Hindawi Publishing Corporation
Abstract and Applied Analysis
Volume 2013, Article ID 561980, 11 pages
http://dx.doi.org/10.1155/2013/561980
Research Article
Approximate Analytic Solutions of Time-Fractional
Hirota-Satsuma Coupled KdV Equation and
Coupled MKdV Equation
Jincun Liu and Hong Li
School of Mathematical Sciences, Inner Mongolia University, 235 West Daxue Road, Hohhot 010021, China
Correspondence should be addressed to Jincun Liu; ljincun@163.com
Received 29 October 2012; Accepted 28 January 2013
Academic Editor: Abdel-Maksoud A. Soliman
Copyright © 2013 J. Liu and H. Li. 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.
By introducing the fractional derivative in the sense of Caputo and combining the pretreatment technique to deal with long
nonlinear items, the generalized two-dimensional differential transform method is proposed for solving the time-fractional
Hirota-Satsuma coupled KdV equation and coupled MKdV equation. The presented method is a numerical method based on
the generalized Taylor series expansion which constructs an analytical solution in the form of a polynomial. The numerical results
show that the generalized two-dimensional differential transform method is very effective for the fractional coupled equations.
1. Introduction
In the last past decade, the fractional differential equations
have been widely used in various fields of physics and engineering. The analytical approximation of such problems has
attracted great attention and became a considbased onerable
interest in mathematical physics. Some powerful methods
including the homotopy perturbation method [1], Adomian
decomposition method [2, 3], variational iteration method
[4, 5], homotopy analysis method [6, 7], fractional complex
transform method [8], and generalized differential transform
method [9] have been developed to obtain exact and approximate analytic solutions. These solution techniques are more
clear and realistic methods for fractional differential equations, because they give the approximate solutions of the considered problems without any linearization or discretization.
The variational iteration method and the homotopy
perturbation method were first proposed by Professor He
in [10, 11], respectively. The idea of the variational iteration
method is to construct correction functionals using general
Lagrange multipliers identified optimally via the variational
theory, and the initial approximations can be freely chosen
with unknown constants. Recently, Wu and Lee proposed a
fractional variational iteration method for fractional differ-
ential equation based on the modified Riemann CLiouville
derivative, which is more effective to solve fractional differential equation [12]. This method has been developed by
many authors, see [13–16] and the references cited therein.
The homotopy perturbation method, which does not require
a small parameter in an equation, has a significant advantage
that it provides an analytical approximate solution to a wide
range of nonlinear problems in applied sciences. Recently,
the fractional complex transform is developed to convert
the fractional differential equation to its differential partner
and gave a geometrical explanation [8]. These methods are
more effective for solving the linear and nonlinear fractional
differential equations.
The differential transform method was used firstly by
Zhou in 1986 to study electric circuits [17]. The differential
transform is an iterative procedure based on the Taylor series
expansion which constructs an analytic solution in the form
of a polynomial. The method is well addressed in [18–23].
Recently, the generalized differential transform method is
developed for obtaining approximate analytic solutions for
some linear and nonlinear differential equations of fractional
order [24, 25].
The aim of this paper is to directly extend the generalized two-dimensional differential transform method to
2
Abstract and Applied Analysis
obtain the approximate analytic solutions of a time-fractional
Hirota-Satsuma coupled KdV equation,
1
𝐷𝑡𝛼 𝑢 = 𝑢𝑥𝑥𝑥 − 3𝑢𝑢𝑥 + 3V𝑤𝑥 + 3V𝑥 𝑤,
2
𝐷𝑡𝛼 V = −V𝑥𝑥𝑥 + 3𝑢V𝑥 ,
𝐷𝑡𝛼 𝑤 = −𝑤𝑥𝑥𝑥 + 3𝑢𝑤𝑥 ,
(1)
𝑡 > 0, 0 < 𝛼 ≤ 1,
Lemma 3 (see [29]). The Caputo fractional derivative of the
power function satisfies
Γ (𝑝 + 1) 𝑝−𝛼
{
𝑡 , 𝑛 − 1 < 𝛼 < 𝑛,
{
{
{
Γ (𝑝 − 𝛼 + 1)
{
{
𝑝 > 𝑛 − 1, 𝑝 ∈ R,
𝐷𝑡𝛼 𝑡𝑝 = {
{
{
{
0,
𝑛 − 1 < 𝛼 < 𝑛,
{
{
𝑝 ≤ 𝑛 − 1, 𝑝 ∈ N.
{
(6)
and a time-fractional coupled MKdV equation,
2. Generalized Two-Dimensional Differential
Transform Method (GDTM)
1
3
𝛽
𝐷𝑡 𝑢 = 𝑢𝑥𝑥𝑥 − 3𝑢2 𝑢𝑥 + VV𝑥 + 3𝑢V𝑥 + 3𝑢𝑥 V − 3𝜆𝑢𝑥 ,
2
2
𝛽
𝐷𝑡 V = − V𝑥𝑥𝑥 − 3VV𝑥 − 3𝑢𝑥 V𝑥 + 3𝑢2 V𝑥 + 3𝜆V𝑥 ,
(2)
𝑡 > 0, 0 < 𝛽 ≤ 1,
where 𝜆 is a constant and 𝛼 and 𝛽 are parameters describing
the order of the time-fractional derivatives of 𝑢(𝑥, 𝑡), V(𝑥, 𝑡),
and 𝑤(𝑥, 𝑡), respectively. The fractional derivatives are considered in the Caputo sense. In the case of 𝛼 = 1 and 𝛽 = 1, (1)
and (2) reduce to the classical Hirota-Satsuma coupled KdV
equation and coupled MKdV equation [26], respectively.
The Caputo fractional derivative is considered here
because it allows traditional initial and boundary conditions
to be included in the formulation of the problem.
Definition 1 (see [27]). The fractional derivative of 𝑓(𝑥) in the
Caputo sense is defined as
𝐷] 𝑓 (𝑥) = 𝐽0𝑚−] 𝐷𝑚 𝑓 (𝑥)
=
𝑥
1
∫ (𝑥 − 𝑡)𝑚−]−1 𝑓(𝑚) (𝑡) 𝑑𝑡,
Γ (𝑚 − ]) 0
(3)
𝑥
1
∫ (𝑥 − 𝑡)𝜇−1 𝑓 (𝑡) 𝑑𝑡,
Γ (𝜇) 0
𝜇 > 0.
