on nonlinear sciencesidosi.org/sns/1(4)10/2.pdfk ∑, n= 0,1,2,k. adomian’s decomposition method...

Studies in Nonlinear Sciences 1 (4): 97-117, 2010 ISSN 2221-3910 © IDOSI Publications, 2010

Corresponding Author: Dr. Ahmet Yildirim, Department of Mathematics, Ege University, 35100 Bornova, Izmir, Turkey 97

On Nonlinear Sciences

1Ahmet Yildirim, 2M.M. Hosseini, 3M. Usman and 4Syed Tauseef Mohyud-Din

1Department of Mathematics, Ege University, 35100 Bornova, Izmir, Turkey 2Faculty of Mathematics, Yazd University, P.O. Box: 89195-74, Yazd, Iran

3Department of Mathematics, University of Dayton, 300 College Park, Dayton, OH 45469-2316, USA 4Department of Mathematics, HITEC University, Taxila Cantt, Pakistan

Abstract: This paper outlines a detailed study of some relatively new techniques which are being used to find appropriate solutions of the physical problems appearing in nonlinear sciences. In particular, we will focus on Variational Iteration (VIM), Homotopy Perturbation (HPM), Homotopy Analysis (HAM), Modified Decomposition (MDM), Variational iteration method using He’s and Adomian’s polynomials (VIMHP & VIMAP), parameter expansion and exp -function method. These relatively new but very reliable techniques proved useful for solving a wide class of nonlinear problems of diversified physical nature and are capable enough to cope with their versatility. Several examples are given to re-confirm the efficiency of these algorithms. Key words: Variational iteration method • homotopy perturbation method • homotopy analysis method •

decomposition • nonlinear problems • he’s polynomials • adomian’s polynomials • exp-function method

INTRODUCTION

The rapid development of nonlinear sciences witnesses number of analytical and numerical techniques developed by various scientists. Most of the previously developed techniques had their limitations like limited convergence, divergent results, linearization, discretization, unrealistic assumptions and non-compatibility with the versatility of physical problems [1-38]. He [14-17] developed a number of efficient and reliable techniques to solve a wide class of nonlinear problems. These relatively new methods proved to be fully synchronized with the complexities of the physical problems [1-17, 20-31, 34-38] and the references therein. The Variational Iteration Method (VIM) [1-17, 14, 15, 25] was developed by He though its origin may be traced back to Inokuti, Sekine and Mura [17] but the real potential of the method was realized and exploited by He. Moreover, He [14-17] introduced another wonderful technique namely Homotopy Perturbation (HPM) by merging the standard homotopy and perturbation. The HPM is independent of the drawbacks of the coupled techniques and absorbs all their positive features. Recently, He and Wu [14, 16, 17] formulated the exp -function method for solving nonlinear problems of versatile nature. It is to be highlighted that the present study would also outline the implementation of He’s parameter-expansion

techniques [14, 15, 25] which comprise the book keeping parameter and modified Lindstedt-Pioncare methods using parameter-expansion for finding the frequency of nonlinear oscillators. These techniques have been applied to a wide class of nonlinear problems [1-17, 19-31, 34-38] and the references therein. With the passage of time some modifications in He’s Variational Iteration Method (VIM) has been introduced by various authors. Abbasbandy [1, 2] made the coupling of Adomian’s polynomials with the correction functional (VIMAP) of the VIM and applied this reliable version for solving Riccati differential and Klein Gordon equations. In a later work, Noor and Mohyud-Din [25] exploited this concept for solving various singular and non singular boundary and initial value problems. Recently, Ghorbani et al. [12] introduced He’s polynomials by splitting the nonlinear term and also proved that He’s polynomials are fully compatible with Adomian’s polynomials but are easier to calculate and are more user friendly. More recently, Noor and Mohyud-Din [25] combined He’s polynomials and correction functional (VIMHP) of the VIM and applied this reliable version to a number of physical problems. It has been observed [25] that the modification based on He’s polynomials (VIMHP) is much easier to implement as compare to the one (VIMAP) where the so-called Adomian’s polynomials along with their complexities are used. The present

Studies in Nonlinear Sci., 1 (4): 97-117, 2010

98

study would also focus on another powerful algorithm which is called the Homotopy Analysis Method (HAM) and was developed by Liao [27]. The HAM contains an auxiliary parameter ,h which provides us with a simple way to adjust and control the convergence region of solution series for large values of t. Unlike some other numerical and analytical techniques Liao’s HAM handles linear and nonlinear problems without any assumption and restriction. Recently, Geijji and Jafari [13] introduced Modified Decomposition Method (MDM) which is independent of the Adomian’s polynomials and is fully capable to grapple complex nonlinear physical problems. The basic motivation of the present study is the review of these reliable techniques which are being used to solve a wide range of problems arising in nonlinear sciences. Several examples are given which reveal the efficiency and reliability of these relatively new techniques.

EXP-FUNCTION METHOD Consider the general nonlinear partial differential equation of the type ( )t x tt xx xxxxP u , u , u , u , u ,u , 0=… (1) Using a transformation kx tη = + ω (2) where k and ω are constants, we can rewrite equation (1) in the following nonlinear ODE;

( )(iv)Q u , u , u , u , u , 0′ ′′ ′′′ =… (3) According to the exp -function method, which was developed by He and Wu [14-17, 25], we assume that the wave solutions can be expressed in the following form

d

nn cq

mm p

a exp[n ]u( )

b exp[m ]=−

=−

ηη =

η∑∑

(4)

where p,q,c and d are positive integers which are known to be further determined, an and bm are unknown constants. We can rewrite equation (4) in the following equivalent form.

c dp q

a exp[c ] ... a exp[ d ]u( )

b exp[p ] ... b exp[ q ]−

−

η + + − ηη =

η + + − η (5)

This equivalent formulation plays an important and fundamental part for finding the analytic solution of

problems [14-17, 25]. To determine the value of c and p, we balance the linear term of highest order of equation (4) with the highest order nonlinear term. Similarly, to determine the value of d and q, we balance the linear term of lowest order of equation (3) with lowest order non linear term.

VARIATIONAL ITERATION METHOD (VIM) To illustrate the basic concept of the He’s VIM, we consider the following general differential equation Lu Nu g(x)+ = (6) where L is a linear operator, N a nonlinear operator and g(x) is the inhomogeneous term. According to variational iteration method [14-17, 25], we can construct a correction functional as follows

x

n 1 n n n0

u (x) u (x) (Lu (s) Nu (s) g(s))ds+ = + λ + −∫ % (7)

where λ is a Lagrange multiplier [14-17], which can be identified optimally via variational iteration method. The subscripts n denote the nth approximation, nu% is considered as a restricted variation. i.e. nu 0;δ =% (7) is called a correction functional. The solution of the linear problems can be solved in a single iteration step due to the exact identification of the Lagrange multiplier. The principles of variational iteration method and its applicability for various kinds of differential equations are given in [14-17]. In this method, it is required first to determine the Lagrange multiplier λ optimally. The successive approximation un+1, n≥0 of the solution u will be readily obtained upon using the determined Lagrange multiplier and any selective function u0, consequently, the solution is given by nnu limu .→∞= We

summarize some useful iteration formulae [14-17] which would be used in the subsequent section:

