Periodic wave solutions to a coupled KdV equations with variable coefficients

Physics Letters A 308 (2003) 31–36 www.elsevier.com/locate/pla

Periodic wave solutions to a coupled KdV equations with variable coefficients Yubin Zhou a,b,∗ , Mingliang Wang a,c , Yueming Wang c a Department of Mathematics, Lanzhou University, Lanzhou 730000, PR China b Department of Mechanics, Lanzhou University, Lanzhou 730000, PR China c Department of Mathematics and Physics, Henan University of Science and Technology, Luoyang 471039, PR China

Received 12 August 2002; accepted 13 December 2002 Communicated by A.R. Bishop

Abstract The periodic wave solutions to a coupled KdV equations with variable coefficients are obtained by using F -expansion method which can be thought of as an over-all generalization of Jacobi elliptic function expansion method. In the limit cases, the solitary wave solutions are obtained as well.  2003 Elsevier Science B.V. All rights reserved. Keywords: Coupled KdV equations with variable coefficients; F -expansion method; Periodic wave solutions; Jacobi elliptic functions; Solitary wave solutions

1. Introduction In this Letter we consider the coupled KdV equations with variable coefficients in the form ut + α(t)uux + β(t)vvx + γ (t)uxxx = 0,

(1.1)

vt + δ(t)uvx + γ (t)vxxx = 0,

(1.2)

where α(t), β(t), δ(t) and γ (t) are all functions of variable t only, and assume that they satisfy the following conditions β(t) = 0,

δ(t) = 0 and δ(t) − α(t) = σ 2 β(t),

γ (t) = kδ(t),

σ, k are constants.

(1.3)

Eqs. (1.1), (1.2) as a simple generalization of Hirota–Satsuma coupled KdV equations [1] are of physically importance, and if v = v(x, t) = 0, then Eqs. (1.1), (1.2) become a single KdV equation with variable coefficients ut + α(t)uux + γ (t)uxxx = 0, * Corresponding author.

E-mail address: [email protected] (Y. Zhou). 0375-9601/03/$ – see front matter  2003 Elsevier Science B.V. All rights reserved. doi:10.1016/S0375-9601(02)01775-9

(1.4)

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Y. Zhou et al. / Physics Letters A 308 (2003) 31–36

which has been payed attention by some authors [2–6]. Using homogeneous balance principle and ε-expansion method, we already obtained the auto-BT and multi-solitary wave solutions of Eqs. (1.1), (1.2). But the periodic wave solutions of Eqs. (1.1), (1.2) is still unknown so far. The aim of this Letter is to obtain the periodic wave solutions of Eqs. (1.1), (1.2) by use of F -expansion method which can be thought of as a succinctly over-all generalization of Jacobi elliptic function expansion method proposed and developed in [7–9]. In Section 2, we describe F -expansion method. In Section 3, we use F -expansion method to obtain the periodic wave solutions of Eqs. (1.1), (1.2).

2. Description of F -expansion method In order to simultaneously obtain more periodic wave solutions expressed by various Jacobi elliptic functions to nonlinear wave equations, we introduce a F -expansion method which can be thought of as a succinctly over-all generalization of Jacobi elliptic function expansion. We briefly show what is F -expansion method and how to use it to obtain various periodic wave solutions to nonlinear wave equations. Suppose a nonlinear wave equation for u(x, t) is given by P (u, ut , ux , ut t , uxt , uxx , uxxx , . . .) = 0,

(2.1)

in which both nonlinear term(s) and higher order derivatives of u(x, t) are all involved. In general, the left-hand side of Eq. (2.1) is a polynomial in u and its various derivatives. The F -expansion method for solving Eq. (2.1) proceeds in the following five steps: Step 1. Look for traveling wave solution of Eq. (2.1) by taking u(x, t) = f (ξ ), ξ = λx + µ(t),

(2.2)

where λ is nonzero constants, µ(t) the function of t satisfying µ (t) = ωµ(t), ¯ both to be determined later, µ(t) ¯ a given function of t. Substituting (2.2) into Eq. (2.1) yields an ordinary differential equation (ode) for f (ξ )  
f, µ f  , λf  , µ f  , µ 2 f  , µ λf  , λ2 f  , λ3 f  , . . . = 0 P (2.3) or  
f, ωµf P ¯  , λf  , ωµ¯  f  , (ωµ) ¯ 2 f  , ωµλf ¯  , λ2 f  , λ3 f  , . . . = 0.

