# Rogue peakon, well-posedness, ill-posedness and blow-up phenomenon for an
integrable Camassa-Holm type equation

Mingxuan Zhu
School of Mathematical Sciences
Qufu Normal University, Qufu, 273100, P. R. China
E-mail<EMAIL_ADDRESS>
Zhenteng Zeng
School of Mathematical and Statistical Sciences
University of Texas Rio Grande Valley, Edinburg, TX 78539, USA
E-mail<EMAIL_ADDRESS>
Zaihong Jiang
Department of Mathematics
Zhejiang Normal University, Jinhua, 321004, P. R. China
E-mail<EMAIL_ADDRESS>
Baoqiang Xia
School of Mathematics and Statistics
Jiangsu Normal University, Xuzhou, 221116, P. R. China
E-mail<EMAIL_ADDRESS>
Zhijun Qiao
School of Mathematical and Statistical Sciences
University of Texas Rio Grande Valley, Edinburg, TX 78539, USA
E-mail<EMAIL_ADDRESS>Corresponding author

###### Abstract

In this paper, we study an integrable Camassa-Holm (CH) type equation with
quadratic nonlinearity. The CH type equation is shown integrable through a Lax
pair, and particularly the equation is found to possess a new kind of peaked
soliton (peakon) solution - called rogue peakon, that is given in a rational
form with some logarithmic function, but not a regular traveling wave. We also
provide multi-rogue peakon solutions. Furthermore, we discuss the local well-
posedness of the solution in the Besov space $B_{p,r}^{s}$ with $1\leq
p,r\leq\infty$, $s>\displaystyle\max\left\\{1+1/p,3/2\right\\}$ or
$B_{2,1}^{3/2}$, and then prove the ill-posedness of the solution in
$B_{2,\infty}^{3/2}$. Moreover, we establish the global existence and blow-up
phenomenon of the solution, which is, if
$m_{0}(x)=u_{0}-u_{0xx}\geq(\not\equiv)0$, then the corresponding solution
exists globally, meanwhile, if $m_{0}(x)\leq(\not\equiv)0$, then the
corresponding solution blows up in a finite time.

Keywords: Camassa-Holm type equation, Lax pair, rogue peakon; well-posedness;
ill-posedness; Besov space; blow-up phenomenon.

Mathematics Subject Classifications (2020): 37K10; 35G25; 35L05

##  1 Introduction and rogue peakons

In this paper, we consider the following Camassa-Holm (CH) type equation that
was proposed recently in [34, 36]

$\displaystyle m_{t}+mu+(mu)_{x}=0,\quad m=u-u_{xx}.$ (1.1)

The above CH type equation (1.1) is actually a reduced case of the following
integrable two-component peakon system [34]

$\displaystyle\begin{cases}m_{t}=F+F_{x}-\displaystyle\frac{1}{2}m(uv-
u_{x}v_{x}+uv_{x}-u_{x}v),\\\ n_{t}=-G+G_{x}+\displaystyle\frac{1}{2}u(uv-
u_{x}v_{x}+uv_{x}-u_{x}v),\\\ m=u-u_{xx},\\\ n=v-v_{xx},\end{cases}$ (1.2)

where $F$ and $G$ are two arbitrary functions of $u$, $v$ and their
derivatives satisfying $mG=nF.$ Let $v=0$, $G=0$, and $F=-mu$ in (1.2), then
Eq. (1.2) is exactly reduced to the scalar model (1.1). Through a
straightforward computation, it is not hard for us to see the scalar CH type
equation (1.1) possesses the following Lax pair
$\psi_{x}=U\psi,\>\psi_{t}=V\psi,$

$U=\displaystyle\frac{1}{2}\begin{pmatrix}-1&\lambda m\\\
0&1\end{pmatrix},\quad
V=-\displaystyle\frac{1}{2}\begin{pmatrix}\lambda^{-2}&-\lambda^{-1}(u-u_{x})+\lambda
mu\\\ 0&-\lambda^{-2}\end{pmatrix},$ (1.3)

where $\psi$ is the 2-dimensional eigenvector, $\lambda$ refers to the
eigenvalue parameter, and $U,V$ satisfy the following compatibility condition
– zero curvature representation $U_{t}-V_{x}+[U,V]=0.$ Therefore, the CH type
equation (1.1) is completely integrable in the sense of Lax pair.

Rogue Peakon

The most attractive property of the CH type equation (1.1) is that it has the
following so-called Rogue Peakon (see the formula given in Eq. (1.4) below and
Remark 1).

Let $u=p(t)e^{-|x-q(t)|}$. One may compute

$\displaystyle u_{x}=-pe^{-|x-q|}sign(x-q),$ $\displaystyle
m=u-u_{xx}=2p\delta(x-q),$ $\displaystyle
m_{t}=2p^{\prime}\delta(x-q)-2pq^{\prime}\delta^{\prime}(x-q)$ $\displaystyle
m_{x}=2p\delta^{\prime}(x-q),$

where $\delta$ the Dirac delta and $\delta^{\prime}$ is its derivative.
Substituting the above formulas to (1.1), we get

$\displaystyle(2p^{\prime}+2p^{2}e^{-|x-q|}-2p^{2}e^{-|x-q|}sign(x-q))\delta(x-q)-2p(q^{\prime}-pe^{-|x-q|})\delta^{\prime}(x-q)=0.$

Evaluating this on a test function $\varphi$ and integratig by parts, we have

$\displaystyle\langle(2p^{\prime}+2p^{2}e^{-|x-q|}-4p^{2}e^{-|x-q|}sign(x-q))\delta(x-q),\varphi\rangle+\langle
2p(q^{\prime}-pe^{-|x-q|})\delta(x-q),\varphi^{\prime}\rangle=0.$

The above equation holds for any $\varphi$. Thus, $p(t)$ and $q(t)$ satisfy

$\displaystyle p^{\prime}=-p^{2},\quad q^{\prime}=p.$

Then, we arrive at

$\displaystyle p(t)$ $\displaystyle=$
$\displaystyle\displaystyle\frac{1}{t-A},\quad q(t)\ =\ \ln|t-A|+B,$
$\displaystyle u(x,t)$ $\displaystyle=$
$\displaystyle\displaystyle\frac{1}{t-A}e^{-|x-\ln|t-A|-B|},$ (1.4)

where $A$, $B$ are two arbitrary constants. The figure of one- rogue peakon
(1.4) is plotted below.

Figure 1: The single rogue peakon solution (1.4)

with $A=1,B=0$ and plotting for $x\in(-3,0),\ t\in(0.6,1.4)$. Apparently, the
solution $u(x,t)$ is continuous everywhere for $x$ and $t$ except $t=1$
because it has a finite jump $2e^{-x}$ at $t=1$ (see Remark 1 below).

Remark 1: Apparently, the single peakon solution given in Eq. (1.4) is not a
traveling solitary wave. The amplitude of the peakon is the rational function
$p(t)=\displaystyle\frac{1}{t-A}$ that has a sole singular point at $t=A$. The
solution $u(x,t)$ is continuous everywhere for both variables $x$ and $t$
except $t=A$ because it has a finite jump at $t=A$ (see the left and right
limits computed below):

$\displaystyle u(A^{-})$ $\displaystyle=\displaystyle\lim_{t\to
A^{-}}u=\displaystyle\lim_{t\to
A^{-}}\displaystyle\frac{e^{-|x-\ln|t-A|-B|}}{t-A},$
$\displaystyle=-e^{-x+B},$ $\displaystyle u(A^{+})$
$\displaystyle=\displaystyle\lim_{t\to A^{+}}u=\displaystyle\lim_{t\to
A^{+}}\displaystyle\frac{e^{-|x-\ln|t-A|-B|}}{t-A},$ $\displaystyle=e^{-x+B}.$

This tells us that the peakon amplitude suddenly changes from negative to
positive nearby $t=A$, that is, right after the crest, it is followed up
immediately by the trough. Since $p(t)$ is a rational function and the peakon
solution is in a non-traveling wave shape, let us call this wave solution
(1.4) as Rogue Peakon. This is the significant difference from all existing
peakon solutions in literature.

Multi-Rogue Peakons

Let us now seek for multi-rogue peakon solutions in the form of

$u(x,t)=\displaystyle\sum_{k=1}^{N}p_{k}(t)e^{-|x-q_{k}(t)|}.$ (1.5)

Substituting Eq. (1.5) to the CH type equation (1.1), we obtain the following
$2N$ ODE system in $t$:

$\displaystyle\left\\{\begin{array}[]{ll}\dot{p}_{k}(t)=-p_{k}(t)\displaystyle\sum_{j=1}^{N}p_{j}(t)e^{-|q_{k}(t)-q_{j}(t)|},\\\
\dot{q}_{k}(t)=\displaystyle\sum_{j=1}^{N}p_{j}(t)e^{-|q_{k}(t)-q_{j}(t)|}.\end{array}\right.$
(1.8)

Solving the above system (1.8) yields solutions

$\displaystyle\left\\{\begin{array}[]{ll}p_{k}(t)=\displaystyle\frac{1}{Nt-
A_{k}}\\\ q_{k}(t)=\ln|Nt-A_{k}|+B_{k},\quad k=1,2,...,N,\end{array}\right.$

where $A_{k}$ and $B_{k}$ are arbitrary constants and $N$ refers to number of
peaks. Therefore, we get the $N$-rogue peakon solutions (1.5) solved with

$\displaystyle u(x,t)$ $\displaystyle=$
$\displaystyle\displaystyle\sum_{k=1}^{N}\displaystyle\frac{e^{-|x-\ln|Nt-
A_{k}|-B_{k}|}}{Nt-A_{k}}.$

As we see the $N$ rogue peakon solution $u(x,t)$ solved above, it can be
understood as a linear superposition of non-traveling peaked waves in $x$ with
coefficients of $t$-related rational form for an nonlinear intgerable PDE.
This property is usually true for a linear differential equation, instead of a
nonlinear PDE. But here in our paper, we find the rogue peakon has the linear
superposition for our integrable nonlinear CH type equation (1.1), which is
the significant difference from other multi-soliton interactions and nonlinear
PDEs. For two- rogue peakons ($N=2$, see Figure 3)

$\displaystyle
u(x,t)=\displaystyle\frac{e^{-|x-\ln|2t-A_{1}|-B_{1}|}}{2t-A_{1}}+\displaystyle\frac{e^{-|x-\ln|2t-A_{2}|-B_{2}|}}{2t-A_{2}},$
(1.10)

and three- rogue peakons ($N=3$, see Figure 3)

$\displaystyle
u(x,t)=\displaystyle\frac{e^{-|x-\ln|3t-A_{1}|-B_{1}|}}{3t-A_{1}}+\displaystyle\frac{e^{-|x-\ln|3t-A_{2}|-B_{2}|}}{3t-A_{2}}+\displaystyle\frac{e^{-|x-\ln|3t-A_{3}|-B_{3}|}}{3t-A_{3}},$
(1.11)

we have the graphs plotted below.

Figure 2: The two-rogue peakon solution (1.10)

with $A_{1}=-1,A_{2}=1,B_{1}=-0.5,B_{2}=0.5$ and plotting for $x\in(-2,1),\
t\in(-1,1)$. Apparently, the solution $u(x,t)$ is continuous everywhere for
$x$ and $t$ except $t=\pm 0.5$ because it has a finite jump $2e^{-x+0.5}$ at
$t=0.5$ and $2e^{-x-0.5}$ at $t=-0.5$, respectively.

