# Applications of Separation Variables Approach in Solving Time-Fractional PDEs

Authors: Yinghui He; Yunmei Zhao
Journal: Mathematical Problems in Engineering
 (2018)

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## Abstract

Based on the homogenous balanced principle and subequation method, an improved separation variables function-expansion method is proposed to seek exact solutions of time-fractional nonlinear PDEs. This method is novel and meaningful without using Leibniz rule and chain rule of fractional derivative which have been proved to be incorrect. By using this method, we studied a nonlinear time-fractional PDE with diffusion term. Some general solutions are obtained which contain many arbitrary parameters. Solutions given in related reference are just our especial case. And we also obtained some new type of solutions.


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## 1. Introduction

It is well known that fractional-order models are more adequate than previously used integer-order models due to an exact description of nonlinear phenomena. Therefore, nonlinear fractional partial differential equations (nfPDEs) have attracted more and more attention. Most recently, FPDEs are increasingly used in mathematical modeling of fluid mechanics, biological and chemical processes, signal processing, and control systems, and they are also used in fractal and differential geometry, and so on(see [1–8] and their references cited). Many natural phenomena associated with real-time problems which depend on both time instant and the previous time history, especially, can be successfully modeled by time-fractional nonlinear partial differential equations. Of course, investigating solutions of nfPDEs plays an important role in a large number of research fields. Until now, there is no general method. Nevertheless, many powerful methods were used for solving fractional differential equations, such as adomian decomposition method [9], symmetry method [10–12], numerical method [13, 14], first integral method [15], the homotopy perturbation method [16], invariant subspace method [17], and fractional variational iteration method [18], and have achieved significant progress.There are many definitions of fractional derivative. The fractional differential equations can be expressed in terms of different differential operators defined by Riemann-Liouville, Caputo, Weyl, and many others. Among these definitions, Caputo definition is most frequently used. Here, let us briefly review this definition of fractional derivative.The Caputo fractional derivative of orderα>0 is defined by the following expression: (1)αCDtαft=1Γn-α∫att-τn-α-1fnτdτ=fnat-an-αΓn-α+1+1Γn-α+1∫att-τn-αfn+1τdτ,where n=[α]+1,n-1<α⩽n,t>a⩾0,n∈N+. In this paper we will adopt Caputo fractional derivative definition to investigate exact solution of the following type of nonlinear time-fractional partial differential equation: (2)CDtαu=Fu,∂u∂x,∂2u∂x2,…,∂nu∂xn,where CDtα=0CDtα is Caputo differential operator, u=u(x,t), and t>0,0<α⩽1, n∈N+, x∈R. Equation (2) denotes a series of nonlinear time-fractional PDEs; these models defined by (2) can be used to accurately describe nonlinear phenomena in connection with real-time problems, which not only depend on time instant but also depend on the previous time history.Recently, Feng [19] introduced a fractional DαG/G method for seeking traveling wave solutions of space-time-fractional partial differential equations under the following modified Riemann-Liouville derivative definition [20]: (3)
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.The following properties for the modified Riemann-Liouville derivative are usually used:(4)Dxαxγ=Γ1+γΓ1+γ-αxγ-α,(5)Dxαfux=f′uDxαux=Dxαfuu′xα.Let us briefly review this method. Using traveling wave transformationξ=kx+ct,u(x,t)=U(ξ) and (4) and (5), they reduced the following fractional partial differential equation: (6)Pu,Dtαu,Dxαu,Dt2αu,Dx2αu,…=0,which can be converted into the following fractional ordinary differential equation with respect to the variable ξ: (7)PU,cαDξαU,kαDξαU,c2αDξ2αU,k2αDξ2αU,…=0.Then, they suppose that the solution of (7) can be expressed by a polynomial in DαG/G as follows: (8)Uξ=∑i=0maiDξαGG,where G=G(ξ) satisfies the following fractional ordinary differential equation: (9)AGξDξ2αGξ-BGξDξαGξ-CDξαGξ2-EG2ξ=0.In order to find exact solutions of (9), a nonlinear fractional complex transformation η=ξα/Γ(1+α) is used. Then, (9) can be turned into the following second ordinary differential equation: (10)AHηH′′η-BHηH′η-CH′η2-EH2η=0.The exact solutions of (10) are known.Finally, substituting (8) into (7), equating each coefficient of this polynomial on DξαG/G to zero, they can obtain a large number of exact solutions of space-time fractional partial differential equations (6).We noticed two problems as follows. At first, in a short communication [21], Tarasov proved that formula (4) and the chain rule (5) cannot be performed together for fractional derivatives of noninteger orders α≠1. Therefore, using traveling wave transformation ξ=kx+ct,u(x,t)=U(ξ), space-time fractional partial differential equations (6) cannot be converted into fractional ordinary differential equation (7). And, (9) cannot convert to (10) under transformation η=ξα/Γ(1+α).In addition, it should be noted that the Leibniz rule in the form(11)Dξαfξgξ=gξDξαfξ+fξDξαgξ.cannot hold for fractional derivatives of order α≠1 for sets of differentiable and nondifferentiable functions [22]. It can be proved easily by using a counterexample. Let (12)fx=xγ1,gx=xγ2,where γ1,γ2∈N. Obviously we have (13)Dxαfx=Γ1+γ1Γ1+γ1-αxγ1-α,Dxαgx=Γ1+γ2Γ1+γ2-αxγ2-α.Then, one has (14)gxDxαfx=Γ1+γ1Γ1+γ1-αxγ1+γ2-α,fxDxαgx=Γ1+γ2Γ1+γ2-αxγ1+γ2-α,Dxαfxgx=Γ1+γ1+γ2Γ1+γ1+γ2-αxγ1+γ2-α.As a result, we have the condition(15)Γ1+γ1+γ2Γ1+γ1+γ2-α=Γ1+γ1Γ1+γ1-α+Γ1+γ2Γ1+γ2-α,which must be performed if the Leibniz rule (11) holds. For example, if we make γ1,γ2∈N+ and take into account Γ(n+1)=nΓ(n), then condition (15) can be represented in the form(16)γ1+k!2γ1-α+k2γ1-α+k-1⋯γ1-α+k-1+γ1!2γ1-α+k2γ1-α+k-1⋯γ1-α+1=2γ1+k!,where k=γ2-γ1. It is not difficult to find that (16) holds only if α=1. That is to say, the Leibniz rule (11) does not hold with 0<α<1. Therefore, the following formula used in references is incorrect: (17)DξαUξ=DξαDξαGξGξ=GξDξ2αGξ-DξαGξ2G2ξ.Therefore, when substituting (8) into (7), the left-hand side of (7) cannot be expressed by the polynomial in DξαG/G.So, this fractionalDαG/G method is not reliable and the obtained results are incorrect. By the way, we should be more cautious when using this kind of methods based on chain rule (5) and Leibniz rule (11), like the subequation method, the G′/G-expansion method, the exp-function method, the functional variable method, the trial equation method, the simple equation method, and so forth. We cannot obtain the exact solutions of compound function type of time-fractional PDEs (2) as in the references. Encouraged by Rui’s work [23], we shall introduce an improved method based on the homogenous balanced principle; by using this improved method we shall investigate exact solutions of a series of nonlinear time-fractional PDEs formed as (2).The rest of this paper is organized as follows. In Section2, we will introduce the improved separation variable expansion method based on the homogenous balanced principle. In Section 3, by using this new method, we will investigate exact solutions of a nonlinear time-fractional PDE with diffusion term discussed in (2).


