ANALYSIS AND APPLICATION OF OPTIMAL HOMOTOPY

ASYMPTOTIC METHOD AND THE USE OF DAFTARDAR-

JAFFERY POLYNOMIALS

Thesis

Presented to the Department of Mathematics

Islamia College Peshawar(Chartered University)

by

Javed Ali

In Partial Fulfillment

of the Requirements for the Degree of

Doctor of Philosophy

In

Mathematics

Supervisor: Dr. Saeed Islam

Co Supervisor: Prof. Dr. Syed Inayat Ali shah

DEPARTMENT OF MATHEMATICS, ISLAMIA COLLEGE PESHAWAR (A

CHARTERED UNIVERSITY), KHYBER PAKHTUNKHWA, PAKISTAN

July 2013

ii

AUTHOR’S DECLARATION

I declare that the work in this thesis was carried out in accordance with the regulation

of the Islamia College Peshawar(a chartered university). The work is original except

where indicated by special reference in the text and no part of the thesis has been

submitted for any other degree either in Pakistan or overseas.

Javed Ali

ICP (CU)

iii

CERTIFICATE

We accept the work contained in this thesis as conforming to the required

standard for the partial fulfillment of the degree of

DOCTOR OF PHILOSOPHY

In the subject of

MATHEMATICS

1.____________________________ 2._____________________________

Supervisor Co-Supervisor

3.___________________________ 4.__________________________

External Examiner I External Examiner II

5.___________________________ 6.___________________________

Chairman Dean

Faculty of Numerical & Physical Sciences

DEPARTMENT OF MATHEMATICS

ISLAMIA COLLEGE PESHAWAR (CHARTERED UNIVERSITY)

KHYBER PAKHTOONKHAWA, PAKISTAN

iv

DEDICATIONS

I dedicate this work to my parents

Hanif Ghulam and Maimoona Jan

v

ACKNOWLEDGEMENT

All praises to Allah that induced the man with intelligence, knowledge and wisdom.

Peace and blessings of Allah be upon the Last Prophet Hazrat Muhammad (S.A.W)

who exhorts his followers to seek for knowledge from cradle to grave.

I acknowledge with immense pleasure and gratitude for the zealous interest and

constant encouragement and inspiration of my supervisor, Dr. Saeed Isalam,

Associate Professor, Department of Mathematics Abdul Wali Khan University

Mardan, Khyber Pakhtoonkhawa, Pakistan and co-supervisor, Prof. Dr. Syed Inayat

Ali Shah, Chairman Department of Mathematics Islamia College Peshawar (Chartered

University), Khyber Pakhtoonkhawa, Pakistan.

I am thankful to them for supervising my research work and giving me a rich

knowledge of this field, of which I had a little knowledge.

I am also thankful to Dr. Siraj-ul-Islam for his endless encouragement and

inspiration.

I would like to express my appreciation to the Ex. Vice Chancellor Ajmal Khan,

present Vice Chancellor, Dean, Faculty of Numerical and Physical Sciences,

Chairman and staff of the Department of Mathematics for their support and dedication

towards my Ph.D. affairs.

Sincere and warm thanks to my friend Hamid Khan for his support and encouragement.

Last but not least, my deepest gratitude goes to my parents for their endless love,

prayers and encouragement that enable me to reach this mile-stone in my life. My

gratitude is also due to my brothers, my sisters and their families for their loving

support

My deepest and special thanks go to my wife and children for their constant support

and patience.

JAVED ALI

vi

ABSTRACT

In this thesis, we use the standard optimal homotopy asymptotic method for the

analytical solution of higher order ordinary differential equations with two or multi-

point boundary conditions, a nonlinear family of partial differential equations with an

initial condition and Integro-differential equations. The results obtained are compared

with the results obtained by the application of Adomain decomposition method,

homotopy perturbation method, variational iteration method, differential transform

method and homotopy analysis method etc. The optimal homotopy asymptotic

method uses a flexible auxiliary function, which controls the convergence of the

solution and allows adjustment in the convergence region where it is needed.

Moreover, the procedure of this method is simple, clearly well-defined and the built-

in recursive relations are explicitly worked out for easy use. Numerical results

obtained by the optimal homotopy asymptotic method reveal high accuracy and

excellent agreement with the exact solutions.

Apart from the application of optimal homotopy asymtotic method, we develop a

new scheme by using Daftardar-Jafari polynomials in the homotopy of optimal

homotopy asymptotic method to solve nonlinear problems more effectively. This

scheme is almost as simple as the optimal homotopy asymtotic method but its results

are more accurate than the usual optimal homotopy asymptotic method. To show the

effectiveness of this scheme, we solve different bench mark problems from literature

and compare the results with those obtained by the standard optimal homotopy

asymptotic method.

To determine the convergence control parameters of the auxiliary function, two well

known methods are followed: The Least squares and the Galerkin methods.

We also explore and use a variety of the forms of auxiliary functions that results

more accuracy and shows flexibility and reliability of the methods. For symbolic

computation, we use Mathematica 7.

The work presented in chapters 2, 3 and 4 of this thesis has been published in

different reputed international journals and the work presented in chapters 3, 5, 6 and

7 is submitted for possible publication. The details of published/submitted work are

included in the list of publications.

vii

CONTENTS

Chapter No. Title Page No.

1. INTRODUCTION AND LITERATURE SURVEY…………… 1

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

1.2 Literature survey…………….…………………………….. 2

1.3 Some basics of homotopy ………………………………... 4

1.4 Intruduction to two well known homotopy methods………. 7

1.4.1 Homotopy Analysis Method………………………… 7

1.4.2 Homotopy Perturbation Method…………………….. 9

1.5 Objective of the thesis….………………………………….. 10

1.6 Organization of the thesis………………………………….. 10

2. NUMERICAL SOLUTION OF THE MULTI-POINT

BOUNDARY VALUE PROBLEMS USING OHAM…………… 11

2.1 Introduction…………………..…………………………… 11

2.2 Analysis of OHAM…………………………………… 11

2.3 Numerical examples……………………………………… 13

3. NUMERICAL SOLUTIONS OF THE TWO POINT

BOUNDARY VALUE PROBLEMS USING OHAM………… 21

3.1 Introduction………………………………………………. 21

3.2 Analysis of the method ………………………………….. 21

3.3 Numerical examples ………………………………. 22

4. NUMERICAL SOLUTIONS OF THE STEADY TWO DIMENSIONAL

RADIAL FLOW OF VISCOUS FLUID USING OHAM & DTM…. 50

4.1 Introduction……………………………………………… 50

4.2 Formulation of the problem……………………………… 50

4.3 Application of OHAM…………………………………… 51

4.4 Differential Transform Method………………………….. 54

4.4.1 Analysis of the method ………………………….. 55

viii

4.4.2 Application of the method……………………….. 56

5. OPTIMAL HOMOTOPY ASYMPTOTIC METHOD FOR THE

SOLUTION OF ZAKHROV-KUZNETSOV EQUATION…. 58

5.1 Introduction……………………………………………….. 58

5.2 Analysis of the method for a PDE…………………………….. 59

5.3 Application…………………………………………......... 60

6. APPLICATION OF OHAM TO INTEGRO-DIFFERENTIAL

EQUATIONS..…………………………………………………. 68

6.1 Introduction……………………………………………….. 68

6.2 Analysis of the method …………………………………… 68

6.3 Numerical examples …………………………………... 71

7. APPLICATION OF OHAM WITH DAFTARDAR-JAFARI

POLYNOMIALS FOR NON-LINEAR PROBLEMS…. 80

7.1 Introduction……………………………………………… 80

7.2 Analysis of the method for differential equations ……… 80

7.3 An illustration about the polynomials of OHAM and DJM 80

7.4 Analysis of the method for nth-order Integro-differential

equations……………………………………………….. 83

7.5 Application of OHAM-DJ……………………………… 86

7.5.1 Ordinary differential equations………………… 86

7.5.2 Partial differential equations…………………… 89

7.5.3 Integro-differential equations…………………... 92

CONCLUSION………………………………………………….. 100

LIST OF PUBLICATION……………………………………… 99

REFERENCES………………………………………………….. 103

1

Chapter 1

INTRODUCTION AND LITERATURE SURVEY

1.1 INTRODUCTION

Most of the physical systems are modeled by either initial or boundary value

problems or both. These problems appear in many areas like physics, astronomy,

population modeling, astrophysics, fluid dynamics, hydrodynamics, biomathematics,

engineering and technology etc. They are also used in finance. Applications of first

and second order ordinary differential equations (ODEs) can be found in many books

of undergraduate level. Third order problems model, Alfven waves in rotating

inhomogeneous plasmas, the shear deformation of a sandwich beams, draining and

coating flows etc. The problems of visco-elastic and inelastic flows, deformations of

elastic beams with simply supported ends in an equilibrium state, transverse

vibrations of hinged beams, plate bending on an elastic foundation and the deflection

of a plate are modeled by fourth order ordinary differential equations. The boundary

value problems (bvps) of fifth order appear in the of visco-elastic flows. Sixth and

eighth-order differential equation arise in stability problems of hydrodynamics. When

a horizontal infinite layer of a fluid is subjected to heat from below then under the

action of rotation, instability sets in. When this instability sets in as ordinary

convection, there arise sixth order ordinary differential equation and when the

instability sets in as overstability, there arise an eighth order ordinary differential

equation. If an infinite horizontal layer of fluid is heated from below, with the

supposition that a uniform magnetic field is also applied across the fluid in the same

direction as gravity and the fluid is subject to the action of rotation, instability sets in.

When instability sets in as ordinary convection, it is modeled by tenth-order bvp and

when instability sets in as oversitbility, it is modeled by twelfth-order bvps [1, 2].

It is fair to say that only the simple physical systems can be modeled by ODEs,

whereas most of the problems are modeled by partial differential equations(PDEs).

The range of application of PDEs is enormous as compared to ODEs. The most

simple PDEs are: the wave equation which governs the motion of a vibrating string or

membrane or a vibrating solid, gas or liquid, the heat equation which gives

temperature, the Laplace equation which arises in the heat flow and electrostatic

2

potential in the steady state etc and the Poisson equation which describes almost the

same physical situations as Laplace equation. The Schrodinger equation which is

second order in three spatial variables describes the quantum mechanical

wavefunction of a non-relativistic particle. The Helmholtz equation governs elastic

waves in solids and electromagnetic waves. In the study of nonlinear dispersive

waves, a famous equation arises which is known as the KdV equation. This equation

was derived in the modeling of shallow water in canal, in 1895 by Korteweg and de

Vries (briefly written as KDV). A special case of the KdV equation is the Burgers

equation which governs the turbulence and it also appears in the approximate theory

of flow through a shock wave traveling in viscous fluid. The generalizations of the

KdV equations yield the two well-known equations: The Kadomtsov–Petviashivilli

(KP) equation and the Zakharov–Kuznetsov (ZK) equation.

1.2 LITERATURE SURVEY

In this modern age of science and engineering, we are not equipped fully to solve

functional equations exactly and as a last option we knock at the door of numerical

analysis. Numerical analysis is a rich subject which can effectively address almost all

the problems of the day. For the functional equations, numerical analysis employ two

approaches i.e, pure numerical approach and semi numerical or analytical approach.

Both of these approaches have their own advantages and disadvantages. Scientists and

engineers follow one or both the approaches to resolve and study their mathematical

models accurately for the better understanding and application. Some well known

pure numerical methods for differential equations are: the classical RK-methods,

shooting method, FDM, FEM, BEM, RBF methods and methods based on wavelets.

Analytical methods contain: Adomian decomposition method (ADM)[3, 4],

variational iteration method (VIM)[5], homotopy perturbation method (HPM)[6,8],

homotopy analysis method(HAM)[9, 10], Differential transform method (DTM)[11,

12] etc. These methods have certain advantages over the commonly used numerical

methods. In numerical methods discretization is used due to which different errors

raise and consequently the accuracy is lost. They also require huge computer memory

and a lot of time. On other hand, the analytical methods do not discretize variables

and such they are free the discritization error. They, also, do not require much cost or

time. Moreover, the analytical methods produce analytical solutions that are

differential over the domain of the problem and they are amenable for mathematical

3

treatments. These methods develope series solution and in certain cases the

developing series leads to the closed form solution.

Perturbation methods assume small/large parameters which restrict their

applications. Consequently, the classical perturbation techniques cannot produce a

general form of an approximate solution and fail to handle properly those problems

which do not have small parameters at all or having strong non-linearity. Some new

perturbation methods like artificial parameter method, perturbation incremental

method, HAM, HPM, parameterized perturbation method, -method and

bookkeeping artificial parameter perturbation method do exist which do not depend

on the small parameter.

The ADM and DTM which are non-perturbation methods, deal strongly

nonlinearity in problems effectively but the region of convergence of the developed

solution is usually small. DTM is very effective in small domains. HPM combines

homotopy from topology and perturbation techniques. It overcomes the problems

related to the existence and size of the small or large parameters in the problems. This

method has been extensively used by many authors for different problems. But there

may arise unexpectedly the problem of convergence as reported in [13]. The HAM

deals nonlinear problems in a more flexible fashion due to the auxiliary parameter

which control the convergence of the solution. An curve is drawn, although

approximately, and the most suitable value of it chosen to control the convergence.

Recently, a new homotopy based method, the optimal homotopy asymptotic method

(OHAM), has been introduced by Vasile Marinca et al. [14-17]. These authors used a

more flexible function, named as auxiliary function, to control the convergence of the

developing series. Unlike HAM, the authors reported straight forward methods for the

determination of auxiliary constants that control the convergence of the solution. They

explicitly defined all the recursive relations that can be handled easily. By the virtue

of the more flexible auxiliary function, this method gives more accurate and reliable

results than HAM and HPM.

In the next lines we mention some important work that has already been done for

the solution of higher order differential equations. Most of this work includes the

analytical approach for the problems.

4

Wazwaz [18-20], used ADM for solving of fifth & sixth order problems. Noor et al.

[21-26] investigated these problems by using VIM, HPM, Variational Iteration

Decomposition Method(VIDM). Modified Variational Iteration Method (MVIM).

Songxin Liang et al. [27], used HAM for solving parameterized sixth-order bvp.

Siraj-ul-Islam et al. [28, 29], used non-polynomial spline approach for solving fifth

and sixth order bvps. S. T. Mohyud-Din et al. [30], used iterative method (ITM) for

such type of problems. K. N. S. Kasi Viswanadham et al. [31], used Quintic B-Spline

Galerkin scheme for fifth order bvps. Che Haziqah et al.[32] and Javed Ali [33], used

DTM for solving fifth and sixth order bvps. Noor et al. [34] used VIM, Inc et al. [35],

used ADM and A. Golbabai et al. [36], used HPM for solving eighth-order bvps.

Ninth, tenth, eleventh and twelfth order bvps have been solved successfully by many

authors using pure numerical or analytical techniques. Detail of these techniques is

referred to [37-40]. Siraj-Ul-Islam et al. [12], applied DTM for solving special twelfth

order boundary value problems.

Biazar et al. [41-43], used VIM to obtain an analytical approximation to the solution

of wave equation. The same authors obtained exact solutions for nonlinear Schrdinger

equations and Helmholtz equation by using HPM. For heat equation Rezanai et al.

[44], used HPM and VIM. D. Ganji studied some nonlinear equations arising in heat

transfer using HPM [45]. To solve electrostatic potential differential equations, Zhang

et al. [46[ used HPM. Biazar et al. [47], obtained exact solutions of Poisson equation

by using VIM and ADM. Coupled Burgers and simple Burgers equations have been

solved by M. A. Abdou et al. [48], using VIM. In [49-51], Wazwaz studied nonlinear

dispersive Zakharov–Kuznetsov(ZK) equations. The same author also used the

extended tanh method [52], for the same equations. In [54], the ZK equations are

solved exactly by using the ADM.

We use the usual (standard) OHAM and OHAM with Daftardar-Jafari polynomials

to find the approximate analytic solutions of; (i) Ordinary differential equations with

two point boundary conditions and multi-point boundary conditions (ii) A partial

differential equation with initial condition.(iii) Integro-differential equations.

1.3 SOME BASICS OF HOMOTOPY

A homotopy between two continuous functions 1f and 2f from a topological space X

to a topological space Y is defined to be a continuous function:

5

H : X × [0,1] → Y,

from the space 0,1X to Y , such that, for all points X , 1( ,0) ( )H f and

2( ,1) ( ).H f

Two spaces are said to be homotopy equivalent if they can be transformed into one

another by the operations of shrinking, expanding and bending.

For ,X Y R and 1 2, : f f R R , are two continuous functions then an another

continuous function : [0, 1]H R R , containing 1f and

2f can be defined by

1 2( , )   (1 ) ( )    ( )H x t t f x t f x , x R which satisfy 1( ,0) ( )H x f x and ( ,1)H x

2 ( )f x , where [0, 1]t . The function ( , )H x t is an example of a homotopy. The

following figure may be helpful to view graphically the function : [0, 1]H R R .

The above fogure Shows that, when 1 2( ,0) ( ,1) then ( ) ( ).H x H x f x f x

From the above definition of homotopy, the continues function ( , )H t associates

with two continuous paths, 1( )f and 2 ( )f , a function of two variables and t

that is equal to 1( )f when t = 0 and equal to 2 ( )f when t = 1. This fact give the

idea of gradual deformation of 1f to 2f without leaving the region as t changes from

0 to 1.

6

This key idea is used to find or approximate 2f when

1f is known in all homotopy

based methods. Great caution should be taken as the deformation of 1f to

2f , when

2f is unknown, does not always goes to 2f . It is the hot issue for all the researchers

and scientists who work in this area.

Many homotopy based computational methods have been developed for algebraic

equations, differential equations, integro-differential equations and integral equations.

The homotopy continuation method is used for algebraic equations and the HAM,

HPM and OHAM are used for differential equations, integro-differential equations

and integral equations. The later three are the most well known methods which

develpos series solutions to the problems. But practically none of these methods,

actually, leads to the exact solutions. However, for some typical examples some

authors have obtained exact solutions by observing the terms of the developing series.

The HAM, HPM, and OHAM can also be used for algebraic equations as well.

As the above methods are homotopy based; it is more important rather than the

obtained results of the methods that weather the deformation of the known 1f (initial

solution) goes to the desired value of 2f (the expected exact solution), which

guarantees the convergence of the developed solution series or else the deformation

goes to some unexpected function . In the terminology of numerical analysis, it is

equivalent to say that weather the homotopy based computational scheme converges

or diverges when p starts from o to 1. In some cases the above discussed deformation

may lead to the exact value of 2f but it is not always the case. It depends upon many

reasons such as the selection of 1f , the type of homotopy used, the iterative scheme

itself, and the problem under investigation.

From physical point of view, the convergence of solution is much more important

than whether or not the used analytic method itself is independent of small/large

physical parameters. If one does not keep this issue in mind, some useless results

might be obtained. This issue arises in HPM while in HAM the convergence is

controlled by the auxiliary function ( )r , where ( )r is a smoothing function of r.

(It has been observed that in almost all research articles/letters ( )r is assumed to be

1. It can lead to more good approximations for other values of 1 as well). We will

7

present a more flexible auxiliary function in OHAM which control the convergence

more sharply than HAM.

In the next section, we describe the two more important and well known homotopy

methods i,e the HAM and the HPM, to equip the readers to compare these methods

with the newly developed method, the optimal homotopy asymptotic method, which

will be presented in chapter 3.

1.4 INTRODUCTION TO TWO WELL-KNOWN HOMOTOPY

METHODS

1.4.1 Homotopy Analysis Method (HAM)

It was introduced by Liao in his Ph.D. thesis, in 1992. This is very effective method in

handling linear and non-linear problems.

To understand the basic idea of HAM, let us consider a differential equation of the

form:

( ) ( ), ,A u f r r (1.1)

with conditions:

( , / ) 0, .B u u n r (1.2)

Here A is considered as a general differential operator, B is taken as a boundary

operator, )(rf is a known analytic function, is the domain’s “ ” boundary.

Generally the operator A is splitted into linear “L” and non-linear “N” parts.

Eqn. (1.1) is then written as:

)()()( rfuNuL (1.3)

According to this method a homotopy is constructed by ( , ) : [0,1]H v p R ,

which satisfy the following equation:

01 ( ( , )) ( ) ( ) ( ( , )) ( )p L v r p L u p r A v r p f r (1.4)

In HAM, the above equation is called zeroth-order deformation equation. The

1,0p is called embedding parameter, is known as convergence control

parameter, ( )r is the smoothing function and 0u is the initial approximation

(solution/guess) satisfying the boundary conditions. When the parameter 0p and

1p , equation (1.4) becomes: 0( ,0) ( )L v L u and ( ,1) ( ) ( ) 0, L v A u f r

8

respectively. So when p starts from 0 to 1, the function ( , )v r p starts tracing

continuously from 0 ( )u r to the desired curve of solution ( ).u r

Next, we use the Taylor series to expand , pv r about p , in the following manner:

0

1

( , ) ( ) ( ) ,m

m

m

v r p u r u r p

(1.5)

where

0

1 ( , )( )

!

m

m m

p

v r pu r

m p

The convergence of (1.5) is dependent upon the parameter . Liao [18–20], found

that there often exists such an effective-region hE that any hE , gives a convergent

series solution. This effective-region is constructed, although approximately, using the

initial/boundary conditions of the problem under consideration.

For example, one may approximately determine hE by plotting curves (0) ,u

(0)u and so on. These curves are called “ -curves” or “curves for convergence-

control parameter”.

We assume that for some value of , the series (1.5), converges at 1p , then

0

1

( ,1) ( ) ( ) ( )m

m

v r u r u r u r

Now at this stage the following vectors are defined:

0 1{ ( ), ( ), ..., ( )}n nu u r u r u r ,

and then the zeroth-order deformation equation (1.3) is differentiated m-times with

respect to p . The obtained result is then divided by m!, which gives the following

mth-order deformation equation for p = 0:

1 1[ ( ) ( )] ( )m m m m mL u r u r u ,

where

1

1 1

0

1 ( ( , ))( )

( 1)!

m

m m m

p

A v r pu

m p

and 0, 1,

1, 1.m

m

m

9

1.4.2 Homotopy Perturbation Method (HPM)

It was proposed by Ji-Huan He in, 1998. Considering again the general equation:

( ) ( ), ,A u f r r (1.6)

with boundary conditions:

( , / ) 0, ,B u u n r (1.7)

where A , B, )(rf , , and have the same meanings as stated earlier in HAM.

We write Eq. (1.6) as:

)()()( rfuNuL (1.8)

According to this method a homotopy is constructed by ( , ) : [0,1]H v p R ,

which satisfy the following equation:

0, 1 ( ) ( ) ( ) ( ) ( ) 0H v p p L v L u p L v N v f r , (1.9)

where 1,0p , is known as embedding parameter and 0 ( )u r is the initial solution

which satisfy the given conditions.

Now from Equation (1.9) we have:

0)()(0, 0 uLvLvH and ,1 ( ) ( ) ( ) 0H v L v N v f r .

or

)()(0, 0uLvLvH and ,1 ( ) ( ) ( ) ( ) ( )H v L v N v f r A u f r

Next, we expand ( , )v r p in the perturbation series in the following manner:

0

( , ) m

m

m

v r p p u

We plug in this series into (1.9) and then equate the like powers of p . In this way we

obtain explicit equations which can be integrated directly.

