# Exact Solutions for(4+1)-Dimensional Nonlinear Fokas Equation Using Extended F-Expansion Method and Its Variant

Authors: Yinghui He
Journal: Mathematical Problems in Engineering
 (2014)

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## Abstract

The construction of exact solution for higher-dimensional nonlinear equation plays an important role in knowing some facts that are not simply understood through common observations. In our work,(4+1)-dimensional nonlinear Fokas equation, which is an important physical model, is discussed by using the extended F-expansion method and its variant. And some new exact solutions expressed by Jacobi elliptic function, Weierstrass elliptic function, hyperbolic function, and trigonometric function are obtained. The related results are enriched.


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## 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 efforts have 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], symmetry reductions method [5–8], Jacobi elliptic function method [9], tanh-function method [10], exp-function method [11–13], simple equation method [14], the meshless methods [15–20], G′/G-expansion method [21–23],F-expansion method [24, 25], improvedF-expansion method [26, 27], and extendedF-expansion method [28].Wang and Li [25] developed a new algebraic method, belonging to the simplest equation method [29–32], 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 extendedF-expansion method is 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 [28].The higher-dimensional integrable model is one of the important problems in mathematical physics which can be obtained from several lower-dimensional integrable equations by extending Lax pairs to higher dimensions. Fokas [33] extended the integrable Kadomtsev-Petviashvili and Davey-Stewartson equation to present a new (4+1)-dimensional (4-dimensional space and one-dimensional time) nonlinear wave equation which is given by
(1)4utx-uxxxy+uxyyy+12uxuy+12uuxy-6uzw=0.
Due to important applications of higher-dimensional equations in real world problems, it is necessary to investigate its analytic solutions. Therefore, (1) is studied by many authors. In [34], Yang and Yan investigated symmetries of (1) including point symmetries and the potential symmetries. In [35], Lee et al. discussed the exact solutions of (1) by modified tanh-coth method, extended Jacobi elliptic function method, and the exp-function method. In [36], Kim obtained a group of exact solutions using G′/G-expansion method.In [37], we have successfully applied the extendedF-expansion method on a higher-order wave equation of KdV type. In this work, we apply this method and its variant on (4+1)-dimensional nonlinear Fokas equation for obtaining new exact traveling solutions. Compared with [37], the subequation is discussed more in detail. Besides the four cases discussed in [37], the other two cases of the subequation are also considered, that is, h0=h4=0 and h2=h4=0, so we can obtain richer results. In addition, in present paper two forms of solution are adopted. They are (4) and (5). Equation (4) is the same with the form in [37]. However, (5) is a new form, by which some new solutions can be obtained. The details can be found in Section 3.The organization of the paper is as follows: in Section2, a brief description of the extendedF-expansion for finding traveling wave solutions of nonlinear equations is given. In Section 3, we will study (1) by the extendedF-expansion methods. Finally conclusions are given in Section 4.


## 2. Description of the ExtendedF-Expansion Methods

Based onF-expansion method, the main procedures of the extendedF-expansion method are as follows [22].Step 1.
Consider a general nonlinear PDE in the form(2)F(u,ux,ut,uxx,uxt,…)=0.
Using u(x,t)=U(ξ),  ξ=x-ct, we can rewrite (2) as the following nonlinear ODE:
(3)F(U,U′,U′′,…)=0,
where the prime denotes differentiation with respect to ξ.Step 2.
Suppose that the solution of ODE (3) can be written as follows:
(4)U(ξ)=A0+∑i=1n‍(AiFi(ξ)+BiF-i(ξ))
or
(5)U(ξ)=A0+∑i=1n‍(AiFi(ξ)+BiFi-1F′(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:
(6)(F′(ξ))2=h0+h1F(ξ)+h2F2(ξ)+h3F3(ξ)+h4F4(ξ),
where h0, h1, h2, h3, h4 are constant.Step 3.
Substituting (4) or (5) along with (6) into (3) and then setting all the coefficients of Fj(ξ)F′k(ξ)(j=1,2,…,k=0,1) of the resulting system to zero yields a set of over-determined nonlinear algebraic equations for A0, Ai, Bi(i=1,2,…n).Step 4.
Assuming that the constantsA0, Ai, 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 (6), depending on the special conditions chosen for the h0, h1, h2, h3, and h4 into (4) (or (5)), we can obtain the explicit solutions of (2) immediately.


## 3. Exact Solutions of (1)

Making a transformationu(x,y,z,t,w)=ϕ(ξ) with ξ=αx+βy+γz+δw+ϵt, (1) can be reduced to the following ODE:
(7)(αβ3-α3β)ϕ(4)+(4αϵ-6γδ)ϕ′′+12αβ(ϕ′)2+12αβϕϕ′′=0,
where α, β, γ, δ, and ɛ are nonzero constants. Balancing ϕ(4) and (ϕ′)2 in (7), we obtain n+4=2n+2 which gives n=2.Suppose that (7) owns the solutions in the form
(8)ϕ(ξ)=A0+A1F(ξ)+A2F2(ξ)+B1F(ξ)+B2F2(ξ).Substituting (8) and (6) into (7) and then setting all the coefficients of Fk(k=-6,…,6) of the resulting system to zero, we can obtain the following results.


