# Further Results about Traveling Wave Exact Solutions of the (2+1)-Dimensional Modified KdV Equation

Authors: Yang Yang; Jian-ming Qi; Xue-hua Tang; Yong-yi Gu
Journal: Advances in Mathematical Physics
 (2019)

---

## Abstract

We used the complex method and the exp(-ϕ
(
z
))-expansion method to find exact solutions of the (2+1)-dimensional mKdV equation. Through the maple software, we acquire some exact solutions. We have faith in that this method exhibited in this paper can be used to solve some nonlinear evolution equations in mathematical physics. Finally, we show some simulated pictures plotted by the maple software to illustrate our results.


---

## Body

## 1. Introduction

Nonlinear science is basic science to study the generality of nonlinear phenomena. It is a comprehensive discipline which has been gradually developed by various branch disciplines characterized by nonlinearity since the 1960s [1]. It was known as the “Third Revolution” of Natural Science in the 20th century. The scientific community believes that the research of nonlinear science has not only great scientific significance but also broad application prospects. It involves almost all fields of natural science and social science, including engineering application, basic physical research, biological research, control theory, and management [2, 3]. And the nonlinear science is changing people’s traditional view of the real world. It is more and more important to find the exact solution of the nonlinear evolution equation. Therefore, a variety of solutions have emerged.In 1967, Gardner and others [4] first proposed scattering inversion method for KdV equation. Since then, many methods and techniques for constructing nonlinear partial differential equations have been gradually proposed for seeking soliton solutions [5–7], such as Bäcklund transform [8], Hirota method [9–11], and Darboux transform [12]. But these methods are complex and difficult to use in solving processes. With the deepening of research and the continuous development of mathematical computing software, in recent years, some analytical tools and direct algebraic methods have gradually emerged, such as fixed point theorems [13–16], variational methods [17–20], topological degree method [21–24], homogeneous balance method [25], tanh function method [26, 27], Jacobi elliptic function method [28, 29], F-expansion method [30, 31], and e
x
p
(
-
ϕ
(
z
)
)-expansion method [32–34].In 2012, Alejo [35] got some numerical results which showed a new family of solutions of the geometric mKdV equation. In 2014, Huang Y, Wu Y, and Meng F, et al. [36] used the complex method to get the meromorphic solutions of complex combined KdV-mKdV equation. Singh [37] used the Jacobian elliptic function expansion method to get the exact solutions of Wick-type stochastic Kersten-Krasil’ shchik coupled KdV-mKdV equations.Consider the following:(1)
u
t
+
τ
u
2
u
x
+
β
u
x
x
x
=
0
.In (1)(see [38]), τ and β are constants.Submittingu
(
x
,
t
)
=
w
(
z
), z
=
k
x
+
w
t, into (1) and after integrating, we get (2)
β
k
3
w
′
′
+
k
τ
3
w
3
-
ϖ
w
+
d
=
0
.In order to get the exact solution of mKdV equation, we use the complex method which was suggested by Yuan et al. [39–42] to get solutions of (2), and then we also get some exact solutions by the exp(
-
ϕ
(
z
)
)-expansion method.Theorem 1.
By using the complex method, we suppose thatβ
k
4
τ
/
3
≠
0 and then all meromorphic solutions w belong to the class W. And we found that there will be three forms of solutions of (2):
(1) The Elliptic Function Solutions
(3)
w
d
z
=
±
1
2
-
β
k
2
τ
-
℘
+
c
4
℘
c
2
+
4
℘
2
c
+
2
℘
′
a
-
℘
g
2
-
c
g
2
12
c
2
-
g
2
℘
+
4
c
3
-
3
c
g
2
℘
′
+
4
℘
3
+
12
c
℘
2
-
3
g
2
℘
-
c
g
2
d
′
,in (3), g
3
=
0, a
2
=
4
c
3
-
g
2
c, and g
2 and c are arbitrary constants.
(2) The Simply Periodic Solutions
(4)
w
s
,
1
z
=
α
-
6
β
k
2
τ
coth
⁡
α
2
z
-
z
0
-
coth
⁡
α
2
z
-
z
0
-
z
1
-
coth
⁡
α
2
z
1
,in (4), z
0
∈
C, ϖ
=
-
β
k
3
α
2
1
/
2
+
3
/
2
s
i
n
h
2
α
/
2
z
1, d
=
-
6
β
k
2
/
τ
t
a
n
h
α
/
2
z
1
/
s
i
n
h
2
α
/
2
z
1, and z
1
≠
0.
And the other solution is(5)
w
s
,
2
z
=
α
-
6
β
k
2
τ
tanh
⁡
α
2
z
-
z
0
,in (5), z
0
∈
C, ϖ
=
-
β
k
3
α
2
/
2
,
d
=
0.
(3) The Rational Function Solutions
(6)
w
r
,
1
z
=
±
-
6
β
k
2
τ
1
z
-
z
0
,and(7)
w
r
,
2
z
=
±
-
6
β
k
2
τ
z
1
2
z
1
z
-
z
0
-
z
1
z
-
z
0
-
z
1
-
1
,in (6), z
0
∈
C
,
ϖ
=
0
,
d
=
0, or in (7), z
0
∈
C, ϖ
=
-
6
β
k
3
/
z
1
2
,
d
=
∓
2
/
3
k
τ
-
6
β
k
2
/
τ
z
1
2
3
/
2.Theorem 2.
By usingexp
⁡
(
-
ϕ
(
z
)
)-expansion method, there will be three forms of solutions of (2).
Ifδ
2
-
4
μ
>
0, μ
≠
0, (8)
u
11
z
=
1
2
λ
-
6
β
k
2
τ
+
-
6
β
k
2
τ
2
μ
δ
2
-
4
μ
tanh
⁡
δ
2
-
4
μ
/
2
z
+
c
+
δ
,
(9)
u
12
z
=
1
2
λ
-
6
β
k
2
τ
+
-
6
β
k
2
τ
2
μ
δ
2
-
4
μ
coth
⁡
δ
2
-
4
μ
/
2
z
+
c
+
δ
,
(10)
u
21
z
=
-
1
2
λ
-
6
β
k
2
τ
-
-
6
β
k
2
τ
2
μ
δ
2
-
4
μ
tanh
⁡
δ
2
-
4
μ
/
2
z
+
c
+
δ
,
(11)
u
22
z
=
-
1
2
λ
-
6
β
k
2
τ
-
-
6
β
k
2
τ
2
μ
δ
2
-
4
μ
coth
⁡
δ
2
-
4
μ
/
2
z
+
c
+
δ
.
Ifδ
2
-
4
μ
<
0, μ
≠
0, (12)
u
13
z
=
1
2
λ
-
6
β
k
2
τ
+
-
6
β
k
2
τ
2
μ
δ
2
-
4
μ
tanh
⁡
δ
2
-
4
μ
/
2
z
+
c
+
δ
,
(13)
u
14
z
=
1
2
λ
-
6
β
k
2
τ
+
-
6
β
k
2
τ
2
μ
δ
2
-
4
μ
coth
⁡
δ
2
-
4
μ
/
2
z
+
c
+
δ
,
(14)
u
23
z
=
-
1
2
λ
-
6
β
k
2
τ
-
-
6
β
k
2
τ
2
μ
δ
2
-
4
μ
t
a
n
h
δ
2
-
4
μ
/
2
z
+
c
+
δ
,
(15)
u
24
z
=
-
1
2
λ
-
6
β
k
2
τ
-
-
6
β
k
2
τ
2
μ
δ
2
-
4
μ
coth
⁡
δ
2
-
4
μ
/
2
z
+
c
+
δ
.
Ifδ
2
-
4
μ
>
0, μ
=
0
,
δ
≠
0, (16)
u
15
z
=
1
2
λ
-
6
β
k
2
τ
+
-
6
β
k
2
τ
δ
exp
⁡
δ
z
+
c
-
1
,
(17)
u
25
z
=
-
1
2
λ
-
6
β
k
2
τ
-
-
6
β
k
2
τ
δ
exp
⁡
δ
z
+
c
-
1
.
Ifδ
2
-
4
μ
=
0, μ
≠
0
,
δ
≠
0, (18)
u
16
z
=
1
2
λ
-
6
β
k
2
τ
+
-
6
β
k
2
τ
δ
2
z
+
c
2
δ
z
+
c
+
2
,
(19)
u
26
z
=
-
1
2
λ
-
6
β
k
2
τ
-
-
6
β
k
2
τ
δ
2
z
+
c
2
δ
z
+
c
+
2
.
Ifδ
2
-
4
μ
=
0, μ
=
0
,
δ
=
0, (20)
u
17
z
=
1
2
λ
-
6
β
k
2
τ
+
-
6
β
k
2
τ
1
z
+
c
,
(21)
u
27
z
=
-
1
2
λ
-
6
β
k
2
τ
-
-
6
β
k
2
τ
1
z
+
c
.