## 2. Introduction of Improved Separation Variable Function-Expansion Method

Although the fractional chain rule (5) and Leibniz rule (11) do not hold, they do not affect the investigation of the exact solutions of nonlinear time-fractional PDE (2) since formulas (4) still hold.Remark 1.
In reference [23], Rui points out that Leibniz rule (11) still holds, which is incorrect. Actually, in our method, it does not matter that Leibniz rule (11) does not hold. Because we just need the following formula: (18)Dtαkfx=kDtαfx,which is easy to be proved by definition for Caputo fractional derivative (1) and for the modified Riemann-Liouville derivative (3).In the following, we introduce main steps of improved function-expansion method of separation variable type as follows.Step 1.
According to the formulas (4) and (18) and characters of nonlinear time-fractional partial differential equation (2), we suppose that (2) has the following exact solutions of the separation variable type: (19)u=∑k=0makφtvxk,where the function φ(t) can be taken as power function tγk, Mittag-Leffler function Eα(λtα), or tβ-1Eα,β(λtα), m is a positive integer, and ak,γk(k=0,1,…,m) are constants to be determined later. The function v(x) satisfies the following subequation: (20)v′x2=r+pvx+qv2x,where r,p,q are constants. Some solutions of (20) are listed as follows.
Whenp=q=0,r≠0, (21)vx=±rx+C,where C is an arbitrary integral constant.
Whenq=0,p≠0, (22)vx=C2p2-2Cp2x+x2p2-4r4p.
Whenq≠0, (23)vx=4pq-p2-4qreqC-x-4qe-qC-x-8q3/2,vx=4pq-p2-4qre-qC-x-4qeqC-x-8q3/2.Taking integral constantC=0, under some conditions, we can obtain many special solutions of (20) which are listed in Table 1.Table 1
Solutions of (20).
Case Conditions Solution of (20) 1 p = q = 0 ,   r = 1 v 1 x = x 2 r = q = 0 ,   p = 4 v 2 x = x 2 3 r = p = 0 ,   q = 1 v 3 x = e x 4 r = 1 ,   p = 0 ,   q = - 1 v 4 x = ± s i n  x 5 r = 1 ,   p = 2 ,   q = 0 v 5 x = x 2 2 - x 6 r = - 1 ,   p = 0 ,   q = 1 v 6 x = cosh ⁡ x 7 r = 1 ,   p = 0 ,   q = - 1 v 7 x = ± c o s  x 8 r = 1 ,   p = 0 ,   q = 1 v 8 x = ± s i n h  xStep 2.
In order to determine the valuem of the function v(x), we balance power of v(x) between the term of the highest order in the right-hand side of (2) and the highest order term in the left-hand side of (2). By the way, the highest order term in the left-hand side of (2) is still vm(x) because we only make derivation for t in the left-hand side of (2). Once the value m has been determined, the expansion expression (19) can be fixed correspondingly.Step 3.
By using specific expansion expression obtained in the Step2, we substitute it into (2) (make fractional-order derivations for t and make integer-order derivations for x); then we balance the power of φ(t) (such as t, Eα(λtα)); thus the values of λ, γk(k=0,1,…,m) can be obtained.Step 4.
In two sides of equation obtained by Step2, we let coefficients of the same order for every term v(x) in two sides of the equation be equal; thus we can determine values of all the coefficients ak(k=0,1,…,m).Step 5.
Finally, substituting the values of all the parametersm,λ,ak,γk(k=0,1,…,m) obtained in above steps into (19), we will obtain exact solutions of the nonlinear time-fractional PDE (2).Remark 2.
In [23], v(x) is taken as one of x,sinx,cos(x), and ex, which is just a special solution of our (20). So our method is more general than Rui’s method [23]. In addition, our method can test a series of functions at one time which satisfy (20). The improved method is more efficient and simple.


## 3. Exact Solutions of a Nonlinear Time-Fractional PDE with Diffusion Term

In this subsection, we will study a nonlinear time-fractional PDE with diffusion term as follows:(24)cDtαu=∂u∂x2∂fu∂u+fu∂2u∂x2+δu2,t⩾0,x∈R,0<α⩽1,where u=u(t,x); the f(u) is diffusion term. When δ=0 and α=1, (24) can be rewritten as the following nonlinear diffusion PDE: (25)∂∂xfu∂u∂x-∂u∂t=0,which have appeared in problems related to plasma and solid state physics; see [24] and references cited therein. In [23], Rui studied the situation of f(u)=βu. Equation (25) becomes the following time-fractional PDE with diffusion term: (26)cDtαu=β∂u∂x2+βu∂2u∂x2+δu2,t⩾0,x∈R,0<α⩽1.And exact solutions of (26) with parameters δ=2β and δ=-2β are given. Here we will investigate more exact solutions by improved method introduced in Section 2.When0<α<1, taking ψ(t)=tγk, we suppose that (26) has an exact solution formed as u(x,t)=∑k=0maktγkv(x). We find that the highest order of v(x) in the term cDtαu is just m. In the right-hand side of (26), the highest order of v(x) in the nonlinear terms ∂u/∂x2 and u∂2u/∂x2 is 2m+2. Considering the relation between parameters β and δ, we can let m be an arbitrary positive integer to test (some terms may be counteracted under certain conditions). Here, for simplicity, we just discuss the situations of m=1 and m=2.


### 3.1. Situation ofm=1

Takingm=1, we first suppose that (26) has an exact solution as the following form: (27)ux,t=a0tγ0+a1tγ1vx,where v(x) satisfies (20) and γ0,γ1,a0,a1≠0 are undetermined constants that can be determined later. Substituting (27) into (26), using (20), it can be reduced to (28)a0Γγ0+1t-α+γ0Γ-α+γ0+1+a1Γγ1+1t-α+γ1Γ-α+γ1+1vx=t2γ1βra12+12tγ1+γ0βpa0a1+t2γ0δa0232t2γ1βpa12+tγ1+γ0βqa0a1+2tγ1+γ0δa0a1vx+2t2γ1βqa12+t2γ1δa12vx2.According to homogenous balanced principle, we let (29)2t2γ1βqa12+t2γ1δa12=0.Solving (29) yields (30)δ=-2qβ.Substituting (30) into (28), it can be reduced to (31)a0Γγ0+1t-α+γ0Γ-α+γ0+1+a1Γγ1+1t-α+γ1Γ-α+γ1+1vx=-12β-3pa12t2γ1+6a1tγ1+γ0qa0vx-12β4qa02t2γ0-2ra12t2γ1-pa0a1tγ1+γ0.


#### 3.1.1.q≠0,p=0,r=0

Equation (31) can be simplified to (32)a0Γγ0+1t-α+γ0Γ-α+γ0+1+a1Γγ1+1t-α+γ1Γ-α+γ1+1vx=-2a02qβt2γ0-3a0a1qβtγ0+γ1vx.According to homogenous balanced principle, we let (33)γ0-α=2γ0,γ1-α=γ0+γ1.Solving (33) yields (34)γ0=-α,γ1=γ1.Substituting (34) into (32), it can be reduced to (35)a0Γ1-αt-2αΓ1-2α+a1Γ1+γ1tγ1-αΓ1+γ1-αvx=-2a02qβt-2α-3a0a1qβtγ1-αvx.Balancing the power of v(x), one has (36)-3a0a1qβ=a1Γ1+γ1Γ1+γ1-α,-2a02qβ=a0Γ1-αΓ1-2α.Solving (36), we obtain the following result: (37)a0=-Γ1-α2βqΓ1-2α,a1=a1,Γ1-α2Γ1-2α=Γ1+γ13Γ1+γ1-α.Substituting (37) into (27), using solution (23) of subequation, we can obtain exact solution of (26) with δ=-2qβ as follows: (38)ux,t=-Γ1-α2βqΓ1-2αt-α+a1tγ1e±qC+x,where α∈(0,1/2)∪(1/2,1) and γ1 is the root of Γ(1-α)/2Γ(1-2α)=Γ(1+γ1)/3Γ(1+γ1-α).


