1 Introduction
In this paper we study the third-order modified equal-width (MEW) equation
(1) u t + 3 α u 2 u x − β u t x x = 0 , α ≠ 0 , β ≠ 0 , $$\begin{equation}\label{mew} u_{t}+3\alpha u^{2}u_{x}-\beta u_{txx}=0,\,\,\, \alpha \neq 0, \, \, \beta \neq 0, \end{equation} $$ where α and β are non-zero real parameters. Equation (1) is used in handling the simulation of a single dimensional wave propagation in nonlinear media with dispersion processes [1 ]. Some researchers have used different techniques and methods to construct travelling wave solutions of (1). Recently MEW Equation (1) was investigated in [1 ], where the researchers employed extended simple equation method and also the exp(-φ (ξ )) expansion method to generate travelling wave solutions of the equation. In [2 ], dynamical system technique for integer order was used and travelling wave solutions of the MEW equation were found, which comprised of solitary, periodic waves and also kink and anti-kink wave solutions. Homotopy perturbation method was applied to (1) and numerical solution of the MEW equation was obtained in [3 ].
In our study we use an entirely different approach to obtain new exact travelling wave solutions, namely cnoidal and snoidal wave solutions of MEW Equation (1) . Moreover, for the first time we derive conservation laws of the MEW equation by employing both the Noether approach as well as the multiplier approach.
2 Exact solutions of (1) constructed on optimal system
In this section, we first compute Lie point symmetries of (1) and then use them to construct an optimal system of one-dimensional subalgebras. Subsequently, we utilise this optimal system of one-dimensional subalgebras to obtain symmetry reductions and group-invariant solutions of (1) [4 , 5 , 6 , 7 , 8 ].
2.1 Lie point symmetries of (1)
The vector field
(2) X = τ ( t , x , u ) ∂ ∂ t + ξ ( t , x , u ) ∂ ∂ x + η ( t , x , u ) ∂ ∂ u , $$\begin{eqnarray} X =\tau (t, x, u)\frac{\partial}{\partial t}+ \xi (t, x, u) \frac{\partial}{\partial x} +\eta (t, x, u) \frac{\partial}{\partial u}, \end{eqnarray}$$ where τ , ξ and η depend on t , x and u is a Lie point symmetry of Equation (1) if
(3) pr ( 3 ) X Δ | Δ = 0 = 0 , $$\begin{eqnarray} \mbox{pr}^{(3)}X\Delta |_{\Delta=0} =0 \label{eq2}, \end{eqnarray}$$ where
Δ ≡ u t + 3 α u 2 u x − β u t x x $$\begin{equation}\nonumber \Delta\equiv u_{t}+3\alpha u^{2}u_{x}-\beta u_{txx} \end{equation}$$ and pr(3) X is the third prolongation [6] of (2) defined as
(4) pr ( 3 ) X = X + ζ t ∂ ∂ u t + ζ x ∂ ∂ u x + ζ t x ∂ ∂ u t x + ζ t x x ∂ ∂ u t x x . $$\begin{eqnarray} \mbox{pr}^{(3)}X =X +\zeta_{t}\frac{\partial }{\partial u_{t}}+\zeta_{x}\frac{\partial }{\partial u_{x}}+\zeta_{tx}\frac{\partial }{\partial u_{tx}}+\zeta_{txx}\frac{\partial }{\partial u_{txx}}. \end{eqnarray}$$ Here ζt , ζx , ζtx and ζtxx are determined by
(5) ζ t = D t ( η ) − u t D t ( τ ) − u x D t ( ξ ) , ζ x = D x ( η ) − u t D x ( τ ) − u x D x ( ξ ) , ζ t x = D x ( ζ t ) − u t t D x ( τ ) − u t x D x ( ξ ) , ζ t x x = D x ( ζ t x ) − u t t x D x ( τ ) − u t x x D x ( ξ ) , $$\begin{equation}\label{pro1} \begin{aligned} &\zeta_{t}= D_{t}(\eta )-u_{t}D_{t}(\tau)-u_{x}D_{t}(\xi), \\ &\zeta_{x}= D_{x}(\eta )-u_{t}D_{x}(\tau)-u_{x}D_{x}(\xi), \\ &\zeta_{tx}=D_{x}(\zeta_t )-u_{tt}D_{x}(\tau)-u_{tx}D_{x}(\xi), \\ &\zeta_{txx}=D_{x}(\zeta_{tx} )-u_{ttx}D_{x}(\tau)-u_{txx}D_{x}(\xi), \end{aligned} \end{equation}$$ where the total derivatives Dt and Dx are defined as
(6) D t = ∂ ∂ t + u t ∂ ∂ u + u t t ∂ ∂ u t + u t x ∂ ∂ u x + ⋯ , D x = ∂ ∂ x + u x ∂ ∂ u + u x x ∂ ∂ u x + u x t ∂ ∂ u t + ⋯ . $$\begin{equation}\label{d} \begin{aligned} & D_t = \frac{\partial}{\partial t} + u_t\frac{\partial}{\partial u}+ u_{tt}\frac{\partial}{\partial u_t}+ u_{tx} \frac{\partial}{\partial u_x} +\cdots, \\ & D_x = \frac{\partial}{\partial x} + u_x\frac{\partial}{\partial u}+ u_{xx} \frac{\partial}{\partial u_x} + u_{xt}\frac{\partial}{\partial u_t} +\cdots. \end{aligned} \end{equation}$$ Expanding (3) and splitting on derivatives of u yields an overdetermined system of linear homogeneous partial differential equations (PDEs). Solving these equations we obtain the values of τ , ξ and η , which lead to three Lie point symmetries of (1) given by
X 1 = ∂ ∂ t , X 2 = ∂ ∂ x , X 3 = 2 t ∂ ∂ t − u ∂ ∂ u . $$\begin{align*} X_1=\frac{\partial}{\partial t},\,\,\, X_2=\frac{\partial}{\partial x},\,\,\, X_3=2t\frac{\partial}{\partial t }-u\frac{\partial}{\partial u}. \end{align*}$$ The infinitesimal generator X 3 represents scaling symmetry whereas the one-parameter groups generated by X 1 and X 2 demonstrate time and space-invariance of the MEW equation.
2.2 Optimal system of one-dimensional subalgebras
We now calculate an optimal system of one-dimensional subalgebras by using Lie point symmetries of (1) obtained in the previous subsection. We employ the method given in [6]. We first construct the commutator table. Thereafter we compute adjoint representation using the Lie series
Ad ( exp ( ε X i ) ) X j = ∑ n = 0 ∞ ε n n ! ( ad X i ) n ( X j ) = X j − ε [ X i , X j ] + ε 2 2 ! [ X i , [ X i , X j ] ] − ⋯ , $$\begin{align*} \mbox{Ad}(\exp (\varepsilon X_i))X_j=\sum_{n=0}^{\infty }\frac{\varepsilon ^n}{n!}(\mbox{ad}X_i)^n(X_j) =X_j-\varepsilon [X_i,X_j]+\frac{\varepsilon ^2}{2!}[X_i,[X_i,X_j]]-\cdots , \end{align*}$$ where ε is a real number and [Xi ,Xj ] denotes the commutator defined by
[ X i , X j ] = X i X j − X j X i . $$\begin{align*} [X_i,X_j]=X_iX_j-X_jX_i. \end{align*}$$ The table of commutators of Lie point symmetries of Equation (1) and adjoint representations of the symmetry group of (1) on its Lie algebra are presented in Table 1 and Table 2 , respectively. Consequently, Table 1 and Table 2 are used to compute an optimal system of one-dimensional subalgebras for Equation (1) .
