Painlevé Analysis, Lie Symmetries and Exact Solutions for Variable Coefficients Benjamin Bona Mahony Burger (BBMB) Equation

Transcription

1 Commun. Theor. Phys. 60 (2013) Vol. 60, No. 2, August 15, 2013 Painlevé Analysis, Lie Symmetries and Exact Solutions for Variable Coefficients Benjamin Bona Mahony Burger (BBMB) Equation Vikas Kumar, R.K. Gupta, and Ram Jiwari School of Mathematics and Computer Applications, Thapar University, Patiala147004, Punjab, India (Received December 21, 2012; revised manuscript received June 19, 2013) Abstract In this paper, a variable-coefficient Benjamin Bona Mahony Burger (BBMB) equation arising as a mathematical model of propagation of small-amplitude long waves in nonlinear dispersive media is investigated. The integrability of such an equation is studied with Painlevé analysis. The Lie symmetry method is performed for the BBMB equation and then similarity reductions and exact solutions are obtained based on the optimal system method. Furthermore different types of solitary, periodic and kink waves can be seen with the change of variable coefficients. PACS numbers: Jb, Sv, Jr, Fg Key words: BBMB equation, Painlevé analysis, Lie symmetric analysis, exact solutions 1 Introduction In recent years, nonlinear partial differential equations (NLPDEs) have been used to model many physical phenomena in various fields such as fluid mechanics, solid state physics, plasma physics, chemical physics, optical fiber, and geochemistry. Thus, it is important to investigate the exact explicit solutions of NLPDEs. To find the symmetry and exact solutions of a nonlinear partial differential equation (NLPDEs), some effective methods have been introduced, such as the classical and non-classical Lie Group method, the CK direct method. [1 3] One such NLPDE is the celebrated Benjamin Bona Mahony Burger (BBMB) equation (4) Y t Y xxt αy xx + Y Y x + Y x = 0, (1) Y (x, t) represents the fluid velocity in the horizontal direction x, α is a positive constant. It has been proposed in [4], as a model to study the unidirectional long waves of small amplitudes in water, which is an alternative to the Korteweg-de Vries equation. For more details on both the mathematical theory and the physical significance of Eq. (1), we refer the reader to Refs. [4 5]. We call Eq. (1) the BBMB Equations, but it is proposed neither by Benjamin, Bona, and Mahony nor by Burgers. Many authors studied Eq. (1) using so many aspects. [6 9] The physical phenomena in which many nonlinear integrable equations with constant coefficient arise tend to highly idealized due the presence of constant coefficients therefore, equations with variable coefficient may provide various models for real physical phenomena, such as, in the propagation of small-amplitude surface waves, which runs on straits or large channels of slowly varying depth variable coefficient of nonlinear integrable evolution equations. [10 12] As there are choices for parameters, the variable-coefficient nonlinear equations can be considered as generalization of the constant coefficients equations. In this paper, we study variable coefficient version of Benjamin Bona Mahony Burger equation Y t e(t)y xxt f(t)y xx + g(t)y Y x + h(t)y x = 0, (2) e(t), f(t), g(t), and h(t) are arbitrary timedependent coefficients. When f(t) = 0, h(t) = 1 and e(t) = 1, Eq. (2) is the alternative regularized long wave equation proposed by Peregrine [5] and Benjamin. [4] In the physical case, the dispersive effect of Eq. (2) is the same as the variable coefficient Benjamin Bona Mahony (BBM) equation Y t e(t)y xxt + g(t)y Y x + h(t)y x = 0, (3) while the dissipative effect is the same as the variable coefficient Burgers equation Y t f(t)y xx + g(t)y Y x + h(t)y x = 0. (4) The study of nonlinear models for integrable properties is one of the important works in nonlinear science. Therefore, in this paper, we check the Painlevé property of the variable coefficient Benjamin Bona Mahony Burger (BBMB) equation (2) firstly, and then the symmetries of Eq. (2) are considered. The exact solutions generated from the symmetries are presented. The outline of this paper is as follows. In Sec. 2, we study the Painlevé property of BBMB equation. In Sec. 3, the classical Lie method is utilized to obtain the optimal system of sub-algebras for Eq. (2), and group invariant solutions of the nonlinear equation (2) are sought. This analysis generates a rich class of solutions. A number of cases arise depending on the nature of the Lie symmetry vikasmath81@gmail.com Corresponding author, rajeshgupta@thapar.edu c 2013 Chinese Physical Society and IOP Publishing Ltd

