# New Exact Solutions for a Higher-Order Wave Equation of KdV Type Using Extended F-Expansion Method

Authors: Yinghui He; Yun-Mei Zhao; Yao Long
Journal: Mathematical Problems in Engineering
 (2013)

---

## Abstract

The F-expansion method is used to find traveling wave solutions to various wave equations. By giving more solutions of the general subequation, an extended F-expansion method is introduced by Emmanuel. In our work, a generalized KdV type equation of neglecting the highest-order infinitesimal term, which is an important water wave model, is discussed by using the extended F-expansion method. And when the parameters satisfy certain relations, some new exact solutions expressed by Jacobi elliptic function, hyperbolic function, and trigonometric function are obtained. The related results are enriched.


---

## Body

## 1. Introduction

It has recently become more interesting to obtain exact solutions of nonlinear partial differential equations. These equations are mathematical models of complex physical phenomena that arise in engineering, applied mathematics, chemistry, biology, mechanics, physics, and so forth. Thus, the investigation of the traveling wave solutions to nonlinear evolution equations (NLEEs) plays an important role in mathematical physics. A lot of physical models have supported a wide variety of solitary wave solutions.In the recent years, much effort has been spent on this task and many significant methods have been established such as inverse scattering transform [1], Backlund and Darboux transform [2], Hirota [3], homogeneous balance method [4], Jacobi elliptic function method [5], tanh-function method [6], exp-function method [7], simple equation method [8], F-expansion method [9, 10], improved F-expansion method [11], and extended F-expansion method [12].Wang and Li [10] developed a new algebraic method, belonging to the simplest equation method [13–16], to seek more new solutions of NLEEs that can be expressed as polynomial in an elementary function which satisfies a more general subequation than other subequations like Riccati equation, auxiliary ordinary equation, elliptic equation, and generalized Riccati equation. The Fans method not only gives a unified formation to construct various traveling wave solutions but also provides a guideline to classify the various types of traveling wave solutions according to five parameters. An extended F-expansion method was proposed by Yomba in 2005 by giving more solutions of the general subequation. Using the new method, exact solutions of many NLEEs are successfully obtained [12].In this work, we apply the extended F-expansion method on a higher-order wave equation of KdV type for obtaining new exact traveling solutions.In 1995, based on the physical and asymptotic considerations, Fokas [17] derived the following generalized KdV equation:
(1)ηt+ηx+αηηx+βηxxx+ρ1α2η2ηx+αβ(ρ2ηηxxx+ρ3ηxηxx)+ρ4α3η3ηx+α2β(ρ5η2ηxxx+ρ6ηηxηxx+ρ7ηx3)=0,
which is an important water wave model, where α=3A/2,β=B/6,ρ1=-1/6,ρ6=5/3,ρ3 = 23/6,ρ4=1/8,ρ5=7/18,ρ6=79/36,  and ρ7=45/36. Regarding the ρ1,  ρ2,  ρ3,  ρ4,  ρ5,  ρ6,  and ρ7 as free parameters and using the ρ~4 to replace the ρ4α2, (1) becomes the following PDE:
(2)ut+ux+αuux+βuxxx+ρ1α2u2ux+αβ(ρ2uuxxx+ρ3uxηxx)+ρ~4αu3ux+α2β(ρ5u2uxxx+ρ6uuxuxx+ρ7ηx3)=0,
which is given by Tzirtzilakis et al. in [18]. They called it high-order wave equation of KdV type. Just as Tzirtzilakis et al. [18] said, these two equations are both water wave equations of KdV type, which are more physically and practically meaningful.Assuming that the waves are unidirectional and neglecting terms ofO(α2,β3,αβ), (1) can be reduced to the classical KdV equation:
(3)ηt+ηx+αηηx+βηxxx=0.
In [17], Fokas assumed that O(β) is less than O(α). According to this assumption, we easily know that O(α2β)<O(α3) and O(α2β)<O(αβ). Neglecting two high-order infinitesimal terms of O(α3,α2β), (1) can be reduced to another high-order wave equation of KdV type [18, 19] as follows:
(4)ηt+ηx+αηηx+βηxxx+ρ1α2η2ηx+αβ(ρ2ηηxxx+ρ3ηxηxx)=0.
Equation (4) is a special case of (1) for ρ4=ρ5=ρ6=ρ7=0. Equation (4) was first derived in [20] by using the method of bi-Hamiltonian systems, whose Lax pair was also given in [21].If only we neglect the highest-order infinitesimal term ofO(α2β), then (1) can be reduced to a new generalized KdV equation as follows:
(5)ηt+ηx+αηηx+βηxxx+ρ1α2η2ηx+αβ(ρ2ηηxxx+ρ3ηxηxx)+ρ4α3η3ηx=0.In fact, (5) is another special case of (1) for ρ5=ρ6=ρ7=0. It is also third-order approximate equation of KdV type. Of course, on describing dynamical behaviors of water waves, (4) is only a rough approximative model of (1) compared with (5); that is, the precision of model (5) is better than that of model (4) on describing dynamical behaviors of water waves. In other words, the model (5) exhibits much richer phenomenology than the model (4). Therefore, the investigation of exact traveling wave solutions for (5) is more practically meaningful than that of (4).Equation (1) is studied by many researchers and some useful results are obtained when ρi takes special values. However, by using the current methods, we cannot obtain exact solutions of (1) in universal conditions. Therefore, the investigation of exact solutions of (5) is necessary and important. Equation (5) is perhaps not integrable. But it would be interesting to check its asymptotic integrability [22]. Equation (5) is studied by Wu et al. in [23] using the integral bifurcation method and some exact solutions in parameter form are given. In this paper, regarding the ρi(i=1,2,3,4) as free parameters and by using the extended F-expansion method [12], we will investigate exact traveling wave solutions of (5).The organization of the paper is as follows. In Section2, a brief description of the extended F-expansion for finding traveling wave solutions of nonlinear equations is given. In Section 3, we will study (5) by the extended F-expansion methods. Finally conclusions are given in Section 4.


