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\AtBeginDocument{{\noindent\small
2005-Oujda International Conference on Nonlinear Analysis.
\newline {\em Electronic Journal of Differential Equations},
Conference 14, 2006, pp. 95--107.\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 2006 Texas State University - San Marcos.}
\vspace{9mm}}
\setcounter{page}{95}

\begin{document}

\title[\hfilneg EJDE/Conf/14 \hfil Maximum and anti-maximum principles]
{Maximum and anti-maximum principles for the p-Laplacian with a
nonlinear boundary condition}

\author[A. Anane, O. Chakrone, N. Moradi \hfil EJDE/Conf/14 \hfilneg]
{Aomar Anane, Omar Chakrone, Najat Moradi}  % in alphabetical order

\address{Aomar Anane \newline
D\'epartement de Math\'ematiques et Informatique,
Facult\'e des Sciences,
Universit\'e Mohammed 1er, Oujda, Maroc}
\email{anane@sciences.univ-oujda.ac.ma}

\address{Omar Chakrone \newline
D\'epartement de Math\'ematiques et Informatique,
Facult\'e des Sciences,
Universit\'e Mohammed 1er, Oujda, Maroc}
\email{chakrone@sciences.univ-oujda.ac.ma}

\address{Najat Moradi \newline
D\'epartement de Math\'ematiques et Informatique,
Facult\'e des Sciences,
Universit\'e Mohammed 1er, Oujda, Maroc}
\email{najat\_moradi@yahoo.fr}


\date{}
\thanks{Published Septebmer 20, 2006.}
\subjclass[2000]{35J65, 35J25}
\keywords{Anti-maximum; p-laplacian; non linear boundary condition}

\begin{abstract}
 In this paper we study the maximum and the anti-maximum principles
 for the problem $\Delta _{p}u=|u|^{p-2}u$ in the bounded smooth domain
 $\Omega \subset \mathbb{R}^{N}$, with
 $|\nabla u|^{p-2}\frac{\partial u}{\partial \nu }=\lambda |u|^{p-2}u+h$
 as a non linear boundary condition on $\partial \Omega $ which
 is supposed $C^{2\beta }$ for some $\beta $ in $]0,1[$, and where
 $h\in L^{\infty }(\partial \Omega )$. We will also examine the existence
 and the non existence of the solutions and their signs.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}{Lemma}[section]
\numberwithin{equation}{section}

\section{Introduction}

In this work we consider the  problem
\begin{equation} \label{e1.1}
\begin{gathered}
\Delta _{p}u=|u|^{p-2}u\quad \text{in } \Omega, \\
|\nabla u|^{p-2}\frac{\partial u}{\partial \nu
}=\lambda |u|^{p-2}u+h\quad \text{on }\partial \Omega,
\end{gathered}
\end{equation}
where $\Omega $ is a bounded domain in $\mathbb{R}^{N}$, with a
$C^{2,\beta } $ boundary where $\beta \in ]0,1[$, $h\in L^{\infty
}(\partial \Omega )$ and $\frac{\partial }{\partial \nu }$ is the outer
normal derivative.

For $h\equiv 0$ in $\partial \Omega $, Fernandez
Bonder, Pinasco and Rossi \cite{F.P.R} proved  that the
problem
\begin{gather*}
\triangle _{p}u=|u|^{p-2}u\quad \text{on } \Omega, \\
|\nabla u|^{p-2}\frac{\partial u}{\partial \nu
}=\lambda |u|^{p-2}u\quad \text{on } \partial \Omega
\end{gather*}
admits an infinite sequence of eigenvalues $(\lambda _{n})$ such that
$\lambda _{n}\to\infty $ as $n\to\infty $.
Martinez and Rossi \cite{M.R}  showed that the first
eigenvalue given by
\begin{equation*}
\lambda _{1}=\inf \big\{ \int_{\Omega }|\nabla u|
^{p}+\int_{\Omega }|u|^{p}: u\in W^{1,p}(\Omega
)\text{ and }\int_{\partial \Omega }|u|^{p}=1\big\}
\end{equation*}
is simple and isolated with the eigenfunctions do not change sign
in $\Omega $.

We will be interested in he case where $h\not\equiv 0$ in
$\partial \Omega $. The case where $h\equiv 0$, is treated by
Godoy, Gossez and Paczka \cite{G.G.P2}; they have proved in
that

\begin{quote}
(i) The maximum principle holds if and only if $\lambda
_{-1}(m)<\lambda <\lambda _{1}(m)$ with $\lambda _{-1}(m)$ and$
\lambda _{1}(m)$ are the principal eigenvalues of the p-Laplacien
with the weight $m$, for the Dirichlet problem, (respectively
$0<\lambda <\lambda ^{\ast }(m)$, for the Neumann problem, with
$\lambda ^{\ast }(m)$ is the nontrivial principal eigenvalue).

\noindent(ii) The anti-maximum principle holds at the right of
 $\lambda _{1}(m)$ and at the left of $\lambda _{-1}(m)$
(resp, at the right of $\lambda ^{\ast }(m)$ and at the left of
$0$. Moreover it is nonuniform when $p\leq N$ and uniform when $p>N$).
\end{quote}

In what follows one supposes that any solution of \eqref{e1.1} is in
$C^{1,\alpha }(\overline{\Omega })$ with $\alpha \in ]0,1[$.

\section{The maximum principle}

The following result will be proven.

\begin{theorem} \label{thm2.1}
The maximum principle holds for problem \eqref{e1.1} if and only if
$\lambda \leq \lambda _{1}$.
\end{theorem}

\begin{proof}
(i) Given $\lambda \leq \lambda_{1}$,
$0\lvertneqq h\in L^{\infty }(\partial \Omega )$ and $u$ be a solution
of \eqref{e1.1}, let us show that $ u\geq 0$ in $ \Omega $.
We recall that $ u$ is a solution of\eqref{e1.1} if and only if
\begin{equation}
 \int_{\Omega }|\nabla u|^{p-2}\nabla u\nabla v+\int_{\Omega }|u|
^{p-2}uv=\lambda \int_{\partial \Omega }|u|
^{p-2}uv+\int_{\partial \Omega }hv \label{e2.1}
\end{equation}
for all $v\in W^{1,p}(\Omega )$. Applying this equality to
$u^{-}$, one finds
$$
 \int_{\Omega }|\nabla u|^{p-2}\nabla u\nabla
u^{-}+\int_{\Omega }|u|^{p-2}uu^{-}=\lambda
\int_{\partial \Omega }|u|^{p-2}uu^{-}+\int_{\partial
\Omega }hu^{-}.
 $$
Then
 $$
\int_{\Omega }|\nabla u^{-}|^{p}+\int_{\Omega }|u^{-}|
^{p}=\lambda \int_{\partial \Omega }|u^{-}|
^{p}-\int_{\partial \Omega }hu^{-},
$$
and
$$
\lambda \int_{\partial \Omega }|u^{-}|^{p}-\Big(
\int_{\Omega }|\nabla u^{-}|^{p}+\int_{\Omega }|
u^{-}|^{p}\Big) =\int_{\partial \Omega }hu^{-}.
$$
However, $ \lambda \leq \lambda _{1}=\inf \{ \int_{\Omega
}|\nabla u|^{p}+\int_{\Omega }|u|^{p}:
 u\in W^{1,p}(\Omega )\text{ and }\int_{\partial \Omega }|u|^{p}=1\}$, so
\begin{align*}
\int_{\partial \Omega }hu^{-}
&= \lambda \int_{\partial \Omega }| u^{-}|^{p}
 -\Big( \int_{\Omega }|\nabla u^{-}|^{p}+\int_{\Omega }|u^{-}|^{p}\Big) \\
&\leq  \lambda _{1}\int_{\partial \Omega }|u^{-}|
^{p}-\Big( \int_{\Omega }|\nabla u^{-}|
^{p}+\int_{\Omega }|u^{-}|^{p}\Big) \leq 0.
\end{align*}
Moreover $h\gneqq 0$ and $u^{-}\geq 0$  thus
$ \int_{\partial \Omega }hu^{-}\geq 0$ from where
$\int_{\partial \Omega }hu^{-}=0$.
Consequently $\lambda \int_{\partial \Omega }|u^{-}|
^{p}=\int_{\Omega }|\nabla u^{-}|^{p}+\int_{\Omega }|
u^{-}|^{p}$; i-e, $u^{-}$ is a positive eigenfunction associated with
$\lambda $.

 If $\lambda \neq \lambda _{1}$ then$ u^{-}$ change sign on $\partial
\Omega $ by  \cite[lemma 2.4]{M.R}, which is not possible, so
$u^{-}\equiv 0$ on $\partial \Omega $. Hence
$0=\lambda \int_{\partial \Omega }|u^{-}|^{p}=\int_{\Omega
}|\nabla u^{-}|
^{p}+\int_{\Omega }|u^{-}|^{p}$ and thus $u^{-}\equiv 0$ in
$\Omega $.

