1. Introduction
Similarity solutions can be obtained in many ways, the most famous methods are the classical Lie group approach, the nonclassical Lie group approach and the direct method [1]. To simplify the calculation, a simple direct method to derive symmetry reductions of a nonlinear system without involving any group theory was introduced by Clarkson and Kruskal (CK). This method has been applied to many nonlinear equations such as the Kadomtsev–Petviashvili equation [2], the Boussinesq equation [3], (2+1)-Dimensional General Nonintegrable KdV equation [4], Jimbo–Miwa equation [5] and the dispersive wave equations in shallow water [6].
The classical Lie group approach can be also important to find the Lie point symmetry groups of a nonlinear system [7]. In [8], the symmetry groups of the KP equation via the traditional Lie group approach had been studied by David, Levi and Winternitz via the traditional Lie group approach. However, the method requires many algebraic operations and cannot obtain all the similar solutions, therefore, the direct method becomes a simple and effective method for us to find symmetry reduction. In [9], two different methods have been applied to the Whitham–Broer–Kaup equation, the results of the classical Lie group approach only got the form of Painlevé but the direct method got five different types of equations, therefore, the results of the classical Lie group approach is only a special case of the direct method. In recent years, similar methods have been applied to some new models, such as the Benney system [10] and Boundary-Layer equations[11].
Recently, a novel (2+1)-dimensional Korteweg–de Vries (KdV) extension, the combined KP3 (Kadomtsev–Petviashvili) and KP4 (CKP34) equation
$\begin{eqnarray}\begin{array}{rcl}{u}_{{xt}} & = & a[{\left(6{{uu}}_{x}+{u}_{{xxx}}\right)}_{x}-3{u}_{{yy}}]\\ & & +b{\left(2{{vu}}_{x}+{v}_{{xxx}}+4{{uu}}_{y}\right)}_{x}-{v}_{{yy}},\end{array}\end{eqnarray}$
$\begin{eqnarray}\,{u}_{y}={v}_{x},\end{eqnarray}$
is proposed by one of the present authors (Lou) [12]. It has been proved that it has the Painlevé property, the Bäcklund/Levi transformations and the residual nonlocal symmetry [12]. It is worth mentioning that the CKP34 equation not only has the same properties as the KP3 equation and the KP4 equation but also has some different properties, such as soliton molecular solutions and D'Alembert solutions, which deserve to be discussed and studied.In section 2 of this paper, the equivalence between the special form and general form of solutions is proved, meanwhile, we get the Lie point group of the CKP34 equation by a direct method. In section 3 , we introduce point symmetry and the Kac-Moody–Virasoro symmetry algebra. In section 4 , we apply the standard Lie point symmetry method to the CKP34 equation. Finally, section 5 is a summary and discussion.
2. Finite symmetry group of the CKP34 equation via a direct method.
The most general finite symmetry transformation of the CKP34 equation read
$\begin{eqnarray}\begin{array}{rcl}u & = & W\left(x,y,t,U(\xi ,\eta ,\tau ),V(\xi ,\eta ,\tau )\right),\\ v & = & Z\left(x,y,t,U(\xi ,\eta ,\tau ),V(\xi ,\eta ,\tau )\right),\end{array}\end{eqnarray}$
where U and V satisfy the same CKP34 equation $\begin{eqnarray*}\begin{array}{rcl}{U}_{\xi \tau } & = & a[{\left(6{{UU}}_{\xi }+{U}_{\xi \xi \xi }\right)}_{\xi }-3{U}_{\eta \eta }]\\ & & +b{\left(2{{VU}}_{\xi }+{V}_{\xi \xi \xi }+4{{UU}}_{\eta }\right)}_{\xi }-{V}_{\eta \eta },\\ {U}_{\eta } & = & {V}_{\xi }.\end{array}\end{eqnarray*}$
Substituting (3 ) into (1 ) and (2 ) and requiring U(ξ, η, τ) and V(ξ, η, τ) also can be the solutions of the CKP34 equation but with different independent variables (eliminating Uξτ and its higher order derivatives by means of the CKP34 equation), we have4 ) and (5 ) are true for an arbitrary solution U and V for all coefficients of the polynomials of the derivatives of U and V being zero.
