A New Fractional Subequation Method and Its Applications for Space-Time Fractional Partial Differential Equations

Abstract

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: 𝐷_𝜉^2𝛼𝐺(𝜉)+𝜆𝐷^𝛼_𝜉𝐺(𝜉)+𝜇𝐺(𝜉)=0,0<𝛼≤1, (1) where𝐷^𝛼_𝜉𝐺(𝜉)denotes the modified Riemann-Liouville derivative of order𝛼for𝐺(𝜉)with respect to𝜉.

The definition and some important properties for Jumarie’s modified Riemann-Liouville derivative of order𝛼are listed as follows (see [13–16]):𝐷^𝛼_𝑡𝑓(𝑡)=1Γ()𝑑1−𝛼∫𝑑𝑡^𝑡0(𝑡−𝜉)^−𝛼𝑓(𝜉)(0)𝑓−𝑓𝑑𝜉,0<𝛼<1,_(𝑛)(𝑡)^{(𝛼−𝑛)}𝐷,𝑛≤𝛼<𝑛+1,𝑛≥1,_𝛼𝑡𝑡^𝑟=Γ(1+𝑟)Γ()𝑡1+𝑟−𝛼^𝑟−𝛼,𝐷^𝛼_𝑡𝑓(𝑡)𝑔(𝑡)=𝑔(𝑡)𝐷^𝛼_𝑡𝑓(𝑡)+𝑓(𝑡)𝐷^𝛼_𝑡𝑔(𝑡),𝐷^𝛼_𝑡𝑓𝑔(𝑡)=𝑓^′_𝑔𝑔(𝑡)𝐷^𝛼_𝑡𝑔(𝑡)=𝐷^𝛼_𝑔𝑓𝑔(𝑡)𝑔^′(𝑡)^𝛼. (2)(3)(4)(5)

We organize this paper as follows. In Section 2, we derive the expression for𝐷^𝛼_𝜉𝐺(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. The General Expression for 𝐷^𝛼_𝜉𝐺(𝜉)/𝐺(𝜉)

In order to obtain the general solutions for (1), we suppose𝐺(𝜉)=𝐻(𝜂) and a nonlinear fractional complex transformation𝜂=𝜉^𝛼/Γ(1+𝛼). Then, by (3) and the first equality in (5), (1) can be turned into the following second ordinary differential equation: 𝐻^′′𝜂+𝜆𝐻^′𝜂+𝜇𝐻𝜂=0. (6) By the general solutions of (6), we have𝐻^′𝜂𝐻𝜂=−𝜆2+𝜆²−4𝜇2𝐶₁sinh𝜆²−4𝜇/2𝜂+𝐶₂cosh𝜆²−4𝜇/2𝜂𝐶₁cosh𝜆²−4𝜇/2𝜂+𝐶₂sinh𝜆²−4𝜇/2𝜂,𝜆²−𝜆−4𝜇>0,2+4𝜇−𝜆²2−𝐶₁sin4𝜇−𝜆²/2𝜂+𝐶₂cos4𝜇−𝜆²/2𝜂𝐶₁cos4𝜇−𝜆²/2𝜂+𝐶₂sin4𝜇−𝜆²/2𝜂,𝜆²−𝜆−4𝜇<0,2+𝐶₂𝐶₁+𝐶₂𝜂,𝜆²−4𝜇=0, (7)where𝐶₁ and 𝐶₂are arbitrary constants.

Since𝐷^𝛼_𝜉𝐺(𝜉)=𝐷^𝛼_𝜉𝐻(𝜂)=𝐻^′(𝜂)𝐷^𝛼_𝜉𝜂=𝐻^′(𝜂), we obtain𝐷^𝛼_𝜉𝐺(𝜉)𝐺(𝜉)=^{−𝜆2+𝜆2−4𝜇2𝐶1sinh𝜆2(𝜉−4𝜇/2Γ1+𝛼)𝛼+𝐶2cosh𝜆2(𝜉−4𝜇/2Γ1+𝛼)𝛼𝐶1cosh𝜆2()𝜉−4𝜇/2Γ1+𝛼𝛼+𝐶2sinh𝜆2()𝜉−4𝜇/2Γ1+𝛼𝛼,𝜆2}_−𝜆^−4𝜇>0,_{2+4𝜇−𝜆}²₂^−𝐶₁^{sin4𝜇−𝜆2()𝜉/2Γ1+𝛼𝛼+𝐶}₂^{cos4𝜇−𝜆2()𝜉/2Γ1+𝛼𝛼}_{𝐶1cos4𝜇−𝜆2(𝜉/2Γ1+𝛼)𝛼+𝐶2sin4𝜇−𝜆2(𝜉/2Γ1+𝛼)𝛼,𝜆2−𝜆−4𝜇<0,2+𝐶2Γ()1+𝛼𝐶1Γ()1+𝛼+𝐶2𝜉𝛼,𝜆2−4𝜇=0.} (8)

