In this paper,we consider the following two-component Camassa-Holm equation

$\left\{ \begin{matrix} {{u}_{i}}-{{u}_{xxt}}+3u{{u}_{x}}-2{{u}_{x}}-u{{u}_{xxx}}+\rho {{\rho }_{x}}=0, \\ {{\rho }_{t}}+{{\rho }_{x}}u+\rho {{u}_{x}}=0, \\ \end{matrix} \right.$

(1)

which was first derived as a bi-Hamiltonian model by Olver and Rosenau^[1]. Eq.(1) includes both velocity and density variables in the dynamics of shallow water waves. In Ref. ^[2],Constantin and Ivanov derived (1) in the context of shallow water waves theory. The variable u(x,t) describes the horizontal velocity of the fluid and the variable ρ(x,t) is in connection with the horizontal deviation of the surface from the equilibrium,all measured in dimensionless units^[2].

For ρ≡0,(1) becomes the Camassa-Holm equation,modeling the unidirectional propagation of shallow water waves over a flat bottom. Here u(x,t) stands for the fluid velocity at time t in the spatial x-direction^[3-8]. The Camassa-Holm equation is also a model for the propagation of axially symmetric waves in hyperelastic rods^[9-10]. It has a bi-Hamiltonian structure^[11-12] and is completely integrable^{[3, 13]}. It was claimed that the equation might be relevant to the modeling of tsunami^[14](see also the discussion in Ref. ^[15]).

The Cauchy problem and initial-boundary value problem for the Camassa-Holm equation have been studied extensively^[16-20]. It has been shown that this equation is locally well-posed for some initial data. More interestingly,it has global strong solutions^{[16-17, 21]} and also finite time blow-up solutions^{[16-17, 21-22]}. For ρ≠0,the Cauchy problems of (1) have been discussed in Refs.^{[2, 23]}.

For the sake of providing more information to understand the shallow water waves moving over a linear shear flow,we will investigate the vector fields,symmetry reductions,and exact solutions to (1) by the Lie symmetry analysis method.

It is known that the Lie symmetry analysis is a powerful and systematic method for dealing with partial differential equations (PDEs)^[24-27]. Moreover,this method has had a profound impact on both pure and applied areas of mathematics,physics,and mechanics,etc. For the PDEs,admitting symmetry is one of the intrinsic properties of the equations. Based on the symmetries of a PDE,many other important properties of the equation such as integrability,conservation laws,reducing equations,and exact solutions can be considered successively^[24-28].

The main purpose of this paper is to apply the Lie group analysis method for dealing with symmetries,symmetry reductions,and exact solutions to the two-component exact Camassa-Holm equation.

1 Lie symmetry analysis for the two-component Camassa-Holm equation

In this section,we perform Lie symmetry analysis for (1),and obtain its infinitesimal generators and commutation table of Lie algebra.

According to the method of determining the infinitesimal generator of PDEs,the infinitesimal generator of (1) can be written as

$V=\xi \left( x,t,u,\rho \right)\frac{\partial }{\partial x}+\tau \left( x,t,u,\rho \right)\frac{\partial }{\partial t}+\phi \left( x,t,u,\rho \right)\frac{\partial }{\partial u}+\psi \left( x,t,u,\rho \right)\frac{\partial }{\partial \rho },$

(2)

where the coefficient functions ξ(x,t,u,ρ),τ(x,t,u,ρ),(x,t,u,ρ),and ψ(x,t,u,ρ) of the vector field are to be determined later.

If the vector field (2) generates a symmetry of (1),then V must satisfy the Lie symmetry condition

$\left\{ \begin{align} & p{{r}^{\left( 3 \right)}}V\left( {{\Delta }_{1}} \right)\left| _{{{\Delta }_{1=0}}}=0, \right. \\ & p{{r}^{\left( 1 \right)}}V\left( {{\Delta }_{2}} \right)\left| _{{{\Delta }_{2=0}}}=0, \right. \\ \end{align} \right.$

(3)

where pr⁽³⁾V and pr⁽¹⁾V denote the 3rd and 1st prolongations of V,respectively,and

${{\Delta }_{1}}={{u}_{t}}-{{u}_{xxt}}+3u{{u}_{x}}-2{{u}_{x}}-u{{u}_{xxx}}+\rho {{\rho }_{x}},$

(4)

