Modified Variational Iteration Method for Goursat and ... - CiteSeerX

World Applied Sciences Journal 4 (4): 487-498, 2008 ISSN 1818-4952 © IDOSI Publications, 2008

Modified Variational Iteration Method for Goursat and Laplace Problems Muhammad Aslam Noor and Syed Tauseef Mohyud-Din Department of Mathematics, COMSATS Institute of Information Technology, Islamabad, Pakistan Abstract: In this paper, we apply the Modified Variational Iteration Method (MVIM) for solving Goursat and Laplace problems which play very important part in applied and engineering sciences. The proposed modification is made by introducing He’s polynomials in the correction functional. Several examples are given to verify the reliability and efficiency of the method. The fact that the proposed MVIM solves nonlinear problems without using Adomian’s polynomials can be considered as a clear advantage of this technique over the decomposition method. Key words : Variational iteration method He’s polynomials Laplace equations •

INTRODUCTION

•

nonlinear problems

•

Goursat problems

•

split the nonlinear term into a series of polynomials calling them as the He’s polynomials. Most recently, Noor and Mohyud-Din used this concept for solving nonlinear boundary value problems; see [23-29]. The basic motivation of this paper is the extension of the Modified Variational Iteration Method (MVIM) which is formulated by the coupling of variational iteration method and He’s polynomials for solving the Goursat and Laplace problems. The proposed MVIM provides the solution in a rapid convergent series which may lead the solution to a closed form. In this technique, the correction functional is developed [7-13] and the Lagrange multipliers are calculated optimally via variational theory. The use of Lagrange multipliers reduces the successive application of the integral operator and the cumbersome of huge computational work while still maintaining a very high level of accuracy. Finally, He’s polynomials are introduced in the correction functional and the comparison of like powers of p gives solutions of various orders. The proposed iterative scheme takes full advantage of variational iteration and the homotopy perturbation methods and absorbs all the positive features of the coupled techniques. It is worth mentioning that the suggested method is applied without any discretization, restrictive assumption or transformation and is free from round off errors. The proposed method work efficiently and the results so far are very encouraging. The MVIM has two major advantages over the decomposition method. Firstly, the calculation of He’s polynomials is much easier than Adomian’s polynomials [21-29]. Secondly, the use of Lagrange multiplier reduces successive application of

This paper is devoted to the study of Goursat and Laplace problems which are known to arise in a variety of physical phenomenon and applied sciences [1-6]. Several techniques including Runge-Kutta, decomposition, finite difference, finite element, geometric mean averaging of the functional values and variational iteration have been used to investigate these problems, see [1-6] and the references therein. He [719] developed the variational iteration and homotopy perturbation methods for solving linear, nonlinear, initial and boundary value problems. It is worth mentioning that the origin of variational iteration method can be traced back to Inokuti, Sekine and Mura [20], but the real potential of this technique was explored by He [7-13]. Moreover, He realized the physical significance of the variational iteration method, its compatibility with the physical problems and applied this promising technique to a wide class of linear and nonlinear, ordinary, partial, deterministic or stochastic differential equation; see [7-13]. The homotopy perturbation method [7, 14-19] was also developed by He by merging two techniques, the standard homotopy and the perturbation. The homotopy perturbation method was formulated by taking full advantage of the standard homotopy and perturbation methods. The variational iteration and homotopy perturbation methods have been applied to a wide class of functional equations; see [7-47] and the references therein. In these methods the solution is given in an infinite series usually converging to an accurate solution [7-47]. In a later work Ghorbani et al. [21, 22]

Corresponding Author: Syed Tauseef Mohyud-Din, Department of Mathematics, COMSATS Institute of Information Technology, Islamabad, Pakistan

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World Appl. Sci. J., 4 (4): 487-498, 2008

the integral operator. Several examples are given to verify the accuracy and efficiency of the proposed algorithm.

where F (u) is a functional operator with known solutions v0 , which can be obtained easily. It is clear that, for

VARIATIO NAL ITERATION METHOD

H(u,p) = 0

we have

To illustrate the basic concept of the technique, we consider the following general differential equation Lu + Nu = g(x)

x

∫

= 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 (v0 , 0) to a solution function H (f, 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 [7, 14-19, 21-29, 35-39, 44, 46]. The homotopy perturbation method uses the homotopy parameter p as an expanding parameter [7, 14-19] to obtain

(1)

where L is a linear operator, N a nonlinear operator and g(x) is the forcing term. According to variational iteration method [7-13, 23-34, 40-43, 45, 47], we can construct a correction functional as follows un +1(x) = u n (x) + λ (Lu n (s) + N u n (s) − g(s))ds