#### 3.1.2.p=0,q≠0,r≠0

Equation (31) can be simplified to (39)a0Γγ0+1t-α+γ0Γ-α+γ0+1+a1Γγ1+1t-α+γ1Γ-α+γ1+1vx=β-2qa02t2γ0+ra12t2γ1-3βa1qa0tγ1+γ0vx.According to homogenous balanced principle, we let (40)γ0-α=γ1+γ0,γ1-α=2γ0=2γ1.Solving (40) yields (41)γ0=-α,γ1=-α.Substituting (41) into (39), it can be reduced to (42)a0Γ1-αΓ1-2α+a1Γ1-αΓ1-2αvx=β-2qa02+ra12-3βa1qa0vx.Balancing the power of v(x), one has (43)-3βa1qa0=a1Γ1-αΓ1-2α,β-2qa02+ra12=a0Γ1-αΓ1-2α.Solving (43), we obtain the following results: (44)a0=-Γ1-α3Γ-2α+1βq,a1=±Γ1-α3Γ-2α+1β-1qr,where α∈(0,1/2)∪(1/2,1),β≠0. Substituting (44) into (27), using solution (23) of subequation, we can obtain exact solution of (26) with δ=-2qβ as follows: (45)ux,t=-Γ1-α3Γ-2α+1βqt-α±Γ1-αe2xq-re2Cqe-qC+x6qβ-rΓ1-2αt-α,ux,t=-Γ1-α3Γ-2α+1βqt-α±Γ1-α-re2xq+e2Cqe-qC+x6qβ-rΓ1-2αt-α.By using Table1, taking parameters p,q,r as some particular values, many specific exact solutions of (26) can be got, parts of which are listed as follows.Whenp=0, q=-ω<0,r=1, then δ=2ωβ, v(x)=±sin⁡ω(C-x)/ω. Equation (45) can be reduced to (46)ux,t=Γ1-α3βωΓ1-2αt-α±sin⁡ωC-x+1,where ω is positive and C is an arbitrary constant.Whenp=0, q=-ω<0,r=-1, then δ=2ωβ, v(x)=-Icos⁡ω(C-x)/ω. Equation (45) can be reduced to (47)ux,t=Γ1-α3βωΓ1-2αt-α±cos⁡ωC-x+1,where I2=-1, ω is positive, and C is an arbitrary constant.If one letsC=0 or π/2,ω=1, then δ=2β,v(x)=±sin0-x=±sin(x), or v(x)=±sinπ/2-x=±cos(x). Therefore, solutions (46) and (47) are reduced to (48)ux,t=Γ1-α3βΓ1-2αt-α1±sin⁡x,ux,t=Γ1-α3βΓ1-2αt-α1±cos⁡x.Remark 3.
Equations (48) are just the solutions (3.37) and (3.39) given in reference [23], so we can say that our solutions are general, including many unreported solutions.
Whenp=0, q=ω>0,r=1, then δ=-2ωβ, v(x)=±sinhω(C-x)/ω. Equation (45) can be reduced to (49)u1x,t=Γ1-α3βωΓ1-2αt-α±sinh⁡ωC-x+1,where ω is positive and C is an arbitrary constant.
Whenp=0, q=ω>0,r=-1, then δ=-2ωβ, v(x)=±coshω(C-x)/ω. Equation (45) can be reduced to(50)ux,t=IΓ1-α3βωΓ1-2αt-α±cosh⁡ωC-x+1,where I2=-1, ω is positive, and C is an arbitrary constant.


#### 3.1.3.p≠0, q=0

Equation (31) can be simplified to (51)a0Γγ0+1t-α+γ0Γ-α+γ0+1+a1Γγ1+1t-α+γ1Γ-α+γ1+1vx=β2a0a1ptγ0+γ1+2ra12t2γ1+32βa12pt2γ1vx.Case 1 (r≠0).
According to homogenous balanced principle, we let(52)2γ1=γ1+γ0=γ0-α,2γ1=γ1-α.Solving (52) yields (53)γ0=-α,γ1=-α.Substituting (53) into (51), it can be reduced to (54)a0Γ1-αΓ1-2α+a1Γ1-αΓ1-2αvx=β2a0a1p+2ra12+32βa12pvx.Balancing the power of v(x), one has (55)32βa12p=a1Γ1-αΓ1-2α,β2a0a1p+2ra12=a0Γ1-αΓ1-2α.Solving (55), we obtain the following results: (56)a0=2rΓ1-α3βp2Γ1-2α,a1=2Γ1-α3βpΓ1-2α.Substituting (56) into (27), using solution (22) of subequation, we can obtain exact solution of (26) with δ=-2qβ=0 as follows: (57)ux,t=2rΓ1-α3βp2Γ1-2αt-α+Γ1-αC2p2-2Cp2+p2x2-4r6βp2Γ1-2αt-α=Γ1-α6βΓ1-2αt-αC-x2,where α∈(0,1/2)∪(1/2,1).Case 2 (r=0).
Equation (51) becomes (58)a0Γγ0+1t-α+γ0Γ-α+γ0+1+a1Γγ1+1t-α+γ1Γ-α+γ1+1vx=β2a0a1ptγ0+γ1+32βa12pt2γ1vx.According to homogenous balanced principle, we let (59)γ0+γ1=γ0-α,2γ1=γ1-α.Solving (59) yields (60)γ0=γ0,γ1=-α.Substituting (60) into (58), it can be reduced to (61)a0Γ1+γ0tγ0-αΓ1+γ0-α+a1Γ1-αt-2αΓ1-2αvx=β2a0a1ptγ0-α+32βa12pt-2αvx.Balancing the power of v(x), one has (62)32βa12p=a1Γ1-αΓ1-2α,β2a0a1p=a0Γ1+γ0Γ1+γ0-α.Solving (62), we obtain the following result: (63)a0=a0,a1=2Γ1-α3βpΓ1-2α,Γ1-α3Γ1-2α=Γ1+γ0Γ1+γ0-α.Substituting (63) into (27), using solution (22) of subequation, we can obtain exact solution of (26) with δ=-2qβ=0 as follows: (64)ux,t=a0tγ0+Γ1-α6βΓ1-2αt-αC-x2,where α∈(0,1/2)∪(1/2,1) and γ0 is the root of Γ(1-α)/3Γ(1-2α)=Γ(1+γ0)/Γ(1+γ0-α).


