The dynamic properties of solutions for a nonlinear shallow water equation

Abstract

The local well-posedness for the Cauchy problem of a nonlinear shallow water equation is established. The wave-breaking mechanisms, global existence, and infinite propagation speed of solutions to the equation are derived under certain assumptions. In addition, the effects of coefficients λ, β, a, b, and index k in the equation are illustrated.

1 Introduction

We aim to consider the problem

$$\begin{aligned} \textstyle\begin{cases} v_{t}-v_{xxt} +\beta (v_{x}-v_{xxx})+\lambda (v-v_{xx})+(a+b)v^{k}v _{x} \\ \quad = bv^{k-1}v_{x}v_{xx}+av^{k }v_{xxx} , \\ v(0,x)=v_{0}(x). \end{cases}\displaystyle \end{aligned}$$

(1.1)

Here $(t,x)\in \mathbb{R}^{+}\times \mathbb{R}$, $v(t,x)$ is fluid velocity of water waves, $\lambda \in \mathbb{R}^{+}$, $\beta \in \mathbb{R}$, $(a,b)\in \mathbb{R}^{2}$, k is a positive integer, $\beta (v-v_{xx})$ is the diffusion term, $\lambda (v-v_{xx})$ is the dissipative term, $v_{0}\in B_{p,r}^{s}(\mathbb{R})\ (s>\operatorname{max}(1+ \frac{1}{p},\frac{3}{2} ))$.

Recently, the Camassa–Holm (CH) equation

$$ v_{t}-v_{xxt}+\beta v_{x} +3vv_{x}=2v_{x}v_{xx}+vv_{xxx} $$

(1.2)

has attracted much attention. Equation (1.2) admits blow-up phenomena. Replacing v with $v+\beta$ in Eq. (1.2), we obtain

$$ v_{t}-v_{xxt}+\beta (v_{x}-v_{xxx}) +3vv_{x}=2v_{x}v_{xx}+vv_{xxx}. $$

(1.3)

Taking $k=1, \lambda =0, a=1, b=2$ in (1.1) gives rise to the Cauchy problem of Eq. (1.3). The solution v to Eq. (1.2) is viewed as a perturbation near β (see [20]). The properties of solutions to the problem with dispersion and dissipative terms are discovered in [15]. Mi et al. [12] investigate the dynamical properties for a generalized CH equation. For a related study of the CH equation and other related partial differential equations, one may refer to references [3, 7, 11, 14, 16].

Taking $k=1, \lambda =\beta =0, a=1, b=3$ in (1.1) yields the Degasperis–Procesi equation

$$\begin{aligned} v_{t}-v_{xxt}+4vv_{x} =3v_{x} v_{xx}+vv_{xxx}. \end{aligned}$$

(1.4)

The formation of singularity for solutions to (1.4) is discovered in [17]. Lai and Wu [10] study the local well-posedness for the Cauchy problem of

$$ v_{t}-v_{xxt}+\beta v_{x}+(a+b)v_{x} =bv_{x} v_{xx}+avv_{xxx}, $$

(1.5)

where $\beta , a, b \in \mathbb{R}$.

Taking $k=2, \lambda =\beta =0, a=1, b=3$ in (1.1), we obtain the Novikov equation

$$\begin{aligned} v_{t}-v_{xxt}+4v^{2}v_{x} =3vv_{x} v_{xx}+v^{2} v_{xxx}. \end{aligned}$$

(1.6)

Guo [4] studies the persistence properties of solutions to the CH-type equation. Fu and Qu [2] discover blow-up of solutions to Eq. (1.6) in $H^{s}(\mathbb{R})\ (s>\frac{5}{2})$. The peakon solutions to the Novikov equation are established in [6].

Himonas and Thompson [8] discover persistence properties for solutions if $\lambda =\beta =0, a=1$ in (1.1). The behaviors of solutions [5], global existence of solutions for $a=1$ [9], and infinite propagation speed of solutions [9, 19] to the problems are investigated. We extend parts of results in [9, 10, 13, 18, 19].

Let $s\in \mathbb{R}, T>0, p\in [1,\infty ]$ and $r\in [1,\infty ]$. Thus we set

$$\begin{aligned} E_{p,r}^{s}(T)= \textstyle\begin{cases} C([0,T];B_{p,r}^{s}(\mathbb{R}))\cap C^{1}([0,T];B_{p,r}^{s-1}( \mathbb{R})),& 1\leq r< \infty , \\ L^{\infty } ([0,T];B_{p,\infty }^{s}(\mathbb{R}))\cap \operatorname{Lip}([0,T];B _{p,\infty }^{s-1}(\mathbb{R})),& r=\infty. \end{cases}\displaystyle \end{aligned}$$

Letting $P_{1}(D)=-\partial _{x}(1-\partial _{x }^{2})^{-1}, P_{2}(D)=(1- \partial _{x }^{2})^{-1}$, problem (1.1) is turned into

$$ \textstyle\begin{cases} v_{t}+(av^{k}+\beta )v_{x}=P_{1}(D)[\frac{b}{k+1}v^{k+1}+ \frac{3ak-b}{2}v^{k-1}v_{x}^{2}] \\ \phantom{v_{t}+(av^{k}+\beta )v_{x}=}{}+P_{2}(D)[\frac{(k-1)(ak-b)}{2} v^{k-2}v_{x}^{3}- \lambda v ], \\ v(0,x)=v_{0}(x). \end{cases} $$

(1.7)

Now we summarize the main results in this paper.

