Hindawi Publishing Corporation
Abstract and Applied Analysis
Volume 2013, Article ID 696074, 7 pages
http://dx.doi.org/10.1155/2013/696074
Research Article
Saddle-Node Heteroclinic Orbit and Exact Nontraveling
Wave Solutions for (2+1)D KdV-Burgers Equation
Da-Quan Xian
School of Sciences, Southwest University of Science and Technology, Mianyang 621010, China
Correspondence should be addressed to Da-Quan Xian; daquanxian@163.com
Received 21 September 2012; Accepted 8 December 2012
Academic Editor: de Dai
Copyright © 2013 Da-Quan Xian. This is an open access article distributed under the Creative Commons Attribution License,
which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
We have undertaken the fact that the periodic solution of (2+1)D KdV-Burgers equation does not exist. The Saddle-node heteroclinic
orbit has been obtained. Using the Lie group method, we get two-(1+1)-dimensional PDE, through symmetric reduction; and by the
direct integral method, spread F-expansion method, and (𝐺󸀠 /𝐺)-expansion method, we obtain exact nontraveling wave solutions,
for the (2+1)D KdV Burgers equation, and find out some new strange phenomenons of sympathetic vibration to evolution of
nontraveling wave.
1. Introduction
We consider the (2+1)-dimensional Korteweg-de Vries Burgers ((2+1)D KdV Burgers) equation
(𝑢𝑡 + 𝑢𝑢𝑥 − 𝛽𝑢𝑥𝑥 + 𝛼𝑢𝑥𝑥𝑥 )𝑥 + 𝛾𝑢𝑦𝑦 = 0,
(1)
where 𝑢 : 𝑅𝑥 × 𝑅𝑦 × 𝑅𝑡+ → 𝑅, 𝛼, 𝛽, and 𝛾 are real parameters.
Equation (1) is model equation for wide class of nonlinear
wave models in an elastic tube, liquid with small bubbles,
and turbulence [1–3]. Much attention has been put on the
study of their exact solutions by some methods [4], such
as, a complex line soliton by extended tanh method with
symbolic computation [5], exact traveling wave solutions
including solitary wave solutions, periodic wave and shock
wave solutions by extended mapping method, and homotopy
perturbation method [6, 7].
It is well known that the investigation of exact solutions
of nonlinear evolution equations plays an important role in
the study of nonlinear physical phenomena. Many effective
methods have been presented [7–22], such as functional
variable separation method [8, 9], homotopy perturbation
method [12], F-expansion method [7, 13], Lie group method
[14, 15], variational iteration method [16], homoclinic test
method [17–19], Exp-function method [20, 21], and homogeneous balance method [22]. Practically, there is no unified
method that can be used to handle all types of nonlinearity.
In this paper, we will discuss the existence of periodic
traveling wave solution and seek the Saddle-Node heteroclinic orbit, and further use the Lie group method with the
aid of the symbolic computation system Maple to construct
the non-traveling wave solutions for (1).
2. Existence of Periodic Traveling
Wave Solution of (1)
Introducing traveling wave transformation in this form
𝑢 (𝑥, 𝑦, 𝑡) = 𝑢 (𝜉) , 𝜉 = 𝑝𝑥 + 𝑞𝑦 − 𝑐𝑡
(2)
permits us to convert (1) into an ODE for 𝑢 = 𝑢(𝜉)
𝑝(𝑝𝑢𝑢𝜉 − 𝛽𝑝2 𝑢𝜉𝜉 + 𝛼𝑝3 𝑢𝜉𝜉𝜉 )𝜉 − 𝑟𝑢𝜉𝜉 = 0,
(3)
where 𝑟 = 𝑝𝑐−𝑞2 𝛾, Integrating (3) with respect to 𝜉 twice and
taking integration constant to 𝐴 yields
2𝛼𝑝4 𝑢𝜉𝜉 − 2𝛽𝑝3 𝑢𝜉 + 𝑝2 𝑢2 − 2𝑟𝑢 = 𝐴.
(4)
Letting 𝑢𝜉 = 𝑣, thus nonlinear ordinary differential equation
(4) is equivalent to the autonomous dynamic system as follows:
𝑑𝑢
= 𝑣,
(5)
𝑑𝜉
𝑑𝑣
1
(2𝛽𝑝3 𝑣 − 𝑝2 𝑢2 + 2𝑟𝑢 + 𝐴) .
