vector y having the same norm is

H = [I − 2wwT ]

with w =

x − y

||x − y||2 =

1

c

(x − y), c = ||x − y||2, ||x|| = ||y|| (P8.4.1)

and that this is an orthonormal symmetric matrix such that HTH = HH = I ; H−1 = H. Note the following facts.

PROBLEMS 393

x − y = c w

m

x y

x = y

w = 1 (x − y) =

c x − y

x − y

2

Figure P8.4 Householder reflection.

(i) y = x − (x − y)

(P8.4.1) = x − cw (P8.4.2a)

(ii) wTw (P8.4.1) = 1 and ||x|| = ||y|| (P8.4.2b)

(iii) m = (x + y)/2 = x − (c/2)w (P8.4.2c)

(iv) The mean vector m of x and y is orthogonal to the difference vector

w = (x − y)/c.

Thus we have

wT (x − (c/2)w) = 0; wT x − (c/2)wT w = wT x − (c/2) = 0

(P8.4.3)

This gives an expression for c = ||x − y||2 as

c = ||x − y||2 = 2wT x (P8.4.4)

We can substitute this into (P8.4.2a) to get the desired result.

y = x − cw = x − 2wwT x = [I − 2wwT ]x ≡ Hx (P8.4.5)

On the other hand, the Householder transform matrix is an orthogonal

matrix, since

HTH = HH = [I − 2wwT ][I − 2wwT ]

= I − 4wwT + 4wwT wwT

= I − 4wwT + 4wwT = I (P8.4.6)

(b) Householder Transform

In order to show that the Householder matrix can be used to zero-out

some part of a vector, let us find the kth Householder matrix Hk transforming

any vector

x = x1 · · · xk−1 xk xk+1 · · · xN (P8.4.7)

394 MATRICES AND EIGENVALUES

into

y = x1 · · · xk−1 −gk 0 · · · 0 (P8.4.8)

where gk is fixed in such a way that the norms of these two vectors are

the same:

gk =

N

n=k

x2

n (P8.4.9)

First, we find the difference vector of unit norm as

wk =

1

c

(x − y)

=

1

c 0 · · · 0 xk + gk xk+1 · · · xN (P8.4.10)

with

c = ||x − y||2 = (xk + gk)2 + x2

k+1 + ·· ·+x2

N (P8.4.11)

Then, one more thing we should do is to substitute this difference vector

into Eq. (P8.4.1).

Hk = [I − 2wkwT

k ] (P8.4.12)

Complete the following routine “Householder()” by permuting the

statements and try it with k = 1, 2, 3, and 4 for a four-dimensional

vector generated by the MATLAB command rand(5,1) to check if it

works fine.

>> x = rand(5,1), for k = 1:4, householder(x,k)*x, end

function H = Householder(x,k)

%Householder transform to zero out tail part starting from k + 1

H = eye(N) – 2*w*w’; %Householder matrix

N = length(x);

w = zeros(N,1);

w(k) =(x(k) + g)/c; w(k + 1:N) = x(k + 1:N)/c; %Eq.(P8.4.10)

tmp = sum(x(k + 1:N).^ 2);

c = sqrt((x(k) + g)^2 + tmp); %Eq.(P8.4.11)

g = sqrt(x(k)^2 + tmp); %Eq.(P8.4.9)

(c) QR Factorization Using Householder Transform

We can use Householder transform to zero out the part under the main

diagonal of each column of an N × N matrix A successively and then

make it a lower triangular matrix R in (N − 1) iterations. The necessary

operations are collectively written as

HN−1HN−2 · · ·H1A =R (P8.4.13)

PROBLEMS 395

which implies that

A = [HN−1HN−2 · · ·H1]−1R = H−1

1 · · ·H−1

N−2H−1

N−1R

= H1 · · ·HN−2HN−1R = QR (P8.4.14)

where the product of all the Householder matrices

Q = H1 · · ·HN−2HN−1 (P8.4.15)

turns out to be not only symmetric, but also orthogonal like each Hk:

QTQ = [H1 · · ·HN−2HN−1]TH1 · · ·HN−2HN−1

= HT

N−1HT

N−2 · · ·HT

1 H1 · · ·HN−2HN−1 = I

This suggests a QR factorization method that is cast into the following

routine “qr_my()”. You can try it for a nonsingular 3 ×3 matrix generated

by the MATLAB command rand(3) and compare the result with

that of the MATLAB built-in routine “qr()”.

function [Q,R] = qr_my(A)

%QR factorization

N = size(A,1); R = A; Q = eye(N);

for k = 1:N – 1

H = Householder(R(:,k),k);

R = H*R; %Eq.(P8.4.13)

Q = Q*H; %Eq.(P8.4.15)

end

8.5 Hessenberg Form Using Householder Transform

function [Hs,HH] = Hessenberg(A)

%Transform into an almost upper triangular matrix

% having only zeros below lower subdiagonal

N = size(A,1); Hs = A; HH = eye(N); %HH*A*HH’ = Hs

for k = 1:N – 2

H = Householder(Hs(:,k), );

Hs = H*Hs*H; HH = H*HH;

end

We can make use of Householder transform (introduced in Problem 8.4) to

zero-out the elements below the lower subdiagonal of a matrix so that it

becomes an upper Hessenberg form which is almost upper-triangular matrix.

Complete the above routine “Hessenberg()” by filling in the second input

argument of the routine “Householder()” and try it for a 5 ×5 matrix

generated by the MATLAB command rand(5) to check if it works.

396 MATRICES AND EIGENVALUES

8.6 QR Factorization of Hessenberg Form Using the Givens Rotation

We can make use of the Givens rotation to get the QR factorization of Hessenberg

form by the procedure implemented in the following routine

“qr_Hessenberg()”, where each element on the lower subdiagonal is zeroed

out at each iteration. Generate a 4 × 4 random matrix A by theMATLAB command

rand(4), transform it into a Hessenberg form Hs by using the routine

“Hessenberg()” and try this routine “qr_Hessenberg()” for the matrix of

Hessenberg form. Check the validity by seeing if norm(Hs-Q*R) ≈ 0 or not.

function [Q,R] = qr_Hessenberg(Hs)

%QR factorization of Hessenberg form by Givens rotation

N = size(Hs,1);

Q = eye(N); R = Hs;

for k = 1:N – 1

x = R(k,k); y = R(k+1,k); r = sqrt(x*x + y*y);

c = x/r; s = -y/r;

R0 = R; Q0 = Q;

R(k,:) = c*R0(k,:) – s*R0(k + 1,:);

R(k + 1,:) = s*R0(k,:) + c*R0(k + 1,:);

Q(:,k) = c*Q0(:,k) – s*Q0(:,k + 1);

Q(:,k + 1) = s*Q0(:,k) + c*Q0(:,k + 1);

end

8.7 Diagonalization by Using QR Factorization to Find Eigenvalues

You will see that a real symmetric matrix A can be diagonalized into a

diagonal matrix having the eigenvalues on its diagonal if we repeat the

similarity transformation by using the orthogonal matrix Q obtained from the

QR factorization. For this purpose, take the following steps.

function [eigs,A] = eig_QR(A,kmax)

%Find eigenvalues by using QR factorization

if nargin < 2, kmax = 200; end

for k = 1:kmax

[Q,R] = qr(A); %A = Q*R; R =Q’*A =Q^-1*A

A = R*Q; %A = Q^ – 1*A*Q

end

eigs = diag(A);

function [eigs,A] = eig_QR_Hs(A,kmax)

%Find eigenvalues by using QR factorization via Hesenberg

if nargin < 2, kmax = 200; end

Hs = hessenberg(A);

for k = 1:kmax

[Q,R] = qr_hessenberg(Hs); %Hs = Q*R; R = Q’*Hs = Q^ – 1*Hs

Hs = R*Q; %Hs = Q^ – 1*Hs*Q

end

eigs = diag(Hs);

PROBLEMS 397

(a) Make the above routine “eig_QR()” that uses the MATLAB built-in

routine “qr()” and then apply it to a 4 × 4 random symmetric matrix A

generated by the following MATLAB statements.

>> A = rand(4); A = A + A’;

(b) Make the above routine “eig_QR_Hs()” that transforms a given matrix

into a Hessenberg form by using the routine “Hessenberg()” (appeared

in Problem 8.5) and then repetitively makes the QR factorization by

using the routine “qr_Hessenberg()” (appeared in Problem 8.6) and

the similarity transformation by the orthogonal matrix Q until the matrix

becomes diagonal. Apply it to the 4 × 4 random symmetric matrix A

generated in (a) and compare the result with those obtained in (a) and

by using the MATLAB built-in routine “eig()” for cross-check.

8.8 Differential/Difference Equation, State Equation, and Eigenvalue

As mentioned in Section 6.5.3, a high-order scalar differential equation such

as

x(3)(t) + a2x(2)(t) + a1x(t) + a0x(t) = u(t) (P8.8.1)

can be transformed into a first-order vector differential equation, called a

state equation, as

x1(t)

x2(t)

x3(t)

=

0 1 0

0 0 1

−a0 −a1 −a2

x1(t)

x2(t)

x3(t)

+

0

0

1

u(t) (P8.8.2a)

x(t) = [1 0 0]

x1(t)

x2(t)

x3(t)

(

P8.8.2b)

The characteristic equation of the differential equation (P8.8.1) is

s3 + a2s2 + a1s + a0 = 0 (P8.8.3)

and its roots are called the characteristic roots.

(a) What is the relationship between these characteristic roots and the eigenvalues

of the system matrix A of the above state equation (P8.8.2)? To

answer this question, write the equation |λI − A| = 0 to solve for the

eigenvalues of A, and show that it is equivalent to Eq. (P8.8.3). To extend

your experience or just for practice, you can try the symbolic computation

of MATLAB by running the following program “nm8p08a.m”.

398 MATRICES AND EIGENVALUES

%nm8p08a

syms a0 a1 a2 s

A =[0 1 0;0 0 1;-a0 -a1 -a2]; %(P8.8.2a)

det(s*eye(size(A))- A) %characteristic polynomial

ch_eq = poly(A) %or, equivalently

(b) Let the input u(t) in the state equation (P8.8.2) be dependent on the state

as

u(t) = K x(t) = [K0x1(t) K1x2(t) K2x3(t) ] (P8.8.4)

Then, the state equation can be written as

x1(t)

x2(t)

x3(t)

=

0 1 0

0 0 1

K0 − a0 K1 − a1 K2 − a2

x1(t)

x2(t)

x3(t)

(P8.8.5)

If the parameters of the original system matrix are a0 = 1, a1 = −2, and

a2 = 3, what are the values of the gain matrix K = [K0 K1 K2] you

will fix so that the virtual system matrix in the state equation (P8.8.5)

has the eigenvalues of λ = −1,−2, and −3? Note that the characteristic

equation of the system whose behavior is described by the state

equation (P8.8.5) is

s3 + (a2 − K2)s2 + (a1 − K1)s + a0 − K0 = 0 (P8.8.6)

and the equation having the roots of λ = −1,−2, and −3 is

(s + 1)(s + 2)(s + 3) = s3 + 6s2 + 11s + 6 = 0 (P8.8.7)

8.9 A Homogeneous Differential Equation—An Eigenvalue Equation

Consider the undamped mass-spring system depicted in Fig. 8.3, where the

masses and the spring constants are m1 = 1,m2 = 1[kg] and k1 = 5, k2 = 10

[N/m], respectively. Complete the following program “nm8p09.m” whose

objective is to solve the second-order differential equation (8.6.1) with the

initial conditions [x1(0), x2(0), x1 (0), x2 (0)] = [1,−0.5, 0, 0] for the time

interval [0,10] in two ways—that is, by using the ODE-solver “ode45()”

(Section 6.5.1) and by using the eigenvalue method (Section 8.6) and plot

the two solutions. Run the completed program to obtain the solution graphs

for x1(t) and x2(t).

(cf) Note that the second-order vector differential equation (8.6.1) can be written as

the following state equation:

x(t )

x(t ) = O I

−A O x(t )

x(t ) (P8.9.1)

PROBLEMS 399

%nm8p09.m solve a set of differential eqs. (a state equation)

clear, clf

global A

df = ’df861’;

k1 = 5; k2 = 10; m1 = 1; m2 = 1; % the spring constants and the masses

A = [(k1 + k2)/m1 -k2/m1; -k2/m2 k2/m2]; NA = size(A,2);

t0 = 0; tf =??; x0 =[? ???? ? ?]; % initial/final time, initial values

[t4,x4] = ode45(df,[t0 tf],x0);

[V,LAMBDA] = eig(A); % modal matrix composed of eigenvectors

w0 = x0(1:NA)*V; w10 = x0(NA+1:end)*V; % Eq.(8.6.8)

omega = ??????????????????;

for n = 1:NA % Eq.(8.6-7)

omegan=omega(n);

w(:,n) = [cos(omega n;*t4) sin(omega n*t4)]*[w0(n);w10(n)/omega n];

end

xE = w*V.’; % Eq.(8.6.3)

for n = 1:NA

subplot(311 + n), plot(t4,x4(:,n),’b’, t4,xE(:,n),’r’)

end

function dx = df861(t,x)

global A

NA = size(A,2);

if length(x) ~= 2*NA, error(’Some dimension problem’); end

dx = [zeros(NA) eye(NA); -A zeros(NA)]*x(:);

if size(x,2) > 1, dx = dx.’; end

9

PARTIAL DIFFERENTIAL

EQUATIONS

What is a partial differential equation (PDE)? It is a class of differential equations

involving more than one independent variable. In this chapter, we consider a general

second-order PDE in two independent variables x and y, which is written as

A(x, y)

∂2u

∂x2 + B(x, y)

∂2u

∂x∂y + C(x, y)

∂2u

∂y2 = f x, y, u,

∂u

∂x

,

∂u

∂y (9.0.1)

for x0 ≤ x ≤ xf , y0 ≤ y ≤ yf

with the boundary conditions given by

u(x, y0) = by0(x), u(x, yf ) = byf (x),

u(x0, y) = bx0(y), and u(xf , y) = bxf (y)

(9.0.2)

These PDEs are classified into three groups:

Elliptic PDE: if B2 − 4AC < 0

Parabolic PDE: if B2 − 4AC = 0

Hyperbolic PDE: if B2 − 4AC > 0

These three types of PDE are associated with equilibrium states, diffusion states,

and oscillating systems, respectively. We will study some numerical methods for

solving these PDEs, since their analytical solutions are usually difficult to find.

Applied Numerical Methods Using MATLAB, by Yang, Cao, Chung, and Morris

Copyr ight 2005 John Wiley & Sons, I nc., ISBN 0-471-69833-4

401

402 PARTIAL DIFFERENTIAL EQUATIONS

9.1 ELLIPTIC PDE

As an example, we will deal with a special type of elliptic equation called

Helmholtz’s equation, which is written as

∇2u(x, y) + g(x, y)u(x, y) =

∂2u(x, y)

∂x2 +

∂2u(x, y)

∂y2 + g(x, y)u(x, y) = f (x, y) (9.1.1)

over a domain D = {(x, y)|x0 ≤ x ≤ xf , y0 ≤ y ≤ yf } with some boundary conditions

of

u(x0, y) = bx0(y), u(xf , y) = bxf (y),

u(x, y0) = by0(x), and u(x, yf ) = byf (x)

(9.1.2)

(cf) Equation (9.1.1) is called Poisson’s equation if g(x, y) = 0 and it is called Laplace’s

equation if g(x, y) = 0 and f (x, y) = 0.

To apply the difference method, we divide the domain into Mx sections, each

of length x = (xf − x0)/Mx along the x-axis and into My sections, each of

length y = (yf − y0)/My along the y-axis, respectively, and then replace the

second derivatives by the three-point central difference approximation (5.3.1)

∂2u(x, y)

∂x2 xj ,yi

∼=

ui,j+1 − 2ui,j + ui,j−1

x2 with xj = x0 + jx, yi = y0 + iy

(9.1.3a)

∂2u(x, y)

∂y2 xj ,yi

∼=

ui+1,j − 2ui,j + ui−1,j

y2 with ui,j = u(xj, yi) (9.1.3b)

so that, for every interior point (xj, yi ) with 1 ≤ i ≤ My − 1 and 1 ≤ j ≤ Mx − 1,

we obtain the finite difference equation

ui,j+1 − 2ui,j + ui,j−1

x2 +

ui+1,j − 2ui,j + ui−1,j

y2 + gi,jui,j = fi,j (9.1.4)

where

ui,j = u(xj, yi ), fi,j = f (xj, yi ), and gi,j = g(xj, yi)

These equations can somehow be arranged into a system of simultaneous

equations with respect to the (My − 1)(Mx − 1) variables {u1,1, u1,2, . . . , u1,Mx−1,

u2,1, . . . , u2,Mx−1, . . . ,uMy−1,1, uMy−1,2, . . . , uMy−1,Mx−1}, but it seems to be

messy to work with and we may be really in trouble as Mx and My become

large. A simpler way is to use the iterative methods introduced in Section 2.5.

To do so, we first need to shape the equations and the boundary conditions into

the following form:

ui,j = ry(ui,j+1 + ui,j−1) + rx(ui+1,j + ui−1,j ) + rxy(gi,jui,j − fi,j ) (9.1.5a)

ELLIPTIC PDE 403

ui,0 = bx0(yi ), ui,Mx = bxf (yi ), u0,j = by0(xj ), uMy ,j = byf (xj ) (9.1.5b)

where

y2

2(x2 + y2) = ry ,

x2

2(x2 + y2) = rx ,

x2y2

2(x2 + y2) = rxy (9.1.6)

How do we initialize this algorithm? If we have no priori knowledge about the

solution, it is reasonable to take the average value of the boundary values as the

initial values of ui,j .

The objective of the MATLAB routine “poisson.m” is to solve the above

equation.

function [u,x,y] = poisson(f,g,bx0,bxf,by0,byf,D,Mx,My,tol,MaxIter)

%solve u_xx + u_yy + g(x,y)u = f(x,y)

% over the region D = [x0,xf,y0,yf] = {(x,y) |x0 <= x <= xf, y0 <= y <= yf}

% with the boundary Conditions:

% u(x0,y) = bx0(y), u(xf,y) = bxf(y)

% u(x,y0) = by0(x), u(x,yf) = byf(x)

% Mx = # of subintervals along x axis

% My = # of subintervals along y axis

% tol : error tolerance

% MaxIter: the maximum # of iterations

x0 = D(1); xf = D(2); y0 = D(3); yf = D(4);

dx = (xf – x0)/Mx; x = x0 + [0:Mx]*dx;

dy = (yf – y0)/My; y = y0 + [0:My]’*dy;

Mx1 = Mx + 1; My1 = My + 1;

%Boundary conditions

for m = 1:My1, u(m,[1 Mx1])=[bx0(y(m)) bxf(y(m))]; end %left/right side

for n = 1:Mx1, u([1 My1],n) = [by0(x(n)); byf(x(n))]; end %bottom/top

%initialize as the average of boundary values

sum_of_bv = sum(sum([u(2:My,[1 Mx1]) u([1 My1],2:Mx)’]));

u(2:My,2:Mx) = sum_of_bv/(2*(Mx + My – 2));

for i = 1:My

for j = 1:Mx

F(i,j) = f(x(j),y(i)); G(i,j) = g(x(j),y(i));

end

end

dx2 = dx*dx; dy2 = dy*dy; dxy2 = 2*(dx2 + dy2);

rx = dx2/dxy2; ry = dy2/dxy2; rxy = rx*dy2;

for itr = 1:MaxIter

for j = 2:Mx

for i = 2:My

u(i,j) = ry*(u(i,j + 1)+u(i,j – 1)) + rx*(u(i + 1,j)+u(i – 1,j))…

+ rxy*(G(i,j)*u(i,j)- F(i,j)); %Eq.(9.1.5a)

end

end

if itr > 1 & max(max(abs(u – u0))) < tol, break; end

u0 = u;

end

%solve_poisson in Example 9.1

f = inline(’0’,’x’,’y’); g = inline(’0’,’x’,’y’);

x0 = 0; xf = 4; Mx = 20; y0 = 0; yf = 4; My = 20;

bx0 = inline(’exp(y) – cos(y)’,’y’); %(E9.1.2a)

bxf = inline(’exp(y)*cos(4) – exp(4)*cos(y)’,’y’); %(E9.1.2b)

by0 = inline(’cos(x) – exp(x)’,’x’); %(E9.1.3a)

byf = inline(’exp(4)*cos(x) – exp(x)*cos(4)’,’x’); %(E9.1.3b)

D = [x0 xf y0 yf]; MaxIter = 500; tol = 1e-4;

[U,x,y] = poisson(f,g,bx0,bxf,by0,byf,D,Mx,My,tol,MaxIter);

clf, mesh(x,y,U), axis([0 4 0 4 -100 100])

404 PARTIAL DIFFERENTIAL EQUATIONS

Example 9.1. Laplace’s Equation—Steady-State Temperature Distribution.

Consider Laplace’s equation

∇2u(x, y) =

∂2u(x, y)

∂x2 +

∂2u(x, y)

∂y2 =0 for0≤ x ≤ 4, 0 ≤ y ≤ 4

(E9.1.1)

with the boundary conditions

u(0, y) = ey − cos y, u(4, y) = ey cos 4 − e4 cos y (E9.1.2)

u(x, 0) = cos x − ex, u(x,4) = e4 cos x − ex cos 4 (E9.1.3)

What we will get from solving this equation is u(x, y), which supposedly

describes the temperature distribution over a square plate having each side 4

units long (Fig. 9.1). We made the MATLAB program “solve_poisson.m” in

order to use the routine “poisson()” to solve Laplace’s equation given above

and run this program to obtain the result shown in Fig. 9.2.

Now, let us consider the so-called Neumann boundary conditions described as

∂u(x, y)

∂x x=x0 = b

x0(y) for x = x0 (the left-side boundary) (9.1.7)

x0 x1 Δx xj xMx

Δy

yi

y1

y0

yMy

ui

+ 1, j

ui, j − 1

ui − 1, j

ui, j ui, j + 1

Dirichlet-type boundary condition (function value fixed)

Neumann-type boundary condition (derivative fixed)

Figure 9.1 The grid for elliptic equations with Dirichlet/Neumann-type boundary condition.

ELLIPTIC PDE 405

100

50

0

−50

−100

4

3

2

1

0 0

1

2

3

4

u(x, y)

y x

Figure 9.2 Temperature distribution over a plate—Example 9.1.

Replacing the first derivative on the left-side boundary (x = x0) by its three-point

central difference approximation (5.1.8)

ui,1 − ui,−1

2x ≈ bx0(yi ), ui,−1 ≈ ui,1 − 2bx0(yi)x for i = 1, 2, . . . , My − 1

(9.1.8)

and then substituting this constraint into Eq. (9.1.5a) at the boundary points, we

have

ui,0 = ry(ui,1 + ui,−1) + rx(ui+1,0 + ui−1,0) + rxy(gi,0ui,0 − fi,0)

= ry(ui,1 + ui,1 − 2bx0(yi)x) + rx(ui+1,0 + ui−1,0) + rxy(gi,0ui,0 − fi,0)

= 2ryui,1 + rx(ui+1,0 + ui−1,0) + rxy(gi,0ui,0 − fi,0 − 2bx0(yi)/x)

for i = 1, 2, . . .,My − 1 (9.1.9)

If the boundary condition on the lower side boundary (y = y0) is also of

Neumann type, then we need to write similar equations for j = 1, 2, . . . , Mx − 1

u0,j = ry(u0,j+1 + u0,j−1) + 2rxu1,j + rxy(g0,ju0,j − f0,j − 2by 0(xj)/y)

(9.1.10)

and additionally for the left-lower corner point (x0, y0),

u0,0 = 2(ryu0,1 + rxu1,0) + rxy(g0,0u0,0 − f0,0 − 2(bx0(y0)/x + 2by0(x0)/y))

(9.1.11)

406 PARTIAL DIFFERENTIAL EQUATIONS

9.2 PARABOLIC PDE

An example of a parabolic PDE is a one-dimensional heat equation describing

the temperature distribution u(x, t) (x is position, t is time) as

A

∂2u(x, t)

∂x2 =

∂u(x, t)

∂t

for 0 ≤ x ≤ xf , 0 ≤ t ≤T (9.2.1)

In order for this equation to be solvable, the boundary conditions u(0, t) = b0(t) & u(xf , t) = bxf (t) as well as the initial condition u(x, 0) = i0(x) should

be provided.

9.2.1 The Explicit Forward Euler Method

To apply the finite difference method, we divide the spatial domain [0, xf ] into

M sections, each of length x = xf/M, and divide the time domain [0, T ] into

N segments, each of duration t = T/N, and then replace the second partial

derivative on the left-hand side and the first partial derivative on the right-hand

side of the above equation (9.2.1) by the central difference approximation (5.3.1)

and the forward difference approximation (5.1.4), respectively, so that we have

A

uk

i+1 − 2uk

i + uk

i−1

x2 =

uk+1

i − uk

i

t

(9.2.2)

This can be cast into the following algorithm, called the explicit forward Euler

method, which is to be solved iteratively:

uk+1

i = r(uk

i+1 + uk

i−1) + (1 − 2r)uk

i with r = A

t

x2 (9.2.3)

for i = 1, 2, . . . ,M − 1

To find the stability condition of this algorithm, we substitute a trial solution

uk

i = λkejiπ/P (P is any nonzero integer) (9.2.4)

into Eq. (9.2.3) to get

λ = r(ejπ/P + e−jπ/P ) + (1 − 2r) = 1 − 2r(1 − cos(π/P )) (9.2.5)

Since we must have |λ| ≤ 1 for nondivergence, the stability condition turns out

to be

r = A

t

x2 ≤

1

2

(9.2.6)

PARABOLIC PDE 407

function [u,x,t] = heat_exp(a,xf,T,it0,bx0,bxf,M,N)

%solve a u_xx = u_t for 0 <= x <= xf, 0 <= t <= T

% Initial Condition: u(x,0) = it0(x)

% Boundary Condition: u(0,t) = bx0(t), u(xf,t) = bxf(t)

% M = # of subintervals along x axis

% N = # of subintervals along t axis

dx = xf/M; x = [0:M]’*dx;

dt = T/N; t = [0:N]*dt;

for i = 1:M + 1, u(i,1) = it0(x(i)); end

for n = 1:N + 1, u([1 M + 1],n) = [bx0(t(n)); bxf(t(n))]; end

r = a*dt/dx/dx, r1 = 1 – 2*r;

for k = 1:N

for i = 2:M

u(i,k+1) = r*(u(i + 1,k) + u(i-1,k)) + r1*u(i,k); %Eq.(9.2.3)

end

end

This implies that as we decrease the spatial interval x for better accuracy, we

must also decrease the time step t at the cost of more computations in order

not to lose the stability.

