Symmetries and conservation laws of the Yao–Zeng two-component short-pulse equation

Abstract

This paper uses the Lie group method to analyze the symmetries of the Yao–Zeng two-component short-pulse equation which describes the propagation of polarized ultrashort light pulses in cubically nonlinear anisotropic optical fibers. Similarity reductions and exact solutions are obtained by constructing an optimal system of one-dimensional subalgebras. Moreover, the explicit solutions are constructed by the power series method and the convergence of power series solutions is proved. In addition, nonlinear self-adjointness and conservation laws of this system are discussed.

1 Introduction

In the last decades, the short pulse equation has attracted much attention. It was introduced as a model equation to describe the propagation of ultrashort optical pules in silica optical fibers [1] which has the form

$$\begin{aligned} u_{xt}=u+\frac{1}{6}\bigl(u^{3} \bigr)_{xx}, \end{aligned}$$

(1)

where u represents the magnitude of the electric field, subscripts denote partial differentiation. In order to obtain more information to study Eq. (1), a number of works have been presented. For instance, its integrability in [2,3,4], solitary wave solutions in [5], periodic and traveling wave solutions in [6], two-loop soliton solutions in [7], periodic and multiloop solitons in [8].

To describe the propagation of polarized ultra-short pulses, the short pulse equation has been generalized to the multi-component integrable systems. Among them, Matsuno presented the two-component system [9]

$$\begin{aligned} \textstyle\begin{cases} u_{xt}=u+\frac{1}{2}(uvu_{x})_{x},\\ v_{xt}=v+\frac{1}{2}(uvv_{x})_{x}, \end{cases}\displaystyle \end{aligned}$$

(2)

which is addressed from its multi-component model.

Another integrable coupled short-pulse equation is given by Feng [10] as follows:

$$\begin{aligned} \textstyle\begin{cases} u_{xt}=u+\frac{1}{6}(u^{3})_{xx}+\frac{1}{2}v^{2}u_{xx},\\ v_{xt}=v+ \frac{1}{6}(v^{3})_{xx}+\frac{1}{2}u^{2}v_{xx}. \end{cases}\displaystyle \end{aligned}$$

(3)

Let $u=v$, Eq. (2) degenerates to Eq. (1); $v=0$, Eq. (3) recasts to Eq. (1).

In this paper, we consider the Yao–Zeng two-component short-pulse equation [11]

$$\begin{aligned} \textstyle\begin{cases} u_{xt}=u+\frac{1}{6}(u^{3})_{xx},\\ v_{xt}=v+\frac{1}{2}(u^{2}v_{x})_{x}, \end{cases}\displaystyle \end{aligned}$$

(4)

which describes the propagation of polarized ultrashort light pulses in cubically nonlinear anisotropic optical fibers. The Hamiltonian structure of Eq. (4) was established in [11], the zero-curvature representation of Eq. (4) was analyzed in [12]. For the sake of getting more information about Eq. (4), the goal of the present paper is to analyze symmetries, invariant solutions, and conservation laws of Eq. (4). The Lie group method [13,14,15] is considered to be one of the most effective methods to obtain exact solutions for lots of nonlinear partial differential equations(PDEs). Another important research topic is related to conservation laws of PDEs. For the PDEs which do not admit a Lagrangian, based on symmetries [16], Ibragimov presented the concept of an adjoint equation to investigate conservation laws by using the conservation law theorem in [17]. There has been a lot of success in this direction to construct conservation laws for PDEs [16, 18, 19]. By means of conservation laws, one can determine exact solutions of PDEs [20, 21].

The rest of the paper is arranged as follows. In Sect. 2, we present the Lie point symmetries of Eq. (4). In Sect. 3, the similarity reductions are made, exact solutions are considered by means of the Lie group method. In Sect. 4, the explicit solutions for the reduced equations are presented via the power series method, and the detailed proof for the convergence of the power series solutions is provided. In Sect. 5, we prove that Eq. (4) is nonlinearly self-adjoint and construct its conservation laws by applying Ibragimov’s method. Finally, we have a summary of the paper.

2 Lie point symmetries

In this section, we apply the Lie point symmetry method to Eq. (4) and determine its infinitesimal generators and the commutation table of Lie algebras.

First of all, let us consider a one-parameter Lie group admitted by Eq. (4) with a generator of the Lie algebras of the form

$$\begin{aligned}& X=\xi \partial _{x}+\tau \partial _{t}+\phi \partial _{u}+\varphi \partial _{v}, \end{aligned}$$

(5)

where ξ, τ, ϕ, φ are functions of x, t, u, v and are described as infinitesimals of the symmetry groups.

The invariance criterion for Eq. (4) with respect to operator (5) is read as [13, 22]

$$\begin{aligned}& \operatorname{pr}X^{(2)} \biggl[u_{xt}-u-\frac{1}{6} \bigl(u^{3}\bigr)_{xx} \biggr]=0, \\& \operatorname{pr}X^{(2)} \biggl[v_{xt}-v-\frac{1}{2} \bigl(u^{2}v_{x}\bigr)_{x} \biggr]=0. \end{aligned}$$

The symbol $\operatorname{pr}X^{(2)}$ is the usual 2th-order prolongation of the operator [13, 22]. In this case,

$$\begin{aligned} &\operatorname{pr}X^{(2)}=X+\phi _{x}^{(1)} \frac{\partial }{\partial {u_{x}}}+ \varphi _{x}^{(1)}\frac{\partial }{\partial {v_{x}}}+\phi _{xx}^{(2)}\frac{ \partial }{\partial {u_{xx}}}+\varphi _{xx}^{(2)} \frac{\partial }{ \partial {v_{xx}}}+\phi _{xt}^{(2)}\frac{\partial }{\partial {u_{xt}}}+ \varphi _{xt}^{(2)}\frac{\partial }{\partial {v_{xt}}}, \end{aligned}$$

where

$$\begin{aligned}& \phi _{x}^{(1)}=D_{x}\phi -u_{x}D_{x} \xi -u_{t}D_{x}\tau , \\& \varphi _{x}^{(1)}=D_{x}\varphi -v_{x}D_{x}\xi -v_{t}D_{x}\tau , \\& \phi _{xx}^{(2)}=D_{x}^{2}(\phi -\xi u_{x}-\tau u_{t})+\xi u_{xxx}+ \tau u_{xxt}, \\& \varphi _{xx}^{(2)}=D_{x}^{2}(\varphi - \xi v_{x}-\tau v_{t})+\xi v _{xxx}+\tau v_{xxt}, \\& \phi _{xt}^{(2)}=D_{x}D_{t}(\phi -\xi u_{x}-\tau u_{t})+\xi u_{xxt}+ \tau u_{xtt}, \\& \varphi _{xt}^{(2)}=D_{x}D_{t}(\varphi - \xi v_{x}-\tau v_{t})+ \xi v_{xxt}+\tau v_{xtt}, \end{aligned}$$

and $D_{x}$, $D_{t}$ are the total derivative operators, e.g.,

$$\begin{aligned}& D_{t}=\frac{\partial }{\partial t}+u_{t} \frac{\partial }{\partial u}+v_{t} \frac{\partial }{\partial v} +u_{tx}\frac{ \partial }{\partial u_{x}} +v_{tx} \frac{\partial }{\partial v_{x}}+u _{tt}\frac{\partial }{\partial u_{t}}+v_{tt} \frac{\partial }{\partial v_{t}}+\cdots . \end{aligned}$$