=
(6)
𝑑𝜉 2𝛼𝑝4
2
Abstract and Applied Analysis
The dynamic system (5) has two balance points:
𝑃1 (𝑢1 , 𝑣1 ) = (
𝑟 + √𝑟2 + 𝑝2 𝐴
𝑝2
Consequently, the system expressed in (12) is not a conservative one, then periodic traveling wave solution of (1) does not
exist.
We conclude the above analysis in the following theorem.
, 0) ,
(7)
𝑃2 (𝑢2 , 𝑣2 ) = (
𝑟−
√𝑟2
+
𝑝2 𝐴
𝑝2
, 0) .
The Jacobi matrixes at the balance points for the right-hand
side of (5) are obtained as follows, respectively:
0
1
),
𝐽1 = ( √𝑟2 + 𝑝2 𝐴
𝛽
−
𝑝4 𝛼
𝑝𝛼
0
𝑝4 𝛼
𝛽
𝑝𝛼
3. Saddle-Node Heteroclinic Orbits of
KdV-Burgers Equation
First, we assume the solutions of (4) in the form
(8)
1
𝐽2 = ( √𝑟2 + 𝑝2 𝐴
Theorem 1. Under the traveling wave transformation, the
periodic solution of (2+1)-dimensional KdV-Burgers equation
does not exist.
But, saddle-node heteroclinic orbits and nontraveling periodic solution do exist, which will be discussed later in this paper.
).
𝑢 (𝜉) =
𝑟 + √𝑟2 + 𝑝2 𝐴
𝑝2
+
𝑏
2
(1 + 𝑒𝑎𝜉 )
.
(14)
Substituting (14) into (4) yields
2 (4𝛼𝑝4 𝑎2 + √𝑟2 + 𝑝2 𝐴 + 2𝛽𝑝3 𝑎) 𝑒2𝑎𝜉
Their latent equations are expressed, respectively, as,
𝑝3 𝜆 (𝑝𝛼𝜆 − 𝛽) + √𝑟2 + 𝑝2 𝐴 = 0,
− 4 (𝛼𝑝4 𝑎2 − √𝑟2 + 𝑝2 𝐴 − 𝛽𝑝3 𝑎) 𝑒𝑎𝜉
(9)
𝑝3 𝜆 (𝑝𝛼𝜆 − 𝛽) − √𝑟2 + 𝑝2 𝐴 = 0.
(15)
+ 2√𝑟2 + 𝑝2 𝐴 + 𝑝2 𝑏 = 0.
Relevant latent roots are as follows respectively:
Then we get
𝜆1 =
𝜆2 =
𝑝𝛽 ±
√ 𝑝2 𝛽2
−
4𝛼√𝑟2
+
𝑝2 𝐴
2𝑝2 𝛼
4𝛼𝑝4 𝑎2 + √𝑟2 + 𝑝2 𝐴 + 2𝛽𝑝3 𝑎 = 0,
,
(10)
𝑝𝛽 ± √ 𝑝2 𝛽2 + 4𝛼√𝑟2 + 𝑝2 𝐴
2𝑝2 𝛼
.
𝑝2 𝛽2 < 4𝛼√𝑟2 + 𝑝2 𝐴, then 𝜆 1 are conjugate complex roots
and real part is positive, so 𝑃1 is a nonsteady focus point. And
𝜆 2 is a positive and minus real root, thus 𝑃2 is a saddle point.
From (5), we know the phase trajectory on the phase plane
satisfies
Solving the system (16) gets
𝑎=−
𝛽
,
5𝛼𝑝
12𝛽2
,
25𝛼
√𝑟2 + 𝑝2 𝐴 =
6𝑝2 𝛽2
.
25𝛼
(17)
Substituting (17) into (14) obtains
(11)
𝑢 (𝜉) =
𝑟 + √𝑟2 + 𝑝2 𝐴
Integrating (11), we can obtain
1
𝐻 (𝑢, 𝑣) = 𝐴𝑢 + 𝑟𝑢2 − 𝑝2 𝑢3 + 2𝛽𝑝3 𝑢𝑣 − 𝛼𝑝4 𝑣2 ,
3
(12)
where 𝐻(𝑢, 𝑣) is a total energy or Hamiliton function of system (4). Apparently
𝜕𝐻
,
𝜕𝑣
𝑏=−
2 2
𝑑𝑣 2𝛽𝑝 𝑣 − 𝑝 𝑢 + 2𝑟𝑢 + 𝐴
=
.
