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International Mathematical Forum, Vol. 9, 2014, no. 3, 137 - 144
HIKARI Ltd, www.m-hikari.com
http://dx.doi.org/10.12988/imf.2014.312238
Exact Two Waves Solutions with Variable
Amplitude to the KdV Equation1
Ying Huang
Department of Mathematics
Chuxiong Normal University
Chuxiong 675000, P.R. China
c 2014 Ying Huang. This is an open access article distributed under the
Copyright Creative Commons Attribution License, which permits unrestricted use, distribution, and
reproduction in any medium, provided the original work is properly cited.
Abstract
By means of the Darboux transformation method, a series of explicit
variable amplitude-two-wave solutions to the Korteweg-de Vries (KdV)
equation are generated from a particular solution with respect to the
independent variable x. Theses new solutions include single soliton solutions, two-soliton solutions, single periodic solutions, doubly periodic
solutions.
Mathematics Subjection Classification: 35Q51; 35A
Keywords: KdV equation; Darboux transformation; soliton solution;periodic
solution
1. Introduction
The KdV equation
ut + 6uux + uxxx = 0,
(1)
is a typical soliton equation, it’s exact expressions of traveling wave solutions
have been studied extensively in many papers[1,2,3,4,5,6]. Especially, the linetwo-soliton solutions obtained by the Inverse scattering method in [7] have
been well known.
1
This work was supported by the Chinese Natural Science Foundation Grant (11261001)
and Yunnan Provincial Department of Education Research Foundation Grant (2012Y130).
138
Ying Huang
We know that, Darboux transformation method is one of the powerful and
direct method for finding exact multiple line-soliton solutions, it is scarcely
used for solving other type of multiple waves solutions because of the matter
of calculation [8,9,10,11,12]. Generating new solutions from the non-trivial
solution amounts to solving the variable coefficient partial differential system
by this method, a major difficult is that for a given partial differential equations there is to date no completely systematic method of solving it, work has
usually relied on some special techniques. In the paper, the author employ a
exploratory method to solve partial differential system, further construct some
two-wave solutions which amplitude depends on the independent variable x.
2. Solution to the lax pair
As described in [13], the Lax pair for equation (1) is given by
⎧
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎨
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎩
⎛
Φx = ⎝
⎛
Φt = ⎝
0
1
λ−u 0
⎞
⎠ Φ,
ux
−(4λ + 2u)
−(4λ + 2u)(λ − u) + uxx
−ux
with the Darboux matrix
⎛
D(x, t, λ) = ⎝
−σ0
1
λ − λ0 + σ02 −σ0
⎞
(2)
⎠ Φ,
⎞
⎠,
(3)
where λ, λ0 are the spectral parameters, in particular, when Φ(x, t, λ) =
(aij (x, t, λ))2×2 is the fundamental solution matrix to the system (2) with u =
u0 , σ0 is defined as
σ0 =
a21 (x, t, λ0 )μ0 + a22 (x, t, λ0 )γ0
,
a11 (x, t, λ0 )μ0 + a12 (x, t, λ0 )γ0
(4)
here, μ0 and γ0 are arbitrary constants, but μ20 + γ02 = 0. Furthermore, if u0 is
a given solution to equation (1), then
u1 = 2λ0 − u0 − 2σ02
(5)
becomes new solution for equation (1), which further concludes that
u2 = 2λ1 − u1 − 2σ12
(6)
is another new solution based on u1 , where
σ1 =
ã21 (x, t, λ1 )μ1 + ã22 (x, t, λ1 )γ1
,
ã11 (x, t, λ1 )μ1 + ã12 (x, t, λ1 )γ1
(7)
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Exact two waves solutions with variable amplitude
and Φ̃(x, t, λ) = D(x, t, λ)Φ(x, t, λ) = (ãij (x, t, λ))2×2 is the fundamental solution matrix of the Lax pair on u1 , μ1 , γ1 are similar to μ0 , γ0 .
