# The Extended Trial Equation Method for Some Time Fractional Differential Equations

Authors: Yusuf Pandir; Yusuf Gurefe; Emine Misirli
Journal: Discrete Dynamics in Nature and Society
 (2013)

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## Abstract

Nonlinear fractional partial differential equations have been  solved with the help of the extended trial equation method. Based on the fractional derivative in the sense of modified Riemann-Liouville derivative and traveling wave transformation, the fractional partial differential equation can be turned into the nonlinear nonfractional ordinary differential equation. For illustrating the reliability of this approach, we apply it to the generalized third order fractional KdV equation and the fractionalKn,n equation according to the complete discrimination system for polynomial method. As a result, some new exact solutions to these nonlinear problems are successfully constructed such as elliptic integral function solutions, Jacobi elliptic function solutions, and soliton solutions.


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## Body

## 1. Introduction

Many physical phenomena in mathematical biology, electromagnetic theory, fluid mechanics, signal processing, engineering, and other fields of science can be modeled using the fractional derivatives. For example, the use of fractional derivative is analyzed in the fields of controller tuning, redundant robots, legged robots, digital circuit synthesis, and heat diffusion [1]. Also, we refer the reader to [1, 2] for details concerning fractional derivative or calculus and their applications in science and engineering. In some sense fractional differential equations could represent various real-life problems [3–5]. However, an effective and general method for solving them cannot be found even in the most useful works. Also, remarkable progress has been made in the construction of the approximate solutions for fractional nonlinear partial differential equations. There are various mathematical methods which can constitute the exact solutions for some time fractional differential equations [6, 7]. Single kink soliton solutions, multiple-soliton solutions, compacton-like solutions, singular solitons, and other solutions have been found by use of these approaches. Apart from all these, some new exact solutions are obtained by using the trial equation methods. Some of them are elliptic integral F, E, and Π functions, and Jacobi elliptic function solutions.In Section2, primarily we give some definitions and properties of the fractional calculus theory and also produce the extended trial equation method for fractional nonlinear evolution equations with higher order nonlinearity. The power of this steerable method showed that this method can be applied to two nonlinear fractional problems. In Section 3, as applications, we solve the fractional nonlinear partial differential equations such as the generalized third order fractional KdV equation [8]
(1)∂αu∂tα+uxxx+upux=0,-∞<x<∞,t>0,
where 0<α≤1 and p≥2 model the subdiffusion and the subdispersion term, and the fractional K(n,n) equation [9]
(2)∂αu∂tα+(un)x+(un)xxx=0,-∞<x<∞,t>0,0<α≤1,
where n>1. Equation (1) has a wide range of applications in plasma physics, fluid physics, and capillary-gravity waves. Although there are a lot of studies for the classical generalized third order KdV and the K(n,n) equation, this is the first study with respect to the fractional versions of these problems. The aim of this paper is to obtain exact solutions of the generalized third order fractional KdV and the fractional K(n,n) equation by extendend trial equation method and to determine the accuracy of extended trial equation method in solving these kinds of problems. The exact solutions obtained by the interesting wave transformation η=kx-(λtα/Γ(1+α)) are used to determine some new phenomena of the fractional nonlinear evolution equations. Based on the derived solutions for (1) and (2), the phenomena of some solutions are shown graphically (see Figures 1, 2, and 3).Graph of the solution (31) corresponding to the values α=0.01,0.50 and α=0.85 from (a) to (c) when τ1=α1=ξ0=ξ1=k=1, p=α1=2.
(a)
(b)
(c)Graph of the solution (33) corresponding to the values α=0.01,0.50 and α=0.85 from (a) to (c) when τ1=α3=ξ0=ξ1=k=1, p=α2=2, and α1=3.
(a)
(b)
(c)Graph of the solution (60) corresponding to the values α=0.01,0.50 and α=0.85 from (a) to (c) when τ0=τ1=ξ1=k=1, n=3, and ξ2=3/2.
(a)
(b)
(c)


## 2. Preliminaries and the Extended Trial Equation Method

In this part of the paper, it would be helpful to give some definitions and properties of the fractional calculus theory. For an introduction to the classical fractional calculus we indicate the reader to [3–5]. Here, we shortly review the modified Riemann-Liouville derivative from the recent fractional calculus proposed by Jumarie [10–12]. Let f:[0,1]→ℝ be a continuous function and α∈(0,1). The Jumarie modified fractional derivative of order α and f may be defined by expression of [13] as follows:
(3)Dxαf(x)={1Γ(-α)∫0x‍(x-ξ)-α-1[f(ξ)-f(0)]dξ,nnnnnnnnnnnnnnnnnnnnnnnnnnα<0,1Γ(1-α)ddx∫0x‍(x-ξ)-α[f(ξ)-f(0)]dξ,nnnnnnnnnnnnnnnnnnnnnnnnn0<α<1,(f(n)(ξ))α-n,nnnnnnn≤α≤n+1,n≥1.In addition to this expression, we may give a summary of the fractional modified Riemann-Liouville derivative properties which are used further in this paper. Some of the useful formulas are given as(4)Dxαk=0,Dxαxμ={0,μ≤α-1,Γ(μ+1)Γ(μ-α+1)xμ-α,μ>α-1.Some new trial equation methods were defined in the literature [14–26]. In this paper, the extended trial equation method will be given. In order to apply this method to fractional nonlinear partial differential equations, we consider the following steps.Step 1.
Consider time fractional partial differential equation in two variables and a dependent variableu(5)P(u,Dtαu,ux,uxx,uxxx,…)=0,
and take the wave transformation
(6)u(x,t)=u(η),η=kx-λtαΓ(1+α),
where λ≠0. Substituting (6) into (5) yields a nonlinear ordinary differential equation
(7)N(u,u′,u′′,…)=0.Step 2.
Take trial equation as follows:(8)u=∑i=0δ‍τiΥi,
where
(9)(Υ′)2=Λ(Υ)=Φ(Υ)Ψ(Υ)=ξθΥθ+⋯+ξ1Υ+ξ0ζϵΥϵ+⋯+ζ1Υ+ζ0.
Using (8) and (9), we can write
(10)(u′)2=Φ(Υ)Ψ(Υ)(∑i=0δ‍iτiΥi-1)2,u′′=Φ′(Υ)Ψ(Υ)-Φ(Υ)Ψ′(Υ)2Ψ2(Υ)(∑i=0δ‍iτiΥi-1)+Φ(Υ)Ψ(Υ)(∑i=0δ‍i(i-1)τiΥi-2),
where Φ(Υ) and Ψ(Υ) are polynomials. Substituting the above relations into (7) yields an equation of polynomial Ω(Υ) of Υ:
(11)Ω(Υ)=ϱsΥs+⋯+ϱ1Υ+ϱ0=0.
According to the balance principle, we can find a relation of θ, ϵ, and δ. We can compute some values of θ, ϵ, and δ.Step 3.
Let the coefficients ofΩ(Υ) all be zero will yield an algebraic equations system:
(12)ϱi=0,i=0,…,s.
Solving this system, we will determine the values of ξ0,…,ξθ, ζ0,…,ζϵ and τ0,…,τδ.Step 4.
Reduce (9) to the elementary integral form
(13)±(η-η0)=∫‍dΥΛ(Υ)=∫‍Ψ(Υ)Φ(Υ)dΥ.
Using a complete discrimination system for polynomial to classify the roots of Φ(Υ), we solve (13) with the help of Mathematica and classify the exact solutions to (7). In addition, we can write the exact traveling wave solutions to (5), respectively.


## 3. Applications to the Extended Trial Equation Method

In this section, we apply the method given in Section2 to the generalized third order fractional KdV and the fractional K(n,n) equation, respectively.


