Well-posedness and Blow-up Phenomena for the Cross-coupled Camassa–Holm System

Abstract

In this paper, we investigate the cross-coupled Camassa–Holm system. First, we prove local well-posedness for the cross-coupled Camassa–Holm equations in $H^s ({\mathbb {R}})$ with $s > 3/2$. Then, a blow-up criterion is established.

1 Introduction

In this paper, we consider the following system of two cross-coupled Camassa–Holm (CCCH, for simplicity) equations, which was proposed by Cotter, Holm, Ivanov and Percival in [4]:

$$\begin{aligned} \left\{ \begin{array}{ll} m_t+(vm)_x+mv_x&{}=0,\\ n_t+(un)_x+nu_x&{}=0, \end{array}\right. \end{aligned}$$

(1.1)

where $m=u-u_{xx}+w$ and $n=v-v_{xx}$. System (1.1) admits peakon solutions, but it is completely different from the two-component Camassa–Holm system, which has been studied in [6, 7] and references therein.

When $v = u$, this coupled system restricts to the standard Camassa–Holm equation, which was first obtained as a bi-Hamiltonian generalization of KdV equation by Fokas and Fuchssteiner [5] and later derived as a model for unidirectional propagation of shallow water over a flat bottom by Camassa and Holm [2]. A lot of works have devoted to the study of the Camassa–Holm equation, because it has both solitary waves like solitons and solutions which blow up in finite time as wave breaking. The well-posedness of the Camassa–Holm equation has been shown in [15, 21] with the initial data $u_0\in H^s({\mathbb {R}}), s>3/2$. Meanwhile, Constantin and Strauss [1] (see also [23]), by using the conservation laws, proved that the Camassa–Holm equation has the stability peaked solitons (peakons) of the form $\varphi _c (t, x) = ce^{-|x-ct|}$ with the traveling speed $c > 0$.

Let $\Lambda =(1-\partial _x^2)^\frac{1}{2},$ then the operator $\Lambda ^{-2}$ can be expressed by its associated Green’s function $G=\frac{1}{2}e^{-|x|}$ as

$$\begin{aligned} \Lambda ^{-2}f(x)=G*f(x)=\frac{1}{2}\int _ {\mathbb {R}}e^{-|x-y|}f(y)\mathrm{d}y. \end{aligned}$$

So the system (1.1) takes the form of a quasilinear evolution equation of hyperbolic type:

$$\begin{aligned} \left\{ \begin{array}{ll} u_t+vu_x+G*(2uv_x+u_xv_{xx})=0,\\ v_t+uv_x+G*(2vu_x+v_xu_{xx})=0. \end{array} \right. \end{aligned}$$

(1.2)

Then, it is easy to verify that this system has the following conserved quantities:

$$\begin{aligned}&\displaystyle E_1=\int _{{\mathbb {R}}}mdx,\quad E_2=\int _{{\mathbb {R}}}ndx,\\&\displaystyle E_3=\int _{{\mathbb {R}}}uv+u_xv_xdx. \end{aligned}$$

Note that when $u=v$, $E_3$ is the useful $H^1$-norm.

Henry, Holm and Ivanov [8] examined whether the solutions of the system (1.1), which initially have compact support, would possess the same persistence properties or not. For the Camassa–Holm equation, a nice description of corresponding solution with compactly supported initial data was given in [9], while the behavior of corresponding solution with algebraic-decaying initial data is investigated in [19]

Very recently, local well-posedness for the system (1.1) with initial data in $H^s({\mathbb {R}})\times H^s({\mathbb {R}}), s > 5/2$ , has been proved in [22] and [16] by Kato’s semigroup theory. One of the purposes of our present paper is to improve the local well-posedness result to $s>3/2$.

The paper is organized as follows. In Sect. 2, we prove local well-posedness for the system (1.1). A blow-up criterion is established in Sect. 3, and the approach is different from that in [22].

2 Local Well-posedness

Let us first state Kato’s theorem in the form suitable for our purpose and present some preliminary results.

Consider the abstract quasilinear evolution equation of the form

$$\begin{aligned} \frac{\mathrm{d}z}{\mathrm{d}t}+A(z)z=f(z),\quad t>0,\quad z(0)=z_0. \end{aligned}$$

(2.1)

Let X and Y be Hilbert spaces such that Y is continuously and densely embedded in X, and let $Q:Y\rightarrow X$ be a topological isomorphism. Let L(Y, X) denote the space of all bounded linear operator from Y to X (L(X),  if $X=Y)$. Assume that:

• (i) $A(y)\in L(Y,X)$ for $Y\in X$ with

  $$\begin{aligned} \Vert (A(y)-A(z))w\Vert _{X}\le \mu _1\Vert y-z\Vert _X\Vert w\Vert _Y,\quad y,~z,~w\in Y. \end{aligned}$$

  and $A(y)\in G(X,1,\beta )$, (i.e., A(y) is quasi-m-accretive), uniformly on bounded sets in Y.

• (ii) $QA(y)Q^{-1}=A(y)+B(y)$, where $B(y)\in L(X)$ is bounded, uniformly on bounded sets in Y. Moreover,

  $$\begin{aligned} \Vert (B(y)-B(Z))w\Vert _X\le \mu _2\Vert y-z\Vert _Y\Vert w\Vert _X,\quad y,z\in Y,~w\in X. \end{aligned}$$

• (iii) $f:Y\rightarrow Y$ and extends also to a map from X into X. f is bounded on bounded sets in Y, and

  $$\begin{aligned} \Vert f(y)-f(z)\Vert _Y\le \mu _3\Vert y-z\Vert _Y,\quad y,z\in Y,\\ \Vert f(y)-f(z)\Vert _X\le \mu _4\Vert y-z\Vert _X,\quad y,z\in Y. \end{aligned}$$

  Here $\mu _1,~\mu _2,~\mu _3,$ and $\mu _4$ depend only on max $\{\Vert y\Vert _Y,\Vert z\Vert _Y\}$.