The MATLAB routine “heat_exp()” has been composed to implement this

algorithm.

9.2.2 The Implicit Backward Euler Method

In this section, we consider another algorithm called the implicit backward Euler

method, which comes out from substituting the backward difference approximation

(5.1.6) for the first partial derivative on the right-hand side of Eq. (9.2.1) as

A

uk

i+1 − 2uk

i + uk

i−1

x2 =

uk

i − uk−1

i

t

(9.2.7)

−ruk

i−1 + (1 + 2r)uk

i − ruk

i+1 = uk−1

i with r = A

t

x2 (9.2.8)

for i = 1, 2, . . . , M − 1

If the values of uk

0 and uk

M at both end points are given from the Dirichlet

type of boundary condition, then the above equation will be cast into a system

of simultaneous equations:

1 + 2r −r 0 · 0 0

−r 1 + 2r −r · 0 0

0 −r 1 + 2r · 0 0

· · · · · · 0 0 0 · 1 + 2r −r

0 0 0 · −r 1 + 2r

uk

1

uk

2

uk

3

·

uk

M−2

uk

M−1

=

uk−1

1 + ruk

0

uk−1

2

uk−1

3

·

uk−1

M−2

uk−1

M−1 + ruk

M

(9.2.9)

408 PARTIAL DIFFERENTIAL EQUATIONS

How about the case where the values of ∂u/∂x|x=0 = b0(t) at one end are

given? In that case, we approximate this Neumann type of boundary condition by

uk

1 − uk

−1

2x = b0 (k) (9.2.10)

and mix it up with one more equation associated with the unknown variable uk

0

−ruk

−1 + (1 + 2r)uk

0 − ruk

1 = uk−1

0 (9.2.11)

to get

(1 + 2r)uk

0 − 2ruk

1 = uk−1

0 − 2rb0 (k)x (9.2.12)

We augment Eq. (9.2.9) with this to write

1 + 2r −2r 0 0 · 0 0

−r 1 + 2r −r 0 · 0 0

0 −r 1 + 2r −r · 0 0

0 0 −r 1 + 2r · 0 0

· · · · · · ·

0 0 0 · · 1 + 2r −r

0 0 0 · · −r 1 + 2r

uk

0

uk

1

uk

2

uk

3

·

uk

M−2

uk

M−1

=

uk−1

0 − 2rb0(k)x

uk−1

1

uk−1

2

uk−1

3

·

uk−1

M−2

uk−1

M−1 + ruk

M

(9.2.13)

Equations such as Eq. (9.2.9) or (9.2.13) are really nice in the sense that they

can be solved very efficiently by exploiting their tridiagonal structures and are

guaranteed to be stable owing to their diagonal dominancy. The unconditional

stability of Eq. (9.2.9) can be shown by substituting Eq. (9.2.4) into Eq. (9.2.8):

−re−jπ/P + (1 + 2r) − rejπ/P = 1/λ, λ =

1

1 + 2r(1 − cos(π/P ))

, |λ| ≤ 1

(9.2.14)

The following routine “heat_imp()” implements this algorithm to solve the

PDE (9.2.1) with the ordinary (Dirichlet type of) boundary condition via Eq. (9.2.9).

function [u,x,t] = heat_imp(a,xf,T,it0,bx0,bxf,M,N)

%solve a u_xx = u_t for 0 <= x <= xf, 0 <= t <= T

% Initial Condition: u(x,0) = it0(x)

% Boundary Condition: u(0,t) = bx0(t), u(xf,t) = bxf(t)

% M = # of subintervals along x axis

% N = # of subintervals along t axis

dx = xf/M; x = [0:M]’*dx;

dt = T/N; t = [0:N]*dt;

for i = 1:M + 1, u(i,1) = it0(x(i)); end

for n = 1:N + 1, u([1 M + 1],n) = [bx0(t(n)); bxf(t(n))]; end

r = a*dt/dx/dx; r2 = 1 + 2*r;

for i = 1:M – 1

A(i,i) = r2; %Eq.(9.2.9)

if i > 1, A(i – 1,i) = -r; A(i,i – 1) = -r; end

end

for k = 2:N + 1

b = [r*u(1,k); zeros(M – 3,1); r*u(M + 1,k)] + u(2:M,k – 1); %Eq.(9.2.9)

u(2:M,k) = trid(A,b);

end

PARABOLIC PDE 409

9.2.3 The Crank–Nicholson Method

Here, let us go back to see Eq. (9.2.7) and try to improve the implicit backward

Euler method. The difference approximation on the left-hand side is taken at

time point k, while the difference approximation on the right-hand side is taken

at the midpoint between time k and k − 1, if we regard it as the central difference

approximation with time step t/2. Doesn’t this seem to be inconsistent?

How about taking the difference approximation of both sides at the same time

point—say, the midpoint between k + 1 and k—for balance? In order to do so,

we take the average of the central difference approximations of the left-hand side

at the two points k + 1 and k, yielding

A

2

uk+1

i+1 − 2uk+1

i + uk+1

i−1

x2 +

uk

i+1 − 2uk

i + uk

i−1

x2 =

uk+1

i − uk

i

t

(9.2.15)

which leads to the so-called Crank–Nicholson method:

−ruk+1

i+1 + 2(1 + r)uk+1

i − ruk+1

i−1 = ruk

i+1 + 2(1 − r)uk

i + ruk

i−1 (9.2.16)

with r = A

t

x2

With the Dirichlet/Neumann type of boundary condition on x0/xM, respectively,

this can be cast into the following tridiagonal system of equations.

2(1 + r) −r 0 · 0 0

−r 2(1 + r) −r · 0 0

0 −r 2(1 + r) · 0 0

· · · · · ·

0 0 0 · 2(1 + r) −r

0 0 0 · −2r 2(1 + r)

uk+1

1

uk+1

2

uk+1

3

·

uk+1

M−1

uk+1

M

=

2(1 − r) r 0 · 0 0

r 2(1 − r) r · 0 0

0 r 2(1 − r) · 0 0

· · · · · ·

0 0 0 · 2(1 − r) r

0 0 0 · 2r 2(1 − r)

uk

1

uk

2

uk

3

·

uk

M−1

uk

M

+

r(uk+1

0 + uk

0)

0

0

·0

2r(bM (k + 1) + bM (k))

(9.2.17)

This system of equations can also be solved very efficiently, and its unconditional

stability can be shown by substituting Eq. (9.2.4) into Eq. (9.2.16):

2λ(1 + r(1 − cos(π/P ))) = 2(1 − r(1 − cos(π/P ))),

λ =

1 − r(1 − cos(π/P ))

1 + r(1 − cos(π/P ))

, |λ| ≤ 1 (9.2.18)

This algorithm is cast into the following MATLAB routine “heat_CN()”.

410 PARTIAL DIFFERENTIAL EQUATIONS

function [u,x,t] = heat_CN(a,xf,T,it0,bx0,bxf,M,N)

%solve a u_xx = u_t for 0 <= x <= xf, 0 <= t <= T

% Initial Condition: u(x,0) = it0(x)

% Boundary Condition: u(0,t) = bx0(t), u(xf,t) = bxf(t)

% M = # of subintervals along x axis

% N = # of subintervals along t axis

dx = xf/M; x = [0:M]’*dx;

dt = T/N; t = [0:N]*dt;

for i = 1:M + 1, u(i,1) = it0(x(i)); end

for n = 1:N + 1, u([1 M + 1],n) = [bx0(t(n)); bxf(t(n))]; end

r = a*dt/dx/dx;

r1 = 2*(1 – r); r2 = 2*(1 + r);

for i = 1:M – 1

A(i,i) = r1; %Eq.(9.2.17)

if i > 1, A(i – 1,i) = -r; A(i,i – 1) = -r; end

end

for k = 2:N + 1

b = [r*u(1,k); zeros(M – 3,1); r*u(M + 1,k)] …

+ r*(u(1:M – 1,k – 1) + u(3:M + 1,k – 1)) + r2*u(2:M,k – 1);

u(2:M,k) = trid(A,b); %Eq.(9.2.17)

end

Example 9.2. One-Dimensional Parabolic PDE: Heat Flow Equation.

Consider the parabolic PDE

∂2u(x, t)

∂x2 =

∂u(x, t)

∂t

for 0 ≤ x ≤ 1, 0 ≤ t ≤ 0.1 (E9.2.1)

with the initial condition and the boundary conditions

u(x, 0) = sin πx, u(0, t) = 0, u(1, t) = 0 (E9.2.2)

We made the MATLAB program “solve_heat.m” in order to use the routines

“heat_exp()”, “heat_imp()”, and “heat_CN()” in solving this equation and ran

this program to obtain the results shown in Fig. 9.3. Note that with the spatial

interval x = xf/M = 1/20 and the time step t = T/N = 0.1/100 = 0.001,

we have

r = A

t

x2 = 1

0.001

(1/20)2 = 0.4 (E9.2.3)

which satisfies the stability condition (r ≤ 1/2) (9.2.6) and all of the three methods

lead to reasonably fair results with a relative error of about 0.013. But,

if we decrease the spatial interval to x = 1/25 for better resolution, we have

r = 0.625, violating the stability condition and the explicit forward Euler method

(“heat_exp()”) blows up because of instability as shown in Fig. 9.3a, while

the implicit backward Euler method (“heat_imp()”) and the Crank–Nicholson

method (“heat_CN()”) work quite well as shown in Figs. 9.3b,c. Now, with the

spatial interval x = 1/25 and the time step t = 0.1/120, the explicit method

as well as the other ones works well with a relative error less than 0.001 in return

PARABOLIC PDE 411

1

0.5

0.5

x

0 0

0.05

t 0.1

1

0

(c) The Crank-Nicholson method

5

0

x

0.05

t

0.1

1

−5 0

1

0.5

0.5

x

0 0

0.05

t

1 0.1

(a) The explicit method (b) The implicit method

0 0

Figure 9.3 Results of various algorithms for a one-dimensional parabolic PDE: heat equation.

for somewhat (30%) more computations, despite that r = 0.5208 doesn’t strictly

satisfy the stability condition.

This implies that the condition (r ≤ 1/2) for stability of the explicit forward

Euler method is not a necessary one, but only a sufficient one. Besides, if it

converges, its accuracy may be better than that of the implicit backward Euler

method, but generally no better than that of the Crank–Nicholson method.

%solve_heat

a = 1; %the parameter of (E9.2.1)

it0 = inline(’sin(pi*x)’,’x’); %initial condition

bx0 = inline(’0’); bxf = inline(’0’); %boundary condition

xf = 1; M = 25; T = 0.1; N = 100; %r = 0.625

%analytical solution

uo = inline(’sin(pi*x)*exp(-pi*pi*t)’,’x’,’t’);

[u1,x,t] = heat_exp(a,xf,T,it0,bx0,bxf,M,N);

figure(1), clf, mesh(t,x,u1)

[u2,x,t] = heat_imp(a,xf,T,it0,bx0,bxf,M,N); %converge unconditionally

figure(2), clf, mesh(t,x,u2)

[u3,x,t] = heat_CN(a,xf,T,it0,bx0,bxf,M,N); %converge unconditionally

figure(3), clf, mesh(t,x,u3)

MN = M*N;

Uo = uo(x,t); aUo = abs(Uo)+eps; %values of true analytical solution

%How far from the analytical solution?

err1 = norm((u1-Uo)./aUo)/MN

err2 = norm((u2-Uo)./aUo)/MN

err3 = norm((u3-Uo)./aUo)/MN

412 PARTIAL DIFFERENTIAL EQUATIONS

9.2.4 Two-Dimensional Parabolic PDE

Another example of a parabolic PDE is a two-dimensional heat equation describing

the temperature distribution u(x, y, t)((x, y) is position, t is time) as

A∂2u(x, y, t)

∂x2 +

∂2u(x, y, t)

∂y2 =

∂u(x, y, t)

∂t

(9.2.19)

for x0 ≤ x ≤ xf , y0 ≤ y ≤ yf , 0 ≤ t ≤ T

In order for this equation to be solvable, we should be provided with the boundary

conditions

u(x0, y, t) = bx0(y, t), u(xf , y, t) = bxf (y, t),

u(x, y0, t) = by0(x, t), and u(x, yf , t) = byf (x, t)

as well as the initial condition u(x, y, 0) = i0(x, y).

We replace the first-order time derivative on the right-hand side by the threepoint

central difference at the midpoint (tk+1 + tk)/2 just as with the Crank–

Nicholson method. We also replace one of the second-order derivatives, uxx and

uyy, by the three-point central difference approximation (5.3.1) at time tk and the

other at time tk+1, yielding

A

uk

i,j+1 − 2uk

i,j + uk

i,j−1

x2 +

uk+1

i+1,j − 2uk+1

i,j + uk+1

i−1,j

y2 =

uk+1

i,j − uk

i,j

t

(9.2.20)

which seems to be attractive, since it can be formulated into a tridiagonal system

of equations with respect to uk+1

i+1,j , uk+1

i,j , and uk+1

i−1,j . But, why do we treat uxx

and uyy with discrimination—that is, evaluate one at time tk and the other at time

tk+1 in a fixed manner? In an alternate manner, we write the difference equation

for the next time point tk+1 as

A

uk+1

i,j+1 − 2uk+1

i,j + uk+1

i,j−1

x2 +

uk

i+1,j − 2uk

i,j + uk

i−1,j

y2 =

uk+2

i,j − uk+1

i,j

t

(9.2.21)

This formulation, proposed by Peaceman and Rachford [P-1], is referred to as the

alternating direction implicit (ADI) method and can be cast into the following

algorithm:

−ry(uk+1

i−1,j + uk+1

i+1,j ) + (1 + 2ry)uk+1

i,j = rx(uk

i,j−1 + uk

i,j+1) + (1 − 2rx)uk

i,j

for 1 ≤ j ≤ Mx − 1 (9.2.22a)

−rx(uk+2

i,j−1 + uk+2

i,j+1) + (1 + 2rx)uk+2

i,j = ry(uk+1

i−1,j + uk+1

i+1,j ) + (1 − 2ry)uk+1

i,j

for 1 ≤ i ≤ My − 1 (9.2.22b)

PARABOLIC PDE 413

with

rx = At/x2, ry = At/y2,

x = (xf − x0)/Mx, y= (yf − y0)/My, t = T/N

The objective of the following MATLAB routine “heat2_ADI()” is to implement

this algorithm for solving a two-dimensional heat equation (9.2.19).

function [u,x,y,t] = heat2_ADI(a,D,T,ixy0,bxyt,Mx,My,N)

%solve u_t = c(u_xx + u_yy) for D(1) <= x <= D(2), D(3) <= y <= D(4), 0 <= t <= T

% Initial Condition: u(x,y,0) = ixy0(x,y)

% Boundary Condition: u(x,y,t) = bxyt(x,y,t) for (x,y)cB

% Mx/My = # of subintervals along x/y axis

% N = # of subintervals along t axis

dx = (D(2) – D(1))/Mx; x = D(1)+[0:Mx]*dx;

dy = (D(4) – D(3))/My; y = D(3)+[0:My]’*dy;

dt = T/N; t = [0:N]*dt;

%Initialization

for j = 1:Mx + 1

for i = 1:My + 1

u(i,j) = ixy0(x(j),y(i));

end

end

rx = a*dt/(dx*dx); rx1 = 1 + 2*rx; rx2 = 1 – 2*rx;

ry = a*dt/(dy*dy); ry1 = 1 + 2*ry; ry2 = 1 – 2*ry;

for j = 1:Mx – 1 %Eq.(9.2.22a)

Ay(j,j) = ry1;

if j > 1, Ay(j – 1,j) = -ry; Ay(j,j-1) = -ry; end

end

for i = 1:My – 1 %Eq.(9.2.22b)

Ax(i,i) = rx1;

if i > 1, Ax(i – 1,i) = -rx; Ax(i,i – 1) = -rx; end

end

for k = 1:N

u_1 = u; t = k*dt;

for i = 1:My + 1 %Boundary condition

u(i,1) = feval(bxyt,x(1),y(i),t);

u(i,Mx+1) = feval(bxyt,x(Mx+1),y(i),t);

end

for j = 1:Mx + 1

u(1,j) = feval(bxyt,x(j),y(1),t);

u(My+1,j) = feval(bxyt,x(j),y(My + 1),t);

end

if mod(k,2) == 0

for i = 2:My

jj = 2:Mx;

bx = [ry*u(i,1) zeros(1,Mx – 3) ry*u(i,My + 1)] …

+rx*(u_1(i-1,jj)+ u_1(i + 1,jj)) + rx2*u_1(i,jj);

u(i,jj) = trid(Ay,bx’)’; %Eq.(9.2.22a)

end

else

for j = 2:Mx

ii = 2:My;

by = [rx*u(1,j); zeros(My-3,1); rx*u(Mx + 1,j)] …

+ ry*(u_1(ii,j-1) + u_1(ii,j + 1)) + ry2*u_1(ii,j);

u(ii,j) = trid(Ax,by); %Eq.(9.2.22b)

end

end

end

414 PARTIAL DIFFERENTIAL EQUATIONS

2

0 0

2

y

4

1

3

x 4

100

0

−100

Figure 9.4 A solution for a two dimensional parabolic PDE obtained using ‘‘heat2 ADI()’’

(Example 9.3).

Example 9.3. A Parabolic PDE: Two-Dimensional Temperature Diffusion.

Consider a two-dimensional parabolic PDE

10−4 ∂2u(x, y, t)

∂x2 +

∂2u(x, y, t)

∂y2 =

∂u(x, y, t)

∂t

(E9.3.1)

for 0 ≤ x ≤ 4, 0 ≤ y ≤ 4, 0 ≤ t ≤ 5000

with the initial conditions and boundary conditions

u(x, y, 0) =0 fort = 0 (E9.3.2a)

u(x, y, t) = ey cos x − ex cos y for x = 0, x = 4, y = 0, y = 4 (E9.3.2b)

We made the following MATLAB program “solve_heat2.m” in order to use

the routine “heat2_ADI()” to solve this equation and ran this program to get the

result shown in Fig. 9.4 at the final time.

%solve_heat2

clear, clf

a = 1e-4;

it0 = inline(’0’,’x’,’y’); %(E9.3.2a)

bxyt = inline(’exp(y)*cos(x)-exp(x)*cos(y)’,’x’,’y’,’t’); %(E9.3.2b)

D = [0 4 0 4]; T = 5000; Mx = 40; My = 40; N = 50;

[u,x,y,t] = heat2_ADI(a,D,T,it0,bxyt,Mx,My,N);

mesh(x,y,u)

9.3 HYPERBOLIC PDE

An example of a hyperbolic PDE is a one-dimensional wave equation for the

amplitude function u(x, t)(x is position, t is time) as

A

∂2u(x, t)

∂x2 =

∂2u(x, t)

∂t2 for 0 ≤ x ≤ xf , 0 ≤ t ≤T (9.3.1)

HYPERBOLIC PDE 415

In order for this equation to be solvable, the boundary conditions u(0, t) = b0(t) and u(xf , t) = bxf (t) as well as the initial conditions u(x, 0) = i0(x) and

∂u/∂t|t=0(x, 0) = i0 (x) should be provided.

9.3.1 The Explicit Central Difference Method

In the same way as with the parabolic PDEs, we replace the second derivatives

on both sides of Eq. (9.3.1) by their three-point central difference approximation

(5.3.1) as

A

uk

i+1 − 2uk

i + uk

i−1

x2 =

uk+1

i − 2uk

i + uk−1

i

t2 with x =

xf

M

,t =

T

N

(9.3.2)

which leads to the explicit central difference method:

uk+1

i = r(uk

i+1 + uk

i−1) + 2(1 − r)uk

i − uk−1

i with r = A

t2

x2 (9.3.3)

Since u−1

i = u(xi ,−t) is not given, we cannot get u1

i directly from this

formula (9.3.3) with k = 0:

u1

i = r(u0

i+1 + u0

i−1) + 2(1 − r)u0

i − u−1

i (9.3.4)

Therefore, we approximate the initial condition on the derivative by the central

difference as

u1

i − u−1

i

2t = i0 (xi) (9.3.5)

and make use of this to remove u−1

i from Eq. (9.3.3):

u1

i = r(u0

i+1 + u0

i−1) + 2(1 − r)u0

i − (u1

i − 2i0 (xi)t)

u1

i =

1

2

r(u0

i+1 + u0

i−1) + (1 − r)u0

i + i0 (xi)t (9.3.6)

We use Eq. (9.3.6) together with the initial conditions to get u1

i and then go

on with Eq. (9.3.3) for k = 1, 2, . . .. Note the following facts:

ž We must have r ≤ 1 to guarantee the stability.

ž The accuracy of the solution gets better as r becomes larger so that x

decreases.

It is therefore reasonable to select r = 1.

The stability condition can be obtained by substituting Eq. (9.2.4) into

Eq. (9.3.3) and applying the Jury test [P-3]:

λ = 2r cos(π/P ) + 2(1 − r) − λ−1, λ2 + 2(r(1 − cos(π/P )) − 1)λ + 1 = 0

416 PARTIAL DIFFERENTIAL EQUATIONS

We need the solution of this equation to be inside the unit circle for stability,

which requires

r ≤

1

1 − cos(π/P )

, r= A

t2

x2 ≤ 1 (9.3.7)

The objective of the following MATLAB routine “wave()” is to implement

this algorithm for solving a one-dimensional wave equation.

Example 9.4. A Hyperbolic PDE: One-Dimensional Wave (Vibration). Consider

a one-dimensional hyperbolic PDE

∂2u(x, t)

∂x2 =

∂u2(x, t)

∂t2 for 0 ≤ x ≤ 2, 0 ≤ y ≤ 2, and 0 ≤ t ≤ 2 (E9.4.1)

with the initial conditions and boundary conditions

u(x, 0) = x(1 − x), ∂u/∂t (x, 0) =0 fort = 0 (E9.4.2a)

u(0, t) =0 forx = 0, u(1, t) =0 forx = 1 (E9.4.2b)

We made the following MATLAB program “solve_wave.m” in order to use

the routine “wave()” to solve this equation and ran this program to get the result

shown in Fig. 9.5 and see a dynamic picture.

function [u,x,t] = wave(a,xf,T,it0,i1t0,bx0,bxf,M,N)

%solve a u_xx = u_tt for 0<=x<=xf, 0<=t<=T

% Initial Condition: u(x,0) = it0(x), u_t(x,0) = i1t0(x)

% Boundary Condition: u(0,t)= bx0(t), u(xf,t) = bxf(t)

% M = # of subintervals along x axis

% N = # of subintervals along t axis

dx = xf/M; x = [0:M]’*dx;

dt = T/N; t = [0:N]*dt;

for i = 1:M + 1, u(i,1) = it0(x(i)); end

for k = 1:N + 1

u([1 M + 1],k) = [bx0(t(k)); bxf(t(k))];

end

r = a*(dt/dx)^ 2; r1 = r/2; r2 = 2*(1 – r);

u(2:M,2) = r1*u(1:M – 1,1) + (1 – r)*u(2:M,1) + r1*u(3:M + 1,1) …

+ dt*i1t0(x(2:M)); %Eq.(9.3.6)

for k = 3:N + 1

u(2:M,k) = r*u(1:M – 1,k – 1) + r2*u(2:M,k-1) + r*u(3:M + 1,k – 1)…

– u(2:M,k – 2); %Eq.(9.3.3)

end

%solve_wave

a = 1;

it0 = inline(’x.*(1-x)’,’x’); i1t0 = inline(’0’); %(E9.4.2a)

bx0t = inline(’0’); bxft = inline(’0’); %(E9.4.2b)

xf = 1; M = 20; T = 2; N = 50;

[u,x,t] = wave(a,xf,T,it0,i1t0,bx0t,bxft,M,N);

figure(1), clf

mesh(t,x,u)

figure(2), clf

for n = 1:N %dynamic picture

plot(x,u(:,n)), axis([0 xf -0.3 0.3]), pause(0.2)

end

HYPERBOLIC PDE 417

1

0.3

0.4 0.6 0.8 x

0.2

0.2

0.1

0

0

−0.1

−0.2

−0.3

0

1

0.5

0 0

1 t

−0.2

0.2

2

(a) (b) A snap shot

Figure 9.5 A solution for a 1-D hyperbolic PDE obtained by using ‘‘wave()’’ (Example 9.4).

