\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
\begin{equation}\label{problema}
\begin{gathered}
-\Delta_p u=\lambda f(u) \quad \mbox{in }  \Omega \\
u=0 \quad\mbox{on }  \partial \Omega,
\end{gathered}
\end{equation}
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
\begin{equation} \label{eP}
\begin{gathered}
-\Delta_p u=\lambda f(u)  \quad\mbox{in }  B\\
u=0  \quad\mbox{on }  \partial B.
\end{gathered}
\end{equation}
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 $\lambda$:
\begin{equation}\label{LP}
\begin{gathered}
-\Delta_p u=\lambda f(\lambda,u) \quad\mbox{in }  B\\
u=0  \quad\mbox{on } \partial B.
\end{gathered}
\end{equation}
The main assumption on $f$ was the existence of a zero
$\bar{u}_0>0$ of order $k<p-1$. It was proved there the existence of
a unique family of solutions $\{u_\lambda\}$ with the property that
$u_\lambda\le \bar{u}_0$ and
\begin{equation}\label{homog}
\lim_{\lambda\to+\infty}u_\lambda=\bar{u}_0, \quad \hbox{uniformly on compacts.}
\end{equation}
It is important to notice that condition (\ref{homog}) was
crucial in \cite{GS2} in order to obtain uniqueness. That is, it
is possible to construct functions $f$ in such a way that problem
\eqref{eP} admits {\it two} families of positive solutions, one of them not
verifying (\ref{homog}) (see \cite{GS3}).

We shall presently consider the complementary case in which $f$
has a positive zero $\bar{u}_0$ of order $k\ge p-1$, therefore
closing the analysis started in \cite{GS2}. It turns out that this
situation is similar to the semilinear phenomenology. Firstly, the
solutions do not have a dead core (see Remark \ref{principal}
(a)). And secondly, it is sufficient to search for families of
solutions $\{u_\lambda\}$ with $\max u_\lambda\to \bar{u}_0$ as
$\lambda\to+\infty$ to obtain uniqueness -- and we have as a
consequence condition (\ref{homog}).

The most important fact in this regard is that we can linearize
around positive solutions, this being a novelty in the context of
the $p$-Laplacian.

We will assume throughout that $f$ satisfies the following
hypotheses, which will be termed as hypotheses (H):
\begin{itemize}
\item[(H1)] $f\in C^1(\mathbb{R})$

\item[(H2)] $f$ has a zero $\bar{u}_0$ of order $k\ge p-1$; that is,
for some positive constant $\gamma$,
$$\lim_{u\to\bar{u}_0-} \frac{f(u)}{(\bar{u}_0-u)^k}=\gamma\,.$$

\item[(H3)] $F(u)<F(\bar{u}_0)$ if $0\le u<\bar{u}_0$, where $F(u)=\int_0^u f(s)ds$.

\item[(H4)] $f'(u)\le 0$ in $[\bar{u}_0-\varepsilon,\bar{u}_0]$ for some $\varepsilon>0$.

\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 $\partial B$:
\begin{equation}\label{BL}
\lim_{\lambda\to+\infty}\lambda^{-1/p}u'_\lambda(1)=-(p'F(\bar{u}_0))^{1/p}\,,
\end{equation}
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 $0\le r_k<R_k$, such that
\begin{equation} \label{dernol}
\begin{gathered}
u(x)=u(\rho),\quad \rho=|x-z_k|  \quad \hbox{in } C_k,\\
\frac{\partial u}{\partial\rho}<0\ \hbox{ in } C_k\,.
\end{gathered}
\end{equation}
Moreover, $u(x)\ge u_{|\partial B_{r_k}(z_k)}$ in $B_{r_k}(z_k)$ and $\nabla u=0$ on
$\partial C_k\cap B$. On the other hand, $f(u)=0$ on $\partial B_{R_k}(z_k)\cap B$ and
if $r_k>0$ 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
$u_1<u(z_1)$. Now we will define an auxiliary function $\bar u$ such
that $\bar u\le u_1$ in $B\setminus B_{R_1}(z_1)$, and show that it
is possible to choose another annulus $C_2$ such that $r_2>R_1$ 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:
$$u^*(x)=\sum_{i=1}^l \bar u(x+z_i)\chi_{\{R_{i-1}\le |x| <R_i\}}\,,$$
where $R_0=0$ (see Figure 1). The main property of the function $u^*$ is the
following:

\begin{lemma}\label{radialidad}
Let $u$ be a positive solution to \eqref{eP}. Then the function $u^*$ defined
above is a radial positive solution to \eqref{eP}. Moreover, if $\nabla u^*\neq 0$
in $B\setminus \{0\}$ then $u^*=u$ and $u$ is a radial function.
\end{lemma}