2 176 Communications in Theoretical Physics Vol. 60 generator. This analysis demonstrates the value of Lie analysis of differential equations for this application. In Sec. 4, we derive some explicit exact solutions of the nonlinear equation (2). In the last section, some conclusions are drawn. 2 Painlevé Analysis Ablowitz, Ramani and Segur [13] stated that when all the ordinary differential equations (ODE) obtained by exact similarity transforms from a given partial differential equation (PDE) have the Painlevé property, then the PDE is integrable. The definition of Painlevé property of the ODE (WTC) was extended to the case of PDE by Weiss, Tabor and Carnevale. [14] Briefly, a partial differential equation has the Painlevé property [15 16] when the solutions of the PDE are single-valued about the movable, and the singularity manifold is non-characteristic. Before finding some exact solutions of Eq. (2), we briefly discuss its Painlevé property by means of the standard WTC approach. The Laurent expansion of the function Y = Y (x, t) about a singular manifold u(x, t) = 0 is Y (x, t) = (u(x, t)) α Y k (x, t)(u(x, t)) k, (5) k=0 by one makes the ansätz Y (x, t) = Y 0 (x, t)[u(x, t)] α. (6) Substitution of Eq. (6) into Eq. (2) and balancing the highest derivative with the nonlinear term gives α = 2, Y 0 (x, t) = 12 e(t) g(t) u xu y. (7) The resonances are determined by setting Y (x, t) = Y 0 (x, t)u 2 (x, t) + Y r (x, t)u r 2 (x, t), (8) and balancing the most singular term in Eq. (2), giving recursion relation Y r [2g(t)Y 0 (x, t)u x (x, t) g(x)y 0 (x, t)(r 2)u x (x, t) + e(t)(r 2)(r 3)(r 3)(r 4)u t (x, t)u 2 x(x, t)] = H(Y 0, Y 1,..., Y r 1, u t, u x,...). (9) Using Eq. (7) into Eq. (9), we find that the resonances for this branch are r = 1, 4, and 6. The resonance at J = 1 corresponds to the arbitrary function defining the singularity manifold for the (BBMB) equation. After detailed calculation, we find that compatibility conditions at J = 4 and 6 are satisfied identically. According to WTC approach, Eq. (2) possesses Painlevé property. 3 Lie Symmetric Analysis for BBMB Equation Let us consider a one-parameter Lie group of infinitesimal transformation, x = x + εη(x, t, Y ) + O(ε 2 ), t = t + εψ(x, t, Y ) + O(ε 2 ), Y = Y + εφ(x, t, Y ) + O(ε 2 ), (10) with a small parameter ε. The partial differential Eq. (2) is said to admit a symmetry generated by the vector field V = η(x, t, Y ) x + ψ(x, t, Y ) + φ(x, t, Y ) t Y, (11) if it is left invariant by the transformation (x, t, Y ) (x, t, Y ). Applying the third prolongation pr (3) V to Eq. (2) results in an overdetermined system of linear partial differential equations. [17 21] After some straightforward and lengthy calculations, the resultant overdetermined system gives the following infinitesimal elements, η = ax + b, φ = cy + d, ψ = 2a e(t) e (t), (12) a, b, c, and d are arbitrary constants and the function e(t), f(t), g(t) and h(t) are governed by the following condition d [h(t)ψ] ah(t) + g(t)d = 0, dt d [g(t)ψ] + (c a)g(t) = 0, dt d [f(t)ψ] 2af(t) = 0. (13) dt The Lie algebra associated with Eq. (2) consists of following four vector fields V 1 = x x + 2 e(t) e (t) t, V 2 = x, V 3 = Y Y, V 4 = Y. (14) The commutator relations between these vector fields is given by [V i, V j ] = V i V j V j V i, i, j = 1,...,4. It immediately follows that [V 1, V 1 ] = [V 2, V 2 ] = [V 3, V 3 ] = [V 4, V 4 ] = 0, [V 1, V 3 ] = [V 1, V 4 ] = [V 2, V 3 ] = [V 2, V 4 ] = [V 3, V 1 ] = [V 3, V 2 ] = [V 4, V 1 ] = [V 4, V 2 ] = 0, [V 1, V 2 ] = [V 2, V 1 ] = V 2, [V 3, V 4 ] = [V 4, V 3 ] = V 4. The adjoint representation of a Lie group on the Lie algebra is given as the Lie series Ad (exp(εv i ))V j = V j ε[v i, V j ] + ε2 2 [V i, [V i, V j ]], ε is a parameter and [V i, V j ] is the commutator of the Lie algebra.