${{\Delta }_{2}}={{\rho }_{t}}+{{\rho }_{x}}u+\rho {{u}_{x}},$

(5)

for Eq.(1). That is

$\left\{ \begin{align} & p{{r}^{\left( 3 \right)}}V=\phi \frac{\partial }{\partial u}+\psi \frac{\partial }{\partial \rho }+{{\psi }^{x}}\frac{\partial }{\partial {{u}_{x}}}+{{\psi }^{x}}\frac{\partial }{\partial \rho }+{{\phi }^{t}}\frac{\partial }{\partial {{u}_{t}}}+{{\phi }^{xx}}\frac{\partial }{\partial {{u}_{xx}}}+{{\phi }^{xxx}}\frac{\partial }{\partial {{u}_{xxx}}}+{{\phi }^{xxt}}\frac{\partial }{\partial {{u}_{xxt}}}, \\ & p{{r}^{\left( 1 \right)}}V=\phi \frac{\partial }{\partial u}+\psi \frac{\partial }{\partial \rho }+{{\phi }^{x}}\frac{\partial }{\partial {{u}_{x}}}+{{\psi }^{x}}\frac{\partial }{\partial {{\rho }_{x}}}+{{\psi }^{t}}\frac{\partial }{\partial {{\rho }_{t}}}, \\ \end{align} \right.$

(6)

where ^x,ψ^x,^t,ψ^t,^xx,^xxx,and ^xxt are the coefficients of pr⁽³⁾V and pr⁽¹⁾V. Furthermore,we have

${{\phi }^{x}}={{D}_{x}}\phi -{{u}_{x}}{{D}_{x}}\xi -{{u}_{t}}{{D}_{x}}\tau ,$

(7)

${{\phi }^{t}}={{D}_{t}}\phi -{{u}_{x}}{{D}_{t}}\xi -{{u}_{t}}{{D}_{t}}\tau ,$

(8)

${{\psi }^{x}}={{D}_{x}}\psi -{{u}_{x}}{{D}_{x}}\xi -{{u}_{t}}{{D}_{x}}\tau ,$

(9)

${{\psi }^{t}}={{D}_{t}}\psi -{{u}_{x}}{{D}_{t}}\xi -{{u}_{t}}{{D}_{t}}\tau ,$

(10)

${{\phi }^{xx}}=D_{x}^{2}\left( \phi -\xi {{u}_{x}}-\tau {{u}_{t}} \right)+\xi {{u}_{xxx}}+\tau {{u}_{xxt}},$

(11)

${{\phi }^{xxt}}={{D}_{t}}D_{x}^{2}\left( \phi -\xi {{u}_{x}}-\tau {{u}_{t}} \right)+\xi {{u}_{xxxt}}+\tau {{u}_{xxtt}},$

(12)

${{\phi }^{xxx}}=D_{x}^{3}\left( \phi -\xi {{u}_{x}}-\tau {{u}_{t}} \right)+\xi {{u}_{xxxx}}+\tau {{u}_{xxxt}},$

(13)

where D_x and D_t are the total derivatives with respect to x and t,respectively.

Substituting (7)-(13) into (6),in terms of the Lie symmetry analysis method,we obtain the following equations for the symmetry group of (1)

$\left\{ \begin{align} & {{\xi }_{x}}={{\xi }_{t}}={{\xi }_{u}}={{\xi }_{\rho }}=0, \\ & {{\tau }_{x}}={{\tau }_{u}}={{\tau }_{\rho }}=0,{{\tau }_{t,t}}=0, \\ & \phi =-u{{\tau }_{t}}, \\ & \psi =-\rho {{\tau }_{t}}. \\ \end{align} \right.$

(14)

Solving (14),we obtain

$\begin{align} & \xi ={{c}_{1}},\tau ={{c}_{2}}t+{{c}_{3}}, \\ & \phi =-{{c}_{2}}u,\psi =-{{c}_{2}}\rho , \\ \end{align}$

(15)

where c₁,c₂,and c₃ are arbitrary constants. Hence the Lie algebra of infinitesimal symmetries of (1) is spanned by the following vector fields

${{V}_{1}}=\frac{\partial }{\partial x},{{V}_{2}}=t\frac{\partial }{\partial t}-u\frac{\partial }{\partial u}-\rho \frac{\partial }{\partial \rho },{{V}_{3}}=\frac{\partial }{\partial t}.$

(16)

It is easy to verify that {V₁,V₂,V₃} is closed under the Lie bracket. In fact,we have

$\begin{align} & \left[ {{V}_{1}},{{V}_{1}} \right]=\left[ {{V}_{2}},{{V}_{2}} \right]=\left[ {{V}_{3}},{{V}_{3}} \right]=0, \\ & \left[ {{V}_{1}},{{V}_{2}} \right]=-\left[ {{V}_{2}},{{V}_{1}} \right]=\left[ {{V}_{1}},{{V}_{3}} \right]=-\left[ {{V}_{3}},{{V}_{1}} \right]=0, \\ & \left[ {{V}_{2}},{{V}_{3}} \right]=-\left[ {{V}_{3}},{{V}_{2}} \right]=-{{V}_{3}}. \\ \end{align}$