#### 3.1.4.p≠0, q≠0

According to (31) and homogenous balanced principle, we let (65)2γ1=γ1+γ0=2γ0=γ0-α,2γ1=γ1+γ0=γ1-α.Solving (65) yields (66)γ0=γ1=-α.Substituting (66) into (31), it can be reduced to (67)a0Γ1-αΓ1-2α+a1Γ1-αΓ1-2αvx=β2pa0a1-4qa02+2ra12+βa12pa1-2qa0vx.Balancing the power of v(x), one has (68)32βa1pa1-2qa0=a1Γ1-αΓ1-2α,β2pa0a1-4qa02+2ra12=a0Γ1-αΓ1-2α.Solving (68) we obtain the following results: (69)a0=ΩΓ1-αΓ1-2αβ,a1=2Γ1-α3qΩ+13βpΓ1-2α,where Ω=-p2+4qr±p4-4p2qr/3q(p2-4qr), p2-4qr≠0. Substituting (69) into (27), using solution (23) of subequation, we can obtain exact solution of (26) with δ=-2qβ as follows: (70)ux,t=Γ1-αp2ΔeC-xqΔ+4e-C-xqq+4pqΔt-α12Γ1-2αΔβpq3/2,ux,t=Γ1-αp2Δe-C-xqΔ+4eC-xqq+4pqΔt-α12Γ1-2αΔβpq3/2,where Δ=p2-4qr.WhenC=0,r=0,q=1,p=4, then v(x)=5/2cosh⁡x±3/2sinh⁡(x)-2. Equation (70) can be reduced to (71)ux,t=-Γ1-α12βΓ1-2αt-α5cosh⁡x±3sinh⁡x+4,ux,t=Γ1-α12βΓ1-2αt-α5cosh⁡x±3sinh⁡x-4,where α∈(0,1/2)∪(1/2,1).WhenC=0,r=0,q=-1,p=4, then v(x)=3/2Icos⁡x±5/2sin⁡(x)+2. Equation (70) can be reduced to (72)ux,t=-Γ1-α12βΓ1-2αt-α-5sin⁡x±3Icosx-4,ux,t=Γ1-α12βΓ1-2αt-α-5sin⁡x±3Icosx+4,where I2=-1.Ifα=1, substituting (27) into (26), using (20), we can obtain the following five families of exact solutions of (26).Family 1.
(73) u x , t = t - 1 2 q β + a 1 t - 3 / 2 e ± q C + x ,where q>0,δ=-2qβ.Family 2.
(74) u x , t = a 0 t - 1 / 3 - 1 6 β t - 1 C - x 2 ,where δ=0, C is an arbitrary constant.Family 3.
(75) u x , t = 1 3 β q t - 1 ± e 2 x q - r e 2 C q e - q C + x 6 q β - r t - 1 , u x , t = 1 3 β q t - 1 ± - r e 2 x q + e 2 C q e - q C + x 6 q β - r t - 1 ,where δ=-2qβ. Taking parameters q,r as suitable values, we can get the following special exact solutions of (26): (76)ux,t=13βωt-1-1±sinωC-x,ux,t=13βωt-1-1±cosωC-x,where δ=2ωβ, ω is positive, and C is an arbitrary constant. (77)ux,t=13βωt-1-1±sinh⁡ωC-x,ux,t=13βωt-1-1±cosh⁡1±ωC-x,where δ=-2ωβ, ω is positive, and C is an arbitrary constant.Family 4.
(78) u x , t = - 2 r 3 β p 2 t - 1 - C 2 p 2 - 2 C p 2 + p 2 x 2 - 4 r 6 β p 2 t - 1 = - 1 6 β t - 1 C - x 2 ,where δ=0 and C is an arbitrary constant.Family 5.
(79) u x , t = p 2 Δ e C - x q Δ + 4 e - C - x q q + 4 p q Δ 12 Δ β p q 3 / 2 t - 1 , u x , t = p 2 Δ e - C - x q Δ + 4 e C - x q q + 4 p q Δ 12 Δ β p q 3 / 2 t - 1 ,where Δ=p2-4qr,p≠0,q≠0,δ=-2qβ and C is an arbitrary constant.


### 3.2. Situation ofm=2

Takingm=2, we suppose that (26) has an exact solution as the following form: (80)ux,t=a0tγ0+a1tγ1vx+a2tγ2v2x,where v(x) satisfies (20) and γ0,γ1,γ2,a0,a1,a2≠0 are constants that can be determined later. Substituting (80) into (26), using (20), balancing the power on t of the reduced equation, we have (81)γ0=γ1=γ2=-α,δ=-8qβ.Substituting (81) and (80) into (26), balancing the power of v(x), yields (82)32βpa12-15βqa0a1+3βpa0a2+6βra1a2=a1Γ1-αΓ-2α+1,-6βqa12+6βra22+152βpa1a2-12βqa0a2=a2Γ1-αΓ-2α+1,-8qβa02+βra12+12βpa0a1+2βra0a2=a0Γ1-αΓ-2α+1,7βa2pa2-qa1=0.Solving the above algebraic equations, we have the following results.Case 1 (p=0,q=0, r≠0, α∈(0,1/2)∪(1/2,1)).
(83) a 0 = 3 β r a 1 2 Γ - 2 α + 1 2 Γ 1 - α , a 1 = a 1 , a 2 = Γ 1 - α 6 Γ - 2 α + 1 β r .
Substituting (83) into (27), using solution (21) of subequation, we can obtain exact solution of (26) with δ=-8qβ=0 as follows: (84)ux,t=3βra12Γ-2α+12Γ1-αt-α+a1±rx+Ct-α+Γ1-α±rx+C26Γ-2α+1βrt-α.Case 2 (q≠0, α∈(0,1/2)∪(1/2,1)).
(85) a 0 = 2 r Γ 1 - α 3 Δ β Γ 1 - 2 α , a 1 = 2 Γ 1 - α p 3 Δ β Γ 1 - 2 α , a 2 = 2 q Γ 1 - α 3 Δ β Γ 1 - 2 α , a 0 = - p 2 Γ 1 - α 6 Δ β q Γ 1 - 2 α , a 1 = - 2 Γ 1 - α p 3 β Δ Γ 1 - 2 α , a 2 = - 2 q Γ 1 - α 3 Δ β Γ 1 - 2 α , where Δ=p2-4qr≠0. Substituting (85) into (27), using solution (23) of subequation, we can obtain the exact solution of (26) with δ=-8qβ as follows: (86)ux,t=Γ1-α96q2ΔβΓ1-2αt-α4qeqC-x-Δe-qC-x2,ux,t=-Γ1-α96q2ΔβΓ1-2αt-α4qeqC-x+Δe-qC-x2.
WhenΔ=4q and q=-ω<0, then δ=8ωβ. Equation (86) can be reduced to (87)ux,t=Γ1-α6βωΓ1-2αt-αsin2⁡wC-x,ux,t=Γ1-α6βωΓ1-2αt-αcos2⁡wC-x,where ω is positive and C is an arbitrary constant.
WhenΔ=4q and q=ω>0, then δ=-8ωβ. Equation (86) can be reduced to (88)ux,t=Γ1-α6βωΓ1-2αt-αsinh2⁡wC-x,ux,t=-Γ1-α6βωΓ1-2αt-αcosh2⁡wC-x,where ω is positive and C is an arbitrary constant.
Ifα=1, substituting (80) into (26), using (20), we can obtain the following exact solutions of (26): (89)ux,t=-196q2Δβt-14qeqC-x-Δe-qC-x2,ux,t=196q2Δβt-14qeqC-x+Δe-qC-x2.When Δ=4q and q=-ω<0, then δ=8ωβ. Equation (89) can be reduced to (90)ux,t=-Γ1-α6βωΓ1-2αt-1sin2⁡wC-x,ux,t=-Γ1-α6βωΓ1-2αt-1cos2⁡wC-x,where ω is positive and C is an arbitrary constant.
WhenΔ=4q and q=ω>0, then δ=-8ωβ. Equation (89) can be reduced to (91)ux,t=-16βωt-1sinh2⁡wC-x,ux,t=16βωt-1cosh2⁡wC-x,where ω is positive and C is an arbitrary constant.


