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\markboth{\hfil Blow-up of radially symmetric solutions \hfil
EJDE--2002/11} {EJDE--2002/11\hfil Dimitrios E. Tzanetis \hfil}
\begin{document}
\title{\vspace{-1in}\parbox{\linewidth}{\footnotesize\noindent
{\sc  Electronic Journal of Differential Equations}, Vol. {\bf
2002}(2002), No. 11, pp. 1--26. \newline ISSN: 1072-6691. URL:
http://ejde.math.swt.edu or http://ejde.math.unt.edu
\newline ftp  ejde.math.swt.edu  (login: ftp)}
 \vspace{\bigskipamount} \\
 %
  Blow-up of radially symmetric solutions of a non-local problem
  modelling Ohmic heating
 %
\thanks{ {\em Mathematics Subject Classifications:}
35B30, 35B40, 35K20, 35K55, 35K99. \hfil\break\indent 
{\em Key words:} Nonlocal parabolic equations, blow-up, global existence,
 steady states.
\hfil\break\indent \copyright 2002 Southwest Texas State
University. \hfil\break\indent 
Submitted October 2, 2001. Published February 1, 2002.
\hfil\break\indent 
Partially supported by the E.C. Human Capital and Mobility Scheme,
 under contract  \hfil\break\indent ERBCHRXCT 93-0409.
} }
\date{}
%
\author{ Dimitrios E. Tzanetis }
\maketitle

\begin{abstract}
  We consider a non-local initial boundary-value problem
  for the equation
   $$ u_t=\Delta u+\lambda f(u)/\Big(\int_{\Omega}f(u)\,dx\Big)^2 ,\quad
      x \in \Omega \subset \mathbb{R}^2 ,\,\;t>0,
   $$
  where $u$ represents a temperature and $f$  is a positive and
  decreasing function. It is shown that for the radially symmetric
  case, if $\int_{0}^{\infty}f(s)\,ds <\infty $
  then there exists a critical value $\lambda^{\ast}>0$
  such that  for $\lambda>\lambda^{\ast}$ there is no stationary
  solution and $u$  blows up, whereas for $\lambda<\lambda^{\ast}$
  there exists at least one stationary solution.
  Moreover, for the Dirichlet problem with
  $-s\,f'(s)<f(s)$ there exists a unique stationary
  solution which is asymptotically stable. For the Robin problem,
  if $\lambda<\lambda^{\ast}$ then there are at least two solutions,
  while if $\lambda=\lambda^{\ast}$ at least one solution.
  Stability and blow-up of these solutions are examined
  in this article.
\end{abstract}

\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{prop}[theorem]{Proposition}
\numberwithin{equation}{section}


\section{Introduction}

In this work we  study the radially symmetric solutions to the
non-local initial boundary-value problem
\begin{subequations} \begin{gather}
u_t=\Delta u+\frac {\lambda f(u)}{(\int_\Omega
f(u)\,dx)^2}\,,\quad x \in \Omega,\;\,t>0\,, \label{enaa}
\\
\mathcal{B} (u):=\frac {\partial u}{\partial n}+\beta
(x)u=0\,,\quad t>0\,,\;x\in\partial\Omega\,, \label{enab}
\\
u(x,0)=u_{0} (x)\,,\quad x\in \Omega\,,   \nonumber
\end{gather} \label{ena}
\end{subequations}
where $u=u(x,t)$,  $\Omega$  is a bounded domain of
$\mathbb{R}^2$, $\lambda$ is a positive parameter, $\partial
\Omega $ and $\beta (x)$ are sufficiently smooth. The function $f$
is continuous, positive and decreasing,
\begin{equation}
f(s)>0\,,\quad f'(s)<0\,,\quad s\geq 0\,. \label{duo}
\end{equation}
We also study, in Section 3, the case of $f$  being the Heaviside
function (which is neither continuous nor always positive). The
equation (\ref{enaa}) arises by reducing the system of two
equations \begin{subequations}
\begin{gather}
u_t=\nabla \cdot (k(u)\nabla u)+\sigma (u)\,|\nabla \phi|^2
\label{tria}\\
\nabla \cdot(\sigma (u)\nabla \phi)=0\,,\label{trib}
\end{gather}
\label{tri} \end{subequations} to a simple, but still realistic
equation. More precisely, (\ref{tria}) is a parabolic equation
while (\ref{trib}) is an elliptic, $u$ represents the temperature
produced by an electric current flowing through a conductor,
$\phi=\phi (x,t)$ is the electric potential, $k(u)$ is the thermal
conductivity and $\sigma (u)$  is the electrical conductivity.
Problem (\ref{tri}) models many physical situations especially in
thermistors \cite{ac1,ac2,all,fow}, fuse wires, electric arcs and
fluorescent lights. The conductivity $\sigma $ may be either
decreasing or increasing in $u $  depending upon the nature of the
conductor. Here we consider materials having constant thermal
conductivity, e.g. $k(u)=1$, and decreasing electrical
conductivity, the latter allowing a thermal runaway to take place
\cite{lac9,lac95}.

Some questions concerning the steady problem to (\ref{tri}) were
investigated by Cimatti \cite{cim89,cim90,cim99}, see also
\cite{all}. A similar problem to (\ref{tri}) with radial symmetry,
Robin boundary conditions of the form $u_{n}+\beta u=0$ and
conductivity $\sigma (u)=\exp(-f(u)/\epsilon)$, $\epsilon\ll 1$
was discussed by Fowler et al. \cite{fow}. Some numerical results
are also given for small $\beta$. See also Howison \cite{how} for
how the steady problem to (\ref{tri}) may
be reduced to one nonlinear o.d.e. and Laplace's equation.\\
Carrillo \cite{car}, has looked at the bifurcation diagram of the
non-local elliptic problem with decreasing nonlinearity and
Dirichlet boundary conditions, in $\Omega \subset \mathbb{R}^N$,
See \cite{beb} for a similar study where $\Omega$  is a unit ball
in $\mathbb{R}^N $.  For an extended study of the structure of
solutions of the non-local elliptic problem see \cite{lo}.

 The two-dimensional mathematical problem for the
single equation can be derived by considering a long and thin
cylindrical conductor $D$, $(x,y,z)\in D \subset \mathbb{R}^3$, of
length $L$, $R\ll L$ where $R$ is the radius of the cross-section
$\Omega$ of  $D$, see Figure \ref{sxh1}.

\begin{figure}
\begin{center}
\includegraphics[width=0.8\textwidth]{fig1.eps}
\end{center}
\caption{A long and thin cylindrical conductor} \label{sxh1}
\end{figure}


The curved surface of $D$ is electrically insulated, i.e. $\phi
_n=0$, and $u=0$ or more generally $u_n+\beta u=0$ for $\beta \in
\left[0,\infty\right]$, ($\beta =0$ gives $u_n=0$ while $\beta
=\infty$ gives $u=0$). Also $\phi (x,y,0,t)=0$, $\phi (x,y,L,t)=V$
at the ends of the conductor, thus $V$ is the potential
difference. Here the temperature $u$ is taken to be initially
independent of $z$ ($u=0$ is likely to be of practical interest).
Also the $z$-derivatives of $u$ are neglected, so the model gives
$z$-independence of $u$ for $t>0$. Moreover we stress that the
model is most definitely only supposed to apply in the bulk of the
device. Thus taking the thermal conductivity to be constant, and
neglecting the end effects, problem (\ref{tri}) can then be
reduced, as in \cite{lac9}, to a single non-local equation
\begin{equation}
u_t =\Delta u + \lambda \sigma (u)/(\int_ \Omega \sigma
(u)\,dx)^2, \label{tes} \end{equation} where $\lambda =I^2
/|\Omega|^2 \geq 0$, $I$  is the electric current which we suppose
to be constant and $|\Omega |$ is the measure of $\Omega $. On the
other hand, by assuming the voltage $V$ to be constant, $\phi_
x=V/L$, problem (\ref{tri}) takes the more standard semi-linear
parabolic form:
\begin{equation}
u_t =\Delta u+ \lambda \sigma
(u)\,,\;\;x\in\Omega,\;\;\mbox{where}\;\;\lambda = V^2/L^2\geq 0.
\label{pen} \end{equation} Finally, taking the more general case
of a conductor connected in series with a resistance $R_0$ under a
constant voltage $E$, then (\ref{tri}) gives, on using
\[
 E=I\,R_0 +V=\Big[I+R_0\, |\Omega|\int_ \Omega \sigma (u)\,dx\Big]\,V\,,
\]
the non-local equation
\begin{equation}
u_ t =\Delta u + \lambda \sigma (u)/\Big[a+b\int_ \Omega \sigma
(u)\,dx\Big]^2 ,\,\; x \in\Omega\,,\;\,t>0,\;\, a,b>0. \label{eks}
\end{equation}

For the derivation of  equations (\ref{tri})-(\ref{eks}), as well
as a complete study of the one-dimensional model for a decreasing
$\rho (u)$, (the electrical resistivity $\sigma (u)=1/\rho (u)$ is
increasing), see \cite{lac9,lac95}. In \cite{lac9,lac95} it was
shown that for $\int_{0}^{\infty} f(s)\,ds<\infty$ there is some
critical value $\lambda ^{\ast}$ such that for $\lambda
>\lambda^*$  there is no steady state and $u$  blows up globally, for
$\lambda =\lambda ^{\ast}$,  and $f(s)=\exp(-s)$, again there is
no steady state and $u$  exists globally in time but is unbounded.
Moreover, for $\lambda <\lambda^* $,   as well as for any $\lambda
>0$ provided now that $\int_{0}^{\infty}f(s)\,ds=\infty$, where a
unique steady state exists, this steady state is  globally
asymptotically stable. A global existence and divergence result
for the solution of (\ref{ept}) (see below), when $f(s)=e^{-s}$,
is also proved in \cite{kth1}.

Chafee \cite{cha} considered a related model
$u_t=u_{xx}-g(u)+\lambda f(u)/ (\int_{-1}^{1} f(u)\,dx)^2$. It was
found that there is a $\lambda^* $  such that for $\lambda
<\lambda ^{\ast}$ there is a homogeneous steady state which is
globally asymptotically stable. There are conditions under which
the homogeneous steady state is unstable and there are then two
stable inhomogeneous steady states. Other works concerning the
blow-up in non-local parabolic problems are \cite{bar,so1,so2}.

