# Exp-Function Method for Solving Fractional Partial Differential Equations

Authors: Bin Zheng
Journal: The Scientific World Journal
 (2013)

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## Abstract

We extend the Exp-function method to fractional partial differential equations in the sense of modified Riemann-Liouville derivative based on nonlinear fractional complex transformation. For illustrating the validity of this method, we apply it to the space-time fractional Fokas equation and the nonlinear fractional Sharma-Tasso-Olver (STO) equation. As a result, some new exact solutions for them are successfully established.


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## Body

## 1. Introduction

Nonlinear differential equations of integer order (NLDEs) can be used to describe many nonlinear phenomena such as fluid mechanics, plasma physics, optical fibers, biology, solid state physics, chemical kinematics, and chemical physics. In the research of the theory of NLDEs, searching for more explicit exact solutions to NLDEs is one of the most fundamental and significant studies in recent years. With the help of computerized symbolic computation, much work has focused on the various extensions and applications of the known algebraic methods to construct the solutions to NLDEs. There have been a variety of powerful methods. For example, these methods include the generalized Riccati subequation method [1, 2], the Jacobi elliptic function expansion [3, 4], the F-expansion method [5], the Exp-function method [6–9], and the (G′/G)-expansion method [10, 11].Fractional differential equations are generalizations of classical differential equations of integer order and have recently proved to be valuable tools to the modeling of many physical phenomena and have been the focus of many studies due to their frequent appearance in various applications in physics, biology, engineering, signal processing, systems identification, control theory, finance, and fractional dynamics. In order to obtain exact solutions for fractional differential equations, many powerful and efficient methods have been proposed so far (e.g., see [12–24]). But nobody has researched the application of the Exp-function method to fractional differential equations so far to our best knowledge.In this paper, we will apply the Exp-function method for solving fractional partial differential equations in the sense of modified Riemann-Liouville derivative by Jumarie [25]. By a certain nonlinear fractional complex transformation ξ=ξ(t,x1,x2,…,xn), a given fractional partial differential equation expressed in independent variables t,x1,x2,…,xn can be turned into another ordinary differential equation of integer order, whose solutions are supposed to have one of the following forms:
(1)u(ξ)=∑p=-m1m2apepξ∑q=-n1n2bqeqξoru(ξ)=∑p=-m1m2apeipξ∑q=-n1n2bqeiqξ,
where m1, m2, n1, and n2 are unknown positive integers that will be determined by using the balance method. Determine the highest order nonlinear term and the linear term of highest order in the ordinary differential equation and express them in terms of (3). Then, in the resulting terms, balance the highest order Exp-function to determine m2 and n2 and balance the lowest order Exp-function to determine m1 and n1.The Jumarie’s modified Riemann-Liouville derivative of orderα is defined by the following expression [25]:
(2)Dtαf(t)={1Γ(1-α)ddt∫0t(t-ξ)-α(f(ξ)-f(0))dξ,0<α<1,(f(n)(t))(α-n),n≤α<n+1,n≥1.
We list some important properties for the modified Riemann-Liouville derivative as follows [25, (3.10)–(3.13)] (see also in [22–24, 26]):
(3)Dtαtr=Γ(1+r)Γ(1+r-α)tr-α,(4)Dtα(f(t)g(t))=g(t)Dtαf(t)+f(t)Dtαg(t),(5)Dtαf[g(t)]=fg′[g(t)]Dtαg(t)=Dgαf[g(t)](g′(t))α.In the following, we will apply the Exp-function method to find exact solutions for the space-time fractional Fokas equation and the nonlinear fractional Sharma-Tasso-Olver (STO) equation.


