Application of Optimal Homotopy Asymptotic Method to

iii

Application of Optimal Homotopy

Asymptotic Method to Initial and Boundary

Value Problems

by

Muhammad Idrees

A thesis submitted for the partial fulfillment of the degree of Doctor of Philosophy to

the Ghulam Ishaq Khan Institute of Engineering Sciences & Technology.

Faculty of Engineering Sciences, Ghulam Ishaq Khan Institute

of Engineering Sciences and Technology, Topi, Pakistan.

May 2011

iv

Application of Optimal Homotopy

Asymptotic Method to Initial and Boundary

Value Problems

By

Muhammad Idrees

Supervised by

Dr. Sirajul Haq

A dissertation submitted in partial fulfilment of the requirement of the Degree of Doctor

of Philosophy in Engineering Sciences (Applied Mathematics)

May 2011

Ghulam Ishaq Khan Institute of Engineering Sciences and Technology, Topi,

Swabi, Pakistan

v

Dedicated to:

Prof. Dr. Syed Ikram Abbas Tirmizi,

Dr. Shaukat Iqbal & Dr. Saeed Islam

vi

CERTIFICATE OF APPROVAL

It is certified that the research work presented in this thesis, entitled “Application of

Optimal Homotopy Asymptotic Method to Initial and Boundary Value Problems” was

conducted by Mr. Muhammad Idrees under the supervision of Dr. Sirajul Haq.

(External Examiner) (Internal Examiner)

(Supervisor) (Dean)

Acknowledgements

vii

Acknowledgements

I would like to thank all people who have helped and inspired me during my doctoral

study. I especially want to thank my supervisor, Dr. Sirajul Haq, for his guidance during

my research and study at GIK Institute. In addition, he was always accessible and willing

to help his students with their research. As a result, research life became smooth and

rewarding for me. I was delighted to interact with Prof. S. I. A. Tirmizi by attending his

classes and having him as my co-advisor. His insights to Numerical Analysis are second

to none. Besides, he sets an example of a world-class researcher for his rigor and passion

on research. I would be failing in my duties if I don’t appreciate the help of professors in

the Faculty of Engineering Sciences and staff for their cooperation. I would like to thank

every one at the FES for day-to-day support, advice, and patience.

Prof. A. M. Siddiqi, Prof. Vasile Marinca, Dr. Nicolae Herisanu, Prof. Aftab Khan, Dr.

Muhammad Sagheer, Dr. Ghulam Shabbir and Dr. Hina Khan deserve special thanks as

my thesis committee members and advisors.

My deepest gratitude goes to my family for their unflagging love and support throughout

my life; this dissertation is simply impossible without them. I am indebted to my parents,

Nawab Gul and Sakeena Bibi, for their care and love. Although they are no longer with

us, they are forever remembered. I am sure they share our joy and happiness in the

heaven. Father and Mother, I love you. I am extremely thankful to my wife Shah

Khurram for her constant support when I encountered difficulties and look after of our

four cute, loveliest but naughty sons viz Hassaan Idrees, Talha Idrees, Mohsin Idrees and

Sarim Idrees. I feel proud of my brothers and sister, Tasleem Begum, for their prayers and

support through out my life. My Uncle Muhammad Iqbal and my ante Mar yam Bibi have

always been a constant source of encouragement during my graduate study. They are like

my spiritual father and mother who drew me close to the Goal.

Dr. Shaukat Iqbal, the former graduate student at the Faculty of Computer Sciences, GIK

Institute has been my friend and mentor for many years. He offers advice and suggestions

whenever I need them. Besides, he has set a role model of a typical friend who cares and

loves their friends as if they were their own family members. Thanks Dr. Shaukat Iqbal.

I thank Dr. Saeed Islam of COMSATS, for hiring me to his team; his friendship and

technical guidance are valued. Furthermore, I am grateful to come across several life-long

friends at work. Thanks to, Dr. Siraj-Ul-Islam, Dr. Gul Zaman, Dr. Abdul Rauf, Arshad

Husain, Javed Ali, Rehan Ali Shah, Meer Aslam and Fazle Mabood.

The generous support from Higher Education Commission (HEC), Pakistan for the

award of the Merit Scholarship for PhD studies in science and technology (200

scholarships) is greatly appreciated. Without their support, my ambition to study at GIK

Institute can hardly be realized.

Last but not least, thanks are to God for my life through all tests in the past five years.

You have made my life more bountiful. May your name be exalted, honoured, and

glorified.

(Muhammad Idrees)

Abstract

viii

Declaration

The contents of this dissertation are my original work except where specific

acknowledgement is given. This dissertation has not been submitted in whole or in part to

any other University. Certain aspects of this dissertation have been

published/accepted/submitted as follows:

1 S. Haq, M. Idrees, S. Islam, Application of Optimal Homotopy Asymptotic Method to

eighth order boundary value problems, International Journal of Applied Mathematics

and Computation, 2(4), 38-47, 2010.

2 M. Idrees, S. Haq, S. Islam, Application of Optimal Homotopy Asymptotic Method to

fourth order boundary value problems, World Applied Sciences Journal, 9(2): 131-

137, 2010.

3 M. Idrees, S. Haq, S. Islam, Application of the Optimal Homotopy Asymptotic

Method to squeezing flow, Computers and Mathematics with Applications, 59(12) ,

3858-3866, 2010.


special sixth order boundary value problems, World Applied Sciences Journal, 9(2):

138-143, 2010.

5 M. Idrees, S. Islam, S. I. A. Tirmizi, S. Haq, Application of Optimal Homotopy

Asymptotic Method to Kortweg-de Vries Equations, Mathematical and Computer

Modeling, In Press, DOI information 10.1016/j.mcm.2011.10.010.


Special fourth order boundary value problems submitted.

7 S. Haq, M. Idrees, Application of Optimal Homotopy Asymptotic Method to Wave

Equations, submitted to Applied Mathematics and Computation.

8 S. Iqbal, M. Idrees, A.M Siddiqi and A. Ansari, Some solutions of the linear and

nonlinear Klein-Gordon equations using the optimal homotopy asymptotic method,

Applied Mathematics and Computation, 216 (2010) 2898-2909.

Abstract

ix

Abstract

The development of nonlinear science has grown an ever-increasing interest among

scientists and engineers for analytical asymptotic techniques for solving nonlinear

problems. Finding solutions to linear problems by means of computer is easier nowadays;

however, it is still difficult to solve nonlinear problems numerically or theoretically. The

reason is the use of iterative techniques in the various discretization methods or numerical

simulations to find numerical solutions to nonlinear problems. Almost all iterative

methods are sensitive to initial solutions; hence, it is very difficult to obtain converging

results in cases of strong nonlinearity.

The objective of this dissertation is to use Optimal Homotopy Asymptotic Method

(OHAM), a new semi-analytic approximating technique, for solving linear and nonlinear

initial and boundary value problems. The semi analytic solutions of nonlinear fourth

order, eighth order, special fourth order and special sixth order boundary-value problems

are computed using OHAM. Successful application of OHAM for squeezing flow is a

major task in this study.

This dissertation also investigates the effectiveness of OHAM formulation for Partial

Differential Equations (Wave Equation and Korteweg de Vries). OHAM is independent

of the free parameter and there is no need of the initial guess as there is in Homotopy

Perturbation Method (HPM), Variational Iteration Method (VIM) and Homotopy

Analysis Method (HAM). OHAM works very well with large domains and provides

better accuracy at lower-order of approximations. Moreover, the convergence domain can

be easily adjusted. The results are compared with other methods like HPM, VIM and

HAM, which reveal that OHAM is effective, simpler, easier and explicit.

Table of Contents

x

Table of Contents

CHAPTER 1. Introduction

1.1 Introduction ................................................................................................................... 1

1.2 Thesis Plan .................................................................................................................... 4

CHAPTER 2. Optimal Homotopy Asymptotic Method

2.1 Introduction ................................................................................................................... 5

2.2 Fundamental Mathematical Theorey of Optimal Homotopy Asymptotic Method ...... 5

2.3 Testing of Optimal Homotopy Asymptotic Method for an ODE (Example 1) ........... 8

2.4 Optimal Homotopy Asymptotic Method for PDEs .................................................... 11

2.5 Testing of Optimal Homotopy Asymptotic Method for a PDE (Example 2) ............ 13

CHAPTER 3. Application of Optimal Homotopy Asymptotic Method for Special and

Nonlinear Boundary Value Problems

3.1. Introduction 17

3.2. Application of OHAM to fourth order nonlinear boundary value problem 17

3.2.1 Model 1 .............................................................................................................. 17

3.2.2 Model 2 .............................................................................................................. 21

3.2.3 Model 3 ............................................................................................................... 25

3.3 Application of OHAM to special fourth order boundary value problem BVPs ........ 27

3.3.1 Model 4 ............................................................................................................... 27

3.3.2 Model 5 .............................................................................................................. 31

3.4 Application of OHAM to special sixth order boundary value problem...................... 34

3.4.1 Model 6 ............................................................................................................... 34

3.5 Application of OHAM to Eighth order boundary value problem ............................... 39

3.5.1 Model 7 ............................................................................................................... 39

CHAPTER 4 Application of Optimal Homotopy Asymptotic Method to Squeezing

Flow

4.1 Introduction .................................................................................................................. 45

4.2 Fundamental Equations ............................................................................................... 45

4.3 Problem Formulations ................................................................................................. 45

4.4 Application of OHAM to squeezing flow problem .................................................... 46

4.5 Solutions by PM and HPM ......................................................................................... 52

Table of Contents

xi

CHAPTER 5 Application of Optimal Homotopy Asymptotic Method to Korteweg de

Vries Equations

5.1 Introduction ................................................................................................................ 57

5.2 Application of OHAM to KdV Equations ................................................................ 57

5.2.1 KdV equation Model 1 ........................................................................................ 57



CHAPTER 6 Application of Optimal Homotopy Asymptotic Method Wave

Equations

6.1 Introduction ................................................................................................................ 69

6.2 Applications of OHAM to Wave Equations ............................................................... 69

6.2.1 Wave equation Model 1 ...................................................................................... 69



CHAPTER 7 Conclusions

Conclusions ....................................................................................................................... 82

References ....................................................................................................................... 84.

Chapter 1 Introduction

1

CHAPTER 1

Introduction

1.1. Literature Survey

In the literature, there are very few exact solutions of the initial and boundary value

problems and even these become rare when problems are nonlinear. Most natural

phenomena can best be described by differential equations with varying boundary

conditions. The roles of analytical and numerical methods have been equally significant

in solving problems of science and engineering. In nonlinear cases, however, these

methods do not always yield exact solutions. Therefore, there is a strong need to build up

other approaches for solving the complicated non-linear problems. Well-known numerical

methods are collocation methods [1-2], finite difference methods [3-4], finite element

methods [3-5], radial basis function collocation method [6-17] etc. Advantage of

numerical methods [18] over analytical methods is that, they can handle nonlinear

problems in simple domains but are costly in complex domains. Great progress has been

made in developing various analytical techniques to ensure economy of cost and time in

solving problems.

Iterative methods [19-20] have a pioneering role but since nearly all iterative methods are

sensitive to initial solutions, they are difficult to yield converged results in cases of strong

nonlinearity. Perturbation methods (PMs) [21-28] provide the most flexible tools for

nonlinear analysis of engineering problems, and they are constantly being developed and

applied to problems that are rather more complex. For nonlinear problems, especially,

PMs hold the sway. The basic theme of the method is to assume the target unknown

function in powers of the parameter . Like other nonlinear asymptotic techniques,

though, PMs have their own merits [29] and limitations:

(1) Almost all PMs depend on assuming the existence of a small parameter in a

mathematical model. This small parameter assumption limits the application of PMs, as

many nonlinear problems, especially those with strong nonlinearity, do not have any

small parameter.

(2) Determining the small parameter in itself is an issue, which seems to be a special

art requiring special techniques. An appropriate choice of small parameter may lead to

ideal results. An improper choice, however, results in seriously adverse effects.


2

(3) Even if a suitable small parameter is determined, the approximate solutions given

by the PMs are valid, in most cases, only for the small values of the parameter. However,

the approximations cannot be relied on fully because there is no criterion on how small

the parameters should be.

Thus, researchers have been trying to improve the approximation of nonlinear

mathematical model of physical phenomena. Liao devised HAM [18, 30-36] using

homotopy [37]. In HAM, the author applies the essential concept of homotopy in

topology to transform a nonlinear problem into many linear sub-problems. HAM

originates from a generalized Taylor series with respect to an embedding parameter.

HAM solves many different types of nonlinear mathematical models. The validity of the

HAM does not depend on whether or not small parameters exist in problems under

consideration. It affords great liberty in choosing auxiliary linear operators.

“It has been successfully applied to many nonlinear problems such as nonlinear

oscillations, boundary layer flows, heat transfer, viscous flows in porous media,

viscous flows of Oldroyd 6-constant fluids, magneto hydrodynamic flows of non-

Newtonian fluids , nonlinear water waves, Thomas-Fermi equation , Lane-Emden

equation , and so on.” [18]

HPM was proposed by He [38]. It is a hybrid of homotopy and perturbation. This method

provides an asymptotic solution with few terms without requiring any convergence

theory. In contrast to the PMs, this technique works independently of small parameters.

This method exploits all the benefits of both PMs and homotopy techniques. This method

converts a given problem to a series of linear equations that are easy to solve. This

method yields a rapid convergence of the solution series and in most cases, only a few

iterations lead to a highly accurate solution. “The applications of the HPM mainly cover

in nonlinear differential equations, nonlinear integral equations, nonlinear differential-

integral equations, difference-differential equations, and fractional differential equations”

[38]. The bottom line is that the method is applicable to a number of nonlinear problems.

Vasile Marinca et al and [39-43] N. Herisanu et-al introduced OHAM for approximate

solution of “nonlinear problems of thin film flow of a fourth grade fluid down a vertical

cylinder” and for the study of behavior of “nonlinear mechanical vibration of electrical

machine.” The same authors used this method for the solution of “nonlinear equations

arising in the steady state flow of a fourth-grade fluid past a porous plate” and for the

solution of “nonlinear equation arising in heat transfer.” Recently, Marinca et-al has

applied OHAM to the “determination of periodic solutions for the motion of a particle on


3

a rotating parabola” [42]. More recently, Javed et-al [44] has applied OHAM to the

“solution of multipoint boundary value problems”. S. Iqbal et-al [45] has done useful

work on “some solutions of linear and nonlinear Klein-Gordon equations using optimal

homotopy asymptotic method.” In [46] OHAM has been applied to “Jeffery-Hamel flow

in which fluid flows between two rigid plane walls, where the angle between them

is 2 .” In [47] problem of “steady incompressible mixed convection flow past vertical

flat plate” has been studied by means of OHAM. Regarding the application of OHAM,

the contributions of Idrees and S. Haq can be seen in [93-96]. Historical hierarchy for the

advancement in the coupling of homotopy with perturbation is summarized in Table 1.

Table 1

Hierarchical coupling of homotopy with perturbation at a glance

Reference Type of Differential

Equation

Family of Homotopy

Liao [31]

1992 0u x N 0(1 ) ; ; 0,p x p u x x p L H N H

J.He [38]

1999 u u f x L N

0(1 )

0,

p u u

p u u f x

L L

L N

Liao [33]

1997 0u x N 0 0(1 ) ; ; ,p x p u x c p x p L H N H

Liao [34]

1999 0u x N

0

0

1 ;

; ,

B p x p u x

c A p x p

L H

N H

V. Marinca

et-al

[39-43] 2008

0

, 0

u x g x

u x

uu

x

L

N

B

2

1 2

(1 ) ( )

( )

where ...

p u x g x

h p u x g x u x

h p pC p C

L

L N

Liao [48]

2009 0u x N

0

2 3

0 1 2

1 ;

; ,

B p x p u x

c p c p c p x p

L H

N H

Z. Niu et-al

[49-51] 2009 0u x N 0 0(1 ) ; ; ,p x p u x c p x p L H N H


4

1.2. Thesis Plan

This dissertation describes the application of OHAM to different linear and non-linear

problems arising in the theory of differential equations and fluid mechanics. It consists of

seven chapters. The material in Chapter 1 covers the historical background of analytic

methods like PM, HPM, HAM and OHAM. The Chapter 2 explores the nature and

application of the basic idea of OHAM for the solution of ordinary differential equations

and partial differential equations. Chapter 3 covers the effectiveness of semi analytic

solutions to nonlinear fourth order, nonlinear eighth order, special fourth order and

special sixth order boundary-value problems. Chapter 4 explains the application of

OHAM formulation to squeezing flow. Chapter 5 and 6 describe the usefulness and

effectiveness of OHAM formulation for Partial Differential Equations, respectively. The

chapters examine both KdV and wave equations extensively. A summary of the results

obtained is presented in Chapter 7. It is observed that the OHAM is better organized and

simpler than the other methods in our problems.

Chapter 2 Optimal Homotopy Asymptotic Method

5

CHAPTER 2

Optimal Homotopy Asymptotic Method

2.1. Introduction

In this chapter, basic steps of OHAM for the solution linear and nonlinear equations

arising in different physical and industrial problems are discussed.

First, homotopy is defined as:

Homotopy [51]:

Homotopy is a branch of geometry concerned with the general properties of shapes and

space. It can be thought of as the study of properties that are not changed by continuous

deformations, such as stretching or twisting. A transformation occurs when there is one-

to-one correspondence between points in one figure and points in another. If one figure

can be transformed into another by such transformation, the figures are said to be

homotopically equivalent. Thus “two continuous functions from one topological space to

another are called homotopic if one can be continuously deformed into the other. Such

deformation is called a homotopy between the two functions. That is,

, : 0,1 ; , 0,1 H and H is the deformation of the original

function f .

1 2[ , ] (1 ) ( ) ( )f f H (2.1)

would be a homotopy between the functions 1( )f and 2 ( )f .Here, is called the

embedding parameter and if 0 , 1( ,0) ( )f H and if 1 , 2( ,1) ( )f H . Thus, as

changes from 0 to 1, 1( )f is gradually transformed into 2 ( )f ”.

Residual [5].

Residual is the difference of true value and the estimated value. For a differential

equation, it is computed by substituting the estimated solution into the original

differential equation.

2.2. Fundamental mathematical theory of OHAM [39-43]

In order to explain the fundamental idea of OHAM, consider the following general

differential equation:


6

( ( )) ( ) ( ( )) 0,w g w L N (2.2)

, 0dw

wd

B (2.3)

where denotes independent variable, ( )w is a required function , ( )g is a given

function, L , N and B are linear, nonlinear and boundary operators.

Applying OHAM to the given problem, a general deformation equation is presented as:

(1 )[ ( ( , ) ( )] ( )[ ( ( , ) ( ) ( ( , ))],g h g L H L H N H (2.4)

,

, , 0,

HB H (2.5)

where 0 1 is an embedding parameter, ( )h is a nonzero auxiliary function for

0 and (0) 0, ,h H is an unknown function. Clearly, when 0 and 1 it

holds 0,0 w H and ,1 w H respectively.

Thus, as changes from 0 to 1 , the solution ( , ) H changes from 0 ( )w to the

solution ( )w , where 0 ( )w is obtained from Eq (2.4) for 0 :

00 0( ) 0, , 0 .

dww g w

d

L B (2.6)

For actual applications iK are finite, say, 1,2,3,..., .i m We propose the auxiliary

function h to be of the form:

2 3

1 2 3 ... m

mh K K K K , (2.7)

where1 2, ,..., mK K K are constants.

For solution, expanding , H in Taylor’s series about , we obtain:

0 1 2

1

, , , ,..., n

k n

n

w w K K K

H . (2.8)

Substituting Eq. (2.8) into Eq. (2.4) and equating the coefficient of like powers of , we

obtained:

Zeroth order problem is given by Eq. (2.6) and the first and second order problems are

given by the Eqs. (2.7)-(2.12) respectively:


7

1 1 0 0w g K w L N (2.9)

11, 0

dww

d

B (2.10)

2 1 2 0 0

1 1 1 0 1, ,

w w K w

K w w w

L L N

L N (2.11)

22 , 0

dww

d

B (2.12)

In general it can be written as:

1 0 0

1

0 1 1

1

, ,..., ,

n n n

n

i n i n i n

i

w w K w

K w w w w

L L N

L N (2.13)

, 0,nn

dww

d

B (2.14)

where 2,3,...,n .

In the above equation 0 1 1, ,...,m nw w w N is the coefficient of m in the

expansion of , N H :

0 0 0 1

1

, , , ,..., m

i m m

m

K w w w w

N H N N (2.15)

Here, convergence of the series (2.8) depends upon the constants , 1,2,3,...iK i .

