Gangwei Wang^,1,∗, Abdul-Majid Wazwaz²¹School of Mathematics and Statistics, Hebei University of Economics and Business, Shijiazhuang, 050061, China
²Department of Mathematics, Saint Xavier University, Chicago, IL 60655, United States of America

First author contact: ∗ Author to whom any correspondence should be addressed.
Received:2020-11-11Revised:2020-12-10Accepted:2021-01-27Online:2021-02-25


Abstract
In this work, we study a generalized double dispersion Boussinesq equation that plays a significant role in fluid mechanics, scientific fields, and ocean engineering. This equation will be reduced to the Korteweg–de Vries equation via using the perturbation analysis. We derive the corresponding vectors, symmetry reduction and explicit solutions for this equation. We readily obtain Bäcklund transformation associated with truncated Painlevé expansion. We also examine the related conservation laws of this equation via using the multiplier method. Moreover, we investigate the reciprocal Bäcklund transformations of the derived conservation laws for the first time.
Keywords： Boussinesq equation;perturbation and symmetry analysis;Bäcklund transformation;conservation laws


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Gangwei Wang, Abdul-Majid Wazwaz. Perturbation, symmetry analysis, Bäcklund and reciprocal transformation for the extended Boussinesq equation in fluid mechanics. Communications in Theoretical Physics, 2021, 73(4): 045003- doi:10.1088/1572-9494/abe03a

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 as(1)$\begin{eqnarray}{u}_{{tt}}-{c}^{2}{u}_{{xx}}-{({u}^{2})}_{{xx}}-{{au}}_{{xxxx}}=0,\end{eqnarray}$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(2)$\begin{eqnarray}{u}_{{tt}}-{c}^{2}{u}_{{xx}}-{({u}^{2})}_{{xx}}-{{bu}}_{{xxtt}}=0,\end{eqnarray}$where the dispersion term u_xxxx is replaced by the term u_xxtt that improves the property of dispersion. Boussinesq equation [1] is commonly used to describe the long waves in shallow water. The classical equation (1) 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.

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 equation(3)$\begin{eqnarray}{u}_{{tt}}-{c}^{2}{u}_{{xx}}-{({u}^{2})}_{{xx}}-{{au}}_{{xxxx}}-{{bu}}_{{xxtt}}=0,\end{eqnarray}$that involves two dispersive terms, namely u_xxxx and u_xxtt. It is clear that if $a\ne 0$, b=0, This equation is reduced to the good Boussinesq equation (1) [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.

