\documentclass[reqno]{amsart}
\usepackage{hyperref}
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\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2015 (2015), No. 117, pp. 1--15.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2015 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2015/117\hfil Global well-posedness]
{Global well-posedness of damped multidimensional generalized 
 Boussinesq equations}

\author[Y. Niu, X. Peng, M. Zhang \hfil EJDE-2015/117\hfilneg]
{Yi Niu, Xiuyan Peng, Mingyou Zhang}

\address{Yi Niu \newline
College of Automation, Harbin Engineering
University, 150001, China}
\email{niuyipde@163.com}

\address{Xiuyan Peng \newline
College of Automation, Harbin Engineering
University, 150001, China}
\email{pxygll@163.com}

\address{Mingyou Zhang \newline
College of Automation, Harbin Engineering
University, 150001, China}
\email{zmy1985624@163.com}

\thanks{Submitted June 3, 2014. Published April 30, 2015.}
\subjclass[2010]{35A01, 35B44, 35L75}
\keywords{Cauchy problem; global solution; finite time blow up; damping term}

\begin{abstract}
 We study the Cauchy problem for a  sixth-order Boussinesq equations
 with the generalized source term and damping term.
 By using Galerkin approximations and potential well methods, we prove
 the existence of a global weak solution. Furthermore, we study the conditions
 for the damped coefficient to obtain the finite time blow up of the solution.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{remark}[theorem]{Remark}
\newtheorem{definition}[theorem]{Definition}
\newtheorem{corollary}[theorem]{Corollary}
\allowdisplaybreaks

\section{Introduction} \label{sec1}

In this article, we consider the Cauchy problem for damped multidimensional 
generalized Boussinesq equations
\begin{gather}
u_{tt}-\Delta u-\Delta u_{tt}+\Delta^2 u_{tt}- k\Delta
u_t=\Delta  f(u), \quad  x\in\mathbb{R}^n, \; t>0,\label{1.1}\\
u(x,0)=u_0(x), \quad u_t(x,0)=u_1(x), \quad  x\in\mathbb{R}^n, \label{1.2}
\end{gather}
where $k$ is a positive constant, and  $f(u)$ satisfies
\begin{itemize}
\item[(A1)]  $f(u)=-|u|^{p-1}u$, 
 $\frac{n+2}{n}\le p<\frac{n+2}{n-2}$ for $n\ge 3$, $1<p<\infty$ for  $n=1,2$.
\end{itemize}

Boussinesq \cite{s1} first derived the equation
\begin{equation}\label{1.3}
u_{tt}=-\gamma u_{xxxx}+ u_{xx}+(u^2)_{xx},
\end{equation}
to describe the propagation of small amplitude long waves on 
the surface of shallow water. Later, Makhankov \cite{s5} obtained that 
the improved Boussinesq  equation (IBq),
\begin{equation} \label{1.6}
u_{tt}-\Delta u-\Delta u_{tt}=\Delta (u^2), \; x\in\mathbb{R}^n, \; t>0,
\end{equation}
which can be derived by using the exact hydro-dynamical set of equations
in plasma. A modification of the IBq equation analogous to the
 modified Korteweg-de Vries equation yields
\begin{equation}\label{1.7}
u_{tt}-\Delta u-\Delta u_{tt}=\Delta (u^3).
\end{equation}
This equation is the so-called IMBq (modified IBq) equation.
Wang and Chen \cite{s6,s7} considered the existence of local and global solutions,
the nonexistence of solutions, and the existence of global small amplitude
solutions for \eqref{1.7} with a general source term $\Delta f(u)$.

Through investigating the water wave problem, Schneider \cite{s3}
 improved the model \eqref{1.3} as follows
\begin{equation}
u_{tt}-u_{xx}-u_{xxtt}- \mu u_{xxxx}+u_{xxxxtt}=(u^2)_{xx}, \label{1.4}
\end{equation}
where $x, t, \mu\in \mathbb{R}$ and $u(x,t)\in \mathbb{R}$. 
This nonlinear wave equation not only models the water wave problem
 with surface tension, but can also be formally derived from the 
two-dimensional water wave problem. 
Because of the linear instability, equation \eqref{1.4} with $\mu>0$ 
is known as the ``bad'' Boussinesq equation. For the case $\mu=-1$, 
Wang and Mu \cite{s14} showed that  \eqref{1.4} has blow up and 
scattering solution. By using contracting mapping principle,
 Wang and Guo \cite{s15} proved the existence and uniqueness for the
 Cauchy problem \eqref{1.4} with $\mu=-1$. 
Furthermore, they gave the sufficient conditions of blowup of the 
solution for the problem in finite time. For the multidimensional 
case \eqref{1.4} and the special case of nonlinear term like $u^p$, 
the Littlewood-Paley dyadic decomposition guarantees the global existence 
and scattering results of solution with the small initial data \cite{s16}.

Xu and Liu \cite{s10} considered the initial boundary value problem 
of the generalized Pochhammer-Chree equation
\begin{align}
u_{tt}-u_{xx}-u_{xxt}-u_{xxtt}=f(u)_{xx},
\end{align}
where $x\in\Omega=(0,1)$. By using the contract mapping principle, 
they established the  existence of local solutions. 
 After modifying the source term $f(u)_{xx}$, they discussed the  $W^{k,p}$  global
solution and global nonexistence of generalized IMBq equations. 
Necat Polat \cite{s13,s20} studied the Cauchy problem of the generalized 
damped multidimensional Boussinesq equation with double dispersive term
\begin{equation}\label{1.10}
u_{tt}-\Delta u-\Delta u_{tt}+\Delta^2 u- k\Delta u_t=\Delta
 f(u).
\end{equation}
First, by starting with the contraction mapping principle, 
the authors pointed out the locally well-posedness of the Cauchy problem. 
Then the authors obtain the necessary a priori bound. 
Thanks to the a priori bound, every local solution is indeed global in time. 
Finally, by using the concavity method, the authors proved that the 
local solution of the Cauchy problem blows up in finite time with negative 
and nonnegative initial energy.
 Unfortunately, it is much less known for the sixth order equations with 
strong damping term.

In this article, we study the Cauchy problem \eqref{1.1}, \eqref{1.2}, 
which is not only the multidimensional generalized sixth order Boussinesq equation,
 but also includes both nondecreasing source term and strong damping term. 
To deal with such problem, we refer to the papers \cite{s8,s9,s10,s17,s18,s19}
especially the work by Xu and Liu \cite{s17} who proved that the Cauchy 
problem \eqref{1.1}, \eqref{1.2} for the multidimensional sixth order 
equation with the generalized source term has the global $H^m$ solution
 and the finite time blow up solution. 
However, the method they employed can not be used directly to solve the 
sixth order Boussinesq equation with strong damping term considered in
this article. So we improve the standard concavity method and exploit 
further the character of the Nehari manifold in order to give a threshold 
result of global existence and nonexistence of solutions, and point that 
the solution blows up in correspondence of the sufficiently small 
damping coefficient. This paper is organized as follows.

In Section 2  we give some preliminary lemmas and local existence theorem.
In Section 3  we give the sufficient conditions for existence and 
nonexistence of global weak solution for problem \eqref{1.1}, \eqref{1.2}, 
and provide the proofs of the main Theorems.
In Section 4  we give some remarks on the main results ofthis article.


