In this section, we discuss the blow-up phenomena of Eq. (4) with g(u)=u. For Eq. (4), which describes shallow water waves, the blow-up occurs only in the form of wave-breaking, *i.e.* the solution remains bounded but its slope becomes unbounded in finite time.

Set p(x)=\frac{1}{2\alpha}{e}^{-|\frac{x}{\alpha}|}, x\in R, then {(1-{\alpha}^{2}{\partial}_{x}^{2})}^{-1}f=p\ast f for all f\in {L}^{2}(R). Using this identity, we can rewrite Eq. (4) as follows:

\{\begin{array}{c}{u}_{t}+(u-\frac{\gamma}{{\alpha}^{2}}){u}_{x}+{\partial}_{x}p\ast k(u)=0,\phantom{\rule{1em}{0ex}}t>0,x\in R,\hfill \\ u(0,x)={u}_{0}(x),\phantom{\rule{1em}{0ex}}x\in R,\hfill \end{array}

(5)

or in the equivalent form:

\{\begin{array}{c}{u}_{t}+(u-\frac{\gamma}{{\alpha}^{2}}){u}_{x}=-{\partial}_{x}{(1-{\alpha}^{2}{\partial}_{x}^{2})}^{-1}k(u),\phantom{\rule{1em}{0ex}}t>0,x\in R,\hfill \\ u(0,x)={u}_{0}(x),\phantom{\rule{1em}{0ex}}x\in R,\hfill \end{array}

(6)

where k(u)=(h(u)+\frac{{\alpha}^{2}}{2}{u}_{x}^{2}-\frac{1}{2}{u}^{2}+\frac{\gamma}{{\alpha}^{2}}u).

We now prove the following result.

**Theorem 3.1** *Let* h\in {C}^{m+3}(R,R), m\ge 2, *and* {u}_{0}\in {H}^{r}, \frac{3}{2}<r\le m. *If* *T* *is the existence time of the corresponding solution of initial data* {u}_{0}, *then the* {H}^{r}-*norm of* u(t,x) *to Eq*. (4) (*or *(6)) *blows up on* [0,T) *if and only if*

\underset{t\to T}{\overline{lim}}\{{\parallel u(t,x)\parallel}_{{L}^{\mathrm{\infty}}}+{\parallel {u}_{x}(t,x)\parallel}_{{L}^{\mathrm{\infty}}}\}=\mathrm{\infty}.

*Proof* Let u(t,x) be the solution of Eq. (4) with the initial data {u}_{0}\in {H}^{r}, \frac{3}{2}<r\le m, which is guaranteed by Theorem 2.1. If

\underset{t\to T}{\overline{lim}}\{{\parallel u(t,x)\parallel}_{{L}^{\mathrm{\infty}}}+{\parallel {u}_{x}(t,x)\parallel}_{{L}^{\mathrm{\infty}}}\}=\mathrm{\infty},

by Sobolev’s embedding theorem, we obtain that the solution u(t,x) will blow up in finite time.

Next, applying the operator {\mathrm{\Lambda}}^{r} to Eq. (6), multiplying by {\mathrm{\Lambda}}^{r}u, and integrating over *R*, we obtain

\frac{d}{dt}{(u,u)}_{r}=-2{(u{u}_{x},u)}_{r}+2{(f(u),u)}_{r}.

(7)

Here, f(u)=-{\partial}_{x}{(1-{\alpha}^{2}{\partial}_{x}^{2})}^{-1}(h(u)+\frac{{\alpha}^{2}}{2}{u}_{x}^{2}-\frac{1}{2}{u}^{2}+\frac{\gamma}{{\alpha}^{2}}u). Assume that there exists R>0 such that

\underset{t\to T}{\overline{lim}}\{{\parallel u(t,x)\parallel}_{{L}^{\mathrm{\infty}}}+{\parallel {u}_{x}(t,x)\parallel}_{{L}^{\mathrm{\infty}}}\}\le R.

Similar to [18], using Lemma 2.1 with s=r, we get

\left|{(u{u}_{x},u)}_{r}\right|\le c{\parallel {u}_{x}\parallel}_{{L}^{\mathrm{\infty}}}{\parallel u\parallel}_{r}^{2}\le cR{\parallel u\parallel}_{r}^{2}.

