1. Introduction
The Boussinesq equation, good or classical and regularized or improved, plays an important role in fluid mechanics fields, scientific applications, and ocean engineering. The good Boussinesq equation given as1 ) is completely integrable that gives multiple soliton solutions, whereas the improved equation (2 ) is not integrable, Both equations (1 ) and (2 ) have been studied intensively in the literatures [1–6] and some of the references therein.
$\begin{eqnarray}{u}_{{tt}}-{c}^{2}{u}_{{xx}}-{({u}^{2})}_{{xx}}-{{au}}_{{xxxx}}=0,\end{eqnarray}$
which describes the gravity induced surface waves as they propagate at constant speed in a canal of uniform depth. However, the regularized Boussinesq equation reads $\begin{eqnarray}{u}_{{tt}}-{c}^{2}{u}_{{xx}}-{({u}^{2})}_{{xx}}-{{bu}}_{{xxtt}}=0,\end{eqnarray}$
where the dispersion term uxxxx is replaced by the term uxxtt that improves the property of dispersion. Boussinesq equation [1] is commonly used to describe the long waves in shallow water. The classical equation (Boussinesq-type equations have attracted many authors to conduct research work on these equations that arise in a variety of scientific and engineering fields. Boussinesq-type equations arise in many physical applications such as, sound waves in plasma, horizontal layer of material, nonlinear string [1–6] etc. It is well know that many nonlinear evolution equations including Boussinesq-type equations have soliton solutions [7–9]. The bilinear method, in general, is a simple and straightforward approach to obtain soliton solutions [10–13]. In addition, symmetry and conservation laws [14–23] play a key role in nonlinear mathematical physics fields. This is due to the fact that symmetry can be used to solve nonlinear evolution equations and find their new solutions. Conservation laws are very useful to interpret and explain many complex natural phenomena, such as momentum conservation, mass conservation, energy conservation and so on. For many nonlinear evolution equations, there are not only rich physical significance, but also elegant mathematical structure.
Recently, the relationship between the symmetry breaking and the energy dissipation of the infinite-dimensional non-conservative dynamic systems was explored [24], which is the theoretical foundation of the structure-preserving approach for the non-conservative dynamic systems and resulted in the broad applications of the structure-preserving method [25, 26]. This achievement established a bridge between the structure-preserving approach and the engineering problems.
In this work, we will study the Boussinesq equation1 ) [2–4, 6, 27–30]. However, for $b\ne 0,a=0$, it is reduced to the improved Boussinesq equation (2 ) [31, 32]. In [27] derived new similarity reductions of the Boussinesq equation (3 ). Moreover, authors of [28] studied non-classical symmetry reduction of the Boussinesq equation. the author in [29] considered the Lie symmetry, conservation laws and solitary wave solutions of the Boussinesq equation.
$\begin{eqnarray}{u}_{{tt}}-{c}^{2}{u}_{{xx}}-{({u}^{2})}_{{xx}}-{{au}}_{{xxxx}}-{{bu}}_{{xxtt}}=0,\end{eqnarray}$
that involves two dispersive terms, namely uxxxx and uxxtt. It is clear that if $a\ne 0$, b = 0, This equation is reduced to the good Boussinesq equation (Concerning the mixed dispersive terms uxxxx and uxxtt, it is worth noting that Wang and Chen [33] studied the existence and uniqueness of the global solution for the Cauchy problem of the generalized double dispersion equation, where mixed dispersive terms, similar to what we used in (3 ), and they proved the blow-up result of the solution by using the concavity method and under some suitable conditions. They also emphasized that when the energy exchange between the surface of nonlinear elastic rod, whose material is hyperplastic (e.g. the Murnagham material), and the medium is considered, the double dispersion Boussinesq equation (3 ) can be derived from Hamilton principle, and it can also be obtained from the Euler equation for surface wave in irrotational motion. Moreover, Schneider and Wayne [34] considered the Boussinesq equation that models the water wave problem with surface tension that involves mixed dispersive terms, where this model can also be derived from the two dimensional water wave problem. In addition, Wang and Xue [35] studied the global solution for a generalized Boussinesq equation with two dispersive terms. More information about the double dispersive terms can be found in [33–35].
This paper is organized as follows, In section 2 , we derive the Korteweg–de Vries (KdV) equation from extended Boussinesq equation (3 ) via using perturbation analysis. Symmetry analysis is employed to investigate this equation in section 3 . In section 4 , Bäcklund transformation associated with truncated Painlevé expansion are addressed. In section 5 , conservation laws of this model (3 ) are presented. Reciprocal Bäcklund transformations of conservation laws are performed in section 6 . Conclusions are displayed in the last section.