Figure 3:  The three-rogue peakon solution (1.11)

with $A_{1}=-1,A_{2}=0,A_{3}=1,B_{1}=-0.5,B_{2}=0,B_{3}=0.5$ and plotting for
$x\in(-2,0),\ t\in(-0.5,0.5)$. Apparently, the solution $u(x,t)$ is continuous
everywhere for $x$ and $t$ except $t=0,\pm 0.33$ because it has a finite jump
$2e^{-x}$ at $t=0$, $2e^{-x+0.33}$ at $t=0.33$ and $2e^{-x-0.33}$ at
$t=-0.33$, respectively.

We already noticed that this is a linear superposition phenomenon with
$t$-fractional coefficients for nonlinear peakons crossing each peak
$x=\ln|Nt-A_{k}|+B_{k}$, but the shape of the rogue peakon between any two
consecutive peaks is still taking on a dynamical interaction, and none of them
are traveling wave type. See this phenomena in the following three figures
plotted for $2-$rogue peakons.

Figure 4:  The two- rogue peakon solution (1.10) with
$A_{1}=-1,A_{2}=1,B_{1}=-0.5,B_{2}=0.5$ and plotting for $x\in(-3,1),\
t\in(-0.4,0.4)$. Apparently, the solution $u(x,t)$ is continuous for
$x\in(-3,1),\ t\in(-0.4,0.4)$ and two peakons (actually peakon and anti-
pekaon) get interacted from positive to negative in this region.

Figure 5:  The two- rogue peakon solution (1.10) with
$A_{1}=-1,A_{2}=1,B_{1}=-0.5,B_{2}=0.5$ and plotting for $x\in(-2,1),\
t\in(-0.49,1)$. Apparently, the solution $u(x,t)$ is continuous everywhere for
$x$ and $t$ except $t=0.5$ because it has a finite jump $2e^{-x+0.5}$ at
$t=0.5$.

But at the moment of going to and departing from $t=0.5$, two rogue peakons
still get interacted.

Figure 6:  The two- rogue peakon solution (1.10) with
$A_{1}=-1,A_{2}=1,B_{1}=-0.5,B_{2}=0.5$ and plotting for $x\in(-2,1),\
t\in(-1,0.49)$. Apparently, the solution $u(x,t)$ is continuous everywhere for
$x$ and $t$ except $t=-0.5$ because it has a finite jump $2e^{-x-0.5}$ at
$t=-0.5$. But at the moment of going to and departing from $t=-0.5$, two rogue
peakons still get interacted.

Background

The system (1.1) is highly related to the celebrated Camassa-Holm(CH) equation

$\displaystyle m_{t}+2mu_{x}+m_{x}u=0,\quad m=u-u_{xx},$

which was introduced in [5] and was also obtained by Fokas and Fuchssteiner
[19] as a nonlinear partial differential equation with bi-Hamiltonian
structures and inherited symmetries. It was shown completely integrable with
the remarkable properties of possessing Lax pair [5], great features on the
circle [12], the integrable hierarchy extension with finite-dimensional
constrained flows and algebro-geometric solutions on a symplectic submanifold
connected to integrable negative order flows [30], inverse scattering
transform [11] and multi-soliton solutions via Darboux transformation [35],
infinitely many conservation laws [25] and geometric background of the CH
equation [32], and explicit solutions including classical soliton, cuspon and
peakon solutions [5, 26, 31].

Local well-posedness for the initial data in $H^{s}$ with $s>3/2$ was proved
in [9, 27], while Danchin have extended the well-posedness to the Besov space
$B_{p,r}^{s}$ ($s>\displaystyle\max\left\\{1+1/p,3/2\right\\}$, $1\leq
p,r\leq\infty$) or $B_{2,1}^{3/2}$ in [14, 15]. There have been extensive
studies on Blow-up phenomenon of solutions, cf. [2, 8, 9, 10, 23, 27, 28] and
references therein. For global existence, McKean [28] (See also [23] for a
simple proof) established a necessary and sufficient condition on the initial
value $u_{0}$, which depends on the shape of $m_{0}=u_{0}-u_{0xx}$. More
precisely, it was proved that the Camassa-Holm equation breaks if and only if
some portion of the positive part of $m_{0}$ lies to the left of some portion
of its negative part. Global conservative solution and global dissipative
solution for the Camassa-Holm equation have been shown in [3, 4]. As for the
asymptotic properties, persistence properties and unique continuation of
solutions of the Camassa-Holm equation were given in [21], the long time
behavior for the support of momentum density of the Camassa-Holm equation was
discussed in [24], and the orbital stability of the peakons was proved in
[13].

Remark 2: An observation with the substitution
$m=u-u_{xx}=\widetilde{m}e^{-x}$, our equation (1.1) is sent to the following
form

$\displaystyle\widetilde{m}_{t}+\widetilde{m}_{x}u+\widetilde{m}u_{x}=0,\quad
m=u-u_{xx}.$

This form exactly falls in the Holm-Staley $b$-family equation [22] when $b=1$

$\displaystyle m_{t}+m_{x}u+bmu_{x}=0,\quad m=u-u_{xx}.$ (1.12)

As shown in [29], the Holm-Staley $b$-family equation (1.12) is integrable
only when $b=2$ and $b=3$, namely, $b=2$ is the case for the integrable
Camassa-Holm equation [5] while $b=3$ is the case for the integrable
Degasperis-Procesi equation [17]. Although our equation (1.1) can be
transformed to the $b$-family with $b=1$, we already show that our equation
(1.1) is completely integrable in the sense of Lax pair (see Eq. (1.3) listed
above). More mathematical properties for the Holm-Staley $b$-family equation
were investigated in [18, 20, 37].

The aim of this paper is to study the Cauchy problem for system (1.1),

$\displaystyle\left\\{\begin{array}[]{ll}&m_{t}+mu+(mu)_{x}=0,\quad
m=u-u_{xx},\quad t\in\mathbb{R}^{+},\quad x\in\mathbb{R}\\\
&u(x,0)=u_{0}(x),\end{array}\right.$ (1.15)

and establish the local and global well-posedness, ill-posedness and blow-up
phenomenon of solutions. Although our system is similar to the Holm-Staley
$b$-family equation, we obtained some completely different results.

The rest of this paper is organized as follows. In Section 2, we discuss the
local well-posedness of the solutions in some Besov spaces. Section 3 aims at
proving the ill-posedness in $B_{2,\infty}^{3/2}$. Sufficient conditions on
the initial data to guarantee global existence and blow-up phenomenon are
established in Section 4. In the Appendix, we recall some conclusions on the
properties of Littlewood-Paley decomposition, the nonhomogeneous Besov spaces
and the theory of the transport equation.

##  2 Local well-posedness

The main goal of this section is to establish the local well-posedness of the
Cauchy problem for (1.1) in the nonhomogeneous Besov spaces, for which we
introduce the following spaces.

###### Definition 2.1

For $T>0$, $s\in\mathbb{R}$ and $1\leq p\leq+\infty$, we set

$\displaystyle E_{p,r}^{s}(T)=:C([0,T];B_{p,r}^{s})\displaystyle\cap
C^{1}([0,T];B_{p,r}^{s-1}),\quad if~{}~{}r<\infty,$ $\displaystyle
E_{p,\infty}^{s}(T)=:L^{\infty}([0,T];B_{p,\infty}^{s})\displaystyle\cap
Lip([0,T];B_{p,\infty}^{s-1}),$

and

$\displaystyle E_{p,r}^{s}=:\bigcap_{T>0}E_{p,r}^{s}(T).$

We rewrite (1.15) as

$\displaystyle u_{t}-u_{xxt}+u^{2}+2uu_{x}=uu_{xx}+u_{x}u_{xx}+uu_{xxx}.$

Denote $P(D)=(1-\partial_{x}^{2})^{-1}$, (1.15) could be reformulated as

$\displaystyle u_{t}+uu_{x}+F(u)=0,$ (2.16)

where

$\displaystyle F(u)=F_{1}(u)+F_{2}(u)+F_{3}(u)$ (2.17)

with $F_{1}=\displaystyle\frac{1}{2}u^{2}$,
$F_{2}=P(D)\left(\displaystyle\frac{1}{2}u^{2}+u_{x}^{2}\right)$ and
$F_{3}=\partial_{x}P(D)\left(\displaystyle\frac{1}{2}u^{2}+u_{x}^{2}\right)$.

###  2.1 Local well-posedness in the Besov spaces

Our main local existence result is given by the following theorem.

###### Theorem 2.1

Suppose that $1\leq p,r\leq\infty$ and
$s>\displaystyle\max\left\\{1+\displaystyle\frac{1}{p},\displaystyle\frac{3}{2}\right\\}$.
If $u_{0}\in B_{p,r}^{s}$, then there exists a time $T>0$ such that the Cauchy
problem (1.15) with initial data $u_{0}$ has a unique solution $u\in
E_{p,r}^{s}(T)$. The map $u_{0}\mapsto u$ is continuous from a neighborhood of
$u_{0}$ in $B_{p,r}^{s}$ into

$C([0,T];B_{p,r}^{s^{\prime}})\displaystyle\cap
C^{1}([0,T];B_{p,r}^{s^{\prime}-1})$

for every $s^{\prime}<s$ when $r=+\infty$ and $s^{\prime}=s$ whereas
$r<+\infty$.

###### Remark 2.1

When $p=r=2$, the Besov space $B_{2,2}^{s}$ coincides with the Sobolev space
$H^{s}$. Theorem 2.1 implies that we can obtain the local well-posedness for
the Cauchy problem (1.15) under the condition $u_{0}\in H^{s}$ with
$s>\displaystyle\frac{3}{2}$.

In what follows, we denote $C>0$ as a generic constant only depending on
$p,r,s$. Uniqueness and continuity with respect to the initial data are an
immediate corollary of the following result.

###### Proposition 2.1

Suppose that $1\leq p,r\leq\infty$ and
$s>\displaystyle\max\left\\{1+\displaystyle\frac{1}{p},\displaystyle\frac{3}{2}\right\\}$.
Let $u,v\in L^{\infty}([0,T);B_{p,r}^{s})\displaystyle\cap
C([0,T];\mathscr{S}^{\prime})$ be two given solutions to the Cauchy problem
(1.15) with the initial data $u_{0},v_{0}\in B_{p,r}^{s}$. Then for every
$t\in[0,T]$, we have

$\displaystyle\|u(t)-v(t)\|_{B_{p,r}^{s-1}}\leq\|u_{0}-v_{0}\|_{B_{p,r}^{s-1}}exp\left\\{C\displaystyle\int_{0}^{t}\|u(\tau)\|_{B_{p,r}^{s}}+\|v(\tau)\|_{B_{p,r}^{s}}d\tau\right\\}.$

Proof : Let $\omega=u-v$, then we have $\omega$ satisfies

$\displaystyle\left\\{\begin{array}[]{ll}\omega_{t}+v\partial_{x}\omega=-\omega\partial_{x}u-(F(u)-F(v)),\\\
\omega_{0}=u_{0}-v_{0}.\end{array}\right.$

By using Lemma A.2, we have

$\displaystyle\|\omega(t)\|_{B_{p,r}^{s-1}}$ $\displaystyle\leq$
$\displaystyle
e^{C\displaystyle\int_{0}^{t}\|\partial_{x}v(\tau)\|_{B_{p,r}^{s-2}}d\tau}\times\left(\|\omega_{0}\|_{B_{p,r}^{s-1}}+C\displaystyle\int_{0}^{t}e^{-C\displaystyle\int_{0}^{\tau}\|\partial_{x}v(s)\|_{B_{p,r}^{s-2}}ds}\|\omega\partial_{x}u+(F(u)-F(v))\|_{B_{p,r}^{s-1}}d\tau\right).$

By using the imbedding $B_{p,r}^{s-1}\hookrightarrow L^{\infty}$ with
$s>1+\displaystyle\frac{1}{p}$, we have