$\begin{eqnarray}\begin{array}{l}4{U}_{{\xi }^{10}}{\tau }_{x}^{3}{\xi }_{x}({W}_{U}{a}^{4}+{Z}_{U}{a}^{3}b)+G(x,y,t,U,{U}_{\xi },\ldots ,\\ {U}_{{\xi }^{9}},{V}_{\xi },\ldots ,{V}_{{\xi }^{9}},{U}_{\eta },{V}_{\eta }...)=0,\end{array}\end{eqnarray}$
$\begin{eqnarray}\begin{array}{l}4{V}_{{\xi }^{10}}{\tau }_{x}^{3}{\xi }_{x}({W}_{U}{a}^{3}b+{Z}_{U}{a}^{2}{b}^{2})+G(x,y,t,U,{U}_{\xi },\ldots ,\\ {U}_{{\xi }^{9}},{V}_{\xi },\ldots ,{V}_{{\xi }^{9}},{U}_{\eta },{V}_{\eta }...)=0,\end{array}\end{eqnarray}$
where ${U}_{{\xi }^{n}}$ ≡ $\tfrac{{\partial }^{n}U}{\partial {\xi }^{n}}$, ${V}_{{\xi }^{n}}$ ≡ $\tfrac{{\partial }^{n}V}{\partial {\xi }^{n}}$, G is a complicated ${U}_{{\xi }^{10}}$, ${V}_{{\xi }^{10}}$ (and Uξτ, Vξτ and the higher order derivatives) independent function. Equations (Obviously, WU and ZU cannot be zero, and if ξx = 0, there are no nontrivial solutions, so causing the coefficients of ${U}_{{\xi }^{10}}$ and ${V}_{{\xi }^{10}}$ to vanish, the only possible case is6 ), (4 ) and (5 ) can be reduced to
$\begin{eqnarray}{\tau }_{x}=0,\qquad \tau \equiv \tau (y,t),\end{eqnarray}$
using conditions ( $\begin{eqnarray}\begin{array}{l}(-{{bZ}}_{U}-3{W}_{U}a){\tau }_{y}^{2}{U}_{{\tau }^{2}}+({{bZ}}_{U}+{{aW}}_{U}){\eta }_{x}^{4}{U}_{{\eta }^{4}}\\ \quad +\,G(x,y,t,U,\ldots ,{U}_{{\xi }^{4}},{U}_{\eta },{U}_{\tau },V,\ldots )=0,\end{array}\end{eqnarray}$
$\begin{eqnarray}\begin{array}{l}(-{{bZ}}_{V}-3{W}_{U}a){\tau }_{y}^{2}{V}_{{\tau }^{2}}+({{bZ}}_{V}+{{aW}}_{V}){\eta }_{x}^{4}{V}_{{\eta }^{4}}\\ \quad +\,G^{\prime} (x,y,t,U,\ldots ,{V}_{{\xi }^{4}},{V}_{\eta },{V}_{\tau },U,\ldots )=0,\end{array}\end{eqnarray}$
vanishing the coefficients of ${U}_{\tau \tau },{U}_{{\eta }^{4}},{V}_{\tau \tau }$ and ${V}_{{\eta }^{4}}$ yields $\begin{eqnarray}{\tau }_{y}={\eta }_{x}=0,\qquad \tau \equiv \tau (t),\qquad \eta \equiv \eta (y,t).\end{eqnarray}$
Equivalent to ξx, we know that ηy ≠ 0 and τt ≠ 0, next, we consider the coefficients of polynomials and get7 ) and (8 ) can be further simplified to3 ).
$\begin{eqnarray*}(3{{bZ}}_{{UU}}{\xi }_{x}^{4}+3{{aW}}_{{UU}}{\xi }_{x}^{4}){U}_{\xi \xi }^{2}=(3{{aW}}_{{VV}}{\xi }_{x}^{4}+3{{bZ}}_{{VV}}{\xi }_{x}^{4}){V}_{\xi \xi }^{2}=0,\end{eqnarray*}$
$\begin{eqnarray*}{W}_{{UU}}={W}_{{VV}}={Z}_{{UU}}={Z}_{{VV}}=0,\end{eqnarray*}$
and $\begin{eqnarray*}-{W}_{V}{\xi }_{x}{\tau }_{t}{V}_{\xi \tau }=0,\qquad {W}_{V}=0,\end{eqnarray*}$
under the above conditions, ( $\begin{eqnarray*}\begin{array}{l}A(x,y,t,U,{U}_{\xi },{U}_{{\xi }^{3}},{U}_{{\xi }^{4}},{U}_{\eta },{U}_{\eta \eta },\ldots )\\ \quad +\,B(x,y,t,U,{U}_{\xi }){U}_{\xi \xi }=0,\\ A^{\prime} (x,y,t,U,{U}_{\xi },{U}_{{\xi }^{3}},{U}_{{\xi }^{4}},V,{V}_{\xi },\ldots )\\ \quad +\,B^{\prime} (x,y,t,V,{V}_{\xi }){V}_{\xi \xi }+C^{\prime} (x,y,t,U,{U}_{\xi }){U}_{\xi \xi }=0,\end{array}\end{eqnarray*}$
then we have come to the conclusion that the form of $\begin{eqnarray}\begin{array}{rcl}u & = & {\alpha }_{1}+{\beta }_{1}U\left(\xi (x,y,t),\eta (y,t),\tau (t)\right),\\ v & = & {\alpha }_{2}+{\beta }_{2}V\left(\xi (x,y,t),\eta (y,t),\tau (t)\right)\\ & & +{\beta }_{3}U\left(\xi (x,y,t),\eta (y,t),\tau (t)\right),\end{array}\end{eqnarray}$
can be equivalent to the general form (By substituting (10 ) into (1 ), we get a complex equation, so it is necessary to find some simple conditions, by combining (2 ) and (10 ), we get
