Hindawi Publishing Corporation
Mathematical Problems in Engineering
Volume 2009, Article ID 737928, 13 pages
doi:10.1155/2009/737928
Research Article
Exact Solutions for a Third-Order KdV Equation
with Variable Coefficients and Forcing Term
Alvaro H. Salas S1, 2 and Cesar A. Gómez S3
1
Department of Mathematics, Universidad de Caldas, Campus La Nubia, Manizales,
Caldas Cll 65 # 26-10, A. A. 275, Colombia
2
Department of Mathematics and Statistics, National University of Colombia,
Manizales Cll 65, Palogrande Stadium, Colombia
3
Department of Mathematics, National University of Colombia, Bogotá, Colombia
Correspondence should be addressed to Alvaro H. Salas S, asalash2002@yahoo.com
Received 20 April 2009; Revised 14 August 2009; Accepted 31 August 2009
Recommended by Ji Huan He
The general projective Riccati equation method and the Exp-function method are used to construct
generalized soliton solutions and periodic solutions to special KdV equation with variable
coefficients and forcing term.
Copyright q 2009 A. H. Salas S and C. A. Gómez S. 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.
1. Introduction
It is well known that in an early phase of the development of the solitons theory, there
were already many applications in physics and engineering. In particular, traveling waves
as solutions of the KdV equation
ut 6uux uxxx 0
1.1
have been of some interest since 150 years. Some generalizations of this last equation have
been studied recently. For instance, the equation
ut k1 tn uux k2 tm uxxx 0,
1.2
where k1 , k2 are arbitrary constants, which have applications in physics, has been analyzed
in 1 from the point of view of its exact solutions. The search of explicit solutions to
2
Mathematical Problems in Engineering
nonlinear partial differential equations NLPDEs using analytic methods is not an easy
task. However, the use of computational methods facilitates this work. Some powerful
computational methods such as the tanh method 2, the generalized tanh method 3, 4,
the extended tanh method 5–10, the improved tanh-coth method 11–13, the Exp-function
method 14–18, the modified Exp-function method 19, the Cole-Hopf transformation 20,
the projective Riccati equation method PREM 21, 22, the generalized projective Riccati
equations method 23–25, the extended hyperbolic function method 26, and many other
methods have been developed in this direction. The PREM and the Exp-function method
have been used in a satisfactory form to solve some NLPDEs 15–17, 24, 25, 27–30. In this
paper, we use this last two methods to obtain soliton and periodic solutions to the following
special KdV equation with variable coefficients and forcing term:
ut αtuux kαtuxxx Ft,
1.3
where Ft is an external forcing function varying with time t, k is a constant, and α αt is a
function of t, αt /
0. Equation 1.3 is a generalization of the following equation 15, 17, 31:
ut αuux βuxxx Ft
α, β const ,
1.4
which results from 1.3 by taking α const and k β/α.
We suppose that the solution to 1.3 has the form
ux, t Ftdt vx, t.
1.5
Ftdtvx kαtvxxx 0.
1.6
Therefore, 1.3 reduces to
vt αtvvx αt
Now we consider the transformation
v V ξ,
ξ λx htdt,
1.7
where λ is a constant and ht is an unknown function of t to be determined later. Substituting
1.7 into 1.6, we obtain
htV ξ αtλ V ξ ft V ξ kλ3 αtV ξ 0,
1.8
where
f ft Ftdt.
1.9
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3
2. The Exp-Function Method
Recently He and Wu 14 have introduced the Exp-function method to solve nonlinear
equations. In particular, the Exp-function method is a useful tool for solving nonlinear
equations with high nonlinearity. The method has been used in a satisfactory way by other
authors to solve a great variety of nonlinear wave equations 14–19. The Exp-function
method is very simple and straightforward and is based on a priori assumption that traveling
wave solutions to a nonlinear partial differential equation in the form
Fu, ux , ut , uxx , uxt , utt , . . . 0
2.1
can be found using the expression
d
an expnξ
,
uξ qn−c
n−p am expmξ
2.2
where c, d, p, and q are positive integers which could be freely chosen; an and bn are unknown
constants to be determined. According to this, we suppose that solutions to 1.8 can be
expressed in the form
d
vξ qn−c
m−p
an expnξ
bm expmξ
a−c exp−cξ · · · ad expdξ
,
b−p exp −pξ · · · bq exp qξ
2.3
where c, d, p, and q are positive integers which are unknown to be determined later; an and
bm are unknown constants. We have the following two cases.