(17)

2 Symmetry groups of the two-component Camassa-Holm equation

In section 1,we have obtained the infinitesimal symmetries of (1). Furthermore,for (1),the one-parameter groups G_i generated by V₁,V₂,and V₃ are given as follows:

$\begin{align} & {{G}_{1}}:\left( x,t,u,\rho \right)\to \left( x+\in ,t,u,\rho \right), \\ & {{G}_{2}}:\left( x,t,u,\rho \right)\to \left( x,t{{e}^{\in }},u{{e}^{-\in }},\rho {{e}^{-\in }} \right), \\ & {{G}_{3}}:\left( x,t,u,\rho \right)\to \left( x+\in ,t,u,\rho \right), \\ \end{align}$

where ∈ is any real number. We note that G₁ is a space translation; G₃ is a time translation; and G₂ is a genuinely local group of transformation. They are very important in our study of the exact solutions of PDEs.

Consequently,if u=f(x,t) and ρ=g(x,t) is a solution of (1),then u_(i) and ρ_(i)(i=1,2,3) given as follows are solutions of (1) as well

${{u}_{\left( 1 \right)}}=f\left( x-\in ,t \right),{{\rho }_{\left( 1 \right)}}g\left( x-\in ,t \right),$

(18)

${{u}_{\left( 2 \right)}}={{e}^{-\in }}f\left( x,t{{e}^{-\in }} \right),{{\rho }_{\left( 1 \right)}}{{e}^{-\in }}g\left( x,t{{e}^{-\in }} \right),$

(19)

${{u}_{\left( 3 \right)}}=f\left( x-\in ,t \right),{{\rho }_{\left( 3 \right)}}g\left( x-\in ,t \right).$

(20)

where ∈ is any real number.

3 Symmetry reductions and exact solutions of the two-component Camassa-Holm equation

Now we deal with the exact solutions for (1) based on the symmetry analysis. To do this,linear combinations of infinitesimals are considered and their corresponding invariants are determined.

(i) For the generator ${{V}_{1}}=\frac{\partial }{\partial x}$,we have the similarity variables

$\xi =t,\omega =u,v=\rho ,$

and the group-invariant solution is ω=f(ξ),ν=g(ξ),that is,

$u=f\left( t \right),\rho =g\left( t \right).$

(21)

Substituting Eq. (21) into Eq.(1),we obtain the reduction equation

$\left\{ \begin{align} & f'=0, \\ & g'=0, \\ \end{align} \right.$

(22)

where $f'=\frac{df}{d\xi },g'=\frac{dg}{d\xi }$. Therefore,(1) has a solution u=c₁,ρ=c₂,where c₁,c₂ are arbitrary constants. Obviously,the solution is not meaningful.

(ii) For the generator ${{V}_{2}}=t\frac{\partial }{\partial t}-u\frac{\partial }{\partial u}-\rho \frac{\partial }{\partial \rho }$,we have the similarity variables

$\xi =x,\omega =ut,v=\rho t,$

and the group-invariant solution is ω=f(ξ),ν=g(ξ),that is,

$u=f\left( x \right){{t}^{-1}},\rho =g\left( x \right){{t}^{-1}}.$

(23)

Substituting (23) into (1),we obtain the reduction equation

$\left\{ \begin{align} & -f+3ff'+f''-2f'f''-f{{f}^{\left( 3 \right)}}+gg'=0, \\ & -g+gf'+g'f=0, \\ \end{align} \right.$

(24)

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

(iii) For the generator ${{V}_{1}}+{{V}_{2}}=\frac{\partial }{\partial x}+t\frac{\partial }{\partial t}-u\frac{\partial }{\partial u}-\rho \frac{\partial }{\partial \rho }$,we have the similarity variables

and the group-invariant solution is $\omega =f\left( \xi \right),\text{ }\nu =g\left( \xi \right)$, that is,

$u={{e}^{-x}}f\left( t{{e}^{-x}} \right),\rho ={{e}^{-x}}g\left( t{{e}^{-x}} \right).$

(25)

Substituting (25) into (1),we obtain the following reduction equation

$\left\{ \begin{align} & -3f'+12\xi ff'+6{{\xi }^{2}}f{{'}^{2}}-5\xi f''+8{{\xi }^{2}}ff''+2{{\xi }^{3}}f'f''-{{\xi }^{2}}{{f}^{\left( 3 \right)}}+{{\xi }^{3}}f{{f}^{\left( 3 \right)}}-{{g}^{2}}-\xi gg'=0, \\ & -2gf+g'-\xi gf'-\xi g'f=0, \\ \end{align} \right.$

(26)

