Some symmetries, similarity solutions and various conservation laws of a type of dispersive water waves

Abstract

We investigate the point symmetries, Lie–Bäcklund symmetries for a type of dispersive water waves. We obtain some Lie transformation groups, various group-invariant solutions, and some similarity solutions. Besides, we produce different formats of conservation laws of the dispersive water waves by using different schemes. Finally, we consider some special solutions of the stationary dispersive water-wave equations.

1 Introduction

The classical dispersiveless long wave equations

$$ u_{t}+uu_{x}+h_{x}, \qquad h_{t}+(uh)_{x}=0 $$

(1)

have a number of dispersive generalizations [1]. Kupershmidt [2] investigated the commuting hierarchy and the Hamiltonian structures of the following generation of (1):

$$ \textstyle\begin{cases} u_{t}=(\frac{1}{2}u^{2}+h+\beta u_{x})_{x}, \\ h_{t}=(uh+\alpha u_{xx}-\beta h_{x})_{x}, \end{cases} $$

(2)

and further turned (2) into the following system:

$$ \textstyle\begin{cases} \bar{u}_{t}=(\frac{1}{2}\bar{u}^{2}+\bar{h}+\mu \bar{u})_{x}, \\ \bar{h}_{t}=(\bar{u}\bar{h}-\mu \bar{h}_{x})_{x},\qquad \mu =\gamma +\beta = \pm \sqrt{(}\alpha ^{2}+\beta ^{2}), \end{cases} $$

(3)

by using the invertible change of variables: $u=\bar{u}$, $h=\bar{h}+ \gamma \bar{u}_{x}$. For $\alpha =\frac{1}{3}$, $\beta =0$, system (2) was given by Broer [1]. For $\beta =0$, in terms of the potential $\varphi :u=\varphi _{x}$, system (2) was derived by Kaup [3], who found its multisoliton solutions. Matveev and Yavor [4] algebro-geometrically found a large class of almost periodic solutions of system (3). In the paper, we want to apply the Lie group analysis method [5] to study the point symmetries and Lie–Bäcklund transformation symmetries of system (2). In fact, many symmetries, similarity reductions, and conservation laws were obtained by Lie group analysis [6,7,8,9,10]. Lou et al. [11, 12] applied the symmetry group method to study some coherent solutions of nonlocal KdV systems and primary branch solutions of a first-order autonomous system. In addition, Ma [13] obtained some new conservation laws of some discrete evolution equations by symmetries and adjoint symmetries. Qu and Ji [14] studied inhomogeneous nonlinear diffusion equations by invariant subspace and conditional Lie–Backlund symmetry methods. It was shown that the equations admit a class of invariant subspaces governed by the nonlinear ordinary differential equations, which is equivalent to a kind of higher-order conditional Lie–Backlund symmetries of the equations. Ji and Qu [15] used the conditional Lie–Backlund symmetry method to study the invariant subspaces of nonlinear diffusion equations with convection and source terms and obtained a complete list of canonical forms for such equations, which admit higher-order conditional Lie–Backlund symmetries and multidimensional invariant subspaces. Ma [16, 17] discussed the conservation laws of differential and discrete equations, respectively. Recent studies by Ma et al. [18,19,20,21,22] also show a remarkable richness of rational function solutions, called lumps, as well as interaction solutions and solutions of other kinds. In addition, the invariant solutions can be formulated from the invariant submanifold method in [23].

Ibragimov and Avdonina [24] applied the Lie group method to propose a new approach for looking for conservation laws and exact solutions to nonlinear self-adjoint differential equations. For a system of m differential equations

$$ F_{\alpha }(x,u,u_{(1)},\ldots ,u_{(s)})=0, \quad \alpha =1,2,\ldots ,m, $$

(4)

where $u_{(1)}=\{u_{i}^{\alpha }\}, \ldots , u_{(s)}=\{u^{\alpha }_{i _{1}\cdots i_{s}}\}$, the steps are as follows.

Step 1: Introducing the adjoint equations of (4) by using the variational derivative:

$$ F^{*}_{\alpha }(x,u,v,u_{(1)},v_{(1)}, \ldots , u_{(s)},v_{(s)})=\frac{ \delta \varphi }{\delta u^{\alpha }}=0, $$

(5)

where $\varphi =\sum_{\beta =1}^{m}v^{\beta }F_{\beta }(x,u,u _{(1)},\ldots , u_{(s)})$.

Step 2: Let

$$ v^{\alpha }=\psi ^{\alpha }(x,u),\quad \alpha =1,2,\ldots ,m, $$

(6)

and require the following relations to be satisfied:

$$ F^{*}_{\alpha }\bigl(x,u,\psi (x,u),\ldots , u_{(s)},\psi _{(s)}\bigr)= \lambda _{\alpha }^{\beta }F_{\beta }(x,u, \ldots ,u_{(s)}). $$

(7)

We say that system (4) is nonlinearly self-adjoint if (7) holds for the solutions of (4), where $\lambda _{\alpha }^{\beta }$ are functions dependent on $x,u,u_{(1)},\ldots$ .

Step 3: Assume that the infinitesimal symmetry of the nonlinear self-adjoint system (4) is given by

$$ X=\xi ^{i}(x,u,u_{(1)},\ldots ) \frac{\partial }{\partial x^{i}}+ \eta ^{\alpha }(x,u,u_{(1)},\ldots) \frac{\partial }{\partial u^{ \alpha }}. $$

(8)

Then a conservation law for the system is presented by

$$ D_{i}\bigl(C^{i}\bigr)=0, $$

(9)

where the components of the conserved vector are the following:

$$\begin{aligned} C^{i} =&W^{\alpha }\biggl[ \frac{\partial \varphi }{\partial u_{i}^{\alpha }}-D_{j} \biggl(\frac{\partial \varphi }{\partial u_{ij}^{\alpha }}\biggr) +D_{j}D_{k}\biggl( \frac{\partial \varphi }{\partial u_{ijk}^{\alpha }}\biggr)-\cdots \biggr] \\ &{}+D_{j}\bigl(W^{\alpha } \bigr)\biggl[\frac{ \partial \varphi }{\partial u_{ij}^{\alpha }}-D_{k}\biggl(\frac{\partial \varphi }{\partial u_{ijk}^{\alpha }}\biggr) + \cdots \biggr]+D_{j}D_{k}\bigl(W^{\alpha }\bigr)\biggl[ \frac{\partial \varphi }{\partial u_{ijk} ^{\alpha }}-\cdots \biggr], \end{aligned}$$

(10)

where $W^{\alpha }=\eta ^{\alpha }-\xi ^{j}u_{j}^{\alpha }$.

By applying (9) and (10) some conservation laws of some nonlinear self-adjoint differential equations were obtained in [5,6,7,8,9]. In addition, Göktas and Hereman [25] proposed a new method for looking for conservation laws of nonlinear differential equations without using the symmetries of differential equations. The explicit steps are as follows:

(1) Consider the form of conservation laws

$$ D_{t}(\rho )+D_{x}(J)=0 $$

(11)

for system (4). Assuming the uniformity in rank in the ith equation, form the linear system

$$ A_{i}=\{r_{i,1},r_{i,2},\ldots , r_{i,\alpha }\} $$

and then gather the $A_{i}$ to form the global linear system $\mathcal{A}=\bigcup_{i=1}^{\alpha }A_{i}$.

(2) Solving for the unknown weights $w(u_{i})$, $w(\partial _{t})$.

(3) Set $\mathcal{V}=\{v_{1},\ldots , v_{Q}\}$ to be the sorted list of all the variables with positive weights, excluding $\partial _{t}$. Form all monomials of rank R or less by taking combinations of the variables in $\mathcal{V}$ and form sets consisting of ordered pairs.

Set $\mathcal{B}_{0}=\{(1;0)\}$. For $q=1, 2,\ldots , Q$, $m=0,1,\ldots , M-1$, where M is the number of pairs in $\mathcal{B}_{q-1}$, form $\mathcal{B}_{q,m}=\bigcup_{i=1}^{p_{q,m}}\{(T_{q,s};W_{q,s})\}$, $T _{q,s}=T_{q-1,m}v_{q}^{s}$, $W_{q,s}=W_{q-1,m}+sw(v_{q})$, $p_{q,m}=[\!\![\frac{R-W _{q-1,m}}{w(v_{q})}]\!\!]$.

Denote $\mathcal{B}_{q}=\bigcup_{m=0}^{M-1}\mathcal{B}_{q,m}$.