Theorem 2.1

[12] Assume that (i), (ii), and (iii) hold. Given $v_0\in Y$, there is a maximal $T>0$ depending only on $\Vert v_0\Vert _Y$ and unique solution v to (2.1) such that

$$\begin{aligned} v=v(\cdot ,v_0)\in C([0,T);Y)\cap C^1([0,T);X). \end{aligned}$$

Moreover, the map $v_0\rightarrow v(\cdot ,v_0)$ is continuous from Y to

$$\begin{aligned} C([0,T);Y)\cap C^1([0,T);X). \end{aligned}$$

By using Theorem 2.1, we will prove the following local well-posedness theorem.

Theorem 2.2

Given $(u_0,v_0)\in H^s\times H^s$, $s>\frac{3}{2}$, then there exists a T and a unique solution (u, v) to (1.2) such that

$$\begin{aligned} (u(x,t),v(x,t))\in C([0,T);H^s\times H^s\cap C^1([0,T);H^{s-1}\times H^{s-1}). \end{aligned}$$

Moreover, the solution depends continuously on the initial data, i.e., the mapping

$$\begin{aligned} z_0 \rightarrow z(\cdot ,z_0):H^s\times H^s\rightarrow C([0,T);H^s\times H^s\cap C^1([0,T);H^{s-1}\times H^{s-1}). \end{aligned}$$

is continuous.

The remainder of this section is devoted to the proof of Theorem 2.2 by applying Kato’s theorem which is similar to that in [3, 18, 21].

Let $z:=\left( \begin{array}{c} u \\ v\\ \end{array} \right)$, $A(z)=\left( \begin{array}{cc} v\partial _x&{} 0\\ 0 &{} u\partial _x\\ \end{array} \right)$, and

$$\begin{aligned} f(z)=\left( \begin{array}{c} -(1-\partial _x)^{-1}(2uv_x+u_xv_{xx})\\ -(1-\partial _x)^{-1}(2vu_x+v_xu_{xx})\\ \end{array} \right) . \end{aligned}$$

Lemma 2.3

[12] Let r, t be real numbers such that $-r<t\le r$. Then,

$$\begin{aligned}&\displaystyle \Vert fg\Vert _{H^t}\le c\Vert f\Vert _{H^r}\Vert g\Vert _{H^t},\quad \mathrm{if} \,\, r>\frac{1}{2},\\&\displaystyle \Vert fg\Vert _{H^{r+t-\frac{1}{2}}}\le c\Vert f\Vert _{H^r}\Vert g\Vert _{H^t},\quad \mathrm{if}\, \, r<\frac{1}{2}, \end{aligned}$$

where c is a positive constant depending on r, t.

Lemma 2.4

[13] Let $f\in H^s$, $s>\frac{3}{2}$. Then,

$$\begin{aligned} \Vert \Lambda ^{-r}[\Lambda ^{r+t+1},M_f]\Lambda ^{-t}\Vert _{L(L^2)}\le \Vert f\Vert _{H^s},\quad |r|,~|t|\le s-1, \end{aligned}$$

where $M_f$ is the operator of multiplication by f and c is a constant depending only on r, t.

Lemma 2.5

[20] Let X and Y be two Banach spaces and Y be continuously and densely embedded in X. Let $-A$ be the infinitesimal generator of the $C_0-$semigroup T(t) on X and S be an isomorphism from Y onto X. Then, Y is A-admissible if and only if $-A_1=-SAS^{-1}$ is the infinitesimal generator of the $C_0-$semigroup $T_1(t)=ST(t)S^{-1}$ on X. Moreover, if Y is $A-$admissible, then the part of $-A$ in Y is the infinitesimal generator of the restriction of T(t) to Y.

Lemma 2.6

The operator $A(z)=\left( \begin{array}{cc} v\partial _x&{} 0\\ 0 &{} u\partial _x\\ \end{array} \right)$ with $z\in H^s\times H^s$, $s\ge \frac{3}{2}$, belongs to $G(L^2\times L^2,1,\beta )$.

Proof

since $L^2\times L^2$ is a Hilbert space, A(z) belongs to $G(L^2\times L^2,1,\beta )$ if and only if there is a real number $\beta$ such that

$(a)(A(z)y,y)_{0\times 0}\ge -\beta \Vert y\Vert _{L^2\times L^2}^2$,

(b) the range of $A+\lambda I$ is all of $L^2\times L^2$, for some (or all) $\lambda >\beta$.