$\begin{eqnarray*}\begin{array}{l}{\beta }_{2}{\xi }_{x}{V}_{\xi }-{\beta }_{1}{\eta }_{y}{U}_{\eta }+({\beta }_{3}{\xi }_{x}-{\beta }_{1}{\xi }_{y}){U}_{\xi }\\ \quad +\,({\beta }_{3x}-{\beta }_{1y})U+{\beta }_{2x}V+{\alpha }_{2x}-{\alpha }_{1y}=0,\end{array}\end{eqnarray*}$
and we know Uη = Vξ, so it is obvious that $\begin{eqnarray}\begin{array}{rcl}{\beta }_{3x} -{\beta }_{1y}=0,\quad {\alpha }_{2x}-{\alpha }_{1y}=0,\quad {\beta }_{2x}=0,\\ {\beta }_{2}{\xi }_{x} -{\beta }_{1}{\eta }_{y}=0,\qquad \qquad {\beta }_{3}{\xi }_{x}-{\beta }_{1}{\xi }_{y}=0.\end{array}\end{eqnarray}$
Now we need to search for some special coefficients in the result of combining (1 ) and (10 ) by the above restrictions, our attention should be focused on the coefficients of polynomials read11 ), (12 ) and analyze different variables in equations, we get11 ), so we know
$\begin{eqnarray*}{\xi }_{{xx}}{\xi }_{x}^{2}{V}_{\xi }{U}_{\xi \xi }=0,\quad 4b{\xi }_{x}^{3}{\beta }_{3x}{U}_{\xi \xi \xi }=0,\end{eqnarray*}$
then we get $\begin{eqnarray}{\xi }_{{xx}}={\beta }_{3x}=0,\end{eqnarray}$
according to ( $\begin{eqnarray*}{\beta }_{1x}={\beta }_{2x}={\beta }_{3x}={\beta }_{1y}=0,\end{eqnarray*}$
assuming ξy is a function of y and t, ξy is a function of x and t, we will find contradictions on both sides of the equation in ( $\begin{eqnarray*}{\beta }_{2y}={\beta }_{3y}={\eta }_{{yy}}=0,\end{eqnarray*}$
under the above conditions, the coefficient of Uη and Vξ read $\begin{eqnarray*}4{\beta }_{1}b{\eta }_{y}{\alpha }_{1x}{U}_{\eta },\qquad {\alpha }_{1x}=0,\end{eqnarray*}$
and $\begin{eqnarray*}-b{\beta }_{2}{\xi }_{{yy}}{V}_{\xi },\qquad {\xi }_{{yy}}=0,\end{eqnarray*}$
therefore, the equation becomes $\begin{eqnarray}\begin{array}{l}{\xi }_{x}({\xi }_{x}^{3}{\beta }_{1}a+{\xi }_{x}^{3}{\beta }_{3}b-{\beta }_{1}a{\tau }_{t}){U}_{\xi \xi \xi \xi }+{\xi }_{x}b({\xi }_{x}^{3}{\beta }_{2}\\ \quad -\,{\beta }_{1}{\tau }_{t}){V}_{\xi \xi \xi \xi }+b({\xi }_{x}{\beta }_{1}{\tau }_{t}-{\eta }_{y}^{2}{\beta }_{2}){V}_{\eta \eta }\\ (4{\beta }_{1}{\alpha }_{1}b{\xi }_{x}{\eta }_{y}-6{\beta }_{1}a{\xi }_{y}{\eta }_{y}-2b{\beta }_{3}{\xi }_{y}{\eta }_{y}-{\beta }_{1}{\xi }_{x}{\tau }_{t}\\ \quad -\,{\beta }_{2}b{\xi }_{y}^{2}){U}_{\xi \eta }+(-2{\beta }_{2}b{\eta }_{y}{\xi }_{y}+3{\beta }_{1}\\ a{\xi }_{x}{\tau }_{t}-3{\beta }_{1}a{\eta }_{y}^{2}-{\beta }_{3}b{\eta }_{y}^{2}){U}_{\eta \eta }+4{\beta }_{1}{\xi }_{x}b({\beta }_{1}{\eta }_{y}-{\tau }_{t}){{UU}}_{\xi \eta }\\ \quad +\,(6{\beta }_{1}{\alpha }_{1}a{\xi }_{x}^{2}+2{\beta }_{1}{\alpha }_{2}b{\xi }_{x}^{2}\\ +\,4{\beta }_{1}{\alpha }_{1}b{\xi }_{y}-3{\beta }_{1}a{\xi }_{y}^{2}-{\beta }_{3}b{\xi }_{y}^{2}-{\beta }_{1}{\xi }_{x}{\xi }_{t}){U}_{\xi \xi }\\ \quad +\,2{\beta }_{1}b{\xi }_{x}({\beta }_{2}{\xi }_{x}-{\tau }_{t}){{VU}}_{\xi \xi }+2{\beta }_{1}{\xi }_{x}(3\\ {\beta }_{1}a{\xi }_{x}+{\beta }_{3}b{\xi }_{x}+2{\beta }_{1}b{\xi }_{y}-3a{\tau }_{t}){{UU}}_{\xi \xi }\\ \quad +\,(2b{\beta }_{1}{\alpha }_{2x}{\xi }_{x}+4{\beta }_{1}b{\alpha }_{1y}{\xi }_{x}+4{\alpha }_{1}{\beta }_{1}b{\xi }_{{xy}}-\\ {\beta }_{1}{\xi }_{{tx}}-{\beta }_{1t}{\xi }_{x}){U}_{\xi }+2{\beta }_{1}{\xi }_{x}b({\beta }_{2}{\xi }_{x}-{\tau }_{t}){U}_{\xi }{V}_{\xi }\\ \quad +\,2{\beta }_{1}{\xi }_{x}(3{\beta }_{1}a{\xi }_{x}+{\beta }_{3}b{\xi }_{x}+2b{\beta }_{1}{\xi }_{y}-\\ 3a{\tau }_{t}){U}_{\xi }^{2}\,+\,4{\beta }_{1}{\xi }_{x}b({\beta }_{1}{\eta }_{y}-{\tau }_{t}){U}_{\xi }{U}_{\eta }+4b{\beta }_{1}^{2}{\xi }_{{xy}}{{UU}}_{\xi }\\ \quad +\,b{\alpha }_{2{xxxx}}-b{\alpha }_{2{yy}}-3a{\alpha }_{1{yy}}=0.\end{array}\end{eqnarray}$