(2)

0

∞

∑

where λ is a Lagrange multiplier [7-13], which can be identified optimally via variational iteration method. The subscripts n denote the nth approximation, u n is

u = p i ui = u0 + p u1 + p 2 u + p 3u 3 +  . 2

i= 0

considered as a restricted variation. i.e. δ u n = 0; (2) is

∞

∑u

= f limu = p →1

(7)

i

i =0

It is well known that series (7) is convergent for most of the cases and also the rate of convergence is dependent on L (u); see [7, 14-19]. We assume that (7) has a unique solution. The comparisons of like powers of p give solutions of various orders. In sum, according to [21, 22], He’s HPM considers the solution, u (x) , of the homotopy equation in a series of p as follows:

n →∞

convergence of variational iteration method, see Ramos [45].

∞

∑ p u =u i

u(x) =

i

0

+ p u1 + p 2 u + ... 2

i= 0

HOMOTOPY PERTURBATION METHOD

and the method considers the nonlinear term N(u) as:

To explain the homotopy perturbation method, we consider a general equation of the type,

∞

∑p H i

N(u) =

i

=H0 + pH 1 + p2H + ... 2

i =0

(3)

where Hn ’s are the so-called He’s polynomials [21, 22], which can be calculated by using the formula

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)

(6)

If p→1, then (6) corresponds to (4) and becomes the approximate solution of the form,

called as 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 [7-13]. In this method, it is required first to determine the Lagrange multiplier λ optimally. The successive approximation u n+1 , n≥0 of the solution u will be readily obtained upon using the determined Lagrange multiplier and any selective function u 0 , consequently, the solution is given by u = lim un . For error estimates and

L(u) = 0

(5)

n

= Hn (u 0,, un )

(4)

488

 1 ∂n   N( p i ui )  , n 0,1,2, . =  n! ∂pn   i= 0  p= 0

∑

World Appl. Sci. J., 4 (4): 487-498, 2008

The correction functional for the above problem is given by:

MODIFIED VARIATIONAL ITERATION METHOD (MVIM) To illustrate the basic concept of the modified variational iteration method (MVIM), we consider the following general differential equation (8)

Lu + N u = g(x)

un ~  − u n  ds.   ∂x∂s 

(s ) ∂

t

u n+1 ( x, t ) = u n ( x, t ) + ∫o

2

Making the correction functional stationary, the Lagrange multiplier can be identified as λ(s) = -1, consequently

where L is a linear operator, N a nonlinear operator and g(x) is the forcing term. According to variational iteration method [7-13, 23-34, 40-43, 45, 47], we can construct a correction functional as follows

un +1 ( x,t ) = u n ( x,t ) −

∫

t

o

 ∂ 2u  n −u   ds, n  ∂x∂ s   

n ≥0

x

∫

un +1= (x) u n (x) + λ (ξ ) ( L u n ( ξ) + N u n ( ξ) − g(ξ ) ) dξ (9)

Applying the modified variational iteration method

0

u0 + pu1 + p2u 2

where λ is a Lagrange multiplier [7-13] , which can be identified optimally via variational iteration method. The subscripts n denote the nth approximation, u n is 0; (9) is considered as a restricted variation. i.e. δ u n = called as a correction functional. Now, we apply the homotopy perturbation method

∞

u ∑ p= (n)

n

x

u0 (x) + p

n=0

∫ 0

−

 ∞  p (n) L(u n )  n =0 λ (ξ )   ∞ + p(n) N(u n )   n=0

∑

∑

 u 0 (x,t) − p +=

∫

t

0

  ∂ 2u  ∂2u1 ∂2 u 2 0  +p + p2 +    ∂x∂s ∂x∂s   ∂x∂s  ds    − u + pu + p2 u +   0 1 2  

(

)

Choosing uo ( x,t = ) Ae x + Be t , as the initial value

    dξ    

and comparing the co-efficient of like powers of p p(0) : uo ( x,t =) Ae x + Be t ,

x

p(1) : u1 ( x,t ) = Ae x +2Bet + Atex − B ,

0

p(2) : u 2 ( x ,t )= Ae x + 4Bet + Ate x − 3B +

∫ λ(ξ) g ( ξ) d ξ

which is the modified variational iteration method (MVIM) [23-29] and is formulated by the coupling of variational iteration method and He’s polynomials. The comparison of like powers of p gives solutions of various orders.

