\documentclass[reqno]{amsart} \usepackage{graphicx} \AtBeginDocument{{\noindent\small {\em Electronic Journal of Differential Equations}, Vol. 2004(2004), No. 110, pp. 1--16.\newline ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu \newline ftp ejde.math.txstate.edu (login: ftp)} \thanks{\copyright 2004 Texas State University - San Marcos.} \vspace{9mm}} \begin{document} \title[\hfilneg EJDE-2004/110\hfil Uniqueness for degenerate elliptic problems] {Uniqueness for degenerate elliptic sublinear problems in the absence of dead cores} \author[J. Garc\'{\i}a-Meli\'an\hfil EJDE-2004/110\hfilneg] {Jorge Garc\'{\i}a-Meli\'an} \address{Jorge Garc\'{\i}a-Meli\'an\hfill\break Departamento de An\'alisis Matem\'atico\\ Universidad de La Laguna \\ 38271 La Laguna - Tenerife, Spain} \email{jjgarmel@ull.es} \date{} \thanks{Submitted July 5, 2004. Published September 21, 2004.} \thanks{Supported by contract PB96-0621 from DGES} \subjclass[2000]{35J60, 35J70} \keywords{p-Laplacian; linearization; uniqueness} \begin{abstract} In this work we study the problem $$-\mathop{\rm div}(|\nabla u|^{p-2}\nabla u)=\lambda f(u)$$ in the unit ball of $\mathbb{R}^N$, with $u=0$ on the boundary, where $p>2$, and $\lambda$ is a real parameter. We assume that the nonlinearity $f$ has a zero $\bar{u}_0>0$ of order $k\ge p-1$. Our main contribution is showing that there exists a unique positive solution of this problem for large enough $\lambda$ and maximum close to $\bar{u}_0$. This will be achieved by means of a linearization technique, and we also prove the new result that the inverse of the $p$-Laplacian is differentiable around positive solutions. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{corollary}[theorem]{Corollary} \newtheorem{lemma}[theorem]{Lemma} \newtheorem{remark}[theorem]{Remark} \section{Introduction} In this paper we are concerned with the nonlinear eigenvalue problem $$\label{problema} \begin{gathered} -\Delta_p u=\lambda f(u) \quad \mbox{in } \Omega \\ u=0 \quad\mbox{on } \partial \Omega, \end{gathered}$$ where $\Delta_p u=\mathop{\rm div}(|\nabla u|^{p-2}\nabla u)$, $p>2$, stands for the $p$-Laplacian operator, $\Omega\subset\mathbb{R}^N$ is a bounded domain, $\lambda$ a real parameter and $f$ a $C^1$ function with a positive zero $\bar{u}_0$ (see hypotheses (H) below). In the semilinear case $p=2$ (where $\Delta_p$ reduces to the usual Laplacian), problems like (\ref{problema}) have been widely considered in the literature. An important number of works (cf. for instance \cite{Ag,CS,SW1,SW2,S} and references therein) deal with nonlinearities $f(u)$ with a positive zero $\bar{u}_0$, and their interest is focused on positive solutions $u$ with $u\le \bar{u}_0$ and $\max u$ close to $\bar{u}_0$. The important matter is then to show that such solutions are unique for large $\lambda$, and to ascertain their qualitative behaviour as $\lambda\to+\infty$. The results obtained in the semilinear case heavily rely on the use of linearization around positive solutions. However, when trying to use the same tools with problems like (\ref{problema}) we encounter an important difficulty: the formal linearization of $\Delta_p$ around a solution $u$ becomes degenerate at points where $\nabla u$ vanishes. Since it is a really hard task for the moment to locate the set of critical points of positive solutions, we are restricting our attention to a symmetric situation, where $\Omega=B$, the unit ball in $\mathbb{R}^N$. That is, we will consider $$\label{eP} \begin{gathered} -\Delta_p u=\lambda f(u) \quad\mbox{in } B\\ u=0 \quad\mbox{on } \partial B. \end{gathered}$$ In this setting -- actually the more general one of rotationally invariant domains of $\mathbb{R}^N$ -- some problems slightly more general than \eqref{eP} were studied in \cite{GS2}. Concretely, the function $f$ was allowed to depend on $\lambda$: $$\label{LP} \begin{gathered} -\Delta_p u=\lambda f(\lambda,u) \quad\mbox{in } B\\ u=0 \quad\mbox{on } \partial B. \end{gathered}$$ The main assumption on $f$ was the existence of a zero $\bar{u}_0>0$ of order $k0$. \item[(H5)] $f$ has a finite number of zeros in the interval $[0,\bar{u}_0]$. \end{itemize} Note that condition (H3) is necessary in order to