Journal of Inequalities and Applications

J Inequal Appl. 2017; 2017(1): 314.
Published online 2017 December 21.
PMCID: PMC5740214

# Symmetry reduction and exact solutions of two higher-dimensional nonlinear evolution equations

## Abstract

In this paper, symmetries and symmetry reduction of two higher-dimensional nonlinear evolution equations (NLEEs) are obtained by Lie group method. These NLEEs play an important role in nonlinear sciences. We derive exact solutions to these NLEEs via the exp(−ϕ(z))-expansion method and complex method. Five types of explicit function solutions are constructed, which are rational, exponential, trigonometric, hyperbolic and elliptic function solutions of the variables in the considered equations.

Keywords: nonlinear evolution equations, symmetry, exp(−ϕ(z))-expansion method, complex method, exact solutions, meromorphic function

## Introduction

In 1998, Yu et al. [1] extended the Bogoyavlenskii Schiff equation

$ut+Φ(u)us=0,Φ(u)=∂x2+4u+2ux∂x−1,$
1

to the (3 + 1)-dimensional NLEE

$(−4ut+Φ(u)us)x+3uyy=0,Φ(u)=∂x2+4u+2ux∂x−1.$
2

Setting u: = ux, equation (2) is changed into the (3 + 1)-dimensional potential Yu-Toda-Sasa-Fukuyama (YTSF) equation

uxxxs − 4uxt + 4uxuxs + 2uxxus + 3uyy = 0.
3

The generalized (3 + 1)-dimensional Zakharov-Kuznetsov (gZK) equation is given by

a1u2uxa2uxxxa3uxyya4uxssa5uuxa6uxxtut = 0.
4

Here ai (i = 1, 2, …, 6) are arbitrary constants.

We note that equation (4) includes many famous NLEEs as its special cases. For instance, if a1a3a4a6 = 0, then equation (4) is the Korteweg-de Vries equation [2, 3]. If a2a4a5 = 0, then equation (4) is the (2 + 1) dimensional ZK-MEW equation [4]. If a3a4a6 = 0, then equation (4) is the Gardner equation [5]. If a4a5a6 = 0, then equation (4) is the modified Zakharov-Kuznetsov equation [6].

In recent years, it has aroused widespread interest in the study of NLEEs [713]. Equations (3) and (4) are very meaningful higher-dimensional NLEEs which can describe many dynamic processes and important phenomena in engineering and physics. The YTSF equation is a mostly used model for investigating the dynamics of solitons and nonlinear waves in fluid dynamics, plasma physics and weakly dispersive media [13]. Zakharov and Kuznetsov [14] proposed the ZK equation to describe nonlinear ion-acoustic waves in a plasma comprised of cold ions and hot isothermal electrons in the presence of a uniform magnetic field. Many physical phenomena, in the purely dispersive limit, are governed by this type of equation, such as the long waves on a thin liquid film [15], the Rossby waves in a rotating atmosphere [16], and the isolated vortex of drift waves in a three-dimensional plasma [17]. The gZK equation is of a generalized setting of ZK equation. Seeking exact solutions of NLEEs is an interesting and significant subject. Over the past few years, many powerful methods for constructing the solutions of NLEEs have been used, for instance, the Bäcklund transform method [18], direct algebraic method [19], modified simple equation method [20], Lie group method [21, 22], exp(−ϕ(z))-expansion method [8, 9, 23, 24], and so on. Recently, Yuan et al. [2527] introduced the complex method to find the exact solutions of NLEEs in mathematical physics. In this paper, we study symmetries, symmetry reduction of the two higher-dimensional NLEEs, and then we obtain their exact solutions via the exp(−ϕ(z))-expansion method and complex method.

## Description of the methods

### Description of the exp(−ϕ(z))-expansion method

Suppose that a nonlinear partial differential equation (PDE) is given by

P(uuxuyutuxxuyyutt, …) = 0,
5

where P is a polynomial of an unknown function u(xyt) and its derivatives in which nonlinear terms and highest order derivatives are involved. The main steps of this method are given in the following.

Step 1. Substituting the traveling wave transform

u(xyt) = w(z),  zkxlyrt

into equation (5) converts it to the following ordinary differential equation (ODE):

F(wwww, …) = 0,
6

in which F is a polynomial of w(z) and its derivatives, while $:′=ddz$.

Step 2. Assume that equation (6) has the following traveling wave solution:

$w(z)=∑j=0nCj(exp(−ϕ(z)))j,$
7

where Cj (0 ≤ j ≤ n) are constants to be determined, such that Cj ≠ 0 and ϕϕ(z) satisfies the ODE as follows:

ϕ(z) = exp(−ϕ(z)) + μexp(ϕ(z)) + δ.
8

Equation (8) has the following solutions.