Substituting $\operatorname{pr}X^{(2)}$ into Eq. (4) yields the following over-determining equations for the unknown functions ξ, τ, ϕ, and φ:

$$\begin{aligned}& \begin{aligned} & \xi _{t}=\xi _{u}=\xi _{v}=0,\qquad \xi _{x}=-\tau _{t}, \\ & \tau _{x}=\tau _{u}=\tau _{v}=0,\qquad \tau _{tt}=0, \\ & \phi =-u\tau _{t}, \\ &\varphi _{x}=\varphi _{t}=0,\qquad \varphi _{vv}=0, \qquad \varphi _{u}=\frac{(1-v) \varphi }{u}. \end{aligned} \end{aligned}$$

(6)

Solving Eq. (6), we get

$$\begin{aligned}& \xi =-c_{1}x+c_{3}, \qquad \tau =c_{1}t+c_{2}, \qquad \phi =-c_{1}u,\qquad \varphi =c _{5}v+c_{4}u, \end{aligned}$$

where $c_{1}$, $c_{2}$, $c_{3}$, $c_{4}$, and $c_{5}$ are arbitrary constants. The Lie algebra of infinitesimal symmetry of Eq. (4) is spanned by the following vector fields:

$$\begin{aligned}& X_{1}=-x\partial _{x}+t\partial _{t}-u\partial _{u}, \qquad X_{2}=\partial _{t},\qquad X_{3}=\partial _{x},\qquad X_{4}=u\partial _{v}, \qquad X_{5}=v\partial _{v}. \end{aligned}$$

Furthermore, in order to classify all the group-invariant solutions, we determine an optimal system of one-dimensional subalgebras of Eq. (4) by using the method in [23, 24], which only relies on the commutator table.

First, the commutator relations of $X_{1}$, $X_{2}$, $X_{3}$, $X_{4}$, $X_{5}$ are represented in Table 1 by applying the commutator operator $[X_{m}, X _{n}]=X_{m}X_{n}-X_{n}X_{m}$.

Table 1 Table of Lie brackets

An arbitrary operator $X\in L_{5}$ is written as

$$\begin{aligned}& X=l_{1}X_{1}+l_{2}X_{2}+l_{3}X_{3}+l_{4}X_{4}+l_{5}X_{5}. \end{aligned}$$

To discuss the linear transformations of the vector $l=(l_{1},l_{2},l _{3},l_{4},l_{5})$, we apply the following generator:

$$\begin{aligned}& E_{i}=c_{ij}^{k}l_{j} \partial _{l_{k}}, \quad i=1,2,3,4,5, \end{aligned}$$

(7)

where $c_{ij}^{k}$ is defined by the formula $[ X_{i},X_{j}]=c_{ij} ^{k}X_{k}$. Based on Eq. (7) and Table 1, $E_{1}$, $E_{2}$, $E_{3}$, $E_{4}$, $E _{5}$ can be represented as

$$\begin{aligned}& E_{1}=-l_{2}\partial _{l_{2}}+l_{3} \partial _{l_{3}}-l_{4}\partial _{l_{4}}, \\& E_{2}=l_{1}\partial _{l_{2}}, \\& E_{3}=-l_{1}\partial _{l_{3}}, \\& E_{4}=(l_{1}+l_{5})\partial _{l_{4}}, \\& E_{5}=-l_{4}\partial _{l_{4}}. \end{aligned}$$

For $E_{1}$, $E_{2}$, $E_{3}$, $E_{4}$, $E_{5}$, the Lie equations with parameters $a _{1}$, $a_{2}$, $a_{3}$, $a_{4}$, $a_{5}$ and the initial condition $\tilde{l}|_{a _{i}=0}=l$, $i=1,2,3,4,5$ are shown as follows:

$$\begin{aligned}& \frac{d\tilde{l_{1}}}{da_{1}}=0,\qquad \frac{d\tilde{l_{2}}}{da_{1}}=- \tilde{l_{2}},\qquad \frac{d\tilde{l_{3}}}{da_{1}}=\tilde{l_{3}},\qquad \frac{d \tilde{l_{4}}}{da_{1}}=- \tilde{l_{4}},\qquad \frac{d\tilde{l_{5}}}{da_{1}}=0, \\& \frac{d\tilde{l_{1}}}{da_{2}}=0,\qquad \frac{d\tilde{l_{2}}}{da_{2}}= \tilde{l_{1}}, \qquad \frac{d\tilde{l_{3}}}{da_{2}}=0,\qquad \frac{d\tilde{l_{4}}}{da _{2}}=0, \qquad \frac{d\tilde{l_{5}}}{da_{2}}=0, \\& \frac{d\tilde{l_{1}}}{da_{3}}=0,\qquad \frac{d\tilde{l_{2}}}{da_{3}}=0, \qquad \frac{d \tilde{l_{3}}}{da_{3}}=- \tilde{l_{1}},\qquad \frac{d\tilde{l_{4}}}{da_{3}}=0,\qquad \frac{d \tilde{l_{5}}}{da_{3}}=0, \\& \frac{d\tilde{l_{1}}}{da_{4}}=0, \qquad \frac{d\tilde{l_{2}}}{da_{4}}=0,\qquad \frac{d \tilde{l_{3}}}{da_{4}}=0, \qquad \frac{d\tilde{l_{4}}}{da_{4}}=\tilde{l_{1}}+ \tilde{l_{5}}, \qquad \frac{d\tilde{l_{5}}}{da_{4}}=0, \\& \frac{d\tilde{l_{1}}}{da_{5}}=0,\qquad \frac{d\tilde{l_{2}}}{da_{5}}=0,\qquad \frac{d \tilde{l_{3}}}{da_{5}}=0, \qquad \frac{d\tilde{l_{4}}}{da_{5}}=- \tilde{l_{4}},\qquad \frac{d\tilde{l_{5}}}{da_{5}}=0. \end{aligned}$$