## 2. Description of the Extended F-Expansion Methods

Based on F-expansion method, the main procedures of the extended F-expansion method are as follows [12].Step 1.
Consider a general nonlinear PDE in the form(6)F(u,ux,ut,uxx,uxt,…)=0.
Using u(x,t)=U(ξ), ξ=x-ct, we can rewrite (6) as the following nonlinear ODE:
(7)F(U,U′,U′′,⋯)=0,
where the prime denotes differentiation with respect to ξ.Step 2.
Suppose that the solution of ODE (7) can be written as follows:
(8)U(ξ)=A0+∑i=1n‍(AiFi(ξ)+BiF-i(ξ)),
where Ai, Bi(i=1,2,…n) are constants to be determined later, n is a positive integer that is given by the homogeneous balance principle, and F(ξ) satisfies the following equation:
(9)(F′(ξ))2=h0+h1F(ξ)+h2F2(ξ)+h3F3(ξ)+h4F4(ξ),
where h0, h1, h2, h3, and h4 are constant.Step 3.
Substituting (8) along with (9) into (7) and then setting all the coefficients of Fj(ξ)(j=0,1,2,…,) of the resulting system’s numerator to zero yield a set of overdetermined nonlinear algebraic equations for A0, Ai, and Bi(i=1,2,…,n).Step 4.
Assuming that the constantsA0, Ai, and Bi(i=1,2,…n) can be obtained by solving the algebraic equations in Step 3 and then substituting these constants and the solutions of (9), depending on the special conditions chosen for the h0, h1, h2, h3, and h4, we can obtain the explicit solutions of (6) immediately.


## 3. Exact Solutions of (5)

Making a transformationη(t,x)=ϕ(ξ) with ξ=x-ct, (5) can be reduced to the following ODE:
(10)-cϕ′+ϕ′+αϕϕ′+βϕ′′′+ρ1α2ϕ2ϕ′+αβ(ρ2ϕϕ′′′+ρ3ϕ′ϕ′′)+ρ4α3ϕ3ϕ′=0,
where c is wave velocity which moves along the direction of x-axis and c≠0. Integrating (10) once and setting the integral constant as R yield
(11)(1-c)ϕ+12αϕ2+βϕ′′+13ρ1α2ϕ3+αβ[ρ2ϕϕ′′+12(ρ3-ρ2)ϕ′2]+14ρ4α3ϕ4+R=0.Suppose that (11) owns the solutions in the form
(12)ϕ(ξ)=A0+A1F(ξ)+B1F(ξ).Substituting (12) and (9) into (11) and then setting all the coefficients of Fk(k=-5,…,5) of the resulting system to zero, we can obtain the following results.


