Exp-Function Method for Solving Fractional Partial Differential Equations

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.

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,…,𝑥_𝑛), a given fractional partial differential equation expressed in independent variables 𝑡,𝑥₁,𝑥₂,…,𝑥_𝑛 can be turned into another ordinary differential equation of integer order, whose solutions are supposed to have one of the following forms: 𝑢(𝜉)=∑^𝑚2_{𝑝=−𝑚1}𝑎_𝑝𝑒^𝑝𝜉∑^𝑛2_{𝑞=−𝑛1}𝑏_𝑞𝑒^𝑞𝜉(𝜉)=∑or𝑢^{𝑚2𝑝=−𝑚1}𝑎_𝑝𝑒^𝑖𝑝𝜉∑^𝑛2_{𝑞=−𝑛1}𝑏_𝑞𝑒^𝑖𝑞𝜉, (1) 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 order 𝛼 is defined by the following expression [25]: 𝐷^𝛼_𝑡𝑓(𝑡)=1Γ()𝑑1−𝛼∫𝑑𝑡^𝑡0(𝑡−𝜉)^−𝛼𝑓(𝜉)(0)𝑓−𝑓𝑑𝜉,0<𝛼<1,_(𝑛)(𝑡)^{(𝛼−𝑛)},𝑛≤𝛼<𝑛+1,𝑛≥1. (2) We list some important properties for the modified Riemann-Liouville derivative as follows [25, (3.10)–(3.13)] (see also in [22–24, 26]): 𝐷^𝛼_𝑡𝑡^𝑟=Γ(1+𝑟)Γ()𝑡1+𝑟−𝛼^𝑟−𝛼,𝐷^𝛼_𝑡𝑓(𝑡)𝑔(𝑡)=𝑔(𝑡)𝐷^𝛼_𝑡𝑓(𝑡)+𝑓(𝑡)𝐷^𝛼_𝑡𝑔(𝑡),𝐷^𝛼_𝑡𝑓𝑔(𝑡)=𝑓^′_𝑔𝑔(𝑡)𝐷^𝛼_𝑡𝑔(𝑡)=𝐷^𝛼_𝑔𝑓𝑔(𝑡)𝑔^′(𝑡)^𝛼. (3)(4)(5)

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] 4𝜕^2𝛼𝑞𝜕𝑡^𝛼𝜕𝑥^𝛼₁−𝜕^4𝛼𝑞𝜕𝑥₁^3𝛼𝜕𝑥^𝛼₂+𝜕^4𝛼𝑞𝜕𝑥₂^3𝛼𝜕𝑥^𝛼₁𝜕+12^𝛼𝑞𝜕𝑥^𝛼₁𝜕^𝛼𝑞𝜕𝑥^𝛼₂𝜕+12𝑞^2𝛼𝑞𝜕𝑥^𝛼₁𝜕𝑥^𝛼₂𝜕−6^2𝛼𝑞𝜕𝑦^𝛼₁𝜕𝑦^𝛼₂=0,0<𝛼≤1. (6) 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=(𝑘₁𝑥^𝛼₁/Γ(1+𝛼))+(𝑘₂𝑥^𝛼₂/Γ(1+𝛼))+(𝑙₁𝑦^𝛼₁/Γ(1+𝛼))+(𝑙₂𝑦^𝛼₂/Γ(1+𝛼))+(𝑐𝑡^𝛼/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 4𝑐𝑘₁𝑈^′′−𝑘³₁𝑘₂𝑈⁽⁴⁾+𝑘³₂𝑘₁𝑈⁽⁴⁾+12𝑘₁𝑘₂𝑈^′2+12𝑘₁𝑘₂𝑈𝑈^′′−6𝑙₁𝑙₂𝑈^′′=0. (7)

