# Exact Solutions for Some Fractional Partial Differential Equations by the(G′/G) Method

Authors: Bin Zheng
Journal: Mathematical Problems in Engineering
 (2013)

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## Abstract

We apply the(G′/G) method to seek exact solutions for several fractional partial differential equations including the space-time fractional (2 + 1)-dimensional dispersive long wave equations,
the (2 + 1)-dimensional space-time fractional Nizhnik-Novikov-Veselov system, and the time fractional fifth-order Sawada-Kotera equation. The fractional derivative is defined in the sense of modified Riemann-liouville derivative. Based on a certain variable transformation, these fractional partial differential equations are transformed into ordinary differential equations of integer order. With the aid of mathematical software, a variety of exact solutions for them are obtained.


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## Body

## 1. Introduction

In recent decades, fractional differential equations have gained much attention as they are widely used to describe various complex phenomena in many fields such as the fluid flow, signal processing, control theory, systems identification, biology, and other areas. Among the investigations for fractional differential equations, research for seeking exact solutions and approximate solutions of fractional differential equations is a hot topic. Many powerful and efficient methods have been proposed so far, for example, the fractional variational iteration method [1–5], the Adomian’s decomposition method [6–8], the homotopy perturbation method [9], the finite difference method [10], the finite element method [11], the fractional subequation method [12–14], and so on. Using these methods, solutions with various forms for some given fractional differential equations have been established.In this paper, we will apply the(G′/G) method [15–21] for solving some fractional partial differential equations in the sense of modified Riemann-Liouville derivative by Jumarie. The definition and some important properties for the Jumarie’s modified Riemann-Liouville derivative of order α are listed as follows [22]:
(1)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,Dtαtr=Γ(1+r)Γ(1+r-α)tr-α,(2)Dtα(f(t)g(t))=g(t)Dtαf(t)+f(t)Dtαg(t),(3)Dtαf[g(t)]=fg′[g(t)]Dtαg(t)=Dgαf[g(t)](g′(t))α.The rest of this paper is organized as follows. In Section2, we give the description of the (G′/G) method for solving fractional partial differential equations. Then in Section 3 we apply this method to establish exact solutions for the space-time fractional (2 + 1)-dimensional dispersive long wave equations, the space-time fractional (2 + 1)-dimensional Nizhnik-Novikov-Veselov system, and the time fractional fifth-order Sawada-Kotera equation. Some conclusions are presented at the end of the paper.


## 2. Description of the(G′/G) Method for Fractional Partial Differential Equations

In this section we give the description of the(G′/G) method for solving fractional partial differential equations.Suppose that a fractional partial differential equation, say in the independent variablest,x1,x2,…,xn, is given by
(4)P(u1,…,uk,Dtαu1,…,Dtαuk,Dx1βu1,…,00Dx1βuk,…,Dxnγu1,…,Dxnγuk,…)=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.Step 1.
Execute a certain variable transformation:(5)ui(t,x1,x2,…,xn)=Ui(ξ),ξ=ξ(t,x1,x2,…,xn),
such that (4) can be turned into the following ordinary differential equation of integer order with respect to the variable ξ:
(6)P~(U1,…,Uk,U1′,…,Uk′,U1′′,…,Uk′′,…)=0.Step 2.
Suppose that the solution of (6) can be expressed by a polynomial in (G′/G) as follows:
(7)Uj(ξ)=∑‍i=0mjaj,i(G′G)i,j=1,2,…,k,
where G=G(ξ) satisfies the second-order ODE in the form
(8)G′′+λG′+μG=0,
and λ, μ, aj,i, i=0,1,…,mj, j=1,2,…,k, are constants to be determined later, aj,m≠0. The positive integer mj can be determined by considering the homogeneous balance between the highest order derivatives and nonlinear terms appearing in (6).
By the generalized solutions of (8), we have
(9)(G′G)={-λ2+λ2-4μ2×((C1sinhλ2-4μ2ξ+C2coshλ2-4μ2ξ)×(C1coshλ2-4μ2ξ+C2sinhλ2-4μ2ξ)-1),λ2-4μ>0,-λ2+4μ-λ22×(+C2sin4μ-λ22ξ)-1(-C1sin4μ-λ22ξ+C2cos4μ-λ22ξ)×(C1cos4μ-λ22ξ+C2sin4μ-λ22ξ)-1),λ2-4μ<0,-λ2+C2C1+C2ξ,λ2-4μ=0,
where C1 and C2 are arbitrary constants.Step 3.
Substituting (7) into (6) and using (8), collecting all terms with the same order of (G′/G) together, the left-hand side of (6) is converted into another polynomial in (G′/G). Equating each coefficient of this polynomial to zero yields a set of algebraic equations for λ, μ, aj,i, i=0,1,…,mj, j=1,2,…,k.Step 4.
Solving the equations system in Step3 and using (9), we can construct a variety of exact solutions for (4).