2. Perturbation analysis to derive KdV equation
Firstly, we use the following transformation6 ) into (3 ) gives7 ) is reduced to
$\begin{eqnarray}\xi ={\varepsilon }^{\alpha }(x-{ct}),\tau ={\varepsilon }^{p\alpha }t,\end{eqnarray}$
where $\alpha ,p$ are need to be fixed, that is to say, $\begin{eqnarray}\displaystyle \frac{\partial }{\partial t}={\varepsilon }^{\alpha }\left(-c\displaystyle \frac{\partial }{\partial \xi }+{\varepsilon }^{(p-1)\alpha }\displaystyle \frac{\partial }{\partial \tau }\right),\displaystyle \frac{\partial }{\partial x}={\varepsilon }^{\alpha }\displaystyle \frac{\partial }{\partial \xi }.\end{eqnarray}$
Assuming that ${u}^{(i)}\to 0(i=1,2,\cdots )$ for $\xi \to 0$, we expand this function into a series by $\begin{eqnarray}\begin{array}{rcl}u(x,t) & = & {u}^{(0)}+\varepsilon {u}^{(1)}+{\varepsilon }^{2}{u}^{(2)}\\ & & +\,{\varepsilon }^{3}{u}^{(3)}+{\varepsilon }^{4}{u}^{(4)}+\cdots .\end{array}\end{eqnarray}$
Substituting ( $\begin{eqnarray}\begin{array}{l}{c}^{2}{\varepsilon }^{2\alpha }\displaystyle \frac{{\partial }^{2}}{\partial {\xi }^{2}}\left({u}^{(0)}+\varepsilon {u}^{(1)}+{\varepsilon }^{2}{u}^{(2)}+\cdots \right)\\ \ \ \ -\,2c{\varepsilon }^{(p+1)\alpha }\displaystyle \frac{{\partial }^{2}}{\partial \xi \partial \tau }\left({u}^{(0)}+\varepsilon {u}^{(1)}+{\varepsilon }^{2}{u}^{(2)}+\cdots \right)\\ +\,{\varepsilon }^{2p\alpha }\displaystyle \frac{{\partial }^{2}}{\partial {\tau }^{2}}\left({u}^{(0)}+\varepsilon {u}^{(1)}+{\varepsilon }^{2}{u}^{(2)}+\cdots \right)\\ \ \ \ -\,{c}^{2}{\varepsilon }^{2\alpha }\displaystyle \frac{{\partial }^{2}}{\partial {\xi }^{2}}\left({u}^{(0)}+\varepsilon {u}^{(1)}+{\varepsilon }^{2}{u}^{(2)}+\cdots \right)\\ \ \ \ -\,{{bc}}^{2}{\varepsilon }^{4\alpha }\displaystyle \frac{{\partial }^{4}}{\partial {\xi }^{4}}\left({u}^{(0)}+\varepsilon {u}^{(1)}+{\varepsilon }^{2}{u}^{(2)}+\cdots \right)\\ \ \ \ -\,a{\varepsilon }^{4\alpha }\displaystyle \frac{{\partial }^{4}}{\partial {\xi }^{4}}\left({u}^{(0)}+\varepsilon {u}^{(1)}+{\varepsilon }^{2}{u}^{(2)}+\cdots \right)\\ +2{bc}{\varepsilon }^{(p+3)\alpha }\displaystyle \frac{{\partial }^{4}}{\partial {\xi }^{3}\partial \tau }\left({u}^{(0)}+\varepsilon {u}^{(1)}+{\varepsilon }^{2}{u}^{(2)}+\cdots \right)\\ \ \ \ -\,{\varepsilon }^{2\alpha }\displaystyle \frac{{\partial }^{2}}{\partial {\xi }^{2}}{\left({u}^{(0)}+\varepsilon {u}^{(1)}+{\varepsilon }^{2}{u}^{(2)}+\cdots \right)}^{2}\\ -\,b{\varepsilon }^{2p\alpha +2\alpha }\displaystyle \frac{{\partial }^{4}}{\partial {\tau }^{2}\partial {\xi }^{2}}\left({u}^{(0)}+\varepsilon {u}^{(1)}+{\varepsilon }^{2}{u}^{(2)}+\cdots \right)=0.\end{array}\end{eqnarray}$
In order to arrive at our goal, we set ${u}^{(0)}={u}^{(1)}=0$, and this gives $(p+1)\alpha =4\alpha ,4\alpha +2=2\alpha +4$, which means $p=3,\alpha =1$. Then ( $\begin{eqnarray}\begin{array}{l}-2c{\varepsilon }^{4\alpha }\displaystyle \frac{{\partial }^{2}}{\partial \xi \partial \tau }\left({\varepsilon }^{2}{u}^{(2)}+\cdots \right)-{\varepsilon }^{2\alpha }\displaystyle \frac{{\partial }^{2}}{\partial {\xi }^{2}}{\left({\varepsilon }^{2}{u}^{(2)}+\cdots \right)}^{2}\\ -\,{{bc}}^{2}{\varepsilon }^{4\alpha }\displaystyle \frac{{\partial }^{4}}{\partial {\xi }^{4}}\left({\varepsilon }^{2}{u}^{(2)}+\cdots \right)\\ \ \ \ -\,a{\varepsilon }^{4\alpha }\displaystyle \frac{{\partial }^{4}}{\partial {\xi }^{4}}\left({\varepsilon }^{2}{u}^{(2)}+\cdots \right)+\cdots \,=\,0.\end{array}\end{eqnarray}$
Therefore, we get $\begin{eqnarray}\begin{array}{l}{\varepsilon }^{6\alpha }\left\{-2\displaystyle \frac{\partial }{\partial \xi }\left(c\displaystyle \frac{\partial {u}^{(2)}}{\partial \tau }\right.\right.\\ \quad \left.\left.+\,{u}^{(2)}\displaystyle \frac{\partial {u}^{(2)}}{\partial \xi }+\displaystyle \frac{a+{{bc}}^{2}}{2}\displaystyle \frac{{\partial }^{3}{u}^{(2)}}{\partial {\xi }^{3}}\right)+o(\varepsilon )\right\}=0.\end{array}\end{eqnarray}$