 If $\lambda =\lambda _{1}$ then $u^{-}$ is a positive eigenfunction
associated to $\lambda _{1}$ and it is in
$C^{1,\alpha }( \overline{\Omega }) $, so $u^{-}>0$ or $u^{-}\equiv 0$ in$
\overline{\Omega }$ by the following maximum principle.

\begin{theorem}[Vasquez \cite{Va}] \label{thm2.2}
Let $u\in C^{1}(\Omega )$ be such that $\Delta
_{p}u\in L_{\rm loc}^{2}(\Omega )$, $u\geq 0$  a.e in
$\Omega $ and $ \Delta _{p}u\leq \beta (u)$ a.e in
$\Omega $, with $\beta : [ 0,+\infty [\to
\mathbb{R}$ is a increasing continuous function,
$\beta (0)=0$ and either $\beta (s)=0$
for some $s>0$ or ($\beta (s)>0$ for all $s>0$ and
$\int_{0}^{1}(j(s)) ^{-\frac{1}{p}}ds=\infty $  with
$j(s)=\int_{0}^{s}\beta (t)dt$). Then if $u $ does not
vanish identically on $\Omega$, it is positive everywhere in
$\Omega $.

Moreover if $u\in C^{1}(\Omega \cup \{x_{0}\})$ for an
$x_{0}\in \partial \Omega $ that satisfies an interior sphere condition
and $u(x_{0})=0$ then $\frac{\partial u}{\partial n}(x_{0})>0$
where $n$ is an interior normal at $x_{0}$.
\end{theorem}

Indeed, ($u^{-}>0$ or $u^{-}=0$) in $ \Omega $, and if there exists
$x_{0}\in \partial \Omega $ such that $u^{-}(x_{0})=0$, then
$\frac{\partial u^{-}}{\partial \nu }(x_{0})<0$. However
$|\nabla u^{-}|^{p-2} \frac{\partial u^{-}}{\partial \nu }(x_{0})
=\lambda _{1}|u^{-}(x_{0})|^{p-2}u^{-}(x_{0})=0$ and thus
$\nabla u^{-}\equiv 0$ in $ \overline{\Omega }$; i.e., $ u^{-}=0$ in
$ \overline{\Omega }$.
Consequently $u\geq 0$ in$ \overline{\Omega }$.

\noindent (ii) Given $\lambda >\lambda _{1}$, and one supposes that
the maximum principle is applicable to the problem \eqref{e1.1}; i.e.,
 for all $h\in L^{\infty}(\partial \Omega )$, if $h\gvertneqq 0$
then any solution is positive on $\Omega  $.

Considering $\Phi $ a positive eigenfunction associated to the
first eigenvalue $\lambda _{1}$, and
$h=(\lambda -\lambda_{1})|\Phi |^{p-2}\Phi \gvertneqq 0$, one checks that
$\Psi =-\Phi $ $\leq 0$ is a solution of \eqref{e1.1}, but this
contradicted the maximum principle.
\end{proof}

\section{Existence and nonexistence of solutions for \eqref{e1.1}}

In this paragraph one shows stronger results existence and
nonexistence. They are stated in the following theorem.

\begin{theorem} \label{thm3.1}
Given $h\in L^{\infty }(\partial \Omega )$ such that $h\geq 0$ in $\partial
\Omega $.
\begin{itemize}
\item[(i)] If$ \lambda <\lambda _{1}$ and $h\gvertneqq 0$ in
 $\partial \Omega $, then the problem admits an unique solution
$u$ which is in $C^{1}(\overline{\Omega })$ and$ u>0$ on
$\overline{\Omega }$.

\item[(ii)] If $\lambda =\lambda _{1}$ and $h\gvertneqq 0$ in
 $\partial \Omega $, then the problem has no solution.

\item[(iii)] If $\lambda >\lambda _{1}$ then the problem does not have any
solution $u$ such that $u\gvertneqq 0$ on $\overline{\Omega }$.
\end{itemize}
\end{theorem}

For the proof of this theorem we need the following lemma.

\begin{lemma} \label{lem3.1}
Let $u$ be a solution of \eqref{e1.1} with $u>0$ on $\overline{\Omega }$, and
$h\geq 0$ in $\partial \Omega $ (or $h\leq 0$ in $\partial \Omega $).
Then $\forall \varphi \in C^{1}(\overline{\Omega })$ with $\varphi \geq 0$ in
$\overline{\Omega }$, one has
\begin{equation}
\lambda \int_{\partial \Omega }\varphi ^{p}+\int_{\partial \Omega
}h\frac{ \varphi ^{p}}{u^{p-1}}\leq \int_{\Omega }|\nabla \varphi
|^{p}+\int_{\Omega }\varphi ^{p}. \label{e3.1}.
\end{equation}
Moreover, the equality holds if and only if $\varphi $ is a
multiple of $u$.
\end{lemma}

\begin{proof}
Given $u$ and $\varphi $ in $C^{1}(\overline{\Omega })$ with $u>0$ and
$\varphi \geq 0$ on  $\overline{\Omega }$, we denote $R(\varphi ,u)=|\nabla
\varphi |^{p}-|\nabla u|^{p-2}\nabla u\nabla (\frac{\varphi ^{p}}{u^{p-1}}
)$, and we prove that $R(\varphi ,u)\geq 0$ and that  $R(\varphi ,u)=0$ if
and only if there exists $c\geq 0$ such that $\varphi =cu$.

One has $R(\varphi ,u)=|\nabla \varphi |^{p}+(p-1)\frac{\varphi ^{p}}{u^{p}}
|\nabla u|^{p}-p\frac{\varphi ^{p-1}}{u^{p-1}}|\nabla u|^{p-2}\nabla
u\nabla \varphi $. Applying Minkovsky inequality  to
$ |\nabla \varphi|$ and $\frac{\varphi ^{p-1}}{u^{p-1}}|\nabla u|^{p-1}$,
one obtains
\begin{equation}
 \frac{\varphi ^{p-1}}{u^{p-1}}|\nabla
u|^{p-1}|\nabla \varphi |\leq \frac{1}{p}|\nabla \varphi
|^{p}+\frac{1}{q}(\frac{\varphi ^{p-1}}{u^{p-1}}|\nabla
u|^{p-1})^{q} \label{e3.2}
\end{equation}
where  $\frac{1}{q}+\frac{1}{p}=1$. This implies
$\frac{\varphi ^{p-1}}{u^{p-1}}|\nabla u|^{p-1}|\nabla
\varphi |\leq \frac{1}{p}|\nabla \varphi |^{p}+\frac{p-1}{q}(\frac{
\varphi ^{p}}{u^{p}}|\nabla u|^{p-1})$. And it follows that
$$
R(\varphi ,u)=|\nabla \varphi |^{p}+(p-1)\frac{\varphi ^{p}}{u^{p}}
|\nabla u|^{p}-p
\frac{\varphi ^{p-1}}{u^{p-1}}|\nabla u|^{p-2}\nabla u\nabla \varphi \geq
0.
$$
The inequality \eqref{e3.2} becomes an equality if and only if
\begin{equation}
|\nabla \varphi |=\frac{\varphi }{u}|\nabla u|.
\label{e3.3}
\end{equation}
 Let $\psi =\frac{\varphi }{u}\geq 0$. Then we have
$\varphi (x)=\psi (x)u(x)$ which implies
$\nabla \varphi =\psi \nabla u+u\nabla \psi $.
Then \eqref{e3.3} is equivalent to
$ |\psi \nabla u+u\nabla \psi |=\psi |\nabla u|$, which is true if and only
if $u^{2}|\nabla \psi |^{2}=-2u\psi \nabla u\nabla \psi $.
We have also:
\begin{align*}
(R(\varphi ,u)=0)
&\Leftrightarrow |\nabla \varphi
|^{p}=|\nabla u|^{p-2}|\nabla u\|\nabla (\frac{\varphi }{u})^{p}u| \\
&\Leftrightarrow |\nabla \varphi |^{p}=|\nabla u|^{p}(\frac{\varphi }{u}
)^{p}+u|\nabla u|^{p-2}|\nabla u|\nabla (\psi ^{p}).
\end{align*}
Since $ |\nabla \varphi |=\psi |\nabla u|$, one has
\begin{equation*}
R(\varphi ,u)=0\Leftrightarrow |\nabla u|^{p-2}|\nabla u|\nabla (\psi
^{p})=|\nabla u|^{p-2}(\nabla u\nabla \psi )p\psi ^{p-1}=0
\end{equation*}

If $ |\nabla u|\equiv 0$, then \eqref{e3.3} implies that
$|\nabla \varphi |=0$. And in this case $u$ and $\varphi $ are
constants; therefore there exists $c\geq 0$ such that
$ \varphi=cu$.