First, let us prove (a). Since $A(z)=\left( \begin{array}{cc} u\\ v\\ \end{array} \right) \in H^s\times H^s,$ $s>\frac{3}{2}$, it follows from the Sobolev embedding theorem that there exists a positive constant c such that

$$\begin{aligned} \Vert u_x\Vert _{L^\infty }\le \Vert u\Vert _{H^s},\quad \Vert v_x\Vert _{L^\infty }\le \Vert v\Vert _{H^s}. \end{aligned}$$

Thus, we have

$$\begin{aligned}&(A(z)y,y)_{0\times 0}\\&\quad =(v\partial _x y_1, y_1)_0+(u\partial _x y_2, y_2)_0\\&\quad =-\frac{1}{2}(v\partial _x y_1, y_1)_0-\frac{1}{2}(u\partial _x y_2, y_2)_0\\&\quad \ge -\frac{1}{2}(\Vert v_x\Vert _{L^\infty }+\Vert u_x\Vert _{L^\infty })(\Vert y_1\Vert _{L^2}^2+\Vert y_2\Vert _{L^2}^2)\\&\quad \ge -\frac{1}{2}(\Vert v\Vert _{H^s}+\Vert u\Vert _{H^s})\Vert y\Vert _{L^2\times L^2}. \end{aligned}$$

Setting $\beta =\frac{1}{2}(\Vert u\Vert _{H^s}+\Vert v\Vert _{H^s})$, we have $(A(z)y,y)_{0\times 0}\ge -\beta \Vert y\Vert _{L^2\times L^2}^2.$

Then, we prove (b). Since A(z) satisfies (a), obviously, A(z) is continuous, then, A(z) is a closed operator. So $(\lambda I+A)$ has closed range in $L^2\times L^2$ forall $\lambda >\beta$. Given $v\in H^s$, $s>\frac{3}{2}$, and $y_1\in L^2$, we have the generalized Leibniz formula,

$$\begin{aligned} \partial _x(vy_1)=v_xy_1+v\partial _xy_1\quad in \, H^{-1}. \end{aligned}$$

Due to $v_x\in L^\infty$, $v_xy_1\in L^2$, so $\partial _x(vy_1)\in L^2$ if and only if $v\partial _x y_1\in L^2$. Integrating by parts, we get

$$\begin{aligned} D(A)&=D\left( \left( \begin{array}{cc} v\partial _x&{} 0\\ 0 &{} u\partial _x\\ \end{array} \right) \right) \\&=\left\{ \left( \begin{array}{cc} y_1\\ y_2\\ \end{array} \right) \in L^2\times L^2, \left( \begin{array}{cc} v\partial _x y_1\\ u\partial _x y_2\\ \end{array} \right) \in L^2\times L^2\right\} \\&=\left\{ \left( \begin{array}{cc} w_1\\ w_2\\ \end{array} \right) \in L^2\times L^2, \left( \begin{array}{cc} -(vw_1)_x\\ -(uw_2)_x\\ \end{array} \right) \in L^2\times L^2\right\} \\&=D\left( \left( \begin{array}{cc} v\partial _x&{} 0\\ 0 &{} u\partial _x\\ \end{array} \right) ^*\right) =D(A^*). \end{aligned}$$

Assume that the range of $A\,+\,\lambda$ is not all of $L^2\times L^2$, then, there exists $w\in L^2\times L^2,$ $w\ne 0,$ such that $((\lambda I+A)y, w)_{0\times 0}=0$, $\forall y\in D(A)$. Since $H^1\times H^1\subset D(A)$, we have that D(A) is dense in $L^2\times L^2$. Therefore, it follows that $w\in D(A^*)$ and $\lambda w+A^*=0$ in $L^2\times L^2$. Note that $D(A)=D(A^*)$. Multiplying by w and integrating by parts, we have

$$\begin{aligned} 0= & {} ((\lambda I+A^*)w, w)_{0\times 0}\\= & {} (\lambda w, w)_{0\times 0}+(w,Aw)_{0\times 0} \ge (\lambda -\beta )\Vert w\Vert _{L^2\times L^2}^2,\quad \forall \lambda >\beta . \end{aligned}$$

Thus, we obtain $w=0$. This contradicts the previous assumption $w\ne 0$. This proves (b) and completes the proof of the lemma. $\square$

We now extend the generation property of A(z) to the full scale $H^{s-1}\times H^{s-1}$ with $s>\frac{3}{2}$.

Lemma 2.7

The operator $A(z)=\left( \begin{array}{cc} v\partial _x&{} 0\\ 0 &{} u\partial _x\\ \end{array} \right)$ with $z\in H^s\times H^s$, $s\ge 2$, belongs to $G(H^{s-1}\times H^{s-1},1,\beta )$.

Proof

Since $H^{s-1}\times H^{s-1}$ is a Hilbert space, A(z) belongs to $G(H^{s-1}\times H^{s-1},1,\beta )$ if and only if there is a real number $\beta$ such that

$(a)(A(z)y,y)_{(s-1)\times (s-1)}\ge -\beta \Vert y\Vert _{H^{s-1}\times H^{s-1}}^2$;

$(b)-A(z)$ is the infinitesimal generator of a $C_0-$ semigroup on $H^{s-1}\times H^{s-1}$ for some $\lambda >\beta$.

First, let us prove (a).

