
\documentclass[twoside]{article}
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\markboth{Stanislav I. Pohozaev \&  Alberto Tesei}
{Instantaneous blow-up of solutions }

\begin{document}
\setcounter{page}{155}
\title{\vspace{-1in}\parbox{\linewidth}{\footnotesize\noindent
2001-Luminy conference on Quasilinear Elliptic and Parabolic Equations
and Systems,\newline
Electronic Journal of Differential Equations,
Conference 08, 2002, pp 155--165. \newline
http://ejde.math.swt.edu or http://ejde.math.unt.edu
\newline ftp  ejde.math.swt.edu (login: ftp)}
 \vspace{\bigskipamount} \\
%
  Instantaneous blow-up of solutions to a class of hyperbolic
  inequalities
%
\thanks{ {\em Mathematics Subject Classifications:} 35L15, 35L70, 35A70.
\hfil\break\indent {\em Key words:} Hyperbolic  inequalities,
instantaneous blow-up, nonnegative solutions, \hfil\break\indent
singular coefficients.
\hfil\break\indent 
\copyright 2002 Southwest Texas State University. \hfil\break\indent
Published December 11, 2002. \hfil\break\indent
Work partially supported through TMR Programme NPE FMRX-CT98-0201.
}}

\date{}
\author{Stanislav I. Pohozaev \&  Alberto Tesei}
\maketitle

\begin{abstract}
 We prove instantaneous blow-up of nonnegative solutions
 for a class of semilinear  hyperbolic  inequalities with first order
 terms and singular coefficients. The present approach relies on a suitable
 choice of the functions used to test the differential inequalities.
\end{abstract}

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


\section{Introduction}

In this paper we investigate {\em instantaneous blow-up} of solutions to
semilinear  hyperbolic  inequalities of the  type
\begin{equation} \label{H.1}
\begin{gathered}
 u_{tt} -\Delta u \ge \lambda | x|^{-\mu} (x,\nabla u )+
|x|^{-\alpha}u^q \quad \hbox{in } Q:=\Omega \times (0,T] \\
 u\ge0 \quad \hbox{in } Q  .
\end{gathered}
\end{equation}
Here $\Omega \subseteq \mathbb{R}^n$,  $n \ge3 $ is a bounded smooth domain
which
contains the origin, $q>1$ and $\lambda, \mu, \alpha$
are real parameters. By $ (\cdot,\cdot )$ we  denote
the scalar product in $\mathbb{R}^n $.

 Instantaneous blow-up can be regarded as {\em nonexistence of  local
 solutions} - namely, nonexistence of
solutions in any neighbourhood of the origin.
Similar nonexistence phenomena were investigated in \cite{BC}
for the  semilinear elliptic  problem:
\begin{equation} \label{0.0}
\begin{gathered}
   -\Delta u \ge | x|^{-2}u^2 \quad\hbox{in } \Omega \\
   u\ge 0 \quad \hbox{in } \Omega
\end{gathered}
\end{equation}
moreover, instantaneous blow-up results were proved
for the companion parabolic inequality (in this connection, see also
 \cite{CM1}).

 A major step of the proof in \cite{BC} was demonstrating  a removable
singularity result of solutions at the origin, then using  comparison
results for the extended solutions. In the present approach we
 take advantage of the local behaviour of
solutions near the origin by a direct bootstrap argument, which relies on
a proper choice of the test functions; no comparison
results are needed. This  allows us to deal not only
with  elliptic or parabolic inequalities (see \cite{PT9}), but also
with the hyperbolic case (\ref{H.1}).  Let us  mention that the same
approach was used elsewhere  (e.g., see \cite{MP}, \cite{PT};
a comprehensive account can be found in  \cite{MP1}).

A specific aim of the present paper is to investigate
how singular first order terms affect  existence of local solutions
to problem (\ref{H.1}). As long as $\mu \le 2$, nonexistence results
are proven only if $\alpha \ge 2$ (see assumptions $(a)$-$(b)$ of Theorem
\ref{tH.1} below); in this case  nonexistence of local solutions depends on
the singularity of the source term as in \cite{BC}. On the other hand,  if
$\mu>{2q  \over q-1}$  nonexistence can be proved also for $\alpha<0$ (see
assumption $(c)$ of Theorem \ref{tH.1}); in this  case the
coefficient of the source term is regular in
$\Omega$, thus the  nonexistence result depends on the singularity of the
first
order term. The same situation occurs for elliptic or parabolic inequalities
analogous to (\ref{H.1}) (see \cite{PT9}).

