Research Article Symmetry Reduced and New Exact Nontraveling Boiti-Leon-Manna-Pempinelli Equation

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Hindawi Publishing Corporation
Abstract and Applied Analysis
Volume 2013, Article ID 784134, 5 pages
http://dx.doi.org/10.1155/2013/784134
Research Article
Symmetry Reduced and New Exact Nontraveling
Wave Solutions of (2+1)-Dimensional Potential
Boiti-Leon-Manna-Pempinelli Equation
Chen Han-Lin and Xian Da-Quan
School of Sciences, Southwest University of Science and Technology, Mianyang, Sichuan 621010, China
Correspondence should be addressed to Xian Da-Quan; daquanxian@163.com
Received 21 September 2012; Accepted 8 December 2012
Academic Editor: Zhengde Dai
Copyright © 2013 C. Han-Lin and X. Da-Quan. 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.
With the aid of Maple symbolic computation and Lie group method, (2+1)-dimensional PBLMP equation is reduced to some (1+1)dimensional PDE with constant coefficients. Using the homoclinic test technique and auxiliary equation methods, we obtain new
exact nontraveling solution with arbitrary functions for the PBLMP equation.
1. Introduction
2. Symmetry of (1)
In this paper, we will consider the potential Boiti-LeonManna-Pempinelli (PBLMP) equation
This section is devoted to Lie point group symmetries of (1).
Let
𝑒𝑦𝑑 + 𝑒π‘₯π‘₯π‘₯𝑦 − 3𝑒π‘₯π‘₯ 𝑒𝑦 − 3𝑒π‘₯ 𝑒π‘₯𝑦 = 0,
𝜎 = 𝜎 (π‘₯, 𝑦, 𝑑, 𝑒, 𝑒π‘₯ , 𝑒𝑦 , 𝑒𝑑 )
(1)
where 𝑒 : 𝑅π‘₯ × π‘…π‘¦ × π‘…π‘‘+ → 𝑅. By some transformations,
the PBLMP equation (1) can be equivalent to the asymmetric Nizhnik-Novikov-Veselov (ANNV) system. In fact, the
ANNV equation can be obtained from the inner parameterdependent symmetry constraint of the KP equation [1] and
may be considered as a model for an incompressible fluid
[2]. The Painlevé analysis, Lax pair, and some exact solutions
have been studied for the PBLMP equation [3]. Tang and
Lou obtained the bilinear form of (1) and variable separation
solutions including two arbitrary functions by the multilinear
variable separation approach [4, 5].
In this paper, by means of Maple symbolic computation,
we will use the Lie group method [6, 7], homoclinic test
technique [8, 9] and so forth to reduce and solve the PBLMP
equation. First, we will derive symmetry of (1). Then we use
the symmetry to reduce (1) to some (1 + 1)-dimensional
PDE with constant coefficients. Finally, solving the reduced
PDE by Homoclinic test technique and auxiliary equation
methods [10, 11] implies abundant exact nontraveling wave
periodic solutions for the PBLMP equation.
(2)
be the symmetry of (1). Based on Lie group theory [6], 𝜎
satisfies the following symmetry equation:
πœŽπ‘¦π‘‘ + πœŽπ‘¦π‘₯π‘₯π‘₯ − 3𝑒π‘₯π‘₯ πœŽπ‘¦ − 3𝑒𝑦 𝜎π‘₯π‘₯ − 3𝑒π‘₯ 𝜎π‘₯𝑦 − 3𝑒π‘₯𝑦 𝜎π‘₯ = 0.