Table 1 Lie brackets for equation (1)
[ , ] X 1 X 2 X 3 X 1 0 0 2X 1 X 2 0 0 0 X 3 - 2X 1 0 0
Table 2 Adjoint representation of subalgebras
Ad X 1 X 2 X 3 X 1 X 1 X 2 - 2ɛ X1 + X 3 X 2 X 1 X 2 X 3 X 3 e 2 ɛ 1 X 2 X 3
Thus following [6] and utilising Tables 1 and 2 we can obtain an optimal system of one-dimensional subalgebras, which is given by {X 1 +cX 2 ,X 3 + aX 2 }, where c and a are arbitrary constants.
2.3 Solutions and symmetry reductions
We now utilise the optimal system of one-dimensional subalgebras obtained above in the previous subsection and find group-invariant solutions and symmetry reductions for Equation (1) .
Consider the first operator X 1 + cX 2 of the optimal system. This operator has two invariants
ξ = x − c t and U = u , $$\begin{align*} & \xi=x-ct\,\,\, \text{and}\,\,\, U = u, \end{align*}$$ which give the group-invariant solution U = U (ξ ). Using ξ as our new independent variable, Equation (1) is transformed into the nonlinear ordinary differential equation (ODE)
(7) c β U ‴ ( ξ ) + 3 α U 2 ( ξ ) U ′ ( ξ ) − c U ′ ( ξ ) = 0. $$\begin{equation} c\beta U'''(\xi)+3\alpha U^{2}(\xi)U'(\xi)-cU'(\xi)=0. \end{equation}$$ We now use the extended Jacobi elliptic function expansion method [9] to obtain travelling wave solutions of (1). We assume that solutions of the third-order nonlinear ODE (7) can be expressed in the form
(8) U ( ξ ) = ∑ i = − M M A i H ( ξ ) i , $$\begin{equation}\label{assum1} U(\xi)= \sum_{i=-M}^{M} A_{i} H(\xi)^{i}, \end{equation}$$ where M is a positive integer obtained by the balancing procedure and Ai are constants to be determined. Here H (ξ ) satisfies the nonlinear first-order ODE
(9) H ′ ( ξ ) = − ( 1 − H 2 ( ξ ) ) ( 1 − ω + ω H 2 ( ξ ) ) $$\begin{equation}\label{simplest} H^{\prime}(\xi)=-\sqrt{(1-H^2(\xi))(1-\omega+\omega H^{2}(\xi))} \end{equation}$$ or
(10) H ′ ( ξ ) = ( 1 − H 2 ( ξ ) ) ( 1 − ω H 2 ( ξ ) ) . $$\begin{equation}\label{simplest1} H^{\prime}(\xi)=\sqrt{(1-H^2(\xi))(1-\omega H^{2}(\xi))} \,. \end{equation}$$ We recall that the Jacobi cosine-amplitude function
(11) H ( ξ ) = cn ( ξ | ω ) $$\begin{align}\label{incognito} & H(\xi) = \text{cn}(\xi|\omega) \end{align}$$ is a solution to (9), whereas the Jacobi sine-amplitude function
(12) H ( ξ ) = sn ( ξ | ω ) $$\begin{align}\label{assum3} H(\xi)=\text{sn}(\xi|\omega) \end{align}$$ is a solution to (10). Here ω is a parameter such that 0 ≤ ω ≤ 1 [9, 10].
We note that when ω → 1, then cn(ξ |w ) → sech(ξ ) and sn(ξ |w ) → tanh(ξ ). Also, when ω → 0, then cn(ξ |w )→ cos(ξ ) and sn(ξ|ω )→ sin(ξ ).
2.3.1 Cnoidal wave solutions
Considering the nonlinear ODE (7), the balancing procedure yields M = 1. Thus (8) takes the form
(13) U ( ξ ) = A − 1 H − 1 ( ξ ) + A 0 + A 1 H ( ξ ) . $$\begin{equation}\label{assum2} U(\xi)=A_{-1}H^{-1}(\xi)+A_{0}+A_{1}H(\xi). \end{equation}$$ Substitution of U from (13) into (7) and utilising (9) we obtain
H ( ξ ) 4 β c A − 1 − H ( ξ ) 6 β c A 1 + H ( ξ ) 4 β c A 1 − 7 H ( ξ ) 2 β c A − 1 − 12 β c ω A − 1 + 6 β c ω 2 A − 1 + 3 H ( ξ ) 10 α ω A 1 3 − 6 H ( ξ ) 8 α ω A 1 3 − H ( ξ ) 8 c ω A 1 + 6 H ( ξ ) 7 α A 0 A 1 2 + 3 H ( ξ ) 6 α ω A 1 3 − 3 α ω A − 1 3 − 3 H ( ξ ) 6 α A 1 3 + 3 H ( ξ ) 8 α A 1 3 − H ( ξ ) 6 c A 1 + H ( ξ ) 4 c A 1 + H ( ξ ) 4 c A − 1 − 3 H ( ξ ) 2 α A − 1 3 − H ( ξ ) 2 c A − 1 − 10 H ( ξ ) 6 β c ω 2 A 1 − 2 H ( ξ ) 6 β c ω 2 A − 1 − 7 H ( ξ ) 8 β c ω A 1 + 14 H ( ξ ) 8 β c ω 2 A 1 − 6 H ( ξ ) 10 β c ω 2 A 1 + 21 H ( ξ ) 2 β c ω A − 1 − 14 H ( ξ ) 2 β c ω 2 A − 1 − 3 H ( ξ ) 4 β c ω A 1 − 10 H ( ξ ) 4 β c ω A − 1 + 2 H ( ξ ) 4 β c ω 2 A 1 + 10 H ( ξ ) 4 β c ω 2 A − 1 + 10 H ( ξ ) 6 β c ω A 1 + H ( ξ ) 6 β c ω A − 1 + 3 H ( ξ ) 8 α ω A 0 2 A 1 + 3 H ( ξ ) 8 α ω A − 1 A 1 2 + 6 H ( ξ ) 9 α ω A 0 A 1 2 − 6 H ( ξ ) α ω A − 1 2 A 0 − 3 H ( ξ ) 2 α ω A − 1 A 0 2 − 3 H ( ξ ) 2 α ω A − 1 2 A 1 + 12 H ( ξ ) 3 α ω A − 1 2 A 0 + 3 H ( ξ ) 4 α ω A 0 2 A 1 + 3 H ( ξ ) 4 α ω A − 1 A 1 2 + 6 H ( ξ ) 4 α ω A − 1 A 0 2 + 6 