\section{Preliminary lemmas and local existence}\label{sec3}

To prove the existence of local solutions, and the main results of this article,
we provide some preliminary lemmas. 
Firstly, we denote $L^p(\mathbb{R}^n)$ and $H^s (\mathbb{R}^n)$ by $L^p$ 
and $H^s$ respectively, with the norm 
$\|\cdot\|_p=\|\cdot\|_{L^p(\mathbb{R}^n)}$, 
$\|\cdot\|=\|\cdot\|_{L^2(\mathbb{R}^n)}$ and the inner product 
$(u,v)=\int_{\mathbb{R}^n} uv {\rm d}x$. We also define the space
$$
H=\{u\in H^1: (-\Delta)^{-1/2} u\in L^2\},
$$
with the norm
$$
\|u\|_{H}^2=\|u\|_{H^1}^2+ \|(-\Delta)^{-1/2} u\|_{L^2}^2,
$$
where $(-\Delta)^{-\alpha}v= \mathscr{F}^{-1} \left(
|\xi|^{-2\alpha} \mathscr{F} v\right)$, $\mathscr{F}$ and
$\mathscr{F}^{-1}$ are the Fourier transformation and the inverse
Fourier transformation respectively.
For problem \eqref{1.1}-\eqref{1.2} we introduce  the  functionals
\begin{gather*}
J(u)= \frac{1}{2} \|u\|^2+
\int_{\mathbb{R}^n} F(u) dx,\quad F(u)=\int_0^u f(s){\rm d} s, \\
I(u)= \|u\|^2+ \int_{\mathbb{R}^n} uf(u) dx, \\
d=\inf_{u\in\mathcal{N}} J(u),\quad
 \mathcal{N}=\{u\in H^1\mid I(u)=0, \|u\|\neq 0\}.
\end{gather*}
And  we define the following subsets of $H^1(\mathbb{R}^n)$:
\begin{gather*}
W= \{u\in H^1: I(u)>0,\;J(u)<d\}\cup\{0\};\\
V= \{u\in H^1: I(u)<0,\;J(u)<d\},
W'= \{u\in H^1: I(u)>0\}\cup\{0\};\\
V'= \{u\in H^1: I(u)<0\}.
\end{gather*}

\begin{definition}\label{def1.1} \rm
We call $u(x,t)$ a weak solution of problem \eqref{1.1}, \eqref{1.2}
on $\mathbb{R}^n\times [0,T)$, if 
$u\in L^\infty(0,T;H^1)$, $u_t \in L^\infty(0,T;H)$
satisfy 
\begin{itemize}
\item[(i)] for all $v\in H$ and all $t\in [0,T)$,
\begin{equation}\label{1.13a}
\begin{split}
&\left( (-\Delta)^{-1/2} u_t, (-\Delta)^{-1/2}
v\right)+(\nabla u_t,\nabla v)+(u_t,v)+ k(u, v)\\
&+ \int_0^t \left((u,v)+ (f(u),v)\right) {\rm d}\tau \\
&=\left( (-\Delta)^{-1/2} u_1, (-\Delta)^{-1/2}
v\right)+(\nabla u_1,\nabla v)+ k(u_0, v)\,.
 \end{split}
\end{equation}

\item[(ii)] There holds $u(x,0)=u_0(x)$ in $H^1$ and
\begin{equation}\label{1.14a}
 u_t(x,0)=u_1(x) \quad \text{in } H.
\end{equation}

\item[(iii)]  for all $t\in [0,T)$,
\begin{equation}\label{1.15a}
E(t)+k\int_0^t\|u_{\tau}\|^2 {\rm d} \tau \le E(0)
\end{equation}
where
$$
E(t)=\frac{1}{2}\|u_t \|_H^2+\frac{1}{2}\|u \|+^2
\int_{\mathbb{R}^n} F(u) {\rm d} x,\quad F(u)=\int_0^u f(s){\rm d} s.
$$
\end{itemize}
\end{definition}

We present the following theorem  about local existence \cite{s3,s13}.

\begin{theorem} \label{thm2.2}
Suppose that $f(x)$ satisfies {\rm (A1)} and $u_0(x),u_1(x)\in H$.
 Then \eqref{1.1}- \eqref{1.2} admits a unique local solution $u(x,t)\in H$.
\end{theorem}

Let $u_0\in H^1$, $u_1\in H$, $\{w_j\}_{j=1}^\infty$ be a basis
function system in $H$. We construct the approximate solutions of
problem \eqref{1.1}, \eqref{1.2}

\begin{equation}\label{4.1}
u_m (x,t)=\sum_{j=1}^m g_{jm} (t) w_j (x), \ m=1,2,\dots,
\end{equation}
satisfying
\begin{gather}\label{4.2}
\begin{aligned}
&\left( (-\Delta)^{-1/2} u_{mtt}, (-\Delta)^{-1/2}
w_s\right)+ (u_m,w_s) + (u_{mtt}, w_s)\\
 &+(\nabla u_{mtt}, \nabla w_s)+ k(u_{mt},w_s)+ (f(u_m),w_s)=0, \quad s=1,2,\dots, m,
\end{aligned} \\
\label{4.3}
u_m (x,0)=\sum_{j=1}^m a_{jm}  w_j (x)\to u_0(x) \quad \text{in } H^1,\\
\label{4.4}
 u_{mt} (x,0)=\sum_{j=1}^m b_{jm}   w_j
(x)\to  u_1(x) \quad \text{in } H.
\end{gather}
Multiplying \eqref{4.2} by $g'_{sm}(t)$ and summing for $s$ we obtain
$$
\frac{\rm d}{{\rm d}t} E_m(t)+k\|u_{mt}\|^2=0
$$
and
\begin{equation}\label{4.5}
E_m (t)+k\int_0^t\|u_{m\tau}\|^2{\rm d}\tau=E_m(0),
\end{equation}
where
\begin{equation}\label{4.6}
E_m(t)=\frac{1}{2}\|u_{mt}\|_H^2+\frac{1}{2}\|u_{m}\|^2
+\int_{\mathbb{R}^n} F(u_m) {\rm d} x,\ \ F(u)=\int_0^u f(s){\rm d} s.
\end{equation}

\begin{lemma}\label{lem2.2}
Let $f(u)$ satisfy {\rm (A1)} and $u\in H^1$. We have
\begin{itemize}
\item[(i)]
$\lim_{\lambda\to 0} J(\lambda u)=0$.

\item[(ii)]
$I(\lambda u)=\lambda\frac{\rm d}{{\rm d}\lambda} J(\lambda u)$,
$\forall\ \lambda >0$.
Furthermore if $\int_{\mathbb{R}^n}  uf(u){\rm d}x<0$ and
$\varphi(\lambda)=-\frac{1}{\lambda} \int_{\mathbb{R}^n}  uf(\lambda
u){\rm d}x$, then $I(\lambda u)>0$ for $\forall \lambda>0$.

\item[(iii)]
$\lim_{\lambda\to +\infty} J(\lambda u)=-\infty$.


\item[(iv)]
$\varphi(\lambda)$ is increasing on $0<\lambda<\infty$.

\item[(v)]
$\lim_{\lambda\to 0}\varphi(\lambda)=0$, $\lim_{\lambda\to
+\infty}\varphi(\lambda)=+\infty$.