The polynomial coefficients in (13 ) related to U and V are zero and the results read
$\begin{eqnarray}\begin{array}{rcl}u & = & \delta \displaystyle \frac{{\tau }_{{tt}}}{8b{\tau }_{t}}y+\displaystyle \frac{1}{4b\sqrt{{\tau }_{t}}}{y}_{0t}+\displaystyle \frac{3{a}^{2}}{4{b}^{2}}(\sqrt{{\tau }_{t}}-1)\\ & & +{C}_{1}^{2}\sqrt{{\tau }_{t}}U(\xi ,\eta ,\tau ),\end{array}\end{eqnarray}$
$\begin{eqnarray}\begin{array}{rcl}v & = & \displaystyle \frac{{\tau }_{{tt}}({bx}-2{{aC}}_{1}y)}{8{b}^{2}{\tau }_{t}}+\displaystyle \frac{{a}^{3}}{4{b}^{3}}{C}_{1}(7-3\sqrt{{\tau }_{t}}-4{\tau }_{t}^{\tfrac{3}{4}})\\ & & +\displaystyle \frac{{{bx}}_{0t}-{{ay}}_{0t}}{2{b}^{2}{\tau }_{t}^{\tfrac{1}{4}}}-\displaystyle \frac{{{ay}}_{0t}}{4{b}^{2}\sqrt{{\tau }_{t}}}\\ & & +{C}_{1}^{2}{\tau }_{t}^{\tfrac{3}{4}}V(\xi ,\eta ,\tau )+\displaystyle \frac{a\sqrt{{\tau }_{t}}}{b}({\tau }_{t}^{\tfrac{1}{4}}-1){C}_{1}^{2}U(\xi ,\eta ,\tau ),\end{array}\end{eqnarray}$
$\begin{eqnarray}\begin{array}{rcl}\xi & = & {C}_{1}{\tau }_{t}^{\tfrac{1}{4}}x-{C}_{1}\displaystyle \frac{a({\tau }_{t}^{\tfrac{1}{4}}-{\tau }_{t}^{\tfrac{1}{2}})}{b}y+{x}_{0}(t),\\ \eta & = & {C}_{1}{\tau }_{t}^{\tfrac{1}{2}}y+{y}_{0}(t),\quad \tau =\tau (t),\end{array}\end{eqnarray}$
the constants δ and C1 possess discrete values determined by $\begin{eqnarray*}\delta =\pm 1,\qquad {C}_{1}=1,\qquad -(1\pm {\rm{i}}\sqrt{3}),\qquad ({\rm{i}}=\sqrt{-1}),\end{eqnarray*}$
in summary, the following theorem reads: If $U=U(x,y,t)$ and $V=V(x,y,t)$ are the solutions of the CKP34 equation, then so is {u, v} with (
According to the symmetry group theorem, we find that for the CKP34 equation, the symmetry group is divided into two sectors: the Lie point symmetry group which reads
$\begin{eqnarray*}\delta ={C}_{1}=1,\end{eqnarray*}$
and a coset of the Lie point group which is related to $\begin{eqnarray*}\delta =-1,\qquad {C}_{1}=1,\end{eqnarray*}$
the coset is equivalent to the reflected transformation of y: y → −y.If we denote by S the Lie point symmetry group of the CKP34 equation, by σy the reflection of y, I is the identity transformation and C2 ≡ {I, σy} is the discrete reflection group, the full Lie symmetry group ${{\mathscr{G}}}_{\mathrm{RCKP}}$ can be expressed as
$\begin{eqnarray*}{{\mathscr{G}}}_{\mathrm{RCKP}}={C}_{2}\otimes S.\end{eqnarray*}$
For the complex cKP3-4 equations, the symmetry group can be divided into six sectors read
$\begin{eqnarray*}\begin{array}{rcl}\delta & = & 1,\qquad \quad {C}_{1}=1,\\ \delta & = & 1,\qquad \quad {C}_{1}={\rm{i}}\sqrt{3}-1,\\ \delta & = & 1,\qquad \quad {C}_{1}={\rm{i}}\sqrt{3}+1,\\ \delta & = & -1,\qquad {C}_{1}=1,\\ \delta & = & -1,\qquad {C}_{1}={\rm{i}}\sqrt{3}-1,\\ \delta & = & -1,\qquad {C}_{1}={\rm{i}}\sqrt{3}+1.\end{array}\end{eqnarray*}$
The full symmetry groups, ${{\mathscr{G}}}_{\mathrm{CCKP}}$, for the complex CKP34 equation are the product of the usual Lie point symmetry group S and the discrete group ${{ \mathcal D }}_{3}$,
$\begin{eqnarray*}{{\mathscr{G}}}_{\mathrm{CCKP}}={{ \mathcal D }}_{3}\otimes S,\end{eqnarray*}$
$\begin{eqnarray*}{{ \mathcal D }}_{3}\equiv \{I,{\sigma }^{y},{R}_{1},{R}_{2},{\sigma }^{y}{R}_{1},{\sigma }^{y}{R}_{2}\},\end{eqnarray*}$