( )

( )( )t

/ /n 1 n n n n

0

u f u,u 0

u (t) u (t) u (s) f u ,u ds+

′ ′+ =

= − +

∫ (3a)

( )

( ) ( )( )t

// / //n 1 n n n n n

0

u f u ,u ,u 0

u (t) u (t) s t u (s) f u ,u ,u ds+

′′ ′ ′′+ =

= + − +

∫ (3b)

( )

( ) ( )( )t

2 /// / // ///n 1 n n n n n n

0

u f u , u , u , u 0

1u (t) u (t) s t u (s) f u ,u ,u ,u ds

2!+

′′′ ′ ′′ ′′′+ =

= − − +

∫ (3c)


99

( )

( ) ( )( )

(iv) (iv)

t3 (iv) / // /// (iv)

n 1 n n n n n n n0

u f u , u , u , u ,u 0

1u (t) u (t) s t u (s) f u ,u ,u , u ,u ds

3!+

′ ′′ ′′′+ =

= + − +

∫ (3d)

( )

( ) ( ) ( ) ( )( )

(n) (iv) (n)

tn n 1 (n) / // /// (iv) (n)

n 1 n n n n n n n n0

u f u , u , u , u ,u , ,u 0

1u (t) u (t) 1 s t u (s) f u ,u ,u ,u , u , , u dsn 1 !

−+

′ ′′ ′′′+ =

= + − − +−

∫

L

L (3e)

HOMOTOPY PERTURBATION METHOD

(HPM) AND HE’S POLYNOMIALS To explain the He’s homotopy perturbation method, we consider a general equation of the type, L(u) 0= (8) where L is any integral or differential operator. We define a convex homotopy H(u,p) by H(u,p) (1 p)F(u) pL(u)= − + (9) where F(u) is a functional operator with known solutions ν0, which can be obtained easily. It is clear that, for H(u,p) 0= (10) we have

H(u,0) F(u)= , H(u,1) L(u)= This shows that H(u,p) continuously traces an implicitly defined curve from a starting point H(ν0,p) to a solution function H(ƒ,1). The embedding parameter monotonically increases from zero to unit as the trivial problem F(u) = 0, is continuously deforms the original problem L(u) = 0. The embedding parameter p∈(0,1] can be considered as an expanding parameter [14-17, 25]. The homotopy perturbation method uses the homotopy parameter p as an expanding parameter [14-17, 25] to obtain

2

i 2 3i 0 1 3

i 0

u p u u p u p u p u∞

=

= = + + + +∑ L (11)

if p→1, then (11) corresponds to (9) and becomes the approximate solution of the form,

ip 1i 0

f limu u∞

→=

= = ∑ (12)

It is well known that series (12) is convergent for most of the cases and also the rate of convergence is

dependent on L (u) [14-17, 25]. We assume that (12) has a unique solution. The comparisons of like powers of p give solutions of various orders. In sum, according to [14-17], He’s HPM considers the nonlinear term N(u) as

2

i 2i 0 1

i 0

N(u) p H H pH p H ...∞

=

= = + + +∑

where Hn’s are the so-called He’s polynomials [11, 25], which can be calculated by using the formula

n ni

n 0 n ini 0 p 0

1H (u , ,u ) N( p u )

n! p = =

∂= ∂

∑… , n 0,1,2,= … .

ADOMIAN’S DECOMPOSITION

METHOD (ADM) Consider the differential equation [1, 2, 32, 33]. Lu R u Nu g+ + = (13) where L is the highest-order derivative which is assumed to be invertible, R is a linear differential operator of order lesser order than L, Nu represents the nonlinear terms and g is the source term. Applying the inverse operator L−1 to both sides of (13) and using the given conditions, we obtain

1 1u f L (Ru) L (Nu)− −= − −

where the function ƒ represents the terms arising from integrating the source term g and by using the given conditions. Adomian’s decomposition method [1, 2, 32, 33] defines the solution u(x) by the series

nn 0

u(x) u ( x )∞

=

=∑ where the components un(x) are usually determined recurrently by using the relation


100

1 1

0 k 1 k ku f , u L ( R u ) L ( N u ) , k 0− −

+= = − ≥ The nonlinear operator F(u) can be decomposed into an infinite series of polynomials

nn 0

F(u) A∞

=

=∑ where An are the so-called Adomian’s polynomials that can be generated for various classes of nonlinearities according to the specific algorithm developed in [1, 2, 32, 33] which yields

( )n n

in in

i 0 0

1 dA N u , n 0,1,2,

n! d = λ =

= λ = λ ∑ L

MODIFIED VARIATIONAL

ITERATION METHODS (MVIMS) The modified variational iteration techniques (MVIMS) are obtained by the elegant coupling of correction functional of VIM with He’s and Adomian’s polynomials. Variational Iteration Method Using He’s Polynomials (VIMHP): This modified version of variational iteration method [25] is obtained by the elegant coupling of correction functional (7) of Variational Iteration Method [25] (VIM) with He’s polynomials [13, 14] and is given by

(n)nx

n 0(n)n 0

n 0 (n)0n

n 0

x

0

p L(u )

p u u (x) p (s) dsp N(u )

(s) g(s)ds

∞

∞=

∞=

=

= + λ

+

− λ

∑∑ ∫

∑

∫

% (14)

Comparisons of like powers of p give solutions of various orders. Variational Iteration Method Using Adomian’s Polynomials (VIMAP): This modified version of VIM is obtained by the coupling of correction functional (3) of VIM with Adomian’s polynomials [25] and is given by

t

n 1 n n nn 00

u (x) u (x) ( L u ( x ) A g(x))dx∞

+=

= + λ + −∑∫ (15) where An are the so-called Adomian’s polynomials and are calculated for various classes of nonlinearities by using the specific algorithm developed in [25].

HOMOTOPY ANALYSIS METHOD (HAM)

Consider the following differential equation ( )N u x,t 0 = (16) where N is a nonlinear operator for this problem, x and t denote an independent variables, u(x,t) is an unknown function. In the frame of HAM [27], we can construct the following zeroth-order deformation: ( ) ( ) ( )( ) ( ) ( )( )01 q L U x,t;q u x,t q H x,t N U x,t;q− − = h (17) where q∈[0,1] is the embedding parameter, 0≠h is an auxiliary parameter, H(x,t)≠0 is an auxiliary function, L is an auxiliary linear operator, u0(x,t) is an initial guess of u(x,t) and U(x,t;q) is an unknown function on the independent variables x, t and q. Obviously, when q = 0 and q = 1, it holds ( ) ( )0U x,t;0 u x,t ,= ( ) ( )U x,t;1 u x,t= (18) respectively. Using the parameter q, we expand U(x,t;q) in Taylor series as follows:

( ) ( ) ( ) m0 mm 1

U x,t;q u x,t u x,t q∞

=

= + ∑ (19)

where

( )m

m m

U t;q1u

q 0m! q∂

==∂

(20)

Assume that the auxiliary linear operator, the initial guess, the auxiliary parameter h and the auxiliary function H(x,t) are selected such that the series (19) is convergent at q = 1, then due to (18) we have