9.3.2 Two-Dimensional Hyperbolic PDE

In this section, we consider a two-dimensional wave equation for the amplitude

function u(x, y, t) ((x, y) is position, t is time) as

A∂2u(x, y, t)

∂x2 +

∂2u(x, y, t)

∂y2 =

∂2u(x, t)

∂t2 (9.3.8)

for 0 ≤ x ≤ xf , 0 ≤ y ≤ yf , 0 ≤ t ≤ T

In order for this equation to be solvable, we should be provided with the boundary

conditions

u(0, y, t) = bx0(y, t), u(xf , y, t) = bxf (y, t),

u(x, 0, t) = by0(x, t), and u(x, yf , t) = byf (x, t)

as well as the initial condition u(x, y, 0) = i0(x, y) and ∂u/∂t|t=0(x, y, 0) = i0(x, y).

In the same way as with the one-dimensional case, we replace the second

derivatives on both sides by their three-point central difference approximation

(5.3.1) as

A

uk

i,j+1 − 2uk

i,j + uk

i,j−1

x2 +

uk

i+1,j − 2uk

i,j + uk

i−1,j

y2 =

uk+1

i,j − 2uk

i,j + uk−1

i,j

t2

(9.3.9)

with x =

xf

Mx

, y=

yf

Ny

, t =

T

N

which leads to the explicit central difference method:

uk+1

i,j = rx(uk

i,j+1 + uk

i,j−1) + 2(1 − rx − ry)uk

i,j + ry(uk

i+1,j + uk

i−1,j ) − uk−1

i,j

(9.3.10)

with rx = A

t2

x2, ry = A

t2

y2

418 PARTIAL DIFFERENTIAL EQUATIONS

Since u−1

i,j = u(xj, yi ,−t) is not given, we cannot get u1

i,j directly from this

formula (9.3.10) with k = 0:

u1

i,j = rx(u0

i,j+1 + u0

i,j−1) + 2(1 − rx − ry)u0

i,j + ry(u0

i+1,j + u0

i−1,j ) − u−1

i,j

(9.3.11)

Therefore, we approximate the initial condition on the derivative by the central

difference as

u1

i,j − u−1

i,j

2t = i0(xj, yi) (9.3.12)

and make use of this to remove u−1

i,j from Eq. (9.3.11) to have

u1

i,j = 12

{rx(u0

i,j+1 + u0

i,j−1) + ry(u0

i+1,j + u0

i−1,j )}

+ 2(1 − rx − ry)u0

i,j + i0(xj, yi)t (9.3.13)

We use this Eq. (9.3.13) together with the initial conditions to get u1

i,j and then

go on using Eq. (9.3.10) for k = 1, 2, . . .. A sufficient condition for stability [S-1,

Section 9.6] is

r =

4At2

x2 + y2 ≤ 1 (9.3.14)

The objective of the MATLAB routine “wave2()” is to implement this algorithm

for solving a two-dimensional wave equation.

Example 9.5. A Hyperbolic PDE: Two-Dimensional Wave (Vibration) Over a

Square Membrane. Consider a two-dimensional hyperbolic PDE

1

4 ∂2u(x, y, t)

∂x2 +

∂2u(x, y, t)

∂y2 =

∂u2(x, y, t)

∂t2

for 0 ≤ x ≤ 2, 0 ≤ y ≤2 and 0≤ t ≤ 2 (E9.5.1)

with the zero boundary conditions and the initial conditions

u(0, y, t) = 0, u(2, y, t) = 0, u(x,0, t) = 0, u(x,2, t) = 0 (E9.5.2)

u(x, y, 0) = 0.1 sin(πx) sin(πy/2), ∂u/∂t (x, y, 0) = 0 for t = 0 (E9.5.3)

We made the following MATLAB program “solve_wave2.m” in order to use

the routine “wave2()” for solving this equation and ran this program to get the

result shown in Fig. 9.6 and see a dynamic picture. Note that we can be sure of

stability, since we have

r =

4At2

x2 + y2 =

4(1/4)(2/20)2

(2/20)2 + (2/20)2 =

1

2 ≤ 1 (E9.5.4)

HYPERBOLIC PDE 419

function [u,x,y,t] = wave2(a,D,T,it0,i1t0,bxyt,Mx,My,N)

%solve a(u_xx + u_yy) = u_tt for D(1) <= x <= D(2), D(3) <= y <= D(4), 0 <= t <= T

% Initial Condition: u(x,y,0) = it0(x,y), u_t(x,y,0) = i1t0(x,y)

% Boundary Condition: u(x,y,t) = bxyt(x,y,t) for (x,y) on Boundary

% Mx/My = # of subintervals along x/y axis

% N = # of subintervals along t axis

dx = (D(2)- D(1))/Mx; x = D(1)+[0:Mx]*dx;

dy = (D(4)- D(3))/My; y = D(3)+[0:My]’*dy;

dt = T/N; t = [0:N]*dt;

%Initialization

u = zeros(My+1,Mx + 1); ut = zeros(My + 1,Mx + 1);

for j = 2:Mx

for i = 2:My

u(i,j) = it0(x(j),y(i)); ut(i,j) = i1t0(x(j),y(i));

end

end

adt2 = a*dt*dt; rx = adt2/(dx*dx); ry = adt2/(dy*dy);

rxy1 = 1- rx – ry; rxy2 = rxy1*2;

u_1 = u;

for k = 0:N

t = k*dt;

for i = 1:My + 1 %Boundary condition

u(i,[1 Mx + 1]) = [bxyt(x(1),y(i),t) bxyt(x(Mx + 1),y(i),t)];

end

for j = 1:Mx + 1

u([1 My + 1],j) = [bxyt(x(j),y(1),t); bxyt(x(j),y(My + 1),t)];

end

if k = = 0

for i = 2:My

for j = 2:Mx %Eq.(9.3.13)

u(i,j) = 0.5*(rx*(u_1(i,j – 1) + u_1(i,j + 1))…

+ ry*(u_1(i – 1,j)+u_1(i + 1,j))) + rxy1*u(i,j) + dt*ut(i,j);

end

end

else

for i = 2:My

for j = 2:Mx %Eq.(<eqnr>9.3.10)</eqnr>

u(i,j) = rx*(u_1(i,j – 1)+ u_1(i,j + 1))…

+ ry*(u_1(i – 1,j) + u_1(i + 1,j)) + rxy2*u(i,j) -u_2(i,j);

end

end

end

u_2 = u_1; u_1 = u; %update the buffer memory

mesh(x,y,u), axis([0 2 0 2 -.1 .1]), pause

end

%solve_wave2

it0 = inline(’0.1*sin(pi*x)*sin(pi*y/2)’,’x’,’y’); %(E9.5.3)

i1t0 = inline(’0’,’x’,’y’); bxyt = inline(’0’,’x’,’y’,’t’); %(E9.5.2)

a = .25; D = [0 2 0 2]; T = 2; Mx = 40; My = 40; N = 40;

[u,x,y,t] = wave2(a,xf,T,it0,i1t0,bxyt,Mx,My,N);

−0.1

0.1

1 1

0 0

0

2

2

(a) At t = 0.1

−0.1

0.1

1 1

0 0

0

2

2

(b) At t = 1.8

Figure 9.6 The solution of a two-dimensional hyperbolic PDE: vibration of a square membrane

(Example 9.5).

420 PARTIAL DIFFERENTIAL EQUATIONS

9.4 FINITE ELEMENT METHOD (FEM) FOR SOLVING PDE

The FEM method is another procedure used in finding approximate numerical

solutions to BVPs/PDEs. It can handle irregular boundaries in the same way as

regular boundaries [R-1, S-2, Z-1]. It consists of the following steps to solve the

elliptic PDE:

∂2u(x, y)

∂x2 +

∂2u(x, y)

∂y2 + g(x, y)u(x, y) = f (x, y) (9.4.1)

for the domain D enclosed by the boundary B on which the boundary condition

is given as

u(x, y) = b(x, y) on the boundaryB (9.4.2)

1. Discretize the (two-dimensional) domain D into, say, Ns subregions

{S1, S2, . . . , SNs} such as triangular elements, neither necessarily of the

same size nor necessarily covering the entire domain completely and

exactly.

2. Specify the positions of Nn nodes and number them starting from the

boundary nodes, say, n = 1, . . .,Nb, and then the interior nodes, say, n = Nb + 1, . . . ,Nn.

3. Define the basis/shape/interpolation functions

φn(x, y) = {φn,s , for s = 1, . . . , Ns} ∀ (x, y) ∈ D (9.4.3a)

φn,s(x, y) = pn,s(1) + pn,s(2)x + pn,s(3)y

for each subregion Ss (9.4.3b)

collectively for all subregions s = 1 : Ns and for each node n = 1 : Nn, so

that φn is 1 only at node n, and 0 at all other nodes. Then, the approximate

solution of the PDE is a linear combination of basis functions

φn(x, y) as

u(x, y) = cT ϕ(x, y) =

Nn

n=1

cnφn(x, y) =

Nb

n=1

cnφn +

Nn

n=Nb+1

cnφn = cT1

ϕ1 + cT2

ϕ2

(9.4.4)

where

ϕ1 = [ φ1 φ2 · φNb ]T , c1 = [ c1 c2 · cNb ]T (9.4.5a)

ϕ2 = [ φNb+1 φNb+2 · φNn ]T , c2 = [ cNb+1 cNb+2 · cNn ]T

(9.4.5b)

FINITE ELEMENT METHOD (FEM) FOR SOLVING PDE 421

For each subregion s = 1, . . .,Ns , this solution can be written as

φs(x, y) =

Nn

n=1

cnφn,s(x, y) =

Nn

n=1

cn(pn,s(1) + pn,s(2)x + pn,s(3)y)

(9.4.6)

4. Set the values of the boundary node coefficients in c1 to the boundary

values according to the boundary condition.

5. Determine the values of the interior node coefficients in c2 by solving the

system of equations

A2c2 = d (9.4.7)

where

A1 =

Ns

s=1

∂

∂x

ϕ2,s ∂

∂x

ϕ1,sT

+ ∂

∂y

ϕ2,s ∂

∂y

ϕ1,sT

− g(xs, ys )ϕ2,sϕT

1,sSs

(9.4.8)

ϕ1,s = [ φ1,s φ2,s · φNb ,s ]T

∂

∂x

ϕ1,s = [ p1,s (2) p2,s (2) · pNb ,s (2) ]T

∂

∂y

ϕ1,s = [ p1,s (3) p2,s (3) · pNb,s (3) ]T

A2 =

Ns

s=1

∂

∂x

ϕ2,s ∂

∂x

ϕ2,sT

+ ∂

∂y

ϕ2,s ∂

∂y

ϕ2,sT

− g(xs, ys )ϕ2,sϕT

2,sSs

(9.4.9)

ϕ2,s = [ φNb+1,s φNb+2,s · φNn,s ]T

∂

∂x

ϕ2,s = [ pNb+1,s (2) φNb+2,s (2) · φNn,s (2) ]T

∂

∂y

ϕ2,s = [ pNb+1,s (3) φNb+2,s (3) · φNn,s (3) ]T

d = −A1c1 −

Ns

s=1

f (xs, ys )ϕ2,sS (9.4.10)

(xs, ys ): the centroid (gravity center) of the sth subregion Ss

The FEM is based on the variational principle that a solution to Eq. (9.4.1)

can be obtained by minimizing the functional

I = R

∂

∂x

u(x, y)2

+ ∂

∂y

u(x, y)2

− g(x, y)u2(x, y) + 2f (x, y)u(x, y)dx dy (9.4.11)

422 PARTIAL DIFFERENTIAL EQUATIONS

which, with u(x, y) = cT ϕ(x, y), can be written as

I = R cT ∂

∂x

ϕ

∂

∂x

ϕT c + cT ∂

∂y

ϕ

∂

∂y

ϕT c

− g(x, y)cT ϕϕT c + 2f (x, y)cT ϕdx dy (9.4.12)

The condition for this functional to be minimized with respect to c is

d

dc2

I = R ∂

∂x

ϕ2

∂

∂x

ϕT c +

∂

∂y

ϕ2

∂

∂y

ϕT c

− g(x, y)ϕ2ϕT c + f (x, y)ϕ2dx dy = 0 (9.4.13)

≈ A1c1 + A2c2 +

Ns

s=1

f (xs, ys)ϕ2,sSs = 0 (9.4.14)

See [R-1] for details.

The objectives of the MATLAB routines “fem_basis_ftn()” and

“fem_coef()” are to construct the basis function φn,s(x, y)’s for each node

n = 1, . . . , Nn and each subregion s = 1, . . . , Ns and to get the coefficient vector

c of the solution (9.4.4) via Eq. (9.4.7) and the solution polynomial φs(x, y)’s

via Eq. (9.4.6) for each subregion s = 1, . . . , Ns , respectively.

Before going into a specific example of applying the FEM method to solve

a PDE, let us take a look at the basis (shape) function φn(x, y) for each node

n = 1, . . . , Nn, which is defined collectively for all of the (triangular) subregions

so that φn is 1 only at node n, and 0 at all other nodes and can be generated by

the routine “fem_basis_ftn()”.

function p = fem_basis_ftn(N,S)

%p(i,s,1:3): coefficients of each basis ftn phi_i

% for s-th subregion(triangle)

%N(n,1:2) : x & y coordinates of the n-th node

%S(s,1:3) : the node #s of the s-th subregion(triangle)

N_n = size(N,1); % the total number of nodes

N_s = size(S,1); % the total number of subregions(triangles)

for n = 1:N_n

for s = 1:N_s

for i = 1:3

A(i,1:3) = [1 N(S(s,i),1:2)];

b(i) = (S(s,i) == n); %The nth basis ftn is 1 only at node n.

end

pnt=A\b’;

for i=1:3, p(n,s,i) = pnt(i); end

end

end

FINITE ELEMENT METHOD (FEM) FOR SOLVING PDE 423

function [U,c] = fem_coef(f,g,p,c,N,S,N_i)

%p(i,s,1:3): coefficients of basis ftn phi_i for the s-th subregion

%c = [ .1 1 . 0 0 .] with value for boundary and 0 for interior nodes

%N(n,1:2) : x & y coordinates of the n-th node

%S(s,1:3) : the node #s of the s-th subregion(triangle)

%N_i : the number of the interior nodes

%U(s,1:3) : the coefficients of p1 + p2(s)x + p3(s)y for each subregion

N_n = size(N,1); % the total number of nodes = N_b + N_i

N_s = size(S,1); % the total number of subregions(triangles)

d=zeros(N_i,1);

N_b = N_n-N_i;

for i = N_b+1:N_n

for n = 1:N_n

for s = 1:N_s

xy = (N(S(s,1),:) + N(S(s,2),:) + N(S(s,3),:))/3; %gravity center

%phi_i,x*phi_n,x + phi_i,y*phi_n,y – g(x,y)*phi_i*phi_n

p_vctr = [p([i n],s,1) p([i n],s,2) p([i n],s,3)];

tmpg(s) = sum(p(i,s,2:3).*p(n,s,2:3))…

-g(xy(1),xy(2))*p_vctr(1,:)*[1 xy]’*p_vctr(2,:)*[1 xy]’;

dS(s) = det([N(S(s,1),:) 1; N(S(s,2),:) 1;N(S(s,3),:) 1])/2;

%area of triangular subregion

if n == 1, tmpf(s) = -f(xy(1),xy(2))*p_vctr(1,:)*[1 xy]’; end

end

A12(i – N_b,n) = tmpg*abs(dS)’; %Eqs. (9.4.8),(9.4.9)

end

d(i-N_b) = tmpf*abs(dS)’; %Eq.(9.4.10)

end

d = d – A12(1:N_i,1:N_b)*c(1:N_b)’; %Eq.(9.4.10)

c(N_b + 1:N_n) = A12(1:N_i,N_b+1:N_n)\d; %Eq.(9.4.7)

for s = 1:N_s

for j = 1:3, U(s,j) = c*p(:,s,j); end %Eq.(9.4.6)

end

Actually, we will plot the basis (shape) functions for the region divided into four

triangular subregions as depicted in Fig. 9.7 in two ways. First, we generate the

basis functions by using the routine “fem_basis_ftn()” and plot one of them

for node 1 by using the MATLAB command mesh(), as depicted in Fig. 9.8a.

Second, without generating the basis functions, we use the MATLAB command

“trimesh()” to plot the shape functions for nodes n = 2, 3, 4, and 5 as depicted

in Figs. 9.8b–e, each of which is 1 only at the corresponding node n and is

0 at all other nodes. Figure 9.8f is the graph of a linear combination of basis

functions

u(x, y) = cT ϕ(x, y) =

Nn

n=1

cnφn(x, y) (9.4.15)

having the given value cn at each node n. This can obtained by using the MATLAB

command “trimesh()” as

>>trimesh(S,N(:,1),N(:,2),c)

where the first input argument S has the node numbers for each subregion, the

second/third input argument N has the x/y coordinates for each node, and the

424 PARTIAL DIFFERENTIAL EQUATIONS

1

1

0.5

0.5

0

0

−0.5

−0.5

−1

−1

n = 2

n = 5

n = 3

n = 4

n = 1

S4

S3

S1

S2

coordinates of

nodes

node numbers

of subregions

N = [−1 1;

1;

−1 −1;

1 −1;

1

1

0.2 0.5]

S = [1

2

2

3

3 4

4 5]

5;

5;

5;

Figure 9.7 A region (domain) divided into four triangular subregions.

fourth input argument c has the function values at each node as follows:

S =

1 2 5

2 3 5

3 4 5

1 4 5

, N=

−1 1

1 1

1 −1

−1 −1

0.2 0.5

, c=

0

1

2

3

0

(9.4.16)

For this job, we make the following program “show_basis.m” and run it to

get Figs. 9.7 and 9.8 together with the coefficients of each basis function as

p(:, :, 1) =

−3/10 0 0 −1/8

−7/10 −7/16 0 0

0 3/16 1/10 0

0 0 7/30 7/24

2 5/4 2/3 5/6

,

p(:, :, 2) =

−1/2 0 0 −5/8

1/2 15/16 0 0

0 5/16 1/2 0

0 0 −1/2 −5/24

0 −5/4 0 5/6

, (9.4.17)

p(:, :, 3) =

4/5 0 0 1/2

6/5 1/2 0 0

0 −1/2 −2/5 0

0 0 −4/15 −1/2

−2 0 2/3 0

The meaning of this Nn (the number of nodes:5) × Ns (the number of subregions:

4) × 3 array p is that, say, the second rows of the three sub-arrays constitute

FINITE ELEMENT METHOD (FEM) FOR SOLVING PDE 425

1

0.5

0

0 0

1 1

−1 −1

(e) f5(x,y )

1

0.5

0

0 0

1 1

−1 −1

(c) f3(x, y )

1

0.5

0

0 0

1 1

−1 −1

(a) f1(x, y )

4

2

0

0 0

1 1

−1 −1

(f) f2(x, y )+2f3(x, y )+3f4(x, y )

1

0.5

0

0 0

1 1

−1 −1

(d) f4(x, y )

1

0.5

0

0 0

1 1

−1 −1

(b) f2(x, y )

(0.2, 0.5)

Figure 9.8 The basis (shape) functions for nodes in Fig. 9.7 and a composite function.

the coefficient vectors of the basis function for node 2 as

φ2(x, y) =

−7/10 + (1/2)x + (6/5)y for subregion S1

−7/16 + (15/16)x + (1/2)y for subregion S2

0 + 0 · x + 0 · y for subregion S3

0 + 0 · x + 0 · y for subregion S4

(9.4.18)

which turns out to be 1 only at node 2 [i.e., (1,1)] and 0 at all other nodes and on

the subregions that do not have node 2 as their vertex, as depicted in Fig. 9.8b.

426 PARTIAL DIFFERENTIAL EQUATIONS

With the program “show_basis.m” in your computer, type the following commands

into the MATLAB command window and see the graphical/textual output.

>>show_basis

>>p

Now, let us see the following example.

%show_basis

clear

N = [-1 1;1 1;1 -1;-1 -1;0.2 0.5]; %the list of nodes in Fig.9.7

N_n = size(N,1); % the number of nodes

S = [1 2 5;2 3 5;3 4 5;1 4 5]; %the list of subregions in Fig.9.7

N_s = size(S,1); % the number of subregions

figure(1), clf

for s = 1:N_s

nodes = [S(s,:) S(s,1)];

for i = 1:3

plot([N(nodes(i),1) N(nodes(i + 1),1)], …

[N(nodes(i),2) N(nodes(i+1),2)]), hold on

end

end

%basis/shape function

p = fem_basis_ftn(N,S);

x0 = -1; xf = 1; y0 = -1; yf = 1; %graphic region

figure(2), clf

Mx = 50; My = 50;

dx = (xf – x0)/Mx; dy = (yf – y0)/My;

xi = x0 + [0:Mx]*dx; yi = y0 + [0:My]*dy;

i_ns = [1 2 3 4 5]; %the list of node numbers whose basis ftn to plot

for itr = 1:5

i_n = i_ns(itr);

if itr == 1

for i = 1:length(xi)

for j = 1:length(yi)

Z(j,i) = 0;

for s = 1:N_s

if inpolygon(xi(i),yi(j), N(S(s,:),1),N(S(s,:),2)) > 0

Z(j,i) = p(i_n,s,1) + p(i_n,s,2)*xi(i) + p(i_n,s,3)*yi(j);

break;

end

end

end

end

subplot(321), mesh(xi,yi,Z) %basis function for node 1

else

c1 = zeros(size(c)); c1(i_n) = 1;

subplot(320 + itr)

trimesh(S,N(:,1),N(:,2),c1) %basis function for node 2-5

end

end

c = [0 1 2 3 0]; %the values for all nodes

subplot(326)

trimesh(S,N(:,1),N(:,2),c) %Fig.9.8f: a composite function

FINITE ELEMENT METHOD (FEM) FOR SOLVING PDE 427

Example 9.6. Laplace’s Equation: Electric Potential Over a Plate with Point

Charge. Consider the following Laplace’s equation:

∇2u(x, y) =

∂2u(x, y)

∂x2 +

∂2u(x, y)

∂y2 = f (x, y) (E9.6.1)

for − 1 ≤ x ≤ +1,−1 ≤ y ≤ +1

where

f (x, y) =

−1 for(x, y) = (0.5, 0.5)

+1 for(x, y) = (−0.5,−0.5)

0 elsewhere

(E9.6.2)

and the boundary condition is u(x, y) = 0 for all boundaries of the rectangular

domain.

In order to solve this equation by using the FEM, we locate 12 boundary

points and 19 interior points, number them, and divide the domain into 36 triangular

subregions as depicted in Fig. 9.9. Note that we have made the size of

the subregions small and their density high around the points (+0.5,+0.5) and

12

11 10

24

25

30 33

36

34 32

23

29

22

8

9

31 21

20

19

18

17

13

14

29 31

35

30

15

1

2 3

28

26

16

27

4 5 6

7

S1 S3

S2

S19

S18

S17

S16

S20

S13

S4

S5

S6

S21

S22

S15

S12

S11

S10

S8 S9

S7

S14

0.8

0.6

0.4

0.2

0

−0.2

−0.4

−0.6

−0.8

−1

−1 −0.5 0 0.5 1

1

y

x

Figure 9.9 An example of triangular subregions for FEM.

428 PARTIAL DIFFERENTIAL EQUATIONS

(−0.5,−0.5), since they are only two points at which the value of the right-hand

side of Eq. (9.6.1) is not zero, and consequently the value of the solution u(x, y)

is expected to change sensitively around them.

We made the following MATLAB program “do_fem.m” in order to use the

routines “fem_basis_ftn()” and “fem_coef()” for solving this equation. For

comparison, we have added the statements to solve the same equation by using the

routine “poisson()” (Section 9.1). The results obtained by running this program

are depicted in Fig. 9.10a–c.

x10−3

5

0.5 0.5

−0.5 −0.5

−5

0

0 0

1

y 1

−1 −1

x

(a) 31-point FEM solution drawn by using trimesh ()

x10−3

5

0.5

0.5

−0.5 −0.5

−5

0

0 0

1

y 1

−1 −1

x

u (x, y )

u (x, y )

u (x, y )

(b) 31-point FEM solution by using mesh ()

x10−3

5

0.5

0.5

−0.5 −0.5

−5

0

0 0

1

y 1

−1 −1

x

(c) 16×15-point FDM (Finite Difference Method) solution

Figure 9.10 Results of Example 9.6.

GUI OF MATLAB FOR SOLVING PDES: PDETOOL 429

%do_fem

% for Example 9.6

clear

N = [-1 0;-1 -1;-1/2 -1;0 -1;1/2 -1; 1 -1;1 0;1 1;1/2 1; 0 1;

-1/2 1;-1 1; -1/2 -1/4; -5/8 -7/16;-3/4 -5/8;-1/2 -5/8;

-1/4 -5/8;-3/8 -7/16; 0 0; 1/2 1/4;5/8 7/16;3/4 5/8;

1/2 5/8;1/4 5/8;3/8 7/16;-9/16 -17/32;-7/16 -17/32;

-1/2 -7/16;9/16 17/32;7/16 17/32;1/2 7/16]; %nodes

N_b = 12; %the number of boundary nodes

S = [1 11 12;1 11 19;10 11 19;4 5 19;5 7 19; 5 6 7;1 2 15; 2 3 15;

3 15 17;3 4 17;4 17 19;13 17 19;1 13 19;1 13 15;7 8 22;8 9 22;

9 22 24;9 10 24; 10 19 24; 19 20 24;7 19 20; 7 20 22;13 14 18;

14 15 16;16 17 18;20 21 25;21 22 23;23 24 25;14 26 28;

16 26 27;18 27 28; 21 29 31;23 29 30;25 30 31;

26 27 28; 29 30 31]; %triangular subregions

f962 = ’(norm([x y]+[0.5 0.5])<0.01)-(norm([x y]-[0.5 0.5]) < 0.01)’;

f=inline(f962,’x’,’y’); %(E9.6.2)

g=inline(’0’,’x’,’y’);

N_n = size(N,1); %the total number of nodes

N_i = N_n – N_b; %the number of interior nodes

c = zeros(1,N_n); %boundary value or 0 for boundary/interior nodes

p = fem_basis_ftn(N,S);

[U,c] = fem_coef(f,g,p,c,N,S,N_i);

%Output through the triangular mesh-type graph

figure(1), clf, trimesh(S,N(:,1),N(:,2),c)

%Output through the rectangular mesh-type graph

N_s = size(S,1); %the total number of subregions(triangles)

x0 = -1; xf = 1; y0 = -1; yf = 1;

Mx = 16; dx = (xf – x0)/Mx; xi = x0+[0:Mx]*dx;

My = 16; dy = (yf – y0)/My; yi = y0+[0:My]*dy;

for i = 1:length(xi)

for j = 1:length(yi)

for s = 1:N_s %which subregion the point belongs to

if inpolygon(xi(i),yi(j), N(S(s,:),1),N(S(s,:),2)) > 0

Z(i,j) = U(s,:)*[1 xi(i) yi(j)]’; %Eq.(9.4.5b)

break;

end

end

end

end

figure(2), clf, mesh(xi,yi,Z)

%For comparison

bx0 = inline(’0’); bxf = inline(’0’);

by0 = inline(’0’); byf = inline(’0’);

[U,x,y] = poisson(f,g,bx0,bxf,by0,byf,[x0 xf y0 yf],Mx,My);

figure(3), clf, mesh(x,y,U)

9.5 GUI OF MATLAB FOR SOLVING PDES: PDETOOL

In this section, we will see what problems can be solved by using the GUI (graphic

user interface) tool of MATLAB for PDEs and then apply the tool to solve the

elliptic/parabolic/hyperbolic equations dealt with in Examples 9.1/9.3/9.5 and 9.6.

