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Discrete Dynamics in Nature and Society
Volume 2012 (2012), Article ID 434976, 15 pages
http://dx.doi.org/10.1155/2012/434976
Research Article

Finite Difference and Iteration Methods for Fractional Hyperbolic Partial Differential Equations with the Neumann Condition

1Department of Mathematics, Fatih University, Buyukcekmece 34500, Istanbul, Turkey
2Department of Mathematics, ITTU, Ashgabad, Turkmenistan
3Department of Mathematics, Ege University, 35100 Izmir, Turkey

Received 19 December 2011; Accepted 18 April 2012

Academic Editor: Chuanxi Qian

Copyright © 2012 Allaberen Ashyralyev and Fadime Dal. 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.

Abstract

The numerical and analytic solutions of the mixed problem for multidimensional fractional hyperbolic partial differential equations with the Neumann condition are presented. The stable difference scheme for the numerical solution of the mixed problem for the multidimensional fractional hyperbolic equation with the Neumann condition is presented. Stability estimates for the solution of this difference scheme and for the first- and second-order difference derivatives are obtained. A procedure of modified Gauss elimination method is used for solving this difference scheme in the case of one-dimensional fractional hyperbolic partial differential equations. He's variational iteration method is applied. The comparison of these methods is presented.

1. Introduction

It is known that various problems in fluid mechanics (dynamics, elasticity) and other areas of physics lead to fractional partial differential equations. Methods of solutions of problems for fractional differential equations have been studied extensively by many researchers (see, e.g., [115] and the references given therein).

The role played by stability inequalities (well posedness) in the study of boundary-value problems for hyperbolic partial differential equations is well known (see, e.g., [1629]).

In the present paper, finite difference and He's iteration methods for the approximate solutions of the mixed boundary-value problem for the multidimensional fractional hyperbolic equation 𝜕2𝑢(𝑡,𝑥)𝜕𝑡2𝑚𝑟=1𝑎𝑟(𝑥)𝑢𝑥𝑟𝑥𝑟+𝐷𝑡1/2𝑥𝑢(𝑡,𝑥)+𝜎𝑢(𝑡,𝑥)=𝑓(𝑡,𝑥),𝑥=1,,𝑥𝑚Ω,0<𝑡<1,𝑢(0,𝑥)=0,𝑢𝑡(0,𝑥)=0,𝑥Ω;𝜕𝑢(𝑡,𝑥)𝜕𝑛=0,𝑥𝑆,(1.1) are studied. Here Ω is the unit open cube in the 𝑚-dimensional Euclidean space: 𝑚{Ω=𝑥=(𝑥1,,𝑥𝑚)0<𝑥𝑗<1,1𝑗𝑚} with boundary 𝑆,Ω=Ω𝑆; 𝑎𝑟(𝑥)(𝑥Ω) and 𝑓(𝑡,𝑥)(𝑡(0,1),𝑥Ω) are given smooth functions and 𝑎𝑟(𝑥)𝑎>0.

1.1. Definition

The Caputo fractional derivative of order 𝛼>0 of a continuous function 𝑢(𝑡,𝑥) is defined by 𝐷𝛼𝑎+1𝑢(𝑡,𝑥)=Γ(1𝛼)𝑡𝑎𝑢(𝑡,𝑥)(𝑡𝑠)𝛼𝑑𝑠,(1.2) where Γ() is the gamma function.

2. The Finite Difference Method

In this section, we consider the first order of accuracy in 𝑡 and the second-orders of accuracy in space variables’ stable difference scheme for the approximate solution of problem (1.1). The stability estimates for the solution of this difference scheme and its first- and second-order difference derivatives are established. A procedure of modified Gauss elimination method is used for solving this difference scheme in the case of one-dimensional fractional hyperbolic partial differential equations.