## 3.1. Situation ofm=1

Takingm=1, we first suppose that (26) has an exact solution as the following form: (27)ux,t=a0tγ0+a1tγ1vx,where v(x) satisfies (20) and γ0,γ1,a0,a1≠0 are undetermined constants that can be determined later. Substituting (27) into (26), using (20), it can be reduced to (28)a0Γγ0+1t-α+γ0Γ-α+γ0+1+a1Γγ1+1t-α+γ1Γ-α+γ1+1vx=t2γ1βra12+12tγ1+γ0βpa0a1+t2γ0δa0232t2γ1βpa12+tγ1+γ0βqa0a1+2tγ1+γ0δa0a1vx+2t2γ1βqa12+t2γ1δa12vx2.According to homogenous balanced principle, we let (29)2t2γ1βqa12+t2γ1δa12=0.Solving (29) yields (30)δ=-2qβ.Substituting (30) into (28), it can be reduced to (31)a0Γγ0+1t-α+γ0Γ-α+γ0+1+a1Γγ1+1t-α+γ1Γ-α+γ1+1vx=-12β-3pa12t2γ1+6a1tγ1+γ0qa0vx-12β4qa02t2γ0-2ra12t2γ1-pa0a1tγ1+γ0.


### 3.1.1.q≠0,p=0,r=0

Equation (31) can be simplified to (32)a0Γγ0+1t-α+γ0Γ-α+γ0+1+a1Γγ1+1t-α+γ1Γ-α+γ1+1vx=-2a02qβt2γ0-3a0a1qβtγ0+γ1vx.According to homogenous balanced principle, we let (33)γ0-α=2γ0,γ1-α=γ0+γ1.Solving (33) yields (34)γ0=-α,γ1=γ1.Substituting (34) into (32), it can be reduced to (35)a0Γ1-αt-2αΓ1-2α+a1Γ1+γ1tγ1-αΓ1+γ1-αvx=-2a02qβt-2α-3a0a1qβtγ1-αvx.Balancing the power of v(x), one has (36)-3a0a1qβ=a1Γ1+γ1Γ1+γ1-α,-2a02qβ=a0Γ1-αΓ1-2α.Solving (36), we obtain the following result: (37)a0=-Γ1-α2βqΓ1-2α,a1=a1,Γ1-α2Γ1-2α=Γ1+γ13Γ1+γ1-α.Substituting (37) into (27), using solution (23) of subequation, we can obtain exact solution of (26) with δ=-2qβ as follows: (38)ux,t=-Γ1-α2βqΓ1-2αt-α+a1tγ1e±qC+x,where α∈(0,1/2)∪(1/2,1) and γ1 is the root of Γ(1-α)/2Γ(1-2α)=Γ(1+γ1)/3Γ(1+γ1-α).


### 3.1.2.p=0,q≠0,r≠0

Equation (31) can be simplified to (39)a0Γγ0+1t-α+γ0Γ-α+γ0+1+a1Γγ1+1t-α+γ1Γ-α+γ1+1vx=β-2qa02t2γ0+ra12t2γ1-3βa1qa0tγ1+γ0vx.According to homogenous balanced principle, we let (40)γ0-α=γ1+γ0,γ1-α=2γ0=2γ1.Solving (40) yields (41)γ0=-α,γ1=-α.Substituting (41) into (39), it can be reduced to (42)a0Γ1-αΓ1-2α+a1Γ1-αΓ1-2αvx=β-2qa02+ra12-3βa1qa0vx.Balancing the power of v(x), one has (43)-3βa1qa0=a1Γ1-αΓ1-2α,β-2qa02+ra12=a0Γ1-αΓ1-2α.Solving (43), we obtain the following results: (44)a0=-Γ1-α3Γ-2α+1βq,a1=±Γ1-α3Γ-2α+1β-1qr,where α∈(0,1/2)∪(1/2,1),β≠0. Substituting (44) into (27), using solution (23) of subequation, we can obtain exact solution of (26) with δ=-2qβ as follows: (45)ux,t=-Γ1-α3Γ-2α+1βqt-α±Γ1-αe2xq-re2Cqe-qC+x6qβ-rΓ1-2αt-α,ux,t=-Γ1-α3Γ-2α+1βqt-α±Γ1-α-re2xq+e2Cqe-qC+x6qβ-rΓ1-2αt-α.By using Table1, taking parameters p,q,r as some particular values, many specific exact solutions of (26) can be got, parts of which are listed as follows.Whenp=0, q=-ω<0,r=1, then δ=2ωβ, v(x)=±sin⁡ω(C-x)/ω. Equation (45) can be reduced to (46)ux,t=Γ1-α3βωΓ1-2αt-α±sin⁡ωC-x+1,where ω is positive and C is an arbitrary constant.Whenp=0, q=-ω<0,r=-1, then δ=2ωβ, v(x)=-Icos⁡ω(C-x)/ω. Equation (45) can be reduced to (47)ux,t=Γ1-α3βωΓ1-2αt-α±cos⁡ωC-x+1,where I2=-1, ω is positive, and C is an arbitrary constant.If one letsC=0 or π/2,ω=1, then δ=2β,v(x)=±sin0-x=±sin(x), or v(x)=±sinπ/2-x=±cos(x). Therefore, solutions (46) and (47) are reduced to (48)ux,t=Γ1-α3βΓ1-2αt-α1±sin⁡x,ux,t=Γ1-α3βΓ1-2αt-α1±cos⁡x.Remark 3.
Equations (48) are just the solutions (3.37) and (3.39) given in reference [23], so we can say that our solutions are general, including many unreported solutions.
Whenp=0, q=ω>0,r=1, then δ=-2ωβ, v(x)=±sinhω(C-x)/ω. Equation (45) can be reduced to (49)u1x,t=Γ1-α3βωΓ1-2αt-α±sinh⁡ωC-x+1,where ω is positive and C is an arbitrary constant.
Whenp=0, q=ω>0,r=-1, then δ=-2ωβ, v(x)=±coshω(C-x)/ω. Equation (45) can be reduced to(50)ux,t=IΓ1-α3βωΓ1-2αt-α±cosh⁡ωC-x+1,where I2=-1, ω is positive, and C is an arbitrary constant.


### 3.1.3.p≠0, q=0

Equation (31) can be simplified to (51)a0Γγ0+1t-α+γ0Γ-α+γ0+1+a1Γγ1+1t-α+γ1Γ-α+γ1+1vx=β2a0a1ptγ0+γ1+2ra12t2γ1+32βa12pt2γ1vx.Case 1 (r≠0).
According to homogenous balanced principle, we let(52)2γ1=γ1+γ0=γ0-α,2γ1=γ1-α.Solving (52) yields (53)γ0=-α,γ1=-α.Substituting (53) into (51), it can be reduced to (54)a0Γ1-αΓ1-2α+a1Γ1-αΓ1-2αvx=β2a0a1p+2ra12+32βa12pvx.Balancing the power of v(x), one has (55)32βa12p=a1Γ1-αΓ1-2α,β2a0a1p+2ra12=a0Γ1-αΓ1-2α.Solving (55), we obtain the following results: (56)a0=2rΓ1-α3βp2Γ1-2α,a1=2Γ1-α3βpΓ1-2α.Substituting (56) into (27), using solution (22) of subequation, we can obtain exact solution of (26) with δ=-2qβ=0 as follows: (57)ux,t=2rΓ1-α3βp2Γ1-2αt-α+Γ1-αC2p2-2Cp2+p2x2-4r6βp2Γ1-2αt-α=Γ1-α6βΓ1-2αt-αC-x2,where α∈(0,1/2)∪(1/2,1).Case 2 (r=0).
Equation (51) becomes (58)a0Γγ0+1t-α+γ0Γ-α+γ0+1+a1Γγ1+1t-α+γ1Γ-α+γ1+1vx=β2a0a1ptγ0+γ1+32βa12pt2γ1vx.According to homogenous balanced principle, we let (59)γ0+γ1=γ0-α,2γ1=γ1-α.Solving (59) yields (60)γ0=γ0,γ1=-α.Substituting (60) into (58), it can be reduced to (61)a0Γ1+γ0tγ0-αΓ1+γ0-α+a1Γ1-αt-2αΓ1-2αvx=β2a0a1ptγ0-α+32βa12pt-2αvx.Balancing the power of v(x), one has (62)32βa12p=a1Γ1-αΓ1-2α,β2a0a1p=a0Γ1+γ0Γ1+γ0-α.Solving (62), we obtain the following result: (63)a0=a0,a1=2Γ1-α3βpΓ1-2α,Γ1-α3Γ1-2α=Γ1+γ0Γ1+γ0-α.Substituting (63) into (27), using solution (22) of subequation, we can obtain exact solution of (26) with δ=-2qβ=0 as follows: (64)ux,t=a0tγ0+Γ1-α6βΓ1-2αt-αC-x2,where α∈(0,1/2)∪(1/2,1) and γ0 is the root of Γ(1-α)/3Γ(1-2α)=Γ(1+γ0)/Γ(1+γ0-α).