### 3.1.h3=h4=0

In this situation, we obtain the following set of nontrivial solutions:(9){A1=A2=B1=0,B2=h0(α2-β2),-2αβ3h2A0α3βh22αnnϵ=-2αβ3h2+3γδ-6αβA0+2α3βh22α},{A1=B1=B2=0,A2=h4(α2-β2),-2αβ3h2A0α3βh22αnnϵ=-2αβ3h2+3γδ-6αβA0+2α3βh22α},{A1=B1=0,A2=h4(α2-β2),B2=h0(α2-β2),-2αβ3h2βA0+2α3βh22αnnϵ=-2αβ3h2+3γδ-6αβA0+2α3βh22α},
where A0, h0, h2, and h4 are arbitrary constants and α, β, γ, and δ are nonzero constants.Substituting (9) into (8), we obtain, respectively, the following formal solution of (1):
(10)u(x,y,z,w,t)=A0+h0(α2-β2)F2(ξ),(11)u(x,y,z,w,t)=A0+h4(α2-β2)F2(ξ),(12)u(x,y,z,w,t)=A0+h4(α2-β2)F2(ξ)+h0(α2-β2)F2(ξ),
where ξ=αx+βy+γz+δw+ϵt and α, β, γ, δ, and ϵ are determined in (9).Whenh1=h3=0, the general elliptic equation (6) is reduced to the auxiliary ordinary equation
(13)F′(ξ)2=h0+h2F2(ξ)+h4F4(ξ).The solutions of (13) are given in Table 1. Combining (10)–(12) with Table 1, many exact solutions of (1) can be obtained. For simplicity, we just give out one case in Table 1; 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), h4=m2, the solution of (13) is F(ξ)=sn(ξ,m) or F(ξ)=cd(ξ,m). Substituting them into (10)–(12), we can obtain the following Jacobi Elliptic function solutions of (1).Form (10), one has
(14)u(x,y,z,w,t)=A0+(α2-β2)sn2(ξ,m)=A0+(α2-β2)ns2(ξ,m),(15)u(x,y,z,w,t)=A0+(α2-β2)cd2(ξ,m)=A0+(α2-β2)dc2(ξ,m),
where ξ=αx+βy+γz+δw+((-2αβ3(-m2-1)+3γδ-6αβA0+2α3β(-m2-1))/2α)t.Whenm→1, sn(ξ,m)→tanh(ξ), solution (14) becomes
(16)u(x,y,z,w,t)=A0+(α2-β2)tanh2⁡(ξ)=A0+(α2-β2)coth⁡2⁡(ξ),
where ξ=αx+βy+γz+δw+((4αβ3+3γδ-6αβA0-4α3β)/2α)t.Whenm→0, dc(ξ,m)→sec(ξ), solution (15) becomes
(17)u(x,y,z,w,t)=A0+(α2-β2)cos⁡2(ξ)=A0+(α2-β2)sec2(ξ),
where ξ=αx+βy+γz+δw+((2αβ3+3γδ-6αβA0-2α3β)/2α)t.From (11), we have
(18)u(x,y,z,w,t)=A0+m2(α2-β2)sn2(ξ,m),(19)u(x,y,z,w,t)=A0+m2(α2-β2)cd2(ξ,m),
where ξ=αx+βy+γz+δw+((-2αβ3(-m2-1)+3γδ-6αβA0+2α3β(-m2-1))/2α)t.Whenm→1, sn(ξ,m)→tanh(ξ), solution (18) becomes
(20)u(x,y,z,w,t)=A0+(α2-β2)tanh2(ξ),
where ξ=αx+βy+γz+δw+((4αβ3+3γδ-6αβA0-4α3β)/2α)t.From (12), we have
(21)u(x,y,z,w,t)=A0+(α2-β2)sn2(ξ,m)+m2(α2-β2)sn2(ξ,m),(22)u(x,y,z,w,t)=A0+(α2-β2)cd2(ξ,m)+m2(α2-β2)cd2(ξ,m).
When m→1, sn(ξ,m)→tanh(ξ), solution (21) becomes
(23)u(x,y,z,w,t)=A0+(α2-β2)coth2(ξ)+(α2-β2)tanh2(ξ),
where ξ=αx+βy+γz+δw+((4αβ3+3γδ-6αβA0-4α3β)/2α)t.Whenm→0, cd(ξ,m)→cos⁡(ξ), solution (22) becomes (17).Remark 1.
Considering the form of (5), we can suppose that (7) owns the solutions in the form
(24)ϕ(ξ)=A0+A1F(ξ)+A2F2(ξ)+B1F′(ξ)+B2F(ξ)F′(ξ).
Substituting (24) and (13) into (7) and then setting all the coefficients of Fj(ξ)F′k(ξ)(j=0,1,…,5,k=0,1) of the resulting system to zero, we can obtain the following result:
(25){A1=B2=0,A2=h42(α2-β2),B1=±h42(α2-β2),nnϵ=α3βh2-αβ3h2-12αβA0+6γδ4α},
where h0, h2, and A0 are arbitrary constants, h4>0, and α, β, γ, and δ are nonzero constants.
Substituting (25) into (24), we obtain the following formal solution of (1):
(26)u(x,y,z,w,t)=A0+h42(α2-β2)F2(ξ)±h42(α2-β2)F′(ξ).
Combining (26) with Table 1, many exact solutions of (1) can be obtained. For simplicity, we just give out one case in Table 1.
Whenh0=1,  h2=-(m2+1),  h4=m2, the solution of (13) is F(ξ)=sn(ξ,m) or F(ξ)=cd(ξ,m). Substituting them into (10)–(12), we can obtain the following Jacobi Elliptic function solutions of (1):
(27)u(x,y,z,w,t)=A0+m22(α2-β2)sn2(ξ,m)±m2(α2-β2)cn(ξ,m)dn(ξ,m),(28)u(x,y,z,w,t)=A0+m22(α2-β2)cd2(ξ,m)±m2(1-m2)(α2-β2)sn(ξ,m)nd2(ξ,m),
where ξ=αx+βy+γz+δw+((α3β(-m2-1)-αβ3(-m2-1)-12αβA0+6δγ)/4α)t.
Whenm→1, sn(ξ,m)→tanh(ξ), cn(ξ,m)→sech(ξ), and dn(ξ,m)→sech(ξ), solution (27) becomes
(29)u(x,y,z,w,t)=A0+12(α2-β2)tanh2(ξ)±12(α2-β2)sech2(ξ)=A0+α2-β22(1-2sech2(ξ)),
where ξ=αx+βy+γz+δw+((-2α3β+2αβ3-12αβA0+6δγ)/4α)t.
Some typical wave figures are given as follows. (14) is an unbounded periodic wave solution, as shown in Figure 1(a). As m increases, the period also increases gradually, as shown in Figure 1(b). When m→1, it becomes unbounded solitary wave solution, as shown in Figure 1(c). (18) is a smooth periodic wave solution, as shown in Figure 2(a). As m increases, the period also increases gradually, as shown in Figure 2(b). When m→1, it becomes a solitary wave solution, as shown in Figure 2(c). The figures of other solutions are similar, so we do not give them out.2-D figures of solution (14) with α=2, β=1, A0=1 in the intervals ξ∈[-6,6].
(a)
m = 0.1
(b)
m = 0.9
(c)
m = 12-D figures of solution (18) with α=2, β=1, A0=1 in the intervals ξ∈[-6,6].
(a)
m = 0.1
(b)
m = 0.9
(c)
m = 1


