DEPARTMENT OF MATHEMATICSijsrme.rdmodernresearch.com/wp-content/uploads/2016/04/TP002.pdfnumerical...

NUMERICAL METHODS FOR SOLVING DEs

WITH MATLAB

This project work is submitted to

ST.JOSEPH’S COLLEGE (AUTONOMOUS)

In partial fulfillment of the requirements of the degree of

MASTER OF PHILOSOPHY IN MATHEMATICS

by

SAHAYA JERNITH.S

(Reg.No:12MMA115)

Under the guidance of

Dr. S. RUBANRAJ, M.Sc., B.Ed., M.Phil., PGDCSA., Ph.D.,

DEPARTMENT OF MATHEMATICS

ST.JOSEPH’S COLLEGE (AUTONOMOUS)

Nationally Re-Accredited (3rd cycle) With A Grade

College With Potential For Excellence

Trichirappalli – 620 002

August – 2013.

Certificate

This is to certify that the project entitled “NUMERICAL METHODS

FOR SOLVING DEs WITH MATLAB” submitted to St.Joseph’s

College (Autonomous), affiliated by Bharathidasan University, Trichy – 24,

by SAHAYA JERNITH.S (12MMA115) for the partial fulfilment of

award of the degree of Master of Philosophy in Mathematics is a record

of bonafide work carried out by her under my guidance and supervision in

the year 2012 - 2013.

HEAD OF THE DEPARTMENT RESEARCH ADVISOR

Dr.S.RUBANRAJ Dr.S.RUBANRAJ

Department of Mathematics Department of Mathematics

St. Joseph’s College (Autonomous) St. Joseph’s College (Autonomous)

Tiruchirappalli – 620 002. Tiruchirappalli – 620 002.

Viva-Voce examination conducted on · · · · · · · · · · · · · · · · · · · · · · · · at St.Joseph’s

College (Autonomous) Tiruchirappalli – 620 002.

Place: Tiruchirappalli EXTERNAL EXAMINER

Date :

Acknowledgement

I would like to express my grateful thanks to Almighty God for giving me

the strength and the wisdom for completing this project.

I would like to convey my heartfelt and sincere thanks to my guide and

Head of the Department of Mathematics Dr. S. RUBANRAJ, M.Sc., B.Ed.,

M.Phil., PGDCSA., Ph.D., Department of Mathematics, St. Joseph’s Col-

lege (Autonomous), Tiruchirappalli-620002, for his unfailing enthusiasm and

endless encouragement in my endeavours. It is not an exaggeration to state

that he is a great source of inspiration and illumination to me.

I am privileged to express my gratitude to Rev.Dr.A.ALBERT MUTHU-

MALAI, SJ, Former Principal, St.Joseph’s College (Autonomous), Trichy

and Rev.Dr.F.ANDREW, SJ, Principal, St.Joseph’s College (Autonomous),

Trichy, for having given me wonderful opportunity to study in this presti-

gious institution.

Last but not the least, I would like to express my gratitude to all mem-

bers, my relations and my friends for their precious support through their

prayers for the successful completion of this dissertation. I whole heartedly

thank my beloved parents for being a great source of inspiration, a role model

and of course, for their invaluable love.

SAHAYA JERNITH.S

Contents

Introduction 1

1 PRELIMINARY 3

1.1 Differential Equations . . . . . . . . . . . . . . . . . . . . . . . 3

1.2 Partial Differential Equations . . . . . . . . . . . . . . . . . . 6

1.3 MATLAB coding for DE . . . . . . . . . . . . . . . . . . . . . 7

2 MATLAB 10

2.1 Basic Steps . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

2.2 Menus and the Toolbar . . . . . . . . . . . . . . . . . . . . . . 13

2.3 Plotting in MATLAB . . . . . . . . . . . . . . . . . . . . . . . 16

3 Numerical Methods for DE 23

3.1 Euler’s Method . . . . . . . . . . . . . . . . . . . . . . . . . . 23

3.2 Heun’s Method . . . . . . . . . . . . . . . . . . . . . . . . . . 25

3.3 Runge-Kutta Method of order 4 . . . . . . . . . . . . . . . . . 26

3.4 Runge-Kutta Fehlberg Method . . . . . . . . . . . . . . . . . 29

4 Solving Differential Equations with MATLAB 34

4.1 First Order Differential Equation . . . . . . . . . . . . . . . . 34

iv

CONTENTS v

4.2 Second Order Differential Equations . . . . . . . . . . . . . . . 36

4.3 System of Differential Equations . . . . . . . . . . . . . . . . . 37

4.4 Runge-Kutta methods in MATLAB . . . . . . . . . . . . . . . 41

4.5 Euler Methods . . . . . . . . . . . . . . . . . . . . . . . . . . 45

5 Comparison of numerical approximations 51

5.1 Comparison for a first order ODE . . . . . . . . . . . . . . . . 51

5.2 Comparison for a system of ODEs . . . . . . . . . . . . . . . . 64

6 Solving Partial Differential Equations with MATLAB 71

6.1 Poisson’s Equation on a Unit Disc . . . . . . . . . . . . . . . . 74

6.2 Helmholtz’s Equation on a Unit Disk with a Square Hole . . . 78

6.3 Solve for Complex Amplitude . . . . . . . . . . . . . . . . . . 79

6.4 Animate Solution to Wave Equation . . . . . . . . . . . . . . 80

Conclusion 82

Bibliography 83

v

List of Figures

1.1 Van Der Pol Equation . . . . . . . . . . . . . . . . . . . . . . 9

2.1 Graph of cosine function . . . . . . . . . . . . . . . . . . . . . 17

2.2 Graph of cosine function from −π to π . . . . . . . . . . . . . 18

2.3 Graph of cosine function in dotted line . . . . . . . . . . . . . 19

2.4 Graph of cosine and sine functions . . . . . . . . . . . . . . . 20

2.5 Graph of meshgrid . . . . . . . . . . . . . . . . . . . . . . . . 21

4.1 Solution of the equation y′ = xy, y(1) = 1 . . . . . . . . . . . 35

4.2 Solution of the DE y + 8y + 2y = cos(x) y(0) = 0 y′(0) = 1 37

4.3 Solutions of x′ = x+ 2y − z, y′ = x+ z, z′ = 4x− 4y + 5z . . 39

4.4 Graph of Lorenz equations . . . . . . . . . . . . . . . . . . . . 40

4.5 Graph of Components for the Lorenz Equations . . . . . . . . 41

4.6 Graph of epidemiology growth of market share . . . . . . . . . 42

4.7 Graph of x(t) vs t for the ODE . . . . . . . . . . . . . . . . . 43

4.8 Graph of y vs t for the ODE . . . . . . . . . . . . . . . . . . . 44

4.9 Graph of y vs x for the ODE . . . . . . . . . . . . . . . . . . . 44

4.10 Graph of the solution by Euler’s methods and the analytical

solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48

vi

LIST OF FIGURES vii

4.11 Solution by Runge-Kutta Method . . . . . . . . . . . . . . . . 50

5.1 Solution of y = y + 2x− x2, y(0) = −1 by Euler’s Method . . 53

5.2 Solution of y = y + 2x− x2, y(0) = −1 by Heun’s Method . . 55

5.3 Solution of y = y + 2x− x2, y(0) = −1 by RK4 Method . . . . 56

5.4 Solution of y = y + 2x− x2, y(0) = −1 by RKF5 Method . . . 58

5.5 Comparison of four methods for an ODE (h = 0.2) . . . . . . 60

5.6 Comparison of four methods for an ODE (h = 0.1) . . . . . . 62

5.7 Comparison of four methods for a system of ODEs . . . . . . 69

6.1 Graph for the linear advection equation . . . . . . . . . . . . . 73

6.2 Graph for Poisson equation . . . . . . . . . . . . . . . . . . . 75

6.3 Graph for Refinement . . . . . . . . . . . . . . . . . . . . . . . 76

6.4 Graph for Exact . . . . . . . . . . . . . . . . . . . . . . . . . . 77

6.5 Graph for FEM Solution . . . . . . . . . . . . . . . . . . . . . 78

6.6 Graph for Helmholtz equation . . . . . . . . . . . . . . . . . . 79

6.7 Graph for Complex Amplitude . . . . . . . . . . . . . . . . . . 80

6.8 Graph for Wave Equation . . . . . . . . . . . . . . . . . . . . 81

vii

Introduction

Numerical methods are commonly used for solving mathematical problems

that are formulated in science and engineering where it is difficult or even

impossible to obtain exact solutions. Only a limited number of differential

equations can be solved analytically. Numerical methods, on the other hand,

can give an approximate solution to (almost) any equation. An ordinary dif-

ferential equation (ODE) is an equation that contains an independent vari-

able, a dependent variable, and derivatives of the dependent variable. Literal

implementation of this procedure results in Euler’s method, which is, how-

ever, not recommended for any practical use. There are other methods more

sophisticated than Euler’s. Among them, there are four types of practical

numerical methods for solving problems of ODEs. They are (i) Runge-Kutta

Methods, (ii) Heun’s Method, (iii) RK4 Method, and (iv)RKF5 Method.

MATLAB is an integrated technical computing environment that com-

bines numeric computation, advanced graphics and visualization and a high-

level programming language. MATLAB was founded by Jack Little Cleve

Moler and Steve Bangert in 1984. Color graphics makes it possible to display

curves, surfaces, and solids in two and three dimensions in a way that it is

both more effective and more engaging. The easy syntax of MATLAB makes

it possible for users to get comfortable results quickly. MATLAB graphics

1

Introduction 2

are excellent and easy to use. It is easy to program in MATLAB, making

it possible to do numerical calculations using simple steps. We can also use

the symbolic capability of MATLAB to get analytical solutions. MATLAB

is becoming the most popular software package in Mathematics.

• In first chapter, we study the preliminaries needed to use MATLAB to

solve problems in ODEs. We see basic definitions based on Differential

Equation and Partial Differential Equation. We introduce MATLAB

plotting to simulate the solution of Differential Equation problems.

• In second chapter, we review the software package MATLAB for high-

performance numerical computation and visualization which provides

an interactive environment with hundreds of built-in-functions for tech-

nical computation, graphics and animation.