When δ2 − 4μ > 0, μ ≠ 0,

$ϕ(z)=ln(−(δ2−4μ)tanh(δ2−4μ2(z+c)−δ)2μ),$
9

$ϕ(z)=ln(−(δ2−4μ)coth(δ2−4μ2(z+c)−δ)2μ).$
10

When δ2 − 4μ < 0, μ ≠ 0,

$ϕ(z)=ln((4μ−δ2)tan((4μ−δ2)2(z+c)−δ)2μ),$
11

$ϕ(z)=ln((4μ−δ2)cot((4μ−δ2)2(z+c)−δ)2μ).$
12

When δ2 − 4μ > 0, μ = 0, δ ≠ 0,

$ϕ(z)=−ln(δexp(δ(z+c))−1).$
13

When δ2 − 4μ = 0, μ ≠ 0, δ ≠ 0,

$ϕ(z)=ln(−2(δ(z+c)+2)δ2(z+c)).$
14

When δ2 − 4μ = 0, μ = 0, δ = 0,

ϕ(z) = ln(zc).
15

Here Cn ≠ 0, δ, μ are constants that will be determined later and c is an arbitrary constant. We take the homogeneous balance between nonlinear terms and highest order derivatives of equation (6) to determine the positive integer n.

Step 3. Substituting equation (7) into equation (6) and accounting the function exp(−ϕ(z)), we obtain a polynomial of exp(−ϕ(z)). Equating all the coefficients of the same power of exp(−ϕ(z)) to zero yields a set of algebraic equations. By solving the algebraic equations, we get the values of Cn ≠ 0, δ, μ, and then we substitute them into equation (7) along with equations (9)-(15) to complete the determination of the solutions of equation (5).

### Description of the complex method

Let m ∈ ℕ: = {1, 2, 3, …}, rj ∈ ℕ = ℕ ∪ {0}, j = 0, 1, …, m, r = (r0r1, …, rm), and

$Kr[w](z):=∏j=0m[w(j)(z)]rj,$

then $d(r):=∑j=0mrj$ is the degree of Kr[w]. Let the differential polynomial be defined by

$F(w,w′,…,w(m)):=∑r∈JarKr[w],$

where J is a finite index set, and ar are constants, then  deg F(ww, …, w(m)): = maxrJ{d(r)} is the degree of F(ww, …, w(m)).

Consider the following differential equation:

F(ww, …, w(m)) = cwnd
16

where n ∈ ℕ, c ≠ 0, d are constants.

Set pq ∈ ℕ, and the meromorphic solutions w of equation (16) have at least one pole. If equation (16) has exactly p distinct meromorphic solutions, and their multiplicity of the pole at z = 0 is q, then equation (16) is said to satisfy the pq condition. It might not be easy to show that the pq condition of equation (16) holds, so we need the weak pq condition as follows.

Inserting the Laurent series

$w(z)=∑λ=−q∞βλzλ,β−q≠0,q>0,$
17

into equation (16), we can determine exactly p different Laurent singular parts:

$∑λ=−q−1βλzλ,$

then equation (16) is said to satisfy the weak pq condition.

Given two complex numbers ν1, ν2 such that $Imν1ν2>0$, and let L be the discrete subset L[2ν1, 2ν2]: = {νν = 2a1ν1 + 2a2ν2a1a2 ∈ ℤ}, and L is isomorphic to ℤ × ℤ. Let the discriminant $Δ=Δ(b1,b2):=b13−27b22$ and

$ln=ln(L):=∑ν∈L∖{0}1νn.$

A meromorphic function ℘(z): = ℘(zg2g3) with double periods 2ν1, 2ν2, which satisfies the following equation:

(℘(z))2 = 4℘(z)3 − g2℘(z)−g3

in which g2 = 60l4, g3 = 140l6, and Δ(g2g3) ≠ 0, is called the Weierstrass elliptic function and satisfies an addition formula [28] as follows:

$℘(z−z0)=14[℘′(z)+℘′(z0)℘(z)−℘(z0)]2−℘(z)−℘(z0).$

If a meromorphic function g is a rational function of z, or a rational function of eαz, α ∈ ℂ, or an elliptic function, then we say that g belongs to the class W.

In 2009, Eremenko et al. [29] studied the mth-order Briot-Bouquet equation (BBEq)

$F(w,w(m))=∑j=0nFj(w)(w(m))j=0,$

where Fj(w) are constant coefficients polynomials, m ∈ ℕ. For the mth-order BBEq, we have the following lemma.

