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Localization of nonlocal symmetries and interaction solutions of the Sawada-Kotera equation

  • Jian-wen Wu ,
  • Yue-jin Cai ,
  • Ji Lin ,
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  • Department of Physics, Zhejiang Normal University, Jinhua, Zhejiang 321004, China

Author to whom any correspondence should be addressed.

Received date: 2021-01-30

  Revised date: 2021-04-06

  Accepted date: 2021-04-07

  Online published: 2021-06-11

Copyright

© 2021 Institute of Theoretical Physics CAS, Chinese Physical Society and IOP Publishing

Abstract

The nonlocal symmetry of the Sawada-Kotera (SK) equation is constructed with the known Lax pair. By introducing suitable and simple auxiliary variables, the nonlocal symmetry is localized and the finite transformation and some new solutions are obtained further. On the other hand, the group invariant solutions of the SK equation are constructed with the classic Lie group method. In particular, by a Galileo transformation some analytical soliton-cnoidal interaction solutions of a asymptotically integrable equation are discussed in graphical ways.

Cite this article

Jian-wen Wu , Yue-jin Cai , Ji Lin . Localization of nonlocal symmetries and interaction solutions of the Sawada-Kotera equation[J]. Communications in Theoretical Physics, 2021 , 73(6) : 065002 . DOI: 10.1088/1572-9494/abf552

1. Introduction

It is well known that certain nonlinear integrable evolution equations, like the Korteweg-de Vries equation, have been widely used in various natural science fields such as nonlinear optics [1, 2], plasma physics [3], ferrimagnetics [4], Bose-Einstein condensation [5] and fluid mechanics [6]. Surface and internal waves in some fields such as fluid can be described by a class of physically important asymptotically integrable equations, like as
$\begin{eqnarray}\begin{array}{l}{v}_{t}+\left[\alpha {v}^{2}+\beta {v}_{{xx}}+\epsilon ({\alpha }_{1}{v}^{3}+{\gamma }_{1}{{vv}}_{{xx}}\right.\\ \,+\,{\left.{\gamma }_{2}{v}_{x}^{2}+{\beta }_{1}{v}_{{xxxx}})\right]}_{x}+o({\epsilon }^{2})=0.\end{array}\end{eqnarray}$

The asymptotically integrable equation was derived by many authors [7, 8]. It was shown in [9] that by constructing a canonical map φ: vu, one can transform the model (1) into a system with same order ε. Further, due to Kodama’s idea [10] by a canonical transformation
$\begin{eqnarray}\begin{array}{rcl}v & = & u+\epsilon \left[\left(\displaystyle \frac{5a\alpha {\beta }_{1}}{3{\beta }^{2}}+{a}_{1}\right){u}^{2}+\left(\displaystyle \frac{5b{\beta }_{1}}{2\beta }+{b}_{1}\right){u}_{{xx}}\right.\\ & & +\left.\left(\displaystyle \frac{5a\alpha {\beta }_{1}}{3{\beta }^{2}}-\displaystyle \frac{{\gamma }_{1}}{3\beta }\right){u}_{x}\int u{\rm{d}}x\right],\end{array}\end{eqnarray}$

with ${a}_{1}\equiv \tfrac{{\gamma }_{1}}{6\beta }-\tfrac{3{\alpha }_{1}}{2\alpha }$,${b}_{1}\equiv \tfrac{{\gamma }_{1}}{4\alpha }+\tfrac{{\gamma }_{2}}{2\alpha }-\tfrac{9\beta {\alpha }_{1}}{4{\alpha }^{2}}$, equation (1) can be solved approximately up to the same order ε system
$\begin{eqnarray}\begin{array}{l}{u}_{t}+\left[\alpha {u}^{2}+\beta {u}_{{xx}}+\epsilon {\beta }_{1}\left({u}_{{xxxx}}+\displaystyle \frac{5a\alpha }{\beta }{{uu}}_{{xx}}\right.\right.\\ \,+\,{\left.\left.\displaystyle \frac{5(a-b)\alpha }{\beta }{u}_{x}^{2}+\displaystyle \frac{5a{\alpha }^{2}}{3{\beta }^{2}}{u}^{3}\right)\right]}_{x}=0,\end{array}\end{eqnarray}$

where a and b are arbitrary constants. In [11], by making a Galileo transformation u → −$\beta$2/(5a$\alpha$$\beta$1ε) + u(x + $\beta$2t/(5a$\beta$1ε), $\beta$1εt), the terms $\alpha$u2 and $\beta$uxx in equation (3) can be eliminated. Thus, equation (3) just is the Sawada-Kotera (SK) equation
$\begin{eqnarray}{u}_{t}+{u}_{{xxxxx}}+5{{uu}}_{{xxx}}+5{u}_{x}{u}_{{xx}}+5{u}^{2}{u}_{x}=0,\end{eqnarray}$

for the suitable parameters a = 1, b = 1.
The SK equation (4) is one of the most important and high order integrable models in mathematical physics, which is firstly derived by Sawada and Kotera [12]. After that, a variety of fruitful algorithmic methods, such as Hirota’s bilinear method [13, 14], recursion operator [15-17], bi-Hamiltonian formulation [18], classical Lie group approach [19-21], inverse recursion operator [22] and function expansion method [23, 24], were used to study abundant properties of equation (4). In addition, abundant traveling wave structures and interaction waves had been separately studied in [25].
Recently, the study of interaction waves between solitons and other nonlinear waves have attracted great attention of many mathematical physicists [26]. So, to search for some interaction waves there were a wealth of approaches, such as the consistent Riccati expansion method [27-29] and the consistent tanh expansion method [30-32]. In addition, Lou and his cooperators produced the localization of the nonlocal symmetry, which can be used to construct much more abundant interaction waves, such as the soliton-Painlev$\acute{e}$ II waves and the soliton-cnoidal waves [33-37]. On the basis of the remarkable method, many new interaction waves for various integrable models have been discussed in detail [38-41]. For instance, by the truncated Painleve analysis and the Lie point symmetry the nonlocal symmetry and its interaction waves for the modified Boussineq equation are studied in the [42]. The motivation of the present paper also stems from the observation of interaction waves in laboratory experiments [43]. In this paper, we employ the nonlocal symmetry for Lax pair to obtain the auto Bäcklund transformation and group invariant solutions. Especially, the analytic soliton-cnoidal wave solution is obtained, which can describe soliton moving on a cnoidal wave background.
The layout of this paper is as follows. Section 2 is devoted to obtain the nonlocal symmetry for the lax pair of equation (4). In section 3 the vector field and the Schwartzian form of equation (4) are composed by the localization of the nonlocal symmetry. In the next section, the auto Bäcklund transformation is found out via solving the initial value problem, while some types of symmetry reduction equations and its group invariant solutions can be given with the help of the standard Lie group approach. The last section contains a conclusion and discussion.

