\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 $$\label{problema} \begin{array}{c} (r^{(k-1)}u')'+r^{(k-1)}a(u')f(r,u)= r^{(k-1)}h(r,u,u'),\\ u'(0)= 0 = u(R) \end{array}$$ ($k>1$). As it is well-known, solutions to $(\ref{problema})$ are radially symmetric solutions to the following elliptic boundary-value problem on a ball ${\cal B}=B(0,R)$ $$\label{BVP} \begin{array}{c} \nabla \cdot (\nabla u)+a(|\nabla u|) f(|x|,u)=h(|x|,u,|\nabla u|)\quad \mbox{in }{\cal B},\\ u=0\quad \mbox{on } \partial {\cal B}. \end{array}$$ 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 $$\lim_{s\to 0}\frac{h(r,s,\xi)}{s}=C(\xi)\mbox{ uniformly in }r\in [0,R].$$ \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 $$\label{auto*} \begin{array}{c} u''+a(u')g(u)=0,\\ u'(0) = 0 = u(R), \end{array}$$ 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 $$\label{phiautonomous} \begin{array}{c} (\phi(u'))' + g(u) = 0,\\ u'(0) = 0 = u(R), \end{array}$$ 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 $$\label{general} \begin{array}{c} (r^{(k-1)} \psi(u'))'+r^{(k-1)}a(u')f(r,u)=r^{(k-1)}h(r,u, u'),\\ u'(0) = 0 = u(R), \end{array}$$ 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 $$\label{autonomous} \begin{array}{c} u''+a(u')g(u)=0,\\ u'(0) = 0 = u(R), \end{array}$$ where $a:\mathbb{R} \to \mathbb{R}$ is defined by $a(\xi):=a_{0} + |\xi|^q, 0 0$. Set $$\label{phi} \phi(s)=\int_0^s \frac{1}{a(x)}\, dx.$$ We assume that $g : [-\varepsilon, \varepsilon] \to \mathbb{R}$ is continuous $(\varepsilon > 0)$ and such that $$\label{refA} \lim_{s\to 0} \frac{g(s)}{\phi(s)}=+\infty.$$ 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 $$\label{equivsys} \begin{array}{c} u' = \phi^{-1}(y),\\ y' = -g(u). \end{array}$$ More precisely, for $$\label{defL} {\cal L}(\xi) =\int_0^{\xi} \frac{x}{a(x)}\, dx,$$ 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 $$\label{tm1} T_1(\alpha)=\int_0^{\alpha} \frac{dx}{{\cal L}^{-1}(G(\alpha)-G(x))}.$$ 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 $$\label{tm2} T_2(\alpha)=\int_{\alpha_1}^0 \frac{dx}{{\cal L}^{-1}(G(\alpha)-G(x))}$$ 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 $$\label{greater} c_1 {\cal L}^{-1}(\xi)\leq {\cal L}^{-1}(c \xi).$$ \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 $$\label{ODE} \begin{array}{c} (r^{(k-1)}u')'+r^{(k-1)}a(u')f(r,u)= r^{(k-1)}h(r,u,u'),\\ u'(0)= 0 = u(R), \end{array}$$ 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 $$\label{refC} \lim_{s\to 0}\frac{h(r,s,\xi)}{s}=C(\xi)\mbox{ uniformly in }r\in [0,R].$$ \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 $$\label{fl} f_{\lambda}(r,s)=\lambda f(r,s)+(1-\lambda) g(s),$$ 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 $\lambda \in [0,1]$, $$\label{Flandapositiva} F_{\lambda}(r,s)>0.$$ 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 $$\label{cauchy} \begin{array}{c} (r^{\lambda (k-1)} u')'+r^{\lambda (k-1)}a(u')f_{\lambda}(r,u)= \lambda r^{\lambda(k-1)}h(r,u,u'),\\ u(0)=d,\ u'(0)=0. \end{array}$$ \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 $$\label{defvlanda} v_{\lambda}(r):=E_{\lambda}(r,u(r),u'(r)).$$ 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 $$\label{vlanda} {\cal L}(u'(r))\leq v_{\lambda}(r)\leq F_{\lambda}(0,d)e^{\int_0^r \tilde\alpha(s)\, ds}.$$ 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 $$\label{aunoadue} a_1+Ra_2\leq {1\over 2} \quad\mbox{and}\quad a_2\leq {1\over 2}$$ (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 $r\in [0,\rho]$, $$\label{uprimo} |u'(r)|\leq {\cal L}^{-1}(F_{\lambda}(0,d)e^{\int_0^r \tilde\alpha(s)\, ds})\leq a_2\varepsilon\leq \frac{\varepsilon}{2}.