where I is the identity transformation, σy is the reflection of y and $\begin{eqnarray*}{R}_{1}:u(x,y,t)\to ({\rm{i}}\sqrt{3}+1)u\left(({\rm{i}}\sqrt{3}-1)x,-(1+{\rm{i}}\sqrt{3})y,t\right),\end{eqnarray*}$
$\begin{eqnarray*}{R}_{2}:u(x,y,t)\to ({\rm{i}}\sqrt{3}-1)u\left(-({\rm{i}}\sqrt{3}+1)x,({\rm{i}}\sqrt{3}-1)y,t\right),\end{eqnarray*}$
and v can be parallel to u.3. Point symmetries and the Kac-Moody–Virasoro symmetry algebra
From the traditional method, we can simply take the arbitrary functions x0, y0 and τ in the forms17 ) into (14 ) and (15 ) and using small parameter expansion method with respect to ε yields
$\begin{eqnarray}\begin{array}{rcl}{x}_{o} & = & \epsilon \alpha (t),\qquad {y}_{0}=\epsilon \beta (t),\\ \tau & = & t+\epsilon \theta (t),\qquad \delta ={C}_{1}=1,\end{array}\end{eqnarray}$
where ε is an infinitesimal parameter. By substituting ( $\begin{eqnarray}\begin{array}{l}\left(\begin{array}{c}u\\ v\end{array}\right)\,=\,\left(\begin{array}{c}U\\ V\end{array}\right)\,+\,\epsilon [{K}_{2}(\theta )+{K}_{0}(\alpha )\\ \quad +\,{K}_{1}(\beta )]+O({\epsilon }^{2}),\end{array}\end{eqnarray}$
and $\begin{eqnarray}\begin{array}{l}\left(\displaystyle \frac{{bx}+3{ay}}{4b}\epsilon {\theta }_{t}+\epsilon \alpha \right){U}_{x}+\left(\displaystyle \frac{\epsilon {\theta }_{t}}{2}y+\epsilon \beta \right){U}_{y}+\epsilon \theta {U}_{t}\\ \quad +\,\left(\displaystyle \frac{\epsilon {\theta }_{t}}{2}+1\right)U+\displaystyle \frac{\epsilon {\theta }_{{tt}}}{8b}y+\displaystyle \frac{\epsilon {\beta }_{t}}{4b}+\displaystyle \frac{3{a}^{2}}{8{b}^{2}}\epsilon {\theta }_{t}=0,\end{array}\end{eqnarray}$
$\begin{eqnarray}\begin{array}{l}\left(\displaystyle \frac{{bx}+3{ay}}{4b}\epsilon {\theta }_{t}+\epsilon \alpha \right){V}_{x}+\left(\displaystyle \frac{\epsilon {\theta }_{t}}{2}y+\epsilon \beta \right){V}_{y}+\epsilon \theta {V}_{t}\\ \quad +\,\left(\displaystyle \frac{3\epsilon {\theta }_{t}}{4}+1\right)V+\displaystyle \frac{3a\epsilon {\theta }_{t}}{4b}U-\displaystyle \frac{\epsilon }{2b}{\alpha }_{t}-\displaystyle \frac{3a\epsilon {\beta }_{t}}{4{b}^{2}}\\ \quad +\,\displaystyle \frac{{bx}-2{ay}}{8{b}^{2}}\epsilon {\theta }_{{tt}}-\displaystyle \frac{9{a}^{3}}{8{b}^{3}}\epsilon {\theta }_{t}=0.\end{array}\end{eqnarray}$
Comparing (18 ), (19 ) and (20 ), we get three symmetry generators,
$\begin{eqnarray}\left(\begin{array}{c}{\sigma }^{u}\\ {\sigma }^{v}\end{array}\right)={K}_{0}(\alpha )=\left(\begin{array}{c}\alpha {u}_{x}\\ \alpha {v}_{x}+\displaystyle \frac{1}{2b}{\alpha }_{t}\end{array}\right),\end{eqnarray}$
$\begin{eqnarray}\left(\begin{array}{c}{\sigma }^{u}\\ {\sigma }^{v}\end{array}\right)={K}_{1}(\beta )=\left(\begin{array}{c}\beta {u}_{y}+\displaystyle \frac{1}{4b}{\beta }_{t}\\ \beta {v}_{y}-\displaystyle \frac{3a}{4{b}^{2}}{\beta }_{t}\end{array}\right),\end{eqnarray}$
and $\begin{eqnarray}{K}_{2}(\theta )=\left(\begin{array}{c}\theta {u}_{t}+\displaystyle \frac{{\theta }_{t}}{4b}({ay}+{bx}){u}_{x}+\displaystyle \frac{1}{2}{\theta }_{t}{\left({yu}\right)}_{y}+\displaystyle \frac{3{a}^{2}}{8{b}^{2}}{\theta }_{t}+\displaystyle \frac{y}{8b}{\theta }_{{tt}}\\ \theta {v}_{t}+\displaystyle \frac{{\theta }_{t}}{4b}({ay}+{bx}){v}_{x}+\displaystyle \frac{1}{2}{\theta }_{t}{\left({yv}\right)}_{y}+\displaystyle \frac{{au}+{bv}}{4b}{\theta }_{t}+\displaystyle \frac{{bx}-2{ay}}{8{b}^{2}}{\theta }_{{tt}}-\displaystyle \frac{9{a}^{3}}{8{b}^{3}}{\theta }_{t}\end{array}\right).\end{eqnarray}$
