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; [email protected] Received 3 February 2013; Revised 22 April 2013; Accepted 23 April 2013 Academic Editor: Dumitru Baleanu
Copyright © 2013 B. Kilic¸ 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 differ- ential 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 B¨acklund transfor- mation, 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
𝑢 (𝑥, 𝑡) = 𝜆∑𝑚
𝑖=0[𝑎𝑖𝐹(𝜉)𝑖+ 𝑎−𝑖𝐹(𝜉)−𝑖] , (2) where,𝜉 = 𝑘(𝑥 − 𝑐𝑡),𝑘 and 𝑐 are the wave number and the wave speed respectively,𝑚is 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 for𝑎0,𝑎1,𝑎2, . . . because all coefficients of 𝐹 have to vanish. From these
4 2
5 0 0
−0.3−5
−0.1−0.20.10
(a)
4 2
5 0 0
−5 0.4
0.2 0
(b) Figure 1: Graph of the solution𝑢(𝑥, 𝑡)from left to the right for (13) and (14).
4 2
5 0 0
−5 0
−0.05
−0.1
(a)
4 2
5 0 0
−5 0.5
0
(b) Figure 2: Graph of the solution𝑢(𝑥, 𝑡)from left to the right for (15) and (16).
relations𝑎0, 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
𝐹 + 𝐵𝐹 + 𝐶𝐹3, (3)
where𝑑𝐹/𝑑𝜉 = 𝐹and𝐴,𝐵,𝐶are constants.
Case 1. If𝐴 = −1/4, 𝐵 = 1/2,𝐶 = −1/2then (3) has the solution𝐹 = 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 +cscℎ(𝜉) +coth(𝜉) or 𝐹 = 1/√1 + 𝑖secℎ(𝜉) +tanh(𝜉).
Case 3. If𝐴 = 1/2,𝐵 = −1,𝐶 = 0then (3) has the solutions 𝐹 = 1/√1 +cotℎ(𝜉)or
𝐹 = 1
√1 +tanh(𝜉). (4)
Remark 1. Depending on the𝐴, 𝐵and𝐶coefficients in the (3), it could be reached only three cases.
In the following we present a new approach to the
“Balance term” definition.
Definition 2. When (1) is transformed with 𝑢(𝑥, 𝑡) = 𝑢(𝜉), 𝜉 = 𝑘(𝑥 − 𝑐𝑡), where𝑘and 𝑐 are real constants, we get a nonlinear ordinary differential equation for𝑢(𝜉)as following 𝑄(𝑢, 𝑘𝑐𝑢, 𝑘𝑢, 𝑘2𝑢, . . .) = 0. (5)
Let𝑢(𝑝)is the highest order derivative linear term and𝑢𝑞𝑢(𝑟) is the highest nonlinear term in (5) and𝐹 = 𝑘0 + 𝑘1𝐹 + 𝑘2𝐹2+ ⋅ ⋅ ⋅ + 𝑘𝑛𝐹𝑛is the auxiliary equation that is used to solve the nonlinear partial differential equation then the “Balance term”𝑚can be decided by the balancing the nonlinear term 𝑢𝑞𝑢(𝑟)and the linear term𝑢(𝑝)with acceptances of𝑢 ≅ 𝜆𝐹𝑖 and𝐹 ≅ 𝐹𝑛 where𝑛is integer(𝑛 ̸= 1)and𝜆is the balance coefficient that can be determined later.