### 3.1. Case ofh1=h3=0

In this situation, we have the following cases.Case 1.
Consider(13)A0=-2βh2α,A1=±2β-6h2h4α,B1=0,ρ1=ρ32+ρ2+14βh2,ρ4=ρ3+3ρ212βh2,c=43β2h22ρ3+1,
where h2≠0 and h0 and h4 are arbitrary constants.Case 2.
Consider(14)A0=-2βh2α,A1=0,B1=±2β-6h0h2α,ρ1=ρ32+ρ2+14βh2,ρ4=ρ3+3ρ212βh2,c=43β2h22ρ3+1,
where h2≠0 and h0 and h4 are arbitrary constants.Case 3.
Also(15)A0=αB1212βh0,A1=wB1,ρ1=-6βh0α2B12-12ρ3,ρ2=-ρ3,ρ4=m4βρ3h0α2B12,c=-ρ3B14α4864β2h02+α2B1224βh0+1-6βh0w+12wB12ρ3α2+βh2-B12ρ3α2h212h0,
where w=±h4/h0, and h0≠0, h2 and h4 are arbitrary constants.Substituting (13)–(15) into (12), we obtain, respectively, the following formal solution of (5):
(16)η(x,t)=2βα(±-6h2h4F(ξ)-h2),
where ξ=x-((4/3)β2h22ρ3+1)t, ρ1=ρ3/2+ρ2+1/4βh2, and ρ4=(ρ3+3ρ2)/12βh2.Moreover,(17)η(x,t)=2βα(±-6h0h2F(ξ)-h2),
where ξ=x-((4/3)β2h22ρ3+1)t, ρ1=ρ3/2+ρ2+1/4βh2, and ρ4=(ρ3+3ρ2)/12βh2.We have(18)η(x,t)=αB1212βh0±h4h0B1F(ξ)+B1F(ξ),
where ξ=x-ct, c is determined in Case 3, ρ1=-(6βh0/α2B12)-(1/2)ρ3, ρ2=-ρ3, and ρ4=4βρ3h0/α2B12.Whenh1=h3=0, the general elliptic equation (9) is reduced to the auxiliary ordinary equation:
(19)F′(ξ)2=h0+h2F2(ξ)+h4F4(ξ).The solutions of (19) are given in Table 1. Combining (16)–(18) with Table 1, many exact solutions of (5) can be obtained. For simplicity, we just give out one case in Table 1, and the other cases can be discussed similarly.Table 1
Solutions ofF(ξ) in F′2=h0+h2F2+h4F4.
Case
h
0
h
2
h
4
F
(
ξ
)
1
1
-
(
m
2
+
1
)
m
2
sn
(
ξ
)
,
cd
(
ξ
)
2
1
-
m
2
m
2
-
1
-
m
2
cn
(
ξ
)
3
m
2
-
1
2
-
m
2
-
1
dn
(
ξ
)
4
m
2
-
(
m
2
+
1
)
1
ns
(
ξ
)
,
dc
(
ξ
)
5
-
m
2
2
m
2
-
1
1
-
m
2
nc
(
ξ
)
6
-
1
2
-
m
2
m
2
-
1
nd
(
ξ
)
7
1
2
-
m
2
1
-
m
2
sc
(
ξ
)
8
1
2
m
2
-
1
-
m
2
(
1
-
m
2
)
sd
(
ξ
)
9
1
-
m
2
2
-
m
2
1
cs
(
ξ
)
10
-
m
2
(
1
-
m
2
)
2
m
2
-
1
1
sd
(
ξ
)
11
1
4
1
-
2
m
2
2
1
4
ns
(
ξ
)
±
cs
(
ξ
)
12
1
-
m
2
4
1
+
m
2
2
1
-
m
2
4
nc
(
ξ
)
±
sc
(
ξ
)
13
m
2
4
m
2
-
2
2
1
4
ns
(
ξ
)
±
ds
(
ξ
)
14
m
2
4
m
2
-
2
2
m
2
4
sn
(
ξ
)
±
i
cn
(
ξ
)Whenh0=1, h2=-(m2+1), and h4=m2, the solution of (16) is F(ξ)=sn(ξ,m) or F(ξ)=cd(ξ,m). Substituting them into (17)–(19), we can obtain the following solutions of (5).From (16), one has
(20)η(x,t)=2βα(m2+1±m6(m2+1)sn(ξ,m)),(21)η(x,t)=2βα(m2+1±m6(m2+1)cd(ξ,m)),
where ξ=x-((4/3)β2ρ3(m2+1)2+1)t, ρ1=ρ3/2+ρ2-(1/4β(m2+1)), and ρ4=-(ρ3+3ρ2)/12β(m2+1).Whenm→1, sn(ξ,m)→tanh(ξ), (20) becomes
(22)η(x,t)=4βα(1±3tanh(ξ)),
where ξ=x-((16/3)β2ρ3+1)t, ρ1=ρ3/2+ρ2-1/8β, and ρ4=-(ρ3+3ρ2)/24β.From (17), we have
(23)η(x,t)=2βα(m2+1±6(m2+1)ns(ξ,m)),(24)η(x,t)=2βα(m2+1±6(m2+1)dc(ξ,m)),
where ξ=x-((4/3)β2ρ3(m2+1)2+1)t, ρ1=ρ3/2+ρ2-(1/4β(m2+1)), ρ4=-(ρ3+3ρ2)/12β(m2+1).Whenm→1, ns(ξ,m)→coth(ξ), (23) becomes
(25)η(x,t)=4βα(1±3coth(ξ)),
where ξ=x-((16/3)β2ρ3+1)t, ρ1=ρ3/2+ρ2-1/8β, and ρ4=-(ρ3+3ρ2)/24β.Whenm→0, ns(ξ,m)→csc(ξ), and dc(ξ,m)→sec(ξ), (23) and (24) become
(26)η(x,t)=2βα(1±6csc(ξ)),η(x,t)=2βα(1±6sec(ξ)),
where ξ=x-((4/3)β2ρ3+1)t, ρ1=ρ3/2+ρ2-1/4β, and ρ4=-(ρ3+3ρ2)/12β.From (18), we have
(27)η(x,t)=αB1212β±mB1sn(ξ,m)+B1sn(ξ,m),(28)η(x,t)=αB1212β±mB1cd(ξ,m)+B1cd(ξ,m),
where ξ=x-ct, c is determined in Case 3, ρ1=-(6β/α2B12)-(1/2)ρ3, ρ2=-ρ3, and ρ4=4βρ3/α2B12.Whenm→1, sn(ξ,m)→tanh(ξ), (27) becomes
(29)η(x,t)=αB1212β±B1tanh(ξ)+B1coth(ξ),
where ξ=x-ct, ρ1=-(6β/α2B12)-(1/2)ρ3, ρ2=-ρ3, and ρ4=4βρ3/α2B12.Whenm→0, sn(ξ,m)→sin(ξ), and dc(ξ,m)→sec(ξ), (27) and (28) become
(30)η(x,t)=αB1212β±B1csc(ξ),η(x,t)=αB1212β±B1sec(ξ),
where ξ=x-ct, ρ1=-6β/α2B12-(1/2)ρ3, ρ2=-ρ3, and ρ4=4βρ3/α2B12.