Consider the boundary condition ${u}^{(2)}\to 0$ for $\xi \to 0$, we get the nonlinear KdV equation $\begin{eqnarray}c\displaystyle \frac{\partial {u}^{(2)}}{\partial \tau }+{u}^{(2)}\displaystyle \frac{\partial {u}^{(2)}}{\partial \xi }+\displaystyle \frac{a+{{bc}}^{2}}{2}\displaystyle \frac{{\partial }^{3}{u}^{(2)}}{\partial {\xi }^{3}}=0.\end{eqnarray}$
It is clear that, from extended Boussinesq equation, we also get the nonlinear KdV equation [6].3. Generalized symmetry analysis, potential symmetry analysis and explicit solution
3.1. Generalized symmetry analysis
Now, we study equation (3 ) by following the generalized symmetry method as presented in the references ([14, 16, 17]). A function σ is a symmetry of equation (3 ), if this function satisfies3 ) as
$\begin{eqnarray}F^{\prime} (u)\sigma =0,\end{eqnarray}$
where $\begin{eqnarray}\begin{array}{rcl}F^{\prime} (u)\sigma & = & \displaystyle \frac{\partial F}{\partial u}\sigma +\displaystyle \frac{\partial F}{\partial {u}_{{tt}}}{\sigma }_{{tt}}+\displaystyle \frac{\partial F}{\partial {u}_{{xx}}}{\sigma }_{{xx}}\\ & & +\,\displaystyle \frac{\partial F}{\partial {u}_{{ttxx}}}{\sigma }_{{ttxx}}+\displaystyle \frac{\partial F}{\partial {u}_{{xxxx}}}{\sigma }_{{xxxx}}+\cdots .\end{array}\end{eqnarray}$
Therefore, we get the symmetry equation of equation ( $\begin{eqnarray}\begin{array}{l}{\sigma }_{{tt}}-{c}^{2}{\sigma }_{{xx}}-4{\sigma }_{x}{u}_{x}-2\sigma {u}_{{xx}}\\ \quad -2u{\sigma }_{{xx}}-a{\sigma }_{{xxxx}}-b{\sigma }_{{xxtt}}=0.\end{array}\end{eqnarray}$
Setting σ equals to $\begin{eqnarray}\sigma ={\beta }_{1}{u}_{x}+{\beta }_{2}{u}_{t}+{\beta }_{3}u+{\beta }_{0},\end{eqnarray}$
where the coefficients ${\beta }_{i}(i=0,1,2,3)$ are functions of $x,t$ and will be solved later.Substituting (14 ) into (13 ) and using (3 ), we obtain the following results
| • | $a\ne 0,b\ne 0$. $\begin{eqnarray}{\beta }_{1}={c}_{1},{\beta }_{2}={c}_{2},{\beta }_{3}={\beta }_{0}=0.\end{eqnarray}$ here ${c}_{1},{c}_{2}$ are arbitrary constants. |
| • | $a=0,b\ne 0$. $\begin{eqnarray}{\beta }_{1}={c}_{3},{\beta }_{2}={c}_{1}t+{c}_{2},{\beta }_{3}=2{c}_{1},{\beta }_{0}={c}^{2}{c}_{1}.\end{eqnarray}$ |
| • | $b=0,a\ne 0$. $\begin{eqnarray}\begin{array}{rcl}{\beta }_{1} & = & {c}_{1}x+{c}_{3},{\beta }_{2}=2{c}_{1}t+{c}_{2},\\ {\beta }_{3} & = & 2{c}_{1},{\beta }_{0}={c}^{2}{c}_{1}.\end{array}\end{eqnarray}$ |
Therefore, we get symmetries for equations (15 )–(17 ) respectively as
$\begin{eqnarray}\sigma ={c}_{1}{u}_{x}+{c}_{2}{u}_{t}.\end{eqnarray}$
$\begin{eqnarray}\sigma ={c}_{3}{u}_{x}+({c}_{1}t+{c}_{2}){u}_{t}+(2{c}_{1})u+{c}^{2}{c}_{1}.\end{eqnarray}$
$\begin{eqnarray}\sigma =({c}_{1}x+{c}_{3}){u}_{x}+(2{c}_{1}t+{c}_{2}){u}_{t}+(2{c}_{1})u+{c}^{2}{c}_{1}.\end{eqnarray}$
The equivalent vector expressions are obtained that read $\begin{eqnarray}V={c}_{1}\displaystyle \frac{\partial }{\partial x}+{c}_{2}\displaystyle \frac{\partial }{\partial t}.\end{eqnarray}$
$\begin{eqnarray}V={c}_{3}\displaystyle \frac{\partial }{\partial x}+({c}_{1}t+{c}_{2})\displaystyle \frac{\partial }{\partial t}-\left(2{c}_{1}u+{c}^{2}{c}_{1}\right)\displaystyle \frac{\partial }{\partial u}.\end{eqnarray}$
$\begin{eqnarray}\begin{array}{rcl}V & = & ({c}_{1}x+{c}_{3})\displaystyle \frac{\partial }{\partial x}+(2{c}_{1}t+{c}_{2})\\ & & \times \,\displaystyle \frac{\partial }{\partial t}-\left(2{c}_{1}u+{c}^{2}{c}_{1}\right)\displaystyle \frac{\partial }{\partial u}.\end{array}\end{eqnarray}$
In fact, for $a\ne 0,b=0$, it is in agreement with the obtained results for the Lie symmetry method reported in [29]. Thus, it is just our special case.3.2. Potential symmetry analysis
To conduct the potential symmetry analysis, we first rewrite (3 ) as conservation laws form ${T}^{t}+{X}^{x}=0$,24 ) we set25 ) let24 ) and (25 )24 ) and (25 ) do not have any potential symmetries.
