\documentclass[reqno]{amsart}
\usepackage{hyperref}

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2012 (2012), No. 119, pp. 1--10.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2012 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2012/119\hfil Non-existence of global solutions]
{Non-existence of global solutions to generalized dissipative
Klein-Gordon equations with positive energy}

\author[M. O. Korpusov \hfil EJDE-2012/119\hfilneg]
{Maxim Olegovich Korpusov}  % in alphabetical order

\address{Maxim Olegovich Korpusov \newline
Chair of Mathematics, Faculty of Physics,
 M. V. Lomonosov Moscow State University, Moscow, Russia}
\email{korpusov@gmail.com}

\thanks{Submitted February 14, 2012. Published July 19, 2012.}
\thanks{Supported by grant N 11-01-12018-ofi-m-2011 from the
Russian Foundation for basic research}
\subjclass[2000]{35B44, 35L05, 35L25, 35L67}
\keywords{Blow-up; wave equation; shocks and singularities }

\begin{abstract}
 In this article the initial-boundary-value problem for generalized
 dissipative high-order equation of Klein-Gordon type is considered.
 We continue our study of nonlinear hyperbolic equations and systems
 with arbitrary positive energy.
 The modified concavity method by Levine is used for proving blow-up
 of solutions.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{remark}[theorem]{Remark}
\newtheorem{definition}[theorem]{Definition}
\allowdisplaybreaks

\section{Introduction}

We consider the  initial-boundary-value problem
\begin{gather}\label{1.1}
u_{tt}+\mu u_t+\Delta h_1(x,\Delta u)
-\operatorname{div}(h_2(x,|\nabla u|)\nabla u)
+\operatorname{div}(h_3(x,|\nabla u|)\nabla u)=0, \\
\label{1.2}
u|_{\partial\Omega}=\frac{\partial u}{\partial n_x}\big|_{\partial\Omega}=0,
\quad u(x,0)=u_0(x),\quad u'(x,0)=u_1(x),\quad\mu\geq  0,
\end{gather}
in a bounded domain $\Omega\subset\mathbb{R}^N$
with smooth boundary $\partial\Omega\in\mathbb{C}^{4,\delta}$ for $\delta\in(0,1]$.

Finite time blow-up of solutions of generalized Klein-Gordon equation have
been studied by many authors; see for example
\cite{anl,guowang,chena,chen2,yang1,yang2,liu1}.
In these references,  the authors considere problems either for
negative energy or for weaker conditions than a condition of negative
initial energy (see \cite{liu1,Wang}).
Other authors have assumed a condition of positive energy under other two
conditions on the initial functions.
However, the mentioned authors have not studied the compatibility of these conditions,
which is come times  hard to understand.
Finally these conditions for any fixed $u_0$ and sufficiently large $u_1$
are not compatible.
These authors have used the classic concavity Levine's method.
In this paper we use a modified method, developed in \cite{Al}, that provide
two conditions for which their compatibility is easily checked.

Let us remember that there are five well-known methods for
studying a blow-up phenomena.
The first method is the concavity method developed by
 Levine \cite{Levine1,Levine2,Levine3,Levine4,Levine5,Levine6}.
The second method is the test functions developed by
 Pokhozhaev, Mitidieri  and Zhang \cite{Po, Pohozaev1,Gal, Zhang}.
And the third method based on different criterion of
comparison and  was developed by Samarskii, Galaktionov,
Kurdyumov, Mikhailov \cite{Galact,Samar}.
The forth method based of positive averageswas developed  by
Keller and  Glassey \cite{Keller,Glassey}.
The fifth method, for nonlinear damping, was developed by  by  Georgiev and
 Todorova \cite{Todorova}.

