\documentclass[twoside]{article} \usepackage{amsfonts} % used for R in Real numbers \pagestyle{myheadings} \markboth{ Existence of infinitely many solutions } { Anna Capietto \& Marielle Cherpion } \begin{document} \setcounter{page}{65} \title{\vspace{-1in}\parbox{\linewidth}{\footnotesize\noindent USA-Chile Workshop on Nonlinear Analysis, \newline Electron. J. Diff. Eqns., Conf. 06, 2001, pp. 65--87.\newline http://ejde.math.swt.edu or http://ejde.math.unt.edu \newline ftp ejde.math.swt.edu or ejde.math.unt.edu (login: ftp)} \vspace{\bigskipamount} \\ % On the existence of infinitely many solutions to a damped sublinear boundary-value problem % \thanks{ {\em Mathematics Subject Classifications:} 34B15. \hfil\break\indent {\em Key words:} Damped sublinear problem, continuation theorem, time-map technique. \hfil\break\indent \copyright 2001 Southwest Texas State University. \hfil\break\indent Published January 8, 2001. \hfil\break\indent A.C. Supported by GNAFA-C.N.R.-Italy, and by EEC grant CHRX-CT94-0555. \hfil\break\indent M.C. Supported by F.N.R.S.-Belgium. } } \date{} \author{ Anna Capietto \& Marielle Cherpion } \maketitle \begin{abstract} We prove the existence of infinitely many solutions (with prescribed nodal properties) to a damped sublinear boundary-value problem. The proofs are performed by means of an abstract continuation theorem and the time-map technique for strongly nonlinear operators. \end{abstract} \newtheorem{Theorem}{Theorem}[section] \newtheorem{proposition}[Theorem]{Proposition} \newtheorem{lemma}[Theorem]{Lemma} \newtheorem{remark}[Theorem]{Remark} \renewcommand{\theequation}{\thesection.\arabic{equation}} \catcode`@=11 \@addtoreset{equation}{section} \catcode`@=12 \section{Introduction} We study the existence and multiplicity of solutions to the boundary-value problem \begin{equation} \label{problema} \begin{array}{c} (r^{(k-1)}u')'+r^{(k-1)}a(u')f(r,u)= r^{(k-1)}h(r,u,u'),\\ u'(0)= 0 = u(R) \end{array} \end{equation} ($k>1$). As it is well-known, solutions to $(\ref{problema})$ are radially symmetric solutions to the following elliptic boundary-value problem on a ball ${\cal B}=B(0,R)$ \begin{equation} \label{BVP} \begin{array}{c} \nabla \cdot (\nabla u)+a(|\nabla u|) f(|x|,u)=h(|x|,u,|\nabla u|)\quad \mbox{in }{\cal B},\\ u=0\quad \mbox{on } \partial {\cal B}. \end{array} \end{equation} We deal with a so-called "sublinear" problem. More precisely, we assume that $a(\xi)=a_0 + |\xi|^q \; (a_0 >0, \, 0 0$ such that for all $(r,s,\xi)\in [0,R]\times [-\varepsilon_{0},\varepsilon_{0}]\times\mathbb{R}$, $$ |h(r,s,\xi)|\leq H|\xi |. $$ Moreover, there exists a continuous function $C:\mathbb{R} \to (0,+\infty)$ such that \begin{equation} \lim_{s\to 0}\frac{h(r,s,\xi)}{s}=C(\xi)\mbox{ uniformly in }r\in [0,R]. \end{equation} \end{enumerate} \noindent We point out that problem $(\ref{problema})$ can be considered ``singular" in a two-fold sense. Indeed, on one hand, under condition $(H_f)$ the uniqueness of the solutions to initial value problems associated to $(\ref{problema})$ must be guaranteed by $(H_F)$; on the other hand, a singularity in the $r$-variable arises because of the boundary condition in zero. For more comments on $(H_F)$ we refer to \cite{CD,CDZ,Ch,WW}. Our main result is the following (cf. Theorem \ref{main}). \paragraph{Theorem A} {\it Assume ($H_f$)-($H_F$)-($H_h$) and let $a:\mathbb{R} \to \mathbb{R}$ be defined by $a(\xi):=a_{0} + |\xi|^q$ with $00$. Then there exists $n_0\in {\mathbb{N}}$ such that for every $n > n_0$ problem $(\ref{problema})$ has at least two solutions $u_n$ and $v_n$ with $u_n(0)>0$ and $v_n(0)<0$, both having exactly $n$ zeros in $[0,R)$. Moreover, we have $$ \lim_{n\to +\infty} |u_n(r)|+|u'_n(r)|=0 =\lim_{n\to +\infty} |v_n(r)|+|v'_n(r)|,\mbox{ uniformly in $r\in [0,R].$} $$ } Multiplicity results for a boundary-value problem of the form $(\ref{problema})$ can be found e.g. in \cite{AGP}, \cite{B}, \cite{CDZ}, \cite{Ch}, \cite{DMS}, \cite{GMZ2}. However, apart from \cite{Ch} where additional regularity conditions are imposed, in those papers the authors considered the case $a \equiv 1$ and/or $h \equiv 0$. In some of the above quoted papers, the differential operator under consideration is strongly nonlinear. We refer to \cite{CDZ} for a more comprehensive list of references. We work in the framework of topological degree methods and use some of the ideas developed in \cite{CDZ} (see also \cite{CK}, \cite{ET}). In this situation, two main tasks have to be accomplished. First, one has to study an autonomous problem \begin{equation} \label{auto*} \begin{array}{c} u''+a(u')g(u)=0,\\ u'(0) = 0 = u(R), \end{array} \end{equation} where $g : [-\varepsilon, \varepsilon] \to \mathbb{R}, \varepsilon >0$, is a continuous function such that \[ \lim_{s\to 0} \frac{g(s)}{s}=+\infty. \] Secondly, suitable estimates on the (possible) solutions to a family of parameter-dependent problems (cf. $(P_{\lambda})$) have to be established. In this paper (cf. Section 2) we overcome the first difficulty by studying problem $(\ref{auto*})$ in the equivalent form \begin{equation} \label{phiautonomous} \begin{array}{c} (\phi(u'))' + g(u) = 0,\\ u'(0) = 0 = u(R), \end{array} \end{equation} where $\phi$ is an odd increasing homeomorphism defined through $a$. In this way, we can use the time-map technique for equations containing the $\phi$-Laplacian (see \cite{CDZ}, \cite{D}, \cite{DMS}, \cite{GMZ3}, \cite{GMZ1}, \cite{GMZ2}, \cite{GU}) and establish a multiplicity result for $(\ref{auto*})$ (cf. Theorem $\ref{autonomo}$ and Theorem 5.4 in \cite{CDZ}). Then, in order to show that the (nodal) properties of the solutions to $(\ref{auto*})$ can be "continued" to problem $(\ref{problema})$, some estimates on the number of zeros of the (possible) solutions to the associated parameter-dependent boundary-value problem (cf. $(P_{\lambda})$) have to be established. To this end, we argue on the lines of \cite{CDZ}; however, some technical difficulties due to the presence in $(\ref{problema})$ of the functions $a$ and $h$ have to be overcome (see in particular the proofs of Lemma 3.1, Lemma 3.3 and Claim 2 in Theorem 4.1). We end this introductory section by observing that a result analogous to Theorem A can be performed for a more general strongly nonlinear boundary-value problem \begin{equation} \label{general} \begin{array}{c} (r^{(k-1)} \psi(u'))'+r^{(k-1)}a(u')f(r,u)=r^{(k-1)}h(r,u, u'),\\ u'(0) = 0 = u(R), \end{array} \end{equation} where $\psi$ is an odd increasing homeomorphism satisfying suitable assumptions. Furthermore, on the lines of \cite{CDZ}, one could prove the existence of an {\sl additional} double sequence of solutions to $(\ref{problema})$ (whose norm tends to infinity) provided that $g$ has a "superlinear" behaviour at infinity and assumption $(H_h)$ is modified accordingly. \smallskip This paper is organized as follows. In Section 2 we study the autonomous problem $(\ref{auto*})$. In Section 3 we introduce a parameter-dependent non-autonomous problem and develop some estimates on its solutions. In Section 4 we recall an abstract continuation theorem which is then applied for the proof of the main result. In what follows, for any Banach space $X$, for any linear compact operator $L:X\to X$ and for any subset $\Omega\subset X$ we will denote by $\deg(I-L,\Omega)$ the Leray-Schauder degree of $I-L$ (if defined). The space $C^1([0,R])$ of the continuously differentiable real functions $u$ on $[0,R]$ will be equipped with the norm \[ ||u||_1=\max\left\{\sqrt{|u(t)|^2+|u'(t)|^2}: t\in [0,R]\right\}. \] Finally, $C^{1}_{\#}([0,R])$ denotes the space of functions $u \in C^{1}([0,R])$ satisfying the boundary condition $u'(0)=0=u(R)$. \section{An autonomous problem} Let us consider the second order ODE \begin{equation} \label{autonomous} \begin{array}{c} u''+a(u')g(u)=0,\\ u'(0) = 0 = u(R), \end{array} \end{equation} where $a:\mathbb{R} \to \mathbb{R}$ is defined by $a(\xi):=a_{0} + |\xi|^q, 0 0$. Set \begin{equation} \label{phi} \phi(s)=\int_0^s \frac{1}{a(x)}\, dx. \end{equation} We assume that $g : [-\varepsilon, \varepsilon] \to \mathbb{R}$ is continuous $(\varepsilon > 0)$ and such that \begin{equation} \label{refA} \lim_{s\to 0} \frac{g(s)}{\phi(s)}=+\infty. \end{equation} We shall also assume (without loss of generality) that $g(s)s>0$ for all $s\in [-\varepsilon, \varepsilon]\setminus\{0\}$ and we set $G(s)=\int_0^s g(\xi)\ d\xi$. We observe that problem $(\ref{autonomous})$ can be written in the form $$ \begin{array}{c} (\phi(u'))' + g(u) = 0,\\ u'(0) = 0 = u(R). \end{array} $$ It is not difficult to check that $\phi$ is an odd increasing homeomorphism. Then, as in \cite{GMZ3}, it is possible to study $(\ref{autonomous})$ with the time-map technique by means of the system \begin{equation} \label{equivsys} \begin{array}{c} u' = \phi^{-1}(y),\\ y' = -g(u). \end{array} \end{equation} More precisely, for \begin{equation} \label{defL} {\cal L}(\xi) =\int_0^{\xi} \frac{x}{a(x)}\, dx, \end{equation} we shall use the fact that if $u$ is a solution of $(\ref{equivsys})$, then $E(r,u(r),u'(r)):=G(u(r))+{\cal L}(u'(r))$ is constant. Observe that our assumptions on $g$ ensure that the orbits of $(\ref{autonomous})$ are closed curves on the phase-plane. Then, denoting by ${\cal L}^{-1}$ the inverse of the restriction to $\mathbb{R}^+$ of