The essential procedure to construct new solutions is to solve the fundamental solution matrix of the lax pair on initial solution u0 . Substituting
u0 = −2x−2 into the system (2) yields
⎧
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎨
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎩
⎛
Φx = ⎝
⎛
Φt = ⎝
0
1
λ + 2x−2 0
⎞
⎠ Φ,
(8a)
4x−3
−4λ + 4x−2
−4λ2 − 4λx−2 − 4x−4
−4x−3
⎞
⎠ Φ,
(8b)
(8a) is an ordinary differential system with variable coefficient, we have no
effective method to deal with it, as for (8b), when x is looked as a parameter,
it is an ordinary differential system with const coefficients. In the case of λ > 0,
with the eigenvalue method, we solve the characteristic equation
4λ − 4x−2 γ − 4x−3
= 0,
γ + 4x−3 √
√
and obtain the characteristic roots γ1 = 4λ λ, γ2 = −4λ λ. After calculating the characteristic vectors associated with γ1 and γ2 , we obtain two basic
solutions to (8b)
⎛
4λ2 + 4λx−2 + 4x−4
√
⎞
⎛
√
⎞
(λ − x−2 )e4λ λt
(λ − x−2 )e−4λ λt
⎝
⎠
⎝
√ 4λ√λt ,
√ −4λ√λt ⎠ .
−3
−3
(x − λ λ)e
(x + λ λ)e
Of course,
⎛
√
⎞
(λ − x−2 )e4λ λt
⎠
k1 (x) ⎝
√
√
(x−3 − λ λ)e4λ λt
is also solution to (8b), where k1 (x) is function to be determined later. In
order to probe whether it is a solution of (8a), substituting it into (8a), we
have
√
k1 (x)(x−2 − λ) = k1 (x)(λ λ + x−3 )
(9a)
√
(9b)
k1 (x)(λ λ − x−3 ) = −k1 (x)(λ2 + λx−2 + x−4 ),
more interesting, (9a) and (9b) have the same solution k1 (x) =
In parallel, we know
⎛
√
⎞
(λ − x−2 )e4λ λt
⎝
k2 (x)
√ 4λ√λt ⎠
−3
(x − λ λ)e
√
x
− λ x
√
e
.
1− λx
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Ying Huang
√
x
is also solution to (8b), where k2 (x) = 1+√
e λ x . In short, we obtain the
λx
fundamental solution matrix to the system (8)
⎛
⎞
√
√
−( λ + x−1 )eξ
( λ − x−1 )e−ξ
⎠,
(10)
Φ(x, t, λ) = ⎝
√ −1
√ −1
−2 ξ
−2 −ξ
(λ + λx + x )e (λ − λx + x )e
√
where ξ = λ (4λt − x).
When λ < 0, the fundamental solution matrix (10) is complex, that is
to say, the system (8) has two linearly independent complex solutions. We
know that the real and imaginary parts of a complex-valued solution are also
solutions, thus taking
⎛ √
⎞
√
−1
−1
−λ
sin
θ
−
x
cos
θ
−(
−λ
cos
θ
+
x
sin
θ)
⎠
Φ (x, t, λ) = ⎝
(11)
I
J
as the
√ fundamental solution matrix to the√system (8), where I = λ√cos θ −
x−1 ( −λ sin θ−x−1 cos θ), J = λ sin θ+x−1 ( −λ cos θ+x−1 sin θ), θ = −λ (4λt−
x).
Similarly, we obtain the fundamental solution matrix to the system (8) for
the case of λ = 0
⎛
Φ (x, t, 0) = ⎝
x2 + 12tx−1
x−1
2x − 12tx−2 −x−2
⎞
⎠.
