The Cauchy problem for the generalized Degasperis-Procesi equation

Abstract

In this paper, we investigate the Cauchy problem for the generalized Degasperis-Procesi equation in a Besov space. Firstly, we prove that the generalized Degasperis-Procesi equation is locally well posed in $B_{p,r}^s$ with $s>1+\frac{1}{p}$ (or $s\ge 1+\frac{1}{p}$ if $r=1$ with $p\in [1,+\mathrm{\infty })$). Secondly, we prove that the generalized Degasperis-Procesi equation possesses the peaked solitary wave which is the weak solution to the generalized Degasperis-Procesi equation. Thirdly, we prove that the data-to-solution map for the generalized Degasperis-Procesi equation is not uniformly continuous in $B_{2,\mathrm{\infty }}^{3/2}$. Fourthly, we prove that the data-to-solution map for the generalized Degasperis-Procesi equation is not uniformly continuous in $H^s(\mathbf{R})$ with $s<3/2$. Finally, we give a blow-up criterion.

MSC:35G25, 35L05.

1 Introduction

In this paper, we consider the Cauchy problem for the following generalized Degasperis-Procesi equation:

$$u_t-u_{txx}+u^k{u}_x-{\left(u^k{u}_x\right)}_{xx}+Q{\left[u^{k+1}\right]}_x=0,$$

(1.1)

$$u(x,0)=u_0(x),$$

(1.2)

where $Q\in \mathbf{R}$ is a constant. When $k=1$ and $Q=\frac{3}{2}$, (1.1) reduces to the Degasperis-Procesi equation

$$u_t-u_{txx}+4u{u}_x=3u_x{u}_{xx}+u{u}_{xxx}.$$

(1.3)

Equation (1.3) possesses the Lax pair and bi-Hamiltonian structures and infinite many conservation laws [1]. The Degasperis-Procesi equation [2] possesses peaked solitons which are stable [3] and shock peakons of the form $u(x,t)=-\frac{1}{t+k}sign(x)e^{-|x|}$, $k>0$. The Degasperis-Procesi equation possesses the global weak solution and blow-up structure [4–6]. Constantin and Lannes studied the relevance between the Camassa-Holm equation and the Degasperis-Procesi equation [7]. The Degasperis-Procesi equation possesses the infinite propagation speed [8]. The Degasperis-Procesi equation possesses multi-peakon solutions [9] and multisoliton [10]. Himonas and his co-authors [11, 12] proved that the data-to-solution for the Camassa-Holm equation, the Degasperis-Procesi equation is not uniformly continuous in $H^s(\mathbf{R})$ with $s>3/2$, respectively. Himonas et al. proved the non-uniform continuity in $H^1$ of the solution map of the CH equation [13]. Recently, Gui and Liu [14] studied the Cauchy problem for the Degasperis-Procesi equation in Besov spaces. Yan et al. [15] studied the Cauchy problem for the Novikov equation in Besov spaces.

Let $P(D)=-{\partial }_x{(1-{\partial }_x^2)}^{-1}$ and $p(x)=\frac{1}{2}{e}^{-|x|}$, $x\in \mathbf{R}$. By using the identity ${(1-{\partial }_x^2)}^{-1}f=p\ast f$ for $f\in L^2$, we can rewrite (1.1)-(1.2) as follows:

$$u_t+u^k{u}_x=QP(D)\left[u^{k+1}\right],$$

(1.4)

$$u(x,0)=u_0(x).$$

(1.5)

In this paper, motivated by [14], we study the Cauchy problem for (1.4) in Besov spaces. Firstly, we use the standard iterative method to prove that the generalized Degasperis-Procesi equation is locally well posed in $B_{p,r}^s$ with $s>1+\frac{1}{p}$ (or $s\ge 1+\frac{1}{p}$ if $r=1$ with $p\in [1,+\mathrm{\infty })$). Secondly, we prove that the generalized Degasperis-Procesi equation possesses the peaked solitary wave which is the weak solution to the generalized Degasperis-Procesi equation. Thirdly, we prove that the data-to-solution map for the generalized Degasperis-Procesi equation is not uniformly continuous in $B_{2,\mathrm{\infty }}^{3/2}$. Fourthly, we prove that the data-to-solution map for the generalized Degasperis-Procesi equation is not uniformly continuous in $H^s(\mathbf{R})$ with $s<3/2$. Finally, we give a blow-up criterion.

Notice that the structure of (1.4) is more complicated than that of the Degasperis-Procesi equation. Thus, to prove that the sequence of smooth solutions ${(u^{(n)})}_{n\in \mathbf{N}}$ is uniformly bounded in $C([0,T];B_{p,r}^s)\cap C^1([0,T];B_{p,r}^{s-1})$ with $s>1+\frac{1}{p}$ (or $s\ge 1+\frac{1}{p}$ if $r=1$ with $p\in [1,+\mathrm{\infty })$), we choose that

$${\parallel u^{(n)}\parallel }_{B_{p,r}^s}\le \frac{{\parallel u_0\parallel }_{B_{p,r}^s}}{{(1-2kC{\parallel u_0\parallel }_{B_{p,r}^s}^{k}t)}^{1/k}},t\in [0,T].$$

(1.6)

In proving Theorem 1.5, we explain why we choose (1.6). It is worthy of pointing out that we use Fatou’s lemma and the upper limit as well as Gronwall’s inequality to prove that ${(u^{(n)})}_{n\in \mathbf{N}}$ is a Cauchy sequence in $C([0,T];B_{p,r}^{s-1})$ with $s>1+\frac{1}{p}$ (or $s\ge 1+\frac{1}{p}$ if $r=1$ with $p\in [1,+\mathrm{\infty })$).

To introduce the main results, we define

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

The main results of this paper are as follows.

Theorem 1.1 Let $u_0(x)\in B_{p,r}^s$ with $s>1+\frac{1}{p}$ (or $s\ge 1+\frac{1}{p}$ if $r=1$ with $p\in [1,+\mathrm{\infty })$). Problem (1.4)-(1.5) is locally well posed. Moreover,

$${\parallel u(t)\parallel }_{B_{p,r}^s}\le \frac{{\parallel u_0\parallel }_{B_{p,r}^s}}{{(1-2kC{\parallel u_0\parallel }_{B_{p,r}^s}^{k}t)}^{1/k}},t\in [0,T].$$

(1.7)

A function $u:[0,T)\times \mathbf{R}$ is called a weak solution to (1.1) (or (1.4)) if u belongs to $L_{\mathrm{loc}}^{\mathrm{\infty }}([0,T);H^1)$ and satisfies the following identity:

$$\begin{array}{r}{\int }_0^T{\int }_{\mathbf{R}}[u{\varphi }_t+\frac{1}{k+1}{u}^{k+1}{\varphi }_x+p\ast \left[\frac{k^2+2k}{k+1}{u}^{k+1}\right]{\varphi }_x]dxdt\\ +{\int }_{\mathbf{R}}u(x,0)\varphi (x,0)dx=0,\end{array}$$

(1.8)

where $p(x)=\frac{1}{2}{e}^{-|x|}$, for any smooth test function $\varphi (x,t)\in C_c^{\mathrm{\infty }}([0,T)\times \mathbf{R})$. If u is a weak solution on $[0,T)$ for every $T>0$, then it is called a global weak solution.