## 2. Preliminary Lemmas, Complex Method andexp
⁡
(
-
ϕ
(
z
)
)-Expansion Method

### 2.1. Introduction of Complex Method

For the introduction of complex method, we have to know some concepts and symbols.Lemma 3 (see [34]).
First we setm
∈
N
≔
{
1,2
,
3
…
}, r
j
∈
N
0
=
N
∪
{
0
}, r
=
(
r
0
,
r
1
,
…
,
r
m
), j
=
0,1
,
…
,
m. Then we can get a differential monomial by (22)
M
r
w
z
≔
w
z
r
0
w
′
z
r
1
w
′
′
z
r
2
⋯
w
m
z
r
m
.
p
(
r
)
≔
r
0
+
2
r
1
+
⋯
+
(
m
+
1
)
r
m and d
e
g
(
M
) are regarded as the weight and degree of M
r
[
w
], separately.
The differential polynomialP
(
w
,
w
′
,
…
,
w
(
m
)
) can be defined as follows: (23)
P
w
,
w
′
,
…
,
w
m
≔
∑
r
∈
I
a
r
M
r
w
.
In (23), a
r are constants, and I is a finite index set.
The total weight and degree ofP
(
w
,
w
′
,
…
,
w
(
m
)
) are marked as W
(
P
)
≔
m
a
x
r
∈
I
{
p
(
r
)
} and d
e
g
(
P
)
≔
m
a
x
r
∈
I
{
d
e
g
(
M
r
)
}, separately.
Considering the complex, ordinary differential equations(24)
P
w
,
w
′
,
…
,
w
m
=
b
w
n
+
c
.
In (24), b
≠
0, c are constants, and n
∈
N.
We takep
,
q
∈
N, and we regard the meromorphic solutions w of (24) to have one or more poles. We can say that (24) is satisfied the 〈
p
,
q
〉 condition, where p means that the equation has p distinct meromorphic solutions and q means that their multiplicity of the pole at z
=
0 is q.
It is difficult for us to find the〈
p
,
q
〉 condition of (24), so we need a method to find the weak 〈
p
,
q
〉 condition showed as follows.
To find out the weak〈
p
,
q
〉 condition of (24), we need to substitute Laurent series (25)
w
z
=
∑
k
=
-
q
∞
c
k
z
k
,
q
>
0
,
c
-
q
≠
0into (24); then we can find out the p distinct Laurent singular parts as below:(26)
∑
k
=
-
q
-
1
c
k
z
k
.Given two complex numbersω
1, ω
2, and I
m
ω
1
/
ω
2
>
0, L
=
L
[
2
ω
1
,
2
ω
2
] are discrete subset L
[
2
ω
1
,
2
ω
2
]
=
{
ω
∣
ω
=
2
n
ω
1
+
2
m
ω
2
,
n
,
m
∈
Z
}, which is isomorphic to Z
×
Z. Let the discriminant Δ
=
Δ
(
c
1
,
c
2
)
≔
c
1
3
-
27
c
2
2 and (27)
s
n
=
s
n
L
≔
∑
ω
∈
L
∖
0
1
ω
n
.A meromorphic functionw
(
z
) means that w
(
z
) is holomorphic in the complex plane C except for poles. ℘
(
z
,
g
2
,
g
3
) is the Weierstrass elliptic function [43, 44] with invariants g
2 and g
3.Iff is an elliptic function, or a rational function of e
α
z
,
α
∈
C, or a rational function of z, then we say that the meromorphic function f belongs to the class W.Weierstrass elliptic function℘
(
z
)
≔
℘
(
z
,
g
2
,
g
3
) is a meromorphic function with double periods ω
1
,
ω
2 and defined as (28)
℘
z
;
w
1
,
w
2
≔
1
z
2
+
∑
μ
,
ν
∈
Z
,
μ
2
+
ν
2
≠
0
1
z
+
μ
w
1
+
ν
w
2
2
-
1
μ
w
1
+
ν
w
2
2
,which satisfies the following:(29)
℘
′
z
2
=
4
℘
z
3
-
g
2
℘
z
-
g
3
,and in (29), g
2
=
60
s
4
,
g
3
=
140
s
6 and Δ
(
g
2
,
g
3
)
≠
0.Or alternating (29) to the form (30)
℘
′
z
2
=
4
℘
z
-
e
1
℘
z
-
e
2
℘
z
-
e
2
,in (30), e
1
=
℘
(
ω
1
)
,
e
2
=
℘
(
ω
2
)
,
e
3
=
℘
(
ω
1
+
ω
2
)
.Contrarily, given two complex numbersg
2 and g
3 and Δ
(
g
2
,
g
3
)
≠
0, then there will have double periods ω
1
,
ω
2 Weierstrass elliptic function ℘
(
z
) which the solutions will possess.In 2009, Eremenko [45] et al. investigated the m-order Briot-Bouquet equation (BBEq) as follows: (31)
F
w
,
w
m
=
∑
j
=
0
m
F
j
w
w
m
j
=
0
.In (31), F
j
(
w
) are constant coefficients polynomials, m
∈
N. For the m order BBEq, there are the following lemmas.Lemma 4 (see [39–42]).
Letp
,
l
,
m
,
n
∈
N, and deg
⁡
P
(
w
,
w
(
m
)
)
<
n. Considering that a m-order Briot-Bouquet equation (32)
P
w
m
,
w
=
b
w
n
+
csatisfies weak 〈
p
,
q
〉 condition, then all the meromorphic solutions w will belong to the class W. For some values of parameters, if the solution w exists, then other meromorphic solutions will form a one-parametric family w
(
z
-
z
0
)
,
z
0
∈
C. Furthermore, it can be written as the following forms of each elliptic solution with pole at z
=
0: (33)
w
z
=
∑
i
=
1
l
-
1
∑
j
=
2
q
i
-
1
j
c
-
i
j
j
-
1
!
d
j
-
2
d
z
j
-
2
1
4
℘
′
z
+
B
i
℘
z
-
A
i
2
-
℘
z
+
∑
i
=
1
l
-
1
c
-
i
1
2
℘
′
z
+
B
i
℘
z
-
A
i
+
∑
j
=
2
q
l
-
1
j
c
-
l
j
j
-
1
!
d
j
-
2
d
z
j
-
2
℘
z
+
c
0
.
In (33), c
-
i
j are given by equation (23), and B
i
2
=
4
A
i
3
-
g
2
A
i
-
g
3, Σ
i
=
1
l
c
-
i
1
=
0.
Each rational function solutionw
≔
R
(
z
) can be show as the following form: (34)
R
z
=
∑
i
=
1
l
∑
j
=
1
q
c
i
j
z
-
z
i
j
+
c
0
,with l
(
≤
p
) distinct poles of multiplicity q.
Every simply periodic solution is a rational functionR
(
ξ
) of ξ
=
e
α
z
(
α
∈
C
). R
(
ξ
) has l
(
≤
p
) distinct poles of multiplicity q and can be show as the following form: (35)
R
ξ
=
∑
i
=
1
l
∑
j
=
1
q
c
i
j
ξ
-
ξ
i
j
+
c
0
.Lemma 5 (see [43, 44]).
Weierstrass elliptic functions℘
(
z
)
≔
℘
(
z
,
g
2
,
g
3
) have two successive degeneracies and addition formula:
(I) Degeneracy to simply periodic functions (i.e., rational functions of one exponentiale
k
z) according to (36)
℘
z
,
3
d
2
,
-
d
3
=
2
d
-
3
d
2
c
o
t
h
2
3
d
2
z
,if one root e
j is double (Δ
(
g
2
,
g
3
)
=
0).
(II) Degeneracy to rational functions ofz according to (37)
℘
z
,
0,0
=
1
z
2
,if one root e
j is triple (g
2
=
g
3
=
0).
(III) Addition formula(38)
℘
z
-
z
0
=
-
℘
z
-
℘
z
0
+
1
4
℘
′
z
+
℘
′
z
0
℘
z
-
℘
z
0
2
.
By the above lemma and results, we introduce complex method to find exact solutions of some PDEs. The detailed five steps are as follows:
(1) Put the transformT
:
u
(
x
,
t
)
→
w
(
z
)
,
(
x
,
t
)
→
z into a given PDE to produce a nonlinear ODE.
(2) Put (25) into (24) or (32) to find out the weak 〈
p
,
q
〉 condition.
(3) By determinant relation equation (33)–(35) we will, respectively, find the elliptic, rational and simply periodic solutions u
(
z
) of (24) or (32) with pole at z
=
0.
(4) By Lemmas3 and 4, we obtain meromorphic solutions and the addition formula.
(5) Put the inverse transformT
-
1 into these meromorphic solutions w
(
z
-
z
0
), all exact solutions u
(
x
,
t
) of the original PDE will be found.More details of complex method can be found in [46–50].