When 1 Eq.(2.8) transforms to:

1 2 0 1 2

1

, , ,... , , ,...,i m

i

w K K w w K K K

(2.16)

In actual application i is finite. Thus Eq.(2.10) reduce to:

1 2 0 1 2

1

, , ,..., , , ,...,m

m i m

i

w K K K w w K K K

(2.17)

Substituting Eq. (2.17) into Eq. (2.4), it results the following residual:

1 2 1 2 1 2, , ,..., ( , , ,..., ) ( ) ( , , ,..., )m m mR K K K w K K K g w K K K L N (2.18)


8

If 0, thenR w yields the exact solution. Generally it doesn’t happen, especially in

nonlinear problems.

For the determinations constants of , 1,2,...,iK i m ,we choose anda b in a manner which

leads to the optimum values of iK s for the convergent solution of the desired problem.

There are various methods like Ritz Method, Galerkin’s Method, and Collocation Method

to find the optimal values of , 1,2,3,...iK i . Here, we apply the Method of Least Squares

as under:

2

1 2 1 2, , ,..., , , ,...,

b

m m

a

S K K K R K K K dx (2.19)

where R is the residual, ( )R w g w L N and

1 2

... 0m

S S S

K K K

(2.20)

where anda b are properly chosen numbers to locate the desired ( 1,2,..., )iK i m . With

these constants known, the approximate solution (of order m ) is well-determined.

Example 1: To see the effectiveness of the above basic idea of OHAM, we are presenting

an example of third order boundary value problem [52]:

3 22 5 3w w e (2.21)

with boundary conditions

0 0, 0 1, 1w w w e . (2.22)

Exact solution of the problem [52] is:

1w e (2.23)

According to OHAM formulation given in Eq. (2.4) for the given problem (2.21), we will

apply OHAM as discussed in Section (2.2); we get the following boundary value

problems generated by OHAM procedure:

Zeroth order boundary value problem is given by:

0 0 0 00; 0 0, 0 1, 1w w w w e (2.24)

The solution of the zeroth order deformation is:


9

2 20

12

2w e (2.25)

The first order deformation problem is given as:

1 1 0 1 0 1

2 31 1 1

1 1 1

1 3

5 2 ;

0 0 1 0

w K w C w e K

e K e K e K

w w w

(2.26)

First order solution is:

51 1

2 2 2 3 6 6

( / 240)(17280 17280 6960 4 10320

3487 1803 2640 240 )

w K e e

e e e e

(2.27)

Second order deformation problem is given as:

2 1 1 2 0 1 1 2

2 32 0 2 2 2

2 2 2

1 3

5 2 ;

0 0 1 0

w K w K w K w e K

K w e K e K e K

w w w

(2.28)

The second order solution can be observed as:

2 5 62 1

2 3 2 4

2 21

4 5 6 9 10

2 3 4

1( 5040 (17280 6960 3487 4

1209600

240 ( 72 43 11 ) ( 1803 ))

( 2206310400 985824000 484667297

3628800 564480 141414 40 7

1209600 (1824 1009 225 24 )

7

w K

e e

K

e

e

2 4 8

2 5 62

2 3 2 4

33730609 11538 )

5040 (17280 6960 3487 4 240

( 72 43 11 ) ( 1803 )))

K e

e

(2.29)

Now we can see the solution as 0 1 2( ) ( ) ( ) ( ) ...w w w w

2 5 61

2 3 2 4

2 21

4 5 6 9 10

2 3 4

1( 10080 (17280 6960 3487 4

1209600

240 ( 72 43 11 ) ( 1803 ))

( 2206310400 985824000 484667297

3628800 564480 141414 40 7

1209600 (1824 1009 225 24 )

7

w K

e e

K

e

e

2 4 8

2 5 62

2 3 2 4

(33730609 11538 )) 5040(120

2 (17280 6960 3487 4

240 ( 72 43 11 ) ( 1803 ))))

e K

e e

(2.30)


10

The residual of the example problem is given as:

2 3 2 31

3 4 2 3 4

2 5 6 21

2 3 4 5

3 4 8

(1/ 2)(( 2 ) (6 10 4 2 )) ( /120)

( ( 1923 ) 240 (3 67 45 12 )

(17280 6720 3367 4 )) ( /1209600)

(1209600 ( 3 1891 1054 237 25 )

7 (35115169 12258 )

R e e K

e e

K

e

e

2 4 5

6 9 10 3 42

2 3 4

2 5 6

( 2293401600

1019692800 501636977 3628800 544320

136374 40 7 )) ( / 240)( ( 1923 )

240 (3 67 45 12 ) (17280 6720

3367 4 ))

K e

e

(2.31)

where 1 20.9873623220546068; and 0.00009539560288304948;K K

We obtain the approximate solution of the second order as:

9

2 4

5 6

10

1636.0020206845597 0.000032238239253077553

116.88592807205187 2.9246530650391955

0.48785969966796544 0.032374372630945884

736.2609681839973 0.000020977400359340836

0.9748843550130653

w

2

6.460588475454943

5.454487464776046

47.62167957157959 10.05922734040209

e

(2.32)

Comparison between the absolute errors of the first order and second order are shown in

table 2.2.

Table 2.2

Comparison of absolute errors of Example 1 at different orders of approximations.

Exact

Solution

First Order Second Order

0.0 0.0 0.0 2.27374×10-13

0.1 9.94654×10-2

9.54924×10-5

7.36235×10-8

0.2 1.95424×10-1

4.11883×10-4

1.44256×10-7

0.3 2.8347×10-1

9.97088×10-4

1.79921×10-8

0.4 3.58038×10-1

1.89223×10-3

5.87428×10-7

0.5 4.1218×10-1

3.11232×10-2

1.60543×10-6

0.6 4.37309×10-1

4.62242×10-2

2.93892×10-6

0.7 4.22888×10-1

6.31257×10-2

4.30696×10-6

0.8 3.56087×10-1

7.97586×10-2

5.38493×10-6

0.9 2.21364×10-1

9.29507×10-2

5.96232×10-6

1.0 0.0 9.84465×10-2

6.09036×10-6


11

2.3. Application of OHAM [45] for PDEs

Let us apply OHAM to the following PDE:

, , , 0,w w L N (2.33)

, / 0,w w B (2.34)

where ,w is an unknown function, L is a linear and N is nonlinear operator, B is

boundary operator, and and represent independent variables, respectively, is the

domain with the boundary and , is a given function.

Constructing the homotopy , ; : H 0,1 which satisfies:

(1 ) ( , ; ) ,

( ) ( , ; ) , 0,h

L H

D H (2.35)

where and 0,1 is an embedding parameter, ( )h is an auxiliary function for

0 , (0) 0h . Clearly, we have:

0 ( , ;0 ) , 0, L H (2.36)

1 (1) , , , 0,h w w L N (2.37)

Obviously, when 0 and 1 it holds that 0, ;0 ,w H and

, ;1 ,w H respectively. Thus, as varies from 0 to 1 , the solution , ; H

varies from 0 ,w to ,w , where 0 ,w is obtained from Eq (2.35) for 0 :

0 , , 0,w L (2.38)

0 0, / 0w w B (2.39)

Next, we choose auxiliary function h in the form

2

1 2 ...h K K (2.40)

Here 1 2, ,...K K are constants to be determined later.

Expanding , ; , iK H in Taylor’s series about as:


12

0

1

, ; , , , ; , 1,2,...k

n n i

n

K w w K n

H (2.41)

Substituting Eq. (2.41) into Eq. (2.35) and comparing the coefficient of like powers of ,

we obtain Zeroth order problem, given by Eq. (2.38), the first and second order problems

are given by Eqs. (2.42)-(2.43) respectively and equations for ,nw are given by Eq.

(2.44):

1 1 0 0 1 1, , ) , , / 0w K w w w L N B (2.42)

2 1 2 0 0

1 1 1 0 1 2 2

, , ,

, ) , , , , , / 0

w w K w

K w w w w w

L L N

L N B (2.43)

1 0 0

1

1 0 1 1

1

, , ,

, , , , ,..., , ,

2,3,..., , / 0

n n n

n

i n n i n

i

n n

w w K w

K w w w w

n w w

L L N

L N

B

(2.44)

where 0 1 1, , , ,..., ,n i nw w w N is the coefficient of k i

in the expansion of

, ; N H with respect to the parameter .

0 0 0 1 2

1

, ; , , , , , ... , k

n n n

n

K w w w w w

N H N N (2.45)

The convergence of the series Eq. (2.44) depends upon the optimal constants 1 2, ,...K K . If

it is convergent at 1 , one has:

0

1

, ; , , ;n k n

k

w K w w K

(2.46)

Substituting Eq. (2.46) into Eq. (2.35), we obtain the residual:

, ; ( , ; ) , ( , ; )n n nR K w K w K L N (2.47)

If , ; 0, then , ;n nR K w K will be the true solution. Usually it doesn’t occur,

especially in case of nonlinear problems.

For the computation of , 1,2,...,nK n m , there are several methods [39-43] . We have

applied the Method of Least Squares as under:

2

0

, ;

t

i iS K R K d d

(2.48)


13

where R is the residual, , ; , ; ( , ) , ;n n nR K w K w K L N and

1 2

... 0m

S S S

K K K

(2.49)

The series solution can be determined by these known constants. The constants iK can

also be found by another method as under:

1 2 1,2; ; ... ; 0, ,..., .i i m iR c K R c K R c K i m (2.50)

at any time , where ic .

Example 2: Consider Cauchy Reaction diffusion equation [53]:

, , ,w w w (2.51)

with the initial conditions

,0w e (2.52)

the boundary conditions are

0, 1, 1, 1w w e (2.53)

Exact solution is:

,w e e (2.54)

Following the OHAM formulation presented in section 2.3, we have

2

2

, ;, ; , ;

HL H H (2.55a)

, ;

, ;

HN H (2.55b)

(2.56)

,0; e H (2.57)

,0; 1 H (2.58)

and

1, ;

1e

H (2.59)


14

Using Eq. (2.36), we obtain a linear sub problems by comparing coefficients of

0 1 2, , ,... called zeroth order, first order, and second order problems.

The zeroth order deformation problem is of the form:

20 0

02

, 0,0, 0, 1, 1

w ww e

(2.60)

having the solution:

0 1 1w e (2.61)

The first order problem with boundary conditions is as:

2 21 0 0

1 1 1 02 2

1

1

, , ,1 , ;

0,0, 0, 0.

w w wK K K w

ww

(2.62)

The solution of first order problem is given as:

3

21 1 1

1,

2 6w K K

(2.63)

Let us see the second order problem with their boundary conditions:

2 2 22 1 0

1 2 1 12 2 2

1 0

1 2 1 1 2 0

2

2

, , ,1 ,

, ,, ,

0,0, 0, 0.

w w wK K K w

w wK K K w K w

ww

(2.64)

The solution of the above problem is as:

2 2 3 22 1 2 1

120 3 60 20 5

120w K K K (2.65)

Now we can see the solution as 0 1 2( ) ( ) ( , ) ( , ) ...w w w w

21 2

2 2 3 21

1, 1 1 3 2

6

160 20 5

120

w e K K

K

(2.66)

The residual of the example is:


15

3 4 5

2 3 2 21 2 1

11 6 6 3 2 1

6 3 24 120R K K K

(2.67)

we will get 1 21.0353022641701395; and 0.002523391938372096;K K and the

final solution of the problem is:

2

2 2 3

1 1 0.3446801894003178 3

0.008932089818298478 60 20 5

w e

(2.68)

Comparison between the absolute errors of the first order and second order are shown in

table 2.3.

Table 2.3

Comparison of absolute errors of Example 2 at different orders of

approximations for 1 .

First Order Second order

0.0 0.0 0.0

0.1 1.66543×10-4

1.79279×10-5

0.2 5.94889×10-4

6.58546×10-5

0.3 1.11152×10-3

1.30847×10-4

0.4 1.46091×10-3

1.96232×10-4

0.5 1.31333×10-3

2.44514×10-4

0.6 2.7189×10-4

2.6103×10-4

0.7 2.12103×10-3

2.38277×10-4

0.8 6.37803×10-3

1.80854×10-4

0.9 1.30615×10-2

1.10942×10-4

1.0 2.27787×10-2

7.42958×10-5

Chapter 3 Application of Optimal Homotopy Asymptotic Method for Special and


16

Chapter 3

Application of Optimal Homotopy Asymptotic Method for

Special and Nonlinear Boundary Value Problems

3.1. Introduction

Accurate solutions to BVPs play a significant role in many scientific and engineering

applications. Fourth-order BVPs have wide application in many areas of pure and applied

sciences. “They arise in the mathematical modeling of visco-elastic and inelastic flows,

deformations of elastic beams in an equilibrium state with simply supported ends,

transverse vibrations of hinged beams, the problem of bending of a plate on an elastic

foundation and plate deflection theory [54-56].” Currently, much attention has been given

to solve the fourth-order BVPs. Numerous analytical and numerical methods are proposed

for solving these BVPs problems. The sixth-order BVPs arise in astrophysics; the narrow

convicting layers bounded by stable layers [57-59]. The past work on the analytical and

numerical solutions to eighth order BVPs is rare. “When an infinite horizontal layer of

fluid is heated from below and is subjected to the action of rotation, instability starts.

When this instability is as ordinary convection, the ordinary differential equation is sixth

order, when the instability sets in as over stability, it is modeled by an eighth order

ordinary differential equation [60].”

3.2. Application of OHAM to Nonlinear Fourth Order BVPs

In this section, we have applied OHAM for the solutions of nonlinear fourth order BVPs.

3.2.1. Model 1:

Solving the following fourth order nonlinear BVPs [61]

2( ) 0, 0 1ivy x g x y x x (3.1)

0 0, 1 sin 1 , 0 1, 1 cos 1y y y y (3.2)

and 2sin sing x x x .



17

Applying (2.4) and comparing the coefficients of 0 1 2, , ,... on both sides, we get the

zeroth order, first order, and second order problem as under:

Zeroth Order Problem:

( )

0 0ivy x g x (3.3)

0 0 0 00 0, 1 sin 1 , 0 1, 1 cos 1y y y y (3.4)

First Order Problem:

2( ) ( )

1 1 1 0 1 1 0, 1 1iv ivy x C C y x C g x C y x (3.5)

1 1 1 10 1 0 1 0y y y y (3.6)

Second Order Problem:

( ) ( )

2 1 2 1 1 1 1 0 1 1

2( )

2 0 2 2 0

, , 1 , 2 ,iv iv

iv

y x C C C y x C C y x y x C

C y x C g x C y x

(3.7)

2 2 2 20 1 0 1 0y y y y (3.8)

Solving problems (3.3)-(3.8) in succession, we obtain

2 3 4 2 2 3 2

0 2 3

3 8 4 9 c 1 6 c 1 3c1

48 6 c 1 s 1 6 c 1 s 1 48 s

x x x x os x os os xy x

x os in x os in in x

(3.8a)



18

2

1 1 1 1 1

3 4 5 6 7

1 1 1 1 1

8 2 3

1 1 1

1 1

, 1/ 92897280 {97823565 -595116480 898780017

-403546230 2146200 -118944 21504 6912

3456 1919877120 cos(1) -1455390720 cos(1)

-278964000cos(2) 33203520 cos(2) 7

y x C C xC x C

x C x C x C x C x C

x C x C x C

C x C

2

1

3 4 5

1 1 1

27 3 2

1 1 1

23 2 3

1 1 1

3

80865056 cos(2)

-530608320 cos(2) - 211680 cos(2) 120960 cos(2)

-20736 cos(2) 522547200 cos(1) (2) 2449440 cos(2)

-2177280 cos(2) - 430080 cos(3) 430080 cos(3)

-15498 cos(4

x C

x C x C x C

x C x Cos C x C

x C x C x C

x 4 5

1 1 1

1 1 1

2

1 1 1

1

) 37800 cos(4) -18144 cos(4)

-92897280cos( ) -371589120 cos( ) 278691840 cos(2) cos( )

-4929120 cos(2 ) 181440 cos(2 ) 362880 cos(2 )

272160 cos(2)cos(2 ) -544320 cos

C x C x C

x C x x C x C

x C x x C x x C

x C x 1 1

2 3 2

1 1 1

2 3

1 1

1 1

(2)cos(2 ) 2835 cos(4 )

-2345656320 sin(1) 1447649280 sin(1) - 6993 cos(4)

-313528320 cos(2) sin(1) 278691840 cos(2)sin(1) -

278873280sin(2) 21591360 sin(2)

x C x C

x C x C x C

x C x C

C x C 2

1

3 4 5

1 1 1

6 7 2

1 1 1

3 3

1

806167872 sin(2)

-547010496 sin(2) - 141120 sin(2) 108864 sin(2)

-8064 sin(2) - 20736 sin(2) - 789626880 cos(1)sin(2)

510935040 cos(1)sin(2) - 725760 cos(2)sin(2

x C

x C x C x C

x C x C x C

x C x

1

1 1

2

1 1

23 2

1 1

23

1 1

)

278691840cos( )sin(2) 181440 cos(2 )sin(2)

-544320 cos(2 )sin(2) - 348364800 sin(1)sin(2)

301916160 sin(1)sin(2) - 2540160 sin(2)

69672960 sin(2)sin( ) 1451520 sin(2)

430080

C

x C x C

x x C x C

x C x C

x x C x C

2 3 2

1 1 1

3 4 5

1 1 1

6 2

1 1 1

2

1 1

sin(3) - 286720 sin(3) 6804 sin(4)

-43092 sin(4) 90720 sin(4) -90720 sin(4)

36288 sin(4) - 23224320 sin( ) - 46448640 sin( )

-801239040 cos(1)cos(2) -34836480cos(2)sin( )

696729

x C x C x C

x C x C x C

x C x x C x x C

x C x C

1 1

1 1 1

1 1

1

60 cos(2)sin( ) - 23224320sin(2)sin( )

-362880sin(2 ) -1451520 sin(2 ) 967680000sin( )

1088640 cos(2) sin(2 ) 1088640sin(2)sin(2 )

-143360sin(3 ) }

x x C x C

x C x x C x C

x C x C

x C

(3.8b)



19

(3.8c)

0 1 1 2 1 2, , ,y y x y x C y x C C (3.9)

1 20.0000418960093 and 0.99071397322C C are calculated by the method of least

squares. Result of Model 1 can be seen in Table 3.1 and Fig. 3.1.