### 3.1. Application to Generalized Fractional KdV Equation

In the case ofα=1, (1) reduces to the classical nonlinear generalized third order KdV equation. Many researchers have tried to get the exact solutions of this equation by using a variety of methods. But, we solve the generalized fractional KdV equation for 0<α≤1. Compactons, solitary patterns, and periodic and solitary traveling plane waves solutions of this equation are found.Let us consider the travelling wave solutions of (1), and we perform the transformation u(x,t)=u(η), η=kx-(λtα/Γ(1+α)), where λ is constant to be determined later. Then, integrating this equation with respect to η and setting the integration constant to zero, we get
(14)-λ(p+1)u+k3(p+1)u′′+kup+1=0.
By use of the following transformation:
(15)u=v1/p,
(14) turns into the equation
(16)k3(p2+p)vv′′+k3(1-p2)(v′)2-λ(p3+p2)v2+kp2v3=0.
From (8)–(10), we can write the following highest order nonlinear terms in order to determine the balance procedure:
(17)v=τδΥδ+⋯,v3=τδ3Υ3δ+⋯,(v′)2=ξθδ2τδ2ζϵΥ2δ-2(Υ′)2+⋯=ξθδ2τδ2ζϵΥ2δ+θ-ϵ-2+⋯,vv′′=ϖΥ2δ+θ-ϵ-2+⋯.
Balancing the highest order nonlinear terms of vv′′ and v3 in (16), we have
(18)2δ+θ-ϵ-2=3δ,
hence
(19)θ=ϵ+δ+2.
This resolution procedure is applied, and we obtain results as follows.Case 1.
If we takeϵ=0, δ=1, and θ=3, then
(20)(v′)2=(τ1)2(ξ3Υ3+ξ2Υ2+ξ1Υ+ξ0)ζ0,v′′=(τ1)(3ξ3Υ2+2ξ2Υ+ξ1)2ζ0,
where ξ3≠0, ζ0≠0. Thus, we have a system of algebraic equations from the coefficients of polynomial of Υ. Solving the algebraic equation system (12), we get
(21)ξ0=ξ0,ξ1=ξ1,ξ2=τ1(2ξ1τ0-3ξ0τ1)τ02,ξ3=τ12(ξ1τ0-2ξ0τ1)τ03,τ0=τ0,τ1=τ1,ζ0=-k2(1+p)(2+p)τ1(ξ1τ0-2ξ0τ1)2p2τ03,λ=2kτ0(ξ1τ0-3ξ0τ1)(2+3p+p2)(ξ1τ0-2ξ0τ1).
Substituting these results into (13), we have
(22)±(η-η0)=-k2(2+3p+p2)2p2τ1×∫‍(nnnnnnnnnnnnnnn+ξ0τ03τ12(ξ1τ0-2ξ0τ1))-1/2dΥ×(Υ3+τ0(2ξ1τ0-3ξ0τ1)τ1(ξ1τ0-2ξ0τ1)Υ2nnnnnnnnnnnnnnn+ξ1τ03τ12(ξ1τ0-2ξ0τ1)Υnnnnnnnnnnnnnnn+ξ0τ03τ12(ξ1τ0-2ξ0τ1))-1/2).
Integrating (22), we procure the solutions to (1) as follows:
(23)±(η-η0)=-2AΥ-α1,±(η-η0)=2Aα2-α1arctanΥ-α2α2-α1,α2>α1,±(η-η0)=Aα1-α2ln|Υ-α2-α1-α2Υ-α2+α1-α2|,α1>α2,±(η-η0)=-2Aα1-α3F(φ,l),α1>α2>α3,
where
(24)A=-k2(2+3p+p2)2p2τ1,F(φ,l)=∫0φ‍dψ1-l2sin2ψ,φ=arcsinΥ-α3α2-α3,l2=α2-α3α1-α3.
Also α1, α2, and α3 are the roots of the polynomial equation
(25)Υ3+ξ2ξ3Υ2+ξ1ξ3Υ+ξ0ξ3=0.
Substituting the solutions (23) into (8) and (15), we have(26)
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(30)u(x,t)=(A~x-λ1tα)2/p,
1-soliton wave solution
(31)u(x,t)=B~cosh2/p(B(x-λ1tα)),
singular soliton solution
(32)u(x,t)=C~sinh2/p(C(x-λ1tα)),whereA~=2Aτ1/k, B~=(τ1(α2-α1))1/p, B=kα1-α2/2A, C~=(τ1(α1-α2))1/p,  C=kα1-α2/2A, and λ1=-(2τ1α1(ξ1α1+3ξ0)/(2+3p+p2)(ξ1α1+2ξ0)Γ(1+α)). Here, B~ and C~ are the amplitude of the soliton, while B and C are the inverse width of the solitons and c is the velocity.On the other hand, if we takeτ0=-τ1α3 and η0=0, then the Jacobi elliptic function solution (29) can be written in the form
(33)u(x,t)=A1sn2/p(Bi(x-λ1tα),α2-α3α1-α3),
where A1=(τ1(α2-α3))1/p and Bi=(-1)ikα1-α3/2A, i=(1,2). Thus, we can say that the solitons exist for τ1>0.Remark 1.
We provide a proof by verifying the exact solution (31) by direct substitution. If we take α=1 and p=2, then (1) becomes
(34)ut+uxxx+u2ux=0,
whereB~=τ1(α2-α1), B=kα1-α2/2A, A=k-(3/2τ1), and λ1=-(τ1α1(ξ1α1+3ξ0)/6(ξ1α1+2ξ0)). Here, B=τ1(α2-α1)/6.
Equation (31) is
(35)u(x,t)=B~sech[B(x-λ1t)].
If ϕ=x-λ1t, then
(36)u(ϕ)=B~sech(Bϕ),ut=B~Bλ1tanh(Bϕ)sech(Bϕ),ux=-B~Btanh(Bϕ)sech(Bϕ),u2ux=-B~3Btanh(Bϕ)sech3(Bϕ),uxxx=B~B3tanh(Bϕ)(6sech3(Bϕ)-sech(Bϕ)).
Substituting the above relations into the partial differential equation, we obtain the following equation:
(37)B~Btanh(Bϕ)sech(Bϕ)[λ1-B2+(6B2-B~2)sech2(Bϕ)]=-τ1(ξ1α1α2+2ξ0α2+ξ0α1)6(ξ1α1+2ξ0)=0.
Otherwise, we can write
(38)-α12α2=ξ0τ03τ12(ξ1τ0-2ξ0τ1),
where τ0=-τ1α1 and (Υ-α1)2(Υ-α2)=Υ3+(ξ2/ξ3)Υ2+(ξ1/ξ3)Υ+(ξ0/ξ3)=0. From this equation, we find ξ1α1α2+2ξ0α2+ξ0α1=0. Substituting this relation into the above equation yields 0=0. Thus, it is seen that the obtained solution (31) is an exact solution of the fractional KdV equation.Remark 2.