### 3.2.  Case ofh0=h1=0

In this situation, we have(31)A0=β(h3w-8h4h2+3h32)4αh4,A1=βwα,B1=0,ρ1=ρ2+ρ32-6h4βw,ρ4=-2h4(3ρ2+ρ3)βw,c=1+43ρ3β2h22-ρ3β2h2h3(4h3+w)4h4+ρ3β2h33(3h3+w)16h42,
where w=9h32-24h2h4 and h2, h3, h4,ρ2, and ρ3 are arbitrary constants.Substituting (31) into (12), we obtain the following formal solution of (5):
(32)η(x,t)=β(h3w-8h4h2+3h32)4αh4+wβαF(ξ),
where ξ=x-ct and c, ρ1, and ρ4 are determined in (31).Whenh0=h1=0, the general elliptic equation (9) is reduced to the auxiliary ordinary equation:
(33)F′(ξ)2=h2F2(ξ)+h3F3(ξ)+h4F4(ξ).The solutions of (33) are given in Table 2.Table 2
Solutions of (33) with Δ=h32-4h2h4, ε=±1.
Case
h
4
,
Δ
F
(
ξ
)
1
h
2
>
0
-
h
2
h
3
sec
h
2
(
(
h
2
/
2
)
ξ
)
h
3
2
-
h
2
h
4
(
1
+
ε
tan
h
(
(
h
2
/
2
)
ξ
)
)
2
h
2
>
0
-
h
2
h
3
csc
h
2
(
(
h
2
/
2
)
ξ
)
h
3
2
-
h
2
h
4
(
1
+
ε
cot
h
(
(
h
2
/
2
)
ξ
)
)
3
h
2
>
0
,
Δ
>
0
2
h
2
sec
h
(
h
2
ξ
)
ε
Δ
-
h
3
sec
h
(
h
2
ξ
)
4
h
2
<
0
,
Δ
>
0
2
h
2
sec
(
-
h
2
ξ
)
ε
Δ
-
h
3
sec
(
-
h
2
ξ
)
5
h
2
>
0
,
Δ
<
0
2
h
2
csc
h
(
h
2
ξ
)
ε
-
Δ
-
h
3
csc
h
(
h
2
ξ
)
6
h
2
<
0
,
Δ
>
0
2
h
2
csc
(
-
h
2
ξ
)
ε
Δ
-
h
3
csc
(
-
h
2
ξ
)
7
h
2
>
0
,
h
4
>
0
-
h
2
sec
h
2
(
(
h
2
/
2
)
ξ
)
h
3
+
2
ε
h
2
h
4
tan
h
(
(
h
2
/
2
)
ξ
)
8
h
2
<
0
,
h
4
>
0
-
h
2
se
c
2
(
(
-
h
2
/
2
)
ξ
)
h
3
+
2
ε
-
h
2
h
4
tan
(
(
-
h
2
/
2
)
ξ
)
9
h
2
>
0
,
h
4
>
0
h
2
csc
h
2
(
(
h
2
/
2
)
ξ
)
h
3
+
2
ε
h
2
h
4
cot
h
(
(
h
2
/
2
)
ξ
)
10
h
2
<
0
,
h
4
>
0
-
h
2
cs
c
2
(
(
-
h
2
/
2
)
ξ
)
h
3
+
2
ε
-
h
2
h
4
cot
(
(
-
h
2
/
2
)
ξ
)
11
h
2
>
0
,
Δ
=
0
-
h
2
h
3
(
1
+
ε
tan
h
(
h
2
2
)
ξ
)
12
h
2
>
0
,
Δ
=
0
-
h
2
h
3
(
1
+
ε
cot
h
(
h
2
2
)
ξ
)
13
h
2
>
0
4
h
2
exp
(
ε
h
2
ξ
)
(
exp
(
ε
h
2
ξ
)
-
h
3
)
2
-
4
h
2
h
4
14
h
2
>
0
,
h
3
=
0
4
h
2
exp
(
ε
h
2
ξ
)
1
-
4
h
2
h
4
exp
(
2
ε
h
2
ξ
)Combining (32) with Table 2, many exact solutions of (5) can be obtained. The following is an example.Ifh2>0, solution of (33) is
(34)F(ξ)=-h2h3sech2((h2/2)ξ)h32-h2h4(1+ɛtanh((h2/2)ξ)).
Therefore, solution of (4) is
(35)η(x,t)=β(h3w-8h2h4+3h32)4αh4+βwα-h2h3sech2((h2/2)ξ)h32-h2h4(1+ɛtanh((h2/2)ξ)),
where ξ=x-ct, w=9h32-24h2h4, and c, ρ1, and ρ4 are determined by (31).