the function ${\cal L}$, we can introduce the function $T_{1}:(0,\varepsilon)\to (0,+\infty)$ by \begin{equation} \label{tm1} T_1(\alpha)=\int_0^{\alpha} \frac{dx}{{\cal L}^{-1}(G(\alpha)-G(x))}. \end{equation} It is straightforward to check that $T_{1}(\alpha)$ represents the time needed for a rotation along the orbit of "energy" $G(\alpha)$ in the upper (resp. lower) half plane from the point $(0,{\cal L}^{-1}(G(\alpha)))$ to the point $(\alpha,0)$ (resp. from $(\alpha,0)$ to $(0,-{\cal L}^{-1}(G(\alpha)))$). Analogously, for $\alpha_1 <0$ s.t. $G(\alpha_1)=G(\alpha)$, the function $T_{2}\, :\, (0,\varepsilon)\to (0,+\infty)$ defined by \begin{equation} \label{tm2} T_2(\alpha)=\int_{\alpha_1}^0 \frac{dx}{{\cal L}^{-1}(G(\alpha)-G(x))} \end{equation} is the time needed for a rotation along the orbit of "energy" $G(\alpha)$ from the point $(\alpha_1,0)$ to the point $(0,{\cal L}^{-1}(G(\alpha)))$ (resp. from $(0,-{\cal L}^{-1}(G(\alpha)))$ to $(\alpha_1,0)$). For a classical reference on this topic, the reader can consult \cite{Op}. See also \cite{D}. For the completion of the study of the autonomous case, we need the following result. \begin{proposition} \label{asint} Let $a:\mathbb{R} \to \mathbb{R}$ be defined by $a(\xi):=a_{0} + |\xi|^q$ with $00$ and $\phi$ given by (\ref{phi}). Assume $g : [-\varepsilon, \varepsilon] \to \mathbb{R}$ is a continuous function such that $g(s)s>0$ for all $s\in [-\varepsilon, \varepsilon]\setminus\{0\}$ and satisfying (\ref{refA}). Then the functions $T_{1}(\alpha)$ and $T_{2}(\alpha)$ defined by (\ref{tm1}) and (\ref{tm2}) are such that for $i=1,\, 2$ we have $$ \lim_{\alpha \to 0} T_i(\alpha)=0. $$ \end{proposition} \noindent{\sc Proof.} Observe that ${\cal L}(s) = (\Phi_{\ast} \circ \phi)(s)$ with $\Phi_{\ast}(s) = \int_0^s \phi^{-1}(x)\, dx$. The proof follows the same arguments as in Lemma 2.1 in \cite{GMZ1} and Theorem 3.2 in \cite{GMZ3} where the assumptions on the function $g$, as well as the result on the asymptotic behaviour of the time-maps, are relative to a neighbourhood of infinity. For a more detailed proof, one can also see Theorem 2.2.8 in \cite{D}. \hfill$\Box$\smallskip Once the time-maps are defined, we can introduce the "generalized Fu$\check c$ik spectrum" as in \cite{CD}, \cite{CDZ}, \cite{D} in order to get a characterization of the existence of solutions with a fixed number of zeros. Indeed, using Proposition \ref{asint}, one gets the following multiplicity result for the autonomous problem (\ref{autonomous}). \begin{Theorem} \label{autonomo} \cite[Th. 5.4]{CDZ} Let $a:\mathbb{R} \to \mathbb{R}$ be defined by $a(\xi):=a_{0} + |\xi|^q$ with $00$ and $\phi$ given by (\ref{phi}). Assume $g : [-\varepsilon, \varepsilon] \to \mathbb{R}$ is a continuous function such that $g(s)s>0$ for all $s\in [-\varepsilon, \varepsilon]\setminus\{0\}$ and satisfying (\ref{refA}). Then there exists $k_0 \in {\mathbb{N}}$ such that for every $k\geq 2k_0$ problem $(\ref{autonomous})$ has at least two solutions $u_k$ and $v_k$ with $u_k(0)>0$ and $v_k(0)<0$, both having exactly $k$ zeros in $[0,R)$. \end{Theorem} We end this section by giving two important properties of ${\cal L}$, which will be crucial in the sequel. \begin{proposition} \label{propL} Let $a:\mathbb{R} \to \mathbb{R}$ be defined by $a(\xi):=a_{0} + |\xi|^q$ with $00$ and ${\cal L}$ given by (\ref{defL}). Then for all $\xi\in\mathbb{R}$, we have $$ \frac{\xi^{2}}{a(\xi)}\leq 2{\cal L}(\xi). $$ \end{proposition} \noindent{\sc Proof.} An easy computation gives $(\frac{s^2}{a(s)})'\leq 2{\cal L}'(s)$, for all $s\geq 0$. Then by integration, we get $$\frac{\xi^2}{a(\xi)}\leq 2{\cal L}(\xi)\,.$$ \begin{proposition} \label{stimastar} Let $a:\mathbb{R} \to \mathbb{R}$ be defined by $a(\xi):=a_{0} + |\xi|^q$ with $00$ and ${\cal L}$ given by (\ref{defL}). Then for any $c_1 >1$, $c\geq c_1^{2}+1$ and $\xi>0$ small enough, we have \begin{equation} \label{greater} c_1 {\cal L}^{-1}(\xi)\leq {\cal L}^{-1}(c \xi). \end{equation} \end{proposition} \noindent{\sc Proof.