(12)
3. Solutions to the KdV equation
Substituting the entries of (10) into (4) shows
−1
σ0 = −x
λ0 x(eξ0 μ0 + e−ξ0 γ0 )
√
− ξ
e 0 μ0 + e−ξ0 γ0 + λ0 x(eξ0 μ0 − e−ξ0 γ0 )
(13)
with ξ0 = ξ(λ0). Then substituting (11) into (5) yields
u1 =
−2λ0 [(eξ0 μ0 + e−ξ0 γ0 )2 + 4λ0 μ0 γ0 x2 ]
√
[eξ0 μ0 + e−ξ0 γ0 + λ0 x(eξ0 μ0 − e−ξ0 γ0 )]2
(14)
in (12), choosing μ0 = ec0 , γ0 = e−c0 and μ0 = ec0 , γ0 = −e−c0 respectively, we
have new soliton solutions
u1−1 =
−2λ0 (cosh2 η0 + λ0 x2 )
√
(cosh η0 + λ0 x sinh η0 )2
u1−2 =
−2λ0 (sinh2 η0 − λ0 x2 )
√
,
(sinh η0 + λ0 x cosh η0 )2
and
141
Exact two waves solutions with variable amplitude
respectively, where c0 is an arbitrary constant, η0 = ξ0 + c0 .
Now we construct the variable amplitude-two-soliton solutions to the KdV
equation generated from u1 . For convenience, we first give the new solution
which is expressed in terms of σ0 rather than u1 , then consider μ0 and γ0 in σ0 .
According to the Darboux matrix (3) and (10), we can obtain the fundamental
matrix to the lax pair associated with the known solution u1 in the following
manner
⎛
Φ̃(x, t, λ) = ⎝
−σ0
1
⎞
P Q
S T
⎠ Φ(x, t, λ) =
,
λ − λ0 + σ02 −σ0
√
√
ξ
−1
)]e
,Q
=
[λ
−
(σ
+
x
)(
λ − x−1 )]e−ξ
where P = [λ + (σ0 + x−1 )( λ + x−1
0
√
2
−1
−1
ξ
2
S = −[λσ
0 + (λ − λ0 + σ0 + σ0 x )( λ + x )]e , T = [−λσ0 + (λ − λ0 + σ0 +
√
σ0 x−1 )( λ − x−1 )]e−ξ , from (7), this implies
σ1 = −σ0 −
√
(λ1 − λ0 )[(eξ1 μ1 + e−ξ1 γ1 ) + λ1 x(eξ1 μ1 − e−ξ1 γ1 )]
√
,
λ1 x(eξ1 μ1 + e−ξ1 γ1 ) + (σ0 + x−1 )[(eξ1 μ1 + e−ξ1 γ1 ) + λ1 x(eξ1 μ1 − e−ξ1 γ1 )]
where ξ1 = ξ(λ1 ). Choosing μ1 = ec1 , γ1 = e−c1 and μ1 = ec1 , γ1 = −e−c1 in
the above formulation, respectively, we obtain
√
(λ1 − λ0 )(cosh ξ1 + λ1 x sinh ξ1 )
√
σ11 = −σ0 −
,
(15)
λ1 x cosh ξ1 + (σ0 + x−1 )(cosh ξ1 + λ1 x sinh ξ1 )
and
σ12
√
(λ1 − λ0 )(sinh ξ1 + λ1 x cosh ξ1 )
√
,
= −σ0 −
λ1 x sinh ξ1 + (σ0 + x−1 )(sinh ξ1 + λ1 x cosh ξ1 )
(16)
Substituting (5) and (15) into (6), we get
u2 = −2x−2 + 2(λ1 − λ0 ) ×
√
λ1 (cosh2 η1 + λ1 x2 ) + (λ0 − σ02 + x−2 )(cosh η1 + λ1 x sinh η1 )2
√
.