20

Table 3.1

Comparison between OHAM solutions with exact solution for Model 1

x Exact OHAM Absolute Error

0.0 0.0 2.0961×10-13

2.0961×10-13

0.1 9.98334×10-2

9.98335×10-2

3.44091×10-8

0.2 1.98669×10-1

1.98669×10-1

1.13845×10-7

0.3 2.9552×10-1

2.9552×10-1

2.06689×10-7

0.4 3.89418×10-1

8.9419×10-1

2.82662×10-7

0.5 4.79426×10-1

4.79426×10-1

3.15173×10-7

0.6 5.64642×10-1

5.64643×10-1

2.90029×10-7

0.7 6.44218×10-1

6.44218×10-1

2.14158×10-7

0.8 7.17356×10-1

7.17356×10-1

1.16136×10-7

0.9 7.83327×10-1

7.83327×10-1

3.40555×10-8

1.0 8.41471×10-1

8.41471×10-1

1.44107×10-13

3.2.2. Model 2:

Consider the BVP [61]:

( ) 2 10 9 8 7 6 44 4 4 8 4 120 48ivy x y x x x x x x x x (3.10)

0 0, 0 0, 1 1, 1 1y y y y (3.11)

The exact solution of this problem is [61]:

5 4 22 2 .y x x x x (3.12)

Let

10 9 8 7 6 44 4 4 8 4 120 48g x x x x x x x x (3.13)

2( ) 0viy x y x g x (3.14)



21

Figure 3.1: Comparison of OHAM solution and exact solution for Model 1

Applying (2.4) and comparing the coefficients of 0 1 2, , ,... on both sides, we obtain:

Zeroth Order Problem:

( )

0 0ivy x g x (3.15)

0 0 0 00 0, 1 sin 1 , 0 1, 1 cos 1y y y y (3.16)


2( ) ( )

1 1 1 0 1 1 0, 1 1iv ivy x C C y x C g x C y x (3.17)

1 1 1 10 1 0 1 0y y y y (3.18)

Second Order Problem

( ) ( )

2 1 2 1 1 1 1 0 1 1

2( )

2 0 2 2 0

, , 1 , 2 ,iv iv

iv

y x C C C y x C C y x y x C

C y x C g x C y x

(3.19)

2 2 2 20 1 0 1 0y y y y (3.20)

Solving problems (3.15)-(3.20) in succession, we obtain:

2 3 4 5

0 8 10 11 12 13 14

2155683 8038 2162160 1081080 (1/1081080)

2574 1716 546 364 252 45

x x x xy x

x x x x x x

(3.20a)



22

1 1

2 3

1 1

8 9

1 1

10

1

,C (1/ 2360410309588661890560000)

14113828503813453911359 C 17512915766704666962322 C

5586404113887189501420 C 23144774000952226800 C

3735433519034813810160 C 1179691619002973294

y x

x x

x x

x

11

1

12 13

1 1

14 15

1 1

16 17

1 1

18

1

400 C

797705495192034374400 C 550212193377310464000 C

97319245468388853600 C 2551023644304960 C

856729299998872080 C 279036341094724200 C

247466505045614400 C 15

x

x x

x x

x x

x

19

1

20 21

1 1

22 23

1 1

24 25 26

1 1 1

27

5275084044250800 C

3650519253533040 C 26398020790188000 C

8405056667479680 C 897927547083840 C

38159719228800 C 21095475553152 C 4049797421376 C

6052906241984 C

x

x x

x x

x x x

x

28 29

1 1 1

30 31 32

1 1 1

1221181635008 C 475889853600 C

295593372480 C 60656299200 C 4738773375 C

x x

x x x

(3.20b)



23

2

1

3

1

8

1

1

2 2

( 32669682535166038177800562717403451774787200

40537647046565498276872580694218893070457600

12931009390154359168924266696920562467136000

5357387038924076963068897

,C ,C

7156787421440000

x C

x C

y

x C

x

9

1

10

1

11

1

12

1

13

1

8646514810996349122418420307721339216128000

2730665933188143522672522412342587755520000

1846471726465979561236202970676720619520000

1273591901712376237386288026597835571200000

x C

x C

x C

x C

x C

14

1

15

1

16

1

17

1

225267641104971069200162200032882362880000

5904927396321076366168727485651968000

1983095815708383589080913455027281664000

645893399999572047403581539117439360000

57281779750567840

x C

x C

x C

x C

18

1

19

1

20

1

21

1

2

3219571120181611520000

359419678365531276056662133636240640000

8449961331839350421935750500018432000

61104253784799540864465470704550400000

19455425077784486158148965087647744000

x C

x C

x C

x C

x 2

1

23

1

24

1

25

1

26

1

2078458576628113491233720925699072000

88329393580142122870344465623040000

48830300656099081258580171252121600

9374181927486220561295860381900800

1401083520940562000888294907678

C

x C

x C

x C

x C

27

1

28

1

29

1

30

1

31

1

7200

2826704059972590662056252066406400

1101555855990308389571183162880000

684218434127313503766165801984000

140402870708442139347389199360000

10968974274097042136514781200000

x C

x C

x C

x C

x C

32

1

42 2

1

43 2

1

44 2

1

2 2

1

3

1

18683464135506877900207872

3279443942818588507522560

3820823675617325912753400

32605172872522266759323921222230868521529727

40451961996974182925822791011410996525862778

x C

x C

x C

x C

x C

x C 2



24

8 2

1

9 2

1

10 2

1

11 2

1

12853457445564057131150272998334822341815742

106873817117301918509679838860690377883320

8620466886846280020965176767572231570851512

2701942454241603867437283593123940978268560

1

x C

x C

x C

x C

12 2

1

13 2

1

14 2

1

15 2

1

853296246170788567680120913554535247983040

1273591901712376237386288026597835571200000

223127551098503747806099303888254244271880

24188753111272816253620159558815649776

3954294

x C

x C

x C

x C

16 2

1

17 2

1

18 2

1

19 2

1

999119289026707499780027151742404

1280476731768992936554600505857043216034

1146299848262837493979094855639474837488

716583570253989121237950089315154553020

18153410829102066767

x C

x C

x C

x C

20 2

1

21 2

1

22 2

1

23 2

1

24

079156365109506436

122325062475997255309755518897728337880

38583478912638559610674295035606805760

4064708510722834491402346035781177920

304944482768990447155003185921595200

x C

x C

x C

x C

x C 2

1

25 2

1

26 2

1

27 2

1

28 2

1

172131824316102518718368542082822400

31225989769228784108742227282106504

48696064947262435404352469277878496

9955954110432674815496294613892416

37680517224202509563809356490

x C

x C

x C

x C

29 2

1

30 2

1

31 2

1

32 2

1

33 2

1

60800

2341220030385245677826756247386880

474526294591146486759733985232768

40917013297075779490615187391192

3867571077895433794192598216016

521615945593005771502089327960

x C

x C

x C

x C

x C

34 2

1

35 2

1

36 2

1

37 2

1

38 2

1

39 2

1

578651890632287750071052046372

265019543551924982873921756160

7980251045802826679782141248

25268923385823287170976713452

8700984701390689923286015560

x C

x C

x C

x C

x C

x C



25

40 2 41 2

1 1

45 2 46 2

1 1

47 2 48 2

1 1

1340309891030782820607242784 135045166608803637462708432

805194420243411190696704 143582548822550108696880

115436320975091250290520 27795688158510545773500

3304269

x C x C

x C x C

x C x C

49 2 50 2

1 1

2

2

3

2

8

2

870772182097500 165213493538609104875

32669682535166038177800562717403451774787200

40537647046565498276872580694218893070457600

12931009390154359168924266696920562467136000

535

x C x C

x C

x C

x C

9

2

10

2

11

2

12

2

73870389240769630688977156787421440000

8646514810996349122418420307721339216128000

2730665933188143522672522412342587755520000

1846471726465979561236202970676720619520000

1273591901

x C

x C

x C

x C

13

2

14

2

15

2

16

2

712376237386288026597835571200000

225267641104971069200162200032882362880000

5904927396321076366168727485651968000

1983095815708383589080913455027281664000

645893399999572047403581

x C

x C

x C

x C

17

2

18

2

19

2

20

2

21

2

539117439360000

572817797505678403219571120181611520000

359419678365531276056662133636240640000

8449961331839350421935750500018432000

61104253784799540864465470704550400000

19

x C

x C

x C

x C

x C

22

2

23

2

24

2

25

2

455425077784486158148965087647744000

2078458576628113491233720925699072000

88329393580142122870344465623040000

48830300656099081258580171252121600

937418192748622056129586038190080

x C

x C

x C

x C

26

2

27

2

28

2

29

2

30

2

0

14010835209405620008882949076787200

2826704059972590662056252066406400

1101555855990308389571183162880000

684218434127313503766165801984000

140402870708442139347389199360000

x C

x C

x C

x C

x C

31

2

32

210968974274097042136514781200000 ) /

5463709258346094058387175634104714600448000000

x C

x C

(3.20c)



26

Adding equations (3.20a, 3.20b & 3.20c) we obtain:

0 1 1 2 1 2, , ,y y x y x C y x C C (3.21)

6

1 21.00113 and 1.07947 10C C are calculated by the method of least squares as

mentioned in Chapter 2.

Table 3.2

Comparison between OHAM solutions with exact solution for Model 2


0.0 0.0 0.0 0.0

0.1 1.981×10-2

1.9809×10-2

7.27908×10-8

0.2 7.712×10-2

7.71198×10-2

2.45216×10-7

0.3 1.6623×10-1

1.6623×10-1

4.48868×10-7

0.4 2.7904×10-1

2.79039×10-1

6.18615×10-7

0.5 4.0625×10-1

4.06249×10-1

6.99951×10-7

0.6 5.3856×10-1

5.38559×10-1

6.6128×10-7

0.7 6.6787×10-1

6.67869×10-1

5.07446×10-7

0.8 7.8848×10-1

7.8848×10-1

2.87104×10-7

0.9 8.9829×10-1

8.9829×10-1

8.56656×10-8

1.0 1.0 1.0 0.0

3.2.3. Model 3:

Consider the non-linear problem [62]:

( ) 2 1, 0 2ivy x y x x (3.22)

0 0 0 00 0, 0 0, 2 0, 2 0y y y y (3.23)

Using (2.4) and equating the coefficients of 0 1 2, , ,... on both sides, we get the

following sub problems:

Zeroth Order Problem

0 1 0iv

y x , (3.24)

0 0 0 00 0 2 2 0y y y y (3.25)



27

First Order Problem

2

1 0 1 0 1 0 11iv iv iv

y x y x C y x C y x C (3.26)

1 1 1 10 0 2 2 0y y y y (3.27)


2 1 2 1 1 1 1 1

2

1 0 1 2 0 2 2 0

, , , ,

2

iv iv iv

iv

y x C C y x C C y x C

C y x y x C y x C C y x

(3.28)

2 2 2 20 0 2 2 0y y y y (3.29)

and so on.

2 3 4

0

1 4 4

24y x x x x (3.30)

2 3 8

1 1 1

1 1 9 10 11 12

1 1 1 1

-7680 5632 - 792 1,

47900160 880 -396 84 -7

x C x C x Cy x C

x C x C x C x C

(3.31)

2

2 1 2 1

3 8 9

1 1 1

10 11 12

1 1 1

2 2

1

, , (1/18246765061324800)(-2925567590400

2145416232960 -301699157760 335221286400

-150849578880 31998395520 - 2666532960

- 2918991347712 214060670

y x C C x C

x C x C x C

x C x C x C

x C

3 2 16 2

1 1

8 2 9 2 10 2

1 1 1

11 2 12 2 14 2

1 1 1

15 2 18 2 19 2

1 1 1

9760 4284918

- 301118688000 334662315520 -150659313792

31975821696 - 2666532960 4186080

- 6480672 328510 - 38220 19

x C x C

x C x C x C

x C x C x C

x C x C x C

20 2

1

2 3 8

2 2 2

9 9 11

2 2 2

12

2

11

- 2925567590400 2145416232960 - 301699157760

335221286400 - 150849578880 31998395520

- 2666532960 )

x C

x C x C x C

x C x C x C

x C

(3.32)

Similarly we can compute 2 1 2, ,y x C C .

Thus the second order solution becomes:

0 1 1 2 1 2, , ,y y x y x C y x C C (3.33)

1 20 1.0022651C and C are calculated be the method of least squares.



28

This problem has no exact solution but from the plot it is clear that the solution given by

OHAM is completely agreed with the analytic solution by L. Xu [62].

Figure 3.2: Comparison of OHAM solution and VIM solution [62] for Model 3

3.3. Application of OHAM to Special Fourth Order Boundary Value Problems

In this section, the technique is applied to special fourth order BVPs (3.34-3.35).

Recently, special fourth order linear BVPs was solved by Liang and Jeffrey [63] by

HAM. Momani and Noor [64] solved the special fourth order linear BVPs by Adomian

Decomposition Method (ADM) , Differential Transform Method (DTM) [64].

3.3.1. Model 4

(4) 211 1, 0 1,

2y x c y x cy x cx x (3.34)

with boundary conditions :

3

0 1, 0 1, 1 sinh 1 , 1 1 cosh 12

y y y y (3.35)

The exact solution of this problem is

211 sinh

2y x x x (3.36)



29

Applying OHAM to (3.34-3.35), we get the sub problems as:

Zeroth order Problem

2

(4)

0 0 0 0 1 0, 0 1,2

c xy x c y x y x c y x x (3.37)

0 0 0 0

30 1, 0 1, 1 sinh 1 , 1 1 cosh 1

2y y y y (3.38)

First order Problem

(4) (4)

1 1 1 0 1 1 1 1 1 0

1 0 0 0 1 1 1 0

2 2

0 1 1

, 1 , ,

,

1 11

2 2

y x C C y x cy x C y x C cC y x

C y x c y x y x c y x C cC y x

c y x c x C C c x

(3.39)

1 1 1 10 0, 0 0, 1 0, 1 0y y y y (3.40)

Solution of Zeroth order Problem is:

(1 ) (2 ) 2 2 (2 )

0

1 2 2

1 2 2 2 1 (2 )

2 2 (2 ) 1

( ((1 6 5 ) ( 1 ) (1 )

10( 1 ) 8 2(1 ) 4( 1 )

4( 2 3) 4( 1 ) 2(2 5 3 )

(1 ) 2( 2 3 ) (1 2 _

x c x c x cx c x

c x c x c x c x

c x c x c x

c x c x

y x e c c e c e c e

c e c e c e c e

e c e c c e

c e c c e c

2

2 2 2 1 2

(1 2 ) 2 2 (1 ) 2 (1 )

1 2 (1 ) 1 (2 ) 2 2 (2 )

2 (2 ) (1 )

3 )

4(1 ) 4( 1 ) 4(2 3 )

2 (5 (5 ( 4 ) )) 4 20

16 2( 2 3 ) ( 1 2 3 )

(1 6 5 ) 8

c x

x cx x c x x cx

x c x c x x c x

x c x c x x c x

x c x c

c e

c e c e c e

e e c e e ce ce

ce c c e c c e

c c e ce

(1 ) 2 1 2 (2 )

2 2 2 (2 ) 2 2 2 (1 )(2 )

2 2 2 2 2 (2 ) 2

2 2

(2 ) 2(2 5 3 )

( 1 ) ( 1 ) (2 ) ( 1 )

(2 ) (1 ) (2 ) (1 ) (2 )))

/(4(4 2 (1 )cosh( ) (1 )( 1 )sinh( )))

x x c x

x c x x cx c x

x cx x c x

x c c e

c e c e x c e

x c e x c e x

ce c e c c e c

(3.41)

Solution of First order Problem is given as under:



30

(1 ) 2 2 1 2 2 2

1 1

1 2 1 2 2 2 1 (2 ) 1 2 (2 )

(2 ) (1 )(2 ) 2 2 2 2

(2 ) 2 (2 )

( , ) ( ( 5 2 4 2

2 2 4 2 2 2

5 (2 ) (2 ) (2 )

(2 ) (2

x c x c x c x c x c x c x

cx x cx x cx x cx c x x c x

x c x c x x cx x cx

x c x c

y x c e e e e e e

e e e e e e

e e x e x e x

e x c e

1 2 1 2

2 2 1 (2 ) 1 2 2 1 2 2 2

1 (1 ) 2 (1 ) 2 (1 ) 1 (1 ) 1 (2 )

2 (2 ) 2 (2 ) 1 2 (2 ) (1 2 )

4 6

2 5 2 2 6 2

4 2 10 4

2 2 5 5 4

x c x c x c x c x

c x c x cx cx x cx x cx x cx

c x c x x c x x c x c x

c x x c x x c x x c x

e e e e

e e e e e e e

e e e e e

e e e e

1 (1 2 )

2 (1 2 ) (1 )(2 ) 2 2 2 2

2 (1 2 ) (1 )(1 ) 2 (2 ) 2 2

2 2 2 2

1

2 (2 ) 2 (2 ) 2 (2 )

8 (2 ) 2 (2 )) ( 1 )

( ( 3 2 ) ( 2 3 ) ( 1 )(2 ))) ) /(2(4 2 (1 )

cosh( ) (

x c x

x c x c x x cx x cx

x c x c x x c x cx c

x x

e

e e x e x e x

e e x e x ae e

e e e e e e x c ce c e

c

21 )( 1 )sinh( )))c e c

(3.42)

Adding Eq. (3.41 & 3.42) in succession we obtain the following, first order solution

0 1 1,y y x y x C (3.43)

1C can be easily calculated by the method of least squares as mentioned in Chapter 2 and

its value is 1. Results of Model 4 can be seen in Table 3.5 for 5c , Table 3.6 for

100c , and Table 3.7 for 810c . In Table 3.8, OHAM solution is analysed at some

further values of c , which are not considered in [63]. We observed that, rapid increase in

the value of c slightly effect the accuracy, so it is concluded that OHAM is independent

of the parameter c .

Table 3.5: Comparison of Absolute error of obtained solution by OHAM at 5c

with HPM and HAM x HPM [63]15thOrder HAM [63]

10th Order

OHAM

1st Order

0.1 7.1×10-15 2.6×10-14 2.44249×10-15

0.2 2.3×10-14 2.0×10-14 3.33067×10-15

0.3 4.0×10-14 1.3×10-14 2.44249×10-15

0.4 4.9×10-14 1.3×10-14 1.77636×10-15

0.5 4.9×10-14 1.3×10-14 2.44249×10-15

0.6 4.0×10-14 1.1×10-14 8.88178×10-16

0.7 2.7×10-14 8.7×10-14 1.77636×10-15

0.8 1.3×10-14 1.1×10-14 2.66454×10-15

0.9 3.2×10-15 1.2×10-15 3.10862×10-15



31

Table 3.6

Comparison of Absolute error of obtained solution by OHAM at 100c with

HPM, DTM, ADM and HAM

x HPM[63]

20th

Order

HAM [63]

20th

Order

DTM [63] ADM & HPM

[63]1st

Order

OHAM

1st

Order

0.1 3.3×106 8.6×10

-9 1.2×10

-11 -6.5×10-4 1.11022×10

-15

0.2 1.1×107 4.6×10

-9 4.0×10

-11 -2.6×10-3 1.77636 ×10

-15

0.3 1.8×107 4.4×10

-9 8.5×10-11 -6.3 ×10

-3 1.11022 ×10-15

0.4 2.3×107 4.3×10

-9 1.6×10-10 -1.2×10

-2 2.22045 ×10-15

0.5 2.3×107 3.9×10

-9 2.9×10-10 -1.8×10

-2 4.44089 ×10-15

0.6 1.9×107 3.5×10

-9 5.1×10

-10 -2.4×10-2 1.11022 ×10

-15

0.7 1.2×107 2.9×10

-9 7.4×10

-10 -2.6×10

-2 0

0.8 5.9×106 2.5×10

-9 8.1×10

-10 -2.0×10

-2 1.33227 ×10

-15

0.9 1.6×106 3.8×10

-9 4.7×10

-10 -8.6×10

-3 1.77636 ×10

-15

Table 3.7


HAM

x HAM[64]

20th Order

OHAM

1st Order

0.1 8.9×10-5

8.88178×10-16

0.2 4.7 ×10-5

2.22045 ×10-15

0.3 3.5 ×10-5

4.44089 ×10-16

0.4 3.4 ×10-5

1.11022 ×10-15

0.5 3.0×10-5

4.44089 ×10-16

0.6 2.6×10-5

1.77636 ×10-15

0.7 2.0 ×10-5

4.44089 ×10-16

0.8 2.6×10-5

1.33227 ×10-15

0.9 4.0 ×10-5

2.22045 ×10-15



32

Table 3.8

Absolute error of obtained solution by OHAM at large values of c .

x 1010c 1210c

1510c 1710c

0.1 1.18807×10-11

1.90092 ×10-10

7.75419 ×10-9

2.40149 ×10-8

0.2 7.43094×10-12

1.18896 ×10-10

1.16605 ×10-8

8.25182 ×10-8

0.3 1.21645×10-11

1.94631 ×10-10

4.90168 ×10-9

7.7014 ×10-9

0.4 1.71392 ×10-11

2.74229 ×10-10

1.00159 ×10-8

6.63313 ×10-9

0.5 2.22045 ×10-16

2.22045 ×10-10

7.01937 ×10-9

1.68309×10-7

0.6 6.90026 ×10-12

1.10405 ×10-10

1.15373 ×10-8

1.0573 ×10-7

0.7 4.37677 ×10-11

4.67041 ×10-10

1.22845 ×10-8

1.45289 ×10-7

0.8 4.77018 ×10-11

2.45057 ×10-10

2.79577×10-9

3.14247 ×10-8

0.9 3.25491×10-11

4.16533 ×10-10

1.87012 ×10-8

4.69593 ×10-8

3.3.2. Model 5

In this example we apply the method to the nonlinear BVPs [64]

2(4) 1, 0 2,y x c y x x (3.44)

with the following boundary conditions :

0 0 2 2 0y y y y (3.45)

Zeroth order problem

(4)

0y 1x (3.46)

0 0 0 00 0 2 2 0y y y y (3.47)

which have the solution:

2 3 4

0

14 4

24y x x x x (3.48)

First Order Problem

(4) (4)

1 1 1 0 1 1y , 1 y 1x C C x cC C (3.49)



33

1 1 1 10 0 2 2 0y y y y (3.50)

and the solution is

2 3 8

1 1 1

1 1 9 10 11 12

1 1 1 1

7680 5632 7921,

47900160 880 396 84 7

c x C c x C c x Cy x C

cx C cx C cx C cx C

(3.51)


(4) (4) (4)

2 1 1 1 2 0 1 0 1 1

2

2 0 2

y , 1 y y 2 y y ,

y

x C C x C x cC x x C

cC x C

(3.52)

2 2 2 20 0 2 2 0y y y y , (3.53)

which have the solution:

2 1 2

2 3

1 1

8 9 10

1 1 1

11 12 2 2

1 1 1

2 2 2

1

, ,

12925567590400 214516232960

18246765061324800

301699157760 335221286400 150849578880

31998395520 2666532960 2925567590400

6576242688 2145416

y x C C

cx C cx C

cx C cx C cx C

cx C cx C cx C

c x C

3 2 2 3 2

1 1

8 2 2 8 2 9 2

1 1 1

2 9 2 10 2 2 10 2

1 1 1

11 2 2 11 2 12 2

1 1 1

232960 4809523200

301699157760 580469760 335221286400

558970880 150849578880 190265088

31998395520 22573824 2666532960

41860

c x C c x C

c x C c x C c x C

c x C cx C c x C

c x C c x C cx C

2 14 2 2 15 2 2 16 2 2 17 2

1 1 1 1

2 18 2 2 19 2 2 20 2 2

1 1 1 2

3 8 9

2 2 2

10

2

80 6480672 4284918 1556100

328510 38220 1911 2925567590400

2145416232960 301699157760 335221286400

150849578880 319983

c x C c x C c x C c x C

c x C c x C c x C cx C

cx C cx C cx C

cx C

11 12

2 295520 2666532960cx C cx C

(3.54)

Adding Eq. (3.48, 3.51 & 3.54) in succession we obtain the following, second order

solution:

0 1 1 2 1 2, , ,y y x y x C y x C C (3.55)

For the determination of the constants 1 2C and C , using the method mentioned in Chapter

2, we have, 1 1.0086569369826812C and 1 0.000040238909386557936C .