(4)
Definition 2 (see [28]). For 𝑚 to be the smallest integer that
exceeds 𝛼, the Caputo time-fractional derivative of order 𝛼 >
0 is defined as
1
{
{
{
Γ (𝑚 − 𝛼)
{
{
{
{
{
𝑡
{
{
{
× ∫ (𝑡 − 𝜉)𝑚−𝛼−1
{
{
𝛼
𝜕 𝑢 { 0
𝐷𝑡𝛼 𝑢 (𝑥, 𝑡) = 𝛼 = { 𝑚
{ 𝜕 𝑢 (𝑥, 𝜉)
𝜕𝑡
{
{
{
×
𝑑𝜉,
{
{
𝜕𝜉𝑚
{
{
{
{
{
{ 𝜕𝑚 𝑢 (𝑥, 𝑡)
{
,
{ 𝜕𝑡𝑚
𝑘𝛼
𝑘=0
∞
ℎ𝛽
ℎ=0
∞ ∞
(7)
𝑘𝛼
ℎ𝛽
= ∑ ∑ 𝑈𝛼,𝛽 (𝑘, ℎ) (𝑥 − 𝑥0 ) (𝑡 − 𝑡0 ) ,
𝑘=0 ℎ=0
where 0 < 𝛼, 𝛽 ≤ 1, 𝑈𝛼,𝛽 (𝑘, ℎ) = 𝐹𝛼 (𝑘)𝐺𝛽 (ℎ) is called the
spectrum of 𝑢(𝑥, 𝑡). The generalized two-dimensional differential transform of the function 𝑢(𝑥, 𝑡) is given by
1
Γ (𝛼𝑘 + 1) Γ (𝛽ℎ + 1)
×
for 𝑚 − 1 < ] < 𝑚 and 𝑚 ∈ 𝑁, 𝑥 > 0. Here,
is the
Riemann-Liouville integral operator of order 𝜇 > 0, defined
by
𝜇
∞
𝑢 (𝑥, 𝑡) = ∑ 𝐹𝛼 (𝑘) (𝑥 − 𝑥0 ) ∑ 𝐺𝛽 (ℎ) (𝑡 − 𝑡0 )
𝑈𝛼,𝛽 (𝑘, ℎ) =
𝜇
𝐽0
𝐽0 𝑓 (𝑥) =
Consider a function of two variables 𝑢(𝑥, 𝑡) and suppose that
it can be represented as a product of two single variable
functions, that is, 𝑢(𝑥, 𝑡) = 𝑓(𝑥)𝑔(𝑡). Based on the properties
of generalized two-dimensional differential transform, the
function 𝑢(𝑥, 𝑡) can be represented as
𝑘
𝛽 ℎ
[(𝐷𝑥𝛼 ) (𝐷𝑡 ) 𝑢 (𝑥, 𝑡)]
(8)
(𝑥0 ,𝑡0 )
,
where (𝐷𝑥𝛼 )𝑘 = 𝐷𝑥𝛼 𝐷𝑥𝛼 ⋅ ⋅ ⋅ 𝐷𝑥𝛼 , 𝑘 is times. In the case of 𝛼 =
1 and 𝛽 = 1, the generalized two-dimensional differential
transform (8) reduces to two-dimensional differential transform.
The fundamental theorems of the generalized twodimensional differential transform are as follows.
Theorem 4. Suppose that 𝑈𝛼,𝛽 (𝑘, ℎ), 𝑉𝛼,𝛽 (𝑘, ℎ), and 𝑊𝛼,𝛽 (𝑘, ℎ)
are the differential transformations of the functions 𝑢(𝑥, 𝑡),
V(𝑥, 𝑡), and 𝑤(𝑥, 𝑡), respectively:
(a) if 𝑢(𝑥, 𝑡) = V(𝑥, 𝑡) ± 𝑤(𝑥, 𝑡), then 𝑈𝛼,𝛽 (𝑘, ℎ) = 𝑉𝛼,𝛽
(𝑘, ℎ) ± 𝑊𝛼,𝛽 (𝑘, ℎ).
for 𝑚 − 1 < 𝛼 < 𝑚,
(b) if 𝑢(𝑥, 𝑡) = 𝑎V(𝑥, 𝑡), 𝑎 ∈ R, then 𝑈𝛼,𝛽 (𝑘, ℎ) = 𝑎𝑉𝛼,𝛽 (𝑘,
ℎ).
(c) if 𝑢(𝑥, 𝑡) = V(𝑥, 𝑡)𝑤(𝑥, 𝑡), then 𝑈𝛼,𝛽 (𝑘, ℎ) =
∑𝑘𝑟=0 ∑ℎ𝑠=0 𝑉𝛼,𝛽 (𝑟, ℎ − 𝑠)𝑊𝛼,𝛽 (𝑘 − 𝑟, 𝑠).
for 𝛼 = 𝑚 ∈ 𝑁,
(5)
(d) if 𝑢(𝑥, 𝑡) = (𝑥− 𝑥0 )𝑛𝛼 (𝑡− 𝑡0 )𝑚𝛽 , then 𝑈𝛼,𝛽 (𝑘, ℎ) = 𝛿(𝑘−
𝑛)𝛿(ℎ − 𝑚).
and the space-fractional derivatives of Caputo type can be
defined analogously.
(e) if 𝑢(𝑥, 𝑡) = 𝐷𝑥𝛼 V(𝑥, 𝑡), 0 < 𝛼 ≤ 1, then 𝑈𝛼,𝛽 (𝑘, ℎ) =
(Γ(𝛼(𝑘 + 1) + 1)/Γ(𝛼𝑘 + 1))𝑉𝛼,𝛽 (𝑘 + 1, ℎ).
Abstract and Applied Analysis
3
Theorem 5. If 𝑢(𝑥, 𝑡) = 𝑓(𝑥)𝑔(𝑡), function 𝑓(𝑥) = 𝑥𝜆 ℎ(𝑥),
where 𝜆 > −1 and ℎ(𝑥) has the generalized Taylor series
𝛼𝑛
expansion ℎ(𝑥) = ∑∞
𝑛=0 𝑎𝑛 (𝑥 − 𝑥0 ) ,
(i) 𝛽 < 𝜆 + 1 and 𝛼 arbitrary, or
(ii) 𝛽 ≥ 𝜆 + 1, 𝛼 arbitrary, and 𝑎𝑛 = 0 for 𝑛 = 0, 1, . . . , 𝑚 −
1, where 𝑚 − 1 < 𝛽 ≤ 𝑚,
then the generalized differential transform (8) becomes
1
𝛽 ℎ
.
[𝐷𝑥𝛼𝑘0 (𝐷𝑡0 ) 𝑢 (𝑥, 𝑡)]
𝑈𝛼,𝛽 (𝑘, ℎ) =
Γ (𝛼𝑘 + 1) Γ (𝛽ℎ + 1)
(𝑥0 ,𝑡0 )
(9)
If 𝑢(𝑥, 𝑡) = 𝐷𝑥𝛾0 V(𝑥, 𝑡), 𝑚 − 1 < 𝛾 ≤ 𝑚, and V(𝑥, 𝑡) = 𝑓(𝑥)
𝑔(𝑡), then
Γ (𝛼𝑘 + 𝛾 + 1)
𝛾
𝑉 (𝑘 + , ℎ) .
