Hindawi Publishing Corporation
Advances in Mathematical Physics
Volume 2013, Article ID 890784, 11 pages
http://dx.doi.org/10.1155/2013/890784
Research Article
A New Method with a Different Auxiliary Equation to
Obtain Solitary Wave Solutions for Nonlinear Partial
Differential Equations
Bülent Kiliç and Hasan Bulut
Department of Mathematics, Firat University, 23119 Elazig, Turkey
Correspondence should be addressed to Hasan Bulut; hbulut@firat.edu.tr
Received 3 February 2013; Revised 22 April 2013; Accepted 23 April 2013
Academic Editor: Dumitru Baleanu
Copyright © 2013 B. Kiliç and H. Bulut. 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.
A new method with a different auxiliary equation from the Riccati equation is used for constructing exact travelling wave solutions
of nonlinear partial differential equations. The main idea of this method is to take full advantage of a different auxilliary equation
from the Riccati equation which has more new solutions. More new solitary solutions are obtained for the RLW Burgers and Hirota
Satsuma coupled equations.
1. Introduction
In the recent years, remarkable progress has been made in
the construction of the exact solutions for nonlinear partial
differential equations, which have been a basic concern for
both mathematicians and physicists [1–3].We do not attempt
to characterize the general form of nonlinear dispersive
wave equations [4, 5]. When an original nonlinear equation
is directly calculated, the solution will preserve the actual
physical characters of solutions [6]. The studies in finding
exact solutions to nonlinear differential equation (NPDE),
when they exist, are very important for the understanding
of most nonlinear physical phenomena. There are many
studies which obtain explicit solutions for nonlinear differential equations. Many explicit exact methods have been
introduced in literature [7–21]. Some of them are generalized
Miura transformation, Darboux transformation, Cole-Hopf
transformation, Hirota’s dependent variable transformation,
the inverse scattering transform and the Bäcklund transformation, tanh method, sine-cosine method, Painleve method,
homogeneous balance method (HB), similarity reduction
method, improved tanh method and so on.
In this article, the first section presents the scope of the
study as an introduction. In the second section contains
analyze of a new method and balance term definition. In the
third section, we will obtain wave solutions of RLW Burgers
and Hirota Satsuma coupled equations by using a new
method. In the last section, we implement the conclusion.
2. Method and Its Applications
Let us simply describe the method [22]. Consider a given
partial differential equation in two variables
𝐻 (𝑢, 𝑢𝑡 , 𝑢𝑥 , 𝑢𝑥𝑥 , . . .) = 0.
(1)
The fact that the solutions of many nonlinear equations can
be expressed as a finite series of solutions of the auxiliary
equation motivates us to seek for the solutions of (1) in the
form
𝑚
𝑢 (𝑥, 𝑡) = 𝜆∑ [𝑎𝑖 𝐹(𝜉)𝑖 + 𝑎−𝑖 𝐹(𝜉)−𝑖 ] ,
(2)
𝑖=0
where, 𝜉 = 𝑘(𝑥 − 𝑐𝑡), 𝑘 and 𝑐 are the wave number and
the wave speed respectively, 𝑚 is a positive integer that can
be determined by balancing the linear term of highest order
with the nonlinear term in (1), 𝜆 is balancing coefficient
that will be defined in a new “Balance term” definition and
𝑎0 , 𝑎1 , 𝑎2 , . . . are parameters to be determined. Substituting
(2) into (1) yields a set of algebraic equations for 𝑎0 , 𝑎1 , 𝑎2 , . . .
because all coefficients of 𝐹 have to vanish. From these
2
Advances in Mathematical Physics
0.4
0.1
0
−0.1
−0.2
−0.3
−5
0.2
4
4
0
−5
2
2
0
0
0
5
0
5
(a)
(b)
Figure 1: Graph of the solution 𝑢(𝑥, 𝑡) from left to the right for (13) and (14).
0
0.5
−0.05
4
−0.1
−5
4
0
−5
2
2
0
0
5
0
5
(a)
0
(b)
Figure 2: Graph of the solution 𝑢(𝑥, 𝑡) from left to the right for (15) and (16).
relations 𝑎0 , 𝑎1 , 𝑎2 , . . . can be determined. The main idea of
our method is to take full advantage of the new auxiliary
equation. The desired auxiliary equation presents as following
𝐹󸀠 =
𝐴
+ 𝐵𝐹 + 𝐶𝐹3 ,
𝐹
In the following we present a new approach to the
“Balance term” definition.
(3)
where 𝑑𝐹/𝑑𝜉 = 𝐹󸀠 and 𝐴, 𝐵, 𝐶 are constants.
Case 1. If 𝐴 = −1/4, 𝐵 = 1/2, 𝐶 = −1/2 then (3) has the
solution 𝐹 = 1/√1 + tan(𝜉) + sec(𝜉).
Case 2. If 𝐴 = 1/4, 𝐵 = −1/2, 𝐶 = 0 then (3)
has the solutions 𝐹 = 1/√1 + csc ℎ(𝜉) + coth(𝜉) or 𝐹 =
1/√1 + 𝑖 sec ℎ(𝜉) + tanh(𝜉).
Case 3. If 𝐴 = 1/2, 𝐵 = −1,𝐶 = 0 then (3) has the solutions
𝐹 = 1/√1 + cot ℎ(𝜉) or
𝐹=
1
√1 + tanh (𝜉)
.
Remark 1. Depending on the 𝐴, 𝐵 and 𝐶 coefficients in the
(3), it could be reached only three cases.
(4)
Definition 2. When (1) is transformed with 𝑢(𝑥, 𝑡) = 𝑢(𝜉),
𝜉 = 𝑘(𝑥 − 𝑐𝑡), where 𝑘 and 𝑐 are real constants, we get a
nonlinear ordinary differential equation for 𝑢(𝜉) as following
𝑄󸀠 (𝑢, 𝑘𝑐𝑢󸀠 , 𝑘𝑢󸀠 , 𝑘2 𝑢󸀠󸀠 , . . .) = 0.
(5)
Let 𝑢(𝑝) is the highest order derivative linear term and 𝑢𝑞 𝑢(𝑟)
is the highest nonlinear term in (5) and 𝐹󸀠 = 𝑘0 + 𝑘1 𝐹 +
𝑘2 𝐹2 + ⋅ ⋅ ⋅ + 𝑘𝑛 𝐹𝑛 is the auxiliary equation that is used to solve
the nonlinear partial differential equation then the “Balance
term” 𝑚 can be decided by the balancing the nonlinear term
𝑢𝑞 𝑢(𝑟) and the linear term 𝑢(𝑝) with acceptances of 𝑢 ≅ 𝜆𝐹𝑖
and 𝐹󸀠 ≅ 𝐹𝑛 where 𝑛 is integer (𝑛 ≠ 1)and 𝜆 is the balance
coefficient that can be determined later.