 Finally we wish to study  problem (\ref{ena}),
which comes from (\ref{tes}) by setting  $\sigma (u)=f(u)$, in the
radially symmetric case. Therefore we take $\Omega $  to be the
unit ball and the initial data are taken to be radially symmetric
and decreasing, $(u(x,0)=u_{0}(r)$ and $u_{0}'(r)
<0\,,\;\,r=|x|)$. Thus our problem, in case of Dirichlet boundary
conditions ($\beta(x)=\infty$), takes the form:
\begin{subequations}
\begin{gather}
u_t=u_{rr}+\frac {u_r}{r}+\frac {\lambda f(u)}{4\pi ^2
(\int_{0}^{1} f(u)\,r\,dr) ^2}\,,\quad\; 0<r<1\,, \quad \: t>0\,,
\label{epta}\\
u(1,t)=u_r (0,t)=0\,,\quad t>0\,,\label{eptb} \\
u(r,0)=u_{0} (r)\,,\quad 0<r<1\,. \label{eptc}
\end{gather} \label{ept}
\end{subequations}
We also consider Neumann or Robin boundary conditions:
\begin{equation}
u_{r}(1,t)+\beta u(1,t)=0\;,\quad \beta\in [0,\infty).
\label{oxto}\end{equation} Note that $u_{r}(0,t)=0$  is a
consequence of boundedness of solutions rather than a specific
constraint upon them.

We note that any solution of (\ref{ena}) with $u_{0}>0$ is
positive,
 moreover if
$u_0(x)=u_0(r)$ and $\Omega $ is a ball, then $u(x,t)=u(r,t) $ is
radially symmetric and satisfies (\ref{ept}) with the proper
boundary conditions ((\ref{eptb}) or (\ref{oxto})). Furthermore if
$u_0'(r) < 0$ then $u_r (r,t)<0$,  $0<r<1$,  $0<t<T$ , i.e. $u$ is
radially decreasing. The same properties hold for the steady
solutions of problem (\ref{ena}), see Gidas et al., \cite{gid}.

The non-local problem under consideration belongs to the class
where the maximum principle holds (due to (\ref{duo})) and
comparison with suitable upper and lower solutions is used to
prove stabilization or blow-up. In the contrary when $f(s)$ is an
increasing function, maximum principle does not hold. Nevertheless
for $f(s)=e^s$, stabilization and blow-up can be studied by using
a Lyapunov functional, \cite{beb,fil}.

 The present work is organized as follows. In
Section 2 the existence and uniqueness of solutions to (\ref{ena})
in $\Omega \subset \mathbb{R}^{N}$ is discussed. In Sections 3, 4,
some particular functions, the Heaviside and the exponential are
studied, while in Section 5 a general decreasing function is
considered. In each of these cases, the critical value $\lambda^*$
is estimated.

 In the rest of this article we mainly follow the
ideas and techniques which have been used in the one-dimensional
case \cite{lac9, lac95}, but have to be modified because of the
extra technical difficulties encountered in this two-dimensional
problem.

\section{Existence, uniqueness and monotonicity}

Problem (\ref{ena}) in $\Omega \subset \mathbb{R}^N$ $N\geq 1$,
for a measurable and bounded $u_{0} (x)$, can be written in a
Green' s integral formulation:
\begin{equation}
u(x,t)=\lambda \int_{0}^{t}\int_\Omega
g(x,y,t-s)\frac{f(u(y,s))}{(\int_\Omega f(u(y,s)) \,dy)^2}\,dy\,ds
+\int_{\Omega}g(x,y,t)u_{0}(y)\,dy\,. \label{dena} \end{equation}
Setting now $v_{n}$ instead of $u$ on the left-hand side and
$v_{n-1}$ instead of $u$ on the right-hand side for $n\geq 1$ and
taking $v_{0} \equiv 0$, we find on passing to the limit and
following on standard Picard - iteration - type arguments, that if
$\lambda
>0$, $f(s)\geq c>0$ and  Lipschitz for $s\in (a,b)$, where
$a<\min \{0,\inf{u_0}\}\leq u\leq \max\{0,\sup u_{0}\}<b\,,$ then
there exists a unique solution $u$ to (\ref{ena}) and (2.1).
Moreover the solution continues to exist as long as it remains
less than or equal to $b\,;$ this implies that it can only cease
to exist due to blow-up.

On the other hand, if we restrict our attention to a decreasing
$f$,  positive and Lipschitz, then we have a sort of comparison.
In particular $\overline u$ is called a strict \bf upper solution
\rm to (\ref{ena}) in $\Omega\subset\mathbb{R}^N,\;\,N\geq1$, if
it satisfies
\begin{subequations}
\begin{gather}
\overline u_{t}(x,t)>\Delta \overline u+\frac{\lambda f(\overline
u)}{(\int_\Omega f(\overline u)\,dx)^2} \,, \quad \mbox{in}
\;\Omega\,, \quad t>0\,, \\
\mathcal{B}(\overline u)>\mathcal{B}(u)=0 \quad \mbox{on} \quad
\partial
\Omega\,,\;\; t>0\,, \\
\overline u_{0} (x)>u_{0}(x)\,,\;\;\mbox{in}\;\; \Omega\,,
\end{gather}
\label{dduo} \end{subequations} while if $\underline u$ satisfies
the reversed inequalities of (\ref{dduo}) it is called a strict
\bf lower solution\rm. Now if we set $v=\overline u-\underline u$
then there exists $T>0$  such that
\begin{equation}
\begin{gathered}
v_{t}>\Delta v+\frac{\lambda f'(s)}{(\int_\Omega f(\underline
u)\,dx)^2}\,v\,,\;\;x\in
 \Omega\,,\;\;0<t<T\,, \\
v>0 \;\;\mbox{at}\;\;\; t=0 \;\;\mbox{in}\;\,\Omega
\;\;\mbox{and}\quad \mathcal{B}
(v)>0\,,\;\;\mbox{on}\;\;\,\partial \Omega\,,\;\;0<t<T\,,
\end{gathered} \label{dtri}
\end{equation}
 which implies, by the
maximum principle, that $v>0$  at $t=T$.  Moreover, if
(\ref{dduo}) holds with $\geq$, then (\ref{dtri}) also holds with
$\geq$ in the place of $>$. As long as $\underline u\,,\;\;
\overline u$ exist and $\overline u\geq \underline u$, with $f$
Lipschitz, we can apply iteration schemes similar to those of
Sattinger \cite{sat}, to show that  there exists a unique solution
$u$ to (\ref{ena}) such that $\underline u\leq u\leq\overline u$.
If now $f$  is increasing then some of the above results can be
adapted by using a pair of upper-lower solutions; see \cite{lac9}.

\section{The Heaviside function}

 We consider now $f(s)$ to
be the Heaviside function (decreasing), $f(s)=H(1-s)$, then
$f(s)=1$  for $s<1$,  and $f(s)=0$  for $s\geq1\,,\:$ which is
neither strictly positive nor Lipschitz continuous. Thus problem
(\ref{ept}) becomes, \begin{subequations}
\begin{gather}
u_t=\Delta_r u+\lambda H(1-u)/4\pi ^2
\Big(\int_0^1H(1-u)r\,dr\Big)^2,\quad 0<r<1 \,,\quad t>0\,,
\label{tenaa}\\
u(1,t)=u_r (0,t)=0\,, \quad t>0\,, \quad u(r,0)=u_0 (r)\,, \quad
0<r<1\,, \label{tenab}
\end{gather} \label{tena} \end{subequations}
where  $(\Delta_r =\partial ^2/\partial r^2 +\frac{1}{r}\partial
/\partial r)$.  In particular equation (\ref{tenaa}) can be
written:
\begin{gather*}
u_t=\Delta_r u\,,\quad \mbox{where} \quad u\geq 1
\,,\;\;\mbox{and} \\
u_t=\Delta_r u+\lambda /m^2 (t)\,,\quad \mbox{where} \quad  u<1\,,
\end{gather*}
writing $m(t)$ the measure of the subset of the unit ball $B(0,1)$
where $u<1$.  The existence and uniqueness of a ``weak" (classical
a.e.) solution to (\ref{tena}) is obtained by using an
approximating regularized version of this problem, see \cite{kth}
and the references therein. Hence, taking into account this
remark, in the following we can use comparison arguments in the
classical sense. We take $u_0 (r)\leq 1$, and for simplicity $u'_0
(r)\leq 0$ and bounded below. With such initial data $v=1$ is an
upper solution to  problem (\ref{tena}), hence $u\leq 1$. Thus
either $u<1$ for $0<r<1$ whereupon  (3.1a) becomes
\begin{equation}
u_t=\Delta_r u +\lambda /\pi ^2,\quad 0<r<1\,,\;\;t>0,
\label{tduo} \end{equation} or there exists an or some
$s=s(t),\;\, 0<s(t)<1$,  such that \begin{subequations}
\begin{gather}
u_t=\Delta_r u+\lambda /\pi ^2 (1-s^2)^2 ,\quad 0\leq u<1,\quad
s<r<1\,,\;\;t>0\,, \\
u=1\,, \quad u_r=0, \quad 0\leq r\leq s\,,\;\; t>0\,,
\end{gather}
\label{ttri} \end{subequations} where $\pi (1-s^2 (t))=m(t)$. Note
that $u$ is continuous and $u_r \leq 0$, the latter follows by
using the maximum principle. The corresponding steady state to
(\ref{tduo}) is
\begin{equation}
\Delta_r w+\lambda /\pi ^2 =0\,, \quad 0<r<1\,, \quad
w'(0)=w(1)=0\,, \label{ttes} \end{equation} which for $\lambda < 4
\pi ^2,$ has a solution $w(r)=\frac {\lambda}{4 \pi ^2}(1-r^2)$.
Also a steady state for (\ref{ttri}) satisfies:
\begin{subequations}
\begin{gather}
\Delta_r w+\frac {\lambda}{\pi ^2 (1-S^2)^2}=0\,,\quad
S<r<1\,,\quad 0\leq w<1\,,\label{tpena} \\
w(r)=1\,,\;w'(r)=0\,,\;0\leq r \leq S \;\;\mbox{for}\quad 0\leq
S<1\,, \quad w(1)=0\,.\label{tpenb}
\end{gather} \label{tpen}\end{subequations}
 Equations (\ref{tpena}), (\ref{tpenb}) give a one-parameter family
 of steady states of the form:
\begin{gather}
w(r;S)=\frac {\hat{\lambda}(S) (1+2S^2 \ln r -r^2)}{4 \pi ^2
(1-S^2)^2}= \frac{1+2S^2 \ln r -r^2}{1+2S^2 \ln S
-S^2}\,,\;\;S<r<1, \label{tpenc}\\
\mbox{where}\quad\lambda =\hat{\lambda} (S)=\frac {4 \pi ^2
(1-S^2)^2}{1+2S^2 \ln S -S^2}.\nonumber
\end{gather}
It is easily seen that $\hat{\lambda} (S)$ is strictly increasing,
$\hat{\lambda} (1-)= 8 \pi ^2\,\; \mbox{and}\,\; \hat{\lambda}
(0+)=4 \pi ^2$. If we note by $\|w'\|=\sup |w'|$, then $\|w'\|
=-w'(1)$ and the following hold: for $0<\lambda < 4 \pi ^2
=\hat{\lambda}(0+)$ there exists a unique steady state $w(r)=\frac
{\lambda}{4 \pi^2}(1-r^2)\,,\;\;\mbox{for}\quad 4\pi ^2\leq \hat
{\lambda} (S)<8\pi ^2,\; S\in [0,1)$, there exists a one-parameter
family of steady states given by (\ref{tpenc}), whereas for
$\lambda \geq \hat{\lambda}(1-)=8\pi^2$ there is no steady
solution. Hence we get the diagram of Figure \ref{sxh2}.

