Int. Journal of Math. Analysis, Vol. 5, 2011, no. 18, 863 - 870
Application of Variational Iteration Method to Partial Differential Equation Systems M. Akbarzade1 Sama Technical and Vocatinal Training college, Islamic Azad University Quchan Branch, Quchan, Iran J. Langari Sama Technical and Vocatinal Training college, Islamic Azad University Quchan Branch, Quchan, Iran Abstract In this article, systems of linear and nonlinear partial differential equations solve with variational iteration method (VIM). The VIM is to construct correction functional using general Lagrange multipliers identified optimally via the variational theory and the initial approximations can be freely chosen with unknown constants. In this article, we used (VIM) to solve partial differential equation system. This method is very simple and reduces computation and readily and fast converges to the answer. The procedure is well organized. Keywords: Variational iteration method (VIM), partial differential equation system, Lagrangian multiplier.
1. Introduction Nonlinear differential equation in engineering and applied mathematics has been a topic to intensive research for many years [3, 4 and 9]. Partial differential equation systems apply in study of very mechanic system and field of another science [7]. For example description wave propagation [2].When several system should be solved existing method for solve of these equations interface with sever computation [5].We solve these equations with Variational iteration method (VIM).This is very important to construct iteration function.
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Corresponding author:
[email protected]
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2. Variational iteration method To clarify the basic ideas of He's VIM, we consider the following differential equation: Lu + Nu = g (t ), (1) Where L is a linear operator, N a nonlinear operator and g(t) an inhomogeneous term. According to VIM, we can write down a correction functional as follows: t
u n +1 (t ) = u n (t ) + ∫ λ ( Lu n (τ ) + N u%n (τ ) − g (τ ))d τ
(2)
0
Where λ is a general Lagrangian multiplier [6, 8 and 9-11] which can be identified optimally via the variational theory. The subscript n indicates the nth approximation and u%n is considered as a restricted variation [5], i.e. δ u% n = 0 .
3. The application of VIM in system of linear and nonlinear partial differential equations Consider the following example with system of linear and nonlinear partial differential equations: 3.1. Example 1 Consider the following linear partial differential equations system: ∂u ( x, t ) ∂v( x, t ) ∂v( x, t ) ∂u ( x, t ) + = 0, + =0 ∂t ∂x ∂t ∂x With the initial condition of:
u (x , 0) = e x ,v (x , 0) = e − x
(3a)
(3b)
3.1.1 Variational iteration method
In order to solve Eq. ((3a) and (3b)) using VIM, we construct a correction functional, as follows: t t ∂u ∂v ∂v ∂u un +1 = un + ∫ λ1 ( n + n )dτ , vn +1 = vn + ∫ λ2 ( n + n +1 )dτ (4) 0 0 ∂τ ∂x ∂τ ∂x Where un = un ( x, t ) , vn = vn ( x, t ). The Lagrangian multiplier can be identified as λ1 = λ2 = -1. As a result, we obtain the following iteration formula: t ∂u t ∂v ∂v ∂u un +1 = un − ∫ ( n + n )dτ , vn +1 = vn − ∫ ( n + n +1 )dτ (5) 0 ∂τ 0 ∂τ ∂x ∂x
