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Latin American applied research
Print version ISSN 0327-0793
Lat. Am. appl. res. vol.34 no.1 Bahía Blanca Jan./Mar. 2004
Exact travelling wave solutions to the generalized Kuramoto-Sivashinsky equation
Changpin Li1, Guanrong Chen2 and Suchuan Zhao3
1 Department of Mathematics, Shanghai University, Shanghai 200436, P. R. China
leecp@online.sh.cn
2 Department of Electronic Engineering, City University of Hong Kong, Hong Kong, P. R. China
gchen@ee.cityu.edu.hk
3 Department of Physics, Shanghai University, Shanghai 200436, P. R. China
Abstract By using a special transformation, the new exact travelling wave solutions to the generalized Kuramoto- Sivashinsky equation are obtained.
Keywords Travelling Wave Solutions. Solitary Wave Solutions. Kuramoto-Sivashinsky Equation.
I. INTRODUCTION
In this paper, we consider the generalized Kuramoto-Sivashinsky equation (Yang, 1994):
ut + βuαux + γuτuxx + δuxxxx = 0, (1)
where α, β, γ, δ, τ ∈ R and αβγδ ≠ 0.
When α = β = 1 and τ = 0, (1) reduces to the original Kuramoto-Sivashinsky (K-S) equation. The K-S equation was derived by Kuramoto (1978) for the study of phase turbulence in the Belousov-Zhabotinsky reaction. An extension of this equation to two or more spatial dimensions was then given by Sivashinsky (1977, 1980) in the study of the propagation of a frame front for the case of mild combustion. The K-S equation represents one class of pattern formation equation (Yang, 1994; Temam, 1988), and it also serves as a good model of bifurcation and chaos (Abdel-Gawad & Abdusalam, 2001; Li and Chen 2001, 2002).
As far as the travelling wave solutions are concerned, one can always use the transform
u(x,t) = u(ξ), ξ = x - ct, (2)
where c is the wave velocity. The travelling wave solutions of (1) satisfy the following ordinary differential equation:
-cu' + βuαu' + γuτu'' + δu'''' = 0 . (3)
In (Yang, 1994), using the ansatz (Bernoulli equation)
u' = au + bun, (4)
where a, b, n ∈ R, ab < 0 and n ≠ 1, the exact travelling wave solution to (1) for a = 3τ = 9 was obtained. In this presentation, we further introduce the following ansatz:
u(ξ) = vh(ξ), v' = av + bvn (5)
where abh ≠ 0, n ≠ l and ab < 0, and obtain a new exact solution for the equation. From (5), one first gets
(6) |
in which c0 is an arbitrary constant. If h/(n - 1) > 0, (6) is the solitary wave solution connecting the two stationary states u = 0 and (Lu, et al., 1993). So, the relative orbit is a heteroclinic orbit.
Repeating some differential calculations, one can obtain the following formulas:
v'' = (a + nbvn-1)v', | (7) |
v'''' = [a3 + a2bn(n2 + n + l)vn-1 + 3ab2n2(2n - l)v2n-2 + b3n(2n - l)(3n - 2)v3n - 3]v'. | (8) |
u' = hvh-lv', | (9) |
u'' = [h2avh-1 + hb(n + h - l)vn+h-2]v', | (10) |
u'''' = {h4a3vh-l + ha2b(n + h- l)[h2 + (n + h - 1)· (n+2h - l)]vn+h-2 + 3hab2(n + h - l)2(2n + h-2)v2n+h-3 + hb3(n + h - l)(2n + h - 2)(3n + h - 3)v3n+h-4} v'. | (11) |
Then, by substituting the first formula of (5) and (9)-(11) into (3), one has
{(-ch + δh4a3)vh-1 + βhvαh+h-1 + γh2avτh+h-1 +γhb(n + h - 1)vn+(τ+1)h-2 + δha2b(n + h - l)· [3h2 + 3h(n - l) + (n - l)2]vn+h-2 + 3δhab2(n + h - l)2· (2n + h - 2)v2n+h-3 + δhb3(n + h - l)(2n + h - 2)· (3n + h - 3)v3n+h-4} v' = 0. | (12) |
By furthermore comparing the same orders of v, one can determine values of the parameters a, b, n and h. However, to consider all possible cases is rather complicated. In order to keep the presentation short, only the following interesting cases with h = 1, n = 0 and n = 2 are considered here.
II. CASE h = l
If h = 1, then u(ξ) = v(ξ) and u' = au + bun, so that (12) is reduced to
(δa3 - c) + βuα + γauτ + γbnuτ+n-1 + δa2bn(n2 + n + l)un-1 + 3δab2n2(2n - l)u2n-2 + δb3n(2n - l)(3n - 2)u3n-3 = 0. | (13) |
By comparing the same orders of u, one finds the following situations.
that is,
In (14), a is a parameter. One should choose a such that ab < 0. The same should be done for the similar cases below.
that is,
(15) |
If , then (13) can be translated into
The following results are immediate.