$$\begin{aligned}&(A(z)y,y)_{(s-1)\times (s-1)}\\&\quad =(\Lambda ^{s-1}(v\partial _x y_1),\Lambda ^{s-1} y_1)_0+(\Lambda ^{s-1}(u\partial _x y_2),\Lambda ^{s-1} y_2)_0\\&\quad =([\Lambda ^{s-1},v]\partial _x y_1,\Lambda ^{s-1} y_1)_0-\frac{1}{2}(v_x\Lambda ^{s-1} y_1),\Lambda ^{s-1} y_1)_0\\&\qquad +\,([\Lambda ^{s-1},u]\partial _x y_1,\Lambda ^{s-1} y_2)_0-\frac{1}{2}(u_x\Lambda ^{s-1} y_2),\Lambda ^{s-1} y_2)_0\\&\quad \ge -\Vert [\Lambda ^{s-1},v]\Lambda ^{2-s}\Vert _{L(L^2)}\Vert \Lambda ^{s-1} y_1\Vert _{L^2}^2-\Vert v_x\Vert _{L^\infty }\Vert \Lambda ^{s-1} y_1\Vert _{L^2}^2\\&\qquad -\,\Vert [\Lambda ^{s-1},u]\Lambda ^{2-s}\Vert _{L(L^2)}\Vert \Lambda ^{s-1} y_2\Vert _{L^2}^2-\Vert u_x\Vert _{L^\infty }\Vert \Lambda ^{s-1} y_2\Vert _{L^2}^2\\&\quad \ge c(\Vert u\Vert _{H^s}+\Vert v\Vert _{H^s})\Vert y\Vert _{H^{s-1}\times H^{s-1}}, \end{aligned}$$

where we use the Lemma 2.4 with $r=0$, $t=s-2$. Setting $\beta =c(\Vert u\Vert _{H^s}+\Vert v\Vert _{H^s})$, we have

$$\begin{aligned} (A(z)y,y)_{(s-1)\times (s-1)}\ge -\beta \Vert y\Vert _{H^{s-1}\times H^{s-1}}^2. \end{aligned}$$

Then, we prove (b). let $S=\left( \begin{array}{cc} \Lambda ^{s-1}&{} 0\\ 0 &{} \Lambda ^{s-1}\\ \end{array} \right)$. Note that S is an isomorphism of $H^{s-1}\times H^{s-1}$ onto $L^2\times L^2$, and $H^{s-1}\times H^{s-1}$ is continuously and densely embedded in $L^2\times L^2$ as $s\ge 2$. Define

$$\begin{aligned} A_1(z):=SA(z)S^{-1},\quad B_1(z)=A_1(z)-A(z). \end{aligned}$$

Note that

$$\begin{aligned} B_1(z)y=[\Lambda ^{s-1},v\partial _x]\Lambda ^{1-s}y_1+[\Lambda ^{s-1},u\partial _x]\Lambda ^{1-s}y_2. \end{aligned}$$

Let $y\in L^2\times L^2$ and $u\in H^s$, $s\ge 2$. Then, we have

$$\begin{aligned}&\Vert B_1(z)y\Vert _{L^2\times L^2}\\&\quad \le \Vert [\Lambda ^{s-1},v\partial _x]\Lambda ^{1-s}y_1\Vert _{L^2}+\Vert [\Lambda ^{s-1},u\partial _x]\Lambda ^{1-s}y_2\Vert _{L^2}\\&\quad \le \Vert [\Lambda ^{s-1},v]\Lambda ^{2-s}\Vert _{L(L^2)}\Vert \Lambda ^{-1}\partial _xy_1\Vert _{L^2} +\Vert [\Lambda ^{s-1},u]\Lambda ^{2-s}\Vert _{L(L^2)}\Vert \Lambda ^{-1}\partial _xy_2\Vert _{L^2}\\&\quad \le c(\Vert u\Vert _{H^s}+\Vert v\Vert _{H^s})\Vert y\Vert _{L^2\times L^2}, \end{aligned}$$

where we applied Lemma 2.4 with $r=0$, $t=s-2$. Thus, we obtain $B_1(z)\in L(L^2\times L^2)$. Note that $A_1(z)=A_(z)+B_1(z)$ and $A(z)\in G(L^2\times L^2,1,\beta )$, see Lemma 2.6. By a perturbation theorem for semigroups (cf.[20]), we obtain $A_1(z)\in G(L^2\times L^2,1,\beta {^\prime })$. By applying Lemma 2.5 with $Y=H^{s-1}\times H^{s-1}$, $X=L^2\times L^2$, and $S=\left( \begin{array}{cc} \Lambda ^{s-1}&{} 0\\ 0 &{} \Lambda ^{s-1}\\ \end{array} \right)$, we conclude that $H^{s-1}\times H^{s-1}$ is A-admissible. Therefore, $-A(z)$ is the infinitesimal generator of a $C_0-$semigroup on $H^{s-1}\times H^{s-1}$. This proves (b) and completes the proof of the lemma. $\square$

Lemma 2.8

Let $S=\left( \begin{array}{cc} v\partial _x&{} 0\\ 0 &{} u\partial _x\\ \end{array} \right)$ with $z\in H^s\times H^s$, $s\ge \frac{3}{2}$, be given. Then, $A(z)\in L(H^s\times H^s,H^{s-1}\times H^{s-1})$ and

$$\begin{aligned} \left\| (A(z)-A(y))w\right\| _{H^{s-1}\times H^{s-1}}\le \mu _1\Vert z-y\Vert _{H^s\times H^s}\Vert w\Vert _{H^s\times H^s}, \end{aligned}$$

for all $z,y,w\in H^s\times H^s$.