have a family of solutions $\{u_\lambda\}$ with $\max u_\lambda\to\bar{u}_0$ as $\lambda\to+\infty$. Our main result can be stated in the following way: \begin{theorem}\label{unicidad} Assume $f$ verifies hypotheses (H). Then there exist $\eta_0>0$, $\lambda^*>0$ such that the problem \eqref{eP} \begin{gather*} -\Delta_p u=\lambda f(u) \quad\mbox{in } B\\ u=0 \quad\mbox{on } \partial B \end{gather*} has a unique positive solution $u_\lambda$ with $\bar{u}_0-\eta_0\le \max u_\lambda<\bar{u}_0$, if $\lambda\ge\lambda^*$. Moreover, $u_\lambda$ is a radial function, the family $\{ u_\lambda\}$ verifies condition (\ref{homog}), and we obtain the following exact estimate for the boundary layer near $\partial B$: $$\label{BL} \lim_{\lambda\to+\infty}\lambda^{-1/p}u'_\lambda(1)=-(p'F(\bar{u}_0))^{1/p}\,,$$ where $F(\bar{u}_0)=\int_0^{\bar{u}_0}f(s)ds$. \end{theorem} \begin{remark}\label{principal} {\rm (a) Condition (H2) on $f$ guarantees that solutions with $\max u\le\bar{u}_0$ also verify $\max u<\bar{u}_0$ (cf. \cite{GS1}), that is, dead cores do not arise even for large $\lambda$. This case is complementary to the one treated in \cite{GS2}. (b) The results in Theorem \ref{unicidad} are also valid when problem \eqref{eP} in considered in an annulus. This will be shown elsewhere. (c) The boundary layer estimate (\ref{BL}) can be shown to be valid even for general domains $\Omega$ (without any knowledge of uniqueness). See \cite{GS1} and \cite{GS2} for related situations. (d) The symmetry of solutions for problems like \eqref{eP} will play an important r\^ole. We refer to \S 2 for details. } \end{remark} This paper is organized as follows: section 2 is devoted to some symmetry considerations. In section 3, after proving the existence of a family of positive solutions, we obtain some precise estimates for all possible solutions. Sections 4 and 5 form the core of the paper: in \S 4 we show that it is possible to linearize problem \eqref{eP} around positive solutions, using this fact in \S 5 to prove uniqueness. \section{Symmetry of solutions} In the semilinear case $p=2$, a well known theorem by Gidas, Ni and Nirenberg (see \cite{GNN}) asserts that positive solutions to \eqref{eP} are radially symmetric. Some attempts to generalize this result to degenerate operators have been made for instance in \cite{BN} and \cite{KP}. However, the possible presence of dead cores in the solutions prevents one to expect a quite general symmetry result. In a recent paper of Brock (\cite{Br}) an almost complete answer to this problem has been given. Nevertheless it is necessary to impose additional assumptions on $f$ in order to obtain symmetry of positive solutions. We are showing in this section how to use the results in \cite{Br} to obtain that all positive solutions to \eqref{eP} in our setting are radially symmetric, without further conditions on the nonlinear term $f$. The following result contains Lemmas 1 and 4, and Remark 1 in \cite{Br}. \begin{lemma}\label{Brock} Let $u>0$ be a solution to \eqref{eP}. Then there exists $m\in \mathbb{N}\cup\{+\infty\}$ so that $B$ admits a decomposition: $$B=\cup_{k=1}^m C_k\cup \{x: \nabla u(x)=0\},$$ where $C_k=B_{R_k}(z_k)\setminus \overline{B_{r_k}(z_k)}$, for certain $z_k\in B$ and $0\le r_k0$ then $f(u)=0$ on $\partial B_{r_k}(z_k)$. \end{lemma} \begin{remark} {\rm This property of the solutions $u$ to \eqref{eP} is called {\it local symmetry.} } \end{remark} By means of this result, we are defining, for an arbitrary positive solution $u$ to \eqref{eP}, a {\it rearrangement} $u^*$, that is, a positive radial solution $u^*$, with $\max u^*=\max u$. Let $u$ be a positive solution to \eqref{eP}. We claim that $\max u$ is attained at some $z_k$ ($z_k$ as in Lemma \ref{Brock}). Indeed, choose $x_0\in B$ such that $u(x_0)=\max u$, and a sequence $x_j\in \partial C_j$ such that ${\rm dist}\ (x_j,x_0)\to \inf \mathop{\rm dist} (C_k,x_0)$. It is easy to see that the $\{x_j\}$ can be chosen in such a way that (passing through a subsequence if necessary) one of the following situations holds: either $x_j\to x_0$ or the segment $[x_j,x_0]$ is contained in the set $\{x:\nabla u(x)=0\}$. In the first case, we have $f(u(x_j))=0$ (since $x_j\in\partial C_j$) and it follows that $u(x_0)$ is an accumulation point of zeros of $f$ unless $u(x_j)=u(x_0)$ for infinitely many $j$'s. Our hypotheses then imply that -- passing through a subsequence -- $u(x_j)=u(x_0)$. If $x_j\in \partial B_{R_j(z_j)}$, we obtain a contradiction to (\ref{dernol}). Hence, $x_j\in \partial B_{r_j}(z_j)$. If $r_j=0$ for some $j$, we have $x_j=z_j$, as was to be proved. If, on the contrary, $r_j>0$, then $u(x)\ge u(x_j)$ in $B_{r_j}(z_j)$, and the maximum is attained in the whole $B_{r_j}(z_j)$, also showing the claim (notice in particular that $\{x_j\}$ reduces to a single point). In the second case, we also arrive at $u(x_0)=u(x_j)$ and the conclusion is the same. \begin{remark} \label{symm} {\rm The above reasoning also shows that if the maximum is achieved in $z_k$, then the solution $u$ is radial in $B_{R_k}(z_k)$, which is not clear from Lemma \ref{Brock} if $r_k>0$.} \end{remark} \begin{figure}[ht] \begin{center} \includegraphics[width=0.8\textwidth]{fig1} \end{center} \caption{A locally symmetric solution $u$ and its rearrangement $u^*$.} \end{figure} Without loss of generality, assume $u(z_1)=\max u$. In virtue of Lemma \ref{Brock}, $u_1=u_{|\partial B_{R_1}(z_1)}$ is a zero of $f$, and $u_1R_1$ and $\bar u=u_1$ on $\partial B_{r_2}(z_2)$. To this aim we are proceeding as follows: if there is a point $x\in C_k$, $k\neq 1$, such that $u(x)>u_1$ then set $\bar u=u_{|\partial B_{R_k}(z_k)}$ in $B_{R_k}(z_k)$. Otherwise define $\bar u=u$. Clearly, $\bar u\le u_1$ in $B\setminus B_{R_1}(z_1)$. Now take the sequences $x_j\in \partial B_{R_1}(z_1)$ and $\hat x_j\in \partial C_j$ in such a way that $\mathop{\rm dist}(x_j,\hat x_j) \to\inf_{k\ne 1} \mathop{\rm dist}\ (C_k,\partial B_{R_1}(z_1))$. As before, it is possible to choose these sequences such that two options may arise: $|x_j-\hat x_j|\to 0$ or $[x_j,\hat x_j]\subset \{x:\nabla u(x)=0\}$. Both of them lead to $\bar u(\hat x_j)=\bar u(x_j)=u_1$ for a subsequence, and $\hat x_j\in \partial B_{r_j}(z_j)$. Thus, $\bar u=u_1$ in $B_{r_j}(z_j) \setminus B_{R_1}(z_1)$. As a conclusion, the sequence $\{\hat x_j\}$ reduces to a point, and $r_j>R_1$, $\bar u=u_1$ on $\partial B_{r_j}(z_j)$. Assume $j=2$. Repeating the above procedure, we arrive, after a finite number of steps, at $u_l=0$ for some $l\in\mathbb{N}$. Denoting by $\chi$ the characteristic function of a set, we define: $$u^*(x)=\sum_{i=1}^l \bar u(x+z_i)\chi_{\{R_{i-1}\le |x| 0, \lambda_0>0 such that for every \lambda\ge \lambda_0, problem \eqref{eP} admits at least a radial positive solution u_\lambda verifying \bar{u}_0-\eta\le\max u_\lambda<\bar{u}_0. In addition the family \{u_\lambda\} satisfies (\ref{homog}) and u_\lambda'(r)<0 for r>0. \end{theorem} \begin{proof} Let us show that there exists \eta>0 such that for the solution u=u(\cdot,u_0) to the problem (\ref{cauchy}) with \bar{u}_0-\eta\le u_0<\bar{u}_0 there exists T>0 with u(T)=0. First we claim that for every \varepsilon_0<\varepsilon (\varepsilon as in hypothesis (H4) on f), and R>0, there exists \eta=\eta(\varepsilon_0,R) such that u(r,u_0) is defined in [0,R], u'(r,u_0)<0 and u(r,u_0)\ge \bar{u}_0-\varepsilon_0 if 0< r\le R whenever \bar{u}_0-\eta\le u_0< \bar{u}_0. To prove the claim, assume first that there exist sequences u_{0n}\to \bar{u}_0- and r_n\le R such that u'(r_n,u_{0n})=0. Since$$ u'(r,u_{0n})=-\varphi_{p'}\Big(\int_0^r \big(\frac \rho r\big)^{N-1} f(u(\rho,u_{0n}))\,d\rho\Big)\,, $$and f>0 in [\bar{u}_0-\varepsilon,\bar{u}_0), it follows easily that u(r_n,u_{0n})<\bar{u}_0-\varepsilon. Thus, there exists \hat r_n0 such that u(r,u_0) is defined in [0,R] and u'(r,u_0)<0 in [0,R] when \bar{u}_0-\eta\le u_0< \bar{u}_0. The remaining part of the claim is proved {\it exactly} in the same way. Now notice that positive solutions to (\ref{cauchy0}) together with their derivatives are uniformly bounded in r\ge 0. Indeed, multiplying the equation by u' and integrating in [0,r] we arrive at the identity$$ |u'(r)|^p+p'({N-1})\int_0^r\frac{|u'(s)|^p}s ds =p'(F(u_0)-F(u(r))), $$so that |u'(r)|^p\le p'(F(u_0)-F(u(r))) if r\ge 0. Choose \tau>0 and R>0 to achieve$$ ({N-1}) \frac{|u'(r)|^{p-1}}r\le \tau,\quad r\ge