\section{Main differential inequality}

Consider the important differential inequality
\begin{equation}\label{2.1}
\Phi\Phi''-\alpha(\Phi')^2+\gamma\Phi'\Phi+\beta\Phi\geq  0,
\quad\alpha>1,\;\beta\geq  0,\;\gamma\geq  0,
\end{equation}
where
$\Phi(t)\in\mathbb{C}^{(2)}([0,T])$,
$\Phi(t)\geq  0$, $\Phi(0)>0$.
Dividing both sides \eqref{2.1} by $\Phi^{1+\alpha}$, we obtain
$$
\Big(\frac{\Phi'}{\Phi^{\alpha}}\Big)'+\gamma\frac{\Phi'}{\Phi^{\alpha}}
+\beta\Phi^{-\alpha}\geq  0.
$$
Therefore,
\begin{equation}\label{2.2}
\frac{1}{1-\alpha}(\Phi^{1-\alpha})''+\frac{\gamma}{1-\alpha}(\Phi^{1-\alpha})'
+\beta\Phi^{-\alpha}\geq  0.
\end{equation}
By definition, put%\label{2.3}
$Z(t)=\Phi^{1-\alpha}(t)$.
Then, from \eqref{2.2} we obtain
\begin{equation}\label{2.4}
Z''+\gamma Z'-\beta(\alpha-1)Z^{\alpha_1}\leq 0,\quad\alpha_1
=\frac{\alpha}{\alpha-1}.
\end{equation}
Also by definition, put %\label{2.5}
$Y(t)=e^{\gamma t}Z(t)$;
hence from \eqref{2.4} we obtain
\begin{equation}\label{2.6}
Y''-\gamma Y'-\beta(\alpha-1)e^{-\delta t}Y^{\alpha_1}\leq 0,\quad
\delta=\frac{\gamma}{\alpha-1}.
\end{equation}
It is easily shown that the following chain of equalities holds:
\begin{equation}\label{2.7}
Y'=\left(\Phi^{1-\alpha}e^{\gamma t}\right)'=
\Phi^{-\alpha}(\alpha-1)e^{\gamma t}
\big[-\Phi'(t)+\frac{\gamma}{\alpha-1}\Phi(t)\big].
\end{equation}
Take the  initial condition
\begin{equation}\label{2.8}
\Phi'(0)>\frac{\gamma}{\alpha-1}\Phi(0);
\end{equation}
then there exists $t_0>0$ such that
\begin{equation}\label{2.9}
\Phi'(t)>\frac{\gamma}{\alpha-1}\Phi(t)\quad\text{for } t\in[0,t_0).
\end{equation}
Combining \eqref{2.9} and \eqref{2.7}, we obtain
$$
Y'(t)<0\quad\text{for}\quad t\in[0,t_0).
$$
Since
$-\gamma Y'(t)\geq  0$, for $t\in[0,t_0)$,
it follows from \eqref{2.6} that
\begin{equation}\label{2.10}
Y''-\beta(\alpha-1)e^{-\delta t}Y^{\alpha_1}\leq 0,\quad\delta
=\frac{\gamma}{\alpha-1}\quad\text{for}\quad t\in[0,t_0).
\end{equation}
Now multiplying both sides \eqref{2.10} by $Y'$, we obtain
\begin{equation}\label{2.11}
Y'Y''-\beta(\alpha-1)e^{-\delta t}Y^{\alpha_1}Y'\geq  0,\quad
\delta=\frac{\gamma}{\alpha-1}\quad\text{for } t\in[0,t_0).
\end{equation}
Let us remark that
\begin{equation*}
e^{-\delta t}Y^{\alpha_1}Y'=
\frac{d}{d t}[e^{-\delta t}Y^{1+\alpha_1}]+\delta e^{-\delta t}Y^{1+\alpha_1}
-\alpha_1 e^{-\delta t}Y^{\alpha_1}Y',
\end{equation*}
Thus we have
\begin{equation}\label{2.12}
e^{-\delta t}Y^{\alpha_1}Y'=
\frac{1}{1+\alpha_1}\frac{d}{d t}[e^{-\delta t}Y^{1+\alpha_1}]
+\frac{1}{1+\alpha_1}\delta e^{-\delta t}Y^{1+\alpha_1}.
\end{equation}
Combining \eqref{2.12} with \eqref{2.11}, we obtain
\begin{equation*}
Y'Y''-\frac{\beta(\alpha-1)}{1+\alpha_1}\frac{d}{d t}
[e^{-\delta t}Y^{1+\alpha_1}]
-\frac{\beta(\alpha-1)\delta}{1+\alpha_1}e^{-\delta t}Y^{1+\alpha_1}
\geq  0\quad\text{for } t\in[0,t_0),
\end{equation*}
clearly, from this inequality we obtain
\begin{equation}\label{2.13}
Y'Y''-\frac{\beta(\alpha-1)}{1+\alpha_1}\frac{d}{d t}[e^{-\delta t}Y^{1+\alpha_1}]
\geq  0\quad\text{for } t\in[0,t_0).
\end{equation}
Integrating the above expression,
\begin{equation}\label{2.14}
(Y')^2\geq  A^2+\frac{2\beta(\alpha-1)^2}{2\alpha-1}e^{-\delta t}Y^{1+\alpha_1}
\geq  A^2,
\end{equation}
where
\begin{equation}\label{2.15}
A^2\equiv (Y'(0))^2-\frac{2\beta(\alpha-1)^2}{2\alpha-1}Y^{1+\alpha_1}(0).
\end{equation}
We assume the  condition
$$
A^2>0.
$$
The reader will have no difficulty in showing that this condition
is equivalent to the condition
\begin{equation}\label{2.16}
A^2=(\alpha-1)^2\Phi^{-2\alpha}(0)[\big(\Phi'(0)-\frac{\gamma}{\alpha-1}\Phi(0)\big)^2
-\frac{2\beta}{2\alpha-1}\Phi(0)\big]>0.
\end{equation}
Therefore, the condition $A^2>0$ and the following condition are equivalent.
\begin{equation}\label{2.17}
\Big(\Phi'(0)-\frac{\gamma}{\alpha-1}\Phi(0)\Big)^2>\frac{2\beta}{2\alpha-1}\Phi(0).
\end{equation}
Thus, combining \eqref{2.14} and \eqref{2.16}, we obtain
$$
Y'(t)\leq -A<0\Rightarrow
\Phi'(t_0)>\frac{\gamma}{\alpha-1}\Phi(t_0).
$$
But now we have that $Y'(t_0)<0$. Therefore, using this algorithm of
``continue in time", we obtain
$$
Y'(t)<0\quad\text{for all } t\in[0,T].
$$
This implies that
\begin{gather*}
|Y'|\geq  A>0\Rightarrow Y'(t)\leq -A\Rightarrow
Y(t)\leq Y(0)-At\Rightarrow\\
\Rightarrow
\Phi^{1-\alpha}(t)\leq e^{-\gamma t}[\Phi^{1-\alpha}(0)-A t]
\Rightarrow
\Phi(t)\geq \frac{e^{\gamma t/(\alpha-1)}}{[\Phi^{1-\alpha}(0)-A t]^{1/(\alpha-1)}}.
\end{gather*}
The result is the following theorem.