Letting then 1p , we have

10

( , ) ( ,1) mp

m

u Lim v r p v r u

.

Practically the truncated series is considered for approximations.

0

M

i

i

u u

This solution is called as Mth-order approximate solution.

10

1.5 OBJECTIVE OF THE THESIS

The objectives of this thesis are many. Firstly, we apply OHAM to various problems

like ODEs, a non-linear family of PDEs and integro differential equations(IDEs).

Secondly, the convergence controlling constants determined by using Least squares

method and the Galerkin’s method are examined for better accuracy. The use of

different auxiliary function to improve the accuracy of the results is also investigated.

Thirdly, we give a new idea to use Daftardar-Jafari polynomials instead of the usual

polynomials of OHAM while dealing non-linear problems. This absolutely new

approach is easily applicable, explicitly defined and provides greater accuracy than

usual OHAM, HPM and the HAM.

1.6 ORGANIZATION OF THE THESIS

The organization of the thesis is as follows: In chapter 2, we use OHAM to multi-

point boundary value problems, in chapter 3, we use OHAM to a variety of higher

order ODEs, in chapter 4, a fluid flow through a wedge shape region(Hamel’s flow) is

presented and the non-linear ODE is solved using OHAM and DTM. In chapter 5, we

use OHAM to solve analytically the nonlinear dispersive Zakharov–Kuznetsov

equation. In chapter 6, the application of OHAM is made to IDEs and finally, in the

last chapter 7, Daftardar-Jaffari polynomials are used in OHAM for the solution of

various functional equations to exhibit the easy applicability and high accuracy of the

new scheme.

11

Chapter 2

NUMERICAL SOLUTION OF THE MULTI VALUED

BOUNDARY PROBLEMS USING OHAM

2.1 INTRODUCTION

This chapter includes the solution of second, third and fourth-order multi-valued

boundary value problems by OHAM. These bvps arise in modeling the variations of a

guy wire of uniform cross-section with different densities at N parts. Multipoint

problems appear in the theory of elastic stability [55]. In [56-58], some theoretical

work has been done regarding these problems. In literature, a little work has been

noted for the numerical solutions of such problems. For numerical work we refer the

reader to see [59-61].

The numerical results obtained by the application of OHAM are compared with the

exact solution where available and the results obtained by ADM and HPM to show

the effectiveness and accuracy of the method.

2.2 THE OPTIMAL HOMOTOPY ASYMPTOTIC METHOD(OHAM)

Consider the following differential equation:

( ( )) ( ) ( ( )) 0, , 0du

L u x g x N u x B udx

. (2.1)

where L is a linear operator, x denotes independent variable, ( )u x is an unknown

function , ( )g x is a known function, N is a nonlinear operator and B is a boundary

operator.

According to OHAM we construct a homotopy ( , ), : 0,1H v x p p which

satisfies

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

,, , 0,

p L v x p g x h p L v x p g x N v x p

v x pB v x p

x

(2.2)

where x and 0,1p is an embedding parameter, ( )h p is a nonzero auxiliary

function for 0p , (0) 0h and ,v x p is an unknown function. Obviously, when

0p and 1p it holds that 0,0 and ,1v x u x v x u x respectively.

12

Thus, as p varies from 0 to 1 , the solution ( , )v x p approaches from 0 ( )u x to ( )u x ,

where 0 ( )u x is obtained from Eq (2.2) for 0p and we have

00 0( ) ( ) 0, , 0

duL u x g x B u

dx

. (2.3)

Next we choose auxiliary function h p in the form

2

1 2 ...h p pC p C (2.4)

where 1 2,, ...C C are parameters to be determined. ( )h p can be expressed in many

forms as reported by V.Marinca et al.[14-17]”.

To get an approximate solution, we expand , , iv x p C in Taylor’s series about p in

the following manner:

0 1 2

1

, , , , ,..., k

i k k

k

v x p C u x u x C C C p

. (2.5)

Substituting Eq. (2.5) into Eq. (2.2) and equating the coefficient of like powers of p ,

we obtain the following linear equations.

Zeroth order problem is given by Eq. (2.3) and the first order problem is given by Eq.

(2.6):

11 1 0 0 1, , 0.

duL u x g x C N u x B u

dx

(2.6)

The general governing equations for ku x are given by:

1 0 0

1

0 1 1

1

, ,..., ,

2,3,..., , 0

k k k

k

i k i k i k

i

kk

L u x L u x C N u x

C L u x N u x u x u x

duk B u

dx

(2.7)

where 0 1, ,...,m mN u x u x u x is the coefficient of mp in the expansion of

,N v x p about the embedding parameter .p

0 0 0 1 2

1

, , , , ,..., m

i m m

m

N v x p C N u x N u u u u p

. (2.8)

It has been observed that the convergence of the series (2.5) depends upon the

auxiliary parameters 1 2,, ...C C .If it is convergent at 1p , one has

13

0 1 2

1

, , , ,..., .i k k

k

v x C u x u x C C C

(2.9)

The result of the mth-order approximations are given by

1 2 0 1 2

1

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

m i i

i

u x C C C u x u x C C C

. (2.10)

Substituting Eq. (2.10) into Eq. (2.1), it results the following residual:

1 2 1 2 1 2, , ,..., ( ( , , ,..., )) ( ) ( ( , , ,..., ))m m mR x C C C L u x C C C g x N u x C C C (2.11)

If 0, thenR u will be the exact solution. Generally, it does not happen, especially

in nonlinear problems.

In order to find the optimal values of , 1,2,3,...iC i , we first construct the

functional,

2

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

b

m m

a

J C C C R x C C C dx (2.12)

and then minimizing it, we have

1 2

... 0.m

J J J

C C C

(2.13)

Here anda b are in the domain of the problem. With these known parameters, the

approximate solution (of order m ) is well-determined.

The method discussed above, for determination of the optimal values of the

auxiliary parameters, is the method of Least Squares. Besides this method, Marinca et

al. [14-17], also reported other methods for this purpose. The other methods contain

the Galerkin’s method, the Ritz’s method and the collocation method.

2.3 NUMERICAL EXAMPLES

Example 1: Third order linear problem

2 0,

(0) (1) 0, (0.5) 0.

u k u a

u u u

(2.14)

Here, the physical constants are 5 and 1, k a [59]. The function ( )u x shows the

shear deformation of sandwich beams. This problem has the following exact solution

3 2 3

1( ) (sinh sinh ) ( ) (cosh cosh ) tanh

2 2 2 2

a k a a k ku x kx x kx

k k k

Applying the method, we obtain:

14

Zero-order problem:

0 0 0( ) 1, (0) (1) 0, (0.5)=0.u x u u u (2.15)

It gives us

2 3

0( ) 0.04167 0. 0.25 0.16667 .u x x x x (2.16)

First order problem:

1 1 1 1 0 1 0

1 1 1

( , ) 1 (1 ) ( ) 25 ( ),

(0) (1) 0, (0.5) 0.

u x C C C u x C u x

u u u

(2.16)

The obtained solution is

2 2

1 1 1

4 5

1 1

1( , ) 0.1041667 0. 0. 0.520833

0.520833 0.208333 .

u x C C x x C x

C x C x

(2.17)

Second order problem:

2 1 2 2 1 1 1 2 0

1 1 1 2 0 2 2 2

( , , ) (1 ) ( , ) 25 C ( )

25 ( , ) ( ), (0) (1) 0, (0.5) 0.

u x C C C C u x C u x

C u x C C u x u u u

(2.18)

We obtain the following solution:

2

2 1 1 2

2 2 2 2 4

1 1 2 1

2 4 4 5 2 5

1 2 1 1

5 2 6 2 7

2 1 1

1 2( , ) 0.104167 0.367684 0.104167

0.520833 1.82292 0.520833 0.520833

1.60590 0.520833 0.208333 0.208333

0.208333 0.434028 0.12400

,

.8

u x C C C C

C x C x C x C x

C x C x C x C x

C x C x C x

C

(2.19)

Third order problem:

3 1 2 3 1 2 1 2 1 3 1 2 3

2 1 1 2 1 1 3 3 0 3 0

( , , , ) (1 ) ( , , ) 25 ( , , , )

( , ) 25 ( , ) ( ) 25 ( ),

u x C C C C u x C C C u x C C C

C u x C C u x C C C u x C u x

(2.20)

3 3 3(0) (1) 0, (0.5) 0. u u u

We obtain the following solution:

2 3

3 1 1 1 2

2 2 2 2

1 2 3 1 1

3 2 2 2 2

1 2 1 2 3

1 1

1 2

4 2

3( , ) 0.104167 0.735367 1.298604 0.104167

0.62066 0.104167 0. 0. 0.520833 3.645833

6.418961 0.520833 3.072917 0.520833

0.52

,

0833 3.2118056

,u x C C C C C

C C C x x C x C x

C x C x C C x

C

C x

x C

C

C

4 3 4 4

1 2

4 4 5 2 5

1 2 3 1 1

3 5 5 5 5

1 2 1 2 3

2 6 3 6 6 2 7

1 1 1 2 1

5.403646 0.520833

2.647569 0.520833 0.208333 0.416667

0.208333 0.208333 0.208333 0.208333

0.868056 1.772280 0.850694 0.248016

0.2

x C x C x

C C x C x C x C x

C x C x C C x C x

C x C x C C x C x

3 7 7 3 8 3 9

1 1 2 1 148016 0.243056 0.193762 0.043058 .C x C C x C x C x

(2.21)

15

Adding up the solutions (2.15), (2.17), (2.19) and (2.21), we get the following third

order approximation/solution, for 1p :

1 2 3 0 1 1 2 1 2 3 1 2 3( , , , ) ( ) ( , ) ( , , ) ( , , , ).u x C C C u x u x C u x C C u x C C C (2.22)

Following the procedure for determining 'iC s , we get the following values for

0, 1a b :

1 2 30.587734742 , 0.140650653 , 0.141306736.C C C

Using these values, solution (2.22) becomes:

2 3 4

5 6 7 8 9

( ) 0.0121071 0.0986628 0.166667 0.205258

0.205679 0.160292 0.0982484 0.0393381 0.0087418 .

u x x x x

x x x x x

(2.23)

Numerical results for this solution are displayed in table 1.

Table 1: In this table, we compare the numerical results of solution (2.23) and the

results of numerical solution based on 'Pade approximants [61].

Table 1

x Exact Sol. OHAM Sol. Error *Error

0.0 -0.0121071 -0.0121071 1.298E-10 6.653E-05

0.1 -0.0112665 -0.0112665 -3.099E-09 6.500E-05

0.2 -0.00922221 -0.00922221 6.959E-09 5.254E-05

0.3 -0.00646687 -0.00646687 1.086E-09 3.630E-05

0.4 -0.00332019 -0.00332018 -1.065E-08 1.875E-05

0.5 0.00 3.70662 E-18 -6.155E-17 ***

0.6 0.00332019 0.00332018 1.065E-08 1.734E-05

0.7 0.00646687 0.00646687 -1.086E-09 3.405E-05

0.8 0.00922221 0.00922221 -6.959E-09 4.980E-05

0.9 0.0112665 0.0112665 3.099E-09 6.201E-05

1.0 0.0121071 0.0121071 -1.298E-10 6.347E-05

Error=Exact-OHAM

Error*=Exact-Pade Approximants[61]

16

Figure 1, shows exact solution represented by solid curve and the OHAM solution

By the dotted curve over the region -2<x<3, enclosing domain of the problem.

Figure 1

Example 2: Fourth order nonlinear problem

Consider the following problem along with four conditions [59]:

(4) 7( ) ( ) ( ) 4 24 0u x u x u x x , (2.24)

(0) 0, (0.25) 6, (0.5) 3, (1) 1.u u u u

The exact solution is 4( )u x x .

By means of OHAM, the following solutions of zeroth, first and second order

problems are obtained:

2 3

0

8

6

4

( ) 0.000451433 0.0000523461 1.27157 10

( ).

u x x

O

x

x

x

x

(2.25)

7 2

1 1 1

7 2 6 3 8 3

1 1

9 5 8

1

( , ) 0.000451433 8.69509 10 0.0000523461

5.0327 10 1.27157 10 7.44941 10

1.69826 10 ( ).

u x C x C x

O x

x

C x x C x

C x

(2.26)

10 2 7

2 1 2 1 1 2

5 2 10 2 2 7 2 6 3

1 1 2 1

9 2 3 8 3 9 5 9 2 5

1 2 1 1

( , , ) 0.000450563 1.06141 10 8.6951 10

5.1843 10 5.3071 10 5.0327 10 1.1981 10

1.0614 10 7.4494 10 3.3965 10 1.6983 10

1.6983

u x C C C x C x C x

C x C x C x C x

C x C x C x C x

9 5 10 6 10 6

2 1 2

12 7 12 7

1 2

8

10 1.9692 10 1.9692 10

9.2576 10 9.2576 ( ).10

C x C x C x

C C xx Ox

(2.27)

17

Adding up the solutions (2.25), (2.26) & (2.27), the following second order solution is

considered for our approximation:

1 2 0 1 1 2 1 2( , , ) ( ) ( , ) ( , , ). u x C C u x u x C u x C C (for p =1)

For 0, 1, a b we follow the procedure stated in section 2.2, and obtain the

following optimal values of the auxiliary parameters:

13 11

1 22.94458 10 and 9.9164 10 .C C

Using these values, the second order solution now becomes:

17 17 2 18 3

4 19 5 20 6 17 7.

( ) 5.24705 10 6.39007 10 7.05627 10

1.6990 10 1.95857 10 9.20741 10

u x x x x

x x x x

(2.28)

Numerical results for this solution are displayed in table 2.

Table 2: In this table, we display the numerical values of the exact solution, OHAM

solution (2.28) and the error between them.

Table 2

E*=Exact-Approx.

x Exact sol. OHAM sol. E*

0.0 0.00 0.00 0.00

0.1 0.0001 0.0001 -4.6151E-18

0.2 0.0016 0.0016 -7.9946E-18

0.3 0.0081 0.0081 -1.0181E-17

0.4 0.0256 0.0256 -1.1218E-17

0.5 0.0625 0.0625 -1.1148E-17

0.6 0.1296 0.1296 -1.0016E-17

0.7 0.2401 0.2401 -7.8692E-18

0.8 0.4096 0.4096 -4.7537E-18

0.9 0.6561 0.6561 -7.1902E-19

1.0 1.00 1.00 4.1836E-18

18

In figure 2, excellent agreement can be seen between the exact solution and the

OHAM solution for the values: -100<x<100.

Figure 2: Red solid curve-exact sol & Dotted curve-OHAM sol.

Example 3: Second order nonlinear problem

We consider second-order nonlinear problem with three conditions [62]:

23 2( ) ( ) ( ) 1 0, 0 1,

8 1089

(0) 0, (1 3) (1).

u x u x u x x

u u u

(2.29)

For 1p , the second order OHAM approximation is:

2

1 1 2 11 2

2 22 31 2 1 1 2

2

1 1

6566 30015838 3283 82 1( , , )

3 88209 864536409 88209 2 9801

104480 4 3235 179618441 3235

288178803 9801 39204 4610860848 78408

3235 31628851

104544 20492714

C C C Cu x C C x

C C C C Cx x

C C

2 24 5 62 1 13235 2030933 2030933

88 209088 2732361984 10929447936.

C C Cx x x

(2.30)

For 0, 1, the following values of ' are obtained:ia b C s

1 21.034937306 , 4.141101662.CC

Knowing these values, the approximate solution now becomes:

2 3

4 5 6

( ) 0.706567121 -0.500456984 0.043730752

0.01551444 0.000796133 0.000199033 .

u x x x x

x x x

(2.31)

We solve the same problem by HPM to compare the results with the results of OHAM.

The obtained HPM solution is:

19

2 3 4

5 6

610695134 288440291 200856379 31783619( ) - -

864536409 576357606 4610860848 2049271488

2030933 2030933 - .

2732361984 10929447936

u x x x x x

x x

(2.32)

Table 3: In this table, we present the numerical values of solution (2.31), HPM solution

(2.32) and solution by ADM[63].

Table 3

x OHAM sol. HPM sol. ADM sol.

0.0 0.00 0.00 0.00

0.1 0.0656 0.0656 0.0656

0.2 0.121 0.1209 0.1209

0.3 0.1659 0.1658 0.1658

0.4 0.2002 0.2001 0.2001

0.5 0.2237 0.2236 0.2236

0.6 0.2364 0.2363 0.2363

0.7 0.2382 0.2381 0.2382

0.8 0.2291 0.2291 0.2291

0.9 0.2092 0.2091 0.2091

Table 4: In this table, we display the numerical values of the residual R using both

the solutions of OHAM and HPM to compare the efficiency of the methods. The

proposed method shows greater accuracy than HPM and ADM.

Table 4

x R(OHAM) R(HPM)

0.0 2.904E-06 9.018E-06

0.1 1.103E-06 1.004E-04

0.2 -7.372E-08 1.860E-04

0.3 -5.032E-07 2.612E-04

0.4 -3.761E-07 3.222E-04

0.5 -9.272E-09 3.659E-04

0.6 2.949E-07 3.901E-04

0.7 3.448E-07 3.936E-04

0.8 1.115E-07 3.762E-04

0.9 -2.641E-07 3.388E-04

20

Figure 3: Solution plots of the exact solution and the solution by OHAM for example 3,

for the values: -4<x<5.

21

Chapter 3

NUMERICAL SOLUTION OF THE TWO POINT BOUNDARY

PROBLEMS USING OHAM

3.1 INTRODUCTION

This chapter includes the solution of the special fifth, sixth, parameterized sixth,

eighth, ninth, tenth, eleventh and twelfth-order bvps by OHAM. The results of

OHAM are more accurate than those obtained using ADM, HPM, HAM, VIM, and

the variational iteration method using He’s polynomials (VIMHP) etc. OHAM

contains the auxiliary function, ( , )iH p C which provides us with a simple way to

adjust and control the convergence region of solution series. It develops a fast

converging series solution to a wide class of differential equations.

The results obtained using OHAM show remarkable low error and have excellent

agreement with exact solutions. In this work, we also use some other forms of the

auxiliary function to show the flexibility and efficiency of the method.

3.2 ANALYSIS OF THE METHOD

We use the same homotopy as stated in chapter 2, and follow the rest of procedure of

OHAM. In this chapter we use a variety of auxiliary functions to examine the

flexibility and reliability of the method. The following forms of the auxiliary function

will be used:

2

1 2

2 3

1 2 3

1 2

1 2

( )( )

( )

x

pC p C

pC p C p Ch p

p C C e

p C C x

By means of OHAM, we obtain linear problems that are directly integrable. To find

the optimal values of iC which minimizes R over the domain of the problem, we

follow two methods: the method of Least Squares that is already stated in chapter 2,

and the Galerekin’s method. In accordance with the Galerekin’s method, we solve the

following system and the values of 'iC s are obtained.

1 2

0, 0, . . . , 0

b b b

ma a a

u u uR dx R dx R dx

C C C

22

3.3 NUMERICAL EXAMPLES

Example 1: Fifth order linear bvp

( ) ( ) 15 10 , 0 1,v x xx u e xe xu

(0) 0, (0) 1, (0) 0, (1) 0, (1) .u u u u u e

Exact solution of this problem is: ( ) (1 ) .xu x x x e

First we use the usually used auxiliary function, 2

1 2( )h p pC p C .

Following the rest of procedure of OHAM, the following linear problems are

obtained:

Zeroth order problem:

(5)

0

0 0 00 0

( ) 0,

(0) 0, (0) 0, (0) 0, (1) 0, (1) . u u u u eu

u x

(3.1)

First order problem:

1

(5) (5)

1 1 1 0 1 0

1 1 1 1

( , (15 10 ( )) (1 ) ( ),

(

)

0) 0, (0) 0, (0) 0, (1) 0, (1) . 0

x xu x C C e xe x C u x

u u

u

u u u

(3.2)

Second order problem:

(5)

2 2

(5)

2 0 1 1 1 1 0

2 2 2 2

1

2

15 10 ( )) ( , ) (1 ) (( , , ) ),

(0) 0, (0) 0, (0) 0, (1) 0

(

, (1) 0.

x xu Cx C C e xe x C x C Cu u

u u

u

u

x

u u

(3.3)

Solutions of the problems (3.1), (3.2) and (3.3), are given by (3.4), (3.5) and (3.6):

3 3 4 4

0( ) .3 2u x ex x x x e x (3.4)

1 1

2 3

4 6 8 9

1 35 35 25 10 7.5 1.051582859

0.59323767 0.001388889 0.00004192 0.00004

( ,

750

)

5

(

)

x xe x e x x x

x x

C

x

C

x

u x

(3.5)

2 3

1

4 6 8 9

2 3 4

2

6 8 2

1 2

9

1

2 35 35 25 10 7.5 1.051583

0.593237685 0.001388889 0.00004192 0.0000475

(35 35 25 10 7.5 1.051583 0.5932377

0.001388889 0.0000419

( , , ) (

2 0.0000475 (50

)

)

x x

x x

C x x e x x

x x x x

C e x

u x

x e x x x

x x x C

C C e

3 4 5 6

7 8 9 8 11

2 10 13 10 14

50

50 8.1151161 2.468193 0.29166667 0.0361111

0.0029761 0.00011456 0.00008674 2.5052 10

25 2.714 10 1.9774 1 ).0

xe

x x x x x

x x x x

x x x

(3.6)

We consider the following second order solution for our approximation:

0 1 1 2 1 2( ) ( ) ( , ) ( , , ). u x u x u x C u x C C (3.7)

For a = 0, b = 1, method of Least squares gives the following optimal values:

23

1 20.912019009, 0.0077753602C C .

By considering these values (3.7) becomes:

3 4 5 6 7

8 9 10 6 11

7 12 8 13 9 14 15

( ) 0.5 0.333334 0.124999 0.0333332 0.00694441

0.00119137 0.000172901 0.0000221613 2.49696 10

2.52 2.563 10 10 10 ( )29636 .2.05622

u x x x x x x x

x x x x

x x x O x

(3.8)

Numerical results for the solution (3.8) are displayed in Table 1.

We now use the auxiliary function, 1 2( ) ( )h p p C C x .

The following problems are obtained:

Zeroth order problem:

(5)

0

0 0 00 0

( ) 0,

(0) 0, (0) 0, (0) 0, (1) 0, (1) . u u u u eu

u x

First order problem:

(5)

1 1 1 1

2 2 2 2

2 2 2 0

2

( , 5 ( ),

(0) 0, (0) 0,

, ) (3 2 )( ) ( )

(0) 0, (1) 0, (1) 0.

xu x C e C C CC x x x u

u u u u u

C x

Second order problem:

(5) (5)

2 2 1 2 1 2 2 2

2 2 2 2

1 1 1 1 1

2

(1 ( ( , ),

(0) 0, (0) 0, (0) 0, (1) 0,

( , , ) ) ( , , ) ) ,

(1) 0 .

u C C C C C C C C x C Cx x u x x u

u u u u u

Adding up the solutions of these problems, the second order approximation,

15

0 1 1 2 2 1 2( ) ( ) ( , , ) ( , , ) ( )u x u x u x C C u x C C O x ,

is determined by knowing the optimal values of the auxiliary parameters 1 2 and C C .