\item[(vi)] In the interval $0<\lambda <\infty$, there exists a
unique $\lambda^\ast=\lambda^\ast(u)$ such that
$$
\frac{\rm d}{{\rm d}\lambda} J(\lambda
u)\Big|_{\lambda=\lambda^\ast}=0.
$$

\item[(vii)]
$J(\lambda u)$ is increasing on $0<\lambda\le\lambda^\ast$,
decreasing on $\lambda^\ast\le \lambda<\infty$ and takes the maximum
at $\lambda=\lambda^\ast$.

\item[(viii)]
$I(\lambda u)>0$ for $0<\lambda<\lambda^\ast$, $I(\lambda u)<0$ for
$\lambda^\ast<\lambda<\infty$ and $I(\lambda^\ast u)=0$.
\end{itemize}
\end{lemma}

\begin{proof}
Parts (i)--(iii) are obvious.
Part (iv) and Part (v) follow from
$$
\varphi(\lambda)=-\frac{1}{\lambda} \int_{\mathbb{R}^n}  uf(\lambda
u){\rm d}x=-\lambda^{p-1} \int_{\mathbb{R}^n}  uf( u){\rm d}x.
$$
Note that $\int_{\mathbb{R}^n}  uf( u){\rm d}x\neq 0$ implies
$\|u\|\neq 0$ and
\begin{equation}\label{2.1o}
\frac{\rm d}{{\rm d}\lambda} J(\lambda u)= \lambda\left(
\|u\|^2-\varphi(\lambda)\right),
\end{equation}
which together with Part (iv) and Part (v) give Part (vi) and Part
(vii).

Part (viii) follows from Part (ii) and \eqref{2.1o}.
\end{proof}


\begin{lemma}\label{lem2.3}
Let $f(u)$ satisfy {\rm (A1)} and $u\in H^1$. We obtain
\begin{itemize}
\item[(i)]
If $0<\|u\|<r_0$, then $I(u)>0$;

\item[(ii)]
If $I(u)<0$, then $\|u\|>r_0$;

\item[(iii)]
If $I(u)=0$ and $\|u\|\neq 0$, i.e. $u\in\mathcal{N}$, then
$\|u\|\ge r_0$, where
$$
r_0=\Big(\frac{1}{a C_\ast^{p+1}}\Big)^\frac{1}{p-1}, \quad
C_\ast=\sup_{u\in H^1, u\neq 0} \frac{\|u\|_{p+1}}{\|u\|}.
$$
\end{itemize}
\end{lemma}

\begin{proof}
(i) If $0<\|u\|<r_0$, then $I(u)>0$ follows from
\begin{align*}
-\int_{\mathbb{R}^n} uf(u) dx 
&\le \int_{\mathbb{R}^n} |uf(u)| dx
=a\|u\|_{p+1}^{p+1}\le a C_\ast^{p+1} \|u\|^{p+1}\\
&=a C_\ast^{p+1} \|u\|^{p-1} \|u\|^{2}< \|u\|^{2}.
\end{align*}

(ii)
If $I(u)<0$, then $\|u\|>r_0$ follows from
$$
\|u\|^{2}< - \int_{\mathbb{R}^n} uf(u) dx\le a C_\ast^{p+1}
\|u\|^{p-1} \|u\|^{2}.
$$

\item[(iii)]
If $I(u)=0$ and $\|u\|\neq 0$, then we have
$$
\|u\|^{2}= - \int_{\mathbb{R}^n} uf(u) dx\le a C_\ast^{p+1}
\|u\|^{p-1} \|u\|^{2},
$$
which together with $\|u\|\neq 0$ gives $\|u\|\ge r_0$.
\end{proof}

\begin{lemma}\label{lem2.4}
Let $f(u)$ satisfy {\rm (A1)}, we have
\begin{itemize}
\item[(i)]
\begin{equation}\label{2.1}
d\ge d_0=\frac{p-1}{2(p+1)} \Big(
\frac{1}{aC_\ast^{p+1}}\Big)^{\frac{2}{p-1}}.
\end{equation}

\item[(ii)]
If  $u\in H^1$ and $I(u)<0$, then
\begin{equation}\label{2.3a}
I(u)< (p+1)\left(J(u)-d\right).
\end{equation}
\end{itemize}
\end{lemma}

\begin{proof}
(i) For any $u\in\mathcal{N}$, by Lemma \ref{lem2.3} we have
$\|u\|\ge r_0$ and
\begin{align*}
J(u)&= \frac{1}{2} \|u\|^2+ \int_{\mathbb{R}^n} F(u) dx=
\frac{1}{2} \|u\|^2+\frac{1}{p+1} \int_{\mathbb{R}^n} u f(u)
dx\\
&= \Big(\frac{1}{2}-\frac{1}{p+1}\Big) \|u\|^2+
\frac{1}{p+1} I(u)=\frac{p-1}{2(p+1)} \|u\|^2 \ge
\frac{p-1}{2(p+1)} r^2_0,
\end{align*}
which gives \eqref{2.1}.

(ii) Let $u\in H^1$ and $I(u)<0$, then from Lemma \ref{lem2.2} it follows
that there exists a $\lambda^\ast$ such that $0<\lambda^\ast<1$ and
$I(\lambda^\ast u)=0$. From the definition of $d$ we have
\begin{equation}\label{2.4}
\begin{split}
d
&\le J(\lambda^\ast u)
= \frac{1}{2}\| \lambda^\ast u \|^2+
\int_{\mathbb{R}^n} F(\lambda^\ast u) dx\\
&= \frac{1}{2}\| \lambda^\ast u \|^2+ \frac{1}{p+1}
\int_{\mathbb{R}^n} \lambda^\ast u f(\lambda^* u)dx\\
&=\Big(\frac{1}{2}-\frac{1}{p+1}\Big) \|\lambda^* u\|^2+
\frac{1}{p+1} I(\lambda^\ast u)\\
&=  \frac{p-1}{2(p+1)} \|\lambda^\ast u\|^2=\lambda^{\ast 2}
\frac{p-1}{2(p+1)} \| u\|^2 \\
&< \frac{p-1}{2(p+1)} \| u\|^2.
\end{split}
\end{equation}
From \eqref{2.4} and
$$
J(u)=\frac{p-1}{2(p+1)} \|  u\|^2+ \frac{1}{p+1} I(u),
$$
we obtain
$$
d<\frac{p-1}{2(p+1)} \|  u\|^2=J(u)- \frac{1}{p+1} I( u),
$$
which gives \eqref{2.3a}.
\end{proof}

\begin{lemma}\label{lem4.1}
Let $f(u)$ satisfy {\rm (A1)},  $u_0\in H^1$ and $u_1\in H $. 
We conclude that $F(u_0)\in L^1$. And for the approximate solutions 
$u_m$ defined by \eqref{4.1}--\eqref{4.4}, there holds $E_m(0)\to E(0)$ 
as $m\to\infty$, where
$$
E(0)=\frac{1}{2}\left(\|u_1\|_H^2 +\|u_0\|^2\right)+\int_{\mathbb{R}^n} F(u_0) {\rm d} x.
$$
\end{lemma}

\begin{proof}
First from the assumptions we have
$$
|F(u)|\le \frac{a }{p +1} |u|^{p +1}, \quad \forall u\in \mathbb{R},
$$
where $\frac{2n+2}{n}\ \le p +1< \frac{2n}{n-2}$ for 
$n\ge 3$ or  $2<p +1 <\infty$ for $n=1,2$. It is obvious that $F(u_0)\in L^1$.

