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Exact Solutions of a Generalized KdV-mKdV Equation - IJNS, The ...

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ISSN 1749-3889 (print), 1749-3897 (online)<br />

International Journal <strong>of</strong> Nonlinear Science<br />

Vol.13(2012) No.1,pp.94-98<br />

<strong>Exact</strong> <strong>Solutions</strong> <strong>of</strong> a <strong>Generalized</strong> <strong>KdV</strong>-m<strong>KdV</strong> <strong>Equation</strong><br />

Cesar A. Gómez Sierra 1 , Motlatsi Molati 2 ∗ , Motlatsi P. Ramollo 2<br />

1 Department <strong>of</strong> Mathematics, Universidad Nacional de Colombia, Bogotá Colombia.<br />

2 Department <strong>of</strong> Mathematics and Computer Science, National University <strong>of</strong> Lesotho, PO Roma 180, Lesotho.<br />

(Received 26 January 2011 , accepted 9 August 2011)<br />

Abstract : We study the combined Korteweg-de Vries (<strong>KdV</strong>) and modified Korteweg-de Vries (m<strong>KdV</strong>) equations.<br />

By using symmetries we construct the exact solutions for the underlying equation. <strong>Exact</strong> solutions in<br />

explicit form are obtained for two particular cases. Other soliton solutions are derived by using the improved<br />

tanh-coth method.<br />

Keywords : <strong>KdV</strong>-m<strong>KdV</strong> equation ; Lie point symmetry ; exact solution ; solitons<br />

1 Introduction<br />

In practical applications finding the exact solutions for nonlinear partial differential equations (NLPDEs) is one <strong>of</strong> the<br />

most important task. Until this end, several direct and computational methods have been implemented. Among the direct<br />

methods <strong>of</strong>ten used are the Hirota method [1], inverse scattering method [2], and analysis by Lie group theory [3],[4], [5].<br />

On the other hand, the tanh-coth [6] method is one <strong>of</strong> the frequently used computational method.<br />

<strong>The</strong> combined <strong>KdV</strong> equation and modified <strong>KdV</strong> equation (<strong>KdV</strong>-m<strong>KdV</strong>) [7] that we study reads<br />

ut + auux + āu 2 ux + buxxx = 0, (1)<br />

with constants a, ā and b. Particular cases <strong>of</strong> (1) and similar models have been studied extensively using various approaches<br />

[8],[9] and the references therein. <strong>The</strong> main goal in this work is obtain the traveling wave solutions to (1) using two<br />

approaches. First, we use Lie point symmetries <strong>of</strong> the underlying equation to obtain the most general traveling wave<br />

solution and then we consider explicit solutions for two particular cases <strong>of</strong> (1). <strong>The</strong>reafter, we use the improved tanh-coth<br />

method to obtain other exact solutions to (1). <strong>The</strong> paper is organized as follows : In Sec. 2, we use the Lie symmetry<br />

approach to derive the traveling wave solution to (1) in an implicit form. As a consequence the soliton solution for two<br />

particular cases are analyzed. In Sec. 3, we use the improved tanh-coth method to obtain the exact traveling wave solutions<br />

to (1) in explicit form. Finally, some conclusions are given.<br />

2 Lie symmetry approach<br />

According to Lie’s algorithm [3],[4], [5], the infinitesimal generator <strong>of</strong> the maximal symmetry group admitted by (1)<br />

is given by<br />

X = ξ 1 (t, x, u) ∂<br />

∂t + ξ2 (t, x, u) ∂<br />

∂<br />

+ η(t, x, u) , (2)<br />

∂x ∂u<br />

if and only if the invariance condition <strong>of</strong> (1) is<br />

X [3] ( ut + auux + āu 2 )<br />

ux + buxxx = 0, (3)<br />

where<br />

X [3] ∂<br />

= X + ζt<br />

∂ut<br />

∗. Corresponding author. E-mail address : m.molati@gmail.com<br />

+ ζx<br />

∂<br />

∂ux<br />

∂<br />

+ ζxxx<br />

∂uxxx<br />

Copyright c⃝World Academic Press, World Academic Union<br />

<strong>IJNS</strong>.2012.02.15/580<br />

, (4)