p(3) : u 3 ( x,t ) = Ae x +8Bet + Atex −7B +

uxt = u ,

1 2 x 1 1 At e − 4Bt + At 3 ex − Bt 2 , 2 6 2



The series solution is given by

NUMERICAL APPLICATIONS In this section, we apply the Modified Variational Iteration Method (MVIM) for solving Goursat and Laplace problems. Numerical results are very encouraging. Example 5.1 [4] Consider the following homogenous Goursat problem

1 2 x At e − Bt , 2

  t 2 t3 t4 t5 un ( x,t= ) A e x 1 + t + 2! + 3! + 4! + 5! + ...     t 1 2 ... + B 8e − 7 − 4t − t +  2  

Imposing

the

boundary

conditions

x = u ( x,0) e= , u ( 0,0) 1, to find the constants A and B

will yield A = 1, and B = 0, consequently

x = u ( x,0) e= , u ( 0,t ) et= , u ( 0,0) 1

u ( x,t ) = e x + t .

489

World Appl. Sci. J., 4 (4): 487-498, 2008 t

 ∂2 u n  un +1 ( x,t ) =u n ( x,t ) − ∫  − 2u n  ds, ∂ x ∂ s   0

n ≥ 0a

Applying the modified variational iteration method u0 + pu 1+ p 2 u 2 +  t

= u 0 (x,t) − p ∫ 0

  ∂ 2u 0  ∂ 2 u1 ∂ 2 u2  +p + p2 +   ∂ x ∂ s ∂ x ∂ s ∂ x ∂ s    ds   2  − 2 u 0 + pu1 + p u 2 +    

(

)

Choosing uo ( x,t = ) Ae x + Be− 2 t , as the initial value and comparing the co-efficient of like powers of p Fig. 1:

p(0 ) : u o ( x,t = ) Ae x + Be −2t , p(1) : u1 ( x,t ) = Ae x + 2Be−2t − 2Ate x − B , p(2 ) : u2 ( x,t ) = Ae x + 4Be−2t − 2Ate x − 3B + At 2e x + Bt ,

p(3) : u 3 ( x,t ) = Ae x + 8Be−2t − 2Ate x − 7B + At 2 e x + 8Bt −

4 3 x At e − 2Bt 2 , 3



The series solution is given by 2 3  ( 2t ) + ( 2t ) + ...  x un ( x,t )= A e  1− ( 2t ) +   2! 3!   2 −2t ... + B 8e − 7 + 8t −2t +

(

Fig. 2: Example 5.2 [4] Consider homogenous Goursat problem

the

following

Imposing

)

the

boundary

conditions

= u ( 0,0) 1,= u( x,0 ) ex , to find the constants A and B

will yield A = 1, and B = 0, consequently u ( x,t ) = e x − 2t .

u xt = 2u,

Example 5.3 [4] Consider the inhomogeneous linear Goursat problem

= u ( x,0) e= , u ( x,t ) e = , u ( 0,0) 1 x

−2 t

The correction functional for the above problem is given by t

u n +1 (x, t ) = u n (x , t ) + ∫ 0

following

u xt= u − t

u ( x,0) = e x x ( x,0) = ex , u ( o,t ) = t + et , u ( 0,0) = 1

 ∂ 2u  (s) n − 2 u~n ds , n ≥ 0.  ∂x∂s 

The correction functional for the above problem is given by

Making the correction functional stationary, the Lagrange multiplier can be identified as λ(s) = -1, consequently

t

u n +1 ( x, t ) = u n (x , t ) + ∫ 0

490

 ∂2 un ~  − u n + s  ds, (s )  ∂x∂s 

n ≥ 0.

World Appl. Sci. J., 4 (4): 487-498, 2008

Making the correction functional stationary, the Lagrange multiplier can be identified as λ(s) = -1, consequently t  ∂ 2u  = un +1 ( x,t ) u n ( x,t ) − ∫  ∂x∂ns − u n + s  ds,  0 

n≥0

Applying the modified variational iteration method u0 + pu 1+ p 2 u 2 +  t

= u 0 (x,t) − p ∫ 0

2   ∂ 2u 0  ∂ 2 u1 2 ∂ u2  +p +p +   ∂x∂s ∂x∂s   ∂x∂s   ds   2  − u 0 + pu 1+ p u 2 +  + s   

(

)