4 2
5 0 0
−5 0
−20
−40
(a)
4
2 0
5 0
−5 20
10 0
(b)
4 2
5 0 0
−5 0
−5
−10
(c)
4 2
5 0 0
−5 2
0
−2
(d)
Figure 3: Graph of the solutions𝑢(𝑥, 𝑡)andV(𝑥, 𝑡)corresponding to the value𝑏0= 1from 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 − 𝑐𝑡we have the ordinary differential equation as following
−𝑐𝑢+ 6𝑢𝑢+ 𝑢= 0. (6) By the balancing linear term𝑢with nonlinear term𝑢𝑢
𝑢= (𝜆𝐹𝑚)= 𝜆𝑚𝐹𝑚−1𝐹= 𝜆𝑚𝐹𝑚−1𝐹𝑛 = 𝜆𝑚𝐹𝑚+𝑛−1, 𝑢= (𝜆𝑚𝐹𝑚+𝑛−1)= 𝜆𝑚 (𝑚 + 𝑛 − 1) 𝐹𝑚+𝑛−2𝐹
= 𝜆𝑚 (𝑚 + 𝑛 − 1) 𝐹𝑚+𝑛−2𝐹𝑛= 𝜆𝑚 (𝑚 + 𝑛 − 1) 𝐹𝑚+2𝑛−2, 𝑢= (𝜆𝑚 (𝑚 + 𝑛 − 1) 𝐹𝑚+2𝑛−2)
= 𝜆𝑚 (𝑚 + 𝑛 − 1) (𝑚 + 2𝑛 − 2) 𝐹𝑚+2𝑛−3𝐹
= 𝜆𝑚 (𝑚 + 𝑛 − 1) (𝑚 + 2𝑛 − 2) 𝐹𝑚+2𝑛−3𝐹𝑛
= 𝜆𝑚 (𝑚 + 𝑛 − 1) (𝑚 + 2𝑛 − 2) 𝐹𝑚+3𝑛−3, 𝑢𝑢= 𝜆𝐹𝑚𝜆𝑚𝐹𝑚+𝑛−1= 𝜆2𝑚𝐹2𝑚+𝑛−1
(7)
we have the equations above and the equating𝑢𝑢to𝑢
𝜆2𝑚𝐹2𝑚+𝑛−1= 𝜆𝑚 (𝑚 + 𝑛 − 1) (𝑚 + 2𝑛 − 2) 𝐹𝑚+3𝑛−3, 𝜆 = (𝑚 + 𝑛 − 1) (𝑚 + 2𝑛 − 2) , 𝑚 = 2 (𝑛 − 1) . (8) If it is noticed that our new balance term𝑚 (𝑚 = 2(𝑛 − 1)) is connected to𝑛. Namely our new balance term definition is connected to chosen auxiliary equation.
3. Application of the Method
Example 2. Let’s consider RLW Burgers equation
𝑢𝑡+ 𝑢𝑥+ 12𝑢𝑢𝑥− 𝑢𝑥𝑥− 𝑢𝑥𝑥𝑡= 0, (9) with the transform𝑢(𝑥, 𝑡) = 𝑢(𝜉), 𝜉 = 𝑥 − 𝑐𝑡we have the following equation
−𝑐𝑢+ 𝑢+ 12𝑢𝑢− 𝑢+ 𝑐𝑢= 0. (10)
4 2
5 0 0
−5 1 0.5
0
(a)
4 2
5 0 0
−5 0.4
0.2 0
−0.2
(b)
4 2
5 0 0
−5 0
−1
−2
−3
(c)
4 2
5 0 0
−5 1 0.5 0
−0.5
(d)
Figure 4: Graph of the solutions𝑢(𝑥, 𝑡)andV(𝑥, 𝑡)corresponding to the value𝑏0= 1from left to the right for (25) and (26).
From theDefinition 2we have the balance term of RLW Burg- ers equation by using the auxiliary equation (3) for “Case1”, is equal to 4. Therefore, we may choose the following ansatz:
𝑢 (𝑥, 𝑡) = 48 (𝑎−4𝐹−4+ 𝑎−3𝐹−3+ 𝑎−2𝐹−2+ 𝑎−1𝐹−1 +𝑎0+ 𝑎1𝐹 + 𝑎2𝐹2+ 𝑎3𝐹3+ 𝑎4𝐹4) .
(11)
Substituting (11) into (10) along with (3) and using Math- ematica yields a system of equations w.r.t. 𝐹𝑖. Setting the coefficients of𝐹𝑖in the obtained system of equations to zero, we can deduce the following set of algebraic polynomials with the respect unknowns𝑎0,𝑎1,𝑎2, . . .for the Case1:
3𝑐𝑎−4+ 12𝑎2−4= 0,
(105𝑐𝑎−3
64 + 21𝑎−4𝑎−3) = 0,
(−3𝑎−4
2 −27𝑐𝑎−4
2 − 24𝑎−42 + 9𝑎2−3 +3𝑐𝑎−2
4 + 18𝑎−4𝑎−2) = 0, (−15𝑎−3
16 −225𝑐𝑎−3
32 − 42𝑎−4𝑎−3+ 15𝑎−3𝑎−2 +15𝑐𝑎−1
64 + 15𝑎−4𝑎−1) = 0, (6𝑎−4+ 28𝑐𝑎−4+ 24𝑎−42 − 18𝑎−32 −𝑎−2
2 − 3𝑐𝑎−2
−36𝑎−4𝑎−2+ 6𝑎2−2+ 12𝑎−3𝑎−1+ 12𝑎−4𝑎0) = 0, (15𝑎−3
4 +429𝑐𝑎−3
32 + 42𝑎−4𝑎−3− 30𝑎−3𝑎−2
−3𝑎−1
16 −27𝑐𝑎−1
32 − 30𝑎−4𝑎−1+ 9𝑎−2𝑎−1 +9𝑎−3𝑎0−3𝑐𝑎1
64 + 9𝑎−4𝑎1) = 0,
(8𝑎−4+ 20𝑐𝑎−4− 3𝑎−2− 4𝑐𝑎−2+ 12𝑎−22 + 24𝑎−3𝑎−1− 6𝑎−12 + 24𝑎−4𝑎0
−12𝑎−2𝑎0− 12𝑎−3𝑎1− 12𝑎−4𝑎2) = 0, (−10𝑎−4− 32𝑐𝑎−4+ 18𝑎−32 + 2𝑎−2
+ 5𝑐𝑎−2+ 36𝑎−4𝑎−2− 12𝑎2−2
− 24𝑎−3𝑎−1+ 3𝑎2−1− 24𝑎−4𝑎0 +6𝑎−2𝑎0+ 6𝑎−3𝑎1+ 6𝑎−4𝑎2) = 0, (−6𝑎−3−27𝑐𝑎−3
2 + 30𝑎−3𝑎−2+3𝑎−1 4 +35𝑐𝑎−1
32 + 30𝑎−4𝑎−1− 18𝑎−2𝑎−1
− 18𝑎−3𝑎0+ 3𝑎−1𝑎0+ 𝑎1 16+3𝑐𝑎1
32
− 18𝑎−4𝑎1+ 3𝑎−2𝑎1+ 3𝑎−3𝑎2 +3𝑐𝑎3
64 + 3𝑎−4𝑎3) = 0, (9𝑎−3
2 +105𝑐𝑎−3
16 − 𝑎−1−𝑐𝑎−1
2 + 18𝑎−2𝑎−1 + 18𝑎−3𝑎0− 6𝑎−1𝑎0−𝑎1
4 +𝑐𝑎1 32 + 18𝑎−4𝑎1− 6𝑎−2𝑎1− 3𝑎0𝑎1
− 6𝑎−3𝑎2− 3𝑎−1𝑎2−3𝑎3 16 +9𝑐𝑎3
32
−6𝑎−4𝑎3− 3𝑎−2𝑎3− 3𝑎−3𝑎4) = 0, (− 2𝑎−4− 6𝑐𝑎−4+ 2𝑎−2+ 𝑐𝑎−2+ 6𝑎−12
+ 12𝑎−2𝑎0+ 12𝑎−3𝑎1− 3𝑎12−𝑐𝑎2 2 + 12𝑎−4𝑎2− 6𝑎0𝑎2− 6𝑎−1𝑎3−𝑎4
2 +3𝑐𝑎4
2 − 6𝑎−2𝑎4) = 0, (−3𝑎−3
4 −9𝑐𝑎−3 8 +𝑎−1
2 −𝑐𝑎−1
16 + 6𝑎−1𝑎0 +𝑐𝑎1
2 + 6𝑎−2𝑎1+ 6𝑎0𝑎1+ 6𝑎−3𝑎2 + 6𝑎−1𝑎2− 9𝑎1𝑎2+3𝑎3
4 −105𝑐𝑎3 32 + 6𝑎−4𝑎3+ 6𝑎−2𝑎3− 9𝑎0𝑎3 +6𝑎−3𝑎4− 9𝑎−1𝑎4) = 0,
(6𝑎21− 𝑎2+ 4𝑐𝑎2+ 12𝑎0𝑎2− 6𝑎22 + 12𝑎−1𝑎3− 12𝑎1𝑎3+ 2𝑎4− 10𝑐𝑎4 +12𝑎−2𝑎4− 12𝑎0𝑎4) = 0,
(−3
8𝑐𝑎−3+𝑎−1
4 −3𝑐𝑎−1 8 +𝑎1
2 −35𝑐𝑎1 16
− 6𝑎0𝑎1− 6𝑎−1𝑎2+ 18𝑎1𝑎2− 3𝑎3 +27𝑐𝑎3
2 − 6𝑎−2𝑎3+ 18𝑎0𝑎3− 15𝑎2𝑎3
−6𝑎−3𝑎4+ 18𝑎−1𝑎4− 15𝑎1𝑎4) = 0, (−6𝑎21+ 2𝑎2− 10𝑐𝑎2− 12𝑎0𝑎2+ 12𝑎22
− 12𝑎−1𝑎3+ 24𝑎1𝑎3− 9𝑎32− 6𝑎4+ 32𝑐𝑎4
−12𝑎−2𝑎4+ 24𝑎0𝑎4− 18𝑎2𝑎4) = 0, (−105𝑐𝑎3
8 − 42𝑎3𝑎4) = 0, (3𝑐𝑎−1
8 −3𝑎1
4 +27𝑐𝑎1
8 − 18𝑎1𝑎2+9𝑎3 2
−429𝑐𝑎3
16 − 18𝑎0𝑎3+ 30𝑎2𝑎3
−18𝑎−1𝑎4+ 30𝑎1𝑎4− 21𝑎3𝑎4) = 0, (−15𝑐𝑎1
8 −15𝑎3
4 +225𝑐𝑎3 8
−30𝑎2𝑎3− 30𝑎1𝑎4+ 42𝑎3𝑎4) = 0, (−24𝑐𝑎4− 24𝑎42) = 0, (−2𝑎2+ 12𝑐𝑎2− 12𝑎22− 24𝑎1𝑎3
+ 18𝑎32+ 8𝑎4− 56𝑐𝑎4− 24𝑎0𝑎4 +36𝑎2𝑎4− 12𝑎42) = 0,
(−6𝑐𝑎2− 18𝑎32− 6𝑎4
+54𝑐𝑎4− 36𝑎2𝑎4+ 24𝑎24) = 0.