First we suppose that the solution of (7) can be expressed by 𝑈(𝜉)=∑^𝑚2_{𝑝=−𝑚1}𝑎_𝑝𝑒^𝑝𝜉∑^𝑛2_{𝑞=−𝑛1}𝑏_𝑞𝑒^𝑞𝜉. (8) 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 𝑈⁽⁴⁾=𝑐₁𝑒^{(15𝑛2+𝑚2)𝑥}+⋯+𝑐₂𝑒^{−(15𝑛1+𝑚1)𝑥}𝑐₃𝑒^16𝑛2𝑥+⋯+𝑐₄𝑒^{−16𝑛1𝑥},𝑈𝑈^′′=𝑐₅𝑒^{(3𝑛2+2𝑚2)𝑥}+⋯+𝑐₆𝑒^{−(3𝑛1+2𝑚1)𝑥}𝑐₇𝑒^5𝑛2𝑥+⋯+𝑐₈𝑒^{−5𝑛1𝑥}, (9) where 𝑐_𝑖, 𝑖=1,2,3,4 are coefficients determined by 𝑎_𝑚2, 𝑎_−𝑚1, 𝑏_𝑛2, 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. Let 𝑚₁=𝑛₁=1, 𝑚₂=𝑛₂=1. Then 𝑈(𝜉)=𝑎₋₁𝑒^−𝜉+𝑎₀+𝑎₁𝑒^𝜉𝑏₋₁𝑒^−𝜉+𝑏₀+𝑏₁𝑒^𝜉. (10) Substituting (10) into (7), eliminating the denominator, and collecting all the terms with the same power of 𝑒^𝑝𝜉 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 𝑎_𝑖,𝑏_𝑖, 𝑖=−1,0,1.

Family  1. Consider 𝑎₁𝑏=−¹𝑘₁𝑘³₂−𝑘₂𝑘³₁+4𝑐𝑘₁−6𝑙₁𝑙₂12𝑘₁𝑘₂,𝑎₀𝑏=−⁰−5𝑘₁𝑘³₂+5𝑘₂𝑘³₁+4𝑐𝑘₁−6𝑙₁𝑙₂12𝑘₁𝑘₂,𝑎₋₁𝑏=−²⁰𝑘₁𝑘³₂−𝑘₂𝑘³₁+4𝑐𝑘₁−6𝑙₁𝑙₂48𝑏₁𝑘₁𝑘₂,𝑏₋₁=𝑏²₀4𝑏₁, (11) where 𝑏₁, 𝑏₀ are arbitrary constants.

Substituting the result above into (10), we can obtain the following exact solutions to (7): 𝑈₁(𝜉)=−𝑘₁𝑘³₂−𝑘₂𝑘³₁+4𝑐𝑘₁−6𝑙₁𝑙₂𝑏²₀𝑒^−𝜉+4𝑏²₁𝑒^𝜉−4𝑏₀𝑏₁−5𝑘₁𝑘³₂+5𝑘₂𝑘³₁+4𝑐𝑘₁−6𝑙₁𝑙₂×12𝑘₁𝑘₂𝑏²₀𝑒^−𝜉+4𝑏²₁𝑒^𝜉+48𝑏₀𝑏₁𝑘₁𝑘₂⁻¹, (12) where 𝜉=(𝑘₁𝑥^𝛼₁/Γ(1+𝛼))+(𝑘₂𝑥^𝛼₂/Γ(1+𝛼))+(𝑙₁𝑦^𝛼₁/Γ(1+𝛼))+(𝑙₂𝑦^𝛼₂/Γ(1+𝛼))+(𝑐𝑡^𝛼/Γ(1+𝛼))+𝜉₀. If we especially take 𝑏₀=2𝑏₁, then we obtain the following hyperbolic function solitary wave solution: 𝑈₂(𝜉)𝑘=−¹𝑘³₂−𝑘₂𝑘³₁+4𝑐𝑘₁−6𝑙₁𝑙₂12𝑘₁𝑘₂+𝑘²₂−𝑘²₁4sech²𝜉2. (13)

Case 2. Let 𝑚₁=𝑛₁=𝑚, 𝑚₂=𝑛₂=𝑚, and for simplicity, we take 𝑈(𝜉)=𝑎_−𝑚𝑒^−𝑚𝜉+𝑎₀+𝑎_𝑚𝑒^𝑚𝜉𝑏_−𝑚𝑒^−𝑚𝜉+𝑏₀+𝑏_𝑚𝑒^𝑚𝜉. (14) Substituting (14) into (7), eliminating the denominator, and collecting all the terms with the same power of 𝑒^𝑝𝑛𝜉 together, equating each coefficient to zero, yield a set of algebraic equations. Solving these equations yields the following result.