(8)

On the other hand, we estimate the second term of the right-hand side of Eq. (7)

\begin{array}{rcl}{(f(u),u)}_{r}& =& {(-{\partial}_{x}{(1-{\alpha}^{2}{\partial}_{x}^{2})}^{-1}(h(u)+\frac{{\alpha}^{2}}{2}{u}_{x}^{2}-\frac{1}{2}{u}^{2}+\frac{\gamma}{{\alpha}^{2}}u),u)}_{r}\\ \le & c{\parallel u\parallel}_{r}({\parallel h(u)\parallel}_{r-1}+{\parallel {u}_{x}^{2}\parallel}_{r-1}+{\parallel {u}^{2}\parallel}_{r-1}+{\parallel u\parallel}_{r-1})\\ \le & c(\tilde{h}\left({\parallel u\parallel}_{{L}^{\mathrm{\infty}}}\right)+{\parallel {u}_{x}\parallel}_{{L}^{\mathrm{\infty}}}+{\parallel u\parallel}_{{L}^{\mathrm{\infty}}}+1){\parallel u\parallel}_{r}^{2}\\ \le & c(\tilde{h}(R)+R+1){\parallel u\parallel}_{r}^{2}.\end{array}

(9)

Here, we applied Lemma 2.2 with F(u)=h(u) and s=r-1, Lemma 2.3 with s=r-1. From (7)-(9), we obtain

\frac{d}{dt}{\parallel u\parallel}_{r}^{2}\le c(\tilde{h}(R)+R+1){\parallel u\parallel}_{r}^{2}.

Thus, by Gronwall’s inequality, we get

{\parallel u(t)\parallel}_{r}^{2}\le {\parallel {u}_{0}\parallel}_{r}^{2}exp(\tilde{h}(R)+R+1)t.

This completes the proof of Theorem 3.1. □

We have the following blow-up scenario for Eq. (4).

**Theorem 3.2** *Assume that* h\in {C}^{m+3}(R,R), m\ge 3. *Given* {u}_{0}\in {H}^{r}, 3\le r\le m, *the solution* u=u({u}_{0},\cdot ) *of Eq*. (4) *is uniformly bounded*. *Blow*-*up in finite time* T<+\mathrm{\infty} *occurs if and only if*

\underset{t\to T}{lim\hspace{0.17em}inf}\left\{\underset{x\in R}{inf}[{u}_{x}(t,x)]\right\}=-\mathrm{\infty}.

*Proof* E(u)=\frac{1}{2}{\int}_{R}({u}^{2}+{\alpha}^{2}{u}_{x}^{2})\phantom{\rule{0.2em}{0ex}}dx is an invariant for Eq. (4). According to the inequalities

\begin{array}{c}{\int}_{R}({u}^{2}+{u}_{x}^{2})\phantom{\rule{0.2em}{0ex}}dx\le {\int}_{R}({u}^{2}+{\alpha}^{2}{u}_{x}^{2})\phantom{\rule{0.2em}{0ex}}dx=2E(u)\phantom{\rule{1em}{0ex}}(\alpha \ge 1),\hfill \\ {\int}_{R}({u}^{2}+{u}_{x}^{2})\phantom{\rule{0.2em}{0ex}}dx\le \frac{1}{{\alpha}^{2}}{\int}_{R}({u}^{2}+{\alpha}^{2}{u}_{x}^{2})\phantom{\rule{0.2em}{0ex}}dx=\frac{2}{{\alpha}^{2}}E(u)\phantom{\rule{1em}{0ex}}(\alpha \le 1),\hfill \end{array}

the invariance of E(u) ensures that the solution *u* is uniformly bounded as long as they exist.

If the slope of the solution u(t,x) becomes unbounded from below in finite time, then by Theorem 2.1 and Sobolev’s embedding theorem, we can see that the solution u(t,x) blows up in finite time.