Theorem 1.2 When $Q=\frac{k(k+2)}{k+1}$ in (1.4), $u_c(x,t)=c^{1/k}{e}^{-|x-ct|}$ with $c>0$ is a weak solution of (1.4) in the sense of (1.8).

Theorem 1.3 When $Q=\frac{k(k+2)}{k+1}$ in (1.4), the data-to-solution map for the generalized Degasperis-Procesi equation is not uniformly continuous in $B_{2,\mathrm{\infty }}^{3/2}$. More precisely, there exists a global solution $u\in L^{\mathrm{\infty }}({\mathbf{R}}^{+};B_{2,\mathrm{\infty }}^{3/2})$ to the Cauchy problem for (1.4) such that for any $T>0$ and $ϵ>0$, there exists a solution $v\in L^{\mathrm{\infty }}(0,T;B_{2,\mathrm{\infty }}^{3/2})$ with

$${\parallel v(0)-u(0)\parallel }_{B_{2,\mathrm{\infty }}^{3/2}}\le ϵ,{\parallel v(t)-u(t)\parallel }_{L^{\mathrm{\infty }}(0,T;B_{2,\mathrm{\infty }}^{3/2})}\ge 1.$$

Theorem 1.4 When $Q=\frac{k(k+2)}{k+1}$ in (1.4), the data-to-solution map for the generalized Degasperis-Procesi equation is not uniformly continuous in $H^s(\mathbf{R})$ with $s<3/2$.

Theorem 1.5 Assume that $T^{\star }$ is the maximal time of existence of the solution to problem (1.4)-(1.5). If $T^{\star }<\mathrm{\infty }$, then

$${\int }_0^{T^{\star }}{\parallel u_x\parallel }_{L^{\mathrm{\infty }}}^{k}d\tau =+\mathrm{\infty }.$$

(1.9)

Moreover, $T^{\star }\ge \frac{1}{Ck{\parallel u_0\parallel }_{B_{p,r}^s}^k}$.

The remainder of this paper is organized as follows. In Section 2, we give some preliminaries. In Section 3, we prove Theorem 1.1. In Section 4, we prove Theorem 1.2. In Section 5, we prove Theorem 1.3. In Section 6, we prove Theorem 1.4. In Section 7, we prove Theorem 1.5.

2 Preliminaries

In this section, we give Lemmas 2.1-2.4. The proof of Lemmas 2.1-2.4 can be seen in [16–21].

Lemma 2.1 (Littlewood-Paley decomposition)

Let $B=\{\xi \in {\mathbf{R}}^n,|\xi |\le \frac{4}{3}\}$ and $C=\{\xi \in {\mathbf{R}}^n,\frac{3}{4}\le |\xi |\le \frac{8}{3}\}$. There exists a couple of smooth radial functions $(\chi ,\varphi )\in (C_c^{\mathrm{\infty }}(B),C_c^{\mathrm{\infty }}(C))$ such that

$$\mathrm{\forall }\xi \in {\mathbf{R}}^n,\chi (\xi )+\sum _{q\in \mathbf{N}}\varphi \left(2^{-q}\xi \right)=1$$

and

$$\begin{array}{c}Supp\varphi (2^{-q}\cdot )\cap Supp\varphi (2^{-q^{\prime }}\cdot )=\mathrm{\varnothing }\mathit{\text{if}}|q-q^{\prime }|\ge 2,\hfill \\ Supp\chi (\cdot )\cap Supp\varphi (2^{-q}\cdot )=\mathrm{\varnothing }\mathit{\text{if}}|q|\ge 1\hfill \end{array}$$

and

$$\frac{1}{3}\le \chi {(\xi )}^2+\sum _{q\ge 0}\varphi {\left(2^{-q}\xi \right)}^2\le 1,\mathrm{\forall }\xi \in {\mathbf{R}}^n.$$

(2.1)

Then, for $u\in {\mathcal{S}}^{\prime }(\mathbf{R})$, the nonhomogeneous dyadic blocks are defined as follows:

$$\begin{array}{c}{\mathrm{\Delta }}_{q}u=0\mathit{\text{if}}q\le -2,\hfill \\ {\mathrm{\Delta }}_{-1}u=\chi (D)u={\mathcal{F}}_x^{-1}\chi {\mathcal{F}}_{x}u,\hfill \\ {\mathrm{\Delta }}_{q}u=\varphi \left(2^{-q}D\right)={\mathcal{F}}_x^{-1}\varphi \left(2^{-q}\xi \right){\mathcal{F}}_{x}u\mathit{\text{if}}q\ge 0.\hfill \end{array}$$

Thus we obtain

$$u=\sum _{q\in \mathbf{Z}}{\mathrm{\Delta }}_{q}u\mathit{\text{in}}{\mathcal{S}}^{\prime }(\mathbf{R}),$$

and the low frequency cut-off $S_q$ is defined by

$$S_{q}u=\sum _{p=-1}^{q-1}{\mathrm{\Delta }}_{p}u=\chi \left(2^{-q}D\right)u={\mathcal{F}}_x^{-1}\chi \left(2^{-q}\xi \right){\mathcal{F}}_{x}u,\mathrm{\forall }q\in \mathbf{N},$$

as well as

$$\begin{array}{c}{\mathrm{\Delta }}_p{\mathrm{\Delta }}_{q}u\equiv 0\mathit{\text{if}}|p-q|\ge 2,\hfill \\ {\mathrm{\Delta }}_q(S_{p-1}u{\mathrm{\Delta }}_{p}v)\equiv 0\mathit{\text{if}}|p-q|\ge 5,\mathrm{\forall }u,v\in {\mathcal{S}}^{\prime }(\mathbf{R}),\hfill \\ {\parallel {\mathrm{\Delta }}_{p}u\parallel }_{L^p}\le C{\parallel u\parallel }_{L^p},\hfill \\ {\parallel S_{q}u\parallel }_{L^p}\le C{\parallel u\parallel }_{L^p},\mathrm{\forall }1\le p\le +\mathrm{\infty },\hfill \end{array}$$

where C is a positive constant independent of q.

Definition (Besov spaces)

Let $s\in \mathbf{R}$ and $1\le p\le +\mathrm{\infty }$. The nonhomogeneous Besov space $B_{p,r}^s({\mathbf{R}}^n)$ is defined by

$$\begin{array}{c}{B}_{p,r}^s\left({\mathbf{R}}^n\right)\hfill \\ =\{f\in {\mathcal{S}}^{\prime }\left({\mathbf{R}}^n\right):{\parallel f\parallel }_{B_{p,r}^s}={\parallel 2^{qs}{\mathrm{\Delta }}_{q}f\parallel }_{l^r(L^p)}={\parallel {\left(2^{qs}{\parallel {\mathrm{\Delta }}_{q}f\parallel }_{L^p}\right)}_{q\ge -1}\parallel }_{l^r}<\mathrm{\infty }\}.\hfill \end{array}$$

In particular, if $s=\mathrm{\infty }$, then $B_{p,r}^s={\bigcap }_{s\in \mathbf{R}}{B}_{p,r}^s$.