Family  2. Consider 𝑎_𝑚𝑏=−^𝑚𝑚²𝑘₁𝑘³₂−𝑚²𝑘₂𝑘³₁+4𝑐𝑘₁−6𝑙₁𝑙₂12𝑘₁𝑘₂,𝑎₀𝑏=−⁰−5𝑚²𝑘₁𝑘³₂+5𝑚²𝑘₂𝑘³₁+4𝑐𝑘₁−6𝑙₁𝑙₂12𝑘₁𝑘₂,𝑎_−𝑚𝑏=−²⁰𝑚²𝑘₁𝑘³₂−𝑚²𝑘₂𝑘³₁+4𝑐𝑘₁−6𝑙₁𝑙₂48𝑏_𝑚𝑘₁𝑘₂,𝑏_−𝑚=𝑏²₀4𝑏_𝑚, (15) where 𝑏_𝑚, 𝑏₀ are arbitrary constants.

Substituting the result above into (14), we can obtain the following exact solutions to (7): 𝑈₃(𝜉)=−𝑚²𝑘₁𝑘³₂−𝑚²𝑘₂𝑘³₁+4𝑐𝑘₁−6𝑙₁𝑙₂𝑏²₀𝑒^−𝜉+4𝑏²_𝑚𝑒^𝜉−4𝑏₀𝑏_𝑚−5𝑚²𝑘₁𝑘³₂+5𝑚²𝑘₂𝑘³₁+4𝑐𝑘₁−6𝑙₁𝑙₂×12𝑘₁𝑘₂(𝑏²₀𝑒^−𝜉+4𝑏²_𝑚𝑒^𝜉)+48𝑏₀𝑏_𝑚𝑘₁𝑘₂⁻¹, (16) where 𝜉=(𝑘₁𝑥^𝛼₁/Γ(1+𝛼))+(𝑘₂𝑥^𝛼₂/Γ(1+𝛼))+(𝑙₁𝑦^𝛼₁/Γ(1+𝛼))+(𝑙₂𝑦^𝛼₂/Γ(1+𝛼))+(𝑐𝑡^𝛼/Γ(1+𝛼))+𝜉₀. If we especially take 𝑏₀=2𝑏_𝑚, then we obtain the following solitary wave solution: 𝑈₄(𝜉)𝑚=−²𝑘₁𝑘³₂−𝑚²𝑘₂𝑘³₁+4𝑐𝑘₁−6𝑙₁𝑙₂12𝑘₁𝑘₂+𝑚²𝑘²₂−𝑘²₁4sech²𝑚2𝜉. (17)

Now we suppose that the solution of (7) can be expressed by 𝑈(𝜉)=∑^𝑚2_{𝑝=−𝑚1}𝑎_𝑝𝑒^𝑖𝑝𝜉∑^𝑛2_{𝑞=−𝑛1}𝑏_𝑞𝑒^𝑖𝑞𝜉. (18) 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. Let 𝑚₁=𝑛₁=1, 𝑚₂=𝑛₂=1. Then 𝑈(𝜉)=𝑎₋₁𝑒^−𝑖𝜉+𝑎₀+𝑎₁𝑒^𝑖𝜉𝑏₋₁𝑒^−𝑖𝜉+𝑏₀+𝑏₁𝑒^𝑖𝜉. (19) Substituting (19) into (7), eliminating the denominator, and collecting all the terms with the same power of 𝑒^𝑖𝑝𝜉 together, equating each coefficient to zero, yield a set of algebraic equations. Solving these equations yields the following.

Family  3. Consider 𝑎₁𝑏=−¹−𝑘₁𝑘³₂+𝑘₂𝑘³₁+4𝑐𝑘₁−6𝑙₁𝑙₂12𝑘₁𝑘₂,𝑎₀𝑏=−⁰5𝑘₁𝑘³₂−5𝑘₂𝑘³₁+4𝑐𝑘₁−6𝑙₁𝑙₂12𝑘₁𝑘₂,𝑎₋₁𝑏=−²⁰−𝑘₁𝑘³₂+𝑘₂𝑘³₁+4𝑐𝑘₁−6𝑙₁𝑙₂48𝑏₁𝑘₁𝑘₂,𝑏₋₁=𝑏²₀4𝑏₁, (20) where 𝑏₁, 𝑏₀ are arbitrary constants.