• In third chapter, we discuss the various Numerical Methods available

for solving DEs and they are (1) Euler’s method, (2) Heun’s method,

(3) RK4 method and (4) RKF5 method.

• In fourth chapter, we study the MATLAB programs for Solving Dif-

ferential Equations. Also we study Euler methods and Runge-Kutta

methods in MATLAB.

• In fifth chapter, we compare various methods and their approximation

for 1st order differential equation and a system of ODEs. We also test

the problem with the above methods.

• In sixth chapter, we attempt to simulate the solution of few partial

differential equations using MATLAB.

2

Chapter 1

PRELIMINARY

In this chapter we see basic definitions based on Differential Equation and

Partial Differential Equation. We introduce MATLAB plotting to simulate

the solution of Differential Equation problems.

1.1 Differential Equations

Definition 1.1.1 (Differential Equation). Any relation involving the depen-

dent variable and the differential coefficient of dependent variable with re-

spect to the independent variable is known as differential equation.

Definition 1.1.2 (Ordinary Differential Equations). An Ordinary Differen-

tial Equation (ODE) is a differential equation that depends on only one

independent variable.

For Example:

dy

dt= 7y

3

1.1. DIFFERENTIAL EQUATIONS 4

is an Ordinary Differential Equation because the independent variable y de-

pends only on the independent variable t.

Definition 1.1.3 (Order of a Differential Equation). The order of a differ-

ential equation is the order of the highest derivative entering the equation.

For Example:

The equation

md2x

dt2= −kx

is called a second-order differential equation because it involves second deriva-

tive. Its order is 2.

Definition 1.1.4 (Degree of a Differential Equation). The degree of a dif-

ferential equation is the degree of the highest derivative occurring in it.

For Example:

dy

dt= t

Its degree is 1.

Definition 1.1.5 (Linear Differential Equation). A first-order differential

equation is linear if it can be written in the form

dy

dt+ g(t)y = r(t)

where g(t) and r(t) are arbitrary functions of t.

4

1.1. DIFFERENTIAL EQUATIONS 5

For Example:

dy

dt+ 4y = cos t

is a first-order linear differential equation where g(t) = 4 and r(t) = cos t.

Definition 1.1.6 (Homogeneous Differential Equation). A linear first-order

differential equation is homogeneous if its right hand side is zero , that is

r(t) = 0. ∴dy

dt+ g(t)y = 0.

For Example:

dy

dt+ ky(t) = 0

, where k is a constant, is homogeneous.

Definition 1.1.7 (Non-homogeneous Differential Equation). A linear first-

order differential equation is non-homogeneous if its right-hand side is non-

zero, that is, r(t) 6= 0.

dy

dt+ g(t)y = r(t)

For Example:

∴

dy

dt+ 2y = sin 2t

is non-homogeneous.

Definition 1.1.8 (Non-linear Differential Equation). It is a differential equa-

tion whose right hand side is not a linear function of the dependent variable.

5

1.2. PARTIAL DIFFERENTIAL EQUATIONS 6

For Example:

dy

dt= k(1− y

n)y

1.2 Partial Differential Equations

Definition 1.2.1 (Partial Differential Equations). A Partial Differential Equa-

tion is differential equation in which the dependent variable depends on two

or more independent variables.

For example

The Laplace’s equation

∂2f

∂x2+

∂2f

∂y2= 0

is a Partial Differential Equation because f depends on two independent

variables x and y.

Definition 1.2.2 (Linear Partial Differential Equation). A linear partial dif-

ferential equation is one in which all of the partial derivatives appears in linear

form and none of the coefficients depends on the dependent variables. The

coefficient may be function of the independent variables.

Definition 1.2.3 (Non-linear Partial Differential Equation). A non-linear

partial differential equation can be described as a partial differential equation

involving non-linear terms.

Definition 1.2.4. Types of Non-Linear Partial Differential Equations

6

1.3. MATLAB CODING FOR DE 7

A non-linear partial differential equation is known as semi-linear if it

is linear in highest order derivatives and the coefficients of highest order

derivatives depend only on independent variables.

A nonlinear partial differential equation is known as quasi-linear if it is

linear in highest order derivatives and the coefficients of highest order deriva-

tives depend on independent variables as well on lesser order derivatives.

A nonlinear partial differential equation is known as fully non-linear if

the coefficients depends on dependent variable or the derivatives appear in

nonlinear form.

For Example:

fxx + fyy + fx + fy = f + xy is linear equation.

afxx + bfyy + f 2x + fy = c is semilinear.

fxfxx + fyfyy + f 2x + f = 0 is quasilinear.

ffxx + ffyy + afx + bfy = 0 is non-linear.

1.3 MATLAB coding for DE

Remark 1.3.1 (Dsolve). Symbolic solution of ordinary differential equa-

tions.

Dsolve(‘eqn1’,‘eqn2’, ...) accepts symbolic equations representing ordi-

nary differential equations and initial conditions. Several equations or initial

conditions may be grouped together, separated by commas, in a single input

argument.

By default, the independent variable is ‘t’. The independent variable

7


may be changed from ‘t’ to some other symbolic variable by including that

variable as the last input argument.

ode45 For solving differential equations, MATLAB provides a powerful

command ode45. ode45 may be invoked from the command line via

[t,y] = ode45(fname, tspan, y0, opts)

where

fname– name of a function Mfile, an inline function object or an anonymous

function used to evaluate the right-hand-side function at a given value of the

independent variable and dependent variable. If an Mfile is used, the func-

tion definition line usually has the form function dydt = fname(t,y) and

the file is stored as fname.m. The output variable (dydt) must be a vector

with the same size as y. Note that the independent variable (t here) must

be included in the input argument list even if it does not explicitly appear

in the expressions used to generate dydt. The variable fname can contain

the name of the Mfile or can be a function handle generated by an inline or

anonymous function.

tspan– 2-element vector defining the range of integration ([to tf ]) or can

be a vector of values for which the solution is desired.

y0– vector of initial conditions for the dependent variable. There should be

as many initial conditions as there are dependent variables.

opts– a MATLAB structure variable (created by odeset) that allows you to

control the details of computation. This argument is optional and, if not

provided, ode45 will use default values.

t– Value of the independent variable at which the solution array (y) is cal-

8


culated. Note that by default this will not be a uniformly distributed set of

values.

y– Values of the solution to the problem (array). Each column of y is a

different dependent variable. The size of the array is length(t)-by-length(y0)

Example for ode45

Consider the second order differential equation known as Van-der-pol

equation x + (x2 − 1)x + x = 0. Rewrite this as a system of coupled first

order differential equations: x1 = x1(1− x22)− x2 and x2 = x1. To simulate

the differential equation defined in vdpol over the interval 0 ≤ t ≤ 20, we

invoke ode23.

function xdot = vdpol(t,x)

xdot = [x(1).*(1-x(2).^2)-x(2); x(1)]

t0 = 0; tf = 20;

x0 = [0 0.25]’;

[t,x] = ode23(‘vdpol’,t0,tf,x0);

plot(t,x)

Figure 1.1: Van Der Pol Equation

0 5 10 15 20−3

−2

−1

0

1

2

3

9

Chapter 2

MATLAB

MATLAB is a software package for high-performance numerical computation

and visualization. It provides an interactive environment with hundreds of

built-in-functions for technical computation graphics and animation. It also

provides easy extensibility with its own high-level programming language.

The name MATLAB stands for MATrix LABoratory.

MATLAB is primarily a numerical computation package, although with

the symbolic MATH toolbox it can so symbolic algebra. MATLAB was pri-

marily designed to do numerical calculations. MATLAB is often much faster

at these calculations. MATLAB deals numerical computation, especially

those use vectors and matrices.

2.1 Basic Steps

The Command Line

First Steps:

10

2.1. BASIC STEPS 11

When we invoke MATLAB to begin a session, we see the prompt

>>

We can give instructions for an operation, or request information,

following this prompt, on the command line.

We can do simple arithmetic operations like (2 + 3.52 − 4.7)/12 on the

command line.

>> (2 + 3.5∧2− 4 ∗ 7)/12

ans =

−1.1458

We can also do this calculation by assigning variable names to the quan-

tities. If we do not wish to see the intermediate results, we can suppress the

numerical output by putting a semicolon at the end of the line. Then the

sequence of commands and output looks like this:

>> x = 2 + 3.5∧2;

>> y = 4 ∗ 7;

>> z = (x− y)/12;

>> z

z =

− 1.1458

Error Message

11

2.1. BASIC STEPS 12

If we enter an expression incorrectly, MATLAB will return an error mes-

sage, which always locates the error. For example, in the following, we left

out the ∗ in 3 ∗ x

>> x = 4;

>> 3x

???3

|

Missing operator, comma, or semicolon.

Making Corrections

To make corrections, we can, of course, retype the expression. But if

the expression is lengthy, we may make more mistakes by typing a second

time. Unfortunately we cannot move the cursor to the line we wish to repair.

Instead we can press the up-arrow key until we reach the desired line and

then the left- and right-arrows until we reach the offending characters. Type

in the correction and enter return.

Exiting

To leave MATLAB session, enter quit. If MATLAB gets hung up in

calculation or is taking a long time, and you want to stop the calculation,

without exiting MATLAB, enter Ctrl+C.

Help

To get information on a particular command or operation, simply enter

help command name. For example, to get information on how to use the

12

2.2. MENUS AND THE TOOLBAR 13

plotting commands, enter help plot.

2.2 Menus and the Toolbar

The Desktop manages the Command window and other MATLAB tools.

Across the top of the Desktop are a row of menu names, and a row of icons

called the toolbar. To the right of the toolbar is a box showing the current

directory, where MATLAB looks for files.

Other windows appear in a MATLAB session, depending on the work to

do. For example, a graphics window containing a plot appears to use the

plotting functions; an editor window, called the Editor/Debugger, appears

for use in creating program files.

Each window type has its own menu bar, with one or more menus, at the

top. Thus the menu bar will change as you change windows. To activate, or

select, a menu, click on it. Each menu has several items. Click on an item to

select it. Keep in mind that menus are context-sensitive. Thus their contents

change, depending on which features you are currently using.