Concerning the mixed dispersive terms u_xxxx and u_xxtt, 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 transformation(4)$\begin{eqnarray}\xi ={\varepsilon }^{\alpha }(x-{ct}),\tau ={\varepsilon }^{p\alpha }t,\end{eqnarray}$where $\alpha ,p$ are need to be fixed, that is to say,(5)$\begin{eqnarray}\displaystyle \frac{\partial }{\partial t}={\varepsilon }^{\alpha }\left(-c\displaystyle \frac{\partial }{\partial \xi }+{\varepsilon }^{(p-1)\alpha }\displaystyle \frac{\partial }{\partial \tau }\right),\displaystyle \frac{\partial }{\partial x}={\varepsilon }^{\alpha }\displaystyle \frac{\partial }{\partial \xi }.\end{eqnarray}$Assuming that ${u}^{(i)}\to 0(i=1,2,\cdots )$ for $\xi \to 0$, we expand this function into a series by(6)$\begin{eqnarray}\begin{array}{rcl}u(x,t) & = & {u}^{(0)}+\varepsilon {u}^{(1)}+{\varepsilon }^{2}{u}^{(2)}\\ & & +\,{\varepsilon }^{3}{u}^{(3)}+{\varepsilon }^{4}{u}^{(4)}+\cdots .\end{array}\end{eqnarray}$Substituting (6) into (3) gives(7)$\begin{eqnarray}\begin{array}{l}{c}^{2}{\varepsilon }^{2\alpha }\displaystyle \frac{{\partial }^{2}}{\partial {\xi }^{2}}\left({u}^{(0)}+\varepsilon {u}^{(1)}+{\varepsilon }^{2}{u}^{(2)}+\cdots \right)\\ \ \ \ -\,2c{\varepsilon }^{(p+1)\alpha }\displaystyle \frac{{\partial }^{2}}{\partial \xi \partial \tau }\left({u}^{(0)}+\varepsilon {u}^{(1)}+{\varepsilon }^{2}{u}^{(2)}+\cdots \right)\\ +\,{\varepsilon }^{2p\alpha }\displaystyle \frac{{\partial }^{2}}{\partial {\tau }^{2}}\left({u}^{(0)}+\varepsilon {u}^{(1)}+{\varepsilon }^{2}{u}^{(2)}+\cdots \right)\\ \ \ \ -\,{c}^{2}{\varepsilon }^{2\alpha }\displaystyle \frac{{\partial }^{2}}{\partial {\xi }^{2}}\left({u}^{(0)}+\varepsilon {u}^{(1)}+{\varepsilon }^{2}{u}^{(2)}+\cdots \right)\\ \ \ \ -\,{{bc}}^{2}{\varepsilon }^{4\alpha }\displaystyle \frac{{\partial }^{4}}{\partial {\xi }^{4}}\left({u}^{(0)}+\varepsilon {u}^{(1)}+{\varepsilon }^{2}{u}^{(2)}+\cdots \right)\\ \ \ \ -\,a{\varepsilon }^{4\alpha }\displaystyle \frac{{\partial }^{4}}{\partial {\xi }^{4}}\left({u}^{(0)}+\varepsilon {u}^{(1)}+{\varepsilon }^{2}{u}^{(2)}+\cdots \right)\\ +2{bc}{\varepsilon }^{(p+3)\alpha }\displaystyle \frac{{\partial }^{4}}{\partial {\xi }^{3}\partial \tau }\left({u}^{(0)}+\varepsilon {u}^{(1)}+{\varepsilon }^{2}{u}^{(2)}+\cdots \right)\\ \ \ \ -\,{\varepsilon }^{2\alpha }\displaystyle \frac{{\partial }^{2}}{\partial {\xi }^{2}}{\left({u}^{(0)}+\varepsilon {u}^{(1)}+{\varepsilon }^{2}{u}^{(2)}+\cdots \right)}^{2}\\ -\,b{\varepsilon }^{2p\alpha +2\alpha }\displaystyle \frac{{\partial }^{4}}{\partial {\tau }^{2}\partial {\xi }^{2}}\left({u}^{(0)}+\varepsilon {u}^{(1)}+{\varepsilon }^{2}{u}^{(2)}+\cdots \right)=0.\end{array}\end{eqnarray}$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 (7) is reduced to(8)$\begin{eqnarray}\begin{array}{l}-2c{\varepsilon }^{4\alpha }\displaystyle \frac{{\partial }^{2}}{\partial \xi \partial \tau }\left({\varepsilon }^{2}{u}^{(2)}+\cdots \right)-{\varepsilon }^{2\alpha }\displaystyle \frac{{\partial }^{2}}{\partial {\xi }^{2}}{\left({\varepsilon }^{2}{u}^{(2)}+\cdots \right)}^{2}\\ -\,{{bc}}^{2}{\varepsilon }^{4\alpha }\displaystyle \frac{{\partial }^{4}}{\partial {\xi }^{4}}\left({\varepsilon }^{2}{u}^{(2)}+\cdots \right)\\ \ \ \ -\,a{\varepsilon }^{4\alpha }\displaystyle \frac{{\partial }^{4}}{\partial {\xi }^{4}}\left({\varepsilon }^{2}{u}^{(2)}+\cdots \right)+\cdots \,=\,0.\end{array}\end{eqnarray}$Therefore, we get(9)$\begin{eqnarray}\begin{array}{l}{\varepsilon }^{6\alpha }\left\{-2\displaystyle \frac{\partial }{\partial \xi }\left(c\displaystyle \frac{\partial {u}^{(2)}}{\partial \tau }\right.\right.\\ \quad \left.\left.+\,{u}^{(2)}\displaystyle \frac{\partial {u}^{(2)}}{\partial \xi }+\displaystyle \frac{a+{{bc}}^{2}}{2}\displaystyle \frac{{\partial }^{3}{u}^{(2)}}{\partial {\xi }^{3}}\right)+o(\varepsilon )\right\}=0.\end{array}\end{eqnarray}$Consider the boundary condition ${u}^{(2)}\to 0$ for $\xi \to 0$, we get the nonlinear KdV equation(10)$\begin{eqnarray}c\displaystyle \frac{\partial {u}^{(2)}}{\partial \tau }+{u}^{(2)}\displaystyle \frac{\partial {u}^{(2)}}{\partial \xi }+\displaystyle \frac{a+{{bc}}^{2}}{2}\displaystyle \frac{{\partial }^{3}{u}^{(2)}}{\partial {\xi }^{3}}=0.\end{eqnarray}$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 satisfies(11)$\begin{eqnarray}F^{\prime} (u)\sigma =0,\end{eqnarray}$where(12)$\begin{eqnarray}\begin{array}{rcl}F^{\prime} (u)\sigma & = & \displaystyle \frac{\partial F}{\partial u}\sigma +\displaystyle \frac{\partial F}{\partial {u}_{{tt}}}{\sigma }_{{tt}}+\displaystyle \frac{\partial F}{\partial {u}_{{xx}}}{\sigma }_{{xx}}\\ & & +\,\displaystyle \frac{\partial F}{\partial {u}_{{ttxx}}}{\sigma }_{{ttxx}}+\displaystyle \frac{\partial F}{\partial {u}_{{xxxx}}}{\sigma }_{{xxxx}}+\cdots .\end{array}\end{eqnarray}$Therefore, we get the symmetry equation of equation (3) as(13)$\begin{eqnarray}\begin{array}{l}{\sigma }_{{tt}}-{c}^{2}{\sigma }_{{xx}}-4{\sigma }_{x}{u}_{x}-2\sigma {u}_{{xx}}\\ \quad -2u{\sigma }_{{xx}}-a{\sigma }_{{xxxx}}-b{\sigma }_{{xxtt}}=0.\end{array}\end{eqnarray}$Setting σ equals to(14)$\begin{eqnarray}\sigma ={\beta }_{1}{u}_{x}+{\beta }_{2}{u}_{t}+{\beta }_{3}u+{\beta }_{0},\end{eqnarray}$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$a\ne 0,b\ne 0$. (15)$\begin{eqnarray}{\beta }_{1}={c}_{1},{\beta }_{2}={c}_{2},{\beta }_{3}={\beta }_{0}=0.\end{eqnarray}$here ${c}_{1},{c}_{2}$ are arbitrary constants.
$a=0,b\ne 0$. (16)$\begin{eqnarray}{\beta }_{1}={c}_{3},{\beta }_{2}={c}_{1}t+{c}_{2},{\beta }_{3}=2{c}_{1},{\beta }_{0}={c}^{2}{c}_{1}.\end{eqnarray}$
$b=0,a\ne 0$. (17)$\begin{eqnarray}\begin{array}{rcl}{\beta }_{1} & = & {c}_{1}x+{c}_{3},{\beta }_{2}=2{c}_{1}t+{c}_{2},\\ {\beta }_{3} & = & 2{c}_{1},{\beta }_{0}={c}^{2}{c}_{1}.\end{array}\end{eqnarray}$