( ) ( ) ( )0 mm 1

u x,t u x,t u x,t∞

=

= + ∑ (21)

Let us define the vector

( ) ( ) ( ) ( )}{n 0 1 nu x,t u x,t , u x,t ,...,u x,t=r (22) Differentiating (17) m times with respect to the embedding parameter q, then setting q = 0 and finally dividing them by m!, we have the so-called mth-order deformation equation ( ) ( ) ( ) ( )m m m 1 m m 1L u x,t u x,t H x,t R u− − − χ =

rh (23) where


101

( ) ( )( )( )m 1

m m 1 m 1

N U t;q1R u

q 0m 1 ! q

−

− −

∂=

=− ∂r

(24)

and

m0 m 11 m 1

≤χ = >

(25)

Finally, for the purpose of computation, we will approximate the HAM solution (21) by the following truncated series:

( ) ( )m 1

m kk 0

t u t−

=

φ = ∑

MODIFIED DECOMPOSITION

METHOD (MDM) Consider the following general functional equations: f(x) 0= (26) To convey the idea of the modified decomposition method [13], we rewrite the above equation as: y N(y) c= + (27)

where N is a nonlinear operator from a banach space B→B and f is a known function. We are looking for a solution of equation (26) having the series form:

ii 0

y y∞

=

= ∑ (28) The nonlinear operator N can be decomposed as

i i 1

i 0 j ji 0 i 0 j 0 j 0

N y N(y ) N y N y∞ ∞ −

= = = =

= + − ∑ ∑ ∑ ∑ (29)

From equations (28) and (29), equation (27) is equivalent to

i i 1

i 0 j ji 0 i 0 j 0 j 0

y c N(y ) N y N y∞ ∞ −

= = = =

= + + − ∑ ∑ ∑ ∑ (30)

We define the following recurrence relation:

0

1 0

m 1 0 m 0 m 1

y cy N(y )

y N(y ... y ) N(y ... y ), m 1,2,3,...+ −

= = = + + − + + =

(31)

then

1 m 1 0 m(y ... y ) N(y ... y ), m 1,2,3,...++ + = + + =

and

ii 1

y f y∞

=

= + ∑

if N is a contraction, i.e. ||N(x) N(y)|| | |x y||, 0 K 1− ≤ − < < , then

m 1 0 m 0 m 1

mm 0

| |y || | |N(y ... y ) N(y ... y )||

K | | y || K | | y ||, m 0,1,2,3,...+ −= + + − + +

≤ ≤ =

and the series ii 1 y∞

==∑ absolutely and uniformly

converges to a solution of equation (26) [13], which is unique, in view of the Banach fixed-point theorem.

NUMERICAL APPLICATIONS In this section, we apply the VIM, HPM, MVIMS, exp-function method, expansion of parameter, HAM and MDM for the solution of initial and boundary value problems of physical nature. Example 9.1: Consider the following “good” Boussinesq equation ( )2tt xxxx xx xxu u u u= − + + (6b) Introducing a transformation as η = kx+ωt, we can covert equation (6b) into ordinary differential equations

( ) ( )iv2 4 2 2 2u k u k u k u 0′′′′ ′′ω + − − = (7b) The solution of the equation (7b) can be expressed in the form, equation (6b) as

( ) [ ] [ ][ ] [ ]c d

p q

a exp c ... a exp du

b exp p ... b exp q−

−

η + + − ηη =

η + + − η

To determine the value of c and p, we balance the linear term of highest order of equation (7b) with the highest order nonlinear term. Proceeding as before, we obtain p = c and d = q. Case 7.2.1: We can freely choose the values of p,c,d but we will illustrate that the final solution does not strongly depends upon the choice of values of c and d. For simplicity, we set p = c = 1 and q = d = 1, hence equation (6b) reduces to the following form:

( ) [ ] [ ][ ] [ ]1 0 1

1 0 1

a exp a a expu

b exp a b exp−

−

η + + −ηη =

η + + −η


102

Substituting equation (7b), we have

( ) ( ) ( ) ( )

( ) ( )( ) ( )

4 3 2 1

0 1 2

3 4

1 [c exp 4 c exp 3 c exp 2 c expA

c c exp c exp 2

c exp 3 c exp 4 ] 0− −

− −

η + η + η + η

+ + −η + − η

+ − η + − η =

(8b)

where

( ) ( )( )51 0 1A= b exp b b exp−η + + −η

( )ic i 4, 3,......,3,4= − − are constants obtained by Maple 11. Equating the coefficients of exp(nη) to be zero, we obtain:

4 3 2 10 1 2 3 4

{ c 0 , c 0, c 0, c 0,c 0 , c 0 , c 0 , c 0 , c 0}

− − − −= = = == = = = =

(9b)

Fig. 1: Depicts the soliton solutions of equation (6b),

when a0 = a−1 = ω = k = 1. In case k is an imaginary number, the obtained soliton solutions can be converted into periodic or compact-like solutions. Therefore, we write k = iK, consequently, equation (11b) becomes

Solution of (9b) will yield

( )

( ) ( )

4 2 2200

1 1 1 0 21

2 4 2 2 2 4 21 0

1 1 0 02 21

b 5k k1 b 1b , b b , a ,4 b 2 k

b k k b k k1 1a , a , b b

2 k 8 k b

− −−

−−

−

+ − ω= = =− ω = ω

ω + − ω + −= = =

(10b)

We, therefore, obtained the following generalized solitary solution u(x,t) of equation (6b):

( ) ( ) ( ) ( ) ( )

( )( )

2 2 2 4 4 2 2 2 2 40 0 12kx 2 t kx t

2 2 21

2kx 2 t2kx t0

0 11

b k k b 5k k b k k1 1 1e e8 k b 2 k 2 ku(x,t)

1 b e b e b4 b

−+ ω + ω

−+ ω

+ ω−

−

ω − + + − ω ω − +− +

=

+ +

or simply, we have

( ) ( )( ) ( )

2 4 2 20

222kx 2 t kx t0

0 11

k k 3 b ku x,t1 b2k e b e b4 b

+ ω +ω−

−

ω + −= −

+ + (11b)

where b0, b−1, ω and k are real numbers.