#### Lemma 2.1

([28, 30, 31])

Let mnph ∈ ℕ,  deg F(ww(m)) < n, and a mth-order BBEq

F(ww(m)) = cwnd

satisfies the weak pq condition, then the meromorphic solutions w ∈ W. Supposing for some values of the parameters the solution w exists, then any other meromorphic solutions will be one parameter family w(zz0), z0 ∈ ℂ. In addition, every elliptic solution w with a pole at z = 0 is expressed as

$w(z)=∑i=1h−1∑j=2q(−1)jβ−ij(j−1)!dj−2dzj−2(14[℘′(z)+Di℘(z)−Bi]2−℘(z))+∑i=1h−1β−i12℘′(z)+Di℘(z)−Bi+∑j=2q(−1)jβ−hj(j−1)!dj−2dzj−2℘(z)+β0,$
18

where βij are determined by (17), $∑i=1hβ−i1=0$ and $Di2=4Bi3−g2Bi−g3$.

Every rational function solution w: = R(z) is expressed as

$R(z)=∑i=1h∑j=1qβij(z−zi)j+β0,$
19

which has h (≤p) distinct poles of multiplicity q.

Every simply periodic solution w: = R(ϑ) is a rational function of ϑeαz (α ∈ ℂ), and is expressed as

$R(ϑ)=∑i=1h∑j=1qβij(ϑ−ϑi)j+β0,$
20

which has h (≤p) distinct poles of multiplicity q.

By the above definitions and lemma, we now present the complex method.

Step 1. Insert the transformation T:u(xyt) → w(z) defined by (xyt) → z into a given PDE to yield a nonlinear ODE.

Step 2. Insert (17) into the ODE to determine whether the weak pq condition holds.

Step 3. Insert the indeterminate solutions introduced in Lemma 2.1 into the ODE, and then get meromorphic solutions of the ODE with a pole at z = 0.

Step 4. Obtain meromorphic solutions w(z − z0) by Lemma 2.1 and the addition formula.

Step 5. Inserting the inverse transformation T−1 into the meromorphic solutions, we get the exact solutions for the original PDE.

## Symmetries and symmetry reduction

### Symmetries

In order to find the symmetry σσ(xystu) of equation (4), we set

21

where u is the solution of equation (4), a, b, c, d, e, f are unknown functions of real variables x, y, s, t. According to Lie group analysis [21, 22], σ satisfies

σta1σ2uxa1u2σxa2σxxxa3σxyya4σxssa5σuxa5uσxa6σxxt = 0.
22

Substituting equation (21) into equation (22), we have a new differential equation, where

a2uxxx = −a1u2ux − a3uxyy − a4uxss − a5uux − a6uxxt − ut.
23

By equation (21), equation (22) and equation (23), we have

$a=c5,b=(c2s+c3),c=(c4−a4a3c2y),d=c1,e=0,f=0,$
24

where ci (i = 2, 3, 4, 5) are real constants. Substituting equations (24) into equation (21), we achieve the symmetry of the gZK equation,

$σ=c5ux+(c2s+c3)uy+(c4−a4a3c2y)us+c1ut.$
25

In order to find the symmetry σσ(xystu) of equation (3), we set

26

Here u is the solution of equation (3), a, b, c, d, e, f are unknown functions of real variables x, y, s, t. According to Lie group analysis, σ satisfies

σxxxs − 4σxt + 4uxσxs + 4uxsσx + 2uxxσs + 2usσxx + 3σyy = 0.
27

Substituting equation (26) into equation (27), we have a new differential equation, where

uxxxs = 4uxt − 4uxuxs − 2uxxus − 3uyy.
28

By equation (26), equation (27) and equation (28), we have

$a=c1x+c2,b=c3y+c4,c=(2c3−3c1)s+ρ(t),d=(2c3−c1)t+c5,e=c1,f=ρ′(t)x+23ρ″(t)y2+τ(t)y+ψ(t),$
29

where ci (i = 1, 2, …, 5) are real constants, ρ(t), τ(t), ψ(t) are arbitrary real functions of t. Substituting equations (29) into equation (26), we achieved the symmetry of YTSF equation

$σ=(c1x+c2)ux+(c3y+c4)uy+((2c3−3c1)s+ρ(t))us+((2c3−c1)t+c5)ut+c1u+ρ′(t)x+23ρ″(t)y2+τ(t)y+ψ(t).$
30

### Symmetry reduction

By solving the characteristic equation (25) of σ

$dxc5=dyc2s+c3=dsc4−a4a3c2y=dtc1=du0,$
31

we find different symmetry reduced equations. Without loss of generality, we have two reduced equations as follows.

