# Construction of New Exact Solutions for the (3 + 1)-Dimensional Burgers System

Authors: Zitian Li
Journal: Mathematical Problems in Engineering
 (2014)

---

## Abstract

By means of a variable separation method and a generalized direct ansätz function approach, new exact solutions including cross kink-wave solutions, doubly periodic kinky-wave solutions, and breather type of two-solitary wave solutions for the (3 + 1)-dimensional Burgers system are obtained. Moreover, the mechanical features are also investigated.


---

## Body

## 1. Introduction

Searching for explicit solutions of nonlinear evolution equations by using various different methods is useful and meaningful in physical science and nonlinear science. Many powerful methods have been presented, such as inverse scattering transform [1], Hirota’s bilinear form method [2], two-soliton method [3], homoclinic test technique [4], Bäcklund transformation method [5], and three-wave type of ansätz approach [6]. And much work has been focused on the various extensions and application of the known algebraic methods to construct the solutions of nonlinear evolution equations [7, 8].In this letter, we will present exact solutions of the (3 + 1)-dimensional Burgers system [9]:
(1)
u
t
=
2
u
u
y
+
2
v
u
x
+
2
w
u
z
+
u
x
x
+
u
y
y
+
u
z
z
,
u
x
=
v
y
,
u
z
=
w
y
,
which was derived from the inverse transformation of heat-conduction equation, where u
(
x
,
y
,
z
,
t
), v
(
x
,
y
,
z
,
t
), w
(
x
,
y
,
z
,
t
) are three physical field function and u
:
R
x
×
R
y
×
R
z
×
R
t
+
→
R. Recently, (1) was widely researched [10–12]. When z
=
x, w
=
u, (1) degenerated into the (
2
+
1
)-dimensional Burgers equation and the soliton-like solutions were obtained by using the extended Riccati equation mapping method [13], and many other results were found in the literature [14, 15]. In our work, we will give some new exact solutions for (1) by using the variable separation hypothesis and the direct ansätz function method all together.