The solutions of these equations give the transformations

$$\begin{aligned}& T_{1}: \tilde{l_{1}}=l_{1},\qquad \tilde{l_{2}}=e^{-a_{1}}l_{2},\qquad \tilde{l_{3}}=e^{a_{1}}l_{3},\qquad \tilde{l_{4}}=e^{-a_{1}}l_{4},\qquad \tilde{l_{5}}=l_{5}, \\& T_{2}: \tilde{l_{1}}=l_{1},\qquad \tilde{l_{2}}=a_{2}l_{1}+l_{2},\qquad \tilde{l_{3}}=l_{3},\qquad \tilde{l_{4}}=l_{4}, \qquad \tilde{l_{5}}=l_{5}, \\& T_{3}: \tilde{l_{1}}=l_{1},\qquad \tilde{l_{2}}=l_{2}, \qquad \tilde{l_{3}}=-a _{3}l_{1}+l_{3},\qquad \tilde{l_{4}}=l_{4}, \qquad \tilde{l_{5}}=l_{5}, \\& T_{4}: \tilde{l_{1}}=l_{1}, \qquad \tilde{l_{2}}=l_{2}, \qquad \tilde{l_{3}}=l _{3},\qquad \tilde{l_{4}}=a_{4}(l_{1}+l_{5})+l_{4}, \qquad \tilde{l_{5}}=l_{5}, \\& T_{5}: \tilde{l_{1}}=l_{1},\qquad \tilde{l_{2}}=l_{2},\qquad \tilde{l_{3}}=l _{3},\qquad \tilde{l_{4}}=e^{-a_{5}}l_{4}, \qquad \tilde{l_{5}}=l_{5}. \end{aligned}$$

The determination of the optimal system demands a simplification of the vector

$$\begin{aligned}& l=(l_{1},l_{2},l_{3},l_{4},l_{5}) \end{aligned}$$

(8)

by using the transformations $T_{1}-T_{5}$. We will obtain the simplest representative of each class of similar vectors (8). Take two cases into account.

Case 2.1 $l_{1}\neq 0$

Let $a_{2}=-\frac{l_{2}}{l_{1}}$ and $a_{3}=\frac{l_{3}}{l_{1}}$ in the transformations $T_{2}$ and $T_{3}$, we have $\tilde{l_{2}}= \tilde{l_{3}}=0$. Vector (8) is simplified as the form

$$\begin{aligned}& l=(l_{1},0,0,l_{4},l_{5}). \end{aligned}$$

(9)

• 2.1.1 $l_{1}+l_{5}\neq 0$

By taking $a_{4}=-\frac{l_{4}}{l_{1}+l_{5}}$ in the transformation $T _{4}$, we have $\tilde{l_{4}}=0$. Vector (9) is hence reduced to the form

$$\begin{aligned}& l=(l_{1},0,0,0,l_{5}). \end{aligned}$$

(10)

We get the following representatives:

$$\begin{aligned}& X_{1},X_{1}+ X_{5}. \end{aligned}$$

(11)

• 2.1.2 $l_{1}+l_{5}=0$

We get the following representatives:

$$\begin{aligned}& X_{1}- X_{5},X_{1}- X_{5}\pm X_{4}. \end{aligned}$$

(12)

Case 2.2 $l_{1}=0$

We obtain vector (8) of the form

$$\begin{aligned}& l=(0,l_{2},l_{3},l_{4},l_{5}). \end{aligned}$$

(13)

• 2.2.1 $l_{5}\neq 0$

Let $a_{4}=-\frac{l_{4}}{l_{5}}$ in the transformations $T_{4}$, we have $\tilde{l_{4}}=0$. Vector (13) is simplified as the form

$$\begin{aligned}& l=(0,l_{2},l_{3},0,l_{5}). \end{aligned}$$

(14)

As a result, we get the following representatives:

$$\begin{aligned}& X_{5},X_{5}\pm X_{2},,X_{5} \pm X_{3},,X_{5}\pm X_{2}\pm X_{3}. \end{aligned}$$

(15)

• 2.2.2 $l_{5}=0$

We obtain vector (13) of the form

$$\begin{aligned}& l=(0,l_{2},l_{3},l_{4},0). \end{aligned}$$

(16)

Considering all the possible combinations, we get the following representatives:

$$\begin{aligned}& X_{2},X_{3},X_{4},X_{2} \pm X_{3},X_{2}\pm X_{4},X_{3}\pm X_{4},X _{2}\pm X_{3}\pm X_{4}. \end{aligned}$$

(17)

Finally, by putting together all the operators (11), (12), (15), and (17), we achieve the following theorem.

Theorem 2.1

The operators in $\{X_{1},X_{2},X_{3},X_{4},X_{5}\}$ create an optimal system:

$$\begin{aligned}& X_{1},X_{1}\pm X_{5},X_{1}-X_{5} \pm X_{4},X_{5},X_{5}\pm X_{2},X _{5}\pm X_{3}, X_{5}\pm X_{2}\pm X_{3}, \\& X_{2},X_{3},X_{4},X_{2}\pm X_{3},X_{2}\pm X_{4},X_{3}\pm X_{4},X _{2}\pm X_{3}\pm X_{4}. \end{aligned}$$

3 Similarity reductions and exact solutions

Based on Theorem 2.1, we will discuss the reductions and exact solutions of Eq. (4) in this section.

Case 3.1 Reduction by $X_{1}$.

Solving the characteristic equation for $X_{1}$, we get similarity variables

$$\begin{aligned}& z=tx,\qquad p=\frac{u}{x},\qquad q=v, \end{aligned}$$

and the group-invariant solution is $p=f(z)$, $q=g(z)$, i.e.,

$$\begin{aligned}& u=xf(z),\qquad v=g(z). \end{aligned}$$

(18)

Using Eq. (18) in Eq. (4), we have

$$ \textstyle\begin{cases} f+f^{3}-2f'+3zf^{2}f'-zf''+z^{2}ff^{\prime 2}+\frac{1}{2}z^{2}f^{2}f''=0, \\ g-g'+zf^{2}g'-zg''+z^{2}ff'g'+\frac{1}{2}z^{2}f^{2}g''=0, \end{cases} $$

(19)

where $f'=\frac{df}{dz}$, $g'=\frac{dg}{dz}$.

Case 3.2 Reduction by $X_{2}+X_{5}$.

Similarly, we have $z=x$, $u=f(z)$, $v=g(z)e^{t}$. The corresponding reduction equation is

$$ \textstyle\begin{cases} f+ff^{\prime 2}+\frac{1}{2}f^{2}f''=0, \\ g-g'+ff'g'+\frac{1}{2}f^{2}g''=0,\end{cases} $$

(20)

where $f'=\frac{df}{dz}$, $g'=\frac{dg}{dz}$. Therefore, Eq. (4) has a solution $u=0$, $v=c _{1}e^{x+t}$, where $c_{1}$ is an arbitrary constant.

Case 3.3 Reduction by $X_{2}+X_{3}+X_{5}$.

