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

Authors: Fanwei Meng; Qinghua Feng
Journal: Journal of Applied Mathematics
 (2013)

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## 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.


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## Body

## 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:(1)Dξ2αG(ξ)+λDξαG(ξ)+μG(ξ)=0,0<α≤1,
whereDξαG(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]):(2)Dtαf(t)={1Γ(1-α)ddt∫0t‍(t-ξ)-α(f(ξ)-f(0))dξ,0<α<1,(f(n)(t))(α-n),n≤α<n+1,n≥1,(3)Dtαtr=Γ(1+r)Γ(1+r-α)tr-α,(4)Dtα(f(t)g(t))=g(t)Dtαf(t)+f(t)Dtαg(t),(5)Dtαf[g(t)]=fg′[g(t)]Dtαg(t)=Dgαf[g(t)](g′(t))α.We organize this paper as follows. In Section2, we derive the expression forDξαG(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 forDξαG(ξ)/G(ξ)

In order to obtain the general solutions for (1), we supposeG(ξ)=H(η) and a nonlinear fractional complex transformationη=ξα/Γ(1+α). Then, by (3) and the first equality in (5), (1) can be turned into the following second ordinary differential equation:
(6)H′′(η)+λH′(η)+μH(η)=0.
By the general solutions of (6), we have(7)H′(η)H(η)={-λ2+λ2-4μ2(C1sinh(λ2-4μ/2)η+C2cosh(λ2-4μ/2)ηC1cosh(λ2-4μ/2)η+C2sinh(λ2-4μ/2)η),λ2-4μ>0,-λ2+4μ-λ22(-C1sin(4μ-λ2/2)η+C2cos(4μ-λ2/2)ηC1cos(4μ-λ2/2)η+C2sin(4μ-λ2/2)η),λ2-4μ<0,-λ2+C2C1+C2η,λ2-4μ=0,whereC1 and C2are arbitrary constants.SinceDξαG(ξ)=DξαH(η)=H′(η)Dξαη=H′(η), we obtain(8)DξαG(ξ)G(ξ)={-λ2+λ2-4μ2(C1sinh(λ2-4μ/2Γ(1+α))ξα+C2cosh(λ2-4μ/2Γ(1+α))ξαC1cosh(λ2-4μ/2Γ(1+α))ξα+C2sinh(λ2-4μ/2Γ(1+α))ξα),λ2-4μ>0,-λ2+4μ-λ22(-C1sin(4μ-λ2/2Γ(1+α))ξα+C2cos(4μ-λ2/2Γ(1+α))ξαC1cos(4μ-λ2/2Γ(1+α))ξα+C2sin(4μ-λ2/2Γ(1+α))ξα),λ2-4μ<0,-λ2+C2Γ(1+α)C1Γ(1+α)+C2ξα,λ2-4μ=0.