## 3. Applications of the Method

### 3.1. Space-Time Fractional (2 + 1)-Dimensional Dispersive Long Wave Equations

We consider the following space-time fractional (2 + 1)-dimensional dispersive long wave equations:(10)∂2αu∂yα∂tα+∂2αv∂x2α+∂α(u(∂αu/∂xα))∂yα=0,∂αv∂tα+∂αu∂xα+∂α(uv)∂xα+∂3αu∂x2α∂yα=0,0<α≤1,
which are a variation of the known (2 + 1)-dimensional dispersive long wave equations [23–36]:
(11)uyt+vxx+(uux)y=0,vt+ux+(uv)x+uxxy=0.
Some types of exact solutions for (11) have been obtained in [23–36] by the use of various methods including the Riccati subequation method [23, 24, 29], the nonlinear transformation method [25], Jacobi function method [27, 28, 36], (G′/G)-expansion method [26], modified CK’s direct method [30], EXP-function method [31], Hopf-Cole transformation method [32], modified extended Fan’s sub-equation method [33, 34], and generalized algebraic method [35]. But we notice so far that no research has been paid for (10). In the following we will apply the described method in Section 2 to (10).To begin with, we supposeu(x,y,t)=U(ξ), v(x,y,t)=V(ξ), where ξ=(c/Γ(1+α))tα+(k1/Γ(1+α))xα+(k2/Γ(1+α))yα+ξ0, k1, k2, c, ξ0 are all constants with k1, k2, c≠0. Then by the use of the first equality in (3), we obtain Dxαu=DxαU(ξ)=U′(ξ)Dxαξ=k1U′(ξ),Dyαu=DyαU(ξ)=U′(ξ)Dyαξ=k2U′(ξ), Dtαu=DtαU(ξ)=U′(ξ)Dtαξ=cU′(ξ), and then (10) can be turned into
(12)ck2U′′+k12V′′+k1k2(UU′′+(U′)2)=0,cV′+k1U′+k1(UV)′+k12k2U′′′=0.
Suppose that the solutions of (12), can be expressed by
(13)U(ξ)=∑‍i=0m1ai(G′G)i,V(ξ)=∑‍i=0m2bi(G′G)i.
Balancing the order of V′′ and UU′′, (UV)′ and U′′′ in (12) we can obtain m1=1, m2=2. So we have
(14)U(ξ)=a0+a1(G′G),V(ξ)=b0+b1(G′G)+b2(G′G)2.
Substituting (14) into (12), using (8), collecting all the terms with the same power of (G′/G) together, and equating each coefficient to zero, yields a set of algebraic equations. Solving these equations yields
(15)a0=±k12λ-ck1,a1=±2k1,b0=-2k1k2μ-1,b1=-2k1k2λ,b2=-2k1k2.
Substituting the result above into (14) and combining it with (9) we can obtain the following exact solutions to (10).Whenλ2-4μ>0,
(16)u1(x,y,t)=±k12λ-ck1000±2k1[-λ2+λ2-4μ200000000000×(00000000000000000+C2sinhλ2-4μ2ξ)-1(C1sinhλ2-4μ2ξ000000000000000000+C2coshλ2-4μ2ξ)00000000000000×(C1coshλ2-4μ2ξ00000000000000000+C2sinhλ2-4μ2ξ)-1)],-32k12[-λ2+λ2-4μ200000000×(+C2sinhλ2-4μ2ξ)-1(C1sinhλ2-4μ2ξ00000000000000+C2coshλ2-4μ2ξ)00000000000×(C1coshλ2-4μ2ξ0000000000000000+C2sinhλ2-4μ2ξ)-1)]2,v1(x,y,t)=-2k1k2μ-1-2k1k2λ0000×[-λ2+λ2-4μ2000000×(+C2sinhλ2-4μ2ξ)-1(C1sinhλ2-4μ2ξ00000000000+C2coshλ2-4μ2ξ)000000000×(C1coshλ2-4μ2ξ000000000000+C2sinhλ2-4μ2ξ)-1)]-2k1k2[-λ2+λ2-4μ2000000×(+C2sinhλ2-4μ2ξ)-1(C1sinhλ2-4μ2ξ0000000000+C2coshλ2-4μ2ξ)000000000×(C1coshλ2-4μ2ξ000000000000+C2sinhλ2-4μ2ξ)-1)]2,
where ξ=(c/Γ(1+α))tα+(k1/Γ(1+α))xα+(k2/Γ(1+α))yα+ξ0.In particular, if we takeC2=0, then we obtain the following solitary wave solutions, which are shown in Figures 1 and 2.Figure 1
The solutionu1 with the positive symbol selected in ± and λ=4, μ=3, c=k1=k2=1, C2=ξ0=0, α=1/2, and t=0.1.Figure 2
The solutionv1 with λ=4, μ=3, c=k1=k2=1, C2=ξ0=0, α=1/2, and t=0.1.Whenλ2-4μ<0,
(17)u2(x,y,t)=±k12λ-ck1000±2k1[-λ2+4μ-λ22000000000×(+C2sin4μ-λ22ξ)-1(-C1sin4μ-λ22ξ0000000000000+C2cos4μ-λ22ξ)000000000000×(C1cos4μ-λ22ξ000000000000000+C2sin4μ-λ22ξ)-1)]-32k12[-λ2+4μ-λ220000000×((-C1sin4μ-λ22ξ0000000000000+C2cos4μ-λ22ξ)000000000×(C1cos4μ-λ22ξ0000000000000+C2sin4μ-λ22ξ))]2,v2(x,y,t)=-2k1k2μ-1000-2k1k2λ[-λ2+4μ-λ220000000000000×(+C2sin4μ-λ22ξ)-1(-C1sin4μ-λ22ξ000000000000000000+C2cos4μ-λ22ξ)0000000000000000×(C1cos4μ-λ22ξ0000000000000000000+C2sin4μ-λ22ξ)-1)]-2k1k2[-λ2+4μ-λ2200000000×((-C1sin4μ-λ22ξ00000000000000+C2cos4μ-λ22ξ)000000000×(C1cos4μ-λ22ξ000000000000+C2sin4μ-λ22ξ))]2,
where ξ=(c/Γ(1+α))tα+(k1/Γ(1+α))xα+(k2/Γ(1+α))yα+ξ0.In Figures3 and 4, the periodic function solutions (17) are demonstrated with some certain parameters.Figure 3
The solutionu2 with the positive symbol selected in ± and λ=2, μ=2, c=k1=k2=1, C2=ξ0=0, α=1/2, and t=0.1.Figure 4
The solutionv2 with λ=2, μ=2, c=k1=k2=1, C2=ξ0=0, α=1/2, and t=0.1.Whenλ2-4μ=0,
(18)u3(x,y,t)=±k12λ-ck1±2k1[-λ2+C2C1+C2ξ]-32k12[-λ2+C2C1+C2ξ]2,v3(x,y,t)=-2k1k2μ-1-2k1k2λ[-λ2+C2C1+C2ξ]-2k1k2[-λ2+C2C1+C2ξ]2,
where ξ=(c/Γ(1+α))tα+(k1/Γ(1+α))xα+(k2/Γ(1+α))yα+ξ0.Remark 1.
The established solutions in (16)–(18) are new exact solutions for the space-time fractional (2 + 1)-dimensional dispersive long wave equations.