H ( ξ ) 4 α ω A − 1 2 A 1 + 6 H ( ξ ) 5 α ω A 0 A 1 2 − 6 H ( ξ ) 5 α ω A − 1 2 A 0 − 6 H ( ξ ) 6 α ω A 0 2 A 1 − 6 H ( ξ ) 6 α ω A − 1 A 1 2 − 3 H ( ξ ) 6 α ω A − 1 A 0 2 − 3 H ( ξ ) 6 α ω A − 1 2 A 1 − 12 H ( ξ ) 7 α ω A 0 A 1 2 + 6 H ( ξ ) α A − 1 2 A 0 + 3 H ( ξ ) 2 α A − 1 A 0 2 + H ( ξ ) 2 c ω A − 1 + 3 H ( ξ ) 2 α A − 1 2 A 1 + 6 H ( ξ ) 2 α ω A − 1 3 − H ( ξ ) 4 c ω A 1 − 6 H ( ξ ) 3 α A − 1 2 A 0 − 2 H ( ξ ) 4 c ω A − 1 − 3 H ( ξ ) 4 α A − 1 A 1 2 − 3 H ( ξ ) 4 α A 0 2 A 1 − 3 H ( ξ ) 4 α A − 1 2 A 1 − 3 H ( ξ ) 4 α A − 1 A 0 2 + 2 H ( ξ ) 6 c ω A 1 − 6 H ( ξ ) 5 α A 0 A 1 2 − 3 H ( ξ ) 4 α ω A − 1 3 + 3 H ( ξ ) 6 α A − 1 A 1 2 + 3 H ( ξ ) 6 α A 0 2 A 1 + H ( ξ ) 6 c ω A − 1 + 6 β c A − 1 + 3 α A − 1 3 = 0. $$\begin{align*} &H(\xi) ^{4}\beta\,cA_{{-1}}- H(\xi) ^{6}\beta\,cA_{{1}}+ H(\xi) ^{4}\beta\,cA_{{1}} -7\,H(\xi)^{2}\beta\,cA_{{-1}}-12\,\beta\,c\omega\,A_{{-1}}\\ & +6\,\beta\,c{\omega}^{2}A_{{-1}}+3\, H(\xi) ^{10}\alpha\,\omega\,{A_{{1}}}^{3} -6\, H(\xi) ^{8}\alpha\,\omega\,{A_{{1}}}^{3}- H(\xi) ^{8}c\omega\,A_{{1}} \\ & +6\, H(\xi) ^{7}\alpha\,A_{{0}}{A_{{1}}}^{2}+3\, H(\xi) ^{6}\alpha\,\omega\,{A_{{1}}}^{3}-3\,\alpha\,\omega\,{A_{{-1}}}^{3}-3\, H(\xi) ^{6}\alpha\,{A_{{1}}}^{3}+3\, H(\xi) ^{8}\alpha\,{A_{{1}}}^{3}\\ & - H(\xi) ^{6}cA_{{1}}+ H(\xi) ^{4}cA_{{1}}+ H(\xi) ^{4}cA_{{-1}}-3\, H(\xi) ^{2}\alpha\,{A_{{-1}}}^{3} - H(\xi)^{2}cA_{{-1}}\\ & -10\,H(\xi)^{6}\beta\,c{\omega}^{2}A_{{1}} -2\, H(\xi) ^{6}\beta\,c{\omega}^{2}A_{{-1}}-7\, H(\xi) ^{8}\beta\,c\omega\,A_{{1}}+14\, H(\xi) ^{8}\beta\,c{\omega}^{2}A_{{1}} \\ & -6\,H(\xi) ^{10}\beta\,c{\omega}^{2}A_{{1}}+21\, H(\xi) ^{2}\beta\,c\omega\,A_{{-1}}-14\, H(\xi) ^{2}\beta\,c{\omega}^{2}A_{{-1}}-3\, H(\xi) ^{4}\beta\,c\omega\,A_{{1}} \\ & -10\, H(\xi) ^{4}\beta\,c\omega\,A_{{-1}} +2\, H(\xi) ^{4}\beta\,c{\omega}^{2}A_{{1}}+10\, H(\xi) ^{4}\beta\,c{\omega}^{2}A_{{-1}}+10\, H(\xi) ^{6}\beta\,c\omega\,A_{{1}} \\ & + H(\xi) ^{6}\beta\,c\omega\,A_{{-1}} +3\, H(\xi) ^{8}\alpha\,\omega\,{A_{{0}}}^{2}A_{{1}}+3\, H(\xi) ^{8}\alpha\,\omega\,A_{{-1}}{A_{{1}}}^{2}+6\, H(\xi) ^{9}\alpha\,\omega\,A_{{0}}{A_{{1}}}^{2} \\ & -6\,H(\xi) \alpha\,\omega\,{A_{{-1}}}^{2}A_{{0}}-3\,H(\xi)^{2}\alpha\,\omega\,A_{{-1}}{A_{{0}}}^{2}-3\, H(\xi) ^{2}\alpha\,\omega\,{A_{{-1}}}^{2}A_{{1}}+12\, H(\xi) ^{3}\alpha\,\omega\,{A_{{-1}}}^{2}A_{{0}} \\ & +3\, H(\xi) ^{4}\alpha\,\omega\,{A_{{0}}}^{2}A_{{1}}+3\, H(\xi) ^{4}\alpha\,\omega\,A_{{-1}}{A_{{1}}}^{2}+6\,H(\xi) ^{4}\alpha\,\omega\,A_{{-1}}{A_{{0}}}^{2}+6\, H(\xi) ^{4}\alpha\,\omega\,{A_{{-1}}}^{2}A_{{1}} \\ & +6\, H(\xi) ^{5}\alpha\,\omega\,A_{{0}}{A_{{1}}}^{2}-6\, H(\xi) ^{5}\alpha\,\omega\,{A_{{-1}}}^{2}A_{{0}} -6\, H(\xi) ^{6}\alpha\,\omega\,{A_{{0}}}^{2}A_{{1}}-6\, H(\xi) ^{6}\alpha\,\omega\,A_{{-1}}{A_{{1}}}^{2} \\ & -3\, H(\xi) ^{6}\alpha\,\omega\,A_{{-1}}{A_{{0}}}^{2} -3\, H(\xi) ^{6}\alpha\,\omega\,{A_{{-1}}}^{2}A_{{1}} -12\, H(\xi) ^{7}\alpha\,\omega\,A_{{0}}{A_{{1}}}^{2}+6\,H(\xi) \alpha\,{A_{{-1}}}^{2}A_{{0}} \\ & +3\, H(\xi) ^{2}\alpha\,A_{{-1}}{A_{{0}}}^{2}+H(\xi) ^{2}c\omega\,A_{{-1}}+3\, H(\xi) ^{2}\alpha\,{A_{{-1}}}^{2}A_{{1}} +6\, H(\xi) ^{2}\alpha\,\omega\,{A_{{-1}}}^{3}\\ & - H(\xi) ^{4}c\omega\,A_{{1}}-6\,H(\xi) ^{3}\alpha\,{A_{{-1}}}^{2}A_{{0}}-2\, H(\xi) ^{4}c\omega\,A_{{-1}} -3\,H(\xi)^{4}\alpha\,A_{{-1}}{A_{{1}}}^{2} \\ & -3\, H(\xi) ^{4}\alpha\,{A_{{0}}}^{2}A_{{1}}-3\, H(\xi) ^{4}\alpha\,{A_{{-1}}}^{2}A_{{1}}-3\, H(\xi) ^{4}\alpha\,A_{{-1}}{A_{{0}}}^{2}+2\, H(\xi) ^{6}c\omega\,A_{{1}} \\ & -6\, H(\xi) ^{5}\alpha\,A_{{0}}{A_{{1}}}^{2}-3\, H(\xi) ^{4}\alpha\,\omega\,{A_{{-1}}}^{3}+3\, H(\xi) ^{6}\alpha\,A_{{-1}}{A_{{1}}}^{2}+3\, H(\xi) ^{6}\alpha\,{A_{{0}}}^{2}A_{{1}} \\ & +H(\xi)^{6}c\omega\,A_{{-1}}+6\,\beta\,cA_{{-1}}+3\,\alpha\,{A_{{-1}}}^{3}=0. \end{align*}$$ The above equation can be separated on like powers of H (ξ ) to obtain an overdetermined system of eleven algebraic equations