The results at 5c are mentioned in Table 3.9 and plotted in Fig.3.5.



34

Table 3.9

Convergence of Absolute error of obtained solution by OHAM at 5c with

respect to increasing order.

x OHAM

(Zeroth order)

OHAM

(First order)

OHAM

(Second order)

0.0 0.0 0.0 0.0

0.2 5.4×10-3

5.4276×10-3

5.42768×10-3

0.4 1.70667×10-2

1.71581×10-2

1.71584×10-2

0.6 2.94×10-2

2.95637×10-2

2.95642×10-2

0.8 3.84×10-2

3.86192×10-2

3.86198×10-2

1.0 4.16667×10-2

4.19066×10-2

4.19073×10-2

1.2 3.84×10-2

3.86192×10-2

3.86198×10-2

1.4 2.94×10-2

2.95637×10-2

2.95642×10-2

1.6 1.70667×10-2

1.71581×10-2

1.71584×10-2

1.8 5.4×10-3

5.4276×10-3

5.42768×10-2

2.0 0.0 0.0 0.0

Fig. 3.5. Plot showing zeroth order, first order and second order

Solutions by OHAM.

3.4. Application of OHAM to Special Sixth Order BVP

3.4.1. Model 6



35

This special sixth order BVP [65-66]:

(6) (4) (2)1 , 0 1 ,y x c y x cy x cx x (3.56)

with boundary conditions :

(1) (1)

(2) (2)

70 1, 1 sinh 1 ,

6

0 1, 1 1 cosh 1 ,

0 0, 1 1 sinh 1 .

y y

y y

y y

(3.57)

The exact solution of this problem is

311 sinh

6y x x x (3.58)

Splitting problem (3.56) as:

6

6

,,

d x pL x p

dx

’ (3.59)

and choosing the nonlinear operator as:

4 2

4 2

, ,, 1

d x p d x pN x p c c

dx dx

, (3.60)

and

g x c x . (3.61)

The boundary conditions are:

(1) (1)

(2) (2)

70, 1, 1, sinh 1 ,

6

0, 1, 1, 1 cosh 1 ,

0, 0, 1, 1 sinh 1 .

(3.62)

Equating coefficient of 0 1 2, , ,... , we get zeroth order, First order and second order

problems as follows:


(6)

0 0,y x c x (3.63)



36

0 0

(1) (1)

0 0

(2) (2)

0 0

70 1, 1 sinh 1 ,

6

0 1, 1 1 cosh 1 ,

0 0, 1 1 sinh 1 .

y y

y y

y y

(3.64)

Its solution is:

3

0

4 7

5

71 ( 4 (1) ( 5 9 (1)))

1680 6

(8 / 630 7 c (1) 16 s (1)) ( / 5040)

( 3 /840 3 c (1) 13 s (1) / 2)

cy x x x cosh sinh

x c osh inh c x

x c osh inh

(3.65)

First order Problem

(6) (6) (4) 2 (2)

1 1 1 0 1 0 1 0 1, 1 1 0y x C C y x c C y x C c y x cC x cx (3.66)

(1) (2) (1) (2)

1 1 1 1 1 10 0 0 1 1 1 0.y y y y y y (3.67)

Its solution is:

3 3 2

1 1

2

2

, (( 1 ) / 279417600 )(( 332640(37 44 21

5(1 )( 19 7 ) ) 22 ( 15120( 37 95 ) (665222

105840 ( 3 5 ) 5 ( 181470 7 (3 )))) 22 ( ( 27662

5 (3558 7 (645 361 ) )) 210 (80 ( 42 ( 57 35 )))

y x C x x e e e

e e x c x e

e x x x x c e

x x x e x x x

3

1

210 (32 ( 150 ( 129 95 )))) (37 (39 (4

( 68 7 (3 )))))))

x x x c e x x

x x x C

(3.68)

Second order Problem

2 (2) 2 (2) (4) (4)

2 1 2 2 2 0 1 1 1 2 0 2 0

(4) (4) (6) (6) (6)

1 1 1 1 1 1 2 0 1 1 1 1 1

, , C ,

, C , , , 0,

y x C C c x c C y x c C y x C C y x cC y x

C y x C c y x C C y x y x C C y x C

(3.69)

(1) (2) (1) (2)

2 2 2 2 2 20 0 0 1 1 1 0.y y y y y y (3.70)

Its solution is:



37

3 3

2 1 2

3 2 3 4 5

2 2

2 3 2

( 1 ) , ,

27461161728000

98280( (37 39 4 68 21 7 ) 332640 (37 (44 60 )

95 7 ( 3 5 )) 22 (105840 ( 3 5 ) 15120 ( 37 95 )

(665222 907350 105 35 ) ) 22 ( (

x xy x C C

e

c e x x x x x e x

x e x c e x x

e x x x c e

2 3 2 2 3

2 3 5 2 3 4 5

1

6 7 8 9 2

3

27662 17790

22575 12635 ) 210 (80 42 57 35 ) 210 (32 150

129 95 ))) ( (965 1011 123 1699 126 630

504 504 63 21 ) 12972960 ( 93526 239160 4515

3325 12

x

x x e x x x x

x x C c e x x x x x

x x x x x x

x e

2 3 2 2

3 2 2 3

2 3 2

3 4 5 2

(9286 12510 315 175 ) 7 (7594 12540 285

175 ) ) 2808 (64680 (3814 6240 285 175 ) 9240(46906

119460 4515 3325 ) ( 517320906 688681610 34846490

19433330 735 245 )) 2808 (105

x x x e x x

x c e x x x

x x x e x x

x x x c e

2

2 3 4 5

2 3 4 5

2 3 4 5 3 2

2 3

( 134608 44488

179123 107387 2604 980 ) 105 ( 61840 221576

402371 293459 6188 2660 ) 2 ( 12013446 4207190

17752070 9668750 252105 88445 )) 3 (98280

( 936 1712 3563 413

x

x x x x x

x x x x e x

x x x x c e

x x x

4 5

2 3 4 5

2 3 4 5 6

7 4 2 2 3 4

5

2604 980 ) 98280 (24 1136

5051 859 6188 2660 ) (164833286 275388546

-615684729 87422539 471965592 165616920 8820

2940 )) 3 (210 ( 8408 6606 3567 22111 19614

630 54

x x x

x x x x e x

x x x x x

x c e x x x x

x

6 7 2 3

4 5 6 7 2

3 4 5 6 7

3 2 3 4 5

18 1470 ) 210( 5080 10350 10551 5683

31710 3906 13146 3990 ) (2678630 1899426 1492689

7576339 7380912 146160 2072700 532140 )))

98280( (37 39 4 68 21 7 ) 332640(3

x x x x x

x x x x e x x

x x x x x

c e x x x x x

2 2

2 3 2 2 3

2 2 3 2 3

2

7 (44 60 )

95 7 ( 3 5 )) 22 (105840 ( 3 5 ) 15120( 37 95 ) (665222

907350 105 35 )) 22 ( ( 27662 17790 22575 12635 )

210 (80 42 57 35 ) 210(32 150 129 95 )))

e x

x e x c e x x e

x x x c e x x x

e x x x x x x C

(3.71)

Now adding the above equations we have:

1 2 0 1 1 2 1 2, , , , ,y x C C y x y x C y x C C (3.72)

Substituting the approximate solution of the second order in equation (3.56), we get the

residual R.

Now to find the optimal values of 1 2andC C , we will apply the method of Least Squares

as mentioned above, we obtain:



38

1 2 0.9939356470997497 , 0.00003879933613304636C C for 1c

Thus the above equation reduce to:

3 9 4

5 8 6

7 7 8

6 9

1 0.33333333875851556 2.6351381677525838 10

0.008333321787 998034 4.583114876711522 10

0.0001985723585094926 1.5801598143928414 10

2.8086542583364374 10 1.760279867970835

y x x x x

x x

x x

x

9 10

8 11 12 12

10 13 13 15

3 10

2.395089749788994 10 -3.3338633864656773 10

4.751453301896842 10 7.554694767677805 10

x

x x

x x

(3.73)

Results can be seen in Table (3.10-3.13) for various values of c by OHAM compared

with HPM [66], ADM [66] and DTM [66]. We observed that OHAM solution is much

better. The beauty of OHAM can be seen from the Table 3.14 which is constructed to

check the convergence of OHAM. Table 3.14, shows the fast convergence of OHAM.

The accuracy of second order OHAM can easily be seen in Fig 3.6 from the plot of the

residual. Moreover unlike DTM, OHAM produced good results at extended domain.

Further, in Table 3.14, we have shown that convergence depends on the number of

convergence constants 1 2C , C , . In zeroth order problem we use no convergence

constant, so we get accuracy up to 710 range. While in the first order problem we

introduced a constant 1C and we see from the Table 3.14 that the accuracy has been

increased. In the second order problem, we introduce another constant C2 and hence the

accuracy increases. Thus we can conclude that the accuracy is directly proportional to the

number of convergence constants iC involving in the auxiliary function H .



39

Table 3.10

Comparison of Absolute error of obtained solution by OHAM at 1c with ADM,

HPM and HAM

x ADM HPM DTM OHAM

0.0 0.0 0.0 0.0 0.0

0.1 7.8×10-10

7.8×10-10

4.5×10-6

5.01488×10-12

0.2 4.7×10-9

4.7×10-9

2.5×10-5

3.4224×10-11

0.3 1.7×10-8

1.7×10-8

5.9×10-5

8.92695×10-11

0.4 1.9×10-8

1.9×10-8

9.1×10-5

1.45841×10-10

0.5 2.4×10-8

2.4×10-8

1.0×10-4

1.7111×10-10

0.6 2.3×10-8

2.3×10-8

9.6×10-5

1.48808×10-10

0.7 1.7×10-8

1.7×10-8

6.6×10-5

9.24671×10-11

0.8 8.6×10-9

8.6×10-9

3.0×10-5

3.56235×10-11

0.9 1.7×10-9

1.7×10-9

5.5×10-6

5.17764×10-12

1. 0.0 0.0 0.0 0.0

Table 3.11


ADM, HPM and HAM

x ADM HPM DTM OHAM

0.0 0.0 0.0 0.0 0

0.1 1.2×10-6

1.2×10-6

2.9×10-5

7.13252×10-8

0.2 7.2×10-6

7.2×10-6

1.6×10-4

3.78362×10-7

0.3 1.7×10-5

1.7×10-5

3.6×10-4

8.04295×10-7

0.4 2.7×10-5

2.7×10-5

5.3×10-4

1.12364×10-6

0.5 3.4×10-5

3.4×10-5

6.0×10-4

1.18255×10-6

0.6 3.2×10-5

3.2×10-5

5.3×10-4

9.68295×10-7

0.7 2.3×10-5

2.3×10-5

3.5×10-4

5.95893×10-7

0.8 1.1×10-5

1.1×10-5

1.5×10-4

2.39795×10-7

0.9 2.2×10-6

2.2×10-6

2.7×10-4

3.83349×10-8

1. 0.0 0.0 0.0 4.44089×10-16



40

Table 3.12


ADM, HPM and HAM

x ADM HPM DTM OHAM

0.0 0.0 0.0 0.0 0

0.1 2.1×10-5

2.1×10-5

4.7×10-5

2.0838×10-7

0.2 1.4×10-4

1.4×10-4

3.2×10-4

8.52908×10-7

0.3 4.1×10-4

4.1×10-4

8.7×10-4

1.21482×10-6

0.4 7.5×10-4

7.5×10-4

1.5×10-3

7.80319×10-7

0.5 1.0×10-3

1.0×10-3

2.1×10-3

2.32466×10-7

0.6 1.1×10-3

1.1×10-3

2.2×10-3

1.14008×10-6

0.7 9.2×10-3

9.2×10-3

1.7×10-3

1.35982×10-6

0.8 4.9×10-3

4.9×10-3

9.1×10-4

8.57547×10-7

0.9 1.0×10-4

1.0×10-4

1.9×10-4

1.98607×10-7

1. 0.0 0.0 0.0 8.88178×10-16

Table 3.13


ADM, HPM and HAM

x ADM HPM DTM OHAM

0.0 0.0 0.0 0.0 0

0.1 1.4×10-3

1.4×10-3

5.9×10-5

2.07403×10-6

0.2 1.0×10-2

1.0×10-2

4.0×10-4

7.95168×10-6

0.3 3.2×10-2

3.2×10-2

1.1×10-3

1.00271×10-5

0.4 6.3×10-2

6.3×10-2

1.9×10-3

4.21226×10-6

0.5 9.3×10-2

9.3×10-2

2.6×10-3

5.78429×10-6

0.6 1.0×10-1

1.0×10-1

2.8×10-3

1.28639×10-5

0.7 8.6×10-2

8.6×10-2

2.2×10-3

1.2749×10-5

0.8 4.7×10-2

4.7×10-2

1.1×10-3

7.06115×10-6

0.9 1.0×10-2

1.0×10-2

2.5×10-4

1.46475×10-6

1. 0.0 0.0 7.1×10-10

1.77636×10-15



41

Table 3.14

Convergence of Absolute error of obtained solution by OHAM at 1c

x Zeroth Order

OHAM

First Order

OHAM

Second Order

OHAM

0.0 0.0 0.0 0.0

0.1 2.17744×10-8

2.87424×10-10

5.01488×10-12

0.2 1.30402×10-7

2.0029×10-9

3.4224×10-11

0.3 3.14964×10-7

5.25813×10-9

8.92695×10-11

0.4 5.03289×10-7

8.67476×10-9

1.45841×10-10

0.5 6.10095×10-7

1.03838×10-8

1.7111×10-10

0.6 5.7986×10-7

9.32644×10-9

1.48808×10-10

0.7 4.17906×10-7

6.04601×10-9

9.24671×10-11

0.8 1.99081×10-7

2.4308×10-9

3.56235×10-11

0.9 3.82003×10-8

3.5721×10-10

5.17764×10-12

1. 0.0 0.0 0.0

0.0 0.2 0.4 0.6 0.8 1.0

0.00006

0.00004

0.00002

0

0.00002

x

Res

idu

al

Fig. 3.6: The accuracy of second order OHAM can easily be seen

from the plot for Residual.



42

3.5. Application of OHAM to Eighth Order Nonlinear BVP

3.5.1 Model 7

Consider the following eighth- order nonlinear problem [67]:

( ) 2 , 0 1,viii xy x e y x x (3.74)

with the following boundary conditions :

( ) 0 1, 0,1,2,...,7.iy i (3.75)

The analytic solution of this problem is .xy x e

Using equation (2.2) and equating the coefficients of 0 1 2, , ,... on both sides, we get the

sub problems as:


( )

0 0, 0 1,viiiy x x (3.76)

( )

0 0 1, 0,1,2,...,7.i

y i (3.77)

First order Problem

2( ) ( )

1 1 1 0 1 0, 1viii viii xy x C C y x C e y x (3.78)

( )

1 0 0, 0,1,2,...,7.i

y i (3.79)

Second order Problem

( ) ( ) ( )

2 1 2 1 1 1 2 0

2

1 0 1 1 2 0

, , 1 ,

2 ,

viii viii viii

x x

y x C C C y x C C y x

C e y x y x C C e y x

(3.80)

( )

2 0 0, 0,1,2,...,7.i

y i (3.81)

Solving problems (3.76)-(3.81) in succession we obtain:

2 3 4 5 6 7

0

15040 5040 2520 840 210 42 7

5040y x x x x x x x x (3.82)



43

2 2

3 3

4

11

22035489982329600 22035489982329600

14335510975833600 7699979006496000

4536322331788800 1218556347120000

927483100185600 113665407408000

137210230886400 678188322

25401600

x

x

x

x

x

e

x xe

x x e

x x e

xe Cy x

4

5 5

6 6

7 7 8

9 10 11 12

13 14

0000

15578144225280 260384392800

1403862888960 6003632880

102343772160 64859760 6087184740

295153320 11537904 354480 8141

126

x

x

x

x

x e

x x e

x x e

x x e x

x x x x

x x

(3.83)



44

(3.84)

For the second order solution, adding eq. (3.82) – eq. (3.84).

1 20.03867627531 and 0.92145807962C C can be easily calculated by the method

of least squares as mentioned in the Chapter 2. Results of Model 7 can be seen in Table

3.15 and Fig. 3.7.



45

Fig. 3.7. Plot showing comparison between exact solutions

and OHAM solution for Model 7.

Table 3.15

Comparison between OHAM solutions with exact

solution for Model 7.

x Exact Solution OHAM Solution Absolute Error

0.0 1.0 1.0 0.0

0.1 1.10517 1.10517 1.45978×10-7

0.2 1.2214 1.2214 1.02754×10-7

0.3 1.34986 1.34986 2.04184×10-7

0.4 1.49182 1.49182 1.49324×10-7

0.5 1.64872 1.64872 1.06158×10-7

0.6 1.82212 1.82212 1.44092×10-7

0.7 2.01375 2.01375 1.05881×10-7

0.8 2.22554 2.22554 1.4508×10-7

0.9 2.4596 2.4596 1.62443×10-10

1.0 2.71828 2.71828 1.65095×10-7

Chapter 4 Application of Optimal Homotopy Asymptotic Method to Squeezing

Flow Problems

46

Chapter 4

Application of Optimal Homotopy Asymptotic Method to

Squeezing Flow Problems

4.1. Introduction [67-81]

The study of squeezing flows has received a great attention due to important applications

in engineering and biophysical sciences. “Squeezing flows are induced by externally

applied normal stresses or vertical velocities by means of moving boundary”. Practical

application of squeezing flows can be seen in “Polymer processing operations such as

sheet forming, compression and injection modeling, bearings, gears, dampers, hydraulic

systems and lubricated parts of living bodies”. “In literature, many studies and

investigations have revealed the importance of squeezing flows. Stefan [67] initiated the

concept of squeezing flow in mechanics. Grim [68] observed squeezing flows of

Newtonian liquid films an analysis include the fluid inertia. Archibald [69] examined load

capacity and time relations in squeeze films and Wolfe [70] observed fluid inertia effects

in squeeze films. Kuzma [71] and Ishizawa [72] determined squeezing flows of

Newtonian liquid films and analysis include the fluid Inertia. Wang and Watson [73],

Jackson [74] used the squeezing of a viscous fluid between elliptic plates. Terrill and

Winner [75], Usha and Sridharan [76] are among those that who have conducted several

investigation on squeezing flows. Huges and Elco [77], Kamiyama [78], Hamza [79],

Bhattacharyya and Pal [80] made studies on lubrication. Halpern and Secomb [81] have

recently studied the squeezing flow of blood”.

4.2. Fundamental Equations [82-83]

Consider governing of an incompressible fluid, neglecting the thermal effects, are:

0v , (4.1)

Dv

divDt

T F , (4.2)

where v is the velocity vector of the fluid, is the constant density, T is the Cauchy

stress tensor, F is the body force per unit mass and tD

D is the material derivative.

The Cauchy stress tensor is:

T P I S , (4.3)

where P is the pressure, I is the identity tensor and the extra stress tensor S is defined

as:


Flow Problems

47

S R (4.4)

where is the viscosity of fluid and R is the Rivlin-Erickson Tensor defined as

T

v v R (4.5)

4.3. Problem Formulation:

Consider an incompressible Newtonian fluid squeezed between two infinite planar

parallel plates, separated by a small distance 2h and the plates approaching each other

with a low constant velocityV , the lower plate is at H and the upper plate is at

H as illustrated in figure 4.1. The coordinate system is chosen at the centre of the

plates in which the radial direction is taken in the direction of fluid flow and is taken

perpendicular to .

