An approximation to the solution of Klein-Gordon equation with initial or boundary value condition

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1 International Mathematical Forum, 1, 6, no 9, An approximation to the solution of Klein-Gordon equation with initial or boundary value condition J Biazar and H Ebrahimi Department of Mathematics Islamic Azad University (Rasht branch) P O Box , Rasht, Iran biazar@guilanacir Abstract Adomian decomposition method has been applied to solve many functional equations so far Some authors have used this method for solving Klien-Gordon equation with initial conditions In this article, Adomian method is applied to solve Klein-Gordon equation with boundary value condition, as well as initial conditionsthree examples are presented to illustrate the method Keywords: Adomian decomposition method, Klein-Gordon 1 Introduction In this work, we will consider the Klein-Gordon equation with initial or boundary conditions and Adomian decomposition method is applied to solve this equationthe Adomian decomposition method has proven to be very effective and results in considerable saving in computation timeklein-gordon equation has the following general form: = c ( u x + u y + u z ) (1) Where c is a constant Let have the following boundary conditions: B 1 u(a 1,y,z,t)=f 1 (y, z, t) () B u(a,y,z,t)=f (y, z, t) (3) Where B 1 and B are identity or any differentiable operators

2 1434 J Biazar and H Ebrahimi The Adomian decomposition method applied to klein-gordon equation For solving this equation by Adomian decomposition method, we can pay attention to initial or boundary conditions using operators L xx =, L x yy =, L y zz = and L z tt = We use the operator L xx = with the inverse x L 1 xx = x x ()dxdx Therefore eq(1)can be written as: L xx u = 1 c g + 1 c L ttu L yy u L zz u (4) By applying the inverse operator L 1 xx to both sides of (4), we have: u(x, y, z, t) =u(,y,z,t)+u x (,y,z,t)x 1 c L 1 xx g +L 1 xx ( 1 c L ttu L yy u L zz u) (5) Let K 1 = u(,y,z,t) and K = u x (,y,z,t) Thus, eq(5) can be written as: u = K 1 + K x 1 c L 1 xx g + L 1 xx ( 1 c L ttu L yy u L zz u) (6) To solve this equation by Adomian decomposition method, as usual in this method, the solution u is considered as the sum of the series u = u n and the integrand on the right side as the sum of a series as: 1 c L ttu L yy u L zz u = A n (u,u 1,,u n ) Where A n (u,u 1,,u n ) are called Adomian polynomials and should be computed By using an Alternate Algorithm for computing Adomian polynomial [4], we have: A n (u,u 1,,u n )= 1 c L ttu n L yy u n L zz u n n =, 1,, Substituting u = u n and A n (u,u 1,,u n )in(6), we derive: u n = K 1 + K x 1 c L 1 xx g + Therefore from (7) the following procedure can be defined: u = K 1 + K x 1 c L 1 xx g L 1 xx ( 1 c L ttu n L yy u n L zz u n ) (7) u n+1 = L 1 xx ( 1 c L ttu n L yy u n L zz u n ) n =, 1,,

3 Klein-Gordon equation 1435 For determine K 1 and K, first we consider one-term approximation ϕ 1 for the exact solution: ϕ 1 = u = K 1 + K x 1 c L 1 xx g (8) By using equations () and (3), we have: B 1 ϕ 1 (a 1,y,z,t)=f 1 (y, z, t) B ϕ 1 (a,y,z,t)=f (y, z, t) Thus [ ][ ] [ 1 a1 K1 B 1 1 f = L 1 ] c xx g x=a1 1 a K B 1 f + 1 L 1 c xx g (9) x=a By solving eq(9), we obtain approximate values K 1 and K 1 and by using Adomian procedure we obtain u 1, and consider two-terms approximated values ϕ = u + u 1 for the exact solution and using equations () and (3), we have: [ 1 a1 1 a ][ K1 K ] = [ B 1 1 f L 1 c xx g x=a 1 u 1 B 1 f + 1 L 1 c xx g x=a u 1 Therefore approximation values K 1 and K from two-terms approximation ϕ will be obtained We can determine the components u n as far as we like to enhance the accuracy of the approximation and similarity we can be obtain approximation values K 1 and K in (n+1)-terms approximation ϕ n+1 = n i= u i Also, we have lim n ϕ n+1 = u 3 Numerical results Example 1 : Consider the Klein-Gordon equation with the following boundary conditions = u x + x t (1) u(,t) = (11) ] u(1,t)= t (1) Regarding boundary conditions we use the operator x Therefore, we have: x = u t + t x (13) Applying the inverse operator L 1 xx = x x ()dxdx to both sides of (13),we get: u(x, t) =u(,t)+ u(,t) x x x x x + (t x )dxdx + x t dxdx