The solutions (30)–(33) obtained by using the extended trial equation method for (1) have been checked by Mathematica. To our knowledge, rational function solution, 1-soliton solution, the singular soliton solution, and the Jacobi elliptic function that we find in this paper are not shown in the previous literature. These results are new traveling wave solutions of (1).Case 2.
If we takeϵ=0, δ=2, and θ=4, then
(39)(v′)2=(τ1+2τ2Υ)2(ξ4Υ4+ξ3Υ3+ξ2Υ2+ξ1Υ+ξ0)ζ0,v′′=((τ1+2τ1Υ)(4ξ4Υ3+3ξ3Υ2+2ξ2Υ+ξ1)nnn+4τ2(ξ4Υ4+ξ3Υ3+ξ2Υ2+ξ1Υ+ξ0))×(2ζ0)-1,
where ξ4≠0, ζ0≠0. Respectively, solving the algebraic equation system (12) yields as follows:
(40)ξ0=(ξ4τ12-ξ2τ12)24ξ4τ24,ξ1=ξ2τ1τ22-ξ4τ13τ23,ξ3=2ξ4τ1τ2,ξ2=ξ2,ξ4=ξ4,τ0=ξ2τ22-ξ4τ122ξ4τ2,τ1=τ1,τ2=τ2,ζ0=-2k2(2+3p+p2)ξ4p2τ2,λ=-k(3ξ4τ12-2ξ2τ22)2(2+3p+p2)ξ4τ2.
Substituting these results into (13), we get
(41)±(η-η0)=-2k2(2+3p+p2)p2τ2×∫(+(ξ4τ12-ξ2τ12)24ξ42τ24)-1/2dΥ‍((ξ4τ12-ξ2τ12)24ξ42τ24Υ4+2τ1τ2Υ3+ξ2ξ4Υ2bbbbbbbbbbbbbbb+ξ2τ1τ22-ξ4τ13τ23ξ4Υbbbbbbbbbbbbbbb+(ξ4τ12-ξ2τ12)24ξ42τ24)-1/2).
Integrating (41), we obtain the solutions to (1) as follows:
(42)±(η-η0)=-A2Υ-α1,±(η-η0)=2A2α1-α2Υ-α2Υ-α1,α1>α2,±(η-η0)=A2α1-α2ln|Υ-α1Υ-α2|,±(η-η0)=2A2(α1-α2)(α1-α3)×ln|(Υ-α2)(α1-α3)-(Υ-α3)(α1-α2)(Υ-α2)(α1-α3)+(Υ-α3)(α1-α2)|,nnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnα1>α2>α3,±(η-η0)=2A2(α1-α3)(α2-α4)F(φ,l),nnnnnnnnnnα1>α2>α3>α4,
where
(43)A2=-2k2(2+3p+p2)p2τ2,φ=arcsin(Υ-α1)(α2-α4)(Υ-α2)(α1-α4),l2=(α2-α3)(α1-α4)(α1-α3)(α2-α4).
Also α1, α2, α3, and α4 are the roots of the polynomial equation
(44)Υ4+ξ3ξ4Υ3+ξ2ξ4Υ2+ξ1ξ4Υ+ξ0ξ4=0.
Substituting the solutions (42) into (8) and (15), we have(45)
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n
n
n
+
τ
2
(
α
1
+
4
A
2
2
(
α
2
-
α
1
)
4
A
2
2
-
[
(
α
1
-
α
2
)
(
k
x
+
(
k
(
3
ξ
4
τ
1
2
-
2
ξ
2
τ
2
2
)
/
2
(
2
+
3
p
+
p
2
)
ξ
4
τ
2
Γ
(
1
+
α
)
)
t
α
-
η
0
)
]
2
)
2
}
1
/
p
,
u
(
x
,
t
)
=
{
(
α
2
+
(
α
2
-
α
1
)
exp
[
(
(
α
1
-
α
2
)
/
A
2
)
(
k
x
+
(
k
(
3
ξ
4
τ
1
2
-
2
ξ
2
τ
2
2
)
/
2
(
2
+
3
p
+
p
2
)
ξ
4
τ
2
Γ
(
1
+
α
)
)
t
α
-
η
0
)
]
-
1
)
2
τ
0
+
τ
1
α
2
+
(
α
2
-
α
1
)
τ
1
exp
[
(
(
α
1
-
α
2
)
/
A
2
)
(
k
x
+
(
k
(
3
ξ
4
τ
1
2
-
2
ξ
2
τ
2
2
)
/
2
(
2
+
3
p
+
p
2
)
ξ
4
τ
2
Γ
(
1
+
α
)
)
t
α
-
η
0
)
]
-
1
u
(
x
,
t
)
n
n
n
n
+
τ
2
(
α
2
+
(
α
2
-
α
1
)
exp
[
(
(
α
1
-
α
2
)
/
A
2
)
(
k
x
+
(
k
(
3
ξ
4
τ
1
2
-
2
ξ
2
τ
2
2
)
/
2
(
2
+
3
p
+
p
2
)
ξ
4
τ
2
Γ
(
1
+
α
)
)
t
α
-
η
0
)
]
-
1
)
2
}
1
/
p
,
u
(
x
,
t
)
=
{
(
α
1
+
(
α
1
-
α
2
)
exp
[
(
(
α
1
-
α
2
)
/
A
2
)
(
k
x
+
(
k
(
3
ξ
4
τ
1
2
-
2
ξ
2
τ
2
2
)
/
2
(
2
+
3
p
+
p
2
)
ξ
4
τ
2
Γ
(
1
+
α
)
)
t
α
-
η
0
)
]
-
1
)
2
τ
0
+
τ
1
α
1
+
(
α
1
-
α
2
)
τ
1
exp
[
(
(
α
1
-
α
2
)
/
A
2
)
(
k
x
+
(
k
(
3
ξ
4
τ
1
2
-
2
ξ
2
τ
2
2
)
/
2
(
2
+
3
p
+
p
2
)
ξ
4
τ
2
Γ
(
1
+
α
)
)
t
α
-
η
0
)
]
-
1
u
(
x
,
t
)
n
n
n
n
+
τ
2
(
α
1
+
(
α
1
-
α
2
)
exp
[
(
(
α
1
-
α
2
)
/
A
2
)
(
k
x
+
(
k
(
3
ξ
4
τ
1
2
-
2
ξ
2
τ
2
2
)
/
2
(
2
+
3
p
+
p
2
)
ξ
4
τ
2
Γ
(
1
+
α
)
)
t
α
-
η
0
)
]
-
1
)
2
(
α
1
-
α
2
)
τ
1
exp
[
(
α
1
-
α
2
/
A
2
)
(
k
x
+
(
k
(
3
ξ
4
τ
1
2
-
2
ξ
2
τ
2
2
)
/
2
(
2
+
3
p
+
p
2
)
ξ
4
τ
2
Γ
(
1
+
α
)
)
t
α
-
η
0
)
]
-
1
}
1
/
p
,
u
(
x
,
t
)
=
{
cosh
[
(
α
1
-
α
2
)
(
α
1
-
α
3
)
A
2
(
k
x
+
k
(
3
ξ
4
τ
1
2
-
2
ξ
2
τ
2
2
)
2
(
2
+
3
p
+
p
2
)
ξ
4
τ
2
Γ
(
1
+
α
)
t
α
-
η
0
)
]
)
-
1
)
2
τ
0
+
τ
1
α
1
-
(
cosh
[
(
α
1
-
α
2
)
(
α
1
-
α
3
)
A
2
(
k
x
+
k
(
3
ξ
4
τ
1
2
-
2
ξ
2
τ
2
2
)
2
(
2
+
3
p
+
p
2
)
ξ
4
τ
2
Γ
(
1
+
α
)
t
α
-
η
0
)
]
)
-
1
(
2
(
α
1
-
α
2
)
(
α
1
-
α
3
)
τ
1
)
n
n
n
n
n
n
n
n
n
n
n
n
n
n
n
n
n
×
(
[
(
α
1
-
α
2
)
(
α
1
-
α
3
)
A
2
(
k
x
+
k
(
3
ξ
4
τ
1
2
-
2
ξ
2
τ
2
2
)
2
(
2
+
3
p
+
p
2
)
ξ
4
τ
2
Γ
(
1
+
α
)
t
α
-
η
0
)
]
2
α
1
-
α
2
-
α
3
+
(
α
3
-
α
2
)
cosh
[
(
α
1
-
α
2
)
(
α
1
-
α
3
)
A
2
(
k
x
+
k
(
3
ξ
4
τ
1
2
-
2
ξ
2
τ
2
2
)
2
(
2
+
3
p
+
p
2
)
ξ
4
τ
2
Γ
(
1
+
α
)
t
α
-
η
0
)
]
)
-
1
)
n
n
n
n
+
τ
2
(
cosh
[
(
α
1
-
α
2
)
(
α
1
-
α
3
)
A
2
(
k
x
+
k
(
3
ξ
4
τ
1
2
-
2
ξ
2
τ
2
2
)
2
(
2
+
3
p
+
p
2
)
ξ
4
τ
2
Γ
(
1
+
α
)
t
α
-
η
0
)
]
)
-
1
α
1
-
(
cosh
[
(
α
1
-
α
2
)
(
α
1
-
α
3
)
A
2
(
k
x
+
k
(
3
ξ
4
τ
1
2
-
2
ξ
2
τ
2
2
)
2
(
2
+
3
p
+
p
2
)
ξ
4
τ
2
Γ
(
1
+
α
)
t
α
-
η
0
)
]
)
-
1
(
2
(
α
1
-
α
2
)
(
α
1
-
α
3
)
)
n
n
n
n
n
n
n
n
n
n
n
n
n
i
n
n
×
(
[
(
α
1
-
α
2
)
(
α
1
-
α
3
)
A
2
(
k
x
+
k
(
3
ξ
4
τ
1
2
-
2
ξ
2
τ
2
2
)
2
(
2
+
3
p
+
p
2
)
ξ
4
τ
2
Γ