From \eqref{4.3} and \eqref{4.4}  we obtain that as $m\to\infty$
\begin{align*}
&\|(-\Delta)^{-1/2} u_{mt}(0)\|^2+
\|u_m(0)\|^2+
\|u_{mt}(0)\|^2+\|\nabla u_{mt}(0)\|^2 \\
&\to  \|(-\Delta)^{-1/2} u_{1}\|^2+ \|u_0\|^2+
\|u_1\|^2+\|\nabla u_1\|^2=\|u_1\|_H^2.
\end{align*}
Next we prove that
$$
\int_{\mathbb{R}^n} F(u_m(0)) {\rm d} x \to \int_{\mathbb{R}^n}
F(u_0) {\rm d} x \ \text{as} \ m\to \infty.
$$
In fact we have
\begin{align*}
\Big| \int_{\mathbb{R}^n} F(u_m(0)) {\rm d} x -
\int_{\mathbb{R}^n} F(u_0) {\rm d} x\Big| 
&\le \int_{\mathbb{R}^n}|f(\varphi_m)| |u_m(0)-u_0| {\rm d} x\\
&\le  \|f(\varphi_m)\|_r \|u_m(0)-u_0\|_q,
\end{align*}
where $1<q$, $r<\infty$, $\frac{1}{q}+\frac{1}{r}=1$, 
$\varphi_m=u_0+\theta\left(u_m(0)-u_0\right)$, $0<\theta<1$.

(i) If $n\ge 3$. Choose $q=\frac{2n}{n-2}$, $r=\frac{2n}{n+2}$. We have
\begin{gather*}
\|u_m(0)-u_0\|_q \le C\|u_m(0)-u_0\|\to 0 \quad \text{as } m\to\infty,\\
\|f(\varphi_m)\|_r^r = \int_{\mathbb{R}^n} \left(a |\varphi_m|^{p }
\right)^r {\rm d}x=A  \|\varphi_m\|_{p r}^{p  r}.
\end{gather*}
From the conditions we have $2\le p  r \le \frac{2n}{n-2}$, hence
$\|f(\varphi_m)\|_r \le C$.

(ii) If $n=1,2$. Choose $q=r=2$, then we have
\begin{gather*}
\|u_m(0)-u_0\|_q \le \|u_m(0)-u_0\| \to 0 \quad \text{as }  m\to\infty.\\
\|f(\varphi_m)\|_r^r=  \|f(\varphi_m)\|^2 \le  A  \|\varphi_m\|_{2 p}^{2 p }.
\end{gather*}
Since $2<2p  <\infty$, we obtain $\|f(\varphi_m)\|_r<C$.

Thus for  two cases above we always have
$$
\int_{\mathbb{R}^n} F(u_m(0)) {\rm d} x \to \int_{\mathbb{R}^n}
F(u_0) {\rm d} x \quad \text{as }  m\to \infty
$$
and $E_m(0)\to E(0)$ as $m\to \infty$.
\end{proof}

\begin{lemma}\label{lem4.3}
Let $f(u)$ satisfy $({\rm A})$, $u_0\in H^1$ and $u_1\in H$, $E(0)<d$. Assume that
 $I(u_0)>0$ or $\|u_0\|= 0$, i.e $u_0\in W'$. Then for the approximate
 solutions $u_m$ defined by
\eqref{4.1}--\eqref{4.4} there holds $u_m\in W'$ for 
$0\le t<\infty$ and sufficiently large $m$. Furthermore we have
\begin{equation}\label{4.8}
\|u_m\|^2\le \frac{2(p+1)}{p-1}d, \quad
\|u_{mt}\|_H^2<2d, \quad 0\le t<\infty,
\end{equation}
for sufficiently large $m$.
\end{lemma}

\begin{proof}
Arguing by contradiction, we assume that there exists a $\bar{t}>0$
such that $u_m (\bar{t})\notin W'$ for some sufficiently large $m$.
Then by the continuity of $I(u_m)$ with respect to $t$ it follows
that there exists a $t_0>0$ such that $u_m(t_0)\in \partial W'$. On
the other hand, from the definition of $W'$ we have $0\notin
\partial W'$. Hence $I(u_m(t_0))=0$ and $\|u_m(t_0)\|\neq 0$ for
some sufficiently large $m$. From the definition of $d$ we obtain
$J(u_m(t_0))\ge d$, which contradicts (by \eqref{4.5})
\begin{equation} \label{4.7}
 E_m(t)=\frac{1}{2}\|u_{mt}\|_H^2+J(u_m)\leq E_m(0)<d, \quad 0\le t<\infty
\end{equation}
for sufficiently large $m$.

On the other hand,
from \eqref{4.7} we obtain that for sufficiently large $m$ there holds
\begin{equation*}
\frac{1}{2}\| u_{mt}\|_H^2+ \frac{p-1}{2(p+1)} \|u_m \|^2+
\frac{1}{p+1} I(u_m )\leq E_m(0)<d, \quad 0\le t<\infty,
\end{equation*}
which together with $u_m(t)\in W'$ gives \eqref{4.8}.
\end{proof}

\section{Existence and nonexistence of global solutions}

We first give the invariance of both subsets $W$ and $V$ of 
$H^1(\mathbb{R}^n)$ under the flow of \eqref{1.1}, \eqref{1.2}.

\begin{theorem}[Invariant sets] \label{thm3.1} 
Let $f(u)$ satisfy {\rm (A1)}, $u_0\in H^1$ and $u_1\in H $. Assume that
$E(0)<d$. Then both sets $W'$ and $V'$ are invariant under the flow
of problem \eqref{1.1}-\eqref{1.2}. Furthermore
\begin{itemize}
\item[(i)]
All weak solutions of problem \eqref{1.1}, \eqref{1.2}  belong to
$W$ provided $u_0\in W'$.

\item[(ii)]
All weak solutions of problem \eqref{1.1}, \eqref{1.2}  belong to
$V$  provided $u_0\in V'$.
\end{itemize}
\end{theorem}

\begin{proof}
We only prove the invariance of $W'$, the proof for the invariance
of $V'$ is similar. Let $u(t)$ be any weak solution of 
\eqref{1.1}, \eqref{1.2} with $u_0\in W'$, 
$T$ be the maximal existence time of $u(t)$. 
 Next we prove that $u(t)\in W'$ for $0<t<T$. Arguing by contradiction 
we assume there is a $\bar{t}\in (0,T)$ such that $u(\bar{t})\notin W'$. 
According to the continuity
of  $I(u(t))$ with respect to $t$, there is a $t_0\in (0,T)$ such
that $u(t_0)\in\partial W'$. From the definition of $W'$ and (i) of
Lemma \ref{lem2.3} we have $B_{r_0}\subset W'$, 
$B_{r_0}=\{u\in H^1: \|u\|<r_0\}$. Hence we know 
$0\notin \partial W'$. So $u(t_0)\in\partial W$ reads
 $I(u(t_0))= 0$ with
$\|u(t_0)\|\neq 0$. The definition of $d$ tells 
$J(u(t_0))\ge d$, which contradicts
\begin{equation}\label{3.1}
\frac{1}{2}\|u_{mt}\|_H^2+k\int_0^t\|u_{\tau}\|^2{\rm d}\tau +
J(u)\le E(0)<d, \quad 0\le t<T.
\end{equation}
So the prove can be completed.
\end{proof}