430 PARTIAL DIFFERENTIAL EQUATIONS

9.5.1 Basic PDEs Solvable by PDETOOL

Basically, the PDE toolbox can be used for the following kinds of PDE.

1. Elliptic PDE

−∇ · (c∇u) + au = f over a domain (9.5.1)

with some boundary conditions like

hu = r (Dirichlet condition)

(9.5.2)

or n · c∇u + qu = g (generalized Neumann condition)

on the boundary ∂, where n is the outward unit normal vector to the boundary.

Note that, in case u is a scalar-valued function on a rectangular domain as

depicted in Fig. 9.1, Eq. (9.5.1) becomes

−c ∂2u(x, y)

∂x2 +

∂2u(x, y)

∂y2 + au(x, y) = f (x, y) (9.5.3)

and if the boundary condition for the left-side boundary segment is of Neumann

type like Eq. (9.1.7), Eq. (9.5.2) can be written as

− i · c ∂u(x, y)

∂x

i +

∂u(x, y)

∂y

j + qu(x, y)

= −c

∂u(x, y)

∂x + qu(x, y) = g(x, y) (9.5.4)

since the outward unit normal vector to the left-side boundary is n = i, where i

and j are the unit vectors along the x axis and y-axis, respectively.

2. Parabolic PDE

−∇ · (c∇u) + au + d

∂u

∂t =f (9.5.5)

over a domain and for a time range 0 ≤ t ≤ T

with boundary conditions like Eq. (9.5.2) and, additionally, the initial condition

u(t0).

3. Hyperbolic PDE

−∇ · (c∇u) + au + d

∂2u

∂t2 =f (9.5.6)

over a domain and for a time range 0 ≤ t ≤ T

GUI OF MATLAB FOR SOLVING PDES: PDETOOL 431

with boundary conditions like Eq. (9.5.2) and, additionally, the initial conditions

u(t0)/u(t0).

4. Eigenmode PDE

−∇ · (c∇u) + au = λdu (9.5.7)

over a domain and for an unknown eigenvalue λ

with some boundary conditions like Eq. (9.5.2).

The PDE toolbox can also deal with a system of PDEs like

−∇ · (c11∇u1)−∇ · (c12∇u2) + a11u1 + a12u2 = f1

−∇ · (c21∇u1)−∇ · (c22∇u2) + a21u1 + a22u2 = f2

over a domain

(9.5.8)

with Dirichlet boundary conditions like

h11 h12

h21 h22 u1

u2 = r1

r2 (9.5.9)

or generalized Neumann boundary conditions like

n · (c11∇u1) + n · (c12∇u2) + q11u1 + q12u2 = g1

n · (c21∇u1) + n · (c22∇u2) + q21u1 + q22u2 = g2

(9.5.10)

or mixed boundary conditions, where

c = c11 c12

c21 c22 , a= a11 a12

a21 a22 , f= f1

f2 , u= u1

u2

h = h11 h12

h21 h22 , r= r1

r2 , q= q11 q12

q21 q22 , g= g1

g2

9.5.2 The Usage of PDETOOL

The PDEtool in MATLAB solves PDEs by using the FEM (finite element method).

We should take the following steps to use it.

0. Type ‘pdetool ’ into the MATLAB command window to have the PDE

toolbox window on the screen as depicted in Fig. 9.11. You can toggle

on/off the grid by clicking ‘Grid’ in the Options pull-down menu

(Fig. 9.12a). You can also adjust the ranges of the x axis and the y axis

in the box window opened by clicking ‘Axes Limits’ in the Options pulldown

menu. If you want the rectangles to be aligned with the grid lines,

click ‘Snap(-to-grid)’ in the Options pull-down menu (Fig. 9.12a). If you

432 PARTIAL DIFFERENTIAL EQUATIONS

Figure 9.11 The GUI (graphical user interface) window of the MATLAB PDEtool.

want to have the x axis and the y axis of equal scale so that a circle/square

may not look like an ellipse/rectangle, click ‘Axes Equal’ in the Options

pull-down menu. You can choose the type of PDE problem you want to

solve in the submenu popped out by clicking ‘Application’ in the Options

pull-down menu (Fig. 9.12a).

(cf) In order to be able to specify the boundary condition for a boundary segment

by clicking it, the segment must be inside in the graphic region of PDEtool.

1. In Draw mode, you can create the two-dimensional geometry of domain

by using the constructive solid geometry (CSG) paradigm, which enables

us to make a set of solid objects such as rectangles, circles/ellipses, and

polygons. In order to do so, click the object that you want to draw in

the Draw pull-down menu (Fig. 9.12b) or click the button with the corresponding

icon ( , , , , ) in the tool-bar just below the top

menu-bar (Fig. 9.11). Then, you can click-and-drag to create/move the

object of any size at any position as you like. Once an object is drawn,

it can be selected by clicking on it. Note that the selected object becomes

surrounded by a black solid line and can be deleted by pressing Delete or

∧R(Ctrl-R) key. The created object is automatically labeled, but it can be

relabeled and resized (numerically) through the Object dialog box opened

GUI OF MATLAB FOR SOLVING PDES: PDETOOL 433

Figure 9.12 Pull-down menu from the top menu and its submenu of the MATLAB PDEtool.

434 PARTIAL DIFFERENTIAL EQUATIONS

by double-clicking the object and even rotated (numerically) through the

box opened by clicking ‘Rotate’ in the Draw pull-down menu. After creating

and positioning the objects, you can make a CSG model by editing

the set formula appropriately in the set formula field of the second line

below the top menu-bar to take the union (by default), the intersection,

and the set difference of the objects to form the shape of the domain

(Fig. 9.11). If you want to see the overall shape of the domain you created,

click ‘Boundary mode’ in the Boundary pull-down menu.

2. In Boundary mode, you can remove the subdomain borders that are

induced by the intersections of the solid objects, but are not between

different materials and also specify the boundary condition for each

boundary segment. First, click the ∂ button in the tool-bar (Fig. 9.11)

or ‘Boundary mode(∧B)’ in the Boundary pull-down menu (Fig. 9.12c),

which will make the boundary segments appear with red/blue/green colors

(indicating Dirichlet(default)/Neumann/mixed type of boundary condition)

and arrows toward its end (for the case where the boundary condition

is parameterized along the boundary). When you want to remove all

the subdomain borders, click ‘Remove All Subdomain Borders’ in the

Boundary pull-down menu. You can set the parameters h, r or g, q

in Eq. (9.5.2) to a constant or a function of x and y specifying the

boundary condition, through the box window opened by double-clicking

each boundary segment. In case you want to specify/change the boundary

condition for multiple segments at a time, you had better use shift-click

the segments to select all of them (which will be colored black) and click

again on one of them to get the boundary condition dialog box.

3. In PDE mode, you can specify the type of PDE (Elliptic/Parabolic/Hyperbolic/

Eigenmode) and its parameters. In order to do so, open the PDE

specification dialog box by clicking the PDE button in the tool-bar or

‘PDE Specification’ in the PDE pull-down menu (Fig. 9.12d), check the

type of PDE, and set its parameters in Eq. (9.5.1)/(9.5.5)/(9.5.6)/(9.5.7).

4. In Mesh mode, you can create the triangular mesh for the domain

drawn in Draw mode by just clicking the button in the tool-bar or

‘Initialize Mesh(∧I)’ in the Mesh pull-down menu (Fig. 9.12e). To improve

the accuracy of the solution, you can refine successively the mesh by

clicking the button in the tool-bar or ‘Refine Mesh(∧M)’ in the Mesh

pull-down menu. You can jiggle the mesh by clicking ‘Jiggle Mesh’ in

expectation of better accuracy. You can also undo any refinement by

clicking ‘Undo Mesh Change’ in the Mesh pull-down menu.

5. In Solve mode, you can solve the PDE and plot the result by just clicking

the = button in the tool-bar or ‘Solve PDE(∧E)’ in the Solve pull-down

(Fig. 9.12f). But, in the case of parabolic or hyperbolic PDE, you must

click ‘Parameters’ in the Solve pull-down menu (Fig. 9.12f) to set up the

initial conditions and the time range before solving the PDE.

GUI OF MATLAB FOR SOLVING PDES: PDETOOL 435

6. In Plot mode, you can change the plot option in the Plot selection dialog

box opened by clicking the button in the tool-bar or ‘Parameters’ in

the Plot pull-down menu (Fig. 9.12g). In the Plot selection dialog box

(Fig. 9.12h), you can set the plot type to, say, Color/Height(3-D) and set

the plot style to, say, interpolated shading and continuous (interpolated)

height. If you want the mesh to be shown in the solution graph, check

the box of Show mesh. In case you want to plot the graph of a known

function, change the option(s) of the Property into ‘user entry’, type in the

MATLAB expression describing the function and click the Plot button. You

can save the plot parameters as the current default by clicking the Done

button. You can also change the color map in the second line from the

bottom of the dialog box.

(cf) We can extract the parameters involved in the domain geometry by clicking ‘Export..’

in the Draw pull-down menu, the parameters specifying the boundary by clicking

‘Export..’ in the Boundary pull-down menu, the parameters specifying the PDE by

clicking ‘Export..’ in the PDE pull-down menu, the parameters specifying the mesh

by clicking ‘Export..’ in the Mesh pull-down menu, the parameters related to the

solution by clicking ‘Export..’ in the Solve pull-down menu, and the parameters

related to the graph by clicking ‘Export..’ in the Plot pull-down menu. Whenever

you want to save what you have worked in PDEtool, you may select File/Save in

the top menu-bar.

(cf) Visit the website “http://www.mathworks.com/access/helpdesk/help/helpdesk.

html” for more details.

9.5.3 Examples of Using PDETOOL to Solve PDEs

In this section, we will make use of PDEtool to solve some PDE problems that

were dealt with in the previous sections.

Example 9.7. Laplace’s Equation: Steady-State Temperature Distribution Over

a Plate. Consider the Laplace’s equation (Example 9.1)

∇2u(x, y) =

∂2u(x, y)

∂x2 +

∂2u(x, y)

∂y2 =0 for0≤ x ≤ 4, 0 ≤ y ≤ 4

(E9.7.1)

with the following boundary conditions.

u(0, y) = ey − cos y, u(4, y) = ey cos 4 − e4 cos y (E9.7.2)

u(x, 0) = cos x − ex, u(x,4) = e4 cos x − ex cos 4 (E9.7.3)

The procedure for using PDEtool to solve this problem is as follows:

0. Type ‘pdetool’ into the MATLAB command window to have the PDE

toolbox window on the screen. Then, adjust the ranges of the x-axis and

436 PARTIAL DIFFERENTIAL EQUATIONS

the y-axis to [0 5] and [0 5], respectively, in the dialog box opened by

clicking ‘Axes Limits’ in the Options pull-down menu. You can also click

‘Axes Equal’ in the Options pull-down menu to have the x axis and

the y axis of equal scale so that a circle/square may not look like an

ellipse/rectangle.

1. Click the button in the tool-bar and click-and-drag on the graphic

region to create a rectangle of domain. Then, in the Object dialog box

opened by double-clicking the rectangle, set the Left/Bottom/Width/Height

to 0/0/4/4. In this case, you don’t have to construct a CSG model by

editing the set formula, because the domain consists of a single object:

a rectangle.

2. Click the button in the tool-bar and double-click each boundary segment

to specify the boundary condition as Eqs. (E9.7.2,3) in the boundary

condition dialog box (see Fig. 9.13a).

3. Open the PDE specification dialog box by clicking the PDE button in the

tool-bar, check the box on the left of Elliptic as the type of PDE, and set

its parameters in Eq. (E9.7.1) as depicted in Fig. 9.13b.

4. Click the button in the tool-bar to divide the domain into a number of

triangular subdomains to get the triangular mesh as depicted in Fig. 9.13c.

You can click the button in the tool-bar to refine the mesh successively

for better accuracy.

5. Click the = button in the tool-bar to plot the solution in the form of twodimensional

graph with the value of u(x, y) shown in color.

6. If you want to plot the solution in the form of a three-dimensional graph

with the value of u(x, y) shown in height as well as color, check the box

before Height on the far-left side of the Plot selection dialog box opened

by clicking the button in the tool-bar. If you want the mesh shown in

the solution plot as Fig. 9.13d, check the box before Show mesh on the

far-left and low side and click the Plot button at the bottom of the Plot

selection dialog box (Fig. 9.12h). You can compare the result with that of

Example 9.3 depicted in Fig. 9.4.

7. If you have the true analytical solution

u(x, y) = ey cos x − ex cosy (E9.7.4)

and you want to plot the difference between the PDEtool (FEM) solution

and the true analytical solution, change the entry ‘u’ into ‘user entry’ in the

Color/Contour row and the Height row of the Property column and write

‘u-(exp(y).*cos(x)-exp(x).*cos(y))’ into the corresponding fields in

the User entry column of the Plot selection dialog box opened by clicking

the button in the tool-bar and click the Plot button at the bottom of the

dialog box.

GUI OF MATLAB FOR SOLVING PDES: PDETOOL 437

(a) Boundary condition for the domain of the PDE

(b) PDE specification (c) Initialize_Mesh

100

−100

u (x, y )

50

−50

0

4

3 4

2 2 3

1 1

0 0

y

x

(d) The mesh plot of solution

4

4

3

3

2

2

1

1

0

0

Figure 9.13 Procedure and results of using PDEtool for Example 9.1/9.7.

438 PARTIAL DIFFERENTIAL EQUATIONS

Example 9.8. A Parabolic PDE: Two-Dimensional Temperature Diffusion Over

a Plate. Consider a two-dimensional parabolic PDE

10−4 ∂2u(x, y, t)

∂x2 +

∂2u(x, y, t)

∂y2 =

∂u(x, y, t)

∂t

for 0 ≤ x ≤ 4, 0 ≤ y ≤4 & 0≤ t ≤ 5000 (E9.8.1)

with the initial conditions and boundary conditions

u(x, y, 0) =0 fort = 0 (E9.8.2a)

u(x, y, t) = ey cos x − ex cos y for x = 0, x = 4, y = 0, y = 4 (E9.8.2b)

The procedure for using the PDEtool to solve this problem is as follows.

0–2. Do exactly the same things as steps 0–2 for the case of an elliptic PDE

in Example 9.7.

3. Open the PDE specification dialog box by clicking the PDE button,

check the box on the left of ‘Parabolic’ as the type of PDE and set its

parameters in Eq. (E9.8.1) as depicted in Fig. 9.14a.

4. Exactly as in step 4 (for the case of elliptic PDE) in Example 9.7, click

the button to get the triangular mesh. You can click the button to

refine the mesh successively for better accuracy.

5. Unlike the case of an elliptic PDE, you must click ‘Parameters’ in

the Solve pull-down menu (Fig. 9.12f) to set the time range, say, as

0:100:5000 and the initial conditions as Eq. (E9.8.2a) before clicking

the = button to solve the PDE. (See Fig. 9.14b.)

6. As in step 6 of Example 9.7, you can check the box before Height in

the Plot selection dialog box opened by clicking the button, check

the box before Show mesh, and click the Plot button. If you want to

plot the solution graph at a time other than the final time, select the time

for plot from

{0, 100, 200, . . . , 500}

in the far-right field of the Plot selection dialog box and click the Plot

button again. If you want to see a movie-like dynamic picture of the

solution graph, check the box before Animation, click Options right after

Animation, fill in the fields of animation rate in fps (i.e., the number of

frames per second and the number of repeats in the Animation Options

dialog box), click the OK button, and then click the Plot button in the

Plot selection dialog box.

(cf) If the dynamic picture is too oblong, you can scale up/down the solution by changing

the Property of the Height row from ‘u’ into ‘user entry’ and filling in the

corresponding field of User entry with, say, ‘u/25’ in the Plot selection dialog box.

GUI OF MATLAB FOR SOLVING PDES: PDETOOL 439

100

50

−50

−100

0

u (x , y )

4

4

Time = 5000, Color: u, Height: u

(d) The mesh plot of solution at t = 5000

(c) The box window popped out by clicking Parameters in the Plot pull-down menu

(a) PDE specification dialog window (b) Solve parameters dialog window

y

2

0 0

1

2

3 x

Figure 9.14 Procedure and results of using PDEtool for Example 9.3/9.8.

440 PARTIAL DIFFERENTIAL EQUATIONS

According to your selection, you will see a movie-like dynamic picture or

the (final) solution graph like Fig. 9.14d, which is the steady-state solution

for Eq. (E9.8.1) with ∂u(x, y, t)/∂t = 0, virtually the same as the elliptic PDE

(E9.7.1) whose solution is depicted in Fig. 9.13d.

Before closing this example, let us have an experience of exporting the values

of some parameters. For example, we extract the mesh data {p, e, t} by clicking

‘Export Mesh’ in the Mesh pull-down menu and then clicking the OK button

in the Export dialog box. Among the mesh data, the matrix p contains the x

and y coordinates in the first and second rows, respectively. We also extract

the solution u by clicking ‘Export Solution’ in the Solve pull-down menu and

then clicking the OK button in the Export dialog box. Now, we can estimate

how far the graphical/numerical solution deviates from the true steady-state solution

u(x, y) = ey cos x − ex cos y by typing the following statements into the

MATLAB command window.

>>x = p(1.:)’; y = p(2.: )’; %x.y coordinates of nodes in column vector

>>err = exp(y).*cos(x) – exp(x).*cos(y) – u(:.end); %deviation from true sol

>>err_max = max(abs(err)) %maximum absolute error

Note that the dimension of the solution matrix u is 177 × 51 and the solution

at the final stage is stored in its last column u(:,end), where 177 is the number

of nodes in the triangular mesh and 51 = 5000/100 + 1 is the number of frames

or time stages.

Example 9.9. A Hyperbolic PDE: Two-Dimensional Wave (Vibration) Over a

Square Membrane. Consider a two-dimensional hyperbolic PDE

1

4 ∂2u(x, y, t)

∂x2 +

∂2u(x, y, t)

∂y2 =

∂u2(x, y, t)

∂t2

for 0 ≤ x ≤ 2, 0 ≤ y ≤ 2, and 0 ≤ t ≤ 2 (E9.9.1)

with the zero boundary conditions and the initial conditions

u(0, y, t) = 0, u(2, y, t) = 0, u(x,0, t) = 0, u(x,2, t) = 0 (E9.9.2)

u(x, y, 0) = 0.1 sin(πx) sin(πy/2), ∂u/∂t (x, y, 0) =0 fort = 0 (E9.9.3)

The procedure for using the PDEtool to solve this problem is as follows:

0–2. Do the same things as steps 0–2 for the case of ellipticPDEinExample 9.7,

except for the following.

ž Set the ranges of the x axis-and the y-axis to [0 3] and [0 3].

ž Set the Left/Bottom/Width/Height to 0/0/2/2 in the Object dialog box

opened by double-clicking the rectangle.

ž Set the boundary condition to zero as specified by Eqs. (E9.9.2) in

the boundary condition dialog box opened by clicking the ∂ button

in the tool-bar, shift-clicking the four boundary segments and doubleclicking

one of the boundary segments.

GUI OF MATLAB FOR SOLVING PDES: PDETOOL 441

(a) PDE specification dialog window

(b) Solve parameters dialog window

(c) The box window popped out by clicking Parameters in the Plot pull-down menu

u(x, y) u(x, y)

0.1 0.1

−0.1 −0.1

0 0

2 2 2 2

1 1 1

0 0 0 0

1

y x y x

(d1) The mesh plot of soution t = 0.1 (d2) The mesh plot of solution at t = 1.7

Figure 9.15 Procedure and results of using PDEtool for Example 9.5/9.9.

3. Open the PDE specification dialog box by clicking the PDE button,

check the box on the left of ‘Hyperbolic’ as the type of PDE, and set

its parameters in Eq. (E9.9.1) as depicted in Fig. 9.15a.

4. Do the same thing as step 4 for the case of elliptic PDE in Example 9.8.

5. Similarly to the case of a parabolic PDE, you must click ‘Parameters’

in the Solve pull-down menu (Fig. 9.12f) to set the time range, say, as

442 PARTIAL DIFFERENTIAL EQUATIONS

0:0.1:2 and the initial conditions as Eq. (E9.9.3) before clicking the = button to solve the PDE. (See Fig. 9.15b.)

6. Do almost the same thing as step 6 for the case of parabolic PDE in

Example 9.8.

Finally, you could see the solution graphs like Figs. 9.15(d1)&(d2), that are

similar to Figs. 9.6(a)&(c).

Example 9.10. Laplace’s Equation: Electric Potential Over a Plate with Point

Charge. Consider the Laplace’s equation (dealt with in Example 9.6)

∇2u(x, y) =

∂2u(x, y)

∂x2 +

∂2u(x, y)

∂y2 = f (x, y)

for −1 ≤ x ≤ +1,−1 ≤ y ≤ +1 (E9.10.1)

where

f (x, y) =

−1 for(x, y) = (0.5, 0.5)

+1 for(x, y) = (−0.5,−0.5)

0 elsewhere

(E9.10.2)

and the boundary condition is u(x, y) = 0 for all boundaries of the rectangular

domain.

The procedure for using the PDEtool to solve this problem is as follows.

0–2. Do the same thing as step 0–2 for the case of elliptic PDE in Example 9.7,

except for the following.

ž Set the Left/Bottom/Width/Height to −1/−1/2/2 in the Object dialog

box opened by double-clicking the rectangle.

ž Set the boundary condition to zero in the boundary condition dialog box

opened by clicking the ∂ button in the tool-bar, shift-clicking the four

boundary segments, and double-clicking one of the boundary segments.

3. Open the PDE specification dialog box by clicking the PDE button, check

the box on the left of ‘Elliptic’ as the type of PDE, and set its parameters

in Eq. (E9.10.1,2) as depicted in Fig. 9.16a.

4. Click the button to initialize the triangular mesh.

5. Click the button to open the Plot selection dialog box, check the box

before ‘Height’, and check the box before ‘Show mesh’ in the dialog box.

6. Click the Plot button to get the solution graph as depicted in Fig. 9.16c.

7. Click ‘Parameters’ in the Solve pull-down menu to open the ‘Solve Parameters’

dialog box depicted in Fig. 9.16b, check the box on the left of

‘Adaptive mode’, and click the OK button in order to activate the adaptive

mesh mode.

8. Click the = button to get a solution graph with the adaptive mesh.

GUI OF MATLAB FOR SOLVING PDES: PDETOOL 443

(c) The mesh plot of the solution with Initialize mesh

× 10−3

5

u (x, y )

0

0

−5

0.5

−0.5

0

1 y 0.5

−0.5

−1 −1

1

x

(d) The mesh plot of the solution obtained by using Adaptive mesh one time

0

1 y

0.5

−0.5

−1

× 10−3

1

u (x, y )

0

0

−1

0.5

−0.5 −1

1

x

(b) Solve parameters dialog window

(a) PDE specification dialog window

Figure 9.16 Procedure and results of using PDEtool for Example 9.6/9.10.

444 PARTIAL DIFFERENTIAL EQUATIONS

9. Noting that the solution is not the right one for the point charge distribution

given by (E9.10.2), reopen the PDE specification dialog box by clicking

the PDE button and rewrite f as below.

, , , ,

10. Noting that the mesh has already been refined in the adaptive way to

yield smaller meshes in the region where the slope of the solution is

steeper, click ‘Parameters’ in the Solve pull-down menu to open the ‘Solve

Parameters’ dialog box, uncheck the box on the left of ‘Adaptive mode’,

and click the OK button in the dialog box in order to inactivate the

adaptive mesh mode.

11. Click the = button to get the solution graph as depicted in Fig. 9.16d.

12. You can click ‘Refine Mesh(∧M)’ in the Mesh pull-down menu and click

the = button to get a more refined solution graph (with higher resolution)

as many times as you want.

PROBLEMS

9.1 Elliptic PDEs: Poisson Equations

Use the routine “poisson()” (in Section 9.1) to solve the following PDEs

and plot the solutions by using the MATLAB command “mesh()”.