2.1. The Difference Scheme: Stability Estimates

The discretization of problem (1.1) is carried out in two steps. In the first step, let us define the grid space Ω=𝑥=𝑥𝑟=1𝑟1,,𝑚𝑟𝑚𝑟,𝑟=1,,𝑟𝑚,0𝑟𝑗𝑁𝑗,𝑗𝑁𝑗,Ω=1,𝑗=1,,𝑚=ΩΩ,𝑆=Ω𝑆.(2.1) We introduce the Banach space 𝐿2=𝐿2(Ω) of the grid functions 𝜑(𝑥)={𝜑(1𝑟1,,𝑚𝑟𝑚)} defined on Ω, equipped with the norm 𝜑𝐿2(Ω)=𝑥Ω||𝜑||(𝑥)21𝑚1/2.(2.2) To the differential operator 𝐴𝑥 generated by problem (1.1), we assign the difference operator 𝐴𝑥 by the formula 𝐴𝑥𝑢=𝑚𝑟=1𝑎𝑟(𝑥)𝑢𝑥𝑟𝑥𝑟,𝑗+𝜎𝑢𝑥(2.3) acting in the space of grid functions 𝑢(𝑥), satisfying the conditions 𝐷𝑢(𝑥)=0 for all 𝑥𝑆. It is known that 𝐴𝑥 is a self-adjoint positive definite operator in 𝐿2(Ω). With the help of 𝐴𝑥 we arrive at the initial boundary value problem 𝑑2𝑣(𝑡,𝑥)𝑑𝑡2+𝐷𝑡1/2𝑣(𝑡,𝑥)+𝐴𝑥𝑣(𝑡,𝑥)=𝑓(𝑡,𝑥),0𝑡1,𝑥Ω,𝑣(0,𝑥)=0,𝑑𝑣(0,𝑥)𝑑𝑡=0,𝑥Ω,(2.4) for an infinite system of ordinary fractional differential equations.

In the second step, we replace problem (2.4) by the first order of accuracy difference scheme 𝑢𝑘+1(𝑥)2𝑢𝑘(𝑥)+𝑢𝑘1(𝑥)𝜏2+1𝜋𝑘𝑚=1Γ(𝑘𝑚+1/2)𝑢(𝑘𝑚)!𝑚𝑢m1𝜏1/2+𝐴𝑥𝑢𝑘+1=𝑓𝑘Ω(𝑥),𝑥,𝑓𝑘𝑡(𝑥)=𝑓𝑘,𝑥,𝑡𝑘Ω=𝑘𝜏,1𝑘𝑁1,𝑁𝜏=1,𝑥,𝑢1(𝑥)𝑢0(𝑥)𝜏=0,𝑢0Ω(𝑥)=0,𝑥.(2.5) Here Γ(𝑘𝑚+1/2)=0𝑡𝑘𝑚1/2𝑒𝑡𝑑𝑡.

Theorem 2.1. Let 𝜏 and ||=21++2𝑚 be sufficiently small numbers. Then, the solutions of difference scheme (2.5) satisfy the following stability estimates: max1𝑘𝑁𝑢𝑘𝐿2+max1𝑘𝑁𝑢𝑘𝑢𝑘1𝜏𝐿2𝐶1max1𝑘𝑁1𝑓𝑘𝐿2,max1𝑘𝑁1𝜏2𝑢𝑘+12𝑢𝑘+𝑢𝑘1𝐿2+max1𝑘𝑁𝑢𝑘𝑥𝑟𝑥𝑟𝐿2𝐶2𝑓1𝐿2+max2𝑘𝑁1𝜏1𝑓𝑘𝑓𝑘1𝐿2.(2.6)
Here 𝐶1 and 𝐶2 do not depend on 𝜏, , and 𝑓𝑘,1𝑘𝑁1.

The proof of Theorem 2.1 is based on the self-adjointness and positive definitness of operator 𝐴𝑥 in 𝐿2 and on the following theorem on the coercivity inequality for the solution of the elliptic difference problem in 𝐿2.

Theorem 2.2. For the solutions of the elliptic difference problem 𝐴𝑥𝑢(𝑥)=𝜔(𝑥),𝑥Ω,𝐷𝑢(𝑥)=0,𝑥𝑆,(2.7) the following coercivity inequality holds [30]: 𝑚𝑟=1𝑢𝑥𝑟𝑥𝑟𝐿2𝜔𝐶𝐿2.(2.8)

Finally, applying this difference scheme, the numerical methods are proposed in the following section for solving the one-dimensional fractional hyperbolic partial differential equation. The method is illustrated by numerical examples.