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

(iv) For the infinitesimal generator $c{{V}_{1}}+{{V}_{3}}=c\frac{\partial }{\partial x}+\frac{\partial }{\partial t}$,we have the similarity variables

and the group-invariant solution is $\omega =f\left( \xi \right),\text{ }\nu =g\left( \xi \right)$,that is,

$u=f\left( x-ct \right),\rho =g\left( x-ct \right).$

(27)

Substituting (27) into (1),we obtain the reduction equation

$\left\{ \begin{align} & \left( 3f-c \right)f'-2f'f''+\left( c-f \right){{f}^{\left( 3 \right)}}+gg'=0, \\ & \left( f-c \right)g'+gf'=0, \\ \end{align} \right.$

(28)

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

Remark 3.1 Noting that the reduced equations such as (24) and (26) are all higher-order nonlinear or nonautonomous ODEs,we will deal with such equations in the next section.

4 The power series solutions

In section 3,we have obtained the reduced equations by using Lie symmetry reductions. In this section,we will treat the nonlinear ODEs (24),(26),and (28). The power series can be used to solve differential equations,including many complicated differential equations with nonconstant coefficients^[29]. Now we consider the power series solutions to the reduced equations.

4.1 The power series solutions to Eq.(24)

Now,we seek a solution of (24) in the form of a power series

$f\left( \xi \right)=\sum\limits_{n=0}^{\infty }{=}{{p}_{n}}{{\xi }^{n}},\text{ }g\left( \xi \right)=\sum\limits_{n=0}^{\infty }{{{q}_{n}}}{{\xi }^{n}},$

(29)

where the coefficients p_n and q_n are all constants to be determined.

Substituting (29) into (24),we have

$\begin{align} & -\sum\limits_{n=0}^{\infty }{{{p}_{n}}}{{\xi }^{n}}+3\sum\limits_{n=0}^{\infty }{\sum\limits_{k=0}^{n}{\left( n+1-k \right)}}{{p}_{k}}{{p}_{n+1-k}}{{\xi }^{n}}+\sum\limits_{n=0}^{\infty }{\left( n+1 \right)\left( n+2 \right)}{{p}_{n+2}}{{\xi }^{n}} \\ & -2\sum\limits_{n=0}^{\infty }{\sum\limits_{k=0}^{n}{\left( k+1 \right)\left( n+1-k \right)\left( n+2-k \right)}}{{p}_{k+1}}{{p}_{n+2-k}}{{\xi }^{n}}- \\ & \sum\limits_{n=0}^{\infty }{\sum\limits_{k=0}^{n}{\left( n+1-k \right)}}\left( n+2-k \right)\left( n+3-k \right){{p}_{k}}{{p}_{n+3-k}}{{\xi }^{n}}+ \\ & \sum\limits_{n=0}^{\infty }{\sum\limits_{k=0}^{n}{\left( n+1-k \right)}}{{q}_{k}}{{q}_{n+1-k}}{{\xi }^{n}}=0,-\sum\limits_{n=0}^{\infty }{{{q}_{n}}{{\xi }^{n}}}+\sum\limits_{n}^{\infty }{=0} \\ & \sum\limits_{k=0}^{n}{\left( n+1-k \right)}{{q}_{k}}{{p}_{n+1-k}}{{\xi }^{n}}+\sum\limits_{n=0}^{\infty }{\sum\limits_{k=0}^{n}{\left( n+1-k \right)}}{{p}_{k}}{{q}_{n+1-k}}{{\xi }^{n}}=0. \\ \end{align}$

(30)

From (30),comparing coefficients,we obtain the recursion formula

$\begin{align} & {{p}_{n+3}}=\frac{1}{\left( n+1 \right)\left( n+2 \right)\left( n+3 \right){{p}_{0}}}\{-{{p}_{n}}+\left( n+1 \right) \\ & [3{{p}_{0}}{{p}_{n+1}}+\left( n+2 \right)(1-2{{p}_{1}}){{p}_{n+2}}+{{q}_{0}}{{q}_{n+1}}]+\sum\limits_{k=1}^{n}{\left( n+1-k \right)} \\ & [3{{p}_{k}}{{p}_{n+1-k}}+{{q}_{k}}{{q}_{n+1-k}}-\left( n+2-k \right)(2\left( k+1 \right){{p}_{k+1}}{{p}_{n+2-k}} \\ & +\left( n+3-k \right){{p}_{k}}{{p}_{n+3-k}})]\},{{q}_{n+1}}=\frac{1}{\left( n+1 \right){{p}_{0}}} \\ & [{{q}_{n}}-\left( n+1 \right){{q}_{0}}{{p}_{n+1}}-\sum\limits_{k=1}^{n}{\left( n+1-k \right)}({{q}_{k}}{{p}_{n+1-k}}+{{p}_{k}}{{q}_{n+1-k}})], \\ \end{align}$