Γ (𝛼𝑘 + 1) 𝛼,𝛽
𝛼
𝑈𝛼,𝛽 (𝑘, ℎ) =
Suppose that the solutions 𝑢(𝑥, 𝑡), V(𝑥, 𝑡), and 𝑤(𝑥, 𝑡) can
be represented as the products of single-valued functions,
respectively. Applying the generalized two-dimensional differential transform to both sides of (11) and using the related
theorems, we have
Γ (𝛼 (ℎ + 1) + 1)
𝑈1,𝛼 (𝑘, ℎ + 1)
Γ (𝛼ℎ + 1)
=
1
(𝑘 + 1) (𝑘 + 2) (𝑘 + 3) 𝑈1,𝛼 (𝑘 + 3, ℎ)
2
𝑘
ℎ
− 3∑ ∑ (𝑘 + 1 − 𝑟) 𝑈1,𝛼 (𝑟, ℎ − 𝑠) 𝑈1,𝛼 (𝑘 + 1 − 𝑟, 𝑠)
𝑟=0 𝑠=0
𝑘+1 ℎ
+ 3 (𝑘 + 1) ∑ ∑𝑉1,𝛼 (𝑟, ℎ − 𝑠) 𝑊1,𝛼 (𝑘 + 1 − 𝑟, 𝑠) ,
𝑟=0 𝑠=0
(10)
Some details of the aformentioned theorems can be found
in [30].
Γ (𝛼 (ℎ + 1) + 1)
𝑉1,𝛼 (𝑘, ℎ + 1)
Γ (𝛼ℎ + 1)
= − (𝑘 + 1) (𝑘 + 2) (𝑘 + 3) 𝑉1,𝛼 (𝑘 + 3, ℎ)
𝑘
ℎ
− 3∑ ∑ (𝑘 + 1 − 𝑟) 𝑈1,𝛼 (𝑟, ℎ − 𝑠) 𝑉1,𝛼 (𝑘 + 1 − 𝑟, 𝑠) ,
3. Applications of GDTM
𝑟=0 𝑠=0
3.1. Fractional Hirota-Satsuma Coupled KdV Equation. Consider the following time-fractional Hirota-Satsuma coupled
KdV equation:
= − (𝑘 + 1) (𝑘 + 2) (𝑘 + 3) 𝑊1,𝛼 (𝑘 + 3, ℎ)
1
𝐷𝑡𝛼 𝑢 = 𝑢𝑥𝑥𝑥 − 3𝑢𝑢𝑥 + 3(V𝑤)𝑥 ,
2
𝐷𝑡𝛼 V
𝐷𝑡𝛼 𝑤 = −𝑤𝑥𝑥𝑥 + 3𝑢𝑤𝑥 ,
𝑘
(11)
= −V𝑥𝑥𝑥 + 3𝑢V𝑥 ,
V (𝑥, 0) =
ℎ
− 3∑ ∑ (𝑘 + 1 − 𝑟) 𝑈1,𝛼 (𝑟, ℎ − 𝑠) 𝑊1,𝛼 (𝑘 + 1 − 𝑟, 𝑠) .
𝑟=0 𝑠=0
(14)
𝑡 > 0, 0 < 𝛼 ≤ 1,
The generalized two-dimensional differential transforms
of the initial conditions can be obtained as follows:
subject to the initial conditions
𝑢 (𝑥, 0) =
Γ (𝛼 (ℎ + 1) + 1)
𝑊1,𝛼 (𝑘, ℎ + 1)
Γ (𝛼ℎ + 1)
1
(𝛽 − 8𝛾2 ) + 4𝛾2 tanh2 (𝛾𝑥) ,
3
−4 (3𝛾4 𝑐0 − 2𝛽𝛾2 𝑐2 + 4𝛾4 𝑐2 )
3𝑐22
𝑈1,𝛼 (𝑘, 0) = 𝑉1,𝛼 (𝑘, 0) = 𝑊1,𝛼 (𝑘, 0) = 0
4𝛾2
+
tanh2 (𝛾𝑥) ,
𝑐2
𝑤 (𝑥, 0) = 𝑐0 + 𝑐2 tanh2 (𝛾𝑥) ,
(12)
where 𝑐0 , 𝑐2 , 𝛽, and 𝛾 are arbitrary constants.
The exact solutions of (11) and (12), for the special case of
𝛼 = 1, given in [26], are
if 𝑘 = 1, 3, 5, . . . ,
𝑈1,𝛼 (0, 0) =
1
(𝛽 − 8𝛾2 ) ,
3
8
𝑈1,𝛼 (4, 0) = − 𝛾6 ,
3
𝑉1,𝛼 (0, 0) = −
1
𝑢 (𝑥, 𝑡) = (𝛽 − 8𝛾2 ) + 4𝛾2 tanh2 [𝛾 (𝑥 + 𝛽𝑡)] ,
3
V (𝑥, 𝑡) =
3𝑐22
+
2
4𝛾
tanh2 [𝛾 (𝑥 + 𝛽𝑡)] ,
𝑐2
𝑤 (𝑥, 𝑡) = 𝑐0 + 𝑐2 tanh2 [𝛾 (𝑥 + 𝛽𝑡)] .
𝑈1,𝛼 (6, 0) =
𝑉1,𝛼 (4, 0) = −
(13)
8𝛾6
,
3𝑐2
𝑊1,𝛼 (0, 0) = 𝑐0 ,
𝑊1,𝛼 (4, 0) = −
68 8
𝛾 ,...,
45
4 (3𝛾4 𝑐0 − 2𝛽𝛾2 𝑐2 + 4𝛾4 𝑐2 )
3𝑐22
𝑉1,𝛼 (2, 0) =
−4 (3𝛾4 𝑐0 − 2𝛽𝛾2 𝑐2 + 4𝛾4 𝑐2 )
𝑈1,𝛼 (2, 0) = 4𝛾4 ,
2𝑐2 4
𝛾,
3
4𝛾4
,
𝑐2
𝑉1,𝛼 (6, 0) =
,
68𝛾8
,...,
45𝑐2
𝑊1,𝛼 (2, 0) = 𝑐2 𝛾2 ,
𝑊1,𝛼 (6, 0) =
17𝑐2 6
𝛾 ,....
45
(15)
4
Abstract and Applied Analysis
Utilizing the recurrence relations (14) and the transformed initial conditions, we can obtain all the 𝑈1,𝛼 (𝑘, ℎ),
𝑉1,𝛼 (𝑘, ℎ), and 𝑊1,𝛼 (𝑘, ℎ) with the help of Mathematica.
Moreover, substituting all 𝑈1,𝛼 (𝑘, ℎ) into (7), we obtain the
series form solution
𝑢 (𝑥, 𝑡) = 𝑢0 (𝑥, 𝑡) + 𝑢1 (𝑥, 𝑡) + 𝑢2 (𝑥, 𝑡) + ⋅ ⋅ ⋅ ,
𝛾6 (1 − 𝛾2 )
𝑢2 (𝑥, 𝑡) =
𝑐2
× {96 (𝑐0 + 2𝑐2 ) sec2 (𝛾𝑥)
(16)
+
where
8𝑐2 𝛽2
(2 − cosh (2𝛾𝑥)) sec4 (𝛾𝑥)
𝛾2 (1 − 𝛾2 )
+ 6 [15𝑐0 + 94𝑐2 − 88𝑐2 cosh (2𝛾𝑥 )
𝑢0 (𝑥, 𝑡)
=
6
8
− (15𝑐0 + 14𝑐2 ) cosh (4𝛾𝑥)] sec8 (𝛾𝑥) }
10
8𝛾 4 68𝛾 6 248𝛾 8
1
(𝛽 − 8𝛾2 ) + 4𝛾4 𝑥2 −
𝑥 +
𝑥 −
𝑥 + ⋅⋅⋅,
3
3
45
315
×
𝑢1 (𝑥, 𝑡)
={
8𝛾4 (𝑐2 𝛽 − 3𝑐0 (−1 + 𝛾2 ))
+
+
𝑐2
3𝑐2
V (𝑥, 𝑡) = V0 (𝑥, 𝑡) + V1 (𝑥, 𝑡) + V2 (𝑥, 𝑡) + ⋅ ⋅ ⋅ ,
𝑥3
8𝛾8 (−51𝑐0 (−1 + 𝛾2 ) + 𝑐2 (17𝛽 + 180 (−1 + 𝛾2 )))
15𝑐2
𝑤 (𝑥, 𝑡) = 𝑤0 (𝑥, 𝑡) + 𝑤1 (𝑥, 𝑡) + 𝑤2 (𝑥, 𝑡) + ⋅ ⋅ ⋅ .