### 3.2. Space-Time Fractional (2 + 1)-Dimensional Nizhnik-Novikov-Veselov System

We will consider the space-time fractional (2 + 1)-dimensional Nizhnik-Novikov-Veselov (NNV) system:(19)Dtαu+aDx3βu+bDy3γu+cDxβu+dDyγu=3aDxβ(uv)+3bDyγ(uw),Dxβu=Dyγv,Dyγu=Dxβw,0<α,β,γ≤1,
which is a variation of the following (2 + 1)-dimensional Nizhnik-Novikov-Veselov (NNV) system [21, 37, 38]:
(20)ut+auxxx+buyyy+cux+duy=3a(uv)x+3b(uw)y,ux=vy,uy=wx.Supposeu(x,y,t)=U(ξ), v(x,y,t)=V(ξ), w(x,y,t)=W(ξ), where ξ=(c/Γ(1+α))tα+(k/Γ(1+β))xβ+(l/Γ(1+γ))yγ+ξ0, c, k, l, ξ0 are all constants with k1,k2, c≠0. Then by the use of the first equality in (3), (19) can be turned into the following forms:
(21)cu′+ak3u′′′+bl3u′′′+cku′+dlu′=3ak(uv)′+3bl(uw)′,ku′=lv′,lu′=kw′.
Integrating (21) once, we have
(22)cu+ak3u′′+bl3u′′+cku+dlu=3akuv+3bluw+g1,ku=lv+g2,lu=kw+g3,
where g1,g2, and g3 are the integration constants.Suppose that the solutions of (22) can be expressed by a polynomial in (G′/G) as follows:
(23)U(ξ)=∑‍i=0m1ai(G′G)i,V(ξ)=∑‍i=0m2bi(G′G)i,W(ξ)=∑‍i=0m3ci(G′G)i,
where ai, bi, and ci are constants. Balancing the order of u′′  and uv, and the order of u and v, the order of u and w in (22), we can obtain m1=m2=m3=2. So we have
(24)U(ξ)=a0+a1(G′G)1+a2(G′G)2,V(ξ)=b0+b1(G′G)1+b2(G′G)2,W(ξ)=c0+c1(G′G)1+c2(G′G)2.Substituting (24) into (22), using (8), collecting all the terms with the same power of (G′/G) together, and equating each coefficient to zero, yields a set of algebraic equations. Solving these equations yields.Case 1.
Consider(25)a2=2kl,a1=2klλ,a0=2klμ,b2=2k2,b1=2k2λ,b0=2k2μ,c2=2l2,c1=2l2λ,c0=2l2μ,c=4ak3μ-ak3λ2-bl3λ2-dl+4bl3μ-ck,g1=g2=g3=0,
where k and l are arbitrary constants.
Substituting the result above into (24) and combining it with (9) we can obtain the corresponding exact solutions for (19).
Whenλ2-4μ>0,
(26)u1(x,y,t)00=2klμ-klλ22+kl(λ2-4μ)20000×((C1sinhλ2-4μ2ξ)-1(C1sinhλ2-4μ2ξ0000000000+C2coshλ2-4μ2ξ)0000000×(C1coshλ2-4μ2ξ0000000000+C2sinhλ2-4μ2ξ)-1)2,v1(x,y,t)00=2k2μ-k2λ22+k22(λ2-4μ)0000×(0000000000+C2sinhλ2-4μ2ξ)-1(C1sinhλ2-4μ2ξ0000000000+C2coshλ2-4μ2ξ)0000000×(C1coshλ2-4μ2ξ0000000000+C2sinhλ2-4μ2ξ)-1)2,w1(x,y,t)=2l2μ-l2λ22+l22(λ2-4μ)×(0000000000+C2sinhλ2-4μ2ξ)-1(C1sinhλ2-4μ2ξ0000000000+C2coshλ2-4μ2ξ)0000000×(C1coshλ2-4μ2ξ0000000000+C2sinhλ2-4μ2ξ)-1)2,
where ξ=((4ak3μ-ak3λ2-bl3λ2-dl+4bl3μ-ck)/Γ(1+α))tα+(k/Γ(1+β))xβ+(l/Γ(1+γ))yγ+ξ0.
Whenλ2-4μ<0,
(27)u2(x,y,t)=2klμ-klλ22+kl(4μ-λ2)2×(0000000000+C2sin4μ-λ22ξ)-1(-C1sin4μ-λ22ξ00000000+C2cos4μ-λ22ξ)0000000×(C1cos4μ-λ22ξ0000000000+C2sin4μ-λ22ξ)-1)2,v2(x,y,t)=2k2μ-k2λ22+k22(4μ-λ2)×(0000000000+C2sin4μ-λ22ξ)-1(-C1sin4μ-λ22ξ0000000000+C2cos4μ-λ22ξ)0000000×(C1cos4μ-λ22ξ0000000000+C2sin4μ-λ22ξ)-1)2,w2(x,y,t)=2l2μ-l2λ22+l22(4μ-λ2)×(0000000000+C2sin4μ-λ22ξ)-1(-C1sin4μ-λ22ξ000000000+C2cos4μ-λ22ξ)0000000×(C1cos4μ-λ22ξ0000000000+C2sin4μ-λ22ξ)-1)2,
where ξ=((4ak3μ-ak3λ2-bl3λ2-dl+4bl3μ-ck)/Γ(1+α))tα+(k/Γ(1+β))xβ+(l/Γ(1+γ))yγ+ξ0.
Whenλ2-4μ=0,
(28)u3(x,y,t)=-klλ22+2klC22(C1+C2ξ)2+2klμ,v3(x,y,t)=-k2λ22+2k2C22(C1+C2ξ)2+2k2μ,w3(x,y,t)=-l2λ22+2l2C22(C1+C2ξ)2+2l2μ,
where ξ=((4ak3μ-ak3λ2-bl3λ2-dl+4bl3μ-ck)/Γ(1+α))tα+(k/Γ(1+β))xβ+(l/Γ(1+γ))yγ+ξ0.Case 2.
Consider(29)a2=2kl,a1=2klλ,a0=13kl(λ2+2μ),b2=2k2,b1=2k2λ,b0=13k2(λ2+2μ),c2=2l2,c1=2l2λ,c0=13l2(λ2+2μ),c=-4ak3μ+ak3λ2+bl3λ2-dl-4bl3μ-ck,g1=g2=g3=0.
Substituting the result above into (24) and combining it with (9) we can obtain the corresponding exact solutions for (19).
Whenλ2-4μ>0,
(30)u4(x,y,t)=13kl(λ2+2μ)-klλ22+kl(λ2-4μ)2×(0000000000+C2sinhλ2-4μ2ξ)-1(C1sinhλ2-4μ2ξ0000000000+C2coshλ2-4μ2ξ)0000000×(C1coshλ2-4μ2ξ0000000000+C2sinhλ2-4μ2ξ)-1)2,v4(x,y,t)=13k2(λ2+2μ)-k2λ22+k22(λ2-4μ)×(0000000000+C2sinhλ2-4μ2ξ)-1(C1sinhλ2-4μ2ξ0000000000+C2coshλ2-4μ2ξ)0000000×(C1coshλ2-4μ2ξ0000000000+C2sinhλ2-4μ2ξ)-1)2,w4(x,y,txi)=13l2(λ2+2μ)-l2λ22+l22(λ2-4μ)×(0000000000+C2sinhλ2-4μ2ξ)-1(C1sinhλ2-4μ2ξ0000000000+C2coshλ2-4μ2ξ)0000000×(C1coshλ2-4μ2ξ0000000000+C2sinhλ2-4μ2ξ)-1)2,
where ξ=((-4ak3μ+ak3λ2+bl3λ2-dl-4bl3μ-ck)/Γ(1+α))tα+(k/Γ(1+β))xβ+(l/Γ(1+γ))yγ+ξ0.
Whenλ2-4μ<0,
(31)u5(x,y,t)=13kl(λ2+2μ)-klλ22+kl(4μ-λ2)2×(00000000000+C2sin4μ-λ22ξ)-1(-C1sin4μ-λ22ξ0000000000+C2cos4μ-λ22ξ)0000000×(C1cos4μ-λ22ξ00000000000+C2sin4μ-λ22ξ)-1)2,v5(x,y,t)=13k2(λ2+2μ)-k2λ22+k22(4μ-λ2)×(0000000000+C2sin4μ-λ22ξ)-1(-C1sin4μ-λ22ξ0000000000+C2cos4μ-λ22ξ)0000000×(C1cos4μ-λ22ξ0000000000+C2sin4μ-λ22ξ)-1)2,w5(x,y,t)=13l2(λ2+2μ)-l2λ22+l22(4μ-λ2)×(0000000000+C2sin4μ-λ22ξ)-1(-C1sin4μ-λ22ξ0000000000+C2cos4μ-λ22ξ)0000000×(C1cos4μ-λ22ξ0000000000+C2sin4μ-λ22ξ)-1)2,
where ξ=((-4ak3μ+ak3λ2+bl3λ2-dl-4bl3μ-ck)/Γ(1+α))tα+(k/Γ(1+β))xβ+(l/Γ(1+γ))yγ+ξ0.
Whenλ2-4μ=0,
(32)u6(x,y,t)=-klλ22+2klC22(C1+C2ξ)2+13kl(λ2+2μ),v6(x,y,t)=-k2λ22+2k2C22(C1+C2ξ)2+13k2(λ2+2μ),w6(x,y,t)=-l2λ22+2l2C22(C1+C2ξ)2+13l2(λ2+2μ),
where ξ=((-4ak3μ+ak3λ2+bl3λ2-dl-4bl3μ-ck)/Γ(1+α))tα+(k/Γ(1+β))xβ+(l/Γ(1+γ))yγ+ξ0.Remark 2.
The established solutions in (26)–(32) for the space-time fractional (2 + 1)-dimensional Nizhnik-Novikov-Veselov system have not been reported by other authors so far.Remark 3.
In [21], Zayed solved the (2 + 1)-dimensional Nizhnik-Novikov-Veselov (NNV) system (20) by the (G′/G)-expansion method. We note that if we put α=1, k=l=1 in (26)–(32), then our results reduce to the results in [21] despite the slight difference of the symbols. So the solutions presented here are of more general forms.