Theorem 1.1

Suppose $1\leq r, p \leq \infty$, $v_{0} \in B_{p,r}^{s}(\mathbb{R})\ (s>\operatorname{max}(1+\frac{1}{p}, \frac{3}{2} ))$. Then solution $v \in E_{p,r}^{s}(T)$ to problem (1.1) is locally well-posed for certain $T>0$.

Theorem 1.2

Suppose $1\leq r, p\leq \infty$, $v_{0} \in B_{p,r}^{s}(\mathbb{R})\ (s>\operatorname{max}(1+\frac{1}{p}, \frac{3}{2} ))$, $t\in [0,T]$. Then a solution v to problem (1.1) blows up in finite time if and only if

$$ \int _{0}^{t} \bigl(1+ \Vert v_{x} \Vert _{L^{\infty }}\bigr)^{k} \,d\tau = \infty. $$

(1.8)

Theorem 1.3

Suppose $b=a(k+1)$ and $v_{0} \in H^{s}( \mathbb{R})\ ( s> \frac{3}{2})$, $t\in [0,T]$. Then a solution v to problem (1.1) blows up in finite time if and only if

$$ \lim_{t\rightarrow T^{-}}\inf_{x\in \mathbb{R}}v_{x}(t,x)=- \infty. $$

(1.9)

Theorem 1.4

Suppose $b=a(k+1)$ and $v_{0} \in H ^{s}( \mathbb{R})\ (s\geq 2)$ satisfies $\Vert v_{0}-v_{0,xx} \Vert _{L^{2}}< \frac{ 4\lambda }{|a|(k+2) \Vert v_{0} \Vert ^{k-1}_{H^{1}}}$. Then there exists a global solution to problem (1.1) in $H ^{s}(\mathbb{R})\ (s\geq 2)$.

Theorem 1.5

Assume $v_{0}\in H^{s}(\mathbb{R})\ (s\geq 2)$, $n_{0}(x)=v_{0}-v_{0,xx}\neq 0$ for all $x\in \mathbb{R}$, $\Vert n_{0} \Vert _{L^{2}}< (\frac{2^{k+1}\lambda }{|ak-2b|})^{ \frac{1}{k}}$ and $b\neq \frac{ak}{2}$. Then a solution v to problem (1.1) is global in $H ^{s}(\mathbb{R})\ (s\geq 2)$.

Theorem 1.6

Assume $a>0$ and let $v_{0}\in H^{s}( \mathbb{R})\ (s>\frac{5}{2})$ be compactly supported in $[a_{0},b_{0}]$, $t\in [0,T]$. Suppose k is a positive odd number and $b=ak$, or $k=1, 0< b<3a$. Then, the solution $v(t,x)$ to (1.1) satisfies

$$ v(t,x)=\frac{1}{2}L_{+}(t)e^{-x}\quad \textit{for } x\geq p(t,b_{0}),\qquad v(t,x)=\frac{1}{2}L_{-}(t)e^{x} \quad\textit{for } x\leq p(t,a_{0}), $$

where $L_{+}(t)$ and $L_{-}(t)$ are continuous non-vanishing functions given in (4.1). What is more, $L_{+}(t)>0$, $L_{-}(t)<0$ for $t\in [0,T]$. In particular, if $k=1$, $b=2a$ or $b=\frac{a}{2}$, then $L_{+}(t)\leq C_{3} e^{(\beta -\lambda )t}$ and $|L_{-}(t)| \leq C_{4} e^{-(\beta +\lambda )t}$.

Remark 1.1

Problem (1.1) is local well-posed in $B_{p,r} ^{s}(\mathbb{R})\ (s>\operatorname{max}(\frac{3}{2},1+\frac{1}{p}))$. $\Vert v(t) \Vert _{H^{1}(\mathbb{R})}$ is bounded if $b=a(k+1)$. Also $\Vert v(t) \Vert _{H^{2}(\mathbb{R})}$ is bounded if $b=\frac{ak}{2}$. Theorem 1.2 improves the result of Theorem 5.1 in [19]. Theorem 1.3 implies that wave-breaking for a solution v occurs if its slope is unbounded. This result improves Theorem 3.1 in [18] and Theorem 5.6 in [19]. From Theorems 1.4, 1.5, and 1.6, we deduce that λ, β, a, b, and k are related to global existence and infinite propagation speed of the solutions. Parts of results in [9, 10, 13, 18, 19] are extended.

2 Proof of Theorem 1.1

We prove Theorem 1.1 in following five steps.

Step 1. Let $v^{0} =0$. Let $(v^{i}) _{i\in \mathbb{N}} \in C(\mathbb{R}^{+};B_{p,r}^{\infty })$ be smooth and satisfy

$$ \textstyle\begin{cases} (\partial _{t}+(a(v^{i})^{k}+\beta )\partial _{x}) v^{i+1}=G , \\ v^{i+1}(0,x)=v^{i+1}_{0}=S_{i+1}v_{0}, \end{cases} $$

(2.1)

and suppose

$$\begin{aligned} G ={}&P_{1}(D)\biggl[\frac{b}{k+1} \bigl(v^{i}\bigr)^{k+1}+\frac{3ak-b}{2} \bigl(v^{i}\bigr)^{k-1}\bigl(v ^{i} \bigr)_{x}^{2}\biggr] \\ &{} +P_{2}(D)\biggl[\frac{(k-1)(ak-b)}{2} \bigl(v^{i} \bigr)^{k-2}\bigl(v^{i}\bigr)_{x}^{3}- \lambda v^{i} \biggr]. \end{aligned}$$

(2.2)

We see $S_{i+1}v_{0} \in B_{p,r}^{\infty }$. Then the solution $v^{i}\in C(\mathbb{R}^{+};B_{p,r}^{\infty })$ in (2.1) is global for all $i\in \mathbb{N}$ by Lemma 2.5 in [13].