$\displaystyle\|\omega\partial_{x}u\|_{B_{p,r}^{s-1}}\leq\|\omega\|_{B_{p,r}^{s-1}}\|u\|_{B_{p,r}^{s}}.$

Direct calculation we obtain

$\displaystyle
F(u)-F(v)=\displaystyle\frac{1}{2}\omega(u+v)+P(D)\left(\displaystyle\frac{1}{2}\omega(u+v)+\omega_{x}(u_{x}+v_{x})\right)+\partial_{x}P(D)\left(\displaystyle\frac{1}{2}\omega(u+v)+\omega_{x}(u_{x}+v_{x})\right),$

then, for
$s>\displaystyle\max\left\\{1+\displaystyle\frac{1}{p},\displaystyle\frac{3}{2}\right\\}$,
we have

$\displaystyle\|F(u)-F(v)\|_{B_{p,r}^{s-1}}\leq
C\|\omega\|_{B_{p,r}^{s-1}}(\|u\|_{B_{p,r}^{s}}+\|v\|_{B_{p,r}^{s}}),$

Hence,

$\displaystyle\|\omega\partial_{x}u+(F(u)-F(v))\|_{B_{p,r}^{s-1}}\leq
C\|\omega\|_{B_{p,r}^{s-1}}(\|u\|_{B_{p,r}^{s}}+\|v\|_{B_{p,r}^{s}}).$ (2.20)

Substituting (2.20) into (2.1), we find that

$\displaystyle\exp\\{{-C\displaystyle\int_{0}^{t}\|v(\tau)\|_{B_{p,r}^{s-1}}d\tau}\\}\|\omega(t)\|_{B_{p,r}^{s-1}}$
$\displaystyle\leq$
$\displaystyle\|\omega_{0}\|_{B_{p,r}^{s-1}}+C\displaystyle\int_{0}^{t}\exp\\{{-C\displaystyle\int_{0}^{\tau}\|v(\tau_{1})\|_{B_{p,r}^{s-1}}d\tau_{1}}\\}\|\omega(\tau)\|_{B_{p,r}^{s-1}}(\|u(\tau)\|_{B_{p,r}^{s}}+\|v(\tau)\|_{B_{p,r}^{s}})d\tau.$

The proof of Proposition 2.1 is completed by using the Gronwall inequality.
$\Box$

Motivated by the local existence theorem for Camassa-Holm equation in [14], we
would like to use the classical Friedrichs regularization method to construct
the approximate solutions to the Cauchy problem (1.15). The following lemma
establishes the existence and compactness of approximate solutions which will
be used in the proof of Theorem 2.1.

###### Lemma 2.1

Let $u_{0},p,r,s$ be the same as the statement of Theorem 2.1 and $u^{(0)}=0$.
There exists a sequence of smooth functions $u^{(n)}\in
C(\mathbb{R}^{+};B_{p,r}^{\infty})$ solving

$\displaystyle\left\\{\begin{array}[]{ll}&u^{(n+1)}_{t}+u^{(n)}u^{(n+1)}_{x}=-F(u^{(n)}),\\\
&u^{(n+1)}(0,x)=S_{n+1}u_{0},\end{array}\right.$ (2.23)

where $F(u)$ is defined in (2.17) and $S_{n+1}$ is the low frequency cut-off
operator defined in Proposition A.1 . Moreover, there is a positive $T$ such
that the solutions satisfying the following properties:

(i) $(u^{(n)})_{n\in\mathbb{N}}$ is uniformly bounded in $E_{p,r}^{s}(T)$,

(ii) $(u^{(n)})_{n\in\mathbb{N}}$ is a Cauchy sequence in
$C([0,T];B_{p,r}^{s-1})$.

Proof : Since all the data $S_{n+1}u_{0}$ belong to $B_{p,r}^{\infty}$, from
Lemma A.3, we obtain that for all $n\in\mathbb{N}$, equation (2.23) has a
global solution $u_{n+1}$ belonging to $C(\mathbb{R}^{+};B_{p,r}^{\infty})$.
Making use of Lemma A.2, we have

$\displaystyle\|u^{(n+1)}(t)\|_{B_{p,r}^{s}}\leq
e^{C\displaystyle\int_{0}^{t}\|u^{(n)}(\tau)\|_{B_{p,r}^{s}}d\tau}\times\left(\|u_{0}^{(n+1)}\|_{B_{p,r}^{s}}+C\displaystyle\int_{0}^{t}e^{-C\displaystyle\int_{0}^{\tau}\|u^{(n)}(s)\|_{B_{p,r}^{s}}ds}\|F(u^{(n)})\|_{B_{p,r}^{s}}d\tau\right).$

By Lemma A.4 and $B_{p,r}^{s-1}\hookrightarrow L^{\infty}$ with
$s>1+\displaystyle\frac{1}{p}$, we estimate $\|F(u^{(n)})\|_{B_{p,r}^{s}}$ as

$\displaystyle\|F(u^{(n)})\|_{B_{p,r}^{s}}$ $\displaystyle\leq$
$\displaystyle\|F_{1}(u^{(n)})\|_{B_{p,r}^{s}}+\|F_{2}(u^{(n)})\|_{B_{p,r}^{s}}+\|F_{3}(u^{(n)})\|_{B_{p,r}^{s}}$
(2.25) $\displaystyle\leq$
$\displaystyle(\|u^{(n)}\|_{L^{\infty}}+\|u^{(n)}_{x}\|_{L^{\infty}})\|u^{(n)}\|_{B_{p,r}^{s}}\leq\|u^{(n)}\|_{B_{p,r}^{s}}^{2}.$

Substituting (2.25) into (2.1), we have

$\displaystyle\|u^{(n+1)}(t)\|_{B_{p,r}^{s}}\leq
e^{C\displaystyle\int_{0}^{t}\|u^{(n)}(\tau)\|_{B_{p,r}^{s}}d\tau}\times\left(\|u_{0}\|_{B_{p,r}^{s}}+C\displaystyle\int_{0}^{t}e^{-C\displaystyle\int_{0}^{\tau}\|u^{(n)}(s)\|_{B_{p,r}^{s}}ds}\|u^{(n)}\|_{B_{p,r}^{s}}^{2}d\tau\right).$

Let $T>0$ such that $2C\|u_{0}\|_{B_{p,r}^{s}}T<1$, we claim that for any
$n\in\mathbb{N}$ and $t\in[0,T]$,

$\displaystyle\|u^{(n)}\|_{B_{p,r}^{s}}\leq\displaystyle\frac{\|u_{0}\|_{B_{p,r}^{s}}}{1-2C\|u_{0}\|_{B_{p,r}^{s}}t}.$
(2.26)

We use mathematical induction method to prove our claim. $u^{(0)}$ satisfies
(2.26) naturally. Assume (2.26) is true for $n$. We now prove that it also
holds true for $n+1$.

$\displaystyle\|u^{(n+1)}(t)\|_{B_{p,r}^{s}}$ $\displaystyle\leq$
$\displaystyle\displaystyle\frac{1}{\left(1-2C\|u_{0}\|_{B_{p,r}^{s}}t\right)^{1/2}}\times\left(\|u_{0}\|_{B_{p,r}^{s}}+C\displaystyle\int_{0}^{t}\displaystyle\frac{\|u_{0}\|_{B_{p,r}^{s}}^{2}}{\left(1-2C\|u_{0}\|_{B_{p,r}^{s}}\tau\right)^{3/2}}d\tau\right)$
$\displaystyle\leq$
$\displaystyle\displaystyle\frac{1}{\left(1-2C\|u_{0}\|_{B_{p,r}^{s}}t\right)^{1/2}}\times\left(\|u_{0}\|_{B_{p,r}^{s}}+\displaystyle\int_{0}^{t}\displaystyle\frac{\|u_{0}\|_{B_{p,r}^{s}}}{2\left(1-2C\|u_{0}\|_{B_{p,r}^{s}}\tau\right)^{3/2}}d\left(1-2C\|u_{0}\|_{B_{p,r}^{s}}\tau\right)\right)$
$\displaystyle\leq$
$\displaystyle\displaystyle\frac{\|u_{0}\|_{B_{p,r}^{s}}}{1-2C\|u_{0}\|_{B_{p,r}^{s}}t}.$

Therefore, (2.26) is true and $(u^{(n)})_{n\in\mathbb{N}}$ is uniformly
bounded in $C([0,T];B_{p,r}^{s})$. It is easy to check that
$u^{(n)}u^{(n)}_{x}$ is uniformly bounded in $C([0,T];B_{p,r}^{s-1})$ and
$F(u^{(n)})$ is uniformly bounded in $C([0,T];B_{p,r}^{s-1})$. We conclude
that the sequence $(u^{(n)})_{n\in\mathbb{N}}$ is uniformly bounded in
$C([0,T];B_{p,r}^{s})\displaystyle\cap C^{1}([0,T];B_{p,r}^{s-1})$.

Now, we turn to show $(u^{(n)})_{n\in\mathbb{N}}$ is a Cauchy sequence in
$C([0,T];B_{p,r}^{s-1})$. For $m,n\in\mathbb{N}$, from (2.23), we have

$\displaystyle\partial_{t}(u^{(n+m+1)}-u^{(n+1)})+u^{(n)}\partial_{x}(u^{(n+m+1)}-u^{(n+1)})=-(u^{(n+m)}-u^{(n)})\partial_{x}u^{(n+m+1)}-(F(u^{n+m})-F(u^{n})).$

Making use of Lemma A.2, we have

$\displaystyle\|(u^{(n+m+1)}-u^{(n+1)})(t)\|_{B_{p,r}^{s-1}}$
$\displaystyle\leq$ $\displaystyle
e^{C\displaystyle\int_{0}^{t}\|\partial_{x}u^{(n)}(\tau)\|_{B_{p,r}^{s-2}}d\tau}\times\left(\|u_{0}^{(n+m+1)}-u_{0}^{(n+1)}\|_{B_{p,r}^{s-1}}+C\displaystyle\int_{0}^{t}e^{-C\displaystyle\int_{0}^{\tau}\|\partial_{x}u^{(n)}(s)\|_{B_{p,r}^{s-2}}ds}\|\tilde{F}\|_{B_{p,r}^{s-1}}d\tau\right),$

where
$\tilde{F}=-(u^{(n+m)}-u^{(n)})\partial_{x}u^{(n+m+1)}-(F(u^{n+m})-F(u^{n}))$.