## 2. Applications of the Exp-Function Method to Fractional Partial Differential Equations

### 2.1. Space-Time Fractional Fokas Equation

We consider the space-time fractional Fokas equation [22]
(6)4∂2αq∂tα∂x1α-∂4αq∂x13α∂x2α+∂4αq∂x23α∂x1α+12∂αq∂x1α∂αq∂x2α+12q∂2αq∂x1α∂x2α-6∂2αq∂y1α∂y2α=0,0<α≤1.
In [22], the author solved (6) by a proposed fractional sub-equation method based on the fractional Riccati equation and established some exact solutions for it. Now we will apply the Exp-function method to this equation. Suppose q(x,y,t)=U(ξ), where ξ=(k1x1α/Γ(1+α))+(k2x2α/Γ(1+α))+(l1y1α/Γ(1+α))+(l2y2α/Γ(1+α))+(ctα/Γ(1+α))+ξ0, k1,k2,l1,l2,c,ξ0 are all constants with k1,k2,l1,l2,c≠0. Then by the use of (5), (6) can be turned into
(7)4ck1U′′-k13k2U(4)+k23k1U(4)+12k1k2U′2+12k1k2UU′′-6l1l2U′′=0.First we suppose that the solution of (7) can be expressed by
(8)U(ξ)=∑p=-m1m2apepξ∑q=-n1n2bqeqξ.
To determine m2, n2, we need to balance the linear term of the highest order with the highest order nonlinear term in (7). By simple calculation, we have
(9)U(4)=c1e(15n2+m2)x+⋯+c2e-(15n1+m1)xc3e16n2x+⋯+c4e-16n1x,UU′′=c5e(3n2+2m2)x+⋯+c6e-(3n1+2m1)xc7e5n2x+⋯+c8e-5n1x,
where ci, i=1,2,3,4 are coefficients determined by am2, a-m1, bn2, b-n1. Balancing the highest order Exp-function in U(4) and UU′′, we have 15n2+m2=14n2+2m2, which implies m2=n2. Similarly, balancing the lowest order Exp-function in U(4) and UU′′, we have 15n1+m1=14n1+2m1, which implies m1=n1. For simplicity, we will proceed in two selected cases.Case 1.
Letm1=n1=1, m2=n2=1. Then
(10)U(ξ)=a-1e-ξ+a0+a1eξb-1e-ξ+b0+b1eξ.
Substituting (10) into (7), eliminating the denominator, and collecting all the terms with the same power of epξ together, equating each coefficient to zero, yield a set of algebraic equations. Solving these equations with the aid of the mathematical software Maple yields the following family of values of ai,bi, i=-1,0,1.Family  1. Consider
(11)a1=-b1(k1k23-k2k13+4ck1-6l1l2)12k1k2,a0=-b0(-5k1k23+5k2k13+4ck1-6l1l2)12k1k2,a-1=-b02(k1k23-k2k13+4ck1-6l1l2)48b1k1k2,b-1=b024b1,
where b1, b0 are arbitrary constants.Substituting the result above into (10), we can obtain the following exact solutions to (7):
(12)U1(ξ)=(-(k1k23-k2k13+4ck1-6l1l2)(b02e-ξ+4b12eξ)-4b0b1(-5k1k23+5k2k13+4ck1-6l1l2)b02e-ξ)×(12k1k2(b02e-ξ+4b12eξ)+48b0b1k1k2)-1,
where ξ=(k1x1α/Γ(1+α))+(k2x2α/Γ(1+α))+(l1y1α/Γ(1+α))+(l2y2α/Γ(1+α))+(ctα/Γ(1+α))+ξ0. If we especially take b0=2b1, then we obtain the following hyperbolic function solitary wave solution:
(13)U2(ξ)=-(k1k23-k2k13+4ck1-6l1l2)12k1k2+(k22-k12)4sech2ξ2.Case 2.
Letm1=n1=m, m2=n2=m, and for simplicity, we take
(14)U(ξ)=a-me-mξ+a0+amemξb-me-mξ+b0+bmemξ.
Substituting (14) into (7), eliminating the denominator, and collecting all the terms with the same power of epnξ together, equating each coefficient to zero, yield a set of algebraic equations. Solving these equations yields the following result.Family  2. Consider