\begin{proof} It is easy to check that $u^*$ is a radial
solution to \eqref{eP} (see Remark \ref{symm}). Thus, assume $\nabla u^*\neq 0$
in $B\setminus\{0\}$.
According to the definition of $u^*$, consider $B_{R_1}(z_1)$. If this
ball does not coincide with $B$ then in virtue of Lemma \ref{Brock}
we obtain $\nabla u=0$ on $\partial B_{R_1}(z_1)$, which is a contradiction.
Thus, $B_{R_1}(z_1)=B$ and $u^*=u$. This proves the lemma.
\end{proof}

\begin{remark}
{\rm If we knew from the beginning that $\nabla u\ne 0$ in $B\setminus
\{0\}$, then Theorem 1 in \cite{BN} could also be applied to conclude
that $u$ is radial.}
\end{remark}

\section{Existence and estimates of solutions}

In this section we are proving that, under the assumptions
(H) on $f$ (see \S 1), we can guarantee the existence of a family
$\{u_\lambda\}$ of positive radial solutions to \eqref{eP} which in addition
verifies (\ref{homog}). The first
important remark is that, due to hypotheses (H)$_2$ on the zero
of $f$, positive solutions $u$ to \eqref{eP} with $u\le \bar{u}_0$ also
satisfy $0<u<\bar{u}_0$ (see \cite{GS1}).

To begin with, notice that if $u$ is a radial solution to \eqref{eP},
then it satisfies (see \S 4):
\begin{gather*}
- (r^{N-1} \varphi_p(u'))'  = r^{N-1} \lambda f(u) \\
 u'(0)=0\,, \quad u(1)=0\,.
\end{gather*}
Setting $v(r)=u(\lambda^{-1/p}r)$, this problem is equivalent to
\begin{equation}\label{cauchy0}
\begin{gathered}
- (r^{N-1} \varphi_p(u'))'  = r^{N-1} f(u) \\
 u'(0)=0\,, \quad u(\lambda^{1/p})=0\,.
\end{gathered}
\end{equation}
Thus, it is apparent that the Cauchy problem
\begin{equation}\label{cauchy}
\begin{gathered}
- (r^{N-1} \varphi_p(u'))'  = r^{N-1} f(u) \\
 u(0)=u_0\,, \quad u'(0)=0\,,
\end{gathered}
\end{equation}
with $u_0$ in a left neighbourhood of $\bar{u}_0$, will play an important
role. Let us just quote that this problem has a unique
solution, denoted henceforth as $u(\cdot,u_0)$, in an interval of
the form $[0,\delta]$ (see Theorem 2.1 in \cite{GS2} and references
therein). Moreover, this solution can be continued as long as
$u'(\cdot,u_0)\ne 0$. It is also worthy of mention that $u(\cdot,\bar{u}_0)
\equiv \bar{u}_0$ (Theorem 2.2 in \cite{GS2}).

Our existence result is the following:

\begin{theorem}\label{existencia} Assume $f$ satisfies hypotheses (H).
Then there exist $\eta>0$, $\lambda_0>0$ 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 $\hat r_n<r_n\le R$ such
that $u(\hat r_n,u_{0n})=\bar{u}_0-\varepsilon_0$. Passing to a subsequence
we can assume $\hat r_n\to \hat r_0$, this also implying that
$u(\hat r_n,u_{0n})\to u(\hat r_0,\bar{u}_0)=\bar{u}_0$. This clear contradiction
proves the existence of $\eta>0$ 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,
\begin{equation}\label{fep}
\left(|u'(r)|^p+p'F_\tau(u(r))\right)'\ge 0, \quad r\ge R\,,
\end{equation}
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 $F(u)<F(\bar{u}_0)$ for
 $-1\le u\le 0$.

Now let $\varepsilon$ be as in hypothesis (H)$_4$. With no loss of
generality, we can assume that $\varepsilon<\eta$. We can find a value
$\lambda_1>\lambda_0$ 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 $0<R<1$ be
such that $u_{\lambda_1}(r)\ge\bar{u}_0- \varepsilon$ if $0\le r\le R$. Then
$u(r)\ge \bar{u}_0-\varepsilon$ for $0\le r \le 1-(1-R)(\lambda/\lambda_1)^{-1/p}$. Thus,
we can take $\Lambda=(1-R) \lambda_1^{1/p}$. \end{proof}

\begin{remark} {\rm Notice that hypothesis (H)$_2$ suffices
to guarantee that every family of positive solutions $\{u_\lambda\}$ to
\eqref{eP} such that $\lim_{\lambda\to+\infty}\max u_\lambda =\bar{u}_0$ verifies
(\ref{homog}). This is in contrast with the case $0<k<p-1$ treated
in \cite{GS2}. } \end{remark}

\section{Differentiability properties}

In this section we are showing some auxiliary results,
which deal with the linearization of the inverse of the $p$-Laplacian
under radial symmetry.