The symmetries K0(α), K1(β) and K0(θ) establish a generalized KMV(Kac-Moody–Virasoro) algebra with the non-zero commutators
$\begin{eqnarray}\begin{array}{rcl}\left[{K}_{2}(\theta ),{K}_{0}(\alpha )\right] & = & {K}_{0}(\theta {\alpha }_{t}),[{K}_{2}(\theta ),{K}_{1}(\beta )]={K}_{1}(\theta {\beta }_{t}),\\ \left[{K}_{2}({\theta }_{1}),{K}_{2}({\theta }_{2})\right] & = & {K}_{2}({\theta }_{1}{\theta }_{2t}-{\theta }_{1t}{\theta }_{2}),\end{array}\end{eqnarray}$
the above commutator [F, G] with $F={\left({F}_{1}(u,v),{F}_{2}(u,v)\right)}^{{\rm{T}}}$ and $G={\left({G}_{1}(u,v),{G}_{2}(u,v)\right)}^{T}$, here the superscript T means the transposition of the matrix, which is defined by $\begin{eqnarray*}[F,G]\equiv \left(\begin{array}{cc}{F}_{1u}^{{\prime} } & {F}_{1v}^{{\prime} }\\ {F}_{2u}^{{\prime} } & {F}_{2v}^{{\prime} }\end{array}\right)G-\left(\begin{array}{cc}{G}_{1u}^{{\prime} } & {G}_{1v}^{{\prime} }\\ {G}_{2u}^{{\prime} } & {G}_{2v}^{{\prime} }\end{array}\right)F,\end{eqnarray*}$
and ${F}_{1u}^{{\prime} },{F}_{1v}^{{\prime} },{F}_{2u}^{{\prime} },{F}_{2v}^{{\prime} },{G}_{1u}^{{\prime} },{G}_{1v}^{{\prime} },{G}_{2u}^{{\prime} }$ and ${G}_{2v}^{{\prime} }$ are partial linearized operators, so $\begin{eqnarray*}{F}_{1u}^{{\prime} }{G}_{1}\equiv {\left.\displaystyle \frac{{\rm{d}}}{{\rm{d}}\epsilon }{F}_{1}(u+\epsilon {G}_{1},v)\right|}_{\epsilon =0}.\end{eqnarray*}$
From (24 ), we know that the Kac-Moody algebra was constituted by K0 and K1Λ, K2 constitutes the Virasoro algebra. When we fix the arbitrary functions α, β and θ as special exponential functions or polynomial functions tm for m = 0, ± 1, ± 2,…, the generalized KMV algebra is reduced to the usual one.
Through the above discussion, we find the series symmetry approach can be equivalent to the Lie point symmetry method to the CKP34 equation and we can achieve three generators, which can be helpful for our following study.
4. Symmetry reductions of the CKP34 equation
In this section, we focus on the symmetry invariant solutions of the CKP34 equation related to the symmetries generated by the following three generators,
$\begin{eqnarray}\left(\begin{array}{c}{\sigma }^{u}\\ {\sigma }^{v}\end{array}\right)={K}_{0}(\alpha )=\left(\begin{array}{c}\alpha {u}_{x}\\ \alpha {v}_{x}+\displaystyle \frac{1}{2b}{\alpha }_{t}\end{array}\right),\end{eqnarray}$
$\begin{eqnarray}\left(\begin{array}{c}{\sigma }^{u}\\ {\sigma }^{v}\end{array}\right)={K}_{1}(\beta )=\left(\begin{array}{c}\beta {u}_{y}+\displaystyle \frac{1}{4b}{\beta }_{t}\\ \beta {v}_{y}-\displaystyle \frac{3a}{4{b}^{2}}{\beta }_{t}\end{array}\right),\end{eqnarray}$
and $\begin{eqnarray}{K}_{2}(\theta )=\left(\begin{array}{c}\theta {u}_{t}+\displaystyle \frac{{\theta }_{t}}{4b}({ay}+{bx}){u}_{x}+\displaystyle \frac{1}{2}{\theta }_{t}{\left({yu}\right)}_{y}+\displaystyle \frac{3{a}^{2}}{8{b}^{2}}{\theta }_{t}+\displaystyle \frac{y}{8b}{\theta }_{{tt}}\\ \theta {v}_{t}+\displaystyle \frac{{\theta }_{t}}{4b}({ay}+{bx}){v}_{x}+\displaystyle \frac{1}{2}{\theta }_{t}{\left({yv}\right)}_{y}+\displaystyle \frac{{au}+{bv}}{4b}{\theta }_{t}+\displaystyle \frac{{bx}-2{ay}}{8{b}^{2}}{\theta }_{{tt}}-\displaystyle \frac{9{a}^{3}}{8{b}^{3}}{\theta }_{t}\end{array}\right),\end{eqnarray}$
where α, β and θ are arbitrary functions of t.Using the Lie point symmetries K0(α), K1(β) and K0(θ) to the cKP3-4 equation (1 ), we can get two nontrivial symmetry reductions.