R. $$Thus, -\varphi_p(u')'=f(u)+({N-1})\varphi_p(u')/r\ge f(u)-\tau  if r\ge R and, as long as u'\le 0, we obtain -\varphi_p(u')'u'\le (f(u)-\tau )u', that is, $$\label{fep} \left(|u'(r)|^p+p'F_\tau(u(r))\right)'\ge 0, \quad r\ge R\,,$$ where F_\tau(u)=F(u)-\tau u, and F(u)=\int_0^u f(s)ds. Notice that f-\tau has a zero \bar{u}_0(\tau)<\bar{u}_0 with \bar{u}_0(\tau)\to\bar{u}_0 as \tau\to 0, verifying in addition the energy condition F_\tau(u)< F_\tau(\bar u), 0\le u<\bar u, if \bar u\in [\bar{u}_0(\tau)-\delta(\tau),\bar{u}_0(\tau)], for some \delta(\tau)>0. Choose \varepsilon_0,\ \tau small so that \bar{u}_0(\tau)-\delta(\tau)<\bar{u}_0-\varepsilon_0. As the claim at the beginning of the proof shows, we have \bar{u}_0(\tau)-\delta(\tau)\le u(R,u_0)<\bar{u}_0. Thus if r\ge R, we have, in virtue of (\ref{fep}), |u'(r)|^p \ge |u'(R)|^p+p' (F_\varepsilon(u(R))-F_\varepsilon(u(r))), for r\ge R. In particular, u'(r)<0 always holds, and then$$ u'(r)\le u'(R), \quad r\ge R\,. $$Integrating this inequality,$$ u(r)\le u(R)+u'(R)(r-R), \quad r\ge R\,, $$and we conclude that u has to vanish for some T=T(u_0). We have defined in this way a continuous mapping T:[\bar{u}_0-\eta,\bar{u}_0) \to R^+. To complete the proof of the Theorem it suffices to show that T(u_0)\to+\infty as u_0\to \bar{u}_0- (then the construction of the family of solutions \{u_\lambda\}_{\lambda\ge\lambda_0} is performed in a standard way). Assume on the contrary that there exists a sequence u_{0n}\to\bar{u}_0 such that T(u_{0n}) is bounded. Without loss of generality, we can assume T(u_{0n}) \to T_0>0, and it follows that u_n:=u(\cdot,u_{0n})\to \bar{u}_0 uniformly in [0,T_0] as seen before. This contradicts the fact that u(T(u_{0n}),u_{0n})=0, finally proving the theorem. \end{proof} \begin{corollary}\label{symmetry} Let \lambda_0, \eta be as in Theorem \ref{existencia}. Then every positive solution u to \eqref{eP} with \bar{u}_0-\eta\le\max u\le\bar{u}_0 and \lambda\ge \lambda_0 is radial. \end{corollary} \begin{proof} Let u be a positive solution to \eqref{eP} with \bar{u}_0-\eta\le\max u\le\bar{u}_0 and \lambda\ge\lambda_0, where \eta, \lambda_0 are as in Theorem \ref{existencia}. Consider the function u^* defined in \S 2. In virtue of Lemma \ref{radialidad}, u^* is a positive radial solution to \eqref{eP} and \max u^*=\max u. Since the function v(r)=u^*(\lambda^{-1/p}r) solves \begin{gather*} - (r^{N-1} \varphi_p(v'))' = r^{N-1} f(v) \\ v(0)=\max u\,, \quad u'(0)=0\,, \end{gather*} it follows by the uniqueness of this Cauchy problem that u^*=u_{\mu} for some \mu\ge \lambda_0, where \{u_\mu\} is the family of functions given by Theorem \ref{existencia}. Thus {u^*}'(r)<0 for r>0, and Lemma \ref{radialidad} implies that u=u^*, and u is a radial function. \end{proof} In the remaining part of the section we are obtaining estimates for the positive solutions u to \eqref{eP} with \max u close to \bar{u}_0 and large \lambda. First of all we are constructing a subsolution of \eqref{eP}, using ideas from \cite{CS}. To this aim, we are redefining f outside [0,\bar{u}_0]. More precisely, we can assume with no loss of generality that f is bounded, f<0 in [\bar{u}_0,+\infty), f=0 in (-\infty,-1] and F(u)\lambda_0 such that the solution u_{\lambda_1} given by Theorem \ref{existencia} satisfies u_{\lambda_1}(0)=\bar{u}_0-\varepsilon/2. Thanks to the condition verified by F (and diminishing \varepsilon again if necessary), we can produce u_{\lambda_1} to reach a value r_0>1 such that u_{\lambda_1}(r_0)=-1, and u_{\lambda_1}'(r)<0 if r\in (0,r_0]. Moreover, since f=0 in (-\infty,-1], u_{\lambda_1} satisfies \begin{gather*} -(r^{N-1}\varphi_p(u'))'= 0,\quad r>r_0\\ u(r_0)=-1\,, \quad u_{\lambda_1}'(r_0)<0\,, \end{gather*} that is$$ u(r)=\begin{cases} -1+u_{\lambda_1}'(r_0)\frac{r_0^\theta r^{1-\theta}-r_0}{1-\theta}, & p\neq N\\ -1+u_{\lambda_1}'(r_0)r_0\log\big(\frac r{r_0}\big), & p=N\,, \end{cases} $$for r\ge r_0, with \theta=(N-1)/(p-1). In particular, u_{\lambda_1}(r) < 0 if r> 1. This function will allow us to obtain a subsolution to problem \eqref{eP}. \begin{lemma} Let u_{\lambda_1} be as before, and define$$ z_\lambda(r)=u_{\lambda_1}\Big(\big(\frac{\lambda}{\lambda_1}\big)^{1/p}r\Big)\,. $$Then z_\lambda is a