V0 (𝑥, 𝑡) =
𝑢2 (𝑥, 𝑡)
+
−
𝑐2
𝑐2
=[
𝑥2
8𝛾8 (744𝑐0 𝛾2 (−1+𝛾2 ) +𝑐2 (−17𝛽2 +2016𝛾2 (−1+𝛾2 )))
3𝑐2
𝑥4
(17)
The closed forms of 𝑢0 (𝑥, 𝑡), 𝑢1 (𝑥, 𝑡), and 𝑢2 (𝑥, 𝑡) are
𝑢0 (𝑥, 𝑡) =
1
(𝛽 − 8𝛾2 ) + 4𝛾2 tanh2 (𝛾𝑥) ,
3
𝑢1 (𝑥, 𝑡) =
𝛾3 sec2 (𝛾𝑥)
𝑐2
4𝛾2
tanh2 (𝛾𝑥) ,
𝑐2
8𝑐0 𝛽𝛾3 sec2 (𝛾𝑥) tanh (𝛾𝑥)
𝑡𝛼
,
𝑐2
Γ (1 + 𝛼)
8𝛽2 𝛾4 (3sec2 (𝛾𝑥) − 2)
𝑐2 cosh2 (𝛾𝑥)
+
𝑡2𝛼
,....
Γ (1 + 2𝛼)
×
576𝛾6 (1 − 𝛾2 ) (𝑐0 + 2𝑐2 tanh2 (𝛾𝑥)) tanh2 (𝛾𝑥)
𝑐22 cosh4 (𝛾𝑥)
𝑡2𝛼
.
Γ (1 + 2𝛼)
]
(20)
The closed forms of 𝑤0 (𝑥, 𝑡), 𝑤1 (𝑥, 𝑡), and 𝑤2 (𝑥, 𝑡) are
𝑤0 (𝑥, 𝑡) = 𝑐0 + 𝑐2 tanh2 (𝛾𝑥) ,
𝑤1 (𝑥, 𝑡) =
2𝑐2 𝛽𝛾sec2 (𝛾𝑥) tanh (𝛾𝑥)
𝑡𝛼
,
𝑐2
Γ (1 + 𝛼)
𝑤2 (𝑥, 𝑡) = [2𝑐2 𝛽2 𝛾2 sec2 (𝛾𝑥) (3sec2 (𝛾𝑥) − 2)
+ 144 (𝑐0 + 2𝑐2 ) 𝛾4 (1 − 𝛾2 )
× [24𝑐0 +8𝑐2 𝛽−24𝑐0 𝛾2 +48𝑐2 (𝛾2 −1) tanh2 (𝛾𝑥)]
×
3𝑐22
+
V2 (𝑥, 𝑡)
8𝛾4 (−12𝑐0 𝛾2 (−1 + 𝛾2 ) + 𝑐2 (𝛽2 − 18𝛾2 (−1 + 𝛾2 )))
+⋅ ⋅ ⋅ )}
−4 (3𝛾4 𝑐0 − 2𝛽𝛾2 𝑐2 + 4𝛾4 𝑐2 )
V1 (𝑥, 𝑡) =
16𝛾6 (51𝑐0 𝛾2 (−1 +𝛾2 )−2𝑐2 (𝛽2 −54𝛾2 (−1 + 𝛾2 )))
(19)
The closed forms of V0 (𝑥, 𝑡), V1 (𝑥, 𝑡) and V2 (𝑥, 𝑡) are
𝑥5
𝑡𝛼
+⋅⋅⋅}
,
Γ (1 + 𝛼)
= {(
(18)
Similarly, substituting all 𝑉1,𝛼 (𝑘, ℎ) and all 𝑊1,𝛼 (𝑘, ℎ) into
(7), respectively, we obtain the series form solutions
𝑥
16𝛾6 (𝑐2 (9 − 2𝛽 − 9𝛾2 ) + 6𝑐0 (−1 + 𝛾2 ))
𝑡2𝛼
.
Γ (1 + 2𝛼)
𝛼
tanh (𝛾𝑥) 𝑡
,
Γ (1 + 𝛼)
× sec4 (𝛾𝑥) tanh2 (𝛾𝑥) − 288𝑐2 𝛾4 (1 − 𝛾2 )
× sec6 (𝛾𝑥) tanh2 (𝛾𝑥)]
𝑡2𝛼
.
Γ (1 + 2𝛼)
(21)
Abstract and Applied Analysis
𝑡
5
0.4
𝑡
0.2
0
0
0.03
0.03
0.02
0.02
0.01
0.01
0
−20
𝑡
−10
0
𝑥
10
0
−20
20
−10
(a)
0.4
0
𝑥
10
20
10
20
(a)
0.4
𝑡
0.2
0.2
0
0
0.03
0.03
0.02
0.02
0.01
0
0.4
0.2
0.01
−20
−10
0
𝑥
10
20
0
−20
0
−10
(b)
0
𝑥
(b)
Figure 1: The surface shows the solution 𝑢(𝑥, 𝑡) of (11): (a) exact
solution; (b) approximate solution of (22) when 𝛼 = 𝑐0 = 𝑐2 = 1,
𝛾 = 0.1, and 𝛽 = 0.08.
Figure 2: The surface shows the solution V(𝑥, 𝑡) of (11): (a) exact
solution; (b) approximate solution of (23) when 𝛼 = 𝑐0 = 𝑐2 = 1,
𝛾 = 0.1, and 𝛽 = 0.08.
The approximate solutions of (11) in finite series forms are
given by
The generalized two-dimensional differential transforms
of the initial conditions of (25) are given by
𝑢 (𝑥, 𝑡) = 𝑢0 (𝑥, 𝑡) + 𝑢1 (𝑥, 𝑡) + 𝑢2 (𝑥, 𝑡) ,
(22)
V (𝑥, 𝑡) = V0 (𝑥, 𝑡) + V1 (𝑥, 𝑡) + V2 (𝑥, 𝑡) ,
(23)
𝑤 (𝑥, 𝑡) = 𝑤0 (𝑥, 𝑡) + 𝑤1 (𝑥, 𝑡) + 𝑤2 (𝑥, 𝑡) .