Proof

Let $z,y,w\in H^s\times H^s$, $s\ge \frac{3}{2}$. Note that $H^{s-1}$ is a Banach algebra and that

$$\begin{aligned} (A(z)-A(y))w=\left( \begin{array}{cc} (v-y_1)\partial _x,&{} 0\\ 0 ,&{} (u-y_2)\partial _x\\ \end{array} \right) \left( \begin{array}{c} w_1\\ w_2\\ \end{array} \right) =\left( \begin{array}{c} (v-y_1)\partial _xw_1\\ (u-y_2)\partial _xw_2\\ \end{array} \right) . \end{aligned}$$

Then, we have

$$\begin{aligned}&\Vert (A(z)-A(y))w\Vert _{H^{s-1}\times H^{s-1}}\\&\quad \le \Vert (v-y_1)\partial _xw_1\Vert _{H^{s-1}}+\Vert (u-y_2)\partial _xw_2\Vert _{H^{s-1}}\\&\quad \le \Vert v-y_1\Vert _{H^{s-1}}\Vert \partial _xw_1\Vert _{H^{s-1}}+\Vert u-y_2\Vert _{H^{s-1}}\Vert \partial _xw_2\Vert _{H^{s-1}}\\&\quad \le c\Vert v-y_1\Vert _{H^{s-1}}\Vert w_1\Vert _{H^s}+c\Vert u-y_2\Vert _{H^{s-1}}\Vert w_2\Vert _{H^s}\\&\quad \le \mu _1\Vert z-y\Vert _{H^s\times H^s}\Vert w\Vert _{H^s\times H^s}, \end{aligned}$$

where we use Lemma 2.3 with $r=s-1$, $t=s-1$. Taking $y=0$ in the above inequality, we obtain that $A(z)\in L(H^s\times H^s,H^{s-1}\times H^{s-1})$. This completes the proof of the lemma. $\square$

Lemma 2.9

Let $B(z):=QA(z)Q^{-1}-A(z)$ with $z\in H^s\times H^s$, $s\ge \frac{3}{2}$. Then, $B(z)\in L(H^{s-1}\times H^{s-1})$ and

$$\begin{aligned} \Vert (B(z)-B(y))w\Vert _{H^s\times H^s}\le \mu _2\Vert z-y\Vert _{H^s\times H^s}\Vert w\Vert _{H^{s-1}\times H^{s-1}}, \end{aligned}$$

for all $z,y\in H^s\times H^s$ and $w\in H^{s-1}\times H^{s-1}$.

Proof

Let $z,y\in H^s\times H^s$ and $w\in H^{s-1}\times H^{s-1}$, $s\ge \frac{3}{2}$. Note that

$$\begin{aligned} (B(z)-B(y))w=[\Lambda ^1,(v-y_1)\partial _x]\Lambda ^{-1}w_1+[\Lambda ^1,(u-y_2)\partial _x]\Lambda ^{-1}w_2. \end{aligned}$$

Then, we have

$$\begin{aligned}&\Vert (B(z)-B(y))w\Vert _{H^{s-1}\times H^{s-1}}\\&\quad \le \Vert \Lambda ^{s-1}[\Lambda ^1,(v-y_1)\partial _x]\Lambda ^{-1}w_1\Vert _{L^2} +\Vert \Lambda ^{s-1}[\Lambda ^1,(u-y_2)\partial _x]\Lambda ^{-1}w_2\Vert _{L^2}\\&\quad \le \Vert \Lambda ^{s-1}[\Lambda ^1,(v-y_1)]\Lambda ^{1-s}\Vert _{L(L^2)}\Vert \Lambda ^{s-2}\partial _xw_1\Vert _{L^2}\\&\qquad +\,\Vert \Lambda ^{s-1}[\Lambda ^1,(u-y_2)]\Lambda ^{1-s}\Vert _{L(L^2)}\Vert \Lambda ^{s-2}\partial _xw_2\Vert _{L^2}\\&\quad \le \mu _2\Vert y-z\Vert _{H^s\times H^s}\Vert w\Vert _{H^{s-1}\times H^{s-1}}, \end{aligned}$$

where we applied Lemma 2.4 with $r=1-s$, $t=s-1$.Taking $y=0$ in the above inequality, we obtain $B(u)\in L(H^{s-1}\times H^{s-1})$. This completes the proof of the lemma. $\square$

Lemma 2.10

Let $z\in H^s\times H^s$, $s\ge \frac{3}{2}$ and let

$$\begin{aligned} f(z)=\left( \begin{array}{c} -(1-\partial _x^2)^{-1}(2uv_x+u_xv_{xx})\\ -(1-\partial _x^2)^{-1}(2vu_x+v_xu_{xx})\\ \end{array} \right) . \end{aligned}$$

Then, f is bounded on bounded sets in $H^s\times H^s$, and satisfies

$$\begin{aligned}&(a)\quad \Vert f(y)-f(z)\Vert _{H^s\times H^s}\le \mu _3\Vert y-z\Vert _{H^s\times H^s},\quad y,z\in H^s\times H^s,\\&(b)\quad \Vert f(y)-f(z)\Vert _{H^{s-1}\times H^{s-1}}\le \mu _4\Vert y-z\Vert _{H^{s-1}\times H^{s-1}},\quad y,z\in H^s\times H^s. \end{aligned}$$

Proof

Let $y,z\in H^s\times H^s$, $s\ge \frac{3}{2}$. Note that $H^s$ is a Banach algebra. Then, we have

