In order to get the energy decay of the solution, we introduce the following set:

${\mathrm{\Sigma}}_{1}=\{(\lambda ,E)\in {R}^{2}|0<\lambda <{\lambda}_{1},0<E<{E}_{1}\},$

(20)

where ${\lambda}_{1}={(p{B}_{1}^{p})}^{-\frac{1}{p-2}}$, ${E}_{1}=(\frac{1}{2}-\frac{1}{p}){\lambda}_{1}^{2}$. Obviously, ${\mathrm{\Sigma}}_{1}\subset {\mathrm{\Sigma}}_{0}$.

Adapting the idea of Vitillaro [35], we have the following lemma.

**Lemma 3.1** *Suppose that* *u* *is the solution of* (1)-(5), ${u}_{0}\in V$, ${u}_{1}\in {L}^{2}$ *and* $(\parallel {u}_{0xx}\parallel ,E(0))\in {\mathrm{\Sigma}}_{1}$, *then* $(\parallel {u}_{xx}(t)\parallel ,E(t))\in {\mathrm{\Sigma}}_{1}$, *for* $t\ge 0$.

**Lemma 3.2** *Under the condition of Lemma * 3.1

*and* $p>2$,

*then*,

*for* $t\ge 0$,

${\parallel {u}_{xx}\parallel}^{2}\ge 2|u(l,t){|}^{p},$

(21)

$E(t)\ge \frac{p-1}{2p}{\parallel {u}_{xx}\parallel}^{2}\ge \frac{p-1}{p}|u(l,t){|}^{p},$

(22)

$E(t)\ge {C}_{3}({\parallel {u}_{xx}\parallel}^{2}+|{u}_{xx}(l,t){|}^{2}).$

(23)

*Proof* By (14) and (18), we have

$\begin{array}{rcl}E(t)& \ge & \frac{1}{2}{\parallel {u}_{xx}\parallel}^{2}-\frac{1}{p}|u(l,t){|}^{p}\ge \frac{1}{2}{\parallel {u}_{xx}\parallel}^{2}-|u(l,t){|}^{p}\\ \ge & \frac{1}{2}{\parallel {u}_{xx}\parallel}^{2}-{B}_{1}^{p}{\parallel {u}_{xx}\parallel}^{p}=G(\parallel {u}_{xx}\parallel ),\end{array}$

(24)

where $G(\lambda )=\frac{1}{2}{\lambda}^{2}-{B}_{1}^{p}{\lambda}^{p}$. Note that $G(\lambda )$ has the maximum at ${\lambda}_{1}={(p{B}_{1}^{p})}^{-\frac{1}{p-2}}$ and the maximum value $G({\lambda}_{1})={E}_{1}$. We see that $G(\lambda )$ is increasing in $(0,{\lambda}_{1})$, decreasing in $({\lambda}_{1},\mathrm{\infty})$ and $G(\lambda )\to 0$ as $\lambda \to \mathrm{\infty}$. Since $\parallel {u}_{xx}\parallel <{\lambda}_{1}$, $E(0)<{E}_{1}$, then $\parallel {u}_{xx}\parallel <{\lambda}_{1}$ for any $t\ge 0$, so $G(\parallel {u}_{xx}\parallel )\ge 0$.

By (24), we have

${\parallel {u}_{xx}\parallel}^{2}-|u(l,t){|}^{p}=\frac{1}{2}{\parallel {u}_{xx}\parallel}^{2}+(\frac{1}{2}{\parallel {u}_{xx}\parallel}^{2}-|u(l,t){|}^{p})\ge \frac{1}{2}{\parallel {u}_{xx}\parallel}^{2}+G(\parallel {u}_{xx}\parallel ),$

then (21) holds since

$G(\parallel {u}_{xx}\parallel )>0$. Furthermore, we have

$E(t)\ge \frac{1}{2}{\parallel {u}_{xx}\parallel}^{2}-\frac{1}{p}|u(l,t){|}^{p}\ge \frac{p-1}{2p}{\parallel {u}_{xx}\parallel}^{2}.$