We have $u=f(z)$, $v=g(z)e^{x}$ in which $z=x-t$. Substituting group invariant solution into Eq. (4), we get

$$ \textstyle\begin{cases} f+ff^{\prime 2}+f''+\frac{1}{2}f^{2}f''=0, \\ g+g'+g''+\frac{1}{2}f^{2}g+ff'g+ff'g'+\frac{1}{2}f^{2}g''+f^{2}g'=0, \end{cases} $$

(21)

where $f'=\frac{df}{dz}$, $g'=\frac{dg}{dz}$.

Case 3.4 Reduction by $X_{3}+X_{4}$.

We have $u=f(z)$, $v=g(z)+xf(z)$ in which $z=t$. Substituting group invariant solution into Eq. (4), we get

$$ \textstyle\begin{cases} f=0, \\ g=0.\end{cases} $$

(22)

Therefore, Eq. (4) has a solution $u=0$, $v=0$. Obviously, the solution is not meaningful.

Some of the similarity reductions for the optimal system of one-dimensional subalgebra are represented in Table 2.

Table 2 Similarity reductions of the Yao–Zeng two-component short-pulse equation

4 The explicit power series solutions

In Sect. 3, we obtained the reduction equations by using symmetry analysis. The power series can be used to treat differential equations, including many complicated differential equations with nonconstant coefficients [25]. In this section, we solve the nonlinear ODE (19) by the power series method. For other reduction equations, power series solutions can also be obtained similarly. For more details on power series solutions, see [15, 26].

Now, we seek a solution of Eq. (19) in a power series of the form

$$\begin{aligned}& f(z)=\sum_{n=0}^{\infty }p_{n}z^{n}, \qquad g(z)=\sum_{n=0}^{\infty }q_{n}z^{n}, \end{aligned}$$

(23)

where the coefficients $p_{n}$ and $q_{n}$ are all constants to be determined.

Substituting (23) into (19), we have

$$\begin{aligned}& \sum_{n=0}^{\infty }p_{n}z^{n}+ \sum_{n=0}^{\infty }\sum _{k=0}^{n} \sum_{l=0}^{k}p_{l}p_{k-l}p_{n-k}z^{n}-2 \sum_{n=0}^{\infty }(n+1)p _{n+1}z^{n} \\& \quad {} +3z\sum_{n=0}^{\infty }\sum _{k=0}^{n}\sum_{l=0}^{k}(n-k+1)p_{l}p_{k-l}p _{n-k+1}z^{n} \\& \quad {} -z\sum_{n=0}^{\infty }(n+1) (n+2)p_{n+2}z^{n} \\& \quad {}+z^{2}\sum_{n=0}^{\infty } \sum_{k=0}^{n}\sum _{l=0}^{k}(k-l+1) (n-k+1)p _{l}p_{k-l+1}p_{n-k+1}z^{n} \\& \quad {}+\frac{1}{2}z^{2}\sum_{n=0}^{\infty } \sum_{k=0}^{n}\sum _{l=0}^{k}(n-k+1) (n-k+2)p _{l}p_{k-l}p_{n-k+2}z^{n}=0, \\& \sum_{n=0}^{\infty }q_{n}z^{n}- \sum_{n=0}^{\infty }(n+1)q_{n+1}z^{n}+z \sum_{n=0}^{\infty }\sum _{k=0}^{n}\sum_{l=0}^{k}(n-k+1)p_{l}p_{k-l}q _{n-k+1}z^{n} \\& \quad {} -z\sum_{n=0}^{\infty }(n+1) (n+2)q_{n+2}z^{n} \\& \quad {}+z^{2}\sum_{n=0}^{\infty } \sum_{k=0}^{n}\sum _{l=0}^{k}(k-l+1) (n-k+1)p_{l}p_{k-l+1}q_{n-k+1}z ^{n} \\& \quad {}+\frac{1}{2}z^{2}\sum_{n=0}^{\infty } \sum_{k=0}^{n}\sum _{l=0}^{k}(n-k+1) (n-k+2)p _{l}p_{k-l}q_{n-k+2}z^{n}=0. \end{aligned}$$

(24)

From (24), comparing coefficients, we obtain

$$\begin{aligned}& \begin{aligned} & p_{1}=\frac{1}{2}p_{0} \bigl(1+p_{0}^{2}\bigr),\qquad q_{1}=q_{0}; \\ & p_{2}=\frac{1}{6}p_{1}\bigl(1+6p_{0}^{2} \bigr),\qquad q_{2}=\frac{1}{4}q_{1}\bigl(1+p _{0}^{2}\bigr). \end{aligned} \end{aligned}$$

(25)

Generally, for $n\geq 0$, we have

$$\begin{aligned}& \begin{aligned} & p_{n+3}=\frac{1}{(n+3)(n+4)}\Biggl\{ p_{n+2}+\sum_{k=0}^{n}\sum _{l=0}^{k}(n-k+1)p _{l} \biggl[(k-l+1)p_{k-l+1}p_{n-k+1} \\ &\hphantom{p_{n+3}=}{} +\frac{1}{2}(n-k+2)p_{k-l}p_{n-k+2} \biggr]+\sum_{k=0}^{n+1}\sum _{l=0}^{k}(3n-3k+7)p _{l}p_{k-l}p_{n-k+2} \\ &\hphantom{p_{n+3}=}{}+\sum_{l=0}^{n+2}p_{l}p_{n+2-l}p_{0} \Biggr\} , \\ & q_{n+3}=\frac{1}{(n+3)^{2}}\Biggl\{ q_{n+2}+\sum _{k=0}^{n}\sum_{l=0}^{k} \biggl[ \frac{1}{2}(n-k+2) (n-k+3)p_{l}p_{k-l}q_{n-k+2} \\ &\hphantom{q_{n+3}=}{} +(k-l+1) (n-k+1)p_{l}p_{k-l+1}q_{n-k+1} \biggr]+\sum_{l=0}^{n+1}p_{l}p_{n+1-l}q _{1}\Biggr\} . \end{aligned} \end{aligned}$$

(26)

In view of Eq. (26), we can obtain all the coefficients $p_{i}$, $q_{i}$ ($i \geq 3$) of the power series (23), e.g.,

$$\begin{aligned}& \begin{aligned} & p_{3}=\frac{1}{12} \bigl(p_{2}+10p_{0}^{2}p_{2}+10p_{0}p_{1}^{2} \bigr), \\ & q_{3}=\frac{1}{9}\bigl(q_{2}+3p_{0}^{2}q_{2}+3p_{0}p_{1}q_{1} \bigr). \end{aligned} \end{aligned}$$

(27)

Therefore, for arbitrary chosen constant numbers $p_{0}$ and $q_{0}$, the other terms of the sequence $\{p_{n}\}_{n=0}^{\infty }$ and $\{q _{n}\}_{n=0}^{\infty }$ can be determined by (25) and (26). This implies that for Eq. (19), there is a power series solution (23) with the coefficients constructed by (25) and (26).