Advances in Mathematical Physics
3
0
20
−20
4
−40
−5
4
10
0
−5
2
2
0
0
0
5
5
(a)
0
(b)
0
2
−5
4
−10
−5
0
4
−2
−5
2
2
0
0
5
0
5
(c)
0
(d)
Figure 3: Graph of the solutions 𝑢(𝑥, 𝑡) and V(𝑥, 𝑡) corresponding to the value 𝑏0 = 1 from left to the right for (22), (3), and (23).
Example 1. For the KdV equation with the transform
𝑢(𝑥, 𝑡) = 𝑢(𝜉), 𝜉 = 𝑥 − 𝑐𝑡 we have the ordinary differential
equation as following
󸀠
󸀠
−𝑐𝑢 + 6𝑢𝑢 + 𝑢
󸀠󸀠󸀠
= 0.
(6)
By the balancing linear term 𝑢󸀠󸀠󸀠 with nonlinear term 𝑢𝑢󸀠
󸀠
𝑢󸀠󸀠 = (𝜆𝑚𝐹𝑚+𝑛−1 ) = 𝜆𝑚 (𝑚 + 𝑛 − 1) 𝐹𝑚+𝑛−2 𝐹󸀠
= 𝜆𝑚 (𝑚 + 𝑛 − 1) 𝐹𝑚+𝑛−2 𝐹𝑛 = 𝜆𝑚 (𝑚 + 𝑛 − 1) 𝐹𝑚+2𝑛−2 ,
𝑚+2𝑛−2 󸀠
)
𝜆 = (𝑚 + 𝑛 − 1) (𝑚 + 2𝑛 − 2) ,
𝑚 = 2 (𝑛 − 1) .
(8)
3. Application of the Method
Example 2. Let’s consider RLW Burgers equation
= 𝜆𝑚 (𝑚 + 𝑛 − 1) (𝑚 + 2𝑛 − 2) 𝐹𝑚+2𝑛−3 𝐹󸀠
= 𝜆𝑚 (𝑚 + 𝑛 − 1) (𝑚 + 2𝑛 − 2) 𝐹
𝜆2 𝑚𝐹2𝑚+𝑛−1 = 𝜆𝑚 (𝑚 + 𝑛 − 1) (𝑚 + 2𝑛 − 2) 𝐹𝑚+3𝑛−3 ,
If it is noticed that our new balance term 𝑚 (𝑚 = 2(𝑛 − 1))
is connected to 𝑛. Namely our new balance term definition is
connected to chosen auxiliary equation.
󸀠
𝑢󸀠 = (𝜆𝐹𝑚 ) = 𝜆𝑚𝐹𝑚−1 𝐹󸀠 = 𝜆𝑚𝐹𝑚−1 𝐹𝑛 = 𝜆𝑚𝐹𝑚+𝑛−1 ,
𝑢󸀠󸀠󸀠 = (𝜆𝑚 (𝑚 + 𝑛 − 1) 𝐹
we have the equations above and the equating 𝑢𝑢󸀠 to 𝑢󸀠󸀠󸀠
𝑢𝑡 + 𝑢𝑥 + 12𝑢𝑢𝑥 − 𝑢𝑥𝑥 − 𝑢𝑥𝑥𝑡 = 0,
(9)
𝑚+2𝑛−3 𝑛
𝐹
with the transform 𝑢(𝑥, 𝑡) = 𝑢(𝜉), 𝜉 = 𝑥 − 𝑐𝑡 we have the
following equation
= 𝜆𝑚 (𝑚 + 𝑛 − 1) (𝑚 + 2𝑛 − 2) 𝐹𝑚+3𝑛−3 ,
𝑢𝑢󸀠 = 𝜆𝐹𝑚 𝜆𝑚𝐹𝑚+𝑛−1 = 𝜆2 𝑚𝐹2𝑚+𝑛−1
(7)
−𝑐𝑢󸀠 + 𝑢󸀠 + 12𝑢𝑢󸀠 − 𝑢󸀠󸀠 + 𝑐𝑢󸀠󸀠󸀠 = 0.
(10)
4
Advances in Mathematical Physics
1
0.4
0.5
4
0
−5
0.2
4
0
−0.2
−5
2
2
0
0
5
0
0
5
(a)
(b)
1
0
0.5
−1
4
−2
−3
−5
0
−0.5
4
−5
2
2
0
0
5
0
5
(c)
0
(d)
Figure 4: Graph of the solutions 𝑢(𝑥, 𝑡) and V(𝑥, 𝑡) corresponding to the value 𝑏0 = 1 from left to the right for (25) and (26).
From the Definition 2 we have the balance term of RLW Burgers equation by using the auxiliary equation (3) for “Case 1”,
is equal to 4. Therefore, we may choose the following ansatz:
(−
3𝑎−4 27𝑐𝑎−4
2
2
+ 9𝑎−3
−
− 24𝑎−4
2
2
+
𝑢 (𝑥, 𝑡) = 48 (𝑎−4 𝐹−4 + 𝑎−3 𝐹−3 + 𝑎−2 𝐹−2 + 𝑎−1 𝐹−1
(11)
(−
+𝑎0 + 𝑎1 𝐹 + 𝑎2 𝐹2 + 𝑎3 𝐹3 + 𝑎4 𝐹4 ) .