Let the flow is axisymmetric and bidirectional, so the velocity field is defined as

= , 0,v v v , , S S (4.6)

It is further assumed that the flow is steady, laminar and isothermal. The gravitational

force is neglected.

The dynamic equation (4.2) in component form in the absence of body force reduces to:

1 d

d

PS (4.7)

0

P (4.8)

1 d

d

PS (4.9)

From equation (4.8), we have , P P .

Substituting equations (4.3-4.5) in equations (4.7) and (4.9) takes the form:


Flow Problems

48

Fig. 4.1. A steady squeezing axisymmetric fluid flow between two parallel plates.

2 2

2 2 2

1,

v v v v v vv v

P (4.10)

2 2

2 2

1.

v v v v vv v

P (4.11)

The boundary conditions are:

0, atv v V H (4.12)

0, 0 at 0v

v

. (4.13)

The continuity equation is:

1

0,v

v

(4.14)

Introducing the stream function:

1 1

, zv v

. (4.15)

The continuity equation is satisfied, substituting Eq. (4.15) in Eq. (4.10-4.11) we obtain:

* 2 2 2

20,

P E E (4.16)


Flow Problems

49

* 2 2 2

20.

P E E (4.17)

where the generalized pressure *P and vorticity , functions are:

* 2 ,2

v

P P (4.18)

v v

(4.19)

Applying integrability conditions:

2 * 2 *

,

P P (4.20)

We get the compatibility equation

2 2 241 ( , / )

,t

E EE (4.21)

where 2 2

2

2 2

1

E (4.22)

From Eq. (4.21) we have

2 24( , / )

,

EE (4.23)

with the boundary conditions

, then 0, ,

0, then 0, 0.

H v v V

vv

(4.24)

The stream function can be expressed as:

2, f . (4.25)

In view of Eq. (4.25), the compatibility Eq. (4.20) and the boundary conditions (4.24)

take the form:


Flow Problems

50

0,iv

f f f

(4.26)

subject to

0 0, 0 0,

, 0.2

f f

Vf H f H

(4.27)

Introducing the following non-dimensional parameters

* *, , ./ 2 2 /

e

f hf R

V H V

(4.28)

Here eR is the well known ratio, called Reynold number.

For simplicity dropping the " " the boundary value problem (4.26) becomes

0,iv

ef R f f (4.29)

with the boundary conditions

0 0, 0 0,

1 1, 1 0.

f f

f f

(4.30)

4.4. Application of OHAM to Squeezing Flow Problem

In this section, we apply OHAM to the following nonlinear problem given in equations

(4.29) and (4.30):

We have:

( )ivf f L (4.31)

0g (4.32)

ef R f f N . (4.33)

Using Eq. (2.2), we construct a family of equations for the given problem (4.29-4.30):

(1 )[ ( ( , )] ( )[ ( ( , ) ( ( , ))], L H L N (4.34)

,

, , 0

B (4.35)


Flow Problems

51

( ) ( )(1 ) ( , ) ( )[ ( , ) ( , )( ( , )],iv iv

eR H (4.36)

,

, , 0

B . (4.37)

Expanding , in a Taylor series with respect to , we obtain:

0

1

, , , , 1,2,...n

i k i

n

K f f K i

. (4.38)

Thus have the following cases:

Zeroth-order problem:

0 0iv

f , 0 0 0 00 1 0 0, 1 1f f f f (4.39)

with solution:

3

0

1 3

2 2f (4.40)

0 ( )f is the initial approximation which satisfies the boundary conditions (4.30), in

literature this is known as Newtonian solution in the absence of inertial terms in the

equations of motion.

First order problem:

( ) ( ) ( )

1 1 0 1 0 1 0 0,iv iv iv

ef K f K f R K f f (4.41)

1 1 1 10 1 1 0 0f f f f (4.42)

with solution is:

3 5 7

1 1 1 1 1 1

1, 19 39 21

560e e e ef K K R K R K R K R . (4.43)

Second-order problem

( ) ( ) ( ) ( )

2 1 2 1 1 1 1 1 2 0

1 1 1 1 1 1 0 2 0 0

, , , ,

, .

iv iv iv iv

e e e

f K K f K K f K K f

R K f f R K f K f R K f f

(4.44)

2 2 2 20 1 1 0 0f f f f . (4.45)


Flow Problems

52

For this problem we have the solution

2

2 1 2 1 1

22 3 2 3

2 1 1 1

23 2 3 5 2 5

2 1 1 1

25 2 5 7

2 1 1

, , (1/ 2587200) 87780 -87780

87780 3288 180180 180180

180180 2215 97020 97020

97020 15708 4620

e e

e e e e

e e e e

e e e

f K K K R K R

K R K R K R K R

K R K R K R K R

K R K R K R

2 7

1

2 2 27 2 7 2 9 2 11

2 1 1 1

4620

4620 11682 1540 63

e

e e e e

K R

K R K R K R K R

(4.46)

For the second order approximation, adding Eqns. (4.40), (4.43) and (4.46), we obtain:

22 2

1

2 222 2 2 2

2 1

2 2 4 6

( / 2587200) 9240 19 1 4620

280 3 19 1 1

4620 19 3288 8791 1414 63

e

e e

e

f K R

K R K R

R

(4.47)

where

0 1 1 2 1 2, , ,f f f K f K K . (4.48)

Substituting Eq. (4.48) in Eq. (4.29), we obtain the residual as:

( )

1 2, , iv

er K K f R f f (4.49)

Using Eq. (4.43), we obtain the expression for the residual r :

2 2

1

24 6 2 2 2 4 6

2

23 2 2 2 4 6

1

2 4

1 (17075520 (39200( 3 ) 560 ( 34 177

223120128000

84 9 ) ( 1 ) ( 741 4029 875 35 )) 1848

( 1 ) ( 9240 ( 741 4029 875 35 ) (86147

2639104 9711786

e e

e

e e

r R C R

K R

K R R

26 8 10

2 4 6 8

10 12 14 2

22 2 2 2 4 6

2 2

3473792 389585 12600 )

(1456584 99401395 7931546 678258231 338212644

60011259 4569642 130977 )) 4268880(78400( 3 )

( 1 ) ( 741 4029 875 35 ) 1120

e

e

R

K R K

2 2 4 6 2 2

1

32 4 6 2 2

2

22 4 6 8 10

2

( 70 ( 3

) ( 34 177 84 9 ))) 924 ( 362208000 ( 3 )

10348800 ( 34 177 84 9 ) ( 1 ) (86147

2639104 9711786 3473792 389585 12600 ) 280

( 52263 839396 2037882

e

e e

e

R K

R K R

R

4 6 8 10

2 2 2 4 6

2

1192356 224455 12768

33 ( 1 ) ( 741 4029 875 35 ))))K

(4.50)


Flow Problems

53

For 1 22, 0.821494, 0.0206056eR K K

21 1 12 2 2 3 19 1

2 280 2587200

2 2 3( 1 ) (14444 28888 309112 106572

4 5 6 7 841914 3640 5040 490 245 ).

f

(4.51)

4.5. Solutions by PM and HPM

For comparison, the reduced model given in Eqs. (4.29-4.30) is solved also by regular PM

and HPM. Second order approximation by PM and HPM are

2 2 2 2

2 2 2 3

4 5 6 7 8

1 1 1( ) ( 3 ) ( 19 )( 1 )

2 560 10348800

( 1 ) (14444 28888 309112 106572

41914 3640 5040 490 245 )

PM e

e

F R

R

(4.52)

and

2 2 2 2

2 2 2 3

4 5 6 7 8

1 1 1( ) ( 3 ) ( 19 )( 1 )

2 560 10348800

( 1 ) (14444 28888 309112 106572

41914 3640 5040 490 245 )

HPM e

e

F R

R

(4.53)

4.6. Comparison analysis

Table 4.1 shows comparison of PM, HPM, HAM, Numerical Method and the present

method (OHAM). For a fix data:

12, 0, 1, 0.821494eR a b K 2and 0.0206056K .

It is obvious that the result of OHAM (second order) is better than the other methods

mention in the Table 4.1.

The plot given Fig. 4.2 shows comparison of zeroth order 0f , first order 1

f and second


Flow Problems

54

order 2f solutions by OHAM. The curve shows that the OHAM results are improving

with the increasing order. In Table 4.2, we analyzed the flow with different Reynolds

numbers 0,1,2,3,4,10eR and observed that we got almost the same solution by our

second order approximation.

Table 4.1

Comparison analysis of OHAM solution with PM, HAM, Numerical

PM HPM HAM[83] Numerical[83] OHAM

0.00 0.0 0 0 0 0

0.15 0.232546 0.232546 0.232179 0.232179 0.231996

0.30 0.45369 0.45369 0.451603 0.451603 0.451343

0.45 0.650927 0.650927 0.646354 0.646353 0.646147

0.60 0.811982 0.811982 0.805988 0.805988 0.805894

0.75 0.926616 0.926616 0.921867 0.921867 0.921852

0.90 0.988531 0.988531 0.987176 0.987176 0.987177

Fig.4.2. Approximation of ( )f by zeroth order solution (dotted lines), first

order solution (dashed lines) and second order solution (dotted-dashed lines) by

OHAM


Flow Problems

55

0.0 0.2 0.4 0.6 0.8 1.0

0.0

0.2

0.4

0.6

0.8

1.0

f

Re4

Re3

Re2

Re1

Fig. 4.3. Analysis of f with increasing Reynolds numbers 1,2,3,4 .eR

We almost get the same behaviour if we slightly increase .eR

Table 4.2

Analysis of f with increasing Reynolds numbers 1,2,3,4 .eR

0eR 1eR 2eR 3eR 4eR 10eR

0.00 0.0 0 0 0 0 0

0.15 0.223312 0.227784 0.231996 0.235947 0.239637 0.256306

0.30 0.4365 0.444185 0.451343 0.457974 0.464078 0.489628

0.45 0.629438 0.638166 0.646147 0.65338 0.659864 0.683059

0.60 0.792 0.799349 0.805894 0.811634 0.816571 0.829301

0.75 0.914063 0.918249 0.921852 0.92487 0.927305 0.929652

0.90 0.9855 0.986419 0.987177 0.987774 0.98821 0.987436


Flow Problems

56

Now let us analyze 0.5, 2.0, 3.0, 4.0.e eR at R We see from Fig. 4.4, that when

eR decreases, the solution of the boundary value problem (4.26-4.27) decreases. Thus, it is

easy to get how , , andV h effect on f

0.0 0.2 0.4 0.6 0.8 1.0

0.0

0.2

0.4

0.6

0.8

1.0

f

Re4.0

Re3.0

Re2.0

Re0.5

Fig.4.4. Analysis of f with Reynolds numbers

0.5, 2.0, 3.0, and 4.0. e e e eR R R R

0.0 0.2 0.4 0.6 0.8 1.0

0.0

0.2

0.4

0.6

0.8

1.0

f

Re0

Re0.5

Re2.0

Re3.0

Fig.4.5. Analysis of f with Reynolds numbers

0.2, 0.5, 2 and 3. e e e eR R R R


Flow Problems

57

0.0 0.2 0.4 0.6 0.8 1.0

0.0

0.2

0.4

0.6

0.8

1.0

1.2

f

Re200

Re30

Re20

Re100

Fig.4.6. Analysis of f for higher Reynolds numbers.

Chapter 5 Application of Optimal Homotopy Asymptotic Method to KdV Equations

58

Chapter 5

Application of Optimal Homotopy Asymptotic Method to

Korteweg de Vries Equations

5.1. Introduction

It is clear that many phenomena in scientific fields such as solid state physics, plasma

physics, fluid dynamics, mathematical biology and chemical kinetics, can be modeled by

systems of linear or nonlinear PDEs. The nonlinear models of real-life problems are still

difficult to solve either numerically or theoretically. A broad class of analytical solutions

methods and numerical solutions methods were used to handle these problems [84-86].

Finding exact or approximate solution of PDEs is interesting and important. A famous

equation which arises in the study of nonlinear dispersive waves is the KdV equation.

KdV equation was derived in 1895 by Korteweg and de Vries to model water in shallow

canal [87]. This equation plays important role in diverse areas of engineering and

scientific applications and therefore, enormous amount of research work has been

invested in the study of such equations. KdV equations have been examined extensively,

analytically and numerically. The existing techniques encountered difficulties in terms of

the size of computational work needed, especially when the system involves several

PDEs.

5.1.1. Model 1 [88]

In this article, we shall deal with the KdV equation in different forms by the analytic

approach. The KdV equation is:

3

3

, , ,6 , 0

u x u x u xw x

x x

(5.1)

where ,u x is the displacement.

Let us consider KdV equation of the form (5.1). The initial condition is 2,0 6 /u x x .

The exact solution is 2

3 3, 6 24 / 12u x x x x which satisfies the KdV

equation (5.1) and the given conditions. Constructing the deformation equations for the

given problem and equating the coefficients of 0 1 2, , ,... on both sides, we get the

zeroth order, first order, and second order problem as under:


59

, ;

, ;x

x

L (5.2)

3

3

, ; , ;, ; 6 , ;

x xx x

x x

N (5.3)

, 0x (5.4)

The initial condition is:

2

6,0;x

x (5.5)


0

0 2

, 60, ,0

u xu x

x

(5.6)

Its solution is

0 2

6u x

x (5.7)

First Order Problem

3

1 0 0 0

1 1 0 1 3

, , , ,1 6 , ,

u x u x u x u xk k u x k

x x

(5.8)

1 ,0 0u x

Its solution is

11 1 5

288, ,

ku x k

(5.9)


3 3

2 1 2 1 1 0 0 1 1

1 2 2 13 3

0 0 1 1

2 0 1 1 1 0

, , , , , , ,1

, ,6 ,

u x k k u x k u x u x u x kk k k k

x x

u x u x u x kk u x k u x k u x

x x x

2 ,0 0u x (5.10)

Its solution is

3 3 2 3 2 3

2 1 2 1 1 1 28

288, , , 21u x k k x k x k x k x k

x

(5.11)


60

Third Order Problem:

3 1 2 3 2 1 2 0 1 1

1 3 2

3 3 3

0 1 1 2 1 2

3 2 13 3 3

0

3 0 2 1 1 1 2 1 2

1 1 1 1 1 1 2

0 2 1 1 0

, , , , , , , , ,1

, , , , ,

6 , , , , ,

, , , , , , ,6 , ,

u x k k k u x k k u x u x kk k k

u x u x k u x k kk k k

x x x

u xk u x k u x k k u x k k

u x k u x k u x k ku x k u x k u x

x x x

(5.12)

3 ,0 0u x

Its solution is:

6 2 3 2 6 2 3 3 2 3 3

1 1 1 1 1

3 1 2 3 11 6 3 6 2 3 6 6

1 2 1 2 1 2 3

78 2 2520 42288, , , ,

42 2

x k x k x k k x ku x k k k

x x k x k x k k x k k x k

(5.13)

Adding equations 5.7, 5.9, 5.11 and 5.13 we obtain:

1 2 0 1 1 2 1 2 3 1 2 3, , , , , , , , , , , , ,u x k k u x u x k u x k k u x k k k (5.14)

For the calculations of the constants 1 2 3, andk k k using the Method Least Squares [39-43]

we have computed 1 2 2 0.52106, 0.54018 and 1.87588.k k k

2 3

5 2 8

3 2 3 6

11

1 6 288, 150.09405 5.70163 0.25239

288356.50877 15.29771 1.77345

u x xx x x

x xx

(5.15)

The series (5.15) is our OHAM solution. The same problem is also solved by VIM [89]

and ADM [88]. Although they get closed form solution but we have taken there, fourth

order expressions to compare with third order OHAM:

2 3

2 5 8 11

6 288 6048 103648, .ADMU x

x x x x

(5.16)

2 3

2 5 8 11

6 288 6048 103648, .V I MU x

x x x x

(5.17)

Table 5.1 shows the effectiveness of OHAM. In comparison with VIM, ADM and

OHAM are much consistent throughout domain. Table 5.2 shows that with the increasing

time, OHAM is very consistent in terms of accuracy and we see marvelous accuracy at


61

large domain. The efficiency of the method (OHAM) is obvious from the residual, which

approaches zero as we move along the large domain shown in Fig. 5.1.

Table 5.1

Comparison of absolute error of obtained solution by OHAM for example 1 at time

1

x VIM [89] ADM [88] OHAM

10 0.000118903 1.56186×10-8

7.68848×10-9

11 0.0000557063 4.10022×10-9

1.51451×10-9

12 0.0000278536 1.206×10-9

2.98386×10-10

13 0.0000147131 3.90184×10-10

5.33781×10-11

14 8.14524×10-6

1.36874×10-10

6.80608×10-12

15 4.69576×10-6

5.14631×10-11

4.56961×10-14

16 2.80456×10-6

2.05464×10-11

6.78659×10-14

17 1.72797×10-6

8.64333×10-12

4.31238×10-13

18 1.09442×10-6

3.80642×10-12

5.97251×10-13

19 7.1044×10-7

1.74522×10-12

5.22402×10-13

20 4.71489×10-7

8.29118×10-13

3.57435×10-13

Table 5.2

Absolute error of obtained solution by OHAM for example 1 at large problem

domain

x 1 10

10 7.68848×10-9

0.000130976

510 2.91379×10-19

2.93412×10-18

1010 8.47033×10-21

9.65618×10-20

1510 1.69407×10-21

1.3129×10-20

2010 4.23516×10-22

2.96462×10-21

2510 2.11758×10-22

9.52912×10-22

3010 0.0 4.23516×10-22

3510 1.05879×10-22

1.05879×10-22

4010 0.0 1.05879×10-22

4510 0.0 5.29396×10-23

5000 5.29396×10-23

0.0


62

100 200 300 400 500

0

2. 1017

4. 1017

6. 1017

8. 1017

1. 1016

1.2 1016

x

Res

idu

al

Residual

Fig. 5.1: Residual of the Model 1.

5.1.2. Model 2 [90]

Let us again consider KdV equation of the form (5.1), now, with initial

2

2,0 2 / 1kx kxw x k e e . The exact solution is 2 2

2

2( , ) 2 / 1k x k k x k

u x k e e

which satisfies the KdV equation (5.1) and the given conditions. Applying the method

formulated in section 2.2 (Chapter 2), with the following initial condition:

2

2,0; 2

1

kx

kx

k ex

e

(5.18)

Constructing the deformation equations for the given problem and equating the

coefficients of 0 1 2, , ,... on both sides, we get the zeroth order, first order, and second

order problem as under:Zeroth Order Problem:

20

0 2

,0, ,0 2

1

kx

kx

u x k eu x

e

(5.19)

Its solution is:

2

0 22

1

kx

kx

k eu x

e

(5.20)


3

1 0 0 0

1 1 0 1 3

, , , ,1 6 , ,

u x u x u x u xk k u x t k

x x

(5.21)

1 ,0 0u x


63

Its solution is:

55

1 1 13

2 1, ,

1kx

kx kxe e ku x k k

e

(5.22)


3 3

2 1 2 1 1 0 0 1 1

1 2 2 13 3

0 0 1 1

2 0 1 1 1 0

, , , , , , ,1

, ,6 ,


x x

u x u x u x kk u x k u x k u x

x x x

(5.23)

2 ,0 0u x

Its solution is

2 2 2 2 3 251 1 1 1 1

2 1 2 4 3 2 2 3 2 2

1 1 2 2

2 2 2 2, , ,

4 2 21

kx kxkx

kx kx kxkx

k e k k e k k ke ku x k k

e k k e k k k e ke

(5.24)

Third Order Problem is:

3 1 2 3 2 1 2 0 1 1

1 3 2

3 3 3

0 1 1 2 1 2

3 2 13 3 3

0

3 0 2 1 1 1 2 1 2

1 1 1 1 1 1 2

0 2 1 1 0

, , , , , , , , ,1

, , , , ,

6 , , , , ,

, , , , , , ,6 , ,



x x x


t


x x x

(5.25)

3 ,0 0u x .