Therefore, we get symmetries for equations (15)–(17) respectively as(18)$\begin{eqnarray}\sigma ={c}_{1}{u}_{x}+{c}_{2}{u}_{t}.\end{eqnarray}$(19)$\begin{eqnarray}\sigma ={c}_{3}{u}_{x}+({c}_{1}t+{c}_{2}){u}_{t}+(2{c}_{1})u+{c}^{2}{c}_{1}.\end{eqnarray}$(20)$\begin{eqnarray}\sigma =({c}_{1}x+{c}_{3}){u}_{x}+(2{c}_{1}t+{c}_{2}){u}_{t}+(2{c}_{1})u+{c}^{2}{c}_{1}.\end{eqnarray}$The equivalent vector expressions are obtained that read(21)$\begin{eqnarray}V={c}_{1}\displaystyle \frac{\partial }{\partial x}+{c}_{2}\displaystyle \frac{\partial }{\partial t}.\end{eqnarray}$(22)$\begin{eqnarray}V={c}_{3}\displaystyle \frac{\partial }{\partial x}+({c}_{1}t+{c}_{2})\displaystyle \frac{\partial }{\partial t}-\left(2{c}_{1}u+{c}^{2}{c}_{1}\right)\displaystyle \frac{\partial }{\partial u}.\end{eqnarray}$(23)$\begin{eqnarray}\begin{array}{rcl}V & = & ({c}_{1}x+{c}_{3})\displaystyle \frac{\partial }{\partial x}+(2{c}_{1}t+{c}_{2})\\ & & \times \,\displaystyle \frac{\partial }{\partial t}-\left(2{c}_{1}u+{c}^{2}{c}_{1}\right)\displaystyle \frac{\partial }{\partial u}.\end{array}\end{eqnarray}$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)$\begin{eqnarray}{\left({u}_{t}-{{bu}}_{{xxt}}\right)}_{t}+{\left(-{c}^{2}{u}_{x}-{({u}^{2})}_{x}-{{au}}_{{xxx}}\right)}_{x}=0,\end{eqnarray}$or(25)$\begin{eqnarray}{\left({u}_{t}\right)}_{t}+{\left(-{c}^{2}{u}_{x}-{({u}^{2})}_{x}-{{au}}_{{xxx}}-{{bu}}_{{xtt}}\right)}_{x}=0.\end{eqnarray}$In order to derive the Lie point symmetry, for (24) we set(26)$\begin{eqnarray}\left\{\begin{array}{l}{u}_{t}-{{bu}}_{{xxt}}={v}_{x},\\ {c}^{2}{u}_{x}+{({u}^{2})}_{x}+{{au}}_{{xxx}}={v}_{t},\end{array}\right.\end{eqnarray}$and for (25) let(27)$\begin{eqnarray}\left\{\begin{array}{l}{u}_{t}={v}_{x},\\ {c}^{2}{u}_{x}+{({u}^{2})}_{x}+{{au}}_{{xxx}}+{{bu}}_{{xtt}}={v}_{t}.\end{array}\right.\end{eqnarray}$Consider the following vector(28)$\begin{eqnarray}\begin{array}{rcl}V & = & \tau (x,t,u,v)\displaystyle \frac{\partial }{\partial t}+\xi (x,t,u,v)\displaystyle \frac{\partial }{\partial x}\\ & & +\,\eta (x,t,u,v)\displaystyle \frac{\partial }{\partial u}+\phi (x,t,u,v)\displaystyle \frac{\partial }{\partial v}.\end{array}\end{eqnarray}$Based on the symmetry method [14, 15, 17–22], we get the same results for (24) and (25)(29)$\begin{eqnarray}\tau ={c}_{3},\xi ={c}_{2},\phi ={c}_{1},\eta =0.\end{eqnarray}$Based on the potential symmetry method [15], we found that these equations (24) and (25) do not have any potential symmetries.

3.3. Explicit solution via traveling wave transformation

Using traveling wave transformation, we set the invariant and invariant function(30)$\begin{eqnarray}\xi =x-\nu t,f(\xi )=u(x-\nu t).\end{eqnarray}$Substituting (30) into (3) leads to(31)$\begin{eqnarray}\left({\nu }^{2}-{c}^{2}\right){f}_{\xi \xi }-{({f}^{2})}_{\xi \xi }-(a+b{\nu }^{2}){f}_{\xi \xi \xi \xi }=0.\end{eqnarray}$Integrating equation (31) once gives(32)$\begin{eqnarray}\left({\nu }^{2}-{c}^{2}\right){f}_{\xi }-{({f}^{2})}_{\xi }-(a+b{\nu }^{2}){f}_{\xi \xi \xi }=B.\end{eqnarray}$Integrating equation (31) again yields(33)$\begin{eqnarray}\left({\nu }^{2}-{c}^{2}\right)f-({f}^{2})-(a+b{\nu }^{2}){f}_{\xi \xi }=A+B\xi ,\end{eqnarray}$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 (33) gives the following form solutions(34)$\begin{eqnarray}\begin{array}{rcl}f(\xi ) & = & {c}_{0}+{c}_{1}\xi +{c}_{2}{\xi }^{2}+\cdots +{c}_{n}{\xi }^{n}\\ & & +\,{c}_{n+1}{\xi }^{n+1}+\cdots =\displaystyle \sum _{n=0}^{\infty }{c}_{n}{\xi }^{n}.\end{array}\end{eqnarray}$

