Numerical Methods for Solving Systems of Ordinary

Differential Equations

Simruy Hürol

Submitted to the

Institute of Graduate Studies and Research

in partial fulfillment of the requirements for the Degree of

Master of Science

in

Applied Mathematics and Computer Science

Eastern Mediterranean University

January 2013

Gazimağusa, North Cyprus

Approval of the Institute of Graduate Studies and Research

Prof. Dr. Elvan Yılmaz

Director

I certify that this thesis satisfies the requirements as a thesis for the degree of Master of

Science in Applied Mathematics and Computer Science.

Prof. Dr. Nazım Mahmudov

Chair, Department of Mathematics

We certify that we have read this thesis and that in our opinion it is fully adequate in

scope and quality as a thesis for the degree of Master of Science in Applied

Mathematics and Computer Science.

Asst. Prof. Dr. Mehmet Bozer

Supervisor

Examining Committee

1. Assoc. Prof. Dr. Derviş Subaşı

2. Asst. Prof. Dr. Mehmet Bozer

3. Asst. Prof. Dr. Mustafa Rıza

ABSTRACT

This thesis concantrates on numerical methods for solving ordinary differential equa-

tions. Firstly, we discuss the concept of convergence, local-truncation error, global-

truncation error, consistency, types of stability, zero-stability, and weak-stability. Af-

terwards, we inform some materials for Euler and Runge-Kutta method. The given

ordinary differential equation is analyzed on Euler and Runge-Kutta method to find

the approximated solution with the given initial conditions. Then, the stability of

each method is examined briefly. We also focus on numerical methods for systems.

Then, the reason of the stiff system is discussed. After investigating the numerical

methods, we gave advantages and disadvantages of Euler method and Fourth Order

Runge-Kutta method. Finally, numerical experiments is applied on Explicit Euler

method and Explicit Fourth Order Runge-Kutta method. The approximated solutions

with different step-size and analytical solutions of methods are computed in Matlab

software. The computation of approximated solutions of methods are compared with

analytical solutions. Then we discussed the accuracy of these methods when they

are applied to the specified system in Chapter 7. Finally, we conclude that Explicit

Fourth Order Runge-Kutta method is more accurate than the Explicit Euler method.

Keywords: Ordinary Differential Equations, Numerical solutions, Euler’s method,

Runge-Kutta method, Stiff System

iii

ÖZ

Bu çalısmada adi diferansiyel denklemlerin çözümü için sayısal yöntemler irdelen-

mistir. Ilk olarak, yakınsama kavramı, yerel kesme hatası, küresel kesme hatası,

tutarlılık, kararlılık türleri, sıfır kararlılık ve zayıf kararlılık kavramları incelenmistir.

Ayrıca, Euler ve Runge-Kutta metodları verilmistir. Verilen bir diferansiyel den-

klemin, yaklasık çözümünü bulmak için Euler ve Runge-Kutta yöntemi verilen bas-

langıç kosulları ile analiz edilmis ve her bir yöntem için kararlılık kısaca ele alın-

mıstır. Daha sonra, verilen sistemler için sayısal yöntemlere odaklanılmıs ve sert

sistemin çıkıs sebebi incelenmistir. Euler yöntemin ve dördüncü dereceden Runge-

Kutta yöntemin avantajları ve dezavantajları verilmistir. Son olarak, sayısal deneyler

üzerinde Açık Euler ve Açık dördüncü dereceden Runge-Kutta yöntemleri uygu-

lanmıstır. Farklı adım büyüklügü ele alınarak yaklasılan çözümler ve yöntemlerin

analitik çözümleri Matlab yazılımıkullanılarak hesaplanmıstır. Elde edilen yaklasılır

çözümler ile analitik çözümler karsılastırılmıstır. Daha sonra yöntemler Bölüm 7’de

belirtilen sistem üzerine uygulanıp yöntemlerin dogrulugu tartısılmıstır. Son olarak,

Açık Runge-Kutta yöntemin yaklastırılmıs çözümünün Açık Euler methoduna göre

daha az hatalı oldugu sonucuna varılmıstır.

Anahtar Kelimeler: Adi Diferensiyel Denklemler, Sayısal Çözümler, Euler Yön-

temi, Runge-Kutta Yöntemi, Sert Sistemi

iv

ACKNOWLEDGMENTS

First of all, I would like to express my gratidute to my supervisor, Dr. Mehmet

Bozer for his recommendation, patient, and encouragement during this thesis. Spe-

cial thanks are extended to Dr. Arif Akkeles and Dr. Mustafa Rıza for assist on

editing my dissertation. I would like to express my sincere thanks to my cousin Dr.

Neyre Tekbıyık Ersoy for her guidance and persistent help on my dissertation. I am

also thankful to my office mates for their help and moral support. Furthermore, I

wish to thank my closest friend for unending spiritual support throught all the hard

times. Last, my thanks to my family for their support during this studies.

v

TABLE OF CONTENTS

ABSTRACT . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . iii

ÖZ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . iv

ACKNOWLEDGMENTS . . . . . . . . . . . . . . . . . . . . . . . . . . . . v

LIST OF TABLES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ix

LIST OF FIGURES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . x

1 INTRODUCTION . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

2 CONCEPT OF CONVERGENCE, LOCAL-GLOBAL TRUNCATION ER-

ROR, CONSISTENCY, STABILITY, WEAK STABILITY AND ZERO STA-

BILITY . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

3 STIFF SYSTEMS AND PROBLEMS OF STABILITY FOR STIFF SYS-

TEM . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19

3.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19

3.2 Definition of Stiffness . . . . . . . . . . . . . . . . . . . . . . . . . 20

3.3 The Problem of Stability for Stiff System . . . . . . . . . . . . . . . 22

4 EULER METHOD . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26

4.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26

4.2 Definition of Euler’s Method . . . . . . . . . . . . . . . . . . . . . 26

4.3 Error Analysis of Euler’s Method . . . . . . . . . . . . . . . . . . . 31

4.4 Numerical Stability of Euler’s Method . . . . . . . . . . . . . . . . 38

vi

4.5 Rounding Error Accumulation of Euler’s Method . . . . . . . . . . 44

4.6 Euler’s Method for Systems . . . . . . . . . . . . . . . . . . . . . . 47

5 RUNGE-KUTTA METHOD . . . . . . . . . . . . . . . . . . . . . . . . 49

5.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49

5.2 Problem of Runge-Kutta Method . . . . . . . . . . . . . . . . . . . 50

5.3 Explicit Runge Kutta Method . . . . . . . . . . . . . . . . . . . . . 51

5.4 Implicit Runge-Kutta Method . . . . . . . . . . . . . . . . . . . . . 59

5.5 Fourth Order Runge-Kutta Method . . . . . . . . . . . . . . . . . . 60

5.6 Stability of Runge-Kutta Method . . . . . . . . . . . . . . . . . . . 64

5.6.1 Absolute Stability of Runge-Kutta Method . . . . . . . . . 70

6 SYSTEM OF ORDINARY DIFFERENTIAL EQUATIONS . . . . . . . 74

6.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 74

6.2 Stability Theory for System . . . . . . . . . . . . . . . . . . . . . . 76

7 NUMERICAL EXPERIMENTS ON SIMPLE SYSTEMS . . . . . . . . 83

7.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83

7.2 Numerical Experiments on Stiff System . . . . . . . . . . . . . . . 84

7.2.1 Numerical Experiments on Explicit Euler method . . . . . . 84

7.2.2 Numerical Experiments on Explicit Fourth Order Runge-

Kutta method . . . . . . . . . . . . . . . . . . . . . . . . . 86

8 COMPARISON OF EULER METHOD AND RUNGE-KUTTA METHOD

WITH ADVANTAGES AND DISADVANTAGES . . . . . . . . . . . . . 88

9 CONCLUSION . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 90

REFERENCES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91

vii

APPENDICES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 96

Appendix A. Explicit Euler method . . . . . . . . . . . . . . . . . . . . . 97

Appendix B. Fourth Order Runge Kutta method . . . . . . . . . . . . . . 112

viii

LIST OF TABLES

Table 1. Explicit Euler method with h=0.1 for n=10 . . . . . . . . . . . 98

Table 2. Explicit Euler method with h=0.05 for n=20 . . . . . . . . . . 100

Table 3. Explicit Euler method with h=0.025 for n=40 . . . . . . . . . 102

Table 4. Explicit Euler method with h=0.0125 for n=80 . . . . . . . . . 104

Table 5. Explicit Euler method with h=0.00625 for n=160 . . . . . . . 106

Table 6. Explicit Euler method with h=0.003125 for n=320 . . . . . . . 108

Table 7. Explicit Euler method with h= 0.0015625 for n=640 . . . . . . 110

Table 8. Fourth Order Runge Kutta method with h=0.1 for n=10 . . . . 113

Table 9. Fourth Order Runge Kutta method with h=0.05 for n=20 . . . 115

Table 10. Fourth Order Runge Kutta method with h=0.025 for n=40 . . . 117

Table 11. Fourth Order Runge Kutta method with h=0.0125 for n=80 . . 119

Table 12. Fourth Order Runge Kutta method with h=0.00625 for n=160 . 121

ix

LIST OF FIGURES

Figure 3.1 The region of A−Stability . . . . . . . . . . . . . . . . . . . 23

Figure 3.2 The region of A (α)−Stability . . . . . . . . . . . . . . . . . 24

Figure 3.3 The region of Stiff Stability . . . . . . . . . . . . . . . . . . 25

Figure 4.1 The region of absolute stability of Forward Euler Method . . . 42

Figure 4.2 The region of absolute stability of Backward Euler Method . . 44

Figure 4.3 Rounding Error Curve for (4.5.7) . . . . . . . . . . . . . . . 47

Figure 1. Graph of approximated solution and exact solution by using

Explicit Euler method with h = 0.1 . . . . . . . . . . . . . . . 99

Figure 2. Graph of approximated solution and exact solution by using

Explicit Euler method with h = 0.05 . . . . . . . . . . . . . . 101

Figure 3. Graph of approximated solution and exact solution by using

Explicit Euler method with h = 0.025 . . . . . . . . . . . . . . 103

Figure 4. Graph of approximated solution and exact solution by using

Explicit Euler method with h = 0.0125 . . . . . . . . . . . . . 105

Figure 5. Graph of approximated solution and exact solution by using

Explicit Euler method with h = 0.00625 . . . . . . . . . . . . 107

Figure 6. Graph of approximated solution and exact solution by using

Explicit Euler method with h = 0.003125 . . . . . . . . . . . . 109

Figure 7. Graph of approximated solution and exact solution by using

Explicit Euler method with h = 0.0015625 . . . . . . . . . . . 111

x

Figure 8. Graph of approximated solution and exact solution by using

Explicit RK4 method with h = 0.1 . . . . . . . . . . . . . . . 114

Figure 9. Graph of approximated solution and exact solution by using

Explicit RK4 method with h = 0.05 . . . . . . . . . . . . . . . 116

Figure 10. Graph of approximated solution and exact solution by using

Explicit RK4 method with h = 0.025 . . . . . . . . . . . . . . 118

Figure 11. Graph of approximated solution and exact solution by using

Explicit RK4 method with h = 0.0125 . . . . . . . . . . . . . 120

Figure 12. Graph of approximated solution and exact solution by using

Explicit RK4 method with h = 0.00625 . . . . . . . . . . . . . 122

xi

Chapter 1

INTRODUCTION

Differential equations are among the most important mathematical tools used in pro-

ducing models in the physical sciences, biological sciences, and engineering. In this

text, we consider numerical methods for solving ordinary differential equations, that

is, those differential equations that have only one independent variable. We consider

the differential equations in the form of

y′ (x) = f (x,y (x))

where y(x) is an unknown function that is being sought. The given function f (x,y)

of two variables defines the differential equation. This equation is called a first order

differential equation because it contains a first order derivative of unknown function.

The numerical methods for a first-order equation can be extended to a system of first

order equations ([12]). The brief of thesis is organized as follows: In the second

chapter, the concept of convergence, local–global truncation error, consistency, zero-

stability, weak-stability are investigated for ordinary differential equations. At this

stage we focused on simple first order differential equations, and multistep method

for the given necessary conditions in Chapter 2. The purpose of this study is to

find the best approximated solution which converge to analytical solution. Chapter

1

3 consist of the problem of stability for stiff system which focuses on the model

problem. It is the instructive form to determine the stiff numerical method. Then, in

the Chapter 4, Euler’s method is reviewed. The derivation of Euler’s method is stated

and using some basic definitions, error and the region of stability is determined. In

the next chapter, we discuss the derivation of the families of Runge-Kutta method.

After completing this, we present some necessary definitions of the stability, and

examined the absolute stability of Fourth Order Runge-Kutta method. In chapter 6,

we give the general form of system for first order equations. The aim of this study

is to compute the approximated solutions of system of differential equations. Then,

it follows the stability theory for systems. In the last chapter, some examples are

given, and each numerical method is solved by using Matlab software. After that,

the results of numerical methods are analyzed. Finally, we concluded by giving the

advantages and disadvantages of the specified numerical methods in Chapter 4 and

Chapter 5. Additionally, in this dissertation, equations and formulas is benefited from

two books.([15]), ([25])

2

Chapter 2

CONCEPT OF CONVERGENCE, LOCAL-GLOBAL

TRUNCATION ERROR, CONSISTENCY, STABILITY, WEAK

STABILITY AND ZERO STABILITY

A simple first order differential equation can be defined as

∂y∂x= y′(x) = f (x,y(x)) (2.1)

where ∂y∂x is describing the first derivative of y with respect to time and f (x,y) is any

function of time and y. We look for a solution to (2.1) for x > x0 as

y(x0) = b (2.2)

where x0 < x < b, x0 and b are two real numbers. The first order differential equation

(2.1) with the condition described in (2.2) is known as the Initial Value Problem.

Analytical solution is defined by y(x), and approximate solution is represented by

yn. Introduce a general multistep method for solving initial value problem. Let the

several nodes given by xn = x0+nh, n = 0,1,2, ...

The general conformation of multistep method can be shown as

yn+1 =

p∑j=0

a jyn− j+hp∑

j=−1

b j f (xn− j,yn− j) (2.3)

3

where h > 0 and a0,a1, .....,ap,b−1,b0, ....,bp are given constants term with∣∣∣ap

∣∣∣+∣∣∣bp∣∣∣ , 0 as p ≥ 0. This is named (p+1) step method. In following chapters, we will

see that some methods can be produced by using multistep method as Euler Method,

The Midpoint Rule, Trapezoidal Rule, and Adams methods. When any numerical

method applied to a system, errors comprise in two forms.

1. Truncation (or discretization) Error: It is caused when approximations are used

to estimate some quantity. Truncation error are composed of two parts.

(a) Local Truncation Error: It is defined by Tn+p, and introduced the local

error at xn+p. It is shown as

Tn+p = y(xn+p)− y(xn)−hΦ(xn,y(xn);h) (2.4)

Local Truncation error arises when a numerical method is used to solve

initial value problem. The error occur after the first step and form in each

step.

Tn+1 = y(xn+1)− yn+1

To obtain the local truncation error, take difference between left hand side

and right hand side of method, and expand by using Taylor series. The

remaining term is called Local Truncation error. Finally, divide the result

4

to step size h. It is shown as

τn(y) =1h

Tn (2.5)

If y(x) is assumed to be sufficiently differentiable, then the local truncation

error for both explicit and implicit method can be written in the form

y(xn+p)− yn+p =Cp+1hp+1yp+1(xn)+O(hp+2)

Tn+p =Cp+1hp+1yp+1(xn)+O(hp+2)

where Cp+1 states the error constant, p indicates the order of the method,

and Cp+1hyp+1 (x) is the principal local truncation error ([27]). Thus, if

the local truncation error is O(hp+1), we can say that p represent the order

of the method. This means that∣∣∣Tn+p

∣∣∣ < Cp+1hp+1. If p is larger, then the

method is more accurate.

(b) Global (or Accumulated) Truncation Error: It is denoted by en which is

expressed

en = y(xn)− yn (2.7)

where y(xn) is exact solution and yn is approximate solution. Global Trun-

cation error is caused by the accumulation of the local error in all of the

iterations.

2. Round-off Error: It originate due to the operations of computer that takes lim-

5

ited edition number of digit. After calculating the approximations of methods,

the result is dropped in specific location. It is denoted by Rn+k which is also

committed at the n th application of the methods.

