Research Article Solution of (3 1)-Dimensional Nonlinear Cubic Schrodinger Equation by Differential Transform Method

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1 Mathematical Problems in Engineering Volume 212, Article ID 5182, 14 pages doi:1.1155/212/5182 Research Article Solution of ( 1)-Dimensional Nonlinear Cubic Schrodinger Equation by Differential Transform Method Hassan A. Zedan 1, 2 and M. Ali Alghamdi 1 1 Mathematics Department, Faculty of Science, King Abdulaziz University, P.O. Box 82, Jeddah 2158, Saudi Arabia 2 Department of Mathematics, Faculty of Science, Kafr El-Sheikh University, Kafr El-Sheikh 516, Egypt Correspondence should be addressed to Hassan A. Zedan, hassanzedan2@yahoo.com Received 15 September 211; Accepted 8 December 211 Academic Editor: Xing-Gang Yan Copyright q 212 H. A. Zedan and M. A. Alghamdi. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. Four-dimensional differential transform method has been introduced and fundamental theorems have been defined for the first time. Moreover, as an application of four-dimensional differential transform, exact solutions of nonlinear system of partial differential equations have been investigated. The results of the present method are compared very well with analytical solution of the system. Differential transform method can easily be applied to linear or nonlinear problems and reduces the size of computational work. With this method, exact solutions may be obtained without any need of cumbersome work, and it is a useful tool for analytical and numerical solutions. 1. Introduction In this paper, we study system of nonlinear partial differential equations PDEs. Inorder to solve systems of differential equations, the commonly used methods are the method of characteristics and the Riemann invariants among other methods. The existing techniques have difficulties in related to the size of computational work, especially when the system has several PDEs. More recently, Adomian decomposition method, were used to handle systems of PDEs by 1, and here we present an alternative method for the system. The concept of differential transform was first introduced by Zhou 2, who solved linear and nonlinear initial value problems in electric circuit analysis. The transformation method, called threedimensional differential transform, is different from the high-order Taylor series method, which consists of computing the coefficients of the Taylor series of the solution using the initial data and the PDE. But the Taylor series method requires more computational work for large orders. The present method is well addressed in 7. Thedifferential transform

2 2 Mathematical Problems in Engineering technique is an iterative procedure for obtaining Taylor series solutions of differential equations and systems of PDEs. This method reduces the size of computational domain and is applicable to many problems easily. In the present work, in order to extend applications of the differential transform method to different problems, four-dimensional differential transform have been defined and new theorems have been prooved. In this paper, Differential Transformation Method DTM is employed to obtain the solution of the 1-dimensional nonlinear cubic Schrödinger equation iψ t ψ xx ψ yy ψ zz ψ( ψ 2 S ( x, y, z, t )). 1.1 We put ψ u iv. 1.2 From 1.1 we obtain that ( Mu, v v t u xx u yy u zz u u 2 v 2 S ( x, y, z, t )), ( Nu, v u t v xx v yy v zz v u 2 v 2 S ( x, y, z, t )) Basic Idea of the Differential Equation 2.1. One-Dimensional Differential Transform Method We introduce in this section the basic definition of the one-dimensional differential transformation. Definition 2.1. If ut is analytic in the domain T, then it will be differentiated continuously with respect to time t: d k ut dt k φt, k, t T, 2.1 for t t i, where φt, k φt i,k, andk and k belong to the set of nonnegative integer denoted as the K domain. Therefore, 1.2 can be written as U i k φt i,k [ ] d k ut where U i k is called spectrum of ut at t t i,inthek domain. dt k tt i, k K, 2.2 Definition 2.2. If ut can be expressed by Taylor s series, then ut can be represented as ut k t t i Uk. k! 2.

3 Mathematical Problems in Engineering Equation 2. is known as the inverts transformation of Uk.IfUkis defined as [ ] d k qtut Uk Mk dt k where k, 1, 2,...,, then the function ut can be described as ut 1 t t i Uk qt k! Mk, k tt i, where Mk /, qt /. The function Mk is called the weighting factor and qt is regarded as kernel corresponding to ut. IfMk 1andqt 1, then 1. and 2.2 are equivalent. In this way, 2.1 can be treated as a special case of 2.. In this paper, the transformation with Mk 1/k! andqt 1isapplied. Then 2.4 becomes Uk [ ] d k ut dt k tt i, where k, 1, 2,...,. 2.6 Using the differential transform, a differential equation in the domain of interest can be transformed to be an algebraic equation in the K domain and ut can be obtained by finiteterm Taylor series plus a remainder, as ut 1 t t i Uk qt k! Mk R n1t t t i k Uk R n1 t. k k 2.7 In order to speed up the convergent rate and the accuracy of calculation, the entire domain of t needs to be split into subdomains Two-Dimensional Differential Transform Method Definition 2.. Given an w function which has two components such as x, t to w-dimensional differential transform of wx, t is defined Wk, h 1 [ ] kh Wx, t k!h! x k y h,, 2.8 where Wx, y the original is function and Wk, h is the transformed function. Again, the transformation can be called T-function, and the lower case and upper case letters represent the original and transformed functions, respectively Table 1. Definition 2.4. The differential inverse transform of Wk, h is defined as Wx, t Wk, hx k t h. k h 2.