𝑑𝑢
2𝛼𝑝4 𝑣
𝑢𝜉 ≠ −
(16)
2√𝑟2 + 𝑝2 𝐴 + 𝑝2 𝑏 = 0.
Obviously, if 𝑝2 𝛽2 > 4𝛼√𝑟2 + 𝑝2 𝐴, then 𝜆 1 are two positive
real roots, therefore 𝑃1 is a nonsteady node point. If 0 <
3
𝛼𝑝4 𝑎2 − √𝑟2 + 𝑝2 𝐴 − 𝛽𝑝3 𝑎 = 0,
𝑣𝜉 ≠
𝜕𝐻
.
𝜕𝑢
(13)
= 𝑢1 −
𝑝2
2
−
12𝛽2
1
25𝛼 (1 + 𝑒−(𝛽/5𝛼𝑝)𝜉 )2
(18)
2
3𝛽
𝛽
(1 + tanh
𝜉) .
25𝛼
20𝛼
Evidently, 𝜉 → −∞ ⇒ 𝑢(𝜉) → 𝑢1 , 𝜉 → +∞ ⇒ 𝑢(𝜉) →
𝑢1 − (6𝛽2 /25𝛼) = 𝑢2 . Thus (18) is a saddle-node heteroclinic
orbit through nonsteady node point 𝑃1 and saddle point 𝑃2
[23].
Abstract and Applied Analysis
3
Ecumenic, taking the Hamiliton function 𝐻(𝑢, 𝑣) = 𝐵, we
obtain
𝑑𝑢
=𝑣
𝑑𝜉
=
3𝑝𝛽𝑢 ± √3𝑢 [3𝐴𝛼 + 3 (𝑝2 𝛽2 + 𝑟𝛼) 𝑢 − 𝑝2 𝛼𝑢2 ] − 9𝐵𝛼
3𝛼𝑝2
,
(19)
−300
−200
−100
u
0
100
200
300
where 𝐵 is an arbitrary constant. Integrating (19) with respect
to 𝜉 we have
∫
2
𝑢(𝜉)
3𝛼𝑝
3𝑝𝛽𝑠 ± √3𝑠 [3𝐴𝛼 +
3 (𝑝2 𝛽2
+ 𝑟𝛼) 𝑠 −
𝑝2 𝛼𝑠2 ]
− 9𝐵𝛼
−10
y
𝑑𝑠
= 𝜉 + 𝜉0 ,
(20)
where 𝜉0 is an arbitrary constant. We can see that (4) has
the general solution (20) and all partial cases as include
above result can be found from the general solution of (20).
Example, take 𝛼√𝑟2 + 𝑝2 𝐴 − 𝑝2 𝛽2 = 0, 3𝐵𝛼 + 𝐴𝛽2 = 0,
𝑟𝛼 + 𝑝2 𝛽2 = 0 in (20), we find a solution of (4) as follows:
2
3𝛽2
𝛽
[1 + tanh (
𝑢 (𝜉) = −
𝜉 + 𝜉0 )] .
4𝛼
4𝑝𝛼
−20
𝜎 = 𝜎 (𝑥, 𝑦, 𝑡, 𝑢, 𝑢𝑡 , 𝑢𝑥 , 𝑢𝑦 , . . .) .
(22)
be the Li symmetry of (1). From Lie group theory, 𝜎 satisfies
the following equation
6
8
t
10
12
14
Based on the integrability of reduced equation of symmetry
(26), we are to consider the following three cases.
Case 1. Taking 𝑘2 (𝑡) = 0 and 𝑐 = 0 in (26) yields
𝜎 = 𝑘1 (𝑡) 𝑢𝑥 − 𝑘1󸀠 (𝑡) .