Setting c1c3c4c5 = 0, c2 = 1, we have the first similarity solution of equation (4)

uφ(ξη),
32

where ξxt, $η=y22a3+s22a4$. Substituting equation (32) into equation (4), we have the first symmetry reduced equation of equation (4)

φξa1φ2φξ + (a2a3)φξξξ + 2φξηηa5φφξ = 0.
33

Setting c1c2 = 0, c3c4c5 = 1, solving σ = 0, we have the second similarity solution of equation (4)

uφ(ξη),
34

where ξxy, ηs. Substituting equation (34) into equation (4), we have the second symmetry reduced equation of equation (4)

a1φ2φξ + (a2a3)φξξξa4φξηηa5φφξ = 0.

By solving the characteristic equation (30) of σ

$dxc1x+c2=dyc3y+c4=ds(2c3−3c1)s+ρ(t)=dt(2c3−c1)t+c5=duc1u+ρ′(t)x+23ρ″(t)y2+τ(t)y+ψ(t),$
35

we obtain symmetry reduction of equation (3). Without loss of generality, we have two reduced equations as follows.

Setting c1c3c4 = 0, c2c5 = 1, ρ(t) = 1, solving σ = 0, we have the first similarity solution of equation (3)

uφ(ξηy)−∫(τ(t)yψ(t)) dt
36

where ξx − t, ηs − t. Substituting equation (36) into equation (3), we have the first symmetry reduced equation of equation (3)

φξξξη + 4φξξ + 4φξη + 4φξφξη + 2φξξφη + 3φyy = 0.
37

Setting c1c2c3c5 = 0, c4 = 1, ρ(t) = τ(t) = 0, solving σ = 0, we have the second similarity solution of equation (3)

uφ(xst)−ψ(t)y.
38

Substituting equation (38) into equation (3), we have the second symmetry reduced equation of equation (3)

φxxxs + 4φxφxs + 2φxxφs − 4φxt = 0.

## Exact solutions

### Exact solutions of gZK equation via the exp(−ϕ(z))-expansion method

Substituting the traveling wave transform

φ(ξη) = w(z),  zkξlη

into equation (33), then integrating it with respect to z, we obtain

$((a2+a3)k2+2l2)w″+w+a52w2+a13w3−γ=0,$
39

where γ is the integration constant which can be determined later.

Taking the homogeneous balance between w and w3 in equation (39) yields

w(z) = C0C1exp(−ϕ(z)),
40

where C1 ≠ 0, C0 are constants to be determined, whereas δ and μ are arbitrary constants.

Substitute w, w2, w3, w into equation (39) and equate the coefficients of exp(−ϕ(z)) to zero, then

$13a1C03+12a5C02+C0+2C1l2δμ+C1k2a2δμ+C1k2a3δμ−γ=0,C1a2k2δ2+C1a3k2δ2+2C1l2δ2+2C1a2k2μ+2C1a3k2μ+C02C1a1+4C1l2μ+C0C1a5+C1=0,12a5C12+a1C0C12+6C1l2δ+3C1k2a2δ+3C1k2a3δ=0,4C1l2+13a1C13+2C1k2a2+2C1k2a3=0.$

Solving the above algebraic equations, we obtain

$γ=−−2a1((δ2−4μ)(a2k2+a3k2+2l2)−2)((δ2−4μ)(a2k2+a3k2+2l2)+1)6a1,C1=−6(a2k2+a3k2+2l2)a1,C0=−6a1(a2k2+a3k2+2l2)δ−2a1(2−(δ2−4μ)(a2k2+a3k2+2l2))2a1,$
41

where μ and δ are arbitrary constants.

Substituting equations (41) into equation (40) yields

$w(z)=−6a1(a2k2+a3k2+2l2)δ−2a1(2−(δ2−4μ)(a2k2+a3k2+2l2))2a1+−6(a2k2+a3k2+2l2)a1exp(−ϕ(z)).$
42

We apply equation (9) to equation (15) into equation (42), respectively, to get traveling wave solutions of the gZK equation as follows.