} Notice that since $\lim_{x \to 0} \frac{{{\cal L}}(c_1 x)}{{{\cal L}}(x)} = c_1^{2}$, then for any $c\geq c_1^{2}+1$ and $x>0$ small enough we have ${\cal L}(c_1 x)\leq c {\cal L}(x)$. As ${\cal L}^{-1}:\mathbb{R}^{+}\to\mathbb{R}^{+}$ is continuous, we have for $\xi>0$ small enough $$ {\cal L}(c_1 {\cal L}^{-1}(\xi))\leq c {\cal L}({\cal L}^{-1}(\xi))=c\xi $$ and since ${\cal L}^{-1}$ is increasing $c_1 {\cal L}^{-1}(\xi)\leq {\cal L}^{-1}(c \xi)$. \hfill$\Box$ \begin{remark} \label{nolower} In general, if one sets $\Phi_{\ast}(s) = \int_0^s \phi^{-1}(x)\, dx$, where $\phi$ is an odd increasing homeomorphism, an inequality like (\ref{greater}) can be proved separately for the functions $\Phi_{\ast}^{-1}$ and $\phi^{-1}$. This is done in \cite{CDZ} under the "{\cal lower} $\sigma$-condition" $$ \liminf_{s\to 0}\frac{{\phi}(\sigma s)}{{\phi}(s)}>1,\quad \forall \sigma>1. $$ In our situation, we observe that ${\cal L}(s) = (\Phi_{\ast} \circ \phi)(s)$, so we could have proved Proposition \ref{stimastar} by combining the inequalities for $\Phi_{\ast}^{-1}$ and $\phi^{-1}$. A direct proof of Proposition $\ref{stimastar}$ is simpler thanks to the fact that we can explicitly use the function ${\cal L}$ and its properties. \end{remark} \section{Preliminary results} We consider the boundary-value problem \begin{equation} \label{ODE} \begin{array}{c} (r^{(k-1)}u')'+r^{(k-1)}a(u')f(r,u)= r^{(k-1)}h(r,u,u'),\\ u'(0)= 0 = u(R), \end{array} \end{equation} where $k>1$, $a:\mathbb{R} \to \mathbb{R}$ is defined by $a(\xi):=a_{0} + |\xi|^q$ with $00$ and for a fixed $\varepsilon_{0} > 0$, the functions $f$ and $h$ satisfy the following properties. \begin{enumerate} \item[($H_f$)] The function $f:[0,R]\times [-\varepsilon_{0},\varepsilon_{0}]\to{\mathbb{R}}$ is continuous and such that $$ f(r,0)=0 $$ and $$ \lim_{s\to 0}\frac{f(r,s)}{s}=+\infty\mbox{ uniformly in }r\in [0,R]. $$ \item[($H_F$)] For $F(r,s):=\int_{0}^s f(r,x)\, dx$, $F$ is differentiable with respect to $r\in [0,R]$ and there exists a continuous positive function $\alpha:[0,R]\to (0,+\infty)$ such that for all $r\in [0,R]$, all $s\in [-\varepsilon_0,\varepsilon_0]$, $$ \left|\frac{\partial F}{\partial r}(r,s) \right|\leq \alpha(r)F(r,s). $$ \item[($H_h$)] The function $h:[0,R]\times [-\varepsilon_{0},\varepsilon_{0}]\times\mathbb{R}\to{\mathbb{R}}$ is continuous and there exists $H>0$ such that for all $(r,s,\xi)\in [0,R]\times [-\varepsilon_{0},\varepsilon_{0}]\times\mathbb{R}$, $$ |h(r,s,\xi)|\leq H|\xi |. $$ Moreover, there exists a continuous function $C:\mathbb{R}\to (0,+\infty)$ such that \begin{equation} \label{refC} \lim_{s\to 0}\frac{h(r,s,\xi)}{s}=C(\xi)\mbox{ uniformly in }r\in [0,R]. \end{equation} \end{enumerate} \noindent A typical example for the function $h$ is $h(r,s,\xi)=\eta(s)|\xi|^{\beta}$ with $\beta > 1$ for $|\xi|<1$, $0<\beta<1$ for $|\xi|\geq1$ and $\eta(s) \sim s$ for $s\to 0$. \hfill$\Box$\smallskip Following a degree approach, problem $(\ref{ODE})$ will be treated by means of the parameter-dependent family of problems ($\lambda \in [0,1]$) $$ \begin{array}{c} (r^{\lambda(k-1)}u')'+r^{\lambda(k-1)}a(u')f_{\lambda}(r,u)= \lambda r^{\lambda(k-1)} h(r,u,u'),\\ u'(0) = 0 = u(R), \end{array} \eqno{(P_{\lambda})} $$ where $f_{\lambda}:[0,R]\times [-\varepsilon_0,\varepsilon_0]\to {\mathbb{R}}$ is defined by \begin{equation} \label{fl} f_{\lambda}(r,s)=\lambda f(r,s)+(1-\lambda) g(s), \end{equation} and $g:[-\varepsilon_0,\varepsilon_0] \to{\mathbb{R}}$ is a continuous nondecreasing function such that $$ \lim_{s\to 0} \frac{g(s)}{s}=+\infty.\leqno{(H_g)} $$ We shall also assume (without loss of generality) that $g(s)s>0$, for every $s \in [-\varepsilon_0,\varepsilon_0]\setminus\{0\}$. Note that our assumptions on $g$ guarantee that condition (\ref{refA}) is satisfied. Set $F_{\lambda}(r,s):=\int_{0}^s f_{\lambda}(r,x)\, dx$. It is immediate to remark that in the situation described above we have \begin{enumerate} \item[($H_{F_{\lambda}}$)] $F_{\lambda}(r,s)$ is differentiable with respect to $r\in [0,R]$ and there exists a continuous positive function $\alpha:[0,R]\to (0,+\infty)$ such that for all $r\in [0,R]$, all $s\in [-\varepsilon_0,\varepsilon_0]$, $$ \left|\frac{\partial F_{\lambda}}{\partial r}(r,s) \right|\leq \alpha(r)F_{\lambda}(r,s). $$ \end{enumerate} Moreover, using ($H_f$) and ($H_g$), we have $$ \lim_{s\to 0} \frac{f_{\lambda}(r,s)}{s}= +\infty\quad \mbox{uniformly in $\lambda \in [0,1]$} $$ and, by ($H_{F_{\lambda}}$), for all $r \in [0,R]$, all $s\in [-\varepsilon_0,\varepsilon_0]\setminus\{0\}$ and all $\lambda \in [0,1]$, \begin{equation} \label{Flandapositiva} F_{\lambda}(r,s)>0. \end{equation} In our main result we will prove the existence of infinitely many solutions of $(P_{1})$ using an abstract continuation theorem. To this end, we need the following lemma concerning the Cauchy problem \begin{equation} \label{cauchy} \begin{array}{c} (r^{\lambda (k-1)} u')'+r^{\lambda (k-1)}a(u')f_{\lambda}(r,u)= \lambda r^{\lambda(k-1)}h(r,u,u'),\\ u(0)=d,\ u'(0)=0. \end{array} \end{equation} \begin{lemma} \label{dipcont} For all $\varepsilon \in (0,\varepsilon_0]$, if $u$ is a (local) solution of problem (\ref{cauchy}) with $d$ small enough, then $u$ can be defined on $[0,R]$ and $||u||_1\leq \varepsilon$. \end{lemma} \noindent{\sc Proof.