(
1
+
α
)
t
α
-
η
0
)
]
2
α
1
-
α
2
-
α
3
+
(
α
3
-
α
2
)
cosh
[
(
α
1
-
α
2
)
(
α
1
-
α
3
)
A
2
(
k
x
+
k
(
3
ξ
4
τ
1
2
-
2
ξ
2
τ
2
2
)
2
(
2
+
3
p
+
p
2
)
ξ
4
τ
2
Γ
(
1
+
α
)
t
α
-
η
0
)
]
)
-
1
)
)
2
}
1
/
p
,
u
(
x
,
t
)
=
{
(
α
2
+
τ
1
(
α
1
-
α
2
)
(
α
4
-
α
2
)
α
4
-
α
2
+
(
α
1
-
α
4
)
s
n
2
[
(
(
α
1
-
α
3
)
(
α
2
-
α
4
)
/
2
A
2
)
(
k
x
+
(
k
(
3
ξ
4
τ
1
2
-
2
ξ
2
τ
2
2
)
/
2
(
2
+
3
p
+
p
2
)
ξ
4
τ
2
Γ
(
1
+
α
)
)
t
α
-
η
0
)
,
(
(
α
2
-
α
3
)
(
α
1
-
α
4
)
/
(
α
1
-
α
3
)
(
α
2
-
α
4
)
)
]
)
2
×
s
n
2
[
(
α
1
-
α
3
)
(
α
2
-
α
4
)
2
A
2
k
(
3
ξ
4
τ
1
2
-
2
ξ
2
τ
2
2
)
2
(
2
+
3
p
+
p
2
)
ξ
4
τ
2
Γ
(
1
+
α
)
,
(
α
2
-
α
3
)
(
α
1
-
α
4
)
(
α
1
-
α
3
)
(
α
2
-
α
4
)
]
)
-
1
)
)
2
τ
0
+
τ
1
α
2
+
(
×
s
n
2
[
(
α
1
-
α
3
)
(
α
2
-
α
4
)
2
A
2
(
k
x
+
k
(
3
ξ
4
τ
1
2
-
2
ξ
2
τ
2
2
)
2
(
2
+
3
p
+
p
2
)
ξ
4
τ
2
Γ
(
1
+
α
)
t
α
-
η
0
)
,
(
α
2
-
α
3
)
(
α
1
-
α
4
)
(
α
1
-
α
3
)
(
α
2
-
α
4
)
]
)
-
1
(
τ
1
(
α
1
-
α
2
)
(
α
4
-
α
2
)
)
n
n
n
n
n
n
n
n
n
n
n
n
n
n
n
n
n
n
×
(
×
s
n
2
[
(
α
1
-
α
3
)
(
α
2
-
α
4
)
2
A
2
(
k
x
+
k
(
3
ξ
4
τ
1
2
-
2
ξ
2
τ
2
2
)
2
(
2
+
3
p
+
p
2
)
ξ
4
τ
2
Γ
(
1
+
α
)
t
α
-
η
0
)
,
(
α
2
-
α
3
)
(
α
1
-
α
4
)
(
α
1
-
α
3
)
(
α
2
-
α
4
)
]
)
α
4
-
α
2
+
(
α
1
-
α
4
)
s
n
2
[
(
α
1
-
α
3
)
(
α
2
-
α
4
)
2
A
2
(
k
x
+
k
(
3
ξ
4
τ
1
2
-
2
ξ
2
τ
2
2
)
2
(
2
+
3
p
+
p
2
)
ξ
4
τ
2
Γ
(
1
+
α
)
t
α
-
η
0
)
n
n
n
n
n
n
n
n
n
n
n
n
n
n
n
n
n
n
n
n
n
n
n
n
n
n
n
n
n
n
n
n
n
n
n
n
n
n
n
n
n
n
×
(
α
2
-
α
3
)
(
α
1
-
α
4
)
(
α
1
-
α
3
)
(
α
2
-
α
4
)
(
α
1
-
α
3
)
(
α
2
-
α
4
)
2
A
2
(
k
x
+
k
(
3
ξ
4
τ
1
2
-
2
ξ
2
τ
2
2
)
2
(
2
+
3
p
+
p
2
)
ξ
4
τ
2
Γ
(
1
+
α
)
t
α
-
η
0
)
]
(
×
s
n
2
[
(
α
1
-
α
3
)
(
α
2
-
α
4
)
2
A
2
(
k
x
+
k
(
3
ξ
4
τ
1
2
-
2
ξ
2
τ
2
2
)
2
(
2
+
3
p
+
p
2
)
ξ
4
τ
2
Γ
(
1
+
α
)
t
α
-
η
0
)
,
(
α
2
-
α
3
)
(
α
1
-
α
4
)
(
α
1
-
α
3
)
(
α
2
-
α
4
)
]
)
-
1
(
τ
1
(
α
1
-
α
2
)
(
α
4
-
α
2
)
)
)
-
1
[
(
α
1
-
α
3
)
(
α
2
-
α
4
)
2
A
2
(
k
x
+
k
(
3
ξ
4
τ
1
2
-
2
ξ
2
τ
2
2
)
2
(
2
+
3
p
+
p
2
)
ξ
4
τ
2
Γ
(
1
+
α
)
t
α
-
η
0
)
,
)
nnnn
+
τ
2
(
×
s
n
2
[
(
α
1
-
α
3
)
(
α
2
-
α
4
)
2
A
2
k
(
3
ξ
4
τ
1
2
-
2
ξ
2
τ
2
2
)
2
(
2
+
3
p
+
p
2
)
ξ
4
τ
2
Γ
(
1
+
α
)
,
(
α
2
-
α
3
)
(
α
1
-
α
4
)
(
α
1
-
α
3
)
(
α
2
-
α
4
)
]
)
-
1
)
α
2
+
(
×
s
n
2
[
(
α
1
-
α
3
)
(
α
2
-
α
4
)
2
A
2
k
(
3
ξ
4
τ
1
2
-
2
ξ
2
τ
2
2
)
2
(
2
+
3
p
+
p
2
)
ξ
4
τ
2
Γ
(
1
+
α
)
,
(
α
2
-
α
3
)
(
α
1
-
α
4
)
(
α
1
-
α
3
)
(
α
2
-
α
4
)
]
)
-
1
(
τ
1
(
α
1
-
α
2
)
(
α
4
-
α
2
)
)
n
n
n
n
n
n
n
n
n
n
n
n
n
n
n
n
n
n
×
(
[
(
α
1
-
α
3
)
(
α
2
-
α
4
)
2
A
2
k
(
3
ξ
4
τ
1
2
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2
ξ
2
τ
2
2
)
2
(
2
+
3
p
+
p
2
)
ξ
4
τ
2
Γ
(
1
+
α
)
,
(
α
2
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α
3
)
(
α
1
-
α
4
)
(
α
1
-
α
3
)
(
α
2
-
α
4
)
]
α
4
-
α
2
+
(
α
1
-
α
4
)
s
n
2
[
(
α
1
-
α
3
)
(
α
2
-
α
4
)
2
A
2
(
k
x
+
k
(
3
ξ
4
τ
1
2
-
2
ξ
2
τ
2
2
)
2
(
2
+
3
p
+
p
2
)
ξ
4
τ
2
Γ
(
1
+
α
)
t
α
-
η
0
)
,
n
n
n
n
n
n
n
n
n
n
n
n
n
n
n
n
n
n
n
n
n
n
n
n
n
n
n
n
n
n
n
n
n
n
n
n
n
n
n
n
n
(
α
2
-
α
3
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+
α
)
t
α
-
η
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,
}
1
/
p
.For simplicity, if we takeη0=0, then we can write the solutions (45) as follows:
(46)u(x,t)=[∑i=02‍τi(α1±A2k(x-λ2tα))i]1/p,u(x,t)=[∑i=02‍τi(α1+4A22(α2-α1)4A22-[(α1-α2)k(x-λ2tα)]2)i]1/p,u(x,t)=[∑i=02‍τi(α2+α2-α1exp[B3(x-λ2tα)]-1)i]1/p,u(x,t)=[∑i=02‍τi(α1+α1-α2exp[B3(x-λ2tα)]-1)i]1/p,u(x,t)=[∑i=02‍τi(×cosh[C(x-λ2tα)])-1((2(α1-α2)(α1-α3)))((2(α1-α2)(α1-α3))α1-(×cosh[C(x-λ2tα)])-1((2(α1-α2)(α1-α3)))((2(α1-α2)(α1-α3))(2(α1-α2)(α1-α3))nnnnnnnnnnnnn×(2α1-α2-α3+(α3-α2)nnnnnnnnnnnnnnnn×cosh[C(x-λ2tα)])-1((2(α1-α2)(α1-α3)))((2(α1-α2)(α1-α3)))i∑i=02‍]1/p,u(x,t)=[∑i=02‍τi(α2+(α1-α2)(α4-α2)α4-α2+(α1-α4)sn2(φ,l))i]1/p,
where B3=k(α1-α2)/A2, C=k(α1-α2)(α1-α3)/A2, φ=(k(α1-α3)(α2-α4)/2A2)(x-λ2tα), l2=(α2-α3)(α1-α4)/(α1-α3)(α2-α4), and λ2=(3ξ4τ12-2ξ2τ22)/2(2+3p+p2)ξ4τ2Γ(1+α). Here, A2 is the amplitude of the soliton, while λ2 is the velocity and B3 and C are the inverse width of the solitons.Remark 3.
To our knowledge, the solutions (46) that we find in this paper have not been found in the literature, and these are new exact solutions of (1).