The general multistep method of initial value problem satisfies

y(xn+1) =p∑

j=0

a jy(xn− j)+hp∑

j=−1

b j f (xn− j,y(xn− j))+Tn+p (2.8)

where Tn+p is the Local Truncation error, and

yn+1 =

p∑j=0

a jyn− j+hp∑

j=−1

b j f (xn− j,yn− j)+Rn+p (2.9)

where {yn} is the solution of multistep method and Rn+p is the Round-off error. Sub-

tracting (2.9) from (2.8) and considering the global error en = y(xn)− yn, we obtain

en+1 =

p∑j=0

a jen− j+hp∑

j=−1

b j[f (xn− j,y(xn− j)− f (xn− j,yn− j)

]+∅n+p

where ∅ is obtained by subtracting round-off error from local truncation error as

∅ = Tn+p−Rn+p. For any differentiable function y(x), define the truncation error

Tn(y) = y(xn+1)−

p∑j=0

a jy(xn− j)+hp∑

j=−1

b jy′(xn− j)

, n ≥ p (2.10)

To state the error of numerical methods, use the Taylor expansion series for y(xn+1).

Then the error will be determined by

Tn(y) = O(hp+1)

6

Consistency is used to indicate the accuracy of the method. Consider the expression

of (2.5) to follow the consistency condition that is satisfied by

τn(y) −→ 0 as h −→ 0

Use the notation (2.5), and rewrite the multistep method. We obtain

y(xn+1) =p∑

j=0

a jy(xn− j)+hp∑

j=−1

b jy′(xn− j)+hτn(y)

Then introduce

τ(h) ≡ maxx0≤xn≤b

|τn(y)| (2.11)

where node points [x0,b] are x0 ≤ x1 ≤ .........≤ xN ≤ b. Multistep method is consistent

if y(x) is continuously differentiable on [x0,b] and satisfied the following status

τ(h) −→ 0 as h −→ 0 (2.12)

Order of the methods have a major role for the speed of the convergence. If the

condition (2.11) goes to zero rapidly, then approximate solution will be closer to the

exact solution. Thus, the speed of the convergence of approximated solution {yn}

to the exact solution y(x) is related with the speed of the convergence of condition

(2.11). When step size h will be chosen small, the intervals will increase. This

provide to be closer to 0. This relation is defined as

τ(h) = O(hp) (2.13)

7

Theorem 2.0.1 ([1])Let m ≥ 1 be a given integer. For (2.11) to hold for all continu-

ously differentiable functions y(x), that is, for the method (2.3) to be consistent, it is

necessary and sufficient that

p∑j=0

a j = 1, (2.14)

p∑j=−1

b j−p∑

j=0

ja j = 1 (2.15)

Further, for τ(h) = O(hm) to be valid for all functions y(x) that are (m+ 1) times

continuously differentiable, it is necessary and sufficient that (2.14)-(2.15) hold and

that

p∑j=0

(− j)ia j+ ip∑

j=−1

(− j)i−1b j = 1, i = 2, ...,m

Proof. ([2])Note that

Tn (αy+βw) = αTn (y)+βTn (w) (2.17)

for all constant α,β and for all differentiable functions y(x),W. To examine the conse-

quences of (2.11)-(2.12) and (2.13), expand y(x) about xn using the Taylor’s theorem

to obtain

y(x) =m∑

i=0

1i!

(x− xn)i y(i) (xn)+Rm+1(x), (2.18)

8

Rm+1(x) =1

m!

∫ x

xn

(x− s)m y(m+1) (s) ds

=(x− xs)m+1

(m+1)!y(m+1) (ξn) (2.19)

with ξn between x and xn. We are assuming that y(x) is m+ 1 times continuously

differentiable on the interval bounded by x and xn. Substituting into (2.10) and using

(2.17), we obtain

Tn(y) =m∑

i=0

1i!

y(i)(xn)Tn((x− xn)i

)+Tn (Rm+1)

It is necessary to calculate Tn((x− xn)i

)for i ≥ 0

• For i = 0,

Tn (1) = c0 ≡ 1−p∑

j=0

a j

• For i ≥ 1,

Tn((x− xn)i

)= (xn+1− xn)i

−

p∑j=0

a j(xn− j− xn

)i+h

p∑j=−1

b ji(xn− j− xn

)i−1

= cihi

ci = 1−p∑

j=0

(− j)i a j+ ip∑

j=−1

(− j)i−1 b j, i ≥ 1 (2.20)

9

This gives us

Tn (y) =m∑

j=0

ci

i!hiy(i) (xn)+Tn (Rm+1) (2.21)

From (2.19), it is straightforward that Tn (Rm+1) = O(hm+1

). If y is m+ 2 times con-

tinuously differentiable, we may write the remainder Rm+1 (x) as

Rm+1(x) =1

(m+1)!(x− xn)(m+1) y(m+1)(xn)+ ...,

and then

Tn (Rm+1) =cm+1

(m+1)!hm+1y(m+1) (xn)+O

(hm+2

)(2.22)

with

cm+1 = 1−

p∑j=0

(− j)m+1a j+ (m+1)p∑

j=−1

(− j)mb j

To obtain the consistency condition (2.11)−(2.12), assuming that y is an arbitrary

twice continuously differentiable function, we need τ (h) = O (h) and this requires

Tn (y) = O(h2

). Using (2.21) with m = 1, we must have c0 = c1 = 0, which gives the

set of equations (2.14)−(2.15). In some texts, these equations are referred to as the

consistency conditions. It can be further shown that (2.14)−(2.15) are the necessary

and sufficient condition for consistency (2.11)−(2.12), even when y is only assumed

to be differentiable. To obtain (2.13) for some m ≥ 1, we must have Tn(y) =O(hm+1).

From (2.21)−(2.20), this will be true if and only if c2 = c3 = . . . = cm = 0. This proves

10

the conditions (2.16).

For the largest value of p, which (2.13) holds, is known as the order or order of

convergence of method, stated in (2.3). Convergence occurs for multistep methods

then approximated solution {yn} tends to analytical solution {y(xn)} as the step size h

goes to zero. This describes that convergence in the limit as h→ 0, n→∞, nh= x− x0

remaining fixed where x is between x0 and b. It can be represented as

limh−→0

yn = y(xn)

To define the convergence of the initial value problem in (2.24) and the general mul-

tistep method in condition (2.3),

y′(x) = f (x,y(x)), x ≥ x0 (2.24)

y(x0) = y0

The following theorem will be examined.

Theorem 2.0.2 ([3])Assume that the derivative function f (t,y) is continuous and

satisfies the Lipschitz Condition

| f (x,y1)− f (x,y2)| ≤ K |y1− y2| (2.25)

for all −∞ < y1, y2 < ∞, x0 ≤ x ≤ b, and for some constant K > 0. Let the initial

11

errors satisfy

η(h) ≡ max0≤ i ≤ p

|y(xi)− yi| −→ 0 as h −→ 0 (2.26)

Assume that the solution y(x) is continuously differentiable and method is consis-

tent, that is, that it satisfies (2.12). Finally, assume that the coefficients a j are all

nonnegative

a j ≥ 0, j = 0,1, ..., p (2.27)

then the multistep method (2.3) is convergent and

maxx0 ≤ xn≤ b

|y(xn)− yn| ≤ c1η(h)+ c2τ(h) (2.28)

for suitable constants c1, c2. If the solution y(x) is (m+1) times continuously differ-

entiable, the method (2.3) is of order m, and the initial errors satisfy η(h) = O(hm),

then the order of convergence of the method is m; that is, the error is of size O(hm).

Proof. ([4])Rewrite (2.10) and use y′(x) = f (x,y(x)) to get

y(xn+1) =

p∑j=0

a jy(xn− j)+hp∑

j=−1

b j f (xn− j,y(xn− j))+hτn(y)

Subtracting (2.3) from this equality and using the notation en = y(xn)− yn, we obtain

en+1 =

p∑j=0

a jen− j+hp∑

j=−1

b j[f (xn− j,y(xn− j)− f (xn− j,yn− j)

]+hτ(h)

12

Apply the Lipschitz condition and the assumption (2.27) to obtain

|en+1| ≤p∑

j=0

a j∣∣∣en− j

∣∣∣+hKp∑

j=−1

∣∣∣b j∣∣∣ ∣∣∣en− j

∣∣∣+hτ(h)

where

τ(h) ≡max |τn(y)|

Introduce the maximum of error

fn = maxi=0,...n

|ei| ,n = 0,1, ...,N (h) .

Using this function, we have

|en+1| ≤p∑

j=0

a j fn+hKp∑

j=−1

∣∣∣b j∣∣∣ fn+1+hτ(h)

and applying (2.14), we obtain

|en+1| ≤ fn+hc fn+1+hτ(h), n ≥ p

c = Kp∑

j=−1

∣∣∣b j∣∣∣

The right hand side is trivially a bound for fn and thus,

fn+1 ≤ fn+hc fn+1+hτ(h)

13

For hc ≤ 12 , which is true as h −→ 0, we obtain

fn+1 ≤fn

1−hc+

h1−hc

τ (h)

≤ (1+2hc) fn+2hτ(h)

Noting that fp = η(h), then

fn ≤ e2c(b−x0)η(h)+[e2c(b−x0)−1

hc

]τ(h), x0 ≤ xn ≤ b (2.29)

The convergence of methods depends on consistency condition and stability. The

stability and convergence of numerical solution is related with the roots of the poly-

nomial

p(r) = rp+1−p∑

j=0

a jrp− j (2.30)

We can see that p(1) = 0 by considering the consistency conditionp∑

j=0a j = 1. Let

r0 = 1 and r1,r2, ...,rp be the roots of polynomial p(r). The method (2.3) satisfies the

root condition if

∣∣∣r j∣∣∣ ≤ 1, j = 0, ..., p (2.31)∣∣∣r j∣∣∣ = 1 =⇒ ρ′(r j) = σ(1) , 0 (2.32)

These says that all roots of ρ(r), which is known first characteristic polynomial, be-

long to the closed unit circle {z : |z| ≤ 1} in complex plane, and some simple roots of

14

p(r) lie on the boundary of the circle. It associated a characteristic polynomial as

π (r,hλ) = ρ(r)−hλσ(r) (2.33)

The general solution of (2.26) can be defined if the roots r j(hλ) are all distinct

yn =p∑

j=0γ j

[r j (hλ)

]n(2.34)

Since hλ = 0, we get from equation (2.33) that ρ(r) = 0. So, roots of equation become

r j(0) = r j, j = 0,1, ..., p.When r0 = 1, then the root of (2.33) is r0(0) = 1.

Definition 2.0.3 ([5])Assume the consistency conditions (2.14)-(2.15). Then the mul-

tistep method (2.3) is stable if and only if the root condition is satisfied.

Stability determines how well the method will perform in finding the approximated

solution in terms of satisfactoriness before actually performing the method. This

means that numerical errors generated during the solution, and these errors should

not be magnified.If Initial condition y0 contains numerical errors, then this will effect

the further approximation yn.

Solving the initial value problem numerically shows that there is a solution y(x) on a

given finite interval x0 ≤ x ≤ b. Accordingly, it is needed to analyze the stability for

methods. We perturb the initial value problem (2.1) with y0+ ϵ. The result is

y′ϵ(x) = f (x,yϵ(x)) (2.35)

yϵ(x) = y0+ ϵ

15

This notations shows that y(x) and yϵ(x) are exist on the given interval for small

values of ε and besides,

∥yϵ − y∥∞ ≡ maxx0≤ x ≤ b

|yϵ(x)− y(x)| ≤ cϵ for c > 0 (2.36)

Similar procedure is analyzed for stability of multistep method. Let state the solution

of (2.4) with initial values y0,y1, ...,yp for some differential equation y′(x)= f (x,y(x))

which perturb to the new values z0,z1, ...,zp. Then,

max0≤n≤p

|yn− zn| ≤ ϵ (2.37)

These initial values are valid for depending on h. In contrast, the numerical solution

{yn : 0 ≤ n ≤ N(h)} is stable if it is independent of h, and there is a constant c with

for all sufficiently small ϵ which is

max0≤n≤N(h)

|yn− zn| ≤ cϵ, 0 ≤ h ≤ h0

and N(h) is accepted the largest subscript N which xN ≤ b. In addition, it is satisfying

the Lipschtz condition. Thus, we can say that numerical solution of multistep method

is stable if all approximate solution {yn} is stable. If the maximum error is less than

the beginning error which is ϵ, then the I.V.P is called well-conditioned. Otherwise,

it is ill-conditioned. Consequently, the small changes in the initial value y0 will affect

the solution of y(x) of the initial value problem for small ϵ.

Theorem 2.0.4 (As described in [32])A linear multistep method is zero stable for

16

any ordinary differential equation of the form (2.1) where f satisfies Lipschitz condi-

tion if and only if its first characteristic polynomial has zeros inside the closed unit

disc with any which lie on the unit circle being simple.

We say that if first characteristic polynomial’s all roots belong to the closed unit

disc, zero stability holds for a linear multistep method. This means that modulus of

first characteristics polynomial p(r) is less than or equal to 1, and satisfies the root

condition. There are two fundamental condition for convergence which is specified

in the following theorem;

Theorem 2.0.5 ([33])To be consistent and zero-stable are the most basic condition

for convergence of linear multistep method.

Definition 2.0.6 ([6])The linear multi-step method (2.3) is said to be relatively sta-

ble for a given hλ if, for a given hλ, the root r j satisfy∣∣∣r j(hλ)

∣∣∣≤ r0(hλ), j= 1,2,3, ..., p

for sufficiently small values of h. If a linear method is stable but not relatively stable,

then it is called weakly stable.

It is clear that a method is relatively stable if the characteristic roots r j(hλ) satisfies

the given definition for nonzero values of |hλ|.

Definition 2.0.7 ([18])A numerical method is said to be absolutely stable in a region

ℜ of complex plane if, for all hλϵℜ, all roots of the stability polynomial π (r,hλ)

17

associated with the method, satisfy

∣∣∣r j∣∣∣ < 1, j = 1,2, . . . , p

Stability also includes some concept of absolute stability. Two parameters are con-

sidering for absolute stability: ”λ” which is the eigenvalues of Jacobian matrix and

”h” which is the step size. Absolute stability depends on value of the product of these

two parameters, is called hλ. Analyzing of the stability at two parameters separately

is insufficient. The region of the stability is considered in a complex plane. Because

λ can be either negative real part or complex. When numerical method is applied to

model equation y′(x) = λy(x) for performing the region of stability, then the modulus

of the nth step iteration should be less than or equal to 1. In system of differential

equations, first , the Jacobian matrix form is created by writing the coefficients of the

matrix variables. Then, the eigenvalues of Jacobian matrix is denoted by λt where

t = 1,2, ... are achieved by taking determinant of the Jacobian matrix which is equal

to zero. Finally, solving the system, we obtain the λt. If all eigenvalues of the matrix

are real, then the matrix is symmetric, and can have complex eigenvalues. Otherwise,

if eigenvalues of matrix are complex, then the parameter hλ will be also complex.

Consequently, the numerical method is absolutely stable if the parameter hλ is in the

region of absolute stability.

18

Chapter 3

STIFF SYSTEMS AND PROBLEMS OF STABILITY FOR STIFF

SYSTEM

3.1. Introduction

Stiff systems arises applications, such as chemical kinetics, mechanical system, con-

trol theory, electronics, and mathematical biology, and numerical solution of partial

differential equations. Stiffness is related with the concept of stability of numerical

methods. There are several definitions of stiff equation.

Let A be the coefficient of matrix of mxm system. Let λµ and λυ be two eigenvalues

of the coefficient matrix. The given problem is stiff if∣∣∣Re(λµ)

∣∣∣≫ |Re(λυ)|. It is re-

quired to integrate numerically over a long range, using the hugely small step length

to the interval.

Let A be the coefficient matrix of linear system. For system of differential equation,

consider the eigenvalues λt of fy(x,y(x)). If eigenvalues λt have negative real part

Re(λt) < 0 (3.1.1)

with the very large magnitude

max1< j<n

|Re(λt)| ≫ min1< j<n

|Re(λt)| (3.1.2)

19

then it make difficult to solve the system. We can say that the given problem is stiff.

By this definition, a stiff problem has a stable point with greatly different magnitudes

of eigenvalues.

To see the large magnitude between the eigenvalues, it is sufficient to look at the ratio

of

ℜ = max |Re(λt)|min |Re(λt)|

(3.1.3)

which provides a measure of stiffness. It is clear that stiffness ratio is calculated by

dividing highest eigenvalue by lowest eigenvalue.