### 3.2.h0=h4=0

In this situation, we have the following result:(30){A2=B1=B2=0,A1=h34(α2-β2),-αβ3h2+α3βh2A04αnnϵ=-αβ3h2+α3βh2+6γδ-12αβA04α},
where A0, h1, h2, and h3 are arbitrary constants.Substituting (30) into (8), we obtain the following formal solution of (1):
(31)u(x,y,z,w,t)=A0+h34(α2-β2)F(ξ),
where ξ=αx+βy+γz+δw+ϵt and α, β, γ, δ, and ϵ are determined in (30).Whenh0=h4=0, the general elliptic equation (6) is reduced to the auxiliary ordinary equation
(32)F′(ξ)2=h1F(ξ)+h2F2(ξ)+h3F3(ξ).Combining (31) with solutions of (32), many exact solutions of (1) can be obtained. The process is similar to the case of h1=h3=0; we omit it.


### 3.3.h0=h1=0

In this situation, we have the following result:(33){B1=B2=0,A1=h32(α2-β2),-h2β3α+A0βα+h2βα34αnnA2=h324h2(α2-β2),h4=h324h2,nnϵ=-h2β3α+6γδ-12A0βα+h2βα34α},
where A0, h2, and h3 are arbitrary constants.Substituting (33) into (8), we obtain the following formal solution of (1):
(34)u(x,y,z,w,t)=A0+h32(α2-β2)F(ξ)+h324h2(α2-β2)F(ξ)2,
where ξ=αx+βy+γz+δw+ϵt and α, β, γ, δ, and ϵ are determined in (33).Whenh0=h1=0, the general elliptic equation (6) is reduced to the auxiliary ordinary equation
(35)F′(ξ)2=h2F(ξ)2+h3F3(ξ)+h4F4(ξ).Ifh2>0, h32-4h2h4=0, the solutions of (35) are given by the following:
(36)F1(ξ)=-h2h3(1±coth⁡(h22ξ+ξ0)),F2(ξ)=-h2h3(1±tanh(h22ξ+ξ0)),
where ξ0 is arbitrary constant.Substituting (36) into (34), we obtain the following hyperbolic function solutions of (1):
(37)u(x,y,z,w,t)=A0-h22(α2-β2)(1±coth⁡⁡(ξ+ξ0))+h24(α2-β2)(1±coth⁡(ξ+ξ0))2,u(x,y,z,w,t)=A0-h22(α2-β2)(1±tanh⁡(ξ+ξ0))+h24(α2-β2)(1±tanh(ξ+ξ0))2,
where ξ=αx+βy+γz+δw+((-h2β3α+6γδ-12A0βα+h2βα3)/4α)t.Remark 2.
Considering the form of (5), we can suppose that (7) owns the solutions in the form
(38)ϕ(ξ)=A0+A1F(ξ)+A2F2(ξ)+B1F′(ξ)+B2F(ξ)F′(ξ).
Substituting (38) and (35) into (7) and then setting all the coefficients of Fj(ξ)F′k(ξ)(j=0,1,…,5,k=0,1) of the resulting system to zero, we can obtain the following result:
(39){α3βh2-αβ3h2-12αβA0+6δγ4αA0=A0,A1=h34(α2-β2),A2=h42(α2-β2),nnB1=±h42(α2-β2),B2=0,nnϵ=α3βh2-αβ3h2-12αβA0+6δγ4α},
where h0, h2, and A0 are arbitrary constants, h4⩾0, and α, β, γ, and δ are nonzero constants.
The solutions of (35) are given in Table 2.
Substituting (39) into (38), we obtain the following formal solution of (1):
(40)u(x,y,z,w,t)=A0+h34(α2-β2)F(ξ)+h42(α2-β2)F2(ξ)±h42(α2-β2)F′(ξ),
where ξ=αx+βy+γz+δw+ϵt and α, β, γ, δ, and ϵ are determined in (39).
Combining (40) with Table 2, many exact solutions of (1) can be obtained. For simplicity, we do not list them carefully. Using case 2 in Table 2, we can obtain the following type of wave figures. Figures 3(a) and 3(b) are kink waves and Figure 3(c) is new type of singular wave. It is interesting that they are dependent on Δ=h32-4h2h4.Table 2
Solutions of (33) with Δ=h32-4h2h4, ε=±1.
Case
h
2
,
Δ
F
(
ξ
)
Case
h
2
,
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
tanh
