1. Introduction
Over the last two decades, more and more scholars have focused on the Camassa–Holm equation (CH) [1],
$\begin{eqnarray*}{m}_{t}+2{u}_{x}m+{{um}}_{x}=0,\ \ m=u-{u}_{{xx}}.\end{eqnarray*}$
This famous equation describes a unidirectional propagation of shallow water waves, which was derived by R. Camassa and D. Holm in 1993 by means of an asymptotic expansion of Euler’s equations. However, it first appeared from the investigation of a recursion operator to a bi-Hamiltonian equation with infinitely many conservation laws in the article of Fokas and Fuchssteiner in 1986 [2]. One of the remarkable properties of the CH equation is that it possesses peaked solutions, i.e. peakons. A peakon is a continuously weak form traveling wave solution, whose profile remains bounded. Meanwhile, it has a sharp angle at the central elevation, and the slope at the critical point changes rapidly. The other property of the CH equation is wave-breaking phenomena [3–5], which means the solution remains bounded, and the slope of the solution becomes unbounded in finite time.Another famous (1+1) dimension CH-type equation is the Fokas–Olver–Resenau–Qiao equation (FORQ), or modified CH equation (mCH) [6–8]
$\begin{eqnarray*}{m}_{t}+{\left({u}^{2}-{u}_{x}^{2})m\right)}_{x}=0,\ \ m=u-{u}_{{xx}},\end{eqnarray*}$
which was derived by the authors, respectively. It should be noted that the nonlinear term in the mCH equation is cubic; this is different to the CH equation. Similarly, the mCH equation also admits peakons and wave-breaking phenomena [9]. A natural thought is to generalize such an integrable equation into multi-component sense. Some two-component extensions of the mCH equation have been derived. Song, Qu and Qiao proposed a two-component modified CH system (SQQ) [10] $\begin{eqnarray*}\begin{array}{rcl}{m}_{t}+{[m(u+{u}_{x})(v-{v}_{x})]}_{x} & = & 0,\\ {n}_{t}+{[n(u+{u}_{x})(v-{v}_{x})]}_{x} & = & 0,\end{array}\end{eqnarray*}$
through a 2 × 2 spectral problem, and the x-part is $\begin{eqnarray*}{\phi }_{x}=\left(\begin{array}{cc}\tfrac{1}{2} & \lambda m\\ \lambda n & -\tfrac{1}{2}\end{array}\right)\phi .\end{eqnarray*}$
Li and Li considered other two-component cases by extending the spectral problem into a 4 × 4 matrix [11]. At the same time, they obtained three two-component generalizations of the mCH equation by taking reduction. The bi-Hamiltonian structures and reciprocal transformations of these systems have also been considered by the authors.The μ-version of the mCH equation was constructed by Qu, Fu and Liu [12]
$\begin{eqnarray*}{m}_{t}+{[(2\mu (u)u-{u}_{x}^{2})m]}_{x}+\gamma {u}_{x}=0,\ \ m=\mu (u)-{u}_{{xx}},\end{eqnarray*}$
where γ is the coefficient of the lower-order term. This equation can be regarded as a Euclidean-invariant version of the μCH equation.All the equations mentioned above are integrable, which means they all admit Lax representations and bi-Hamiltonian structures. To the best of our knowledge, there are less multi-component cases of the μmCH equation. Thus, the aim of this paper is to propose some two-component generalizations of the μmCH equation. The outline of this paper is as follows: in section 2 , we consider an extended spectral problem and three two-component reductions. It is proved that both reductions admit bi-Hamiltonian structures in section 3 . As an example, we derive two blow-up criteria for the 2μmCH system in section 4 , and the periodic peakons are derived in section 5 . Finally, some conclusions are given in the last section.
2. Extension of Lax pair
In this section, we derive some two-component generalizations of the μmCH equation. Recall the Lax pair1 ). Now we extend the spectral problem of the μmCH to a 4 × 4 matrix sense,
$\begin{eqnarray}{\left(\begin{array}{c}{\phi }_{1}\\ {\phi }_{2}\end{array}\right)}_{x}=P(m,\lambda )\left(\begin{array}{c}{\phi }_{1}\\ {\phi }_{2}\end{array}\right),\ \ {\left(\begin{array}{c}{\phi }_{1}\\ {\phi }_{2}\end{array}\right)}_{t}=Q(m,u,\lambda ,\gamma )\left(\begin{array}{c}{\phi }_{1}\\ {\phi }_{2}\end{array}\right),\end{eqnarray}$
where $\begin{eqnarray}\begin{array}{l}P(m,\lambda )\\ =\,\left(\begin{array}{cc}-\sqrt{-\tfrac{\gamma }{2}}\lambda & \lambda m\\ -\lambda m & \sqrt{-\tfrac{\gamma }{2}}\lambda \end{array}\right),\end{array}\end{eqnarray}$
$\begin{eqnarray}Q(m,u,\lambda ,\gamma )=\left(\begin{array}{cc}\sqrt{-\tfrac{\gamma }{2}}\left(\tfrac{1}{2}{\lambda }^{-1}+\lambda B(u\right) & -L(\lambda ,u)+\sqrt{-\tfrac{\gamma }{2}}{u}_{x}\\ L(\lambda ,u)+\sqrt{-\tfrac{\gamma }{2}}{u}_{x} & -\sqrt{-\tfrac{\gamma }{2}}\left(\tfrac{1}{2}{\lambda }^{-1}+\lambda B(u\right)\end{array}\right),\end{eqnarray}$
in which $B(u)=2\mu (u)u-{u}_{x}^{2}$ and $L(\lambda ,u)=\tfrac{1}{2}{\lambda }^{-1}\mu (u)\,+\lambda B(u)m$. It is easy to check the μmCH equation arises as a zero-curvature equation ${P}_{t}-{Q}_{x}+[P,Q]=0$, which is the compatibility condition of a linear system ( $\begin{eqnarray}P(M,\lambda )=\left(\begin{array}{cc}-\sqrt{-\tfrac{\gamma }{2}}\lambda I & \lambda M\\ -\lambda M & \sqrt{-\tfrac{\gamma }{2}}\lambda I\end{array}\right),\end{eqnarray}$