$\begin{eqnarray}{\left({u}_{t}-{{bu}}_{{xxt}}\right)}_{t}+{\left(-{c}^{2}{u}_{x}-{({u}^{2})}_{x}-{{au}}_{{xxx}}\right)}_{x}=0,\end{eqnarray}$
or $\begin{eqnarray}{\left({u}_{t}\right)}_{t}+{\left(-{c}^{2}{u}_{x}-{({u}^{2})}_{x}-{{au}}_{{xxx}}-{{bu}}_{{xtt}}\right)}_{x}=0.\end{eqnarray}$
In order to derive the Lie point symmetry, for ( $\begin{eqnarray}\left\{\begin{array}{l}{u}_{t}-{{bu}}_{{xxt}}={v}_{x},\\ {c}^{2}{u}_{x}+{({u}^{2})}_{x}+{{au}}_{{xxx}}={v}_{t},\end{array}\right.\end{eqnarray}$
and for ( $\begin{eqnarray}\left\{\begin{array}{l}{u}_{t}={v}_{x},\\ {c}^{2}{u}_{x}+{({u}^{2})}_{x}+{{au}}_{{xxx}}+{{bu}}_{{xtt}}={v}_{t}.\end{array}\right.\end{eqnarray}$
Consider the following vector $\begin{eqnarray}\begin{array}{rcl}V & = & \tau (x,t,u,v)\displaystyle \frac{\partial }{\partial t}+\xi (x,t,u,v)\displaystyle \frac{\partial }{\partial x}\\ & & +\,\eta (x,t,u,v)\displaystyle \frac{\partial }{\partial u}+\phi (x,t,u,v)\displaystyle \frac{\partial }{\partial v}.\end{array}\end{eqnarray}$
Based on the symmetry method [14, 15, 17–22], we get the same results for ( $\begin{eqnarray}\tau ={c}_{3},\xi ={c}_{2},\phi ={c}_{1},\eta =0.\end{eqnarray}$
Based on the potential symmetry method [15], we found that these equations (3.3. Explicit solution via traveling wave transformation
Using traveling wave transformation, we set the invariant and invariant function30 ) into (3 ) leads to31 ) once gives31 ) again yields33 ) gives the following form solutions
$\begin{eqnarray}\xi =x-\nu t,f(\xi )=u(x-\nu t).\end{eqnarray}$
Substituting ( $\begin{eqnarray}\left({\nu }^{2}-{c}^{2}\right){f}_{\xi \xi }-{({f}^{2})}_{\xi \xi }-(a+b{\nu }^{2}){f}_{\xi \xi \xi \xi }=0.\end{eqnarray}$
Integrating equation ( $\begin{eqnarray}\left({\nu }^{2}-{c}^{2}\right){f}_{\xi }-{({f}^{2})}_{\xi }-(a+b{\nu }^{2}){f}_{\xi \xi \xi }=B.\end{eqnarray}$
Integrating equation ( $\begin{eqnarray}\left({\nu }^{2}-{c}^{2}\right)f-({f}^{2})-(a+b{\nu }^{2}){f}_{\xi \xi }=A+B\xi ,\end{eqnarray}$
where $A,B$ are the integral constants. If $A=B=0$, it is a reduced equation as given in [29]. On the other hand, equation ( $\begin{eqnarray}\begin{array}{rcl}f(\xi ) & = & {c}_{0}+{c}_{1}\xi +{c}_{2}{\xi }^{2}+\cdots +{c}_{n}{\xi }^{n}\\ & & +\,{c}_{n+1}{\xi }^{n+1}+\cdots =\displaystyle \sum _{n=0}^{\infty }{c}_{n}{\xi }^{n}.\end{array}\end{eqnarray}$
Substituting (34 ) into (33 ), one gets35 ), we have35 ), if $n\geqslant 2$, we obtain38 ) gives
$\begin{eqnarray}\begin{array}{l}-A-B\xi -(a+b{\nu }^{2})\\ \ \ \times \,\left(2{c}_{2}+6{c}_{3}\xi +\displaystyle \sum _{n=2}^{\infty }(n+1)(n+2){c}_{n+2}{\xi }^{n}\right)\\ \ \ -\,\left({c}_{0}^{2}+2{c}_{0}{c}_{1}\xi +\displaystyle \sum _{n=2}^{\infty }\displaystyle \sum _{k=0}^{n}{c}_{k}{c}_{n-k}{\xi }^{n}\right)\\ \ \ +\,\left({\nu }^{2}-{c}^{2}\right)\left({c}_{0}+{c}_{1}\xi +\displaystyle \sum _{n=2}^{\infty }{c}_{n}{\xi }^{n}\right)=0.\end{array}\end{eqnarray}$
Comparing coefficients of ( $\begin{eqnarray}\begin{array}{rcl}{c}_{2} & = & \displaystyle \frac{A+{c}_{0}^{2}-{\nu }^{2}{c}_{0}+{c}^{2}{c}_{0}}{-2a-2b{\nu }^{2}},\\ {c}_{3} & = & \displaystyle \frac{B+2{c}_{0}{c}_{1}-{\nu }^{2}{c}_{1}+{c}^{2}{c}_{1}}{-6a-6b{\nu }^{2}}.\end{array}\end{eqnarray}$
By considering the other coefficients of ( $\begin{eqnarray}\begin{array}{rcl}{c}_{n+2} & = & \displaystyle \frac{1}{(n+1)(n+2)(a+b{\nu }^{2})}\\ & & \times \,\left(\left({\nu }^{2}-{c}^{2}\right){c}_{n}-\displaystyle \sum _{k=0}^{n}{c}_{k}{c}_{n-k}\right).\end{array}\end{eqnarray}$