(4) Let $\mathcal{G}=\mathcal{B}_{Q}$, which consists of all possible combinations of powers of the variables that produce rank R or less. For each pair $(T_{Q,s};W_{q,s})$ in $\mathcal{G}$, apply $\frac{ \partial ^{l}}{\partial x^{l}}$ to the term $T_{Q,s}$, where $l=R-W_{Q,s}$. Set $\mathcal{H}$ to contain the terms that result from computing the various $\frac{\partial ^{l}}{\partial x^{l}}(T_{Q,s})$.

(5) Removing those terms in $\mathcal{H}$ that can be written as a total derivative with respect to x,or as a derivative up to terms kept previously in the set, we denote such a set by I.

(6) For all terms from I with desired rank R, let

$$ \rho =\sum_{i=1}^{\sigma }c_{i}I(i), $$

(12)

where $I(i)$ is the ith element in I, σ is the number of the terms in I, and $c_{i}$ are constants to be determined later.

(7) Computing $D_{t}(\rho )$ and replacing the terms with derivatives with respect to $\partial _{t}$ by using the original system (4), we denote the result by E. Then act on E by the variational derivatives $\frac{\delta E}{\delta u^{\alpha }}$ ($\alpha =1,\ldots ,m$). Next, solving the system $\frac{\delta E}{\delta u^{\alpha }}=0$ ($\alpha =1, \ldots ,m$), we obtain the relations among the coefficients $c_{i}$. Thus we get the density ρ in (12). Substituting $c_{i}$ into the expression E, we can obtain the resulting fluxes of differential equations.

In the paper, we apply the Lie group analysis method and the above approach to investigate the conservation laws of system (2).

2 Point symmetries, Lie–Bäcklund symmetries, and conservation laws

By using the Lie-group analysis method we easily get the point symmetry of the system (2):

$$ \textstyle\begin{cases} V_{1}=\partial _{t},\qquad V_{2}=\partial _{x},\qquad V_{3}=t\partial _{x}-\partial _{u}, \\ V_{4}=t\partial _{t}+\frac{x}{2}\partial _{x}-\frac{u}{2}\partial _{u}-h \partial _{h}. \end{cases} $$

(13)

Set

$$ \mathcal{L}=p(u_{t}-uu_{x}-h_{x}-\beta u_{xx})+q(h_{t}-u_{x}h-uh_{x}- \alpha u_{xxx}+\beta h_{xx}). $$

Then the adjoint equations of system (2) are given by

$$ \textstyle\begin{cases} \frac{\delta \mathcal{L}}{\delta u}=-p_{t}+p_{x}u-\beta p_{xx}+q_{x}h+ \alpha q_{xxx}=0, \\ \frac{\delta \mathcal{L}}{\delta h}=-q_{t}+p_{x}+uq_{x}+\beta q_{xx}=0, \end{cases} $$

(14)

where u, h are solutions to system (2). System (14) has the solutions

$$ p=h,\qquad q=u. $$

Hence system (2) is strictly self-adjoint. Besides, system (14) has some special solutions for given u and h. For example, when $u=h$, (14) has the solution

$$ p=-c_{1}x+c_{2},\qquad q=c_{1}(x-t)+c_{3}, $$

where $c_{1}$, $c_{2}$, $c_{3}$ are constants. In particular, for $\beta =0$, system (2) reduces to

$$ \bigl(u^{\prime }\bigr)^{2}=u^{2}- \frac{1}{3}u^{3}. $$

(15)

Obviously, (15) is solvable. Since system (2) is nonlinearly self-adjoint, we can look for the conservation laws by using the Lie group method. For system (2), the conservation laws are of the following form:

$$\begin{aligned}& C^{1}=W^{u}\frac{\partial \mathcal{L}}{\partial u_{t}}+W^{h} \frac{ \partial \mathcal{L}}{\partial h_{t}}, \end{aligned}$$

(16)

$$\begin{aligned}& \begin{aligned}[b] C^{2}&=W^{u}\biggl[ \frac{\partial \mathcal{L}}{\partial u_{x}}-D_{x}\biggl(\frac{ \partial \mathcal{L}}{\partial u_{xx}}\biggr)+D_{x}^{2} \biggl(\frac{\partial \mathcal{L}}{\partial u_{xxx}}\biggr)\biggr]+D_{x}\bigl(W^{u} \bigr)\biggl[\frac{\partial \mathcal{L}}{\partial u_{xx}}-D_{x}\biggl(\frac{\partial \mathcal{L}}{ \partial u_{xxx}}\biggr) \biggr] \\ &\quad {}+D_{x}^{2}\bigl(W^{u}\bigr) \frac{\partial \mathcal{L}}{ \partial u_{xxx}}+W^{h}\biggl[\frac{\partial \mathcal{L}}{\partial h_{x}}-D_{x} \biggl(\frac{ \partial \mathcal{L}}{\partial h_{xx}}\biggr)\biggr]+D_{x}\bigl(W^{h} \bigr)\frac{\partial \mathcal{L}}{\partial h_{xx}}. \end{aligned} \end{aligned}$$

(17)

For the vector $V_{3}$, the conservation law is

$$ D_{t}\bigl(C^{1}\bigr)+D_{x} \bigl(C^{2}\bigr)=0. $$

(18)

Note that $W^{u}=-1-tu_{t}$ and $W^{h}=-th_{t}$. Then we have

$$\begin{aligned}& C^{1}=(1+tu_{t})p+th_{t}q=h+D_{t}(thu)-uh, \\& \begin{aligned} C^{2}&=(1+tu_{t}) (2hu-\beta h_{x}+\beta u_{x}+\alpha u_{xxx})-tu_{xt}(- \beta h+\beta u+\alpha u)+\alpha tu_{xxt}u \\ &\quad {}+th_{t}\bigl(h+u^{2}+\beta u_{x}\bigr)- \beta th_{xt}u. \end{aligned} \end{aligned}$$

To cancel the trivial operation in computing conservation laws, we may assume that

$$ C^{1}|_{(2)}=\tilde{C}^{1}+D_{x} \bigl(H^{2}\bigr)+D_{z}\bigl(H^{3}\bigr)+\cdots . $$

(19)

Then the conserved vector $C=(X^{1},C^{2},\ldots , c^{m})=0$ can be written as

$$ \tilde{C}=\bigl(\tilde{C}^{1},\ldots , \tilde{C}^{m}\bigr)=0 $$

(20)

with the components

$$ \tilde{C}^{1},\qquad \tilde{C}^{2}=C^{2}+D_{1} \bigl(H^{2}\bigr),\qquad \ldots ,\qquad \tilde{C} ^{m}=C^{m}+D_{1} \bigl(H^{m}\bigr). $$

(21)

Based on versions (18)–(21), we get the reduced forms of the components of the conserved density:

$$\begin{aligned}& \tilde{C}^{1}=h-uh, \\& \begin{aligned} \tilde{C}^{2}&=C^{2}+D_{t}(thu) \\ &=(1+tu_{t}) (2hu-\beta h_{x}+\beta u _{x}+ \alpha u_{xxx})-tu_{xt}(-\beta h+\beta u+\alpha u)+\alpha tu_{xxt}u \\ &\quad {}+th _{t}\bigl(h+u^{2}+\beta u_{x} \bigr)-\beta th_{xt}u. \end{aligned} \end{aligned}$$

In particular, when $u=h$, $p=-c_{1}x+c_{2}$, and $q=c_{1}(x-t)+c_{3}$, we can obtain the special components of the conserved density:

$$\begin{aligned}& \bar{C}^{1}=-c_{1}x+c_{2}+tu_{t}(-c_{1}t+c_{2}+c_{3}), \\& \begin{aligned} \bar{C}^{2}&=(1+tu_{t}) \bigl[(-c_{1}t+c_{2}+c_{3})u+2c_{1} \beta \bigr]-tu_{xt}\bigl[(2c _{1}x-c_{1}t+c_{3}-c_{2}) \beta \\ &\quad {}+\alpha c_{1}(x-t)+\alpha c_{3}\bigr]+ \alpha tu_{xxt}\bigl[c_{1}(x-t)+c_{3}\bigr] \\ &\quad {}+tu_{t}\bigl[-c_{1}x+c_{2}+ \bigl(c_{1}(x-t)+c _{3}\bigr)u+\beta c_{1}\bigr]- \beta tu_{xt}\bigl[c_{1}(x-t)+c_{3}\bigr]. \end{aligned} \end{aligned}$$