### 3.2. Application to the FractionalK(n,n) Equation

In the case ofα=1, (2) reduces to the classical nonlinear generalized K(n,n) equation. Many authors have studied to find the exact solutions of this equation by using a variety of methods [27, 28]. In this paper, we consider the generalized fractional K(n,n) equation for the case of 0<α≤1. Compactons, solitary patterns, and periodic and solitary traveling plane waves solutions of this equation are obtained.Let us consider the travelling wave solutions of (2), and we perform the transformation u(x,t)=u(η), η=kx-(λtα/Γ(1+α)), where λ is constant to be determined later. Then, integrating this equation with respect to η and setting the integration constant to zero, we get
(47)k3(un)′′+kun-λu=0.
By use of the following transformation:
(48)u=v1/(n-1),
(47) turns into the equation
(49)k3(n2-n)vv′′+k3n(v′)2-λ(n-1)2v+k(n-1)2v2=0.
From (8)–(10), we can write the following highest order nonlinear terms in order to determine the balance procedure:
(50)v=τδΥδ+⋯,v2=τδ2Υ2δ+⋯,vv′′=ϖΥ2δ+θ-ϵ-2+⋯.
Balancing the highest order nonlinear terms of vv′′ and v2 in (49), we have
(51)2δ+θ-ϵ-2=2δ,
hence
(52)θ=ϵ+2.
This resolution procedure is applied and we obtain results as follows.Case 1.
If we takeϵ=0,  θ=2, and δ=1, then
(53)(v′)2=τ12(ξ2Υ2+ξ1Υ+ξ0)ζ0,v′′=τ1(2ξ2Υ+ξ1)2ζ0,
where ξ2≠0, ζ0≠0. Thus, we have a system of algebraic equations from the coefficients of polynomial of u. Solving the algebraic equation system (12), we get
(54)ξ0=τ1(ξ1τ1-ξ2τ0)τ12,ξ1=ξ1,ξ2=ξ2,ζ0=-k2n2ξ2(n-1)2,τ0=τ0,τ1=τ1,λ=k(n+1)(2ξ2τ0-ξ1τ1)2nξ2.
Substituting these results into (13), we have
(55)±(η-η0)=-k2n2(n-1)2∫‍dΥΥ2+(ξ1/ξ2)Υ+τ1(ξ1τ1-ξ2τ0)/τ12ξ2.
Integrating (55), we procure the solutions to (2) as follows:
(56)±(η-η0)=A3ln|Υ-α1|,±(η-η0)=2A3ln|Υ-α1+Υ-α2|,
where A3=-(k2n2/(n-1)2). Also α1, α2 are the roots of the polynomial equation
(57)Υ2+ξ1ξ2Υ+ξ0ξ2=0.
Substituting the solutions (56) into (8) and (48), we have(58)
u
(
x
,
t
)
=
[
τ
0
+
τ
1
(
exp
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1
A
3
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(
x
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)
=
[
τ
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exp
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(
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n
n
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n
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n
n
n
n
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+
(
α
1
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2
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2
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[
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3
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k
x
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(
4
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-
1
exp
[
1
A
3
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k
x
-
k
(
n
+
1
)
(
2
ξ
2
τ
0
-
ξ
1
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t
α
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+
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1
/
(
n
-
1
)
.If we takeη0=α2=0 and α1=1, then the solutions (58) can reduce to single kink solution and the hyperbolic function solution, respectively,
(59)u(x,t)=(τ0+τ1(exp[B4(x-λ3tα)]+1))1/(n-1),(60)u(x,t)=(τ0+τ1(12(cosh[B4(x-λ3tα)]+1)))1/(n-1),whereB4=k/A3, λ3=(n+1)(2ξ2τ0-ξ1τ1)/2nξ2Γ(1+α). Here, B4 is the inverse width of the solitons.Remark 4.
The solutions (59)-(60) obtained by using the extended trial equation method for (2) have been checked by Mathematica. To our knowledge, the hyperbolic function solution and single kink solution, that we find in this paper, are not shown in the previous literature. These results are new traveling wave solutions of (2).Case 2.
If we takeϵ=0, θ=2, and δ=2, then
(61)(v′)2=(τ1+2τ2Υ)2(ξ2Υ2+ξ1Υ+ξ0)ζ0,v′′=4τ2(ξ2Υ2+ξ1Υ+ξ0)+(τ1+2τ1Υ)(2ξ2Υ+ξ1)2ζ0,
where ξ2≠0, ζ0≠0. Thus, we have a system of algebraic equations from the coefficients of polynomial of u. Solving the algebraic equation system (12), we get
(62)ξ0=ξ0,ξ1=ξ2τ1τ2,ξ2=ξ2,ζ0=-4k2n2ξ2(n-1)2,τ0=τ124τ2,τ1=τ1,τ2=τ2,λ=k(n+1)(ξ2τ12-4ξ0τ22)8nξ2τ2.
Substituting these results into (13), we have
(63)±(η-η0)=-4k2n2(n-1)2∫‍dΥΥ2+(τ1/τ2)Υ+(ξ0/ξ2).
Integrating (63), we procure the solutions to (2) as follows:
(64)±(η-η0)=A4ln|Υ-α1|,±(η-η0)=2A4ln|Υ-α1+Υ-α2|,
where A4=-(4k2n2/(n-1)2). Also α1, α2 are the roots of the polynomial equation
(65)Υ2+ξ1ξ2Υ+ξ0ξ2=0.
Substituting the solutions (64) into (8) and (48), we have(66)
u
(
x
,
t
)
=
{
(
exp
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4
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n
n
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n
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n
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+
(
α
1
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exp
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4
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k
x
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(
4
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4
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k
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2
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exp
[
1
A
4
(
k
x
-
k
(
n
+
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ξ
2
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1
2
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4
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0
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t
α
8
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ξ
2
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2
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(
1
+
α
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n
n
n
n
n
+
(
α
1
-
α
2
)
2
exp
[
-
1
A
4
(
k
x
-
k
(
n
+
1
)
(
ξ
2
τ
1
2
-
4
ξ
0
τ
2
2
)
t
α
8
n
ξ
2
τ
2
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(
1
+
α
)
-
η
0
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+
2
(
α
1
+
α
2
)
)
×
(
4
)
-
1
)
2
}
1
/
(
n
-
1
)
.For simplicity,η0=α2=0 and α1=1, then we can write the solutions (66) as follows:
(67)u(x,t)=[∑i=02‍τi(exp[B5(x-λ4tα)+1])i]1/(n-1),u(x,t)=[∑i=02‍τi(12(cosh[B5(x-λ4tα)]+1))i]1/(n-1),whereB5=k/A4, λ4=(n+1)(ξ2τ12-4ξ0τ22)/8nξ2τ2Γ(1+α). Here, B5 is the inverse width of the solitons.Case 3.
If we takeϵ=1, θ=3, and δ=1, then
(68)(v′)2=τ12(ξ3Υ3+ξ2Υ2+ξ1Υ+ξ0)ζ0+ζ1Υ,v′′=(-ζ1(ξ3Υ3+ξ2Υ2+ξ1Υ+ξ0))τ1((ζ0+ζ1Υ)(3ξ3Υ2+2ξ2Υ+ξ1)nnnnnn-ζ1(ξ3Υ3+ξ2Υ2+ξ1Υ+ξ0)))×(2(ζ0+ζ1Υ)2)-1,
where ξ3≠0, ζ1≠0. Respectively, solving the algebraic equation system (12) yields
(69)ξ0=-(n-1)2ζ0τ0((n-1)2ζ0τ0+k2n2ξ1τ1)k2n2τ1(k2n2ξ3τ0-(n-1)2ζ0τ1),ξ1=ξ1,ξ2=((k2τ1(k2n4ξ3τ0-(n-1)2n2ζ0τ1))-1k4n4ξ32τ02+(n-1)4ζ02τ12nnn+k2ξ3τ1(k2n4ξ1τ1-(n-1)2n2ζ0τ0)(k2τ1(k2n4ξ3τ0-(n-1)2n2ζ0τ1))-1)×(k2τ1(k2n4ξ3τ0-(n-1)2n2ζ0τ1))-1,ξ3=ξ3,ζ0=ζ0,ζ1=-k2n2ξ3(n-1)2,τ0=τ0,τ1=τ1,λ=k(n+1)(k2n2ξ3τ02-τ1(2(n-1)2ζ0τ0+k2n2ξ1τ1))2k2n3ξ3τ0-2(n-1)2nζ0τ1.
Substituting these results into (13), we have
(70)±(η-η0)=-k2n2(n-1)2×∫‍Υ+(ζ0/ζ1)Υ3+(ξ2/ξ3)Υ2+(ξ1/ξ3)Υ+(ξ0/ξ3)dΥ.
Integrating (70), we obtain the solutions to (2) as follows:
(71)±(η-η0)=2A3(ln|ζ0+ζ1Υζ1+Υ-α1|-ζ0+ζ1Υζ1(Υ-α1)),±(η-η0)=2A3ζ1(α2-α1)×(ζ0+ζ1α1arctan[(ζ0+ζ1α1)(Υ-α2)(ζ0+ζ1Υ)(α2-α1)]nnnnnnn+α2-α1ln|ζ0+ζ1Υζ1+Υ-α2|),±(η-η0)=2(α1-α2)ζ1(ζ0+ζ1α3)×((ζ0+ζ1α1)F(φ,l)+(α3-ζ1α1)π(φ,n,l)),
where
(72)F(φ,l)=∫0φ‍dψ1-l2sin2ψ,π(φ,n,l)=∫0φ‍dψ(1+nsin2ψ)1-l2sin2ψ,φ=arcsin(Υ-α3)(α2-α1)(Υ-α1)(α2-α3),l2=(ζ0+ζ1α1)(α3-α2)(ζ0+ζ1α3)(α1-α2),n=α3-α2α1-α2.
Also α1, α2, and α3 are the roots of the polynomial equation
(73)Υ3+a2a3Υ2+a1a3Υ+a0a3=0.