(24)
In order to verify whether the approximate solutions of
(22)–(24) lead to higher accuracy, we draw the figures of the
approximate solutions of (22)–(24) with 𝛼 = 1, as well as the
exact solutions (13) when 𝑐0 = 𝑐2 = 1, 𝛾 = 0.1, and 𝛽 =
0.08. It can be seen from Figures 1(a) to 3(b) that the solutions
obtained by the presented method is nearly identical with the
exact solutions. So, we conclude that a good approximation is
achieved by using the GDTM.
In the following, we will construct an approximate solution of (11) with the new initial conditions,
𝛽 − 2𝛾2
+ 2𝛾2 tanh2 (𝛾𝑥) ,
3
4𝛾2 𝑐0 (𝛽 + 𝛾2 ) 4𝛾2 (𝛽 + 𝛾2 )
V (𝑥, 0) = −
+
tanh (𝛾𝑥) ,
3𝑐1
3𝑐12
𝑢 (𝑥, 0) =
𝑈1,𝛼 (0, 0) =
1
(𝛽 − 2𝛾2 ) ,
3
4
𝑈1,𝛼 (4, 0) = − 𝛾6 ,
3
𝑉1,𝛼 (1, 0) =
𝑐
𝑊1,𝛼 (3, 0) = − 1 𝛾3 ,
3
𝑈1,𝛼 (2, 0) = 2𝛾4 ,
𝑈1,𝛼 (6, 0) =
𝑉1,𝛼 (0, 0) = −
𝑊1,𝛼 (0, 0) = 𝑐0 ,
(25)
if 𝑘 = 1, 3, 5, . . . ,
𝑉1,𝛼 (𝑘, 0) = 𝑊1,𝛼 (𝑘, 0) = 0 if 𝑘 = 2, 4, 6, . . . ,
𝑉1,𝛼 (3, 0) = −
𝑤 (𝑥, 0) = 𝑐0 + 𝑐1 tanh (𝛾𝑥) ,
where 𝑐0 , 𝑐1 , 𝛾, and 𝛽 are arbitrary constants.
𝑈1,𝛼 (𝑘, 0) = 0,
4𝑐0 𝛾2 (𝛽 + 𝛾2 )
3𝑐12
4𝛾3 (𝛽 + 𝛾2 )
3𝑐1
4𝛾5 (𝛽 + 𝛾2 )
9𝑐1
34 8
𝛾 ,...,
45
,
,
,...,
𝑊1,𝛼 (1, 0) = 𝑐1 𝛾,
𝑊1,𝛼 (5, 0) =
2𝑐1 5
𝛾 ,....
15
(26)
6
Abstract and Applied Analysis
Utilizing the recurrence relations in (14) and the transformed initial conditions, we can obtain the following
approximate solutions:
𝑢 (𝑥, 𝑡) = 𝑢0 (𝑥, 𝑡) + 𝑢1 (𝑥, 𝑡) + 𝑢2 (𝑥, 𝑡) + 𝑢3 (𝑥, 𝑡) ,
× (3𝑐1 )
−1
Γ (1 + 2𝛼)
𝑡3𝛼
,
}
Γ2 (1 + 𝛼) Γ (1 + 3𝛼)
𝑤0 (𝑥, 𝑡) = 𝑐0 + 𝑐1 tanh (𝛾𝑥) ,
V (𝑥, 𝑡) = V0 (𝑥, 𝑡) + V1 (𝑥, 𝑡) + V2 (𝑥, 𝑡) + V3 (𝑥, 𝑡) ,
𝑤 (𝑥, 𝑡) = 𝑤0 (𝑥, 𝑡) + 𝑤1 (𝑥, 𝑡) + 𝑤2 (𝑥, 𝑡) + 𝑤3 (𝑥, 𝑡) ,
(27)
where the closed forms of 𝑢𝑖 (𝑥, 𝑡), V𝑖 (𝑥, 𝑡), and 𝑤𝑖 (𝑥, 𝑡) (𝑖 =
0, 1, 2, 3) are
𝛽 − 2𝛾2
+ 2𝛾2 tanh2 (𝛾𝑥) ,
3
𝑡𝛼
𝑢1 (𝑥, 𝑡) = 4𝛽𝛾3 sech2 (𝛾𝑥) tanh (𝛾𝑥)
,
Γ (1 + 𝛼)
𝑡2𝛼
𝑢2 (𝑥, 𝑡) = −4𝛽2 𝛾4 (cosh (2𝛾𝑥) − 2) sech4 (𝛾𝑥)
,
Γ (1 + 2𝛼)
𝑢0 (𝑥, 𝑡) =
2
× (5 + cosh (2𝛾𝑥)) sech4 (𝛾𝑥) tanh3 (𝛾𝑥))
𝑤1 (𝑥, 𝑡) = 𝑐1 𝛽𝛾 sech2 (𝛾𝑥)
𝑤2 (𝑥, 𝑡) = −2𝑐1 𝛽2 𝛾2 sech2 (𝛾𝑥) tanh (𝛾𝑥)
𝑤3 (𝑥, 𝑡) = {
− 15 (3𝛽 + 56𝛾2 ) cosh (2𝛾𝑥)
+ 576𝛾2 cosh (4𝛾𝑥)
+ (5𝛽 − 56𝛾2 ) cosh (6𝛾𝑥))
𝑢3 (𝑥, 𝑡) = {2𝛾 sech (𝛾𝑥) tanh (𝛾𝑥)
× sech8 (𝛾𝑥) +
× (64𝛽2 𝛾5 sech2 (𝛾𝑥) + 144𝛽2 𝛾5 tanh4 (𝛾𝑥))
2𝑐1 𝛽2 5
𝛾
5
× (7 cosh (4𝛾𝑥) − 9 − 58 cosh (2𝛾𝑥))
− 2𝛾5 (144𝛽2 𝛾2 − 8𝛽3 ) sech2 (𝛾𝑥) tanh3 (𝛾𝑥)
+ (179𝛽 + 2921𝛾2 ) cosh (2𝛾𝑥)
𝑡2𝛼
,
Γ (1 + 2𝛼)
𝑐1 𝛽2 𝛾3
(40 (8𝛾2 − 𝛽)
40
2
1
− 𝛽2 𝛾5 (878𝛽 + 3626𝛾2
88
𝑡𝛼
,
Γ (1 + 𝛼)
× sech6 (𝛾𝑥) tanh2 (𝛾𝑥)
×
𝑡3𝛼
.