Using Galerkin’s method, we obtain:

1 21.000245451, 0.000124615C C .

By considering these values the second order becomes:

3 4 5 6 7

8 9 10 6 11

7 12 8 13 9 14 15

0.333333 0.125 0.0333333 0.00694444

0.00119049 0.00017360

( ) 0.5

1 0.0000220495 2.48013

2.50501 2.2

1

7501 2.0

0

10 10 10 ( ).326

u x x x x x x x

x x x x

x x x O x

(3.9)

Numerical results for the solution (3.9) are displayed in Table 1.

24

Table 1: We compare the errors of the second order solution (3.8) and (3.9) with the

errors obtained (see [21] and references therein) by using HPM, VIM, ADM, Iterative

method (ITM), and the variational iteration method using He’s polynomials

(VIMHP). We find that both the OHAM solution (3.8) & (3.9) are superior to the

solutions obtained by the above named methods. The use of new auxiliary function

produced much more accurate results than the results obtained by the use of

commonly used auxiliary function.

Table 1

x E* (3.8) E* (3.9) E*([21])

0.0 0.0000 0.0000 0.0000

0.1 -9E-11 -1E-12 -3E-11

0.2 -4E-10 -7E-12 -2E-10

0.3 -5E-10 -1E-11 -4E-10

0.4 -2E-11 -1E-11 -8E-10

0.5 1E-09 -8E-12 -1E-09

0.6 2E-09 3E-14 -2E-09

0.7 2E-09 6E-12 -2E-09

0.8 1E-09 2E-12 -2E-09

0.9 4E-10 -3E-11 -1E-09

1.0 0.000 0.000 0.000

E*=Exact-Approx.

Example 2: Another Fifth order linear bvp

3(5) 219 2( ) 41( 2) x x Cos x x Cou x u x s x Sin x x Sin x

( 1) (1) (1),

( 1) (1) 4 (1) (1),

( 1) 3 (1) 8 (1).

u u Cos

u u Cos Sin

u Cos Sin

Exact solution of this problem is: 2( ) (2 1) ( )u x x Cos x .

We use the auxiliary function 2

1 2( )h p pC p C and consider the second order

solution:

13

0 1 1 2 1 2( ) ( ) ( , ) ( , , ) ( )u x u x u x C u x C C O x .

Using the method of Least Squares, we obtain:

1 20.9940605306, 3.9762851376C C .

25

Having these values, our second order approximation becomes:

2 4 10

8 6 7 2 11 3

0.999978 2.49992 1.04155 0.0000420239

0.00276866 0.0846365 3.38286 10

( )

( ).

x x

x

x

O

xu

x xx

(3.10)

Numerical results for the solution (3.10) are displayed in Table 2.

Table 2: The maximum absolute error as reported in [31] is 1.8775 × 10 -5

, while in

our case it is 88.5757 10 .

Table 2

X Exact Sol. OHAM Sol. E*(3.10)

1.0 0.540302306 0.540302301 3.90E-09

0.8 0.195077879 0.195077879 8.44E-09

0.6 -0.231093972 -0.231094010 3.74E-08

0.4 -0.626321476 -0.626321543 6.70E-08

0.2 -0.901661252 -0.901661334 8.22E-08

0.0 -1.000000000 -1.000000086 8.58E-08

0.2 -0.901661252 -0.901661334 8.22E-08

0.4 -0.626321476 -0.626321543 6.70E-08

0.6 -0.231093972 -0.231094010 3.74E-08

0.8 0.195077879 0.195077879 8.44E-09

1.0 0.540302306 0.540302301 3.90E-09

E*=Exact-Approx.

Example 3: Fifth order non-linear bvp

( ) 2( ) ( ) , 0 1,v xx u x e xu

(0) 1, (0) 1, (0) 1, (1) , (1) .u u u u e u e

Exact solution of this problem is: ( ) .xu x e

For 1p , we consider the second order solution:

13

0 1 1 2 1 2( ) ( ) ( , ) ( , , ) ( ).u x u x u x C u x C C O x

Using Least Squares procedure, we obtain the following values of ,

iC s for 0a

and 1:b

1 20.97647733 0.031756606.,C C

The second order solution thus becomes:

26

2 3 4 5

6 7 5 8 6 9

7 10 9 11 8 12 13

0.166666439 0.041667135 0.0083327684

0.001389521 0.00019810282 2.4837 2.6754

3.4474 4.29

(

26 1

) 1 / 2

10 10

10 10 10 (.462 )6

u x x x x x x

x x x x

x x x O x

(3.11)

Numerical results for the solution (3.11) are displayed in Table 3.

Table 3: We compare the second order solution (3.11) with the exact solution and the

error is compared with error of HPM, VIM, ADM, ITM and VIMHP[21].

Table 3

x Exact Sol. OHAM Sol. E*(3.11) E*([21])

0.0 0.00000 0.0000 0.0000 0.0000

0.1 1.105170918 1.105170918 1.9E-10 1.0E-09

0.2 1.221402758 1.221402757 1.2E-09 2.0E-09

0.3 1.349858808 1.349858804 3.3E-09 1.0E-08

0.4 1.491824698 1.491824691 6.3E-09 2.0E-08

0.5 1.648721271 1.648721261 9.3E-09 3.1E-08

0.6 1.822118800 1.822118789 1.1E-08 3.7E-08

0.7 2.013752707 2.013752696 1.1E-08 4.1E-08

0.8 2.225540928 2.225540920 8.2E-09 3.1E-08

0.9 2.459603111 2.459603109 1.9E-09 1.4E-08

1.0 2.718281828 2.718281834 0.000 0.000

E*=Exact-Approx.

Example 4: Another fifth order non-linear bvp[32]

Consider the following problem

3( ) ( ) ( ) , 0 1,v xx u x e xu

1/2 1/2(0) 1, (0) 1/ 2, (0) 1/ 4, (1) , (1) 1/ 2 .u u u u e u e

Exact solution of this problem is: 2( ) .xu x e

We consider the following second order solution:

15

0 1 1 2 1 2( ) ( ) ( , ) ( , , ) ( ).u x u x u x C u x C C O x

Using Galerkin’s procedure, we obtain the following values of auxiliary parameters

1 20.010868466 0., 029423113C C .

The second order solution thus becomes:

27

2 3 4 5 5

6 6 7 7 8 8 9 9

11 10 12 11 14 12

15 13 17 14

0.0205993 0.0030533 6.3067

5.2556 3.754 2.29401

( ) 1 0.5 0.

1.74892

3.2692 1.4

125 10

10 1

8596 6.20243

2.58495 2.75

0 10 10

10 10 10

10 10287

u x x x x x x

x x x x

x x x

x x

15( )O x

(3.12)

Numerical results for the solution (3.12) are displayed in Table 4.

Table 4

x Exact Sol. OHAM Sol. E*(3.12) E*(DTM [32])

0.0 0.00000 0.0000 0.0000 0.0000

0.1 1.105170918 1.105170919 -9.2E-10 1.0E-09

0.2 1.221402758 1.221402763 -5.0E-09 2.0E-09

0.3 1.349858808 1.349858818 -1.1E-08 1.0E-08

0.4 1.491824698 1.491824713 -1.5E-08 2.0E-08

0.5 1.648721271 1.648721287 -1.6E-08 3.1E-08

0.6 1.822118800 1.822118814 -1.4E-08 3.7E-08

0.7 2.013752707 2.013752717 -9.9E-09 4.1E-08

0.8 2.225540928 2.225540934 -5.6E-09 3.1E-08

0.9 2.459603111 2.459603112 -1.1E-09 1.4E-08

1.0 2.718281828 2.718281824 0.000 0.000

E*=Exact-Approx.

Example 5: Sixth order linear bvp

( ) ( ) ( ) 6 , 0 1,vi xx u x e xu

( ) ( )(0) 1, (0) 1, (0) 3, (1) 0, (1) 2 , (1) 4 . iv ivu uu u u e u e

The exact solution is: ( ) (1 ) .xu x x e

Now following the OHAM procedure, the second order solution,

13

0 1 1 2 1 2( ) ( ) ( , ) ( , , ) ( ),u x u x u x C u x C C O x

is obtained by determining the values of , .iC s Following the procedure of Least

Squares, we get:

1 2 0.99900, .35462C C

The OHAM second order solution thus becomes:

7 2 3 4 5

6 7 8 9

6 10 7 11 8 12 13

( ) 1 2.062484 10 / 2 0.333333 / 8 0.0333325

0.00694578 0.00118984 0.000173587 0.0000221313

2.47936 10 2.39498 10 2.504 10 ( ).

u x x x x x x

x x x x

x x x O x

(3.13)

28

Numerical results for the solution (3.13) are displayed in Table 5.

Table 5: In this table, we compare the errors obtained by ADM[19], VIM[23] and

HPM [24].

Table 5

x E*(OHAM) E*(ADM) E*(VIM) E*(HPM)

0.0 0 0 0 0

0.1 2.05E-08 -4.09E-04 -4.09E-04 -4.09E-04

0.2 4.00E-08 -7.78E-04 -7.78E-04 -7.78E-04

0.3 5.69E-08 -1.07E-03 -1.07E-03 -1.07E-03

0.4 6.95E-08 -1.26E-03 -1.26E-03 -1.26E-03

0.5 7.59E-08 -1.32E-03 -1.32E-03 -1.32E-03

0.6 7.46E-08 -1.26E-03 -1.26E-03 -1.26E-03

0.7 6.52E-08 -1.07E-03 -1.07E-03 -1.07E-03

0.8 4.82E-08 -4.09E-04 -4.09E-04 -4.09E-04

0.9 2.52E-08 -7.78E-04 -7.78E-04 -7.78E-04

1.0 -2.07E-09 0 0 0

E*=Exact-Approx.

Example 6: Sixth order non-linear bvp

( ) 2( ) ( ), 0 1,vi xu ux e x x

with two set of boundary conditions:

(i) (0) 1, (0) 1, ''''(0) 1, (1) , ''(1) , ''''(1) .u u u u e u e u e

(ii) 1 1 1(0) 1, '(0) 1, ''(0) 1, (1) , '(1) , ''(1) .u u u u e u e u e

Exact solution of the given problem using the first (i) set of boundary conditions is:

( ) .xu x e

Exact solution in the case of the second (ii) set of boundary conditions is:

( ) .xu x e

(i) Consider the given problem with conditions:

(0) 1, (0) 1, ''''(0) 1, (1) , ''(1) , ''''(1) .u u u u e u e u e

Let us the auxiliary function 1 2( ) ( )xh p p C C e and consider the second order

solution:

13

0 1 1 2 2 1 2( ) ( ) ( , , ) ( , , ) ( ).u x u x u x C C u x C C O x

29

For 0a and 1b , Least Squares gives the following optimal values ,

iC s :

1 20.99142072, 0.015535939.C C

Knowing these values, the approximate solution is:

2 3 4 5

6 7 5 8 6 9

7 10 8 11 9 12 13

( ) 0.166666497 041666667 0.0083334658

0013888217 0001984188 2.4802 2.7589

2.74

1 0.5

27 2.4703 +2.1384

0. 0.

0. 0. 10 10

1 1 10 )0 0 ( .

x x x x x

x x x x

x

u x

O xx x

(3.14)

Numerical results for the solution (3.14) are displayed in Table 6.

(ii) Now we use the second set of conditions:

1 1 1(0) 1, '(0) 1, ''(0) 1, (1) , '(1) , ''(1) .u u u u e u e u e

We consider the same the auxiliary function and the same order of solution. i.e.

1 2( ) ( )xh p p C C e and

13

0 1 1 2 2 1 2( ) ( ) ( , , ) ( , , ) ( ).u x u x u x C C u x C C O x

Using Galerkin’s method, we obtain: 1 20.41243798998, 0.0014069149.C C

OHAM solution in this case is:

2 3 4 5

6 7 5 8 9

7 10 9 11 9 12 10 13

11 14 15

0.166666775 0.008332465

0.001387441 0.000197784 2.5071 3.002

2.918 6.7 2.257 1.806

1.974

( ) 1 / 2 / 24

10

10 10 10 10

10 ( ).

u x x x x x x

x x x x

x x x x

x O x

(3.15)

Numerical results of the solution (3.15) are displayed in Table 6.

30

Table 6: In this table, we compare the errors obtained by OHAM with the errors

obtained by ADM [19], VIM [23] and HPM [24]. Numerical results show the

superiority of the proposed method. The results also reveal that the Galerkin’s method

produced much better values for the auxiliary constants.

Table 6

x E*(OHAM)

(3.14)

E*(OHAM)

(3.15) E*(ADM) E*(VIM) E*(HPM)

0.0 0 0 0 0 0

0.1 -9.4E-09 -4.82E-10 -1.2E-04

-1.2E-04

-1.2E-04

0.2 -1.8E-08 -4.92E-10 -2.3E-04

-2.3E-04

-2.3E-04

0.3 -2.4E-08 -2.37E-11 -3.2E-04

-3.2E-04

-3.2E-04

0.4 -2.9E-08 5.11E-10 -3.8E-04

-3.8E-04

-3.8E-04

0.5 -3.0E-08 6.42E-10 -4.0E-04

-4.0E-04

-4.0E-04

0.6 -2.8E-08 2.02E-10 -3.9E-04

-3.9E-04

-3.9E-04

0.7 -2.4E-08 -5.37E-10 -3.3E-04

-3.3E-04

-3.3E-04

0.8 -1.7E-08 -1.02E-09 -2.4E-04

-2.4E-04

-2.4E-04

0.9 -9.1E-09 -8.23E-10 -1.2E-04

-1.2E-04

-1.2E-04

1.0 2.5E-10 -2.05E-12 2.0E-09

2.0E-09

2.0E-09

E*=Exact-Approx.

Example 7: Parameterized sixth order bvp

(6) (4)( ) (1 ) ( ) ( )u x c u x cu x cx , 0 1x

(0) 1 , (0) 1 , (0) 0,

7 1(1) sinh(1) , (1) cosh(1) , (1) 1 sinh(1).

6 2

u u u

u u u

The problem has the exact solution: 31( ) 1 sinh( ).

6u x x x

It is interesting to note that the exact solution is free from the parameter c but the

problem depends upon this parameter. This can be viewed by rewriting the given

problem as

(6) (4) (4) (2){ } { } 0.u u c u u x

It shows that, no matter what the value of c is, the solution of sixth-order problem is

actually the solution of the fourth-order problem.

31

For the solution of the problem under discussion, we follow the procedure of OHAM

with the following auxiliary function:

2 3

1 2 3( )H p pC p C p C .

Accordingly, the following problems are obtained:

Zeroth-order problem:

(6)

0 ( )u x cx ,

(1) (2)

0 0 0

(1) (2)

0 0 0

(0) 1 , (0) 1 , (0) 0,

7 1(1) sinh(1) , (1) cosh(1) , (1) 1 sinh(1).

6 2

u u u

u u u

First order problem:

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

1 1 0 1 0 1 0 1( ) (1 ) ( ) (1 ) ( ) ( ) (1 ),u x C u x C c u x C cu x c C

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

1 1 1 1 1 1(0) 0, (0) 0, (0) 0, (1) 0, (1) 0, (1) 0.u u u u u u

Second order problem:

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

2 1 1 2 0 1 1

(4) (2) (2)

2 0 1 1 2 0

( ) (1 ) ( ) ( ) (1 ) ( )

(1 ) ( ) ( ) ( ( ) ),

u x C u x C u x C c u x

C c u x C c u x C c u x x

(1) (2)

2 2 2

(1) (2)

2 2 2

(0) 0 , (0) 0 , (0) 0,

(1) 0 , (1) 0 , (1) 0.

u u u

u u u

Third order problem:

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

3 1 2 2 1 3 0 1 2

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

2 1 3 0 1 2 2 1

(2)

3 0

( ) (1 ) ( ) ( ) ( ) (1 ) ( )

(1 ) ( ) (1 ) ( ) ( ) ( )

( ( ) ),

u x C u x C u x C u x C c u x

C c u x C c u x C c u x C c u x

C c u x x

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

3 3 3 3 3 3(0) 0, (0) 0, (0) 0, (1) 0, (1) 0, (1) 0.u u u u u u

Solutions to the zeroth and the first order problems are

5 3 4

0

3 4 5 7

( ) 1 0.00119048 0.000595238 0.00165465

0.333957 0.0015873 0.00956585 0.00019841 .3

u x x c x c x x

x c x x c x

4 4 6 2 4

1 1 1

5 3 3 6 2 3

5 5 2 5 5 6 6 6

5 2 6 7

( , ) ( 0.0016566 0.00158697 2.14732 10

0.00120109 0.00062812 0.000594872 4.69906 10

0.00119192 3.0005 10 5.5155 10 2.2451 10

5.291 10 9.8491 10

u x C C x c x c x

x x c x c x

c x c x x c x

c x c

2 8 5 2 7 7 2 8

7 7 7 9 6 2 9

8 2 11

2.7636 10 9.4482 10

0.000227758 0.000198154 4.0758 10 3.1494 10

2.5052 10 ).

x c x c x

x c x c x c x

c x

Solutions to the second and third order problems can be obtained easily.

32

For 1p , we consider the third order solution for our approximation:

0 1 1 2 1 2 3 1 2 3( ) ( ) ( , ) ( , , ) ( , , , ).u x u x u x C u x C C u x C C C

For this approximation the residual becomes:

(6) (4)( ) (1 ) ( ) ( )R u x c u x cu x cx .

Now using the Least Squares with 0 and 1a b , we obtain the following values of

the 'iC s for 10c :

1

2

3

0.935780212,

0.001555044,

0.0000438473.

C

C

C

The third order solution thus becomes:

3 6 4 5 5 6

5 8 9 5 10

5 11 6 12 6 13 8 14

8 15

7

( ) 1 0.333333 1.29887 10 0.00832295 4.16114 10

9.6176 10 0.000162708 0.000162731 9.92639 10

2.67093 10 2.10774 10 2.11295 10 1.05849 10

2.3803 10

u x x x x x x

x x x x

x x x x

x

11 16 11 17 14 191.33648 10 7.66735 10 6.73639 10 .x x x

(3.16)

For 1000c , the following values of the 'iC s are obtained:

1

2

3

0.135830834,

0.025540511,

0.0020075724.

C

C

C

In this case our solution is

3 4 5 6

7 8 9 10 11

12 13 14 15

6 16 17 8 19

( ) 1 0.274438 0.833256 4.9068 16.3271

34.4276 482972 44.7369 25.2817 6.55572

0.59442 0.552725 0.00328518 0.00652957

4.5582 10 0.0000221568 2.06016 10 .

u x x x x x x

x x x x x

x x x x

x x x

(3.17)

33

In Table 7 and 8, numerical results for the third-order solution are compared with the

numerical results of exact solution and the error is compared with the error of the

fifth-order HAM solution. It is clear from this table that OHAM is more effective and

explicit than HAM.

Table 7: Numerical Results for 10c

x E*(HAM[27]) E*(OHAM)

0.1 6.9E-11 -2.6E-11

0.2 2.6E-10 -1.9E-10

0.3 1.1E-09 -3.4E-10

0.4 1.7E-09 -2.9E-10

0.5 1.9E-09 -7.0E-11

0.6 1.5E-09 1.4E-10

0.7 7.6E-10 1.8E-10

0.8 1.6E-10 8.8E-11

0.9 3.5E-11 2.9E-11

E* = Exact Sol. – Approx. Sol.

Table 8: Numerical Results for 1000c

E*(HAM[27]) E*(OHAM)

0.1 9.1E-06 1.1E-05

0.2 1.6E-04 3.3E-06

0.3 4.4E-04 -1.4E-05

0.4 6.8E-04 5.2E-06

0.5 7.3E-04 4.2E-05

0.6 5.8E-04 5.7E-05

0.7 3.2E-04 4.9E-05

0.8 9.8E-05 4.5E-05

0.9 4.7E-06 2.4E-05

E* = Exact Sol. – Approx. Sol.

Example 8: Eighth order nonlinear bvp

(8) 2( ) ( ), 0 1,xu ux e x x

'(0) 1, '(0) 1, ''(0) 1, '''(0) 1, (1) , '(1) , ''(1) .u u u u u e u e u e

We consider the second order approximation:

34

13

0 1 1 2 1 2( ) ( ) ( , ) ( , , ) ( ).u x u x u x C u x C C O x

The following values of the convergence control parameters are obtained by using

Galerkin’s method:

11

1 21.451894673 10 , 0.000647581.C C

The approximate solution in this case is:

2 3 4 5

6 7 8 8 9 9

10 10 11 11 12 12 31

1 0.5 0.166667 0.0416275 0.00884857

0.00117027 0.00033161 .6061 10 1.78456 10

1.78456 10 1.62233 10 1

( )

1

( ).3 .494 10

x x xu x x

x x x

x x

x

x

x O x

(3.18)

If the method of Least-Squares is used to determine C’s, we have then:

8

1 21.793 10 , 1.001347284C C .

The approximate solution in this case is:

2 3 4 5

6 4 7 5 8 6 9

7 10 8 11 9 12 13

1 0.5 0.166667 0.0416667 0.00833313

0.00138918 1.9823 10 2.4835 10 2.75944 10

2.75944 10 2.50859 10 2.08656 10

( )

( ).

x x x x x

x x x

x x x

u x

x

O x

(3.19)

Let us use the auxiliary function 1 2( ) ( )xh p p C C e and consider the second order

approximation:

13

0 1 1 2 2 1 2( ) ( ) ( , , ) ( , , ) ( ).u x u x u x C C u x C C O x

Using Galerkin’s method, we obtain

1 20.9993171458, 0.0012314995C C .

The OHAM approximation now becomes

2 34 7 5 8 6 9

7 10 4 5 6

8 11 9 12 13

( ) 1.984 2.480 2.756

2.756 0.041666667

1 10 10 10

+0.00

2! 3!

10 833333

10 10

3 0.001388889

2.505 2.088 ( ).

x xx x x

x x x

u x x

x O x

x

x

(3.20)

Numerical results for the solutions (3.18), (3.19) and (3.20) are displayed in table 9.

35

Table 9

x E*(3.18) E*(3.19) E*(3.20) E*([24])

0.0 0 0 0 0

0.1 2.60E-09 -3.5E-12 1.39E-16 1.27E-05

0.2 2.62E-08 -3.5E-11 5.32E-16 2.43E-05

0.3 7.86E-08 -1.1E-10 1.48E-15 3.35E-05

0.4 1.36E-07 -1.8E-10 3.36E-15 3.94E-05

0.5 1.61E-07 -2.2E-10 2.28E-14 4.16E-05

0.6 1.39E-07 -1.9E-10 2.21E-13 3.96E-05

0.7 8.22E-08 -1.1E-10 1.64E-12 3.38E-05

0.8 2.80E-08 -2.8E-11 9.36E-12 2.45E-05

0.9 2.84E-09 4.1E-11 4.36E-11 1.29E-05

1.0 1.14E-13 1.8E-10 1.73E-10 1.00E-09

E*=Exact-Approx.