Therefore, one can easily find that
(16) |
so,
(17) |
so,
Besides and , one has the following case.
(3) τ = n - 1, α = 3n - 3
For this case,
δa3 - c = 0, γa + δa2bn(n2 + n + 1) = 0,
γbn + 3δab2n2(2n - l) = 0, β + δb3n(2n - l)(3n - 2) = 0
From the second and the third equations, it follows that n = 4. So, if and only if n = 4, α = 3τ = 9, there exist real number solutions for a, b and c, as
(19) |
This result is the same as that obtained in Yang (1994).
III. CASE n = 0
For n = 0, (12) is reduced to
(δa3h3-c)vh-1 + βvαh+h-1 + γahvτh+h-1 + γb(h - 1)· vτh+h-2 + δa2b(h - l)(3h2 - 3h + l)vh-2+3δab2(h - l)2· (h - 2)vh-3 + δb3(h - l)(h - 2)(h - 3)vh-4 = 0. | (20) |
After considering the coefficients of some orders of v, one has the following cases, with h = 1, h = 2 and h = 3, respectively.
(1) h = 1.
There exists one and only one sub-case with α = τ ≠ 0 for h = 1 (Note: α ≠ 0), as follows.
(1a) α = τ ≠ 0
(21) |
(2) h = 2
For h = 2, one also has a sub-case.
For this sub-case, one has
7δa2b + γb + β = 0, 8δa3 - c + 2γa = 0.
Thus,
(22) |
(3) h = 3
For h = 3, (20) can be changed to
(27δa3 - c)v2 + βv3α+2 + 3γav3τ+2 + 2γbv3τ+1 + 38δa2bv + 12δab2 = 0, | (23) |
and only three cases exist, as follows.
Here,
27δa3 - c = 0, 3γa + 38δa2b = 0, β + 2γb + 12βab2 = 0.
Therefore,
(24) |
It is clear that
27δa3 - c = 0, β + 3γa + 38δa2b = 0,2γb + 12δab2 = 0.
Hence,
(25) |
By the same reasoning, one has
27δa3 - c + 3γa = 0, 2γb + 38δa2b = 0, β + 12δab2 = 0,
so a, b, c are as below:
(26) |
For h ≠ 1, h ≠ 2, and h ≠ 3, a comparison between the corresponding terms in (20) gives only one case, as follows.
It follows that
δa3h3 - c = 0 , (δa2b(h - 1) (3h2 - 3h + 1) + γah = 0 ,
3δab2(h - l)2(h - 2) + γb(h - 1) = 0 ,
δb3(h - l)(h - 2)(h - 3) + β = 0 .
The second and the third equations of the above system give . So, if and only if , α = 3τ = 9, the above system has real number solutions for a, b and c, as
(27) |
IV. CASE n = 2
Substituting n = 2 into (12) gives
(-c + δh3a3)vh-l + (δa2b(h + l)(3h2 + 3h + 1)vh + 3δab2(h + l)2(h + 2)vh+1 + δb3(h + l)(h + 2)(h + 3)vh+2 +γhavτh+h-1 + γb(h + l)vτh+h + βvαh+h-1 = 0. | (28) |
(1) h = -1
For h = - 1, there exist only one sub-case.
(la) γ = α ≠ 0
(29) |
(2) h = -2
For h = - 2, there exist two sub-cases.
By the same reason, one has
-c - 8δa3 - 2γa = 0 , -7δa2b - γb + β = 0,
so,
(30) |
Similarly, one has
-c - 8δa3 = 0, -7δa2b - 2γa = 0 , -γb + β = 0,
so,
(31) |
(3) h = -3
For h = -3, there exist three sub-cases.
For this sub-case,
c + 27δa3 + 3γa = 0, 38δa2b + 2γb = 0, 12γab2 - β = 0.
The solutions are
(32) |
Similarly,
c + 27δa3 = 0, 38δa2b + 3γa - β = 0, 12δab2 + 2γb = 0.
The solutions are
(33) |
The following system is determined by the same reasoning:
c + 27δa3 = 0, 38δa2b + 3γa = 0, 12δab2 + 2γb - β = 0,
so,
(34) |
Besides h = - 1, h = - 2 and h = -3, there exists only one case, with , α = 3γ.
(4) , α = 3γ.
It follows that
-c + δh3a3 = 0, δa2b(h + l)(3h2 + 3h + l) + γha = 0 ,
3δab2(h + l)2(h + 2) + γb(h + 1) = 0 ,
δb3(h + l)(h + 2)(h + 3) + β = 0.
The second and third equations in the above system give . So, if and only if and α = 3τ = 9, parameters a, b and c are given by
(35) |
ACKNOWLEDGEMENTS
This research was supported by the NSF of China (Grant No. 19971057) and the Hong Kong CERG (Grant No. 9040579).
REFERENCES
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Received: April 30, 2002.
Accepted for publication: July 25, 2002.
Recommended by Subject Editor Jorge L. Moiola.