### 3.3. Time Fractional Fifth-Order Sawada-Kotera Equation

We consider the time fractional fifth-order Sawada-Kotera equation:(33)Dtαu+uxxxxx+45uxu2+15(uxuxx+uuxxx)=0,0<α≤1,
which is the variation of the fifth-order Sawada-Kotera equation [39]:
(34)ut+uxxxxx+45uxu2+15(uxuxx+uuxxx)=0.
Now we apply the proposed method to (33). To begin with, we suppose u(x,t)=U(ξ), where ξ=kx+(c/Γ(1+α))tα+ξ0, k, c, ξ0 are all constants with k, c≠0. Then by the use of the first equality in (3), (33) can be turned into
(35)cU′+k5U(5)+45kUU′+15k3(U′U′′+UU′′′)=0.Suppose that the solution of (35) can be expressed by
(36)U(ξ)=∑‍i=0mai(G′G)i.
Balancing the order of U(5) and UU′′′ in (36) we have m=2. So
(37)U(ξ)=a0+a1(G′G)+a2(G′G)2.Substituting (37) into (35), using (8), collecting all the terms with the same power of (G′/G) together, and equating each coefficient to zero, yields a set of algebraic equations. Solving these equations yields
(38)a0=-((80k6μ2-40k6μλ2+5k6λ4-20kc40k3μ+5k3λ2±80k6μ2-40k6μλ2+5k6λ4-20kc)×(30k)-1(80k6μ2-40k6μλ2+5k6λ4-20kc40k3μ+5k3λ2),a1=-2k2λ,a2=-2k2.Substituting the result above into (37) and combining it with (9) we can obtain the corresponding exact solutions to (33).Whenλ2-4μ>0,
(39)u1(x,t)=-((80k6μ2-40k6μλ2+5k6λ4-20kc40k3μ+5k3λ2000000±80k6μ2-40k6μλ2+5k6λ4-20kc)00000×(30k)-1(80k6μ2-40k6μλ2+5k6λ4-20kc40k3μ+5k3λ2)-2k2λ[-λ2+λ2-4μ2000000000000×(0000000000000000000+C2sinhλ2-4μ2ξ)-1(C1sinhλ2-4μ2ξ00000000000000000+C2coshλ2-4μ2ξ)0000000000000000×(C1coshλ2-4μ2ξ0000000000000000000+C2sinhλ2-4μ2ξ)-1)]-2k2[-λ2+λ2-4μ200000000000×(0000000000000000+C2sinhλ2-4μ2ξ)-1(C1sinhλ2-4μ2ξ00000000000000+C2coshλ2-4μ2ξ)0000000000000×(C1coshλ2-4μ2ξ0000000000000000+C2sinhλ2-4μ2ξ)-1)]2,
where ξ=kx+(c/Γ(1+α))tα+ξ0.Whenλ2-4μ<0,
(40)u2(x,t)=-((80k6μ2-40k6μλ2+5k6λ4-20kc40k3μ+5k3λ20±80k6μ2-40k6μλ2+5k6λ4-20kc)00×(30k)(80k6μ2-40k6μλ2+5k6λ4-20kc40k3μ+5k3λ2)-1-2k2λ[-λ2+4μ-λ22000000000000×(000000000000000000+C2sin4μ-λ22ξ)-1(-C1sin4μ-λ22ξ0000000000000000+C2cos4μ-λ22ξ)000000000000000×(C1cos4μ-λ22ξ000000000000000000+C2sin4μ-λ22ξ)-1)]-2k2[-λ2+4μ-λ2200000000000×(00000000000000000+C2sin4μ-λ22ξ)-1(-C1sin4μ-λ22ξ000000000000000+C2cos4μ-λ22ξ)00000000000000×(C1cos4μ-λ22ξ00000000000000000+C2sin4μ-λ22ξ)-1)]2,
where ξ=kx+(c/Γ(1+α))tα+ξ0.Whenλ2-4μ=0,
(41)u3(x,t)=-((80k6μ2-40k6μλ2+5k6λ4-20kc40k3μ+5k3λ2±80k6μ2-40k6μλ2+5k6λ4-20kc)×(30k)-1(80k6μ2-40k6μλ2+5k6λ4-20kc40k3μ+5k3λ2)-2k2λ[-λ2+C2C1+C2ξ]-2k2[-λ2+C2C1+C2ξ]2,
where ξ=kx+(c/Γ(1+α))tα+ξ0.Remark 4.
The established solutions in (39)–(41) are new exact solutions for the time fractional fifth-order Sawada-Kotera equation.