Substituting the result above into (19), we can obtain the following exact solutions to (7): 𝑈₅(𝜉)=−−𝑘₁𝑘³₂+𝑘₂𝑘³₁+4𝑐𝑘₁−6𝑙₁𝑙₂𝑏²₀𝑒^−𝑖𝜉+4𝑏²₁𝑒^𝑖𝜉−4𝑏₀𝑏₁5𝑘₁𝑘³₂−5𝑘₂𝑘³₁+4𝑐𝑘₁−6𝑙₁𝑙₂×12𝑘₁𝑘₂𝑏²₀𝑒^−𝑖𝜉+4𝑏²₁𝑒^𝑖𝜉+48𝑏₀𝑏₁𝑘₁𝑘₂⁻¹, (21) where 𝜉=(𝑘₁𝑥^𝛼₁/Γ(1+𝛼))+(𝑘₂𝑥^𝛼₂/Γ(1+𝛼))+(𝑙₁𝑦^𝛼₁/Γ(1+𝛼))+(𝑙₂𝑦^𝛼₂/Γ(1+𝛼))+(𝑐𝑡^𝛼/Γ(1+𝛼))+𝜉₀. If we especially take 𝑏₀=2𝑏₁, then we obtain the following trigonometric function solution: 𝑈₆(𝜉)=−−𝑘₁𝑘³₂+𝑘₂𝑘³₁+4𝑐𝑘₁−6𝑙₁𝑙₂12𝑘₁𝑘₂−𝑘²₂−𝑘²₁4sec²𝜉2. (22)

Case 2. Let 𝑈(𝜉)=𝑎_−𝑚𝑒^{−𝑖𝑚𝜉}+𝑎₀+𝑎_𝑚𝑒^𝑖𝑚𝜉𝑏_−𝑚𝑒^{−𝑖𝑚𝜉}+𝑏₀+𝑏_𝑚𝑒^𝑖𝑚𝜉. (23) Substituting (23) into (7), eliminating the denominator, and collecting all the terms with the same power of 𝑒^{𝑖𝑝𝑛𝜉} together, equating each coefficient to zero, yield a set of algebraic equations. Solving these equations, we obtain another family of values of 𝑎_𝑖,𝑏_𝑖, 𝑖=−𝑚,0,𝑚.

Family  4. Consider 𝑎_𝑚𝑏=−^𝑚−𝑚²𝑘₁𝑘³₂+𝑚²𝑘₂𝑘³₁+4𝑐𝑘₁−6𝑙₁𝑙₂12𝑘₁𝑘₂,𝑎₀𝑏=−⁰5𝑚²𝑘₁𝑘³₂−5𝑚²𝑘₂𝑘³₁+4𝑐𝑘₁−6𝑙₁𝑙₂12𝑘₁𝑘₂,𝑎_−𝑚𝑏=−²⁰−𝑚²𝑘₁𝑘³₂+𝑚²𝑘₂𝑘³₁+4𝑐𝑘₁−6𝑙₁𝑙₂48𝑏_𝑚𝑘₁𝑘₂,𝑏_−𝑚=𝑏²₀4𝑏_𝑚. (24) Substituting the result above into (23), we can obtain the following exact solutions to (7): 𝑈₇(𝜉)=−−𝑚²𝑘₁𝑘³₂+𝑚²𝑘₂𝑘³₁+4𝑐𝑘₁−6𝑙₁𝑙₂𝑏²₀𝑒^−𝜉+4𝑏²_𝑚𝑒^𝜉−4𝑏₀𝑏_𝑚5𝑚²𝑘₁𝑘³₂−5𝑚²𝑘₂𝑘³₁+4𝑐𝑘₁−6𝑙₁𝑙₂×12𝑘₁𝑘₂𝑏²₀𝑒^−𝜉+4𝑏²_𝑚𝑒^𝜉+48𝑏₀𝑏_𝑚𝑘₁𝑘₂⁻¹, (25) where 𝜉=(𝑘₁𝑥^𝛼₁/Γ(1+𝛼))+(𝑘₂𝑥^𝛼₂/Γ(1+𝛼))+(𝑙₁𝑦^𝛼₁/Γ(1+𝛼))+(𝑙₂𝑦^𝛼₂/Γ(1+𝛼))+(𝑐𝑡^𝛼/Γ(1+𝛼))+𝜉₀. If we especially take 𝑏₀=2𝑏_𝑚, then we obtain the following solitary wave solution: 𝑈₈(𝜉)=−−𝑚²𝑘₁𝑘³₂+𝑚²𝑘₂𝑘³₁+4𝑐𝑘₁−6𝑙₁𝑙₂12𝑘₁𝑘₂−𝑚²𝑘²₂−𝑘²₁4sec²𝑚2𝜉. (26)