𝑘1󸀠 (𝑡)
𝑥 + 𝐹 (𝑦, 𝑡) ,
𝑘1 (𝑡)
(27)
(23)
We take the function 𝜎 in the form
𝜎 = 𝑎1 𝑢𝑥 + 𝑎2 𝑢𝑦 + 𝑎3 𝑢𝑡 + 𝑎4 𝑢 + 𝑎5 ,
(24)
where 𝑎𝑖 = 𝑎𝑖 (𝑥, 𝑦, 𝑡) : 𝑅𝑥 × 𝑅𝑦 × 𝑅𝑡+ → 𝑅 (𝑖 = 1, . . . , 5)
are functions to be determined later. Substituting (3) into (2)
yields
1 󸀠
𝑘 (𝑡) 𝑦 + 𝑘1 (𝑡) ,
2𝛾 2
𝑎5 =
𝑎2 = 𝑘2 (𝑡) ,
1 󸀠󸀠
𝑘 (𝑡) 𝑦 − 𝑘1󸀠 (𝑡) ,
2𝛾 2
(25)
where 𝑘𝑗 (𝑡) (𝑗 = 1, 2) are arbitrary functions of 𝑡, 𝑐 is an
arbitrary constant. Substituting (25) into (24), we obtain the
Li symmetries of (1) as follows:
1 󸀠
𝑘 (𝑡) 𝑦 + 𝑘1 (𝑡)] 𝑢𝑥 + 𝑘2 (𝑡) 𝑢𝑦
2𝛾 2
1
+ 𝑐𝑢𝑡 + 𝑘2󸀠󸀠 (𝑡) 𝑦 − 𝑘1󸀠 (𝑡) .
2𝛾
(26)
𝐹 (𝑦, 𝑡) : 𝑅𝑦 × 𝑅𝑡+ → 𝑅.
(28)
Substituting (28) into (1) yields the function 𝐹(𝑦, 𝑡) which
satisfies the following linear PDE:
𝑘1󸀠󸀠
𝜕2 𝐹
+ 𝛾 2 = 0.
𝑘1
𝜕𝑦
𝜎𝑥𝑡 + 2𝑢𝑥 𝜎𝑥 + 𝑢𝜎𝑥𝑥 + 𝜎𝑢𝑥𝑥 − 𝛽𝜎𝑥3 + 𝛼𝜎𝑥4 + 𝛾𝜎𝑦𝑦 = 0.
𝜎 = [−
4
5. Symmetry Reduction and Solutions of (1)
𝑢=
This section devotes to Li symmetry of (1) [14, 15]. Let
𝑎4 = 0,
20 0
2
The solution of the differential equation 𝜎 = 0 is
4. Li Symmetry of (1)
𝑎3 = 𝑐,
10
Figure 1: The strange phenomenon which is a sympathetic vibration
of periodicity on the 𝑡-axis and paraboloid on 𝑦-axis for 𝑢1 (𝑥, 𝑦, 𝑡)
as 𝑥 = 1.
(21)
It is a heteroclinic orbit too.
𝑎1 = −
0
(29)
By integrating both sides, we find out the following result:
𝐹 (𝑦, 𝑡) = −
𝑘1󸀠󸀠 2
𝑦 + 𝑘3 (𝑡) 𝑦 + 𝑘4 (𝑡) ,
2𝛾𝑘1
(30)
where 𝑘3 (𝑡), 𝑘4 (𝑡) are new arbitrary functions of 𝑡. Substituting (30) into (28), we can get the solutions of (1) as follows:
𝑢1 (𝑥, 𝑦, 𝑡) =
𝑘1󸀠 (𝑡)
𝑘󸀠󸀠
𝑥 − 1 𝑦2 + 𝑘3 (𝑡) 𝑦 + 𝑘4 (𝑡) .
𝑘1 (𝑡)
2𝛾𝑘1
(31)
(1) Given 𝑘𝑖 (𝑡) = cn(𝑡, 0.95) (𝑖 = 1, 3, 4), 𝑥 = 1, 𝛾 = 0.6 in
(31), the local structure of 𝑢1 is obtained (Figure 1). Where
cn(𝑡, 0.95) is an Jacobian elliptic cosine function.
(2) Given 𝑘1 (𝑡) = sech(𝑡), 𝑘3 (𝑡) = sin(𝑡), 𝑘4 (𝑡) = cn(𝑡,
0.1), 𝑦 = 1, 𝛾 = 0.6 in (31), the local structure of 𝑢1 is obtained
(Figure 2).
Case 2. Take 𝑘1 (𝑡) = 𝑡, 𝑘2 (𝑡) = 1 and 𝑐 = 0 in (26), then
𝜎 = 𝑡𝑢𝑥 + 𝑢𝑦 − 1.