2. Nonlocal symmetry of the SK equation

The Lie’s symmetry analysis theory [44] of differential equation (DE) have been widely applied to explore the exact solution, which is based on the algebraic symmetry of DE. Furthermore, abundant new explicit exact solutions can be constructed with using a nonlocal symmetry. Thus, how to construct a new nonlocal symmetry of equation (4) is interesting problem. Previously, the Lie point symmetries of the potential system, namely the nonlocal symmetries, were derived by the conservation law [20]. In recent years, various nonlocal symmetries were constructed by Bäcklund transformation and Darboux transformation (Lax pair) [45, 33]. For instance, starting from a known Lax pair, one can get some infinitely many coupled Lax pairs and infinitely many nonlocal symmetries [46]. In [47], the nonlocal symmetry and some interaction solutions are studied by a Lax pair. However, it is known that for an integrable model, its Lax pair is not unique. The SK equation has another Lax pair [45]
$\begin{eqnarray}{\psi }_{{xxx}}+u{\psi }_{x}-\lambda \psi =0,\end{eqnarray}$


$\begin{eqnarray}{\psi }_{t}-{(9{\psi }_{{xxxx}}+15u{\psi }_{{xx}})}_{x}-(10{u}_{{xx}}+5{u}^{2}){\psi }_{x}=0,\end{eqnarray}$

with spectral function ψ and arbitrary parameter λ. Fortunately, when λ = 0 the compatibility condition ψxxxt = ψtxxx of Lax pair (5) and (6) is just the SK equation (4).
A symmetry σu of the SK equation (4) is as a solution of its linearized equation
$\begin{eqnarray}\begin{array}{l}{\sigma }_{t}^{u}+5{\sigma }_{x}^{u}{u}_{{xx}}+5{u}_{x}{\sigma }_{{xx}}^{u}+{\sigma }_{{xxxxx}}^{u}\\ \,+\,5{\sigma }^{u}{u}_{{xxx}}+5u{\sigma }_{{xxx}}^{u}+10{\sigma }^{u}{{uu}}_{x}+5{u}^{2}{\sigma }_{x}^{u}=0,\end{array}\end{eqnarray}$

which means equation (4) is form invariant under the transform uu + εσu. The symmetry can be written in the form [20]
$\begin{eqnarray}\begin{array}{l}{\sigma }^{u}=X(x,t,u,\psi ,{\psi }_{x},{\psi }_{{xx}}){u}_{x}+T(x,t,u,\psi ,{\psi }_{x},{\psi }_{{xx}}){u}_{t}\\ \qquad -\ U(x,t,u,\psi ,{\psi }_{x},{\psi }_{{xx}}),\end{array}\end{eqnarray}$

where X, T, U are the functions of the variables (x, t, u, ψ, ψx, ψxx). Substituting equation (8) into (7) and eliminating ut, ψxxx and ψt through equation (4) and the Lax pair (5)-(6) with λ = 0, we get the symmetry determining equations for the functions X, T, U. Then, after solving this symmetry determining equations, the symmetry is given
$\begin{eqnarray}\begin{array}{l}{\sigma }^{u}=({C}_{1}x+{C}_{2}){u}_{x}+(5{C}_{1}t+{C}_{3}){u}_{t}\\ \,+(2{C}_{1}u-{C}_{4}{\psi }_{{xx}}).\end{array}\end{eqnarray}$

If we set ${C}_{1}={C}_{2}={C}_{3}=0$ and ${C}_{4}=-1$ in equation (9), equation (4) will have a simple nonlocal symmetry

$\begin{eqnarray}{\sigma }^{u}={\psi }_{{xx}}\end{eqnarray}$

with ψ being the solution of equations (5) and (6) with $\lambda =0$.

3. Localization of nonlocal symmetry

Utilizing the standard Lie algebra theory, the homologous finite transformation will be obtained by working out the solution of the initial value problem
$\begin{eqnarray}\displaystyle \frac{{\rm{d}}\bar{u}(\epsilon )}{{\rm{d}}\epsilon }={\psi }_{{xx}},\bar{u}(0)=u.\end{eqnarray}$