$\begin{eqnarray}\begin{array}{l}Q(M,U,\lambda ,\gamma )\\ =\,\left(\begin{array}{cc}\sqrt{-\tfrac{\gamma }{2}}\left(\tfrac{1}{2}{\lambda }^{-1}I+\lambda B(U\right) & -L(\lambda ,U)+\sqrt{-\tfrac{\gamma }{2}}{U}_{x}\\ L(\lambda ,U)+\sqrt{-\tfrac{\gamma }{2}}{U}_{x} & -\sqrt{-\tfrac{\gamma }{2}}\left(\tfrac{1}{2}{\lambda }^{-1}I+\lambda B(U\right)\end{array}\right),\end{array}\end{eqnarray}$
where φ1 and φ2 are two-dimension column vectors, M, U, μ(U) and I are 2 × 2 matrices and I is an identity matrix. Taking P and Q into the zero-curvature equation, we have $\begin{eqnarray}\begin{array}{rcl} & & {{MU}}_{x}+{U}_{x}M=B{(U)}_{x},\\ & & M\mu (U)=\mu (U)M,\ \ {MB}(U)=B(U)M,\\ & & {M}_{t}+{\left(B(U)M\right)}_{x}+\gamma {U}_{x}=0,\\ & & M=\mu (U)-{U}_{{xx}}={AU}.\end{array}\end{eqnarray}$
To obtain two-component generalizations of the μmCH equation, we must restrict the choice of matrix U. It is sufficient to choose $U=\left(\begin{array}{cc}a & b\\ c & d\end{array}\right)$ by taking
$\begin{eqnarray*}a=d,\ \ {{bc}}_{x}={b}_{x}c.\end{eqnarray*}$
Under this assumption, all the matrices M, U and μ(U) are commutable. Here are three possible choices for U, which can be written as $\begin{eqnarray}U=\left(\begin{array}{cc}u & v\\ v & u\end{array}\right),\ \ U=\left(\begin{array}{cc}u & v\\ -v & u\end{array}\right),\ \ U=\left(\begin{array}{cc}u & v\\ 0 & u\end{array}\right),\end{eqnarray}$
and the corresponding expressions of M and μ(U) are $\begin{eqnarray}M=\left(\begin{array}{cc}m & n\\ n & m\end{array}\right),\ \ M=\left(\begin{array}{cc}m & n\\ -n & m\end{array}\right),\ \ M=\left(\begin{array}{cc}m & n\\ 0 & m\end{array}\right),\end{eqnarray}$
and $\begin{eqnarray}\begin{array}{rcl}\mu (U) & = & \left(\begin{array}{cc}\mu (u) & \mu (v)\\ \mu (v) & \mu (u)\end{array}\right),\ \ \mu (U)=\left(\begin{array}{cc}\mu (u) & \mu (v)\\ -\mu (v) & \mu (u)\end{array}\right),\\ \mu (U) & = & \left(\begin{array}{cc}\mu (u) & \mu (v)\\ 0 & \mu (u)\end{array}\right).\end{array}\end{eqnarray}$
Under the assumptions of (7 )–(9 ), we finally obtain three two-component generalizations of the μmCH equation (for simplicity, we only consider γ = 0)
$\begin{eqnarray}\begin{array}{l}{m}_{t}+\left[\left(2\mu (u)u+2\mu (v)v-({u}_{x}^{2}+{v}_{x}^{2}\right)m\right.\\ \quad {\left.+2\left(\mu (u)v+\mu (v)u-{u}_{x}{v}_{x}\right)n\right]}_{x}=0,\\ {n}_{t}+\left[\left(2\mu (u)u+2\mu (v)v-({u}_{x}^{2}+{v}_{x}^{2}\right)n\right.\\ \quad {\left.+2\left(\mu (u)v+\mu (v)u-{u}_{x}{v}_{x}\right)m\right]}_{x}=0,\end{array}\end{eqnarray}$
$\begin{eqnarray}\begin{array}{l}{m}_{t}+\left[\left(2\mu (u)u-2\mu (v)v-({u}_{x}^{2}-{v}_{x}^{2}\right)m\right.\\ \quad {\left.-2\left(\mu (u)v+\mu (v)u-{u}_{x}{v}_{x}\right)n\right]}_{x}=0,\\ {n}_{t}+\left[\left(2\mu (u)u-2\mu (v)v-({u}_{x}^{2}-{v}_{x}^{2}\right)n\right.\\ \quad {\left.+2\left(\mu (u)v+\mu (v)u-{u}_{x}{v}_{x}\right)m\right]}_{x}=0,\end{array}\end{eqnarray}$
$\begin{eqnarray}\begin{array}{rcl} & & {m}_{t}+{\left[\left(2\mu (u)u-{u}_{x}^{2}\right)m\right]}_{x}=0,\\ & & {n}_{t}+\left[\left(2\mu (u)u-{u}_{x}^{2}\right)n\right.\\ & & {\left.+2\left(\mu (u)v+\mu (v)u-{u}_{x}{v}_{x}\right)m\right]}_{x}=0.\end{array}\end{eqnarray}$
3. Bi-Hamiltonian structures of reductions
In this section, we show that systems (10 ), (11 ) and (12 ) admit bi-Hamiltonian structures. This property is very important for integrable systems. It is already known that the μmCH equation admits a bi-Hamiltonian structure10 ) and (11 ), inspired by Li [11], we denote $\bar{u}=u+v,\bar{v}=u-v$ and $\bar{u}=u+{\rm{i}}v,\bar{v}=u-{\rm{i}}v$, then (10 ) and (11 ) can be rewritten as two decoupled μmCH equations10 ) can be reformulated as a bi-Hamiltonian system11 ) can also be reformulated as a bi-Hamiltonian system. The Hamiltonian functionals are19 ), and J is12 ) admits the following two conserved quantities12 ) into the bi-Hamiltonian form, the associated Hamiltonian operator K is the same as (10 ) and (11 ), and J is
$\begin{eqnarray*}{m}_{t}=K\displaystyle \frac{\delta {H}_{2}}{\delta m}=J\displaystyle \frac{\delta {H}_{1}}{\delta m},\end{eqnarray*}$
in which the Hamiltonian functionals are $\begin{eqnarray}\begin{array}{rcl}{H}_{1} & = & -{\int }_{{\mathbb{S}}}{mu}{\rm{d}}x,\\ {H}_{2} & = & -{\int }_{{\mathbb{S}}}\left({\mu }^{2}(u){u}^{2}+\mu (u){{uu}}_{x}^{2}-\displaystyle \frac{1}{12}{u}_{x}^{4}\right){\rm{d}}x,\end{array}\end{eqnarray}$
and the associated Hamiltonian operators are $\begin{eqnarray}J=\partial m{\partial }^{-1}m\partial ,\ \ K=\partial A=-{\partial }^{3},\end{eqnarray}$
where ∂ = ∂x. To derive the Hamiltonian structure of ( $\begin{eqnarray}\begin{array}{rcl} & & {\bar{m}}_{t}+{\left[(2\mu (\bar{u})\bar{u}-{\bar{u}}_{x}^{2})\bar{m}\right]}_{x}=0,\\ & & {\bar{n}}_{t}+{\left[(2\mu (\bar{v})\bar{v}-{\bar{v}}_{x}^{2})\bar{n}\right]}_{x}=0.\end{array}\end{eqnarray}$
Thus, the Hamiltonian functionals are $\begin{eqnarray}\begin{array}{rcl}\bar{{H}_{1}} & = & \displaystyle \frac{1}{2}{\displaystyle \int }_{{\mathbb{S}}}\left(\bar{m}\bar{u}+\bar{n}\bar{v}\right){\rm{d}}x,\\ \bar{{H}_{2}} & = & {\displaystyle \int }_{{\mathbb{S}}}\left(4{\mu }^{2}(\bar{u}){\bar{u}}^{2}+4\mu (\bar{u})\bar{u}{\bar{u}}_{x}^{2}-\displaystyle \frac{1}{3}{\bar{u}}_{x}^{4})\right.\\ & & \left.-(4{\mu }^{2}(\bar{v}){\bar{v}}^{2}+4\mu (\bar{v})\bar{v}{\bar{v}}_{x}^{2}-\displaystyle \frac{1}{3}{\bar{v}}_{x}^{4}\right){\rm{d}}x.\end{array}\end{eqnarray}$