(15)am=-bm(m2k1k23-m2k2k13+4ck1-6l1l2)12k1k2,a0=-b0(-5m2k1k23+5m2k2k13+4ck1-6l1l2)12k1k2,a-m=-b02(m2k1k23-m2k2k13+4ck1-6l1l2)48bmk1k2,b-m=b024bm,
where bm, b0 are arbitrary constants.Substituting the result above into (14), we can obtain the following exact solutions to (7):
(16)U3(ξ)=(-(m2k1k23-m2k2k13+4ck1-6l1l2)(b02e-ξ+4bm2eξ)-4b0bm(-5m2k1k23+5m2k2k13+4ck1-6l1l2)k13)×(12k1k2(b02e-ξ+4bm2eξ)+48b0bmk1k2)-1,
where ξ=(k1x1α/Γ(1+α))+(k2x2α/Γ(1+α))+(l1y1α/Γ(1+α))+(l2y2α/Γ(1+α))+(ctα/Γ(1+α))+ξ0. If we especially take b0=2bm, then we obtain the following solitary wave solution:
(17)U4(ξ)=-(m2k1k23-m2k2k13+4ck1-6l1l2)12k1k2+m2(k22-k12)4sech2(m2ξ).Now we suppose that the solution of (7) can be expressed by
(18)U(ξ)=∑p=-m1m2apeipξ∑q=-n1n2bqeiqξ.
Similar to the balancing process for (8), we have m1=n1 and m2=n2. Similar to the above, we only consider the following two cases.Case 1.
Letm1=n1=1, m2=n2=1. Then
(19)U(ξ)=a-1e-iξ+a0+a1eiξb-1e-iξ+b0+b1eiξ.
Substituting (19) into (7), eliminating the denominator, and collecting all the terms with the same power of eipξ together, equating each coefficient to zero, yield a set of algebraic equations. Solving these equations yields the following.Family  3. Consider
(20)a1=-b1(-k1k23+k2k13+4ck1-6l1l2)12k1k2,a0=-b0(5k1k23-5k2k13+4ck1-6l1l2)12k1k2,a-1=-b02(-k1k23+k2k13+4ck1-6l1l2)48b1k1k2,b-1=b024b1,
where b1, b0 are arbitrary constants.Substituting the result above into (19), we can obtain the following exact solutions to (7):
(21)U5(ξ)=(-(-k1k23+k2k13+4ck1-6l1l2)(b02e-iξ+4b12eiξ)-4b0b1(5k1k23-5k2k13+4ck1-6l1l2)4b12eiξ)×(12k1k2(b02e-iξ+4b12eiξ)+48b0b1k1k2)-1,
where ξ=(k1x1α/Γ(1+α))+(k2x2α/Γ(1+α))+(l1y1α/Γ(1+α))+(l2y2α/Γ(1+α))+(ctα/Γ(1+α))+ξ0. If we especially take b0=2b1, then we obtain the following trigonometric function solution:
(22)U6(ξ)=-(-k1k23+k2k13+4ck1-6l1l2)12k1k2-(k22-k12)4sec2ξ2.Case 2.
Let(23)U(ξ)=a-me-imξ+a0+ameimξb-me-imξ+b0+bmeimξ.
Substituting (23) into (7), eliminating the denominator, and collecting all the terms with the same power of eipnξ together, equating each coefficient to zero, yield a set of algebraic equations. Solving these equations, we obtain another family of values of ai,bi, i=-m,0,m.Family  4. Consider
(24)am=-bm(-m2k1k23+m2k2k13+4ck1-6l1l2)12k1k2,a0=-b0(5m2k1k23-5m2k2k13+4ck1-6l1l2)12k1k2,a-m=-b02(-m2k1k23+m2k2k13+4ck1-6l1l2)48bmk1k2,b-m=b024bm.
Substituting the result above into (23), we can obtain the following exact solutions to (7):
(25)U7(ξ)=(-(-m2k1k23+m2k2k13+4ck1-6l1l2)(b02e-ξ+4bm2eξ)-4b0bm(5m2k1k23-5m2k2k13+4ck1-6l1l2))4bm2eξ×(12k1k2(b02e-ξ+4bm2eξ)+48b0bmk1k2)-1,
where ξ=(k1x1α/Γ(1+α))+(k2x2α/Γ(1+α))+(l1y1α/Γ(1+α))+(l2y2α/Γ(1+α))+(ctα/Γ(1+α))+ξ0. If we especially take b0=2bm, then we obtain the following solitary wave solution:
(26)U8(ξ)=-(-m2k1k23+m2k2k13+4ck1-6l1l2)12k1k2-m2(k22-k12)4sec2(m2ξ).Remark 1.
Compared with the results in [22], the established solutions by (12), (16), (21), and (25) are new exact solutions to the space-time fractional Fokas equation, while the solutions by (13), (17), (22), and (26) are new solitary wave solutions.