(a) ∇2u(x, y) =

∂2u(x, y)

∂x2 +

∂2u(x, y)

∂y2 = x +y (P9.1.1)

for 0 ≤ x ≤ 1, 0 ≤ y ≤ 1

with the boundary conditions

u(0, y) = y2, u(1, y) = 1,

u(x, 0) = x2, u(x,1) = 1 (P9.1.2)

Divide the solution region (domain) into Mx × My = 5 × 10 sections.

(b)

∂2u(x, y)

∂x2 +

∂2u(x, y)

∂y2 − 12.5π2u(x, y)

= −25π2 cos5π

2

xcos5π

2

y for 0 ≤ x, y ≤ 0.4 (P9.1.3)

with the boundary conditions

u(0, y) = cos5π

2

y, u(0.4, y) = −cos5π

2

y (P9.1.4)

u(x, 0) = cos5π

2

x, u(x,0.4) = −cos5π

2

x (P9.1.5)

PROBLEMS 445

Divide the solution region into Mx ×My = 40 × 40 sections.

(c)

∂2u(x, y)

∂x2 +

∂2u(x, y)

∂y2 + 4π(x2 + y2)u(x, y)

= 4π cos(π(x2 + y2)) for 0 ≤ x ≤ 1, 0 ≤ y ≤ 1 (P9.1.6)

with the boundary conditions

u(0, y) = sin(πy2), u(1, y) = sin(π(y2 + 1)) (P9.1.7)

u(x, 0) = sin(πx2), u(x, 1) = sin(π(x2 + 1)) (P9.1.8)

Divide the solution region into Mx ×My = 40 × 40 sections.

(d)

∂2u(x, y)

∂x2 +

∂2u(x, y)

∂y2 = 10e2x+y for 0 ≤ x ≤ 1, 0 ≤ y ≤ 2 (P9.1.9)

with the boundary conditions

u(0, y) = 2ey, u(1, y) = 2e2x+y,

u(x, 0) = 2e2x, u(x,2) = 2e2x+2 (P9.1.10)

Divide the solution region into Mx ×My = 20 × 40 sections.

(e)

∂2u(x, y)

∂x2 +

∂2u(x, y)

∂y2 =0 for0≤ x ≤ 1, 0 ≤ y ≤ π/2 (P9.1.11)

with the boundary conditions

u(0, y) = 4 cos(3y), u(1, y) = 4e−3 cos(3y),

u(x, 0) = 4e−3x, u(x,π/2) = 0 (P9.1.12)

Divide the solution region into Mx ×My = 20 × 20 sections.

9.2 More General PDE Having Nonunity Coefficients

Consider the following PDE having nonunity coefficients.

A

∂2u(x, y)

∂x2 + B

∂2u(x, y)

∂x∂y + C

∂2u(x, y)

∂y2 + g(x, y)u(x, y) = f (x, y)

(P9.2.1)

Modify the routine “poisson()” so that it can solve this kind of PDEs and

declare it as

function [u,x,y] = poisson_abc(ABC,f,g,bx0,bxf,by0,…,Mx,My,tol,imax)

446 PARTIAL DIFFERENTIAL EQUATIONS

where the first input argument ABC is supposed to carry the vector containing

three coefficients A,B, and C. Use the routine to solve the following PDEs

and plot the solutions by using the MATLAB command “mesh()”.

(a)

∂2u(x, y)

∂x2 + 2

∂2u(x, y)

∂y2 = 10 for 0 ≤ x ≤ 1, 0 ≤ y ≤ 1 (P9.2.2)

with the boundary conditions

u(0, y) = y2, u(1, y) = (y + 2)2,

u(x, 0) = 4×2, u(1, y) = (2x + 1)2

(P9.2.3)

Divide the solution region (domain) into Mx ×My = 20 × 40 sections.

(b)

∂2u(x, y)

∂x2 + 3

∂2u(x, y)

∂x∂y + 2

∂2u(x, y)

∂y2 =0 for0≤ x ≤ 1, 0 ≤ y ≤ 1

(P9.2.4)

with the boundary conditions

u(0, y) = ey + cos y, u(1, y) = ey−1 + cos(y − 2) (P9.2.5)

u(x, 0) = e−x + cos(−2x), u(x, 1) = e1−x + cos(1 − 2x) (P9.2.6)

Divide the solution region into Mx × My = 40 × 40 sections.

(c)

∂2u(x, y)

∂x2 + 3

∂2u(x, y)

∂x∂y + 2

∂2u(x, y)

∂y2 = x sin y

for 0 ≤ x ≤ 2, 0 ≤ y ≤ π

(P9.2.7)

with the boundary conditions

u(0, y) = (3/4) cos y, u(2, y) = −sin(y) + (3/4) cos y (P9.2.8)

u(x, 0) = 3/4, u(x,π)= −3/4 (P9.2.9)

Divide the solution region into Mx × My = 20 × 40 sections.

(d) 4

∂2u(x, y)

∂x2 − 4

∂2u(x, y)

∂x∂y +

∂2u(x, y)

∂y2 =0 for0≤ x ≤ 1, 0 ≤ y ≤ 1

(P9.2.10)

with the boundary conditions

u(0, y) = ye2y, u(1, y) = (1 + y)e1+2y,

u(x, 0) = xex, u(x,1) = (x + 1)ex+2

(P9.2.11)

Divide the solution region into Mx × My = 40 × 40 sections.

PROBLEMS 447

function [u,x,y] = poisson_Neuman(f,g,bx0,bxf,by0,byf,x0,xf,y0,yf,…)

.. .. .. .. .. .. .. ..

Neum = zeros(1,4); %Not Neumann, but Dirichlet condition by default

if length(x0) > 1, Neum(1) = x0(2); x0 = x0(1); end

if length(xf) > 1, Neum(2) = xf(2); xf = xf(1); end

if length(y0) > 1, Neum(3) = y0(2); y0 = y0(1); end

if length(yf) > 1, Neum(4) = yf(2); yf = yf(1); end

.. .. .. .. .. .. .. ..

dx_2 = dx*dx; dy_2 = dy*dy; dxy2=2*(dx_2 + dy_2);

rx = dx_2/dxy2; ry = dy_2/dxy2; rxy = rx*dy_2; rx = rx;

dx2 = dx*2; dy2 = dy*2; rydx = ry*dx2; rxdy = rx*dy2;

u(1:My1,1:Mx1) = zeros(My1,Mx1);

sum_of_bv = 0; num = 0;

if Neum(1) == 0 %Dirichlet boundary condition

for m = 1:My1, u(m,1) = bx0(y(m)); end %side a

else %Neumann boundary condition

for m = 1:My1, duxa(m) = bx0(y(m)); end %du/dx(x0,y)

end

if Neum(2) == 0 %Dirichlet boundary condition

.. .. .. .. .. .. .. .. .. .. ..

end

if Neum(3) == 0 %Dirichlet boundary condition

n1 = 1; nM1 = Mx1;

if Neum(1) == 0, u(1,1)=(u(1,1) + by0(x(1)))/2; n1 = 2; end

if Neum(2) == 0, u(1,Mx1)=(u(1,Mx1) + by0(x(Mx1)))/2; nM1 = Mx; end

for n = n1:nM1, u(1,n) = by0(x(n)); end %side c

else %Neumann boundary condition

for n = 1:Mx1, duyc(n) = by0(x(n)); end %du/dy(x,y0)

end

if Neum(4) == 0 %Dirichlet boundary condition

.. .. .. .. .. .. .. .. .. .. ..

end

for itr = 1:imax

if Neum(1) %Neumann boundary condition

for i = 2:My

u(i,1) = 2*ry*u(i,2) + rx*(u(i + 1,1) + u(i-1,1)) …

+ rxy*(G(i,1)*u(i,1) – F(i,1)) – rydx*duxa(i); %(9.1.9)

end

if Neum(3), u(1,1) = 2*(ry*u(1,2) +rx*u(2,1)) …

+ rxy*(G(1,1)*u(1,1) – F(1,1)) – rydx*duxa(1) – rxdy*duyc(1);%(9.1.11)

end

if Neum(4), u(My1,1) = 2*(ry*u(My1,2) +rx*u(My,1)) …

+ rxy*(G(My1,1)*u(My1,1)- F(My1,1))+rxdy*duyd(1)- rydx*duxa(My1);

end

end

if Neum(2) %Neumann boundary condition

.. .. .. .. .. .. .. .. .. .. .. .. ..

end

if Neum(3) %Neumann boundary condition

for j = 2:Mx

u(1,j) = 2*rx*u(2,j)+ry*(u(1,j+1) + u(1,j-1)) …

+rxy*(G(1,j)*u(1,j) – F(1,j)) – rxdy*duyc(j); %(9.1.10)

end

end

if Neum(4) %Neumann boundary condition

.. .. .. .. .. .. .. .. .. .. .. .. ..

end

.. .. .. .. .. .. .. .. .. .. .. .. ..

end

448 PARTIAL DIFFERENTIAL EQUATIONS

9.3 Elliptic PDEs with Neumann Boundary Condition

Consider the PDE (E9.1.1) (dealt with in Example 9.1)

∂2u(x, y)

∂x2 +

∂2u(x, y)

∂y2 =0 for0≤ x ≤ 4, 0 ≤ y ≤ 4 (P9.3.1)

with different boundary conditions of Neumann type, which was discussed

in Section 9.1. Modify the routine “poisson()” so that it can deal with the

Neumann boundary condition and declare it as

function [u,x,y] = poisson_Neuman(f,g,bx0,bxf,by0,byf,x0,xf,y0,yf,…)

where the third/fourth/fifth/sixth input arguments are supposed to carry the

functions of

u(x0, y)/u(xf , y)/u(x, y0)/u(x, yf )

or

∂u(x, y)/∂x|x=x0/∂u(x, y)/∂x|x=xf /∂u(x, y)/∂y|y=y0/∂u(x, y)/∂y|y=yf

and the seventh/eighth/ninth/tenth input arguments are to carry x0/xf /y0/yf

or [x0 1]/[xf 1]/[y0 ]/[yf 1] depending on whether each boundary condition

is of Dirichlet or Neumann type. Use it to solve the PDE with the

following boundary conditions and plot the solutions by using the MATLAB

command “mesh()”. Divide the solution region (domain) into Mx ×My = 20 × 20 sections.

(cf) You may refer to the related part of the program in the previous page.

(a) ∂u(x, y)/∂x|x=0 = −cos y, u(4, y) = ey cos 4 − e4 cosy (P9.3.2)

∂u(x, y)/∂y|y=0 = cos x, u(x, 4) = e4 cos x − ex cos 4 (P9.3.3)

(b) u(0, y) = ey − cos y, ∂u(x, y)/∂x|x=4 = −ey sin 4 − e4 cosy (P9.3.4)

u(x, 0) = cos x − ex, ∂u(x, y)/∂y|y=4 = e4 cos x + ex sin 4 (P9.3.5)

(c) ∂u(x, y)/∂x|x=0 = −cos y, u(4, y) = ey cos 4 − e4 cosy (P9.3.6)

u(x, 0) = cos x − ex, ∂u(x, y)/∂yy=4 = e4 cos x + ex sin 4 (P9.3.7)

(d) u(0, y) = ey − cos y, ∂u(x, y)/∂xx=4 = −ey sin 4 − e4 cosy (P9.3.8)

∂u(x, y)/∂yy=0 = cos x, u(x, 4) = e4 cos x − ex cos 4 (P9.3.9)

PROBLEMS 449

(e) ∂u(x, y)/∂xx=0 = −cos y, ∂u(x, y)/∂xx=4 = −ey sin 4 − e4 cos y

(P9.3.10)

∂u(x, y)/∂yy=0 = cos x, u(x, 4) = e4 cos x − ex cos 4 (P9.3.11)

(f) ∂u(x, y)/∂xx=0 = −cos y, u(4, y) = ey cos 4 − e4 cos y(P9.3.12)

∂u(x, y)/∂yy=0 = cos x, ∂u(x, y)/∂yy=4 = e4 cos x + ex sin 4

(P9.3.13)

(g) u(0, y) = ey − cos y, ∂u(x, y)/∂xx=4 = −ey sin 4 − e4 cos y(P9.3.14)

∂u(x, y)/∂yy=0 = cos x, ∂u(x, y)/∂yy=4 = e4 cos x + ex sin 4

(P9.3.15)

(h) ∂u(x, y)/∂xx=0 = −cos y, ∂u(x, y)/∂xx=4 = −ey sin 4 − e4 cos y

(P9.3.16)

∂u(x, y)/∂yy=0 = cos x, ∂u(x, y)/∂yy=4 = e4 cos x + ex sin 4

(P9.3.17)

9.4 Parabolic PDEs: Heat Equations

Modify the program “solve_heat.m” (in Section 9.2.3) so that it can solve

the following PDEs by using the explicit forward Euler method, the implicit

backward Euler method, and the Crank–Nicholson method.

(a)

∂2u(x, t)

∂x2 =

∂u(x, t)

∂t

for 0 ≤ x ≤ 1, 0 ≤ t ≤ 0.1 (P9.4.1)

with the initial/boundary conditions

u(x, 0) = x4, u(0, t) = 0, u(1, t) = 1 (P9.4.2)

(i) With the solution region divided into M × N = 10 × 20 sections,

does the explicit forward Euler method converge? What is the value

of r = At/(x)2?

(ii) If you increase M and N to make M × N = 20 × 40 for better

accuracy, does the explicit forward Euler method still converge?

What is the value of r = At/(x)2?

(iii) What is the number N of subintervals along the t axis that we should

choose in order to keep the same value of r for M = 20? With that

value of r, does the explicit forward Euler method converge?

450 PARTIAL DIFFERENTIAL EQUATIONS

(b) 10−5 ∂2u(x, t)

∂x2 =

∂u(x, t)

∂t

for 0 ≤ x ≤ 1, 0 ≤ t ≤ 6000 (P9.4.3)

with the initial/boundary conditions

u(x, 0) = 2x + sin(2πx), u(0, t) = 0, u(1, t) = 2 (P9.4.4)

(i) With the solution region divided into M × N = 20 × 40 sections,

does the explicit forward Euler method converge? What is the value

of r = At/(x)2? Does the numerical stability condition (9.2.6)

seem to be so demanding?

(ii) If you increase M and N to make M × N = 40 × 160 for better

accuracy, does the explicit forward Euler method still converge?

What is the value of r = At/(x)2? Does the numerical stability

condition (9.2.6) seem to be so demanding?

(iii) With the solution region divided into M × N = 40 × 200 sections,

does the explicit forward Euler method converge? What is the value

of r = At/(x)2?

(c) 2

∂2u(x, t)

∂x2 =

∂u(x, t)

∂t

for 0 ≤ x ≤ π, 0 ≤ t ≤ 0.2 (P9.4.5)

with the initial/boundary conditions

u(x, 0) = sin(2x), u(0, t) = 0, u(π,t)= 0 (P9.4.6)

(i) By substituting

u(x, t) = sin(2x)e−8t (P9.4.7)

into the above equation (P9.4.5), verify that this is a solution to

the PDE.

(ii) With the solution region divided into M × N = 40 × 100 sections,

does the explicit forward Euler method converge? What is the value

of r = At/(x)2?

(iii) If you increase N (the number of subintervals along the t-axis) to

125 for improving the numerical stability, does the explicit forward

Euler method converge? What is the value of r = At/(x)2? Use

the MATLAB statements in the following box to find the maximum

absolute errors of the numerical solutions obtained by the three

methods. Which method yields the smallest error?

uo = inline(’sin(2*x)*exp(-8*t)’,’x’,’t’); %true analytical solution

Uo = uo(x,t);

err = max(max(abs(u1 – Uo)))

PROBLEMS 451

(iv) If you increase N to 200, what is the value of r = At/(x)2? Find

the maximum absolute errors of the numerical solutions obtained

by the three methods as in (iii). Which method yields the smallest

error?

(d)

∂2u(x, t)

∂x2 =

∂u(x, t)

∂t

for 0 ≤ x ≤ 1, 0 ≤ t ≤ 0.1 (P9.4.8)

with the initial/boundary conditions

u(x, 0) = sin(πx) + sin(3πx), u(0, t) = 0, u(1, t) = 0 (P9.4.9)

(i) By substituting

u(x, t) = sin(πx)e−π2t + sin(3πx)e−(3π)2t (P9.4.10)

into Eq. (P9.4.5), verify that this is a solution to the PDE.

(ii) With the solution region divided into M × N = 25 × 80 sections,

does the explicit forward Euler method converge? What is the value

of r = At/(x)2?

(iii) If you increase N (the number of subintervals along the t axis) to

100 for improving the numerical stability, does the explicit forward

Euler method converge? What is the value of r = At/(x)2? Find

the maximum absolute errors of the numerical solutions obtained by

the three methods as in (c)(iii).

(iv) If you increase N to 200, what is the value of r = At/(x)2? Find

the maximum absolute errors of the numerical solutions obtained by

the three methods as in (c)(iii). Which one gained the accuracy the

most of the three methods through increasing N?

9.5 Parabolic PDEs with Neumann Boundary Conditions

Let us modify the routines “heat_exp()”, “heat_imp()”, and “heat_cn()”

(in Section 9.2) so that they can accommodate the heat equation (9.2.1) with

Neumann boundary conditions

∂u(x, t)/∂xx=x0 = bx0 (t), ∂u(x, t)/∂xx=xf = bxf (t) (P9.5.1)

(a) Consider the explicit forward Euler algorithm described by Eq. (9.2.3)

uk+1

i = r(uk

i+1 + uk

i−1) + (1 − 2r)uk

i

for i = 1, 2, . . .,M − 1 with r = A

t

x2 (P9.5.2)

452 PARTIAL DIFFERENTIAL EQUATIONS

In the case of Dirichlet boundary condition, we don’t need to get uk+1

0

and uk+1

M , because they are already given. But, in the case of the Neumann

boundary condition, we must get them by using this equation for

i = 0 and M as

uk+1

0 = r(uk

1 + uk

−1) + (1 − 2r)uk

0 (P9.5.3a)

uk+1

M = r(uk

M+1 + uk

M−1) + (1 − 2r)uk

M (P9.5.3b)

and the boundary conditions approximated as

uk

1 − uk

−1

2x = b0(k), uk

−1 = uk

1 − 2b0 (k)x (P9.5.4a)

uk

M+1 − uk

M−1

2x = bM (k), uk

M+1 = uk

M−1 + 2bM (k)x (P9.5.4b)

Substituting Eqs. (P9.5.4a,b) into Eq. (P9.5.3) yields

uk+1

0 = 2r(uk

1 − b0(k)x) + (1 − 2r)uk

0 (P9.5.5a)

uk+1

M = 2r(uk

M−1 + bM (k)x) + (1 − 2r)uk

M (P9.5.5b)

Modify the routine “heat_exp()” so that it can use this scheme to deal

with the Neumann boundary conditions for solving the heat equation and

declare it as

function [u,x,t] = heat_exp_Neuman(a,xfn,T,it0,bx0,bxf,M,N)

where the second input argument xfn and the fifth and sixth input arguments

bx0,bxf are supposed to carry [xf 0 1] and bx0 (t), bxf

(t), respectively,

if the boundary condition at x0/xf is of Dirichlet/Neumann type and

they are also supposed to carry [xf 1 1] and bx0 (t), bxf

(t), respectively,

if both of the boundary conditions at x0/xf are of Neumann type.

(b) Consider the implicit backward Euler algorithm described by Eq. (9.2.13),

which deals with the Neumann boundary condition at the one end for

solving the heat equation (9.2.1). With reference to Eq. (9.2.13), modify

the routine “heat_imp()” so that it can solve the heat equation with the

Neumann boundary conditions at two end points x0 and xf and declare

it as

function [u,x,t] = heat_imp_Neuman(a,xfn,T,it0,bx0,bxf,M,N)

(c) Consider theCrank–Nicholson algorithm described by Eq. (9.2.17),which

deals with the Neumann boundary condition at the one end for solving the

heat equation (9.2.1). With reference to Eq. (9.2.17), modify the routine

“heat_cn()” so that it can solve the heat equation with the Neumann

boundary conditions at two end points x0 and xf and declare it as

function [u,x,t] = heat_cn_Neuman(a,xfn,T,it0,bx0,bxf,M,N)

PROBLEMS 453

(d) Solve the following heat equation with three different boundary conditions

by using the three modified routines in (a), (b), (c) with M = 20,N

= 100 and find the maximum absolute errors of the three solutions as in

Problem 9.4(c)(iii).

∂2u(x, t)

∂x2 =

∂u(x, t)

∂t

for 0 ≤ x ≤ 1, 0 ≤ t ≤ 0.1 (P9.5.6)

with the initial/boundary conditions

(i) u(x, 0) = sin(πx), ∂u(x, t)/∂xx=0 = π e−π2t, u(x,t)x=1 = 0

(P9.5.7)

(ii) u(x, 0) = sin(πx), u(x, t)x=0 = 0, ∂u(x,t)/∂xx=1 = −π e−π2t

(P9.5.8)

(iii) u(x, 0) = sin(πx), ∂u(x, t)/∂xx=0 = π e−π2t ,

∂u(x, t)/∂xx=1 = −π e−π2t (P9.5.9)

Note that the true analytical solution is

u(x, t) = sin(πx)e−π2t (P9.5.10)

9.6 Hyperbolic PDEs: Wave Equations

Modify the program “solve_wave.m” (in Section 9.3) so that it can solve

the following PDEs by using the explicit forward Euler method, the implicit

backward Euler method, and the Crank–Nicholson method.

(a) 4

∂2u(x, t)

∂x2 =

∂2u(x, t)

∂t2 for 0 ≤ x ≤ 1, 0 ≤ t ≤ 1 (P9.6.1)

with the initial/boundary conditions

u(x, 0) = 0, ∂u(x,t)/∂tt=0 = 5 sin(πx),

u(0, t) = 0, u(1, t) = 0 (P9.6.2)

Note that the true analytical solution is

u(x, t) =

2.5

π

sin(πx) sin(2πt) (P9.6.3)

(i) With the solution region divided into M × N = 20 × 50 sections,

what is the value of r = A(t)2/(x)2? Use the MATLAB statements

in Problem 9.4(c)(iii) to find the maximum absolute error of

the solution obtained by using the routine “wave()”.

454 PARTIAL DIFFERENTIAL EQUATIONS

(ii) With the solution region divided into M × N = 40 × 100 sections,

what is the value of r? Find the maximum absolute error of the

numerical solution.

(iii) If we increase M (the number of subintervals along the x axis) to

50 for better accuracy, what is the value of r? Find the maximum

absolute error of the numerical solution and determine whether it

has been improved.

(iv) If we increase the number M to 52, what is the value of r? Can

we expect better accuracy in the light of the numerical stability

condition (9.3.7)? Find the maximum absolute error of the numerical

solution and determine whether it has been improved or not.

(v) What do you think the best value of r is?

(b) 6.25

∂2u(x, t)

∂x2 =

∂2u(x, t)

∂t2 for 0 ≤ x ≤ π, 0 ≤ t ≤ 0.4π (P9.6.4)

with the initial/boundary conditions

u(x, 0) = sin(2x), ∂u(x, t)/∂tt=0 = 0,

u(0, t) = 0, u(1, t) = 0 (P9.6.5)

Note that the true analytical solution is

u(x, t) = sin(2x) cos(5t) (P9.6.6)

(i) With the solution region divided into M × N = 50 × 50 sections,

what is the value of r = A(t)2/(x)2? Find the maximum absolute

error of the solution obtained by using the routine “wave()”.

(ii) With the solution region divided into M × N = 50 × 49 sections,

what is the value of r? Find the maximum absolute error of the

numerical solution.

(iii) If we increase N (the number of subintervals along the t axis) to

51 for better accuracy, what is the value of r? Find the maximum

absolute error of the numerical solution.

(iv) What do you think the best value of r is?

(c)

∂2u(x, t)

∂x2 =

∂2u(x, t)

∂t2 for 0 ≤ x ≤ 10, 0 ≤ t ≤ 10 (P9.6.7)

with the initial/boundary conditions

u(x, 0) = (x − 2)(3 − x) for 2 ≤ x ≤ 3

0 elsewhere

(P9.6.8)

∂u(x, t)/∂t |t=0 = 0, u(0, t) = 0, u(10, t) = 0 (P9.6.9)

PROBLEMS 455

(i) With the solution region divided into M × N = 100 × 100 sections,

what is the value of r = A(t)2/(x)2?

(ii) Noting that the initial condition (P9.6.8) can be implemented by the

MATLAB statement as

>>it0 = inline(’(x-2).*(3-x).*(2<x&x<3)’,’x’);

solve the PDE (P9.6.7) in the same way as in “solve_wave.m” and

make a dynamic picture out of the numerical solution, with the current

time printed on screen. Estimate the time when one of the two separated

pulses propagating leftwards is reflected and reversed. How about the

time when the two separated pulses are reunited?

9.7 FEM (Finite Element Method)

In expectation of better accuracy/resolution, modify the program

“do_fem.m” (in Section 9.6) by appending the following lines

;-17/32 -31/64; -1/2 -17/32;-15/32 -31/64

;17/32 31/64; 1/2 17/32; 15/32 31/64

to the last part of the Node array N and replacing the last line of the subregion

array S with

26 32 33; 27 33 34; 28 32 34; 29 35 36;

30 36 37; 31 35 37; 32 33 34; 35 36 37

This is equivalent to refining the triangular mesh in the subregions nearest

to the point charges at (0.5, 0.5) and (−0.5, −0.5) as depicted in Fig. P9.7.

Plot the new solution obtained by running the modified program. You may

have to change a statement of the program as follows.

f962 =’(norm([x y]+[0.5 0.5])<1e-3)-(norm([x y]-[0.5 0.5])<1e-3)’;

S37

S41

S35 S36

S39

S42

S40

S38

28

34

27

37

30

36

35

29

31

33

26

32

(−1/2,−7/16)

(−17/32,−31/64)

(−7/16,−17/32)

(7/16,17/32) (9/16,17/32)

(−9/16,−17/32) (−1/2,−17/32)

(1/2,17/32)

(15/32,31/64)

(−15/32,−31/64)

(17/32,31/64)

(1/2,7/16)

Figure P9.7 Refined triangular meshes.