3.2. Definition of Stiffness

Given the mxm linear systems

y′= Ay+Φ(x) (3.2.1)

where A is the coefficient of matrix that has different eigenvalues λt and correspond-

ing eigenvectors ct, t = 1,2, ..,m has general solution in the form of

y(x) =m∑

t=1

kteλt xct +Ψ(x) (3.2.2)

where kt is a constant term. Suppose that Re(λt) < 0, then the first term

m∑t=1

kteλt xct −→ 0 as x −→∞ (3.2.3)

20

which includes exponential terms, is called "transient part" and the remaining term

Ψ(x) is called "steady state part". In this system, to reach the accurate approximation,

the step size h is taken extremely small in our choice. It follows that the transient part

approach to zero as x −→∞ and λ is real and negative. Steady state part will attain

to exact solution. Thus, the aim of numerical solution is to find the approximate so-

lution in steady state part and to ignore the transient part that includes the slowest

decaying exponential eλt . If the eigenvalues lie outside the region of stability, then

step length should be chosen exceedingly small to satisfy hλ. Hence hλ will lie in

region of the absolute stability. In order to recognize the necessary condition, step

size is to be excessively small for stability and it has negative real eigenvalues or

negative real part of complex eigenvalues. Under this consideration system of differ-

ential equation is referred as stiff system.

In general, stiffness is affected from stiffness ratio which resulted in enormous nu-

meral. It arises when the huge difference observed in the modulus of real part of

eigenvalues.

Definition 3.2.1 ([19])The linear system y′= Ay+Φ(x) is said to be stiff if

1. Re(λt) < 0, t = 1,2, . . . ,m and

2. maxt=1,2,...,m

|Re(λt)| ≫ mint=1,2,...,m

|Re(λt)|, where λt,t = 1,2, . . . ,m are the eigenvalues

of A. The ratio

[max

t=1,2,...,m|Re(λt)|

]:[

mint=1,2,...,m

|Re(λt)|]

(3.2.4)

21

is called the stiffness ratio.

3.3. The Problem of Stability for Stiff System

We have seen the difficulties of the numerical solution of stiff system. It is essential to

consider the absolute stability of methods. There are various definitions of stability

for stiff systems that is proposed. These are applicable to any numerical method

which involves discretization with associated step length h.

Definition 3.3.1 ([20])(Dahlquist37) A numerical method is said to be A-stable if its

region of absolute stability contains the whole of the left-hand half-plane Re(hλ) < 0.

It is shown in figure

Figure 3.1. The region of A−Stability

When A−stable is considered on stiff system that Re(λ j) < 0, then step size can be

taken without any restriction, and the magnitude of real part of eigenvalues, which

represent maxt=1,2,...,m

|Re(λt)| , disregard.

22

Definition 3.3.2 ([21])([35])(Widlund 181) A numerical method is said to be A(α)−stable,

αϵ (0,π/2), if its region of absolute stability contains the infinite wedge

Wα ={hλ| −α < π− arghλ < α

}

it is said to be A(0)−stable if it is A(α)−stable for some (sufficiently small) αϵ (0,π/2).

A numerical method is A0−stable if its region of absolute stability includes the nega-

tive real axis in the complex plane.

If for a given λ, Re(λ)< 0, then the point hλ either will lie inside or outside the wedge

Wα for all h > 0. Eventually, eigenvalues of a stiff system lie in a some wedge Wβ;

so, A(β)− stable can be considered on the numerical solution of initial value problem

without any restriction on step size. Especially, A(0)−stable can be used for real and

nonnegative eigenvalues of matrix A. It is shown as

Figure 3.2. The region of A (α)−Stability

Definition 3.3.3 ([22])(Gear52,53) A numerical method is said to be stiffly stable iff

1. its region of absolute stability containsℜ1 andℜ2,

23

2. it is accurate for all hǫℜ2 when applied to the scalar test equationy′= λy, λ

complex constant withRe(λ) < 0, whereℜ1 = {hλ|Re(hλ) < −a}, and

ℜ2= {hλ| −a≤ Re(hλ) < b,−c≤ Im(hλ) < c} and a, c are positive real numbers.

It is shown as ([30])

Figure 3.3. The region of Stiff Stability

Sinceλ is real and negative, the regionR1 has no restriction onh. Eigenvalues decays

rapidly in the transient part. The valuehλ in ℜ1 will lie in stability region by elimi-

nating the step size. Nevertheless, in the regionℜ2, the step sizeh should be chosen

excessively small to satisfyhλ. Hereby, the valuehλ will lie in a stable region that is

shown in the above figure.

Definition 3.3.4 ([23])A one step numerical method is said to beL-Stable if it is A-

stable and, in addition, when applied to the scalar test equation y′= λy, λ a complex

constant withRe(λ) < 0, it yields yn+1 =ℜ(hλ)yn, where∣

∣

∣ℜ(hλ)∣

∣

∣ −→ 0 asRe(hλ) −→

−∞.

24

It is clear that A−stable is not enough to get accurate approximation, L− stabil-

ity is also required which can also be called as left stability. Definition of L−

stability needed, the ℜ(hλ) tends to zero along with the real part of λ tends to

negative infinity. Obviously, A−stability=⇒stiff-stability=⇒ A(α)−stability=⇒ A(0)-

stability=⇒ A0−stability ([34]).

25

Chapter 4

EULER METHOD

4.1. Introduction

Many differential equations in engineering are so intricate. It is inapplicable to have

solution. Numerical methods provide ease for solving the differential equations. The

simplest method to solve initial value problem is Euler method which have one step

process for each equation before move on next step. This method is not an adequate

method to get the certain approximation. Differential equation can be solved simply

even though it is rather rough and least accurate. It is restricted to utilize because

each successive step during the process accumulates large errors. It has slow of

convergence which means a method of order 1. So the error is O(h). In contrast, the

remainder term and error analysis in Euler method provide convenience to state the

difference between the approximate and exact solutions.

4.2. Definition of Euler’s Method

Assume that initial value problem (2.1)−(2.2) is applied to numerical method on

the specified interval: [x0,b] . We can create nodes with equi-spaced subinterval for

26

simplicity as

x0 < x1 < ... < xN < b

If nodes is taken to be evenly spaced, numerical methods will generate an approxi-

mate solution yn. It is written simply which is called mesh point as follows

xn = x0+nh, n = 0,1, ...,N

where

h =(b− x0)

N

and h is defined as step-size or (step of integration), a positive real number N. For

each n, the numerical approximation yn at a mesh points xn can be smoothly obtained.

The initial condition is known as

y(x0) = y0

Assume that we have already calculated yn up to some n. This represent

yn+1 = yn+h f (xn,yn) (4.2.1)

27

It is known as Euler method with the initial condition. To attain Euler’s method,

consider a forward difference approximation to the derivative

y′(x) ≈ 1

h[y(x+h)− y(x)

]

Equalize this to the initial value problem (2.1−2.2) at xn.We obtain

1h[y(xn+h)− y(xn)

]= f (xn,y(xn))

y(xn+1) = y(xn)+h f (xn,y(xn))

Euler’s method can be represented by considering the approximated values

yn+1 = yn+h f (xn,yn), 0 ≤ n ≤ N −1

The other way to derive the Euler’s method is to integrate the differential equation

(2.1) between two consecutive mesh points xn and xn+1.We conclude that

xn+1∫xn

y′(x)dx =

xn+1∫xn

f (x,y(x))dx

y(xn+1)− y(xn) =xn+1∫xn

f (x,y(x))dx, n = 0,1, ...,N −1

y(xn+1) = y(xn)+xn+1∫xn

f (x,y(x))dx (4.2.2)

28

Here, we cannot integrate f (x,y(x)) without knowing y(x). Hence we must approxi-

mate the integral. Apply the numerical integration rule

xn+1∫xn

g(x)dx ≈ hg(x)

which is knowns as the rectangle rule with g(x)= f (x,y(x)). This means the simplest

approach is to approximate the area under function f (x,y(x)) by the rectangle with

base [xn, xn+1] and height f (x,y(x)).We can define the rectangle rule

xn+1∫xn

g(x)dx ≈ h[(1−Θ)g(xn)+Θg(xn+1)

](4.2.3)

with Θϵ [0,1] . Then, substitute (4.2.3) into (4.2.2) by considering g(xn) = f (xn,y(xn)

to obtain

y(xn+1) = y(xn)+h[(1−Θ)g(xn)+Θg(xn+1)

]y(xn+1) ≈ y(xn)+h

[(1−Θ) f (xn,y (xn))+Θ f (xn+1,y (xn+1))

]y(x0) = y0

Then, supply the initial conditions to the one parameter method that is mentioned

above. This gives us the following method where Θϵ [0,1]

yn+1 = yn+h[(1−Θ) f (xn,yn)+Θ f (xn+1,yn+1)

]

This definition motivatesΘ−method by considering approximate values. TheΘ−method

referred Euler’s method as Θ = 0 which yn+1 must be found merely left hand side.

29

This definition that give yn+1 directly is called explicit methods. For Θ = 1 we re-

cover the Implicit (backward) Euler Method.

yn+1 = yn+h f (xn+1,yn+1), n = 0, ..,N −1 (4.2.4)

In order to identify yn+1 (4.2.4) need the solution of an implicit equation. Euler’s

method is also referred as Explicit Euler Method in order to pick out the difference.

This scheme gives a result for the value of Θ = 12 which is denominated Trapezium

Rule Method. It is shown as

yn+1 = yn+12

h[ f (xn,yn)+ f (xn+1,yn+1)], n = 0, ..,N −1

The other way to obtain numerical method is using multistep method. In general

form of multistep method, it is easier to achieve any numerical methods. Consider

the given form of (2.3) to obtain Euler method. As p = 0, we get that

yn+1 = a0yn+h[b−1 f (xn+1,yn+1)+b0 f (xn,yn)

]

If we give values instead of a0,b−1,b0 kind of

a0 = 1,b−1 = 0,b0 = 1

gives Euler method formulae. In contrast, if the values are taken as

a0 = 1,b−1 = 1,b0 = 0

30

, we get Implicit Euler method

yn+1 = yn+h f (xn+1,yn+1)

which yn+1 will occur in both sides.

4.3. Error Analysis of Euler’s Method

The purpose of examining error of Euler’s method is to see how the approximated

solution works. In Euler’s method the slope of function affects the accuracy of meth-

ods. In order to minimize the error that occur, the step size should be chosen very

small. In other words, number of point need to be taken enormous between the given

interval. Thus, when step size is chosen excessively small, error is minimized and

approximated solution will be more better.

For error analysis, we consider the differential equation y(x) that satisfies

y′(x) = f (x,y(x))

Assume that the solution of initial value problem is unique on x0 ≤ x ≤ b, and this

solution has bounded second derivative y′′(x) over given interval. To state the error

of Euler method, we begin by applying Taylor expansion series for y (xn+1) .

y (xn+1) = y(xn)+hy′(xn)+

h2

2y′′(ξn)+O(h3)

31

for some xn ≤ ξn ≤ xn+1. Taylor approximation becomes

y (xn+1) = y(xn)+h f (xn,y(xn))+h2

2y′′(ξn)+O(h3) (4.3.1)

If we compare the eguations (4.2.1) and (4.3.1), it yields

yn+1 = yn+h f (xn,yn)+O(h2)

It is shown that local truncation error of forward Euler method is O(h2

). The other

way to find truncation error; for any differentiable function y(x), we can define the

truncation error as follows

Tn(y) = y (xn+1)− y(xn)−h f (xn,y(xn)) (4.3.2)

and the term

Tn =h2

2y′′(ξn)+O

(h3

)(4.3.3)

is called truncation error for Euler method. By considering the 2.5 in Chapter 2,

we have the local truncation error of Euler method

τn =1h

Tn

τn =1h

[y (xn+1)− y(xn)−h f (xn,y(xn))], as y′(xn) = f (xn,y(xn))

32

Then substitute the Taylor expansion about y(xn+1) to get

τn =1h

[y(xn)+hy

′(xn)+O(h2)− y(xn)−h f (xn,y(xn))

]τn =

1h

[O(h2)

]τn = O(h1)

From the above explanation, It is clear that the truncation error is defined Tn =O(h2)

for Euler’s method and in general if Tn = O(hp+1) which p indicates the order of

method, then method is the pth order. That is; Euler method is the first order method.

It can be explained that exact solution is equal to the approximate solution addedly to

local truncation error: y(xn) = yn+Tn. As the order of method is 1, the method is too

small. For this reason, it is not efficient method to obtain accuracy. To comprehend

the error in Euler method, subtract

yn+1 = yn+h f (xn,yn)

from (4.3.1) getting

y (xn+1)− yn+1 = y(xn)− yn+h f (xn,y(xn))−h f (xn,yn)+h2

2y′′(ξn) (4.3.4)

From the above explanation, the propagated error arises as following

y(xn)− yn+h[f (xn,y(xn))− f (xn,yn)

](4.3.5)

33

Let en = y(xn)− yn to rewrite the (4.3.4)

en+1 = en [1+hK]+h2

2y′′(ξn)

These results give a general error of Euler Methods. Now, we investigate the con-

vergence of Euler’s method for solving the general initial value problem on a given

interval [x0,b] .

y′(x) = f (x,y(x)), x0 ≤ x ≤ b (4.3.6)

y(x0) = y0

For complete the error analysis the following Lemma and Theorem will be consid-

ered.

Lemma 4.3.1 ([9])For any real t,

1+ t ≤ et, (4.3.7)

and for any t ≥ 1, any m ≥ 0,

0 ≤ (1+ t)m ≤ emt (4.3.8)

Theorem 4.3.2 ([7])Let f (x,y) be a continuous function for x0 ≤ x≤ b and −∞< y<

∞, and further assume that f (x,y) satisfies the Lipschitz condition. Assume that the

solution y(x) of the general solution of initial value problem has a continuous second

34

derivative on [x0,b] . Then the solution {y(xn)| x0 ≤ x ≤ b} obtained by Euler’s method

satisfies

maxx0≤xn≤b

|y(xn)− yn| ≤ e(b−x0)K |e0|+[e(b−x0)K −1

K

]τ(h), (4.3.9)

where

τ(h) =12

h∥∥∥∥y′′∥∥∥∥∞ = 1

2h max

x0≤xn≤b

∣∣∣∣y′′(x)∣∣∣∣ (4.3.10)

and e0 = y(x0)− y0. If, in addition, we have

|y(x0)− y0| ≤ c1h as h −→ 0 (4.3.11)

for some c1 ≥ 0 (e.g., if y(x0) = y0 for all h, then c1 = 0), then there is a constant

B ≥ 0 for which

maxx0≤xn≤b

|y(xn)− yn| ≤ Bh (4.3.12)

Proof. ([8])Let en = y(xn)− yn,n ≥ 0. Let N ≡ N(h) be integer for which

xN ≤ b, xN+1 > b.

Define

τn =12

hy′′(ξn), 0 ≤ n ≤ N(h)−1,

35

based on the truncation error. Easily, we obtain

max0≤n≤N−1

|τn| ≤ τ(h)

using (4.3.10). Recalling (4.3.5), we have

en+1 = en+h[f (xn,y(xn))− f (xn,yn)

]+hτn (4.3.13)

Taking bounds using Lipschtz condition, we obtain

|en+1| ≤ |en|+hK |y(xn)− yn|+h |τn|

|en+1| ≤ (1+hK) |en|+hτ(h), 0 ≤ n ≤ N(h)−1 (4.3.14)

Apply this recursively to obtain

∣∣∣en| ≤ (1+hK)n∣∣∣e0|+

[1+ (1+hK)+ . . .+ (1+hK)n−1

]hτ(h)

Using the formula for the sum of a finite geometric series,

1+ r+ r2+ ...+ rn−1 =rn−1r−1

, r , 1 (4.3.15)

we obtain

∣∣∣en| ≤ (1+hK)n∣∣∣e0|+

[(1+hK)n−1

K

]τ(h) (4.3.16)

36

Using the Lemma 1, we obtain

(1+hK)n ≤ enhK = e(xn−x0)K ≤ e(b−x0)K

and this with (4.3.16 ) implies the main result (4.3.9).

The remaining result (4.3.12) is a trivial corollary of (4.3.9) with the constant B given

by

B = c1e(b−x0)K +12

[e(b−x0)K −1

K

]∥∥∥y”∥∥∥∞

We can say that the result (4.3.12) is consistent. When step size is halved, Bh is also

halved. Euler’s method is convergent with order 1, because that is the power of h that

occurs in the error bound. In general, if we have

|y(xn)− yn| ≤ chp, x0 ≤ xn ≤ b

for some constant p ≥ 0, then numerical methods is convergent with order p ([14]).

