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i

OPTIMAL HOMOTOPY ASYMPTOTIC METHOD TO

SOLUTION OF PARTIAL DIFFERENTIAL EQUATIONS

Submitted By

RASHID NAWAZ

Supervisor: Prof. Dr. Syed Inayath Ali Shah

Co Supervisor: Dr. Saeed Islam

Thesis Submitted in Fulfillment For The Degree of Ph.D.

DEPARTMENT OF MATHEMATICS, ISLAMIA COLLEGE

PESHAWAR (CHARTERED UNIVERSITY), KHYBER

PAKHTUNKHAWA, PAKISTAN

SEPTEMBER 2014

iii

AUTHOR’S DECLARATION

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

Islamia College Peshawar (a chartered university). The contents of this Thesis are my

original work except where specific acknowledgement is given. This dissertation has not

been submitted in whole or in part to any other University.

Rashid Nawaz

ICUP Peshawar

iv

Dedicated to:

My Family and Parents

v

Acknowledgement

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

study. I especially want to thank my supervisor, Prof. Dr. Syed Inayath Ali Shah, for his

guidance during my research and study at Islamia College University. In addition, he was

always accessible and willing to help his students with their research. As a result, research

life became charming and rewarding for me. I was delighted to interact with Dr. Saeed

Islam by attending his classes and having him as my co-supervisor. His insights to fluid

dynamics are second to none. Besides, he sets an example of a world-class researcher for

his rigor and passion on research. I would be failing in my duties if I don’t appreciate the

help of professors in the Faculty of physical and numerical Sciences and staff for their

cooperation.

My deepest appreciation goes to my family for their love and support throughout my life.

This dissertation is simply impossible without them. I am indebted to my parents for their

care, love and prayers. They have always been a constant source of encouragement during

my graduate study. Father and Mother, I love you. I feel proud of my brothers and sisters,

for their prayers and support throughout my life.

Dr. Adam khan, Sharif khan, Dr. Fazal Haq and Dr. Arshid Ali have been my friend and

mentor for many years. They offer advice and suggestions whenever I need them.

Besides, they have set a role model of typical friends who cares and loves their friends as

if they were their own family members.

I thank Dr. Muhammad Idrees of Islamia College University; his friendship and technical

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

at work. Thanks to Hakeem Ullah, Dr. Siraj-Ul-Islam, Dr. Rehan Ali Shah.

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

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

glorified.

Rashid Nawaz

vi

List of Abbreviations

Words Abbreviations

Ordinary Differential Equation ODEs

Partial Differential Equations PDEs

Differential Difference Equations DDEs

Equal Width Wave EW

Modified Equal Width Wave MEW

Regularized Long-Wave RLW

Damped Generalized Regularized Long-Wave DGRLW

Witham Broer Kaup WBK

Approximate Long Wave ALW

Modified Boussinesq MB

Kortweg-de Vries KdV

Modified Kortweg-de Vries MKdV

Differential Transform Method DTM

Variational iteration method VIM

Adomian Decomposition Method ADM

Homotopy Analysis Method HAM

Homotopy Perturbation Method HPM

Optimal Homotopy Asymptotic Method OHAM

Modified Variational iteration method MVIM

vii

Abstract

Nonlinear Differential equations are of major importance in different fields of science and

engineering. For complicated nonlinear problems exact solutions are not available and

alternate way is to use numerical methods, Iterative methods or analytical techniques of

perturbation. Numerical methods use discretization a have slow rate of convergence.

Iterative methods are sensitive to initial conditions and in case of high nonlinearity they

do not yield converged results. In perturbation methods small parameter is applied on the

equation and hence cannot be applied for high nonlinear problems as they do not have

small parameter. One of domain type methods is known as OHAM. This method is free

from small parameter assumption and do not need the initial guess. The proposed method

provides better accuracy at lower-order of approximations. Moreover the convergence

domain can be easily adjusted.

In this thesis OHAM is implemented for solution linear and nonlinear tenth order ODEs.

Then its effectiveness and generalization is shown to a nonlinear family of PDEs,

including Burger, Fisher, Burger’s–Huxley, Burger’s–Fisher, MEW and DGRLW

equations. The results of the proposed method are compared with that of DTM, VIM,

ADM, HAM and HPM, which reveal that OHAM is effective, simpler, easier and

explicit.

Apart from application to PDEs, OHAM is applied to couple system of PDEs. The

coupled WBK, ALW, MB systems are used as test examples and results are compared

with those obtained by HPM.

OHAM is implemented to DDEs as well, and solution of MKdV lattice equation is

presented for the illustration of proposed technique. The results are compared with HAM

and HPM. In all cases the results obtained by OHAM are in close agreement with the

exact solution and reveal high accuracy.

viii

Table of Contents

CHAPTER 1. Introductionand Literature Survey

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

1.2 Literature Survey………………………………………………………………………3

1.3 Basics of Homotopy ………………………………………………………………….4

1.4 Introduction to Well known Homotopy Mathods……………………………………..5

1.4.1 Homotopy Analysis Method………………………………………………..5

1.4.2 Homotopy Perturbation Method …………………………………………....6

1.5 Thesis Plan………………………………………………………………………….....7

CHAPTER 2. Application of OHAM to Tenth Order Boundary Value Problems

2.1 Introduction ................................................................................................................... 9

2.2 Fundamental Mathematical Theory of OHAM for (ODE)……………………………9

2.3 Application of OHAM to Tenth Order Boundary Value Problems ………………....12

2.3.1 Model 1……………………………………………………………………12

2.3.2 Model 2…………………………………………………………………….15

2.3.3. Model 3…………………………………………………………………….20

CHAPTER 3. Implementation of OHAM to Family of Burger’s Equation

3.1. Introduction 23

3.2. Analysis of OHAM for (PDE)………………………………………………………23

3.3. Implementaion OHAM to Burger’s Family ………………………………………..24

3.3.1 Model 1 (Burger Equation) .......................................................................... 24

3.3.2. Model 2 (Burger’s–Huxley Equation)……………………………………. .30

3.3.3. Model 3 (Burger’s–Fisher Equation)……………………………………...35

CHAPTER 4 Application of OHAM to MEW and Fisher’s Equation

4.1 Introduction……………………………………………………………..……………38

4.2 Application of OHAM to Fisher’s and MEWequation ....................................... …… …39

4.2.1 Model 1 (Fisher’s equation)………………………………………………..39

4.2.2 Model 2 (Defusion equation of Fisher’s equation) .......................................47

4.2.3 Model 2 (MEWequation) ............................................................................. 52

CHAPTER 5 Application of OHAM to DGRLW Equation

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

5.2 Applications of OHAM to (DGRLW Equation) ......................................................... 59

ix

5.2.1 Model 1 ............................................................................................................... 59

5.2.2 Model 2 ............................................................................................................... 63

5.2.3 Model 3 ............................................................................................................... 66

5.2.4 Model 4 ............................................................................................................... 70

CHAPTER 6 Analysis of OHAM to Coupled System of PDEs

6.1 Introduction………………………………………………………………………….75

6.2 Solution of coupled system of PDE’s by OHAM...............................................……76

6.2.1 Model 1. (WBK Equation)………………………………………………....76

6.2.2 Model 2 (MB equation)………………………………………………….....79

6.2.3 Model 3 (ALW Equation)………………………………………………….82

CHAPTER 7 Analysis of OHAM to Differential-Difference Equations

7.1 Introduction…………………………………………………………………………..86

7.2 Analysis of OHAM to MKdV lattice equation………………………………………86

7.2.1 Model 1 (Mkdv lattice equation)…………………………………………………...87

CHAPTER 8 Conclusion

........................................................................................................................................... 91

List of publications…………………………………………………………………….93

Appendix A…………………………………………………………………………….94

Appendix B…………………………………………………………………………….96

Appendix C…………………………………………………………………………….99

References ….............................................................................................................10

1

Chapter 1

Introduction and Literature Survey

1.1 Introduction

Differential equations play a vital role in modeling of different physical problems. Variety of

these problems can be modeled by either ODEs or PDEs. ODEs along with initial and boundary

conditions have numerous applications in physics, biology, astronomy, fluid dynamics, chemical

reactions and in engineering sciences. The newton law of cooling, nuclear decay and population

growth is modeled by first order ODEs. The flow of current in RLC circuits, the motion of free

falling body and motion of mass attached to a string are modeled by second order ODEs. The

third order boundary value problems included draining flow, coating flow and shear deformation

of sandwich beams. The fourth order ODEs are used to model variety of the physical problems

such as plate binding on an elastic foundation, inelastic flows and transverse vibration of hinged

beams. The viscous-elastic flows are modeled by fifth order ODEs. An infinite horizontal layer

of fluid when heated and subjected to the action of rotation results in instability and cannot be

modeled by ODEs up to fifth order. Hence, for systems having instability as ordinary convection

can be solved by using sixth order ODEs. However systems resulting in over stability can be

solved by using eight order ODEs. The tenth order boundary value problem arises when a

uniform magnetic field is also applied across the fluid in the same direction as gravity and fluid

is subjected to the action of rotation [1].

ODEs are used in modeling of simple physical problems whereas PDEs are used in modeling of

complex problems. The simplest PDEs are wave equation, heat equation, Laplace equation and

Poisson equation. The wave equation models the vibration of string or membrane. The heat

equation governs heat flow. Laplace and Poisson equations models electrostatic potential in

steady state. The KdV equation arises in the study of nonlinear dispersive waves. It was derived

by Korteweg and de Vries in 1895 for modeling of shallow water waves in canal. A special case

of the KdV equation is burger equation. This equation is similar to the one dimension Navies-

Stokes equation without the stress term, and was presented for the first time in a paper in 1940

from Burger. It is the model for the solution of Navier-Stokes equation and is applied to laminar

2

and turbulence flows as well. It can be used as a model for any nonlinear wave transmission

problem subject to dissipation [2-6].

The Differential Difference equations models complicated physical phenomena such as particle

vibrations in lattices, currents flow in electrical networks, and pulses in biological chains. The

solutions of these DDEs can provide numerical simulations of nonlinear partial differential

equations, queuing problems, and discretization in solid state and quantum physics.

After modeling a physical system into mathematical form, two approaches are used for its

solution, one is analytical and other is numerical approach. Both of these have their own

advantages and disadvantages. Complicated nonlinear problems are solved either by using

numerical methods or by analytical techniques of perturbation as exact solutions are rare. In the

numerical methods, stability and convergence should be considered so as to elude divergence or

unsuitable results. Numerical methods use discretization technique which affects the accuracy.

Well-known numerical methods are collocation methods [7-8], finite difference methods [9-10],

finite element methods [9-11], radial basis function, and collocation method [11-23].

Iterative methods have a revolutionary role but in presence of strong nonlinearity they do not

yield converged results as they are sensitive to initial conditions [24-25]. Researchers usually

use perturbation methods for the analytical solution of nonlinear problems [26-33]. These

methods play an important role in development of engineering sciences but have some

limitations as well. In the perturbation techniques, a small parameter is applied on the equation.

Defining this small parameter in itself is an issue, and its improper choice can affect the results

badly. Another drawback of perturbation methods is that they cannot be applied for high

nonlinear problems as they do not have small parameter. Due to these restrictions implementing

the method for different applications is too difficult. To avoid such type of difficulty, different

methods such as artificial small parameter, expension method and ADM have been recently

introduced to eliminate the small parameter assumption. However, all these methods cannot

provide larger flexibility to control the convergence region of series solution. Hence, it is

necessary to develop more effective analytical methods. Liao formulated a kind of analytic

method namely HAM by applying the important concept of homotopy in topology [34-42]. In

contrast to the perturbation methods, this technique is free of small parameters assumption and

transforms the nonlinear problem into many linear sub-problems. HAM provides larger

flexibility to control the convergence region of series solution.

3

HPM was proposed by He [43]. It is a combination of homotopy and perturbation. This method

provides an asymptotic solution with few terms without requiring any convergence theory. The

validity of HPM does not require the assumption of small parameter in the problem. This method

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

transforms a given problem into a series of linear equations that are easy to solve. This method is

rapid convergent, but requires that the nonlinearity must be sufficiently differentiable.

In this work, OHAM is proposed for solving nonlinear problems. It uses more flexible function

called auxiliary function to control the convergence region of series solution and is free from the

small parameter assumption. The convergence criterion of proposed method is similar to that of

HAM and HPM, but this method is more efficient and flexible.

1.2 Literature Survey

The OHAM has been introduced by Marinca and Herisanu et al. [44-47], for approximate

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

They used OHAM for understanding the behavior of nonlinear mechanical vibration of an

electrical machine, nonlinear equation arising in heat transfer, for investigation of solution of

nonlinear equations arising in the study of state flow of a fourth grade fluid past a porous plate

and for ―determination of periodic solutions for the motion of a particle on a rotating parabola‖

[44-47]. The authors made newest advances in OHAM by including functions of a physical

parameter in the auxiliary function beside the convergence-control parameters. All these

measures were intended to increase the efficiency of the procedure and the accuracy of the

results for complicated problems [48-49].

Idrees et al. have applied OHAM for solution of fourth order, special sixth order, eight order

boundary value problems, sin Gordon equation, KdV equation and for squeezing flow [50-56].

Javed Ali et al. has applied OHAM for solution of fifth order, sixth order, parameterized sixth

order and multipoint boundary value problems [57-61]. Shah et al. has applied OHAM for

solution of Couette and Poiseuille flows for fourth grade fluids, tagnation point flow and Couette

and Poiseuille flows for third grade fluids with heat transfer analysis [62-64]. Iqbal et al. has

applied OHAM for analytic solution of singular Lane–Emden type equation, heat transfer flow of

a third grade fluid between parallel plates and weakly singular Volterra Integral equations [65-

67]. Ganji et al. has applied OHAM for the solution of the Jeffery–Hamel flow problem [68].

http://www.sciencedirect.com/science/article/pii/S0898122109007494



4

Ghoreishi et al. has applied OHAM for nonlinear age-structured population models [69].

Sheikholeslami et al. has investigated the laminar viscous flow in a semi-porous channel in the

presence of uniform magnetic field using OHAM [70]. Farrokhzad et al. has investigated beam

deformation equation using OHAM [71]. Ene et al. has applied OHAM for solving a nonlinear

problem in Elasticity [72]. Pandey et al. used OHAM for the electro hydrodynamic flow [73].

Han et al. applied OHAM for solving Integro-Differential equations [74]. Anakira et al. has

applied OHAM for delay Differential equations [75]. Ghazanfari et al. applied OHAM for

solving system of Fredholm Integral equations [76]. Mabood et al. has applied OHAM for heat

transfer in hollow sphere with robin boundary conditions, Riccati equation, convection heat

transfer flow, Analytical solution for radiation effects on heat transfer in blasius flow, nonlinear

boundary layer equation for flat plate, steady heat transfer in a heat-generating fin with

Convection and Radiation, Painleve equation, one dimensional heat and Advection-Diffusion

equations and for a water quality model in a uniform stream [77-88]. Abdou et al. has applied

OHAM for Quantum Zakharov-Kuznetsov equation in Ion Acoustic Waves [89]. Kashefi et al.

has applied OHAM to Fredholm Fuzzy Integral equations [90]. Regarding the application of

OHAM, the contributions of Rashid Nawaz can be seen in [91-95].

1.3 Basics of Homotopy.

1.3.1 Homotopy

Homotopy is the study of properties that are not changed by continuous distortions, such as

stretching or twisting. If one mathematical object continuously distorts into the other then these

objects are homotopic and distortion is called a homotopy between the two objects. The

elementary concept of homotopy is between mapping of functions i.e.

, : 0,1 ; , 0,1 , H

where is problem domain andH is the deformation of the original function f .

1 2[ , ] (1 ) ( ) ( ),g g H (1.1)

http://link.springer.com/search?facet-author=%22Ram+K+Pandey%22

http://www.hindawi.com/25082947/

http://scholar.google.com.my/citations?view_op=view_citation&hl=en&user=xy71xwEAAAAJ&citation_for_view=xy71xwEAAAAJ:u5HHmVD_uO8C


http://scholar.google.com.my/citations?view_op=view_citation&hl=en&user=xy71xwEAAAAJ&citation_for_view=xy71xwEAAAAJ:YsMSGLbcyi4C

http://scholar.google.com.my/citations?view_op=view_citation&hl=en&user=xy71xwEAAAAJ&citation_for_view=xy71xwEAAAAJ:d1gkVwhDpl0C


http://scholar.google.com.my/citations?view_op=view_citation&hl=en&user=xy71xwEAAAAJ&citation_for_view=xy71xwEAAAAJ:_FxGoFyzp5QC

5

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

parameter if 0 , 1( ,0) ( )g H and if 1 , then 2( ,0) ( )g H . Thus, as changes from 0

to 1, 1( )g is gradually transformed into 2 ( )g ‖.

3.2.2 Residual.

Residual is the difference of exact and the estimated value. For a differential equation, it is

calculated by substituting the approximate solution into the original differential equation.

1.4. Introduction to Well Known Homotopy Methods

1.4.1. Homotopy Analysis Method

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

equation:

( ( )) ( ), ,A f w (1.2)

, 0, ,d

d

wB w (1.3)

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

,A B are the differential and boundary operators respectively. The operator A is split into linear

L and nonlinearN parts. Eq. (1.3) is then written as

( ( )) ( ( )) ( ).f L w N w

Applying HAM [36] to the given problem, a general deformation equation is presented as:

0(1 )[ ( ( , ) ( ( ))] ( )[ ( ( , ) ( ( , )) ( )],f L H L w L H N H (1.4)

where 0 1 is an embedding parameter, is convergence control parameter, ( )

smoothing function, 0 ( )w is an initial guess approximation. Clearly, when 0 and 1 , it

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

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

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

6

0

1

, n

k

n

w w

H , (1.5)

where

0

,1|

!

k

kwk

H

. (1.6)

With proper selection of auxiliary linear operator, the initial guess, the auxiliary parameter , and

the auxiliary function the series (1.5) converges at 1, then we have

0

1

,1 .k

n

w w

H

(1.7)

Define the vector

0 1, ,..., .m mw w w w

Differentiating Eq. (1.7) m times with respect to the embedding parameter and then setting

0, and finally dividing them by m ! we have the so-called mth-order deformation equation.

For 0

1 1( )) ( ( )]m m m m m L[w N w wX R , (1.8)

where

1

1

1

1 ( ( , ))

1 !

m

m m

m

A p

m p

R

ww ,

and

0, 1

1, 1{ .m

m m

X

1.4.2. Homotopy Perturbation Method

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

equation:

( ( )) ( ),A f w (1.9)

, 0,d

d

wB w (1.10)

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

and B are differential and boundary operators. The operator A is split into linearL and nonlinear

7

N parts. Eq. (1.6) is then written as

( ( )) ( ( )) ( ).f L w N w

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

0(1 )[ ( ( , ) ( ( ))] [ ( ( , ) ( ( , )) ( )],f L H L w L H N H

(1.11)

where 0 1 is an embedding parameter, is convergence control parameter, ( ) smoothing

function, 0 ( )w is an initial guess approximation. Clearly, when 0 and 1 it holds

0,0 H w and ,1 H w

respectively. Thus, as changes from 0 to 1 , the

solution ( , ) H changes from 0 ( )w to the solution ( )w . For solution, expanding , H in

Taylor’s series about , we obtain:

0

1

, n

k

n

H w w . (1.12)

Substituting this series into (1.11) and the equating the like power of and letting 1 , we have

10

lim , ,1 .n

n

w H H w

(1.13)

The mth order approximate solution is obtain by truncating the series up to mth term and is given

by

0

.ii

w w (1.14)

1.5 THESIS PLAN

This thesis describes the application of OHAM to different linear and non-linear problems. The

material in Chapter 1 covers the historical background of analytic methods like Perturbation

methods, HPM, HAM and OHAM. The Chapter 2 explores the nature and the basic idea of

OHAM for ODEs and its application to tenth order ordinary differential equations. Chapter 3, 4

and 5 describes the effectiveness of OHAM formulation for time dependent nonlinear PDEs.

These chapters include Burger, Fisher, Burger’s–Huxley, Burger’s–Fisher, MEW and DGRLW

equations. In chapter 6 OHAM is applied to coupled system of PDEs. This chapter describes

WBK, MB and ALW equations extensively. Chapter 7 describes the implementation of OHAM

8

for DDEs and the solution of MKdV lattice equation is presented for the illustration of technique.

Application of OHAM to coupled system of PDEs and DDEs is major task of this thesis. A

summary of the results obtained is presented in Chapter 8. It is observed that the OHAM is better

organized and simpler and efficient than the other methods.

9

Chapter 2

Application of OHAM to Tenth Order Boundary Value Problems

2.1. Introduction

Ultrasonically assisted development of resists feature on semiconductor substrate is a popular

development technique, but is difficult to understand. During this development process the

developer i.e. 1:3 MythleIso Butyle Ketone and Isopropanal Alcohol is heated because of

ultrasonic agitation. During heating an infinite horizontal layer of fluid and then subjecting to the

action of rotation, instability sets in. Tenth order boundry value problems arise when instability

is as ordinary convection and a uniform magnetic field is also applied across the fluid in the

same direction as gravity. In literature little work has been done regarding these problems.

Siddiqi and Twizell found the solutions of these problems using thirteen, eleventh, tenth and

nonic degree splines [96- 99]. Wazwaz implemented ADM for linear and nonlinear boundary

value problems of tenth and twelfth-order [100]. Erturk and Momani applied DTM to linear and

nonlinear tenth-order boundary value problems with two-point boundary condition [101]. VIM

was used by Siddiqi et al., for the solution of tenth order boundary value problems [102]. Barari

et al. applied HPM for solving tenth order boundary value problems [103].

This chapter includes the solution of linear and nonlinear tenth order boundary value problems

using OHAM. The 2D images of exact, OHAM and DTM solutions are drawn and absolute error

by OHAM is compared with that of DTM to assess the efficiency of OHAM.

2.2 Fundamental Mathematical Theory of OHAM for (ODE) [44-49]

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

equation:

( ( )) ( ) ( ( )) 0,w g w L N (2.1)

, 0.dw

wd

B (2.2)

Where is an n tuples, ( )w is a required function, ( )g is a given function,L , N and B are

linear, nonlinear and boundary operators respectively.