Substituting (11) into (10) along with (3) and using Mathematica yields a system of equations w.r.t. 𝐹𝑖 . Setting the
coefficients of 𝐹𝑖 in the obtained system of equations to zero,
we can deduce the following set of algebraic polynomials with
the respect unknowns 𝑎0 , 𝑎1 , 𝑎2 , . . . for the Case 1:
(
105𝑐𝑎−3
+ 21𝑎−4 𝑎−3 ) = 0,
64
15𝑎−3 225𝑐𝑎−3
−
− 42𝑎−4 𝑎−3 + 15𝑎−3 𝑎−2
16
32
+
15𝑐𝑎−1
+ 15𝑎−4 𝑎−1 ) = 0,
64
2
2
(6𝑎−4 + 28𝑐𝑎−4 + 24𝑎−4
− 18𝑎−3
−
𝑎−2
− 3𝑐𝑎−2
2
2
−36𝑎−4 𝑎−2 + 6𝑎−2
+ 12𝑎−3 𝑎−1 + 12𝑎−4 𝑎0 ) = 0,
(
2
3𝑐𝑎−4 + 12𝑎−4
= 0,
3𝑐𝑎−2
+ 18𝑎−4 𝑎−2 ) = 0,
4
15𝑎−3 429𝑐𝑎−3
+
+ 42𝑎−4 𝑎−3 − 30𝑎−3 𝑎−2
4
32
−
3𝑎−1 27𝑐𝑎−1
−
− 30𝑎−4 𝑎−1 + 9𝑎−2 𝑎−1
16
32
+9𝑎−3 𝑎0 −
3𝑐𝑎1
+ 9𝑎−4 𝑎1 ) = 0,
64
Advances in Mathematical Physics
5
2
(8𝑎−4 + 20𝑐𝑎−4 − 3𝑎−2 − 4𝑐𝑎−2 + 12𝑎−2
(6𝑎12 − 𝑎2 + 4𝑐𝑎2 + 12𝑎0 𝑎2 − 6𝑎22
2
+ 24𝑎−3 𝑎−1 − 6𝑎−1
+ 24𝑎−4 𝑎0
+ 12𝑎−1 𝑎3 − 12𝑎1 𝑎3 + 2𝑎4 − 10𝑐𝑎4
−12𝑎−2 𝑎0 − 12𝑎−3 𝑎1 − 12𝑎−4 𝑎2 ) = 0,
+12𝑎−2 𝑎4 − 12𝑎0 𝑎4 ) = 0,
2
+ 2𝑎−2
(−10𝑎−4 − 32𝑐𝑎−4 + 18𝑎−3
2
+ 5𝑐𝑎−2 + 36𝑎−4 𝑎−2 − 12𝑎−2
− 24𝑎−3 𝑎−1 +
2
3𝑎−1
𝑎
3𝑐𝑎−1 𝑎1 35𝑐𝑎1
3
(− 𝑐𝑎−3 + −1 −
+
−
8
4
8
2
16
− 6𝑎0 𝑎1 − 6𝑎−1 𝑎2 + 18𝑎1 𝑎2 − 3𝑎3
− 24𝑎−4 𝑎0
27𝑐𝑎3
− 6𝑎−2 𝑎3 + 18𝑎0 𝑎3 − 15𝑎2 𝑎3
2
+
+6𝑎−2 𝑎0 + 6𝑎−3 𝑎1 + 6𝑎−4 𝑎2 ) = 0,
(−6𝑎−3 −
27𝑐𝑎−3
3𝑎
+ 30𝑎−3 𝑎−2 + −1
2
4
35𝑐𝑎−1
+
+ 30𝑎−4 𝑎−1 − 18𝑎−2 𝑎−1
32
− 18𝑎−3 𝑎0 + 3𝑎−1 𝑎0 +
𝑎1 3𝑐𝑎1
+
16
32
−6𝑎−3 𝑎4 + 18𝑎−1 𝑎4 − 15𝑎1 𝑎4 ) = 0,
(−6𝑎12 + 2𝑎2 − 10𝑐𝑎2 − 12𝑎0 𝑎2 + 12𝑎22
− 12𝑎−1 𝑎3 + 24𝑎1 𝑎3 − 9𝑎32 − 6𝑎4 + 32𝑐𝑎4
−12𝑎−2 𝑎4 + 24𝑎0 𝑎4 − 18𝑎2 𝑎4 ) = 0,
− 18𝑎−4 𝑎1 + 3𝑎−2 𝑎1 + 3𝑎−3 𝑎2
+
(
(−
3𝑐𝑎3
+ 3𝑎−4 𝑎3 ) = 0,
64
(
9𝑎−3 105𝑐𝑎−3
𝑐𝑎
+
− 𝑎−1 − −1 + 18𝑎−2 𝑎−1
2
16
2
𝑎
𝑐𝑎
+ 18𝑎−3 𝑎0 − 6𝑎−1 𝑎0 − 1 + 1
4
32
+ 18𝑎−4 𝑎1 − 6𝑎−2 𝑎1 − 3𝑎0 𝑎1
−6𝑎−4 𝑎3 − 3𝑎−2 𝑎3 − 3𝑎−3 𝑎4 ) = 0,
(− 2𝑎−4 − 6𝑐𝑎−4 + 2𝑎−2 + 𝑐𝑎−2 +
2
6𝑎−1
𝑐𝑎2
2
𝑎4
+ 12𝑎−4 𝑎2 − 6𝑎0 𝑎2 − 6𝑎−1 𝑎3 −
2
+ 12𝑎−2 𝑎0 + 12𝑎−3 𝑎1 − 3𝑎12 −
+
3𝑐𝑎4
− 6𝑎−2 𝑎4 ) = 0,
2
3𝑎
9𝑐𝑎−3 𝑎−1 𝑐𝑎−1
(− −3 −
+
−
+ 6𝑎−1 𝑎0
4
8
2
16
𝑐𝑎
+ 1 + 6𝑎−2 𝑎1 + 6𝑎0 𝑎1 + 6𝑎−3 𝑎2
2
3𝑎
105𝑐𝑎3
+ 6𝑎−1 𝑎2 − 9𝑎1 𝑎2 + 3 −
4
32
+ 6𝑎−4 𝑎3 + 6𝑎−2 𝑎3 − 9𝑎0 𝑎3
+6𝑎−3 𝑎4 − 9𝑎−1 𝑎4 ) = 0,
3𝑐𝑎−1 3𝑎1 27𝑐𝑎1
9𝑎
−
+
− 18𝑎1 𝑎2 + 3
8
4
8
2
429𝑐𝑎3
− 18𝑎0 𝑎3 + 30𝑎2 𝑎3
16
−
−18𝑎−1 𝑎4 + 30𝑎1 𝑎4 − 21𝑎3 𝑎4 ) = 0,
(−
3𝑎
9𝑐𝑎3
− 6𝑎−3 𝑎2 − 3𝑎−1 𝑎2 − 3 +
16
32
105𝑐𝑎3
− 42𝑎3 𝑎4 ) = 0,
8
15𝑐𝑎1 15𝑎3 225𝑐𝑎3
−
+
8
4
8
−30𝑎2 𝑎3 − 30𝑎1 𝑎4 + 42𝑎3 𝑎4 ) = 0,
(−24𝑐𝑎4 − 24𝑎42 ) = 0,
(−2𝑎2 + 12𝑐𝑎2 − 12𝑎22 − 24𝑎1 𝑎3
+ 18𝑎32 + 8𝑎4 − 56𝑐𝑎4 − 24𝑎0 𝑎4
+36𝑎2 𝑎4 − 12𝑎42 ) = 0,
(−6𝑐𝑎2 − 18𝑎32 − 6𝑎4
+54𝑐𝑎4 − 36𝑎2 𝑎4 + 24𝑎42 ) = 0.