With a direct calculation, we find ( $\begin{eqnarray}\left(\begin{array}{c}{m}_{t}\\ {n}_{t}\end{array}\right)=K\left(\begin{array}{c}\delta {H}_{2}/\delta m\\ \delta {H}_{2}/\delta n\end{array}\right)=J\left(\begin{array}{c}\delta {H}_{1}/\delta m\\ \delta {H}_{1}/\delta n\end{array}\right),\end{eqnarray}$
by using Hamiltonian functionals $\begin{eqnarray}\begin{array}{rcl}{H}_{1} & = & -{\displaystyle \int }_{{\mathbb{S}}}\left({mu}+{nv}\right){\rm{d}}x,\\ {H}_{2} & = & -{\displaystyle \int }_{{\mathbb{S}}}\left(2[\mu (u)u+\mu (v)v][\mu (u)v+\mu (v)u+{u}_{x}{v}_{x}]\right.\\ & & \left.+\,[\mu (u)v+\mu (v)u-\displaystyle \frac{1}{3}{u}_{x}{v}_{x}]({u}_{x}^{2}+{v}_{x}^{2})\right){\rm{d}}x,\end{array}\end{eqnarray}$
and associated Hamiltonian operators $\begin{eqnarray}\begin{array}{rcl}J & = & \left(\begin{array}{cc}\partial m{\partial }^{-1}m\partial +\partial n{\partial }^{-1}n\partial \ \ & \ \ \partial m{\partial }^{-1}n\partial +\partial n{\partial }^{-1}m\partial \\ \partial m{\partial }^{-1}n\partial +\partial n{\partial }^{-1}m\partial \ \ & \ \ \partial m{\partial }^{-1}m\partial +\partial n{\partial }^{-1}n\partial \end{array}\right),\\ K & = & \left(\begin{array}{cc}0 & \partial A\\ \partial A & 0\end{array}\right).\end{array}\end{eqnarray}$
Similarly, ( $\begin{eqnarray}\begin{array}{rcl}{H}_{1} & = & -{\displaystyle \int }_{{\mathbb{S}}}\left({mu}-{nv}\right){\rm{d}}x,\\ {H}_{2} & = & -{\displaystyle \int }_{{\mathbb{S}}}\left(2[\mu (u)u-\mu (v)v][\mu (u)v+\mu (v)u+{u}_{x}{v}_{x}]\right.\\ & & \left.+[\mu (u)v+\mu (v)u-\displaystyle \frac{1}{3}{u}_{x}{v}_{x}]({u}_{x}^{2}-{v}_{x}^{2})\right){\rm{d}}x.\end{array}\end{eqnarray}$
The Hamiltonian operator K is defined in ( $\begin{eqnarray}J=\left(\begin{array}{cc}\partial m{\partial }^{-1}m\partial -\partial n{\partial }^{-1}n\partial \ \ & \ \ \partial m{\partial }^{-1}n\partial +\partial n{\partial }^{-1}m\partial \\ \partial m{\partial }^{-1}n\partial +\partial n{\partial }^{-1}m\partial \ \ & \ \ -\partial m{\partial }^{-1}m\partial +\partial n{\partial }^{-1}n\partial \end{array}\right).\end{eqnarray}$
It is not difficult to check that ( $\begin{eqnarray}\begin{array}{rcl}{H}_{1} & = & -{\displaystyle \int }_{{\mathbb{S}}}\left({mv}+{nu}\right){\rm{d}}x,\\ {H}_{2} & = & -{\displaystyle \int }_{{\mathbb{S}}}\left(2\mu (u)u[\mu (u)v+\mu (v)u+{u}_{x}{v}_{x}]\right.\\ & & \left.+[\mu (u)v+\mu (v)u-\displaystyle \frac{1}{3}{u}_{x}{v}_{x}]{u}_{x}^{2}\right){\rm{d}}x.\end{array}\end{eqnarray}$
Thus, we can also rewrite ( $\begin{eqnarray}J=\left(\begin{array}{cc}0 & \partial m{\partial }^{-1}m\partial \\ \partial m{\partial }^{-1}m\partial \ \ & \ \ \partial m{\partial }^{-1}n\partial +\partial n{\partial }^{-1}m\partial \end{array}\right).\end{eqnarray}$
4. Blow-up criteria
In this section, we provide a criterion for the blow up of solutions to the Cauchy problem of system (10 ) (2μmCH)
$\begin{eqnarray}\begin{array}{l}{m}_{t}+\left[\left(2\mu (u)u+2\mu (v)v-({u}_{x}^{2}+{v}_{x}^{2}\right)m\right.\\ \quad {\left.+2\left(\mu (u)v+\mu (v)u-{u}_{x}{v}_{x}\right)n\right]}_{x}=0,\\ {n}_{t}+\left[\left(2\mu (u)u+2\mu (v)v-({u}_{x}^{2}+{v}_{x}^{2}\right)n\right.\\ \quad {\left.+2\left(\mu (u)v+\mu (v)u-{u}_{x}{v}_{x}\right)m\right]}_{x}=0,\\ u(0,x)={u}_{0}(x),\ \ u(t,x+1)=u(t,x),\\ \ \ m=\mu (u)-{u}_{{xx}},\\ v(0,x)={u}_{0}(x),\ \ v(t,x+1)=u(t,x),\\ \ \ n=\mu (v)-{v}_{{xx}}.\end{array}\end{eqnarray}$
Firstly, we offer the local well-posedness of (24 ) in Sobolev space. The process of the proof is similar to the paper [8]. Therefore, we only show the result here.
Let $({u}_{0},{v}_{0})\in {\left({H}^{s}({\mathbb{S}}\right)}^{2}$ with $s\gt \tfrac{5}{2}$. Then a time $T\gt 0$ exists such that the initial-value problem (
Now we introduce the following transport estimation lemma, which is crucial to the proof of the blow-up criterion for (24 ).
Consider the one-dimensional linear transport equation
$\begin{eqnarray}{\partial }_{t}f+v{\partial }_{x}f=g,\ \ f{| }_{t=0}={f}_{0}.\end{eqnarray}$
Let 0 ≤ σ < 1, and suppose that $\begin{eqnarray*}\begin{array}{rcl}{f}_{0}\in {H}^{\sigma },\ & & g\in {L}^{1}([0,T];{H}^{\sigma }),\\ {v}_{x}\in {L}^{1}([0,T];{L}^{\infty }),\ & & f\in {L}^{\infty }([0,T];{H}^{\sigma })\cap C([0,T];{ \mathcal S }^{\prime} ).\end{array}\end{eqnarray*}$
Then $f\in C([0,T];{H}^{\sigma })$. More precisely, there exists a constant $C$ depending only on $\sigma $ such that, for every $0\lt t\leqslant T$, $\begin{eqnarray*}\parallel f(t){\parallel }_{{H}^{\sigma }}\leqslant \parallel {f}_{0}{\parallel }_{{H}^{\sigma }}+C{\int }_{0}^{t}\left(\parallel g(\tau ){\parallel }_{{H}^{\sigma }}+\parallel f(\tau ){\parallel }_{{H}^{\sigma }}V^{\prime} (\tau \right){\rm{d}}\tau ,\end{eqnarray*}$
and hence $\begin{eqnarray*}\begin{array}{l}\parallel f(t){\parallel }_{{H}^{\sigma }}\leqslant {{\rm{e}}}^{{CV}(t)}\left(\parallel {f}_{0}{\parallel }_{{H}^{\sigma }}+C{\displaystyle \int }_{0}^{t}\parallel g(\tau ){\parallel }_{{H}^{\sigma }}{\rm{d}}\tau \right)\\ \quad {with}\ \ \ V(t)={\displaystyle \int }_{0}^{t}\parallel {\partial }_{x}v(\tau ){\parallel }_{{L}^{\infty }}{\rm{d}}\tau .\end{array}\end{eqnarray*}$
The blow-up criterion for (24 ) is as follows.