Step 2. It is derived from Lemma 2.4 in [13] that

$$\begin{aligned} \bigl\Vert v^{i+1} \bigr\Vert _{B_{p,r}^{s}}\leq{}& e^{C_{1}\int _{0}^{t} \Vert (v^{i}(\tau ))^{k} \Vert _{B_{p,r}^{s}} \,d\tau } \\ &{} \times \biggl[ \Vert v_{0} \Vert _{B_{p,r}^{s}}+ \int _{0}^{t} e^{-C_{1} \int _{0}^{\tau } \Vert (v^{i}(\xi ))^{k} \Vert _{B_{p,r}^{s}} \,d \xi } \bigl\Vert G(\tau ,\cdot ) \bigr\Vert _{B_{p,r}^{s}} \,d\tau \biggr]. \end{aligned}$$

(2.3)

The notation $a\lesssim b$ means $a\leq Cb$ for a certain positive constant C. We acquire the estimates

$$\begin{aligned} \bigl\Vert G(t,x) \bigr\Vert _{B_{p,r}^{s}} \lesssim \bigl( \bigl\Vert v^{i} \bigr\Vert _{B_{p,r}^{s}}+ 1 \bigr)^{k} \bigl\Vert v^{i} \bigr\Vert _{B_{p,r}^{s}} . \end{aligned}$$

(2.4)

That is,

$$\begin{aligned} \bigl\Vert v^{i+1} \bigr\Vert _{B_{p,r}^{s}} \leq{}& C_{2}\cdot e^{C_{2} \int _{0}^{t} ( \Vert v^{i}(\tau ) \Vert _{B_{p,r}^{s}}+1)^{k} \,d \tau } \biggl[ \Vert v_{0} \Vert _{B_{p,r}^{s}} \\ &{}+ \int _{0}^{t}e^{-C_{2} \int _{0}^{\tau }( \Vert v^{i}(\xi ) \Vert _{B_{p,r}^{s}}+1)^{k} \,d\xi }\bigl( \bigl\Vert v^{i} \bigr\Vert _{B_{p,r}^{s}}+ 1\bigr)^{k} \bigl\Vert v^{i} \bigr\Vert _{B_{p,r}^{s}} \,d \tau \biggr]. \end{aligned}$$

(2.5)

One may find certain $T>0$ which satisfies $2kC_{2}^{k+1}(1+ \Vert v_{0} \Vert _{B_{p,r}^{s}} )^{k}T<1$ and

$$ \bigl( 1+ \bigl\Vert v^{i}(t) \bigr\Vert _{B_{p,r}^{s}} \bigr)^{k}\leq \frac{C_{2} ^{k}( 1+ \Vert v_{0} \Vert _{B_{p,r}^{s}} )^{k}}{1-2kC_{2}^{k+1}( 1+ \Vert v_{0} \Vert _{B_{p,r}^{s}} )^{k}t}. $$

(2.6)

Further, we deduce

$$\begin{aligned} \bigl( 1+ \bigl\Vert v^{i+1}(t) \bigr\Vert _{B_{p,r}^{s}} \bigr)^{k}\leq \frac{C_{2} ^{k}( 1+ \Vert v_{0} \Vert _{B_{p,r}^{s}} )^{k}}{1-2kC_{2}^{k+1}( 1+ \Vert v_{0} \Vert _{B_{p,r}^{s}} )^{k}t}, \end{aligned}$$

which implies that $(v^{i})_{i\in \mathbb{N}}$ is uniformly bounded in $E_{p,r}^{s}(T)$.

Step 3. Let $m, n\in \mathbb{N}$. From (2.1), we deduce that

$$\begin{aligned} &\bigl(\partial _{t}+\bigl(a\bigl(v^{m+n} \bigr)^{k}+\beta \bigr)\partial _{x}\bigr) \bigl(v^{m+n+1}-v^{m+1}\bigr) \\ &\quad{}=-a\bigl(\bigl(v ^{m+n}\bigr)^{k}-\bigl(v^{m} \bigr)^{k}\bigr)\partial _{x}v^{m+1} \\ &\qquad{} +P_{1}(D)\biggl[\frac{b}{k+1}\bigl(\bigl(v^{m+n} \bigr)^{k+1}-\bigl(v^{m}\bigr)^{k+1}\bigr) \biggr] \\ &\qquad{} +P_{1}(D)\biggl[\frac{3ak-b}{2}\bigl(\bigl(v^{m+n} \bigr)^{k-1}\bigl(v^{m+n}\bigr)^{2}_{x}- \bigl(v ^{m}\bigr)^{k-1}\bigl(v^{m} \bigr)_{x}^{2}\bigr)\biggr] \\ &\qquad{} +P_{2}(D)\biggl[\frac{(k-1)(ak-b )}{2} \bigl(\bigl(v^{m+n} \bigr)^{k-2} \bigl(v^{m+n}\bigr)_{x} ^{3}-\bigl(v^{m }\bigr)^{k-2} \bigl(v^{m}\bigr)_{x}^{3}\bigr)\biggr] \\ &\qquad{} +P_{2}(D)\bigl[-\lambda \bigl(v^{m+n}-v^{m} \bigr)\bigr]. \end{aligned}$$