It is known that for general point symmetry the initial value problem can be solved, but in this paper the symmetry is a nonlocal symmetry which cannot be solved. However, recent study [33] presents that the general Lie point symmetry can be obtained by the localization procedure. The original system is extended to a suitable prolonged system as follow
$\begin{eqnarray}\begin{array}{l}{u}_{t}+{u}_{{xxxxx}}+5{{uu}}_{{xxx}}+5{u}_{x}{u}_{{xx}}+5{u}^{2}{u}_{x}=0,\\ {\psi }_{{xxx}}+u{\psi }_{x}=0,g={\psi }_{x},h={g}_{x},\\ {\psi }_{t}-{(9{\psi }_{{xxxx}}+15u{\psi }_{{xx}})}_{x}-(10{u}_{{xx}}+5{u}^{2}){\psi }_{x}=0,\end{array}\end{eqnarray}$

by introducing the auxiliary functions g and h which are functions of x and t. The symmetries σk(k = u, ψ, g, h) are the solutions of the linearized equations of the prolonged system (12)
$\begin{eqnarray}\begin{array}{l}{\sigma }_{t}^{u}-5{\sigma }_{x}^{u}{u}_{{xx}}-5{u}_{x}{\sigma }_{{xx}}^{u}-{\sigma }_{{xxxxx}}^{u}-5{\sigma }^{u}{u}_{{xxx}}\\ \,-5u{\sigma }_{{xxx}}^{u}-10{\sigma }^{u}{{uu}}_{x}-5{u}^{2}{\sigma }_{x}^{u}=0,\\ {\sigma }_{{xxx}}^{\psi }+{\sigma }^{u}{\psi }_{x}+u{\sigma }_{x}^{\psi }=0,{\sigma }^{g}={\sigma }_{x}^{\psi },{\sigma }^{h}={\sigma }_{x}^{g},\\ {\sigma }_{t}^{\psi }-9{\sigma }_{{xxxxx}}^{\psi }-15({\sigma }^{u}{\psi }_{{xxx}}+u{\sigma }_{{xxx}}^{\psi }+{\sigma }_{x}^{u}{\psi }_{{xx}}+{u}_{x}{\sigma }_{{xx}}^{\psi })\\ \,-10({\sigma }_{{xx}}^{u}{\psi }_{x}+u{\sigma }^{u}{\psi }_{x}+{u}_{{xx}}{\sigma }_{x}^{\psi })-5{u}^{2}{\sigma }^{\psi }=0,\end{array}\end{eqnarray}$

and these localized symmetries are constructed
$\begin{eqnarray}{\sigma }^{u}=h,{\sigma }^{\psi }=-\displaystyle \frac{1}{6}{\psi }^{2},{\sigma }^{g}=-\displaystyle \frac{1}{3}\psi g,{\sigma }^{h}=-\displaystyle \frac{1}{3}({g}^{2}+\psi h).\end{eqnarray}$

Consequently, the corresponding vector field of the prolonged system (12) is
$\begin{eqnarray}V=h\displaystyle \frac{\partial }{\partial u}-\displaystyle \frac{1}{6}{\psi }^{2}\displaystyle \frac{\partial }{\partial \psi }-\displaystyle \frac{1}{3}\psi g\displaystyle \frac{\partial }{\partial g}-\displaystyle \frac{1}{3}({g}^{2}+\psi h)\displaystyle \frac{\partial }{\partial h}.\end{eqnarray}$

Fortunately, the auxiliary spectral function ψ satisfies the Schwartzian form of equation (4):
$\begin{eqnarray}{\psi }_{t}=-4\{\psi ;x\}{}^{2}{\psi }_{x}-\{\psi ;x\}{}_{{xx}}{\psi }_{x},\end{eqnarray}$

where $\{\psi ;x\}=({\psi }_{{xxx}}/{\psi }_{x})-\tfrac{3}{2}{\left({\psi }_{{xx}}/{\psi }_{x}\right)}^{2}$ is the Schwartzian derivative.

4. Explicit solutions from nonlocal symmetry

4.1. Finite symmetry transformation

Thanks to the Lie’s first fundamental theorem [48], by solving the following initial value problem for the prolonged system (12)
$\begin{eqnarray}\begin{array}{l}\displaystyle \frac{{\rm{d}}\bar{u}(\epsilon )}{{\rm{d}}\epsilon }=\bar{h}(\epsilon ),\bar{u}(\epsilon ){| }_{\epsilon =0}=u,\\ \displaystyle \frac{{\rm{d}}\bar{\psi }(\epsilon )}{{\rm{d}}\epsilon }=-\displaystyle \frac{1}{6}\bar{\psi }{\left(\epsilon \right)}^{2},\bar{\psi }(\epsilon ){| }_{\epsilon =0}=\psi ,\\ \displaystyle \frac{{\rm{d}}\bar{g}(\epsilon )}{{\rm{d}}\epsilon }=-\displaystyle \frac{1}{3}\bar{\psi }(\epsilon )\bar{g}(\epsilon ),\bar{g}(\epsilon ){| }_{\epsilon =0}=g,\\ \displaystyle \frac{{\rm{d}}\bar{h}(\epsilon )}{{\rm{d}}\epsilon }=-\displaystyle \frac{1}{3}[\bar{g}{\left(\epsilon \right)}^{2}+\bar{\psi }(\epsilon )\bar{h}(\epsilon )],\bar{h}(\epsilon ){| }_{\epsilon =0}=h,\end{array}\end{eqnarray}$

the finite symmetry transformation is obtained.

If $\{u,\psi ,g,h\}$ is a solution of the prolonged system (12), another solution $\{\bar{u},\bar{\psi },\bar{g},\bar{h}\}$ will be obtained via the following finite symmetry transformation.

$\begin{eqnarray}\begin{array}{l}\bar{u}=u-\displaystyle \frac{6{\epsilon }^{2}{g}^{2}}{{\left(\psi \epsilon +6\right)}^{2}}+\displaystyle \frac{6h\epsilon }{\psi \epsilon +6},\\ \bar{\psi }=\displaystyle \frac{6\psi }{\psi \epsilon +6},\bar{g}=\displaystyle \frac{36g}{{\left(\psi \epsilon +6\right)}^{2}},\\ \bar{h}=\displaystyle \frac{36h}{{\left(\psi \epsilon +6\right)}^{2}}-\displaystyle \frac{72\epsilon {g}^{2}}{{\left(\psi \epsilon +6\right)}^{3}}.\end{array}\end{eqnarray}$