### 2.2. Introduction of exp(
-
ϕ
(
z
)
)-Expansion Method

Consider that a nonlinear partial differential equation (PDE) in the following form:(39)
P
μ
,
μ
x
,
μ
y
,
μ
t
,
μ
x
x
,
μ
y
y
,
μ
t
t
…
=
0In (39), P is a polynomial with an unknown function μ
(
x
,
y
,
t
) and its derivatives in which nonlinear terms and highest order derivatives are involved. And it can be processed as follows.Step 1.
Insert the traveling wave transformμ
(
x
,
y
,
t
)
=
w
(
z
)
,
z
=
k
x
+
l
y
+
r
t into (39) alternating it to the following ordinary differential equation (ODE): (40)
K
w
,
w
′
,
w
′
′
,
w
′
′
′
,
…
=
0
,and in (40), K is a polynomial of w
(
z
) and its derivatives.Step 2.
Regarding that (40) has the following traveling wave solution: (41)
w
z
=
∑
j
=
0
n
C
j
exp
⁡
-
ϕ
z
j
,in (41), C
j
(
0
≤
j
≤
n
) are constants and will be determined later and C
j
≠
0 and ϕ
=
ϕ
(
z
) satisfies the ODE as follows: (42)
ϕ
′
z
=
exp
⁡
-
ϕ
z
+
μ
exp
⁡
ϕ
z
+
δ
,Equation (42) has different style solutions as follows:Ifδ
2
-
4
μ
>
0
,
μ
≠
0, then (43)
ϕ
z
=
ln
⁡
-
δ
2
-
4
μ
tanh
⁡
δ
2
-
4
μ
/
2
z
+
c
-
δ
2
μ
,
(44)
ϕ
z
=
ln
⁡
-
δ
2
-
4
μ
coth
⁡
δ
2
-
4
μ
/
2
z
+
c
-
δ
2
μ
.Ifδ
2
-
4
μ
<
0
,
μ
≠
0, then (45)
ϕ
z
=
ln
⁡
-
4
μ
-
δ
2
tan
⁡
4
μ
-
δ
2
/
2
z
+
c
-
δ
2
μ
,
(46)
ϕ
z
=
ln
⁡
-
4
μ
-
δ
2
cot
⁡
4
μ
-
δ
2
/
2
z
+
c
-
δ
2
μ
.Ifδ
2
-
4
μ
<
0
,
μ
=
0
,
δ
≠
0, then (47)
ϕ
z
=
-
ln
⁡
δ
exp
⁡
δ
z
+
c
-
1
.Ifδ
2
-
4
μ
=
0
,
μ
≠
0
,
δ
≠
0, then (48)
ϕ
z
=
ln
⁡
-
2
δ
z
+
c
+
2
δ
2
z
+
c
.Ifδ
2
-
4
μ
=
0
,
μ
=
0
,
δ
=
0, then (49)
ϕ
z
=
ln
⁡
z
+
c
.In above equations,C
n
≠
0, δ, and μ are constants and will be determined later and c is an arbitrary constant. We consider the homogeneous balance between nonlinear terms and highest order derivatives of (40), so we can find the positive integer n.Step 3.
Putting (41) into (40) and accounting the function e
x
p
(
-
ϕ
(
z
)
), we get a polynomial of e
x
p
(
-
ϕ
(
z
)
). Calculating all the coefficients of the same power of e
x
p
(
-
ϕ
(
z
)
) to zero and then we get a set of algebraic equations. By solving the algebraic equations, we get the values of C
n
≠
0, δ, and μ, and then we put these into (34) along with(43)-(49) to get the determination of the solutions of (39).