Now we show the convergence of the power series solution (23) of Eq. (19). In fact, from (26), we have

$$\begin{aligned} \vert p_{n+3} \vert \leq& M\Biggl[ \vert p_{n+2} \vert +\sum_{k=0}^{n}\sum _{l=0}^{k} \vert p_{l} \vert \bigl( \vert p _{k-l+1} \vert \vert p_{n-k+1} \vert + \vert p_{k-l} \vert \vert p_{n-k+2} \vert \bigr) \\ &{} +\sum_{k=0}^{n+1}\sum _{l=0}^{k} \vert p_{l} \vert \vert p_{k-l} \vert \vert p_{n-k+2} \vert +\sum _{l=0} ^{n+2} \vert p_{l} \vert \vert p_{n+2-l} \vert \Biggr],\quad n=0,1,\ldots , \end{aligned}$$

where $M=\max\{1,\frac{1}{2},3,p_{0}\}$. Similarly, from (26), we have

$$\begin{aligned} \vert q_{n+3} \vert \leq& N\Biggl[ \vert q_{n+2} \vert +\sum_{k=0}^{n} \sum_{l=0}^{k}\bigl( \vert p_{l} \vert \vert p _{k-l} \vert \vert q_{n-k+2} \vert + \vert p_{l} \vert \vert p_{k-l+1} \vert \vert q_{n-k+1} \vert \bigr) \\ &{} +\sum_{l=0}^{n+1} \vert p_{l} \vert \vert p_{n+1-l} \vert \Biggr],\quad n=0,1, \ldots , \end{aligned}$$

(28)

where $N=\max\{1,\frac{1}{2},q_{1}\}$.

Now, we define two power series $R=R(z)=\sum_{n=0}^{\infty }r_{n}z ^{n}$ and $S=S(z)=\sum_{n=0}^{\infty }s_{n}z^{n}$ by

$$\begin{aligned}& r_{i}= \vert p_{i} \vert ,\qquad s_{j}= \vert q_{j} \vert ,\quad i,j=0,1,2, \end{aligned}$$

and

$$\begin{aligned}& \begin{aligned} & r_{n+3}=M\Biggl[r_{n+2}+\sum _{k=0}^{n}\sum_{l=0}^{k}(r_{l}r_{k-l+1}r_{n-k+1}+r _{l}r_{k-l}r_{n-k+2}) \\ &\hphantom{r_{n+3}=}{} +\sum_{k=0}^{n+1}\sum _{l=0}^{k}r_{l}r_{k-l}r_{n-k+2}+ \sum_{l=0}^{n+2}r _{l}r_{n+2-l} \Biggr], \\ & s_{n+3}=N\Biggl[s_{n+2}+\sum_{k=0}^{n} \sum_{l=0}^{k}(r_{l}r_{k-l}s_{n-k+2}+r _{l}r_{k-l+1}s_{n-k+1}) \\ &\hphantom{s_{n+3}=}{} +\sum_{l=0}^{n+1}r_{l}r_{n+1-l} \Biggr], \end{aligned} \end{aligned}$$

(29)

where $n=0,1,\ldots$ . Then, it is easily seen that

$$\begin{aligned}& \vert p_{n} \vert {\leq }r_{n},\qquad \vert q_{n} \vert {\leq }s_{n},\quad n=0,1,2,\ldots . \end{aligned}$$

Thus, the two series $R=R(z)=\sum_{n=0}^{\infty }r_{n}z^{n}$ and $S=S(z)= \sum_{n=0}^{\infty }s_{n}z^{n}$ are majorant series of (23), respectively. Next, we show that the series $R = R(z)$ and $S = S(z)$ have positive radius of convergence.

$$\begin{aligned} R(z) =&r_{0}+r_{1}z+r_{2}z^{2}+\sum _{n=0}^{\infty }r_{n+3}z^{n+3}=r _{0}+r_{1}z+r_{2}z^{2} \\ &{} +M\Biggl[\sum_{n=0}^{\infty }r_{n+2}z^{n+3}+ \sum_{n=0}^{\infty }\sum _{k=0} ^{n}\sum_{l=0}^{k}(r_{l}r_{k-l+1}r_{n-k+1}+r_{l}r_{k-l}r_{n-k+2})z ^{n+3} \\ &{} +\sum_{n=0}^{\infty }\sum _{k=0}^{n+1}\sum_{l=0}^{k}r_{l}r_{k-l}r_{n-k+2}z ^{n+3}+\sum_{n=0}^{\infty }\sum _{l=0}^{n+2}r_{l}r_{n+2-l}z^{n+3} \Biggr] \\ =&r_{0}+r_{1}z+r_{2}z^{2}+M \bigl[z(R-r_{0}-r_{1}z)+zR(R-r_{0})^{2}+zR^{2}(R-r _{0}-r_{1}z) \\ &{} +z\bigl(R^{2}(R-r_{0})-r_{0}^{2}r_{1}z \bigr)+z\bigl((R+r_{0}) (R-r_{0}-r_{1}z)+r_{1}z(R-r _{0})\bigr)\bigr], \end{aligned}$$

and

$$\begin{aligned} S(z) =&s_{0}+s_{1}z+s_{2}z^{2}+\sum _{n=0}^{\infty }s_{n+3}z^{n+3} \\ =&s_{0}+s_{1}z+s_{2}z^{2}+N\Biggl[ \sum_{n=0}^{\infty }s_{n+2}z^{n+3}+ \sum_{n=0}^{\infty }\sum _{k=0}^{n}\sum_{l=0}^{k}(r_{l}r_{k-l}s_{n-k+2}+r _{l}r_{k-l+1}s_{n-k+1})z^{n+3} \\ &{} +\sum_{n=0}^{\infty }\sum _{l=0}^{n+1}r_{l}r_{n+1-l}z^{n+3} \Biggr] \\ =&s_{0}+s_{1}z+s_{2}z^{2}+N \bigl[z(S-s_{0}-s_{1}z) \\ &{} +zR^{2}(S-s_{0}-s_{1}z)+zR(RS-r_{0}S-Rs_{0}+r_{0}s_{0})+z^{2} \bigl(R^{2}-r _{0}^{2}\bigr)\bigr]. \end{aligned}$$

Consider now the implicit functional system with respect to the independent variable z:

$$\begin{aligned}& F(z,R,S)=R-r_{0}-r_{1}z-r_{2}z^{2}-M \bigl[z(R-r_{0}-r_{1}z)+zR(R-r_{0})^{2}+zR ^{2}(R-r_{0}-r_{1}z) \\& \hphantom{F(z,R,S)=}{} +z\bigl(R^{2}(R-r_{0})-r_{0}^{2}r_{1}z \bigr)+z\bigl((R+r_{0}) (R-r_{0}-r_{1}z)+r_{1}z(R-r _{0})\bigr)\bigr], \\& G(z,R,S)=S-s_{0}-s_{1}z-s_{2}z^{2}-N \bigl[z(S-s_{0}-s_{1}z)+zR^{2}(S-s _{0}-s_{1}z) \\& \hphantom{G(z,R,S)=}{} +zR(RS-r_{0}S-Rs_{0}+r_{0}s_{0})+z^{2} \bigl(R^{2}-r_{0}^{2}\bigr)\bigr]. \end{aligned}$$