3 No. 2 Communications in Theoretical Physics 177 It follows that the adjoint representations of the vector fields for the BBMB equation is i = 1,...,4 and Ad (exp(εv i ))V i = V i, Ad (exp(εv 1 ))V 2 = e ε V 2, Ad (exp(εv 1 ))V 3 = V 3, Ad (exp(εv 1 ))V 4 = V 4, Ad (exp(εv 2 ))V 1 = V 1 εv 2, Ad (exp(εv 2 ))V 3 = V 3, Ad (exp(εv 2 ))V 4 = V 4, Ad (exp(εv 3 ))V 1 = V 1, Ad (exp(εv 3 ))V 2 = V 2, Ad (exp(εv 3 ))V 4 = e ε V 4, Ad (exp(εv 4 ))V 1 = V 1, Ad (exp(εv 4 ))V 2 = V 2, Ad (exp(εv 4 ))V 3 = V 3 εv 4. The classification of one-dimensional subalgebras of the whole symmetry algebra (14) is done by an inductive approach. [19] If V = 4 i=1 a iv i is none zero vector field, we will simplify as many of the coefficients a i as possible by acting it to various adjoint transformations to obtain the subalgebra. The types of one-dimensional subalgebras of Eq. (2) is as follows, (i) V 1 + αv 3, (ii) V 2 + βv 3, (iii) V 3, (iv) V 4, (15) α and β are arbitrary constants. 4 Reduced Systems and Exact Solutions In the following, some exact solutions for the reduced systems are studied for each type of subalgebra. These solutions are obtained by solving the associated Lagrange equations, [22 28] 4.1 Subalgebra V 1 + αv 3 dx η = dt ψ = dy φ. (16) The invariants associated with the symmetry operators, can be obtained by integrating the characteristic equation (16). In this type, the form of invariant solution and coefficient functions are as follows Y (x, t) = x α F(ξ), h(t) = [e(t)] ( 1/2 )e (t) 2 g(t) = k 2 [e(t)] ( 1 α/2 )e (t) 2, f(t) = k 3 e (t) 2, (17) ξ = x 2 /e(t) is an invariant of the symmetry. Substitution of Eq. (17) into Eq. (2) gives the following ODE, [( α 2 + α)k 3 + αξ (1/2 ) ]F + [ 2ξ 2 + (2α 2 + 6α 4αk 3 2k 3 + 4)ξ + 2 ξ 3/2 ]F + [(4k 3 + 8α + 20)]ξ 2 F + αk 3 ξ (α+1/2 )F 2 + 2k 2 ξ (α+3/2) FF + 8ξ 3 F = 0. (18) To solve Eq. (18), we consider two cases Case (i) α = 0, the reduced ODE (18) is as follows, [ 2ξ 2 + 4ξ 2k 3 ξ + 2 ξ 3/2 ]F +[20 + 4k 3 ]ξ 2 F + 2k 2 ξ 3/2 FF + 8ξ 3 F = 0. (19) Thus, by solving Eq. (19) and reverting back to the original variables, we obtain the following group-invariant solutions of the BBMB equation (2) as follows. Solution 1, Y (x, t) = k 3 + (x 2 /e(t)) + 3k3 2 (x 2 /e(t)) 1/2 k 2 (x 2 /e(t)) 1/2 + 6 ( + 10k 3 + k3 2)tanh( C 1 + 1/4 + (1/20)k 3 ) 3 ( + 10k 3 + k3 2)tanh( C 1 + 1/4 + (1/20)k 3 ) 2, (20) k 2 (x 2 /e(t)) 1/2 k 2 (x 2 /e(t)) 1/2, k 2, k 3, and C 1 are arbitrary constant and e(t) is an arbitrary function of t, see Figs Fig. 1 The inverted two solitary solution (20) in form of 3D plot at different times t = 0, 1, 2, 3, 4, 5, 6 with = k 2 = k 3 = C 1 and e(t) = sec 2 (t). Fig. 2 The contour plot of solution (20) (projection of some of its level curves) at different times t = 0,1, 2,3, 4,5, 6 with = k 2 = k 3 = C 1 = 1, and e(t) = sec 2 (t).