$$\begin{aligned}&\Vert f(y)-f(z)\Vert _{H^s\times H^s}\\&\quad \le \Vert -(1-\partial _x^2)^{-1}(2y_2y_{1x}-2uv_x+y_{2x}y_{1xx}-u_xv_{xx})\Vert _{H^s}\\&\qquad +\,\Vert -(1-\partial _x^2)^{-1}(2y_1y_{2x}-2vu_x+y_{1x}y_{2xx}-v_xu_{xx})\Vert _{H^s}\\&\quad \le \Vert (2y_2y_{1x}-2uv_x+y_{2x}y_{1xx}-u_xv_{xx})\Vert _{H^{s-2}}\\&\qquad +\,\Vert (2y_1y_{2x}-2vu_x+y_{1x}y_{2xx}-v_xu_{xx})\Vert _{H^{s-2}}\\&\quad \le 2\Vert y_{1x}(y_2-u)+u(y_{1x}-v_x)\Vert _{H^{s-2}}+\Vert y_{1xx}(y_{2x}-u_x)+u_x(y_{1xx}{-}v_{xx})\Vert _{H^{s-2}}\\&\qquad +\,2\Vert y_{2x}(y_1-v)+v(y_{2x}-u_x)\Vert _{H^{s-2}}+\Vert y_{2xx}(y_{1x}-v_x){+}v_x(y_{2xx}{-}u_{xx})\Vert _{H^{s-2}}\\&\quad \le c(\Vert y_1\Vert _{H^{s-1}}\Vert (y_2-u)\Vert _{H^s}+\Vert u\Vert _{H^{s-2}}\Vert (y_1-v)\Vert _{H^{s}}\\&\qquad +\,\Vert y_1\Vert _{H^s}\Vert y_2-u\Vert _{H^s}+\Vert u\Vert _{H^{s-1}}\Vert y_1-v\Vert _{H^s}\\&\qquad +\,\Vert y_2\Vert _{H^{s-1}}\Vert y_1-v\Vert _{H^s}+\Vert v\Vert _{H^{s-2}}\Vert y_2-u\Vert _{H^s}\\&\qquad +\,\Vert y_2\Vert _{H^s}\Vert y_1-v\Vert _{H^s}+\Vert v\Vert _{H^{s-1}}\Vert y_2-u\Vert _{H^s})\\&\quad \le c(\Vert y\Vert _{H^s\times H^s}+\Vert z\Vert _{H^s\times H^s})\Vert y-z\Vert _{H^s\times H^s}. \end{aligned}$$

This proves (a). Taking $y=0$ in the above inequality, we obtain that f is bounded on bounded set in $H^s\times H^s$. Next, we prove (b). Note that $H^{s-1}$ is a Banach algebra. Then, we have

$$\begin{aligned}&\Vert f(y)-f(z)\Vert _{H^{s-1}\times H^{s-1}}\\&\quad \le \Vert -(1-\partial _x^2)^{-1}(2y_2y_{1x}-2uv_x+y_{2x}y_{1xx}-u_xv_{xx})\Vert _{H^{s-1}}\\&\qquad +\,\Vert -(1-\partial _x^2)^{-1}(2y_1y_{2x}-2vu_x+y_{1x}y_{2xx}-v_xu_{xx})\Vert _{H^{s-1}}\\&\quad \le \Vert (2y_2y_{1x}-2uv_x+y_{2x}y_{1xx}-u_xv_{xx})\Vert _{H^{s-3}}\\&\qquad +\,\Vert (2y_1y_{2x}-2vu_x+y_{1x}y_{2xx}-v_xu_{xx})\Vert _{H^{s-3}}\\&\quad \le 2\Vert y_{1x}(y_2-u)+u(y_{1x}-v_x)\Vert _{H^{s-3}}+\Vert y_{1xx}(y_{2x}-u_x)+u_x(y_{1xx}-v_{xx})\Vert _{H^{s-3}}\\&\qquad +\,2\Vert y_{2x}(y_1-v)+v(y_{2x}-u_x)\Vert _{H^{s-3}}+\Vert y_{2xx}(y_{1x}-v_x){+}v_x(y_{2xx}{-}u_{xx})\Vert _{H^{s-3}}\\&\quad \le c(\Vert y_1\Vert _{H^{s-2}}\Vert y_2-u\Vert _{H^{s-1}}+\Vert u\Vert _{H^{s-3}}\Vert y_1-v\Vert _{H^{s-1}}\\&\qquad +\,\Vert y_1\Vert _{H^{s-1}}\Vert y_2-u\Vert _{H^{s-1}}+\Vert u\Vert _{H^{s-2}}\Vert y_1-v\Vert _{H^{s-1}}\\&\qquad +\,\Vert y_2\Vert _{H^{s-2}}\Vert y_1-v\Vert _{H^{s-1}}+\Vert v\Vert _{H^{s-3}}\Vert y_2-u\Vert _{H^{s-1}}\\&\qquad +\,\Vert y_2\Vert _{H^{s-1}}\Vert y_1-v\Vert _{H^{s-1}}+\Vert v\Vert _{H^{s-2}}\Vert y_2-u\Vert _{H^{s-1}})\\&\quad \le c(\Vert y\Vert _{H^s\times H^s}+\Vert z\Vert _{H^s\times H^s})\Vert y-z\Vert _{H^{s-1}\times H^{s-1}}. \end{aligned}$$

This proves (b) and completes the proof of the lemma. $\square$

Proof of Theorem 2.2

Using Theorem 2.1 and Lemmas 2.6–2.10, we arrive at the conclusion of Theorem 2.2. $\square$

3 Blow-up Scenario

The following lemmas are useful for our discussion.

Lemma 3.1

[14] If $r>0$, then,

$$\begin{aligned} \Vert [\Lambda ,f]g\Vert _{L^2}\le c(\Vert \partial _x f\Vert _{L^\infty }\Vert \Lambda ^{r-1}g\Vert _{L^2}+\Vert \Lambda ^rf\Vert _{L^2}\Vert g\Vert _{L^\infty }), \end{aligned}$$

where c is a constant depending only on r.

In this section, we present the precise blow-up scenarios for the solutions to (1.1).