When δ2 − 4μ > 0, μ ≠ 0,

$w11(z)=−6a1(a2k2+a3k2+2l2)δ−2a1(2−(δ2−4μ)(a2k2+a3k2+2l2))2a1−−6(a2k2+a3k2+2l2)a12μ(δ2−4μ)tanh(δ2−4μ2(z+c)+δ),w12(z)=−6a1(a2k2+a3k2+2l2)δ−2a1(2−(δ2−4μ)(a2k2+a3k2+2l2))2a1−−6(a2k2+a3k2+2l2)a12μ(δ2−4μ)coth(δ2−4μ2(z+c)+δ).$

When δ2 − 4μ < 0, μ ≠ 0,

$w13(z)=−6a1(a2k2+a3k2+2l2)δ−2a1(2−(δ2−4μ)(a2k2+a3k2+2l2))2a1+−6(a2k2+a3k2+2l2)a12μ(4μ−δ2)tan(4μ−δ22(z+c)−δ),w14(z)=−6a1(a2k2+a3k2+2l2)δ−2a1(2−(δ2−4μ)(a2k2+a3k2+2l2))2a1+−6(a2k2+a3k2+2l2)a12μ(4μ−δ2)cot(4μ−δ22(z+c)−δ).$

When δ2 − 4μ > 0, μ = 0, δ ≠ 0,

$w15(z)=−6a1(a2k2+a3k2+2l2)δ−2a1(2−δ2(a2k2+a3k2+2l2))2a1+−6(a2k2+a3k2+2l2)a1δexp(δ(z+c))−1.$

When δ2 − 4μ = 0, μ ≠ 0, δ ≠ 0,

$w16(z)=−3(a2k2+a3k2+2l2)2a1δ−1a1−−6(a2k2+a3k2+2l2)a1δ2(z+c)2(δ(z+c)+2).$

When δ2 − 4μ = 0, μ = 0, δ = 0,

$w17(z)=−1a1+−6(a2k2+a3k2+2l2)a11z+c.$

### Exact solutions of gZK equation via the complex method

Inserting (17) into equation (39) we have p = 2, q = 1, $β−1=±−6(a2k2+a3k2+2l2)a1$, $β0=−a52a1$, $β1=−a5224a12−6a1a2k2+a3k2+2l2$, $β2=−12a12γ−a53+6a1a548a12(a2k2+a3k2+2l2)$ and β3 is an arbitrary constant.

Therefore, equation (39) is a second order BBEq and satisfies the weak 〈2, 1〉 condition. Hence, by Lemma 2.1, we see that meromorphic solutions of equation (39) belong to W. We will show meromorphic solutions of equation (39) in the following.

By (19), we infer that the indeterminate rational solutions of equation (39) are

$R1(z)=β11z+β12z−z1+β10,$

with a pole at z = 0.

Substituting R1(z) into equation (39), we have

$R1,1(z)=±−6(a2k2+a3k2+2l2)a11z−a52a1,$

where $a52=4a1$ and 9a1γ2 = 1;

$R1,2(z)=±−6(a2k2+a3k2+2l2)a1(1z−1z−z1−1z1)−a52a1,$

where $k=4a1z12−a52z12−48l2a124a1(a2+a3)$ and $γ=(a53−6a1a5+(a52−4a1)33)z13$.

So the rational solutions of equation (39) are

$wr,1(z)=±−6(a2k2+a3k2+2l2)a11z−z0−a52a1$

and

$wr,2(z)=±−6(a2k2+a3k2+2l2)a1(1z−z0−1z−z0−z1−1z1)−a52a1,$

where z0 ∈ ℂ, z1 ≠ 0. $a52=4a1$, 9a1γ2 = 1 in the former case, or $k=4a1z12−a52z12−48l2a124a1(a2+a3)$, $γ=(a53−6a1a5+(a52−4a1)33)z13$ in the latter case.

To obtain simply periodic solutions, let ϑeαz, and substitute wR(ϑ) into equation (39), then

$((a2+a3)k2+2l2)α2(ϑR′+ϑ2R″)+R+a52R2+a13R3−γ=0.$
43

Substituting

$R2(z)=β21ϑ−1+β22(ϑ−ϑ1)+β20$

into equation (43), we obtain

$R2,1(z)=±−6(a2k2+a3k2+2l2)a1α(1ϑ−1+12)−a52a1$
44

and

$R2,2(z)=±−6(a2k2+a3k2+2l2)a1α(1ϑ−1−ϑ1ϑ−ϑ1−ϑ1+12(ϑ1−1))−a52a1,$
45

where $γ=a5(a52−6a1)12a12$, $l=12α4a1−a52−2a1k2α2(a2+a3)a1$ in the former case, or $γ=3z1(z1+1)(4a1−a52)32(z12+10z1+1)32a12+a5(a52−6a1)12a12$, $k=−(4a1−a52−4a1l2α2)(z12+1)+2(a52−4a1−20a1l2α2)z12a1(z12+10z1+1)(a2+a3)α2$ in the latter case.