### 3.3.  Case ofh0≠0, h1≠0, h2≠0, h3≠0, and h4≠0

In this case, there exist three parametersr, p, and q such that
(36)(F′(ξ))2=h0+h1F(ξ)+h2F2(ξ)+h3F3(ξ)+h4F4(ξ)=(r+pF(ξ)+qF2(ξ))2.
Equation (36) is satisfied only if the following relations hold:
(37)h0=r2,h1=2rp,h2=2rq+p2,h3=2pq,h4=q2.Equation (36) is the general Riccati equation. The solutions of (36) are listed in [12]. There are 24 group solutions named ϕiI(i=1,2,…24), which we do not list for simplicity.Substituting (36) and (9) into (11) and then setting all the coefficients of Fk(k=-4,…,4) of the resulting system to zero, we can obtain the following results.Case 1.
Consider(38)A0=β(wp+2p2-8qr)2α,A1=0,B1=wrβα,c=1+13β2ρ3(p2-4qr)2,ρ1=ρ32+ρ2-12β(p2-4qr),ρ2=ρ2,ρ3=ρ3,ρ4=-ρ3+3ρ26β(p2-4qr),
where w=±23p2-4qr and p2-4qr>0.Case 2.
Consider(39)A0=β(wp+2p2-8qr)2α,A1=wrβα,B1=0,c=1+13β2ρ3(p2-4qr)2,ρ1=ρ32+ρ2-12β(p2-4qr),ρ2=ρ2,ρ3=ρ3,ρ4=-ρ3+3ρ26β(p2-4qr),
where w=±23p2-4qr and p2-4qr>0.Case 3.
Consider(40)A0=β(δp+16qr+2p2)2α,A1=δβqα,B1=rδβα,c=1-13ρ3β2(6rpδq-(8qr+p2)2),ρ1=ρ32+ρ2-12β(8qr+p2),ρ2=ρ2,ρ3=ρ3,ρ4=-ρ3+3ρ26β(8qr+p2),
where δ=±23p2+8qr and p2+8qr>0.Substituting (38)–(40) into (12), we obtain, respectively, the following formal solution of (5):
(41)η(x,t)=β(wp+2p2-8qr)2α+wrβαF(ξ),
where ξ=x-(1+(1/3)β2ρ3(p2-4qr)2)t, ρ1=(ρ3/2)+ρ2-1/2β(p2-4qr), and ρ4=-(ρ3+3ρ2)/6β(p2-4qr).Also(42)η(x,t)=β(wp+2p2-8qr)2α+wrβα1F(ξ),
where ξ=x-(1+(1/3)β2ρ3(p2-4qr)2)t, ρ1=(ρ3/2)+ρ2-1/2β(p2-4qr), and ρ4=-(ρ3+3ρ2)/6β(p2-4qr).Moreover,(43)η(x,t)=β(δp+16qr+2p2)2α+δβqαF(ξ)+rδβα1F(ξ),
where ξ=x-(1-(1/3)ρ3β2(6rpδq-(8qr+p2)2))t, ρ1=(ρ3/2)+ρ2-1/2β(8qr+p2), and ρ4=-(ρ3+3ρ2)/6β(8qr+p2).Substituting solutions of (36) ϕiI(i=1,2,…,24) into (41)–(43), we can obtain a lot of solutions of (5). We just give one example.Whenp2-4pq>0 and pq≠0, ϕ1I=-(1/2q)(p+p2-4qrtanh((p2-4qr/2)ξ)). Substituting ϕ1I into (41) and (42), we have
(44)η(x,t)=β(wp+2p2-8qr)2α+wrβαϕ1I,η(x,t)=β(wp+2p2-8qr)2α+wrβα1ϕ1I,
where ξ=x-ct and c, w, ρ1, and ρ4 are determined by (38).


### 3.4.  Case ofh0≠0, h1≠0, h2=0, h3≠0, and h4≠0

In this case, there exist three parametersr, p, and q such that
(45)(F′(ξ))2=h0+h1F(ξ)+h3F3(ξ)+h4F4(ξ)=(r+pF(ξ)+qF2(ξ))2.
Equation (45) is satisfied only if the following relations hold:
(46)h0=r2,h1=2rp,h3=2pq,h4=q2.
The following constraint should exist between r, p, and q parameters:
(47)p2=-2qr,qr<0.
Therefore, we can discuss the solution of (5) similarly as in the case shown in Section 3.3 under condition (47). Here, we omit it.