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]: 𝐷^𝛼_𝑡𝑢+3𝑎𝑢²_𝑥+3𝑎𝑢²𝑢_𝑥+3𝑎𝑢𝑢_𝑥𝑥+𝑎𝑢_𝑥𝑥𝑥=0,0<𝛼≤1. (27) 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+(𝑐𝑡^𝛼/Γ(1+𝛼))+𝜉₀, 𝑘,𝑐,𝜉₀ are all constants with 𝑘,𝑐≠0. Then by use of (5), (27) can be turned into 𝑐𝑈^′+3𝑎𝑘²𝑈^′2+3𝑎𝑘𝑈²𝑈^′+3𝑎𝑈𝑈^′′+𝑎𝑈^′′′=0. (28)

First we suppose that the solution of (28) can be expressed by 𝑈(𝜉)=∑^𝑚2_{𝑝=−𝑚1}𝑎_𝑝𝑒^𝑝𝜉∑^𝑛2_{𝑞=−𝑛1}𝑏_𝑞𝑒^𝑞𝜉. (29) 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. Let 𝑚₁=𝑛₁=1, 𝑚₂=𝑛₂=1. Then 𝑈(𝜉)=𝑎₋₁𝑒^−𝜉+𝑎₀+𝑎₁𝑒^𝜉𝑏₋₁𝑒^−𝜉+𝑏₀+𝑏₁𝑒^𝜉. (30) Substituting (30) into (28), eliminating the denominator, and collecting all the terms with the same power of 𝑒^𝑝𝜉 together, equating each coefficient to zero, yield a set of algebraic equations. Solving these equations yields two families of values for 𝑎_𝑖,𝑏_𝑖, 𝑖=−1,0,1.

Family  1. Consider 𝑎₁=𝑏₁3𝑎𝑘²∓−3𝑎²𝑘⁴−3𝑎𝑘𝑐3𝑎𝑘,𝑎₀=0,𝑏₀𝑎=0,₋₁𝑏=−⁻¹2𝑎𝑘³±2𝑘−3𝑎²𝑘⁴−3𝑎𝑘𝑐−𝑐3𝑎𝑘²±−3𝑎²𝑘⁴,−3𝑎𝑘𝑐 (31) where 𝑏₁, 𝑏₋₁ are arbitrary constants.

Family  2. Consider 𝑎₁=𝑏₁3𝑎𝑘²∓−3𝑎²𝑘⁴−3𝑎𝑘𝑐,𝑎3𝑎𝑘₀=𝑏₀3𝑎𝑘²∓−3𝑎²𝑘⁴−3𝑎𝑘𝑐,𝑎3𝑎𝑘₋₁=𝑏₋₁9𝑎𝑘³±𝑘−3𝑎²𝑘⁴−3𝑎𝑘𝑐+3𝑐6𝑎𝑘²±3−3𝑎²𝑘⁴,−3𝑎𝑘𝑐 (32) where 𝑏₁, 𝑏₀, 𝑏₋₁ are arbitrary constants.

Substituting the results above into (30), we can obtain the following exact solutions to (28): 𝑈₁(𝜉)=−𝑏₋₁2𝑎𝑘³±2𝑘−3𝑎²𝑘⁴−3𝑎𝑘𝑐−𝑐3𝑎𝑘²±−3𝑎²𝑘⁴𝑒−3𝑎𝑘𝑐^−𝜉+𝑏₁3𝑎𝑘²∓−3𝑎²𝑘⁴−3𝑎𝑘𝑐𝑒3𝑎𝑘^𝜉×𝑏₋₁𝑒^−𝜉+𝑏₁𝑒^𝜉−1,𝑈₂(𝜉)=𝑏₋₁9𝑎𝑘³±𝑘−3𝑎²𝑘⁴−3𝑎𝑘𝑐+3𝑐6𝑎𝑘²±3−3𝑎²𝑘⁴𝑒−3𝑎𝑘𝑐^−𝜉+𝑏₀3𝑎𝑘²∓−3𝑎²𝑘⁴−3𝑎𝑘𝑐+𝑏3𝑎𝑘₁3𝑎𝑘²∓−3𝑎²𝑘⁴−3𝑎𝑘𝑐𝑒3𝑎𝑘^𝜉×𝑏₋₁𝑒^−𝜉+𝑏₀+𝑏₁𝑒^𝜉−1, (33)(34) where 𝜉=𝑘𝑥+(𝑐𝑡^𝛼/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: 𝑈₃(𝜉)𝑈=𝑘tanh𝜉,₄(𝜉)=𝑘coth𝜉. (35)