(12)
From the system of (12) we have
(i)
𝑎0 = −
11
𝑖
− ,
60 12
𝑎1 = 𝑎2 = 𝑎3 = 𝑎4 = 𝑎−1 = 𝑎−3 = 0,
6
Advances in Mathematical Physics
𝑎−2 =
𝑎−4
1
𝑖
+ ,
10 10
𝑖
=− ,
20
(
𝑖
𝑐= ,
5
9𝑎−3 135𝑐𝑎−3
−
+ 30𝑎−3 𝑎−2
4
16
𝑢 (𝑥, 𝑡)
= −(
11 + 5𝑖
1+𝑖
)+(
)
60
10
−9𝑎−2 𝑎−1 − 9𝑎−3 𝑎0 ) = 0,
(−2𝑎−4 + 6𝑐𝑎−4 + 𝑎−2 − 3𝑐𝑎−2
−2
1
)
×(
√1 + sec [𝑥 − (𝑖/5) 𝑡] + tan [𝑥 − (𝑖/5) 𝑡]
2
2
+ 12𝑎−2
+ 24𝑎−3 𝑎−1 − 3𝑎−1
−4
−
3𝑎−1 27𝑐𝑎−1
+
+ 30𝑎−4 𝑎−1
16
32
−
+24𝑎−4 𝑎0 − 6𝑎−2 𝑎0 ) = 0,
𝑖
1
(
) .
20 √1 + sec [𝑥 − (𝑖/5) 𝑡] + tan [𝑥 − (𝑖/5) 𝑡]
(−
(13)
3𝑎−3 15𝑐𝑎−3 𝑎−1 9𝑐𝑎−1
+
+
−
4
8
4
16
+18𝑎−2 𝑎−1 + 18𝑎−3 𝑎0 − 3𝑎−1 𝑎0 ) = 0,
(ii)
𝑎0 =
1 + 5𝑖
,
60
𝑐 = −𝑎4 ,
(−
𝑎1 = 𝑎−2 = 𝑎3 = 𝑎−4 = 𝑎−1 = 𝑎−3 = 0,
𝑎2 = −
1+𝑖
,
5
2
+ 12𝑎−2 𝑎0 ) = 0,
(6𝑎−1
𝑖
𝑎4 = ± ,
5
(
1 + 5𝑖 1 + 𝑖
−
60
5
𝑢 (𝑥, 𝑡) =
(15)
From the system of (15) we have
(i)
2
𝑎0 = 𝑎−1 = 0,
4
𝑖
𝑖
𝑖
+ (√ 1 + sec [𝑥 + 𝑡] + tan [𝑥 + 𝑡]) .
5
5
5
From the Definition 2 we have the balance term of RLW
Burgers equation by using the auxiliary equation (3) for “Case
2”, is equal to −4 then we have the following system of
equations
𝑎−3 = 0,
1
,
20
1
𝑐=− ,
5
−2
1
1
=− (
)
5 √1 + 𝑖 sech [𝑥 + (1/5) 𝑡] + tanh [𝑥 + (1/5) 𝑡]
−4
3𝑎
27𝑐𝑎−4
2
2
(− −4 +
− 9𝑎−3
+ 24𝑎−4
2
2
3𝑐𝑎−2
− 18𝑎−4 𝑎−2 ) = 0,
4
15𝑎−3 225𝑐𝑎−3
+
+ 42𝑎−4 𝑎−3 − 15𝑎−3 𝑎−2
16
32
+
1
1
(
) ,
20 √1 + 𝑖 sech [𝑥 + (1/5) 𝑡] + tanh [𝑥 + (1/5) 𝑡]
(16)
or
𝑢 (𝑥, 𝑡)
1
1
=− (
)
5 √1 + csch [𝑥 + (1/5) 𝑡] + coth [𝑥 + (1/5) 𝑡]
15𝑐𝑎−1
−
− 15𝑎−4 𝑎−1 ) = 0,
64
2
(4𝑎−4 − 18𝑐𝑎−4 + 18𝑎−3
−
𝑎−4 =
1
𝑎−2 = − ,
5
𝑢 (𝑥, 𝑡)
2
= 0,
−3𝑐𝑎−4 − 12𝑎−4
(−
𝑎−1 3𝑐𝑎−1
−
+ 6𝑎−1 𝑎0 ) = 0.
4
8
(14)
𝑖
𝑖
× (√ 1 + sec [𝑥 + 𝑡] + tan [𝑥 + 𝑡])
5
5
−
105
𝑐𝑎 − 21𝑎−4 𝑎−3 ) = 0,
64 −3
𝑎−2
2
2
+ 3𝑐𝑎−2 + 36𝑎−4 𝑎−2 − 6𝑎−2
− 12𝑎−3 𝑎−1 −12𝑎−4 𝑎0 ) = 0,
−2
−4
1
1
+ (
) .
20 √1 + csch [𝑥 + (1/5) 𝑡] + coth [𝑥 + (1/5) 𝑡]
(17)
Advances in Mathematical Physics
7
Example 3. Let’s consider Hirota Satsuma coupled equation
𝑎1 𝑐𝑎1 3𝑎0 𝑎1 9𝑎1 𝑎2
−
−
+
2
2
2
4
(−
1
𝑢𝑡 − 3𝑢𝑢𝑥 + 6VV𝑥 − 𝑢𝑥𝑥𝑥 = 0,
2
+
(18)
V𝑡 + 3𝑢V𝑥 + V𝑥𝑥𝑥 = 0,
129𝑎3 3𝑐𝑎3 9𝑎0 𝑎3
+
+
64
4
4
+ 3𝑏0 𝑏1 −
(−
with the transform 𝑢(𝑥, 𝑡) = 𝑢(𝜉), V(𝑥, 𝑡) = V(𝜉), 𝜉 = 𝑥 − 𝑐𝑡 we
have
3𝑎12 5𝑎2
−
− 𝑐𝑎2 − 3𝑎0 𝑎2
2
2
+
󸀠
−𝑐V + 3𝑢V + V
󸀠󸀠󸀠
(19)
−6𝑏1 𝑏3 − 6𝑏0 𝑏4 ) = 0,
= 0.