\begin{figure}
\begin{center}
\includegraphics[width=0.6\textwidth]{fig2.eps}
\end{center}
\caption{Response diagram for  (\ref{ttes}), (\ref{tpen}),
  $f(s)=H(1-s)$.}\label{sxh2}
\end{figure}

We wish now to study the stability of the steady solutions for
$\lambda <8\pi ^2=\hat{\lambda} (1-)$. Therefore we construct an
upper solution $\overline v$  (lower solution $\underline v$) to
problem (\ref{tena}), decreasing (increasing) in time, of a form
similar to the steady state, i.e. $w(r;s(t))$. Namely for
$\lambda<4\pi ^2$  we take, \begin{subequations}
\begin{gather}
\overline v(r,t)=1\,,\;\;\overline v_r
(r,t)=0\;\;\mbox{for}\;\;0\leq r\leq s(t)\,,\;\;\, \mbox{and} \\
\overline v(r,t)=w(r;s(t))=\frac{\hat{\lambda} (s)}{4\pi ^2
(1-s^2)^2}\left(1+2s^2 \ln r-r^2\right)=\frac{1+2s^2 \ln r-r^2
}{1+2s^2 \ln s-s^2}\,,
\end{gather} \label{texi} \end{subequations}
$s(t)<r<1\,,\quad 0\leq t<t_1$  where $s=s(t)\,\in
(0,1)\,,\;\;s(0)=s_0$. For any initial data $u_0 (r)\leq 1$, we
choose $s_0$  so that $u_0 (r) \leq 1$  for $0\leq r\leq s_0$  and
$u_0 (r)\leq w(r;s_0)$, $0<s_0 <r<1$,  i.e. $\overline v(r,0)=
w(r;s_0)\geq u_0 (r)$.  Then we have,
\begin{eqnarray*}
\mathcal{E}(\overline v)&:=&\overline v_t-\Delta_r \overline
v-\lambda H(1-\overline v)/4\pi ^2 \Big(\int_0^1H (1-\overline
v)r\,dr\Big)^2\\
 &=&\left\{\begin{array}{ll}\displaystyle
0\,, & 0\leq r\leq s(t)\,,\\
\displaystyle \overline v_t+\frac{\overline \lambda(s)
-\lambda}{\pi ^2 (1-s^2)^2}\geq0\,, & s(t)\leq r\leq 1\,,
\end{array}\right.
\end{eqnarray*}
provided that $s(t)$  satisfies:
\[
0<-\dot{s}=h(s)\equiv \frac{(\overline \lambda(s) -\lambda)(1+2s^2
\ln s-s^2)}{4\pi ^2(1-s^2 )^2 (1-s)}\,,
\]
giving $\dot{s}(t)<0\,,\;\;\dot{\overline \lambda}=\overline
\lambda '(s)\dot{s}<0$  for $\overline \lambda (s)>\lambda \;\, $
and $s(t_1 )=0$,  for $t_1 <\infty$.

Hence $\overline v(0,t_1)=1$, and  $w(r;s(t))\to 1-r^2$ as $t\to
t_1 -$. Again for $t\geq t_1$  we can take \begin{subequations}
\begin{gather}
\overline v (r,t)=a(t)(1-r^2)\,,\;\; a_1
=a(t_1)=1\,,\;\mbox{for}\;\;t\geq t_1\,, \\
\overline v_r (0,t)=\overline v(1,t)=0\,,\;\;\, t\geq t_1\,,
\end{gather}
\label{tept} \end{subequations} giving
\[
\mathcal{E}(\overline v)=\dot{a}(1-r^2)+4a-\frac{\lambda}{\pi
^2}\geq\dot{a}+4(a- \frac{\lambda}{4\pi ^2})=0\,,
\]
on taking $\dot{a}=-4(a-\frac{\lambda}{4\pi ^2})<0$,
$\frac{\lambda}{4\pi ^2}<a<1$, since $\dot{a}(t)<0$.  Then
\[
a(t)=\frac{\lambda}{4\pi ^2}+(1-\frac{\lambda}{4\pi ^2})e^{4(t_1
-t)}\to \frac{\lambda}{4\pi ^2} \;\;\,\mbox{as}\;\;t\to\infty\,.
\]
Hence $\overline v$ is an upper solution to $u$-problem which
exists for all time, in particular $u\leq \overline v$  and
$\overline v\to w=\frac{\lambda}{4\pi ^2}(1-r^2 )$ as
$t\to\infty$.  On the other hand for $\underline v(r,t)$  we take
\begin{equation}
\begin{gathered}
\underline v(r,t)=b(t)(1-r^2)\,,\;\;0<r<1\,,\quad t>0\,,\\
b(0)=0\,,\quad \mbox{where}\;\; 0\leq b\leq \min\{1,\lambda /4\pi
^2 \}\,.
\end{gathered}
\label{toct} \end{equation} Then we have,
\[
\mathcal{E}(\underline v)=\dot{b}(t)(1-r^2)+4b-\frac{\lambda}{4\pi
^2}\leq \dot{b}(t)+4b- \frac{\lambda}{4\pi ^2}=0\,,
\]
on taking $\dot{b}(t)+4b-\frac{\lambda}{4\pi ^2}=0$,  since
$\dot{b}(t)>0$,  giving $b(t)=\frac{\lambda}{4\pi
^2}(1-e^{-4t})\to \frac{\lambda}{4\pi ^2}$  as $ t\to \infty$.
Thus for $0<\lambda<4\pi ^2$  we have that $\underline v(r,t)\leq
u\leq \overline v(r,t)$, $\overline v(r,t)$, $\underline v(r,t)\to
w(r)=\frac{\lambda}{4\pi ^2}(1-r^2)$ as $t\to\infty$  uniformly in
$r$, which implies that $u$  is bounded for all time and
$u(r,t)\to w(r)$  as $t\to \infty$ ($w$ is the unique steady state
for $\lambda <4\pi ^2$).  Hence $w$, is globally asymptotically
stable solution \cite{lac9,lac95,sat}. Moreover, if $4\pi ^2 \leq
\lambda <8\pi ^2$,  then we can proceed in a similar way. In fact,
we construct an upper solution decreasing in time as (\ref{texi}),
then $\mathcal{E}(\overline v)\geq 0$,  provided that
\[
0<-\dot{s}=h(s)\equiv \frac{(\overline \lambda(s) -\lambda)(1+2s^2
\ln s-s^2)} {4\pi ^2(1-s^2 )^2 (1-s)}\,,
\]
giving now $\dot{s}(t)<0$  for $\overline \lambda (s)>\lambda $
and $s\to S_0 +\,, \;\;\overline \lambda (s)\to \lambda =\lambda
(S_0)$,  as $t\to \infty$.
  Also we construct a lower solution $\underline v$
increasing in time, having a similar form to that as in the proof
of the blow-up (see below), in particular like (\ref{toct})
followed by a complementary version of (\ref{texi}). But now
$\dot{s}(t)>0\,,\;\; \dot{\underline \lambda}=\underline \lambda
'(s)\dot{s}>0$  for $t>t_1\,,\;\;s(t_1)=0$  and $s(t)\to S_0
-\,,\;\;\underline \lambda (s) \to \lambda = \lambda (S_0) $, as
$t\to \infty$. Hence $\underline v\leq u\leq\overline v,u$  exists
for all time and $u\to w(r;S_0)$  the unique solution for $4\pi ^2
\leq\lambda =\lambda (S_0)<8\pi ^2$,  which is globally
asymptotically stable.

We  show now that the solution $u$,  ``blows up" (it ceases to be
less than $1$  in $[0,1)$,  we recall that $u\leq1$  in $(0,1)$
as long as $u_0(r)\leq1$ ) in the sense that it becomes $1$ in
$[0,1)$  in finite time, for $\lambda
>8\pi ^2$.  Therefore we get a lower solution $\underline v(r,t)$  of the
form:\hspace{.3cm} $\underline v(r,t)=b(t)(1-r^2)\,,\quad
0<r<1\,,\quad 0\leq t\leq t_1$, which satisfies (\ref{toct}) (note
that $u_0 (r)\leq 1$ ), $b(t_1)=1\,, \;\;u(r,t_1)\geq \underline
v(r,t_1)= 1-r^2 $,  provided that $u$  still exists ($u\leq 1$ )
up to $t_1 $.  Also we take $\underline v(r,t)$ to satisfy:
\begin{gather*}
\underline v(r,t)=1\,,\quad \underline v_r (r,t)=0\quad
\mbox{for}\;\;0\leq r\leq s(t)\,,\;\;
\,t>t_1\,,\;\;\,\mbox{and} \\
\underline v(r,t)=\frac{1+2s^2 \ln r-r^2}{1+2s^2 \ln
s-s^2}\,,\quad s(t)<r<1\,,\; \;t>t_1\,,
\end{gather*}
$b(t_1)=1$, $s(t_1)=0$, $\underline v(r,t_1)=1-r^2$. Then we have,
$$\mathcal{E}(\underline v)=\left\{\begin{array}{ll}
\displaystyle 0\,, & 0\leq r\leq
s(t)\,,\; t\geq t_1\,,\\
\displaystyle \underline v_t +\frac{\underline \lambda -
\lambda}{\pi ^2 (1-s^2)^2}\leq 0\,, & s(t)\leq r\leq
1\,,\;\;t>t_1\,,
\end{array}\right.
$$
provided that  $s(t)$  satisfies
\[
0<\dot{s}=-h(s)\equiv \frac{(\lambda -\underline \lambda(s)
)(1+2s^2 \ln s-s^2)} {4\pi ^2(1-s^2 )^2 (1-s)}\,,
\]
for $\underline \lambda (s)<8\pi ^2 <\lambda $.  This implies that
$\dot{\underline \lambda }=\underline \lambda '(s)\dot{s}>0$, and
$s(t)\to 1-\,,$ $\underline \lambda (s) \to 8\pi ^2 -$,  as $t\to
T^{\ast}<\infty$.  Hence $\underline v$ becomes $1$  in $[0,1)$
at $T^{\ast}$  and a form of `` blow-up" $(\,u\to 1$ in $[0,1)$ as
$t\to t^{\ast}-$, $t^{\ast}\leq T^{\ast})$  has been established
for $u$, with derivative $u_r(1,t)$ becoming unbounded as $t\to
t^{\ast}-$.

 For the critical value $\lambda =8\pi ^2 $,
again we construct an upper solution
 $\overline v$  but now increasing in time; indeed
$\mathcal{E}(\overline v)\geq 0$,  provided that $s(t)$ satisfies
\[
0<\dot{s}=-h(s)\equiv \frac{(8\pi ^2 -\overline \lambda(s))(1+2s^2
\ln s-s^2)}{4\pi ^2(1-s^2 )^2 (1-s)}\,.
\]
For $4\pi ^2<\overline \lambda (s)<8\pi ^2 $,  we get $s\to 1$  as
$t \to \infty$,  which implies that $u$  exists for all time but
becomes $1$  in $[0,1)$  as $t\to \infty$.