If $|\nabla u|\neq 0$, on the set $\{|\nabla u|\neq 0\}$, \eqref{e3.3}
is equivalent to $\nabla u\nabla \psi =0$. Therefore, one has also \eqref{e3.3}
is equivalent to
$-2\psi \nabla u\nabla (\varphi \psi )=u|\nabla \psi |^{2}$
Thus $\nabla u\nabla \psi =0$ if and only if $ |\nabla \psi |=0$,
which is equivalent to $ \psi$ equals a constant.

Consequently $R(\varphi ,u)=0$ if and only there exists $c\geq 0$ such that
$ \varphi =cu$.
Conclusions:
\begin{itemize}
\item[(i)] $0\leq \int_{\Omega }R(\varphi ,u)=\int_{\Omega }|\nabla \varphi
|^{p}-\int_{\Omega }|\nabla u|^{p-2}\nabla u\nabla (\frac{\varphi ^{p}}{
u^{p-1}})$ for all$ (u,\varphi )\in (C^{1}(\overline{\Omega }))^{2}$ with
 $u>0$ and $\varphi \geq 0$ on $\overline{\Omega }$.

\item[(ii)] $\int_{\Omega }R(\varphi ,u)=\int_{\Omega }|\nabla \varphi
|^{p}-\int_{\Omega }|\nabla u|^{p-2}\nabla u\nabla (\frac{\varphi ^{p}}{
u^{p-1}})=0 $ if and only there exists $c\geq 0$ such that $\varphi =cu$.
\end{itemize}
Then if $u$ is a solution of \eqref{e1.1} and
$\varphi \in C^{1}( \overline{\Omega })$ such that  $\varphi \geq 0$ on
$\overline{\Omega }$, we get
\begin{equation*}
\int_{\Omega }|\nabla u|^{p-2}\nabla u\nabla (\frac{\varphi ^{p}}{u^{p-1}}
)+\int_{\Omega }(u)^{p-1}(\frac{\varphi ^{p}}{u^{p-1}})=\lambda
\int_{\partial \Omega }(u)^{p-1}(\frac{\varphi ^{p}}{u^{p-1}}
)+\int_{\partial \Omega }h\frac{\varphi ^{p}}{u^{p-1}}.
\end{equation*}
Consequently,
$$
 0\leq \int_{\Omega }|\nabla \varphi |^{p}-\lambda \int_{\partial
\Omega }(u)^{p-1}(\frac{\varphi ^{p}}{u^{p-1}})-\int_{\partial
\Omega }h \frac{\varphi ^{p}}{u^{p-1}}+\int_{\Omega }(u)^{p-1}(\frac{\varphi ^{p}}{
u^{p-1}}),
$$
then
$$
0\leq \int_{\Omega }|\nabla \varphi |^{p}-\lambda \int_{\partial
\Omega }\varphi ^{p}-\int_{\partial \Omega }h\frac{\varphi
^{p}}{u^{p-1}}+\int_{\Omega }\varphi ^{p},
$$
and
\begin{equation}
\int_{\Omega }|\nabla \varphi |^{p}+\int_{\Omega }\varphi ^{p}\geq
\lambda \int_{\partial \Omega }\varphi ^{p}+\int_{\partial \Omega
}h\frac{\phi^p}{u^{p-1}}. \label{e3.1b}
\end{equation}
If there is equality in \eqref{e3.1b} then $\int_{\Omega }R(\varphi ,u)=0 $
which is equivalent to $R(\varphi ,u)=0$ almost everywhere in
$\Omega $. So there exists $c\geq 0$ such that
$ \varphi=cu$.
\end{proof}


\begin{proof}[Proof of Theorem \ref{thm3.1}]
The case of $\lambda <\lambda _{1}$:
(i) Existence of the solutions for the problem \eqref{e1.1}:
One considers the function $\Phi :W^{1,p}(\Omega )\to \mathbb{R}$
defined by
\begin{equation*}
\Phi (v)=\frac{1}{p}(\int_{\Omega }|\nabla v|
^{p}+\int_{\Omega }|v|^{p})-\frac{\lambda }{p}\int_{\partial
\Omega }|v|^{p}-\int_{\partial \Omega }hv\,.
\end{equation*}
(a) $\Phi $ is $C^{1}$ and weakly lower semi-continuous.

\noindent(b) Let us show that $\Phi $ is coercive:
Let $v\in W^{1,p}(\Omega )$ such that $\|v\|_{W^{1,p}(\Omega
)}\neq 0$, and we consider $u=\frac{v}{\|v\|_{W^{1,p}(\Omega )}}$
and $s=\|v\|_{W^{1,p}(\Omega )}$. So we have
\begin{align*}
\Phi (v) &=\Phi (su)\\
&=\frac{1}{p}s^{p}\Big[ \Big(\int_{\Omega }|
\nabla u|^{p}+\int_{\Omega }|u|^{p}\Big)-\lambda
\int_{\partial \Omega }|u|^{p}\Big] -s\int_{\partial \Omega }hu \\
&\geq \frac{1}{p}s^{p}(1-\frac{\lambda }{\lambda _{1}})
\|u\|_{W^{1,p}(\Omega )}^{p}-s\|h\|
_{L^{q}(\partial \Omega )}\|u\|_{L^{p}(\partial \Omega )} \\
&\geq \frac{1}{p}s^{p}(1-\frac{\lambda }{\lambda _{1}})-
\frac{s}{(\lambda _{1})^{\frac{1}{p}}}\|h\|
_{L^{q}(\partial \Omega )}
\end{align*}
Thus as $s\to +\infty $ one has $\Phi (su)\to +\infty $;  i. e.
$\Phi (v)\to +\infty $ as $\|v\|_{W^{1,p}(\Omega )}\to +\infty $,
so $\Phi $ is coercive.
The function $\Phi $ is weakly lower semi-continuous coercive,
so it admits a critical point $u$ which is solution of \eqref{e1.1}.

\noindent (ii) One shows that $u>0$ on $\overline{\Omega }$.
According to the regularity  and the theorem \ref{thm2.1}, one has
$u\in C^{1}(\overline{\Omega })$ and $u\geqslant 0$ on $\Omega $.
And with the maximum principle by Vazquez \cite{Va}, one gets $u>0$ on
$\Omega $. Moreover $\partial \Omega $ is $C^{2,\beta }$, so $u>0$ on
$\overline{\Omega }$, because if there exists $x_{0}\in \partial \Omega $
such that $u(x_{0})=0$, then
$\frac{\partial u}{\partial n}(x_{0})>0$; i.e.
$\frac{\partial u}{\partial \nu }(x_{0})<0$. However,
$|\nabla u|^{p-2}\frac{\partial u}{\partial \nu }
(x_{0})=\lambda |u|^{p-2}u(x_{0})+h(x_{0})=h(x_{0})$, then
$|\nabla u|^{p-2}\frac{\partial u}{\partial \nu }(x_{0})=h(x_{0})<0$.
That is impossible since $h\geqslant 0$ in $\partial \Omega $.
So $u>0$ on $\overline{\Omega }$.