Lemma 2.2 Let $s\in \mathbf{R}$, $1\le p,r,p_j,r_j\le \mathrm{\infty }$, $j=1,2$, then:

• (1) $B_{p,r}^s$ is a Banach space and is continuously embedded in ${\mathcal{S}}^{\prime }({\mathbf{R}}^n)$.

• (2) $B_{p_1,r_1}^{s_1}\hookrightarrow B_{p_2,r_2}^{s_2}$, if $p_1\le p_2$ and $r_1\le r_2$ and $s_2=s_1-n(\frac{1}{p_1}-\frac{1}{p_2})$

  $$B_{p,r_2}^{s_1}\hookrightarrow B_{p,r_1}^{s_2}\mathit{\text{locally compact if}}{s}_2<s_1.$$

• (3) $\mathrm{\forall }s>0$, $B_{p,r}^s\cap L^{\mathrm{\infty }}$ is a Banach algebra. $B_{p,r}^s$ is a Banach algebra iff $B_{p,r}^s\hookrightarrow L^{\mathrm{\infty }}$ and iff $s>\frac{1}{p}$ or ($s\ge \frac{1}{p}$ and $r=1$). (4)

• (i) For $s>0$,

  $${\parallel fg\parallel }_{B_{p,r}^s}\le C({\parallel f\parallel }_{B_{p,r}^s}{\parallel g\parallel }_{L^{\mathrm{\infty }}}+{\parallel f\parallel }_{L^{\mathrm{\infty }}}{\parallel g\parallel }_{B_{p,r}^s}),\mathrm{\forall }f,g\in B_{p,r}^s\cap L^{\mathrm{\infty }}.$$

• (ii) $\mathrm{\forall }{s}_1\le \frac{1}{p}<s_2$ ($s_2\ge \frac{1}{p}$ if $r=1$) and $s_1+s_2>0$,

  $${\parallel fg\parallel }_{B_{p,r}^{s_1}}\le C{\parallel f\parallel }_{B_{p,r}^{s_1}}{\parallel g\parallel }_{B_{p,r}^{s_2}},\mathrm{\forall }f\in B_{p,r}^{s_1},g\in B_{p,r}^{s_2}.$$

• (5) $\mathrm{\forall }\theta \in [0,1]$ and $s=\theta s_1+(1-\theta )s_2$,

  $${\parallel f\parallel }_{B_{p,r}^s}\le C{\parallel f\parallel }_{B_{p,r}^{s_1}}^{\theta }{\parallel f\parallel }_{B_{p,r}^{s_2}}^{1-\theta },\mathrm{\forall }f\in B_{p,r}^{s_1}\cap B_{p,r}^{s_2}.$$

• (6) If ${(u_n)}_{n\in \mathbf{N}}$ is bounded in $B_{p,r}^s$ and $u_n\to u$ in ${\mathcal{S}}^{\prime }({\mathbf{R}}^n)$, then $u\in B_{p,r}^s$ and

  $${\parallel u\parallel }_{B_{p,r}^s}\le \underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}inf}{\parallel u_n\parallel }_{B_{p,r}^s}.$$

• (7) Let $m\in \mathbf{R}$ and Ψ be an $S^m$-multiplier. Then the operator $\mathrm{\Psi }(D)$ is continuous from $B_{p,r}^s$ into $B_{p,r}^{s-m}$. In particular, $-{\partial }_x{(1-{\partial }_x^2)}^{-1}$ is continuous from $B_{p,r}^s$ into $B_{p,r}^{s-1}$.

Lemma 2.3 (A priori estimates in Besov spaces)

Let $1\le p,r\le \mathrm{\infty }$ and $s>-min\{\frac{1}{p},1-\frac{1}{p}\}$. Assume that $f_0\in B_{p,r}^s$, $F\in L^1(0,T;B_{p,r}^s)$ and that ${\partial }_{x}v$ belongs to $L^1(0,T;B_{p,r}^{s-1})$ if $s>1+\frac{1}{p}$ or to $L^1(0,T;B_{p,r}^{1/p}\cap L^{\mathrm{\infty }})$ otherwise. If $f\in L^{\mathrm{\infty }}(0,T;B_{p,r}^s)\cap C([0,T];{\mathcal{S}}^{\prime }(\mathbf{R}))$ solves the following 1-D linear transport equation:

$$f_t+v{f}_x=F,$$

(2.2)

$$f(x,0)=f_0,$$

(2.3)

then there exists a constant C depending only on s, p, r such that the following statements hold:

• (1) If $r=1$ or $s\ne 1+\frac{1}{p}$, then

  $${\parallel f\parallel }_{B_{p,r}^s}\le {\parallel f_0\parallel }_{B_{p,r}^s}+{\int }_0^t{\parallel F(\tau )\parallel }_{B_{p,r}^s}d\tau +C{\int }_0^t{V}^{\prime }(\tau ){\parallel f(\tau )\parallel }_{B_{p,r}^s}d\tau$$

or

$${\parallel f\parallel }_{B_{p,r}^s}\le e^{CV(t)}({\parallel f_0\parallel }_{B_{p,r}^s}+{\int }_0^t{e}^{-CV(\tau )}{\parallel F(\tau )\parallel }_{B_{p,r}^s}d\tau )$$

(2.4)

with $V(t)={\int }_0^t{\parallel v_x(\tau )\parallel }_{B_{p,r}^{1/p}\cap L^{\mathrm{\infty }}}d\tau$ if $s<1+\frac{1}{p}$ and $V(t)={\int }_0^t{\parallel v_x(\tau )\parallel }_{B_{p,r}^{s-1}}d\tau$ else.

• (2) If $s\le 1+\frac{1}{p}$, $f_0^{\prime }\in L^{\mathrm{\infty }}$ and $f_x\in L^{\mathrm{\infty }}((0,T)\times \mathbf{R})$ and $F_x\in L^1(0,T;L^{\mathrm{\infty }})$, then

  $$\begin{array}{c}{\parallel f(t)\parallel }_{B_{p,r}^s}+{\parallel f_x(t)\parallel }_{L^{\mathrm{\infty }}}\hfill \\ \le e^{CV(t)}({\parallel f_0\parallel }_{B_{p,r}^s}+{\parallel f_0^{\prime }\parallel }_{L^{\mathrm{\infty }}}+{\int }_0^t{e}^{-CV(\tau )}[{\parallel F(\tau )\parallel }_{B_{p,r}^s}+{\parallel F_x(\tau )\parallel }_{L^{\mathrm{\infty }}}]d\tau )\hfill \end{array}$$

with

$$V(t)={\int }_0^t{\parallel {\partial }_{x}v(\tau )\parallel }_{B_{p,r}^{1/p}\cap L^{\mathrm{\infty }}}d\tau .$$

• (3) If $f=v$, then for all $s>0$, (1) holds true when $V(t)={\int }_0^t{\parallel v_x(\tau )\parallel }_{L^{\mathrm{\infty }}}d\tau$.