### 2.2. Nonlinear Fractional Sharma-Tasso-Olver (STO) Equation

We consider the nonlinear fractional Sharma-Tasso-Olver (STO) equation with time-fractional derivative [26, 27]:
(27)Dtαu+3aux2+3au2ux+3auuxx+auxxx=0,0<α≤1.
In [26], the author solved (27) by the first integral method and obtained some hyperbolic function and trigonometric function solutions, while in [27], the variational iteration method, the Adomian decomposition method, and the homotopy perturbation method are used for obtaining a rational approximation solution for (27). Now we will apply the Exp-function method to (27). To begin with, we suppose u(x,t)=U(ξ), where ξ=kx+(ctα/Γ(1+α))+ξ0, k,c,ξ0 are all constants with k,c≠0. Then by use of (5), (27) can be turned into
(28)cU′+3ak2U′2+3akU2U′+3aUU′′+aU′′′=0.First we suppose that the solution of (28) can be expressed by
(29)U(ξ)=∑p=-m1m2apepξ∑q=-n1n2bqeqξ.
Similar to the previous subsection, by balancing the highest order Exp-function in U′′′ and U2U′, we have m2=n2, while we obtain m1=n1 by balancing the lowest order Exp-function in U′′′ and U2U′. Similar to the solving process for the space-time fractional Fokas equation, we will also proceed the computation under two selected cases.Case 1.
Letm1=n1=1, m2=n2=1. Then
(30)U(ξ)=a-1e-ξ+a0+a1eξb-1e-ξ+b0+b1eξ.
Substituting (30) into (28), eliminating the denominator, and collecting all the terms with the same power of epξ together, equating each coefficient to zero, yield a set of algebraic equations. Solving these equations yields two families of values for ai,bi, i=-1,0,1.Family  1. Consider
(31)a1=b1(3ak2∓-3a2k4-3akc)3ak,a0=0,b0=0,a-1=-b-1(2ak3±2k-3a2k4-3akc-c)3ak2±-3a2k4-3akc,
where b1, b-1 are arbitrary constants.Family  2. Consider
(32)a1=b1(3ak2∓-3a2k4-3akc)3ak,a0=b0(3ak2∓-3a2k4-3akc)3ak,a-1=b-1(9ak3±k-3a2k4-3akc+3c)6ak2±3-3a2k4-3akc,
where b1, b0, b-1 are arbitrary constants.Substituting the results above into (30), we can obtain the following exact solutions to (28):
(33)U1(ξ)=(-b-1(2ak3±2k-3a2k4-3akc-c)3ak2±-3a2k4-3akce-ξ+b1(3ak2∓-3a2k4-3akc)3akeξb-1(2ak3±2k-3a2k4-3akc-c)3ak2±-3a2k4-3akc)×(b-1e-ξ+b1eξ)-1,(34)U2(ξ)=(b-1(9ak3±k-3a2k4-3akc+3c)6ak2±3-3a2k4-3akce-ξ+b0(3ak2∓-3a2k4-3akc)3ak+b1(3ak2∓-3a2k4-3akc)3akeξb-1(9ak3±k-3a2k4-3akc+3c)6ak2±3-3a2k4-3akc)×(b-1e-ξ+b0+b1eξ)-1,
where ξ=kx+(ctα/Γ(1+α))+ξ0. If we especially take c=-ak3, b1=b-1 or c=-ak3, b1=-b-1 in (33), then we obtain the hyperbolic function solitary wave solutions:
(35)U3(ξ)=ktanhξ,U4(ξ)=kcothξ.Case 2.
Let(36)U(ξ)=a-me-mξ+a0+amemξb-me-mξ+b0+bmemξ.
Substituting (36) into (28), eliminating the denominator, and collecting all the terms with the same power of epnξ together, equating each coefficient to zero, yield a set of algebraic equations. Solving these equations yields the following results.Family  3. Consider
(37)am=bm(3amk2∓-3a2m2k4-3akc)3ak,a0=0,b0=0,a-m=-b-m(2am2k3±2mk-3a2m2k4-3akc-c)3amk2±-3a2m2k4-3akc,
where bm, b-m are arbitrary constants.Family  4. Consider
(38)am=bm(3amk2∓-3a2k4-3akc)3ak,a0=b0(3amk2∓-3a2m2k4-3akc)3ak,a-m=b-m(9am2k3±mk-3a2m2k4-3akc+3c)6amk2±3-3a2m2k4-3akc,
where bm, b0, b-m are arbitrary constants.Substituting the results above into (36), we can obtain the following exact solutions to (28):
(39)U5(ξ)=(-b-m(2am2k3±2mk-3a2m2k4-3akc-c)3amk2±-3a2m2k4-3akce-mξ+bm(3amk2∓-3a2m2k4-3akc)3akemξb-m(2am2k3±2mk-3a2m2k4-3akc-c)3amk2±-3a2m2k4-3akc)×(b-me-mξ+bmemξ)-1,(40)U6(ξ)=(b-m(9am2k3±mk-3a2m2k4-3akc+3c)6amk2±3-3a2m2k4-3akce-ξ+bm(3amk2∓-3a2k4-3akc)3akemξ+b0(3amk2∓-3a2m2k4-3akc)3akb-m(9am2k3±mk-3a2m2k4-3akc+3c)6amk2±3-3a2m2k4-3akc)×(b-me-mξ+b0+bmemξ)-1,