\noindent (iii) Uniqueness of the solution:
By contradiction one supposes that there exist two nontrivial
solutions of \eqref{e1.1} $u$ and $v$ with $h\gvertneqq 0$ in
$\partial \Omega $. According to what precedes one has $u>0$ and $v>0$
on $\overline{\Omega }$. And Applying  lemma \ref{lem3.1} at $u$ and $ v$ one
gets
\begin{equation}
\lambda \int_{\partial \Omega }v^{p}+\int_{\partial \Omega }h(\frac{v^{p}}{
u^{p-1}})\leq \int_{\Omega }|\nabla v|^{p}+\int_{\Omega
}|v|^{p}=\lambda \int_{\partial \Omega }v^{p}+\int_{\partial
\Omega }hv.  \label{e3.4}
\end{equation}
Then
\begin{equation}
\lambda \int_{\partial \Omega }h(\frac{v^{p}}{u^{p-1}}
)-\int_{\partial \Omega }hv\leq 0,  \mbox{ i-e }  \int_{\partial
\Omega }hv(\frac{ v^{p-1}-u^{p-1}}{u^{p-1}})\leq 0.
\label{e3.5}
\end{equation}
By exchanging the roles of $u$ and $v$ one has
\begin{equation}
\int_{\partial \Omega }hu( \frac{u^{p-1}-v^{p-1}}{u^{p-1}})\leq
0. \label{e3.6}
\end{equation}
Inequalities \eqref{e3.5} and \eqref{e3.6} imply
$$
\int_{\partial \Omega }h\big[ u((\frac{u^{p-1}-v^{p-1}}{u^{p-1}})+v((
\frac{v^{p-1}-u^{p-1}}{u^{p-1}})\big] \leq 0;
$$
i.e.,
$\int_{\partial \Omega }h\big[(u^{p-1}-v^{p-1})(\frac{v}{u^{p-1}}
- \frac{u}{v^{p-1}})\big]\leq 0$.
Consequently
$$
 \int_{\partial \Omega}h\frac{(v^{p-1}-u^{p-1})(v^{p}-u^{p})}
{ u^{p-1}v^{p-1}}\leq 0.
$$
However,
$$
\frac{(v^{p-1}-u^{p-1})(v^{p}-u^{p})}{u^{p-1}v^{p-1}}\geq 0
$$
and
$h\geq 0$ in $\partial \Omega $. Then  \eqref{e3.5} and \eqref{e3.6} are
equalities, and one has equality in \eqref{e3.4}; that is true
if and only if $u$ is a multiple of $v$. Consequently there
exists $c\geq 0$ such that $ u=cv$ on $\Omega $ and since
$u>0$ and$ v>0$ we have $c>0$.
Replacing $u$ by $cv$ in  equation \eqref{e1.1}, one obtains
\begin{gather*}
\Delta _{p}(cv)=|cv|^{p-2}cv \quad \text{in } \Omega, \\
|\nabla cv|^{p-2}\frac{\partial (cv)
}{\partial \nu }=\lambda |cv|^{p-2}cv+h\quad  \text{on }
\partial \Omega.
\end{gather*}
This is equivalent to
\begin{gather*}
\Delta _{p}(v)=|v|^{p-2}v \quad \text{in } \Omega, \\
|\nabla v|^{p-2}\frac{\partial (v)}{\partial \nu } =\lambda |v|
^{p-2}v+\frac{h}{c^{p-1}}\quad \text{on }\partial \Omega.
\end{gather*}
 But $v$ is a solution of \eqref{e1.1} then
$\frac{h}{c^{p-1}}=h$, and since$ h\neq 0$ one has $c=1$ and
$u=v$.
So the uniqueness of the solution of \eqref{e1.1} in the case
$\lambda <\lambda_{1} $ and  $h\gvertneqq 0$ is proved.

The case $\lambda =\lambda _{1}$:
Suppose that there exists $u\in W^{1,p}(\Omega )$ a nontrivial
solution of  \eqref{e1.1}. One has $u\in C^{1}(\overline{\Omega }) $
and $u>0$ on $\overline{\Omega }$. And let $\varphi _{1}$ be a positive
eigenfunction associated to $\lambda _{1}$, then
$\varphi _{1}\in C^{1}(\overline{\Omega })$ and
$ \varphi _{1}>0 $ on $ \overline{\Omega}$. Applying
 lemma \ref{lem3.1} at $u$ and $ \varphi _{1}$ one gets
\[
\lambda _{1}\int_{\partial \Omega }\varphi _{1}^{p}+\int_{\partial \Omega }h(
\frac{\varphi _{1}^{p}}{u^{p-1}})\leq \int_{\Omega }|\nabla \varphi
_{1}|^{p}+\int_{\Omega }|\varphi _{1}|^{p}.
\]
However $\int_{\Omega }|\nabla \varphi _{1}|^{p}+\int_{\Omega }|\varphi
_{1}|^{p}=\lambda _{1}\int_{\partial \Omega }\varphi _{1}^{p}$, then
$\int_{\partial \Omega }h\frac{\varphi _{1}^{p}}{u^{p-1}}=0$ and
$ h\frac{\varphi _{1}^{p}}{u^{p-1}}=0\sigma $ a.e. in $\partial \Omega $. But
$\varphi _{1}>0$ and$  u>0$ on $\overline{\Omega }$ and  $h\neq 0$ in
$\partial \Omega $ a contradiction.
Then if $\lambda =\lambda _{1}$ and $h\gvertneqq 0$  in
$\partial \Omega $,  problem \eqref{e1.1} has no solution in
$W^{1,p}(\Omega )$.

The case of $\lambda >\lambda _{1}$:
By contradiction we suppose that there is $u\gvertneqq 0$ a solution of
\eqref{e1.1}. Then $u\in C^{1}(\overline{\Omega })$ and
$ u>0$ on $\overline{\Omega }$. Lemma \ref{lem3.1} implies
\begin{equation*}
\lambda \int_{\partial \Omega }\varphi ^{p}+\int_{\partial \Omega }h\frac{
\varphi ^{p}}{u^{p-1}}\leq \int_{\Omega }|\nabla \varphi |^{p}+\int_{\Omega
}|\nabla \varphi |^{p}
\end{equation*}
This implies that $\lambda \int_{\partial \Omega }\varphi ^{p}\leq
\int_{\Omega }|\nabla \varphi |^{p}+\int_{\Omega }|\nabla \varphi |^{p}$ for
all $ \varphi \in C^{1}(\overline{\Omega })$ with $\varphi \geq 0$. Also
by density $\lambda \int_{\partial \Omega }\varphi ^{p}\leq \int_{\Omega
}|\nabla \varphi |^{p}+\int_{\Omega }|\nabla \varphi |^{p}|^{p}$
for all $\varphi \in W^{1,p}(\Omega )$. Then
$$
\lambda \leq \frac{\int_{\Omega }|\nabla \varphi
|^{p}+\int_{\Omega }|\nabla \varphi |^{p}|^{p}}{\int_{\partial
\Omega }\varphi ^{p}} \quad   \forall \varphi \in W^{1,p}(\Omega )
 \mbox{ with }  \int_{\partial \Omega }\varphi ^{p}\neq 0
$$
Consequently $ \lambda \leq \lambda _{1}$, which is a contradiction, and the
result follows.
\end{proof}

\section{The anti-maximum principle }

In this part we study the anti-maximum principle for the
problem \eqref{e1.1}.

\begin{theorem} \label{thm4.1}
Given $h\in L^{\infty }(\partial \Omega )$ with $h\gneqq 0$, there exists
$\delta =\delta (h)>0$ such that if
$\lambda _{1}<\lambda <\lambda_{1}+\delta $,  and $u$ is a solution
of \eqref{e1.1}, then
\begin{equation}
 u<0 \quad \mbox{in }
\overline{\Omega } \quad \mbox{and}\quad  \frac{\partial u}{\partial v}<0
\quad\mbox{on }
\partial \Omega   \label{e4.1}
\end{equation}
\end{theorem}

It will thus be said that the antimaximum principle (AMP) holds
on the right  of $ \lambda _{1}$.

\begin{proof}
 By contradiction we suppose that for all
$k\in N^{\ast}$ there exists $(\mu _{k})_{k}\subset \mathbb{R})$
such that
$$
\lambda _{1}<\mu _{k}<\lambda _{1}+\frac{1}{k}
$$
and there exists $(u_{k})_{k}\subset W^{1,p}(\Omega )$ so that
\begin{equation} \label{Pk}
\begin{gathered}
\Delta _{p}u_{k}=|u_{k}|^{p-2}u_{k}+h \quad \text{in } \Omega, \\
|\nabla u_{k}|^{p-2}\frac{\partial u_{k}}{\partial
\nu }=\mu _{k}|u_{k}|^{p-2}u_{k}+h  \quad \text{on } \partial \Omega
\end{gathered}
\end{equation}
 and  $u_{k}$ does not check \eqref{e4.1}.
One has according to the regularity result
$ u_{k}\in C^{1,\alpha }(\overline{\Omega })$ for some
$\alpha \in ]0,1[$. And $\|u_{k}\|_{\infty }$ does not remain bounded.
Indeed, if there exists $M>0 $
such that $\|u_{k}\|_{\infty }\leq M$  for all $k$, then
$\|\Delta_{p}u_{k}\|=\||u_{k}|^{p-1}\|_{\infty }\leq M^{p-1}=
M'$, and we get also $\|u_{k}\|_{C^{1,\alpha }(\overline{\Omega })}\leq
K_{1} $ independently of $k$.

So since $ (u_{k})\subset C^{1,\alpha }(\overline{\Omega }){
\hookrightarrow }C^{1}(\overline{\Omega })$, compact, then for a subsequence
$u_{k}\to u$ in $ C^{1}(\overline{\Omega })$.
Moreover, if
\begin{gather*}
\Delta _{p}u_{k}=|u_{k}|^{p-2}u_{k}+h \quad \text{in} \Omega, \\
|\nabla u_{k}|^{p-2}\frac{\partial u_{k}}{\partial
\nu }=\mu _{k}|u_{k}|^{p-2}u_{k}+h \quad\text{on }\partial \Omega,
\end{gather*}
 then
\begin{gather*}
\Delta _{p}u=|u|^{p-2}u \quad \text{in }  \Omega, \\
|\nabla u|^{p-2}\frac{\partial u}{\partial \nu
}=\lambda _{1}|u|^{p-2}u+h \quad \text{on }\partial \Omega.
\end{gather*}
This contradicts the result of  theorem \ref{thm3.1} which ensures the
nonexistence of solution for \eqref{e1.1} when
$\lambda =\lambda _{1}$ and $h\gvertneqq 0$. Consequently
$ \|u_{k}\|_{\infty }\to +\infty $.