\section{Mathematical background and results}

Solutions to problem (\ref{H.1}) are meant in the following sense.


\begin{definition}\label{dH.1} \rm
By a {\rm solution } to problem (\ref{H.1})  in $Q:=\Omega \times (0,T]$ we
mean any function $u \in C([0,T];H^1_{\rm loc}(\Omega \setminus \{0\})) \cap
C^1([0,T]; L^q_{\rm loc}(\Omega \setminus \{0\}))$   such that:
 \item (i) $u \ge 0$ almost everywhere in $Q$;

\item (ii)  for any test function $\zeta  \in C^{\infty,2}_{x,t}(\bar Q)$,
$\zeta \ge 0$, $\zeta(\cdot,t) \in C^{\infty}_{0}(\Omega \setminus \{0\})$
($t\in [0,T]$), $\zeta(\cdot,T)=\zeta_t(\cdot,T)=0$ there holds:
\begin{multline} \label{2.2}
\iint_{Q} (\nabla u,\nabla \zeta) +
\lambda \iint_{Q}u   \mathop{\rm div}\Big (| x|^{-\mu} x \zeta \Big)\\
\ge \iint_{Q} | x|^{-\alpha}u^q \zeta  +
\int_{\Omega} u_t(x,0)\zeta(x,0) -\int_{\Omega} u(x,0)\zeta_t(x,0)
- \iint_{Q} u \zeta_{tt}  .
\end{multline}
\end{definition}
The following nonexistence result will be proven.

\begin{theorem}\label{tH.1} Let either of the following assumptions
be satisfied:
\begin{enumerate}
\item[(a)] $\mu<2$, $\alpha \ge 2$

\item[(b)] $\mu=2$ and either
$\alpha >2, \lambda \ge 2-n$, or $\alpha =2, \lambda > 2-n$

\item[(c)] $\mu>2$, $\alpha > \mu + (2-\mu)q$,
and $ \lambda > 0$.
\end{enumerate}
Moreover, let
\begin{equation}\label{H}
 \lim \inf_{x \to 0} | x|^{-\gamma}   u(x,0) >0
\end{equation}
for some $\gamma < 0$ and
\begin{equation}\label{H1}
  u_t(x,0) \ge 0
\end{equation}
in some neighbourhood of the origin.
Then the only solution to problem (\ref{H.1}) in any cylinder
$\Omega_1 \times (0, \tau]$, $\Omega_1
\subseteq \Omega$  containing the origin and $\tau \in (0,T)$,
is trivial.
\end{theorem}

\begin{remark}\label{r3.1} \rm
With the exception of the case $\alpha =2$,
 Theorem \ref{tH.1} still holds if we assume
$\gamma < {\alpha \over q-1}$ in (\ref{H}).
\end{remark}

The proof of Theorem \ref{tH.1}
relies on a proper choice of the test function in inequality (\ref{2.2}).
  For this purpose we introduce some preliminary material.

Let $\Omega_1\subseteq \Omega$ be any neighbourhood containing the origin,
 $0<\epsilon<\eta$, $\eta>2\epsilon$ so small that
$A_{\epsilon,\eta}:= \{x \in \mathbb{R}^n : \epsilon<| x|<\eta\}
\subseteq \Omega_1 \setminus \{0\}$.
 For $r \in [\epsilon,\eta]$ define
$$ \phi_0(r):=r^{\sigma}-\eta^{\sigma}   ,$$
where $\sigma<0$ will be fixed later. Define also
$$ \phi_1(r):= \bar \phi({r \over \epsilon})\quad (r \in
[\epsilon,\eta]) ,$$
where $\bar \phi \in C^{\infty}([0,{\eta \over \epsilon}])$ is
nondecreasing, such that
 $$
 \bar \phi(s):= \begin{cases} 0 &\mbox{if }  s\in (0,1)\\
     1   &\mbox{if } s \in (2, \eta/\epsilon) .
\end{cases}
$$
Finally, set
$$ \bar \zeta (r):= r^{\rho}\phi_0(r)\phi_1(r) \quad (r \in
[\epsilon,\eta])   ,$$
where $\rho$ is a real parameter to be chosen later.

\begin{remark}\label{l3.1} \rm
The following  properties of the function $ \bar
\zeta$ are easily checked. \\
(i) There holds:
$$
\bar \zeta (\epsilon)=\bar \zeta (\eta)=0; \quad {d \bar\zeta \over dr}
(\epsilon) \ge 0, \quad {d \bar\zeta \over dr} (\eta)\le 0   .
$$
(ii) There exists a sequence $\{\zeta_k\} \subseteq C^{\infty}_{0}
(A_{\epsilon,\eta})$, $\zeta_k\ge0$ for any $k$, such that $\zeta_k
\rightarrow \tilde \zeta$
in $W^{1,p}_0(A_{\epsilon,\eta})$ $(p \in (1, \infty))$, where
\begin{equation}\label{3.0} \tilde \zeta (x):= \bar \zeta (| x|)
\quad (x \in \bar A_{\epsilon,\eta})   .
\end{equation}
\end{remark}