The Desktop Menus

Most of the interaction will be in the Command window. When the Com-

mand window is active, the default MATLAB has six menus: File, Edit,

Debug, Desktop, Window, and Help. These menus change depending

on what window is active. Every item on a menu can be selected with the

menu open either by clicking on the item or by typing its underlined letter.

The three most useful menus are the File, Edit, and Help menus.

The File menu in MATLAB contains the following items, which perform

the indicated actions when we select them.

13


The File Menu

New:Opens a dialog box that allow to create a new program file, called an

M-file, using a text editor called the Editor/Debugger, or a new Figure or

Model file.

Open: Opens a dialog box that allows you to select a file for editing.

Close Command Window: Closes the Command window.

Import Data: Starts the Import Wizard which enables you to import data

easily.

Save Workspace As: Opens a dialog box that enables you to save a file.

Set Path: Opens a dialog box that enables you to set the MATLAB search

path.

Preferences: Opens a dialog box that enables you to set preferences for

such items as fonts, colors, tab spacing, and so forth.

Page Setup: Opens a dialog box that enables you to format printed output.

Print: Opens a dialog box that enables you to print all of the Command

window.

Print Selection: Opens a dialog box that enables you to print selected

portions of the Command window.

File List: Contains a list of previously used files, in order of most recently

used.

Exit MATLAB: Closes MATLAB.

TheEdit menu contains the following items.

The Edit Menu

Undo: Reverses the previous editing operation.

Redo: Reverses the previous Undo operation.

14


Cut: Removes the selected text and stores it for pasting later.

Copy: Copies the selected text for pasting later, without removing it.

Paste: Inserts any text on the clipboard at the current location of the cursor.

Paste to Workspace: Inserts the contents of the clipboard into the workspace

as one or more variables.

Select All: Highlights all text in the Command window.

Delete: Clears the variable highlighted in the Workspace Browser.

Find: Finds and replaces phrases.

Find Files: Finds files.

Clear Command Window: Removes all text from the Command window.

Clear Command History: Removes all text from the Command History

window.

ClearWorkspace: Removes the values of all variables from the workspace.

Use the Copy and Paste selections to copy and paste commands ap-

pearing on the Command window.

Use the Debug menu to access the Debugger.

Use the Desktop menu to control the configuration of the Desktop and

to display toolbars.

The toolbar, which is below the menu bar, provides buttons as short cuts

to some of the features on the menus. Clicking on the button is equivalent

to clicking on the menu, then clicking on the menu item; thus the button

eliminates one click of the mouse.

The first seven buttons from the left correspond to the New M-File,

Open File, Cut, Copy, Paste, Undo, and Redo. The eighth button

activates Simulink, which is a program built on top of MATLAB. The ninth

15

2.3. PLOTTING IN MATLAB 16

button activates the Profile which can be used to optimize program perfor-

mance. The tenth button activates the GUIDE Quick Start, which is used to

create and edit graphical user interfaces (GUIs). The eleventh button (the

one with the question mark) accesses the Help system.

2.3 Plotting in MATLAB

Remark 2.3.1 (Plotting). If x and y are two vectors of the same length then

plot (x,y) plots x versus y. Commands for data Visualization that exists in

MATLAB include

Subplot Create an array of (tiled) plots in the same window.

loglog Plot using log-log scales.

semilog x Plot using log scale on the x-axis.

semilog y Plot using log scale on the y-axis.

surf 3-D shaded surface graph.

surfl 3-D shaded surface graph with lighting.

mesh 3-D mesh surface.

Remark 2.3.2 (Legend). Legend(string1,string2,string3,....) puts a legend

on the current plot using the specified strings as label. Legend works on line

graphs, bar graphs, pie graphs, ribbon plots etc., We can label any solid-

colored patch or surface object. The fontsize and fontname for the legend

string matches the axes fontsize and fontname.

16


Example for Plotting.

To obtain the graph of y = cos(x) from -π to π, we can first define the

vector x with components equally spaced numbers between π and π, with

increment, say 0.01

>> x=-pi:0.01:pi;

>>y=cos(x);

>>plot(x,y)

>>xlabel(‘x’)

>>ylabel(‘y’)

>>ylabel(‘y=cos(x)’)

Figure 2.1: Graph of cosine function

−4 −3 −2 −1 0 1 2 3 4−1

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1

x

y=co

s(x)

Graph of cosine from −π to π

So, to obtain the same graph but in green, we type

17


>>title(‘Graph of cosine from - \pi to \pi’)

>>plot(x,y,‘g’)

Figure 2.2: Graph of cosine function from −π to π

−4 −3 −2 −1 0 1 2 3 4−1

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1

x

y=co

s(x)


where the third argument indicating the color, appears within single quotes.

We could get a dashed line instead of a solid one by typing

>> plot(x,y,‘--’)

or even a combination of line type and color, say a blue dotted line by typing

>> plot(x,y,‘b:’)

Multiple curves can appear on the same graph. If for example we define

another vector

>> z = sin(x);

18


Figure 2.3: Graph of cosine function in dotted line

−4 −3 −2 −1 0 1 2 3 4−1

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1

x

y=cos(

x)


we can get both graphs on the same axis, distinguished by their line type,

using

>> plot(x,y,‘r--’,x,z,‘m:’)

The resulting graph can be seen below, with the red dashed line representing

y = cos(x) and the blue dotted line representing z = sin(x)

When multiple curves appear on the same axis, it is a good idea to create

a legend to label and distinguish them. The command legend does exactly

this.

>> legend(‘cos(x)’,‘sin(x)’)

Remark 2.3.3 (Meshgrid). X and Y arrays for 3-D plots.

[X,Y]=meshgrid(x,y)

19


Figure 2.4: Graph of cosine and sine functions

−4 −3 −2 −1 0 1 2 3 4−1

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1cos(x)sin(x)

transforms the domain specified by vectors x and y into arrays X and Y that

can be used for the evaluation of function of two variables and 3D surface

plots.

The row of the output array X are copies of the vector x and the columns

of the output array Y are copies of the vector y.

Example for Meshgrid.

>> [x,y] = meshgrid(-3:.1:3,-3:.1:3);

>> z = 3*(1-x).^2.*exp(-(x.^2) - (y+1).^2) ...

- 10*(x/5 - x.^3 - y.^5).*exp(-x.^2-y.^2) ...

- 1/3*exp(-(x+1).^2 - y.^2);

>> surf(z)

>> xlabel(‘x’)

>> ylabel(‘y’)

>> zlabel(‘z’)

>> title(‘Peaks’)

20


Figure 2.5: Graph of meshgrid

020

4060

80

0

20

40

60

80−10

−5

0

5

10

x

peaks

y

z

Remark 2.3.4 (Peaks). A sample function of two variables. PEAKS is

a function of two variables, obtained by translating and scaling Gaussian

distributions, which is useful for demonstrating MESH, SURF, PCOLOR,

CONTOUR, etc.

Remark 2.3.5 (Initmesh). Initmesh build an initial PDE triangular mesh.

[P,E,T]=Initmesh(G)

returns a triangular mesh using the geometry specification function G. It

uses a Delaunay triangulation algorithm. The mesh size is determined from

the shape of the geometry. The outputs P, E, and T are the mesh data.

Remark 2.3.6 (Pdemesh). Plot a PDE triangular mesh. Pdemesh(P,E,T)

plots the mesh specified by P, E,T. Pdemesh(P,E,T,U) plots the solution

column vector U using a mesh plot. If U is a column vector, node data is

assumed. If U is a row vector, triangle data is assumed.

21


Remark 2.3.7 (Pdesurf). Surface plot of PDE solution. Pdesurf(P,T,U)

plots a 3-D surface of PDE node or triangle data. If U is a column vector,

node data is assumed, and continuous style and interpolated shading are

used. If U is a row vector, triangle data is assumed, and discontinuous style

and flat shading are used.

Remark 2.3.8 (Assempde). Assemble the stiffness matrix and right hand

side a PDE problem. U=ASSEMPDE(B,P,E,T,C,A,F) assembles and solves

the PDE problem on a mesh described by P, E, and T, with boundary con-

ditions given by the function name B.

It eliminates the Dirichlet boundary conditions from the system of linear

equations when solving for U.

Remark 2.3.9 (Refinemesh). Refine a triangular mesh.

[P1,E1,T1]=REFINEMESH(G,P,E,T)

returns a refined version of the triangular mesh specified by the geometry G,

point matrix P, edge matrix E, and triangle matrix T.

22

Chapter 3

Numerical Methods for DE

3.1 Euler’s Method

Obtaining Approximations:

The object of Euler’s method is to obtain approximations to the well-posed

initial-value problem

dy

dt= f(t, y), a ≤ t ≤ b, y(a) = α

A continuous approximation to the solution y(t) will not be obtained; In-

stead, approximations to y will be generated at various values, called mesh

points, in the interval [a, b]. Once the approximate solution is obtained at

the points, the approximate solution at other points in the interval can

be found by interpolation. We first make the stipulation that the mesh

points are equally distributed throughout the interval [a, b]. This condition

is ensured by choosing a positive integer N and selecting the mesh points

ti = a + ih, for each i = 0, 1, 2, . . . , N . The common distance between the

points h =b− a

N= ti+1 − ti is called the step size.

23

3.1. EULER’S METHOD 24

Use Taylor’s Theorem to derive Euler method: Suppose that y(t),

the unique solution tody

dt= f(t, y), a ≤ t ≤ b, y(a) = α has two continu-

ous derivatives on [a, b], so that for each i = 0, 1, 2, . . . , N − 1,

y(ti+1) = y(ti) + (ti+1 − ti)y′(ti) +

(ti+1 − ti)2

2y′′(ǫi)

for some number ǫi ∈ (ti, ti+1).

Since h = ti+1 − ti, we have

y(ti+1) = y(ti) + h y′(ti) +h2

2y′′(ǫi)

Because y(t) satisfies the differential equation

y′ = f(t, y)

we write

y(ti+1) = y(ti) + hf(ti, y(ti)) +h2

2y′′(ǫi)

Euler’s method constructs wi ≈ y(ti), for each i = 0, 1, 2, . . . , N , by deleting

the remainder term. Thus Euler’s method is

w0 = α

wi+1 = wi + hf(ti, wi), for each i = 0, 1, 2, . . . , N − 1.