In fact, the transformation (18) is just the auto Bäcklund transformation. As a example, taking the solution $u=-2{k}^{2}$ of equation (4), we obtain the corresponding special solutions for the auxiliary variables from equation (12):
$\begin{eqnarray}\begin{array}{rcl}\psi & = & {\kappa }_{1}{{\rm{e}}}^{{\rm{\Sigma }}}+{\kappa }_{2}{{\rm{e}}}^{-{\rm{\Sigma }}}+{\kappa }_{3},\ g=\sqrt{2}{\kappa }_{2}{{k}{\rm{e}}}^{-{\rm{\Sigma }}}\\ & & -\sqrt{2}{\kappa }_{1}{{k}{\rm{e}}}^{{\rm{\Sigma }}},\ h=2{\kappa }_{1}{k}^{2}{{\rm{e}}}^{{\rm{\Sigma }}}+2{\kappa }_{2}{k}^{2}{{\rm{e}}}^{-{\rm{\Sigma }}},\end{array}\end{eqnarray}$

where ${\kappa }_{1}$, ${\kappa }_{2},\,{\kappa }_{3}$ and k are arbitrary constants, and ${\rm{\Sigma }}=\sqrt{2}k(4{k}^{4}t-x)$. Substituting equation (19) into (18), the new solution of the SK equation (4) is constructed
$\begin{eqnarray}\begin{array}{l}\bar{u}=-2{k}^{2}\\ \displaystyle \,+\,\frac{12{k}^{2}\epsilon {{\rm{e}}}^{{\rm{\Sigma }}}\left[{\kappa }_{1}({\kappa }_{3}\epsilon +6){{\rm{e}}}^{2{\rm{\Sigma }}}+4{\kappa }_{1}{\kappa }_{2}\epsilon {{\rm{e}}}^{{\rm{\Sigma }}}+{\kappa }_{2}({\kappa }_{3}\epsilon +6)\right]}{{\left[{\kappa }_{1}\epsilon {{\rm{e}}}^{2{\rm{\Sigma }}}+({\kappa }_{3}\epsilon +6){{\rm{e}}}^{{\rm{\Sigma }}}+{\kappa }_{2}\epsilon \right]}^{2}}.\end{array}\end{eqnarray}$

4.2. Soliton and Jacobi periodic wave solutions

Not only the finite transformation but also the symmetry reduction can be applied to get the explicit solutions by reducing dimensions of a partial DE. Firstly, to obtain the Lie point symmetries of the whole prolonged system (12), we assume that the prolonged system (12) is invariant under the infinitesimal transformations
$\begin{eqnarray}\begin{array}{l}u\to u+\epsilon {\sigma }^{u},\psi \to \psi +\epsilon {\sigma }^{\psi },\\ g\to g+\epsilon {\sigma }^{g},h\to h+\epsilon {\sigma }^{h},\end{array}\end{eqnarray}$

with
$\begin{eqnarray}\begin{array}{rcl}{\sigma }^{u} & = & X(x,t,u,\psi ,g,h){U}_{x}+T(x,t,u,\psi ,g,h){U}_{t}\\ & & -U(x,t,u,\psi ,g,h),\\ {\sigma }^{\psi } & = & X(x,t,u,\psi ,g,h){{\rm{\Psi }}}_{x}+T(x,t,u,\psi ,g,h){{\rm{\Psi }}}_{t}\\ & & -{\rm{\Psi }}(x,t,u,\psi ,g,h),\\ {\sigma }^{g} & = & X(x,t,u,\psi ,g,h){G}_{x}+T(x,t,u,\psi ,g,h){G}_{t}\\ & & -G(x,t,u,\psi ,g,h),\\ {\sigma }^{h} & = & X(x,t,u,\psi ,g,h){H}_{x}+T(x,t,u,\psi ,g,h){H}_{t}\\ & & -H(x,t,u,\psi ,g,h).\end{array}\end{eqnarray}$

Now substituting the formula (22) into the symmetry definition equations (13) and collecting together the coefficients of partial differential of dependent variables, a set of overdetermined linear equations for the infinitesimals X, T, U, Ψ, G and H are obtained. Fortunately, by solving the overdetermined linear equations with Maple the corresponding solutions can be obtained as
$\begin{eqnarray}\begin{array}{rcl}X & = & \displaystyle \frac{1}{5}{c}_{1}x+{c}_{4},\ T={c}_{1}t+{c}_{2},\\ U & = & {c}_{3}h-\displaystyle \frac{2}{5}{c}_{1}u,\,{\rm{\Psi }}=-\displaystyle \frac{1}{6}{c}_{3}{\psi }^{2}+{c}_{5}\psi +{c}_{6},\\ G & = & -\displaystyle \frac{1}{3}{c}_{3}g\psi +g\left({c}_{5}-\displaystyle \frac{1}{5}{c}_{1}\right),\\ H & = & h({c}_{5}-\displaystyle \frac{1}{3}{c}_{3}\psi -\displaystyle \frac{2}{5}{c}_{1})-\displaystyle \frac{1}{3}{c}_{3}{g}^{2},\end{array}\end{eqnarray}$

where ci (i = 1, 2,..,6) are arbitrary constants. Obviously the finite symmetry transformation (14) is a very special case of the general infinitesimal operator (23) when ci = 0 (i = 1, 2, 4, 5, 6) and c3 = −1. To find the symmetry reduction equations, we will solve equation (22) with (23) by the symmetry constraint situations σk = 0(k = u, ψ, g, h). Namely, by explicitly solving the corresponding characteristic equations [48]
$\begin{eqnarray}\begin{array}{rcl}\displaystyle \frac{{\rm{d}}x}{\tfrac{1}{5}{c}_{1}x+{c}_{4}} & = & \displaystyle \frac{{\rm{d}}t}{{c}_{1}t+{c}_{2}}=\displaystyle \frac{{\rm{d}}\psi }{-\tfrac{1}{6}{c}_{3}{\psi }^{2}+{c}_{5}\psi +{c}_{6}}\\ & = & \displaystyle \frac{{\rm{d}}u}{{c}_{3}h-\tfrac{2}{5}{c}_{1}u}=\displaystyle \frac{{\rm{d}}g}{-\tfrac{1}{3}{c}_{3}g\psi +g\left({c}_{5}-\tfrac{1}{5}{c}_{1}\right)}\\ & = & \displaystyle \frac{{\rm{d}}h}{h\left({c}_{5}-\tfrac{1}{3}{c}_{3}\psi -\tfrac{2}{5}{c}_{1}\right)-\tfrac{1}{3}{c}_{3}{g}^{2}},\end{array}\end{eqnarray}$