Γ (1 + 3𝛼)
Γ (1 + 2𝛼)
}
Γ2 (1 + 𝛼)
(28)
+ (386𝛽 − 922𝛾2 ) cosh (4𝛾𝑥)
+7 (−5𝛽 + 𝛾2 ) cosh (6𝛾𝑥))
× sech8 (𝛾𝑥) tanh (𝛾𝑥)
1
3
𝛽
𝐷𝑡 𝑢 = 𝑢𝑥𝑥𝑥 − 3𝑢2 𝑢𝑥 + V𝑥𝑥 + 3𝑢V𝑥 + 3𝑢𝑥 V − 3𝜆𝑢𝑥 ,
2
2
𝛽
𝐷𝑡 V = − V𝑥𝑥𝑥 − 3VV𝑥 − 3𝑢𝑥 V𝑥 + 3𝑢2 V𝑥
𝑡3𝛼
Γ (1 + 2𝛼)
}
× 2
,
Γ (1 + 𝛼) Γ (1 + 3𝛼)
V0 (𝑥, 𝑡) = −
4𝛾2 𝑐0 (𝛽 + 𝛾2 )
V1 (𝑥, 𝑡) =
V2 (𝑥, 𝑡) = −
3𝑐12
+
4𝛾2 (𝛽 + 𝛾2 )
3𝑐1
4𝛽𝛾3 (𝛽 + 𝛾2 ) sech2 (𝛾𝑥)
3𝑐1
tanh (𝛾𝑥) ,
𝑡𝛼
,
Γ (1 + 𝛼)
8𝛽2 𝛾4 (𝛽 + 𝛾2 ) sech2 (𝛾𝑥) tanh (𝛾𝑥)
3𝑐1
V3 (𝑥, 𝑡) = {
𝑡2𝛼
,
Γ (1 + 2𝛼)
𝛽2 𝛾5
(𝛽 + 𝛾2 )
30𝑐1
× (40 (8𝛾2 − 𝛽) − 15 (3𝛽 + 56𝛾2 ) cosh (2𝛾𝑥)
+ 576𝛾2 cosh (4𝛾𝑥)
2
8
+ (5𝛽 − 56𝛾 ) cosh (6𝛾𝑥)) sech (𝛾𝑥)
2 6
2
− (16𝛽 𝛾 (𝛽 + 𝛾 )
3.2. Fractional Coupled MKdV Equation. Consider the following time-fractional coupled MKdV equation:
+ 3𝜆V𝑥 ,
𝑡 > 0, 0 < 𝛽 ≤ 1,
(29)
subject to the initial conditions
𝑢 (𝑥, 0) = 𝛾 tanh (𝛾𝑥) ,
V (𝑥, 0) =
1
(4𝛾2 + 𝜆) − 2𝛾2 tanh2 (𝛾𝑥) ,
2
(30)
where 𝛾 is an arbitrary constant.
The exact solutions of (29) and (30), for the special case
of 𝛽 = 1, given in [26], are
𝑢 (𝑥, 𝑡) = 𝛾 tanh (𝛾𝜉) ,
3
𝜉 = 𝑥 − (𝛾2 + 𝜆) 𝑡,
2
1
V (𝑥, 𝑡) = (4𝛾2 + 𝜆) − 2𝛾2 tanh2 (𝛾𝜉) .
2
(31)
Abstract and Applied Analysis
𝑡
7
𝐹1,𝛽 (𝑘, ℎ) = (𝑘 + 1) 𝑉1,𝛽 (𝑘 + 1, ℎ) ,
0.4
0.2
𝐺1,𝛽 (𝑘, ℎ) = (𝑘 + 1) 𝑈1,𝛽 (𝑘 + 1, ℎ) ,
0
𝑃1,𝛽 (𝑘, ℎ) = 𝐺1,𝛽 (𝑘, ℎ) + 3𝑉1,𝛽 (𝑘, ℎ) ,
1.8
2
𝑄1,𝛽 (0, 0) = 𝑈1,𝛽
(0, 0) − 𝑉1,𝛽 (0, 0) + 𝜆,
1.6
𝑘
1.4
ℎ
𝑄1,𝛽 (𝑘, ℎ) = ∑ ∑𝑈1,𝛽 (𝑟, ℎ − 𝑠) 𝑈1,𝛽 (𝑘 − 𝑟, 𝑠)
1.2
𝑟=0 𝑠=0
1
−20
− 𝑉1,𝛽 (𝑘, ℎ) ,
−10
0
𝑥
10
Γ (𝛽 (ℎ + 1) + 1)
𝑈1,𝛽 (𝑘, ℎ + 1)
Γ (𝛽ℎ + 1)
20
(a)
𝑡
𝑘2 + ℎ2 ≠ 0,
=
0.4
1
(𝑘 + 1) (𝑘 + 2) 𝑃1,𝛽 (𝑘 + 2, ℎ)
2
𝑘
0.2
ℎ
− 3∑ ∑𝑄1,𝛽 (𝑟, ℎ − 𝑠) 𝐺1,𝛽 (𝑘 − 𝑟, 𝑠)
𝑟=0 𝑠=0
0
𝑘+1 ℎ
1.8
+ 3 ∑ ∑𝑈1,𝛽 (𝑟, ℎ − 𝑠) 𝐹1,𝛽 (𝑘 − 𝑟, 𝑠) ,
1.6
𝑟=0 𝑠=0
1.4
Γ (𝛽 (ℎ + 1) + 1)
𝑉1,𝛽 (𝑘, ℎ + 1)
Γ (𝛽ℎ + 1)
1.2
1
−20
−10
𝑘
0
𝑥
10
ℎ
= −∑ ∑ (𝑘 + 1 − 𝑟) 𝐺1,𝛽 (𝑟, ℎ − 𝑠) 𝑃1,𝛽
20
𝑟=0 𝑠=0
(b)
Figure 3: The surface shows the solution 𝑤(𝑥, 𝑡) of (11): (a) exact
solution; (b) approximate solution (24) when 𝛼 = 𝑐0 = 𝑐2 = 1, 𝛾 =
0.1 and 𝛽 = 0.08.
× (𝑘 + 1 − 𝑟, 𝑠) − (𝑘 + 1) (𝑘 + 2) 𝐹1,𝛽 (𝑘 + 2, ℎ)
𝑘
𝑟=0 𝑠=0
𝑘
Equation system of (29) is more complex. In order to
obtain an explicit iteration scheme, we follow the pretreatment technique introduced recently by Chang [20, 21] to deal
with long nonlinear items. Firstly, we suppose that
𝑓 = V𝑥 ,
𝑔 = 𝑢𝑥 ,
𝑝 = 𝑔 + 3V,
𝑞 = 𝑢2 − V + 𝜆.
(32)
According to (32), (29) can be equivalently written as the
following form:
𝑓 = V𝑥 ,
𝑝 = 𝑔 + 3V,
𝑔 = 𝑢𝑥 ,
2
𝑞 = 𝑢 − V + 𝜆,
1
𝛽
𝐷𝑡 𝑢 = 𝑝𝑥𝑥 − 3𝑞𝑔 + 3𝑢𝑓,
2
(33)
𝛽
𝐷𝑡 V = −𝑔𝑝𝑥 + 3𝑞𝑓 − 𝑓𝑥𝑥 + 𝑔𝑔𝑥 .
Suppose that the solutions 𝑢(𝑥, 𝑡) and V(𝑥, 𝑡) can be represented as the products of single-valued functions, respectively. Applying the generalized two-dimensional differential
transform to both sides of (33) and using the related theorems, we have
ℎ
+ 3∑ ∑𝑄1,𝛽 (𝑟, ℎ − 𝑠) 𝐹1,𝛽 (𝑘 − 𝑟, 𝑠)
ℎ
+ ∑ ∑ (𝑘 + 1 − 𝑟) 𝐺1,𝛽 (𝑟, ℎ − 𝑠) 𝐺1,𝛽
𝑟=0 𝑠=0
× (𝑘 + 1 − 𝑟, 𝑠) .