(
From the Definition 2 we have the balance term of Hirota
Satsuma coupled equation by using the auxiliary equation (3)
for “Case 1”, is equal to 4 for 𝑢 and V. Therefore, we may choose
the following ansatz:
2
3
43𝑎1 𝑐𝑎1 3𝑎0 𝑎1 9𝑎1 𝑎2
+
+
−
32
2
2
2
−
15𝑎3 3𝑐𝑎3 9𝑎0 𝑎3 15𝑎2 𝑎3
−
−
+
2
2
2
4
+
15𝑎1 𝑎4
− 3𝑏0 𝑏1 + 9𝑏1 𝑏2
4
+9𝑏0 𝑏3 −
4
𝑢 (𝑥, 𝑡) = 48 (𝑎0 + 𝑎1 𝐹 + 𝑎2 𝐹 + 𝑎3 𝐹 + 𝑎4 𝐹 ) ,
V (𝑥, 𝑡) = 48 (𝑏0 + 𝑏1 𝐹 + 𝑏2 𝐹2 + 𝑏3 𝐹3 + 𝑏4 𝐹4 ) .
(20)
(
Substituting (20) into (19) along with (3) and using Mathematica yields a system of equations w.r.t. 𝐹𝑖 . Setting the
coefficients of 𝐹𝑖 in the obtained system of equations to zero,
we can deduce the following set of algebraic polynomials with
the respect unknowns 𝑎0 , 𝑎1 , 𝑎2 , . . . for the Case 1:
3𝑎
3𝑎
(− 1 − 3 ) = 0,
64 128
3𝑎 𝑎
9𝑎
3𝑏 𝑏
7𝑎
𝑐𝑎
( 1 + 1 + 0 1 − 3 − 0 1 ) = 0,
64
4
4
64
2
3𝑎12 𝑎2 𝑐𝑎2 3𝑎0 𝑎2
+
+
+
4
2
2
2
−
3𝑎4 3𝑏12
−
− 3𝑏0 𝑏2 ) = 0,
4
2
15𝑏2 𝑏3 15𝑏1 𝑏4
−
) = 0,
2
2
3𝑎12 11𝑎2
+
+ 𝑐𝑎2 + 3𝑎0 𝑎2 − 3𝑎22
2
2
− 6𝑎1 𝑎3 +
9𝑎32
− 17𝑎4 − 2𝑐𝑎4
4
− 6𝑎0 𝑎4 +
9𝑎2 𝑎4
− 3𝑏12 − 6𝑏0 𝑏2
2
+ 6𝑏22 + 12𝑏1 𝑏3 −
9𝑏32
2
+12𝑏0 𝑏4 − 9𝑏2 𝑏4 ) = 0,
3𝑎1
= 0,
128
(
3𝑎22
11𝑎4
+ 3𝑎1 𝑎3 +
+ 𝑐𝑎4
2
2
+ 3𝑎0 𝑎4 + 3𝑏12 + 6𝑏0 𝑏2 − 3𝑏22
1
−𝑐𝑢󸀠 − 3𝑢𝑢󸀠 + 6VV󸀠 − 𝑢󸀠󸀠󸀠 = 0,
2
󸀠
9𝑏1 𝑏2 9𝑏0 𝑏3
−
) = 0,
2
2
(
105𝑎3 21𝑎3 𝑎4
+
− 21𝑏3 𝑏4 ) = 0,
16
2
(−
27𝑎1 9𝑎1 𝑎2 453𝑎3 3𝑐𝑎3
+
+
+
16
2
32
2
+
9𝑎0 𝑎3 15𝑎2 𝑎3 15𝑎1 𝑎4
−
−
2
2
2
+
21𝑎3 𝑎4
− 9𝑏1 𝑏2 − 9𝑏0 𝑏3
4
+15𝑏2 𝑏3 + 15𝑏1 𝑏4 −
21𝑏3 𝑏4
) = 0,
2
8
Advances in Mathematical Physics
(
15𝑎1 225𝑎3 15𝑎2 𝑎3
−
+
16
16
2
+
−2𝑐𝑏4 + 6𝑎0 𝑏4 − 3𝑎2 𝑏4 ) = 0,
15𝑎1 𝑎4 21𝑎3 𝑎4
−
− 15𝑏2 𝑏3
2
2
−15𝑏1 𝑏4 + 21𝑏3 𝑏4 ) = 0,
(12𝑎4 + 6𝑎42 − 12𝑏42 ) = 0,
(−6𝑎2 +
3𝑎22
9𝑎2
+ 6𝑎1 𝑎3 − 3
2
+ 29𝑎4 + 2𝑐𝑎4 + 6𝑎0 𝑎4
− 9𝑎2 𝑎4 + 3𝑎42 − 6𝑏22
(−
(
27𝑏1 3𝑎2 𝑏1 3𝑎4 𝑏1
−
+
− 3𝑎1 𝑏2
8
2
2
+ 3𝑎3 𝑏2 −
+
(−
− 12𝑏1 𝑏3 + 9𝑏32 − 12𝑏0 𝑏4
+18𝑏2 𝑏4 −
6𝑏42 )
3𝑏
= 0 − 1 = 0,
64
−
𝑐𝑏1 3𝑎0 𝑏1 3𝑎2 𝑏1 3𝑎1 𝑏2
+
−
−
2
2
4
2
(−6𝑏2 − 3𝑎4 𝑏2 −
9𝑎3 𝑏3
+ 54𝑏4
2
−6𝑎2 𝑏4 + 6𝑎4 𝑏4 ) = 0,
3𝑎1 𝑏1 3𝑎3 𝑏1
−
+ 5𝑏2 − 𝑐𝑏2
2
4
+ 3𝑎0 𝑏2 −
9𝑎1 𝑏3 9𝑎3 𝑏3
+
− 58𝑏4 + 2𝑐𝑏4
2
2
−6𝑎0 𝑏4 + 6𝑎2 𝑏4 − 3𝑎4 𝑏4 ) = 0,
129𝑏3 3𝑐𝑏3 9𝑎0 𝑏3
+
−
) = 0,
32
4
4
(
9𝑎2 𝑏3 9𝑎4 𝑏3
+
− 6𝑎1 𝑏4 + 6𝑎3 𝑏4 ) = 0,
2
2
3
(− 𝑎3 𝑏1 + 12𝑏2 − 3𝑎2 𝑏2 + 3𝑎4 𝑏2
2
3𝑎 𝑏
𝑐𝑏
3𝑏
3
(− 𝑎1 𝑏1 − 𝑏2 + 2 − 0 2 + 4 ) = 0,
4
2
2
2
−
9𝑎2 𝑏3 9𝑎4 𝑏3
−
+ 6𝑎1 𝑏4 − 3𝑎3 𝑏4 ) = 0,
2
4
(−24𝑏4 − 6𝑎4 𝑏4 ) = 0,
7𝑏1 𝑐𝑏1 3𝑎0 𝑏1 9𝑏3
+
−
+
) = 0,
32
4
4
32
(𝑏1 −
453𝑏3 3𝑐𝑏3 9𝑎0 𝑏3
+
−
16
2
2
225𝑏3
15𝑏1 3𝑎4 𝑏1
−
− 3𝑎3 𝑏2 +
8
2
8
−
3𝑏
3𝑏
( 1 + 3 ) = 0,
32
64
(−
105𝑏3 9𝑎4 𝑏3
−
− 6𝑎3 𝑏4 ) = 0,
8
2
(3𝑎2 +
3𝑎2 𝑏2 9𝑎1 𝑏3
−
2
4
9𝑎32
− 27𝑎4 + 9𝑎2 𝑎4 − 6𝑎42
2
−9𝑏32 − 18𝑏2 𝑏4 + 12𝑏42 ) = 0.