the symmetry reductions of the prolong system are given. Here, we focus on c3 ≠ 0, corresponding to the nonlocal symmetry circumstance. Without loss of generality, we rewrite the parameter ${\rm{\Delta }}=\sqrt{6{c}_{3}{c}_{6}+9{c}_{5}^{2}}$ and there are four cases to consider.
Case I: c1 ≠ 0, Δ ≠ 0
After solving the characteristic equations (24), the similarity solution is obtained as follow:
$\begin{eqnarray}\begin{array}{rcl}u & = & -\displaystyle \frac{1}{{\left({c}_{1}t+{c}_{2}\right)}^{\tfrac{2}{5}}}\left[\displaystyle \frac{6{c}_{3}H}{{\rm{\Delta }}}\tanh ({\rm{\Gamma }})\right.\\ & & \left.+\displaystyle \frac{6{c}_{3}^{2}{G}^{2}}{{{\rm{\Delta }}}^{2}}{\tanh }^{2}({\rm{\Gamma }})-U\right],\\ h & = & -\displaystyle \frac{H}{{\left({c}_{1}t+{c}_{2}\right)}^{\tfrac{2}{5}}}{{\rm{sech}} }^{2}({\rm{\Gamma }})\\ & & -\displaystyle \frac{2{c}_{3}{G}^{2}}{{\rm{\Delta }}{\left({c}_{1}t+{c}_{2}\right)}^{\tfrac{2}{5}}}\tanh ({\rm{\Gamma }}){{\rm{sech}} }^{2}({\rm{\Gamma }}),\\ g & = & \displaystyle \frac{G}{{\left({c}_{1}t+{c}_{2}\right)}^{\tfrac{1}{5}}}{{\rm{sech}} }^{2}({\rm{\Gamma }}),\\ \psi & = & \displaystyle \frac{{\rm{\Delta }}}{{c}_{3}}\tanh ({\rm{\Gamma }})+\displaystyle \frac{3{c}_{5}}{{c}_{3}},\end{array}\end{eqnarray}$

with $\xi =\tfrac{{c}_{1}x+5{c}_{4}}{{c}_{1}{\left({c}_{1}t+{c}_{2}\right)}^{\tfrac{1}{5}}}$ and ${\rm{\Gamma }}=\tfrac{{\rm{\Delta }}}{6{c}_{1}}[{c}_{1}{\rm{\Psi }}+\mathrm{ln}({c}_{1}t+{c}_{2})]$.
Here, U(ξ), Ψ(ξ), G(ξ) and H(ξ) represent four group invariant functions. The ξ is the similarity variable. By substituting equations (25) into the prolonged system (12) we have the reduced equations
$\begin{eqnarray}U=\displaystyle \frac{{{\rm{\Delta }}}^{2}}{18}{{\rm{\Psi }}}_{\xi }^{2}-\displaystyle \frac{{{\rm{\Psi }}}_{\xi \xi \xi }}{{{\rm{\Psi }}}_{\xi }},H=\displaystyle \frac{{{\rm{\Delta }}}^{2}}{6{c}_{3}}{{\rm{\Psi }}}_{\xi \xi },G=-\displaystyle \frac{{{\rm{\Delta }}}^{2}}{6{c}_{3}}{{\rm{\Psi }}}_{\xi },\end{eqnarray}$

where Ψ(ξ) satisfies a five-order ordinary differential equation (ODE)
$\begin{eqnarray}\begin{array}{l}{{\rm{\Psi }}}_{\xi \xi \xi \xi \xi }+5\displaystyle \frac{{\psi }_{\xi \xi }}{{\psi }_{\xi }}{\psi }_{\xi \xi \xi \xi }-5\displaystyle \frac{{\psi }_{\xi \xi }^{2}}{{\psi }_{\xi }^{2}}{\psi }_{\xi \xi \xi }+\displaystyle \frac{5{\rm{\Delta }}}{9}{\psi }_{\xi }^{2}{\psi }_{\xi \xi \xi }\\ -\displaystyle \frac{5{{\rm{\Delta }}}^{2}}{9}{\psi }_{\xi }{\psi }_{\xi \xi }^{2}-\displaystyle \frac{{{\rm{\Delta }}}^{4}}{81}{\psi }_{\xi }^{5}+\displaystyle \frac{{c}_{1}}{5}\xi {\psi }_{\xi }-1=0.\end{array}\end{eqnarray}$

It is obvious that U, Ψ, G, H can been solved from equation (26), and the exact solution of the SK equation will be obtained by equation (25).
Case II: c1 ≠ 0, Δ = 0
If Ψ ≡ Ψ($\eta$) is a solution of the ODE
$\begin{eqnarray}{{\rm{\Psi }}}_{\eta \eta \eta \eta \eta }=\displaystyle \frac{5{{\rm{\Psi }}}_{\eta \eta }}{{{\rm{\Psi }}}_{\eta }}{{\rm{\Psi }}}_{\eta \eta \eta \eta }-\displaystyle \frac{5{{\rm{\Psi }}}_{\eta \eta }^{2}}{{{\rm{\Psi }}}_{\eta }^{2}}{{\rm{\Psi }}}_{\eta \eta \eta }+\displaystyle \frac{{c}_{1}}{5}\eta {{\rm{\Psi }}}_{\eta }-1,\end{eqnarray}$