Case 2. Let 𝑈(𝜉)=𝑎_−𝑚𝑒^−𝑚𝜉+𝑎₀+𝑎_𝑚𝑒^𝑚𝜉𝑏_−𝑚𝑒^−𝑚𝜉+𝑏₀+𝑏_𝑚𝑒^𝑚𝜉. (36) Substituting (36) into (28), eliminating the denominator, and collecting all the terms with the same power of 𝑒^𝑝𝑛𝜉 together, equating each coefficient to zero, yield a set of algebraic equations. Solving these equations yields the following results.

Family  3. Consider 𝑎_𝑚=𝑏_𝑚3𝑎𝑚𝑘²∓−3𝑎²𝑚²𝑘⁴−3𝑎𝑘𝑐3𝑎𝑘,𝑎₀=0,𝑏₀𝑎=0,_−𝑚𝑏=−^−𝑚2𝑎𝑚²𝑘³±2𝑚𝑘−3𝑎²𝑚²𝑘⁴−3𝑎𝑘𝑐−𝑐3𝑎𝑚𝑘²±−3𝑎²𝑚²𝑘⁴,−3𝑎𝑘𝑐 (37) where 𝑏_𝑚, 𝑏_−𝑚 are arbitrary constants.

Family  4. Consider 𝑎_𝑚=𝑏_𝑚3𝑎𝑚𝑘²∓−3𝑎²𝑘⁴−3𝑎𝑘𝑐,𝑎3𝑎𝑘₀=𝑏₀3𝑎𝑚𝑘²∓−3𝑎²𝑚²𝑘⁴−3𝑎𝑘𝑐,𝑎3𝑎𝑘_−𝑚=𝑏_−𝑚9𝑎𝑚²𝑘³±𝑚𝑘−3𝑎²𝑚²𝑘⁴−3𝑎𝑘𝑐+3𝑐6𝑎𝑚𝑘²±3−3𝑎²𝑚²𝑘⁴,−3𝑎𝑘𝑐 (38) where 𝑏_𝑚, 𝑏₀, 𝑏_−𝑚 are arbitrary constants.

Substituting the results above into (36), we can obtain the following exact solutions to (28): 𝑈₅(𝜉)=−𝑏_−𝑚2𝑎𝑚²𝑘³±2𝑚𝑘−3𝑎²𝑚²𝑘⁴−3𝑎𝑘𝑐−𝑐3𝑎𝑚𝑘²±−3𝑎²𝑚²𝑘⁴𝑒−3𝑎𝑘𝑐^−𝑚𝜉+𝑏_𝑚3𝑎𝑚𝑘²∓−3𝑎²𝑚²𝑘⁴−3𝑎𝑘𝑐𝑒3𝑎𝑘^𝑚𝜉×𝑏_−𝑚𝑒^−𝑚𝜉+𝑏_𝑚𝑒^𝑚𝜉−1,𝑈₆(𝜉)=𝑏_−𝑚9𝑎𝑚²𝑘³±𝑚𝑘−3𝑎²𝑚²𝑘⁴−3𝑎𝑘𝑐+3𝑐6𝑎𝑚𝑘²±3−3𝑎²𝑚²𝑘⁴𝑒−3𝑎𝑘𝑐^−𝜉+𝑏_𝑚3𝑎𝑚𝑘²∓−3𝑎²𝑘⁴−3𝑎𝑘𝑐𝑒3𝑎𝑘^𝑚𝜉+𝑏₀3𝑎𝑚𝑘²∓−3𝑎²𝑚²𝑘⁴−3𝑎𝑘𝑐×𝑏3𝑎𝑘_−𝑚𝑒^−𝑚𝜉+𝑏₀+𝑏_𝑚𝑒^𝑚𝜉−1, (39)(40) where 𝜉=𝑘𝑥+(𝑐𝑡^𝛼/Γ(1+𝛼))+𝜉₀. If we especially take 𝑐=−𝑎𝑚²𝑘³, 𝑏_𝑚=𝑏_−𝑚 or 𝑐=−𝑎𝑚²𝑘³, 𝑏_𝑚=−𝑏_−𝑚 in (39), then we obtain the hyperbolic function solitary wave solutions: 𝑈₇(𝜉)𝑈=𝑚𝑘tanh𝜉,₈(𝜉)=𝑚𝑘coth𝜉. (41)