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

10

(1 )[ ( ( , ) ( )] ( )[ ( ( , ) ( ) ( ( , ))],g h g L H L H N H (2.3)

,

, , 0,

HB H (2.4)

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

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

0,0 w H and ,1 w H respectively. Thus, as changes from 0 to 1 , the solution

( , ) H changes from 0 ( )w to the solution ( )w , where 0 ( )w is obtained from Eq. (2.3) for

0 :

00 0( ) 0, , 0 .

dww g w

d

L B (2.5)

For actual applications iC are finite, say, 1,2,3,... ,i m we propose the auxiliary function h

to be of the form:

2 3

1 2 3 ... m

mh C C C C , (2.6)

where 1 2, ,..., mC C C are constants. For solution, expanding , H in Taylor’s series about ,

we obtain:

0 1 2

1

, , , ,..., n

k n

n

w w C C C

H . (2.7)

Substituting Eq. (2.7) into Eq. (2.3) and equating the coefficient of like powers of , we

obtained the Zeroth order problem given by Eq. (2.5), the first and second order problems are

given by the eqs. (2.8)- (2.11) respectively:

1 1 0 0 ,w g C w L N (2.8)

11, 0,

dww

d

B (2.9)

2 1 2 0 0

1 1 1 0 1, ,

w w C w

C w w w

L L N

L N (2.10)

11

22 , 0,

dww

d

B (2.11)

In general it can be written as:

1 0 0

1

0 1 1

1

, ,..., ,

n n n

n

i n i n i n

i

w w C w

C w w w w

L L N

L N (2.12)

, 0,nn

dww

d

B where 2,3,...,n . (2.13)

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

, N H :

0 0 0 1

1

, , , ,..., .m

i m m

m

C w w w w

N H N N (2.14)

Here, convergence of the series (2.3) depends upon the constants , 1,2,3,...iC i . When 1

Eq. (2.8) transforms to:

1 2 0 1 2

1

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

i

w C C w w C C C

(2.15)

In actual application i is finite. Thus eq. (2.10) reduces to:

1 2 0 1 2

1

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

m i m

i

w C C C w w C C C

(2.16)

Substituting Eq. (2.16) into Eq. (2.1), it results the following residual:

1 2 1 2 1 2, , ,..., ( , , ,..., ) ( ) ( , , ,..., ).m m mR C C C w C C C g w C C C L N (2.17)

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

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

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

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

the optimal values of , 1,2,3,...iC i . Here, we apply the Method of Least Squares as under:

12

2

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

b

m m

a

S C C C R C C C dx (2.18)

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

1 2

... 0.m

S S S

C C C

(2.19)

the numbers anda b are properly chosen to locate the desired , 1,2,...,iC i m . With these

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

2.3 Application of OHAM to Tenth Order Boundary Value Problems

For implementation of method three examples are considered

2.3.1. Model 1

The following linear boundary value problem is considered

0)2189()()( 32)10( xxxexxuxu x

,[ 1, 1]x , (2.20)

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

(3) (3) (4) (4)

2 2( 1) 0, (1) 0, ( 1) , (1) 2 , ( 1) , (1) 6 ,

4( 1) 0, (1) 12 , ( 1) , (1) 20 .

u u u u e u u ee e

u u e u u ee

The exact solution is given by

2( ) (1 ) .xu x x e (2.21)

Here (10) 2 3( ( , )) ( , ), ( ( , )) ( , ) and (89 21 )xx x x x x g e x x x L H H N H H .

Using basic idea of OHAM we have

Zeroth order problem

2 3 (10)

089 21 ( ) 0,x x x xe xe x e e x u x (2.22)

with conditions

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

0 0 0 0 0 0

(3) (3) (4) (4)

0 0 0 0

2 2(1) 0, (1) 0, (1) 2 , ( 1) , (1) 6 , ( 1) ,

4(1) 12 , ( 1) 0, (1) 20 , ( 1) ,

u u u e u u e ue e

u e u u e ue

13

It is obtained that

2 (1 ) 2 (1 )

0

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

2 4 5 2 5 6 2 6

1( ) (545372 111250 502656 1221081 303837 126336

384

394320 103656 11904 2080060 269804 384 256152

32724 1921662 260310 105584 14312

x x

x x

u x e e x e x e xe

x e x e x x e x e x x

e x x e x x e x

7 2 7

8 2 8 9 2 9

921900 124764

19020 2574 179857 24341 . (2.23)

x e x

x e x x e x

First order problem

2 3 2 3 (10)

1 1 1 1 1 0 0

(10) (10)

1 0 1

89 21 89 21 ( ) ( )

( ) ( ) 0,

x x x x x x x x xe e x e x e x e C e C e x C e x C e C u x u x

C u x u x

(2.24)

with conditions

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

1 1 1 1 1 1

(3) (3) (4) (4)

1 1 1 1

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

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

u u u u u u

u u u u

Its solution is

17 21 20 (1 )

1 1 1 1

21 20 (1 ) 20 2 19 (1 ) 2

1 1 1 1

20 3 17 (1 ) 3 19 4

1 1 1

( , ) 6.80446 10 (1.57779 10 5.80436 10 1.23391

10 1.26505 10 4.73219 10 1.0375 10

1.18185 10 3.83858 10 2.15006 10 5.40645

x

x x

x

u x c C e C

xc e C x C e x C

x C e x C x C

15 (1 ) 4 18 5 17 6 16 7

1 1 1 1

15 8 13 9 11 11 10 12

1 1 1 1

9 13 8 14 7 15 16

1 1 1 1

10 3.01334 10 3.34258 10 2.94362 10

1.98866 10 8.81679 10 4.82305 10 6.01961 10

5.04113 10 3.35131 10 1.85435 10 863957

3

xe x C x C x C x C

x C x C x C x C

x C x C x C x C

17 18 19 20

1 1 1 13550.1 1046.7 23.9233 0.304585 .x C x C x C x C

(2.25)

Using Eqs. (2.5) and (2.6), the first order approximate solution by OHAM is

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

(2.26)

Now using the method of least square the following residuals

(10) 2 3( ) (89 21 ).xR u x xu e x x x (2.27)

is minimized for 1a and 1b . We obtain 1 0.008297818711013732C . The first-order

approximate solution is:

14

4 1 6 (1 )

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

6 7 8 9

17

( ) 0.000958019(1.3674 10 50256 1.02399 10 126336

371600 11904 86463.1 384 14352.5 1783.19

168.171 11.8049 0.569601 0.0145041 )

6.080446 10 ( 1.30922 1

x x

x x

u x e x e x

x e x x e x x x

x x x x

19 18 (1 ) 19

18 (1 ) 18 2 16 (1 ) 17 3

15 (1 ) 3 17 4 (1 ) 4

16 5 15 6

0 ) 4.81636 10 1.02388 10

1.04972 10 3.92669 10 8.60897 10 9.80676 10

3.18518*10 1.78408 10 4.48617 1013

2.50043 10 2.77361 10 2.4425

x

x x

x x

e x

e x e x

e x x e x

x x

14 7 13 8

11 9 9 11 8 12 7 13

6 14 15 16 17 18

19 20

7 10 1.65015 10

7.31601 10 4.00208 10 4.99479 10 4.18303 10

2.78086 10 153870 7168.96 278.393 8.68532

0.198511 0.0025274 . 2.28

x x

x x x x

x x x x x

x x

Table 2.1 displays values of the exact solution (2.21), OHAM solution (2.28) and comparison of

absolute error of OHAM with DTM [101]. 1st order OHAM solution gives very encouraging

results after comparing with DTM.

Table 2.1

Comparison of Absolute errors of OHAM with DTM

x Exact

Solution

OHAM

First Order

Error

DTM [101]

Error OHAM

-1 0

5.2398× 1310 2.49×10-7

5.2398× 1310

-0.8 0.161758 0.161758 1.66145×10-8

1.08119×10-9

-0.6 0.351239 0.351239 3.79826×10-9

5.56501×10-9

-0.4 0.563069 0.563069 1.41344×10-8

3.71371××10-8

-0.2 0.785982 0.785982 2.62138×10-8

1.04842×10-8

0 1.00000 1.00000 3.2×10-8

1.70981×10-8

0.2 1.17255 1.17255 2.68453×10-8

1.75885×10-8

0.4 1.25313 1.25313 1.46553×10-8

1.0973×10-9

0.6 1.16616 1.16616 3.64376×10-9

3.33432×10-9

0.8 0.801195 0.801195 1.95265×10-8

2.17311×10-10

1.0 0.000000 0.0000 2.97×10-7

7.84106×10-13

Error OHAM=Exact-OHAM, Error DTM=Exact-DTM

15

Fig. 2.1 shows the plots of exact solution (blue) OHAM solution (red) & DTM solution (Yellow)

for the domain 1815 x

Figure 2.1

2.3.2. Model 2

Consider the following tenth order nonlinear differential equation

(10) (3) 2( ) ( ) 2 ( ),xu x u x e u x 10 x (2.29)

with the following boundary conditions

(2) (2) (4)

(4) (6) (6) (8) (8)

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

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

u u u u ue e

u u u u ue e e

The analytic solution of this problem is

(10) ( ) .xu x e (2.30)

Here (10) (3) 2( ( , )) ( , ) ( , ), ( ( , )) 2 ( , ) and 0.xx x x x e x g L H H H N H H

Applying the proposed method, we have

Zero-Order Problem

(10)

0 ( ) 0,u x (2.31)

with conditions

(2) (2) (4)

0 0 0 0 0

(4) (6) (6) (8) (8)

0 0 0 0 0

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

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

u u u u ue e

u u u u ue e e

16

Its solution is

2 3

0

3 4 5 6 7 7 8

9 9

1( ) (1814400 1543941 -2382336 907200 257260 -

(1814400 )

397120 75600 12894 -19824 5 2520 300 -480 45

5 -5 .

u x e x e x e x xe

ex e x x e x e x x e x e x

x e x

(2.32)

First order problem

2 (3) (3) (3) (10) (10) (10)

1 0 0 1 0 1 0 1 0 12 ( ) ( ) ( ) ( ) ( ) ( ) ( ) 0xe C u x u x C u x u x u x C u x u x , (2.33)

with conditions

(2) (2) (4)

1 1 1 1 1

(4) (6) (6) (8) (8)

1 1 1 1 1

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

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

u u u u u

u u u u u

its solution is

17

-21 15 15 3 15 5

1

14 7 14 9 13 10 12 11

11 12 8 15 6 16 31

1

( ) 5.70399 10 (-4.65271 10 7.66574 10 -3.81067 10

9.31126 10 -1.77757 10 4.83251 10 -4.39204 10

3.65048 10 -1.34067 10 5.29666 10 9.96692 10 -

u x x x x

x x x x

x x x C

31 31 1 31 2. 31

1 1 1 1

30 31 1 31 2

1 1 1

30 2 30 2 30 1

1 1

1.29092 10 3.57953 10 -2.491 10 3.49096 10

8.54154 10 -2.35326 10 1.62654 10

5.67024 10 -2.77036 10 7.57977 10

x x x

x x x

x

e C e C e C xC

e xC e x C e xC

x C e x C e

2

1

30 2 2 30 2 2 29 3

1 1 1

29 3 30 1 3 30 2 3

1 1 1

28 4 28 29 1

1 4 1

-

5.20077 10 5.20077 10 5.61832 10

5.86092 10 -1.59155 10 1.0834 10

3.75206 10 -9.07735 10 e 2.44496 10

x

x x

x x x

x x

x C

e x C e x C x C

e x C e x C e x C

x C x C e

4

1

29 2 4 27 5 28 5

1 1 1

28 1 5 28 2. 5 25 6

1 1 1

27 6 27 1 6 27 2

1 1

-

1.65009 10 1.75658 10 1.09485 10 -

2.92287 10 1.95425 10 5.7851 10 -

1.06775 10 2.82296 10 -1.86827 10

x x

x x

x x x

x C

e x C x C e x c

e x C e x C x C

e x C e x c e x

6

1

24 7 25 7 26 1 7

1 1 1

26 2 7 22 8 24 8

1 1 1

25 1 8 24 2 8 20 9

1 1 1

1.29785 10 8.62687 10 -2.2566 10

1.47682 10 1.80918 10 -5.86693 10

1.51671 10 -9.80447 10 1.20224 10

3.39207

x x

x x

x x

C

x C e x C e x C

e x C x C e x C

e x C e x C x C

23 9 23 1 9 23 2 9

1 1 1

13 10 22 10 22 1 10

1 1 1

22 2 10 12 11 20 11

1 1 1

10 -8.6555 10 5.51949 10

4.83251 10 -1.67593 10 4.21474 10

-2.64732 10 -4.39204 10 7.08135 10

-1.75208

x x x

x x

x x

e x C e x C e x C

x C e x C e x C

e x C x C e x C

21 1 11 21 2. 11 11 12

1 1 1

19 12 19 1 12 19 2 12

1 1 1

10 13 17 13 18 1 13

1 1

10 1.08206 10 3.65048 10

-2.54894 10 6.19159 10 -3.75189 10

-2.81541 10 7.74164 10 -1.84152 10

x x

x x x

x x

e x C e x C x C

e x C e x C e x C

x C e x C e x C

1

18 2 13 9 14 16 14

1 1 1

16 1 14 16 2 14 8 15

1 1 1

14 15 14 1 15 14

1 1

1.09217 10 2.06483 10 -1.94963 10

4.52728 10 -2.62006 10 -1.34067 10

3.9507 10 -8.92081 10 5.01915 10

x x

x x

x x

e x C x C e x C

e x C e x C x C

e x C e x C e

2 15

1

6 16 12 16 13 1 16

1 1 1

12 2 16 10 17 11 1 17

1 1 1

10 2 17 8 18

1 1

5.29666 10 -6.1081 10 1.33449 10

-7.26594 10 6.48648 10 -1.36216 10

7.13513 10 -3.6036 10

x

x x

x x x

x x

x C

x C e x C e x C

e x C e x C e x C

e x C e x C

8 1 18

1

8 2 18

1

7.2072 10 -

3.6036 10 ).

x

x

e x C

e x C

2.34

Using Eqs. (2.13) and (2.15), the first order approximate solution by OHAM is;

0 1 1( ) ( ) ( , ).u x u x u x C 2.35

Now using the method of least square the following residuals is minimized for 0a and 1b .

(10) (3) 2( ) ( ) 2 ( ) 0,xu x u x e u x

18

Hence -7

1 -1.7374791232536796 10C is selected. The approximate solution is given by

2

3 3 4 5 5

6 7 7 8 9 9 25

24

1( ) (1814400 1543941 -2382336 907200

1814400

257260 -397120 75600 12894 -19824

2520 300 -480 45 5 -5 ) (-1.73173 10

2.24294 10 -6.21935 1024x

u x e x e x exe

x e x e x x e x

ex x ex e x x ex

e

1 2

24 24 1

24 2 23 2 23 2

1 2 2 2 22 3

23

4.32807 1024 -

6.06547 10 -1.48407 1024 4.08874 10

-2.82609 10 -9.85193 10 4.81345 10

-1.31697 1024 9.03623 1023 -9.76172 10

-1.01832 10

x x

x x

x x

x x

e e

x e x e x

e x x e x

e x e x x

3 23 1 3 23 2 3

21 4 22 4 1 4

2 4 20 5 21 5

21 1 5 21 2 5

2.76529 10 -1.88239 10

-6.51912 10 1.57717 10 -4.24807 1022

2.86699 1022 -3.05206 10 -1.90228 10

5.07842 10 -3.39547 10 -1.0051

x x x

x x

x x

x x

e x e x e x

x e x e x

e x x e x

e x e x

19 6

20 6 20 1 6 20 2 6

17 7 19 7 19 1 7

19 2 7 15 8 8

1 8

5 10

1.8552 10 -4.90484 10 3.24608 10

-2.24567 10 -1.4989 10 3.9208 10 -

2.56594 10 -3.14341 10 1.01937 1018

-2.63525 1018 1.70351

x x x

x x

x x

x

x

e x e x e x

x e x e x

e x x e x

e x

18 2 8 14 9

16 9 1 9 16 2 9

13 10 10 15 1 10

15 2 10 11 2 11

10 -1.98646 10

-5.89364 10 1.50388 1017 -9.59 10

4.83251 10 2.91189 1015 -7.32303 10

4.59967 10 763107. -594397440000 -

1.23037

x

x x x

x x

x

e x x

e x e x e x

x e x e x

e x x e x

11 14 1 11 14 2 11

11 12 12 12 13 1 12

12 2 12 13 2 13

11 13 11 1 13

1014 3.0442 10 -1.88005 10

3.65048 10 4.42872 10 -1.07578 10

6.51884 10 4891.71 -3810240000 -

1.34509 10 3.1996 10 -1.89

x x x

x x

x

x x

e x e x e x

x e x e x

e x x e x

e x e x

11 2 13

9 14 14 9 1 14

9 2 14 15 7 15

1 15 7 2 15 6 16

6 16

763 10

2.06483 10 3.38745 109 -7.86606 10

4.55229 10 23.2939 -6.86426 10

1.54997 108 -8.72067 10 5.29666 10

1.06127 10 -2.31864 1

x

x x

x x

x x

x

e x

x e x e x

e x x e x

e x e x x

e x

6 1 16 6 2 16

17 1 17 2 17 18

1 18 2 18 2

0 1.26244 10

-11270.1 23667.3 -12397.1 62.6118

-125.224 62.6118 )/ 23726443732992000000 .

x x

x x x x

x x

e x e x

e x e x e x e x

e x e x e

2.36

19

The following table displays values of the exact solution (2.30), OHAM solution (2.36) and

comparison of absolute error of OHAM with DTM [101].

Table 2.2

Comparison of Absolute errors of OHAM with DTM

x Exact

Solution

OHAM

First order

=

Error DTM Error OHAM

0.0 1.000000 1.000000 2.48699×10-11

2.22045×10-16

0.1 0.904837 0.904837 1.06434×10-6

2.14132×10-11

0.2 0.818731 0.818731 2.02411×10-6

9.42912×10-13

0.3 0.740818 0.740818 2.78518×10-6

1.03221×10-10

0.4 0.67032 0.67032 3.27306×10-6

4.62774×10-12

0.5 0.606531 0.606531 3.44023×10-6

6.16236×10-11

0.6 0.548812 0.548812 3.27066×10-6

3.40034×10-11

0.7 0.496585 0.496585 2.78129×10-6

1.19528×10-11

0.8 0.449329 0.449329 2.02022×10-6

8.85421×10-11

0.9 0.40657 0.40657 1.06193×10-6

4.02983×10-11

1.0 0.367879 0.367879 2.85577×10-6

4.85739×10-11

Error OHAM=Exact-OHAM, Error DTM=Exact- DTM

Fig. 2.2 shows exact solution (Yellow) OHAM solution (red) & DTM solution (blue) for the

domain 80 x .

Figure 2.2

20

2.2.3. Model 3

For ]1,1[x , the following linear boundary value problem is considered

(10) 2 3( ) ( ) (89 21 ),xu x xu x e x x x

with change initial conditions

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

(6) (6) (8) (8)

2 4( 1) 0, (1) 0, ( 1) , (1) 6 , ( 1) , (1) 20

18 40( 1) , (1) 42 , ( 1) , (1) 72 .

u u u u e u u ee e

u u e u u ee e

The exact solution is given by

2( ) (1 ) .xu x x e

(2.38)

Zeroth order problem

2 3 (10)

089 21 ( ) 0.x x x xe xe x e e x u x (2.39)

The initial conditions are given by

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

0 0 0 0 0 0

(6) (6) (8) (8)

0 0 0 0

2 4( 1) 0, (1) 0, ( 1) (1) 6 , ( 1) , (1) 20

18 40( 1) , (1) 42 , ( 1) , (1) 72 .

u u u u e u u ee e

u u e u u ee e

It is obtained that

2 (1 )

0

2 (1 ) 2

2 2 (1 ) 2 3 2 3

(1 ) 3 4

1( ) [1159766595 717443145 -2375049600 -

(181440 )

1379102043 841439505 59693760x 340322220

191564100 -56246400 -132536060 73226420

1814400 14726250 718

x

x

x

x

u x e ee

x e x e x

e x e x x e x

e x x

2 4 5 2 5

6 2 6 7 2 7 8 2 8 9 2 9

5150 -3353322 1594110

207900 79380 -32460 11940 1035 225 -115 25 .

e x x e x

x e x x e x x e x x e x

(2.40)

First order problem

2 3 2 3

1 1 1 1 1 0

(10) (10) (10)

0 1 0 1

89 21 89 21 ( )

( ) ( ) ( ) 0,

x x x x x x x x xe e x e x e x e C e C e x C e x C e C u x

u x C u x u x

(2.41)

with conditions

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

1 1 1 1 1 1

(3) (3) (4) (4)

1 1 1 1

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

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

u u u u u u

u u u u

21

Its solution is

-20 24 24 1

1 1 1

24 23 1 24 2 22

1 1 1

1 2 23 3 21 1 3

1 1 1

23 4

1

( ) 1.4401 10 (7.45506 10 -2.74256 10 5.83023

10 5.97738 10 2.23596 10 -4.90218 10

5.58423 10 1.81373 10 1.0159

10 -2.55455

x

x

x x

u x C e C

xC e xC x C

e x C x C e x C

x C

19 1 4 22 5

1 1

21 6 20 7 18 8 17 9

1 1 1 1

15 11 14 12 13 13

1 1 1

12 14 10 15

1 1

10 1.42381 10 1.57937

10 1.39086 10 9.39651 10 4.1661 10

-2.2789 10 -2.84427 10 -2.38192 10 -1.58349

10 -8.76205 10 -4

xe x C x C

x C x C x C x C

x C x C x C

x C x C

9 16 8 17

1 1

6 18 19 20

1 1 1

.08223 10 -1.58491 10

-4.94453 10 -113297 -1464.25 .

x C x C

x C x C x C

(2.42)

Approximate solution of OHAM using (2.40) and (2.42) is

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

For 1a and 1b , using method of least square we obtain

1 0.9983163864661831.C

With this value our solution is

2 1

2 1 2 2 2

1 2 3 2 3 1 3

4

1( ) (1159766595 717443145 -2375049600 -1379102043

1814400

841439505 596937600 340322220 191564100 -

56246400 -132536060 73226420 1814400

14726250 7185

x

x

x x

u x e e xe

e x e x x e x

e x x e x e x

x

2 4 5 2 5 6

2 6 7 2 7 8 2 8 9

2 9 24 24 1

24 23 1

150 -3353322 1594110 207900

79380 -32460 11940 1035 225 -115

125 (-7.4425 10 2.73794 10

(25545471085854720000 )

-5.82042 10 -5.96732 10 -

x

x

e x x e x x

e x x e x x e x x

e x ee

x e x

24 2 22 1 2

23 3 21 1 3 23 4

19 1 4 22 5 21 6 20 7

18 8 17 9 15 11

2.2322 10 4.89392 10

-5.57483 10 -1.81067 10 -1.01419 10 2.55025

10 -1.42141 10 -1.57671 10 -1.38852 10 -

9.38069 10 -4.15909 10 2.27506 10

x

x

x

x e x

x e x x

e x x x x

x x x

14 12

13 12 14 10 15 16

17 18 19 20

2.83948 10

2.37791 1013 1.58083 10 8.7473 10 4.07535 109

1.58225 108 4.9362 106 113106 1461.79 .

x

x x x x

x x x x

(2.43)

22

The following table displays values of the exact solution (2.38), OHAM solution (2.43) and

absolute error.

Table 2.3

Abosute errors of 1st order OHAM solution

x Exact Solution OHAM First

Order Solution

Absolute error

OHAM

-1.0 0.000000 0.000000 1.10934×10-11

-0.8 0.161759 0.161759 8.39316×10-7

-0.6 0.351241 0.351239 1.59673×10-6

-0.4 0.563071 0.563069 2.1983×10-6

-0.2 0.785984 0.785982 2.58517×10-6

0.0 1.000000 1.000000 2.71929×10-6

0.2 1.17255 1.17255 2.58731×10-6

0.4 1.25313 1.25313 2.20175×10-6

0.6 1.16616 1.16616 1.60013×10-6

0.8 0.801196 0.801195 8.41367×10-7

1.0 0.000000 -4.85494×10-11

4.85494×10-11

Fig.2.3. Shows OHAM solution (red) & exact solution (blue) for the domain 3222 x .

Figure 2.3

23

Chapter 3

Implementation of OHAM to Family of Burger Equations

3.1. Introduction

This chapter is devoted to analysis and application of OHAM to family of Burger equations. In

particular we will focus on Burger, Burger’s–Huxley and Burger’s–Fisher equations. These

equations arise in modeling of different physical problems. The Burger equation can be used as a

model for different problems of a fluid flow nature, where shocks or viscous dissipation is a

major factor. The first theoretical solution of Burger equation has been given by Cole [104],

which was based on Fourier series analysis using the appropriate initial and boundary conditions.

Another theoretical solution based on the ―test and trial‖ method has been given by Madsen and

Sincovec[105], using the appropriate initial and boundary conditions.