\begin{theorem} \label{thm1}
Suppose $\Phi(t)\in\mathbb{C}^{(2)}([0,T])$, satisfies inequality \eqref{2.1} and
\begin{gather}\label{2.18}
\Phi'(0)>\frac{\gamma}{\alpha-1}\Phi(0), \\
\label{2.19}
\big(\Phi'(0)-\frac{\gamma}{\alpha-1}\Phi(0)\big)^2
>\frac{2\beta}{2\alpha-1}\Phi(0)
\end{gather}
where $\Phi(t)\geq  0$, $\Phi(0)>0$, then the time $T>0$ can not be arbitrarily large,
 but the following inequality holds
\begin{gather*}
T\leq T_{\infty}\leq\Phi^{1-\alpha}(0)A^{-1}, \\
A^2\equiv(\alpha-1)^2\Phi^{-2\alpha}(0)
\big[\big(\Phi'(0)-\frac{\gamma}{\alpha-1}\Phi(0)\big)^2
-\frac{2\beta}{2\alpha-1}\Phi(0)\big],
\end{gather*}
where $\limsup_{t\uparrow T}\Phi(t)=+\infty$.
\end{theorem}


\section{Conditions}


We begin with conditions on the functions $h_1(x,s)$, $h_2(x,s)$, and $h_3(x,s)$.

\noindent\textbf{Conditions on $h_1(x,s)$:}
\begin{itemize}
\item[(H1.1)] $h_1(x,s):\Omega\times\mathbb{R}^1\to\mathbb{R}^1$
is a Caratheodory function;
\item[(H1.2)] for almost all $x\in\Omega$ the function
$h_1(x,s)\in\mathbb{C}^{(1)}(\mathbb{R}^1)$
and a ``growth conditions" take place
\begin{equation}\label{3.1}
|h_1(x,s)|\leq c_1+c_2|s|^{p_1-1},\quad
|h'_{1s}(x,s)|\leq c_1+c_2|s|^{p_1-2} \quad
\text{for } p_1\geq  2;
\end{equation}