2.1. Case 1: p c 1 and d q 1
In this case, the trial solution to 1.8 becomes
V ξ a1 expξ a0 a−1 exp−ξ
.
expξ b0 b−1 exp−ξ
2.4
Substituting 2.4 into 1.8 we obtain a polynomial equation in the variable η expξ.
Equating to zero the coefficients of all powers of η yields a set of algebraic equations. Solving
it with the aid of a computer, we get a0 b0 a1 6kλ2 , a−1 1/4a1 b02 , b−1 b02 /4, and
ht −λαta1 ft kλ2 , and, from 1.5, 1.7, and 2.3, one solution to 1.3 is given
by
2
a1 b0 2eξ 24b0 kλ2 eξ
Ftdt,
ux, t 2
b0 2eξ
2.5
4
Mathematical Problems in Engineering
where
ξ ξx, t λx htdt λx −
λαt a1 kλ2 Ftdt dt,
2.6
and a1 , b0 , and λ are arbitrary real or complex numbers.
2.2. Case 2: p c 2 and d q 2
In this case, the trial solution to 1.8 becomes
a2 exp2ξ a1 expξ a0 a−1 exp−ξ a−2 exp−2ξ
.
exp2ξ b1 expξ b0 b−1 exp−ξ b−2 exp−2ξ
V ξ 2.7
As in the first case, we obtain an algebraic system. Solving it gives
−2a1 b1 a2 3kλ2 a2 a2 b12 6 b12 − b0 kλ2 a21
a0 −
,
6kλ2
a1 − a2 b1 2
2
2 4
a−1 −a
27a
b
kλ
2a
72k
λ
1
1
2
2
432k3 λ6
a2 6kλ2 12a2 b12 kλ2 a22 b12 72b0 k2 λ4 a21 a2 9kλ2 ,
1
2
2
2
2
2 2
2
2 4
8a
a
a
,
a
−
a
b
b
kλ
−
2a
b
4kλ
b
a
48b
k
λ
a
2
1
2
1
2
1
1
2
0
2 1
1
1
6912k4 λ8
1
b−1 a1 − a2 b1 9a2 b12 kλ2 − a1 b1 2a2 9kλ2 a22 b12 a21 72b0 k2 λ4 ,
3
6
432k λ
1
2
2
2
2
2 2
2
2 4
b−2 8a
a
a
,
−
a
b
b
kλ
−
2a
b
4kλ
b
a
48b
k
λ
a
1
2
1
2
1
1
2
0
2 1
1
1
6912k4 λ8
ht −λαt a2 ft kλ2 .
a−2 2.8
From 1.5, 1.7, and 2.7 we may verify that to this set of values corresponds the solution
1
ux, t a1 − a2 b1 12kλ2 e
Ftdt,
a21 a2 a1 −2a22 b1 24a2 kλ2 eξ 144k2 λ4 eξ
2
ξ
a2 a22 b12 − 24a2 b1 kλ2 eξ 144k2 λ4 eξ eξ − b1
2.9
Mathematical Problems in Engineering
5
where
ξ ξx, t λx htdt λx −
λαt a2 kλ 2
Ftdt dt,
2.10
and a1 , a2 , b1 , and λ are arbitrary real or complex numbers.
3. General Projective Riccati Equation Method
The projective Riccati equation method was introduced initially in 21 and generalizations
of this method have been used in a satisfactory way by several authors to solve nonlinear
partial equations 22–25. Using this last method 24, 25, 27–30, we seek solutions to 1.8 in
the form
V ξ a0 m
σ i−1 ξai σξ bi τξ,
3.1
i1
where a0 , a1 , b1 , . . . are constants and σξ and τξ satisfy the system
σ ξ σξτξ,
τ ξ R τ 2 ξ − μσξ.