3. Description of the Fractional Subequation Method

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 variables𝑡,𝑥₁,𝑥₂,…,𝑥_𝑛, is given by 𝑃𝑢₁,…,𝑢_𝑘,𝐷^𝛼_𝑡𝑢₁,…,𝐷^𝛼_𝑡𝑢_𝑘,𝐷^𝛼_𝑥1𝑢₁𝐷,…,_𝛼𝑥1𝑢_𝑘,…,𝐷^𝛼_𝑥𝑛𝑢₁,…,𝐷^𝛼_𝑥𝑛𝑢_𝑘,𝐷_𝑡^2𝛼𝑢₁𝐷,…,_𝑡2𝛼𝑢_𝑘,𝐷_𝑥^2𝛼₁𝑢₁,…=0, (9) where𝑢_𝑖=𝑢_𝑖(𝑡,𝑥₁,𝑥₂,…,𝑥_𝑛), 𝑖=1,…,𝑘, are unknown functions and 𝑃 is a polynomial in 𝑢_𝑖 and their various partial derivatives including fractional derivatives.

Step 1. Suppose that 𝑢_𝑖𝑡,𝑥₁,𝑥₂,…,𝑥_𝑛=𝑈_𝑖(𝜉),𝜉=𝑐𝑡+𝑘₁𝑥₁+𝑘₂𝑥₂+⋯+𝑘_𝑛𝑥_𝑛+𝜉₀. (10) Then, by the second equality in (5), (9) can be turned into the following fractional ordinary differential equation with respect to the variable𝜉: 𝑃𝑈₁,…,𝑈_𝑘,𝑐^𝛼𝐷^𝛼_𝜉𝑈₁,…,𝑐^𝛼𝐷^𝛼_𝜉𝑈_𝑘,𝑘^𝛼₁𝐷^𝛼_𝜉𝑈₁,…,𝑘^𝛼₁𝐷^𝛼_𝜉𝑈_𝑘,…,𝑘^𝛼_𝑛𝐷^𝛼_𝜉𝑈₁𝑘,…,_𝛼𝑛𝐷^𝛼_𝜉𝑈_𝑘,𝑐^2𝛼𝐷_𝜉^2𝛼𝑈₁𝑐,…,_2𝛼𝐷_𝜉^2𝛼𝑈_𝑘,𝑘₁^2𝛼𝐷_𝜉^2𝛼𝑈₁,…=0. (11)

Step 2. Suppose that the solution of (11) can be expressed by a polynomial in(𝐷^𝛼_𝜉𝐺/𝐺)as follows: 𝑈_𝑗(𝜉)=𝑎_𝑗,0+^𝑚𝑗_𝑖=1𝑎_𝑗,𝑖𝐷^𝛼_𝜉𝐺𝐺^𝑖+𝑏_𝑗,𝑖𝐷^𝛼_𝜉𝐺𝐺^𝑖−1𝜎11+𝜇𝐷^𝛼_𝜉𝐺𝐺²,𝑗=1,2,…,𝑘, (12) where𝐺=𝐺(𝜉)satisfies (1),𝜎is a constant, and 𝑎_𝑗,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(𝐷^𝛼_𝜉𝐺/𝐺)𝜎(1+(1/𝜇)(𝐷^𝛼_𝜉𝐺/𝐺)²)together, the left-hand side of (11) is converted into another polynomial in(𝐷^𝛼_𝜉𝐺/𝐺). Equating each coefficient of this polynomial to zero yields a set of algebraic equations for𝑎_𝑗0,𝑎_𝑗,𝑖,𝑏_𝑗,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