Example 9: Ninth order linear bvp

(9) ( ) ( ) 9 ,xu x u x e

0) 1, ' 0) 0, '' 0) 1, ''' 0) 2, '''' 0( ( ( ( (

(1 (1

) 3,

) 0, (1 (1' ) , '' ) 2 , ''' ) 3 .

u u u u

u u e

u

u e u e

Exact solution of this problem is: ( ) (1 ) xu x x e .

For this linear problem, we take 1 2( ) ( )h p p C C x .

We consider the following second order solution:

13

0 1 1 2 2 1 2( ) ( ) ( , , ) ( , , ) ( ).u x u x u x C C u x C C O x

We follow the Galerkin’s method to find the values of ' .iC s For 0a . and 1b , we

obtain 1 21 and 0.C C

The second order solution now becomes:

2 35 6 7

4 8

4

1

5 9 6 10

2 3

7 11

8 1

0.033333333 0.006944444 0.001190476

1

( ) 12 3 4

10 10 10 10

1

.7361 2.2046 2.4802 2.5052

2.296 )04 ( .

x x xu x x x x

x x x

O

x

x x

(3.21)

Numerical results for the solution (3.21) are displayed in table 10.

36

Table 10

x Exact Sol. OHAM Sol. E*(3.21) E*([40])

0.0 1 1 0.0000 0.0000

0.1 0.994653826 0.994653826 1.03E-16 -2.0E-10

0.2 0.977122206 0.977122206 1.33E-16 -2.0E-10

0.3 0.944901165 0.944901165 -2.12E-16 -2.0E-10

0.4 0.895094818 0.895094818 -1.30E-14 -2.0E-10

0.5 0.824360635 0.824360635 -2.44E-13 -2.0E-10

0.6 0.728847520 0.728847520 -2.64E-12 -6.0E-10

0.7 0.604125812 0.604125812 -1.97E-11 -1.0E-09

0.8 0.445108186 0.445108186 -1.13E-10 -2.0E-09

0.9 0.245960311 0.245960311 -5.26E-10 -3.4E-09

1.0 0.0000 2.09E-09 -2.09E-09 0.000

E*=Exact-Approx

Example 10: Tenth order nonlinear

(10) 2( ) ( ), 0 1, xu ux e x x

( )

( )

(0) 1, '(0) 1, ''(0) 1, '''(0) 1, (0) 1,

(1) , '(1) , ''(1) , '''(1) , (1) .

iv

iv

u u u u u

u e u e u e u e u e

We consider the second order approximation:

13

0 1 1 2 1 2( ) ( ) ( , ) ( , , ) ( ).u x u x u x C u x C C O x

To find the values of iC , we apply the Galarkin’s method. So solving the system

1 2

0, 0,

b b

a a

u uR dr R dr

C C

we obtain:

1 20 & 1.023966086C C .

In this case the approximate solution is:

2 3 4 5 6

7 8 6 9 7 1

8 1 9

0

1 2 131

1 2 6 24 0.008333323 0.00138894

0.000198312 0.00002

( ) / /

4898 2.712 10 2.8218 10

2.5652 10 2.1377 10

/

( ).

x x x x x x

x x x x

u

x x

x

O x

(3.22)

Numerical results of the solution (3.22) are displayed in table 11.

37

Table 11

x Exact Sol. OHAM Sol. E*(3.22) E*([40])

0.0 1 1 0 0

0.1 1.105170918 1.105170918 8.12E-17 -1.41E-06

0.2 1.221402758 1.221402758 2.18E-18 -2.69E-06

0.3 1.349858808 1.349858808 -2.84E-18 -3.70E-06

0.4 1.491824698 1.491824698 7.95E-16 -4.35E-06

0.5 1.648721271 1.648721271 1.91E-14 -4.58E-06

0.6 1.822118800 1.822118800 2.15E-13 -4.36E-06

0.7 2.013752707 2.013752707 1.63E-12 -3.71E-06

0.8 2.225540928 2.225540928 9.33E-12 -2.69E-06

0.9 2.459603111 2.459603111 4.36E-11 -1.42E-06

1.0 2.718281828 2.718281828 1.73E-10 2.00E-09

E*=Exact-Approx.

Example 11: Eleventh order linear bvp

(11) ( ) ( ) 22(5 ) ,0 1,xu x u x x e x

( ) ( )

( )

(0) 1, (0) 1, (0) 1, (0) 5, (0) 11, (0) 19,

(1) 0, (1) 2 , (1) 6 , (1) 12 , (1) 20.

iv v

iv

u u u u u u

u u e u e u e u

Exact solution of this problem is: 2( ) (1 ) xu x x e .

According to the procedure of OHAM, we obtain:

Zeroth order problem:

(11)

0

( ) ( )

0 0 0 0 0 0

( )

0 0 0 0 0

( ) 0,

(0) 1, (0) 1, (0) 1, (0) 5, (0) 11, (0) 19,

(1) 0, (1) 2 , (1) 6 , (1) 12 , (1) 20.

iv v

iv

u x

u u u u u u

u u e u e u e u

First order problem:

(11)

1 1 1 0

( ) ( )

1 1 1 1 1 1

( )

1 1 1 1 1

( ) 22 ( 5) ( ),

(0) 0, (0) 0, (0) 0, (0) 0, (0) 0, (0) 0,

(1) 0, (1) 0, (1) 0, (1) 0, (1) 0.

x

iv v

iv

u x c e x c u x

u u u u u u

u u u u u

Second order problem:

(11) (11) (11)

2 2 0 0 1 1 1 1

( ) ( )

2 2 2 2 2 2

( )

2 2 2 2 2

( ) (22 ( 5) ( ( ) ( ))) ( ) (1 ) ( ),

(0) 0, (0) 0, (0) 0, (0) 0, (0) 0, (0) 0,

(1) 0, (1) 0, (1) 0, (1) 0, (1) 0.

x

iv v

iv

u x c e x u x u x c u x c u x

u u u u u u

u u u u u

38

We add up solutions of the problems, to obtain the second order approximation:

16

0 1 2( ) ( ) ( ) ( ) ( ).u x u x u x u x O x

To find the value of 'iC s , we follow the Galerkin’s method.

For 0, and 1a b , we obtain: 1 20, 1.0008235064. C C

Using these values, we obtain

2 3 4 56 7

8 9 10 6 11

7 12 8 13 9 14 10 15

16

5 11 19( ) 1 0.040277774 0.008134939

2 6 24 120

0.001364052 0.000195691 0.0000245105 2.7329 10

2.7371 10 2.4912 10 2.0779 10 1.5996 10

( ).

x x x xu x x x x

x x x x

x x x x

O x

(3.23)

Numerical results for the solution (3.23) are displayed in table 12.

Table 12: In this table, we display numerical results of example 11. The error is

denoted by E* and k is the number of iterations. OHAM produced very accurate

results for only two iterations. In columns 3 & 4 of this table, we present numerical

results for the solutions obtained by using ADM & VIM [39].

Table 12

x E*(OHAM k=2) E*(ADM k=2) E*(VIM k=4)

0.0 0.0000 0.0000 0.0000

0.1 -2.66E-15 -8.12E-13 6.44E-15

0.2 -8.04E-14 -2.94E-11 2.40E-13

0.3 -4.74E-13 -1.74E-10 1.44E-12

0.4 -1.24E-12 -4.62E-10 3.85E-12

0.5 -1.92E-12 -7.20E-10 6.06E-12

0.6 -1.90E-12 -7.16E-10 6.09E-12

0.7 -1.19E-12 -4.35E-10 3.75E-12

0.8 -6.80E-13 -1.30E-10 1.14E-12

0.9 -2.27E-12 -8.34E-12 9.95E-14

1.0 -1.22E-11 8.86E-18 6.37E-14

E*=Exact-Approx.

39

Example 12: Another eleventh order linear bvp

(11) 2( ) ( ) 22( cos 5sin ) (1 )(sin ), 0 1,u x u x x x x x x cosx x

( ) ( )(0) 1, (0) 0, (0) 3, (0) 0, (0) 13, (0) 0,iv vu u u u u u

( )

(1) 0, (1) 2cos1, (1) 2cos1 4sin1, (1) 6(cos1 sin1),

(1) 12cos1 8sin1.iv

u u u u

u

Exact solution of this problem is: 2( ) (1 )cosu x x x .

We follow the OHAM procedure and consider the second order solution:

16

0 1 2( ) ( ) ( ) ( ) ( )u x u x u x u x O x .

For 0, and 1a b , we obtain: 1 0,C & 2 1.0023268176,C by Galerkin’s

method. Inserting these values our solution becomes:

2 46 8 7 8

8 9 5 10 7 12 9 14 16

3 13( ) 1 0.0430555586 1.4739 10 0.00141366

2 24

+2.4768 10 2.5086 10 +2.7831 10 2.104 10 ( ).

x xu x x x x

x x x x O x

(3.24)

Numerical results for solution (3.24) are displayed in table 13.

Table 13: In this table, we display numerical results of example 12. The error is

denoted by E* and k is the number of iterations. OHAM produced very accurate

results for only two iterations. In column 3 of this table, we present numerical results

for the solutions obtained by using VIM [39].

Table 13

x E*(OHAM k=2) E*(VIM k=4)

0.0 0.0000 0.0000

0.1 2.00E-15 3.89E-15

0.2 6.66E-14 1.46E-13

0.3 3.96E-13 8.81E-13

0.4 1.05E-12 2.36E-12

0.5 1.63E-12 3.80E-12

0.6 1.63E-12 5.15E-12

0.7 1.02E-12 1.56E-11

0.8 6.18E-13 8.99E-11

0.9 2.15E-12 4.70E-10

1.0 1.15E-11 2.06E-09

E*=Exact-Approx.

40

Example 13: Eleventh order nonlinear bvp

(11) 2 2( ) ( ) ( 11)(sin cos ) (1 2sin cos ), 0 1,u x u x x x x x x x x

( ) ( )

( )

(0) 0, (0) 1, (0) 2, (0) 3, (0) 4, (0) 5,

(1) sin1 cos1, (1) 3cos1 sin1, (1) 2(cos1 2sin1),

(1) 5cos1 3sin1, (1) 2sin1.

iv v

iv

u u u u u u

u u u

u u

Exact solution of this problem is: ( ) (sin cos )u x x x x .

We follow the OHAM procedure and consider the second order solution:

16

0 1 2( ) ( ) ( ) ( ) ( )u x u x u x u x O x .

For 0, and 1a b , we obtain 22

1 22.5961 10 & 1.0001860659C C , by

Galerkin’s method.

Our solution in this case is:

2 3 4 5 6 7

8 5 9 6 10 7 11

8 12 9 13 11 14 11 15 16

( ) 3 6 24 0.008333805 0.0013865074

0.0001935835 2.9733 10 5.3105 10 2.7562 10

2.088 10 2.088 10 9.1783 10 1.1473 10 ( ).

u x x x x x x x x

x x x x

x x x x O x

(3.25)

Numerical results for the solution (3.25) are displayed in table 14.

Table 14: In this table, we display numerical results of example 11. The error is

denoted by E* and k is the number of iterations. OHAM produced very accurate

results for only two iterations. In column 3 of this table, we present numerical results

for the solutions obtained by using VIM [39].

Table 14

x E*(OHAM k=2) E*(VIM k=3)

0.0 0 0

0.1 -2.77E-13 1.03E-15

0.2 -9.79E-12 5.34E-14

0.3 -5.69E-11 5.59E-13

0.4 -1.47E-10 3.32E-12

0.5 -2.24E-10 1.48E-11

0.6 -2.18E-10 5.38E-11

0.7 -1.30E-10 1.66E-10

0.8 -3.78E-11 4.42E-10

0.9 -2.42E-12 1.04E-09

1.0 -2.38E-13 2.19E-09

E*=Exact-Approx.

41

Example 14: Twelfth order linear bvp

(12) ( ) ( ) 24 cos 132sinu x u x x x x , 11 x

( 1) 0u , (1) ( 1) 2sin(1)u , (2) ( 1) 4cos(1) 2sin(1)u ,

(3) ( 1) 6cos(1) 6sin(1)u , (4) ( 1) 8cos(1) 12sin(1)u ,

(5) ( 1) 20cos(1) 10sin(1)u ,

(1) 0u , (1) (1) 2sin(1)u , (2) (1) 4cos(1) 2sin(1)u ,

(3) (1) 6cos(1) 6sin(1)u , (4) (1) 8cos(1) 12sin(1)u ,

(5) (1) 20cos(1) 10sin(1)u .

Exact solution of this problem is: 2( ) ( 1) u x x Sin x .

Here we take the auxiliary function 2

1 2( )H p pC p C , and according to OHAM the

following order problems are obtained:

Zeroth-oder problem:

(12)

0

(1) (2)

0 0 0

(3) (4)

0 0

(5)

0 0

( ) 24 c ( ) 132 ( ),

( 1) 0 , ( 1) 2sin(1) , ( 1) 4cos(1) 2sin(1),

( 1) 6cos(1) 6sin(1) , ( 1) 8cos(1) 12sin(1),

( 1) 20cos(1) 10sin(1) , (1)

u x x os x sin x

u y u

u u

u u (1)

(2) (3)

0 0

(4) (5)

0 0

0, (1) 2sin(1),

(1) 4cos(1) 2sin(1) , (1) 6cos(1) 6sin(1),

(1) 8cos(1) 12sin(1) , (1) 20cos(1) 10sin(1)

y

u u

u u

First order oroblem:

(12) (12)

1 1 0 1 1 1 0( ) (1 ) ( ) 24(1 ) cos( ) 132(1 )sin( ) C u x C y x C x x C x y

( ) ( )

1 1( 1) (1) 0, 0(1)5.k ku u k

Solutions to these problems are:

3 5

0

7 4 9 6 11

( ) 396.999999959 59.166666418 2.674999379

0.058530916 7.627137352 10 6.437881583 10

24 cos( ) 420 sin( )

u x x x x

x x x

x x x

42

3 5

1 1 1 1 1

7 7 6

1 1

6 11 8 13 10 15

1 1 1

13 17

1

( , ) (420 264) (44 62 ) (2.7 2.2) (0.0523801

0.0547627 ) (0.000628931 0.000726891) (6.36358 10

4.76041 10 ) 6.37544 10 2.71474 10

9.02478 10 2.4250

u x C C x C x C x

C x C x

C x C x C x

C x

15 19 18 21

1 1

21 23

1 1 1

5 10 5.41727 10

9.94041 10 24 cos( ) (264 444 )sin( ).

C x C x

C x C x x C x

For 1p , we consider only the first order approximation:

1 0 1 1( , ) ( ) ( , ).u x C u x u x C

For 0 and 1a b , using the residual,

(12)

1 1( , ) ( , ) 24 cos 132sin ,R u x C u x C x x x

we get, 1 1.9984,C by using Least Squares.

In this case, our first order approximation becomes:

3 5 7 9

5 11 7 13 10 15 12 17

15 19 17 21 20 23

( ) 1500.33 227.067 10.2707 0.220349 0.00274646

2.23146 10 1.27406 10 5.42513 10 1.80351 10

4.84621 10 1.08259 10 1.98649 10 71.9615 cos( )

1571

u x x x x x x

x x x x

x x x x x

.29 sin( )x

(3.26)

The second order OHAM solution is given by

1 0 1 1 2 1 2( , ) ( ) ( , ) ( , , )u x C u x u x C u x C C ,

Using this solution, we obtain:

1 22.000000019 and 0.999999940C C .

Having these values, the second-order solution is determined.

Fig.1: Dotted curve represents the OHAM solution while the solid curve is for the exact solution.

43

Table 15: In this table, numerical results for the first-order solution are compared

with the exact solution. The error of the first and second-order solutions is compared

with the error of DTM solution. It is clear from the table that the error of the DTM

grows as we depart from zero while this is not the case with the OHAM.

Table 15

x E*(DTM[12]) E*(OHAM)

Ist-Order

E*(OHAM)

2nd-Order

0.0 0.0000 0.0000 0.0000

0.1 1.6376E-15 5.9944E-12 -3.9288E-14

0.2 2.0797E-13 9.9150E-12 1.7736E-14

0.3 3.4360E-12 1.0666E-11 -3.5250E-13

0.4 2.4556E-11 8.8775E-12 1.3883E-13

0.5 1.1021E-10 5.5447E-12 -6.3283E-14

0.6 3.6677E-10 2.4814E-12 -2.6101E-13

0.7 9.8945E-10 7.4829E-13 -2.4547E-13

0.8 2.2839E-09 2.7400E-13 1.2379E-14

0.9 4.6760E-09 5.6399E-14 -1.8097E-13

1.0 8.7157E-09 0.0000 -4.5474E-13

E* = Exact Sol.– Approx.Sol.

Fig.2: Error graph for the OHAM Ist-order solution.

44

Fig. 3: Error graph for the OHAM 2nd-order solution.

Example 15: Another twelfth order linear bvp

(12) 3( ) (120 23 ) xu x x x e

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

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

(0) 0, (0) 1, (0) 0, (0) 3, (0) 8, (0) 15,

(1) 0, (1) , (1) 4 , (1) 9 , (1) 16 , (1) 25 .

u u u u u u

u u e u e u e u e u e

The exact solution of this problem is: ( ) (1 ) xu x x x e

We consider only the zeroth and first-order problems.

Zeroth-order problem:

(12) 3

0 ( ) (120 23 ) xu x x x e ,

(1) (2) (3)

0 0 0 0(0) 0, (0) 1, (0) 0, (0) 3,u u u u (4) (5)

0 0(0) 8, (0) 15,u u

(1)

0 0(1) 0, (1) ,u u e (2)

0 (1) 4 ,u e (3)

0 (1) 9 ,u e (4) (5)

0 0(1) 16 , (1) 25 .u e u e

First order problem:

(12) 3 (12)

1 1 1 0 1 0( ) (1 )(120 23 ) ( ) (1 ) ( )xu x C x x e C xu x C u x ,

( ) ( )

1 1(0) (1) 0, 0(1)5.k ku u k

Solutions to these problems are

2

0

2 3 3 4 5 6

6 7 7 8

8 9 9

1( ) ( 280800 280800 221760 58920 85800

120

4320 21600 120 3960 560 197720040

72737161 836896800 307877125 1451896600

534122910 1282301040 471732190 5741817

x x

x x

u x e x e x x

e x x e x x x x

e x x e x x

e x x e x

10

10 11 11

60

211229665 103986080 38254341 )

x

e x x e x

45

1 1 1

2 3 4 10

6 7 8 9 5

11 7 13

( , ) (216060 216060 (176352 39708 ) (71045

2723 ) (18795 84 ) (3663 ) 0.00237 265

69.1417 7.07738 0.603671 0.0425263 558.833

0.0000892391 3.75782 10 4.23959 10

x x

x x x

u x C C e e x

e x e x e x x

x x x x x

x x

8 14

9 15 10 16 11 17

13 18 14 19 16 20

17 21 19 22 20 23

22 24

3.28063 10 2.06473 10 1.11334 10

5.24805 10 2.17518 10 7.89181 10

2.46619 10 6.40591 10 1.27589 10

1.55619 10 )

x

x x x

x x x

x x x

x

(3.27)

We consider the first order approximation:

1 0 1 1( , ) ( ) ( , ).u x C u x u x C

Using the Least Square procedure with 0.5 and 1a b , we get,

1 0.00260417 C .

The above first order approximation has now determined.

Numerical results are of the first order solution are displayed in table 16.

Table 16

X Exact Sol. OHAM Sol. E*(DTM[12]) E*(OHAM)

0.0 0.000000000 0.000000000 0.0000 0.0000

0.1 0.099465383 0.099465383 -2.1410E-13 -7.5065E-14

0.2 0.195424441 0.195424441 -2.4416E-13 -2.7686E-12

0.3 0.283470350 0.283470350 6.5645E-13 -1.7271E-11

0.4 0.358037927 0.358037927 -1.3413E-12 -5.0232E-11

0.5 0.412180318 0.412180318 -2.7926E-12 -9.3401E-11

0.6 0.437308512 0.437308512 4.1875E-11 -1.2791E-10

0.7 0.422888068 0.422888069 8.0884E-11 -1.3917E-10

0.8 0.356086548 0.356086549 -2.7222E-10 -1.2278E-10

0.9 0.221364280 0.221364280 -3.3264E-10 -7.4997E-11

1.0 0.000000000 -0.000000000 1.0680E-09 1.9454E-11

E* = Exact Sol.–Approx. Sol.

46

Fig. 4: Dotted Curve represents OHAM solution (3.27) while the solid Curve is for the exact

solution.

Fig. 5: Error graph for the OHAM Ist-order solution (3.27).

Example 16: Twelfth order nonlinear bvp

Now let us consider a nonlinear boundary value problem

(12) 2 (3)( ) 2 ( ) ( )xu x e u x u x , 10 x (3.28)

with two sets of boundary conditions (a) and (b):

(a) kku )1()0()( , 1)( )1()1( eu kk , .5)1(0k

(b) 1)0()2( ku , 1)2( )0( eu k , .5)1(0k

Exact solution of this problem is: ( ) xu x e .

For the first-order OHAM solutions of both the problems (3.28a) and (3.28b), we use

Least Squares with 0 and 1.a b For the problem (3.28a), we get,

34

1 8.143655212 10 . C

47

Solution of the problem (3.28a) is:

2 3 4 5

6 7 8 6 9

7 10 8 11 42 12 43 13

45 14 46 15

( ) 1 0.5 0.166666667 0.041666667 0.008333333

0.001388888 0.000198405 0.000024781 2.7276 10

2.5343 10 1.5342 10 1.7 10 1.3 10

9.3 10 6.2 10 3.9

u x x x x x x

x x x x

x x x x

x x

47 16 48 17

49 18 51 19 52 20

10 2.3 10

1.2 10 6.2 10 2.0 10 .

x x

x x x

(3.29)

For the problem (3.28b), we get, 1C = 123.285859812 10 .