(31)

for all n=0,1,2,…. Thus,for arbitrarily chosen constants p₀≠0,p₁,p₂,and q₀,from (31),we obtain

$\begin{align} & {{p}_{3}}=\frac{{{p}_{0}}(-1+3{{p}_{1}})+2{{p}_{2}}(1-2{{p}_{1}})+{{q}_{0}}{{q}_{1}}}{6{{p}_{0}}}, \\ & {{q}_{1}}=\frac{{{q}_{0}}(1-{{p}_{1}})}{{{p}_{0}}}. \\ \end{align}$

(32)

then,we have

$\begin{align} & {{p}_{4}}=\frac{{{p}_{1}}(3{{p}_{1}}-1)+2{{p}_{2}}(3{{p}_{0}}-4{{p}_{2}})}{24{{p}_{0}}}+\frac{6{{p}_{3}}(1-3{{p}_{1}})+2{{q}_{0}}{{q}_{2}}+{{q}^{2}}_{1}24{{p}_{0}}}{24{{p}_{0}}}, \\ & {{q}_{2}}=\frac{{{q}_{1}}-2{{q}_{0}}{{p}_{2}}-2{{p}_{1}}{{q}_{1}}}{2{{p}_{0}}}, \\ \end{align}$

(33)

and so on.

Thus,the other terms of the sequence {p_n}^∞_n=0 and {q_n}^∞_n=0 can be determined successively from (31) in a unique manner. This implies that,for (24),there exists a power series solution (29) with the coefficients given in (31). Furthermore,it is easy to prove the convergence of the power series (29) with the coefficients given in (31). As an example,we consider the convergence of the power series solution (29) of (24). From (31),we have

$\begin{align} & |{{p}_{n+3}}|\le M[|{{p}_{n}}\left| + \right|{{p}_{n+1}}\left| + \right|{{p}_{n+2}}\left| + \right|{{q}_{n+1}}|+ \\ & \sum\limits_{k+1}^{n}{(|{{p}_{k}}||{{p}_{n+1-k}}\left| + \right|{{q}_{k}}||{{q}_{n+1-k}}|+}|{{p}_{k+1}}||{{p}_{n+2-k}}\left| + \right|{{p}_{k}}||{{p}_{n+3-k}}|)], \\ & n=0,1,2,\ldots , \\ \end{align}$

where $M=max\left\{ \frac{1}{\left| {{p}_{0}} \right|} \right\},1,\frac{|1-2{{p}_{1}}|}{\left| {{p}_{0}} \right|},\left. \frac{\left| {{q}_{0}} \right|}{\left| {{p}_{0}} \right|} \right\}$. Similarly,from (31),we have

$|{{q}_{n+1}}\left| \le N[ \right|{{q}_{n}}\left| + \right|\left. {{p}_{n+1}} \right|\sum\limits_{k=1}^{n}{\left( \left| {{q}_{k}} \right| \right)}|{{p}_{n+1-k}}|+|{{p}_{k}}||{{q}_{n+1-k}}|)],n=0,1,2,\ldots ,$

where $N=max\left\{ \frac{1}{\left| {{p}_{0}} \right|},\frac{\left| {{q}_{0}} \right|}{\left| {{p}_{0}} \right|} \right\}$.

Now,we define two power series $R=R\left( \xi \right)=\sum\limits_{n=0}^{\infty }{{{r}_{n}}{{\xi }^{n}}}$ and $S=S\left( \xi \right)=\sum\limits_{n=0}^{\infty }{{{s}_{n}}{{\xi }^{n}}}$,by taking

${{r}_{i}}=|{{p}_{i}}|,\text{ }{{s}_{j}}=|{{q}_{j}}|,\text{ }i=0,1,2,\text{ }j=0,$

and

$\begin{align} & {{r}_{n+3}}=M[|{{r}_{n}}\left| + \right|{{r}_{n+1}}\left| + \right|{{r}_{n+2}}\left| + \right|{{s}_{n+1}}|+ \\ & \sum\limits_{k=1}^{n}{(|{{r}_{k}}||{{r}_{n+1-k}}\left| + \right|}\left. {{s}_{k}} \right|{{s}_{n+1-k}}|+{{r}_{k+1}}|{{r}_{n+2-k}}\left| + \right|\left. {{r}_{k}} \right||{{r}_{n+3-k}}|)], \\ & {{s}_{n+1}}=N[|{{s}_{n}}\left| + \right|{{r}_{n+1}}+\sum\limits_{k=1}^{n}{\left( \left| {{s}_{k}} \right| \right)}|{{r}_{n+1-k}}|+|{{r}_{k}}||{{s}_{n+1-k}}|)], \\ \end{align}$

where n=0,1,2,…. Then,it is easily seen that

$|{{p}_{n}}|\le {{r}_{n}},\text{ }|{{q}_{n}}|\le {{s}_{n}},\text{ }n=0,1,2,\cdots .$