## 3.1. Space-Time Fractional (2 + 1)-Dimensional Dispersive Long Wave Equations

We consider the following space-time fractional (2 + 1)-dimensional dispersive long wave equations:(10)∂2αu∂yα∂tα+∂2αv∂x2α+∂α(u(∂αu/∂xα))∂yα=0,∂αv∂tα+∂αu∂xα+∂α(uv)∂xα+∂3αu∂x2α∂yα=0,0<α≤1,
which are a variation of the known (2 + 1)-dimensional dispersive long wave equations [23–36]:
(11)uyt+vxx+(uux)y=0,vt+ux+(uv)x+uxxy=0.
Some types of exact solutions for (11) have been obtained in [23–36] by the use of various methods including the Riccati subequation method [23, 24, 29], the nonlinear transformation method [25], Jacobi function method [27, 28, 36], (G′/G)-expansion method [26], modified CK’s direct method [30], EXP-function method [31], Hopf-Cole transformation method [32], modified extended Fan’s sub-equation method [33, 34], and generalized algebraic method [35]. But we notice so far that no research has been paid for (10). In the following we will apply the described method in Section 2 to (10).To begin with, we supposeu(x,y,t)=U(ξ), v(x,y,t)=V(ξ), where ξ=(c/Γ(1+α))tα+(k1/Γ(1+α))xα+(k2/Γ(1+α))yα+ξ0, k1, k2, c, ξ0 are all constants with k1, k2, c≠0. Then by the use of the first equality in (3), we obtain Dxαu=DxαU(ξ)=U′(ξ)Dxαξ=k1U′(ξ),Dyαu=DyαU(ξ)=U′(ξ)Dyαξ=k2U′(ξ), Dtαu=DtαU(ξ)=U′(ξ)Dtαξ=cU′(ξ), and then (10) can be turned into
(12)ck2U′′+k12V′′+k1k2(UU′′+(U′)2)=0,cV′+k1U′+k1(UV)′+k12k2U′′′=0.
Suppose that the solutions of (12), can be expressed by
(13)U(ξ)=∑‍i=0m1ai(G′G)i,V(ξ)=∑‍i=0m2bi(G′G)i.
Balancing the order of V′′ and UU′′, (UV)′ and U′′′ in (12) we can obtain m1=1, m2=2. So we have
(14)U(ξ)=a0+a1(G′G),V(ξ)=b0+b1(G′G)+b2(G′G)2.
Substituting (14) into (12), using (8), collecting all the terms with the same power of (G′/G) together, and equating each coefficient to zero, yields a set of algebraic equations. Solving these equations yields
(15)a0=±k12λ-ck1,a1=±2k1,b0=-2k1k2μ-1,b1=-2k1k2λ,b2=-2k1k2.
Substituting the result above into (14) and combining it with (9) we can obtain the following exact solutions to (10).Whenλ2-4μ>0,
(16)u1(x,y,t)=±k12λ-ck1000±2k1[-λ2+λ2-4μ200000000000×(00000000000000000+C2sinhλ2-4μ2ξ)-1(C1sinhλ2-4μ2ξ000000000000000000+C2coshλ2-4μ2ξ)00000000000000×(C1coshλ2-4μ2ξ00000000000000000+C2sinhλ2-4μ2ξ)-1)],-32k12[-λ2+λ2-4μ200000000×(+C2sinhλ2-4μ2ξ)-1(C1sinhλ2-4μ2ξ00000000000000+C2coshλ2-4μ2ξ)00000000000×(C1coshλ2-4μ2ξ0000000000000000+C2sinhλ2-4μ2ξ)-1)]2,v1(x,y,t)=-2k1k2μ-1-2k1k2λ0000×[-λ2+λ2-4μ2000000×(+C2sinhλ2-4μ2ξ)-1(C1sinhλ2-4μ2ξ00000000000+C2coshλ2-4μ2ξ)000000000×(C1coshλ2-4μ2ξ000000000000+C2sinhλ2-4μ2ξ)-1)]-2k1k2[-λ2+λ2-4μ2000000×(+C2sinhλ2-4μ2ξ)-1(C1sinhλ2-4μ2ξ0000000000+C2coshλ2-4μ2ξ)000000000×(C1coshλ2-4μ2ξ000000000000+C2sinhλ2-4μ2ξ)-1)]2,
where ξ=(c/Γ(1+α))tα+(k1/Γ(1+α))xα+(k2/Γ(1+α))yα+ξ0.In particular, if we takeC2=0, then we obtain the following solitary wave solutions, which are shown in Figures 1 and 2.Figure 1
The solutionu1 with the positive symbol selected in ± and λ=4, μ=3, c=k1=k2=1, C2=ξ0=0, α=1/2, and t=0.1.Figure 2
The solutionv1 with λ=4, μ=3, c=k1=k2=1, C2=ξ0=0, α=1/2, and t=0.1.Whenλ2-4μ<0,
(17)u2(x,y,t)=±k12λ-ck1000±2k1[-λ2+4μ-λ22000000000×(+C2sin4μ-λ22ξ)-1(-C1sin4μ-λ22ξ0000000000000+C2cos4μ-λ22ξ)000000000000×(C1cos4μ-λ22ξ000000000000000+C2sin4μ-λ22ξ)-1)]-32k12[-λ2+4μ-λ220000000×((-C1sin4μ-λ22ξ0000000000000+C2cos4μ-λ22ξ)000000000×(C1cos4μ-λ22ξ0000000000000+C2sin4μ-λ22ξ))]2,v2(x,y,t)=-2k1k2μ-1000-2k1k2λ[-λ2+4μ-λ220000000000000×(+C2sin4μ-λ22ξ)-1(-C1sin4μ-λ22ξ000000000000000000+C2cos4μ-λ22ξ)0000000000000000×(C1cos4μ-λ22ξ0000000000000000000+C2sin4μ-λ22ξ)-1)]-2k1k2[-λ2+4μ-λ2200000000×((-C1sin4μ-λ22ξ00000000000000+C2cos4μ-λ22ξ)000000000×(C1cos4μ-λ22ξ000000000000+C2sin4μ-λ22ξ))]2,
where ξ=(c/Γ(1+α))tα+(k1/Γ(1+α))xα+(k2/Γ(1+α))yα+ξ0.In Figures3 and 4, the periodic function solutions (17) are demonstrated with some certain parameters.Figure 3
The solutionu2 with the positive symbol selected in ± and λ=2, μ=2, c=k1=k2=1, C2=ξ0=0, α=1/2, and t=0.1.Figure 4
The solutionv2 with λ=2, μ=2, c=k1=k2=1, C2=ξ0=0, α=1/2, and t=0.1.Whenλ2-4μ=0,
(18)u3(x,y,t)=±k12λ-ck1±2k1[-λ2+C2C1+C2ξ]-32k12[-λ2+C2C1+C2ξ]2,v3(x,y,t)=-2k1k2μ-1-2k1k2λ[-λ2+C2C1+C2ξ]-2k1k2[-λ2+C2C1+C2ξ]2,
where ξ=(c/Γ(1+α))tα+(k1/Γ(1+α))xα+(k2/Γ(1+α))yα+ξ0.Remark 1.
The established solutions in (16)–(18) are new exact solutions for the space-time fractional (2 + 1)-dimensional dispersive long wave equations.