(32)
4
Abstract and Applied Analysis
−8
Solving the system of function equations with the aid of
Maple, we obtain
−6
−4
𝑎0 =
u −2
0
3𝛽2 − 25𝛼
,
25𝛼
𝑎1 =
6𝛽2 𝑞
,
25𝑠𝛼
𝑎2 =
3𝛽2 𝑞
.
25𝛼𝑝
(40)
2
−5
4
x 0
5
14
12
10
8
6
4
when 𝑘 = 𝛽/10𝑠𝛼, 𝑝𝑞 < 0, 𝐴 = (625𝛼2 − 36𝛽4 )/625𝛼2 , where
𝑠 = √−𝑝𝑞.
It is known that solutions of (38) are as follows [24]:
2
t
Figure 2: The periodic solution which is a periodic nontraveling
wave traveling on the 𝑡-axis for 𝑢1 (𝑥, 𝑦, 𝑡) as 𝑦 = 1.
𝜉 = 𝑥 − 𝑡𝑦.
(33)
Substituting (33) into (1) and integrating once with respect to
𝜉 yield
𝐹𝑡 + 𝐹𝐹𝜉 + 𝛾𝑡2 𝐹𝜉 − 𝛽𝐹𝜉𝜉 + 𝛼𝐹𝜉𝜉𝜉 = 0.
=
1
{3𝛽2 − 25𝛼 − 3𝑞𝛽2
25𝛼
(34)
× [ tanh (
(35)
(36)
2
𝑤 = 𝑝 + 𝑞𝑤 .
𝑢3 (𝑥, 𝑦, 𝑡)
=
1
{3𝛽2 − 25𝛼 − 3𝑞𝛽2
25𝛼
× [coth (
𝛽
𝛾
(𝑥 − 𝑡𝑦 + 𝑡 − 𝑡3 ))
10𝑠𝛼
3
(37)
where 𝑎𝑖 (𝑖 = 0, 1, 2) are constants to be determined later,
𝑤(𝜃) satisfies the following auxiliary equation
󸀠
𝛽
𝛾
(𝑥 − 𝑡𝑦 + 𝑡 − 𝑡3 ))]}
10𝑠𝛼
3
+ 𝑦,
where 𝐴 is an integration constant, 𝐹󸀠 = 𝑑𝐹(𝜃)/𝑑𝜃. We
assume that the solution of (36) can be expressed in the form
𝐹 (𝜃) = 𝑎0 + 𝑎1 𝑤 (𝜃) + 𝑎2 𝑤(𝜃)2 ,
𝛽
𝛾
(𝑥 − 𝑡𝑦 + 𝑡 − 𝑡3 ))
10𝑠𝛼
3
− 2𝑝𝑞tanh2 (
Substituting (35) into (34) and integrating once with respect
to 𝜃 yield
2𝑘2 𝛼𝐹󸀠󸀠 − 2𝑘𝛽𝐹󸀠 + 𝐹2 + 2𝐹 + 𝐴 = 0,
(41)
𝑢2 (𝑥, 𝑦, 𝑡)
Again, further using the transformation of dependent variable to (34),
1
𝐹 (𝑡, 𝜉) = 𝐹 (𝜃) , 𝜃 = 𝑘 (𝑡 − 𝛾𝑡3 + 𝜉) .
3
𝑤 (𝜃) = −𝑠 coth (𝑠𝜃) .
Substituting (41), (40), (37), and (35) into (33), we obtain solutions of (1) as follows:
Solving the differential equation 𝜎 = 0, we can get
𝑢 = 𝑦 + 𝐹 (𝑡, 𝜉) ,
𝑤 (𝜃) = −𝑠 tanh (𝑠𝜃) ,
− 2𝑝𝑞 coth2 (
𝛽
𝛾
(𝑥 − 𝑡𝑦 + 𝑡 − 𝑡3 ))]}
10𝑠𝛼
3
+ 𝑦.
(42)
(38)
Substituting (37) and (38) into (36) and equating the coefficients of all powers of 𝑤 to zero yield a set of algebra equations
for 𝑎0 , 𝑎1 , 𝑎2 , and 𝐴 as follows.
(see Figures 3 and 4).