(2.7)

Using Lemma 2.4 in [13] yields

$$\begin{aligned} & \bigl\Vert v^{m+n+1}-v^{m+1} \bigr\Vert _{B_{p,r}^{s-1}} \\ &\quad\leq e^{C \int _{0}^{t} \Vert v^{m+n} \Vert _{B_{p,r}^{s}} ^{k} \,d \tau }\biggl[ \bigl\Vert v_{0}^{m+n+1}-v_{0}^{m+1} \bigr\Vert _{B_{p,r} ^{s-1}}+C\times \int _{0}^{t} e^{-C \int _{0}^{\tau } \Vert v^{m+n} \Vert _{B_{p,r}^{s}}^{k} \,d\xi } \\ &\qquad{}\times \bigl( \bigl\Vert v^{m+n}-v^{m} \bigr\Vert _{B_{p,r}^{s-1}}\bigl( \bigl\Vert v^{m} \bigr\Vert _{B_{p,r}^{s}}+ \bigl\Vert v^{m+n} \bigr\Vert _{B_{p,r}^{s}}+ \bigl\Vert v^{m+1} \bigr\Vert _{B_{p,r}^{s}}+1\bigr)^{k }\bigr) \,d \tau \biggr]. \end{aligned}$$

(2.8)

We note that the initial values satisfy

$$\begin{aligned} v_{0}^{m+n+1}-v_{0}^{m+1}=\sum _{q=m+1}^{m+n}\Delta _{q} v_{0}. \end{aligned}$$

One may find a constant $C_{T_{1}}$ independent of m to satisfy

$$\begin{aligned} \bigl\Vert v^{m+n+1}-v^{m+1} \bigr\Vert _{L^{\infty }([0,T];B_{p,r}^{s-1})} \leq C_{T_{1}}2^{-m}. \end{aligned}$$

We obtain the desired results.

Step 4. Following the discussions in Step 4 in Sect. 3.1 in [13], one derives that $v \in E_{p,r}^{s}(T)$, which is continuous.

Step 5. (Proof of the uniqueness). Suppose $1\leq r, p \leq \infty , s>\operatorname{max}(\frac{3}{2}, 1+\frac{1}{p} )$. Assume $v^{1}$ and $v^{2}$ satisfy (1.7) with $v_{0}^{1}, v_{0}^{2} \in B_{p,r}^{s}$, $v^{1},v^{2}\in L^{\infty }([0,T];B_{p,r}^{s}) \cap C([0,T];B_{p,r}^{s-1})$. We write $v^{12}=v^{1}-v^{2}$. Then

$$\begin{aligned} v^{12}\in L^{\infty }\bigl([0,T];B_{p,r}^{s} \bigr) \cap C\bigl([0,T];B_{p,r}^{s-1}\bigr), \end{aligned}$$

which results in

$$ \textstyle\begin{cases} \partial _{t}v^{12}+(a(v^{1})^{k}+\beta )\partial _{x}v^{12}=-a((v^{ 1})^{k}-(v ^{ 2})^{k})\partial _{x}v^{2}+G_{1}, \\ v^{12}(0,x)=v_{0}^{12}=v_{0}^{1}-v_{0}^{2}, \end{cases} $$

(2.9)

where

$$\begin{aligned} G_{1} ={}&P_{1}(D)\biggl[\frac{b}{k+1}\bigl( \bigl(v^{1}\bigr)^{k+1}-\bigl(v^{2} \bigr)^{k+1}\bigr)\biggr] \\ &{} +P_{1}(D)\biggl[\frac{3ak-b}{2}\bigl(\bigl(v^{1} \bigr)^{k-1}\bigl(v^{1}\bigr)^{2}_{x}- \bigl(v^{2}\bigr)^{k-1}\bigl(v ^{2} \bigr)_{x}^{2}\bigr)\biggr] \\ &{}+P_{2}(D)\biggl[\frac{(k-1)(ak-b)}{2} \bigl(\bigl(v^{1} \bigr)^{k-2} \bigl(v^{1}\bigr)_{x}^{3}- \bigl(v ^{2 }\bigr)^{k-2}\bigl(v^{2} \bigr)_{x}^{3}\bigr)-\lambda v^{12}\biggr]. \end{aligned}$$

Using Lemma 2.4 in [13], we derive the estimates

$$\begin{aligned} &e^{-C\int _{0}^{t} \Vert v^{1} \Vert _{B_{p,r}^{s }}^{k} \,d \tau } \bigl\Vert v^{12} \bigr\Vert _{B_{p,r}^{s-1}} \\ &\quad \leq \bigl\Vert v^{12} _{0} \bigr\Vert _{B_{p,r}^{s-1}} \\ &\qquad{}+ C \int _{0}^{t} e^{-C\int _{0}^{\tau } \Vert v^{1} \Vert _{B_{p,r} ^{s }} ^{k} \,d\xi } \bigl\Vert v^{12} \bigr\Vert _{B_{p,r}^{s-1}} \bigl( \bigl\Vert v^{1} \bigr\Vert _{B_{p,r}^{s}}+ \bigl\Vert v^{2} \bigr\Vert _{B_{p,r}^{s}}+1\bigr)^{k } \,d \tau , \end{aligned}$$

which finishes the proof of the uniqueness.