Application of variational iteration method
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Now we start with an arbitrary initial approximation that satisfies the initial condition: u0 ( x, t ) = e x , v0 ( x, t ) = e− x Using the above variational formula (5), we have: t ∂u t ∂v ∂v ∂u u1 = u0 − ∫ ( 0 + 0 )dτ , v1 = v0 − ∫ ( 0 + 1 )dτ 0 ∂τ 0 ∂τ ∂x ∂x Substituting Eq.(6) into Eq. (7)and after simplifications, we have: 1 u1 = e x + te − x , v1 = e− x − te x + t 2 e− x 2 In the same way, we obtain u 2 (x , t ),v 2 (x , t ), u 3 (x , t ),v 3 (x , t ) as follows:
(6)
(7)
(8)
1 1 1 1 1 u2 = e x + te − x + t 2 e x + t 3e− x , v2 = e− x − te x + t 2 e − x − t 3e x + t 4 e− x , 2 6 2 6 24 1 1 1 1 5 −x u3 = e x + te − x + t 2 e x + t 3e − x + t 4 e x + te (9) 2 6 24 120 1 1 1 1 5 x 1 6 −x v3 = e − x − te x + t 2e − x − t 3e x + t 4 e − x − te + t e 2 6 24 120 720 And so on. In the same manner the rest of the components of the iteration formula can be obtained: 1 1 1 1 5 u n (x , t ) = (1 + t 2 + t 4 + ...)e x + (t + t 3 + t + ...)e − x 2 24 6 120 (10) 1 2 1 4 1 6 1 3 1 5 x −x v n (x , t ) = (1 + t + t + t + ...)e + (t + t + t + ...)e 2 24 720 6 120 We know that (1 + 1 t 2 + 1 t 4 + ...) is Taylor series of cosh(t ) and (t + 1 t 3 + 1 t 5 + ...) 2
24
6
120
is Taylor series of sinh(t ) . In order to compare with the decomposition method solution, we write Abdul-Majid Wazwaz’s result [1]: u ( x, t ) = cosh(t )e x + sinh(t )e− x , v( x, t ) = cosh(t )e- x − sinh(t )e x (11) 3.2 Example 2
Now we solve the following nonlinear partial differential equations system: ∂u ( x, t ) ∂u ( x, t ) ∂v( x, t ) ∂v( x, t ) + v ( x, t ) + u ( x, t ) = 1, − u ( x, t ) − v( x, t ) = 1 (12a) ∂t ∂x ∂t ∂x Initial condition: u (x , 0) = e x , v (x , 0) = e − x
(12b)
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3.2.1 Variational iteration method
In this system we combine two equation then select specific figure of the iteration function: t ∂v ∂v ∂u ∂u u n +1 = u n + ∫ λ1 ( n + v 0 n + u n − 0 + u n 0 + v 0 )d τ 0 ∂τ ∂x ∂τ ∂x (13) t ∂u 0 ∂u 0 ∂v 0 ∂v n v n +1 = v n + ∫ λ2 (− −v n −u0 + −u0 − v n )d τ 0 ∂τ ∂x ∂τ ∂x Where un = un ( x, t ) , vn = vn ( x, t ). The Lagrangian multiplier can be identified as λ1 = λ2 = -1. As a result, we obtain the following iteration formula: t
u n +1 = u n − ∫ ( 0
∂v ∂v ∂u n ∂u + v 0 n + u n − 0 + u n 0 + v 0 )d τ ∂τ ∂x ∂τ ∂x
∂u 0 ∂u ∂v ∂v − v n 0 − u 0 + n − u 0 0 − v n )d τ ∂τ ∂x ∂τ ∂x Now we start with an arbitrary initial approximation as follows: u 0 (x , t ) = e x ,v 0 (x , t ) = 0
(14a)
t
v n +1 = v n − ∫ (− 0
By the above variational formula (14a), we can obtain the following result: t ∂u ∂u ∂v ∂v u1 = u 0 − ∫ ( 0 + v 0 0 + u 0 − 0 + u 0 0 + v 0 )d τ 0 ∂τ ∂x ∂τ ∂x Substituting Eq. (15) into Eq. (16) , we have: u1 (x , t ) = (1 − t )e x Now we start with an arbitrary initial approximation as follows: u 0 (x , t ) = 0,v 0 (x , t ) = e − x By the above variational formula (14b), we can obtain the following result: t ∂u ∂u ∂v ∂v v 1 = − ∫ (− 0 − v 0 0 − u 0 + 0 − u 0 0 − v 0 )d τ 0 ∂τ ∂x ∂τ ∂x Substituting Eq. (18) into Eq. (19) , we have: v1 = (1 + t ) e − x