### 3.1.4.p≠0, q≠0

According to (31) and homogenous balanced principle, we let (65)2γ1=γ1+γ0=2γ0=γ0-α,2γ1=γ1+γ0=γ1-α.Solving (65) yields (66)γ0=γ1=-α.Substituting (66) into (31), it can be reduced to (67)a0Γ1-αΓ1-2α+a1Γ1-αΓ1-2αvx=β2pa0a1-4qa02+2ra12+βa12pa1-2qa0vx.Balancing the power of v(x), one has (68)32βa1pa1-2qa0=a1Γ1-αΓ1-2α,β2pa0a1-4qa02+2ra12=a0Γ1-αΓ1-2α.Solving (68) we obtain the following results: (69)a0=ΩΓ1-αΓ1-2αβ,a1=2Γ1-α3qΩ+13βpΓ1-2α,where Ω=-p2+4qr±p4-4p2qr/3q(p2-4qr), p2-4qr≠0. Substituting (69) into (27), using solution (23) of subequation, we can obtain exact solution of (26) with δ=-2qβ as follows: (70)ux,t=Γ1-αp2ΔeC-xqΔ+4e-C-xqq+4pqΔt-α12Γ1-2αΔβpq3/2,ux,t=Γ1-αp2Δe-C-xqΔ+4eC-xqq+4pqΔt-α12Γ1-2αΔβpq3/2,where Δ=p2-4qr.WhenC=0,r=0,q=1,p=4, then v(x)=5/2cosh⁡x±3/2sinh⁡(x)-2. Equation (70) can be reduced to (71)ux,t=-Γ1-α12βΓ1-2αt-α5cosh⁡x±3sinh⁡x+4,ux,t=Γ1-α12βΓ1-2αt-α5cosh⁡x±3sinh⁡x-4,where α∈(0,1/2)∪(1/2,1).WhenC=0,r=0,q=-1,p=4, then v(x)=3/2Icos⁡x±5/2sin⁡(x)+2. Equation (70) can be reduced to (72)ux,t=-Γ1-α12βΓ1-2αt-α-5sin⁡x±3Icosx-4,ux,t=Γ1-α12βΓ1-2αt-α-5sin⁡x±3Icosx+4,where I2=-1.Ifα=1, substituting (27) into (26), using (20), we can obtain the following five families of exact solutions of (26).Family 1.
(73) u x , t = t - 1 2 q β + a 1 t - 3 / 2 e ± q C + x ,where q>0,δ=-2qβ.Family 2.
(74) u x , t = a 0 t - 1 / 3 - 1 6 β t - 1 C - x 2 ,where δ=0, C is an arbitrary constant.Family 3.
(75) u x , t = 1 3 β q t - 1 ± e 2 x q - r e 2 C q e - q C + x 6 q β - r t - 1 , u x , t = 1 3 β q t - 1 ± - r e 2 x q + e 2 C q e - q C + x 6 q β - r t - 1 ,where δ=-2qβ. Taking parameters q,r as suitable values, we can get the following special exact solutions of (26): (76)ux,t=13βωt-1-1±sinωC-x,ux,t=13βωt-1-1±cosωC-x,where δ=2ωβ, ω is positive, and C is an arbitrary constant. (77)ux,t=13βωt-1-1±sinh⁡ωC-x,ux,t=13βωt-1-1±cosh⁡1±ωC-x,where δ=-2ωβ, ω is positive, and C is an arbitrary constant.Family 4.
(78) u x , t = - 2 r 3 β p 2 t - 1 - C 2 p 2 - 2 C p 2 + p 2 x 2 - 4 r 6 β p 2 t - 1 = - 1 6 β t - 1 C - x 2 ,where δ=0 and C is an arbitrary constant.Family 5.
(79) u x , t = p 2 Δ e C - x q Δ + 4 e - C - x q q + 4 p q Δ 12 Δ β p q 3 / 2 t - 1 , u x , t = p 2 Δ e - C - x q Δ + 4 e C - x q q + 4 p q Δ 12 Δ β p q 3 / 2 t - 1 ,where Δ=p2-4qr,p≠0,q≠0,δ=-2qβ and C is an arbitrary constant.


## 3.1.1.q≠0,p=0,r=0

Equation (31) can be simplified to (32)a0Γγ0+1t-α+γ0Γ-α+γ0+1+a1Γγ1+1t-α+γ1Γ-α+γ1+1vx=-2a02qβt2γ0-3a0a1qβtγ0+γ1vx.According to homogenous balanced principle, we let (33)γ0-α=2γ0,γ1-α=γ0+γ1.Solving (33) yields (34)γ0=-α,γ1=γ1.Substituting (34) into (32), it can be reduced to (35)a0Γ1-αt-2αΓ1-2α+a1Γ1+γ1tγ1-αΓ1+γ1-αvx=-2a02qβt-2α-3a0a1qβtγ1-αvx.Balancing the power of v(x), one has (36)-3a0a1qβ=a1Γ1+γ1Γ1+γ1-α,-2a02qβ=a0Γ1-αΓ1-2α.Solving (36), we obtain the following result: (37)a0=-Γ1-α2βqΓ1-2α,a1=a1,Γ1-α2Γ1-2α=Γ1+γ13Γ1+γ1-α.Substituting (37) into (27), using solution (23) of subequation, we can obtain exact solution of (26) with δ=-2qβ as follows: (38)ux,t=-Γ1-α2βqΓ1-2αt-α+a1tγ1e±qC+x,where α∈(0,1/2)∪(1/2,1) and γ1 is the root of Γ(1-α)/2Γ(1-2α)=Γ(1+γ1)/3Γ(1+γ1-α).