## 2. Exact Solutions to the (3 + 1)-Dimensional Burgers System

We firstly suppose the solution of system (1) has the following ansätz:
(2)
u
=
u
0
+
(
ln
⁡
ϕ
)
y
,
v
=
v
0
+
(
ln
⁡
ϕ
)
x
,
w
=
w
0
+
(
ln
⁡
ϕ
)
z
,
where u
0
,
v
0, and w
0 are real constants and ϕ
=
ϕ
(
x
,
y
,
z
,
t
) has the following variable separation form:
(3)
ϕ
(
x
,
y
,
z
,
t
)
=
a
0
+
a
1
f
(
y
,
t
)
+
a
2
g
(
x
,
z
)
,
with a
i  (
i
=
0,1
,
2
) constants and f
(
y
,
t
)
,
g
(
x
,
z
) are unknown function about {
y
,
t
} and {
x
,
z
}, respectively. Substituting (2) and (3) into (1), we have
(4)
a
0
(
f
y
t
-
2
u
0
f
y
y
-
f
y
y
y
)
+
a
1
[
f
(
f
y
t
-
2
u
0
f
y
y
-
f
y
y
y
)
+
+
a
1
-
f
y
(
f
t
-
2
u
0
f
y
-
f
y
y
)
]
=
0
,
(5)
f
y
(
g
x
x
+
g
z
z
)
+
2
f
y
(
v
0
g
x
+
w
0
g
z
)
+
g
(
f
y
t
-
2
u
0
f
y
y
-
f
y
y
y
)
=
0
.Now let us consider (5), since g
(
x
,
z
) is only a function about variables x and z, letting
(6)
f
y
t
-
2
u
0
f
y
y
-
f
y
y
y
f
y
=
C
(
constant
)
,
and by using the wave transformation ξ
=
a
x
+
b
z
+
ξ
0, (5) can be changed into
(7)
g
′′
(
ξ
)
+
2
(
a
v
0
+
b
w
0
)
a
2
+
b
2
g
′
(
ξ
)
+
C
a
2
+
b
2
g
(
ξ
)
=
0
,
where a
,
b, and ξ
0 are arbitrary constants. And when a
0 and C are taken as a
0
=
0, C
≠
0 or a
0
≠
0, C
=
0, we find that (4) is automatically satisfied.Whena
v
0
+
b
w
0
=
0, solutions of (7) are well known as follows:
(8)
(
i
)
g
(
x
,
z
)
=
c
1
cos
⁡
[
τ
a
2
+
b
2
(
a
x
+
b
z
+
ξ
0
)
]
(
i
)
g
(
x
,
z
)
=
+
c
2
sin
[
τ
a
2
+
b
2
(
a
x
+
b
z
+
ξ
0
)
]
,
C
>
0
,
(
ii
)
g
(
x
,
z
)
=
c
1
(
a
x
+
b
z
+
ξ
0
)
+
c
2
,
C
=
0
,
(
iii
)
g
(
x
,
z
)
=
c
1
cosh
⁡
[
-
τ
a
2
+
b
2
(
a
x
+
b
z
+
ξ
0
)
]
(
i
i
i
)
g
(
x
,
z
)
=
+
c
2
sinh
⁡
[
-
τ
a
2
+
b
2
(
a
x
+
b
z
+
ξ
0
)
]
,
(
i
i
i
)
g
(
x
,
z
)
=
sin
[
-
τ
a
2
+
b
2
(
a
x
+
b
z
+
ξ
0
)
]
C
<
0
,
where c
1
,
c
2 are arbitrary constants.Therefore, we only need to solve (6) from which we can obtain the solutions of system (1).Example 1.
Let the test function be(9)
f
(
y
,
t
)
=
b
1
e
k
y
+
l
t
+
b
0
+
b
2
e
-
(
k
y
+
l
t
)
,
where k, l, b
0, b
1, and b
2 are constants to be determined.
By introducing means of computer algebra (i.e., Maple), we obtain(10)
u
1
=
u
0
+
(
(
2
k
a
1
b
1
b
2
sinh
⁡
⁡
(
k
y
+
2
k
u
0
t
+
r
)
)
b
1
b
2
)
-
1
)
u
1
+
+
=
u
0
×
(
a
1
b
0
+
2
a
1
b