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]: 𝐷^𝛼_𝑡1𝑢−2𝐷_𝑥^3𝛼𝑢+3𝑢𝐷^𝛼_𝑥𝑢−3𝐷^𝛼_𝑥()𝐷𝑣𝑤=0,_𝛼𝑡𝑣+𝐷_𝑥^3𝛼𝑣−3𝑢𝐷^𝛼_𝑥𝐷𝑣=0,_𝛼𝑡𝑤+𝐷_𝑥^3𝛼𝑤−3𝑢𝐷^𝛼_𝑥𝑤=0,0<𝛼≤1. (13) 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), where𝜉=𝑘𝑥+𝑐𝑡+𝜉₀, 𝑘,𝑐,𝜉₀ are all constants with𝑘,𝑐≠0. Then, by usinge the second equality in (4), we obtain 𝐷^𝛼_𝑥𝑢=𝐷^𝛼_𝑥𝑈(𝜉)=𝐷^𝛼_𝜉𝑈𝜉^′_𝑥^𝛼=𝑘^𝛼𝐷^𝛼_𝜉𝐷𝑈,_𝛼𝑡𝑢=𝐷^𝛼_𝑡𝑈(𝜉)=𝐷^𝛼_𝜉𝑈𝜉^′_𝑡^𝛼=𝑐^𝛼𝐷^𝛼_𝜉𝑈, (14) and similarly we have 𝐷^𝛼_𝑥𝑣=𝑘^𝛼𝐷^𝛼_𝜉𝑉,𝐷^𝛼_𝑡𝑣=𝑐^𝛼𝐷^𝛼_𝜉𝐷𝑉,_𝛼𝑥𝑤=𝑘^𝛼𝐷^𝛼_𝜉𝑊,𝐷^𝛼_𝑡𝑤=𝑐^𝛼𝐷^𝛼_𝜉𝑊; (15) then (11) can be turned into the following fractional ordinary differential equations with respect to the variable𝜉: 𝑐^𝛼𝐷^𝛼_𝜉1𝑈−2𝑘^3𝛼𝐷_𝜉^3𝛼𝑈+3𝑘^𝛼𝑈𝐷^𝛼_𝜉𝑈−3𝑘^𝛼𝐷^𝛼_𝜉()𝑐𝑉𝑊=0,_𝛼𝐷^𝛼_𝜉𝑉+𝑘^3𝛼𝐷_𝜉^3𝛼𝑉−3𝑘^𝛼𝑈𝐷^𝛼_𝜉𝑐𝑉=0,_𝛼𝐷^𝛼_𝜉𝑊+𝑘^3𝛼𝐷_𝜉^3𝛼𝑊−3𝑘^𝛼𝑈𝐷^𝛼_𝜉𝑊=0. (16)

Suppose that the solutions of (16) can be expressed by 𝑈(𝜉)=𝑎₀+^𝑚1_𝑖=1𝑎_𝑖𝐷^𝛼_𝜉𝐺𝐺^𝑖+𝑏_𝑖𝐷^𝛼_𝜉𝐺𝐺^𝑖−1𝜎11+𝜇𝐷^𝛼_𝜉𝐺𝐺²,𝑉(𝜉)=𝑐₀+^𝑚2_𝑖=1𝑐_𝑖𝐷^𝛼_𝜉𝐺𝐺^𝑖+𝑑_𝑖𝐷^𝛼_𝜉𝐺𝐺^𝑖−1𝜎11+𝜇𝐷^𝛼_𝜉𝐺𝐺²,𝑊(𝜉)=𝑒₀+^𝑚3_𝑖=1𝑒_𝑖𝐷^𝛼_𝜉𝐺𝐺^𝑖+𝑓_𝑖𝐷^𝛼_𝜉𝐺𝐺^𝑖−1𝜎11+𝜇𝐷^𝛼_𝜉𝐺𝐺². (17) Balancing the order of 𝐷_𝜉^3𝛼𝑈 and 𝐷^𝛼_𝜉(𝑉𝑊), 𝐷_𝜉^3𝛼𝑉 and 𝑈𝐷^𝛼_𝜉𝑉, and 𝐷_𝜉^3𝛼𝑊 and 𝑈𝐷^𝛼_𝜉𝑊 in (16), we have  𝑚₁=𝑚₂=𝑚₃=2. So 𝑈(𝜉)=𝑎₀+𝑎₁𝐷^𝛼_𝜉𝐺𝐺+𝑎₂𝐷^𝛼_𝜉𝐺𝐺²+𝑏₁𝜎11+𝜇𝐷^𝛼_𝜉𝐺𝐺²+𝑏₂𝐷^𝛼_𝜉𝐺𝐺𝜎11+𝜇𝐷^𝛼_𝜉𝐺𝐺²,𝑉(𝜉)=𝑐₀+𝑐₁𝐷^𝛼_𝜉𝐺𝐺+𝑐₂𝐷^𝛼_𝜉𝐺𝐺²+𝑑₁𝜎11+𝜇𝐷^𝛼_𝜉𝐺𝐺²+𝑑₂𝐷^𝛼_𝜉𝐺𝐺𝜎11+𝜇𝐷^𝛼_𝜉𝐺𝐺²,𝑊(𝜉)=𝑒₀+𝑒₁𝐷^𝛼_𝜉𝐺𝐺+𝑒₂𝐷^𝛼_𝜉𝐺𝐺²+𝑓₁𝜎11+𝜇𝐷^𝛼_𝜉𝐺𝐺²+𝑓₂𝐷^𝛼_𝜉𝐺𝐺𝜎11+𝜇𝐷^𝛼_𝜉𝐺𝐺². (18)