subsolution to \eqref{eP} for \lambda > \lambda_1. \end{lemma} \begin{proof} Clearly, z_\lambda verifies the equation. Moreover, z_\lambda(1)=u_{\lambda_1}((\lambda/\lambda_1)^{1/p})<0, since u_{\lambda_1}(r) < 0 if r > 1. \end{proof} The existence of this subsolution is essential. Indeed, for large enough \lambda, every positive solution with maximum close to \bar{u}_0 lies above it. \begin{lemma}\label{estim} There exist 0<\eta_0\le\eta, \lambda_2>\lambda_1 such that every positive solution u to \eqref{eP} with \lambda\ge \lambda_2 and \bar{u}_0-\eta_0\le \max u<\bar{u}_0 verifies u\ge z_\lambda. \end{lemma} \begin{proof} In virtue of Corollary \ref{symmetry}, u is a radial solution. Moreover, as seen there, u=u_\mu for some \mu=\mu(\lambda). It is not hard to show that \mu(\lambda)\to +\infty as \lambda\to+\infty and \max u\to\bar{u}_0-. Thus, for \delta>0 fixed, there exist \bar\lambda, 0<\eta_0\le\eta such that \lambda\ge\bar\lambda and \bar{u}_0-\eta_0\le \max u<\bar{u}_0 imply u(r)\ge \bar{u}_0-\varepsilon/2, for every r\in [0,1-\delta] (this is a consequence of condition (\ref{homog})). Since \max z_\lambda= \bar{u}_0-\varepsilon/2, we obtain u\ge z_\lambda in [0,1-\delta]. In addition,$$ z_\lambda(r)\le z_\lambda(1-\delta)=u_{\lambda_1}\Big(\big(\frac \lambda{\lambda_1}\big)^{1/p} (1-\delta)\Big)\le 0\,, $$if \lambda \ge \lambda_1/(1-\delta)^p, r\ge 1-\delta. Hence, u_\lambda\ge z_\lambda in [1-\delta,1], and we can take \lambda_2=\max\{\lambda_1/(1-\delta)^p,\bar\lambda\}. This concludes the proof of the Lemma. \end{proof} \begin{lemma}\label{estim2} Let \eta_0, \lambda_2 be as in Lemma \ref{estim}. Then there exists \Lambda>0 such that every positive solution u to \eqref{eP} with \lambda\ge\lambda_2 and \bar{u}_0-\eta_0\le \max u<\bar{u}_0 verifies u(r)\ge \bar{u}_0-\varepsilon if 0\le r \le 1-\Lambda \lambda^{-1/p}. \end{lemma} \begin{proof} Choose e\in\mathbb{R}^N with |e|=1 and define the family of subsolutions$$ u_{t,\lambda}(x)=u_{\lambda_1}\Big(\big(\frac{\lambda}{\lambda_1} \big)^{1/p}|x-te|\Big) \,,$$for x\in B and t\in [0,1-(\lambda/\lambda_1)^{-1/p}). Since, in virtue of Lemma \ref{estim}, u\ge u_{t\big|t=0}, we are in a position to apply the {\it sweeping principle} of the Appendix to conclude that u\ge u_{t\big| t=1-(\lambda/\lambda_1)^{-1/p}}. Let 02, it is well known that for every m\ge 0 there exists a unique weak solution u to the equation \begin{gather*} -\Delta_p u+mu=f \quad\mbox{in } B\\ u=0 \quad\mbox{on } \partial B\,, \end{gather*} which is C^{1,\beta}(\overline{B}) for some 0<\beta<1 (cf. \cite{T}). Moreover, u is a radial function, so that u\in C^1[0,1], r^{N-1}\varphi_p (u')\in C^1[0,1] and u solves \begin{gather*} - (r^{N-1} \varphi_p(u'))' + m r^{N-1} u = r^{N-1} f(r) \quad 0 < r < 1\\ u'(0)= 0\,,\quad u(1) = 0\,, \end{gather*} where '=d/dr. Thus we can define an operator K_m:C[0,1]\to C^1[0,1] given by u=K_m(f), which is compact. For m=0 it is easily seen that$$ K(f)(r):=K_0(f)(r)=\int_r^1\varphi_{p'}\Big(\int_0^s(\frac{\rho}{s})^{N-1} f(\rho)\,d\rho\Big)\,ds\,. $$For simplicity, we still denote K_m to the restriction of those operators to C^1[0,1]. Back to problem \eqref{eP}, taking m>0 such that f'(u)+m>0 in [0,\bar{u}_0], then radial solutions to \eqref{eP} coincide with fixed points of the operator equation$$u=K_{\lambda m}(\lambda f(u)+\lambda mu)\,.$$Denote T_\lambda(u)=K_{\lambda m}(\lambda f(u)+\lambda mu). T_\lambda is a compact operator in C^1[0,1], and it is increasing in the order interval [0,\bar{u}_0]. Our first objective is to show that T_\lambda is differentiable in a neighbourhood of its fixed points in the interval [z_\lambda,\bar{u}_0] (see \S 3). With this in mind, it is convenient to consider first the case m=0. See Theorem 2.1 in \cite{GS3} for a related result. \begin{theorem}\label{diff} Assume f\in C^1[0,1] verifies f(0)\ne 0 and u'(r)\neq 0 if 00. Thanks to Theorem 2.1 in \cite{GS2}, we have$$ u'(r)\sim -Cr^{\frac 1{p-1}}, \quad r\to 0+ $$with a certain constant C>0. Thus there exists a constant c>0 such that |u'(s)|/s^\frac 1{p-1} \ge c for 00$$ in $(0,1]$, where $v=K(f+g)$. As the proof of Theorem \ref{diff} shows, this condition turns out to be sufficient for the differentiability