In order to help the readers quickly get the main idea of the present paper, we show the main theorem in the beginning of this section.

**Theorem 3.1** *Let* $2\le s<p+1$,

${u}_{0}(x),{u}_{1}(x)\in {H}^{s}$ *be given and let* (3.1)

*hold*.

*Assume that* $E(0)>0$,

${u}_{0}\in \mathcal{W}$ *and the initial data satisfy* ${\parallel {u}_{0}\parallel}^{2}+{\parallel {u}_{0xx}\parallel}^{2}+2({u}_{0},{u}_{1})+2({u}_{0xx},{u}_{1xx})+\frac{2(p+1)}{p+3}E(0)<0.$

(3.1)

*Then the existence time of a global solution for problem* (1.1)-(1.2) *is infinite*.

In what follows, we show a preliminary lemma about the monotonicity of the functional ${\parallel u(x,t)\parallel}^{2}+{\parallel {u}_{xx}(x,t)\parallel}^{2}$, which will be used to prove the invariance of the new stable set
under the flow of problem (1.1)-(1.2).

**Lemma 3.2** *Let* ${u}_{0}(x),{u}_{1}(x)\in {H}^{2}$ *be given and* $u(x,t)$ *be the solution of Equation* (1.1) *with initial data* $({u}_{0},{u}_{1})$. *Assume that* $E(0)>0$ *and the initial data satisfy Equation* (3.1), *then the map* $\{t\mapsto {\parallel u(t)\parallel}^{2}+{\parallel {u}_{xx}(t)\parallel}^{2}\}$ *is strictly decreasing as long as* $u(x,t)\in \mathcal{W}$.

*Proof* Let

$F(t)={\parallel u(t)\parallel}^{2}+{\parallel {u}_{xx}(t)\parallel}^{2},$

(3.2)

then we get

${F}^{\prime}(t)=2(u,{u}_{t})+2({u}_{xx},{u}_{xxt})$

(3.3)

and

${F}^{\u2033}(t)=2(u,{u}_{tt})+2{\parallel {u}_{t}\parallel}^{2}+2({u}_{xx},{u}_{xxtt})+2{\parallel {u}_{xxt}\parallel}^{2}.$

(3.4)

Note that by testing Equation (

1.1) with

*u* we have

$({u}_{tt},u)+({u}_{xxxxtt},u)+a{\parallel {u}_{x}\parallel}^{2}+{\parallel {u}_{xx}\parallel}^{2}=-(f({u}_{x}),{u}_{x})\phantom{\rule{1em}{0ex}}\text{for}t\in [0,\mathrm{\infty}).$

Then, Equation (

3.4) becomes

${F}^{\u2033}(t)=2{\parallel {u}_{t}\parallel}^{2}+2{\parallel {u}_{xxt}\parallel}^{2}-2I(u).$

(3.5)

Furthermore, from

$u(t)\in \mathcal{W}$ we have

${F}^{\u2033}(t)<0$ for

$t\in [0,{T}_{max})$. Obviously from

$E(0)>0$ and (3.1) we can get

$2{\int}_{\mathbb{R}}{u}_{0}{u}_{1}\phantom{\rule{0.2em}{0ex}}dx+2{\int}_{\mathbb{R}}{u}_{0xx}{u}_{1xx}\phantom{\rule{0.2em}{0ex}}dx<0,$

which implies ${F}^{\prime}(0)=({u}_{0},{u}_{1})+({u}_{0xx},{u}_{1xx})<0$. It is easy to see that ${F}^{\prime}(t)<{F}^{\prime}(0)<0$, namely, ${F}^{\prime}(t)<0$. Therefore, we find that the map $\{t\mapsto {\parallel u(t)\parallel}^{2}+{\parallel {u}_{xx}(t)\parallel}^{2}\}$ is strictly decreasing. □

Subsequently we show the invariance of the new stable set
under the flow of problem (1.1)-(1.2), which plays a key role in proving existence of global solutions for problem (1.1)-(1.2) at high initial energy level $E(0)>0$.