• (4) If $r<\mathrm{\infty }$, then $f\in C([0,T];B_{p,r}^s)$. If $r=\mathrm{\infty }$, then $f\in C([0,T];B_{p,1}^{s^{\prime }})$ for all $s^{\prime }<s$.

Lemma 2.4 (Existence and uniqueness)

Let p, r, s, $f_0$ and F be as in the statement of Lemma  2.3. Assume that $v\in L^{\rho }(0,T;B_{\mathrm{\infty },\mathrm{\infty }}^{-M})$ for some $\rho >1$ and $M>0$ and $v_x\in L^1(0,T;B_{p,r}^{s-1})$ if $s>1+\frac{1}{p}$ or $s=1+\frac{1}{p}$ and $r=1$ and $v_x\in L^1(0,T;B_{p,\mathrm{\infty }}^{1/p}\cap L^{\mathrm{\infty }})$ if $s<1+\frac{1}{p}$. Then problem (2.1)-(2.2) has a unique solution $f\in L^{\mathrm{\infty }}(0,T;B_{p,r}^s)\cap ({\bigcap }_{s^{\prime }<s}C([0,T];B_{p,1}^{s^{\prime }}))$ and the inequalities of Lemma  2.3 can hold true. Moreover, if $r<\mathrm{\infty }$, then $f\in C([0,T];B_{p,r}^s)$.

3 Proof of Theorem 1.1

In this section, we complete the proof of Theorem 1.1 and suppose that $s>1+\frac{1}{p}$ (or $s\ge 1+\frac{1}{p}$ if $r=1$ with $p\in [1,+\mathrm{\infty })$).

First step: approximate solution

By using the standard iterative process, we construct a sequence of smooth solutions ${(u^{(n)})}_{n\in \mathbf{N}}\in C({\mathbf{R}}^{+};B_{p,r}^{\mathrm{\infty }})$. Assume that $u^{(0)}:=0$, by induction we define a sequence of smooth functions ${(u^{(n)})}_{n\in \mathbf{N}}$ by solving the following linear transport equation:

$$u_t^{(n+1)}+{\left[u^{(n)}\right]}^k{u}_x^{(n+1)}=QP(D)\left[{\left(u^{(n)}\right)}^{k+1}\right],$$

(3.1)

$$u^{(n+1)}(x,0)=u_0^{(n+1)}(x)=S_{n+1}{u}_0(x).$$

(3.2)

By using the fact that $S_{n+1}{u}_0$ belong to $B_{p,r}^{\mathrm{\infty }}$, from Lemma 2.4, for all $n\in \mathbf{N}$, we can show by induction that problem (3.1)-(3.2) has a global solution ${(u^{(n)})}_{n\in \mathbf{N}}\in C({\mathbf{R}}^{+},B_{p,r}^{\mathrm{\infty }})$.

Second step: uniform bounds

For $s>1+\frac{1}{p}$ (or $s\ge 1+\frac{1}{p}$ if $r=1$ with $p\in [1,+\mathrm{\infty })$) and $n\in \mathbf{N}$, we prove that

$$\begin{array}{r}{\parallel u^{(n+1)}\parallel }_{B_{p,r}^s}\\ \le e^{C{U}^n(t)}({\parallel u_0\parallel }_{B_{p,r}^s}+C{\int }_0^t{e}^{-C{U}^n(\tau )}{\parallel u^{(n)}\parallel }_{B_{p,r}^s}^{k+1}d\tau ),\end{array}$$

(3.3)

with $U^n={\int }_0^t{\parallel u^{(n)}\parallel }_{B_{p,r}^s}^{k}d\tau$.

From (2.3) of Lemma 2.3 and (3.1), we derive that

$$\begin{array}{rl}{\parallel u^{(n+1)}(t)\parallel }_{B_{p,r}^s}\le & e^{C{\int }_0^t{\parallel u_x^{(n)}(t^{\prime })\parallel }_{B_{p,r}^{s-1}}d{t}^{\prime }}{\parallel u_0\parallel }_{B_{p,r}^s}\\ +C{\int }_0^t{e}^{C{\int }_{\tau }^t{\parallel u_x^{(n)}(t^{\prime })\parallel }_{B_{p,r}^{s-1}}d{t}^{\prime }}{\parallel F\left(u^{(n)}\right)\parallel }_{B_{p,r}^s}d\tau ,\end{array}$$

(3.4)

where

$$F\left(u^{(n)}\right)=P(D)\left[{\left(u^{(n)}\right)}^{k+1}\right].$$

(3.5)

By using the $S^{-1}$ multiplier property of $P(D)$ and the fact $B_{p,r}^s$ with $s>1+\frac{1}{p}$ (or $s\ge 1+\frac{1}{p}$ if $r=1$ with $p\in [1,+\mathrm{\infty })$) is a Banach algebra, we have

$$\begin{array}{rcl}{\parallel F\left(u^{(n)}\right)\parallel }_{B_{p,r}^{s-1}}& \le & C{\parallel P(D)\left[{\left(u^{(n)}\right)}^{k+1}\right]\left(t^{\prime }\right)\parallel }_{B_{p,r}^s}\\ \le & C{\parallel {\left(u^{(n)}\right)}^{k+1}\left(t^{\prime }\right)\parallel }_{B_{p,r}^{s-1}}\le C{\parallel u^{(n)}\left(t^{\prime }\right)\parallel }_{B_{p,r}^s}^{k+1}.\end{array}$$

(3.6)

Inserting (3.6) into (3.4) yields (3.3).

Let us fix $T>0$ such that

$$T\le \frac{1}{4kC{\parallel u_0\parallel }_{B_{p,r}^s}^k},$$

(3.7)

and suppose that

$${\parallel u^{(n)}(t)\parallel }_{B_{p,r}^s}\le \frac{{\parallel u_0\parallel }_{B_{p,r}^s}}{{(1-2kC{\parallel u_0\parallel }_{B_{p,r}^s}^{k}t)}^{1/k}},t\in [0,T].$$

(3.8)

Since $U^n(t)={\int }_0^t{\parallel u^{(n)}\parallel }_{B_{p,r}^s}^{k}d\tau$, by using (3.8), we have

$$\begin{array}{rl}{e}^{C(U^n(t)-U^n(\tau ))}& =e^{C{\int }_{\tau }^t{\parallel u^{(n)}(t^{\prime })\parallel }_{B_{p,r}^s}^{k}d{t}^{\prime }}\le e^{-\frac{1}{2k}{\int }_{\tau }^t\frac{d(1-2kC{\parallel u_0\parallel }_{B_{p,r}^s}^k{t}^{\prime })}{(1-2kC{\parallel u_0\parallel }_{B_{p,r}^s}^k{t}^{\prime })}}\\ ={\left(\frac{1-2kC{\parallel u_0\parallel }_{B_{p,r}^s}^k\tau }{1-2kC{\parallel u_0\parallel }_{B_{p,r}^s}^{k}t}\right)}^{\frac{1}{2k}}.\end{array}$$