## 2.1. Introduction of Complex Method

For the introduction of complex method, we have to know some concepts and symbols.Lemma 3 (see [34]).
First we setm
∈
N
≔
{
1,2
,
3
…
}, r
j
∈
N
0
=
N
∪
{
0
}, r
=
(
r
0
,
r
1
,
…
,
r
m
), j
=
0,1
,
…
,
m. Then we can get a differential monomial by (22)
M
r
w
z
≔
w
z
r
0
w
′
z
r
1
w
′
′
z
r
2
⋯
w
m
z
r
m
.
p
(
r
)
≔
r
0
+
2
r
1
+
⋯
+
(
m
+
1
)
r
m and d
e
g
(
M
) are regarded as the weight and degree of M
r
[
w
], separately.
The differential polynomialP
(
w
,
w
′
,
…
,
w
(
m
)
) can be defined as follows: (23)
P
w
,
w
′
,
…
,
w
m
≔
∑
r
∈
I
a
r
M
r
w
.
In (23), a
r are constants, and I is a finite index set.
The total weight and degree ofP
(
w
,
w
′
,
…
,
w
(
m
)
) are marked as W
(
P
)
≔
m
a
x
r
∈
I
{
p
(
r
)
} and d
e
g
(
P
)
≔
m
a
x
r
∈
I
{
d
e
g
(
M
r
)
}, separately.
Considering the complex, ordinary differential equations(24)
P
w
,
w
′
,
…
,
w
m
=
b
w
n
+
c
.
In (24), b
≠
0, c are constants, and n
∈
N.
We takep
,
q
∈
N, and we regard the meromorphic solutions w of (24) to have one or more poles. We can say that (24) is satisfied the 〈
p
,
q
〉 condition, where p means that the equation has p distinct meromorphic solutions and q means that their multiplicity of the pole at z
=
0 is q.
It is difficult for us to find the〈
p
,
q
〉 condition of (24), so we need a method to find the weak 〈
p
,
q
〉 condition showed as follows.
To find out the weak〈
p
,
q
〉 condition of (24), we need to substitute Laurent series (25)
w
z
=
∑
k
=
-
q
∞
c
k
z
k
,
q
>
0
,
c
-
q
≠
0into (24); then we can find out the p distinct Laurent singular parts as below:(26)
∑
k
=
-
q
-
1
c
k
z
k
.Given two complex numbersω
1, ω
2, and I
m
ω
1
/
ω
2
>
0, L
=
L
[
2
ω
1
,
2
ω
2
] are discrete subset L
[
2
ω
1
,
2
ω
2
]
=
{
ω
∣
ω
=
2
n
ω
1
+
2
m
ω
2
,
n
,
m
∈
Z
}, which is isomorphic to Z
×
Z. Let the discriminant Δ
=
Δ
(
c
1
,
c
2
)
≔
c
1
3
-
27
c
2
2 and (27)
s
n
=
s
n
L
≔
∑
ω
∈
L
∖
0
1
ω
n
.A meromorphic functionw
(
z
) means that w
(
z
) is holomorphic in the complex plane C except for poles. ℘
(
z
,
g
2
,
g
3
) is the Weierstrass elliptic function [43, 44] with invariants g
2 and g
3.Iff is an elliptic function, or a rational function of e
α
z
,
α
∈
C, or a rational function of z, then we say that the meromorphic function f belongs to the class W.Weierstrass elliptic function℘
(
z
)
≔
℘
(
z
,
g
2
,
g
3
) is a meromorphic function with double periods ω
1
,
ω
2 and defined as (28)
℘
z
;
w
1
,
w
2
≔
1
z
2
+
∑
μ
,
ν
∈
Z
,
μ
2
+
ν
2
≠
0
1
z
+
μ
w
1
+
ν
w
2
2
-
1
μ
w
1
+
ν
w
2
2
,which satisfies the following:(29)
℘
′
z
2
=
4
℘
z
3
-
g
2
℘
z
-
g
3
,and in (29), g
2
=
60
s
4
,
g
3
=
140
s
6 and Δ
(
g
2
,
g
3
)
≠
0.Or alternating (29) to the form (30)
℘
′
z
2
=
4
℘
z
-
e
1
℘
z
-
e
2
℘
z
-
e
2
,in (30), e
1
=
℘
(
ω
1
)
,
e
2
=
℘
(
ω
2
)
,
e
3
=
℘
(
ω
1
+
ω
2
)
.Contrarily, given two complex numbersg
2 and g
3 and Δ
(
g
2
,
g
3
)
≠
0, then there will have double periods ω
1
,
ω
2 Weierstrass elliptic function ℘
(
z
) which the solutions will possess.In 2009, Eremenko [45] et al. investigated the m-order Briot-Bouquet equation (BBEq) as follows: (31)
F
w
,
w
m
=
∑
j
=
0
m
F
j
w
w
m
j
=
0
.In (31), F
j
(
w
) are constant coefficients polynomials, m
∈
N. For the m order BBEq, there are the following lemmas.Lemma 4 (see [39–42]).
Letp
,
l
,
m
,
n
∈
N, and deg
⁡
P
(
w
,
w
(
m
)
)
<
n. Considering that a m-order Briot-Bouquet equation (32)
P
w
m
,
w
=
b
w
n
+
csatisfies weak 〈
p
,
q
〉 condition, then all the meromorphic solutions w will belong to the class W. For some values of parameters, if the solution w exists, then other meromorphic solutions will form a one-parametric family w
(
z
-
z
0
)
,
z
0
∈
C. Furthermore, it can be written as the following forms of each elliptic solution with pole at z
=
0: (33)
w
z
=
∑
i
=
1
l
-
1
∑
j
=
2
q
i
-
1
j
c
-
i
j
j
-
1
!
d
j
-
2
d
z
j
-
2
1
4
℘
′
z
+
B
i
℘
z
-
A
i
2
-
℘
z
+
∑
i
=
1
l
-
1
c
-
i
1
2
℘
′
z
+
B
i
℘
z
-
A
i
+
∑
j
=
2
q
l
-
1
j
c
-
l
j
j
-
1
!
d
j
-
2
d
z
j
-
2
℘
z
+
c
0
.
In (33), c
-
i
j are given by equation (23), and B
i
2
=
4
A
i
3
-
g
2
A
i
-
g
3, Σ
i
=
1
l
c
-
i
1
=
0.
Each rational function solutionw
≔
R
(
z
) can be show as the following form: (34)
R
z
=
∑
i
=
1
l
∑
j
=
1
q
c
i
j
z
-
z
i
j
+
c
0
,with l
(
≤
p
) distinct poles of multiplicity q.
Every simply periodic solution is a rational functionR
(
ξ
) of ξ
=
e
α
z
(
α
∈
C
). R
(
ξ
) has l
(
≤
p
) distinct poles of multiplicity q and can be show as the following form: (35)
R
ξ
=
∑
i
=
1
l
∑
j
=
1
q
c
i
j
ξ
-
ξ
i
j
+
c
0
.Lemma 5 (see [43, 44]).
Weierstrass elliptic functions℘
(
z
)
≔
℘
(
z
,
g
2
,
g
3
) have two successive degeneracies and addition formula:
(I) Degeneracy to simply periodic functions (i.e., rational functions of one exponentiale
k
z) according to (36)
℘
z
,
3
d
2
,
-
d
3
=
2
d
-
3
d
2
c
o
t
h
2
3
d
2
z
,if one root e
j is double (Δ
(
g
2
,
g
3
)
=
0).
(II) Degeneracy to rational functions ofz according to (37)
℘
z
,
0,0
=
1
z
2
,if one root e
j is triple (g
2
=
g
3
=
0).
(III) Addition formula(38)
℘
z
-
z
0
=
-
℘
z
-
℘
z
0
+
1
4
℘
′
z
+
℘
′
z
0
℘
z
-
℘
z
0
2
.
By the above lemma and results, we introduce complex method to find exact solutions of some PDEs. The detailed five steps are as follows:
(1) Put the transformT
:
u
(
x
,
t
)
→
w
(
z
)
,
(
x
,
t
)
→
z into a given PDE to produce a nonlinear ODE.
(2) Put (25) into (24) or (32) to find out the weak 〈
p
,
q
〉 condition.
(3) By determinant relation equation (33)–(35) we will, respectively, find the elliptic, rational and simply periodic solutions u
(
z
) of (24) or (32) with pole at z
=
0.
(4) By Lemmas3 and 4, we obtain meromorphic solutions and the addition formula.
(5) Put the inverse transformT
-
1 into these meromorphic solutions w
(
z
-
z
0
), all exact solutions u
(
x
,
t
) of the original PDE will be found.More details of complex method can be found in [46–50].