1
b
2
cosh
⁡
⁡
(
k
y
+
2
k
u
0
t
+
r
)
b
1
b
2
u
1
+
+
=
×
i
i
u
0
+
a
2
[
c
1
cosh
⁡
B
(
a
x
+
b
z
+
ξ
0
)
u
1
+
+
=
×
i
i
u
0
+
+
a
2
+
c
2
sinh
⁡
B
(
a
x
+
b
z
+
ξ
0
)
]
b
1
b
2
)
-
1
)
,
v
1
=
v
0
+
(
(
a
a
2
B
[
c
1
sinh
⁡
B
(
a
x
+
b
z
+
ξ
0
)
b
1
b
2
)
-
1
)
k
k
k
k
k
k
k
k
k
k
k
k
k
k
+
c
2
cosh
⁡
B
(
a
x
+
b
z
+
ξ
0
)
]
)
v
1
=
+
v
0
+
×
(
a
1
b
0
+
2
a
1
b
1
b
2
cosh
⁡
(
k
y
+
2
k
u
0
t
+
r
)
v
1
=
+
+
+
v
0
+
+
a
2
[
c
1
cosh
⁡
B
(
a
x
+
b
z
+
ξ
0
)
v
1
=
+
+
+
+
i
+
v
0
+
b
1
b
2
+
c
2
sinh
⁡
B
(
a
x
+
b
z
+
ξ
0
)
]
)
-
1
)
,
w
1
=
w
0
+
(
(
b
a
2
B
[
c
1
sinh
⁡
B
(
a
x
+
b
z
+
ξ
0
)
c
2
sinh
⁡
B
(
a
x
+
b
z
+
ξ
0
)
]
b
1
b
2
)
-
1
)
k
k
k
k
k
k
k
k
k
k
k
k
k
k
+
c
2
cosh
⁡
B
(
a
x
+
b
z
+
ξ
0
)
]
)
w
1
=
w
0
+
+
×
(
a
1
b
0
+
2
a
1
b
1
b
2
cosh
⁡
(
k
y
+
2
k
u
0
t
+
r
)
b
1
b
2
kkkkkkkkkkkkkkkk
+
a
2
[
c
1
cosh
⁡
B
(
a
x
+
b
z
+
ξ
0
)
kkkkkkkkkkkkkkkkkkkk
+
c
2
sinh
⁡
B
(
a
x
+
b
z
+
ξ
0
)
]
b
1
b
2
)
-
1
)
,
which is a cross kink-wave solution, where r
=
ln
⁡
⁡
|
b
1
/
b
2
|, B
=
k
2
/
(
a
2
+
b
2
) (see Figure 1: F
1).F
1: mechanical feature of u
1: (
a
1
=
a
2
=
1
,
b
1
=
b
2
=
1
,
k
=
u
0
=
1
,
c
1
=
c
2
=
B
=
1
,
t
=
0
.
), F
6: mechanical feature of u
2
:
(
a
1
=
a
2
=
1
,
h
0
=
h
1
=
1
,
c
1
=
c
2
=
1
,
u
0
=
A
=
1
,
k
=
0.1
,
t
=
1
).
(a)
(b)Example 2.
Similarly, iff
(
y
,
t
) was taken as
(11)
f
(
y
,
t
)
=
(
h
1
e
k
y
+
l
t
+
h
0
+
h
2
e
-
(
k
y
+
l
t
)
)
2
,
where k
,
l
,
h
0
,
h
1, and h
2 are constants to be determined later, then, we derived a periodic soliton solution and a multiple-soliton solution as follows:
(12)
u
2
=
u
0
+
(
2
a
1
h
1
k
[
h
1
e
k
y
+
(
2
k
u
0
+
3
k
2
)
t
+
h
0
]
e
k
y
+
(
2
k
u
0
+
3
k
2
)
t
kkkkkkkkkkkkkkkkk
+
c
2
sin
[
A
(
a
x
+
b
z
+
ξ
0
)
]
)
[
h
1
e
k
y
+
(
2
k
u
0
+
3
k
2
)
t
+
h
0
]
2
)
-
1
)
u
2
=
u
0
+
+
×
(
a
1
[
h
1
e
k
y
+
(
2
k
u
0
+
3
k
2
)
t
+
h
0
]
2
u
2
=
u
0
+
+
×
+
a
2
(
c
1
cos
⁡
⁡
[
A
(
a
x
+
b
z
+
ξ
0
)
]
2
a
1
h
1
k
[
h
1
e
k
y
+
(
2
k
u
0
+
3
k
2
)
t
+
h
0
]
e
k
y
+
(
2
k
u
0
+
3
k
2
)
t
kkkkkkkkkkkkkkkkk
+
c
2
sin
[
A
(
a
x
+
b
z
+
ξ
0
)
]
)
[
h
1
e
k
y
+
(
2
k
u
0
+
3
k
2
)
t
+
h
0
]
2
)
-
1
)
,
v
2
=
v
0
+
(
A
a
a
2
[
-
c
1
sin
[
A
(
a
x
+
b
z
+
ξ
0
)
]
(
a
1
[
h
1
e
k
y
+
(
2
k
u
0
+
3
k
2
)
t
+
h
0
]
2
)
-
1
)
k
k
k
k
k
k
k
k
k
k
+
c
2
cos
⁡
[
A