Let ${({u}_{0},{v}_{0})\in ({H}^{s})}^{2}$ be as in theorem
$\begin{eqnarray}{\int }_{0}^{{T}^{* }}{\left(\parallel m(\tau ){\parallel }_{{L}^{\infty }}+\parallel n(\tau ){\parallel }_{{L}^{\infty }}\right)}^{2}{\rm{d}}\tau =\infty .\end{eqnarray}$
In order to apply lemma
$\begin{eqnarray}\begin{array}{rcl} & & {\bar{m}}_{t}+{\left[(2\mu (\bar{u})\bar{u}-{\bar{u}}_{x}^{2})\bar{m}\right]}_{x}=0,\\ & & {\bar{n}}_{t}+{\left[(2\mu (\bar{v})\bar{v}-{\bar{v}}_{x}^{2})\bar{n}\right]}_{x}=0,\\ & & \bar{m}(0,x)={\bar{m}}_{0},\ \ \bar{n}(0,x)={\bar{n}}_{0},\ \ \end{array}\end{eqnarray}$
where $\bar{m}=m+n$ and $\bar{n}=m-n$. Now we break the proof into three steps. Step 1. For $s\in \left(\tfrac{1}{2},1\right)$, applying lemma
$\begin{eqnarray}\begin{array}{rcl}\parallel \bar{m}{\parallel }_{{H}^{s}} & \leqslant & \parallel {\bar{m}}_{0}{\parallel }_{{H}^{s}}+C{\displaystyle \int }_{0}^{t}\left(\parallel {\left(2\mu (\bar{u})\bar{u}-{\bar{u}}_{x}^{2}\right)}_{x}(\tau )\right.\\ & & \left.\times {\parallel }_{{L}^{\infty }}\parallel \bar{m}(\tau ){\parallel }_{{H}^{s}}+\parallel ({\bar{u}}_{x}{\bar{m}}^{2}(\tau ){\parallel }_{{H}^{s}}\right){\rm{d}}\tau ,\\ \parallel \bar{n}{\parallel }_{{H}^{s}} & \leqslant & \parallel {\bar{n}}_{0}{\parallel }_{{H}^{s}}+C{\displaystyle \int }_{0}^{t}\left(\parallel {\left(2\mu (\bar{v})\bar{v}-{\bar{v}}_{x}^{2}\right)}_{x}(\tau )\right.\\ & & \left.\times {\parallel }_{{L}^{\infty }}\parallel \bar{n}(\tau ){\parallel }_{{H}^{s}}+\parallel ({\bar{v}}_{x}{\bar{n}}^{2}(\tau ){\parallel }_{{H}^{s}}\right){\rm{d}}\tau ,\end{array}\end{eqnarray}$
for all $0\lt t\lt {T}^{* }$. Recall the 1-D Moser-type estimates [13], for s ≥ 0, the following estimates hold: $\begin{eqnarray*}\begin{array}{rcl}\parallel {fg}{\parallel }_{{H}^{s}} & \leqslant & C\left(\parallel f{\parallel }_{{H}^{s}}\parallel g{\parallel }_{{L}^{\infty }}+\parallel f{\parallel }_{{L}^{\infty }}\parallel g{\parallel }_{{H}^{s}}\right),\\ \parallel f{\partial }_{x}g{\parallel }_{{H}^{s}} & \leqslant & C\left(\parallel f{\parallel }_{{H}^{s+1}}\parallel g{\parallel }_{{L}^{\infty }}+\parallel f{\parallel }_{{L}^{\infty }}\parallel {\partial }_{x}g{\parallel }_{{H}^{s}}\right),\end{array}\end{eqnarray*}$
where the constant C is independent of f and g. Owing to the first Moser-type estimate, we have $\begin{eqnarray}\begin{array}{rcl}\parallel {\bar{u}}_{x}{\bar{m}}^{2}{\parallel }_{{H}^{s}} & \leqslant & C\left(\parallel {\bar{u}}_{x}{\parallel }_{{H}^{s}}\parallel {\bar{m}}^{2}{\parallel }_{{L}^{\infty }}+\parallel {\bar{u}}_{x}{\parallel }_{{L}^{\infty }}\parallel {\bar{m}}^{2}{\parallel }_{{H}^{s}}\right)\\ & \leqslant & C\left(\parallel {\bar{u}}_{x}{\parallel }_{{H}^{s}}\parallel \bar{m}{\parallel }_{{L}^{\infty }}^{2}+\parallel {\bar{u}}_{x}{\parallel }_{{L}^{\infty }}\parallel \bar{m}{\parallel }_{{L}^{\infty }}\parallel \bar{m}{\parallel }_{{H}^{s}}\right).\end{array}\end{eqnarray}$
As A is invertible, the inverse ${A}^{-1}m=u$ is given explicitly by $\begin{eqnarray*}\begin{array}{rcl}u(x) & = & \left[\displaystyle \frac{1}{2}{\left(x-\displaystyle \frac{1}{2}\right)}^{2}+\displaystyle \frac{23}{24}\right]\mu (m)\\ & & +\left(x-\displaystyle \frac{1}{2}\right){\displaystyle \int }_{{\mathbb{S}}}{\displaystyle \int }_{0}^{y}m(s){\rm{d}}s{\rm{d}}y\\ & & -{\displaystyle \int }_{0}^{x}{\displaystyle \int }_{0}^{y}m(s){\rm{d}}s{\rm{d}}y+{\displaystyle \int }_{{\mathbb{S}}}{\displaystyle \int }_{0}^{y}{\displaystyle \int }_{0}^{s}m(r){\rm{d}}r{\rm{d}}s{\rm{d}}y,\end{array}\end{eqnarray*}$
and the Green function of the operator ${A}^{-1},u={A}^{-1}m=G* m$ is given by $\begin{eqnarray*}G(x)=\displaystyle \frac{1}{2}{\left(x-[x]-\displaystyle \frac{1}{2}\right)}^{2}+\displaystyle \frac{23}{24}.\end{eqnarray*}$
With an application of Young’s inequality, we obtain $\begin{eqnarray}\parallel {\bar{u}}_{x}{\parallel }_{{L}^{\infty }}\leqslant \parallel {G}_{x}{\parallel }_{{L}^{1}}\parallel \bar{m}{\parallel }_{{L}^{\infty }}\leqslant \parallel \bar{m}{\parallel }_{{L}^{\infty }}.\end{eqnarray}$
Combining the fact $\parallel {\bar{u}}_{x}{\parallel }_{{H}^{s}}\leqslant C\parallel \bar{m}{\parallel }_{{H}^{s}}$ with ( $\begin{eqnarray}\parallel {\bar{u}}_{x}{\bar{m}}^{2}{\parallel }_{{H}^{s}}\leqslant C\parallel \bar{m}{\parallel }_{{H}^{s}}\parallel \bar{m}{\parallel }_{{L}^{\infty }}^{2},\end{eqnarray}$
and $\begin{eqnarray}\begin{array}{l}\parallel {\left(2\mu (\bar{u})\bar{u}-{\bar{u}}_{x}^{2}\right)}_{x}{\parallel }_{{L}^{\infty }}\\ \quad =\,2\parallel \bar{m}{\bar{u}}_{x}{\parallel }_{{L}^{\infty }}\leqslant C\parallel \bar{m}{\parallel }_{{L}^{\infty }}\parallel {\bar{u}}_{x}{\parallel }_{{L}^{\infty }}\leqslant C\parallel \bar{m}{\parallel }_{{L}^{\infty }}^{2}.\end{array}\end{eqnarray}$
Plugging ( $\begin{eqnarray}\parallel \bar{m}(t){\parallel }_{{H}^{s}}\leqslant \parallel {\bar{m}}_{0}{\parallel }_{{H}^{s}}+C{\int }_{0}^{t}\parallel \bar{m}(\tau ){\parallel }_{{L}^{\infty }}^{2}\parallel \bar{m}(\tau ){\parallel }_{{H}^{s}}{\rm{d}}\tau ,\end{eqnarray}$