(12)
From the system of (12) we have (i)
𝑎0= −11 60− 𝑖
12,
𝑎1= 𝑎2= 𝑎3= 𝑎4= 𝑎−1 = 𝑎−3 = 0,
𝑎−2= 1 10+ 𝑖
10, 𝑎−4= − 𝑖
20, 𝑐 = 𝑖 5, 𝑢 (𝑥, 𝑡)
= − (11 + 5𝑖
60 ) + (1 + 𝑖 10 )
× ( 1
√1 +sec[𝑥 − (𝑖/5) 𝑡] +tan[𝑥 − (𝑖/5) 𝑡])
−2
− 𝑖
20( 1
√1 +sec[𝑥 − (𝑖/5) 𝑡] +tan[𝑥 − (𝑖/5) 𝑡])
−4
. (13) (ii)
𝑎0= 1 + 5𝑖
60 , 𝑐 = −𝑎4, 𝑎1= 𝑎−2= 𝑎3= 𝑎−4= 𝑎−1 = 𝑎−3= 0,
𝑎2= −1 + 𝑖
5 , 𝑎4= ±𝑖 5, 𝑢 (𝑥, 𝑡) = 1 + 5𝑖
60 −1 + 𝑖 5
× (√1 +sec[𝑥 + 𝑖
5𝑡] +tan[𝑥 + 𝑖 5𝑡])
2
+ 𝑖
5(√1 +sec[𝑥 + 𝑖
5𝑡] +tan[𝑥 + 𝑖 5𝑡])
4
. (14)
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
−3𝑐𝑎−4− 12𝑎−42 = 0, (−3𝑎−4
2 +27𝑐𝑎−4
2 + 24𝑎−42 − 9𝑎2−3
−3𝑐𝑎−2
4 − 18𝑎−4𝑎−2) = 0, (−15𝑎−3
16 +225𝑐𝑎−3
32 + 42𝑎−4𝑎−3− 15𝑎−3𝑎−2
−15𝑐𝑎−1
64 − 15𝑎−4𝑎−1) = 0, (4𝑎−4− 18𝑐𝑎−4+ 18𝑎2−3−𝑎−2
2 + 3𝑐𝑎−2+ 36𝑎−4𝑎−2− 6𝑎−22
− 12𝑎−3𝑎−1−12𝑎−4𝑎0) = 0,
(9𝑎−3
4 −135𝑐𝑎−3
16 + 30𝑎−3𝑎−2
−3𝑎−1
16 +27𝑐𝑎−1
32 + 30𝑎−4𝑎−1
−9𝑎−2𝑎−1− 9𝑎−3𝑎0) = 0, (−2𝑎−4+ 6𝑐𝑎−4+ 𝑎−2− 3𝑐𝑎−2
+ 12𝑎−22 + 24𝑎−3𝑎−1− 3𝑎−12 +24𝑎−4𝑎0− 6𝑎−2𝑎0) = 0, (−3𝑎−3
4 +15𝑐𝑎−3 8 +𝑎−1
4 −9𝑐𝑎−1 16 +18𝑎−2𝑎−1+ 18𝑎−3𝑎0− 3𝑎−1𝑎0) = 0,
(−105
64 𝑐𝑎−3− 21𝑎−4𝑎−3) = 0, (6𝑎−12 + 12𝑎−2𝑎0) = 0, (𝑎−1
4 −3𝑐𝑎−1
8 + 6𝑎−1𝑎0) = 0.
(15) From the system of (15) we have
(i)
𝑎0= 𝑎−1= 0, 𝑎−2= −1 5, 𝑎−3= 0, 𝑎−4= 1
20, 𝑐 = −1 5, 𝑢 (𝑥, 𝑡)
= −1
5( 1
√1 + 𝑖sech[𝑥 + (1/5) 𝑡] +tanh[𝑥 + (1/5) 𝑡])
−2
+ 1
20( 1
√1 + 𝑖sech[𝑥 + (1/5) 𝑡] +tanh[𝑥 + (1/5) 𝑡])
−4
, (16) or
𝑢 (𝑥, 𝑡)
= −1
5( 1
√1 +csch[𝑥 + (1/5) 𝑡] +coth[𝑥 + (1/5) 𝑡])
−2
+ 1
20( 1
√1 +csch[𝑥 + (1/5) 𝑡] +coth[𝑥 + (1/5) 𝑡])−4.