Let us consider the consistency on the examples:

1. For Euler methods: The numerical solution is

yn+1 = yn+h f (xn+1,yn+1)

37

Tn(y) = y(xn+1)− y(xn)−hy′(xn)+O(h2)

τn(y) =1h[y(xn+1)− y(xn)−hy′(xn)

]

This says that

y(xn+1)− y(xn)h

≈ y′(xn)

and the order of the method is 1, O(h1).

2. ([31])For midpoint method:

yn+1 = yn−1+2h f (xn,yn)

Here

Tn(y) = y(xn+1)− y(xn−1)−2hy′(xn)

τn(y) = 2[y(xn+1)− y(xn)

2h− y

′(xn)

]

Thusly,

y(xn+1)− y(xn)2h

≈ y′(xn)

4.4. Numerical Stability of Euler’s Method

Recollect the definition of stability for initial value problem. Small changes in initial

value y0 will perturbed the solution y(x) by ϵ. Similar analysis is valid for Euler’s

38

method. Introduce a new numerical solution {zn} by

zn+1 = zn+h f (xn,zn), n = 0,1, ...,N −1

with initial values z0 = y0 + ϵ. It is perturbed to the initial value problem. In order

to inspect the stability, compare two numerical solution {yn} in (2.1) and {zn} by

subtracting. We get

zn+1− yn+1 = zn− yn+h[f (xn,zn)− f (xn,yn)

]

with initial value z0− y0 = ϵ. Introduce en = zn− yn to obtain

en+1 = en+h[f (xn,zn)− f (xn,yn)

]

To be more straightforward, use the Lipschtz condition and apply theorem (4.3.2) to

get

max0≤n≤N

|zn− yn| ≤ e(b−x0)K |ϵ |

There is a constant c ≥ 0 such that

max0≤n≤N

|zn− yn| ≤ c |ϵ|

This analog shows that error must be less than the received perturbed error in startup

for getting stable method. Thus, Euler method is a stable under the solution of the

initial value problem Now, we turn to examine the performance of Euler’s method on

39

model problem. We have

y′(x) = λy(x), x ≥ 0 (4.4.1)

y(0) = 1

When it is applied to the model problem, the numerical method is displayed

yn+1 = yn+hλyn = (1+hλ)yn, n ≥ 0 (4.4.2)

y0 = 1

We investigate the case λ to find the region of stability of Euler methods. Recursively,

it is obtained

yn = (1+hλ)n y0 (4.4.3)

It can be written for a fixed node point xn = nh ≡ −x, as n −→∞

yn =

1+ λ−xn

n

−→ eλ−x

Since exact solution is decaying exponentially, numerical method also has same be-

havior. To show the stability of Explicit Euler method, we need that

|1+hλ| < 1

for λ is real and negative. Thus, in the complex plane radius is 1 with centre 1. The

point hλ is in region of absolutely stable under specified interval with Re(λ). Then

40

the condition indicates the interval that Euler method is stable. It becomes

−2 < hλ < 0

namely

0 < h <2λ

This also satisfied the convergence of Euler method since in the notation (4.4.3),

yn −→ 0 as n −→∞ for any choice of step size holds. Thus, in Explicit Euler method,

step size should be chosen excessively small to provide the stability. The stability

region of Explicit Euler method is straightforward in the figure 4.1

Figure 4.1. The region of absolute stability of Forward Euler Method

41

On the other hand, we show the implicit Euler method to understand how methods

behaves in the model problem. Implicit Euler method states that

yn+1 = yn+h f (xn+1,yn+1)

After that substitute model problem to numerical method to obtain

yn+1 = yn+hλyn+1

yn+1 =1

1−hλyn

and by using induction

yn =

(1

1−hλ

)n

y0

Therefore, it is clear that for Re(λ) < 0 implicit Euler method is stable if and only if

it satisfies

(1

1−hλ

)< 1

So, |1−hλ| > 1 for any step size h > 0. Thus, we should choose step size h large to

ensure the stability. It is straightforward in figure (4.2)

42

Figure 4.2. The region of absolute stability of Backward Euler Method

Consequently, when numerical method is absolutely stable, there is no restriction on

h. If the values of λ that is real and negative has the large magnitude, then step size

should be chosen excessively small to satisfies hλ. Hereby, it belong to the region of

absolute stability. This also affect the approximation solution. It provides the trun-

cation error to be small. Even though Euler’s method is absolute stable, we cannot

make any comment for Euler Method about weak-stable. Consider root condition

(2.30) and take p = 0

ρ (r) = r−a0

and root is

r0 = a0

From the condition (2.14) of theorem 2.0.1, manifestly r0 = 1. This shows that Euler

method is absolute stable, but we cannot apply the definition 2.0.5 which is related

with relatively stable and weak stable. Because it has one root.

43

4.5. Rounding Error Accumulation of Euler’s Method

"Integer mode" and "floating point mode" provide to represent the numbers. Integer

mode is used for integer numbers and floating point mode is used for real numbers.

We concern with floating point mode. In numerical methods, numbers have greatly

varying size. In order to facilitate the operations, the limitations on the numbers of

digit are required. Rounding error is caused due to this modes. It affect the accuracy

in the numerical solution. Now, we analyze the error for Euler’s method. After

numerical method applied the to any differential equation, rounding error is being

performed.

Denote δn which is called rounding error. It is the precision of the arithmetic and

affected the approximate solution {yn}. We have

yn+1 = yn+h f (xn,yn)+δn, n = 0,1, . . . ,N(h)−1 (4.5.1)

Assume that

|δn| ≤ cu. maxx0≤x≤b

|y(x)| = cu.∥y∥∞ (4.5.2)

where u = 2.2x10−16 is the machine epsilon of the computer and the magnitude of c

is 1 or larger 1. Let δ(h)be a bound on the rounding errors

δ(h) = max0≤n≤N(h)−1

|δn| (4.5.3)

To see the effect of rounding error subtract the approximate solution (4.5.1) from the

44

exact solution to have

y(xn+1)− yn+1 = y(xn)− yn+h[f (xn,y(xn))− f (xn,yn)

]+

12

h2y′′(ξn)−δn (4.5.4)

y(x0)− y0 = 0

This is equivalent to the form

y(xn+1)− yn+1 = y(xn)− yn+h[f (xn,y(xn))− f (xn,yn)

]+hτ(h)−δn

where Tn+1 =12h2y”(ξn) is the local truncation error. It is same as the notation

τ(h) =1h

Tn+1

The error term of (4.5.4) can be written as follows

hτ(h)−δ(h) = h[τ(h)− δn

h

](4.5.5)

Setting en = y(xn)− yn, take bounds of notation (4.5.4) using the Lipschtz condition

|en+1| ≤ |en|+hK |en|+hτ(h)−δ(h) (4.5.6)

|en+1| ≤ (1+hK) |en|+hτ(h)−δ(h) (4.5.1)

By an inductive argument and applying the proof of the theorem (4.3.2) we obtain,

|en| ≤ e(b−x0)K |e0|+e(b−x0)K

hKh[τ(h)− δn

h

]

45

where e0 = 0 is getting from (4.5.4). So, we reach the last form as

|en| ≤e(b−x0)K

K

[12

h∥∥∥∥y′′∥∥∥∥∞− cu

h∥y∥∞

]|y(xn)− yn| ≤ c1

{12

h∥∥∥∥y′′∥∥∥∥∞− cu

h∥y∥∞

}= E(h) (4.5.7)

It is clear from the above operations that error will decrease as h decrease at the

beginning; but at the critical point the optimum value of h∗, the error will increase.

At the second term in brackets on the right hand side of (4.5.7) affect the error despite

the fact that machine epsilon u is small. The figure (4.3) also demonstrate the above

explanation.

Figure 4.3. Rounding Error Curve for (4.5.7)

46

4.6. Euler’s Method for Systems

The numerical solution for system is same as for a single equation. Numerical

method is applied to each equation in the system. It is shown in detailed in Chapter

6 that is written in a matrix form. To be more clarity, we consider Euler’s method for

a system of two differential equations. Let

y′1(x) = f1(x,y1,y2)

y′2(x) = f2(x,y1,y2)

with initial conditions

y1(x0) = y1,0

y2(x0) = y2,0

In order to obtain system of Euler’s method, consider the Taylor expansion for each

equation.

y1(xn+1) = y1(xn)+h f1(xn,y1,n,y2,n)+h2

2y′′1 (ξn)

y2(xn+1) = y2(xn)+h f2(xn,y1,n),y2,n)+h2

2y′′2 (ζn)

47

for some ξn, ζn in [xn, xn+1]. We get Euler’s method for a system of two equation by

ignoring the error terms and considering the initial conditions

y1,n+1 = y1,n+h f1(xn,y1,n,y2,n)

y2,n+1 = y2,n+h f2(xn,y1,n,y2,n)

This indicates the general form of system in matrices format as

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

48

Chapter 5

RUNGE-KUTTA METHOD

5.1. Introduction

The Runge-Kutta method is popular method for solving initial value problem. It is

most accurate and stable method. It arise when Leonhard Euler have made improve-

ments on Euler method to produce Improved Euler method. Then, Runge is realized

this method which is similar method with the second order Runge Kutta method. A

few years later in 1989 Runge acquired Fourth Order of Runge Kutta method and af-

terwards, it is developed by Heun(1900) and Kutta(1901). Fourth Order Runge-Kutta

method intend to increase accuracy to get better approximated solution. This means

that the aim of this method is to achieve higher accuracy and to find explicit method

of higher order. In this section, we discuss the formulation of method, concept of

convergence, stability, consistency for RK4 method. In spite of the fact that Runge

Kutta methods are all explicit, implicit Runge Kutta method is also observed. It has

the same idea of Euler method. Euler method is the first order accurate; in addition it

require only a single evaluation of f (xn,yn) to obtain yn+1 from yn. In contrast, Runge

Kutta method has higher accuracy. It re-evaluates the function f at two consecutive

points (xn,yn) and (xn+1,yn+1). It requires four evaluations per step. Due to this,

Runge-Kutta method is quite accurate, and it has faster rates of convergence.

49

5.2. Problem of Runge-Kutta Method

Analyze the initial value problem to find the solution of ordinary differential equa-

tions

y′(x) = f (x,y(x)) (5.2.1)

y(x0) = y0

The general form Runge-Kutta method is

yn+1 = yn+hF (xn,yn;h) , n = 0, . . . ,N −1 (5.3.2)

where F(., .; .) is called an increment function on the interval [xn, xn+1] . It can be

defined in general form as

F (x,y;h) = b1k1+b2k2+ . . .+bnkn (5.2.3)

where bn’s are constant and kn’s are

k1 = f (xn,yn) (5.2.4)

k2 = f (xn+ c2h,yn+a21k1h)

k3 = f (xn+ c3h,yn+a31k1h+a32k2h)

...

ks = f (xn+ csh,yn+as,1k1h+as,2k2h+ ...+as,s−1ks−1h)

50

where c’s and a’s are constants. Here, each of function k’s are represent slope of the

solution which are approximated to y(x). These coefficients ai j,c j,b j must be satisfied

the system of Runge-Kutta method. These are usually arranged in Butcher tableau or

(partitioned tableau)

c A

bT

The coefficient bT is a vector of quadrature weights, and ai j (i, j = 1, . . . , s) indicates

the matrix A.

5.3. Explicit Runge Kutta Method

The family of s−stage Explicit Runge Kutta method can be formed as shown in the

following:

yn+1 = yn+h

b1 f (xn,yn)+b2 f (xn+ c2h, f (xn,yn+h [a21k1]))+ . . .

+bs−1 f (xn+ cs−1h,yn+h[as−1,1k1+as−1,2k2+ . . .+as−1,s−2

])

+bs f (xn+ csh,yn+h[as,1k1+as,2k2+ . . .+as,s−1

])

(5.3.1)

yn+1 = yn+hs∑

i=1

biki

51

where

k1 = f (xn,yn),

ki = f (xn+ cih,yn+hs−1∑j=1

ai, jk j), i = 1,2, . . . , s (5.3.2)

where h = xn+1 − xn. The coefficients{ci,ai, j,b j

}determine the numerical method.

This form was developed for higher accuracy methods. However, formulas can be

defined different notations as

zi = yn+hi−1∑j=1

ai, j f (xn+ c jh,z j), i = 1,2, ..., s (5.3.4)

yn+1 = yn+hs∑

j=1

b j f(xn+ c jh,z j

)(4.3.5)

The coefficients{ci,ai, j,b j

}for Explicit Runge Kutta methods can be simplified by

using the following Butcher Tableau

0

c2 a21

c3 a31 a32

......

.... . .

cs as1 as2 . . . . . . as,s−1

b1 b2 . . . bs−1 bs

52

If Runge-Kutta method is consistent, then the coefficients{ai, j

}and

{c j

}must satisfy

the condition

i−1∑j=1

ai, j = ci, i = 2, . . . , s (5.3.5)

The simplest Runge–Kutta method is the (Explicit) Euler method. The formula of

Euler method which is the first order, is mentioned previous chapter as yn+1 = yn +

f (xn,yn). The corresponding tableau is

0 0

1

The other example is given the midpoint method which is a second-order method

yn+1 = yn+h f (xn+12

h,yn+12

h f (xn,yn)

and its Butcher tableau is shown as

0 0

12

12 0

0 1

In order to evaluate y(xn+1), each of higher order derivatives is required to expand at

the point xi in the Taylor series. It is convenient to analyze any Runge-Kutta method

53

by using some formulae of derivatives which is

y′(x) = f (y(x))

y′′(x) = f ′(y(x)) y′(x) =⇒ f ′(y(x)) f (y(x))

y′′′(x) = f′′(y(x))

(f (y(x)),y

′(x)

)+ f

′(y(x)) f ′(y(x))y

′(x)

=⇒ f ′′ (y (x)) ( f (y(x)), f (y(x)))+ f ′(y(x)) f ′(y(x)) f (y (x))

It can be simplified to avoid complication by taking f = f (y (x)), fy = f ′ (y(x)), fyy =

f′′

(y (x)). We obtain

y′(x) = f

y′′(x) = fy f (5.3.6)

y′′′

(x) = fyy f 2+ f 2y f

y′′′′

(x) = fyyyy f 3+4 fyy fy f 2+ f 3y f

In order to generate the second order Runge Kutta method, we observe Taylor expan-

sions for y(xn+1) which has the form

y(xn+1) = y(xn)+hy′(xn)+

h2

2!y′′(xn)+

h3

3!y′′′

(xn)+ . . .+hp

p!y(p)(xn+) (5.3.7)

So as to attain the second order Runge Kutta method, we should take the derivation

of f (x,y) for the functions k1,k2. Consider the second order Taylor series and use

notations of (5.3.5)−(5.3.6) to construct the Runge Kutta method with order 2. Thus,

instead of y′(xn), we can write f (x,y) and we need to evaluate y

′′(xn) with respect to

54

y. After substituting, we takes form

y(xn+1) = y(xn)+h f +12

h2 fy f +O(h3) (5.3.8)

The approximate solution of second order Runge Kutta method can be obtained

yn+1 = yn+h [b1k1+b2k2] (5.3.9)

where

k1 = f (x,y) = f (5.3.10)

k2 = f (xn+ c2h,yn+ha21k1) = f +ha21 f fy+h2a12 f fx

Then, replace the slope of solution to form of the approximation solution yn+1 and

neglect function of f that depend on x to obtain

yn+1 = yn+h[b1 f +b2

(f +ha21 f fy

)](5.3.11)

Now, we match (5.3.8) with Taylor series (5.3.7). This satisfy the following equations

b1+b2 = 1 (5.3.12)

b2a21 =12

55

So, we have two conditions with three unknowns values. If we choose b1 = 0, b2 = 1

and a21 =12 , Runge Kutta method with order 2

yn+1 = yn+h f(xn+

12

h,yn+12

h f (xn,yn))

This is called Modified Euler method. Another particular solution of (4.3.12) is

Heun’s method since b1 =12 ,b2 =

12 and a21 = 1. The resulting method is

yn+1 = yn+h2

f (xn,yn+ f (xn+h,yn+h f (xn,yn)))

It is possible to acquire third order Runge Kutta method by using the same construc-

tion. Expansion of the term will include the order h3.Thus, for the Runge Kutta

method of order third, we consider third order Taylor expansion which has the form

y(xn+1) = y(xn)+hy′(xn)+

h2

2!y′′(xn)+

h3

3!y′′′

(xn)+O(h4)