456 PARTIAL DIFFERENTIAL EQUATIONS

9.8 PDEtool: GUI (Graphical User Interface) of MATLAB for Solving PDEs

(a) Consider the PDE

4

∂2u(x, y)

∂x2 − 4

∂2u(x, y)

∂x∂y +

∂2u(x, y)

∂y2 =0 for0≤ x ≤ 1, 0 ≤ y ≤ 1

(P9.8.1)

with the boundary conditions

u(0, y) = ye2y, u(1, y) = (1 + y)e1+2y ,

u(x, 0) = xex, u(x,1) = (x + 1)ex+2

(P9.8.2)

Noting that the field of coefficient c should be filled in as

Elliptic

c 4 -2 -2 1 4 -2 1

in the PDE specification dialog box and the true analytical solution is

u(x, y) = (x + y)ex+2y (P9.8.3)

use the PDEtool to solve this PDE and fill in Table P9.8.1 with the

maximum absolute error and the number of nodes together with those of

Problem 9.2(d) for comparison.

You can refer to Example 9.8 for the procedure to get the numerical

value of the maximum absolute error. Notice that the number of nodes is

the number of columns of p, which is obtained by clicking ‘Export Mesh’

in the Mesh pull-down menu and then, clicking the OK button in the

Export dialog box. You can also refer to Example 9.10 for the usage

of ‘Adaptive Mesh’, but in this case you only have to check the box

on the left of ‘Adaptive Mode’ and click the OK button in the ‘Solve

Parameters’ dialog box opened by clicking ‘Parameters’ in the Solve

pull-down menu, and then the mesh is adaptively refined every time you

click the = button in the tool-bar to get the solution. With the box on the

left of ‘Adaptive Mode’ unchecked in the ‘Solve Parameters’ dialog box,

Table P9.8.1 The Maximum Absolute Error and the Number of Nodes

The Maximum

Absolute Error

The Number

of Nodes

poisson() 1.9256 41 × 41

PDEtool with Initialize Mesh 0.1914 177

PDEtool with Refine Mesh

PDEtool with second Refine Mesh

PDEtool with Adaptive Mesh

PDEtool with second Adaptive Mesh

PROBLEMS 457

the mesh is nonadaptively refined every time you click ‘Refine Mesh’

in the Mesh pull-down menu. You can restore the previous mesh by

clicking ‘Undo Mesh Change’ in the Mesh pull-down menu.

(b) Consider the PDE

∂2u(x, y)

∂x2 +

∂2u(x, y)

∂y2 =0 for0≤ x ≤ 4, 0 ≤ y ≤ 4 (P9.8.4)

with the Dirichlet/Neumann boundary conditions

u(0, y) = ey − cos y, ∂u(x, y)/∂x|x=4 = −ey sin 4 − e4 cos y (P9.8.5)

∂u(x, y)/∂y|y=0=cos x, ∂u(x, y)/∂y|y=4=e4 cos x + ex sin 4 (P9.8.6)

Noting that the true analytical solution is

u(x, y) = ey cos x − ex cosy (P9.8.7)

use the PDEtool to solve this PDE and fill in Table P9.8.2 with the

maximum absolute error and the number of nodes together with those of

Problem 9.3(g) for comparison.

(c) Consider the PDE

2

∂2u(x, t)

∂x2 =

∂u(x, t)

∂t

for 0 ≤ x ≤ π, 0 ≤ t ≤ 0.2 (P9.8.8)

with the initial/boundary conditions

u(x, 0) = sin(2x), u(0, t) = 0, u(π,t)= 0 (P9.8.9)

Noting that the true analytical solution is

u(x, t) = sin(2x)e−8t (P9.8.10)

Table P9.8.2 The Maximum Absolute Error and the Number of Nodes

The Maximum

Absolute Error

The Number

of Nodes

poisson() 0.2005 21 × 21

PDEtool with Initialize Mesh 0.5702 177

PDEtool with Refine Mesh

PDEtool with second Refine Mesh

PDEtool with Adaptive Mesh

PDEtool with second Adaptive Mesh

458 PARTIAL DIFFERENTIAL EQUATIONS

Table P9.8.3 The Maximum Absolute Error and the Number of Nodes

The Maximum

Absolute Error

The Number

of Nodes

poisson() 7.5462 × 10−4 41 × 101

PDEtool with Initialize Mesh

PDEtool with Refine Mesh

PDEtool with second Refine Mesh

use the PDEtool to solve this PDE and fill in Table P9.8.3 with the

maximum absolute error and the number of nodes together with those

obtained with the MATLAB routine ‘heat_CN()’ in Problem 9.4(c) for

comparison. In order to do this job, take the following steps.

(1) Click the button in the tool-bar and click-and-drag on the

graphic region to create a rectangular domain. Then, doubleclick

the rectangle to open the Object dialog box and set

the Left/Bottom/Width/Height to 0/0/pi/0.01 to make a long

rectangular domain.

(cf) Even if the PDEtool is originally designed to deal with only 2-D PDEs,

we can use it to solve 1-D PDEs like (P9.8.8) by proceeding in this way.

(2) Click the ∂ button in the tool-bar, double-click the upper/lower

boundary segments to set the homogeneous Neumann boundary condition

(g = 0, q = 0) and double-click the left/right boundary segments

to set the Dirichlet boundary condition (h = 1, r = 0) as given

by Eq. (P9.8.9).

(3) Open the PDE specification dialog box by clicking the PDE button,

check the box on the left of ‘Parabolic’ as the type of PDE, and set

its parameters in Eq. (9.5.5) as c = 2, a = 0, f = 0 and d = 1, which

corresponds to Eq. (P9.8.8).

(4) Click ‘Parameters’ in the Solve pull-down menu to set the time range,

say, as 0:0.002:0.2 and to set the initial conditions as Eq. (P9.8.9).

(5) In the Plot selection dialog box opened by clicking the button,

check the box before Height and click the Plot button. If you

want to plot the solution graph at a time other than the final time,

select the time for plot from {0, 0.002, 0.004, . . . , 0.2} in the farright

field of the Plot selection dialog box and click the Plot button

again.

(6) If you want to see a movie-like dynamic picture of the solution graph,

check the box before Animation and then click the Plot button in the

Plot selection dialog box.

(7) Click ‘Export Mesh’ in the Mesh pull-down menu, and then click the

OK button in the Export dialog box to extract the mesh data {p,e,t }.

PROBLEMS 459

Also click ‘Export Solution’ in the Solve pull-down menu, and then

click the OK button in the Export dialog box to extract the solution

u. Now, you can estimate how far the graphical/numerical solution

deviates from the true solution (P9.8.10) by typing the following

statements into the MATLAB command window:

>>x = p(1,:)’; y = p(2,: )’; %x,y coordinates of nodes in columns

>>tt = 0:0.01:0.2; %time vector in row

>>err = sin(2*x)*exp(-8*tt)-u; %deviation from true sol.(P9.8-10)

>>err_max = max(abs(err)) %maximum absolute error

(d) Consider the PDE

∂2u(x, t)

∂x2 =

∂2u(x, t)

∂t2 for 0 ≤ x ≤ 10, 0 ≤ t ≤ 10 (P9.8.11)

with the initial/boundary conditions

u(x, 0) = (x − 2)(3 − x) for 2 ≤ x ≤ 3

0 elsewhere ,

∂u

∂t |t=0 = 0 (P9.8.12)

u(0, t) = 0, u(10, t) = 0 (P9.8.13)

Use the PDEtool to make a dynamic picture out of the solution for

this PDE and see if the result is about the same as that obtained in

Problem 9.6(c) in terms of the time when one of the two separated pulses

propagating leftward is reflected and reversed and the time when the two

separated pulses are reunited.

(cf) Even if the PDEtool is originally designed to solve only 2-D PDEs, we can

solve 1-D PDE like (P9.8.11) by proceeding as follows:

(0) In the PDE toolbox window, adjust the ranges of the x axis and the

y axis to [−0.5 10.5] and [−0.01 +0.01], respectively, in the box

opened by clicking ‘Axes Limits’ in the Options pull-down menu.

(1) Click the button in the tool-bar and click-and-drag on the graphic

region to create a long rectangle of domain ranging from x0 = 0 to

xf = 10. Then, double-click the rectangle to open the Object dialog

box and set the Left/Bottom/Width/Height to 0/−0.01/10/0.02.

(2) Click the ∂ button in the tool-bar, double-click the upper/lower

boundary segments to set the homogeneous Neumann boundary condition

(g = 0, q = 0) and double-click the left/right boundary segments

to set the Dirichlet boundary condition (h = 1, r = 0) as given

by Eq. (P9.8.13).

(3) Open the PDE specification dialog box by clicking the PDE button,

check the box on the left of ‘Hyperbolic’ as the type of PDE, and

set its parameters in Eq. (P9.8.11) as c = 1, a = 0, f = 0 and d = 1.

(See Fig. 9.15a.)

460 PARTIAL DIFFERENTIAL EQUATIONS

(4) Click ‘Parameters’ in the Solve pull-down menu to set the time range

to, say, as 0:0.2:10, the boundary condition as (P9.8.13) and the initial

conditions as (P9.8.12). (See Fig. 9.15b and Problem 9.6(c)(ii).)

(5) In the Plot selection dialog box opened by clicking the button,

check the box before ‘Height’ and the box before ‘Animation’ and

then click the Plot button in the Plot selection dialog box to see a

movie-like dynamic picture of the solution graph.

(6) If you want to have better resolution in the solution graph, click Mesh

in the top menu bar and click ‘Refine Mesh’ in the Mesh pull-down

menu. Then, select Plot in the top menu bar or type CTRL + P(∧P)

on the keyboard and click ‘Plot Solution’ in the Plot pull-down menu

to see a smoother animation graph.

(7) In order to estimate the time when one of the two separated pulses

propagating leftward is reflected and reversed and the time when

the two separated pulses are reunited, count the flickering frame

numbers, noting that one flickering corresponds to 0.2 s according

to the time range set in step (4).

(8) If you want to save the PDEtool program, click File in the top menu

bar, click ‘Save As’ in the File pull-down menu, and input the file

name of your choice.

APPENDIX A

MEAN VALUE THEOREM

Theorem A.1. Mean Value Theorem1. Let a function f (x) be continuous on

the interval [a, b] and differentiable over (a, b). Then, there exists at least one

point ξ between a and b at which

f (ξ ) =

f (b) − f (a)

b − a

, f(b)= f (a) + f (ξ )(b − a) (A.1)

In other words, the curve of a continuous function f (x) has the same slope as

the straight line connecting the two end points (a, f (a)) and (b, f (b)) of the

curve at some point ξ ∈ [a, b], as in Fig. A.1.

f (a)

f'(x) = f'(x)

f (b)

f (b) − f (a)

b − a

a x b

Figure A.1 Mean value theorem.

1 See the website @http://www.maths.abdn.ac.uk/∼igc/testing/tch/ma2001/notes/notes.html

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462 MEAN VALUE THEOREM

Theorem A.2. Taylor Series Theorem1. If a function f (x) is continuous and

its derivatives up to order (K + 1) are also continuous on an open interval D

containing some point a, then the value of the function f (x) at any point x ∈ D

can be represented by

f (x) =

K

k=0

f k(a)

k!

(x − a)k + RK+1(x) (A.2)

where the first term of the right-hand side is called the Kth-degree Taylor polynomial,

and the second term called the remainder (error) term is

RK+1(x) =

f (K+1)(ξ )

(K + 1)!

(x − a)K+1 for some ξ between a andx (A.3)

Moreover, if the function f (x) has continuous derivatives of all orders on D,

then the above representation becomes

f (x) =

∞

k=0

f k(a)

k!

(x − a)k (A.4)

which is called the (infinite) Taylor series expansion of f (x) about a.

APPENDIX B

MATRIX

OPERATIONS/PROPERTIES

B.1 ADDITION AND SUBTRACTION

A + B =

a11 a12 · a1N

a21 a22 · a2N

· · · · aM1 aM2 · aMN

+

b11 b12 · b1N

b21 b22 · b2N

· · · · bM1 bM2 · bMN

=

c11 c12 · c1N

c21 c22 · c2N

· · · · cM1 cM2 · cMN

= C (B.1.1)

with

amn + bmn = cmn (B.1.2)

B.2 MULTIPLICATION

AB =

a11 a12 · a1K

a21 a22 · a2K

· · · · aM1 aM2 · aMK

b11 b12 · b1N

b21 b22 · b2N

· · · · bK1 bK2 · bKN

=

c11 c12 ·

c1N

c21 c22 · c2N

· · · · cM1 cM2 · cMN

= C (B.2.1)

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463

464 MATRIX OPERATIONS/PROPERTIES

with

cmn =

K

k=1

amkbkn (B.2.2)

(cf) For this multiplication to be done, the number of columns of A must equal the

number of rows of B.

(cf) Note that the commutative law does not hold for the matrix multiplication, that is,

AB = BA.

B.3 DETERMINANT

The determinant of a K × K (square) matrix A = [amn] is defined by

det(A) = |A| =

K

k=0

akn(−1)k+nMkn or

K

k=0

amk(−1)m+kMmk (B.3.1)

for any fixed 1 ≤ n ≤ K or 1 ≤ m ≤ K

where the minor Mkn is the determinant of the (K − 1) × (K − 1) (minor)

matrix formed by removing the kth row and the nth column from A and Akn = (−1)k+nMkn is called the cofactor of akn.

In particular, the determinants of a 2 ×2 matrix A2×2 and a 3 ×3 matrix

A3×3 are

det(A2×2) =

a11 a12

a21 a22

=

2

k=1

akn(−1)k+nMkn = a11a22 − a12a21 (B.3.2)

det(A3×3) =

a11 a12 a13

a21 a22 a23

a31 a32 a33

= a11

a22 a23

a32 a33

− a12

a21 a23

a31 a33

+ a13

a21 a22

a31 a32

= a11(a22a33 − a23a32) − a12(a21a33 − a23a31) + a13(a21a32 − a22a31)

(B.3.3)

Note the following properties.

ž If the determinant of a matrix is zero, the matrix is singular.

ž The determinant of a matrix equals the product of the eigenvalues of a

matrix.

ž If A is upper/lower triangular having only zeros below/above the diagonal

in each column, its determinant is the product of the diagonal elements.

ž det(AT ) = det(A); det(AB) = det(A)det(B); det(A−1) = 1/det(A)

INVERSE MATRIX 465

B.4 EIGENVALUES AND EIGENVECTORS OF A MATRIX2

The eigenvalue or characteristic value and its corresponding eigenvector or characteristic

vector of an N × N matrix A are defined to be a scalar λ and a nonzero

vector v satisfying

Av = λv ⇔ (A − λI)v = 0 (v = 0) (B.4.1)

where (λ, v) is called an eigenpair and there are N eigenpairs for an N × N

matrix A.

The eigenvalues of a matrix can be computed as the roots of the characteristic

equation

|A − λI| = 0 (B.4.2)

and the eigenvector corresponding to an eigenvalue λi can be obtained by substituting

λi into Eq. (B.4.1) and solve it for v.

Note the following properties.

ž If A is symmetric, all the eigenvalues are real-valued.

ž If A is symmetric and positive definite, all the eigenvalues are real and

positive.

ž If v is an eigenvector of A, so is cv for any nonzero scalar c.

B.5 INVERSE MATRIX

The inverse matrix of a K × K (square) matrix A = [amn] is denoted by A−1

and defined to be a matrix which is premultiplied/postmultiplied by A to form

an identity matrix—that is, satisfies

A × A−1 = A−1 × A =I (B.5.1)

An element of the inverse matrix A−1 = [αmn] can be computed as

αmn =

1

det(A)

Amn =

1

|A|

(−1)m+nMmn (B.5.2)

where Mkn is the minor of akn and Akn = (−1)k+nMkn is the cofactor

of akn.

2 See the website @http://www.sosmath.com/index.html or http://www.psc.edu/∼burkardt/papers/

linear glossary.html.)

466 MATRIX OPERATIONS/PROPERTIES

Note that a square matrix A is invertible/nonsingular if and only if

ž No eigenvalue of A is zero, or equivalently,

ž The rows (and the columns) of A are linearly independent, or equivalently,

ž The determinant of A is nonzero.

B.6 SYMMETRIC/HERMITIAN MATRIX

A square matrix A is said to be symmetric if it is equal to its transpose, that is,

AT ≡A (B.6.1)

A complex-valued matrix is said to be Hermitian if it is equal to its complex

conjugate transpose, that is,

A ≡ A∗T where ∗ means the conjugate. (B.6.2)

Note the following properties of a symmetric/Hermitian matrix.

ž All the eigenvalues are real.

ž If all the eigenvalues are distinct, the eigenvectors can form an orthogonal/

unitary matrix U.

B.7 ORTHOGONAL/UNITARY MATRIX

A nonsingular (square) matrix A is said to be orthogonal if its transpose is equal

to its inverse, that is,

AT A ≡ I, AT ≡ A−1 (B.7.1)

A complex-valued (square) matrix is said to be unitary if its conjugate transpose

is equal to its inverse, that is,

A∗T A ≡ I, A∗T ≡ A−1 (B.7.2)

Note the following properties of an orthogonal/unitary matrix.

ž The magnitude (absolute value) of every eigenvalue is one.

ž The product of two orthogonal matrices is also orthogonal; (AB)∗T (AB) = B∗T (A∗T A)B ≡ I .

ROW ECHELON FORM 467

B.8 PERMUTATION MATRIX

A matrix P having only one nonzero element of value 1 in each row and column

is called a permutation matrix and has the following properties.

ž Premultiplication/postmultiplication of a matrix A by a permutation matrix

P (i.e., PA or AP) yields the row/column change of the matrix A, respectively.

ž A permutation matrix A is orthogonal, that is, AT A ≡ I .

B.9 RANK

The rank of an M × N matrix is the number of linearly independent

rows/columns and if it equals min(M,N), then the matrix is said to be of

maximal or full rank; otherwise, the matrix is said to be rank-deficient or to

have rank-deficiency.

B.10 ROW SPACE AND NULL SPACE

The row space of an M × N matrix A, denoted by R(A), is the space spanned

by the row vectors—that is, the set of all possible linear combinations of row

vectors of A that can be expressed by AT α with an M-dimensional column vector

α. On the other hand, the null space of the matrix A, denoted by N(A), is the

space orthogonal (perpendicular) to the row space—that is, the set of all possible

linear combinations of the N-dimensional vectors satisfying Ax = 0.

B.11 ROW ECHELON FORM

A matrix is said to be of row echelon form if

ž Each nonzero row having at least one nonzero element has a 1 as its first

nonzero element.

ž The leading 1 in a row is in a column to the right of the leading 1 in the

upper row.

ž All-zero rows are below the rows that have at least one nonzero element.

A matrix is said to be of reduced row echelon form if it satisfies the above

conditions and, additionally, each column containing a leading 1 has no other

nonzero elements.

468 MATRIX OPERATIONS/PROPERTIES

Any matrix, singular or rectangular, can be transformed into this form through

the Gaussian elimination procedure (i.e., a series of elementary row operations)

or, equivalently, by using the MATLAB built-in routine “rref()”. For example,

we have

A =

0 0 1 3

2 4 0 −8

1 2 1 −1

row

−−−→ change

2 4 0 −8

0 0 1 3

1 2 1 −1

row

division

−−−−−row →

subtraction

1 2 0 −4

0 0 1 3

0 0 1 3

row

−−−−−→ subtraction

1 2 0 −4

0 0 1 3

0 0 0 0

= rref(A)

Once this form is obtained, it is easy to compute the rank, the determinant and

the inverse of the matrix, if only the matrix is invertible.

B.12 POSITIVE DEFINITENESS

A square matrix A is said to be positive definite if

x∗T Ax > 0 for any nonzero vector x (B.12.1)

A square matrix A is said to be positive semidefinite if

x∗TAx ≥ 0 for any nonzero vector x (B.12.2)

Note the following properties of a positive definite matrix A.

ž A is nonsingular and all of its eigenvalues are positive.

ž The inverse of A is also positive definite.

There are similar definitions for negative definiteness and negative semidefiniteness.

Note the following property, which can be used to determine if a matrix

is positive (semi-) definite or not. A square matrix is positive definite if and

only if:

(i) Every diagonal element is positive.

(ii) Every leading principal minor matrix has positive determinant.

On the other hand, a square matrix is positive semidefinite if and only if:

(i) Every diagonal element is nonnegative.

(ii) Every principal minor matrix has nonnegative determinant.

MATRIX INVERSION LEMMA 469

Note also that the principal minor matrices are the submatrices taking the diagonal

elements from the diagonal of the matrix A and, say for a 3 ×3 matrix, the

principal minor matrices are

a11, a22, a33, a11 a12

a21 a22

, a22 a23

a32 a33

, a11 a13

a31 a33

,

a11 a12 a13

a21 a22 a23

a31 a32 a33

among which the leading ones are

a11, a11 a12

a21 a22

,

a11 a12 a13

a21 a22 a23

a31 a32 a33

B.13 SCALAR (DOT) PRODUCT AND VECTOR (CROSS) PRODUCT

A scalar product of two N-dimensional vectors x and y is denoted by x · y and

is defined by

x · y =

N

n=1

xnyn = xT y (B.13.1)

An outer product of two three-dimensional column vectors x = [x1 x2 x3]T and

y = [y1 y2 y3]T is denoted by x × y and is defined by

x × y =

x2y3 − x3y2

x3y1 − x1y3

x1y2 − x2y1

(B.13.2)

B.14 MATRIX INVERSION LEMMA

Matrix Inversion Lemma. Let A,C, and [C−1 + DA−1B] be well-defined with

nonsingularity as well as compatible dimensions. Then we have

[A + BCD]−1 = A−1 − A−1B[C−1 + DA−1B]−1DA−1 (B.14.1)

Proof. We will show that postmultiplying Eq. (B.14.1) by [A + BCD] yields an

identity matrix.

[A−1 − A−1B[C−1 + DA−1B]−1DA−1][A + BCD]

= I + A−1BCD − A−1B[C−1 + DA−1B]−1D

− A−1B[C−1 + DA−1B]−1DA−1BCD

470 MATRIX OPERATIONS/PROPERTIES

= I + A−1BCD − A−1B[C−1 + DA−1B]−1C−1CD

− A−1B[C−1 + DA−1B]−1DA−1BCD

= I + A−1BCD − A−1B[C−1 + DA−1B]−1[C−1 + DA−1B]CD

= I + A−1BCD − A−1BCD ≡ I

APPENDIX C

DIFFERENTIATION WITH

RESPECT TO A VECTOR

The first derivative of a scalar-valued function f (x) with respect to a vector

x = [x1 x2]T is called the gradient of f (x) and defined as

∇f (x) =

d

dx

f (x) = ∂f/∂x1

∂f/∂x2 (C.1)

Based on this definition, we can write the following equation.

∂

∂x

xT y =

∂

∂x

yT x =

∂

∂x

(x1y1 + x2y2) = y1

y2 = y (C.2)

∂

∂x

xT x =

∂

∂x

(x2

1 + x2

2 ) = 2 x1

x2 = 2x (C.3)

Also with an M × N matrix A, we have

∂

∂x

xTAy =

∂

∂x

yTAT x = Ay (C.4a)

∂

∂x

yTAx =

∂

∂x

xTAT y = AT y (C.4b)

where

xT Ay =

M

m=1

N

n=1

amnxmyn (C.5)

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472 DIFFERENTIATION WITH RESPECT TO A VECTOR

Especially for a square, symmetric matrix A with M = N, we have

∂

∂x

xTAx = (A + AT )x

if A is symmetric

−−−−−−−−−→ 2Ax (C.6)

The second derivative of a scalar function f (x) with respect to a vector x = [x1 x2]T is called the Hessian of f (x) and is defined as

H(x) = ∇2f (x) =

d2

dx2 f (x) = ∂2f/∂x2

1 ∂2f/∂x1∂x2

∂2f/∂x2∂x1 ∂2f/∂x2

2 (C.7)

Based on this definition, we can write the following equation:

d2

dx2 xT Ax = A + AT

if A is symmetric

−−−−−−−−−→ 2A (C.8)

On the other hand, the first derivative of a vector-valued function f(x) with

respect to a vector x = [x1 x2]T is called the Jacobian of f (x) and is defined as

J(x) =

d

dx

f(x) = ∂f1/∂x1 ∂f1/∂x2

∂f2/∂x1 ∂f2/∂x2 (C.9)

APPENDIX D

LAPLACE TRANSFORM

Table D.1 Laplace Transforms of Basic Functions

x(t) X(s) x(t) X(s) x(t) X(s)

(1) δ(t) 1 (5)e−atus (t)

1

s + a

(9) e−at sin ωt us (t)

ω

(s + a)2 + ω2

(2) δ(t − t1) e−t1s (6) tme−atus (t)

m!