The Generalized Burger’s–Huxley equation introduced by Satsuma shows a prototype model for

describing the communication between reaction mechanisms, convection effects and diffusion

transports [106]. Burger-Fisher equation has significant applications in various fields of applied

mathematics and has physical applications such as gas dynamic, traffic flow, convection effect

and diffusion transport [107-113]. In section 3.3.1 comparison is made between the results of the

proposed method and ADM for Burger equation [115]. In section 3.3.2 the results obtained by

proposed method for Burger’s–Huxley equation are compared to that of ADM [115]. In section

3.3.3 comparison is made between the results of the proposed method and ADM for Burger’s

fisher equation [115]. As a result it is concluded that the method is explicit, effective, and simple

to use.

3.2 Analysis of OHAM for PDE

Consider the following general PDE:

( ) ( ) ( ) 0,w g w L N (3.1)

, 0,dw

wd

B (3.2)

where is n-tupple, w is required function, ( )g is a given function,L , N and B are linear,

24

nonlinear and boundary operators respectively. The basic idea of OHAM is same to that defined

in section 2.2. The method of least square for above purpose is as follows.

We solve the system

2

0

, ,

t

i iS K R C d

(3.3)

where R is the residual,

( ) ,R w g w L N (3.4)

and

1 2

... 0.m

S S S

C C C

(3.5)

The values of constants iC can also be found by collocation method by taking points ip in

problem domain and then solving the system

1 2 1,2; ; ... ; 0, ,..., .i i m iR p C R p C R p C i m

(3.6)

3.3 Implementation of OHAM to Burger’s Family

In this section, we have applied OHAM for finding approximate solutions of Burger, Burger’s–

Huxley and Burger’s–Fisher equations.

3.3.1. Model 1. (Burger Equation)

Let us consider Burger Equation

2

2

( , ) ( , ) ( , )( , ) 0,δu x t u x t u x t

αu x tt x x

for all 0 1x and 0t . (3.7)

with initial condition given by

( ,0) 0.5 0.5 tanh .2 1

αδu x

δ

(3.8)

Case 1: when 1α and 1δ

for 1α and 1δ Eq. (3.7) takes the following form

2

2

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

u x t u x t u x tu x t

t x x

(3.9)

subject to constant initial condition

25

( ,0) 0.5 0.5tanh(0.25 ).u x x (3.10)

The exact solution of equation (3.10) with given condition is given by

)]5.0(25.0tanh[5.05.0),( txtxu , (3.11)

here

2

2

( , , )( ( , , )) ,

( , , ) ( , , )( ( , , )) ( , , ) ,

and 0.

x tx t

t

x t x tx t t

x x

g

HL H

H HN H H

Following the basic idea of OHAM presented in preceding section we start with

Zeroth Order Problem

0),(0

t

txu, 0( ,0) 0.5 0.5tanh(0.25 ).u x x (3.12)

Its solution is

).25.0tanh(5.05.0),(0 xtxu (3.13)

First Order Problem

0),(),(

),(),(

)1(),(

2

0

2

10

010

11

x

txuC

x

txutxuC

t

txuC

t

txu (3.14)

.0)0,(1 xu

Its solution is

2

1 1 1( , , ) ( 0.0625 sech (0.25 )).u x t C t C x (3.15)

Second Order Problem

2

0 0 02 11 2 0 2 22

2

0 1 11 1 1 0 1 22

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

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

u x t u x t u x tu x t u x tC C u x t C C

t t x x t

u x t u x t u x tC u x t C u x t C u x

x x x

(3.16)

Its solution is

2 5

2 1 1 1 2

2 2 4 2 2

1 1

( , , ) (( 0.0625C (1 ) 0.0625 ) sec (0.25 ) sec (0.25 )

(0.00195313 sinh(0.75 )) sec (0.25 )( 0.00195313 tanh(0.75 ))).

u x t C C C t h x h x

C t x h x C t x

(3.17)

Third Order Problem

26

2

3 0 0 021 3 0 3 32

2

0 1 1 12 1 2 0 2 22

2

0 011 2 1 1 1 2

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

( , ) ( , ) ( , ) ( , )( , ) ( , )

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

u x t u x t u x t u x tu x tC C u x t C C

t t x x t

u x t u x t u x t u x tC u x t C u x t C C

x x x t

u x t u x tu x tC u x t C u x t C

x x x

3, ,0 0.u x

(3.18)

Its solution is

2

3 1 2 3 2 30.5 2

1 1 2

2

1

2 3 1 2 1

1( , , , , ) ( sec (0.25 )(-0.0625 -0.0625

(1 )

( -0.0625 (-0.125-0.015625 ) (-0.125

-0.015625 )-0.000651042 C (5.0718 ) (18.9282 ))

((288 288 (288 576 (576-2

xu x t C C C t h x C C

e

C C t C

t t t

C C C C C

2

2 1

2 3 1 2 1

2

2 1

2

2 1

4 )

-24 -3 (-6.58301 ) (14.583 ) cosh(0.25 )

(-96 - 96 + (-96- 192 + (-192- 24 )

-24 +5 (-7.396+ ) (2.596+ ))) sinh(0.25 ))

-24 + 5 (-7.396 + ) (2.596 + ))) sinh(0.25 )

(

t

C t C t t x

C C C C C t

C t C t t x

C t C t t x

-0.000651042 cosh(0.75 ) - 0.000651042 sinh(0.75 )))).x x

(3.19)

Adding equations (3.13, 3.15, 3.17, 3.19) we obtain:

1 2 0 1 1 2 1 2 3 1 2 3, , , , , , , , , , , , , .u x t C C u x t u x t C u x t C C u x t C C C

(3.20)

For the calculations of the constants 1 2 3, andC C C using the Method Least Squares we have

computed that

1

-8

2

-11

3

1.000084070827354,

9.465713851467995 10 ,

1.98940934674601 10 .

C

C

C

Putting the values of these constants into equation (3.20) the third order approximate solution

using of OHAM is

27

-6 2

0.5 2

-9 -6

-8

2

1( , ) (0.5-5.2607845247854 10 sech (0.25 ) +

(1 )

(6.360090415451543 10 (1.3153067008227217 10

0.0006512058813062292 ) (1.2720180853076355 10 -

0.002604823525224917 ) cosh(0.5 )+

xu x t t x

e

t t

t x

-9

-6

-8 2

-9 -9

(6.36009042653818 10

(-1.3153067008227217 10 0.0006512058813062292 ) ) cosh( )

(1.2720180853076355 10 -0.002604823525224917 ) sinh(0.5 )

(6.36009042653818 10 ( 1.3153067008227217 10

0.0006

t t x

t x

-

2 2 2

512058813062292 ) )sinh( ) -0.5 tanh(0.25 )

0.007813813661895406 sech (0.25 ) tanh(0.25 ) sech (0.25 )

(0.06250525442670962 )).

t t x x

t x x x

t

(3.21)

The 3rd

order OHAM solution yields very encouraging results after comparing with 4th

order

ADM solution [115].

Table 3.1

Comparison of absolute errors of 3rd

order OHAM solution and 4th

order ADM solution for

Burger equation for 0.1x and [0, 2]t

t Exact solution Absolute Error ADM Absolute Error OHAM

0.5 0.518741 6.34216 ×10-8

5.32631×10-8

1.0 0.549834 2.02886 ×10-6

7.98928×10-8

2.0 0.610639 6.42801×10-5

3.2441×10-5

Table 3.2


order OHAM solution and 4th

order ADM solution for

Burger equation for 0.5x and [0, 2]t

t Exact solution Absolute Error ADM Absolute Error OHAM

0.5 0.468791 5.66705×10-8

5.82744×10-8

1.0 0.5 1.8471×10-6

3.89112×10-6

2.0 0.562177 6.06928×10-5

7.40943×10-5

28

Table 3.3

Absolute errors of 3rd

order OHAM solution for various values of x and t

x 003.0t 1.0t 5.0t 1t

-4 7.88258×10-15

3.00377×10-10

1.86591×10-7

2.68667×10-6

-2 8.54872×10-15

2.023×10-9

1.2447×10-6

1.96202×10-5

0 4.55191×10-15

1.02224×10-10

7.37402×10-8

2.10382×10-6

2 1.11022×10-15

1.98526×10-9

1.26475×10-6

2.0391×10-5

4 9.57567×10-16

3.6574×10-10

2.37125×10-7

4.14971×10-6

Case 2. When 1α and 2δ

For 1α and 2δ Eq. (3.7) takes the following form

22

2

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

u x t u x t u x tu x t

t x x

(3.22)

subject to constant initial condition

( ,0) 0.5 0.5 tanh( ).3

xu x

(3.23)


)]3

1(

4

1tanh[5.05.0),( txtxu , (3.24)

Using same lines as above the third order approximate solution using OHAM is obtained and

absolute errors for various values of x and t are given in Table (3.4-3.5).

29

Table 3.4

Absolute error of the solution of Burger equation by OHAM at 0.1x and various values of t

t Exact solution OHAM solution Absolute error

0.1 0.6992207 0.6992207 1.026602×10-11

0.2 0.703168 0.703168 3.52315×10-9

0.3 0.707107 0.707107 2.39034×10-8

0.4 0.711024 0.711024 8.5422×10-8

0.5 0.714919 0.714919 2.23819×10-7

0.6 0.718791 0.718791 4.86719×10-7

0.7 0.722639 0.722639 9.34107×10-7

0.8 0.726464 0.726464 1.63865×10-6

0.9 0.730263 0.730263 2.86617×10-6

1.0 0.734037 0.734037 4.17601×10-6

Table 3.5

Absolute error of the solution of Burger equation by OHAM at 0.5x and various values of t

t Exact solution OHAM solution Absolute error

0.1 0.650264 0.650264 3.16105×10-10

0.2 0.654428 0.654428 1.15677×10-9

0.3 0.658578 0.658578 7.62659×10-10

0.4 0.662715 0.662715 1.33499×10-8

0.5 0.666837 0.666837 4.97044×10-8

0.6 0.670944 0.670944 1.28683×10-7

0.7 0.675035 0.675035 2.7546×10-7

0.8 0.679109 0.679109 5.22082×10-7

0.9 0.683166 0.683165 9.08017×10-7

1.0 0.687205 0.687204 1.48069×10-6

30

3.3.2 Model 2 (Burger’s–Huxley Equation).

Consider Burger’s–Huxley equation of the form

2

2

( , ) ( , ) ( , )( , ) ( , )(1 ( , ))( ( , ) ) 0,δ δ δu x t u x t u x t

αu x t βu x t u x t u x t γt x x

(3.25)

for all 0 1x and 0,t Here , , and are parameters where 0, 0 and (0,1)

the initial condition is

1

( ,0) (0.5 0.5 tanh( ))2( 1)

u x x


1

1 2( ,0) (0.5 0.5 tanh( ( - )))δu x γ γ A x A t , (3.26)

where

2

1

- 4 (1 )

4(1 )

αδ δ α β δA γ

δ

,

and

2

2

(1 - )(- 4 (1 ))- .

(1 ) 2(1 )

δ γ α α β δαγA

δ δ

For computational work, we have taken 0,1, 1, 1,2α β δ and 0.001γ for various values of

x and t .We take

2

2

( , , )( ( , , )) ,

( , , ) ( , , )( ( , , )) ( , , )

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

x tx t

t

x t x tx t x t

x x

x t x t x t g

HL H

H HN H H

H H H

Following the basic idea of OHAM we start with


0),(0

t

txu, 0( ,0) (0.0005+0.0005tanh (0.00025 ))u x x .

Its solution is

0 (0.0005+0.0005tanh (0.00025 ))u (x,t) x .

(3.27)

31

First Order Problem

0 0 01

1 1 1

1 0 1 0 1 0

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

0.001 ( , ) 0.001 ( , ) ( , ) 0,

2

0 2

3 3

u x t u x t u x tu x t( C ) C u (x,t) C

t t x x

- C u x t + C u x t - C u x t

(3.28)

.0)0,(1 xu

Its solution is

-7 -11 2

1 1 1

-10

1

-10 2

1

-7 2 -10

1 1

( , ) -2.4987500000000003×10 6.25×10 sech (0.00025 )

1.249999999999317×10 tanh(0.00025 )

1.25×10 sech (0.00025 ) tanh(0.00025 )

2.49875×10 tanh (0.00025 ) -1.25×10 t

u x t -t ( C - C x

+ C x

- C x x

+ C x C

3anh (0.00025 )x .

(3.29)


0 0 02 11 2 0 2 2

20 1 11 1 1 0 1 2 0 2 0

3

2 0 1 1 1 0

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

( , ) ( , ) ( , )( , ) ( , ) 0.001 ( , ) 1.001 ( , )

( , ) 0.001 ( , ) 2.002 ( ,

2

2

2

2

u x t u x t u x tu x t u x t( C ) C u x t C C

t t x x t

u x t u x t u x tC u x t C u x t C C u x t + C u x t

x x x

C u x t C u x t + C u x

2

1 1 0 1) ( , ) 3 ( , ) ( , ) 0,t u x t C u x t u x t

2 ,0 0.u x (3.30)

Its solution is

-7 -11 2

1 2 1 1

-10

1

-10 2

1

-7 2

1

( , , , ) -2.4987500000000003×10 6.25×10 sech (0.00025 )

1.249999999999317×10 tanh(0.00025 )

1.25×10 sech (0.00025 ) tanh(0.00025 )

2.49875×10 tanh (0.00025 ) 1.25×10

2u x t C C t ( C C x

C x

C x x

C x

-10 3

1 tanh (0.00025 )C x .

(3.31)

32

Third Order Problem

3 0 0 021 0

0 1 1 11 0

0 1 22 1

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

( ) ( ) ( ) ( )( ) ( )

( ) ( ) ( )( ) ( )

2

3 3 32

2

2 2 2 22

2

1 1 1 2

u x,t u x,t u x,t u x,tu x,tC C u x,t C C

t t x x t

u x,t u x,t u x,t u x,tC u x,t C u x,t C C

x x x t

u x,t u x,t u x,tC u x,t C u x,t C C

x x x

0

2 3

2 0 0 0

2 2 2

0 1 1 1 2

2

0 2 0 2 0

( )

0.001 ( ) 1.001 ( ) ( ) 0.001 ( )

3 ( ) ( ) 1.001 ( ) 3 ( ) 0.001 ( )

2.002 ( ) ( ) 3 ( ) ( ) 2.002 ( )

21

3 3 3 2

2 1 1 0 1

1 1 2

u (x,t)u x,t

x

C u x,t + C u x,t C u x,t C u x,t

C u x,t u x,t C u x,t C u (x,t)u x,t C u x,t

+ C u x,t u x,t C u x,t u x,t + C u x,t

1( ) 0,u x,t

3 ,0 0.u x (3.32)

Its solution is

2

3 1 2 3 2 30.5 2

1 1 2

2

1

2 3 1 2 1

1( , , , , ) ( sec (0.25 )( - 0.0625 - 0.0625

(1 )

(-0.0625 (-0.125 - 0.015625 ) (-0.125

- 0.015625 ) - 0.000651042 (5.0718 ) (18.9282 ))

((288 288 (288 576 (576 - 2

xu x t C C C t h x C C

e

C C t C

t C t t

C C C C C

2

2 1

2 3 1 2 1

2

2 1

2

2 1

4 )

- 24 -3 (-6.58301 ) (14.583 ) cosh(0.25 )

(-96 - 96 (-96 - 192 (-192 - 24 )

- 24 5 (-7.396 ) (2.596 ))) sinh(0.25 ))

- 24 5 (-7.396 ) (2.596 ))) sinh(0.25 )

t

C t C t t x

C C C C C t

C t C t t x

C t C t t x

(-0.000651042 cosh(0.75 ) - 0.000651042 sinh(0.75 )))).x x

(3.33)


1 2 0 1 1 2 1 2 3 1 2 3, , , , , , , , , , , , , .u x t C C u x t u x t C u x t C C u x t C C C (3.34)

For the calculations of the constants 1 2 3, andC C C using the Collocation method, we have

computed that

1

-7

2

-12

3

-1.0000010231545267,

-9.98159444155818 10 ,

- 2.041939789528322 10 .

C

C

C

Putting the values of these constants into Eq. (3.34) the third order approximate solution using

OHAM is

33

-6 2

3 0.5 2

-9 -6

-8

2

1( ) = (0.5 -5.2607845247854×10 sech (0.25 ) +

(1+ e )

(6.360090415451543×10 + (1.3153067008227217×10 +

0.0006512058813062292 ) + (1.2720180853076355×10 -

0.002604823525224917 ) cosh(0.5 ) +

xu x,t t x

t t

t x -9

-6

-8 2

-9 -9

(6.36009042653818×10 +

(-1.3153067008227217×10 + 0.0006512058813062292 ) ) cosh( ) +

(1.2720180853076355×10 - 0.002604823525224917 ) sinh(0.5 ) +

(6.36009042653818×10 + (-1.3153067008227217×10 +

0.0006

t t x

t x

2 2 2

512058813062292 ) )sinh( ) - 0.5 tanh(0.25 ) +

0.007813813661895406 sech (0.25 ) tanh(0.25 ) + sech (0.25 )

(0.06250525442670962 )).

t t x x

t x x x

t

(3.35)

Table 3.6 shows a comparison between OHAM solution and ADM solution for

1, 1, 1and 0.001α . For 0α Eq. (3.25) is reduced to the Generalize Huxley

equation which describes nerve pulse propagation in nerve fibers and wall motion in liquid

crystals [06]. Table 3.7 and 3.8 shows a comparison between ADM [115] solution and OHAM

solution for 0, 1, 0.001 and 1,2α respectively. Table 3.9 shows absolute errors of

OHAM solution for larger domain for 0,1, 1, 1,2 and 0.001α β δ γ .

Table 3.6

Comparison of absolute errors of OHAM and ADM for various values of x and t , where

1, 1, 1and 0.001α

t ADM

1.0x

OHAM

1.0x

ADM

5.0x

OHAM

5.0x

ADM

9.0x

OHAM

9.0x

0.05 1.93715×10-7

1.87406×10-8

1.9373×10-7

1.87406×10-8

1.93745×10-7

1.87406×10-8

0.1 3.87434×10-7

3.74812×10-8

3.87464×10-7

3.74812×10-8

3.87494×10-7

3.74812×10-8

1 3.87501×10-6

3.74812×10-7

3.87531×10-6

3.74812×10-7

3.87561×10-6

3.74812×10-7

34

Table 3.7

Comparison of absolute errors of OHAM and ADM for various values of x and t and

0, 1, 0.001and 1α

t ADM

1.0x

OHAM

1.0x

ADM

5.0x

OHAM

5.0x

ADM

9.0x

OHAM

9.0x

0.05 1.93715×10-7

2.49875×10-8

1.9373×10-7

2.49875×10-8

1.93745×10-7

2.49875×10-8

0.1 3.87434×10-7

4.9975×10-8

3.87464×10-7

4.9975×10-8

3.87494×10-7

4.9975×10-8

1 3.87501×10-6

4.9975×10-7

3.87531×10-6

4.9975×10-7

3.87561×10-6

4.9975×10-7

Table 3.8

Comparison of absolute errors of OHAM and ADM for 0.5x 001.02,1,0 andα

Table 3.9

Absolute errors of OHAM for 0,1, 1, 1, 0.001 2α β δ γ and x

t 001.0,1,1,1 α 001.0,1,1,0 α 0, 1, 0.001, 2α β γ δ

0.1 3.74812×10-8

2.49875×10-8

2.23403×10-6

0.2 7.49625×10-8

4.9975×10-8

4.46806×10-6

0.3 1.12444×10-7

7.49625×10-8

6.70209×10-6

0.4 1.49925×10-7

9.995×10-8

8.93612×10-6

0.5 1.87406×10-7

1.24937×10-7

1.11702×10-5

0.6 2.24887×10-7

1.49925×10-7

1.34042×10-5

0.7 2.62369×10-7

1.74912×10-7

1.56382×10-5

0.8 2.9985×10-7

1.999×10-7

1.78722×10-5

0.9 3.37331×10-7

2.24887×10-7

2.01063×10-5

1.0 3.74812×10-7

2.49875×10-7

2.23403×10-5

t Error ADM Error OHAM

0.05 5.58836×10-7

2.7938×10-7

0.1 1.11766×10-6

5.58771×10-7

1 1.00741×10-5

5.5896×10-6

35

3.3.3. Model 3 (Burger’s–Fisher Equation).

Consider the Burger’s-Fisher equation of the form

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

2δ δ

2

u x,t u x,t u x,tαu x,t βu x,t u x,t

t x x

(3.36)

for all 0 1x and 0t . Here , and are parameters where 0 and 0.

Subject to constant initial condition

( ,0) (0.5+ 0.5 tanh( ))1

δαδ

u x x2(δ 1)

,

with exact solution given by

( 1)

( ) = (0.5+ 0.5 tanh( ( ( ) )) .2( ( 1)

1

δαδ α β δ

u x,t x tδ 1) δ α

(3.37)

For computational work, we have taken 0.001, 0.001 and 1,2α and various values of x

and t . Following the basic idea of OHAM with

( , , )( ( , , )) ,

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

and 0, we have

2δ

2

x tx t

t

x t x tx t α x t β x t x t

x x

g

HL H

H HN H H H H


0 ( )0

u x,t

t

,

0

-0.001( ) = (0.5+ 0.5 tanh( )).

4u x,0 x

Its solution is

0

0.001( ) = (0.5-0.5 tanh( )).

4u x,t x (3.38)

First Order Problem

0 011 1 0 1 0 0

01 1

( ) ( )( )(1+ ) 0.001 ( ) 0.001 ( )(1- ( ))

( )0, ( ,0) = 0.

2

2

u x,t u x,tu x,tC C u x,t C u x,t u x,t

t t x

u x,tC u x

x

(3.39)

Its solution is

-8 2

1 1 1 1

2

1

( ) = - (0.00025 + 6.25×10 sech (0.00025 ) -

0.00025 tanh (0.00025 )).

u x,t,C t C C x

C x

(3.40)

36


02 11 2 0 0 2 2

0 0 12 0 1 1 1 0

11 1 1 0

( )( ) ( )(1+ ) 0.001 ( )(1 ( ))

( ) ( ) ( )0.001 ( ) 0.001 ( ) 0.001 ( )

( )0.001 ( ) - 0.002 (

2

0

2

2

1 2

u x,t u (x,t)u x,t u x,tC C u x,t u x,t C C

t t x t

u x,t u x,t u x,tC u x,t C u x,t C u x,t

x x x

u x,tC C u x,t C u x,t

x

1) ( ) = 0,u x,t

2 ,0 0.u x (3.41)

Its solution is

5

2 1 2 1 2

2

1 1

2

2 1

-20

1

( , ) = sech (0.00025 ) ((-0.000187547 - 0.000187547 +

(-0.000187547 )) cosh(0.00025 ) + (-0.0000625156

- 0.0000625156 + (-0.0000625156 )

+ 2.71051×10 sinh(0.00025

u x,t,C C x C t C t

C t x C t

C t C t

C t -20 2

1

-20 -8 2 2

2 1

2 2

1

) + 2.71051×10 sinh(0.00025 )

+ 2.71051×10 sinh(0.00025 ) + 3.12656×10 sinh(0.00025 )

+ 3.12656×10 -8 sinh(0.00075 ).

x C t x

C t x C t x

C t x

The third order approximate solution using OHAM is given by


Where 3 1 2 3( , , , , )u x t C C C is obtained in same lines.

For the calculations of the constants 1 2 3, andC C C using the collocation method we have

computed that -7

1

2

3

= -5.928318703338053×10 ,

= -465.9630543691778 and

= 1.8651679832921486.

C

C

C

The third order OHAM solution yields very encouraging results after comparing with fourth

order approximate solution by ADM [115].