(17)
Example 3. Let’s consider Hirota Satsuma coupled equation
𝑢𝑡− 3𝑢𝑢𝑥+ 6VV𝑥−1
2𝑢𝑥𝑥𝑥= 0, V𝑡+ 3𝑢V𝑥+V𝑥𝑥𝑥= 0,
(18)
with the transform𝑢(𝑥, 𝑡) = 𝑢(𝜉),V(𝑥, 𝑡) =V(𝜉),𝜉 = 𝑥 − 𝑐𝑡we have
−𝑐𝑢− 3𝑢𝑢+ 6VV−1 2𝑢= 0,
−𝑐V+ 3𝑢V+V= 0.
(19)
From theDefinition 2 we have the balance term of Hirota Satsuma coupled equation by using the auxiliary equation (3) for “Case1”, is equal to 4 for𝑢andV. Therefore, we may choose the following ansatz:
𝑢 (𝑥, 𝑡) = 48 (𝑎0+ 𝑎1𝐹 + 𝑎2𝐹2+ 𝑎3𝐹3+ 𝑎4𝐹4) , V(𝑥, 𝑡) = 48 (𝑏0+ 𝑏1𝐹 + 𝑏2𝐹2+ 𝑏3𝐹3+ 𝑏4𝐹4) .
(20)
Substituting (20) into (19) along with (3) and using Math- ematica yields a system of equations w.r.t. 𝐹𝑖. Setting the coefficients of𝐹𝑖in the obtained system of equations to zero, we can deduce the following set of algebraic polynomials with the respect unknowns𝑎0,𝑎1,𝑎2, . . .for the Case1:
3𝑎1 128= 0, (−3𝑎1
64 −3𝑎3 128) = 0, (7𝑎1
64 +𝑐𝑎1
4 +3𝑎0𝑎1 4 −9𝑎3
64 −3𝑏0𝑏1 2 ) = 0, (3𝑎12
4 +𝑎2 2 +𝑐𝑎2
2 +3𝑎0𝑎2 2
−3𝑎4 4 −3𝑏12
2 − 3𝑏0𝑏2) = 0,
( −𝑎1 2 −𝑐𝑎1
2 −3𝑎0𝑎1
2 +9𝑎1𝑎2 4 +129𝑎3
64 +3𝑐𝑎3
4 +9𝑎0𝑎3 4 + 3𝑏0𝑏1 −9𝑏1𝑏2
2 −9𝑏0𝑏3 2 ) = 0, (−3𝑎12
2 −5𝑎2
2 − 𝑐𝑎2− 3𝑎0𝑎2 +3𝑎22
2 + 3𝑎1𝑎3+11𝑎4 2 + 𝑐𝑎4 + 3𝑎0𝑎4+ 3𝑏12+ 6𝑏0𝑏2− 3𝑏22
−6𝑏1𝑏3− 6𝑏0𝑏4) = 0, (43𝑎1
32 +𝑐𝑎1
2 +3𝑎0𝑎1
2 −9𝑎1𝑎2 2
−15𝑎3 2 −3𝑐𝑎3
2 −9𝑎0𝑎3
2 +15𝑎2𝑎3 4 +15𝑎1𝑎4
4 − 3𝑏0𝑏1+ 9𝑏1𝑏2 +9𝑏0𝑏3−15𝑏2𝑏3
2 −15𝑏1𝑏4 2 ) = 0, (3𝑎12
2 +11𝑎2
2 + 𝑐𝑎2+ 3𝑎0𝑎2− 3𝑎22
− 6𝑎1𝑎3+9𝑎32
4 − 17𝑎4− 2𝑐𝑎4