Reduction I: θ ≠ 0. For θ ≠ 0, we replace the arbitrary functions in the form:28 ), the group invariant condition becomes28 ) into (29 ) we get28 ), (29 ), (32 ) and vanishing the coefficients of Uξ and Uη, we get
$\begin{eqnarray}\begin{array}{rcl}\theta & \equiv & {\rho }^{4}\ne 0,\qquad \alpha \equiv {\rho }^{5}{\alpha }_{1t},\\ \beta & \equiv & {\rho }^{6}{\beta }_{1t},\qquad {\theta }_{1t}\equiv \rho {\beta }_{1t},\end{array}\end{eqnarray}$
with the new definitions ( $\begin{eqnarray}{K}_{0}(\alpha )+{K}_{1}(\beta )+{K}_{2}(\theta )=0,\end{eqnarray}$
because the process of solving v can be parallel to u, so we only solve u here. Substituting ( $\begin{eqnarray}\begin{array}{l}{\rho }^{3}{\alpha }_{1t}{u}_{x}+{\rho }^{4}{\beta }_{1t}{u}_{y}+\displaystyle \frac{1}{4b}(6{\rho }^{3}{\beta }_{1t}+{\rho }^{4}{\beta }_{1{tt}})\\ \quad +\,{\rho }^{2}{u}_{t}+\displaystyle \frac{\rho {\rho }_{t}}{b}({ay}+{bx}){u}_{x}\\ \quad +\,2\rho {\rho }_{t}(u+{{yu}}_{y})\displaystyle \frac{3{a}^{2}}{2{b}^{2}}\rho {\rho }_{t}+\displaystyle \frac{y}{2b}(3{\rho }_{t}+\rho {\rho }_{{tt}})=0,\end{array}\end{eqnarray}$
here we give some equivalent transformations: $\begin{eqnarray}\begin{array}{rcl}x & \to & \xi (x,y,t),\quad y\to \eta (y,t),\\ {u}_{x} & \to & {U}_{\xi }{\xi }_{x},\qquad {u}_{y}\to {U}_{\xi }{\xi }_{y}+{U}_{\eta }{\eta }_{y},\\ {v}_{x} & \to & {V}_{\xi }{\xi }_{x},\qquad {v}_{y}\to {V}_{\xi }{\xi }_{y}+{V}_{\eta }{\eta }_{y},\end{array}\end{eqnarray}$
and setting $\begin{eqnarray}\xi ={A}_{1}x+{B}_{1}y+{C}_{1},\qquad \eta ={B}_{2}y+{C}_{2},\end{eqnarray}$
where U(ξ, η) ≡ U and V(ξ, η) ≡ V are invariant functions of the group with invariant variables ξ and η, combining ( $\begin{eqnarray}\xi =\displaystyle \frac{x}{\rho }-\displaystyle \frac{{ay}}{b\rho }-{\alpha }_{1}+\displaystyle \frac{a{\theta }_{1}}{b},\quad \eta =\displaystyle \frac{y}{{\rho }^{2}}-{\beta }_{1},\end{eqnarray}$
so the first type of group invariant solutions become $\begin{eqnarray}u=\displaystyle \frac{U(\xi ,\eta )}{{\rho }^{2}}-\displaystyle \frac{{\rho }_{t}y}{2b\rho }-\displaystyle \frac{3{a}^{2}}{4{b}^{2}}-\displaystyle \frac{{\beta }_{1t}\rho }{4b},\end{eqnarray}$
$\begin{eqnarray}\begin{array}{rcl}v & = & \displaystyle \frac{V(\xi ,\eta )}{{\rho }^{3}}-\displaystyle \frac{{aU}(\xi ,\eta )}{b{\rho }^{2}}+\displaystyle \frac{(2{ay}-{bx}){\rho }_{t}}{2{b}^{2}\rho }\\ & & +\displaystyle \frac{7{a}^{3}}{4{b}^{3}}-\displaystyle \frac{\rho {\alpha }_{1t}}{2b}+\displaystyle \frac{3a\rho {\beta }_{1t}}{4{b}^{2}}.\end{array}\end{eqnarray}$
By substituting (34 ) and (35 ) into (1 ) and (2 ), we can achieve the group invariant reduction equations about the group invariant functions U and V,36 ) is Lax integrable with the fourth order spectral problem
$\begin{eqnarray}\begin{array}{rcl}{U}_{\eta } & = & {V}_{\xi },\\ {V}_{\eta \eta } & = & {\left({V}_{\xi \xi \xi }+4{{UV}}_{\xi }+2{{VU}}_{\xi }\right)}_{\xi },\end{array}\end{eqnarray}$
it is worth mentioning that the reduction system ( $\begin{eqnarray}\begin{array}{rcl}\lambda {\rm{\Psi }} & = & 2{{\rm{\Psi }}}_{\xi \xi \xi \xi }+4U{{\rm{\Psi }}}_{\xi \xi }+2(2{U}_{\xi }-{\rm{i}}V){{\rm{\Psi }}}_{\xi }\\ & & -\left(\displaystyle \int {V}_{\eta }{\rm{d}}\xi -2{U}^{2}-2{U}_{\xi \xi }+{\rm{i}}{V}_{\xi }\right){\rm{\Psi }},\end{array}\end{eqnarray}$
$\begin{eqnarray}{{\rm{\Psi }}}_{\eta }={\rm{i}}({{\rm{\Psi }}}_{\xi \xi }+U{\rm{\Psi }}).\end{eqnarray}$
On the other hand, in the reduction system (36 ), we can make the following transformation:36 ) can be equivalent to the Boussinesq equation.