Solution of the problem (3.28b) is:

2

3 4 5 6

4 7 5 8 6 9 7 10

8 11 21 12 2

( ) 1126.713391869 389.738335887 64.27317 7328

6.302588356 0.495553602 0.014829714 0.002268134

1.7302 10 2.5317 10 2.8220 10 2.7564 10

1.5836 10 6.8 10 5.2 10

u x x x

x x x x

x x x x

x x

2 13 23 14 24 15

25 16 27 17 28 18 29 19 31 20

2

3 4 5

3.8 10 2.5 10

1.6 10 9.2 10 5.3 10 2.7 10 8.5 10

( 1125.713391869 734.97505329 4 235.891534688

49.553656086 7.652893232 0.925140116 0.091002769

x

x x x

x x x x x

e x x

x x x

6

7 4 8 5 9 6 10

8 11 9 12 10 13 12 14

14 15 15 16 17 17 19 18

21 19

0.007474311 2.1817 10 3.1357 10 1.6360 10

7.4518 10 2.9717 10 1.0379 10 3.1671 10

8.398 10 1.918 10 3.716 10 5.975 10

7.7 10 7.5 1

x

x x x x

x x x x

x x x x

x

23 20 25 21 27 220 4.9 10 1.6 10 )x x x

(3.30)

Table 17: Comparison of numerical results for problem (3.28b)

x Exact Sol. OHAM Sol. E*(DTM[12]) E*(OHAM)

0.0 1.000000000 1.000000000 0.0000 0.0000

0.1 0.904837418 0.904837418 -4.1078E-15 7.9308E-16

0.2 0.818730753 0.818730753 -1.3023E-13 2.2086E-14

0.3 0.740818221 0.740818221 -6.7535E-13 1.1141E-13

0.4 0.670320046 0.670320046 -1.5278E-12 2.4591E-13

0.5 0.606530660 0.606530660 -1.9817E-12 3.1165E-13

0.6 0.548811636 0.548811636 -1.5745E-12 2.4304E-13

0.7 0.496585304 0.496585304 -7.1704E-13 1.1459E-13

0.8 0.449328964 0.449328964 -1.4222E-13 1.4041E-14

0.9 0.406569660 0.406569660 -4.1633E-15 1.9715E-14

1.0 0.367879441 0.367879441 1.2212E-15 2.2560E-13

E*=Exact Sol.-Approx. Sol.

48

Fig. 6: Dotted Curve represents OHAM solution (3.28b) while the solid Curve is for the exact

solution.

Fig. 7: Error graph for the OHAM Ist-order solution (3.28b).

Table 18: Comparison of numerical results for problem (3.28a)

X E*(ADM [3]) E*(DTM[12]) E*(OHAM)

0.0 0.0000 0.0000 0.0000

0.1 -1.61E-07

-1.61E-07

2.64E-07

0.2 -3.07E-07

-3.07E-07

5.03E-07

0.3 -4.22E-07

-4.22E-07

6.92E-07

0.4 -4.97E-07

-4.97E-07

8.14E-07

0.5 -5.22E-07

-5.22E-07

8.55E-07

0.6 -4.97E-07

-4.96E-07

8.14E-07

0.7 -4.22E-07

-4.22E-07

6.92E-07

0.8 -3.07E-07

-3.07E-07

5.03E-07

0.9 -1.61E-07

-1.61E-07

2.64E-07

1.0 2.00E-10

1.11E-16

-9.5E-15

E* =Exact Sol.-Approx. Sol.

49

Fig. 8: Dotted Curve represents OHAM solution (3.28a) while the solid Curve is for the exact

solution.

Fig. 9: Error graph for the OHAM Ist-order solution (3.28a).

50

Chapter 4

NUMERICAL SOLUTION OF THE STEADY TWO DIMENTIONAL

RADIAL FLOW OF VISCOUS FLUID USING OHAM & DTM

4.1 INTRODUCTION

In this chapter, we present the problem of steady two dimensional radial flow is

presented and the optimal homotopy asymptotic method and differential transform

method are employed to obtain approximate analytical solution to the fourth order

nonlinear boundary value problem governing the problem. The numerical results

reveal that the proposed methods are very efficient and accurate.

4.2 FORMULATION OF THE PROBLEM

Consider an incompressible fluid that flows outward in a trough between two

nonparallel plane walls, as shown in Fig. (1). We look for a solution such that the

flow is purely radial.

Fig.1

For such flows the velocity components and Zu u are zero in and Z direction. The

Navier-Stokes equations then become

Continuity: ( )

0rru

r

(4.1)

Momentum: 2

2 2

r rr

u upu

r r r

(4.2)

2

1 20 rup

r r

(4.3)

51

Equation (4.1) means that the product rr u is independent of r . For two dimensional

flows, this condition implies that

( )rr u (4.4)

where the right hand side is multiplied by the viscosity so that ( ) is

dimensionless.

Substitute the expression for ru into Eq.(4.2) and Eq.(4.3), we get

2 2 2

2

3 3 2

p d

r r r d

(4.5)

2

2

20

p d

r r d

(4.6)

Eliminating the pressure p from Eq.(4.5) and Eq.(4.6), we get

3

32 4 0.

d d dF

d d d

The boundary conditions are:

: 0 (no slip boundary)

0 : 0 (symmetry)

r

r

u

u

From Eq.(4.2), we have

2

2 2: 0rup

r r

The boundary conditions in terms of ( )F become:

2 2: ( ) 0, ( ) (the drag coefficient)

0 : (0) 0.

d d K

d d

(4.7)

Introducing the change of variable in Eq.(4.6), we get

3

3(2 4) 0

d d

d d

, (4.8)

and the associated boundary conditions are

2 20 : (0)=0, (0)

: ( ) 0.

d d K

d d

In the next section, we solve this using OHAM.

4.3 APPLICATION OH OHAM

Now we solve Eq.(4.8). Setting 0.087(angle of 5 degrees) , and 0.10K (the

drag coefficient) the boundary conditions of Eq.(4.8) become

52

(2) (1)(0) 0, (0) 0.10, (0.087) 0 .

According to the procedure of OHAM, the following zeroth, first and second-order

problems are formed:

Zeroth-order problem:

3

0

3

(2) (1)

0 0 0

0,

(0) 0, (0) 0.10, (0.087) 0.

d

d

(4.9)

First-order problem:

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

1 1 0 1 0 1 0 0( ) (1 ) ( ) 4 ( ) 2 ( ) ( ),C C C (4.10)

(2) (1)

0 0 0(0) 0, (0) 0, (0.087) 0.

Second-order problem:

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

2 2 0 2 0 0 1 1 0

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

1 1 1 0 1 2 0 1 1

(2) (1)

0 0 0

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

4 ( ) 2 ( ) ( ) ( ) (1 ) ( ),

(0) 0, (0) 0, (0.087) 0.

C C C

C C C C

(4.11)

Solutions of (4.9), (4.10) and (4.11) are respectively given by (4.12), (4.13) and

(4.14):

2

0( ) 0.0087 0.05 (4.12)

5 6 5 5 3

1 1

4 5

( ) (8.3333 10 4.35 10 0.0058

0.0166603592 8.7807 10 ).

C

(4.13)

5 5 2 5

2 1 1 2

3 2

1 1 2

4 2

1 1 2

5 5 2 5

1 1 2

6

( ) (8.7807 10 8.674 10 8.7807 10 )

(0.0057999999 0.0057414620 0.0057999999 )

(0.0166603592 0.0166604865 0.0166603592 )

(4.35 10 0.0011169390 4.35 10 )

(

C C C

C C C

C C C

C C C

5 2 5

1 1 2

5 2 7 5 2 8 8 2 9

1 1 1

8 2 10

1

8.3333 10 0.0021346839 8.3333 10 )

2.4854 10 0.00003.569 10 8.0556 10

9.2592 10 .

C C C

C C C

C

(4.14)

Using (4.12), (4.13) and (4.14), the second-order approximate OHAM solution is:

0 1 2( ) ( ) ( ) ( ). (4.15)

For 0 and 0.087a b , Least Squares gives us:

1

2

1.006793718.

0.0000313319.

C

C

Knowing 1 2 and C C the solution (4.15) is well determined.

53

Residual of the solution is:

3

3(2 4)

d dR

d d

. (4.16)

Table 1: In this table, we display numerical results for the OHAM solution (4.15) and

the residual (4.16) for different domain points. Results for the finite deference

solution, given in T.Y.Na. [64] are very poor to be compared.

Table 1

Fig. 2: Plot of the OHAM solution for the domain 0 0.087 .

x OHAM Sol. Residual

0.0 0.000 3.4310E-08

0.01 0.0000828831 -4.4662E-09

0.02 0.000155733 -1.4392E-08

0.03 0.000218522 -9.6683E-09

0.04 0.000271223 -1.4560E-10

0.05 0.000313816 8.0324E-09

0.06 0.000346284 1.1779E-08

0.07 0.000368614 1.0417E-08

0.08 0.000380797 5.0250E-09

54

Fig.3: Plot of the residual (4.16) in the domain 0 0.087

Fig. 4: Plot of the OHAM solution for the domain 0.087 0.087.

4.4 DIFFERENTIAL TRANSFORM METHOD(DTM)

If ( ) is a given function, its differential transform is defined as

0

1 ( )( )

!

k

k

dk

k d

. (4.17)

The inverse transform of ( )k is defined by

0

( ) ( )k

k

k

. (4.18)

In actual application, the function ( ) is expressed by a finite series

0

( ) ( )N

k

k

k

. (4.19)

Equation (4.18) implies that: 1

( ) ( )k

k N

k

, is negligibly small.

The basic operations of the DTM are given in table 2.

55

Table 2

Function Transformed function

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

( )( ) ( )mg ( )!

( ) ( )!

m kk G m k

k

1 2( ) ( ) ( ) 1 2

0

( ) ( ) ( )k

m

k m k m

( ) m ( ) ( )k k m 1,

0,

k m

k m

1 2( ) ( ) ( )... ( )m 1 3 2

1 2 2 1

1 1 2 2 1

0 0 0 0 1

( ) ( )...( ) ...

( )

m

m m

k k kk

k k k k m m

k k kk

k k

4.4.1 Analysis of the Method

In this section, analysis of the DTM is given for a general third order boundary

problem. For this purpose, consider the following bvp:

(3) ( ) ( , , , )g , a b (4.20)

with conditions:

0( )a , 1( )b , 2( )a , (4.21)

where : 0 , 1 , 2.i i are finite real constants and ( , , , )g is a continuous

function on interval ba , .

Differential transform of equation (4.20) gives

!

( 3) ( )( 3)!

kk G k

k

, (4.22)

where ( ) is the differential transform of ( , , , ).G k g

The differential transform of the boundary conditions (4.21) are:

0

0

( )N

k

k

a k

, 1

1

0

( )N

k

k

k b k

, 2

2

0

( 1) ( )N

k

k

k k a k

. (4.23)

Using equations (4.22) and (4.23), values for ( )k are obtained which give the

following series solution of 1( )NO .

0

( ) ( )N

k

k

k

+ 1( )NO .

The approximate solution is:

56

0

( ) ( )N

k

k

k

(4.24)

4.4.2 Application of the Method

The non-linear third order problem (4.8) for our study is,

3

3(2 4) 0

d d

d d

,

with the boundary conditions: (For 0.087 and 0.10K )

(2) (1)(0) 0, (0) 0.10, (0.087) 0

The differential transformations of the first two boundary conditions are:

0.10(0) 0 , (2)2!

. (4.25)

The differential transform of the third boundary condition is:

1

0

(0.087) ( ) 0N

k

k

k k

. (4.26)

The differential transform of the differential equation is:

0

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

( 3)!

k

m

kk k k m m k m

k

. (4.27)

Using Eqs.(4.25)-(4.27) and (4.19), for 10N , the following series solution of

11( )O is obtained.

2 3 4

5 6 7 8

9 10

( ) 0.008788884 0.05 0.005859256 0.01666023

0.001215796 0.002301264 0.000136713 0.000193132

0.000012557 0.00001404

(4.28)

Residual of the solution is: 3

3(2 4)

d dR

d d

. (4.29)

Fig. 5: Plot of the residual (4.29) in the domain 0 0.087 .

57

Table 3: In this table, we display numerical results for the DTM solution (4.28) and

the residual (4.29) for different domain points. Results via finite deference scheme,

given in T.Y.Na. [64] are very poor to be compared. From this table, it is clear that

DTM is very effective in small domains.

Table 3

x OHAM Sol. Residual

0.0 0.000 6.9389E-18

0.01 0.000082883 1.3878E-17

0.02 0.000155733 4.1633E-17

0.03 0.000218522 7.7369E-16

0.04 0.000271223 7.5669E-15

0.05 0.000313816 4.4466E-14

0.06 0.000346284 1.8865E-13

0.07 0.000368614 6.3879E-13

0.08 0.000380797 1.8337E-12

58

Chapter 5

OPTIMAL HOMOTOPY ASYMPTOTIC METHOD FOR THE

SOLUTION OF ZAKHAROV-KUZNETSOV EQUATION

In this chapter we solve a nonlinear dispersive Zakharov-Kuznetsov equation by

optimal homotopy asymptotic method. It is initial value problem and we to test the

proposed method in this case. The beauty of this method lies in the auxiliary function,

2

1 2 ...H p pC p C . in which 1 2, ,...C C are used to control and adjust the region

of convergence of the solution.

The obtained results are compared with those obtained by HPM and ADM. The

results reveal that in small domains, the optimal homotopy asymptotic method is

effective and reliable for initial value problems.

5.1 INTRODUCTION

The discovery of solitary waves by John Scott Russell [65] in 1834, attracted many

scientists to work on this concept. Zabuski and Kruskal[66] named these solitary

waves as solitons. Due to its significance in the scientific fields such as fluid

dynamics, astrophysics and plasma physics substantial work has been done in this

area. The KdV equation governs the height of the surface of shallow water in the

presence of solitary waves. It was discovered that solitons are solutions to the

Korteweg-de Vries (KdV) equation [67]. In simplest form, the KdV equation is

0. t x xxxu auu u

This equation appears in many areas like plasma physics, acoustic waves, heat

pulses etc.

Rosenau and Hyman [68] introduced a family of full non-linear partial differential

equations

( , ) : ( ) ( ) 0, 0, 1 3,m n

t x xxxK m n u a u u m n

to study the role of non-linear dispersion in the formation of patterns in liquid drops.

They discovered a class of solitary waves with compact support and finite wavelength

and they named these waves as compactons.

The best known two-dimensional generalizations of the KdV equations are the

Kadomtsov–Petviashivilli (KP) equation, and the Zakharov–Kuznetsov (ZK)

equation. The KP equation is given by

59

{ } 0,t x xxx x yyu auu u u

which characterizes small-amplitude, weakly dispersive waves on a fluid sheet, and it

accounts for slowly varying transverse perturbations of unidirectional KdV solitons

moving along the x-direction [45].The ZK equation in (2+1) dimensions, is given by

( ) 0.t x xx yy xu auu u u

The ZK equation governs the behavior of weakly nonlinear ion-acoustic waves in a

plasma comprising cold ions and hot isothermal electrons in the presence of a uniform

magnetic field.

Recently Ma Hong-Cai et al. [69], solved ZK equation by using the generalized

auxiliary equation method and new solitary pattern, solitary wave and singular

solitary wave solutions are found. B. Ganjavi et al. [70], used HPM and for solving

ZK equation.

We apply OHAM to the ZK equation. The main goal of this work is to solve the ZK

equation of the form ZK(m, n, k ):

( ) ( ) ( ) 0, ( , , ) 0,                       m n k

t x xxx yyxu a u b u c u m n k

where , , a b c are arbitrary constants and , , m n k are integers.

5.2 ANALYSIS OF THE METHOD FOR A PDE

We consider a general PDE

( ) ( ) ( ) 0,

, 0,

L u g r N u r

uB u n

n

(5.1)

where L is a linear operator, u is the unknown function of r which is to be

determined, r is an n-tupple, ( )g r is a source function, N is a nonlinear and B is a

boundary operator.

According to the OHAM procedure stated in section of chapter 2.2, we obtain the

following problems:

0 0 0( ) ( ) 0, , 0L u r g r B u u (5.2)

1 1 0 0 1 1, , 0L u r g r C N u r B u u (5.3)

The mth-order equation is given by:

60

1 0 0

1

0 1

1

( ( )) ( ( )) ( ( ))

( ( )) ( ), ( ),..., ( ) ,

2,3,..., , 0

k k k

k

i k i k i k i

i

k k

L u r L u r C N u r

C L u r N u r u r u r

k m B u u

(5.4)

The mth-order approximate solution is:

0 1 2

1

, ( ) , , ,...,m

i i i

i

u r C u r u r C C C

. (5.5)

Residual in this case is:

, ( ( , )) ( ) ( ( , ))i i iR r C L u r C g r N u r C . (5.6)

We will have the exact solution, u if 0R . Otherwise the residual is minimized

over the domain of the problem using the methods: Least squares, Ritz or Galerkin’s.

This minimization gives us the optimal values of , 1,2,3,...iC i .

We apply the Galerkin’s method for the above purpose for a fixed t and

( , )u u x y . In accordance with this method, we solve the following system:

1 2

0, 0, ... , 0

b d b d b d

ma c a c a c

u u uR dr R dr R dr

C C C

The values , 1,2,3,...iC i can also be obtained viva collocation by taking fixed

points iP in the domain of the problem using (5.6) and then solving ( 0, )iR C .

5.3 APPLICATION

We consider the following ZK (3, 3, 3) equation with an initial value

3 3 3( ) 2( ) 2( ) 0,t x xxx yyxu u u u

3

, ,0 sinh ,2 6

x yu x y

(5.7)

where the arbitrary constant is assumed to be 1.

By means of OHAM, taking 2

1 2( )H p pC p C , we obtain the following zero, first

and 2nd

order problems:

Zeroth-Order:

00

30, with initial condition ( , , 0) sinh

2 6

u x yu x y

t

First-Order:

61

2 220 0 0 0 0 01

1 1 0 1 1 0 2

3 2 3 220 0 0 0 0 0

1 1 0 1 0 1 02 2

32 0

1 0 13

(1 ) 3 12 12

12 24 6 36

6 , ( , ,0) 0.

u u u u u uuC C u C C u

t t x y x y x

u u u u u uC C u C u C u

x y x y x y x xu

C u u x yx

Second-Order:

220 0 0 0 02 1 1

2 1 2 0 1 0 1 2

2 2 2

0 0 0 0 0 0 01 11 2 0 1 1 1 02 2 2

3 2 220 0 01 1

2 1 0 1 1 0 2

3 6 12

24 12 12 12

12 3 12 12

u u u u uu u uC C C u C u u C

t t t t x x y xu u u u u u uu u

C C u C u C uy y x y x y x y xu u uu u u

C C u C C ux x y x y

2

01 11

2 2 2 2

0 0 0 0 0 01 12 0 1 1 1 0 1 0

3 3 3 22 20 0 0 0 0

2 0 1 0 1 1 0 2 02 2 2 2

2

0 0 11 1 1 02

36

24 24 24 24

6 12 6 36

36 36

u uC

x x xu u u u u uu u

C u C u C u C uy x y y x y y x y y x y

u u u u uC u C u u C u C u

x y x y x y x xu u u

C u C ux x x

2 3220 0 01

1 0 2 02 2 3

3 320 0

1 0 1 1 0 23 3

36 6

12 6 , ( , ,0) 0.

u u uuC u C u

x x x xu u

C u u C u u x yx x

The obtained solutions are:

Solution of zeroth-order problem:

0

3( , , )

2 6

x yu x y t Sinh

Solution of first-order problem:

11

3( , , ) 9 5

32 2 6

C x y x yu x y t t Cosh Cosh

Solution of second-order problem:

2

2 1 1

2

2 1 1

2 2

2 1

15 15 ( , , )

32 6 32 615 27 27

32 6 32 2 32 227 105 1863

32 2 512 6 1024

x y x yu x y t C t Cosh C t Cosh

x y x y x yC t Cosh C t Cosh C t Cosh

x y x yC t Cosh C t Sinh C

2 2

1

2 2

1

22295

1024 6 / 5

x yt Sinh

x yC t Sinh

We consider the following second order approximation:

0 1 2( , , ) ( , , ) ( , , ) ( , , )u x y t u x y t u x y t u x y t .

62

1

2

1 1 2

2

1 1 2

3( , , ) 5 9

32 6 215 15 15

32 6 32 6 32 627 27 27

32 2 32 2 32 23

2 6

C x y x yu x y t Cosh t Cosh

x y x y x yC t Cosh C t Cosh C t Cosh

x y x y x yC t Cosh C t Cosh C t Cosh

x ySinh

2 2 2 2

1 1

2 2

1

105 1863

512 6 1024 22295

1024 6 / 5

x y x yC t Sinh C t Sinh

x yC t Sinh

(5.8)

Residual of this solution is: 3 3 3( ) 2( ) 2( )t x xxx yyxR u u u u .

For 0.001,t we solve

0.5 0.5 0.5 0.5

1 20 0 0 0

0, 0,u u

R dydx R dydxC C

(5.9)

and obtain:

1 21.0279768796 & 2.0002481934.C C

Using these values, the second order approximation (3.2) becomes:

6

6

( , ) 1.5000002061 1.8287 10

6 2

0.00046875496 0.0008437589

6 2

2.2528 10 .

6 / 5

x y x yu x y Sinh Sinh

x y x yCosh Cosh

x ySinh

(5.10)

Figure 1: Plot of the solution (5.10)

For 0.1,y we display the numerical results for solution (5.10) in table 1.

63

Table1: In this table, we compare OHAM solution (5.10) with results obtained by

applying ADM and HPM. The residual for OHAM solution is computed at different

mesh points indicating sufficiently good accuracy although only two iterations are

performed.

Table 1

x OHAM Sol. ADM HPM Residual

0.00 0.025377251 0.024751123 0.024625267 -1.0552E-06

0.05 0.037881288 0.037253829 0.037126828 -6.6948E-07

0.10 0.050388426 0.049759121 0.049630498 -3.3234E-07

0.15 0.062899536 0.062267870 0.062137143 -4.3465E-08

0.20 0.075415486 0.074780942 0.074647632 1.9726E-07

0.25 0.087937148 0.087299208 0.087162832 3.8965E-07

0.30 0.100465394 0.099823536 0.099683610 5.3322E-07

0.35 0.113001096 0.112354796 0.112210833 6.2721E-07

0.40 0.125545128 0.124893859 0.124745368 6.7053E-07

0.45 0.138098363 0.137441595 0.137288085 6.6182E-07

0.50 0.150661676 0.149998876 0.149839851 5.9938E-07

Using again Eq.(5.9), with the interval of integration from 0 to 1, we obtain

1 21.005832689 & 2.023377226786,C C for 0.001.t For these values, the

second order approximation (5.8) becomes:

6 6 ( , ) 2.2674 10 1.8406 10

6 / 5 2

0.0008437395 0.0004687442

2 6

1.500000207 .

6

x y x yu x y Sinh Sinh

x y x yCosh Cosh

x ySinh

(5.11)

For 0.1,y we display the numerical results for the solution (5.11) in table 2.

64

Table 2

x OHAM Sol. Residual

0.0 0.025377243 -8.4394E-06

0.1 0.050388419 -6.4705E -06

0.2 0.075415479 -4.7138E -06

0.3 0.100465388 -3.1555E -06

0.4 0.125545122 -1.7864E -06

0.5 0.150661670 -6.0192E -07

0.6 0.175822037 3.9804E -07

0.7 0.201033244 1.2086E -06

0.8 0.226302331 1.8196E -06

0.9 0.251636360 2.2154E -06

1.0 0.277042417 2.3742E -06

Using again Eq.(5.9), we obtain 1 21.0279768796 & 2.0002481934,C C for

0.01.t For these values, Eq.(5.8) becomes:

( , ) 0.0084462 1.50002

2 6

0.00469233 0.000192256

6 2

0.000236837 .

6 / 5

x y x yu x y Cosh Sinh

x y x yCosh Sinh

x ySinh

(5.12)

For 0.1,y we display the numerical results for the solution (5.12) in table 3.

