# Exact Solutions with Variable Coefficient Function Forms for Conformable Fractional Partial Differential Equations by an Auxiliary Equation Method

Authors: Fanwei Meng; Qinghua Feng
Journal: Advances in Mathematical Physics
 (2018)

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## Abstract

In this paper, an auxiliary equation method is introduced for seeking exact solutions expressed in variable coefficient function forms for fractional partial differential equations, where the concerned fractional derivative is defined by the conformable fractional derivative. By the use of certain fractional transformation, the fractional derivative in the equations can be converted into integer order case with respect to a new variable. As for applications, we apply this method to the time fractional two-dimensional Boussinesq equation and the space-time fractional (2+1)-dimensional breaking soliton equation. As a result, some exact solutions including variable coefficient function solutions as well as solitary wave solutions for the two equations are found.


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## Body

## 1. Introduction

Fractional differential equations are the generalizations of classical differential equations with integer order derivatives. It is well known that fractional partial differential equations have proved to be very useful in many research fields such as physics, mathematical biology, engineering, fluid mechanics, plasma physics, optical fibers, neural physics, solid state physics, viscoelasticity, electromagnetism, electrochemistry, signal processing, control theory, chaos, finance, and fractal dynamics. In the last few decades, research on various aspects for fractional partial differential equations has received more and more attention by many authors, such as the oscillation [1, 2] and existence and uniqueness [3–5]. In order to better understand the physical process described by fractional partial differential equations, one usually needs to obtain exact solutions or numerical solutions for fractional partial differential equations. And so far a lot of effective methods have been developed and used by many authors. For example, these methods include the finite difference method [6], the (G′/G) method [7–10], the Jacobi elliptic function method [11], the projective Riccati equation method [12], the modified Kudryashov method [13–19], the exp method [20–25], the ansatz method [26], the first integral method [27–29], and the subequation method [30–34]. Based on these methods, a lot of fractional partial differential equations have been investigated.In this paper, we develop an auxiliary equation method for solving fractional partial differential equations, where the fractional derivative is defined in the sense of the conformable fractional derivative. Our aim is to seek exact solutions with variable coefficient function forms for some certain fractional partial differential equations.The conformable fractional derivative is defined by [35](1)Dtαft=limε→0⁡ft+εt1-α-ftε.The following properties for the conformable fractional derivative are well known, which can be easily proved due to the definition of the conformable fractional derivative. (i)Dtα[af(t)+bg(t)]=aDαf(t)+bDαg(t).(ii)Dtα(tγ)=γtγ-α.(iii)Dtα[f(t)g(t)]=f(t)Dαg(t)+g(t)Dαf(t).(iv)DtαC=0, where C is a constant.(v)Dtαf[g(t)]=fg′[g(t)]Dtαg(t).(vi)Dtα(f/g)(t)=(gtDαft-ftDαgt)/g2(t).(vii)Dtαf(t)=t1-αf′(t).The paper is organized as follows. In Section2, we propose the description of the auxiliary equation method for seeking exact solutions with variable coefficient function forms for fractional partial differential equations. Then in Section 3, we apply the method to solve the time fractional two-dimensional Boussinesq equation and the space-time fractional (2+1)-dimensional breaking soliton equation. In Section 4, we present some concluding statements.