(
a
x
+
b
z
+
ξ
0
)
]
]
×
(
a
1
[
h
1
e
k
y
+
(
2
k
u
0
+
3
k
2
)
t
+
h
0
]
2
+
+
+
+
+
a
2
(
c
1
cos
⁡
[
A
(
a
x
+
b
z
+
ξ
0
)
]
+
+
+
+
+
+
a
2
+
c
2
sin
⁡
[
A
(
a
x
+
b
z
+
ξ
0
)
]
)
(
a
1
[
h
1
e
k
y
+
(
2
k
u
0
+
3
k
2
)
t
+
h
0
]
2
)
-
1
)
,
w
2
=
w
0
+
(
A
b
a
2
[
-
c
1
sin
[
A
(
a
x
+
b
z
+
ξ
0
)
]
[
h
1
e
k
y
+
(
2
k
u
0
+
3
k
2
)
t
+
h
0
]
2
)
-
1
k
k
k
k
k
k
k
k
k
+
c
2
cos
⁡
[
A
(
a
x
+
b
z
+
ξ
0
)
]
]
×
(
a
1
[
h
1
e
k
y
+
(
2
k
u
0
+
3
k
2
)
t
+
h
0
]
2
kk
+
kkkk
+
a
2
(
kkkkkkk
+
a
2
[
h
1
e
k
y
+
(
2
k
u
0
+
3
k
2
)
t
+
h
0
]
2
+
c
2
sin
⁡
[
A
(
a
x
+
b
z
+
ξ
0
)
]
)
c
1
cos
⁡
[
A
(
a
x
+
b
z
+
ξ
0
)
]
kkkk
+
kkk
+
a
2
[
h
1
e
k
y
+
(
2
k
u
0
+
3
k
2
)
t
+
h
0
]
2
+
c
2
sin
⁡
[
A
(
a
x
+
b
z
+
ξ
0
)
]
)
)
-
1
)
,
u
3
=
u
0
+
(
4
k
h
1
h
2
sinh
⁡
2
[
(
k
y
+
2
k
u
0
t
)
+
ln
⁡
h
1
h
2
]
[
(
k
y
+
2
k
u
0
t
)
+
ln
⁡
h
1
h
2
]
)
-
1
×
(
4
a
1
h
1
h
2
cosh
⁡
2
[
(
k
y
+
2
k
u
0
t
)
+
ln
⁡
h
1
h
2
]
[
(
k
y
+
2
k
u
0
t
)
+
ln
⁡
h
1
h
2
]
)
-
1
kkkkkk
+
a
2
(
c
1
cosh
⁡
⁡
[
B
(
a
x
+
b
z
+
ξ
0
)
]
k
k
k
k
k
k
k
k
k
k
k
[
(
k
y
+
2
k
u
0
t
)
+
ln
⁡
h
1
h
2
]
+
c
2
sinh
⁡
⁡
[
B
(
a
x
+
b
z
+
ξ
0
)
]
)
4
a
1
h
1
h
2
cosh
⁡
2
[
(
k
y
+
2
k
u
0
t
)
+
ln
⁡
h
1
h
2
]
)
-
1
)
,
v
3
=
v
0
+
(
B
a
a
2
(
c
1
sinh
⁡
⁡
[
B
(
a
x
+
b
z
+
ξ
0
)
]
[
(
k
y
+
2
k
u
0
t
)
+
ln
⁡
h
1
h
2
]
)
-
1
kkkkkkkkkkk
+
c
2
cosh
⁡
⁡
[
B
(
a
x
+
b
z
+
ξ
0
)
]
)
×
(
4
a
1
h
1
h
2
cosh
⁡
2
[
(
k
y
+
2
k
u
0
t
)
+
ln
⁡
h
1
h
2
]
+
c
2
sinh
⁡
⁡
[
B
(
a
x
+
b
z
+
ξ
0
)
]
)
)
-
1
kkkkkk
+
a
2
(
c
1
cosh
⁡
⁡
[
B
(
a
x
+
b
z
+
ξ
0
)
]
k
k
k
k
k
k
k
k
k
k
[
(
k
y
+
2
k
u
0
t
)
+
ln
⁡
h
1
h
2
]
+
c
2
sinh
⁡
⁡
[
B
(
a
x
+
b
z
+
ξ
0
)
]
)
)
-
1
)
,
w
3
=
w
0
+
(
B
b
a
2
(
c
1
sinh
⁡
[
B
(
a
x
+
b
z
+
ξ
0
)
]
[
(
k
y
+
2
k
u
0
t
)
+
ln
⁡
h
1
h
2
]
)
-
1
k
k
k
k
k
k
k
k
k
k
+
c
2
cosh
⁡
[
B
(
a
x
+
b
z
+
ξ
0
)
]
)
×
(
4
a
1
h
1
h
2
cosh
⁡
2
[
(
k
y
+
2
k
u
0
t
)
+
ln
⁡
h
1
h
2
]
[
(
k
y
+
2
k
u
0
t
)
+
ln
⁡
h
1
h
2
]
)
-
1
kkkkkk
+
a
2
(
c
1
cosh
⁡
[
B
(
a
x
+
b
z
+
ξ
0
)
]
kkkkkk
+
a
2
+
c
2
sinh
⁡
⁡
[
B
(
a
x
+
b
z
+
ξ
0
)
]
)
[
(
k
y
+
2
k
u
0
t
)
+
ln
⁡
h
1
h
2
]
)
-
1
)
,
where A
=
2
k
2
/
(
a
2
+
b
2
) and B
=
4
k
2
/
(
a
2
+
b
2
) (see Figure 1: F
6, Figure 2: F
7).F
7: mechanical feature of u
3: (
a
0
=
1
,
a
2
=
2
,
u
0
=
1
,
c
1
=
2
,
c
2
=
1
,
B
=