( ) ( )( ) ( )

2 4 2 20

22IKx t IKx t0

0 11

K K 3 b Ku x,t1 b2K e b b e4 b

+ω − −ω−

−

ω + += +

− + + (12b)

( )( )

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

2 4 2 20

22t t0

0 11

K K 3 b Ku x,t

1 b2K e cos Kx Isin Kx b b e cos Kx Isin Kx4 b

ω −ω−

−

ω + += +

− + + + + (13c)

Now, to obtain periodic solutions from soliton solutions, we put imaginary part of equation (13b) which is sin kx equal to zero, hence equation (13b) becomes

( )( )

( ) ( ) ( ) ( )

2 4 2 20

22t t0

0 11

K K 3 b Ku x,t

1 b2K e cos Kx b b e cos Kx4 b

ω −ω−

−

ω + += +

− + + (14b)


103

Fig. 2: Depicts the periodic solutions of equation (6b)

drawn on the whole domain when b−1 = b0 = k = ω = 1


drawn on the whole domain when b−1 = b0 = k = ω = 1


when b−1 = b0 = ω = 1 and K = 2 Case 7. 2.2: If p = c = 2 and q = d = 1, than equation (6b) reduces to

( ) [ ] [ ] [ ][ ] [ ] [ ]2 1 0 1

2 1 0 1

a exp 2 a exp a a expu

b exp 2 b exp b b exp−

−

η + η + + −ηη =

η + η + + −η (15b)

Proceeding as before, we obtain

( )

( )

( )

4 2 2211

2 1 1 0 0 1 20

2 4 20

1 0 2

2 2 4 21

2 0 0 120

b 5k k1 b 1b , b b , b b , a

4 b 2 k

b k k1, b 0 , a

2 kb k k1

a ,b b ,a 08 k b

−

−

+ − ω= = = = −

ω + −ω = ω = =

ω + −= = =

(16b)

Hence, we get the generalized solitary solution u(x,t) of equation (6b) as follows

( ) ( ) ( ) ( ) ( )

( )( )

2 2 2 4 4 2 2 2 2 41 1 02kx 2 t kx t

2 2 20

2kx 2 t2kx t1

1 00

b k k b 5k k b k k1 1 1e e8 k b 2 k 2 ku(x,t)

1 b e b e b4 b

+ ω + ω

+ ω+ ω

ω − + + − ω ω − +− +

=

+ +

or simply, we have

( ) ( )( ) ( )

2 4 2 21

222kx 2 t kx t1

1 00

k k 3 b ku x,t1 b2k e b e b4 b

+ ω +ω

ω + −= −

+ + (17b)

where b0, b1, ω and k are real numbers. Example 9.2: Consider the following nonlinear boundary value problem of twelfth order

(xii) x 21y (x) e y (x), 0 x 12

−= < <

with boundary conditions (iv) (vi) (viii) ( x )

(iv) (vi) (viii) (x)

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

y(1) y (1) y (1) y (1) y (1) y (1) 2e

′′= = = = = =

′′= = = = = =


104

The exact solution of the problem is

xy(x) 2e= The correction functional is given as

x 12x 2n

n 1 n n120

d y 1y (x) y (x) (s) e y (x) ds

dx 2−

+

= + λ −

∫ %

Making the correction functional stationary, the Lagrange multiplier is identified as

( ) ( )111s s x11!

λ = −

we get the following iterative formula

( )x 12

11 x 2nn 1 n n12

0

1 d y 1y (x) y (x) s x e y (x) ds

11! d x 2−

+

= + − −

∫

where

(3) (5)

(7) (9) (11)

A y(0) ,B y (0),C y (0),

D y (0),E y (0),F y (0)

′= = =

= = =

Consequently, following approximants are obtained

0y ( x ) 2=

2 3 4 5 6 7 8 9 10 11 12 13

1

1 2 1 2 1 2 1 2 1 2 2y (x) 2 Ax x Bx x Cx x Dx x Ex x Fx x x

3! 4! 5! 6! 7! 8! 9! 10! 11! 12! 13!= + + + + + + + + + + + + − + L

2 3 4 5 6 7 8 9 10 11 12 13 13

2

1 2 1 2 1 2 1 2 1 2 2 2y ( x ) 2 Ax x Bx x Cx x Dx x Ex x Fx x x Ax

3! 4! 5! 6! 7! 8! 9! 10! 11! 12! 13! 13!= + + + + + + + + + + + + − + +L

M

The series solution is given as

2 3 4 5 6 7 8 9 10 11 12 132 1 2 1 1 1 2 1 2 1 2 2y(x) 2 Ax x Bx x Cx x Dx x Ex x Fx x (A 1)x2! 3! 4! 5! 6! 7! 8! 9! 10! 11! 12! 13!

= + + + + + + + + + + + + + − +L

Imposing the boundary conditions at x = 1, will yield

A 2.002114383, B 1.986107510, C 2.333176702, D 3.001515917, E 1.986107510, F 2.102525395= = = = = = The series solution is given as

2 3 4 5 6 7 8

5 9 10 7 11 12 9 13

1 1 1y(x) 2 2.002114783x x 0.329322880x x 0.01944313918x x 0.0005955388724x x12 720 20160

1 10.5473179867 10 x x 0.5267269408 10 x x 0.3218601046 10 x

1814400 239500800− − −

= + + + + + + + +

+ × + + × + + × +L

Fig. 5: Clearly indicates the accuracy of the proposed VIM


105

Table 1: Error estimates

x Exact solution Series solution *Errors

0.0 2.0000000000 2.0000000000 0.0000000000

0.1 2.2103418362 2.2105496398 0.0002078037

0.2 2.4428055163 2.4431996667 0.0003941504

0.3 2.6997176152 2.7002578996 0.0005402845

0.4 2.9836493953 2.9842815093 0.0006321141

0.5 3.2974425414 3.2981039180 0.0006613766

0.6 3.6442376008 3.6448637056 0.0006261048

0.7 4.0275054149 4.0280359083 0.0005304933

0.8 4.4510818570 4.4514661221 0.0003842651

0.9 4.9192062223 4.9194078617 0.0002016394

1.0 5.4365636569 5.4365636559 0.0002016394

*Error = Exact solution-Series solution

Table 1 exhibits the exact solution and the series solution along with the errors obtained by using the Variational Iteration Method (VIM). It is obvious that the errors can be reduced further and higher accuracy can be obtained by evaluating more components of y(x). Example 9.3: Consider the following one-dimensional Burger’s equation

t xx xxu u u vu 0+ − =

with initial condition

( ) ( )( )

expu(x,0) , t 0

1 expα + β + β − α γ

= ≥+ γ

where

( )( )xvαγ = − λ and the parameters α,β,λ re the arbitrary constants. The correction functional is given by

( )t 2 2

n n nn 1 n n 2 2

0

u u uu (x,t) u (x,t) s u v ds

t x x+ ∂ ∂ ∂

= + λ + − ∂ ∂ ∂ ∫

% %%

Making the above functional stationary, the Lagrange multiplier can be determined as λ(s) = -1, we get the following iterative scheme

t 2 2n n n

n 1 n n 2 20

u u uu (x,t) u (x,t) u v ds

t x x+ ∂ ∂ ∂

= − + − ∂ ∂ ∂ ∫

Applying the Modified Variational Iteration Method (MVIM)

( ) ( )( )

( )

20 1 2

t20 1 2

0

2 22 0 1

0 1 2 2 2t

2 20 0 1

2 2

expu pu p u

1 exp

u u up p p dst t t

u uu pu p u p

x xp ds

u uv p

x x

α + β + β − α γ+ + + = + γ

∂ ∂ ∂ − + + + ∂ ∂ ∂

∂ ∂+ + + + + ∂ ∂ − ∂ ∂ − + + ∂ ∂

∫

∫

L

L

L L

L

Comparing the co-efficient of like powers of p

( ) ( )( )

(0)0

expp : u (x,t)

1 expα + β + β − α γ

=+ γ

( ) ( )( )

( )( )( )

2(1)

1

exp 2 expp : u ( x , t ) t

1 exp v 1 exp

α + β + β − α γ αβ γ= +

+ γ + γ

( ) ( )( )

( )( )( )