Then, we estimate $\|\tilde{F}\|_{B_{p,r}^{s-1}}$ as

$\displaystyle\|\tilde{F}\|_{B_{p,r}^{s-1}}$ $\displaystyle\leq$
$\displaystyle\|(u^{(n)}-u^{(n+m)})\partial_{x}u^{(n+m+1)}\|_{B_{p,r}^{s-1}}+\|F(u^{n+m})-F(u^{n})\|_{B_{p,r}^{s-1}}.$

By using the embedding $B_{p,r}^{s-1}\hookrightarrow L^{\infty}$ with
$s>1+\displaystyle\frac{1}{p}$, we have

$\displaystyle\|(u^{(n+m)}-u^{(n)})\partial_{x}u^{(n+m+1)}\|_{B_{p,r}^{s-1}}\leq\|(u^{(n+m)}-u^{(n)})\|_{B_{p,r}^{s-1}}\|u^{(n+m+1)}\|_{B_{p,r}^{s}}.$

Direct calculation, we obtain

$\displaystyle F(u^{n+m})-F(u^{n})$ $\displaystyle=$
$\displaystyle\displaystyle\frac{1}{2}(u^{(n+m)}-u^{(n)})(u^{(n+m)}+u^{(n)})$
$\displaystyle+P(D)\left(\displaystyle\frac{1}{2}(u^{(n+m)}-u^{(n)})(u^{(n+m)}+u^{(n)})+(u_{x}^{(n+m)}-u_{x}^{(n)})(u_{x}^{(n+m)}+u_{x}^{(n)})\right)$
$\displaystyle+\partial_{x}P(D)\left(\displaystyle\frac{1}{2}(u^{(n+m)}-u^{(n)})(u^{(n+m)}+u^{(n)})+(u_{x}^{(n+m)}-u_{x}^{(n)})(u_{x}^{(n+m)}+u_{x}^{(n)})\right).$

If
$s>\displaystyle\max\\{1+\displaystyle\frac{1}{p},\displaystyle\frac{3}{2}\\}$,
then

$\displaystyle\|F(u^{n+m})-F(u^{n})\|_{B_{p,r}^{s-1}}\leq
C\|u^{(n+m)}-u^{(n)}\|_{B_{p,r}^{s-1}}(\|u^{(n+m)}\|_{B_{p,r}^{s}}+\|u^{(n)}\|_{B_{p,r}^{s}}).$

Hence,

$\displaystyle\|\tilde{F}\|_{B_{p,r}^{s-1}}\leq
C\|u^{(n+m)}-u^{(n)}\|_{B_{p,r}^{s-1}}(\|u^{(n+m)}\|_{B_{p,r}^{s}}+\|u^{(n)}\|_{B_{p,r}^{s}}+\|u^{(n+m+1)}\|_{B_{p,r}^{s}}).$

We also have

$\displaystyle\|u_{0}^{(n+m+1)}-u_{0}^{(n+1)}\|_{B_{p,r}^{s-1}}$
$\displaystyle=$
$\displaystyle\|S_{n+m+1}u_{0}-S_{n+1}u_{0}\|_{B_{p,r}^{s-1}}$ (2.28)
$\displaystyle=$
$\displaystyle\left\|\displaystyle\sum_{q=n+1}^{n+m}\Delta_{q}u_{0}\right\|_{B_{p,r}^{s-1}}$
$\displaystyle=$
$\displaystyle\left(\displaystyle\sum_{k\geq-1}2^{rk(s-1)}\|\Delta_{k}\displaystyle\sum_{q=n+1}^{n+m}\Delta_{q}u_{0}\|_{L^{p}}^{r}\right)^{\displaystyle\frac{1}{r}}$
$\displaystyle\leq$ $\displaystyle\left(\displaystyle\sum_{n\leq k\leq
m+n+1}2^{rk(s-1)}(\|\Delta_{k}\Delta_{k-1}u_{0}\|_{L^{p}}+\|\Delta_{k}\Delta_{k+1}u_{0}\|_{L^{p}}+\|\Delta_{k}\Delta_{k}u_{0}\|_{L^{p}})^{r}\right)^{\displaystyle\frac{1}{r}}$
$\displaystyle\leq$ $\displaystyle C\left(2^{-rn}\displaystyle\sum_{n\leq
k\leq
m+n+1}2^{rks}\|\Delta_{k}u_{0}\|_{L^{p}}^{r}\right)^{\displaystyle\frac{1}{r}}$
$\displaystyle\leq$ $\displaystyle C2^{-n}\|u_{0}\|_{B_{p,r}^{s}}.$

Since $u^{(n)}\in C([0,T];B_{p,r}^{s})$, consequently, for all $t\in[0,T]$,

$\displaystyle\|(u^{(n+m+1)}-u^{(n+1)})(t)\|_{B_{p,r}^{s-1}}\leq
C\left(2^{-n}+\displaystyle\int_{0}^{t}\|(u^{(n+m)}-u^{(n)})\|_{B_{p,r}^{s-1}}d\tau\right).$

Arguing by induction with respect to the index $n$, one can easily prove that

$\displaystyle\|(u^{(n+m+1)}-u^{(n+1)})(t)\|_{B_{p,r}^{s-1}}\leq
2^{-n}\left(C\displaystyle\sum_{k=0}^{n}\displaystyle\frac{(2TC)^{k}}{k!}\right)+C\displaystyle\frac{(TC)^{n+1}}{(n+1)!}.$

Hence, $(u^{(n)})_{n\in\mathbb{N}}$ is a Cauchy sequence in
$C([0,T];B_{p,r}^{s-1})$. $\Box$
_Proof of Theorem 2.1_ From Lemma 2.1, $(u^{(n)})_{n\in\mathbb{N}}$ is a
Cauchy sequence in $C([0,T];B_{p,r}^{s-1})$. Then,
$(u^{(n)})_{n\in\mathbb{N}}$ converges to a function $u\in
C([0,T];B_{p,r}^{s-1})$. Indeed, by using the locally compact embedding
$B_{p,r}^{s}\hookrightarrow B_{p,r}^{s-1}$, the Arzela-Ascoli theorem and
Cantor’s diagonal process, we could extract a subsequence of
$(u^{(n)})_{n\in\mathbb{N}}$ (we still use $(u^{(n)})_{n\in\mathbb{N}}$ to
denote it.) and there exists $u\in Lip([0,T],B_{p,r}^{s-1})$, such that, for
any $\phi\in C_{0}^{\infty}(\mathbb{R})$

$\displaystyle\displaystyle\lim_{n\to\infty}\|u^{(n)}(t)\phi-u(t)\phi\|_{B_{p,r}^{s-1}}=0,$

uniformly for $t\in[0,T]$. On the other hand, since
$(u^{(n)})_{n\in\mathbb{N}}$ is uniformly bounded in
$L^{\infty}([0,T];B_{p,r}^{s})$, it follows from the Fatou property for the
Besov spaces(Lemma A.7) that $u\in L^{\infty}([0,T];B_{p,r}^{s})$. By
interpolation, we could deduce for any $s^{\prime}<s$

$\displaystyle\displaystyle\lim_{n\to\infty}\displaystyle\max_{t\in[0,T]}\|u^{(n)}(t)\phi-u(t)\phi\|_{B_{p,r}^{s^{\prime}}}=0.$

Passing to the limit in (2.23) in the sense of distributions, we see that $u$
is a solution to (2.16) in the sense of distributions, one can easily get that
$u_{t}\in L^{\infty}([0,T];B_{p,r}^{s-1})$. Lemma A.3 guarantees that $u\in
C([0,T];B_{p,r}^{s})$. From the equation (2.16), one can also check $u_{t}\in
C([0,T];B_{p,r}^{s-1})$. Thus, $u\in E_{p,r}^{s}(T)$. Uniqueness and
continuity with respect to the initial data are an immediate consequence of
Proposition 2.1.

###  2.2 Local well-posedness in the critical Besov space

This section will deal with the local existence result in the critical Besov
space $B_{2,1}^{3/2}$, which is given in the following theorem.

###### Theorem 2.2

Let $u_{0}\in B_{2,1}^{3/2}$, there exists a time $T>0$ such that the Cauchy
problem (1.1) has a unique solution $u\in E_{2,1}^{3/2}(T)$. The map
$u_{0}\mapsto u$ is continuous from a neighborhood of $u_{0}$ in
$B_{2,1}^{3/2}$ into $C([0,T];B_{2,1}^{3/2})\displaystyle\cap
C^{1}([0,T];B_{2,1}^{1/2}).$

Proof : Let $(u^{(n)})_{n\in\mathbb{N}}$ be given by (2.23). Since the proof
is very long, we split it into four steps.

Step 1. By replacing $B_{p,r}^{s}$ with $B_{2,1}^{3/2}$ in Lemma 2.1, we could
prove that there exits a time $T>0$, such that the sequence
$(u^{(n)})_{n\in\mathbb{N}}$ is uniformly bounded in
$C([0,T];B_{2,1}^{3/2})\displaystyle\cap C^{1}([0,T];B_{2,1}^{1/2})$.

Step 2. In this step, we prove $(u^{(n)})_{n\in\mathbb{N}}$ is a Cauchy
sequence in $C([0,T];B_{2,1}^{1/2})$.

From (2.1) and Lemma A.2, we have

$\displaystyle\|(u^{(n+m+1)}-u^{(n+1)})(t)\|_{B_{2,\infty}^{1/2}}$
$\displaystyle\leq$ $\displaystyle
e^{C\displaystyle\int_{0}^{t}\|u^{(n)}(\tau)\|_{B_{2,\infty}^{3/2}}d\tau}\times\left(\|u_{0}^{(n+m+1)}-u_{0}^{(n+1)}\|_{B_{2,\infty}^{1/2}}+C\displaystyle\int_{0}^{t}e^{-C\displaystyle\int_{0}^{\tau}\|u^{(n)}(s)\|_{B_{2,\infty}^{3/2}}ds}\|\tilde{F}\|_{B_{2,\infty}^{1/2}}d\tau\right),$

where
$\tilde{F}=-(u^{(n+m)}-u^{(n)})\partial_{x}u^{(n+m+1)}-(F(u^{n+m})-F(u^{n}))$.
Since $(u^{(n)})_{n\in\mathbb{N}}$ is uniformly bounded in
$C([0,T];B_{2,1}^{3/2})$, there exists an $M>0$ such that
$\|u^{(n)}(t)\|_{B_{2,\infty}^{3/2}}\leq M$ for $t\in[0,T]$, which implies

$\displaystyle
e^{C\displaystyle\int_{0}^{t}\|u^{(n)}(\tau)\|_{B_{2,\infty}^{3/2}}d\tau}\leq
e^{CMT}:=M_{1}.$

From Lemma A.4 and A.5, we obtain

$\displaystyle\|\tilde{F}\|_{B_{2,\infty}^{1/2}}$ $\displaystyle\leq$
$\displaystyle\|(u^{(n+m)}-u^{(n)})\partial_{x}u^{(n+m+1)}\|_{B_{2,\infty}^{1/2}}+\|(F(u^{n+m})-F(u^{n}))\|_{B_{2,\infty}^{1/2}}$
$\displaystyle\leq$ $\displaystyle
C\|(u^{(n+m)}-u^{(n)})\|_{B_{2,1}^{1/2}}(\|u^{(n+m+1)}\|_{B_{2,1}^{3/2}}+\|u^{(n)}\|_{B_{2,1}^{3/2}}+\|u^{(n+m)}\|_{B_{2,1}^{3/2}})$
$\displaystyle\leq$ $\displaystyle
C\|(u^{(n+m)}-u^{(n)})\|_{B_{2,\infty}^{1/2}}\ln\left(e+\displaystyle\frac{\|(u^{(n+m)}-u^{(n)})\|_{B_{2,\infty}^{3/2}}}{\|(u^{(n+m)}-u^{(n)})\|_{B_{2,\infty}^{1/2}}}\right).$

Consequently, for all $t\in[0,T]$,

$\displaystyle\|(u^{(n+m+1)}-u^{(n+1)})(t)\|_{B_{2,\infty}^{1/2}}$
$\displaystyle\leq$ $\displaystyle
M_{1}\left(\|u_{0}^{(n+m+1)}-u_{0}^{(n+1)}\|_{B_{2,\infty}^{1/2}}+C\displaystyle\int_{0}^{t}{\|(u^{(n+m)}-u^{(n)})\|_{B_{2,\infty}^{1/2}}}\ln\left(e+\displaystyle\frac{\|(u^{(n+m)}-u^{(n)})\|_{B_{2,\infty}^{3/2}}}{\|(u^{(n+m)}-u^{(n)})\|_{B_{2,\infty}^{1/2}}}\right)d\tau\right).$

By using similar calculation as (2.28) and Lemma A.1
($B_{2,1}^{3/2}\hookrightarrow B_{2,\infty}^{3/2}$)