⁡
(
(
h
2
/
2
)
ξ
)
2
2
h
2
>
0
-
h
2
h
3
csc
h
2
(
(
h
2
/
2
)
ξ
)
h
3
2
-
h
2
h
4
(
1
+
ε
coth
⁡
(
(
h
2
/
2
)
ξ
)
2
3
h
2
>
0, Δ>0
2
h
2
sech
⁡
(
h
2
ξ
)
ε
Δ
-
h
3
sech
⁡
(
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
csch
⁡
(
h
2
ξ
)
ε
-
Δ
-
h
3
csch
⁡
(
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, h4>0
-
h
2
sec
h
2
(
(
h
2
/
2
)
ξ
)
h
3
+
2
ε
h
2
h
4
tanh
⁡
(
(
h
2
/
2
)
ξ
)
8
h
2
<
0, h4>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, h4>0
h
2
csc
h
2
(
(
h
2
/
2
)
ξ
)
h
3
+
2
ε
h
2
h
4
coth
⁡
(
(
h
2
/
2
)
ξ
)
10
h
2
<
0, h4>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
tanh
⁡
(
h
2
2
)
ξ
)
12
h
2
>
0, Δ=0
-
h
2
h
3
(
1
+
ε
coth
⁡
(
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, h3=0
4
h
2
exp
⁡
(
ε
h
2
ξ
)
1
-
4
h
2
h
4
exp
⁡
(
2
ε
h
2
ξ
)2-D figures of solution (40) with α=2, β=1, A0=0, h2=h4=1 in the intervals ξ∈[-6,6].
(a)
h
3
=
1
(
Δ
<
0
)
(b)
h
3
=
2
(
Δ
=
0
)
(c)
h
3
=
3
(
Δ
>
0
)


### 3.4.h0≠0,  h1≠0,  h2≠0,  h3≠0,  h4≠0

In this case, there exists three parametersr, p, and q such that
(41)(F′(ξ))2=h0+h1F(ξ)+h2F2(ξ)+h3F3(ξ)+h4F4(ξ)=(r+pF(ξ)+qF2(ξ))2.
Equation (41) is satisfied only if the following relations hold:
(42)h0=r2,h1=2rp,h2=2rq+p2,h3=2pq,h4=q2.Equation (33) is the general Riccati equation. The solutions of (41) are listed in [12]. There are 24 group solutions named ϕiI, (i=1,2,…24), which we do not list for simplicity.Substituting (41) and (8) into (7) and then setting all the coefficients of Fk(k=-5,…,5) of the resulting system to zero, we can obtain the following results:
(43){A0=A0,B1=B2=0,A1=pq(α2-β2)(4α)-1},A2=q2(α2-β2),ϵ=(8α3βqr+α3βp2-αβ3p2nnnnnn-12αβA0-8αβ3qr+6γδ)×(4α)-1}(44){×(4α)-1A0=A0,A1=A2=0,B1=rp(α2-β2),nnB2=r2(α2-β2),nnϵ=(8α3βqr+α3βp2-αβ3p2nnnnnn-12αβA0-8αβ3qr+6γδ)nn×(4α)-1},
where r, p, and q are arbitrary constants, and
(45){A0=A0,A1=B1=p=0,α3βA0β3γδ2αnnA2=q2(α2-β2),B2=r2(α2-β2),nnϵ=4α3βqr-6αβA0-4αβ3qr+3γδ2α},
where r and q are arbitrary constants.Substituting (43)–(45) into (8), we obtain, respectively, the following formal solutions of (1):
(46)u(x,y,z,w,t)=A0+pq(α2-β2)F(ξ)+q2(α2-β2)F(ξ)2,u(x,y,z,w,t)=A0+rp(α2-β2)F(ξ)-1+r2(α2-β2)F(ξ)-2,
where ξ=αx+βy+γz+δw+((8α3βqr+α3βp2-αβ3p2-12αβA0-8αβ3qr+6γδ)/4α)t, and
(47)u(x,y,z,w,t)=A0+q2(α2-β2)F(ξ)2+r2(α2-β2)F(ξ)-2,
where ξ=αx+βy+γz+δw+((4α3βqr-6αβA0-4αβ3qr+3γδ)/2α)t.Substituting solutions of (41) F(ξ)=ϕiI, (i=1,2,…24) into (46)–(47), we can obtain a lot of solutions of (1). We just give one example.Whenp2-4pq>0 and pq≠0, ϕ1I=-(1/2q)(p+p2-4qrtanh((p2-4qr/2)ξ)). Substituting ϕ1I into (46) and (51), we have
(48)u(x,y,z,w,t)=A0+pq(α2-β2)ϕ1I+q2(α2-β2)ϕ1I2,u(x,y,z,w,t)=A0+rp(α2-β2)ϕ1I-1+r2(α2-β2)ϕ1I-2,
where ξ=αx+βy+γz+δw+((8α3βqr+α3βp2-αβ3p2-12αβA0-8αβ3qr+6γδ)/4α)t.Whenp=0, qr<0, ϕ=rtanh(-qrξ). Substituting ϕ into (47), we have
(49)u(x,y,z,w,t)=A0+q2(α2-β2)ϕ2+r2(α2-β2)ϕ-2,
where ξ=αx+βy+γz+δw+((4α3βqr-6αβA0-4αβ3qr+3γδ)/2α)t.