Next we show the existence  of global solution for  \eqref{1.1}, \eqref{1.2}.
 And we  give some sufficient conditions for global well-posedness and 
finite time blow up. What is more, these results are independent of the 
local existence theory, so they are not restricted by the conditions for 
the local solution.

\begin{theorem}\label{thm4.6}
Let $f(u)$ satisfy {\rm (A1)}, $u_0\in H^1$, $u_1\in H$ and
$E(0)<d$. Then  problem
\eqref{1.1}, \eqref{1.2} admits a global weak solution 
$u(t)\in L^\infty (0,\infty; H^1)$  with 
$u_t (t)\in L^\infty (0,\infty; H)$
and $u(t)\in W$ for $0\le t<\infty$ provided $u_0\in W'$.
\end{theorem}

\begin{proof}
First let us turn to the existence of global solution
for problem \eqref{1.1}, \eqref{1.2}.

For problem \eqref{1.1}, \eqref{1.2}, construct the approximate
solutions $u_m(x,t)$ by \eqref{4.1}-\eqref{4.4}. From Lemma
\ref{lem4.3}, it follows that $\{u_m\}$ in $L^\infty (0,\infty;
H^1)$ and $\{\nabla u_{mt}\}$ in $L^\infty (0,\infty; H)$ 
are bounded respectively.
Moreover by an argument similar to that in the proof of Lemma
\ref{lem4.1} we can get $\{f(u_m)\}$ are bounded in 
$L^\infty (0,\infty; L^r)$, where $r$ is defined in the proof of Lemma
\ref{lem4.1}. Hence there exists a $u$ and a subsequence $\{u_\nu\}$
of $\{u_m\}$ such that as
 $\nu\to \infty$;
$u_\nu\to u$ in $L^\infty\left(0,\infty; H^1\right)$ weakly star and
a.e. in $Q=\mathbb{R}^n\times [0,\infty)$;
$\nabla u_{\nu t}\to \nabla u_t$ in $L^\infty\left(0,\infty;
H\right)$ weakly star;
$f(u_\nu)\to \chi=f(u)$ in $L^\infty\left(0,\infty;L^r\right)$
weakly star.

Integrating \eqref{4.2} with respect to $t$ from $0$ to $t$ we obtain
\begin{equation}\label{4.9}
\begin{split}
&\left( (-\Delta)^{-1/2} u_{mt}, (-\Delta)^{-1/2}
w_s\right) +(\nabla u_{mt}, \nabla w_s)+(u_{mt},w_s)\\
&+ k(u_m, w_s)+\int_0^t \left( (u_m,w_s)+(f(u_m),w_s)\right){\rm d}\tau\\
&=  \left( (-\Delta)^{-1/2} u_{mt}(0),
(-\Delta)^{-1/2} w_s\right)+(\nabla
u_{mt}(0), \nabla w_s)\\
&\quad +(u_{mt}(0),w_s)+ k(u_m(0), w_s).
\end{split}
\end{equation}
Let $m=\nu\to\infty$ in \eqref{4.9} we obtain
\begin{align*}
&\left( (-\Delta)^{-1/2} u_{t}, (-\Delta)^{-1/2}
w_s\right) +(\nabla u_{t}, \nabla w_s)+(u_t, w_s)\\
&+ k(u, w_s)+\int_0^t \left( (u,w_s)+ (f(u),w_s)\right){\rm d}\tau\\
&=\left( (-\Delta)^{-1/2} u_{1}, (-\Delta)^{-1/2}
w_s\right)+(\nabla u_{1}, \nabla w_s)+(u_{1},w_s)+k(u_0, w_s) ,\
\forall s
\end{align*}
and
\begin{align*}
&\left( (-\Delta)^{-1/2} u_{t}, (-\Delta)^{-1/2}
v \right) +(\nabla u_{t}, \nabla v)+(u_t,v)\\
&+ k(u, v)+\int_0^t \left( (u,v)+(f(u),v)\right){\rm d}\tau\\
&=  \left( (-\Delta)^{-1/2} u_{1}, (-\Delta)^{-1/2}
v\right)+(\nabla u_{1}, \nabla v)+(u_1,v)+k(u_0, v),
  \end{align*}
for all $v\in H$ and all $t\in[0,\infty)$.
On the other hand, from \eqref{4.3}, \eqref{4.4} we obtain
\begin{gather*}
u(x,0)=u_0(x)\quad \text{in } H^1, \\
u_t (x,0)=u_1(x)\quad \text{in } H.
\end{gather*}

Next we prove that above $u$ satisfies \eqref{1.15a}.
Note that the embedding $H^1 \hookrightarrow L^{p+1}$ is compact
under the condition $2\left(1+\frac{1}{n}\right)\le p+1<
\frac{2n}{n-2}$ for $n\ge 3$ or  $2<p+1<\infty$ for
$n=1,2$. Thus from $\{u_m\}$ is bounded in $L^\infty (0,\infty;
H^1)$ it follows that there exists a subsequence $\{u_\nu\}$ of
$\{u_m\}$ such that as $\nu\to\infty$
$u_\nu\to u$ in $L^{p+1}$ strongly for each $t>0$.
Hence
\begin{equation*}
\Big| \int_{\mathbb{R}^n} F(u_\nu) {\rm d} x - \int_{\mathbb{R}^n}
F(u) {\rm d} x\Big| 
\le \int_{\mathbb{R}^n} |f(v_\nu)| |u_\nu-u| {\rm d} x
\le  \|f(v_\nu)\|_{\bar{r}} \|u_\nu-u\|_{\bar{q}},
\end{equation*}
where $\bar{q}=p+1$, $\bar{r}=\frac{p+1}{p}$, $u_\nu=u+\theta
(u_\nu-u)$, $0<\theta<1$. From
$\|u_\nu-u\|_{\bar{q}}\to 0$  as $\nu\to\infty$
and
$$
\|f(v_\nu)\|_{\bar{r}}^{\bar{r}}= \int_{\mathbb{R}^n}
\left(a|v_\nu|^p\right)^{\bar{r}} {\rm d}x= a^{\frac{p+1}{p}}
\|v_\nu\|_{p+1}^{p+1}\le C,
$$
we obtain
$$
\int_{\mathbb{R}^n} F(u_\nu) {\rm d} x \to \int_{\mathbb{R}^n} F(u)
{\rm d} x \quad \text{as }  \nu\to\infty.
$$
Hence from \eqref{4.5} we obtain
\begin{align*}
&\frac{1}{2}\left(\|(-\Delta)^{-1/2} u_{t}\|^2+\|\nabla u_t\|^2+\|u_t\|^2+ \|u\|^2
\right)+k\int_0^t\|u_{\tau}\|^2\rm d\tau\\
&\le  \frac{1}{2}\Big(\liminf_{\nu\to\infty}
\|(-\Delta)^{-1/2} u_{\nu t}\|^2+\liminf_{\nu\to\infty} \|\nabla u_{\nu t}\|^2
+\liminf_{\nu\to\infty} \|u_{\nu t}\|^2+
\liminf_{\nu\to\infty} \|u_\nu\|^2 \Big)\\
&\quad +k\liminf_{\nu\to\infty}\int_0^t\|u_{\tau}\|^2\rm d\tau\\
&\le  \liminf_{\nu\to\infty} \Big(
\frac{1}{2}\|(-\Delta)^{-1/2} u_{\nu t}\|^2+
\frac{1}{2}\|\nabla u_{\nu t}\|^2+\frac{1}{2}\|u_{\nu t}\|^2
+\frac{1}{2}\|u_\nu\|^2 +k\int_0^t\|u_{\tau}\|^2\rm d\tau\Big)\\
&=  \liminf_{\nu\to\infty} \Big( E_\nu(0)- \int_{\mathbb{R}^n}
F(u_\nu) {\rm d} x\Big)\\
&=\lim_{\nu\to\infty} \Big( E_\nu(0)-
\int_{\mathbb{R}^n} F(u_\nu) {\rm d} x\Big)\\
&=E (0)- \int_{\mathbb{R}^n} F(u ) {\rm d} x,
\end{align*}
which gives $E(t)\le E(0)$ for $0\le t<\infty$. Therefore 
$u(x)$ is a global weak solution of problem \eqref{1.1},
\eqref{1.2}. 
Finally from Theorem \ref{thm3.1} we obtain $u(t)\in W$
for $0\le t<\infty$.
\end{proof}