Remark 2.1

Suppose $b=a(k+1), 1\leq r, p \leq \infty$, $v_{0} \in B_{p,r}^{s} (\mathbb{R})\ (s>\operatorname{max}(1+\frac{1}{p}, \frac{3}{2} ))$, $t\in [0,T]$. Then, the solution v to (1.1) satisfies

$$\begin{aligned} \bigl\Vert v(t) \bigr\Vert _{H^{1}}\leq \Vert v_{0} \Vert _{H^{1}}. \end{aligned}$$

3 Proofs of Theorems 1.2, 1.3, 1.4, and 1.5

3.1 Proof of Theorem 1.2

Taking advantage of the operator $\Delta _{q}$ to (1.7) yields

$$\begin{aligned} \bigl(\partial _{t}+\bigl(av^{k}+\beta \bigr)\partial _{x}\bigr)\Delta _{q} v=a \bigl[v^{k},\Delta _{q}\bigr]\partial _{x}v+ \Delta _{q} G_{2}(t,x), \end{aligned}$$

(3.1)

where

$$\begin{aligned} G_{2}(t,x)={}&P_{1}(D)\biggl[\frac{b}{k+1}v^{k+1}+ \frac{3ak-b}{2}v^{k-1}v _{x}^{2} \biggr] \\ &{} +P_{2}(D)\biggl[\frac{(k-1)(ak-b)}{2} v^{k-2}v_{x}^{3}- \lambda v \biggr]. \end{aligned}$$

Applying Lemma 2.3 in [13] gives rise to the estimates

$$\begin{aligned} \bigl\Vert a \bigl[v^{k},\Delta _{q}\bigr]\partial _{x}v \bigr\Vert _{B_{p,r}^{s }} \lesssim \Vert v_{x} \Vert _{L^{\infty }}^{k} \Vert v \Vert _{B_{p,r}^{s }} \end{aligned}$$

and

$$\begin{aligned} \bigl\Vert G_{2}(t,x) \bigr\Vert _{B_{p,r}^{s }} \lesssim \bigl( \Vert v _{x} \Vert _{L^{\infty }}^{k} +1 \bigr) \Vert v \Vert _{B_{p,r}^{s }}. \end{aligned}$$

We derive that

$$\begin{aligned} \bigl\Vert v(t) \bigr\Vert _{B_{p,r}^{s }} \lesssim \Vert v_{0} \Vert _{B_{p,r}^{s }} + \int _{0}^{t} \bigl(1+ \bigl\Vert v_{x}(\tau ) \bigr\Vert _{L^{\infty }}\bigr)^{k} \bigl\Vert v(\tau ) \bigr\Vert _{B_{p,r} ^{s }} \,d\tau. \end{aligned}$$

That is,

$$\begin{aligned} \bigl\Vert v(t) \bigr\Vert _{B_{p,r}^{s }} \lesssim \Vert v_{0} \Vert _{B_{p,r}^{s }} e^{\int _{0}^{t} (1+ \Vert v_{x}(\tau ) \Vert _{L^{\infty }})^{k} \,d\tau }. \end{aligned}$$

(3.2)

Letting $t\in [0,T^{\ast } ], T^{\ast } <\infty$ and

$$\begin{aligned} \int _{0}^{t} \bigl(1+ \bigl\Vert v_{x}(\tau ) \bigr\Vert _{L^{\infty }}\bigr)^{k} \,d \tau < \infty , \end{aligned}$$

(3.3)

we see that $\Vert v(T^{\ast }) \Vert _{B_{p,r}^{s }}$ is bounded by using (3.2). It yields a contradiction, ending the proof.

From Remark 2.1, we obtain a blow-up result.

Remark 3.1

If assumption $b=a(k+1)$ is added into Theorem 1.2, then condition in (1.8) is changed into

$$\begin{aligned} \int _{0}^{t} \bigl(1+ \Vert v_{x} \Vert _{L^{\infty }}\bigr)^{2} \,d\tau = \infty. \end{aligned}$$

3.2 Proof of Theorem 1.3

We only need to prove Theorem 1.3 with $s=2$ by density argument. Take $b=a(k+1)$. It is deduced from (1.1) that

$$\begin{aligned} \frac{1}{2}\frac{ d}{ dt} \int _{\mathbb{R}}\bigl(v^{2}+v_{x}^{2} \bigr) \,dx+ \int _{\mathbb{R}} \lambda \bigl(v^{ 2}+ v_{x}^{2}\bigr) \,dx=0, \end{aligned}$$

(3.4)

which results in

$$\begin{aligned} \frac{1}{2}\frac{ d}{ dt} \int _{\mathbb{R}}\bigl(v^{2}+v_{x}^{2} \bigr) \,dx\leq 0. \end{aligned}$$

(3.5)

A direct calculation shows that

$$\begin{aligned} & \frac{1}{2}\frac{ d}{ dt} \int _{\mathbb{R}}\bigl(v_{x}^{2}+v^{2}_{xx} \bigr) \,dx \\ &\quad=a(k+2) \int _{\mathbb{R}}v^{k}v_{x}v_{xx} \,dx - \int _{\mathbb{R}} \lambda \bigl(v_{x}^{ 2} + v _{xx}^{2}\bigr) \,dx \\ &\qquad{} - \int _{\mathbb{R}}\bigl[a(k+1) v^{k-1}v_{x}v_{xx}^{2}+av^{k}v _{xxx}v_{xx} \bigr]\,dx. \end{aligned}$$

(3.6)

Let $T<\infty$ and $v_{x}(t,x)\geq -M$ for a certain $M>0$. We come to the estimate

$$\begin{aligned} \bigl\Vert v(t) \bigr\Vert _{H^{2}}\leq \Vert v_{0} \Vert _{H^{2}}e ^{(1+M+ \Vert v_{0} \Vert _{H^{1}})^{k}t}, \quad\text{for all } t\in [0,T], \end{aligned}$$

which yields a contradiction.