2
,
k
=
1.5
,
t
=
2.), F
8: mechanical feature of u
4
:
(
k
1
=
a
1
=
c
2
=
1
,
a
2
=
k
2
=
h
1
=
c
1
=
H
0
=
u
0
=
r
=
2
,
B
=
L
=
3
,
t
=
1).
(a)
(b)Example 3.
Furthermore, when choosingf
(
y
,
t
) to be
(13)
f
(
y
,
t
)
=
L
cos
⁡
[
k
1
y
+
l
1
t
]
+
e
k
2
x
+
l
2
t
+
H
0
,
with L
,
H
0
,
k
1
,
k
2
,
l
1, and l
2 constants, we obtained a breather type of two-solitary wave solution as follows:
(14)
u
4
=
u
0
+
(
(
a
1
k
2
[
sinh
⁡
⁡
(
ξ
1
)
+
cosh
⁡
⁡
(
ξ
1
)
]
-
a
1
k
1
L
sin
(
ξ
2
)
)
k
k
k
k
k
k
+
a
2
[
c
1
cos
⁡
⁡
(
ξ
3
)
+
c
2
sin
⁡
(
ξ
3
)
]
)
-
1
×
(
a
1
H
0
+
a
1
[
cosh
⁡
⁡
(
ξ
1
)
+
sinh
⁡
⁡
(
ξ
1
)
]
+
a
1
L
cos
⁡
(
ξ
2
)
k
k
k
k
k
k
+
a
2
[
c
1
cos
⁡
⁡
(
ξ
3
)
+
c
2
sin
⁡
(
ξ
3
)
]
)
-
1
)
,
v
4
=
v
0
-
(
a
2
a
B
[
c
1
sin
⁡
[
B
(
a
x
+
b
z
+
ξ
0
)
]
+
a
2
[
c
1
cos
⁡
⁡
(
ξ
3
)
+
c
2
sin
⁡
(
ξ
3
)
]
)
-
1
k
k
k
k
k
k
k
k
k
-
c
2
cos
⁡
⁡
[
B
(
a
x
+
b
z
+
ξ
0
)
]
]
×
(
a
1
H
0
+
a
1
[
cosh
⁡
⁡
(
ξ
1
)
+
sinh
⁡
⁡
(
ξ
1
)
]
+
a
1
L
cos
⁡
(
ξ
2
)
k
k
k
k
k
k
+
a
2
[
c
1
cos
⁡
⁡
(
ξ
3
)
+
c
2
sin
⁡
(
ξ
3
)
]
)
-
1
)
,
w
4
=
w
0
-
(
[
c
1
cos
⁡
⁡
(
ξ
3
)
+
c
2
sin
⁡
(
ξ
3
)
]
)
-
1
a
2
b
B
[
c
1
sin
[
B
(
a
x
+
b
z
+
ξ
0
)
]
k
k
k
k
k
k
k
k
-
c
2
cos
⁡
[
B
(
a
x
+
b
z
+
ξ
0
)
]
]
×
(
a
1
H
0
+
a
1
[
cosh
⁡
(
ξ
1
)
+
sinh
⁡
(
ξ
1
)
]
+
a
1
L
cos
⁡
(
ξ
2
)
kkkkkk
+
a
2
[
c
1
cos
⁡
⁡
(
ξ
3
)
+
c
2
sin
⁡
(
ξ
3
)
]
a
1
H
0
+
a
1
[
cosh
⁡
(
ξ
1
)
+
sinh
⁡
(
ξ
1
)
]
+
a
1
L
cos
⁡
(
ξ
2
)
)
-
1
)
,
where ξ
1
=
k
2
y
+
(
k
1
2
+
k
2
2
+
2
k
2
u
0
)
t, ξ
2
=
k
1
y
+
2
k
1
u
0
t, ξ
3
=
B
(
a
x
+
b
z
+
ξ
0
).The solution represented by (14) is breather type of two-solitary wave solution which contains a periodic wave and two solitary waves, whose amplitude periodically oscillates with the evolution of time (see Figure 2: F
8).


## 3. Conclusion

In this paper, by using the variable separation method and the generalized direct ansätz method the(
3
+
1
)-dimensional Burgers system is investigated. New exact solutions including cross kink-wave solution, doubly kinky-wave solution, and breather type of two-solitary wave solutions are obtained; these solutions enrich the structures of solutions of the (
3
+
1
)-dimensional Burgers system. Moreover, the mechanical features are also investigated.



---

*Source: 390742-2014-10-12.xml*