Let $f\in C(\overline{B})$ be radially symmetric. Since $p>2$,
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 $0<r\le 1$, where $u=Kf$. Then  $K$, as an operator
defined in $C^1[0,1]$, is Fr\'echet-differentiable
on $f$, and
\begin{equation}\label{difk}
DK(f)g=\frac 1{p-1}\int_r^1\frac 1{|u'(s)|^{p-2}}\int_0^s
(\frac{\rho}{s})^{N-1} g(\rho)\,d\rho\,ds
\end{equation}
for every $g\in C^1[0,1]$. In particular, $w=DK(f)g$  is a solution to
the equation
\begin{equation}\label{difk2}
\begin{gathered}
- (r^{N-1} |u'|^{p-2} w')'  = \frac{r^{N-1}}{p-1} g(r) \quad 0 < r < 1\\
 w'(0)=0\,, \quad w(1)=0\,.
\end{gathered}
\end{equation}
\end{theorem}

\begin{proof} Without loss of generality, assume
$f(0)>0$. 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 $0<s\le 1$, and the expression (\ref{difk}) makes sense.
Denote it by $R(g)$. Then
\begin{align*}
&|(K(f+g)-K(f)-R(g))'(s)|\\
&\le \Big| \varphi_{p'}\Big(\int_0^s (\frac{\rho}{s})^{N-1} (f(\rho)+g(\rho))\,d\rho\Big)
-\varphi_{p'}\Big(\int_0^s(\frac{\rho}{s})^{N-1} f(\rho)\,d\rho\Big)\\
&\quad -\frac 1{p-1} \frac 1{|u'(s)|^{p-2}}\int_0^s(\frac{\rho}{s})^{N-1} g(\rho)
\, d\rho\Big| \\
&\le\frac 1{p-1}\Big| |\xi(s)|^{p'-2}-\frac 1
{|u'(s)|^{p-2}}\Big|\int_0^s(\frac{\rho}{s})^{N-1} |g(\rho)|\,d\rho\\
&\le \frac 1{p-1}\Big| s|\xi(s)|^{p'-2}-\frac s{|u'(s)|^{p-2}}\Big|
|g|_1\,,
\end{align*}
where $\xi(s)$ is an intermediate function comprised between
$\int_0^s(\frac{\rho}{s})^{N-1} (f(\rho)+g(\rho))\,d\rho$ and $\int_0^s (\frac{\rho}{s})^{N-1}
f(\rho)\,d\rho$. Thus, as $|g|_1\to 0$ we obtain
$$
\xi(s)\to \int_0^s(\frac{\rho}{s})^{N-1} f(\rho)\,d\rho=-|u'(s)|^{p-1}
$$
uniformly in $[0,1]$. Let us see that $\xi(s)/s\to
-|u'(s)|^{p-1}/s$ uniformly in $[0,1]$ as $|g|_1\to 0$. Indeed,
\begin{equation}\label{chap}
\Big|\frac 1s\int_0^s(\frac{\rho}{s})^{N-1} (f(\rho)+g(\rho))\,d\rho-\frac 1s
\int_0^s (\frac{\rho}{s})^{N-1} f(\rho)\,d\rho\Big|\le |g|_1\,,
\end{equation}
and since $\xi(s)/s$ is an intermediate value, we are done.
This implies
\begin{align*}
&|(K(f+g)-K(f)-R(g))'(s)|\\
&\le \frac 1{p-1} s^\frac 1{p-1}
\Big| \big(\frac{|\xi(s)|}s\big)^{p'-2}
- \big( \frac {|u'(s)|^{p-1}}s\big)^{p'-2}\Big| |g|_1 = o(|g|_1)\,,
\end{align*}
uniformly as $|g|_1\to 0$. In the same way $|K(f+g)-K(f)-R(g)|_\infty =
o(|g|_1)$ as $|g|_1\to 0$. This proves the differentiability
assertion. That $w=R(g)$ is a solution to (\ref{difk2}) is a direct
consequence of expression (\ref{chap}).
\end{proof}

\begin{corollary} \label{diff1}
Assume $f\in C^1[0,1]$ verifies $f(0)\ne 0$ and
$u'(r)\neq 0$ if $0<r\le 1$, where $u=Kf$. Then $K$ is $C^1$ in
a neighbourhood of $f$ in $C^1[0,1]$.
\end{corollary}