A 0 A 1 2 = 0 , A − 1 2 A 0 − ω A − 1 2 A 0 = 0 , 2 ω A − 1 2 A 0 − A − 1 2 A 0 = 0 , A 0 A 1 2 − 2 ω A 0 A 1 2 = 0 , α A 1 3 − 2 β c ω A 1 = 0 , ω A 0 A 1 2 − ω A − 1 2 A 0 − A 0 A 1 2 = 0 , 2 β c ω 2 A − 1 − α ω A − 1 3 + α A − 1 3 − 4 β c ω A − 1 + 2 β c A − 1 = 0 , 3 α ω A − 1 A 1 2 + 3 α ω A 0 2 A 1 − 6 α ω A 1 3 + 14 β c ω 2 A 1 + 3 α A 1 3 − 7 β c ω A 1 − c ω A 1 = 0 , 6 α ω A − 1 3 − 3 α ω A − 1 2 A 1 − 3 α ω A − 1 A 0 2 − 14 β c ω 2 A − 1 − 3 α A − 1 3 + 3 α A − 1 2 A 1 + 3 α A − 1 A 0 2 + 21 β c ω A − 1 − 7 β c A − 1 + c ω A − 1 − c A − 1 = 0 , 3 α ω A 1 3 − 3 α ω A − 1 2 A 1 − 3 α ω A − 1 A 0 2 − 6 α ω A − 1 A 1 2 − 6 α ω A 0 2 A 1 − 2 β c ω 2 A − 1 + 3 α A − 1 A 1 2 − 10 β c ω 2 A 1 + 3 α A 0 2 A 1 − 3 α A 1 3 + β c ω A − 1 + 10 β c ω A 1 − β c A 1 + c ω A − 1 + 2 c ω A 1 − c A 1 = 0 , 6 α ω A − 1 2 A 1 − 3 α ω A − 1 3 + 6 α ω A − 1 A 0 2 + 3 α ω A − 1 A 1 2 + 3 α ω A 0 2 A 1 + 10 β c ω 2 A − 1 + 2 β c ω 2 A 1 − 3 α A − 1 2 A 1 − 3 α A − 1 A 0 2 − 3 α A − 1 A 1 2 − 3 α A 0 2 A 1 − 10 β c ω A − 1 − 3 β c ω A 1 + β c A − 1 + β c A 1 − 2 c ω A − 1 − c ω A 1 + c A − 1 + c A 1 = 0. $$\begin{align*} & A_{{0}}{A_{{1}}}^{2}=0,\\ & {A_{{-1}}}^{2}A_{{0}}-\omega\,{A_{{-1}}}^{2}A_{{0}}=0, \\ & 2\,\omega\,{A_{{-1}}}^{2}A_{{0}}-{A_{{-1}}}^{2}A_{{0}}=0, \\ & A_{{0}}{A_{{1}}}^{2}-2\omega\,A_{{0}}{A_{{1}}}^{2}=0, \\ & \alpha\,{A_{{1}}}^{3}-2\,\beta\,c{\omega}A_{{1}}=0, \\ & \omega\,A_{{0}}{A_{{1}}}^{2}-\omega\,{A_{{-1}}}^{2}A_{{0}}-A_{{0}}{A_{{1}}}^{2}=0, \\ & 2\,\beta\,c{\omega}^{2}A_{{-1}}-\alpha\,\omega\,{A_{{-1}}}^{3}+\alpha\,{A_{{-1}}}^{3}-4\,\beta\,c\omega\,A_{{-1}}+2\,\beta\,cA_{{-1}}=0, \\ & 3\,\alpha\,\omega\,A_{{-1}}{A_{{1}}}^{2}+3\,\alpha\,\omega\,{A_{{0}}}^{2}A_{{1}}-6\,\alpha\,\omega\,{A_{{1}}}^{3}+14\,\beta\,c{\omega}^{2}A_{{1}} +3\,\alpha\,{A_{{1}}}^{3}-7\,\beta\,c\omega\,A_{{1}}-c\omega\,A_{{1}}=0, \\ & 6\,\alpha\,\omega\,{A_{{-1}}}^{3}-3\,\alpha\,\omega\,{A_{{-1}}}^{2}A_{{1}}-3\,\alpha\,\omega\,A_{{-1}}{A_{{0}}}^{2}-14\,\beta\,c{\omega}^{2}A_{{-1}} -3\,\alpha\,{A_{{-1}}}^{3}+3\,\alpha\,{A_{{-1}}}^{2}A_{{1}} \\ & +3\,\alpha\,A_{{-1}}{A_{{0}}}^{2}+21\,\beta\,c\omega\,A_{{-1}}-7\,\beta\,cA_{{-1}}+c\omega\,A_{{-1}}-cA_{{-1}}=0, \\ & 3\,\alpha\,\omega\,{A_{{1}}}^{3}-3\,\alpha\,\omega\,{A_{{-1}}}^{2}A_{{1}}-3\,\alpha\,\omega\,A_{{-1}}{A_{{0}}}^{2}-6\,\alpha\,\omega\,A_{{-1}}{A_{{1}}}^{2}-6\,\alpha\,\omega\,{A_{{0}}}^{2}A_{{1}} -2\,\beta\,c{\omega}^{2}A_{{-1}} \\ & +3\,\alpha\,A_{{-1}}{A_{{1}}}^{2}-10\,\beta\,c{\omega}^{2}A_{{1}}+3\,\alpha\,{A_{{0}}}^{2}A_{{1}} -3\,\alpha\,{A_{{1}}}^{3}+\beta\,c\omega\,A_{{-1}}+10\,\beta\,c\omega\,A_{{1}}-\beta\,cA_{{1}}+c\omega\,A_{{-1}} \\ & +2\,c\omega\,A_{{1}}-cA_{{1}}=0, \\ & 6\,\alpha\,\omega\,{A_{{-1}}}^{2}A_{{1}}-3\,\alpha\,\omega\,{A_{{-1}}}^{3}+6\,\alpha\,\omega\,A_{{-1}}{A_{{0}}}^{2}+3\,\alpha\,\omega\,A_{{-1}}{A_{{1}}}^{2} +3\,\alpha\,\omega\,{A_{{0}}}^{2}A_{{1}}+10\,\beta\,c{\omega}^{2}A_{{-1}} \\ & +2\,\beta\,c{\omega}^{2}A_{{1}}-3\,\alpha\,{A_{{-1}}}^{2}A_{{1}}-3\,\alpha\,A_{{-1}}{A_{{0}}}^{2}-3\,\alpha\,A_{{-1}}{A_{{1}}}^{2}-3\,\alpha\,{A_{{0}}}^{2}A_{{1}}-10\,\beta\,c\omega\,A_{{-1}} \\ & -3\,\beta\,c\omega\,A_{{1}}+\beta\,cA_{{-1}}+\beta\,cA_{{1}}-2\,c\omega\,A_{{-1}}-c\omega\,A_{{1}}+cA_{{-1}}+cA_{{1}}=0. \end{align*}$$ Solving the above system of equations we obtain
ω = 8 β + 3 k − 1 16 β , A 0 = 0 , A 1 = ± c ( 3 k + 8 β − 1 ) 8 α , A − 1 = − 3 β ± k β + 1 A 1 $$\begin{align*} \omega=\frac{8\beta+3k-1}{16\beta},\,\, A_{0}=0,\,\, A_{1}=\pm\sqrt{\frac{c(3k+8\beta-1)}{8\alpha}},\,\, A_{-1}=-\frac{3\beta\pm k}{\beta+1}A_1 \end{align*}$$ with k = 8 β 2 + 1 $k=\sqrt{8\beta^2+1}$
Thus reverting to the original variables the solutions of (1) are
(14) u ( t , x ) = ± c ( 3 k + 8 β − 1 ) 8 α cn ξ ω − 3 β ± k β + 1 nc ξ ω , $$\begin{equation} u(t,x)=\pm\sqrt{\frac{c(3k+8\beta-1)}{8\alpha}}\left\{\text{cn}\left(\xi\left|\omega \right.\right)-\left(\frac{3\beta\pm k}{\beta+1}\right)\text{nc}\left(\xi\left|\omega\right.\right)\right\}, \end{equation}$$ where nc = 1/cn.