( ) ( )( )( )( )

2(2)

2

3 2

332


1 exp v 1 exp

exp 1 expt

v 1 exp

α + β + β − α γ αβ γ = + + γ + γ

α β γ − + γ + + γ

( ) ( )( )

( )( )( )

( ) ( )( )( )( )

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

2(3)

3

3 2

332

24 3

443


1 exp v 1 exp

exp 1 expt

v 1 exp

exp 1 4exp expt

3v 1 exp

α + β + β − α γ αβ γ = + + γ + γ α β γ − + γ + + γ α β γ − γ + γ +

+ γ

M The series solution is given by

( ) ( )( )

( )( )( )

( ) ( )( )( )( )

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

2

3 2

332

24 3

443

exp 2 expu(x,t) t

1 exp v 1 exp

exp 1 expt

v 1 exp

exp 1 4exp expt

3v 1 exp

α + β + β − α γ αβ γ = + + γ + γ α β γ − + γ + + γ α β γ − γ + γ + +

+ γ

L

and in a closed form by

( ) ( )( )( )( )

xexp x tvu(x,t)x1 exp x tv

α + β + β − α −β − λ = + − β − λ


106

Fig. 6: α = β = ν = λ = 1

Fig. 7: α = ν = 1, β = λ = 2

Fig. 8: α = 1, β = 2, ν = 4, λ = 3 Example 9.4: Consider the following singularly perturbed fourth-order Boussinesq equation

2tt xx xx xxxxu u 3(u ) u= + +

with initial conditions

2 kx

k x 2

2 a k eu(x,0)(1 a e )

=+

Fig. 9: α = 1, β = 0.2, ν = 0.5, λ = 1

Fig. 10: α = 1, β = 0.2, ν = 0.5, λ = 0.4

3 2 kx kx

t kx 3

2ak 1 k (1 a e )eu(x,0)

(1 a e )+ −

=+

where a and k are arbitrary constants. The exact solution u(x,t) of the above initial value problem is given as

)(2 2

22

a k exp(kx k 1 k t)u(x,t) 2

1 aexp(kx k 1 k t )

+ +=

+ + +

The correction functional is given as

( ) ( )

2

2 kx 3 2 kx kx

n 1 kx 3kx

t 2n

n n n2 xx xxxxn 00

2ak e 2ak 1 k (1 ae )eu (x,t) t(1 ae )(1 a e )

uu u 3 B dt

t

+

∞

=

+ −= +++

∂+ λ − − − ∂

∑∫ % %

where Bn are Adomian’s polynomials for nonlinear operator F(u) = u2(x) and can be generated for all type of nonlinearities according to the algorithm developed in [1, 2, 32, 33] which yields the following


107

( )( ) ( )

20 0 1 0 1xx 0 x 1x 0xx 1,x x

2

2 0 2xx 0 x 2x 0 x x 2 1x 1xx 1x 3 0 3 1 2, xx

B u ,B 2 u u 4u u 2u u

B 2 u u 4u u 2 u u 2u u 2 u B 2 u u 2 u u

= = + +

= + + + + = +

M

Making the correction functional stationary, the Lagrange multiplier can be identified as λ(s) = (s-t), consequently

( ) ( ) ( )2t2 k x 3 2 k x k x 2

nn 1 n n nkx 3 2 xx xxxxkx

n 00

2 a k e 2ak 1 k (1 a e )e uu (x,t) t s t u u 3 B ds

(1 ae ) t(1 a e )

∞

+=

+ − ∂= + + − − − − + ∂+

∑∫ (7)

As per the traditional approach, we are to use

2

2 kx 3 2 kx kx

kx 3kx

2ak e 2ak 1 k (1 ae )et

(1 ae )(1 a e )

+ −+

++

as the initial approximation, but following the modified version, we selected 22 kx

kx

2ak e(1 a e )+

as the initial value.

Consequently, following approximants are obtained

2 k x

0 k x 2

2 a k eu (x,t)(1 ae )

=+

( )( )

x x 2 x2 k x 3 2 kx kx2

1 4kx 2 kx 3 x

2e 1 4e e2 a k e 2ak 1 k (1 a e )eu (x,t) t t

(1 ae ) (1 a e ) 1 e

− ++ −= + +

+ + +

( )( )

( )( )( )

( )( )( )

x x 2x x x x 2x2 k x 3 2 kx kx2 3

2 4 5k x 2 kx 3 x x

x x 2x x 2x 3x 4x4

8x

2e 1 4e e 2 2 e 1 e 1 10e e2 a k e 2ak 1 k (1 a e )eu ( x , t ) t t t(1 ae ) (1 a e ) 1 e 3 1 e

e 1 4e e 1 44e 78e 44e et

3 1 e

− + − + − ++ −= + + −+ + + +

− + − + − ++

+

( )( )

( )( )( )

( )( )( )

( )( )( )

x x 2x x x x 2x2 k x 3 2 kx kx2 3

3 4 5k x 2 kx 3 x x

x x 2x x 2x 3x 4x4

8x

x x x 2x 3x 4x

57x

x x

2e 1 4e e 2 2 e 1 e 1 10e e2 a k e 2ak 1 k (1 a e )eu (x,t) t t t

(1 ae ) (1 a e ) 1 e 3 1 e

e 1 4e e 1 44e 78e 44e et

3 1 e

2 e 1 e 1 56e 246e 56e et

1 5 1 e

e 1 452e 1914

− + − + − ++ −= + + −

+ + + +

− + − + − ++

+

− + − + − +−

+

− ++

( )( )

( )( )

2 x 3x 4x 5x

612x

x 6 x 7x 8x 9x 1 0 x

612x

9e 207936e 807378e 1256568et

4 5 1 e

e 807378e 207936e 19149e 452e et

4 5 1 e

− + −

+

− + − ++

+

M

The series solution is given by


108

Table 2: Error estimates

t j

---------------------------------------------------------------------------------------------------------------------------------------------------------------- xi 0.01 0.02 0.04 0.1 0.2 0.5

-1 2.80886 E-14 1.79667 E-12 1.15235 E-10 2.83355 E-8 1.83899 E-6 4.74681 E-4 -0.8 6.27276 E-14 4.01362 E-12 2.57471 E-10 6.33178 E-8 4.10454 E-6 1.04489 E-3

-0.6 6.08402 E-14 3.90188 E-12 2.25663 E-10 6.18024 E-8 4.02299 E-6 1.03093 E-3 -0.4 1.16573 E-14 7.41129 E-13 4.82756 E-11 1.23843 E-8 8.53800 E-6 2.46302 E-4

-0.2 5.53446 E-14 3.53395 E-12 2.25663 E-10 5.47485 E-8 3.47264 E-6 8.35783 E-4

0.0 8.63198 E-14 5.53357 E-12 2.54174 E-10 8.65197 E-8 5.54893 E-6 1.37353 E-3 0.2 5.56222 E-14 3.55044 E-12 2.27779 E-10 5.60362 E-8 3.63600 E-6 9.29612 E-4

0.4 1.14353 E-14 7.14928 E-13 4.49107 E-11 1.03370 E-8 5.93842 E-7 9.61260 E-5 0.6 6.06182 E-14 3.87551 E-12 2.47218 E-10 5.97562 E-8 3.76275 E-6 8.79002 E-4

