Symmetry Analysis and Conservation Laws of a Generalized Two-Dimensional Nonlinear KP-MEW Equation

Lie symmetry analysis is performed on a generalized two-dimensional nonlinear Kadomtsev-Petviashvili-modified equal width equation. The symmetries and adjoint representations for this equation are given and an optimal system of one-dimensional subalgebras is derived. The similarity reductions and exact solutions with the aid of (G/G)-expansion method are obtained based on the optimal systems of one-dimensional subalgebras. Finally conservation laws are constructed by using the multiplier method.

The purpose of this paper is to study one such NLEE, namely, the generalized two-dimensional nonlinear Kadomtsev-Petviashvili-modified equal width (KP-MEW) equation [26] that is given by Here, in (1) , ,  and  > 1 are real valued constants.The solutions of (1) have been studied in various aspects.See, for example, the recent papers [26][27][28].Wazwaz [26] used the tanh method and the sine-cosine method, for finding solitary waves and periodic solutions.Saha [27] used the theory of bifurcations of planar dynamical systems to prove the existence of smooth and nonsmooth travelling wave solutions.Wei et al. [28] used the qualitative theory of differential equations and obtained peakon, compacton, cuspons, loop soliton solutions, and smooth soliton solutions.
In this paper we obtain symmetry reductions of (1) using Lie group analysis [19][20][21][22][23][24] and based on the optimal systems of one-dimensional subalgebras.Furthermore, the (  /)expansion method is employed to obtain some exact solutions of (1).In addition to this conservation laws will be derived for (1) using the multiplier method [29].

Symmetry Reductions and Exact
Solutions of (1) The vector field of the form where   ,  = 1, 2, 3, and  depend on , , , and , is a Lie point symmetry of (1) if pr (4)  [(  +  (  )  +   )  +   ] = 0 whenever (  + (  )  +   )  +   = 0.Here pr (4)  [20] denotes the fourth prolongation of .Expanding (3) and splitting on the derivatives of , we obtain an overdetermined system of linear partial differential equations.Solving this system one obtains the following four Lie point symmetries: 2.1.One-Dimensional Optimal System of Subalgebras.We now calculate the optimal system of one-dimensional subalgebras for (1) and use it to find the optimal system of groupinvariant solutions for (1).We follow the method given in [20].Recall that the adjoint transformations are given by where [  ,   ] is the commutator defined by We present the commutator table of the Lie symmetries and the adjoint representations of the symmetry group of (1) on its Lie algebra in Tables 1 and 2, respectively.These two tables are then used to construct the optimal system of one-dimensional subalgebras for (1).As a result, after some calculations, one can obtain an optimal system of onedimensional subalgebras given by { 1 + 2 + 3 ,  1 + 4 }, where ,  ∈ R, ,  = 0, ±1.

Symmetry Reductions and Exact Solutions of (1).
In this subsection we use the optimal system of one-dimensional subalgebras calculated above to obtain symmetry reductions and exact solutions of the KP-MEW equation.
Case 1.Consider the following: The symmetry  1 +  2 +  3 gives rise to the following three invariants: Table 1: Commutator table of the Lie algebra of equation (1).
Table 2: Adjoint table of the Lie algebra of equation (1).
Now treating  as the new dependent variable and  and  as new independent variables, the KP-MEW equation (1) transforms to which is a nonlinear PDE in two independent variables.We now use the Lie point symmetries of (8) and transform it to an ordinary differential equation (ODE).Equation ( 8) has the two translational symmetries; namely, The combination Γ 1 +Γ 2 of the two symmetries Γ 1 and Γ 2 yields the two invariants which gives rise to a group-invariant solution  = ().Consequently using these invariants, ( 8) is transformed into the fourth-order nonlinear ODE: Integrating the above equation twice and taking the constants of integration to be zero we obtain a second-order ODE: Multiplying ( 12) by   , integrating once and taking the constant of integration to be zero, we obtain the first-order ODE: One can integrate the above equation by separating the variables.After integrating and reverting back to the original variables, we obtain the following group-invariant solutions of the KP-MEW equation ( 1) for arbitrary values of  in the following form: where and  is a constant of integration.By taking  = 2,  = 1/2,  = 1,  = 1,  = 1,  = 1,  = 1,  = 0, and  = 1 in ( 14), the profile of the solution is given in Figure 1.
The symmetry  1 +  4 gives rise to the three invariants: By treating  as the new dependent variable and  and  as new independent variables, the KP-MEW equation ( 1) transforms to Equation ( 17) has a single Lie point symmetry; namely, and this symmetry yields the two invariants which gives rise to a group-invariant solution  = () and consequently, using these invariants, ( 17) is then transformed to a second-order Cauchy-Euler ODE: where  = / 2 and  1 and  2 are constants of integration.
Let us consider the solutions of (11) in the form where () satisfies and  and  are constants.The homogeneous balance method between the highest order derivative and highest order nonlinear term appearing in (11) determines the value of  and A 0 , . . ., A  are constants to be determined.
Consider  = 2. Application of the balancing procedure to fourth-order ODE (11) yields  = 2, so the solution of ( 11) is of the form Substituting ( 23) and ( 24) into (11) leads to an overdetermined system of algebraic equations.Solving this system of algebraic equations with the aid of Maple, we obtain Now using the general solution of ( 23) in ( 24), we have the following three types of travelling wave solutions of the KP-MEW equation ( 1).When  2 − 4 > 0, we obtain the hyperbolic function solution: where  =  −  − (( + )/),  1 = (1/2)√ 2 − 4, and  1 and  2 are arbitrary constants.The profile of the solution ( 26) is given in Figure 2. When  2 − 4 < 0, we obtain the trigonometric function solution: (, , )    Consider  = 3. Again the application of the balancing procedure to fourth-order ODE yields  = 1, so the solution of ( 11) is of the form  () = A 0 + A 1 (   ()  () ) . (29) Solving this system of algebraic equations with the aid of Maple, we obtain