C. A. Gómez Sierra, M. Molati, M. P. Ramollo: <strong>Exact</strong> <strong>Solutions</strong> <strong>of</strong> a <strong>Generalized</strong> <strong>KdV</strong>-m<strong>KdV</strong> <strong>Equation</strong> 95<br />

is the prolongation <strong>of</strong> the vector field (2). <strong>The</strong> variables ζis are given by the formulae :<br />

⎧<br />

ζ<br />

⎪⎨<br />

α i = Di(ηα ) − uα j Di(ξj ),<br />

ζα ij = Dj(ζ α i ) − uα ilDj(ξ l ),<br />

.<br />

⎪⎩<br />

ζα i1···ik = Dik (ζα i1···ik−1 ) − uα i1···ik−1lDik (ξl ).<br />

Executing the Lie’s algorithm, we obtain the Lie point symmetries <strong>of</strong> (1) given by<br />

X1 = ∂<br />

∂t , X2 = ∂<br />

∂x , X3 = −3t ∂<br />

∂t +<br />

( ) 2 a t ∂<br />

− x<br />

2ā ∂x +<br />

We look for solutions invariant under the linear combination X1 + λX2 where λ is a constant.<br />

Solving the characteristic system for the invariants <strong>of</strong> the linear combination, we obtain<br />

where v is an arbitrary function <strong>of</strong> ξ. <strong>The</strong> substitution <strong>of</strong> (7) into (1) yields<br />

(5)<br />

(<br />

a<br />

)<br />

∂<br />

+ u . (6)<br />

2ā ∂u<br />

ξ = x − λt, u(t, x) = v(ξ), (7)<br />

−λ v ′ (ξ) + av(ξ)v ′ (ξ) + āv 2 (ξ)v ′ (ξ) + bv ′′′ (ξ) = 0, (8)<br />

where a prime denotes differentiation with respect to ξ and λ represents the wave speed. Integration <strong>of</strong> (8) once leads to<br />

Multiplying both sides <strong>of</strong> (9) by v ′ and integrating once more we obtain<br />

−λv + a<br />

2 v2 + ā<br />

3 v3 + bv ′′ = k1. (9)<br />

− λ<br />

2 v2 + a<br />

6 v3 + ā<br />

12 v4 + b<br />

2 (v′ ) 2 = k1v + k2. (10)<br />

Finally, from (10) we have<br />

√<br />

√<br />

b dv<br />

2k1v + 2k2 + λv2 − a<br />

3 v3 − ā<br />

6 v4<br />

= dξ (11)<br />

where k1 and k2 are arbitrary constants.<br />

Integrating both sides <strong>of</strong> this last equation we obtain the general traveling wave solution to (1) in implicit form<br />

∫<br />

√<br />

b dv<br />

√<br />

2k1v + 2k2 + λv2 − a<br />

3 v3 − ā<br />

6 v4<br />

= ξ + k3 = x − λ t + k3<br />

for an arbitrary constant k3.<br />

2.1 First particular case<br />

Setting a = 0 in (1) we obtain the following <strong>KdV</strong> <strong>of</strong> higher order<br />

(12)<br />

ut + āu 2 ux + buxxx = 0. (13)<br />

If we assume that v → 0, v ′ → 0, v ′′ → 0 when ξ → ∞ in the analysis presented in the previous section, then (12)<br />

reduces to<br />

With the substitution v =<br />

√ 6λ<br />

ā<br />

where k4 is a constant <strong>of</strong> integration.<br />

√ ∫<br />

b<br />

λ<br />

dv<br />

√<br />

v 1 − ā<br />

6λv2 = ξ + k3. (14)<br />

sechw finally we obtain the following soliton solution to (13)<br />

√<br />

6λ<br />

u(t, x) =<br />

ā sech<br />

[√ ]<br />

λ<br />

(x − λ t + k4) , (15)<br />

b<br />

<strong>IJNS</strong> homepage: http://www.nonlinearscience.org.uk/


96 International Journal <strong>of</strong> Nonlinear Science, Vol.13(2012), No.1, pp. 94-98<br />

2.2 Second particular case<br />

If we set ā = 0, then the equation under study (1) reduces to<br />

ut + auux + buxxx = 0, (16)<br />

which is the generalized <strong>KdV</strong> equation. Proceeding as in the first case, from (12) we obtain<br />