2.3.2 Snoidal wave solutions
We now obtain snoidal wave solutions for Equation (1) . Here again M = 1. Substituting the value of U from (13) into (7) and making use of (10) we obtain
3 H ( ξ ) 6 α A 1 3 − H ( ξ ) 4 c A 1 + H ( ξ ) 2 c A − 1 + 3 H ( ξ ) 2 α A − 1 3 + 3 H ( ξ ) 4 α A − 1 2 A 1 + 3 H ( ξ ) 4 α A − 1 A 0 2 + H ( ξ ) 6 β c A 1 − H ( ξ ) 4 β c A − 1 − H ( ξ ) 4 β c A 1 + 7 H ( ξ ) 2 β c A − 1 + 3 H ( ξ ) 4 α A − 1 A 1 2 + 3 H ( ξ ) 4 α A 0 2 A 1 + 3 H ( ξ ) 10 α ω A 1 3 − 3 H ( ξ ) 8 α ω A 1 3 − H ( ξ ) 8 c ω A 1 − 6 H ( ξ ) 7 α A 0 A 1 2 − 3 H ( ξ ) 6 α A − 1 A 1 2 − 3 H ( ξ ) 6 α A 0 2 A 1 + H ( ξ ) 6 c ω A − 1 + H ( ξ ) 6 c ω A 1 + 6 H ( ξ ) 5 α A 0 A 1 2 − 3 H ( ξ ) 4 α ω A − 1 3 − 3 α A − 1 3 − 6 H ( ξ ) 7 α ω A 0 A 1 2 − 3 H ( ξ ) 6 α ω A − 1 2 A 1 − 3 H ( ξ ) 6 α ω A − 1 A 0 2 − 3 H ( ξ ) 6 α ω A − 1 A 1 2 − 3 H ( ξ ) 6 α ω A 0 2 A 1 − 6 H ( ξ ) 5 α ω A − 1 2 A 0 + 3 H ( ξ ) 4 α ω A − 1 2 A 1 + 3 H ( ξ ) 4 α ω A − 1 A 0 2 + 6 H ( ξ ) 9 α ω A 0 A 1 2 + 3 H ( ξ ) 8 α ω A − 1 A 1 2 + 3 H ( ξ ) 8 α ω A 0 2 A 1 − 3 H ( ξ ) 8 α A 1 3 + H ( ξ ) 6 c A 1 − H ( ξ ) 4 c A − 1 − 6 β c A − 1 + 7 H ( ξ ) 2 β c ω A − 1 − H ( ξ ) 4 β c ω A 1 − 8 H ξ 4 β c ω A − 1 − H ( ξ ) 4 β c ω 2 A − 1 + 8 H ( ξ ) 6 β c ω A 1 + H ( ξ ) 6 β c ω A − 1 + H ( ξ ) 6 β c ω 2 A 1 + H ( ξ ) 6 β c ω 2 A − 1 − 7 H ( ξ ) 8 β c ω A 1 − 7 H ( ξ ) 8 β c ω 2 A 1 − 3 H ( ξ ) 2 α A − 1 A 0 2 − 6 H ξ α A − 1 2 A 0 + 3 H ( ξ ) 2 α ω A − 1 3 − 3 H ( ξ ) 2 α A − 1 2 A 1 − H ( ξ ) 4 c ω A − 1 + 6 H ( ξ ) 3 α A − 1 2 A 0 + 6 H ( ξ ) 10 β c ω 2 A 1 + 6 H ( ξ ) 3 α ω A − 1 2 A 0 = 0. $$\begin{align*} & 3\,H(\xi) ^{6}\alpha\,{A_{{1}}}^{3}-H(\xi) ^{4}cA_{{1}}+H(\xi) ^{2}cA_{{-1}} +3\,H(\xi) ^{2}\alpha\,{A_{{-1}}}^{3}+3\,H(\xi) ^{4}\alpha\,{A_{{-1}}}^{2}A_{{1}} \\ & +3\, H(\xi) ^{4}\alpha\,A_{{-1}}{A_{{0}}}^{2}+H(\xi) ^{6}\beta\,cA_{{1}}- H(\xi) ^{4}\beta\,cA_{{-1}}-H(\xi) ^{4}\beta\,cA_{{1}}+7\,H(\xi) ^{2}\beta\,cA_{{-1}} \\ & +3\,H(\xi) ^{4}\alpha\,A_{{-1}}{A_{{1}}}^{2}+3\,H(\xi) ^{4}\alpha\,{A_{{0}}}^{2}A_{{1}} +3\,H(\xi) ^{10}\alpha\,\omega\,{A_{{1}}}^{3}-3\,H(\xi) ^{8}\alpha\,\omega\,{A_{{1}}}^{3} -H(\xi) ^{8}c\omega\,A_{{1}} \\ & -6\, H(\xi) ^{7}\alpha\,A_{{0}}{A_{{1}}}^{2}-3\, H(\xi) ^{6}\alpha\,A_{{-1}}{A_{{1}}}^{2}-3\,H(\xi) ^{6}\alpha\,{A_{{0}}}^{2}A_{{1}} +H(\xi) ^{6}c\omega\,A_{{-1}}+ H(\xi) ^{6}c\omega\,A_{{1}} \\ & +6\,H(\xi)^{5}\alpha\,A_{{0}}{A_{{1}}}^{2}-3\, H(\xi) ^{4}\alpha\,\omega\,{A_{{-1}}}^{3}-3\,\alpha\,{A_{{-1}}}^{3} -6\,H(\xi) ^{7}\alpha\,\omega\,A_{{0}}{A_{{1}}}^{2}-3\,H(\xi) ^{6}\alpha\,\omega\,{A_{{-1}}}^{2}A_{{1}} \\ & -3\, H(\xi) ^{6}\alpha\,\omega\,A_{{-1}}{A_{{0}}}^{2}-3\,H(\xi) ^{6}\alpha\,\omega\,A_{{-1}}{A_{{1}}}^{2} -3\,H(\xi) ^{6}\alpha\,\omega\,{A_{{0}}}^{2}A_{{1}}-6\,H(\xi) ^{5}\alpha\,\omega\,{A_{{-1}}}^{2}A_{{0}} \\ & +3\, H(\xi) ^{4}\alpha\,\omega\,{A_{{-1}}}^{2}A_{{1}}+3\,H(\xi) ^{4}\alpha\,\omega\,A_{{-1}}{A_{{0}}}^{2} +6\,H(\xi) ^{9}\alpha\,\omega\,A_{{0}}{A_{{1}}}^{2}+3\,H(\xi) ^{8}\alpha\,\omega\,A_{{-1}}{A_{{1}}}^{2} \\ & +3\, H(\xi) ^{8}\alpha\,\omega\,{A_{{0}}}^{2}A_{{1}}-3\,H(\xi) ^{8}\alpha\,{A_{{1}}}^{3} + H(\xi) ^{6}cA_{{1}}-H(\xi) ^{4}cA_{{-1}}-6\,\beta\,cA_{{-1}} \\ & +7\,H(\xi) ^{2}\beta\,c\omega\,A_{{-1}}-H(\xi) ^{4}\beta\,c\omega\,A_{{1}}-8\, H \left( \xi \right) ^{4}\beta\,c\omega\,A_{{-1}}-H(\xi) ^{4}\beta\,c{\omega}^{2}A_{{-1}}+8\, H(\xi) ^{6}\beta\,c\omega\,A_{{1}} \\ & + H(\xi) ^{6}\beta\,c\omega\,A_{{-1}}+ H(\xi) ^{6}\beta\,c{\omega}^{2}A_{{1}} + H(\xi) ^{6}\beta\,c{\omega}^{2}A_{{-1}}-7\, H(\xi) ^{8}\beta\,c\omega\,A_{{1}} -7\, H(\xi) ^{8}\beta\,c{\omega}^{2}A_{{1}} \\ & -3\,H(\xi) ^{2}\alpha\,A_{{-1}}{A_{{0}}}^{2}-6\,H \left( \xi \right) \alpha\,{A_{{-1}}}^{2}A_{{0}}+3\, H(\xi) ^{2}\alpha\,\omega\,{A_{{-1}}}^{3} -3\,H(\xi) ^{2}\alpha\,{A_{{-1}}}^{2}A_{{1}} \\ & -H(\xi) ^{4}c\omega\,A_{{-1}}+6\,H(\xi) ^{3}\alpha\,{A_{{-1}}}^{2}A_{{0}}+6\,H(\xi) ^{10}\beta\,c{\omega}^{2}A_{{1}}+6\,H(\xi) ^{3}\alpha\,\omega\,{A_{{-1}}}^{2}A_{{0}}=0. \end{align*}$$ Splitting on powers of H (ξ ) yields the following overdetermined system of algebraic equations:
A − 1 2 A 0 = 0 , A 0 A 1 2 = 0 , α A − 1 3 + 2 β c A − 1 = 0 , ω A − 1 2 A 0 + A − 1 2 A 0 = 0 , ω A 0 A 1 2 + A 0 A 1 2 = 0 , α A 1 3 + 2 β c ω A 1 = 0 , ω A − 1 2 A 0 − A 0 A 1 2 = 0 , 3 α ω A − 1 A 1 2 + 3 α ω A 0 2 A 1 − 3 α ω A 1 3 − 7 β c ω 2 A 1 − 3 α A 1 3 − 7 β c ω A 1 − c ω A 1 = 0 , 3 α ω A − 1 3 + 3 α A − 1 3 − 3 α A − 1 2 A 1 − 3 α A − 1 A 0 2 + 7 β c ω A − 1 + 7 β c A − 1 + c A − 1 = 0 , β c ω 2 A − 1 − 3 α ω A − 1 2 A 1 − 3 α ω A − 1 A 0 2 − 3 α ω A − 1 A 1 2 − 3 α ω A 0 2 A 1 + β c ω 2 A 1 + 3 α A 1 3 − 3 α A − 1 A 1 2 − 3 α A 0 2 A 1 + β c ω A − 1 + 8 β c ω A 1 + β c A 1 + c ω A − 1 + c ω A 1 + c A 1 = 0 , 3 α ω A − 1 2 A 1 − 3 α ω A − 1 3 + 3 α ω A − 1 A 0 2 − β c ω 2 A − 1 + 3 α A − 1 2 A 1 + 3 α A − 1 A 0 2 + 3 α A − 1 A 1 2 + 3 α A 0 2 A 1 − 8 β c ω A − 1 − β c ω A 1 − β c A − 1 − β c A 1 − c ω A − 1 − c A − 1 − c A 1 = 0. $$\begin{align*} & {A_{{-1}}}^{2}A_{{0}}=0,\\ & A_{{0}}{A_{{1}}}^{2}=0,\\ & \alpha\,{A_{{-1}}}^{3}+2\,\beta\,cA_{{-1}}=0,\\ & \omega\,{A_{{-1}}}^{2}A_{{0}}+{A_{{-1}}}^{2}A_{{0}}=0, \\ & \omega\,A_{{0}}{A_{{1}}}^{2}+A_{{0}}{A_{{1}}}^{2}=0, \\ & \alpha\,{A_{{1}}}^{3}+2\,\beta\,c{\omega}A_{{1}}=0, \\ & \omega\,{A_{{-1}}}^{2}A_{{0}}-A_{{0}}{A_{{1}}}^{2}=0, \\ & 3\,\alpha\,\omega\,A_{{-1}}{A_{{1}}}^{2}+3\,\alpha\,\omega\,{A_{{0}}}^ {2}A_{{1}}-3\,\alpha\,\omega\,{A_{{1}}}^{3}-7\,\beta\,c{\omega}^{2}A_{{1}}-3\,\alpha\,{A_{{1}}}^{3}-7\,\beta\,c\omega\,A_{{1}}-c\omega\,A_{{1}}=0, \\ & 3\,\alpha\,\omega\,{A_{{-1}}}^{3}+3\,\alpha\,{A_{{-1}}}^{3}-3\, \alpha\,{A_{{-1}}}^{2}A_{{1}}-3\,\alpha\,A_{{-1}}{A_{{0}}}^{2}+7\, \beta\,c\omega\,A_{{-1}}+7\,\beta\,cA_{{-1}}+cA_{{-1}}=0, \\ & \beta\,c{\omega}^{2}A_{{-1}}-3\,\alpha\,\omega\,{A_{{-1}}}^{2}A_{{1}}-3\,\alpha\,\omega\,A_{{-1}}{A_{{0}}}^{2} -3\,\alpha\,\omega\,A_{{-1}}{A_{{1}}}^{2}-3\,\alpha\,\omega\,{A_{{0}}}^{2}A_{{1}}+\beta\,c{\omega}^{2}A_{{1}}\\ & +3\,\alpha\,{A_{{1}}}^{3}-3\,\alpha\,A_{{-1}}{A_{{1}}}^{2}-3\,\alpha\,{A_{{0}}}^{2}A_{{1}}+\beta\,c\omega\,A_{{-1}}+8\,\beta\,c\omega\,A_{{1}}+\beta\,cA_{{1}}+c\omega\,A_{{-1}}+c\omega\,A_{{1}}+cA_{{1}}=0, \\ & 3\,\alpha\,\omega\,{A_{{-1}}}^{2}A_{{1} }-3\, \alpha\,\omega\,{A_{{-1}}}^{3}+3\,\alpha\,\omega\,A_{{-1}}{A_{{0}}}^{2}-\beta\,c{\omega}^{2}A_{{-1} }+3\,\alpha\,{A_{{-1}}}^{2}A_{{1}}+3\,\alpha\,A_{{-1}}{A_{{0}}}^{2}+3 \,\alpha\,A_{{-1}}{A_{{1}}}^{2}\\ & +3\,\alpha\,{A_{{0}}}^{2}A_{{1}}-8\, \beta\,c\omega\,A_{{-1}}-\beta\,c\omega\,A_{{1}}-\beta\,cA_{{-1}}- \beta\,cA_{{1}}-c\omega\,A_{{-1}}-cA_{{-1}}-cA_{{1}}=0. \end{align*}$$ Solving the above system of equations we get
β = − 1 1 + ω , A − 1 = A 0 = 0 , A 1 = ± 2 c ( β + 1 ) α . $$\begin{align*} \beta =-\frac{1}{1 + \omega},\,\, A_{-1}=A_{0}=0,\,\, A_{1}=\pm \sqrt{\frac{2c(\beta+1)}{\alpha}}. \end{align*}$$ Reverting to original variables we obtain solutions of (1) as
(15) u ( t , x ) = ± 2 c ( β + 1 ) α sn ξ ω . $$\begin{equation} u(t,x)=\pm \sqrt{\frac{2c(\beta+1)}{\alpha}}\, \text{sn}\left(\xi\left|\omega\right.\right). \end{equation}$$ We now consider the second operator X 3 + aX 2 of the optimal system. This symmetry operator yields two invariants J 1 = ex t-a / 2 J 2 = ω 1/ 2 . Thus J 2 = f (J 1 ) provides a group-invariant solution to (1). That is
u = t − 1 / 2 f ( e x t − a / 2 ) . $$\begin{equation*} u=t^{-1/2} \, f(e^x \, t^{-a/2}). \end{equation*} $$ Substituting the above value of u in (1), we obtain the third-order nonlinear ODE
a β z 3 f ′ ′ ′ ( z ) + β 3 a + 1 z 2 f ′ ′ ( z ) + ( a β − a + β ) z f ′ ( z ) + 6 α z f ( z ) 2 f ′ ( z ) − f ( z ) = 0 , $$\begin{equation*} a \beta z^3 f^{\prime \prime \prime }(z)+ \beta\left(3 a+1\right) z^2f^{\prime \prime }(z)+ (a \beta -a +\beta )zf'(z)+6 \alpha z f(z)^2 f'(z)-f(z)=0, \end{equation*}$$ where z = e x t − a / 2 $z=e^x \, t^{-a/2}$
3 In the section we derive conservation laws for (1) by employing two different techniques, namely the multiplier method and Noether approach.