(25)

So (22) holds. Similar to (25), the above equality becomes

$E(t)\ge \frac{1}{2}{\parallel {u}_{xx}\parallel}^{2}+\frac{1}{2}|{u}_{xx}(l,t){|}^{2}-\frac{1}{p}|u(l,t){|}^{p}\ge \frac{p-1}{2p}{\parallel {u}_{xx}\parallel}^{2}+\frac{1}{2}|{u}_{xx}(l,t){|}^{2},$

so (23) holds. □

**Theorem 3.3** *Let* $p>2$,

$(\parallel {u}_{0xx}\parallel ,E(0))\in {\mathrm{\Sigma}}_{1}$,

*and* *u* *be the solution of problem* (1)-(5),

*then there exist two positive constants* *l* *and* *θ* *independent of* *t* *such that* $E(t)\le l{e}^{-\theta t},\phantom{\rule{1em}{0ex}}t\ge 0.$

*Proof* From (16), we have

${\int}_{t}^{t+1}[{\parallel {u}_{t}(s)\parallel}^{2}+|{u}_{t}(l,s){|}^{2}]\phantom{\rule{0.2em}{0ex}}ds=E(t)-E(t+1).$

(26)

Now, for the above estimate and the mean value theorem, we choose

${t}_{1}\in [t,t+\frac{1}{4}]$ and

${t}_{2}\in [t+\frac{3}{4},t+1]$ such that, for

$i=1,2$,

$\parallel {u}_{t}({t}_{i})\parallel \le {\left({\int}_{t}^{t+1}{\parallel {u}_{t}(s)\parallel}^{2}\phantom{\rule{0.2em}{0ex}}ds\right)}^{\frac{1}{2}}\le C{(E(t)-E(t+1))}^{\frac{1}{2}},$

(27)

$|u(l,{t}_{i})|\le {\left({\int}_{t}^{t+1}\right|u(l,s){|}^{2}\phantom{\rule{0.2em}{0ex}}ds)}^{\frac{1}{2}}\le C{(E(t)-E(t+1))}^{\frac{1}{2}}.$

(28)

Multiplying equation (

1) by

*u* and integrating over

$[0,l]\times [{t}_{1},{t}_{2}]$, by the boundary conditions (2) and (3), we have

$\begin{array}{r}{\int}_{{t}_{1}}^{{t}_{2}}[{\parallel {u}_{xx}\parallel}^{2}+{\parallel {u}_{x}\parallel}^{2}+k{\parallel {u}_{x}\parallel}^{2(m+1)}+|{u}_{xx}(l,s){|}^{2}-|u(l,s){|}^{p}]\phantom{\rule{0.2em}{0ex}}ds\\ \phantom{\rule{1em}{0ex}}={\int}_{{t}_{1}}^{{t}_{2}}[({u}_{tt},u)+{u}_{t}(l,s)u(l,s)+({u}_{t},u)+{u}_{tt}(l,s)u(l,s)]\phantom{\rule{0.2em}{0ex}}ds\\ \phantom{\rule{1em}{0ex}}\le \sum _{i=1}^{2}[\parallel {u}_{t}({t}_{i})\parallel \parallel u({t}_{i})\parallel +|{u}_{t}(l,{t}_{i})\left|\right|u(l,{t}_{i})\left|\right]+{\int}_{{t}_{1}}^{{t}_{2}}[{\parallel {u}_{t}\parallel}^{2}+{u}_{t}^{2}(l,s)]\phantom{\rule{0.2em}{0ex}}ds\\ \phantom{\rule{2em}{0ex}}+{\int}_{{t}_{1}}^{{t}_{2}}[\parallel {u}_{t}\parallel \parallel u\parallel +|{u}_{t}(l,s)\left|\right|u(l,s)\left|\right]\phantom{\rule{0.2em}{0ex}}ds.\end{array}$