In other words,the two series $ are majorant series of (29),respectively. Next,we show that the series R=R(ξ) and S=S(ξ) have positive radius of convergence. Indeed,by formal calculation,we have

$\begin{align} & R\left( \xi \right)={{r}_{0}}+{{r}_{1}}\xi +{{r}_{2}}{{\xi }^{2}}+\sum\limits_{n=0}^{\infty }{{{r}_{n+3}}}{{\xi }^{n+3}}={{r}_{0}}+{{r}_{1}}\xi +{{r}_{2}}{{\xi }^{2}}+ \\ & M[\sum\limits_{n=0}^{\infty }{{{r}_{n}}}{{\xi }^{n+3}}+\sum\limits_{n=0}^{\infty }{{{r}_{n+1}}}{{\xi }^{n+3}}+\sum\limits_{n=0}^{\infty }{{{r}_{n+2}}}{{\xi }^{n+3}}+\sum\limits_{n=0}^{\infty }{{{s}_{n+1}}} \\ & {{\xi }^{n+3}}+\sum\limits_{n=0}^{\infty }{\sum\limits_{k=1}^{n}{{{r}_{k}}}}{{r}_{n+1-k}}{{\xi }^{n+3}}+\sum\limits_{n=0}^{\infty }{\sum\limits_{k=1}^{n}{{{s}_{k}}{{s}_{n+1-k}}{{\xi }^{n+3}}}} \\ & +\sum\limits_{n=0}^{\infty }{\sum\limits_{k=1}^{n}{{{r}_{k+1}}{{r}_{n+2-k}}{{\xi }^{n+3}}}}+\sum\limits_{n=0}^{\infty }{\sum\limits_{k=1}^{n}{{{r}_{k}}}}{{r}_{n+3-k}}{{\xi }^{n+3}}]={{r}_{0}}+{{r}_{1}}\xi +{{r}_{2}}{{\xi }^{2}}+ \\ & M[{{\xi }^{3}}R+({{\xi }^{2}}-{{r}_{1}}\xi -{{r}_{0}}{{\xi }^{2}}+{{\xi }^{2}}R-{{\xi }^{2}}{{r}_{2}})(R-{{r}_{0}})+ \\ & 2{{R}^{2}}-3{{r}_{0}}R+{{r}^{2}}_{0}+(\xi -{{r}_{1}}\xi -{{r}_{0}})(R-{{r}_{0}}- \\ & {{r}_{1}}\xi )+({{\xi }^{2}}-{{s}_{0}}{{\xi }^{2}}+{{\xi }^{2}}S)(S-{{s}_{0}})-\xi {{r}_{1}}R], \\ \end{align}$

and

S(ξ)=s₀+N[ξS+(1+2S-2s₀)(R-r₀)].

Consider now the implicit functional system with respect to the independent variable ξ,

$S\left( \xi \right)={{s}_{0}}+N[\xi S+(1+2S-2{{s}_{0}})(R-{{r}_{0}})].$

Since F,G are analytic in the neighborhood of (0,r₀,s₀),F(0,r₀,s₀)=0,G(0,r₀,s₀)=0.

Furthermore,the Jacobian determinant

$\frac{\partial \left( F,G \right)}{\partial \left( R,S \right)}\left| _{(0,{{r}_{0}},{{s}_{0}})} \right.=1\ne 0,$

if we choose the parameters r₀=|p₀| and s₀=|q₀|[JP4] properly. By the implicit function theorem^[30],we see that R=R(ξ) and S=S(ξ) are analytic in a neighborhood of the point (0,r₀,s₀) and with the positive radius. This implies that the two power series (29) converge in a neighborhood of the point (0,r₀,s₀).