3.3 Proof of Theorem 1.4

We take $n =v-v_{xx}$. The first equation in (1.1) is written in the form

$$\begin{aligned} n_{t}+\beta n_{x}+\lambda n +bv^{k-1}v_{x}n +av^{k} n_{x}=0. \end{aligned}$$

(3.7)

We see $b=a(k+1)$ in Theorem 1.4. Multiplying (3.7) by n and applying (3.6) gives rise to

$$\begin{aligned} \frac{1}{2}\frac{ d}{ dt} \int _{\mathbb{R}}n^{2} \,dx+\lambda \int _{\mathbb{R}}n^{2} \,dx\lesssim \frac{ \vert a \vert (k+2)}{4} \Vert v_{0} \Vert _{H^{1}}^{k-1} \Vert n \Vert _{L^{2}}^{3}. \end{aligned}$$

Taking $\lambda _{1}=2\lambda$ and $M_{1}=\frac{|a|(k+2)}{2} \Vert v_{0} \Vert _{H^{1}}^{k-1}$, we have

$$\begin{aligned} \frac{ d}{ dt} \Vert n \Vert _{L^{2}}^{2}+ \lambda _{1} \Vert n \Vert _{L^{2}}^{2}\leq M_{1}\bigl( \Vert n \Vert _{L^{2}}^{2} \bigr)^{ \frac{3}{2}}. \end{aligned}$$

It follows that $\Vert n \Vert _{L^{2}}\leq e^{-\frac{1}{2} \lambda _{1} t}(\frac{1}{ \Vert n_{0} \Vert _{L^{2}}}-\frac{M _{1}}{ \lambda _{1} })^{-1}$ if $\Vert n_{0} \Vert _{L^{2}}<\frac{ \lambda _{1}}{M_{1}}$. Then

$$\begin{aligned} \Vert v_{x} \Vert _{L^{\infty }}\leq \Vert n \Vert _{L ^{2}}\leq C_{2}(T). \end{aligned}$$

Using Theorem 1.3, we end the proof.

3.4 Proof of Theorem 1.5

We investigate problem

$$ \textstyle\begin{cases} \frac{d}{dt}p(t,x)=av^{k}(t,p(t,x))+\beta , \\ p(0,x)=x, \end{cases} $$

(3.8)

where $(t,x)\in (0,T)\times \mathbb{R}$.

Lemma 3.1

([1])

Let $v\in C([0,T];H^{s}(\mathbb{R})) \cap C^{1}([0,T];H^{s-1}(\mathbb{R}))\ (s\geq 2)$, $(t,x)\in [0,T] \times \mathbb{R}$. It follows that $p\in C^{1}([0,T]\times \mathbb{R}, \mathbb{R})$ to (3.8) is unique and

$$\begin{aligned} p_{x}(t,x)=e^{\int _{0}^{t}akv^{k-1}v_{x}(\tau ,p(\tau ,x)) \,d\tau }. \end{aligned}$$

(3.9)

Lemma 3.2

Let $v_{0}\in H^{s}(\mathbb{R})\ (s\geq 2)$, $(t,x)\in [0,T]\times \mathbb{R}$. Then

$$\begin{aligned} n(t,p ) (p_{x})^{\frac{b}{ak}}(t,x)=n_{0} e^{-\lambda t}. \end{aligned}$$

(3.10)

Moreover, $\Vert n \Vert _{L^{\frac{ak}{b}}}=e^{-\lambda t} \Vert n_{0} \Vert _{L^{\frac{ak}{b}}}$. If $b=\frac{ak}{2}$, it holds that

$$\begin{aligned} \Vert n \Vert _{L^{2}}=e^{-\lambda t} \Vert n_{0} \Vert _{L^{2}}. \end{aligned}$$

(3.11)

Proof

From (3.10), we acquire that

$$\begin{aligned} \frac{d}{dt}\bigl[n(t,p ) (p_{x})^{\frac{b}{ak}} \bigr]=-\lambda n (p_{x})^{ \frac{b}{ak}}. \end{aligned}$$

(3.12)

That is,

$$\begin{aligned} n(t,p ) (p_{x})^{\frac{b}{ak}}=e^{-\lambda t}n_{0}(x). \end{aligned}$$

A direct computation gives rise to

$$\begin{aligned} \bigl\Vert e^{-\lambda t}n_{0}(x) \bigr\Vert _{L^{\frac{ak}{b}}}= \Vert n \Vert _{L^{\frac{ak}{b}}}. \end{aligned}$$

We note $b=\frac{ak}{2}$. Thus we get (3.11). □

Proof of Theorem 1.5

Multiplying (3.7) by $ne^{2\lambda t}$, we come to

$$\begin{aligned} \frac{d}{dt}\biggl(e^{2\lambda t} \int _{\mathbb{R}}n^{2} \,dx\biggr)=(ak-2b)e^{2 \lambda t} \int _{\mathbb{R}}n^{2}v^{k-1}v_{x} \,dx. \end{aligned}$$