Table 3.10 shows a comparison between OHAM solution and ADM solution for

0.001, 0.001 and 1α [115]. In table 3.11 comparisons is given between OHAM

solution and ADM solution for 0.001, 0.001 and 2α [115]. Table 3.12 shows the

reliability of OHAM for larger domain.

37

Table 3.10

Comparison of absolute errors by OHAM and ADM [115] for 0.001, 0.001 and 1α

t ADM for

1.0x

OHAM

1.0x

ADM

5.0x

OHAM

5.0x

ADM

9.0x

OHAM

9.0x

0.005 9.68763×10-6

1.12257×10-7

9.68691×10-6

2.28888×10-7

9.68619×10-6

2.28888×10-7

0.001 1.93753×10-6

2.24513×10-8

1.93738×10-6

4.57775×10-8

1.93724×10-6

4.57775×10-8

0.01 1.93752×10-5

2.24514×10-7

1.93738×10-5

4.57777×10-7

1.93724×10-5

4.57777×10-7

Table 3.11

Comparison of absolute errors obtained by OHAM and ADM [115] for

0.001, 0.001 and 2α

t ADM for

1.0x

OHAM

1.0x

ADM

5.0x

OHAM

5.0x

ADM

9.0x

OHAM

9.0x

0.0005 1.40177×10-3

5.87633×10-5

1.34526×10-3

×10-5

1.27699×10-3

4.64718×10-5

0.0001 2.80396×10-4

1.17539×10-5

2.69094×10-4

5.33686×10-5

2.55438×10-4

9.29303×10-6

0.001 2.80301×10-3

1.17512×10-4

2.69000×10-3

×10-4

2.55346×10-3

9.296×10-4

Table 3.12

Absolute errors of OHAM for 2x and [0.1, 1]t

t .1001.0,001.0 andα .2001.0,001.0 andα

0.1 1.98526×10-9

1.09926×10-5

0.2 3.20807×10-8

2.19856×10-5

0.3 1.63084×10-7

2.9789×10-5

0.4 5.16881×10-7

4.39726×10-5

0.5 1.26475×10-6

5.49666×10-5

0.6 2.62763×10-6

6.5961×10-5

0.7 4.87621×10-6

7.69557×10-5

0.8 8.33106×10-6

8.79507×10-5

0.9 1.33626×10-5

9.89461×10-5

38

Chapter 4

Application of OHAM to MEW and Fisher’s Equation

4.1 Introduction

Fisher’s equation occur in chemical kinetics and population dynamics which include problems

such as neutron population in a nuclear reaction, nonlinear evolution of a population in a one-

dimensional habitat, logistic population growth models, flame propagation, neurophysiology,

autocatalytic chemical reactions, and branching Brownian motion processes [116-117].

Wazwaz et al. used ADM for the exact solutions of Fisher’s equation and to a nonlinear diffusion

equation of the Fisher's type [118]. Matinfar et al. used HPM, VIM and MVIM for Fisher’s

equation, Generalized Fisher’s equation and nonlinear diffusion equation of the Fisher's type

[119-122]. The MEW equation was suggested by Morrison et al, to use as a model for the

simulation of one-dimensional wave propagation in nonlinear media with dispersion processes

[123].

The objective of this chapter is to show the effectiveness of OHAM for the solution of Fisher’s

equation, nonlinear diffusion equation of the Fisher's type and MEW equation.

In Section 4.2.1 numerical solution of Fisher's equation of the form:

2

2

, ,, (1 , )

u x t u x tu x t u x t

t x

, (4.1)

is presented by OHAM, and results are compared with ADM [118], HPM [119] and VIM [121].

In Section 4.2.2 nonlinear diffusion equation of the Fisher type

2

2

, ,, (1 , )( , ),

u x t u x tu x t u x t u x t a

t x

(4.2)

is considered and results are compared with ADM [118] and HPM [119]..

In Section 4.2.3 numerical solution of nonlinear MEW equation of the form:

0),(),(

),(),(

2

32

tx

txu

x

txutxu

t

txu , (4.3)

is presented by OHAM, and results are compared with HAM [124]. Here is a positive

parameter. The obtained results reveal the reliability and efficiency of the given method.

39

4.2. Application of OHAM to Fisher’s and MEW Equation

Here we apply OHAM to Fisher’s and Diffusion equation of Fisher’s type

4.2.1 Model 1 (Fisher’s equation)

Let us consider Fisher’s equation of the form (4.1)

2

2

, ,, (1 , )

u x t u x tu x t u x t

t x

, (4.4)


( ,0) .u x f x

Case 1:

In Eqs. (4.3) and (4.4), set 1 and ( ,0)u x f x , where is constant. The exact solution

is given by

, .

1

t

t

eu x t

e

(4.5)

We take

( , , )( ( , , )) ,

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

2

2

x tx t

t

x tx t α x t x t g

x

HL H

HN H H H

Following the basic idea of OHAM presented in preceding section we start with


0

0

,0, ,0 .

u x tu x

t

(4.6)

Its solution is

0 .u x

(4.7)

First Order Problem

221 0 0

1 1 0 1 0 1 2

, , ,1 , , ,

u x t u x t u x tC C u x t C u x t C

t t x

(4.8)

1 ,0 0.u x

40

Its solution is

1 1 1( , , ) (1 ) .u x t C C t (4.9)


2 2

1 1 0 0 1 1

1 2 2 12 1 2 2 2

2

1 1 1 1 0 1 1 2 0 2 0

, , , ,1, , ,

,

, , 2 , ,

u x t C u x u x u x t CC C C Cu x t C C

t t x xt

C u x t C C u x u x t C C u x C u x

2 ,0 0.u x (4.10)

Its solution is

2 2 2 2 2

2 1 2 1 1 1 1 1

1, , , 1 2 2 2 2 .

2u x t C C C t C t C t C t C t

. (4.11)

Third Order Problem

2

3 0 3 0 2 1 1 2 0 1 1

2

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

3 1 2 30 1 1 2 1 2 2 1 2

3 2 1

2 2 2

0 1 1 2

3 2 12 2

, , 2 , ,

, , , , , 2 , , ,

, , , , , , , , , , , ,

, , ,

C u x C u x C u x t C C u x u x t C

C u x t C C u x t C C C u x u x t C C

u x t C C C u x u x t C u x t C C u x t C CC C Ct t t t t

u x u x t C u x tC C C

x x

1 2

2

,

, ,C C

x

3 ,0 0.u x (4.12)

Its solution is

2 3

1 1 1 2 1 2 3

2 2 2 3 2 2 2 3

3 1 2 3 1 1 1 2 1 1

2 3 3 3 3 3 2 3

1 2 1 1 1

6 12 6 6 12 61

, , , , 1 6 6 6 12 12 .6

12 6 6

t C tC t C t C C C t tC

u x t C C C t C t C t C C t C t C

t C C C t t C t C

(4.13)


1 2 0 1 1 2 1 2 3 1 2 3, , , , , , , , , , , , , .u x t C C u x t u x t C u x t C C u x t C C C

. (4.14)


computed that

41

1 2

-7

3

0.9835299805207357, 0.0000802958639931476

and 8.730218357092782 10 .

C C

C

Putting the values of these constants into Eq. (4.14) the approximate solution using of

OHAM is

2

2

2

2

2

, 0.9835299805207357 0.9835299805207357

0.03223692414724755 0.96733122258311881

2 1.9346624451662375

0.00206143051225105 0.095117944135773241

+0.19023588827154647 0.96

u x t t

t t

t

t t

t

3

3 3 2

513992585042742 .

5.708395551025645 5.708395551025645

t

t t

The same problem is also solved by ADM [118], HPM [119] and VIM [121]. Although they get

closed form solution but we have compared the third order OHAM solution with their fourth

order expressions, which yields very encouraging results

2 32, 1 1 1 2 1 1 6 6 .

2 6NIM

t tU x t t

(4.15)

2 32, 1 1 1 2 1 1 6 6 .

2 6HPM

t tU x t t (4.16)

2 32, 1 1 1 2 1 1 6 6 .

2 6ADM

t tU x t t (4.17)

Tables (4.1-4.3) shows the effectiveness of OHAM for 0.6 , 0.8 and 1.2 . The

results of OHAM are very consistent in comparison with ADM [118], HPM [119] and VIM

[121].

42

Table 4.1

Comparison of Absolute errors of OHAM with VIM, HPM and ADM and for model 4.2.1 case 1

with 0.6

t VIM HPM ADM OHAM

0 0. 0. 0. 0.

0.2 6.43922×10-6

6.43922×10-6

6.43922×10-6

9.44737×10-7

0.4 1.08705×10-4

1.08705×10-4

0.000108705 51031×10-5

0.6 5.73326×10-4

5.73326×10-4

5.73326×10-4

3.81397×10-4

0.8 1.86677×10-3

1.86677×10-3

1.86677×10-3

1.41556×10-3

1 4.64969×10-3

4.64969×10-3

4.64969×10-3

3.77312×10-3

Table 4.2

Comparison of Absolute errors of OHAM with VIM, HPM and ADM for model 4.2.1 case 1

with 0.8

t VIM HPM ADM OHAM

0 0. 0. 0. 0.

0.2 5.57339×10-6

5.57339×10-6

5.57339×10-6

4.39509×10-6

0.4 8.40865×10-5

8.40865×10-5

8.40865×10-5

8.05211×10-5

0.6 4.00147×10-4

4.00147×10-4

4.00147×10-4

3.95474×10-4

0.8 1.18584×10-3

1.18584×10-3

1.18584×10-3

1.018382×10-3

1 2.70952×10-3

2.70952×10-3

2.70952×10-3

2.71643×10-3

43

Table 4.3

Comparison of Absolute errors of OHAM with VIM, HPM and ADM for model 1 case 1 with

1.2

t VIM HPM ADM OHAM

0 0. 0. 0. 0.

0.2 7.82112×10-5

7.82112×10-5

7.82112×10-5

4.63504×10-5

0.4 1.13756×10-3

1.13756×10-3

1.13756×10-3

8.59097×10-3

0.6 5.27903×10-2

5.27903×10-2

5.27903×10-2

4.31153×10-3

0.8 1.54016×10-2

1.54016×10-2

1.54016×10-2

1.30749×10-2

1 3.49181×10-2

3.49181×10-2

3.49181×10-2

3.345×10-2

Case 2:

Now consider the case for 6 and

2

1( ,0)

1 xu x f x

e

. (4.18)

The exact solution of (4.3) and (4.4) along with (4.18) is given as

5 2

1, .

(1 )x tu x t

e

(4.19)


0

0 2

, 10, ,0 .

1 x

u x tu x

t e

(4.20)

Its solution is

0 2

1.

1 xu x

e

(4.21)

First Order Problem

221 0 0

1 1 0 1 0 1 2

, , ,1 6 , 6 , ,

u x t u x t u x tC C u x t C u x t C

t t x

1 ,0 0.u x

44

Its solution is

11 1 3

10( , , ) .

1

x

x

C e tu x t C

e

(4.22)


2 2

1 1 0 0 1 1

1 2 2 12 1 2 2 2

2

1 1 1 1 0 1 1 2 0 2 0

, , , ,1, , ,

,

6 , , 12 , , 6 6

u x t C u x u x u x t CC C C Cu x t C C

t t x xt

C u x t C C u x u x t C C u x C u x

2 ,0 0u x

. (4.23)

Its solution is

2

1 1 2 1

2 1 2 4 2 2 2

1 2 1 1

2 2 2 25( , , , ) .

2 2 5 101

xx

x x xx

C C C C ee tu x t C C

C e C e C t C e te

(4.24)

Third Order Problem

2

3 0 3 0 2 1 1 2 0 1 1

2

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

3 1 2 30 1 1 2 1 2 2 1 2

3 2 1

2 2 2

0 1 1 2

3 2 12 2

6 6 6 , , 12 , ,

6 , , 6 , , , 12 , , ,

, , , , , , , , , , , ,

, , ,

C u x C u x C u x t C C u x u x t C

C u x t C C u x t C C C u x u x t C C

u x t C C C u x u x t C u x t C C u x t C CC C Ct t t t t

u x u x t C u x tC C C

x x

1 2

2

, ,.

C C

x

3 ,0 0.u x

(4.25)

Its solution is

2 3

1 1 1 2 1 2 3

2 3

1 1 1 2

2 2 2

1 2 3 1 1

3 2 2 2 2

1 2 1 2 3

3 1 2 3 2 2 2 3 2 2 2

1 1 1 2 1

6 12 6 6 12 6

12 24 12 12

24 12 6 12

6 6 12 65, , , ,

3 1 30 30 30 30

x x x x

x x x x

x x x xx

x x


t C e t C e t C e tC e

t C C e t C e t C e tC e

t C e t C e C C te tC eeu x t C C C

e t C t C C C t t C e

2 2 2 3 2 2 2 2

1 1 1 2 1

2 3 2 2 2 3 3 3 3

1 1 2 1 1

3 3 2

1

30 30 30 60

60 60 25 175

100 .

x x x x

x x x

x

t C e t C e C C t e t C e

t C e t C C e t C t C e

t C e

(4.26)

45




computed that

-8

1 2

-11

3

1.0002004146462424, 7.918538356949736 10

and 4.930081333209121 10 .

C C

C

(4.28)

Putting the values of these constants into Eq. (4.27), the approximate solution using of

OHAM is

2 2

5

3

2 3

1, [3 (-9 - 60.00000000534542 -74.99997908124764

3 1

+875.5261938892598 (-12 - 60.00000000534542

-74.99997908124764 -625.3758527780427 )cosh( )

(6 - 75.01503109846819(-4.751123014169842 1

x

xu x t e t t

e

t t

t t x

-7

2

-7

2

0

-1.0002004146462424(-2.998797512122545- 5.001002073231212 )) sinh( )))

(6 - 75.01503109846819(-4.751123014169842 10 -1.0002004146462424

(-2.998797512122545 - 5.001002073231212 )) )sinh( ))].

t t x

t t x

(4.29)

The third order OHAM solution yields very encouraging results after comparing with fourth

order expressions of ADM [118] and HPM [119].

2 32

2 3 4 5

125 1 7 41 10 25(2 1), .

1 1 1 3 1

x x xx x x

HPMx x x x

e e e te e e tU x t t

e e e e

(4.30)

2 32

2 3 4 5

125 1 7 41 10 25(2 1), .

1 1 1 3 1

x x xx x x

ADMx x x x

e e e te e e tU x t t

e e e e

(4.31)

Table 2.1, 2.2, 2.3 and 2.4 shows the effectiveness of OHAM for 0.1t , 0.05t 0.01t

and 0.005t . The results of OHAM are very consistent in comparison with ADM [118], HPM

[119].

46

Table 4.4

Comparison of Absolute errors of OHAM with HPM and ADM for model 1 case 2 with 0.1t

x HPM ADM OHAM

-0.4 8.18403×10-5

8.18403×10-5

7.93903×10-5

-0.2 3.46596×10-4

3.46596×10-4

3.44479×10-4

0 5.65214×10-4

5.65214×10-4

5.63644×10-4

0.2 6.92342×10-4

6.92342×10-4

6.91442×10-4

0.4 7.07282×10-4

7.07282×10-4

7.07066×10-4

Table 4.5

Comparison of Absolute errors of OHAM with HPM and ADM for model 1 case 2 with

0.05t

x HPM ADM OHAM

-0.4 9.31454×10-6

9.31454×10-6

9.00804×10-6

-0.2 2.56797×10-5

2.56797×10-5

2.54148×10-5

0 3.831×10-5

3.831×10-5

3.81142×10-5

0.2 4.46248×10-5

4.46248×10-5

4.45116×10-5

0.4 4.37787×10-5

4.37787×10-5

4.3751×10-5

Table 4.6

Comparison of absolute errors of OHAM with HPM and ADM for model 1 case 2 with 0.01t

x HPM ADM OHAM

-0.4 2.03807×10-8

2.03807×10-8

1.79136×10-8

-0.2 4.59065×10-8

4.59065×10-8

4.37612×10-8

0 6.44302×10-8

6.44302×10-8

6.28217×10-8

0.2 7.22659×10-8

7.22659×10-8

7.1321×10-8

0.4 6.86949×10-8

6.86949×10-8

6.84319×10-8

47

Table 4.7

Comparison of Absolute errors of OHAM with HPM and ADM for model 1 case 2 with

0.005t

x HPM ADM OHAM

-0.4 1.00597×10-9

1.00597×10-9

1.00597×10-9

-0.2 2.63313×10-9

2.63313×10-9

2.90499×10-9

0 3.84221×10-9

3.84221×10-9

4.04831×10-9

0.2 4.3959×10-9

4.3959×10-9

4.51994×10-9

0.4 4.24073×10-9

4.24073×10-9

4.27997×10-9

4.2.2. Model 2 (Diffusion Equation of Fisher’s type)

In this section we consider diffusion Equation of Fisher type given by

2

2

, ,, (1 , )( , ).

u x t u x tu x t u x t u x t a

t x

(4.32)


( ,0) .u x f x

(4.33)

In Eq. (4.33) we set 2

1( ,0)

1

xu x f x

e

, then its exact solution is given by

2 0.5

2

1( , ) .

1

x a tu x t

e

(4.34)

The initial condition is:

2

1( ,0)

1

xu x f x

e

(4.35)

48

Following the basic idea of OHAM we start with


0

0

2

, 10, ,0 .

1

x

u x tu x

te

Its solution is

2

0

2

.

1

x

x

eu x

e

(4.36)

First Order Problem

22 3 21 0 0

1 1 0 1 0 1 0 1 0 1 2

, , ,1 , , , , ,

u x t u x t u x tC aC u x t C u x t C u x t aC u x t C

t t x

1 ,0 0.u x (4.37)

Its solution is

21

1 1 2

2

2 1( , , ) .

2 1

x

x

a C e tu x t C

e

(4.38)


2 2

1 1 0 0 1 1

1 2 2 12 2

2 22 1 2

1 1 1 1 0 1 1 2 0 2 0

2 3

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

, , , ,1

, , ,, , 3 , ,

2 , , 2 , , .

u x t C u x u x u x t CC C C C

t t x xu x t C C

aC u x t C C u x u x t C aC u x C u xt

aC u x C u x C u x u x t C aC u x u x t C

2 ,0 0.u x (4.39)

49

Its solution is

2 21 1 2 1

22 2 22 2

2 1 2 1 2 1 13

22 22 2

1 1

4 4 4 4

2 1( , , , ) 4 4 2 .

8 12

x

x

x x

xx x

C C C C e

a e tu x t C C C e C e C t aC t

eC e t aC e t

(4.40)

Third Order Problem

2 2 3

3 0 3 0 3 0 3 0 2 1 1

2

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

2 2 2

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

3 1 2 31 0 2 1 2 1 0 2 1 2

, ,

2 , , 2 , , 3 , ,

, , , , 3 , , , , ,

, , , ,2 , , , 2 , , , 3

aC u x C u x aC u x C u x aC u x t C

C u x u x t C aC u x u x t C C u x u x t C

C u x t C aC u x t C C u x u x t C aC u x t C C

u x t C C CC u x u x t C C aC u x u x t C C

t

2

1 0 1 1

0 1 1 2 1 2 2 1 2

3 2 1

2 2 2

0 1 1 2 1 2

3 2 12 2 2

, ,

, , , , , , , ,

, , , , ,.

C u x u x t C

u x u x t C u x t C C u x t C CC C C

t t t t

u x u x t C u x t C CC C C

x x x

3 ,0 0.u x (4.41)

Its solution is

2 3

1 1 1 2 1 2 3

2 32 2 2 21 1 1 2

2 2 22 21 2 3 1 1

3 2 2 2 2

1 2 1 2 3

2 221

3 1 2 3 4

2

24 48 24 24 48 24

48 96 48 48

96 48 24 48

24 24 48 24

122 1, , , ,

48 1

x x x x

x x

x x

x x x x

x

x


t C e t C e t C e tC e

t C C e t C e t C e tC e

t C e t C e C C te tC e

t Ca eu x t C C C

e

2 2 3 2 3 2 2

1 1 1 1 2

2 2 2 2 2 2 2 2 3 2

1 2 1 1 1

2 3 2 2 2 2 2 3 3

1 1 2 1 2 1

3 3 2 3 3 3 3 3 32 21 1 1 1

2 3 3 3 3 2 3 3 221 1 1

24 12 24 12

24 12 24 12

24 12 24

4 4 16 4

16 100 .

4

x x x

x x x

x x

x

x x

at C C t aC t C C t

aC C t t C e at C e t C e

at C e C C t e aC C t e C t

aC t a C t at C e t C e

a t C e t C e s t C e

a

3 3 2 2 3 3 2

1 1

4.42

4x xC t e a C t e

50

Adding equations (4.38, 4.40, 4.42) we obtain:


For the calculations of the constants 1 2 and C C using the collocation method we have computed

-8

1 21.0002004146462424, and 7.918538356949736 10 .C C

Putting the values of these constants into Eq. (4.43) the approximate solution using of OHAM is

22

3 1 2 3 2

2 2

2

2

3

2

0.5000089104949663 2a-1, , , ,

1 1

0.00007130017186333967 0.000071300171863519782a-1

1.000035642297453 2.000071284594906a -

8 11.0000356422974

xx

x x

x

x

x

e teu x t C C C

e e

ee t

t t

e

2 2

-8 -7 2

-8 2

2

4

2

53 +2.000071284594906a

9.72801870840521 10 1.9456037136933446 10

9.72801870840521 10 0.00021390432750933996 -

0.00042780865501867993a 0.000212a-1

48 1

x x

x

x

x

x

te te

e

e t

te t

e

2

2 2

2 2 2

2 2 2 22

2 2

390432750933996

+0.00042780865501867993a 1.0000534639225667

+4.000213855690267a 16.000855422761067a

+16.000855422761067a 1.0000534639225667

+4.000213855690267a 4.000213855

x

x

x

x

x

x

te

te t

t t e

t e t e

t e

2 2 2

.

690267a xt e

.

The third order OHAM solution yields very encouraging results after comparing with ADM

[118] and HPM [119] fourth order expressions. Table 4.8shows the effectiveness of OHAM for

a=0.8 t=0.4 . Table 4.9 shows comparison for a=0.8 t=0.8 . Table 4.10 shows reliability of

OHAM for a=0.6 t=0.8 and table 4.11 shows comparison for a=0.6 t=0.05. The results of

OHAM are very consistent in comparison with ADM [118] and HPM [119].

51

Table 4.8

Comparison of Absolute errors of OHAM with HPM and ADM for model 4.2.2 for 0.1t

x HPM ADM OHAM

-0.4 3.25805×10-5

3.25805×10-5

6.21342×10-7

-0.2 349349×10-5

349349×10-5

3.52046×10-7

0 3.59482×10-5

3.59482×10-5

5.36893×10-8

0.2 3.5536×10-5

3.5536×10-5

2.49086×10-7

0.4 3.37328×10-5

3.37328×10-5

5.31016×10-7

Table 4.9


x HPM ADM OHAM

-0.4 2.55052×10-4

2.55052×10-4

1.05625×10-5

-0.2 2.7592×10-4

2.7592×10-4

6.37554×10-6

0 2.86351×10-4

2.86351×10-4

1.66466×10-6

0.2 2.8548×10-4

2.8548×10-4

3.18475×10-6

0.4 2.73382×10-4

2.73382×10-4

7.76774×10-6

Table 4.10


x HPM ADM OHAM

-0.4 9.71739×10-6

9.71739×10-6

1.2018×10-7

-0.2 1.03889×10-5

1.03889×10-5

6.65322×10-8

0 1.06598×10-5

1.06598×10-5

7.39252×10-9

0.2 1.05077×10-5

1.05077×10-5

5.23425×10-8

0.4 9.94527×10-6

9.94527×10-6

1.07693×10-7

52

Table 4.11


4.2.3. Model 3 (MEW Equation).

Consider the MEW equation of the form

0),(),(

),(),(

2

32

tx

txu

x

txutxu

t

txu , (4.44)

where is a positive parameter. The initial condition is

)(sec)0,( kxhxu , (4.45)

its exact solution given by

0( , ) sec [ ( )]u x t h k x x ct ,

where 2

2 1, .