Inserting ϑeαz into equation (44) and equation (45), we can get simply periodic solutions to equation (39) with a pole at z = 0,

$ws0,1(z)=±−3(a2k2+a3k2+2l2)2a1αcothα2z−a52a1,ws0,2(z)=±−3(a2k2+a3k2+2l2)2a1α(cothα2z−cothα2(z−z1)−cothα2z1)−a52a1,$

where $γ=a5(a52−6a1)12a12$, $l=12α4a1−a52−2a1k2α2(a2+a3)a1$ in the former case, or $γ=3z1(z1+1)(4a1−a52)32(z12+10z1+1)32a12+a5(a52−6a1)12a12$, $k=−(4a1−a52−4a1l2α2)(z12+1)+2(a52−4a1−20a1l2α2)z12a1(z12+10z1+1)(a2+a3)α2$ in the latter case.

So simply periodic solutions of equation (39) are

$ws,1(z)=±−3(a2k2+a3k2+2l2)2a1αcothα2(z−z0)−a52a1$

and

$ws,2(z)=±−3(a2k2+a3k2+2l2)2a1⋅α(cothα2(z−z0)−cothα2(z−z0−z1)−cothα2z1)−a52a1,$

where z0 ∈ ℂ, z1 ≠ 0. $l=12α4a1−a52−2a1k2α2(a2+a3)a1$, $γ=a5(a52−6a1)12a12$ in the former case, or $k=−(4a1−a52−4a1l2α2)(z12+1)+2(a52−4a1−20a1l2α2)z12a1(z12+10z1+1)(a2+a3)α2$, $γ=3z1(z1+1)(4a1−a52)32(z12+10z1+1)32a12+a5(a52−6a1)12a12$ in the latter case.

From (18), we have the indeterminate relations for the elliptic solutions of equation (39) with a pole at z = 0,

$wd1(z)=β−12℘′(z)+D1℘(z)−B1+β0,$

where $D12=4B13−g2B1−g3$. Considering the results obtained above, we infer that $β0=−a52a1$, g3 = 0, D1B1 = 0. So we obtain

$wd1(z)=±−3(a2k2+a3k2+2l2)2a1℘′(z)℘(z)−a52a1,$

where g3 = 0.

Thus, the elliptic function solutions of equation (39) are

$wd(z)=±−3(a2k2+a3k2+2l2)2a1℘′(z−z0,g2,0)℘(z−z0,g2,0)−a52a1,$

where z0 ∈ ℂ, g3 = 0, g2 is arbitrary. Applying the addition formula, we can rewrite it as

$wd(z)=±−3(a2k2+a3k2+2l2)2a1⋅(−℘+E)(4E℘2+(4E2−g2)℘+2F℘′−Eg2)((12E2−g2)℘+4E3−3Eg2)℘′+(4℘3+12E℘2−3g2℘−Eg2)F−a52a1,$

where g3 = 0, F2 = 4E3 − g2E, E and g2 are arbitrary.

### Exact solutions of YTSF equation via the exp(−ϕ(z))-expansion method

Substituting the traveling wave transform

φ(ξηy) = v(z),  zkξlηry

into equation (37), then integrating it with respect to z, we obtain

k3lv + (4k2 + 4kl + 3r2)v + 3k2l(v)2γ = 0,
46

where γ is the integration constant which can be determine later.

Setting wv, equation (46) becomes

k3lw + (4k2 + 4kl + 3r2)w + 3k2lw2γ = 0.
47

Taking the homogeneous balance between w and w2 in equation (47) yields

w(z) = C0C1exp(−ϕ(z)) + C2(exp(ϕ(z)))2
48

where C2 ≠ 0, Ci (i = 0, 1, 2) are constants to be determined, whereas δ and μ are arbitrary constants.

Substitute w, w2, w into equation (47) and equate the coefficients of exp(−ϕ(z)) to zero, then

$k3lC1δμ+2k3lC2μ2+3k2lC02+4C0k2+4C0kl+3C0r2+γ=0,C1lk3δ2+6C2lk3δμ+2C1lk3μ+6C0C1lk2+4C1k2+4C1lk+3C1r2=0,4C2lk3δ2+3C1lk3δ+8C2lk3μ+6C0C2lk2+3C12lk2+4C2k2+4C2lk+3C2r2=0,10C2lk3δ+6C1C2lk2+2C1lk3=0,3C22lk2+6C2lk3=0.$

Solving the above algebraic equations, we obtain

$γ=−(δ2−4μ)2l2k6−(4lk+4k2+3r2)212k2l,C2=−2k,C1=−2kδ,C0=−lk3δ2+8lk3μ+4lk+4k2+3r26k2l,$
49

where μ and δ are arbitrary constants.