of $K$ on $f+g$. Moreover, for $h\in C^1[0,1]$, \begin{align*} &|(DK(f+g)h-DK(f)h)'(s)|\\ &\le \frac 1{p-1} \Big|\frac 1{|u'(s)|^{p-2}}-\frac 1{|v'(s)|^{p-2}}\Big| \int_0^s (\frac{\rho}{s})^{N-1} |h(\rho)|d\rho \\ &\le \frac 1{p-1} s\Big|\frac 1{|u'(s)|^{p-2}}-\frac 1{|v'(s)|^{p-2}}\Big| |h|_1 \\ &\le \frac 1{p-1}\Big| \big(\frac {s^\frac 1{p-1}}{|u'(s)|}\big)^{p-2} -\big(\frac{s^\frac 1{p-1}}{|v'(s)|}\big)^{p-2}\Big| |h|_1 \end{align*} and we obtain, in virtue of (\ref{chap}), $$\sup_h \frac{|DK(f+g)h-DK(f)h|_1}{|h|_1}\to 0$$ as $g\to 0$ in $C^1[0,1]$. This proves the corollary. \end{proof} As a consequence of the implicit function theorem, we can now obtain the differentiability of the operator $T_\lambda$ on its fixed points in the interval $[z_\lambda,\bar{u}_0]$. \begin{corollary}\label{diff2} Let $u$ be a fixed point of the operator $T_\lambda$ in the interval $[z_\lambda,\bar{u}_0]$, with $f(u(0))\ne 0$, and $u'(r)\ne 0$ if $00$ for $0\le r<1$. \end{lemma} \begin{proof} Let us see that the operator $DT_\lambda(u)$ is positive. That is, $g\ge 0$ implies $DT_\lambda(u)g\ge 0$. Let $w=DT_\lambda(u)g$. Then \begin{gather*} -(r^{N-1}|u'|^{p-2}w')'+\frac{r^{N-1}}{p-1}\lambda mw= \frac{r^{N-1}}{p-1}(\lambda f'(u)+\lambda m)g\\ w'(0)=0\,, \quad w(1)=0\,. \end{gather*} Multiplying this equation by $w^-=\max\{0,-w\}$, integrating in $(0,1)$ and performing an integration by parts in the left-hand side, we obtain $$\int_{w\le 0}\big(r^{N-1} |u'|^{p-2}(w')^2+\frac{r^{N-1}}{p-1} \lambda mw^2\big)dr =\int_{w\le 0}\frac{r^{N-1}}{p-1}(\lambda f'(u)+ \lambda m)wg\,dr\le 0\,,$$ hence $w^-\equiv 0$. Thus $w\ge 0$ follows. Since $DT_\lambda(u)$ is a compact operator, Krein-Rutman's theorem \cite[Theorem 3.1]{Am} guarantees the existence of an eigenfunction $v$ associated to $\sigma$ such that $v\ge 0$. Let us show that $v>0$. Assume on the contrary that $v(r_0)=0$ for some $0\le r_0<1$. Since $v\ge 0$, we have $v'(r_0)=0$. Moreover, $v$ satisfies $$-v'(r)=\frac 1{p-1}\frac \lambda\sigma\frac 1{|u'(r)|^{p-2}} \int_{r_0}^r\left(\frac{\rho}{r}\right)^{N-1} (f'(u)+m(1-\sigma)) v(\rho) d\rho\,.$$ Then, letting $|v|_{\infty,\delta}=\sup_{|r-r_0|\le\delta}|v(r)|$, we obtain $$|v'(r)|\le C |v|_{\infty,\delta}\,.$$ for a certain constant $C>0$. After an integration we arrive at $|v|_{\infty,\delta}\le C\delta |v|_{\infty,\delta}$, and thus $v\equiv 0$ in $|r-r_0|\le\delta$ if $\delta$ is small. A continuation argument gives $v\equiv 0$ in $[0,1]$, which is clearly impossible. Thus $v(r)>0$ if $r\in [0,1)$. \end{proof} \begin{remark}{\rm Note that the conclussion of Lemma \ref{eig} cannot be achieved by means of the strong maximum principle, since the operator becomes degenerate for $r=0$.} \end{remark} \begin{proof}[Proof of Theorem \ref{unicidad}] As seen in \S 3, every positive solution to \eqref{eP} with large $\lambda$ and maximum close to $\bar{u}_0$ lies in the ordered interval $[z_\lambda,\bar{u}_0]$. Since the operator $T_\lambda$ is increasing, $z_\lambda\le T_\lambda(z_\lambda)$ and $T_\lambda(\bar{u}_0)\le \bar{u}_0$, it follows that $T_\lambda$ leaves the interval $[z_\lambda,\bar{u}_0]$ invariant. Furthermore, $T_\lambda$ is compact, and does not have fixed points in the boundary of the interval. Thus, the Leray-Schauder degree of $I-T_\lambda$ makes sense. We will denote it by $d(I-T_\lambda, (z_\lambda,\bar{u}_0),0)$. As usual, the local index of a fixed point $u$ will be denoted by $i(I-T_\lambda,u,0)$. Since $(z_\lambda,\bar{u}_0)$ is convex, we have (\cite{Am}) $$d(I-T_\lambda, (z_\lambda,\bar{u}_0),0)=1\,.$$ Let us show that, for large enough $\lambda$, every fixed point of $T_\lambda$ in the interval $[z_\lambda,\bar{u}_0]$ is isolated, and has index $1$. This will conclude the proof of the uniqueness assertion in Theorem \ref{unicidad}. \begin{lemma} \label{nondeg} There exists $\lambda^*>0$ such that for $\lambda\ge \lambda^*$, every fixed point $u$ of $T_\lambda$ in the interval $[z_\lambda,\bar{u}_0]$ is isolated, and $i(I-T_\lambda, u,0)=1$. \end{lemma} \begin{proof} Let $u\in (z_\lambda,\bar{u}_0)$ be a fixed point of $T_\lambda$. In virtue of Corollary \ref{diff2}, $T_\lambda$ is differentiable on $u$. To prove the theorem it will suffice