4 1436 J Biazar and H Ebrahimi Let K 1 = u(,t) and K = u(,t) Therefore, the Adomian scheme would be x as follows: u = K 1 + K x + x t x4 1 x x n u n+1 = dxdx n =, 1,, (14) t By using (11) we have K 1 = To find K, we consider the following one-term approximation: ϕ 1 = u = K x + x t x4 1 By using (1) we have: ϕ 1 (1,t)= t K = 1 1 Therefore, u = x + x t x4 By using (14) u would be derived as: x x u x x u 1 = t dxdx = x dxdx = x4 1 To improve the value of K, let us consider two-terms approximation ϕ = u + u = K x + x t for solution u Regarding (1), we obtain K = Therefore ϕ = x t Again by using (14), we get: x x u x x 1 u = t dxdx = ()dxdx = x x u x x u = t dxdx = ()dxdx = u n = Therefore n-terms approximation is ϕ n = x t and solution is: x t u(x, t) = lim ϕ n = lim n n This solution is the exact solution Example : Consider the following Klein-Gordon = x t = u x + u + xy (15) y

5 Klein-Gordon equation 1437 u(x,,t) = (16) u(x, 1,t)= xt (17) u(,y,t) = (18) We can use one of the operators x operator we have: y u(x, y, t) =u(x,,t)+ u(1,y,t)= yt and y (19) for obtain the solution By using u(x,,t) y y y xydydy + ( u y t u x )dydy Consider K 1 = u(x,,t) and K = u(x,,t) Then the Adomian scheme would y be as follows: u = K 1 + K y xy3 6 u n+1 = ( u n u n )dydy n =, 1,, () x By using (16), we have K 1 = and to obtain K we consider the following one-term approximation: ϕ 1 = u = K y xy3 6 By Considering (17) we have K = xt + x 6 Therefore u = xyt + xy 6 xy3 6 Also u 1 would be as u 1 = ( u t u xy3 )dydy = x 6 Similarity by using two-terms approximation ϕ = u + u 1 and (17), we obtain k = xt Therefore ϕ = xyt and from () we have: u = ( u 1 t u 1 )dydy = x u 3 =

6 1438 J Biazar and H Ebrahimi u n = Then n-terms approximation is ϕ n = xyt and the exact solution is: xyt u(x, y, t) = lim ϕ n = lim n n = xyt Also by using operator derive the same solution x Example 3 : Consider the following Klein-Gordon equation with the initial conditions: By using operator t t =4( u x + u y + u 1 x )+t x u(x, y, z, ) = sin y u(x, y, z, ) = z t Adomian procedure would be as follows: u = 1 1 x t 4 + z t + sin y u n+1 =4 ( u n x + u n y First few terms would be as follows: u = 1 1 x t 4 + z t + sin y + u n )dtdt n =, 1,, z u 1 = 1 45 t t3 sin y (t)! u = sin y (t)4 4! u 3 = sin y (t)6 6! Therefore, the general term would be as: u n =( 1) n sin y (t)n (n)! n =, 3, 4 Then, the solution is: u(x, y, z, t) = u n = 1 45 t x t t3 + z t + sin y(1 (t) + (t)4 )! 4! = 1 45 t x t t3 + z t + sin y cos t

7 Klein-Gordon equation Conclusions and Discussion The Adomian decomposition method is a powerful method, which has provided an efficient potential for the solution of physical applications modeled by linear and nonlinear differential equations [1,,3] The main goal of this work has been to derive an approximation for solution of Klein-Gordon equation We have achieved this goal by applying Adomian decomposition method We can be understood from the Examples to solve the equation we have different choices, in Example the choices are and, and using each of them leads to the x y same solution For computations we used the package Maple 9 5 References [1] Adomian, G, Nonlinear Stochastic Systems Theory and Applications to Physics, Kluwer Academic Press, (1989) [] Adomian, G, Solving Frontier Problems of Physics The Decomposition Method, Kluwer Academic Press, (1994) [3] Adomian, G, Bellman, R, Partial Differential Equations, Reidel Publishing, (1985) [4] Biazar, J, Babolian, E, Nouri, A,Islam, R, An Alternate Algorithm for Computing Adomian Polynomial in Special Cases, Applied Mathematics and Computing, 8(-3),PP53-59,(3) Received: November 7, 5