− 6𝑎0𝑎4+9𝑎2𝑎4
2 − 3𝑏12− 6𝑏0𝑏2 + 6𝑏22+ 12𝑏1𝑏3−9𝑏32
2 +12𝑏0𝑏4− 9𝑏2𝑏4) = 0,
(105𝑎3
16 +21𝑎3𝑎4
2 − 21𝑏3𝑏4) = 0, ( − 27𝑎1
16 +9𝑎1𝑎2
2 +453𝑎3 32 +3𝑐𝑎3
2 +9𝑎0𝑎3
2 −15𝑎2𝑎3
2 −15𝑎1𝑎4 2 +21𝑎3𝑎4
4 − 9𝑏1𝑏2− 9𝑏0𝑏3 +15𝑏2𝑏3+ 15𝑏1𝑏4−21𝑏3𝑏4
2 ) = 0,
(15𝑎1
16 −225𝑎3
16 +15𝑎2𝑎3 2 +15𝑎1𝑎4
2 −21𝑎3𝑎4
2 − 15𝑏2𝑏3
−15𝑏1𝑏4+ 21𝑏3𝑏4) = 0, (12𝑎4+ 6𝑎42− 12𝑏42) = 0, (−6𝑎2+ 3𝑎22+ 6𝑎1𝑎3−9𝑎23
2 + 29𝑎4+ 2𝑐𝑎4+ 6𝑎0𝑎4
− 9𝑎2𝑎4+ 3𝑎24− 6𝑏22
− 12𝑏1𝑏3+ 9𝑏32− 12𝑏0𝑏4 +18𝑏2𝑏4− 6𝑏42) = 0 − 3𝑏1
64 = 0, (3𝑏1
32 +3𝑏3 64) = 0, (−7𝑏1
32 +𝑐𝑏1
4 −3𝑎0𝑏1 4 +9𝑏3
32) = 0, (−3
4𝑎1𝑏1− 𝑏2+𝑐𝑏2
2 −3𝑎0𝑏2 2 +3𝑏4
2 ) = 0, (𝑏1−𝑐𝑏1
2 +3𝑎0𝑏1 2 −3𝑎2𝑏1
4 −3𝑎1𝑏2 2
−129𝑏3 32 +3𝑐𝑏3
4 −9𝑎0𝑏3 4 ) = 0, (3𝑎1𝑏1
2 −3𝑎3𝑏1
4 + 5𝑏2− 𝑐𝑏2 + 3𝑎0𝑏2−3𝑎2𝑏2
2 −9𝑎1𝑏3 4
−11𝑏4+ 𝑐𝑏4− 3𝑎0𝑏4) = 0, (−43𝑏1
16 +𝑐𝑏1 2 −3𝑎0𝑏1
2 +3𝑎2𝑏1 2
−3𝑎4𝑏1
4 + 3𝑎1𝑏2−3𝑎3𝑏2 2 + 15𝑏3
−3𝑐𝑏3 2 +9𝑎0𝑏3
2 −9𝑎2𝑏3
4 − 3𝑎1𝑏4) = 0, (−3
2𝑎1𝑏1+3𝑎3𝑏1
2 − 11𝑏2+ 𝑐𝑏2
− 3𝑎0𝑏2+ 3𝑎2𝑏2−3𝑎4𝑏2 2 +9𝑎1𝑏3
2 −9𝑎3𝑏3 4 + 34𝑏4
−2𝑐𝑏4+ 6𝑎0𝑏4− 3𝑎2𝑏4) = 0,
(−105𝑏3 8 −9𝑎4𝑏3
2 − 6𝑎3𝑏4) = 0, (27𝑏1
8 −3𝑎2𝑏1
2 +3𝑎4𝑏1 2 − 3𝑎1𝑏2 + 3𝑎3𝑏2−453𝑏3
16 +3𝑐𝑏3
2 −9𝑎0𝑏3 2 +9𝑎2𝑏3
2 −9𝑎4𝑏3
4 + 6𝑎1𝑏4− 3𝑎3𝑏4) = 0, (−15𝑏1
8 −3𝑎4𝑏1
2 − 3𝑎3𝑏2+225𝑏3 8
−9𝑎2𝑏3 2 +9𝑎4𝑏3
2 − 6𝑎1𝑏4+ 6𝑎3𝑏4) = 0, (−24𝑏4− 6𝑎4𝑏4) = 0,
(−3
2𝑎3𝑏1+ 12𝑏2− 3𝑎2𝑏2+ 3𝑎4𝑏2
−9𝑎1𝑏3 2 +9𝑎3𝑏3
2 − 58𝑏4+ 2𝑐𝑏4
−6𝑎0𝑏4+ 6𝑎2𝑏4− 3𝑎4𝑏4) = 0, (−6𝑏2− 3𝑎4𝑏2−9𝑎3𝑏3
2 + 54𝑏4
−6𝑎2𝑏4+ 6𝑎4𝑏4) = 0,
(3𝑎2+9𝑎32
2 − 27𝑎4+ 9𝑎2𝑎4− 6𝑎42
−9𝑏32− 18𝑏2𝑏4+ 12𝑏42) = 0.
