\documentclass[twoside]{article} \usepackage{amssymb, amsmath} % font used for R in Real numbers \usepackage{graphicx} % For including postscript figures \pagestyle{myheadings} \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 $-s\,f'(s)0\,, \label{enaa} \\ \mathcal{B} (u):=\frac {\partial u}{\partial n}+\beta (x)u=0\,,\quad t>0\,,\;x\in\partial\Omega\,, \label{enab} \\ u(x,0)=u_{0} (x)\,,\quad x\in \Omega\,, \nonumber \end{gather} \label{ena} \end{subequations} where$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, $$f(s)>0\,,\quad f'(s)<0\,,\quad s\geq 0\,. \label{duo}$$ 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 $$u_t =\Delta u + \lambda \sigma (u)/(\int_ \Omega \sigma (u)\,dx)^2, \label{tes}$$ 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: $$u_t =\Delta u+ \lambda \sigma (u)\,,\;\;x\in\Omega,\;\;\mbox{where}\;\;\lambda = V^2/L^2\geq 0. \label{pen}$$ 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 $$u_ t =\Delta u + \lambda \sigma (u)/\Big[a+b\int_ \Omega \sigma (u)\,dx\Big]^2 ,\,\; x \in\Omega\,,\;\,t>0,\;\, a,b>0. \label{eks}$$ 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$), takes the form: \begin{subequations} \begin{gather} u_t=u_{rr}+\frac {u_r}{r}+\frac {\lambda f(u)}{4\pi ^2 (\int_{0}^{1} f(u)\,r\,dr) ^2}\,,\quad\; 00\,, \label{epta}\\ u(1,t)=u_r (0,t)=0\,,\quad t>0\,,\label{eptb} \\ u(r,0)=u_{0} (r)\,,\quad 00$ 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 $u_r (r,t)<0$, $00$, $f(s)\geq c>0$ and Lipschitz for $s\in (a,b)$, where $a<\min \{0,\inf{u_0}\}\leq u\leq \max\{0,\sup u_{0}\}\Delta \overline u+\frac{\lambda f(\overline u)}{(\int_\Omega f(\overline u)\,dx)^2} \,, \quad \mbox{in} \;\Omega\,, \quad t>0\,, \\ \mathcal{B}(\overline u)>\mathcal{B}(u)=0 \quad \mbox{on} \quad \partial \Omega\,,\;\; t>0\,, \\ \overline u_{0} (x)>u_{0}(x)\,,\;\;\mbox{in}\;\; \Omega\,, \end{gather} \label{dduo} \end{subequations} while if$\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$such that \begin{gathered} v_{t}>\Delta v+\frac{\lambda f'(s)}{(\int_\Omega f(\underline u)\,dx)^2}\,v\,,\;\;x\in \Omega\,,\;\;00 \;\;\mbox{at}\;\;\; t=0 \;\;\mbox{in}\;\,\Omega \;\;\mbox{and}\quad \mathcal{B} (v)>0\,,\;\;\mbox{on}\;\;\,\partial \Omega\,,\;\;00$ 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 00\,, \label{tenaa}\\ u(1,t)=u_r (0,t)=0\,, \quad t>0\,, \quad u(r,0)=u_0 (r)\,, \quad 00, \label{tduo} or there exists an or some $s=s(t),\;\, 00\,, \\ u=1\,, \quad u_r=0, \quad 0\leq r\leq s\,,\;\; t>0\,, \end{gather} \label{ttri} \end{subequations} where$\pi (1-s^2 (t))=m(t)$. Note that$u$is continuous and$u_r \leq 0$, the latter follows by using the maximum principle. The corresponding steady state to (\ref{tduo}) is \Delta_r w+\lambda /\pi ^2 =0\,, \quad 0\lambda \;\,$ 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}0\,,\\ b(0)=0\,,\quad \mbox{where}\;\; 0\leq b\leq \min\{1,\lambda /4\pi ^2 \}\,. \end{gathered} \label{toct} 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 0t_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)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 w''(r)+\frac {1}{r}w'(r)+\mu e ^{-w(r)} =0\,,\quad 01$,$M=\sup\|w\|=w(0)=2 \ln \alpha$. The parameter$\lambda$is given by $$\lambda =4\pi ^2 \mu \Big(\int_0^1 e^{-w} r\,dr\Big)^2 =8\big(1-\frac {1}{\alpha} \big)\pi ^2=8\pi ^2 \left(1-e^{-M/2}\right)<8\pi ^2\,, \label{Ttes}$$ 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 $$\dot{\alpha}+2 \frac{\lambda^{\ast} -\lambda}{\lambda^{\ast}}\alpha -2=0\,,\;\; \alpha (0)=\alpha_0\; \label{Tpen}$$ and$\alpha_0$is sufficiently large. Hence for$\lambda \in (0,\lambda^{\ast})$the solution of (\ref{Tpen}) is $$\alpha (t)=\frac{\lambda^{\ast}}{\lambda^* -\lambda}+\big[\alpha_0 -\frac{\lambda^{\ast}}{\lambda^* -\lambda}\big] \exp\big(-2\frac{\lambda^{\ast} -\lambda}{\lambda^{\ast}}\,t\big), \label{Texi}$$ 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$,$01$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$11$,$2(p-1)^2/p <1$, while$\frac{2(p-1)^2}{p}\to 1$as$p\to 2- $, so we can choose$p\in (1,2)$: $$\frac{\lambda}{\lambda^*}\frac{2(p-1)^2}{p}>1 \,. \label{Tokt}$$ Now for a fixed$\lambda >\lambda^*$we can choose suitable$p$and$k$so that both (\ref{Tokt}) and the following hold: $$\frac{\lambda}{\lambda^*}\frac{2(p-1)^2}{p}>k^2 >1\;\;\mbox{or}\quad \Lambda=\frac{\lambda}{\lambda^*} \frac{2(p-1)^2}{pk^2}>1\;. \label{Tene}$$ 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, $$\dot{\alpha}-4(\Lambda -1)\alpha ^{3-p} =0\;,\quad \alpha (0)=\alpha_0>0 \,. \label{Tdek}$$ 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 10. Furthermore from (\ref{Tdek}) we obtain, $$4(\Lambda -1)(t-t_1)=\int_{\alpha (t_1)}^{\alpha (t)}s^{p-3}\,ds<\int_{\alpha_1}^{\infty} s^{p-3}\,ds=\frac{\alpha_1 ^{p-2}}{2-p}<\infty\,, \label{Tent}$$ 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 \[ T^{\ast}=t_1+\frac{\alpha_1 ^{p-2}}{4(\Lambda -1)(2-p)}<\infty$ and, since$Z(r,t)$is a lower solution for$u$, this means that$u$blows up at$t^{\ast}\leq T^{\ast}<\infty$. This completes the proof of the blow-up of$u$. In the next section, for general decreasing functions, we shall show that this blow-up is global, this means that$ u(r,t)\to \infty$as$t\to t^{\ast}- $for every$r\in [0,1)$. \section{General decreasing functions} \subsection{ Stationary Solutions} We consider an arbitrary decreasing function$f$satisfying (\ref{duo}). Again we may use comparison techniques due to the monotonicity of$f$as in Section 2. We follow the same procedure as in the previous section; see also \cite{beb,car}. For the moment we suppose that$\int_0^\infty f(s)\,ds<\infty \,,$unless otherwise stated. The corresponding steady problem of (\ref{ept}) is \begin{subequations}\begin{gather} w''(r)+\frac{1}{r}w'(r)+\mu f(w(r))=0\,,\;\;\; 00\;\;\mbox{in}\;\;0\leq r \leq 1-\delta$. Also taking into account (\ref{aux})-(\ref{pept}), \begin{eqnarray} \lefteqn{ z''(r)+\frac{1}{r}z'+\mu f(z)}\nonumber\\ &=&\frac{z'}{r}+\mu(f(z)-g(z))>\frac{z'(r)} {1-\delta}+\mu (f(z)-g(z)) \label{pokt}\\ &=&-\frac{\sqrt{2\mu}\;[G(z)-G(M)]^{1/2}}{1-\delta}+\mu (f(z)-g(z))\,,\;\;\mbox{in} \;\;1-\delta\mu (f(z)-g(z))-\mu (f(z)-g(z))=0\,. \] In addition $z'(0)=z(1)=w'(0)=w(1)=0$, hence $z$ is a lower solution to $w$-problem. This implies $$z(r)\leq w(r)\;\;\mbox{and}\;\; w'(1)\leq z'(1)<0\,, \label{pdek}$$ (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 $01-2\epsilon$, $\epsilon >0$, 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 $$\int_0^\infty f(s)\,ds=\infty \label{pent}$$ 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 $00\,, \label{pdod} then problem (\ref{pena}) has a unique solution. \end{prop} \paragraph{Proof:} From (\ref{pduo}) we get$\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 $$\mathcal{E}(v)=\overline w_{\overline \mu}\:\dot{\overline \mu}+[4\pi ^2 \overline \mu \:I^2 (\overline w)-\lambda ] f(\overline w) /4\pi ^2 I^2 (\overline w)\;,\smallskip \label{pdtr}$$ 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 $$0<-\dot{\overline \mu }=h(\overline \mu)\equiv\left(\overline \lambda (t) -\lambda \right)\; I^{-2} (\overline w)\;\inf_{r} \{\frac{f(\overline w)}{\overline w_{\overline \mu}}\}, \label{pdte}$$ 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$01/[(\lambda /8\pi ^2)^{1/2}-1]$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$00\,,\;\;\beta >0\,. \label{sunor} The corresponding steady problem is \begin{subequations}\begin{gather} w''+\frac{1}{r}w'+\mu f(w)=0\,,\quad 00\;,\quad w'(0)=0\,. \end{gather} \label{exis} \end{subequations} Multiplying again by $w'$ and integrating we obtain $$\frac{(w'(1))^2}{2}+\int_0^1 \frac{(w'(r))^2}{r}\,dr-\mu \int_m^M f(s)\,ds=0\,, \label{sunth}$$ where $0\lambda^*$. 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 $w(r;\overline \mu (0))>u_0(r)$, $w(r;\underline \mu (0))0$, 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 $w_1(r;\underline \mu (0))w^{\ast}$. 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\,,\;\;00 \\ v(1,t)=v_r (0,t)=0\,,\;\;t\geq t_k \\ v(r,t_k)=0\,,\;\;0\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. 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