\documentclass[twoside]{article} \usepackage{amssymb} % font used for R in Real numbers \pagestyle{myheadings} \markboth{\hfil Bifurcations for semilinear elliptic equations\hfil EJDE--1999/43} {EJDE--1999/43\hfil J. Kar\'atson \& P. L. Simon \hfil} \begin{document} \title{\vspace{-1in}\parbox{\linewidth}{\footnotesize\noindent {\sc Electronic Journal of Differential Equations}, Vol. {\bf 1999}(1999), No.~43, pp. 1--16. \newline ISSN: 1072-6691. URL: http://ejde.math.swt.edu or http://ejde.math.unt.edu \newline ftp ejde.math.swt.edu \quad ftp ejde.math.unt.edu (login: ftp)} \vspace{\bigskipamount} \\ % Bifurcations for semilinear elliptic equations with convex nonlinearity \thanks{ {\em 1991 Mathematics Subject Classifications:} 35J60. \hfil\break\indent {\em Key words and phrases:} semilinear elliptic equations, time-map, bifurcation diagram. \hfil\break\indent \copyright 1999 Southwest Texas State University and University of North Texas. \hfil\break\indent Submitted June 22, 1999. Published October 18, 1999.} } \date{} % \author{ J. Kar\'atson \& P. L. Simon} \maketitle \begin{abstract} We investigate the exact number of positive solutions of the semilinear Dirichlet boundary value problem $\Delta u+f(u) = 0$ on a ball in ${\mathbb R}^n$ where $f$ is a strictly convex $C^2$ function on $[0,\infty)$. For the one-dimensional case we classify all strictly convex $C^2$ functions according to the shape of the bifurcation diagram. The exact number of positive solutions may be 2, 1, or 0, depending on the radius. This full classification is due to our main lemma, which implies that the time-map cannot have a minimum. For the case $n>1$ we prove that for sublinear functions there exists a unique solution for all $R$. For other convex functions estimates are given for the number of positive solutions depending on $R$. The proof of our results relies on the characterization of the shape of the time-map. \end{abstract} \newtheorem{theorem}{Theorem} \newtheorem{prop}{Proposition} \newtheorem{cor}{Corollary} \newtheorem{lem}{Lemma} \newtheorem{defi}{Definition} \newtheorem{rem}{Remark} \section{Introduction} We investigate the exact number of positive solutions of the semilinear boundary value problem \begin{eqnarray} & \Delta u+f(u) = 0 & \label{BVP} \\ & u|_{\partial B_R}=0 & \nonumber \end{eqnarray} where $B_R \subset {\mathbb R}^n$ is the ball centered at the origin with radius $R$ and $f$ is a strictly convex $C^2$ function on $[0,\infty)$ (i.e. $f''\ge 0$ and $f''$ does not vanish identically on any interval). According to the well-known result of Gidas, Ni and Nirenberg \cite{GNN} every positive solution of (\ref{BVP}) is radially symmetric, hence it satisfies \begin{eqnarray} & ru''(r) + (n-1)u'(r) + rf(u(r)) = 0 & \label{RBVP} \\ & u'(0) =0, \ u(R) =0 ; & \nonumber \end{eqnarray} further, there holds \begin{equation} u'(r) < 0 \qquad (00$ \cite{SW84}. The time-map, associating the first zero of $u(\cdot ,p)$ to $p$, determines the number of positive solutions of (\ref{RBVP}). There are several results concerning the number of positive solutions of (\ref{RBVP}) for special convex functions $f$, but the problem is far from being solved for a general convex function. Joseph and Lundgren \cite{JL} determined the number of solutions in the case $f(u) = \mbox{e}^u$ and $f(u) = (1+ \alpha u)^{\beta}$ for $\alpha ,\ \beta >0$. They used Emden's transformation, because for these special functions (\ref{RBVP}) can be transformed to a two-dimensional autonomous system and phase plane techniques can be applied. It can be shown by Pohozaev's formula \cite{Poh} that for $f(u) = u^p$ there exists a positive solution (on a star-shaped domain) if and only if $p<(n+2)/(n-2)$, and by Emden's transformation or Sturm's theorem that it is unique (on the ball). Several authors studied the case $f(u) = u^p +\lambda u^q$, especially when $p$ is near to the critical value $(n+2)/(n-2)$ \cite{AP, BN, Budd, CK, MP}. Mc Leod's result \cite{McL} on the uniqueness of positive