Open Access

Analysis of fractional partial differential equations by Taylor series expansion

Boundary Value Problems20132013:68

DOI: 10.1186/1687-2770-2013-68

Received: 19 December 2012

Accepted: 11 March 2013

Published: 30 March 2013

Abstract

We develop a formulation for the analytic or approximate solution of fractional differential equations (FDEs) by using respectively the analytic or approximate solution of the differential equation, obtained by making fractional order of the original problem integer order. It is shown that this method works for FDEs very well. The results reveal that it is very effective and simple in determination of solutions of FDEs.

1 Introduction

Fractional differential equations (FDEs) are obtained by generalizing differential equations to an arbitrary order. Since fractional differential equations are used to model complex phenomena, they play a crucial role in engineering, physics and applied mathematics. Therefore they have been generating increasing interest from engineers and scientist in recent years. Since FDEs have memory, nonlocal relations in space and time, complex phenomena can be modeled by using these equations. Due to this fact, materials with memory and hereditary effects, fluid flow, rheology, diffusive transport, electrical networks, electromagnetic theory and probability, signal processing, and many other physical processes are diverse applications of FDEs [17].

In [8], the solutions for some nonlinear fractional differential equations are constructed by using symmetry analysis. But, in general, FDEs do not have exact analytic solutions, hence the approximate and numerical solutions of these equations are studied. Analytical approximations of linear and nonlinear FDEs are obtained by the variational iteration method, Adomian’s decomposition method, the homotopy perturbation method and the Lagrange multiplier method [924].

In the present paper, we use the Taylor series of an analytical solution for the differential equations which is obtained from FDEs by making the fractional order of the derivative integer, to obtain the analytical or approximate solution of FDEs. We can obtain the exact or approximate solution of FDEs by changing the terms of Taylor series expansion for a solution of a differential equation in such a way that the relationship among the terms of Taylor series expansion in the sense of derivative and fractional derivative remains the same. Applications of this method show that it is easy and effective when applied to any FDEs as long as the differential equation obtained from FDEs has an analytical or approximate solution. We take the fractional derivative in the Caputo sense.

The structure of this article is as follows. In Section 2, we give the construction of analytical or approximate solutions for FDEs including fractional derivative with respect to time. In the same manner, we obtain the analytical or approximate solution of FDEs with fractional derivative with respect to space variable in Section 3. In Section 4, we take the combination of previous two sections, and we get the analytical or approximate solution of FDEs with fractional derivative with respect to time and space variable. Finally, we give some illustrative examples of this method for all cases in Section 5.