## 2. Description of the Auxiliary Equation Method

In this section, we give the description of the auxiliary equation method for solving fractional partial differential equations.Suppose that a fractional partial differential equation in the independent variablest,x1,x2,…,xn is given by (2)Pu,ut,ux1,…,uxi,…,Dtαu,Dx1γu…,Dxiβu,…=0,where u is an unknown function, the orders of the fractional derivatives are, for example, α,β,γ∈(0,1], and P is a polynomial in u and its various partial derivatives including fractional derivatives. Without loss of generality, next we may assume that the fractional partial derivatives are related to the variables t,xi, while the other variables are related to integer order derivatives.Step 1.
For those variables involving fractional derivatives, fulfil corresponding fractional transformations so that the fractional partial derivatives can be converted into integer order partial derivatives with respect to new variables.
Taking the expressionsDtαu and Dxiβu, for example, one can use two fractional transformations T=c(tα/α) and Xi=k(xiβ/β) and denote u(t,x1,…,xi,…,xn)=u~(T,x1,…,Xi,…,xn). Then due to the properties (ii) and (v) of the conformable fractional derivative, one can obtain that Dtαu=(∂u~/∂T)DtαT=(∂u~/∂T)c=cu~T, Dxiβu=(∂u~/∂Xi)DxiβXi=(∂u~/∂Xi)k=ku~Xi. Therefore, the original fractional partial differential equation can be converted into another partial differential equation of integer order as follows: (3)P~u~,u~T,u~x1,…,u~Xi,…=0.Step 2.
Suppose that the solution of (3) can be expressed by a polynomial in ψ′(ξ)/ψ(ξ) as follows: (4)u~T,x1,…,Xi,…,xn=∑i=0maiT,x1,…,Xi,…,xnψ′ξψξi,where ξ=ξ(T,x1,…,Xi,…,xn),am(T,x1,…,Xi,…,xn), am-1(T,x1,…,Xi,…,xn),…,a0(T,x1,…,Xi,…,xn) are all unknown functions to be determined later with am(T,x1,…,Xi,…,xn)≠0, and ψ=ψ(ξ) satisfies some certain auxiliary equation with the following form: (5)Fψ,ψ′,ψ′′,…=0whose solutions are known. Furthermore, as [ψ′ξ/ψξ)m′=m(ψ′(ξ)/ψ(ξ))m-1[(ψξψ′′(ξ)-(ψ′ξ)2)/ψ2(ξ)], then the degree of ψ′ξ/ψξ′ is usually one more than [ψ′(ξ)/ψ(ξ)]. The positive integer m can be determined by considering the homogeneous balance of the degrees between the highest order derivatives and nonlinear terms appearing in (3).Step 3.
Substituting (4) into (3), using the relation between ψ′(ξ) and ψ(ξ) derived from (5), and collecting all terms with the same order of ψ(ξ) together, the left-hand side of (3) is converted to another polynomial in ψ(ξ). Equating each coefficient of this polynomial to zero yields a set of partial differential equations for amT,x1,…,Xi,…,xn,am-1T,x1,…,Xi,…,xn…, a0(T,x1,…,Xi,…,xn),ξ(T,x1,…,Xi,…,xn).Step 4.
Solving the equations yielded in Step3 and using the solutions of (5), together with the fractional transformations introduced in Step 1, one can obtain exact solutions for (2).Remark 1.
In the present method, the expression of the transformation denoted byξ is underdetermined, and the coefficients ai in (4) are variable coefficient functions, which may contribute to the seeking of exact solutions with variable coefficient function forms. Furthermore, if (5) are selected for some different forms, such as the Riccati equation, Bernoulli equation, and Jacobi elliptic equation, then different exact solutions for (2) can be obtained correspondingly.Remark 2.
As the partial differential equations yielded in Step3 are usually overdetermined, we may choose some special forms of am,am-1,…,a0 as in the following.