Since F, G are analytic in the neighborhood of $(0,r_{0}, s_{0})$ and $F(0,r_{0}, s_{0}) = 0$, $G(0,r_{0}, s_{0}) = 0$. Furthermore, the Jacobian determinant

$$\begin{aligned}& \frac{\partial (F,G)}{\partial (R,S)}\bigg|_{(0,r_{0},s _{0})}=1\neq 0, \end{aligned}$$

if we choose the parameters $r_{0} = |p_{0}|$ and $s_{0} = |q_{0}|$ properly. By the implicit function theorem [27], we see that $R=R(z)$ and $S=S(z)$ are analytic in a neighborhood of the point $(0,r_{0},s_{0})$ and with the positive radius. This implies that the two power series (23) converge in a neighborhood of the point $(0,r_{0},s_{0})$. This completes the proof.

Hence, the power series solution (23) for Eq. (19) is an analytic solution. The power series solution of Eq. (19) can be written as the following:

$$\begin{aligned}& f(z)=p_{0}+p_{1}z+p_{2}z^{2}+\sum _{n=0}^{\infty }p_{n+3}z^{n+3} \\& \hphantom{f(z)} =p_{0}+\frac{1}{2}p_{0} \bigl(1+p_{0}^{2}\bigr)z+\frac{1}{6}p_{1} \bigl(1+6p_{0}^{2}\bigr)z ^{2} \\& \hphantom{f(z)=}{} +\sum_{n=0}^{\infty } \frac{1}{(n+3)(n+4)}\Biggl\{ p_{n+2}+\sum_{k=0}^{n} \sum_{l=0}^{k}(n-k+1)p_{l} \biggl[(k-l+1)p_{k-l+1}p_{n-k+1} \\& \hphantom{f(z)=}{} +\frac{1}{2}(n-k+2)p_{k-l}p_{n-k+2} \biggr]+\sum_{k=0}^{n+1}\sum _{l=0}^{k}(3n-3k+7)p _{l}p_{k-l}p_{n-k+2} \\& \hphantom{f(z)=}{}+\sum_{l=0}^{n+2}p_{l}p_{n+2-l}p_{0} \Biggr\} z^{n+3}, \\& g(z)=q_{0}+q_{1}z+q_{2}z^{2}+\sum _{n=0}^{\infty }q_{n+3}z^{n+3} \\& \hphantom{g(z)} =q_{0}+q_{0}z+\frac{1}{4}q_{1} \bigl(1+p_{0}^{2}\bigr)z^{2} \\& \hphantom{g(z)=}{} +\sum_{n=0}^{\infty } \frac{1}{(n+3)^{2}}\Biggl\{ q_{n+2}+\sum_{k=0}^{n} \sum_{l=0}^{k}\biggl[\frac{1}{2}(n-k+2) (n-k+3)p_{l}p_{k-l}q_{n-k+2} \\& \hphantom{g(z)=}{} +(k-l+1) (n-k+1)p_{l}p_{k-l+1}q_{n-k+1} \biggr]+\sum_{l=0}^{n+1}p_{l}p_{n+1-l}q _{1}\Biggr\} z^{n+3}. \end{aligned}$$

Thus, the explicit power series solution of Eq. (4) is

$$\begin{aligned}& u(x,t) =p_{0}x+\frac{1}{2}p_{0}\bigl(1+p_{0}^{2} \bigr)tx^{2}+\frac{1}{6}p_{1}\bigl(1+6p _{0}^{2}\bigr)t^{2}x^{3} \\& \hphantom{u(x,t) =}{} +\sum_{n=0}^{\infty } \frac{1}{(n+3)(n+4)}\Biggl\{ p_{n+2}+\sum_{k=0}^{n} \sum_{l=0}^{k}(n-k+1)p_{l} \biggl[(k-l+1)p_{k-l+1}p_{n-k+1} \\& \hphantom{u(x,t) =}{} +\frac{1}{2}(n-k+2)p_{k-l}p_{n-k+2} \biggr]+\sum_{k=0}^{n+1}\sum _{l=0}^{k}(3n-3k+7)p _{l}p_{k-l}p_{n-k+2} \\& \hphantom{u(x,t) =}{}+\sum_{l=0}^{n+2}p_{l}p_{n+2-l}p_{0} \Biggr\} t^{n+3}x ^{n+4}, \\& v(x,t)=q_{0}+q_{0}tx+\frac{1}{4}q_{1} \bigl(1+p_{0}^{2}\bigr)t^{2}x^{2} \\& \hphantom{v(x,t) =}{} +\sum_{n=0}^{\infty } \frac{1}{(n+3)^{2}}\Biggl\{ q_{n+2}+\sum_{k=0}^{n} \sum_{l=0}^{k}\biggl[\frac{1}{2}(n-k+2) (n-k+3)p_{l}p_{k-l}q_{n-k+2} \\& \hphantom{v(x,t) =}{} +(k-l+1) (n-k+1)p_{l}p_{k-l+1}q_{n-k+1} \biggr]+\sum_{l=0}^{n+1}p_{l}p_{n+1-l}q _{1}\Biggr\} t^{n+3}x^{n+3}, \end{aligned}$$

where $p_{0}$ and $q_{0}$ are arbitrary constants, the other coefficients $p _{n}$, $q_{n}$ ($n\geq 1$) depend on (25) and (26) completely.

Remark 4.1

The power series solutions can greatly enrich the solutions of Eq. (4) and converge quickly, so it is convenient for computations in both theory and applications.

5 Nonlinear self-adjointness and conservation law

In this section, we prove the nonlinear self-adjointness of Eq. (4) and determine its conservation laws.

5.1 Preliminaries

Consider the rth-order system of m PDEs:

$$\begin{aligned}& E_{\alpha }(x,u,u_{(1)},\ldots ,u_{(r)})=0, \quad \alpha =1,2,\ldots ,m, \end{aligned}$$

(30)

where $x=(x^{1},\dots ,x^{n})$, $u=(u^{1},\dots ,u^{m})$, $u^{\alpha } _{i}=\partial u^{\alpha }/\partial x^{i}$, $u^{\alpha }_{ij}=\partial ^{2} u^{\alpha }/\partial x^{i}\partial x^{j}$, and $u_{(i)}$ denotes the collection of all ith-order partial derivatives of u with respect to x.