For the symmetry vector $V_{4}$, we have that

$$ W^{u}=-\frac{1}{2}u-tu_{t}-\frac{1}{2}xu_{x}, \qquad W^{h}=-h-th_{t}-\frac{1}{2}xh_{x}. $$

Similarly as before, we can obtain the components of the density of system (2):

$$\begin{aligned}& \begin{aligned} C^{1}&=\biggl(\frac{1}{2}u+tu_{t}+ \frac{1}{2}xu_{x}\biggr)p+\biggl(h+th_{t}+ \frac{1}{2}xh _{x}\biggr)q=\biggl(\frac{1}{2}u+tu_{t}+ \frac{1}{2}xu_{x}\biggr)h \\ &\quad {}+\biggl(h+th_{t} + \frac{1}{2}xh_{x} \biggr)u, \end{aligned} \\& \begin{aligned} C^{2}&=\biggl(\frac{1}{2}u+tu_{t}+ \frac{1}{2}xu_{x}\biggr) (pu+qh-\beta p_{x}+ \beta q_{x}+\alpha u_{xxx}) \\ &\quad {}+\biggl(tu_{xt}+\frac{1}{2}xu_{xx}\biggr) (- \beta p+ \beta q+\alpha q_{xxx})-\alpha \biggl(tu_{xxt}+ \frac{1}{2}u_{xx}+ \frac{1}{2}xu_{xxx} \biggr)q_{xxx} \\ &\quad {}+\biggl(h+th_{t}+\frac{1}{2}xh_{x}\biggr) (p+qu+\beta q _{xx})-\beta \biggl(th_{xt}+ \frac{3}{2}h_{x}+\frac{1}{2}xh_{xx}\biggr)q \\ &=\biggl( \frac{1}{2}u+tu_{t}+\frac{1}{2}xu_{x} \biggr) (2hu-\beta h_{x}+\beta u_{x}+ \alpha u_{xxx}) \\ &\quad {}+\biggl(tu_{xt}+\frac{1}{2}xu_{xx}\biggr) (- \beta h+\beta u+\alpha u_{xxx})-\alpha \biggl(tu_{xxt}+ \frac{1}{2}u_{xx}+\frac{1}{2}xu_{xxx} \biggr)u_{xxx} \\ &\quad {}+\biggl(h+th _{t}+\frac{1}{2}xh_{x}\biggr) \bigl(h+u^{2}+\beta u_{xx}\bigr)-\beta \biggl(th_{xt}+ \frac{3}{2}h_{x}+\frac{1}{2}xh_{xx}\biggr)u. \end{aligned} \end{aligned}$$

In what follows, we investigate the Lie–Bäcklund symmetries of system (2) and the resulting conservation laws.

Set

$$\begin{aligned} X =&\eta ^{u}(x,t,u,h,u_{x},h_{x},u_{xx},h_{xx},u_{xxx},h_{xxx}) \partial _{u} \\ &{}+\eta ^{h}(x,t,u,h,u_{x},h_{x},u_{xx},h_{xx},u_{xxx}, h_{xxx}) \partial _{h}. \end{aligned}$$

(22)

Substituting (22) into system (2), we infer the following Lie-Bäcklund symmetries by using the software Maple:

$$\begin{aligned}& X_{1}=h_{x}\partial _{h}+u_{x} \partial _{u},\qquad X_{2}=h_{x}\partial _{h}+(1+tu _{x})\partial _{u}, \\& X_{3}=(6u_{x}+uh_{x}+\alpha u_{xxx}- \beta h_{xx}\partial _{h}+(uu_{x}+ \beta u_{xx}+h_{x})\partial _{u}, \\& \begin{aligned} X_{4}&=(2thu_{x}+2tuh_{x}+2 \alpha tu_{xxx}-2\beta th_{xx}+h_{xx}+2h) \partial _{h} \\ &\quad {}+(2tuu_{x}+2\beta tu_{xx}+2th_{x}+u_{xx}+u) \partial _{u} , \end{aligned} \\& \begin{aligned} X_{5}&=\bigl(6huu_{x}+3u^{2}h_{x}-6 \beta u_{x}h_{x}+6\alpha uu_{xxx}-6 \beta uh_{xx}+12\alpha u_{x}h_{xx}+4\beta ^{2}h_{xxx} \\ &\quad {}+6hh_{x}+4\alpha h_{xxx}\bigr)\partial _{h}+\bigl(3u^{2}u_{x}+6\beta uu_{xx}+6 \beta u_{x}^{2}+4 \beta ^{2}u_{xxx} \\ &\quad {}+6hu_{x}+6uh_{x}+4\alpha u_{xxx}\bigr) \partial _{u} , \end{aligned} \\& \begin{aligned} X_{6}&=\bigl(6tuhu_{x}+3tu^{2}h_{x}-6 \beta th_{x}u_{x}+6\alpha tuu_{xxx}-6 \beta tuh_{xx}+12\alpha tu_{x}u_{xx}+4\beta ^{2}th_{xxx} \\ &\quad {}+6thh_{x}+2xhu _{x}+2xuh_{x}+4\alpha th_{xxx}+2\alpha xu_{xxx}-2\beta xh_{xx}+4hu \\ &\quad {}-10 \beta h_{x}+6\alpha u_{xx}\bigr)\partial _{h}+\bigl(3tu^{2}u_{x}+6\beta tuu_{xx}+6 \beta th_{x}^{2}+4\beta ^{2}tu_{xxx}+6thu_{x}+6tuh_{x} \\ &\quad {}+4\alpha tu_{xxx}+2xuu _{x}+2\beta xu_{xx}+2xh_{x}+u^{2}+4h\bigr)\partial _{u} . \end{aligned} \end{aligned}$$

Applying (10), we can deduce the components of the conserved density for system (2). For $X_{1}$, we have

$$\begin{aligned}& C^{1}=u_{x}p+h_{x}q=u_{x}h+h_{x}u=D_{x}(uh), \quad \mbox{hence } \tilde{C}_{1}=0, \\& \begin{aligned} C^{2}&=-2huu_{x}+\beta uh_{x}- \beta u_{x}^{2}-\beta hu_{xx}+\beta uu _{xx}-\alpha uu_{xxx}-hh_{x}-u^{2}h_{x} \\ &\quad {}-\beta u_{x}h_{x}+\beta uh_{x}h _{xx}+\beta uh_{xx}. \end{aligned} \end{aligned}$$

For $X_{2}$, $W^{u}=\eta ^{u}=1+tu_{x}$ and $W^{h}=\eta ^{h}=h_{x}$; substituting into (16) and (17), we have that

$$\begin{aligned}& C^{1}=W^{u}p+W^{h}q=(1+tu_{x})p+h_{x}q=(1+tu_{x})h+uh_{x}, \\& \begin{aligned} C^{2}&=(\beta p_{x}-pu-qh-\alpha q_{xx})W^{u}+(\alpha q_{x}-\beta p)D _{x}\bigl(W^{u}\bigr)-\alpha qD_{x}^{2} \bigl(W^{u}\bigr) \\ &\quad {}-(p+qu+\beta q_{x})W^{h}+\beta qD _{x} \bigl(W^{h}\bigr) \\ &=(\beta p_{x}-pu-qh-\alpha q_{xx}) (1+tu_{x})+( \alpha q_{x}- \beta p)tu_{xx}-\alpha qtu_{xxx} \\ &\quad {}-(p+qu+\beta q_{x})h_{x}+\beta qh_{xx}. \end{aligned} \end{aligned}$$

Similarly, for $X_{3}$, we get that

$$\begin{aligned}& C^{1}=W^{u}p+W^{h}q=p(uu_{x}+\beta u_{xx}+h_{x})+q(6u_{x}+uh_{x}+ \alpha u_{xxx}-\beta h_{xx}) , \\& \begin{aligned} C^{2}&=(\beta p_{x}-pu-qh-\alpha q_{xx}) (uu_{x}+\beta u_{xx}+h_{x})+( \alpha q_{x}-\beta p)D_{x}(uu_{x}+\beta u_{xx}+h_{x}) \\ &\quad {}-\alpha qD_{x} ^{2}(uu_{x}+\beta u_{xx}+h_{x})-(p+qu+\beta q_{x}) (6u_{x}+uh_{x}+ \alpha u_{xxx}-\beta h_{xx}) \\ &\quad {}+\beta qD_{x}(6u_{x}+uh_{x}+\alpha u_{xxx}- \beta h_{xx}) . \end{aligned} \end{aligned}$$