Substituting (34) into (33), one gets(35)$\begin{eqnarray}\begin{array}{l}-A-B\xi -(a+b{\nu }^{2})\\ \ \ \times \,\left(2{c}_{2}+6{c}_{3}\xi +\displaystyle \sum _{n=2}^{\infty }(n+1)(n+2){c}_{n+2}{\xi }^{n}\right)\\ \ \ -\,\left({c}_{0}^{2}+2{c}_{0}{c}_{1}\xi +\displaystyle \sum _{n=2}^{\infty }\displaystyle \sum _{k=0}^{n}{c}_{k}{c}_{n-k}{\xi }^{n}\right)\\ \ \ +\,\left({\nu }^{2}-{c}^{2}\right)\left({c}_{0}+{c}_{1}\xi +\displaystyle \sum _{n=2}^{\infty }{c}_{n}{\xi }^{n}\right)=0.\end{array}\end{eqnarray}$Comparing coefficients of (35), we have(36)$\begin{eqnarray}\begin{array}{rcl}{c}_{2} & = & \displaystyle \frac{A+{c}_{0}^{2}-{\nu }^{2}{c}_{0}+{c}^{2}{c}_{0}}{-2a-2b{\nu }^{2}},\\ {c}_{3} & = & \displaystyle \frac{B+2{c}_{0}{c}_{1}-{\nu }^{2}{c}_{1}+{c}^{2}{c}_{1}}{-6a-6b{\nu }^{2}}.\end{array}\end{eqnarray}$By considering the other coefficients of (35), if $n\geqslant 2$, we obtain(37)$\begin{eqnarray}\begin{array}{rcl}{c}_{n+2} & = & \displaystyle \frac{1}{(n+1)(n+2)(a+b{\nu }^{2})}\\ & & \times \,\left(\left({\nu }^{2}-{c}^{2}\right){c}_{n}-\displaystyle \sum _{k=0}^{n}{c}_{k}{c}_{n-k}\right).\end{array}\end{eqnarray}$Therefore, we rewrite the power series solution as follows(38)$\begin{eqnarray}\begin{array}{rcl}f(\xi ) & = & {c}_{0}+{c}_{1}\xi +{c}_{2}{\xi }^{2}++{c}_{3}{\xi }^{3}+\displaystyle \sum _{n=2}^{\infty }{c}_{n+2}{\xi }^{n+2}\\ & = & {c}_{0}+{c}_{1}\xi +\displaystyle \frac{A+{c}_{0}^{2}-{\nu }^{2}{c}_{0}+{c}^{2}{c}_{0}}{-2a-2b{\nu }^{2}}{\xi }^{2}\\ & & +\,\displaystyle \frac{B+2{c}_{0}{c}_{1}-{\nu }^{2}{c}_{1}+{c}^{2}{c}_{1}}{-6a-6b{\nu }^{2}}{\xi }^{3}\\ & & +\,\displaystyle \sum _{n=2}^{\infty }\displaystyle \frac{1}{(n+1)(n+2)(a+b{\nu }^{2})}\\ & & \times \,\left(\left({\nu }^{2}-{c}^{2}\right){c}_{n}-\displaystyle \sum _{k=0}^{n}{c}_{k}{c}_{n-k}\right){\xi }^{n+2}.\end{array}\end{eqnarray}$Thus, substituting $\xi =x-\nu t$ into (38) gives(39)$\begin{eqnarray}\begin{array}{rcl}f(\xi ) & = & {c}_{0}+{c}_{1}\xi +{c}_{2}{\xi }^{2}++{c}_{3}{\xi }^{3}+\displaystyle \sum _{n=2}^{\infty }{c}_{n+2}{\xi }^{n+2}\\ & = & {c}_{0}+{c}_{1}\left(x-\nu t\right)+\displaystyle \frac{A+{c}_{0}^{2}-{\nu }^{2}{c}_{0}+{c}^{2}{c}_{0}}{-2a-2b{\nu }^{2}}{\left(x-\nu t\right)}^{2}\\ & & +\,\displaystyle \frac{B+2{c}_{0}{c}_{1}-{\nu }^{2}{c}_{1}+{c}^{2}{c}_{1}}{-6a-6b{\nu }^{2}}{\left(x-\nu t\right)}^{3}\\ & & +\,\displaystyle \sum _{n=2}^{\infty }\displaystyle \frac{\left(\left({\nu }^{2}-{c}^{2}\right){c}_{n}-{\displaystyle \sum }_{k=0}^{n}{c}_{k}{c}_{n-k}\right)}{(n+1)(n+2)(a+b{\nu }^{2})}{\left(x-\nu t\right)}^{n+2},\end{array}\end{eqnarray}$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 derive(40)$\begin{eqnarray}\begin{array}{l}\displaystyle \frac{1}{2}\left({\nu }^{2}-{c}^{2}\right){f}^{2}-\displaystyle \frac{1}{3}({f}^{3})\\ \quad -\,\displaystyle \frac{1}{2}(a+b{\nu }^{2}){f}_{\xi }^{2}={Af}+M,\end{array}\end{eqnarray}$where M is the integral constant. Rewrite equation (40)(41)$\begin{eqnarray}{f}_{\xi }^{2}=-\displaystyle \frac{2}{3(a+b{\nu }^{2})}\left({f}^{3}-\displaystyle \frac{3\left({\nu }^{2}-{c}^{2}\right)}{2}{f}^{2}+3{Af}+3M\right).\end{eqnarray}$Consider the equation(42)$\begin{eqnarray}{f}^{3}-\displaystyle \frac{3\left({\nu }^{2}-{c}^{2}\right)}{2}{f}^{2}+3{Af}+3M=0.\end{eqnarray}$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,(43)$\begin{eqnarray}{u}_{\xi }^{2}=-\displaystyle \frac{2}{3(a+b{\nu }^{2})}\left((u-{u}_{1})(u-{u}_{2})(u-{u}_{3})\right).\end{eqnarray}$As a result we obtain the solution as(44)$\begin{eqnarray}\begin{array}{rcl}u & = & {u}_{1}+({u}_{1}-{u}_{2}){\mathrm{cn}}^{2}\left(\sqrt{\frac{{u}_{1}-{u}_{3}}{6(a+b{\nu }^{2})}}\xi ,k\right),\\ & & \,{u}_{2}\leqslant u\leqslant {u}_{1},\end{array}\end{eqnarray}$where(45)$\begin{eqnarray}k=\sqrt{\displaystyle \frac{{u}_{1}-{u}_{2}}{{u}_{1}-{u}_{3}}}.\end{eqnarray}$If $k\to 0$, we get trigonometric solution. However, for $k\to 1$, we get a solitary wave solution(46)$\begin{eqnarray}u={u}_{2}+({u}_{1}-{u}_{2}){{\rm{sech}} }^{2}\left(\sqrt{\displaystyle \frac{{u}_{1}-{u}_{2}}{6(a+b{\nu }^{2})}}\xi \right).\end{eqnarray}$In particular, set $A=M=0$, ${u}_{2}={u}_{3}=0$, ${u}_{1}=\tfrac{3\left({\nu }^{2}-{c}^{2}\right)}{2},$ we have(47)$\begin{eqnarray}u=\displaystyle \frac{3\left({\nu }^{2}-{c}^{2}\right)}{2}{{\rm{sech}} }^{2}\left(\sqrt{\displaystyle \frac{{\nu }^{2}-{c}^{2}}{4(a+b{\nu }^{2})}}\xi \right).\end{eqnarray}$