\section{ Proofs}

Let us prove the following result.

\begin{proposition}\label{p5.1}
Let $u$ be a solution to problem (\ref{H.1})
in some cylinder $\Omega_1 \times (0, \tau]\subseteq Q$,
$\Omega_1$  containing the origin and $\tau \in (0,T)$.
Then for any $0<\epsilon<\eta$, $\eta$ sufficiently small
and any $\tau \in (0,T)$ there holds:
\begin{equation}\label{5.1}
\begin{aligned}
\int_0^{\tau}& (\tau-t)^{\beta}  \,dt  \int_{A_{\epsilon,\eta}}
| x|^{-\alpha}      u^q(x,t)  \tilde \zeta (x) \\
 \le& - \int_0^{\tau} (\tau-t)^{\beta}  \,dt
 \int_{A_{\epsilon,\eta}}
| x|^{-(n-1)}   {d \psi \over dr} (| x|)   u(x,t)  \\
&+   \beta ( \beta-1)\int_0^{\tau} (\tau-t)^{\beta-2}  \,dt
\int_{A_{\epsilon,\eta}}u(x,t)   \tilde \zeta (x)   \\
&-  {\tau}^{\beta}
\int_{A_{\epsilon,\eta}} u_t(x,0)   \tilde \zeta (x)   -
  \beta{\tau}^{\beta-1}
\int_{A_{\epsilon,\eta}} u(x,0)   \tilde \zeta (x) ,
\end{aligned}
\end{equation}
where  $\beta>  {q+1 \over q-1}$,
$ \tilde \zeta $ is the function (\ref{3.0}) and
\begin{equation}\label{3.2}
\psi(r):= r^{n-1} {d \bar\zeta \over dr} (r) - \lambda r^{n-\mu}
\bar\zeta(r)
 \quad  (r \in [\epsilon,\eta])     .
\end{equation}
\end{proposition}

\paragraph{Proof.}
Let $\tau \in (0,T)$; set
 $$
\hat \phi(t):= \begin{cases} (\tau-t)^{\beta} &\mbox{if } t \in (0,\tau)\\
  0   &\mbox{if } t \in (\tau,T)\,.
\end{cases}
$$
 Let $\{\zeta_k\}
\subseteq C^{\infty}_{0}(A_{\epsilon,\eta})$ be the
approximating sequence in Remark \ref{l3.1}-$(ii)$; set
$\zeta(x,t)=\zeta_k(x)\hat \phi(t)$ in inequality (\ref{2.2}).
Letting $k \to \infty$ we obtain easily:
\begin{align*}
\int_0^{\tau}& (\tau-t)^{\beta}  \,dt  \int_{A_{\epsilon,\eta}} |
x|^{-\alpha}
 u^q(x,t)  \tilde \zeta (x)   \\
 \le&  \int_0^{\tau} (\tau-t)^{\beta}  \,dt
  \Big(\int_{A_{\epsilon,\eta}} (\nabla u(x,t),\nabla \tilde \zeta) +
\lambda \int_{A_{\epsilon,\eta}}u(x,t)
 \mathop{\rm div} \big (| x|^{-\mu} x \zeta \big) \Big) \\
&+   \beta ( \beta-1)\int_0^{\tau} (\tau-t)^{\beta-2}  \,dt
\int_{A_{\epsilon,\eta}} u(x,t)   \tilde \zeta (x)  \\
&- {\tau}^{\beta} \int_{A_{\epsilon,\eta}} u_t(x,0)   \tilde \zeta (x)   -
  \beta{\tau}^{\beta-1}
\int_{A_{\epsilon,\eta}} u(x,0)   \tilde \zeta (x) \,.
\end{align*}
On the other hand, for any $t \in (0,\tau)$ there holds:
\begin{multline*}
\int_{A_{\epsilon,\eta}} (\nabla u(x,t),\nabla \tilde \zeta) +
\lambda \int_{A_{\epsilon,\eta}}u(x,t)
 \mathop{\rm div} \Big (| x|^{-\mu} x \zeta \Big) \\
\le - \int_{A_{\epsilon,\eta}}  u(x,t)
\Big \{\Delta \tilde \zeta - \lambda \
\mathop{\rm div} \big (| x|^{-\mu} x \zeta \Big) \big \} \,.
\end{multline*}
An elementary calculation shows that
$$
\Delta \tilde \zeta - \lambda \
\mathop{\rm div} \Big (| x|^{-\mu} x \zeta \Big) =
| x|^{-(n-1)}   {d \psi \over dr} (| x|)   ;
$$
Then from the above inequalities the conclusion follows.
\hfill$\square$