For $X_{4}$, we have

$$\begin{aligned}& \begin{aligned} C^{1}&=p(2tuu_{x}+2\beta tu_{xx}+2th_{x}+u_{xx}+u)\partial _{u} \\ &\quad {}+q(2thu _{x}+2tuh_{x}+2\alpha tu_{xxx}-2 \beta th_{xx}+h_{xx}+2h), \end{aligned} \\& \begin{aligned} C^{2}&=(\beta p_{x}-pu-qh-\alpha q_{xx}) (2tuu_{x}+2\beta tu_{xx}+2th _{x}+u_{xx}+u) \\ &\quad {}+(\alpha q_{x}-\beta p)D_{x}(2tuu_{x}+2 \beta tu_{xx}+2th _{x}+u_{xx}+u) \\ &\quad {}-\alpha qD_{x}^{2}(2tuu_{x}+2\beta tu_{xx}+2th_{x}+u _{xx}+u) \\ &\quad {}-(p+qu+\beta q_{x}) (2thu_{x}+2tuh_{x}+2 \alpha tu_{xxx}-2 \beta th_{xx}+h_{xx}+2h) \\ &\quad {}+\beta qD_{x}(2thu_{x}+2tuh_{x}+2 \alpha tu _{xxx}-2\beta th_{xx}+h_{xx}+2h) . \end{aligned} \end{aligned}$$

For $X_{5}$, we have

$$\begin{aligned}& \begin{aligned} C^{1}&=p\bigl(3u^{2}u_{x}+6\beta uu_{xx}+6\beta u_{x}^{2}+4\beta ^{2}u_{xxx}+6hu _{x}+6uh_{x}+4\alpha u_{xxx}\bigr) \\ &\quad {}+q\bigl(6huu_{x}+3u^{2}h_{x}-6\beta u_{x}h _{x}+6\alpha uu_{xxx}-6\beta uh_{xx}+12\alpha u_{x}h_{xx} \\ &\quad {}+4\beta ^{2}h _{xxx}+6hh_{x}+4\alpha h_{xxx}\bigr) , \end{aligned} \\& \begin{aligned} C^{2}&=(\beta p_{x}-pu-qh-\alpha q_{xx})+(\alpha q_{x}-\beta p)D_{x}\bigl(3u ^{2}u_{x}+6\beta uu_{xx}+6\beta u_{x}^{2}+4\beta ^{2}u_{xxx} \\ &\quad {}+6hu_{x}+6uh _{x}+4\alpha u_{xxx}\bigr)- \alpha qD_{x}^{2}\bigl(3u^{2}u_{x}+6 \beta uu_{xx}+6 \beta u_{x}^{2}+4\beta ^{2}u_{xxx}+6hu_{x} \\ &\quad {}+6uh_{x}+4\alpha u_{xxx}\bigr)-(p+qu+ \beta q_{x}) \bigl(6huu_{x}+3u^{2}h_{x}-6 \beta u_{x}h_{x}+6\alpha uu_{xxx}-6 \beta uh_{xx} \\ &\quad {}+12\alpha u_{x}h_{xx}+4\beta ^{2}h_{xxx}+6hh_{x}+4 \alpha h_{xxx}\bigr)+\beta qD_{x}\bigl(6huu_{x}+3u^{2}h_{x}-6 \beta u_{x}h_{x} \\ &\quad {}+6\alpha uu_{xxx}-6\beta uh_{xx}+12\alpha u_{x}h_{xx}+4\beta ^{2}h_{xxx}+6hh _{x}+4\alpha h_{xxx}\bigr) . \end{aligned} \end{aligned}$$

For $X_{6}$, we infer that

$$\begin{aligned}& \begin{aligned} C^{1}&=p\bigl(3tu^{2}u_{x}+6 \beta tuu_{xx}+6\beta th_{x}^{2}+4\beta ^{2}tu _{xxx}+6thu_{x}+6tuh_{x}+4\alpha tu_{xxx} \\ &\quad {}+2xuu_{x}+2\beta xu_{xx}+2xh _{x}+u^{2}+4h \bigr)+q\bigl(6tuhu_{x}+3tu^{2}h_{x}-6\beta th_{x}u_{x} \\ &\quad {}+6\alpha tuu _{xxx}-6\beta tuh_{xx}+12\alpha tu_{x}u_{xx}+4\beta ^{2}th_{xxx}+6thh _{x}+2xhu_{x}+2xuh_{x} \\ &\quad {}+4\alpha th_{xxx}+2\alpha xu_{xxx}-2\beta xh _{xx}+4hu-10\beta h_{x}+6\alpha u_{xx}\bigr) , \end{aligned} \\& \begin{aligned} C^{2}&=(\beta p_{x}-pu-qh-\alpha q_{xx}) \bigl(3tu^{2}u_{x}+6\beta tuu_{xx}+6 \beta th_{x}^{2}+4\beta ^{2}tu_{xxx}+6thu_{x}+6tuh_{x} \\ &\quad {}+4\alpha tu_{xxx}+2xuu _{x}+2\beta xu_{xx}+2xh_{x}+u^{2}+4h\bigr)+(\alpha q_{x}-\beta p)D_{x}\bigl(3tu ^{2}u_{x}+6 \beta tuu_{xx} \\ &\quad {}+6\beta th_{x}^{2}+4\beta ^{2}tu_{xxx}+6thu _{x}+6tuh_{x}+4\alpha tu_{xxx}+2xuu_{x}+2 \beta xu_{xx}+2xh_{x} \\ &\quad {}+u^{2}+4h\bigr)- \alpha qD_{x}^{2} \bigl(3tu^{2}u_{x}+6\beta tuu_{xx}+6\beta th_{x}^{2}+4 \beta ^{2}tu_{xxx}+6thu_{x}+6tuh_{x} \\ &\quad {}+4\alpha tu_{xxx}+2xuu_{x}+2\beta xu_{xx}+2xh_{x}+u^{2}+4h\bigr)-(p+qu+\beta q_{x}) \bigl(6tuhu_{x}+3tu^{2}h_{x} \\ &\quad {}-6 \beta th_{x}u_{x}+6\alpha tuu_{xxx} -6\beta tuh_{xx}+12\alpha tu_{x}u _{xx}+4\beta ^{2}th_{xxx}+6thh_{x} \\ &\quad {}+2xhu_{x}+2xuh_{x}+4\alpha th_{xxx}+2 \alpha xu_{xxx}-2\beta xh_{xx}+4hu-10\beta h_{x}+6 \alpha u_{xx}\bigr) \\ &\quad {}+ \beta qD_{x}\bigl(6tuhu_{x}+3tu^{2}h_{x}-6 \beta th_{x}u_{x}+6\alpha tuu _{xxx}-6\beta tuh_{xx}+12\alpha tu_{x}u_{xx} \\ &\quad {}+4\beta ^{2}th_{xxx}+6thh _{x}+2xhu_{x}+2xuh_{x}+4 \alpha th_{xxx}+2\alpha xu_{xxx}-2\beta xh _{xx} \\ &\quad {}+4hu-10\beta h_{x}+6\alpha u_{xx}\bigr) , \end{aligned} \end{aligned}$$

where $p=h$ and $q=u$, which can substituted into the above conserved densities so that we obtain more explicit formulas for $c^{1}$ and $C^{2}$. Here we omit the computations.