2 Solution of FDEs including fractional derivative with respect to time

Let us consider the following FDE:
D t α u ( x , t ) = F ( u , u x n u x n , x , t ) , m 1 < α m , t > 0 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-68/MediaObjects/13661_2012_Article_317_Equ1_HTML.gif
(1)
In order to determine the solution of this equation, we first need to determine the solution of the following differential equation:
D t m u ( x , t ) = F ( u , u x n u x n , x , t ) , t > 0 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-68/MediaObjects/13661_2012_Article_317_Equ2_HTML.gif
(2)
which is obtained by taking α = m https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-68/MediaObjects/13661_2012_Article_317_IEq1_HTML.gif. After finding an analytic or approximate solution of equation (2), we can obtain the exact or approximate solution of equation (1) by changing the terms of Taylor series expansion for the solution of differential equation (2) in such a way that the relationship among the terms of Taylor series expansion in the sense of derivative and fractional derivative with respect to time remains the same. In other words, we expand the exact or approximate solution into its Taylor series with respect to t. Then we replace the derivatives with respect to t by fractional derivatives with respect to t in such a way that the relation among the terms of Taylor series is preserved. Moreover, we leave the first m terms of Taylor series fixed since α = m https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-68/MediaObjects/13661_2012_Article_317_IEq1_HTML.gif. This also allows the solution of the fractional differential equation to satisfy the initial conditions of the problem. Let us assume that the solution of equation (2) is expanded into its Taylor series with respect to t as follows:
u ( x , t ) = n = 0 n u ( x , 0 ) t n t n n ! . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-68/MediaObjects/13661_2012_Article_317_Equ3_HTML.gif
(3)
Then the solution of equation (1) becomes
u ( x , t ) = n = 0 m 1 n u ( x , 0 ) t n t n n ! + n = 1 i = 0 m 1 m n + i u ( x , 0 ) t m n + i t n α + i Γ ( n α + i + 1 ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-68/MediaObjects/13661_2012_Article_317_Equ4_HTML.gif
(4)
For instance, let us assume that 0 < α 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-68/MediaObjects/13661_2012_Article_317_IEq2_HTML.gif and the solution of equation (2) is u ( x , t ) = e x e t https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-68/MediaObjects/13661_2012_Article_317_IEq3_HTML.gif. In order to find the solution of equation (1), we expand this solution into Taylor series with respect to t as follows:
u ( x , t ) = n = 0 e x t n n ! . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-68/MediaObjects/13661_2012_Article_317_Equa_HTML.gif
Based on (4), the solution of equation (1) can be written in the following form:
u ( x , t ) = e x + e x t α Γ ( α + 1 ) + e x t 2 α Γ ( 2 α + 1 ) + e x t 3 α Γ ( 3 α + 1 ) + . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-68/MediaObjects/13661_2012_Article_317_Equ5_HTML.gif
(5)
Figures 1-3 show the evolution results for the expansion (5) obtained for different values of α.
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-68/MediaObjects/13661_2012_Article_317_Fig1_HTML.jpg
Figure 1

The surface shows the expansion ( 5 ) for α = 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-68/MediaObjects/13661_2012_Article_317_IEq4_HTML.gif .

https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-68/MediaObjects/13661_2012_Article_317_Fig2_HTML.jpg
Figure 2

The surface shows the expansion ( 5 ) for α = 0.75 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-68/MediaObjects/13661_2012_Article_317_IEq5_HTML.gif .

https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-68/MediaObjects/13661_2012_Article_317_Fig3_HTML.jpg
Figure 3

The surface shows the expansion ( 5 ) for α = 0.5 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-68/MediaObjects/13661_2012_Article_317_IEq6_HTML.gif .

3 Solution of FDEs including fractional derivative with respect to space variable

Let us consider the following FDE:
D x α u ( x , t ) = F ( u , u t n u t n , x , t ) , m 1 < α m , t > 0 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-68/MediaObjects/13661_2012_Article_317_Equ6_HTML.gif
(6)
In order to determine the solution of this equation, we first need to determine the solution of the following differential equation:
D x m u ( x , t ) = F ( u , u t n u t n , x , t ) , t > 0 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-68/MediaObjects/13661_2012_Article_317_Equ7_HTML.gif
(7)

which is obtained by taking α = m https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-68/MediaObjects/13661_2012_Article_317_IEq1_HTML.gif.

After finding the analytic or approximate solution of equation (7), we can obtain the exact or approximate solution of equation (6) by changing the terms of Taylor series expansion for the solution of differential equation (7) in such a way that the relationship among the terms of Taylor series expansion in the sense of derivative and fractional derivative with respect to space remains the same. In other words, we expand the exact or approximate solution into its Taylor series with respect to x. Then we replace the derivatives with respect to x by fractional derivatives with respect to x in such a way that the relation among the terms of Taylor series is preserved. Moreover, we leave the first m terms of Taylor series fixed since α = m https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-68/MediaObjects/13661_2012_Article_317_IEq1_HTML.gif. This also allows the solution of the fractional differential equation to satisfy the boundary conditions of the problem.