## 3.1. Case ofh1=h3=0

In this situation, we have the following cases.Case 1.
Consider(13)A0=-2βh2α,A1=±2β-6h2h4α,B1=0,ρ1=ρ32+ρ2+14βh2,ρ4=ρ3+3ρ212βh2,c=43β2h22ρ3+1,
where h2≠0 and h0 and h4 are arbitrary constants.Case 2.
Consider(14)A0=-2βh2α,A1=0,B1=±2β-6h0h2α,ρ1=ρ32+ρ2+14βh2,ρ4=ρ3+3ρ212βh2,c=43β2h22ρ3+1,
where h2≠0 and h0 and h4 are arbitrary constants.Case 3.
Also(15)A0=αB1212βh0,A1=wB1,ρ1=-6βh0α2B12-12ρ3,ρ2=-ρ3,ρ4=m4βρ3h0α2B12,c=-ρ3B14α4864β2h02+α2B1224βh0+1-6βh0w+12wB12ρ3α2+βh2-B12ρ3α2h212h0,
where w=±h4/h0, and h0≠0, h2 and h4 are arbitrary constants.Substituting (13)–(15) into (12), we obtain, respectively, the following formal solution of (5):
(16)η(x,t)=2βα(±-6h2h4F(ξ)-h2),
where ξ=x-((4/3)β2h22ρ3+1)t, ρ1=ρ3/2+ρ2+1/4βh2, and ρ4=(ρ3+3ρ2)/12βh2.Moreover,(17)η(x,t)=2βα(±-6h0h2F(ξ)-h2),
where ξ=x-((4/3)β2h22ρ3+1)t, ρ1=ρ3/2+ρ2+1/4βh2, and ρ4=(ρ3+3ρ2)/12βh2.We have(18)η(x,t)=αB1212βh0±h4h0B1F(ξ)+B1F(ξ),
where ξ=x-ct, c is determined in Case 3, ρ1=-(6βh0/α2B12)-(1/2)ρ3, ρ2=-ρ3, and ρ4=4βρ3h0/α2B12.Whenh1=h3=0, the general elliptic equation (9) is reduced to the auxiliary ordinary equation:
(19)F′(ξ)2=h0+h2F2(ξ)+h4F4(ξ).The solutions of (19) are given in Table 1. Combining (16)–(18) with Table 1, many exact solutions of (5) can be obtained. For simplicity, we just give out one case in Table 1, and the other cases can be discussed similarly.Table 1
Solutions ofF(ξ) in F′2=h0+h2F2+h4F4.
Case
h
0
h
2
h
4
F
(
ξ
)
1
1
-
(
m
2
+
1
)
m
2
sn
(
ξ
)
,
cd
(
ξ
)
2
1
-
m
2
m
2
-
1
-
m
2
cn
(
ξ
)
3
m
2
-
1
2
-
m
2
-
1
dn
(
ξ
)
4
m
2
-
(
m
2
+
1
)
1
ns
(
ξ
)
,
dc
(
ξ
)
5
-
m
2
2
m
2
-
1
1
-
m
2
nc
(
ξ
)
6
-
1
2
-
m
2
m
2
-
1
nd
(
ξ
)
7
1
2
-
m
2
1
-
m
2
sc
(
ξ
)
8
1
2
m
2
-
1
-
m
2
(
1
-
m
2
)
sd
(
ξ
)
9
1
-
m
2
2
-
m
2
1
cs
(
ξ
)
10
-
m
2
(
1
-
m
2
)
2
m
2
-
1
1
sd
(
ξ
)
11
1
4
1
-
2
m
2
2
1
4
ns
(
ξ
)
±
cs
(
ξ
)
12
1
-
m
2
4
1
+
m
2
2
1
-
m
2
4
nc
(
ξ
)
±
sc
(
ξ
)
13
m
2
4
m
2
-
2
2
1
4
ns
(
ξ
)
±
ds
(
ξ
)
14
m
2
4
m
2
-
2
2
m
2
4
sn
(
ξ
)
±
i
cn
(
ξ
)Whenh0=1, h2=-(m2+1), and h4=m2, the solution of (16) is F(ξ)=sn(ξ,m) or F(ξ)=cd(ξ,m). Substituting them into (17)–(19), we can obtain the following solutions of (5).From (16), one has
(20)η(x,t)=2βα(m2+1±m6(m2+1)sn(ξ,m)),(21)η(x,t)=2βα(m2+1±m6(m2+1)cd(ξ,m)),
where ξ=x-((4/3)β2ρ3(m2+1)2+1)t, ρ1=ρ3/2+ρ2-(1/4β(m2+1)), and ρ4=-(ρ3+3ρ2)/12β(m2+1).Whenm→1, sn(ξ,m)→tanh(ξ), (20) becomes
(22)η(x,t)=4βα(1±3tanh(ξ)),
where ξ=x-((16/3)β2ρ3+1)t, ρ1=ρ3/2+ρ2-1/8β, and ρ4=-(ρ3+3ρ2)/24β.From (17), we have
(23)η(x,t)=2βα(m2+1±6(m2+1)ns(ξ,m)),(24)η(x,t)=2βα(m2+1±6(m2+1)dc(ξ,m)),
where ξ=x-((4/3)β2ρ3(m2+1)2+1)t, ρ1=ρ3/2+ρ2-(1/4β(m2+1)), ρ4=-(ρ3+3ρ2)/12β(m2+1).Whenm→1, ns(ξ,m)→coth(ξ), (23) becomes
(25)η(x,t)=4βα(1±3coth(ξ)),
where ξ=x-((16/3)β2ρ3+1)t, ρ1=ρ3/2+ρ2-1/8β, and ρ4=-(ρ3+3ρ2)/24β.Whenm→0, ns(ξ,m)→csc(ξ), and dc(ξ,m)→sec(ξ), (23) and (24) become
(26)η(x,t)=2βα(1±6csc(ξ)),η(x,t)=2βα(1±6sec(ξ)),
where ξ=x-((4/3)β2ρ3+1)t, ρ1=ρ3/2+ρ2-1/4β, and ρ4=-(ρ3+3ρ2)/12β.From (18), we have
(27)η(x,t)=αB1212β±mB1sn(ξ,m)+B1sn(ξ,m),(28)η(x,t)=αB1212β±mB1cd(ξ,m)+B1cd(ξ,m),
where ξ=x-ct, c is determined in Case 3, ρ1=-(6β/α2B12)-(1/2)ρ3, ρ2=-ρ3, and ρ4=4βρ3/α2B12.Whenm→1, sn(ξ,m)→tanh(ξ), (27) becomes
(29)η(x,t)=αB1212β±B1tanh(ξ)+B1coth(ξ),
where ξ=x-ct, ρ1=-(6β/α2B12)-(1/2)ρ3, ρ2=-ρ3, and ρ4=4βρ3/α2B12.Whenm→0, sn(ξ,m)→sin(ξ), and dc(ξ,m)→sec(ξ), (27) and (28) become
(30)η(x,t)=αB1212β±B1csc(ξ),η(x,t)=αB1212β±B1sec(ξ),
where ξ=x-ct, ρ1=-6β/α2B12-(1/2)ρ3, ρ2=-ρ3, and ρ4=4βρ3/α2B12.