Theorem 3.2

Let $z:=\left( \begin{array}{c} u_0 \\ v_0\\ \end{array} \right) \in H^s\times H^s$, $s>\frac{5}{2}$, be given and assume that T is the existence time of the corresponding solution $z:=\left( \begin{array}{c} u \\ v\\ \end{array} \right)$ to (1.1) with the initial data $Z_0$. If there exists $M>0$ such that

$$\begin{aligned} \Vert u_x\Vert _{L^\infty }+\Vert v_x\Vert _{L^\infty }+\Vert u_{xx}\Vert _{L^\infty }+\Vert u_{xx}\Vert _{L^\infty }\le M,\quad t\in [0,T), \end{aligned}$$

then, the $H^s\times H^s$-norm of $z(\cdot ,t)$ does not blow up on [0, T).

Proof

Throughout the proof, $c>0$ stands for a generic constant depending only on s. Applying the operator $\Lambda ^s$ to the first and second equation in (1.2), multiplying by $\Lambda ^su$ and $\Lambda ^sv$, respectively, and integrating over ${\mathbb {R}}$, we obtain

$$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}t}\Vert u\Vert _{H^s}^2=-(vu_x,u)_s-(u,f_1(u,v))_s,\end{aligned}$$

(3.1)

$$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}t}\Vert v\Vert _{H^s}^2=-(uv_x,v)_s-(v,f_2(u,v))_s, \end{aligned}$$

(3.2)

where

$$\begin{aligned} f_1(u,v)=(1-\partial _x^2)^{-1}(2uv_x+u_xv_{xx}),\\ f_2(u,v)=(1-\partial _x^2)^{-1}(2vu_x+v_xu_{xx}). \end{aligned}$$

Let us estimate the right-hand side of (3.1) and (3.2).

$$\begin{aligned} |(vu_x,u)_s|&=|(\Lambda ^svu_x,\Lambda ^su)_0|\\&=|([\Lambda ^s,v]u_x,\Lambda ^su)_0+(v\Lambda ^su_x,\Lambda ^su)_0|\\&\le \Vert [\Lambda ^s,v]u_x\Vert _{L^2}\Vert \Lambda ^su\Vert _{L^2}+\frac{1}{2}|(v_x\Lambda ^su,\Lambda ^su)_0|\\&\le c(\Vert v_x\Vert _{L^\infty }\Vert u\Vert _{H^s}+\Vert u_x\Vert _{L^\infty }\Vert v\Vert _{H^s})\Vert u\Vert _{H^s} +\frac{1}{2}\Vert v^x\Vert _{L^\infty }\Vert u\Vert _{H^s}^2\\&\le c(\Vert u_x\Vert _{L^\infty }+\Vert v_x\Vert _{L^\infty })(\Vert u\Vert _{H^s}^2+\Vert v\Vert _{H^s}^2). \end{aligned}$$

In the above inequality, we used Lemma 3.1 with $r=s$. Similarly, we have

$$\begin{aligned} |(uv_x,v)_s| \le c(\Vert u_x\Vert _{L^\infty }+\Vert v_x\Vert _{L^\infty })(\Vert u\Vert _{H^s}^2+\Vert v\Vert _{H^s}^2). \end{aligned}$$

For the other two terms, we have

$$\begin{aligned} |(u,f_1(u,v))_s|= & {} |((1-\partial _x^2)^{-1}(2u_xv+v_xu_{xx}),u)_s|\\\le & {} |(\Lambda ^{s-1}(2u_xv),\Lambda ^{s_1}u)_0|+|(\Lambda (v_xu_{xx}),\Lambda ^su)_0|\\\le & {} 2\Vert [\Lambda ^{s-1},v]u_x\Vert _{L^2}\Vert \Lambda ^{s-1}u\Vert _{L^2}+2|(v\Lambda ^{s-1}u_x,\Lambda ^{s-1}u)_0|\\&+\,\Vert [\Lambda ^{s-2},v_x]u_{xx}\Vert _{L^2}\Vert \Lambda ^su\Vert _{L^2}+2|(v_x\Lambda ^{s-2}u_{xx},\Lambda ^su)_0|\\\le & {} c(\Vert v_{xx}\Vert _{L^\infty }\Vert \Lambda ^{s-2}u_{xx}\Vert _{L^2} +\,\Vert u_{xx}\Vert _{L^\infty }\Vert \Lambda ^{s-2}v_x\Vert _{L^2})\Vert \Lambda ^su\Vert _{L^2}\\&c(\Vert v_x\Vert _{L^\infty }\Vert \Lambda ^{s-2}u_x\Vert _{L^2} +\Vert u_x\Vert _{L^\infty }\Vert \Lambda ^{s-1}v_x\Vert _{L^2})\Vert \Lambda ^{s-1}u\Vert _{L^2}\\&+\Vert v_x\Vert _{L^\infty }\Vert u\Vert _{H^s}^2+\frac{1}{2}\Vert v\Vert _{L^\infty }\Vert u\Vert _{H^{s-1}}^2\\\le & {} c(\Vert u_x\Vert _{L^\infty }+\Vert v_x\Vert _{L^\infty }+\Vert u_{xx}\Vert _{L^\infty } +\Vert v_{xx}\Vert _{L^\infty })(\Vert u\Vert _{H^s}^2+\Vert v\Vert _{H^s}^2), \end{aligned}$$