3.2
In 3.2, ±1 and R and μ are certain constants. These equations have following solutions.
Case 1. When −1 and R /
0,
√ R sech Rξ
σ1 ξ ,
√ μ sech Rξ 1
√
σ2 ξ R csch Rξ
,
√ μ csch Rξ 1
√ R tanh Rξ
τ1 ξ ,
√ μ sech Rξ 1
√
√ √
R coth Rξ
τ2 ξ .
√ μ csch Rξ 1
3.3
Case 2. When 1 and R /
0,
√ R sec Rξ
,
σ3 ξ √ μ sec Rξ 1
√ R tan Rξ
τ3 ξ ,
√ μ sec Rξ 1
√ R csc Rξ
σ4 ξ ,
√ μ csc Rξ 1
√ R cot Rξ
τ4 ξ .
√ μ csc Rξ 1
√
√
3.4
6
Mathematical Problems in Engineering
Case 3. When R μ 0,
σ5 ξ C
,
ξ
τ5 ξ 1
.
ξ
3.5
In this last case, we seek solutions to 1.8 in the form
V ξ m
ai τ i ξ,
3.6
i0
where τ ξ τ 2 ξ.
For any pair σξ, τξ of functions given by 3.3 or 3.4 the following equation
holds:
μ2 − 1 2
σ ξ .
τ ξ − R − 2μσξ R
2
3.7
3.1. Periodic and Soliton Solutions
Periodic and soliton solutions are obtained when R /
0 and ±1 and it corresponds to the
first two cases. Substituting 3.1, along with 3.2 and 3.7 into the left hand of 1.8 and
collecting all terms with the same power in σ i ξτ j ξ, we get a polynomial in the variables
σ σξ and τ τξ. We may choose m 2. Thus, we seek solutions to 1.8 in the form
V ξ a0 a1 σξ b1 τξ σξa2 σξ b2 τξ.
3.8
We equate each coefficient of the polynomial to zero. This will give an overdetermined system
of algebraic equations involving the parameters ai , bi i 0, . . . , m, λ and μ, R, and the
unknown function ht. Having determined these parameters, we may determine V ξ, and
using 1.5 we obtain an exact solution ux, t in a closed form. The corresponding system
reads.
i b12 R3 αλ − b12 R3 α2 λ 0.
ii 6a1 kRαλ3 4 − 6a1 kRαλ3 μ2 4 48a2 kR2 αλ3 μ4 − 3b1 b2 Rαλμ2 2 3b1 b2 Rαλ2 −
18a2 kR2 αλ3 μ2 4b22 R2 αλμ2 3a1 a2 R2 αλ − b22 R2 αλμ 0.
iii 24a2 kRαλ3 4 − 24a2 kRαλ3 μ2 4 − 2b22 Rαλμ2 2 2b22 Rαλ2 2a22 R2 αλ 0.
iv 6b1 kαλ3 μ4 5 6b1 kαλ3 5 − 96b2 kRαλ3 μ3 5 − 12b1 kαλ3 μ2 5 96b2 kRαλ3 μ5 36b2 kRαλ3 μ3 3 − 36b2 kRαλ3 μ3 − 3a2 b1 Rαλμ2 2 − 3a1 b2 Rαλμ2 2 3a2 b1 Rαλ2 3a1 b2 Rαλ2 8a2 b2 R2 αλμ2 − a2 b2 R2 αλμ 0.
v 24b2 kαλ3 μ4 5 24b2 kαλ3 5 − 48b2 kαλ3 μ2 5 − 4a2 b2 Rαλμ2 2 4a2 b2 Rαλ2 0,
−6a1 kR3 αλ3 4 5a1 kR3 αλ3 2 − 3b1 b2 R3 αλ2 2b12 R2 αλμ2 a0 a1 R2 αλ a1 fR2 αλ a1 R2 h 2b1 b2 R3 αλ − b12 R2 αλμ 0.