It is easily checked that the function $\psi$ defined in (\ref{3.2}) reads:
\begin{equation}\label{3.3}
\psi=\phi_1  \psi_1+ r^{n-1+\rho}\phi_0{d \phi_1 \over dr}  ,
\end{equation}
where
$$
\psi_1=\psi_1(r):= r^{n-1} {d  \over dr} [r^{\rho}\phi_0(r)] -
\lambda r^{n-\mu+\rho} \phi_0(r) \quad (r \in
[\epsilon,\eta])     .
$$
The following technical lemma plays an important role in the sequel.
\begin{lemma}\label{l3.2}
Let any of the following assumptions be satisfied:
\begin{itemize}
 \item (i) $\mu<2$,  $\rho\le 2-n$, $\sigma<0$
 \item (ii) $\mu<2$, $\rho \in (2-n,0)$,
 $\rho \le 4-n-\mu$, $\sigma=-\rho+2-n$
 \item (iii) $\mu=2$, $\rho\le 2-n \le \lambda$,  $\sigma<0$
 \item (iv) $\mu=2$, $\rho=\lambda>2-n$,  $\sigma=-\rho+2-n$
\item (v) $\mu>2$, $\rho \le \mu-n$, $\lambda>0$, $\sigma<0$.
\end{itemize}
Then there exists $\eta_0>0$ (depending on
$n,\lambda,\mu,\rho,\sigma$) such that for any $\eta<\eta_0$ there holds
\begin{equation}\label{3.4}
{d  \psi_1\over dr} \ge 0 \quad \mbox{in } (\epsilon ,\eta) \,.
\end{equation}
\end{lemma}

\paragraph{Proof .}
We deal only with cases $(i)$-$(ii)$ for shortness.
Observe that
$$
{d  \psi_1\over dr}= r^{n-3+\rho+\sigma}\psi_0({r \over \eta})   ,
$$
where
\begin{align*}
\psi_0(s):=&(\rho+\sigma)(n-2+\rho+\sigma)-\rho(n-2+\rho)s^{-\sigma}\\
&-\lambda \Big [(n-\mu+\rho+\sigma)-(n-\mu+\rho)
s^{-\sigma} \Big] \eta^{2-\mu} s^{2-\mu} \quad (s \in [0,1])   ,
\end{align*}
as an elementary calculation shows.

\noindent $(i)$ Since $\rho \le2-n\le 0$ and $\sigma<0$, there holds
\begin{align*}
(\rho+\sigma)(n-2+\rho+\sigma)-\rho(n-2+\rho)s^{-\sigma} &\\
\geq (\rho+\sigma)(n-2+\rho+\sigma)-\rho(n-2+\rho) &>0\,;
\end{align*}
in fact, it is easily seen that the function $f(s):=s(n-2+s)$ is strictly
increasing in the interval $[\rho+\sigma, \rho]$. Since by assumption
$2-\mu>0$, choosing $\eta$ sufficiently small, we  prove the claim.

\noindent $(ii)$ In this case  $n-2+\rho+\sigma=0$ and $-\rho(n-2+\rho)>0$;
since $2-\mu+\sigma=4-n-\mu -\rho \ge0$, the claim follows as in $(i)$.

In the remaining cases we can argue similarly; hence the conclusion
follows. \hfill$\square$

Then we have the following result.

\begin{proposition}\label{p5.2}
Let $u$ be a solution to problem (\ref{H.1})
in some cylinder $\Omega_1 \times (0, \tau]\subseteq Q$,
$\Omega_1$  containing the origin and $\tau \in (0,T)$. Let the assumptions
 of Lemma \ref{l3.2} be satisfied; choose
 $\eta<\eta_0$ accordingly. Moreover, let
\begin{equation}\label{5.2}
 \rho+ \sigma > - {\alpha \over q-1} - n \,.
\end{equation}
Then for any $\epsilon>0$ sufficiently small
and any $\tau \in (0,T)$ there holds:
\begin{equation}\label{5.3}
\begin{aligned}
\int_0^{\tau}& (\tau-t)^{\beta}  \,dt
  \Big(\int_{A_{\epsilon,\eta}} u(x,t) \tilde \zeta(x) \Big)^q \\
\leq&  M   \tau^{\beta-1}C_1(\epsilon,\eta) \Big(C_1(\epsilon,\eta)^
{1 \over q-1}  \tau^{-{2 \over q-1}} + C_2(\epsilon,\eta)   \tau^2\\
&-  {\tau}\int_{A_{\epsilon,\eta}} u_t(x,0)   \tilde \zeta (x)
- \int_{A_{\epsilon,\eta}} u(x,0) \tilde \zeta(x) \Big) \,,
\end{aligned}
\end{equation}
for some constant $M=M(\beta,q)>0$. Here
\begin{gather} \label{4.4}
C_1(\epsilon,\eta):= \Big(\int_{\epsilon}^{\eta}
r^{{\alpha  \over q-1}+n-1}   \bar \zeta (r)   dr\Big)
^{ q-1}  \,,\\
\label{4.5}
C_2(\epsilon,\eta):=  \int_{A_{\epsilon,\eta}} | x|^{-(n-1)q'}
\Big [| x|^{-\alpha}   \tilde \zeta (x)
\Big ] ^{-(q'-1)}  \chi (| x|)^{q'}   dx \,.
\end{gather}
\end{proposition}

