\documentclass[reqno]{amsart}
\usepackage{hyperref}

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2008(2008), No. 78, pp. 1--13.\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 2008 Texas State University - San Marcos.}
\vspace{8mm}}

\begin{document}
\title[\hfilneg EJDE-2008/78\hfil Boundary eigencurve problems]
{Boundary eigencurve problems involving
the p-Laplacian operator}

\author[A. El Khalil, M. Ouanan\hfil EJDE-2008/78\hfilneg]
{Abdelouahed El Khalil, Mohammed Ouanan}  % in alphabetical order

\address{Abdelouahed El Khalil  \newline
Department of mathematics, College of Sciences\\
Al-Imam Muhammad Ibn Saud Islamic University\\
P. Box 90950, Riyadh 11623 Saudi Arabia}
\email{alakhalil@imamu.edu.sa \quad lkhlil@hotmail.com}

\address{Mohammed Ouanan \newline
University Moulay Ismail\\ 
Faculty of  Sciences et Technology\\
Department of informatics\\
P.O. Box 509, Boutalamine\\
52000 Errachidia, Morocco} 
\email{m\_ouanan@hotmail.com}

\thanks{Submitted September 16, 2007. Published May 27, 2008.}
\subjclass[2000]{35P30, 35J20, 35J60}
\keywords{p-Laplacian operator; nonlinear boundary conditions;
\hfill\break\indent principal eigencurve; Sobolev trace embedding}

\begin{abstract}
 In this paper, we show that for each $\lambda \in \mathbb{R}$, there is
 an increasing  sequence of eigenvalues for the
 nonlinear boundary-value problem
 \begin{gather*}
 \Delta_pu=|u|^{p-2}u \quad \text{in }   \Omega\\
 |\nabla u|^{p-2}\frac{\partial u}{\partial \nu}=\lambda
 \rho(x)|u|^{p-2}u+\mu|u|^{p-2}u \quad \text{on } \partial \Omega\,;
 \end{gather*}
 also we show that the first eigenvalue is simple and isolated.
 Some results about their variation, density,  and continuous
 dependence on the parameter $\lambda$ are obtained.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{proposition}[theorem]{Proposition}
\newtheorem{corollary}[theorem]{Corollary}
\newtheorem{remark}[theorem]{Remark}
\allowdisplaybreaks

\textbf{Editor's note:}
After publication, we learned that this article is an unauthorized
copy of 
``On the principal eigencurve of the p-Laplacian related to the 
Sobolev trace embedding'', Applicationes Mathematicae, 32, 1 (2005), 1-16.
The authors alone are responsible for this action which 
may be in violation of the Copyright Laws.

\section{Introduction and Notation}

Let $\Omega$ be a smooth bounded domain in $\mathbb{R}^N$, with $N\geq 1$.
Let $\rho$ be a function in $L^{\infty}(\partial\Omega)$ with
$\rho\not\equiv 0$ and that can change sign.
Let $\lambda, p, \mu$ be real numbers, with $1<p<\infty$.
 We are interested in the  nonlinear boundary-value problem
\begin{gather}
\Delta_{p} u=|u|^{p-2}u \quad \text{in } \Omega \label{e1.1} \\
|\nabla u|^{p-2}\frac{\partial u}{\partial \nu}=\lambda
\rho(x)|u|^{p-2}u+\mu|u|^{p-2}u \quad \text{on }  \partial\Omega.
\label{e1.2}
\end{gather}
Here $\Delta_{p}u=\nabla (|\nabla u|^{p-2}\nabla u)$, which is
known as the $p$-Laplacian and has attracted a lot of attention
because of its applications. It appears in mathematical models for
subject such as glaciology, nonlinear diffusion, filtration
problem \cite{p2}, power-low materials \cite{g1},  non-Newtonian
fluids \cite{a4}, reaction-diffusion problems, flow through porous
media, nonlinear elasticity, petroleum extraction, torsional creep
problems, etc. For a discussion and some physical background, we
refer the reader to \cite{d1}. The nonlinear boundary condition
\eqref{e1.2} describes a flux through the boundary
$\partial\Omega$ which depends on the solution itself. For a
physical motivation of such conditions, see for example \cite{p1}.

Observe that in the restrictive cases $\mu=0$ or $p=2$,
\eqref{e1.1}--\eqref{e1.2} becomes linear and it is known as the
Steklov problem \cite{b3}.