## 3.2. Space-Time Fractional (2 + 1)-Dimensional Nizhnik-Novikov-Veselov System

We will consider the space-time fractional (2 + 1)-dimensional Nizhnik-Novikov-Veselov (NNV) system:(19)Dtαu+aDx3βu+bDy3γu+cDxβu+dDyγu=3aDxβ(uv)+3bDyγ(uw),Dxβu=Dyγv,Dyγu=Dxβw,0<α,β,γ≤1,
which is a variation of the following (2 + 1)-dimensional Nizhnik-Novikov-Veselov (NNV) system [21, 37, 38]:
(20)ut+auxxx+buyyy+cux+duy=3a(uv)x+3b(uw)y,ux=vy,uy=wx.Supposeu(x,y,t)=U(ξ), v(x,y,t)=V(ξ), w(x,y,t)=W(ξ), where ξ=(c/Γ(1+α))tα+(k/Γ(1+β))xβ+(l/Γ(1+γ))yγ+ξ0, c, k, l, ξ0 are all constants with k1,k2, c≠0. Then by the use of the first equality in (3), (19) can be turned into the following forms:
(21)cu′+ak3u′′′+bl3u′′′+cku′+dlu′=3ak(uv)′+3bl(uw)′,ku′=lv′,lu′=kw′.
Integrating (21) once, we have
(22)cu+ak3u′′+bl3u′′+cku+dlu=3akuv+3bluw+g1,ku=lv+g2,lu=kw+g3,
where g1,g2, and g3 are the integration constants.Suppose that the solutions of (22) can be expressed by a polynomial in (G′/G) as follows:
(23)U(ξ)=∑‍i=0m1ai(G′G)i,V(ξ)=∑‍i=0m2bi(G′G)i,W(ξ)=∑‍i=0m3ci(G′G)i,
where ai, bi, and ci are constants. Balancing the order of u′′  and uv, and the order of u and v, the order of u and w in (22), we can obtain m1=m2=m3=2. So we have
(24)U(ξ)=a0+a1(G′G)1+a2(G′G)2,V(ξ)=b0+b1(G′G)1+b2(G′G)2,W(ξ)=c0+c1(G′G)1+c2(G′G)2.Substituting (24) into (22), using (8), collecting all the terms with the same power of (G′/G) together, and equating each coefficient to zero, yields a set of algebraic equations. Solving these equations yields.Case 1.
Consider(25)a2=2kl,a1=2klλ,a0=2klμ,b2=2k2,b1=2k2λ,b0=2k2μ,c2=2l2,c1=2l2λ,c0=2l2μ,c=4ak3μ-ak3λ2-bl3λ2-dl+4bl3μ-ck,g1=g2=g3=0,
where k and l are arbitrary constants.
Substituting the result above into (24) and combining it with (9) we can obtain the corresponding exact solutions for (19).
Whenλ2-4μ>0,
(26)u1(x,y,t)00=2klμ-klλ22+kl(λ2-4μ)20000×((C1sinhλ2-4μ2ξ)-1(C1sinhλ2-4μ2ξ0000000000+C2coshλ2-4μ2ξ)0000000×(C1coshλ2-4μ2ξ0000000000+C2sinhλ2-4μ2ξ)-1)2,v1(x,y,t)00=2k2μ-k2λ22+k22(λ2-4μ)0000×(0000000000+C2sinhλ2-4μ2ξ)-1(C1sinhλ2-4μ2ξ0000000000+C2coshλ2-4μ2ξ)0000000×(C1coshλ2-4μ2ξ0000000000+C2sinhλ2-4μ2ξ)-1)2,w1(x,y,t)=2l2μ-l2λ22+l22(λ2-4μ)×(0000000000+C2sinhλ2-4μ2ξ)-1(C1sinhλ2-4μ2ξ0000000000+C2coshλ2-4μ2ξ)0000000×(C1coshλ2-4μ2ξ0000000000+C2sinhλ2-4μ2ξ)-1)2,
where ξ=((4ak3μ-ak3λ2-bl3λ2-dl+4bl3μ-ck)/Γ(1+α))tα+(k/Γ(1+β))xβ+(l/Γ(1+γ))yγ+ξ0.
Whenλ2-4μ<0,
(27)u2(x,y,t)=2klμ-klλ22+kl(4μ-λ2)2×(0000000000+C2sin4μ-λ22ξ)-1(-C1sin4μ-λ22ξ00000000+C2cos4μ-λ22ξ)0000000×(C1cos4μ-λ22ξ0000000000+C2sin4μ-λ22ξ)-1)2,v2(x,y,t)=2k2μ-k2λ22+k22(4μ-λ2)×(0000000000+C2sin4μ-λ22ξ)-1(-C1sin4μ-λ22ξ0000000000+C2cos4μ-λ22ξ)0000000×(C1cos4μ-λ22ξ0000000000+C2sin4μ-λ22ξ)-1)2,w2(x,y,t)=2l2μ-l2λ22+l22(4μ-λ2)×(0000000000+C2sin4μ-λ22ξ)-1(-C1sin4μ-λ22ξ000000000+C2cos4μ-λ22ξ)0000000×(C1cos4μ-λ22ξ0000000000+C2sin4μ-λ22ξ)-1)2,