4 4 Mathematical Problems in Engineering Table 1: The fundamental theorems. Original function 1 Wx, t ux, t ± vx, t, 2 Wx, t λux, t, Wx, t ux, t/ x, 4 Wx, t ux, t/ t 5 Wx, t rs ux, t/ x r t s 6 Wx, t ux, tvx, t 7 Wx, t ux, t/ x vx, t/ x, 8 Wx, t ux, t ux, t/ x, Transformed function Wk, h Uk, h ± V k, h. Wk, h λuk, h, λ is constant Wk, h k 1Uk 1,h Wk, h h1uk, h 1 Wk, h k1k 2 k rh 1 h 2 h suk r, h s Wk, h k k r s Ur, h sv k r, s Wk, h k hr 1k r 1Ur 1,h s Uk r 1,s Wk, h k hk r 1Ur, h suk r 1,s and from 2.8 and 2. can be concluded as follows: [ ] 1 kh Wx, t Wx, t k!h! x k t h k h, x k t h Three-Dimensional Differential Transform Definition 2.5. Given an w function which has three components such as x, y, t. Threedimensional differential transform of wx, y, t is defined as [ 1 hkm w ( x, y, t ) ] Wk, h, m k!h!m! x k y h t m,,, 2.11 where wx, y, t is the original function and Wk, h, m is the transformed function. Again, the transformation can be called T-function, and the lower case and upper case letters represent the original and transformed functions, respectively Table 2. Definition 2.6. The differential inverse transform of Wk, h, m is defined as w ( x, y, t ) Wk, h, mx k y h t m k h m 2.12 and from 2.11 and 2.12 can be concluded as w ( x, y, t ) [ 1 hkm w ( x, y, t ) ] k!h!m! x k y h t m k h m,, x k y h t m. 2.1

5 Mathematical Problems in Engineering 5 Original function Table 2: Fundamental theorems for three-dimensional case. Transformed function 1 wx, y, t ux, y, t ± vx, y, t Wh, k, m Uk, h, m ± V k, h, m 2 wx, y, t cux, y, t Wh, k, m cuk, h, m wx, y, t ux, y, t/ x, Wh, k, m k1uk1,h,m 4 wx, y, t ux, y, t/ y Wh, k, m h1uk, h 1,m 5 wx, y, t ux, y, t/ t Wh, k, m m1uk, h, m 1 6 wx, y, t rsp ux, y, t/ x r y s t p Wh, k, m k1k 2 k r h 2 h sm 1m 2 m p 7 wx, y, t ux, y, tvx, y, t Wh, k, m k h m r s p Ur, h s, V k r, s, p 8 wx, y, t ux, y, t/ x vx, y, t/ y Wh, k, m k h m r s pk r 1h s 1 Uk r 1,s,p V r, h s 1,m p Table : Fundamental theorems for four-dimensional case. Original function Transformed function 1 wx, y, z, t ux, y, z, t ± vx, y, z, t Wh, k, m, n Uk, h, m, n ± V k, h, m, n 2 wx, y, z, t cux, y, z, t Wh, k, m, n cuk, h, m, n wx, y, z, t rspq Wh, k, m, n k1k 2 k rh 1h 2 ux, y, z, t/ x r y s z p t q hsm1m2 mpn1n2 n quk r, h s, m p, n q 4 wx, y, z, t ux, y, z, t/ t Wh, k, m, n m1uk, h, m, n 1 Wh, k, m, n k h m n r s p q 5 wx, y, z, t ux, y, z, tvx, y, z, t Ur, h s, m p, n qv k r, s, p, q 2.4. Four-Dimensional Differential Transform Definition 2.7. Given an w function which has four components such as x, y, z, t. Fourdimensional differential transform of wx, y, z, t is defined as [ 1 hkmn w ( x, y, z, t ) ] Wk, h, m, n k!h!m!n! x k y h z m t n,,,, 2.14 where wx, y, z, t is the original function and Wk, h, m, n is the transformed function. Again, the transformation can be called T-function, and the lower case and upper case letters represent the original and transformed functions, respectively Table. Definition 2.8. The differential inverse transform of Wk, h, m, n is defined as w ( x, y, z, t ) Wk, h, m, nx k y h z m t n k h m n 2.15 and from 2.14 and 2.15 can be concluded as w ( x, y, z, t ) [ 1 hkm w ( x, y, t ) ] k!h!m!n! x k y h t m k h m n,,, x k y h z m t n. 2.16