\section{The exponential function}

\subsection{Stationary solutions}
We now consider $f(s)=e ^{-s}$, so $ f(s)>0$, $f'(s)<0$ for $s\geq
0$  and $\int_0^\infty f(s)\,ds=1$.  The corresponding steady
problem to (\ref{ept}) for $f(s)=e ^{-s}$ is
\begin{equation}
w''(r)+\frac {1}{r}w'(r)+\mu e ^{-w(r)} =0\,,\quad 0<r<1\,, \quad
w(1)=w' (0)=0, \label{Tena} \end{equation} where
\begin{equation}
\mu =\frac {\lambda }{4 \pi ^2 (\int_0^1 e ^{-w(r)} r\,dr)^2}\,.
\label{Tduo} \end{equation} The solution of (\ref{Tena}) is
\begin{equation}
w(r)=2\ln [\alpha (1-r^2)+r^2]\,,\quad  \mu =8 \alpha (\alpha
-1)\,, \label{Ttri} \end{equation} where $\alpha >1$,
$M=\sup\|w\|=w(0)=2 \ln \alpha$. The parameter $\lambda$ is given
by
\begin{equation}
\lambda =4\pi ^2 \mu \Big(\int_0^1 e^{-w} r\,dr\Big)^2
=8\big(1-\frac {1}{\alpha} \big)\pi ^2=8\pi ^2
\left(1-e^{-M/2}\right)<8\pi ^2\,, \label{Ttes}
\end{equation}
so $\alpha=\frac{\lambda^* }{\lambda^* -\lambda }$, $\lambda
=\lambda (M) \to 8\pi ^2=\lambda ^{\ast}$ as $M\to \infty$, or
equivalently as $\alpha \to \infty $,  and $\lambda '(M)>0$. For
each $M$ there is a corresponding unique solution $w(r)$(this
follows from a shooting argument). Finally from the above we get
the diagram of Figure \ref{sxh3}.

\begin{figure}
\begin{center}
\includegraphics[width=0.5\textwidth]{fig3.eps}
\end{center}
\caption{Response diagram for (4.1), $f(s)=\exp(-s)$.}
\label{sxh3}
\end{figure}

If $\lambda \to \lambda^{\ast} -$, which implies that $\alpha \to
\infty$, then the solution $w(r)= 2\ln [\alpha (1-r^2)+r^2)] \to
\infty$, for every compact subset of $[0,1)$. We  see below that
this also holds for a general decreasing $f$.

\subsection{Stability for {\boldmath $\lambda <\lambda^{\ast} $}}

  To study  the stability, we use upper solutions which are
decreasing in time and lower solutions which are increasing in
time to  problem (\ref{ept}) with $f(s)=\exp (-s)$.

 We first note that $v(r,t)=w(r;\mu (t))=2\ln[\alpha
(t)(1-r^2)+r^2] $  is an upper solution, provided that
\begin{equation}
\dot{\alpha}+2 \frac{\lambda^{\ast}
-\lambda}{\lambda^{\ast}}\alpha -2=0\,,\;\; \alpha (0)=\alpha_0\;
\label{Tpen} \end{equation} and $\alpha_0$  is sufficiently large.
Hence for $\lambda \in (0,\lambda^{\ast})$  the solution of
(\ref{Tpen}) is
\begin{equation}
\alpha (t)=\frac{\lambda^{\ast}}{\lambda^* -\lambda}+\big[\alpha_0
-\frac{\lambda^{\ast}}{\lambda^* -\lambda}\big]
\exp\big(-2\frac{\lambda^{\ast} -\lambda}{\lambda^{\ast}}\,t\big),
\label{Texi}
\end{equation}
and $\dot{\alpha} (t)<0$ for $\alpha_0
>\frac{\lambda^{\ast}}{\lambda^{\ast} -\lambda}$. Furthermore we require
$v(r,0)=2 \ln [\alpha_0 (1-r^2)+r^2] \geq u_0(r)$. It is
sufficient to choose
$$\alpha_0 =\max\{\frac{\lambda^*}{\lambda^*
-\lambda}\,,\,\sup_{r} \frac{\exp(u_0 (r)/2) -r^2}{1-r^2}\}\,.
$$
  Also   $\mathcal{B}(v)\geq\mathcal{B}(u)\;\mbox{on}\;\partial \Omega $,
in fact it is $v(1,t)=u(1,t)=0$. The calculations are like these
of the one-dimensional case \cite{lac9,lac95}; we find an upper
solution $v$ decreasing in time, $v\geq u$ and $v(r,t)\to 2\ln
[A(1-r^2)+r^2]=w(r;\lambda )$ as $t \to \infty\,,\; \alpha (t) \to
A =\frac{\lambda^*}{\lambda^* -\lambda}$ as $t \to \infty$ (see
(4.6)).

 In a similar way we construct a lower solution $z(r,t)$ increasing in
time. Again $z(r,t)=2\ln [\alpha (t)(1-r^2)+r^2]$ is a lower
solution provided that $\alpha(t)$ satisfies (\ref{Tpen}) and
$\alpha _0 -\frac{\lambda^*}{\lambda^* -\lambda}<0$,  $\alpha (t)$
is of the form of (\ref{Texi}). Also we require $z(r,0)\leq u_0
(r)$. It is sufficient to choose $\alpha_0
=\min\{\frac{\lambda^*}{\lambda^* -\lambda}\,,\inf_{r} \frac{\exp
(u_0 (r)/2)-r^2} {1-r^2}\}$. But on $\partial \Omega \;\;
z(r,t)=u(r,t)=0$,  which finally implies that $z\leq u$. Hence for
$0<\lambda <\lambda^* =8\pi ^2$ we find an upper solution $v$ and
a lower solution $z$ such that $z\leq u\leq v$ with $v(r,t)\to
w+\,,\quad z(r,t) \to w-$, as $t \to \infty$. Thus the solution
$u$ is global  and $u(r,t) \to w(r;\lambda )=2\ln
\left[\frac{\lambda^*}{\lambda^* -\lambda} (1-r^2)+r^2\right]$ as
$t\to \infty$,  where $w(r;\lambda)$ is the unique steady state.
The above procedure holds for any (admissible) initial data $u_0
(r)$, from which it follows that the solution $w$ is globally
asymptotically stable.


\subsection{Blow-up for {\boldmath $\lambda >\lambda^{\ast} $} }
To prove that the solution $u(r,t)$ blows up for $\lambda
>\lambda^* =8\pi ^2$, we construct a lower solution which blows up.
Again we take as a lower solution a function with a similar form
to the steady state $w(r):$  $z(r,t)=w(r;\mu (t))=2\ln [\alpha
(t)(1-r^2)+r^2]$. We first note that if $\alpha (t)$ satisfies
(\ref{Tpen}) and $\alpha_0 <\frac{\lambda^*}{\lambda^* -\lambda}$,
then $\dot{\alpha}(t)>0$,  moreover $z(r,t)$ is an unbounded lower
solution to (\ref{ept})  and $z(r,t) \to \infty \;\;\mbox{as}\; t
\to \infty$ for any $r\in [0,1)$. This implies that $u(r,t)$ is
unbounded, more precisely $\limsup_{t\to t^{\ast}}\|u(\cdot,t)\|
\to\infty\,,\quad t^{\ast}\leq\infty$. To prove that
$t^{\ast}<\infty$  we take a modified comparison function,
$Z(r,t)=p\ln [\alpha (t)(1-r^2)+r^2]$. We show that $Z(r,t)$ is a
lower solution to (\ref{ept}) and blows up for a certain value of
$p$. Thus we have
\begin{eqnarray}
\mathcal{E}(Z)&:=& Z_t-\Delta_r Z-\lambda e^{-Z}/4\pi^2(\int_0^1
e^{-Z} r\,dr)^2  \nonumber\\
&\leq& \frac{p}{(\alpha -\beta r^2)}
\Big\{\dot{\alpha}(1-r^2)(\alpha -\beta r^2) \label{Tept}\\
&&- 4(\alpha-\beta r^2)^{2-p}
\big[\frac{\lambda}{\lambda^*}\frac{2(p-1)^2}{p\,k^2}-1\big]\alpha^2
\Big\} \nonumber
\end{eqnarray}
where   $\beta (t)=\alpha (t)-1$, $\dot{\alpha}(t)>0$, $0<p<2$,
$k>1$ and $\alpha-1\geq \alpha /k$.  The last condition is
satisfied for $t\geq t_1$,  for some $t_1$  since the use of the
lower solution $z$ above guarantees unboundedness of $u$ and
allows $Z$ to be large for $t\geq t_1$.  For $\lambda >\lambda^* $
and $1<p<2$,  we have $\lambda/\lambda^* >1$, $2(p-1)^2/p <1$,
while $\frac{2(p-1)^2}{p}\to 1$ as $p\to 2- $, so we can choose
$p\in (1,2)$:
\begin{equation}
\frac{\lambda}{\lambda^*}\frac{2(p-1)^2}{p}>1 \,. \label{Tokt}
\end{equation}
Now for a fixed $\lambda >\lambda^*$ we can choose suitable $p$
and $k$ so that both (\ref{Tokt}) and the following hold:
\begin{equation}
\frac{\lambda}{\lambda^*}\frac{2(p-1)^2}{p}>k^2
>1\;\;\mbox{or}\quad \Lambda=\frac{\lambda}{\lambda^*}
\frac{2(p-1)^2}{pk^2}>1\;. \label{Tene} \end{equation} The
inequalities (\ref{Tept}), (\ref{Tene}) imply
\[
\mathcal{E}(Z)\leq \frac{p(1-r^2)}{(\alpha -\beta
r^2)}\left[\dot{\alpha}-4(\alpha -\beta r^2)^{1-p} (\Lambda
-1)a^2\right]\leq 2\left[\dot{a} -4(\Lambda
-1)\alpha^{3-p}\right]=0\,,
\]
by taking $\alpha (t)$ to satisfy,
\begin{equation}
\dot{\alpha}-4(\Lambda -1)\alpha ^{3-p} =0\;,\quad \alpha
(0)=\alpha_0>0 \,. \label{Tdek} \end{equation} We also require
$Z(r,0)\leq u_0 (r)$,   for which it is sufficient to take
\[
\alpha_0 \leq \inf_{r} \frac{\exp (u_0 (r)/p) -r^2}{1-r^2}\,,\quad
1<p<2\,,
\]
and $Z(1,t)=u(1,t)=0$  holds on $\partial \Omega\:$. Hence
$Z(r,t)$, is a lower solution to (\ref{ept}) i.e. $Z(r,t)\leq
u(r,t)$ and $Z(r,t)\:$ is increasing in time since
$\dot{\alpha}(t)>0$. Furthermore from (\ref{Tdek}) we obtain,
\begin{equation}
4(\Lambda -1)(t-t_1)=\int_{\alpha (t_1)}^{\alpha
(t)}s^{p-3}\,ds<\int_{\alpha_1}^{\infty}
s^{p-3}\,ds=\frac{\alpha_1 ^{p-2}}{2-p}<\infty\,, \label{Tent}
\end{equation}
where $\alpha_1 =\alpha (t_1)=k/(k-1)<\alpha (t)$, since we have
used that $\alpha -1> \alpha /k$. The relation (\ref{Tent})
implies that $\alpha (t)\:$ blows up at
\[
T^{\ast}=t_1+\frac{\alpha_1 ^{p-2}}{4(\Lambda -1)(2-p)}<\infty
\]
and, since $Z(r,t)$ is a lower solution for $u$,  this means that
$u$  blows up at $t^{\ast}\leq T^{\ast}<\infty$. This completes
the proof of the blow-up of $u$. In the next section, for general
decreasing functions, we shall show that this blow-up is global,
this means that $ u(r,t)\to \infty$ as $t\to t^{\ast}- $ for every
$r\in [0,1)$.