(3.9)

When $\tau =0$ in (3.9), we have

$$e^{C{U}^n(t)}\le {\left(\frac{1}{1-2kC{\parallel u_0\parallel }_{B_{p,r}^s}^{k}t}\right)}^{\frac{1}{2k}}.$$

(3.10)

Inserting (3.9), (3.10) into (3.3) yields

$$\begin{array}{rl}{\parallel u^{(n+1)}(t)\parallel }_{B_{p,r}^s}& \le \frac{{\parallel u_0\parallel }_{B_{p,r}^s}}{{(1-2kC{\parallel u_0\parallel }_{B_{p,r}^s}^{k}t)}^{1/2k}}[1-\frac{1}{2k}{\int }_0^t\frac{d(1-2kC{\parallel u_0\parallel }_{B_{p,r}^s}^{k}t)}{{(1-2kC{\parallel u_0\parallel }_{B_{p,r}^s}^{k}t)}^{1+\frac{1}{2k}}}]\\ \le \frac{{\parallel u_0\parallel }_{B_{p,r}^s}}{{(1-2kC{\parallel u_0\parallel }_{B_{p,r}^s}^{k}t)}^{1/k}}.\end{array}$$

(3.11)

Thus, ${(u^{(n)})}_{n\in \mathbf{N}}$ is uniformly bounded in $C([0,T];B_{p,r}^s)$. By using (3.9) and the fact that $B_{p,r}^s$ is a Banach algebra and (3.5) and using the $S^{-1}$-multiplier property of $P(D)$, we have that

$${\parallel {\left[u^{(n)}\right]}^k{u}_x^{(n+1)}\parallel }_{B_{p,r}^s}\le C{\parallel u^{(n)}\parallel }_{B_{p,r}^s}^k{\parallel u^{(n+1)}\parallel }_{B_{p,r}^s}\le \frac{C{\parallel u_0\parallel }_{B_{p,r}^s}^{k+1}}{{(1-2C{\parallel u_0\parallel }_{B_{p,r}^s}^{k}t)}^{\frac{k+1}{k}}}$$

(3.12)

and

$$\begin{array}{rcl}{\parallel P(D)\left[{\left(u^{(n)}\right)}^{k+1}\right]\parallel }_{B_{p,r}^{s-1}}& \le & C{\parallel {\left(u^{(n)}\right)}^{k+1}\parallel }_{B_{p,r}^{s-1}}\le C{\parallel u^{(n)}\parallel }_{B_{p,r}^s}^{k+1}\\ \le & \frac{C{\parallel u_0\parallel }_{B_{p,r}^s}^{k+1}}{{(1-2kC{\parallel u_0\parallel }_{B_{p,r}^s}t)}^{\frac{k+1}{k}}}.\end{array}$$

(3.13)

Consequently,

$${\left(u^{(n)}\right)}_n\subset C([0,T];B_{p,r}^s)\cap C^1([0,T];B_{p,r}^{s-1}).$$

(3.14)

Remark Inserting (3.7) into (3.8) yields

$${\parallel u^{(n)}\parallel }_{B_{p,r}^s}\le 4{\parallel u_0\parallel }_{B_{p,r}^s}$$

(3.15)

for $n\in \mathbf{N}$. From (3.11) and (3.7), $\mathrm{\forall }n\in {\mathbf{N}}^{+}$, we have that

$$e^{C{U}^n}\le exp\left[{\int }_0^t\frac{C{\parallel u_0\parallel }_{B_{p,r}^s}^k}{1-2kC{\parallel u_0\parallel }_{B_{p,r}^s}^{k}t}d\tau \right]\le 4$$

(3.16)

and

$${\parallel u(t)\parallel }_{B_{p,r}^s}\le \frac{{\parallel u_0\parallel }_{B_{p,r}^s}}{{(1-2kC{\parallel u_0\parallel }_{B_{p,r}^s}^{k}t)}^{1/k}}$$

(3.17)

with the aid of Fatou’s lemma. We define

$$L=4({\parallel u_0\parallel }_{B_{p,r}^s}+1).$$

(3.18)

Thus,

$${\parallel u^{(n)}\parallel }_{B_{p,r}^s}+1\le L$$

(3.19)

for $n\in {\mathbf{N}}^{+}$.

Third step: convergence

We prove that ${(u^{(n)})}_{n\in \mathbf{N}}$ is a Cauchy sequence in $C([0,T];B_{p,r}^{s-1})$. For $(m,n)\in {\mathbf{N}}^2$, we have

$$\begin{array}{r}[{\partial }_t+{\left(u^{(n+m)}\right)}^k{\partial }_x](u^{(n+1+m)}-u^{(n+1)})\\ =({\left(u^{(n)}\right)}^k-{\left(u^{(n+m)}\right)}^k){\partial }_x{u}^{(n+1)}\\ +QP(D)\left[(u^{(n+m)}-u^{(n)})\sum _{j=0}^k{\left(u^{(n+m)}\right)}^{k-j}{\left(u^{(n)}\right)}^j\right].\end{array}$$

(3.20)

Combining (2.3) with (3.20), we have that

$$\begin{array}{r}{\parallel [u^{(n+1+m)}-u^{(n+1)}](t)\parallel }_{B_{p,r}^{s-1}}\\ \le e^{U^n(t)}\left({\parallel u_0^{(n+1+m)}-u_0^{(n+1)}\parallel }_{B_{p,r}^{s-1}}\right)\\ +{\int }_0^t{e}^{U^n(t)-U^n(\tau )}{\parallel F(u^{(n)},u^{(n+m)},{\partial }_x{u}^{(n+1)})\parallel }_{B_{p,r}^{s-1}}d\tau ,\end{array}$$

(3.21)

where

$$\begin{array}{c}{U}^n(t)={\int }_0^t{\parallel {\left(u^{(n+m)}\right)}^k\parallel }_{B_{p,r}^s}d\tau ,\hfill \\ F(u^{(n)},u^{(n+m)},{\partial }_x{u}^{(n+1)})\hfill \\ =({\left(u^{(n)}\right)}^k-{\left(u^{(n+m)}\right)}^k){\partial }_x{u}^{(n+1)}+QP(D)\left[(u^{(n+m)}-u^{(n)})\sum _{j=0}^k{\left(u^{(n+m)}\right)}^{k-j}{\left(u^{(n)}\right)}^j\right].\hfill \end{array}$$