(29)

Now, we estimate the terms of the right-hand side of (29). By (10), (23), (27) and the Young inequality, we have

$\begin{array}{rl}{I}_{1}& =\sum _{i=1}^{2}\parallel {u}_{t}({t}_{i})\parallel \parallel u({t}_{i})\parallel \le C{(E(t)-E(t+1))}^{\frac{1}{2}}\sum _{i=1}^{2}\parallel u({t}_{i})\parallel \\ \le \epsilon E(t)+C(\epsilon )(E(t)-E(t+1)).\end{array}$

(30)

It follows from (9), (10), (23), (28) and the Young inequality that

$\begin{array}{rl}{I}_{2}& =\sum _{i=1}^{2}|{u}_{t}(l,{t}_{i})||u(l,{t}_{i})|\le C{(E(t)-E(t+1))}^{\frac{1}{2}}\sum _{i=1}^{2}|u(l,{t}_{i})|\\ \le \epsilon E(t)+C(\epsilon )(E(t)-E(t+1)).\end{array}$

(31)

From (26), we get

${I}_{3}={\int}_{{t}_{1}}^{{t}_{2}}[{\parallel {u}_{t}\parallel}^{2}+{u}_{t}^{2}(l,s)]\phantom{\rule{0.2em}{0ex}}ds\le C(E(t)-E(t+1)).$

(32)

From the Young inequality, (10), (23), (26) and from the fact that

$E(t)$ is non-increasing, we arrive at

${I}_{4}={\int}_{{t}_{1}}^{{t}_{2}}\parallel {u}_{t}\parallel \parallel u\parallel \phantom{\rule{0.2em}{0ex}}ds\le \epsilon E(t)+C(\epsilon )(E(t)-E(t+1)).$

(33)

By the Young inequality, (9), (10), (23), (26) and the fact that

$E(t)$ is non-increasing, we obtain

${I}_{5}={\int}_{{t}_{1}}^{{t}_{2}}|u(l,s)||{u}_{t}(l,s)|\phantom{\rule{0.2em}{0ex}}ds\le \epsilon E(t)+C(\epsilon )(E(t)-E(t+1)).$

(34)

Substituting (30)-(34) into (29), we get the estimate

$\begin{array}{r}{\int}_{{t}_{1}}^{{t}_{2}}[{\parallel {u}_{xx}\parallel}^{2}+{\parallel {u}_{x}\parallel}^{2}+k{\parallel {u}_{x}\parallel}^{2(m+1)}+|{u}_{xx}(l,s){|}^{2}-|u(l,s){|}^{p}]\phantom{\rule{0.2em}{0ex}}ds\\ \phantom{\rule{1em}{0ex}}\le \epsilon E(t)+C(\epsilon )(E(t)-E(t+1)).\end{array}$

(35)

On the other hand, it follows from (22) that

$\begin{array}{rcl}E(t)& =& \frac{1}{2}({\parallel {u}_{xx}\parallel}^{2}-|u(l,t){|}^{p})+\frac{1}{2}{\parallel {u}_{x}\parallel}^{2}+\frac{k}{2(m+1)}{\parallel {u}_{x}\parallel}^{2(m+1)}+\frac{1}{2}{\parallel {u}_{t}\parallel}^{2}\\ +\frac{1}{2}|{u}_{t}(l,t){|}^{2}+\frac{1}{2}|{u}_{xx}(l,t){|}^{2}+\frac{p-2}{2p}|u(l,t){|}^{p}\\ \le & \frac{1}{2}({\parallel {u}_{xx}\parallel}^{2}-|u(l,t){|}^{p})+\frac{1}{2}{\parallel {u}_{x}\parallel}^{2}+\frac{k}{2(m+1)}{\parallel {u}_{x}\parallel}^{2(m+1)}+\frac{1}{2}{\parallel {u}_{t}\parallel}^{2}\\ +\frac{1}{2}|{u}_{t}(l,t){|}^{2}+\frac{1}{2}|{u}_{xx}(l,t){|}^{2}+\frac{p-2}{2(p-1)}E(t).\end{array}$