Figure 2: Plot for the solution (5.12) at y=0.1

65

Table 3

x OHAM Sol. Residual

0.00 0.028775435 -1.3400E-04

0.05 0.041295863 -9.3262E-05

0.10 0.053823905 -5.7657E-05

0.15 0.066360457 -2.7151E-05

0.20 0.078906416 -1.7412E-06

0.25 0.091462684 1.8542E-05

0.30 0.104030166 3.3638E-05

0.35 0.116609772 4.3456E-05

0.40 0.129202415 4.7871E-05

0.45 0.141809019 4.6727E-05

0.50 0.154430487 3.9832E-05

0.1 0.2 0.3 0.4 0.5

0.0001

0.00005

0.00005

Figure 3: Plot of the residual(for solution 5.12) at t=0.01 & y=0.1

For 0.25,y we display the numerical results for the solution (5.12) in table 4.

66

Table 4

x OHAM Sol. Residual

0.00 0.066360457 -2.7151E-05

0.05 0.078906416 -1.7412E-06

0.10 0.091462684 1.8542E-05

0.15 0.104030166 3.3638E-05

0.20 0.116609772 4.3456E-05

0.25 0.129202415 4.7871E-05

0.30 0.141809013 4.6727E-05

0.35 0.154430487 3.9832E-05

0.40 0.167067765 2.6959E-05

0.45 0.179721776 7.8447E-06

0.50 0.192393456 -1.7812E-05

For 0.5,y we display the numerical results for the solution (5.12) in table 5.

Table 5

x OHAM Sol. Residual

0.00 0.129202415 4.7871E-05

0.05 0.141809013 4.6727E-05

0.10 0.154430487 3.9832E-05

0.15 0.167067765 2.6959E-05

0.20 0.179721776 7.8447E-06

0.25 0.192393456 -1.7812E-05

0.30 0.205083745 -5.0354E-05

0.35 0.217793589 -9.0163E-05

0.40 0.230523937 -1.3767E-04

0.45 0.243275745 -1.9334E-04

0.50 0.256049972 -2.5771E-04

For 1,y we display the numerical results for the solution (5.12) in table 6.

67

Table 6

x OHAM Sol. Residual

0.0 0.256046827 2.5264E-04

0.1 0.281667078 2.7162E-04

0.2 0.307388716 2.6025E-04

0.3 0.333219638 2.1427E-04

0.4 0.359167841 1.2842E-04

0.5 0.385241425 -3.6642E-06

0.6 0.411448602 -1.8962E-04

0.7 0.437797698 -4.3853E-04

0.8 0.464297162 -7.6110E-04

0.9 0.490955568 -1.1699E-03

1.0 0.517781630 -1.6797E-03

68

Chapter 6

APPLICATION OF OHAM TO INTEGRO-DIFFERENTIAL

EQUATIONS

6.1 INTRODUCTION

This chapter includes the application of optimal homotopy asymptotic method to

integro-differential equations. Many mathematical formulations of physical

phenomena contain integro-differential equations. These equations arise in fluid

dynamics, biological models, chemical kinetics, population dynamics, nuclear

reactors, viscoelasticity, wave propagation, image processing, economics and

engineering systems. Exact solutions of integro-differential equations(IDEs) are rare

and thus it is required to obtain an efficient approximate solution. There has recently

been much attention paid to the search for better and more efficient solution methods

for determining approximate solutions. Theorems which list the conditions for the

existence and uniqueness of solutions of such problems are contained in a book by

Agarwal [73] though no numerical method is presented therein. Two point boundary

value problems for integro-differential equation of second order is discussed by J.

Morchalo in [74]. Also, J. Morchalo [75] studied two point boundary value problems

for integro-differential equation of higher order. A reliable algorithm for solving

boundary value problems for higher-order integro-differential equation has been

proposed by A. M-Wazwaz [76].

We develop OHAM for nth-order integro-differential equation and is applied to two

first-order and two fourth order problems.

Numerical results show excellent accuracy and efficiency of the proposed

algorithm.

6.2 ANALYSIS METHOD

We consider the nth order integro-differential equation of the form:

( ) ( 1)

0( ) ( ) ( ) ( , ) ( ( )) ,

xn nu x f x u x K x t F u t dt (6.1)

with boundary conditions:

69

( )

( )

(0) , 0,1,2,..., ( -1),

( ) , , 1,..., ( -1),

i

i

i

i

u i j

u b i j j n

where ( ) ( )nu x is nth derivative of the unknown function ( ),u x ( ( ))F u x is a nonlinear

function. The kernel ( , )K x t and ( )f x are real and derivable functions on [0, ]b and

and i i are real finite constants.

We develop OHAM for (6.1). In the first step, we construct a family of curves

defined by the homotopy equation;

0(1 ) ( ( , ) ( )) ( , )( ( ( , )) ( ) ( , ) ( ( , )) ),

x

ip L x p f x H p C L x p f x K x t F t p dt (6.2)

along with the boundary conditions:

( ( , ), ) 0.B x px

(6.3)

Where L is a linear, F is a nonlinear and B is a boundary operator. ( , )iH p C is

auxiliary function with [0,1]p as an embedding parameter and 'iC s are

convergence control parameters. For 0p , ( , ) 0iH p C , and for 0p ,

( , ) 0iH p C .

Eq.(6.2) satisfies

( ( ,0)) ( ), for 0,L x f x p (6.4)

0( ( ,1)) ( ) ( , ) ( ( ,1)) , for 1.

x

L x f x K x t F t dt p (6.5)

Solution of Eq.(6.4) is denoted by 0 ( )u x and is called initial solution which satisfy the

boundary conditions. It serves as an initial or starting point for the scheme. As p

moves from 0 to 1, the initial solution 0 ( )u x traces continuously the solution curve

( ) ( ,1)u x x , of the given problem (6.1).

For 0p , we have the linear ODE, ( ( )) ( )L u x f x , and for 1p , we get the

complete problem, 0

( ( )) ( ) ( , ) ( ( )) .x

L u x f x K x t F u t dt

Now we go to the auxiliary function ( , )iH p C . It is interesting to note that the shape

of this function is not fixed and one can choose the best which suits the given

problem. Some of its forms have been discussed by Marinca et.al [14-17]. The most

famous and simple form of this auxiliary function is

2

1 2( , ) ...iH p C pC p C (6.6)

70

Next, the unknown function ( , )x p is expanded in usual way as

0 1

1

( , , ) ( ) ( , ,..., ) .

k

i k k

k

x p C u x u x C C p (6.7)

Plugging in Eq.(6.6) and Eq.(6.7) into Eqs.(6.2, 6.3), and then equating the like

powers of p , we get the following linear problems which are directly integrable:

Zeroth-order problem:

( )

0

( )

0

( )

0

( ) ( ),

(0) , 0,1,2,..., ( -1),

( ) , , 1,..., ( -1).

n

i

i

i

i

u x f x

u i j

u b i j j n

(6.8)

First-order problem:

( ) ( ) ( )

1 1 0 1 0

( 1)

0 1 0 00

( )

1

( )

1

( ) ( 1) ( ) ( ) ( ( )

( )) ( , ) ( ( )) ,

(0) 0, 0,1,2,..., ( -1),

( ) 0, , 1,..., ( -1).

n n n

xn

i

i

u x C f x u x C u x

u x C K x t F u t dt

u i j

u b i j j n

(6.9)

Second-order problem:

( ) ( ) ( 1) ( ) ( 1)

2 1 1 1 1 2 0 0 2

1 1 0 1 2 0 00 0

( ) ( )

2 2

( ) (1 ) ( ) ( ) ( ( ) ( )) ( )

( , ) ( ( ), ( )) ( , ) ( ( )) ,

(0) 0, 0,1,2,..., ( -1), ( ) 0, , 1,..., ( -1).

n n n n n

x x

i i

u x C u x C u x C u x u x C f x

C K x t F u t u t dt C K x t F u t dt

u i j u b i j j n

(6.10)

Where the nonlinear term ( ( ))F u x is decomposed in the following manner:

2

0 0 1 0 1 1 0 1 2( ( )) ( ( )) ( ( ), ( )) ( ( ), ( ), ( )) ...F u x F u x p F u x u x p F u x u x u x

Higher order problems can be constructed easily in the similar way.

For 1p , if the series (6.7) converges, then

0 1

1

( ,1, ) ( , ) ( ) ( , ,..., ).i i k k

k

x C u x C u x u x C C

In actual application, the truncated series is taken and the approximate solution then

becomes:

0 1

1

( , ) ( ) ( , ,..., ),M

i k k

k

u x C u x u x C C

(6.11)

where M is the order of approximation or solution and 'iC s are to be determined.

In most of the cases approximate solution upto the second order i.e,

1 2 0 1 1 2 1 2( , , ) ( ) ( , ) ( , , )u x C C u x u x C u x C C , (6.12)

71

is enough to produce excellent results. We confine ourselves to this order of solution

for the rest of OHAM procedure.

To determine 'iC s , we plug in Eq.(6.11) into Eq.(6.1), and obtain the following

residual:

( ) ( 1)

0( , ) ( , ) ( , ) ( ) ( , ) ( ( , )) .

xn n

i i i iR x C u x C u x C f x K x t F u t C dt

Generally, 0R , but we can minimize it over the domain of the problem. Marinca et

al. [14-17] reported many methods to achieve this goal. We use the Galerkin’s method

for this purpose. According to this method the following system is to be solved for

1 2 and .C C

1 2

0, 0.

b b

a a

u uR dx R dx

C C

Having these values the second order approximate solution is determined.

6.3 NUMERICAL EXAMPLES

Example 1: First order non-linear problem

2

0

( ) 1 ( )

x

u x u t dt ,

where [0,1] and (0) 0.x u

According to the procedure discussed in section 6.2, the following problems are

obtained:

Zeroth-order problem:

0 0( ) 1, (0) 0.u x u (6.13)

First-order problem:

2

1 1 0 1 0 1

0

( ) (1 )(1 ( )) ( ) , (0) 0.

x

u x C u x C u t dt u (6.14)

Second-order problem:

2

2 2 2 0 1 0 1 2 0

0 0

1 1 2

( ) ( ) 2 ( ) ( ) ( )

(1 ) ( ), (0) 0.

x x

u x C C u t dt C u t u t dt C u x

C u x u

(6.15)

Solutions of the problems (6.13), (6.14) and (6.15) are:

0( ) .u x x

72

4

11 1( , ) .

12

C xu x C

4 2 4 4 2 7

2 1 2 1 1 2 1( , , ) 1/ 252 (21 21 21 ).u x C C C x C x C x C x

The second-order approximate solution is:

2 1 2 0 1 1 2 1 2

4 2 4 4 2 7

1 1 2 1

( , , ) ( ) ( , ) ( , , )

.6 12 12 252

u x C C u x u x C u x C C

C x C x C x C xx

Using the Galerkin’s method, we find

1 20.962737702, 3.850657281.C C

For these values our solution is:

4 7( ) 0.0831932 0.00367803 .u x x x x (6.16)

Numerical results for the solution (6.16) are displayed in table 1. Accuracy of the

results can be improved by increasing the order of the solution

Table 1

x OHAM Sol. Residual

0.0 0 0

0.1 -0.09999168 -5.5868E-07

0.2 -0.199866938 -4.3591E-06

0.3 -0.299326940 -1.3722E-05

0.4 -0.397876281 -2.8168E-05

0.5 -0.494829162 -4.2156E-05

0.6 -0.589321127 -4.4376E-05

0.7 -0.680328222 -2.2109E-05

0.8 -0.766695418 2.3807E-05

0.9 -0.847176155 4.5728E-05

1.0 -0.920484865 -1.1209E-04

73

Figure 1: Graph of the residual for example 1.

Example 2: Another first order non-linear problem

0

( ) 1 ( )

xdu

u x u t dtdt

,

where [0,1] and (0) 0.x u

According to the procedure discussed in section 6.2, the following problems are

obtained:

Zeroth-order problem:

0 0( ) 1, (0) 0.u x u (6.17)

First-order problem:

1 1 0 1 0 0 1

0

( ) (1 )( ( ) 1) ( ) ( ) , (0) 0

x

u x C u x C u t u t dt u . (6.18)

Second-order problem:

2 2 0 1 1 2 0 0

0

1 0 1 0 1 2

0

( ) ( ( ) 1) (1 ) ( ) ( ) ( )

( ( ) ( ) ( ) ( )) , (0) 0.

x

x

u x C u x C u x C u t u t dt

C u t u t u t u t dt u

(6.19)

Solutions of the problems (6.17), (6.18) and (6.19) are

0 ( ) .u x x

3

11 1( , ) .

6

C xu x C

3 2 3 3 2 5

2 1 2 1 1 2 1( , , ) 1/ 30 (5 5 5 ).u x C C C x C x C x C x

We consider the second-order solution:

74

2 1 2 0 1 1 2 1 2

3 2 3 3 2 5

1 1 2 1

( , , ) ( ) ( , ) ( , , )

.3 6 6 30

u x C C u x u x C u x C C

C x C x C x C xx

Using the Galerkin’s method, we find

1 21.233570579, 0.006543424.C C

For these values our solution is:

3 5( ) 0.156484 +0.0507232 .u x x x x (6.20)

Numerical results for the solution (6.20) are displayed in table 2. Accuracy of the

results can be improved by increasing the order of the solution.

Figure 2: Graph of the residual for example 2.

Table 2

x OHAM Sol. Residual

0.0 0 0

0.1 0.100156991 -2.9584E-04

0.2 0.201268100 -1.0706E-03

0.3 0.304348314 -2.0091E-03

0.4 0.410534354 -2.6645E-03

0.5 0.521145545 -2.5827E-03

0.6 0.637744686 -1.4882E-03

0.7 0.762198911 4.5046E-04

0.8 0.896740565 2.2578E-03

0.9 1.044028065 1.6552E-03

1.0 1.207206773 -5.6074E-03

75

Example 3: Linear problem of order four

( )

0( ) (1 exp( )) 3exp( ) ( ) ( ) , 0 1,

xivu x x x x u x u t dt x

(0) 1, (0) 1, (1) 1 , (1) 2 .u u u e u e

The exact solution is: ( ) 1 .xu x xe

According to the procedure discussed in section 6.2, the following problems are

obtained:

Zeroth-order problem:

(4)

0

0 0 0 0

( ) 1,

(0) 1, (0) 1, (1) 1 , (1) 2 .

u x

u u u e u e

(6.21)

First-order problem:

(4) (4)

1 1 0 1 0 1 0 1

0

1 1 1 1

( ) (1 ) ( ) ( ) ( ) ( 3),

(0) 0, (0) 0, (1) 0, (1) 0.

x

xu x C u x C u x C u t dt C xe x

u u u u

(6.22)

Second-order problem:

(4) (4) (4)

2 1 1 2 0 1 1 2 0

2 2 0 1 1

0 0

( ) (1 ) ( ) ( ) ( ) ( )

( 3) ( ) ( ) ,

x x

x x

u x C u x C u x C u x C u x

C e xe x C u t dt C u t dt

(6.23)

with boundary conditions: 2 2 2 2(0) 0, (0) 0, (1) 0, (1) 0.u u u u

The problems (6.21), (6.22) and (6.23) are directly integrable and their solutions are

obtained easily.

We consider the second order approximations given by

2 1 2 0 1 1 2 1 2( , , ) ( ) ( , ) ( , , )u x C C u x u x C u x C C .

Using the Galerkin’s method, we solve the system

1 2

0, 0.

b b

a a

u uR dx R dx

C C

and find 1 21.000753860 and 4.003015722.C C

The solution thus becomes:

76

2 3 4

5 6 7 8

5 9 6 10 7 11 9 12

9 13 14

( ) 1 0.999999988 0.500000048 0.166666619

0.041666655 0.008333381 0.001388768 0.000198576

2.4806 10 2.6006 10 3.7915 10 2.969 10

3.0529 10 ( ).

u x x x x x

x x x x

x x x x

x O x

(6.24)

Numerical results for the solution (6.24) are displayed in table 3.

Table 3

x OHAM Sol. E*

0.0 1.0000 0.0000

0.1 1.110517092 7.7102E-11

0.2 1.244280551 1.7138E-10

0.3 1.404957642 1.6548E-10

0.4 1.596729879 5.3070E-11

0.5 1.824360635 -8.8474E-11

0.6 2.093271280 -1.5771E-10

0.7 2.409626895 -1.0851E-10

0.8 2.780432743 -1.4894E-12

0.9 3.213642800 3.8564E-11

1.0 3.718281828 -1.7647E-15

E*=Exact-Approx.

Figure 3: Solid curve-Exact sol. Dotted curve-OHAM sol.

77

Figure 4: Graph of the residual for example 3.

Example 4: Nonlinear problem of order four

( ) 2

0( ) 1 ( ) , 0 1,

xiv tu x e u t dt x

(0) 1, (0) 1, (1) , (1) .u u u e u e

The exact solution of this problem is ( ) xu x e .

According to the procedure discussed in section 6.2, the following problems are

obtained:

Zeroth-order problem:

(4)

0

0 0 0 0

( ) 1,

(0) 1, (0) 1, (1) , (1) .

u x

u u u e u e

(6.25)

First-order problem:

(4) (4) 2

1 1 0 1 0

0

1 1 1 1

( ) (1 )( ( ) 1) ( ) ,

(0) 0, (0) 0, (1) 0, (1) 0.

x

tu x C u x C e u t dt

u u u u

(6.26)

Second-order problem:

(4) (4) (4)

2 2 0 1 1

2

2 0 1 0 1

0 0

2 2 2 2

( ) ( ( ) 1) (1 ) ( )

( ) 2 ( ) ( ) ,

(0) 0, (0) 0, (1) 0, (1) 0.

x x

t t

u x C u x C u x

C e u t dt C e u t u t dt

u u u u

(6.27)

The problems (6.25), (6.26) and (6.27) are directly integrable and their solutions are

obtained easily.

We consider the second-order solution

15

2 1 2 0 1 1 2 1 2( , , ) ( ) ( , ) ( , , ) ( )u x C C u x u x C u x C C O x .

Using the Galerkin’s method, we find

78

1 20 and 1.001331910.C C

The solution becomes:

2 3 4 5

6 7 8 6 9

7 10 7 11 8 12 10 13

9 14

( ) 1 0.500003 0.166659 / 24 0.00834443

0.00139074 0.000181376 0.0000342871 2.7908 10

-3.3706 10 2.21015 10 1.68469 10 7.34336 10

1.43975 10

u x x x x x x

x x x x

x x x x

x

(6.28)

Numerical results for the solution (6.28) are displayed in table 4.

If the method of Least squares is used to determine 'iC s , then

1 20.998739362, 3.994966024.C C

For these values the approximate solution is:

2 3 4 5

6 7 8 6 9

7 10 8 11 8 12 10 13

9 14

( ) 1 0.500000041 0.166666591 / 24 0.008333392

0.0013888986 0.0001983497 0.0000248359 2.7246 10

3.3507 10 2.2059 10 1.6466 10 7.8934 10

1.3966 10 .

u x x x x x x

x x x x

x x x x

x

(6.29)

Numerical results for the solution (6.29) are displayed in table 4.

Table 4: In this table, we compare the errors obtained by using Galerkin’s method

and the method of Least squares for determining the convergence control parametrs

1C and 2C .

Table 4

x E*(6.28) E*(6.29)

0.0 0 0

0.1 -2.0165E-08 -3.3887E-10

0.2 -5.3455E-08 -1.0685E- 09

0.3 -6.9629E-08 -1.8172E-09

0.4 -5.7490E-08 -2.3265E-09

0.5 -2.4575E-08 -2.4695E-09

0.6 1.0375E-08 -2.2363E-09

0.7 2.9168E-08 -1.7013E-09

0.8 2.5253E-08 -9.9204E-10

0.9 8.4974E-09 -2.7455E-10

1.0 -1.2808E-09 2.6119E-10

E*=Exact-Approx.

79

Fig. 5: Graph of the errors for example 4.

Solid curve-E*(6.28) & Dotted curve- E*(6.29)

80

Chapter 7

APPLICATION OF OHAM WITH DAFTARDAR-JAFARI

POLYNOMAILS FOR NON-LINEAR PROBLEMS

7.1 INTRODUCTION

In this chapter we present a new idea of using Daftardar-Jafari polynomials in the

optimal homotopy asumptotic method (OHAM) for the analytical solution of non-

linear functional equations. These polynomials appeared for the first time in DJM [77]

(Daftardar-Gejji and Jafari method) in 2006. We name these polynomials as DJ

Polynomails. For non-linear function we plug in these polynomials in the homotopy

used in OHAM and then follow the rest of procedure of OHAM. It brings more

accurate results than using the built-in polynomials used in OHAM for non-linear

functions. This new approach is almost as simple as OHAM and leads to explicit

equations that are directly integrable. We name this new procedure as OHAM with

DJ-Polynomials and abbreviate it as OHAM-DJ.

To test this method, we solve two non-linear ordinary differential equations of order

fifth and sixth, one nonlinear partial differential equation namely the ZK(3, 3, 3)

equation and three non-linear integro-differential equations; two of order one and one

of order fourth.

The obtained results are compared with results of the usual or standard OHAM.

7.2 ANALYSIS OF THE METHOD FOR DIFFERENTIAL EQUATIONS

Consider the problem

Lu Fu f , r (7.1)

, 0,u

B un

where L , F , B and f appear with the usual meanings. According to OHAM, we

construct a homotopy defined by

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

,, , 0

q L v r q f r h q L v r q f r F v r q

v r qB v r q

n

(7.2)

81

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

0q , and it is 0 for 0q . The unknown function ,v r q starts form 0( ,0) ( )v r u r

to ( ,1) ( )v r u r as q starts from 0 to 1.

The auxiliary function ( )h q is chosen in the form (it is a commonly used form),

1

( )m

i

i

i

h q q C

, (7.3)

where 1 2, ,...C C , are the convergence control parameters which are to be determined.

We will use this function unless otherwise stated.

Next, we use Taylor’s series to expand the function ( , )v r q about q .

0 1 2

1

( , ) ( ) ( , , ,..., ) m

m m

m

v r q u r u r C C C q

(7.4)

The nonlinear function ( ( , ))F v r p is decomposed as

2

0 0 1 0 0 1 2 0 1( ( , )) ( ) [ ( ) ( )] [ ( ) ( )] ...F v r p F u q F u u F u q F u u u F u u

The expressions on the right hand side

0 0 1 0 0 1 2 0 1( ), [ ( ) ( )], [ ( ) ( )], . . .F u F u u F u F u u u F u u

are DJ-Polynomials. Actually, these polynomials are the terms of the Taylor’s series

of the nonlinear term. Convergence of these polynomials has been proved by Sachin

Bhalekar and Varsha Daftardar-Gejji [78].

For simplicity and convenience these polynomials are expressed as:

0 0

1

0 0

( )

m m

m i i

i i

F F u

F F u F u

We can now express

0

1

( ( , )) k

k

k

F v r q F q F

(7.5)

Substituting, Eqs. (7.3), (7.4) and (7.5) into Eq. (7.2) and equating the coefficient of

like powers of q , we obtain the following linear equations which can be integrated

directly:

Zeroth order problem:

0 ( ),Lu f r 00 , 0.

uB

nu

(7.6)

First order problem:

82

11 1 0 1 0 1(1 ) ( ), , 0.

uLu C Lu C F f B u

n

(7.7)

Second order problem:

2 1 1 1 1 2 0 0(1 ) ( )Lu C Lu C F C Lu F f , 22 , 0

uB u

n

. (7.8)

Though we can construct higher order problems easily but solutions upto the second

order problems are enough to produce excellent results.