Substituting (18) into (16), using (1), and collecting all the terms with the same power of(𝐷^𝛼_𝜉𝐺/𝐺)  𝜎(1+(1/𝜇)(𝐷^𝛼_𝜉𝐺/𝐺)²)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 1𝜇=4𝑘^−4𝛼𝜎𝑏²₂,𝜆=0,𝑎₀=13𝑘^−𝛼𝑐^𝛼+5𝑘12^−2𝛼𝜎𝑏²₂,𝑎₁=0,𝑎₂=2𝑘^2𝛼,𝑏₁=0,𝑏₂=𝑏₂,𝑐₀=1𝑏24²16𝑘^−𝛼𝑐^𝛼𝑓₂+5𝑘^−2𝛼𝑓₂𝜎𝑏²₂−6𝑒₀𝑏₂𝑓²₂,𝑐₁𝑐=0,₂=𝑘^2𝛼𝑏₂2𝑓₂,𝑑₁=0,𝑑₂=𝑏²₂4𝑓₂,𝑒₀=𝑒₀,𝑒₁=0,𝑒₂=2𝑘^2𝛼𝑓₂𝑏₂,𝑓₁=0,𝑓₂=𝑓₂. (19)

Case 2. One has 1𝜇=𝜇,𝜆=4𝑘^−2𝛼𝑎₁,𝑎₀=1𝑘48^−2𝛼𝑎²₁+83𝑘^2𝛼1𝜇+3𝑘^−𝛼𝑐^𝛼,𝑎₁=𝑎₁,𝑎₂=4𝑘^2𝛼,𝑏₁=0,𝑏₂𝑐=0,₀=1𝑐96²𝑘^−4𝛼𝑎²₁+128𝜇+64𝑐^𝛼𝑘^−3𝛼−24𝑘^−4𝛼𝑐₂𝑒₀,𝑐₁=14𝑐₂𝑘^−2𝛼𝑎₁,𝑐₂=𝑐₂,𝑑₁𝑑=0,₂=0,𝑒₀=𝑒₀,𝑒₁=𝑘^2𝛼𝑎₁𝑐₂,𝑒₂=4𝑘^4𝛼𝑐₂,𝑓₁=0,𝑓₂=0. (20)

Case 3. One has 1𝜇=𝑘16^−4𝛼𝑎²₁+𝑐^𝛼𝑘^−3𝛼−34𝑘^−4𝛼𝑐₁𝑒₁,1𝜆=2𝑘^−2𝛼𝑎₁,𝑎₀=18𝑘^−2𝛼𝑎²₁+𝑘^−𝛼𝑐^𝛼−12𝑘^−2𝛼𝑐₁𝑒₁,𝑎₁=𝑎₁,𝑎₂=2𝑘^2𝛼,𝑏₁𝑏=0,₂=0,𝑐₀=𝑐₀,𝑐₁=𝑐₁,𝑐₂=0,𝑑₁=0,𝑑₂𝑒=0,₀=𝑒₁𝑘^−2𝛼𝑎₁𝑐₁−2𝑐₀2𝑐₁,𝑒₁=𝑒₁,𝑒₂=0,𝑓₁=0,𝑓₂=0. (21)