3 Lie symmetry groups and similarity solutions

In this section, we apply the point symmetries (13) to consider the Lie symmetry groups and some similarity solutions to system (2). Denote the Lie symmetry groups generated by $V_{1}$, $V_{2}$, $V_{3}$, $V_{4}$ by $g_{1}$, $g_{2}$, $g_{3}$, $g_{4}$, respectively. It is easy to see that

$$\begin{aligned}& g_{1}:(x,t,u,h)\rightarrow (x,t+\epsilon ,u,h), \\& g_{2}:(x,t,u,h)\rightarrow (x+\epsilon ,t,u,h), \\& g_{3}: (x,t,u,h)\rightarrow \bigl(x, e^{\epsilon }t,u-\epsilon ,h \bigr), \\& g_{4}:(x,t,u,h)\rightarrow \bigl(e^{\frac{1}{2}\epsilon }x,e^{\epsilon }t,e ^{-\frac{1}{2}\epsilon }u,e^{-\epsilon }h\bigr). \end{aligned}$$

If $u=f(x,t)$, $h=g(x,t)$ are solutions to e system (2), then we can get the following new solutions based on these symmetry groups:

$$\begin{aligned}& u^{(1)}=u(x,t-\epsilon )=f(x,t-\epsilon ), \qquad h^{(1)}=g(x,t- \epsilon ), \\& u^{(2)}=f(x-\epsilon ,t), \qquad h^{(2)}=g(x-\epsilon ,t), \\& u^{(3)}=f\bigl(x,e^{-\epsilon }t\bigr)-\epsilon , \qquad h^{(3)}=g\bigl(x,e^{-\epsilon }t\bigr), \\& u^{(4)}=e^{-\frac{1}{2}\epsilon }f\bigl(e^{-\frac{1}{2}\epsilon }x,e^{- \epsilon }t \bigr), \qquad h^{(4)}=e^{-\epsilon }g\bigl(e^{-\frac{1}{2}\epsilon }x, e^{-\epsilon }t\bigr). \end{aligned}$$

Of course, we can go on getting some iteration solutions following the work [8]:

$$\begin{aligned}& \textstyle\begin{cases} u^{(3,1)}=f(x,e^{-\epsilon }t)-2\epsilon , \\ h^{(3,1)}=g(x,e^{-\epsilon }t), \end{cases}\displaystyle \\& \textstyle\begin{cases} u^{(4,1)}=e^{-\epsilon }f(e^{-\frac{1}{2}\epsilon } x,e^{-\epsilon }t), \\ h^{(4,1)}=e^{-2\epsilon }g(e^{-\frac{1}{2}\epsilon }x,e^{-\epsilon }t), \end{cases}\displaystyle \\& \ldots , \\& \textstyle\begin{cases} u^{(4,n)}=e^{-\frac{2n-1}{n}\epsilon }f{-\frac{1}{2}\epsilon } x,e ^{-\epsilon }t), \\ h^{(4,n)}=e^{-n\epsilon }g(e^{-\frac{1}{2}\epsilon } x,e^{-\epsilon }t), \end{cases}\displaystyle \end{aligned}$$

where $n\in N^{+}$.

For the transformation groups $g_{1}$ and $g_{2}$, the invariant solutions are traveling wave solutions. Indeed, set

$$ u=U(\xi ),\qquad h=H(\xi ),\qquad \xi =x-ct. $$

(23)

Substituting (23) into system (2), we have

$$ \textstyle\begin{cases} -cU^{\prime }=UU^{\prime }+H^{\prime }+\beta U^{\prime \prime }, \\ -cH^{\prime }=(UH)^{\prime }+\alpha U^{\prime \prime \prime }-\beta H ^{\prime \prime }. \end{cases} $$

Integrating gives rise to

$$ \textstyle\begin{cases} -cU=\frac{1}{2}U^{2}+H+\beta U^{\prime }, \\ -cH=UH+\alpha U^{\prime \prime }-\beta H^{\prime }. \end{cases} $$

(24)

From (24) we find that

$$ \bigl(\alpha +\beta ^{2}\bigr)U^{\prime \prime }- \frac{1}{2}U^{3}-\frac{3}{2}cU ^{2}-c^{2}U=0. $$

(25)

Let $U^{\prime }=y(\xi )$. Then (25) turns to

$$ \bigl(U^{\prime }\bigr)^{2}=\frac{1}{\alpha +\beta ^{2}} \biggl(\frac{1}{4}U^{u}+cU^{3}+c ^{2}U^{2} \biggr). $$

(26)

Suppose $U^{\prime }=aU+U^{2}$. Then inserting this into (26), we have that

$$ \alpha ^{2}+\beta ^{2}=\frac{1}{4}, \qquad a=2\pm c. $$

Thus we get

$$ U^{\prime }=2\pm cU+U^{2}=:2\epsilon cU+U^{2}. $$

(27)

Solving (27) yields

$$ U=\frac{2\epsilon c\bar{c}e^{2\epsilon c\xi }}{1-\bar{c}e^{2\epsilon c\xi }}, $$

(28)

and hence

$$ H=-\frac{2\epsilon c\bar{c}(\epsilon c-2\epsilon c\beta )e^{2\epsilon c\xi }}{1-\bar{c}e^{2\epsilon c\xi }}-\biggl(\frac{1}{2} +\beta \biggr) \frac{4c ^{2}\bar{c}^{2}e^{4\epsilon c\xi }}{(1-\bar{c}e^{2\epsilon \xi })^{2}}, $$

(29)

where c̄ is an integral constant, which does not vanish. Again applying (23) and (28)–(29), we have

$$ \textstyle\begin{cases} u(x,t)=\frac{2\epsilon c\bar{c}e^{2\epsilon c(x-ct)}}{1-\bar{c}e^{2 \epsilon c(x-ct)}}, \\ h(x,t)=-\frac{2\epsilon c\bar{c}(\epsilon c-2\epsilon c\beta )e^{2 \epsilon c(x-ct)}}{1-\bar{c}e^{2\epsilon c(x-ct)}}-(\frac{1}{2} + \beta )\frac{4c^{2}\bar{c}^{2}e^{4\epsilon c(x-ct)}}{(1-\bar{c}e^{2 \epsilon (x-ct)})^{2}}. \end{cases} $$

The characteristic equation of the vector field $V_{3}$ presents

$$ \frac{dt}{0}=\frac{dx}{t}=\frac{du}{-1}, $$

which gives

$$ w=\xi x+u, \qquad x\xi =t. $$

The resulting group-invariant solution reads

$$ u=f(t)-tx, \qquad h=g(t). $$

(30)

Substituting (30) into system (2), we infer

$$ \textstyle\begin{cases} \frac{df}{dt}+tf=x+t^{2}x, \\ \frac{dg}{dt}=-tg(t), \end{cases} $$

which has the following set of solutions:s

$$ \textstyle\begin{cases} f(t)=\exp(-\frac{1}{2}t^{2})[x\int ^{t}\exp(\frac{1}{2}t ^{2})\,dt+\frac{x}{3}t^{3}e^{\frac{t^{2}}{2}} -\frac{2}{3}xe^{\frac{t ^{2}}{2}}+\tilde{c}], \\ g(t)=c_{0}\exp(-\frac{1}{2}t^{2}), \end{cases} $$

(31)

where c̃, $c_{0}$ are constants. Substituting (31) into (30), we obtain the similarity solutions of system (2):

$$ \textstyle\begin{cases} u(x,t)=xe^{-\frac{1}{2}t^{2}}\int ^{t}e^{\frac{1}{2}t^{2}}\,dt+ \frac{1}{3}x(t^{3}-2)+\tilde{c}e^{-\frac{1}{2}t^{2}}-tx, \\ h(x,t)=c_{0}e^{-\frac{1}{2}}t^{2}. \end{cases} $$

Similarly, for $g_{4}$, the characteristic equation reads

$$ \frac{dt}{t}=\frac{dx}{\frac{1}{2}x}=\frac{du}{-\frac{1}{2}u}= \frac{dh}{-h}, $$

from which we get the invariants

$$ \xi =x^{2}t^{-1}, \qquad u=x^{-1}f( \xi ), \qquad h=x^{-\frac{1}{2}}g(\xi ), $$

(32)

where $f(\xi )$, $g(\xi )$ are arbitrary invariant functions with respect to the variable ξ. Inserting (32) into the system, we get the following ordinary differential equations with variable coefficients:

$$ \textstyle\begin{cases} -f^{\prime }(\xi )=(2\beta -1)\xi ^{-2}f^{2}(\xi )+2(1-\beta )\xi ^{-1}f ^{\prime }(\xi )-2\xi ^{-2}g(\xi ) \\ \hphantom{-f^{\prime }(\xi )={}}{}+2\xi ^{-1}g^{\prime }(\xi )+2\beta f ^{\prime \prime }(\xi ), \\ -g^{\prime }(\xi )=-3\xi ^{2}f(\xi )g(\xi )+2\xi ^{-1}f^{\prime }( \xi )g(\xi )+2\xi ^{-1}f(\xi )g^{\prime }(\xi )+6\alpha \xi ^{-2}f( \xi ) \\ \hphantom{-g^{\prime }(\xi )={}}{}-2\alpha \xi ^{-1}f^{\prime }(\xi )-2\alpha f^{\prime \prime }(\xi )+4\alpha \xi f^{\prime \prime \prime }(\xi )-6\beta \xi ^{-2}g(\xi ) \\ \hphantom{-g^{\prime }(\xi )={}}{}+6\beta \xi ^{-1}g^{\prime }(\xi )-4 \beta g^{\prime \prime }(\xi ). \end{cases} $$

(33)