where ξ=kx+(ctα/Γ(1+α))+ξ0. If we especially take c=-am2k3, bm=b-m or c=-am2k3, bm=-b-m in (39), then we obtain the hyperbolic function solitary wave solutions:
(41)U7(ξ)=mktanhξ,U8(ξ)=mkcothξ.Now we suppose that the solution of (28) can be expressed by
(42)U(ξ)=∑p=-m1m2apeipξ∑q=-n1n2bqeiqξ.
By the balancing process for (28), we have m1=n1 and m2=n2. Similar to the above, we will also select two cases for computation.Case 1.
Letm1=n1=1, m2=n2=1. Then
(43)U(ξ)=a-1e-iξ+a0+a1eiξb-1e-iξ+b0+b1eiξ.
Substituting (43) into (28), eliminating the denominator, and collecting all the terms with the same power of eipξ together, equating each coefficient to zero, yield a set of algebraic equations. Solving these equations, we obtain the following two families of values.Family  5. Consider
(44)a1=b1(3iak2∓3a2k4-3akc)3ak,a0=0,a-1=-b-1(-2ak3±2ik3a2k4-3akc-c)3iak2±3a2k4-3akc,
where b1, b-1 are arbitrary constants.Family  6. Consider
(45)a1=b1(3iak2∓3a2k4-3akc)3ak,a0=b0(3iak2∓3a2k4-3akc)3ak,a-1=b-1(-9ak3±ik3a2k4-3akc+3c)6iak2±33a2k4-3akc,
where b1, b0, b-1 are arbitrary constants.Substituting the results above into (43), we can obtain the following exact solutions to (28):
(46)U9(ξ)=(-b-1(-2ak3±2ik3a2k4-3akc-c)3iak2±3a2k4-3akce-iξ+b1(3iak2∓3a2k4-3akc)3akeiξb-1(-2ak3±2ik3a2k4-3akc-c)3iak2±3a2k4-3akc)×(b-1e-iξ+b1eiξ)-1,(47)U10(ξ)=(b-1(-9ak3±ik3a2k4-3akc+3c)6iak2±33a2k4-3akce-iξ+b0(3iak2∓3a2k4-3akc)3ak+b1(3iak2∓3a2k4-3akc)3akeiξb-1(-9ak3±ik3a2k4-3akc+3c)6iak2±33a2k4-3akc)×(b-1e-iξ+b0+b1eiξ)-1If we especially takec=-ak3, b1=b-1 or c=-ak3, b1=-b-1 in (46), then we obtain the trigonometric function solutions:
(48)U11(ξ)=-ktanξ,U12(ξ)=-kcotξ.Case 2.
Let(49)U(ξ)=a-me-imξ+a0+ameimξb-me-imξ+b0+bmeimξ.
Substituting (49) into (28), eliminating the denominator, and collecting all the terms with the same power of eipmξ together, equating each coefficient to zero, yield a set of algebraic equations. Solving these equations, we obtain the following two families of values.Family  7. Consider
(50)am=bm(3iamk2∓3a2m2k4-3akc)3ak,a0=0,a-m=-b-m(-2am2k3±2imk3a2m2k4-3akc-c)3iamk2±3a2m2k4-3akc,
where bm, b-m are arbitrary constants.Family  8. Consider
(51)am=bm(3iamk2∓3a2k4-3akc)3ak,a0=b0(3iamk2∓3a2m2k4-3akc)3ak,a-m=b-m(-9am2k3±mk3a2m2k4-3akc+3c)6iamk2±33a2m2k4-3akc,
where bm, b0, b-m are arbitrary constants.Substituting the results above into (49), we can obtain the following exact solutions to (28):
(52)U13(ξ)=(-b-m(-2am2k3±2imk3a2m2k4-3akc-c)3iamk2±3a2m2k4-3akce-imξ+bm(3iamk2∓3a2m2k4-3akc)3akeimξ)×(b-me-imξ+bmeimξ)-1,(53)U14(ξ)=(b-m(-9am2k3±mk3a2m2k4-3akc+3c)6iamk2±33a2m2k4-3akce-imξ+bm(3iamk2∓3a2k4-3akc)3akeimξ+b0(3iamk2∓3a2m2k4-3akc)3ak)×(b-me-imξ+b0+bmeimξ)-1.
If we especially take c=-am2k3, bm=b-m or c=-am2k3, bm=-b-m in (53), then we obtain the trigonometric function solutions:
(54)U11(ξ)=-mktanξ,U12(ξ)=-mkcotξ.Remark 2.
If we letm=2, then the solutions by (41), (54) reduce to the solutions established by Lu [26, equations (55), (56), (53), (54)]. So in this way, our solutions (41), (54) are of more general forms. Moreover, we note that the solutions denoted by (33), (34), (39), (40), (46), (47), (52), and (53) are essentially different from the results in [26, 27] and are new exact solutions to the nonlinear fractional Sharma-Tasso-Olver equation so far in the literature, while the solutions (35), (48) are new solitary wave solutions to it.