2.2. Numerical Results

For the numerical result, the mixed problem 𝐷2𝑡𝑢(𝑡,𝑥)+𝐷𝑡1/2𝑢(𝑡,𝑥)𝑢𝑥𝑥(𝑡,𝑥)+𝑢(𝑡,𝑥)=𝑓(𝑡,𝑥),𝑓(𝑡,𝑥)=2+𝑡2+8𝑡3/23𝜋+(𝜋𝑡)2cos(𝜋𝑥),0<𝑡,𝑥<1,𝑢(0,𝑥)=0,𝑢𝑡(𝑢0,𝑥)=0,0𝑥1,𝑥(𝑡,0)=𝑢𝑥(𝑡,1)=0,0𝑡1(2.9) for solving the one-dimensional fractional hyperbolic partial differential equation is considered. Applying difference scheme (2.5), we obtained 𝑢𝑛𝑘+12𝑢𝑘𝑛+𝑢𝑛𝑘1𝜏2+1𝜋𝑘𝑚=1Γ(𝑘𝑚+1/2)𝑢(𝑘𝑚)!𝑚𝑛𝑢𝑛𝑚1𝜏1/2𝑢𝑘+1𝑛+12𝑢𝑛𝑘+1+𝑢𝑘+1𝑛12+𝑢𝑘𝑛=𝜑𝑘𝑛,𝜑𝑘𝑛𝑡=𝑓𝑘,𝑥𝑛𝑢,1𝑘𝑁1,1𝑛𝑀1,0𝑛=0,𝜏1𝑢1𝑛𝑢0𝑛𝑢=0,0𝑛𝑀,𝑘1𝑢𝑘0=𝑢𝑘𝑀𝑢𝑘𝑀1=0,0𝑘𝑁.(2.10) We get the system of equations in the matrix form: 𝐴𝑈𝑛+1+𝐵𝑈𝑛+𝐶𝑈𝑛1=𝐷𝜑𝑛𝑈,1𝑛𝑀1,1=𝑈0,𝑈𝑀=𝑈𝑀1,(2.11) where 𝐴=00000000000000000000𝑎000000𝑎000000𝑎0000000(𝑁+1)×(𝑁+1),𝑏𝐵=1,1𝑏000002,1𝑏2,2𝑏00003.1𝑏3,2𝑏3,3𝑏0004,1𝑏4,2𝑏4,3𝑏4,4𝑏00𝑁,1𝑏𝑁,2𝑏𝑁,3𝑏𝑁,4𝑏𝑁,𝑁0𝑏𝑁+1,1𝑏𝑁+1,2𝑏𝑁+1,3𝑏𝑁+1,4𝑏𝑁+1,𝑁𝑏𝑁+1,𝑁+1(𝑁+1)×(𝑁+1),𝐶=𝐴,𝐷=000000000000000000001000000100000010000001(𝑁+1)×(𝑁+1),𝑈𝑠=𝑈0𝑠𝑈1𝑠𝑈2𝑠𝑈3𝑠𝑈𝑠𝑁1𝑈𝑁𝑠(𝑁+1)×(1),𝑠=𝑛1,𝑛,𝑛+1.(2.12) Here 1𝑎=2,𝑏1,1=1,𝑏2,1=1,𝑏2,2=1,𝑏3,1=1𝜏21𝜏1/2,𝑏3,22=𝜏2+1𝜏1/2,𝑏3,31=1+𝜏2+22,𝑏𝑘+2,11=𝜋Γ(𝑘1+1/2)Γ(𝑘)𝜏1/2𝑏,2𝑘𝑁1,𝑘+2,𝑘+12=𝜏2+1𝜏1/2𝑏,1𝑘𝑁1,𝑘+2,𝑘=1𝜏2+1𝜋Γ(1+0.5)Γ(2)Γ(0.5)1Γ(1)𝜏1/2𝑏,2𝑘𝑁1,𝑘+2,𝑘+21=1+𝜏2+22𝑏,1𝑘𝑁1,𝑘+2,𝑖+1=1𝜋Γ(𝑘𝑖+1/2)Γ(𝑘(𝑖1))Γ(𝑘(𝑖+1)+1/2)1Γ(𝑘(𝑖1)1)𝜏1/2𝜑,3𝑘𝑁1,1𝑖𝑘2,𝑘𝑛=2+(𝑘𝜏)2+8(𝑘𝜏)3/23𝜋+(𝜋𝑘𝜏)2𝜑cos𝜋(𝑛),𝑛=𝜑0𝑛𝜑1𝑛𝜑2𝑛𝜑𝑁𝑛(𝑁+1)×1.(2.13) So, we have the second-order difference equation with respect to 𝑛 matrix coefficients. To solve this difference equation, we have applied a procedure of modified Gauss elimination method for difference equation with respect to 𝑘 matrix coefficients. Hence, we seek a solution of the matrix equation in the following form: 𝑈𝑗=𝛼𝑗+1𝑈𝑗+1+𝛽𝑗+1,(2.14)𝑛=𝑀1,,2,1,  𝛼𝑗(𝑗=1,,𝑀) are (𝑁+1)×(𝑁+1) square matrices, and 𝛽𝑗(𝑗=1,,𝑀) are (𝑁+1)×1 column matrices defined by 𝛼𝑛+1=𝐵+𝐶𝛼𝑛1𝛽(𝐴),𝑛+1=𝐵+𝐶𝛼𝑛1𝐷𝜑𝑛𝐶𝛽𝑛,𝑛=2,3,,𝑀,(2.15) where 𝛼1=1000010000100001(𝑁+1)×(𝑁+1),𝛽1=0000(𝑁+1)×1.(2.16)