−11𝑏4 + 𝑐𝑏4 − 3𝑎0 𝑏4 ) = 0,
(21)
3𝑎 𝑏
43𝑏
𝑐𝑏
3𝑎 𝑏
(− 1 + 1 − 0 1 + 2 1
16
2
2
2
−
3𝑎 𝑏
3𝑎4 𝑏1
+ 3𝑎1 𝑏2 − 3 2 + 15𝑏3
4
2
−
3𝑐𝑏3 9𝑎0 𝑏3 9𝑎2 𝑏3
+
−
− 3𝑎1 𝑏4 ) = 0,
2
2
4
3𝑎 𝑏
3
(− 𝑎1 𝑏1 + 3 1 − 11𝑏2 + 𝑐𝑏2
2
2
− 3𝑎0 𝑏2 + 3𝑎2 𝑏2 −
+
3𝑎4 𝑏2
2
9𝑎1 𝑏3 9𝑎3 𝑏3
−
+ 34𝑏4
2
4
From the system of (21) we have
(i)
𝑐=
1
(5 − 6𝑏0 ) ,
4
𝑎1 = 𝑎3 = 0,
𝑎0 =
1
(−5 − 2𝑏0 ) ,
4
𝑏1 = 𝑏3 = 0,
𝑎2 = 4,
𝑏2 = −2,
𝑎4 = −4,
𝑏4 = 2,
Advances in Mathematical Physics
𝑢 (𝑥, 𝑡) =
9
1
(−5 − 2𝑏0 )
4
𝑏2
2
V (𝑥, 𝑡) = −
+ 4 (((1 + sec [𝑥 −
+ tan [𝑥 −
− 𝑏2 (((1 + sec [𝑥 −
−1
2
−1
4
−1/2
1
(5 − 6𝑏0 ) 𝑡]) ) )
4
− 4 (((1 + sec [𝑥 −
+ tan [𝑥 −
1
(5 − 6𝑏0 ) 𝑡]
4
−1
1
(5 − 6𝑏0 ) 𝑡]
4
From the Definition 2 we have the balance term of Hirota
Satsuma coupled equation by using the auxiliary equation (3)
for “Case 2”, is equal to −4 then we have the following system
of equations
V (𝑥, 𝑡) = 𝑏0
3𝑎−4
2
2
− 6𝑏−4
= 0,
+ 3𝑎−4
2
1
(5 − 6𝑏0 ) 𝑡]
4
(
−1
−1/2
1
+ tan [𝑥 − (5 − 6𝑏0 ) 𝑡]) ) )
4
+ 2 (((1 + sec [𝑥 −
2
105𝑎−3 21
21
+ 𝑎−4 𝑎−3 − 𝑏−4 𝑏−3 ) = 0,
128
4
2
(−
1
(5 − 6𝑏0 ) 𝑡]
4
(−
4
−1
𝑎−1 𝑐𝑎−1 3
−
− 𝑎−1 𝑎0 + 3𝑏−1 ) = 0,
16
2
2
9𝑎2
27𝑎−4
3𝑎
9
2
+ −3 + −2 + 𝑎−4 𝑎−2
− 6𝑎−4
4
4
8
2
2
+12𝑏−4
−
−1/2
1
+ tan [𝑥 − (5 − 6𝑏0 ) 𝑡]) ) ) .
4
(22)
(−
(ii)
𝑐=
1
(−1 − 3𝑏22 ) ,
4
𝑎1 = 𝑎3 = 0,
𝑎2 = 2,
𝑎0 =
1
(−3 − 𝑏22 ) ,
4
𝑏1 = 𝑏3 = 𝑏4 = 0,
𝑏2 ≠ 0,
(
𝑎4 = −2,
1
𝑢 (𝑥, 𝑡) = (−3 − 𝑏22 )
4
+ 2 (((1 + sec [𝑥 −
225𝑎−3 21
15
− 𝑎−4 𝑎−3 + 𝑎−3 𝑎−2
64
2
4
+
15𝑎−1 15
+ 𝑎−4 𝑎−1 + 21𝑏−4 𝑏−3
128
4
−
15
15
𝑏 𝑏 − 𝑏 𝑏 ) = 0,
2 −3 −2 2 −4 −1
2
3𝑎−2
+ 3𝑎−3 𝑎−1 + 3𝑎−4 𝑎0 − 6𝑏−4
2
2
2
+9𝑏−3
+ 18𝑏−4 𝑏−2 − 3𝑏−2
− 6𝑏−3 𝑏−1 ) = 0,
1
(−1 − 3𝑏22 ) 𝑡]
4
−1
− 2 (((1 + sec [𝑥 −
2
9𝑏−3
− 9𝑏−4 𝑏−2 ) = 0,
2
9𝑎2
19𝑎−4
3𝑎
+ 𝑐𝑎−4 − −3 − −2 − 9𝑎−4 𝑎−2
2
2
2
+
−1/2
1
+ tan [𝑥 − (−1 − 3𝑏22 ) 𝑡]) ) )
4
2
(
147𝑎−3 3𝑐𝑎−3 15
+
− 𝑎−3 𝑎−2
32
4
2
−
1
(−1 − 3𝑏22 ) 𝑡]
4
−1
2
−1/2
1
+ tan [𝑥 − (−1 − 3𝑏22 ) 𝑡]) ) ) .