4 178 Communications in Theoretical Physics Vol. 60 Fig. 3 The inverted three solitary solution (20) in form of 3D plot at different times t = 0, 1, 2, 3, 4, 5, 6 with = k 2 = k 3 = C 1 = 1 and e(t) = cos 2 (t). Fig. 4 The contour plot of solution (20) (projection of some of its level curves) at different times t = 0, 1, 2, 3, 4, 5, 6 with = k 2 = k 3 = C 1 = 1 and e(t) = cos 2 (t). Solution 2 Y (x, t) = C 3 96(x2 /e(t)) 3/2 C2 2 tanh(c 1 + C 2 (x 2 /e(t))) 48(x2 /e(t)) 3/2 C2 2 tanh(c 1 + C 2 (x 2 /e(t))) 2, (21) k 2 k 2 = C 3k2 2(x2 /e(t)) 1/ C 2 x 2 /e(t) x 2 /e(t) 48(x 2 /e(t)) 2 C2 2 x 2, k (x 2 /e(t)) 1/2 3 = 20C 2 e(t) 5 and k 2, C 1, C 2, C 3 are arbitrary constants and e(t) be an arbitrary function of t Solution 3 Fig. 5 The distribution of some singularities of the solution (22) in form of 3D plot at t = 0, 1, 2, 3, 4, 5, 6 with k 2 = 0.5, C 1 = C 2 = C 3 = 1 and e(t) = sec 2 (t). Fig. 7 The damped oscillatory kink wave solution (22) in form of 3D plot at t = 0, 1, 2, 3, 4, 5, 6 with k 2 = 0.5, C 1 = C 2 = C 3 = 1 and e(t) = cos 2 (t). Fig. 8 The contour plot of solution (22) (projection of some of its level curves) at t = 0, 1, 2, 3, 4, 5, 6 with k 2 = 0.5, C 1 = C 2 = C 3 = 1 and e(t) = cos 2 (t). Fig. 6 The contour plot of solution (22) (projection of some of its level curves) at t = 0, 1, 2, 3, 4, 5, 6 with k 2 = 0.5, C 1 = C 2 = C 3 = 1 and e(t) = sec 2 (t). Y (x, t)=c 3 48(x2 /e(t)) 3/2 C 2 2 tanh(c 1 +C 2 x 2 /e(t)) 2 k 2, (22)

5 No. 2 Communications in Theoretical Physics 179 = C 3k 2 (x 2 /e(t)) 1/2 32(x 2 /e(t)) 2 C 2 2 x2 /e(t) + 7 (x 2 /e(t)) 1/2, k 3 = 5 and k 2, C 1, C 2 are arbitrary constants also e(t) be an arbitrary function of t, see Figs Case (ii) α = 1 The reduced ODE is as follows, ξ 1/2 F +[(12 6k 3 )ξ 2ξ ξ 3/2 ]F +[28 +4k 3 ]ξ 2 F +k 3 ξf 2 + 2k 2 ξ 2 FF + 8ξ 3 F = 0. (23) Solving Eq. (23) and reverting back to the original variables, we obtain the following group-invariant solutions of the BBMB equation (2) as follows Solution 1 Y (x, t) = x 2 /e(t) 294 k 2 tanh(c 1 7/20) k 2 x 2 /e(t) 147 tanh(c 1 7/20) 2 k 2 x 2, (24) /e(t) = 0, k 3 = 0 and k 2, C 1 are arbitrary constants and e(t) is arbitrary function of t, see Figs Fig. 11 The three solitary solution (24) in form of 3D plot at different times t = 0, 1, 2, 3, 4, 5, 6 with k 2 = 0.5, C 1 = 1 and e(t) = cos 2 (t). Fig. 12 The contour plot of solution (24) (projection of some of its level curves) at t = 0, 1, 2, 3, 4, 5, 6 with k 2 = 0.5, C 1 = 1 and e(t) = cos 2 (t). Fig. 9 The two solitary solution (24) in form of 3D plot at different times t = 0, 1, 2, 3, 4, 5, 6 with k 2 = 0.5, C 1 = 1 and e(t) = sec 2 (t). Fig. 10 The contour plot of solution (24) (projection of some of its level curves) at t = 0, 1, 2, 3, 4, 5, 6 with k 2 = 0.5, C 1 = 1 and e(t) = sec 2 (t). Solution Y (x, t) = 50x 