Conservation laws have several important uses in the study of partial differential equations, especially for determining conserved quantities and constants of motion, detecting integrability and linearizations, finding potentials and nonlocally-related systems, as well as checking the accuracy of numerical solution methods [11 , 12 , 13 , 14 , 15 , 16 , 17 ].
3.1 Conservation laws of (1) using multiplier approach
We look for zeroth-order multiplier 𝛬=𝛬(t,x,u ). Thus, the determining equation for this multiplier is stated as
(16) δ δ u { Λ ( t , x , u ) ( u t + 3 α u 2 u x − β u t x x ) } = 0 , $$\begin{equation} \frac{\delta}{\delta u}\Big\{\Lambda(t,x,u)\Big(u_{t}+3\alpha u^{2}u_{x}-\beta u_{txx}\Big)\Big\}=0, \end{equation}$$ where δ/δu is the Euler-Lagrange operator defined as
δ δ u = ∂ ∂ u − D t ∂ ∂ u t − D x ∂ ∂ u x − D t D x 2 ∂ ∂ u t x x $$\begin{equation*} \frac{\delta }{\delta u}=\frac{\partial }{\partial u}-D_{t}\frac{\partial}{\partial u_{t}} - D_{x}\frac{\partial}{\partial u_{x}} - D_{t} D^{2}_{x}\frac{\partial}{\partial u_{txx}} \end{equation*}$$ and the total derivatives Dt and Dx are defined as in (6). The above equation yields
u t Λ u + 3 α u 2 u x Λ u − β u t x x Λ u + 6 α u u x Λ − D t ( Λ ) − D x ( 3 α u 2 Λ ) + β D x 2 D t ( Λ ) = 0 , $$\begin{equation*} u_t\Lambda_u+3\alpha u^{2}u_x\Lambda_u-\beta u_{txx}\Lambda_u+6\alpha uu_x\Lambda-D_t(\Lambda)-D_x(3\alpha u^{2}\Lambda)+\beta D^{2}_xD_t(\Lambda)=0, \end{equation*}$$ which on expanding gives
u t Λ u + 3 α u 2 u x Λ u − β u x x t Λ u + 6 α u u x Λ − Λ t − u t Λ u − 3 α u 2 Λ x − 6 α u u x Λ − 3 α u 2 u x Λ u + β Λ t x x + β u t Λ u x x + β u x Λ t u x + β u t u x Λ u u x + β u t x Λ u x + β u x Λ t u x + β u t u x Λ u u x + β u x 2 Λ t u u + β u t u x 2 Λ u u u + β u t x u x Λ u u + β u t x Λ u x + β u t x u x Λ u u + β u x x Λ t u + β u x x u t Λ u u + β u x x t Λ u = 0. $$\begin{align*} &u_t\Lambda_u+3\alpha u^{2}u_x\Lambda_u-\beta u_{xxt}\Lambda_u+6\alpha uu_x\Lambda-\Lambda_t-u_t\Lambda_u-3\alpha u^{2}\Lambda_x-6\alpha uu_x\Lambda-3\alpha u^2u_x\Lambda_u+\beta\Lambda_{txx} \\ & +\beta u_t\Lambda_{uxx}+\beta u_x \Lambda_{tux}+\beta u_tu_x\Lambda_{uux}+\beta u_{tx}\Lambda_{ux}+\beta u_x\Lambda_{tux}+\beta u_tu_x\Lambda_{uux}+\beta u^2_x\Lambda_{tuu}+\beta u_tu^{2}_x\Lambda_{uuu} \\ & +\beta u_{tx}u_x\Lambda_{uu}+\beta u_{tx}\Lambda_{ux}+\beta u_{tx}u_x\Lambda_{uu}+\beta u_{xx}\Lambda_{tu}+\beta u_{xx}u_t \Lambda_{uu}+\beta u_{xxt}\Lambda_u=0. \end{align*}$$ Splitting the above equation on derivatives of u , we obtain
Λ u u = 0 , Λ u x = 0 , Λ t u = 0 , β Λ t x x − 3 α u 2 Λ x − Λ t = 0. $$\begin{align*} \Lambda_{uu}=0,\,\, \Lambda_{ux}=0,\,\, \Lambda_{tu}=0,\,\, \beta\Lambda_{txx}-3\alpha u^{2}\Lambda_x-\Lambda_t=0. \end{align*}$$ By solving the above equations we get two multipliers given by
Λ 1 ( t , x , u ) = u $$\begin{equation*} \Lambda _1(t,x,u)=u \end{equation*}$$ and
Λ 2 ( t , x , u ) = 1. $$\begin{equation*} \Lambda _2(t,x,u)=1. \end{equation*}$$ Corresponding to these two multipliers, we obtain the following two conservation laws:
T 1 t = 1 2 u 2 + 1 2 β u x 2 , T 1 x = 3 4 α u 4 − β u u t x $$\begin{eqnarray*} T^{t}_{1}&=& \frac{1}{2}u^{2} + \frac{1}{2}\beta u_x ^{2},\\ T^{x}_{1}&=& \frac{3}{4} \alpha u^4 - \beta u u_{tx} \end{eqnarray*}$$ and
T 2 t = u , T 2 x = α u 3 − β u t x . $$\begin{eqnarray*} T^{t}_{2}&=&u,\\ T^{x}_{2}&=&\alpha u^3 - \beta u_{tx}. \end{eqnarray*}$$ 3.2 Conservation laws of (1) using Noether’s theorem
In this subsection we derive conservation laws for the modified equal-width Equation (1) using Noether’s theorem [18 , 19 ]. This equation as it is does not have a Lagrangian. In order to apply Noether’s theorem we transform Equation (1) to a fourth-order equation which will have a Lagrangian. Thus using the transformation u =Vx , Equation (1) becomes
(17) V t x + 3 α V x 2 V x x − β V t x x x = 0. $$\begin{equation}\label{noe1} V_{tx}+3\alpha V_{x}^{2}V_{xx}-\beta V_{txxx}=0. \end{equation}$$ It can readily be verified that a Lagrangian for equation (17) is given by
(18) L = − 1 2 V x V t − 1 4 α V x 4 − 1 2 β V x x V t x $$\begin{equation}\label{noe5} \mathcal{L}=-\frac{1}{2}V_xV_{t}-\frac{1}{4}\alpha V_{x}^{4}-\frac{1}{2}\beta V_{xx}V_{tx} \end{equation}$$ because δℒ/δV = 0 on (17). Here δ/δV is the Euler-Lagrange operator defined as