## 3. Applications of the Auxiliary Equation Method

### 3.1. Time Fractional Two-Dimensional Boussinesq Equation

We consider the time fractional two-dimensional Boussinesq equation with the following form:(6)DtαDtαu-uxx-uyy-u2xx-uxxxx=0,0<α≤1.The fractional Boussinesq equation is used in the analysis of long waves in shallow water and also used in the analysis of many other physical applications, such as the percolation of water in a porous subsurface of a horizontal layer of material. Tasbozan et al. [36] solved (7) using the Jacobi elliptic function expansion method and obtained a series of exact solutions with Jacobi elliptic function forms.Now we use the proposed auxiliary equation method to solve (6). First we let T=c(tα/α) and u(x,y,t)=u~(x,y,T). Then Dtαu=cu~T, and (6) can be converted into the following form: (7)c2u~TT-u~xx-u~yy-u~2xx-u~xxxx=0. Suppose that the solution of (7) can be expressed by a polynomial in ψ′(ξ)/ψ(ξ) as follows: (8)u~x,y,T=∑i=0maiy,Tψ′ξψξi,where ai(y,T),i=0,1,…,m, are underdetermined functions and ψ=ψ(ξ) satisfies (5). Balancing the degrees of u~xxxx and (u~2)xx in (7), one can obtain m+4=2m+2, which means m=2. Thus, one has (9)u~x,y,T=a2y,Tψ′ξψξ2+a1y,Tψ′ξψξ+a0y,T.Next we will discuss two cases, in whichψ satisfies two certain auxiliary equations.Case 1.
ψ = ψ ( ξ ) satisfies the following Riccati equation: (10)ψ′ξ+λψξ=ψ2ξ,where λ≠0.
Substituting (9) into (7), using (10), collecting all the terms with the same power of ψ together, and equating each coefficient to zero yield a set of underdetermined partial differential equations a0(y,T),a1(y,T),a2(y,T), and ξ(x,y,T). Solving these equations yields the following families of results, where C1,C2 are arbitrary constants and Fi,i=1,2,…,5, are arbitrary functions.Family 1 (11) a 0 y , T = - 1 2 F 1 c y + T 2 λ 2 - 1 2 , a 1 y , T = - 6 F 1 c y + T 2 λ , a 2 y , T = - 6 F 1 c y + T 2 , ξ x , y , T = F 1 c y + T x + F 2 c y + T .Family 2 (12) a 0 y , T = F 3 - T - c y + F 4 T - c y , a 1 y , T = F 2 c y + T λ , a 2 y , T = F 2 c y + T , ξ x , y , T = F 1 c y + T .Family 3 (13) a 0 y , T = - 1 2 C 1 2 λ 2 - 1 2 , a 1 y , T = - 6 C 1 2 λ , a 2 y , T = - 6 C 1 2 , ξ x , y , T = C 1 x + F 1 c y + T .Family 4 (14) a 0 y , T = - C 1 2 + c 2 C 4 2 - C 2 4 λ 2 - C 2 2 2 C 2 2 , a 1 y , T = - 6 C 2 2 λ , a 2 y , T = - 6 C 2 2 , ξ x , y , T = C 1 y + C 4 T + C 2 x + C 3 .Family 5 (15) a 0 y , T = F 4 - T - c y + F 5 T - c y , a 1 y , T = F 3 c y + T , a 2 y , T = F 2 c y + T , ξ x , y , T = F 1 c y + T .Remark 3.
It is obvious that if we letF1(cy+T)=k1, F2(cy+T)=cy+T in Family 1, then the transformation denoted by ξ becomes ξ(x,y,T)=k1x+cy+T=k1x+cy+c(tα/α), which has been used by many authors so far in the literature.
On the other hand, it is well known that the solution of (10) is denoted by (16)ψξ=λ1+λdeλξ, where d is an arbitrary constant, so (17)ψ′ξψξ=-dλ2eλξ1+λdeλξ.In particular, when d=1/λ, one can obtain (18)ψ′ξψξ=-λ21+tanh⁡λξ2.By a combination of the results denoted in Families 1–5 and (17), together with the expression of ξ, one can obtain a series of exact solutions for the time fractional two-dimensional Boussinesq equation as follows: (19)u1x,y,t=-6F1cy+T2dλ2eλξ1+λdeλξ2+6F1cy+T2λdλ2eλξ1+λdeλξ-12F1cy+T2λ2-12,where T=c(tα/α),ξ=F1(cy+T)x+F2(cy+T), and F1,F2 are arbitrary functions. (20)u2x,y,t=F2cy+Tdλ2eλξ1+λdeλξ2-F2cy+Tλdλ2eλξ1+λdeλξ+F3-T-cy+F4T-cy,where