Hindawi Publishing Corporation
Journal of Applied Mathematics
Volume 2013, Article ID 481729, 10 pages
http://dx.doi.org/10.1155/2013/481729
Research Article
A New Fractional Subequation Method and Its Applications for
Space-Time Fractional Partial Differential Equations
Fanwei Meng1 and Qinghua Feng2
1
2
School of Mathematical Sciences, Qufu Normal University, Qufu, Shandong 273165, China
School of Science, Shandong University of Technology, Zibo, Shandong 255049, China
Correspondence should be addressed to Qinghua Feng; fqhua@sina.com
Received 22 February 2013; Accepted 1 July 2013
Academic Editor: Magdy A. Ezzat
Copyright © 2013 F. Meng and Q. Feng. 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.
A new fractional subequation method is proposed for finding exact solutions for fractional partial differential equations (FPDEs).
The fractional derivative is defined in the sense of modified Riemann-Liouville derivative. As applications, abundant exact solutions
including solitary wave solutions as well as periodic wave solutions for the space-time fractional generalized Hirota-Satsuma
coupled KdV equations are obtained by using this method.
1. Introduction
Fractional differential equations are generalizations of classical differential equations of integer order. Recently, fractional
differential equations have been the focus of many studies
due to their frequent appearance in various applications
in physics, biology, engineering, signal processing, systems
identification, control theory, finance, and fractional dynamics. Among the investigations for fractional differential equations, research for seeking exact solutions and approximate
solutions of fractional differential equations is a hot topic.
New exact solutions for fractional differential equations
may help to understand better corresponding nonlinear
wave phenomena they describe. Some powerful methods
have been proposed so far (e.g., see [1–12]). Using these
methods, a variety of fractional differential equations have
been investigated.
In this paper, we propose a new fractional subequation
method to establish exact solutions for fractional partial differential equations (FPDEs), which is based on the following
fractional ordinary differential equation:
𝐷𝜉2𝛼 𝐺 (𝜉) + 𝜆𝐷𝜉𝛼 𝐺 (𝜉) + 𝜇𝐺 (𝜉) = 0,
0 < 𝛼 ≤ 1,
(1)
where 𝐷𝜉𝛼 𝐺(𝜉) denotes the modified Riemann-Liouville
derivative of order 𝛼 for 𝐺(𝜉) with respect to 𝜉.
The definition and some important properties for
Jumarie’s modified Riemann-Liouville derivative of
order 𝛼 are listed as follows (see [13–16]):
𝐷𝑡𝛼 𝑓 (𝑡)
𝑑 𝑡
1
{
{
∫ (𝑡 − 𝜉)−𝛼 (𝑓 (𝜉) − 𝑓 (0)) 𝑑𝜉,
{
{ Γ (1 − 𝛼) 𝑑𝑡 0
={
0 < 𝛼 < 1,
{
{
(𝛼−𝑛)
{ (𝑛)
,
𝑛 ≤ 𝛼 < 𝑛 + 1, 𝑛 ≥ 1,
{ (𝑓 (𝑡))
(2)
𝐷𝑡𝛼 𝑡𝑟 =
Γ (1 + 𝑟) 𝑟−𝛼
𝑡 ,
Γ (1 + 𝑟 − 𝛼)
(3)
𝐷𝑡𝛼 (𝑓 (𝑡) 𝑔 (𝑡)) = 𝑔 (𝑡) 𝐷𝑡𝛼 𝑓 (𝑡) + 𝑓 (𝑡) 𝐷𝑡𝛼 𝑔 (𝑡) ,
(4)
𝛼
𝐷𝑡𝛼 𝑓 [𝑔 (𝑡)] = 𝑓𝑔󸀠 [𝑔 (𝑡)] 𝐷𝑡𝛼 𝑔 (𝑡) = 𝐷𝑔𝛼 𝑓 [𝑔 (𝑡)] (𝑔󸀠 (𝑡)) .
(5)
We organize this paper as follows. In Section 2, we derive
the expression for 𝐷𝜉𝛼 𝐺(𝜉)/𝐺(𝜉) related to (1). In Section 3,
we give the description of the fractional subequation method
for solving FPDEs. Then in Section 4 we apply this method
to establish exact solutions for the space-time fractional
generalized Hirota-Satsuma coupled KdV equations. Some
conclusions are presented at the end of the paper.
2
Journal of Applied Mathematics
2. The General Expression for 𝐷𝜉𝛼 𝐺(𝜉)/𝐺(𝜉)
equality in (5), (1) can be turned into the following second
ordinary differential equation:
In order to obtain the general solutions for (1), we
suppose 𝐺(𝜉) = 𝐻(𝜂) and a nonlinear fractional complex
transformation 𝜂 = 𝜉𝛼 /Γ(1 + 𝛼). Then, by (3) and the first
𝐻󸀠󸀠 (𝜂) + 𝜆𝐻󸀠 (𝜂) + 𝜇𝐻 (𝜂) = 0.
By the general solutions of (6), we have
2
𝐶1 sinh (√𝜆2 − 4𝜇/2) 𝜂 + 𝐶2 cosh (√𝜆2 − 4𝜇/2) 𝜂
{
{
𝜆 √𝜆 − 4𝜇
{
{
(
),
{
{− 2 +
{
2
{
𝐶1 cosh (√𝜆2 − 4𝜇/2) 𝜂 + 𝐶2 sinh (√𝜆2 − 4𝜇/2) 𝜂
{
{
{
{
{
{
𝐻󸀠 (𝜂) {
= { 𝜆 √4𝜇 − 𝜆2 −𝐶1 sin (√4𝜇 − 𝜆2 /2) 𝜂 + 𝐶2 cos (√4𝜇 − 𝜆2 /2) 𝜂
{− +
𝐻 (𝜂) {
),
(
{
{
2
2
{
{
𝐶1 cos (√4𝜇 − 𝜆2 /2) 𝜂 + 𝐶2 sin (√4𝜇 − 𝜆2 /2) 𝜂
{
{
{
{
{
{
𝐶2
{
{− 𝜆 +
,
{ 2 𝐶1 + 𝐶2 𝜂
𝜆2 − 4𝜇 > 0,
𝜆2 − 4𝜇 < 0,
(7)
𝜆2 − 4𝜇 = 0,
Since 𝐷𝜉𝛼 𝐺(𝜉) = 𝐷𝜉𝛼 𝐻(𝜂) = 𝐻󸀠 (𝜂)𝐷𝜉𝛼 𝜂 = 𝐻󸀠 (𝜂), we
obtain
where 𝐶1 and 𝐶2 are arbitrary constants.