Let us consider $v_{k}=\frac{u_{k}}{\|u_{k}\|_{\infty }}$, so
 $\|v_{k}\|_{\infty }=1$ and as previously for a
subsequence $ v_{k}\to v$ in $ C^{1}(\overline{\Omega })$
with $\|v\|_{\infty }=1$. Also $v_{k}$ solves
\begin{equation} \label{Pk'}
\begin{gathered}
\Delta _{p}v_{k}=\frac{\Delta _{p}u_{k}}{\|u_{k}\|_{\infty
}^{p-1}}=|v_{k}|^{p-2}v_{k} \quad \text{in }  \Omega, \\
|\nabla v_{k}|^{p-2}\frac{\partial v_{k}}{\partial
\nu }= \frac{1}{\|u_{k}\|_{\infty }^{p-1}}|\nabla u_{k}|^{p-2}\frac{
\partial u_{k}}{\partial \nu }=\mu _{k}|v_{k}|^{p-2}v_{k}+
\frac{h}{\|u_{k}\|_{\infty }^{p-1}} \quad \text{on }  \partial
\Omega.
\end{gathered}
\end{equation}
Then
\begin{gather*}
\Delta _{p}v=|v|^{p-2}v \quad \text{in }  \Omega, \\
|\nabla v|^{p-2}\frac{\partial v}{\partial \nu
}=\lambda _{1}|v|^{p-2}v \quad \text{on } \partial \Omega.
\end{gather*}
So $v$ is a eigenfunction associated to $\lambda _{1}$ and
$v\neq 0$. Applying again the maximum principle one has ($v>0$ in
$\overline{\Omega }$)  or ($v<0$ in $ \overline{\Omega}$).

\noindent(i) If $v>0$ in $\overline{\Omega }$, then for $k$
sufficiently large we have $v_{k}>0$ in$ \overline{\Omega }$,
but $ v_{k}$ is a solution
of \eqref{e1.1} with $\lambda =\mu _{k}>\lambda _{1}$ and
$h'=\frac{h}{\|u_{k}\|_{\infty }^{p-1}}\gvertneqq 0$ on
$ \partial \Omega $ which
leads to a  contradiction with the theorem \ref{thm3.1}.

\noindent(ii) If $ v<0$ in $\overline{\Omega }$, one has
$|\nabla v|^{p-2}\frac{\partial v}{\partial \nu }=\lambda _{1}|v|
^{p-2}v<0 $ in $ \partial \Omega $, then
$\frac{\partial v}{\partial \nu }<0$ in $ \partial \Omega $.
So for $k$ sufficiently large$, (v_{k}<0$ in
$\overline{\Omega }$ and $\frac{\partial v_{k}}{\partial \nu }<0$ in
$\partial \Omega $).
This means that $u_{k}$ checks \eqref{e4.1}. Contradiction
with the assumption. This completes the proof of the
anti-maximum principle.
\end{proof}

Now we study the uniformity of this principle. One will show
that the AMP is nonuniform on the right of $\lambda _{1}$ when
$p\leq N$, and uniform when $p>N$. In the latter case one will
characterize the interval of uniformity. Moreover one shows that
the AMP still holds on the right of this interval but not
uniformly. For this, one shows accesses the following result.

\begin{lemma} \label{lem4.1}
If $p\leq N$ then $\lambda _{1}=\overline{\lambda _{1}}$.
If  $p>N$ then  $\lambda _{1}<\overline{\lambda _{1}}$, and the interval
$]\lambda _{1},\overline{\lambda _{1}}] $ does not
contain any eigenvalue where
$\overline{\lambda _{1}}=\inf_{u\in P}(\int_{\Omega }|
\nabla u|^{p}+\int_{\Omega }|u|^{p})$ and
$$
P=\big\{ u\in W^{1,p}(\Omega ): \int_{\partial \Omega }|u|
^{p}=1\text{ and $u$ vanishes in some ball in
}\overline{\Omega }\big\}.
$$
\end{lemma}

\begin{proof}
It is clear that $\lambda _{1}\leq \overline{\lambda _{1}}$.

\noindent (i) If  $p<N$: As in   Godoy,
Gossez and  Paczka \cite{G.G.P1},  one defines a sequence
$(v_{k})_{k}$ as follows: For $k\in \mathbb{N}^*$,
\[
\begin{cases}
1&\text{if }  |x|\geq 1/k \\
2k|x|-1 &\text{if } 1/(2k) <|x|< 1/k, \\
0 & \text{if } |x|\leq 1/2k
\end{cases}
\]
Note that $v_{k}$converges to the constant function 1 as $k\to+\infty $
in $W_{\rm loc}^{1,p}(R^{N})$. Given $x_{0}\in \Omega $, and $u$
a eigenfunction associated to $\lambda _{1}$; i.e.,
 $\lambda _{1}=\frac{\int_{\Omega }|\nabla u|^{p}+\int_{\Omega }|
u|^{p}}{\int_{\partial \Omega }|u|^{p}}$, then the
sequence ($w_{k}$) defined by $w_{k}(x)=u(x) v_{k}(x-x_{0})$ vanishes
in the ball $B(x_{0},\frac{1}{2k})$ and
\begin{equation*}
\overline{\lambda _{1}}\leq \frac{\int_{\Omega }|\nabla
w_{k}|^{p}+\int_{\Omega }|w_{k}|^{p}}{\int_{\partial
\Omega }|w_{k}|^{p}}\to\frac{\int_{\Omega }|
\nabla u|^{p}+\int_{\Omega }|u|^{p}}{
\int_{\partial \Omega }|u|^{p}} \text{as } k\to+\infty
\end{equation*}
So $\overline{\lambda _{1}}\leq \lambda _{1}$, then
$\overline{\lambda _{1}}=\lambda _{1}$.

\noindent(ii) If $p=N$: One defines a sequence $(v_{k})_{k}$ as follows:
For $k\in \mathbb{N}^*$,
\[
v_k=\begin{cases}
1-\frac{1}{2k} &\text{if } |x|\geq \frac{1}{k} \\
|x|^{\delta _{k}}-\frac{1}{k} &\text{if }(\frac{1}{k}
)^{1/\delta _{k}}<|x|<\frac{1}{k} \\
0 & \text{if } |x|\leq (\frac{1}{k})^{1/\delta _{k}},
\end{cases}
\]
where $\delta _{k}$ satisfies $(\frac{1}{k})^{\delta _{k}}=1-
\frac{1}{k}$. ($\delta _{k}=1-\frac{\ln (k+1)}{\ln (k)}\to 0$ as
$k\to+\infty $). And  $(w_{k})$ is as previously, and we show that
$$
\frac{\int_{\Omega }|\nabla w_{k}|
^{p}+\int_{\Omega }|w_{k}|^{p}}{\int_{\partial \Omega
}|w_{k}|^{p}} \to\frac{\int_{\Omega }|
\nabla u|^{p}+\int_{\Omega }|u|
^{p}}{\int_{\partial \Omega }|u|^{p}}\quad \text{as }
k\to+\infty
$$
Indeed
$$
\int_{\Omega }|w_{k}|^{p}=\int_{|x-x_{0}|\geq
\frac{1}{k}}|1-\frac{1}{2k}|^{p}|u(x)|
^{p}+\int_{(\frac{1}{k})^{\frac{1}{\delta _{k}}}\leq |
x-x_{0}|\leq \frac{1}{k}}\|x-x_{0}|^{\delta _{k}}-
\frac{1}{k}|^{p}|u(x)|^{p},
$$
and
$$
\int_{(\frac{1}{k})^{\frac{1}{\delta _{k}}}\leq
|x-x_{0}|\leq \frac{1}{k}}\|
x-x_{0}|^{\delta _{k}}- \frac{1}{k}|^{p}|
u(x)|^{p}\leq (1-\frac{2}{k} )^{p}\int_{(
\frac{1}{k})^{\frac{1}{\delta _{k}}}\leq |
x-x_{0}|\leq \frac{1}{k}}|u(x)|^{p}\to 0,
$$
consequently
$$
\overline{\lambda _{1}}\leq \frac{\int_{\Omega }|
\nabla w_{k}|^{p}+\int_{\Omega }|w_{k}|^{p}}{
\int_{\partial \Omega }|w_{k}|^{p}}\to\frac{
\int_{\Omega }|\nabla u|^{p}+\int_{\Omega
}|u|^{p}}{\int_{\partial \Omega }|u|
^{p}}=\lambda _{1},
$$
then $\overline{\lambda _{1}}=\lambda_{1}$.