Case 4. One has 𝜇=𝑘^−4𝛼𝜎𝑏²₂𝑎,𝜆=0,₀=23𝑘^−2𝛼𝜎𝑏²₂+13𝑘^−𝛼𝑐^𝛼,𝑎₁𝑎=0,₂=𝑘^2𝛼,𝑏₁=0,𝑏₂=𝑏₂,𝑐₀=𝑐₀,𝑐₁=𝑘^2𝛼𝑑₁𝑏₂,𝑐₂𝑑=0,₁=𝑑₁,𝑑₂𝑒=0,₀=𝑐₀𝑏²₂𝑘^−4𝛼𝜎𝑏²₂−4𝑐^𝛼𝑘^−3𝛼12𝑑²₁,𝑒₁=𝑏₂−𝑘^−2𝛼𝜎𝑏²₂+4𝑘^−𝛼𝑐^𝛼12𝑑₁,𝑒₂𝑓=0,₁𝑏=−²²𝑘^−4𝛼𝜎𝑏²₂−4𝑐^𝛼𝑘^−3𝛼12𝑑₁,𝑓₂=0. (22)

Case 5. One has 𝜇=4𝑐^𝛼𝑘^−3𝛼,𝜆=0,𝑎₀=3𝑘^−𝛼𝑐^𝛼,𝑎₁=0,𝑎₂=𝑘^2𝛼,𝑏₁𝑏=0,₂=±𝑘^−𝛼𝑐^𝛼𝑘^3𝛼𝜎,𝑐₀=0,𝑐₁𝑐=0,₂=0,𝑑₁=0,𝑑₂=0,𝑒₀=𝑒₀,𝑒₁𝑘=±^3𝛼𝑓₁2𝜎𝑐^𝛼𝑘^3𝛼,𝑒₂𝑓=0,₁=𝑓₁,𝑓₂=0. (23)

Case 6. One has 𝜇=𝜇,𝜆=0,𝑎₀=23𝑘^2𝛼1𝜇+3𝑘^−𝛼𝑐^𝛼,𝑎₁=0,𝑎₂=2𝑘^2𝛼,𝑏₁𝑏=0,₂=0,𝑐₀=𝑐₀,𝑐₁=𝑐₁,𝑐₂=0,𝑑₁=0,𝑑₂𝑒=0,₀=−4𝑐₀−𝑘^4𝛼𝜇+𝑘^𝛼𝑐^𝛼3𝑐²₁,𝑒₁=4−𝑘^4𝛼𝜇+𝑘^𝛼𝑐^𝛼3𝑐₁,𝑒₂=0,𝑓₁=0,𝑓₂=0. (24)

Case 7. One has 𝜇=𝜇,𝜆=0,𝑎₀=53𝑘^2𝛼1𝜇+3𝑘^−𝛼𝑐^𝛼,𝑎₁=0,𝑎₂=2𝑘^2𝛼,𝑏₁𝑏=0,₂=0,𝑐₀=𝑐₀,𝑐₁𝑐=0,₂=0,𝑑₁=𝑑₁,𝑑₂𝑒=0,₀=−2𝑐₀𝜇𝑘^4𝛼𝜇+2𝑘^𝛼𝑐^𝛼3𝜎𝑑²₁,𝑒₁=0,𝑒₂𝑓=0,₁=2𝜇𝑘^4𝛼𝜇+2𝑘^𝛼𝑐^𝛼3𝜎𝑑₁,𝑓₂=0. (25)
Substituting the previous results into (18) and combining with (8), we can obtain a series of exact solutions for (13).

From Case 1, we obtain the following exact solutions.