Let us assume that the solution of equation (7) is expanded into its Taylor series with respect to x as follows:
u ( x , t ) = n = 0 n u ( 0 , t ) x n x n n ! . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-68/MediaObjects/13661_2012_Article_317_Equ8_HTML.gif
(8)
Then the solution of equation (6) becomes
u ( x , t ) = n = 0 m 1 n u ( 0 , t ) x n x n n ! + n = 1 i = 0 m 1 m n + i u ( 0 , t ) x m n + i x n α + i Γ ( n α + i + 1 ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-68/MediaObjects/13661_2012_Article_317_Equ9_HTML.gif
(9)
For instance, let us assume that 0 < α 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-68/MediaObjects/13661_2012_Article_317_IEq2_HTML.gif and the solution of equation (7) is u ( x , t ) = e t e x https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-68/MediaObjects/13661_2012_Article_317_IEq7_HTML.gif. In order to find the solution of equation (6), we expand this solution into Taylor series with respect to x as follows:
u ( x , t ) = n = 0 e t x n n ! . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-68/MediaObjects/13661_2012_Article_317_Equb_HTML.gif
Based on (9), the solution of equation (6) can be written in the following form:
u ( x , t ) = e t + e t x α Γ ( α + 1 ) + e t x 2 α Γ ( 2 α + 1 ) + e t x 3 α Γ ( 3 α + 1 ) + . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-68/MediaObjects/13661_2012_Article_317_Equ10_HTML.gif
(10)
Figures 4-6 show the evolution results for the expansion (10) obtained for different values of α.
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-68/MediaObjects/13661_2012_Article_317_Fig4_HTML.jpg
Figure 4

The surface shows the expansion ( 10 ) for α = 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-68/MediaObjects/13661_2012_Article_317_IEq4_HTML.gif .

https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-68/MediaObjects/13661_2012_Article_317_Fig5_HTML.jpg
Figure 5

The surface shows the expansion ( 10 ) for α = 0.75 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-68/MediaObjects/13661_2012_Article_317_IEq5_HTML.gif .

https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-68/MediaObjects/13661_2012_Article_317_Fig6_HTML.jpg
Figure 6

The surface shows the expansion ( 10 ) for α = 0.5 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-68/MediaObjects/13661_2012_Article_317_IEq6_HTML.gif .