### 3.5.h0≠0,  h1≠0,  h2=0,  h3≠0,  h4≠0

In this case, there exists three parametersr, p, and q such that
(50)(F′(ξ))2=h0+h1F(ξ)+h3F3(ξ)+h4F4(ξ)=(r+pF(ξ)+qF2(ξ))2.
Equation (50) is satisfied only if the following relations hold:
(51)h0=r2,h1=2rp,h3=2pq,h4=q2.
The following constraint should exist between r, p, and q parameters:
(52)p2=-2qr,qr<0.Therefore, we can discuss the solution of (1) similarly as 3.4 under the condition (52). Here, we omit it.


### 3.6.h2=0,  h4=0

In this situation, we have the following result:(53)A2=B1=B2=0,A1=h34(α2-β2),ϵ=-32-γδ+2A0βαα.
Substituting (53) into (8), we obtain the following formal solution of (1):
(54)u(x,y,z,w,t)=A0+h34(α2-β2)F(ξ),
where ξ=αx+βy+γz+δw+ϵt and α, β, γ, δ, and ϵ are determined in (53).Whenh2=h3=0, the general elliptic equation (6) is reduced to the auxiliary ordinary equation
(55)F′(ξ)2=h0+h1F(ξ)+h3F3(ξ).
The solution of (55) is the Weierstrass elliptic doubly periodic type solution is as follows:
(56)F(ξ)=℘(h32ξ,g2,g3),h3>0.
Substituting (56) into (54), the solution of (1) is
(57)u(x,y,z,w,t)=A0+h34(α2-β2)℘(h32ξ,g2,g3),h3>0,
where g2=-4h1/h3 and g3=-4h0/h3.