Choosing uo ( x,t ) = t + Ae x + B et , as the initial value

Fig. 3:

and comparing the co-efficient of like powers of p


p(0 ) : u o ( x,t ) = t + Ae x + Be t , p(1) : u1 ( x,t ) = t + Ae x + 2Be t + 2Ate x − Bt ,

t



 ∂ x∂s

0

p(2 ) : u2 ( x,t ) = t + Ae x +4Bet + Ate x − 3B +

1 2 x At e − Bt , 2

1 2 x 1 1 At e − 8Bt − At 3 e x − Bt 2 , 2 6 2

t  ∂ 2u  = un +1 ( x,t ) u n ( x,t ) − ∫  ∂x∂ns − u n − 4xs − x 2 s2 ds,  0 




n≥ 0

u0 + pu 1+ p 2 u 2 +  = u 0 (x,t) − p ∫

t

0

the

boundary conditions = u ( 0,0) 1,= u( x,0 ) ex , to find the constants A and B

  ∂2 u 0   ∂ 2 u1  +p +  ∂x∂s   ∂x ∂s   ds    − u 0 + pu1 + p2 u 2 +  − 4xs − x 2s2   

(

)

Choosing uo ( x,t ) = t 2 x 2 + Ae2 + Bet , as the initial


value and comparing the co-efficient of like powers of p

u ( x,t )= t + e x + t .

Example 5.4 [4] Consider the inhomogeneous linear Goursat problem

n ≥ 0.


  t 2 t 3 t4 un ( x,t ) = t + Ae x 1 + t + + + + ... 2! 3! 4!   1 2 ...   t + B  8e − 7 − 4t − t +  2  

Imposing

 − u~n − 4 xs − x 2 s 2  ds, 

Making the correct functional stationary, the Lagrange multiplier can be identified as λ(s) = -1 consequently

p(3) : u 3 ( x,t ) = t + Ae x +8Bet + Ate x − 7B +

2

(s ) ∂ un

u n+1 ( x,t ) = u n (x, t ) + ∫

p(0 ) : u o ( x,t ) = t 2 x 2 + Ae x + Be t ,

following

p(1) : u1 ( x,t ) = t 2 x 2 + Ae x +2Be t + 2Ate x − B ,

u xt =u + 4xt − x2 t 2 ,

p(2 ) : u2 ( x,t ) = t 2 x 2 + Ae x +4Bet + Ate x − 3B +

x = u ( x,0) e= , u ( 0,t ) et= , u ( 0,0) 1

491

1 2 x At e − Bt , 2

World Appl. Sci. J., 4 (4): 487-498, 2008

Fig. 4: Fig. 5:

p(3) : u 3 ( x,t ) = t 2 x 2 + Ae x +8Be t + Ate x − 7B +

1 2 x 1 1 At e − 4Bt + At 3 e x − Bt 2 , 2 6 2

Applying the method, we have




Imposing

2

x

the

boundary

= u 0( x , t ) − p ∫

conditions

p(0 ) : u 0 (x,t) = Ax + Bt,

Example 5.5 [4] Consider the following nonlinear inhomogeneous Goursat problem 2

1 p(1) : u1( x , t ) = Ax + Bt − (B3 − 1)t 4 − (AB2 − 1)xt3 4 3 2 − ( A B − 1)x 2 t 2 − A3 x 3 t + x3t , 2 

3

u xt = −u + x + 3x t + 3xt + t , = u(x,0) x,= u(0,t) t,= u(0,0) 0

The correction functional for the above problem is given by  ∂ un  t + u 3n − x3   un +1 ( = x,t ) u n ( x,t ) + ∫ λ (s )  ∂x∂ s  ds, n ≥ 0  − 3 x2s −3xs2 − s 3  0   2

Imposing

the

boundary

conditions

= u ( x,0 ) x,= u ( 0,t ) t, to find the constants A and B will

yield A = 1, and B = 1, consequently u(x,t)= x + t .

Making the correction functional stationary, the Lagrange multiplier can be identified as λ(s) = -1 consequently 2  t ∂ u n u n +1 (x, t ) = u n (x, t ) − ∫  + u 3n − x3 − 3 x 2s − 3 xs 2 − s 3  ds, o ∂ ∂ x s  

2   ∂ 2 u0   ∂ u1 +p +    ∂x∂s   ∂x∂s    3 2  + ( u 0 + pu1 + p u 2 + )  ds  3  2 2 3  − x −3x s −3xs −s     

comparing the co-efficient of like powers of p

u ( x,t = ) x 2 t 2 + e x +t

2

iteration

Choosing u0 (x,t) = Ax + Bx, as the initial value and


3

t

0

= u ( 0,0) 1,= u( x,0 ) ex , to find the constants A and B

3

variational

u0 + pu 1+ p 2 u 2 + 

2 3 4   t t t un ( x,t ) = t x + Ae 1 + t + + + + ... 2! 3! 4!   1 2 ...   t + B  8e − 7 − 4t − t +  2   2

modified

Example 5.6 [4] Consider the following nonlinear inhomogeneous Goursat problem 2

2x

2t

u xt = −u + e + e t + 2e

n ≥ 0.