\section{General decreasing functions}

\subsection{ Stationary Solutions}
We consider an arbitrary decreasing function $f$ satisfying
(\ref{duo}). Again we may use comparison techniques due to the
monotonicity of $f$  as in Section 2. We follow the same procedure
as in the previous section; see also \cite{beb,car}. For the
moment we suppose that $\int_0^\infty f(s)\,ds<\infty \,,$ unless
otherwise stated. The corresponding steady problem of (\ref{ept})
is
\begin{subequations}\begin{gather}
w''(r)+\frac{1}{r}w'(r)+\mu
f(w(r))=0\,,\;\;\; 0<r<1\,,\label{penaa}\\
w(1)=w'(0)=0\,, \label{penab}
\end{gather}
\label{pena} \end{subequations} where $\lambda =4\pi ^2 \mu
(\int_0^1 f(w)r\,dr)^2$. Multiplying (\ref{penaa}) by $r$  and
integrating,
\begin{equation}
\lambda =\frac{4\pi ^2}{\mu}(w'(1))^2\;. \label{pduo}
\end{equation}
Again multiplying (\ref{penaa}) by $w'$ and integrating as before
we get
\begin{equation}
\frac{(w'(1))^2}{2}+\int_0^1\frac{(w' (r))^2}{r}\;dr -\mu \int_0^M
f(s)\,ds =0\,, \label{ptri} \end{equation} which implies
$\frac{(w'(r))^2}{\mu}<2\int_0^M f(s)\,ds$. By rescaling the
problem we may assume that
\begin{equation}
\int_0^\infty f(s)\,ds=1, \label{ptes} \end{equation} then from
(\ref{pduo}) and (\ref{ptri}) we get
\[
\lambda <8\pi ^2 \int_0^M f(s)\,ds <8 \pi ^2 \,\quad
\mbox{and}\quad \frac{(w'(r))^2}{\mu}<2\,.
\]

\begin{lemma}\label{PrENA}
For the Dirichlet problem (\ref{pena}), if (\ref{ptes}) holds,
then $\frac{(w'(1))^2}{\mu}\to 2$  as $\mu \to \infty$.
\end{lemma}

\paragraph{Proof:} We consider the auxiliary problem:
\begin{subequations}
\begin{gather}
z''(r)+\mu g(z(r))=0\,,\quad 1-\delta <r<1, \label{auxa}\\
z(r)=\sup_r z(r)=M\,,\quad z'(r)=0\,,\quad 0\leq r\leq 1-\delta
\,,\quad z(1)=0\,,
\end{gather} \label{aux} \end{subequations}
where $0<g(s)<f(s)$,  and $ z, z_r$, are continuous at $ 1-\delta
$. Multiplying (\ref{auxa}) by $z'$ and integrating we obtain
\begin{equation}
(z'(r))^2=2\mu \int_{z(r)}^M g(s)\,ds=2\mu [G(z)-G(M)]\,,
\label{pexi}
\end{equation}
where $G(z)=\int_z^\infty g(s)\,ds$.  Then
\begin{equation}
\int_0^M [G(z)-G(M)]^{-1/2}\,dz=\delta \sqrt{2\mu}\,, \label{pept}
\end{equation}
since $z'(r)<0$,  and $(z'(1))^2=2\mu [G(0)-G(M)]=2\mu \int_0^M
g(s)\,ds$. We  prove now that the solution to problem (\ref{aux})
is a lower solution to problem (\ref{pena}). Indeed
$z''(r)+\frac{1}{r}z'+\mu f(z)=\mu f(z)>0\;\;\mbox{in}\;\;0\leq r
\leq 1-\delta$.  Also taking into  account
(\ref{aux})-(\ref{pept}),
\begin{eqnarray}
\lefteqn{ z''(r)+\frac{1}{r}z'+\mu f(z)}\nonumber\\
&=&\frac{z'}{r}+\mu(f(z)-g(z))>\frac{z'(r)} {1-\delta}+\mu
(f(z)-g(z))
\label{pokt}\\
&=&-\frac{\sqrt{2\mu}\;[G(z)-G(M)]^{1/2}}{1-\delta}+\mu
(f(z)-g(z))\,,\;\;\mbox{in} \;\;1-\delta<r<1\,. \nonumber
\end{eqnarray}
Now choosing $\mu$ large enough such that
\begin{equation}
\mu \geq \mu _0=\sup_{z\in (0,M)}\frac{2[G(z)-G(M)]}{(1-\delta)^2
\,[f(z)-g(z)]^2} \,, \label{pene} \end{equation} and $\delta <1$,
relations (\ref{pokt}), (\ref{pene}) give
\[
z''(r)+\frac{1}{r}z'+\mu f(z)>\mu (f(z)-g(z))-\mu (f(z)-g(z))=0\,.
\]
In addition $z'(0)=z(1)=w'(0)=w(1)=0$,  hence $z$ is a lower
solution to $w$-problem. This  implies
\begin{equation}
z(r)\leq w(r)\;\;\mbox{and}\;\; w'(1)\leq z'(1)<0\,, \label{pdek}
\end{equation} (if the latter inequality were $w'(1)>z'(1)$ it would give
$z(r)>w(r)$ for some $r,\,$ which would be a contradiction).\\
Now taking:
\begin{enumerate}
\item[(a)] $g$, such that
$0<g(s)<f(s)$  and $1-\epsilon< G(0)=\int_0^\infty g(s)\,ds\leq 1$

\item[(b)] $M$ such that
$[G(0)-G(M)]>1-2\epsilon$, $\epsilon >0$,  from the definition of
$G$,

\item[(c)] $\mu$ to satisfy (\ref{pene}).
\end{enumerate}
Note that $G'(z)=-g(z) <0$,  $G(z)$ is decreasing and $G(0)\leq
1$); from (\ref{ptri}), (\ref{pexi}) and (\ref{pdek}) we obtain
\[
2>\frac{(w'(1))^2}{\mu }\geq
\frac{(z'(1))^2}{\mu}=2[G(0)-G(M)]>2(1-2\epsilon )\,.
\]
This relation holds for every $\epsilon >0$, as far as $\mu \gg
1$, hence this proves the lemma.\quad\hfill$\Box$\medskip


\begin{prop}\label{PrDUO}
If (\ref{ptes}) holds then $\,\lambda <\lambda^* =8\pi ^2$ and
$\lambda \to 8\pi ^2-$ as $M\to \infty$ $(\lambda \to \lambda^*
\int_0^\infty f(s)\,ds=\lambda^* $  as $M\to \infty).$
\end{prop}

\paragraph{Proof:}
  The first relation is obtained by (\ref{ptri}), (\ref{ptes}).  For
the second, using (\ref{pduo}) and Lemma \ref{PrENA} we obtain,
\[
\lambda =4\pi ^2 \frac{(w'(1))^2}{\mu} \to 8\pi ^2 \quad \mbox{as
}
 \mu \to \infty\, \;(\mbox{or equivalently as }
M\to\infty) .
\]
\quad\hfill $\Box$ \medskip

Also from Lemma  \ref{PrENA} and relation (\ref{ptri}) we deduce
that
\begin{gather*}
\lim_{\mu \to \infty}\frac{2}{\mu}\int_0^1
\frac{(w'(r))^2}{r}\,dr=0\,\;\;\mbox{and} \\
\Big[\int_0^1 \frac{(w'(r))^2}{r}\,dr\Big]/(w'(1))^2
=\Big[\frac{4\pi ^2} {\lambda}\int_0^Mf(s)\,ds-1/2\Big]\to 0\;\;
\mbox{as}\;\; \mu \to \infty\,.
\end{gather*}
Now we assume  that
\begin{equation}
\int_0^\infty f(s)\,ds=\infty \label{pent} \end{equation} holds
instead of (\ref{ptes}), so we have the following statement.

\begin{prop}\label{PrTRI}
Let  (\ref{pent}) hold and $w$ is the solution to problem
(\ref{pena}) then $(w'(1))^2/\mu \to \infty$ as $\mu \to\infty$
and $\lambda (M) \to \infty$ as $M\to \infty$.
\end{prop}

\paragraph{Proof:}
  Again the solution $z(r)$ to problem (\ref{aux}) is a lower
solution to $w$-problem, provided that (\ref{pene}) holds. Thus we
have $z(r)\leq w(r)$ which implies that $w'(1)\leq z'(1)<0$,  or
$(w'(1))^2\geq (z'(1))^2$, but now from (\ref{pexi}) at $r=1$  we
get,
\[
\frac{(w'(1))^2}{\mu}\geq \frac{(z'(1))^2}{\mu}=2\int_0^M
g(s)\,ds\to \infty \quad \mbox{as} \quad M\to \infty \,,
\]
provided that we take $g$ such that $0<g(s)=\gamma f(s)<f(s)$, for
$0<\gamma<1$, then $\int_0^{\infty}g(s)\,ds= \infty$.
\hfill$\Box$\medskip


We now obtain a uniqueness result for the steady problem.

\begin{prop}\label{PrTES}
Let $f$, satisfy
\begin{equation}
-sf'(s)<f(s)\,,\quad s>0\,, \label{pdod} \end{equation} then
problem (\ref{pena}) has a unique solution.
\end{prop}

\paragraph{Proof:}
 From (\ref{pduo}) we get
 $\lambda (\mu)=4\pi ^2 \frac{(w'(1))^2}{\mu}=4\pi ^2 (W'
(1))^2$ by writing $w(r)=\nu W(r)$, $\nu =\sqrt \mu$. Then
(\ref{pena}) gives $W'_{\nu}(0)=W_{\nu}(1)=W'(0)=W(1)=0$ and
\[
W''+\frac{W'}{r}+\nu f(w)=0\quad \mbox{or}\quad
W''_{\nu}+\frac{W'_{\nu}}{r}+\nu ^2
f'(w)W_{\nu}=-f(w)-f'(w)w\,.\quad
\]
If (\ref{pdod}) holds, then using maximum principle and  Hopf's
boundary lemma, we get that $W_{\nu} (r)>0$  and $W'_{\nu}(1)<0$
or $\frac{d}{d\nu} (W'(1))^2 >0 $,  since also $W'(1)<0\,.$ Then
$\lambda '(\mu)=\frac{2\pi ^2}{\nu}\frac{d}{d\nu}(W'(1))^2
>0\;\; $ which implies uniqueness.\quad\hfill$\Box$ \medskip


\paragraph{Remark:} Proposition \ref{PrTES}  is also true for a
general domain $\Omega$.  The relation (\ref{pdod}) implies
(\ref{pent}).