4
(23)
−1/2
1
(5 − 6𝑏0 ) 𝑡]) ) ) ,
4
− 2 (((1 + sec [𝑥 −
1
(−1 − 3𝑏22 ) 𝑡]
4
4
−1/2
1
+ tan [𝑥 − (−1 − 3𝑏22 ) 𝑡]) ) ) ,
4
27𝑎−1 15
9
− 𝑎−4 𝑎−1 + 𝑎−2 𝑎−1
64
2
4
9𝑏
9
+ 𝑎−3 𝑎0 − −3 + 15𝑏−3 𝑏−2
4
2
9
+15𝑏−4 𝑏−1 − 𝑏−2 𝑏−1 ) = 0,
2
10
Advances in Mathematical Physics
(−
9
(8𝑏−4 − 2𝑐𝑏−4 + 6𝑎0 𝑏−4 + 𝑎−1 𝑏−3
2
27𝑎−3 3𝑐𝑎−3 13𝑎−1 𝑐𝑎−1
−
+
+
16
2
32
4
9
9
3
− 𝑎−2 𝑎−1 − 𝑎−3 𝑎0 + 𝑎−1 𝑎0
2
2
4
+9𝑏−3 −
−
3𝑏−1
+ 9𝑏−2 𝑏−1 ) = 0,
2
(− 4𝑎−4 − 2𝑐𝑎−4 +
3
3
3
− 𝑎0 𝑏−2 + 𝑎−3 𝑏−1 − 𝑎−1 𝑏−1 ) = 0,
2
2
4
7𝑎−2 𝑐𝑎−2
+
4
2
2
− 3𝑎−2
− 6𝑎−3 𝑎−1 +
3
(𝑏−2 − 𝑐𝑏−2 + 3𝑎0 𝑏−2 + 𝑎−1 𝑏−1 ) = 0,
2
2
3𝑎−1
4
(
3
− 6𝑎−4 𝑎0 + 𝑎−2 𝑎0 + 12𝑏−4
2
27𝑏−3 3𝑐𝑏−3 9
−
+ 𝑎0 𝑏−3 + 3𝑎−1 𝑏−2
8
2
2
−
3𝑏2
2
−3𝑏−2 + 6𝑏−2
+ 12𝑏−3 𝑏−1 − −1 ) = 0,
2
(−
𝑎−2
2
) = 0,
− 𝑐𝑎−2 −
− 3𝑎−2 𝑎0 + 6𝑏−2 + 3𝑏−1
2
2
(−3𝑎−3 𝑏−4 −
(
105𝑏−3 9
− 𝑎−4 𝑏−3 ) = 0,
64
4
From the system of (3) we have
(i)
𝑎−1 = 𝑎−3 = 𝑏−1 = 𝑏−3 = 0,
3𝑏
9
3
− 𝑎−3 𝑏−3 − −2 − 𝑎−4 𝑏−2 ) = 0,
4
4
2
𝑏−2 = ±1,
225𝑏−3
32
15𝑏−1 3
− 𝑎−4 𝑏−1 ) = 0,
64
4
( − 19𝑏−4 + 𝑐𝑏−4 + 6𝑎−2 𝑏−4 − 3𝑎0 𝑏−4
9
9
+ 𝑎−3 𝑏−3 − 𝑎−1 𝑏−3 + 3𝑏−2
2
4
3
3
+3𝑎−4 𝑏−2 − 𝑎−2 𝑏−2 − 𝑎−3 𝑏−1 ) = 0,
2
4
(6𝑎−1 𝑏−4 −
147𝑏−3 3𝑐𝑏−3
+
16
4
9
9
+ 𝑎−2 𝑏−3 − 𝑎0 𝑏−3
2
4
27𝑏−1
3
+ 3𝑎−3 𝑏−2 − 𝑎−1 𝑏−2 +
2
32
3
3
+ 𝑎−4 𝑏−1 − 𝑎−2 𝑏−1 ) = 0,
2
4
𝑏−4 = −
𝑎−4 = −1,
9
9
3
+ 𝑎−4 𝑏−3 − 𝑎−2 𝑏−3 − 𝑎−3 𝑏−2
2
4
2
−
𝑏−1 𝑐𝑏−1 3
−
+ 𝑎0 𝑏−1 ) = 0.
8
2
2
(24)
27𝑏−4
+ 6𝑎−4 𝑏−4 − 3𝑎−2 𝑏−4
2
(6𝑎−3 𝑏−4 − 3𝑎−1 𝑏−4 +
13𝑏−1 𝑐𝑏−1 3
3
+
+ 𝑎−2 𝑏−1 − 𝑎0 𝑏−1 ) = 0,
16
4
2
4
(
2
3𝑎−1
−3𝑏−4 − 3𝑎−4 𝑏−4 = 0,
7𝑏−2 𝑐𝑏−2
+
+ 3𝑎−2 𝑏−2
2
2
𝑎0 =
𝑐=
𝑏−2
,
2
𝑎−2 = 2,
1
(1 + 6𝑏−2 ) ,
4
1
(−1 + 2𝑏−2 ) ,
4
𝑢 (𝑥, 𝑡)
1
1
= +2 (
)
4
√1 + 𝑖 sech [𝑥 − (7/4) 𝑡] + tanh [𝑥 − (7/4) 𝑡]
−2
−4
1
−(
) ,
√1 + 𝑖 sech [𝑥 − (7/4) 𝑡] + tanh [𝑥 − (7/4) 𝑡]
V (𝑥, 𝑡)
−2
= 𝑏0 + (
1
)
√1 + 𝑖 sech [𝑥 − (7/4) 𝑡] + tanh [𝑥 − (7/4) 𝑡]
−4
1
1
− (
) ,
2 √1 + 𝑖 sech [𝑥 − (7/4) 𝑡] + tanh [𝑥 − (7/4) 𝑡]
(25)
Advances in Mathematical Physics
11
or
𝑢 (𝑥, 𝑡)
=
1
1
)
+2(
4
√1 + csch [𝑥 − (7/4) 𝑡] + coth[𝑥 − (7/4) 𝑡]
−2
−4
1
−(
) ,
√1 + csch [𝑥 − (7/4) 𝑡] + coth [𝑥 − (7/4) 𝑡]
V (𝑥, 𝑡)
−2
= 𝑏0 + (
1
)
√1 + csch [𝑥 − (7/4) 𝑡] + coth [𝑥 − (7/4) 𝑡]
−4
1
1
− (
) .
2 √1 + csch [𝑥 − (7/4) 𝑡] + coth [𝑥 − (7/4) 𝑡]
(26)
Figures 1, 2, 3 and 4 gives to us 3D graphics for RLW Burgers
and RLW Burgers and Hirota Satsuma coupled equations.
4. Conclusion
We have presented a new method and balance term definition
and used it to solve the RLW Burgers and Hirota Satsuma
coupled equations. In fact, this method is readily applicable
to a large variety of nonlinear PDEs. First, all the nonlinear
PDEs which can be solved by the other methods can be
solved by our method. Second, we used only the special
solutions of (3). If we use other solutions of (3), we can obtain
more travelling wave solutions. Third it is a computerizable
method, which allow us to perform complicated and tedious
algebraic calculation on computer and so our balance term
definition is effectively useful for any to chosen auxiliary
equation.