When𝜇<0, 𝑈₁(𝜉)=13𝑘^−𝛼𝑐^𝛼+5𝑘12^−2𝛼𝜎𝑏²₂−2𝑘^2𝛼𝜇×𝐶₁√sinh−𝜇𝜉^𝛼Γ(1+𝛼)+𝐶₂√cosh−𝜇𝜉^𝛼Γ()×𝐶1+𝛼₁√cosh−𝜇𝜉^𝛼Γ(1+𝛼)+𝐶₂√sinh−𝜇𝜉^𝛼Γ(1+𝛼)⁻¹²+𝑏₂√−𝜇𝐶₁√sinh−𝜇𝜉^𝛼Γ(1+𝛼)+𝐶₂√cosh−𝜇𝜉^𝛼Γ()×𝐶1+𝛼₁√cosh−𝜇𝜉^𝛼Γ(1+𝛼)+𝐶₂√sinh−𝜇𝜉^𝛼Γ(1+𝛼)⁻¹×𝜎1−𝐶₁√sinh−𝜇𝜉^𝛼Γ(1+𝛼)+𝐶₂√cosh−𝜇𝜉^𝛼Γ()×𝐶1+𝛼₁√cosh−𝜇𝜉^𝛼Γ(1+𝛼)+𝐶₂√sinh−𝜇𝜉^𝛼Γ(1+𝛼)^−121/2,𝑉₁(𝜉)=1𝑏24²16𝑘^−𝛼𝑐^𝛼𝑓₂+5𝑘^−2𝛼𝑓₂𝜎𝑏²₂−6𝑒₀𝑏₂𝑓²₂−𝑘^2𝛼𝑏₂2𝑓₂𝜇𝐶₁√sinh−𝜇𝜉^𝛼Γ(1+𝛼)+𝐶₂√cosh−𝜇𝜉^𝛼Γ()×𝐶1+𝛼₁√cosh−𝜇𝜉^𝛼Γ(1+𝛼)+𝐶₂√sinh−𝜇𝜉^𝛼Γ(1+𝛼)⁻¹²+𝑏²₂4𝑓₂√−𝜇𝐶₁√sinh−𝜇𝜉^𝛼Γ(1+𝛼)+𝐶₂√cosh−𝜇𝜉^𝛼Γ()×𝐶1+𝛼₁√cosh−𝜇𝜉^𝛼Γ(1+𝛼)+𝐶₂√sinh−𝜇𝜉^𝛼Γ(1+𝛼)⁻¹×𝜎1−𝐶₁√sinh−𝜇𝜉^𝛼Γ(1+𝛼)+𝐶₂√cosh−𝜇𝜉^𝛼Γ()×𝐶1+𝛼₁√cosh−𝜇𝜉^𝛼Γ(1+𝛼)+𝐶₂√sinh−𝜇𝜉^𝛼Γ(1+𝛼)^−121/2,𝑊₁(𝜉)=𝑒₀−2𝑘^2𝛼𝑓₂𝑏₂𝜇𝐶₁√sinh−𝜇𝜉^𝛼Γ(1+𝛼)+𝐶₂√cosh−𝜇𝜉^𝛼Γ()×𝐶1+𝛼₁√cosh−𝜇𝜉^𝛼Γ(1+𝛼)+𝐶₂√sinh−𝜇𝜉^𝛼Γ(1+𝛼)⁻¹²+𝑓₂√−𝜇𝐶₁√sinh−𝜇𝜉^𝛼Γ(1+𝛼)+𝐶₂√cosh−𝜇𝜉^𝛼Γ()×𝐶1+𝛼₁√cosh−𝜇𝜉^𝛼Γ(1+𝛼)+𝐶₂√sinh−𝜇𝜉^𝛼Γ(1+𝛼)⁻¹×𝜎1−𝐶₁√sinh−𝜇𝜉^𝛼Γ(1+𝛼)+𝐶₂√cosh−𝜇𝜉^𝛼Γ()×𝐶1+𝛼₁√cosh−𝜇𝜉^𝛼Γ(1+𝛼)+𝐶₂√sinh−𝜇𝜉^𝛼Γ(1+𝛼)^−121/2, (26)(27)(28) where𝜉=𝑘𝑥+𝑐𝑡+𝜉₀,𝜇=(1/4)𝑘^−4𝛼𝜎𝑏²₂.