\begin{theorem}\label{thm4.5}
Let $f(u)$ satisfy {\rm (A1)}, $u_0\in H^1$, $u_1\in H$, 
$(-\Delta)^{-1/2} u_0\in L^2$ and
$E(0)<d$. Then the solution$u(t)$  of \eqref{1.1}, \eqref{1.2} 
belongs to $L^\infty (0,\infty; H^1)$,  with $u_t (t)\in L^\infty (0,\infty; H)$
 blows up in finite time when $I(u_0)<0$, and $k$ satisfies
\begin{equation} \label{Hk} 
\begin{gathered}  0<k<p-1, \quad \text{if }  E(0)\le0;\\
0<k<(p-1)\sqrt{1-\frac{E(0)}{d_0}}, \quad \text{if } 0<E(0) < d_0,
\end{gathered}
\end{equation}
where
$$
 d_0=\frac{p-1}{2(p+1)} \Big(
\frac{1}{aC_\ast^{p+1}}\Big)^{\frac{2}{p-1}},\quad 
 C_*=\sup_{u\in H^1,\ u\neq0}\frac{\|u\|_{p+1}}{\|u\|}.
$$
\end{theorem}

\begin{proof}
Let  $u(t)\in L^\infty (0,\infty; H^1)$  with $u_t (t)\in L^\infty
(0,\infty; H)$ be any weak solution of  \eqref{1.1},
\eqref{1.2}, $T$ be the maximal existence time of $u(t)$. Now we
need to show $T<\infty$. Arguing by contradiction, we suppose that
$T=+\infty$. Let $\phi(t)=\|u\|_H^2,$ then
\begin{equation*}
\dot{\phi}(t)=2((-\Delta)^{-1/2}
u_t,(-\Delta)^{-1/2} u)+2(\nabla u_t,\nabla u)+2(u_t, u).
\end{equation*}
From Schwartz inequality we obtain
\begin{align*}
& \left(((-\Delta)^{-1/2}u_t,(-\Delta)^{-1/2} u)+ (\nabla u_t,\nabla u)
 +(u_t, u)\right)^2\\
&= ((-\Delta)^{-1/2}u_t, (-\Delta)^{-1/2} u)^2+(\nabla u_t,\nabla u)^2+(u_t, u)^2\\
&\quad+2((-\Delta)^{-1/2} u_t,(-\Delta)^{-1/2}u)(\nabla u_t,\nabla u)
 +2(\nabla u,\nabla u_t)(u,u_t)\\
&\quad+((-\Delta)^{-1/2} u_t,(-\Delta)^{-1/2}u)(u_t, u)\\
&\le \|(-\Delta)^{-1/2} u_t\|^2
\|(-\Delta)^{-1/2}u\|^2+\|\nabla u_t\|^2\|\nabla u\|^2+\|u\|^2\|u_t\|^2\\
&\quad +\|(-\Delta)^{-1/2} u_t\|^2\|\nabla u\|^2
 +\|(-\Delta)^{-1/2} u\|^2\|\nabla u_t\|^2+\|\nabla u\|^2\|u_t\|^2\\
&\quad +\|\nabla u_t\|^2\|u\|^2+\|(-\Delta)^{-1/2} u_t\|^2\|u\|^2
 +\|(-\Delta)^{-1/2} u\|^2\|u_t\|^2\\
&=\|(-\Delta)^{-1/2} u\|^2\|u_t\|_H^2+\|\nabla u\|^2\|u_t\|_H^2
 +\|u\|^2\|u_t\|_H^2.
\end{align*}
we have
\begin{equation}\label{4.10}
\big(\dot{\phi}(t)\big)^2\le 4 \phi (t)\|u_t\|_H^2,
\end{equation}
and
\begin{equation}\label{4.11}
\begin{split}
\ddot{\phi}(t)
&=2\|(-\Delta)^{-1/2} u_t\|^2+2\|\nabla u_{t}\|^2+2\|u_t\|^2\\ 
&\quad +2((-\Delta)^{-1/2}u_{tt},(-\Delta)^{-1/2} u)
 +2(\nabla u_{tt},\nabla u)+2(u,u_{tt})\\
&= 2\|u_t\|_H^2-2((-\Delta)^{-1}u_{tt},u)-2(\Delta u_{tt},u)+2(u_{tt},u)\\
&= 2\|u_t\|_H^2-2(u,u)-2(f(u),u)-2k(u_t,u)\\
&= 2\|u_t\|_H^2-2(u,u)-2(f(u),u)-2k(u_t,u)\\
&= 2\|u_t\|_H^2-2\Big( \|u\|^2+\int_{\mathbb{R}^n} uf(u){\rm d}x\Big)-2k(u_t,u)\\
&= 2\|u_t\|_H^2-2I(u)-2k(u_t,u).
\end{split}
\end{equation}
From \eqref{1.15a} we have
\begin{gather*}
\frac{1}{2}\|u_t\|_H^2+\frac{p-1}{2(p+1)}\|u\|^2+ \frac{1}{p+1} I(u)
\le\frac{1}{2}\|u_t\|_H^2+ J(u)\le E(0)<d, \\
-2I(u)\ge (p+1)\|u_t\|_H^2+(p-1)\|u\|^2-2(p+1)E(0).
\end{gather*}
 Hence we have
 \begin{equation}\label{4.12}
\ddot{\phi}(t)\ge(p+3)\|u_t\|_H^2+(p-1)\|u\|^2-2k(u_t,u)-2(p+1)E(0).
\end{equation}