$\displaystyle\|u_{0}^{(n+m+1)}-u_{0}^{(n+1)}\|_{B_{2,\infty}^{1/2}}=\|S_{n+m+1}u_{0}-S_{n+1}u_{0}\|_{B_{2,\infty}^{1/2}}\leq
C2^{-n}\|u_{0}\|_{B_{2,1}^{3/2}}.$

Setting $\omega_{n,m}(t)=\|(u^{(n+m)}-u^{(n)})(t)\|_{B_{2,\infty}^{1/2}}$,
since $u^{(n)}\in C([0,T];B_{2,1}^{3/2})$ and the function
$x\ln(e+\displaystyle\frac{2M}{x})$ is nondecreasing, then we have

$\displaystyle\omega_{n+1,m}(t)\leq
M_{1}\left(2^{-n}+C\displaystyle\int_{0}^{t}\omega_{n,m}(\tau)\ln\left(e+\displaystyle\frac{2M}{\omega_{n,m}(\tau)}\right)d\tau\right).$

Setting $\omega_{n}(t)=\displaystyle\sup_{m\in\mathbb{N}}\omega_{n,m}(t)$ and
$\tilde{\omega}(t)=\limsup_{n\rightarrow\infty}\omega_{n}(t)$. Taking supremum
with respect to $m$ in $\mathbb{N}$, $\omega_{n}(t)$ satisfies

$\displaystyle\omega_{n+1}(t)\leq
CM_{1}\left(2^{-n}+\displaystyle\int_{0}^{t}\omega_{n}(\tau)\ln\left(e+\displaystyle\frac{2M}{\omega_{n}(\tau)}\right)d\tau\right).$

Which together with the Fatou–Lebesgue theorem implies

$\displaystyle\tilde{\omega}(t)\leq
CM_{1}\displaystyle\int_{0}^{t}\tilde{\omega}(\tau)\ln\left(e+\displaystyle\frac{2M}{\tilde{\omega}(\tau)}\right)d\tau.$

Making use of the Lemma A.6 with $a=0$ and
$\mu(r)=r\ln\left(e+\displaystyle\frac{2M}{r}\right)$, we have
$\tilde{\omega}(t)=0$, for $\forall t\in[0,T]$, that is to say
$(u^{(n)})_{n\in\mathbb{N}}$ is a Cauchy sequence in
$C([0,T];B_{2,\infty}^{1/2})$. By Lemma 2.1, we see

$\displaystyle\|(u^{(n+m)}-u^{(n)})\|_{B_{2,1}^{1/2}}$ $\displaystyle\leq$
$\displaystyle\|(u^{(n+m)}-u^{(n)})\|_{B_{2,1}^{1}}$ $\displaystyle\leq$
$\displaystyle
C\|(u^{(n+m)}-u^{(n)})\|_{B_{2,\infty}^{1/2}}^{1/2}\|(u^{(n+m)}-u^{(n)})\|_{B_{2,\infty}^{3/2}}^{1/2}$
$\displaystyle\leq$ $\displaystyle
C(2M)^{1/2}\|(u^{(n+m)}-u^{(n)})\|_{B_{2,\infty}^{1/2}}^{1/2}.$

Hence, $(u^{(n)})_{n\in\mathbb{N}}$ is a Cauchy sequence in
$C([0,T];B_{2,1}^{1/2})$.

Step 3. In this step we prove the existence and uniqueness of the solution.

We now prove the existence of solutions. From step 2,
$(u^{(n)})_{n\in\mathbb{N}}$ is a Cauchy sequence in $C([0,T];B_{2,1}^{1/2})$.
Then, $(u^{(n)})_{n\in\mathbb{N}}$ converges to a function $u\in
C([0,T];B_{2,1}^{1/2})$. By using the same method that is used in the proof of
Theorem 2.1 we could obtain $u\in E_{2,1}^{3/2}(T)$.

For the uniqueness, assume that $u_{1}$ and $u_{2}$ are two solutions with the
same initial datum $u_{0}$ to the system (1.15). If we set
$\varpi=u_{1}-u_{2}$, then we have $\varpi$ satisfies

$\displaystyle\partial_{t}(\varpi)+u_{2}\partial_{x}\varpi=-\varpi\partial_{x}u_{1}-(F(u_{1})-F(u_{2}))$

with initial data $\varpi(x,0)=0$. Using the similar method that is used in
the Step 2, we obtain

$\displaystyle\|\varpi(t)\|_{B_{2,1}^{1/2}}=0.$

This complete the proof of uniqueness.

Step 4.  Continuity with respect to the initial data in $B_{2,1}^{3/2}$ can be
obtained as in [15], we omit the details in this paper.

$\Box$

##  3 Ill-posedness in $B_{2,\infty}^{3/2}$

###### Theorem 3.1

When $u_{0}\in{B_{2,\infty}^{3/2}}$, the Cauchy problem (1.15) is ill-
posedness. More precisely, there is a solution $u\in
L^{\infty}([0,\infty];{B_{2,\infty}^{3/2}})$ of (1.15) such that for any $T$,
$\varepsilon>0$, there exists a solution $v\in
L^{\infty}([0,\infty];{B_{2,\infty}^{3/2}})$ with

$\displaystyle\|v(0)-u(0)\|_{B_{2,\infty}^{3/2}}\leq\varepsilon\quad\emph{and}\quad\|v(t)-u(t)\|_{L([0,T];B_{2,\infty}^{3/2})}\geq
1.$

###### Remark 3.1

The proof of ill-posedness of CH equation is based on the solitary wave
solution $u_{c}=ce^{-|x-ct|}$. However, such special solution is not a weak
solution of (1.15). Our proof relies on the single rogue peakon (1.4) which is
a solution in the sense of distribution. Moreover, the result of Theorem 3.1
also indicates that $B_{2,1}^{3/2}$ is a critical space to consider the local
well-posedness.

Proof : Let $A=\displaystyle\frac{1}{c}$ and $B=\ln|c|$ with $c\neq 0$ in
(1.4), we define $u_{c}(x,t):=\displaystyle\frac{c}{ct-1}e^{-|x-\ln|ct-1||}$,
$c\neq 0$, $t\geq 0$. Its Fourier transform in $x$ is

$\hat{u}_{c}(x,t):=\displaystyle\frac{2c}{ct-1}\displaystyle\frac{e^{-i\xi\ln|ct-1|}}{1+\xi^{2}}.$

When $p=2$, and $r\in[0,+\infty]$, the Definition A.1 is equivalent to

$\displaystyle\|u\|_{B_{2,r}^{s}}=\left\\{\begin{array}[]{ll}\left[\left(\displaystyle\int_{-1}^{1}(1+\xi^{2})^{s}|\hat{u}(\xi)|^{2}d\xi\right)^{\displaystyle\frac{r}{2}}+\displaystyle\sum_{q\in\mathbb{N}}\left(\displaystyle\int_{2^{q}\leq|\xi|\leq
2^{q+1}}(1+\xi^{2})^{s}|\hat{u}(\xi)|^{2}d\xi\right)^{\displaystyle\frac{r}{2}}\right]^{\displaystyle\frac{1}{r}},&r\leq\infty,\\\
\displaystyle\max\left\\{\left(\displaystyle\int_{-1}^{1}(1+\xi^{2})^{s}|\hat{u}(\xi)|^{2}d\xi\right)^{\displaystyle\frac{1}{2}},\,\displaystyle\sup_{q\in\mathbb{N}}\left(\displaystyle\int_{2^{q}\leq|\xi|\leq
2^{q+1}}(1+\xi^{2})^{s}|\hat{u}(\xi)|^{2}d\xi\right)^{\displaystyle\frac{1}{2}}\right\\},&r=\infty.\end{array}\right.$

For the initial data, we compute
$\|u_{c_{2}}(0)-u_{c_{1}}(0)\|_{B_{2,\infty}^{3/2}}$ according to the above
Definition.

$\displaystyle\|u_{c_{2}}(0)-u_{c_{1}}(0)\|_{B_{2,\infty}^{3/2}}^{2}$
$\displaystyle=$ $\displaystyle
8(c_{2}-c_{1})^{2}\displaystyle\max\left\\{\displaystyle\int_{0}^{1}\displaystyle\frac{1}{\sqrt{1+\xi^{2}}}d\xi,\displaystyle\sup_{q\in\mathbb{N}}\displaystyle\int_{2^{q}}^{2^{q+1}}\displaystyle\frac{1}{\sqrt{1+\xi^{2}}}d\xi\right\\}$
(3.30) $\displaystyle=$ $\displaystyle 8(c_{2}-c_{1})^{2}\ln(1+\sqrt{2}).$

Direct calculation, we have

$\displaystyle|\hat{u}_{c_{2}}(t)-\hat{u}_{c_{1}}(t)|^{2}=4\left|\displaystyle\frac{c_{2}}{c_{2}t-1}e^{-i\xi\ln|c_{2}t-1|}-\displaystyle\frac{c_{1}}{c_{1}t-1}e^{-i\xi\ln|c_{1}t-1|}\right|^{2}$
(3.31) $\displaystyle=$ $\displaystyle
4\left(\displaystyle\frac{c_{2}}{c_{2}t-1}e^{-i\xi\ln|c_{2}t-1|}-\displaystyle\frac{c_{1}}{c_{1}t-1}e^{-i\xi\ln|c_{1}t-1|}\right)\left(\displaystyle\frac{c_{2}}{c_{2}t-1}e^{i\xi\ln|c_{2}t-1|}-\displaystyle\frac{c_{1}}{c_{1}t-1}e^{i\xi\ln|c_{1}t-1|}\right)$
$\displaystyle=$ $\displaystyle
4\left(\displaystyle\frac{c_{2}}{c_{2}t-1}-\displaystyle\frac{c_{1}}{c_{1}t-1}\right)^{2}+\displaystyle\frac{8c_{1}c_{2}}{(c_{2}t-1)(c_{1}t-1)}\left(1-\cos((\ln|c_{2}t-1|-\ln|c_{1}t-1|)\xi)\right),$

For $T>0$, we choose $c_{1}=\displaystyle\frac{1}{1+T}$,
$c_{2}=\displaystyle\frac{1+T-e^{2^{-q}\pi}}{(1+T)T}$ and $q$ large enough,
then $c_{1}c_{2}>0$ and

$\displaystyle\ln|c_{2}T-1|-\ln|c_{1}T-1|=2^{-q}\pi.$ (3.32)

Substituting (3.31) and (3.32) into above definition of $B_{2,\infty}^{3/2}$,
we have