By using (3) and (7) of Lemma 2.2, we have that

$$\begin{array}{r}{\parallel F(u^{(n)},u^{(n+m)},{\partial }_x{u}^{(n+1)})\parallel }_{B_{p,r}^{s-1}}\\ \le C{\parallel {\left(u^{(n)}\right)}^k-{\left(u^{(n+m)}\right)}^k\parallel }_{B_{p,r}^{s-1}}{\parallel {\partial }_x{u}^{(n+1)}\parallel }_{B_{p,r}^{s-1}}\\ +C{\parallel P(D)\left[(u^{(n+m)}-u^{(n)})\sum _{j=0}^k{\left(u^{(n+m)}\right)}^{k-j}{\left(u^{(n)}\right)}^j\right]\parallel }_{B_{p,r}^{s-1}}\\ \le C{\parallel u^{(n)}-u^{(n+m)}\parallel }_{B_{p,r}^{s-1}}{[{\parallel u^{(n)}\parallel }_{B_{p,r}^s}+{\parallel u^{(n+1)}\parallel }_{B_{p,r}^s}+{\parallel u^{(n+m)}\parallel }_{B_{p,r}^s}+1]}^k\\ \le C{L}^k{\parallel u^{(n)}-u^{(n+m)}\parallel }_{B_{p,r}^{s-1}}.\end{array}$$

(3.22)

Inserting (3.22) into (3.21) yields

$$\begin{array}{r}{\parallel [u^{(n+1+m)}-u^{(n+1)}](t)\parallel }_{B_{p,r}^{s-1}}\\ \le e^{U^n(t)}\left({\parallel u_0^{(n+1+m)}-u_0^{(n+1)}\parallel }_{B_{p,r}^{s-1}}\right)+C{L}^k{\int }_0^t{e}^{U^n(t)-U^n(\tau )}{\parallel u^{(n)}-u^{(n+m)}\parallel }_{B_{p,r}^{s-1}}d\tau .\end{array}$$

(3.23)

From (3.2) and Lemma 2.1, we can easily obtain that

$${\parallel u_0^{(n+1+m)}-u_0^{(n+1)}\parallel }_{B_{p,r}^{s-1}}\le C{2}^{-n}.$$

(3.24)

Obviously,

$$e^{U^n(t)}\le 4,e^{U^n(t)-U^n(\tau )}\le 4.$$

(3.25)

Inserting (3.24) and (3.25) into (3.23) leads to

$$\begin{array}{r}{\parallel (u^{(n+1+m)}-u^{(n+1)})(t)\parallel }_{B_{p,r}^{s-1}}\\ \le C{2}^{-n}+C{L}^k{\int }_0^t{\parallel (u^{(n)}-u^{(n+m)})(\tau )\parallel }_{B_{p,r}^{s-1}}d\tau .\end{array}$$

(3.26)

We define

$$A_{(n,m)}(t)={\parallel (u^{(n+m)}-u^{(n)})\parallel }_{B_{p,r}^{s-1}}.$$

(3.27)

Inserting (3.27) into (3.26) leads to

$$A_{(n+1,m)}(t)\le C{2}^{-n}+C{L}^k{\int }_0^t{A}_{(n,m)}(\tau )d\tau .$$

(3.28)

We define

$${\rho }_n(t)=\underset{m\in {\mathbf{N}}^{+}}{sup}{A}_{(n,m)}(t)=\underset{m\in {\mathbf{N}}^{+}}{sup}{\parallel (u^{(n+m)}-u^{(n)})(t)\parallel }_{B_{p,r}^s}$$

(3.29)

and

$$\tilde{\rho }(t)=\underset{n\to +\mathrm{\infty }}{lim\hspace{0.17em}sup}{\rho }_n(t).$$

(3.30)

Combining (3.30) with (3.29), (3.28), by using Fatou’s lemma, we have that

$$\tilde{\rho }(t)=\underset{n\to +\mathrm{\infty }}{lim\hspace{0.17em}sup}{\rho }_{n+1}(t)\le C^k{\int }_0^t\tilde{\rho }(\tau )d\tau .$$

(3.31)

Applying Gronwall’s inequality to (3.31) yields

$$\tilde{\rho }(t)\le e^{t{C}^k}\tilde{\rho }(0)$$

(3.32)

for $t\in [0,T]$. According to the definition of $\tilde{\rho }(t)$, we can easily obtain that

$$\tilde{\rho }(0)=0.$$

(3.33)

Combining (3.32) with (3.33), we have that

$$\tilde{\rho }(t)=0.$$

(3.34)

Hence, ${(u^n)}_n$ is a Cauchy sequence in $C([0,T];B_{p,r}^{s-1})$.

Fourth step: existence in $E_{p,r}^s(T)$

Now we prove that $u\in E_{p,r}^s(T)$ and satisfies (1.4)-(1.5) since ${(u^n)}_{n\in \mathbf{N}}$ is uniformly bounded in $L^{\mathrm{\infty }}(0,T;B_{p,r}^s)$. From (6) in Lemma 2.2, we have that $u\in L^{\mathrm{\infty }}(0,T;B_{p,r}^s)$. From (1.4), we can easily prove that $u_t\in L^{\mathrm{\infty }}(0,T;B_{p,r}^{s-1})$. It is easily checked that u is indeed a solution to (1.4)-(1.5) by passing to the limit in (3.1)-(3.2).

Now we prove that $u\in E_{p,r}^s(T)$. Since $u\in B_{p,r}^s$, $\mathrm{\forall }ϵ>0$, there exists $q_0\in {\mathbf{N}}^{+}$ such that

$$\sum _{q\ge q_0}{2}^{qsr}{\parallel {\mathrm{\Delta }}_{q}u\parallel }_{L^{\mathrm{\infty }}(0,T;L^p)}^r\le \frac{ϵ}{4}.$$

(3.35)

From the definition of Besov spaces, we have that

$$\begin{array}{r}{\parallel u(t+\delta )-u(t)\parallel }_{B_{p,r}^s}^r\\ =\sum _{q<q_0}{2}^{qsr}{\parallel {\mathrm{\Delta }}_{q}u(t)-{\mathrm{\Delta }}_{q}u(t+\delta )\parallel }_{L^p}^r+\sum _{q\ge q_0}{2}^{qsr}{\parallel {\mathrm{\Delta }}_{q}u-{\mathrm{\Delta }}_{q}u(t+\delta )\parallel }_{L^p}^r\\ \le \sum _{q<q_0}{2}^{qsr}{\parallel {\mathrm{\Delta }}_{q}u(t)-{\mathrm{\Delta }}_{q}u(t+\delta )\parallel }_{L^p}^r+2\sum _{q\ge q_0}{2}^{qsr}{\parallel {\mathrm{\Delta }}_{q}u(t)\parallel }_{L^{\mathrm{\infty }}(0,T;L^p)}^r\\ \le \sum _{q<q_0}{2}^{qsr}{\parallel {\mathrm{\Delta }}_{q}u(t)-{\mathrm{\Delta }}_{q}u(t+\delta )\parallel }_{L^p}^r+\frac{ϵ}{2}.\end{array}$$

(3.36)

By using the mean value theorem, we have that

$${\parallel {\mathrm{\Delta }}_{q}u(t)-{\mathrm{\Delta }}_{q}u(t+\delta )\parallel }_{L^p}={\parallel {\mathrm{\Delta }}_q{u}_t(t+\theta \delta )\parallel }_{L^p}|\delta |\le {\parallel {\mathrm{\Delta }}_q{u}_t(t)\parallel }_{L^{\mathrm{\infty }}(0,T;L^p)}|\delta |,$$