4 Solution of FDEs including fractional derivative with respect to space variable and time

Let us consider the following linear FDE with constant coefficients:
F ( u , D t β u , D x α u , x , t ) = 0 , k 1 < β k , m 1 < α m , t > 0 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-68/MediaObjects/13661_2012_Article_317_Equ11_HTML.gif
(11)
In order to determine the solution of this equation, we first need to determine the solution of the following differential equation:
F ( u , D t k u , D x m u , x , t ) = 0 , t > 0 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-68/MediaObjects/13661_2012_Article_317_Equ12_HTML.gif
(12)
which is obtained by taking β = k https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-68/MediaObjects/13661_2012_Article_317_IEq8_HTML.gif, α = m https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-68/MediaObjects/13661_2012_Article_317_IEq1_HTML.gif. Since it is a linear partial differential equation, its solution can be written as
u ( x , t ) = u 0 ( x ) u 1 ( t ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-68/MediaObjects/13661_2012_Article_317_Equ13_HTML.gif
(13)
After finding the analytic or approximate solution of equation (12), we expand u 0 ( x ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-68/MediaObjects/13661_2012_Article_317_IEq9_HTML.gif into its Taylor series with respect to x and u 1 ( t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-68/MediaObjects/13661_2012_Article_317_IEq10_HTML.gif into its Taylor series with respect to t. Then we replace the derivatives with respect to x by fractional derivatives with respect to x in u 0 ( x ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-68/MediaObjects/13661_2012_Article_317_IEq9_HTML.gif in such a way that the relation among the terms of Taylor series is preserved. Moreover, we leave the first m terms of Taylor series fixed since α = m https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-68/MediaObjects/13661_2012_Article_317_IEq1_HTML.gif. Similarly, we do the same thing for u 1 ( t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-68/MediaObjects/13661_2012_Article_317_IEq10_HTML.gif. This also allows the solution of the fractional differential equation to satisfy the initial and boundary conditions of the problem. Let us assume that u 0 ( x ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-68/MediaObjects/13661_2012_Article_317_IEq9_HTML.gif of equation (12) is expanded into its Taylor series with respect to x as follows:
u 0 ( x ) = n = 0 n u 0 ( 0 ) x n x n n ! . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-68/MediaObjects/13661_2012_Article_317_Equc_HTML.gif
Then u 0 ( x ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-68/MediaObjects/13661_2012_Article_317_IEq9_HTML.gif of solution (13) becomes
u 0 ( x ) = n = 0 m 1 n u 0 ( 0 ) x n x n n ! + n = 1 i = 0 m 1 m n + i u 0 ( 0 ) x m n + i x n α + i Γ ( n α + i + 1 ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-68/MediaObjects/13661_2012_Article_317_Equ14_HTML.gif
(14)
Let us assume that u 1 ( t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-68/MediaObjects/13661_2012_Article_317_IEq10_HTML.gif of equation (12) is expanded into its Taylor series with respect to t as follows:
u 1 ( t ) = n = 0 n u 1 ( 0 ) t n t n n ! . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-68/MediaObjects/13661_2012_Article_317_Equd_HTML.gif
Then u 1 ( t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-68/MediaObjects/13661_2012_Article_317_IEq10_HTML.gif of solution (13) becomes
u 1 ( t ) = n = 0 m 1 n u 1 ( 0 ) t n t n n ! + n = 1 i = 0 m 1 m n + i u 1 ( 0 ) t m n + i t n α + i Γ ( n α + i + 1 ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-68/MediaObjects/13661_2012_Article_317_Equ15_HTML.gif
(15)
For instance, let us assume that 0 < β 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-68/MediaObjects/13661_2012_Article_317_IEq11_HTML.gif, 0 < α 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-68/MediaObjects/13661_2012_Article_317_IEq2_HTML.gif, and that the solution of equation (12) is u ( x , t ) = e x e t https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-68/MediaObjects/13661_2012_Article_317_IEq3_HTML.gif. In order to find the solution of equation (11), we first expand u 0 ( x ) = e x https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-68/MediaObjects/13661_2012_Article_317_IEq12_HTML.gif into Taylor series with respect to x as follows:
u 0 ( x ) = n = 0 x n n ! . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-68/MediaObjects/13661_2012_Article_317_Eque_HTML.gif
Similarly,
u 1 ( t ) = n = 0 t n n ! . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-68/MediaObjects/13661_2012_Article_317_Equf_HTML.gif
Based on (4) and (9), the expansions of u 0 ( x ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-68/MediaObjects/13661_2012_Article_317_IEq9_HTML.gif and u 1 ( t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-68/MediaObjects/13661_2012_Article_317_IEq10_HTML.gif can be written in the following form:
u 0 ( x ) = 1 + x α Γ ( α + 1 ) + x 2 α Γ ( 2 α + 1 ) + x 3 α Γ ( 3 α + 1 ) + , u 1 ( t ) = 1 + t β Γ ( β + 1 ) + t 2 β Γ ( 2 β + 1 ) + t 3 β Γ ( 3 β + 1 ) + https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-68/MediaObjects/13661_2012_Article_317_Equg_HTML.gif
and then the solution of equation (11) can be written in the following form:
u ( x , t ) = ( 1 + x α Γ ( α + 1 ) + x 2 α Γ ( 2 α + 1 ) + x 3 α Γ ( 3 α + 1 ) + ) × ( 1 + t β Γ ( β + 1 ) + t 2 β Γ ( 2 β + 1 ) + ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-68/MediaObjects/13661_2012_Article_317_Equ16_HTML.gif
(16)

5 Illustrative applications

Example 1

Let us consider the following time-fractional initial boundary value problem:
D t α u ( x , t ) = 1 2 x 2 u x x ( x , t ) , 0 < α 1 , t > 0 , u ( x , 0 ) = x 2 , u ( 0 , t ) = 0 , u ( 1 , t ) = e t . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-68/MediaObjects/13661_2012_Article_317_Equh_HTML.gif
The exact solution, for the special case α = 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-68/MediaObjects/13661_2012_Article_317_IEq13_HTML.gif, is given by
u ( x , t ) = x 2 e t . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-68/MediaObjects/13661_2012_Article_317_Equi_HTML.gif
Now we can apply series (4) to construct the solution u ( x , t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-68/MediaObjects/13661_2012_Article_317_IEq14_HTML.gif for 0 < α 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-68/MediaObjects/13661_2012_Article_317_IEq2_HTML.gif, and we have
u ( x , t ) = x 2 [ 1 + t α Γ ( α + 1 ) + t 2 α Γ ( 2 α + 1 ) + t 3 α Γ ( 3 α + 1 ) + ] , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-68/MediaObjects/13661_2012_Article_317_Equj_HTML.gif

which is exactly the same solution as in [18].