(21)
From the system of (21) we have (i)
𝑐 = 1
4(5 − 6𝑏0) , 𝑎0=1
4(−5 − 2𝑏0) , 𝑎1= 𝑎3= 0, 𝑏1= 𝑏3= 0,
𝑎2= 4, 𝑏2= −2, 𝑎4= −4, 𝑏4= 2,
𝑢 (𝑥, 𝑡) = 1
4(−5 − 2𝑏0)
+ 4 (((1 +sec[𝑥 −1
4(5 − 6𝑏0) 𝑡]
+tan[𝑥 −1
4(5 − 6𝑏0) 𝑡])−1/2)
−1
)
2
− 4 (((1 +sec[𝑥 −1
4(5 − 6𝑏0) 𝑡]
+tan[𝑥 −1
4(5 − 6𝑏0) 𝑡])−1/2)
−1
)
4
, V(𝑥, 𝑡) = 𝑏0
− 2 (((1 +sec[𝑥 −1
4(5 − 6𝑏0) 𝑡]
+tan[𝑥 −1
4(5 − 6𝑏0) 𝑡])−1/2)
−1
)
2
+ 2 (((1 +sec[𝑥 −1
4(5 − 6𝑏0) 𝑡]
+tan[𝑥 −1
4(5 − 6𝑏0) 𝑡])−1/2)
−1
)
4
. (22) (ii)
𝑐 = 1
4(−1 − 3𝑏22) , 𝑎0=1
4(−3 − 𝑏22) , 𝑎1= 𝑎3= 0, 𝑏1= 𝑏3= 𝑏4= 0,
𝑎2= 2, 𝑏2 ̸= 0, 𝑎4= −2, 𝑢 (𝑥, 𝑡) = 1
4(−3 − 𝑏22)
+ 2 (((1 +sec[𝑥 −1
4(−1 − 3𝑏22) 𝑡]
+tan[𝑥 −1
4(−1 − 3𝑏22) 𝑡])−1/2)
−1
)
2
− 2 (((1 +sec[𝑥 −1
4(−1 − 3𝑏22) 𝑡]
+tan[𝑥 −1
4(−1 − 3𝑏22) 𝑡])−1/2)
−1
)
4
,
V(𝑥, 𝑡) = −𝑏2 2
− 𝑏2(((1 +sec[𝑥 −1
4(−1 − 3𝑏22) 𝑡]
+tan[𝑥 −1
4(−1 − 3𝑏22) 𝑡])−1/2)
−1
)
2
. (23)
From theDefinition 2 we have the balance term of Hirota Satsuma coupled equation by using the auxiliary equation (3) for “Case2”, is equal to−4 then we have the following system of equations
3𝑎−4
2 + 3𝑎2−4− 6𝑏−42 = 0, (105𝑎−3
128 +21
4𝑎−4𝑎−3−21
2 𝑏−4𝑏−3) = 0, (−𝑎−1
16 −𝑐𝑎−1 2 −3
2𝑎−1𝑎0+ 3𝑏−1) = 0, (−27𝑎−4
4 − 6𝑎−42 +9𝑎−32 4 +3𝑎−2
8 +9 2𝑎−4𝑎−2 +12𝑏−42 −9𝑏−32
2 − 9𝑏−4𝑏−2) = 0, (−225𝑎−3
64 −21
2𝑎−4𝑎−3+15 4 𝑎−3𝑎−2 +15𝑎−1
128 +15
4 𝑎−4𝑎−1+ 21𝑏−4𝑏−3
−15
2 𝑏−3𝑏−2−15
2𝑏−4𝑏−1) = 0, (19𝑎−4
2 + 𝑐𝑎−4−9𝑎−32 2 −3𝑎−2
2 − 9𝑎−4𝑎−2 +3𝑎−22
2 + 3𝑎−3𝑎−1+ 3𝑎−4𝑎0− 6𝑏−4 +9𝑏−32 + 18𝑏−4𝑏−2− 3𝑏−22 − 6𝑏−3𝑏−1) = 0,
(147𝑎−3
32 +3𝑐𝑎−3 4 −15
2 𝑎−3𝑎−2
−27𝑎−1 64 −15
2 𝑎−4𝑎−1+9 4𝑎−2𝑎−1 +9
4𝑎−3𝑎0−9𝑏−3
2 + 15𝑏−3𝑏−2 +15𝑏−4𝑏−1−9
2𝑏−2𝑏−1) = 0,
(−27𝑎−3
16 −3𝑐𝑎−3
2 +13𝑎−1 32 +𝑐𝑎−1
4
−9
2𝑎−2𝑎−1−9
2𝑎−3𝑎0+3 4𝑎−1𝑎0 +9𝑏−3−3𝑏−1
2 + 9𝑏−2𝑏−1) = 0, (− 4𝑎−4− 2𝑐𝑎−4+7𝑎−2
4 +𝑐𝑎−2 2
− 3𝑎−22 − 6𝑎−3𝑎−1+3𝑎2−1 4
− 6𝑎−4𝑎0+3
2𝑎−2𝑎0+ 12𝑏−4
−3𝑏−2+ 6𝑏−22 + 12𝑏−3𝑏−1−3𝑏−12 2 ) = 0, (−𝑎−2
2 − 𝑐𝑎−2−3𝑎−12
2 − 3𝑎−2𝑎0+ 6𝑏−2+ 3𝑏−12 ) = 0,
−3𝑏−4− 3𝑎−4𝑏−4= 0, (−3𝑎−3𝑏−4−105𝑏−3
64 −9
4𝑎−4𝑏−3) = 0, (27𝑏−4
2 + 6𝑎−4𝑏−4− 3𝑎−2𝑏−4