## 2.2. Introduction of exp(
-
ϕ
(
z
)
)-Expansion Method

Consider that a nonlinear partial differential equation (PDE) in the following form:(39)
P
μ
,
μ
x
,
μ
y
,
μ
t
,
μ
x
x
,
μ
y
y
,
μ
t
t
…
=
0In (39), P is a polynomial with an unknown function μ
(
x
,
y
,
t
) and its derivatives in which nonlinear terms and highest order derivatives are involved. And it can be processed as follows.Step 1.
Insert the traveling wave transformμ
(
x
,
y
,
t
)
=
w
(
z
)
,
z
=
k
x
+
l
y
+
r
t into (39) alternating it to the following ordinary differential equation (ODE): (40)
K
w
,
w
′
,
w
′
′
,
w
′
′
′
,
…
=
0
,and in (40), K is a polynomial of w
(
z
) and its derivatives.Step 2.
Regarding that (40) has the following traveling wave solution: (41)
w
z
=
∑
j
=
0
n
C
j
exp
⁡
-
ϕ
z
j
,in (41), C
j
(
0
≤
j
≤
n
) are constants and will be determined later and C
j
≠
0 and ϕ
=
ϕ
(
z
) satisfies the ODE as follows: (42)
ϕ
′
z
=
exp
⁡
-
ϕ
z
+
μ
exp
⁡
ϕ
z
+
δ
,Equation (42) has different style solutions as follows:Ifδ
2
-
4
μ
>
0
,
μ
≠
0, then (43)
ϕ
z
=
ln
⁡
-
δ
2
-
4
μ
tanh
⁡
δ
2
-
4
μ
/
2
z
+
c
-
δ
2
μ
,
(44)
ϕ
z
=
ln
⁡
-
δ
2
-
4
μ
coth
⁡
δ
2
-
4
μ
/
2
z
+
c
-
δ
2
μ
.Ifδ
2
-
4
μ
<
0
,
μ
≠
0, then (45)
ϕ
z
=
ln
⁡
-
4
μ
-
δ
2
tan
⁡
4
μ
-
δ
2
/
2
z
+
c
-
δ
2
μ
,
(46)
ϕ
z
=
ln
⁡
-
4
μ
-
δ
2
cot
⁡
4
μ
-
δ
2
/
2
z
+
c
-
δ
2
μ
.Ifδ
2
-
4
μ
<
0
,
μ
=
0
,
δ
≠
0, then (47)
ϕ
z
=
-
ln
⁡
δ
exp
⁡
δ
z
+
c
-
1
.Ifδ
2
-
4
μ
=
0
,
μ
≠
0
,
δ
≠
0, then (48)
ϕ
z
=
ln
⁡
-
2
δ
z
+
c
+
2
δ
2
z
+
c
.Ifδ
2
-
4
μ
=
0
,
μ
=
0
,
δ
=
0, then (49)
ϕ
z
=
ln
⁡
z
+
c
.In above equations,C
n
≠
0, δ, and μ are constants and will be determined later and c is an arbitrary constant. We consider the homogeneous balance between nonlinear terms and highest order derivatives of (40), so we can find the positive integer n.Step 3.
Putting (41) into (40) and accounting the function e
x
p
(
-
ϕ
(
z
)
), we get a polynomial of e
x
p
(
-
ϕ
(
z
)
). Calculating all the coefficients of the same power of e
x
p
(
-
ϕ
(
z
)
) to zero and then we get a set of algebraic equations. By solving the algebraic equations, we get the values of C
n
≠
0, δ, and μ, and then we put these into (34) along with(43)-(49) to get the determination of the solutions of (39).