solutions is valid for a certain class of convex functions, typically for $f(u) = u^p -u$ ( if $11$. \subsection*{The main results and the method} For the one-dimensional case we classify all strictly convex $C^2$ functions according to the shape of the bifurcation diagram and thus determine the exact number of positive solutions. We prove that for asymptotically linear or superlinear functions (at infinity) the bifurcation diagram is determined by the behaviour of $f$ at infinity and at $0$, further, by the sign changes of $f$. We essentially prove that there exist numbers $0\le R_{\infty} < R_{sup}\le \infty$, depending on $\lim\limits_{u\to \infty} {f(u) \over u}$ and the number of zeros of $f$, respectively, which divide the interval $(0,\infty)$ according to the number of positive solutions of (\ref{BVP}). As it is sketched in Table 1, these numbers are influenced by the sign of $f(0)$. (The exact formulation of these results is given in Theorems 1 and 2.) \bigskip \begin{center} \begin{tabular}{|c|c|c|c|} \hline & $R\leq R_{\infty}$ & $R_{\infty}< R < R_{sup}$ & $R_{sup}< R$ \\ \hline $f(0)>0$ & 1 & 2 & 0 \\ \hline $f(0) \leq 0$ & 0 & 1 & 0 \\ \hline \end{tabular} \\[6pt] Table 1. Number of positive solutions \end{center} The above table shows the number of positive solutions of (\ref{BVP}) in the superlinear and asymptotically linear case for $n=1$. In the superlinear case $R_{\infty}=0$. If $f$ has two zeros or a double zero, then $R_{sup} = \infty$. \bigskip For sublinear functions there exists a unique solution for all $R$; this result is considered in the general case $n>1$. This full classification in one dimension is due to our main lemma (Lemma 1), which implies that the time-map cannot have a minimum. Otherwise, where it is possible, we prove our results for any dimension $n\geq 1$. Especially, we avoid the application of the integral formula for the time-map, except for determining the limit of the time map at infinity, where the result is indeed not true for all $n$ (see e.g. \cite{JL}). For the case $n>1$ we prove that for sublinear functions there exists a unique solution for all $R$ (Theorem 3). For other convex functions estimates are given for the number of positive solutions (Theorem 4). The proof of our results relies on the characterization of the shape of the time-map. The following three characteristic properties determine the exact number of positive solutions of (\ref{RBVP}): \begin{itemize} \item the domain of the time-map \item the limit of the time-map at the boundary points of its domain \item the monotonicity of the time-map on the maximal subintervals of its domain. \end{itemize} The main tools during our studies are Sturm's theorems, Pohozaev's formula and the application of auxiliary functions, e.g. the Hamiltonian of the 1-dimensional case as a Liapunov function for the case $n>1$. \begin{defi} \rm The {\it time-map } associated to the above initial value problem is the following function $T$: $$ T(p) = \min \{ r>0\ : u(r,p) =0 \} \ ; \qquad D(T) = \{ p> 0 \ : \exists r>0 \ u(r,p)=0 \}. $$ \end{defi} The function $T$ is determined by the implicit equation \begin{equation} u(T(p),p) \equiv 0 \label{T0} \end{equation} and the assumption $u(r,p)>0$ if $r\in [0,T(p))$. Differentiating (\ref{T0}) one gets the following equations for the derivatives of $T$: \begin{equation} \partial_r u(T(p),p) T'(p) + \partial_p u(T(p),p) \equiv 0, \label{T1} \end{equation} \begin{equation} \partial_r^2 u(T(p),p) T'(p)^2 + 2 \partial_{rp} u(T(p),p) T'(p) + \partial_r u(T(p),p) T''(p) + \partial_p^2 u(T(p),p) \equiv 0. \label{T2} \end{equation} Differentiating (\ref{IVP}) with respect to $p$ and introducing the notations $h(r,p)=\partial_p u(r,p)$ and $z(r,p)=\partial_p^2 u(r,p)$ we get \begin{eqnarray} & rh''(r,p) + (n-1)h'(r,p) + rf'(u(r,p))h(r,p) = 0 & \label{T3} \\ & h(0,p) =1, \ h'(0,p) =0; & \nonumber \end{eqnarray} \begin{eqnarray} & rz''(r,p) + (n-1)z'(r,p) + rf'(u(r,p))z(r,p) + rf''(u(r,p)) h^2(r,p) = 0 & \label{T4} \\ & z(0,p) =0, \ z'(0,p) =0. & \nonumber \end{eqnarray} Differentiating (\ref{IVP}) with respect to $r$ and introducing the notation $v(r,p)=\partial_r u(r,p)$ we get \begin{eqnarray} & v''(r,p) + \frac{n-1}{r} v'(r,p) + (f'(u(r,p)) -\frac{n-1}{r^2} )v(r,p) = 0 & \label{T5} \\ & v(0,p) =0, \ v'(0,p) = -\frac{f(p)}{n}. & \nonumber \end{eqnarray} Using the above notations, (\ref{T1}) is written as \begin{equation} v(T(p),p) T'(p) + h(T(p),p) \equiv 0\, . \label{T6} \end{equation} \section{General results on the shape of the time-map} In this section we formulate results on the domain $D(T)$ of the time-map , the limit of the time-map at the boundary points of $D(T)$ and the monotonicity of the time-map. These will enable us to cover the possible cases of the shape of the time-map for strictly convex $f$. The results are proved for all dimensions $n$ where it is possible. Most of the one-dimensional results exploit the first main result (Lemma 1). This is given in the first subsection. \subsection*{Main lemma in one dimension} \begin{lem} Let $n=1$, $f\in C^2$ be strictly convex. If $T'(p)=0$, then $T''(p)<0$. \end{lem} {\sc Proof.} Let $ T'(p)=0$. Then (\ref{T6}) implies $h(T(p),p)=0$. This means that $h>0$ in the interval $[0,T(p))$. Otherwise, using Sturm comparison, $v$ should have a root between the previous root of $h$ and $T(p)$ since $v$ and $h$ are two solutions of the same equation (\ref{T3}). This cannot occur by (\ref{u'<0}). From this we obtain $z(T(p),p)<0$ for the solution of (\ref{T4}). Namely, assume first that $f'' (p)>0$. Then $z'' (0)=-f'' (p)<0$ implies $z<0$ in some right neighbourhoodof $0$. Assume that $z(r_1)=0$ for some $r_1\in (0,T(p)]$. Then comparison of the equations $$ h'' + f'(u)h=0 \quad \hbox{and} \quad z'' + \left( f'(u)+ \frac{f'' (u) h^2}{z}\right) z =0 $$ shows that $h$ should have a root in $(0,r_1)$. I.e. $z<0$ in $(0,T(p)]$. If $f'' (p)=0$ then we can use the above argument for all other values $\tilde p$ where $f'' (\tilde p)>0$ holds to show that $z(r,\tilde p)<0$ until the first root of $h(r,\tilde p)$. Then by continuity this extends to $p$. Finally, $z(T(p),p)<0$ yields $T'' (p)<0$, using (\ref{T2}) (and $v(T(p),p)\le 0$). \hfill$\diamondsuit$ \begin{cor} If $n=1$ and $f\in C^2$ is strictly convex then $T$ may have at most one local extremum in any subinterval of $D(T)$, namely, a local maximum. \end{cor} \subsection*{The domain $D(T)$ of the time-map } In this subsection we do not assume first that $f$ is convex. Necessary conditions for $p\in D(T)$ can be obtained using the Hamiltonian of the 1-dimensional case \begin{equation} H(r):=\, { u'(r)^2 \over 2} + F(u(r)) \label{H} \end{equation} (where $F(u):=\int_0^u f$) as a Liapunov function, because $H'(r) = -\frac{n-1}{r}u'^2(r)$ is negative (see (\ref{u'<0})). Namely: \begin{prop} \begin{enumerate} \item If $p\in D(T)$, then $f(p)>0$. \item $D(T)\subset \{ p>0 \ :\ F(p) > F(s) \ \forall \ s\in I_p \mbox{ and } f(p)\neq 0 \}=:{\cal P}_f$ where $I_p:=(0,p)$ if both $n=1$ and $f(0)<0$ and $I_p:=[0,p)$ otherwise. \end{enumerate} \end{prop} {\sc Proof.} 1. If $f(p) = 0$, then $u(r) \equiv p$. In case $f(p)<0$ we have $u''(0) >0$, hence $u$ initially increases. This implies $u(r) \geq u(0)$ for all $r>0$, because $u(r_0) = u(0)$ would imply $H(r_0) \geq H(0)$. This is impossible for $n>1$, and in the case $n=1$ there holds $u''(r_0) > 0$. (We note that the end of the proof also follows from (\ref{u'<0}).) 2. Let $p\in D(T)$. According to 1. we have $f(p) > 0$. Let $s\in (0,p)$, then there exists $r\in (0, T(p))$ with $u(r) = s$. Then $$ F(s) = F(u(r)) = H(r) - \frac{u'(r)^2}{2} < H(r) \leq H(0) = F(p). $$ It is easy to see that if $n>1$ or $f(0) \geq 0$, then the above inequality also holds for $s=0$ (i.e. $r=T(p)$). \hfill$\diamondsuit$\medskip The sufficient condition may be a difficult question (cf. e.g. Pohozaev's identity \cite{Poh}). The characterization of $D(T)$ can only be given for $n=1$. Besides this, we mention special cases for $n>1$. \begin{prop} If $n=1$ then $D(T)={\cal P}_f$. \label{domT1} \end{prop} {\sc Proof.