0.8 6.23945 E-14 3.99519 E-12 2.55127 E-10 6.18881 E-8 3.92220 E-6 9.36404 E-4 1.0 2.79776 E-14 1.78946 E-12 1.14307 E-10 2.77684 E-8 1.76607 E-6 4.28986 E-4

( )( )

( )( )( )

( )( )( )

( )( )

( )

x x 2x x x x 2xx 3 2 kx kx2 3

4 5x 2 kx 3 x x

x x 2x x 2x 3x 4x 2x x 2x 3x 4x4 4

8 8x x

x x x 2x 3x

2 e 1 4e e 2 2 e 1 e 1 10e e2e 2 a k 1 k (1 a e )eu(x,t) t t t

(1 e ) (1 a e ) 1 e 3 1 e

e 1 4e e 1 44e 78e 44e e 8e 1 10e 20e 10e et t

3 1 e 1 e

2 e 1 e 1 56e 246e 56e

− + − + − ++ −= + + −

+ + + +

− + − + − + − + − ++ +

+ +

− + − + − +−

( )( )

( )( )

( )( )

4x5

7x

x x 2x 3x 4x 5x6

12x

x 6x 7 x 8 x 9x 10x6

12x

et

1 5 1 e

e 1 452e 19149e 207936e 807378e 1256568et

4 5 1 e

e 807378e 207936e 19149e 452e et

45 1 e

+

− + − + −+

+

− + − ++ +

+L

Fig. 11: Table 2 exhibits the absolute error between the exact and the series solutions. Higher accuracy can be obtained by introducing some more components of the series solution. Example 9.5: Consider the Fisher’s equation of the following form

t xxu(x, t ) u (x,t) u(x,t)(1 u(x,t)) 0− − − = with initial conditions

u(x,0) = β The correction functional is given by

( ) ( )( ) ( )

( ) ( )( )

2n nt

2n 1 n

0n n

u x,s u x,s

s xu x,t u x,t dsu x, 1 u x,s

+

∂ ∂−

∂ ∂= + λ − τ −

∫%

% %

Making the above correction functional stationary, the Lagrange multiplier can be identified as λ(s) = -1, consequently, we obtain the following iteration formula

( ) ( )( ) ( )

( ) ( )( )

2n nt

2n 1 n

0n n

u x,s u x,s

s xu x,t u x,t dsu x,s 1 u x,s

+

∂ ∂−

∂ ∂= − − −

∫


109

Table 3 Numerical results for Fisher’s equation

β = 0.2 β = 0.8

-------------------------------------------------------------------------- ------------------------------------------------------------

t *E (ADM) *E(VIMHP) (VIMHPDM) *E (ADM) *E(VIMHP)

0 0 0 0 0

0.2 6.19201E-06 1.96407E-06 5.57339E-06 2.63705E-06

0.4 1.03635E-04 2.60211E-05 1.13137E-04 4.7283E-05

0.6 5.45505E-04 1.05333E-04 4.00147E-04 2.63379E-04

0.8 1.78050E-03 2.54998E-04 1.18584E-03 9.01304E-04

1 4.45699E-03 4.515414E-04 1.99502E-03 1.60455E-05

*Error = Exact solution-series solution Applying the variational iteration method using He’s polynomials (VIMHP), we get

( ) ( ) ( ) ( )

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

t0 1 22 2

0 1 2 00

2 2t0 1

0 1 0 12 20

u x,s u x,s u x,su pu p u u x,t p p p ds

s s s

u x,s u x,sp p u pu 1 u pu ds

x x

∂ ∂ ∂ + + + = − + + + ∂ ∂ ∂

∂ ∂ + + + + + + − + + ∂ ∂

∫

∫

L L

L L L

Comparing the co-efficient of like powers of p, following approximants are obtained

(0)0p : u (x,t) = β ,

(1)1p : u ( x , t ) (1 )t= β + β − β

2 3

(2) 2 2 22

t tp : u (x,t) (1 ) t (1 3 2 ) ( 1 )

2! 3= β + β −β + β − β + β − β − + β

3 4

(3) 2 2 23

5 6 72 2 2 3 3 4 4

t tp : u ( x , t ) (1 ) t (1 ) (1 6 6 ) ( 1 ) ( 1 2 )

3! 3t t t( 1 ) (3 20 20 ) ( 1 ) ( 1 2 ) ( 1 )60 18 63

= β + β − β + β − β − β + β + − +β β − + β

− − + β β − β + β + − + β β − + β − − + β β

M

The series solution is given by

3 42 2 2

5 6 72 2 2 3 3 4 4

t tu(x,t) (1 ) t (1 ) ( 1 6 6 ) ( 1 ) ( 1 2 )

3! 3t t t( 1 ) (3 20 20 ) ( 1 ) ( 1 2 ) ( 1 )60 18 63

= β + β − β + β − β − β + β + − +β β − + β

− − + β β − β + β + − + β β − + β − − +β β + L

and in a closed form by expt

u(x,t)1 expt

β=

− β + β

Table 3 exhibits errors obtained by applying Adomian’s Decomposition Method (ADM) and the VIMHP Example 9.6: Consider the following coupled nonlinear Klein-Gordon-Schrödinger equations

2tt xx t xxu u u v 0; i v v uv 0− + − = + + =

with the initial conditions

( )

( )

2 2

2 2t

2 idx

u x,0 6B sech(Bx)

u(x,0) 12Bcsech (Bx)tanh(Bx)

v x,0 3Bsech (Bx)e

=

= −

=

where 1

B2

≥ , c and d are arbitrary constants.


110

24B 1 c

c , d2 2B

− −= =

for the above coupled K-G-S. We can construct following homotopies :

2 220 0

2 2

u u u up u v

t t x t ∂ ∂ ∂ ∂

− = − + − ∂ ∂ ∂ ∂

2_

2 0 02

v v v vv v v p i iuv

t t x t ∂ ∂ ∂ ∂

= ⋅ − = + − ∂ ∂ ∂ ∂

We get the following set of differential equations

0 0

0

0 0

u u0

t tp :v v 0t t

∂ ∂ − = ∂ ∂

∂ ∂ − = ∂ ∂

2 2 _

1 00 0 02 2

1

21 0

0 02

u u u v vt x

p :v v

i iu vt x

∂ ∂= − + ⋅ ∂ ∂

∂ ∂ = + ∂ ∂

2 2 _ _

2 11 01 0 12 2

22 2 1

0 1 1 02

u uu v v v v

t x

v vp : u v u v

t x

∂ ∂= − + +

∂ ∂

∂ ∂= + +

∂ ∂

M

Solving the above equations using the initial conditions yield

0

0

u 0v 0

= =

( )

( )

2 2 2 21

2 idx1

u 6B sech Bx 12Bctsech (Bx)tanh(Bx)

v 3Btanh Bx e

= −

=

( ) ( )( ) ( )2 2 2 4 4 3 4 5 2 22idx 2 2 2 2 2

2

u t ( 3B sech Bx 9B sech Bx t 2Bcsech(Bx)( 11sinh(Bx) sinh(3Bx)) 2Bcsech(Bx)tanh(Bx)

v 3iBe tsech (Bx)(2B sech (Bx) (d 2iBtanh (Bx))

= − − + − − + + = − + +

M

The series solution is give by

2 2 2 2

2 4 5 2 2

2 idx idx 2 2 2 2 2

u(x,t) 6B sech (Bx) 12Bctsech (Bx)tanh(Bx)

t ( 2Bcsech(Bx)( 11sinh(Bx) sinh(3Bx) 2Bcsech (Bx)tanh(Bx)) ....

v(x,t) 3Bsech ( Bx)e 3iBe tsech (Bx)(2B sech (Bx)) (d 2iBtanh (Bx)) ....