(34)
Herein 𝐹1,𝛽 (𝑘, ℎ), 𝐺1,𝛽 (𝑘, ℎ), 𝑃1,𝛽 (𝑘, ℎ), and 𝑄1,𝛽 (𝑘, ℎ) denote
the differential transformations of the functions 𝑓(𝑥, 𝑡),
𝑔(𝑥, 𝑡), 𝑝(𝑥, 𝑡), and 𝑞(𝑥, 𝑡), respectively.
The generalized two-dimensional differential transforms
of the initial conditions of (30) can be obtained as follows:
𝑈1,𝛽 (𝑘, 0) = 0,
if 𝑘 = 0, 2, 4, . . . ,
𝑉1,𝛽 (𝑘, 0) = 0,
if 𝑘 = 1, 3, 5, . . . ,
𝑈1,𝛽 (1, 0) = 𝛾2 ,
𝑈1,𝛽 (5, 0) =
𝑉1,𝛽 (0, 0) =
2𝛾6
,
15
𝛾4
,
3
17𝛾8
𝑈1,𝛽 (7, 0) = −
,...,
315
𝑈1,𝛽 (3, 0) = −
1
(4𝛾2 + 𝜆) ,
2
𝑉1,𝛽 (4, 0) =
4𝛾6
,
3
𝑉1,𝛽 (2, 0) = −2𝛾4 ,
𝑉1,𝛽 (6, 0) = −
34𝛾8
,....
45
(35)
8
Abstract and Applied Analysis
Utilizing the recurrence relations of (34) and the transformed initial conditions, we can obtain all the 𝑈1,𝛽 (𝑘, ℎ)
and 𝑉1,𝛽 (𝑘, ℎ) with the help of Mathematica. Through the
complex calculation which is similar to the solving process
in Section 3.1, we have the approximate solutions of (29) in
finite series,
𝑢 (𝑥, 𝑡) = 𝑢0 (𝑥, 𝑡) + 𝑢1 (𝑥, 𝑡) + 𝑢2 (𝑥, 𝑡) + 𝑢3 (𝑥, 𝑡) ,
(36)
V (𝑥, 𝑡) = V0 (𝑥, 𝑡) + V1 (𝑥, 𝑡) + V2 (𝑥, 𝑡) + V3 (𝑥, 𝑡) ,
(37)
0
𝑡
0.5
0.1
1
0.05
0
−0.05
−0.1
where the closed forms of 𝑢𝑖 (𝑥, 𝑡) and V𝑖 (𝑥, 𝑡) (𝑖 = 0, 1, 2, 3)
are given by
40
20
0
𝑢0 (𝑥, 𝑡) = 𝛾 tanh (𝛾𝑥) ,
−40
(a)
𝑡
sec2 (𝛾𝑥) 𝑡𝛼
1
,
𝑢1 (𝑥, 𝑡) = − 𝛾2 (2𝛾2 + 3𝜆)
2
Γ (1 + 𝛼)
1
0.5
0.52
1
𝑢2 (𝑥, 𝑡) = − 𝛾3 sec2 (𝛾𝑥)
2
0
0.515
0.51
2
× [(2𝛾2 + 3𝜆) − 144𝛾2 𝜆sec2 (𝛾𝑥)]
×
−20
𝑥
0.505
2𝛼
0.5
−40
tanh (𝛾𝑥) 𝑡
,
Γ (1 + 2𝛼)
−20
0
20
𝑥
1
𝑢3 (𝑥, 𝑡) = { 𝛾4 sec2 (𝛾𝑥)
4
40
(b)
3
Figure 4: The surface shows the exact solutions (31): (a) 𝑢(𝑥, 𝑡); (b)
V(𝑥, 𝑡) when 𝜆 = 1 and 𝛾 = 0.1.
× [−2(2𝛾2 + 3𝜆) + 3sec2 (𝛾𝑥)
× (−152𝛾6 + 1092𝛾4 𝜆 + 270𝛾2 𝜆2 + 27𝜆3
+ (200𝛾6 − 4776𝛾4 𝜆 − 270𝛾2 𝜆2 )
V3 (𝑥, 𝑡) = {2𝛾5 sec8 (𝛾𝑥)
× [92𝛾6 + 3420𝛾4 𝜆 + 351𝛾2 𝜆2 + 54𝜆3
× sec2 (𝛾𝑥)
− 9 (12𝛾6 + 1456𝛾4 𝜆 + 63𝛾2 𝜆2 − 9𝜆3 )
𝑡3𝛼
+ 4032𝛾 𝜆sec (𝛾𝑥))] }
Γ (1 + 3𝛼)
4
4
× sinh2 (𝛾𝑥)
3
+ 𝛾6 (10𝛾2 − 9𝜆) (2𝛾2 + 3𝜆)
4
− 96𝛾2 (2𝛾4 − 48𝛾2 𝜆 + 9𝜆2 ) sinh4 (𝛾𝑥)
3
×
sec6 (𝛾𝑥) Γ (1 + 2𝛼) 𝑡3𝛼
Γ(1 + 𝛼)2 Γ (1 + 3𝛼)
V0 (𝑥, 𝑡) =
×
,
1
(4𝛾2 + 𝜆) − 2𝛾2 tanh2 (𝛾𝑥) ,
2
tanh (𝛾𝑥) 𝑡𝛼
V1 (𝑥, 𝑡) = 2𝛾 (2𝛾 − 3𝜆) sec (𝛾𝑥)
,
Γ (1 + 𝛼)
3
2
V2 (𝑥, 𝑡) = {2(2𝛾2 − 3𝜆) − 3sec2 (𝛾𝑥)
2
2
2
2
× [4𝛾 − 60𝛾 𝜆 + 9𝜆 + 48𝛾 𝜆sec (𝛾𝑥)] }
×
4
2
2𝛼
𝛾 sec (𝛾𝑥) 𝑡
,
Γ (1 + 2𝛼)
tanh (𝛾𝑥) 𝑡3𝛼
Γ (1 + 3𝛼)
2
− [27(𝜆 − 2𝛾2 ) sinh (𝛾𝑥)
−96𝛾2 (2𝛾2 − 3𝜆) sinh3 (𝛾𝑥) ]
2
2
4
6
+(2𝛾2 − 3𝜆) sinh (𝛾𝑥) ]}
× [2 cosh (2𝛾𝑥) − 3]
×
𝛾7 sec7 (𝛾𝑥) Γ (1 + 2𝛼) 𝑡3𝛼
Γ(1 + 𝛼)2 Γ (1 + 3𝛼)
.
(38)
The effectiveness and accuracy of the approximate solutions can be seen from the comparison figures.
Figures 4 and 5 show the approximate solutions of (36)
and (37) and the exact ones of (31) with 𝛼 = 𝜆 = 1 and
Abstract and Applied Analysis
9
𝛾 = 0.1, respectively. Comparing Figures 1(a) and 1(b) with
Figures 2(a) and 2(b), we can see that the solutions obtained
by different methods are nearly identical. From these figures,
we can know that the series solutions converge rapidly, so a
good approximation has been achieved.