𝛼
𝛼
{
√𝜆2 − 4𝜇 𝐶1 sinh (√𝜆2 − 4𝜇/2Γ (1 + 𝛼)) 𝜉 + 𝐶2 cosh (√𝜆2 − 4𝜇/2Γ (1 + 𝛼)) 𝜉
{
𝜆
{
{
(
− +
) , 𝜆2 − 4𝜇 > 0,
{
{
2
2
{
2
𝛼
2
𝛼
{
𝐶1 cosh (√𝜆 − 4𝜇/2Γ (1 + 𝛼)) 𝜉 + 𝐶2 sinh (√𝜆 − 4𝜇/2Γ (1 + 𝛼)) 𝜉
{
{
{
{
{
{
{
𝛼
𝛼
2
2
{
𝐷𝜉𝛼 𝐺 (𝜉) {
{ 𝜆 √4𝜇 − 𝜆2 −𝐶1 sin (√4𝜇 − 𝜆 /2Γ (1 + 𝛼)) 𝜉 + 𝐶2 cos (√4𝜇 − 𝜆 /2Γ (1 + 𝛼)) 𝜉
) , 𝜆2 − 4𝜇 < 0,
(
= {− +
2
2
{
𝐺 (𝜉)
2
𝛼
2
𝛼
{
√
√
𝐶1 cos ( 4𝜇 − 𝜆 /2Γ (1 + 𝛼)) 𝜉 + 𝐶2 sin ( 4𝜇 − 𝜆 /2Γ (1 + 𝛼)) 𝜉
{
{
{
{
{
{
𝐶2 Γ (1 + 𝛼)
𝜆
{
{
{
,
𝜆2 − 4𝜇 = 0.
− +
{
𝛼
{
2
𝐶
Γ
+
𝛼)
+
𝐶
𝜉
(1
{
1
2
{
{
{
{
3. Description of the Fractional
Subequation Method
(8)
Step 1. Suppose that
In this section, we give the main steps of the fractional
subequation method for finding exact solutions for FPDEs.
Suppose that an FPDE, say in the independent variables 𝑡, 𝑥1 , 𝑥2 , . . . , 𝑥𝑛 , is given by
𝑃 (𝑢1 , . . . , 𝑢𝑘 , 𝐷𝑡𝛼 𝑢1 , . . . , 𝐷𝑡𝛼 𝑢𝑘 , 𝐷𝑥𝛼1 𝑢1 , . . . ,
𝐷𝑥𝛼1 𝑢𝑘 , . . . , 𝐷𝑥𝛼𝑛 𝑢1 , . . . , 𝐷𝑥𝛼𝑛 𝑢𝑘 , 𝐷𝑡2𝛼 𝑢1 , . . . ,
(6)
(9)
𝐷𝑡2𝛼 𝑢𝑘 , 𝐷𝑥2𝛼1 𝑢1 , . . .) = 0,
where 𝑢𝑖 = 𝑢𝑖 (𝑡, 𝑥1 , 𝑥2 , . . . , 𝑥𝑛 ), 𝑖 = 1, . . . , 𝑘, are unknown
functions and 𝑃 is a polynomial in 𝑢𝑖 and their various partial
derivatives including fractional derivatives.
𝑢𝑖 (𝑡, 𝑥1 , 𝑥2 , . . . , 𝑥𝑛 ) = 𝑈𝑖 (𝜉) ,
𝜉 = 𝑐𝑡 + 𝑘1 𝑥1 + 𝑘2 𝑥2 + ⋅ ⋅ ⋅ + 𝑘𝑛 𝑥𝑛 + 𝜉0 .
(10)
Then, by the second equality in (5), (9) can be turned into
the following fractional ordinary differential equation with
respect to the variable 𝜉:
𝑃̃ (𝑈1 , . . . , 𝑈𝑘 , 𝑐𝛼 𝐷𝜉𝛼 𝑈1 , . . . , 𝑐𝛼 𝐷𝜉𝛼 𝑈𝑘 ,
𝑘1𝛼 𝐷𝜉𝛼 𝑈1 , . . . , 𝑘1𝛼 𝐷𝜉𝛼 𝑈𝑘 , . . . , 𝑘𝑛𝛼 𝐷𝜉𝛼 𝑈1 , . . . ,
𝑘𝑛𝛼 𝐷𝜉𝛼 𝑈𝑘 , 𝑐2𝛼 𝐷𝜉2𝛼 𝑈1 , . . . ,
𝑐2𝛼 𝐷𝜉2𝛼 𝑈𝑘 , 𝑘12𝛼 𝐷𝜉2𝛼 𝑈1 , . . .) = 0.