(36)

Then we have

$\begin{array}{rcl}\frac{p}{2(p-1)}E(t)& \le & \frac{1}{2}[{\parallel {u}_{xx}\parallel}^{2}+{\parallel {u}_{x}\parallel}^{2}+k{\parallel {u}_{x}\parallel}^{2(m+1)}+|{u}_{xx}(l,t){|}^{2}-|u(l,t){|}^{p}]\\ +\frac{1}{2}{\parallel {u}_{t}\parallel}^{2}+\frac{1}{2}|{u}_{t}(l,t){|}^{2}.\end{array}$

(37)

Therefore, by (37), (35) and (26), we arrive at

$\begin{array}{rcl}{\int}_{{t}_{1}}^{{t}_{2}}E(s)\phantom{\rule{0.2em}{0ex}}ds& \le & C{\int}_{{t}_{1}}^{{t}_{2}}[{\parallel {u}_{xx}\parallel}^{2}+{\parallel {u}_{x}\parallel}^{2}+k{\parallel {u}_{x}\parallel}^{2(m+1)}\\ +|{u}_{xx}(l,s){|}^{2}-|u(l,s){|}^{p}]\phantom{\rule{0.2em}{0ex}}ds+C{\int}_{{t}_{1}}^{{t}_{2}}({\parallel {u}_{t}\parallel}^{2}+|{u}_{t}(l,s){|}^{2})\phantom{\rule{0.2em}{0ex}}ds\\ \le & {\epsilon}_{1}E(t)+C({\epsilon}_{1})(E(t)-E(t+1))\\ \le & {\epsilon}_{1}\underset{s\in [t,t+1]}{sup}E(s)+C({\epsilon}_{1})(E(t)-E(t+1)).\end{array}$

(38)

Since

$E(t)$ is non-increasing, we choose

${t}_{3}\in [{t}_{1},{t}_{2}]$ such that

$E({t}_{3})\le C{\int}_{{t}_{1}}^{{t}_{2}}E(s)\phantom{\rule{0.2em}{0ex}}ds.$

(39)

Then, using (26),

${t}_{3}<t+1$, and the fact that

$E(t)$ is non-increasing, we have

$\begin{array}{rcl}E(t)& =& E(t+1)+{\int}_{t}^{t+1}[{\parallel {u}_{t}(s)\parallel}^{2}+|{u}_{t}(l,s){|}^{2}]\phantom{\rule{0.2em}{0ex}}ds\le E({t}_{3})+{\int}_{t}^{t+1}[{\parallel {u}_{t}(s)\parallel}^{2}+|{u}_{t}(l,s){|}^{2}]\phantom{\rule{0.2em}{0ex}}ds\\ \le & C{\int}_{{t}_{1}}^{{t}_{2}}E(s)\phantom{\rule{0.2em}{0ex}}ds+{\int}_{t}^{t+1}[{\parallel {u}_{t}(s)\parallel}^{2}+|{u}_{t}(l,s){|}^{2}]\phantom{\rule{0.2em}{0ex}}ds.\end{array}$

(40)

Since

$E(t)$ is non-increasing, combining this with (40), (38) and (26), we have

$\underset{s\in [t,t+1]}{sup}E(s)\le {\epsilon}_{1}\underset{s\in [t,t+1]}{sup}E(s)+C({\epsilon}_{1})(E(t)-E(t+1)).$

(41)

Choosing

${\epsilon}_{1}$ sufficiently small, (41) leads to

$\underset{s\in [t,t+1]}{sup}E(s)\le C(E(t)-E(t+1)),$

(42)

then, applying Lemma 2.2, we obtain the energy decay. □