Now we look for the formal series solutions to (33), so we assume that

$$ f(\xi )=\sum_{n=0}^{\infty }a_{n}\xi ^{n}, \qquad g(\xi )=\sum_{n=0}^{\infty }b_{n} \xi ^{n}. $$

(34)

Substituting (34) into system (2) and comparing the coefficients of both sides in the system (33), one infers that

$$\begin{aligned}& 2b_{0}=(2\beta -1)a_{0}^{2} , \\& (4\beta -2)a_{0}a_{1}+(2-2\beta )a_{1}=0 , \\& (4\beta -2)a_{0}a_{2}+4a_{2}+2b_{2}+(2 \beta -1)a_{1}^{2}+a_{1}=0 , \\& -2a_{2}+(4\beta -2)a_{0}a_{3}+(4\beta -2)a_{1}a_{2}+(6\beta +6)a_{3}+4b _{3}+2a_{2}=0 , \\& \ldots , \\& -\sum_{k=0}^{4}a_{k}a_{4-k}+8(4- \beta )a_{4}+8b_{4}+2\beta \sum _{k=0}^{4}a_{k}a_{4-k}=0 , \\& (1-n)a_{n-1}=2nb_{n}+(2\beta -1)\sum _{k=0}^{n}a_{k}a_{n-k}+\bigl(2n-4 \beta n+2\beta n^{2}\bigr)a_{n} , \quad n\geq 5 , \\& a_{0}b_{0}-2\alpha a_{0}+2\beta b_{0}=0 , \\& a_{0}b_{1}+a_{1}b_{0}-4\alpha a_{1}=0 , \\& a_{2}b_{0}-a_{1}b_{1}+a_{0}b_{2}+2a_{0}b_{1}-2 \alpha a_{2}-14\beta b _{2}+b_{1}+6\beta b_{1}=0 , \\& -3a_{3}b_{0}+2a_{1}b_{2}+3a_{0}b_{3}+2b_{2}+12 \alpha a_{3}+12 \beta b_{3}=0 , \\& \ldots , \\& \begin{aligned} -3a_{3}&=-3\sum_{k=0}^{4}a_{k}b_{4-k}+2 \sum_{k=0}^{3}(3-k)a _{3-k}+2\sum _{k=0}^{4}(4-k)a_{4-k}+2\sum _{k=0}^{3}(3-k)a _{k}b_{3-k} \\ &\quad {}+2\sum_{k=0}^{4}(4-k)a_{k}b_{4-k}-8 \alpha a_{4}-24a _{4}-30\beta b_{4} , \end{aligned} \\& -4a_{4}=-3\sum_{k=0}^{5}a_{k}b_{5-k}+2 \sum_{k=0}^{5}(5-k)a _{5-k}b_{k}+2 \sum_{k=0}^{5}(5-k)a_{k}b_{5-k}+190 \alpha a_{5}-56 \beta b_{5} , \\& \ldots , \\& \begin{aligned} (1-n)a_{n-1}&=-3\sum_{k=0}^{n}a_{k}b_{n-k}+2 \sum_{k=0} ^{n}(n-k)a_{n-k}b_{k}+2 \sum_{k=0}^{n}(n-k)a_{k}b_{n-k}+6 \alpha a_{n-2} \\ &\quad {}+\bigl[-2\alpha n-2\alpha n(n-1)+4\alpha n(n-1) (n-2) \bigr]a_{n} \\ &\quad {}+\bigl[6\beta n-6\beta -4\beta n(n-1)\bigr]b_{n} , \quad n \geq 6 . \end{aligned} \end{aligned}$$

When $a_{1}\neq 0$, there exists a constraint between α and β:

$$ 5\beta ^{2}-8\alpha \beta -4\beta +4\alpha +1=0. $$

In terms of the above relations among $a_{i}$, $b_{i}$ ($i=0,1,2,\ldots ,n$), we can write out the formal series solutions where $a_{1}$ is a free parameter. As for the convergence of the series, we skip the discussion. However, we can follow the way presented in [8] to proceed.

4 A new scheme for seeking conservation laws of system (2)

In this section, we adopt the approach given by Göktas and Hereman to investigate the conserved densities and the fluxes of the system for the given rank of ever term in the equations. We rewrite the system (2) as follows

$$ \textstyle\begin{cases} u_{1,t}=u_{1}u_{1,x}+u_{2,x}+\beta u_{1,2x}, \\ u_{2,t}=(u_{1}u_{2})_{x}+\alpha u_{1,3x}-\beta u_{2,2x}, \end{cases} $$

(35)

where $u_{i,nx}$ denotes the nth derivative with respect to x for the variables $u_{i}$ ($i=1,2$). We denote the weight of the variables $u_{i}$ and derivative $\partial _{t}$ by $w(u_{i})$ and $w(\partial _{t})$, respectively. We easily find that

$$\begin{aligned}& r_{1,1}=w(u_{1})+w(\partial _{t}), \qquad r_{1,2}=1+w(u_{1})+w(u_{2}), \\& r_{1,3}=3+w(u_{1}), \qquad r_{1,4}=2+w(u_{2}), \\& r_{2,1}=w(u_{2})+w(\partial _{t}), \qquad r_{2,2}=1+w(u_{1})+w(u_{2}), \\& r_{2,3}=3+w(u_{1}), \qquad r_{2,4}=2+w(u_{2}). \end{aligned}$$

Solving

$$ \textstyle\begin{cases} r_{1,1}=r_{1,2}=r_{1,3}=r_{1,4}, \\ r_{2,1}=r_{2,2}=r_{2,3}=r_{2,4}, \end{cases} $$

we get

$$ w(u_{1})=1,\qquad w(u_{2})=2,\qquad w(\partial _{t})=2. $$

List of the variables in system (2):

$$ \mathcal{V}=\{v_{1},v_{2}\},\qquad v_{1}=u_{2}, \qquad v_{2}=u_{1}. $$

In what follows, we consider the case where $\operatorname{rank}=4$ and

$$ \mathcal{B}_{0}=\bigl\{ (1;0)\bigr\} . $$

For $q=1$ and $m=0$, a direct calculation gives

$$ T_{1,s}=v_{1}^{s}=u_{2}^{s}, \qquad W_{1,s}=sw(v_{1})=2s,\qquad p_{1,0}=\biggl[\!\!\biggl[\frac{4-0}{w(v _{1})}\biggr]\!\!\biggr]=2. $$

Hence $s=0,1,2$, and

$$ \mathcal{B}_{1}=B_{1,0}=\bigl\{ (1;0),(u_{2};2), \bigl(u_{2}^{2};4\bigr)\bigr\} . $$

For $q=2$ and $m=0$, we have

$$ T_{2,s}=u_{1}^{s},\qquad W_{2,s}=W_{1,0}+sw(v_{2})=s, \qquad p_{2,0}=\biggl[\!\!\biggl[\frac{4-0}{1}\biggr]\!\!\biggr]=4. $$

Hence $s=0,1,2,3,4$, and

$$ B_{2,0}=\bigl\{ (1;0),(u_{1};1),\bigl(u_{1}^{2};2 \bigr),\bigl(u_{1}^{3};3\bigr),\bigl(u_{1}^{4};4 \bigr)\bigr\} . $$

For $q=2$ and $m=1$, we have that

$$ T_{2,s}=u_{2}u_{1}^{s},\qquad W_{2,s}=2+s,\qquad p_{2,1}=\biggl[\!\!\biggl[\frac{4-2}{1} \biggr]\!\!\biggr]=2. $$

Thus $s=0,1,2$. It follows that

$$ B_{2,1}=\bigl\{ (u_{2};2),(u_{1}u_{2};3), \bigl(u_{2}u_{1}^{2};4\bigr)\bigr\} . $$

For $q=2$ and $m=2$, we find that

$$ T_{2,s}=u_{2}u_{1}^{s},\qquad W_{2,s}=4+s,\qquad p_{2,2}=\biggl[\!\!\biggl[\frac{4-W_{1,2}}{w(v_{2})} \biggr]\!\!\biggr]=\biggl[\!\!\biggl[\frac{4-4}{1}\biggr]\!\!\biggr]=0. $$