(3)
To get some symmetries of (1), we take the function 𝜎 in the
form
𝜎 = π‘Ž (π‘₯, 𝑦, 𝑑) 𝑒π‘₯ + 𝑏 (π‘₯, 𝑦, 𝑑) 𝑒𝑦 + 𝑐 (π‘₯, 𝑦, 𝑑) 𝑒𝑑
+𝑑 (π‘₯, 𝑦, 𝑑) 𝑒 + 𝑒 (π‘₯, 𝑦, 𝑑) ,
(4)
where π‘Ž, 𝑏, 𝑐, 𝑑, 𝑒 are functions of π‘₯, 𝑦, 𝑑 to be determined, and
𝑒(π‘₯, 𝑦, 𝑑) satisfies (1). Substituting (4) and (1) into (3), one can
get
1
π‘Ž = π‘˜1 π‘₯ + πœ† (𝑑) ,
3
𝑐 = π‘˜1 𝑑 + π‘˜2 ,
1
𝑑 = π‘˜1 ,
3
𝑏 = πœ‡ (𝑦) ,
1
𝑒 = πœ†σΈ€  (𝑑) π‘₯ + πœ‰ (𝑑) ,
3
(5)
2
Abstract and Applied Analysis
where π‘˜1 , π‘˜2 are arbitrary constants. πœ†(𝑑), πœ‰(𝑑) are arbitrary
functions of t. πœ‡(𝑦) is a arbitrary function of y. Substituting
(5) into (4), we obtain the symmetries of (1) as follows:
Substituting (15) into (1) yields a reduced PDE of (1) with
constant coefficients:
1
1
𝜎 = ( π‘˜1 π‘₯ + πœ† (𝑑)) 𝑒π‘₯ + πœ‡ (𝑦) 𝑒𝑦 + (π‘˜1 𝑑 + π‘˜2 ) 𝑒𝑑 + π‘˜1 𝑒
3
3
π‘€πœƒπœƒπœƒπœƒ − 6π‘€πœƒπœƒ π‘€πœƒ + π‘€πœƒπ‘‘ = 0.
1
+ πœ†σΈ€  (𝑑) π‘₯ + πœ‰ (𝑑) .
3
(6)
(16)
Integrating (16) once with respect to πœƒ and taking integration
constant to zero yield
π‘€πœƒπœƒπœƒ − 3π‘€πœƒ2 + 𝑀𝑑 = 0.
(17)
3. Symmetry Reduction of (1)
Based on the integrability of reduced equation of the symmetry (6), we consider the following three cases.
Case 1. Let π‘˜1 = π‘˜2 = 0, πœ†(𝑑) = π‘Ÿ, πœ‰(𝑑) = 1, πœ‡(𝑦) = −1/𝜏(𝑦)
in (6), then
−1
𝜎 = 𝜏(𝑦) (π‘Ÿπœ (𝑦) 𝑒π‘₯ − 𝑒𝑦 + 𝜏 (𝑦)) ,
(7)
where π‘Ÿ is an arbitrary nonzero constant, 𝜏(𝑦) =ΜΈ 0. Solving the
differential equation for 𝜎 = 0, one gets
Combining the above results, we obtain some reduced
equations of (1) expressed by (10), (13), and (17), respectively.
Meanwhile many new explicit solutions of (1) from these
reduced Equations. can be achieved. We omit other cases
based on symmetries (6) here.
4. Solve Reduced PDE and Get Exact
Nontraveling Wave Solutions of (1)
(8)
In this section, we seek exact nontraveling wave solutions
of (1) by using some appropriate methods to solve reduced
equations (10), (13), and (17).
Substituting (8) into (1), we get the following (1 + 1)-dimensional nonlinear PDE with constant coefficients:
4.1. Solve Reduced PDE (10). Now, we seek solutions of (10) by
auxiliary equation method. Make transformation as follows:
𝑒 = ∫ 𝜏 (𝑦) 𝑑𝑦 + 𝑀 (πœƒ, 𝑑) ,
πœƒ = π‘₯ + ∫ π‘Ÿπœ (𝑦) 𝑑𝑦.
π‘Ÿπ‘€πœƒπœƒπœƒπœƒ − 6π‘Ÿπ‘€πœƒ π‘€πœƒπœƒ + π‘Ÿπ‘€πœƒπ‘‘ − 3π‘€πœƒπœƒ = 0.
(9)
Integrating (9) once with respect to πœƒ and taking integration
constant to zero yield
π‘Ÿπ‘€πœƒπœƒπœƒ − 3π‘Ÿπ‘€πœƒ2 + π‘Ÿπ‘€π‘‘ − 3π‘€πœƒ = 0.
𝜎 = 𝜏(𝑦) (𝑒𝑦 + π‘Ÿπœ (𝑦) 𝑒𝑑 ) .
πœƒ = 𝑑 − ∫ 𝜏 (𝑦) 𝑑𝑦.
(12)
(13)
Case 3. Let π‘˜1 = π‘˜2 = 0, πœ†(𝑑) = 1, πœ‰(𝑑) = 0, πœ‡(𝑦) = −1/𝜏(𝑦)
in (6), then
−1
𝜎 = 𝜏(𝑦) (𝜏 (𝑦) 𝑒π‘₯ − 𝑒𝑦 ) .