and taking advantage of Gronwall’s inequality yields $\begin{eqnarray}\parallel \bar{m}(t){\parallel }_{{H}^{s}}\leqslant \parallel {\bar{m}}_{0}{\parallel }_{{H}^{s}}{{\rm{e}}}^{C{\int }_{0}^{t}\parallel \bar{m}(\tau ){\parallel }_{{L}^{\infty }}^{2}{\rm{d}}\tau }.\end{eqnarray}$
With a similar computation, we have $\begin{eqnarray}\parallel \bar{n}(t){\parallel }_{{H}^{s}}\leqslant \parallel {\bar{n}}_{0}{\parallel }_{{H}^{s}}{{\rm{e}}}^{C{\int }_{0}^{t}\parallel \bar{n}(\tau ){\parallel }_{{L}^{\infty }}^{2}{\rm{d}}\tau }.\end{eqnarray}$
Squaring both sides of ( $\begin{eqnarray*}\begin{array}{l}2(\parallel m(t){\parallel }_{{H}^{s}}^{2}+\parallel n(t){\parallel }_{{H}^{s}}^{2})\\ \quad \leqslant \,2(\parallel {m}_{0}{\parallel }_{{H}^{s}}^{2}+\parallel {n}_{0}{\parallel }_{{H}^{s}}^{2}){{\rm{e}}}^{2C{\displaystyle \int }_{0}^{t}{\left(\parallel m(\tau ){\parallel }_{{L}^{\infty }}+\parallel n(\tau ){\parallel }_{{L}^{\infty }}\right)}^{2}{\rm{d}}\tau },\end{array}\end{eqnarray*}$
which means $\begin{eqnarray}\begin{array}{l}\parallel m(t){\parallel }_{{H}^{s}}+\parallel n(t){\parallel }_{{H}^{s}}\\ \leqslant \,\sqrt{2(\parallel {m}_{0}{\parallel }_{{H}^{s}}^{2}+\parallel {n}_{0}{\parallel }_{{H}^{s}}^{2})}{{\rm{e}}}^{C{\displaystyle \int }_{0}^{t}{\left(\parallel m(\tau ){\parallel }_{{L}^{\infty }}+\parallel n(\tau ){\parallel }_{{L}^{\infty }}\right)}^{2}{\rm{d}}\tau }.\end{array}\end{eqnarray}$
Thus, if the maximal existence time ${T}^{* }\lt \infty $ satisfies $\begin{eqnarray*}{\int }_{0}^{{T}^{* }}{\left(\parallel m(\tau ){\parallel }_{{L}^{\infty }}+\parallel n(\tau ){\parallel }_{{L}^{\infty }}\right)}^{2}{\rm{d}}\tau \lt \infty ,\end{eqnarray*}$
throw inequality ( $\begin{eqnarray}\mathop{\mathrm{lim}\,\sup }\limits_{t\to {T}^{* }}\left(\parallel m(t){\parallel }_{{H}^{s}}+\parallel n(t){\parallel }_{{H}^{s}}\right)\lt \infty ,\end{eqnarray}$
which contradicts the assumption that the maximal existence time T* is finite. Step 2. For s ∈ [1,2), differentiating (
$\begin{eqnarray}\begin{array}{rcl}{\bar{m}}_{{xt}}+(2\mu (\bar{u})\bar{u}-{\bar{u}}_{x}^{2}){\bar{m}}_{{xx}} & = & -2{\bar{u}}_{{xx}}{\bar{m}}^{2}-3{\bar{u}}_{x}{\left({\bar{m}}^{2}\right)}_{x},\\ {\bar{n}}_{{xt}}+(2\mu (\bar{v})\bar{v}-{\bar{v}}_{x}^{2}){\bar{n}}_{{xx}} & = & -2{\bar{v}}_{{xx}}{\bar{n}}^{2}-3{\bar{v}}_{x}{\left({\bar{n}}^{2}\right)}_{x}.\end{array}\end{eqnarray}$
Applying lemma $\begin{eqnarray}\begin{array}{rcl}\parallel {\bar{m}}_{x}{\parallel }_{{H}^{s-1}} & \leqslant & \parallel {\bar{m}}_{0x}{\parallel }_{{H}^{s-1}}\\ & & +C{\displaystyle \int }_{0}^{t}\left(\parallel {\left(2\mu (\bar{u})\bar{u}-{\bar{u}}_{x}^{2}\right)}_{x}{\parallel }_{{L}^{\infty }}\parallel {\bar{m}}_{x}{\parallel }_{{H}^{s-1}}\right.\\ & & \left.+\parallel {\bar{u}}_{x}{\left({\bar{m}}^{2}\right)}_{x}{\parallel }_{{H}^{s-1}}+\parallel {\bar{u}}_{{xx}}{\bar{m}}^{2}{\parallel }_{{H}^{s-1}}\right){\rm{d}}\tau .\end{array}\end{eqnarray}$
With a similar computation to ( $\begin{eqnarray*}\begin{array}{rcl}\parallel {\bar{u}}_{{xx}}{\parallel }_{{L}^{\infty }} & = & \parallel \mu (\bar{m})-\bar{m}{\parallel }_{{L}^{\infty }}\\ & \leqslant & C\left(\parallel \bar{m}{\parallel }_{{L}^{\infty }}+\parallel G* \bar{m}{\parallel }_{{L}^{\infty }}\right)\leqslant C\parallel \bar{m}{\parallel }_{{L}^{\infty }},\end{array}\end{eqnarray*}$
we have $\begin{eqnarray}\begin{array}{rcl}\parallel {\bar{u}}_{x}{\left({\bar{m}}^{2}\right)}_{x}{\parallel }_{{H}^{s-1}} & \leqslant & C\parallel \bar{m}{\parallel }_{{L}^{\infty }}^{2}\parallel \bar{m}{\parallel }_{{H}^{s}},\\ \parallel {\bar{u}}_{{xx}}{\bar{m}}^{2}{\parallel }_{{H}^{s-1}} & \leqslant & C\parallel \bar{m}{\parallel }_{{L}^{\infty }}^{2}\parallel \bar{m}{\parallel }_{{H}^{s-1}}.\end{array}\end{eqnarray}$
Plugging (Step 3. For $s\in [2,\infty )$, assume $2\leqslant k\in {\mathbb{N}}$ is a positive integer. We prove that by induction, if (
$\begin{eqnarray}\begin{array}{l}{\left({\partial }_{x}^{k}\bar{m}\right)}_{t}+(2\mu (\bar{u})\bar{u}-{\bar{u}}_{x}^{2}){\left({\partial }_{x}^{k}\bar{m}\right)}_{x}\\ \quad =\,-\sum _{j=0}^{k-1}{C}_{k}^{j}{\partial }_{x}^{k-j}(2\mu (\bar{u})\bar{u}-{\bar{u}}_{x}^{2}){\partial }_{x}^{j+1}\bar{m}-2{\partial }_{x}^{k}({\bar{u}}_{x}{\bar{m}}^{2}),\\ {\left({\partial }_{x}^{k}\bar{n}\right)}_{t}+(2\mu (\bar{v})\bar{v}-{\bar{v}}_{x}^{2}){\left({\partial }_{x}^{k}\bar{n}\right)}_{x}\\ \quad =\,-\sum _{j=0}^{k-1}{C}_{k}^{j}{\partial }_{x}^{k-j}(2\mu (\bar{v})\bar{v}-{\bar{v}}_{x}^{2}){\partial }_{x}^{j+1}\bar{n}-2{\partial }_{x}^{k}({\bar{v}}_{x}{\bar{n}}^{2}).\end{array}\end{eqnarray}$
Taking advantage of lemma $\begin{eqnarray}\begin{array}{rcl}\parallel {\partial }_{x}^{k}\bar{m}(t){\parallel }_{{H}^{s-k}} & \leq & \parallel {\partial }_{x}^{k}{\bar{m}}_{0}{\parallel }_{{H}^{s-k}}\\ & & +C\displaystyle \int \left(\parallel {\partial }_{x}^{k}\bar{m}(\tau ){\parallel }_{{H}^{s-k}}\parallel (\bar{m}{\bar{u}}_{x})(\tau ){\parallel }_{{L}^{\infty }}\right.\\ & & +\parallel (\sum _{j=0}^{k-1}{C}_{k}^{j}{\partial }_{x}^{k-j}(2\mu (\bar{u})\bar{u}-{\bar{u}}_{x}^{2}){\partial }_{x}^{j+1}\bar{m}\\ & & \left. {+2{\partial }_{x}^{k}({\bar{u}}_{x}{\bar{m}}^{2}))(\tau )\parallel }_{{H}^{s-k}}\right){\rm{d}}\tau .\end{array}\end{eqnarray}$