(11)
Journal of Applied Mathematics
3
Step 2. Suppose that the solution of (11) can be expressed by
a polynomial in (𝐷𝜉𝛼 𝐺/𝐺) as follows:
constants with 𝑘, 𝑐 ≠ 0. Then, by usinge the second equality in
(4), we obtain
𝛼
𝐷𝑥𝛼 𝑢 = 𝐷𝑥𝛼 𝑈 (𝜉) = (𝐷𝜉𝛼 𝑈) (𝜉𝑥󸀠 ) = 𝑘𝛼 𝐷𝜉𝛼 𝑈,
𝑈𝑗 (𝜉)
𝑚𝑗
= 𝑎𝑗,0 + ∑ [𝑎𝑗,𝑖 (
𝐷𝜉𝛼 𝐺
𝑖=1
+ 𝑏𝑗,𝑖 (
𝐺
)
𝐷𝜉𝛼 𝐺
𝐺
and similarly we have
𝑖−1
)
(14)
𝛼
𝐷𝑡𝛼 𝑢 = 𝐷𝑡𝛼 𝑈 (𝜉) = (𝐷𝜉𝛼 𝑈) (𝜉𝑡󸀠 ) = 𝑐𝛼 𝐷𝜉𝛼 𝑈,
𝑖
𝐷𝜉𝛼 𝐺
2
]
√ 𝜎 (1 + 1 (
) )] ,
𝜇
𝐺
]
𝑗 = 1, 2, . . . , 𝑘,
(12)
where 𝐺 = 𝐺(𝜉) satisfies (1), 𝜎 is a constant, and 𝑎𝑗,𝑖 , 𝑖 =
0, 1, . . . , 𝑚, 𝑗 = 1, 2, . . . , 𝑘, are constants to be determined
later. The positive integer 𝑚 can be determined by considering the homogeneous balance between the highest-order
derivatives and nonlinear terms appearing in (11).
Step 3. Substituting (12) into (11), using (1), and
collecting all terms with the same order of (𝐷𝜉𝛼 𝐺/
𝐺)√𝜎(1 + (1/𝜇)(𝐷𝜉𝛼 𝐺/𝐺)2 ) together, the left-hand side
of (11) is converted into another polynomial in (𝐷𝜉𝛼 𝐺/𝐺).
Equating each coefficient of this polynomial to zero yields
a set of algebraic equations for 𝑎𝑗0 , 𝑎𝑗,𝑖 , 𝑏𝑗,𝑖 , 𝑖 = 1, . . . , 𝑚,
𝑗 = 1, 2, . . . , 𝑘.
Step 4. Solving the equations in Step 3 and using (8), we can
construct a variety of exact solutions for (9).
4. Application of
the Method to Space-Time Fractional
Generalized Hirota-Satsuma Coupled
KdV Equations
𝐷𝑥𝛼 V = 𝑘𝛼 𝐷𝜉𝛼 𝑉,
𝐷𝑡𝛼 V = 𝑐𝛼 𝐷𝜉𝛼 𝑉,
𝐷𝑥𝛼 𝑤 = 𝑘𝛼 𝐷𝜉𝛼 𝑊,
𝐷𝑡𝛼 𝑤 = 𝑐𝛼 𝐷𝜉𝛼 𝑊;
then (11) can be turned into the following fractional ordinary
differential equations with respect to the variable 𝜉:
1
𝑐𝛼 𝐷𝜉𝛼 𝑈 − 𝑘3𝛼 𝐷𝜉3𝛼 𝑈 + 3𝑘𝛼 𝑈𝐷𝜉𝛼 𝑈 − 3𝑘𝛼 𝐷𝜉𝛼 (𝑉𝑊) = 0,
2
𝑐𝛼 𝐷𝜉𝛼 𝑉 + 𝑘3𝛼 𝐷𝜉3𝛼 𝑉 − 3𝑘𝛼 𝑈𝐷𝜉𝛼 𝑉 = 0,
Suppose that the solutions of (16) can be expressed by
𝑚1
𝑈 (𝜉) = 𝑎0 + ∑ [𝑎𝑖 (
𝐷𝜉𝛼 𝐺
𝐺
𝑖=1
+ 𝑏𝑖 (
𝑚2
𝑉 (𝜉) = 𝑐0 + ∑ [𝑐𝑖 (
𝐺
𝐷𝜉𝛼 𝐺
𝐺
𝑖=1
𝑚3
𝑊 (𝜉) = 𝑒0 + ∑ [𝑒𝑖 (
(13)
𝐷𝑡𝛼 𝑤 + 𝐷𝑥3𝛼 𝑤 − 3𝑢𝐷𝑥𝛼 𝑤 = 0,
0 < 𝛼 ≤ 1.
Equations (13) can be used to describe the interaction of two
long waves with different dispersion relations [17]. In [15],
the authors solved equations (13) by a proposed fractional
subequation method based on the fractional Riccati equation,
while in [16], (13) are solved by the known (𝐺󸀠 /𝐺)-expansion
method. Now we apply the described method in Section 3 to
solve (13). To begin with, suppose that 𝑢(𝑥, 𝑡) = 𝑈(𝜉), V(𝑥, 𝑡) =
𝑉(𝜉), 𝑤(𝑥, 𝑡) = 𝑊(𝜉), where 𝜉 = 𝑘𝑥 + 𝑐𝑡 + 𝜉0 , 𝑘, 𝑐, 𝜉0 are all
𝑖
)
𝐷𝜉𝛼 𝐺
)
𝐺
𝐺
+ 𝑓𝑖 (
)
2
𝛼
𝐷 𝐺
√ 𝜎 (1 + 1 ( 𝜉 ) )]
],
𝜇
𝐺
]
𝑖
𝐷𝜉𝛼 𝐺
𝑖=1
𝑖−1
𝐷𝜉𝛼 𝐺
𝑖−1
)
2
𝐷𝛼 𝐺
√ 𝜎 (1 + 1 ( 𝜉 ) )]
],
𝜇
𝐺
]
𝑖
)
𝐷𝜉𝛼 𝐺
𝐺
1
𝐷𝑡𝛼 𝑢 − 𝐷𝑥3𝛼 𝑢 + 3𝑢𝐷𝑥𝛼 𝑢 − 3𝐷𝑥𝛼 (V𝑤) = 0,
2
𝐷𝑡𝛼 V + 𝐷𝑥3𝛼 V − 3𝑢𝐷𝑥𝛼 V = 0,
(16)
𝑐𝛼 𝐷𝜉𝛼 𝑊 + 𝑘3𝛼 𝐷𝜉3𝛼 𝑊 − 3𝑘𝛼 𝑈𝐷𝜉𝛼 𝑊 = 0.