6 6 Mathematical Problems in Engineering Theorem 2.. If wx, y, z, t = ux, y, z, t)vx, y, z, tωx, y, z, t, then n n q Wh, k, m, n U [ r, s, p, n q a ] V [ t, i, j, q ] r t s i p j q a W [ k r t, h s i, m p j, a ] Proof. From the definition of transform, W,,, U,,, V,,, ω,,,, W1,,, U1,,, V,,, ω,,, U,,, V 1,,, ω,,, U,,, V,,, ω1,,,, W2,,, U2,,, V,,, ω,,, U1,,, V 1,,, ω,,, U,,, V 2,,, ω,,, U1,,, V,,, ω1,,, U,,, V 1,,, ω1,,, U,,, V,,, ω2,,,, W, 1,, U, 1,, V,,, ω,,, U,,, V, 1,, ω,,, U,,, V,,, ω, 1,,, W, 2,, U, 2,, V,,, ω,,, U, 1,, V, 1,, ω,,, U,,, V, 2,, ω,,,, W,, 1, U,, 1, V,,, ω,,, U,,, V,, 1, ω,,, U,,, V,,, ω,, 1, In general, we have n n q Wh, k, m, n U [ r, s, p, n q a ] V [ t, i, j, q ] r t s i p j q a W [ k r t, h s i, m p j, a ] Analysis of Method To investigate the solution of 1., we first construct a system by differential transformation method as follows: n 1V k, h, m, n 1 k 1k 2Uk 2,h,m,n h 1h 2Uk, h 2,m,n m 1m 2Uk, h, m 2,n U [ r, s, p, n q a ] U [ t, i, j, q ] r t s i p j q a U [ k r t, h s i, m p j, a ]

7 Mathematical Problems in Engineering 7 U [ r, s, p, n q a ] V [ t, i, j, q ] r t s i p j q a V [ k r t, h s i, m p j, a ] k h p U [ r, h s, m p, n q ] S [ k r, s, p, q ], r s p q n 1Uk, h, m, n 1 k 1k 2V k 2,h,m,n h 1h 2V k, h 2,m,n m 1m 2V k, h, m 2,n U [ r, s, p, n q a ] U [ t, i, j, q ] r t s i p j q a V [ k r t, h s i, m p j, a ] V [ r, s, p, n q a ] V [ t, i, j, q ] r t s i p j q a V [ k r t, h s i, m p j, a ] k h p V [ r, h s, m p, n q ] S [ k r, s, p, q ]. r s p q.1 Subject to Taylor s expand of initial condition u ( x, y, z, ) G ( x, y, z ) F,, v ( x, y, z, ) G ( x, y, z ) G,, k h m k h m Uk, h, mx k y h z m k h m k h m F khm,, x k y h z m, k!h!m! V k, h, mx k y h z m G khm,, x k y h z m. k!h!m!.2 Substituting.2 into.1 and using operation of Table, weget V k, h, m, n 1 1 k 1k 2Uk 2,h,m,n h 1h 2Uk, h 2,m,n n 1 m 1m 2Uk, h, m 2,n U [ r, s, p, n q a ] U [ t, i, j, q ] r t s i p j q a