= −

+ − − + + + = − + + +

Some numerical results of these solutions are presented in Fig. 12-15.

(a) (b) Fig. 12: (a) The homotopy perturbation solution for u(x,t), (b) Corresponding solution for u(x,t) when t=0


111

(a) (b) Fig. 13: (a) The homotopy perturbation solution for re(v(x,t)), (b) Corresponding solution for re (v(x,t)) when t=0

(a) (b) Fig 14: (a) The homotopy perturbation solution for im(v(x,t)), (b) Corresponding solution for im(v(x,t)) when t=0

(a) (b) Fig. 15: (a) The homotopy perturbation solution for abs(v(x,t)), (b) Corresponding solution for abs (v(x,t)) when t=0


112

Example 9.6: Consider the following Duffing harmonic oscillator

( ) ( )3

tt t2

uu , u 0 A, u 0 0

1 u= − = =

+

The above problem can be re-written as

2 3tt ttu o.u 1.u u 1.u 0+ + + =

Assuming that the solution of above equation and co-efficient 0, 1 can be expressed as power series in p as follows

2 21 10 pa p a= ω + + +L

21 21 pb p b= + +L

Consequently, we have

( ) ( )20 0 tu u 0 , u 0 A , u 0 0′′ + ω = = =

( ) ( )2 2 31 1 1 0 1 0 0 1 0 tu u a u b u u b u 0,u 0 0,u 0 0′′ ′′+ ω + + + = = = Solving the above, we have u0 = A cox ωt, substituting u0 gives

( )

( )

2 2 21 1 1 1

3 21

3u u Acos t a b A 14

1b A 1 cos3 t 0

4

′′ + ω + ω + − ω

+ − ω ω =

Elimination of the secular term requires

( )2 21 13a b A 14= − − ω If only the first-order approximation is searched for, then a1 = -ω

2b1, is obtained which leads to

22

2

3A4 3A

ω =+

Example 9.7: Consider the initial value problem B (3, 3)

( ) ( )3 3tt xx xxxxu u u 0+ − = (23a)

( ) 6 xu x,0 vsinh2 3

=

( ) 2t6 xu x,0 v cosh

6 3 = −

(23b)

where v is an arbitrary constant. According to (5), the zeroth-order deformation can ve given by

( ) ( ) ( )( )

( ) ( ) ( )( )0

3 3tt xx xxxx

1 q L U x,t;q u x, t

q H x,t U U U

− −

= + −h (24)

We can start with an initial approximation

( ) 206 x 6 xu x,t vsinh vtcosh

2 3 6 3 = −

and we choose the auxiliary linear operator

( )( ) ( )2

2

U x,t;qL U x,t;q

t∂

=∂

with the property ( )L Cx D 0+ =

where C and D are integral constants. We also choose the auxiliary function to be

( )H x,t 1= Hence, the mth-order deformation can be given by

( ) ( ) ( ) ( )m m m 1 m m 1L u x,t u x,t H x,t R u− − − χ = rh

where

( ) ( )

2 2 m 1 m 1 im 1

m m 1 i k m 1 i k2 2i 0 k 0

4 m 1 m 1 i

i k m 1 i k4i 0 k 0

uR u u u u

t x

u u ux

− − −−

− − − −= =

− − −

− − −= =

∂ ∂ = + ∂ ∂ ∂ − ∂

∑ ∑

∑ ∑

r

(25)

Now the solution of the mth-order deformation equations (25) for m≥1 become ( ) ( ) ( )1m m m 1 m m 1u x,t u x,t L R u−− − = χ +

rh (26) Consequently, the first few terms of the HAM series solution are as follows:

( ) 206 x 6 xu x,t vsinh vtcosh

2 3 6 3 = −

( ) 3 2 4 31

5 4 6 5

6 x 6 xu x,t v t sinh v t cosh36 3 324 3

6 x 6 xv t sinh v t cosh

1944 3 9720 3

= − +

+ −

h h

h h


113

(a) (b) Fig. 16: The surface shows the solution u(x,t): (a) exact solution for v=1, (b) exact solution for v=2

(a) (b) Fig. 17: The surface shows the solution u(x,t) for Eq.(29a-b): (a) exact solution for v=1, (b) exact solution for v=2 and so on. Hence, the HAM series solution (for 1)= −h i

( ) ( ) ( )0 1u x,t u x,t u x,t .....= + +

( ) 21

3 2 4 3

5 4 6 5

6 x 6 xu x,t vsinh vtcosh

2 3 6 3

6 x 6 xv t sinh v t cosh

36 3 324 3

6 x 6 xv t sinh v t cosh .....1944 3 9720 3

= −

+ −

− + +

h (27)

Using Taylor series into (27), we find the closed form solution

( ) 6 x vtu x,t vsinh2 3

− = (28)

To obtain another exact solution for B(3, 3), we consider the initial value problem of B(3, 3) equation

( ) ( )3 3tt xx xxxxu u u 0+ − = (29a)

( ) 6 xu x,0 vsinh2 3

= −

( ) 2t6 xu x,0 v cosh

6 3 =

(29b)

According to the similar steps as discussed above, we have another exact solution given by

( ) 6 x vtu x,t vsinh2 3

− = − (30)

Example 9.7: Consider the following Whitham-Broer-Kaup (WBK) equations which arise quite frequently in mathematical physics, nonlinear sciences and is of the form


114

t x x xxu u u v u 0+ + + β = (1)

t x xxx xxv (uv) u v 0+ + α − β =

with initial conditions

2 2u(x,0) 2Bkcoth(k ),v(x,0) 2B(B ) k csch ( k )= λ − ξ = − + β ξ

where

20 0B and x x and x ,k,= α + β ξ = + λ

are arbitrary constants. Applying modified decomposition method (MDM), consequently, following approximants are obtained

0

0

0

2 20

u ( x , t ) c

u ( x , t ) 2Bkcoth(k )v ( x , t ) c

v ( x , t ) 2B(B )k csch ( k )

= = λ − ξ = = − + β ξ

1 0

2 21 0

1 0

2 2 2 21

2 4 40

u(x , t ) Nu(x , t )

u (x , t ) 2Bkcoth(k ) 2Bk tcsch (k(x x ))v (x,t) N u ( x , t )

v (x,t) 2B(B )k csch (k ) 2B(B )k t( 2Bkcoth(k ))csc h ( k )

4B( B )k t(2 cosh(2k(x x )))csch (k )