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 set(48)$\begin{eqnarray}u(x,t)={u}_{0}(x,t)+\displaystyle \frac{{u}_{1}(x,t)}{f}+\displaystyle \frac{{u}_{2}(x,t)}{{f}^{2}}={u}_{0}+\displaystyle \frac{{u}_{1}}{f}+\displaystyle \frac{{u}_{2}}{{f}^{2}}.\end{eqnarray}$Inserting equation (48) into (3), we get polynomials of different orders of f, all coefficients of different orders should be zero. Therefore, from coefficient: ${f}^{-6}$, one gets(49)$\begin{eqnarray}{u}_{2}=-6{{af}}_{x}^{2}-6{{bf}}_{t}^{2}.\end{eqnarray}$And then, substitute equation (49) into coefficient ${f}^{-5}$, one finds(50)$\begin{eqnarray}{u}_{1}=\displaystyle \frac{6\left({f}_{{tt}}{f}_{x}^{2}{ab}+5{a}^{2}{f}_{x}^{2}{f}_{{xx}}{ab}+8{{abf}}_{x}{f}_{t}{f}_{{tx}}+5{f}_{{tt}}{f}_{t}^{2}{b}^{2}+{f}_{t}^{2}{f}_{{xx}}{ab}\right)}{5{{af}}_{x}^{2}+5{{bf}}_{t}^{2}}.\end{eqnarray}$Substituting equations (49) and (50) into coefficient of ${f}^{-4}$, one obtains(51)$\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 (49)–(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).

Therefore, we get the following theorems is a solution of (3), where u₀, u₁ and u₂ are given by equations (49)–(51) respectively.

Theorem 1. If $f$ satisfies the equation of the coefficient ${f}^{-3}$, then ${u}_{0}$ is a solution of (3).

Theorem 2. If $f$ satisfies the equation of the coefficient ${f}^{-3}$, then(52)$\begin{eqnarray}u(x,t)={u}_{0}+\displaystyle \frac{{u}_{1}}{f}+\displaystyle \frac{{u}_{2}}{{f}^{2}},\end{eqnarray}$

Theorem 3. ${u}_{0}$ is a auto Bäcklund transformation.