the solution of the SK equation (4) is given
$\begin{eqnarray}\begin{array}{rcl}u & = & \displaystyle \frac{1}{{\left({c}_{1}t+{c}_{2}\right)}^{\tfrac{2}{5}}}\left(\displaystyle \frac{3{c}_{5}^{2}}{2{c}_{1}{c}_{6}{\rm{\Omega }}}H-\displaystyle \frac{3{c}_{5}^{4}}{8{c}_{1}^{2}{c}_{6}^{2}{{\rm{\Omega }}}^{2}}{G}^{2}+U\right),\\ U & = & -\displaystyle \frac{{{\rm{\Psi }}}_{\eta \eta \eta }}{{{\rm{\Psi }}}_{\eta }},H=\displaystyle \frac{4{c}_{6}{c}_{1}^{2}}{{c}_{5}^{2}}{{\rm{\Psi }}}_{\eta \eta },G=\displaystyle \frac{4{c}_{6}{c}_{1}^{2}}{{c}_{5}^{2}}{{\rm{\Psi }}}_{\eta },\end{array}\end{eqnarray}$

where $\eta =\tfrac{{c}_{1}x+5{c}_{4}}{{c}_{1}{\left({c}_{1}t+{c}_{2}\right)}^{\tfrac{1}{5}}}$ and ${\rm{\Omega }}={c}_{1}{\rm{\Psi }}+\mathrm{ln}({c}_{1}t+{c}_{2})$. Similarly, U($\eta$), Ψ($\eta$), G($\eta$) and H($\eta$) represent four group invariant functions. The $\eta$ is the similarity variable.
Case III: c1 = 0, Δ ≠ 0
By solving equation (24), we obtain
$\begin{eqnarray}\begin{array}{rcl}u & = & -\displaystyle \frac{6{c}_{3}}{{{\rm{\Delta }}}^{2}}\tanh ({\rm{\Theta }})[{c}_{3}{G}^{2}\tanh ({\rm{\Theta }})+{\rm{\Delta }}H]+U,\\ h & = & -\displaystyle \frac{2{c}_{3}}{{\rm{\Delta }}}{G}^{2}\tanh ({\rm{\Theta }}){{\rm{sech}} }^{2}({\rm{\Theta }})-H{{\rm{sech}} }^{2}({\rm{\Theta }}),\\ \psi & = & \displaystyle \frac{{\rm{\Delta }}}{{c}_{3}}\tanh ({\rm{\Theta }})+\displaystyle \frac{3{c}_{5}}{{c}_{3}},\\ g & = & -G{{\rm{sech}} }^{2}({\rm{\Theta }}),\end{array}\end{eqnarray}$

with ${\rm{\Theta }}=\tfrac{{\rm{\Delta }}}{6{c}_{2}}(\int W(\tau ){\rm{d}}\tau +t)$ and $\tau =x-\tfrac{c4}{{c}_{2}}t$.
Substituting equation (30) into the prolonged system (13), the reduced equations are obtained as follows:
$\begin{eqnarray}\begin{array}{rcl}U & = & \displaystyle \frac{{{\rm{\Delta }}}^{2}}{18{c}_{2}^{2}}W{\left(\tau \right)}^{2}-\displaystyle \frac{{W}_{\tau \tau }(\tau )}{W(\tau )},\\ H & = & -\displaystyle \frac{{{\rm{\Delta }}}^{2}}{6{c}_{2}{c}_{3}}{W}_{\tau }(\tau ),\ G=\displaystyle \frac{{{\rm{\Delta }}}^{2}}{6{c}_{2}{c}_{3}}W(\tau ),\end{array}\end{eqnarray}$

where W(τ) satisfies
$\begin{eqnarray}\begin{array}{rcl}{W}_{\tau }^{2}(\tau ) & = & {a}_{0}+{a}_{1}W(\tau )+{a}_{2}{W}^{2}(\tau )\\ & & +{a}_{3}{W}^{3}(\tau )+{a}_{4}{W}^{4}(\tau ),\end{array}\end{eqnarray}$

with ${a}_{0}=\tfrac{3{c}_{2}}{{\rm{\Delta }}}({c}_{2}{a}_{2}^{2}+{c}_{4})$, a1 = 0, ${a}_{3}=-\tfrac{2{{\rm{\Delta }}}^{2}}{27{c}_{2}({c}_{2}{a}_{2}^{2}+{c}_{4})}$ and ${a}_{4}=\tfrac{{{\rm{\Delta }}}^{2}}{9{c}_{2}^{2}}$.
Then summarizing the above expressions, we obtain the explicit solution of the SK equation (4):
$\begin{eqnarray}\begin{array}{rcl}u & = & \displaystyle \frac{{{\rm{\Delta }}}^{2}}{18{c}_{2}^{2}}{W}^{2}(\tau )-\displaystyle \frac{{W}_{\tau \tau }(\tau )}{W(\tau )}+\displaystyle \frac{{\rm{\Delta }}}{{c}_{2}}{W}_{\tau }(\tau )\\ & & \times \tanh \left[\displaystyle \frac{{\rm{\Delta }}}{6{c}_{2}}\left(\int W(\tau ){\rm{d}}\tau +t\right)\right]\\ & & -\displaystyle \frac{{{\rm{\Delta }}}^{2}}{6{c}_{2}^{2}}{W}^{2}(\tau ){\tanh }^{2}\left[\displaystyle \frac{{\rm{\Delta }}}{6{c}_{2}}\left(\int W(\tau ){\rm{d}}\tau +t\right)\right],\end{array}\end{eqnarray}$

where W(τ) satisfies equation (32).
We know that the general solution of equation (32) can be written in terms of Jacobi elliptic functions. In order to show more detail for this kind of solution, we offer two special case of equation (33) by solving equation (32).
Type 1:The solution of the equation (32) is given as
$\begin{eqnarray}W(\tau )={b}_{1}+{b}_{0}\mathrm{sn}(K\tau ,m),\end{eqnarray}$