} Let $\varepsilon>0$ be fixed and $u$ be a solution of $(\ref{cauchy})$. Assume that there exists $\rho\in (0,R]$ such that for all $r\in [0,\rho]$, $$ |u(r)|\leq \varepsilon\hspace{1cm}\mbox{ and }\hspace{1cm}|u'(r)|\leq \varepsilon. $$ Let $$ E_{\lambda}(r,s,\xi):=F_{\lambda}(r,s)+{\cal L}(\xi) $$ where ${\cal L}$ is given in (\ref{defL}) and for all $r\in [0,\rho]$, we consider the function \begin{equation} \label{defvlanda} v_{\lambda}(r):=E_{\lambda}(r,u(r),u'(r)). \end{equation} We have, using ($H_{h}$), ($H_{F_{\lambda}}$) and Proposition \ref{propL} \begin{eqnarray*} v'_{\lambda}(r) &=&\frac{\partial F_{\lambda}}{\partial r}(r,u(r))+\frac{u'(r)}{a(u'(r))} \left( \lambda h(r,u(r),u'(r))-\frac{\lambda (k-1)}{r} u'(r)\right)\\ &\leq&\frac{\partial F_{\lambda}}{\partial r}(r,u(r))+\lambda h(r,u(r),u'(r))\frac{u'(r)}{a(u'(r))}\\ &\leq&\frac{\partial F_{\lambda}}{\partial r}(r,u(r))+H\frac{(u'(r))^{2}}{a(u'(r))}\\ &\leq&\alpha(r) F_{\lambda}(r,u(r))+2H{\cal L}(u'(r))\\ &\leq&\tilde\alpha(r) v_{\lambda}(r), \end{eqnarray*} where $\tilde\alpha:[0,R]\to (0,+\infty)$ is a continuous function. Integrating on $(0,r)$, we get $$ v_{\lambda}(r)\leq v_{\lambda}(0) e^{\int_0^r \tilde\alpha(s)\, ds}=F_{\lambda}(0,d)e^{\int_0^r \tilde\alpha(s)\, ds} $$ and by definition of $v_{\lambda}$, we have \begin{equation} \label{vlanda} {\cal L}(u'(r))\leq v_{\lambda}(r)\leq F_{\lambda}(0,d)e^{\int_0^r \tilde\alpha(s)\, ds}. \end{equation} For the rest of the proof, we argue as in the proof of Lemma 2.3 in \cite{CDZ}; however, we give the details for the reader's convenience. Consider $(a_1,a_2)\in (0,1)^2$ such that \begin{equation} \label{aunoadue} a_1+Ra_2\leq {1\over 2} \quad\mbox{and}\quad a_2\leq {1\over 2} \end{equation} (observe that, for every $R>0$, a similar choice of $a_1$ and $a_2$ is always possible). Since $\lim_{d\to 0} {\cal L}^{-1}(F_{\lambda}(0,d)e^{\int_0^r \tilde\alpha(s)\, ds})=0$ uniformly in $\lambda \in [0,1]$, for every $\varepsilon\leq\varepsilon_0$ there exists $d_{\varepsilon}>0$ such that $d_{\varepsilon}\leq a_1\varepsilon$ and for all $0<|d|\leq d_{\varepsilon}$, all $\lambda\in [0,1]$, \[ {\cal L}^{-1}(F_{\lambda}(0,d)e^{\int_0^r \tilde\alpha(s)\, ds})\leq a_2\varepsilon. \] Then, for $0<|d|\leq d_{\varepsilon}$, we deduce from $(\ref{vlanda})$ that for all $r\in [0,\rho]$, \begin{equation} \label{uprimo} |u'(r)|\leq {\cal L}^{-1}(F_{\lambda}(0,d)e^{\int_0^r \tilde\alpha(s)\, ds})\leq a_2\varepsilon\leq \frac{\varepsilon}{2}. \end{equation} The above estimate implies that for all $r\in [0,\rho]$, \begin{equation} \label{ustima} \begin{array}{rcl} |u(r)|&\leq&\displaystyle d+\int_0^r |u'(s)|\, ds\leq d+R {\cal L}^{-1}( F_{\lambda}(0,d)e^{\int_0^r \tilde\alpha(s)\, ds})\\ &\leq& a_1\varepsilon + R a_2\varepsilon=(a_1+R a_2)\varepsilon\leq\frac{\varepsilon}{2}. \end{array} \end{equation} Since $(\ref{uprimo})$ and $(\ref{ustima})$ hold independently on $\rho$, we can extend $u$ on $[0,R]$ as a $C^1$-function. Finally, $(\ref{uprimo})$ and $(\ref{ustima})$ imply that $$ ||u||_1=\max_{r\in [0,R]} \sqrt{|u(r)|^2+|u'(r)|^2}\leq\varepsilon. $$ \begin{remark} \label{rem1} We deduce from Lemma \ref{dipcont} that if $u$ is a solution of (\ref{cauchy}) with $d$ small enough then $u'$ is bounded. Hence condition (\ref{refC}) in ($H_{h}$) implies that there exists $\tilde\delta>0$ such that for all $r\in [0,R]$, $$ 0<|u(r)|\leq\tilde\delta \Longrightarrow|h(r,u(r),u'(r))|\leq C|u(r)|, $$ with $C$ independent of $u'$. \end{remark} \begin{lemma} \label{energia} There exists $\bar\delta>0$ such that if $u$ is a solution of $(P_{\lambda})$ with $|u(0)|=\bar d$ small enough then for all $r\in [0,R]$, $$ |u(r)|^2+|u'(r)|^2\geq\bar\delta. $$ \end{lemma} \noindent{\sc Proof.} {\bf Step 1 - } Let $0<\tilde\varepsilon0. \end{equation} Hence if $d>0$ and $r$ are small enough, for every $s\in [0,r]$ we have $00\,, \end{eqnarray*} which proves that $u$ is decreasing in $[0,r]$. In the same way, if $d<0$ is small enough then $$ f_{\lambda}(s,u(s))a(u'(s))-\lambda h(s,u(s),u'(s))\leq\lambda C u(s)\left(\frac{a(u'(s))}{\tilde\varepsilon}-1\right)<0 $$ and $u$ is increasing in $[0,r]$. Arguing as in \cite{CK}, for every $\theta \in (0,1)$ we can consider the first point $r_{\theta}(d)$ such that $$ u(r_{\theta}(d);d)=\theta d. $$ Moreover, we denote by $r_0(d)$ the first zero of $u(\cdot;d)$. \noindent {\bf Step 2 - }There exists $\bar\delta>0$ such that if $u$ is a solution of $(P_{\lambda})$ with $|u(0)|=\bar d$ small enough then for all $r\in [0,R]$ we have $|u(r)|^2+|u'(r)|^2\geq\bar\delta$. Let $u$ be a solution of $(P_{\lambda})$ with $|u(0)|=\bar d$ sufficiently small. To prove this Step we will need