with showing that ${\rm spr}(DT_\lambda(u))<1$, for large $\lambda$ (this implies in particular the isolation of $u$). Then since $$i(I-T_\lambda,u,0)=(-1)^\chi\,,$$ where $\chi$ stands for the sum of multiplicities of the eigenvalues of $DT_\lambda(u)$ greater than $1$ (cf. \cite[Theorem 11.4]{Am}), the conclusion follows. Assume on the contrary that there exist sequences $\lambda_n\to+\infty$, $u_n>0$ in such a way that $\sigma_n={\rm spr}(DT_{\lambda_n}(u_n))\ge 1$. In virtue of Lemma \ref{eig}, $\sigma_n$ has an associated eigenfunction $v_n>0$, which will be normalized by $|v_n|_\infty=1$. Notice that $f'(u_n)\le 0$ in $[0,1-\Lambda \lambda_n^{-1/p}]$, in virtue of Lemma \ref{estim2}. Then $v_n$ verifies $$-(r^{N-1} |u_n'|^{p-2}v_n')'\le \frac{r^{N-1}}{p-1} f'(u_n)v_n\le 0$$ in $[0,1-\Lambda \lambda_n^{-1/p}]$. The maximum principle is then applicable to conclude that $v_n$ attains its maximum in $[1-\Lambda \lambda_n^{-1/p},1]$. Choose $r_n\in [1-\Lambda \lambda_n^{-1/p},1]$ such that $v_n(r_n)=1=\max v_n$. We introduce the functions \begin{gather*} U_n(x)=u_n(1-\lambda_n^{-1/p}x) \\ V_n(x)=v_n(1-\lambda_n^{-1/p}x)\,, \end{gather*} if $0\le x\le\lambda_n^{1/p}$. Since $u_n$ is a solution to \eqref{eP}, we have that $$U_n(x)=\int_0^x\varphi_{p'} \Big(\int_s^{\lambda_n^{1/p}} \Big(\frac{\lambda_n^{1/p}-\rho }{\lambda_n^{1/p}-s}\Big)^{N-1} f(U_n(\rho ))\,d\rho \Big)\,ds\,.$$ Similarly, \begin{align*} V_n(x)&=\frac 1{p-1}\int_0^x\frac 1{|U_n'(s)|^{p-2}}\int_s^{\lambda_n^{1/p}} \Big(\frac{\lambda_n^{1/p}-\rho }{\lambda_n^{1/p}-s}\Big)^{N-1}\\ &\quad\times \Big[ \frac 1{\sigma_n}f'(U_n(\rho ))+m\big(\frac 1{\sigma_n}-1\big) \Big] V_n(\rho )\,d\rho \,ds\,. \end{align*} Now note that $U_n'\ne 0$ in $[0,\lambda_n^{1/p})$. Hence $U_n,\ V_n\in C^3[0,\lambda_n^{1/p})$. Moreover, $\{U_n\}$, $\{V_n\}$ are precompact in $C^2[0,T]$ for every $T>0$. Thus, we can assume $U_n\to \overline{U}$, $V_n\to \overline{V}$ in $C^2_{\rm loc}[0,+\infty)$, where \begin{gather*} \overline{U}(x)=\int_0^x\varphi_{p'} \Big(\int_0^{+\infty}f(\overline{U}(\rho ))\,d\rho \Big)\,ds,\\ \overline{V}(x)=\frac 1{p-1}\int_0^x\frac 1{|\overline{U}'(s)|^{p-2}}\int_s^{+\infty} \left[ \frac 1{\overline{\sigma}} f'(\overline{U}(\rho ))+m\big(\frac 1{\overline{\sigma}}-1\big)\right]\overline{V}(\rho )\,d\rho \,ds\,, \end{gather*} and $\overline{\sigma}=\lim_{n\to\infty}\sigma_n$ ($\overline{\sigma}=+\infty$ is not excluded and then we should set $1/\overline{\sigma}=0$). Thus, $\overline{U}$, $\overline{V}$ are solutions to the one-dimensional problems $$\label{unid1} \begin{gathered} -\varphi_p(\overline{U}')'= f(\overline{U})\\ \overline{U}(0)=\overline{U}'(+\infty)=0\,, \end{gathered}$$ and $$\label{unid2} \begin{gathered} -(|\overline{U}'|^{p-2}\overline{V}')'=\frac 1{p-1} \left(\frac 1{\overline{\sigma}} f'(\overline{U})+m\big(\frac 1{\overline{\sigma}}-1\big)\right)\overline{V} \\ \overline{V}(0)=\overline{V}'(+\infty)=0\,, \end{gathered}$$ while $0\le \overline{U}\le \bar{u}_0$, $0\le \overline{V}\le 1$. Since the functions $V_n$ attain their maxima in $x_n=\lambda_n^{1/p}(1-r_n)\le \Lambda$, we obtain that $\overline{V}\not\equiv 0$. Thus $\overline{V}>0$. On the other hand, notice that $\overline{U}$ verifies $|\overline{U}'|^p=p'(F(\bar{u}_0)-F(\overline{U}))$, together with $\overline{U}'>0$ in $[0,+\infty)$. Taking derivatives in (\ref{unid1}), it follows that $\overline{U}'$ solves the second order equation $$-(|\overline{U}'|^{p-2}(\overline{U}')')'=\frac 1{p-1} f'(\overline{U})\overline{U}'\,,$$ and consequently $\overline{U}'\in C^2[0,+\infty)$. Choose the least $C>0$ such that $W:=C\overline{U}'-\overline{V}\ge 0$ in $[0,\Lambda+1]$. $W$ has to vanish in some point of $[0,\Lambda+1]$. Furthermore, $W\in C^2[0,+\infty)$ and satisfies the equation $$-(|\overline{U}'|^{p-2}W')'=\frac 1{p-1}\left( f'(\overline{U})C\overline{U}'-\frac 1{\overline{\sigma}} f'(\overline{U})\overline{V}-m\big(\frac 1{\overline{\sigma}}-1\big)\overline{V}\right)\,.