(2.4)

Step 2. Suppose that f (ξ ) can be expressed by a finite power series of F (ξ ) f (x, t) =

n 

ak F k (ξ ),

an = 0,

(2.5)

k=0

instead of any kind of Jacobi elliptic function in Jacobi elliptic function expansion method [7–9] where a0 , a1 , . . . , an are constants to be determined later, F (ξ ) satisfies the first order nonlinear ode in the form F  2 (ξ ) = q0 + q2 F 2 (ξ ) + q4 F 4 (ξ ), and hence holds for F (ξ )     3   F F = q2 F F + 2q4F F (ξ ), F  = q2 F + 2q4 F 3 ,   F = q2 F  + 6q4 F 2 F  .

(2.6)

(2.7)

Y. Zhou et al. / Physics Letters A 308 (2003) 31–36

33

Table 1 Relations between values of (q0 , q2 , q4 ) and corresponding F (ξ ) in ode F  2 = q0 + q2 F 2 + q4 F 4 q0

q2

q4

1

  − 1 + m2

m2

1 − m2

2m2 − 1

−m2

m2 − 1

2 − m2

−1

m2

  − 1 + m2

−m2

2m2 − 1

1 − m2

−1

2 − m2

m2 − 1

1

2 − m2

1 − m2

1

2m2 − 1

1 − m2   −m2 1 − m2

2 − m2

1

2m2 − 1

1

1

 2

−m 1 − m

 2

F  2 = q0 + q2 F 2 + q4 F 4    F  2 = 1 − F 2 1 − m2 F 2    F  2 = 1 − F 2 m2 F 2 + 1 − m2    F  2 = 1 − F 2 F 2 + m2 − 1    F  2 = 1 − F 2 m2 − F 2    F  2 = 1 − F 2 m2 − 1 F 2 − m2    F  2 = 1 − F 2 1 − m2 F 2 − 1    F  2 = 1 + F 2 1 − m2 F 2 + 1     F  2 = 1 + m2 F 2 1 + m2 − 1 F 2    F  2 = 1 + F 2 F 2 + 1 − m2    F  2 = F 2 + m2 F 2 + m2 − 1

F cn ξ sn ξ, cd ξ = dn ξ cn ξ

dn ξ ξ ns ξ = (sn ξ )−1 , dc ξ = dn cn ξ nc ξ = (cn ξ )−1

nd ξ = (dn ξ )−1 sn ξ sc ξ = cn ξ sn ξ sd ξ = dn ξ cn ξ cs ξ = sn ξ ξ ds ξ = dn sn ξ

Integer n in (2.5) can be determined by considering homogeneous balance between the nonlinear terms and the highest order derivatives of f (ξ ) in Eq. (2.3). Step 3. Substituting the F -expansion (2.5) into ode (2.3) and using ode (2.6) and (2.7), then the left-hand side of Eq. (2.3) can be converted into a polynomial in F (ξ ). Setting each coefficient of the polynomial to zero yields equations for a0 , a1 , . . . , an , ω and λ. Step 4. Solving the equations obtained in Step 3, probably with the aid of Mathematica and using Wu elimination method, then a0 , a1 , . . . , aN , ω and λ can be expressed by q0 , q2 , q4 . Substituting these results into F -expansion (2.5), then a general form of traveling wave solution of Eq. (2.1) can be obtained. Step 5. Choose properly (q0 , q2 , q4 ) in ode (2.6) such that the corresponding solution F (ξ ) of it is one of Jacobi elliptic functions. The relations between values of (q0 , q2 , q4 ) and corresponding Jacobi elliptic function solution F (ξ ) of Eq. (2.6) are given in Table 1 which involves twelve relations that has been examined by Maple. Substitute the values of (q0 , q2 , q4 ) and the corresponding Jacobi elliptic function solution F (ξ ) chosen from Table 1 into the general form of solution obtained in Step 4, then an ideal periodic wave solution expressed by Jacobi elliptic function can be obtained. In this way we can simultaneously obtained more periodic wave solutions to Eq. (2.1).

3. Periodic wave solutions and solitary wave solutions Now we discuss the Eqs. (1.1), (1.2) with (1.3) using the method described in the previous section. Step 1. Suppose the solution of Eqs. (1.1), (1.2) is of the form u = f (ξ ),

v = g(ξ ),

ξ = λx + µ(t), λ = 0.