Now, we will give the results of the numerical analysis. First, we give an estimate for the constants 𝐶1 and 𝐶2 figuring in the stability estimates of Theorem 2.1. We have 𝐶1=max𝑓,𝑢𝐶𝑡1,𝐶2=max𝑓,𝑢𝐶𝑡2,𝐶𝑡1=max1𝑘𝑁𝑢𝑘𝐿2+max1𝑘𝑁𝜏1𝑢𝑘𝑢𝑘1𝐿2×max1𝑘𝑁1𝑓𝑘𝐿21,𝐶𝑡2=max1𝑘𝑁1𝜏2𝑢𝑘+12𝑢𝑘+𝑢𝑘1𝐿2+max𝑛1𝑘𝑁𝑟=1𝑢𝑘𝑥𝑟,𝑥𝑟𝐿2×max2𝑘𝑁1𝜏1𝑓𝑘𝑓𝑘1𝐿2+𝑓1𝐿21.(2.17) The constants 𝐶𝑡1 and 𝐶𝑡2 in the case of numerical solution of initial-boundary value problem (2.9) are computed. The constants 𝐶𝑡1 and 𝐶𝑡2 are given in Table 1 for 𝑁=20,40,80, and 𝑀=80, respectively.

tab1
Table 1: Stability estimates for (2.9).

Second, for the accurate comparison of the difference scheme considered, the errors computed by 𝐸0=max1𝑘𝑁1𝑀1𝑛=1||𝑢(𝑡𝑘,𝑥𝑛)𝑢𝑘𝑛||21/2,𝐸1=max1𝑘𝑁1𝑀1𝑛=1||||𝑢𝑡(𝑡𝑘,𝑥𝑛)(𝑢𝑛𝑘+1𝑢𝑛𝑘1)||||2𝜏21/2,𝐸2=max1𝑘𝑁1𝑀1𝑛=1||||𝑢𝑡𝑡(𝑡𝑘,𝑥𝑛)(𝑢𝑛𝑘+12𝑢𝑘𝑛+𝑢𝑛𝑘1)𝜏2||||21/2(2.18) of the numerical solutions are recorded for higher values of 𝑁=𝑀, where 𝑢(𝑡𝑘,𝑥𝑛) represents the exact solution and 𝑢𝑘𝑛 represents the numerical solution at (𝑡𝑘,𝑥𝑛). The errors 𝐸0,𝐸1 and 𝐸2 results are shown in Table 2 for 𝑁=20,40,60 and 𝑀=60, respectively.

tab2
Table 2: Comparison of the errors for the difference scheme.

The figure of the difference scheme solution of (2.9) is given by the Figure 2. The exact solution of (2.9) is given by as follows: 𝑢(𝑡,𝑥)=𝑡2cos(𝜋𝑥).(2.19) The figure of the exact solution of (2.9) is shown by the Figure 1.

434976.fig.001
Figure 1: The surface shows the exact solution 𝑢(𝑡,𝑥) for (2.9).
434976.fig.002
Figure 2: Difference scheme solution for (2.9).