## 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,…,xn, is given by
(9)P(u1,…,uk,Dtαu1,…,Dtαuk,Dx1αu1,…,Dx1αuk,…,Dxnαu1,…,Dxnαuk,Dt2αu1,…,Dt2αuk,Dx12αu1,…)=0,
whereui=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.Step 1.
Suppose that(10)ui(t,x1,x2,…,xn)=Ui(ξ),ξ=ct+k1x1+k2x2+⋯+knxn+ξ0.
Then, by the second equality in (5), (9) can be turned into the following fractional ordinary differential equation with respect to the variableξ:
(11)P~(U1,…,Uk,cαDξαU1,…,cαDξαUk,k1αDξαU1,…,k1αDξαUk,…,knαDξαU1,…,knαDξαUk,c2αDξ2αU1,…,c2αDξ2αUk,k12αDξ2αU1,…)=0.Step 2.
Suppose that the solution of (11) can be expressed by a polynomial in(DξαG/G)as follows:
(12)Uj(ξ)=aj,0+∑‍i=1mj[aj,i(DξαGG)i+bj,i(DξαGG)i-1σ(1+1μ(DξαGG)2)],j=1,2,…,k,
whereG=G(ξ)satisfies (1),σis 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(DξαG/G)σ(1+(1/μ)(DξαG/G)2)together, the left-hand side of (11) is converted into another polynomial in(DξαG/G). Equating each coefficient of this polynomial to zero yields a set of algebraic equations foraj0,aj,i,bj,i, i=1,…,m, j=1,2,…,k.Step 4.
Solving the equations in Step3 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 Section3 to solve the space-time fractional generalized Hirota-Satsuma coupled KdV equations [15, 16]:
(13)Dtαu-12Dx3αu+3uDxαu-3Dxα(vw)=0,Dtαv+Dx3αv-3uDxαv=0,Dtαw+Dx3αw-3uDxαw=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), whereξ=kx+ct+ξ0, k,c,ξ0 are all constants withk,c≠0. Then, by usinge the second equality in (4), we obtain
(14)Dxαu=DxαU(ξ)=(DξαU)(ξx′)α=kαDξαU,Dtαu=DtαU(ξ)=(DξαU)(ξt′)α=cαDξαU,
and similarly we have
(15)Dxαv=kαDξαV,Dtαv=cαDξαV,Dxαw=kαDξαW,Dtαw=cαDξαW;
then (11) can be turned into the following fractional ordinary differential equations with respect to the variableξ:
(16)cαDξαU-12k3αDξ3αU+3kαUDξαU-3kαDξα(VW)=0,cαDξαV+k3αDξ3αV-3kαUDξαV=0,cαDξαW+k3αDξ3αW-3kαUDξαW=0.Suppose that the solutions of (16) can be expressed by
(17)U(ξ)=a0+∑‍i=1m1[ai(DξαGG)i+bi(DξαGG)i-1σ(1+1μ(DξαGG)2)],V(ξ)=c0+∑‍i=1m2[ci(DξαGG)i+di(DξαGG)i-1σ(1+1μ(DξαGG)2)],W(ξ)=e0+∑‍i=1m3[ei(DξαGG)i+fi(DξαGG)i-1σ(1+1μ(DξαGG)2)].
Balancing the order of Dξ3αU and Dξα(VW), Dξ3αV and UDξαV, and Dξ3αW and UDξαW in (16), we have  m1=m2=m3=2. So
(18)U(ξ)=a0+a1(DξαGG)+a2(DξαGG)2+b1σ(1+1μ(DξαGG)2)+b2(DξαGG)σ(1+1μ(DξαGG)2),V(ξ)=c0+c1(DξαGG)+c2(DξαGG)2+d1σ(1+1μ(DξαGG)2)+d2(DξαGG)σ(1+1μ(DξαGG)2),W(ξ)=e0+e1(DξαGG)+e2(DξαGG)2+f1σ(1+1μ(DξαGG)2)+f2(DξαGG)σ(1+1μ(DξαGG)2).Substituting (18) into (16), using (1), and collecting all the terms with the same power of(DξαG/G)  σ(1+(1/μ)(DξαG/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(19)μ=14k-4ασb22,λ=0,a0=13k-αcα+512k-2ασb22,a1=0,a2=2k2α,b1=0,b2=b2,c0=124b2(16k-αcαf2+5k-2αf2σb22-6e0b2)f22,c1=0,c2=k2αb22f2,d1=0,d2=b224f2,e0=e0,e1=0,e2=2k2αf2b2,f1=0,f2=f2.Case 2.