## 3. Proof of Theorems1 and 2

### 3.1. Proof of Theorem1

Putting (25) into (2), we have q
=
1, p
=
2, c
-
1
=
±
-
6
β
k
2
/
τ, c
0
=
0, c
1
=
ϖ
/
k
τ
c
-
1, c
2
=
d
/
β
k
3
-
4
k
τ, and c
4
=
-
ϖ
d
/
24
β
2
k
6 and c
3 is arbitrary constant.Because (2) satisfies weak 〈
2,1
〉 condition and is a two-order mKdv equation, (2) satisfies the dominant condition. By Lemma 4, we know that all meromorphic solutions of (2) belong to W. Now we will give the forms of all meromorphic solutions of (2).By (2), we infer the indeterminant rational solutions of (2) with pole at z
=
0 that (50)
u
r
z
=
c
11
z
+
c
12
z
-
z
1
+
c
10
.Puttingu
r
(
z
) into (2), we get two classes: one is (51)
R
1,1
z
=
±
-
6
β
k
2
τ
1
z
,in (51), ϖ
=
0
,
d
=
0.And the other one is(52)
R
1,2
z
=
±
-
6
β
k
3
k
τ
z
1
2
z
1
z
-
z
1
z
-
z
1
-
1
,in (52), ϖ
=
-
6
β
k
3
/
z
1
2
,
d
=
∓
2
/
3
k
τ
-
6
β
k
2
/
τ
z
1
2
3
/
2.All rational solutions of (2) are as follows:(53)
w
r
,
1
z
=
±
-
6
β
k
2
τ
1
z
-
z
0
,and (54)
w
r
,
2
z
=
±
-
6
β
k
2
τ
z
1
2
z
1
z
-
z
0
-
z
1
z
-
z
0
-
z
1
-
1
.In (53)-(54), z
0
∈
C
,
ϖ
=
0
,
d
=
0, or in (54), ϖ
=
-
6
β
k
3
/
z
1
2
,
d
=
∓
2
/
3
k
τ
-
6
β
k
2
/
τ
z
1
2
3
/
2.In order to get simply periodic solutions, we setξ
=
e
x
p
(
α
z
) and then put u
=
u
(
ξ
) into (2). We get (55)
β
k
3
α
2
ξ
R
′
+
ξ
2
R
′
′
-
ϖ
R
+
k
τ
3
R
3
+
d
=
0
.By putting(56)
u
2
ξ
=
c
2
ξ
-
1
+
c
1
ξ
-
ξ
1
+
c
20into (55), we get (57)
R
2,1
ξ
=
α
1
-
ξ
1
-
6
β
k
2
τ
-
2
ξ
1
-
1
1
ξ
-
1
-
1
ξ
1
-
1
+
ξ
1
+
1
,in (57), ϖ
=
-
1
/
2
+
6
ξ
1
/
1
-
ξ
1
2
β
k
3
α
2
,
d
=
8
k
τ
ξ
1
(
ξ
1
+
1
)
α
3
/
3
(
1
-
ξ
1
)
3
-
6
β
k
2
/
τ
z
1
2
3
/
2.And the other solution is(58)
R
2,2
ξ
=
α
-
6
β
k
2
τ
1
ξ
-
1
+
1
,in (58), ϖ
=
-
β
k
3
α
2
/
2
,
d
=
0
.So forz
=
0, all simply periodic solutions of (2) are gotten, which are (59)
w
s
0,1
z
=
α
-
6
β
k
2
τ
coth
⁡
α
2
z
-
coth
⁡
α
2
z
-
z
1
-
coth
⁡
α
2
z
1
,and(60)
w
s
0,2
z
=
α
-
6
β
k
2
τ
tanh
⁡
α
2
z
.So all simply periodic solutions of (2) are gotten: (61)
w
s
,
1
z
=
α
-
6
β
k
2
τ
coth
⁡
α
2
z
-
z
0
-
coth
⁡
α
2
z
-
z
0
-
z
1
-
coth
⁡
α
2
z
1
,in (61), z
0
∈, ϖ
=
-
β
k
3
α
2
1
/
2
+
3
/
2
s
i
n
h
2
α
/
2
z
1, d
=
-
6
β
k
2
/
τ
t
a
n
h
α
/
2
z
1
/
s
i
n
h
2
α
/
2
z
1, and z
1
≠
0.And the other solution is(62)
w
s
,
2
z
=
α
-
6
β
k
2
τ
tanh
⁡
α
2
z
-
z
0
,in (62), ϖ
=
-
β
k
3
α
2
/
2
,
d
=
0.By (33) of Lemma 4, we have indeterminant relations of elliptic solutions of (2) with the pole at z
=
0
(63)
u
d
0
z
=
c
-
1
2
℘
′
z
+
F
℘
z
-
E
+
c
30
,in (63), F
2
=
4
E
3
-
g
2
E
-
g
3. Applying the conclusion II of Lemma 5 to u
d
0
(
z
) and noting that the results of rational solutions obtained above, we deduce that c
30
=
0, E
=
F
=
0, and g
3
=
0. Then we get that (64)
u
d
0
z
=
±
1
2
-
6
β
k
2
τ
℘
′
z
℘
z
,in (64), g
3
=
0.So all elliptic function solutions of (2) are (65)
u
d
0
z
=
±
1
2
-
6
β
k
2
τ
℘
′
z
-
z
0
℘
z
-
z
0
,in (65), z
0
∈
C
,
g
3
=
0. Making use the addition of Lemma 5, we can rewrite it to the form (66)
w
d
z
=
±
1
2
-
6
β
k
2
τ
-
℘
+
c
4
℘
c
2
+
4
℘
2
c
+
2
℘
′
a
-
℘
g
2
-
c
g
2
12
c
2
-
g
2
℘
+
4
c
3
-
3
c
g
2
℘
′
+
4
℘
3
+
12
c
℘
2
-
3
g
2
℘
-
c
g
2
d
′
,In (66), g
3
=
0, a
2
=
4
c
3
-
g
2
c, and g
2 and c are arbitrary constants.


### 3.2. Proof of Theorem2

Taking the homogeneous balance betweenu
′
′ and u
3 in (2) we get(67)
u
z
=
C
0
+
C
1
exp
⁡
-
ϕ
z
,in (67), C
1
≠
0 and C
0 are constants which need to be determined, and ϕ
(
z
) satisfies equation ϕ
′
(
z
)
=
exp
⁡
(
-
ϕ
(
z
)
)
+
μ
exp
⁡
(
ϕ
(
z
)
)
+
δ, whereas δ and μ are arbitrary constants.From (67), we insert u, u
3, u
′
′ into (2) and sort out the coefficients of exp
⁡
(
-
ϕ
(
z
)
) to zero, and then we obtain (68)
c
1
=
±
-
6
β
k
2
τ
,
c
0
=
±
1
2
λ
-
6
β
k
2
τ
,in (68), λ
=
±
2
2
β
k
3
μ
-
ϖ
/
β
k
3 and d
=
0.By putting (68) into (67), we get(69)
u
1
z
=
1
2
λ
-
6
β
k
2
τ
+
-
6
β
k
2
τ
exp
⁡
-
ϕ
z
,or (70)
u
2
z
=
-
1
2
λ
-
6
β
k
2
τ
-
-
6
β
k
2
τ
exp
⁡
-
ϕ
z
.We apply (43)–(49) into (69) and (70), respectively, to obtain traveling wave solutions of the mKdV equation as follows.Ifδ
2
-
4
μ
>
0, μ
≠
0, (71)
u
11
z
=
1
2
λ
-
6
β
k
2
τ
+
-
6
β
k
2
τ
2
μ
δ
2
-
4
μ
tanh
⁡
δ
2
-
4
μ
/
2
z
+
c
+
δ
,
(72)
u
12
z
=
1
2
λ
-
6
β
k
2
τ
+
-
6
β
k
2
τ
2
μ
δ
2
-
4
μ
coth
⁡
δ
2
-
4
μ
/
2
z
+
c
+
δ
,
(73)
u
21
z
=
-
1
2
λ
-
6
β
k
2
τ
-
-
6
β
k
2
τ
2
μ
δ
2
-
4
μ
tanh
⁡
δ
2
-
4
μ
/
2
z
+
c
+
δ
,
(74)
u
22
z
-
1
2
λ
-
6
β
k
2
τ
-
-
6
β
k
2
τ
2
μ
δ
2
-
4
μ
coth
⁡
δ
2
-
4
μ
/
2
z
+
c
+
δ
.Ifδ
2
-
4
μ
<
0, μ
≠
0, (75)
u
13
z
=
1
2
λ
-
6
β
k
2
τ
+
-
6
β
k
2
τ
2
μ
δ
2
-
4
μ
tanh
⁡
δ
2
-
4
μ
/
2
z
+
c
+
δ
,
(76)
u
14
z
=
1
2
λ
-
6
β
k
2
τ
+
-
6
β
k
2
τ
2
μ
δ
2
-
4
μ
coth
⁡
δ
2
-
4
μ
/
2
z
+
c
+
δ
,
(77)
u
23
z
=
-
1
2
λ
-
6
β
k
2
τ
-
-
6
β
k
2
τ
2
μ
δ
2
-
4
μ
tanh
⁡
δ
2
-
4
μ
/
2
z
+
c
+
δ
,
(78)
u
24
z
=
-
1
2
λ
-
6
β
k
2
τ
-
-
6
β
k
2
τ
2
μ
δ
2
-
4
μ
coth
⁡
δ
2
-
4
μ
/
2
z
+
c
+
δ
.Ifδ
2
-
4
μ
>
0, μ
=
0
,
δ
≠
0, (79)
u
15
z
=
1
2
λ
-
6
β
k
2
τ
+
-
6
β
k
2
τ
δ
exp
⁡
δ
z
+
c
-
1
,
(80)
u
25
z
=
-
1
2
λ
-
6
β
k
2
τ
-
-
6
β
k
2
τ
δ
exp
⁡
δ
z
+
c
-
1
.Ifδ
2
-
4
μ
=
0, μ
≠
0
,
δ
≠
0, (81)
u
16
z
=
1
2
λ
-
6
β
k
2
τ
+
-
6
β
k
2
τ
δ
2
z
+
c
2
δ
z
+
c
+
2
,
(82)
u
26
z
=
-
1
2
λ
-
6
β
k
2
τ
-
-
6
β
k
2
τ
δ
2
z
+
c
2
δ
z
+
c
+
2
.Ifδ
2
-
4
μ
=
0, μ
=
0
,
δ
=
0, (83)
u
17
z
=
1
2
λ
-
6
β
k
2
τ
+
-
6
β
k
2
τ
1
z
+
c
,
(84)
u
27
z
=
-
1
2
λ
-
6
β
k
2
τ
-
-
6
β
k
2
τ
1
z
+
c
.