## 3.1. Application to Generalized Fractional KdV Equation

In the case ofα=1, (1) reduces to the classical nonlinear generalized third order KdV equation. Many researchers have tried to get the exact solutions of this equation by using a variety of methods. But, we solve the generalized fractional KdV equation for 0<α≤1. Compactons, solitary patterns, and periodic and solitary traveling plane waves solutions of this equation are found.Let us consider the travelling wave solutions of (1), and we perform the transformation u(x,t)=u(η), η=kx-(λtα/Γ(1+α)), where λ is constant to be determined later. Then, integrating this equation with respect to η and setting the integration constant to zero, we get
(14)-λ(p+1)u+k3(p+1)u′′+kup+1=0.
By use of the following transformation:
(15)u=v1/p,
(14) turns into the equation
(16)k3(p2+p)vv′′+k3(1-p2)(v′)2-λ(p3+p2)v2+kp2v3=0.
From (8)–(10), we can write the following highest order nonlinear terms in order to determine the balance procedure:
(17)v=τδΥδ+⋯,v3=τδ3Υ3δ+⋯,(v′)2=ξθδ2τδ2ζϵΥ2δ-2(Υ′)2+⋯=ξθδ2τδ2ζϵΥ2δ+θ-ϵ-2+⋯,vv′′=ϖΥ2δ+θ-ϵ-2+⋯.
Balancing the highest order nonlinear terms of vv′′ and v3 in (16), we have
(18)2δ+θ-ϵ-2=3δ,
hence
(19)θ=ϵ+δ+2.
This resolution procedure is applied, and we obtain results as follows.Case 1.
If we takeϵ=0, δ=1, and θ=3, then
(20)(v′)2=(τ1)2(ξ3Υ3+ξ2Υ2+ξ1Υ+ξ0)ζ0,v′′=(τ1)(3ξ3Υ2+2ξ2Υ+ξ1)2ζ0,
where ξ3≠0, ζ0≠0. Thus, we have a system of algebraic equations from the coefficients of polynomial of Υ. Solving the algebraic equation system (12), we get
(21)ξ0=ξ0,ξ1=ξ1,ξ2=τ1(2ξ1τ0-3ξ0τ1)τ02,ξ3=τ12(ξ1τ0-2ξ0τ1)τ03,τ0=τ0,τ1=τ1,ζ0=-k2(1+p)(2+p)τ1(ξ1τ0-2ξ0τ1)2p2τ03,λ=2kτ0(ξ1τ0-3ξ0τ1)(2+3p+p2)(ξ1τ0-2ξ0τ1).
Substituting these results into (13), we have
(22)±(η-η0)=-k2(2+3p+p2)2p2τ1×∫‍(nnnnnnnnnnnnnnn+ξ0τ03τ12(ξ1τ0-2ξ0τ1))-1/2dΥ×(Υ3+τ0(2ξ1τ0-3ξ0τ1)τ1(ξ1τ0-2ξ0τ1)Υ2nnnnnnnnnnnnnnn+ξ1τ03τ12(ξ1τ0-2ξ0τ1)Υnnnnnnnnnnnnnnn+ξ0τ03τ12(ξ1τ0-2ξ0τ1))-1/2).
Integrating (22), we procure the solutions to (1) as follows:
(23)±(η-η0)=-2AΥ-α1,±(η-η0)=2Aα2-α1arctanΥ-α2α2-α1,α2>α1,±(η-η0)=Aα1-α2ln|Υ-α2-α1-α2Υ-α2+α1-α2|,α1>α2,±(η-η0)=-2Aα1-α3F(φ,l),α1>α2>α3,
where
(24)A=-k2(2+3p+p2)2p2τ1,F(φ,l)=∫0φ‍dψ1-l2sin2ψ,φ=arcsinΥ-α3α2-α3,l2=α2-α3α1-α3.
Also α1, α2, and α3 are the roots of the polynomial equation
(25)Υ3+ξ2ξ3Υ2+ξ1ξ3Υ+ξ0ξ3=0.
Substituting the solutions (23) into (8) and (15), we have(26)
u
(
x
,
t
)
=
[
τ
0
+
τ
1
α
1
+
4
A
2
τ
1
(
k
x
-
(
2
k
τ
0
(
ξ
1
τ
0
-
3
ξ
0
τ
1
)
/
(
2
+
3
p
+
p
2
)
(
ξ
1
τ
0
-
2
ξ
0
τ
1
)
Γ
(
1
+
α
)
)
t
α
-
η
0
)
2
]
1
/
p
,
(27)
u
(
x
,
t
)
=
[
τ
0
+
τ
1
α
1
+
τ
1
(
α
2
-
α
1
)
sec
h
2
(
α
1
-
α
2
2
A
(
k
x
-
2
k
τ
0
(
ξ
1
τ
0
-
3
ξ
0
τ
1
)
(
2
+
3
p
+
p
2
)
(
ξ
1
τ
0
-
2
ξ
0
τ
1
)
Γ
(
1
+
α
)
t
α
-
η
0
)
)
]
1
/
p
,
(28)
u
(
x
,
t
)
=
[
τ
0
+
τ
1
α
1
+
τ
1
(
α
1
-
α
2
)
cosec
h
2
(
α
1
-
α
2
2
A
(
k
x
-
2
k
τ
0
(
ξ
1
τ
0
-
3
ξ
0
τ
1
)
(
2
+
3
p
+
p
2
)
(
ξ
1
τ
0
-
2
ξ
0
τ
1
)
Γ
(
1
+
α
)
t
α
-
η
0
)
)
]
1
/
p
,
(29)
u
(
x
,
t
)
=
[
×
(
k
x
-
2
k
τ
0
(
ξ
1
τ
0
-
3
ξ
0
τ
1
)
(
2
+
3
p
+
p
2
)
(
ξ
1
τ
0
-
2
ξ
0
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1
)
Γ
(
1
+
α
)
t
α
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(
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2
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1
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α
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)
)
-
1
)
τ
0
+
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1
α
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+
(
×
(
k
x
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2
k
τ
0
(
ξ
1
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0
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3
ξ
0
τ
1
)
(
2
+
3
p
+
p
2
)
(
ξ
1
τ
0
-
2
ξ
0
τ
1
)
Γ
(
1
+
α
)
t
α
-
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0
,
(
α
2
-
α
3
)
(
α
1
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α
3
)
)
)
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-
1
(
τ
1
(
α
2
-
α
3
)
)
n
n
n
n
n
n
n
n
n
n
n
n
n
n
n
n
n
n
×
(
s
n
2
(
±
(
α
1
-
α
3
)
2
A
(
k
x
-
2
k
τ
0
(
ξ
1
τ
0
-
3
ξ
0
τ
1
)
(
2
+
3
p
+
p
2
)
(
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1
τ
0
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2
ξ
0
τ
1
)
Γ
(
1
+
α
)
t
α
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η
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(
α
2
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α
3
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1
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α
3
)
)
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-
1
)
]
1
/
p
.If we takeτ0=-τ1α1 and η0=0, then the solutions (26)–(28) can reduce to rational function solution
(30)u(x,t)=(A~x-λ1tα)2/p,
1-soliton wave solution
(31)u(x,t)=B~cosh2/p(B(x-λ1tα)),
singular soliton solution
(32)u(x,t)=C~sinh2/p(C(x-λ1tα)),whereA~=2Aτ1/k, B~=(τ1(α2-α1))1/p, B=kα1-α2/2A, C~=(τ1(α1-α2))1/p,  C=kα1-α2/2A, and λ1=-(2τ1α1(ξ1α1+3ξ0)/(2+3p+p2)(ξ1α1+2ξ0)Γ(1+α)). Here, B~ and C~ are the amplitude of the soliton, while B and C are the inverse width of the solitons and c is the velocity.On the other hand, if we takeτ0=-τ1α3 and η0=0, then the Jacobi elliptic function solution (29) can be written in the form
(33)u(x,t)=A1sn2/p(Bi(x-λ1tα),α2-α3α1-α3),
where A1=(τ1(α2-α3))1/p and Bi=(-1)ikα1-α3/2A, i=(1,2). Thus, we can say that the solitons exist for τ1>0.Remark 1.
We provide a proof by verifying the exact solution (31) by direct substitution. If we take α=1 and p=2, then (1) becomes
(34)ut+uxxx+u2ux=0,
whereB~=τ1(α2-α1), B=kα1-α2/2A, A=k-(3/2τ1), and λ1=-(τ1α1(ξ1α1+3ξ0)/6(ξ1α1+2ξ0)). Here, B=τ1(α2-α1)/6.
Equation (31) is
(35)u(x,t)=B~sech[B(x-λ1t)].
If ϕ=x-λ1t, then
(36)u(ϕ)=B~sech(Bϕ),ut=B~Bλ1tanh(Bϕ)sech(Bϕ),ux=-B~Btanh(Bϕ)sech(Bϕ),u2ux=-B~3Btanh(Bϕ)sech3(Bϕ),uxxx=B~B3tanh(Bϕ)(6sech3(Bϕ)-sech(Bϕ)).
Substituting the above relations into the partial differential equation, we obtain the following equation:
(37)B~Btanh(Bϕ)sech(Bϕ)[λ1-B2+(6B2-B~2)sech2(Bϕ)]=-τ1(ξ1α1α2+2ξ0α2+ξ0α1)6(ξ1α1+2ξ0)=0.
Otherwise, we can write
(38)-α12α2=ξ0τ03τ12(ξ1τ0-2ξ0τ1),
where τ0=-τ1α1 and (Υ-α1)2(Υ-α2)=Υ3+(ξ2/ξ3)Υ2+(ξ1/ξ3)Υ+(ξ0/ξ3)=0. From this equation, we find ξ1α1α2+2ξ0α2+ξ0α1=0. Substituting this relation into the above equation yields 0=0. Thus, it is seen that the obtained solution (31) is an exact solution of the fractional KdV equation.Remark 2.
The solutions (30)–(33) obtained by using the extended trial equation method for (1) have been checked by Mathematica. To our knowledge, rational function solution, 1-soliton solution, the singular soliton solution, and the Jacobi elliptic function that we find in this paper are not shown in the previous literature. These results are new traveling wave solutions of (1).Case 2.
If we takeϵ=0, δ=2, and θ=4, then