It is required to find y′′(xn), y

′′′(xn) and also we need to expand k1, k2, k3 by using

derive expansion. The numerical solution is written as

yn+1 = yn+h [b1k1+b2k2+b3k3] (5.3.13)

56

where

k1 = f (x,y)

k2 = f (xn+ c2h,yn+ha21k1) (5.3.14)

k3 = f (xn+ c3h,yn+h [a31k1+a32k2])

Expanding k2 as a Taylor series we obtain

k2 = f +hc2 fx+ha21k1 fy+h2a12 f fx+12

h2a221

[fxx+2k1 fxy+ fyyk2

1

]+O(h3) (5.3.15)

and substitute k1 to equation (5.3.15) and ignore the function which is depending of

x to obtain

k2 = f +ha21 f fy+12

h2a221 fyy f 2+O(h3)

Similarly, in order to obtain k3, same process is applied in turn

k3 = f +h (a31k1+a32k2) fy+12

h2(a2

31k21 +a2

32k22

)fyy

Then by substituting ki’s (i = 1,2,3) to the numerical solution, we obtain

yn+1 = yn+h (b1+b2+b3) f +h2 (b2a21+b3a31+b3a32) f fy

+h3

2

(b2a2

21+b3a231+b3a2

32

)f 2 fyy+h3b3a32a21 f fy fy

57

Ultimately, we match the Taylor expansion of exact solution with the expansion of

solution. We have

y (xn+1)− yn+1 = h (b1+b2+b3−1) f +h2(b2a21+b3a31+b3a32−

12

)f fy (5.3.16)

+h3

2

(b2a2

21+b3a231+b3a2

32−13

)f 2 fyy+h3(b3a32a21−

16

) f fy fy

It is clear that set of functions

b1+b2+b3 = 1

b2a21+b3a31+b3a32 =12

(5.3.17)

b2a221+b3a2

31+b3a232 =

13

b3a32a21 =16

There are four equations with six unknowns. By taking convenient value for un-

knowns, we get through two particular solution of third order Runge Kutta method:

1. ([28])Since b1 =14 ,b2 = 0,b3 =

34 ,a21 =

13 ,a31 =

23 ,a32 =

23 , the result gives

Heun’s third order formula

yn+1 = yn+h4

(k1+3k3)

k1 = f (x,y)

k2 = f (x+13

h,y+13

hk1)

k3 = f (x+23

h,y+23

hk2)

58

2. ([29])Since b1 =16 ,b2 =

23 ,b3 =

16 ,a21 =

12 ,a31 = 1,a32 = 2, the result gives

Kutta’s third order formula

yn+1 = yn+h6

(k1+4k2+ k3)

k1 = f (x,y)

k2 = f (x+12

h,y+12

hk1)

k2 = f (x+h,y−hk1+2hk2)

5.4. Implicit Runge-Kutta Method

The implicit method is more different than explicit methods. It is such a complicated

method; but in solving differential equation, it has helpful numerical stability. The

form of s- stage Runge Kutta method is defined as

yn+1 = yn+hs∑

i=1

biki

where

ki = f (xn+ cih,yn+hs∑

j=1

ai jk j), i = 1, . . . , s (5.4.2)

It has a Butcher tableau

59

c1 a11 a12 . . . a1s

c2 a21 a22 . . . a2s

......

.... . .

...

cs as1 as2 . . . ass

b1 b2 ... bs

If the equation is a system of m differential equation, then the coefficients will be

system of sm nonlinear equation. It is clear that s nonlinear equations of system has

s unknowns ki.We now introduce a famous Runge Kutta method.

5.5. Fourth Order Runge-Kutta Method

The most popular method Fourth Order Runge-Kutta method is obtained as

yn+1 = yn+16

(k1+2k2+2k3+ k4)

where

k1 = f (xn,yn)

k2 = f (xn+12

h,yn+12

k1h) (5.5.2)

k3 = f (xn+12

h,yn+12

k2h)

k4 = f (xn+h,yn+ k3h)

Fourth Order Runge Kutta method is the well-known sample of all Runge Kutta

methods. Its tableau is

60

0

12

12

12 0 1

2

1 0 0 1

16

13

13

16

It is clear that RK4 method is consistent over the condition (5.3.8). The local trun-

cation error is O(h5). Also, it has four evaluation of function f in each step. We can

construct the function of f same as the second order method. First we begin with

Taylor expansion of exact solution

y(xn+1) = y(xn)+hy′(xn)+12

h2y′′(xn)+13!

h3y′′′(xn)+14!

h4y′′′′(xn)+O(h5) (5.5.3)

Then we replaced the first derivative, second derivative, third derivative, and fourth

derivative by using differentiation of (5.2.1). The expression will be interpreted

y(xn+1) = y(xn)+h f +12

h2 f fy+13!

h3[fyy f 2+ f 2

y f]

(5.5.4)

+14!

h4[fyyyy f 3+4 fyy fy f 2+ f 3

y f]+O(h5)

To contrast the approximate and exact solution, we need to indicate the expansion of

numerical method. By assuming

yn+1 = yn+h [b1k1+b2k2+b3k3+b4k4] (5.5.5)

61

with

k1 = f (x,y)

k2 = f (xn+ c2h,yn+ha21k1) (5.5.6)

k3 = f (xn+ c3h,yn+h [a31k1+a32k2])

k4 = f (xn+ c4h,yn+h [a41k1+a42k2+a43k3]

In order to extend ki’s we use again Taylor expansion formulae

∞∑m=0

m∑j=0

1j!(m− j)!

D j1D(m− j)

2 f (x,y)h jk(m− j) (5.5.7)

Here, expansion is only depending on y, so we neglect x which appears in argument

of f . After expansion is carry out, substitute ki’s to numerical method. Afterwards,

we have to match the expansions (5.5.4) and (5.5.5). We achieve ([16])

b1+b2+b3+b4 = 1 (5.5.8a)

b2c2+b3c3+b4c4 =12

(5.5.8b)

b2c22+b3c2

3+b4c24 =

13

(5.5.8c)

b3a32c2+b4a42c2+b4a43c3 =16

(5.5.8d)

b2c32+b3c3

3+b4c34 =

14

(5.5.8e)

b3c3a32c2+b4c4a42c2+b4c4a43c3 =18

(5.5.8f)

b3a32c22+b4a42c2

2+b4a43c23 =

112

(5.5.8g)

b4a43a32c2 =124

(5.5.8h)

62

In place of above a system of 8 equations in 10 unknowns is obtained. To show the

application is so complicated for Fourth Order Runge Kutta method. In contrast, It

can be solved by taking some appropriate value. Thus, since c4 = 1 is taken, b2,b3,b4

obtain for equation (5.5.8b),(5.5.8c) and (5.5.8c) ; then solve for a32,a42,a43 from

equation (5.5.8d), (5.5.8f), (5.5.8g); finally substituting to h. Many solution and s−

stages families of solution is arise these condition. There are two well known fourth

order methods

yn+1 = yn+h6

(k1+2k2+2k3+ k4)

k1 = f (xn,yn)

k2 = f (xn+12

h,yn+12

k1h)

k3 = f (xn+12

h,yn+12

k2h)

k4 = f (xn+h,yn+ k3h)

and

yn+1 = yn+h8

(k1+3k2+3k3+ k4)

63

k1 = f (xn,yn)

k2 = f (xn+13

h,yn+13

hk1)

k3 = f (xn+23

h,yn−13

hk1+hk2)

k4 = f (xn+h,yn+hk1−hk2+hk3)

5.6. Stability of Runge-Kutta Method

Runge-Kutta method has the same property of stability. Observe that the model

problem

y′ = λy (5.6.1)

y(0) = y0

with real and negative λ. Denote zT = [z1,z2, . . . ,zs] and eT = [1,1, . . . ,1] is the

s−dimensional vector. Apply the general form of s−stages to model problem to

get

zn = yne+hλAzn (5.6.2)

yn+1 = yn+hλbT zn

After some operations applied, it gives

yn+1 = yn+hλbT (1−hλA)−1eyn =(1+hλbT (1−hλA)−1e

)yn (5.6.3)

64

Then the method has stability function if

R(η) =(1+hλbT (1−hλA)−1e

)(5.6.4)

The Runge-Kutta method is A−Stable if modulus of function is less than 1 for Re(hλ)<0

which is

|R(η)| < 1 (5.6.5)

for all complex hλ.

Definition 5.6.1 ([24]) The method (5.3.2) is said to have order p if p is the largest

integer for which

y(x+h)− y(x)−hF(x,y(x);h) = O(hp+1) (5.6.6)

holds, where y(x) is known as the theoretical solution of the initial value problem.

Definition 5.6.2 ([26]) The method (5.3.2) is said to be consistent with initial value

problem if

F(x,y,0) ≡ f (x,y) (5.6.7)

The method is zero stable if F(x,y(x);h) = 0 as h −→ 0. It can be shown as follows:

yn+1 = yn+16

(k1+2k2+2k3+ k4) (5.6.8)

65

This equation becomes

yn+1− yn = 0 (5.6.9)

If we apply the root condition, the characteristic equation is satisfied

r−1 = 0

r = 1

Then root conditions hold when the modulus of root is less than or equal to 1 which

shown as |r| ≤ 1. We can say that the fourth order of Runge Kutta method is zero

stable.

To confirm that Runge Kutta method have greater accuracy than Euler method, we

need to calculate the local truncation error. Assume that

F(yn, xn;h) =s∑

i=1

biki

RK4 is consistent if F(yn, xn;0) = f (xn,yn)) then we must approve this rule. Replac-

ing the consistency condition into F(yn, xn;h) =∑s

i=1 biki, we get

s∑i=1

bi = 1

66

which is the necessary condition to satisfy the consistency condition for Runge-Kutta

method. Furthermore, we can define the local truncation error is

Tn+1 = y(xn+1)− y(xn)−hF(y(xn), xn;h)

The comparison of definition (5.6.2) and the local truncation error gives us

Tn+1 = O(hp+1)

where p is the order of the method. It is clear that Tn+1 =O(h5) is the local truncation

error of Fourth Order Runge Kutta method. Order of method assure the accuracy of

the method. For RK4 method, the local truncation error is appeared O(h5). This

means that the order of RK4 method is four. In case that RK4 is consistent and zero

stable, RK4 converges to analytical solution.

The definition of convergence is given in Chapter 2. To investigate the convergence

of Runge-Kutta method, consistency and stability condition must be hold. Consider

the form of method

yn+1 = yn+hF(xn,yn;h), n ≥ 0 (5.6.10)

which refers to the numerical solution of initial value problem (5.2.1). Using the

truncation error

Tn+1 = y(xn+1)− y(xn)−hF(x,y(x),h; f ) (5.6.11)

67

we define

τn(y) =1h

Tn+1(y)

In order to show the convergence of the solution (5.6.10), we need to have τn(y)−→ 0

as h −→ 0 since

τn(y) =y(xn+1)− yn

h−F(x,y(x),h; f ) (5.6.12)

We need that

F (xn,y(xn),h; f ) −→ y′(x) = f (x,y(x)) as h −→ 0

This shows that

η(h) = sup x0≤x≤b, −∞<y<∞| f (x,y)−F(x,y,h; f )| (5.6.1)

and since

η(h) −→ 0 as h −→ 0 (5.6.14)

the consistency condition is hold. We can also show the consistency result by other

way. Rewrite (5.6.12) in the form

y(xn+1) = y(xn)+hF(xn,y(xn),h; f )+hτn(y) (5.6.15)

68

then define

τ(h) = maxx0≤x≤b

|τn(y)|

The condition can be used for consistency condition while τ(h) −→ 0 as h −→ 0.

Apart from that we need to check the Lipschitz condition on F;

|F(x,y,h; f )−F(x,z,h; f )| ≤ L |y− z| (5.6.16)

for all x0 ≤ x ≤ b, −∞ < y, z <∞ and h > 0.

Theorem 5.6.3 ([10])Assume that the Runge-Kutta method satisfies the Lipschitz

condition. Then, for the initial value problem, the solution {yn} satisfies

maxx0≤xn≤b

|y(xn)− yn| ≤ e(b−x0)L |y(x0)− y0|+[e(b−x0)L−1

L

]τ(h) (5.6.17)

where

τ(h) ≡ maxx0≤xn≤b

|τn(y)| (5.6.18)

If the consistency condition (5.6.14) is also satisfied, then the numerical solution {yn}

converges to y(x).

Proof. ([11])Subtract (5.6.15) from (5.6.10) to obtain

en+1 = en+h[F(xn,y(xn),h; f )−F(xn,yn,h; f )

]+hτn(y) (5.6.19)

69

in which en = y(xn)− yn. Apply the Lipschitz condition (5.6.16) and use (5.6.18) to

get

|en+1| ≤ (1+hL) |en|+hτ(h), x0 ≤ xN ≤ b (5.6.20)

As with the convergence proof in Theorem 4.3.2 for the Euler method, given in

Section 4.3 of Chapter 3, this leads easily to the result (5.6.9). In most cases, it is

known by direct computation that τ (h)−→ 0 as h−→ 0, and in that case, convergence

of {yn} to y(x) is immediately proved. But all that we need to know is that (5.6.14) is

satisfied. To see this, write

hτn(y) = y(xn+1)− y(xn)−hF(xn,y(xn),h; f )

= hy′(xn)+12

h2y′′(ξn)−hF(xn,y(xn);h), y′(xn) = f (xn,y(xn))

h |τn(y)| ≤ hη(h)+12

h2∥∥∥y′′

∥∥∥∞τ(h) ≤ η(h)+

12

h∥∥∥y′′

∥∥∥∞Thus τ(h) −→ 0 as h −→ 0, completing the proof.

5.6.1. Absolute Stability of Runge-Kutta Method

The adequate technique is to consider model problem

y′ = λy (5.7.1)

y (0) = y0

70

The analytical solution of initial value problem is y′(x)= y0 exp(λx) where λ negative

real number. Here, analytical solution converge to 0 as x −→ +∞ at the exponen-

tial part. Note that this condition also is necessary for the solution of Runge Kutta

method. We investigate that what condition on step size the solution of Runge Kutta

method will satisfy the same behavior. By choosing the appropriate step size in the

numerical solution of initial value problem on the interval [x0, xm] with xm≫ x0, the

numerical solution reproduce desired status. When the numerical solution is applied

to the model problem, the interval of line hλ must satisfy the absolute stability con-

dition. This shows that solution of Runge Kutta method tends to zero as x −→ ∞.

Now, perform the Runge Kutta fourth order method to model problem to obtain

k1 = f (x,y) = λy

k2 = f (xn+ c2h,yn+ha21k1) = λ (y+ha21λy) =

= λy (1+a21hλ)

k3 = f (xn+ c3h,yn+h [a31k1+a32k2]) (5.7.2)

= λ(y+h

[a31λy+a32λy (1+a21hλ)

])= λy

(1+hλ [a31+a32]+h2λ2a32a21

)k4 = f (xn+ c4h,yn+h [a41k1+a42k2+a43k3] =

= λ(y+h

[a41λy

]+ha42λy [1+a21hλ]+ha43λy

[1+hλ [a31+a32]+h2λ2a32a21

])= λy

(1+hλa41+hλa42+h2λ2a42a21+hλa43+h2λ2a43 [a31+a32]+h3λ3a43a32a21

)

71

Assuming−h = hλ and

i−1∑j=1

ai j = ci, (i = 1, .., s), then rewrite (5.7.2)

k1 = λy

k2 = λy(1+a21

−h)

k3 = λy(1+−hc3+

−h2a32a21

)(5.7.3)

k4 = λy(1+−hc4+

−h2a42a21+

−h2a43c3+

−h3a43a32a21

)

Afterwards, substitute into (5.2.3) to obtain

F(x,y;h) = b1k1+b2k2+b3k3+b4k4

and

yn+1 = yn+−h

b1+b2

(1+a21

−h)+b3

(1+−hc3+

−h2a32a21

)+b4

(1+−hc4+

−h2a42a21+

−h2a43c3+

−h3a43a32a21

) yn

yn+1− yn =−h

(b1+b2+b3+b4)+ (b2a21+b3c3+b4c4)

−h

+ (b3a32a21+b4a42a21+b4a43c3)−h2+b4a43a32a21

−h3

yn (5.7.4)

yn+1/yn = 1+ (b1+b2+b3+b4)−h+ (b2a21+b3c3+b4c4)

−h2

+ (b3a32a21+b4a42a21+b4a43c3)−h3+b4a43a32a21

−h4

72

Now, we turn to the conditions (5.5.8a), (5.5.8b), (5.5.8d), (5.5.8h). It is evident that

the difference equation satisfies these conditions; so it is obtained

yn+1 =

(1+−h+

12

−h2+

16

−h3+

124

−h4

)yn (5.7.5)

Consequently, method is absolutely stable, that is, yn tends to 0 if and only if

∣∣∣∣∣∣1+ −h+ 12

−h2+

16

−h3+

124

−h4

∣∣∣∣∣∣ < 1 (5.7.6)

The plot of this function towards−h pose that the interval of absolute stability in

−hϵ [−2.78,0] .

