Physics Letters A 372 (2008) 3240–3242 www.elsevier.com/locate/pla
An effective method for finding special solutions of nonlinear differential equations with variable coefficients ✩ Maochang Qin a,∗ , Guihong Fan b a Mathematics and Statistics College, Chongqing Technology and Business University, Chongqing 400067, China b Mathematics and Statistics Department, McMaster University, Hamilton Ontario, L8S 4K1, Canada
Received 27 June 2007; received in revised form 16 January 2008; accepted 30 January 2008 Available online 5 February 2008 Communicated by A.B Bishop
Abstract There are many interesting methods can be utilized to construct special solutions of nonlinear differential equations with constant coefficients. However, most of these methods are not applicable to nonlinear differential equations with variable coefficients. A new method is presented in this Letter, which can be used to find special solutions of nonlinear differential equations with variable coefficients. This method is based on seeking appropriate Bernoulli equation corresponding to the equation studied. Many well-known equations are chosen to illustrate the application of this method. © 2008 Elsevier B.V. All rights reserved. PACS: 02.30.Jr; 04.20.Jb; 04.20.Jb Keywords: Special solutions; Nonlinear differential equations; Variable coefficients
1. Introduction One of the most interesting problems of nonlinear model analysis is the construction of partial solutions. This is important because a lot of mathematical-physics models are described by nonlinear differential equations. When nonlinear differential equations belong to the class of exactly solvable equations, the inverse scattering transformation [1,2] and the Hirota method [3,4] are very efficient tools in seeking their special solutions. While most of the nonlinear differential equations, which are used to describe various phenomena or courses in engineering, physics, biology, economics, etc., are not belong to the exactly solvable class. Some effective methods containing special functions were proposed to find partial solutions of this kind of differential equations with constant coefficients. The typical methods are given as follows: the trigonometric function method in [5,6], the Jacobian elliptic function method in [7,8], the tanh function method in [9,10], and the Weisertrass function method in [11,12]. However, no analogous methods can be applied to nonlinear differential equations with variable coefficients. In this Letter, we will consider following ordinary differential equation n k=1
an−k (x)
dky + an (x)y + b(x)y m = 0, dx k
(1)
✩ This project supported by the Youth Foundation of Chongqing Technology & Business University under No. 0652002 and National Natural Sciences Foundation of China under No. 10572021. * Corresponding author. E-mail address:
[email protected] (M. Qin).
0375-9601/$ – see front matter © 2008 Elsevier B.V. All rights reserved. doi:10.1016/j.physleta.2008.01.058
M. Qin, G. Fan / Physics Letters A 372 (2008) 3240–3242
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where ai (x), i = 1, 2, . . . , n, and b(x) are continuous functions, and m = 0 a constant. We intend to find special solution of Eq. (1), which satisfies Bernoulli equation dy (2) = a(x)y α + c(x)y, dx where a(x) and c(x) are undetermined continuous functions, and α = 0, 1 is a constant to be determined. Substituting (2) in Eq. (1) and expanding it as a polynomial of y. Assuming the highest-order term is y m , we determine α. Letting the coefficients equal to zero, we can solve coefficient functions a(x) and c(x). For example, for n = 2, Eq. (1) can be written as d 2y dy + a1 (x) + a2 (x)y + b(x)y m = 0. dx dx 2 Substituting (2) in Eq. (3), we have αa0 (x)a 2 (x)y 2α−1 + a0 a (x) + (α + 1)a(x)c(x) + a1 (x)a(x) y α + a0 c (x) + c2 (x) + a1 (x)c(x) + a2 (x) y + b(x)y m = 0.