+ 𝑑𝑖 (
In this section, we will apply the described method in
Section 3 to solve the space-time fractional generalized
Hirota-Satsuma coupled KdV equations [15, 16]:
(15)
𝑖−1
)
2
𝐷𝛼 𝐺
√ 𝜎 (1 + 1 ( 𝜉 ) )]
].
𝜇
𝐺
]
(17)
Balancing the order of 𝐷𝜉3𝛼 𝑈 and 𝐷𝜉𝛼 (𝑉𝑊), 𝐷𝜉3𝛼 𝑉 and 𝑈𝐷𝜉𝛼 𝑉,
and 𝐷𝜉3𝛼 𝑊 and 𝑈𝐷𝜉𝛼 𝑊 in (16), we have 𝑚1 = 𝑚2 = 𝑚3 = 2.
So
𝑈 (𝜉) = 𝑎0 + 𝑎1 (
𝐷𝜉𝛼 𝐺
𝐺
) + 𝑎2 (
𝐷𝜉𝛼 𝐺
𝐺
2
)
2
𝛼
1 𝐷𝜉 𝐺
√
+ 𝑏1 𝜎 (1 + (
))
𝜇
𝐺
+ 𝑏2 (
𝐷𝜉𝛼 𝐺
𝐺
2
) √ 𝜎 (1 +
𝛼
1 𝐷𝜉 𝐺
(
) ),
𝜇
𝐺
4
Journal of Applied Mathematics
𝑉 (𝜉) = 𝑐0 + 𝑐1 (
𝐷𝜉𝛼 𝐺
𝐺
) + 𝑐2 (
𝐷𝜉𝛼 𝐺
𝐺
𝐷𝜉𝛼 𝐺
2
𝑐0 =
)
1
𝑐1 = 𝑐2 𝑘−2𝛼 𝑎1 ,
4
2
1
+ 𝑑1 √ 𝜎 (1 + (
))
𝜇
𝐺
𝑑2 = 0,
𝐷𝜉𝛼 𝐺
2
𝛼
1 𝐷𝜉 𝐺
+ 𝑑2 (
) √ 𝜎 (1 + (
) ),
𝐺
𝜇
𝐺
𝑊 (𝜉) = 𝑒0 + 𝑒1 (
𝐷𝜉𝛼 𝐺
𝐺
) + 𝑒2 (
𝐷𝜉𝛼 𝐺
𝐺
𝑒2 =
)
𝐺
𝜇=
𝛼
1 𝐷𝜉 𝐺
(
) ).
𝜇
𝐺
(1/𝜇)(𝐷𝜉𝛼 𝐺/𝐺)2 )
√𝜎(1 +
together, equating each coefficient
to zero yields a set of algebraic equations. Solving these
equations, with the aid of the mathematical software Maple,
yields the following seven groups of values.
Case 1. One has
𝑐2 =
𝑘 𝑏2
,
2𝑓2
1
5
𝑎0 = 𝑘−𝛼 𝑐𝛼 + 𝑘−2𝛼 𝜎𝑏22 ,
3
12
𝜆 = 0,
2𝛼
𝑎2 = 2𝑘 ,
𝑏1 = 0,
𝑑1 = 0,
𝑑2 =
2𝑘2𝛼 𝑓2
𝑒2 =
,
𝑏2
𝑒1 = 0,
𝑎1 = 𝑎1 ,
𝑎2 = 2𝑘2𝛼 ,
𝑏1 = 0,
𝑏2 = 0,
𝑐0 = 𝑐0 ,
𝑐1 = 𝑐1 ,
𝑐2 = 0,
𝑑1 = 0,
𝑑2 = 0,
𝑒0 =
𝑒1 (𝑘−2𝛼 𝑎1 𝑐1 − 2𝑐0 )
2𝑐1
𝑒2 = 0,
,
(21)
𝑒1 = 𝑒1 ,
𝑓1 = 0,
𝑏2 = 𝑏2 ,
𝑏22
4𝑓2
𝑓2 = 0.
𝑐0 = 𝑐0 ,
𝑒0 = 𝑒0 ,
𝑓1 = 0,
𝑒0 =
(19)
𝜇 = 𝜇,
𝑓1 = −
1
8
1
𝑎0 = 𝑘−2𝛼 𝑎12 + 𝑘2𝛼 𝜇 + 𝑘−𝛼 𝑐𝛼 ,
48
3
3
𝑎2 = 4𝑘2𝛼 ,
𝑏1 = 0,
𝑏2 = 0,
𝑐1 =
𝑏2 = 𝑏2 ,
𝑘2𝛼 𝑑1
,
𝑏2
𝑐2 = 0,
𝑑2 = 0,
(22)
𝑐0 𝑏22 (𝑘−4𝛼 𝜎𝑏22 − 4𝑐𝛼 𝑘−3𝛼 )
12𝑑12
𝑏2 (−𝑘−2𝛼 𝜎𝑏22 + 4𝑘−𝛼 𝑐𝛼 )
Case 2. One has
𝑎1 = 0,
𝑏1 = 0,
𝑑1 = 𝑑1 ,
𝑓2 = 𝑓2 .