Its solution is


64

5 7 2 3 4

3 1 1 1 1 1

5 2 2 2 2 3 2 4 2 5 2

1 1 1 1 1 1 1

3 2 2 3 2 2 2 3 2 2 3 3 2 2 4 3 2

1 1 1 1 1

, ( / 3(1 ) ) 6 18 12 12 18

6 12 36 24 24 36 12

6 30 84 84 30

kx kx kx kx kx kx

kx kx kx kx kx kx

kx kx kx kx

u x e k e k e k e k e k e tk

e k k e k e k e k e k e k

k k e k k e k k e k k e k k

2

5 3 2 2 3 3 2 3 3 3 4 3 5 3

1 1 1 1 1 1 1

3 2 3 3 2 3 2 3 2 3 3 3 2 3 4 3 2 3 5 3 2 3

1 1 1 1 1 1

6 3 3 6 3 3 2 6 3 3 3 6 3 3

1 1 1 1

6 6 18 12 12 18 6

6 6 48 48 6 6

33 110 110 3

kx kx kx kx kx kx

kx kx kx kx kx

kx kx kx

e k k k e k e k e k e k e k

k k e k k e k k e k k e k k e k k

k t k e k k e k k e k k

4 6 3 3 5 6 3 3

1 1

2 3 4 5

2 2 2 2 2 2 1 2

2 3 4 5 3 2

1 2 1 2 1 2 1 2 1 2 1 2

3 2 2 3 2 3 3

1 2 1 2

3

6 18 12 12 18 6 12

36 24 24 36 12 6

6 48 48

kx kx

kx kx kx kx kx

kx kx kx kx kx

kx kx kx

e k k e k k

tk e k e k e k e k e k k k

e k k e k k e k k e k k e k k k k k

e k k k e k k k e k

2 4 3 2 5 3 2

1 2 1 2 1 2

2 3 4 5

3 3 3 3 3 3

6 6

6 18 12 12 18 6

kx kx

kx kx kx kx kx

k k e k k k e k k k

k e k e k e k e k e k

(5.26)

Adding the Equations (5.20), (5.22), (5.24) and (5.26), we obtain:


Substituting equation Eq. (5.27), in Eq. (5.1) and calculating the constants 1 2 3, andk k k

using the Method Least Squares, we obtain:

16

1 2 3 4.66455 10 , 6.03026, 13.58218k k k

2 16 5

2 3

52 31 3

4

31 3 31 2 3

5

7

(2 ) 9.32911 10 ( 1 ) ,

(1 ) (1 )

(12.06051 12.0605 2.17581 10(1 )

8.70322 10 2.17581 10 )

k(45.3115

3(1 )

kx kx kx

kx kx

kxkx

kx

kx kx

kx

kx

e k e e ku x

e e

e ke k

e

e k e k

e

e

2

3 4 5

14 3 2 14 3 2

13 2 3 2 13 3 3 2

14 4 3 2 14 5 3 2

4 135.93462 90.62308

90.62308 135.93462 45.31154

1.68771 10 1.68771 10

1.35017 10 1.35017 10

1.68771 10 1.68771 10

kx kx

kx kx kx

kx

kx kx

kx kx

e e

e e e

k e k

e k e k

e k e k

46 6 3 45 6 3

44 2 6 3 44 3 6 3

45 4 6 3 46 5 6 3

1.01492 10 3.34922 10

1.11641 10 1.11641 10

3.34922 10 1.01492 10 )

kx

kx kx

kx kx

k e k

e k e k

e k e k

(5.28)


65

10 5 0 5 10

0.0

0.1

0.2

0.3

0.4

0.5

x

ux

Exact

3 rd Order OHAM

Fig. 5.2. OHAM and Exact Solutions for Example 2.


66

Table 5.3

Absolute error of obtained solution by OHAM for example 2 at time

0.001 and 1k


-10 -0.0000907009 -0.0000906535 4.7404×10-8

-9 -0.000246512 -0.000246383 1.28817×10-7

-8 -0.000669806 -0.000669456 3.49865×10-7

-7 -0.00181863 -0.00181768 9.48839×10-7

-6 -0.00492811 -0.00492555 2.5631×10-6

-5 -0.013283 -0.0132762 6.84968×10-6

-4 -0.0352914 -0.0352736 0.000017781

-3 -0.0902716 -0.0902289 0.0000426965

-2 -0.209827 -0.209744 0.0000834664

-1 -0.393042 -0.392947 0.0000947597

0 -0.5 -0.5 0.0

1 -0.393406 -0.3935 0.0000948303

2 -0.210147 -0.210231 0.0000833887

3 -0.0904351 -0.0904778 0.0000426306

4 -0.0353595 -0.0353772 0.0000177494

5 -0.0133092 -0.0133161 6.83691×10-6

6 -0.00493793 -0.00494049 2.55824×10-6

7 -0.00182226 -0.00182321 9.47029×10-7

8 -0.000671146 -0.000671495 3.49196×10-7

9 -0.000247006 -0.000247134 1.28571×10-7

10 -0.0000908824 -0.0000909298 4.73133×10-8


67

10 5 0 5 10

0.10

0.05

0.00

0.05

0.10

x

Res

idu

al

Residual

Fig. 5.3. Residual (error) of OHAM solution for Model 2.

5.1.3. Model 3 [90]

Now let us consider KdV equation (5.1) with initial condition 2,0 6secu x h x . The

exact solution is

8 2 24 4 72 6

8 2 64 4 72 6

2( , ) 12

1 3 3

x x x

x x x

e e eu x

x e e e

which satisfies the KdV

equation (5.1) and the given conditions. Applying the method formulated in Section 2.2

(Chapter 2), leads to the same equations the following equation:

2

2,0; 2

1

kx

kx

k ex

e

(5.29)

Constructing the deformation equations for the given problem and equating the

coefficients of 0 1 2, , ,... on both sides, we get the zeroth order, first order, and second

order problem as under:


20

0 2

,0, ,0 2

1

kx

kx

u x k eu x

e

(5.30)

Its solution is

20 6secu x h x (5.31)

First order problem


68

3

1 0 0 0

1 1 0 1 13

, , , ,1 6 , , ,0 0

u x u x u x u xk k u x k u x

x x

(5.32)

Its solution is

4 2 3

1 1 1 1, , 48 7 sec tan sec tanu x k h x k h x h x k h x (5.33)

Second order problem

3 3

2 1 2 1 1 0 0 1 1

1 2 2 13 3

0 0 1 1

2 0 1 1 1 0 2

, , , , , , ,1

, ,6 , , ,0 0


x x

u x u x u x kk u x k u x k u x u x

x x x

(5.34)

Its solution is:

2 8 2 4

2 1 2 1 1

4 2 4

1 2

2 6 2 2 2 3

1 1

2 2 3 2 3

1 2

2 4 2 4 2

1

, , , 48(80 sec 7 sec tanh

7 sec tanh 7 sec tanh

288 sec tanh sec tanh

sec tanh sec tanh

246 sec tanh 4 sec

u x k k h x k h x k x

h x k x h x k x

h x k x h x k x

h x k x h x k x

h x k x

2 2 6

1 tanh )h x k x

(5.35)

Third order problem is:

3 1 2 3 2 1 2 0 1 1

1 3 2

3 3 3

0 1 1 2 1 2

3 2 13 3 3

0

3 0 2 1 1 1 2 1 2

1 1 1 1 1 1 2

0 2 1 1 0

, , , , , , , , ,1

, , , , ,

6 , , , ,

, , , , , , ,6 , ,



x x x



x x x

(5.36)

3 ,0 0u x ;

Its solution can be obtained as usual.

Adding the Equations (5.31), (5.33), (5.35) and (5.37), we obtain:


Substituting equation Eq. (5.38) in the residual equation and calculating the constants

1 2 3, andk k k using the Method Least Squares, we obtain:


69

1 0.0161,k 2 30.965326,k 3 59.97914.k

2 2 3

2 8 4

2 6 2 2 3

2 4 4 2 2 6

, sec ( ) 6 48 0.11295sec ( ) tanh( ) 0.01613tanh ( )

48(0.02082 sec 16.86839 sec tanh

0.07496 sec tanh 30.98119 sec tanh

0.06403 sec tanh 0.00104 sec tanh )

16( 239.92

u x h x h x x x

h x h x x

h x x h x x

h x x h x x

2 8 4

3 10 2 6 2

2 3 3 8 3

2 4 4 3 6 5

2 2 6

565 sec 629.94482 sec tanh




11.99628 sec tanh 0

h x h x x

h x x h x x

h x x h x x

h x x h x x

h x x

3 4 7

3 2 9

.06544 sec tanh

0.00013 sec tanh )

h x x

h x x

(5.39)

Tables (5.1-5.4) show the absolute errors between OHAM and the exact solution. It is

observed that as we move along the domain, we get consistent accuracy. This method

accurately approximates the solutions of partial differential equations.

4 2 0 2 4

0

1

2

3

4

5

6

x

ux

Exact

OHAM

Fig. 5.4 :OHAM and Exact Solutions for Model 3.


70

Table 5.4

Absolute error of obtained solution by OHAM for example 5 at time

0.001 and 1k


-5.0 -0.00108081 -0.00108081 4.43069×10-10

-4.5 -0.00293745 -0.00293745 1.24451×10-9

-4. -0.00798114 -0.00798114 3.67919×10-9

-3.5 -0.0216679 -0.0216679 1.21755×10-8

-3. -0.0587 -0.0587001 4.88768×10-8

-2.5 -0.158106 -0.158106 2.43777×10-7

-2. -0.419315 -0.419316 1.37153×10-6

-1.5 -1.06821 -1.06822 6.97502×10-6

-1. -2.46698 -2.467 0.0000167845

-0.5 -4.61928 -4.61923 0.000047836

0. -5.99616 -5.99616 2.14761×10-6

0.5 -4.81869 -4.81875 0.0000508896

1. -2.57509 -2.57507 0.0000165221

1.5 -1.10095 -1.10094 7.13239×10-6

2. -0.428628 -0.428626 1.40907×10-6

2.5 -0.161027 -0.161026 2.50052×10-7

3. -0.0596984 -0.0596983 4.99005×10-8

3.5 -0.0220244 -0.0220244 1.23646×10-8

4. -0.00811082 -0.00811082 3.72275×10-9

4.5 -0.00298496 -0.00298495 1.257×10-9

5.0 -0.00109826 -0.00109826 4.47158×10-10

1.0 0.5 0.0 0.5 1.0

0.2

0.1

0.0

0.1

0.2

x

Res

idu

al

Residual

Fig. 5.5: Residual (error) of OHAM solution for Model 3.

Chapter 6 Application of Optimal Homotopy to Wave Equations

71

Chapter 6

Application of Optimal Homotopy to Wave Equations

6.1. Introduction

OHAM, a latest semi-analytic approximate technique for treatment of wave equations,

has been used in this chapter. To see the usefulness of the method, we take up wave

equations in different forms.

The wave equations perform a great function in several areas of technology. A great

amount of research work has been done in the study of wave equations [91-92]. In this

article, we shall deal with the wave equations [91-92] in different forms by the latest

numerical analytic approach. The governing equations are:

, ,0, 0

w wL

(6.1)

2 2

2

2 2

, ,, 0

w wL

(6.2)

and

2 22

2 2

, ,, , , , 0

w ww w L

(6.3)

where ,w is the displacement of a string, is a positive constant showing the wave

speed and , is a source function.

The motivation of this part of the work is to extend the use of OHAM to the solution of

wave equations. Here, we have proved that OHAM is effective and reliable for solving of

wave equations, hence, showing its capacity for the solution of time depending physical

phenomena.

To demonstrate the effectiveness of the extended formulation of OHAM three models of

wave equations of the type (6.1-6.3) is studied.

Example 6.1

Here, we consider wave equation of the form (6.1) [91], with the initial conditions


72

,0 sin / , ,0 / cos /w L w L L , (6.4)

and boundary conditions

0, sin / , 0, cos / ,w L w L (6.5)

The exact solution is

( , ) sin /w L . (6.6)

Applying the method formulated in section Chapter 2, leads to the following:

, ;

, ;

L , (6.7)

, ;

, ;

N . (6.8)

The initial conditions are:

,0; sinL

(6.9)

and

,0;

cos .L L

(6.10)


0 ,0,

w

(6.11)

0 ,0w sinL

(6.12)

Its solution is

0w sinL

(6.13)

First order problem

1 0 0

1 1

, , ,1

w w wk k

, (6.14)

1 ,0 0w (6.15)

Its solution is


73

1

1

cos

,

kL

wL

. (6.16)


1 0

1 2

2

1 0

1 2

, ,1

,

, ,

w wk k

w

w wk k

(6.17)

2 ,0 0w

Its solution is:

2

1 1

2 2

2 2 2 2

1 2

2 cos 2 cos1

,2

sin 2 cos

L k L kL L

wL

k L kL L

(6.18)

Adding equations (6.13) (6.16), and (6.18), we obtain:

1 2 0 1 1 2 1 2, , , , , , , , ,w k k w w k w k k (6.19)

For the calculations of the constants 1k and 2k , using (6.19) in Eq.(2.17) and applying

the procedure mentioned in (2.19-2.20), we get

1 0.999708426k and 2 0.000416477k when =0.1 , 2L

2 2

0.049985421 / 2 / 2

0.000493435 c / 2, 8

0.098638498 s / 2

cos sin

osw

in

(6.20)

Example 6.2 [91]

Now we consider the second order wave equation in one dimension of the form (6.2).


,0 sin / , ,0 0w L w , (6.21)

where as the boundary conditions are given by

0, 0,w 0, / cos /w L L (6.22)

The exact solution is


74

, sin / cos /w L L (6.22a)

Applying the method discussed in section 2.3, we have:

2

2

, ;, ;

L (6.23)

2

2

2

, ;, ;

N (6.24)


,0; sin / L , and ,0;

0

. (6.25)

Zeroth order problem is

2

0 0

02

, ,00, ,0 , 0

w ww sin

L

(6.26)

Its solution is

0 ,w sinL

(6.27)

First order problem

2 2 2

1 0 02

1 12 2 2

, , ,1

w w wk k

(6.28)

1

1

,0,0 0 , 0

ww

(6.29)

Whose solution is

2 2 2

1

1 2,

2

k sinL

wL

(6.30)


2 2

1 0

2 1 22 22

2 2 2

1 02 2

1 22 2

, ,1

,

, ,

w wk k

w

w wk k

(6.31)

2

2

,0,0 0, 0

ww

(6.32)

Solution of the above problem is given by:


75

2 2 2

2

2 1 1 22

4 4 4 2

1

4

,2

24

w k sin k sin k sinL L L L

ksin

L L

(6.33)

From equations (6.27), (6.30), and (6.33), we obtain:

1 2 0 1 1 2 1 2, , , , , , , , ,w k k w w k w k k (6.34)

Using equation (6.34) in (2.17) and applying the technique as discussed in equations

(2.19) and (2.20) , we obtain:

1 0.999910714k and -8

2 = 1.434657458 10k for 2L and 0.1

Substituting values in equation (6.34), we have:

2 -7 4, sin / 2 - 0.00125 sin / 2 2.6037 10 sin / 2w (6.35)

Example 6.3

Finally we take the second order inhomogeneous nonlinear wave equation of the form

(6.3) [92], where 0 and 0 . The initial and boundary conditions are

,0 0, ,0 /w w (6.36)

and

0, 0, ,w w (6.37)

respectively. The exact solution is

( , ) ,w (6.38)

which satisfies the wave equation (6.3) and the given conditions. Apply the same method

as in Examples 6.1 and 6.2 we leads to the following:

2

2

, , ;, , ;

L (6.39)

2 2, (6.40)

2

2

2

, ;, ; , ; , ;

N (6.41)


,0; 0 and ,0;

(6.42)


76


2

0

2

,, 0,

w

(6.43)

0

0

,0,0 0,

ww

(6.44)

having the solution

3 4 2

0

1, 12 2

12w . (6.45)

First order problem

2 2

0 021 1 1 02 21

22

1 0 1

, ,1 ,,

, 1 ,

w wk k k ww

k w k

(6.46)

1

1

,0,0 0, 0

ww

(6.47)

Its solution is as follow

6 3 5

1 1 1

4 2 6 2 8 2

1 1 1 1 1

7 3 9 3 10 4

1 1 1

504 15120 756 1

, , 7560 1260 45 .90720

360 35 7

k k k

w k k k k

k k k

(6.48)


2 2

1 1

1 1 1 12 2

2 2 2

2 0 0

1 0 1 2 22 2 2

2

2 0 2 0 2

, ,1 ,

, , ,2 , ,

, , ,

w wk k k w

w w wk w w k k

k w k w k

(6.49)

2

2

,0,0 0, 0

ww

(6.50)

Solution of the above problem is given by:


77

2 1 2

6 2 6 6 2 8

1 1 2 1

2 10 3 2 3 2 16 6

1 1 1 1

3 5 2 5

2 1 1

1, , ,

130767436800

726485760 + 1452971520 726485760 77837760

1441440 21794572800 21794572800 7007

21794572800 1089728640 2179457280

1

w k k

k k k k

k k k k

k k k

5 2 7 2 9

2 1 1

2 11 4 2 2 4 2

1 1 1

4 2 6 2 2 6 2

2 1 1

6 2 8 2 2 8 2

2 1 1

089728640 25945920 63423360

4953312 10897286400 10897286400

10897286400 1816214400 3632428800

1816214400 64864800 265945680

k k k

k k k

k k k

k k k

8 2 2 10 2 2 12 2 2 13 5

2 1 1 1

7 3 2 7 3 2 12 4

1 1 1

7 3 9 3 2 9 3

2 1 1

9 3 2 11 3

2 1

64864800 4756752 1834560 683760

518918400 1037836800 4444440

518918400 50450400 209008800

50450400 8792784 1386

k k k k

k k k

k k k

k k

2 13 3 2 15 5

1 1

10 4 2 10 4 10 4 2 14 4

1 1 2 1

00 56056

10090080 41801760 10090080 151800

k k

k k k k

(6.51)

From equations (6.45), (6.48) and (6.51), we obtain:

1 2 0 1 1 2 1 2, , , , , , , , ,w k k w w k w k k (6.52)

Following similar procedure as discussed in previous examples, we get:

1 1.0009996521882438k and 6

2 1.4704901755719086 10k .

Therefore, the final solution is of example 3 is:

3 4 2

1 2

6 3 5 4 2

1 1 1 1

6 2 8 2 7 3 9 3 10 4

1 1 1 1 1

6 2 6

1 1

, , , (1/12) 12 2 (1/ 90720)

504 15120 756 7560(1/130767436800)

1260 45 360 35 7

726485760 + 1452971520 726485760

w k k

k k k k

k k k k k

k k

6 2 8

2 1

2 10 3 2 3 2 16 6

1 1 1 1

3 5 2 5

2 1 1

5 2 7 2 9 10 4

2 1 1 1

77837760

1441440 21794572800 21794572800 7007

21794572800 1089728640 2179457280

1089728640 25945920 63423360 10090080

495331

k k

k k k k

k k k

k k k k

2 11 4 2 2 4 2

1 1 1

4 2 6 2 2 6 2

2 1 1

6 2 8 2 2 8 2 2 14 4

2 1 1 1

8 2 2 10 2

2 1

2 10897286400 10897286400

10897286400 1816214400 3632428800

1816214400 64864800 265945680 151800

64864800 4756752 18345

k k k

k k k

k k k k

k k

2 12 2 2 13 5

1 1

7 3 2 7 3 2 12 4 10 4

1 1 1 2

7 3 9 3 2 9 3 2 10 4

2 1 1 1

9 3 2 11 3 2 13

2 1 1

60 683760

518918400 1037836800 4444440 10090080

518918400 50450400 209008800 41801760

50450400 8792784 138600

k k

k k k k

k k k k

k k k

3 2 15 5

156056k

(6.53)


78

Results and Discussions

The formulation presented in this chapter, provides highly accurate solutions for the

problems. In Table 6.1, we have presented absolute errors, for example 6.1, at a spatial

domain [0,6] and 0.1, 0.5, 1, 1.5and 2 . It can be seen from Table

6.2 that the results are remarkable with just second order OHAM approximation. In Table

6.2, convergence of the OHAM solution is given, for example 6.1, through zeroth order,

first order and second order absolute error at time 1 and 0 6 . Here we

observe that the OHAM solution converges rapidly with the increase in the order of

approximation. Table 6.3 and Table 6.4 show much better results in case of example 6.2.

In Table 6.3, we have computed second order OHAM approximation for the spatial

domain [0,6] and 0.1, 0.5, 1, 1.5and 2 . Here the results are very

consistent with the increasing time. In case of example 6.2, Table 6.4 shows the faster

convergence of the OHAM solution by computing zeroth order, first order and second

order absolute errors at time 1 and 0 6 . The results for example 6.3 are

demonstrated in Tables 6.5-6.10. This example has been taken from Wazwaz paper [92],

which is a second order inhomogeneous nonlinear wave equation. Although we get good

results for small time in the time interval (0,1) but as the time increases, the accuracy of

OHAM decreases for the wave equation. We notice, from Table 6.6

( 0.01 and 0 6 ) and 8 ( 0.1 and 0 6 ), that the order of

convergence is reasonable. On the other hand, Tables 6.8-6.10

( 0.5, 1,2 and 0 6 ) show that the convergence of OHAM is not satisfactory.