$\displaystyle\|u_{c_{2}}(T)-u_{c_{1}}(T)\|_{B_{2,\infty}^{3/2}}$
$\displaystyle\geq$
$\displaystyle\displaystyle\frac{16c_{1}c_{2}}{(c_{2}T-1)(c_{1}T-1)}\displaystyle\int_{2^{q}}^{2^{q+1}}\displaystyle\frac{1-\cos(2^{-q}\pi\xi)}{\sqrt{1+|\xi|^{2}}}d\xi$
$\displaystyle\geq$
$\displaystyle\displaystyle\frac{4c_{1}c_{2}}{\sqrt{2}(c_{2}T-1)(c_{1}T-1)},$

where we have used

$\displaystyle\displaystyle\int_{2^{q}}^{2^{q+1}}\displaystyle\frac{\cos(2^{-q}\pi\xi)}{\sqrt{1+|\xi|^{2}}}d\xi$
$\displaystyle=$
$\displaystyle\displaystyle\int_{\pi}^{2\pi}\displaystyle\frac{\cos
t}{\sqrt{\pi^{2}+|2^{q}t|^{2}}}dt$ $\displaystyle=$
$\displaystyle\displaystyle\int_{\pi}^{\displaystyle\frac{3\pi}{2}}\displaystyle\frac{\cos
t}{\sqrt{\pi^{2}+|2^{q}t|^{2}}}dt+\displaystyle\int_{\displaystyle\frac{3\pi}{2}}^{2\pi}\displaystyle\frac{\cos
t}{\sqrt{\pi^{2}+|2^{q}t|^{2}}}dt$ $\displaystyle=$
$\displaystyle-\displaystyle\int_{0}^{\displaystyle\frac{\pi}{2}}\displaystyle\frac{\cos\theta}{\sqrt{\pi^{2}+|2^{q}(\theta+\pi)|^{2}}}d\theta+\displaystyle\int_{0}^{\displaystyle\frac{\pi}{2}}\displaystyle\frac{\cos\theta}{\sqrt{\pi^{2}+|2^{q}(2\pi-\theta)|^{2}}}d\theta$
$\displaystyle=$
$\displaystyle\displaystyle\int_{0}^{\displaystyle\frac{\pi}{2}}[\displaystyle\frac{1}{\sqrt{\pi^{2}+|2^{q}(2\pi-\theta)|^{2}}}-\displaystyle\frac{1}{\sqrt{\pi^{2}+|2^{q}(\theta+\pi)|^{2}}}]\cos\theta
d\theta$ $\displaystyle\leq$ $\displaystyle 0,$

and

$\displaystyle\displaystyle\int_{2^{q}}^{2^{q+1}}\displaystyle\frac{1}{\sqrt{1+|\xi|^{2}}}d\xi\geq\displaystyle\frac{2^{q}}{\sqrt{1+2^{2(q+1)}}}\geq\displaystyle\frac{1}{4\sqrt{2}}.$

From (3.30), we have

$\displaystyle\|u_{c_{2}}(0)-u_{c_{1}}(0)\|_{B_{2,\infty}^{3/2}}$
$\displaystyle=$ $\displaystyle
8\left(\displaystyle\frac{1+T-e^{2^{-q}\pi}}{(1+T)T}-\displaystyle\frac{1}{1+T}\right)^{2}\ln(1+\sqrt{2})$
$\displaystyle=$ $\displaystyle
8\left(\displaystyle\frac{1-e^{2^{-q}\pi}}{(1+T)T}\right)^{2}\ln(1+\sqrt{2}),$

which implies $\|u_{c_{2}}(0)-u_{c_{1}}(0)\|_{B_{2,\infty}^{3/2}}$ may be
arbitrary small for $q$ large enough, however,

$\displaystyle\|u_{c_{2}}(T)-u_{c_{1}}(T)\|_{B_{2,\infty}^{3/2}}\geq 2.$

$\Box$

##  4 Global existence and blow-up phenomenon

Suppose $u(x,t)$ is a solution to the Cauchy problem (1.15) in its lifespan.
Let us now consider the following differential equation:

$\displaystyle\left\\{\begin{array}[]{ll}\displaystyle\frac{dq(x,t)}{dt}=u(q,t),\qquad
0<t<T,x\in\mathbb{R},\\\ q(x,0)=x,\qquad x\in\mathbb{R},\end{array}\right.$
(4.35)

where $T$ is the lifespan of the solution. Taking derivative (4.35) with
respect to $x$, we obtain

$\displaystyle\frac{dq_{t}}{dx}=q_{tx}=u_{x}(q,t)q_{x},\qquad t\in(0,T).$

Therefore

$\displaystyle\left\\{\begin{array}[]{ll}q_{x}=\exp\left(\displaystyle\int^{t}_{0}(u_{x}(q,s))ds\right),\qquad
0<t<T,x\in\mathbb{R},\\\ q_{x}(x,0)=1,\qquad
x\in\mathbb{R},\end{array}\right.$

which is always positive before the blow-up time. Therefore, the function
$q(x,t)$ is an increasing diffeomorphism of the line before blow-up. In fact,
direct calculation yields

$\displaystyle\displaystyle\frac{d}{dt}(e^{q(x,t)}m(q,t)q_{x})=e^{q(x,t)}[m_{t}(q,t)+m(q,t)u(q,t)+(m(q,t)u(q,t))_{x}]q_{x}=0.$

Hence, the following identity can be proved:

$\displaystyle e^{q(x,t)}m(q,t)q_{x}=e^{x}m_{0}(x),$ (4.37)

which implies that $m(x,t)$ keeps the sign with respect to the initial datum.

By (4.37), we have

$\displaystyle\int_{\mathbb{R}}e^{q(x,t)}m(q,t)q_{x}dx=\displaystyle\int_{\mathbb{R}}e^{q(x,t)}m(q,t)dq=\displaystyle\int_{\mathbb{R}}e^{x}m(x,t)dx.$

It follows a conserved quantity as

$\displaystyle\int_{\mathbb{R}}e^{x}m(x,t)dx=\displaystyle\int_{\mathbb{R}}e^{x}m_{0}dx.$

We establish the blow-up scenario for the Cauchy problem (1.15) in the
following Theorem.

###### Theorem 4.1

Let $u_{0}(x)\in H^{s}(\mathbb{R})$, $s>\displaystyle\frac{3}{2}$ and let $T$
be the maximal existence time of the solution $u(x,t)$ to (1.15) with the
initial data $u_{0}(x)$. Then the corresponding solution blows up in finite
time if and only if

$\displaystyle\liminf_{t\rightarrow
T-}\displaystyle\inf_{x\in\mathbb{R}}\left(u+\displaystyle\frac{1}{2}u_{x}\right)=-\infty.$

Proof : On one hand, suppose for any $t\in(0,T]$,
$\displaystyle\inf_{x\in\mathbb{R}}\left(u+\displaystyle\frac{1}{2}u_{x}\right)\geq-M$
with $M>0$. Multiplying (1.15) by $m$, after integrating by parts, we have

$\displaystyle\displaystyle\frac{1}{2}\displaystyle\frac{d}{dt}\|m\|_{L^{2}}^{2}$
$\displaystyle=$
$\displaystyle-\displaystyle\int_{\mathbb{R}}um^{2}+(um)_{x}m)dx$
$\displaystyle=$
$\displaystyle-\displaystyle\int_{\mathbb{R}}\left(u+\displaystyle\frac{1}{2}u_{x}\right)m^{2}dx.$

By using Gronwall’s inequality, we have

$\displaystyle\|m(t)\|_{L^{2}}^{2}\leq\|m_{0}\|_{L^{2}}^{2}e^{2MT},$

then $\|u(t)\|_{H^{2}}^{2}\leq\|m(t)\|_{L^{2}}^{2}$ is bounded. This
contradicts with the fact that $T$ is the maximal time of existence.

On the other hand, the solution does not blow up in
$H^{s},s>\displaystyle\frac{3}{2}$, that is $\|u\|_{H^{s}}$ is bounded, by
Morrey’s inequality, we have

$\displaystyle\|u+\displaystyle\frac{1}{2}u_{x}\|_{L^{\infty}}\leq
C\|u\|_{H^{s}}<+\infty.$

This completes the proof of Theorem 4.1. $\Box$

The operator $(1-\partial_{x}^{2})^{-1}$ can be expressed by its associated
Green’s function as

$u(x,t)=(1-\partial_{x}^{2})^{-1}m(t,x)=G*m,\quad
G=\displaystyle\frac{1}{2}e^{-|x|}.$

More precisely,

$\displaystyle u(x,t)=G\ast
m(x,t)=\displaystyle\frac{1}{2}e^{-x}\displaystyle\int_{-\infty}^{x}e^{\xi}m(\xi,t)d\xi+\displaystyle\frac{1}{2}e^{x}\displaystyle\int_{x}^{\infty}e^{-\xi}m(\xi,t)d\xi.$
(4.38) $\displaystyle
u_{x}(x,t)=-\displaystyle\frac{1}{2}e^{-x}\displaystyle\int_{-\infty}^{x}e^{\xi}m(\xi,t)d\xi+\displaystyle\frac{1}{2}e^{x}\displaystyle\int_{x}^{\infty}e^{-\xi}m(\xi,t)d\xi.$
(4.39)

###### Remark 4.1

We know that for CH type equations, when $m_{0}(x)$ does not change sign, the
solution usually exists globally. But for our system, we find a new a new
phenomenon, which is if $m_{0}(x)\geq(\not\equiv)0$, then the corresponding
solution $u(x,t)$ exists globally, while, if $m_{0}(x)\leq(\not\equiv)0$, then
the corresponding solution $u(x,t)$ blows up in finite time.

###### Theorem 4.2

Suppose that $u_{0}(x)\in H^{s}(\mathbb{R})$, $s>\displaystyle\frac{3}{2}$ ,
$m_{0}(x)\geq(\not\equiv)0$. Then the corresponding solution $u(x,t)$ to the
Cauchy problem (1.15) with $u_{0}$ as the initial datum exists globally.

Proof : By the blow-up scenario, we only need to show
$u+\displaystyle\frac{1}{2}u_{x}$ has a lower bound. From the initial data
condition $m_{0}(x)\geq(\not\equiv)0$ and (4.37), we have

$\displaystyle m(x,t)\geq(\not\equiv)0.$

By using (4.38) and (4.39), we have

$\displaystyle
u+\displaystyle\frac{1}{2}u_{x}=\displaystyle\frac{1}{4}e^{-x}\displaystyle\int_{-\infty}^{x}e^{\xi}m(\xi,t)d\xi+\displaystyle\frac{3}{4}e^{x}\displaystyle\int_{x}^{\infty}e^{-\xi}m(\xi,t)d\xi>0.$

$\Box$

In the remainder of this section, we would like to establish a sufficient
condition for the blow-up of solutions, which is given in the following
Theorem.

###### Theorem 4.3

Suppose that $u_{0}(x)\in H^{s}(\mathbb{R})$, $s>\displaystyle\frac{3}{2}$ ,
$m_{0}(x)\leq(\not\equiv)0$. Then, the corresponding solution $u(x,t)$ to the
Cauchy problem (1.15) with $u_{0}$ as the initial datum blows up in finite
time.

Proof : From the condition $m_{0}(x)\leq(\not\equiv)0$, we have, there exist a
point $x_{0}\in\mathbb{R}$, such that

$\displaystyle(u_{0}+\displaystyle\frac{1}{2}u_{0x})(x_{0})=\displaystyle\frac{1}{4}e^{-x_{0}}\displaystyle\int_{-\infty}^{x_{0}}e^{\xi}m_{0}(\xi)d\xi+\displaystyle\frac{3}{4}e^{x_{0}}\displaystyle\int_{x_{0}}^{\infty}e^{-\xi}m_{0}(\xi)d\xi<0.$
(4.40)

Rewrite (2.16) as

$\displaystyle
u_{t}+uu_{x}+\displaystyle\frac{1}{2}u^{2}+G\ast\left(\displaystyle\frac{1}{2}u^{2}+u_{x}^{2}\right)+\partial_{x}G\ast\left(\displaystyle\frac{1}{2}u^{2}+u_{x}^{2}\right)=0.$
(4.41)

Differentiating (4.41) with respect to $x$, we obtain

$\displaystyle
u_{xt}+uu_{xx}+u_{x}^{2}+uu_{x}+\partial_{x}G\ast\left(\displaystyle\frac{1}{2}u^{2}+u_{x}^{2}\right)+\partial_{x}^{2}G\ast\left(\displaystyle\frac{1}{2}u^{2}+u_{x}^{2}\right)=0,$

it follows

$\displaystyle
u_{xt}+uu_{xx}-\displaystyle\frac{1}{2}u^{2}+uu_{x}+G\ast\left(\displaystyle\frac{1}{2}u^{2}+u_{x}^{2}\right)+\partial_{x}G\ast\left(\displaystyle\frac{1}{2}u^{2}+u_{x}^{2}\right)=0.$

Then, we have
$\displaystyle\frac{d\left(u+\displaystyle\frac{1}{2}u_{x}\right)}{dt}$ at the
point $q(x_{0},t)$.