2 2
Remark 2. If we direct assume that the solution of (34) can
be expressed in the form
𝑤3 : − 4𝛽𝑘𝑎2 𝑞 + 2𝑎1 𝑎2 + 4𝛼𝑘2 𝑎1 𝑞2 = 0,
𝐹 (𝑡, 𝜉) = 𝑎0 (𝑡) + 𝑎1 (𝑡) 𝑤 (𝜃) + 𝑎2 (𝑡) 𝑤(𝜃)2 ,
4
𝑤 : 𝑎2 (𝑎2 + 12𝛼𝑘 𝑞 ) = 0,
𝑤2 : 𝑎12 + 16𝛼𝑘2 𝑎2 𝑞𝑝 − 2𝛽𝑘𝑎1 𝑞 + 2𝑎2 + 2𝑎2 𝑎0 = 0,
𝑤1 : 2𝑎1 𝑎0 − 4𝛽𝑘𝑎2 𝑝 + 2𝑎1 + 4𝛼𝑘2 𝑎1 𝑞𝑝 = 0,
𝑤0 : 2𝑎0 + 𝐴 + 4𝛼𝑘2 𝑎2 𝑝2 + 𝑎02 − 2𝛽𝑘𝑎1 𝑝 = 0.
(39)
(43)
where 𝜃 = 𝑓(𝑡)𝜉 + 𝑔(𝑡), 𝑓(𝑡), and 𝑔(𝑡) are continuous functions of 𝑡 to be determined later. 𝑤(𝜃) satisfies the auxiliary
equation (38). Substituting (43) and (38) into (34), equating
the coefficients of all powers of 𝑤 to zero yields a set of
Abstract and Applied Analysis
5
Solving the system of function equations, we obtain
𝑎0 (𝑡) =
−20
u
𝑎2 (𝑡) =
0
3𝛽2 𝑞
,
25𝛼𝑝
3𝛽2
,
25𝛼
𝑓 (𝑡) = ±
𝑎1 (𝑡) = ±
6𝛽2 𝑞
,
25𝑠𝛼
𝛽
,
10𝑠𝛼
𝑔 (𝑡) = ∓
20
−2
−1
t
0
1
2
6
4
2
−2
0
y
−4
−6
𝛽𝛾 3
𝑡.
30𝑠𝛼
(45)
This result indicate the idea is equivalent to idea of Case 2
above.
Case 3. Take 𝑘2 (𝑡) = 0 and 𝑐 = 1 in (26), then
𝜎 = 𝑘1 (𝑡) 𝑢𝑥 + 𝑢𝑡 − 𝑘1 (𝑡) .
Figure 3: Local structure of 𝑢2 (𝑥, 𝑦, 𝑡) is shown as 𝑥 = 1, 𝛼 = 1, 𝛽 =
10, 𝑝 = −1, 𝑞 = 1, and 𝛾 = 6.
(46)
Solving the differential equation 𝜎 = 0, we obtain
𝑢 = 𝑘1 (𝑡) + 𝐹 (𝜉, 𝑦) , 𝜉 = 𝑥 − ∫ 𝑘1 (𝑡) 𝑑𝑡.
(47)
Substituting (47) into (1) yield
𝛼𝐹𝜉𝜉𝜉𝜉 − 𝛽𝐹𝜉𝜉𝜉 + 𝐹𝐹𝜉𝜉 + 𝐹𝜉2 + 𝛾𝐹𝑦𝑦 = 0.
−20
u
(48)
Using the transformation 𝐹(𝜉, 𝑦) = 𝐹(𝜂), 𝜂 = 𝑘𝜉 − 𝑐𝑦 and
integrating the resulting equation with respect to 𝜂 we have
0
20
𝑘2 𝐹2 + 2𝛾𝑐2 𝐹 + 2𝑘4 𝛼𝐹󸀠󸀠 − 2𝑘3 𝛽𝐹󸀠 + 𝐴 = 0,
−2
−1
t
0
1
2
6
4
2
0
−2
−4
−6
y
(49)
󸀠
where 𝐴 is an arbitrary constant, 𝐹 = 𝑑𝐹/𝑑𝜂. Suppose that
the solution of ODE (49) can be expressed by a polynomial in
(𝐺󸀠 /𝐺) as follows:
𝑛
Figure 4: Local structure of 𝑢3 (𝑥, 𝑦, 𝑡) is shown as 𝑥 = 1, 𝛼 = 1, 𝛽 =
10, 𝑝 = −1, 𝑞 = 1, 𝛾 = 6.
function equations for 𝑎0 (𝑡), 𝑎1 (𝑡), 𝑎2 (𝑡), 𝑓(𝑡), and 𝑔(𝑡) as
follows:
𝐹 (𝜂) = 𝑏𝑛 (
𝐺󸀠
) + ⋅⋅⋅,
𝐺
(50)
where 𝐺 = 𝐺(𝜂) satisfies the second-order LODE in the form
[25]
𝐺󸀠󸀠 + 𝜆𝐺󸀠 + 𝜇𝐺 = 0.