## 3.2.  Case ofh0=h1=0

In this situation, we have(31)A0=β(h3w-8h4h2+3h32)4αh4,A1=βwα,B1=0,ρ1=ρ2+ρ32-6h4βw,ρ4=-2h4(3ρ2+ρ3)βw,c=1+43ρ3β2h22-ρ3β2h2h3(4h3+w)4h4+ρ3β2h33(3h3+w)16h42,
where w=9h32-24h2h4 and h2, h3, h4,ρ2, and ρ3 are arbitrary constants.Substituting (31) into (12), we obtain the following formal solution of (5):
(32)η(x,t)=β(h3w-8h4h2+3h32)4αh4+wβαF(ξ),
where ξ=x-ct and c, ρ1, and ρ4 are determined in (31).Whenh0=h1=0, the general elliptic equation (9) is reduced to the auxiliary ordinary equation:
(33)F′(ξ)2=h2F2(ξ)+h3F3(ξ)+h4F4(ξ).The solutions of (33) are given in Table 2.Table 2
Solutions of (33) with Δ=h32-4h2h4, ε=±1.
Case
h
4
,
Δ
F
(
ξ
)
1
h
2
>
0
-
h
2
h
3
sec
h
2
(
(
h
2
/
2
)
ξ
)
h
3
2
-
h
2
h
4
(
1
+
ε
tan
h
(
(
h
2
/
2
)
ξ
)
)
2
h
2
>
0
-
h
2
h
3
csc
h
2
(
(
h
2
/
2
)
ξ
)
h
3
2
-
h
2
h
4
(
1
+
ε
cot
h
(
(
h
2
/
2
)
ξ
)
)
3
h
2
>
0
,
Δ
>
0
2
h
2
sec
h
(
h
2
ξ
)
ε
Δ
-
h
3
sec
h
(
h
2
ξ
)
4
h
2
<
0
,
Δ
>
0
2
h
2
sec
(
-
h
2
ξ
)
ε
Δ
-
h
3
sec
(
-
h
2
ξ
)
5
h
2
>
0
,
Δ
<
0
2
h
2
csc
h
(
h
2
ξ
)
ε
-
Δ
-
h
3
csc
h
(
h
2
ξ
)
6
h
2
<
0
,
Δ
>
0
2
h
2
csc
(
-
h
2
ξ
)
ε
Δ
-
h
3
csc
(
-
h
2
ξ
)
7
h
2
>
0
,
h
4
>
0
-
h
2
sec
h
2
(
(
h
2
/
2
)
ξ
)
h
3
+
2
ε
h
2
h
4
tan
h
(
(
h
2
/
2
)
ξ
)
8
h
2
<
0
,
h
4
>
0
-
h
2
se
c
2
(
(
-
h
2
/
2
)
ξ
)
h
3
+
2
ε
-
h
2
h
4
tan
(
(
-
h
2
/
2
)
ξ
)
9
h
2
>
0
,
h
4
>
0
h
2
csc
h
2
(
(
h
2
/
2
)
ξ
)
h
3
+
2
ε
h
2
h
4
cot
h
(
(
h
2
/
2
)
ξ
)
10
h
2
<
0
,
h
4
>
0
-
h
2
cs
c
2
(
(
-
h
2
/
2
)
ξ
)
h
3
+
2
ε
-
h
2
h
4
cot
(
(
-
h
2
/
2
)
ξ
)
11
h
2
>
0
,
Δ
=
0
-
h
2
h
3
(
1
+
ε
tan
h
(
h
2
2
)
ξ
)
12
h
2
>
0
,
Δ
=
0
-
h
2
h
3
(
1
+
ε
cot
h
(
h
2
2
)
ξ
)
13
h
2
>
0
4
h
2
exp
(
ε
h
2
ξ
)
(
exp
(
ε
h
2
ξ
)
-
h
3
)
2
-
4
h
2
h
4
14
h
2
>
0
,
h
3
=
0
4
h
2
exp
(
ε
h
2
ξ
)
1
-
4
h
2
h
4
exp
(
2
ε
h
2
ξ
)Combining (32) with Table 2, many exact solutions of (5) can be obtained. The following is an example.Ifh2>0, solution of (33) is
(34)F(ξ)=-h2h3sech2((h2/2)ξ)h32-h2h4(1+ɛtanh((h2/2)ξ)).
Therefore, solution of (4) is
(35)η(x,t)=β(h3w-8h2h4+3h32)4αh4+βwα-h2h3sech2((h2/2)ξ)h32-h2h4(1+ɛtanh((h2/2)ξ)),
where ξ=x-ct, w=9h32-24h2h4, and c, ρ1, and ρ4 are determined by (31).