(3.13)

We derive that

$$\begin{aligned} \frac{d}{dt}\biggl(e^{2\lambda t} \int _{\mathbb{R}}n^{2} \,dx\biggr)\leq \frac{ \vert ak-2b \vert }{2^{k}}e^{-k\lambda t}\biggl[e^{2\lambda t} \int _{\mathbb{R}}n ^{2} \,dx\biggr]^{\frac{k+2}{2}}. \end{aligned}$$

(3.14)

Let $h(t)=e^{2\lambda t}\int _{\mathbb{R}}n^{2} \,dx$. Bearing in mind that $n_{0}(x)\neq 0$, $x\in \mathbb{R}$ and (3.10), one deduces that $h(t)$ is positive. Then

$$\begin{aligned} \frac{d}{dt}\bigl[h(t)\bigr]^{-\frac{k}{2}}\geq - \frac{k}{2} \frac{ \vert ak-2b \vert }{2^{k}}e^{-k\lambda t}. \end{aligned}$$

(3.15)

Using the assumption $n_{0}(x)\neq 0, b\neq \frac{ak}{2}$, $\Vert n_{0} \Vert _{L^{2}}< ({\frac{2^{k+1}\lambda }{|ak-2b|}})^{ \frac{1}{k}}$, we have $[h(0)]^{-\frac{k}{2}} - \frac{|ak-2b|}{2^{k+1} \lambda }> 0$. We obtain the inequality

$$\begin{aligned} \biggl(e^{2\lambda t} \int _{\mathbb{R}} n^{2} \,dx\biggr)^{\frac{k}{2}}\leq \biggl[ \Vert n_{0} \Vert _{L^{2}}^{-k}- \frac{ \vert ak-2b \vert }{2^{k+1}\lambda }\biggr]^{-1}. \end{aligned}$$

Consequently, we have the estimate

$$\begin{aligned} \Vert v_{x} \Vert _{L^{\infty }} \leq \Vert n \Vert _{L ^{2}}\leq e^{-\lambda t}\biggl[ \Vert n_{0} \Vert _{L^{2}}^{-k}-\frac{ \vert ak-2b \vert }{2^{k+1} \lambda } \biggr]^{-\frac{1}{k}}. \end{aligned}$$

Applying Theorem 1.3, we complete the proof. □

We give a global existence result.

Lemma 3.3

Let $b=a(k+1)$ or $b=\frac{ak}{2}$, $v_{0} \in H^{s}(\mathbb{R})\ (s\geq 2)$. Assume $n_{0}=v_{0}-v_{0,xx}$ does not change sign. It holds that a solution $v(t,x)$ to problem (1.1) exists globally.

Proof

One may assume $n_{0}(x)> 0$. We use Lemma 3.2 to derive that $n > 0$. Thus

$$\begin{aligned} v(t,x)= \int _{\mathbb{R}}\frac{1}{2}e^{-|x-\xi |} n(t,\xi ) \,d\xi \geq 0. \end{aligned}$$

That is,

$$\begin{aligned} v(t,x)=\frac{1}{2}e^{-x} \int _{-\infty }^{x} e^{\xi }n(t,\xi ) \,d \xi +\frac{1}{2}e^{x} \int _{x}^{\infty } e^{-\xi }n(t,\xi ) \,d \xi. \end{aligned}$$

(3.16)

We conclude that

$$\begin{aligned} v_{x}(t,x)=-\frac{1}{2}e^{-x} \int _{-\infty }^{x} e^{\xi }n(t,\xi ) \,d \xi +\frac{1}{2}e^{x} \int _{x}^{\infty } e^{-\xi }n(t,\xi ) \,d \xi. \end{aligned}$$

(3.17)

Hence $|v_{x} |\leq v$.

Applying $b=a(k+1)$ and recalling Remark 2.1, we derive

$$\begin{aligned} \vert v_{x} \vert \leq \vert v \vert \lesssim \bigl\Vert v(t) \bigr\Vert _{H^{1}}\lesssim \Vert v_{0} \Vert _{H^{1}}. \end{aligned}$$

(3.18)

Taking advantage of $b=\frac{ak}{2}$ and using Lemma 3.2 results in

$$\begin{aligned} \vert v_{x} \vert \leq \vert v \vert \lesssim \Vert n \Vert _{L^{2}}\lesssim \Vert n_{0} \Vert _{L^{2}}. \end{aligned}$$

(3.19)

Combining (3.18) or (3.19) with Theorem 1.2, we obtain the desired results. □

4 Proof of Theorem 1.6

Note that $a>0$. Using $\operatorname{supp} v_{0}(x)\subset [a_{0},b_{0}]$, we derive that $\operatorname{supp} v_{0}(x)\subset [p(t,a_{0}),p(t,b_{0})]$. Applying Lemma 3.2 yields that $\operatorname{supp} n(t,x)\subset [p(t,a _{0}),p(t,b_{0})]$, $t\in [0,T]$.