1
𝜆 = 𝑘−2𝛼 𝑎1 ,
4
𝜆 = 0,
2
1
𝑎0 = 𝑘−2𝛼 𝜎𝑏22 + 𝑘−𝛼 𝑐𝛼 ,
3
3
𝑎2 = 𝑘2𝛼 ,
𝑐1 = 0,
,
𝜇 = 𝑘−4𝛼 𝜎𝑏22 ,
𝑒1 =
𝑎1 = 𝑎1 ,
1 −4𝛼 2 𝛼 −3𝛼 3 −4𝛼
𝑘 𝑎1 + 𝑐 𝑘 − 𝑘 𝑐1 𝑒1 ,
16
4
Case 4. One has
−𝛼 𝛼
−2𝛼
2
1 𝑏2 (16𝑘 𝑐 𝑓2 + 5𝑘 𝑓2 𝜎𝑏2 − 6𝑒0 𝑏2 )
,
𝑐0 =
24
𝑓22
2𝛼
𝑓2 = 0.
1
1
𝑎0 = 𝑘−2𝛼 𝑎12 + 𝑘−𝛼 𝑐𝛼 − 𝑘−2𝛼 𝑐1 𝑒1 ,
8
2
Substituting (18) into (16), using (1), and collecting
all the terms with the same power of (𝐷𝜉𝛼 𝐺/𝐺)
𝑎1 = 0,
𝑓1 = 0,
𝑘2𝛼 𝑎1
,
𝑐2
1
𝜆 = 𝑘−2𝛼 𝑎1 ,
2
(18)
1
𝜇 = 𝑘−4𝛼 𝜎𝑏22 ,
4
4𝑘4𝛼
,
𝑐2
𝑒1 =
Case 3. One has
2
) √ 𝜎 (1 +
𝑒0 = 𝑒0 ,
𝑑1 = 0,
(20)
2
𝐷𝜉𝛼 𝐺
𝑐2 = 𝑐2 ,
2
𝛼
1 𝐷𝜉 𝐺
+ 𝑓1 √ 𝜎 (1 + (
))
𝜇
𝐺
+ 𝑓2 (
1
𝑐 (𝑘−4𝛼 𝑎12 + 128𝜇 + 64𝑐𝛼 𝑘−3𝛼 − 24𝑘−4𝛼 𝑐2 𝑒0 ) ,
96 2
12𝑑1
,
𝑏22 (𝑘−4𝛼 𝜎𝑏22 − 4𝑐𝛼 𝑘−3𝛼 )
12𝑑1
,
,
𝑒2 = 0,
𝑓2 = 0.
Case 5. One has
𝜇 = 4𝑐𝛼 𝑘−3𝛼 ,
𝑎1 = 0,
𝜆 = 0,
𝑎2 = 𝑘2𝛼 ,
𝑎0 = 3𝑘−𝛼 𝑐𝛼 ,
𝑏1 = 0,
Journal of Applied Mathematics
𝑏2 = ±𝑘
𝛼 3𝛼
−𝛼 √ 𝑐 𝑘
,
𝑐0 = 0,
𝑐1 = 0,
𝑑1 = 0,
𝑑2 = 0,
𝑒0 = 𝑒0 ,
𝜎
𝑐2 = 0,
𝑒1 = ±
5
𝑘3𝛼 𝑓1
𝜎
√ 𝛼 3𝛼 ,
2
𝑐 𝑘
𝑓1 = 𝑓1 ,
+ 𝐶2 cosh (
𝛼
√−𝜇𝜉
))
Γ (1 + 𝛼)
× (𝐶1 cosh (
𝛼
√−𝜇𝜉
)
Γ (1 + 𝛼)
−1 2
𝑒2 = 0,
+ 𝐶2 sinh (
𝑓2 = 0.
(23)
𝛼
√−𝜇𝜉
)) ]
Γ (1 + 𝛼)
+ 𝑏2 √−𝜇 [ (𝐶1 sinh (
Case 6. One has
𝜇 = 𝜇,
2
1
𝑎0 = 𝑘2𝛼 𝜇 + 𝑘−𝛼 𝑐𝛼 ,
3
3
𝜆 = 0,
2𝛼
𝑎1 = 0,
𝑎2 = 2𝑘 ,
𝑐0 = 𝑐0 ,
𝑐1 = 𝑐1 ,
𝑐2 = 0,
𝑑1 = 0,
𝑑2 = 0,
𝑒1 =
4𝑐0 (−𝑘4𝛼 𝜇 + 𝑘𝛼 𝑐𝛼 )
3𝑐12
4 (−𝑘4𝛼 𝜇 + 𝑘𝛼 𝑐𝛼 )
3𝑐1
𝑒2 = 0,
𝑓1 = 0,
𝛼
√−𝜇𝜉
))
Γ (1 + 𝛼)
× (𝐶1 cosh (
𝑏1 = 0,
𝑏2 = 0,
𝑒0 = −
+ 𝐶2 cosh (
𝛼
√−𝜇𝜉
)
Γ (1 + 𝛼)
+ 𝐶2 sinh (
(24)
𝛼
√−𝜇𝜉
)
Γ (1 + 𝛼)
𝛼
{
√−𝜇𝜉
)
× (𝜎 {1 − [ (𝐶1 sinh (
Γ (1 + 𝛼)
{
,
,
+ 𝐶2 cosh (
𝑓2 = 0.
𝛼
√−𝜇𝜉
))
Γ (1 + 𝛼)
× (𝐶1 cosh (
Case 7. One has
𝜇 = 𝜇,
𝑏2 = 0,
𝑐2 = 0,
𝑒0 = −
𝑐1 = 0,
𝑑1 = 𝑑1 ,
𝑑2 = 0,
2𝑐0 𝜇 (𝑘4𝛼 𝜇 + 2𝑘𝛼 𝑐𝛼 )
3𝜎𝑑12
𝑒1 = 0,
𝑉1 (𝜉) =
(25)
,
−𝛼 𝛼
−2𝛼
2
1 𝑏2 (16𝑘 𝑐 𝑓2 + 5𝑘 𝑓2 𝜎𝑏2 − 6𝑒0 𝑏2 )
24
𝑓22
−
𝑒2 = 0,
2𝜇 (𝑘4𝛼 𝜇 + 2𝑘𝛼 𝑐𝛼 )
3 (𝜎𝑑1 )
,
𝑓2 = 0.