## 2.1. Space-Time Fractional Fokas Equation

We consider the space-time fractional Fokas equation [22]
(6)4∂2αq∂tα∂x1α-∂4αq∂x13α∂x2α+∂4αq∂x23α∂x1α+12∂αq∂x1α∂αq∂x2α+12q∂2αq∂x1α∂x2α-6∂2αq∂y1α∂y2α=0,0<α≤1.
In [22], the author solved (6) by a proposed fractional sub-equation method based on the fractional Riccati equation and established some exact solutions for it. Now we will apply the Exp-function method to this equation. Suppose q(x,y,t)=U(ξ), where ξ=(k1x1α/Γ(1+α))+(k2x2α/Γ(1+α))+(l1y1α/Γ(1+α))+(l2y2α/Γ(1+α))+(ctα/Γ(1+α))+ξ0, k1,k2,l1,l2,c,ξ0 are all constants with k1,k2,l1,l2,c≠0. Then by the use of (5), (6) can be turned into
(7)4ck1U′′-k13k2U(4)+k23k1U(4)+12k1k2U′2+12k1k2UU′′-6l1l2U′′=0.First we suppose that the solution of (7) can be expressed by
(8)U(ξ)=∑p=-m1m2apepξ∑q=-n1n2bqeqξ.
To determine m2, n2, we need to balance the linear term of the highest order with the highest order nonlinear term in (7). By simple calculation, we have
(9)U(4)=c1e(15n2+m2)x+⋯+c2e-(15n1+m1)xc3e16n2x+⋯+c4e-16n1x,UU′′=c5e(3n2+2m2)x+⋯+c6e-(3n1+2m1)xc7e5n2x+⋯+c8e-5n1x,
where ci, i=1,2,3,4 are coefficients determined by am2, a-m1, bn2, b-n1. Balancing the highest order Exp-function in U(4) and UU′′, we have 15n2+m2=14n2+2m2, which implies m2=n2. Similarly, balancing the lowest order Exp-function in U(4) and UU′′, we have 15n1+m1=14n1+2m1, which implies m1=n1. For simplicity, we will proceed in two selected cases.Case 1.
Letm1=n1=1, m2=n2=1. Then
(10)U(ξ)=a-1e-ξ+a0+a1eξb-1e-ξ+b0+b1eξ.
Substituting (10) into (7), eliminating the denominator, and collecting all the terms with the same power of epξ together, equating each coefficient to zero, yield a set of algebraic equations. Solving these equations with the aid of the mathematical software Maple yields the following family of values of ai,bi, i=-1,0,1.Family  1. Consider
(11)a1=-b1(k1k23-k2k13+4ck1-6l1l2)12k1k2,a0=-b0(-5k1k23+5k2k13+4ck1-6l1l2)12k1k2,a-1=-b02(k1k23-k2k13+4ck1-6l1l2)48b1k1k2,b-1=b024b1,
where b1, b0 are arbitrary constants.Substituting the result above into (10), we can obtain the following exact solutions to (7):
(12)U1(ξ)=(-(k1k23-k2k13+4ck1-6l1l2)(b02e-ξ+4b12eξ)-4b0b1(-5k1k23+5k2k13+4ck1-6l1l2)b02e-ξ)×(12k1k2(b02e-ξ+4b12eξ)+48b0b1k1k2)-1,
where ξ=(k1x1α/Γ(1+α))+(k2x2α/Γ(1+α))+(l1y1α/Γ(1+α))+(l2y2α/Γ(1+α))+(ctα/Γ(1+α))+ξ0. If we especially take b0=2b1, then we obtain the following hyperbolic function solitary wave solution:
(13)U2(ξ)=-(k1k23-k2k13+4ck1-6l1l2)12k1k2+(k22-k12)4sech2ξ2.Case 2.
Letm1=n1=m, m2=n2=m, and for simplicity, we take
(14)U(ξ)=a-me-mξ+a0+amemξb-me-mξ+b0+bmemξ.
Substituting (14) into (7), eliminating the denominator, and collecting all the terms with the same power of epnξ together, equating each coefficient to zero, yield a set of algebraic equations. Solving these equations yields the following result.Family  2. Consider
(15)am=-bm(m2k1k23-m2k2k13+4ck1-6l1l2)12k1k2,a0=-b0(-5m2k1k23+5m2k2k13+4ck1-6l1l2)12k1k2,a-m=-b02(m2k1k23-m2k2k13+4ck1-6l1l2)48bmk1k2,b-m=b024bm,
where bm, b0 are arbitrary constants.Substituting the result above into (14), we can obtain the following exact solutions to (7):
(16)U3(ξ)=(-(m2k1k23-m2k2k13+4ck1-6l1l2)(b02e-ξ+4bm2eξ)-4b0bm(-5m2k1k23+5m2k2k13+4ck1-6l1l2)k13)×(12k1k2(b02e-ξ+4bm2eξ)+48b0bmk1k2)-1,
where ξ=(k1x1α/Γ(1+α))+(k2x2α/Γ(1+α))+(l1y1α/Γ(1+α))+(l2y2α/Γ(1+α))+(ctα/Γ(1+α))+ξ0. If we especially take b0=2bm, then we obtain the following solitary wave solution:
(17)U4(ξ)=-(m2k1k23-m2k2k13+4ck1-6l1l2)12k1k2+m2(k22-k12)4sech2(m2ξ).Now we suppose that the solution of (7) can be expressed by
(18)U(ξ)=∑p=-m1m2apeipξ∑q=-n1n2bqeiqξ.
Similar to the balancing process for (8), we have m1=n1 and m2=n2. Similar to the above, we only consider the following two cases.Case 1.
Letm1=n1=1, m2=n2=1. Then
(19)U(ξ)=a-1e-iξ+a0+a1eiξb-1e-iξ+b0+b1eiξ.
Substituting (19) into (7), eliminating the denominator, and collecting all the terms with the same power of eipξ together, equating each coefficient to zero, yield a set of algebraic equations. Solving these equations yields the following.Family  3. Consider
(20)a1=-b1(-k1k23+k2k13+4ck1-6l1l2)12k1k2,a0=-b0(5k1k23-5k2k13+4ck1-6l1l2)12k1k2,a-1=-b02(-k1k23+k2k13+4ck1-6l1l2)48b1k1k2,b-1=b024b1,
where b1, b0 are arbitrary constants.Substituting the result above into (19), we can obtain the following exact solutions to (7):
(21)U5(ξ)=(-(-k1k23+k2k13+4ck1-6l1l2)(b02e-iξ+4b12eiξ)-4b0b1(5k1k23-5k2k13+4ck1-6l1l2)4b12eiξ)×(12k1k2(b02e-iξ+4b12eiξ)+48b0b1k1k2)-1,
where ξ=(k1x1α/Γ(1+α))+(k2x2α/Γ(1+α))+(l1y1α/Γ(1+α))+(l2y2α/Γ(1+α))+(ctα/Γ(1+α))+ξ0. If we especially take b0=2b1, then we obtain the following trigonometric function solution:
(22)U6(ξ)=-(-k1k23+k2k13+4ck1-6l1l2)12k1k2-(k22-k12)4sec2ξ2.Case 2.
Let(23)U(ξ)=a-me-imξ+a0+ameimξb-me-imξ+b0+bmeimξ.
Substituting (23) into (7), eliminating the denominator, and collecting all the terms with the same power of eipnξ together, equating each coefficient to zero, yield a set of algebraic equations. Solving these equations, we obtain another family of values of ai,bi, i=-m,0,m.Family  4. Consider
(24)am=-bm(-m2k1k23+m2k2k13+4ck1-6l1l2)12k1k2,a0=-b0(5m2k1k23-5m2k2k13+4ck1-6l1l2)12k1k2,a-m=-b02(-m2k1k23+m2k2k13+4ck1-6l1l2)48bmk1k2,b-m=b024bm.
Substituting the result above into (23), we can obtain the following exact solutions to (7):
(25)U7(ξ)=(-(-m2k1k23+m2k2k13+4ck1-6l1l2)(b02e-ξ+4bm2eξ)-4b0bm(5m2k1k23-5m2k2k13+4ck1-6l1l2))4bm2eξ×(12k1k2(b02e-ξ+4bm2eξ)+48b0bmk1k2)-1,
where ξ=(k1x1α/Γ(1+α))+(k2x2α/Γ(1+α))+(l1y1α/Γ(1+α))+(l2y2α/Γ(1+α))+(ctα/Γ(1+α))+ξ0. If we especially take b0=2bm, then we obtain the following solitary wave solution:
(26)U8(ξ)=-(-m2k1k23+m2k2k13+4ck1-6l1l2)12k1k2-m2(k22-k12)4sec2(m2ξ).Remark 1.
Compared with the results in [22], the established solutions by (12), (16), (21), and (25) are new exact solutions to the space-time fractional Fokas equation, while the solutions by (13), (17), (22), and (26) are new solitary wave solutions.