Therefore, we rewrite the power series solution as follows $\begin{eqnarray}\begin{array}{rcl}f(\xi ) & = & {c}_{0}+{c}_{1}\xi +{c}_{2}{\xi }^{2}++{c}_{3}{\xi }^{3}+\displaystyle \sum _{n=2}^{\infty }{c}_{n+2}{\xi }^{n+2}\\ & = & {c}_{0}+{c}_{1}\xi +\displaystyle \frac{A+{c}_{0}^{2}-{\nu }^{2}{c}_{0}+{c}^{2}{c}_{0}}{-2a-2b{\nu }^{2}}{\xi }^{2}\\ & & +\,\displaystyle \frac{B+2{c}_{0}{c}_{1}-{\nu }^{2}{c}_{1}+{c}^{2}{c}_{1}}{-6a-6b{\nu }^{2}}{\xi }^{3}\\ & & +\,\displaystyle \sum _{n=2}^{\infty }\displaystyle \frac{1}{(n+1)(n+2)(a+b{\nu }^{2})}\\ & & \times \,\left(\left({\nu }^{2}-{c}^{2}\right){c}_{n}-\displaystyle \sum _{k=0}^{n}{c}_{k}{c}_{n-k}\right){\xi }^{n+2}.\end{array}\end{eqnarray}$
Thus, substituting $\xi =x-\nu t$ into ( $\begin{eqnarray}\begin{array}{rcl}f(\xi ) & = & {c}_{0}+{c}_{1}\xi +{c}_{2}{\xi }^{2}++{c}_{3}{\xi }^{3}+\displaystyle \sum _{n=2}^{\infty }{c}_{n+2}{\xi }^{n+2}\\ & = & {c}_{0}+{c}_{1}\left(x-\nu t\right)+\displaystyle \frac{A+{c}_{0}^{2}-{\nu }^{2}{c}_{0}+{c}^{2}{c}_{0}}{-2a-2b{\nu }^{2}}{\left(x-\nu t\right)}^{2}\\ & & +\,\displaystyle \frac{B+2{c}_{0}{c}_{1}-{\nu }^{2}{c}_{1}+{c}^{2}{c}_{1}}{-6a-6b{\nu }^{2}}{\left(x-\nu t\right)}^{3}\\ & & +\,\displaystyle \sum _{n=2}^{\infty }\displaystyle \frac{\left(\left({\nu }^{2}-{c}^{2}\right){c}_{n}-{\displaystyle \sum }_{k=0}^{n}{c}_{k}{c}_{n-k}\right)}{(n+1)(n+2)(a+b{\nu }^{2})}{\left(x-\nu t\right)}^{n+2},\end{array}\end{eqnarray}$
where ${c}_{i}(i=0,1)$ are arbitrary constants.Let the integration constant B equal to zero and multiply ${f}_{\xi }$, integrate equation (33 ) once, we derive40 )
$\begin{eqnarray}\begin{array}{l}\displaystyle \frac{1}{2}\left({\nu }^{2}-{c}^{2}\right){f}^{2}-\displaystyle \frac{1}{3}({f}^{3})\\ \quad -\,\displaystyle \frac{1}{2}(a+b{\nu }^{2}){f}_{\xi }^{2}={Af}+M,\end{array}\end{eqnarray}$
where M is the integral constant. Rewrite equation ( $\begin{eqnarray}{f}_{\xi }^{2}=-\displaystyle \frac{2}{3(a+b{\nu }^{2})}\left({f}^{3}-\displaystyle \frac{3\left({\nu }^{2}-{c}^{2}\right)}{2}{f}^{2}+3{Af}+3M\right).\end{eqnarray}$
Consider the equation $\begin{eqnarray}{f}^{3}-\displaystyle \frac{3\left({\nu }^{2}-{c}^{2}\right)}{2}{f}^{2}+3{Af}+3M=0.\end{eqnarray}$
Consider the three solutions to this equation justify the inequality ${u}_{1}\geqslant {u}_{2}\geqslant {u}_{3}$, therefore, one gets ${u}_{1}+{u}_{2}\,+{u}_{3}=\tfrac{3\left({\nu }^{2}-{c}^{2}\right)}{2}$. Also, $\begin{eqnarray}{u}_{\xi }^{2}=-\displaystyle \frac{2}{3(a+b{\nu }^{2})}\left((u-{u}_{1})(u-{u}_{2})(u-{u}_{3})\right).\end{eqnarray}$
As a result we obtain the solution as $\begin{eqnarray}\begin{array}{rcl}u & = & {u}_{1}+({u}_{1}-{u}_{2}){\mathrm{cn}}^{2}\left(\sqrt{\frac{{u}_{1}-{u}_{3}}{6(a+b{\nu }^{2})}}\xi ,k\right),\\ & & \,{u}_{2}\leqslant u\leqslant {u}_{1},\end{array}\end{eqnarray}$
where $\begin{eqnarray}k=\sqrt{\displaystyle \frac{{u}_{1}-{u}_{2}}{{u}_{1}-{u}_{3}}}.\end{eqnarray}$
If $k\to 0$, we get trigonometric solution. However, for $k\to 1$, we get a solitary wave solution $\begin{eqnarray}u={u}_{2}+({u}_{1}-{u}_{2}){{\rm{sech}} }^{2}\left(\sqrt{\displaystyle \frac{{u}_{1}-{u}_{2}}{6(a+b{\nu }^{2})}}\xi \right).\end{eqnarray}$
In particular, set $A=M=0$, ${u}_{2}={u}_{3}=0$, ${u}_{1}=\tfrac{3\left({\nu }^{2}-{c}^{2}\right)}{2},$ we have $\begin{eqnarray}u=\displaystyle \frac{3\left({\nu }^{2}-{c}^{2}\right)}{2}{{\rm{sech}} }^{2}\left(\sqrt{\displaystyle \frac{{\nu }^{2}-{c}^{2}}{4(a+b{\nu }^{2})}}\xi \right).\end{eqnarray}$
It is clear that equation (41 ) includes a great number of solutions, such as trigonometric solutions, Jacobi elliptic solutions, hyperbolic solutions and so on. Here, we do not list them in details.