Next we consider the following three cases:
\smallskip

(i) We consider $E(0)<0$. In this case from $0<k<p-1$ it follows that there exists a
$\varepsilon$ such that $0<\varepsilon<p-1$ and $k^2<(p-1-\varepsilon)(p-1)$.
And \eqref{4.12} gives
\begin{equation}\label{4.13}
 \begin{split}
\ddot{\phi}(t) 
&\ge(4+\varepsilon)\|u_t\|_H^2+(p-1-\varepsilon)
\left(\|(-\Delta)^{-1/2} u_t\|^2+ \|\nabla u_t\|^2\right)\\
&\quad +(p-1-\varepsilon)\|u_t\|^2+(p-1)\|u\|^2-
2k(u_t,u)-2(p+1)E(0).
\end{split}
\end{equation}
And from
\begin{align*}
2k|(u_t,u)|
&\le (p-1-\varepsilon)\|u_t\|^2+\frac{k^2}{p-1-\varepsilon}\|u\|^2\\
&\le (p-1-\varepsilon)\|u_t\|^2+(p-1)\|u\|^2,
\end{align*}
we have
\begin{equation}\label{4.14}
\ddot{\phi}(t) \ge(4+\varepsilon)\|u_t\|_H^2-2(p+1)E(0).
\end{equation}
\smallskip

(ii) Suppose $E(0)=0$. In this case from $0<k<p-1$ it follows that there exists a
$\varepsilon$ such that $0<\varepsilon<p-1$ and $k<p-1-\varepsilon$.
And \eqref{4.12} gives
\begin{equation}\label{4.15}
 \begin{split}
\ddot{\phi}(t) 
&\ge(4+\varepsilon)\|u_t\|_H^2+(p-1-\varepsilon)
\left(\|(-\Delta)^{-1/2} u_t\|^2+ \|\nabla u_t\|^2\right)\\
&\quad +(p-1-\varepsilon)\|u_t\|^2+(p-1-\varepsilon)\|u\|^2
+ \varepsilon\|u\|^2- 2k(u_t,u),
\end{split}
\end{equation}
and from
\begin{align*}
2k|(u_t,u)|
&\le (p-1-\varepsilon)\|u_t\|^2+\frac{k^2}{p-1-\varepsilon}
\|u\|^2\\
&\le (p-1-\varepsilon)\|u_t\|^2+(p-1-\varepsilon)\|u\|^2,
\end{align*}
we have
\begin{equation}\label{4.16}
\ddot{\phi}(t) \ge(4+\varepsilon)\|u_t\|_H^2+\varepsilon\|u\|^2
\ge(4+\varepsilon)\|u_t\|_H^2+\varepsilon r_0^2.
\end{equation}
\smallskip

(iii) We consider $0<E(0)<d_0$. In this case from 
$0<k<(p-1)\sqrt{1-\frac{E(0)}{d_0}}$, it follows that there exists a 
$\varepsilon$ such that $0<\varepsilon<(p-1)\big(1-\frac{E(0)}{d_0}\big)$ 
and $k^2<(p-1-\varepsilon)\big((p-1)\big(1-\frac{E(0)}{d_0}\big)-\varepsilon\big)$. 
And \eqref{4.12} gives
\begin{equation}\label{4.17}
 \begin{split}
\ddot{\phi}(t) 
&\ge (4+\varepsilon)\|u_t\|_H^2+(p-1-\varepsilon)
\Big(\|(-\Delta)^{-1/2} u_t\|^2+ \|\nabla u_t\|^2+\|u_t\|^2\Big)\\
&\quad+\Big((p-1)\big(1-\frac{E(0)}{d_0}\big)-\varepsilon\Big)\|u\|^2
+\varepsilon\|u\|^2\\
&\quad +(p-1)\Big(\frac{E(0)}{d_0}\Big)\|u\|^2
-2k(u_t,u)-2(p+1)E(0).
\end{split}
\end{equation}
And from Theorem \ref{thm3.1} we have $u(t) \in V$ for $0\le t<\infty$. 
By Lemma \ref{lem2.3}, we obtain $\|u\|>r_0$. 
From $d_0=\frac{p-1}{2(p+1)}r_0^2$, we obtain
\begin{equation}\label{4.18}
(p-1)\Big(\frac{E(0)}{d_0}\Big)\|u\|^2\ge
(p-1)\Big(\frac{E(0)}{d_0}\Big)r_0^2=2(p+1)E(0).
\end{equation}
We can derive
\begin{align*}
\ddot{\phi}(t) 
&\ge (4+\varepsilon)\|u_t\|^2+(p-1-\varepsilon)
 \left(\|(-\Delta)^{-1/2} u_t\|^2+\|\nabla u_t\|^2\right)\\
&\quad +(p-1-\varepsilon)\|u_t\|^2+\Big((p-1)\Big(1-\frac{E(0)}{d_0}\Big)
-\varepsilon\Big)\|u\|^2 +\varepsilon\|u\|^2\\
&\quad +(p-1)\Big(\frac{E(0)}{d_0}\Big)\|u\|^2 -2k(u_t,u)-2(p+1)E(0).
\end{align*}
On the other hand, 
\begin{align*}
2k|(u_t,u)|
&\le (p-1-\varepsilon)\|u_t\|^2+\frac{k^2}{(p-1-\varepsilon)}\|u\|^2\\
&\le (p-1-\varepsilon)\|u_t\|^2+\Big((p-1)\Big(1-\frac{E(0)}{d_0}\Big)
-\varepsilon\Big)\|u\|^2.
\end{align*}
 Hence we have
\begin{equation}\label{4.19}
\ddot{\phi}(t) \ge(4+\varepsilon)\|u_t\|_H^2+\varepsilon\|u\|^2>(4+\varepsilon)\|u_t\|_H^2+\varepsilon r_0^2.
\end{equation}

From \eqref{4.14}, \eqref{4.16} and \eqref{4.19}, it follows that there
exists a $\delta_0$ such that for all above cases there holds
\begin{equation}\label{4.20}
\ddot{\phi}(t) \ge (4+\varepsilon)\|u_t\|_H^2+\delta_0.
\end{equation}
Hence
\begin{equation}\label{4.21}
\phi(t)\ddot{\phi}(t)-\frac{\varepsilon+4}{4}\big(\dot{\phi}(t)\big)^2
\ge \delta_0\|u_t\|_H^2\ge 0,
\end{equation}
and
\begin{equation}\label{4.22}
\big(\phi^{-\alpha}(t)\big)''
=\frac{-\alpha}{\phi(t)^{\alpha+2}}
\Big(\phi(t)\ddot{\phi}(t)-(\alpha+1)\big(\dot{\phi}(t)\big)^2\Big)\le
0,
\end{equation}
$$
\alpha=\frac{\varepsilon}{4}, \quad  0< t<\infty.
$$
On the other hand, from  \eqref{4.20}  we obtain
$$
\dot{\phi}(t)\ge \delta_0 t+\dot{\phi}(0), \quad 0< t<\infty.
$$
Hence there exists a $t_0\ge 0$ such that $\dot{\phi}(t)>
\dot{\phi}(t_0)>0$ for $t> t_0$ and
$$
\phi(t)> \dot{\phi}(t_0)(t-t_0)+\phi(t_0)\ge \dot{\phi}(t_0)(t-t_0),
\ t_0<t<\infty.
$$
Therefore there exits a $t_1>0$ such that $\phi(t_1)>0$ and
$\dot{\phi}(t_1)>0$. From this and \eqref{4.22} it follows that
there exists a $T_1>0$ such that
\[
\lim_{t\to T_1} \phi^{-\alpha}(t)=0,
\]
and
\begin{equation}\label{4.23}
\lim_{t\to T_1}\phi(t)=+\infty,
\end{equation}
which contradicts $T=+\infty$. So we prove the nonexistence of
global weak solutions.
\end{proof}