The adjoint equations of Eq. (30) are defined by [28]

$$\begin{aligned}& E_{\alpha }^{\ast }(x,u,v,u_{(1)},v_{(1)}, \ldots ,u_{(r)},v_{(r)})=\frac{ \delta \mathcal{L}}{\delta u^{\sigma }}=0, \end{aligned}$$

(31)

where $v=(v^{1},\dots ,v^{m})$, $\mathcal{L}=v^{\beta }E_{\beta }(x,u,u _{(1)},\ldots ,u_{(r)})$ is the formal Lagrangian, and $\delta / \delta u^{\sigma }$ is the Euler–Lagrange operator defined by [28]

$$\begin{aligned}& \frac{\delta }{\delta u^{\sigma }}=\frac{\partial }{\partial u^{ \sigma }}+\sum_{s=1}^{\infty }(-1)^{s}D_{i_{1}} \dots D_{i_{s}}\frac{ \partial }{\partial u^{\sigma }_{i_{1}\dots i_{s}}}, \end{aligned}$$

where $D_{i}$ denotes the total derivative operators with respect to $x_{i}$.

Definition 5.1

([16])

System (30) is said to be nonlinearly self-adjoint if the adjoint system (31) is satisfied for all solutions u of Eq. (30) upon a substitution

$$\begin{aligned}& v^{\alpha }=\varphi ^{\alpha }(x,u), \quad \alpha =1, \dots ,m, \end{aligned}$$

(32)

such that $\varphi (x,u)=(\varphi ^{1},\dots ,\varphi ^{m})\neq 0$.

Definition 5.1 is identical to the following identities:

$$\begin{aligned}& E_{\alpha }^{\ast }(x,u,v,u_{(1)},v_{(1)}, \ldots ,u_{(r)},v_{(r)}) = \lambda _{\alpha }^{\beta }E_{\beta }(x,u,u_{(1)}, \ldots ,u_{(r)}), \alpha ,\quad \beta =1,\dots ,m, \end{aligned}$$

(33)

where $\lambda _{\alpha }^{\beta }$ is a certain function.

The following theorem will be used to obtain conservation laws [17].

Theorem 5.2

Any infinitesimal symmetry (local and nonlocal)

$$\begin{aligned}& X=\xi ^{i}(x,u,u_{(1)},\dots )\frac{\partial }{\partial x^{i}}+ \eta ^{\alpha }(x,u,u_{(1)},\dots )\frac{\partial }{\partial u^{\alpha }} \end{aligned}$$

admitted by Eq. (30) gives rise to a conservation law $D_{i}(C^{i})=0$, where $C^{i}$ is constructed by the formula

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

(34)

with $W^{\alpha }=\eta ^{\alpha }-\xi ^{j}u_{j}^{\alpha }$. The Lagrangian $\mathcal{L}$ should be written in the symmetric form with respect to all mixed derivatives $u_{ij}^{\alpha },u_{ijk}^{\alpha },\dots$.

5.2 Nonlinear self-adjointness

Following Definition 5.1 and Theorem 5.2, we can prove the nonlinear self-adjointness of Eq. (4) and thus obtain its conservation laws.

Theorem 5.3

Equation (4) is nonlinearly self-adjoint under the substitution

$$ \varLambda _{1}=c_{1}v,\qquad \varLambda _{2}=-c_{1}u, $$

where $c_{1}$ is an arbitrary constant.

Proof

Let the formal Lagrangian of Eq. (4) be of the form

$$ \mathcal{L}=\varLambda _{1}\biggl(u_{xt}-u- \frac{1}{6}\bigl(u^{3}\bigr)_{xx}\biggr)+\varLambda _{2}\biggl(v _{xt}-v-\frac{1}{2}\bigl(u^{2}v_{x} \bigr)_{x}\biggr), $$

where $\varLambda _{1}$, $\varLambda _{2}$ are two new dependent variables.

Using the equivalent formula (33) of the definition of nonlinear self-adjointness, the identities

$$\begin{aligned}& \begin{aligned} &\frac{\delta \mathcal{L}}{\delta u}=\lambda _{1}^{1} \biggl(u_{xt}-u- \frac{1}{6}\bigl(u^{3} \bigr)_{xx}\biggr)+\lambda _{1}^{2} \biggl(v_{xt}-v-\frac{1}{2}\bigl(u^{2}v _{x} \bigr)_{x}\biggr), \\ &\frac{\delta \mathcal{L}}{\delta v}=\lambda _{2}^{1}\biggl(u_{xt}-u- \frac{1}{6}\bigl(u^{3}\bigr)_{xx}\biggr)+\lambda _{2}^{2}\biggl(v_{xt}-v-\frac{1}{2} \bigl(u^{2}v _{x}\bigr)_{x}\biggr), \end{aligned} \end{aligned}$$

(35)

are established under the substitution $\varLambda _{1}=\varLambda _{1}(x,t,u,v)$, $\varLambda _{2}=\varLambda _{2}(x,t,u,v)$.

By splitting Eq. (35) with respect to the coefficients of different order derivatives of u and v, we obtain a system in the unknown variables $\varLambda _{1}$, $\varLambda _{2}$ whose solutions are

$$ \varLambda _{1}=c_{1}v,\qquad \varLambda _{2}=-c_{1}u, $$

where $c_{1}$ is an arbitrary constant. This completes the proof. □

5.3 Construction of conservation laws

For the infinitesimal operator $X=\xi \partial _{x}+\tau \partial _{t}+\phi \partial _{u}+\varphi \partial _{v}$, by Theorem 5.2, the conservation law of Eq. (4) is represented in the form $D_{x} C^{x}+D _{t} C^{t}=0$. Moreover, we have

$$\begin{aligned}& C^{x}=W^{u}\biggl[\frac{\partial \mathcal{L}}{\partial u_{x}}-D_{t} \biggl(\frac{ \partial \mathcal{L}}{\partial u_{xt}}\biggr)-D_{x}\biggl(\frac{\partial \mathcal{L}}{\partial u_{xx}}\biggr) \biggr]+D_{t}\bigl(W^{u}\bigr)\frac{\partial \mathcal{L}}{\partial u_{xt}}+D_{x} \bigl(W^{u}\bigr)\frac{\partial \mathcal{L}}{ \partial u_{xx}} \\& \hphantom{C^{x}=}{} +W^{v}\biggl[\frac{\partial \mathcal{L}}{\partial v_{x}}-D_{t} \biggl(\frac{\partial \mathcal{L}}{\partial v_{xt}}\biggr)-D_{x}\biggl(\frac{\partial \mathcal{L}}{ \partial v_{xx}}\biggr) \biggr]+D_{t}\bigl(W^{v}\bigr)\frac{\partial \mathcal{L}}{\partial v _{xt}}+D_{x} \bigl(W^{v}\bigr)\frac{\partial \mathcal{L}}{\partial v_{xx}}, \\& C^{t}=W^{u}\biggl[-D_{x}\biggl( \frac{\partial \mathcal{L}}{\partial u_{xt}}\biggr)\biggr]+D _{x}\bigl(W^{u}\bigr) \frac{\partial \mathcal{L}}{\partial u_{xt}} \\& \hphantom{C^{t}=}{}+W^{v}\biggl[-D_{x}\biggl( \frac{ \partial \mathcal{L}}{\partial v_{xt}}\biggr)\biggr]+D_{x}\bigl(W^{v}\bigr) \frac{\partial \mathcal{L}}{\partial v_{xt}}, \end{aligned}$$

where $W^{u}=\phi -\xi u_{x}-\tau u_{t}$, $W^{v}=\varphi -\xi v_{x}- \tau v_{t}$, the Lagrangian $\mathcal{L}=\varLambda _{1}(u_{xt}-u- \frac{1}{6}(u^{3})_{xx})+\varLambda _{2}(v_{xt}-v-\frac{1}{2}(u^{2}v_{x})_{x})=c _{1}v(u_{xt}-u-\frac{1}{6}(u^{3})_{xx})-c_{1}u(v_{xt}-v-\frac{1}{2}(u ^{2}v_{x})_{x})$.