(14)
Solving the equation for 𝜎 = 0, we obtain
𝑒 = 𝑀 (πœƒ, 𝑑) ,
πœƒ = π‘₯ + ∫ 𝜏 (𝑦) 𝑑𝑦.
where 𝑝, π‘ž are nonzero constants. Substituting (18) into (10)
obtains an ordinary differential equation for πœ‘(πœ‰) as follows:
𝑝3 π‘Ÿπœ‘σΈ€ σΈ€ σΈ€  − 3π‘Ÿπ‘2 πœ‘σΈ€ 2 + (π‘žπ‘Ÿ − 3𝑝) πœ‘σΈ€  = 0,
(19)
where πœ‘σΈ€  = π‘‘πœ‘/π‘‘πœ‰. Let πœ‘σΈ€  = 𝑓, then (19) can be written as
𝑝3 π‘Ÿπ‘“σΈ€ σΈ€  − 3π‘Ÿπ‘2 𝑓2 + (π‘žπ‘Ÿ − 3𝑝) 𝑓 = 0.
Substituting (12) into (1), we have the function 𝑀(π‘₯, πœƒ) which
must satisfy the following PDE:
𝑀π‘₯π‘₯π‘₯πœƒ − 3𝑀π‘₯π‘₯ π‘€πœƒ − 3𝑀π‘₯ 𝑀π‘₯πœƒ + π‘€πœƒπœƒ = 0.
(18)
(11)
Solving the differential equation for 𝜎 = 0, one gets
𝑒 = 𝑀 (π‘₯, πœƒ) ,
πœ‰ = π‘πœƒ + π‘žπ‘‘,
(10)
Case 2. Taking π‘˜1 = 0, π‘˜2 = 1, πœ†(𝑑) = 0, πœ‰(𝑑) = 0, πœ‡(𝑦) =
1/𝜏(𝑦) in (6) yields
−1
𝑀 (πœƒ, 𝑑) = πœ‘ (πœ‰) ,
(15)
(20)
This is the fourth type of ellipse equation (12), its solutions are
as follows:
3𝑝 − π‘žπ‘Ÿ
3𝑝 − π‘žπ‘Ÿ
{
{
sech2 [√
(πœ‰ − πœ‰0 )] ,
−
{
2
{
4𝑝3 π‘Ÿ
{
{ 2𝑝 π‘Ÿ
{
{
{
π‘π‘Ÿ (3𝑝 − π‘žπ‘Ÿ) > 0,
{
{
{
{
{
{
{
{
{
3𝑝 − π‘žπ‘Ÿ
3𝑝 − π‘žπ‘Ÿ
{
{
csch2 [√
(πœ‰ − πœ‰0 )] ,
2
𝑓 = { 2𝑝 π‘Ÿ
4𝑝3 π‘Ÿ
{
{
{
π‘π‘Ÿ (3𝑝 − π‘žπ‘Ÿ) > 0,
{
{
{
{
{
{
{
{
{
3𝑝 − π‘žπ‘Ÿ
3𝑝 − π‘žπ‘Ÿ 2
{
{
−
sec [√−
(πœ‰ − πœ‰0 )] ,
{
2π‘Ÿ
{
{
2𝑝
4𝑝3 π‘Ÿ
{
{
π‘π‘Ÿ (3𝑝 − π‘žπ‘Ÿ) < 0,
{
(21)
Abstract and Applied Analysis
3
where πœ‰0 is the integration constant. From the result of
(21), some new exact solutions 𝑒1 through 𝑒3 of (1) can be
obtained:
1.5
u
3𝑝 − π‘žπ‘Ÿ
𝑒1 = ∫ 𝜏 (𝑦) 𝑑𝑦 − √
π‘π‘Ÿ
0.5
−0.5
−1.5
−15
−10
3𝑝 − π‘žπ‘Ÿ
× tanh [√
(𝑝 (π‘₯ + π‘Ÿ ∫ 𝜏 (𝑦) 𝑑𝑦) + π‘žπ‘‘ − πœ‰0 )] ,
4𝑝3 π‘Ÿ
−5
t
0
5
10
× coth [√
3𝑝 − π‘žπ‘Ÿ
π‘π‘Ÿ
3𝑝 − π‘žπ‘Ÿ
(𝑝 (π‘₯ + π‘Ÿ ∫ 𝜏 (𝑦) 𝑑𝑦) + π‘žπ‘‘ − πœ‰0 )] ,
4𝑝3 π‘Ÿ
π‘π‘Ÿ (3𝑝 − π‘žπ‘Ÿ) > 0,
𝑒3 = ∫ 𝜏 (𝑦) 𝑑𝑦 − √
π‘žπ‘Ÿ − 3𝑝
π‘π‘Ÿ
π‘π‘Ÿ (π‘žπ‘Ÿ − 3𝑝) < 0.