Now we suppose that the solution of (28) can be expressed by 𝑈(𝜉)=∑^𝑚2_{𝑝=−𝑚1}𝑎_𝑝𝑒^𝑖𝑝𝜉∑^𝑛2_{𝑞=−𝑛1}𝑏_𝑞𝑒^𝑖𝑞𝜉. (42) 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. Let 𝑚₁=𝑛₁=1, 𝑚₂=𝑛₂=1. Then 𝑈(𝜉)=𝑎₋₁𝑒^−𝑖𝜉+𝑎₀+𝑎₁𝑒^𝑖𝜉𝑏₋₁𝑒^−𝑖𝜉+𝑏₀+𝑏₁𝑒^𝑖𝜉. (43) Substituting (43) into (28), eliminating the denominator, and collecting all the terms with the same power of 𝑒^𝑖𝑝𝜉 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 𝑎₁=𝑏₁3𝑖𝑎𝑘²∓3𝑎²𝑘⁴−3𝑎𝑘𝑐3𝑎𝑘,𝑎₀𝑎=0,₋₁𝑏=−⁻¹−2𝑎𝑘³±2𝑖𝑘3𝑎²𝑘⁴−3𝑎𝑘𝑐−𝑐3𝑖𝑎𝑘²±3𝑎²𝑘⁴,−3𝑎𝑘𝑐 (44) where 𝑏₁, 𝑏₋₁ are arbitrary constants.

Family  6. Consider 𝑎₁=𝑏₁3𝑖𝑎𝑘²∓3𝑎²𝑘⁴−3𝑎𝑘𝑐,𝑎3𝑎𝑘₀=𝑏₀3𝑖𝑎𝑘²∓3𝑎²𝑘⁴−3𝑎𝑘𝑐,𝑎3𝑎𝑘₋₁=𝑏₋₁−9𝑎𝑘³±𝑖𝑘3𝑎²𝑘⁴−3𝑎𝑘𝑐+3𝑐6𝑖𝑎𝑘²±33𝑎²𝑘⁴,−3𝑎𝑘𝑐 (45) where 𝑏₁, 𝑏₀, 𝑏₋₁ are arbitrary constants.

Substituting the results above into (43), we can obtain the following exact solutions to (28): 𝑈₉(𝜉)=−𝑏₋₁−2𝑎𝑘³±2𝑖𝑘3𝑎²𝑘⁴−3𝑎𝑘𝑐−𝑐3𝑖𝑎𝑘²±3𝑎²𝑘⁴𝑒−3𝑎𝑘𝑐^−𝑖𝜉+𝑏₁3𝑖𝑎𝑘²∓3𝑎²𝑘⁴−3𝑎𝑘𝑐𝑒3𝑎𝑘^𝑖𝜉×𝑏₋₁𝑒^−𝑖𝜉+𝑏₁𝑒^𝑖𝜉−1,𝑈₁₀(𝜉)=𝑏₋₁−9𝑎𝑘³±𝑖𝑘3𝑎²𝑘⁴−3𝑎𝑘𝑐+3𝑐6𝑖𝑎𝑘²±33𝑎²𝑘⁴𝑒−3𝑎𝑘𝑐^−𝑖𝜉+𝑏₀3𝑖𝑎𝑘²∓3𝑎²𝑘⁴−3𝑎𝑘𝑐+𝑏3𝑎𝑘₁3𝑖𝑎𝑘²∓3𝑎²𝑘⁴−3𝑎𝑘𝑐𝑒3𝑎𝑘^𝑖𝜉×𝑏₋₁𝑒^−𝑖𝜉+𝑏₀+𝑏₁𝑒^𝑖𝜉−1 (46)(47)

If we especially take 𝑐=−𝑎𝑘³, 𝑏₁=𝑏₋₁ or 𝑐=−𝑎𝑘³, 𝑏₁=−𝑏₋₁ in (46), then we obtain the trigonometric function solutions: 𝑈₁₁(𝜉)𝑈=−𝑘tan𝜉,₁₂(𝜉)=−𝑘cot𝜉. (48)