\paragraph{Proof.}
$(i)$ Due to Lemma \ref{l3.2} and the choice
$\eta<\eta_0$, for any $t \in (0,\tau)$ we have
\begin{equation}\label{3.666}
 - \int_{A_{\epsilon,\eta}}
| x|^{-(n-1)}   {d \psi \over dr} (| x|)   u(x,t)   \le
- \int_{A_{\epsilon,\eta}} | x|^{-(n-1)}   \chi (| x|)   u(x,t)  ,
\end{equation}
where
\begin{equation}\label{3.6}
\chi(r):= {d \phi_1 \over dr}  \psi_1+
{d  \over dr}[r^{n-1+\rho}\phi_0{d \phi_1 \over dr}]
 \quad (r \in
[\epsilon,\eta])      .
\end{equation}
Using H\"older inequality, the right-hand side of inequality (\ref{3.666})
can be estimated as follows
\begin{align}
&\big |\int_{A_{\epsilon,\eta}} | x|^{-(n-1)}   \chi (| x|)
  u(x,t)  \big | \nonumber\\
&\le
 \Big(\int_{A_{\epsilon,\eta}} | x|^{-\alpha}    u^q(x,t)
   \tilde \zeta (x)  \Big)^{1/q}
\Big(\int_{A_{\epsilon,\eta}} | x|^{-(n-1)q'}\big[| x|^{-\alpha}
\tilde \zeta (x) \big] ^{-(q'-1)}  \chi (| x|)^{q'} \Big)^{1/q'} \nonumber
\\
&=   \Big(\int_{A_{\epsilon,\eta}} | x|^{-\alpha}    u^q
   \tilde \zeta (x)  \Big)^{1/q}
\Big(C_2(\epsilon,\eta) \Big)^{1/q'} \label{3.7}
\end{align}
(see definition (\ref{4.5}); here $q':={q \over q-1}$). Using
inequality (\ref{3.7}) and Young inequality we obtain
\begin{equation}\label{4.6a}
\Big | \int_{A_{\epsilon,\eta}} | x|^{-(n-1)}   \chi (| x|)
  u(x,t) \Big |
\le   { 1 \over q} \Big | \int_{A_{\epsilon,\eta}}
| x|^{-\alpha} u^q(x,t) \tilde \zeta(x) \Big |
+ { 1 \over q'} C_2(\epsilon,\eta)
\end{equation}
for any $t \in (0,\tau)$. Then from
 inequalities (\ref{4.6a}), (\ref{3.666})  we obtain
for any $t \in (0,\tau)$:
\begin{align*}
- \int_0^{\tau}& (\tau-t)^{\beta}  \,dt
 \int_{A_{\epsilon,\eta}}
| x|^{-(n-1)}   {d \psi \over dr} (| x|)   u(x,t)   \\
&\le   {1 \over q}\int_0^{\tau} (\tau-t)^{\beta}  \,dt
 \int_{A_{\epsilon,\eta}}  | x|^{-\alpha}      u^q(x,t)  \tilde \zeta
(x)   +  {1 \over q'} {{\tau}^{\beta+1} \over \beta+1} C_2(\epsilon,
\eta).
\end{align*}
Substituting the above inequality in (\ref{5.1}) easily gives
\begin{equation} \label{5.4}
\begin{aligned}
\int_0^{\tau} &(\tau-t)^{\beta}  \,dt  \int_{A_{\epsilon,\eta}}
| x|^{-\alpha}      u^q(x,t)  \tilde \zeta (x)  \,dt   \\
\leq& {{\tau}^{\beta+1} \over \beta+1} C_2(\epsilon,\eta)
+   \beta(\beta-1)   q'  \int_0^{\tau} (\tau-t)^{\beta -2}  \,dt
\int_{A_{\epsilon,\eta}} u(x,t) \tilde \zeta(x)  \\
& - q'   {\tau}^{\beta} v'(0) - q' \beta   {\tau}^{\beta-1}v(0) \,.
\end{aligned}
\end{equation}