This equation is called the difference equation associated with Euler’s method.

Merits of Euler’s Method

• Euler’s Method is simple and direct.

• This method is used for non-linear IVPs.

24

3.2. HEUN’S METHOD 25

Demerits of Euler’s Method

• It is less accurate and numerically unstable.

• Approximation error is proportional to the step size h. Hence, good

approximation is obtained with a very small h.

• This method requires a large number of time discretization leading to

a larger computation time.

• Usually applicable to explicit differential equations.

3.2 Heun’s Method

This method introduces a new idea for constructing an algorithm to solve

the I.V.P.

(1) y′(t) = f(t, y(t)) over [a, b] with y(t0) = y0.

To obtain the solution point (t1, y1), use the fundamental theorem of calculus

and integrate y′(t) over [t0, t1] to get,

(2)t1∫t0

f(t, y(t))dt =t1∫t0

y′(t)dt = y(t1)− y(t2)

the antiderivative of y′(t) is the desired function y(t). When equation (2) is

solved for y(t1), the result is

(3) y(t1) = y(t0) +t1∫t0

f(t, y(t))dt

A numerical integration method can be used to approximate the definite

integral in (3). If the trapezoidal rule is used with the step size h = t1 − t0,

then the result is

(4) y(t1) ≈ y(t0) +h

2(f(t0, y(t0))) + f(t1, y(t1))

25

3.3. RUNGE-KUTTA METHOD OF ORDER 4 26

Here the formula on the right-hand side of (4) involves the yet to be

determined value y(t1). To proceed, we use an estimate for y(t1). Euler’s

solution will suffice for this purpose.

After it is substituted into (4), the resulting formula for finding (t1, y1) is

called Heun’s method

(5) y1 = y(t0) +h

2(f(t0, y0) + f(t1, y0 + hf(t0, y0))).

The process is repeated and generates a sequence of points that approx-

imates the solution curve y = y(t). At each step, Euler’s method is used as

a prediction, and then the trapezoidal rule is used to make a correction to

obtain the final value. The general step for Heun’s method is

pk+1 = yk + hf(tk, yk);

yk+1 = yk +h

2(f(tk, yk) + f(tk+1, pk+1));

tk+1 = tk + h.

Note the role played by differentiation and integration in Heun’s Method.

3.3 Runge-Kutta Method of order 4

The Runge-Kutta methods are developed to show the derivation of a simple

second-order method. Here, the increment to the y is a weighted average of

two estimates of the increment, which it call k1 and k2. Thus for the equation

dy

dx= f(x, y),

yn+1 = yn + ak1 + bk2,

k1 = hf(xn, yn),

26


k2 = hf(xn + α, yn + βk1)

The values of k1 and k2 as estimates of changes in y when x advances

by h, because they are the product of the change in x. The Runge-Kutta

method is always the simple Euler estimate as the first estimate of δy; the

other estimate is taken with x and y stepped up by the fractions α and β of

h and of the earlier estimate of δy, k1. The y-derivatives are written in terms

of f , fromdy

dx= f(x, y).

yn+1 = yn + hf(xn, yn) +h2

2f ′(xn, yn) + · · ·

An equivalent form,

df

dx= fx + fy,

dy

dx= fx + fyf, is

yn+1 = yn + hfn + h2(1

2fx +

1

2fyf)n (1)

Now rewrite equation (1) by substituting the definitions of k1 and k2

yn+1 = yn + ahf(xn, yn) + bhf [xn + αh, yn + βhf(xn, yn)] (2)

To make the last term of Eq.(2) comparable to Eq.(1), we expand f(x,y)

in a Taylor series in terms of xn, yn remembering that f is a function of two

variables, retaining only first derivative terms

f [xn + αh,+βhf(xn, yn)] = (f + fxαh+ fyβhf)n. (3)

On the right side of both equation (1) and (3), f and its partial derivatives

are all to be evaluated at (xn, yn). Substituting from Eq.(3) into Eq.(2), we

get

yn+1 = yn + ahfn + bh(f + fxαh+ fyβhf)n

27


or rearranging

yn+1 = yn + (a + b)hfn + h2(αbfx + βbfyf)n (4)

Equation (4) will be identical to equation (1) if

a + b = 1, αb =1

2, βb =

1

2.

Note that only three equations need to be satisfied by the four unknowns.

This last set of parameters gives the modified Euler algorithm. This is a

special case of a second-order Runge-Kutta method.

Fourth-order Runge-Kutta methods are most widely used and are

derived similar. The set of 11 equations can be solved with 2 unknowns

being chosen arbitrarily. The most commonly used set of values lead to the

algorithm.

yn+1 = yn +1

6(k1 + 2k2 + 3k3 + k4),

k1 = hf(xn, yn),

k2 = hf(xn +1

2h, yn +

1

2k1),

k3 = hf(xn +1

2h, yn +

1

2k2),

k4 = hf(xn + h, yn + k2).

The local error term for the fourth order Runge-Kutta method is O(h5);

The global error would be O(h4). It is computationally more than the mod-

ified Euler Method.

Merits of Runge-Kutta Method

• This method is easy to implement.

• This method is stable.

28

3.4. RUNGE-KUTTA FEHLBERG METHOD 29

Demerits of Runge-Kutta Method

• This method requires relatively large computer time.

• Error estimation are not easy to be done.

• The Runge-Kutta methods do not work well for simple differential equa-

tions (eg. linear differential equations with widely spread eigenvalues).

• They are not good for systems of differential equations with a mix of

fast and slow state dynamics.

3.4 Runge-Kutta Fehlberg Method

The appropriate use of varying step size produce computationally efficient

integral approximating methods. In itself, this might not be sufficient to favor

these methods due to the increased complication of applying them. However,

they have another feature that makes them worthwhile. They incorporate

in the step-size procedure an estimate of the truncation error that does not

require the approximation of the higher derivatives of the function. These

methods are called adaptive because they adapt the number and position of

the nodes used in the approximation to ensure that the truncation error is

kept within a specified bound.

There is a close connection between the problem of approximating the

value of a definite integral and that of approximating the solution to an

initial-value problem. It is not surprising, then, that there are adaptive

methods for approximating the solutions to initial-value problems and that

these methods are not only efficient, but also incorporate the control of error.

29


An ideal difference-equation method

wi+1 = wi + hiφ(ti, wi, hi), i = 0, 1, 2, . . . , N − 1.

for approximating the solution, y(t), to the initial-value problem

y′ = f(t, y), a < t < b, y(a) = α,

would have the property that, given a tolerance ǫ > 0, the minimal number

of mesh points would be used to ensure that the global error,| y(ti) − wi |,

would not exceed ǫ for any i = 0, 1, . . . , N . Having a minimal number of

mesh points and also controlling the global error of a difference method is,

not surprisingly, inconsistent with the points being equally spaced in the

interval. Now examine techniques used to control the error of a difference-

equation method in an efficient manner by the appropriate choice of mesh

points.

Although we cannot generally determine the global error of a method,

that there is a close connection between the local truncation error and the

global error. By using methods of differing order we can predict the local

truncation error and, using this prediction, choose a step size that will keep

it and the global error in check.

Suppose that we have two approximation techniques. The first is an

nth-order method obtained from an nth-order Taylor method of the form

y(ti+l) = y(ti) + hφ(ti, y(ti, h) +O(hn+1),

producing approximations

W0 = α,

Wi+l = Wi + hφ(ti,Wi, h), for i > 0,

30


with local truncation error τi+1(h) = O(hn). In general, the method is gen-

erated by applying a Runge-Kutta modification to the Taylor method, but

the specific derivation is unimportant.

The second method is similar but one order higher; it comes from an

(n + l)st-order Taylor method of the form

y(ti+l) = y(ti) + hφ(ti, y(ti), h) +O(hn+2),

producing approximations

W0 = α,

Wi+l = Wi + hφ(ti, Wi, h), for i > 0,

with local truncation error τi+l(h) = O(hn+1).

Now first make the assumption that Wi ≈ y(ti) ≈ Wi and choose a fixed

step size h to generate the approximations Wi+1 and Wi+1 to y(ti+1). Then

τ i + 1 =y(ti+1)− y(ti)

h− φ(ti, y(ti), h)

=y(ti+1)− wi

h− φ(ti, wi, h)

=y(ti+1)− [wi + hφ(ti, wi, h)]

h

=1

h(y(ti+1)− wi+1)

In a similar manner,

τi+l(h) =1

h(y(ti+1)− wi+1)

and, as a consequence,

τi+1(h) =1

h(y(ti+1)−Wi+1)

=1

h[(y(ti+1 − Wi+1) + (Wi+1 −Wi+1)]

31


= τi+1(h) +1

h(Wi+1 −Wi+1).

But τi+1(h) is O(hn) and τi+1(h) is O(hn+1), so the significant portion of

τi+1(h) must come from

1

h(wi+1 − wi+1)

This gives us an easily computed approximation for the local truncation error

of the 0(hn) method

τi+1(h) ≈1

h(wi+1 − wi+1)

The object is not simply to estimate the local truncation error but to adjust

the step size to keep it within a specified bound. We now assume that since

τi+1(h) is O(hn), a number K, independent of h, exists with

τi+1(h) ≈ Khn

Then the local truncation error produced by applying the nth-order method

with a new step size qh can be estimated using the original approximations

Wi+1 and Wi+1

τi+1(qh) ≈ K(qh)n = qn(khn) ≈ qnτi+1(h) ≈qn

h(Wi+1 −Wi+1)

To bound τi+1(qh) by ǫ, we choose q so that

qn

h| Wi+1 −Wi+1 |≈| τi+1(qh) |≤ ǫ

that is, so that

q ≤ (ǫh

| wi+1 − wi+1 |)1/n

One popular technique that uses this inequality for error control is the

Runge-Kutta-Fehlberg method. This technique uses a Runge-Kutta

method with local truncation error of order five,

Wi+1 = Wi +16

135k1 +

6656

12825k3 +

28561

56430k4 −

9

50k5 +

2

55k6

32


to estimate the local error in a Runge-Kutta method of order four given by

Wi+1 = Wi +25

216k1 +

1408

2565k3 +

2197

4104k4 −

1

5k5

where,

k1 = hf(ti,Wi),

k2 = hf(ti +h

4,Wi +

1

4k1),

k3 = hf(ti +3h

8,Wi +

3

32k1 +

9

32k2),

k4 = hf(ti +12h

13,Wi +

1932

2197k1 +

7200

2197k2 +

7296

2197k3),

k5 = hf(ti + h,Wi +439

216k1 − 8k2 +

3680

513k3 −

845

4104k4),

k6 = hf(ti +h

2,Wi −

8

27k1 + 2k2 −

3544

2565k3 −

1859

4104k4 −

11

40k5).

with global error O(h5)

Error, E =k1360

− 128

4275k3 −

2197

75240k4 +

k550

+2

55k6

Merits of Runge-kutta-Fehlberg Method

• In this method, only six evaluations of f are required per step.