(51)
Balancing 𝐹󸀠󸀠 with 𝐹2 in (49) gives 𝑛 = 2. So that
2
𝐹 (𝜂) = 𝑏2 (
𝑤5 : 2𝑓𝑎2 𝑞 (12𝑓2 𝑞2 𝛼 + 𝑎2 ) = 0,
𝐺󸀠
𝐺󸀠
) + 𝑏1 ( ) + 𝑏0 ,
𝐺
𝐺
𝑏2 ≠ 0,
(52)
where 𝑏𝑖 (𝑖 = 0, 1, 2) and 𝜇 are constants to be determined
later. Substituting (52) and (51) into (49). Setting these coefficients of the 𝐺󸀠 /𝐺 to zero, yields a set of algebraic equations
as follows:
𝑤4 : − 3𝑓𝑞 (−2𝑎1 𝑞2 𝑓2 𝛼 + 2𝑞𝑓𝑎2 𝛽 − 𝑎1 𝑎2 ) = 0,
𝑤3 : − 2𝛽𝑎1 𝑓2 𝑞2 + 2𝑎22 𝑓𝑝 + 40𝛼𝑎2 𝑓3 𝑞2 𝑝 + 2𝑎2 𝑔󸀠 𝑞
𝑘2 𝑏2 (12𝛼𝑘2 + 𝑏2 ) = 0,
+ 2𝑎0 𝑎2 𝑓𝑞 + 𝑎12 𝑓𝑞 + 2𝑎2 𝑓󸀠 𝜉𝑞 + 2𝛾𝑡2 𝑎2 𝑓𝑞 = 0,
2𝑘2 (10𝛼𝑘2 𝑏2 𝜆 + 𝑏1 𝑏2 + 2𝛼𝑘2 𝑏1 + 2𝛽𝑘𝑏2 ) = 0,
𝑤2 : − 8𝛽𝑎2 𝑓2 𝑝𝑞 + 𝑎1 𝑔󸀠 𝑞 + 𝑎2󸀠 + 𝑎0 𝑎1 𝑓𝑞 + 𝛾𝑡2 𝑎1 𝑓𝑞
+ 8𝛼𝑎1 𝑓3 𝑝𝑞2 + 𝑎1 𝑓󸀠 𝜉𝑞 + 3𝑎1 𝑎2 𝑓𝑞 = 0,
8𝛼𝑘4 𝑏2 𝜆2 + 2𝛽𝑘3 𝑏1 + 𝑘2 𝑏12 + 16𝛼𝑘4 𝑏2 𝜇 + 2𝑘2 𝑏2 𝑏0
𝑤1 : 𝑎12 𝑓𝑝 + 16𝛼𝑎2 𝑓3 𝑝2 𝑞 + 𝑎1󸀠 + 2𝛾𝑡2 𝑎2 𝑓𝑝 + 2𝑎0 𝑎2 𝑓𝑝
+ 6𝛼𝑘4 𝑏1 𝜆 + 4𝛽𝑘3 𝑏2 𝜆 + 2𝛾𝑐2 𝑏2 = 0,
+ 2𝑎2 𝑔󸀠 𝑝 + 2𝑎2 𝑓󸀠 𝜉𝑝 − 2𝛽𝑎1 𝑓2 𝑝𝑞 = 0,
2𝑘2 𝑏1 𝑏0 + 4𝛽𝑘3 𝑏2 𝜇 + 4𝛼𝑘4 𝑏1 𝜇 + 2𝛾𝑐2 𝑏1 + 2𝛼𝑘4 𝑏1 𝜆2
𝑤0 : 𝑎1 𝑔󸀠 𝑝 + 𝑎1 𝑓󸀠 𝜉𝑝 + 𝑎0 𝑎1 𝑓𝑝 − 2𝛽𝑎2 𝑓2 𝑝2 + 𝑎0󸀠
+ 2𝛽𝑘3 𝑏1 𝜆 + 12𝛼𝑘4 𝑏2 𝜆𝜇 = 0,
+ 2𝛼𝑎1 𝑓3 𝑝2 𝑞 + 𝛾𝑡2 𝑎1 𝑓𝑝 = 0.