Next, if the slope of the solution is bounded from below in finite time, then we deduce that the solution will not blow up in finite time. Differentiating Eq. (5) with respect to *x*, in view of the identity {\partial}_{x}^{2}(p\ast f)=\frac{1}{{\alpha}^{2}}(p\ast f-f), we have

\begin{array}{rcl}{u}_{tx}& =& -\frac{1}{2}{u}_{x}^{2}-u{u}_{xx}+\frac{\gamma}{{\alpha}^{2}}{u}_{xx}+\frac{1}{{\alpha}^{2}}h(u)-\frac{1}{2{\alpha}^{2}}{u}^{2}+\frac{\gamma}{{\alpha}^{4}}u\\ -\frac{1}{{\alpha}^{2}}p\ast (h(u)+\frac{{\alpha}^{2}}{2}{u}_{x}^{2}-\frac{1}{2}{u}^{2}+\frac{\gamma}{{\alpha}^{2}}u).\end{array}

(10)

Note that p\ast (\frac{1}{2}{u}_{x}^{2})\ge 0 and

{\parallel u\parallel}_{{L}^{\mathrm{\infty}}}^{2}\le \frac{1}{2}{\parallel u\parallel}_{1}^{2}\le max(1,\frac{1}{{\alpha}^{2}})E({u}_{0})\le max(1,\frac{1}{{\alpha}^{4}}){\parallel {u}_{0}\parallel}_{1}^{2}.

By Young’s inequality, we get

\begin{array}{c}{\parallel p\ast h(u)\parallel}_{{L}^{\mathrm{\infty}}}\le {\parallel p\parallel}_{{L}^{1}}{\parallel h(u)\parallel}_{{L}^{\mathrm{\infty}}}\le {\parallel h(u)\parallel}_{{L}^{\mathrm{\infty}}}\le \underset{|v|\le max(1,\frac{1}{{\alpha}^{2}}){\parallel {u}_{0}\parallel}_{1}}{sup}|h(v)|,\hfill \\ {\parallel p\ast {u}^{2}\parallel}_{{L}^{\mathrm{\infty}}}\le {\parallel p\parallel}_{{L}^{1}}{\parallel {u}^{2}\parallel}_{{L}^{\mathrm{\infty}}}\le {\parallel u\parallel}_{{L}^{\mathrm{\infty}}}^{2}\le max(1,\frac{1}{{\alpha}^{4}}){\parallel {u}_{0}\parallel}_{1}^{2},\hfill \end{array}

and

{\parallel p\ast u\parallel}_{{L}^{\mathrm{\infty}}}\le {\parallel p\parallel}_{{L}^{1}}{\parallel u\parallel}_{{L}^{\mathrm{\infty}}}\le {\parallel u\parallel}_{{L}^{\mathrm{\infty}}}\le max(1,\frac{1}{{\alpha}^{2}}){\parallel {u}_{0}\parallel}_{1}.

Define M(t)={u}_{x}(t,\zeta (t))={sup}_{x\in R}[{u}_{x}(t,x)]. Since {u}_{xx}(t,\zeta (t))=0 for all t\in [0,T), it follows that a.e. on [0,T)

\begin{array}{rcl}{M}^{\prime}(t)& =& -\frac{1}{2}{M}^{2}(t)+\frac{1}{{\alpha}^{2}}h\left(u(t,\zeta (t))\right)-\frac{1}{2{\alpha}^{2}}{u}^{2}(t,\zeta (t))+\frac{\gamma}{{\alpha}^{4}}u(t,\zeta (t))\\ -\frac{1}{{\alpha}^{2}}p\ast (h(u)+\frac{{\alpha}^{2}}{2}{u}_{x}^{2}-\frac{1}{2}{u}^{2}+\frac{\gamma}{{\alpha}^{2}}u).\end{array}

Then

{M}^{\prime}(t)\le -\frac{1}{2}{M}^{2}(t)+{A}_{0}^{2},

where

{A}_{0}={(\frac{2\gamma}{{\alpha}^{4}}max(1,\frac{1}{{\alpha}^{2}}){\parallel {u}_{0}\parallel}_{1}+\frac{1}{2{\alpha}^{2}}max(1,\frac{1}{{\alpha}^{4}}){\parallel {u}_{0}\parallel}_{1}^{2}+\frac{2}{{\alpha}^{2}}{G}_{0})}^{\frac{1}{2}}

by

{G}_{0}=\underset{|v|\le max(1,\frac{1}{{\alpha}^{2}}){\parallel {u}_{0}\parallel}_{1}}{sup}|h(v)|.