√ ∫<br />

b<br />

λ<br />

dv<br />

v √ 1 − a<br />

3λ v = ξ + k3. (17)<br />

Using the substitution v = 3λ<br />

a sech 2 w, and after simplifications, finally we obtain the following traveling wave<br />

solution to (16)<br />

u(t, x) = 3λ<br />

a sech2<br />

3 <strong>The</strong> improved tanh-coth method<br />

[ √ ]<br />

1 λ<br />

(x − λ t + k4) . (18)<br />

2 b<br />

In this section we consider the equation (1) with variable coefficients (depending <strong>of</strong> t), that is,<br />

ut + a(t)uux + ā(t)u 2 ux + b(t)uxxx = 0. (19)<br />

Our goal is to obtain the exact solutions <strong>of</strong> (19) using the improved tanh-coth method [10]. As before, we seek a solution<br />

in the form u(t, x) = v(ξ), being ξ = x − λ t. In this case, (19) reduces to<br />

−λ v ′ (ξ) + a(t)v(ξ)v ′ (ξ) + ā(t)v 2 (ξ)v ′ (ξ) + b(t)v ′′′ (ξ) = 0. (20)<br />

Using the idea <strong>of</strong> the tanh-coth method introduced by Wazwaz [6], we assume the solution <strong>of</strong> (20) in the form<br />

where ϕ(ξ) satisfies the Riccati equation<br />

M∑<br />

ai(t)ϕ(ξ) i +<br />

i=0<br />

2M∑<br />

M+1<br />

ai(t)ϕ(ξ) M−i , (21)<br />

ϕ ′ (ξ) = γ(t)ϕ 2 (ξ) + β(t)ϕ(ξ) + α(t). (22)<br />

<strong>The</strong> coefficients ai(t) are unknown functions to be determined later and α(t), γ(t), β(t) are independent <strong>of</strong> variable ξ. It<br />

is well known that the solution <strong>of</strong> (22) is given by [11]<br />

⎧<br />

1<br />

(<br />

−<br />

⎪⎨ γ(t)<br />

ϕ(ξ) =<br />

⎪⎩<br />

1<br />

)<br />

β(t)<br />

ξ − 2 , β2 = 4γ(t)α(t)<br />

( √ [√ ]<br />

1 β2 (t)−4α(t)γ(t)<br />

β2 (t)−4α(t)γ(t)<br />

−<br />

γ(t)<br />

2 tanh<br />

2 ξ − β(t)<br />

)<br />

2 , β2 ̸= 4γ(t)α(t)<br />

(23)<br />

Balancing the linear term <strong>of</strong> highest order with the nonlinear term in (20) we have<br />

so that M = 1. <strong>The</strong>refore, (21) takes the form<br />

M + 3 = 2M + (M + 1),<br />

a0(t) + a1(t)ϕ(ξ) + a2(t)ϕ(ξ) −1 . (24)<br />

Substitution <strong>of</strong> (24) along with (22) into (20) yields a set <strong>of</strong> algebraic equations for λ, ai(t), α(t), β(t) and γ(t).<br />

Solving the algebraic system with aid <strong>of</strong> Mathematica we consider only the following non-trivial solutions :<br />

1. ⎧⎨<br />

λ = −a(t)2 +4ā(t) 2 √<br />

+a1(t)a2(t)<br />

ā(t)<br />

6ā(t) , α(t) = ±i a2(t) 6b(t)<br />

⎩a0(t)<br />

= − a(t)<br />

ā(t) .<br />

<strong>IJNS</strong> email for contribution: editor@nonlinearscience.org.uk<br />

√<br />

ā(t)<br />

ia(t)<br />

, γ(t) = ±i a1(t) 6b(t) , β(t) = ± √ ,<br />

6ā(t)b(t)


C. A. Gómez Sierra, M. Molati, M. P. Ramollo: <strong>Exact</strong> <strong>Solutions</strong> <strong>of</strong> a <strong>Generalized</strong> <strong>KdV</strong>-m<strong>KdV</strong> <strong>Equation</strong> 97<br />