Proof: From coefficient f⁰, we can find that(53)$\begin{eqnarray}{u}_{0{tt}}-{c}^{2}{u}_{0{xx}}-{({u}^{2})}_{0{xx}}-{{au}}_{0{xxxx}}-{{bu}}_{0{xxtt}}=0.\end{eqnarray}$

5. Conservation laws

In this section, we study conservation laws of equation (3) by using the multipliers method [15]. One should get the multipliers,(54)$\begin{eqnarray}\begin{array}{l}{\rm{\Lambda }}1\left(t,x,u,{\text{}}{u}_{x},{\text{}}{u}_{{xx}},{\text{}}{u}_{{xxx}},{\text{}}{u}_{{xxxx}}\right)\\ \quad =\,\left({c}_{1}x+{c}_{2}\right)t+{c}_{3}x+{c}_{4}.\end{array}\end{eqnarray}$Therefore, we get

Theorem 4. For the multiplier ${xt}$, we have(55)$\begin{eqnarray}\left\{\begin{array}{l}{T}^{t}={{xtu}}_{t}-{{bxtu}}_{{xxt}}-{xu}+{{bxu}}_{{xx}},\\ {T}^{x}={{tu}}^{2}-2{{xtuu}}_{x}-{c}^{2}{{xtu}}_{x}-{{axtu}}_{{xxx}}+{c}^{2}{tu}+{{atu}}_{{xx}};\end{array}\right.\end{eqnarray}$On the basis of the multiplier $t$, we get(56)$\begin{eqnarray}\left\{\begin{array}{l}{T}^{t}={{tu}}_{t}-{{btu}}_{{xxt}}-u+{{bu}}_{{xx}},\\ {T}^{x}=-2{{tuu}}_{x}-{c}^{2}{{tu}}_{x}-{{atu}}_{{xxx}};\end{array}\right.\end{eqnarray}$For the multiplier $x$, we have(57)$\begin{eqnarray}\left\{\begin{array}{l}{T}^{t}={{xu}}_{t}-{{bxu}}_{{xxt}},\\ {T}^{x}={u}^{2}-2{{uu}}_{{xx}}-{c}^{2}{{xu}}_{x}-{{axu}}_{{xxx}}+{c}^{2}u+{{au}}_{{xx}};\end{array}\right.\end{eqnarray}$As to the multiplier 1, we get(58)$\begin{eqnarray}\left\{\begin{array}{l}{T}^{t}={u}_{t}-{{bu}}_{{xxt}},\\ {T}^{x}=-2{{uu}}_{x}-{c}^{2}{u}_{x}-{{au}}_{{xxx}},\end{array}\right.\end{eqnarray}$or(59)$\begin{eqnarray}\left\{\begin{array}{l}{T}^{t}={u}_{t},\\ {T}^{x}=-2{{uu}}_{x}-{c}^{2}{u}_{x}-{{au}}_{{xxx}}-{{bu}}_{{xtt}}.\end{array}\right.\end{eqnarray}$

6. Reciprocal Bäcklund transformations of conservation laws

Based on the method addressed in [36], we consider the reciprocal Bäcklund transformations of conservation laws.

Theorem 5. [36] The conservation law(60)$\begin{eqnarray}\begin{array}{l}\displaystyle \frac{\partial }{{\partial }_{t}}\left\{T\left(\partial /{\partial }_{x};\partial /{\partial }_{t};u\right)\right\}\\ \quad +\displaystyle \frac{\partial }{{\partial }_{x}}\left\{F\left(\partial /{\partial }_{x};\partial /{\partial }_{t};u\right)\right\}=0\end{array}\end{eqnarray}$is transformed to the reciprocally associated conservation law(61)$\begin{eqnarray}\begin{array}{l}\displaystyle \frac{\partial }{{\partial }_{t^{\prime} }}\left\{T^{\prime} \left(\partial /{\partial }_{x^{\prime} };\partial /{\partial }_{t^{\prime} };u\right)\right\}\\ \quad +\,\displaystyle \frac{\partial }{{\partial }_{x^{\prime} }}\left\{F^{\prime} \left(\partial /{\partial }_{x^{\prime} };\partial /{\partial }_{t^{\prime} };u\right)\right\}=0\end{array}\end{eqnarray}$by the reciprocal transformation(62)$\begin{eqnarray}{\rm{d}}x^{\prime} =T{\rm{d}}x-F{\rm{d}}t,t^{\prime} =t,\end{eqnarray}$where(63)$\begin{eqnarray}T^{\prime} T=1,F^{\prime} T=-F.\end{eqnarray}$

From equations (62) and (63), we get(64)$\begin{eqnarray}\displaystyle \frac{\partial }{{\partial }_{t^{\prime} }}=\displaystyle \frac{F}{T}\displaystyle \frac{\partial }{{\partial }_{x}}+\displaystyle \frac{\partial }{{\partial }_{t}},\displaystyle \frac{\partial }{{\partial }_{x^{\prime} }}=\displaystyle \frac{1}{T}\displaystyle \frac{\partial }{{\partial }_{x}}.\end{eqnarray}$Now, based on the Theorem 5, we will deal with reciprocal Bäcklund transformations of conservation laws.

As to the multiplier 1, we obtain(65)$\begin{eqnarray}\left\{\begin{array}{l}{T}^{t}={u}_{t}-{{bu}}_{{xxt}},\\ {T}^{x}=-2{{uu}}_{x}-{c}^{2}{u}_{x}-{{au}}_{{xxx}}.\end{array}\right.\end{eqnarray}$Therefore, conservation laws (65) transform to the following form:(66)$\begin{eqnarray}\left\{\begin{array}{l}({T}^{t})^{\prime} =\tfrac{1}{T}=\tfrac{1}{{u}_{t}-{{bu}}_{{xxt}}},\\ ({T}^{x})^{\prime} =-\tfrac{F}{T}=-\tfrac{-2{{uu}}_{x}-{c}^{2}{u}_{x}-{{au}}_{{xxx}}}{{u}_{t}-{{bu}}_{{xxt}}},\end{array}\right.\end{eqnarray}$or equivalently(67)$\begin{eqnarray}({T}^{t}){{\prime} }_{t^{\prime} }+({T}^{x}){{\prime} }_{x^{\prime} }=0.\end{eqnarray}$And similarly, we can get the remaining reciprocal Bäcklund transformations of conservation laws.