(3.37)

where $0<\theta <1$. Inserting (3.37) into (3.36) yields

$$\begin{array}{r}{\parallel u(t+\delta )-u(t)\parallel }_{B_{p,r}^s}^r\\ =\sum _{q<q_0}{2}^{qsr}{\parallel {\mathrm{\Delta }}_{q}u(t)-{\mathrm{\Delta }}_{q}u(t+\delta )\parallel }_{L^p}^r+\sum _{q\ge q_0}{2}^{qsr}{\parallel {\mathrm{\Delta }}_{q}u(t)-{\mathrm{\Delta }}_{q}u(t+\delta )\parallel }_{L^p}^r\\ \le |\delta |\sum _{q<q_0}{\parallel {\mathrm{\Delta }}_q{u}_t(t)\parallel }_{L^{\mathrm{\infty }}(0,T;L^p)}^r+\frac{ϵ}{2}\\ \le |\delta |{\parallel u_t\parallel }_{L^{\mathrm{\infty }}(0,T;B_{p,r}^{s-1})}^r+\frac{ϵ}{2}.\end{array}$$

(3.38)

We may choose δ sufficiently small such that

$$|\delta |{\parallel u_t\parallel }_{L^{\mathrm{\infty }}(0,T;B_{p,r}^{s-1})}^r\le \frac{ϵ}{2}.$$

(3.39)

Inserting (3.39) into (3.38) leads to

$${\parallel u(t+\delta )-u(t)\parallel }_{B_{p,r}^s}\le {ϵ}^{1/r}.$$

(3.40)

Thus, we derive that

$$u\in C([0,T];B_{p,r}^s).$$

(3.41)

Combining (3.41) with (1.4), we can easily obtain

$$u_t\in C([0,T];B_{p,r}^{s-1}).$$

(3.42)

From (3.41) and (3.42), we have that $u\in E_{p,r}^s(T)$.

Fifth step: uniqueness of solution

The uniqueness of the solution to the Cauchy problem for (1.4) can be proved similarly to Proposition 3.1 of [14].

Sixth step: continuity in $E_{p,r}^s(T)$ with $s>1+\frac{1}{p}$ (or $s\ge 1+\frac{1}{p}$ if $r=1$ with $p\in [1,+\mathrm{\infty })$)

The continuity of the solution to the Cauchy problem for (1.4) can be proved similarly to the continuity of the solution to the Degasperis-Procesi equation which can be seen in [14].

4 Proof of Theorem 1.2

From (6.5) and (6.7) of [22], we have that

$${\partial }_x{u}_c(x,t)=-sign(x-ct)u_c(x,t),{\partial }_t{u}_c(x,t)=csign(x-ct)u_c(x,t),$$

(4.1)

where $u_c(x,t)=c^{1/k}{e}^{-|x-ct|}$. By using integration by parts and (4.1), we have that

$$\begin{array}{r}{\int }_0^{+\mathrm{\infty }}{\int }_{\mathbf{R}}(u_c{\partial }_t\varphi +\frac{1}{k+1}{u}^{k+1}{\partial }_x\varphi )dxdt+{\int }_{\mathbf{R}}{u}_c(x,0)\varphi (x,0)dx\\ =-{\int }_0^{+\mathrm{\infty }}{\int }_{\mathbf{R}}\varphi [{\partial }_t{u}_c+u_c^k{\partial }_x{u}_c]dxdt\\ =-{\int }_0^{+\mathrm{\infty }}{\int }_{\mathbf{R}}\varphi sign(x-ct)[c{u}_c-u_c^{k+1}]dxdt.\end{array}$$

(4.2)

Since $u_c(x,t)=c^{1/k}{e}^{-|x-ct|}$, we have that when $x>ct$,

$$sign(x-ct)[c{u}_c-u_c^{k+1}]=c^{1+\frac{1}{k}}[e^{ct-x}-e^{(k+1)(ct-x)}]$$

(4.3)

and when $x\le ct$,

$$sign(x-ct)[c{u}_c-u_c^{k+1}]=c^{1+\frac{1}{k}}[e^{(x-ct)}-e^{(k+1)(x-ct)}].$$

(4.4)

By using ${(1-{\partial }_x^2)}^{-1}f=p\ast f$ and (4.1), we have that

$$\begin{array}{r}{\int }_0^{+\mathrm{\infty }}{\int }_{\mathbf{R}}\varphi {\partial }_x{(1-{\partial }_x^2)}^{-1}\left[\frac{k^2+2k}{k+1}{u}_c^{k+1}\right]dxdt\\ ={\int }_0^{+\mathrm{\infty }}{\int }_{\mathbf{R}}\varphi {\partial }_{x}p\ast \left[\frac{k^2+2k}{k+1}{u}_c^{k+1}\right]dxdt\\ =-\frac{k^2+2k}{2(k+1)}{c}^{1+\frac{1}{k}}{\int }_0^{+\mathrm{\infty }}{\int }_{{\mathbf{R}}_x}{\int }_{{\mathbf{R}}_y}\varphi sign(x-y)e^{-|x-y|}{e}^{-(k+1)|y-ct|}dydxdt.\end{array}$$

(4.5)

Now, we compute

$$\begin{array}{r}-\frac{k^2+2k}{2(k+1)}{c}^{1+\frac{1}{k}}{\int }_{{\mathbf{R}}_y}sign(x-y)e^{-|x-y|}{e}^{-(k+1)|y-ct|}dy\\ :=I_1+I_2+I_3.\end{array}$$

(4.6)

When $x>ct$, we have that

$$\begin{array}{rl}{I}_1& =-\frac{k^2+2k}{2(k+1)}{c}^{1+\frac{1}{k}}{\int }_{-\mathrm{\infty }}^{ct}{e}^{-(x-y)}{e}^{(k+1)(y-ct)}dy\\ =-\frac{k^2+2k}{2(k+1)}{c}^{1+\frac{1}{k}}{e}^{-(x+(k+1)ct)}{\int }_{-\mathrm{\infty }}^{ct}{e}^{(k+2)y}dy\\ =-\frac{k}{2(k+1)}{c}^{1+\frac{1}{k}}{e}^{ct-x}\end{array}$$

(4.7)

and

$$\begin{array}{rcl}{I}_2& =& -\frac{k^2+2k}{2(k+1)}{c}^{1+\frac{1}{k}}{\int }_{ct}^x{e}^{-(x-y)}{e}^{-(k+1)(y-ct)}dy\\ =& -\frac{k^2+2k}{2(k+1)}{c}^{1+\frac{1}{k}}{e}^{-(x-(k+1)ct)}{\int }_{ct}^x{e}^{-\frac{k(k+2)}{k+1}y}dy\\ =& -\frac{k+2}{2(k+1)}{c}^{1+\frac{1}{k}}[e^{ct-x}-e^{(k+1)(ct-x)}]\end{array}$$