## 3.1.2.p=0,q≠0,r≠0

Equation (31) can be simplified to (39)a0Γγ0+1t-α+γ0Γ-α+γ0+1+a1Γγ1+1t-α+γ1Γ-α+γ1+1vx=β-2qa02t2γ0+ra12t2γ1-3βa1qa0tγ1+γ0vx.According to homogenous balanced principle, we let (40)γ0-α=γ1+γ0,γ1-α=2γ0=2γ1.Solving (40) yields (41)γ0=-α,γ1=-α.Substituting (41) into (39), it can be reduced to (42)a0Γ1-αΓ1-2α+a1Γ1-αΓ1-2αvx=β-2qa02+ra12-3βa1qa0vx.Balancing the power of v(x), one has (43)-3βa1qa0=a1Γ1-αΓ1-2α,β-2qa02+ra12=a0Γ1-αΓ1-2α.Solving (43), we obtain the following results: (44)a0=-Γ1-α3Γ-2α+1βq,a1=±Γ1-α3Γ-2α+1β-1qr,where α∈(0,1/2)∪(1/2,1),β≠0. Substituting (44) into (27), using solution (23) of subequation, we can obtain exact solution of (26) with δ=-2qβ as follows: (45)ux,t=-Γ1-α3Γ-2α+1βqt-α±Γ1-αe2xq-re2Cqe-qC+x6qβ-rΓ1-2αt-α,ux,t=-Γ1-α3Γ-2α+1βqt-α±Γ1-α-re2xq+e2Cqe-qC+x6qβ-rΓ1-2αt-α.By using Table1, taking parameters p,q,r as some particular values, many specific exact solutions of (26) can be got, parts of which are listed as follows.Whenp=0, q=-ω<0,r=1, then δ=2ωβ, v(x)=±sin⁡ω(C-x)/ω. Equation (45) can be reduced to (46)ux,t=Γ1-α3βωΓ1-2αt-α±sin⁡ωC-x+1,where ω is positive and C is an arbitrary constant.Whenp=0, q=-ω<0,r=-1, then δ=2ωβ, v(x)=-Icos⁡ω(C-x)/ω. Equation (45) can be reduced to (47)ux,t=Γ1-α3βωΓ1-2αt-α±cos⁡ωC-x+1,where I2=-1, ω is positive, and C is an arbitrary constant.If one letsC=0 or π/2,ω=1, then δ=2β,v(x)=±sin0-x=±sin(x), or v(x)=±sinπ/2-x=±cos(x). Therefore, solutions (46) and (47) are reduced to (48)ux,t=Γ1-α3βΓ1-2αt-α1±sin⁡x,ux,t=Γ1-α3βΓ1-2αt-α1±cos⁡x.Remark 3.
Equations (48) are just the solutions (3.37) and (3.39) given in reference [23], so we can say that our solutions are general, including many unreported solutions.
Whenp=0, q=ω>0,r=1, then δ=-2ωβ, v(x)=±sinhω(C-x)/ω. Equation (45) can be reduced to (49)u1x,t=Γ1-α3βωΓ1-2αt-α±sinh⁡ωC-x+1,where ω is positive and C is an arbitrary constant.
Whenp=0, q=ω>0,r=-1, then δ=-2ωβ, v(x)=±coshω(C-x)/ω. Equation (45) can be reduced to(50)ux,t=IΓ1-α3βωΓ1-2αt-α±cosh⁡ωC-x+1,where I2=-1, ω is positive, and C is an arbitrary constant.


## 3.1.3.p≠0, q=0

Equation (31) can be simplified to (51)a0Γγ0+1t-α+γ0Γ-α+γ0+1+a1Γγ1+1t-α+γ1Γ-α+γ1+1vx=β2a0a1ptγ0+γ1+2ra12t2γ1+32βa12pt2γ1vx.Case 1 (r≠0).
According to homogenous balanced principle, we let(52)2γ1=γ1+γ0=γ0-α,2γ1=γ1-α.Solving (52) yields (53)γ0=-α,γ1=-α.Substituting (53) into (51), it can be reduced to (54)a0Γ1-αΓ1-2α+a1Γ1-αΓ1-2αvx=β2a0a1p+2ra12+32βa12pvx.Balancing the power of v(x), one has (55)32βa12p=a1Γ1-αΓ1-2α,β2a0a1p+2ra12=a0Γ1-αΓ1-2α.Solving (55), we obtain the following results: (56)a0=2rΓ1-α3βp2Γ1-2α,a1=2Γ1-α3βpΓ1-2α.Substituting (56) into (27), using solution (22) of subequation, we can obtain exact solution of (26) with δ=-2qβ=0 as follows: (57)ux,t=2rΓ1-α3βp2Γ1-2αt-α+Γ1-αC2p2-2Cp2+p2x2-4r6βp2Γ1-2αt-α=Γ1-α6βΓ1-2αt-αC-x2,where α∈(0,1/2)∪(1/2,1).Case 2 (r=0).
Equation (51) becomes (58)a0Γγ0+1t-α+γ0Γ-α+γ0+1+a1Γγ1+1t-α+γ1Γ-α+γ1+1vx=β2a0a1ptγ0+γ1+32βa12pt2γ1vx.According to homogenous balanced principle, we let (59)γ0+γ1=γ0-α,2γ1=γ1-α.Solving (59) yields (60)γ0=γ0,γ1=-α.Substituting (60) into (58), it can be reduced to (61)a0Γ1+γ0tγ0-αΓ1+γ0-α+a1Γ1-αt-2αΓ1-2αvx=β2a0a1ptγ0-α+32βa12pt-2αvx.Balancing the power of v(x), one has (62)32βa12p=a1Γ1-αΓ1-2α,β2a0a1p=a0Γ1+γ0Γ1+γ0-α.Solving (62), we obtain the following result: (63)a0=a0,a1=2Γ1-α3βpΓ1-2α,Γ1-α3Γ1-2α=Γ1+γ0Γ1+γ0-α.Substituting (63) into (27), using solution (22) of subequation, we can obtain exact solution of (26) with δ=-2qβ=0 as follows: (64)ux,t=a0tγ0+Γ1-α6βΓ1-2αt-αC-x2,where α∈(0,1/2)∪(1/2,1) and γ0 is the root of Γ(1-α)/3Γ(1-2α)=Γ(1+γ0)/Γ(1+γ0-α).