(44)
2𝛾𝑐2 𝑏0 + 2𝛼𝑘4 𝑏1 𝜆𝜇 + 𝐴 + 2𝛽𝑘3 𝑏1 𝜇 + 4𝛼𝑘4 𝑏2 𝜇2 + 𝑘2 𝑏02 = 0.
(53)
6
Abstract and Applied Analysis
Solving the algebraic equations above yields
𝑏0 =
3
2 2
2
15𝑘 𝜆𝛼 (5𝑘𝜆𝛼 + 2𝛽) − 3𝑘 𝛽 + 25𝑐 𝛼𝛾
,
25𝑘2 𝛼
12𝑘 (5𝑘𝛼𝜆 + 𝛽)
𝑏1 = −
,
5
(54)
2
𝑏2 = −12𝑘 𝛼.
when 25𝑘2 𝛼2 (4𝜇 − 𝜆2 ) + 𝛽2 = 0 and 625𝛼2 (𝐴𝑘2 − 𝑐2 𝛾2 ) +
36𝑘4 𝛽4 = 0. Consequently, we obtain the following solution
of (1) for 𝜆2 − 4𝜇 > 0:
𝑢4 (𝑥, 𝑦, 𝑡) = −12𝑘2 𝛼𝜏2
× [(𝐶1 sinh 𝜏 (𝑘 (𝑥 − ∫ 𝑘1 (𝑡) 𝑑𝑡) − 𝑐𝑦)
+ 𝐶2 cosh 𝜏 (𝑘 (𝑥 − ∫ 𝑘1 (𝑡) 𝑑𝑡) − 𝑐𝑦))
× (𝐶1 cosh 𝜏 (𝑘 (𝑥 − ∫ 𝑘1 (𝑡) 𝑑𝑡) − 𝑐𝑦)
+ 𝐶2 sinh 𝜏
−1 2
× (𝑘 (𝑥 − ∫ 𝑘1 (𝑡) 𝑑𝑡) − 𝑐𝑦)) ]
+ (12𝑘2 𝜆𝛼𝜏 −
12𝑘 (5𝑘𝛼𝜆 + 𝛽)
)
5
× [(𝐶1 sinh 𝜏 (𝑘 (𝑥 − ∫ 𝑘1 (𝑡) 𝑑𝑡) − 𝑐𝑦)
+ 𝐶2 cosh 𝜏 (𝑘 (𝑥 − ∫ 𝑘1 (𝑡) 𝑑𝑡) − 𝑐𝑦))
× (𝐶1 cosh 𝜏 (𝑘 (𝑥 − ∫ 𝑘1 (𝑡) 𝑑𝑡) − 𝑐𝑦)
+ 𝐶2 sinh 𝜏
−1 2
× (𝑘 (𝑥 − ∫ 𝑘1 (𝑡) 𝑑𝑡) − 𝑐𝑦)) ]
+
15𝑘3 𝜆𝛼 (5𝑘𝜆𝛼 + 2𝛽) − 3𝑘2 𝛽2 + 25𝑐2 𝛼𝛾
25𝑘2 𝛼
+
𝜆2
,
4
(55)
where 𝜏 = (1/2)√𝜆2 − 4𝜇.
6. Conclusions
Based on the fact that the periodic solution of (2+1)D KdVBurgers equation does not exist, we have obtained Saddlenode Heteroclinic Orbits. By applying the Lie group method,
we reduce the (2+1)D KdV Burgers equation to (1+1)-dimensional equations including the (1+1)-dimensional linear partial differential equation with constants coefficients (29), (48)
and (1+1)-dimensional nonlinear partial differential equation
with variable coefficients (34). By solving the equations (29),
(34), and (48), we obtain some new exact solutions and
discover the strange phenomenon of sympathetic vibration
to evolution of nontraveling wave soliton for the (2+1)D KdV
Burgers equation. Our results show that the unite of Lie group
method with others is effective to search simultaneously exact
solutions for nonlinear evolution equations. Other structures
of solutions with symmetry (26) are to be further studied.
Acknowledgments
The authors would like to thank professor S. Y. Lou for the
helpful discussions. This work was supported by key research
projects of Sichuan Provincial Educational Administration
no. 10ZA021 and Chinese Natural Science Foundation Grant
no. 10971169.
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