(s + a)m+1 (10) e−at cos ωt us (t)

s + a

(s + a)2 + ω2

(3) us (t)

1

s

(7) sin ωt us (t)

ω

s2 + ω2

(4) tmus (t)

m!

sm+1 (8) cos ωt us (t)

s

s2 + ω2

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474 LAPLACE TRANSFORM

Table D.2 Properties of Laplace Transform

(0) Definition X(s) = L{x(t)} = ∞

0

x(t)e−st dt

(1) Linearity αx(t) + βx(t) → αX(s) + βY(s)

(2) Time shifting x(t − t1)us (t − t1), t1 > 0 → e−st1 X(s) + 0

−t1

x(τ)e−sτ dτ

(3) Frequency shifting es1tx(t) → X(s − s1)

(4) Real convolution g(t) ∗ x(t) → G(s)X(s)

(5) Time derivative x(t ) → sX(s) − x(0)

(6) Time integral t

−∞

x(τ) dτ →

1

s

X(s) +

1

s 0

−∞

x(τ) dτ

(7) Complex derivative t x(t)→−

d

ds

X(s)

(8) Complex convolution x(t)y(t) →

1

2πj σ0+∞

σ0−∞

X(v)Y(s − v) dv

(9) Initial value theorem x(0) → lim

s→∞

sX(s)

(10) Final value theorem x(∞) → lim

s→0

sX(s)

APPENDIX E

FOURIER TRANSFORM

Table E.1 Properties of CtFT (Continuous-Time Fourier Transform)

(0) Definition X(ω) = F{x(t)} = ∞

−∞

x(t)e−jωt dt

(1) Linearity αx(t) + βx(t) → αX(ω) + βY(ω)

(2) Symmetry x(t) = xe(t) + xo(t): real → X(ω) ≡ X∗(−ω)

xe(t): real and even → Xe(ω) = Re{X(ω)}

xo(t): real and odd → Xo(ω) = jIm{X(ω)}

x(−t) → X(−ω)

(3) Time shifting x(t − t1)→ e−jωt1X(ω)

(4) Frequency shifting ejω1tx(t) → X(ω − ω1)

(5) Real convolution g(t) ∗ x(t) = ∞

−∞

g(τ)x(t − τ) dτ → G(ω)X(ω)

(6) Time derivative x(t) → jωX(ω)

(7) Time integral t

−∞

x(τ) dτ →

1

jω

X(ω) + πX(0)δ(ω)

(8) Complex derivative t x(t) → j

d

dω

X(ω)

(9) Complex convolution x(t)y(t) →

1

2π

X(ω) ∗ Y(ω)

(10) Scaling x(at) →

1

|a|

X(ω/a)

(11) Duality g(t) → f (ω) ⇔ f (t) → 2πg(ω)

(12) Parseval’s relation ∞

−∞ |x(t)|2 dt →

1

2π ∞

−∞ |x(ω)|2 dω

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475

476 FOURIER TRANSFORM

Table E.2 Properties of DtFT (Discrete-Time Fourier Transform)

(0) Definition X() =

∞

n=−∞

x[n]e−jn

(1) Linearity αx[n] + βx[n]→ αX() + βY()

(2) Symmetry x[n] = xe[n] + xo[n]: real → X() ≡ X∗(−)

xe[n]: real and even → Xe() = Re{X()}

xo[n]: real and odd → Xo() = jIm{X()}

x[−n]→ X(−)

(3) Time shifting x[n − n1]→ e−jn1X()

(4) Frequency shifting ej1nx[n]→ X( − 1)

(5) Real convolution g[n] ∗ x[n] =

∞

m=−∞

g[m]x[n − m]→ G()X()

(6) Complex derivative n x[n] → j

d

d

X()

(7) Complex convolution x[n]y[n] →

1

2π

X() ∗ Y() (periodic/circular convolution)

(8) Scaling x[n/M] ifn = mM(m : an integer)

0, otherwise → X(M)

(9) Parseval’s relation

∞

n=−∞

|x[n]|2 =

1

2π 2π |x()|2 d

APPENDIX F

USEFUL FORMULAS

Formulas for Summation of Finite Number of Terms

N

n=0

an =

1 − aN+1

1 − a

(F.1)

N

n=0

nan = a

1 − (N + 1)aN + NaN+1

(1 − a)2 (F.2)

N

n=0

n =

N(N + 1)

2

(F.3)

N

n=0

n2 =

N(N + 1)(2N + 1)

6

(F.4)

N

n=0

n(n + 1) =

N(N + 1)(N + 2)

3

(F.5)

(a + b)N =

N

n=0

NCnaN−nbn with NCn = NCN−n =

NPn

n! =

N!

(N − n)!n!

(F.6)

Formulas for Summation of Infinite Number of Terms

∞

n=0

xn =

1

1 − x

, |x| < 1 (F.7)

∞

n=0

nxn =

x

(1 − x)2 , |x| <1 (F.8)

∞

n=0

nkxn = lim

a→0

(−1)k ∂k

∂ak x

x − e−a , |x| <1 (F.9)

∞

n=0

(−1)n

2n + 1 = 1 −

1

3 +

1

5 −

1

7 +· · · =

1

4

π (F.10)

(continued overleaf )

Applied Numerical Methods Using MATLAB, by Yang, Cao, Chung, and Morris

Copyr ight 2005 John Wiley & Sons, I nc., ISBN 0-471-69833-4

477

478 USEFUL FORMULAS

∞

n=0

1

n2 = 1 +

1

22 +

1

32 +

1

42 +· · · =

1

6

π2 (F.11)

ex =

∞

n=0

1

n!

xn = 1 +

1

1!

x +

1

2!

x2 +

1

3!

x3 +· · · (F.12)

ax =

∞

n=0

(ln a)n

n!

xn = 1 +

ln a

1!

x +

(ln a)2

2!

x2 +

(ln a)3

3!

x3 +· · · (F.13)

ln(1 ± x) = −

∞

n=1

(±1)n 1

n

xn = ±x −

1

2

x2 ±

1

3

x3 −· · · , |x| < 1 (F.14)

sin x =

∞

n=0

(−1)n

(2n + 1)!

x2n+1 = x −

1

3!

x3 +

1

5!

x5 −

1

7!

x7 +· · · (F.15)

cos x =

∞

n=0

(−1)n

(2n)!

x2n = 1 −

1

2!

x2 +

1

4!

x4 −

1

6!

x6 +· · · (F.16)

tan x = x +

1

3

x3 +

2

15

x5 +· · · , |x| <

π

2

(F.17)

tan−1 x =

∞

n=0

(−1)n

2n + 1

x2n+1 = x −

1

3

x3 +

1

5

x5 −

1

7

x7 +· · · (F.18)

Trigonometric Formulas

sin(A ± B) = sinAcosB ± cosAsinB (F.19) tan(A ± B) =

tanA ± tan B

1 ∓ tanAtanB

(F.21)

cos(A ± B) = cosAcosB ∓ sinAsinB (F.20)

sinAsinB =

1

2 {cos(A − B) − cos(A + B)} (F.22)

sinAcosB =

1

2 {sin(A + B) + sin(A − B)} (F.23)

cosAsin B =

1

2 {sin(A + B) − sin(A − B)} (F.24)

cosAcos B =

1

2 {cos(A + B) + cos(A − B)} (F.25)

sinA + sin B = 2 sin A + B

2 cosA − B

2 (F.26)

cosA + cos B = 2 cos A + B

2 cosA − B

2 (F.27)

a cosA − b sinA = √a2 + b2 cos(A + θ), θ = tan−1 b

a (F.28)

a sinA + b cosA = √a2 + b2 sin(A + θ), θ = tan−1 b

a (F.29)

sin2 A =

1

2

(1 − cos 2A) (F.30) cos2 A =

1

2

(1 + cos 2A) (F.31)

USEFUL FORMULAS 479

sin3 A =

1

4

(3 sinA − sin 3A) (F.32) cos3 A =

1

4

(3 cosA + cos 3A) (F.33)

sin 2A = 2 sinAcosA (F.34) sin 3A = 3 sinA − 4 sin3 A (F.35)

cos 2A = cos2 A − sin2 A = 1 − 2 sin2 A = 2 cos2 A − 1 (F.36)

cos 3A = 4 cos3 A − 3 sinA (F.37)

a

sinA =

b

sin B =

c

sinC

(F.38) e±jθ = cos θ ± j sin θ (F.40)

a2 = b2 + c2 − 2bc cosA (F.39a) sin θ =

1

j2

(ejθ − e−jθ ) (F.41a)

b2 = c2 + a2 − 2ca cosB (F.39b) cos θ =

1

2

(ejθ + e−jθ ) (F.41b)

c2 = a2 + b2 − 2ab cosC (F.39c) tan θ =

1

j

ejθ − e−jθ

ejθ + e−jθ

(F.41c)

APPENDIX G

SYMBOLIC COMPUTATION

G.1 HOW TO DECLARE SYMBOLIC VARIABLES AND HANDLE

SYMBOLIC EXPRESSIONS

To declare any variable(s) as a symbolic variable, you should use the sym or

syms command as below.

>>a = sym(’a’); t = sym(’t’); x = sym(’x’);

>>syms a x y t %or, equivalently and more efficiently

Once the variables have been declared as symbolic, they can be used in expressions

and as arguments to many functions without being evaluated as numeric.

>>f = x^2/(1 + tan(x)^2);

>>ezplot(f,-pi,pi)

>>simplify(cos(x)^2+sin(x)^2) %simplify an expression

ans = 1

>>simplify(cos(x)^2 – sin(x)^2) %simplify an expression

ans = 2*cos(x)^2-1

>>simple(cos(x)^2 – sin(x)^2) %simple expression

ans = cos(2*x)

>>simple(cos(x) + i*sin(x)) %simple expression

ans = exp(i*x)

>>eq1 = expand((x + y)^3 – (x + y)^2) %expand

eq1 = x^3 + 3*x^2*y + 3*x*y^2 + y^3 – x^2 – 2*x*y – y^2

>>collect(eq1,y) %collect similar terms in descending order with respect to y

ans = y^3 + (3*x – 1)*y^2 + (3*x^2 – 2*x)*y + x^3 – x^2

Applied Numerical Methods Using MATLAB, by Yang, Cao, Chung, and Morris

Copyright 2005 John Wiley & Sons, Inc., ISBN 0-471-69833-4

481

482 SYMBOLIC COMPUTATION

>>factor(eq1) %factorize

ans = (x + y – 1)*(x + y)^2

>>horner(eq1) %nested multiplication form

ans = (-1 + y)*y^2 + ((- 2 + 3*y)*y + (-1 + 3*y + x)*x)*x

>>pretty(ans) %pretty form

2

(-1 + y) y + ((-2 + 3 y) y + (-1 + 3 y + x) x) x

If you need to substitute numeric values or other expressions for some symbolic

variables in an expression, you can use the subs function as below.

>>subs(eq1,x,0) %substitute numeric value

ans = -y^2 + y^3

>>subs(eq1,{x,y},{0,x – 1}) %substitute numeric values

ans = (x – 1)^3 – (x – 1)^2

The sym command allows you to declare symbolic real variables by using the

‘real’ option as illustrated below.

>>x = sym(’x’,’real’); y = sym(’y’,’real’);

>>syms x y real %or, equivalently

>>z = x + i*y; %declare z as a symbolic complex variable

>>conj(z) %complex conjugate

ans = x – i*y

>>abs(z)

ans = (x^2 + y^2)^(1/2) %equivalently

The sym function can be used to convert numeric values into their symbolic

expressions.

>>sym(1/2) + 0.2

ans = 7/10 %symbolic expression

On the other hand, the double command converts symbolic expressions into

their numeric (double-precision floating-point) values and the vpa command finds

the variable-precision arithmetic (VPA) expression (as a symbolic representation)

of a numeric or symbolic expression with d significant decimal digits, where d

is the current setting of DIGITS that can be set by the digits command. Note

that the output of the vpa command is a symbolic expression even if it may look

like a numeric value. Let us see some examples.

>>f = sym(’exp(i*pi/4)’)

f = exp(i*pi/4)

>>double(f)

ans = 0.7071 + 0.7071i %numeric value

>>vpa(ans,2)

ans = .71 + .71*i %symbolic expression with 2 significant digits

CALCULUS 483

G.2 CALCULUS

G.2.1 Symbolic Summation

We can use the symsum() function to obtain the sum of an indefinite/definite

series as below.

>>syms x n N %declare x,n,N as symbolic variables

>>simple(symsum(n,0,N))

ans = 1/2*N*(N + 1) %Nn

=0 n =

N(N + 1)

2

>>simple(symsum(n^2,0,N))

ans = 1/6*N*(N + 1)*(2*N + 1) %Nn

=0 n2 =

N(N + 1)(2N + 1)

6

>>symsum(1/n^2,1,inf))

ans = 1/6*pi^2 %N

n=0

1

n2 =

π2

6

>>symsum(x^n,n,0,inf))

ans = -1/(-1 + x) %Nn

=0 x n =

1

1 − x

under the assumption that |x| < 1

G.2.2 Limits

We can use the limit() function to get the (two-sided) limit and the right/leftsided

limits of a function as below.

>>syms h n x

>>limit(sin(x)/x,x,0) % lim

x→0

sin x

x = 1

ans = 1

>>limit(x/abs(x),x,0,’right’) % lim

x→0+

x

| x| = 1

ans = 1

>>limit(x/abs(x),x,0,’left’) % lim

x→0−

x

|x| = −1

ans = -1

>>limit(x/abs(x),x,0) % lim

x→0

x

|x| = ?

ans = NaN %Not a Number

>>limit((cos(x+h)-cos(x))/h,h,0) % lim

h→0

cos(x + h) − cos(x)

h =

d

dx

cosx = −sinx

ans = -sin(x)

>>limit((1 + x/n)^n,n,inf) % lim

n→∞1 +

x

n n

= ex

ans = exp(x)

G.2.3 Differentiation

The diff() function differentiates a symbolic expression w.r.t. the variable given

as one of its 2nd or 3rd input arguments or its free variable which might be

determined by using the findsym function.

484 SYMBOLIC COMPUTATION

>>syms a b x n t

>>diff(x^n))

ans = x^n*n/x

>>simplify(ans)

ans = x^(n – 1)*n

>>f = exp(a*x)*cos(b*t)

>>diff(f) %equivalently diff(f,x)

ans = a*exp(a*x)*cos(b*t) %

d

dx

f =

d

dx

eax cos (bt ) = aeax cos (bt )

>>diff(f,t)

ans = -exp(a*x)*sin(b*t)*b %

d

dt

f =

d

dt

eax cos (bt ) = −beax sin (bt )

>>diff(f,2) %equivalently diff(f,x,2)

ans = a^2*exp(a*x)*cos(b*t) %

d2

dx2 f = a2eax cos (bt )

>>diff(f,t,2)

ans = -exp(a*x)*cos(b*t)*b^2 %

d2

dt 2 f = −eax cos (bt )b2

>>g = [cos(x)*cos(t) cos(x)*sin(t)];

>>jacob g = jacobian(g,[x t])

jacob g = [ -sin(x)*cos(t), -cos(x)*sin(t)]

[ -sin(x)*sin(t), cos(x)*cos(t)]

Note that the jacobian() function finds the jacobian defined by (C.9)—that is,

the derivative of a vector function [g1 g2]T with respect to a vector variable

[x t]T —as

J = ∂g1/∂x ∂g1/∂t

∂g2/∂x ∂g2/∂t (G.1)

G.2.4 Integration

The int() function returns the indefinite/definite integral (anti-derivative) of a

function or an expression with respect to the variable given as its second input

argument or its free variable which might be determined by using the findsym

function.

>>syms a x y t

>>int(x^n)

ans = x^(n + 1)/(n + 1) % xn dx =

1

n + 1

xn + 1

>>int(1/(1 + x^2))

ans = atan(x) %

1

1 + x2 dx = tan− 1 x

>>int(a^x) %equivalently diff(f,x,2)

ans = 1/log(a)*a^x % ax dx =

1

log a

ax

>>int(sin(a*t),0,pi) %equivalently int(sin(a*t),t,0,pi)

ans = -cos(pi*a)/a + 1/a %π

0 sin (at ) dt = −

1

a

cos (at )

π

0 = −

1

a

cos (aπ ) +

1

a

CALCULUS 485

>>int(exp(-(x – a)^2),a,inf) %equivalently int(exp(-(x – a)^2),x,0,inf)

ans = 1/2*pi^(1/2) % ∞ a e − (x − a)2 dx = ∞ 0 e − x2 dx =

1

2

√π

G.2.5 Taylor Series Expansion

We can use the taylor() function to find the Taylor series expansion of a

function or an expression with respect to the variable given as its second or

third input argument or its free variable that might be determined by using the

findsym function.

One may put ‘help taylor’ into the MATLAB command window to see its

usage, which is restated below. Let us try applying it.

>>syms x t; N = 3;

>>Tx0 = taylor(exp(-x),N + 1) %f (x) ˜= N

n = 0

1

n!

f (n)(0) xn

Tx0 = 1-x + 1/2*x^2 – 1/6*x^3

>>sym2poly(Tx0) %extract the coefficients of Taylor series polynomial

ans = -0.1667 0.5000 -1.0000 1.0000

>>xo = 1; Tx1 = taylor(exp(-x),N + 1,xo) %f (x) ˜= Nn

= 0

1

n !

f (n)(x0) (x − x0)n

Tx1 = exp(-1) – exp(-1)*(x – 1) + 1/2*exp(-1)*(x – 1)^2 – 1/6*exp(-1)*(x – 1)^3

>>pretty(Tx1)

2 3

exp(-1) -exp(-1)(x – 1) +1/2 exp(-1)(x – 1) -1/6 exp(-1)(x – 1)

>>f = exp(-x)*sin(t);

>>Tt = taylor(f,N + 1,t) %f (x) ˜= N

n = 0

1

n!

f (n)(0)t n

Tt = exp(-x)*t – 1/6*exp(-x)*t^3

ž taylor(f) gives the fifth-order Maclaurin series expansion of f.

ž taylor(f,n+1) with an integer n > 0 gives the nth-order Maclaurin series

expansion of f.

ž taylor(f,a) with a real number (a) gives the fifth-order Taylor series

expansion of f about a.

ž taylor(f,n + 1,a) gives the nth-order Taylor series expansion of f about

default variable=a.

ž taylor(f,n + 1,a,y) gives the nth-order Taylor series expansion of f(y)

about y = a.

(cf) The target function f must be a legitimate expression given directly as the first

input argument.

(cf) Before using the command “taylor()”, one should declare the arguments of the

function as symbols by putting, say, “syms x t”.

(cf) In case the function has several arguments, it is a good practice to put the independent

variable as the last input argument of “taylor()”, though taylor() takes

486 SYMBOLIC COMPUTATION

one closest (alphabetically) to ‘x’ as the independent variable by default only if

it has been declared as a symbolic variable and is contained as an input argument

of the function f.

(cf) One should use the MATLAB command “sym2poly()” if he wants to extract the

coefficients from the Taylor series expansion obtained as a symbolic expression.

G.3 LINEAR ALGEBRA

Several MATLAB commands and functions can be used to manipulate the vectors

or matrices consisting of symbolic expressions as well as those consisting

of numerics.

>>syms a11 a12 a21 a22

>>A = [a11 a12; a21 a22];

>>det(A)

ans = a11*a22 – a12*a21

>>AI = A^ – 1

AI = [ a22/(a11*a22 – a12*a21), -a12/(a11*a22 – a12*a21)]

[ -a21/(a11*a22 – a12*a21), a11/(a11*a22 – a12*a21)]

>>A*AI

ans = [ a11*a22/(a11*a22 – a12*a21)-a12*a21/(a11*a22 – a12*a21), 0]

[ 0, a11*a22/(a11*a22 – a12*a21) – a12*a21/(a11*a22 – a12*a21)]

>>simplify(ans) %simplify an expression

ans = [ 1, 0]

[ 0, 1]

>>syms x t;

>>G = [cos(t) sin(t); -sin(t) cos(t)] %The Givens transformation matrix

G = [ cos(t), sin(t)]

[ -sin(t), cos(t)]

>>det(G), simple(ans)

ans = cos(t)^2 + sin(t)^2

ans = 1

>>G2 = G^2, simple(G2)

G2 = [ cos(t)^2 – sin(t)^2, 2*cos(t)*sin(t)]

[ -2*cos(t)*sin(t), cos(t)^2 – sin(t)^2]

ans = [ cos(2*t), sin(2*t)]

[ -sin(2*t), cos(2*t)]

>>GTG = G.’*G, simple(GTG)

GTG = [ cos(t)^2 + sin(t)^2, 0]

[ 0, cos(t)^2 + sin(t)^2]

ans = [ 1, 0]

[ 0, 1]

>>simple(G^ – 1) %inv(G) for the inverse of Givens transformation matrix

G = [ cos(t), -sin(t)]

[ sin(t), cos(t)]

>>syms b c

>>A = [0 1; -c -b];

>>[V,E] = eig(A)

V = [ -(1/2*b + 1/2*(b^2 – 4*c)^(1/2))/c, -(1/2*b – 1/2*(b^2 – 4*c)^(1/2))/c]

[ 1, 1]

E = [ -1/2*b + 1/2*(b^2 – 4*c)^(1/2), 0]

[ 0, -1/2*b – 1/2*(b^2 – 4*c)^(1/2)]

>> solve(poly(A))%another way to get eigenvalues(characteristic roots)

ans = [ -1/2*b+1/2*(b^2 – 4*c)^(1/2)]

[ -1/2*b-1/2*(b^2 – 4*c)^(1/2)]

SOLVING DIFFERENTIAL EQUATIONS 487

Besides, other MATLAB functions such as jordan(A) and svd(A) can be

used to get the Jordan canonical form together with the corresponding similarity

transformation matrix and the singular value decomposition of a symbolic matrix.

G.4 SOLVING ALGEBRAIC EQUATIONS

We can use the backslash (\) operator to solve a set of linear equations written

in a matrix–vector form.

>>syms R11 R12 R21 R22 b1 b2

>>R = [R11 R12; R21 R22]; b = [b1; b2];

>>x = R\b

x = [ (R12*b2 – b1*R22)/(-R11*R22 + R21*R12)]

[ (-R11*b2 + R21*b1)/(-R11*R22 + R21*R12)]

We can also use the MATLAB function solve() to solve symbolic algebraic

equations.

>>syms a b c x

>>fx = a*x^2+b*x+c;

>>solve(fx) %formula for roots of 2nd-order polynomial eq

ans = [ 1/2/a*(-b + (b^2 – 4*a*c)^(1/2))]

[ 1/2/a*(-b – (b^2 – 4*a*c)^(1/2))]

>>syms x1 x2 b1 b2

>>fx1 = x1 + x2 – b1; fx2 = x1 + 2*x2 – b2; %a system of simultaneous algebraic eq.

>>[x1o,x2o] = solve(fx1,fx2) %

x1o = 2*b1 – b2

x2o = -b1 + b2

G.5 SOLVING DIFFERENTIAL EQUATIONS

We can use the MATLAB function dsolve() to solve symbolic differential

equations.

>>syms a b c x

>>xo = dsolve(’Dx + a*x = 0’) % a differential eq.(d.e.) w/o initial condition

xo = exp(-a*t)*C1 % a solution with undetermined constant

>>xo = dsolve(’Dx + a*x = 0’,’x(0) = 2’) % a d.e. with initial condition

xo = 2*exp(-a*t) % a solution with undetermined constant

>>xo = dsolve(’Dx=1+x^2’) % a differential eq. w/o initial condition

xo = tan(t – C1) % a solution with undetermined constant

>>xo = dsolve(’Dx = 1 + x^2’,’x(0) = 1’) % with the initial condition

xo = tan(t + 1/4*pi) % a solution with determined constant

>>yo = dsolve(’D2u = -u’,’t’) % a 2nd-order d.e. without initial condition

yo = C1*sin(t) + C2*cos(t)

>>xo = dsolve(’D2u = -u’,’u(0) = 1,Du(0) = 0’,’t’) % with the initial condition

xo = cos(t))

>>yo = dsolve(’(Dy)^2 + y^2 = 1’,’y(0) = 0’,’x’) % a 1st-order nonlinear d.e.(nlde)

yo = [ sin(x)] %two solutions

[ -sin(x)]

>>yo = dsolve(’D2y = cos(2*x) – y’,’y(0) = 1,Dy(0) = 0’,’x’) % a 2md-order nlde

yo = 4/3*cos(x) – 2/3*cos(x)^2 + 1/3

>>S = dsolve(’Df=3*f + 4*g’,’Dg=-4*f + 3*g’);

488 SYMBOLIC COMPUTATION

>>f = S.f, g = S.g

f = exp(3*t)*(C1*sin(4*t) + C2*cos(4*t))

g = exp(3*t)*(C1*cos(4*t) – C2*sin(4*t))

>>[f,g] = dsolve(’Df = 3*f + 4*g,Dg = -4*f + 3*g’,’f(0) = 0,g(0) = 1’)

f = exp(3*t)*sin(4*t)

g = exp(3*t)*cos(4*t)

APPENDIX H

SPARSE MATRICES

A matrix is said to be sparse if it has a large portion of zero elements. MATLAB

has some built-in functions/routines that enable us to exploit the sparsity of a

matrix for computational efficiency.

The MATLAB routine sparse() can be used to convert a (regular) matrix

into a sparse form by squeezing out any zero elements and to generate a sparse

matrix having the elements of a vector given together with the row/column index

vectors. On the other hand, the MATLAB routine full() can be used to convert

a matrix of sparse form into a regular one.

>>row_index = [1 1 2 3 4]; col_index = [1 2 2 3 4]; elements = [1 2 3 4 5];

>>m = 4; n = 4; As = sparse(row_index,col_index,elements,m,n)

As = (1,1) 1

(1,2) 2

(2,2) 3

(3,3) 4

(4,4) 5

>>Af = full(As)

Af= 1 2 0 0

0 3 0 0

0 0 4 0

0 0 0 5

We can use the MATLAB routine sprandn(m,n,nzd) to generate an m × n

sparse matrix having the given non-zero density nzd. Let us see how efficient

the operations can be on the matrices in sparse forms.