(14b)
(15)
(16)
(17) (18)
(19)
(20)
In the same way, we obtain u 2 (x , t ),u 3 (x , t ),u 4 (x , t ),v 2 (x , t ),v 3 (x , t ),v 4 (x , t ) as:
1 2 x 1 1 1 1 1 t ) e , v 2 = (1 + t + t 2 ) e − x , u 3 = (1 − t + t 2 − t 3 ) e x , v3 = (1 + t + t 2 + t 3 ) e − x 2 2 2 6 2 6 (21) 1 2 1 3 1 4 x 1 2 1 3 1 4 −x u 4 = (1 − t + t − t + t ) e , v 4 = (1 + t + t + t + t )e 2 6 24 2 6 24 With continue these stages u(x,t) , v(x,t) converge to the below functions: u 2 = (1 − t +
Application of variational iteration method
u n ( x , t ) = (1 − t +
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1 2 1 3 1 4 1 1 1 4 t − t + t − ...) e x , v n ( x , t ) = (1 + t + t 2 + t 3 + t + ...) e − x (22) 2 6 24 2 6 24
We know that (1 − t + 1 t 2 − 1 t 3 + ...) is Taylor series of e − t and (1 + t + 1 t 2 + 1 t 3 + ...) 2
6
2
6
t
is Taylor series of e . (23)
u n ( x , t ) = e − t e x , vn ( x , t ) = e t e − x
In order to compare with the decomposition method solution, we write Abdul Majid Wazwaz’s result [1]:
u ( x, t ) = e x −t , v ( x, t ) = et − x
(24)
3.3 Example 3
Consider the following partial differential equation, with specified initial conditions: ∂u (x , y , t ) ∂v (x , y , t ) ∂w (x , y , t ) ∂v (x , y , t ) ∂w (x , y , t ) + − = −u (x , y , t ) ∂t ∂x ∂y ∂y ∂x ∂v (x , y , t ) ∂w ( x , y , t ) ∂u ( x , y , t ) ∂w ( x , y , t ) ∂u ( x , y , t ) + − = v (x , y , t ) (25a) ∂t ∂x ∂y ∂y ∂x ∂w (x , y , t ) ∂u ( x , y , t ) ∂v (x , y , t ) ∂u (x , y , t ) ∂v ( x , y , t ) + − = w (x , y , t ) ∂t ∂x ∂y ∂y ∂x and: u (x , y , 0) = e x + y ,v (x , y , 0) = e x − y ,w (x , y , 0) = e − x + y
(25b)
3.3.1 Variational iteration method
In this system we combine two equation then select specific figure of the iteration function:
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∂un ∂v0 ∂w0 ∂v0 ∂w0 ∂v ∂w ∂u ∂w ∂u + − + un + 0 + 0 n + 0 n − v0 ∂τ ∂x ∂y ∂y ∂x ∂τ ∂x ∂y ∂y ∂x ∂w ∂u ∂v ∂u ∂v + 0 + n 0 + n 0 − w0 )dτ , ∂τ ∂x ∂y ∂y ∂x t ∂u ∂v ∂w ∂v ∂w ∂v ∂w ∂u ∂w ∂u vn +1 = vn + ∫ λ2 ( 0 + n 0 − n 0 + u0 + n + 0 0 + 0 0 − vn 0 ∂τ ∂x ∂y ∂y ∂x ∂τ ∂x ∂y ∂y ∂x (26) ∂w0 ∂u0 ∂vn ∂u0 ∂vn + − w0 )dτ + + ∂τ ∂x ∂y ∂y ∂x t ∂u ∂v ∂w ∂v ∂w ∂v ∂w ∂u ∂w ∂u wn +1 = wn + ∫ λ3 ( 0 + 0 n − 0 n + u0 + 0 + n 0 + n 0 − vn 0 ∂τ ∂x ∂y ∂y ∂x ∂τ ∂x ∂y ∂y ∂x ∂w ∂u ∂v ∂u ∂v + n + 0 n + 0 0 − wn )dτ ∂τ ∂x ∂y ∂y ∂x t
un +1 = un + ∫ λ1 ( 0
The Lagrangian multiplier can be identified as λ1 =λ2 = λ3 =-1. As a result, we obtain the following iteration formula: t ∂v ∂w 0 ∂v 0 ∂w 0 ∂v ∂w 0 ∂u n ∂w 0 ∂u n ∂u u n +1 = u n + ∫ −( n + 0 − +un + 0 + + −v 0 0 ∂τ ∂x ∂y ∂y ∂x ∂τ ∂x ∂y ∂y ∂x ∂w 0 ∂u n ∂v 0 ∂u n ∂v 0 + + + − w 0 )d τ ∂τ ∂x ∂y ∂y ∂x t ∂u ∂w 0 ∂u 0 ∂w 0 ∂u 0 ∂v ∂w 0 ∂v n ∂w 0 ∂v v n +1 = v n + ∫ −( 0 + n − +u0 + n + + −v n 0 ∂τ ∂x ∂y ∂y ∂x ∂τ ∂x ∂y ∂y ∂x (27) ∂w 0 ∂u 0 ∂v n ∂u 0 ∂v n − w 0 )d τ + + + ∂y ∂x ∂τ ∂x ∂y t ∂u ∂v ∂w n ∂v 0 ∂w n ∂v ∂w n ∂u 0 ∂w n ∂u 0 w n +1 = w n + ∫ −( 0 + 0 − +u0 + 0 + + −v 0 0 ∂τ ∂x ∂y ∂y ∂x ∂τ ∂x ∂y ∂y ∂x ∂w n ∂u 0 ∂v 0 ∂u 0 ∂v 0 + + + − w n )d τ ∂τ ∂x ∂y ∂y ∂x Now we start with an arbitrary initial approximation to obtain u(x, y, t) : u 0 (x , y , t ) = e x + y ,v 0 (x , y , t ) = 0,w 0 (x , y , t ) = 0 Similary, we can obtain the following result: u1 = (1 − t )e x + y , v1 = (1 + t )e x − y , w1 = (1 + t )e − x + y In the same way, we obtain: u2 ( x, y, t ), u3 ( x, y, t ), u4 ( x, y, t ), v2 ( x, y, t ), v3 ( x, y, t ), v4 ( x, y, t ), w2 ( x, y, t ), w3 ( x, y, t ), w4 ( x, y, t ) :