T=c(tα/α),ξ=F1(cy+T), and F1,F2,F3,F4 are arbitrary functions. (21)u3x,y,t=-6C12dλ2eλξ1+λdeλξ2+6C12λdλ2eλξ1+λdeλξ-12C12λ2-12,where T=c(tα/α),ξ=C1x+F1(cy+T), C1 is an arbitrary constant, and F1 is an arbitrary function. (22)u4x,y,t=-6C22dλ2eλξ1+λdeλξ2+6C22λdλ2eλξ1+λdeλξ+-C12+c2C42-C24λ2-C222C22,where T=c(tα/α),ξ=C1y+C4T+C2x+C3, and C1,C2,C3,C4 are arbitrary constants. (23)u5x,y,t=F2cy+Tdλ2eλξ1+λdeλξ2-F3cy+Tdλ2eλξ1+λdeλξ+F4-T-cy+F5T-cy,where T=c(tα/α),ξ=F1(cy+T), F1,F2,F3,F4,F5 are arbitrary functions.Similarly, by a combination of the results in Families 1–5 and (18), one can obtain the following solitary wave solutions, in which T=c(tα/α), Ci,i=1,2,3,4, are arbitrary constants, and Fj,j=1,2,3,4,5, are arbitrary functions. (24)u6x,y,t=-3λ22F1cy+T21+tanh⁡λξ22+3F1cy+T2λ21+tanh⁡λξ2-12F1cy+T2λ2-12,where ξ=F1(cy+T)x+F2(cy+T).(25)u7x,y,t=λ24F2cy+T1+tanh⁡λξ22-λ22F2cy+T1+tanh⁡λξ2+F3-T-cy+F4T-cy,where ξ=F1(cy+T).(26)u8x,y,t=-3C12λ221+tanh⁡λξ22+3C12λ21+tanh⁡λξ2-12C12λ2-12,where ξ=C1x+F1(cy+T).(27)u9x,y,t=-3C22λ221+tanh⁡λξ22+3C22λ21+tanh⁡λξ2+-C12+c2C42-C24λ2-C222C22,where ξ=C1y+C4T+C2x+C3.(28)u10x,y,t=λ24F2cy+T1+tanh⁡λξ22-λ2F3cy+T1+tanh⁡λξ2+F4-T-cy+F5T-cy,where ξ=F1(cy+T).In Figure1, the solitary wave solution u6(x,y,t) in (24) with some special parameters is demonstrated.Figure 1
The solitary wave solutionu6 with α=0.5, F1cy+T=1, F2cy+T=cy+T, c=λ=1, t=3.Case 2.
ψ = ψ ( ξ ) satisfies the following equation: (29)ψ′+λψ=μψ3,where λ,μ are arbitrary constants with λ≠0.Substituting (9) into (7), using (29), collecting all the terms with the same power of ψ together, and equating each coefficient to zero yield a set of underdetermined partial differential equations. Solving these equations yields the following results, where Ci,i=1,2,3,4, are arbitrary constants and Fi,i=1,2,3,4, are arbitrary functions.Family 1 (30) a 0 y , T = - 2 F 1 c y + T 2 λ 2 - 1 2 , a 1 y , T = - 24 F 1 c y + T 2 λ , a 2 y , T = - 24 F 1 c y + T 2 , ξ x , y , T = F 1 c y + T x + F 2 c y + T .Family 2 (31) a 0 y , T = - C 1 2 - 4 C 2 4 λ 2 - C 2 2 + c 2 C 4 2 2 C 2 2 , a 1 y , T = - 24 C 2 2 λ , a 2 y , T = - 24 C 2 2 , ξ x , y , T = C 1 y + C 4 T + C 2 x + C 3 .Family 3 (32) a 0 y , T = - 2 C 1 2 λ 2 - 1 2 , a 1 y , T = - 24 C 1 2 λ , a 2 y , T = - 24 C 1 2 , ξ x , y , T = C 1 x + F 1 c y + T .Family 4 (33) a 0 y , T = F 3 - T - c y + F 4 T - c y , a 1 y , T = F 2 c y + T λ , a 2 y , T = F 2 c y + T , ξ x , y , T = F 1 c y + T .For the solutions of (29), one has (34)ψξ=±1μ/λ+Ae2λξ, where A is an arbitrary constant and μ2+A2≠0, so (35)ψ′ξψξ=-Aλe2λξμ/λ+Ae2λξ.Substituting the results denoted by Families 1–4 into (9) and combining them with (35), one can obtain the following exact solutions for the time fractional two-dimensional Boussinesq equation, in which T=c(tα/α). (36)u11x,y,t=-24F1cy+T2Aλe2λξμ/λ+Ae2λξ2+24F1cy+T2λAλe2λξμ/λ+Ae2λξ-2F1cy+T2λ2-12,where ξ=F1(cy+T)x+F2(cy+T). (37)u12x,y,t=-24C22Aλe2λξμ/λ+Ae2λξ2+24C22λAλe2λξμ/λ+Ae2λξ+-C12-4C24λ2-C22+c2C422C22,where ξ=C1y+C4T+C2x+C3. (38)u13x,y,t=-24C12Aλe2λξμ/λ+Ae2λξ2+24C12λAλe2λξμ/λ+Ae2λξ-2C12λ2-12,where ξ=C1x+F1(cy+T). (39)u14x,y,t=F2cy+TAλe2λξμ/λ+Ae2λξ2-F2cy+TλAλe2λξμ/λ+Ae2λξ+F3-T-cy+F4T-cy,where ξ=F1(cy+T).If we setμ=λA in (35), then we obtain the following solitary wave solutions: (40)ψ′ξψξ=-λ21+tanh⁡λξ.By a combination of the results in Families 1–4 and (40), one can obtain corresponding solitary wave solutions, which are omitted here.