\item[(H1.3)] for any $v(x)\in\mathbb{W}_0^{2,p_1}(\Omega)$ there exist the
inequalities
\begin{equation}\label{3.2}
0\leq\int_{\Omega}h_1(x,\Delta v(x))\Delta v(x)\,dx\leq
\theta_1\int_{\Omega}H_1(x,\Delta v(x))\,dx,
\end{equation}
where $\theta_1>0$ and
$H_1(x,s)=\int_0^sd\sigma\,h_1(x,\sigma)$;
\end{itemize}


\noindent\textbf{conditions on $h_2(x,s)$:}
\begin{itemize}
\item[(H2.1)] $h_2(x,s):\Omega\times\mathbb{R}_+^1\to\mathbb{R}_+^1$
is a Caratheodory function;

\item[(H2.2)] for almost all $x\in\Omega$ the function
 $h_2(x,s)\in\mathbb{C}^{(1)}(\mathbb{R}^1_+)$
and we suppose the following inequalities
\begin{equation}\label{3.4}
0\leq h_2(x,s)\leq c_3+c_4 s^{p_2-2},\quad
|h'_{2s}(x,s)s|\leq c_3+c_4 s^{p_2-2} \quad\text{for } p_2\geq  2;
\end{equation}

\item[(H2.3)] for any $v(x)\in\mathbb{W}_0^{1,p_2}(\Omega)$ an inequality holds
\begin{equation}\label{3.5}
0\leq\int_{\Omega}h_2(x,|\nabla v|)|\nabla v|^2\,dx
\leq\theta_2\int_{\Omega}dx\,H_2(x,|\nabla v|)\quad
\text{for } \theta_2>0,
\end{equation}
where $H_2(x,s)=\int_0^sd\sigma\, h_2(x,\sigma)\sigma$.
\end{itemize}


\noindent\textbf{Conditions on $h_3(x,s)$:}
\begin{itemize}
\item[(H3.1)] $h_3(x,s):\Omega\times\mathbb{R}_+^1\to\mathbb{R}_+^1$
is a Caratheodory function;

\item[(H3.2)] for almost all $x\in\Omega$ the function
$h_3(x,s)\in\mathbb{C}^{(1)}(\mathbb{R}^1_+)$
and
\begin{equation}\label{3.6}
0\leq h_3(x,s)\leq c_5+c_6 s^{p_3-2},\quad
|h'_{3s}(x,s)s|\leq c_5+c_6 s^{p_3-2},\quad p_3>2;
\end{equation}

\item[(H3.3)] for all $v(x)\in\mathbb{W}_0^{1,p_3}(\Omega)$ we assume that
\begin{equation}\label{3.7}
\int_{\Omega}h_3(x,|\nabla v|)|\nabla v|^2\,dx
\geq \theta_3\int_{\Omega}dx\,H_3(x,|\nabla v|)\quad\text{for }\theta_3>2,
\end{equation}
 where $H_3(x,s)=\int_0^sd\sigma\, h_3(x,\sigma)\sigma$.
\end{itemize}


We define
$$
p^{\ast}=\begin{cases}
Np/(N-p),&\text{for $N>p$;}\\
+\infty,&\text{for $N\leq p$.}
\end{cases}
$$
It can easily be checked that from the conditions on the functions
 $h_1(x,s)$, $h_2(x,s)$, and $h_3(x,s)$ we have
\begin{gather*}
\Delta h_1(x,\Delta v):\mathbb{W}_0^{2,p_1}(\Omega)\to\mathbb{W}^{-2,p_1'}(\Omega),
\quad p'_1=p_1/(p_1-1),
\\
\operatorname{div}(h_2(x,|\nabla v|)\nabla v):\mathbb{W}_0^{1,p_2}(\Omega)
\to\mathbb{W}^{-1,p_2'}(\Omega),\quad p_2=p_2/(p_2-1),
\\
\operatorname{div}(h_3(x,|\nabla v|)\nabla v):\mathbb{W}_0^{1,p_3}(\Omega)
\to\mathbb{W}^{-1,p_3'}(\Omega),\quad p_3=p_3/(p_3-1),
\end{gather*}
and this operators are continuous in the corresponding topologies.