8 8 Mathematical Problems in Engineering U [ k r t, h s i, m p j, a ] U [ r, s, p, n q a ] V [ t, i, j, q ] r t s i p j q a V [ k r t, h s i, m p j, a ] k h p U [ r, h s, m p, n q ] S [ k r, s, p, q ], r s p q Uk, h, m, n 1 1 k 1k 2V k 2,h,m,n h 1h 2V k, h 2,m,n n 1 m 1m 2V k, h, m 2,n U [ r, s, p, n q a ] U [ t, i, j, q ] r t s i p j q a V [ k r t, h s i, m p j, a ] V [ r, s, p, n q a ] V [ t, i, j, q ] r t s i p j q a V [ k r t, h s i, m p j, a ] k h p V [ r, h s, m p, n q ] S [ k r, s, p, q ]. r s p q. In order to obtain the unknowns of Uk, h, m, n and V k, h, m, n, k, h, m, n, 1, 2,...,we must construct and solve the above equation and substitute in Application In this section, the differential transformation technique is applied to solve 1.. This method only needs the initial condition of covering PDEs. Firstly, we consider the solution of 1. with the initial conditions: u ( x, y, z, ) 1 Tanh ( x y z ), v ( x, y, z, ) 1 Tanh ( x y z ), S ( x, y, z, ) 4 8Tanh ( x y z )

9 Mathematical Problems in Engineering Suppose that x y z t, in Definition 2.7, then we have U,,, 1, U1,,, 1, U2,,,, U,,, 1, U4,,,, 2 U5,,, 15, U, 1,, 1, U,, 2,, U,,, 1, U2,,, 4 2, U,, 4,, U,, 5,, U2, 1,, 1, 15 U4, 1,, 2, U1, 2,, 1, U, 2,,, U,,, 1, U, 4,,, 2 U, 5,, 15, U,, 1, 1, U4,,, 17, U1, 4,, 2, U, 4,, 17, U2,, 1, 1, U4,, 1, 2, 4.2 U,, 2, 4, U4,,, 17, U1,, 4, 2, U,, 4, 17, U, 2, 1, 1,... Using. and 4.2 and by recursive method, we get U,,, 1 6, U,, 1, 1, U, 1,, 1, U1,,, 1, U, 1, 1, 1 12, U1,, 1, 1 12, U1, 1,, 1 12, U1, 1, 1, 1, U,, 2, 1 6, U, 2,, 1 6, U2,,, 1 6, U,,, 2, U2, 1,, 1, U, 1, 2, 1, U, 2, 1, 1, U2,, 1, 1, U1,, 2, 1,..., V,,, 1 6, V,, 1, 1, V, 1,, 1, V 1,,, 1, V, 1, 1, 1 12, V1,, 1, 1 12, V 1, 1,, 1 12, V1, 1, 1, 1, V,, 2, 1 6, V, 2,, 1 6, V 2,,, 1 6, V,,, 2, V 2, 1,, 1, V, 1, 2, 1, V, 2, 1, 1, V1,, 2, 1, V 1, 2,, 1, V1, 2, 1, 1 48, V2, 1, 1, 1 16, V 2, 2,, 1 24, V2,, 2, 1 24, V1, 1, 2, 1 16, V, 2, 2, 1 24, V2, 2, 1, 1 64, V1, 2, 2, 1 64,... 4.

10 1 Mathematical Problems in Engineering Substituting all Uk, h, m, n and V k, h, m, n into 1.2 yields u ( x, y, z, t ) 1 6t x 2t 2 x 6tx 2 56t 2 x 2 x 4tx4 y 2t 2 y 12txy x 2 y 2x4 y 6ty 2 56t 2 y 2 xy 2 24tx 2 y 2 4x y 2 17x y 4 z 2t 2 z 12txz x 2 z 2x4 z y 4x2 y 17x4 y 4ty 4 2xy4 12tyz 48t 2 yz 2xyz 16tx 2 yz 8 x yz y 2 z 48txy 2 z 4x 2 y 2 z 64tx 2 y 2 z 17 x4 y 2 z 16ty z 8 xy z 2y4 z 17 x2 y 4 z 6tz 2 56t 2 z 2 xz 2 24tx 2 z 2 4x z 2 yz 2 16txyz 2 4x 2 yz 2 128tx 2 yz 2 17 x4 yz 2 24ty 2 z 2 4xy 2 z 2 64txy 2 z 2 24tx 2 y 2 z 2 4 x y 2 z 2 4y z 2 4y2 z 4 x2 y 2 z 17y4 z 17 xy2 z 4 17y z 4, 4 x2 y z 2 17 xy4 z 2 z 4tz 4 2xz4 4x2 z 17x z 4 17x4 z 16tyz 8 xyz 2yz4 17 x2 yz v ( x, y, z, t ) 16t x 2t 2 x 6tx 2 56t 2 x 2 x 4tx4 y 2t 2 y 12txy x 2 y 2x4 y 6ty2 56t 2 y 2 xy 2 24tx 2 y 2 4x y 2 y 4x2 y 17x4 y 4ty 4 2xy4 17x y 4 z 2t 2 z 12txz x 2 z 2x4 z 12tyz 48t 2 yz 2xyz 16tx 2 yz 8 x yz y 2 z 48txy 2 z 4x 2 y 2 z 64tx 2 y 2 z 17 x4 y 2 z 16ty z 8 xy z 2y4 z 17 x2 y 4 z 6tz 2 56t 2 z 2 xz 2 24tx 2 z 2 4x z 2 yz 2 16txyz 2 4x 2 yz 2 128tx 2 yz 2 17 x4 yz 2 24ty 2 z 2 4xy 2 z 2 64txy 2 z 2 24tx 2 y 2 z 2 4 x y 2 z 2 4 x2 y z 2 17 xy4 z 2 z 4x2 z 4y z 2 4 x2 y 2 z 17y4 z 17 xy2 z 4 17y z 4, 4tz 4 2xz4 17x z 4 17x4 z 16tyz 8 xyz 4y2 z 2yz4 17 x2 yz 4 4.5