For the multiplier xt, we derive(68)$\begin{eqnarray}\left\{\begin{array}{l}({T}^{t})^{\prime} =\tfrac{1}{T}=\tfrac{1}{{{xtu}}_{t}-{{bxtu}}_{{xxt}}-{xu}+{{bxu}}_{{xx}}},\\ ({T}^{x})^{\prime} =-\tfrac{F}{T}=-\tfrac{{{tu}}^{2}-2{{xtuu}}_{x}-{c}^{2}{{xtu}}_{x}-{{axtu}}_{{xxx}}+{c}^{2}{tu}+{{atu}}_{{xx}}}{{{xtu}}_{t}-{{bxtu}}_{{xxt}}-{xu}+{{bxu}}_{{xx}}}.\end{array}\right.\end{eqnarray}$

As to the multiplier t, we find(69)$\begin{eqnarray}\left\{\begin{array}{l}({T}^{t})^{\prime} =\tfrac{1}{T}=\tfrac{1}{{{tu}}_{t}-{{btu}}_{{xxt}}-u+{{bu}}_{{xx}}},\\ ({T}^{x})^{\prime} =-\tfrac{F}{T}=-\tfrac{-2{{tuu}}_{x}-{c}^{2}{{tu}}_{x}-{{atu}}_{{xxx}}}{{{tu}}_{t}-{{btu}}_{{xxt}}-u+{{bu}}_{{xx}}}.\end{array}\right.\end{eqnarray}$For the multiplier x, we have(70)$\begin{eqnarray}\left\{\begin{array}{l}({T}^{t})^{\prime} =\tfrac{1}{T}=\tfrac{1}{{{xu}}_{t}-{{bxu}}_{{xxt}}},\\ ({T}^{x})^{\prime} =-\tfrac{F}{T}=-\tfrac{{u}^{2}-2{{uu}}_{{xx}}-{c}^{2}{{xu}}_{x}-{{axu}}_{{xxx}}+{c}^{2}u+{{au}}_{{xx}}}{{{xu}}_{t}-{{bxu}}_{{xxt}}}.\end{array}\right.\end{eqnarray}$As to the multiplier 1, one can get(71)$\begin{eqnarray}\left\{\begin{array}{l}({T}^{t})^{\prime} =\tfrac{1}{T}=\tfrac{1}{{u}_{t}},\\ ({T}^{x})^{\prime} =-\tfrac{F}{T}=-\tfrac{-2{{uu}}_{x}-{c}^{2}{u}_{x}-{{au}}_{{xxx}}-{{bu}}_{{xtt}}}{{u}_{t}}.\end{array}\right.\end{eqnarray}$

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.

Acknowledgments

This work is supported by Natural Science Foundation of Hebei Province, China (Grant No. A2018207030), Youth Key Program of Hebei University of Economics and Business (2018QZ07), Key Program of Hebei University of Economics and Business (2020ZD11), Youth Team Support Program of Hebei University of Economics and Business. National Natural Science Foundation of China (Grant No. 11 801 133).

Reference View Option By original order
By published year
By cited within times
By Impact factor

[1]Boussinesq M J 1871Theorie de lintumescence liquide appelee onde solitaire ou de translation se propageant dans un canal rectangulaire
C. R. Math. Acad. Sci. 72 755 759
[Cited within: 3]

[2]Benjamin T B 1986On the Boussinesq model for two-dimensional wave motions in heterogeneous fluids
J. Fluid Mech. 165 445 474
DOI:10.1017/S0022112086003178 [Cited within: 1]

[3]Madsen P A Bingham H B Liu H 2002A new Boussinesq method for fully nonlinear waves from shallow to deep water
J. Fluid Mech. 462 1 30
DOI:10.1017/S0022112002008467

[4]Wazwaz A M 2007Multiple-soliton solutions for the Boussinesq equation
Appl. Math. Comput. 192 479 486
DOI:10.1016/j.amc.2007.03.023 [Cited within: 1]

[5]Hirota R 1977J. Satsuma, Nonlinear evolution equations generated from the Backlund transformation for the Boussinesq equation
Prog. Theor. Phys. 57 797 807
DOI:10.1143/PTP.57.797

[6]Koji M et al. 1976A perturbation method and its application to obliquely propagating nonlinear alfven wave
J. Phys. Soc. Japan 41 2114 2120
DOI:10.1143/JPSJ.41.2114 [Cited within: 4]

[7]Ablowitz M J et al. 1973Nonlinear evolution equations of physical significance
Phys. Rev. Lett. 31 125 127
DOI:10.1103/PhysRevLett.31.125 [Cited within: 1]

[8]Newell A C 1985Solitons in Mathematics and Physics
CBMS Lectures vol 48 Philadelphia, PA SIAM


[9]Li H Lou S Y 2019Multiple soliton solutions of alice-bob Boussinesq equations
Chin. Phys. Lett. 36 050501
DOI:10.1088/0256-307X/36/5/050501 [Cited within: 1]

[10]Hirota R 2004 The Direct Method in Soliton Theory Cambridge Cambridge University Press
[Cited within: 1]

[11]Guo H D Xia T C Hu B B 2020Dynamics of abundant solutions to the (3+1)-dimensional generalized Yu-Toda-Sasa-Fukuyama equation
Appl. Math. Lett. 105 106301
DOI:10.1016/j.aml.2020.106301