In this chapter, we have seen the effectiveness of enhanced OHAM [39-43] to wave

equations. By applying the basic idea of OHAM to wave equations, we found it simpler

in applicability, more convenient to control convergence and involved less computational

overhead. Therefore, extended OHAM shows its validity and great potential for the

solution of time dependent problems in science and engineering.


79

Table 6.1

Absolute error of the obtained solution by OHAM for example 6.1 when time

0.1, 0.5, 1, 1.5and 2, and 0 6

0.1 0.5 1 1.5 2.0

0.0 2.06198×10-6

7.80998×10-6

2.60562×10-9

3.90505×10-5

1.24927×10-4

0.5 1.99967×10-6

7.60824×10-6

1.13368×10-7

3.7757×10-5

1.21353×10-4

1.0 1.81304×10-6

6.93345×10-6

2.22292×10-7

3.41158×10-5

1.10233×10-4

1.5 1.51368×10-6

5.82758×10-6

3.17396×10-7

2.83535×10-5

9.22598×10-5

2.0 1.1202×10-6

4.35937×10-6

3.92765×10-7

2.08284×10-5

6.85502×10-5

2.5 6.5708×10-7

2.62013×10-6

4.43714×10-7

1.20082×10-5

4.05785×10-5

3.0 1.53103×10-7

7.17972×10-7

4.67076×10-7

2.44138×10-6

1.00838×10-5

3.5 3.60394×10-7

1.22882×10-6

4.61396×10-7

7.2772×10-6

2.10378×10-5

4.0 8.51483×10-7

3.09922×10-6

4.27029×10-7

1.65433×10-5

5.08514×10-5

4.5 1.28963×10-6

4.77691×10-6

3.66112×10-7

2.47809×10-5

7.75033×10-5

5.0 1.6476×10-6

6.15761×10-6

2.82432×10-7

3.14777×10-5

9.93365×10-5

5.5 1.90312×10-6

7.15545×10-6

1.81191×10-7

3.62173×10-5

1.14993×10-4

6.0 2.04032×10-6

7.7084×10-6

6.86849×10-8

3.87051×10-5

1.23501×10-4

Table 6.2

Comparison of zeroth order, first order and second order absolute error

corresponding to example 1 at time 1, and 0 6

Zeroth Order

Absolute Error

First Order

Absolute Error

Second Order

Absolute Error

0.0 4.99792×10-2

6.25204×10-6

2.60562×10-9

0.5 4.87346×10-2

3.03133×10-4

1.13368×10-7

1.0 4.446×10-2

5.9367×10-4

2.22292×10-7

1.5 3.74211×10-2

8.47296×10-4

3.17396×10-7

2.0 2.80555×10-2

1.04824×10-3

3.92765×10-7

2.5 1.69455×10-2

1.18401×10-3

4.43714×10-7

3.0 4.782×10-3

1.24617×10-3

4.67076×10-7

3.5 7.67886×10-3

1.23084×10-3

4.61396×10-7

4.0 1.96623×10-2

1.13899×10-3

4.27029×10-7

4.5 3.04232×10-2

9.76316×10-4

3.66112×10-7

5.0 3.92926×10-2

7.52943×10-4

2.82432×10-7

5.5 4.57189×10-2

4.82756×10-4

1.81191×10-7

6.0 4.93026×10-2

1.82553×10-4

6.86849×10-8


80

Table 6.3

Absolute error of the obtained solution by OHAM for example 6.2 at time

0.1, 0.5, 1, 1.5, 2, and 0 6

0.1 0.5 1 1.5 2

0.0 0.0 0.0 0.0 0.0 0.0

0.5 6.7862×10-14

1.0904×10-12

7.6630×10-13

1.8438×10-11

1.8710×10-10

1.0 1.3156×10-13

2.1130×10-12

1.4850×10-12

3.5730×10-11

3.6255×10-10

1.5 1.8696×10-13

3.0042×10-12

2.1113×10-12

5.0800×10-11

5.1547×10-10

2.0 2.3070×10-13

3.7086×10-12

2.6064×10-12

6.2712×10-11

6.3633×10-10

2.5 2.6035×10-13

4.1824×10-12

2.9394×10-12

7.0724×10-11

7.1764×10-10

3.0 2.7367×10-13

4.3962×10-12

3.0898×10-12

7.4340×10-11

7.5432×10-10

3.5 2.6990×10-13

4.3365×10-12

3.0477×10-12

7.3333×10-11

7.4410×10-10

4.0 2.4958×10-13

4.0075×10-12

2.8164×10-12

6.7766×10-11

6.8762×10-10

4.5 2.1350×10-13

3.4292×10-12

2.4100×10-12

5.7987×10-11

5.8839×10-10

5.0 1.6420×10-13

2.6377×10-12

1.8537×10-12

4.4602×10-11

4.5257×10-10

5.5 1.0469×10-13

1.6822×10-12

1.1822×10-12

2.8444×10-11

2.8862×10-10

6.0 3.8747×10-14

6.2196×10-13

4.3710×10-13

1.0517×10-11

1.0672×10-10

Table 6.4

Comparison of zeroth, first and second order absolute error corresponding to

example 6.2 when time 1, and 0 6

Zeroth Order

Absolute Error

First Order

Absolute Error

Second Order

Absolute Error

0.0 0.0 0.0 0.0

0.5 3.09191×10-4

3.68105×10-8

7.66304×10-13

1.0 5.99157×10-4

7.13323×10-8

1.48503×10-12

1.5 8.51871×10-4

1.01419×10-7

2.11131×10-12

2.0 1.05162×10-3

1.252×10-7

2.60636×10-12

2.5 1.18598×10-3

1.41197×10-7

2.93943×10-12

3.0 1.24661×10-3

1.48414×10-7

3.08975×10-12

3.5 1.22973×10-3

1.46404×10-7

3.04767×10-12

4.0 1.13639×10-3

1.35292×10-7

2.81641×10-12

4.5 9.72389×10-4

1.15767×10-7

2.40996×10-12

5.0 7.47934×10-4

8.90449×10-8

1.85374×10-12

5.5 4.76977×10-4

5.67862×10-8

1.18222×10-12

6.0 1.76363×10-4

2.09968×10-8

4.37095×10-13


81

Table 6.5

Absolute error of the obtained solution by OHAM for example 6.3 when time

0.1, 0.5, 1, 1.5, 2 , and 0 6 .

0.1 0.5 1 1.5 2

0.0 5.16334×10-12

2.14516×10-6

5.74257×10-4

1.45222×10-2

1.40664×10-1

0.5 1.28195×10-10

4.25725×10-6

1.23371×10-3

3.30734×10-2

3.20379×10-1

1.0 2.63197×10-10

9.45678×10-6

3.05451×10-3

8.65331×10-2

8.25718×10-1

1.5 3.99907×10-10

1.92301×10-5

6.87982×10-3

2.03228×10-1

1.88292×100

2.0 5.3841×10-10

3.52893×10-5

1.37175×10-2

4.14331×10-1

3.66171×100

2.5 6.7882×10-10

5.95703×10-5

2.47234×10-2

7.5174×10-1

6.26611×100

3.0 8.21273×10-10

9.42299×10-5

4.11858×10-2

1.24634×100 9.76478×10

0

3.5 9.65931×10-10

1.41644×10-4

6.45095×10-2

1.92668×100 1.4261×10

1

4.0 1.11298×10-9

2.04405×10-4

9.62018×10-2

2.81803×100 2.00023×10

1

4.5 1.26263×10-9

2.8532×10-4

1.37858×10-1

3.94184×100 2.75294×10

1

5.0 1.41513×10-9

3.87408×10-4

1.91148×10-1

5.31559×100 3.7865×10

1

5.5 1.57072×10-9

5.13897×10-4

2.57802×10-1

6.95309×100 5.27424×10

1

6.0 1.7297×10-9

6.68224×10-4

3.39602×10-1

8.86505×100 7.48729×10

1

Table 6.6

Comparison of zeroth order, first order and second order absolute

error corresponding to example 6.3 at time 0.01, and 0 6

Zeroth Order

Absolute Error

First Order

Absolute Error

Second Order

Absolute Error

0.0 0.0 5.56111×10-15

1.10669×10-17

0.5 8.35417×10-8

8.30865×10-11

2.0548×10-13

1.0 1.675×10-7

1.66588×10-10

4.11985×10-13

1.5 2.51875×10-7

2.50499×10-10

6.19506×10-13

2.0 3.36667×10-7

3.3482×10-10

8.28042×10-13

2.5 4.21875×10-7

4.1955×10-10

1.0376×10-12

3.0 5.075×10-7

5.04689×10-10

1.24816×10-12

3.5 5.93542×10-7

5.90238×10-10

1.45974×10-12

4.0 6.8×10-7

6.76196×10-10

1.67234×10-12

4.5 7.66875×10-7

7.62564×10-10

1.88594×10-12

5.0 8.54167×10-7

8.49341×10-10

2.10058×10-12

5.5 9.41875×10-7

9.36527×10-10

2.31622×10-12

6.0 1.03×10-6

1.02412×10-9

2.53286×10-12


82

Table 6.7


corresponding to example 6.3 at time 0.1, and 0 6

Zeroth Order

Absolute Error

First Order

Absolute Error

Second Order

Absolute Error

0.0 0.0 5.56111×10-9

5.16334×10-12

0.5 8.54167×10-5

3.45935×10-8

1.28195×10-10

1.0 1.75×10-4

7.16668×10-8

2.63197×10-10

1.5 2.6875×10-4

1.05361×10-7

3.99907×10-10

2.0 3.66667×10-4

1.35379×10-7

5.3841×10-10

2.5 4.6875×10-4

1.61423×10-7

6.7882×10-10

3.0 5.75×10-4

1.83195×10-7

8.21273×10-10

3.5 6.85417×10-4

2.00398×10-7

9.65931×10-10

4.0 8.0×10-4

2.12734×10-7

1.11298×10-9

4.5 9.1875×10-4

2.19905×10-7

1.26263×10-9

5.0 1.04167×10-3

2.21615×10-7

1.41513×10-9

5.5 1.16875×10-3

2.17566×10-7

1.57072×10-9

6.0 1.3×10-3

2.07459×10-7

1.7297×10-9

Table 6.8



Zeroth Order

Absolute Error

First Order

Absolute Error

Second Order

Absolute Error

0.0 0.0 8.68923×10-5

2.14516×10-6

0.5 1.17188×10-2

2.63119×10-4

4.25725×10-6

1.0 2.60417×10-2

5.67031×10-4

9.45678×10-6

1.5 4.29688×10-2

1.02117×10-3

1.92301×10-5

2.0 6.25×10-2

1.64796×10-3

3.52893×10-5

2.5 8.46354×10-2

2.46971×10-3

5.95703×10-5

3.0 1.09375×10-1

3.50862×10-3

9.42299×10-5

3.5 1.36719×10-1

4.78679×10-3

1.41644×10-4

4.0 1.66667×10-1

6.32618×10-3

2.04405×10-4

4.5 1.99219×10-1

8.14865×10-3

2.8532×10-4

5.0 2.34375×10-1

1.0276×10-2

3.87408×10-4


83

Table 6.9



Zeroth Order

Absolute Error

First Order

Absolute Error

Second Order

Absolute Error

0.0 0.0 5.56111×10-3

5.74257×10-4

0.5 1.04167×10-1

1.34228×10-2

1.23371×10-3

1.0 2.5×10-1

3.05679×10-2

3.05451×10-3

1.5 4.375×10-1

5.95122×10-2

6.87982×10-3

2.0 6.66667×10-1

1.02655×10-1

1.37175×10-2

2.5 9.375×10-1

1.62282×10-1

2.47234×10-2

3.0 1.25×100 2.40559×10

-1 4.11858×10

-2

3.5 1.60417×100 3.39541×10

-1 6.45095×10

-2

4.0 2.0×100 4.61162×10

-1 9.62018×10

-2

4.5 2.4375×100 6.07243×10

-1 1.37858×10

-1

5.0 2.91667×100 7.7949×10

-1 1.91148×10

-1

5.5 3.4375×100 9.79492×10

-1 2.57802×10

-1

6.0 4.0×100 1.20872×10

0 3.39602×10

-1

Table 6.10



Zeroth Order

Absolute Error

First Order

Absolute Error

Second Order

Absolute Error

0.0 0.0 3.55911×10-1

1.40664×10-1

0.5 1. ×100 7.12941×10

-1 3.20379×10

-1

1.0 2.66667×100 1.61447×10

0 8.25718×10

-1

1.5 5.0×100 3.11558×10

0 1.88292×10

0

2.0 8.0 ×100 5.15271×10

0 3.66171×10

0

2.5 1.16667×101 7.54368×10

0 6.26611×10

0

3.0 1.6×101 9.98765×10

0 9.76478×10

0

3.5 2.1×101 1.20652×10

1 1.4261×10

1

4.0 2.66667×101 1.32381×10

1 2.0023×10

1

4.5 3.3×101 1.28497×10

1 2.5294×10

1

5.0 4.0×101 1.1247×10

1 3.7865×10

1

5.5 4.76667×101 4.16891×10

0 5.27424×10

1

6.0 5.6×101 6.0302×10

0 7.48729×10

1

Chapter 7 Conclusions

84

Chapter 7

Conclusions

OHAM [39-43] is applied to nonlinear initial and boundary value problems of fourth

order, eighth order, special fourth order, special sixth order, squeezing flow problem,

KdV and wave equations . Special boundary value problems involve a constant c . For the

case special fourth order BVPs, it has been observed that an increasing value c effect the

results badly while solving it analytically by Momani and Noor [64]. By using OHAM,

we get accurate results up to 1710c . Similarly for the case special sixth order BVPs, we

compared OHAM with HPM [65] , ADM [65] and DTM [65] . We observed that OHAM

solution is much better than HPM [65] , ADM [65] and DTM [65].

Further, OHAM formulation is tested on time dependent wave equations and KdV

equations. By applying the basic idea of OHAM we found it simpler in applicability,

more convenient to control convergence and involve less computational overhead.

Therefore, OHAM shows its validity and great potential for the solution of various initial

and boundary value problems in science and engineering. This method is very useful for

problems with large domain. This approach does not require discretization or perturbation

like other numerical and approximate methods. This confirms that the efficiency of the

OHAM gives it much wider applicability. Results show that with the increase in the order

of approximations, convergence of the OHAM solutions get faster and faster. This

method accurately approximates the solutions of partial differential equations at less

computational cost.

Further more we have the following observations with the application of OHAM:

1. OHAM is very efficient for solving a variety of linear and non-linear differential

equations.

2. OHAM fastly converges to the exact solution.

3. Marinca and Herisanu [39-43] provides a variety of options for computing siC in

the auxiliary function H p e-g Galerkin’s Method, Collocation Method, Method

of Least squares, Ritz Method. We can easily select a method, a convenient one,

for the desired goal.

Chapter 7 Conclusions

85

4. OHAM works very well with some of time dependent partial differential

equations using boundary conditions [45].

5. OHAM works very well for simple forcing function and initial conditions. But if

someone working with forcing functions/initial conditions involving

trigonometric, logarithmic or exponential functions, then it is extremely time

consuming to evaluate the residual for the increasing number of convergence

constants. Hence computing of more than three or four convergence constants is

not practical. Although OHAM can results best approximation but it will hardly

give the closed form solution because of the involvement of the convergence

constants siC in the auxiliary function ( )H p .

6. While solving problems to the corresponding initial conditions, the accuracy of

OHAM decreases for the increasing temporal domain.

7. HPM is a special case of OHAM when H p p with zero initial conditions

and HAM is a special case for H p p .

References

86

References

[1] R. Alt and J. Vignes, Validation of results of collocation methods for ODEs with the CADNA

library Original Research Article, Applied Numerical Mathematics, 21 (2) (1996), 119-139.

[2] H. Y. Hu, Z. C. Li ,Collocation methods for Poisson‟s equation Original Research Article,

Compute Methods in Applied Mechanics and Engineering, 195 (33, 36) (2006), 4139-4160.

[3] V. Thomée, From finite differences to finite elements: A short history of numerical analysis of

partial differential equations Original Research Article, Journal of Computational and Applied

Mathematics, 128(1-2) (2001) 1-54.

[4] J.H. Ferziger and M. Peric, Computational Methods for Fluid Dynamics, Springer, 1996.

[5] W Young, Hwon and Hyohoong Bank, The finite element method using MATLAB, CRC Press,

New York, 1996.

[6] M. Sharan, E.J. Kansa, S. Gupta, Application of the multiquadric method for numerical solution of

elliptic partial differential equations, Appl. Math. Comp. 1 (1994) 146–171.

[7] G. Fasshauer, Solving partial differential equations by collocation with radial basis functions,

Chamonix Proceedings, Vanderbilt University Press, Nashville, TN, 1996.

[8] C. Franke and R. Schaback, Solving partial differential equations by collocation using radial basis

functions, Appl. Math. Comput. 93 (1998) 73–82.

[9] C. Franke and R. Schaback, Convergence order estimates of meshless collocation methods using

radial basis functions, Adv. Comput. Math. 8 (1998) 381–399.

[10] G. Fasshauer, Solving differential equations with radial basis functions: multilevel methods and

smoothing, Adv. Comput. Math. 11 (1999) 139–159.

[11] Siraj-ul-Islam, Arshed Ali and Sirajul Haq , A computational modeling of the behavior of the two-

dimensional reaction–diffusion Brusselator system Applied Mathematical Modeling, 34(12) ,

(2010), 3896-3909.

[12] Siraj-ul-Islam, Sirajul Haq and Arshed Ali, A meshfree method for the numerical solution of the

RLW equation, Journal of Computational and Applied Mathematics, 223( 2), (2009), 997-1012.

[13] Marjan Uddin, Sirajul Haq and Siraj-ul-Islam, Numerical solution of complex modified Korteweg–

de Vries equation by mesh-free collocation method, Computers & Mathematics with Applications,

58( 3), (2009), 566-578.

[14] Marjan Uddin, Sirajul Haq and Siraj-ul-Islam, A mesh-free numerical method for solution of the

family of Kuramoto–Sivashinsky, equations Applied Mathematics and Computation, 212( 2)

, 2009, 458-469.

[15] Sirajul Haq, Siraj-ul-Islam and Marjan Uddin, A mesh-free method for the numerical solution of

the KdV–Burgers equation, Applied Mathematical Modelling, 33(8), (2009), 3442-3449.

[16] Sirajul Haq, Siraj-ul-Islam and Marjan Uddin, A meshfree interpolation method for the numerical

solution of the coupled nonlinear partial differential equations, Engineering Analysis with

Boundary Elements, 33( 3), (2009), 99-409.

[17] Siraj-Ul Islam, Sirajul Haq and Javid Ali , Numerical solution of special 12th-order boundary

value problems using differential transform method, Communications in Nonlinear Science and

Numerical Simulation, 14, (4), (2009), 1132-1138.