2k and c


0),(0

t

txu, 0( ,0) sec ( ).u x α h kx

Its solution is

0( , ) sec ( ).u x t α h k x

(4.46)

First Order Problem

,0

),(),(),(3

),()1(

),(2

0

3

10

02

10

11

tx

txuC

x

txutxuC

t

txuC

t

txu

1( ,0) 0.u x

Its solution is

3 2

1 1 1( , ,C ) -3 sec ( ) tanh( ).u x t α C k t h kx kx (4.47)

x HPM ADM OHAM

-0.4 2.39988×10-9

2.39988×10-9

1.86984×10- 12

-0.2 2.55154×10-9

2.55154×10-9

1.04816×10-12

0 2.60416×10-9

2.60416×10-9

1.45217×10-13

0.2 2.55335×10-9

2.55335×10-9

7.635×10-13

0.4 2.40336×10-9

2.40336×10-9

1.60283×10-12

53


22 0 0 02 1

1 2 0 2 22

320 0 1 1

2 1 1 0 1 0 1 2

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

( , ) ( , ) ( , ) ( , )6 ( , ) ( , ) 3 ( , ) 0,

u x t u x t u x tu x t u x tC C u x t C C

t t x x t t

u x t u x t u x t u x tC C u x t u x t C u x t C

x x x x t

2 ,0 0.u x

Its solution is

3 7 2 2 2 2

2 1 1 1

2

2 1 1 1

2

2 1 1 1

3( , , ) sech ( ) (-42 30 cosh(2 )-

8

2 ( (1 31 )) sinh(2 )-

( (1 -9 )) sinh(4 )).

u x t C k t kx C k t C k t kx

C C C C k kx

C C C C k kx

(4.48)

Third Order Problem

323 0 0 02

1 3 0 3 32

32 0 0 1 1

1 1 2 0 1 2 22

01 0 2 1 0 1

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

( , ) ( , ) ( , ) ( , )3 ( , ) 6 ( , ) ( , )

( , )6 ( , ) ( , ) 6 ( , )

u x t u x t u x t u x tu x tC C u x t C C

t t x x t t

u x t u x t u x t u x tC u x t C u x t u x t C C

x x x t t

u x tC u x t u x t C u x t u

x

21 12 0

32 02

1 0 1 32

( , ) ( , )( , ) 3 ( , )

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

u x t u x tx t C u x t

x x

u x tu x tC u x t C u x

x x t

Its solution is

3 4

3 1 2 3 2 3 1 1 1 2

2 2 2 4 2 2 2

1 1 2 1 1 1 1 2 1

2 2 2 5

1

3( , , , , ) sech ( ) (-2 ( + + (1+ (2+ )+2 -

2

18 ( + + ) +81 ))sinh( ) 10C (- - - 17 )

sech( )(-3 8 tanh( )) 24 3 tech ( ) (-62

u x t C C C k t kx C C C C C C

C C C k C k kx k C C C C k

kx t k kx C k t kx k

2 3 2 2 2 2

1 1 1 2 1

2 2 4 2

1

9 tanh( )) 3 sech ( ) (2 (-6 ( ) 319C )

-7 (80 7 )tanh( )).

t kx C k kx C C C k t

C k k t kx

(4.49)


1 2 3 0 1 1 2 1 2 3 1 2 3, , , , , , , , , , , , , , .u x t C C C u x t u x t C u x t C C u x t C C C (4.50)

For the calculations of the constants 1 2 3, andC C C using the Method Least Squares we have

computed that

54

1

2

3

-0.041314777653485436,

0.02625035702785775 and

0.22457035785347018.

C

C

C

Putting the values of these constants into Eq. (4.50) the approximate solution using of

OHAM is

). tanh()) 0.0189612-77(-0.005949

0.0277842(-0.049378 ) 0.03927123(-0.006829 (

) (sech ) Tanh( ) (sech 0.123944

)Sinh(7 ) (7.00454 ) (2.29468 ) (-2.29468 80.00026775

)sinh(5 ) 0.00765259- 0.0018108-2(-0.049378)sinh(3

)) 0.003887452(-0.008226 0.01986377(-0.088880

)cosh(5 ) 0.00337177 1(0.0015521

))cosh(3 0.0337573- 23(0.0003104 ( )(sech

)sinh(4 k2) 0.00576082 7(0.0050090)sinh(2 ) (0.502431

)(-0.502431 0.0396857-)cosh(2 0.0192027

8(-0.026883 ) (sech )sech( ),(

2 42

422

9333

2

42

2424

22

22103

2

273

3

kxtk

ktkk

xkktxkxktk

kxkkk

ktkkkx

tkk

kxktk

ktktkkxtk

kxkxk

kkxtk

tkxktkkxtxu

(4.51)

In Tables (4.12.-4.16) we compare third order OHAM solution with fifth order HAM [124]

solution for 10x , 5x , 0,x 5x , 10x , 0.25 1.and k These results

shows that third order OHAM gives the same error as fifth order HAM solution.

Table 4.12

Comparison of absolute errors of the obtained solution by third order OHAM solution with fifth

order HAM solution [124] for MEW equation for 0.25 and 1,k and

10x

t HAM OHAM

0.01 7.098400×10-9

7.09263×10-9

0.03 2.127125×10-8

2.12712×10-8

0.05 3..544099×10-8

3.5441×10-8

0.07 4.960190×10-8

4.960190×10-8

0.09 6.375394×10-8

6.3754×10-8

0.1 7.082664×10-8

7.08267×10-8

55

Table 4.13

Comparison of absolute errors of the obtained solution by third order OHAM

solution with fifth order HAM solution [124] for MEW equation for 0.25 and 1,k and

5x

t HAM OHAM

0.01 1.045271×10-6

1.05264×10-6

0.03 3.134831×10-6

3.15693×10-6

0.05 5.223074×10-6

5.25991×10-6

0.07 7.310005×10-6

7.36158×10-6

0.09 9.395622×10-6

9.46193×10-6

0.1 1.043794×10-5

1.05116×10-5

Table 4.14



0x

t HAM OHAM

0.01 9.77214×10-7

1.18555×10-8

0.03 8.79492 ×10-6

1.067×10-7

0.05 1.56356×10-5

2.96388×10-7

0.07 4.78842 ×10-5

5.80919×10-7

0.09 7.91551 ×10-5

9.60294×10-7

0.1 9.77226×10-5

1.18555×10-6

56

Table 4.15



5x

t HAM OHAM

0.01 1.04560×10-6

1.05297×10-6

0.03 3.13779×10-6

3.15989×10-6

0.05 5.23130×10-6

5.26813×10-6

0.07 7.32612×10-6

7.37769×10-6

0.09 9.42226×10-6

9.48857×10-6

0.1 1.04708×10-5

1.05445×10-5

Table 4.16



10x

t HAM OHAM

0.01 7.09484×10-9

7.09485×10-9

0.03 2.12912×10-8

2.12912×10-8

0.05 3.54964 ×10-8

3.54964×10-8

0.07 4.97105×10-8

4.97105×10-8

0.09 6.39335 ×10-8

6.39335×10-8

0.1 7.10483×10-8

7.10483×10-8

For the MEW equation we discuss the following three invariant conditions which, respectively

correspond to conservation of mass, momentum and energy and are given as

∫

∫

∫

57

These invariants will be used to check the conservation properties of the numerical solutions.

The analytical values of the invariants are

In Table 4.17 we used absolute errors to compare the invariants obtained by proposed method

with the analytical values of the invariants for 0.05, 1, 0 0x and [ 30,30]x . The

results obtained are in good agreement with analytical values of the invariants.

Table 4.17

Invariants for single solitary wave with, for 0.05, 1, 0 0x and [ 30,30]x in third

order approximation of OHAM with absolute errors

t Error Error Error

0.0 0.15708 1.86795×10-14 0.00666667 0.0000000 8.33333×10-6 1.69407×10

-21

0.1 0.15708 1.86795×10-14

0.00666667 3.15379×10-11

8.33333×10-6 5.591×10-14

0.2 0.15708 1.86795×10-14

0.00666667 1.26152×10-10

8.33333×10-6 2.2364×10-13

0.3 0.15708 1.86795×10-14

0.00666667 2.83841×10-10

8.33333×10-6 5.0319×10-13

0.4 0.15708 1.86795×10-14

0.00666667 5.04607×10-10

8.33333×10-6 8.9456×10-13

0.5 0.15708 1.86795×10-14

0.00666667 7.88448×10-10

8.33333×10-6 1.39775×10-12

0.6 0.15708 1.86795×10-14

0.00666667 1.13537×10-9

8.33334×10-6 2.01276×10-12

0.7 0.15708 1.86795×10-14

0.00666667 1.54536×10-9

8.33334×10-6 3.57824×10-12

0.8 0.15708 1.86795×10-14

0.00666667 2.01843×10-9

8.33334×10-6 2.73959×10-12

0.9 0.15708 1.86795×10-14

0.00666667 2.55457×10-9

8.33334×10-6 4.52871×10-12

1.0 0.15708 1.86795×10-14

0.00666667 3.15379×10-9

8.33334×10-6 5.591×10-12

58

Chapter 5

Application of OHAM to DGRLW Equation

5.1 Introduction

The DGRLW equation is a partial differential equation that describes the amplitude of long-

wave, which takes the following for

2 2

2, 0.pu u u u u

φ x t α ut x x t x x x

(5.1)

Here 0α , 0p is an integer, ,φ x t is known function and ,u x t

is the amplitude of the

long-wave at the position x and time t . For 0α in Eq. (5.1), features a balance between

nonlinear and dispersive effects, but also takes into account mechanisms of dissipation. In the

physical sense, Eq. (5.1) with the dissipative term xxαu is suggested if the good predictive power

is preferred; such type of problem arises in the bore propagation as well as for water waves

[125]. The EW, RLW and GRLW equations are special cases of the DGRLW equation [126–

128]. The EW equation corresponds to 0α , , 1φ x t whereas the RLW equation corresponds

to 0α , , 1φ x t and . On the other hand, the GRLW equation corresponds 0α ,

, 1φ x t and 1p . Different techniques are available in the literature to solve these equations.

Basak et al. have used fully Galerkin method for DGRLW equation [129]. Demir et al has

presented VIM for numerical solutions for the DGRLW equation [130]. Yousefi, et al. has used

Bernstein Ritz-Galerkin Method for solving the DGRLW equation [131].

In this chapter, the approximate solution of the DGRLW equation with a variable coefficient is

presented using OHAM. Absolute errors by OHAM for homogeneous and non-homogenous

DGRLW equations are compared with VIM [130]. The 3D and 2D images of the approximate

solution and exact solution are also drawn. In all cases, the proposed method yields very

encouraging results.

59

5.2 Application of OHAM to DGRLW Equation

5.2.1. Model 1

Consider Eq. (5.1) with 1, 2α p and 2 41, ,

6

x tφ x t e

which in simplest form is given as

2 2

2 4 2

2

11 0.

6

x tu u u u ue u

t x x t x x x

(5.2)

The initial condition is ,0 xu x e and exact solution given by

2, .

x tu x t e

We take

2 2

2 4 2

2

( , , )( ( , , )) ,

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

6

and 0,

x t

x tx t

t

x t x t x t x tx t e x t

x x t x x x

g

HL H

H H H HN H H


000, ,0 .xu

u x et

Its solution is given as under

0 , .xu x t e (5.3)

First Order Problem

2 2 3

4 2 4 220 0 0 0 0 0 011 1 1 0 1 1 12 2

10.

3

t x t xu u u u u u uuC C C u C e C C e

t t t x x x t x x t

Its solution is as follows

3 2

1 1 1, , 1 2 .x xu x t C C e e t (5.4)


2 20 0 0 01 1 1 1 12 1 2 2 0 1 0 1 1 1 0

2 24 2 4 20 1

2 1

2

1 10.

3 3

t x t x

u u u uu u u u uC C C C u C u u C C u

t t t t x x x x x

u uC e C e

x t x t

Its solution is as under

60

4 22 2 4 2 2 2 2

1 1 1 1 1

5 2 2 2 4 2 4 4

2 1 2 1 2 1 1 2

2 2 2 2 2 2 4 2

1 1 1

5 5 2 2 81

, , , 8 8 16 16 16 .8

20 72 16

t xt x x

x x x x x x

x x

C C e C e C e C e t

u x t C C e C e t C e t C e t C e t C e t

C t C t e C e t

(5.5)

Third Order Problem

20 3 0 0 01 2 23 2 1 3 3 0 2 0 1

2 20 01 1 1 11 1 1 1 0 2 2 2 0 1 0 1

2 24 2 4 22 02 2 1

1 1 0 3 2

4 2

1

2

2 2

1 1

3 3

1

3

t x t x

t x

u u u u uu u uC C C C C u C u u

t t t t t x x x

u uu u u uC C u C u u C C u C u u

x x x x x x

uu u uC C u C e C e

x x x t x t

C e

32 2 2 24 2 02 2 1 2

3 2 1 32 2 2

3 34 2 4 21 2

2 12 2

1

6

1 10.

6 6

t x

t x t x

uu u u uC C C C e

x t x t x x x t

u uC e C e

x t x t

(5.6)

Its solution is given as follows

3 3 4 3 8 2 2 3 2 2

1 1 1 1 1 1 2

4 2 4 22 4 3 4 4 2 3

1 1 1 2 1 1

4 2 8 2 4 4 4 43 2 3

1 2 1 1 1

4 4 3

1 2 1

7

3 1 2 3

245 70 175 120 60 120

48 12 48 120 90

120 30 48 12

48 420

1, , , ,

96

t t x x x

t x t xx x x

t x t x t x t x

t x

x

C C e C e C e C e C C e

C e C e C C e C e C e

C C e C e C e C e

C C e C

u x t C C C e

3 4 3 2 4

1 1 1

2 4 3 4 4 4 4

1 1 2 1 2 3

6 2 6 3 6 6 6

1 1 1 2 1 2

4 2 4 46 3 3 2 2 2

3 1 1 1

3 2

1

700 1920 96

192 384 96 192 96

192 384 192 192 384

192 1080 48 480

480

t x x

x x x x x

x x x x x

t x t xx x

x

t C e t C e t C e t

C e t C e t C e t C C e t C e t

C e t C e t C e t C e t C C e t

C e t C e t C e t C e t

C e t

2 2 2 2 4 2 3 4 2

1 2 1 1

4 2 2 6 2 3 6 2 6 2

1 2 1 1 1 2

3 3 3 2 3 3 4 3 3 6 3

1 1 1 1

.

480 1728 1728

1728 384 384 384

784 4480 4032 128 .

x x x

x x x x

x x x

C C e t C e t C e t

C C e t C e t C e t C C e t

C t C e t C e t C e t

(5.7)

61

The third order approximate solution is given by following equation

1 2 3 0 1 1 2 1 2 3 1 2 3

3 2

1

4 22 2 4 2 2 2 2 2 2 2

1 1 1 1 1 1 25

4 2 4 4 2 2 2 2 2 2 4 2

1 1 2 1 1 1

, , , , , , , , , , , , , ,

1 2

5 5 2 2 8 8 81

8 16 16 16 20 72 16

x x x

t xt x x x x

x

x x x x x

u x t C C C u x t u x t C u x t C C u x t C C C

e C e e t

C C e C e C e C e t C e t C e te

C e t C e t C e t C t C t e C e t

3 3 4 3 8 2 2 3 2 2

1 1 1 1 1 1 2

4 2 4 22 4 3 4 4 2 3

1 1 1 2 1 1

4 2 8 2 4 4 4 43 2 3

1 2 1 1 1

4 4 3 3 4

1 2 1 1

7

245 70 175 120 60 120

48 12 48 120 90

120 30 48 12

48 420 700

1

96

t t x x x

t x t xx x x

t x t x t x t x

t x t

x

C C e C e C e C e C C e

C e C e C C e C e C e

C C e C e C e C e

C C e C t C e t

e

3 2 4

1 1

2 4 3 4 4 4 4

1 1 2 1 2 3

6 2 6 3 6 6 6

1 1 1 2 1 2

4 2 4 46 3 3 2 2 2

3 1 1 1

3 2 2 2

1 1 2

1920 96

192 384 96 192 96

192 384 192 192 384

192 1080 48 480

480 480

x x

x x x x x

x x x x x

t x t xx x

x x

C e t C e t

C e t C e t C e t C C e t C e t

C e t C e t C e t C e t C C e t

C e t C e t C e t C e t

C e t C C e

2 2 4 2 3 4 2

1 1

4 2 2 6 2 3 6 2 6 2

1 2 1 1 1 2

3 3 3 2 3 3 4 3 3 6 3

1 1 1 1

.

1728 1728

1728 384 384 384

784 4480 4032 128 .

x x

x x x x

x x x

t C e t C e t

C C e t C e t C e t C C e t

C t C e t C e t C e t

(5.8)

The constant 1 2,C C and 3C are calculated using the collocation method, their optimal values are

as follows

1

2

3

0.8410486261349961,

0.0226537091096181,

0.00319308340395879.

C

C

C

Figures (5.1-5.4) show plots of exact and approximate solution for Eq. (5.2)

Figure 5.1: Plot of 3rd

order approximate solution Figure 5.2: Plot exact solution

62

Figure 5.3: Approximate solution plot for t=0.04 Figure 5.4: Exact solution plot for t=0.04

The 3rd

order OHAM solution yields very encouraging results after comparing with 3rd

order

VIM solution [130]. Tables (5.1—5.3) show the effectiveness of OHAM for 1.2, 1.4x x and

1.6.x

Table 5.1


order OHAM solution and 3rd

order VIM solution

for model 1 at 1.2x and various values of t

t Exact solution OHAM solution Absolute error VIM Absolute error OHAM

0.02 0.313486 0.313486 1.92728 ×10-5

3.146×10-7

0.04 0.32628 0.32627 3.39912 ×10-5

1.01891×10-5

0.06 0.339596 0.339571 1.301192 ×10-4

2.46667×10-5

0.08 0.353455 0.353416 2.401071 ×10-4

3.87876×10-5

0.10 0.367879 0.367829 3.356920 ×10-4

5.08783×10-5

Table 5.2



order VIM solution



0.02 0.256661 0.256661 1.331×10-7

3.29357×10-8

0.04 0.267135 0.267135 2.11230 ×10-5

5.45591×10-7

0.06 0.278037 0.278037 4.57127 ×10-5

1.19768×10-7

0.08 0.289384 0.289387 5.71818 ×10-5

2.34162×10-6

0.10 0.301194 0.301199 3.93977 ×10-5

4.79793×10-6

1.0 1.5 2.0 2.5 3.0

0.10

0.15

0.20

0.25

0.30

0.35

0.40

1.0 1.5 2.0 2.5 3.0

0.10

0.15

0.20

0.25

0.30

0.35

0.40

63

Table 5.3



order VIM solution



0.02 0.210136 0.210136 1.4946×10-6

4.27298×10-7

0.04 0.218712 0.218712 7.2701×10-6

1.00227×10-7

0.06 0.227638 0.227635 7.6219×10-6

2.32488×10-6

0.08 0.236928 0.23692 6.6969×10-6

7.99082×10-6

0.10 0.246597 0.246577 4.44429×10-5

1.99024×10-5

5.2.2. Model 2

Consider Eq. (5.1) with 1, 2α p and 2 41,

6

x tφ x t e

which in simplest form is given as

2 2

2 4 2

2

11 0.

6

x tu u u u ue u

t x x t x x x

(5.9)

The initial condition is 2, sec4

xu x t h

and exact solution given by

2, sec .4 3

x tu x t h

(5.10)


2000, ,0 sec .

4

u xu x h

t


2

0 , sec .4

xu x t h

(5.11)

First Order Problem

2 2 3

4 2 4 220 0 0 0 0 0 011 1 1 0 1 1 12 2

10.

3

t x t xu u u u u u uuC C C u C e C C e

t t t x x x t x x t

64


4 2

1 1

1 1 1

6 2 2

1 1

sec 4 sec tanh4 4 41

, , .8

4 sec tanh 2 sec tanh4 4 4 4

x x xC h C h

u x t C C tx x x x

C h C h

(5.12)


2 20 0 0 01 1 1 1 12 1 2 2 0 1 0 1 1 1 0

2 24 2 4 20 1

2 1

2

1 10.

3 3

t x t x

u u u uu u u u uC C C C u C u u C C u

t t t t x x x x x

u uC e C e

x t x t

Its approximate solution 2 1 2, , ,u x t C C is obtained in similar manner. The second order

approximate solution is given by

1 2 0 1 1 2 1 2

4 2

1 1

2

1 2 1 2

6 2 2

1 1

, , , , , , , , ,


sec , , ,4 8


u x t C C u x t u x t C u x t C C

x x xC h C h

xh C t u x t C C

x x x xC h C h

(5.13)

Using collocation method the optimal values of constants 1C and 2C are computed and are given

as under

1

2

0.8122876966282708,

0.0115021833336221.

C

C


Figure 5.5: plot of 2nd order approximate solution Figure 5.6: Exact solution plot

65


Table (5.4) shows the effectiveness of OHAM for 15, 20x x and 25x , while table (5.5)

shows the effectiveness of OHAM for various values of x and t

Table 5.4

Absolute error of the solution of model 2 by OHAM at 15, 20x x and 25x

And various values of t

t 15x 20x 25x

0.01 2.7905×10-6

2.30504×10-7

1.89307×10-8

0.02 2.83577×10-6

2.35074×10-7

1.93117×10-8

0.03 8.32571×10-6

6.87803×10-7

5.6488×10-8

0.04 5.60061×10-6

4.64385×10-7

3.81506×10-8

0.05 ×10-5

1.13979×10-6

9.36098×10-8

0 2 4 6 8 10

0.2

0.4

0.6

0.8

1.0

2 4 6 8 10

0.2

0.4

0.6

0.8

1.0

66

Table 5.5

Absolute error of the solution of model 2 by optimal homotopy asymptotic method (OHAM) at

various values of and t

0.1t 0.2t 0.3t 0.4t 0.5t

15 2.24995×10-5

4.36594×10-5

6.27619×10-5

9.29866×10-5

9.16758×10-5

16 1.36916×10-

5

26576×10-5

3.82161×10-5

5.66063×10-5

5.58636×10-5

17 8.32101×10-6

1.61544×10-5

2.32343×10-5

3.441×10-5

3.39789×10-5

18 5.05308×10-6

9.81109×10-6

1.41127×10-5

2.0899×10-5

2.06447×10-5

19 3.0671×10-6

5.95551×10-6

8.56725×10-6

1.26863×10-5

1.25347×10-5

20 1.86112×10-6

3.61396×10-6

5.19906×10-6

7.69844×10-6

7.60747×10-6

21 1.12913×10-6

2.19263×10-6

3.1544×10-6

4.67075×10-6

4.61593×10-6

22 6.84967×10-7

1.33013×10-6

1.91361×10-6

2.83347×10-6

2.80036×10-6

23 4.15495×10-7

8.06854×10-7

1.1608×10-6

1.71878×10-6

1.69874×10-6

24 2.52026×10-7

4.89414×10-7

7.04113×10-7

1.04256×10-6

1.03043×10-6

25

1.52867×10-7

2.96856×10-7

4.27085×10-7

6.32372×10-7

6.25018×10-7

5.2.3. Model 3

Consider equation 1 with 1, 1α p and 2

3cosh sinh4 3 4 3

, 1 ,

3 2cosh4 3

x t x t

φ x tx t

which in

simplest form is given as

2 2

22

3cosh sinh4 3 4 3

1 0

3 2cosh4 3

x t x t

u u u u uu

x tt x x t x x x

(5.14)

67

The initial condition is 2, sec4

xu x t h


2, sec .4 3

x tu x t h

(5.15)


2000, ,0 sec .