} Let $p\in {\cal P}_f$, $u(r)=u(r,p)$. We argue by contradiction, so assume that $u(r)>0$ for all $r>0$. Using that $H$ is constant and $p\in {\cal P}_f$, (\ref{H}) shows that $u'(r)<0$ ($r>0$). Hence there exists $\lim_{\infty}u = :c\in [0,p]$, and (\ref{H}) implies $F(c)=F(p)$. In case $f(0)\geq 0$ this contradicts $F(s)0$, which contradicts $\lim_{\infty}u = 0$. \hfill$\diamondsuit$\medskip If $f$ is strictly convex, then ${\cal P}_f$ clearly consists of at most two intervals (around 0 and $\infty$). First we consider the unbounded component. (The bounded one can be studied for any $n$.) \begin{cor} Let $n=1$ and $\alpha>0$. Let $f$ be strictly convex, $f(\alpha)=0$, $f>0$ on $(\alpha, +\infty)$. Then there exists $\beta \ge \alpha$ such that $(\beta, +\infty)$ or $[\beta, +\infty)$ is a connected component of $D(T)$ (in the cases $f(0)\ge 0$ and $f(0)<0$, respectively). Further, $\beta$ is the solution of $F(\beta) = F(\alpha_1)$ in $[\alpha, \infty)$, where $\alpha_1$ is the first root of $f$ in $[0,\alpha]$ if $f(0)\geq 0$ and $\alpha_1 =0$ otherwise. Therefore, if $f'(\alpha)=0$ ($f'(\alpha)>0$), then $\beta=\alpha$ ($\beta>\alpha$), resp. \label{cordomT1} \end{cor} \begin{prop} Let $n\le 2$. If $f(u)>0 \ (u\in (0,+\infty))$, then $D(T)= (0,+\infty)$. \label{domT2} \end{prop} {\sc Proof.} It is a consequence of the integral form of (\ref{IVP}) \begin{equation} -r^{n-1} u'(r) = \int_0^r\rho^{n-1}f(u(\rho))d\rho \label{I} \end{equation} since this implies $u'(r)\le - {K \over r^{n-1} }$ ($r>r_1$) for some $r_1>0$ and $K>0$, hence $u$ attains $0$ if $n\le 2$. \hfill$\diamondsuit$ \begin{rem} \rm We obtain similarly that for $n\le 2$ and for nonnegative $f$ we have $D(T)={\cal P}_f$. \end{rem} \begin{rem} \rm Propositions \ref{domT1} and \ref{domT2} cannot be extended for all $n\geq 1$. An important restriction is imposed by Pohozaev's formula \cite{Poh} on the growth of $f$ depending on $n$. \end{rem} \begin{prop} Let $n\geq 1$, $\alpha\in (0, +\infty]$. If $f>0$ in $(0,\alpha)$ and $\liminf\limits_{u\to 0} {f(u) \over u} >0$, then $(0,\alpha)\subset D(T)$. Consequently, if $f(\alpha)=0$ then $(0,\alpha)$ is a maximal subinterval of $D(T)$. \label{domTn} \end{prop} {\sc Proof.} \begin{enumerate} \item[(i)] In case $f(0)>0$ the statement is an easy consequence of (\ref{I}). Namely, let $p\in (0,\alpha )$ and let $M:=\min f|_{[0,p]}$. Then (\ref{I}) yields $u(r,p)\leq p-{M \over 2n} r^2$, hence $u(\cdot ,p)$ has a root, i.e. $p\in D(T)$. \item[(ii)] In case $f(0)=0$ we use a linear lower estimate for $f$, namely, we choose $\delta >0$ and $m>0$ such that $f(u)\ge mu $ for $u\in [0,\delta )$. Let $p\in (0,\alpha )$. If $p>\delta $, then let $M:=\min f|_{[\delta ,p]}>0$. Then applying (\ref{I}) again we get that there exists $r_1>0$ such that $u(r_1)=\delta$. In case $p\leq \delta $ let $r_1:=0$. Comparison of $$ u'' + {n-1 \over r}u' + {f(u) \over u}u =0 \quad \hbox{and} \quad w'' + {n-1 \over r}w' + mw =0 $$ with $w(r_1)=\delta$, $w'(r_1)=0$ shows that $u$ attains 0 before the first root of $w$ which exists since the equation for $w$ is linear and $m>0$. \end{enumerate} \hfill$\diamondsuit$ \begin{cor} Let $n\geq 1$. Let $f(u)\ge mu \ (u\in [0,+\infty))$ for some $m>0$. Then $D(T)= (0,+\infty)$, and $T$ is bounded. \label{cordomTn} \end{cor} \subsection{The limit of the time-map at the boundary points of $D(T)$} \begin{prop} Let $n\geq 1$. Let $0\in \partial D(T)$. \begin{enumerate} \item[(a)] If $f(0)>0$, then $\lim\limits_{0} T = 0$. \item[(b)] If $f(0)=0$ and $f'(0)>0$, then $\lim\limits_{0} T \in (0,+\infty)$. \item[(c)] If $f(0)=0$ and $f'(0)=0$, then $\lim\limits_{0} T = +\infty $. \end{enumerate} \label{limT0} \end{prop} {\sc Proof.