(22)
Particularly, we assume 𝑝 = π‘ž = 1, π‘Ÿ = 2, 𝜏(𝑦) = sin(𝑦),
πœ‰0 = 0, π‘₯ = sech(𝑑), then the solution 𝑒1 can be depicted by
Figure 1(a). If 𝑝 = −1, π‘ž = 1, π‘Ÿ = −1, 𝜏(𝑦) = βˆ“ cos(𝑦), πœ‰0 =
0, π‘₯ = sin(t), then 𝑒3 can be depicted by Figures 1(b) and 2(a).
4.2. Solve Reduced PDE (13). Make transformation to (13) as
follows:
𝑀 (π‘₯, πœƒ) = πœ‘ (πœ‰) ,
πœ‰ = π‘˜π‘₯ + π‘πœƒ,
−1.5
−1
−0.5 0
0.5
1
1.5
−15
−10
−5
0
5
10
15
y
(b)
Figure 1: (a) The figure of 𝑒1 as 𝑝 = 1, π‘ž = 1, π‘Ÿ = 2, 𝜏(𝑦) =
sin(𝑦), πœ‰0 = 0, π‘₯ = sech(𝑑). (b) The figure of 𝑒3 as 𝑝 = −1, π‘ž =
1, π‘Ÿ = −1, 𝜏(𝑦) = − cos(𝑦), πœ‰0 = 0, π‘₯ = sin(𝑑).
0.2
0.1
0
u −0.1
−0.2
−0.1
−0.05
t 0
0.05
0.1
(23)
where π‘˜, 𝑐 are non-zero constants. Substituting (23) into (13)
then we have
π‘πœ‘σΈ€  + π‘˜3 πœ‘σΈ€ σΈ€ σΈ€  − 3𝐾2 πœ‘σΈ€ 2 = 0.
y
6
4
2
u 0
−2
−4
−6
t
π‘žπ‘Ÿ − 3𝑝
× tan [√
(𝑝 (π‘₯ + π‘Ÿ ∫ 𝜏 (𝑦) 𝑑𝑦) + π‘žπ‘‘ − πœ‰0 )] ,
4𝑝3 π‘Ÿ
−5
(a)
π‘π‘Ÿ (3𝑝 − π‘žπ‘Ÿ) > 0,
𝑒2 = ∫ 𝜏 (𝑦) 𝑑𝑦 − √
−10
15
10
5
0
−6 −4 −2
0
2 4
y
6
8 10
(a)
(24)
It is equivalent to (19). Based on the above accordant idea, we
can get
−0.16
−0.12
𝑐
𝑒4 = √−
2π‘˜
𝑐
×tanh [√− 3 (π‘˜π‘₯+𝑐 (𝑑−∫𝜏 (𝑦) 𝑑𝑦)− πœ‰0)] ,
2π‘˜
𝑒5 = √−
u −0.08
−0.04
π‘˜π‘ < 0,
𝑐
2π‘˜
𝑐
×coth [√− 3 (π‘˜π‘₯+𝑐(𝑑−∫𝜏 (𝑦) 𝑑𝑦)−πœ‰0)] ,
2π‘˜
0
−10
−5
0
y
5
−1
−2
−3 t
−4
10 −5
(b)
π‘˜π‘ < 0,
Figure 2: (a) The figure of 𝑒3 as 𝑝 = −1, π‘ž = 1, π‘Ÿ = −1, 𝜏(𝑦) =
cos(𝑦), πœ‰0 = 0, π‘₯ = sin(𝑑). (b) The figure of 𝑒9 as 𝑝1 = 1, 𝑐1 =
1, 𝑝2 = 1, 𝜏(𝑦) = sin(𝑦), π‘₯ = sin(𝑑).
4
Abstract and Applied Analysis
𝑒6 = √
𝑐
2π‘˜
×tan[√
where 𝑖2 = −1. Substituting (30)–(33) into (28) yields the
solutions 𝑒7 through 𝑒11 of (1) as follows:
𝑐
(π‘˜π‘₯+𝑐 (𝑑−∫ 𝜏 (𝑦) 𝑑𝑦)−πœ‰0 )] ,
2π‘˜3
𝑒7 = −2𝑝1 tanh (𝑝1 (π‘₯ + ∫ 𝜏 (𝑦) 𝑑𝑦) − 4𝑝12 𝑑 +
π‘˜π‘ > 0.