3. He's Variational Iteration Method

In the present paper, the mixed boundary value problem for the multidimensional fractional hyperbolic equation (1.1) is considered. The correction functional for (1.1) can be approximately expressed as follows: 𝑢𝑛+1(𝑡,𝑥)=𝑢𝑛+(𝑡,𝑥)𝑡0𝜆𝜕2𝑢(𝑠,𝑥)𝜕𝑠2𝑚𝑟=1𝑎𝑟(𝑥)̃𝑢𝑥𝑟𝑥𝑟+D𝑠1/2̃𝑢(𝑠,𝑥)+𝜎̃𝑢(𝑠,𝑥)𝑓(𝑠,𝑥)𝑑𝑠,(3.1) where 𝜆 is a general Lagrangian multiplier (see, e.g., [31]) and ̃𝑢 is considered as a restricted variation as a restricted variation (see, e.g., [32]); that is, 𝛿̃𝑢=0, 𝑢0(𝑡,𝑥) is its initial approximation. Using the above correction functional stationary and noticing that 𝛿̃𝑢=0, we obtain 𝛿𝑢𝑛+1(𝑡,𝑥)=𝛿𝑢𝑛(𝑡,𝑥)+𝑡0𝛿𝜆𝜕𝑢2𝑛(𝑡,𝑥)𝜕𝑠2𝑑𝑠,𝛿𝑢𝑛+1(𝑡,𝑥)=𝛿𝑢𝑛(𝑡,𝑥)𝜕𝜆𝜕𝑠𝛿𝑢𝑛|||(𝑠,𝑥)𝑠=𝑡𝜕+𝜆𝜕𝑠𝛿𝑢𝑛|||(𝑠,𝑥)𝑠=𝑡+𝑡0𝜕2𝜆(𝑡,𝑠)𝜕𝑠2𝛿𝑢𝑛(𝑠,𝑥)𝑑𝑠=0.(3.2) From the above relation for any 𝛿𝑢𝑛, we get the Euler-Lagrange equation: 𝜕𝜆2(𝑡,𝑠)𝜕𝑠2=0,(3.3) with the following natural boundary conditions: 1𝜕𝜆(𝑡,𝑠)|||𝜕𝑠𝑠=𝑡||=0,𝜆(𝑡,𝑠)𝑠=𝑡=0.(3.4) Therefore, the Lagrange multiplier can be identified as follows: 𝜆(𝑡,𝑠)=𝑠𝑡.(3.5) Substituting the identified Lagrange multiplier into (3.1), the following variational iteration formula can be obtained: 𝑢𝑛+1(𝑡,𝑥)=𝑢𝑛(𝑡,𝑥)+𝑡0𝜕(𝑠𝑡)2𝑢𝑛(𝑠,𝑥)𝜕𝑠2𝑚𝑟=1𝑎𝑟(𝑥)𝑢𝑛𝑥𝑟𝑥𝑟+𝐷𝑠1/2𝑢𝑛(𝑠,𝑥)+𝑢𝑛(𝑠,𝑥)𝑓(𝑠,𝑥)𝑑𝑠.(3.6)

In this case, let an initial approximation 𝑢0(𝑡,𝑥)=𝑢(0,𝑥)+𝑡𝑢𝑡(0,𝑥). Then approximate solution takes the form 𝑢(𝑡,𝑥)=lim𝑛u𝑛(𝑡,𝑥).

3.1. Variational Iteration Solution 1

For the numerical result, the mixed problem 𝐷2𝑡𝑢(𝑡,𝑥)+𝐷𝑡1/2𝑢(𝑡,𝑥)𝑢𝑥𝑥(𝑡,𝑥)+𝑢(𝑡,𝑥)=𝑓(𝑡,𝑥),𝑓(𝑡,𝑥)=2+𝑡2+8𝑡3/23𝜋+(𝜋𝑡)2cos(𝜋𝑥),0<𝑡,𝑥<1,𝑢(0,𝑥)=0,𝑢𝑡(𝑢0,𝑥)=0,0𝑥1,𝑥(𝑡,0)=𝑢𝑥(𝑡,1)=0,0𝑡1(3.7) for solving the one-dimensional fractional hyperbolic partial differential equation is considered.

According to formula (3.6), the iteration formula for (3.7) is given by 𝑢𝑛+1(𝑡,𝑥)=𝑢𝑛+(𝑡,𝑥)𝑡0(𝑠𝑡)𝜕𝑢2𝑛(𝑠,𝑥)𝜕𝑠2+𝐷𝑠1/2𝑢𝑛(𝑠,𝑥)𝜕𝑢2𝑛(𝑠,𝑥)𝜕𝑥2+𝑢𝑛(𝑠,𝑥)𝑓(𝑠,𝑥)𝑑𝑠.(3.8) Now we start with an initial approximation 𝑢0(𝑡,𝑥)=𝑢(0,𝑥)+𝑡𝑢𝑡(0,𝑥).(3.9) Using the above iteration formula (3.8), we can obtain the other components as 𝑢0𝑢(𝑡,𝑥)=0,11(𝑡,𝑥)=420𝜋128𝑡7/2+35𝑡4𝜋+35𝑡4𝜋5/2+420𝑡2𝜋𝑢cos(𝜋𝑥),21(𝑡,𝑥)=420𝜋cos(𝜋𝑥)128𝑡7/2+35𝑡4𝜋+35𝑡4𝜋5/2+420𝑡2𝜋[+cos(𝜋𝑥)0.9058003666𝑡40.1510268880𝑡11/20.1719434921𝑡7/20.3281897218𝑡60.01666666667𝑡5](3.10) The figure of (3.10) is given by the Figure 3.