} \begin{enumerate} \item[(a)] The proof of Proposition \ref{domTn} (i) is used: with suitable $M>0$ for small enough $p$ we have $f(u(r))\ge M$ in $[0,T(p)]$, hence $u$ attains 0 before $w(r)= p-{M \over 2n} r^2$ does. The root of the latter tends to $0$ as $p\to 0$. \item[(b)] Similarly, for any $\varepsilon >0$ comparison to $u_{\pm}'' + {n-1 \over r}u_{\pm}' + (f'(0)\pm \varepsilon)u_{\pm} =0 $ shows that for small enough $p$ the first root of $u$ lies between those of $u_{\pm}$, hence $\lim\limits_{0} T$ coincides with the root of the linearized equation at $0$. \item[(c)] Similarly, for any $\delta>0$ comparison to $w_{\delta}'' + {n-1 \over r}w_{\delta}' + \delta w_{\delta} =0 $ shows that for small enough $p$ the first root of $u$ lies after that of $w_{\delta}$ which tends to $+\infty$ as $\delta\to 0$. \end{enumerate} \hfill$\diamondsuit$ \begin{prop} Let $n\geq 1$. Let $c>0$ belong to $\partial D(T)\setminus D(T)$. Then $\lim\limits_{c} T= +\infty$. \label{limTc} \end{prop} {\sc Proof.} Let $R>0$. Let $\varepsilon >0$ such that $u(r,c)\ge \varepsilon$ ($r\in [0,R]$). The continuous dependence of $u(r,p)$ on $p$ is uniform on compact intervals, hence for small enough $\delta >0$ we have $u(r,p)>0$ ($r\in [0,R]$) for all $|p-c|< \delta$. \hfill$\diamondsuit$ \begin{prop} Let $n=1$. Let $+\infty\in \partial D(T)$. \begin{enumerate} \item[(a)] If $\lim\limits_{u\to +\infty} {f(u) \over u} = +\infty$ then $\lim\limits_{+\infty} T = 0$. \item[(b)] If $\lim\limits_{u\to +\infty} {f(u) \over u}=L \in (0, +\infty)$ then $\lim\limits_{+\infty} T = {\pi \over 2\sqrt{L}}$. \item[(c)] If $\lim\limits_{u\to +\infty} {f(u) \over u} = 0$ then $\lim\limits_{+\infty} T = +\infty$. \end{enumerate} \label{limTinfty1} \end{prop} {\sc Proof.} \begin{enumerate} \item[(a)] Let $k(u)=\inf \{ {f(s) \over s} : s\ge u \}$ \ $(u>0)$ and $K(u)= \int_0^u k(s)s\, ds$. Then $k$ increases to $+\infty$ and $f(u)\ge k(u)u$. For large enough $p$ we have \begin{eqnarray} T(p)&=& {1 \over \sqrt{2} } \int_0^p \frac{1}{ \sqrt{F(p)-F(s)} }\, ds \nonumber\\ &\le& {1 \over \sqrt{2} } \int_0^p \frac{1}{ \sqrt{K(p)-K(s)} }\, ds \label{tbecs}\\ &=& {p \over \sqrt{2K(p)} } \int_0^1 \frac{1}{ \sqrt{1-{K(pt) \over K(p)} } } dt\, .\nonumber \end{eqnarray} Here $K(pt)= t^2 \int_0^p k(vt)v\, dv \le t^2 K(p)$, hence the integral is bounded by $\int_0^1 \frac{1}{ \sqrt{1-t^2 } } dt\, ={\pi \over 2}$. Thus \begin{equation} T(p) \le {p \over \sqrt{2K(p)} } \cdot {\pi \over 2} . \label{kpiper2} \end{equation} Further, $K(p)\ge \int_{p/2}^p k(s)s\, ds \ge k({p/2}){3p^2 \over 8}$, hence we have $$ T(p)\, \le \, const.\cdot \frac{1}{ \sqrt{k(p/2) }} \to 0 \quad (p\to +\infty)\, . $$ \item[(b)] Let $\varepsilon, \delta >0$ be fixed. Then for sufficiently large $p$ $$ K(p)\ge \int_{\delta p }^p k(s)s\, ds \ge k(\delta p){p^2 \over 2}(1-\delta^2). $$ Here $\lim\limits_{p\to +\infty} k(\delta p) = L$, hence for sufficiently large $p$ we have \begin{equation} {p \over \sqrt{2K(p)} } \le \frac{1}{\sqrt{L(1-\delta^2)}} + \varepsilon . \label{pkp} \end{equation} In order to obtain a lower estimate for $T(p)$, we also introduce $g(u)=\sup \{ {f(s) \over s} : s\ge u \}$ \ $(u>0)$ and $G(u)= \int_0^u g(s)s\, ds$. Then $g$ decreases to $L$. Exchanging $K$ to $G$ in (\ref{tbecs}), the estimate is reversed, and similarly we obtain \begin{equation} T(p) \ge {p \over \sqrt{2G(p)} } \cdot {\pi \over 2} . \label{gpiper2} \end{equation} Let ${p_\delta}>0$ such that there holds $g(p)< L+\delta$ \, $(p>{p_\delta})$. Then $$ G(p) = \int_0^{p_\delta} g(s)s\, ds + \int_{p_\delta}^p g(s)s\, ds \le g(0){{p_\delta}^2 \over 2} + (L+\delta){p^2 \over 2} . $$ Hence for sufficiently large $p$ we have \begin{equation} {p \over \sqrt{2G(p)} } \ge \frac{1}{\sqrt{L+\delta}} - \varepsilon . \label{pgp} \end{equation} Summarizing (\ref{kpiper2})--(\ref{pgp}), we obtain for sufficiently large $p$ $$ \left( \frac{1}{\sqrt{L+\delta}} - \varepsilon \right) {\pi \over 2} \le T(p) \le \left( \frac{1}{\sqrt{L(1-\delta^2)}} + \varepsilon \right) {\pi \over 2}. $$ \item[(c)] Similar to (b), now using $g$ and $G$ only to get the suitable lower estimate for $T(p)$. \hfill$\diamondsuit$\medskip \end{enumerate} \begin{rem} \rm \begin{enumerate} \item[(a)] Proposition \ref{limTinfty1} is proved for $f>0$ in \cite{Lae70}. \item[(b)] The whole proposition cannot be extended to all $n\geq 1$ as the examples $f(u) = \mbox{e}^u$ or $(1+\alpha u)^\beta$ show (see \cite{JL}). It is possible for last part. \end{enumerate} \end{rem} \begin{prop} Part {\it (c)} of Proposition \ref{limTinfty1} holds for all $n\geq 1$. \label{limTinftyn} \end{prop} {\sc Proof.