Finally we obtain the response diagram of Figure \ref{sxh4}.

\begin{figure}
\begin{center}
\includegraphics[width=0.5\textwidth]{fig4.eps}
\end{center}
\caption{Response diagram for the Dirichlet problem,
$\int_0^\infty f(s)\,ds=\infty$.}\label{sxh4}
\end{figure}


\subsection {Stability where a unique steady state exists}
We follow the same procedure as in the previous section, we seek
for a decreasing (increasing)-in-time upper (lower) solution to
problem (\ref{ept}). We first look for an upper solution of a form
similar to the steady state: $v(r,t)=w(r;\overline
{\mu}(t))=\overline w$, where $\overline w$ is a steady state.
Then
\begin{equation}
\mathcal{E}(v)=\overline w_{\overline \mu}\:\dot{\overline
\mu}+[4\pi ^2 \overline \mu \:I^2 (\overline w)-\lambda ]
f(\overline w) /4\pi ^2 I^2 (\overline w)\;,\smallskip
\label{pdtr}
\end{equation}
where $I(w)=\int_0^1 f(w)r\,dr$ and $-\Delta _r w=\mu f(w)$
($\Delta_r =\partial ^2/\partial r^2 +\frac{1}{r}\partial
/\partial r$). Also $\overline w_{\overline \mu}=\frac{\partial
\overline w}{\partial \overline \mu }>0$, $w(r;\overline \mu (t))$
is increasing with respect to $\overline \mu$, by using the
maximum principle. Taking now any $\lambda>0 $, so that
$w(x;\lambda)$  is the unique stationary solution, we can choose
$\overline \mu (0)$ such that $w(r;\overline \mu (0))=v(r,0)\geq
u_0 (r)\,;$ this can be done since $u_0 (r),\; u'_0 (r)\:$ are
bounded. Furthermore, since a unique steady state exists (see
Proposition \ref{PrTES} ) for these  values of $\lambda$, there
exists a $\mu$ such that $\lambda =4\pi ^2 \mu I^2 (w)$,
$M=w(0;\mu )=w(0)$, where $w(r)= w(r;\mu )$ is the unique steady
state of problem (\ref{pena}), and as long as $\overline \mu >\mu$
then $\overline w>w$ and $\overline \lambda=\overline \lambda (t)
=4\pi ^2 \overline \mu (t) I^2 (\overline w)>\lambda$. As before
$w(r;\overline \mu)$ is an upper solution, decreasing in time,
which tends to the stationary solution $w$ provided that
\begin{equation}
0<-\dot{\overline \mu }=h(\overline \mu)\equiv\left(\overline
\lambda (t) -\lambda \right)\; I^{-2} (\overline w)\;\inf_{r}
\{\frac{f(\overline w)}{\overline w_{\overline \mu}}\},
\label{pdte}
\end{equation}
and $\overline \mu (0)>\mu $  (note that $f(s)$ is bounded away
from zero and $w_{\mu}$  is also finite). Hence $ u=u(r,t)\leq
v(r,t)=w(r;\overline \mu (t))$  and $\dot{\overline \mu}<0$ which
implies that $v_t =\overline w_{\overline \mu}\:\dot{\overline
\mu} <0$. In a similar way we can construct a lower solution
$z(r,t)=w(r;\underline \mu (t))$ which is increasing in time  and
tends to the steady state $w$. Finally we obtain, $
z(r,t)=w(r;\underline \mu (t))\leq u(r,t)\leq v(r,t)=
w(r;\overline \mu (t))$, and $\overline \mu (t)\to \mu
+\,,\;\underline \mu (t) \to \mu-\,,\; v(r,t)\to w+\,, \; z(r,t)
\to w-$, as $t\to\infty$. This implies that $u(\cdot ,t) \to
w(\cdot )$  uniformly as $t\to \infty$, and that $w$ is globally
asymptotically stable.\smallskip\\ For the case of $\lambda \geq
\lambda^*$ and $\int_0^\infty f(s)\,ds=1$, there is no steady
solution to (\ref{pena}), then $\underline \lambda =\underline
\lambda(t) =4\pi ^2 \underline \mu(t) I^2 (\underline w) <\lambda$
for every $\underline \mu
>0$. Taking the above lower solution $z(r,t)$ we obtain that
$\underline \mu (t)\to \infty $  $(\dot{\underline \mu}>0)$  and $
u(r,t)\geq z(r,t)=w(r;\underline \mu(t)) \to \infty$  as $t\to
\infty$  (note that $w_{\mu} >0$  and $\mu (t) \to \infty$  as
$t\to \infty$  then $z \to \infty$ as $t\to \infty)$. In
particular $\sup_{r}u(r,t)\to \infty$ as $t\to t^*\leq\infty$,
hence $u$  is unbounded.

\subsection{Blow-up for {\boldmath $\lambda >\lambda^* $}}

We can now prove that the solution $u$ to  problem (\ref{ept})
blows up in finite time if $\lambda >\lambda^* =8\pi ^2. $  To
prove this we use similar methods to those in the previous
sections, (or see \cite{kth,lac9, lac95}). We look for a lower
solution $z(r,t)$ to the $u$-problem which itself blows up. We try
to find a lower solution with a similar form to the steady state.
We take into account the form of blow-up in the one-dimensional
case, therefore we consider $z(r,t)$ to satisfy:
\begin{subequations}
\begin{gather}
z(r,t)=M(t)=\sup_{r}z(r,t)\,,\quad z_r (r,t)=0\,,\quad 0\leq r\leq
1-\delta (t)\, ,\;\;t>0\,, \label{depena}\\
z_{rr}+\mu (t) g(z)=0\;,\quad1-\delta (t)\leq r\leq 1\,,\;\;z(1,t)
=0\,,\;\; t>0\,, \label{depenb}
\end{gather} \label{depen}\end{subequations}
where $0<g(s)=\gamma f(s)<f(s)$, $0<\gamma <1$,  and $z,z_r$  are
continuous at $1-\delta (t)$. Multiplying (\ref{depenb}) by $z_r$,
and integrating in $(1-\delta ,r)$ we obtain
\begin{equation}
\frac{z^2_{r}}{2}+\mu (t)\int_M^z g(s)\,ds =0\,, \label{pdpe}
\end{equation} which gives
\begin{equation}
z^2_{r}=2\mu (t)\,[G(z)-G(M)]\quad \mbox{and}\quad
\frac{z^2_{r}}{\mu}<2\,, \label{pdex}
\end{equation}
on writing $G(z)=\int_z^\infty g(s)\,ds$ (then $G'(z)=-g(z)$ and
$G(0)=\gamma)$). The relation (\ref{pdex}) implies
\begin{equation}
\delta \sqrt{2\mu} =\int_0^M \left[G(s)-G(M)\right]^{-1/2}\,ds\,.
\label{pdep} \end{equation} Again integrating (\ref{pdex}) we
obtain
\begin{equation}
(1-r)\sqrt{2\mu}=\int_0^z \left[G(s)-G(M)\right]^{-1/2}\,ds\,.
\label{pdok} \end{equation} On the other hand, we can get
\begin{eqnarray*}
\int_0^1 g(z)r\,dr&=&\int_0^{1-\delta} g(z)r\,dr+\int_{1-\delta}^1
g(z)r\,dr \leq g(M)\frac{(1-\delta)^2}{2}-\frac{1}{\mu}z_r (1,t)\\
&=& g(M)\frac{(1-\delta
)^2}{2}+\sqrt{\frac{2}{\mu}}\,\Big(\int_0^M g(s)\,ds
\Big)^{1/2}\\
&\sim& g(M)/2+\sqrt{2\gamma /\mu}\quad \mbox{for}\quad\delta \ll
1\ll M\;\; \mbox{and}\;\;1\ll \mu\,,
\end{eqnarray*}
 by using (\ref{pdex}) at $r=1$, $\int_0^M g(s)\,ds\sim
\gamma$, $1- \delta \sim 1$, for $\delta\ll 1\ll M$. Finally we
get \begin{subequations}
\begin{equation}
\int_0^1 g(z)r\,dr\;\,\lesssim \,\;\frac{g(M)}{2}(\alpha
+1)\,,\;\;\quad\delta \ll 1\ll M\,,
\end{equation}
on taking
\begin{equation}
\sqrt{2\gamma /\mu }=\alpha g(M)/2, \end{equation} \label{pden}
\end{subequations}
where $\alpha $ is a suitable chosen constant; in particular
choose $\alpha >1/[(\lambda /8\pi ^2)^{1/2}-1]$ for
$\lambda>\lambda^* =8\pi ^2 $. Such an $\alpha$  gives $\Lambda
=\frac{1}{3}[\lambda /\pi ^2 (1+\alpha)^2 -\frac{8}{\alpha
^2}]>0$.

\begin{lemma}\label{PPEN}
$ M\,f(M)\to 0 $  as $M\to \infty$.
\end{lemma}
For the proof of this lemma, see \cite{lac95}.

\begin{lemma}\label{PEXI}
$\delta \to 0$  as $ \mu \to \infty$.
\end{lemma}
\paragraph{Proof:}
 From the previous lemma we have that $f(M)\to 0$  and $g(M)\to 0$
as $M\to \infty$  and for fixed $\alpha$, (\ref{pden}) implies
that $\mu \to \infty$ as $M\to \infty\,.$ For $0<s<M$, we have,
$\,(M-s)g(M)\leq G(s)-G(M)\leq (M-s)g(s)$  and (\ref{pdep}),
(\ref{pden}) give:
\begin{eqnarray*}
\delta &\leq &\frac{\alpha g(M)}{4\,\sqrt{\gamma}}\int_0^M
[(M-s)g(M)]^{-1/2}\,ds= \frac{\alpha
g^{1/2}(M)}{4\,\sqrt{\gamma}}\int_0^M (M-s)^{-1/2}\,ds\\
&=&\frac{\alpha}{2\,\sqrt{\gamma}}[g(M)M]^{1/2},\;\;\;for\;\;
\delta \ll1\ll M\;.
\end{eqnarray*}
The above relation implies, $0<\delta \to 0$  since $g(M)M<f(M)M
\to 0$, as $M\to\infty$. \quad\hfill$\Box$


\begin{prop}\label{PrEPT}
 The solution $z$, to  (\ref{depen}) is a lower solution to the
$u$-problem; moreover the solution $u$ blows up in finite time.
\end{prop}
\paragraph{Proof:}
  For $0<r<1-\delta (t)$,
\begin{eqnarray}
\mathcal{E}(z)&=&\dot{M}-\lambda f(M)/4\pi ^2 \Big(\int_0^1
f(z)r\,dr\Big)^2 =\dot{M}-\frac{\lambda g(M)/ \gamma}{4\pi^2
(\int_0^1 g(z)r/\gamma\,dr)^2}\nonumber \\
&\lesssim&\,\dot{M}-\lambda \gamma /\pi ^2 (\alpha +1)^2
g(M)\leq\dot{M}-\gamma \Lambda /g(M)=0\,, \label{peien}
\end{eqnarray}
on choosing  $\dot{M}-\gamma \Lambda /g(M)=0$,  where $\Lambda $
is taken as $3\Lambda =(\lambda /\pi ^2 (\alpha +1)^2)-8/\alpha ^2
<(\lambda /\pi^2 (\alpha +1)^2)$. Hence $\mathcal{E}(z)\lesssim 0$
for $ M \gg 1$.