434976.fig.003
Figure 3: Variational iteration method for (3.10).
3.2. Variational Iteration Solution 2

For the numerical result, the mixed problem 𝐷2𝑡𝑢(𝑡,𝑥,𝑦)+𝐷𝑡1/2𝑢(𝑡,𝑥,𝑦)𝑢𝑥𝑥(𝑡,𝑥,𝑦)𝑢𝑦𝑦(𝑡,𝑥,𝑦)+𝑢(𝑡,𝑥,𝑦)=𝑓(𝑡,𝑥,𝑦),𝑓(𝑡,𝑥,𝑦)=2+𝑡2+8𝑡3/23𝜋+(𝜋𝑡)2cos(𝜋𝑥)cos(𝜋𝑦),0<𝑡,𝑥<1,𝑦<1,𝑢(0,𝑥,𝑦)=0,𝑢𝑡𝑢(0,𝑥,𝑦)=0,0𝑥1,0𝑦1,𝑥(𝑡,0,𝑦)=𝑢𝑥(𝑢𝑡,1,𝑦)=0,0𝑡1,0𝑦1,𝑦(𝑡,𝑥,0)=𝑢𝑦(𝑡,𝑥,1)=0,0𝑡1,0𝑥1(3.11) for solving the two-dimensional fractional hyperbolic partial differential equation is considered.

According to formula (3.6), the iteration formula for (3.11) is given by 𝑢𝑛+1(𝑡,𝑥,𝑦)=𝑢𝑛+(𝑡,𝑥,𝑦)𝑡0𝜕(𝑠𝑡)2𝑢𝑛(𝑠,𝑥,𝑦)𝜕𝑠2+𝐷𝑠1/2𝑢𝑛(𝑠,𝑥,𝑦)𝜕𝑢2𝑛(𝑠,𝑥,𝑦)𝜕𝑥2𝜕𝑢2𝑛(𝑠,𝑥,𝑦)𝜕𝑦2+𝑢𝑛(𝑠,𝑥,𝑦)𝑓(𝑠,𝑥,𝑦)𝑑𝑠;(3.12) we start with an initial approximation 𝑢0(𝑡,𝑥,𝑦)=𝑢(0,𝑥,𝑦)+𝑡𝑢𝑡(0,𝑥,𝑦).(3.13) Using the above iteration formula (3.12), we can obtain the other components as 𝑢0𝑢(𝑡,𝑥,𝑦)=0,11(𝑡,𝑥,𝑦)=420𝜋128𝑡7/2+35𝑡4𝜋+35𝑡4𝜋5/2+420𝑡2𝜋𝑢cos(𝜋𝑥)cos(𝜋𝑦),2(𝑡,𝑥,𝑦)=cos(𝜋𝑥)cos(𝜋𝑦)321105𝑡5/2𝑡7/2+1112𝑡5/2𝑡4𝜋+1112𝑡5/2𝑡4𝜋5/2+1𝑡5/2𝑡2𝜋×+cos(𝜋𝑥)cos(𝜋𝑦)0.2195931203𝑡11/2𝜋1𝑡5/20.1719434921𝜋𝑡7/2×1𝑡5/20.6261860981𝑡6𝜋1𝑡5/21.728267400𝜋𝑡41𝑡5/20.01666666667𝜋𝑡5/2(3.14) The exact solution of (3.11) is given by as follows: 𝑢(𝑡,𝑥,𝑦)=𝑡2cos(𝜋𝑥)cos(𝜋𝑦).(3.15) The figure of the exact solution of (3.11) is shown by the Figure 5.

The figure of (3.14) is given by the Figure 4, and so on; in the same manner the rest of the components of the iteration formula (3.12) can be obtained using the Maple package.

434976.fig.004
Figure 4: Variational iteration method for (3.14).
434976.fig.005
Figure 5: The surface shows the exact solution 𝑢(𝑡,𝑥,𝑦) for (3.11).

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