4

u xu x h

t


2

0 , sec .4

xu x t h

(5.16)

First Order Problem

2

0 0 0 0 011 1 1 0 1 2

2 2 2 2

1 22 22

3cosh 3cosh sinh 3sinh3 4 3 4 3 4 3 4

4 3 2cosh 4 3 2cosh3 2cosh3 4 3 43 4

u u u u uuC C C u C

t t t x x x

t x t x t x t x

Ct x t xt x

2

0

3

01 2

2

3cosh sinh3 4 3 4

1 0.

3 2cosh3 4

u

x t

t x t x

uC

x tt x

(5.17)


4 2

1 1

1 1

4 2 2

1 1


, , .8


x x xC h C h

u x t C tx x x x

C h C h

(5.18)

The first order approximate solution is given by

1 0 1 1, , , , , ,u x t C u x t u x t C ,

68

4 2

1 1

2

1

4 2 2

1 1


, , sec .4 8


x x xC h C h

xu x t C h t

x x x xC h C h

(5.19)

The constants 1C is calculated using the Least Squares we have its optimal value as follows

1 0.802767563787412.C

The first order optimum solution using OHAM is as follows

4 2

2

4 2 2

0.8027675637 3.211070255

3.211070255 1.6055351

sec sec tanh4 4 41

, sec .4 8

sec tanh sec tanh4 4 4 4

27

x x xh h

xu x t h t

x x x xh h

(6.20)


Figure 5.9: 1st order approximate solution plot Figure 5.10: Exact solution plot

Figure 5.11: approximate solution plot for t=0.1 Figure 5.12: Exact solution plot for t=0.1

20 10 10 20

0.2

0.4

0.6

0.8

1.0

20 10 10 20

0.2

0.4

0.6

0.8

1.0

69

The first order OHAM solution yields very encouraging results after comparing with 2nd

order

VIM solution [130]. Tables (5.6-5.9) show the effectiveness of OHAM for

2.5,x 0,x 2.5x 5.and x

Table 5.6

Comparison of absolute errors of 1st order OHAM solution and 2

nd order VIM solution



0.02 0.687297 0.687094 1.510466×10-3

2.02246×10-4

0.04 0.682171 0.68177 3.017151×10-3

4.0071×10-4

0.06 0.677041 0.676445, 4.521014×10-3

5.96353×10-4

0.08 0.671911 0.67112 6.023003×10-3

7.90121×10-4

0.10 0.666779 0.665796 7.524051×10-3

9.82948×10-4

Table 5.7



for model 3 at 0x and various values of t


0.02 0.999956 0.997993 2.455556×10-3

1.96248×10-3

0.04 0.999822 0.995986 4.822243××10-3

3.83608×10-3

0.06 0.9996 0.993979 7.100106×10-3

5.62086×10-3

0.08 0.999289 0.991972 9.289225 ××10-3

7.3169×10-3

0.10 0.99889 0.989965 1.138971×10-3

8.92431×10-3

Table 5.8



for example 3 at 2.5x and various values of t


0.02 0.697537 0.697529 1.247750×10-3

7.71606×10-6

0.04 0.702649 0.702639 2.501243×10-3

9.6889×10-6

0.06 0.707754 0.707749 3.761480×10-3

4.91862×10-6

,

0.08 0.712851 0.712859 5.029472×10-3

7.60727×10-6

0.10 0.71794 0.717969 6.306245×10-3

2.89136×10-5

70

Table 5.9



for model 3 at 5x and various values of t


0.02 0.283601 0.283512 6.719964×10-4

889401×10-5

0.04 0.286816 0.286609 1.315023×10-3

20685×10-4

0.06 0.29006 0.289706 1.928998×10-3

353812×10-4

0.08 0.293333 0.292803 2.513842×10-3

529904×10-4

0.10 0.296636 0.295901 3.069485×10-3

735198×10-4

5.2.4. Model 4:

Let us consider the inhomogeneous DGRLW equation

2 2

2, ( , ).

u u u u uφ x t u f x t

t x x t x x x

(5.21)

Where

,φ x t xt and ( , ) cos sin cos sin cos .t tf x t t x xt x x x x e e

The initial condition is

,0 sinu x x

(5.22)


, sin .tu x t x e

(5.23)


000, ,0 sin .

uu x x

t


0 , sin .u x t x (5.24)

First Order Problem

2 2 3

0 0 0 0 0 0 011 1 1 0 1 1 12 2

1 cos sin cos sin cos 0.t t

u u u u u u uuC C C u C C C xt

t t t x x x t x x t

C e t x xt x x x x e

71


2 2 2

2

1 1 1 2 2

2 cos 2 cos 2 sin 2 sin 2 sin1, , .

2 2 sin sin cos sin cos 2 sin cos

t t t t t

t

t t t

te x te x te x xe x xe xu x t C C e

xte x x x e x x te x x

(5.26)


2 22

0 0 0 0 0 02 1 1 1 12 2 2 0 1 1 1 1 0 2 1 22 2

3 32

0 011 2 1 22 2 2

cos sin cos sin cos 0.t t

u u u u u uu u u u uC C C u C u C C u C t C t C

t t t x x x x x x t x x

u uuC C xt C xt C e t x xt x x x x e

x x t x t

Its approximate solution as under

2 1 2

2 2 2 2 2 2 2 2 2

1 1 1 1 1 1

2 2

1 1 2 1

2 2 2 2 2 2 2

1 2 1 1

2

1

2

1

, , ,

16 16 8 8 80 cos 80 cos

16 cos 32 cos 16 cos 16 cos

48 cos 16 cos 8 cos 144 cos

144 cos

1

16

t t t t t t

t t t t

t t t t

t

t

u x t C C

C e C e C te C te C te x C te x

C te x C te x C te x C te x

C te x C te x C t e x C xe x

C xe

C e

2 2 2

1 1

2 2 2 2 2 2 2

1 1 1 1

2 2 2 2

1 1 1

2 2 2 2 2 2 2

1 1 1 1

2 2

1 1

128 cos 16 cos

64 cos 8 cos 8 cos 2 328 cos 2

248 cos 2 8 cos 2 24 cos 2

8 cos 2 sin 96 sin 95 sin

80 sin 16 sin 2

t t

t t t

t t

t t t

t t

x C xte x C xte x

C xt e x C xt e x C x C e x

C e x C t x C te x

C t e x C x C e x C e x

C e x C te x

2 2 2

1 2

2 2 2 2 2 2

1 1 1 1

2 2 2 2

2 1 1 2

2 2 2 2 2

1 2 1 1

2 2

1

sin 16 sin

32 sin 6 sin 16 sin 16 sin



32 sin

t t

t t t t

t t t t

t t t t

t

C te x C te x

C t e x C t e x C xe x C xe x

C xe x C xe x C xe x C xe x

C xte x C xte x C xte x C xt e x

C x e x

2 2 2 2 2 2 2 2

1 1 1

2 2 2 2

1 1 2 1 1

2 2 2 2 2

2 1 1 2

2 2 2 2 2 2

1 1 1 1

32 sin 32 sin 16 sin

4 sin 2 4 sin 2 4 sin 2 4 sin 2 4 sin 2

4 sin 2 8 sin 2 4 sin 2 8 sin 2

24 sin 2 8 sin 2 32 sin 2 24

t t t

t t

t t t t

t t t

C x e x C x te x C x t e x

C x C x C x C e x C e x

C e x C te x C te x C te x

C t e x C x x C xe x C xe

2 2 2 2 2 2 2

1 1 1 1

2 2 2 2 2 2 2 2

1 1 1 1

sin 2

16 sin 2 16 sin 2 16 sin 2 16 sin 2

3 sin 3 3 sin 3 6 sin 3 6 sin 3 .

t t t

t t t

x

C xt x C xte x C xte x C xt e x

C x C e x C te x C t e x

(5.27)

72

The second order approximate solution is given by following equation

1 2 0 1 1 2 1 2, , , , , , , , , .u x t C C u x t u x t C u x t C C

Using method of collocation method the optimum values of constants 1C and 2C are computed

which are as follows

1

2

1.0433989069917953,

0.0018859250375959.

C

C


Figure 5.13: 2nd order approximate solution Figure 5.14: The surface shows exact solution


The 2nd

order OHAM solution yields very encouraging results after comparing with 2nd

order

VIM solution [130]. Table (5.10—5.13) shows the effectiveness of OHAM for 0.2,x 0.4,x

0.6x and 1.x

3 2 1 1 2 3

0.5

0.5

3 2 1 1 2 3

0.5

0.5

73

Table 5.10

Comparison of absolute errors of 2nd

order OHAM solution and 2

nd order VIM solution for

model 4 at 0.2x and various values of t


0.02 0.194735 0.194388 3.596760×10-4

3.46922×10-4

0.04 0.190879 0.189563 1.374338×10-3

1.31649×10-3

0.06 0.1871 0.184295 2.950802×10-3

2.8045×10-3

0.08 0.183395 0.178685 5.000174×10-3

4.71013×10-3

0.10 0.179763 0.172828 7.437789×10-3

6.93584×10-3

Table 5.11


order OHAM solution and 2nd

order VIM solution for



0.02 0.381707 0.381449 2.832281×10-4

2.58242×10-4

0.04 0.374149 0.373216 1.043581×10-3

9.32938×10-4

0.06 0.36674 0.364862 2.151953×10-3

1.87854×10-3

0.08 0.359478 0.356523 3.485485×10-3

2.95526×10-3

0.10 0.35236 0.348331 4.927456×10-3

4.02886×10-3

Table 5.12


order OHAM solution and 2nd

order VIM solution for



0.02 0.553462 0.55333 1.676249×10-4

1.31821×10-4

0.04 0.542503 0.542094 5.651230×10-4

4.08832×10-4

0.06 0.53176 0.531101 1.040306×10-3

6.59364×10-4

0.08 0.521231 0.520511 1.4483802×10-3

7.194×10-4

0.10 0.51091 0.510477 1.651669×10-3

4.32267×10-4

74

Table 5.13


order OHAM solution and 2

nd order VIM solution for

model 4 at 1x and various values of t


0.02 0.703151 0.703174 1.439370×10-4

2.23769×10-4

0.04 0.689228 0.689438 6.672725×10-4

2.10203×10-4

0.06 0.675581 0.676318 1.702709×10-3

7.37797×10-4

0.08 0.662203 0.663974 3.376802×10-3

1.77096×10-3

0.10 0.649091 0.652558 5.810408×10-3

3.46734×10-3

75

Chapter 6

Analysis of OHAM to Coupled System of PDEs

In this chapter, for the first time OHAM is introduced for deriving approximate solution of

coupled system of PDEs. The WBK, ALW and MB equations are taken as tests examples. It has

been attempted to show the reliability, efficiency and wide rang applications of OHAM in

comparison with HPM [132].

The WBK equations which are introduced by Witham, Broer and Kaup, describes the

propagation of shallow water waves with different dispersion relation and takes the following

form

2

2

3 2

3 2

0,

0.

u u v uu

t x x x

v u u vuv

t x x x x

(6.1)

Where ( , )u u x y is the field of horizontal velocity and ( , )v v x y is the height that deviate

from equilibrium position of liquid. The constants , represent different diffusion power. The

initial conditions for Eq. (6.1) are

0

2 2

0

( ,0) 2 coth ,

( ,0) 2 cosech ,

u x Bk k x x

v x B B k k x x

(6.2)

and exact solution is given by

0

2 2

0

( ,0) 2 coth ,

( ,0) 2 cosech ,

u x Bk k x x t

v x B B k k x x t

(6.3)

With 1, 0 Eq. (6.1) reduces to the MB equation which models shallow water waves

motion under gravity and in a one dimension nonlinear lattice. For 0, 0.5 then Eq. (6.1)

reduces to the ALW equation which describes shallow water wave with dispersive.

Mohyud-Din et al. has found traveling wave solutions of WBK equations by HPM [132]. Xie et

al. has found some new solitary wave solutions for the WBK equation using hyperbolic function

method [133]. El-Sayed et al. has used ADM for solving the governing WBK problem [134].

76

6.2 Solution of coupled system of PDE’s by OHAM

In this section, OHAM is applied Coupled system of PDEs including WBK, ALW, MB systems.

6.2.1 Model 1. (WBK Equation)

With 00.1, 0.005, 1.5, 1.5and 10.k x Eq. (6.1) takes the following form

1.5 0,

( ) 1.5 1.5 0,

t x x xx

t x x xxx xx

u uu v u

v u uv u v

(6.4)

with initial conditions

2

( ,0) 0.005 0.387298coth 0.1 10 ,

( ,0) 0.1330947502cosech 0.1 10 ,

u x x

v x x


2

( , ) 0.005 0.387298coth 0.1 10 0.005 ,

( , ) 0.1330947502cosech 0.1 10 0.005 .

u x t x t

v x t x t

(6.5)

According to OHAM, homotopy for Eq. (6.4) takes the following form

1

2

(1 ) ( ) ( )[ 1.5 ] 0,

(1 )( ) ( )[ ( ) 1.5 1.5 ] 0,

t t x x xx

t t x x xxx xx

u h u uu v u

v h v u uv u v

(6.6)

We consider 1 2, , ( ), ( )u v h h as

2 3

0 1 2 3

3

0 1 2 3

1 11 12

2 21 22

...,

...,

( ) ...,

( ) ....

u u u u u

v v v v v

h C C

h C C

(6.7)

The constants in 1 2( )are different then theconstantsin ( )h h . Substituting 1 2, , ( ), ( )u v h h

from Eq. (6.7) into Eq. (6.6) and some simplification and rearranging based on powers of -

terms, we have:


00

200

( , )( ,0) 0.005 0.387298coth 0.1 10 ,

( , )( ,0) 0.1330947502cosech 0.

0,

0, 1 10 .

u x tu x x

t

v x tv x x

t

(6.8)

77


0

2

0

( , ) 0.005 0.387298coth 0.1 10 ,

( , ) 0.1330947502cosech 0.1 10 .

u x t x

v x t x

(6.9)

First Order Problem

2

11 11 11 11 2

2

21 21 21 2

0 0 0 0 010

1

0 0 0 0 010 0 1 2

, ) , ) , ) , ) , ), ), ) 1.5 0,

,0) 0,

, ) , ) , ) ,

( ( ( ( (((

(

( ( ( ( ((( (

) , ), ), ) , ) 1.5

u x t u x t u x t v x t u x tu x tC C u x t C C

t t t x x x

u x

v x t v x t u x t v x t v x tv x tC C v x t C u x t C

t t t x x x

3

11 3

02

, )1.5 0, ,0) 0( .

(u x tC v x

x


2

11

1 11 2

11

2

21

2 2

1 11 21

18

19

1

0.000193649 cosech

3.4694469 10 cosech

0.000133094 cosec

0.1 10, , .

coth 0.1 10 0.1 10

coth 0.1 10 0.1h

4.3368087 10

10

, , coth 0.1 10 0cose .1 10ch

2.16840 10

C xu x t C t

C x x

C x x

v x t C t C x x

49

21cosech

.

0.1 10C x

(6.10)


2 11 12 11

2 2

1 12 11

0 0 01 1 20

12 1

1

0 01 1 10

2

2

2

20

1 2

, ) , ) , ), ) , ) , ), ) , )

, ) , ), ) , ) , ), ) 1.5 1.5 0,

,0) 0,

( ( (( ( (( (

( (( ( ((

(

( ,

u x t u x t u x tu x t u x t u x tC C C u x t C u x t

t t t t x x

v x t u x tu x t v x t u x tC u x t C C C C

x x x x x

u x

v xC

21 22 21

21 22 21

0 01 1 20 1

0 01 10 0 21 0

0 1

1

2 2

22 212

) , ) , ), ) , ) , ), ) , )

, ) , ), ) , ), ) , ) , ) , )

, ) ,1.

( (( ( (( (

( (( (( ( ( (

5 1.5( (

t u x t u x tv x t v x t v x tC C v x t C v x t

t t t t x x

v x t v x tu x t v x tC v x t C u x t C u x t C u x t

x x x x

v x t v x tC C

x

3 3

22 212 3

1

3

1) , ) , )1.5 1.5

(0.

(u x t u x tC C

x x x

78

Its solutions 2 11 12 21 22( , , , , , )u x t C C C C and 2 11 12 21 22( , , , , , )v x t C C C C are given in appendix A.

The final expression for 2nd

order approximate solution using OHAM for 11 12 21 22( , , , , , )u x t C C C C

and 11 12 21 22( , , , , , )v x t C C C C is

11 12 21 22 0 1 11 21 2 11 12 21 22

11 12 21 22 0 1 11 21 2 11 12 21 22

( , , , , , ) ( , ) ( , , , ) ( , , , , , ),

( , , , , , ) ( , ) ( , , , ) ( , , , , , ),

u x t C C C C u x t u x t C C u x t C C C C

v x t C C C C v x t v x t C C v x t C C C C

(6.11)

the constants are calculated using collocation method, but residuals for Eq. (6.4) we will take the

following form

1 1 2 3 4 11 12 21 22 11 12 21 22

11 12 21 22 11 12 21 22

2 1 2 3 4 11 12 21 22 11 12 21 22

11 12

11 12, , , , , , , , , , ,

, , , 1.5 , , ,

, , , , , , , , , , , , , ,

, , , , , , ,

,

, , , ,

( , ,

,

,

0

,

t x

x xx

t x

x t C C C C u u C C C C u C C C C

v C C C C u C C C C

x t C C C C v x t C C C C u x t C C C C

x t C C x t x

u x t C C

t

x t x t

R

R

21 22 11 12 21 22

11 12 21 22 11 12 21 22

, , , , , , )

1.5 , , , , , 1.5 , , , , , 0.

x

xxx xx

C C v x t C C C C

u x t C C C C v x t C C C C

(6.12)

The optimal values of the constants are

11 12

21 22

0.0199759697279381, 0.9605137555212906,

0.5086085429455561, 0.2490893184736806.

C C

C C

The 2nd

order OHAM solution gives very encouraging results after comparing with 2nd

order

HPM solution. Tables (6.1-6.2) shows the effectiveness of OHAM for WBK equation with

00.1, 0.005, 1.5, 1.5and 10.k x

Table 6.1

Comparison of absolute errors of OHAM with HPM for ( , )u x t of Eq. (6.4)

absolute error 0.1x absolute error 0.3x absolute error 0.5x

t HPM[132] OHAM HPM[132] OHAM HPM[132] OHAM

0.1 1.04892×10-4

1.90473×10-9

9.64474×10-5

1.7706×10-9

8.88312×10-5

1.64829×10-9

0.2 4.25408×10-4

5.79784×10-9

3.91098×10-4

5.35353×10-9

3.60161×10-4

4.95094×10-9

0.3 9.71992×10-4

1.16791×10-8

8.93309×10-4

1.07485×10-8

8.22452×10-4

9.90771×10-9

0.4 1.75596×10-3

1.95481×10-8

1.61430×10-3

1.79553×10-8

1.48578×10-3

1.65184×10-8

0.5 2.79519×10-3

2.94046×10-8

2.56714×10-3

2.69737×10-8

2.36184×10-3

2.47827×10-8

79

Table 6.2

Comparison of OHAM absolute errors with HPM for ( , )v x t of Eq. (6.4)



0.1 6.41419×10-3

8.17374×10-8

5.99783×10-3

8.17374×10-8

5.61507×10-3

8.17374×10-8

0.2 1.33181×10-3

1.61843×10-7

1.24441×10-2

1.61843×10-7

1.16416×10-2

1.61843×10-7

0.3 2.07641×10-2

2.40316×10-7

1.93852×10-2

2.40316×10-7

1.81209×10-2

2.40316×10-7

0.4 2.88100×10-2

3.17156×10-7

2.68724×10-2

3.17156×10-7

2.50985×10-2

3.17156×10-7

0.5 3.75193×10-2

3.92363×10-7

3.49617×10-2

3.92363×10-7

3.26239×10-2

3.92363×10-7

6.2.2 Model 2 (MB) equation

With 1, 0 the WBK equation is be reduced to the MB equations. For 0 10.x 0.1k and

0.005 Eq. (6.1) takes the following form

0,

( ) 0,

t x x

t x x xxx

u uu v

v u uv u

(6.13)


2

( ,0) 0.005 0.2coth 0.1 10 ,

( ,0) 0.02cosech 0.1 10 ,

u x x

v x x


2

( ,0) 0.005 0.2coth 0.1 10 0.005 ,

( ,0) 0.02 cosech 0.1 10 0.005 .

u x x t

v x x t

(6.14)

Homotopy for Eq. (6.13) takes the following form

1

2

(1 ) ( ) ( )[ ] 0,

(1 )( ) ( )[ ( ) ] 0,

t t x x

t t x x xxx

u h u uu v

v h v u uv u

(6.15)

we consider 1 2, , ( ), ( )u v h h as the following

2 3

0 1 2 3

3

0 1 2 3

1 11 12

2 21 22

...,

...,

( ) ...,

( ) ....

u u u u u

v v v v v

h C C

h C C

(6.16)

Substituting 1 2, , ( ), ( )u v h h from (6.16) into (6.15) and some simplification and rearranging

based on powers of -terms, we have:

80


00

200

( , )( ,0) 0.005 0.2coth 0.1 10 ,

( , )( ,0) 0.02cosech 0.1 .

0,

0, 10

u x tu x x

t

v x tv x x

t

(6.17)


0

2

0

( , ) 0.005 0.2coth 0.1 10 ,

( , ) 0.02cosech 0.1 10 .

u x t x

v x t x

(6.18)

First Order Problem

11 11 11

21 21

0 0 1 0 00

1

0 0 1 0

3

21 21

0

0 00 13

( ( ( ( ((

(

(

, ) , ) , ) , ) , ), ) 0,

,0) 0,

, ) , ) , ) , ), )

, ) , ), ) 0, ,0) 0.

( ( ((

( (( (

u x t u x t u x t u x t v x tC C u x t C

t t t x x

u x

v x t v x t v x t u x tC C v x t

t t t x

v x t u x tC u x t C v x

x x


23

2

2

1

1 11 2

11

2

3

1 21 40

21

0.1 10, , .

coth 0.1 10 0.1 10

coth 0.1 10 0.1 10,

0.0001 cosech

0.4694469 10 cosech

0.000020 cosech

5.4210108624 1, .

0 cos .1 10ech 0

C xu x t C t

C x x

C x xv x t C t

C x

(6.19)

81


12 11 12

11 11 12

0 01 1 20

0 01 11 11

22 21

0

2

0 1 1 2

, ) , ), ) , ) , ), )

, ) , ), ) , ), ) ,

( (( ( ((

( (( (( (

(

(

) 0,

,0) 0,

, ) ( (, ) () , ,

u x t u x tu x t u x t u x tC C C u x t

t t t t x

u x t v x tu x t v x tC u x t C u x t C C

x x x x

u x

v x t v x t v x t v xC C

t t t

00

0 01

22

21 21 22

3 3

21 21 22 21

1 0 0

0 01

3 3

11 0

2

, )), )

, ) , ), ), ) , ) , )

, ) , ), ) , ), ) , ) 0,

,0) 0

((

( ((( ( (

( (

.