where ξ=((4ak3μ-ak3λ2-bl3λ2-dl+4bl3μ-ck)/Γ(1+α))tα+(k/Γ(1+β))xβ+(l/Γ(1+γ))yγ+ξ0.
Whenλ2-4μ=0,
(28)u3(x,y,t)=-klλ22+2klC22(C1+C2ξ)2+2klμ,v3(x,y,t)=-k2λ22+2k2C22(C1+C2ξ)2+2k2μ,w3(x,y,t)=-l2λ22+2l2C22(C1+C2ξ)2+2l2μ,
where ξ=((4ak3μ-ak3λ2-bl3λ2-dl+4bl3μ-ck)/Γ(1+α))tα+(k/Γ(1+β))xβ+(l/Γ(1+γ))yγ+ξ0.Case 2.
Consider(29)a2=2kl,a1=2klλ,a0=13kl(λ2+2μ),b2=2k2,b1=2k2λ,b0=13k2(λ2+2μ),c2=2l2,c1=2l2λ,c0=13l2(λ2+2μ),c=-4ak3μ+ak3λ2+bl3λ2-dl-4bl3μ-ck,g1=g2=g3=0.
Substituting the result above into (24) and combining it with (9) we can obtain the corresponding exact solutions for (19).
Whenλ2-4μ>0,
(30)u4(x,y,t)=13kl(λ2+2μ)-klλ22+kl(λ2-4μ)2×(0000000000+C2sinhλ2-4μ2ξ)-1(C1sinhλ2-4μ2ξ0000000000+C2coshλ2-4μ2ξ)0000000×(C1coshλ2-4μ2ξ0000000000+C2sinhλ2-4μ2ξ)-1)2,v4(x,y,t)=13k2(λ2+2μ)-k2λ22+k22(λ2-4μ)×(0000000000+C2sinhλ2-4μ2ξ)-1(C1sinhλ2-4μ2ξ0000000000+C2coshλ2-4μ2ξ)0000000×(C1coshλ2-4μ2ξ0000000000+C2sinhλ2-4μ2ξ)-1)2,w4(x,y,txi)=13l2(λ2+2μ)-l2λ22+l22(λ2-4μ)×(0000000000+C2sinhλ2-4μ2ξ)-1(C1sinhλ2-4μ2ξ0000000000+C2coshλ2-4μ2ξ)0000000×(C1coshλ2-4μ2ξ0000000000+C2sinhλ2-4μ2ξ)-1)2,
where ξ=((-4ak3μ+ak3λ2+bl3λ2-dl-4bl3μ-ck)/Γ(1+α))tα+(k/Γ(1+β))xβ+(l/Γ(1+γ))yγ+ξ0.
Whenλ2-4μ<0,
(31)u5(x,y,t)=13kl(λ2+2μ)-klλ22+kl(4μ-λ2)2×(00000000000+C2sin4μ-λ22ξ)-1(-C1sin4μ-λ22ξ0000000000+C2cos4μ-λ22ξ)0000000×(C1cos4μ-λ22ξ00000000000+C2sin4μ-λ22ξ)-1)2,v5(x,y,t)=13k2(λ2+2μ)-k2λ22+k22(4μ-λ2)×(0000000000+C2sin4μ-λ22ξ)-1(-C1sin4μ-λ22ξ0000000000+C2cos4μ-λ22ξ)0000000×(C1cos4μ-λ22ξ0000000000+C2sin4μ-λ22ξ)-1)2,w5(x,y,t)=13l2(λ2+2μ)-l2λ22+l22(4μ-λ2)×(0000000000+C2sin4μ-λ22ξ)-1(-C1sin4μ-λ22ξ0000000000+C2cos4μ-λ22ξ)0000000×(C1cos4μ-λ22ξ0000000000+C2sin4μ-λ22ξ)-1)2,
where ξ=((-4ak3μ+ak3λ2+bl3λ2-dl-4bl3μ-ck)/Γ(1+α))tα+(k/Γ(1+β))xβ+(l/Γ(1+γ))yγ+ξ0.
Whenλ2-4μ=0,
(32)u6(x,y,t)=-klλ22+2klC22(C1+C2ξ)2+13kl(λ2+2μ),v6(x,y,t)=-k2λ22+2k2C22(C1+C2ξ)2+13k2(λ2+2μ),w6(x,y,t)=-l2λ22+2l2C22(C1+C2ξ)2+13l2(λ2+2μ),
where ξ=((-4ak3μ+ak3λ2+bl3λ2-dl-4bl3μ-ck)/Γ(1+α))tα+(k/Γ(1+β))xβ+(l/Γ(1+γ))yγ+ξ0.Remark 2.
The established solutions in (26)–(32) for the space-time fractional (2 + 1)-dimensional Nizhnik-Novikov-Veselov system have not been reported by other authors so far.Remark 3.
In [21], Zayed solved the (2 + 1)-dimensional Nizhnik-Novikov-Veselov (NNV) system (20) by the (G′/G)-expansion method. We note that if we put α=1, k=l=1 in (26)–(32), then our results reduce to the results in [21] despite the slight difference of the symbols. So the solutions presented here are of more general forms.