• Arbitrary Runge-Kutta methods of orders four and five used together

require at least four evaluations of f for the fourth-order method.

• In additional six for the fifth-order method, for a total of at least ten

functional evaluations.

33

Chapter 4

Solving Differential Equations

with MATLAB

MATLAB has an extensive library of functions for solving ordinary differ-

ential equations. Though MATLAB is primarily a numerics package, it can

certainly solve straight forward differential equations.

4.1 First Order Differential Equation

Example:

Solve the first order differential equation:

dy

dx(x) = xy; y(1) = 1

Solution:

>>y=dsolve(‘Dy=y*x’,‘x’)

y =

C2*exp(x^2/2)

34

4.1. FIRST ORDER DIFFERENTIAL EQUATION 35

>>eqn1=‘Dy=y*x’

eqn1 =

Dy=y*x

>>y=dsolve(eqn1,‘x’)

y =

C2*exp(x^2/2)

To solve an initial value problem

Solution:

>>inits=‘y(1)=1’;

>>y=dsolve(eqn1,inits,‘x’)

y =

exp(x^2/2)/exp(1)^(1/2)

>>x = linspace(0,1,20);

>>z = eval(vectorize(y));

>>plot(x,z)

Figure 4.1: Solution of the equation y′ = xy, y(1) = 1

0 0.2 0.4 0.6 0.8 1

0.65

0.7

0.75

0.8

0.85

0.9

0.95

1

35

4.2. SECOND ORDER DIFFERENTIAL EQUATIONS 36

Now to use the same equation a number of times, we might choose to define

it as a variable, say eqn1.

There are two minor difficulties.

• Our expression for y(x) is not suited for array operations (.∗, ./, .∧)

• y, as MATLAB returns it, is actually a symbol.

To overcome these two difficulties, use Vectorize(). Vectorize() converts

symbolic objects into strings.

And use the command eval(), which evaluates or executes text strings.

4.2 Second Order Differential Equations

Example:

Solve and plot the solution of the second order equation:

d2ydx

(x) + 8 dydx(x) + 2y(x) = cos(x); y(0) = 0; y′(0) = 1.

Solution:

>>eqn2=‘D2y+8*Dy+2*y=cos(x)’;

>>inits2=‘y(0)=0,Dy(0)=1’;

>>y=dsolve(eqn2,inits2,‘x’)

y =

(14^(1/2)*exp(4*x - 14^(1/2)*x)*exp(x*(14^(1/2)- 4))

*(sin(x)-cos(x)*(14^(1/2)- 4)))/(28*((14^(1/2) - 4)2 +

1))-(98*14^(1/2)+378)/(exp(x*(14^(1/2)+4))

*(868*14^(1/2)+3136)) -(14^(1/2)*exp(4*x

36

4.3. SYSTEM OF DIFFERENTIAL EQUATIONS 37

+ 14^(1/2)*x)*(sin(x) +cos(x)*(14^(1/2)+ 4)))/

(28*exp(x*(14^(1/2)+4))*((14^(1/2)+4)^2 +1))

- (exp(x*(14^(1/2)- 4))*(98*14^(1/2)-378))

/(868*14^(1/2)-3136)

>>x=linspace(0,1,20);

>>z=eval(vectorize(y));

>>plot(x,z)

Figure 4.2: Solution of the DE y + 8y + 2y = cos(x) y(0) = 0 y′(0) = 1

0 0.2 0.4 0.6 0.8 10

0.02

0.04

0.06

0.08

0.1

0.12

0.14

0.16

0.18

0.2

4.3 System of Differential Equations

(1) Solve and plot solutions of the system of three ordinary differential equa-

tions:

x′(t) = x(t) + 2y(t)− z(t)

y′(t) = x(t) + z(t)

z′(t) = 4x(t)− 4y(t) + 5z(t)

37


Solution:

>>[x,y,z]=dsolve(‘Dx=x+2*y-z’,‘Dy=x+z’,‘Dz=4*x-4*y+5*z’)

x =

-(C3*exp(t))/2 -(C4*exp(2*t))/2 -(C5*exp(3*t))/4

y=

(C3*exp(t))/2 + (C4*exp(2*t))/4 + (C5*exp(3*t))/4

z =

C3*exp(t) + C4*exp(2*t) + C5*exp(3*t)

>>inits=‘x(0)=1,y(0)=2,z(0)=3’;

>>[x,y,z]=dsolve(‘Dx=x+2*y-z’,‘Dy=x+z’,‘Dz=4*x-4*y+5*z’,inits)

x =

6*exp(2*t) - (5*exp(3*t))/2 - (5*exp(t))/2

y =

(5*exp(3*t))/2 - 3*exp(2*t) + (5*exp(t))/2

z =

10*exp(3*t) - 12*exp(2*t) + 5*exp(t)

>>t=linspace(0,.5,25);

>>xx=eval(vectorize(x));

>>yy=eval(vectorize(y));

>>zz=eval(vectorize(z));

>>plot(t, xx, t, yy, t, zz)

To solve the Equation

dy

dx= xy2 + y; y(0) = 1.

We write the M-file firstode.m as follows

38


Figure 4.3: Solutions of x′ = x+ 2y − z, y′ = x+ z, z′ = 4x− 4y + 5z

0 0.1 0.2 0.3 0.4 0.50

5

10

15

20

25

function yprime = firstode(x,y);

yprime = x*y^2 + y;

>>xspan = [0,.5];

>>y0 = 1;

>>[x,y]=ode45(@firstode,xspan,y0);

(2) Solve the system of Lorenz Equations:

dx

dt= −σx + σy

dy

dt= ρx− y − xz

dz

dt= −βx+ xy

Solution:

For the purpose of this problem we will take σ = 10, β = 83and ρ = 28

as well as x(0) = −8, y(0) = 8, z(0) = 27

function xprime = lorenz(t,x);

sig=10;

39


beta=8/3;

rho=28;

xprime=[-sig*x(1) + sig*x(2); rho*x(1) - x(2) - x(1)*x(3);

-beta*x(3) + x(1)*x(2)];

>>x0=[-8 8 27];

>>tspan=[0,20];

>>[t,x]=ode45(@lorenz,tspan,x0);

>>plot(x(:,1),x(:,3))

Figure 4.4: Graph of Lorenz equations

−20 −15 −10 −5 0 5 10 15 205

10

15

20

25

30

35

40

45

Observe that x is stored as x(1), y as x(2) and z as x(3). xprime is a

column vector.

Components for the Lorenz Equations

>>subplot(3,1,1)

>>plot(t,x(:,1))

>>subplot(3,1,2)

40

4.4. RUNGE-KUTTA METHODS IN MATLAB 41

>>plot(t,x(:,2))

>>subplot(3,1,3)

>>plot(t,x(:,3))

Figure 4.5: Graph of Components for the Lorenz Equations

0 5 10 15 20−20

0

20

0 5 10 15 20−50

0

50

0 5 10 15 200

20

40

60

4.4 Runge-Kutta methods in MATLAB

The Runge-Kutta methods are designed to give greater accuracy and they

possess the advantage of requiring only the function values at some selected

points on the interval.

The Runga-Kutta method is halfway between a 4th order and 5th order

method. Consequently, the inbuilt MATLAB program to run the method is

called ODE45.

(1) Use ODE45 to find a numerical solution todu

dt= ku(1− u), over the

time interval [0, 100] where k = 0.2 and u(0) = 0.001

This equation is a common modelling tool for epidemiology growth of

market share and population dynamics.

41


Solution:

We construct a function representing the right-hand-side of the equation by

function output=RHS(t,u)

k=0.2;

output=k*u*(1-u);

>>ode45(‘RHS’,[0,100],0.001)

The procedure for ode45 to produce a graph automatically is

>>[t,u]=ode45(‘RHS’,[0,100],0.001);

>>plot(t,u)

Figure 4.6: Graph of epidemiology growth of market share

0 20 40 60 80 1000

0.005

0.01

0.015

0.02

0.025

Runge-Kutta method for systems of ODEs:

dx

dt= x+ y

dy

dt= y − 3x x(0) = 1, y(0) = −1, 0 ≤ t ≤ 10

42


function output = systemRHS(t,v)

x=v(1);

y=v(2);

output1=x+y;

output2=y-3*x;

output=[output1;output2];

>>[t,u]=ode45(‘systemRHS’,[0,10],[1,-1]);

>>plot(t,v(:,1))

Figure 4.7: Graph of x(t) vs t for the ODE

0 2 4 6 8 10−1

−0.5

0

0.5

1

1.5x 10

4

>>plot(t,v(:,2))

>>plot(v(:,1),v(:,2))

43


Figure 4.8: Graph of y vs t for the ODE

0 2 4 6 8 10−1

−0.5

0

0.5

1

1.5

2

2.5

3

3.5

4x 10

4

Figure 4.9: Graph of y vs x for the ODE

−1 −0.5 0 0.5 1 1.5

x 104

−1

−0.5

0

0.5

1

1.5

2

2.5

3

3.5

4x 10

4

44

4.5. EULER METHODS 45

4.5 Euler Methods

The Euler methods are simple methods of solving the first-order ODE, par-

ticularly suitable for quick programming because of their great simplicity,

although their accuracy is not high. Euler methods include three versions,

namely,

• Forward Euler method

• Modified Euler method

• Backward Euler method

Forward Euler Mehtod

If y′ = f(y, x) is derived by rewriting the forward difference approxima-

tion,

yn+1 − ynh

≈ y′n (4.1)

in to

yn+1 = yn + hf(yn, xn) (4.2)

where y′n = f(yn, xn) is used. In order to advance time steps, Eq. 4.2 is

recursively applied as

y1 = y0 + hy′0

y1 = y0 + hf(y0, x0)

y2 = y1 + hf(y1, x1)

y3 = y2 + hf(y2, x2)

. . . . . . . . .

yn = yn−1 + hf(yn−1, xn−1)

45


Modifed Euler method

The modifed Euler method is more accurate than the forward Euler

method. It is more stable. It is derived by applying the trapezoidal rule

to the solution of y′ = f(y, x)

yn+1 = yn +h

2[f(yn+1, xn+1) + f(yn, xn)] (4.3)

Backward Euler Method

The backward Euler method is based on the backward difference approx-

imation and written as

yn+1 = yn + hf(yn+1, xn+1) (4.4)

The accuracy of this method is quite the same as that of the forward Euler

method.