## 2.2. Nonlinear Fractional Sharma-Tasso-Olver (STO) Equation

We consider the nonlinear fractional Sharma-Tasso-Olver (STO) equation with time-fractional derivative [26, 27]:
(27)Dtαu+3aux2+3au2ux+3auuxx+auxxx=0,0<α≤1.
In [26], the author solved (27) by the first integral method and obtained some hyperbolic function and trigonometric function solutions, while in [27], the variational iteration method, the Adomian decomposition method, and the homotopy perturbation method are used for obtaining a rational approximation solution for (27). Now we will apply the Exp-function method to (27). To begin with, we suppose u(x,t)=U(ξ), where ξ=kx+(ctα/Γ(1+α))+ξ0, k,c,ξ0 are all constants with k,c≠0. Then by use of (5), (27) can be turned into
(28)cU′+3ak2U′2+3akU2U′+3aUU′′+aU′′′=0.First we suppose that the solution of (28) can be expressed by
(29)U(ξ)=∑p=-m1m2apepξ∑q=-n1n2bqeqξ.
Similar to the previous subsection, by balancing the highest order Exp-function in U′′′ and U2U′, we have m2=n2, while we obtain m1=n1 by balancing the lowest order Exp-function in U′′′ and U2U′. Similar to the solving process for the space-time fractional Fokas equation, we will also proceed the computation under two selected cases.Case 1.
Letm1=n1=1, m2=n2=1. Then
(30)U(ξ)=a-1e-ξ+a0+a1eξb-1e-ξ+b0+b1eξ.
Substituting (30) into (28), eliminating the denominator, and collecting all the terms with the same power of epξ together, equating each coefficient to zero, yield a set of algebraic equations. Solving these equations yields two families of values for ai,bi, i=-1,0,1.Family  1. Consider
(31)a1=b1(3ak2∓-3a2k4-3akc)3ak,a0=0,b0=0,a-1=-b-1(2ak3±2k-3a2k4-3akc-c)3ak2±-3a2k4-3akc,
where b1, b-1 are arbitrary constants.Family  2. Consider
(32)a1=b1(3ak2∓-3a2k4-3akc)3ak,a0=b0(3ak2∓-3a2k4-3akc)3ak,a-1=b-1(9ak3±k-3a2k4-3akc+3c)6ak2±3-3a2k4-3akc,
where b1, b0, b-1 are arbitrary constants.Substituting the results above into (30), we can obtain the following exact solutions to (28):
(33)U1(ξ)=(-b-1(2ak3±2k-3a2k4-3akc-c)3ak2±-3a2k4-3akce-ξ+b1(3ak2∓-3a2k4-3akc)3akeξb-1(2ak3±2k-3a2k4-3akc-c)3ak2±-3a2k4-3akc)×(b-1e-ξ+b1eξ)-1,(34)U2(ξ)=(b-1(9ak3±k-3a2k4-3akc+3c)6ak2±3-3a2k4-3akce-ξ+b0(3ak2∓-3a2k4-3akc)3ak+b1(3ak2∓-3a2k4-3akc)3akeξb-1(9ak3±k-3a2k4-3akc+3c)6ak2±3-3a2k4-3akc)×(b-1e-ξ+b0+b1eξ)-1,
where ξ=kx+(ctα/Γ(1+α))+ξ0. If we especially take c=-ak3, b1=b-1 or c=-ak3, b1=-b-1 in (33), then we obtain the hyperbolic function solitary wave solutions:
(35)U3(ξ)=ktanhξ,U4(ξ)=kcothξ.Case 2.
Let(36)U(ξ)=a-me-mξ+a0+amemξb-me-mξ+b0+bmemξ.
Substituting (36) into (28), eliminating the denominator, and collecting all the terms with the same power of epnξ together, equating each coefficient to zero, yield a set of algebraic equations. Solving these equations yields the following results.Family  3. Consider
(37)am=bm(3amk2∓-3a2m2k4-3akc)3ak,a0=0,b0=0,a-m=-b-m(2am2k3±2mk-3a2m2k4-3akc-c)3amk2±-3a2m2k4-3akc,
where bm, b-m are arbitrary constants.Family  4. Consider
(38)am=bm(3amk2∓-3a2k4-3akc)3ak,a0=b0(3amk2∓-3a2m2k4-3akc)3ak,a-m=b-m(9am2k3±mk-3a2m2k4-3akc+3c)6amk2±3-3a2m2k4-3akc,
where bm, b0, b-m are arbitrary constants.Substituting the results above into (36), we can obtain the following exact solutions to (28):
(39)U5(ξ)=(-b-m(2am2k3±2mk-3a2m2k4-3akc-c)3amk2±-3a2m2k4-3akce-mξ+bm(3amk2∓-3a2m2k4-3akc)3akemξb-m(2am2k3±2mk-3a2m2k4-3akc-c)3amk2±-3a2m2k4-3akc)×(b-me-mξ+bmemξ)-1,(40)U6(ξ)=(b-m(9am2k3±mk-3a2m2k4-3akc+3c)6amk2±3-3a2m2k4-3akce-ξ+bm(3amk2∓-3a2k4-3akc)3akemξ+b0(3amk2∓-3a2m2k4-3akc)3akb-m(9am2k3±mk-3a2m2k4-3akc+3c)6amk2±3-3a2m2k4-3akc)×(b-me-mξ+b0+bmemξ)-1,
where ξ=kx+(ctα/Γ(1+α))+ξ0. If we especially take c=-am2k3, bm=b-m or c=-am2k3, bm=-b-m in (39), then we obtain the hyperbolic function solitary wave solutions:
(41)U7(ξ)=mktanhξ,U8(ξ)=mkcothξ.Now we suppose that the solution of (28) can be expressed by