δ δ V = ∂ ∂ V − D t ∂ ∂ V t − D x ∂ ∂ V x + D x 2 ∂ ∂ V x x + D t D x ∂ ∂ V t x . $$\begin{equation*} \frac{\delta }{\delta V}=\frac{\partial }{\partial V}-D_{t}\frac{\partial}{\partial V_{t}} - D_{x}\frac{\partial}{\partial V_{x}} +D^{2}_{x}\frac{\partial}{\partial V_{xx}} + D_{t}D_{x} \frac{\partial}{\partial V_{tx}}. \end{equation*}$$ Consider the vector field
(19) X = τ ( t , x , V ) ∂ ∂ t + ξ ( t , x , V ) ∂ ∂ x + η ( t , x , V ) ∂ ∂ V , $$\begin{equation} X=\tau(t,x,V)\frac{\partial}{\partial t}+\xi(t,x,V)\frac{\partial}{\partial x}+\eta(t,x,V) \frac{\partial}{\partial V}, \end{equation}$$ where τ , ξ and η depend on t , x and V . To determine Noether point symmetries X of (17) we insert the value of L from (18) in the determining equation
(20) pr [ 2 ] X ( L ) + L D t τ + D x ξ = D t B t + D x B x , $$\begin{equation}\label{noe8} \mbox{pr}^{[2]}X (\mathcal{L}) + \mathcal{L} \left [D_{t}\left( \tau \right) +D_{x}\left( \xi \right) \right ]=D_{t}\left( B^{t} \right) +D_{x}\left( B^{x}\right) , \end{equation}$$ where Bt = Bt (t,x,V ) and Bx = Bx (t,x,V ) are gauge terms and pr[2] X is the second prolongation of X defined as
(21) pr [ 2 ] X = X + ζ t ∂ ∂ V t + ζ x ∂ ∂ V x + ζ x x ∂ ∂ V x x + ζ t x ∂ ∂ V t x $$\begin{equation} \mbox{pr}^{[2]}X=X+\zeta_{t}\frac{\partial}{\partial V_{t}}+\zeta_{x}\frac{\partial}{\partial V_{x}} +\zeta_{xx}\frac{\partial}{\partial V_{xx}}+\zeta_{tx}\frac{\partial}{\partial V_{tx}} \end{equation}$$ with ζt , ζx , ζxx and ζtx as defined in (5). Expansion of equation (20) and separating with respect to derivatives of V yields an overdetermined system of linear PDEs. Thereafter solving these PDEs we obtain the following Noether point symmetries together with their gauge functions:
X 1 = ∂ ∂ t , B t = 0 , B x = 0 , X 2 = ∂ ∂ x , B t = 0 , B x = 0 , X f = f ( t ) ∂ ∂ V , B t = 0 , B x = − 1 2 f ′ ( t ) V . $$\begin{eqnarray*} X_{1}&=&\frac{\partial}{\partial t}, \,\, B^{t}=0, \,\, B^{x}=0,\\ X_{2}&=&\frac{\partial}{\partial x}, \,\, B^{t}=0, \,\, B^{x}=0,\\X_{f}&=&f(t)\frac{\partial}{\partial V}, \,\, B^{t}=0, \,\, B^{x}=-\frac{1}{2}f'(t)V. \end{eqnarray*}$$ Next, we use the above results to compute conserved vectors of the fourth-order equation (17) . Using formulae for the conserved vector (Tt ,Tx ) [20 ]
F k = L τ k + ( ξ α − ψ x j α τ j ) ( ∂ L ∂ ψ x k α − ∑ l = 1 k D x l ( ∂ L ∂ ψ x l x k α ) ) + ∑ l = k n ( η l α − ψ x l x j α τ j ) ∂ L ∂ ψ x k x l α − f k $$\begin{equation*} F^{k}={\mathcal{L}}\tau^{k}+(\xi^{\alpha}-\psi^{\alpha}_{x^{j}}\tau^{j})\Bigg(\frac{\partial {\mathcal{L}}}{\partial \psi^{\alpha}_{x^{k}}}-\sum_{l=1}^{k}D_{x^{l}}\Big(\frac{\partial {\mathcal{L}}}{\partial \psi^{\alpha}_{x^{l}x^{k}}}\Big)\Bigg)+\sum_{l=k}^{n}(\eta^{\alpha}_{l}-\psi^{\alpha}_{x^{l}x^{j}}\tau^{j})\frac{\partial {\mathcal{L}}}{\partial \psi^{\alpha}_{x^{k}x^{l}}}-f^{k} \end{equation*}$$ we obtain three conserved vectors associated with three Noether point symmetries X 1 , X 2 and X f . Then reverting to the original variable u , we have
T 1 t = − 1 4 α u 4 − 1 2 β u x u t − 1 2 β u x x ∫ u t d x , T 1 x = 1 2 ∫ u t d x 2 + α u 3 ∫ u t d x − 1 2 β u x t ∫ u t d x + 1 2 β u t 2 + 1 2 β u x ∫ u t t d x ; T 2 t = 1 2 u 2 − 1 2 β u u x x , T 2 x = 3 4 α u 4 − 1 2 β u u x t + 1 2 β u x u t ; T f t = − 1 2 f ( t ) u + 1 2 β f ( t ) u x x , T f x = − 1 2 f ( t ) ∫ u t d x − α f ( t ) u 3 + 1 2 β f ( t ) u x t − 1 2 β f ′ ( t ) u x + 1 2 f ′ ( t ) ∫ u d x . $$\begin{eqnarray*} T^{t}_1&=&-\frac{1}{4}\alpha u^{4}-\frac{1}{2}\beta u_{x}u_{t}-\frac{1}{2}\beta u_{xx}\int u_{t}dx,\\ T^{x}_1&=&\frac{1}{2}\left( \int u_{t}dx\right) ^{2}+\alpha u^{3}\int u_{t}dx-\frac{1}{2}\beta u_{xt}\int u_{t}dx+\frac{1}{2}\beta u^{2}_{t}+\frac{1}{2}\beta u_{x}\int u_{tt}dx; \\ \\ T^{t}_2&=&\frac{1}{2} u^{2}-\frac{1}{2}\beta uu_{xx},\\ T^{x}_2&=&\frac{3}{4}\alpha u^{4}-\frac{1}{2}\beta uu_{xt}+\frac{1}{2}\beta u_{x}u_{t}; \\ \\ T^{t}_f&=&-\frac{1}{2}f(t)u+\frac{1}{2}\beta f(t)u_{xx},\\ T^{x}_f&=&-\frac{1}{2}f(t)\int u_tdx-\alpha f(t)u^{3} + \frac{1}{2} \beta f(t)u_{xt}-\frac{1}{2}\beta f'(t)u_{x}+\frac{1}{2}f'(t)\int udx. \end{eqnarray*}$$ Remark: It should be noted that due to the presence of arbitrary function f (t ) we have infinitely many nonlocal conservation laws.