Euler Methods implemented in MATLAB

function [x,y]=euler_forward(f,xinit,yinit,xfinal,n)

h=(xfinal-xinit)/n;

x=[xinit zeros(1,n)]; y=[yinit zeros(1,n)];

for k=1:n

x(k+1)=x(k)+h;

y(k+1)=y(k)+h*f(x(k),y(k));

end

end

function [x,y]=euler_modified(f,xinit,yinit,xfinal,n)

h=(xfinal-xinit)/n;

46


x=[xinit zeros(1,n)];

y=[yinit zeros(1,n)];

for l=1:n

x(l+1)=x(l)+h;

ynew=y(l)+h*f(x(l),y(l));

y(l+1)=y(l)+(h/2)*(f(x(l),y(l))+f(x(l+1),ynew));

end

end

function [x,y]=euler_backward(f,xinit,yinit,xfinal,n)

h=(xfinal-xinit)/n;

x=[xinit zeros(1,n)];

y=[yinit zeros(1,n)];

for r=1:n

x(r+1)=x(r)+h;

ynew=y(r)+h*(f(x(r),y(r)));

y(r+1)=y(r)+h*f(x(r+1),ynew);

end

end

Solve the problem using Euler’s methods

dy

dx=

x

yy(0) = 1

and compare with its exact or analytical solution

y(x) =√x2 + 1

Solution:

We solve this problem by calling the three Euler’s methods.

47


f=@(x,y) x./y;

g=@(x) sqrt(x.^2+1);

xe=[0:0.01:0.3];

ye=g(xe);

% Call functions

[x1,y1]=euler_forward(f,0,1,0.3,6);

[x2,y2]=euler_modified(f,0,1,0.3,6);

[x3,y3]=euler_backward(f,0,1,0.3,6);

% Plot

plot(xe,ye,‘r-’,x1,y1,‘m-.’,x2,y2,‘g:’,x3,y3,‘b--’)

xlabel(‘x’)

ylabel(‘y’)

legend(‘Analytical’,‘Forward’,‘Modified’,‘Backward’)

axis([0 0.3 1 1.07])

Figure 4.10: Graph of the solution by Euler’s methods and the analytical

solution

0 0.05 0.1 0.15 0.2 0.25 0.31

1.01

1.02

1.03

1.04

1.05

1.06

1.07

x

y

AnalyticalForwardModifiedBackward

48


% Estimate errors

error1=[‘Forward error:

‘ num2str(-100*(ye(end)-y1(end))/ye(end)) ‘%’]

error1 =

Forward error: -0.6751%

error2=[‘Modified error:


error2 =

Modified error: 0.0003983%

error3=[‘Backward error:


error3 =

Backward error: 0.67244%

Runge-Kutta Methods

There are many variants of the Runge-Kutta methods, but the most

widely used one is the following equation,

y′ = f(x, y)

y(xn) = yn

Solution

Now use this script file

f=@(x,y) x./y;

% Calculate exact solution

49


g=@(x) sqrt(x.^2+1);

xe=[0:0.01:0.3];

ye=g(xe);

% Call function

[x4,y4]=ode45(f,[0,0.3],1);

% Plot

plot(xe,ye,‘b-’,x4,y4,‘mo’)

xlabel(‘x’)

ylabel(‘y’)

legend(‘Analytical’,‘ode45’)

axis([0 0.3 1 1.05])

Figure 4.11: Solution by Runge-Kutta Method

0 0.05 0.1 0.15 0.2 0.25 0.31

1.005

1.01

1.015

1.02

1.025

1.03

1.035

1.04

1.045

1.05

x

y

Analyticalode45

50

Chapter 5

Comparison of numerical

approximations

We will now compare various methods and their approximation for 1st order

differential equation and a system of ODEs. We will test the problem with

the following methods: (1) Euler’s method, (2) Heun’s method, (3) RK4

method and (4) RKF5 method.

5.1 Comparison for a first order ODE

Consider the first order differential equation

y′(x) = f(x, y) = y + 2x− x2, y(0) = −1

The exact solution is given by

y(x) = x2 − ex

Now this problem is tested by various methods.

51

5.1. COMPARISON FOR A FIRST ORDER ODE 52

• Euler’s Method

• Heun’s Method

• RK4 Method

• RKF5 Method

with the interval x ∈ [0, 1]

case(i):

Step size h = 0.2.

Euler Method

MATLAB code for Euler Method is

function[y,x]=my_Euler(x0,xN,N,y0)

h=(xN-x0)/(N);

x=[x0:h:xN];

y=zeros(size(x));

y(1)=y0;

for n=1:1:N,

k=h*sysfun(x(n),y(n));

y(n+1)=y(n)+k;

end

end

The function f(x, y) is defined in the file sysfun.m as follows

function f=sysfun(x,y)

f=y+2*x-x*x;

end

52


The exact solution is defined in a function as

function y=exact(x)

y=x.*x-exp(x);

end

The program should be called by the commands:

>>[y,x]=my_Euler(0,1,5,-1);

>>plot(x,exact(x),x,y,‘.’)

Figure 5.1: Solution of y = y + 2x− x2, y(0) = −1 by Euler’s Method

0 0.2 0.4 0.6 0.8 1−1.8

−1.7

−1.6

−1.5

−1.4

−1.3

−1.2

−1.1

−1

%Error tabulation

>>[x’,y’,exact(x’),abs(y’-exact(x’))]

ans =

0 -1.0000 -1.0000 0

0.2000 -1.2000 -1.1814 0.0186

53


0.4000 -1.3680 -1.3318 0.0362

0.6000 -1.5136 -1.4621 0.0515

0.8000 -1.6483 -1.5855 0.0628

1.0000 -1.7860 -1.7183 0.0677

Heun’s Method

MATLAB code for Heun’s Method is

function[y,x]=my_Heun(x0,xN,N,y0)

h=(xN-x0)/(N);

x=[x0:h:xN];

y=zeros(size(x));

y(1)=y0;

for n=1:1:N,

fn=sysfun(x(n),y(n));

k=y(n)+h*fn;

y(n+1)=y(n)+h/2*(fn+sysfun(x(n+1),k));

end

end


>>[y2,x]=my_Heun(0,1,5,-1);

>>plot(x,exact(x),x,y2,‘*’);

%Error tabulation

>>[x’,y2’,exact(x’),abs(y2’-exact(x’))]

54


Figure 5.2: Solution of y = y + 2x− x2, y(0) = −1 by Heun’s Method

0 0.2 0.4 0.6 0.8 1−1.8

−1.7

−1.6

−1.5

−1.4

−1.3

−1.2

−1.1

−1

ans=

0 -1.0000 -1.0000 0

0.2000 -1.1840 -1.1814 0.0026

0.4000 -1.3373 -1.3318 0.0055

0.6000 -1.4707 -1.4621 0.0086

0.8000 -1.5974 -1.5855 0.0119

1.0000 -1.7337 -1.7183 0.0154

RK4 Method

MATLAB code for RK4 Method is

function[y,x]=my_RK4(x0,xN,N,y0)

h=(xN-x0)/(N);

x=[x0:h:xN];

y=zeros(size(x));

y(1)=y0;

for n=1:1:N,

55


k1=h*sysfun(x(n),y(n));

k2=h*sysfun(x(n)+h/2,y(n)+k1/2);


k4=h*sysfun(x(n)+h,y(n)+k3);

y(n+1)=y(n)+k1/6+k2/3+k3/3+k4/6;

end

end


>>[y4,x]=my_RK4(0,1,5,-1);

>>plot(x,exact(x),x,y4,‘o’);

Figure 5.3: Solution of y = y + 2x− x2, y(0) = −1 by RK4 Method

0 0.2 0.4 0.6 0.8 1−1.8

−1.7

−1.6

−1.5

−1.4

−1.3

−1.2

−1.1

−1

%Error tabulation


ans =

56


0 -1.0000 -1.0000 0

0.2000 -1.1814 -1.1814 0.0000

0.4000 -1.3318 -1.3318 0.0000

0.6000 -1.4621 -1.4621 0.0000

0.8000 -1.5856 -1.5855 0.0000

1.0000 -1.7183 -1.7183 0.0000

RKF5 Method

MATLAB code for RKF5 method is

function[y,x]=my_RKF5(x0,xN,N,y0)

h=(xN-x0)/(N);

x=[x0:h:xN];

y=zeros(size(x));

y(1)=y0;

R=[16/135,0,6656/12825,28561/56430,-9/50,2/55];

for n=1:1:N,

k1=h*sysfun(x(n),y(n));


k3=h*sysfun(x(n)+3*h/8,y(n)+3*k1/32+9*k2/32);

k4=h*sysfun(x(n)+12*h/13,y(n)+1932*k1/2197-7200*k2/2197

+7296*k3/2197);

k5=h*sysfun(x(n)+h,y(n)+439*k1/216-8*k2+3680*k3/513

-845*k4/4104);

k6=h*sysfun(x(n)+h/2,y(n)-8*k1/27+2*k2-3544*k3/2565

+1859*k4/4104-11*k5/40);

57


y(n+1)=y(n)+R(1)*k1+R(2)*k2+R(3)*k3+R(4)*k4+R(5)*k5

+R(6)*k6;

end

end

It should be called by the commands:

>>[y5,x]=my_RKF5(0,1,5,-1);

>>plot(x,exact(x),x,y5,‘+’);

Figure 5.4: Solution of y = y + 2x− x2, y(0) = −1 by RKF5 Method

0 0.2 0.4 0.6 0.8 1−1.8

−1.7

−1.6

−1.5

−1.4

−1.3

−1.2

−1.1

−1

%Error tabulation


ans =

0 -1.0000 -1.0000 0

0.2000 -1.1814 -1.1814 0.0000

0.4000 -1.3318 -1.3318 0.0000

58


0.6000 -1.4621 -1.4621 0.0000

0.8000 -1.5855 -1.5855 0.0000

1.0000 -1.7183 -1.7183 0.0000

Comparison of Four Methods

Now we plot the solutions together in one figure, where the straight line

is the exact solution.