} The integral formula (\ref{I}) implies $u(r,p)\ge p-{M_p \over 2n} r^2$ where $M_p = \max f_{\mid [0,p]}$. Then $T(p)\ge \sqrt{ 2np \over M_p } \to +\infty$ since $\lim_{p\to +\infty} {M_p \over p} = 0$. \hfill$\diamondsuit$ \subsection{The monotonicity of the time-map } \begin{prop} Let $n\geq 1$. Let $\alpha\in (0, +\infty]$, $f\in C^2$ be convex and positive on $(0,\alpha )$. Then for any $p\in (0,\alpha )$ $T'(p)=0$ implies $f'(p)\ge 0$. \label{monTn} \end{prop} {\sc Proof. } (\ref{T3}), (\ref{u'<0}) and the convexity of $f$ imply that the function $$ H(r) = f'(u(r))\frac{h^2(r)}{2} + \frac{h'^2(r)}{2} $$ is strictly decreasing (here $h=\partial_pu$). According to (\ref{T6}) $T'(p)=0$ implies $h(T(p))=0$. Hence $$ 0 \leq \frac{h'^2(T(p))}{2} = H(T(p)) \le H(0) = \frac{f'(p)}{2} . $$ \hfill$\diamondsuit$ \begin{cor} Let $n\geq 1$. Let $\alpha\in (0, +\infty]$, $f\in C^2$ be convex and positive on $(0,\alpha )$. If $\lim_{\alpha} f=0$ then $T$ is strictly increasing on $(0,\alpha)$. \label{cormonTn} \end{cor} \begin{prop} Let $n=1$ and $f\in C^2$ be strictly convex. If $f(0)\le 0$ then $T$ is strictly decreasing. \label{monTf(0)neg} \end{prop} {\sc Proof.} Let $l(u)=uf'(u)-f(u)$ ($u>0$). The strict convexity of $f$ implies that $l$ is strictly increasing, thus, owing to $l(0)=-f(0)\ge 0$, we have $l(u)>0$ ($u>0$), i.e. ${f(u) \over u}1$. Under a certain set of assumptions it follows from \cite{McL} in the case when $f$ has a positive root and $f(0)=0$. \end{rem} \begin{prop} Let $n=1$ and $f\in C^2$ be strictly convex. If one of the endpoints of a maximal subinterval of $D(T)$ does not belong to $D(T)$, then $T$ is strictly monotonic on this subinterval. \label{monTsubint} \end{prop} {\sc Proof.} It is the consequence of Lemma 1 and Proposition \ref{limTc} \hfill$\diamondsuit$ \section{Classification of the number of positive solutions in one dimension} The results of Section 2 enable us to give a complete classification of strictly convex $C^2$ functions according to the number of positive solutions. Thus we obtain the extension of the results in \cite{Lae70} concerning the case $f(u)>0$ ($u>0$). It is easy to see that for $u$ large enough the function ${f(u) \over u}$ is monotone, therefore the limit $\lim\limits_{u\to +\infty} {f(u) \over u} $ exists. The value of this limit serves as the first level of our classification. The superlinear ($\lim\limits_{u\to +\infty} {f(u) \over u} = +\infty$), aymptotically linear ($\lim\limits_{u\to +\infty} {f(u) \over u} \in (0, +\infty)$) and sublinear ($\lim\limits_{u\to +\infty} {f(u) \over u} \leq 0$) cases are considered in Theorems 1, 2, 3, respectively. \begin{theorem} Let $n=1$, $f:[0,\infty)\to{\mathbb R}$, $f\in C^2$ be strictly convex, $\lim\limits_{u\to +\infty} {f(u) \over u} = +\infty$. \begin{enumerate} \item[(i)] If $f(u)>0 \ (u\in[0,\infty))$ then there exists $R_{sup} >0$ such that (\ref{BVP}) has two solutions for $RR_{sup}$. \item[(ii)] If $f(0)>0$ and $f$ has a root in $(0,\infty)$ then (\ref{BVP}) has two solutions for all $R>0$. \item[(iii)] If $f(0)=0$ and $f'(0)> 0$ then there exists $R_{sup} >0$ such that (\ref{BVP}) has one solution for $R0$. \item[(v)] If $f(0)<0$ then there exists $R_{sup} >0$ such that (\ref{BVP}) has one solution for $R\le R_{sup}$ and no solution for $R> R_{sup}$. \end{enumerate} \end{theorem} {\sc Proof.