For 1q , if the series (7.4) converges, then

0 1 1 2 1 2( ) ( ) ( ) ( , ) ( , , ).v r u r u r u r C u r C C (7.9)

By substituting Eq. (7.9) into Eq. (7.1), the resulting residual is

1 2 ( ( ) ( ( )) ( )( , , ) L u r F u r f rR r C C . (7.10)

We have the exact solution u if 0R . Otherwise, we minimize R over domain of

the problem. We follow two methods: the method of Least Squares and the

Galerekin’s method.

According to the method of Least Squares, we first construct the functional

1 2

1 2

2

1 2( , ) ...n

n

bb b

a a a

J C C R dr , (7.11)

and then minimizing it, we have

1 2

0J J

C C

, (7.12)

and according to the Galerekin’s method, we solve the following system:

1 2 1 2

1 2 1 21 2

... 0, ... 0 n n

n n

b bb b b b

a a a a a a

u uR dr R dr

C C

(7.13)

7.3 AN ILLUSTRATION ABOUT THE POLYNOMIALS OF OHAM AND

DJM

Suppose we have a nonlinear term 2u in a functional equation. If this function is

expanded in perturbation series, we then have:

2 2 3 2

0 1 2 3

2 2 2 2

0 1 0 1 2 0 2 1 2

3 2

0 3 1 3 2 3 3

( )

( 2 ) ( 2 2 )

(2 2 2 )

u u qu q u q u

u q u u u q u u u u u

q u u u u u u u

(7.14)

Relating to the power of q , the OHAM built-in polynomials for this nonlinear term

are:

83

2

0 0 0( )F u u

1 0 1 0 1( , ) 2F u u u u

2

2 0 1 2 1 0 2( , , ) 2F u u u u u u

3 0 1 2 3 1 2 0 3( , , , ) 2 2F u u u u u u u u

We notice that these polynomials took only six terms of the series (7.14) while the

series (7.14) is consisting of ten terms.

Relating to the power of q , the polynomials in DJM for this nonlinear term are:

2

0 0 0( )F u u

2

1 0 1 1 0 1( , ) 2F u u u u u

2

2 0 1 2 2 0 2 1 2( , , ) 2 2F u u u u u u u u

2

3 0 1 2 3 0 3 1 3 2 3 3( , , , ) 2 2 2 F u u u u u u u u u u u

We notice that these polynomials took all the terms of the expansion (7.14). It clearly

indicates the superiority of DJM-polynomials over the OHAM polynomials.

7.4 ANALYSIS OF THE METHOD FOR Nth-ORDER INTEGRO-

DIFFERENTIAL EQUATIONS

Consider the nth order integro-differential equation of the form:

( ) ( 1)

0( ) ( ) ( ) ( , ) ( ( )) .

xn nu x f x u x K x t F u t dt (7.15)

with boundary conditions:

( )

( )

(0) , 0,1,2,..., ( -1),

( ) , , 1,..., ( -1),

i

i

i

i

u i j

u b i j j n

where ( ) ( )nu x is nth derivative of the unknown function ( ),u x ( ( ))F u x is a nonlinear

function. The kernel ( , )K x t and ( )f x are real and derivable functions on [0, ]b and

and i i are real finite constants.

We develop OHAM with Daftardar-Jafari Polynomials for (7.15).

As a first step, we construct a family of curves defined by the homotopy equation

0(1 ) ( ( , ) ( )) ( , )( ( ( , )) ( ) ( , ) ( ( , )) ),

x

iq L v x q f x H q C L v x q f x K x t F v t q dt (7.16)

with

( , ), 0.v

B v x qx

(7.17)

84

where L is a linear, F is a nonlinear and B is a boundary operator. ( , )iH q C is

auxiliary function with [0,1]q as an embedding parameter and 'iC s are

convergence control parameters. For 0q , ( , ) 0iH q C , otherwise ( , ) 0iH q C .

Eq.(7.16) satisfies:

( ( ,0) ( ), for 0,L v x f x q (7.18)

0

( ( ,1)) ( ( )) ( ) ( , ) ( ( )) . for 1.x

L v x L u x f x K x t F u t dt q (7.19)

Solution of Eq.(7.18) is denoted by 0 ( )u x and is called initial solution which satisfy

the boundary conditions. It serves as an initial or starting point. As q moves from 0 to

1, the initial solution 0 ( )u x traces continuously the solution curve ( ) ( ,1).u x v x

For 0q , we have the linear part ( ( )) ( )L u x f x of our problem (7.15) and for

1q , we get our complete problem 0

( ( )) ( ) ( , ) ( ( )) .x

L u x f x K x t F u t dt

Now we go to the auxiliary function ( , )iH q C . It is interesting to note that the shape

of this function is not fixed and one can choose the best which suits the given

problem.

The most famous and simple form is

2

1 2( , ) ...iH q C qC q C (7.20)

Next, ( , )v r q is expanded in the following manner

0 1 2

1

( , ) ( ) ( , , ,..., ) .m

m m

m

v x q u x u x C C C q

(7.21)

The nonlinear function ( ( , ))F v x q is expanded in Taylor’s series about q in the

following manner

2

0 0 1 0 0 1 2 0 1( ( , )) ( ) [ ( ) ( )] [ ( ) ( )] ...F v x p F u q F u u F u q F u u u F u u

This can be expressed as discussed in above section as

0

1

( ( , )) k

k

k

F v x q F q F

(7.22)

where

0 0

1

0 0

( )

m m

m i i

i i

F F u

F F u F u

85

Plugging in Eq.(7.20), Eq.(7.21) and Eq.(7.22) into Eqs.(7.16, 7.17), and then

equating the like powers of q , we get the following linear problems which are

directly integrable.

Zeroth-order problem:

( )

0

( )

0

( )

0

( ) ( ),

(0) , 0,1,2,..., ( -1),

( ) , , 1,..., ( -1).

n

i

i

i

i

u x f x

u i j

u b i j j n

(7.23)

First-order problem:

( ) ( )

1 1 0 1 1 00

( )

1

( )

1

( ) (1 ) ( ) ( ) ( , ) ,

(0) 0, 0,1,2,..., ( -1),

( ) 0, , 1,..., ( -1).

xn n

i

i

u x C u x C f x C K x t F dt

u i j

u b i j j n

(7.24)

Second-order problem:

( ) ( ) ( )

2 1 1 2 0 2

1 0 2 10 0

( ) (1 ) ( ) ( ) ( )

( , ) ( , ) ,

n n n

x x

u x C u x C u x C f x

C K x t F dt C K x t F dt

(7.25)

( )

2

( )

2

(0) 0, 0,1,2,..., ( -1),

( ) 0, , 1,..., ( -1).

i

i

u i j

u b i j j n

Higher order problems can be constructed in the similar way.

If the series (7.21) is convergent at 1q , then

0

1

( ,1, ) ( , ) ( ) ( , ).i i k k

k

v x C u x C u x u x C

In actual application, we consider the following truncated series

0

1

( , ) ( ) ( , ),M

i k k

k

u x C u x u x C

(7.26)

where M is the order of the solution and 'iC s are to be determined. We confine our

study to the following second order OHAM approximation

1 2 0 1 1 2 1 2( , , ) ( ) ( , ) ( , , ).u x C C u x u x C u x C C

To determine 'iC s , we plug in Eq.(7.26) into Eq.(7.15), and obtain the following

residual

( )

0( , ) ( , ) ( ) ( , ) ( ( , )) .

xn

i i iR x C u x C f x K x t F u t C dt

86

Many methods can be applied to determine 'iC s . Here we apply the Galerkin’s

method. In accordance with this method, the following system is to be solved for

1 2 and C C :

1 2

0, 0.

b b

a a

u uR dx R dx

C C

7.5 APPLICATION OF OHAM-DJ

7.5.1 Ordinary Differential Equations

Example 1: Fifth order non-linear bvp

( ) 2( ) ( ) , 0 1,v xx u x e xu

(0) 1, (0) 1, (0) 1, (1) , (1) .u u u u e u e

The exact solution for this problem is ( ) .xu x e

We consider the following second order approximation

13

0 1 1 2 1 2( ) ( ) ( , ) ( , , ) ( ).u x u x u x C u x C C O x

Using the Galerkin’s method, we obtain the following values of ,

iC s

for 0a and 1:b

1 20.945336440 0.003029798.,C C

The second order approximate solution is

2 3 4 5

6 7 5 8 6 9

7 10 8 11 9 12 13

0.1666668 0.041666251 0.008333681

0.001388947 0.000198421

( ) 1 / 2

10 10

1

2.4427 3.0627

2.1695 +1.9868 1.240 10 1073 ( )

u x x x x x x

x x x x

x x x O x

(7.27)

Numerical results for the solution (7.27) are displayed in Table 1.

Table 1: In this table, we compare the results of the second order solution (7.27) of

OHAM-DJ method with the second order OHAM solution by computing the error at

different points of the problem domain. In both the methods the same homotopy, the

same auxiliary function, the same method (Galerkin’s) for determining convergence

controlling parameters and the same order solutions are used. Numerical results show

that this method provides more accuracy than the standard OHAM.

87

Table 1

x E*(OHAM) E*(OHAM-DJ)

0.0 0.0000 0.0000

0.1 -9.2E-10 -9.5E-11

0.2 -5.0E-09 -5.2E-10

0.3 -1.1E-08 -1.1E-09

0.4 -1.5E-08 -1.5E-09

0.5 -1.6E-08 -1.6E-09

0.6 -1.4E-08 -1.4E-09

0.7 -9.9E-09 -9.5E-10

0.8 -5.6E-09 -5.3E-10

0.9 -1.0E-09 -1.1E-10

1.0 4.2E-09 3.7E-10

E*=Exact-Approx.

Figure 1: Solid curve-E*(OHAM) & Dotted curve-E*(OHAM-DJ)

Example 2: Sixth order non-linear bvp

( ) 2( ) ( ), 0 1,vi xu ux e x x

(0) 1, (0) 1, ''''(0) 1, (1) , ''(1) , ''''(1) .u u u u e u e u e

Exact solution of this problem is ( ) .xu x e

Let us try a different auxiliary function 1 2( ) ( )xh q q C C e and consider the second

order OHAM approximation

13

0 1 1 2 2 1 2( ) ( ) ( , , ) ( , , ) ( ).u x u x u x C C u x C C O x

For 0a and 1b , we obtain the following values of ,

iC s by Galerkin’s method:

88

1 20.999015489, 0.001829698.C C

Knowing these values the approximate solution is

2 3 4 5

6 7 5 8 6 9

7 10 8 11 9 12 13

( ) 0.166666667 041666667 0.008333338

001388878 000198421 2.4800 2.7549

2.758

1 0.5

2 2.5106 +2.0414 ( )

0. 0.

0. 0. 10 10

10 10 1 .0

x x x x x

x x x x

x

u x

O xx x

(7.28)

Numerical results of the solution (7.28) are displayed in Table 2.

Table 2: In this table, we compare the results of the second order solution (7.28) of

OHAM-DJ method with the second order OHAM solution by computing the error at

different points of the problem domain. In both the methods the same homotopy, the

same auxiliary function, the same method (Galerkin’s) for determining convergence

controlling parameters and the same order solutions are used. Numerical results show

that this method provides more accuracy than the standard OHAM.

Table 2

x *E (OHAM) *E (OHAM-DJ)

0.0 0 0

0.1 8.1E-10 2.6E-11

0.2 1.4E-09 5.0E-11

0.3 2.1E-09 7.0E-11

0.4 2.5E-09 8.0E-11

0.5 2.6E-09 7.8E-11

0.6 2.5E-09 6.6E-11

0.7 2.1E-09 5.0E-11

0.8 1.5E-09 3.8E-11

0.9 7.3E-10 5.9E-11

1.0 3.3E-10 1.8E-10

E*=Exact-Approx.

89

Figure 2: Solid curve-E*(OHAM) & Dotted curve-E*(OHAM-DJ)

7.5.2 Partial differential Equations

Example 3: We consider the following ZK (3, 3, 3) equation with an initial value

3 3 3( ) 2( ) 2( ) 0,t x xxx yyxu u u u

3

, ,0 sinh2 6

x yu x y

.

where is an arbitrary constant. We assume =1.

By means of OHAM, taking 2

1 2( )H q qC q C , we obtain the following zeroth, first

and second order problems:

Zeroth-order problem:

00

30, with initial condition ( , , 0) sinh

2 6

u x yu x y

t

First order problem:

2 3220 0 0 0 0 0 01

1 1 0 1 1 0 12

2 3 2 32 20 0 0 0 0 0

1 0 1 0 1 0 1 0 12 2 3

(1 ) 3 12 12 12

24 6 36 6 , ( , ,0) 0.

u u u u u u uuC C u C C u C

t t x y x y x x

u u u u u uC u C u C u C u u x y

y x y x y x x x

Second order problem:

90

22 20 0 0 0 0 02 1 1

2 1 2 0 1 0 1 1 1 2

2 2 32

0 0 0 0 0 0 0 01 11 1 2 0 1 1 22 2

2 2

01 11 0 1 12 2

3 6 3 12

24 12 12 12 12

12 12

u u u u u uu u uC C C u C u u C u C

t t t t x x x y x

u u u u u u u uu uC C C u C u C

y y x y x y x y x x

u uu uC u C u

y x y

20 01 1 1 11 0 1 0 1 1

2 2 222 0 0 01 1 1 1 1 1

1 1 1 1 1 0 1 12 2

22 2 2

0 01 1 1 1 1 11 0 1 1 1 12 2

3 6 24

3 12 12 12 12

12 12 36 36

uu u u uC u C u u C

x x x y y x

u u uv u u u u uC u C C C u C u

x y x y x y x y x

u uu u u u u uC u C u C C

y x y x x x x

3

11

2 2 2 2

0 0 0 0 0 01 12 0 1 1 1 0 1 1

2 2 2 2

0 01 1 1 1 1 11 0 1 1 1 0 1 1

3 32 0

2 0 1 0 12

12

24 24 24 24

24 24 24 24

6 12

uC

x x

u u u u u uu uC u C u C u C u

y x y y x y y x y y x y

u uu u u u u uC u C u C u C u

y x y y x y y x y y x y

u uC u C u u

x y

3 3 32 2 20 0 0 1

1 1 1 0 1 12 2 2 3

3 3 33 32 2 20 0 01 1

1 0 1 1 1 2 0 1 0 1 1 12 2 3 3 3

2 2 2

0 0 0 0 01 12 0 1 1 1 0 1 12 2 2

6 6 6

12 6 6 12 6

36 36 36 36

u u uC u C u C u

x y x y x y x

u u uu uC u u C u C u C u u C u

x y x y x x x

u u u u uu uC u C u C u C u

x x x x x x x

2

0

2

2 2 2 2

0 01 1 1 1 1 11 0 1 1 1 0 1 12 2 2 2

3 32 0 1

1 0 1 0 13 3

36 36 36 36

6 12 .

u

x

u uu u u u u uC u C u C u C u

x x x x x x x x

u uC u C u u

x x

Solution of Zeroth-order problem is

0

3( , , )

2 6

x yu x y t Sinh

Solution of first-order problem is

11

3( , , ) 9 5

32 2 6

C x y x yu x y t t Cosh Cosh

Solution of second-order problem is

91

2

2 1 1

3 3 3 3

1 1

2 2 4 4

1 1

( , , ) 0.46875 0.46875

6 6

1.774291992 2.0072021486 / 5 6

0.205078125 0.020313263 [6 6

x y x yu x y t C tCosh C tCosh

x y x yC t Cosh C t Cosh

x y x yC t Sinh C t Sinh

2 2 4 4

1 1

4 4 4 4

1 1

2 2 4 4

1 1

2.2412109375 0.087547302

6 / 5 6 / 5

0.4704380035 0.5631351476 / 7 2 / 3

1.819335938 0.175712585

2

x y x yC t Sinh C t Sinh

x y x yC t Sinh C t Sinh

x y xC t Sinh C t Sinh

3 3

2 1

3 3

2 1

2

1 1

2

0.84375 0.0823974612 2

0.46875 0.169067383

6 6

0.84375 0.84375 .2 2

y

x y x yC t Cosh C t Cosh

x y x yC t Cosh C t Cosh

x y x yC t Cosh t C Cosh

The second-order approximate solution by OHAM-DJ is now ready to be obtained

0 1 2( , , ) ( , , ) ( , , ) ( , , )u x y t u x y t u x y t u x y t . (7.29)

Residual of this solution is 3 3 3( ) 2( ) 2( )t x xxx yyxR u u u u .

For 0.001,t we solve

1 1 1 1

1 20 0 0 0

0, 0,u u

R dydx R dydxC C

(7.30)

and obtain 1 21.004720177 & 2.018914023.C C

Using these values the second order solution (7.29) becomes

6

9 9

6

( , ) 1.5 1.8366 10

2 2

0.000843741 0.000468745

2 6

1.7995 10 2.0358 10

6 / 5 6 / 7

2.2624 10

6 / 5

x y x yu x y Sinh Sinh

x y x yCosh Cosh

x y x yCosh Cosh

x ySinh

13

13

4.7938 10

6 / 7

5.7384 10 .

2 / 3

x ySinh

x ySinh

(7.31)

For 0.1,y we display the numerical results for the solution (7.31) in table 3.

92

Table 3: In this table, we compare the results of the second order solution (7.31) of

OHAM-DJ method with the second order OHAM solution by computing the error at

different points of the problem domain. In both the methods, the same homotopy, the

same auxiliary function, the same method (Least squares) for determining

convergence controlling parameters and the same order solutions are used. Numerical

results show that this method provides more accuracy than the standard OHAM.

Table 3

x OHAM Residual OHAM-DJ Residual

0.0 -8.4394E-06 -6.8472E-06

0.1 -6.4705E -06 -5.2545E-06

0.2 -4.7138E -06 -3.8322E-06

0.3 -3.1555E -06 -2.5694E-06

0.4 -1.7864E -06 -1.4588E-06

0.5 -6.0192E -07 -4.9675E-07

0.6 3.9804E -07 3.1656E-07

0.7 1.2086E -06 9.7695E-07

0.8 1.8196E -06 1.4760E-06

0.9 2.2154E -06 1.8006E-06

1.0 2.3742E -06 1.9327E-06

Figure 3: Dotted curve OHAM residual & solid curve OHAM-DJ residual

7.5.3 Integro-Differential Equations

Example 4: First order non-linear problem

93

2

0

( ) 1 ( )

x

u x u t dt ,

where [0,1] and (0) 0.x u

According to the procedure discussed in section 7.3, the following problems are

obtained.

Zeroth-order problem:

0 0( ) 1, (0) 0.u x u (7.32)

First-order problem:

2

1 1 0 1 0 1

0

( ) (1 )(1 ( )) ( ) , (0) 0.

x

u x C u x C u t dt u (7.33)

Second-order problem:

2 2

2 2 2 0 1 0 1 1 2 0

0 0

1 1 2

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

(1 ) ( ), (0) 0.

x x

u x C C u t dt C u t u t u t dt C u x

C u x u

(7.34)

Solutions of the problems (7.32), (7.33) and (7.34) are

0( ) .u x x

4

11 1( , ) .

12

C xu x C

4 2 4

2 1 2 1 1

4 2 7 3 10

2 1 1

( , , ) 1/ 90720 (7560 7560

7560 360 7 ).

u x C C C x C x

C x C x C x

The second-order approximate solution is

2 1 2 0 1 1 2 1 2

4 2 4 4 2 7 3 10

1 1 2 1 1

( , , ) ( ) ( , ) ( , , )

.6 12 12 252 12960

u x C C u x u x C u x C C

C x C x C x C x C xx

Using the Galerkin’s method, we find

1 20.980179091 and 0.000494614.C C

For these values our solution is

4 7 10( ) 0.083259376 0.0038125042 0.00007266266 .u x x x x x (7.35)

Numerical results for the solution (7.35) are displayed in table 4.

94

Table 4: In this table, we compare the results of the second order solution (7.35) of

OHAM-DJ method with the second order OHAM solution by computing the error at

different points of the problem domain. In both the methods the same homotopy, the

same auxiliary function, the same method (Galerkin’s) for determining convergence

controlling parameters and the same order solutions are used. Numerical results show

that this method provides more accuracy than the standard OHAM.

Table 4

x Residual

(OHAM)

Residual

(OHAM-DJ)

0.0 0 0

0.1 -5.9E-07 -2.9E-07

0.2 -4.4E-06 -2.3E-06

0.3 -1.4E-05 -7.2E-06

0.4 -2.8E-05 -1.5E-05

0.5 -4.2E-05 -2.2E-05

0.6 -4.4E-05 -2.3E-05

0.7 -2.2E-05 -1.1E-05

0.8 2.4E-05 1.3E-05

0.9 4.6E-05 2.3E-05

1.0 -1.1E-04 -5.8E-05

Figure 4: Solid curve-OHAM residual & Dotted curve-OHAM-DJ residual

95

Example 5: Another first order non-linear problem

0

( ) 1 ( )

xdu

u x u t dtdt

,

where [0,1] and (0) 0.x u

According to the procedure discussed in section 7.3, the following problems are

obtained:

Zeroth-order problem:

0 0( ) 1, (0) 0.u x u (7.36)

First-order problem:

1 1 0 1 0 0 1

0

( ) (1 )( ( ) 1) ( ) ( ) , (0) 0

x

u x C u x C u t u t dt u . (7.37)

Second-order problem:

2 2 0 1 1 2 0 0

0

1 0 1 0 1 1 1 2

0

( ) ( ( ) 1) (1 ) ( ) ( ) ( )

( ( ) ( ) ( ) ( ) ( ) ( )) , (0) 0.

x

x

u x C u x C u x C u t u t dt

C u t u t u t u t u t u t dt u

(7.38)

Solutions of the problems (7.36), (7.37) and (7.38) are

0 ( ) .u x x

3

11 1( , ) .

6

C xu x C

3 2 3 3 2 5 3 7

2 1 2 1 1 2 1 1( , , ) 1/ 2520 (420 420 420 84 5 ).u x C C C x C x C x C x C x

The second-order approximate solution is

2 1 2 0 1 1 2 1 2

3 2 3 3 2 5 3 7

1 1 2 1 1

( , , ) ( ) ( , ) ( , , )

.3 6 6 30 504

u x C C u x u x C u x C C

C x C x C x C x C xx

Using the Galerkin’s method, we find

1 21.161089512 and 0.017064207.C C

For these values, our solution is

3 5 7( ) 0.159497660 0.0449376285 0.0031057505 .u x x x x x (7.39)

Numerical results for the solution (7.39) are displayed in table 5.

96

Table 5: In this table, we compare the results of the second order solution (7.39) of

OHAM-DJ method with the second order OHAM solution by computing the error at

different points of the problem domain. In both the methods the same homotopy, the

same auxiliary function, the same method (Galerkin’s) for determining convergence

controlling parameters and the same order solutions are used. Numerical results show

that this method provides more accuracy than the standard OHAM.