The power series solution of (1) can be written as

$\begin{align} & u\left( x,t \right)={{p}_{0}}{{t}^{-1}}+{{p}_{1}}x{{t}^{-1}}+{{p}_{2}}{{x}^{2}}{{t}^{-1}}+{{p}_{3}}{{x}^{3}}{{t}^{-1}}+ \\ & \sum\limits_{n=1}^{\infty }{{{p}_{n+3}}}{{x}^{n+3}}{{t}^{-1}}={{p}_{0}}{{t}^{-1}}+{{p}_{1}}x{{t}^{-1}}+{{p}_{2}}{{x}^{2}}{{t}^{-1}}+ \\ & \frac{{{p}_{0}}(-1+3{{p}_{1}})+2{{p}_{2}}(1-2{{p}_{1}})+{{q}_{0}}{{q}_{1}}}{6{{p}_{0}}}{{x}^{3}}{{t}^{-1}}+ \\ & \sum\limits_{n=1}^{\infty }{\frac{1}{\left( n+1 \right)\left( n+2 \right)\left( n+3 \right){{p}_{0}}}}\{-{{p}_{n}}+\left( n+1 \right) \\ & \left[ 3{{p}_{0}}{{p}_{n+1}}+\left( n+2 \right)(1-2{{p}_{1}}){{p}_{n+2}}+{{q}_{0}}{{q}_{n+1}} \right]+\sum\limits_{k=1}^{n}{\left( n+1-k \right)} \\ & \left[ 3{{p}_{k}}{{p}_{n+1-k}}+{{q}_{k}}{{q}_{n+1-k}}-\left( n+2-k \right)(2\left( k+1 \right){{p}_{k+1}}{{p}_{n+2-k}}+\left( n+3-k \right){{p}_{k}}{{p}_{n+3-k}}) \right] \\ & {{x}^{n+3}}{{t}^{-1}},\rho \left( x,t \right)={{q}_{0}}{{t}^{-1}}+{{q}_{1}}x{{t}^{-1}}+\sum\limits_{n=1}^{\infty }{{{q}_{n+1}}{{x}^{n+1}}{{t}^{-1}}}= \\ & {{q}_{0}}{{t}^{-1}}+{{q}_{0}}(1-{{p}_{1}}){{p}_{0}}x{{t}^{-1}}+\sum\limits_{n=1}^{\infty }{\frac{1}{\left( n+1 \right){{p}_{0}}}} \\ & [{{q}_{n}}-\left( n+1 \right){{q}_{0}}{{p}_{n+1}}-\sum\limits_{k=1}^{n}{n\left( n+1-k \right)}({{q}_{k}}{{p}_{n+1-k}}+{{p}_{k}}{{q}_{n+1-k}})]{{x}^{n+1}}{{t}^{-1}}, \\ \end{align}$

(34)

where p₀≠0,p₁,p₂,and q₀ are arbitrary constants. The other coefficients p_n(n≥3) and q_n(n≥1) can be determined successively from (31).

In physical applications,it will be convenient to write the solution of (1) in the approximate form

$\begin{align} & u\left( x,t \right)={{p}_{0}}{{t}^{-1}}+{{p}_{1}}x{{t}^{-1}}+{{p}_{2}}{{x}^{2}}{{t}^{-1}}+ \\ & \frac{{{p}_{0}}(-1+3{{p}_{1}})+2{{p}_{2}}(1-2{{p}_{1}})+{{q}_{0}}{{q}_{1}}}{6{{p}_{0}}}{{x}^{3}}{{t}^{-1}}+ \\ & \frac{{{p}_{1}}(3{{p}_{1}}-1)+2{{p}_{2}}(3{{p}_{0}}-4{{p}_{2}})}{24{{p}_{0}}}{{x}^{4}}{{t}^{-1}}+ \\ & \frac{6{{p}_{3}}(1-3{{p}_{1}})+2{{q}_{0}}{{q}_{2}}+{{q}^{2}}_{1}}{24{{p}_{0}}}{{x}^{4}}{{t}^{-1}}+\cdots , \\ & \rho \left( x,t \right)={{q}_{0}}{{t}^{-1}}+\frac{{{q}_{0}}(1-{{p}_{1}})}{{{p}_{0}}}x{{t}^{-1}}+\frac{{{q}_{1}}-2{{q}_{0}}{{p}_{2}}-2{{p}_{1}}{{q}_{1}}}{2{{p}_{0}}}{{x}^{2}}{{t}^{-1}}+\cdots , \\ \end{align}$

in terms of the above computation.

4.2 The power series solutions to (26)

Now,we seek a solution of (26) in the form of the power series (29). Substituting (29) into (26) and comparing the coefficients,we obtain

$\begin{align} & {{p}_{1}}=-\frac{{{q}^{2}}_{0}}{3},~{{q}_{1}}=-2{{q}_{0}}{{p}_{0}}, \\ & {{p}_{2}}=\frac{3(-{{q}_{0}}{{q}_{1}}+4{{p}_{0}}{{p}_{1}})}{16}, \\ & {{p}_{3}}=\frac{2(-2{{q}_{0}}{{q}_{2}}-{{q}^{2}}_{1}+20{{p}_{0}}{{p}_{2}}+9{{p}^{2}}_{1})}{45}. \\ \end{align}$

(35)