Using Moser-type estimates and the Sobolev embedding inequality, it is easy to check that $\begin{eqnarray}\begin{array}{l}\parallel {\partial }_{x}^{k}({\bar{u}}_{x}{\bar{m}}^{2}){\parallel }_{{H}^{s-k}}\leqslant C\parallel \bar{m}{\parallel }_{{L}^{\infty }}^{2}\parallel \bar{m}{\parallel }_{{H}^{s}},\\ {\parallel \sum _{j=0}^{k-1}{C}_{k}^{j}{\partial }_{x}^{k-j}(2\mu (\bar{u})\bar{u}-{\bar{u}}_{x}^{2}){\partial }_{x}^{j+1}\bar{m}\parallel }_{{H}^{s-k}}\\ \quad \leqslant \,C\parallel \bar{m}{\parallel }_{{H}^{s}}\parallel \bar{m}{\parallel }_{{H}^{k-\tfrac{1}{2}+\epsilon }}^{2},\end{array}\end{eqnarray}$
where ϵ is a sufficiently small positive number such that ${H}^{\tfrac{1}{2}+\epsilon }\hookrightarrow {L}^{\infty }$ holds. Taking ( $\begin{eqnarray}\parallel \bar{m}(t){\parallel }_{{H}^{s}}\leqslant \parallel {\bar{m}}_{0}{\parallel }_{{H}^{s}}+C{\int }_{0}^{t}\parallel \bar{m}(\tau ){\parallel }_{{H}^{k-\tfrac{1}{2}+\epsilon }}^{2}\parallel \bar{m}(\tau ){\parallel }_{{H}^{s}}{\rm{d}}\tau ,\end{eqnarray}$
and applying Gronwall’s inequality, we obtain $\begin{eqnarray}\parallel \bar{m}(t){\parallel }_{{H}^{s}}\leqslant \parallel {\bar{m}}_{0}{\parallel }_{{H}^{s}}{{\rm{e}}}^{C{\int }_{0}^{t}\parallel \bar{m}(\tau ){\parallel }_{{H}^{k-\tfrac{1}{2}+\epsilon }}^{2}{\rm{d}}\tau }.\end{eqnarray}$
Similarly, we have the estimation for $\bar{n}$ $\begin{eqnarray}\parallel \bar{n}(t){\parallel }_{{H}^{s}}\leqslant \parallel {\bar{n}}_{0}{\parallel }_{{H}^{s}}{{\rm{e}}}^{C{\int }_{0}^{t}\parallel \bar{n}(\tau ){\parallel }_{{H}^{k-\tfrac{1}{2}+\epsilon }}^{2}{\rm{d}}\tau }.\end{eqnarray}$
By the definition of the Sobolev norm, and squaring both sides of ( $\begin{eqnarray}\begin{array}{l}\parallel m(t){\parallel }_{{H}^{s}}+\parallel n(t){\parallel }_{{H}^{s}}\\ \leqslant \,\sqrt{2(\parallel {m}_{0}{\parallel }_{{H}^{s}}^{2}+\parallel {n}_{0}{\parallel }_{{H}^{s}}^{2})}{{\rm{e}}}^{C{\displaystyle \int }_{0}^{t}{\left(\parallel m(\tau ){\parallel }_{{H}^{k-\tfrac{1}{2}+\epsilon }}+\parallel n(\tau ){\parallel }_{{H}^{k-\tfrac{1}{2}+\epsilon }}\right)}^{2}{\rm{d}}\tau }.\end{array}\end{eqnarray}$
If the maximal existence time T* is finite, and $\begin{eqnarray*}{\int }_{0}^{{T}^{* }}{\left(\parallel m(\tau ){\parallel }_{{L}^{\infty }}+\parallel n(\tau ){\parallel }_{{L}^{\infty }}\right)}^{2}{\rm{d}}\tau \lt \infty ,\end{eqnarray*}$
we find that $\parallel m(\tau ){\parallel }_{{H}^{k-\tfrac{1}{2}+\epsilon }}+\parallel n(\tau ){\parallel }_{{H}^{k-\tfrac{1}{2}+\epsilon }}$ is uniformly bounded in $(0,{T}^{* })$ by the induction assumption. Throw ( In theorem
$\begin{eqnarray*}{\int }_{0}^{{T}^{* }}{\left(\parallel m(\tau ){\parallel }_{{L}^{\infty }}+\parallel n(\tau ){\parallel }_{{L}^{\infty }}\right)}^{2}{\rm{d}}\tau =\infty ,\end{eqnarray*}$
which means $\begin{eqnarray}\mathop{\mathrm{lim}\,\sup }\limits_{t\to {T}^{* }}\parallel m(t){\parallel }_{{L}^{\infty }}=\infty ,\ {\rm{or}}\ \mathop{\mathrm{lim}\,\sup }\limits_{t\to {T}^{* }}\parallel n(t){\parallel }_{{L}^{\infty }}=\infty .\end{eqnarray}$
On the other hand, if (
$\begin{eqnarray*}\mathop{\mathrm{lim}\,\sup }\limits_{t\to {T}^{* }}\parallel m(t){\parallel }_{{H}^{s}}=\infty ,\ \ {or}\ \ \ \mathop{\mathrm{lim}\,\sup }\limits_{t\to {T}^{* }}\parallel n(t){\parallel }_{{H}^{s}}=\infty .\end{eqnarray*}$
Thus we have the following corollary. Let ${({u}_{0},{v}_{0})\in ({H}^{s})}^{2}$ be as in theorem
The following theorem shows that the wave-breaking only depends on the infimum of ${{mu}}_{x},{{mv}}_{x},{{nu}}_{x}$ and nvx.
Let ${({u}_{0},{v}_{0})\in ({H}^{s})}^{2}$ be as in theorem
$\begin{eqnarray}\mathrm{lim}\,{\inf }_{t\to {T}^{* }}[\mathop{\inf }\limits_{x\in {\mathbb{S}}}\left({ab}(t,x\right)]=-\infty ,\end{eqnarray}$
where $a\in \{m,n\}$ and $b\in \{{u}_{x},{v}_{x}\}$. Arguing by density, it suffices to prove the case s = 3. Multiplying the first equation of (
$\begin{eqnarray}\begin{array}{rcl}\displaystyle \frac{{\rm{d}}}{{\rm{d}}t}{\displaystyle \int }_{{\mathbb{S}}}{m}^{2}{\rm{d}}x & = & -{\displaystyle \int }_{{\mathbb{S}}}2m\left[\left(2\mu (u)u+2\mu (v)v\right.\right.\\ & & -\ \left.({u}_{x}^{2}+{v}_{x}^{2}\right)m\\ & & {\left.+2\left(\mu (u)v+\mu (v)u-{u}_{x}{v}_{x}\right)n\right]}_{x}{\rm{d}}x,\\ \displaystyle \frac{{\rm{d}}}{{\rm{d}}t}{\displaystyle \int }_{{\mathbb{S}}}{n}^{2}{\rm{d}}x & = & -{\displaystyle \int }_{{\mathbb{S}}}2n\left[\left(2\mu (u)u+2\mu (v)v\right.\right.\\ & & -\ \left.({u}_{x}^{2}+{v}_{x}^{2}\right)n\\ & & {\left.+2\left(\mu (u)v+\mu (v)u-{u}_{x}{v}_{x}\right)m\right]}_{x}{\rm{d}}x.\end{array}\end{eqnarray}$
Plugging the two equations of ( $\begin{eqnarray}\begin{array}{l}\displaystyle \frac{{\rm{d}}}{{\rm{d}}t}{\displaystyle \int }_{{\mathbb{S}}}({m}^{2}+{n}^{2}){\rm{d}}{x}\\ =\,-{\displaystyle \int }_{{\mathbb{S}}}\left({m}^{2}+{n}^{2})(2{{mu}}_{x}+2{{nv}}_{x})-2{mn}(2{{mv}}_{x}+2{{nu}}_{x}\right){\rm{d}}x.\end{array}\end{eqnarray}$
Differentiating ( $\begin{eqnarray}\begin{array}{l}{m}_{{xt}}+\left[\left(2\mu (u)u+2\mu (v)v-({u}_{x}^{2}+{v}_{x}^{2}\right)m\right.\\ \quad {\left.+2\left(\mu (u)v+\mu (v)u-{u}_{x}{v}_{x}\right)n\right]}_{{xx}}=0,\\ {n}_{{xt}}+\left[\left(2\mu (u)u+2\mu (v)v-({u}_{x}^{2}+{v}_{x}^{2}\right)n\right.\\ \quad {\left.+2\left(\mu (u)v+\mu (v)u-{u}_{x}{v}_{x}\right)m\right]}_{{xx}}=0.\end{array}\end{eqnarray}$