(3.1)

We can write the µ(t) as

t µ(t) = ωµ(t) ¯ + µ0 ,

µ(t) ¯ =

δ(τ ) dτ, 0

where µ0 is an arbitrary constant and ω undetermined parameter.

(3.2)

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Y. Zhou et al. / Physics Letters A 308 (2003) 31–36

Substituting (3.1) and (3.2) into Eqs. (1.1), (1.2), we have ωδf  + λαff  + λβgg  + λ3 γf  = 0, 



(3.3)



ωδg + λδfg + λ γ g = 0. 3

(3.4)

Step 2. Substituting (2.5) and (2.6) into Eqs. (3.3), (3.4), using relations (2.7), considering homogeneous balance between ff  , gg  and f  in (3.3) as well as fg  and g  in (3.4), we suppose that the solution of odes (3.3), (3.4) is of the form f (ξ ) = a0 + a1 F (ξ ) + a2 F 2 (ξ ) = 0,

a2 = 0,

(3.5)

g(ξ ) = b0 + b1 F (ξ ) + b2 F (ξ ) = 0,

b2 = 0,

(3.6)

2

where F (ξ ) satisfies ode (2.6), a0 , a1 , a2 , b0 , b1 , b2 are constants to be determined later. Step 3. Now substituting (3.5), (3.6) into (3.3), (3.4), and using (2.6) and (2.7) yields     2λ 12γ (t)a2 q4 λ2 + b22 β(t) + α(t)a22 F  F 3 + 3λ 2γ (t)a1 q4 λ2 + α(t)a1 a2 + b2 b1 β(t) F  F 2   + 2a2 ωδ(t) + β(t)b12 λ + 8γ (t)a2 q2 λ3 + 2α(t)a0 a2 λ + α(t)a12 λ + 2β(t)b0 b2 λ F  F   + a1 ωδ(t) + α(t)a0 a1 λ + β(t)b0 b1 λ + γ (t)a1 q2 λ3 F  = 0, (3.7)     2b2λ 12q4γ (t)λ2 + δ(t)a2 F  F 3 + λ 6γ (t)b1 q4 λ2 + 2δ(t)a1 b2 + δ(t)a2 b1 F  F 2     + 2b2 ωδ(t) + 2δ(t)a0 b2 λ + 8γ (t)b2 q2 λ3 + δ(t)a1 b1 λ F  F + b1 ωδ(t) + δ(t)a0 λ + λ3 q2 γ (t) F  = 0. (3.8) Canceling F 

(k = 0, 1, 2, 3) to zero yields a set of equations for a0 , a1 , a2 , b0 , b1 , b2 , λ and ω   2λ 12γ (t)a2 q4 λ2 + b22 β(t) + α(t)a22 = 0, (3.9)   2 3λ 2γ (t)a1 q4 λ + α(t)a1 a2 + b2 b1 β(t) = 0, (3.10) and setting each coefficient of F k

2a2ωδ(t) + β(t)b12 λ + 8γ (t)a2 q2 λ3 + 2α(t)a0 a2 λ + α(t)a12 λ + 2β(t)b0 b2 λ = 0,

(3.11)

a1 ωδ(t) + α(t)a0 a1 λ + β(t)b0 b1 λ + γ (t)a1 q2 λ = 0,   2b2λ 12q4γ (t)λ2 + δ(t)a2 = 0,   λ 6γ (t)b1 q4 λ2 + 2δ(t)a1 b2 + δ(t)a2 b1 = 0,

(3.12)

3

(3.13) (3.14)

2b2ωδ(t) + 2δ(t)a0 b2 λ + 8γ (t)b2 q2 λ + δ(t)a1 b1 λ = 0,   b1 ωδ(t) + δ(t)a0 λ + λ3 q2 γ (t) = 0. 3

(3.15) (3.16)

Step 4. The solution of Eqs. (3.9)–(3.16) under condition (1.3) and a2 = 0,

b2 = 0,

ω = 0,

λ = 0,

is only λ, a0 = arbitrary constant, b0 = −σ a0 ,   2 a2 = −12kλ2 q4 , ω = −λ a0 + 4kλ q2 ,

a1 = 0,

b1 = 0,

b2 = 12kσ λ q4 . 2

(3.17) (3.18)