2. ⎧ ⎨<br />

3. ⎧ ⎨<br />

4. ⎧ ⎨<br />

5. ⎧ ⎨<br />

6. ⎧ ⎨<br />

7. ⎧ ⎨<br />

λ = −a(t)2<br />

√<br />

ā(t)<br />

6ā(t) , α(t) = 0, γ(t) = ±i a1(t) 6b(t)<br />

⎩<br />

a0(t) = 0, a2(t) = 0.<br />

λ = −a(t)2<br />

√<br />

ā(t)<br />

6ā(t) , α(t) = 0, γ(t) = ±i a1(t) 6b(t)<br />

⎩a0(t)<br />

= − a(t)<br />

ā(t) , a2(t) = 0.<br />

λ = −a(t)2<br />

√<br />

ā(t)<br />

6ā(t) , α(t) = ±i a2(t) 6b(t)<br />

⎩<br />

a0(t) = 0, a1(t) = 0.<br />

λ = −a(t)2<br />

√<br />

ā(t)<br />

6ā(t) , α(t) = ±i a2(t) 6b(t)<br />

⎩a0<br />

= − a(t)<br />

ā(t) , a1(t) = 0.<br />

λ = −a(t)2<br />

6ā(t)<br />

, α(t) = ±<br />

⎩<br />

a0(t) = − a(t)<br />

2ā(t) , a2(t) =<br />

λ = −a(t)2<br />

6ā(t)<br />

, α(t) = ±<br />

⎩<br />

a0 = − a(t)<br />

2ā(t) , a2(t) = 0.<br />

ia1(t) 2<br />

16a1(t) 5 2<br />

2<br />

a1(t)<br />

16ā(t) 2a1(t) .<br />

ia1(t) 2<br />

8a1(t)ā(t) 3 2<br />

8.<br />

⎧<br />

⎨λ<br />

=<br />

⎩<br />

−a(t)2 +a(t)ā(t)a0(t)+2ā(t) 2 a0(t) 2<br />

6ā(t) , α(t) =<br />

a2(t) = − a(t)a0(t)<br />

4ā(t)a1(t) .<br />

ia(t)<br />

, β(t) = ± √ ,<br />

6ā(t)b(t)<br />

ia(t)<br />

, β(t) = ∓√ ,<br />

6ā(t)b(t)<br />

ia(t)<br />

, γ(t) = 0, β(t) = ± √ ,<br />

6ā(t)b(t)<br />

ia(t)<br />

, γ(t) = 0, β(t) = ∓√ ,<br />

6ā(t)b(t)<br />

√ 6b(t) , γ(t) = ∓i a1(t)<br />

√ 6b(t) , γ(t) = ∓i a1(t)<br />

−i a(t)a0(t)<br />

4 √ 6ā(t)b(t)a1(t)<br />

√<br />

ā(t)<br />

6b(t)<br />

√<br />

ā(t)<br />

6b(t)<br />

, β(t) = 0,<br />

, β(t) = 0,<br />

√<br />

ā(t)<br />

i[a(t)+2ā(t)a0(t)]<br />

, γ(t) = i a1(t) 6b(t) , β(t) = √ ,<br />

6ā(t)b(t)<br />

In all cases, substituting the respective values <strong>of</strong> α(t), β(t), γ(t) into (23) and using (24) and taking into account that<br />

u(t, x) = v(ξ) with ξ = x − λ t, we obtain traveling wave solutions to (19). By conserving the space, we consider the<br />

solutions for the values given by (1), (2) and (8) :<br />

From (1) we have<br />

i sech<br />

u(t, x) = −<br />

2 ( 1<br />

2ξ √ (<br />

a(t) 2−4ā(t) 2a1(t)a2(t) 1<br />

2i<br />

ā(t) tanh<br />

[ ]<br />

2 2 −a(t) + 4ā(t) + a1(t)a2(t)<br />

ξ = x −<br />

t.<br />

6ā(t)<br />

According to (2) we obtain<br />

Finally, from (8) we have the following solution<br />

√<br />

2ā(t)a1(t)a2(t)<br />

3b(t) − a(t)2<br />

6ā(t)b(t)<br />

2ξ √<br />

2ā(t)a1(t)a2(t)<br />

3b(t) − a(t)2<br />

6ā(t)b(t)<br />

u(t, x) = − [<br />

a(t)<br />

] ,<br />

ξ =<br />

ia(t)ξ<br />

exp √<br />

6ā(t) b(t)<br />

[ ] 2 −a(t)<br />

x − t.<br />

6ā(t)<br />

ā(t) + ā(t)<br />

<strong>IJNS</strong> homepage: http://www.nonlinearscience.org.uk/<br />

) [a(t) 2 − 4ā(t) 2 a1(t)a2(t) ]<br />

)<br />

ā(t) 3/2 ,<br />

+ 2ia(t)ā(t)