(42)U(ξ)=∑p=-m1m2apeipξ∑q=-n1n2bqeiqξ.
By the balancing process for (28), we have m1=n1 and m2=n2. Similar to the above, we will also select two cases for computation.Case 1.
Letm1=n1=1, m2=n2=1. Then
(43)U(ξ)=a-1e-iξ+a0+a1eiξb-1e-iξ+b0+b1eiξ.
Substituting (43) into (28), eliminating the denominator, and collecting all the terms with the same power of eipξ together, equating each coefficient to zero, yield a set of algebraic equations. Solving these equations, we obtain the following two families of values.Family  5. Consider
(44)a1=b1(3iak2∓3a2k4-3akc)3ak,a0=0,a-1=-b-1(-2ak3±2ik3a2k4-3akc-c)3iak2±3a2k4-3akc,
where b1, b-1 are arbitrary constants.Family  6. Consider
(45)a1=b1(3iak2∓3a2k4-3akc)3ak,a0=b0(3iak2∓3a2k4-3akc)3ak,a-1=b-1(-9ak3±ik3a2k4-3akc+3c)6iak2±33a2k4-3akc,
where b1, b0, b-1 are arbitrary constants.Substituting the results above into (43), we can obtain the following exact solutions to (28):
(46)U9(ξ)=(-b-1(-2ak3±2ik3a2k4-3akc-c)3iak2±3a2k4-3akce-iξ+b1(3iak2∓3a2k4-3akc)3akeiξb-1(-2ak3±2ik3a2k4-3akc-c)3iak2±3a2k4-3akc)×(b-1e-iξ+b1eiξ)-1,(47)U10(ξ)=(b-1(-9ak3±ik3a2k4-3akc+3c)6iak2±33a2k4-3akce-iξ+b0(3iak2∓3a2k4-3akc)3ak+b1(3iak2∓3a2k4-3akc)3akeiξb-1(-9ak3±ik3a2k4-3akc+3c)6iak2±33a2k4-3akc)×(b-1e-iξ+b0+b1eiξ)-1If we especially takec=-ak3, b1=b-1 or c=-ak3, b1=-b-1 in (46), then we obtain the trigonometric function solutions:
(48)U11(ξ)=-ktanξ,U12(ξ)=-kcotξ.Case 2.
Let(49)U(ξ)=a-me-imξ+a0+ameimξb-me-imξ+b0+bmeimξ.
Substituting (49) into (28), eliminating the denominator, and collecting all the terms with the same power of eipmξ together, equating each coefficient to zero, yield a set of algebraic equations. Solving these equations, we obtain the following two families of values.Family  7. Consider
(50)am=bm(3iamk2∓3a2m2k4-3akc)3ak,a0=0,a-m=-b-m(-2am2k3±2imk3a2m2k4-3akc-c)3iamk2±3a2m2k4-3akc,
where bm, b-m are arbitrary constants.Family  8. Consider
(51)am=bm(3iamk2∓3a2k4-3akc)3ak,a0=b0(3iamk2∓3a2m2k4-3akc)3ak,a-m=b-m(-9am2k3±mk3a2m2k4-3akc+3c)6iamk2±33a2m2k4-3akc,
where bm, b0, b-m are arbitrary constants.Substituting the results above into (49), we can obtain the following exact solutions to (28):
(52)U13(ξ)=(-b-m(-2am2k3±2imk3a2m2k4-3akc-c)3iamk2±3a2m2k4-3akce-imξ+bm(3iamk2∓3a2m2k4-3akc)3akeimξ)×(b-me-imξ+bmeimξ)-1,(53)U14(ξ)=(b-m(-9am2k3±mk3a2m2k4-3akc+3c)6iamk2±33a2m2k4-3akce-imξ+bm(3iamk2∓3a2k4-3akc)3akeimξ+b0(3iamk2∓3a2m2k4-3akc)3ak)×(b-me-imξ+b0+bmeimξ)-1.
If we especially take c=-am2k3, bm=b-m or c=-am2k3, bm=-b-m in (53), then we obtain the trigonometric function solutions:
(54)U11(ξ)=-mktanξ,U12(ξ)=-mkcotξ.Remark 2.
If we letm=2, then the solutions by (41), (54) reduce to the solutions established by Lu [26, equations (55), (56), (53), (54)]. So in this way, our solutions (41), (54) are of more general forms. Moreover, we note that the solutions denoted by (33), (34), (39), (40), (46), (47), (52), and (53) are essentially different from the results in [26, 27] and are new exact solutions to the nonlinear fractional Sharma-Tasso-Olver equation so far in the literature, while the solutions (35), (48) are new solitary wave solutions to it.


## 3. Conclusions

We have extended the Exp-function method to solve fractional partial differential equations successfully. As applications, some generalized and new exact solutions for the space-time fractional Fokas equation and the nonlinear fractional Sharma-Tasso-Olver (STO) equation have been successfully found. As one can see, this method is based on the homogeneous balancing principle. So it can also be applied to other fractional partial differential equations where the homogeneous balancing principle is satisfied.



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*Source: 465723-2013-05-28.xml*