\begin{definition} \label{def1} \rm
A strong generalized solution of {\rm\eqref{1.1}, \eqref{1.2}} is a
function $u(x)(t)$ in the class
$$
u(x)(t)\in\mathbb{C}^{(2)}([0,T];\mathbb{W}_0^{2,p_1}(\Omega)),\quad T>0,
$$
for some large $s>0$, if the following condition hold:
\begin{equation} \label{3.8}
\begin{aligned}
&\int_{\Omega}u''(x)(t)w(x)\,dx+\mu\int_{\Omega}u'(x)(t)w(x)\,dx+
\int_{\Omega}h_1(x,\Delta u)\Delta w(x)\,dx\\
&+ \int_{\Omega}h_2(x,|\nabla u|)(\nabla u,\nabla w)\,dx-
\int_{\Omega}h_3(x,|\nabla u|)(\nabla u,\nabla w)\,dx=0,\quad t\in[0,T]
\end{aligned}
\end{equation}
for all $w(x)\in\mathbb{W}_0^{2,p_1}(\Omega)$; and
\begin{equation}\label{3.9}
u(x)(0)=u_0(x)\in\mathbb{W}_0^{2,p_1}(\Omega),\quad
u'(x)(0)=u_1(x)\in\mathbb{L}^2(\Omega).
\end{equation}
\end{definition}