98 International Journal <strong>of</strong> Nonlinear Science, Vol.13(2012), No.1, pp. 94-98<br />

u(t, x) =<br />

a(t)a0(t)<br />

2a(t) + 4ā(t)a0(t) + 2 √ [a(t) + ā(t)a0(t)][a(t) + 4ā(t)a0(t)] tanh<br />

a(t) +<br />

−<br />

√ [a(t) + ā(t)a0(t)][a(t) + 4ā(t)a0(t)] tanh<br />

2ā(t)<br />

[ 2 −a(t) + a(t)ā(t)a0(t) + 2ā(t)<br />

ξ = x −<br />

2a0(t) 2 ]<br />

t.<br />

6ā(t)<br />

4 Conclusion<br />

( iξ<br />

2<br />

( √<br />

iξ [a(t)+ā(t)a0(t)][a(t)+4ā(t)a0(t)]<br />

2<br />

6ā(t) b(t)<br />

√ )<br />

[a(t)+ā(t)a0(t)][a(t)+4ā(t)a0(t)]<br />

6ā(t) b(t)<br />

We have used two approaches to determine the exact solutions <strong>of</strong> the equation under study (combined <strong>KdV</strong>-m<strong>KdV</strong><br />

equation). <strong>The</strong> first direct method was based upon the use <strong>of</strong> Lie symmetries. <strong>The</strong> most general traveling wave solution in<br />

implicit form was derived. With the aim to obtain solutions in explicit form we have used the improved tanh-coth method<br />

and derived new traveling wave solutions.It should be noted that other solutions beside the travelling wave solutions can<br />

be constructed using Lie point symmetry X3 and the linear combination <strong>of</strong> X3 with Lie point symmetries X1 and X2.<br />

Acknowledgments<br />

M. Molati acknowledges financial support by the National University <strong>of</strong> Lesotho and IMU.<br />

Références<br />

[1] R. Hirota. Direct methods in Soliton <strong>The</strong>ory, Springer, 1980.<br />

[2] M.J Ablowitz and P.A Clarkson. Soliton, Nonlinear Evolution <strong>Equation</strong>s and Inverse Scattering, Cambridge University<br />

Press, 1991.<br />

[3] G. W. Bluman and S. Kumei. Symmetries and Differential <strong>Equation</strong>s, Springer, 1989.<br />

[4] P. J. Olver. Applications <strong>of</strong> Lie Groups to Differential <strong>Equation</strong>s, Springer, 1986.<br />

[5] L. V. Ovsiannikov. Group Analysis <strong>of</strong> Differential <strong>Equation</strong>s, Academic, 1982.<br />

[6] A. M. Wazwaz. <strong>The</strong> extended tanh method for new solitons solutions for many forms <strong>of</strong> the fifth-order <strong>KdV</strong> equations,<br />

Appl. Math. Comput, 84 (2007) :1002-1014.<br />

[7] M. P. Ramollo. <strong>The</strong> influence <strong>of</strong> obstacles on non-linear waves in stratified fluids, Ph.D thesis : University <strong>of</strong> Newcastle<br />

Upon Tyne, 1994.<br />

[8] E. Yasar. On the conservation laws and invariant solutions <strong>of</strong> the m<strong>KdV</strong> equation, J. Math. Anal. Appl. 363<br />

(2010) :174-181.<br />

[9] A. M. Wazwaz. Solitons and singular solitons for the Gardner–KP equation, Appl. Math. Comput. 204 (2008) :162-<br />

169.<br />

[10] C.A. Gómez and A. Salas. <strong>The</strong> Cole Hopf transformation and improved tanh-coth method applied to new integrable<br />

system (<strong>KdV</strong>6), Appl.Math and Comp. 204 (2008) :957-962.<br />

[11] C.A.Gómez and A.H.Salas. Special symmetries to Riccati equation and applications, Appl. Math.and Comp. 57<br />

(2010) :692-700.<br />

<strong>IJNS</strong> email for contribution: editor@nonlinearscience.org.uk<br />

,<br />

)

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