[12]Guo H D Xia T C Hu B B 2020High-order lumps, high-order breathers and hybrid solutions for an extended (3+1)-dimensional Jimbo-Miwa equation in fluid dynamics
Nonlinear Dyn. 100 601 614
DOI:10.1007/s11071-020-05514-9

[13]Guo H D Xia T C Ma W X 1919Localized waves and interaction solutions to an extended (3+1)-dimensional Kadomtsev-Petviashvili equation
Mod. Phys. Lett. B 34 2050076
DOI:10.1142/S0217984920500761 [Cited within: 1]

[14]Olver P J 1986 Application of Lie Group to Differential Equation New York Springer
[Cited within: 3]

[15]Bluman G W Cheviakov A Anco S 2010 Applications of Symmetry Methods to Partial Differential Equations New York Springer
[Cited within: 3]

[16]Tian C 2001 Lie Groups and its Applications to Differential Equations Beijing Science Press (in Chinese)
[Cited within: 1]

[17]Wang G W Liu X Q Ying Y Y 2013Symmetry reduction, exact solutions and conservation laws of a new fifth-order nonlinear integrable equation
Commun. Nonlinear Sci. Numer. Simul. 18 2313 2320
DOI:10.1016/j.cnsns.2012.12.003 [Cited within: 2]

[18]Wang G W 2016Symmetry analysis and rogue wave solutions for the (2+1)-dimensional nonlinear Schrödinger equation with variable coefficients
Appl. Math. Lett. 56 56 64
DOI:10.1016/j.aml.2015.12.011

[19]Wang G W et al. 2020A (2+1)-dimensional sine-Gordon and sinh-Gordon equations with symmetries and kink wave solutions
Nucl. Phys. B 953 114956
DOI:10.1016/j.nuclphysb.2020.114956

[20]Wang G W Kara A H 2019A (2+1)-dimensional KdV equation and mKdV equation: symmetries, group invariant solutions and conservation laws
Phys. Lett. A 383 728 731
DOI:10.1016/j.physleta.2018.11.040

[21]Wang G W et al. 2020(2+1)-dimensional Boiti–Leon–Pempinelli equation-domain walls, invariance properties and conservation laws
Phys. Lett. A 384 126255
DOI:10.1016/j.physleta.2020.126255

[22]Wang G W et al. 2020Symmetry analysis for a seventh-order generalized KdV equation and its fractional version in fluid mechanics
Fractals 28 2050044
DOI:10.1142/S0218348X20500449 [Cited within: 1]

[23]Wang G W 2021A novel (3+1)-dimensional sine-Gorden and sinh-Gorden equation: Derivation, symmetries and conservation laws
Appl. Math. Lett. 113 106768
DOI:10.1016/j.aml.2020.106768 [Cited within: 1]

[24]Hu W et al. 2020Symmetry breaking of infinite-dimensional dynamic system
Appl. Math. Lett. 103 106207
DOI:10.1016/j.aml.2019.106207 [Cited within: 1]

[25]Hu W Zhang C Deng Z 2020Vibration and elastic wave propagation in spatial flexible damping panel attached to four special springs
Commun. Nonlinear Sci. Numer. Simul. 84 105199
DOI:10.1016/j.cnsns.2020.105199 [Cited within: 1]

[26]Hu W et al. 2021Coupling dynamic behaviors of flexible stretching hub-beam system
Mech. Syst. Signal Process. 151 107389
DOI:10.1016/j.ymssp.2020.107389 [Cited within: 1]

[27]Clarkson P A Kruskal M D 1989New similarity reductions of the Boussinesq equation
J. Math. Phys. 30 2201 2213
DOI:10.1063/1.528613 [Cited within: 2]

[28]Levi D Winternitz P 1989Non-classical symmetry reduction: example of the Boussinesq equation
J. Phys. A: Math. Gen. 22 2915 2924
DOI:10.1088/0305-4470/22/15/010 [Cited within: 1]

[29]Tian S F 2020Lie symmetry analysis, conservation laws and solitary wave solutions to a fourth-order nonlinear generalized Boussinesq water wave equation
Appl. Math. Lett. 100 106056
DOI:10.1016/j.aml.2019.106056 [Cited within: 3]

[30]Yang B Yang J K 2020General rogue waves in the Boussinesq equation
J. Phys. Soc. Jpn. 89 024003
DOI:10.7566/JPSJ.89.024003 [Cited within: 1]

[31]Boyd J 1998 Weakly Nonlocal Solitary Waves and Beyond-All-Orders Asymptotics Boston Kluwer
[Cited within: 1]

[32]Wazwaz A M 2009 Partial Differential Equations and Solitary Waves Theory New York Springer
[Cited within: 1]

[33]Wang S Chen G 2006Cauchy problem of the generalized double dispersion equation
Nonlinear Anal. 64 159 173
DOI:10.1016/j.na.2005.06.017 [Cited within: 2]

[34]Schneider G Wayne C E 2001Kawahara dynamics in dispersive media
Physica D 152 384 394
DOI:10.1016/S0167-2789(01)00181-6 [Cited within: 1]

[35]Wang S Xue H 2008Global solution for a generalized Boussinesq equation
Appl. Math. Comput. 204 130 136
DOI:10.1016/j.amc.2008.06.059 [Cited within: 2]

[36]Kingston J G Rogers C 1982Reciprocal Bäcklund transformations of conservation laws
Phys. Lett. A 92 261 264
DOI:10.1016/0375-9601(82)90081-0 [Cited within: 2]