## 3.3. Time Fractional Fifth-Order Sawada-Kotera Equation

We consider the time fractional fifth-order Sawada-Kotera equation:(33)Dtαu+uxxxxx+45uxu2+15(uxuxx+uuxxx)=0,0<α≤1,
which is the variation of the fifth-order Sawada-Kotera equation [39]:
(34)ut+uxxxxx+45uxu2+15(uxuxx+uuxxx)=0.
Now we apply the proposed method to (33). To begin with, we suppose u(x,t)=U(ξ), where ξ=kx+(c/Γ(1+α))tα+ξ0, k, c, ξ0 are all constants with k, c≠0. Then by the use of the first equality in (3), (33) can be turned into
(35)cU′+k5U(5)+45kUU′+15k3(U′U′′+UU′′′)=0.Suppose that the solution of (35) can be expressed by
(36)U(ξ)=∑‍i=0mai(G′G)i.
Balancing the order of U(5) and UU′′′ in (36) we have m=2. So
(37)U(ξ)=a0+a1(G′G)+a2(G′G)2.Substituting (37) into (35), using (8), collecting all the terms with the same power of (G′/G) together, and equating each coefficient to zero, yields a set of algebraic equations. Solving these equations yields
(38)a0=-((80k6μ2-40k6μλ2+5k6λ4-20kc40k3μ+5k3λ2±80k6μ2-40k6μλ2+5k6λ4-20kc)×(30k)-1(80k6μ2-40k6μλ2+5k6λ4-20kc40k3μ+5k3λ2),a1=-2k2λ,a2=-2k2.Substituting the result above into (37) and combining it with (9) we can obtain the corresponding exact solutions to (33).Whenλ2-4μ>0,
(39)u1(x,t)=-((80k6μ2-40k6μλ2+5k6λ4-20kc40k3μ+5k3λ2000000±80k6μ2-40k6μλ2+5k6λ4-20kc)00000×(30k)-1(80k6μ2-40k6μλ2+5k6λ4-20kc40k3μ+5k3λ2)-2k2λ[-λ2+λ2-4μ2000000000000×(0000000000000000000+C2sinhλ2-4μ2ξ)-1(C1sinhλ2-4μ2ξ00000000000000000+C2coshλ2-4μ2ξ)0000000000000000×(C1coshλ2-4μ2ξ0000000000000000000+C2sinhλ2-4μ2ξ)-1)]-2k2[-λ2+λ2-4μ200000000000×(0000000000000000+C2sinhλ2-4μ2ξ)-1(C1sinhλ2-4μ2ξ00000000000000+C2coshλ2-4μ2ξ)0000000000000×(C1coshλ2-4μ2ξ0000000000000000+C2sinhλ2-4μ2ξ)-1)]2,
where ξ=kx+(c/Γ(1+α))tα+ξ0.Whenλ2-4μ<0,
(40)u2(x,t)=-((80k6μ2-40k6μλ2+5k6λ4-20kc40k3μ+5k3λ20±80k6μ2-40k6μλ2+5k6λ4-20kc)00×(30k)(80k6μ2-40k6μλ2+5k6λ4-20kc40k3μ+5k3λ2)-1-2k2λ[-λ2+4μ-λ22000000000000×(000000000000000000+C2sin4μ-λ22ξ)-1(-C1sin4μ-λ22ξ0000000000000000+C2cos4μ-λ22ξ)000000000000000×(C1cos4μ-λ22ξ000000000000000000+C2sin4μ-λ22ξ)-1)]-2k2[-λ2+4μ-λ2200000000000×(00000000000000000+C2sin4μ-λ22ξ)-1(-C1sin4μ-λ22ξ000000000000000+C2cos4μ-λ22ξ)00000000000000×(C1cos4μ-λ22ξ00000000000000000+C2sin4μ-λ22ξ)-1)]2,
where ξ=kx+(c/Γ(1+α))tα+ξ0.Whenλ2-4μ=0,
(41)u3(x,t)=-((80k6μ2-40k6μλ2+5k6λ4-20kc40k3μ+5k3λ2±80k6μ2-40k6μλ2+5k6λ4-20kc)×(30k)-1(80k6μ2-40k6μλ2+5k6λ4-20kc40k3μ+5k3λ2)-2k2λ[-λ2+C2C1+C2ξ]-2k2[-λ2+C2C1+C2ξ]2,
where ξ=kx+(c/Γ(1+α))tα+ξ0.Remark 4.
The established solutions in (39)–(41) are new exact solutions for the time fractional fifth-order Sawada-Kotera equation.


## 4. Conclusions

We have solved the space-time fractional (2 + 1)-dimensional dispersive long wave equations, the space-time fractional (2 + 1)-dimensional Nizhnik-Novikov-Veselov system, and the time fractional fifth-order Sawada-Kotera equation by the known(G′/G) method successfully. In the process, the transformation for ξ is very important, which ensures that certain fractional partial differential equations can be turned into ordinary differential equations of integer order. With the aid of mathematical software, a variety of exact solutions for these equations are obtained, which may help to understand better physical phenomenon related to these equations. Finally, we note that this approach can also be applied to solve other fractional partial differential equations.



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*Source: 826369-2013-10-31.xml*