4. Bäcklund transformation associated with truncated Painlevé expansion
Based on the assumption presented earlier in section 2 , we set48 ) into (3 ), we get polynomials of different orders of f, all coefficients of different orders should be zero. Therefore, from coefficient: ${f}^{-6}$, one gets49 ) into coefficient ${f}^{-5}$, one finds49 ) and (50 ) into coefficient of ${f}^{-4}$, one obtains49 )–(51 ) into coefficient of ${f}^{-3}$, one derives an equation with regard to f. In other words, f needs to satisfy the equation of this coefficient ${f}^{-3}$ (see appendix ).
$\begin{eqnarray}u(x,t)={u}_{0}(x,t)+\displaystyle \frac{{u}_{1}(x,t)}{f}+\displaystyle \frac{{u}_{2}(x,t)}{{f}^{2}}={u}_{0}+\displaystyle \frac{{u}_{1}}{f}+\displaystyle \frac{{u}_{2}}{{f}^{2}}.\end{eqnarray}$
Inserting equation ( $\begin{eqnarray}{u}_{2}=-6{{af}}_{x}^{2}-6{{bf}}_{t}^{2}.\end{eqnarray}$
And then, substitute equation ( $\begin{eqnarray}{u}_{1}=\displaystyle \frac{6\left({f}_{{tt}}{f}_{x}^{2}{ab}+5{a}^{2}{f}_{x}^{2}{f}_{{xx}}{ab}+8{{abf}}_{x}{f}_{t}{f}_{{tx}}+5{f}_{{tt}}{f}_{t}^{2}{b}^{2}+{f}_{t}^{2}{f}_{{xx}}{ab}\right)}{5{{af}}_{x}^{2}+5{{bf}}_{t}^{2}}.\end{eqnarray}$
Substituting equations ( $\begin{eqnarray}\begin{array}{rcl}{u}_{0} & = & -\displaystyle \frac{1}{50{f}_{x}^{2}\left({f}_{x}^{6}{a}^{3}+3{f}_{x}^{4}{f}_{t}^{2}{a}^{2}b+3{f}_{x}^{2}{f}_{t}^{4}{{ab}}^{2}+{f}_{t}^{6}{b}^{3}\right)}\\ & & \times \left(25{f}_{x}^{2}{f}_{t}^{6}{b}^{3}{c}^{2}\right.\\ & & +\,25{f}_{x}^{8}{a}^{3}{c}^{2}+100{f}_{x}^{7}{f}_{{xxx}}{a}^{4}\\ & & -\,75{f}_{x}^{6}{f}_{{xx}}^{2}{a}^{4}-25{f}_{x}^{6}{f}_{t}^{2}{a}^{3}\\ & & +\,104{f}_{x}^{5}{f}_{t}{f}_{{tt}}{f}_{{tx}}{a}^{2}{b}^{2}-600{f}_{x}^{5}{f}_{t}{f}_{{tx}}{f}_{{xx}}{a}^{3}b\\ & & -\,412{f}_{x}^{4}{f}_{t}^{2}{f}_{{tt}}{f}_{{xx}}\\ & & -\,600{f}_{x}^{3}{f}_{t}^{3}{f}_{{tt}}{f}_{{tx}}{{ab}}^{3}\\ & & +\,104{f}_{x}^{3}{f}_{t}^{3}{f}_{{tx}}{f}_{{xx}}{a}^{2}{b}^{2}\\ & & -\,30{f}_{x}^{2}{f}_{t}^{4}{f}_{{tt}}{f}_{{xx}}{{ab}}^{3}-25{f}_{t}^{8}{b}^{3}\\ & & +\,160{f}_{x}^{4}{f}_{t}^{3}{f}_{{ttt}}{{ab}}^{3}\\ & & +\,240{f}_{x}^{4}{f}_{t}^{3}{f}_{{txx}}{a}^{2}{b}^{2}+150{f}_{x}^{4}{f}_{t}^{2}{f}_{{tt}}^{2}{{ab}}^{3}\\ & & -\,344{f}_{x}^{4}{f}_{t}^{2}{f}_{{tx}}^{2}{{ab}}^{2}+150{f}_{x}^{4}{f}_{t}^{2}{f}_{{xx}}^{2}{a}^{3}b\\ & & +\,180{f}_{x}^{3}{f}_{t}^{4}{f}_{{ttx}}{{ab}}^{3}\\ & & +\,60{f}_{x}^{3}{f}_{t}^{4}{f}_{{xxx}}{a}^{2}{b}^{2}+60{f}_{x}^{2}{f}_{t}^{5}{f}_{{txx}}{{ab}}^{3}\\ & & +\,180{f}_{x}^{2}{f}_{t}^{4}{f}_{{tx}}^{2}{{ab}}^{3}\\ & & +\,49{f}_{x}^{2}{f}_{t}^{4}{f}_{{xx}}^{2}{a}^{2}{b}^{2}+60{f}_{x}^{6}{f}_{t}{f}_{{ttt}}{a}^{2}{b}^{2}\\ & & +\,180{f}_{x}^{6}{f}_{t}{f}_{{xxt}}{a}^{3}b-30{f}_{x}^{6}{f}_{{tt}}{f}_{{xx}}{a}^{3}b\\ & & +\,240{f}_{x}^{5}{f}_{t}^{2}{f}_{{ttx}}{a}^{2}{b}^{2}\\ & & +\,160{f}_{x}^{5}{f}_{t}^{2}{f}_{{xxx}}{a}^{3}b+75{f}_{x}^{6}{f}_{t}^{2}{a}^{2}{{bc}}^{2}\\ & & +\,75{f}_{x}^{4}{f}_{t}^{4}{{ab}}^{2}{c}^{2}+60{f}_{x}^{7}{f}_{{xtt}}{a}^{3}b\\ & & +\,49{f}_{x}^{6}{f}_{{tt}}^{2}{a}^{2}{b}^{2}+180{f}_{x}^{6}{f}_{{tx}}^{2}{a}^{3}b+100{f}_{x}^{2}{f}_{t}^{5}{f}_{{ttt}}{b}^{4}\\ & & \left.-75{f}_{x}^{2}{f}_{t}^{4}{f}_{{tt}}^{2}{b}^{4}-75{f}_{x}^{4}{f}_{t}^{4}{a}^{2}b-75{f}_{x}^{2}{f}_{t}^{6}{{ab}}^{2}\right).\end{array}\end{eqnarray}$
Substitute equations (Therefore, we get the following theorems is a solution of (3 ), where u0, u1 and u2 are given by equations (49 )–(51 ) respectively.