\section{Blow-up of solutions}

Assume that there is a weak solution $u(x)(t)$ in the class
$\mathbb{C}^{(2)}([0,T];\mathbb{W}_0^{2,p_1}(\Omega))$ for some
$T>0$.
Let us put $w=u(x)(t)$ in equation \eqref{3.8}, then we obtain
the first energy equality
\begin{equation} \label{4.1}
\begin{split}
&\frac{1}{2}\frac{d^2\Phi}{d t^2}+\frac{\mu}{2}\frac{d\Phi}{d t}- J
+\int_{\Omega}h_1(x,\Delta u)\Delta u\,dx
+\int_{\Omega}h_2(x,|\nabla u|)|\nabla u|^2\,dx\\
&=\int_{\Omega}h_3(x,|\nabla u|)|\nabla u|^2\,dx,
\end{split}
\end{equation}
where we denote
$$
\Phi(t)\equiv\int_{\Omega}|u|^2\,dx,\quad
 J(t)\equiv\int_{\Omega}|u'|^2\,dx.
$$
Let us put  $w=u'(x)(t)$ in  \eqref{3.8}, we obtain the
second energy equality
\begin{equation}\label{4.2}
\begin{split}
&\frac{d}{d t}\Big(\frac{1}{2} J+\int_{\Omega}H_1(x,\Delta u)\,dx+
\int_{\Omega}H_2(x,|\nabla u|)\,dx\Big)+\mu J\\
&=\frac{d}{d t}\int_{\Omega}H_3(x,|\nabla u|)\,dx.
\end{split}
\end{equation}
Furthermore, integrating \eqref{4.2} over time, we obtain the inequality
\begin{equation}\label{4.3}
\frac{1}{2} J+\int_{\Omega}H_1(x,\Delta u)\,dx+
\int_{\Omega}H_2(x,|\nabla u|)\,dx-E(0)\leq\int_{\Omega}H_3(x,|\nabla u|)\,dx,
\end{equation}
where
\begin{equation} \label{4.4}
\begin{split}
E(0)&\equiv\frac{1}{2}\int_{\Omega}|u_{1}|^2\,dx+\int_{\Omega}H_1(x,\Delta u_{0})\,dx\\
&\quad + \int_{\Omega}H_2(x,|\nabla u_{0}|)\,dx-\int_{\Omega}H_3(x,|\nabla u_{0}|)\,dx.
\end{split}
\end{equation}
By \eqref{4.3} we obtain the  inequality
\begin{align*}
&\frac{\theta_3}{2} J+\theta_3\int_{\Omega}H_1(x,\Delta u)\,dx+
\theta_3\int_{\Omega}H_2(x,|\nabla u|)\,dx-\theta_3E(0)\\
&\leq \theta_3\int_{\Omega}H_3(x,|\nabla u|)\,dx,
\end{align*}
combining this with the condition (H3.3), we obtain
\begin{align*}
&\frac{\theta_3}{2} J+\theta_3\int_{\Omega}H_1(x,\Delta u)\,dx+
\theta_3\int_{\Omega}H_2(x,|\nabla u|)\,dx-\theta_3E(0)\\
&\leq \int_{\Omega}h_3(x,|\nabla u|)|\nabla u|^2\,dx.
\end{align*}
Using the above inequality and \eqref{4.1}, we obtain
\begin{equation} \label{4.5}
\begin{split}
&\frac{1}{2}\frac{d^2\Phi}{d t^2}+\frac{\mu}{2}\frac{d\Phi}{d t}- J+\int_{\Omega}h_1(x,\Delta u)\Delta u\,dx+
\int_{\Omega}h_2(x,|\nabla u|)|\nabla u|^2\,dx \\
&\geq \frac{\theta_3}{2} J+\theta_3\int_{\Omega}H_1(x,\Delta u)\,dx+
\theta_3\int_{\Omega}H_2(x,|\nabla u|)\,dx-\theta_3E(0).
\end{split}
\end{equation}
Now taking into account the conditions (H3.1) and (H3.2), from \eqref{4.5} we obtain
\begin{equation} \label{4.6}
\begin{split}
&\frac{1}{2}\frac{d^2\Phi}{d t^2}+\frac{\mu}{2}\frac{d\Phi}{d t}
- J+\theta_1\int_{\Omega}H_1(x,\Delta u)\,dx+
\theta_2\int_{\Omega}H_2(x,|\nabla u|)\,dx \\
&\geq
\frac{\theta_3}{2} J+\theta_3\int_{\Omega}H_1(x,\Delta u)\,dx+
\theta_3\int_{\Omega}H_2(x,|\nabla u|)\,dx-\theta_3E(0).
\end{split}
\end{equation}
Under the conditions
$\theta_3\geq \theta_1$, $\theta_3\geq \theta_2$
using the inequalities
$$
\int_{\Omega}H_1(x,\Delta u)\,dx\geq  0,\quad
\int_{\Omega}H_2(x,|\nabla u|)\,dx\geq  0,
$$
 from \eqref{4.6}, we obtain
\begin{equation}\label{4.7}
\frac{1}{2}\frac{d^2\Phi}{d t^2}+\frac{\mu}{2}\frac{d\Phi}{d t}+\theta_3E(0)
\geq \big(1+\frac{\theta_3}{2}\big) J(t).
\end{equation}
Using the Cauchy-Bunyakovsky-Schwarz inequality, it is easily shown the
 differential inequality
\begin{equation}\label{4.8}
(\Phi')^2\leq 4 J\Phi.
\end{equation}
Combining \eqref{4.7} and \eqref{4.8}, we obtain the important differential
inequality
\begin{equation}\label{4.9}
\Phi\Phi''-\frac{1}{2}\big(1+\frac{\theta_3}{2}\big)(\Phi')^2+\mu\Phi\Phi'
+2\theta_3 E(0)\Phi\geq  0.
\end{equation}
Comparing this differential inequality with \eqref{2.1}, we obtain that
\begin{gather*}
\alpha=\frac{1}{2}\big(1+\frac{\theta_3}{2}\big)>1\quad\text{for }\theta_3>2,\\
\beta=2\theta_3 E(0),\quad \gamma=\mu,\quad \frac{2\beta}{2\alpha-1}=8 E(0),\quad
\frac{\gamma}{\alpha-1}=\frac{4\mu}{\theta_3-2}.
\end{gather*}
We assume the following conditions
\begin{gather}\label{4.10}
\Phi'(0)>\frac{4\mu}{\theta_3-2}\Phi(0)>0, \\
\label{4.11}
\Big(\Phi'(0)-\frac{4\mu}{\theta_3-2}\Phi(0)\Big)^2>8 E(0)\Phi(0), \\
\label{4.12}
\begin{aligned}
E(0)&\equiv\frac{1}{2}\int_{\Omega}|u_{1}|^2\,dx+\int_{\Omega}H_1(x,\Delta u_{0})\,dx\\
&\quad + \int_{\Omega}H_2(x,|\nabla u_{0}|)\,dx
-\int_{\Omega}H_3(x,|\nabla u_{0}|)\,dx>0,
\end{aligned}
\end{gather}
Under conditions \eqref{4.10}--\eqref{4.12}, the time $T>0$ of existence of $u(x)(t)$
is bounded from above
\begin{gather*}
T\leq\Phi^{(2-\theta_3)/4}(0)A^{-1}, \\
A^2\equiv\big(\frac{\theta_3-2}{4}\big)^2\Phi^{-1-\theta_3/2}(0)
\Big[\Big(\Phi'(0)-\frac{4\mu}{\theta_3-2}
\Phi(0)\Big)^2-8 E(0)\Phi(0)\Big],
\end{gather*}
at the same time
\begin{equation}\label{4.13}
\Phi(t)\geq
\frac{e^{4\mu t/(\theta_3-2)}}{[\Phi^{(2-\theta_3)/4}(0)-A t]^{4/(\theta_3-2)}},
\end{equation}
where
$$
\Phi'(0)=2\int_{\Omega}u_{1}(x)u_{0}(x)\,dx,\quad
\Phi(0)=\int_{\Omega}|u_{0}|^2\,dx.
$$
Therefore, our main result of is the following theorem.