 For the interval $1-\delta <r<1$,  we first differentiate
(\ref{pdok}) with respect to $t$ and get
\begin{eqnarray*}
z_t &=& (1-r)\frac{\dot{\mu}}{\sqrt{2\mu}}[G(z)-G(M)]^{1/2}\\
&&+\frac{1}{2}g(M) \dot{M}[G(z)-G(M)]^{1/2} \int_0^z
[G(s)-G(M)]^{-3/2}\,ds\\ &=&A+B.
\end{eqnarray*}
For $A$ we have:
\begin{eqnarray*}
A&=&(1-r)\frac{\dot{\mu}}{\sqrt{2\mu}}[G(z)-G(M)]^{1/2}\\
&\leq& -\frac{g'(M)\dot{M}} {g(M)}[g(z)(M-z)]^{1/2}\int_0^z
[G(s)-G(M)]^{1/2}\,ds\\ &\leq&\frac{-g'(M)\dot{M}
g^{1/2}(z)M}{2g^{3/2}(M)}\leq\frac{\gamma \Lambda g(z)}{g^2
(M)}\quad\mbox{for}\;\; M\gg 1\,,
\end{eqnarray*}
provided that
$$\dot{M}\leq -\frac{\gamma \Lambda
g^{1/2}(z)}{Mg'(M)g^{1/2}(M)}$$
 which certainly holds if $0\leq
\dot{M}\leq -\gamma\Lambda /Mg'(M)$  since $g'(s)\leq 0$  so
$g(z)/g(M)\geq 1$  for $z\leq M$.

For B we have:
\begin{eqnarray*}
B&=&\frac{1}{2}g(M)\dot{M}[G(z)-G(M)]^{1/2} \int_0^z
[G(s)-G(M)]^{-3/2}\,ds \leq\frac{\dot{M}g^{1/2}(z)}{g^{1/2}(M)}\\
&\leq& \frac{\gamma\Lambda g(z)}{g^2 (M)}\quad \mbox{for} \;\;
M\gg 1\,,
\end{eqnarray*}
provided that  $\dot{M}\leq \frac{\gamma \Lambda
g^{1/2}(z)}{g^{3/2}(M)}$, which holds if
$0\leq\dot{M}\leq\frac{\gamma\Lambda }{g (M)}$,  since
$g(z)/g(M)\geq 1$  for $z\leq M$.  \\ Also, on using (\ref{pdex}),
we have the estimate,
\[
- \frac{z_r}{r}\leq
\frac{4\,\sqrt{\gamma}}{\alpha}(g(M)M)^{1/2}\frac{g(z)}{g^2 (M)}
\;\lesssim \;\gamma\Lambda \frac{g(z)}{g^2
(M)}\;\,\mbox{since}\;g(M)M\to 0\;\,\mbox{for}\; M\gg
1\,.\nonumber\\
\]
Thus for $1-\delta <r<1$  if $\,0\leq \dot{M}=\min\{\gamma\Lambda
/g(M)\,,-\gamma\Lambda /Mg'(M)\}$ and using the previous estimate
we obtain,
\begin{eqnarray*}
\mathcal{E}(z)&=& A+B-\frac{z_r}{r}+\mu g(z)-\lambda f(z)/4\pi ^2
(\int_0^1 f(z)r\,dr) ^2 \\ &=& \frac{2\gamma \Lambda g(z)}{g^2
(M)}-\frac{z_r}{r}+\mu g(z)-\frac{\lambda g(z)/\gamma}{4\pi ^2
(\int_0^1 g(z)r/\gamma\,dr)^2}\\ &\lesssim &
2\gamma\Lambda\frac{g(z)}{g^2 (M)}+\gamma\Lambda\frac{g(z)}{g^2
(M)}+\mu g(z)- \frac{\lambda\gamma g(z)}{\pi ^2 (a+1)^2 g^2
(M)}\\&=& \left[3\gamma \Lambda +8\gamma /\alpha ^2 -\gamma
\lambda /\pi ^2 (a +1)^2 \right]\frac{g(z)}{g^2 (M)} =(3\gamma
\Lambda -3\gamma \Lambda)=0\,,
\end{eqnarray*}
for $M\gg 1$. Also $z(1,t)=u(1,t)=z_r (0,t)=u_r (0,t)=0$  on the
boundary and taking $z(r,0)\geq u_0 (r),\,$ the function $z(r,t)$
is a lower solution to the $u$-problem, hence $u(r,t)\geq
z(r,t),\,$ for $M$, large enough (after some time at which $u$, is
sufficiently large).

 Now we  show that $u$  blows up. Indeed
\[
\dot{M}=\min\{\Lambda\gamma /g(M),-\Lambda\gamma /Mg'(M)\}
\]
which implies
\[
\Lambda\gamma \frac{dt}{dM}=\max\{g(M)\,,-Mg'(M)\}\leq
g(M)-Mg'(M)\quad \mbox{or}
\]
\begin{eqnarray*}
\Lambda\gamma t&\leq& \int^M (g(s)-sg'(s))\,ds\\ &=&-Mg(M) +\int^M
g(s)\,ds<\int^M f(s)\,ds<\infty\,,
\end{eqnarray*}
since $Mg(M)\to 0$ as $M\to \infty$, and (\ref{ptes}) holds.
Hence, $z$, blows up at $T^{\ast}<\infty $,  and $u$, must also
blows  up at some $t^{\ast}\leq T^{\ast}<\infty$. This completes
the proof of the proposition.\quad\hfill$\Box$\medskip


As in the one-dimensional case \cite{lac9,lac95}, the blow-up is
global, i.e. $u(r,t)\to \infty$ as $t\to t^{\ast}-$  for all $r\in
[0,1)$.  Since $f(u)$  is bounded, then $u$
 blows up only by having $\int_0^1 f(u)r\,dr\to 0$  as $t\to
t^{\ast}-$.  Indeed,
\[
\dot{M}\,\leq\, \frac{\lambda f(0)}{4\pi ^2 (\int_0^1
f(u)r\,dr)^2}=h(t)\,,
\]
giving
\[
M(t)-M(0)\leq \int_0^t h(s)\,ds\to \infty\quad \mbox{as }t\to t^
{\ast}\,.
\]
This implies $\int_0^1 f(u)rdr\to 0$ as $t\to t^{\ast}$, but then
$u$  blows up globally and $u_r (1,t)\to \infty$  as $t\to
t^{\ast}-$.

\subsection{The Robin Problem }

We consider again $u$  to satisfy (\ref{epta}), (\ref{eptc}) but
now we take boundary conditions of Robin type,
\begin{equation}
u_r (1,t)+\beta u(1,t)=0\,,\quad t>0\,,\;\;\beta >0\,.
\label{sunor}
\end{equation}
The corresponding steady problem is
\begin{subequations}\begin{gather}
w''+\frac{1}{r}w'+\mu f(w)=0\,,\quad 0<r<1\,,\label{exisa}\\
w'(1)+\beta w(1)=0\;,\quad \beta >0\;,\quad w'(0)=0\,.
\end{gather}
\label{exis} \end{subequations} Multiplying again by $w'$ and
integrating we obtain
\begin{equation}
\frac{(w'(1))^2}{2}+\int_0^1 \frac{(w'(r))^2}{r}\,dr-\mu \int_m^M
f(s)\,ds=0\,, \label{sunth} \end{equation} where $0<m=w(1)<M=\max
w= w(0)\,,\;\;(m=\min w$,  by using the maximum principle). We
also consider the auxiliary problem,
\begin{equation}
\begin{gathered}
z(r)=\sup_r z(r)=N\,,\quad z'(r)=0\quad 0\leq r\leq
1-\delta\,,\\
z''(r)+\mu g(z(r))=0\;,\quad 1-\delta <r<1\,, \\
z'(1-\delta)=0\quad z'(1)+\beta z(1)=0\,,
\end{gathered}\label{auxil}
\end{equation}
where $0<g(s)<f(s)$,  and $ z, z_r$, are continuous at $ 1-\delta
$.

The following result is  similar to the one of Lemma \ref{PrENA}.

\begin{lemma}\label{PrOKT}
Let (\ref{ptes})  hold, then the solution to (\ref {auxil}) is a
lower solution to the Robin problem (\ref{exis}); moreover
$\frac{(w'(1))^2}{\mu}\to 0$  as $\mu \to \infty$.
\end{lemma}

\paragraph{Proof:} As in  the proof of Lemma \ref{PrENA} we
again have $(z'(r))^2=2\mu [G(z)-G(N)]$, and  $z'(r)<0$, where
$G(z)=\int_z^\infty g(s)\,ds$.  This implies
\[
\delta
=\frac{1}{2\mu}\int_{z(1)}^{N}[G(s)-G(N)]^{-1/2}\,ds<\frac{1}{2\mu}
\int_{0}^{N}[G(s)-G(N)]^{-1/2}\,ds\,.
\]
Taking $\mu \geq \mu _0 $, where
\[
\mu _0=\sup_{z\in (0,N)}\frac{2[G(z)-G(N)]}{(1-\delta )^2
[f(z)-g(z)]^2}\,,
\]
$z$ is a lower solution to problem (\ref{exis}), $z(r)<w(r)$  and
$N\leq M$. Note that $N \to \infty$, and $M\to \infty$, as $\mu\to
\infty$. Moreover $z''(r)<0$  in $(1-\delta ,1)$,  then
$0<-z'(r)<-z'(1)$, $z(r)<z(1)[1+\beta (1-r)]$, and for
$r=1-\delta$, $N<z(1)(1+\beta \delta )<w(1) (1+\beta \delta
)=m(1+\beta \delta)$, which implies that $m \to \infty$ as $N\to
\infty$. Since $\int_0^\infty f(s)\,ds<\infty$, $\int_{m(\mu
)}^{M(\mu )} f(s)\,ds\to 0$  as $\mu\to \infty$.
 From (\ref{sunth}) we get $\frac{(w'(1))^2} {\mu}\to
0$  as $\mu \to \infty$  and $\frac{1}{\mu}\int_0^1
\frac{(w'(r))^2}{r}\, dr\to 0$  as $\mu \to \infty$.
\quad\hfill$\Box$\medskip