>>As = sprandn(10,10,0.2); %a sparse matrix and

>>Af = full(As); its full version

Applied Numerical Methods Using MATLAB, by Yang, Cao, Chung, and Morris

Copyr ight 2005 John Wiley & Sons, I nc., ISBN 0-471-69833-4

489

490 SPARSE MATRICES

>>flops(0), AsA = As*As; flops %in sparse forms

ans = 50

>>flops(0), AfA = Af*Af; flops %in full(regular) forms

ans = 2000

>>b = ones(10,1); flops(0), x = As\b; flops

ans = 160

>>flops(0), x = Af\b; flops

ans = 592

>>flops(0), inv(As); flops

ans = 207

>>flops(0), inv(Af); flops

ans = 592

>>flops(0), [L,U,P] = lu(As); flops

ans = 53

>>flops(0), [L,U,P] = lu(Af); flops

ans = 92

Additionally, the MATLAB routine speye(n) is used to generate an n × n

identity matrix and the MATLAB routine spy(n) is used to visualize the sparsity

pattern. The computational efficiency of LU factorization can be upgraded if

one pre-orders the sparse matrix by the symmetric minimum degree permutation,

which is cast into the MATLAB routine symmmd().

Interest readers are welcome to run the following program “do_sparse” to

figure out the functions of several sparsity-related MATLAB routines.

%do_sparse

clear, clf

%create a sparse mxn random matrix

m = 4; n = 5; A1 = sprandn(m,n,.2)

%create a sparse symmetric nxn random matrix with non-zero density nzd

nzd = 0.2; A2 = sprandsym(n,nzd)

%create a sparse symmetric random nxn matrix with condition number r

r = 0.1; A3 = sprandsym(n,nzd,r)

%a sparse symmetric random nxn matrix with the set of eigenvalues eigs

eigs = [0.1 0.2 .3 .4 .5]; A4=sprandsym(n,nzd,eigs)

eig(A4)

tic, A1A = A1*A1’, time_sparse = toc

A1f = full(A1); tic, A1Af = A1f*A1f’; time_full = toc

spy(A1A), full(A1A), A1Af

sparse(A1Af)

n = 10; A5 = sprandsym(n,nzd)

tic, [L,U,P] = lu(A5); time_lu = toc

tic, [L,U,P] = lu(full(A5)); time_full = toc

mdo = symmmd(A5); %symmetric minimum degree permutation

tic, [L,U,P] = lu(A5(mdo,mdo)); time_md=toc

(cf) The command ‘flops’ is not available in MATLAB of version 6.x and that is why we

use ‘tic’ and ‘toc’ to count the process time instead of the number of floating-point

operations.

APPENDIX I

MATLAB

First of all, the following should be noted:

1. The index of an array in MATLAB starts from 1, not 0.

2. A dot(.) must be put before an operator to make a termwise (element-byelement)

operation.

Some of useful MATLAB commands are listed in Table I.1.

Table I.1 Commonly Used Commands and Functions in MATLAB

General Commands

break to exit from a for or while loop

fprintf fprintf(‘\n x(%d) = %6.4f \a’,ind,x(ind))

keyboard stop execution until the user types any key

return terminate a routine and go back to the calling routine

load *** x y read the values of x and y from the MATLAB file

***.mat

load x.dat read the value(s) of x from the ASCII file x.dat

save *** x y save the values of x and y into the MATLAB file

***.mat

save x.dat x save the value(s) of x into the ASCII file x.dat

clear remove all or some variables/functions from memory

Two-Dimensional Graphic Commands

bar(x,y),plot(x,y),stairs(x,y)

stem(x,y),loglog(x,y)

semilogx(x,y),semilogy(x,y)

plot the values of y versus x in a bar\continuous

\stairs\discrete\xy-log\x-log\y-log graph

Applied Numerical Methods Using MATLAB, by Yang, Cao, Chung, and Morris

Copyright 2005 John Wiley & Sons, Inc., ISBN 0-471-69833-4

491

492 MATLAB

Table I.1 Commonly Used Commands and Functions in MATLAB

plot(y) (y: read-valued) plot the values of vector\array over the index

plot(y) (y: complex-valued) plot the imaginary part versus the real part:

plot(real(y),imag(y))

bar(y, s1s2s3)

plot(y, s1s2s3)

The string of three characters s1s2s3, given as one of the

input arguments to these graphic commands specifies the

color, the symbol, and the line types:

stairs(y, s1s2s3)

stem(y, s1s2s3)

s1(color): y(ellow), m(agenta), c(yan), r(ed), g(reen),

b(lue), w(hite), (blac)k

loglog(y, s1s2s3)

semilogx(y, s1s2s3)

s2(symbol):.(point), o,x,+,*, s(quare: ), d(iamond:♦),

v(), ˆ(), <(), >(), p(entagram:✩), h(exagram)

semilogy(y, s1s2s3)

plot(y1, s1s2s3, y2, s1s2s3)

s3(line symbol): -(solid, default), :(dotted),

-.(dashdot),–(dashed)

(ex) plot(x,’b+:’) plots x(n) with the + symbols on

a blue dotted line

polar(theta,r) plot the graph in polar form with the phase theta and

magnitude r

Auxiliary Graphic Commands

axis([xmin xmax ymin ymax]) specify the ranges of graph on horizontal/vertical axes

clf(clear figure) clear the existent graph(s)

grid on/off draw/remove the grid lines

hold on/off keep/remove the existent graph(s)

subplot(ijk) divide the screen into i × j sections and use the kth one

text(x,y,plot(y,‘***’) print the string ‘***’ in the position (x,y) on the graph

title(‘**’), xlabel(‘**’),

ylabel(‘**’)

print the string ‘**’ into the top/low/left side of graph

Three-Dimensional Graphic Commands

mesh(X,Y, Z) connect the points of height Z at points (X,Y) where

X,Y and Z are the matrices of the same dimension

mesh(x, y, Z) connect the points of height Z(j, i) at points specified by

the two vectors (x(i),y(j))

mesh(Z), surf(), plot3(),

contour()

connect the points of height Z(j, i) at points specified by

(i, j)

Once you installed MATLAB, you can click the icon like the one in the left side

to run MATLAB. Then you will see the MATLAB command window

on your monitor as depicted in Fig. I.1, where a cursor appears

(most likely blinking) to the right of the prompt like ‘>>’ or

‘?’ waiting for you to type in a command. If you are running

MATLAB of version 6.x, the main window has not only the command window,

but also the workspace box and the command history box on the left-up/down

side of the command window, in which you can see the contents of MATLAB

MATLAB 493

Figure I.1 The MATLAB command window with the workspace box and the command box.

memory and the commands you have typed into the Command window up to

the present time, respectively. You might clear the boxes by clicking the corresponding

submenu under the ‘Edit’ menu and even remove/restore them by

un-checking/checking the corresponding submenu under the ‘View’ menu.

How do we work with the MATLAB command window?

ž By clicking ‘File’ on the top menu and then ‘New’/‘Open’ in the File pulldown

menu, you can create/edit any file with the MATLAB editor.

ž By clicking ‘File’ on the top menu and then ‘Set Path’ in the File pull-down

menu, you can make the MATLAB search path include/exclude the paths

containing the files you want to be run.

ž If you are a beginner in MATLAB, then it may be worthwhile to click ‘Help’

on the top menu, click ‘Demos’ in the Help pull-down menu, (double-)click

any topic that you want to learn, and watch the visual explanation about it.

ž By typing any MATLAB commands/statements in the MATLAB command

window, you can use various powerful mathematic/graphic functions

of MATLAB.

ž If you have an m-file that contains a series of commands/statements composed

for performing your job, you can type in the file name (without the

extension ‘.m’) to make it run.

It is helpful to know the procedure of debugging in MATLAB, which is

summarized below.

1. With the program (you want to edit) loaded into the MATLAB Editor/

Debugger window, set breakpoint(s) at any statement(s) which you think

494 MATLAB

Figure I.2 The MATLAB file editor/debugger window.

is (are) suspicious to be the source(s) of error, by clicking the pertinent

statement line of the program with the left mouse button and pressing the

F12 key or clicking ‘Set/Clear Breakpoint’ in the ‘Breakpoints’ pull-down

menu of the Editor/Debugger window. Then, you will see a small red disk

in front of every statement at which you set the breakpoint.

2. Going to the MATLAB Command window, type in the name of the file

containing the main program to try running the program. Then, go back to

the Editor/Debugger window and you will see the cursor blinking just after

a green arrow between the red disk and the first statement line at which

you set the breakpoint.

3. Determining which variable to look into, go to the Command window

and type in the variable name(s) (just after the prompt ‘K>>’) or whatever

statement you want to run for debugging.

4. If you want to proceed to the next statement line in the program, go back

to the Editor/Debugger window and press the F10 (single step) key or the

F11 (step in) key to dig into a called routine. If you want to jump to the

next breakpoint, press F5 or click ‘Run (Continue)’ in the Debug pull-down

menu of the Editor/Debugger window. If you want to run the program until

just before a statement, move the cursor to the line and click ‘Go Until

Cursor’ in the Debug pull-down menu (see Fig. I.2).

5. If you have figure out what is wrong, edit the pertinent part of the program,

save the edited program in the Editor/Debugger window, and then go to the

Command window, typing the name of the file containing the main program

to try running the program for test. If the result seems to reflect that the

program still has a bug, go back to step 1 and restart the whole procedure.

MATLAB 495

If you use the MATLAB of version 5.x, you can refer to the usage of the

constrained minimization routine ‘constr()’, which is summarized in the box

below.

USAGE OF THE MATLAB 5.X BUILT-IN FUNCTION “CONSTR()”

FOR CONSTRAINED OPTIMIZATION

[x,options] = constr(’ftn’,x0,options,l,u)

ž Input arguments (only ‘ftn’ and x0 required, the others optional)

‘ftn’ : usually defined in an m-file and should return two output

arguments, one of which is a scalar value (f (x)) of the

function (ftn) to be minimized and the other is a vector

(g(x)) of constraints such that g(x) ≤ 0.

x0 : the initial guess of solution

options: is used for setting the termination tolerance on x, f (x), and

constraint violation through options(2)/(3)/(4), the number of

the (leading) equality constraints among g(x) ≤ 0 through

options (13), etc.

(For more details, type ‘help foptions’ into the MATLAB

command window)

l,u : lower/upper bound vectors such that l ≤ x ≤ u.

ž Output arguments

x : minimum point reached in the permissible region satisfying

the constraints.

options: outputs some information about the search process and the

result like the function value at the minimum point (x)

reached through options (8).

REFERENCES

[B-1] Burden, Richard L., and Fairs, J. Douglas, Numerical Analysis, 7th ed., Brooks/

Cole Thomson, Pacific Grove, CA, 2001.

[B-2] Bell, H. E., Gerschgorin’s theorem and the zeros of polynomials, Am. Math.

Monthly 72, 292–295 (1965).

[C-1] Canale, Raymond, and Chapra, Steven, Numerical Methods for Engineers: with

Software and Programming Applications, McGraw-Hill, New York, 2002.

[F-1] Fausett, L. V., Applied Numerical Analysis Using MATLAB, Prentice-Hall, Upper

Saddle River, NJ, 1999.

[H-1] Hamming, R. W., Numerical Methods for Scientists and Engineers, 2nd ed.,

McGraw-Hill, New York, 1973.

[K-1] Kreyszig, Erwin, Advanced Engineering Mathematics, 8th ed., John Wiley & Sons,

New York, 1999.

[K-2] Kreyszig, Erwin, Introductory Functional Analysis with Applications, John Wiley

& Sons, New York, 1978.

[L-1] Lindfield, G. R., and Peny, J. E. T., Numerical Methods Using MATLAB, 8th ed.,

Prentice-Hall, Upper Saddle River, NJ, 2000.

[L-2] Luenberger, D. G., Linear and Nonlinear Programming, 2nd ed., Addison-Wesley

Publishing Company, Reading, MA, 1984.

[M-1] Mathews, J. H., and Fink, K. D., Numerical Methods Using MATLAB, Prentice-

Hall, Upper Saddle River, NJ, 1999.

[M-2] Maron, Melvin J., Numerical Analysis, Macmillan, Inc., New York, 1982.

[N-1] Nakamura, Shoichiro, Numerical Analysis and Graphic Visualizationwith MATLAB,

2nd ed., Prentice-Hall, Upper Saddle River, NJ, 2002.

[O-1] Oppenheim, Alan V., and Schafer, Ronald W., Discrete-Time Signal Processing,

Prentice-Hall, Englewood Cliffs, NJ, 1989.

[P-1] Peaceman, D. W., and Rachford, H. H., The numerical solution of parabolic and

elliptic differential equations, J. Soc. Ind. Appl. Math. 3, 28–41 (1955).

[P-2] Pham, D. T., and Karaboga, D., Intelligent Optimization Techniques, Springer-

Verlag, London, 1998.

Applied Numerical Methods Using MATLAB, by Yang, Cao, Chung, and Morris

Copyr ight 2005 John Wiley & Sons, I nc., ISBN 0-471-69833-4

497

498 REFERENCES

[P-3] Phillips, C. L., and Nagle, H. T., Jr., Digital Control System Analysis and Design,

Prentice-Hall, Upper Saddle River, NJ, 2002.

[R-1] Rao, S. S., The Finite Element Method in Engineering, 3rd ed., Butterworth Heinemann,

Boston, 1999.

[R-2] Recktenwald, G. W., Numerical Methods with MATLAB, Prentice-Hall, Upper Saddle

River, NJ, 2000.

[S-1] Schilling, R. J., and Harris, S. L., Applied Numerical Methods for Engineers Using

MATLAB and C, Brooks/Cole, Pacific Grove, CA, 2000.

[S-2] Silvester, P. P., and Ferrari, R. L., Finite Elements for Electrical Engineers, 3rd

ed., Cambridge University Press, Cambridge, U.K., 1996.

[S-3] Stoer, J., and Bulirsch, R., Introduction to Numerical Analysis, Springer-Verlag,

New York, 1980.

[W-1] Website <http://www.maths.abdn.ac.uk/igc/testing/tch/ma2001/notes/node79

.html>

[W-2] Website <http://mathworld.wolfram.com/PoissonDistribution.html>

[W-3] Website <http://www-ccrma.stanford.edu/ jos/mdft/>

[W-4] Website <http://mathworld.wolfram.com/FourierTransform.html

[W-5] Website <http://www.psc.edu/∼burkardt/papers/linear glossary.html>

[W-6] Website <http://www.mathworks.com/access/helpdesk/help/helpdesk.html>

[W-7] Website <http://csep1.phy.ornl.gov/CSEP/MO/NODE27.html>

[W-8] Website <http://mathews.ecs.fullerton.edu/numerical.html>

[Z-1] Zienkiewicz, O. C., and Taylor, R. L., The Finite Element Method, 4th ed., Vol. 1,

McGraw-Hill, London, 1989.

SUBJECT INDEX

A

Absolute error, 33

Acceleration of Aitken, 201

Adams–Bashforth–Mou1ton (ABM) method,

269

Adaptive input argument, 46

Adaptive quadrature, 231

Alignment, 30

alternating direction imp1icit (ADI) method,

417

Animation, 302, 438

Apostrophe, 15

Approximation, 124, 209, 212, 323

B

backslash, 19, 59, 60, 76, 109, 110

backward difference approximation, 210

backward substitution, 82

basis function, 420

bilinear interpolation, 142

bisection method, 183

Boltzmann, 335

boundary condition, 134, 401, 404, 420,

430–432

Boundary mode, 434

boundary node, 420

boundary value problem (BVP), 287, 305–319

bracketing method, 188

breakpoint, 493

Bulirsch–Stoer, 161

C

case, 24

catastrophic cancellation, 32

central difference approximation, 211, 212

characteristic equation, 371, 465

characteristic value, 371, 465

characteristic vector, 371, 465

Chebyshev coefficient polynomial, 126

Chebyshev node, 125, 160

Chebyshev polynomial, 124, 127, 240

chemical reactor, 297

Cholesky decomposition (factorization), 97

circulant matrix, 391

conjugate gradient, 332

constrained linear least squares (LLS), 354

constrained optimization, 343, 350, 352

constructive solid geometry (CSG), 432

contour, 11, 295, 345, 349

convergence, 103, 378–379

covariance matrix, 386

Crank–Nicholson method, 409, 452

CtFT, 68, 475

cubic spline, 133, 162–164

curve fitting, 143, 147, 165, 167

D

damped Newton method, 193

data file, 47

dat-file, 2

dc motor, 298

debugging, 493

decoupling, 374, 376

determinant, 464

DFT, 151–156, 171–175

diagonalization, 374–376

difference approximation, 209, 211, 216, 218

differential equation, 263, 487

Dirichlet boundary condition, 404, 430, 434,

452

discretization, 281

distinct eigenvalues, 373

divided difference, 120–122

Applied Numerical Methods Using MATLAB, by Yang, Cao, Chung, and Morris

Copyright 2005 John Wiley & Sons, Inc., ISBN 0-471-69833-4

499

500 SUBJECT INDEX

double integral/integration, 241, 259

Draw mode, 432

DtFT (Discrete-time Fourier Transform), 476

E

eigenmode PDE, 431

eigenpair, 371

eigenvalue, 371, 377, 385, 389, 465

eigenvalue problem, 314, 389

eigenvector, 371, 377, 385, 465

electric potential, 427, 442

element-by-element operation, 15, 52

elliptic PDE, 401, 402, 420, 430

eps, 13

error, 31, 33, 35, 40, 213, 226, 274

error analysis, 159, 225

error estimate, 226

error magnification, 31

error propagation, 33

errorbar, 148

Euler’s method, 263

explicit central difference method, 415, 417

explicit forward Euler method, 406, 410

exponent field, 28

F

factoriaL, 40

false position, 185

FFT (Fast Fourier Transform), 151

finite difference method (FDM), 290

finite element method (FEM), 420, 431, 455

fixed-point, 99, 179, 197

Fletcher–Reeves (FR), 332, 333

forward difference approximation, 209, 218,

406

Fourier series/transform, 150, 475

full pivoting, 85

G

Gauss elimination, 79

Gauss quadrature, 234

Gauss–Chebyshev, 240

Gauss–Hermite, 238, 251, 253

Gauss–Jordan elimination, 89, 106

Gauss–Laguerre, 239, 254, 255

Gauss–Legendre, 235, 251, 255

Gauss–Seidel iteration, 100, 103, 115

Gaussian distribution, 24

genetic algorithm, 338, 340

Gerschgorin’s Disk Theorem, 380

golden search, 321, 322

gradient, 294, 328, 330, 471

graphic command, 491

H

Hamming method, 273

heat equation, 406, 412

heat flow equation, 410

Helmholtz’s equation, 402

Hermite interpolating polynomial l39,

Hermite polynomial, 66, 238

Hermitian, 466

Hessenberg form, 395–397

Hessian, 330, 472

Heun’s method, 266

hidden bit, 28

Hilbert matrix, 88

histogram, 23

Householder, 392–395

hyperbolic PDE, 401, 414, 430, 440, 453

I

IDFT, 151

IEEE 64-bit floating-point number, 28

ilaplace, 280

ill-condition, 88

implicit backward Euler method, 407, 452

improper integral, 248, 249

inconsistency, 83, 85

independent eigenvectors, 373

input, 2,4

interior node, 425

interpolation, 117, 119, 133, 141, 161

2-dimensional 141

interpolation by using DFS, 155

interpolation function, 420

inverse matrix, 92, 465

inverse power method, 380, 381

IVP (initial value problem), 263, 284

J

Jacobi iteration, 98

Jacobi method, 381–384

Jacobian, 191, 472, 484

K

keyboard input, 2

L

Lagrange coefficient polynomial, 118

Lagrange multiplier method, 74, 343, 344

Lagrange polynomial, 117, 118

Laguerre polynomial, 239

Laplace transform, 278, 280, 473

Laplace’s Equation, 402, 404, 427, 435, 442

largest number in MATLAB, 27

leakage, 155, 174

least squares (LS), 144, 165, 169, 171, 351, 354

SUBJECT INDEX 501

Legendre polynomial, 236

length of arc/curve, 257

limit, 483

linear equation, 71, 79

linear programming (LP), 355, 361

logical operator, 25

loop, 26

loop iteration, 39

Lorenz equation, 297

loss of significance, 31, 32

LSE (least squares error), 75

LU decomposition (factorization), 92

M

mantissa field, 28

mat-file, 2

mathematical functions, 10

matrix, 15, 463

matrix inversion lemma, 78, 469

mean value theorem, 461

mesh, 11, 48, 49, 431–444

midpoint rule, 222

minimum-norm solution, 73

mixed boundary condition, 287, 306, 308

modal matrix, 373–376

mode, 285, 377–378, 386, 432, 434

modification formula, 272, 274

mu law, mu-inverse law, 53, 335

N

negligible addition, 31

Nelder-Mead Algorithm, 325

nested computing, 38, 121

nested (calling) routine, 40

Neumann boundary condition, 404, 431, 447,

448, 451

Newton method, 186, 188, 191, 330, 332

Newton polynomial, 119

nonlinear BVP, 312

nonlinear least squares (NLLS), 352

nonnegative least squares (NLS), 355

norm, 58

normal (Gaussian) distribution, 24

normalized range, 29

null space, 73, 467

numerical differentiation, 209, 244

numerical integration, 222, 247, 249

O

on-line recursive computation of DFT, 176

orthogonal, 382, 395, 466

orthonormal, 382, 385

over-determined, 75

overflow, 34, 64

P

Pade approximation, 129, 160

parabolic PDE, 406, 410, 412, 414, 430, 438,

449

two-dimensional PDE, 412

parallelepiped, 389

parallelogram, 388

parameter passing through VARARGIN, 45

parameter sharing via GLOBAL, 44

partial differential equation (PDE), 401

partial pivoting, 81,85,105

path, 1

PDE mode, 434

PDEtool, 429–431, 435, 456

penalty, 346–349, 362, 366

permutation, 94, 467

persistent excitation, 169

physical meaning of eigenvalues and

eigenvectors, 385

pivoting, 85–88, 105–106

plot, 6–11

Plot mode, 440

Polak-Ribiere (PR) method, 332, 333

polynomial approximation, 124

polynomial curve fitting by least squares, 146,

169

polynomial wiggle, 124

positive definite, 468

predictor/corrector errors, 272

projection operator, 74

pseudo (generalized) inverse, 17, 73, 76

Q

QR decomposition (factorization), 97, 392–396

quadratic approximation method, 323–325

quadratic interpo1ation, 157

quadratically convergent, 188

quadrature, 222, 231, 234

quantization error, 63, 212

quenching factor, 335

R

rank, 467

recursive, 40, 66, 176, 201, 228, 231

recursive least square estimation (RLSE), 76,

104

redundancy, 83, 85

regula falsi, 185

relational operators, 25

relative error, 33

relaxation, 104, 115

reserved constants/variables, 13

Richardson’s extrapolation, 211, 216

RLSE (Recursive Least Squares Estimation), 76

502 SUBJECT INDEX

robot path planning, 164

Romberg integration, 228–230

rotation matrix, 382, 384

round-off error, 31, 35, 212, 213

row echelon form, 467

row space, 467

row switching, 91, 105

Runge Phenomenon, 124

Runge-Kutta (RK4), 267

runtime error, 40

S

saddle point, 358

sampling period, 151, 153, 172

scalar product, 469

scaled partial pivoting, 85, 105

scaled power method, 378–379

Schroder method, 202

secant method, 189, 201

self-calling, 201

shifted inverse power method, 380

shooting method, 287, 305, 309, 312

shooting position, 288, 307

similarity transformation, 373

Simpson’s rule, 222, 226

simulated annealing, 334, 336

sinc function, 41, 51

single step, 494

smallest positive number in MATLAB,

27

Solve mode, 434

SOR (successive over-relaxation), 104

sparse, 489

stability, 378, 386, 406–410, 415–416, 418,

450

state equation, 277, 281, 283, 295, 299

steepest descent, 328, 330

Steffensen method, 201

step-size, 212–215, 264–265, 269, 286, 328,

332

step-size dilemma, 213

step in, 494

stiff, 284–286, 298–299, 386

Sturm-Liouville (BVP) equation, 319

surface area of revolutionary object, 258

SVD (singular value decomposition), 98, 112

symbolic, 193, 233, 280, 481

symbolic variable, 194, 481

symmetric matrix, 381–382, 466

Symmetric Diagonalization Theorem, 382

T

Taylor series theorem, 462, 485

temperature, 404, 406, 412, 435, 438

term-wise operation, 15, 52

Toeplitz matrix, 390

trapezoidal rule, 222, 225, 226

tri-diagonal, 107, 108

truncation error, 31, 212, 213

two-dimensional interpolation, 141, 168

U

unconstrained optimization, 321, 350

unconstrained least squares, 355

underdetermined, 72

underflow, 34, 64

uniform probabilistic distribution, 22

unitary, 466

un-normalized range, 29

V

Van der Pol equation, 285, 296

vector, 15, 469

vector differential equation, 277, 284

vector operation, 39

vector product, 469

vibration, 416, 418, 440

volume, 243, 258

W

wave equation, 414, 416–418, 453

weight least-squares (WLS), 145, 147, 171

X

zero-padding, 151

INDEX FOR MATLAB

ROUTINES

(cf) A/C/E/P/S/T stand for Appendix/Chapter/Example/Problems/Section/Table, respectively.

(cf) The routines whose name starts with a capital letter are constructed in this book.

(cf) A program named “nmijk.m” can be found in Section i.j-k.

Name Place Description

abmc S6.4-1 Predictor/Corrector coefficients in Adams-Bashforth-

Moulton ODE solver

adapt

−

Smpsn() S5.8 ntegration by the adaptive Simpson method

adc1() P1.10 AD conversion

adc2() P1.10 AD conversion

axis() S1.1-4 specify axis limits or appearance

backslash(\) P1.14 left matrix division

backsubst() S2.4-1 backward substitution for