$$ The choice of $m$ implies $f'(\overline{U})+m>0$, and, since $\overline{\sigma}\ge 1$, $$-(|\overline{U}'|^{p-2}W')'\ge\frac 1{p-1}f'(\overline{U})W$$ in $[0,\Lambda+1]$. This implies $$\label{quec} -(|\overline{U}'|^{p-2}W')'+\frac m{p-1}W\ge \frac 1{p-1}(f'(\overline{U})+m)W\ge 0$$ in $[0,\Lambda+1]$. Notice that $\overline{U}'\neq 0$ in $[0,\Lambda+1]$, and the operator in (\ref{quec}) becomes nondegenerate. The strong maximum principle gives us that $W>0$ in $(0,\Lambda+1)$. Moreover, $W(0)>0$ and we obtain $W(\Lambda+1)=0$. Hopf's boundary lemma then provides with $W'(\Lambda+1)<0$. Thus, we can choose $\delta>0$ small so that $W<0$ in $(\Lambda+1,\Lambda+1+\delta)$. We claim that this inequality holds in $(\Lambda+1,+\infty)$. If, on the contrary we have $\delta_0=\sup\{\delta>0:W<0\ \quad\mbox{in }\ (\Lambda+1,\Lambda+1+\delta)\}<+\infty$, we obtain \begin{gather*} -(|\overline{U}'|^{p-2}W')'\ge\frac 1{p-1} f'(\overline{U})W \\ W(\Lambda+1)=W(\Lambda+1+\delta_0)=0\,. \end{gather*} Since $f'(\overline{U})\le 0$ in $[\Lambda+1,+\infty)$, maximum principle implies $W\ge 0$ in $[\Lambda+1,\Lambda+1+\delta]$ -- impossible. Thus $W<0$ in $[\Lambda+1,+\infty)$. This leads us to $$-(|\overline{U}'|^{p-2}W')'\ge 0$$ in $[\Lambda+1,+\infty)$. Integrating this inequality we arrive at $$\label{contrad} W(x)\le W'(\Lambda+1)|\overline{U}'(\Lambda+1)|^{p-2} \int_{\Lambda+1}^x \frac {ds}{|\overline{U}'(s)|^{p-2}} \,,$$ and since $p>2$, we have $\lim_{x\to+\infty} W(x)=-\infty$, contradicting $W\ge -1$. This finishes the proof. \end{proof} It only remains to prove estimate (\ref{BL}). Let $\lambda_n\to +\infty$ be an arbitrary sequence and define $U_n$ as in Lemma \ref{nondeg}. As already seen, we can assume $U_n\to \overline{U}$ in $C^2_{\rm loc} [0,+\infty)$. Thus, $U_n'(0)\to \overline{U}'(0)$. The proof is concluded by noticing that $\overline{U}'(0)=(p'F(\bar{u}_0))^{1/p}$ and $U_n'(0)=- \lambda_n^{1/p} u_n'(1)$. \end{proof} \section{Appendix} In this Appendix we are providing a generalization of Serrin's sweeping principle adequate for our purposes. \begin{theorem}[Sweeping principle] Let $\{u_t\}_{t\in [0,a]} \subset W_0^{1,p}(\Omega) \cap C^{1,\beta}(\overline{\Omega})$ be a family of subsolutions to the problem $$\label{eA} \begin{gathered} -\Delta_p u=f(u) \quad\mbox{in } \Omega\\ u=0 \quad \mbox{on } \partial\Omega\,, \end{gathered}$$ where $f$ is a $C^1$ function and $\Omega$ a smooth bounded domain of $\mathbb{R}^N$. Let $u>0$ be a solution to \eqref{eA}. Assume that $\{u_t\}$ verifies \begin{enumerate} \item[(i)] $u_t<0$ on $\partial\Omega$. \item[(ii)] The mapping $t\to u_t\in C(\overline{\Omega})$ is continuous. \item[(iii)] The set $\{x:\nabla u_t(x)=0\}$ reduces to a single point $x_t$, and for every $t$ we have $\nabla u(x_t)\ne 0$ or $u(x_t)>u_t(x_t)$. \item[(iv)] $u\ge u_{t\big|t=0}$. \end{enumerate} Then $u\ge u_{t\big|t=a}$. \end{theorem} \begin{proof} Consider the set $E=\{t\in [0,a]:u\ge u_t\ \quad\mbox{in }\ \Omega\}$. Hypotheses (ii) and (iv) imply that $E$ is closed and nonempty. Let us show that it is also open. Indeed, assume $t_0\in E$, and define $B_{t_0}:=\{ x\in \Omega\setminus\{x_{t_0}\}:u(x)=u_{t_0}(x)\}$. The set $B_{t_0}$ is closed with respect to $\Omega\setminus\{x_{t_0}\}$. To prove it is also open, let $x_0\in B_{t_0}$. Since $u\ge u_{t_0}$, $u(x_0)=u_{t_0}(x_0)$ and $x_0\ne x_{t_0}$, we obtain $\nabla u(x_0)=\nabla u_{t_0}(x_0)\ne 0$. Thus, choosing $m>0$ so that $f(u)+mu$ is increasing in a neighbourhood of $u(x_0)$, $$-\Delta_p u+\Delta_p u_{t_0}+ m(u-u_{t_0})\ge 0 \quad\mbox{in }\ \Omega\,,$$ and since the gradients of $u$ and $u_{t_0}$ do not vanish, we arrive at $L(u-u_{t_0})\ge 0$, where $L$ is an uniformly elliptic operator in a neighbourhood of $x_0$ (cf. Appendix in \cite{Sk}). This implies $u\equiv u_{t_0}$ in that neighbourhood, and $B_{t_0}$ is open. Since $\Omega\setminus\{x_{t_0}\}$ is connected, we should have $B_{t_0}=\Omega\setminus\{x_{t_0}\}$ or $B_{t_0}=\emptyset$. The first possibility implies $u\equiv u_{t_0}$ in $\overline{\Omega}$, which is impossible since $u_{t_0}<0$ on $\partial\Omega$. The second leads to $u>u_{t_0}$ in $\Omega\setminus\{x_{t_0}\}$. Hypothesis (iii) then gives $u>u_{t_0}$ in $\Omega$, and then $u>u_t$ in $\Omega$ for $t\sim t_0$, that is, $E$ is open. Finally, the connectedness of $[0,a]$ implies $E=[0,a]$, and $u\ge u_{t\big| t=a}$ follows. 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