\noindent (iii) When $p>N$, we have
\begin{equation*}
\overline{\lambda _{1}}=\inf_{u\in G} \Big(\int_{\Omega }|
\nabla u|^{p}+\int_{\Omega }|u|^{p}\Big)
\end{equation*}
where $G=\big\{ u\in W^{1,p}(\Omega ):\int_{\partial
\Omega }|u|^{p}=1\text{ and }\exists x_{0}\in \overline{
\Omega }: u(x_{0})=0\big\} $. Since
$W^{1,p}(\Omega)\underset{cpt}{\hookrightarrow }C(\overline{\Omega })$,
the minimum is achieved in a certain function
$\overline{u}$, that one can suppose positive on $\overline{\Omega }$. And
there exists $x_{0}\in \overline{\Omega }$ such that
$\overline{u}(x_{0})=0$. Let us show that it vanishes only
once on $\overline{\Omega}$.

One assumes, by contradiction, that there exists
$x_{1}\neq x_{0}\in \overline{\Omega }$ with  $\overline{u}(x_{1})=0$.
Setting $F=\{ u\in W^{1,p}(\Omega )
: \int_{\partial \Omega }|u|^{p}=1\;\text{and}
\;u(x_{0})=0\} $ and
$\mu =\inf_{u\in F}(\int_{\Omega }|\nabla u|^{p}+\int_{\Omega}|u|^{p})$,
one gets $\overline{\lambda _{1}}\leq \mu $ , $\overline{u}\in F$ and
$\overline{\lambda _{1}}=\int_{\Omega }|\nabla \overline{u}
|^{p}+\int_{\Omega }|\overline{u}|^{p}$ then $\overline{
\lambda _{1}}=\mu $.

Applying the standard theorem on Lagrange multipliers, we ensure
the existence of $(\alpha _{1},\alpha _{2})\in R^{2}$ such that
for all $w\in W^{1,p}(\Omega)$ with  $w(x_{0})=0$ one
has
$$
\Phi '(\overline{u}).w=\alpha _{1}\Psi _{1}'(\overline{u}).w
+\alpha _{2}\Psi _{2}'(\overline{u}).w=\alpha _{1}\Psi
_{1}'(\overline{u}).w
$$
where $\Psi _{1}(u)=\int_{\partial \Omega }|u|^{p}-1$,
$\Psi_{2}(u)=u(x_{1})$ and $ \Phi (u)=\int_{\Omega }|\nabla u|
^{p}+\int_{\Omega }|u|^{p}$. Then for all
$w\in W^{1,p}(\Omega ): w(x_{0})=0$ one has
$\int_{\Omega }|\nabla \overline{u}|^{p-2}\nabla \overline{u}\nabla w
+\int_{\Omega }\overline{u}^{p-1}w=\alpha_{1}\int_{\partial \Omega }
\overline{u}^{p-1}w$. Taking $w=\overline{u}$
one gets $\alpha _{1}=\overline{\lambda _{1}}$ and
\begin{equation}
\int_{\Omega }|\nabla \overline{u}|^{p-2}\nabla \overline{u}\nabla
w+\int_{\Omega }\overline{u}^{p-1}w=\overline{\lambda
_{1}}\int_{\partial \Omega }\overline{u}^{p-1}w \label{e4.2}
\end{equation}
 for all $w\in W^{1,p}(\Omega)$ with $w(x_{0})=0$.
By the same process, one has
$$
\int_{\Omega }|\nabla \overline{u} |^{p-2}\nabla
\overline{u}\nabla w+\int_{\Omega }\overline{u}^{p-1}w=
\overline{\lambda _{1}}\int_{\partial \Omega }\overline{u}^{p-1}w
$$
for all
$w\in W^{1,p}(\Omega)$  with $w(x_{1})=0$. Knowing
that for all $w\in W^{1,p}(\Omega )$, there exists
$(w_{0},w_{1})\in (W^{1,p}(\Omega ))^{2}$
such that $w=w_{0}+w_{1}$, $ w_{0}(x_{0})=0$ and
$w_{1}(x_{1})=0$, one arrives at the result:
For all $w\in W^{1,p}(\Omega )$,
\begin{equation}
\int_{\Omega }|\nabla
\overline{u} |^{p-2}\nabla \overline{u}\nabla w+\int_{\Omega
}\overline{u}^{p-1}w= \overline{\lambda _{1}}\int_{\partial \Omega
}\overline{u}^{p-1}w  \label{e4.3}
\end{equation}
Then $\overline{\lambda _{1}}$ is a eigenvalue and
$\overline{u}$ is a positive eigenfunction associated. However by
\cite[Lemma 2.4]{M.R}, $\overline{ \lambda _{1}}=\lambda _{1}$
and $\overline{u}>0$ on $\overline{\Omega }$
by the maximum principle. But this contradicted the assumption that
$\overline{u}$ vanishes in $\overline{\Omega }$. Consequently
$\overline{u}$ vanishes only once on $\overline{\Omega }$, and
$$
\overline{\lambda _{1}}=\int_{\Omega }|\nabla
\overline{u} |^{p}+\int_{\Omega }|\overline{u}|
^{p}=\inf_{u\in F}\Big(\int_{\Omega }|\nabla
u|^{p}+\int_{\Omega }|u|^{p}\Big)
$$
with $\overline{u}>0$ in $\overline{\Omega }\backslash \{
x_{0}\} $ and $\overline{u}(x_{0})=0$.

Now one considers $u_{\epsilon }(x)=\max (\overline{u}(x),\epsilon )$ and
$B_{\epsilon }=\{ x\in \overline{\Omega }:\overline{u}(x)<\epsilon
\} $ where $\epsilon >0$. One has $u_{\epsilon }\to \overline{u}$,
as $\epsilon\to 0$,  in $W^{1,p}(\Omega )$, and
\begin{equation*}
\int_{\partial \Omega }|u_{\epsilon }|^{p}\partial \sigma
=\int_{\partial \Omega \backslash B_{\epsilon }}|\overline{u}|
^{p}\partial \sigma +\epsilon ^{p}\int_{\partial \Omega \cap B_{\epsilon
}}\partial \sigma \underset{\epsilon \to 0}{\to}
\int_{\partial \Omega }|\overline{u}|^{p}\partial \sigma =1
\end{equation*}
Then $\int_{\partial \Omega }|u_{\epsilon }|^{p}\partial \sigma >0$
for $\epsilon $ enough small.
One hopes to show that
\begin{equation*}
\lambda _{1}\leq \frac{\int_{\Omega }|\nabla u_{\epsilon
}|^{p}+\int_{\Omega }|u_{\epsilon }|^{p}}{\int_{\partial
\Omega }|u_{\epsilon }|^{p}}<\overline{\lambda _{1}}
=\int_{\Omega }|\nabla \overline{u}|^{p}+\int_{\Omega
}|\overline{u}|^{p}
\end{equation*}
 for $\epsilon $ enough small.
We notice that
\begin{gather*}
\int_{\Omega }|u_{\epsilon }|^{p} =\int_{\Omega
}\overline{u} ^{p}-\int_{\Omega \cap B_{\epsilon
}}\overline{u}^{p}+\epsilon
^{p}\mathop{\rm meas}(\Omega \cap B_{\epsilon }), \\
\int_{\partial \Omega }|u_{\epsilon }|^{p}
=\int_{\partial \Omega }\overline{u}^{p}-\int_{B_{\epsilon }\cap
\partial \Omega }\overline{u
}^{p}+\epsilon ^{p}\mathop{\rm meas}_{\sigma }(B_{\epsilon }\cap \partial \Omega ), \\
\int_{\Omega }|\nabla u_{\epsilon }|^{p} =\int_{\Omega
}|\nabla \overline{u}|^{p}-\int_{\Omega \cap B_{\epsilon
}}|\nabla \overline{u}|^{p}
\end{gather*}
and we conclude that
\begin{align*}
A_{\varepsilon }
&=\frac{\int_{\Omega }|\nabla u_{\epsilon
}|^{p}+\int_{\Omega }|u_{\epsilon }|^{p}}{\int_{\partial
\Omega }|u_{\epsilon }|^{p}}-\int_{\Omega }|\nabla
\overline{u}|^{p}-\int_{\Omega }|\overline{u}|^{p} \\
&=\frac{\int_{\Omega }|\nabla u_{\epsilon }|
^{p}+\int_{\Omega }|u_{\epsilon }|^{p}-\int_{\partial \Omega
}|u_{\epsilon }|^{p}(\int_{\Omega }|\nabla
\overline{u}|^{p}+\int_{\Omega }\overline{u}^{p})}{
\int_{\partial \Omega }|u_{\epsilon }|^{p}} \\
&=\Big(\int_{\Omega }|\nabla \overline{u}|^{p}+\int_{\Omega
}|\overline{u}|^{p}\Big)C_{\epsilon }
\end{align*}
where
$$
C_{\epsilon }=\frac{\big(\int_{B_{\epsilon }\cap \partial \Omega }
\overline{u}^{p}-\epsilon ^{p}\mathop{\rm meas}_{\sigma }(B_{\epsilon }\cap \partial
\Omega )\big)-\big(\int_{\Omega \cap B_{\epsilon }}|\nabla
\overline{u}|^{p}+\int_{\Omega \cap B_{\epsilon }}\overline{u}
^{p}-\epsilon ^{p}mes(B_{\epsilon })\big)}{1-\int_{B_{\epsilon }\cap
\partial \Omega }|\overline{u}|^{p}+\epsilon ^{p}\mathop{\rm meas}_{\sigma
}(B_{\epsilon }\cap \partial \Omega )}.
$$
We will prove that
$A_{\varepsilon }/\varepsilon $ converges towards a non
positive value. For that we recall that $\overline{u}$ checks the
property \eqref{e4.2}, and considering the sequence
$v_{\epsilon }=\min (\overline{u},\epsilon )$ we get
\begin{equation*}
\int_{\Omega \cap B_{\epsilon }}|\nabla \overline{u}|^{p}+\int_{\Omega \cap
B_{\epsilon }}\overline{u}^{p}=\overline{\lambda _{1}}\int_{\partial \Omega
\cap B_{\epsilon }}\overline{u}^{p}+\epsilon (P_{\varepsilon }),
\end{equation*}
where $P_{\varepsilon }=\overline{\lambda _{1}}\int_{\partial
\Omega \backslash B_{\epsilon }}\overline{u}^{p-1}-\int_{\Omega
\backslash
B_{\epsilon }}\overline{u}^{p-1}\to\overline{\lambda _{1}}
\int_{\partial \Omega }\overline{u}^{p-1}-\int_{\Omega
}\overline{u}^{p-1}$ as $\varepsilon \to 0$.