### 3.2. Space-Time Fractional (2+1)-Dimensional Breaking Soliton Equation

We consider the space-time fractional (2+1)-dimensional breaking soliton equation [37] of the form (41)DtαDxαu-4DxαuDyαDxαu-2DxαDxαuDyαu+DyαDxαDxαDxαu=0,0<α≤1.Now we solve (41) using the introduced auxiliary equation method.LetT=c(tα/α),X=k1(xα/α),Y=k1(yα/α), and u(x,y,t)=u~(X,Y,T). Then Dtαu=cu~T, Dxαu=k1u~X, Dyαu=k2u~Y, and (41) can be converted into the following equation: (42)ck1u~XT-4k12k2u~Xu~XY-2k12k2u~XXu~Y+k13k2u~XXXY=0. Suppose that the solutions of (42) can be expressed by a polynomial in ψ′(ξ)/ψ(ξ) as follows: (43)u~X,Y,T=∑i=0maiY,Tψ′ξψξi, where ai(Y,T) are underdetermined functions. Balancing the degrees of u~XXXY and u~XXu~Y in (42), one has m=1. Therefore, (44)u~X,Y,T=a1Y,Tψ′ξψξ+a0Y,T.Case 1.
Ifψ(ξ) satisfies (10), then substituting (44) into (42), collecting all the terms with the same power of ψ together, and equating each coefficient to zero yield a set of underdetermined partial differential equations for a0(Y,T), a1(Y,T), ξ(X,Y,T). Solving these equations yields several families of results as follows.Family 1 (45) a 0 Y , T = 1 2 C 3 λ T + c C 3 4 k 2 C 1 k 1 Y + C 5 , a 1 Y , T = C 3 T + C 4 , ξ X , Y , T = C 1 X + C 2 .Family 2 (46) a 0 Y , T = 1 2 C 1 k 1 λ 2 F Y + C 2 , a 1 Y , T = 2 C 1 k 1 , ξ X , Y , T = C 1 X + F Y ,where F is an arbitrary function.By a combination of (17), (44), and the results above, one can obtain the following exact solutions for the space-time fractional (2+1)-dimensional breaking soliton equation, in which T=c(tα/α),X=k1(xα/α),Y=k1(yα/α). (47)u1x,y,t=-C3T+C4dλ2eλξ1+λdeλξ+12C3λT+cC34k2C1k1Y+C5,where ξ=C1X+C2. (48)u2x,y,t=-2C1k1dλ2eλξ1+λdeλξ+12C1k1λ2FY+C2,where ξ=C1X+F(Y).By a combination of (18) and the results above, one can obtain the following solitary wave solutions: (49)u3x,y,t=-λ2C3T+C41+tanhλξ2+12C3λT+cC34k2C1k1Y+C5,where ξ=C1X+C2. (50)u4x,y,t=-C1k1λ1+tanh⁡λξ2+12C1k1λ2FY+C2,where ξ=C1X+F(Y).Case 2.
Ifψ(ξ) satisfies (29), then substituting (44) into (42), using (29), collecting all the terms with the same power of ψ together, and equating each coefficient to zero yield a set of underdetermined partial differential equations. Solving these equations yields the following results.Family 1 (51) a 0 Y , T = 2 C 1 k 1 λ 2 F 2 Y + C 2 , a 1 Y , T = 4 C 1 k 1 , ξ X , Y , T = C 1 X + F 2 Y .Family 2 (52) a 0 Y , T = 1 2 C 3 λ T + c C 3 8 k 2 C 1 k 1 Y + C 5 , a 1 Y , T = C 3 T + C 4 , ξ x , y , T = C 1 X + C 2 .By a combination of the results above and (35), one can obtain the following exact solutions for (41), where T=c(tα/α),X=k1(xα/α),Y=k1(yα/α). (53)u5x,y,t=-4C1k1dλ2eλξ1+λdeλξ+2C1k1λ2F2Y+C2,where ξ=C1X+F2(Y). (54)u6x,y,t=-C3T+C4dλ2eλξ1+λdeλξ+12C3λT+cC38k2C1k1Y+C5,where ξ=C1X+C2.Similarly, by a combination of (40) and the results above, one can obtain corresponding solitary wave solutions, which are omitted here.