} \begin{enumerate} \item[(i)] Propositions \ref{limT0} and \ref{limTinfty1} imply that $ \lim_0 T = \lim_{+\infty} T = 0$. The maximum of $T$ is a unique extremum owing to Corollary 1, hence $T$ increases from 0 to some $R_{sup}>0$, then it decreases again to 0. \item[(ii)] Proposition \ref{domTn} and Corollary \ref{cordomT1} imply that $D(T)$ consists of two maximal subintervals $(0,\alpha)$ and $(\beta,\infty)$ where the endpoint(s) $\alpha$ and $\beta$ are not in $D(T)$. Using Propositions \ref{limT0}, \ref{limTc}, and \ref{limTinfty1} we conclude that $ \lim_0 T = \lim_{+\infty} T = 0$ and $ \lim_{\alpha^-} T = \lim_{\beta^+} T = +\infty$. From Proposition \ref{monTsubint} $T$ is strictly monotonic on both subintervals, hence $T$ attains each value twice. \item[(iii)] Corollary \ref{cordomTn} and Propositions \ref{limTinfty1} and \ref{limT0} yield $D(T)= (0,\infty )$, further, that $\lim_{+\infty} T = 0$ and $\lim_0 T = R_{sup}$ where $R_{sup}>0$ is the first root of the linearized equation $u'' + f'(0)u = 0$. Proposition \ref{monTf(0)neg} implies that $T$ is strictly decreasing, hence it attains each value $R\in (0, R_{sup})$ once. \item[(iv)] If $f'(0)=0$, then propositions \ref{domT2} and \ref{limT0} yield that $D(T)= (0,\infty )$ and $\lim_0 T = +\infty$. In the case $f'(0)<0$ Corollary \ref{cordomT1} and Proposition \ref{limTc} yield that for some $\beta>0$ we have $D(T)= (\beta,\infty )$ and $\lim_{\beta^+} T = +\infty$. In both cases we have $\lim_{+\infty} T = 0$. Proposition \ref{monTf(0)neg} implies that $T$ strictly decreases to 0. \item[(v)] Corollary \ref{cordomT1} yields that $D(T)= [\beta, \infty)$ for some $\beta >0$. Proposition \ref{monTf(0)neg} and $\lim_{+\infty} T = 0$ imply again that $T$ strictly decreases to 0. \end{enumerate} \hfill$\diamondsuit$ \begin{theorem} Let $n=1$, $f:[0,\infty)\to{\mathbb R}$, $f\in C^2$ be strictly convex, $\lim\limits_{u\to +\infty} {f(u) \over u}=L\in (0, +\infty)$ and $R_{\infty}:=\frac{\pi}{2 \sqrt{L}}$. \begin{enumerate} \item[(i)] If $f(u)>0 \ (u\in[0,\infty))$ then there exists $R_{sup} >R_{\infty}$ such that (\ref{BVP}) has one solution for $R\le R_{\infty}$ and $R=R_{sup}$, two solutions for $R_{\infty}R_{sup}$. \item[(ii)] If $f(0)>0$ and $f$ has a root in $(0,\infty)$ then (\ref{BVP}) has one solution for $R\le R_{\infty}$ and two solutions for $R>R_{\infty}$. \item[(iii)] If $f(0)=0$ and $f'(0)> 0$ then there exists $R_{sup} >R_{\infty} $ such that (\ref{BVP}) has no solution for $R\le R_{\infty}$, one solution for $R_{\infty}< R R_{\infty}$. \item[(v)] If $f(0)<0$ then there exists $R_{sup} >R_{\infty}$ such that (\ref{BVP}) has no solution for $R\le R_{\infty}$, one solution for $R_{\infty}< R \le R_{sup}$ and no solution for $R> R_{sup}$. \end{enumerate} \end{theorem} {\sc Proof.} The proof proceeds just in the same way as in Theorem 1, now using $\lim_{+\infty} T = R_{\infty}$ from Proposition \ref{limTinfty1}. (The existence of the maximum of $T$ in case {\it (i) } can be found in \cite{Lae70}).) \hfill$\diamondsuit$\medskip In the sublinear case the result is independent of $n$, therefore it is dealt with in the next section. \section{The $n$-dimensional case} For the general case ($n\geq 1$) the whole classification of bifurcation diagrams cannot be extended from the case $n=1$. In the sublinear case the next theorem gives the answer. \begin{theorem} Let $n\ge 1$, $f:[0,\infty)\to{\mathbb R}$, $f\in C^2$ be strictly convex, and $\lim\limits_{u\to +\infty} {f(u) \over u} \le 0$. If $f$ has a positive value then for any $R>0$ (\ref{BVP}) has a unique solution. (If $f\le 0$ then (\ref{BVP}) has obviously no positive solution for any $R>0$.) \end{theorem} {\sc Proof.} Due to its convexity, $f$ is strictly decreasing. In the nontrivial case $f(0)>0$ there are two possibilities: either $f$ has a single positive root $\alpha >0$ (and then $u>0$ on $(0,\alpha)$ and $u<0$ on $(\alpha, \infty)$) or $f>0$ on $(0, \infty)$ (in which case we define $\alpha:= \infty$). Proposition \ref{domTn} implies $D(T)=(0,\alpha)$. Propositions \ref{limT0}, \ref{limTc} and \ref{limTinftyn} imply that $ \lim_0 T = 0$ and $ \lim_{\alpha} T = +\infty $, and from Corollary \ref{cormonTn} $T$ is strictly increasing, hence it attains each value once. \hfill$\diamondsuit$ \begin{rem} \rm This result can be