One has(20)μ=μ,λ=14k-2αa1,a0=148k-2αa12+83k2αμ+13k-αcα,a1=a1,a2=4k2α,b1=0,b2=0,c0=196c2(k-4αa12+128μ+64cαk-3α-24k-4αc2e0),c1=14c2k-2αa1,c2=c2,d1=0,d2=0,e0=e0,e1=k2αa1c2,e2=4k4αc2,f1=0,f2=0.Case 3.
One has(21)μ=116k-4αa12+cαk-3α-34k-4αc1e1,λ=12k-2αa1,a0=18k-2αa12+k-αcα-12k-2αc1e1,a1=a1,a2=2k2α,b1=0,b2=0,c0=c0,c1=c1,c2=0,d1=0,d2=0,e0=e1(k-2αa1c1-2c0)2c1,e1=e1,e2=0,f1=0,f2=0.Case 4.
One has(22)μ=k-4ασb22,λ=0,a0=23k-2ασb22+13k-αcα,a1=0,a2=k2α,b1=0,b2=b2,c0=c0,c1=k2αd1b2,c2=0,d1=d1,d2=0,e0=c0b22(k-4ασb22-4cαk-3α)12d12,e1=b2(-k-2ασb22+4k-αcα)12d1,e2=0,f1=-b22(k-4ασb22-4cαk-3α)12d1,f2=0.Case 5.
One has(23)μ=4cαk-3α,λ=0,a0=3k-αcα,a1=0,a2=k2α,b1=0,b2=±k-αcαk3ασ,c0=0,c1=0,c2=0,d1=0,d2=0,e0=e0,e1=±k3αf12σcαk3α,e2=0,f1=f1,f2=0.Case 6.
One has(24)μ=μ,λ=0,a0=23k2αμ+13k-αcα,a1=0,a2=2k2α,b1=0,b2=0,c0=c0,c1=c1,c2=0,d1=0,d2=0,e0=-4c0(-k4αμ+kαcα)3c12,e1=4(-k4αμ+kαcα)3c1,e2=0,f1=0,f2=0.Case 7.
One has(25)μ=μ,λ=0,a0=53k2αμ+13k-αcα,a1=0,a2=2k2α,b1=0,b2=0,c0=c0,c1=0,c2=0,d1=d1,d2=0,e0=-2c0μ(k4αμ+2kαcα)3σd12,e1=0,e2=0,f1=2μ(k4αμ+2kαcα)3(σd1),f2=0.
Substituting the previous results into (18) and combining with (8), we can obtain a series of exact solutions for (13).From Case1, we obtain the following exact solutions.Whenμ<0,
(26)U1(ξ)=13k-αcα+512k-2ασb22-2k2αμ×[C2sinh(-μξαΓ(1+α)))-1(C1sinh(-μξαΓ(1+α))+C2cosh(-μξαΓ(1+α)))×(C1cosh(-μξαΓ(1+α))+C2sinh(-μξαΓ(1+α)))-1]2+b2-μ[C2sinh(-μξαΓ(1+α)))-1(C1sinh(-μξαΓ(1+α))+C2cosh(-μξαΓ(1+α)))×(C1cosh(-μξαΓ(1+α))+C2sinh(-μξαΓ(1+α)))-1]×(σ{C2sinh(-μξαΓ(1+α)))-1]21-[C2sinh(-μξαΓ(1+α)))-1(C1sinh(-μξαΓ(1+α))+C2cosh(-μξαΓ(1+α)))×(C1cosh(-μξαΓ(1+α))+C2sinh(-μξαΓ(1+α)))-1]2})1/2,(27)V1(ξ)=124b2(16k-αcαf2+5k-2αf2σb22-6e0b2)f22-k2αb22f2μ[C2sinh(-μξαΓ(1+α)))-1(C1sinh(-μξαΓ(1+α))+C2cosh(-μξαΓ(1+α)))×(C1cosh(-μξαΓ(1+α))+C2sinh(-μξαΓ(1+α)))-1]2+b224f2-μ[C2sinh(-μξαΓ(1+α)))-1(C1sinh(-μξαΓ(1+α))+C2cosh(-μξαΓ(1+α)))×(C1cosh(-μξαΓ(1+α))+C2sinh(-μξαΓ(1+α)))-1]×(σ{C2sinh(-μξαΓ(1+α)))-1]21-[C2sinh(-μξαΓ(1+α)))-1(C1sinh(-μξαΓ(1+α))+C2cosh(-μξαΓ(1+α)))×(C1cosh(-μξαΓ(1+α))+C2sinh(-μξαΓ(1+α)))-1]2})1/2,(28)W1(ξ)=e0-2k2αf2b2μ[C2sinh(-μξαΓ(1+α)))-1(C1sinh(-μξαΓ(1+α))+C2cosh(-μξαΓ(1+α)))×(C1cosh(-μξαΓ(1+α))+C2sinh(-μξαΓ(1+α)))-1]2+f2-μ[(C1sinh(-μξαΓ(1+α))+C2cosh(-μξαΓ(1+α)))×(C1cosh(-μξαΓ(1+α))+C2sinh(-μξαΓ(1+α)))-1]×(σ{C2sinh(-μξαΓ(1+α)))-1]21-[C2sinh(-μξαΓ(1+α)))-1(C1sinh(-μξαΓ(1+α))+C2cosh(-μξαΓ(1+α)))×(C1cosh(-μξαΓ(1+α))+C2sinh(-μξαΓ(1+α)))-1]2})1/2,