If M(t)>\sqrt{2}{A}_{0}, then {M}^{\prime}(t)<0 and M(t) is decreasing. Otherwise, M(t)\le \sqrt{2}{A}_{0}. Thus we get

m(t)\le M(t)\le max\{M(0),\sqrt{2}{A}_{0}\},\phantom{\rule{1em}{0ex}}t\in [0,T).

By Theorem 3.1 and the above inequality, we have that if the slope of the solution is bounded from below in finite time, then the solution will not blow up in finite time. □

Next, we present the following blow-up result.

**Theorem 3.3** *Assume that* h\in {C}^{m+3}(R,R), m\ge 3. *Given* \alpha >0, {u}_{0}\in {H}^{r}, 3\le r\le m, *assume that we can find* {x}_{0}\in R *with*

\begin{array}{rcl}{u}_{0}^{\prime}({x}_{0})& <& -{(\frac{4\gamma}{{\alpha}^{4}}max(1,\frac{1}{{\alpha}^{2}}){\parallel {u}_{0}\parallel}_{1}+\frac{1}{{\alpha}^{2}}max(1,\frac{1}{{\alpha}^{4}}){\parallel {u}_{0}\parallel}_{1}^{2}+\frac{4}{{\alpha}^{2}}{G}_{0})}^{\frac{1}{2}}\\ =& -\sqrt{{B}_{0}},\end{array}

*where*

{G}_{0}=\underset{|v|\le max(1,\frac{1}{{\alpha}^{2}}){\parallel {u}_{0}\parallel}_{1}}{sup}|h(v)|.

*Then the corresponding solution to Eq*. (4) *for* g(u)=u *blows up in finite time*. *Moreover*, *the maximal time of existence* *T* *satisfies the inequality*

T\le 2c{({c}^{2}-{B}_{0})}^{-1},\phantom{\rule{1em}{0ex}}\mathit{\text{where}}\phantom{\rule{0.3em}{0ex}}c=-{u}_{0}^{\prime}({x}_{0}).

*Proof* Now define m(t)={inf}_{x\in R}[{u}_{x}(t,x)]={u}_{x}(t,\xi (t)) by Lemma 2.4, and let \xi (t)\in R be a point where this infimum is attained. From Eq. (10), we have

\begin{array}{rcl}{m}^{\prime}(t)& =& -\frac{1}{2}{m}^{2}(t)+\frac{1}{{\alpha}^{2}}h\left(u(t,\xi (t))\right)-\frac{1}{2{\alpha}^{2}}{u}^{2}(t,\xi (t))+\frac{\gamma}{{\alpha}^{4}}u(t,\xi (t))\\ -\frac{1}{{\alpha}^{2}}p\ast (h(u)+\frac{{\alpha}^{2}}{2}{u}_{x}^{2}-\frac{1}{2}{u}^{2}+\frac{\gamma}{{\alpha}^{2}}u).\end{array}

For x=\xi (t), since {u}_{xx}(t,\xi (t))=0, we arrive at

{m}^{\prime}(t)\le -\frac{1}{2}{m}^{2}(t)+K(u)\phantom{\rule{1em}{0ex}}\text{a.e. on}(0,T),

where

K(u)=\frac{\gamma}{{\alpha}^{4}}{\parallel u\parallel}_{{L}^{\mathrm{\infty}}}+\frac{1}{{\alpha}^{2}}\left(\underset{|v|\le {\parallel u\parallel}_{{L}^{\mathrm{\infty}}}}{sup}\right|h(v)\left|\right)+{\parallel \frac{1}{{\alpha}^{2}}p\ast (h(u)+\frac{1}{2}{u}^{2}+\frac{\gamma}{{\alpha}^{2}}u)\parallel}_{{L}^{\mathrm{\infty}}}.

Note that

{\parallel u\parallel}_{{L}^{\mathrm{\infty}}}^{2}\le \frac{1}{2}{\parallel u\parallel}_{1}^{2}\le max(1,\frac{1}{{\alpha}^{2}})E({u}_{0})\le max(1,\frac{1}{{\alpha}^{4}}){\parallel {u}_{0}\parallel}_{1}^{2}.