$$ The above estimate implies that for all $r\in [0,\rho]$, $$\label{ustima} \begin{array}{rcl} |u(r)|&\leq&\displaystyle d+\int_0^r |u'(s)|\, ds\leq d+R {\cal L}^{-1}( F_{\lambda}(0,d)e^{\int_0^r \tilde\alpha(s)\, ds})\\ &\leq& a_1\varepsilon + R a_2\varepsilon=(a_1+R a_2)\varepsilon\leq\frac{\varepsilon}{2}. \end{array}$$ 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 $0<\tilde\varepsilon0. Hence if$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 $${\hat f}(s)=\left\{ \begin{array}{ll} \sup\{f(r,x),\, r\in [0,R],\, x\in [0,s]\}&\mbox{if }00. If \bar d is small enough, we have for every s\in [0,r_0(\bar d)] that 00 such that if u is a solution of (P_{\lambda}) with |u(0)|=\bar d sufficiently small, we have E_{\lambda}(r,u(r),u'(r))\geq \delta_0 for every r\in [0,R]. First, we observe that, by (H_{F_{\lambda}}), there is a constant \gamma such that for all r\in (0,R], all s\in [-\epsilon_0,\epsilon_0] and all \lambda \in[0,1], $$\label{gammasuper} {{\partial F_{\lambda}}\over {\partial r}}(r,s)+{{\gamma }\over r}F_{\lambda}(r,s)\geq 0.$$ Recall that E_{\lambda}(r,s,\xi):=F_{\lambda}(r,s)+{\cal L}(\xi). Using (H_{h}) and Proposition \ref{propL}, \begin{eqnarray*} \lefteqn{ \frac{d}{dr} E_{\lambda}(r,u(r),u'(r))+\frac{\gamma}{r}E_{\lambda}(r,u(r),u'(r)) }\\ &=& \frac{\partial F_{\lambda}}{\partial r}(r,u(r))+\frac{\gamma}{r}F_{\lambda}(r,u(r))+\frac{u'(r)}{a(u'(r))}\lambda h(r,u(r),u'(r))+\frac{\gamma}{r}{\cal L}(u'(r))\\ &&-\lambda\frac{(k-1)}{r} \frac{(u'(r))^{2}}{a(u'(r))}\\ &\geq&\frac{u'(r)}{a(u'(r))}\lambda h(r,u(r),u'(r))+\frac{\gamma}{r}{\cal L}(u'(r))-\frac{2(k-1)}{r}{\cal L}(u'(r))\\ &\geq& -H\frac{(u'(r))^{2}}{a(u'(r))}+\frac{\gamma}{r}{\cal L}(u'(r))-\frac{2(k-1)}{r}{\cal L}(u'(r))\\ &\geq& {\cal L}(u'(r))\left(-2H+\frac{\gamma}{r}-\frac{2(k-1)}{r}\right) \geq 0 \end{eqnarray*} if \gamma\geq 2HR+2(k-1). Multiplying by r^{\gamma} and integrating from r_{\theta}(\bar d) to r, we obtain$$ 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 $$\label{gap1} n \notin {\bf k}(\partial (\bar B \setminus A))$$ and $$\label{gap2} {\bf k}^{-1}(n) \quad \mbox{is bounded}.$$ 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 $$\label{gradonozero} \deg (I-{\cal N}_0,U^n_0)\not= 0,$$ 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. $$\noindent Let us consider u\in S_{d,\lambda}. We observe that for every N>0 there exists M(N)> 2^{2(2k-1)} such that for all |s|\leq \varepsilon_0 $$\label{asterisco1} \frac{1}{\sqrt{M(N)}} \int_0^s \frac{du}{{\cal L}^{-1}(s^2-u^2)} < \frac{1}{N}.$$ Let \tilde M(N):=M(N)+\frac{C}{a_{0}}. By (H_f), there is \eta:=\eta_{\tilde M(N)} such that for all r\in [0,R], all 0<|s|\leq \eta and all \lambda \in [0,1], $$\label{asterisco4} |f_{\lambda}(r,s)|>\tilde M(N) |s|.$$ Let \varepsilon_N\leq\min\{\eta,\bar d\}. From Lemma \ref{dipcont}, we can consider d_N>0 small enough such that$$ |u(0)|\leq d_N\quad \Longrightarrow \quad ||u||_1\leq \varepsilon_N. $$\noindent Now, let (u,\lambda) \in \Sigma with |u(0)|\leq d_N. The equation in (P_{\lambda}) can be written as $$\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}$$ We shall be concerned with the zeros \{r_i\}_{i=1,\ldots ,I} of u in the interval [R/2,R]. More precisely, we first estimate the distance between two successive zeros r_{i} and r_{i+1} of u in the case when$$ 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. $$Hence for r\in (r_i,r^*)\subset [R/2,R], we get $$\label{sistema2} \begin{array}{c} u'\leq \frac{y}{A},\\ \\ y'\leq -A M(N) u a\left(\frac{y}{B}\right). \end{array}$$ Multiplying the first inequality in (\ref{sistema2}) by \frac{A}{B}M(N) u and the second one by \frac{y}{AB a(\frac{y}{B})} and adding up, we obtain $\frac{A}{B}M(N)uu' + \frac{yy'}{a(\frac{y}{B}) AB}\leq\frac{M(N)uy}{B} - \frac{M(N)uy}{B} = 0.$ This means that the function \frac{A}{B}M(N)\frac{u^2}{2} + \frac{B}{A}{\cal L}(\frac{y}{B}) is non-increasing in (r_i,r^*). Hence, setting u^*:=u(r^*), we obtain $\frac{A}{B}M(N)\frac{u(r)^2}{2} + \frac{B}{A}{\cal L}\left(\frac{y(r)}{B}\right) \geq \frac{A}{B}M(N)\frac{(u^*)^{2}}{2}$ which implies for all r\in (r_i,r^*), \begin{eqnarray*} u'(r) &\geq& \frac{B}{r^{\lambda (k-1)}}{\cal L}^{-1} \left(\frac{M(N)A^2}{2B^{2}}((u^*)^2 - u^2(r))\right)\\ &\geq& {\cal L}^{-1} \left(\frac{M(N)A^2}{2B^{2}}((u^*)^2 - u^2(r))\right). \end{eqnarray*} Finally, using Proposition \ref{stimastar} with c=\frac{M(N)A^2}{2B^2}, c_1=(1/2)^{2k-1}\sqrt{M(N)}, we have for all r\in (r_i,r^*),$$ 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. 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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}