 If $\overline{\lambda _{1}}\int_{\partial \Omega }\overline{u}
^{p-1}=\int_{\Omega }\overline{u}^{p-1}$, then  \eqref{e4.2} holds for
$w\equiv 1 $; consequently it holds also for all $\varphi $ in
$W^{1,p}(\Omega )$ because $\varphi =\psi +\varphi
(x_{0})w$ with $\psi =\varphi -\varphi (
x_{0})$. So $\overline{u}$ is a positive solution of the
problem
\begin{gather*}
\Delta _{p}v=|v|^{p-2}v \quad \text{in } \Omega,  \\
|\nabla v|^{p-2}\frac{\partial v}{\partial \nu
}=\overline{ \lambda _{1}}|v|^{p-2}v \quad \text{on }
\partial \Omega,
\end{gather*}
then $\overline{u}>0$ on $\overline{\Omega }$, absurdity since
$\overline{u}(x_{0})=0$. So we deduce that
$\overline{\lambda _{1}}\int_{\partial \Omega }\overline{u}^{p-1}
\neq \int_{\Omega }\overline{u}^{p-1}$.

 If $\overline{\lambda _{1}}\int_{\partial \Omega }\overline{u}
^{p-1}<\int_{\Omega }\overline{u}^{p-1}$, then since
$\int_{\partial \Omega \cap B_{\epsilon }}\overline{u}^{p}=o(\epsilon ^{p})$
one has
\begin{equation*}
\frac{1}{\epsilon }(\int_{\Omega \cap B_{\epsilon }}|\nabla
\overline{u}|^{p}+\int_{\Omega \cap B_{\epsilon }}\overline{u}
^{p})\underset{\epsilon \to0}{\to}\overline{\lambda
_{1}}\int_{\partial \Omega }\overline{u}^{p-1}-\int_{\Omega }\overline{u}
^{p-1}<0.
\end{equation*}
This implies that for $\epsilon $ enough small
$\int_{\Omega \cap B_{\epsilon }}|\nabla \overline{u}|^{p}
+\int_{\Omega \cap B_{\epsilon }}\overline{u}^{p}<0$, which is not true.
Then $\overline{\lambda _{1}}\int_{\partial \Omega }\overline{u}^{p-1}
>\int_{\Omega } \overline{u}^{p-1}$. Moreover
$1-\int_{B_{\epsilon }\cap \partial \Omega }
\overline{u}^{p}+\epsilon ^{p}\mathop{\rm meas}_{\sigma }(B_{\epsilon }
\cap \partial \Omega )\to 1$ as $\epsilon \to0$ and
$\int_{B_{\epsilon }\cap \partial \Omega }\overline{u}^{p}-\epsilon
^{p}\mathop{\rm meas}_{\sigma }(B_{\epsilon }\cap \partial \Omega )=o(\epsilon
^{p})$, then
\begin{equation*}
\frac{1}{\epsilon }(\frac{\int_{\Omega }|\nabla
u_{\epsilon }|^{p}+\int_{\Omega }|u_{\epsilon }|^{p}}{
\int_{\partial \Omega }|u_{\epsilon }|^{p}}-\int_{\Omega
}|\nabla \overline{u}|^{p}-\int_{\Omega }|\overline{u}
|^{p})\underset{\epsilon \to0}{\to}\overline{
\lambda _{1}}\int_{\partial \Omega }\overline{u}^{p-1}-\int_{\Omega }
\overline{u}^{p-1}<0
\end{equation*}
So for $\epsilon $ enough small
$\lambda _{1}\leq \frac{\int_{\Omega }|
\nabla u_{\epsilon }|^{p}+\int_{\Omega }|u_{\epsilon
}|^{p}}{\int_{\partial \Omega }|u_{\epsilon }|^{p}}
<\int_{\Omega }|\nabla \overline{u}|^{p}+\int_{\Omega
}|\overline{u}|^{p}=\overline{\lambda _{1}}$.
Consequently, $\lambda _{1}<\overline{\lambda _{1}}$. To complete
the proof, one shows that there is no eigenvalue in
$]\lambda _{1},\overline{\lambda _{1}}]$. Let us suppose by
absurdity that there exists an eigenvalue $\mu $ in
$]\lambda _{1},\overline{\lambda _{1}}] $, with associated eigenfunction $v$.
 By \cite[Lemma 2.4]{M.R}, $v$ changes sign on $\partial \Omega $,
consequently it vanishes somewhere in $\overline{\Omega }$. Then
$v\in G$, so
\begin{equation*}
\overline{\lambda _{1}}\int_{\partial \Omega }|v|^{p}\leq
\|\nabla v\|_{L^{p}(\Omega )}^{p}+\|v\|
_{L^{p}(\Omega )}^{p}=\mu \int_{\partial \Omega }|v|^{p}
\end{equation*}
Since $\int_{\partial \Omega }|v|^{p}\neq 0$, one
concluded that $\overline{\lambda _{1}}=\mu $ . In this case $v$
satisfies \eqref{e4.3},
and it vanishes only once on $\overline{\Omega }$. Applying \eqref{e4.3} to
$w=v^{-}$ one obtains
\begin{equation*}
\overline{\lambda _{1}}\int_{\partial \Omega }|v^{-}|
^{p}=\|\nabla v^{-}\|_{L^{p}(\Omega
)}^{p}+\|v^{-}\|_{L^{p}(\Omega )}^{p},
\end{equation*}
then $v^{-}$ vanishes only once on $\overline{\Omega }$. Contradiction
since $v$ changes sign on $\partial \Omega $.
\end{proof}

\begin{theorem} \label{thm4.2}
If  $p>N$
\begin{itemize}
\item[(i)] For all $h\in L^{\infty }(\partial \Omega )$, such that
$h\gneqq 0$ in $ \overline{\Omega }$, if $\lambda \in ]
\lambda _{1},\overline{\lambda _{1}}]$, then any solution $u$
 of \eqref{e1.1} satisfies $u<0$ in
$\overline{\Omega }$.
\item[(ii)] $\overline{\lambda _{1}}$ is the largest number such that the
preceding implication holds.
\item[(iii)] Given $h\in L^{\infty }(\partial \Omega )$, $h\gneqq 0$ in $\partial
\Omega $, there exists $\delta =\delta (h)>0$ such that if
$\overline{\lambda _{1}}<\lambda <\overline{\lambda _{1}}+\delta $, and $u$
is a solution of \eqref{e1.1} then $u<0$ on $\overline{\Omega }$.
\end{itemize}
\end{theorem}

\begin{theorem} \label{thm4.3}
The AMP is not uniform on the right of $\overline{\lambda _{1}}$
for all $1<p<\infty $; i.e., there is no $\delta $ independent of
$h$ satisfying (i).
\end{theorem}