## 3.3.  Case ofh0≠0, h1≠0, h2≠0, h3≠0, and h4≠0

In this case, there exist three parametersr, p, and q such that
(36)(F′(ξ))2=h0+h1F(ξ)+h2F2(ξ)+h3F3(ξ)+h4F4(ξ)=(r+pF(ξ)+qF2(ξ))2.
Equation (36) is satisfied only if the following relations hold:
(37)h0=r2,h1=2rp,h2=2rq+p2,h3=2pq,h4=q2.Equation (36) is the general Riccati equation. The solutions of (36) are listed in [12]. There are 24 group solutions named ϕiI(i=1,2,…24), which we do not list for simplicity.Substituting (36) and (9) into (11) and then setting all the coefficients of Fk(k=-4,…,4) of the resulting system to zero, we can obtain the following results.Case 1.
Consider(38)A0=β(wp+2p2-8qr)2α,A1=0,B1=wrβα,c=1+13β2ρ3(p2-4qr)2,ρ1=ρ32+ρ2-12β(p2-4qr),ρ2=ρ2,ρ3=ρ3,ρ4=-ρ3+3ρ26β(p2-4qr),
where w=±23p2-4qr and p2-4qr>0.Case 2.
Consider(39)A0=β(wp+2p2-8qr)2α,A1=wrβα,B1=0,c=1+13β2ρ3(p2-4qr)2,ρ1=ρ32+ρ2-12β(p2-4qr),ρ2=ρ2,ρ3=ρ3,ρ4=-ρ3+3ρ26β(p2-4qr),
where w=±23p2-4qr and p2-4qr>0.Case 3.
Consider(40)A0=β(δp+16qr+2p2)2α,A1=δβqα,B1=rδβα,c=1-13ρ3β2(6rpδq-(8qr+p2)2),ρ1=ρ32+ρ2-12β(8qr+p2),ρ2=ρ2,ρ3=ρ3,ρ4=-ρ3+3ρ26β(8qr+p2),
where δ=±23p2+8qr and p2+8qr>0.Substituting (38)–(40) into (12), we obtain, respectively, the following formal solution of (5):
(41)η(x,t)=β(wp+2p2-8qr)2α+wrβαF(ξ),
where ξ=x-(1+(1/3)β2ρ3(p2-4qr)2)t, ρ1=(ρ3/2)+ρ2-1/2β(p2-4qr), and ρ4=-(ρ3+3ρ2)/6β(p2-4qr).Also(42)η(x,t)=β(wp+2p2-8qr)2α+wrβα1F(ξ),
where ξ=x-(1+(1/3)β2ρ3(p2-4qr)2)t, ρ1=(ρ3/2)+ρ2-1/2β(p2-4qr), and ρ4=-(ρ3+3ρ2)/6β(p2-4qr).Moreover,(43)η(x,t)=β(δp+16qr+2p2)2α+δβqαF(ξ)+rδβα1F(ξ),
where ξ=x-(1-(1/3)ρ3β2(6rpδq-(8qr+p2)2))t, ρ1=(ρ3/2)+ρ2-1/2β(8qr+p2), and ρ4=-(ρ3+3ρ2)/6β(8qr+p2).Substituting solutions of (36) ϕiI(i=1,2,…,24) into (41)–(43), we can obtain a lot of solutions of (5). We just give one example.Whenp2-4pq>0 and pq≠0, ϕ1I=-(1/2q)(p+p2-4qrtanh((p2-4qr/2)ξ)). Substituting ϕ1I into (41) and (42), we have
(44)η(x,t)=β(wp+2p2-8qr)2α+wrβαϕ1I,η(x,t)=β(wp+2p2-8qr)2α+wrβα1ϕ1I,
where ξ=x-ct and c, w, ρ1, and ρ4 are determined by (38).


## 3.4.  Case ofh0≠0, h1≠0, h2=0, h3≠0, and h4≠0

In this case, there exist three parametersr, p, and q such that
(45)(F′(ξ))2=h0+h1F(ξ)+h3F3(ξ)+h4F4(ξ)=(r+pF(ξ)+qF2(ξ))2.
Equation (45) is satisfied only if the following relations hold:
(46)h0=r2,h1=2rp,h3=2pq,h4=q2.
The following constraint should exist between r, p, and q parameters:
(47)p2=-2qr,qr<0.
Therefore, we can discuss the solution of (5) similarly as in the case shown in Section 3.3 under condition (47). Here, we omit it.


## 4. Conclusions

The investigation of the exact solutions of (5) is very important. And (5) is just studied by Wu et al. [23] using the integral bifurcation method. In our work, (5) is studied by extended F-expansion method and some new exact solutions expressed by Jacobi elliptic function, hyperbolic function, and trigonometric function are obtained. We believe that the results we obtained are useful in describing related physical phenomena. The correctness of all the solutions is verified by substituting them into original equation (5). Comparing with [23], it is easy to see that our method is more straightforward and the forms of the solutions obtained in our paper are also more simple. The related results are enriched.



---

*Source: 128970-2013-12-19.xml*