(39)(v′)2=(τ1+2τ2Υ)2(ξ4Υ4+ξ3Υ3+ξ2Υ2+ξ1Υ+ξ0)ζ0,v′′=((τ1+2τ1Υ)(4ξ4Υ3+3ξ3Υ2+2ξ2Υ+ξ1)nnn+4τ2(ξ4Υ4+ξ3Υ3+ξ2Υ2+ξ1Υ+ξ0))×(2ζ0)-1,
where ξ4≠0, ζ0≠0. Respectively, solving the algebraic equation system (12) yields as follows:
(40)ξ0=(ξ4τ12-ξ2τ12)24ξ4τ24,ξ1=ξ2τ1τ22-ξ4τ13τ23,ξ3=2ξ4τ1τ2,ξ2=ξ2,ξ4=ξ4,τ0=ξ2τ22-ξ4τ122ξ4τ2,τ1=τ1,τ2=τ2,ζ0=-2k2(2+3p+p2)ξ4p2τ2,λ=-k(3ξ4τ12-2ξ2τ22)2(2+3p+p2)ξ4τ2.
Substituting these results into (13), we get
(41)±(η-η0)=-2k2(2+3p+p2)p2τ2×∫(+(ξ4τ12-ξ2τ12)24ξ42τ24)-1/2dΥ‍((ξ4τ12-ξ2τ12)24ξ42τ24Υ4+2τ1τ2Υ3+ξ2ξ4Υ2bbbbbbbbbbbbbbb+ξ2τ1τ22-ξ4τ13τ23ξ4Υbbbbbbbbbbbbbbb+(ξ4τ12-ξ2τ12)24ξ42τ24)-1/2).
Integrating (41), we obtain the solutions to (1) as follows:
(42)±(η-η0)=-A2Υ-α1,±(η-η0)=2A2α1-α2Υ-α2Υ-α1,α1>α2,±(η-η0)=A2α1-α2ln|Υ-α1Υ-α2|,±(η-η0)=2A2(α1-α2)(α1-α3)×ln|(Υ-α2)(α1-α3)-(Υ-α3)(α1-α2)(Υ-α2)(α1-α3)+(Υ-α3)(α1-α2)|,nnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnα1>α2>α3,±(η-η0)=2A2(α1-α3)(α2-α4)F(φ,l),nnnnnnnnnnα1>α2>α3>α4,
where
(43)A2=-2k2(2+3p+p2)p2τ2,φ=arcsin(Υ-α1)(α2-α4)(Υ-α2)(α1-α4),l2=(α2-α3)(α1-α4)(α1-α3)(α2-α4).
Also α1, α2, α3, and α4 are the roots of the polynomial equation
(44)Υ4+ξ3ξ4Υ3+ξ2ξ4Υ2+ξ1ξ4Υ+ξ0ξ4=0.
Substituting the solutions (42) into (8) and (15), we have(45)
u
(
x
,
t
)
=
[
(
α
1
±
A
2
k
x
+
(
k
(
3
ξ
4
τ
1
2
-
2
ξ
2
τ
2
2
)
/
2
(
2
+
3
p
+
p
2
)
ξ
4
τ
2
Γ
(
1
+
α
)
)
t
α
-
η
0
)
2
τ
0
+
τ
1
α
1
±
τ
1
A
2
k
x
+
(
k
(
3
ξ
4
τ
1
2
-
2
ξ
2
τ
2
2
)
/
2
(
2
+
3
p
+
p
2
)
ξ
4
τ
2
Γ
(
1
+
α
)
)
t
α
-
η
0
u
(
x
,
t
)
n
n
n
n
+
τ
2
(
α
1
±
A
2
k
x
+
(
k
(
3
ξ
4
τ
1
2
-
2
ξ
2
τ
2
2
)
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2
(
2
+
3
p
+
p
2
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ξ
4
τ
2
Γ
(
1
+
α
)
)
t
α
-
η
0
)
2
]
1
/
p
,
u
(
x
,
t
)
=
{
(
α
1
+
4
A
2
2
(
α
2
-
α
1
)
4
A
2
2
-
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(
α
1
-
α
2
)
(
k
x
+
(
k
(
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ξ
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τ
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2
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2
ξ
2
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2
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3
p
+
p
2
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τ
2
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1
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t
α
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2
τ
0
+
τ
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1
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4
A
2
2
(
α
2
-
α
1
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τ
1
4
A
2
2
-
[
(
α
1
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k
x
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1
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t
α
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2
u
(
x
,
t
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n
n
n
n
+
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-
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t
α
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]
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)
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}
1
/
p
,
u
(
x
,
t
)
=
{
(
α
2
+
(
α
2
-
α
1
)
exp
[
(
(
α
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2
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1
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1
/
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,
u
(
x
,
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=
{
(
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1
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α
1
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[
(
(
α
1
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2
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1
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[
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(
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1
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k
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1
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(
x
,
t
)
n
n
n
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+
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2
(
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1
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1
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)
exp
[
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(
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1
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2
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k
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t
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2
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2
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k
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4
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α
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1
}
1
/
p
,
u
(
x
,
t
)
=
{
cosh
[
(
α
1
-
α
2
)
(
α
1
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α
3
)
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2
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k
x
+
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2
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p
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2
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0
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1
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1
-
(
cosh
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(
α
1
-
α
2
)
(
α
1
-
α
3
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}
1
/
p
.For simplicity, if we takeη0=0, then we can write the solutions (45) as follows:
(46)u(x,t)=[∑i=02‍τi(α1±A2k(x-λ2tα))i]1/p,u(x,t)=[∑i=02‍τi(α1+4A22(α2-α1)4A22-[(α1-α2)k(x-λ2tα)]2)i]1/p,u(x,t)=[∑i=02‍τi(α2+α2-α1exp[B3(x-λ2tα)]-1)i]1/p,u(x,t)=[∑i=02‍τi(α1+α1-α2exp[B3(x-λ2tα)]-1)i]1/p,u(x,t)=[∑i=02‍τi(×cosh[C(x-λ2tα)])-1((2(α1-α2)(α1-α3)))((2(α1-α2)(α1-α3))α1-(×cosh[C(x-λ2tα)])-1((2(α1-α2)(α1-α3)))((2(α1-α2)(α1-α3))(2(α1-α2)(α1-α3))nnnnnnnnnnnnn×(2α1-α2-α3+(α3-α2)nnnnnnnnnnnnnnnn×cosh[C(x-λ2tα)])-1((2(α1-α2)(α1-α3)))((2(α1-α2)(α1-α3)))i∑i=02‍]1/p,u(x,t)=[∑i=02‍τi(α2+(α1-α2)(α4-α2)α4-α2+(α1-α4)sn2(φ,l))i]1/p,
where B3=k(α1-α2)/A2, C=k(α1-α2)(α1-α3)/A2, φ=(k(α1-α3)(α2-α4)/2A2)(x-λ2tα), l2=(α2-α3)(α1-α4)/(α1-α3)(α2-α4), and λ2=(3ξ4τ12-2ξ2τ22)/2(2+3p+p2)ξ4τ2Γ(1+α). Here, A2 is the amplitude of the soliton, while λ2 is the velocity and B3 and C are the inverse width of the solitons.Remark 3.
To our knowledge, the solutions (46) that we find in this paper have not been found in the literature, and these are new exact solutions of (1).