Example 2

Let us consider the following time-fractional initial boundary value problem:
D t α u = u x x ( x , t ) + x u x ( x , t ) + u ( x , t ) , 0 < α 1 , t > 0 , u ( x , 0 ) = x , u x ( x , 0 ) = 1 , u ( 0 , t ) = 0 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-68/MediaObjects/13661_2012_Article_317_Equk_HTML.gif
The exact solution, for the special case α = 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-68/MediaObjects/13661_2012_Article_317_IEq13_HTML.gif, is given by
u ( x , t ) = x e 2 t . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-68/MediaObjects/13661_2012_Article_317_Equl_HTML.gif
Now we can apply series (4) to u ( x , t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-68/MediaObjects/13661_2012_Article_317_IEq14_HTML.gif for 0 < α 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-68/MediaObjects/13661_2012_Article_317_IEq2_HTML.gif, and we have
u ( x , t ) = x 2 [ 1 + 2 t α Γ ( α + 1 ) + 4 t 2 α Γ ( 2 α + 1 ) + 8 t 3 α Γ ( 3 α + 1 ) + ] , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-68/MediaObjects/13661_2012_Article_317_Equm_HTML.gif

which is exactly the same solution as in [19].

Example 3

Let us consider the nonlinear time-fractional Fisher’s equation
D t α u = u x x ( x , t ) + 6 u ( x , t ) ( 1 u ( x , t ) ) , 0 < α 1 , t > 0 , u ( x , 0 ) = 1 ( 1 + e x ) 2 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-68/MediaObjects/13661_2012_Article_317_Equn_HTML.gif
The exact solution, for the special case α = 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-68/MediaObjects/13661_2012_Article_317_IEq13_HTML.gif, is given by
u ( x , t ) = 1 ( 1 + e x 5 t ) 2 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-68/MediaObjects/13661_2012_Article_317_Equo_HTML.gif
As in the previous examples, we can apply series (4) to obtain the solution u ( x , t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-68/MediaObjects/13661_2012_Article_317_IEq14_HTML.gif for 0 < α 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-68/MediaObjects/13661_2012_Article_317_IEq2_HTML.gif, and we get
u ( x , t ) = 1 ( 1 + e x ) 2 + 10 e x ( 1 + e x ) 3 t α Γ ( α + 1 ) + 50 e x ( 1 + 2 e x ) ( 1 + e x ) 4 t 2 α Γ ( 2 α + 1 ) + , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-68/MediaObjects/13661_2012_Article_317_Equp_HTML.gif

which is totaly the same solution as in [20].

Example 4

Let us consider the following time-fractional initial boundary value problem:
D t α u = ± u x x ( x , t ) , 0 < α 1 , t > 0 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-68/MediaObjects/13661_2012_Article_317_Equ17_HTML.gif
(17)
u ( x , 0 ) = x 2 , u ( 0 , t ) = 2 k t α Γ ( α + 1 ) , u ( , t ) = 2 + 2 k t α Γ ( α + 1 ) , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-68/MediaObjects/13661_2012_Article_317_Equ18_HTML.gif
(18)

where the boundary conditions are given in fractional terms.