Now multiplying  by $r$, (\ref{exisa}) gives, $(\int_0^1
f(w)r\,dr)^2= \frac{(w'(1))^2}{\mu ^2}$  and $\lambda =4\pi ^2
\frac{(w'(1))^2}{\mu}$. From Lemma \ref{PrOKT} we obtain that
$\lambda (M) \to 0$ as $M\to \infty$  and from (\ref{sunth}) that
$\frac{(w'(1))^2}{\mu}<2\int_0^M f(s)\,ds\to 2$  as $M\to \infty$.
Hence $\lambda (M)<8\pi ^2$  which means that there exists a
$0<\lambda^* \leq 8\pi ^2$  such that for $0<\lambda < \lambda^* $
problem (\ref{exis}) has at least two solutions whereas it has no
solutions for $\lambda >\lambda^*$. Thus we have the diagrams of
Figure \ref{sxh5}.

\begin{figure}
\begin{center}
\includegraphics[width=0.9\textwidth]{fig5.eps}
\end{center}
\caption{Possible non-local response diagrams for the Robin
problem.}\label{sxh5}
\end{figure}


\subsection{Stability} We consider,  as in the Dirichlet
problem,  an upper solution $v(r,t)=w(r;\overline \mu (t))$ which
is decreasing in time and a lower solution $z(r,t)= w(r;\underline
\mu (t))$ which is increasing in time,   to the $u$-problem. More
precisely we have $\mathcal{E}(v)\geq 0$,  and $\mathcal{E}(z)\leq
0$  provided  that
\begin{gather*}
0<-\dot{\overline \mu}=h(\overline \mu)\equiv [4\pi ^2 \overline
\mu \,I^2 (\overline w)-\lambda]\inf_{(0,1)} \{\frac{f(\overline
w)}{\overline w_\mu }\}/ I^2 (\overline w)\,,
\\
0<\dot{\underline \mu}=h(\underline \mu)\equiv[\lambda -4\pi ^2
\underline \mu \,I^2 (\underline w)]\inf_{(0,1)}
\{\frac{f(\underline w)}{\underline w_\mu }\}/ I^2 (\underline
w)\,,
\end{gather*}
respectively, with $\overline \mu (0)$, and $\underline\mu (0)$
chosen so that $w(r;\overline \mu (0))>u_0(r)$, $w(r;\underline
\mu (0))<u_0 (r)$,  and $\lambda =\lambda (\mu)= 4\pi ^2\mu
(\int_0^1 f(w)r\,dr)^2 =4\pi ^2 \mu I^2 (w)$, where $w(r;\mu)$  is
the steady solution. To each $\mu$ corresponds a unique $M$ but to
each $\lambda \in (0,\lambda^* )$ corresponds more than one $M$
and hence many solutions $w(r;\lambda )$,  see Figure \ref{sxh5}.
Following the same procedure as in the one-dimensional case we
know that the quantity $\Phi (\hat{\mu},\lambda )=4\pi ^2
\hat{\mu} I^2 (w)-\lambda =\hat{\lambda}(\hat {\mu})-\lambda $, is
either greater than, or equal to, or less than zero, as this is
the key term for the construction of upper and lower solutions.

 Thus if we consider
the left response diagram of Figure \ref{sxh5}, then $\Phi
(\overline \mu,\lambda )=\overline \lambda -\lambda
>0$, if $\mu_1<\overline \mu <\mu_2$,
$\Phi (\underline \mu ,\lambda)=\underline\lambda -\lambda <0$ if
$\underline\mu<\mu_1$ or $\underline \mu>\mu_2$ and $\Phi(\mu
,\lambda )=0$ if $\mu=\mu_1$ or $\mu =\mu_2 $.

 For $\lambda =\lambda^*$, $\Phi (\underline \mu ,\lambda )<0$,
 hence $z_1(r,t)\to w^ {\ast}$  as $t\to \infty$, provided
that $w_1(r;\underline \mu (0))<w^{\ast}$  and $z_2 (r,t)\to
\infty$  as $t\to \infty$ provided now that $w_2 (r;\underline \mu
(0))>w^{\ast}$. This means that $w^{\ast}$  is unstable. More
precisely it is unstable from above and stable from below. For
$\lambda>\lambda^* $  again $\Phi(\underline \mu ,\lambda )<0$,
$z(r,t) \to \infty$  as $t\to t^{\ast}\leq \infty$, hence $u$ is
unbounded for any initial data; this also holds even for
$\lambda<\lambda^* $  provided that the initial data are greater
than the largest steady state.
 The above procedure can
also be applied  to the rest of the response diagrams of Figure
\ref{sxh5}.

\subsection{Blow-up of unbounded solutions} We consider now the
unbounded solutions appearing for $\lambda>\lambda^* $  or for
$\lambda\leq \lambda^*$  but with initial conditions larger than
the greatest steady state. Following the same method as in the
one-dimensional case \cite{lac95}, if $u$  fails to blow-up, then
for any given $k$,  there must be a $t_k>0$  such that $u\geq k$,
for $t_k\geq t$ (this is due to the use of the lower solutions,
note that $m=\min w(r)=w(1)\to \infty$  as $M\to \infty )$. Then
we consider the problem,
\begin{gather*}
v_t=\Delta _r v+\lambda f_k (v)/4\pi ^2 (\int_0^1 f_k
(v)r\,dr)^2\,,\;\;0<r<1\,,\;\; t>0 \\
v(1,t)=v_r (0,t)=0\,,\;\;t\geq t_k \\
v(r,t_k)=0\,,\;\;0<r<1\,,
\end{gather*}
where $f_k (s)=f(s+k)$.   Thus it can easily seen that $v+k$ is a
lower solution to the $u$-problem, hence $u\geq v+k$  for $t\geq
t_k$.  But from the Dirichlet problem (see Proposition
\ref{PrDUO}, we have that if $\lambda>\lambda^* \int_0^\infty f_k
(s)\,ds= \lambda^* \int_0^\infty f(s+k)\,ds=\lambda^*
\int_k^\infty f(S)\,dS$,  then $v$  blows up at finite time. Hence
choosing $k$  sufficiently large so that the previous inequality
holds, we get that $u$  blows up. This blow-up is global.

 Finally, carrying over the analysis similar to the
Dirichlet problem and the one-dimensional case \cite{lac9,lac95},
we obtain Figure \ref{sxh6}. We use the notation: $(\to \cdot
\leftarrow )$,  for stable stationary solutions, $(\leftarrow
\cdot \to )$  for unstable, and the double arrows  $(\to \to )$
for solutions  $u$ which blow up. If we lie in the region where
$\Phi (\overline \mu (t),t)>0$  then the arrows point downwards
while where $\Phi (\underline \mu (t),t)<0$  the arrows point
upwards.

\begin{figure}
\begin{center}
\includegraphics[width=\textwidth]{fig6.eps}
\end{center}
\caption{Stability and blow-up of solutions for the Robin
problem.}\label{sxh6}
\end{figure}

Any solution which corresponds to a point of the curves of this
type $ (\to \cdot \leftarrow)$  is stable while all others are
unstable. More precisely this $(\to \cdot \to)$,  is stable from
one side and unstable from the other whereas this $(\leftarrow
\cdot \to)$,  is unstable from both sides.


 For the Neumann problem  (the boundary conditions
are  $u_r (0,t)=u_r (1,t)=0)$ there is no positive steady state
for any $\lambda>0$. Concerning the solution $u(r,t)$, this
behaves as in the one-dimensional case  \cite{lac9,lac95};  if
$\int_0^\infty f(s)\,ds<\infty$  then $u$,  blows up globally at
$t^{\ast}=\frac{2\pi ^2}{\lambda}\int_0^\infty f(s)\,ds<\infty$
for $u(0,t)=0$, whereas if $\int_0^\infty f(s)\,ds =\infty$, then
blow-up does not occur but the solution tend to $\infty$,
uniformly as $t\to \infty$,


\section{Discussion}

 In the present work we have studied the non-local, two-dimensional, radially
 symmetric problem of the form:
\[
u_t=\Delta _r u+\lambda f(u)/4\pi ^2 \Big(\int_0^1
f(u)r\,dr\Big)^2
\]
where $f(u)>0$, $f'(u)<0$, $u$, represents the temperature which
is produced in a conductor having fixed electric current $I$,
i.e. $\lambda =I^2 /\pi ^2$.   The function $f$ represents
electrical conductivity $(f(u)=\sigma (u))$,  in contrast to the
one-dimensional model where it represents electrical resistivity
$(f(u)=\rho (u)=1/\sigma (u))$.  This work extends the results of
the one-dimensional problem, and the methods used are similar to
the one-dimensional case and are based on comparison techniques.

  We find similar behaviour in both problems, and it is rather like that of
the standard reaction-diffusion model, $u_t=\Delta u+\lambda
f(u)$, $f(u)>0$, $f'(u)>0$, $f''(u)>0$, see \cite{lac} and the
references therein. More precisely,  for the Dirichlet problem, if
$\int_0^\infty f(s)\,ds <\infty$  we find a critical value
$\lambda^*$  such that for $\lambda >\lambda^* $ there is no
steady state and $u$,  blows up globally. Also in case  we have a
unique steady solution, this solution is asymptotically stable.

For the Robin problem, provided that $\int_0^\infty
f(s)\,ds<\infty$,  again there exists a critical value $\lambda^*
\leq \lambda^* _D$ ($\lambda^* _D$  refers to  the Dirichlet
problem). If $0<\lambda <\lambda^*$  then there exists at least
one stationary solution and no solution for $\lambda>\lambda^*$.
Concerning the stability, the minimal solution is stable, the next
greater one is unstable, the next  stable and so on. On the other
hand, if $\lambda>\lambda^* $  then $u$ blows up; $u$  also blows
up for $\lambda \in (0,\lambda^* )$ and for sufficiently large
initial data . The solution(s) at $\lambda=\lambda^* $  is(are)
unstable.\\ For the Neumann problem there is no steady state and
the solution $u$  blows up in finite time if $\int_0^\infty
f(s)\,ds<\infty$, whereas $u\to \infty$  uniformly as $t\to
\infty$  if $\int_0^\infty f(s)\,ds=\infty$.

It is an interesting question whether or not a similar behaviour
occurs for asymmetric problems for dimensions greater than or
equal to two.


\paragraph{Acknowledgments}
The author would like to thank A. A. Lacey for several fruitful
discussions. Part of this work was done during the author's visit
to the Departments of Mathematics of Heriot-Watt University,
Edinburgh, and Universidad Complutense, Madrid.


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\noindent\textsc{Dimitrios E. Tzanetis}\\
Department of Mathematics,
Faculty of Applied Sciences,\\
National Technical University of Athens,\\
Zografou Campus,  157 80 Athens, Greece \\
e-mail: dtzan@math.ntua.gr

\end{document}