By simplifying $C^{x}$, $C^{t}$, we have

$$\begin{aligned}& C^{x}=\biggl(-c_{1}uvu_{x}+ \frac{3}{2}c_{1}u^{2}v_{x}-c_{1}v_{t} \biggr)W^{u}+c _{1}vD_{t}\bigl(W^{u} \bigr)-\frac{1}{2}c_{1}u^{2}vD_{x} \bigl(W^{u}\bigr) \\& \hphantom{C^{x}=}{}+\biggl(-\frac{1}{2}c_{1}u^{2}u_{x}+c_{1}u_{t} \biggr)W^{v}-c_{1}uD_{t}\bigl(W^{v} \bigr)+ \frac{1}{2}c_{1}u^{3}D_{x} \bigl(W^{v}\bigr), \end{aligned}$$

(36)

$$\begin{aligned}& C^{t}=-c_{1}v_{x}W^{u}+c_{1}vD_{x} \bigl(W^{u}\bigr)+c_{1}u_{x}W^{v}-c_{1}uD _{x}\bigl(W^{v}\bigr). \end{aligned}$$

(37)

Case 5.1 For the generator $X_{1}=-x\partial _{x}+t\partial _{t}-u\partial _{u}$, we obtain $W^{u}=-u+xu_{x}-t u_{t}$, $W^{v}=xv_{x}-t v_{t}$. According to Eqs. (36) and (37), we have

$$\begin{aligned}& C^{x}=\biggl(-c_{1}uvu_{x}+ \frac{3}{2}c_{1}u^{2}v_{x}-c_{1}v_{t} \biggr) (-u+xu _{x}-tu_{t})+c_{1}v(-2u_{t}+xu_{tx}-tu_{tt}) \\& \hphantom{C^{x}=}{} -\frac{1}{2}c_{1}u^{2}v(xu_{xx}-tu_{tx})+ \biggl(-\frac{1}{2}c_{1}u^{2}u_{x}+c _{1}u_{t}\biggr) (xv_{x}-tv_{t})-c_{1}u(-v_{t}-tv_{tt}+xv_{xt}) \\& \hphantom{C^{x}=}{} +\frac{1}{2}c_{1}u^{3}(v_{x}-tv_{tx}+xv_{xx}), \\& C^{t}=-c_{1}v_{x}(-u+xu_{x}-tu_{t})+c_{1}v(xu_{xx}-tu_{tx})+c_{1}u _{x}(xv_{x}-tv_{t})-c_{1}u(v_{x}-tv_{tx}+xv_{xx}). \end{aligned}$$

Case 5.2 For the generator $X_{2}=\partial _{t}$, we obtain $W ^{u}=-u_{t}$, $W^{v}=-v_{t}$. According to Eqs. (36) and (37), we have

$$\begin{aligned}& C^{x}=-\biggl(-c_{1}uvu_{x}+ \frac{3}{2}c_{1}u^{2}v_{x}-c_{1}v_{t} \biggr)u_{t}-c _{1}vu_{tt}+\frac{1}{2}c_{1}u^{2}vu_{tx} \\& \hphantom{C^{x}=}{} -\biggl(-\frac{1}{2}c_{1}u^{2}u_{x}+c_{1}u_{t} \biggr)v_{t}+c_{1}uv_{tt}- \frac{1}{2}c_{1}u^{3}v_{tx}, \\& C^{t}=c_{1}v_{x}u_{t}-c_{1}vu_{tx}-c_{1}u_{x}v_{t}+c_{1}uv_{tx}. \end{aligned}$$

Case 5.3 For the generator $X_{3}=\partial _{x}$, we obtain $W ^{u}=-u_{x}$, $W^{v}=-v_{x}$. According to Eqs. (36) and (37), we have

$$\begin{aligned}& C^{x}=-\biggl(-c_{1}uvu_{x}+ \frac{3}{2}c_{1}u^{2}v_{x}-c_{1}v_{t} \biggr)u_{x}-c _{1}vu_{xt}+\frac{1}{2}c_{1}u^{2}vu_{xx} \\& \hphantom{C^{x}=}{} -\biggl(-\frac{1}{2}c_{1}u^{2}u_{x}+c_{1}u_{t} \biggr)v_{x}+c_{1}uv_{xt}- \frac{1}{2}c_{1}u^{3}v_{xx}, \\& C^{t}=c_{1}v_{x}u_{x}-c_{1}vu_{xx}-c_{1}u_{x}v_{x}+c_{1}uv_{xx}. \end{aligned}$$

Case 5.4 For the generator $X_{4}=u\partial _{v}$, we obtain $W ^{u}=0$, $W^{v}=u$. According to Eqs. (36) and (37), we have

$$\begin{aligned}& C^{x}=\biggl(-\frac{1}{2}c_{1}u^{2}u_{x}+c_{1}u_{t} \biggr)u-c_{1}uu_{t}+ \frac{1}{2}c_{1}u^{3}u_{x}, \\& C^{t}=c_{1}u_{x}u-c_{1}uu_{x}. \end{aligned}$$

Case 5.5 For the generator $X_{5}=v\partial _{v}$, we obtain $W ^{u}=0$, $W^{v}=u$. According to Eqs. (36) and (37), we have

$$\begin{aligned}& C^{x}=\biggl(-\frac{1}{2}c_{1}u^{2}u_{x}+c_{1}u_{t} \biggr)v-c_{1}uv_{t}+ \frac{1}{2}c_{1}u^{3}v_{x}, \\& C^{t}=c_{1}u_{x}v-c_{1}uv_{x}. \end{aligned}$$

6 Conclusions

In this paper, the Lie symmetry analysis is applied for the Yao–Zeng two-component short-pulse equation. New invariant solutions are constructed based on the optimal system. Moreover, we use the properties of nonlinear self-adjointness of Eq. (4) to obtain general formulae of conservation law. Our results can be applied to describe the propagation of polarized ultrashort light pulses in cubically nonlinear anisotropic optical fibers.