When𝜇>0, 𝑈₂(𝜉)=13𝑘^−𝛼𝑐^𝛼+5𝑘12^−2𝛼𝜎𝑏²₂+2𝑘^2𝛼𝜇×−𝐶₁√sin𝜇Γ()𝜉1+𝛼^𝛼+𝐶₂√cos𝜇Γ()𝜉1+𝛼^𝛼×𝐶₁√cos𝜇Γ()𝜉1+𝛼^𝛼+𝐶₂√sin𝜇Γ()𝜉1+𝛼^𝛼−12+𝑏₂√𝜇−𝐶₁√sin𝜇Γ()𝜉1+𝛼^𝛼+𝐶₂√cos𝜇Γ()𝜉1+𝛼^𝛼×𝐶₁√cos𝜇Γ()𝜉1+𝛼^𝛼+𝐶₂√sin𝜇Γ()𝜉1+𝛼^𝛼−1×𝜎1+−𝐶₁√sin𝜇Γ()𝜉1+𝛼^𝛼+𝐶₂√cos𝜇Γ()𝜉1+𝛼^𝛼×𝐶₁√cos𝜇Γ()𝜉1+𝛼^𝛼+𝐶₂√sin𝜇Γ()𝜉1+𝛼^𝛼−121/2,𝑉₂(𝜉)=1𝑏24²16𝑘^−𝛼𝑐^𝛼𝑓₂+5𝑘^−2𝛼𝑓₂𝜎𝑏²₂−6𝑒₀𝑏₂𝑓²₂+𝑘^2𝛼𝑏₂2𝑓₂𝜇−𝐶₁√sin𝜇Γ()𝜉1+𝛼^𝛼+𝐶₂√cos𝜇Γ()𝜉1+𝛼^𝛼×𝐶₁√cos𝜇Γ()𝜉1+𝛼^𝛼+𝐶₂√sin𝜇Γ()𝜉1+𝛼^𝛼−12+𝑏²₂4𝑓₂√𝜇−𝐶₁√sin𝜇Γ()𝜉1+𝛼^𝛼+𝐶₂√cos𝜇Γ()𝜉1+𝛼^𝛼×𝐶₁√cos𝜇Γ()𝜉1+𝛼^𝛼+𝐶₂√sin𝜇Γ()𝜉1+𝛼^𝛼−1×𝜎1+−𝐶₁√sin𝜇Γ()𝜉1+𝛼^𝛼+𝐶₂√cos𝜇Γ()𝜉1+𝛼^𝛼×𝐶₁√cos𝜇Γ()𝜉1+𝛼^𝛼+𝐶₂√sin𝜇Γ()𝜉1+𝛼^𝛼−121/2,𝑊₂(𝜉)=𝑒₀+2𝑘^2𝛼𝑓₂𝑏₂𝜇×−𝐶₁√sin𝜇Γ()𝜉1+𝛼^𝛼+𝐶₂√cos𝜇Γ()𝜉1+𝛼^𝛼×𝐶₁√cos𝜇Γ()𝜉1+𝛼^𝛼+𝐶₂√sin𝜇Γ()𝜉1+𝛼^𝛼−12+𝑓₂√𝜇−𝐶₁√sin𝜇Γ()𝜉1+𝛼^𝛼+𝐶₂√cos𝜇Γ()𝜉1+𝛼^𝛼×𝐶₁√cos𝜇Γ()𝜉1+𝛼^𝛼+𝐶₂√sin𝜇Γ()𝜉1+𝛼^𝛼−1×𝜎1+−𝐶₁√sin𝜇Γ()𝜉1+𝛼^𝛼+𝐶₂√cos𝜇Γ()𝜉1+𝛼^𝛼×𝐶₁√cos𝜇Γ()𝜉1+𝛼^𝛼+𝐶₂√sin𝜇Γ()𝜉1+𝛼^𝛼−121/2, (29)(30)(31) where𝜉=𝑘𝑥+𝑐𝑡+𝜉₀ and  𝜇=(1/4)𝑘^−4𝛼𝜎𝑏²₂.