## 3.1.h3=h4=0

In this situation, we obtain the following set of nontrivial solutions:(9){A1=A2=B1=0,B2=h0(α2-β2),-2αβ3h2A0α3βh22αnnϵ=-2αβ3h2+3γδ-6αβA0+2α3βh22α},{A1=B1=B2=0,A2=h4(α2-β2),-2αβ3h2A0α3βh22αnnϵ=-2αβ3h2+3γδ-6αβA0+2α3βh22α},{A1=B1=0,A2=h4(α2-β2),B2=h0(α2-β2),-2αβ3h2βA0+2α3βh22αnnϵ=-2αβ3h2+3γδ-6αβA0+2α3βh22α},
where A0, h0, h2, and h4 are arbitrary constants and α, β, γ, and δ are nonzero constants.Substituting (9) into (8), we obtain, respectively, the following formal solution of (1):
(10)u(x,y,z,w,t)=A0+h0(α2-β2)F2(ξ),(11)u(x,y,z,w,t)=A0+h4(α2-β2)F2(ξ),(12)u(x,y,z,w,t)=A0+h4(α2-β2)F2(ξ)+h0(α2-β2)F2(ξ),
where ξ=αx+βy+γz+δw+ϵt and α, β, γ, δ, and ϵ are determined in (9).Whenh1=h3=0, the general elliptic equation (6) is reduced to the auxiliary ordinary equation
(13)F′(ξ)2=h0+h2F2(ξ)+h4F4(ξ).The solutions of (13) are given in Table 1. Combining (10)–(12) with Table 1, many exact solutions of (1) can be obtained. For simplicity, we just give out one case in Table 1; 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), h4=m2, the solution of (13) is F(ξ)=sn(ξ,m) or F(ξ)=cd(ξ,m). Substituting them into (10)–(12), we can obtain the following Jacobi Elliptic function solutions of (1).Form (10), one has
(14)u(x,y,z,w,t)=A0+(α2-β2)sn2(ξ,m)=A0+(α2-β2)ns2(ξ,m),(15)u(x,y,z,w,t)=A0+(α2-β2)cd2(ξ,m)=A0+(α2-β2)dc2(ξ,m),
where ξ=αx+βy+γz+δw+((-2αβ3(-m2-1)+3γδ-6αβA0+2α3β(-m2-1))/2α)t.Whenm→1, sn(ξ,m)→tanh(ξ), solution (14) becomes
(16)u(x,y,z,w,t)=A0+(α2-β2)tanh2⁡(ξ)=A0+(α2-β2)coth⁡2⁡(ξ),
where ξ=αx+βy+γz+δw+((4αβ3+3γδ-6αβA0-4α3β)/2α)t.Whenm→0, dc(ξ,m)→sec(ξ), solution (15) becomes
(17)u(x,y,z,w,t)=A0+(α2-β2)cos⁡2(ξ)=A0+(α2-β2)sec2(ξ),
where ξ=αx+βy+γz+δw+((2αβ3+3γδ-6αβA0-2α3β)/2α)t.From (11), we have
(18)u(x,y,z,w,t)=A0+m2(α2-β2)sn2(ξ,m),(19)u(x,y,z,w,t)=A0+m2(α2-β2)cd2(ξ,m),
where ξ=αx+βy+γz+δw+((-2αβ3(-m2-1)+3γδ-6αβA0+2α3β(-m2-1))/2α)t.Whenm→1, sn(ξ,m)→tanh(ξ), solution (18) becomes
(20)u(x,y,z,w,t)=A0+(α2-β2)tanh2(ξ),
where ξ=αx+βy+γz+δw+((4αβ3+3γδ-6αβA0-4α3β)/2α)t.From (12), we have
(21)u(x,y,z,w,t)=A0+(α2-β2)sn2(ξ,m)+m2(α2-β2)sn2(ξ,m),(22)u(x,y,z,w,t)=A0+(α2-β2)cd2(ξ,m)+m2(α2-β2)cd2(ξ,m).
When m→1, sn(ξ,m)→tanh(ξ), solution (21) becomes
(23)u(x,y,z,w,t)=A0+(α2-β2)coth2(ξ)+(α2-β2)tanh2(ξ),
where ξ=αx+βy+γz+δw+((4αβ3+3γδ-6αβA0-4α3β)/2α)t.Whenm→0, cd(ξ,m)→cos⁡(ξ), solution (22) becomes (17).Remark 1.
Considering the form of (5), we can suppose that (7) owns the solutions in the form
(24)ϕ(ξ)=A0+A1F(ξ)+A2F2(ξ)+B1F′(ξ)+B2F(ξ)F′(ξ).
Substituting (24) and (13) into (7) and then setting all the coefficients of Fj(ξ)F′k(ξ)(j=0,1,…,5,k=0,1) of the resulting system to zero, we can obtain the following result:
(25){A1=B2=0,A2=h42(α2-β2),B1=±h42(α2-β2),nnϵ=α3βh2-αβ3h2-12αβA0+6γδ4α},
where h0, h2, and A0 are arbitrary constants, h4>0, and α, β, γ, and δ are nonzero constants.
Substituting (25) into (24), we obtain the following formal solution of (1):
(26)u(x,y,z,w,t)=A0+h42(α2-β2)F2(ξ)±h42(α2-β2)F′(ξ).
Combining (26) with Table 1, many exact solutions of (1) can be obtained. For simplicity, we just give out one case in Table 1.
Whenh0=1,  h2=-(m2+1),  h4=m2, the solution of (13) is F(ξ)=sn(ξ,m) or F(ξ)=cd(ξ,m). Substituting them into (10)–(12), we can obtain the following Jacobi Elliptic function solutions of (1):
(27)u(x,y,z,w,t)=A0+m22(α2-β2)sn2(ξ,m)±m2(α2-β2)cn(ξ,m)dn(ξ,m),(28)u(x,y,z,w,t)=A0+m22(α2-β2)cd2(ξ,m)±m2(1-m2)(α2-β2)sn(ξ,m)nd2(ξ,m),
where ξ=αx+βy+γz+δw+((α3β(-m2-1)-αβ3(-m2-1)-12αβA0+6δγ)/4α)t.
Whenm→1, sn(ξ,m)→tanh(ξ), cn(ξ,m)→sech(ξ), and dn(ξ,m)→sech(ξ), solution (27) becomes
(29)u(x,y,z,w,t)=A0+12(α2-β2)tanh2(ξ)±12(α2-β2)sech2(ξ)=A0+α2-β22(1-2sech2(ξ)),
where ξ=αx+βy+γz+δw+((-2α3β+2αβ3-12αβA0+6δγ)/4α)t.
Some typical wave figures are given as follows. (14) is an unbounded periodic wave solution, as shown in Figure 1(a). As m increases, the period also increases gradually, as shown in Figure 1(b). When m→1, it becomes unbounded solitary wave solution, as shown in Figure 1(c). (18) is a smooth periodic wave solution, as shown in Figure 2(a). As m increases, the period also increases gradually, as shown in Figure 2(b). When m→1, it becomes a solitary wave solution, as shown in Figure 2(c). The figures of other solutions are similar, so we do not give them out.2-D figures of solution (14) with α=2, β=1, A0=1 in the intervals ξ∈[-6,6].
(a)
m = 0.1
(b)
m = 0.9
(c)
m = 12-D figures of solution (18) with α=2, β=1, A0=1 in the intervals ξ∈[-6,6].
(a)
m = 0.1
(b)
m = 0.9
(c)
m = 1


## 3.2.h0=h4=0

In this situation, we have the following result:(30){A2=B1=B2=0,A1=h34(α2-β2),-αβ3h2+α3βh2A04αnnϵ=-αβ3h2+α3βh2+6γδ-12αβA04α},
where A0, h1, h2, and h3 are arbitrary constants.Substituting (30) into (8), we obtain the following formal solution of (1):
(31)u(x,y,z,w,t)=A0+h34(α2-β2)F(ξ),
where ξ=αx+βy+γz+δw+ϵt and α, β, γ, δ, and ϵ are determined in (30).Whenh0=h4=0, the general elliptic equation (6) is reduced to the auxiliary ordinary equation
(32)F′(ξ)2=h1F(ξ)+h2F2(ξ)+h3F3(ξ).Combining (31) with solutions of (32), many exact solutions of (1) can be obtained. The process is similar to the case of h1=h3=0; we omit it.