\section{Remarks}

In this section, we  give some remarks on the main results of this paper. 
First Theorem \ref{thm4.5} can be written  as follows

\begin{theorem}\label{thm4.10}
Let $f(u)$, $u_0$ and $u_1$  be same as those in Theorem \ref{thm4.6} 
and \ref{thm4.5}. Assume that $E(0)<d_0$, where $d_0$ is defined in 
Lemma \ref{lem2.4}, i.e.
$$
d_0=\frac{p-1}{2(p+1)} \Big(
\frac{1}{aC_\ast^{p+1}}\Big)^{\frac{2}{p-1}}.
$$
Then when $\|u_0\|<r_0$  problem \eqref{1.1}, \eqref{1.2}
admits a global weak solution; and when $\|u_0\|\ge r_0$,
and $k$ satisfies
\begin{equation} \label{Hk2} 
\begin{gathered}  
0<k<p-1, \quad \text{if }  E(0)\le0;\\
0<k<(p-1)\sqrt{1-\frac{E(0)}{d_0}}, \quad \text{if }  0<E(0) < d_0.
\end{gathered}
\end{equation}
Then problem \eqref{1.1}, \eqref{1.2} does not admit any global weak
solution, where $r_0$ is defined in Lemma \ref{lem2.3}; i.e.,
$$
r_0=\Big(\frac{1}{a C_\ast^{p+1}}\Big)^\frac{1}{p-1}, \quad
C_\ast=\sup_{u\in H^1, u\neq 0} \frac{\|u\|_{p+1}}{\|u\|}.
$$
\end{theorem}

\begin{proof}
We will complete this proof by considering case $\|u_0\|<r_0$
and   case $\|u_0\|\ge r_0$  separately.
\smallskip

(i) Since $\|u_0\|<r_0$ implies $0<\|u_0\|<r_0$ or
$\|u_0\|=0$. If  $0<\|u_0\|<r_0$, from Lemma
\ref{lem2.3} we can derive $I(u_0)>0$. Hence the weak solution 
exists globally.

(ii) If  $\|u_0\|\ge r_0$, then from
\begin{align*}
&\frac{1}{2}\|u_1\|_H^2+k\int_0^t\| u_1\|^2{\rm d}\tau
+ \frac{p-1}{2(p+1)} \|u_0 \|^2+ \frac{1}{p+1}
I(u_0 )\\
&\le E(0)<d_0= \frac{p-1}{2(p+1)} \Big(
\frac{1}{aC_\ast^{p+1}}\Big)^{\frac{2}{p-1}}
= \frac{p-1}{2(p+1)} r_0^2,
\end{align*}
we obtain $I(u_0)<0$. Then  Theorem \ref{thm4.5} gives that there is no
global weak solution for problem \eqref{1.1}, \eqref{1.2}.
\end{proof}

So the results of Theorem \ref{thm4.10} show that the space
 $H=\{u\in H^1\mid (-\Delta)^{-1/2} u\in L^2\}$ is divided into two subspaces:
 $\|u\|<r_0$ and $\|u\|>r_0$ by  the surface $\|u\|=r_0$.
 Furthermore, we have all  weak solutions $u(t)$ of problem 
\eqref{1.1}, \eqref{1.2}
 with $E(0)<d_0$ belong to $B_{r_0} =\{u\in H \mid \|u\|<r_0\}$,
 and problem \eqref{1.1}, \eqref{1.2} does not admit any global weak
 solutions if  $u_0\in \bar{B}_{r_0}^c = \{u\in H \mid \|u\|\geq r_0\}$ 
and $k$ satisfies \eqref{Hk2}.

In the case  $E(0)\le 0$, which is a special case of
the energy restriction $E(0)<d$. We have the following result.

\begin{theorem}\label{thm3.3}
Let $f(u)$ satisfy {\rm (A1)} and $u_0,u_1\in H$. Assume that
$E(0)<0$ or $E(0)=0$, $\|u_0\|\neq 0$.  Then all weak
solutions of problem \eqref{1.1}-\eqref{1.2} belong to $V$.
\end{theorem}

\begin{proof}
Let $u(t)$ be any weak solution of problem \eqref{1.1}-\eqref{1.2}
with $E(0)<0$ or $E(0)=0$, $\|u_0\|\neq 0$, $T$ be the maximal
existence time of $u(t)$. From
\begin{equation*}
\frac{1}{2}\|u_1\|_H^2+k\int_0^t\| u_1\|^2{\rm} d\tau
+\frac{p-1}{2(p+1)} \|u_0\|^2+ \frac{1}{p+1} I(u_0) \le E(0),
\end{equation*}
we see that if $E(0)<0$ or $E(0)=0$ with $\|u_0\|\neq 0$, then
$I(u_0)<0$. Hence from Theorem \ref{thm3.1} we obtain $u(t)\in V$
for $0\le t<T$.
\end{proof}

Furthermore  from  Theorem \ref{thm4.5} and Theorem \ref{thm3.3} we can
conclude the following corollary.

\begin{corollary}\label{coro4.8}
Let $f(u)$ satisfy  {\rm (A1)} and $u_0,u_1\in H$. Assume that
$E(0)<0$ or $E(0)=0$, $\|u_0\|\neq 0$ and $k$ satisfies
\begin{gather*}
0<k<p-1, \quad \text{if }  E(0)\le0;\\
0<k<(p-1)\sqrt{1-\frac{E(0)}{d_0}},\quad\text{if }  0<E(0) < d_0.
\end{gather*}
 Then problem \eqref{1.1}, \eqref{1.2} does not admit any
global weak solution.
\end{corollary}

\begin{corollary}\label{coro4.6a}
Under the conditions of Theorem \ref{thm4.5}, for the global weak
solution of problem \eqref{1.1}-\eqref{1.2} given in Theorem
\ref{thm4.5} we further have
$$
u(t)\in L^\infty (0,T; H), \quad \forall T>0.
$$
\end{corollary}

\begin{proof}
From
$$
(-\Delta)^{-1/2} u= \int_0^t (-\Delta)^{-1/2} u_\tau
{\rm d}\tau+ (-\Delta)^{-1/2} u_0, \quad 0\le t<\infty,
$$
we obtain
\begin{align*}
\|(-\Delta)^{-1/2} u\|
&\le \int_0^t
\|(-\Delta)^{-1/2} u_\tau\| {\rm d}\tau+
\|(-\Delta)^{-1/2} u_0\|\\
&\le T \max_{0\le t\le T} \left(\|(-\Delta)^{-1/2}
u_t\|\right)+ \|(-\Delta)^{-1/2} u_0\|, \quad  0\le t\le T,
\end{align*}
which gives
\begin{gather*}
(-\Delta)^{-1/2} u\in L^\infty (0,T; L^2), \quad \forall T>0, \\
u(t)\in L^\infty (0,T; H), \quad \forall T>0.
\end{gather*}
\end{proof}

\subsection*{Acknowledgements}
This work was supported by the National Natural Science Foundation 
of China (11471087, 41306086).


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