\noindent $(ii)$ Observe that for any $t \in (0,\tau)$,
\begin{align*}
\int_{A_{\epsilon,\eta}} u(x,t) \tilde \zeta(x)
\le & \Big(\int_{A_{\epsilon,\eta}}
| x|^{-\alpha} u^q(x,t) \tilde \zeta(x) \Big)^{1/q}
\Big(\int_{A_{\epsilon,\eta}}
| x|^{\alpha  \over q-1}  \tilde \zeta(x) \Big)^{1/q'}\\
 =& \Big(\int_{A_{\epsilon,\eta}}
| x|^{-\alpha} u^q(x,t) \tilde \zeta(x) \Big)^{1/q}
\Big(\int_{\epsilon}^{\eta}
r^{{\alpha  \over q-1}+n-1}   \bar \zeta (r)   dr\Big)
^{1/q'} \,.
\end{align*}
 Set
$$
v(t):= \int_{A_{\epsilon,\eta}} u(x,t) \tilde \zeta(x),
\quad t \in (0,\tau) \,.
$$
Then by definition (\ref{4.4}) the above inequality reads
\begin{equation}\label{4.6}
v^q(t)   \le   C_1(\epsilon,\eta)
\int_{A_{\epsilon,\eta}}
| x|^{-\alpha} u^q(x,t) \tilde \zeta(x)
\end{equation}
for any $t \in (0,\tau)$. Then from
 inequalities (\ref{4.6}), (\ref{5.4})  we get
\begin{equation}\label{5.4Z}
\begin{aligned}
\int_0^{\tau} (\tau-t)^{\beta} v^q(t) \,dt
 \leq&  C_1(\epsilon,\eta)   \Big[
{{\tau}^{\beta+1} \over \beta+1} C_2(\epsilon,\eta)
+ \beta(\beta-1)   q'  \int_0^{\tau} (\tau-t)^{\beta -2}  \,dt\\
&\times \int_{A_{\epsilon,\eta}} u(x,t) \tilde \zeta(x)
 -  q' {\tau}^{\beta} v'(0)- q' \beta   {\tau}^{\beta-1}v(0)  \Big]\,.
\end{aligned} \end{equation}

\noindent$(iii)$
Due to Young inequality,
\begin{align*}
&\beta (\beta-1)  q' C_1(\epsilon,\eta) \int_0^{\tau} (\tau-t)^{\beta -2}
v(t)   \,dt  \\
&\leq { 1 \over q}
\int_0^{\tau} (\tau-t)^{\beta} v^q(t)   \,dt   +
{[\beta(\beta-1)] ^{  q' } { (q') } ^{ q'-1} \over \beta - 2q' +1}
 C_1^{  q' }(\epsilon,\eta)   {\tau}^{\beta - 2q' +1 }
\end{align*}
(here we used  the assumption $\beta>  {q+1 \over q-1}$).
From the previous inequality and  (\ref{5.4}) we obtain
\begin{align*}
(1- &{ 1 \over q}) \int_0^{\tau} (\tau-t)^{\beta} v^q(t)   \,dt\\
 \le&
{{\tau}^{\beta+1} \over \beta+1} C_1(\epsilon,\eta) C_2(\epsilon,\eta)
+ {[\beta(\beta-1)] ^{  q' } { (q') } ^{ q'-1} \over \beta - 2q' +1}
 C_1^{  q' }(\epsilon,\eta)   {\tau}^{\beta - 2q' +1 }\\
&  -   q' C_1(\epsilon,\eta)  {\tau}^{\beta} v'(0)
  -   q' C_1(\epsilon,\eta)  \beta   {\tau}^{\beta-1} v(0)\,.
\end{align*}
Then the conclusion follows. \hfill$\square$\smallskip

Let us now proceed to prove Theorem \ref{H.1}. Observe  that, due to
assumption (\ref{H1}), from inequality (\ref{5.3}), we obtain
\begin{align} \label{5.5}
&\int_0^{\tau} (\tau-t)^{\beta}  \,dt
  \Big(\int_{A_{\epsilon,\eta}} u(x,t) \tilde \zeta(x) \Big)^q\\
&\leq   M   \tau^{\beta-1}C_1(\epsilon,\eta) \Big\{C_1(\epsilon,\eta)^
{1 \over q-1}  \tau^{-{2 \over q-1}} + C_2(\epsilon,\eta)   \tau^2   -
   \int_{A_{\epsilon,\eta}} u(x,0) \tilde \zeta(x) \Big\} \,. \nonumber
\end{align}
We will estimate the various terms in inequality (\ref{5.5})
 as $\epsilon \to 0^+$. This motivates the following considerations:

\noindent$\mathbf{(\alpha)}$
Observe that for any $t \in [0, \tau]$,
$$
\lim_{\epsilon \to 0^+} \int_{A_{\epsilon,\eta}} u(x,t) \tilde \zeta(x)
  =
\int_{B_{\eta}}
| x|^{\rho+\sigma} \Big [1- \big ({ \eta \over | x|} \big)^{\sigma}
 \Big ]  u(x,t)
$$
by monotonicity, due to the choice of the function $ \tilde \zeta$.
Moreover, due to
assumption (\ref{H}), there exist $k>0$ and $\eta_1>0$ such that for any
$|x|< \eta<\eta_1$ there holds: $u(x,0) \ge k|x|^{\gamma}$. Hence
$$
\int_{B_{\eta}}
| x|^{\rho+\sigma} \Big [1- \Big ({ \eta \over | x|} \Big)^{\sigma}
 \Big ]  u(x,0)    dx   \ge
k \int_0^{\eta} r^{\gamma+\rho+\sigma+n-1} \Big [1-
\Big ({ \eta \over r} \Big)^{\sigma} \Big ]   dr \,.
$$
 If $\gamma \le -2$, the integral in the right-hand side of the above
inequality diverges. On the other hand, if $\gamma > -2$ we obtain:
$$
\int_{B_{\eta}}
| x|^{\rho+\sigma} \Big [1- \Big ({ \eta \over | x|} \Big)^{\sigma}
 \Big ]  u(x,0)    dx   \ge   K   {\eta}^{\gamma+\rho+\sigma+n}
$$
for some $K>0$.

\noindent$\mathbf{(\beta)}$
Concerning the coefficient $C_1(\epsilon,\eta)$ we
have (see definition (\ref{4.4}))
$$
C_1(\epsilon,\eta)   \le    \Big(\int_{\epsilon}^{\eta}
r^{{\alpha  \over q-1}+\rho+\sigma+n-1}    dr \Big )^{ q-1} \,;
$$
thus, by monotonicity
$$
\lim_{\epsilon \to 0^+} C_1(\epsilon,\eta)   \le
L {\eta}^{\alpha +(\rho+\sigma+n)(q-1)}
$$
for some $L>0$, provided that condition (\ref{5.2}) is satisfied.

\noindent$\mathbf{(\gamma)}$
As for the coefficient $C_2(\epsilon,\eta)$,
we claim that
\begin{equation}\label{5.44}
 C_2(\epsilon,\eta)   =   C_2(\epsilon,2\epsilon)
 \le \bar C \epsilon ^{\theta}
\end{equation}
for some constant $\bar C>0$, where
\begin{equation}\label{5.55}
\theta:= n-\alpha +\rho+\sigma +(\alpha -2){q  \over q-1}    .
\end{equation}
In fact, let us  estimate the integral in the right-hand side
of inequality (\ref{4.5}). To this purpose, set
$s:={r  \over\epsilon}\in [1,2]$. It is easily seen that
\begin{gather*}
\bar \zeta (\epsilon s) \ge c_1\epsilon^{\rho+\sigma} \bar \phi(s)  ,\\
\chi (\epsilon s) \le c_2\epsilon^{n-3+\rho+\sigma} [\bar \phi'(s)
+\bar \phi''(s)]  ,
\end{gather*}
for some $c_1,c_2>0$ and any $s \in [1,2]$; here we used the equalities
$$
\phi_1(\epsilon s)=\bar \phi(s), \quad {d \phi_1  \over dr}
 (\epsilon s)={\bar \phi'(s) \over \epsilon},
\quad {d^2 \phi_1  \over dr^2}
 (\epsilon s)={\bar \phi''(s) \over \epsilon^2}.
$$
 Moreover, choosing $\bar \phi(s)=O((s-1)^{\gamma})$
with $\gamma> \max \{2,{q+1  \over q-1}\}$ as $s \to 1^+$, we obtain
$$
\int_1^2 \Big [{\bar \phi'(s)^q \over \bar \phi(s)}\Big]
^{1  \over q-1} < \infty,
\quad
\int_1^2 \Big [{\bar \phi''(s)^q \over \bar \phi(s)}\Big]
^{1  \over q-1} < \infty   .
$$
It follows that
\begin{equation}\label{3.13}
\int_{\epsilon}^{2\epsilon}
\Big [r^{n-\alpha-1}   \bar \zeta (r)\Big ]
^{-{1  \over q-1}}  \chi (r)^{q  \over q-1}   dr   \le \tilde C
\epsilon ^{\theta}
\end{equation}
for some  $\tilde C>0$, where
\begin{align*}
\theta:=& (n-3+\rho+\sigma){q  \over q-1}
-(n-\alpha -1+\rho+\sigma){1  \over q-1} +1\\
=& n-\alpha +\rho+\sigma +(\alpha -2){q  \over q-1}    .
\end{align*}
This proves the claim. \hfill$\square$


\paragraph{Proof of Theorem \ref{H.1}}
Suppose that
assumption $(a)$ is satisfied. (the proof cases $(b)$-$(c)$ being
the same we omit them).