## 3.1. Time Fractional Two-Dimensional Boussinesq Equation

We consider the time fractional two-dimensional Boussinesq equation with the following form:(6)DtαDtαu-uxx-uyy-u2xx-uxxxx=0,0<α≤1.The fractional Boussinesq equation is used in the analysis of long waves in shallow water and also used in the analysis of many other physical applications, such as the percolation of water in a porous subsurface of a horizontal layer of material. Tasbozan et al. [36] solved (7) using the Jacobi elliptic function expansion method and obtained a series of exact solutions with Jacobi elliptic function forms.Now we use the proposed auxiliary equation method to solve (6). First we let T=c(tα/α) and u(x,y,t)=u~(x,y,T). Then Dtαu=cu~T, and (6) can be converted into the following form: (7)c2u~TT-u~xx-u~yy-u~2xx-u~xxxx=0. Suppose that the solution of (7) can be expressed by a polynomial in ψ′(ξ)/ψ(ξ) as follows: (8)u~x,y,T=∑i=0maiy,Tψ′ξψξi,where ai(y,T),i=0,1,…,m, are underdetermined functions and ψ=ψ(ξ) satisfies (5). Balancing the degrees of u~xxxx and (u~2)xx in (7), one can obtain m+4=2m+2, which means m=2. Thus, one has (9)u~x,y,T=a2y,Tψ′ξψξ2+a1y,Tψ′ξψξ+a0y,T.Next we will discuss two cases, in whichψ satisfies two certain auxiliary equations.Case 1.
ψ = ψ ( ξ ) satisfies the following Riccati equation: (10)ψ′ξ+λψξ=ψ2ξ,where λ≠0.
Substituting (9) into (7), using (10), collecting all the terms with the same power of ψ together, and equating each coefficient to zero yield a set of underdetermined partial differential equations a0(y,T),a1(y,T),a2(y,T), and ξ(x,y,T). Solving these equations yields the following families of results, where C1,C2 are arbitrary constants and Fi,i=1,2,…,5, are arbitrary functions.Family 1 (11) a 0 y , T = - 1 2 F 1 c y + T 2 λ 2 - 1 2 , a 1 y , T = - 6 F 1 c y + T 2 λ , a 2 y , T = - 6 F 1 c y + T 2 , ξ x , y , T = F 1 c y + T x + F 2 c y + T .Family 2 (12) a 0 y , T = F 3 - T - c y + F 4 T - c y , a 1 y , T = F 2 c y + T λ , a 2 y , T = F 2 c y + T , ξ x , y , T = F 1 c y + T .Family 3 (13) a 0 y , T = - 1 2 C 1 2 λ 2 - 1 2 , a 1 y , T = - 6 C 1 2 λ , a 2 y , T = - 6 C 1 2 , ξ x , y , T = C 1 x + F 1 c y + T .Family 4 (14) a 0 y , T = - C 1 2 + c 2 C 4 2 - C 2 4 λ 2 - C 2 2 2 C 2 2 , a 1 y , T = - 6 C 2 2 λ , a 2 y , T = - 6 C 2 2 , ξ x , y , T = C 1 y + C 4 T + C 2 x + C 3 .Family 5 (15) a 0 y , T = F 4 - T - c y + F 5 T - c y , a 1 y , T = F 3 c y + T , a 2 y , T = F 2 c y + T , ξ x , y , T = F 1 c y + T .Remark 3.
It is obvious that if we letF1(cy+T)=k1, F2(cy+T)=cy+T in Family 1, then the transformation denoted by ξ becomes ξ(x,y,T)=k1x+cy+T=k1x+cy+c(tα/α), which has been used by many authors so far in the literature.
On the other hand, it is well known that the solution of (10) is denoted by (16)ψξ=λ1+λdeλξ, where d is an arbitrary constant, so (17)ψ′ξψξ=-dλ2eλξ1+λdeλξ.In particular, when d=1/λ, one can obtain (18)ψ′ξψξ=-λ21+tanh⁡λξ2.By a combination of the results denoted in Families 1–5 and (17), together with the expression of ξ, one can obtain a series of exact solutions for the time fractional two-dimensional Boussinesq equation as follows: (19)u1x,y,t=-6F1cy+T2dλ2eλξ1+λdeλξ2+6F1cy+T2λdλ2eλξ1+λdeλξ-12F1cy+T2λ2-12,where T=c(tα/α),ξ=F1(cy+T)x+F2(cy+T), and F1,F2 are arbitrary functions. (20)u2x,y,t=F2cy+Tdλ2eλξ1+λdeλξ2-F2cy+Tλdλ2eλξ1+λdeλξ+F3-T-cy+F4T-cy,where T=c(tα/α),ξ=F1(cy+T), and F1,F2,F3,F4 