Hence $s=0$. Thus we get

$$\begin{aligned}& B_{2,2}=\bigl\{ \bigl(u_{2}^{2};4\bigr)\bigr\} , \\& \begin{aligned} \mathcal{B}_{2}&=B_{2,0}\cup B_{2,1}\cup B_{2,2} \\ &=\bigl\{ (1;0),(u_{1};1),\bigl(u _{1}^{2};2 \bigr),\bigl(u_{1}^{3};3\bigr),\bigl(u_{1}^{4};4 \bigr),(u_{2};2),(u_{1}u_{2};3),\bigl(u _{2}i_{1}^{2};4\bigr),\bigl(u_{2}^{2};4 \bigr)\bigr\} . \end{aligned} \end{aligned}$$

Computation of l for each pair of $\mathcal{B}_{2}$ reads

$$ l=4,3,2,1,0,2,1,0,0. $$

Gathering the terms by applying the number l of partial derivatives with respect to x, we get that

$$ \mathcal{H}=\bigl\{ 0,u_{1,3x},u_{1,x}^{2},u_{1}u_{1,2x},u_{1}^{2}u_{1,x},u _{1}^{4},u_{2,2x},u_{1,x}u_{2},u_{1}u_{2,x},u_{2}u_{1}^{2},u_{2}^{2} \bigr\} . $$

Removing from $\mathcal{H}$ the constant terms and the terms that can be written as an x-derivative, or an x-derivative up to terms retained earlier in the set $\mathcal{H}$, we have that

$$ \mathcal{I}=\bigl\{ u_{1,x}^{2},u_{1}^{4},u_{1}u_{2,x},u_{2}u_{1}^{2},u_{2} ^{2}\bigr\} . $$

Let

$$ \rho =c_{1}u_{1,x}^{2}+c_{2}u_{1}^{4}+c_{3}u_{1}u_{2,x}+c_{4}u_{2}u _{1}^{2}+c_{5}u_{2}^{2}, $$

where $c_{i}$ ($i=1,2,3,4$) are constants to be determined. Then we have

$$\begin{aligned} D_{t}(\rho ) =&2c_{1}u_{1,x}u_{1,xt}+4c_{2}u_{1}^{3}u_{1,t}+c_{3}(u_{1,t}u _{2,x}+u_{1}u_{2,xt}) \\ &{}+c_{4}u_{2,t}u_{1}^{2}+2c_{4}u_{1}u_{2}u_{1,t}+2c _{5}u_{2}u_{2,t}. \end{aligned}$$

(36)

Substituting (35) into (36) to cancel the terms with t-derivatives, we get

$$\begin{aligned} D_{t}(\rho )|_{\text{(35)}} =&2c_{1}u_{1,x} \bigl(u_{1,x}^{2}+u_{1}u_{1,2x}+u_{2,2x}+ \beta u_{1,3x}\bigr)+4c_{2}u_{1}^{3}(u_{1}u_{1,x}+u_{2,x}+ \beta u_{1,2x}) \\ &{}+c _{3}\bigl[u_{2,x}(u_{1}u_{1,x}+u_{2,x}+ \beta u_{1,2x})+u_{1}(u_{2}u_{1,2x}+2u _{1,x}u_{2,x} \\ &{}+u_{1}u_{2,2x}+\alpha u_{1,3x}-\beta u_{2,3c})\bigr]+2c_{4}u _{1}u_{2}(u_{1}u_{1,x}+u_{2,x}+ \beta u_{1,2x}) \\ &{}+c_{4}u_{1}^{2}(u_{1,x}u _{2}+u_{1}u_{2,x}+\alpha u_{1,4x}-\beta u_{2,3x}) \\ &{}+2c_{5}u_{2}(u_{1,x}u _{2}+u_{1}u_{2,x}+ \alpha u_{1,3x}-\beta u_{2,2x}) . \end{aligned}$$

Acting on the variational derivative to $D_{t}(\rho )|_{\text{(35)}}$ and comparing the coefficients of the same terms, we get the following relations:

$$\begin{aligned} S =&\{\alpha c_{4}+c_{1}=0, 2\beta c_{4}+c_{3}=0, 2\beta c_{4}+c_{3}=0,2 \alpha c_{3}-4\beta c_{1}=0,\alpha c_{5}+c_{1}=0, \\ &2\beta c_{5}+c_{3}=0,c _{4}-c_{5}=0 \} . \end{aligned}$$

Solving S gives that

$$ c_{1}=-\alpha c_{4},\qquad c_{3}=-2\beta c_{4},\qquad c_{5}=c_{4}, $$

(37)

where $c_{4}$ is a free parameter, and $c_{2}$ can be an arbitrary constant.

Suppose $c_{4}=1$ and $c_{2}=0$. Then $c_{1}=-\alpha$, $c_{3}=-2\beta$, $c_{5}=1$. Thus, we have the conserved density:

$$ \rho =-\alpha u-1, x^{2}-2\beta u_{1}u_{2,x}+u_{2}u_{1}^{2}+u_{2} ^{2}. $$

The resulting flux of system (2) when $\operatorname{rank}=4$ is given by

$$\begin{aligned} J =&\alpha \bigl(2u_{1}u_{1,x}^{2}-u_{1}^{2}u_{1,2x} \bigr)+2\alpha (u_{1,x}u_{2,x}-u _{2}u_{1,2x})+2 \alpha \beta u_{1}u_{1,3x} \\ &{}-2\beta ^{2}u_{1}u_{2,2x}+2 \beta u_{2}u_{2,x}+3\beta u_{1}^{2}u_{2,x}+u_{1}^{3}u_{2}+2u_{2}^{2}u _{1} . \end{aligned}$$

If $c_{1}=c_{3}=c_{4}=c_{5}=0$ and $c_{2}\neq 0$, then the conserved density is of the form

$$ \rho =c_{2}u_{1}^{4}, $$

whereas

$$\begin{aligned} D_{t}(\rho ) =&4c_{2}u_{1}^{3}(u_{1}u_{1,x}+u_{2,x}+ \beta u_{1,2x}) \\ =&D _{x}\biggl[\frac{1}{5}c_{2}u_{1}^{5}+4c_{2}u_{1}^{3}u_{1,x}+4c_{2} \beta u _{1}^{3}u_{2}-12c_{2}D_{x}^{-1} \bigl(u_{1}^{2}u_{1,x}^{2} \bigr)-12c_{2}\beta D _{x}^{-1} \bigl(u_{1}^{2}u_{2}u_{1,x}\bigr)\biggr] . \end{aligned}$$

Hence the flux has nonlocal differential form

$$ J=c_{2}\biggl[-\frac{1}{5}u_{1}^{5}-4u_{1}^{3}u_{1,x}-4 \beta u_{1}^{3}u_{2}+12D _{x}^{-1} \bigl(u_{1}^{2}u_{1,x}^{2}\bigr)+12\beta D_{x}^{-1}\bigl(u_{1}^{2}u_{2}u _{1,x}\bigr)\biggr]. $$

When $\operatorname{rank}=5,6,\ldots$ , we can obtain the resulting conserved densities and fluxes of system (2), and we omit the complicated computations.

5 Special solutions of the stationary system (2)

An important application of the conservation laws of evolution equations is the study of nonvariant solutions of symmetry groups. In this section, we want to apply the conservation laws of system (2) to investigating some noninvariant solutions of the symmetries. For simplicity, we only choose simpler conservation laws and only consider the solutions of the stationary system (2). System (2) can be written as

$$ u_{t}=\biggl(\frac{1}{2}u^{2}+h+\beta u_{x}\biggr)_{x}, \qquad h_{t}=(uh+\alpha u_{xx}-\beta h_{x})_{x}, $$

(38)

where the conserved densities are u and h, and the corresponding fluxes are $\frac{1}{2}u^{2}+h+\beta u_{x}$ and $uh+\alpha u_{xx}- \beta h_{x}$, which are the simplest conservation laws. We further want to use them to deduce some stationary solutions of system (2). Taking

$$ u=g(x), \qquad \frac{1}{2}u^{2}+h+\beta u_{x}=f(t), $$

and substituting into the second conserved equation in (38), we have that

$$ f^{\prime }(t)=g^{\prime }(x)f(t)-\frac{3}{2}g^{2}g^{\prime }(x)-2 \beta g^{\prime }(x)^{2}-2\beta gg^{\prime \prime }+\alpha g^{\prime \prime \prime }+\beta ^{2}gg^{\prime \prime \prime }. $$

(39)

Assume that $f(t)=0$ and then integrate (39):

$$ \bigl(\beta ^{2}g+\alpha \bigr)g^{\prime \prime }- \frac{1}{2}\beta ^{2}\bigl(g^{\prime }\bigr)^{2}-2 \beta gg^{\prime }-\frac{1}{2}g^{3}=0. $$