extended to general bounded domains using monotone operator theory, because $f$ is strictly decreasing. \end{rem} In the superlinear and asymptotically linear case ($\lim\limits_{u\to \infty} \frac{f(u)}{u} > 0$) the problem is much more complicated. The known results thus usually concern given special functions, as described in the Introduction. First, we briefly list some of the reasons for the difficulties concerning the domain, the limits and the monotonicity of the time-map. The domain of the time-map depends sensitively on the dimension $n$, as is shown by example $f(u) = u^k$: according to the Pohozaev identity \cite{Poh} $D(T)=\emptyset$ if $k\geq \frac{n+2}{n-2}$, and using variational methods \cite{Stu} or Emden's transformation \cite{JL} one can prove that $D(T)= (0,\infty )$ if $k< \frac{n+2}{n-2}$. The limit of the time map at infinity may also change as the dimension $n$ increases: in \cite{JL} it is shown for the example $f(u) =\mbox{e}^u$ that for $n\geq 3$ $\lim_{\infty} T \neq 0$. We note that in case $n\leq 2$, $\lim_{u\to \infty} \frac{f(u)}{u} = \infty$ implies $\lim_{\infty} T = 0$. The monotonicity properties of the time-map also change for higher dimensions. Our main lemma (Lemma 1) is not true generally, since e.g. for $f(u) =\mbox{e}^u$ and for $3\leq n \leq 10$ the time-map has infinitely many maxima and minima, see \cite{JL}. For the case $f(0)\leq 0$ McLeod proved a uniqueness result for a sophisticatedly chosen function class with main typical member $f(u)=u^p-u$. Unfortunately, many convex functions are not contained in this class, e.g. $f(u) = u^2 +au - b$ for $a,b>0$. In the class $f(0)=0$, $f'(0)>0$ Srikanth \cite{Sri} and Zhang \cite{Zhang} have proved uniqueness for $f(u)=u^p+u$ for $n\ge 3$. The difficulties concerning the monotonicity of the time-map generally arise from the fact that in case $n>1$ the Sturm comparison of $h$ and $v$ is not possible; from equations (\ref{T3}) and (\ref{T5}) one can see that $v$ oscillates more slowly than $h$. Now we summarize the bifurcation results that follow from our investigations on the domain, limits and monotonicity of the time-map. Of course, this theorem for a general convex function $f$ does not contain the strong results concerning the special functions mentioned above. \begin{theorem} Let $n\ge 1$, $f:[0,\infty)\to{\mathbb R}$, $f\in C^2$ be strictly convex, $\lim\limits_{u\to +\infty} {f(u) \over u} > 0$. \begin{enumerate} \item[(i)] If $f(u)>0 \ (u\in[0,\infty))$ then there exists $R_{sup} >0$ such that (\ref{BVP}) has at least one solution for $R\leq R_{sup}$ and no solution for $R>R_{sup}$. \item[(ii)] If $f(0)>0$ and $f$ has a root in $(0,\infty)$ then (\ref{BVP}) has at least one solution for all $R>0$. \item[(iii)] If $f(0)=0$ and $f'(0)> 0$ then there exist $R_{sup} >0$ and $R_{inf}\in [0,R_{sup}]$ such that (\ref{BVP}) has no solution for $R>R_{sup}$ or $R1$, $f\in C^2$ be strictly convex. If $f(0)\leq 0$, then (\ref{BVP}) has at most one solution for any $R>0$. \end{enumerate} \noindent {\sc Open Problems} \begin{enumerate} \item Let $n>2$ and $f\geq 0$. Determine the domain $D(T)$. \item Let $n>2$, $f(0)\leq 0$. Assume that $f$ has at most one positive root and $f$ is positive after this root, and $\lim \limits_{u\to \infty} \frac{f(u)}{u^p} =0$ for some $p<\frac{n+2}{n-2}$. Prove that there exists $b$ such that $D(T)=(b,\infty)$. \item Let $n>2$, $f(0)\leq 0$. Determine the limit $\lim\limits_{\infty} T$. \end{enumerate} \paragraph{Acknowledgement} The authors are grateful to Prof. L.A. Peletier for the useful discussion. This work was supported by OTKA grants F-022228 and T-019460. \begin{thebibliography}{99} \bibitem{Amb} Ambrosetti, A., On the exact number of positive solutions of convex nonlinear problems, {\it Boll. U.M.I.} {\bf 15-A}, 610-615, (1978). \bibitem{ABC} Ambrosetti, A., Brezis, H., Cerami, G., Combined effects of concave and convex nonlinearities in some elliptic problems, {\it J. Func. 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