are arbitrary functions. (21)u3x,y,t=-6C12dλ2eλξ1+λdeλξ2+6C12λdλ2eλξ1+λdeλξ-12C12λ2-12,where T=c(tα/α),ξ=C1x+F1(cy+T), C1 is an arbitrary constant, and F1 is an arbitrary function. (22)u4x,y,t=-6C22dλ2eλξ1+λdeλξ2+6C22λdλ2eλξ1+λdeλξ+-C12+c2C42-C24λ2-C222C22,where T=c(tα/α),ξ=C1y+C4T+C2x+C3, and C1,C2,C3,C4 are arbitrary constants. (23)u5x,y,t=F2cy+Tdλ2eλξ1+λdeλξ2-F3cy+Tdλ2eλξ1+λdeλξ+F4-T-cy+F5T-cy,where T=c(tα/α),ξ=F1(cy+T), F1,F2,F3,F4,F5 are arbitrary functions.Similarly, by a combination of the results in Families 1–5 and (18), one can obtain the following solitary wave solutions, in which T=c(tα/α), Ci,i=1,2,3,4, are arbitrary constants, and Fj,j=1,2,3,4,5, are arbitrary functions. (24)u6x,y,t=-3λ22F1cy+T21+tanh⁡λξ22+3F1cy+T2λ21+tanh⁡λξ2-12F1cy+T2λ2-12,where ξ=F1(cy+T)x+F2(cy+T).(25)u7x,y,t=λ24F2cy+T1+tanh⁡λξ22-λ22F2cy+T1+tanh⁡λξ2+F3-T-cy+F4T-cy,where ξ=F1(cy+T).(26)u8x,y,t=-3C12λ221+tanh⁡λξ22+3C12λ21+tanh⁡λξ2-12C12λ2-12,where ξ=C1x+F1(cy+T).(27)u9x,y,t=-3C22λ221+tanh⁡λξ22+3C22λ21+tanh⁡λξ2+-C12+c2C42-C24λ2-C222C22,where ξ=C1y+C4T+C2x+C3.(28)u10x,y,t=λ24F2cy+T1+tanh⁡λξ22-λ2F3cy+T1+tanh⁡λξ2+F4-T-cy+F5T-cy,where ξ=F1(cy+T).In Figure1, the solitary wave solution u6(x,y,t) in (24) with some special parameters is demonstrated.Figure 1
The solitary wave solutionu6 with α=0.5, F1cy+T=1, F2cy+T=cy+T, c=λ=1, t=3.Case 2.
ψ = ψ ( ξ ) satisfies the following equation: (29)ψ′+λψ=μψ3,where λ,μ are arbitrary constants with λ≠0.Substituting (9) into (7), using (29), collecting all the terms with the same power of ψ together, and equating each coefficient to zero yield a set of underdetermined partial differential equations. Solving these equations yields the following results, where Ci,i=1,2,3,4, are arbitrary constants and Fi,i=1,2,3,4, are arbitrary functions.Family 1 (30) a 0 y , T = - 2 F 1 c y + T 2 λ 2 - 1 2 , a 1 y , T = - 24 F 1 c y + T 2 λ , a 2 y , T = - 24 F 1 c y + T 2 , ξ x , y , T = F 1 c y + T x + F 2 c y + T .Family 2 (31) a 0 y , T = - C 1 2 - 4 C 2 4 λ 2 - C 2 2 + c 2 C 4 2 2 C 2 2 , a 1 y , T = - 24 C 2 2 λ , a 2 y , T = - 24 C 2 2 , ξ x , y , T = C 1 y + C 4 T + C 2 x + C 3 .Family 3 (32) a 0 y , T = - 2 C 1 2 λ 2 - 1 2 , a 1 y , T = - 24 C 1 2 λ , a 2 y , T = - 24 C 1 2 , ξ x , y , T = C 1 x + F 1 c y + T .Family 4 (33) a 0 y , T = F 3 - T - c y + F 4 T - c y , a 1 y , T = F 2 c y + T λ , a 2 y , T = F 2 c y + T , ξ x , y , T = F 1 c y + T .For the solutions of (29), one has (34)ψξ=±1μ/λ+Ae2λξ, where A is an arbitrary constant and μ2+A2≠0, so (35)ψ′ξψξ=-Aλe2λξμ/λ+Ae2λξ.Substituting the results denoted by Families 1–4 into (9) and combining them with (35), one can obtain the following exact solutions for the time fractional two-dimensional Boussinesq equation, in which T=c(tα/α). (36)u11x,y,t=-24F1cy+T2Aλe2λξμ/λ+Ae2λξ2+24F1cy+T2λAλe2λξμ/λ+Ae2λξ-2F1cy+T2λ2-12,where ξ=F1(cy+T)x+F2(cy+T). (37)u12x,y,t=-24C22Aλe2λξμ/λ+Ae2λξ2+24C22λAλe2λξμ/λ+Ae2λξ+-C12-4C24λ2-C22+c2C422C22,where ξ=C1y+C4T+C2x+C3. (38)u13x,y,t=-24C12Aλe2λξμ/λ+Ae2λξ2+24C12λAλe2λξμ/λ+Ae2λξ-2C12λ2-12,where ξ=C1x+F1(cy+T). (39)u14x,y,t=F2cy+TAλe2λξμ/λ+Ae2λξ2-F2cy+TλAλe2λξμ/λ+Ae2λξ+F3-T-cy+F4T-cy,where ξ=F1(cy+T).If we setμ=λA in (35), then we obtain the following solitary wave solutions: (40)ψ′ξψξ=-λ21+tanh⁡λξ.By a combination of the results in Families 1–4 and (40), one can obtain corresponding solitary wave solutions, which are omitted here.