73

Chapter 6

SYSTEM OF ORDINARY DIFFERENTIAL EQUATIONS

6.1. Introduction

More application problems include a system of several equation. The solution of a

system of ordinary differential equation is required in engineering and science that

has more complicated situations. The initial value problem of m differential equa-

tion’s system can be put into a form as

y′(x) = f(x,y(x)), y(x0) = y0 (6.1)

We can write as follows

y′1(x) = f1(x,y1(x),y2(x), ...,ym(x)), y1(x0) = y1,0

y′2(x) = f2(x,y1(x),y2(x), ...,ym(x)), y2(x0) = y2,0

...

y′m(x) = fm(x,y1(x),y2(x), ...,ym(x)), ym(x0) = ym,0

74

with the given some interval x0 ≤ x ≤ b. The general form of system can be repre-

sented the solution and the differential equation by using the column vector. Indicate,

y(x) =

y1(x)

y2(x)

...

ym(x)

, y0 =

y1,0

y2,0

...

ym,0

, f(x,y) =

f1(x,y1,y2, ...,ym)

f2(x,y1,y2, ...,ym)

...

fm(x,y1,y2, ...,ym)

with y =

[y1,y2, ...,ym

]T .

Example 6.1.1 ([13])The initial value problem

y′1 = y1(x)−2y2(x)+4cos(x)−2sin(x) , y1(0) = 1,

y′2 = 3y1(x)−4y2(x)+5cos(x)−5sin(x), y2(0) = 2

has the solution

y1(x) = cos(x)+ sin(x), y2(x) = 2cos(x)

System can be written as

y′(x) = Ay(x)+Φ(x), y(0) = y0

75

with

y =

y1

y2

, A =

1 −2

3 −4

,

Φ(x) =

4cos(x)−2sin(x)

5cos(x)−5sin(x)

, y0 =

1

2

In the notation above it can be represented as

f(x,y) =Ay+Φ(x), y =[y1,y2

]T

6.2. Stability Theory for System

Stability means that small disturbance in the initial value problems causes a small

change in the solution. Here, we consider the numerical methods for solving the

initial value problems that are numerically stable. Small change in initial value prob-

lem will cause a small change in the numerical solution for any sufficiently small

step size.

To examine the stability of system keep in view the initial value problem

y′(x) = f (x,y(x)),

y(x0) = y0

76

Instead of this, we consider the stability of numerical method for the model problem

y′(x) = λy(x)+g(x)

y(0) = 1

Stability and convergence can be answered for this problem. Expand y′(x) = f (x,y)

to get

f (x,y(x)) = f (x0,y0)+ fx(x0,y0)(x− x0)+ fy(x0,y0)(y(x)− y0)

Thus,

y′(x) = f (x,y(x))

= λ(y(x)− y0)+g(x)

with

g(x) = f (x0,y0)+ fx(x0,y0)(x− x0),

λ = fy(x0,y0)

Let V(x) = y(x)− y0, then we obtain

V ′(x) = λV(x)+g(x)

which is called model equation for the initial value problem. Make small change in

77

the initial value problem and see the difference in the solution

V ′ϵ(x) = λVϵ(x)+g(x)

Vϵ(x0) = V0+ ϵ

Here, g(x) will be cancelled. Because we are interested in the differ of the solutions

and we get model problem. Subtracting them we get

V ′ϵ(x)−V′(x) = λ(Vϵ(x)−V(x))

Vϵ(x0)−V(x0) = ϵ

Take W = V′ϵ(x)−V ′(x) to obtain

W′ = λW

W(x0) = ϵ

These operations demonstrate the stability and convergence of model equation. Now,

we examine the more general problem for system by considering the model equation.

It is similar with the above analysis.

The initial value problem of m differential equations of system describe as

y′ = f(x,y), x0 ≤ x ≤ b

y(x0)= y0

78

The final version of the model problem is defined

y′(x) = ∧y(x)+g(x)

with ∧ = fy(x0,y0). if f is differentiable, then fy(x,y) indicate a Jacobian matrix.

fy(x,y)i, j =∂ fi(x,y1,y2, ...,ym)

∂y j, 1 ≤ i, j ≤ m

This system can be written as

y′ = ∧y+g(x)

which can be reduced to equivalent system

z′i = λizi+γi(x), 1 ≤ i ≤ m

with λ1,λ2, ...,λm the eigenvalues of ∧ = fy(x0,y(x0)).

To investigate the stability of multistep method (3), examine the special model equa-

tion case

y′(x) = λy(x),

y0 = 1

79

After applied it to the multistep method, it becomes

yn+1 =

p∑j=0

a jyn− j+hλp∑

j=−1

b jyn− j , n ≥ p

yn+1 =

p∑j=0

a jyn− j+hλb−1yn+1+hλp∑

j=0

b jyn− j

we get

(1−hλb−1)yn+1 =

p∑j=0

(a j+hλb j

)yn− j

This is called a linear difference equation of (p+1) .We investigate a general solution

in a form

yn = rn , n ≥ 0

Substitute this form to the method and multiply by rp−n

rp+1 =

p∑j=0

a jrp− j+hλp∑

j=−1

b jrp− j

which is called characteristic equation. Denote the second characteristic roots,

σ(r) =p∑

j=−1

b jrp− j = b−1rp+1+

p∑j=0

rp− j

and remind that the first root of polynomials

ρ(r) = rp+1−p∑

j=0

a jrp− j

80

Thus first and second characteristic polynomial of (2.3) can be defined by

π(r,hλ) = ρ(r)−hλσ(r) = 0

The stability of numerical methods will be analyzed, when the step size is not chosen

very small. The model problem is the instructive technique to state the region of the

stability for any methods. We will investigate the case λ for this.

In general multistep methods (2.3), the characteristic equation is represented as

rp+1−p∑

j=0

a jrp− j−hλp∑

j=−1

b jrp− j = 0

for the determined model equation.

All roots of characteristic equation have magnitude 1 to satisfy the absolute stability.

We can find hλ from the characteristic equation.

rp+1−p∑

j=0

a jrp− j = hλp∑

j=−1

b jrp− j

hλ =rp+1−∑p

j=0 a jrp− j∑pj=−1 b jrp− j

This shows the region of the stability of method. When the constant λ is real, then λ

is negative (λ < 0) ; or when λ is complex, assume that Re(λ) < 0 in stable differential

equation problems. The true solution of the given model problem is

y(x) = eλx

Considering the above cases, the solution of model problem tends to zero as x tends

81

to infinity.

y(x) −→ 0 as x −→∞

When the any numerical method is applied to model problem, the approximated

solution satisfied

y(xn) −→ 0 as xn −→∞

If hλ is satisfied for above definition in any numerical method, then it is called region

of absolute stability of the numerical method.

82

Chapter 7

NUMERICAL EXPERIMENTS ON SIMPLE SYSTEMS

7.1. Introduction

Numerical experiments is mentioned to prove which numerical methods converge

faster to analytic solutions. Numerical experiments are investigated over an exam-

ple. In comparing different numerical methods, numerical experiments must be done

using the same of ordinary differential equations.([17]). For comparison bases, Hull

and Enright (1976) pointed out some assumptions that must be undertaken. Assump-

tions include assuming that the method is modelled to integrate between initial values

specified. Another assumption is assuming that local error is observed by keeping the

absolute error under the specified error tolerance. Differential equation is inspected

two varied numerical methods.

The given differential equations are analyzed for Explicit Euler method, Explicit

Runge-Kutta method. Analytical solution of ODE is calculated as well. Each one

of all is examined at varied step sizes. Step size is started with 0.1 and continued

with halved. Afterward, the absolute error is identified.

Initially, exact solution of DE is computed and then approximated solution is cal-

culated by using Matlab software for each numerical method at different step size.

Afterwards, absolute error are computed by taking difference analytical solution

83

to approximated solutions and presented in the same table at each step size. Ulti-

mately,errors at each step size of method are compared with other numerical meth-

ods. All numbers in the table. The computation process, errors and according to

graphs of each steps can be found in Appendix A and Appendix B.

7.2. Numerical Experiments on Stiff System

7.2.1. Numerical Experiments on Explicit Euler method

Consider the following example of stiff system;

dudx= 8u (x)−5v (x)+10w (x)

dvdx= 2u (x)+1v (x)+2w (x)

dwdx= −4u (x)+4v (x)+6w (x)

where initial conditions are u (0) = 2,v (0) = 2,w (0) = −3. It can be shown in general

form as y (0) = [2,2,−3]T . Its theoretical solution is given

u (x) = 6e2x−4e3x,

v (x) = −4e3x+6e2x,

w (x) = −e−2x+3e2x

where y (x) = [u (x) ,v (x) ,w (x)]T .[36]

Explicit Euler method and Explicit Fourth Order Runge Kutta method of numerical

methods is considered to solve system of ordinary differential equations. The system

84

is computed in Matlab software to evaluate values of the approximated and analytical

solution for Explicit Euler method and Explicit Fourth Order Runge Kutta method.

Initially, Explicit Euler method and Explicit Fourth Order Runge Kutta method is

computed for the given system at different step size. The system and analytical solu-

tions is entered into Matlab software for each method separately. Then analytical so-

lution and approximated solution is evaluated with step sizes 0.1,0.05,0.025,0.0125,

0.00625 and 0.003125. Afterwards, the absolute errors which is obtain by differ ap-

proximated solution from analytical solution, are calculated at specified step size.

Error tables with exact solution, approximated solution, and graphs, which is related

with exact and approximated solution, are shown in Appendix 2. In this table, we can

see how the approximated solutions behave when the step size is reduced by half. We

can compare the exact solution with approximate solution and identify the safe step

size ,which the approximated solution tends to analytical solution, through this table.

It can be determined the proximity of the approximated solution to exact solution

since the step size is changed.

In order to compute the approximated solution, step size is started with 0.1 for Ex-

plicit Euler method. As can be seen from the table in step size 0.1,0.05, 0.025 and

0.0125, the resulted of approximated solution is not closeness to exact solution. So,

these step sizes do not lie in the region of absolute stability. The difference between

exact solution and approximated solution must be excessively small. The computed

approximated solutions have much difference from analytical solutions as it seen in

the table. Nevertheless, the step size will lie inside the region of absolute stability

85

since h < 0.003125. The approximated solution is closer to exact solution. This

shows that the small step size provides the better approximation.

Application of Explicit Euler method is not occasionally preferred to use in stiff

systems. The reason is that cannot give the accurate approximated solution in fixed

error tolerance well. In order to achieve certain approximation, the step size must be

taken very small. This cause more iterations and high computation.

7.2.2. Numerical Experiments on Explicit Fourth Order Runge-Kutta method

The same system is computed in Matlab software. It is taken for comparing the

Explicit Euler method and Explicit Runge Kutta method. The same technique which

is implemented for Explicit Euler method, is applied to RK4. The computation of

Explicit Runge Kutta of fourth order method in same stiff system is also given at

different step sizes in Table for given systems. Step size is started from 0.1 and com-

putation progress is proceed of half. Evaluation of analytical solution is same and the

approximated solution is computed by using the step sizes 0.1,0.05,0.025,0.0125

respectively. In addition, local error of numerical method is generated by taking

modulus. Error tables with graphic can be found in Appendix 2. As it is seen in

table, at the step size h = 0.1, errors in system lie in the region of absolute stability.

Thus,absolute error is started to behave as approximate solutions But, the approxi-

mation of u(x),v(x),w(x) is not close to analytical solution. It is also appreciable in

absolute error which is not small absolute error. Again, when step sizes are taken

0.05 and 0.025, we can see in the table that absolute error is started to behave as

86

approximate solutions. These results indicate that the acceptable approximations,

and step sizes are in the region of the absolute stability. if the step size is used less

than 0.0125, the approximate solution will have same solutions with analytical so-

lutions. It is also noticeable from the error evaluation that have excessively small

difference. This shows that approximated solution turn into the exact solution at step

size h = 0.0125.

87

Chapter 8

COMPARISON OF EULER METHOD AND RUNGE-KUTTA

METHOD WITH ADVANTAGES AND DISADVANTAGES

1. EULER METHOD:

Advantages:

Euler’s method is the simplest of all linear multi-step method to obtain the ap-

proximated solution of the specified initial value problem. It has the one-step

techniques, and it can be easily programmed. Most of the ordinary differen-

tial equations can be solved conjecturally with numerical method and approxi-

mated solution yn+1 can be obtained from yn. The derivation of Euler’s method

can be revealed by constructing Taylor series. Approximated solution can be

acquired at each step before progressing to the next step. In addition, it has

merely single computation of function f in each time (step-size). The error

analysis, which involve local and global truncation error, and remainder term

can be obtained smoothly. Consequently, Euler’s method has the practical so-

lution techniques in order to solve the complicated differential equations.

Disadvantages:

In spite of the simplicity, it is restricted to use. The reason is; it generate large

error in each successive step during the computation which is the accumulated

error. In order to avoid the formation of larger error, step-size should be taken

88

excessively small.Therefore, it needs high computation of time. Additively,

approximated solution converge slower to analytical solution. This means that

the order of method is 1 and the error is observable O(h2

). It is a slow rate of

convergence.

2. RUNGE-KUTTA METHOD:

Advantages:

The idea of families of Runge Kutta method is too complicated, but higher or-

der provides much better approximated solutions than Euler method. The most

popular Runge-Kutta method is the method of order four. It is good choice to

get more accurate and more efficient solutions for solving the specified ordi-

nary differential equations.The approximated solution converge faster to exact

solution and the order of RK4 is 4 and the truncation error is O(h5

).

Disadvantages:

Method is re-evaluating the function f at each time to obtain the predictable

solution. It requires four evaluation per step. So, the computation of function

may take long time. The derivation of Runge-Kutta method is obtained from

Taylor series, but it is tedious to calculate higher derivative. To avoid this, the

function f is evaluated at more points.

89

Chapter 9

CONCLUSION

In this thesis, we have discussed numerical methods for solving systems of ordi-

nary differential equations. Some necessary conditions and definitions are given to

examine the numerical methods. After that, by considering these conceptions and

definitions, Euler methods and Runge Kutta method of order 4 are developed and,

derivations and stabilities of both methods are discussed. Afterwards, the system of

ODE is given and, we discussed the efficiency of two methods at the different step

size for systems using Tables of approximated solutions with exact solutions. At the

end, we gave information about the advantages and disadvantages of Euler’s method

and Runge Kutta method. Consequently, we see that in the Euler’s method exces-

sively small step size converge to analytical solution. Therefore, large number of

computation is needed. In contrast, Runge Kutta method gives better results and it

converge faster to analytical solution and has less iteration to get accuracy solution.