Multiplying the first equation of ( $\begin{eqnarray}\begin{array}{rcl}\displaystyle \frac{{\rm{d}}}{{\rm{d}}t}{\displaystyle \int }_{{\mathbb{S}}}{m}_{x}^{2}{\rm{d}}x & = & -{\displaystyle \int }_{{\mathbb{S}}}2{m}_{x}\left[\left(2\mu (u)u+2\mu (v)v-({u}_{x}^{2}+{v}_{x}^{2}\right)m\right.\\ & & {\left.+2\left(\mu (u)v+\mu (v)u-{u}_{x}{v}_{x}\right)n\right]}_{{xx}}{\rm{d}}x,\\ \displaystyle \frac{{\rm{d}}}{{\rm{d}}t}{\displaystyle \int }_{{\mathbb{S}}}{n}_{x}^{2}{\rm{d}}x & = & -{\displaystyle \int }_{{\mathbb{S}}}2{n}_{x}\left[\left(2\mu (u)u+2\mu (v)v-({u}_{x}^{2}+{v}_{x}^{2}\right)n\right.\\ & & {\left.+2\left(\mu (u)v+\mu (v)u-{u}_{x}{v}_{x}\right)m\right]}_{{xx}}{\rm{d}}x.\end{array}\end{eqnarray}$
Plugging the two equations of ( $\begin{eqnarray}\begin{array}{l}\displaystyle \frac{{\rm{d}}}{{\rm{d}}t}{\displaystyle \int }_{{\mathbb{S}}}({m}_{x}^{2}+{n}_{x}^{2}){\rm{d}}{x}\\ =\,-{\displaystyle \int }_{{\mathbb{S}}}\left(8({m}_{x}^{2}+{n}_{x}^{2})({{mu}}_{x}+{{nv}}_{x})+16{m}_{x}{n}_{x}({{mv}}_{x}+{{nu}}_{x}\right){\rm{d}}x.\end{array}\end{eqnarray}$
By virtue of ( $\begin{eqnarray}\begin{array}{l}\displaystyle \frac{{\rm{d}}}{{\rm{d}}t}(\parallel m(t){\parallel }_{{H}^{1}}^{2}+\parallel n(t){\parallel }_{{H}^{1}}^{2})\\ \quad =\,-{\displaystyle \int }_{{\mathbb{S}}}\left({m}^{2}+{n}^{2})(2{{mu}}_{x}+2{{nv}}_{x})-2{mn}(2{{mv}}_{x}+2{{nu}}_{x})\right.\\ \qquad \left.+\,8({m}_{x}^{2}+{n}_{x}^{2})({{mu}}_{x}+{{nv}}_{x})+16{m}_{x}{n}_{x}({{mv}}_{x}+{{nu}}_{x}\right){\rm{d}}x.\end{array}\end{eqnarray}$
Thus, if a positive constant C > 0 exists such that ab≥ − C on $[0,{T}^{* })\times {\mathbb{S}}$, then by the above estimation, we have $\begin{eqnarray}\begin{array}{l}\displaystyle \frac{{\rm{d}}}{{\rm{d}}t}(\parallel m(t){\parallel }_{{H}^{1}}^{2}+\parallel n(t){\parallel }_{{H}^{1}}^{2})\\ \quad \leqslant \,32C{\int }_{{\mathbb{S}}}({m}^{2}+{m}_{x}^{2}+{n}^{2}+{n}_{x}^{2}){\rm{d}}x.\end{array}\end{eqnarray}$
Applying Gronwall’s inequality finally yields $\begin{eqnarray}\parallel m(t){\parallel }_{{H}^{1}}^{2}+\parallel n(t){\parallel }_{{H}^{1}}^{2}\leqslant {{\rm{e}}}^{16{Ct}}(\parallel {m}_{0}{\parallel }_{{H}^{1}}^{2}+\parallel {n}_{0}{\parallel }_{{H}^{1}}^{2})\end{eqnarray}$
for $t\in [0,{T}^{* })$, which contradicts the fact that ${T}^{* }\lt \infty $ is the maximal existence time. On the other hand, if ( $\begin{eqnarray*}\mathop{\mathrm{lim}\,\sup }\limits_{t\to {T}^{* }}\parallel m(t){\parallel }_{{L}^{\infty }}=\infty ,\ \ {or}\ \ \ \mathop{\mathrm{lim}\,\sup }\limits_{t\to {T}^{* }}\parallel n(t){\parallel }_{{L}^{\infty }}=\infty .\end{eqnarray*}$
Then, by the Sobolev embedding theorem, we claim that the solution of (5. Periodic peakons
A peakon is a continuously weak form traveling wave solution, whose profile remains bounded. A big feature of CH-type equations is that they possess peaked traveling wave solutions. As we have just mentioned that A is invertible and the Green function of operator ${A}^{-1},u={A}^{-1}m=G* m$ is given by10 )–(12 ) admit periodic peakons with the form10 ). Due to the fact that G(ξ) only admits a first order partial derivative, we have to give the definition of weak solutions for (10 ) first.
$\begin{eqnarray*}G(x)=\displaystyle \frac{1}{2}{\left(x-[x]-\displaystyle \frac{1}{2}\right)}^{2}+\displaystyle \frac{23}{24}.\end{eqnarray*}$
Thus, similarly to the CH equation, it is natural to assume both ( $\begin{eqnarray}{\phi }_{u}(x,t)=p(t)G(\xi ),\ \ {\phi }_{v}(x,t)=q(t)G(\xi ),\end{eqnarray}$
where $\xi =x-s(t),p(t),q(t)$ and s(t) are functions of t. As an example, we calculate the peakons for ( Solutions u and v are called weak for (
$\begin{eqnarray}\begin{array}{l}{\displaystyle \int }_{0}^{T}{\displaystyle \int }_{{\mathbb{S}}}\left({u}_{t}{\phi }_{{xx}}+\left(-\displaystyle \frac{1}{3}{u}_{x}^{3}-{u}_{x}{v}_{x}^{2}+2\mu (u){{uu}}_{x}\right.\right.\\ \quad \left.+\,2\mu (u){{vv}}_{x}+2\mu (v){{uv}}_{x}+2\mu (v){u}_{x}v,),\,\right){\phi }_{{xx}}\\ \quad +\,\left(2{\mu }^{2}(u)u+2{\mu }^{2}(v)u+4\mu (u)\mu (v)v\right.\\ \quad +\,\mu (u){u}_{x}^{2}+\mu (u){v}_{x}^{2}\\ \quad \left.\left.+\,2\mu (v){u}_{x}{v}_{x},),\,\right),),\,\right){\phi }_{x}{\rm{d}}x{\rm{d}}t=0,\\ {\displaystyle \int }_{0}^{T}{\displaystyle \int }_{{\mathbb{S}}}\left({v}_{t}{\phi }_{{xx}}+\left(-\displaystyle \frac{1}{3}{v}_{x}^{3}-{u}_{x}^{2}{v}_{x}\right.\right.\\ \quad \left.+\,2\mu (v){{uu}}_{x}+2\mu (v){{vv}}_{x}+2\mu (u){u}_{x}v+2\mu (u){{uv}}_{x},),\,\right){\phi }_{{xx}}\\ \quad +\,\left(2{\mu }^{2}(u)v+2{\mu }^{2}(v)v+4\mu (u)\mu (v)u\right.\\ \quad \left.\left.+\,\mu (v){u}_{x}^{2}+\mu (v){v}_{x}^{2}+2\mu (u){u}_{x}{v}_{x},),\,\right),),\,\right){\phi }_{x}{\rm{d}}x{\rm{d}}t=0.\end{array}\end{eqnarray}$
Denoting $\zeta =\xi -[\xi ]-\tfrac{1}{2}$, and taking (58 ) into the first equation of (59 ), we have61 )–(63 ) into (60 ), we obtain the following ordinary differential equations (ODEs)59 ), we have