whereξ=kx+ct+ξ0,μ=(1/4)k-4ασb22.Whenμ>0,
(29)U2(ξ)=13k-αcα+512k-2ασb22+2k2αμ×[C2sinh(-μξαΓ(1+α)))-1(-C1sin(μΓ(1+α))ξα+C2cos(μΓ(1+α))ξα)×(C1cos(μΓ(1+α))ξα+C2sin(μΓ(1+α))ξα)-1]2+b2μ[C2sinh(-μξαΓ(1+α)))-1(-C1sin(μΓ(1+α))ξα+C2cos(μΓ(1+α))ξα)×(C1cos(μΓ(1+α))ξα+C2sin(μΓ(1+α))ξα)-1]×(σ{+C2sin(μΓ(1+α))ξα)-1]21+[C2sinh(-μξαΓ(1+α)))-1(-C1sin(μΓ(1+α))ξα+C2cos(μΓ(1+α))ξα)×(C1cos(μΓ(1+α))ξα+C2sin(μΓ(1+α))ξα)-1]2})1/2,(30)V2(ξ)=124b2(16k-αcαf2+5k-2αf2σb22-6e0b2)f22+k2αb22f2μ[C2sinh(-μξαΓ(1+α)))-1(-C1sin(μΓ(1+α))ξα+C2cos(μΓ(1+α))ξα)×(C1cos(μΓ(1+α))ξα+C2sin(μΓ(1+α))ξα)-1]2+b224f2μ[C2sinh(-μξαΓ(1+α)))-1(-C1sin(μΓ(1+α))ξα+C2cos(μΓ(1+α))ξα)×(C1cos(μΓ(1+α))ξα+C2sin(μΓ(1+α))ξα)-1]×(σ{+C2sin(μΓ(1+α))ξα)-1]21+[C2sinh(-μξαΓ(1+α)))-1(-C1sin(μΓ(1+α))ξα+C2cos(μΓ(1+α))ξα)×(C1cos(μΓ(1+α))ξα+C2sin(μΓ(1+α))ξα)-1]2})1/2,(31)W2(ξ)=e0+2k2αf2b2μ×[C2sinh(-μξαΓ(1+α)))-1(-C1sin(μΓ(1+α))ξα+C2cos(μΓ(1+α))ξα)×(C1cos(μΓ(1+α))ξα+C2sin(μΓ(1+α))ξα)-1]2+f2μ[C2sinh(-μξαΓ(1+α)))-1(-C1sin(μΓ(1+α))ξα+C2cos(μΓ(1+α))ξα)×(C1cos(μΓ(1+α))ξα+C2sin(μΓ(1+α))ξα)-1]×(σ{+C2sin(μΓ(1+α))ξα)-1]21+[C2sinh(-μξαΓ(1+α)))-1(-C1sin(μΓ(1+α))ξα+C2cos(μΓ(1+α))ξα)×(C1cos(μΓ(1+α))ξα+C2sin(μΓ(1+α))ξα)-1]2})1/2,
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:
(32)U3(ξ)=13k-αcα+512k-2ασb22-2k2αμ[tanh(-μξαΓ(1+α))]2+b2-μ[tanh(-μξαΓ(1+α))]×σ{1-[tanh(-μξαΓ(1+α))]2},(33)V3(ξ)=124b2(16k-αcαf2+5k-2αf2σb22-6e0b2)f22-k2αb22f2μ[tanh(-μξαΓ(1+α))]2+b224f2-μ[tanh(-μξαΓ(1+α))]×σ{1-[tanh(-μξαΓ(1+α))]2},(34)W3(ξ)=e0-2k2αf2b2μ[tanh(-μξαΓ(1+α))]2+f2-μ[tanh(-μξαΓ(1+α))]×σ{1-[tanh(-μξαΓ(1+α))]2}.Figure 1
The solutionU3 in (32) with γ=4/5, k=c=σ=b2=1, μ=-1, ξ0=0.Figure 2
The solutionV3 in (33) with γ=4/5, k=c=σ=b2=f2=1, μ=-1, ξ0=0, e0=1.Figure 3
The solutionW3 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:
(35)U4(ξ)=13k-αcα+512k-2ασb22+2k2αμ[tanμΓ(1+α)ξα]2-b2μ[tanμΓ(1+α)ξα]×σ{1+[tanμΓ(1+α)ξα]2},(36)V4(ξ)=124b2(16k-αcαf2+5k-2αf2σb22-6e0b2)f22+k2αb22f2μ[tanμΓ(1+α)ξα]2-b224f2μ[tanμΓ(1+α)ξα]×σ{1+[tanμΓ(1+α)ξα]2},(37)W4(ξ)=e0+2k2αf2b2μ[tanμΓ(1+α)ξα]2-f2μ[tanμΓ(1+α)ξα]×σ{1+[tanμΓ(1+α)ξα]2}.Figure 4
The solutionU4 in (35) with γ=4/5, k=c=σ=b2=1, μ=1, ξ0=0.Figure 5
The solutionV4 in (36) with γ=4/5, k=c=σ=b2=f2=1, μ=1, ξ0=0,e0=1.Figure 6
The solutionW4 in (37) with γ=4/5, k=c=σ=b2=f2=1, μ=-1, ξ0=0,e0=1.Similar to the established solutions from Case1, 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+⋯+knxn+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.



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