−9
4𝑎−3𝑏−3−3𝑏−2 4 −3
2𝑎−4𝑏−2) = 0, (6𝑎−3𝑏−4− 3𝑎−1𝑏−4+225𝑏−3
32 +9
2𝑎−4𝑏−3−9
4𝑎−2𝑏−3−3 2𝑎−3𝑏−2
−15𝑏−1 64 −3
4𝑎−4𝑏−1) = 0, ( − 19𝑏−4+ 𝑐𝑏−4+ 6𝑎−2𝑏−4− 3𝑎0𝑏−4
+9
2𝑎−3𝑏−3−9
4𝑎−1𝑏−3+ 3𝑏−2 +3𝑎−4𝑏−2−3
2𝑎−2𝑏−2−3
4𝑎−3𝑏−1) = 0, (6𝑎−1𝑏−4−147𝑏−3
16 +3𝑐𝑏−3 4 +9
2𝑎−2𝑏−3−9 4𝑎0𝑏−3 + 3𝑎−3𝑏−2−3
2𝑎−1𝑏−2+27𝑏−1 32 +3
2𝑎−4𝑏−1−3
4𝑎−2𝑏−1) = 0,
(8𝑏−4− 2𝑐𝑏−4+ 6𝑎0𝑏−4+9 2𝑎−1𝑏−3
−7𝑏−2 2 +𝑐𝑏−2
2 + 3𝑎−2𝑏−2
−3
2𝑎0𝑏−2+3
2𝑎−3𝑏−1−3
4𝑎−1𝑏−1) = 0, (𝑏−2− 𝑐𝑏−2+ 3𝑎0𝑏−2+3
2𝑎−1𝑏−1) = 0, (27𝑏−3
8 −3𝑐𝑏−3 2 +9
2𝑎0𝑏−3+ 3𝑎−1𝑏−2
−13𝑏−1 16 +𝑐𝑏−1
4 +3
2𝑎−2𝑏−1−3
4𝑎0𝑏−1) = 0, (𝑏−1
8 −𝑐𝑏−1 2 +3
2𝑎0𝑏−1) = 0.
(24)
From the system of (3) we have (i)
𝑎−1= 𝑎−3= 𝑏−1= 𝑏−3= 0, 𝑏−2= ±1, 𝑏−4 = −𝑏−2
2 , 𝑎−2= 2, 𝑎−4 = −1, 𝑐 = 1
4(1 + 6𝑏−2) , 𝑎0=1
4(−1 + 2𝑏−2) , 𝑢 (𝑥, 𝑡)
= 1
4+2 ( 1
√1 + 𝑖sech[𝑥 − (7/4) 𝑡] +tanh[𝑥 − (7/4) 𝑡])−2
− ( 1
√1 + 𝑖sech[𝑥 − (7/4) 𝑡] +tanh[𝑥 − (7/4) 𝑡])
−4
, V(𝑥, 𝑡)
= 𝑏0+ ( 1
√1 + 𝑖sech[𝑥 − (7/4) 𝑡] +tanh[𝑥 − (7/4) 𝑡])
−2
−1
2( 1
√1 + 𝑖sech[𝑥 − (7/4) 𝑡] +tanh[𝑥 − (7/4) 𝑡])
−4
, (25)
or 𝑢 (𝑥, 𝑡)
= 1
4+2( 1
√1 +csch[𝑥 − (7/4) 𝑡] +coth[𝑥 − (7/4) 𝑡])
−2
− ( 1
√1 +csch[𝑥 − (7/4) 𝑡] +coth[𝑥 − (7/4) 𝑡])
−4
, V(𝑥, 𝑡)
= 𝑏0+ ( 1
√1 +csch[𝑥 − (7/4) 𝑡] +coth[𝑥 − (7/4) 𝑡])
−2
−1
2( 1
√1 +csch[𝑥 − (7/4) 𝑡] +coth[𝑥 − (7/4) 𝑡])
−4
.
(26) Figures1,2,3and4gives 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.
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Probability and Statistics
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Journal of Function Spaces
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The Scientific World Journal
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Discrete Dynamics in Nature and Society
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Decision Sciences
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Discrete Mathematics
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Stochastic Analysis
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