the following two claims. \smallskip \noindent {\it Claim 1 : }For every $\theta \in (0,1)$ and for every $\lambda \in (0,1)$, we have $$ r_{\theta}(\bar d)\geq \sqrt{\frac{2\bar d(1-\theta)(1+\lambda (k-1))}{({\hat f}(\bar d)+g(\bar d)) a(\varepsilon_{0}) + H\varepsilon_{0}}}=:\beta(\bar d)>0, $$ where ${\hat f}$ is defined by $$ {\hat f}(s)=\left\{ \begin{array}{ll} \sup\{f(r,x),\, r\in [0,R],\, x\in [0,s]\}&\mbox{if }00$. If $\bar d$ is small enough, we have for every $s\in [0,r_0(\bar d)]$ that $00$ such that if $u$ is a solution of $(P_{\lambda})$ with $|u(0)|=\bar d$ sufficiently small, we have $E_{\lambda}(r,u(r),u'(r))\geq \delta_0$ for every $r\in [0,R]$. First, we observe that, by $(H_{F_{\lambda}})$, there is a constant $\gamma$ such that for all $r\in (0,R]$, all $s\in [-\epsilon_0,\epsilon_0]$ and all $\lambda \in[0,1]$, \begin{equation} \label{gammasuper} {{\partial F_{\lambda}}\over {\partial r}}(r,s)+{{\gamma }\over r}F_{\lambda}(r,s)\geq 0. \end{equation} Recall that $E_{\lambda}(r,s,\xi):=F_{\lambda}(r,s)+{\cal L}(\xi)$. Using $(H_{h})$ and Proposition \ref{propL}, \begin{eqnarray*} \lefteqn{ \frac{d}{dr} E_{\lambda}(r,u(r),u'(r))+\frac{\gamma}{r}E_{\lambda}(r,u(r),u'(r)) }\\ &=& \frac{\partial F_{\lambda}}{\partial r}(r,u(r))+\frac{\gamma}{r}F_{\lambda}(r,u(r))+\frac{u'(r)}{a(u'(r))}\lambda h(r,u(r),u'(r))+\frac{\gamma}{r}{\cal L}(u'(r))\\ &&-\lambda\frac{(k-1)}{r} \frac{(u'(r))^{2}}{a(u'(r))}\\ &\geq&\frac{u'(r)}{a(u'(r))}\lambda h(r,u(r),u'(r))+\frac{\gamma}{r}{\cal L}(u'(r))-\frac{2(k-1)}{r}{\cal L}(u'(r))\\ &\geq& -H\frac{(u'(r))^{2}}{a(u'(r))}+\frac{\gamma}{r}{\cal L}(u'(r))-\frac{2(k-1)}{r}{\cal L}(u'(r))\\ &\geq& {\cal L}(u'(r))\left(-2H+\frac{\gamma}{r}-\frac{2(k-1)}{r}\right) \geq 0 \end{eqnarray*} if $\gamma\geq 2HR+2(k-1)$. Multiplying by $r^{\gamma}$ and integrating from $r_{\theta}(\bar d)$ to $r$, we obtain $$ E_{\lambda}(r,u(r),u'(r))r^{\gamma}-E_{\lambda}(r_{\theta}(\bar d),u(r_{\theta}(\bar d)), u'(r_{\theta}(\bar d)))r_{\theta}(\bar d)^{\gamma}\geq 0 $$ and \begin{eqnarray*} E_{\lambda}(r,u(r),u'(r))&\geq& E_{\lambda}(r_{\theta}(\bar d),u(r_{\theta}(\bar d)), u'(r_{\theta}(\bar d)))r_{\theta}(\bar d)^{\gamma}R^{-\gamma}\\ &=&R^{-\gamma}\left({\cal L}(u'(r_{\theta}(\bar d)))+F_{\lambda}(r_{\theta}(\bar d),u(r_{\theta}(\bar d)))\right) r_{\theta}(\bar d)^{\gamma}\\ &\geq& R^{-\gamma}F^0(\theta {\bar d}) r_{\theta}(\bar d)^{\gamma}\\ &\geq&R^{-\gamma}F^0(\theta {\bar d})(\beta(\bar d))^{\gamma}, \end{eqnarray*} where $F^0(\theta {\bar d})= \min\{F_{\lambda}(r,\theta {\bar d}): r\in [0,R],\ \lambda\in [0,1]\}>0$. We finish the proof of the Claim by setting $\delta_{0}=R^{-\gamma}F^0(\theta {\bar d})(\beta(\bar d))^{\gamma}$. \smallskip If the claim in Step 2 were not true then for every $\bar\delta>0$, there exists $\bar r\in [0,R]$ such that $|u(\bar r)|^2+|u'(\bar r)|^2<\bar\delta$, which contradicts Claim 2. \hfill$\Box$\smallskip Now, for every $d\in {\mathbb{R}_{0}}$ and for every $\lambda \in [0,1]$ we can define $$ {\bf n}:S_{d,\lambda}\to {\mathbb{N}}: u\mapsto {\bf n}(u), $$ where $$ S_{d,\lambda}=\{u:u\mbox{ is a solution of }(P_{\lambda})\mbox{ and } u(0) > d\mbox{ if }d>0,\, u(0)1$), using an abstract theorem. Note that solutions to $(P_{1})$ are solutions to $(\ref{ODE})$, while solutions to $(P_{0})$ are solutions to the autonomous problem $$ \begin{array}{c} u''+ a(u')g(u)=0,\\ u'(0) = 0 = u(R). \end{array} $$ Our main result is the following. \begin{Theorem} \label{main} Assume ($H_f$)-($H_F$)-($H_h$) and let $a:\mathbb{R} \to \mathbb{R}$ be defined by $a(\xi):=a_{0} + |\xi|^q$ with $00$. Then there exists $n_0\in {\mathbb{N}}$ such that for every $n > n_0$ problem $(\ref{ODE})$ has at least two solutions $u_n$ and $v_n$ with $u_n(0)>0$ and $v_n(0)<0$, both having exactly $n$ zeros in $[0,R)$. Moreover, we have $$ \lim_{n\to +\infty} |u_n(r)|+|u'_n(r)|=0 =\lim_{n\to +\infty} |v_n(r)|+|v'_n(r)|,\mbox{ uniformly in $r\in [0,R].$} $$ \end{Theorem} To prove this theorem, we will need an abstract continuation result. In order to state this theorem, let us consider a Banach space $X$ and a completely continuous operator ${\cal N}: {\rm dom}\ {\cal N}\subset X\times [0,1] \to X$. Moreover, let $A$, $B$ be two open sets such that $A \subset \bar A \subset B \subset \bar B$ and $(\bar B \setminus A)\subset {\rm dom}\ {\cal N}$. Let $\Sigma$ be the set of the solutions of the abstract equation $u={\cal N}(u,\lambda)$, i.e. \[ \Sigma=\{(u,\lambda): u={\cal N}(u,\lambda)\}. \] For any subset $D\subset X\times [0,1]$, we denote the section of $D$ at $\lambda\in [0,1]$ by $D_{\lambda}=\{x\in X:(x,\lambda)\in D\}$ and we also set ${\cal N}_{\lambda}={\cal