## 3.2. Space-Time Fractional (2+1)-Dimensional Breaking Soliton Equation

We consider the space-time fractional (2+1)-dimensional breaking soliton equation [37] of the form (41)DtαDxαu-4DxαuDyαDxαu-2DxαDxαuDyαu+DyαDxαDxαDxαu=0,0<α≤1.Now we solve (41) using the introduced auxiliary equation method.LetT=c(tα/α),X=k1(xα/α),Y=k1(yα/α), and u(x,y,t)=u~(X,Y,T). Then Dtαu=cu~T, Dxαu=k1u~X, Dyαu=k2u~Y, and (41) can be converted into the following equation: (42)ck1u~XT-4k12k2u~Xu~XY-2k12k2u~XXu~Y+k13k2u~XXXY=0. Suppose that the solutions of (42) can be expressed by a polynomial in ψ′(ξ)/ψ(ξ) as follows: (43)u~X,Y,T=∑i=0maiY,Tψ′ξψξi, where ai(Y,T) are underdetermined functions. Balancing the degrees of u~XXXY and u~XXu~Y in (42), one has m=1. Therefore, (44)u~X,Y,T=a1Y,Tψ′ξψξ+a0Y,T.Case 1.
Ifψ(ξ) satisfies (10), then substituting (44) into (42), collecting all the terms with the same power of ψ together, and equating each coefficient to zero yield a set of underdetermined partial differential equations for a0(Y,T), a1(Y,T), ξ(X,Y,T). Solving these equations yields several families of results as follows.Family 1 (45) a 0 Y , T = 1 2 C 3 λ T + c C 3 4 k 2 C 1 k 1 Y + C 5 , a 1 Y , T = C 3 T + C 4 , ξ X , Y , T = C 1 X + C 2 .Family 2 (46) a 0 Y , T = 1 2 C 1 k 1 λ 2 F Y + C 2 , a 1 Y , T = 2 C 1 k 1 , ξ X , Y , T = C 1 X + F Y ,where F is an arbitrary function.By a combination of (17), (44), and the results above, one can obtain the following exact solutions for the space-time fractional (2+1)-dimensional breaking soliton equation, in which T=c(tα/α),X=k1(xα/α),Y=k1(yα/α). (47)u1x,y,t=-C3T+C4dλ2eλξ1+λdeλξ+12C3λT+cC34k2C1k1Y+C5,where ξ=C1X+C2. (48)u2x,y,t=-2C1k1dλ2eλξ1+λdeλξ+12C1k1λ2FY+C2,where ξ=C1X+F(Y).By a combination of (18) and the results above, one can obtain the following solitary wave solutions: (49)u3x,y,t=-λ2C3T+C41+tanhλξ2+12C3λT+cC34k2C1k1Y+C5,where ξ=C1X+C2. (50)u4x,y,t=-C1k1λ1+tanh⁡λξ2+12C1k1λ2FY+C2,where ξ=C1X+F(Y).Case 2.
Ifψ(ξ) satisfies (29), then substituting (44) into (42), using (29), collecting all the terms with the same power of ψ together, and equating each coefficient to zero yield a set of underdetermined partial differential equations. Solving these equations yields the following results.Family 1 (51) a 0 Y , T = 2 C 1 k 1 λ 2 F 2 Y + C 2 , a 1 Y , T = 4 C 1 k 1 , ξ X , Y , T = C 1 X + F 2 Y .Family 2 (52) a 0 Y , T = 1 2 C 3 λ T + c C 3 8 k 2 C 1 k 1 Y + C 5 , a 1 Y , T = C 3 T + C 4 , ξ x , y , T = C 1 X + C 2 .By a combination of the results above and (35), one can obtain the following exact solutions for (41), where T=c(tα/α),X=k1(xα/α),Y=k1(yα/α). (53)u5x,y,t=-4C1k1dλ2eλξ1+λdeλξ+2C1k1λ2F2Y+C2,where ξ=C1X+F2(Y). (54)u6x,y,t=-C3T+C4dλ2eλξ1+λdeλξ+12C3λT+cC38k2C1k1Y+C5,where ξ=C1X+C2.Similarly, by a combination of (40) and the results above, one can obtain corresponding solitary wave solutions, which are omitted here.


## 4. Conclusions

Based on the properties of conformable fractional calculus, we have proposed an auxiliary equation method for seeking exact solutions with variable coefficient function forms for fractional partial differential equations and applied it to the time fractional two-dimensional Boussinesq equation and the space-time fractional (2+1)-dimensional breaking soliton equation. As a result, some exact solutions including variable coefficient function solutions as well as solitary wave solutions for them have been successfully found. These solutions are new exact solutions so far in the literature to the best of our knowledge. We note that, by a combination of other auxiliary equations different from the two equations used here, more exact solutions can be found subsequently.



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