Boundary value problem (17)-(18), for the special case α = 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-68/MediaObjects/13661_2012_Article_317_IEq13_HTML.gif, becomes as follows:
u t = ± u x x ( x , t ) , 0 < α 1 , t > 0 , u ( x , 0 ) = x 2 , u ( 0 , t ) = 2 k t , u ( , t ) = 2 + 2 k t , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-68/MediaObjects/13661_2012_Article_317_Equq_HTML.gif
and its analytic solution is obtained as follows:
u ( x , t ) = x 2 + 2 k t . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-68/MediaObjects/13661_2012_Article_317_Equr_HTML.gif
Now we can apply series (4) to u ( x , t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-68/MediaObjects/13661_2012_Article_317_IEq14_HTML.gif for 0 < α 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-68/MediaObjects/13661_2012_Article_317_IEq2_HTML.gif, and we have
u ( x , t ) = x 2 + 2 k t α Γ ( α + 1 ) , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-68/MediaObjects/13661_2012_Article_317_Equs_HTML.gif

which is exactly the same solution as in [21].

Example 5

Let us consider the following space-fractional initial boundary value problem:
u t ( x , t ) = D x β u ( x , t ) ( 1 + tan t ) u ( x , t ) , 1 < β 2 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-68/MediaObjects/13661_2012_Article_317_Equ19_HTML.gif
(19)
u ( x , 0 ) = e x , u ( 0 , t ) = cos t , u x ( 0 , t ) = cos t . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-68/MediaObjects/13661_2012_Article_317_Equ20_HTML.gif
(20)
The exact solution, for the special case β = 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-68/MediaObjects/13661_2012_Article_317_IEq15_HTML.gif, is given by
u ( x , t ) = e x cos t . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-68/MediaObjects/13661_2012_Article_317_Equt_HTML.gif
Now we can apply series (9) to u ( x , t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-68/MediaObjects/13661_2012_Article_317_IEq14_HTML.gif for 1 < β 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-68/MediaObjects/13661_2012_Article_317_IEq16_HTML.gif, then we have
u ( x , t ) = cos t ( 1 + x + x β Γ ( β + 1 ) + x β + 1 Γ ( β + 2 ) + x 2 β Γ ( 2 β + 1 ) + x 2 β + 1 Γ ( 2 β + 2 ) + ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-68/MediaObjects/13661_2012_Article_317_Equu_HTML.gif

Example 6

Let us consider the following space and time-fractional initial boundary value problem:
D t α u ( x , t ) = 2 D x β u ( x , t ) + cos t , 0 < α 1 , 1 < β 2 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-68/MediaObjects/13661_2012_Article_317_Equ21_HTML.gif
(21)
u ( x , 0 ) = e x , u ( 0 , t ) = e 2 t + sin t , u x ( 0 , t ) = e 2 t + sin t . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-68/MediaObjects/13661_2012_Article_317_Equ22_HTML.gif
(22)
The exact solution, for the special case α = 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-68/MediaObjects/13661_2012_Article_317_IEq13_HTML.gif and β = 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-68/MediaObjects/13661_2012_Article_317_IEq15_HTML.gif, is given by
u ( x , t ) = e 2 t e x + sin t . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-68/MediaObjects/13661_2012_Article_317_Equv_HTML.gif
Now we can apply the series (4)-(9) to u ( x , t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-68/MediaObjects/13661_2012_Article_317_IEq14_HTML.gif for 0 < α 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-68/MediaObjects/13661_2012_Article_317_IEq2_HTML.gif and 1 < β 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-68/MediaObjects/13661_2012_Article_317_IEq16_HTML.gif, then we have
u ( x , t ) = ( 1 + 2 t α Γ ( α + 1 ) + 4 t 2 α Γ ( 2 α + 1 ) + ) ( 1 + x + x β Γ ( β + 1 ) + x β + 1 Γ ( β + 2 ) + ) + ( t α Γ ( α + 1 ) t 3 α Γ ( 3 α + 1 ) + t 5 α Γ ( 5 α + 1 ) ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-68/MediaObjects/13661_2012_Article_317_Equw_HTML.gif