x

x+ t

, t

u(x,0) = 1+ e , u(0,t) = 1+ e , u(0,0) = 2

492

World Appl. Sci. J., 4 (4): 487-498, 2008

The correction functional for the above problem is given by t

un +1 ( = x,t ) u n ( x,t ) + ∫ 0

 ∂ 2 un  + u 2n − e 2x   λ (s )  ∂x∂ s  ds, n ≥ 0  − e2s s− 2e x + s   

Making the correction functional stationary, the Lagrange multiplier can be identified as λ(s) = -1, consequently  ∂2 u n  + un2 − e2x  un +1 ( x,t ) = u n ( x,t ) − ∫  ∂x∂s  ds , n ≥ 0 2s x +s  0  − − e s 2e   t

Fig. 6:


with Dirichlet boundary conditions

u0 + pu 1 + p 2 u 2 + 

t

= u 0( x , t ) − p ∫ 0

u(0,y) = 0,

2   ∂ 2u 0   ∂ u1 +p +    ∂ x∂ s   ∂x∂s    2 2 u pu p u  )  ds  +( 0 + 1 + 2 +  2x  2s x+s  −e + e s − 2e     

x  ∂ 2 n ∂ 2u~  u n+1 ( x, y ) = u n ( x, y ) + ∫0 ( s ) 2 + 2n  ds. ∂y   ∂s

and comparing the co-efficient of like powers of p

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

p(0 ) : u 0 (x,t) = Ae x + Be t , 1 2 2t B e − A2t e 2 x 2 1 + 2ABe x − 2ex + (B2 − 1) , 2

p(1) : u1( x , t ) = Aex + Be t − 2ABe x + t −

x  ∂2 u ∂ 2u n  = u n (x,y) + ∫ (s − x)  2n + un +1 (x,y)  ds ∂y2   ∂s 0





u0 + pu 1 + p 2 u 2 + 

1 u(x,t) = Aex + Be t − 2ABe x + t − B 2e2 t − A 2te 2x 2 1 2t 1 2x x+t x + e + te +2e + 2ABe − 2e x + (B 2 − 1) 2 2

Imposing

u(x, π ) =− sinhx.


Choosing u0 (x,t) = Ae x + Be x , as the initial value

1 + e 2t + te 2x +2e x + t 2

u( π,y) = sinh πcosy,

u(x,0) =sinhx,

  ∂2 u  ∂2 u ∂ 2 u2   20 + p 21 + p2 +   2 ∂s ∂s   ∂s  = u 0 (x,y) + p∫ (s − x)   ds 2 2  +  ∂ u 0 + p ∂ u1 +   0  2   ∂y 2  ∂ y    x

the

boundary conditions u ( 0,0) = 2, u ( x,0) = 1+e , to find the constants A and B x

Choosing x cos y as the initial value and comparing the co-efficient of like powers of p

will yield A = 1, and B = 1, consequently u(x,t) = ex + et .

Example 5.7 [5] Consider the following Laplace problem = u xx + u yy 0,

0 < x,

p(0 ) : u 0 (x,y) = x co s y ,  1  p(1) : u1( x , y= )  x + x 3  cosy , 3!  

y < π,

493

World Appl. Sci. J., 4 (4): 487-498, 2008

Fig. 7:

Fig. 8: x  ∂2u ∂ 2 u~n un +1 ( x, y) = un ( x, y ) + ∫ (s − x) 2n + ∂y 2 0  ∂s

 1 3 1 5 p(2 ) : u 2 (x,y) =  x + x + x  cos y , 3! 5!    1 3 1 5 1 7 p(3) : u 3( x , y ) = x + x + x  cosy , x + 3! 5! 7!  

  ds. 




u0 + pu 1 + p 2 u 2 +  2   ∂2 u 0  ∂2 u 2 ∂ u2   2 + p 21 + p +   2 ∂s ∂s   ∂s  = u 0 (x,y) + p∫ (s − x)   ds 2 2 0  +  ∂ u 0 + p ∂ u1 +      ∂y 2  ∂y2    


x

 1 3 1 5 1 7  un (x,y) =  x + x + x + x +  cos y , 3! 5! 7!  

and in a closed form by



Choosing  1 + 

u(x,y) = sinhxcosy .

comparing the co-efficient of like powers of p  1  p(0) : u 0 (x,y)=  1 + x 2  si ny , 2!  

Example 5.7 [5] Consider the following Laplace problem = uxx + uyy 0,

0 < x,

y