• the dot ‘.’ is the solution for Euler’s Method,

• the star ‘*’ is the solution for Heun’s Method,

• the circle ‘o’ is the solution for RK4 Method,

• the plus ‘+’ is the solution for RKF5 method.

>>x0=0;

>>x1=1;

>>N=5;

>>y0=-1;

>>[y,x]=my_Euler(x0,x1,N,y0);

>>[y2,x]=my_Heun(x0,x1,N,y0);

>>[y4,x]=my_RK4(x0,x1,N,y0);

>>[y5,x]=my_RKF5(x0,x1,N,y0);

>>ex=exact(x);

>>plot(x,ex,x,y,‘.’,x,y2,‘*’,x,y4,‘o’,x,y5,‘+’)

59


Figure 5.5: Comparison of four methods for an ODE (h = 0.2)

0 0.2 0.4 0.6 0.8 1−1.8

−1.7

−1.6

−1.5

−1.4

−1.3

−1.2

−1.1

−1

60


%Error tabulation

>>format short e

>>[x’,(y-ex)’,(y2-ex)’,(y4-ex)’,(y5-ex)’]

ans =

0 0 0 0 0

2.0000e-001 -1.8597e-002 -2.5972e-003 -3.9085e-006 -3.3122e-008

4.0000e-001 -3.6175e-002 -5.4553e-003 -8.0717e-006 -6.0133e-008

6.0000e-001 -5.1481e-002 -8.5628e-003 -1.2411e-005 -7.6702e-008

8.0000e-001 -6.2779e-002 -1.1891e-002 -1.6799e-005 -7.6884e-008

1.0000e+000 -6.7702e-002 -1.5385e-002 -2.1047e-005 -5.2608e-008

case:2

Step size h = 0.1

>>x0=0;

>>x1=1;

>>N=10;

>>y0=-1;

>>[y,x]=my_Euler(x0,x1,N,y0);

>>[y2,x]=my_Heun(x0,x1,N,y0);

>>[y4,x]=my_RK4(x0,x1,N,y0);

>>[y5,x]=my_RKF5(x0,x1,N,y0);

>>ex=exact(x);

>>plot(x,ex,x,y,‘.’,x,y2,‘*’,x,y4,‘o’,x,y5,‘+’)

61


Figure 5.6: Comparison of four methods for an ODE (h = 0.1)

0 0.2 0.4 0.6 0.8 1−1.8

−1.7

−1.6

−1.5

−1.4

−1.3

−1.2

−1.1

−1

62


%Error tabulation

>>format short e

>>[x’,(y-ex)’,(y2-ex)’,(y4-ex)’,(y5-ex)’]

ans =......x_n......Euler......Heun.........RK4.......... RKF5

0 0 0 0 0

1.0000e-001 -4.8291e-003 -3.2908e-004 -1.2359e-007 -5.2611e-010

2.0000e-001 -9.5972e-003 -6.7474e-004 -2.5127e-007 -1.0099e-009

3.0000e-001 -1.4241e-002 -1.0368e-003 -3.8252e-007 -1.4368e-009

4.0000e-001 -1.8685e-002 -1.4150e-003 -5.1670e-007 -1.7892e-009

5.0000e-001 -2.2840e-002 -1.8086e-003 -6.5295e-007 -2.0470e-009

6.0000e-001 -2.6598e-002 -2.2167e-003 -7.9024e-007 -2.1863e-009

7.0000e-001 -2.9836e-002 -2.6380e-003 -9.2727e-007 -2.1792e-009

8.0000e-001 -3.2407e-002 -3.0708e-003 -1.0625e-006 -1.9935e-009

9.0000e-001 -3.4139e-002 -3.5128e-003 -1.1940e-006 -1.5918e-009

1.0000e+000 -3.4835e-002 -3.9613e-003 -1.3194e-006 -9.3046e-010

Now we compare the errors at the point xn = 0.1. We see that when we

half the step size,

• the error for Euler’s Method becomes circa 1/2.

• the error for Heun’s Method, it is circa 1/4 = 1/22.

• the error for RK4 Method, it is circa 1/16 = 1/24.

• the error for RKF5 Method, it si circa 1/32 = 1/25

This shows that the Euler’s Method is 1st order, Heun’s Method is 2nd

order, RK4 Method is 4th order and RKF5 is a 5th order method.

63

5.2. COMPARISON FOR A SYSTEM OF ODES 64

5.2 Comparison for a system of ODEs

We solve the problem:

d2

dx+ 2

dy

dx+ 0.75y = 0, y(0) = 3, y′(0) = −2.5.

The exact solution is

y′(x) = y1(x) = 2e−0.5x + e−1.5x;

y′(x) = y2(x) = −e−0.5x − 1.5e−1.5x

Here, we define the system in a function as follows

function f=sysfunsys(x,y)

f=zeros(2,1);

f(1)=y(2);

f(2)=-2*y(2)-0.75*y(1);

end

The exact solution is defined in a function

function y=exactsys(x)

y1=2*exp(-0.5*x)+exp(-1.5*x);

y2=-exp(-0.5*x)-1.5*exp(-1.5*x);

y=[y1;y2];

end

Euler Method

The MATLAB code for Euler method is

64


function [y,x]=sys_Euler(x0,xN,N,y0)

h=(xN-x0)/(N);

x=[x0:h:xN];

y=zeros(length(y0),length(x));

y(:,1)=y0;

for n=1:1:N,

k=h*sysfunsys(x(n),y(:,n));

y(:,n+1)=y(:,n)+k;

end

end

Heun’s Method

The MATLAB code for Heun’s Method is

function [y,x]=sys_Heun(x0,xN,N,y0)

h=(xN-x0)/(N);

x=[x0:h:xN];


y(:,1)=y0;

for n=1:1:N,

fn=sysfunsys(x(n),y(:,n));

k=y(:,n)+h*fn;

y(:,n+1)=y(:,n)+h/2*(fn+sysfunsys(x(n+1),k));

end

end

RK4 Method

The MATLAB code for RK4 Method is

65


function[y,x]=sys_RK4(x0,xN,N,y0)

h=(xN-x0)/(N);

x=[x0:h:xN];


y(:,1)=y0;

for n=1:1:N,

k1=h*sysfunsys(x(n),y(:,n));

k2=h*sysfunsys(x(n)+h/2,y(:,n)+k1/2);


k4=h*sysfunsys(x(n)+h,y(:,n)+k3);

y(:,n+1)=y(:,n)+k1/6+k2/3+k3/3+k4/6;

end

end

RKF5 Method

The MATLAB code for RKF5 Method is

function [y,x]=sys_RKF5(x0,xN,N,y0)

h=(xN-x0)/(N);

x=[x0:h:xN];


y(:,1)=y0;

R=[16/135, 0, 6656/12825, 28561/56430, -9/50, 2/55];

for n=1:1:N,

k1=h*sysfunsys(x(n),y(:,n));


k3=h*sysfunsys(x(n)+3*h/8,y(:,n)+3*k1/32+9*k2/32);

66


k4=h*sysfunsys(x(n)+12*h/13,y(:,n)+1932*k1/2197

-7200*k2/2197+7296*k3/2197);

k5=h*sysfunsys(x(n)+h,y(:,n)+439*k1/216-8*k2

+3680*k3/513-845*k4/4104);

k6=h*sysfunsys(x(n)+h/2,y(:,n)-8*k1/27+2*k2

-3544*k3/2565+1859*k4/4104-11*k5/40);

y(:,n+1)=y(n)+R(1)*k1+R(2)*k2+R(3)*k3+R(4)*k4

+R(5)*k5+R(6)*k6;

end

end

Comparison of Four Methods

>>x0=0;

>>x1=1;

>>N=5;

>>y0=[3;-2.5];

>>[y1,x]=sys_Euler(x0,x1,N,y0);

>>[y2,x]=sys_Heun(x0,x1,N,y0);

>>[y4,x]=sys_RK4(x0,x1,N,y0);

>>[y5,x]=sys_RKF5(x0,x1,N,y0);

>>ex=exactsys(x);

>>subplot(2,1,1);

>>plot(x,ex(1,:),x,y1(1,:),‘+’,x,y2(1,:),‘*’,x,y4(1,:),‘o’,

x,y5(1,:),‘.’),ylabel(‘y1’)

>>subplot(2,1,2);

>>plot(x,ex(2,:),x,y1(2,:),‘+’,x,y2(2,:),‘*’,x,y4(2,:),‘o’,

67


x,y5(2,:),‘.’),ylabel(‘y2’)

Now we plot the solution together in a figure, where the straight line is

the exact solution.

• the dot ‘+’ is the solution for Euler’s Method,

• the star ‘o’ is the solution for Heun’s Method,

• the circle ‘*’ is the solution for RK4 Method,

• the plus ‘.’ is the solution for RKF5 method.