\begin{proof} Let us show first that $K$ is
differentiable in a neighbourhood of $f$. Notice that, in virtue of
(\ref{chap}), if $|g|_1$ is small, we have that
$$
\frac{|v'(s)|^{p-1}}s\ge c>0
$$
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
$0<r\le 1$. Then $T_\lambda$ is $C^1$ in a neighbourhood of $u$ and
we have, for every $g\in C^1[0,1]$, that $w=DT_\lambda(u)g$ solves
\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*}
\end{corollary}

\begin{proof} Let us prove that $K_{\lambda m}$ is differentiable
in a neighbourhood of $\lambda f(u)+\lambda mu$.
Denote $\hat f=\lambda f(u)+\lambda mu$. Solving the equation $-\Delta_p v+\lambda mv=f$
is equivalent to solving
$$
\mathcal{F}(v,f):=v-K(f-\lambda mv)=0\,.
$$
We will show that the implicit function theorem can be applied in this
case. Indeed, $\mathcal{F}(u,\hat f)=0$, and in virtue of Corollary
\ref{diff1},
$K$ is $C^1$ in a neighbourhood of $\hat f-\lambda mu=\lambda f(u)$
(since $f(u(0))\ne 0$ and $u'(r)\ne 0$ for $0<r\le 1$). Thus,
$\mathcal{F}$ is $C^1$ with respect to both variables. Assume there
exists $g$ such that $D_v\mathcal{F}(u,\hat f)g=g +DK(\hat f-\lambda mu)\lambda mg
=0$. Then $g$ solves
\begin{gather*}
-(r^{N-1}|u'|^{p-2}g')' +\frac{r^{N-1}}{p-1}\lambda mg= 0\\
 g'(0)=0\,, \quad g(1)=0\,,
\end{gather*}
and we obtain $g\equiv 0$. Hence, $D_v\mathcal{F}(u,\hat f)$ is an isomorphism
(notice that $D_v\mathcal{F}(u,\hat f)$ is a compact perturbation of the
identity).

Implicit function theorem guarantees the existence of a unique $C^1$
function $R$, such that $\mathcal{F}(v,f)=0$ implies $v=R(f)$ in a
 neighbourhood of $(u,\hat f)$. Uniqueness yields  $R(f)=K_{\lambda m}(f)$,
hence $K_{\lambda m}$ is $C^1$. In addition,
$$
D_v\mathcal{F}(u,\hat f) DK_{\lambda m}(\hat f) + D_f\mathcal{F}(u,\hat f)=0\,,
$$
so that $w=DK_{\lambda m}(\hat f)g$ is a solution to the equation
\begin{gather*}
-(r^{N-1}|u'|^{p-2}w')'+\frac{r^{N-1}}{p-1}\lambda mw= \frac{r^{N-1}}{p-1} g\\
 w'(0)=0\,, \quad w(1)=0\,.
\end{gather*}
The conclusion of this theorem is then a direct consequence of the
chain rule. \end{proof}

\section{Uniqueness}

 This section is devoted to the proof of Theorem \ref{unicidad}.
First of all, we need a result about the eigenvalues of the
linearization of the operator $T_\lambda$. The spectral radius of a
bounded linear operator $L$ will be denoted by spr($L$). We
recall that $|\mu|\le {\rm spr}(L)$ for every spectral
value of $L$, in particular for possible eigenvalues
$\mu$ (cf. \cite{Yo}). We have the following lemma.

\begin{lemma}\label{eig}
Let $u$ be a fixed point of $T_\lambda$ in the interval
$[z_\lambda,\bar{u}_0]$. If $\sigma=\mathop{\rm spr}(DT_\lambda(u))$
is positive,
then $\sigma$ is an eigenvalue of $DT_\lambda(u)$ which admits
an eigenfunction $v$ such that $v(r)>0$
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
\begin{equation}\label{unid1}
\begin{gathered}
-\varphi_p(\overline{U}')'= f(\overline{U})\\
 \overline{U}(0)=\overline{U}'(+\infty)=0\,,
\end{gathered}
\end{equation}
and
\begin{equation}\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}
\end{equation}
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
\begin{equation}\label{quec}
-(|\overline{U}'|^{p-2}W')'+\frac m{p-1}W\ge \frac 1{p-1}(f'(\overline{U})+m)W\ge 0
\end{equation}
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
\begin{equation}\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}} \,,
\end{equation}
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
\begin{equation} \label{eA}
\begin{gathered}
-\Delta_p u=f(u) \quad\mbox{in } \Omega\\
u=0 \quad \mbox{on } \partial\Omega\,,
\end{gathered}
\end{equation}
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. This proves the sweeping principle.
\end{proof}


\subsection*{Acknowledgements}
The author would like to thank Professor Jos\'e Sabina de Lis for
his careful reading of this paper and also for some very helpful
comments.

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\end{document}