where $\mathrm{sn}(K\tau ,m)$ is the usual Jacobi elliptic sine function. Substituting equation (34) into (32) and vanishing the coefficients of ${\rm{d}}{\rm{n}}(\tau ,m)$, $\mathrm{sn}(\tau ,m)$, the solution of equation (3) are constructed
$\begin{eqnarray}\begin{array}{rcl}u & = & -\displaystyle \frac{{{\rm{\Delta }}}^{2}}{6{c}_{2}^{2}}{\left({b}_{1}+{b}_{0}\mathrm{sn}(K\tau ,m)\right)}^{2}\tanh {\left(\theta \right)}^{2}\\ & & +\displaystyle \frac{K{\rm{\Delta }}{b}_{0}}{{c}_{2}}\mathrm{cn}(K\tau ,m)\mathrm{dn}(K\tau ,m)\tanh (\theta )\\ & & +\displaystyle \frac{1}{18{c}_{2}^{2}({b}_{1}+{b}_{0}\mathrm{sn}(K\tau ,m))}(-(36{K}^{2}{c}_{2}^{2}{m}^{2}{b}_{0}\\ & & -{{\rm{\Delta }}}^{2}{b}_{0}^{3})\mathrm{sn}{\left(K\tau ,m\right)}^{3}+3{{\rm{\Delta }}}^{2}{b}_{0}^{2}{b}_{1}\mathrm{sn}{\left(K\tau ,m\right)}^{2}\\ & & +3{b}_{0}(6{K}^{2}{c}_{2}^{2}{m}^{2}+6{K}^{2}{c}_{2}^{2}+{{\rm{\Delta }}}^{2}{b}_{1}^{2})\\ & & \times \mathrm{sn}(K\tau ,m)+{{\rm{\Delta }}}^{2}{b}_{1}^{3})-\displaystyle \frac{{\beta }^{2}}{5a\alpha {\beta }_{1}\epsilon },\end{array}\end{eqnarray}$

where
$\begin{eqnarray}\begin{array}{rcl}\theta & = & \displaystyle \frac{{\rm{\Delta }}({b}_{1}{Km}\tau +{Kmt}+{b}_{0}\mathrm{ln}(\mathrm{dn}(K\tau ,m)-m\mathrm{cn}(K\tau ,m)))}{6{{Kmc}}_{2}},\\ \tau & = & x+\displaystyle \frac{{\beta }^{2}t}{a{\beta }_{1}\epsilon }-\displaystyle \frac{{c}_{4}\epsilon {\beta }_{1}t}{{c}_{2}},\end{array}\end{eqnarray}$

and
$\begin{eqnarray}\begin{array}{l}{c}_{4}=\displaystyle \frac{{a}_{2}^{2}{c}_{2}(19{m}^{4}+26{m}^{2}+19)}{16{\left({m}^{2}+1\right)}^{2}},{b}_{1}=\displaystyle \frac{8}{9}\displaystyle \frac{{\left({m}^{2}+1\right)}^{2}}{{a}_{2}^{2}{\left({m}^{2}-1\right)}^{2}},\\ K=\sqrt{\displaystyle \frac{{a}_{2}}{2{m}^{2}+2}},{a}_{0}=-\displaystyle \frac{16}{81}\displaystyle \frac{{\left({m}^{2}+1\right)}^{2}}{{a}_{2}^{3}{\left({m}^{2}-1\right)}^{2}},{a}_{1}=0,\\ {a}_{3}=-\displaystyle \frac{9}{8}\displaystyle \frac{{a}_{2}^{3}{\left({m}^{2}-1\right)}^{2}}{{\left({m}^{2}+1\right)}^{2}},\\ {a}_{4}=\displaystyle \frac{81}{256}\displaystyle \frac{{a}_{2}^{5}{\left({m}^{2}-1\right)}^{4}}{{\left({m}^{2}+1\right)}^{4}},{\rm{\Delta }}=\displaystyle \frac{27}{16}\displaystyle \frac{{a}_{2}^{\tfrac{5}{2}}{c}_{2}{\left({m}^{2}-1\right)}^{2}}{{\left({m}^{2}+1\right)}^{2}},\\ {b}_{0}=\displaystyle \frac{8\sqrt{2}}{9}\displaystyle \frac{m{\left({m}^{2}+1\right)}^{\tfrac{3}{2}}}{{a}_{2}^{2}{\left({m}^{2}-1\right)}^{2}}.\end{array}\end{eqnarray}$

Figure 1(a) shows the interaction of a soliton with cnoidal wave solution (35) by fixing parameters
$\begin{eqnarray}a=\alpha =1,\epsilon =\frac{1}{100},\beta =1,{\beta }_{1}=20,{a}_{2}=\frac{1}{2},m=\frac{3}{100}.\end{eqnarray}$

figures 1(b)-(d) clearly display the evolutions of the soliton-cnoidal wave solution (35) along the time with t = −10, t = 0 and t = 10, respectively.
Figure 1. The type of a bright soliton on a cnoidal wave background expressed by equation (35) with equations (36) and (37). (a) The evolution of u, (b) t = −10, (c) t = 0, (d) t = 10.
Type 2: Another special solution of equation (32) reads
$\begin{eqnarray}W(\tau )={b}_{4}+{b}_{3}\mathrm{cn}(K\tau ,m),\end{eqnarray}$