(25)
4.3. Solve Reduced PDE (17). In this section, we use homoclinic test technique [8, 9] to (17) and transform the unknown
function as follows:
𝑀 (πœƒ, 𝑑) = −2(ln 𝑓 (πœƒ, 𝑑))πœƒ .
(π·πœƒ 𝐷𝑑 +
π·πœƒ4 ) (𝑓
when 𝑐2 > 0 in (30);
𝑒8 = −2𝑝1 coth (𝑝1 (π‘₯ + ∫ 𝜏 (𝑦) 𝑑𝑦) − 4𝑝12 𝑑 +
(26)
Substituting (26) into (17) and using the bilinear form, we can
get
⋅ 𝑓) = 0,
𝑒9 = −2𝑝1 𝑝2
(27)
𝑓=𝑒
𝑝1 (πœƒ−πœ”1 𝑑)
+ 𝑐1 cos (𝑝2 (πœƒ + πœ”2 𝑑)) + 𝑐2 𝑒
,
× (coth (𝑝1 (π‘₯ + ∫ 𝜏 (𝑦) 𝑑𝑦)
− (𝑝12 − 3𝑝22 ) 𝑑 + ln
(28)
𝑐1 𝑝2
)
𝑝1
+ sin (𝑝2 (π‘₯ + ∫ 𝜏 (𝑦) 𝑑𝑦) − (3𝑝12 − 𝑝22 ) 𝑑))
where 𝑝2 , πœ”1 , πœ”2 , 𝑐1 , and 𝑐2 are real constants to be determined. Substituting (28) into (27) yields a set of algebraic
equations as follows:
× (𝑝2 sinh (𝑝1 (π‘₯ + ∫ 𝜏 (𝑦) 𝑑𝑦)
𝑝1 𝑐1 𝑝2 (4 (𝑝12 − 𝑝22 ) + πœ”2 − πœ”1 ) = 0,
− (𝑝12 − 3𝑝22 ) 𝑑 + ln
𝑐1 ((𝑝14 + 𝑝44 − 6𝑝12 𝑝22 ) − 𝑝12 πœ”1 − 𝑝22 πœ”2 ) = 0,
𝑝1 𝑝2 𝑐1 𝑐2 (4 (𝑝12 − 𝑝22 ) + πœ”2 − πœ”1 ) = 0,
1
ln (−𝑐2 )) ,
2
(35)
when 𝑐2 < 0 in (30);
where the Hirota operator 𝐷 is defined in [12]. In this case we
choose extended homoclinic test function
−𝑝1 (πœƒ−πœ”1 𝑑)
1
ln 𝑐2 ) ,
2
(34)
𝑐1 𝑝2
)
𝑝1
−1
(29)
+ 𝑝1 cos (𝑝2 (π‘₯ + ∫ 𝜏 (𝑦) 𝑑𝑦) − (3𝑝12 − 𝑝22 ) 𝑑)) ,
𝑐1 𝑐2 ((𝑝14 + 𝑝44 − 6𝑝12 𝑝22 ) − 𝑝12 πœ”1 − 𝑝22 πœ”2 ) = 0,
(36)
when 𝑐1 𝑝1 𝑝2 > 0 in (31) (see Figure 2(b));
4 (4𝑝14 𝑐2 + 𝑐12 𝑝24 ) − 4𝑝12 πœ”1 𝑐2 − 𝑐12 𝑝22 πœ”2 = 0.
𝑒10 (π‘₯, 𝑦, 𝑑) = 𝑝2 tan (𝑝2 (π‘₯ + ∫ 𝜏 (𝑦) 𝑑𝑦) + 4𝑝22 𝑑) ,
Solving the above equations (29) yields
(37)
𝑝 = 𝑝1 ,
(1) { 1
πœ”1 = 4𝑝12 ,
𝑝2 = 𝑝2 ,
πœ”2 = πœ”2 ,
𝑐1 = 0,
when 𝑐2 = 1 in (32);
𝑐2 = 𝑐2 ,
𝑒11 (π‘₯, 𝑦, 𝑑)
(30)
{
{𝑝 = 𝑝1 ,
𝑝2 = 𝑝2 ,
𝑐1 = 𝑐1 ,
𝑐2 =
(2){ 1
{
2
2
πœ”2 = −3𝑝12 + 𝑝22 ,
{πœ”1 = −3𝑝2 + 𝑝1 ,
𝑝 = 𝑝2 𝑖, 𝑝2 = 𝑝2 ,
𝑐1 = 𝑐1 ,
(3) { 1
πœ”1 = −4𝑝22 ,
πœ”2 = 4𝑝22 ,
𝑐2 𝑝2
− 1 22 ,
𝑝1
(31)
𝑐2 = 𝑐2 ,
×
sin (𝑝2 (π‘₯+∫ 𝜏 (𝑦) 𝑑𝑦)+(8𝑝22 −πœ”2 ) 𝑑)+sin (𝑝2 (π‘₯+∫ 𝜏 (𝑦) 𝑑𝑦)+πœ”2 𝑑)
,
cos (𝑝2 (π‘₯+∫ 𝜏 (𝑦) 𝑑𝑦)+(8𝑝22 −πœ”2 ) 𝑑)+cos (𝑝2 (π‘₯+∫ 𝜏 (𝑦) 𝑑𝑦)+πœ”2 𝑑)
(38)
when 𝑐1 = 2 in (33).