## 3.3.h0=h1=0

In this situation, we have the following result:(33){B1=B2=0,A1=h32(α2-β2),-h2β3α+A0βα+h2βα34αnnA2=h324h2(α2-β2),h4=h324h2,nnϵ=-h2β3α+6γδ-12A0βα+h2βα34α},
where A0, h2, and h3 are arbitrary constants.Substituting (33) into (8), we obtain the following formal solution of (1):
(34)u(x,y,z,w,t)=A0+h32(α2-β2)F(ξ)+h324h2(α2-β2)F(ξ)2,
where ξ=αx+βy+γz+δw+ϵt and α, β, γ, δ, and ϵ are determined in (33).Whenh0=h1=0, the general elliptic equation (6) is reduced to the auxiliary ordinary equation
(35)F′(ξ)2=h2F(ξ)2+h3F3(ξ)+h4F4(ξ).Ifh2>0, h32-4h2h4=0, the solutions of (35) are given by the following:
(36)F1(ξ)=-h2h3(1±coth⁡(h22ξ+ξ0)),F2(ξ)=-h2h3(1±tanh(h22ξ+ξ0)),
where ξ0 is arbitrary constant.Substituting (36) into (34), we obtain the following hyperbolic function solutions of (1):
(37)u(x,y,z,w,t)=A0-h22(α2-β2)(1±coth⁡⁡(ξ+ξ0))+h24(α2-β2)(1±coth⁡(ξ+ξ0))2,u(x,y,z,w,t)=A0-h22(α2-β2)(1±tanh⁡(ξ+ξ0))+h24(α2-β2)(1±tanh(ξ+ξ0))2,
where ξ=αx+βy+γz+δw+((-h2β3α+6γδ-12A0βα+h2βα3)/4α)t.Remark 2.
Considering the form of (5), we can suppose that (7) owns the solutions in the form
(38)ϕ(ξ)=A0+A1F(ξ)+A2F2(ξ)+B1F′(ξ)+B2F(ξ)F′(ξ).
Substituting (38) and (35) into (7) and then setting all the coefficients of Fj(ξ)F′k(ξ)(j=0,1,…,5,k=0,1) of the resulting system to zero, we can obtain the following result:
(39){α3βh2-αβ3h2-12αβA0+6δγ4αA0=A0,A1=h34(α2-β2),A2=h42(α2-β2),nnB1=±h42(α2-β2),B2=0,nnϵ=α3βh2-αβ3h2-12αβA0+6δγ4α},
where h0, h2, and A0 are arbitrary constants, h4⩾0, and α, β, γ, and δ are nonzero constants.
The solutions of (35) are given in Table 2.
Substituting (39) into (38), we obtain the following formal solution of (1):
(40)u(x,y,z,w,t)=A0+h34(α2-β2)F(ξ)+h42(α2-β2)F2(ξ)±h42(α2-β2)F′(ξ),
where ξ=αx+βy+γz+δw+ϵt and α, β, γ, δ, and ϵ are determined in (39).
Combining (40) with Table 2, many exact solutions of (1) can be obtained. For simplicity, we do not list them carefully. Using case 2 in Table 2, we can obtain the following type of wave figures. Figures 3(a) and 3(b) are kink waves and Figure 3(c) is new type of singular wave. It is interesting that they are dependent on Δ=h32-4h2h4.Table 2
Solutions of (33) with Δ=h32-4h2h4, ε=±1.
Case
h
2
,
Δ
F
(
ξ
)
Case
h
2
,
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
tanh
⁡
(
(
h
2
/
2
)
ξ
)
2
2
h
2
>
0
-
h
2
h
3
csc
h
2
(
(
h
2
/
2
)
ξ
)
h
3
2
-
h
2
h
4
(
1
+
ε
coth
⁡
(
(
h
2
/
2
)
ξ
)
2
3
h
2
>
0, Δ>0
2
h
2
sech
⁡
(
h
2
ξ
)
ε
Δ
-
h
3
sech
⁡
(
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
csch
⁡
(
h
2
ξ
)
ε
-
Δ
-
h
3
csch
⁡
(
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, h4>0
-
h
2
sec
h
2
(
(
h
2
/
2
)
ξ
)
h
3
+
2
ε
h
2
h
4
tanh
⁡
(
(
h
2
/
2
)
ξ
)
8
h
2
<
0, h4>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, h4>0
h
2
csc
h
2
(
(
h
2
/
2
)
ξ
)
h
3
+
2
ε
h
2
h
4
coth
⁡
(
(
h
2
/
2
)
ξ
)
10
h
2
<
0, h4>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
tanh
⁡
(
h
2
2
)
ξ
)
12
h
2
>
0, Δ=0
-
h
2
h
3
(
1
+
ε
coth
⁡
(
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
>
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(a)
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(b)
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## 3.4.h0≠0,  h1≠0,  h2≠0,  h3≠0,  h4≠0