(4.8)

and

$$\begin{array}{rcl}{I}_3& =& \frac{k^2+2k}{2(k+1)}{c}^{1+\frac{1}{k}}{\int }_x^{\mathrm{\infty }}{e}^{(x-y)}{e}^{-(k+1)(y-ct)}dy\\ =& \frac{k^2+2k}{2(k+1)}{c}^{1+\frac{1}{k}}{e}^{(x+(k+1)ct)}{\int }_x^{\mathrm{\infty }}{e}^{-(k+2)y}dy\\ =& \frac{k}{2(k+1)}{c}^{1+\frac{1}{k}}{e}^{(k+1)(ct-x)}.\end{array}$$

(4.9)

Thus, when $x>ct$, from (4.7)-(4.9), we have that

$$\begin{array}{r}{I}_1+I_2+I_3\\ =-\frac{k}{2(k+1)}{c}^{1+\frac{1}{k}}{e}^{ct-x}-\frac{k+2}{2(k+1)}{c}^{1+\frac{1}{k}}[e^{ct-x}-e^{(k+1)(ct-x)}]\\ +\frac{k}{2(k+1)}{c}^{1+\frac{1}{k}}{e}^{(k+1)(ct-x)}\\ =-c^{1+\frac{1}{k}}[e^{(k+1)(ct-x)}-e^{ct-x}].\end{array}$$

(4.10)

Similarly, when $x<ct$, we can obtain

$$I_1+I_2+I_3=c^{1+\frac{1}{k}}[e^{(k+1)(x-ct)}-e^{x-ct}].$$

(4.11)

From (4.3), (4.4) and (4.10) as well as (4.11), we have that

$$\begin{array}{r}{\int }_0^{+\mathrm{\infty }}{\int }_{\mathbf{R}}[u_c{\varphi }_t+\frac{1}{k+1}{u}_c^{k+1}{\varphi }_x+p\ast \left[\frac{k^2+2k}{k+1}{u}_c^{k+1}\right]{\varphi }_x]dxdt\\ +{\int }_{\mathbf{R}}{u}_c(x,0)\varphi (x,0)dx=0.\end{array}$$

(4.12)

Thus, $u_c=c^{1/k}{e}^{-|x-ct|}$ is the solution in the sense of (4.12). For $c>0$, let $u_c(x,t)=c^{1/k}{e}^{-|x-ct|}$. Thus $u_c(x,t)$ is the solitary wave for (1.1) (or (1.4)).

5 Proof of Theorem 1.3

Since when $Q=\frac{k(k+2)}{k+1}$ in (1.4), (1.4) possesses the peaked solitary wave $c^{1/k}{e}^{|x-ct|}$, Theorem 1.3 can be proved similarly to Proposition 4 of [19].

6 Proof of Theorem 1.4

Since when $Q=\frac{k(k+2)}{k+1}$ in (1.4), (1.4) possesses the peaked solitary wave $c^{1/k}{e}^{|x-ct|}$, Theorem 1.4 can be proved similarly to Theorem 3 of [23].

7 Proof of Theorem 1.5

In this section, we always assume that $s>1+\frac{1}{p}$ (or $s\ge 1+\frac{1}{p}$ if $r=1$ with $p\in [1,+\mathrm{\infty })$).

Proof of Theorem 1.5 Applying ${\mathrm{\Delta }}_q$ to (1.4) yields

$$({\partial }_t+u^k{\partial }_x){\mathrm{\Delta }}_{q}u=[u^k,{\mathrm{\Delta }}_q]{\partial }_{x}u+QP(D){\mathrm{\Delta }}_q\left[\frac{k^2+2k}{k+1}{u}^{k+1}\right].$$

(7.1)

From (2.54) of page 112 in [24], since $s>1+\frac{1}{p}$ (or $s\ge 1+\frac{1}{p}$ if $r=1$ with $p\in [1,+\mathrm{\infty })$), we have that

$${\parallel 2^{sq}{\parallel [u^k,{\mathrm{\Delta }}_q]{\partial }_{x}u\parallel }_{L^p}\parallel }_{{\ell }^r}\le C{\parallel u_x\parallel }_{L^{\mathrm{\infty }}}^k{\parallel u\parallel }_{B_{p,r}^s}\le C{\parallel u\parallel }_{B_{p,r}^s}^{k+1}.$$

(7.2)

By using (4) of Lemma 2.2 and $P(D)$ is an $S^{-1}$-multiplier, since $s>1+\frac{1}{p}$ (or $s\ge 1+\frac{1}{p}$ if $r=1$ with $p\in [1,+\mathrm{\infty })$), we have that

$${\parallel P(D)\left[\frac{k^2+2k}{k+1}{u}^{k+1}\right]\parallel }_{B_{p,r}^s}\le C{\parallel u_x\parallel }_{L^{\mathrm{\infty }}}^k{\parallel u\parallel }_{B_{p,r}^s}\le C{\parallel u\parallel }_{B_{p,r}^s}^{k+1}.$$

(7.3)

Going along the lines of the proof of Proposition A.1 of [18], from (7.2) and (7.3), we have that

$${\parallel u\parallel }_{B_{p,r}^s}\le {\parallel u_0\parallel }_{B_{p,r}^s}+C{\int }_0^t{\parallel u_x\parallel }_{L^{\mathrm{\infty }}}^k{\parallel u\parallel }_{B_{p,r}^s}d\tau$$

(7.4)

$$\le {\parallel u_0\parallel }_{B_{p,r}^s}+C{\int }_0^t{\parallel u\parallel }_{B_{p,r}^s}^{k+1}d\tau .$$

(7.5)

Solving (7.4) yields

$${\parallel u\parallel }_{B_{p,r}^s}\le e^{c{\int }_0^t{\parallel u_x\parallel }_{L^{\mathrm{\infty }}}^{k}d\tau }{\parallel u_0\parallel }_{B_{p,r}^s}.$$

(7.6)

Solving (7.5) yields

$${\parallel u\parallel }_{B_{p,r}^s}\le \frac{{\parallel u_0\parallel }_{B_{p,r}^s}}{{(1-Ckt{\parallel u_0\parallel }_{B_{p,r}^s}^k)}^{1/k}}.$$

(7.7)

Assume that $T^{\star }$ is the maximal time of existence of the solution to problem (1.4)-(1.5). If $T^{\star }<\mathrm{\infty }$, we claim that

$${\int }_0^{T^{\star }}{\parallel u_x\parallel }_{L^{\mathrm{\infty }}}^{k}d\tau =+\mathrm{\infty }.$$

(7.8)

We prove the claim (7.8) by contradiction. If (7.8) is untrue, then from (7.8), we have that

$${\parallel u\left(T^{\star }\right)\parallel }_{B_{p,r}^s}<\mathrm{\infty },$$

(7.9)

which contradicts the fact that $T^{\star }$ is the maximal time of existence of the solution to problem (1.4)-(1.5). Consequently, (7.8) is true. From (7.7), we know that $T^{\star }\ge \frac{1}{Ck{\parallel u_0\parallel }_{B_{p,r}^s}^k}$. Moreover, (7.7) ensures the validity of (3.8).

The proof of Theorem 1.5 is completed. □