Substituting equations (49) into equation (48), yields

$w(z)=−lk3δ2+8lk3μ+4lk+4k2+3r26k2l−2kδexp(−ϕ(z))−2k(exp(ϕ(z)))2.$
50

We apply equation (9) to equation (15) into equation (50), respectively, to get traveling wave solutions of the YTSF equation as follows.

When δ2 − 4μ > 0, μ ≠ 0,

$w21(z)=−lk3δ2+8lk3μ+4lk+4k2+3r26k2l+4kδμ(δ2−4μ)tanh(δ2−4μ2(z+c)+δ)−8kμ2((δ2−4μ)tanh(δ2−4μ2(z+c)+δ))2,w22(z)=−lk3δ2+8lk3μ+4lk+4k2+3r26k2l+4kδμ(δ2−4μ)coth(δ2−4μ2(z+c)+δ)−8kμ2((δ2−4μ)coth(δ2−4μ2(z+c)+δ))2.$

When δ2 − 4μ < 0, μ ≠ 0,

$w23(z)=−lk3δ2+8lk3μ+4lk+4k2+3r26k2l−4kδμ(δ2−4μ)tan(δ2−4μ2(z+c)−δ)−8kμ2((δ2−4μ)tan(δ2−4μ2(z+c)−δ))2,w24(z)=−lk3δ2+8lk3μ+4lk+4k2+3r26k2l−4kδμ(δ2−4μ)cot(δ2−4μ2(z+c)−δ)−8kμ2((δ2−4μ)cot(δ2−4μ2(z+c)−δ))2.$

When δ2 − 4μ > 0, μ = 0, δ ≠ 0,

$w25(z)=−lk3δ2+4lk+4k2+3r26k2l−2kδ2exp(δ(z+c))−1−2kδ2(exp(δ(z+c))−1)2.$

When δ2 − 4μ = 0, μ ≠ 0, δ ≠ 0,

$w26(z)=−12lk3μ+4lk+4k2+3r26k2l+kδ3(z+c)(δ(z+c)+2)−kδ4(z+c)22((δ(z+c)+2))2.$

When δ2 − 4μ = 0, μ = 0, δ = 0,

$w27(z)=−4lk+4k2+3r26k2l−2k(z+c)2.$

### Exact solutions of YTSF equation via the complex method

Inserting (17) into equation (47) we have p = 1, q = 2, β−2 = −2k, β−1 = 0, $β0=−4lk+4k2+3r26k2l$, β1 = 0, $β2=−16k4+32lk3+(16l2−12lγ+24r2)k2+24lkr2+9r4120k5l2$, and β3 is an arbitrary constant.

Therefore, equation (47) is a second order BBEq and satisfies the weak 〈1, 2〉 condition. Hence, by Lemma 2.1, we see that meromorphic solutions of equation (47) belong to W. We will show meromorphic solutions of equation (47) in the following.

By (19), we deduce the indeterminacy rational solutions of equation (47) are

$R1(z)=β32z2+β31z+β30,$

with a pole at z = 0.

Substituting R1(z) into equation (47), we get the following form:

$R1(z)=−2kz2−4lk+4k2+3r26k2l,$

where $γ=16k4+32lk3+(16l2+24r2)k2+24lkr2+9r412k2l$.

So the rational solutions of equation (47) are

$wr(z)=−2k(z−z0)2−4lk+4k2+3r26k2l,$

where $γ=16k4+32lk3+(16l2+24r2)k2+24lkr2+9r412k2l$, z0 ∈ ℂ.

To obtain simply periodic solutions, let ϑeαz, and substitute wR(ϑ) into equation (47), then we get

k3lα2(ϑRϑ2R) + (4k2 + 4kl + 3r2)R + 3k2lR2γ = 0.
51

Substituting

$R2(z)=β42(ϑ−1)2+β41(ϑ−1)+β40,$

into equation (51), we obtain

$R2(z)=−2kα2(ϑ−1)2−2kα2(ϑ−1)−kα26−4lk+4k2+3r26k2l,$
52

where $γ=(4lk+4k2+3r2)2−(lα2k3)212k2l$. Substituting ϑeαz into equation (52), we can obtain simply periodic solutions of equation (47),

$ws0(z)=−2kα2(eαz−1)2−2kα2(eαz−1)−kα26−4lk+4k2+3r26k2l=−2kα2eαz(eαz−1)2−kα26−4lk+4k2+3r26k2l=−kα22coth2αz2+kα23−4lk+4k2+3r26k2l,$

with a pole at z = 0.