Remark 1. If one lets π‘€πœƒ = 𝑣 in (16), then (16) can be written
as
𝑣𝑑 − 6π‘£π‘£πœƒ + π‘£πœƒπœƒπœƒ = 0.
(32)
{𝑝 = 𝑝2 𝑖, 𝑝2 = 𝑝2 ,
𝑐1 = 𝑐1 ,
(4) { 1
2
πœ”2 = πœ”2 ,
{πœ”1 = πœ”2 − 8𝑝2 ,
= −2𝑝2
1
𝑐2 = 𝑐12 ,
4
(39)
This is the famous KdV equation.
5. Conclusions
(33)
In this paper, a combination of Lie group method and
homoclinic test technique and so forth is applied and thus the
Abstract and Applied Analysis
symmetries (6) are obtained. The (2+1)-dimensional potential Boiti-Leon-Manna-Pempinelli equation (1) is reduced to
(1 + 1)-dimensional nonlinear PDE of constant coefficients
(10), (13), and (17). Further auxiliary equation method and
homoclinic test technique are used and some new exact nontraveling wave solutions are obtained. And they include some
special and strange structures to be further studied and other
relevant solutions about symmetry (6) will be discussed later
in another paper. Our results show that combining the Lie
group method with homoclinic test technique and so forth
is effective in finding nontraveling wave exact solutions of
nonlinear evolution equations.
Acknowledgments
The authors would like to thank Professor Dai Zhengde
for his helpful discussions. This work was supported by the
Chinese Natural Science Foundation Grant no. 10971169 and
the key research projects of Sichuan Provincial Educational
Administration, no. 10ZA021.
References
[1] S.-Y. Lou and X.-B. Hu, “Infinitely many Lax pairs and symmetry constraints of the KP equation,” Journal of Mathematical
Physics, vol. 38, no. 12, pp. 6401–6427, 1997.
[2] M. J. Ablowitz and P. A. Clarkson, Solitons, Nonlinear Evolution
Equations and Inverse Scattering, vol. 149, Cambridge University
Press, Cambridge, Mass, USA, 1991.
[3] P. G. Estévez and S. Leble, “A wave equation in 2 + 1 Painlevé
analysis and solutions,” Inverse Problems, vol. 11, no. 4, pp. 925–
937, 1995.
[4] X. Y. Tang, Locallized Excitations and Symmetries of (2+1)Dimensional Nonlinear Systems, Shanghai Physics Department,
Shanghai Jiao Tong University, Shanghai, China, 2004.
[5] S. Y. Lou and X. Y. Tang, Methods of Nonlinear Mathematical
Physics, Beling Science Press of China, Beijing, China, 2006.
[6] P. J. Olver, Applications of Lie Groups to Differential Equations,
vol. 107, Springer, New York, NY, USA, 1986.
[7] D. Q. Xian, “New exact solutions to a class of nonlinear
wave equations,” Journal of University of Electronic Science and
Technology of China, vol. 35, no. 6, pp. 977–980, 2006.
[8] Z. Dai and D. Xian, “Homoclinic breather-wave solutions for
Sine-Gordon equation,” Communications in Nonlinear Science
and Numerical Simulation, vol. 14, no. 8, pp. 3292–3295, 2009.
[9] Z.-D. Dai, D.-Q. Xian, and D.-L. Li, “Homoclinic breather-wave
with convective effect for the (1+1)-dimensional boussinesq
equation,” Chinese Physics Letters, vol. 26, no. 4, Article ID
040203, 2009.
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805–809, 1973.
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