vi −24a2 kR3 αλ3 4 12a1 kR2 αλ3 μ4 16a2 kR3 αλ3 2 −b12 Rαλμ2 2 −2b22 R3 αλ2 b12 Rαλ2 −
6a1 kR2 αλ3 μ2 6b1 b2 R2 αλμ2 a21 R2 αλ 2a0 a2 R2 αλ 2a2 fR2 αλ 2a2 R2 h b22 R3 αλ − 2b1 b2 R2 αλμ 0.
Mathematical Problems in Engineering
7
vii 6b1 kR4 αλ3 5 − 8b1 kR4 αλ3 3 − a0 b1 R3 αλ2 − b1 fR3 αλ2 − b1 R3 h2 2b1 kR4 αλ3 a0 b1 R3 αλ b1 fR3 αλ b1 R3 h 0.
viii 24b2 kR4 αλ3 5 − 24b1 kR3 αλ3 μ5 − 28b2 kR4 αλ3 3 28b1 kR3 αλ3 μ3 − 2a1 b1 R3 αλ2 −
2a0 b2 R3 αλ2 − 2b2 fR3 αλ2 2a0 b1 R2 αλμ2 2b1 fR2 αλμ2 − 2b2 R3 h2 2b1 R2 μh2 5b2 kR4 αλ3 −5b1 kR3 αλ3 μa1 b1 R3 αλa0 b2 R3 αλb2 fR3 αλ−a0 b1 R2 αλμ−b1 fR2 αλμ
b2 R3 h − b1 R2 μh 0.
ix −12b1 kR2 αλ3 5 36b1 kR2 αλ3 μ2 5 −96b2 kR3 αλ3 μ5 8b1 kR2 αλ3 3 −32b1 kR2 αλ3 μ2 3 92b2 kR3 αλ3 μ3 − a0 b1 Rαλμ2 2 − b1 fRαλμ2 2 − 3a2 b1 R3 αλ2 − 3a1 b2 R3 αλ2 a0 b1 Rαλ2 b1 fRαλ2 4a1 b1 R2 αλμ2 4a0 b2 R2 αλμ2 4b2 fR2 αλμ2 − b1 Rμ2 h2 b1 Rh2 4b2 R2 μh2 3b1 kR2 αλ3 μ2 − 11b2 kR3 αλ3 μ a2 b1 R3 αλ a1 b2 R3 αλ −
a1 b1 R2 αλμ − a0 b2 R2 αλμ − b2 fR2 αλμ − b2 R2 μh 0.
x −48b2 kR2 αλ3 5 −24b1 kRαλ3 μ3 5 144b2 kR2 αλ3 μ2 5 24b1 kRαλ3 μ5 28b2 kR2 αλ3 3 12b1 kRαλ3 μ3 3 − 100b2 kR2 αλ3 μ2 3 − 12b1 kRαλ3 μ3 − 2a1 b1 Rαλμ2 2 − 2a0 b2 Rαλμ2 2 −
2b2 fRαλμ2 2 −4a2 b2 R3 αλ2 2a1 b1 Rαλ2 2a0 b2 Rαλ2 2b2 fRαλ2 6a2 b1 R2 αλμ2 6a1 b2 R2 αλμ2 − 2b2 Rμ2 h2 2b2 Rh2 6b2 kR2 αλ3 μ2 a2 b2 R3 αλ − a2 b1 R2 αλμ −
a1 b2 R2 αλμ 0.
In the equations above, f ft Ftdt, α αt, and h ht. Solving the previous
system with the aid of a computer, we obtain many solutions to 1.3. These solutions
may
be obtained from 1.5 and are given by 3.9–3.19. In√these formulas, Ht ht dt and
λ, a0 are arbitrary parameters. In a formula containing R, we suppose that
R > 0 and if the
√
√
expression −R appears, we choose R < 0. If a formula involves −R and 1 − μ2 see, e.g.,
3.17 we consider that R < 0 and |μ| ≤ 1.