Substituting (3.17) and (3.18) into (3.5) and (3.6) yields general form solutions to (1.1), (1.2) u(x, t) = f (ξ ) = a0 − 12kλ2 q4 F 2 (ξ ),

(3.19)

v(x, t) = g(ξ ) = −σ a0 + 12kσ λ q4 F (ξ ),

(3.20)

2

2

Y. Zhou et al. / Physics Letters A 308 (2003) 31–36

  with ξ = λx − λ a0 + 4kλ2 q2

35

t δ(τ ) dτ + µ0 ,

(3.21)

0

where k, a0 , λ are arbitrary constants. Step 5. From Table 1, if we take q0 = 1, q2 = −(1 + m2 ), q4 = m2 , then F (ξ ) = sn ξ , thus u(x, t) = f (ξ ) = a0 − 12kλ2 m2 sn2 ξ,

(3.22)

v(x, t) = g(ξ ) = −σ a0 + 12kσ λ m sn ξ,

t    2 2 with ξ = λx − λ a0 − 4kλ 1 + m δ(τ ) dτ + µ0 . 2

2

2

(3.23) (3.24)

0

In the limit case when m → 1, then sn ξ → tanh ξ , thus (3.22)–(3.24) become u(x, t) = f (ξ ) = a0 − 12kλ2 tanh2 ξ = a0 − 12kλ2 + 12kλ2 sech2 ξ,

(3.25)

v(x, t) = g(ξ ) = −σ a0 + 12kσ λ tanh ξ = −σ a0 + 12kσ λ − 12kσ λ sech ξ,

t   2 with ξ = λx − λ a0 − 8kλ δ(τ ) dτ + µ0 , 2

2

2

2

2

(3.26) (3.27)

0

which is solitary wave solution of Eqs. (1.1), (1.2). If q0 = m2 , q2 = −(1 + m2 ), q4 = 1, then F (ξ ) = dc ξ , thus u(x, t) = f (ξ ) = a0 − 12kλ2 dc2 ξ,

(3.28)

v(x, t) = g(ξ ) = −σ a0 + 12kσ λ dc ξ,

t    2 2 with ξ = λx − λ a0 − 4kλ 1 + m δ(τ ) dτ + µ0 . 2

2

(3.29) (3.30)

0

In the limit case when m → 0, then (3.28)–(3.30) become u(x, t) = f (ξ ) = a0 − 12kλ2 sec2 ξ = a0 − 12kλ2 − 12kλ2 tan2 ξ,

(3.31)

v(x, t) = g(ξ ) = −σ a0 + 12kσ λ sec ξ = −σ a0 + 12kσ λ + 12kσ λ tan ξ,

t   2 with ξ = λx − λ a0 − 4kλ δ(τ ) dτ + µ0 . 2

2

2

2

2

(3.32) (3.33)

0

If q0 = 1 − m2 , q2 = 2m2 − 1, q4 = −m2 (0 < m < 1), then F (ξ ) = cn ξ , thus we have the periodic wave solution expressed by Jacobi cn-function to Eqs. (1.1), (1.2) u(x, t) = f (ξ ) = a0 + 12kλ2 m2 cn2 ξ,

(3.34)

v(x, t) = g(ξ ) = −σ a0 − 12kσ λ m cn ξ,

t   2  2 with ξ = λx − λ a0 − 4kλ 2m − 1 δ(τ ) dτ + µ0 . 2

2

2

0

(3.35) (3.36)

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Y. Zhou et al. / Physics Letters A 308 (2003) 31–36

In the limit case when m → 1, then cn ξ → sech ξ , thus (3.34)–(3.36) becomes the solitary wave solution to Eqs. (1.1), (1.2) u(x, t) = f (ξ ) = a0 + 12kλ2 sech2 ξ,

(3.37)

v(x, t) = g(ξ ) = −σ a0 − 12kσ λ sech ξ,

t   2 with ξ = λx − λ a0 − 4kλ δ(τ ) dτ + µ0 . 2

2

(3.38) (3.39)

0

We can obtain the other solutions to Eqs. (3.1), (3.2) in terms of Table 1, but we omit them here for simplicity.

Acknowledgements This work is suported in part by the National Natural Science Foundation of China, the Natural Science Foundation of Henan province of China and the Natural Science Foundation of Education Committee of Henan province of China.

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