## 3.1. Proof of Theorem1

Putting (25) into (2), we have q
=
1, p
=
2, c
-
1
=
±
-
6
β
k
2
/
τ, c
0
=
0, c
1
=
ϖ
/
k
τ
c
-
1, c
2
=
d
/
β
k
3
-
4
k
τ, and c
4
=
-
ϖ
d
/
24
β
2
k
6 and c
3 is arbitrary constant.Because (2) satisfies weak 〈
2,1
〉 condition and is a two-order mKdv equation, (2) satisfies the dominant condition. By Lemma 4, we know that all meromorphic solutions of (2) belong to W. Now we will give the forms of all meromorphic solutions of (2).By (2), we infer the indeterminant rational solutions of (2) with pole at z
=
0 that (50)
u
r
z
=
c
11
z
+
c
12
z
-
z
1
+
c
10
.Puttingu
r
(
z
) into (2), we get two classes: one is (51)
R
1,1
z
=
±
-
6
β
k
2
τ
1
z
,in (51), ϖ
=
0
,
d
=
0.And the other one is(52)
R
1,2
z
=
±
-
6
β
k
3
k
τ
z
1
2
z
1
z
-
z
1
z
-
z
1
-
1
,in (52), ϖ
=
-
6
β
k
3
/
z
1
2
,
d
=
∓
2
/
3
k
τ
-
6
β
k
2
/
τ
z
1
2
3
/
2.All rational solutions of (2) are as follows:(53)
w
r
,
1
z
=
±
-
6
β
k
2
τ
1
z
-
z
0
,and (54)
w
r
,
2
z
=
±
-
6
β
k
2
τ
z
1
2
z
1
z
-
z
0
-
z
1
z
-
z
0
-
z
1
-
1
.In (53)-(54), z
0
∈
C
,
ϖ
=
0
,
d
=
0, or in (54), ϖ
=
-
6
β
k
3
/
z
1
2
,
d
=
∓
2
/
3
k
τ
-
6
β
k
2
/
τ
z
1
2
3
/
2.In order to get simply periodic solutions, we setξ
=
e
x
p
(
α
z
) and then put u
=
u
(
ξ
) into (2). We get (55)
β
k
3
α
2
ξ
R
′
+
ξ
2
R
′
′
-
ϖ
R
+
k
τ
3
R
3
+
d
=
0
.By putting(56)
u
2
ξ
=
c
2
ξ
-
1
+
c
1
ξ
-
ξ
1
+
c
20into (55), we get (57)
R
2,1
ξ
=
α
1
-
ξ
1
-
6
β
k
2
τ
-
2
ξ
1
-
1
1
ξ
-
1
-
1
ξ
1
-
1
+
ξ
1
+
1
,in (57), ϖ
=
-
1
/
2
+
6
ξ
1
/
1
-
ξ
1
2
β
k
3
α
2
,
d
=
8
k
τ
ξ
1
(
ξ
1
+
1
)
α
3
/
3
(
1
-
ξ
1
)
3
-
6
β
k
2
/
τ
z
1
2
3
/
2.And the other solution is(58)
R
2,2
ξ
=
α
-
6
β
k
2
τ
1
ξ
-
1
+
1
,in (58), ϖ
=
-
β
k
3
α
2
/
2
,
d
=
0
.So forz
=
0, all simply periodic solutions of (2) are gotten, which are (59)
w
s
0,1
z
=
α
-
6
β
k
2
τ
coth
⁡
α
2
z
-
coth
⁡
α
2
z
-
z
1
-
coth
⁡
α
2
z
1
,and(60)
w
s
0,2
z
=
α
-
6
β
k
2
τ
tanh
⁡
α
2
z
.So all simply periodic solutions of (2) are gotten: (61)
w
s
,
1
z
=
α
-
6
β
k
2
τ
coth
⁡
α
2
z
-
z
0
-
coth
⁡
α
2
z
-
z
0
-
z
1
-
coth
⁡
α
2
z
1
,in (61), z
0
∈, ϖ
=
-
β
k
3
α
2
1
/
2
+
3
/
2
s
i
n
h
2
α
/
2
z
1, d
=
-
6
β
k
2
/
τ
t
a
n
h
α
/
2
z
1
/
s
i
n
h
2
α
/
2
z
1, and z
1
≠
0.And the other solution is(62)
w
s
,
2
z
=
α
-
6
β
k
2
τ
tanh
⁡
α
2
z
-
z
0
,in (62), ϖ
=
-
β
k
3
α
2
/
2
,
d
=
0.By (33) of Lemma 4, we have indeterminant relations of elliptic solutions of (2) with the pole at z
=
0
(63)
u
d
0
z
=
c
-
1
2
℘
′
z
+
F
℘
z
-
E
+
c
30
,in (63), F
2
=
4
E
3
-
g
2
E
-
g
3. Applying the conclusion II of Lemma 5 to u
d
0
(
z
) and noting that the results of rational solutions obtained above, we deduce that c
30
=
0, E
=
F
=
0, and g
3
=
0. Then we get that (64)
u
d
0
z
=
±
1
2
-
6
β
k
2
τ
℘
′
z
℘
z
,in (64), g
3
=
0.So all elliptic function solutions of (2) are (65)
u
d
0
z
=
±
1
2
-
6
β
k
2
τ
℘
′
z
-
z
0
℘
z
-
z
0
,in (65), z
0
∈
C
,
g
3
=
0. Making use the addition of Lemma 5, we can rewrite it to the form (66)
w
d
z
=
±
1
2
-
6
β
k
2
τ
-
℘
+
c
4
℘
c
2
+
4
℘
2
c
+
2
℘
′
a
-
℘
g
2
-
c
g
2
12
c
2
-
g
2
℘
+
4
c
3
-
3
c
g
2
℘
′
+
4
℘
3
+
12
c
℘
2
-
3
g
2
℘
-
c
g
2
d
′
,In (66), g
3
=
0, a
2
=
4
c
3
-
g
2
c, and g
2 and c are arbitrary constants.