References
[1] B. R. Duffy and E. J. Parkes, “Travelling solitary wave solutions
to a seventh-order generalized KdV equation,” Physics Letters A,
vol. 214, no. 5-6, pp. 271–272, 1996.
[2] E. J. Parkes and B. R. Duffy, “Travelling solitary wave solutions
to a compound KdV-Burgers equation,” Physics Letters A, vol.
229, no. 4, pp. 217–220, 1997.
[3] E. J. Parkes, “Exact solutions to the two-dimensional Kortewegde Vries-Burgers equation,” Journal of Physics A, vol. 27, no. 13,
pp. L497–L501, 1994.
[4] L. Debnath, Nonlinear Partial Differential Equations for Scientists and Engineers, Birkhäuser, Boston, Mass, USA, 1997.
[5] A. M. Wazwaz, Partial Differential Equations: Methods and
Applications, Balkema, Rotterdam, The Netherlands, 2002.
[6] W. Hereman, P. P. Banerjee, A. Korpel, G. Assanto, A. van
Immerzeele, and A. Meerpoel, “Exact solitary wave solutions of
nonlinear evolution and wave equations using a direct algebraic
method,” Journal of Physics A, vol. 19, no. 5, pp. 607–628, 1986.
[7] X.-B. Hu and W.-X. Ma, “Application of Hirota’s bilinear
formalism to the Toeplitz lattice—some special soliton-like
solutions,” Physics Letters A, vol. 293, no. 3-4, pp. 161–165, 2002.
[8] A. M. Abourabia and M. M. El Horbaty, “On solitary wave solutions for the two-dimensional nonlinear modified Kortweg-de
Vries-Burger equation,” Chaos, Solitons & Fractals, vol. 29, no.
2, pp. 354–364, 2006.
[9] T. L. Bock and M. D. Kruskal, “A two-parameter Miura
transformation of the Benjamin-Ono equation,” Physics Letters
A, vol. 74, no. 3-4, pp. 173–176, 1979.
[10] V. B. Matveev and M. A. Salle, Darboux Transformations and
Solitons, Springer Series in Nonlinear Dynamics, Springer,
Berlin, Germany, 1991.
[11] F. Cariello and M. Tabor, “Painlevé expansions for nonintegrable evolution equations,” Physica D, vol. 39, no. 1, pp. 77–94,
1989.
[12] E. Fan and H. Zhang, “New exact solutions to a solutions to a
system of coupled KdV equations,” Physics Letters A, vol. 245,
no. 5, pp. 389–392, 1998.
[13] E. Fan, “Auto-Bäcklund transformation and similarity reductions for general variable coefficient KdV equations,” Physics
Letters A, vol. 294, no. 1, pp. 26–30, 2002.
[14] E.-G. Fan, “Traveling wave solutions for nonlinear equations
using symbolic computation,” Computers & Mathematics with
Applications, vol. 43, no. 6-7, pp. 671–680, 2002.
[15] Y. Ugurlu and D. Kaya, “Solutions of the Cahn-Hilliard equation,” Computers & Mathematics with Applications, vol. 56, no.
12, pp. 3038–3045, 2008.
[16] I. E. Inan, “Exact solutions to the various nonlinear evolution
equations,” Physics Letters A, vol. 371, no. 1-2, pp. 90–95, 2007.
[17] E. Fan and H. Zhang, “A note on homogeneous balance
method,” Physics Letters A, vol. 246, no. 5, pp. 403–406, 1998.
[18] L. Yang, Z. Zhu, and Y. Wang, “Exact solutions of nonlinear
equations,” Physics Letters A, vol. 260, no. 1-2, pp. 55–59, 1999.
[19] E. Fan, “Two new applications of the homogeneous balance
method,” Physics Letters A, vol. 265, no. 5-6, pp. 353–357, 2000.
[20] M. L. Wang, “Solitary wave solutions for variant Boussinesq
equations,” Physics Letters A, vol. 199, no. 3-4, pp. 169–172, 1995.
[21] M. Wang, Y. Zhou, and Z. Li, “Application of a homogeneous
balance method to exact solutions of nonlinear equations in
mathematical physics,” Physics Letters A, vol. 216, no. 1–5, pp.
67–75, 1996.
[22] B. Kiliç, Some methods for traveling wave solutions of the
nonlinear partial differential equations and numerical analysis of
these solutions [Ph.D. thesis], Firat University, 2012.
Advances in
Hindawi Publishing Corporation
http://www.hindawi.com
Algebra
Advances in
Operations
Research
Decision
Sciences
Volume 2014
Hindawi Publishing Corporation
http://www.hindawi.com
Journal of
Probability
and
Statistics
Mathematical Problems
in Engineering
Volume 2014
Hindawi Publishing Corporation
http://www.hindawi.com
Volume 2014
Hindawi Publishing Corporation
http://www.hindawi.com
Volume 2014
The Scientific
World Journal
Hindawi Publishing Corporation
http://www.hindawi.com
Hindawi Publishing Corporation
http://www.hindawi.com
Volume 2014
International Journal of
Differential Equations
Hindawi Publishing Corporation
http://www.hindawi.com
Volume 2014
Volume 2014
Submit your manuscripts at
http://www.hindawi.com
International Journal of
Advances in
Combinatorics
Mathematical Physics
Volume 2014
Hindawi Publishing Corporation
http://www.hindawi.com
Journal of
Complex Analysis
Hindawi Publishing Corporation
http://www.hindawi.com
Volume 2014
International
Journal of
Mathematics and
Mathematical
Sciences
Hindawi Publishing Corporation
http://www.hindawi.com
Journal of
Hindawi Publishing Corporation
http://www.hindawi.com
Stochastic Analysis
Abstract and
Applied Analysis
Hindawi Publishing Corporation
http://www.hindawi.com
Hindawi Publishing Corporation
http://www.hindawi.com
International Journal of
Mathematics
Volume 2014
Volume 2014
Journal of
Volume 2014
Discrete Dynamics in
Nature and Society
Volume 2014
Hindawi Publishing Corporation
http://www.hindawi.com
Journal of
Applied Mathematics
Optimization
Hindawi Publishing Corporation
http://www.hindawi.com
Hindawi Publishing Corporation
http://www.hindawi.com
Volume 2014
Journal of
Discrete Mathematics
Journal of Function Spaces
Hindawi Publishing Corporation
http://www.hindawi.com
Volume 2014
Hindawi Publishing Corporation
http://www.hindawi.com
Volume 2014
Hindawi Publishing Corporation
http://www.hindawi.com
Volume 2014
Volume 2014
Volume 2014