N}(\cdot,\lambda)$. We are now in position to state the following \begin{Theorem} \label{continuazione} \cite[Th. 2.1]{CDZ} Let ${\bf k}:\Sigma \cap (\bar B \setminus A) \to {\mathbb{N}}$ be a continuous function; suppose that there exists a positive integer $n$ satisfying the following conditions \begin{equation} \label{gap1} n \notin {\bf k}(\partial (\bar B \setminus A)) \end{equation} and \begin{equation} \label{gap2} {\bf k}^{-1}(n) \quad \mbox{is bounded}. \end{equation} Then, for an open set $U^n_0$ such that $({\bf k}^{-1}(n))_0\subset U^n_0\subset {\overline {U^n_0}}\subset (\bar B \setminus A)_0$ and $\Sigma_0 \cap U^n_0 = ({\bf k}^{-1}(n))_0$, the Leray-Schauder degree $\deg (I-{\cal N}_0,U^n_0)$ is defined. If \begin{equation} \label{gradonozero} \deg (I-{\cal N}_0,U^n_0)\not= 0, \end{equation} then there is a continuum $C_n\subset \Sigma$ with \[ \{\lambda\in [0,1]: \exists u\in X: (u,\lambda)\in C_n\}=[0,1] \] and such that \[ (u,\lambda)\in C_n\quad \Longrightarrow \quad (u,\lambda)\in (B \setminus \bar A) \quad \mbox{and} \quad {\bf k}(u,\lambda)=n. \] In particular there is at least one solution $\tilde u \in (B \setminus \bar A)_1$ of the operator equation \[ u={\cal N}(u,1) \] with \[ {\bf k}(\tilde u,1)=n. \] \end{Theorem} \noindent {\sc Proof of Theorem \ref{main}.} First note that problem $(P_{\lambda})$ can be put into the form $u={\cal N}(u,\lambda)$ where $$ {\cal N}: {\rm dom}\ {\cal N}\subset C^{1}_{\#}([0,R])\times [0,1] \to C^{1}_{\#}([0,R]) $$ is a completely continuous operator (see for example \cite{Maw}). In order to give the appropriate definition for the sets $A$ and $B$, we need some estimates on {\bf n}. Let $\delta$ be given by (\ref{delta}) and $\bar d\leq\delta$ sufficiently small. \smallskip \noindent {\it Claim 1 : }There exists $n^*\in {\mathbb{N}}$ such that for any solution $u$ of $(P_{\lambda})$ we have $$ |u(0)|={\bar d}\quad \Longrightarrow \quad {\bf n}(u)0$ there exists $d_N>0$, $d_N< {\bar d}$, such that for any solution $u\in S_{d,\lambda}$ (for some $d$) we have $$ |u(0)|\leq d_N\Longrightarrow \quad {\bf n}(u)>N. $$ \noindent Let us consider $u\in S_{d,\lambda}$. We observe that for every $N>0$ there exists $M(N)> 2^{2(2k-1)}$ such that for all $|s|\leq \varepsilon_0$ \begin{equation} \label{asterisco1} \frac{1}{\sqrt{M(N)}} \int_0^s \frac{du}{{\cal L}^{-1}(s^2-u^2)} < \frac{1}{N}. \end{equation} Let $\tilde M(N):=M(N)+\frac{C}{a_{0}}$. By $(H_f)$, there is $\eta:=\eta_{\tilde M(N)}$ such that for all $r\in [0,R]$, all $0<|s|\leq \eta$ and all $\lambda \in [0,1]$, \begin{equation} \label{asterisco4} |f_{\lambda}(r,s)|>\tilde M(N) |s|. \end{equation} Let $\varepsilon_N\leq\min\{\eta,\bar d\}$. From Lemma $\ref{dipcont}$, we can consider $d_N>0$ small enough such that $$ |u(0)|\leq d_N\quad \Longrightarrow \quad ||u||_1\leq \varepsilon_N. $$ \noindent Now, let $(u,\lambda) \in \Sigma$ with $|u(0)|\leq d_N$. The equation in $(P_{\lambda})$ can be written as \begin{equation} \label{sistema} \begin{array}{c} u'=\displaystyle\frac{y}{r^{\lambda (k-1)}},\\ \\ y'=-r^{\lambda (k-1)}(f_{\lambda}(r,u)a(u')-\lambda h(r,u,u')). \end{array} \end{equation} We shall be concerned with the zeros $\{r_i\}_{i=1,\ldots ,I}$ of $u$ in the interval $[R/2,R]$. More precisely, we first estimate the distance between two successive zeros $r_{i}$ and $r_{i+1}$ of $u$ in the case when $$ u'(r_i)>0,\quad u'(r_{i+1})<0,\quad\mbox{ and }\quad u(r)>0,\, \forall r\in (r_i,r_{i+1}). $$ From $(\ref{sistema})$ we infer that $y'(r)<0$ for every $r\in (r_i,r_{i+1})$. Since $y(r_i)>0$ and $y(r_{i+1})<0$, we deduce that there exists exactly one point $r^*\in (r_i,r_{i+1})$ such that $y(r^*)=0$ and again from $(\ref{sistema})$ it follows that \[ u'(r)>0,\quad \forall r\in (r_i,r^*),\quad u'(r)<0,\quad \forall r\in (r^*,r_{i+1})\quad \mbox{ and } \quad u'(r^*)=0. \] Let $A=(R/2)^{\lambda (k-1)}$ and $B=R^{\lambda (k-1)}$. Using $(H_h)$, (\ref{delta}) and (\ref{asterisco4}), we observe that $$ f_{\lambda}(r,u)-\frac{\lambda h(r,u,u')}{a(u')}\geq \left(\tilde M(N)-\frac{C}{a_{0}}\right)u=M(N)u. $$ Hence for $r\in (r_i,r^*)\subset [R/2,R]$, we get \begin{equation} \label{sistema2} \begin{array}{c} u'\leq \frac{y}{A},\\ \\ y'\leq -A M(N) u a\left(\frac{y}{B}\right). \end{array} \end{equation} Multiplying the first inequality in $(\ref{sistema2})$ by $\frac{A}{B}M(N) u$ and the second one by $\frac{y}{AB a(\frac{y}{B})}$ and adding up, we obtain \[ \frac{A}{B}M(N)uu' + \frac{yy'}{a(\frac{y}{B}) AB}\leq\frac{M(N)uy}{B} - \frac{M(N)uy}{B} = 0. \] This means that the function $\frac{A}{B}M(N)\frac{u^2}{2} + \frac{B}{A}{\cal L}(\frac{y}{B})$ is non-increasing in $(r_i,r^*)$. Hence, setting $u^*:=u(r^*)$, we obtain \[ \frac{A}{B}M(N)\frac{u(r)^2}{2} + \frac{B}{A}{\cal L}\left(\frac{y(r)}{B}\right) \geq \frac{A}{B}M(N)\frac{(u^*)^{2}}{2} \] which implies for all $r\in (r_i,r^*)$, \begin{eqnarray*} u'(r) &\geq& \frac{B}{r^{\lambda (k-1)}}{\cal L}^{-1} \left(\frac{M(N)A^2}{2B^{2}}((u^*)^2 - u^2(r))\right)\\ &\geq& {\cal L}^{-1} \left(\frac{M(N)A^2}{2B^{2}}((u^*)^2 - u^2(r))\right). \end{eqnarray*} Finally, using Proposition $\ref{stimastar}$ with $c=\frac{M(N)A^2}{2B^2}$, $c_1=(1/2)^{2k-1}\sqrt{M(N)}$, we have for all $r\in (r_i,r^*)$, $$ u'(r)\geq (1/2)^{2k-1}\sqrt{M(N)}{\cal L}^{-1}((u^*)^2 - u^2(r))) $$ (notice that with the above choices $c_1>1$ and $c>c_1^2+1$). Integrating from $r_i$ to $r^*$, we get \[ \int_{r_i}^{r^*} \frac{u'(r)}{(1/2)^{2k-1}\sqrt{M(N)}{\cal L}^{-1}((u^*)^2 - u(r)^2)}\, dr \geq r^*-r_i. \] If we set $u(r)=u$, then using (\ref{asterisco1}) we obtain $$ r^*-r_i\leq \int_{0}^{u^*} \frac{du}{(1/2)^{2k-1}\sqrt{M(N)}{\cal L}^{-1}((u^*)^2 - u^2)}<\frac{2^{2k-1}}{N}. $$ For the completion of the proof of the Claim, it is now sufficient to observe that a computation analogous to the one developed above can be performed if we consider the interval $(r^*,r_{i+1})$ or an interval $(r_i,r_{i+1})$ where $u$ is negative. Now, let $n_0=\max(n^*,2 k_0)$ (for the definition of $k_0$, see Theorem $\ref{autonomo}$). Next, let us consider $n > n_0$ and the number $d_n$ arising from Claim 2. In order to prove the existence of the solutions with exactly $n$ zeros by an application of Theorem $\ref{continuazione}$, we introduce the sets $$ B = \{(u,\lambda) \in {\rm dom}\ {\cal N} : u(0) < \bar d \} $$ ($\bar d$ as in Lemma \ref{energia} and Claim 1) and $$ A_n = \{(u,\lambda) \in {\rm dom}\ {\cal N} : u(0) < d_n \}. $$ Moreover, the functional $$ {\bf k}: \Sigma \cap (\bar B \setminus A_n) \to {\mathbb{N}} $$ will be defined by $$ {\bf k}(u,\lambda)={\bf n}(u). $$ Let us now prove that conditions $(\ref{gap1})$ and $(\ref{gap2})$ are satisfied. We observe that \[ \partial(\bar B \setminus A_n)= \{(u,\lambda): u(0)=\bar d \} \cup \{(u,\lambda): u(0)=d_n \}. \] If $(u,\lambda) \in \Sigma$ and $u(0)=d_n$ then, by Claim 2, we get ${\bf n}(u) > n$; on the other hand, if $(u,\lambda) \in \Sigma$ and $u(0)=\bar d$ then, by Claim 1, we have ${\bf n}(u) < n^*$. Hence, being $n^* < n$, condition $(\ref{gap1})$ is satisfied. As far as the boundedness of ${\bf k}^{-1}(n)$ is concerned, if $(u,\lambda) \in {\bf k}^{-1}(n) \subset \Sigma \cap (\bar B \setminus A_n)$, then $u(0) < \bar d$ and we deduce from Lemma $\ref{dipcont}$ that $u$ is bounded, hence $(\ref{gap2})$ is fulfilled. Now we have to choose an open set on which to compute the degree. From Theorem \ref{autonomo}, we know that problem $(\ref{autonomous})$ has solutions with exactly $n$ zeros in $[0,R)$. These solutions enable us to determine an open bounded set $\Omega_0$ such that $$ ({\bf k}^{-1}(n))_0\subset \Omega_0. $$ We define $$ U^n_0=\Omega_0 \cap (\bar B \setminus A_n). $$ Arguing as in \cite{CDZ}, we can prove that the degree $\deg(I-{\cal N}_0,U^n_0)$ is well defined and $\deg(I-{\cal N}_0,U^n_0) \not= 0$. Hence, an application of Theorem $\ref{continuazione}$ provides the existence of a solution $u_n$ of problem $(\ref{ODE})$ with $$ {\bf n}(u_n)= n\quad \mbox{and}\quad u_n(0)>0. $$ Moreover, this solution $u_n$ is such that $||u_n||_1\leq \varepsilon_0$. A similar argument, considering the sets $$ B = \{(u,\lambda) \in {\rm dom}\ N : u(0) > -\bar d \} $$ and $$ A_n = \{(u,\lambda) \in {\rm dom}\ N : u(0) > -d_n \} $$ shows that there exists at least one solution $v_n$ of $(\ref{ODE})$ such that $$ {\bf n}(v_n)= n\quad \mbox{and} \quad v_n(0)<0. $$ The last statement in Theorem $\ref{main}$ follows from the properties of ${\bf n}$. \hfill$\Box$\smallskip \paragraph{Acknowledgments} The first author wishes to thank the Organizers of the meeting ``USA-Chile Workshop on Nonlinear Analysis'' for the invitation and the hospitality. \begin{thebibliography}{00} \bibitem{AGP} {\sc A. Ambrosetti - J. Garcia Azorero - I. 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Walter}, {\it Radial solutions of equations and inequalities involving the $p$-laplacian}, J. Inequal. and Appl., {\bf 1} (1997), pp. 47-71. \end{thebibliography} \noindent{\sc Anna Capietto }\\ Dipartimento di Matematica - Universit\`a di Torino \\ Via Carlo Alberto 10 - 10123 Torino - Italy \\ e-mail: capietto@dm.unito.it \smallskip \noindent{\sc Marielle Cherpion }\\ D\'epartement de Math\'ematique - Universit\'e Catholique de Louvain \\ Chemin du Cyclotron, 2 - B-1348 Louvain-la-Neuve - Belgium \\ e-mail: cherpion@amm.ucl.ac.be \end{document}