**Lemma 3.3** (Invariant set)

*Let* ${u}_{0}(x),{u}_{1}(x)\in {H}^{2}$ *be given and* $u(x,t)$ *be a weak solution of problem* (1.1)-(1.2) *with maximal existence time interval* $[0,{T}_{0})$, ${T}_{0}\le +\mathrm{\infty}$. *Assume that the initial data satisfy* (3.1). *Then all solutions of problem* (1.1)-(1.2) *with* $E(0)>0$ *belong to*
, *provided* ${u}_{0}\in \mathcal{W}$.

*Proof* We prove

$u(t)\in \mathcal{W}$. If it is false, let

${t}_{0}\in (0,{T}_{0})$ be the first time that

$I(u({t}_{0}))={\parallel {u}_{t}({t}_{0})\parallel}^{2}+{\parallel {u}_{xxt}({t}_{0})\parallel}^{2},$

(3.6)

namely,

$I(u(t))>{\parallel {u}_{t}\parallel}^{2}+{\parallel {u}_{xxt}\parallel}^{2}\phantom{\rule{1em}{0ex}}\text{for}t\in [0,{t}_{0}).$

Let

$F(t)$ be defined as (3.2) above. Hence by Lemma 3.2, we see that

$F(t)$ and

${F}^{\prime}(t)$ are strictly decreasing on the interval

$[0,{t}_{0})$. And then by (3.1), for all

$t\in (0,{t}_{0})$, we have

$F(t)<{\parallel {u}_{0}\parallel}^{2}+{\parallel {u}_{0xx}\parallel}^{2}<-2({u}_{0},{u}_{1})-2({u}_{0xx},{u}_{1xx})-\frac{2(p+1)}{p+3}E(0).$

Therefore from the continuity of

${\parallel u(t)\parallel}^{2}+{\parallel {u}_{xx}(t)\parallel}^{2}$ in

*t* we get

$F({t}_{0})<{\parallel {u}_{0}\parallel}^{2}+{\parallel {u}_{0xx}\parallel}^{2}<-2({u}_{0},{u}_{1})-2({u}_{0xx},{u}_{1xx})-\frac{2(p+1)}{p+3}E(0).$

(3.7)

On the other hand, by (2.1) and (2.2) we can obtain

$\begin{array}{rl}E(0)=& E({t}_{0})=\frac{1}{2}({\parallel {u}_{t}({t}_{0})\parallel}^{2}+a{\parallel {u}_{x}({t}_{0})\parallel}^{2}+{\parallel {u}_{xx}({t}_{0})\parallel}^{2}+{\parallel {u}_{xxt}({t}_{0})\parallel}^{2})\\ +\frac{b}{p+1}{\int}_{\mathbb{R}}{|{u}_{x}({t}_{0})|}^{p}{u}_{x}({t}_{0})\phantom{\rule{0.2em}{0ex}}dx\\ =& \frac{1}{2}({\parallel {u}_{t}({t}_{0})\parallel}^{2}+{\parallel {u}_{xxt}({t}_{0})\parallel}^{2})+\frac{1}{p+1}I(u({t}_{0}))\\ +(\frac{1}{2}-\frac{1}{p+1})(a{\parallel {u}_{x}({t}_{0})\parallel}^{2}+{\parallel {u}_{xx}({t}_{0})\parallel}^{2}).\end{array}$

Recalling (3.6) we have

$\begin{array}{rl}E(0)=& (\frac{1}{2}+\frac{1}{p+1})({\parallel {u}_{t}({t}_{0})\parallel}^{2}+{\parallel {u}_{xxt}({t}_{0})\parallel}^{2})\\ +(\frac{1}{2}-\frac{1}{p+1})(a{\parallel {u}_{x}({t}_{0})\parallel}^{2}+{\parallel {u}_{xx}({t}_{0})\parallel}^{2})\\ \ge & \frac{p+3}{2(p+1)}({\parallel {u}_{t}({t}_{0})\parallel}^{2}+{\parallel {u}_{xxt}({t}_{0})\parallel}^{2}).\end{array}$

(3.8)

Then from the following equalities:

$\begin{array}{r}{\parallel {u}_{t}({t}_{0})\parallel}^{2}={\parallel {u}_{t}({t}_{0})+u({t}_{0})\parallel}^{2}-{\parallel u({t}_{0})\parallel}^{2}-2(u({t}_{0}),{u}_{t}({t}_{0})),\\ {\parallel {u}_{xxt}({t}_{0})\parallel}^{2}={\parallel {u}_{xxt}({t}_{0})+{u}_{xx}({t}_{0})\parallel}^{2}-{\parallel {u}_{xx}({t}_{0})\parallel}^{2}-2({u}_{xx}({t}_{0}),{u}_{xxt}({t}_{0}))\end{array}$

and (3.8), we can derive

$\begin{array}{rl}E(0)\ge & A{\parallel {u}_{t}({t}_{0})+u({t}_{0})\parallel}^{2}+A{\parallel {u}_{xxt}({t}_{0})+{u}_{xx}({t}_{0})\parallel}^{2}-A{\parallel u({t}_{0})\parallel}^{2}\\ -A{\parallel {u}_{xx}({t}_{0})\parallel}^{2}-2A(u({t}_{0}),{u}_{t}({t}_{0}))-2A({u}_{xx}({t}_{0}),{u}_{xxt}({t}_{0}))\\ \ge & -A({\parallel u({t}_{0})\parallel}^{2}+{\parallel {u}_{xx}({t}_{0})\parallel}^{2})-2A(({u}_{0},{u}_{1})+({u}_{0xx},{u}_{1xx})),\end{array}$

where

$A=\frac{p+3}{2(p+1)}$. Or equivalently

$F({t}_{0})\ge -2({u}_{0},{u}_{1})-2({u}_{0xx},{u}_{1xx})-\frac{2(p+1)}{p+3}E(0),$

(3.9)

which contradicts the first inequality of (3.7). This completes the proof. □

At this point we can prove the global existence for the solution of problem (1.1)-(1.2) with arbitrarily positive initial energy.

*Proof of Theorem 3.1* From Theorem 2.1 there exists a unique local solution of problem (1.1)-(1.2) defined on a maximal time interval

$[0,{T}_{0})$. Let

$u(t)$ be the weak solution of problem (1.1)-(1.2) with

$E(0)>0$,

${u}_{0}\in \mathcal{W}$ and (3.1). Then from Lemma 3.3 we have

$u(x,t)\in \mathcal{W}$, namely,

$I(u(t))>{\parallel {u}_{t}\parallel}^{2}+{\parallel {u}_{xxt}\parallel}^{2}\phantom{\rule{1em}{0ex}}\text{for}t\in [0,{T}_{0}).$

(3.10)

Therefore from (2.1)-(2.2) and (3.6), we can obtain

$\begin{array}{rl}E(0)& =E(t)=\frac{1}{2}({\parallel {u}_{t}\parallel}^{2}+a{\parallel {u}_{x}\parallel}^{2}+{\parallel {u}_{xx}\parallel}^{2}+{\parallel {u}_{xxt}\parallel}^{2})+\frac{b}{p+1}{\int}_{\mathbb{R}}{|{u}_{x}|}^{p}{u}_{x}\phantom{\rule{0.2em}{0ex}}dx\\ =\frac{1}{2}({\parallel {u}_{t}\parallel}^{2}+{\parallel {u}_{xxt}\parallel}^{2})+\frac{p-1}{2(p+1)}(a{\parallel {u}_{x}\parallel}^{2}+{\parallel {u}_{xx}\parallel}^{2})+\frac{1}{p+1}I(u)\\ >\frac{p+3}{2(p+1)}({\parallel {u}_{t}\parallel}^{2}+{\parallel {u}_{xxt}\parallel}^{2})+\frac{p-1}{2(p+1)}(a{\parallel {u}_{x}\parallel}^{2}+{\parallel {u}_{xx}\parallel}^{2}),\end{array}$

which implies

$u(x,t),{u}_{t}(x,t)\text{are bounded in}{C}^{1}(0,{T}_{0};{H}^{s}).$

Hence from Theorem 2.1 it follows that ${T}_{0}=\mathrm{\infty}$ and the solution of problem (1.1)-(1.2) exists globally. This completes the proof of Theorem 3.1. □