$\square$

where Lemma 3.1 with $r=s-1$ and $r=s-2$ was used. Similarly, we have

$$\begin{aligned} |(u,f_2(u,v))_s| \le c(\Vert u_x\Vert _{L^\infty }+\Vert v_x\Vert _{L^\infty }+\Vert u_{xx}\Vert _{L^\infty } +\Vert v_{xx}\Vert _{L^\infty })(\Vert u\Vert _{H^s}^2+\Vert v\Vert _{H^s}^2). \end{aligned}$$

Therefore,

$$\begin{aligned}&\frac{\mathrm{d}}{\mathrm{d}t}(\Vert u\Vert _{H^s}^2+\Vert v\Vert _{H^s}^2)\\&\quad \le c(\Vert u_x\Vert _{L^\infty }+\Vert v_x\Vert _{L^\infty }+\Vert u_{xx}\Vert _{L^\infty } +\Vert v_{xx}\Vert _{L^\infty })(\Vert u\Vert _{H^s}^2+\Vert v\Vert _{H^s}^2). \end{aligned}$$

An application of Gronwall’s inequality and the assumption of the theorem yield

$$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}t}(\Vert u\Vert _{H^s}^2+\Vert v\Vert _{H^s}^2) \le \exp ^{cMt}(\Vert u_0\Vert _{H^s}^2+\Vert v_0\Vert _{H^s}^2). \end{aligned}$$

This completes the proof of the theorem.

Theorem 3.3

Let $z_0:=\left( \begin{array}{c} u_0 \\ v_0\\ \end{array} \right) \in H^s\times H^s$, $s>\frac{5}{2}$, be given and assume that T is the existence time of the corresponding solution $z:=\left( \begin{array}{c} u \\ v\\ \end{array} \right)$ to(1.1) with the initial data $z_0$. Then, the corresponding solution blows up in finite time if and only if

$$\begin{aligned} \liminf _{t\rightarrow T}\{\inf _{x\in {\mathbb {R}}}[u_x(x,t)]\}=-\infty \quad or\quad \liminf _{t\rightarrow T}\{\inf _{x\in {\mathbb {R}}}[v_x(x,t)]\}=-\infty . \end{aligned}$$

Proof

By direct computation, one has

$$\begin{aligned} \left\| m\right\| _{L^2}^2=\int _{{\mathbb {R}}}(u-u_{xx})^2\mathrm{d}x =\int _{{\mathbb {R}}}u^2+2u_x^2+u_{xx}^2\mathrm{d}x. \end{aligned}$$

and

$$\begin{aligned} \Vert n\Vert _{L^2}^2=\int _{{\mathbb {R}}}(v-v_{xx})^2\mathrm{d}x =\int _{{\mathbb {R}}}v^2+2v_x^2+v_{xx}^2\mathrm{d}x. \end{aligned}$$

Hence,

$$\begin{aligned} \Vert z\Vert _{H^2\times H^2}^2\le \Vert m\Vert _{L^2}^2+\Vert n\Vert _{L^2}^2 \le 2\Vert z\Vert _{H^2\times H^2}^2. \end{aligned}$$

(3.3)

Applying y on (1.1) and integration by parts, we obtain

$$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}t}\int _{{\mathbb {R}}}m^2\mathrm{d}x=\int _{{\mathbb {R}}}2mm_t\mathrm{d}x =-2\int _{{\mathbb {R}}}m(2mv_x+m_xv)\mathrm{d}x =-3\int _{{\mathbb {R}}}v_xm^2\mathrm{d}x.\qquad \end{aligned}$$

(3.4)

and

$$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}t}\int _{{\mathbb {R}}}n^2\mathrm{d}x=\int _{{\mathbb {R}}}2nn_t\mathrm{d}x =-2\int _{{\mathbb {R}}}m(2nu_x+n_xu)\mathrm{d}x =-3\int _{{\mathbb {R}}}u_xn^2\mathrm{d}x. \end{aligned}$$

(3.5)

Combining (3.4) and (3.5) together, we get

$$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}t}\int _{{\mathbb {R}}}m^2+n^2\mathrm{d}x =-3\int _{{\mathbb {R}}}v_xm^2+u_xn^2\mathrm{d}x. \end{aligned}$$

(3.6)

Due to Gronwall’s inequality, it is clear that, from (3.6), $u_x$ and $v_x$ are bounded from below on [0, T ) then, the $H_2\times H_2$-norm of the solution z is also bounded on [0, T ). On the other hand,

$$\begin{aligned} u(x,t)=(1-\partial _x^2)^{-1}y(x,t)=\int _{{\mathbb {R}}}G(x-\xi )y(\xi )\mathrm{d}\xi . \end{aligned}$$

Therefore,

$$\begin{aligned}&\Vert u_x\Vert _{L^\infty }+\Vert v_x\Vert _{L^\infty }\le \left| \int _{{\mathbb {R}}}G_x(x-\xi )(m+n)(\xi )\mathrm{d}\xi \right| \nonumber \\&\quad \le \Vert G_x\Vert _{L^2}(\Vert m\Vert _{L^2}+\Vert n\Vert _{L^2})=\frac{1}{2}(\Vert m\Vert _{L^2}+\Vert n\Vert _{L^2})\le \Vert z\Vert _{H^2\times H^2}, \end{aligned}$$

(3.7)

where we use (3.3). Hence, (3.7) tells us if $H_2$-norm of the solution is bounded then the $L_\infty$-norm of the first derivative is bounded. $\square$

Comparing with the standard Camassa–Holm equation, it seems very difficult to establish a necessary and sufficient condition on the initial data to guarantee blow-up in finite time. For the standard Camassa–Holm equation, we refer to [17] and [10] for details. Studies on large time behavior is given in [11].