Generally,for n≥0,we have

$\begin{align} & {{p}_{n+4}}=\frac{1}{{{\left( n+4 \right)}^{2}}\left( n+6 \right)} \\ & \{\sum\limits_{k=0}^{n}{[-\left( n-k+4 \right)}{{q}_{k}}{{q}_{n-k+3}}+\left( n-k+3 \right) \\ & (12+\left( n-k+2 \right)\left( n-k+9 \right)){{p}_{k}}{{p}_{n-k+3}}+ \\ & 2\left( k+1 \right)(n-k+2)\left( n-k+4 \right){{p}_{k+1}}{{p}_{n-k+2}}]- \\ & 3{{q}_{n+1}}{{q}_{2}}-2{{q}_{n+2}}{{q}_{1}}-{{q}_{n+3}}{{q}_{0}}+40{{p}_{n+1}}{{p}_{2}}+6\left( n+4 \right){{p}_{n+2}}{{p}_{1}}\}, \\ & {{q}_{n+2}}=\frac{1}{\left( n+2 \right)}\{\sum\limits_{k=0}^{n}{[\left( n-k+3 \right)}{{q}_{k}}{{p}_{n-k+1}}+\left( n-k+1 \right){{p}_{k}}{{q}_{n-k+1}}]+2{{q}_{n+1}}{{p}_{0}}\}. \\ \end{align}$

(36)

Thus,for arbitrarily chosen constants p₀ and q₀,from (36) we have

$\begin{align} & {{p}_{4}}=\frac{-5{{q}_{0}}{{q}_{3}}+90{{p}_{0}}{{p}_{3}}+80{{p}_{1}}{{p}_{2}}-5{{q}_{1}}{{q}_{2}}}{96}, \\ & {{q}_{2}}=\frac{3}{2}({{q}_{0}}{{p}_{1}}+{{q}_{1}}{{p}_{0}}), \\ \end{align}$

(37)

and so on.

Thus,for arbitrary chosen constants p₀ and q₀,the other terms of the sequence {p_n}^∞_n=0 and {q_n}^∞_n=0 can be determined successively from (35) and (36) in a unique manner. This implies that,for (26),there exists a power series solution (29) with the coefficients given by (35) and (36).

4.3 The power series solutions to (28)

Similarly,we seek a solution of (28) in the form of the power series (29). Substituting (29) into (28) and comparing the coefficients,we obtain recursion formula

$\begin{align} & {{p}_{n+3}}=\frac{1}{c\left( n+1 \right)\left( n+2 \right)\left( n+3 \right)}\{c\left( n+1 \right){{p}_{n+1}}+ \\ & \sum\limits_{k=0}^{n}{\left( n+1-k \right)}[-3{{p}_{k}}{{p}_{n+1-k}}-{{q}_{k}}{{q}_{n+1-k}}+ \\ & 2\left( k+1 \right)\left( n+2-k \right){{p}_{k+1}}{{p}_{n+2-k}}+\left( n+3-k \right)\left( n+2-k \right){{p}_{k}}{{p}_{n+3-k}}]\}, \\ & {{q}_{n+1}}=\frac{1}{c\left( n+1 \right)}[\sum\limits_{k=0}^{n}{\left( n+1-k \right)}({{q}_{k}}{{p}_{n+1-k}}+{{p}_{k}}{{q}_{n+1-k}})], \\ \end{align}$

(38)

for all n=0,1,2,….

Thus,for arbitrarily chosen constants p₀≠0,p₁,p₂,and q₀,from (38),we obtain

$\begin{align} & {{p}_{3}}=\frac{c{{p}_{1}}-3{{p}_{0}}{{p}_{1}}+4{{p}_{1}}{{p}_{2}}+6{{p}_{0}}{{p}_{3}}-{{q}_{0}}{{q}_{1}}}{6c}, \\ & {{q}_{1}}=\frac{{{q}_{0}}{{p}_{1}}+{{q}_{1}}{{p}_{0}}}{c}, \\ \end{align}$

(39)

and so on.

Thus,for arbitrary chosen constants c≠0,p₀≠0,p₁,p₂,and q₀,the other terms of the sequence {p_n}^∞_n=0 and {q_n}^∞_n=0 can be determined successively from (38) in a unique manner. This implies that,for (28),there exists a power series solution (29) with the coefficients given by (38).

The approximate solution of (1) can be obtained in terms of the above computation. The details are omitted here.

Remark 4.1 The proofs for the convergence of the power series solutions to (26) and (28) are similar to the one for (24). The details are omitted here. We reiterate that the power series solutions which have been obtained in this paper are approximate solutions. Furthermore,such power series solutions converge quickly. So it is convenient for computations in both theory and application.

5 Conclusions

In this paper,the exact solutions of the two-component Camassa-Holm equation have been studied by using Lie symmetry analysis and the power series method. The similarity reductions and exact solutions are given for the first time. These similarity solutions possess significant features in physical systems. The results of this paper show that the Lie symmetry analysis is a very powerful method.