The proof of this result is a consequence of the following
theorem.

 \begin{theorem} \label{thm4.4}
Given $\epsilon >0$, there exists $h\in L^{\infty }(\partial \Omega )$, such
that $h\gneqq 0$ in $\partial \Omega $ and if
$\overline{\lambda _{1}} +\epsilon <\lambda $ then  problem \eqref{e1.1}
does not admit any non-positive solution.
\end{theorem}

\textbf{Proof of theorem \ref{thm4.2}}
(i) Given $h\in L^{\infty }(\partial \Omega )$, such that $h\gneqq 0$ in
$\partial \Omega $, and $\lambda \in ]\lambda _{1},\overline{\lambda
_{1}}] $, then if $u$ is a solution of \eqref{e1.1}, it does not satisfy
$u\gvertneqq 0$ on $\overline{\Omega }$ by the Theorem \ref{thm3.1}. Taking
$u^{-}\neq 0$ as testing function in \eqref{e1.1}, one gets
\begin{equation*}
\int_{\Omega }|\nabla u^{-}|^{p}+\int_{\Omega }|
u^{-}|^{p}=\lambda \int_{\partial \Omega }|
u^{-}|^{p}-\int_{\partial \Omega }hu^{-}\leq \lambda
\int_{\partial \Omega }|u^{-}|^{p}.
\end{equation*}
If $\int_{\partial \Omega }|u^{-}|^{p}=0$, then
$\int_{\Omega }|\nabla u^{-}|^{p}+\int_{\Omega
}|u^{-}|^{p}=0$, consequently $u^{-}\equiv 0$, this
is not true, so $\int_{\partial \Omega }|u^{-}|
^{p}>0$. One deduces that
\begin{equation*}
\frac{\int_{\Omega }|\nabla u^{-}|^{p}+\int_{\Omega
}|u^{-}|^{p}}{\int_{\partial \Omega }|
u^{-}|^{p}}\leq \lambda.
\end{equation*}
If $\lambda <\overline{\lambda _{1}}$ then
\begin{equation*}
\frac{\int_{\Omega }|\nabla u^{-}|^{p}+\int_{\Omega
}|u^{-}|^{p}}{\int_{\partial \Omega }|
u^{-}|^{p}}\leq \lambda <\overline{\lambda
_{1}}=\underset{u\in G}{\inf }\Big(\int_{\Omega }|
\nabla u|^{p}+\int_{\Omega }|u|^{p}\Big),
\end{equation*}
so $u^{-}\notin G$ i-e $u^{-}$ does not vanish any where on
$\overline{ \Omega }$. Consequently $u<0$ on $\overline{\Omega
}$.
If $\lambda =\overline{\lambda _{1}}$ then
$\frac{\int_{\Omega }|\nabla u^{-}|^{p}+\int_{\Omega }|u^{-}|
^{p}}{ \int_{\partial \Omega }|u^{-}|^{p}}\leq
\overline{\lambda _{1}}$. Two cases are distinguished: If
$u^{-}\notin G$ then $u<0$ on $\overline{ \Omega }$. But if
$u^{-}\in G$ then $\frac{\int_{\Omega }|\nabla u^{-}|
^{p}+\int_{\Omega }|u^{-}|^{p}}{\int_{\partial \Omega
}|u^{-}|^{p}}=\overline{\lambda _{1}}$, and $u^{-}$
vanishes only once on $\overline{\Omega }$. In this case one has
the equality
\begin{equation*}
\int_{\partial \Omega }hu^{-}=\lambda \int_{\partial \Omega
}|u^{-}|^{p}-\int_{\Omega }|\nabla u^{-}|
^{p}+\int_{\Omega }|u^{-}|^{p}=0,
\end{equation*}
so $u^{-}$ vanishes on the set where $h\neq 0$ which is of
positive measure. Absurdity. So $u^{-}\notin G$ and as previously
$u<0$ on $\overline{\Omega }$.

\noindent (ii) This part is a consequence of Theorem \ref{thm4.4}.

\noindent(iii) This result is similar as Theorem \ref{thm4.1}.
As in his proof we suppose, by contradiction, that for all
$k\in N^{\ast}$ there exists $(\mu _{k})_{k}\subset \mathbb{R})$
such that $\overline{ \lambda _{1}}<\mu _{k}<\overline{\lambda
_{1}}+\frac{1}{k}$, and there exists $(u_{k})_{k}\subset
W^{1,p}(\Omega )$ holding
\begin{gather*}
\Delta _{p}u_{k}=|u_{k}|^{p-2}u_{k}+h\quad \text{in }\Omega,\\
|\nabla u_{k}|^{p-2}\frac{\partial u_{k}}{\partial
\nu }=\mu _{k}|u_{k}|^{p-2}u_{k}+h\quad  \text{in }
\partial \Omega,
\end{gather*}
where $u_{k}$ is not non-positive on $\overline{\Omega }$,
and $ u_{k}\in C^{1,\alpha }(\overline{\Omega })$ for
some $\alpha \in ]0,1 [$.
One distinguishes two cases, either $\|u_{k}\|_{\infty }$ remains
bounded, or $\|u_{k}\|_{\infty }\to+\infty $. In the
first case, one has for a subsequence $u_{k}\to u$ in
$C^{1}(\overline{\Omega })$, and $u$  solves
\begin{gather*}
\Delta _{p}u=|u|^{p-2}u \quad \text{in }  \Omega, \\
|\nabla u|^{p-2}\frac{\partial u}{\partial \nu
}=\overline{ \lambda _{1}}|u|^{p-2}u+h \quad \text{on }
\partial \Omega.
\end{gather*}
 By part (i) one gets $u<0$ on $\overline{\Omega }$.
Contradiction. In the second case one considers
$v_{k}=\frac{u_{k}}{\|u_{k}\|_{\infty }}$, so
$\|v_{k}\|_{\infty }=1$ and as previously for a subsequence
$v_{k}\to v$ in $ C^{1}(\overline{\Omega })$ with $\|v\|_{\infty }=1$,
and $v$ solves
\begin{gather*}
\Delta _{p}v=|v|^{p-2}v \quad \text{in } \Omega,  \\
|\nabla v|^{p-2}\frac{\partial v}{\partial \nu
}=\overline{ \lambda _{1}}|v|^{p-2}v \quad \text{on }
\partial \Omega.
\end{gather*}
But this yields a contradiction since by Lemma \ref{lem4.1},
$\overline{\lambda_{1}}$ is not an eigenvalue.


\begin{proof}[Proof of theorem \ref{thm4.4}]
 Assume by contradiction that
there exists $\varepsilon >0$ such that for any $h\gneqq 0$ there
exists $\lambda (h)$ with $\lambda (h)\geq \overline{ \lambda
_{1}}+\varepsilon $ such that \eqref{e1.1} has a solution $u(h)<0$ in
$\overline{\Omega }$. We consider $\varphi \in C^{1}(\overline{\Omega })$
satisfying $ \int_{\partial \Omega }|\varphi |^{p}\neq 0$ and $\varphi $
vanishes in some ball in $\overline{\Omega }$, and we choose
$h\in L^{\infty }(\partial \Omega )$, such that $h\gneqq 0$ in
$\partial \Omega $ and
$ \mathop{\rm supp}\varphi \cap \mathop{\rm supph}\cap \partial \Omega
 =\emptyset $.
Applying Lemma \ref{lem3.1} to $|\varphi |$ and $v=-u(h)>0$
in $\overline{\Omega }$, we obtain
\begin{equation*}
\lambda (h)\int_{\partial \Omega }|\varphi |^{p}\leq
\int_{\Omega }|\nabla \varphi |^{p}+\int_{\Omega }|\varphi
|^{p},
\end{equation*}
Then
\begin{equation*}
\overline{\lambda _{1}}+\varepsilon \leq \lambda (h)\leq
\frac{\int_{\Omega }|\nabla \varphi |^{p}+\int_{\Omega }|
\varphi |^{p}}{ \int_{\partial \Omega }|\varphi
|^{p}},
\end{equation*}
for all $\varphi \in W^{1,p}(\Omega )$ satisfying
$\int_{\partial \Omega }|\varphi |^{p}\neq 0$ and $\varphi $
vanishes in some ball in $\overline{\Omega }$, which contradicts
the definition of $\overline{\lambda _{1}}$.
\end{proof}

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\end{document}