If $f$ satisfies the equation of the coefficient ${f}^{-3}$, then ${u}_{0}$ is a solution of (
If $f$ satisfies the equation of the coefficient ${f}^{-3}$, then
$\begin{eqnarray}u(x,t)={u}_{0}+\displaystyle \frac{{u}_{1}}{f}+\displaystyle \frac{{u}_{2}}{{f}^{2}},\end{eqnarray}$
${u}_{0}$ is a auto Bäcklund transformation.
Proof: From coefficient f0, we can find that
$\begin{eqnarray}{u}_{0{tt}}-{c}^{2}{u}_{0{xx}}-{({u}^{2})}_{0{xx}}-{{au}}_{0{xxxx}}-{{bu}}_{0{xxtt}}=0.\end{eqnarray}$
5. Conservation laws
In this section, we study conservation laws of equation (3 ) by using the multipliers method [15]. One should get the multipliers,
$\begin{eqnarray}\begin{array}{l}{\rm{\Lambda }}1\left(t,x,u,{\text{}}{u}_{x},{\text{}}{u}_{{xx}},{\text{}}{u}_{{xxx}},{\text{}}{u}_{{xxxx}}\right)\\ \quad =\,\left({c}_{1}x+{c}_{2}\right)t+{c}_{3}x+{c}_{4}.\end{array}\end{eqnarray}$
Therefore, we getFor the multiplier ${xt}$, we have
$\begin{eqnarray}\left\{\begin{array}{l}{T}^{t}={{xtu}}_{t}-{{bxtu}}_{{xxt}}-{xu}+{{bxu}}_{{xx}},\\ {T}^{x}={{tu}}^{2}-2{{xtuu}}_{x}-{c}^{2}{{xtu}}_{x}-{{axtu}}_{{xxx}}+{c}^{2}{tu}+{{atu}}_{{xx}};\end{array}\right.\end{eqnarray}$
On the basis of the multiplier $t$, we get $\begin{eqnarray}\left\{\begin{array}{l}{T}^{t}={{tu}}_{t}-{{btu}}_{{xxt}}-u+{{bu}}_{{xx}},\\ {T}^{x}=-2{{tuu}}_{x}-{c}^{2}{{tu}}_{x}-{{atu}}_{{xxx}};\end{array}\right.\end{eqnarray}$
For the multiplier $x$, we have $\begin{eqnarray}\left\{\begin{array}{l}{T}^{t}={{xu}}_{t}-{{bxu}}_{{xxt}},\\ {T}^{x}={u}^{2}-2{{uu}}_{{xx}}-{c}^{2}{{xu}}_{x}-{{axu}}_{{xxx}}+{c}^{2}u+{{au}}_{{xx}};\end{array}\right.\end{eqnarray}$
As to the multiplier 1, we get $\begin{eqnarray}\left\{\begin{array}{l}{T}^{t}={u}_{t}-{{bu}}_{{xxt}},\\ {T}^{x}=-2{{uu}}_{x}-{c}^{2}{u}_{x}-{{au}}_{{xxx}},\end{array}\right.\end{eqnarray}$
or $\begin{eqnarray}\left\{\begin{array}{l}{T}^{t}={u}_{t},\\ {T}^{x}=-2{{uu}}_{x}-{c}^{2}{u}_{x}-{{au}}_{{xxx}}-{{bu}}_{{xtt}}.\end{array}\right.\end{eqnarray}$
6. Reciprocal Bäcklund transformations of conservation laws
Based on the method addressed in [36], we consider the reciprocal Bäcklund transformations of conservation laws.