From Case 1, we obtain the following exact solutions.
When 𝜇 < 0,
× [ (𝐶1 sinh (
𝛼
𝑘2𝛼 𝑏2
√−𝜇𝜉
)
𝜇 [ (𝐶1 sinh (
2𝑓2
Γ (1 + 𝛼)
+𝐶2 cosh (
× (𝐶1 cosh (
Substituting the previous results into (18) and combining
with (8), we can obtain a series of exact solutions for (13).
1 −𝛼 𝛼 5 −2𝛼 2
𝑘 𝑐 + 𝑘 𝜎𝑏2 − 2𝑘2𝛼 𝜇
3
12
𝛼
√−𝜇𝜉
)
Γ (1 + 𝛼)
+𝐶2 sinh (
+
𝛼
√−𝜇𝜉
))
Γ (1 + 𝛼)
−1 2
𝛼
√−𝜇𝜉
)) ]
Γ (1 + 𝛼)
𝛼
𝑏22
√−𝜇𝜉
)
√−𝜇 [ (𝐶1 sinh (
4𝑓2
Γ (1 + 𝛼)
+ 𝐶2 cosh (
𝛼
√−𝜇𝜉
)
Γ (1 + 𝛼)
,
(26)
𝑏1 = 0,
𝑐0 = 𝑐0 ,
1/2
−1
𝛼
}
√−𝜇𝜉
)) ] })
+ 𝐶2 sinh (
Γ (1 + 𝛼)
}
5
1
𝑎0 = 𝑘2𝛼 𝜇 + 𝑘−𝛼 𝑐𝛼 ,
3
3
𝑎2 = 2𝑘2𝛼 ,
𝑎1 = 0,
𝑈1 (𝜉) =
𝛼
√−𝜇𝜉
)
Γ (1 + 𝛼)
2
𝜆 = 0,
𝑓1 =
−1
𝛼
√−𝜇𝜉
)) ]
Γ (1 + 𝛼)
× (𝐶1 cosh (
𝛼
√−𝜇𝜉
))
Γ (1 + 𝛼)
𝛼
√−𝜇𝜉
)
Γ (1 + 𝛼)
6
Journal of Applied Mathematics
−1
+ 𝐶2 sinh (
𝛼
√−𝜇𝜉
)) ]
Γ (1 + 𝛼)
When 𝜇 > 0,
5
1
𝑈2 (𝜉) = 𝑘−𝛼 𝑐𝛼 + 𝑘−2𝛼 𝜎𝑏22 + 2𝑘2𝛼 𝜇
3
12
𝛼
{
√−𝜇𝜉
× (𝜎 {1 − [ (𝐶1 sinh (
)
Γ (1 + 𝛼)
{
+ 𝐶2 cosh (
× (𝐶1 cosh (
× [ (−𝐶1 sin (
𝛼
√−𝜇𝜉
))
Γ (1 + 𝛼)
+𝐶2 cos (
𝛼
√−𝜇𝜉
)
Γ (1 + 𝛼)
× (𝐶1 cos (
1/2
2
−1
𝛼
}
√−𝜇𝜉
)) ] }) ,
+ 𝐶2 sinh (
Γ (1 + 𝛼)
}
(27)
−1 2
+𝐶2 cos (
𝛼
× (𝐶1 cos (
𝛼
√−𝜇𝜉
)
× (𝐶1 cosh (
Γ (1 + 𝛼)
√−𝜇𝜉
)) ]
Γ (1 + 𝛼)
𝛼
𝛼
√𝜇
) 𝜉𝛼
Γ (1 + 𝛼)
+𝐶2 sin (
𝑉2 (𝜉) =
𝛼
√−𝜇𝜉
))
Γ (1 + 𝛼)
+
2
−1 2
𝑘2𝛼 𝑏2
√𝜇
𝜇 [ (−𝐶1 sin (
) 𝜉𝛼
2𝑓2
Γ (1 + 𝛼)
+𝐶2 cos (
1/2
−1
𝛼
}
√−𝜇𝜉
+𝐶2 sinh (
)) ] }) ,
Γ (1 + 𝛼)
}
(28)
× (𝐶1 cos (
√𝜇
) 𝜉𝛼 )
Γ (1 + 𝛼)
√𝜇
) 𝜉𝛼
Γ (1 + 𝛼)
−1 2
√𝜇
+𝐶2 sin (
) 𝜉𝛼 ) ]
Γ (1 + 𝛼)
+
1/2
√𝜇
) 𝜉𝛼 ) ] }) ,
Γ (1 + 𝛼)
(29)
−𝛼 𝛼
−2𝛼
2
1 𝑏2 (16𝑘 𝑐 𝑓2 + 5𝑘 𝑓2 𝜎𝑏2 − 6𝑒0 𝑏2 )
24
𝑓22
𝛼
√−𝜇𝜉
)
Γ (1 + 𝛼)
where 𝜉 = 𝑘𝑥 + 𝑐𝑡 + 𝜉0 , 𝜇 = (1/4)𝑘−4𝛼 𝜎𝑏22 .