## 3.1.4.p≠0, q≠0

According to (31) and homogenous balanced principle, we let (65)2γ1=γ1+γ0=2γ0=γ0-α,2γ1=γ1+γ0=γ1-α.Solving (65) yields (66)γ0=γ1=-α.Substituting (66) into (31), it can be reduced to (67)a0Γ1-αΓ1-2α+a1Γ1-αΓ1-2αvx=β2pa0a1-4qa02+2ra12+βa12pa1-2qa0vx.Balancing the power of v(x), one has (68)32βa1pa1-2qa0=a1Γ1-αΓ1-2α,β2pa0a1-4qa02+2ra12=a0Γ1-αΓ1-2α.Solving (68) we obtain the following results: (69)a0=ΩΓ1-αΓ1-2αβ,a1=2Γ1-α3qΩ+13βpΓ1-2α,where Ω=-p2+4qr±p4-4p2qr/3q(p2-4qr), p2-4qr≠0. Substituting (69) into (27), using solution (23) of subequation, we can obtain exact solution of (26) with δ=-2qβ as follows: (70)ux,t=Γ1-αp2ΔeC-xqΔ+4e-C-xqq+4pqΔt-α12Γ1-2αΔβpq3/2,ux,t=Γ1-αp2Δe-C-xqΔ+4eC-xqq+4pqΔt-α12Γ1-2αΔβpq3/2,where Δ=p2-4qr.WhenC=0,r=0,q=1,p=4, then v(x)=5/2cosh⁡x±3/2sinh⁡(x)-2. Equation (70) can be reduced to (71)ux,t=-Γ1-α12βΓ1-2αt-α5cosh⁡x±3sinh⁡x+4,ux,t=Γ1-α12βΓ1-2αt-α5cosh⁡x±3sinh⁡x-4,where α∈(0,1/2)∪(1/2,1).WhenC=0,r=0,q=-1,p=4, then v(x)=3/2Icos⁡x±5/2sin⁡(x)+2. Equation (70) can be reduced to (72)ux,t=-Γ1-α12βΓ1-2αt-α-5sin⁡x±3Icosx-4,ux,t=Γ1-α12βΓ1-2αt-α-5sin⁡x±3Icosx+4,where I2=-1.Ifα=1, substituting (27) into (26), using (20), we can obtain the following five families of exact solutions of (26).Family 1.
(73) u x , t = t - 1 2 q β + a 1 t - 3 / 2 e ± q C + x ,where q>0,δ=-2qβ.Family 2.
(74) u x , t = a 0 t - 1 / 3 - 1 6 β t - 1 C - x 2 ,where δ=0, C is an arbitrary constant.Family 3.
(75) u x , t = 1 3 β q t - 1 ± e 2 x q - r e 2 C q e - q C + x 6 q β - r t - 1 , u x , t = 1 3 β q t - 1 ± - r e 2 x q + e 2 C q e - q C + x 6 q β - r t - 1 ,where δ=-2qβ. Taking parameters q,r as suitable values, we can get the following special exact solutions of (26): (76)ux,t=13βωt-1-1±sinωC-x,ux,t=13βωt-1-1±cosωC-x,where δ=2ωβ, ω is positive, and C is an arbitrary constant. (77)ux,t=13βωt-1-1±sinh⁡ωC-x,ux,t=13βωt-1-1±cosh⁡1±ωC-x,where δ=-2ωβ, ω is positive, and C is an arbitrary constant.Family 4.
(78) u x , t = - 2 r 3 β p 2 t - 1 - C 2 p 2 - 2 C p 2 + p 2 x 2 - 4 r 6 β p 2 t - 1 = - 1 6 β t - 1 C - x 2 ,where δ=0 and C is an arbitrary constant.Family 5.
(79) u x , t = p 2 Δ e C - x q Δ + 4 e - C - x q q + 4 p q Δ 12 Δ β p q 3 / 2 t - 1 , u x , t = p 2 Δ e - C - x q Δ + 4 e C - x q q + 4 p q Δ 12 Δ β p q 3 / 2 t - 1 ,where Δ=p2-4qr,p≠0,q≠0,δ=-2qβ and C is an arbitrary constant.


## 3.2. Situation ofm=2

Takingm=2, we suppose that (26) has an exact solution as the following form: (80)ux,t=a0tγ0+a1tγ1vx+a2tγ2v2x,where v(x) satisfies (20) and γ0,γ1,γ2,a0,a1,a2≠0 are constants that can be determined later. Substituting (80) into (26), using (20), balancing the power on t of the reduced equation, we have (81)γ0=γ1=γ2=-α,δ=-8qβ.Substituting (81) and (80) into (26), balancing the power of v(x), yields (82)32βpa12-15βqa0a1+3βpa0a2+6βra1a2=a1Γ1-αΓ-2α+1,-6βqa12+6βra22+152βpa1a2-12βqa0a2=a2Γ1-αΓ-2α+1,-8qβa02+βra12+12βpa0a1+2βra0a2=a0Γ1-αΓ-2α+1,7βa2pa2-qa1=0.Solving the above algebraic equations, we have the following results.Case 1 (p=0,q=0, r≠0, α∈(0,1/2)∪(1/2,1)).
(83) a 0 = 3 β r a 1 2 Γ - 2 α + 1 2 Γ 1 - α , a 1 = a 1 , a 2 = Γ 1 - α 6 Γ - 2 α + 1 β r .
Substituting (83) into (27), using solution (21) of subequation, we can obtain exact solution of (26) with δ=-8qβ=0 as follows: (84)ux,t=3βra12Γ-2α+12Γ1-αt-α+a1±rx+Ct-α+Γ1-α±rx+C26Γ-2α+1βrt-α.Case 2 (q≠0, α∈(0,1/2)∪(1/2,1)).
(85) a 0 = 2 r Γ 1 - α 3 Δ β Γ 1 - 2 α , a 1 = 2 Γ 1 - α p 3 Δ β Γ 1 - 2 α , a 2 = 2 q Γ 1 - α 3 Δ β Γ 1 - 2 α , a 0 = - p 2 Γ 1 - α 6 Δ β q Γ 1 - 2 α , a 1 = - 2 Γ 1 - α p 3 β Δ Γ 1 - 2 α , a 2 = - 2 q Γ 1 - α 3 Δ β Γ 1 - 2 α , where Δ=p2-4qr≠0. Substituting (85) into (27), using solution (23) of subequation, we can obtain the exact solution of (26) with δ=-8qβ as follows: (86)ux,t=Γ1-α96q2ΔβΓ1-2αt-α4qeqC-x-Δe-qC-x2,ux,t=-Γ1-α96q2ΔβΓ1-2αt-α4qeqC-x+Δe-qC-x2.
WhenΔ=4q and q=-ω<0, then δ=8ωβ. Equation (86) can be reduced to (87)ux,t=Γ1-α6βωΓ1-2αt-αsin2⁡wC-x,ux,t=Γ1-α6βωΓ1-2αt-αcos2⁡wC-x,where ω is positive and C is an arbitrary constant.
WhenΔ=4q and q=ω>0, then δ=-8ωβ. Equation (86) can be reduced to (88)ux,t=Γ1-α6βωΓ1-2αt-αsinh2⁡wC-x,ux,t=-Γ1-α6βωΓ1-2αt-αcosh2⁡wC-x,where ω is positive and C is an arbitrary constant.
Ifα=1, substituting (80) into (26), using (20), we can obtain the following exact solutions of (26): (89)ux,t=-196q2Δβt-14qeqC-x-Δe-qC-x2,ux,t=196q2Δβt-14qeqC-x+Δe-qC-x2.When Δ=4q and q=-ω<0, then δ=8ωβ. Equation (89) can be reduced to (90)ux,t=-Γ1-α6βωΓ1-2αt-1sin2⁡wC-x,ux,t=-Γ1-α6βωΓ1-2αt-1cos2⁡wC-x,where ω is positive and C is an arbitrary constant.
WhenΔ=4q and q=ω>0, then δ=-8ωβ. Equation (89) can be reduced to (91)ux,t=-16βωt-1sinh2⁡wC-x,ux,t=16βωt-1cosh2⁡wC-x,where ω is positive and C is an arbitrary constant.


## 4. Conclusions

In this work, we proved that the fractional Leibniz rule that appeared in many references does not hold under Riemann-Liouville definition and Caputo definition of fractional derivative. Based on the homogenous balanced principle, we introduced a general method for investigating exact solution of nonlinear time-fractional PDEs. By using this method called improved separation variable function-expansion method, we studied a nonlinear time-fractional PDE with diffusion term. Some new results are obtained.Firstly, compared with Rui’s method [23], it is easy to find that our method is more general. All solutions given in reference [23] can be obtained by taking special parameters in our results. For example, taking ω=1,C=0, our solutions (46) and (47) become solutions (3.40) and (3.41) in [23], respectively. Taking q=1,C=0, our solutions (38) become solution (3.47) in [23]. Taking r=1,C=0, our solutions (84) become solution (3.60) in [23]. Other solutions obtained in our work are new, such as solutions (49), (50), (57), (64), (71), (72), (87), and (88) which are not reported in related references. In addition, we should adopt different subequation for other time-fractional PDEs, such as v′(x)=r+pv2(x)+qv4(x). Finally, our method is simple and efficient for application without any skill.According to symmetrical characteristic, this method also can be used to investigate exact solutions of space fractional PDEs which are formed as follows:(92)CDxαu=Fu,∂u∂t,∂2u∂t2,…,∂nu∂tn.



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