In this case, there exists three parametersr, p, and q such that
(41)(F′(ξ))2=h0+h1F(ξ)+h2F2(ξ)+h3F3(ξ)+h4F4(ξ)=(r+pF(ξ)+qF2(ξ))2.
Equation (41) is satisfied only if the following relations hold:
(42)h0=r2,h1=2rp,h2=2rq+p2,h3=2pq,h4=q2.Equation (33) is the general Riccati equation. The solutions of (41) are listed in [12]. There are 24 group solutions named ϕiI, (i=1,2,…24), which we do not list for simplicity.Substituting (41) and (8) into (7) and then setting all the coefficients of Fk(k=-5,…,5) of the resulting system to zero, we can obtain the following results:
(43){A0=A0,B1=B2=0,A1=pq(α2-β2)(4α)-1},A2=q2(α2-β2),ϵ=(8α3βqr+α3βp2-αβ3p2nnnnnn-12αβA0-8αβ3qr+6γδ)×(4α)-1}(44){×(4α)-1A0=A0,A1=A2=0,B1=rp(α2-β2),nnB2=r2(α2-β2),nnϵ=(8α3βqr+α3βp2-αβ3p2nnnnnn-12αβA0-8αβ3qr+6γδ)nn×(4α)-1},
where r, p, and q are arbitrary constants, and
(45){A0=A0,A1=B1=p=0,α3βA0β3γδ2αnnA2=q2(α2-β2),B2=r2(α2-β2),nnϵ=4α3βqr-6αβA0-4αβ3qr+3γδ2α},
where r and q are arbitrary constants.Substituting (43)–(45) into (8), we obtain, respectively, the following formal solutions of (1):
(46)u(x,y,z,w,t)=A0+pq(α2-β2)F(ξ)+q2(α2-β2)F(ξ)2,u(x,y,z,w,t)=A0+rp(α2-β2)F(ξ)-1+r2(α2-β2)F(ξ)-2,
where ξ=αx+βy+γz+δw+((8α3βqr+α3βp2-αβ3p2-12αβA0-8αβ3qr+6γδ)/4α)t, and
(47)u(x,y,z,w,t)=A0+q2(α2-β2)F(ξ)2+r2(α2-β2)F(ξ)-2,
where ξ=αx+βy+γz+δw+((4α3βqr-6αβA0-4αβ3qr+3γδ)/2α)t.Substituting solutions of (41) F(ξ)=ϕiI, (i=1,2,…24) into (46)–(47), we can obtain a lot of solutions of (1). We just give one example.Whenp2-4pq>0 and pq≠0, ϕ1I=-(1/2q)(p+p2-4qrtanh((p2-4qr/2)ξ)). Substituting ϕ1I into (46) and (51), we have
(48)u(x,y,z,w,t)=A0+pq(α2-β2)ϕ1I+q2(α2-β2)ϕ1I2,u(x,y,z,w,t)=A0+rp(α2-β2)ϕ1I-1+r2(α2-β2)ϕ1I-2,
where ξ=αx+βy+γz+δw+((8α3βqr+α3βp2-αβ3p2-12αβA0-8αβ3qr+6γδ)/4α)t.Whenp=0, qr<0, ϕ=rtanh(-qrξ). Substituting ϕ into (47), we have
(49)u(x,y,z,w,t)=A0+q2(α2-β2)ϕ2+r2(α2-β2)ϕ-2,
where ξ=αx+βy+γz+δw+((4α3βqr-6αβA0-4αβ3qr+3γδ)/2α)t.


## 3.5.h0≠0,  h1≠0,  h2=0,  h3≠0,  h4≠0

In this case, there exists three parametersr, p, and q such that
(50)(F′(ξ))2=h0+h1F(ξ)+h3F3(ξ)+h4F4(ξ)=(r+pF(ξ)+qF2(ξ))2.
Equation (50) is satisfied only if the following relations hold:
(51)h0=r2,h1=2rp,h3=2pq,h4=q2.
The following constraint should exist between r, p, and q parameters:
(52)p2=-2qr,qr<0.Therefore, we can discuss the solution of (1) similarly as 3.4 under the condition (52). Here, we omit it.


## 3.6.h2=0,  h4=0

In this situation, we have the following result:(53)A2=B1=B2=0,A1=h34(α2-β2),ϵ=-32-γδ+2A0βαα.
Substituting (53) into (8), we obtain the following formal solution of (1):
(54)u(x,y,z,w,t)=A0+h34(α2-β2)F(ξ),
where ξ=αx+βy+γz+δw+ϵt and α, β, γ, δ, and ϵ are determined in (53).Whenh2=h3=0, the general elliptic equation (6) is reduced to the auxiliary ordinary equation
(55)F′(ξ)2=h0+h1F(ξ)+h3F3(ξ).
The solution of (55) is the Weierstrass elliptic doubly periodic type solution is as follows:
(56)F(ξ)=℘(h32ξ,g2,g3),h3>0.
Substituting (56) into (54), the solution of (1) is
(57)u(x,y,z,w,t)=A0+h34(α2-β2)℘(h32ξ,g2,g3),h3>0,
where g2=-4h1/h3 and g3=-4h0/h3.


## 4. Conclusions

The investigation of the exact solutions of higher-dimensional integrable models is one of the important problems in mathematical physics. In our work, (4 + 1)-dimensional nonlinear Fokas equation (1), which is an important physical model, is discussed by using the extendedF-expansion method and its variant. Some new exact solutions expressed by Jacobi elliptic function, Weierstrass elliptic function, hyperbolic function, and trigonometric function are obtained and some typical wave figures are given including periodic wave, solitary wave, kink wave, and some new types. The correctness of all the solutions is verified by substituting them into original equation (1). Comparing with [34–36], it is easy to see that our method is more straightforward and the form of the solutions obtained in our paper is also more simple and many solutions are new. The related results are enriched.



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*Source: 972519-2014-04-17.xml*