√𝜇
) 𝜉𝛼 )
Γ (1 + 𝛼)
−1
𝛼
{
√−𝜇𝜉
× (𝜎 {1 − [ (𝐶1 sinh (
)
Γ (1 + 𝛼)
{
× (𝐶1 cosh (
−1
√𝜇
) 𝜉𝛼 ) ]
Γ (1 + 𝛼)
× (𝐶1 cos (
√−𝜇𝜉
)) ]
Γ (1 + 𝛼)
+𝐶2 cosh (
√𝜇
) 𝜉𝛼
Γ (1 + 𝛼)
+𝐶2 cos (
𝛼
√−𝜇𝜉
)
Γ (1 + 𝛼)
𝛼
√𝜇
) 𝜉𝛼 )
Γ (1 + 𝛼)
{
√𝜇
) 𝜉𝛼
× (𝜎 {1 + [ (−𝐶1 sin (
Γ (1 + 𝛼)
{
√−𝜇𝜉
))
Γ (1 + 𝛼)
+𝐶2 sinh (
√𝜇
) 𝜉𝛼
Γ (1 + 𝛼)
+𝐶2 sin (
−1 2
𝛼
√−𝜇𝜉
+ 𝑓2 √−𝜇 [(𝐶1 sinh (
)
Γ (1 + 𝛼)
× (𝐶1 cosh (
√𝜇
) 𝜉𝛼
Γ (1 + 𝛼)
√𝜇
+𝐶2 sin (
) 𝜉𝛼 ) ]
Γ (1 + 𝛼)
√−𝜇𝜉
))
+ 𝐶2 cosh (
Γ (1 + 𝛼)
+𝐶2 cosh (
√𝜇
) 𝜉𝛼 )
Γ (1 + 𝛼)
+ 𝑏2 √𝜇 [ (−𝐶1 sin (
𝛼
2𝑘2𝛼 𝑓2
√−𝜇𝜉
𝑊1 (𝜉) = 𝑒0 −
𝜇 [ (𝐶1 sinh (
)
𝑏2
Γ (1 + 𝛼)
+𝐶2 sinh (
√𝜇
) 𝜉𝛼
Γ (1 + 𝛼)
𝑏22
√𝜇
) 𝜉𝛼
√𝜇 [ (−𝐶1 sin (
4𝑓2
Γ (1 + 𝛼)
Journal of Applied Mathematics
+𝐶2 cos (
× (𝐶1 cos (
7
√𝜇
) 𝜉𝛼 )
Γ (1 + 𝛼)
× (𝐶1 cos (
√𝜇
) 𝜉𝛼
Γ (1 + 𝛼)
+𝐶2 sin (
+𝐶2 sin (
× (𝜎 {1 + [ (−𝐶1 sin (
+𝐶2 cos (
√𝜇
) 𝜉𝛼 )
Γ (1 + 𝛼)
5
1
𝑈3 (𝜉) = 𝑘−𝛼 𝑐𝛼 + 𝑘−2𝛼 𝜎𝑏22
3
12
√𝜇
) 𝜉𝛼
Γ (1 + 𝛼)
− 2𝑘2𝛼 𝜇[tanh (
2𝑘2𝛼 𝑓2
𝜇
𝑏2
× (𝐶1 cos (
−1 2
√𝜇
) 𝜉𝛼 ) ]
Γ (1 + 𝛼)
+𝐶2 cos (
√𝜇
) 𝜉𝛼
Γ (1 + 𝛼)
√𝜇
) 𝜉𝛼 )
Γ (1 + 𝛼)
× (𝐶1 cos (
√𝜇
) 𝜉𝛼
Γ (1 + 𝛼)
+ 𝐶2 sin (
−1
√𝜇
) 𝜉𝛼 ) ]
Γ (1 + 𝛼)
× (𝜎 {1 + [ (−𝐶1 sin (
(32)
2
𝛼
√−𝜇𝜉
)] },
Γ (1 + 𝛼)
−𝛼 𝛼
−2𝛼
2
1 𝑏2 (16𝑘 𝑐 𝑓2 + 5𝑘 𝑓2 𝜎𝑏2 − 6𝑒0 𝑏2 )
24
𝑓22
𝛼
𝑘2𝛼 𝑏2
√−𝜇𝜉
)]
𝜇[tanh (
2𝑓2
Γ (1 + 𝛼)
2
(33)
𝛼
𝑏2
√−𝜇𝜉
)]
+ 2 √−𝜇 [tanh (
4𝑓2
Γ (1 + 𝛼)
√𝜇
) 𝜉𝛼
Γ (1 + 𝛼)
+ 𝑓2 √𝜇 [ (−𝐶1 sin (
2
𝛼
√−𝜇𝜉
)]
Γ (1 + 𝛼)
× √ 𝜎 {1 − [tanh (
−
√𝜇
) 𝜉𝛼 )
Γ (1 + 𝛼)
+𝐶2 sin (
+ 𝑏2 √−𝜇 [tanh (
𝑉3 (𝜉) =
√𝜇
) 𝜉𝛼
Γ (1 + 𝛼)
𝛼
√−𝜇𝜉
)]
Γ (1 + 𝛼)
1/2
(30)
+𝐶2 cos (
,
where 𝜉 = 𝑘𝑥 + 𝑐𝑡 + 𝜉0 and 𝜇 = (1/4)𝑘−4𝛼 𝜎𝑏22 .
In particular, if we let 𝐶2 = 0 in (26)–(28), then we obtain
the following solitary wave solutions, which are shown in
Figures 1, 2, and 3:
√𝜇
) 𝜉𝛼
Γ (1 + 𝛼)
−1 2
× [ (−𝐶1 sin (
1/2
(31)
√𝜇
+𝐶2 sin (
) 𝜉𝛼 ) ] }) ,
Γ (1 + 𝛼)
𝑊2 (𝜉) = 𝑒0 +
−1 2
√𝜇
) 𝜉𝛼 ) ] })
Γ (1 + 𝛼)
−1
√𝜇
) 𝜉𝛼 ) ]
Γ (1 + 𝛼)
× (𝐶1 cos (
√𝜇
) 𝜉𝛼
Γ (1 + 𝛼)
√𝜇
) 𝜉𝛼
Γ (1 + 𝛼)
√𝜇
+ 𝐶2 cos (
) 𝜉𝛼 )
Γ (1 + 𝛼)
× √ 𝜎 {1 − [tanh (
𝑊3 (𝜉) = 𝑒0 −
2
𝛼
√−𝜇𝜉
)] },
Γ (1 + 𝛼)
𝛼
2𝑘2𝛼 𝑓2
√−𝜇𝜉
𝜇[tanh (
)]
𝑏2
Γ (1 + 𝛼)
+ 𝑓2 √−𝜇 [tanh (
2
𝛼
√−𝜇𝜉
)]
Γ (1 + 𝛼)
× √ 𝜎 {1 − [tanh (
(34)
2
𝛼
√−𝜇𝜉
)] }.