[36] The conservation law
$\begin{eqnarray}\begin{array}{l}\displaystyle \frac{\partial }{{\partial }_{t}}\left\{T\left(\partial /{\partial }_{x};\partial /{\partial }_{t};u\right)\right\}\\ \quad +\displaystyle \frac{\partial }{{\partial }_{x}}\left\{F\left(\partial /{\partial }_{x};\partial /{\partial }_{t};u\right)\right\}=0\end{array}\end{eqnarray}$
is transformed to the reciprocally associated conservation law $\begin{eqnarray}\begin{array}{l}\displaystyle \frac{\partial }{{\partial }_{t^{\prime} }}\left\{T^{\prime} \left(\partial /{\partial }_{x^{\prime} };\partial /{\partial }_{t^{\prime} };u\right)\right\}\\ \quad +\,\displaystyle \frac{\partial }{{\partial }_{x^{\prime} }}\left\{F^{\prime} \left(\partial /{\partial }_{x^{\prime} };\partial /{\partial }_{t^{\prime} };u\right)\right\}=0\end{array}\end{eqnarray}$
by the reciprocal transformation $\begin{eqnarray}{\rm{d}}x^{\prime} =T{\rm{d}}x-F{\rm{d}}t,t^{\prime} =t,\end{eqnarray}$
where $\begin{eqnarray}T^{\prime} T=1,F^{\prime} T=-F.\end{eqnarray}$
From equations (62 ) and (63 ), we get
$\begin{eqnarray}\displaystyle \frac{\partial }{{\partial }_{t^{\prime} }}=\displaystyle \frac{F}{T}\displaystyle \frac{\partial }{{\partial }_{x}}+\displaystyle \frac{\partial }{{\partial }_{t}},\displaystyle \frac{\partial }{{\partial }_{x^{\prime} }}=\displaystyle \frac{1}{T}\displaystyle \frac{\partial }{{\partial }_{x}}.\end{eqnarray}$
Now, based on the Theorem 5, we will deal with reciprocal Bäcklund transformations of conservation laws.As to the multiplier 1, we obtain65 ) transform to the following form:
$\begin{eqnarray}\left\{\begin{array}{l}{T}^{t}={u}_{t}-{{bu}}_{{xxt}},\\ {T}^{x}=-2{{uu}}_{x}-{c}^{2}{u}_{x}-{{au}}_{{xxx}}.\end{array}\right.\end{eqnarray}$
Therefore, conservation laws ( $\begin{eqnarray}\left\{\begin{array}{l}({T}^{t})^{\prime} =\tfrac{1}{T}=\tfrac{1}{{u}_{t}-{{bu}}_{{xxt}}},\\ ({T}^{x})^{\prime} =-\tfrac{F}{T}=-\tfrac{-2{{uu}}_{x}-{c}^{2}{u}_{x}-{{au}}_{{xxx}}}{{u}_{t}-{{bu}}_{{xxt}}},\end{array}\right.\end{eqnarray}$
or equivalently $\begin{eqnarray}({T}^{t}){{\prime} }_{t^{\prime} }+({T}^{x}){{\prime} }_{x^{\prime} }=0.\end{eqnarray}$
And similarly, we can get the remaining reciprocal Bäcklund transformations of conservation laws.For the multiplier xt, we derive
$\begin{eqnarray}\left\{\begin{array}{l}({T}^{t})^{\prime} =\tfrac{1}{T}=\tfrac{1}{{{xtu}}_{t}-{{bxtu}}_{{xxt}}-{xu}+{{bxu}}_{{xx}}},\\ ({T}^{x})^{\prime} =-\tfrac{F}{T}=-\tfrac{{{tu}}^{2}-2{{xtuu}}_{x}-{c}^{2}{{xtu}}_{x}-{{axtu}}_{{xxx}}+{c}^{2}{tu}+{{atu}}_{{xx}}}{{{xtu}}_{t}-{{bxtu}}_{{xxt}}-{xu}+{{bxu}}_{{xx}}}.\end{array}\right.\end{eqnarray}$
As to the multiplier t, we find
$\begin{eqnarray}\left\{\begin{array}{l}({T}^{t})^{\prime} =\tfrac{1}{T}=\tfrac{1}{{{tu}}_{t}-{{btu}}_{{xxt}}-u+{{bu}}_{{xx}}},\\ ({T}^{x})^{\prime} =-\tfrac{F}{T}=-\tfrac{-2{{tuu}}_{x}-{c}^{2}{{tu}}_{x}-{{atu}}_{{xxx}}}{{{tu}}_{t}-{{btu}}_{{xxt}}-u+{{bu}}_{{xx}}}.\end{array}\right.\end{eqnarray}$
For the multiplier x, we have $\begin{eqnarray}\left\{\begin{array}{l}({T}^{t})^{\prime} =\tfrac{1}{T}=\tfrac{1}{{{xu}}_{t}-{{bxu}}_{{xxt}}},\\ ({T}^{x})^{\prime} =-\tfrac{F}{T}=-\tfrac{{u}^{2}-2{{uu}}_{{xx}}-{c}^{2}{{xu}}_{x}-{{axu}}_{{xxx}}+{c}^{2}u+{{au}}_{{xx}}}{{{xu}}_{t}-{{bxu}}_{{xxt}}}.\end{array}\right.\end{eqnarray}$
As to the multiplier 1, one can get $\begin{eqnarray}\left\{\begin{array}{l}({T}^{t})^{\prime} =\tfrac{1}{T}=\tfrac{1}{{u}_{t}},\\ ({T}^{x})^{\prime} =-\tfrac{F}{T}=-\tfrac{-2{{uu}}_{x}-{c}^{2}{u}_{x}-{{au}}_{{xxx}}-{{bu}}_{{xtt}}}{{u}_{t}}.\end{array}\right.\end{eqnarray}$
7. Conclusions
In the present work, the extended Boussinesq equation (3 ), that involves two dispersive terms, is investigated. From this equation, we derived the KdV equation using the perturbation analysis. Then, based on the generalized and potential symmetry, the corresponding symmetries and vector fields are performed. Some explicit solutions are also given. In addition, Bäcklund transformation associated with truncated Painlevé expansion are studied. Meanwhile, conservation laws also presented. After this, reciprocal Bäcklund transformations of conservation laws are performed for the first time. In this paper, we just considered the constant coefficients case. The obtained results will be employed for further works in the future.