Example 7

Let us consider the following space and time-fractional initial boundary value problem:
D t α u ( x , t ) = 2 D x β u ( x , t ) + 2 sin x , 0 < α 1 , 1 < β 2 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-68/MediaObjects/13661_2012_Article_317_Equ23_HTML.gif
(23)
u ( x , 0 ) = e x + sin x , u ( 0 , t ) = e 2 t , u x ( 0 , t ) = e 2 t + 1 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-68/MediaObjects/13661_2012_Article_317_Equ24_HTML.gif
(24)
The exact solution, for the special case α = 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-68/MediaObjects/13661_2012_Article_317_IEq13_HTML.gif and β = 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-68/MediaObjects/13661_2012_Article_317_IEq15_HTML.gif, is given by
u ( x , t ) = e 2 t e x + sin x . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-68/MediaObjects/13661_2012_Article_317_Equx_HTML.gif
Now we can apply the series (4)-(9) to u ( x , t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-68/MediaObjects/13661_2012_Article_317_IEq14_HTML.gif for 0 < α 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-68/MediaObjects/13661_2012_Article_317_IEq2_HTML.gif and 1 < β 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-68/MediaObjects/13661_2012_Article_317_IEq16_HTML.gif, then we have
u ( x , t ) = ( 1 + 2 t α Γ ( α + 1 ) + 4 t 2 α Γ ( 2 α + 1 ) + ) ( 1 + x + x β Γ ( β + 1 ) + x β + 1 Γ ( β + 2 ) + ) + ( x x β + 1 Γ ( β + 2 ) + x 2 β + 1 Γ ( 2 β + 2 ) x 3 β + 1 Γ ( 3 β + 2 ) ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-68/MediaObjects/13661_2012_Article_317_Equy_HTML.gif
Figures 7-9 show the evolution results for the approximate solutions of problem (23)-(24) obtained for different values of α and β.
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-68/MediaObjects/13661_2012_Article_317_Fig7_HTML.jpg
Figure 7

The surface shows the approximate solution u ( x , t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-68/MediaObjects/13661_2012_Article_317_IEq17_HTML.gif of problem ( 23 )-( 24 ) for α = 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-68/MediaObjects/13661_2012_Article_317_IEq4_HTML.gif , β = 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-68/MediaObjects/13661_2012_Article_317_IEq18_HTML.gif .

https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-68/MediaObjects/13661_2012_Article_317_Fig8_HTML.jpg
Figure 8

The surface shows the approximate solution u ( x , t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-68/MediaObjects/13661_2012_Article_317_IEq17_HTML.gif of problem ( 23 )-( 24 ) for α = 0.75 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-68/MediaObjects/13661_2012_Article_317_IEq5_HTML.gif , β = 1.5 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-68/MediaObjects/13661_2012_Article_317_IEq19_HTML.gif .

https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-68/MediaObjects/13661_2012_Article_317_Fig9_HTML.jpg
Figure 9

The surface shows the approximate solution u ( x , t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-68/MediaObjects/13661_2012_Article_317_IEq17_HTML.gif of problem ( 23 )-( 24 ) for α = 0.5 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-68/MediaObjects/13661_2012_Article_317_IEq6_HTML.gif , β = 1.75 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-68/MediaObjects/13661_2012_Article_317_IEq20_HTML.gif .

Declarations

Acknowledgements

Dedicated to Professor Hari M Srivastava.

The research was supported by parts by the Scientific and Technical Research Council of Turkey (TUBITAK).

Authors’ Affiliations

(1)
Department of Mathematics, Kocaeli University, Umuttepe
(2)
Ardahan University