90

REFERENCES

[1] Atkinson, K. E., and Han, W., and Steward, D.,(1961), Numerical Solutions

of Ordinary Differential Equations, John Wiley & Sons, Inc., Hoboken, New

Jersey, Canada

[2] Atkinson, K. E., and Han, W., and Steward, D.,(1961), Numerical Solutions

of Ordinary Differential Equations, John Wiley & Sons, Inc., Hoboken, New

Jersey, Canada

[3] Atkinson, K. E., and Han, W., and Steward, D.,(1961), Numerical Solutions

of Ordinary Differential Equations, John Wiley & Sons, Inc., Hoboken, New

Jersey, Canada

[4] Atkinson, K. E., and Han, W., and Steward, D.,(1961), Numerical Solutions

of Ordinary Differential Equations, John Wiley & Sons, Inc., Hoboken, New

Jersey, Canada

[5] Atkinson, K. E., and Han, W., and Steward, D.,(1961), Numerical Solutions

of Ordinary Differential Equations, John Wiley & Sons, Inc., Hoboken, New

Jersey, Canada

[6] Atkinson, K. E., and Han, W., and Steward, D.,(1961), Numerical Solutions

of Ordinary Differential Equations, John Wiley & Sons, Inc., Hoboken, New

Jersey, Canada

91

[7] Atkinson, K. E., and Han, W., and Steward, D.,(1961), Numerical Solutions

of Ordinary Differential Equations, John Wiley & Sons, Inc., Hoboken, New

Jersey, Canada

[8] Atkinson, K. E., and Han, W., and Steward, D.,(1961), Numerical Solutions

of Ordinary Differential Equations, John Wiley & Sons, Inc., Hoboken, New

Jersey, Canada

[9] Atkinson, K. E., and Han, W., and Steward, D.,(1961), Numerical Solutions

of Ordinary Differential Equations, John Wiley & Sons, Inc., Hoboken, New

Jersey, Canada

[10] Atkinson, K. E., and Han, W., and Steward, D.,(1961) Numerical Solutions

of Ordinary Differential Equations, John Wiley & Sons, Inc., Hoboken, New

Jersey, Canada

[11] Atkinson, K. E., and Han, W., and Steward, D.,(1961), Numerical Solutions

of Ordinary Differential Equations, John Wiley & Sons, Inc., Hoboken, New

Jersey, Canada

[12] Atkinson, K. E., and Han, W., and Steward, D.,(1961), Numerical Solutions

of Ordinary Differential Equations, John Wiley & Sons, Inc., Hoboken, New

Jersey, Canada

[13] Atkinson, K. E., and Han, W., and Steward, D.,(1961) Numerical Solutions

of Ordinary Differential Equations, John Wiley & Sons, Inc., Hoboken, New

Jersey, Canada

92

[14] Atkinson, K. E., and Han, W., and Steward, D.,(1961) Numerical Solutions

of Ordinary Differential Equations, John Wiley & Sons, Inc., Hoboken, New

Jersey, Canada

[15] Atkinson, K. E., and Han, W., and Steward, D.,(1961), Numerical Solutions

of Ordinary Differential Equations, John Wiley & Sons, Inc., Hoboken, New

Jersey, Canada

[16] Butcher, J., C., (1933), Numerical Methods for Ordinary Differential Equa-

tions,John Wiley & Sons, The University of Auckland, New Zealand

[17] Hull, T.E. and Enright,W.H., (1976), The Results on Initial Value Methods for

Non-Stiff Ordinary Differential Equations, Society for Industrial and Applied

Mathematics Journal on Numerical Analysis, 13(6), pp,944-961.

[18] Lambert J.D., (1973), Computational Methods in Ordinary Differential Equa-

tions, London

[19] Lambert J.D., (1973), Computational Methods in Ordinary Differential Equa-

tions, London

[20] Lambert J.D., (1973), Computational Methods in Ordinary Differential Equa-

tions, London

[21] Lambert J.D., (1973), Computational Methods in Ordinary Differential Equa-

tions, London

93

[22] Lambert J.D., (1973), Computational Methods in Ordinary Differential Equa-

tions, London

[23] Lambert J.D., (1973), Computational Methods in Ordinary Differential Equa-

tions, London

[24] Lambert J.D., (1973), Computational Methods in Ordinary Differential Equa-

tions, London

[25] Lambert J.D., (1973), Computational Methods in Ordinary Differential Equa-

tions, London

[26] Lambert J.D., (1973), Computational Methods in Ordinary Differential Equa-

tions, London

[27] Lambert J.D., (1973), Computational Methods in Ordinary Differential Equa-

tions, London

[28] Lambert J.D., (1973), Computational Methods in Ordinary Differential Equa-

tions, London

[29] Lambert J.D., (1973), Computational Methods in Ordinary Differential Equa-

tions, London

[30] Lambert J.D., (1973), Computational Methods in Ordinary Differential Equa-

tions, London

[31] http://www.math.uiowa.edu/~atkinson/m171.dir/sec_6-3.pdf

[32] http://people.maths.ox.ac.uk/suli/nsodes.pdf

94

[33] http://people.maths.ox.ac.uk/suli/nsodes.pdf

[34] http://people.maths.ox.ac.uk/suli/nsodes.pdf

[35] http://people.maths.ox.ac.uk/suli/nsodes.pdf

[36] http://plato.asu.edu/MAT275/odes_matlab/Chapter12.pdf

[37] http://www.siam.org/books/ot98/sample/OT98Chapter7.pdf

95

96

APPENDICES

Appendix A. Explicit Euler method

.

97

Tabl

e1.

Exp

licit

Eul

erm

etho

dw

ithh=

0.1

forn=

10

ih

u app

(x)

v app

(x)

wap

p(x)

u(x)

v(x)

w(x

)∣ ∣ ∣ u( x)

−u a

pp(x

)∣ ∣ ∣∣ ∣ ∣ v(x)−

v app

(x)∣ ∣ ∣∣ ∣ ∣ w(x)

−w

app(

x)∣ ∣ ∣

00

22

−3

10.

1−0.4

2−1.2

−0.4

8705

11.

9289

8−1.2

4818

0.08

7050

70.

0710

187

0.04

8176

2

20.

2−2.9

21.

880.

48−3.2

6655

1.66

247

0.45

3554

0.34

6555

0.21

7527

0.02

6446

2

30.

3−5.7

161.

582.

112

−6.5

4554

1.09

432.

1734

90.

8295

430.

4857

0.06

1486

6

40.

4−8.9

668

1.01

723.

7632

−10.

5845

0.07

2777

93.

9806

51.

6176

90.

9444

220.

2174

49

50.

5−1

2.88

560.

0782

5.49

888−1

5.71

95−1.6

1707

5.94

757

2.83

384

1.69

527

0.44

8689

60.

6−1

7.73

44−1.3

9133

7.38

509−2

2.39

14−4.2

7789

8.15

319

4.65

705

2.88

656

0.76

8097

70.

7−2

3.84

11−3.6

0032

9.49

125−3

1.18

51−8.3

3348

10.6

867.

3439

84.

7331

61.

1947

7

80.

8−3

1.62

26−6.8

3033

11.8

928−4

2.88

13−1

4.37

4513.6

477

11.2

587

7.54

418

1.75

49

90.

9−4

1.61

27−1

1.45

9314.6

74−5

8.52

71−2

3.22

117.1

571

16.9

144

11.7

617

2.48

311

101−5

4.49

92−1

7.99

317.9

31−7

9.53

01−3

6.00

7821.3

552

25.0

3118.0

148

3.42

419

0 0.2 0.4 0.6 0.8 1−80

−60

−40

−20

0

20

40

h

u,v,

w

App1App2App3 Exact1Exact2Exact3

Figure 1. Graph of approximated solution and exact solution by using Explicit Eulermethod with h = 0.1

99

Tabl

e2.

Exp

licit

Eul

erm

etho

dw

ithh=

0.05

forn=

20

ih

u app

(x)

v app

( x)

wap

p( x

)u

( x)

v(x)

w( x

)∣ ∣ ∣ u( x)

−u a

pp( x

)∣ ∣ ∣∣ ∣ ∣ v( x)−

v app

( x)∣ ∣ ∣∣ ∣ ∣ w( x)

−w

app

( x)∣ ∣ ∣

00

22

−3

20.

1−0.4

31.

97−1.2

3−0.4

8705

11.

9289

8−1.2

4818

0.05

7050

70.

0410

187

0.01

8176

2

40.

2−3.0

5943

1.78

857

0.45

57−3.2

6655

1.66

247

0.45

3554

0.20

713

0.12

6102

0.00

2146

18

60.

3−6.0

636

1.37

712

2.12

604−6.5

4554

1.09

432.

1734

90.

4819

460.

2828

230.

0474

496

80.

4−9.6

5329

0.62

5441

3.84

796−1

0.58

450.

0727

779

3.98

065

0.93

1206

0.55

2664

0.13

2686

100.

5−1

4.09

02−0.6

1977

65.

6891

6−1

5.71

95−1.6

1707

5.94

757

1.62

932

0.99

7289

0.25

8412

120.

6−1

9.70

64−2.5

7043

7.72

071−2

2.39

14−4.2

7789

8.15

319

2.68

51.

7074

60.

4324

78

140.

7−2

6.93

02−5.5

1783

10.0

199−3

1.18

51−8.3

3348

10.6

864.

2548

82.

8156

50.

6661

3

160.

8−3

6.31

87−9.8

6065

12.6

731−4

2.88

13−1

4.37

4513.6

477

6.56

266

4.51

387

0.97

4611

180.

9−4

8.60

12−1

6.14

2315.7

792−5

8.52

71−2

3.22

117.1

571

9.92

589

7.07

873

1.37

796

201−6

4.73

67−2

5.10

1119.4

53−7

9.53

01−3

6.00

7821.3

552

14.7

934

10.9

067

1.90

212

0 0.2 0.4 0.6 0.8 1−80

−60

−40

−20

0

20

40

h

u,v,

w

App1App2App3 Exact1Exact2Exact3

Figure 2. Graph of approximated solution and exact solution by using Explicit Eulermethod with h = 0.05

101

Tabl

e3.

Exp

licit

Eul

erm

etho

dw

ithh=

0.02

5fo

rn=

40

ih

u app

( x)

v app

( x)

wap

p( x

)u(

x)v(

x)w

( x)

∣ ∣ ∣ u( x)−

u app

( x)∣ ∣ ∣∣ ∣ ∣ v( x)

−v a

pp( x

)∣ ∣ ∣∣ ∣ ∣ w( x)−

wap

p( x

)∣ ∣ ∣0

22

2−3

40.

1−0.4

548

1.95

116−1.2

405−0.4

871

1.92

898−1.2

482

0.03

221

0.02

218

0.00

766

80.

2−3.1

534

1.73

082

0.45

184−3.2

666

1.66

247

0.45

355

0.11

317

0.06

835

0.00

171

120.

3−6.2

851.

2480

22.

1454

1−6.5

455

1.09

432.

1734

90.

2605

90.

1537

20.

0280

8

160.

4−1

0.08

20.

3740

83.

9078

6−1

0.58

50.

0727

83.

9806

50.

5020

80.

3013

0.07

279

200.

5−1

4.84

1−1.0

716

5.80

898

−15.

72−1.6

171

5.94

757

0.87

899

0.54

545

0.13

859

240.

6−2

0.94−3.3

409

7.92

337−2

2.39

1−4.2

779

8.15

319

1.45

186

0.93

699

0.22

982

280.

7−2

8.87

7−6.7

8310.3

334−3

1.18

5−8.3

335

10.6

862.

3082

71.

5504

60.

3525

9

320.

8−3

9.30

8−1

1.88

13.1

326−4

2.88

1−1

4.37

513.6

477

3.57

381

2.49

438

0.51

516

360.

9−5

3.1−1

9.29

516.4

288−5

8.52

7−2

3.22

117.1

571

5.42

767

3.92

580.

7283

8

401−7

1.40

6−2

9.93

720.3

489−7

9.53−3

6.00

821.3

552

8.12

425

6.07

079

1.00

626

0 0.2 0.4 0.6 0.8 1−80

−60

−40

−20

0

20

40

h

u,v,

w

App1App2App3 Exact1Exact2Exact3

Figure 3. Graph of approximated solution and exact solution by using Explicit Eulermethod with h = 0.025

103

Tabl

e4.

Exp

licit

Eul

erm

etho

dw

ithh=

0.01

25fo

rn=

80

ih

u app

( x)

v app

( x)

wap

p( x

)u

( x)

v(x)

w( x

)∣ ∣ ∣ u( x)

−u a

pp( x

)∣ ∣ ∣∣ ∣ ∣ v( x)−

v app

( x)∣ ∣ ∣∣ ∣ ∣ w( x)

−w

app

( x)∣ ∣ ∣

00

22

−3

80.

1−0.4

6997

1.94

053

−1.2

447−0.4

8705

1.92

898−1.2

4818

0.01

7078

0.01

1553

0.00

3474

160.

2−3.2

0739

1.69

812

0.45

1996

−3.2

6655

1.66

247

0.45

3554

0.05

9165

0.03

565

0.00

1558

240.

3−6.4

099

1.17

462.

1583

3−6.5

4554

1.09

432.

1734

90.

1356

390.

0803

030.

0151

58

320.

4−1

0.32

340.

2304

413.

9425

8−1

0.58

450.

0727

783.

9806

50.

2610

890.

1576

630.

0380

73

400.

5−1

5.26

21−1.3

3113

5.87

58−1

5.71

95−1.6

1707

5.94

757

0.45

7359

0.28

5933

0.07

1772

480.

6−2

1.63

49−3.7

8579

8.03

466−2

2.39

14−4.2

7789

8.15

319

0.75

6507

0.49

2101

0.11

8523

560.

7−2

9.98

01−7.5

1763

10.5

045−3

1.18

51−8.3

3348

10.6

861.

205

0.81

5851

0.18

1523

640.

8−4

1.01

17−1

3.05

9413.3

826−4

2.88

13−1

4.37

4513.6

477

1.86

965

1.31

511

0.26

5073

720.

9−5

5.68

11−2

1.14

7116.7

823−5

8.52

71−2

3.22

117.1

571

2.84

601

2.07

393

0.37

4821

801

−75.

26−3

2.79

4220.8

371−7

9.53

01−3

6.00

7821.3

552

4.27

015

3.21

361

0.51

808

0 0.2 0.4 0.6 0.8 1−80

−60

−40

−20

0

20

40

h

u,v,

w

App1App2App3 Exact1Exact2Exact3

Figure 4. Graph of approximated solution and exact solution by using Explicit Eulermethod with h = 0.0125

105

Tabl

e5.

Exp

licit

Eul

erm

etho

dw

ithh=

0.00

625

forn=

160

ih

u app

( x)

v app

( x)

wap

p( x

)u

( x)

v(x)

w( x

)∣ ∣ ∣ u( x)

−u a

pp( x

)∣ ∣ ∣∣ ∣ ∣ v( x)−

v app

( x)∣ ∣ ∣∣ ∣ ∣ w( x)

−w

app

( x)∣ ∣ ∣

00

22

−3

160.

1−0.4

7826

11.

9348

8−1.2

4653−0.4

8705

11.

9289

8−1.2

4818

0.00

8789

670.

0058

9864

0.00

1648

6

320.

2−3.2

363

1.68

069

0.45

2598

−3.2

6655

1.66

247

0.45

3554

0.03

0253

40.

0182

149

0.00

0955

953

480.

3−6.4

7632

1.13

536

2.16

562

−6.5

4554

1.09

432.

1734

90.

0692

190.

0410

640.

0078

6322

640.

4−1

0.45

130.

1534

743.

9611

8−1

0.58

450.

0727

779

3.98

065

0.13

3188

0.08

0695

70.

0194

665

800.

5−1

5.48

61−1.4

7058

5.91

105

−15.

7195

−1.6

1707

5.94

757

0.23

3399

0.14

6485

0.03

6522

9

960.

6−2

2.00

51−4.0

2554

8.09

299

−22.

3914

−4.2

7789

8.15

319

0.38

6359

0.25

2353

0.06

0194

8

112

0.7−3

0.56

91−7.9

1468

10.5

939

−31.

1851

−8.3

3348

10.6

860.

6160

250.

4187

960.

0921

156

128

0.8−4

1.92

44−1

3.69

8713.5

132

−42.

8813

−14.

3745

13.6

477

0.95

6895

0.67

5778

0.13

4482

144

0.9−5

7.06

88−2

2.15

4216.9

67−5

8.52

71−2

3.22

117.1

571

1.45

837

1.06

683

0.19

0177

160

1−7

7.33

93−3

4.35

321.0

922

−79.

5301

−36.

0078

21.3

552

2.19

089

1.65

485

0.26

2934

0 0.2 0.4 0.6 0.8 1−80

−60

−40

−20

0

20

40

h

u,v,

w

App1App2App3 Exact1Exact2Exact3

Figure 5. Graph of approximated solution and exact solution by using Explicit Eulermethod with h = 0.00625

107

Tabl

e6.

Exp

licit

Eul

erm

etho

dw

ithh=

0.00

3125

forn=

320

ih

u app

( x)

v app

( x)

wap

p( x

)u

( x)

v(x)

w( x

)∣ ∣ ∣ u( x)

−u a

pp( x

)∣ ∣ ∣∣ ∣ ∣ v( x)−

v app

( x)∣ ∣ ∣∣ ∣ ∣ w( x)

−w

app

( x)∣ ∣ ∣

00

22

−3

320.

1−0.4

8259

21.

9319

6−1.2

4737−0.4

8705

11.

9289

8−1.2

4818

0.00

4458

430.

0029

8071

0.00

0802

203

640.

2−3.2

5126

1.67

168

0.45

3031

−3.2

6655

1.66

247

0.45

3554

0.01

5297

90.