( (( (

(

u x ttC v x t

t x

u x t v x tu x tC v x t C v x t C u x t

x x x

v x t u x tv x t u x tC u x t C u x t C C

x x x x

v x

Its solutions 2 11 12 21 22 2 11 12 21 22, , an, , , , ,d , , ,u x t vC xC C C C C C Ct are given in appendix B.

Therefore final expression for 11 12 21 22 11 12 21 22, , an, , , ,d , ,, ,C Cu C C Cv C Cx t Cx t is

11 12 21 22 0 1 11 21 2 11 12 21 22

11 12 21 22 0 1 11 21 2 11 12 21 22

( , , , , , ) ( , ) ( , , , ) ( , , , , , ),

( , , , , , ) ( , ) ( , , , ) ( , , , , , ),



(6.20)

The constants are calculated using collocation method, but the residuals for Eq. (6.13) we will

take the following form

1 11 12 21 22 11 12 21 22 11 12 21 22 11 12 21 22

11 12 21 22

2 11 12 21 22 11 12 21 22 11 12 21 22

11 12 21 22

, , , , , ,

,

, , , , , , , , , , , , , ,

, , ,

, , , , , , , , , , , , , , ,

, , , , ,

,

,

,

,

t x

x

t x

x t C C C C u C C C C u C C C C u C C C C

v C C C C

x t C C C C v x t C C C C u x t C C C C

u x t

x t x t x

C C C C

v

t

x

x t C

t

R

R

11 12 21 22

11 12 21 22

, , , , , 0., , ,

xxx

x

u x t C C C CC C C


11 12

21 22

0.2713077090827, 1.616535607314,

0.130949863039, 0.63817921763.

C C

C C

The 2nd


order

HPM solution. Table 6.3-6.4 shows the effectiveness of OHAM for MB equation with

00.1, 0.005, 1.5, 1.5 10.k and x

82

Table 6.3

Comparison of OHAM absolute errors with HPM for ( , )u x t of MB Eq. (6.13)


t HPM[132] OHAM HPM[132] OHAM HPM[132] 0HAM

0.1 8.16297×10-7

1.39222×10-9

7.64245×10-7

1.23917×10-9

7.16083×10-7

1.10332×10-9

0.2 3.26243×10-6

9.9748×10-9

3.05458×10-6

9.13951×10-9

2.86226×10-6

8.38644×10-9

0.3 7.33445×10-6

2.57479×10-8

6.86758×10-6

2.37012×10-8

6.43557×10-6

2.18495×10-8

0.4 1.30286×10-5

4.87116×10-8

1.22000×10-5

4.49242×10-8

1.14333×10-5

4.14926×10-8

0.5 2.03415×10-5

7.88661×10-8

1.90489×10-5

7.28089×10-8

1.78528×10-5

6.73159×10-8

Table 6.4

Comparison of OHAM absolute errors with HPM for ( , )v x t of Eq. (6.13)



0.1 5.88676×10-5

2.16276×10-7

5.56914×10-5

2.03149×10-7

5.27169×10-5

1.91004×10-7

0.2 1.18213×10-4

4.33772×10-7

1.11833×10-4

4.07414×10-7

1.05858×10-4

3.83031×10-7

0.3 1.78041×10-4

6.5249×10-7

1.68429×10-4

6.12797×10-7

1.59428×10-4

5.76081×10-7

0.4 2.38356×10-4

8.72429×10-7

2.25483×10-4

8.19296×10-7

2.13430×10-4

7.70155×10-7

0.5 2.99162×10-4

1.09359×10-6

2.83001×10-4

1.02691×10-6

2.67868×10-4

9.65252×10-7

6.2.3 Model 3 (ALW Equation)

With 0, 0.5 the WBK equation is reduced to the ALW equation. For 0 10,x 0.1k

and 0.005 Eq. (6.1) takes the following form

0.5 0,

( ) 0.5 0,

t x x xx

t x x xx

u uu v u

v u uv v

(6.21)


2

( ,0) 0.005 0.141421356coth 0.1 10 ,

( ,0) 0.0170710678cosech 0.1 10 ,

u x x

v x x


2

( , ) 0.005 0.141421356coth 0.1 10 0.005 ,

( , ) 0.0170710678cosech 0.1 10 0.005 .

u x t x t

v x t x t

(6.22)

Homotopy for Eq. (6.21) takes the following form

1

2

(1 ) ( ) ( )[ 0.5 ] 0,

(1 )( ) ( )[ ( ) 0.5 ] 0,

t t x x xx

t t x x xxx

u h u uu v u

v h v u uv v

(6.23)

83

we consider 1 2, , ( ), ( )u v h h as the following

2 3

0 1 2 3

3

0 1 2 3

1 11 12

2 21 22

...,

...,

( ) ...,

( ) ....

u u u u u

v v v v v

h C C

h C C

(6.24)

Substituting 1 2, , ( ), ( )u v h h from Eq. (6.24) into Eq. (6.23) and some simplification and

rearranging based on powers of -terms, we have:


00

200

( , )( , ) 0.005 0.141421356 0.1 10 0.005 ,

( , )( ,0) 0.0

0

170710678 0.1 10 0.005 .

,

0,

u x tu x t Coth x t

t

v x tv x Csch x t

t

(6.25)


0

2

0

( , ) 0.005 0.141421356coth 0.1 10 ,

( , ) 0.0170710678cosech 0.1 10 .

u x t x

v x t x

(6.26)

First Order Problem

2

11 11 11 11 2

2

21 21 21 2

0 0 0 0 010

1

0 0 0 0 010 0 1 2

, ) , ) , ) , ) , ), ), ) 0.5 0,

,0) 0,

, ) , ) , ) ,

( ( ( ( (((

(

( ( ( ( ((( (

) , ), ), ) , ) 0.5

u x t u x t u x t v x t u x tu x tC C u x t C C

t t t x x x

u x

v x t v x t u x t v x t v x tv x tC C v x t C u x t C

t t t x x x

1

0,

0)( , 0v x


2

11

1 11 21 2

11

2

21

2 2

1 11

4

21 1

2

2

0.1 10, , , .

coth 0.1 10 0.1 10

coth 0.1 10 0.1 10

, , , c

0.000070710678 cosech

0.064469 10 cosech

0

oth 0.1 10

.0000170710678 cosech

0.000141421356 cosec 0.1h

C xu x t C C t

C x x

C x x

v x t C C t C x

4

210.000070710678 cos

10 .

0.1 1h 0ec

x

C x

(6.27)

84


12 11 12 11

2 2

11 12 11

0 0 01 1 20 1

0 01 1 10

2

12 112 2

022

, ) , ) , ), ) , ) , ), ) , )

, ) , ), ) , ) , ),

( ( (( ( (( (

( (( () 0.5 0.5 0

(,

(

(

,0)

(

0,

u x t u x t u x tu x t u x t u x tC C C u x t C u x t

t t t t x x

v x t u x tu x t v x t u x tC u x t C C C C

x x x x x

u x

vC

0 01 1 2

0 1

0

3 22 3

3 401 1

0 0 1 0

0

3 3

2 2

2

14 32

( (( ( (( (

( ((

, ) , ) , ), ) , ) , ), ) , )

, ) , ), ) ,(( ( ( (

), ) , ) , ) , )

, ) , )0.5 0.5 0

( (

x t u x t u x tv x t v x t v x tC C v x t C v x t

t t t t x x

v x t v x tu x t v x tC v x t C u x t C u x t C u x t

x x x x

v x t v x tC C

x x

2, ,( 0) 0,v x

The terms 1 2 3 4 1 22 2 3 4, , , , , and , , , , ,u x t C C C C v x t C C C C are given in appendix C. Therefore

final expression for , ,u x t and v x t is

1 2 3 4 0 1 1 3 2 1 2 3 4

1 2 3 4 0 1 1 3 2 1 2 3 4

( , , , , , ) ( , ) ( , , , ) ( , , , , , ),

( , , , , , ) ( , ) ( , , , ) ( , , , , , ),



(6.28)

The constants are calculated using collocation method, but for Eq. (6.21) the residuals we will

take the following form

1 1 2 3 4 1 2 3 4 1 2 3 4 1 2 3 4

1 2 3 4 1 2 3 4

2 1 2 3 4 1 2 3 4 1 2 3 4

1 2 3 4

1 2 3 4

, , , , , , , , , , , , , ,

, , , 0.5 , , ,

, , , , , , ,

, , , , , ,

, , , ,

, , , , , , , ,

, , , , ,

, , , , ,

,

t x

x xx

t x

x t C C C C u C C C C u C C C C u C C C C

v C C C C u C C C C

x t C C C C v x t C C C C u x

x t x t x t

x t

t C C C C

u x t C C C C

v x t C C

t

C C

x

R

R

1 2 3 40.5 , , , , , 0,xx

x

v x t C C C C


11 12

21 22

0.2671352269414923, 0.5377914218783247,

2.4059074418238753, 1.0265395222595246.

C C

C C

The 2nd


order

HPM solution. Table 6.5-6.6 shows the effectiveness of OHAM for ALW equation with

00.1, 0.005, 1.5, 1.5 10.k and x

85

Table 6.5

Comparison of OHAM absolute errors with HPM for ( , )u x t of ALW equation



0.1 8.02989×10-6

2.43527×10-7

7.38281×10-6

2.2051×10-7

6.79923×10-6

2.00051×10-7

0.2 3.23228×10-5

9.81096×10-7

2.97172×10-5

8.88677×10-7

2.73673×10-5

8.06506×10-7

0.3 7.32051×10-5

2.21271×10-6

6.73006×10-5

2.0045×10-6

6.19760×10-5

1.81937×10-6

0.4 1.31032×10-4

3.93836×10-6

1.20455×10-4

3.56798×10-6

1.10919×10-4

3.23863×10-6

0.5 2.06186×10-4

6.15806×10-6

1.89528×10-4

5.57911×10-6

1.74510×10-4

5.0643×10-6

Table 6.6

Comparison of OHAM absolute errors with HPM for ( , )u x t of ALW equation


t HPM[132] OHAM HPM [132] OHAM HPM[132] OHAM

0.1 4.81902×10-4

2.61984×10-6

4.50818×10-4

2.43827×10-6

4.22221×10-5

2.27251×10-6

0.2 9.76644×10-4

5.44025×10-6

9.13502×10-4

5.04965×10-6

8.55426×10-4

4.69456×10-6

0.3 1.48482×10-3

8.46123×10-6

1.38858×10-3

7.83414×10-6

1.30009×10-4

7.26613×10-6

0.4 2.00705×10-3

1.16828×10-5

1.87661×10-3

1.07917×10-5

1.75670×10-4

9.98723×10-6

0.5 2.54396×10-3

1.51049×10-5

2.37815×10-3

1.39225×10-5

2.22578×10-4

1.28579×10-5

86

Chapter 7

Analysis of OHAM to DDEs

7.1 Introduction

In this chapter we implemented OHAM for nonlinear DDEs. We consider the following

discretized MKdV lattice equation as test example.

2

1 1

( , )( ( ( , )) )( ( , ) ( , )).n

n n n

u n tu n t u n t u n t

dt

(7.1)

Here is a constant. This equation arises in problems where Nano technology has been used,

e.g. heat conduction, electronic current and flow in carbon nanotubes [135]. The same lattice

equation has been solved by HPM [136] and HAM [137]. Our 2nd

order approximate solution to

the problem of Mkdv lattice equation yield very encouraging results after comparing with third

order approximate solution by HPM and HAM.

In literature, only a few papers are available for solution of DDEs. The fundamental paper of

Baldwin et al. [138], resulted in numerous publications for finding some exact solutions of DDEs

[159- 142]. The exp-function method for approximate solution of DDEs has been reported in

[156]. Zhu has found the exact traveling wave solutions with the exp-function method for

Hybrid- lattice system, discrete MKdV lattice, and discrete (2 + 1)-dimensional toda lattice

equation [144 -146]. Moreover, Zhu et al. used HPM for discontinued problems arising in

nanotechnology [136]. Nik et al. used HAM for solving discontinued problems arising in

nanotechnology [137]. Mokhtari et al. used VIM for finding approximate solution of

discontinued problems arising in nanotechnology [147].

7.2 Analysis of OHAM to Mkdv lattice equation

Consider Eq. (7.1) with initial condition given by

( ,0) ( ) ( ).n tanh d tanhu n dn

(7.2)

Where d is any arbitrary constant.

The homotopy using OHAM for Eq. (7.1) is as follows

87

20 01 1

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

n n n

u n t u n tu n t u n t u n t

t t

H

(7.3)

where

2 3

1 2 3 ... ,m

mC C C C H (7.4)

2

0 1 3

3

1 2, , , , , , ...,n n n n nu n t u n t u n t C u n t u n t (7.5)

1 ( 1)0 ( 1)1 1 (

2 3

1)2 ( 1)3, , , , , , ...,n n n n nu n t u n t u n t C u n t u n t (7.6)

1 ( 1)0 ( 1)1 1 (

2 3

1)2 ( 1)3, , , , , , ....n n n n nu n t u n t u n t C u n t u n t (7.7)

Residuals for Eq. (7.1) we will take the following form

1 11 2 1 2 1 2, ,, ; ( )( )., , , , , , ,i n n nn n t C C nt C u u t C C n tu C C R

(7.9)

11 2, , ,n x t Cu C and 11 2, , ,n x t Cu C are obtained by replacing n by 1 1n and n respectively

in equation (7.7).

7.2.1 Model 1 (Mkdv lattice equation)

From Eq. (7.3) we get the following problem


00

( , )( ,00, tanh( ) tanh( )) .n

n

u n tu n

td d n

(7.10)


0 tanh( ( ) tanh( )., )n du n t d n (7.11)

First Order Problem

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

0 0

2 2

1 1 1

11

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

, ) ,

( ( (

) , )

(

( ( (0.

(n n n n n n

n n n

C u n t C u n t C u n t u n t C u n t u n t

u n t u n t u n tC

t t t


3/2 3/2 3 2

1 1 1 1

3/2 3/2 3 2

1 1

, , ( tanh( ) tanh( ) tanh ( ) tanh ( ) tanh( )

tanh( ) tanh( ) tanh ( ) tanh ( ) tanh( )).

nu n t C t C d d nd C d nd d nd

C d d nd C d nd d nd

(7.12)

88


1 ( 1)1 2 ( 1)0 2 ( 1)0 1 ( 1)1

2 2 2

1 ( 1)1 2 ( 1)0 2 ( 1)0

2

1 ( 1)1 1 ( 1)

0 0 0

0 0 10

( , ) ( , ) ( , ) ( , )

( , )( ( , )) ( , )( ( , )) ( , )( ( , ))

( , )( ( , )) 2 ( , ) ( , ) ( , )

n n n n

n n n

n n

n n n

n n n

C u n t C u n t C u n t C u n t

C u n t u n t C u n t u n t C u n t u n t

C u n t u n t C u n t u n t u n t

0 11 ( 1)0

0 11

12

2

2 ( , ) ( , ) ( , )

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

n n

n n n

n

n

C u n t u n t u n t

u n t u n t u n t u n tC C

t t t t


2 2 5/2 2 2 5/2

2 1 2 1 1

2 2 5/2 3 2

1

2 2 5/2 3 2

1

2 2 5/2 3

1

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

2

tanh ( ) tanh( ( 1) ) tanh (( 1) )

tanh ( ) tanh( ( 1) ) tanh (( 1) )

tanh ( ) tanh( ( 1)

nu n t C C C t d h d n d C t d d n d

C t d d n d h n d

C t d d n d n d

C t d d n

2 3

2 2 5/2 3 2 3

1

2 2 5/2 5 2 2

1

2 2 5/2 5 2 2

1

5/2

1

) tanh (( 1) ) ( )

( ) ( ( 1) ) tanh (( 1) ) ( )

tanh ( ) tanh( ( 1) ) tanh (( 1) ) ( )

tanh ( ) tanh( ( 1) ) tanh (( 1) ) tanh ( )

2 tanh(

d n d tanh nd

C t tanh d tanh d n d n d tanh nd

C t d d n d n d tanh nd

C t d d n d n d nd

C t

2 3/2

1

5/2 3/2 3 2

2 1

2 3/2 3 2 3/2 3 2

1 2

2 2 5/2

1

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

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

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

2

d n d C t d n d

C t d n d C t nd nd n d

C t nd nd n d C t nd nd n d

C t

3 2 2 2 5/2 5 3 2

1

3/2 2 3/2 3/2

1 1 2

3/2 3 2 2

1 1

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

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

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

nd nd n d C t d nd n d

C t d n d C t d n d C t d n d

C t nd nd n d C

3/2 3 2

3/2 3 2 2 2 5/2 3

2 1

2 2 5/2 5 3 2 2 5/2 3

1 1

tanh ( ) tanh ( ) tanh((1 ) )

2 tanh ( ) tanh ( ) tanh((1 ) ) 4 tanh ( ) tanh( ) tanh((1 ) ) tanh((1 ) )

4 tanh ( ) tanh ( ) tanh((1 ) ) tanh((1 ) ) 2 tanh ( ) tanh

t nd nd n d

C t nd nd n d C t d nd n d n d

C t d nd n d n d C t d

2

2 2 5/2 5 3 2 2 2 5/2

1 1

2 2 5/2 3 2 2 2 5/2 3 2

1 1

2 2 5/2 3 2

1

( ) tanh ((1 ) )

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

tanh ( ) tanh ( ) tanh( (1 )) tanh ( ) tanh ( (1 )) tanh( (1 ))

tanh ( ) tanh (

nd n d

C t d nd n d C t d d d n

C t d nd d d n C t d d n d d n

C t d n

2 2

2 2 5/2 2 2 5/2 3 2

1 1

2 2 5/2 3 2

1

2 2 5/2 5 2 2

1

) tanh[ (1 )] ( (1 ))

tanh( ) tanh( (1 )) tanh ( ) tanh ( ) tanh( (1 ))

tanh ( ) tanh ( (1 )) tanh( (1 ))

tanh ( ) tanh ( ) tanh ( (1 )) tanh( (1 )))

d d n tanh d n

C t d d d n C t d nd d d n

C t d d n d d n

C t d nd d n d d n

.

(7.13)

89

The 2nd

order approximate solution is given by following equation

1 2 0 1 1 2 1 2

3/2

1

3/2 3 2 3/2

1 1

3/2 3 2

1

2 2 5/2

1

, , , , , , , , ,

tanh( ) tanh( ) ( tanh( ) tanh( )

tanh ( ) tanh ( ) tanh( ) tanh( ) tanh( )

tanh ( ) tanh ( ) tanh( ))

1( tanh( ) t

2

n n n nu n t C C u n t u n t C u n t C C

d d n t C d d nd

C d nd d nd C d d nd

C d nd d nd

C t d

2 2 5/2

1

2 2 5/2 3 2

1

2 2 5/2 3 2

1

2 2 5/2 3 2 3

1

2 2 5/2

1

anh ( ( 1) ) tanh( ) tanh( ( 1) )

tanh ( ) tanh( ( 1) ) tanh (( 1) )

tanh ( ) tanh( ( 1) ) tanh (( 1) )

tanh ( ) tanh( ( 1) ) tanh (( 1) ) ( )

h d n d C t d d n d

C t d d n d h n d

C t d d n d n d


C t

3 2 3

2 2 5/2 5 2 2

1

2 2 5/2 5 2 2

1

5/2 2 3/2

1 1

( ) ( ( 1) ) tanh (( 1) ) ( )

tanh ( ) tanh( ( 1) ) tanh (( 1) ) ( )

tanh ( ) tanh( ( 1) ) tanh (( 1) ) tanh ( )

2 tanh( ) tanh((1 ) ) 2 tanh( ) tan

tanh d tanh d n d n d tanh nd


C t d d n d n d nd

C t d n d C t d

5/2 3/2 3 2

2 1

2 3/2 3 2 3/2 3 2

1 2

2 2 5/2 3 2

1 1

h((1 ) )

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

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

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

n d

C t d n d C t nd nd n d


C t nd nd n d C

2 2 5/2 5 3 2

3/2 2 3/2 3/2

1 1 2

3/2 3 2 2 3/2 3 2

1 1

tanh ( ) tanh ( ) tanh ((1 ) )

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

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

t d nd n d

C t d n d C t d n d C t d n d


3/2 3 2 2 2 5/2 3

2 1

2 2 5/2 5 3 2 2 5/2 3 2

1 1

2 2 5/2 5

1

)

2 tanh ( ) tanh ( ) tanh((1 ) ) 4 tanh ( ) tanh( ) tanh((1 ) ) tanh((1 ) )

4 tanh ( ) tanh ( ) tanh((1 ) ) tanh((1 ) ) 2 tanh ( ) tanh( ) tanh ((1 ) )

2 tanh (

C t nd nd n d C t d nd n d n d

C t d nd n d n d C t d nd n d

C t

3 2 2 2 5/2

1

2 2 5/2 3 2 2 2 5/2 3 2

1 1

2 2 5/2 3 2 2 2

1

2 2

1

) tanh ( ) tanh ((1 ) ) tanh( ) tanh( (1 ))

tanh ( ) tanh ( ) tanh( (1 )) tanh ( ) tanh ( (1 )) tanh( (1 ))

tanh ( ) tanh ( ) tanh[ (1 )] ( (1 ))

d nd n d C t d d d n

C t d nd d d n C t d d n d d n

C t d nd d n tanh d n

C t

5/2 2 2 5/2 3 2

1

2 2 5/2 3 2

1

2 2 5/2 5 2 2

1

tanh( ) tanh( (1 )) tanh ( ) tanh ( ) tanh( (1 ))

tanh ( ) tanh ( (1 )) tanh( (1 ))

tanh ( ) tanh ( ) tanh ( (1 )) tanh( (1 ))).

d d d n C t d nd d d n

C t d d n d d n

C t d nd d n d d n

The constants are calculated using collocation method. The optimal values of constants are

1 0.9984520064122604,C

2 0.0000378857828636.C

We take 1, 0.1d and 1t . In table 7.1 we have presented 2nd

order approximate solution

by OHAM and absolute error of OHAM are compared with 3rd

order HAM and HPM solution

90

[136, 137]. The 2nd

order OHAM solution yield vary encouraging results after comparing with

3rd

order HAM and HPM solution. Results can be further improved if we take higher order

approximations of OHAM. If the proposed equation defines the flow in a carbon nanotube, then

the velocity of the flow have a tendency to the maximum when 20n , this can explain some

attractive phenomena of Nano hydrodynamics. Figure 7.1 presents the 2D plot of the

approximate solution.

Figure 7.1 Approximate solution graph for 1, 0.1 and 1d t

Table 7. 1



order HAM and HPM

solution for 1, 0.1 and 1d t

n OHAM

solution

Exact

solution

Absolute error

HAM [137]

Absolute error

HPM [136]

Absolute error

OHAM

-15 -0.0859782 -0.08590325 6.870534×10-5

6.535234×10-4

7.50032×10-5

-5 -0.0289982 -0.02909509 9.392570×10-5

4.958405×10-5

9.68803×10-5

-4 -0.0195853 -0.0197355 1.4563×10-4

1.0105422×10-4

1.5028×10-4

-3 -0.00979998 -0.0099992 1.9388×10-4

1.5394854×10-4

1.9925×10-4

3 0.0461641 0.0460062 1.6340×10-4

2.0187103×10-4

1.5786×10-4

4 0.0535852 0.0534795 1.1127×10-4

1.5366629×10-4

1.0564×10-4

5 0.0602487 0.060194 5.918924×10-5

1.0078519×10-4

5.45908×10-5

15 0.0931588 0.0932221 6.920412×10-5

7.3120040×10-5

6.32132×10-5

91

Chapter 8

Conclusion

OHAM [44-49] is implemented to linear and nonlinear tenth order BVPs, nonlinear time

dependent PDEs, coupled system of PDEs and DDE’s. For tenth order BVPs the absolute error

of OHAM were compared to that of DTM [101]. The nonlinear time dependent PDEs in this

thesis are Burger’s family, fisher’s equation, nonlinear diffusion equation fisher’s type, MEW

equation and DGRLW equation. In Burger’s family, we studied Burger, Burger-Huxley and

Burger-fisher equations and the results of OHAM were compared to that of ADM [115]. For

fisher’s equation and nonlinear diffusion equation of fisher’s type, the absolute errors of OHAM

were compared to that of ADM, HPM and VIM [118-120]. For MEW equation the results of

proposed method were compared to HAM [124]. For DGRLW equation the results of proposed

method were compared to VIM [124]. In all these cases the results of OHAM were found to be

very encouraging. We observed that OHAM is simpler in applicability, more efficient and

involve less computational work. This technique is free from small parameter assumption and

does not require discretization or perturbation. The convergence of proposed method do not

depends upon the initial solution. By taking higher order of approximations the OHAM solution

can get closer to exact solution. The auxiliary function ( )h involves convergence constants siC ,

due to which OHAM cannot give a closed form solution, but gives more accurate results while

analyzing any physical problem.