$\begin{eqnarray*}U={w}_{\xi },\qquad V={w}_{\eta },\end{eqnarray*}$
with respect to η, the system (Reduction II: θ = 0, β ≠ 0. In this case, the group invariant condition becomes:
$\begin{eqnarray*}{K}_{0}(\alpha )+{K}_{1}(\beta )=0,\end{eqnarray*}$
and $\begin{eqnarray*}{v}_{x}={u}_{y},\end{eqnarray*}$
then we can get the usual KdV reduction $\begin{eqnarray}\begin{array}{l}{\left({U}_{T}+{U}_{{XXX}}+6{{UU}}_{X}\right)}_{X}=0,X=\displaystyle \frac{1}{\sqrt{\beta }}\left(\displaystyle \frac{\alpha }{\beta }y-x\right),\\ T=\displaystyle \int {\beta }^{-5/2}(a\beta -b\alpha ){\rm{d}}t,\end{array}\end{eqnarray}$
with $\begin{eqnarray}u=\displaystyle \frac{U(X,T)}{\beta }-\displaystyle \frac{{\beta }_{t}y}{4b\beta }+\displaystyle \frac{{\alpha }^{2}(3a\beta -b\alpha )}{6{\beta }^{2}(a\beta -b\alpha )},\end{eqnarray}$
$\begin{eqnarray}\begin{array}{rcl}v & = & \displaystyle \frac{-\alpha U(X,T)}{{\beta }^{2}}-\displaystyle \frac{{\beta }_{t}x}{4b\beta }+\left[\displaystyle \frac{(3a\beta +b\alpha ){\beta }_{t}}{4{b}^{2}{\beta }^{2}}-\displaystyle \frac{{\alpha }_{t}}{2b\beta }\right]y\\ & & -\displaystyle \frac{{\alpha }^{3}(3a\beta -b\alpha )}{6{\beta }^{3}(a\beta -b\alpha )}.\end{array}\end{eqnarray}$
5. Summary and discussion
In this paper, a direct method was applied to find the finite point symmetry groups of the CKP34 equation, and whence the finite Lie point symmetry group is obtained, to find its related Lie symmetry algebra is quite straightforward. However, the simple ansatz (10 ) is not a universal formal for some other (2+1)-dimensional integrable systems. For example, in the Ablowitz–Kaup–Newell–Segur system, in order to find its generalized finite symmetry groups, the form of solutions should be modified to another form mentioned in [13], therefore, it is worthwhile for us to find the more general method.
Another method to find similarity reductions is the CK direct method. For the CKP34 equation, the results of the CK direct method can be equivalent to the results of (36 ) and this process can be divided into two parts; the one case can be proved that U and V are not related to τ, and the other case is that U and V are not related to η. We can substitute1 ) and (2 ), then we can select a standardization coefficient which can be convenient for us to use the method mentioned in [1] to achieve the purpose of taking a special value and simplifying the equations and the standardization coefficient reads: ${\beta }_{2}{\xi }_{x}^{4}.$
$\begin{eqnarray*}u={\alpha }_{1}+{\beta }_{1}U(\xi ,\eta ),\end{eqnarray*}$
and $\begin{eqnarray*}v={\alpha }_{2}+{\beta }_{2}V(\xi ,\eta )+{\beta }_{3}U(\xi ,\eta ),\end{eqnarray*}$
into (During our computational process, we found that the form of the CK direct method seems can be related to the Bäcklund transformation of the CKP34 equation because both of them have the same coefficients and similar forms, which can be worthwhile for further research.
Recently, linear superposition in the general heavenly equation [14] had been found, the approximate method involves constructing Poisson brackets and using hodographtransformation. Whether this method can be applied to find symmetry reductions in different nonlinear systems is intriguing.
By using the Lie approach, we can reduce the KP equation to the Boussinesq equation [15], and then use the nonclassical symmetry reduction method to reduce the Boussinesq equation to the ordinary differential equation [16]. Therefore, the CKP34 equation has many remaining problems for us to do further study.