## 3.2. Proof of Theorem2

Taking the homogeneous balance betweenu
′
′ and u
3 in (2) we get(67)
u
z
=
C
0
+
C
1
exp
⁡
-
ϕ
z
,in (67), C
1
≠
0 and C
0 are constants which need to be determined, and ϕ
(
z
) satisfies equation ϕ
′
(
z
)
=
exp
⁡
(
-
ϕ
(
z
)
)
+
μ
exp
⁡
(
ϕ
(
z
)
)
+
δ, whereas δ and μ are arbitrary constants.From (67), we insert u, u
3, u
′
′ into (2) and sort out the coefficients of exp
⁡
(
-
ϕ
(
z
)
) to zero, and then we obtain (68)
c
1
=
±
-
6
β
k
2
τ
,
c
0
=
±
1
2
λ
-
6
β
k
2
τ
,in (68), λ
=
±
2
2
β
k
3
μ
-
ϖ
/
β
k
3 and d
=
0.By putting (68) into (67), we get(69)
u
1
z
=
1
2
λ
-
6
β
k
2
τ
+
-
6
β
k
2
τ
exp
⁡
-
ϕ
z
,or (70)
u
2
z
=
-
1
2
λ
-
6
β
k
2
τ
-
-
6
β
k
2
τ
exp
⁡
-
ϕ
z
.We apply (43)–(49) into (69) and (70), respectively, to obtain traveling wave solutions of the mKdV equation as follows.Ifδ
2
-
4
μ
>
0, μ
≠
0, (71)
u
11
z
=
1
2
λ
-
6
β
k
2
τ
+
-
6
β
k
2
τ
2
μ
δ
2
-
4
μ
tanh
⁡
δ
2
-
4
μ
/
2
z
+
c
+
δ
,
(72)
u
12
z
=
1
2
λ
-
6
β
k
2
τ
+
-
6
β
k
2
τ
2
μ
δ
2
-
4
μ
coth
⁡
δ
2
-
4
μ
/
2
z
+
c
+
δ
,
(73)
u
21
z
=
-
1
2
λ
-
6
β
k
2
τ
-
-
6
β
k
2
τ
2
μ
δ
2
-
4
μ
tanh
⁡
δ
2
-
4
μ
/
2
z
+
c
+
δ
,
(74)
u
22
z
-
1
2
λ
-
6
β
k
2
τ
-
-
6
β
k
2
τ
2
μ
δ
2
-
4
μ
coth
⁡
δ
2
-
4
μ
/
2
z
+
c
+
δ
.Ifδ
2
-
4
μ
<
0, μ
≠
0, (75)
u
13
z
=
1
2
λ
-
6
β
k
2
τ
+
-
6
β
k
2
τ
2
μ
δ
2
-
4
μ
tanh
⁡
δ
2
-
4
μ
/
2
z
+
c
+
δ
,
(76)
u
14
z
=
1
2
λ
-
6
β
k
2
τ
+
-
6
β
k
2
τ
2
μ
δ
2
-
4
μ
coth
⁡
δ
2
-
4
μ
/
2
z
+
c
+
δ
,
(77)
u
23
z
=
-
1
2
λ
-
6
β
k
2
τ
-
-
6
β
k
2
τ
2
μ
δ
2
-
4
μ
tanh
⁡
δ
2
-
4
μ
/
2
z
+
c
+
δ
,
(78)
u
24
z
=
-
1
2
λ
-
6
β
k
2
τ
-
-
6
β
k
2
τ
2
μ
δ
2
-
4
μ
coth
⁡
δ
2
-
4
μ
/
2
z
+
c
+
δ
.Ifδ
2
-
4
μ
>
0, μ
=
0
,
δ
≠
0, (79)
u
15
z
=
1
2
λ
-
6
β
k
2
τ
+
-
6
β
k
2
τ
δ
exp
⁡
δ
z
+
c
-
1
,
(80)
u
25
z
=
-
1
2
λ
-
6
β
k
2
τ
-
-
6
β
k
2
τ
δ
exp
⁡
δ
z
+
c
-
1
.Ifδ
2
-
4
μ
=
0, μ
≠
0
,
δ
≠
0, (81)
u
16
z
=
1
2
λ
-
6
β
k
2
τ
+
-
6
β
k
2
τ
δ
2
z
+
c
2
δ
z
+
c
+
2
,
(82)
u
26
z
=
-
1
2
λ
-
6
β
k
2
τ
-
-
6
β
k
2
τ
δ
2
z
+
c
2
δ
z
+
c
+
2
.Ifδ
2
-
4
μ
=
0, μ
=
0
,
δ
=
0, (83)
u
17
z
=
1
2
λ
-
6
β
k
2
τ
+
-
6
β
k
2
τ
1
z
+
c
,
(84)
u
27
z
=
-
1
2
λ
-
6
β
k
2
τ
-
-
6
β
k
2
τ
1
z
+
c
.


## 4. Computer Simulations

In this section, we will show some computer simulation pictures to illustrate some results. Consideringk
=
1, l
=
1, r
=
1, the simply periodic solutions are shown in Figure 1, and the rational function solutions w
r
,
1 are shown in Figure 2. And through the exp(-ϕ
(
z
))-expansion method, we get some other simply periodic solutions. We take the solutions u
s
,
11
(
z
) to further analyze their properties by Figure 3.Figure 1
The solution of mKdV equation corresponding tow
s
,
1, take α
=
2, -
6
β
k
2
/
τ
=
1, and z
0
=
1, z
1
=
2; from left to right, t take the following three different values: (a) t
=
-
5, (b) t
=
0, and (c) t
=
5.Figure 2
The solution of mKdV equation corresponding tow
r
,
1, take 6
β
k
2
/
τ
=
1; from left to right, t take the following three different values: (a) t
=
-
5, (b) t
=
0, and (c) t
=
5.Figure 3
The solution of mKdV equation corresponding tou
s
,
11, take λ
=
2, β
k
2
/
τ
=
-
1
/
6, μ
=
2, and δ
2
-
μ
=
2; from left to right, t take the following three different values: (a) t
=
-
5, (b) t
=
0, and (c) t
=
5.


### 4.1. The Physical Significance of the Figures

Figures1, 2, and 3 shows Waveform Graphs of several functions at different times. We can see that there will be some distinct generation poles shown in Figures.


## 4.1. The Physical Significance of the Figures

Figures1, 2, and 3 shows Waveform Graphs of several functions at different times. We can see that there will be some distinct generation poles shown in Figures.


## 5. Conclusions

It can be seen from the above analysis that the complex method and exp(
-
ϕ
(
z
)
)-expansion method are powerful tools for solving the exact solutions of nonlinear evolution equations. The general meromorphic solutions of (2+1)-dimensional mKdV equation are obtained by the complex method, and we found eight solutions of (2+1)-dimensional mKdV equation. Using exp(
-
ϕ
(
z
)
)-expansion method, we also find fourteen solutions of (2+1)-dimensional mKdV equation. By comparing with the two methods, we find more solutions by exp(
-
ϕ
(
z
)
)-expansion method, while we can say that the solutions of the elliptic function are only obtained by the complex method.



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*Source: 3053275-2019-05-16.xml*