In particular, if we let𝐶₂=0in (26)–(28), then we obtain the following solitary wave solutions, which are shown in Figures 1, 2, and 3: 𝑈₃(𝜉)=13𝑘^−𝛼𝑐^𝛼+5𝑘12^−2𝛼𝜎𝑏²₂−2𝑘^2𝛼𝜇√tanh−𝜇𝜉^𝛼Γ(1+𝛼)²+𝑏₂√√−𝜇tanh−𝜇𝜉^𝛼Γ()×1+𝛼𝜎√1−tanh−𝜇𝜉^𝛼Γ(1+𝛼)²,𝑉₃(𝜉)=1𝑏24²16𝑘^−𝛼𝑐^𝛼𝑓₂+5𝑘^−2𝛼𝑓₂𝜎𝑏²₂−6𝑒₀𝑏₂𝑓²₂−𝑘^2𝛼𝑏₂2𝑓₂𝜇√tanh−𝜇𝜉^𝛼Γ(1+𝛼)²+𝑏²₂4𝑓₂√√−𝜇tanh−𝜇𝜉^𝛼Γ()×1+𝛼𝜎√1−tanh−𝜇𝜉^𝛼Γ(1+𝛼)²,𝑊₃(𝜉)=𝑒₀−2𝑘^2𝛼𝑓₂𝑏₂𝜇√tanh−𝜇𝜉^𝛼Γ(1+𝛼)²+𝑓₂√√−𝜇tanh−𝜇𝜉^𝛼Γ()×1+𝛼𝜎√1−tanh−𝜇𝜉^𝛼Γ(1+𝛼)². (32)(33)(34)

Figure 1 

The solution 𝑈3 in (32) with 𝛾=4/5, 𝑘=𝑐=𝜎=𝑏2=1, 𝜇=−1, 𝜉0=0.

Figure 2 

The solution 𝑉3 in (33) with 𝛾=4/5, 𝑘=𝑐=𝜎=𝑏2=𝑓2=1, 𝜇=−1, 𝜉0=0, 𝑒0=1.

Figure 3 

The solution 𝑊3 in (34) with 𝛾=4/5, 𝑘=𝑐=𝜎=𝑏2=𝑓2=1, 𝜇=−1, 𝜉0=0, 𝑒0=1.

If we let𝐶₂=0in (29)–(31), then we obtain the following periodic wave solutions, which are shown in Figures 4, 5, and 6: 𝑈₄(𝜉)=13𝑘^−𝛼𝑐^𝛼+5𝑘12^−2𝛼𝜎𝑏²₂+2𝑘^2𝛼𝜇√tan𝜇Γ()𝜉1+𝛼^𝛼2−𝑏₂√𝜇√tan𝜇Γ()𝜉1+𝛼^𝛼×𝜎√1+tan𝜇Γ()𝜉1+𝛼^𝛼2,𝑉₄(𝜉)=1𝑏24²16𝑘^−𝛼𝑐^𝛼𝑓₂+5𝑘^−2𝛼𝑓₂𝜎𝑏²₂−6𝑒₀𝑏₂𝑓²₂+𝑘^2𝛼𝑏₂2𝑓₂𝜇√tan𝜇Γ()𝜉1+𝛼^𝛼2−𝑏²₂4𝑓₂√𝜇√tan𝜇Γ()𝜉1+𝛼^𝛼×𝜎√1+tan𝜇Γ()𝜉1+𝛼^𝛼2,𝑊₄(𝜉)=𝑒₀+2𝑘^2𝛼𝑓₂𝑏₂𝜇√tan𝜇Γ()𝜉1+𝛼^𝛼2−𝑓₂√𝜇√tan𝜇Γ()𝜉1+𝛼^𝛼×𝜎√1+tan𝜇Γ()𝜉1+𝛼^𝛼2. (35)(36)(37)

Figure 4 

The solution 𝑈4 in (35) with 𝛾=4/5, 𝑘=𝑐=𝜎=𝑏2=1, 𝜇=1, 𝜉0=0.

Figure 5 

The solution 𝑉4 in (36) with 𝛾=4/5, 𝑘=𝑐=𝜎=𝑏2=𝑓2=1, 𝜇=1, 𝜉0=0,𝑒0=1.

Figure 6 

The solution 𝑊4 in (37) with 𝛾=4/5, 𝑘=𝑐=𝜎=𝑏2=𝑓2=1, 𝜇=−1, 𝜉0=0,𝑒0=1.

Similar to the established solutions from Case 1, we can construct corresponding exact solutions to (13) from Cases 2–7, which are omitted here.

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.

5. Conclusions

Based on the concept of the modified Riemann-Liouville derivative and a variable transformation𝜉=𝑐𝑡+𝑘₁𝑥₁+𝑘₂𝑥₂+⋯+𝑘_𝑛𝑥_𝑛+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 better the nonlinear wave phenomena. It is supposed that this method can be further applied to solve other FPDEs.

Acknowledgments

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.