In this paper, combining the Caputo fractional derivative,
the GDTM was applied to derive approximate analytical
solutions of the time-fractional Hirota-Satsuma coupled KdV
equation and coupled MKdV equation with initial conditions. The numerical solutions obtained from the GDTM
are shown graphically. The obtained results demonstrate the
reliability of the algorithm and its wider applicability to
nonlinear fractional coupled partial differential equations.
In [8] and the references cited therein, the so-called
fractional complex transform (FCT) is suggested to convert
a fractional differential equation with Jumarie’s modification
of Riemann-Liouville derivative into its classical differential
partner. According to the idea of FCT, for some fractional
differential equations with Caputo time-fractional derivative
𝛽
𝐷𝑡 , we assume that
+∞
𝑚=0
𝑡𝑚𝛽
V (𝑥, 𝑡) = ∑ V𝑚 (𝑥)
,
Γ (1 + 𝑚𝛽)
𝑚=0
+∞
𝑔 (𝑥, 𝑡) = ∑ 𝑔𝑚 (𝑥)
𝑚=0
𝑡𝑚𝛽
𝑝 (𝑥, 𝑡) = ∑ 𝑝𝑚 (𝑥)
,
Γ (1 + 𝑚𝛽)
𝑚=0
+∞
𝑞 (𝑥, 𝑡) = ∑ 𝑞𝑚 (𝑥)
𝑚=0
𝑡𝑚𝛽
.
Γ (1 + 𝑚𝛽)
𝑢0 (𝑥) = 𝑢 (𝑥, 0) ,
V0 (𝑥) = V (𝑥, 0) ,
𝑓0 (𝑥) = V0󸀠 (𝑥) ,
𝑔0 (𝑥) = 𝑢0󸀠 (𝑥) ,
𝑝0 (𝑥) = 𝑔0 (𝑥) + 3V0 (𝑥) ,
𝑞0 (𝑥) = 𝑢02 (𝑥) − V0 (𝑥) + 𝜆,
󸀠
𝑓𝑚 (𝑥) = V𝑚
(𝑥) ,
(39)
𝑞𝑚 (𝑥) = ∑𝑢𝑟 (𝑥) 𝑢𝑚−𝑟 (𝑥)
𝑟=0
󸀠
𝑔𝑚 (𝑥) = 𝑢𝑚
(𝑥) ,
Γ (1 + 𝑚𝛽)
Γ (1 + 𝑟𝛽) Γ (1 + (𝑚 − 𝑟) 𝛽)
− V𝑚 (𝑥) ,
𝑢𝑚+1 (𝑥) =
𝑚
1 󸀠󸀠
𝑝𝑚 (𝑥) − 3∑𝑞𝑟 (𝑥) 𝑔𝑚−𝑟 (𝑥)
2
𝑟=0
×
+∞
𝑡𝑚𝛽
= ∑ 𝑢𝑚+1 (𝑥)
,
Γ (1 + 𝑚𝛽)
𝑚=0
(42)
From (33), (40), and (41), we have the following iteration
formulae:
𝑚
consequently, from Lemma 3,
𝛽
𝐷𝑡 𝑢 (𝑥, 𝑡)
𝑡𝑚𝛽
,
Γ (1 + 𝑚𝛽)
𝑝𝑚 (𝑥) = 𝑔𝑚 (𝑥) + 3V𝑚 (𝑥) ,
𝑡𝑚𝛽
,
Γ (1 + 𝑚𝛽)
+∞
𝑚=0
𝑡𝑚𝛽
,
Γ (1 + 𝑚𝛽)
+∞
4. Summary and Discussion
𝑢 (𝑥, 𝑡) = ∑ 𝑢𝑚 (𝑥)
+∞
𝑓 (𝑥, 𝑡) = ∑ 𝑓𝑚 (𝑥)
(40)
Γ (1 + 𝑚𝛽)
Γ (1 + 𝑟𝛽) Γ (1 + (𝑚 − 𝑟) 𝛽)
𝑚
+ 3∑𝑢𝑟 (𝑥) 𝑓𝑚−𝑟 (𝑥)
𝑟=0
and it can be seen easily that
+∞
𝑚
𝑚=0
𝑟=0
×
𝑢 (𝑥, 𝑡) V (𝑥, 𝑡) = ∑ (∑𝑢𝑟 (𝑥) V𝑚−𝑟 (𝑥)
×
×
Γ (1 + 𝑚𝛽)
,
Γ (1 + 𝑟𝛽) Γ (1 + (𝑚 − 𝑟) 𝛽)
𝑚
Γ (1 + 𝑚𝛽)
)
Γ (1 + 𝑟𝛽) Γ (1 + (𝑚 − 𝑟) 𝛽)
𝑡𝑚𝛽
.
Γ (1 + 𝑚𝛽)
󸀠
V𝑚+1 (𝑥) = − 𝑓𝑚󸀠󸀠 (𝑥) − ∑𝑔𝑟 (𝑥) 𝑝𝑚−𝑟
(𝑥)
𝑟=0
×
Γ (1 + 𝑚𝛽)
Γ (1 + 𝑟𝛽) Γ (1 + (𝑚 − 𝑟) 𝛽)
𝑚
(41)
+3∑𝑞𝑟 (𝑥) 𝑓𝑚−𝑟 (𝑥)
𝑟=0
𝑚
We can point out that the approximate solutions of (11)
and (29) by GDTM can be derived by the similar method
compared with FCT. Without loss of generality, we only
consider (29) with the initial conditions of (30). In fact,
suppose that
󸀠
+ ∑𝑔𝑟 (𝑥) 𝑔𝑚−𝑟
(𝑥)
𝑟=0
for 𝑚 = 0, 1, 2, 3, . . ..
Γ (1+𝑚𝛽)
Γ (1+𝑟𝛽) Γ (1+(𝑚−𝑟) 𝛽)
Γ (1+𝑚𝛽)
,
Γ (1+𝑟𝛽) Γ (1+(𝑚−𝑟) 𝛽)
(43)
10
Abstract and Applied Analysis
0
𝑡
0.5
0.1
[5]
1
0.05
[6]
0
−0.05
−0.1
40
[7]
20
0
−20
𝑥
−40
(a)
𝑡
1
0.5
0.52
[8]
[9]
0
0.515
0.51
[10]
0.505
0.5
−40
−20
[11]
0
20
𝑥
40
(b)
Figure 5: The surface shows the approximate solutions of (29): (a)
𝑢(𝑥, 𝑡) of (36); (b) V(𝑥, 𝑡) of (37) when 𝛼 = 𝜆 = 1 and 𝛾 = 0.1.
Setting 𝑢𝑚 (𝑥, 𝑡) = 𝑢𝑚 (𝑥)(𝑡𝑚𝛽 /Γ(1 + 𝑚𝛽)) and V𝑚 (𝑥, 𝑡) =
V𝑚 (𝑥)(𝑡𝑚𝛽 /Γ(1 + 𝑚𝛽)), and using (43), we can obtain the 3order approximate solutions as (36) and (37) by using the
symbol computational software Mathematica.
Acknowledgments
The project is supported by the NNSF of China (11061021),
the NSF-IMU of China (2012MS0105, 2012MS0106), and the
YSF-IMU of China (ND0811).
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