(3)
a0 (x)
Letting exponent 2α − 1 = m and coefficients equal to zero, we have ⎧ m+1 2 ⎪ ⎨ 2 a0 (x)a (x) + b(x) = 0, (4) a0 (a (x) + m+3 2 a(x)c(x)) + a1 (x)a(x) = 0, ⎪ ⎩ 2 a0 (c (x) + c (x)) + a1 (x)c(x) + a2 (x) = 0. That is to say coefficient functions a(x) and c(x) satisfying (4), then Eq. (3) has a special solution which satisfying following Bernoulli equation m+1 dy = a(x)y 2 + c(x)y. dx Combining Eqs. (3) and (4), we know our method has an effect upon second-order nonlinear differential equation
a0 (x)
m+1 d 2y dy + a1 (x) + b2 (x)y + b1 (x)y 2 + b0 (x)y m = 0. 2 dx dx
2. Applications Example 1. Consider second-order nonlinear differential equation y −
m+1 2 m (m + 5)x + 2 (m + 5)x + 2 y− y + x y = 0, 2 2(x + 1)x 2 2(x + 1) x
which is difficult to solve in general. Substituting (2) in Eq. (5), we have α = special solution y
1−m 2
=
1−m 1−m (x + 1) + (x + 1) 2 , m+1
(5) m+1 2 ,
a(x) = x, and c(x) =
1 x+1 .
Hence Eq. (5) has a
m = ±1.
Example 2. Consider third-order nonlinear differential equation
(m + 3)x + 2 (m + 8)((m + 3)x + 2) (m + 2)(m + 5) y − y + − y x2 3x 3 9x 2
(m + 8)((m + 3)x + 2) (m + 2)(m + 5) (2m + 1)(m + 2) 3 m − − x y = 0. y− 4 3 9 3x 9x This equation is difficult to solve in general. Substituting (2) in Eq. (6), we have α = Eq. (6) has following special solution
m+2 3 ,
(6) a(x) = x, and c(x) = x1 . Consequently
1−m 1−m 2 x + x 3 , m = ±1, m = −2, m = −5. m+5 Our method is valid for following third-order differential equations
y
1−m 3
=
2m+1 m+2 d 3y d 2y dy + a (x) + a2 (x) + b3 (x)y + b2 (x)y 3 + b1 (x)y 3 + b0 (x)y m = 0. 1 dx dx 3 dx 2 In particular, if coefficient functions ai (x), i = 1, 2, . . . , n, and b(x) in Eq. (1) are constants, a(x) and c(x) in (2) are also constants. Then calculation becomes much simpler.
a0 (x)
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M. Qin, G. Fan / Physics Letters A 372 (2008) 3240–3242
Example 3. The generalized Korteveg–deVries–Burgers equation (see [1,4,13]) ut − βuδ ux + γ uxxx = κuxx .
(7)
Use the travelling wave transformation u(x, t) = u(ξ ),
ξ = x − c0 t.
Integrating and letting the integration constant equal to zero, we have γ uξ ξ − κuξ − c0 u −
β δ+1 = 0. u δ+1
2β 2κ , a = ± Substituting (2) in above equation, we have α = δ+1 2 γ (δ+1)(δ+2) , c = γ (δ+4) , and c0 = solution
− 2
δ βγ κδ 2κ 2 (δ + 2)t δ+4 u(x, t) = c1 exp − . x+ ± 2 γ (δ + 4) κ 2(δ + 1)(δ + 2) γ (δ + 4)
−2κ 2 (δ+2) . γ (δ+4)2
Eq. (7) has a special
Example 4. The generalized Kuramoto–Sivashinsky equation (see [14,15]) ut + βuδ ux + κuxx + γ uxxx + σ uxxxx = 0.
(8)
Use the travelling wave transformation u(x, t) = u(ξ ),
ξ = x − c0 t.
Integrating and letting the integration constant equal to zero, we have σ uξ ξ ξ + γ uξ ξ + κuξ − c0 u +
β δ+1 = 0. u δ+1
Substituting (2) in above equation, we have α = Therefore Eq. (8) has special solution u(x, t) = c1 exp −
δ+3 3 ,
a=
3
−9β σ (δ+1)(δ+3)(2δ+3) ,
c=
γ σ (δ+3) ,
κ=
γ 2 (27−δ 2 ) , σ (δ+3)2
and c0 =
(9−δ 2 )γ 3 . σ 2 (δ+3)3
3 γδ 1 3 −9βσ 2 (δ + 3)2 − δ (9 − δ 2 )γ 3 t − . x− 2 3σ (δ + 3) γ (δ + 1)(2δ + 3) σ (δ + 3)3
3. Conclusion In summary, basing on Bernoulli equation, an efficient method, which can be used to find special solutions of nonlinear differential equation with variable coefficients as well as constant coefficients, is present in this Letter. This method can be applied to a more general nonlinear differential equations. References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11] [12] [13] [14] [15]
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