Let

$$\begin{aligned} L_{+}(t)= \int _{p(t,a_{0})}^{p(t,b_{0})} e^{\xi }n(t,\xi ) \,d \xi ,\qquad L_{-}(t)= \int _{p(t,a_{0})}^{p(t,b_{0})} e^{-\xi }n(t,\xi ) \,d \xi. \end{aligned}$$

(4.1)

From (3.16) and (4.1), we have

$$\begin{aligned} v(t,x)={}&\frac{1}{2}e^{-x}\biggl( \int _{-\infty }^{p(t,a_{0})}+ \int _{p(t,a _{0})}^{p(t,b_{0})}+ \int _{p(t,b_{0})}^{x}\biggr) e^{\xi }n(t,\xi ) \,d \xi \\ &{} +\frac{1}{2}e^{x} \int _{x}^{\infty } e^{-\xi }n(t,\xi ) \,d\xi \\ ={}&\frac{1}{2}e^{-x}L_{+}(t),\quad x>p(t,b_{0}). \end{aligned}$$

(4.2)

We derive $v =\frac{1}{2}e^{x}L_{-}(t)$ if $x< p(t,a_{0})$. Combining (3.17) with (4.2) gives rise to

$$\begin{aligned} v =-v_{x} =v_{xx} = \frac{1}{2}e^{-x}L_{+}(t),\quad x>p(t,b_{0}) \end{aligned}$$

(4.3)

and

$$\begin{aligned} v =v_{x} =v_{xx} = \frac{1}{2}e^{x}L_{-}(t), \quad x< p(t,a_{0}). \end{aligned}$$

(4.4)

An application of (4.1) leads to the identity

$$\begin{aligned} L_{+}(0)= \int _{a_{0}}^{b_{0}} e^{\xi }n_{0}( \xi ) \,d\xi =0. \end{aligned}$$

(4.5)

A direct calculation shows

$$\begin{aligned} \frac{ d }{ dt} L_{+}(t) ={}& \int _{-\infty }^{\infty } e^{\xi }n_{t}(t, \xi ) \,d\xi \\ ={}&{-} \int _{-\infty }^{\infty } e^{\xi }(\lambda -\beta )n\,d\xi + \int _{-\infty }^{\infty } e^{\xi } \frac{b}{k+1} v^{k+1} \,d\xi \\ &{} +\frac{3ak-b}{2} \int _{-\infty }^{\infty } e^{\xi }v_{x}^{2} v^{k-1} \,d \xi + \frac{ (k-1)(ak-b)}{2} \int _{-\infty }^{\infty } e^{\xi }v_{x}^{3} v ^{k-2} \,d\xi. \end{aligned}$$

(4.6)

If $b=ak$ and k is a positive odd number, we obtain

$$\begin{aligned} \frac{ d }{ dt} L_{+}(t)+(\lambda -\beta )L_{+}(t) >0, \end{aligned}$$

(4.7)

which is equivalent to the inequality

$$\begin{aligned} \frac{ d [L_{+}(t)e^{(\lambda -\beta )t}]}{ dt}>0. \end{aligned}$$

(4.8)

Hence $L_{+}(t)>0$, $t\in [0,T)$.

Similarly, we have

$$\begin{aligned} \frac{ d [-L_{-}(t)e^{(\lambda +\beta )t}]}{ dt}>0. \end{aligned}$$

(4.9)

Thus, $L_{-}(t)<0$, $t\in [0,T)$.

If $k=1, 0< b<3a$, we derive that (4.8) and (4.9) still hold true.

We give the estimates for curve $p(t,b_{0})$. Using the assumption $k=1, b=2a$ and (3.4) yields

$$\begin{aligned} \Vert v \Vert _{L^{\infty }}\leq \Vert v \Vert _{H^{1}} \leq e^{-\lambda t} \Vert v_{0} \Vert _{H^{1}}. \end{aligned}$$

(4.10)

Taking $x=b_{0}$ in (3.8) and integrating (3.8) on $[0,t]$, we come to the estimate

$$\begin{aligned} p(t,b_{0})&=b_{0}+ \int _{0}^{t} a v(\tau ,p ) \,d \tau +\beta t \\ &\leq \frac{1}{\lambda }C_{5}+b_{0}+\beta t. \end{aligned}$$

(4.11)

We conclude from (4.2) that

$$\begin{aligned} L_{+}(t)=2e^{p(t,b_{0})}v\bigl(t,p(t,b_{0}) \bigr)\leq C_{3}e^{(\beta -\lambda )t}. \end{aligned}$$

(4.12)

Similar to the derivation in (4.11), we have

$$\begin{aligned} p(t,a_{0})&=a_{0}+ \int _{0}^{t} av(\tau ,p ) \,d \tau +\beta t \\ &\geq -\frac{1}{\lambda }C_{5}+a_{0}+\beta t, \end{aligned}$$

(4.13)

which, combining with (4.4), implies

$$\begin{aligned} \bigl\vert L_{-}(t) \bigr\vert \leq C_{4}e^{-(\beta +\lambda )t}. \end{aligned}$$

(4.14)

If $k=1, b=\frac{a}{2}$, it is deduced from (3.11) that $\Vert v \Vert _{L^{\infty }} \leq e^{-\lambda t} \Vert v_{0} \Vert _{H^{2}}$. Similarly, we establish (4.12) and (4.14).

Remark 4.1

If $\operatorname{supp} v_{0}(x)\subset [a_{0},b_{0}]$ in (1.1), then $n =(1-\partial _{x}^{2})v(t,x)$ satisfies $\operatorname{supp} n\subset [p(t,a_{0}),p(t,b_{0})]$. Indeed, v does not have compact support. Also $v(t,x)$ is positive if $x\rightarrow \infty$ and $v(t,x)$ is negative if $x\rightarrow -\infty$.