\begin{theorem} \label{thm2}
  Assume all conditions on $h_1$, $h_2$ and $h_3$ hold.
Under the following conditions
\begin{gather}\label{4.17}
\Phi'(0)>\frac{4\mu}{\theta_3-2}\Phi(0)+(8 E(0)\Phi(0))^{1/2}>0,\quad E(0)>0,\\
\label{4.18}
\theta_3\geq \theta_1,\quad \theta_3\geq \theta_2,
\end{gather}
there exists the estimation from above for the time of solution existence $T$,
$$
T_1\leq T_{\infty}=\Phi^{(2-\theta_3)/4}(0)A^{-1};
$$
i. e.,  we have that
$\limsup_{t\uparrow T_1}\Phi(t)=+\infty$,
where
\begin{gather*}
\Phi(0)=\int_{\Omega}|u_0|^2\,dx,\quad \Phi'(0)=2\int_{\Omega}u_1u_0\,dx,\\
\begin{aligned}
E(0)&\equiv\frac{1}{2}\int_{\Omega}|u_{1}|^2\,dx+\int_{\Omega}H_1(x,\Delta u_{0})\,dx\\
&\quad +\int_{\Omega}H_2(x,|\nabla u_{0}|)\,dx-\int_{\Omega}H_3(x,|\nabla u_{0}|)\,dx,
\end{aligned}\\
A^2\equiv\big(\frac{\theta_3-2}{4}\big)^2\Phi^{-1-\theta_3/2}(0)
\big[\big(\Phi'(0)-\frac{4\mu}{\theta_3-2}
\Phi(0)\big)^2-8 E(0)\Phi(0)\big].
\end{gather*}
\end{theorem}