which yields the interaction solution of the equation (3)
$\begin{eqnarray}\begin{array}{l}u=-\displaystyle \frac{{{\rm{\Delta }}}^{2}}{6{c}_{2}^{2}}{\left({b}_{4}+{b}_{3}\mathrm{cn}(K\tau ,m)\right)}^{2}\tanh {\left(\theta \right)}^{2}\\ \,+\,\displaystyle \frac{K{\rm{\Delta }}{b}_{3}}{{c}_{2}}\mathrm{dn}(K\tau ,m)\mathrm{sn}(K\tau ,m)\tanh (\theta )\\ \,+\,\displaystyle \frac{1}{18{c}_{2}^{2}({b}_{4}+{b}_{3}\mathrm{cn}(K\tau ,m))}((36{K}^{2}{c}_{2}^{2}{m}^{2}{b}_{3}+{{\rm{\Delta }}}^{2}{b}_{3}^{3})\\ \,\times \,\mathrm{cn}{\left(K\tau ,m\right)}^{3}+3{{\rm{\Delta }}}^{2}{b}_{3}^{2}{b}_{4}\mathrm{cn}{\left(K\tau ,m\right)}^{2}\\ \,-\,3{b}_{3}(12{K}^{2}{c}_{2}^{2}{m}^{2}-6{K}^{2}{c}_{2}^{2}-{{\rm{\Delta }}}^{2}{b}_{4}^{2})\mathrm{sn}(K\tau ,m)\\ \,+\,{{\rm{\Delta }}}^{2}{b}_{4}^{3})-\displaystyle \frac{{\beta }^{2}}{5a\alpha {\beta }_{1}\epsilon },\end{array}\end{eqnarray}$

where
$\begin{eqnarray}\begin{array}{rcl}\theta & = & \displaystyle \frac{{\rm{\Delta }}}{6{c}_{2}}({b}_{4}\tau +\displaystyle \frac{{b}_{3}}{{Km}}\arctan \left[m\displaystyle \frac{\mathrm{sn}(K\tau ,m)}{\mathrm{dn}(K\tau ,m)}\right]+t),\\ \tau & = & x+\displaystyle \frac{{\beta }^{2}t}{a{\beta }_{1}\epsilon }-\displaystyle \frac{{c}_{4}\epsilon {\beta }_{1}t}{{c}_{2}},K=\sqrt{-\displaystyle \frac{{a}_{2}}{4{m}^{2}-2}},\\ {\rm{\Delta }} & = & \displaystyle \frac{27}{16}\displaystyle \frac{{a}_{2}^{\tfrac{5}{2}}{c}_{2}}{{\left(2{m}^{2}-1\right)}^{2}},{b}_{3}=\displaystyle \frac{8}{9{a}_{2}^{2}}\sqrt{4{m}^{2}-2}(2{m}^{2}-1)m,\\ {b}_{4} & = & \displaystyle \frac{8}{9{a}_{2}^{2}}{\left(2{m}^{2}-1\right)}^{2},{c}_{4}={c}_{2}{a}_{2}^{2}+\displaystyle \frac{3{c}_{2}{a}_{2}^{2}}{16{\left(2{m}^{2}-1\right)}^{2}},\\ {a}_{0} & = & -\displaystyle \frac{16}{81}\displaystyle \frac{{\left(2{m}^{2}-1\right)}^{2}}{{a}_{2}^{3}},{a}_{3}=-\displaystyle \frac{9}{8}\displaystyle \frac{{a}_{2}^{3}}{{\left(2{m}^{2}-1\right)}^{2}},\\ {a}_{4} & = & \displaystyle \frac{81}{256}\displaystyle \frac{{a}_{2}^{5}}{{\left(2{m}^{2}-1\right)}^{4}}.\end{array}\end{eqnarray}$

Case IV: c1 = 0, Δ = 0
The explicit solution of the SK equation (4) is given as
$\begin{eqnarray}u=\displaystyle \frac{6{{\rm{\Psi }}}_{\zeta }}{\int {\rm{\Psi }}{\rm{d}}\zeta +t}-\displaystyle \frac{6{{\rm{\Psi }}}^{2}}{{\left(\int {\rm{\Psi }}{\rm{d}}\zeta +t\right)}^{2}}-\displaystyle \frac{{{\rm{\Psi }}}_{\zeta \zeta }}{{\rm{\Psi }}},\end{eqnarray}$

where $\zeta =x-\tfrac{{c}_{4}}{{c}_{2}}t$ and Ψ(ζ) satisfies an ODE:
$\begin{eqnarray}{c}_{2}{{\rm{\Psi }}}^{2}{{\rm{\Psi }}}_{\zeta \zeta \zeta \zeta }-5{c}_{2}{\rm{\Psi }}{{\rm{\Psi }}}_{\zeta }{{\rm{\Psi }}}_{\zeta \zeta \zeta }+5{{\rm{\Psi }}}_{\zeta }^{2}{{\rm{\Psi }}}_{\zeta \zeta }+{c}_{4}{{\rm{\Psi }}}^{3}+{c}_{2}{{\rm{\Psi }}}^{2}=0,\end{eqnarray}$

with the arbitrary constants c2 and c4.

5. Conclusion and discussion

In this paper, we have studied the nonlocal symmetry of the SK equation (4) by using a symmetry assumption method with the Lax pair. The nonlocal symmetry can be localized by introducing the three potential functions, namely, the spectral function ψ, the x-derivatives of the spectral function g = ψx and h = ψxx. On the one hand, considering to the Lie’s first fundamental theorem, the initial value problem have been solved, which lead to a finite symmetry transformation, and with the help of the finite symmetry transform one can find some new solutions from the given ones. On the other hand, by the means of the symmetry reductions, several classes of new interaction solutions are obtained. Especially, the interaction solutions for soliton-cnoidal wave are unearthed, which can be explicitly constructed by the Jacobi elliptic functions.
Using the Lax pair to search for nonlocal symmetries of integrable systems and then considering the nonlocal symmetries to come out the explicit solutions could be of particular interest. To get interaction solutions among different waves, one has to treat the method to more nonlinear evolution equations and it needs further investigate in the future.

The authors extend their gratitude to Professors S Y Lou for helpful discussions. This work was supported by the National Natural Science Foundation of China, Grant No. 11835011 and Grant No. 11675146.

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