[18] S.J. Liao. Beyond perturbation. CRC Press, Boca Raton, 2003.

http://www.sciencedirect.com/science?_ob=ArticleURL&_udi=B6TYD-3VTJGW2-1&_user=4335646&_coverDate=06%2F30%2F1996&_alid=1473133559&_rdoc=9&_fmt=high&_orig=search&_origin=search&_zone=rslt_list_item&_cdi=5616&_st=13&_docanchor=&view=c&_ct=14426&_acct=C000062921&_version=1&_urlVersion=0&_userid=4335646&md5=1e5bee9432054f59760abbe49607ffcd&searchtype=a

http://www.sciencedirect.com/science?_ob=ArticleURL&_udi=B6TYD-3VTJGW2-1&_user=4335646&_coverDate=06%2F30%2F1996&_alid=1473133559&_rdoc=9&_fmt=high&_orig=search&_origin=search&_zone=rslt_list_item&_cdi=5616&_st=13&_docanchor=&view=c&_ct=14426&_acct=C000062921&_version=1&_urlVersion=0&_userid=4335646&md5=1e5bee9432054f59760abbe49607ffcd&searchtype=a

http://www.sciencedirect.com/science?_ob=ArticleURL&_udi=B6V29-4HG6H81-1&_user=4335646&_coverDate=07%2F01%2F2006&_alid=1473133559&_rdoc=20&_fmt=high&_orig=search&_origin=search&_zone=rslt_list_item&_cdi=5697&_st=13&_docanchor=&view=c&_ct=14426&_acct=C000062921&_version=1&_urlVersion=0&_userid=4335646&md5=639239cda3f86b5a6f02565245ec6885&searchtype=a

http://www.sciencedirect.com/science?_ob=ArticleURL&_udi=B6TYH-42DP215-3&_user=4335646&_coverDate=03%2F01%2F2001&_alid=1473149822&_rdoc=1&_fmt=high&_orig=search&_origin=search&_zone=rslt_list_item&_cdi=5619&_sort=r&_st=13&_docanchor=&view=c&_ct=51663&_acct=C000062921&_version=1&_urlVersion=0&_userid=4335646&md5=8d9065e515a72e4d2a402703ff78c9a1&searchtype=a



http://www.sciencedirect.com/science?_ob=ArticleURL&_udi=B6TYC-4YX002B-1&_user=10&_coverDate=12%2F31%2F2010&_alid=1389820549&_rdoc=1&_fmt=high&_orig=search&_cdi=5615&_sort=r&_docanchor=&view=c&_ct=3&_acct=C000050221&_version=1&_urlVersion=0&_userid=10&md5=109b8dd6e83939d5d4d219a467458cb5

http://www.sciencedirect.com/science?_ob=ArticleURL&_udi=B6TYC-4YX002B-1&_user=10&_coverDate=12%2F31%2F2010&_alid=1389820549&_rdoc=1&_fmt=high&_orig=search&_cdi=5615&_sort=r&_docanchor=&view=c&_ct=3&_acct=C000050221&_version=1&_urlVersion=0&_userid=10&md5=109b8dd6e83939d5d4d219a467458cb5

http://www.sciencedirect.com/science?_ob=ArticleURL&_udi=B6TYH-4S6G8YV-1&_user=10&_coverDate=01%2F15%2F2009&_alid=1389820549&_rdoc=2&_fmt=high&_orig=search&_cdi=5619&_sort=r&_docanchor=&view=c&_ct=3&_acct=C000050221&_version=1&_urlVersion=0&_userid=10&md5=b4169b3482dd2d4a97300425befb9e14



http://www.sciencedirect.com/science?_ob=ArticleURL&_udi=B6TYJ-4W7B53Y-1&_user=10&_coverDate=08%2F31%2F2009&_alid=1389824754&_rdoc=2&_fmt=high&_orig=search&_cdi=5620&_sort=r&_docanchor=&view=c&_ct=8&_acct=C000050221&_version=1&_urlVersion=0&_userid=10&md5=31f50a6397bee3b2e428cfcd68da3c80

http://www.sciencedirect.com/science?_ob=ArticleURL&_udi=B6TYJ-4W7B53Y-1&_user=10&_coverDate=08%2F31%2F2009&_alid=1389824754&_rdoc=2&_fmt=high&_orig=search&_cdi=5620&_sort=r&_docanchor=&view=c&_ct=8&_acct=C000050221&_version=1&_urlVersion=0&_userid=10&md5=31f50a6397bee3b2e428cfcd68da3c80

http://www.sciencedirect.com/science?_ob=ArticleURL&_udi=B6TY8-4VPM59K-4&_user=10&_coverDate=06%2F15%2F2009&_alid=1389824754&_rdoc=5&_fmt=high&_orig=search&_cdi=5612&_sort=r&_docanchor=&view=c&_ct=8&_acct=C000050221&_version=1&_urlVersion=0&_userid=10&md5=c49a822762dea2b786f5d4f5a8dadf9c

http://www.sciencedirect.com/science?_ob=ArticleURL&_udi=B6TY8-4VPM59K-4&_user=10&_coverDate=06%2F15%2F2009&_alid=1389824754&_rdoc=5&_fmt=high&_orig=search&_cdi=5612&_sort=r&_docanchor=&view=c&_ct=8&_acct=C000050221&_version=1&_urlVersion=0&_userid=10&md5=c49a822762dea2b786f5d4f5a8dadf9c

http://www.sciencedirect.com/science?_ob=ArticleURL&_udi=B6TYC-4V3546B-1&_user=10&_coverDate=08%2F31%2F2009&_alid=1389824754&_rdoc=6&_fmt=high&_orig=search&_cdi=5615&_sort=r&_docanchor=&view=c&_ct=8&_acct=C000050221&_version=1&_urlVersion=0&_userid=10&md5=9d824979cd6e1033cddeb3770b65f2a2

http://www.sciencedirect.com/science?_ob=ArticleURL&_udi=B6TYC-4V3546B-1&_user=10&_coverDate=08%2F31%2F2009&_alid=1389824754&_rdoc=6&_fmt=high&_orig=search&_cdi=5615&_sort=r&_docanchor=&view=c&_ct=8&_acct=C000050221&_version=1&_urlVersion=0&_userid=10&md5=9d824979cd6e1033cddeb3770b65f2a2

http://www.sciencedirect.com/science?_ob=ArticleURL&_udi=B6V2N-4T6CF17-3&_user=10&_coverDate=03%2F31%2F2009&_alid=1389824754&_rdoc=8&_fmt=high&_orig=search&_cdi=5707&_sort=r&_docanchor=&view=c&_ct=8&_acct=C000050221&_version=1&_urlVersion=0&_userid=10&md5=e21eac8af61a38875074d86d1e012325

http://www.sciencedirect.com/science?_ob=ArticleURL&_udi=B6V2N-4T6CF17-3&_user=10&_coverDate=03%2F31%2F2009&_alid=1389824754&_rdoc=8&_fmt=high&_orig=search&_cdi=5707&_sort=r&_docanchor=&view=c&_ct=8&_acct=C000050221&_version=1&_urlVersion=0&_userid=10&md5=e21eac8af61a38875074d86d1e012325

http://www.sciencedirect.com/science?_ob=ArticleURL&_udi=B6X3D-4S26651-1&_user=10&_coverDate=04%2F30%2F2009&_alid=1389824754&_rdoc=4&_fmt=high&_orig=search&_cdi=7296&_sort=r&_docanchor=&view=c&_ct=8&_acct=C000050221&_version=1&_urlVersion=0&_userid=10&md5=146babd4bfb005defe29664e15c97116

http://www.sciencedirect.com/science?_ob=ArticleURL&_udi=B6X3D-4S26651-1&_user=10&_coverDate=04%2F30%2F2009&_alid=1389824754&_rdoc=4&_fmt=high&_orig=search&_cdi=7296&_sort=r&_docanchor=&view=c&_ct=8&_acct=C000050221&_version=1&_urlVersion=0&_userid=10&md5=146babd4bfb005defe29664e15c97116

References

87

[19] F.B. Bao and J.Z. Lin. Linear stability analysis for various forms of one-dimensional Burnett

equations, Int. J. Non-linear Sci. Num. Simul. 6(3), (2005), 295-305.

[20] X.P. Chen, W.X. He and B.T. Jin, Symmetric boundary knots method for membrane vibrations

under mixed-type boundary conditions, Int. J. Nonl. Sci. Num. Simul., 6(4), (2005) 421-429.

[21] J.R. Acton and P.T. Squire, Solving equations with physical understanding, Adam Hilger Ltd,

Boston, 1985.

[22] J. Awrejcewicz, I.V. Andrianov and L.I. Manevitch, Asymptotic approaches in nonlinear

dynamics: New trends and applications, Springer-Verlag, Heidelberg, 1998.

[23] R. Bellman, Perturbation techniques in mathematics, physics and engineering, Holt Rinehart and

Winston, Inc. New York, 1964.

[24] M.N. Bogoliubov and A. Yu, Mitropolsky, Asymptotic method in the theory of nonlinear

oscillations, Gordon and Breach, London, 1985.

[25] R.E. Mickens, An introduction to nonlinear oscillations, Cambridge University Press, Cambridge,

1981

[26] A.H. Nayfeh, Introduction to perturbation methods, John Wiley, New York, 1981.

[27] A.H. Nayfeh, Problems in perturbation, John Wiley, New York, 1985.

[28] R.H. Rand and D. Armbruster, Perturbation methods, bifurcation theory and computer algebraic.

Springer; 1987.

[29] J.H. He. Some asymptotic methods for strongly non linear equations, Int. J. Mod. Phys., B20 (10),

(2006), 1141-1199.

[30] S.J. Liao, A second-order approximate analytical solution of a simple pendulum by the process

analysis method. ASME J. Appl. Mech. 59 (1992,) 970-975.

[31] S.J. Liao, The proposed homotopy analysis technique for the solution of nonlinear problems. PhD

thesis, Shanghai Jiao Tong University, 1992.

[32] S.J. Liao, A kind of linearity-invariance under homotopy and some simple applications of it in

mechanics. Technical Report 520, Institute of Shipbuilding, University of Hamburg, Jan. 1992.

[33] S.J. Liao, A kind of approximate solution technique, which does not depend upon small parameters

(II): an application in fluid mechanics. Int.J. of Non-Linear Mech., 32 (1997), 815–822.

[34] S.J. Liao, An explicit, totally analytic approximation of Blasius viscous flow problems. Int. J. of

Non-Linear Mech., 34(4):759–778, 1999, (reprinted with permission from Elsevier).

[35] S.J. Liao, A new analytic algorithm of Lane-Emden equation. Applied Mathematics and

Computation, 142(1), (2003) 1–16.

[36] S.J. Liao, A short review on the homotopy analysis method in fluid mechanics, Journal of

Hydrodynamics, 22(5) (2010), 882-884.

[37] P.J. Hilton, An introduction to homotopy theory, Cambridge University Press, 1953.

[38] J.H. He. Homotopy perturbation technique. Comput. Meth. Appl. Mech. Eng., 178 (1999), 257-

262.

[39] V. Marinca, N.Herisanu, I. Nemes, An Optimal Homotopy Asymptotic Method with application to

thin film flow, Central European Journal of Physics 6(3), (2008), 648-653.

[40] V. Marinca, N. Herisanu, C. Bota and B. Marinca, An Optimal Homotopy Asymptotic Method applied to

steady flow of a fourth-grade fluid past a porous plate, Applied Mathematics Letters 22(2009) 245-251.

http://www.sciencedirect.com/science?_ob=ArticleURL&_udi=B8CX5-51FPWXC-50&_user=4335646&_coverDate=10%2F31%2F2010&_alid=1661640750&_rdoc=56&_fmt=high&_orig=search&_origin=search&_zone=rslt_list_item&_cdi=40077&_sort=r&_st=13&_docanchor=&view=c&_ct=8446&_acct=C000062921&_version=1&_urlVersion=0&_userid=4335646&md5=15cc1d4b521d3437208180f0fa59ba67&searchtype=a

References

88

[41] V. Marinca and N. Herisanu, Application of Optimal Homotopy Asymptotic Method for solving

nonlinear equations arising in heat transfer, International Communications in Heat and Mass

Transfer 35, (2008), 710-715.

[42] V. Marinca and N. Herisanu, Determination of periodic solutions for the motion of a particle on a

rotating parabola by means of the Optimal Homotopy Asymptotic Method , Journal of Sound and

Vibration , doi: 10. 1016 / j. jsv. 2009. 11. 005.

[43] N. Herisanu, V. Marinca, T. Dordea and G. Madescu, A new analytical approach to nonlinear

vibration of an electric machine, Proc. Romanian. Acad. Series A: Mathematics, Physics,

Technical Sciences, Information Science, 9(3) (2008)229-236.

[44] Javed Ali, S. Islam, Sirajul Islam, Gul Zaman, The solution of multipoint boundary value problems

by the Optimal Homotopy Asymptotic Method, Computers and Mathematics with Application,

59(6), (2010), 2000-2006.

[45] S. Iqbal, M. Idrees, A.M Siddiqi and A. Ansari, Some solutions of the linear and nonlinear Klein-

Gordon equations using the optimal homotopy asymptotic method, Applied Mathematics and

Computation, 216 (2010) 2898-2909.

[46] M. Esmaeilpour , D.D. Ganji, Solution of the Jeffery-Hamel flow problem by optimal homotopy

asymptotic method, Computers and Mathematics with Applications, 59 (2010) 3405-3411.

[47] M. Babaelahi, M., D. D. Ganji, A. A. Jonidi, Analytical treatment of mixed convection flow past

vertical flat plate, Journal of Thermal Science, 14 (2), (2010), 409-416.

[48] S. J. Liao, An optimal homotopy-analysis approach for strongly nonlinear differential equations.

Commun Nonlinear Sci Numer Simul 2009; doi:10.1016/j.cnsns.2009 .09.002, Online.

[49] Z. Niu and C. Wang. A one-step optimal homotopy analysis method for nonlinear differential

equations. Commun Nonlinear Sci Numer Simul. Online. doi:10.1016/j.cnsns.2009.08.014.

[50] C. Wang and Z. Niu, Reply to “Comments on „A one-step optimal homotopy analysis method for

nonlinear differential equations‟ Commun Nonlinear Sci Numer Simul. Online. doi

10.1016/j.cnsns.2009.12.026.

[51] Erik Sweet, A dissertation submitted in partial fulfillment of the requirements for the degree of

Doctor of Philosophy in the Department of Mathematics in the College of Sciences at the

University of Central Florida Orlando, Florida, 2009.

[52] S Talaat and El-Danaf, Quartic Nonpolynomial Spline Solutions for Third Order Two-Point

Boundary Value Problem, World Academy of Science, Engineering and Technology, (45) 2008.E.

[53] Babolian, A. Azizi, J. Saeidian, Some notes on using the homotopy perturbation method for

solving time-dependent differential equations, Mathematical and Computer Modelling,

doi:10.1016/j.mcm.2009.03.003.

[54] S.M. Momani, Some problems in non-Newtonian fluid mechanics, Ph.D. Thesis, Walse University,

United Kingdom, 1991.

[55] T.F. Ma, J. Silva, Iteration solution for a beam equation with nonlinear boundary conditions of

third order, Applied Mathematics and Computation 159 (1) (2004) 11–18.

[56] M.M. Chawla, C.P. Katti, Finite difference methods for two-point boundary-value problems

involving higher order differential equation, BIT 19 (1979) 27–33.

References

89

[57] A Boutayeb, H Twizell E, Numerical methods for the solution of special sixth-order boundary

value problems, Int J Comput Math, 45 (1992), 207-233.

[58] Aslam Noor and S. T. Mohyud-Din, Variational iteration technique for solving higher order

boundary value problems, Appl Math Comp 189(2007), 1929-1942.

[59] G.A. Glatzmaier, Numerical simulations of stellar convection dynamics at the convection zone,

Fluid Dyn, 31(1985), 137-150.

[60] S. Chandrasekhar, Hydrodynamic hydromagnetic stability, Clarendon Press, Oxford, 1961.

[61] M.A. Noor and S.T. Mohyud-Din, An efficient method for fourth- order boundary value problems,

Comput. and Math. with applications 54(2007) 1101-1111.

[62] L. Xu, The variational method for fourth order boundary value problems, Chaos, Solitons and

Fractals 39(2009) 1386-1394.

[63] S. Liang and D.J. Jeffrey, An efficient analytical approach for solving fourth order boundary value

problems, Computer Physics Communications 10.1016/j.cpc.2009.06.006.

[64] S. Momani and M.A. Noor, Numerical comparison of methods for solving a special fourth-order

boundary value problem, Appl. Math. Comput. 191(2007) 218-224.

[65] S. Momani, M.A. Noor and S. T. Mohyud-Din, Numerical methods for solving a special sixth-

order boundary value problem, Nonlinear Analysis forum, 13(2), (2008), 125-134.

[66] Golbabai and M. Javidi, Application of homotopy perturbation method for solving eighth-order

boundary value problems, Appl. Math. Comput. 191(2007) 334-346.

[67] J. Stefan, Akad and Wiss. Math. Natur. Wien 69(1874), 713-735.

[68] R.J. Grimm. Squeezing flows of Newtonian liquid films an analysis include the fluid inertia. Appl.

Sci. Res. 32 (1976), 149-146.

[69] F.R. Archibald. Load capacity and time relations in squeeze films. Trans. ASME J. Lub. Tech.

78(1956), 29-35.

[70] W.A. Wolfe, Squeeze film pressures, Appl. Sci. Res, 14 (1964), 77-90.

[71] D.C. Kuzma, Fluid inertia effects in squeeze films. Appl. Sci. Res, 18(1967), 15-20.

[72] S. Ishizawa, Squeezing flows of Newtonian liquid films an analysis include the fluid Inertia. Appl.

Sci. Res. 32(1976), 149-166.

[73] C.Y. Wang and L.T. Watson. Squeezing of a viscous fluid between elliptic plates. Appl. Sci. Res.

35(1979), 195-207.

[74] J.D. Jackson, A study of squeezing flow, Appl. Sci. Res, 11(1962), 148-152.

[75] R.M. Terrill and W.O. Winer, Inertial considerations in parallel circular squeeze film bearings,

Trans. ASME J. Lub, 92(1970), 588-592.

[76] R. Usha and R. Sridharan, Arbitrary squeezing of a viscous fluid between elliptic plates. Fluid

Dynamic Res. 18(1996), 35-51.

[77] W.F. Hughes and R.A. Elco. Magneto hydrodynamic lubrication flow between parallel rotating

disks, J. Fluid Mech, 13 (1962), 21-32.

[78] S. Kamiyama, Inertia effects in MUD hydrostatic thrust bearing. ASME J. Lub. Tech., 91(4)

(1969), 589-596.

[79] E.A. Hamza. The magneto hydrodynamic squeeze film. ASME J. Trib., 110(2) (1988), 375-377.

[80] S. Bhattacharya and A. Pal, Unsteady MHD squeezing flow between two parallel rotating disks,

References

90

Mech. Research. Commun. 24(1997), 615-623.

[81] D. Halpern and T.W. Secomb, The squeezing of red blood cells through parallel sided channels

with near minimal widths, J.Fluid Mech. 244 (1992), 307-322.

[82] Q.K.Ghori,M.Ahmed and A.M.Siddiqui, Application of Perturbation Method to Squeezing Flow of

a Newtonian Fluid, International Journal of Nonlinear sciences and Numerical

Simulation,8(2)(2007),179-184.

[83] X.J. Ran, Q.Y. Zhu and Y. Li, An explicit series solution of the squeezing flow between two

infinite parallel plates, Comm. Nonlinear. Sci. Num. Simul doi:10.1016/j.cnsns. (2007).

[84] X.Y. Wang, Exact and explicit solitary wave solutions for the generalized Fisher equation, Physics

Letters A, 131 (4/5) (1988) 277-279.

[85] Jeffrey, M.N.B. Mohammad, Exact solutions to the KdV-Burgers equation, Wave Motion 14

(1991) 369-375.

[86] M. Wadati, The exact solution of the modified Korteweg-de Vries equation, J. Phys. Soc. Jpn. 32

(1972) 1681-1687.

[87] D.J. Korteweg and G. de Vries, On the change of form of long waves advancing in a rectangular

canal, and on a new type of long stationary wave, Philos. Mag. 39 (1895) 422–443.

[88] Dogan Kaya and Mohammed Aassila, Application for a generalized KdV equation by the

decomposition method, Physics Letters A, 299(2002) 201-206.

[89] T.A. Abassy,A.Magdy, El-Tawil and H. El-Zoheiry, Exact solutions of some nonlinear partial

differential equations using the variational iteration method linked with Laplace transforms

andthe Pad´e technique, Computers and Mathematics with Applications, doi:10.1016/j. camwa.

2006.12.067

[90] F. Kangalgil and F. Ayaz, Solitary wave solutions for the KdV and mKdV equationsby differential

transform method, Chaos, Solitons and Fractals, doi: 10.1016/j. chaos. 2008. 02. 009.

[91] A. A. Hemeda. Variational Iteration Method for solving wave equation Computer and

Mathematics with Application, 56(2008), 1948-1953.

[92] M. Wazwaz. The Variational Iteration Method: A reliable analytic tool for solving linear and

nonlinear wave equations, Computer and Mathematics with Application 54 (2007), 926-932.

[93] S. Haq, M. Idrees, S. Islam, Application of Optimal Homotopy Asymptotic Method to eighth order

boundary value problems, International Journal of Applied Mathematics and Computation, 2(4),

38-47, 2010.

[94] M. Idrees, S. Haq, S. Islam, Application of Optimal Homotopy Asymptotic Method to fourth order

boundary value problems, World Applied Sciences Journal, 9(2): 131-137, 2010.

[95] M. Idrees, Sirajul Haq, S. Islam, Application of the Optimal Homotopy Asymptotic Method to

squeezing flow, Computers and Mathematics with Applications, 59(12) , 3858-3866, 2010.

[96] M. Idrees, S. Haq, S. Islam, Application of Optimal Homotopy Asymptotic Method to special sixth

order boundary value problems, World Applied Sciences Journal, 9(2): 138-143, 2010.

Documents

Application of Optimal Homotopy Asymptotic Method to