$\displaystyle\displaystyle\frac{d\left(u+\displaystyle\frac{1}{2}u_{x}\right)(q(x_{0},t),t)}{dt}$
$\displaystyle=$ $\displaystyle
u_{t}(q(x_{0},t),t)+uu_{x}(q(x_{0},t),t)+\displaystyle\frac{1}{2}u_{xt}(q(x_{0},t),t)+\displaystyle\frac{1}{2}uu_{xx}(q(x_{0},t),t)$
$\displaystyle=$
$\displaystyle-\displaystyle\frac{1}{4}u^{2}(q(x_{0},t),t)-\displaystyle\frac{1}{2}uu_{x}(q(x_{0},t),t)-\displaystyle\frac{3}{2}G\ast\left(\displaystyle\frac{1}{2}u^{2}+u_{x}^{2}\right)(q(x_{0},t),t)-\displaystyle\frac{3}{2}\partial_{x}G\ast\left(\displaystyle\frac{1}{2}u^{2}+u_{x}^{2}\right)(q(x_{0},t),t).$

Note that,

$\displaystyle
G\ast\left(\displaystyle\frac{1}{2}u^{2}+u_{x}^{2}\right)(q(x_{0},t),t)$
(4.43) $\displaystyle=$
$\displaystyle\displaystyle\frac{1}{2}e^{-q(x_{0},t)}\displaystyle\int_{-\infty}^{q(x_{0},t)}e^{\xi}\left(\displaystyle\frac{1}{2}u^{2}+u_{\xi}^{2}\right)d\xi+\displaystyle\frac{1}{2}e^{q(x_{0},t)}\displaystyle\int_{q(x_{0},t)}^{\infty}e^{-\xi}\left(\displaystyle\frac{1}{2}u^{2}+u_{\xi}^{2}\right)d\xi$

and

$\displaystyle\partial_{x}G\ast\left(\displaystyle\frac{1}{2}u^{2}+u_{x}^{2}\right)(q(x_{0},t),t)$
(4.44) $\displaystyle=$
$\displaystyle-\displaystyle\frac{1}{2}e^{-q(x_{0},t)}\displaystyle\int_{-\infty}^{q(x_{0},t)}e^{\xi}\left(\displaystyle\frac{1}{2}u^{2}+u_{\xi}^{2}\right)d\xi+\displaystyle\frac{1}{2}e^{q(x_{0},t)}\displaystyle\int_{q(x_{0},t)}^{\infty}e^{-\xi}\left(\displaystyle\frac{1}{2}u^{2}+u_{\xi}^{2}\right)d\xi.$

Substituting (4.43) and (4.44) into (4), we obtain

$\displaystyle\displaystyle\frac{d\left(u+\displaystyle\frac{1}{2}u_{x}\right)(q(x_{0},t),t)}{dt}$
$\displaystyle=$
$\displaystyle-\displaystyle\frac{1}{4}u^{2}(q(x_{0},t),t)-\displaystyle\frac{1}{2}uu_{x}(q(x_{0},t),t)-\displaystyle\frac{3}{2}e^{q(x_{0},t)}\displaystyle\int_{q(x_{0},t)}^{\infty}e^{-\xi}\left(\displaystyle\frac{1}{2}u^{2}+u_{\xi}^{2}\right)d\xi$
(4.45) $\displaystyle\leq$
$\displaystyle-\displaystyle\frac{5}{8}u^{2}(q(x_{0},t),t)-\displaystyle\frac{1}{2}uu_{x}(q(x_{0},t),t)-\displaystyle\frac{9}{8}e^{q(x_{0},t)}\displaystyle\int_{q(x_{0},t)}^{\infty}e^{-\xi}u_{\xi}^{2}d\xi$
$\displaystyle\leq$
$\displaystyle-\displaystyle\frac{5}{8}u^{2}(q(x_{0},t),t)-\displaystyle\frac{1}{2}uu_{x}(q(x_{0},t),t),$

where, we have used

$\displaystyle\displaystyle\int_{x}^{\infty}e^{-\xi}\left(u^{2}+u_{x}^{2}\right)d\xi\geq\displaystyle\int_{x}^{\infty}e^{-\xi}\left(-2uu_{\xi}\right)d\xi\geq
e^{-x}u^{2}(x,t)-\displaystyle\int_{x}^{\infty}e^{-\xi}u^{2}d\xi$

yields

$\displaystyle
e^{x}\displaystyle\int_{x}^{\infty}e^{-\xi}\left(2u^{2}+u_{\xi}^{2}\right)d\xi\geq
u^{2}(x,t).$

Under the initial data condition $m_{0}(x,t)\leq(\not\equiv)0$ and (4.37), we
have

$\displaystyle m(x,t)\leq(\not\equiv)0,$

which implies that for any $x\in\mathbb{R}$,

$\displaystyle(u^{2}-u_{x}^{2})(x,t)=\displaystyle\int_{-\infty}^{x}e^{\xi}m(\xi,t)d\xi\displaystyle\int_{x}^{\infty}e^{-\xi}m(\xi,t)d\xi>0.$
(4.46)

Then, from (4.45) and (4.46), we have

$\displaystyle\displaystyle\frac{d\left(u+\displaystyle\frac{1}{2}u_{x}\right)(q(x_{0},t),t)}{dt}$
$\displaystyle\leq$
$\displaystyle-\displaystyle\frac{1}{2}\left(u+\displaystyle\frac{1}{2}u_{x}\right)^{2}(q(x_{0},t),t)-\displaystyle\frac{1}{8}u^{2}(q(x_{0},t),t)+\displaystyle\frac{1}{8}u_{x}^{2}(q(x_{0},t),t)$
$\displaystyle\leq$
$\displaystyle-\displaystyle\frac{1}{2}\left(u+\displaystyle\frac{1}{2}u_{x}\right)^{2}(q(x_{0},t),t).$

By (4.40) and using standard arguments for ordinary differential equations, it
is easy to conclude that there exists a finite time $T$, such that

$\displaystyle\displaystyle\lim_{t\rightarrow
T}\left(u+\displaystyle\frac{1}{2}u_{x}\right)(q(x_{0},t),t)=-\infty\quad
with\quad
T\leq\displaystyle\frac{-2}{(u_{0}+\displaystyle\frac{1}{2}u_{0x})(x_{0})}.$

This completes the proof of Theorem 4.3. $\Box$

##  5 Conclusion and open problem

In this paper, we develop a new type of soliton solution – called rogue peakon
in the integrable peakon theory. The rogue peakon takes on a rational form at
the peak of a logarithmic function, instead of a regular traveling wave, and
is found in the integrable Camassa-Holm (CH) type equation (1.1). We also
provide multi-rogue peakon solutions with some interesting interactional
dynamics as shown in Figures 2 - 6. Moreover, we discuss the local well-
posedness of the solution in the Besov space $B_{p,r}^{s}$ with $1\leq
p,r\leq\infty$, $s>\displaystyle\max\left\\{1+1/p,3/2\right\\}$ or
$B_{2,1}^{3/2}$, and then prove the ill-posedness of the solution in
$B_{2,\infty}^{3/2}$. It is worth to point out that the proof of ill-posedness
of relies on the single rogue peakon (1.4) which is a solution in the sense of
distribution. For the classical Camassa-Holm equation, Degasperis-Procesi
equation and Holm-Staley $b$-family equation, when $m_{0}(x)$ does not change
sign, the solution usually exists globally. But for our system, we find a new
phenomenon, which is if $m_{0}(x)\geq(\not\equiv)0$, then the corresponding
solution $u(x,t)$ exists globally, while if $m_{0}(x)\leq(\not\equiv)0$, then
the corresponding solution $u(x,t)$ blows up in a finite time. But the global
rogue peakon stability keeps open.

##  6 Acknowledgments

This work was partially supported by the National Natural Science Foundation
of China (Grant No.12071439 and Grant No. 11971475). The authors really very
much appreciate the fruitful discussions with Prof. Senyue Lou (NBU), Prof.
Hongyu Liu (CUHK), Prof. Jingsong He (SZU), Prof. Ruguang Zhou (JSNU), Prof.
Zuonong Zhu (SHJT), Prof. Yufeng Zhang (CUMT), and Prof. Jifeng Chu (SHNU).

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## Appendix

For consistence of our paper, we recall some basic results on the Littlewood-
Paley decomposition theory and Besov spaces which required in the proof of our
main theorems.

######  Proposition A.1 (Littlewood–Paley decomposition)

_[14, 16]_ Let
$\mathcal{B}:=\\{\xi\in\mathbb{R}^{d};|\xi|\leq\displaystyle\frac{4}{3}\\}$
and
$\mathcal{C}:=\\{\xi\in\mathbb{R}^{d};\displaystyle\frac{8}{3}\leq|\xi|\leq\displaystyle\frac{8}{3}\\}$.
Then there exist two smooth radial functions $\chi\in
C_{c}^{\infty}(\mathcal{B})$ and $\varphi\in C_{c}^{\infty}(\mathcal{C})$
valued in $[0,1]$, such that

$\displaystyle\chi(\xi)+\displaystyle\sum_{q\geq
0}\varphi(2^{-q}\xi)=1,\quad\forall\xi\in\mathbb{R}^{d},$
$\displaystyle\mathrm{supp}\varphi(2^{-q}\cdot)\displaystyle\cap\mathrm{supp}\varphi(2^{-q^{\prime}}\cdot)=\varnothing,\,\,if\,\,|q-q^{\prime}|\geq
2,$
$\displaystyle\mathrm{supp}\chi(\cdot)\displaystyle\cap\mathrm{supp}\varphi(2^{-q}\cdot)=\varnothing,\,\,if\,\,q\geq
1,$
$\displaystyle\displaystyle\frac{1}{2}\leq\chi^{2}(\xi)+\displaystyle\sum_{q\geq
0}\varphi^{2}(2^{-q}\xi)\leq 1,\quad\forall\xi\in\mathbb{R}^{d}.$

We denote the Fourier transform and its inverse by $\mathcal{F}$ and
$\mathcal{F}^{-1}$. Let $h:=\mathcal{F}^{-1}\varphi$ and
$\tilde{h}:=\mathcal{F}^{-1}\chi$. Then for all
$f\in\mathscr{S}^{\prime}(\mathbb{R}^{d})$, the dyadic operators $\Delta_{q}$
and $S_{q}$ can be defined as follows

$\displaystyle\Delta_{q}f:=\varphi(2^{-q}D)f=2^{qd}\displaystyle\int_{\mathbb{R}^{d}}h(2^{q}y)f(x-y)dy\quad
if\quad q\geq 0,$ $\displaystyle
S_{q}f:=\chi(2^{-q}D)f=\displaystyle\sum_{-1\leq k\leq
q-1}\Delta_{k}f=2^{qd}\displaystyle\int_{\mathbb{R}^{d}}\tilde{h}(2^{q}y)f(x-y)dy,$
$\displaystyle\Delta_{-1}f:=S_{0}f\quad and\quad\Delta_{q}f:=0\quad if\quad
q\leq-2.$

Hence,

$\displaystyle f=\displaystyle\sum_{q\in\mathbb{Z}}\Delta_{q}f$

holds in $\mathscr{S}^{\prime}(\mathbb{R}^{d})$ and is called the Littlewood-
Paley decomposition.