First Group. 1:
i a2 b1 b2 0, a1 6kλ2 , ht −λαta0 Ftdt − Rkλ2 , μ −1:
u1 x, t Ftdt a0 −
6kλ2 R
,
√
1 − sin Rλx Ht
Ftdt a0 −
6kλ2 R
,
√
1 − cos Rλx Ht
Ftdt a0 −
6kλ2 R
.
√
1 − cosh −Rλx Ht
u2 x, t u3 x, t ii a2 b1 b2 0, a1 −6kλ2 , ht −λαta0 Ftdt − Rkλ2 , μ 1:
u4 x, t Ftdt a0 −
6kλ2 R
,
√
1 sin Rλx Ht
Ftdt a0 −
6kλ2 R
,
√
1 cos Rλx Ht
u5 x, t 3.9
8
Mathematical Problems in Engineering
u6 x, t Ftdt a0 −
6kλ2 R
.
√
1 cosh −Rλx Ht
3.10
iii a1 −6kλ2 μ, b1 0, a2 6kλ2 μ2 −1/R, b2 6kλ2 R1 − μ2 /R, ht −λαta0 Ftdt − Rkλ2 , μ μ:
u7 x, t Ftdt a0
√
√
6kλ2 R 1 μ cos Rλx Ht − 1 − μ2 sin Rλx Ht
,
−
√
√
μ2 2μ cos Rλx Ht cos2 Rλx Ht
u8 x, t Ftdt a0
√
√
6kλ2 R 1 μ cosh −Rλx Ht − μ2 − 1 sinh −Rλx Ht
−
,
2
√
μ cosh −Rλx Ht
u9 x, t Ftdt a0
√
√
6λ2 kR 1 μ sin Rλx Ht − 1 − μ2 cos Rλx Ht
.
−
√
√
1 μ2 2μ sin Rλx Ht − cos2 Rλx Ht
3.11
iv a1 −6kλ2 μ, b1 0, a2 6kλ2 μ2 − 1/R, b2 −6kλ2 R1 − μ2 /R, ht −λαta0 Ftdt − Rkλ2 , μ μ:
u10 x, t Ftdt a0
√
√
6kλ2 R 1 μ cos Rλx Ht 1 − μ2 sin Rλx Ht
,
−
√
√
μ2 2μ cos Rλx Ht cos2 Rλx Ht
u11 x, t Ftdt a0
√
√
6kλ2 R 1 μ cosh −Rλx Ht μ2 − 1 sinh −Rλx Ht
−
.
2
√
μ cosh −Rλx Ht
3.12
Mathematical Problems in Engineering
9
Figure 1 shows solution u11 x, t for the choices μ 1.3, R −2.3, k 2.3, λ 1.9,
a0 1, Ft sint, αt t, −3 ≤ x ≤ 4, and −1 ≤ t ≤ 1.1:
u12 x, t Ftdt a0
√
√
6kλ2 R 1 μ sin Rλx Ht 1 − μ2 cos Rλx Ht
.
−
√
√
1 μ2 2μ sin Rλx Ht − cos2 Rλx Ht
3.13
Figure 2 shows solution u12 x, t for the choices μ 0.3, R 2.3, k 2.3, λ 1.9, a0 1,
Ft sint, αt t, −3 ≤ x ≤ 4, and −1 ≤ t ≤ 1.1.
Second Group. −1:
√
v a1 b1 0, a2 6kλ2 /R, b2 −6kλ2 / −R, ht −λαta0 Ftdt Rkλ2 ,
μ 0:
u13 x, t √
6kλ2 R 1 sin −Rλx Ht
.
Ftdt a0 √
cos2 −Rλx Ht
3.14
√
vi a1 b1 0, a2 6kλ2 /R, b2 6kλ2 / −R, ht −λαta0 Ftdt Rkλ2 , μ 0:
u14 x, t √
6kλ2 R 1 − sin −Rλx Ht
.