(40)

Case 1: When $\alpha >0$, $\beta =0$, Eq. (39) reduces to

$$ \alpha g^{\prime \prime }-\frac{1}{2}g^{3}=0, $$

which has a special solution

$$ g=-\frac{2\sqrt{\alpha }}{\epsilon x},\quad \epsilon =\pm 1. $$

Thus we obtain a set of special solutions to system (2):

$$ u=-\frac{2\sqrt{\alpha }}{\epsilon x},\qquad h=-\frac{2\alpha }{x^{2}}. $$

(41)

Case 2: When $\beta \neq 0$, assume a formal solution of (40)

$$ g=\frac{A}{B+Cx}. $$

(42)

Then we get

$$ g^{\prime }=\frac{AC}{(B+Cx)^{2}},\qquad g^{\prime \prime }= \frac{2A^{2}C}{(B+Cx)^{3}}. $$

(43)

Inserting (42) and (43) into (40) and comparing the coefficients at the powers of x, we infer that

$$ A=-\frac{\alpha }{\beta }C, \quad \beta =4. $$

(44)

Taking $B=1$, we have the special solution to system (2)

$$ \textstyle\begin{cases} u=-\frac{\alpha C}{4(1+Cx)}, \\ h=-\frac{\alpha ^{2}C^{2}}{32(1+Cx)^{2}}+\frac{\alpha C}{(1+Cx)^{2}}, \end{cases} $$

(45)

where $C\neq 0$ is a parameter.

In the case $f(t)=0$, we find that $u_{t}=h_{t}=0$. Hence the system becomes the following ordinary differential equations:

$$ \textstyle\begin{cases} D_{x}(\frac{1}{2}u^{2}+h+\beta u_{x})=0, \\ D_{x}(hu+\alpha u_{xx}-\beta h_{x})=0, \end{cases} $$

which can be written as

$$ \frac{1}{2}u^{2}+h+\beta u_{x}=c_{1}, \qquad hu+\alpha u_{xx}-\beta h_{x}=c _{2}, $$

(46)

where $c_{1}$, $c_{2}$ are integral constants. It is easy to rewrite (46) as the following differential equation with respect to the variable u:

$$ \frac{1}{2}u^{3}-\bigl(\alpha +\beta ^{2}\bigr)u_{xx}=c_{1}u-c_{2}. $$

(47)

When $c_{1}=c_{2}=0$, solving (47) yields

$$ u=-\frac{2\sqrt{\alpha +\beta ^{2}}}{\epsilon x}. $$

(48)

Substituting (48) into the first equation in (46), we get

$$ h=-\beta u_{x}-\frac{1}{2}u^{2}= \frac{2\sqrt{\alpha +\beta ^{2}}}{ \epsilon x^{2}}-\frac{2(\alpha +\beta ^{2})}{x^{2}}. $$

(49)

When $c_{1}\neq 0$ and $c_{2}\neq 0$, (47) becomes

$$ u_{xx}+\frac{1}{2(\alpha +\beta ^{2})}u^{3}- \frac{c_{1}u}{\alpha +\beta ^{2}}+\frac{c_{2}}{\alpha +\beta ^{2}}=0. $$

(50)

Let $u_{x}=y$. Then (50) can be written as

$$ y\frac{dy}{du}+\frac{u^{3}}{2(\alpha +\beta ^{2})}-\frac{c_{1}u}{ \alpha +\beta ^{2}}+ \frac{c_{2}}{\alpha +\beta ^{2}}=0. $$

(51)

Integrating (5) with respect to u, we get

$$ \frac{1}{2}y^{2}+\frac{1}{8(\alpha +\beta ^{2})}u^{4}- \frac{c_{1}u^{2}}{2( \alpha +\beta ^{2})}+\frac{c_{2}u}{\alpha +\beta ^{2}}+c_{0}=0, $$

(52)

where $c_{0}$ is an integral constant. Thus (52) can be written as

$$ u^{\prime }=\sqrt{\frac{c_{1}u^{2}}{\alpha +\beta ^{2}}-\frac{1}{4( \alpha +\beta ^{2})}u^{4}- \frac{2c_{2}u}{\alpha +\beta ^{2}}-2c_{0}}. $$

(53)

Fan [26] studied the solutions to the formal ODE (53) taking different parameters. Now we follow his method to give some exact solutions to (53).

(i) When $c_{0}=0$, (53) has the following solutions:

$$\begin{aligned}& u=\sqrt{4c_{1}}\operatorname{sech}\biggl(\sqrt{\frac{c_{1}}{\alpha +\beta ^{2}}}x \biggr),\quad c _{1}>0,\alpha +\beta _{2}>0, \\& u=\sqrt{4c_{1}}\operatorname{sec}\biggl(\sqrt{-\frac{c_{1}}{\alpha +\beta ^{2}}}x \biggr),\quad c _{1}>0,\alpha +\beta _{2}< 0, \\& u=-\frac{1}{\sqrt{-4(\alpha +\beta ^{2})}x},\quad c_{1}=0,\alpha +\beta ^{2}< 0. \end{aligned}$$

(ii) Equation (53) still has the following three Jacobi elliptic function solutions: when $\frac{c_{1}}{\alpha +\beta ^{2}}>0$, $c_{0}=\frac{2c_{1}^{2}m ^{2}(m^{2}-1)}{(\alpha +\beta ^{2})(2m^{2}-1)^{2}}$,

$$ u=\sqrt{\frac{4c_{1}m^{2}}{2m^{2}-1}}\operatorname{cn}\biggl(\sqrt{\frac{c_{1}}{( \alpha +\beta ^{2})(2m^{2}-1)}}x \biggr); $$

when $\frac{c_{1}}{\alpha +\beta ^{2}}<0$, $c_{0}=\frac{2c_{1}^{2}m ^{2}}{(\alpha +\beta ^{2})(m^{2}+1)}$,

$$ u=\sqrt{\frac{4c_{1}m^{2}}{m^{2}+1}}\operatorname{sn}\biggl(\sqrt{\frac{-c_{1}}{( \alpha +\beta ^{2})(m^{2}+1)}}x \biggr); $$

when $\frac{c_{1}}{\alpha +\beta ^{2}}>0$, $c_{0}=\frac{2c_{1}^{2}(1-m ^{2})}{(\alpha +\beta ^{2})(2-m^{2})^{2}}$,

$$ u=\sqrt{\frac{4c_{1}}{2-m^{2}}}\operatorname{dn}\biggl(\sqrt{\frac{c_{1}}{( \alpha +\beta ^{2})(2-m^{2})}}x \biggr), $$

where m denotes the module of Jacobi elliptic functions. Substituting all these u into (46), we can obtain the resulting h.

From the above discussion we find that we indeed obtain some new special solutions of system (2). However, there are two questions to further consider. One is that when $f(t)\neq 0$, can we obtain solutions dependent on the variable t? Of course, we can. Because we may investigate the traveling-wave solutions of system (2) by setting $\xi =x-ct$, which can transform system (2) to the ODEs with respect to the variable ξ. As for this, we do not further discuss them. Another one is that when $f(t)=0$, can we obtain solutions only dependent on the variable x and not on the variable t? We do not think so. In fact, if we replace system (2) with the system

$$ \textstyle\begin{cases} u_{t}=(\frac{1}{2}u^{2}+h+\beta h_{x})_{x}, \\ h_{t}=(uh+\alpha u_{xx}-\beta h_{x})_{x}, \end{cases} $$

(54)

then we can get a set of solutions dependent on the variable t. In fact, taking $u_{t}=g(x)$, $\frac{1}{2}u^{2}+h+\beta h_{x}=0$, we have

$$ h_{x}+\frac{1}{\beta }h=-\frac{1}{2\beta }g^{2}(x). $$

Substituting this into the second equation in (54) gives rise to

$$ G_{x}+\frac{1}{\beta }G=0, $$

where $G(x,t)=hg(x)+\alpha g_{xx}-\beta h_{x}$, which has the solution

$$ G=\sigma (t)e^{\frac{-1}{\beta }x}, $$

from which we have

$$ hg(x)+\alpha g_{xx}-\beta h_{x}=\sigma (t)e^{\frac{-1}{\beta }x}, $$

where $\sigma (t)$ is a function to be determined. A direct verification indicates that

$$ g(x)=-1, \qquad h=-\frac{1}{2}+e^{(-\frac{1}{\beta }x+\frac{1}{2}t)} $$

is a set of solutions of (54). We see that the function h is dependent on the variable t.