Table 5

x Residual.

(OHAM)

Residual

(OHAM-DJ)

0.0 0 0

0.1 -3.0E-04 -2.1E-04

0.2 -1.1E-03 -7.6E-04

0.3 -2.0E-03 -1.4E-03

0.4 -2.7E-03 -1.9E-03

0.5 -2.6E-03 -1.9E-03

0.6 -1.5E-03 -1.2E-03

0.7 4.5E-04 2.5E-04

0.8 2.2E-03 1.6E-03

0.9 1.7E-03 1.3E-03

1.0 -5.6E-03 -4.2E-03

Figure 5: Solid curve-OHAM residual & Dotted curve-OHAM-DJ residual

97

Example 6: Nonlinear problem of order four

Consider the nonlinear boundary value problem for the fourth-order integro

differential equation

( ) 2

0( ) 1 ( ) , 0 1,

xiv tu x e u t dt x

with the boundary conditions:

(0) 1, (0) 1, (1) , (1) .u u u e u e

The exact solution of this problem is ( ) xu x e .

According to the procedure discussed in section 7.3, the following problems are

obtained:

Zeroth-order problem:

(4)

0

0 0 0 0

( ) 1,

(0) 1, (0) 1, (1) , (1) .

u x

u u u e u e

(7.40)

First-order problem:

(4) (4) 2

1 1 0 1 0

0

( ) (1 )( ( ) 1) ( ) ,

x

tu x C u x C e u t dt (7.41)

1 1 1 1(0) 0, (0) 0, (1) 0, (1) 0.u u u u

Second-order problem:

(4) (4) (4)

2 2 0 1 1

2 2

2 0 1 0 1 1

0 0

2 2 2 2

( ) ( ( ) 1) (1 ) ( )

( ) (2 ( ) ( ) ( )) ,

(0) 0, (0) 0, (1) 0, (1) 0.

x x

t t

u x C u x C u x

C e u t dt C e u t u t u t dt

u u u u

(7.42)

The problems (7.40), (7.41) and (7.42) are directly integrable and their solutions are

obtained easily.

We consider the second-order approximation

15

2 1 2 0 1 1 2 1 2( , , ) ( ) ( , ) ( , , ) ( )u x C C u x u x C u x C C O x .

Using the Galerkin’s method, we find

1 21.00235514 and 4.0094258598.C C

By using these values the approximate solution becomes

98

2 3 4 5

6 7 8 6 9

7 10 7 11 8 12 10 13

9 14

( ) 1 0.500003 0.166659 / 24 0.00834443

0.00139074 0.000181376 0.0000342871 2.7908 10

-3.3706 10 2.21015 10 1.68469 10 7.34336 10

1.43975 10 .

u x x x x x x

x x x x

x x x x

x

(7.43)

Numerical results for the solution (7.43) are displayed in table 6.

Table 6: In this table, we compare the results of the second order solution (7.43) of

OHAM-DJ method with the second order OHAM solution by computing the error at

different points of the problem domain. In both the methods, the same homotopy, the

same auxiliary function, the same method (Galerkin’s) for determining convergence

controlling parameters and the same order solutions are used. Numerical results show

that this method provides more accuracy than the standard OHAM.

Table 6

x E*(OHAM) E*(OHAM-DJ)

0.0 0 0

0.1 -3.3887E-10 -9.0694E-13

0.2 -1.0685E-09 -3.3683E-12

0.3 -1.8172E-09 -6.5234E-12

0.4 -2.3265E-09 -8.9932E-12

0.5 -2.4695E-09 -4.4978E-12

0.6 -2.2363E-09 -7.6340E-12

0.7 -1.7013E-09 -4.2673E-12

0.8 -9.9204E-10 -1.2165E-12

0.9 -2.7455E-10 -3.1821E-13

1.0 2.6119E-10 -2.4673E-12

E*=Exact-Approx.

99

Figure 6: Solid cure-OHAM error & Dotted curve-error of the new method

100

CONCLUSION

In this thesis, we have used OHAM for the solution of different functional equations.

We found that this recently developed method has great potential to handle a wide

class of functional equations. In this method, all the recursive relations have been

clearly and explicitly defined and for the determination of auxiliary parameters which

control the convergence, a straight forward approach has been used. The residual is

forced to minimize or vanish over the domain of the problem and by virtue of this

procedure we find the optimal values for the developed solution. OHAM gives a very

good response to boundary value problems but it response to initial problems is still

awaited. In chapter 5, we have solved an initial value problem i.e, the ZK-equation

with an initial condition to examine the efficiency of the method. We found that the

method works very well at the small domains. By small domain we mean a domain

whose length does not exceed one.

We also extended OHAM to integro-differential equations. The obtained results are

very much encouraging and one can fully rely over the obtained solution as the

convergence of the developed solution is control viva the flexible auxiliary function.

In the application of OHAM, no restrictive assumptions are needed and one feels very

comfortable as the convergence of the method is not dependent on the initial solution

or guess.

We also observed that for determination of auxiliary constants, the method of Least

Squares and the Galerkin’s method provide the best optimal values but the Least

Squares requires more CPU time and working memory storage than the other. The

efficiency of both the methods is problem dependent and one can easily choose the

one which best leads to the optimal values of the auxiliary constants for more

accuracy.

Generally, a closed form solution cannot be obtained by using OHAM but it gives

more accurate results to analyze and interpret any physical model.

In this thesis, we have also developed a new scheme which provides more accurate

results than the standard OHAM and is applicable to a more wide class of nonlinear

101

functional equations. We have tested this new scheme to different functional

equations i.e, ODEs, a PDE and some integro-differential equations successfully and

compared the results obtained by using the standard OHAM. We found better results

than the standard OHAM.

The newly developed method i.e, the OHAM-DJ is more accurate than the standard

OHAM and as such it will be more appealing for the scientists and researchers to

apply and extend its application to more complex problems arising in science and

technology. It is hoped that it will attract both students/scholars of every field of

application.

102

LIST OF PUBLICATIONS

1. Siraj-Ul-Islam, Sirajul-Haq and Javed Ali, Numerical solution of special 12th

-

order boundary value problems using differential transform method, Comm.

Nonl.Sc.Nu.Sim.14(4)(2009)1132-1138.

2. Javed Ali, S.Islam, Siraj-UL-Islam, Gulzaman, The solution of multipoint

boundary value problems by the Optimal Homotopy Asymptotic

Method,.Computer Math.Appl.CAMWA-D-09-00585RI(2009).

3. Javed Ali, S. Islam, M. Tariq Rahim and Gul Zaman, The Solution of Special

Twelfth OrderBoundary Value Problems by the Optimal Homotopy

Asymptotic Method,.World Applied Sciences Journal 11 (3): 371-378, 2010.

4. Javed Ali, S. Islam, Inayat Ali, Hamid Khan, Application of Optimal

Homotopy Asymptotic Method to higher order boundary value problems,

Volume 2012 (2012), Article ID 401217, 14 pages, doi:10.1155/2012/401217,

Abstract and Applied Analysis.

5. Javed Ali, Saeed Islam, Hamid Khan, Gul Zaman, Solution of a Parameterized

Sixth-Order Boundary Value Problem by the Optimal Homotopy Asymptotic

Method, Volume 12 , Number 3/2011, pp. 167–172, Proceedings of the

Romanian Academy, Series A.

6. Javed Ali, S. Islam, Inayat Ali, Hamid Khan, Application of Optimal

Homotopy Asymptotic Method to 5th

& 6th

order boundary value problems,

Volume 15(8):1120-1126, 2011, World Applied Science Journal(WASJ).

7. Javed Ali, One Dimensional Differential Transform Method for Some Higher

Order Boundary Value Problems in Finite Domain, Int. J. Contemp. Math.

Sciences, Vol. 7, 2012, no. 6, 263 – 272.

8. Yasir Khan, S. Islam, Javed Ali, Sirajul Islam, QingbiaoWu, Numerical

solution of the steady two-dimensional radial flow of viscous fuid via

differential transform method Vol. 2, Issue. 2, 2010, pp. 13-20, Journal of

Advanced Research in Scientific Computing.

9. Hamid Khan, S. Islam, Javed Ali, Inayat Ali Shah, Comparison of Different

Analytical Solutions to Axisymmetric Squeezing fluid between two Infinite

Parallel Plates with Slip boundary conditions, Abstract and Applied Analysis,

Published Volume 2012 (2012), Article ID 835268, 18 pages

doi:10.1155/2012/835268.

103

REFERENCES

[1] S. Chandrasekhar, Hydrodynamic hydromegnatic stability, Clarendon Press,

Oxford, 1961(reprinted: dover Books, New York 1981).

[2] R. P. Agarwal, Boundary value problems for higher order differential

equations, world scientific, Singapore(1986).

[3] G. Adomian, Solving Frontier Problems of Physics: The Decomposition

method, vol. 60 of Fundamental Theories of Physics, Kluwer Academic

Publishers, Boston, mass, USA, 1994.

[4] G. Adomian, A review of the decomposition method and some recent results

for nonlinear equations, Computer Math, Appl.,21(5)(1991)101-127.

[5] Ji-Huan He, Variational approach to the sixth-order boundary value

problems, Appl. Math. Comput.143(2-3)(2003)537-538.

[6] Ji-Huan He, Homotopy perturbation technique, Computer Methods in Applied

Mechanics and Engineering, 170(3-4)(1999)257-262.

[7] Ji-Huan He, Homotopy perturbation method for solving boundary value

problems, Phys. Lett.A. 350(2006)87-88.

[8] Ji-Huan He, A note on the homotopy perturbation method, Thermal

Sc.14(2)(2010)565-568.

[9] Shijun Liao, Beyond Perturbation: Introduction to the Homotopy Analysis

Method. Chapman & Hall/CRC, Florida, 2004.

[10] Shijun Liao, Notes on the homotopy analysis method: Some definitions and

theorems, Communications in Nonlinear Science and Numerical Simulation,

14(4)(2008)983-997.

[11] J. K. Zhou, Deferential Transformation and its Application for Electrical

Circuits,1986, Huazhong University Press, Wuhan China (in Chinese).

[12] Siraj-Ul-Islam, Sirajul-Haq and Javed Ali, Numerical solution of special 12th

-

order boundary value problems using differential transform method, Comm.

Nonl.Sc.Nu.Sim.14(4)(2009)1132-1138.

[13] Songxin Liang, David J. Jeffrey, Comparison of homotopy analysis method

and homotopy perturbation method through an evolution equation, Preprint

submitted to Elsevier, 23 February 2009.

104

[14] V. Marinca, N.Herisanu, I. Nemes, Optimal homotopy asymptotic method with

application to thin film flow, Cent. Eur. J. Phys., 6(3)(2008)648-653.

[15] N.Herisanu, V. Marinca, T.Dordea,G.Madescu,A new analytical approach to

nonlinear vibration of an electric machine, Proce. Romanian. Acad.

9(3)(2008).

[16] V. Marinca, N. Herisanu, C. Bota, B.Marinca, An optimal homotopy

asymptotic method applied to the steady flow of fourth-grade fluid past a

porous plate,Appl.Math.Lett.22(2),(2009)245-251.

[17] V. Marinca, N. Herisanu, An optimal homotopy asymptotic method for solving

Nonlinear equations arising in heat transfer, Int. Comm. Heat and Mass

Transfer,5(2008)710-715.

[18] A. M. Wazwaz, The numerical solution of fifth-order boundary value

problems by decomposition method, J. Comp. and Appl. Math., Vol. 136,

(2001), p. 259-270.

[19] A. M. Wazwaz, The numerical solution of sixth-order boundary value

problems by the modified decomposition method, Appl. Math. Comput.,

Vol.118, (2001), p. 311-325.

[20] A. M. Wazwaz, Approximate solutions to boundary value problems of higher-

order by the modified decomposition method, Computer Math, Appl.,

40(2000)679-691.

[21] Muhammad Aslam Noor and Syed Tauseef Mohyud-Din, Variational

Iteration Method for Fifth-Order Boundary Value Problems Using He’s

Polynomials, Mathematical Problems in Engineering,

doi:10.1155/2008/954794.

[22] M.A. Noor, S.T. Mohyud-Din, Variational iteration method for solving sixth

order boundary value problems, pre-print (2007).

[23] M. A. Noor and S. T.Mohyud-Din, Variational iteration technique for solving

higher order boundary value problems, Applied Mathematics and

Computation, vol. 189, no. 2, pp. 1929–1942, 2007.

[24] Muhammad Aslam Noor_, Syed Tauseef Mohyud-Din, Homotopy

perturbation method for solving sixth-order boundary value problems,

Computer Math.Appl., 55:2953–2972,2008.

105

[25] M. A. Noor, S. T. Mohyud-Din. An efficient algorithm for solving fifth-order

boundary value problems, Mathematical and Computer Modeling, Vol.45

(2007), p. 954-964.

[26] Muhammad Aslam Noor and Syed Tauseef Mohyud-Din, A Reliable

Approach for Solving Linear and Nonlinear Sixth-order Boundary Value

Problems, International Journal of Computational and Applied Mathematics.,2

(2):163–172(2007).

[27] Songxin Liang , David J. Jeffrey, Approximate solutions to a parameterized

sixth order boundary value problem, Comput.Math.Appl., 59(1):247-

253(2010).

[28] Siraj-ul-Islam, M. A. Khan, A numerical method based on nonpolynomial

sextic spline function for solution of special fifth order boundary value

problems, Appl. Math. Comput., Vol. 181(2006), p. 356-361.

[29] Siraj-ul-Islam, I. A Tirmizi Fazal-i-Haq and M. A. khan, Non-polynomial

splines approach to the solution of sixth-order boundary-value problems,

Appl. Math. Comput. Vol.195 (2008), p.270-284.

[30] Syed Tauseef Mohyud-Din, Ahmet Yildirim and M. M. Hosseini, An

Iterative Algorithm for Fifth-order Boundary Value Problems, World Applied

Sciences Journal 8 (5): 531-535, 2010.

[31] K. N. S. Kasi Viswanadham and P. Murali Krishna, Quintic B-Spline

Galerkin method for fifth order boundary value problems, ARPN Journal of

Engineering and Applied Sciences, 5(2):2010.

[32] Che Haziqah Che Hussin and Adem Kilic¸man, On the Solutions of Nonlinear

Higher-Order Boundary Value Problems by Using Differential

Transformation Method and Adomian Decomposition Method: Mathematical

Problems in Engineering Volume 2011, doi:10.1155/2011/724927.

[33] Javed Ali, One Dimensional Differential Transform Method for Some Higher

Order Boundary Value Problems in Finite Domain, Int. J. Contemp. Math.

Sciences, Vol.7, 2012, no. 5, 263-272.

[34] Muhammad Aslam Noor and Syed Tauseef Mohyud-Din, Variational

Iteration Decomposition Method for Solving Eighth-Order Boundary Value

Problems, Differential Equations and Nonlinear Mechanics Volume 2007,

Article ID 19529, 16 pages,doi:10.1155/2007/19529.

106

[35] Inc M., Evans D.J., An efficient approach to approximate solutions of eighth

order boundary-value problems, Intel. J. Comput. Math, 81, 6, 685-692(2004).

[36] A. Golbabai and M. Javidi, “Application of homotopy perturbation method for

solving eighth-order boundary value problems,” Applied Math. Comput., vol.

191, pp. 334–346, 2007.

[37] A. Barari, M. Omidvar, T. Najafi, Abdoul R. Ghotbi, Homotopy Perturbation

Method for Solving Tenth Order Boundary Value Problems, Intel, J.Math.

Comput, 3(9):8-14(2009).

[38] Wazwaz, A.M. (2000). The Modified Decomposition Method for Solving

Linear and Nonlinear Boundary Value Problems of Tenth-Order and Twelfth

Order. Int. J. Nonlinear Sci. Numer. Simul., 1: 17-24.

[39] Shahid S. Siddiqi, Ghazala Akram, Imran Zulfiqar, Solution of Eleventh

Order Boundary Value Problems Using Variational Iteration Technique.,

European Journal of Scientific Research,Vol.30 No.4 (2009), pp.505-525.

[40] Syed Tauseef Mohyud-Din and Ahmet Yildirim. Solutions of Tenth and

Ninth- OrderBoundary Value Problems by Modified Variational Iteration

Method. Applications and Applied Mathematics, 5(1):11-25(2010).

[41] J. Biazar, Ghazvini H. (2008) "An analytical approximation to the solution of

a wave equation by a variational iteration method," Appl, Math. Lett.,21, 780-

785.

[42] J. Biazar, H. Ghazvini, Exact solutions for non-linear Schrdinger equations by

He’s homotopy perturbation method, Phys. Lett. A 366 (2007)79-84.

[43] Biazar, J., 2008. Int. J. Contemp. "He’s Homotopy Perturbation Method for

Solving Helmholtz Equation", Math. Sciences, 3 (15) , pp. 739 – 744.

[44] Rezania, A., GHorbali, A.R., Ganji, D.D., and Bararnia, H., 2009. Application

on Homotopy perturbation and Variational Iteration Methods for Heat

Equation, Auturalian journal of Basic and Applied Sciences, 3, pp. 1863-1874.

[45] Ganji, D.D., 2006. "The application of He’s homotopy perturbation method to

Nonlinear equations arising in heat transfer", Physics Letters A, 355(337–

341).

[46] Zhang, Li-Na., and Ji-Huan, He., 2006. "Homotopy Perturbation method for

the solution of the electrostatic potential differential equation", Mathematical

Problems in Engineering-Hindawi, Volume (2008), Article ID 83878, 6 pages.

107

[47] Biazar J., Gholamin P., Hosseini K. (2006) "Exact solutions of Poisson

equation by using variational iteration and Adomian decomposition methods,"

Journal of Applied Mathematics, Vol.3, No.10

[48] Abdou, M.A. and A.A. Soliman, 2005. Variational iteration method for

solving Burger's and coupled Burger's equations, J. Comput. Appl. Math.,

181: 245–251.

[49] A.M. Wazwaz: “Special types of the nonlinear dispersive Zakharov–

Kuznetsov equations with compactons, solitons, and periodic solutions”, Int. J.

Computer Math.,Vol. 81, (2004), pp. 1107–1119.

[50] A.M. Wazwaz: “Nonlinear dispersive special type of the Zakharov–Kuznetsov

equation ZK(n,n) with compact and noncompact structures”, Appl. Math.

Comput., Vol.161, (2005), pp. 577–590.

[51] A.M. Wazwaz: “Exact solutions with solitons and periodic structures for the

Zakharov–Kuznetsov (ZK) equation and its modified form”, Commun.

Nonlinear Sci. Numer. Simul., Vol. 10, (2005), pp. 597–606.

[52] Wazwaz, A.M. 2005.The extended tanh method for the Zakharov–Kuznetsov

(ZK) equation, the modified ZK equation, and its generalized forms. Commun.

Nonlinear Sci. Numer. Simul., Volume 13, Issue 6, August 2008(1039-1047).

[53] Zhao, X., Zhou, H., Tang and Jia, H. 2006.Travelling wave solutions for

modified Zakharov–Kuznetsov equation. Appl. Math.Comput.181(1), 634–

648.

[54] Mustafa. Exact solutions with solitary patterns for the Zakharov–Kuznetsov

equations with fully nonlinear dispersion. Chaos, Solitons & Fractals Volume

33, Issue 5, August 2007, Pages

[55] S. Timoshenko. Theory of Elastic stability, McGraw-Hill, New York, 1961.

[56] C. Bai and J. Fang, Existence of multiple positive solutions for nonlinear m-

point boundary value problems .J. Math. Anal. Appl.,281(2003)76-85.

[57] J. Henderson, Five-point boundary value problems for third-order differential

equations by solution matching. Math. Comput. Model,42(2005)133-137.

[58] R. Ma. Existence theorems for a second-order three-point boundary value

problem, J. Math. Anal. Appl.,212(1997)430-442.

[59] M.Haque, M. H. Baluch and M. F. N. Mohsen, Solution of multiple point,

nonlinear boundary value problems by method of weighted residuals. Int. J

.Comput. Math.,19 1986)69-84.

108

[60] G. R. Liu , T Y Wu, Multiple boundary value problems by differential

quadrature method, Math.Comput.Model,35(2002)215-227.

[61] Ikram. A. Tirmizi., E. H. Twizell, and Siraj-Ul-Islam, A numerical method for

thir -order non-linear boundaryvalue problems in engineering, International

Journal of Computer Mathematics,82(1)(2005)103 -109.

[62] Yao. Q. Successive iteration and positive solution for nonlinear second-order

three-point boundary value problems Comput. Math. Appl. 50(2005) 433–44.

[63] M. Tatari , M. Dehgan, The use of the Adomian decomposition method for

solving multipoint boundary value problems, Phys. Scr.,73(2006)672-676.

[64] T.Y.Na.,Computional Methods in Engineering Boundary Value Problems,

pp.122-125. Acdemic Press, New York, 1979.

[65] J. S. Russell, Report on Waves, in: Report of the Fourteenth Meeting of the

British Association for the Advancement of Science, John Murray, London,

1844.

[66] N. J. Zabusky and M. D. Kruskal, Interaction of "Solitons" in a Collisionless

Plasma and the Recurrence of Initial States., Phys. Rev. Lett. 15, 240–243

(1965).

[67] D. J. Korteweg, G. de Vries, Phil. Mag. 39 (1895) 422.

[68] Rosenau, P., Hyman, J.M., 1993. Compactons: solitons with finite

wavelengths, Physics Review Letters. 70(5), 564-567.

[69] Ma Hong-Cai, Yu Yao-Dong and Ge Dong-Jie, New Exact Travelling Wave

Solutions for Zakharov–Kuznetsov Equation. Commun. Theor. Phys. 2009. 51

609-612.

[70] B. Ganjavi, D.D. Ganji, Homotopy Perturbation Method and Variational

Iteration Method for Solving Zakharov-Kuznetsov Equation. American Journal

of Applied Sciences 5 (7): 811-817, 2008.

[73] R. P. Agarwal, Boundary value problems for higher order integro-differential

equations, Nonlinear Analysis, Theory, Meth. Appl., 7, pp. 259-270, (1983)

[74] J. Morchalo, On two point boundary value problem for integro-differential

equation of second order, Fasc. Math., 9, pp. 51-56, (1975),

[75] J. Morchalo, On two point boundary value problem for integro-differential

equation of higher order, Fasc. Math., 9, pp. 77-96, (1975)

109

[76] A. Wazwaz, A reliable algorithm for solving boundary value problems for

higher-order integro-differential equations, Appl. Math. Comp., 118, pp. 327-

342, (2001).

[77] V. Daftardar-Gejji and H. Jafari, An iterative method for solving nonlinear

functional equations, Journal of Mathematical Analysis and Applications, vol.

316, no. 2, pp. 753–763, 2006.

[78] Sachin Bhalekar and Varsha Daftardar-Gejji, Convergence of the New

Iterative Method, International Journal of Differential Equations, vol. 2011,

Article ID 989065,10pages doi:10.1155/2011/989065.