Ftdt a0 √
cos2 −Rλx Ht
3.15
vii a1 6kλ2 μ, a2 6kλ2 1 − μ2 /R, b1 0, b2 6kλ2 Rμ2 − 1/R, ht −λαta0 Ftdt Rkλ2 , μ μ:
u15 x, t 6kλ2 μR
√
μ cosh Rλx Ht
√
6kλ2 R 1 − μ2 μ2 − 1 sinh Rλx Ht
,
2
√
μ cosh Rλx Ht
Ftdt a0 3.16
u16 x, t 6kλ2 μR
√
μ cos −Rλx Ht
√
6kλ2 R 1 − μ2 1 − μ2 sin −Rλx Ht
.
√
√
μ2 2μ cos −Rλx Ht cos2 −Rλx Ht
Ftdt a0 3.17
10
Mathematical Problems in Engineering
Figure 1: Graphic of solution u11 x, t, −3 ≤ x ≤ 4 and −1 ≤ t ≤ 1.1.
viii a1 6kλ2 μ, a2 −6kλ2 1 − μ2 /R, b1 0, b2 −6kλ2 Rμ2 − 1/R, ht −λαta0 Ftdt Rkλ2 , μ μ:
Ftdt a0 u17 x, t 6kλ2 μR
√
μ cosh Rλx Ht
√
6kλ2 R 1 − μ2 − μ2 − 1 sinh Rλx Ht
,
2
√
μ cosh Rλx Ht
u18 x, t Ftdt a0 3.18
6kλ2 μR
√
μ cos −Rλx Ht
√
6kλ2 R 1 − μ2 − 1 − μ2 sin −Rλx Ht
.
√
√
μ2 2μ cos −Rλx Ht cos2 −Rλx Ht
3.19
3.2. Rational Solutions
We seek rational solutions to 1.8 R μ 0 in the form given by 3.6 with m 2,
V ξ 2
i0
ai τ i ξ,
3.20
Mathematical Problems in Engineering
11
Figure 2: Graphic of solution u12 x, t, −3 ≤ x ≤ 4 and −1 ≤ t ≤ 1.1.
where
τ ξ τ 2 ξ.
3.21
Note that this last equation has the general solution
τξ −
1
.
ξC
3.22
We now substitute 3.20 into 1.8, and using the relation τ ξ τ 2 ξ we obtain an equation
whose left-hand side is a polynomial in the variable τ τξ. We equate each coefficient of
this polynomial to zero and we get the following algebraic system.
i hta1 αtλ Ftdta1 αtλa0 a1 0.
ii αtλa21 2hta2 2αtλ Ftdta2 2αtλa0 a2 0.
iii 6ka1 λ3 3a1 a2 λ 0.
iv 24ka2 λ3 2a22 λ 0.
Solving this system gives a0 a0 , a1 0, a2 −12kλ2 and
ht −λαt
Ftdt a0 .
3.23
A rational solution of 1.8 is given by
V ξ a0 −
12kλ2
ξ C2
.
3.24
12
Mathematical Problems in Engineering
According to 1.5, 1.6, and 1.7 we obtain the following rational solution to 1.3:
u19 x, t Ftdt a0 −
λx − λαt
12kλ2
.
2
Ftdt a0 dt C
3.25
4. Conclusions
In this paper, by using the projective Riccati equation and the Exp-function methods, with
the help of a symbolic computation engine, we obtain exact solutions for a generalized KdV
equation with forcing term 1.3. The methods certainly works well for a large class of very
interesting nonlinear equations. The main advantage of these methods is their capability of
greatly reducing the size of computational work compared to existing techniques such as
the pseudospectral method, the inverse scattering method, Hirota’s bilinear method, and the
truncated Painlevé expansion. The Exp-function method gives us more general solutions with
some free parameters than the projective Riccati equation method. It also has other interesting
applications. For instance, the Exp-function method may be applied not only to differential
equations but also to differential-difference equations or Stochastic equations 32–35.
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