%Error tabulation for $y_1$

>>format short e

>>[x’,ex(1,:)’-y(1,:)’,ex(1,:)’-y2(1,:)’,

ex(1,:)’-y4(1,:)’,ex(1,:)’-y5(1,:)’]

ans =......x_n ....Euler.....Heun........RK4.......... RKF5

0 0 0 0 0

2.0000e-001 5.0493e-002 -4.5069e-003 -1.9443e-005 6.2198e-007

4.0000e-001 7.6273e-002 -6.8019e-003 -2.8862e-005 4.1484e+000

6.0000e-001 8.7206e-002 -7.7228e-003 -3.2146e-005 -3.1614e-002

8.0000e-001 8.9534e-002 -7.8223e-003 -3.1841e-005 -2.4988e+000

1.0000e+000 8.7141e-002 -7.4594e-003 -2.9585e-005 3.1527e+000

%Error tabulation for $y_2$

>>[x’,ex(2,:)’-y(2,:)’,ex(2,:)’-y2(2,:)’,

ex(2,:)’-y4(2,:)’,ex(2,:)’-y5(2,:)’]

ans =......x_n ....Euler.....Heun........RK4.......... RKF5

0 0 0 0 0

68


Figure 5.7: Comparison of four methods for a system of ODEs

0 0.2 0.4 0.6 0.8 1−2

0

2

4

6

y1

0 0.2 0.4 0.6 0.8 1−5

0

5

y2

69


2.0000e-001 -6.6065e-002 6.4353e-003 2.9001e-005 -5.5000e+000

4.0000e-001 -9.6948e-002 9.6143e-003 4.2996e-005 2.3605e+000

6.0000e-001 -1.0717e-001 1.0785e-002 4.7816e-005 -5.5081e+000

8.0000e-001 -1.0586e-001 1.0770e-002 4.7275e-005 -2.9514e+000

1.0000e+000 -9.8631e-002 1.0099e-002 4.3828e-005 2.1544e+000

Now we see that errors are much smaller with higher order methods.

When h is halved, errors are smaller for all methods.

• For Euler’s Method it is circa 12.

• For Heun’s Method it is circa 14.

• For RK4 Method it is circa 116

• For RKF5 Method it is circa 1256

.

This agree with the theoretical result that the Euler’s Method is 1st order,

Heun’s Method is 2nd order, RK4Method is 4th order and RKF5 is 5th order.

70

Chapter 6

Solving Partial Differential

Equations with MATLAB

We solve the linear advection equation for u(x, t),

∂u

∂t= c

∂u

∂x

defined on (−∞,∞),t ∈ [0,∞) and with the boundary condition

u(x, 0) = f(x).

% Define and initialize the variables

for i = 1:10

xi(i) = i;

for j = 1:3

uij(i,j) = 0;

end

end

% The boundary condition (which essentially defines the "initial state")

71

72

uij(4,1) = 1;

% integration to t = 0.1 and 0.2

for j = 1:2

for i = 2:9

uij(i,j+1) = -0.25*uij(i-1,j)+uij(i,j)+0.25*uij(i+1,j);

end

end

% making the plots for t = 0, 0.1, and 0.2

for i = 1:10

u0(i) = uij(i,1);

u1(i) = uij(i,2);

u2(i) = uij(i,3);

z(i) = 0;

end

>>subplot(3,1,1);

>>plot(xi,z,‘k--’,xi,u0,‘r-o’);

>>axis([1 10 -0.5 1])

>>subplot(3,1,2);


>>axis([1 10 -0.5 1])

>>subplot(3,1,3);


>>axis([1 10 -0.5 1])

72

73

Figure 6.1: Graph for the linear advection equation

1 2 3 4 5 6 7 8 9 10−0.5

0

0.5

1

1 2 3 4 5 6 7 8 9 10−0.5

0

0.5

1

1 2 3 4 5 6 7 8 9 10−0.5

0

0.5

1

73

6.1. POISSON’S EQUATION ON A UNIT DISC 74

6.1 Poisson’s Equation on a Unit Disc

(1) Poisson’s Equation on a Unit Disk

The particular PDE is

−△ u = 1,

on the unit disk with zero-Dirichlet boundary conditions. The exact solution

is

u(x, y) =1− x2 − y2

4

Problem Definition:

The following variables will define our problem

• g: A specification function that is used by initmesh.

• b: A boundary file used by assempde.

• c,a,f: The coefficients and inhomogeneous term.

>>g=‘circleg’;

>>b=‘circleb1’;

>>c=1;

>>a=0;

>>f=1;

Generate Initial Mesh

The function initmesh takes a geometry specification function and returns

a discretization of that domain. The ‘hmax’ option lets the user specify the

maximum edge length. In this case, the domain is a unit disk, a maximum

edge length of one creates a coarse discretization.

74


[p,e,t]=initmesh(g,‘hmax’,1);

figure;

pdemesh(p,e,t);

axis equal

Figure 6.2: Graph for Poisson equation

−1 −0.5 0 0.5 1−1

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1

Refinement

We repeatedly refine the mesh until the infinity-norm of the error vector

is less than a 10−3.

er = Inf;

while er > 0.001

[p,e,t]=refinemesh(g,p,e,t);

u=assempde(b,p,e,t,c,a,f);

exact=(1-p(1,:).^2-p(2,:).^2)’/4;

er=norm(u-exact,‘inf’);

fprintf(‘Error: %e. Number of nodes: %d\n’,er,size(p,2));

end

75


Error: 1.292265e-02. Number of nodes: 25




Plot Final Mesh

>>figure;

>>pdemesh(p,e,t);

>>axis equal

Figure 6.3: Graph for Refinement

−1 −0.5 0 0.5 1−1

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1

Plot Error

76


The scale of the vertical axis shows that the error is small and of the

order 10−4.

>>figure;

>>pdesurf(p,t,u-exact);

Figure 6.4: Graph for Exact

−1−0.5

00.5

1

−1

−0.5

0

0.5

1−1

0

1

2

3

4

x 10−4

Plot FEM Solution

>>figure;

>>pdesurf(p,t,u);

77

6.2. HELMHOLTZ’S EQUATION ON A UNIT DISK WITH A SQUAREHOLE 78

Figure 6.5: Graph for FEM Solution

−1−0.5

00.5

1

−1

−0.5

0

0.5

10

0.1

0.2

0.3

0.4

6.2 Helmholtz’s Equation on a Unit Disk with

a Square Hole

(2) The Helmholtz equation, an elliptic equation, ie., the time-independent

form of wave equation is

−△ u− k2u = 0

Problem Definition

The following variables will define our problem:

• g: A specification function that is used by initmesh.

• b: A boundary file used by assempde.

• c,a,f: The coefficients and inhomogeneous term.

>>g=‘scatterg’;

>>b=‘scatterb’;

>>c=1;

>>k=60;

78

6.3. SOLVE FOR COMPLEX AMPLITUDE 79

>>a=-k^2;

>>f=0;

Create Mesh

We need a fine mesh to resolve the waves. To achieve this, we refine the

initial mesh twice.

>>[p,e,t]=initmesh(g);

>>[p,e,t]=refinemesh(g,p,e,t);

>>[p,e,t]=refinemesh(g,p,e,t);

>>pdemesh(p,e,t);

axis equal

Figure 6.6: Graph for Helmholtz equation

0.4 0.6 0.8 1 1.2

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

6.3 Solve for Complex Amplitude

The real part of the vector u stores an approximation to a real-valued solution

of the Helmholtz equation.

>>u=assempde(b,p,e,t,c,a,f);

79

6.4. ANIMATE SOLUTION TO WAVE EQUATION 80

Plot FEM Solution

>>figure;

>>h = newplot;

>>set(get(h,‘Parent’),‘Renderer’,‘zbuffer’)

>>pdeplot(p,e,t,‘xydata’,real(u),‘zdata’,real(u),‘mesh’,‘off’);

>>colormap(cool)

Figure 6.7: Graph for Complex Amplitude

6.4 Animate Solution to Wave Equation

Using the solution to the Helmholtz equation, we construct an animation

showing the corresponding solution to the time-dependent wave equation.

>>figure;

>>m=10;

>>h = newplot;

>>hf=get(h,‘Parent’);

>>set(hf,‘Renderer’,‘zbuffer’)

80

6.4. ANIMATE SOLUTION TO WAVE EQUATION 81

>>axis tight, set(gca,‘DataAspectRatio’,[1 1 1]);

>>axis off

>>M=moviein(m,hf);

>>maxu=max(abs(u));

>>for j=1:m

>>uu=real(exp(-j*2*pi/m*sqrt(-1))*u);

>>pdeplot(p,e,t,‘xydata’,uu,‘colorbar’,‘off’,‘mesh’,‘off’);

>>caxis([-maxu maxu]);

>>axis tight, set(gca,‘DataAspectRatio’,[1 1 1]);

>>axis off;

>>M(:,j)=getframe(hf);

>>end

>>movie(hf,M,2);

Figure 6.8: Graph for Wave Equation

81

Conclusion

In this project we demonstrated that the different types of differential equa-

tion could be solved with MATLAB.

• Ordinary differential equation and partial differential equation could

be easily solved by using MATLAB coding and get the figure for any

complicated equations.

• By comparing Four numerical types of methods, we arrived at the the-

oretical result that Eulers method is 1st order, Heuns method is 2nd

order, RK4 method is 4th order and RKF5 is 5th order method.

• We attempted to simulate the solution of few partial differential equa-

tions.

82

Bibliography

[1] Curtis F. Gerald, Patrick O. Wheatley, “Applied Numerical Analysis”,

Addision-Wesley, Fifth Edition

[2] Wen Shen, “Journal on Introduction to Numerical computation”,

xuru.org, 2012.

[3] P.Howard, “Solving Ordinary Equations in MATLAB”, Fall, 2009.

[4] David Houcque, “Applications of MATLAB on Ordinary Differential

Equations”, Robert R. McCormick School of Engineering and Applied

Science, Northwestern University.

[5] Rudra Pratap, “Getting Started with MATLAB”, Department of Me-

chanical Engineering, Indian Institute of Science, Bangalore, 2010.

[6] Hari Kishan, “Differential Equations”, Atlantiv Publishers And Distrib-

utors, 2006.

83

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