[λ1 x cosh η1 + (σ0 + x−1 )(cosh η1 + λ1 x sinh η1 )]2
(17)
where η1 = ξ(λ1 ) + c1 , c1 is an arbitrary constant. Substituting (13) into (17),
at the same time, choosing μ0 = ec0 , γ0 = e−c0 and μ0 = ec0 , γ0 = −e−c0 ,
respectively, we obtain the two-soliton solutions
u2−1 = 2
(λ1 − λ0 )x2 (λ21 A21 − λ20 B12 ) − λ0 λ1 (cosh η1 A1 − cosh η0 B1 )2
x2 (λ1 cosh η1 A1 − λ0 cosh η0 B1 )2
and
u2−2 = 2
(λ1 − λ0 )x2 (λ21 A22 + λ20 B12 ) − λ0 λ1 (cosh η1 A2 − sinh η0 B1 )2
,
x2 (λ1 cosh η1 A2 − λ0 sinh η0 B1 )2
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Ying Huang
√
√
respectively, where
A
=
cosh
η
+
λ
x
sinh
η
,
B
=
cosh
η
+
λ1 x sinh η1 ,
1
0
0
0
1
1
√
A2 = sinh η0 + λ0 x cosh η0 .
In the same way, compound (5), (6), (13) and (16), we obtain new solutions
u2−3 = 2
(λ0 − λ1 )x2 (λ21 A21 + λ20 B22 ) − λ0 λ1 (sinh η1 A1 − cosh η0 B2 )2
x2 (λ1 sinh η1 A1 − λ0 cosh η0 B2 )2
and
(λ1 − λ0 )x2 (λ20 B22 − λ21 A22 ) − λ0 λ1 (sinh η1 A2 − sinh η0 B2 )2
,
x2 (λ1 sinh η1 A2 − λ0 sinh η0 B2 )2
√
with B2 = sinh η1 + λ1 x cosh η1 .
Next, the same calculation as the case λ > 0 can be done in λ < 0,
combining (11) with (4), we obtain
u2−4 = 2
u1−1 =
−2λ0 (cos2 θ0 + λ0 x2 )
√
(cos θ0 − −λ0 x sin θ0 )2
(18)
u1−2 =
−2λ0 (sin2 θ0 + λ0 x2 )
√
(sin θ0 + −λ0 x cos θ0 )2
(19)
and
with θ0 = θ(λ0 ). In exactly the same manner, we further obtain new doubly
periodic solutions
u2−1 = 2
2 2
2
(λ1 − λ0 )x2 (λ21 A2
1 − λ0 B1 ) − λ0 λ1 (cos θ1 A1 − cos θ0 B1 )
,
x2 (λ1 cos θ1 A1 − λ0 cos θ0 B1 )2
u2−2 = 2
2 2
2
(λ1 − λ0 )x2 (λ21 A2
2 − λ0 B1 ) − λ0 λ1 (cos θ1 A2 − sin θ0 B1 )
,
x2 (λ1 cos θ1 A2 − λ0 sin θ0 B1 )2
u2−3 = 2
2 2
2
(λ1 − λ0 )x2 (λ21 A2
1 − λ0 B2 ) − λ0 λ1 (sin θ1 A1 − cos θ0 B2 )
x2 (λ1 sin θ1 A1 − λ0 cos θ0 B2 )2
together with
2 2
2
(λ1 − λ0 )x2 (λ21 A2
2 − λ0 B2 ) − λ0 λ1 (sin θ1 A2 − sin θ0 B2 )
,
x2 (λ1 sin θ1 A2 − λ0 sin θ0 B2 )2
√
√
sin θ0 , B1 = cos θ1 − −λ1 x sin θ1 , A2 = sin θ0 +
where
A1 = cos θ0 − −λ0 x√
√
−λ0 x cos θ0 , B2 = sin θ1 + −λ1 x cos θ1 .
u2−4 = 2
Exact two waves solutions with variable amplitude
143
Finally, from (4) and (12), we get nonlinear rational solution
u =
8μ0 x(12μ0 t + γ0 )
.
(μ0 x3 + 12μ0 t + γ0 )2
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Received: December 7, 2013