=

= λ − ξ − λ + = = − +β ξ − + β −λ+ ξ ξ − α + β + β + + ξ

( )2 0 1 03 2

5 2 2 2 2 22

2 2 2 2 2 2

2

2

u ( x , t ) N u (x,t) u ( x , t ) Nu (x,t)

Bk tu ( x , t ) csch (k )(( 44 k 44B k 44 k B )cosh(k )

2(4 k 4B k 4 k B )cosh(3k ) (6Bk 6B k 6Bk 6 k )sinh(k )

(2Bk 2B k 2Bk 2 k )sinh(3k )

v ( x , t

= + −

= ξ − α − β − β − λ − β λ + λ ξ

− α + β + β − λ − β λ + λ ξ − + β − λ − β λ ξ

− + β − λ − β λ ξ

( )0 1 02 2

6 3 2 2 2 4 2 4 3 4 2 2 22

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

2 2 2 2 2

) N u (x,t) u(x , t ) N u ( x , t )

Bk tv ( x , t ) csch ( k ) ( 4 B k 4B k 528 k 528B k 528 k 12 k 8 B k8

16B k 24 k 3B 3 (416 k 416B k 416 k 8 k 8 B k

8B k 16 k 4B 4 )cosh(2k

= + −

= ξ + β − αβ − β − β − α λ + λ

− β λ − β λ − λ − βλ − αβ + β + β − α λ + λ

− β λ − β λ − λ − βλ 3 2 2 2 4 2 4 3 4

2 2 2 2 2 2 3 2 3 2 3

2 3 3 2 3 2 3

2 3

) (4Bk 4B k 16 k 16B k 16 k

4 k cosh(4k ) 8B k 8 k B )cosh(4k ) (32 Bk 112B k 112B k

8B k 8B k 80 k )sinh(2k ) (8 Bk 16B k 16B k

4 B k 4B k 8 k )sinh(4k ))

ξ − + β + αβ + β + β

− α λ ξ + β λ − β λ + λ +βλ ξ − α + β + β

+ λ + β λ + α λ ξ − α + β + β

− λ − β λ + α λ ξM

Hence, the closed form solutions are given as

u(x,t) 2Bkcoth(k( t))= λ − ξ − λ , 2 2v(x,t) 2B(B )k csch (k( t))= − +β ξ − λ

where 2B = α + β and ξ = x+x0 and x0, k, λ are arbitrary constants. As a special case, if α=1 and β = 0, WBK equations can be reduced to the modified Boussinesq (MB) equations.We shall second consider the initial conditions of the MB equations

u(x,0) 2kcoth(k ),= λ − ξ 2 2v(x,0) 2k csch (k )= − ξ

where ξ = x+x0 being arbitrary constant. Procedding as before, we obtain exact solution as follows

u(x,t) 2kcoth(k t),= λ − ξ − λ 2 2v(x,t) 2k csch ( k t)= − ξ − λ


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Table 4:The numerical results for φn(x,t) and ϕn(x,t) in comparison with the exact solution for u(x,t) and ν(x,t) when k = 0.1, λ = 0.005, α = 1.5,

β = 1.5 and x0 = 10, for the approximate solution of the WBK equation

t i/x i 0.1 0.2 0.3 0.4 0.5

|u-φn|

0.1 1.04892E-04 9.64474E-05 8.88312E-05 4.25408E-04 3.91098E-04

0.3 3.60161E-04 9.71992E-04 8.93309E-04 8.22452E-04 1.75596E-03

0.5 1.61430E-03 1.48578E-03 2.79519E-03 2.56714E-03 2.36184E-03

|ν-φn|

0.1 6.41419E-03 5.99783E-03 5.61507E-03 1.33181E-03 1.24441E-02 0.3 1.16416E-02 2.07641E-02 1.93852E-02 1.81209E-02 2.88100E-02

0.5 2.68724E-02 2.50985E-02 3.75193E-02 3.49617E-02 3.26239E-02

Table 5:The numerical results for φn(x,t) and ϕn(x,t) in comparison with the analytical solution for u(x,t) and ν(x,t) when k = 0.1, λ = 0.005, α =

1.5, β = 0 and x0 = 10 for the approximate solution of the MB equation

t i/x i 0.1 0.2 0.3 0.4 0.5

|u-φn| 0.1 8.16297E-07 7.64245E-07 7.16083E-07 3.26243E-06 3.05458E-06

0.3 2.86226E-06 7.33445E-06 6.86758E-06 6.43557E-06 1.30286E-05

0.5 1.22000E-05 1.14333E-05 2.03415E-05 1.90489E-05 1.78528E-05

|ν-φn|

0.1 5.88676E-05 5.56914E-05 5.27169E-05 1.18213E-04 1.11833E-04

0.3 1.05858E-04 1.78041E-04 1.68429E-04 1.59428E-04 2.38356E-04 0.5 2.25483E-04 2.13430E-04 2.99162E-04 2.83001E-04 2.67868E-04

Table 6:The numerical results for φn(x,t) and ϕn(x,t) in comparison with the analytical solution for u(x,t) and ν(x,t) when k = 0.1, λ = 0.005, α =

0, β = 0.5 and x0 = 10 for the approximate solution of the ALW equation

t i/x i 0.1 0.2 0.3 0.4 0.5

|u-φn|

0.1 8.02989E-06 7.38281E-06 6.79923E-06 3.23228E-05 2.97172E-05 0.3 2.73673E-05 7.32051E-05 6.73006E-05 6.19760E-05 1.31032E-04

0.5 1.20455E-04 1.10919E-04 2.06186E-04 1.89528E-04 1.74510E-04

|ν-φn|

0.1 4.81902E-04 4.50818E-04 4.22221E-04 9.76644E-04 9.13502E-04

0.3 8.55426E-04 1.48482E-03 1.38858E-03 1.30009E-03 2.00705E-03

0.5 1.87661E-03 1.75670E-03 2.54396E-03 2.37815E-03 2.22578E-03

where k, λ are constants to be determined and x0 is an arbitrary constant. In the last example, if α = 0 and β = 1/2, WBK equations can be reduced to the approximate long wave (ALW) equation in shallow water. We can compute the ALW equation with the initial conditions

u(x,0) kcoth(k ),= λ − ξ 2 2v(x,0) k csch (k )= − ξ

where k is constant to be determined and ξ = x+x0. Procedding as before, we obtain exact solution as follows

u(x,t) kcoth(k t),= λ − ξ − λ 2 2v(x,t) 2k csch ( k t)= − ξ − λ

where k,λ are constants to be determined and ξ = x+x0, x0 is an arbitrary constant. In order to verify numerically whether the proposed methodology lead to higher accuracy, we evaluate the numerical solutions using the n-term approximation. Table 4-6 show the difference of analytical solution and numerical solution of the absolute error. We achieved a very good approximation with the actual solution of the equations by using 5 terms only of the proposed MDM.

CONCLUSION In this paper, we made a detailed study of some relatively new techniques along with some of their


116

modifications. These proposed methods and their modifications are employed without using linearization, discretization, transformation or restrictive assumptions, absorb the positive features of the coupled techniques and hence are very much compatible with the diversified and versatile nature of the physical problems. It may be concluded that theses relatively new techniques can be treated as alternatives for solving a wide class of nonlinear problems.

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