Case 2. Let 𝑈(𝜉)=𝑎_−𝑚𝑒^{−𝑖𝑚𝜉}+𝑎₀+𝑎_𝑚𝑒^𝑖𝑚𝜉𝑏_−𝑚𝑒^{−𝑖𝑚𝜉}+𝑏₀+𝑏_𝑚𝑒^𝑖𝑚𝜉. (49) Substituting (49) into (28), eliminating the denominator, and collecting all the terms with the same power of 𝑒^{𝑖𝑝𝑚𝜉} 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 𝑎_𝑚=𝑏_𝑚3𝑖𝑎𝑚𝑘²∓3𝑎²𝑚²𝑘⁴−3𝑎𝑘𝑐3𝑎𝑘,𝑎₀𝑎=0,_−𝑚𝑏=−^−𝑚−2𝑎𝑚²𝑘³±2𝑖𝑚𝑘3𝑎²𝑚²𝑘⁴−3𝑎𝑘𝑐−𝑐3𝑖𝑎𝑚𝑘²±3𝑎²𝑚²𝑘⁴,−3𝑎𝑘𝑐 (50) where 𝑏_𝑚, 𝑏_−𝑚 are arbitrary constants.

Family  8. Consider 𝑎_𝑚=𝑏_𝑚3𝑖𝑎𝑚𝑘²∓3𝑎²𝑘⁴−3𝑎𝑘𝑐,𝑎3𝑎𝑘₀=𝑏₀3𝑖𝑎𝑚𝑘²∓3𝑎²𝑚²𝑘⁴−3𝑎𝑘𝑐,𝑎3𝑎𝑘_−𝑚=𝑏_−𝑚−9𝑎𝑚²𝑘³±𝑚𝑘3𝑎²𝑚²𝑘⁴−3𝑎𝑘𝑐+3𝑐6𝑖𝑎𝑚𝑘²±33𝑎²𝑚²𝑘⁴,−3𝑎𝑘𝑐 (51) where 𝑏_𝑚, 𝑏₀, 𝑏_−𝑚 are arbitrary constants.

Substituting the results above into (49), we can obtain the following exact solutions to (28): 𝑈₁₃(𝜉)=−𝑏_−𝑚−2𝑎𝑚²𝑘³±2𝑖𝑚𝑘3𝑎²𝑚²𝑘⁴−3𝑎𝑘𝑐−𝑐3𝑖𝑎𝑚𝑘²±3𝑎²𝑚²𝑘⁴𝑒−3𝑎𝑘𝑐^{−𝑖𝑚𝜉}+𝑏_𝑚3𝑖𝑎𝑚𝑘²∓3𝑎²𝑚²𝑘⁴−3𝑎𝑘𝑐𝑒3𝑎𝑘^𝑖𝑚𝜉×𝑏_−𝑚𝑒^{−𝑖𝑚𝜉}+𝑏_𝑚𝑒^{𝑖𝑚𝜉−1},𝑈₁₄(𝜉)=𝑏_−𝑚−9𝑎𝑚²𝑘³±𝑚𝑘3𝑎²𝑚²𝑘⁴−3𝑎𝑘𝑐+3𝑐6𝑖𝑎𝑚𝑘²±33𝑎²𝑚²𝑘⁴𝑒−3𝑎𝑘𝑐^{−𝑖𝑚𝜉}+𝑏_𝑚3𝑖𝑎𝑚𝑘²∓3𝑎²𝑘⁴−3𝑎𝑘𝑐𝑒3𝑎𝑘^𝑖𝑚𝜉+𝑏₀3𝑖𝑎𝑚𝑘²∓3𝑎²𝑚²𝑘⁴−3𝑎𝑘𝑐×𝑏3𝑎𝑘_−𝑚𝑒^{−𝑖𝑚𝜉}+𝑏₀+𝑏_𝑚𝑒^{𝑖𝑚𝜉−1}. (52)(53) If we especially take 𝑐=−𝑎𝑚²𝑘³, 𝑏_𝑚=𝑏_−𝑚 or 𝑐=−𝑎𝑚²𝑘³, 𝑏_𝑚=−𝑏_−𝑚 in (53), then we obtain the trigonometric function solutions: 𝑈₁₁(𝜉)𝑈=−𝑚𝑘tan𝜉,₁₂(𝜉)=−𝑚𝑘cot𝜉. (54)

Remark 2. If we let 𝑚=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.