11 Mathematical Problems in Engineering 11 Table 4: Exact and approximate solutions and absolute errors of ux, y, z, t, y z.1. x t Exact solutions Approximate solutions with DTM Absolute errors Table 5: Exact and approximate solutions and absolute errors of vx, y, z, t, y z.1. x t Exact solutions Approximate solutions with DTM Absolute errors Figure 1: The analytic solution of ux, y, z, t. and the analytical solution of the problem by Tanh method is u ( x, y, z, t ) 1 Tanh ( x y z 6t ), v ( x, y, z, t ) 1 Tanh ( x y z 6t ). 4.6

12 12 Mathematical Problems in Engineering Figure 2: The approximate solution with DTM of u x, y, z, t Figure : The analytic solution of v x, y, z, t Figure 4: The approximate solution with DTM of v x, y, z, t.

13 Mathematical Problems in Engineering t (x, y, z) = (.1,.1,.2) (x, y, z) = (.2,.2,.1) (x, y, z) = (.,.1,.1) Figure 5: The error of u exact -u DTM t v(x, y, z) = (.1,.1,.2) v(x, y, z) = (.2,.2,.1) v(x, y, z) = (.,.1,.1) Figure 6: The error of v exact -v DTM. When 4.4, 4.5, and4.6 are compared, it can be seen that these two results are quite compatible for small values of t. For example, for x.1,y.1, z.1, and t.1, analytic solution of ux, y, z, t is and transform solution is , and analytic solution of vx, y, z, t is and transform solution is For x.1, y.1, z.1, and t.1, analytic solution of ux, y, z, t is and transform solution is , and analytic solution of vx, y, z, t is and transform solution is Tables 4 and 5, Figures 1 to 6.

14 14 Mathematical Problems in Engineering 5. Conclusion Four-dimensional differential transform has been applied to nonlinear system of PDEs. Analytic solution and transform solution are compared; the results are quite compatible for small values of t. The present method reduces the computational difficulties of the other methods, and all the calculations can be made by simple manipulations. On the other hand, the results are quite reliable. Therefore, this method can be applied to many complicated linear and nonlinear PDEs and system of PDEs and does not require linearization, discretization, or perturbation. Acknowledgments This paper was funded by the Deanship of Scientific Research DSR, King Abdulaziz University, Jeddah, under Grant no D142. The authors, therefore, acknowledge with thanks DSR technical and financial support. References 1 A. M. Wazwaz, The decomposition method applied to systems of partial differential equations and to the reaction-diffusion Brusselator model, Applied Mathematics and Computation, vol. 11, no. 2-, pp , 2. 2 J. K. Zhou, Differential Transformation and Its Application for Electrical Circuits, Huazhong University Press, Wuhan, China, 186. C. Kuang Chen and S. Huei Ho, Solving partial differential equations by two-dimensional differential transform method, Applied Mathematics and Computation, vol. 16, no. 2-, pp , 1. 4 M. J. Jang, C. L. Chen, and Y. C. Liu, Two-dimensional differential transform for partial differential equations, Applied Mathematics and Computation, vol. 121, no. 2-, pp , F. Ayaz, On the two-dimensional differential transform method, Applied Mathematics and Computation, vol. 14, pp , 2. 6 F. Ayaz, Applications of differential transform method to differential-algebraic equations, Applied Mathematics and Computation, vol. 152, no., pp , F. Ayaz, Solutions of the system of differential equations by differential transform method, Applied Mathematics and Computation, vol. 147, no. 2, pp , 24.

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