Γ (1 + 𝛼)
If we let 𝐶2 = 0 in (29)–(31), then we obtain the following
periodic wave solutions, which are shown in Figures 4, 5, and
6:
1
5
𝑈4 (𝜉) = 𝑘−𝛼 𝑐𝛼 + 𝑘−2𝛼 𝜎𝑏22
3
12
+ 2𝑘2𝛼 𝜇[tan
√𝜇
𝜉𝛼 ]
Γ (1 + 𝛼)
2
8
Journal of Applied Mathematics
1.15
1.14
2.8
1.13
1.12
1.11
2.7
1.10
1.09
2.6
1.08
1.07
1.06
2.5
0
2
4
6
8
t
Figure 1: The solution 𝑈3 in (32) with 𝛾 = 4/5, 𝑘 = 𝑐 = 𝜎 = 𝑏2 = 1,
𝜇 = −1, 𝜉0 = 0.
− 𝑏2 √𝜇 [tan
√𝜇
𝜉𝛼 ]
Γ (1 + 𝛼)
× √ 𝜎 {1 + [tan
−
(36)
2
4
6
8
7
10 9 8
4
6 5
x
1
3 2
Figure 3: The solution 𝑊3 in (34) with 𝛾 = 4/5, 𝑘 = 𝑐 = 𝜎 = 𝑏2 =
𝑓2 = 1, 𝜇 = −1, 𝜉0 = 0, 𝑒0 = 1.
2
√𝜇
𝜉𝛼 ] },
Γ (1 + 𝛼)
√𝜇
𝜉𝛼 ]
Γ (1 + 𝛼)
× √ 𝜎 {1 + [tan
8
2.9
t
Remark 1. We note that the solutions obtained here are of new
forms compared with the solutions obtained in [15, 16] since
a fully new method is used here.
2
2𝑘2𝛼 𝑓2
√𝜇
𝜇[tan
𝜉𝛼 ]
𝑏2
Γ (1 + 𝛼)
− 𝑓2 √𝜇 [tan
6
Figure 2: The solution 𝑉3 in (33) with 𝛾 = 4/5, 𝑘 = 𝑐 = 𝜎 = 𝑏2 =
𝑓2 = 1, 𝜇 = −1, 𝜉0 = 0, 𝑒0 = 1.
0
𝑏22
√𝜇
𝜉𝛼 ]
√𝜇 [tan
4𝑓2
Γ (1 + 𝛼)
𝑊4 (𝜉) = 𝑒0 +
t
2 1
4 3
5
6
8 7
10 9
x
2.8
2
𝑘2𝛼 𝑏2
√𝜇
𝜇[tan
𝜉𝛼 ]
2𝑓2
Γ (1 + 𝛼)
× √ 𝜎 {1 + [tan
4
3
2
√𝜇
𝜉𝛼 ] },
Γ (1 + 𝛼)
−𝛼 𝛼
−2𝛼
2
1 𝑏2 (16𝑘 𝑐 𝑓2 + 5𝑘 𝑓2 𝜎𝑏2 − 6𝑒0 𝑏2 )
24
𝑓22
+
2
3.1
(35)
𝑉4 (𝜉) =
0
2 1
4 3
5
6
8 7
x
10 9
(37)
2
√𝜇
𝜉𝛼 ] }.
Γ (1 + 𝛼)
Similar to the established solutions from Case 1, we can
construct corresponding exact solutions to (13) from Cases
2–7, which are omitted here.
5. Conclusions
Based on the concept of the modified Riemann-Liouville
derivative and a variable transformation 𝜉 = 𝑐𝑡 + 𝑘1 𝑥1 +
𝑘2 𝑥2 + ⋅ ⋅ ⋅ + 𝑘𝑛 𝑥𝑛 + 𝜉0 , we have proposed a new fractional
subequation method for solving fractional partial differential
equations (FPDEs). By using this method, the space-time
fractional generalized Hirota-Satsuma coupled KdV equations are solved successfully, and, as a result, some exact
solutions are established, which may help to understand
Journal of Applied Mathematics
9
better the nonlinear wave phenomena. It is supposed that this
method can be further applied to solve other FPDEs.
25000
20000
Acknowledgments
15000
This work is partially supported by the National Natural
Science Foundation of China (11171178) and the Program
for Scientific Research Innovation Team in Colleges and
Universities of Shandong Province. The authors would like to
thank the reviewers very much for their valuable suggestions
in the paper.
10000
5000
0
0
2
4
t
6
8
8 7
10 9
6
5 4
1
3 2
x
Figure 4: The solution 𝑈4 in (35) with 𝛾 = 4/5, 𝑘 = 𝑐 = 𝜎 = 𝑏2 = 1,
𝜇 = 1, 𝜉0 = 0.
20000
15000
10000
5000
0
0
2
4
t
6
8
8
10 9
7 6
5
2
4 3
1
x
Figure 5: The solution 𝑉4 in (36) with 𝛾 = 4/5, 𝑘 = 𝑐 = 𝜎 = 𝑏2 =
𝑓2 = 1, 𝜇 = 1, 𝜉0 = 0, 𝑒0 = 1.
80000
70000
60000
50000
40000
30000
20000
10000
0
0
2
4
t
6
8
10
9 8
7 6
5
4 3
2
1
x
Figure 6: The solution 𝑊4 in (37) with 𝛾 = 4/5, 𝑘 = 𝑐 = 𝜎 = 𝑏2 =
𝑓2 = 1, 𝜇 = −1, 𝜉0 = 0, 𝑒0 = 1.
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