References

  1. Oldham KB, Spanier J: The Fractional Calculus. Academic Press, New York; 1974.
  2. Podlubny I: Fractional Differential Equations. Academic Press, San Diego; 1999.
  3. Kilbas AA, Srivastava HM, Trujillo JJ: Theory and Applications of Fractional Differential Equations. Elsevier, Amsterdam; 2006.
  4. He JH: Nonlinear oscillation with fractional derivative and its applications. International Conference on Vibrating Engineering’98 1998, 288-291.
  5. He JH: Some applications of nonlinear fractional differential equations and their approximations. Bull Sci Technol 1999, 15: 86-90.
  6. He JH: Approximate analytical solution for seepage flow with fractional derivatives in porous media. Comput Methods Appl Mech Eng 1998, 167: 57-68. 10.1016/S0045-7825(98)00108-XView Article
  7. Metzler R, Klafter J: The random walk’s guide to anomalous diffusion: a fractional dynamics approach. Phys Rep 2000, 339: 1-77. 10.1016/S0370-1573(00)00070-3MathSciNetView Article
  8. Gazizov RK, Kasatkin AA, Lukashchuk SY: Group-invariant solutions of fractional differential equations. 1. In Nonlinear Science and Complexity. 1st edition. Edited by: Tenreiro Machado JA, Luo ACJ, Barbosa RS, Silva MF, Figueiredo LB. Springer, Dordrecht; 2011:51-61.View Article
  9. Huang F, Liu F: The time-fractional diffusion equation and fractional advection-dispersion equation. ANZIAM J. 2005, 46: 1-14.View Article
  10. Huang F, Liu F: The fundamental solution of the space-time fractional advection-dispersion equation. J. Appl. Math. Comput. 2005, 18(2):339-350. 10.1007/BF02936577MathSciNetView Article
  11. Momani S: Non-perturbative analytical solutions of the space- and time-fractional Burgers equations. Chaos Solitons Fractals 2006, 28(4):930-937. 10.1016/j.chaos.2005.09.002MathSciNetView Article
  12. Odibat Z, Momani S: Application of variational iteration method to nonlinear differential equations of fractional order. Int. J. Nonlinear Sci. Numer. Simul. 2006, 1(7):15-27.
  13. Das S: Solution of fractional vibration equation by the variational iteration method and modified decomposition method. Int. J. Nonlinear Sci. Numer. Simul. 2008., 9: Article ID 361
  14. Momani S, Odibat Z: Numerical comparison of methods for solving linear differential equations of fractional order. Chaos Solitons Fractals 2007, 31(5):1248-1255. 10.1016/j.chaos.2005.10.068MathSciNetView Article
  15. Odibat Z, Momani S: Approximate solutions for boundary value problems of time-fractional wave equation. Appl. Math. Comput. 2006, 181(1):767-774. 10.1016/j.amc.2006.02.004MathSciNetView Article
  16. Yildirim A: An algorithm for solving the fractional nonlinear Schrödinger equation by means of the homotopy perturbation method. Int. J. Nonlinear Sci. Numer. Simul. 2009, 10: 445-451.
  17. Ganji ZZ, Ganji DD, Jafari H, Rostamian M: Application of the homotopy perturbation method to coupled system of partial differential equations with time fractional derivatives. Topol. Methods Nonlinear Anal. 2008., 31: Article ID 341
  18. Hang X, Shi-Yun L, Xiang-Cheng Y: Analysis of nonlinear fractional partial differential equations with the homotopy analysis method. Commun. Nonlinear Sci. Numer. Simul. 2009, 14: 1152-1156. 10.1016/j.cnsns.2008.04.008MathSciNetView Article
  19. Saha RS, Bera RK: The random walk’s guide to anomalous diffusion: a fractional dynamics approach. Phys. Rep. 2000, 339: 1-77. 10.1016/S0370-1573(00)00070-3View Article
  20. Odibat Z, Momani S: Numerical methods for nonlinear partial differential equations of fractional order. Appl. Math. Model. 2008, 32: 28-39. 10.1016/j.apm.2006.10.025View Article
  21. El-Sayed AMA, Gaber M: The Adomian decomposition method for solving partial differential equations of fractal order in finite domains. Phys. Lett. A 2006, 359: 175-182. 10.1016/j.physleta.2006.06.024MathSciNetView Article
  22. Sheu LJ, Tam LM, Lao SK: Parametric analysis and impulsive synchronization of fractional-order Newton-Leipnik systems. Int. J. Nonlinear Sci. Numer. Simul. 2009, 10: 33-44.View Article
  23. Xu C, Wu G, Feng JW, Zhang WQ: Synchronization between two different fractional-order chaotic systems. Int. J. Nonlinear Sci. Numer. Simul. 2008, 9: 89-95.
  24. Yildirim A: Homotopy perturbation method for solving the space-time fractional advection-dispersion equation. Adv. Water Resour. 2009, 32: 1711-1716. 10.1016/j.advwatres.2009.09.003View Article

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