(28) (29)
Application of variational iteration method
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1 2 x+ y 1 1 t ) e , v 2 = (1 + t + t 2 ) e x − y , w2 = (1 + t + t 2 ) e − x + y , 2 2 2 (30) 1 1 1 1 1 1 u 3 = (1 − t + t 2 − t 3 ) e x + y , v3 = (1 + t + t 2 + t 3 ) e x − y , w3 = (1 + t + t 2 + t 3 ) e − x + y 2 6 2 6 2 6 With continue these stages u(x,y,t) , v(x,y,t) ,w(x,y,t) converge to the below functions: 1 1 1 1 u n ( x , y , t ) = (1 − t + t 2 − t 3 + ...) e x + y , v n ( x , y , t ) = (1 + t + t 2 + t 3 + ...) e x − y 2 6 2 6 (31) 1 2 1 3 − x+ y wn ( x , y , t ) = (1 + t + t + t + ...) e 2 6 u 2 = (1 − t +
We know that (1 − t + 1 t 2 − 1 t 3 + ...) is Taylor series of e − t and (1 + t + 1 t 2 + 1 t 3 + ...) 2
6
2
6
t
is Taylor series of e . In order to compare with the decomposition method solution, we write Abdul-Majid Wazwaz’s result [1]:
u ( x, y , t ) = e −t e x+ y , v ( x, y , t ) = et e x− y , w( x, y, t ) = et e − x+ y
(32)
4. Conclusions Variational iteration method (VIM), for the first time, was applied to partial differential equation system. We demonstrated the accuracy and efficiency of the method by solving some examples. Moreover, we showed that the obtained solutions are valid for the whole domain. It suggests that the variational iteration method is accurate, reliable and easy to use.
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[4] H. Pashaei, D. D. Ganji, and M. Akbarzade, APPLICATION OF THE ENERGY BALANCE METHOD FOR STRONGLY NONLINEAR OSCILLATORS, Progress In Electromagnetics Research M, (2008): Vol. 2, 4756. [5] J.H. He, Variational iteration method for autonomous ordinary differential systems, Appl. Math. Comput. 114 (2000), pp. 115–123. [6] L. Xu, Variational principles for coupled nonlinear Schrodinger equations, Physics Letters A 359 (6) (2006) 627–629. [7] L.Debnath, Nonlinear Partial Differential Equations for Scientists and Engineers, Birkhauser, Boston 1997. [8] M. Akbarzade, D. D. Ganji, Coupled Method of Homotopy Perturbation Method and Variational Approach for Solution to Nonlinear Cubic-Quintic Duffing Oscillator, Adv. Theor. Appl. Mech., Vol. 3, (2010), no. 7, 329 - 337. [9] M. Akbarzade, D. D. Ganji, and H. Pashaei, ANALYSIS OF NONLINEAR n OSCILLATORS WITH u FORCE BY HE’S ENERGY BALANCE METHOD, Progress In Electromagnetics Research C, (2008): Vol. 3, 57–66. [10] N. Bildik and A. Konuralp, The use of variational iteration method, differential transform method and adomian decomposition method for solving different types of nonlinear partial differential equations, Int. J. Nonlinear Sci. Numer. Simul. 7 (2006), pp. 65–70. [11] Z.M. Odibat and S. Momani, Application of variational iteration method to nonlinear differential equations of fractional order, Int. J. Nonlinear Sci. Numer. Simul. 7 (2006), pp. 27–34. Received: October, 2010