2 /e(t) 147 = tanh(c 1 7/20) x 2 /e(t) 147 tanh(c 1 7/20) 2 50 x 2, () /e(t) x 2 /e(t) + 144, k (x 2 /e(t)) 1/2 3 = 44 e(t) , k 2 = 2 and C 1 is an arbitrary constant and e(t) is an arbitrary function of t. Solution Y (x, t) = 50x 2 /e(t) 147 tanh(c 1 7/20) x 2 /e(t) = x tanh(c 1 7/20) 2 50 x 2, (26) /e(t) x 2 /e(t) + 144, k (x 2 /e(t)) 1/2 3 = 544 x 2 e(t) , k 2 = 2 and C 1 is an arbitrary constant and e(t) is an arbitrary function of t. 4.2 Subalgebra V 2 + βv 3 To solve Eq. (2) we consider the two cases under this type of subalgebra.

6 180 Communications in Theoretical Physics Vol. 60 Case (i) e(t) = constant and ψ =arbitrary The symmetry in this case gives the group invariant solution and form of the coefficient functions as follows ( β ) Y (x, t) = exp h(t)dt F(ξ), g(t) = k 2 h(t)exp ( β ) h(t)dt, h(t) = arbitrary and f(t) = k 3 h(t), (27) ξ = x (1/ ) h(t)dt is an invariant of the symmetry. Substitution of Eq. (27) into Eq. (2) gives the ODE F satisfies C 1 F + C 2 F C 3 F + C 4 FF + C 5 F = 0, (28) C 1 = β, C 2 =1 1, C 3 = eβ + k 3, C 4 =k 2, C 5 = e, 0 and k 2, k 3, β, e are arbitrary constants. The solutions of Eq. (2) corresponding to the solutions of Eq. (28) are the following. Solution 1 Y (x, t) = C 9 tanh (C 6 + C 7 (x 1 )) h(t)dt, (29) C 1 = 0, C 2 = 0, C 3 = (1/2)C 9 C 4 /C 7, C 5 = 0 and C 4, C 6, C 7, C 9 are arbitrary constants and h(t) is an arbitrary function of t, see Figs Fig. 15 The periodic solution (29) in form of 3D plot at different times t = 0, 1, 2, 3, 4, 5, 6 with = 0.5, C 6 = C 7 = C 9 = 1 and h(t) = sec 2 (t). Fig. 16 The contour plot of solution (29) (projection of some of its level curves) at t = 0, 1, 2, 3, 4, 5, 6 with = 0.5, C 6 = C 7 = C 9 = 1 and h(t) = sec 2 (t). Solution 2 Y (x, t) = C 8 +C 9 tanh (C 6 + C 7 (x 1 )) h(t)dt, (30) Fig. 13 The Singularity at t = 0 of solution (29) in form of 3D plot at t = 0, 1, 2, 3, 4, 5, 6 with = 0.5, C 6 = C 7 = C 9 = 1 and h(t) = cosh(t). C 1 = 0, C 2 = C 4 C 8, C 3 = (1/2)C 9 C 4 /C 7, C 5 = 0 and C 4, C 6, C 7, C 8, C 9 are arbitrary constants and h(t) is an arbitrary function of t. Solution 3 Y (x, t) = C 8 + C 9 tanh (C 6 + C 7 (x 1 )) h(t)dt + C 10 tanh (C 6 + C 7 (x 1 )) h(t)dt, (31) C 1 = 0, C 2 = 0, C 3 = 0, C 4 = 0, C 5 = 0 and C 6, C 7, C 8, C 9, C 10 are arbitrary constants and h(t) is an arbitrary function of t. Solution 4 Y (x, t)=c 8 +C 10 tanh (C 6 +C 7 (x 1 2, h(t)dt)) (32) C 1 = 0, C 2 = C 4 C C 4C 10, Fig. 14 The contour plot of solution (29) (projection of some of its level curves) at t = 0, 1, 2, 3, 4, 5, 6 with = 0.5, C 6 = C 7 = C 9 = 1 and h(t) = cosh(t). C 3 = 0, C 5 = 1 12 C 10 C 4 C 2 7 and C 4, C 6, C 7, C 8, C 10 are arbitrary constants and h(t) is an arbitrary function of t.