Therefore the simply periodic solutions of equation (47) are

$ws(z)=−kα22coth2α(z−z0)2+kα23−4lk+4k2+3r26k2l,$

where $γ=(4lk+4k2+3r2)2−(lα2k3)212k2l$, z0 ∈ ℂ.

From (18), we can express the elliptic solutions of equation (47) as

wd0(z) = β−2℘(z) + β0

with a pole at z = 0.

Substituting wd0(z) into equation (47), we obtain

$wd0(z)=−2k℘(z)−4lk+4k2+3r26k2l,$

where $g2=16k4+32lk3+(16l2−12lγ+24r2)k2+24lkr2+9r412k6l2$, g3 is arbitrary.

Therefore, the elliptic solutions of equation (47) are

$wd(z)=−2k℘(z−z0)−4lk+4k2+3r26k2l,$

in which z0 ∈ ℂ. Applying the addition formula, we can rewrite it as

$wd(z)=−2k(−℘(z)+14(℘′(z)+C℘(z)−D)2)+2kD−4lk+4k2+3r26k2l,$

where $g2=16k4+32lk3+(16l2−12lγ+24r2)k2+24lkr2+9r412k6l2$, C2 = 4D3 − g2D − g3, g3 is arbitrary.

### Comparison

Implementing the exp(−ϕ(z))-expansion method, we found seven solutions for the gZK and YSFT equation, respectively. Using the complex method, we found five solutions for the gZK equation and three solutions for the YSFT equation. Rational solutions w17(z) and w27(z) are obtained via the exp(−ϕ(z))-expansion method, and Wr,1(z) and Wr(z) are obtained via the complex method. If we let c = −z0, then w17(z) is equivalent to Wr,1(z), and w27(z) is equivalent to Wr(z). For getting rational solutions, these two methods are in good agreement. Rational solutions Wr,2(z) and simply periodic solutions Ws,2(z) and Ws(z) are new and cannot be degenerated successively through elliptic function solutions. From the results, we can find more solutions by the exp(−ϕ(z))-expansion method, whereas we can obtain elliptic function solutions just by the complex method. These two methods are very useful tools in finding the exact solutions of NLEEs.

## Computer simulations

In this section, we illustrate some results by the computer simulations. We carry out further analysis to the properties of simply periodic solutions Ws,2(z) and Ws(z) as in Figures 1 and and2.2.

1. By employing the complex method, we are capable to obtain simply periodic solutions Ws,1(z) and Ws,2(z) of the gZK equation. The solutions Ws,1(z) and Ws,2(z) come from hyperbolic function. Figure 1 shows the shape of solutions Ws,2(z) for k = 1, l = 1, α = 1, a1 = −6, a2 = 1, a3 = 1, a5 = −24, and z1 = 1 within the interval −2π ≤ ξη ≤ 2π. Note that they have two distinct generation poles which are showed by Figure 1.
2. By using the complex method, we achieve to obtain simply periodic solutions Ws(z) of the YSTF equation. The solutions Ws(z) are in terms of the hyperbolic function solution. The solutions Ws(z) in Figure 2 of the YSTF equation are represented the singular soliton solution for the parameters k = 1, l = 1, r = 1, α = 1 and y = 0 within the interval −2π ≤ ξη ≤ 2π.

The solution of the gZK equation corresponding to Ws,2(z) . (a) z0 = −8, (b) z0 = 0, (c) z0 = 8.
The solution of the YSTF equation corresponding to Ws(z) . (a) z0 = −8, (b) z0 = 0, (c) z0 = 8.

## Conclusions

In this article, we utilize Lie group analysis to obtain symmetries and symmetry reduction for two higher-dimensional NLEEs. In this way, we can reduce the dimension of the NLEEs, which is relevant in the fields of mathematical physics and engineering. Five types of explicit function solutions are constructed by the exp(−ϕ(z))-expansion method and complex method. It demonstrates these methods are very efficient and powerful to seek the exact solutions of NLEEs. We can apply the idea of this study to other NLEEs.

## Acknowledgements

This work was supported by the NSF of China (11271090, 11701111); the NSF of Guangdong Province (2016A030310257); the Foundation for Young Talents in Educational Commission of Guangdong Province (2015KQNCX116). Thanks to the Joint PHD Program of Guangzhou University and Curtin University. Thanks to the editors and referees with their very useful suggestions and helpful comments.

## Authors’ contributions

Authors’ contributions

All authors typed, read and approved the final manuscript.

## Notes

### Competing interests

The authors declare that they have no competing interests.

## Footnotes

Publisher’s Note

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## Contributor Information

Yongyi Gu, moc.361@iygnoyugdg.

Jianming Qi, moc.361@ujdsgnimnaijiq.

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