######  Definition A.1 (Besov spaces)

_[14, 16]_ Let $1\leq p,r\leq\infty$ and $s\in\mathbb{R}$. The nonhomogeneous
Besov space $B_{p,r}^{s}(\mathbb{R}^{d})$ is defined by

$\displaystyle
B_{p,r}^{s}(\mathbb{R}^{d}):=\\{f\in\mathscr{S}^{\prime}(\mathbb{R}^{d});\|f\|_{B_{p,r}^{s}}<\infty\\},$

where

$\displaystyle\|f\|_{B_{p,r}^{s}}:=\left\\{\begin{array}[]{ll}\left(\displaystyle\sum_{q\in\mathbb{Z}}2^{qsr}\|\Delta_{q}f\|_{L^{p}}^{r}\right)^{\displaystyle\frac{1}{r}},&for\quad
r<\infty,\\\
\displaystyle\sup_{q\in\mathbb{Z}}2^{qs}\|\Delta_{q}f\|_{L^{p}},&for\quad
r=\infty.\end{array}\right.$

If $s=\infty$,
$B_{p,r}^{\infty}:=\displaystyle\cap_{s\in\mathbb{R}}B_{p,r}^{s}$.

Then, we present some useful properties of $B_{p,r}^{s}$.

###### Lemma A.1

_[14, 16, 33]_ Let $s\in\mathbb{R}$ and $1\leq p,r,p_{j},r_{j}\leq\infty$,
$j=1,2$. Then the following properties hold:
_(1)_ $B_{p,r}^{s}(\mathbb{R})$ is a Banach space and is continuously embedded
in $\mathscr{S}^{\prime}(\mathbb{R})$.
_(2)_ $C_{c}^{\infty}(\mathbb{R})$ is dense in
$B_{p,r}^{s}(\mathbb{R})\Leftrightarrow 1\leq p,r<\infty$.
_(3)_ $B_{p_{1},r_{1}}^{s_{1}}(\mathbb{R})\hookrightarrow
B_{p_{2},r_{2}}^{s_{2}}(\mathbb{R})$, if $p_{1}\leq p_{2}$, $r_{1}\leq r_{2}$
and
$s_{2}=s_{1}-\left(\displaystyle\frac{1}{p_{1}}-\displaystyle\frac{1}{p_{2}}\right)$.

$B_{p,r_{1}}^{s_{1}}(\mathbb{R})\hookrightarrow
B_{p,r_{2}}^{s_{2}}(\mathbb{R})$, if $s_{2}<s_{1}$ or $s_{2}=s_{1}$,
$r_{1}\leq r_{2}$.

$B_{p_{1},r_{1}}^{s_{1}}(\mathbb{R})\hookrightarrow
B_{p_{2},r_{2}}^{s_{2}}(\mathbb{R})$ is locally compact if $s_{2}\leq s_{1}$;
$r_{1}\leq r_{2}$.
_(4)_ $\forall s>0$, $B_{p,r}^{s}(\mathbb{R})\displaystyle\cap
L^{\infty}(\mathbb{R})$ is a Banach algebra. Moreover,
$B_{p,r}^{s}(\mathbb{R})$ is a Banach algebra$\Leftrightarrow$
$B_{p,r}^{s}(\mathbb{R})\hookrightarrow L^{\infty}(\mathbb{R})$
$\Leftrightarrow$ $s>\displaystyle\frac{1}{p}$(or
$s\geq\displaystyle\frac{1}{p}$ and $r=1$).
_(5)_ $\forall$ $\theta\in[0,1]$, $s=\theta s_{1}+(1-\theta)s_{2}$,

$\|f\|_{B_{p,r}^{s}}\leq
C\|f\|_{B_{p,r}^{s_{1}}}^{\theta}\|f\|_{B_{p,r}^{s_{2}}}^{1-\theta},\quad\forall
f\in B_{p,r}^{s_{1}}\displaystyle\cap B_{p,r}^{s_{2}}.$

_(6)_ $\forall$ $\theta\in(0,1)$, $s_{1}>s_{2}$, $s=\theta
s_{1}+(1-\theta)s_{2}$,

$\|f\|_{B_{p,1}^{s}}\leq\displaystyle\frac{C(\theta)}{s_{1}-s_{2}}\|f\|_{B_{p,\infty}^{s_{1}}}^{\theta}\|f\|_{B_{p,\infty}^{s_{2}}}^{1-\theta},\quad\forall
f\in B_{p,\infty}^{s_{1}}.$

###### Lemma A.2

_[1, 14, 15]_Let $1\leq p,r\leq\infty$ and
$s>-\displaystyle\min\left\\{\displaystyle\frac{1}{p},1-\displaystyle\frac{1}{p}\right\\}$.
Assume that $f_{0}\in B_{p,r}^{s}$, $F\in L^{1}([0,T];B_{p,r}^{s})$ and
$\partial_{x}v\in L^{1}([0,T];B_{p,r}^{s-1})$ if
$s>1+\displaystyle\frac{1}{p}$ or $\partial_{x}v\in
L^{1}([0,T];B_{p,r}^{1/p}\displaystyle\cap L^{\infty})$ otherwise. If $f\in
L^{\infty}([0,T];B_{p,r}^{s})\displaystyle\cap C([0,T];\mathscr{S}^{\prime})$
solves the following 1D linear transport equation:

$\displaystyle\left\\{\begin{array}[]{ll}f_{t}+vf_{x}=F,t>0,x\in\mathbb{R},\\\
f(0,x)=f_{0},x\in\mathbb{R},\end{array}\right.$ (.50)

then there exists a constant C depending only on $s,p,r$ such that the
following statements hold:
_(1)_ For all $t\in[0,T]$,

$\displaystyle\|f\|_{B_{p,r}^{s}}\leq\|f_{0}\|_{B_{p,r}^{s}}+\displaystyle\int_{0}^{t}\|F(\tau)\|_{B_{p,r}^{s}}d\tau+C\displaystyle\int_{0}^{t}V^{\prime}(\tau)\|f(\tau)\|_{B_{p,r}^{s}}d\tau,$

and hence,

$\displaystyle\|f\|_{B_{p,r}^{s}}\leq
e^{CV(t)}\left(\|f_{0}\|_{B_{p,r}^{s}}+\displaystyle\int_{0}^{t}e^{-CV(\tau)}\|F(\tau)\|_{B_{p,r}^{s}}d\tau\right),$

where

$\displaystyle
V(t)=\left\\{\begin{array}[]{ll}\displaystyle\int_{0}^{t}\|v_{x}(\tau)\|_{B_{p,r}^{1/p}\displaystyle\cap
L^{\infty}}d\tau,\quad if~{}~{}s<1+\displaystyle\frac{1}{p},\\\
\displaystyle\int_{0}^{t}\|v_{x}(\tau)\|_{B_{p,r}^{s-1}}d\tau,\quad
if~{}~{}s>1+\displaystyle\frac{1}{p}(or~{}~{}s=1+\displaystyle\frac{1}{p}~{}~{}and~{}~{}r=1).\end{array}\right.$

_(2)_ If $f=v$, then (1) holds true for all $s>0$ with
$V(t)=\displaystyle\int_{0}^{t}\|v_{x}(\tau)\|_{L^{\infty}}d\tau$.

###### Lemma A.3

_[1, 16]_ Let $p,r,s,f_{0}$ and $F$ be as in the statement of Lemma A.2.
Assume that $v\in L^{\rho}([0,T];B_{\infty,\infty}^{-M})$ for some $\rho>1$,
$M>0$ and $v_{x}\in L^{1}([0,T];B_{p,r}^{s-1})$ if
$s>1+\displaystyle\frac{1}{p}$ or $s=1+\displaystyle\frac{1}{p}$, $r=1$ and
$v_{x}\in L^{1}([0,T];B_{p,\infty}^{1/p}\displaystyle\cap L^{\infty})$ if
$s<1+\displaystyle\frac{1}{p}$. Then, equation (.50) have a unique solution
$f\in L^{\infty}([0,T];B_{p,r}^{s})\displaystyle\cap
C([0,T];B_{p,1}^{s^{\prime}})$ for any $s^{\prime}<s$ and the inequalities of
Lemma A.2 hold true. Moreover, if $r<\infty$, then $f\in
C([0,T];B_{p,r}^{s})$.

######  Lemma A.4 (Moser-type estimates)

_[6, 15, 16]_ Let $1\leq p,r\leq\infty$, then we have the following estimates:
_(1)_ For $s>0$, $\|fg\|_{B_{p,r}^{s}}\leq
C(\|f\|_{B_{p,r}^{s}}\|g\|_{L^{\infty}}+\|f\|_{L^{\infty}}\|g\|_{B_{p,r}^{s}})$,
$\forall f,g\in B_{p,r}^{s}\displaystyle\cap L^{\infty}$.
_(2)_ For all $s_{1}\leq\displaystyle\frac{1}{p}<s_{2}$
($s_{2}\geq\displaystyle\frac{1}{p}$ if $r=1$) and $s_{1}+s_{2}>0$,

$\|fg\|_{B_{p,r}^{s_{1}}}\leq
C\|f\|_{B_{p,r}^{s_{1}}}\|g\|_{B_{p,r}^{s_{2}}},\forall f\in
B_{p,r}^{s_{1}},g\in B_{p,r}^{s_{2}}.$

###### Lemma A.5

_[15, 16]_ There is a constant $C>0$ such that for $s\in\mathbb{R}$,
$\varepsilon>0$ and $1\leq p\leq\infty$,

$\|f\|_{B_{p,1}^{s}}\leq
C\displaystyle\frac{1+\varepsilon}{\varepsilon}\|f\|_{B_{p,\infty}^{s}}\ln\left(e+\displaystyle\frac{\|f\|_{B_{p,\infty}^{s+\varepsilon}}}{\|f\|_{{B_{p,\infty}^{s}}}}\right),\quad
f\in B_{p,\infty}^{s+\varepsilon}.$

######  Lemma A.6 (Osgood Lemma)

_[1, 7]_ Let $\rho$ be a measurable function from $[t_{0},T]$ to $[0,c]$,
$\gamma$ be a locally integrable function from $[t_{0},T]$ to
$\mathbb{R}^{+}$, $\mu$ be a continuous and nondecreasing function from
$[0,c]$ to $\mathbb{R}^{+}$, and $a\geq 0$ be a real number. Assume that
$\rho$ satisfies

$\displaystyle\rho(t)\leq
a+\displaystyle\int_{t_{0}}^{t}\gamma(t^{\prime})\mu(\rho(t^{\prime}))dt^{\prime},\quad
a.e.\quad t\in[t_{0},T].$

If $a>0$, then for a.e. $t\in[t_{0},T]$ we have

$\displaystyle-\mathcal{M}(\rho(t))+\mathcal{M}\mathcal{(}a)\leq\displaystyle\int_{t_{0}}^{t}\gamma(t^{\prime})dt^{\prime},\quad
with\quad\mathcal{M}(x)=\displaystyle\int_{x}^{c}\displaystyle\frac{dr}{\mu(r)}.$

If $a=0$ and $\mu$ satisfies the condition
$\displaystyle\int_{0}^{1}\displaystyle\frac{dr}{\mu(r)}=+\infty$, then
$\rho=0$ a.e. $t\in[t_{0},T]$.

######  Lemma A.7 (Fatou-type lemma)

_[15, 16]_ If $\\{u^{(k)}\\}$ is a bounded sequence in $B_{p,r}^{s}$, which
tends to $u$ in $\mathscr{S}^{\prime}$, then $u\in B_{p,r}^{s}$ and

$\displaystyle\|u\|_{B_{p,r}^{s}}\leq\liminf_{k\rightarrow\infty}\|u^{(k)}\|_{B_{p,r}^{s}}.$