7 No. 2 Communications in Theoretical Physics 181 Solution 5 Y (x, t) = C 8 6 C3 2 tanh( C 6 + (1/10)[C 3 (x (1/k1) h(t)dt)/c 5 ) C 5 C 4 3 C3 2 tanh( C 6 + (1/10)C 3 (x (1/ ) h(t)dt)c 5 ) 2, (33) C 5 C 4 C 1 = 0, C 2 = 1 C 8 C 5 C 4 + 3C3 2 C 5 and C 3, C 4, C 5, C 6, and C 8 are arbitrary constants and h(t) is an arbitrary function of t. Case (ii) e(t) = arbitrary and ψ = 0 Suitable invariants and form of the coefficient functions of this case for reduction are as follows, Y (x, t) = F(ξ), g(t) = arbitrary, h(t) = arbitrary and g(t) = arbitrary, (34) ξ = t represents the invariant of the symmetry, which reduce Eq. (2) to F = 0. (35) The obvious solution of Eq. (2), in this case, is Y = constant. 4.3 Subalgebra V 3 Under this type to find the solutions of equation we consider the following cases. Case (i) e(t) = constant and ψ = arbitrary In this case we derive the following expression of Y by solving the corresponding characteristic equation and form of coefficient functions from Eq. (13), ( 1 ) Y (x, t) = exp h(t)dt F(ξ), g(t) = k 2 h(t)exp ( 1 ) h(t)dt, h(t) = arbitrary and f(t) = k 3 h(t), (36) ξ = x and F must satisfy C 1 F + F + C 2 FF C 3 F = 0, (37) C 1 = 1/, C 2 = k 2, C 3 = e/ + k 3, 0, and k 2, k 3, e are arbitrary constants. The solution of the main Eq. (2) is given by Y (x, t) = 1 C 2 + C 7 tanh(c 4 + C 5 x), (38) C 1 = 0, C 3 = (1/2)C 7 /C 5, C 5 0, and C 2, C 4 are arbitrary constants, see Figs Fig. 17 The kink wave solution (38) in form of 3D plot at different times t = 0, 1, 2, 3, 4, 5, 6 with C 2 = 1, C 4 = C 5 = C 7 = 1. Case (ii) e(t) = arbitrary and ψ = 0 This gives a constant solution. 4.4 Subalgebra V 4 It leads to a constant solution. 5 Discussion and Concluding Remarks We have performed Painlevé analysis to cheek the integrability of the nonlinear variable coefficient BBMB equation. The Lie symmetry method is utilized to investigate the symmetries and invariant solutions of the BBMB equations. The vector fields of the optimal system lead to reduction of the nonlinear PDE to system of ODEs. The infinitesimal generators in the optimal system are used Fig. 18 The contour plot of solution (38) (projection of some of its level curves) at t = 0, 1, 2, 3, 4, 5, 6 with C 2 = 1, C 4 = C 5 = C 7 = 1. for reductions and exact solutions. By using these vector fields, we have found group-invariant solutions. In almost all the cases, one can choose the arbitrary function e(t) or h(t) along with various other arbitrary parameters, in a suitable manner, to simulate physical situations with the help of Figures 1 18 governed by Eq. (2). Acknowledgments Vikas Kumar would like to express his thanks to Prof. Dinesh Kumar (Director U.I.E.T. Kurukhetra University Kurukshetra) and Prof. Pawan Diwan (Head Dept of Applied Science U.I.E.T. Kurukhetra University Kurukshetra) for providing help and necessary facilities and direct or indirect encouragement during this work.