\begin{remark} \label{rmk1} \rm
Now we shall see that all conditions \eqref{4.17} are compatible for enough
small $\mu\geq  0$.
Indeed, first we shall choose $u_0\in\mathbb{W}_0^{2,p_1}(\Omega)$ enough
 large to satisfy the  inequality
\begin{equation} \label{4.19}
\begin{split}
&\int_{\Omega}H_3(x,|\nabla u_{0}|)\,dx\\
&>\int_{\Omega}H_1(x,\Delta u_{0})\,dx
+ \int_{\Omega}H_2(x,|\nabla u_{0}|)\,dx
+\frac{2}{\theta_3-2}\int_{\Omega}|u_0|^2\,dx.
\end{split}
\end{equation}
Secondly we fix $u_0$ and choose
$u_1=\lambda u_0$ for $\lambda>2\mu/(\theta_3-2)$.
In this case we have
\begin{equation}\label{4.20}
\Phi'(0)-\frac{4\mu}{\theta_3-2}\Phi(0)
=2\Big(\lambda-\frac{2\mu}{\theta_3-2}\Big)\Phi(0)>0.
\end{equation}
Finally we choose $\lambda>2\mu/(\theta_3-2)$ enough large to satisfy
the inequality
\begin{align*}
E(0)&=\frac{\lambda^2}{2}\int_{\Omega}|u_{0}|^2\,dx
+\int_{\Omega}H_1(x,\Delta u_{0})\,dx\\
&\quad + \int_{\Omega}H_2(x,|\nabla u_{0}|)\,dx-\int_{\Omega}H_3(x,|\nabla u_{0}|)
\,dx>0.
\end{align*}
Under  condition \eqref{4.20},   inequality \eqref{4.17} is equivalent to
\begin{equation}\label{4.21}
\Big(\Phi'(0)-\frac{4\mu}{\theta_3-2}\Phi(0)\Big)^2>8 E(0)\Phi(0),
\end{equation}
by our substitution we obtain from the left and right side of this inequality
\begin{equation} \label{4.22}
\begin{split}
\Big(\Phi'(0)-\frac{4\mu}{\theta_3-2}\Phi(0)\Big)^2
&= 4\Big(\lambda-\frac{2\mu}{\theta_3-2}\Big)^2\Phi^2(0)\\
&=\Big(4\lambda^2-\frac{16\mu\lambda}{\theta_3-2}+\frac{16\mu^2}{(\theta_3-2)^2}\Big)
 \Phi^2(0),
\end{split}
\end{equation}
and
\begin{equation} \label{4.23}
\begin{split}
8 E(0)\Phi(0)
&=4\lambda^2\Phi^2(0)+8\Phi(0)\Big[\int_{\Omega}H_1(x,\Delta u_{0})\,dx\\
&\quad +\int_{\Omega}H_2(x,|\nabla u_{0}|)\,dx
 -\int_{\Omega}H_3(x,|\nabla u_{0}|)\,dx\Big]
\end{split}
\end{equation}
Combining \eqref{4.21} with \eqref{4.22} and \eqref{4.23}, we obtain that
\begin{equation} \label{4.24}
\begin{aligned}
\Big(\Phi'(0)-\frac{4\mu}{\theta_3-2}\Phi(0)\Big)^2
&= \Big(4\lambda^2-\frac{16\mu\lambda}{\theta_3-2}
 +\frac{16\mu^2}{(\theta_3-2)^2}\Big)\Phi^2(0)
> 8 E(0)\Phi(0)\\
&=4\lambda^2\Phi^2(0)+8\Phi(0)\Big[\int_{\Omega}H_1(x,\Delta u_{0})\,dx \\
&\quad +\int_{\Omega}H_2(x,|\nabla u_{0}|)\,dx
-\int_{\Omega}H_3(x,|\nabla u_{0}|)\,dx\Big],
\end{aligned}
\end{equation}
Now it is not hard to prove that
\begin{equation} \label{4.25}
\begin{aligned}
&\int_{\Omega}H_3(x,|\nabla u_{0}|)\,dx
+\frac{2\mu^2}{(\theta_3-2)^2}\int_{\Omega}|u_0|^2\,dx\\
&> \frac{2\mu\lambda}{\theta_3-2}\int_{\Omega}|u_0|^2\,dx+
\int_{\Omega}H_1(x,\Delta u_{0})\,dx+
\int_{\Omega}H_2(x,|\nabla u_{0}|)\,dx.
\end{aligned}
\end{equation}
Moreover, we choose
$$
\lambda=\frac{1}{\mu}\quad\text{for }\mu\in\big(0,(\frac{\theta_3-2}{2})^{1/2}\big),
$$
and if $\mu=0$, then $\lambda>0$ and large enough. We see that all foregoing
conditions are satisfied for small enough
$\mu\geq  0.$ Now combining  this large enough $\lambda$ and \eqref{4.25},
we obtain the inequality
\begin{align*}
&\int_{\Omega}H_3(x,|\nabla u_{0}|)\,dx
+\frac{2\mu^2}{(\theta_3-2)^2}\int_{\Omega}|u_0|^2\,dx\\
&> \frac{2}{\theta_3-2}\int_{\Omega}|u_0|^2\,dx+
\int_{\Omega}H_1(x,\Delta u_{0})\,dx+
\int_{\Omega}H_2(x,|\nabla u_{0}|)\,dx.
\end{align*}
Obviously, this inequality holds by \eqref{4.19}. Therefore, we have to
prove \eqref{4.19} for some functions on  $h_1(x,s)$, $h_2(x,s)$, and $h_3(x,s)$.
At the same time we check the condition \eqref{4.18}.
Suppose
$$
h_1(x,s)=|s|^{p_1-2}s,\quad h_2(x,s)=s^{p_2-2},\quad h_3(x,s)=s^{p_3-2},
$$
where
$p_3>p_1>2$, $p_3>p_2\geq  2$.
Then
$$
H_1(x,s)=\frac{|s|^{p_1}}{p_1},\quad
H_2(x,s)=\frac{|s|^{p_2}}{p_2},\quad
H_3(x,s)=\frac{|s|^{p_3}}{p_3},
$$
and $\theta_3=p_3>\theta_1=p_1>2$, $\theta_3=p_3>\theta_2=p_2$.
Therefore, first note that the condition \eqref{4.18} holds, and further note
that for large enough
$u_0(x)\in\mathbb{W}_0^{2,p_1}(\Omega)$ the condition \eqref{4.19} also holds.
\end{remark}


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\end{document}