\noindent  $(i)$  Let $\mu<2$,  $\alpha>2$. In this case  we can choose the
parameters $\rho$, $\sigma$ so that both assumption $(i)$ of Lemma
\ref{l3.2}
and condition (\ref{5.2}) are satisfied, and moreover the exponent $\theta$
defined in (\ref{5.55}) is positive .

Due to the above remarks $(\alpha)$--$(\gamma)$, taking the limit of
inequality  (\ref{5.5}) as $\epsilon \to 0^+$ gives
\begin{multline*}
\int_0^{\tau} (\tau-t)^{\beta}    \,dt   \Big \{
\int_{B_{\eta}}
| x|^{\rho+\sigma} \Big [1- \Big ({ \eta \over | x|} \Big)^{\sigma}
 \Big ]  u(x,t)    dx \Big \} ^q\\
\le   M   \tau^{\beta} {\eta}^{{\alpha  \over q-1}+\rho+\sigma+n}
\Big\{\tau^{-{2 \over q-1}}
- K  {\eta}^{\gamma -{\alpha  \over q-1}} \Big\}
\end{multline*}
for any $\tau \in (0,T)$, if $\gamma > -2$.
In this case the
right-hand side of the above inequality is negative for any
$\tau> \tau_*=\tau_*(\eta):=K^{-(q-1)}{\eta}^{ \alpha - \gamma (q-1)}$;
since
$\tau_*(\eta) \to 0^+$ as $\eta \to 0^+$, the conclusion follows in this
case.

On the other hand, if $\gamma \le -2$  the right-hand side of
inequality (\ref{5.5}) tends to $-\infty$ as $\epsilon \to 0^+$,
thus a contradiction follows in this case, too. This proves the
result in the case $\alpha>2$.

\noindent $(ii)$  Let us assume $\mu<2,  \alpha=2$. In this case  we make
the choice
$\rho+\sigma+n=2$, so that both  assumption $(ii)$ of Lemma \ref{l3.2}
and condition (\ref{5.2}) are satisfied; the above choice gives  $\theta=0$.
Taking the limit of inequality (\ref{5.5})
as $\epsilon \to 0^+$, we obtain
$$
\int_0^{\tau} (\tau-t)^{\beta}    \,dt   \Big \{
\int_{B_{\eta}}
| x|^{\rho+\sigma} \Big [1- \Big ({ \eta \over | x|} \Big)^{\sigma}
 \Big ]  u(x,t)    dx \Big \} ^q   \le
 M   \tau^{\beta} \Big\{ g(\eta, \tau)
- K  {\eta}^{\gamma +2} \Big\} \,,
$$
where
$$
g(\eta, \tau):= {\eta}^{{2q \over q-1}}
\tau^{-{2 \over q-1}} +  \bar C   \tau^2 \,.
$$
It is easily seen that the function $g(\eta, \cdot)$ has a unique
minimum $ \tau_*=\tau_*(\eta):=[(q-1)\bar C]^{-{q-1 \over 2q}}{\eta}$
in $[0,T]$; moreover, $g(\eta, \tau_*)=q\bar C\tau_*^2$. Then by
the above inequality %(\ref{4.9})
there holds:
$$
\int_0^{\tau} (\tau-t)^{\beta}    \,dt   \Big \{
\int_{B_{\eta}}
| x|^{\rho+\sigma} \Big [1- \Big ({ \eta \over | x|} \Big)^{\sigma}
 \Big ]  u(x,t)  \Big \} ^q   \le
 M'   \tau^{\beta+2} \Big\{ \bar C- K  {\eta}^{\gamma } \Big\}
$$
for some $M'>0$. Since $\gamma<0$ and $\tau_*(\eta) \to 0^+$
as $\eta \to 0^+$, the conclusion follows in this case, too.
This completes the proof. \hfill$\square$


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\noindent\textsc{Stanislav I. Pohozaev}\\
Steklov Mathematical Institute, Russian Academy of Sciences,\\
Gubkina 8, 117966 Moscow, Russia \\
e-mail: pohozaev@mi.ras.ru \smallskip

\noindent\textsc{Alberto Tesei} \\
Dipartimento di Matematica ``G. Castelnuovo", \\
Universit\`a  di Roma ``La Sapienza'', P.le A. Moro 5, \\
I-00185 Roma, Italia\\
e-mail: tesei@mat.uniroma1.it
\end{document}