## 3.2. Application to the FractionalK(n,n) Equation

In the case ofα=1, (2) reduces to the classical nonlinear generalized K(n,n) equation. Many authors have studied to find the exact solutions of this equation by using a variety of methods [27, 28]. In this paper, we consider the generalized fractional K(n,n) equation for the case of 0<α≤1. Compactons, solitary patterns, and periodic and solitary traveling plane waves solutions of this equation are obtained.Let us consider the travelling wave solutions of (2), and we perform the transformation u(x,t)=u(η), η=kx-(λtα/Γ(1+α)), where λ is constant to be determined later. Then, integrating this equation with respect to η and setting the integration constant to zero, we get
(47)k3(un)′′+kun-λu=0.
By use of the following transformation:
(48)u=v1/(n-1),
(47) turns into the equation
(49)k3(n2-n)vv′′+k3n(v′)2-λ(n-1)2v+k(n-1)2v2=0.
From (8)–(10), we can write the following highest order nonlinear terms in order to determine the balance procedure:
(50)v=τδΥδ+⋯,v2=τδ2Υ2δ+⋯,vv′′=ϖΥ2δ+θ-ϵ-2+⋯.
Balancing the highest order nonlinear terms of vv′′ and v2 in (49), we have
(51)2δ+θ-ϵ-2=2δ,
hence
(52)θ=ϵ+2.
This resolution procedure is applied and we obtain results as follows.Case 1.
If we takeϵ=0,  θ=2, and δ=1, then
(53)(v′)2=τ12(ξ2Υ2+ξ1Υ+ξ0)ζ0,v′′=τ1(2ξ2Υ+ξ1)2ζ0,
where ξ2≠0, ζ0≠0. Thus, we have a system of algebraic equations from the coefficients of polynomial of u. Solving the algebraic equation system (12), we get
(54)ξ0=τ1(ξ1τ1-ξ2τ0)τ12,ξ1=ξ1,ξ2=ξ2,ζ0=-k2n2ξ2(n-1)2,τ0=τ0,τ1=τ1,λ=k(n+1)(2ξ2τ0-ξ1τ1)2nξ2.
Substituting these results into (13), we have
(55)±(η-η0)=-k2n2(n-1)2∫‍dΥΥ2+(ξ1/ξ2)Υ+τ1(ξ1τ1-ξ2τ0)/τ12ξ2.
Integrating (55), we procure the solutions to (2) as follows:
(56)±(η-η0)=A3ln|Υ-α1|,±(η-η0)=2A3ln|Υ-α1+Υ-α2|,
where A3=-(k2n2/(n-1)2). Also α1, α2 are the roots of the polynomial equation
(57)Υ2+ξ1ξ2Υ+ξ0ξ2=0.
Substituting the solutions (56) into (8) and (48), we have(58)
u
(
x
,
t
)
=
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+
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.If we takeη0=α2=0 and α1=1, then the solutions (58) can reduce to single kink solution and the hyperbolic function solution, respectively,
(59)u(x,t)=(τ0+τ1(exp[B4(x-λ3tα)]+1))1/(n-1),(60)u(x,t)=(τ0+τ1(12(cosh[B4(x-λ3tα)]+1)))1/(n-1),whereB4=k/A3, λ3=(n+1)(2ξ2τ0-ξ1τ1)/2nξ2Γ(1+α). Here, B4 is the inverse width of the solitons.Remark 4.
The solutions (59)-(60) obtained by using the extended trial equation method for (2) have been checked by Mathematica. To our knowledge, the hyperbolic function solution and single kink solution, that we find in this paper, are not shown in the previous literature. These results are new traveling wave solutions of (2).Case 2.
If we takeϵ=0, θ=2, and δ=2, then
(61)(v′)2=(τ1+2τ2Υ)2(ξ2Υ2+ξ1Υ+ξ0)ζ0,v′′=4τ2(ξ2Υ2+ξ1Υ+ξ0)+(τ1+2τ1Υ)(2ξ2Υ+ξ1)2ζ0,
where ξ2≠0, ζ0≠0. Thus, we have a system of algebraic equations from the coefficients of polynomial of u. Solving the algebraic equation system (12), we get
(62)ξ0=ξ0,ξ1=ξ2τ1τ2,ξ2=ξ2,ζ0=-4k2n2ξ2(n-1)2,τ0=τ124τ2,τ1=τ1,τ2=τ2,λ=k(n+1)(ξ2τ12-4ξ0τ22)8nξ2τ2.
Substituting these results into (13), we have
(63)±(η-η0)=-4k2n2(n-1)2∫‍dΥΥ2+(τ1/τ2)Υ+(ξ0/ξ2).
Integrating (63), we procure the solutions to (2) as follows:
(64)±(η-η0)=A4ln|Υ-α1|,±(η-η0)=2A4ln|Υ-α1+Υ-α2|,
where A4=-(4k2n2/(n-1)2). Also α1, α2 are the roots of the polynomial equation
(65)Υ2+ξ1ξ2Υ+ξ0ξ2=0.
Substituting the solutions (64) into (8) and (48), we have(66)
u
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k
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/
(
n
-
1
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.For simplicity,η0=α2=0 and α1=1, then we can write the solutions (66) as follows:
(67)u(x,t)=[∑i=02‍τi(exp[B5(x-λ4tα)+1])i]1/(n-1),u(x,t)=[∑i=02‍τi(12(cosh[B5(x-λ4tα)]+1))i]1/(n-1),whereB5=k/A4, λ4=(n+1)(ξ2τ12-4ξ0τ22)/8nξ2τ2Γ(1+α). Here, B5 is the inverse width of the solitons.Case 3.
If we takeϵ=1, θ=3, and δ=1, then
(68)(v′)2=τ12(ξ3Υ3+ξ2Υ2+ξ1Υ+ξ0)ζ0+ζ1Υ,v′′=(-ζ1(ξ3Υ3+ξ2Υ2+ξ1Υ+ξ0))τ1((ζ0+ζ1Υ)(3ξ3Υ2+2ξ2Υ+ξ1)nnnnnn-ζ1(ξ3Υ3+ξ2Υ2+ξ1Υ+ξ0)))×(2(ζ0+ζ1Υ)2)-1,
where ξ3≠0, ζ1≠0. Respectively, solving the algebraic equation system (12) yields
(69)ξ0=-(n-1)2ζ0τ0((n-1)2ζ0τ0+k2n2ξ1τ1)k2n2τ1(k2n2ξ3τ0-(n-1)2ζ0τ1),ξ1=ξ1,ξ2=((k2τ1(k2n4ξ3τ0-(n-1)2n2ζ0τ1))-1k4n4ξ32τ02+(n-1)4ζ02τ12nnn+k2ξ3τ1(k2n4ξ1τ1-(n-1)2n2ζ0τ0)(k2τ1(k2n4ξ3τ0-(n-1)2n2ζ0τ1))-1)×(k2τ1(k2n4ξ3τ0-(n-1)2n2ζ0τ1))-1,ξ3=ξ3,ζ0=ζ0,ζ1=-k2n2ξ3(n-1)2,τ0=τ0,τ1=τ1,λ=k(n+1)(k2n2ξ3τ02-τ1(2(n-1)2ζ0τ0+k2n2ξ1τ1))2k2n3ξ3τ0-2(n-1)2nζ0τ1.
Substituting these results into (13), we have
(70)±(η-η0)=-k2n2(n-1)2×∫‍Υ+(ζ0/ζ1)Υ3+(ξ2/ξ3)Υ2+(ξ1/ξ3)Υ+(ξ0/ξ3)dΥ.
Integrating (70), we obtain the solutions to (2) as follows:
(71)±(η-η0)=2A3(ln|ζ0+ζ1Υζ1+Υ-α1|-ζ0+ζ1Υζ1(Υ-α1)),±(η-η0)=2A3ζ1(α2-α1)×(ζ0+ζ1α1arctan[(ζ0+ζ1α1)(Υ-α2)(ζ0+ζ1Υ)(α2-α1)]nnnnnnn+α2-α1ln|ζ0+ζ1Υζ1+Υ-α2|),±(η-η0)=2(α1-α2)ζ1(ζ0+ζ1α3)×((ζ0+ζ1α1)F(φ,l)+(α3-ζ1α1)π(φ,n,l)),
where
(72)F(φ,l)=∫0φ‍dψ1-l2sin2ψ,π(φ,n,l)=∫0φ‍dψ(1+nsin2ψ)1-l2sin2ψ,φ=arcsin(Υ-α3)(α2-α1)(Υ-α1)(α2-α3),l2=(ζ0+ζ1α1)(α3-α2)(ζ0+ζ1α3)(α1-α2),n=α3-α2α1-α2.
Also α1, α2, and α3 are the roots of the polynomial equation
(73)Υ3+a2a3Υ2+a1a3Υ+a0a3=0.


## 4. Conclusions

In this study, the extended trial equation method is applied successfully for solving the nonlinear fractional differential equations. We used it to obtain some soliton and elliptic function or elliptic integral function solutions, which are one-wave solutions, to the generalized third order fractional KdV equation and the fractionalK(n,n) equation. The performance of this method is reliable and effective, and also this method gives more general solutions. We think that the proposed method can also be applied to other fractional nonlinear differential equations with nonlinear evolution. In our future studies, we will define the multiple trial equation method for the fractional or nonfractional nonlinear partial differential equations.



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*Source: 491359-2013-06-10.xml*