8 182 Communications in Theoretical Physics Vol. 60 References [1] X. Hu, S. Lou, and Y. Chen, Phys. Rev. E 85 (2012) [2] Z.Z. dong, Y. Chen, and L. Wang, Commun. Theor. Phys. 50 (2008) 803. [3] C.C. Cheng and Y. Chen, Commun. Theor. Phys. 51 (2009) 973. [4] T.B. Benjamin, J. Bona, and J. Mahony, Philos. Trans. R. Soc. Lond. A 272 (1972) 47. [5] D.H. Peregrine, J. Fluid Mech. (1996) 321. [6] K. Omrani and M. Ayadi, Numer. Methods Partial Differ. Equat. 24 (2008) 239. [7] K.Al. Khaled, S. Momani, and A. Alawneh, Appl. Math. Comput. 171 (2005) 281. [8] G. Fakhari, Domairry, and Ebrahimpour, Phys. Lett. A 368 (2007) 64. [9] H. Tari and D.D. Ganji, Phys. Lett. A 367 (2007) 95. [10] K. Singh and R.K. Gupta, Int. J. Eng. Sci. 44 (2006) 241. [11] K. Singh and R.K. Gupta, Int. J. Eng. Sci. 44 (2006) 16. [12] S. Kumar, K. Singh, and R.K. Gupta, Commun. Nonl. Sci. Numer. Simul. 17 (2012) [13] M.J. Ablowitz, A. Ramani, and H. Segur, J. Math. Phys. 21 (1980) 715. [14] J. Weiss, M. Tabor, and G. Carnevale, J. Math. Phys. 24 (1983) 522. [15] A. Pickering, J. Phys. A: Math. Theor. 26 (1993) [16] D. Rollin, J. Math. Phys. 32 (1991) [17] Y.Q. Yang and Y. Chen, Commun. Theor. Phys. 56 (2011) 463. [18] T. Chou, Lie Group and Its Applications in Differential Equations, Science Press, Beijing (2001). [19] P.J. Olver, Applications of Lie Groups to Differential Equations, Springer, New York (1993). [20] G. Bluman and S. Anco, Symmetry and Integration Methods for Differential Equations, Springer-Verlag, New York (2002). [21] L.V. Ovsyannikov, Group Analysis of Differential Equations, Academic, New York (1982) [22] L. Kaur and R.K. Gupta, Phys. Scr. 87 (2013) [23] H.Z. Liu, J.B. Li, and L. Liu, Commun. Theor. Phys. 57 (2012) 737. [24] A.G. Johnpillai, A.H. Kara, and A. Biswas, Appl. Math. Lett. 26 (2013) 376. [] N. Goyal and R.K. Gupta, Phys. Scr. 85 (2012) [26] V. Kumar, R.K. Gupta, and RamJiwari, Chin. Phys. B 22 (2013) [27] A. Bansal and R.K. Gupta, Phys. Scr. 86 (2012) [28] C.M. Khalique and K.R. Adem, Math. Comput. Model 54 (2011) 184.