0092

0782

0.00

0522

396

960.

3−6.5

1057

1.11

507

2.16

948

−6.5

4554

1.09

432.

1734

90.

0349

677

0.02

0767

10.

0040

0324

128

0.4−1

0.51

720.

1136

063.

9708

1−1

0.58

450.

0727

779

3.98

065

0.06

7272

20.

0408

286

0.00

9842

160

0.5−1

5.60

16−1.5

4291

5.92

915

−15.

7195

−1.6

1707

5.94

757

0.11

7913

0.07

4151

10.

0184

23

192

0.6−2

2.19

62−4.1

5008

8.12

285

−22.

3914

−4.2

7789

8.15

319

0.19

5268

0.12

7806

0.03

0334

6

224

0.7−3

0.87

36−8.1

2127

10.6

396

−31.

1851

−8.3

3348

10.6

860.

3115

030.

2122

120.

0464

025

256

0.8−4

2.39

72−1

4.03

1913.5

8−4

2.88

13−1

4.37

4513.6

477

0.48

4152

0.34

2612

0.06

7736

5

288

0.9−5

7.78

88−2

2.67

9917.0

614

−58.

5271

−23.

221

17.1

571

0.73

8337

0.54

1163

0.09

5793

8

320

1−7

8.42

02−3

5.16

7921.2

227

−79.

5301

−36.

0078

21.3

552

1.10

991

0.83

9907

0.13

2461

0 0.2 0.4 0.6 0.8 1−80

−60

−40

−20

0

20

40

h

u,v,

w

App1App2App3 Exact1Exact2Exact3

Figure 6. Graph of approximated solution and exact solution by using Explicit Eulermethod with h = 0.003125

109

Tabl

e7.

Exp

licit

Eul

erm

etho

dw

ithh=

0.00

1562

5fo

rn=

640

ih

u app

( x)

v app

( x)

wap

p( x

)u

( x)

v(x)

w( x

)∣ ∣ ∣ u( x)

−u a

pp( x

)∣ ∣ ∣∣ ∣ ∣ v( x)−

v app

( x)∣ ∣ ∣∣ ∣ ∣ w( x)

−w

app

( x)∣ ∣ ∣

00

22

−3

640.

1−0.4

8480

51.

9304

8−1.2

4778−0.4

8705

11.

9289

8−1.2

4818

0.00

2245

240.

0014

9832

0.00

0395

578

128

0.2−3.2

5886

1.66

710.

4532

81−3.2

6655

1.66

247

0.45

3554

0.00

7692

180.

0046

2937

0.00

0272

326

192

0.3−6.5

2797

1.10

474

2.17

147

−6.5

4554

1.09

432.

1734

90.

0175

745

0.01

0443

20.

0020

1961

256

0.4−1

0.55

070.

0933

143

3.97

57−1

0.58

450.

0727

779

3.98

065

0.03

3808

0.02

0536

40.

0049

4835

320

0.5−1

5.66

02−1.5

7976

5.93

832

−15.

7195

−1.6

1707

5.94

757

0.05

9264

20.

0373

064

0.00

9252

19

384

0.6−2

2.29

33−4.2

1357

8.13

796

−22.

3914

−4.2

7789

8.15

319

0.09

8163

90.

0643

174

0.01

5227

1

448

0.7−3

1.02

85−8.2

2666

10.6

627

−31.

1851

−8.3

3348

10.6

860.

1566

380.

1068

220.

0232

882

512

0.8−4

2.63

78−1

4.20

213.6

137

−42.

8813

−14.

3745

13.6

477

0.24

3526

0.17

2509

0.03

3993

4

576

0.9−5

8.15

56−2

2.94

8517.1

091

−58.

5271

−23.

221

17.1

571

0.37

1497

0.27

2556

0.04

8075

1

640

1−7

8.97

15−3

5.58

4721.2

887

−79.

5301

−36.

0078

21.3

552

0.55

8639

0.42

3136

0.06

6481

9

0 0.2 0.4 0.6 0.8 1−80

−60

−40

−20

0

20

40

h

u,v,

w

App1App2App3 Exact1Exact2Exact3

Figure 7. Graph of approximated solution and exact solution by using Explicit Eulermethod with h = 0.0015625

111

Appendix B. Fourth Order Runge Kutta method

.

112

Tabl

e8.

Four

thO

rder

Run

geK

utta

met

hod

with

h=0.

1fo

rn=

10

ih

u app

( x)

v app

( x)

wap

p( x

)u

( x)

v(x)

w( x

)∣ ∣ ∣ u( x)

−u a

pp( x

)∣ ∣ ∣∣ ∣ ∣ v( x)−

v app

( x)∣ ∣ ∣∣ ∣ ∣ w( x)

−w

app

( x)∣ ∣ ∣

00

22

−3

10.

1−0.4

8695

1.92

905

−1.2

482−0.4

8705

1.92

9−1.2

4818

0.00

0101

6.87

E−

052.

38E−

05

20.

2−3.2

663

1.66

2663

0.45

3508

−3.2

6656

1.66

250.

4535

540.

0002

550.

0001

94.

56E−

05

30.

3−6.5

4505

1.09

4692

2.17

3418

−6.5

4554

1.09

432.

1734

870.

0004

970.

0003

926.

82E−

05

40.

4−1

0.58

360.

0734

963.

9805

55−1

0.58

450.

0727

783.

9806

490.

0008

720.

0007

189.

43E−

05

50.

5−1

5.71

8−1.6

1584

5.94

7442

−15.

7195

−1.6

171

5.94

7569

0.00

145

0.00

1231

0.00

0127

60.

6−2

2.38

91−4.2

7587

8.15

3016

−22.

3914

−4.2

779

8.15

3185

0.00

2326

0.00

2022

0.00

0169

70.

7−3

1.18

15−8.3

3026

10.6

8579−3

1.18

51−8.3

335

10.6

8602

0.00

3642

0.00

3225

0.00

0225

80.

8−4

2.87

57−1

4.36

9513.6

4742−4

2.88

13−1

4.37

513.6

4772

0.00

5598

0.00

5031

0.00

0299

90.

9−5

8.51

87−2

3.21

3317.1

5675−5

8.52

71−2

3.22

117.1

5715

0.00

8483

0.00

7717

0.00

0397

101−7

9.51

74−3

5.99

6121.3

5463−7

9.53

01−3

6.00

821.3

5516

0.01

2707

0.01

168

0.00

0526

0 0.2 0.4 0.6 0.8 1−80

−60

−40

−20

0

20

40

h

u,v,

w

App1App2App3 Exact1Exact2Exact3

Figure 8. Graph of approximated solution and exact solution by using Explicit RK4method with h = 0.1

114

Tabl

e9.

Four

thO

rder

Run

geK

utta

met

hod

with

h=0.

05fo

rn=

20

ih

u app

( x)

v app

( x)

wap

p( x

)u

( x)

v(x)

w( x

)∣ ∣ ∣ u( x)

−u a

pp( x

)∣ ∣ ∣∣ ∣ ∣ v( x)−

v app

( x)∣ ∣ ∣∣ ∣ ∣ w( x)

−w

app

( x)∣ ∣ ∣

00

22

−3

20.

1−0.4

8704

381.

9289

86−1.2

4817

8−0.4

8705

071.

929

−1.2

4817

66.

92E−

064.

91E−

061.

45E−

06

40.

2−3.2

6653

71.

6624

870.

4535

51−3.2

6655

51.

6625

0.45

3553

81.

77E−

051.

35E−

052.

83E−

06

60.

3−6.5

4550

81.

0943

282.

1734

82−6.5

4554

31.

0943

2.17

3487

3.48

E−

052.

79E−

054.

30E−

06

80.

4−1

0.58

443

0.07

2829

033.

9806

43−1

0.58

449

0.07

2778

3.98

0649

6.13

E−

055.

12E−

056.

05E−

06

100.

5−1

5.71

938−1.6

1697

85.

9475

61−1

5.71

948−1.6

171

5.94

7569

0.00

0102

135

8.76

E−

058.

25E−

06

120.

6−2

2.39

126−4.2

7774

48.

1531

74−2

2.39

142−4.2

779

8.15

3185

0.00

0164

166

0.00

0143

872

1.11

E−

05

140.

7−3

1.18

484−8.3

3325

110.6

86−3

1.18

51−8.3

335

10.6

8602

0.00

0257

318

0.00

0229

322

1.49

E−

05

160.

8−4

2.88

093−1

4.37

415

13.6

477

−42.

8813

3−1

4.37

513.6

4772

0.00

0395

824

0.00

0357

608

2.00

E−

05

180.

9−5

8.52

653−2

3.22

049

17.1

5712

−58.

5271

3−2

3.22

117.1

5715

0.00

0600

045

0.00

0548

329

2.67

E−

05

201

−79.

5292

4−3

6.00

698

21.3

5512

−79.

5301

4−3

6.00

821.3

5516

0.00

0899

018

0.00

0829

558

3.55

E−

05

0 0.2 0.4 0.6 0.8 1−80

−60

−40

−20

0

20

40

h

u,v,

w

App1App2App3 Exact1Exact2Exact3

Figure 9. Graph of approximated solution and exact solution by using Explicit RK4method with h = 0.05

116

Tabl

e10

.Fou

rth

Ord

erR

unge

Kut

tam

etho

dw

ithh=

0.02

5fo

rn=

40

ih

u app

( x)

v app

( x)

wap

p( x

)u

( x)

v(x)

w( x

)∣ ∣ ∣ u( x)

−u a

pp( x

)∣ ∣ ∣∣ ∣ ∣ v( x)−

v app

( x)∣ ∣ ∣∣ ∣ ∣ w( x)

−w

app

( x)∣ ∣ ∣

00

22

−3

40.

1−0.4

871

1.92

898−1.2

482−0.4

871

1.92

9−1.2

482

4.55

E−

073.

28E−

079.

00E−

08

80.

2−3.2

666

1.66

247

0.45

355−3.2

666

1.66

250.

4535

51.

17E−

069.

04E−

071.

77E−

07

120.

3−6.5

455

1.09

432.

1734

9−6.5

455

1.09

432.

1734

92.

30E−

061.

87E−

062.

71E−

07

160.

4−1

0.58

40.

0727

83.

9806

5−1

0.58

40.

0727

83.

9806

54.

06E−

063.

41E−

063.

84E−

07

200.

5−1

5.71

9−1.6

171

5.94

757−1

5.71

9−1.6

171

5.94

757

6.78

E−

065.

85E−

065.

27E−

07

240.

6−2

2.39

1−4.2

779

8.15

319−2

2.39

1−4.2

779

8.15

319

1.09

E−

059.

60E−

067.

15E−

07

280.

7−3

1.18

5−8.3

335

10.6

86−3

1.18

5−8.3

335

10.6

861.

71E−

051.

53E−

059.

63E−

07

320.

8−4

2.88

1−1

4.37

413.6

477−4

2.88

1−1

4.37

513.6

477

2.63

E−

052.

38E−

051.

29E−

06

360.

9−5

8.52

7−2

3.22

117.1

572−5

8.52

7−2

3.22

117.1

572

3.99

E−

053.

65E−

051.

73E−

06

401

−79.

53−3

6.00

821.3

552−7

9.53

−36.

008

21.3

552

5.98

E−

055.

53E−

052.

30E−

06

0 0.2 0.4 0.6 0.8 1−80

−60

−40

−20

0

20

40

h

u,v,

w

App1App2App3 Exact1Exact2Exact3

Figure 10. Graph of approximated solution and exact solution by using Explicit RK4method with h = 0.025

118

Tabl

e11

.Fou

rth

Ord

erR

unge

Kut

tam

etho

dw

ithh=

0.01

25fo

rn=

80

ih

u app

( x)

v app

( x)

wap

p( x

)u

( x)

v(x)

w( x

)∣ ∣ ∣ u( x)

−u a

pp( x

)∣ ∣ ∣∣ ∣ ∣ v( x)−

v app

( x)∣ ∣ ∣∣ ∣ ∣ w( x)

−w

app

( x)∣ ∣ ∣

00

22

−3

80.

1−0.4

8705

071.

9289

81−1.2

4817

6−0.4

8705

071.

929

−1.2

4817

62.

91E−

082.

12E−

085.

60E−

09

160.

2−3.2

6655

51.

6624

730.

4535

538−3.2

6655

51.

6625

0.45

3553

87.

52E−

085.

84E−

081.

11E−

08

240.

3−6.5

4554

21.

0943

2.17

3487

−6.5

4554

31.

0943

2.17

3487

1.48

E−

071.

21E−

071.

70E−

08

320.

4−1

0.58

449

0.07

2778

13.

9806

49−1

0.58

449

0.07

2778

3.98

0649

2.62

E−

072.

20E−

072.

42E−

08

400.

5−1

5.71

948−1.6

1706

55.

9475

69−1

5.71

948−1.6

171

5.94

7569

4.37

E−

073.

78E−

073.

33E−

08

480.

6−2

2.39

142−4.2

7788

88.

1531

85−2

2.39

142−4.2

779

8.15

3185

7.03

E−

076.

20E−

074.

53E−

08

560.

7−3

1.18

51−8.3

3347

910.6

8602

−31.

1851

−8.3

335

10.6

8602

1.10

E−

069.

87E−

076.

12E−

08

640.

8−4

2.88

132−1

4.37

451

13.6

4772

−42.

8813

3−1

4.37

513.6

4772

1.70

E−

061.

54E−

068.

22E−

08

720.

9−5

8.52

713−2

3.22

104

17.1

5715

−58.

5271

3−2

3.22

117.1

5715

2.57

E−

062.

36E−

061.

10E−

07

801

−79.

5301

3−3

6.00

781

21.3

5516

−79.

5301

4−3

6.00

821.3

5516

3.86

E−

063.

57E−

061.

47E−

07

0 0.2 0.4 0.6 0.8 1−80

−60

−40

−20

0

20

40

h

u,v,

w

App1App2App3 Exact1Exact2Exact3

Figure 11. Graph of approximated solution and exact solution by using Explicit RK4method with h = 0.0125

120

Tabl

e12

.Fou

rth

Ord

erR

unge

Kut

tam

etho

dw

ithh=

0.00

625

forn=

160

ih

u app

( x)

v app

( x)

wap

p( x

)u

( x)

v(x)

w( x

)∣ ∣ ∣ u( x)

−u a

pp( x

)∣ ∣ ∣∣ ∣ ∣ v( x)−

v app

( x)∣ ∣ ∣∣ ∣ ∣ w( x)

−w

app

( x)∣ ∣ ∣

00

22

−3

160.

1−0.4

8705

071.

9289

81−1.2

4817

6−0.4

8705

071.

929

−1.2

4817

61.

84E−

091.

35E−

093.

50E−

10

320.

2−3.2

6655

51.

6624

730.

4535

538−3.2

6655

51.

6625

0.45

3553

84.

77E−

093.

71E−

096.

91E−

10

480.

3−6.5

4554

31.

0943

2.17

3487

−6.5

4554

31.

0943

2.17

3487

9.38

E−

097.

66E−

091.

07E−

09

640.

4−1

0.58

449

0.07

2777

893.

9806

49−1

0.58

449

0.07

2778

3.98

0649

1.66

E−

081.

40E−

081.

52E−

09

800.

5−1

5.71

948−1.6

1706

55.

9475

69−1

5.71

948−1.6

171

5.94

7569

2.77

E−

082.

40E−

082.

10E−

09

960.

6−2

2.39

142−4.2

7788

88.

1531

85−2

2.39

142−4.2

779

8.15

3185

4.46

E−

083.

94E−

082.

85E−

09

112

0.7

−31.

1851

−8.3

3348

10.6

8602

−31.

1851

−8.3

335

10.6

8602

7.00

E−

086.

27E−

083.

86E−

09

128

0.8−4

2.88

133−1

4.37

451

13.6

4772

−42.

8813

3−1

4.37

513.6

4772

1.08

E−

079.

77E−

085.

19E−

09

144

0.9−5

8.52

713−2

3.22

104

17.1

5715

−58.

5271

3−2

3.22

117.1

5715

1.63

E−

071.

50E−

076.

94E−

09

160

1−7

9.53

014−3

6.00

781

21.3

5516

−79.

5301

4−3

6.00

821.3

5516

2.45

E−

072.

27E−

079.

26E−

09

0 0.2 0.4 0.6 0.8 1−80

−60

−40

−20

0

20

40

h

u,v,

w

App1App2App3 Exact1Exact2Exact3

Figure 12. Graph of approximated solution and exact solution by using Explicit RK4method with h = 0.00625

122