By Young’s inequality, we get

\begin{array}{c}{\parallel p\ast h(u)\parallel}_{{L}^{\mathrm{\infty}}}\le \underset{|v|\le max(1,\frac{1}{{\alpha}^{2}}){\parallel {u}_{0}\parallel}_{1}}{sup}|h(v)|,\hfill \\ {\parallel p\ast {u}^{2}\parallel}_{{L}^{\mathrm{\infty}}}\le max(1,\frac{1}{{\alpha}^{4}}){\parallel {u}_{0}\parallel}_{1}^{2}\hfill \end{array}

and

{\parallel p\ast u\parallel}_{{L}^{\mathrm{\infty}}}\le max(1,\frac{1}{{\alpha}^{2}}){\parallel {u}_{0}\parallel}_{1}.

So, it follows that

{m}^{\prime}(t)\le -\frac{1}{2}{m}^{2}(t)+{K}_{0},

(11)

where (note that {A}_{0}^{2}={K}_{0})

\begin{array}{c}{G}_{0}=\underset{|v|\le max(1,\frac{1}{{\alpha}^{2}}){\parallel {u}_{0}\parallel}_{1}}{sup}|h(v)|,\hfill \\ {K}_{0}=\frac{2\gamma}{{\alpha}^{4}}max(1,\frac{1}{{\alpha}^{2}}){\parallel {u}_{0}\parallel}_{1}+\frac{1}{2{\alpha}^{2}}max(1,\frac{1}{{\alpha}^{4}}){\parallel {u}_{0}\parallel}_{1}^{2}+\frac{2}{{\alpha}^{2}}{G}_{0}.\hfill \end{array}

The absolute continuity of the locally Lipschitz function m(t) allows us to perform an integration over [0,t] and to have

m(t)\le m(0)-\frac{1}{2}{\int}_{0}^{t}{m}^{2}(\tau )\phantom{\rule{0.2em}{0ex}}d\tau +{K}_{0}t,\phantom{\rule{1em}{0ex}}t\in [0,T).

We claim now that m(t)<-c for all t\in (0,T), where c>\sqrt{2{K}_{0}} is fixed arbitrarily provided that m(0)<-c. In fact, assuming the contrary, in view of m(t) being continuous, ensure the existence of {t}_{0}\in (0,T) such that m(t)<-c in (0,{t}_{0}) and m({t}_{0})=c. Then we deduce that

m(t)\le m(0)-{\int}_{0}^{t}{K}_{0}\phantom{\rule{0.2em}{0ex}}d\tau +{K}_{0}t=m(0)<-c,\phantom{\rule{1em}{0ex}}t\in [0,{t}_{0}],

and a contradiction appears as we take t={t}_{0}. Using (11), we get

\begin{array}{rcl}{m}^{\prime}(t)& \le & -\frac{1}{2}{m}^{2}(t)+{K}_{0}\le -\frac{1}{2}{m}^{2}(t)+(\frac{1}{2}-\epsilon ){c}^{2}\\ \le & -\frac{1}{2}{m}^{2}(t)+(\frac{1}{2}-\epsilon ){m}^{2}(t)\\ \le & -\epsilon {m}^{2}(t)\phantom{\rule{1em}{0ex}}\text{a.e. on}(0,T),\end{array}

where \epsilon \in (0,\frac{1}{2}-\frac{{K}_{0}}{{c}^{2}}). Since m(t)<-c, and m(t) is locally Lipschitz, it follows that \frac{1}{m(t)} is locally Lipschitz as well. This gives

\frac{d}{dt}\left(\frac{1}{m(t)}\right)=-\frac{{m}^{\prime}(t)}{{m}^{2}(t)}\ge \epsilon \phantom{\rule{1em}{0ex}}\text{a.e. on}(0,T).

Integration of this inequality yields

-\frac{1}{m(t)}+\frac{1}{m(0)}\le -\epsilon t.

Since m(t)<0, we obtain

0\le t<\frac{1}{\epsilon (-m(0))},\phantom{\rule{1em}{0ex}}t\in [0,T).

In fact, as a consequence of these considerations, we obtain that the maximal existence time

T\le \frac{1}{\epsilon (-m(0))}\phantom{\rule{1em}{0ex}}\text{for all}\epsilon \in (0,\frac{1}{2}-\frac{{K}_{0}}{{c}^{2}}).

An estimation from the above for *T* is obtained immediately, namely

T\le \frac{2{c}^{2}}{(-m(0))({c}^{2}-2{K}_{0})}.

The conclusion is reached by letting c\to -m(0). □