$\begin{eqnarray}\begin{array}{l}{\displaystyle \int }_{{\mathbb{S}}}\left(\left[\dot{p}\left(\displaystyle \frac{1}{2}{\zeta }^{2}+\displaystyle \frac{23}{24}\right)-p\dot{s}\zeta \right]{\phi }_{{xx}}\right.\\ \quad +\,\left[\left(\displaystyle \frac{2}{3}{p}^{3}+2{{pq}}^{2}\right){\zeta }^{3}+\left(\displaystyle \frac{23}{12}{p}^{3}+\displaystyle \frac{23}{4}{{pq}}^{2}\right)\zeta \right]{\phi }_{{xx}}\\ \quad \left.+\,\left[(2{p}^{3}+6{{pq}}^{2}){\zeta }^{2}+\left(\displaystyle \frac{23}{12}{p}^{3}+\displaystyle \frac{23}{4}{{pq}}^{2}\right)\right]{\phi }_{x}\right){\rm{d}}x=0,\end{array}\end{eqnarray}$
where μ(G(x)) = 1. For convenience, we denote $\dot{f}=\tfrac{{\rm{d}}f}{{\rm{d}}t}$, and with a direct calculation, we have $\begin{eqnarray}\begin{array}{l}{\displaystyle \int }_{{\mathbb{S}}}[\dot{p}(\displaystyle \frac{1}{2}{\zeta }^{2}+\displaystyle \frac{23}{24})-p\dot{s}\zeta ]{\phi }_{{xx}}{\rm{d}}x\\ \quad =\,{\displaystyle \int }_{0}^{s}\left[\displaystyle \frac{\dot{p}}{2}{\left(x-s+\displaystyle \frac{1}{2}\right)}^{2}-p\dot{s}\left(x-s+\displaystyle \frac{1}{2}\right)\right]{\phi }_{{xx}}{\rm{d}}x\\ \qquad +\,{\displaystyle \int }_{s}^{1}\left[\displaystyle \frac{\dot{p}}{2}{\left(x-s-\displaystyle \frac{1}{2}\right)}^{2}-p\dot{s}\left(x-s-\displaystyle \frac{1}{2}\right)\right]{\phi }_{{xx}}{\rm{d}}x\\ \quad =\,{\left.\left(\left[\displaystyle \frac{\dot{p}}{2}{\left(x-s+\displaystyle \frac{1}{2}\right)}^{2}-p\dot{s}\left(x-s+\displaystyle \frac{1}{2}\right)\right]{\phi }_{x}\right)\right|}_{0}^{s}\\ \qquad +\,{\left.\left(\left[\displaystyle \frac{\dot{p}}{2}{\left(x-s-\displaystyle \frac{1}{2}\right)}^{2}-p\dot{s}\left(x-s-\displaystyle \frac{1}{2}\right)\right]{\phi }_{x}\right)\right|}_{s}^{1}\end{array}\end{eqnarray}$
$\begin{eqnarray}\begin{array}{l}\qquad -\,{\displaystyle \int }_{0}^{s}\left[\dot{p}\left(x-s+\displaystyle \frac{1}{2}\right)-p\dot{s}\right]{\phi }_{x}{\rm{d}}x\\ \qquad -\,{\displaystyle \int }_{s}^{1}\left[\dot{p}\left(x-s-\displaystyle \frac{1}{2}\right)-p\dot{s}\right]{\phi }_{x}{\rm{d}}x\\ \quad =\,-p\dot{s}{\phi }_{x}(s)-\dot{p}\phi (s),\end{array}\end{eqnarray}$
$\begin{eqnarray}\begin{array}{l}{\displaystyle \int }_{{\mathbb{S}}}\left(\displaystyle \frac{2}{3}{p}^{3}+2{{pq}}^{2}\right){\left({\zeta }^{3}{\phi }_{x}\right)}_{x}{\rm{d}}x\\ \quad =\,\left(\displaystyle \frac{2}{3}{p}^{3}+2{{pq}}^{2}\right)\left[{\left.{\left(x-s+\displaystyle \frac{1}{2}\right)}^{3}{\phi }_{x}\right|}_{0}^{s}\right.\\ \qquad +\ \left.{\left.{\left(x-s-\displaystyle \frac{1}{2}\right)}^{3}{\phi }_{x}\right|}_{s}^{1}\right]\\ \quad =\,\left(\displaystyle \frac{1}{6}{p}^{3}+\displaystyle \frac{1}{2}{{pq}}^{2}\right){\phi }_{x}(s),\end{array}\end{eqnarray}$
$\begin{eqnarray}\begin{array}{l}{\displaystyle \int }_{{\mathbb{S}}}\left(\displaystyle \frac{23}{12}{p}^{3}+\displaystyle \frac{23}{4}{{pq}}^{2}\right){\left(\zeta {\phi }_{x}\right)}_{x}{\rm{d}}x\\ \quad =\,\left(\displaystyle \frac{23}{12}{p}^{3}+\displaystyle \frac{23}{4}{{pq}}^{2}\right)\left[{\left.\left(x-s+\displaystyle \frac{1}{2}\right){\phi }_{x}\right|}_{0}^{s}\right.\\ \qquad +\ \left.{\left.\left(x-s-\displaystyle \frac{1}{2}\right){\phi }_{x}\right|}_{s}^{1}\right]\\ \quad =\,\left(\displaystyle \frac{23}{12}{p}^{3}+\displaystyle \frac{23}{4}{{pq}}^{2}\right){\phi }_{x}(s).\end{array}\end{eqnarray}$
Plugging ( $\begin{eqnarray}\left\{\begin{array}{l}\dot{p}=0,\\ \dot{s}=\tfrac{25}{12}{p}^{2}+\tfrac{25}{4}{q}^{2}.\end{array}\right.\end{eqnarray}$
Similarly, from the second equation of ( $\begin{eqnarray}\left\{\begin{array}{l}\dot{q}=0,\\ \dot{s}=\tfrac{25}{12}{q}^{2}+\tfrac{25}{4}{p}^{2}.\end{array}\right.\end{eqnarray}$
Thus $s(t)={ct},{p}^{2}\,=\,{q}^{2}=\tfrac{3c}{25},c\gt 0$, $\begin{eqnarray}{\phi }_{u}(\xi )=\pm {\phi }_{v}(\xi )=\pm \displaystyle \frac{\sqrt{3c}}{5}\left(\displaystyle \frac{1}{2}{\left(\xi -[\xi ]-\displaystyle \frac{1}{2}\right)}^{2}+\displaystyle \frac{23}{24}\right).\end{eqnarray}$
With a similar technique, one can verify that (11 ) and (12 ) also admit periodic peakons with form (58 ).
6. Conclusions
In this paper, we derived some two-component generalizations of the μmCH equation by extending the Lax pair through a zero-curvature equation, and we established bi-Hamiltonian structures of these systems. As an example, we presented two blow-up criteria and periodic peakons for system (10 ). The remaining problem is whether the peaked solutions are orbital stable, and how to prove the stability if such an assumption holds by taking advantage of the conserved quantities. In addition, the symmetry analysis method is one of the effective methods used to investigate the properties of partial differential equations [14–17]. Based on the conservation laws, one can construct the nonlocally related systems to search the nonlocal symmetries. One of our authors investigated the conservation laws and nonlocally related systems of the Hunter–Saxton equation [18], which is a short-wave limit of the CH equation. Many new local and nonlocal symmetries were obtained, and the inverse potential system was also constructed. It has been proved that the symmetry analysis method is an effective method for studying the CH-type equation. In the forthcoming days, we will study the symmetry properties of the μmCH equation presented in this paper.