While solving tenth order boundary value problems the optimal values of constants were

calculated using method of least squares. The method of least squares requires more CPU timing

and working storage memory, so for the complicated problems in this thesis we have used

collocation method.

Apart from application to ODEs and PDEs, we implemented OHAM to time dependent coupled

system of PDEs including WBK, MB and ALW systems and to DDEs including MKdV lattice

equation. The homotopy using OHAM for WBK, MB and ALW systems with 0 10,x 0.1k

and 0.005 were given by Eqs. (6.6), (6.15) and (6.23) respectively and the optimal values of

constants were calculated using collocation method. The homotopy using OHAM for MKdV

lattice equation was given by Eq. (7.3). For MKdV lattice equation we first approximated the

92

solution at three mesh points 1, , 1n n n , found residual using Eq. (7.9) and then used

collocation method for computation of optimal values of constants.

The implantation of OHAM to coupled system of PDEs and DDEs were more accurate and as

such it will be more appealing for the researchers to apply it to more complex problems in

science and technology especially in fluid dynamics.

93

LIST OF PUBLICATIONS

[1] R. Nawaz, S. Islam, M. N. Khalid and S. Naeem, Solution of Tenth Order Boundary Value Problems Using Optimal

Homotopy Asymptotic Method Canadian Journal on Computing in Mathematics, Natural Sciences, Engineering &

Medicine, 2(4)2010, 38-47.

[2] R. Nawaz, S. Islam, M. Idrees, and H. Ullah, Application of Optimal Homotopy Asymptotic Method to Burger

Equations (Burger Huxley and burger fisher), J. Appl. Math. volume (2013), Article ID 387478, 8 pages.

[3] R. Nawaz, S. Islam, M. Idrees, I.A. Shah, and H. Ullah, Optimal Homotopy Asymptotic Method to Nonlinear

Damped Generalized Regularized Long-Wave Equation, J. problems in engineering (2013).

[4] S. Islam, R. Nawaz, M. Arif, I.A. Shah, Application of Optimal Homotopy Asymptotic Method to the Equal Width

Wave and Burger Equations, Life Science Journal 2012 9(4).

[5] H. Ullah, M. Idrees, Saeed Islam, and R. Nawaz, Application of Optimal Homotopy Asymptotic Method to Doubly

Wave Solutions of the Couple Drinfel’d-Sokolv-Wilson Equations. J. problems in eng., vol. 2013, Article ID 362816, 8

pages, 2013. doi:10.1155/2013/362816

[6] R. Nawaz, S. Islam, I.A. Shah, M. Idrees and H. Ullah, Application of Optimal Homotopy Asymptotic Method

to Fisher’s Equation. Submitted.

[7] R. Nawaz, S. Islam, I.A. Shah, M. Idrees and H. Ullah, Modification of optimal homotopy asymptotic method for

discontinued problems arising in nanotechnology. Submitted.

[8] R. Nawaz, S. Islam, Syed Inayat Ali Shah,Idrees Muhammad and H. Ullah, Modification of optimal homotopy

asymptotic method for coupled system of partial differential equation. Submitted.

[9] H. Ullah, S. Islam, R. Nawaz, and Mehreen Fiza, The Optimal Homotopy Asymptotic Method with Applications

to steady three dimensional problem of condensation film on inclined rotating disk. Submitted.

[10] H. Ullah, S. Islam, M. Idrees, R. Nawaz, and Mehreen Fiza, The Optimal Homotopy Asymptotic Method with

application to Fornberg-Whitham Equation. Submitted.

[11] H. Ullah, S. Islam, M. Idrees, R. Nawaz, Optimal Homotopy Asymptotic Method to nonlinear Advection

equations. Submitted

[12] H. Ullah, R. Nawaz ,S. Islam, M. Idrees, The optimal homotopy asymptotic method for modified Khwara Equation.

Submitted.

http://www.informatik.uni-trier.de/~ley/db/journals/jam/jam2013.html#NawazUII13

94

Appendix A

2 11 12 21 22 12

11 11 21

12 11 11

( , , , , , ) (0.00290669 cosh 0.4 0.00290669sinh 0.4 ((5.35254

(5.35254 (5.35254 0.365203 ) 0.367879 )) cosh 0.1

(-86.5355 (-86.5355 - 86.5355 5.96199

u x t C C C C t x x C

C C t C t x

C C C

11 21

2

11 11 12

2

11 11 21 11

2

11 12

5.94757 )) cosh 0.1

63.0429 cosh 0.1 63.0429 cosh 0.1 63.0429 cosh 0.1

6.49798 cosh 0.1 6.5402 cosh 0.1 284.995 cosh 0.3

284.995 cosh 0.1 284.995 cosh 0.

C t C t x

C x C x C x

C t x C C t x C x

C x C

2

11

2

11 21 11 11

2

12 11 11 21

11

3 20.337 cosh 0.3

20.4838 cosh 0.3 0.490176 cosh 0.5 0.490176 cosh 0.5

0.490176 cosh 0.5 0.00659089 cosh 0.5 0.00673795 cosh 0.5

0.0132676 cosh 0.7 0.01

x C t x

C C t x C x C x

C x C t x C C t x

C x

2

11 12

2

11 11 21 11

2 2

11 12 11

11 21

32676 cosh 0.7 0.0132676 cosh 0.7

0.000918516 cosh 0.7 0.000911882 cosh 0.7 5.35254 sinh 0.1

5.35254 sinh 0.1 5.35254 sinh 0.1 0.365203 sinh 0.1

0.367879 s

C x C x

C t x C C t x C x

C x C x C t x

C C t

2

11 11

2

12 11 11 21

2 2

11 11 12 11

inh 0.4 150.766 sinh 0.1 150.766 sinh 0.1

150.766 sinh 0.1 10.3872 sinh 0.1 10.3621 sinh 0.1

95.1581 sinh 0.1 95.1581 sinh 0.1 95.1581 sinh 0.1 4.29605 sinh

x C x C x

C x C t x C C t x

C x C x C x C t

2

11 21 11 11

2

12 11 11 21

2

11

0.1

4.33293 sinh 0.1 299.483 sinh 0.3 299.483 sinh 0.3

299.483 sinh 0.3 19.5418 sinh 0.3 19.6872 sinh 0.3

0.490176 1sinh 0.5 - 0.490176 sinh 0.5 0.490176

x

C C t x C x C x

C x C t x C C t x

c x C x

12

2

11 11 21 12

11 11 11 21

2

sinh 0.5

0.00659089 sinh 0.5 0.00673795 sinh 0.5 (0.0132676

(0.0132676 0.0132676 0.000918516 0.000911882 ))sinh 0.7 ))

1 c 2. 0.2 s 2. 0.2 2.3504/

C x

C t x C C t x C

C C C t C t x

osh x inh x

5

[0.1 ] 3.08616 sinh 0.5 Cosh x x

95

7 7

2 11 12 21 22 22

21 21 11 21 22

21 21 11

( , , , , , ) ( 6.28982 10 cosh 0.2 6.28982 10 sinh 0.2 (801.848

(801.848 801.848 21.3649 20.3912 ) ( 846.414

( 846.414 ( 846.414 7.38906 ) 7.38906 ))

v x t C C C C t x x C

C C C t C t C

C C t C t

22 21 21 11 21

2

21 21 22

2

11 21 21

cosh 0.2

(7493.71 (7493.71 7493.71 90.7483 110.317 )) cosh 0.2

6864.33 cosh 0.2 6864.33 cosh 0.2 6864.33 cosh 0.2

88.3645 cosh 0.2 86.7104 cosh 0.2 8754

x

C C C C t C t x

C x C x C x

C C t x C t x

21

2

21 22 11 21

2 2

21 21 21

22 11 21

6.3 cosh 0.2

87546.3 cosh 0.4 87546.3 cosh 0.4 2332.75 cosh 0.4

2439.07 cosh 0.4 646977 cosh 0.6 646977 cosh 0.6

646977 cosh 0.6 14920.4 cosh 0.6 15105.2

C x

C x C x C C t x

C t x C x C x

C x C C t x C

2

21

6 6 2 6

21 21 22

2 6

11 21 21 21

6 2 6

21 22

cosh 0.6

2.04881 10 cosh 0.8 2.04881 10 cosh 0.8 2.04881 10 cosh 0.8

35014.2 cosh 0.8 34672.8 cosh 0.8 2.52313 10 cosh

2.52313 10 cosh 2.5231 10 cosh

t x

C x C x C x

C C t x C t x C x

C x C

11 21

2 2

21 21 21

2

22 11 21 21

2

21 21

22026.5 cosh

20764.9 cosh 846.414 sinh 0.2 846.414 sinh 0.2

846.414 sinh 0.2 7.38906 sinh 0.2 7.38906 sinh 0.2

7741.75 sinh 0.2 7741.75 sinh 0.2

x C C t x

C t x C x C x

C x C C t x C t x

C x C x

22

2

11 21 21 21

2

21 22 11 21

2

21 21

7741.75 sinh 0.2

93.8589 sinh 0.2 113.226 sinh 0.2 6678.3 sinh 0.2

6678.3 sinh 0.2 6678.3 sinh 0.2 81.4646 sinh 0.2

79.733 sinh 0.2 87571.4 sinh 0.4 87571.

C x

C C t x C t x C x

C x C x C C t x

C t x C x

2

21

2

22 11 21 21

2

21 21 22

2

11 21 21

4 sinh 0.4

87571.4 sinh 0.4 2333.18 sinh 0.4 2439.51 sinh 0.4

646977 sinh 0.6 646977 sinh 0.6 646977 sinh 0.6

14920.4 sinh 0.6 15105.2 sinh 0.6 2.04881 1

C x

C x C C t x C t x

C x C x C x

C C t x C t x

6

21

6 2 6

21 22 11 21

2 6 6

21 22 21

6

21 11 21

0 sinh 0.8

2.04881 10 sinh 0.8 2.04881 10 sinh 0.8 35014.2 sinh 0.8

34672.8 sinh 0.8 (2.52313 2.52313 10 (2.52313 10

2.52313 10 22026.5 20764.9

C x

C x C x C C t x

C t x C C

C C t C t

10

)) sinh ))

/ 2.3504 cosh 0.1 3.08616 sinh 0.1 .

x

x x

96

Appendix B

2

11 12 21 22 12 11 11

11 21 12 11 11

21

( , , , , , ) (0.25 cos 0.2 (-0.602647 (-0.602647 - 0.602647

0.00154044 0.00135337 ) (0.160242 (0.160242 (0.160242 0.00266232 )

- 0.00273932 )) c 5.55112 1

u x t C C C C t ech x C C C

C t C t C C C t

C t osh

17

12 11 11

2

11 21 11 11

2

12 11 11 21

0 (-0.0803693 (-0.0803693- 0.0803693

-0.0015672 0.00160739 )) cos 0.2 1.1723 cos 0.2 1.1723 cos 0.2

1.1723 cos 0.2 0.0176596 cos 0.2 - 0.0182169 cos 0.2

-0.3

x C C C

C t C t x C x C x

C x C t x C C t x

2

11 11 12

2

11 11 21 11

2 2

11 21 11

76939 cos 0.2 - 0.376939 cos 0.2 - 0.376939 cos 0.2

-0.00500628 cos 0.2 0.00517685 cos 0.2 0.554572 cos 0.4

0.554572 cos 0.4 0.554572 cos 0.4 - 0.00814981 cos 0

C x C x C x

C t x C C t x C x

C x C x C t

2

11 21 11 11

2

12 11 11 21

2

11 11 12

.4

0.00803925 cos 0.4 -1.49116 cos 0.6 -1.49116 cos 0.6

- 1.49116 cos 0.6 - 0.0161196 cos 0.6 0.016766 cos 0.6

0.690331 cos 0.8 0.690331 cos 0.8 0.690331 cos 0.

x

C C t x C x C x

C x C t x C C t x

C x C x C

2

11 11 21 11

2 2

11 12 11 11 21

2

11 11

8

0.0202007 cos 0.8 - 0.0206143 cos 0.8 0.591951 cos ...

+ 0.591951 cos +0.591951 cos 0.00144118 cos - 0.0016 cos

-0.677329 cos 1.2 - 0.677329 cos 1.2 - 0.677

x

C t x C C t x C x

C x C x C t x C C t x

C x C x

12

2

11 11 21 11

2 2

11 12 11

11 21 11

329 cos 1.2

-0.010241 cos 1.2 0.0105547 cos 1.2 0.184223 cos 1.4

0.184223 cos 1.4 0.184223 cos 1.4 0.00359235 cos 1.4

-0.00368446 cos 1.4 - 0.160242 sinh 5.55112

C x

C t x C C t x C x

C x C x C t x

C C t x C

17

2 17 17

11 12

2 17 17

11 11 21

2

11 11

10

-0.160242 sinh 5.55112 10 - 0.160242 sinh 5.55112 10

-0.00266232 sinh 5.55112 10 0.00273932 sinh 5.55112 10

-0.0229423 sinh 0.2 - 0.0229423 sinh 0.2 - 0.0229423

x

C x C x

C t x C C t x

C x C x

12

2

11 11 21 11

2 2

11 12 11

11 21 1

sinh 0.2

-0.000447374 sinh 0.2 0.000458845 sinh 0.2 0.423974 sinh 0.2

0.423974 sinh 0.2 0.423974 sinh 0.2 0.0208896 sinh 0.2

-0.0212515 sinh 0.2 - 0.143773

C x

C t x C C t x C x

C x C x C t x

C C t x C

2

1 11

2

12 11 11 21

2

11 11 12

2

11

sinh 0.2 - 0.143773 sinh 0.2

- 0.143773 sinh 0.2 - 0.00488911 sinh 0.2 0.00498801 sinh 0.2

0.649989 sinh 0.4 0.649989 sinh 0.4 0.649989 sinh 0.4

- 0.00967593 sinh 0.4 0

x C x

C x C t x C C t x

C x C x C x

C t x

11 21 11

2 2

11 12 11

.00956151 sinh 0.4 -1.50216 sinh 0.6

-1.50216 sinh 0.6 -1.50216 sinh 0.6 - 0.0159679 sinh 0.6

C C t x C x

C x C x C t x

97

2

11 21 11 11

2

12 11 11 21

2

11 11

... 0.0166128 sinh 0.6 0.690999 sinh 0.8 0.690999 sinh 0.8

0.690999 sinh 0.8 0.0201995 sinh 0.8 - 0.0206129 sinh 0.8

0.591935 sinh 0.591935 sinh

C C t x C x C x

C x C t x C C t x

C x C x

2

12 11

2

11 21 11 11

2

12 11 11 21

12 11

0.591935 sinh 0.00144083 sinh

-0.00164852 sinh - 0.677329 sinh 1.2 - 0.677329 sinh 1.2

-0.677329 sinh 1.2 - 0.010241 sinh 1.2 0.0105547 sinh 1.2

(0.184223 (0.

C x C t x

C C t x C x C x

C x C t x C C t x

C C

11 11 21

2 3

184223 0.184223 0.00359235 - 0.00368446 )) sinh 1.4 )) /

-1. + cos 2+0.2 + Sinh 2+0.2 2.3504cos 0.1 + 3.08616 sinh 0.1

(6.09275 cos 0.1 - 9.13913cos 0.3 + 3.04638cos 0.5

+ 17.6799sinh 0.1

C C t C t x

x x x x

x x x

- 11.1601sinh 0.1 + 3.16005sinh 0.1 )x x x

98

2 11 21 22

13 13 13

21 22 22

21 21 22

( , , , , ) ( (0.0000968229cosh 334.875 - 0.0000968229sinh 334.875 )

((-2.86663 10 - 2.86663 10 )cosh 333.875 2.86663 10 cosh 334.875

- 6.10521 (1 )cosh 335.075 - 6.10521

v x t C C C t x x

C C x C x

C C x C

2

11 21 21

2

21 21 22

2

11 21 21

cosh 335.075

0.122104 cosh 335.075 - 0.125157 cosh 335.075

90.2235 cosh 335.275 90.2235 cosh 335.275 90.2235 cosh 335.275

5.41341 cosh 335.275 -5.45852 cosh 335.275

x

C C t x C t x

C x C x C x

C C t x C t x

2

21

2

21 21 22

2

11 21 21 21

2

21

cosh 335.475

- 4926.04 cosh 335.675 - 4926.04 cosh 335.675 - 4926.04 cosh 335.675

-295.562 cosh 335.675 293.099 cosh 335.675 18199.4 cosh 335.875

18199.4 cosh 335.875 181

C t x

C x C x C x

C C t x C t x C x

C x

22 11 21

2 13 13

21 21 22

13

22 21

99.4 cosh 335.875 -363.988 cosh 335.875

354.888 t cosh 335.875 -2.86663 10 sinh 335.875 -2.86663 10 sinh 335.875

+2.86663 10 sinh 334.875 - 6.10521 sinh 335.075 - 6.10521

C x C C t x

C x C x C x

C x C x C

2

21

2

22 11 21 21

2

21 21 22

11 21

sinh 335.075

-6.10521 sinh 335.075 0.122104 sinh 335.075 - 0.125157 sinh 335.075

90.2235 sinh 335.275 90.2235 sinh 335.275 90.2235 sinh 335.275

5.41341 sinh 335.275 -5.

x

C x C C t x C t x

C x C x C x

C C t x

2 2

21 21

2

21 21 22

2

11 21 21 22

21

45852 sinh 335.275 sinh 335.475

-4926.04 sinh 335.675 - 4926.04 sinh 335.675 - 4926.04 sinh 335.275

-295.562 sinh 335.675 293.099 sinh 335.675 (18199.4

(18199.4 -36

C t x C t x

C x C x C x

C C t x C t x C

C

11 21

6

3.988 (18199.4 354.888 )))sinh 335.875 ))

/ 1- 7.38906 cosh 0.2 - 7.38906 sinh 0.2

C t C t x

x x

99

Appendix C

11 12 21 22 12 11 11

21 12 11 11 11

21

( , , , , , ) ( ((0.00230654 (0.00230654 (0.00230654 - 0.000018946 )

-0.000442746 )) cosh 0.1 (0.008914 (0.008914 0.008914 - 0.000148854

0.00342988 ))cosh 0.1 - 0

u x t C C C C t C C C t

C t x C C C C t

C t x

2

11 11

6 2

12 11 11 21

2

11 11 12

.000312156 cosh 0.1 - 0.000312156 cosh 0.1

-0.000312156 cosh 0.1 2.46001 10 cosh 0.1 - 0.0000649432 cosh 0.1

-0.113607 cosh 0.3 - 0.113607 cosh 0.3 - 0.113607 cosh 0.3

0.0021

C x C x

C x C t x C C t x

C x C x C x

2

11 11 21 11

2 2

11 12 11

2

11 21 11 11

9479 cosh 0.3 - 0.0021976 cosh 0.3 0.629664 cosh 0.5

+0.629664 cosh 0.5 0.629664 cosh 0.5 - 0.0092258 cosh 0.5

-0.150394 cosh 0.5 -1.86105 cosh 0.7 -1.86105 cosh 0.

C t x C C t x C x

C x C x C t x

C C t x C x C

2

12 11 11 21

2 2

11 11 12 11

11 21 11

7

-1.86105 cosh 0.7 0.000620349 cosh 0.7 0.49628 cosh 0.7

2.2919 cosh 0.9 2.2919 cosh 0.9 2.2919 cosh 0.9 0.0541854 cosh 0.9

0.403048 cosh 0.9 0.00230654 sinh 0

x

C x C t x C C t x

C x C x C x C t x

C C t x C

2

11

2

12 11 11 21

2

11 11 12

2

11

.1 0.00230654 sinh 0.1

0.00230654 sinh 0.1 - 0.000018946 sinh 0.1 - 0.000442746 sinh 0.1

0.00953831 sinh 0.1 0.00953831 sinh 0.1 0.00953831 sinh 0.1

-0.000143518 sinh 0.1

x C x

C x C t x C C t x

C x C x C x

C t

11 21 11

2 6 2

11 12 11

2

11 21 11 11

0.00363296 sinh 0.1 0.000312156 sinh 0.1

0.000312156 sinh 0.1 0.000312156 sinh 0.1 - 2.46001 10 sinh 0.1

0.0000649432 sinh 0.3 - 0.113635 sinh 0.3 - 0.113635 sinh

x C C t x C x

C x C x C t x

C C t x C x C

2

12 11 11 21

2

11 11 12

2

11 11 21

0.3

-0.113635 sinh 0.3 0.00219409 sinh 0.3 - 0.00219129 sinh 0.3

0.629664 sinh 0.5 0.629664 sinh 0.5 0.629664 sinh 0.5

-0.0092258 sinh 0.5 - 0.150394 sinh 0.5 - 1

x

C x C t x C C t x

C x C x C x

C t x C C t x

11

2 2

11 12 11

11 21 12 11 11 11

21

.86105 sinh 0.7

-1.86105 sinh 0.7 -1.86105 sinh 0.7 0.000620349 sinh 0.7

0.49628 sinh 0.7 (2.2919 (2.2919 2.2919 0.0541854

0.403048 ))sinh 0.9 ))

( -1 cosh 2 0.2 /

C x

C x C x C t x

C C t x C C C C t

C t x

x

3 3

sinh 2 0.2 2.3504 cosh 0.1 3.08616 sinh 0.1

-2. 7.52439 cosh 0.2 7.25372 sinh 0.2

x x x

x x

100

-7 2 2 3 2

2 11 21 22 21

-7 2 2 4 2

21 22

-9 4

21 21 21

( , , , , ) 1.41421 10 coth 1 0.1 cos 1 0.1

-5.85786 10 coth 1 0.1 cos 1 0.1 (-0.0000707107

(-0.0000707107 - 0.0000707107 - 4.26777 10 )) cos 1 0.1

(

v x t C C C C t x ch x

C t x ch x t C

C C C t ch x

2 -7 2 2 6

11 21 21

2

21 21 22

-7 -8 2

11 21 21

3

22 21

0. 9.14214 10 )cos 1 0.1

coth 1 0.1 cos 1 0.1 (0.0000170711 (1 ) 0.0000170711

(2.41421 10 4.14214 10 ) cos 1 0.1

coth 1 0.1 ( (-0.000141421

C C t C t ch x

t x ch x C C C

C C C t ch x

x t C C

21

-9 2 2 -7 2 2 4

21 11 21 21

(-0.000141421- 0.000141421

-8.53553 10 )) cos 1 0.1 (0. 2.77817 10 )cos 1 0.1 )

C

C t ch x C C t C t ch x

101

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