Classical Dirichlet problems involving the p-Laplacian have been
extensively  studied by various authors in the cases:
$\lambda=0$ and $\mu=0$; we cite the works \cite{a1,a2,a3,d1,s1,s2}.
For the nonlinear boundary condition  \eqref{e1.2},
recently the authors in \cite{b4} studied
the case when $\mu=0$ and $\rho$ belongs to  $L^s(\partial\Omega)$, which is
not necessary essentially bounded, with an additional  condition
on its sign.

We set
\begin{equation}
\mu_{1}(\lambda)=\inf\big\{\|v\|_{1,p}^p-\lambda\int_{\partial
\Omega}\rho(x)|v|^{p}d\sigma: v\in W^{1,p}(\Omega),\int_{\partial
\Omega}|u|^pd\sigma=1\big\}, \label{e1.3}
\end{equation}
where $\|\cdot\|_{1,p}$ denotes the $W^{1,p}(\Omega)$-norm; i.e.,
$\|v\|_{1,p}=(\|\nabla v\|_p^p+\|v\|_p^p)^{1/p}$ and $\|\cdot\|_p$
is the $L^p$-norm, with $\sigma$ is the Lebesgue measure of
$\mathbb{R}^{N-1}$. We understand by the principal (or first)
eigencurve of the $p$-Laplacian related to Sobolev trace
embedding, the graph of the map
$\mu_{1}:\lambda\to\mu_{1}(\lambda)$ from $\mathbb{R}$ into
$\mathbb{R}$. In \cite{e1}, the authors proved the simplicity and
isolation of the first eigencurve of Dirichlet p-Laplacian by
extending a similar result shown by Binding and Huang in
\cite{b2}.

Our purpose of this paper is to extend some of the results
known in the ordinary Dirichlet p-Laplacian, by using suitable
Sobolev trace embeddings which lead to a nonlinear eigenvalue
problem where the two parameter eigenvalues appear at the
nonlinear boundary condition. We show that $\mu_{1}(\lambda)$
is simple and isolated for any $\lambda\in\mathbb{R}$.
Note that to show the simplicity (uniqueness) result,
we use a simple convexity argument, by remarking that the
energy functional associated with
 \eqref{e1.1}--\eqref{e1.2}  is convex in $u^{p}$ for nonnegative
 functions $u$, without use in any way
$C^1(\Omega)$ and $L^{\infty}(\Omega)$ regularities of the eigenfunctions.
Here our process is new.

Remark that $\mu_{1}(0)=\lambda_{1}$ the optimal reciprocal
constant of the Sobolev embedding
$W^{1,p}(\Omega)\hookrightarrow L^p(\partial\Omega)$.
For the particular case $\mu=0$ and
$\rho \in L^{s}(\partial \Omega)$ (for a suitable $s$), the isolation
and simplicity of the first
eigenvalue of \eqref{e1.1}--\eqref{e1.2} are studied by \cite{b4}.

The main objective of our work is to extend this result to any
$\lambda\in\mathbb{R}$, by using new technical methods.

The rest of the paper is organized as follows. In Section 2, we
establish some definitions and preliminaries. In Section 3, we use
a variational method to prove the existence of a sequence of
eigencurves of \eqref{e1.1}--\eqref{e1.2}. In Section 4, we prove the
simplicity and the isolation results of each point of the first
eigencurve. Finally, in Section 5, we show some results about
variations of the weight as a direct application of the simplicity
result.

\subsection*{Definitions}
In this paper, all solutions are weak solutions; i.e.,
$u\in W^{1,p}(\Omega)$ is a solution of \eqref{e1.1}--\eqref{e1.2},
if for all $v\in W^{1,p}(\Omega)$,
\begin{equation}
\int_{\Omega}|\nabla u|^{p-2}\nabla u\nabla vdx+\int_\Omega |u|^{p-2}uvdx=
\int_{\partial\Omega}(\lambda \rho(x)+\mu)|u|^{p-2}uvd\sigma.\label{e2.1}
\end{equation}
If $u\in W^{1,p}(\Omega)\backslash \{0\}$, then $u$ shall be called an
eigenfunction of
\eqref{e1.1}--\eqref{e1.2} associated with the eigenpair
$(\lambda,\mu)$.

Set
\begin{equation}
\mathcal{M}=\big\{u\in W^{1,p}(\Omega): \int_{\partial
\Omega } |u|^pd\sigma=1\big\}.\label{e2.2}
\end{equation}
We say that a principal eigenfunction of \eqref{e1.1}--\eqref{e1.2},
an any eigenfunction $u\in \mathcal{M}$, $u\geq 0$ a.e. on
$\bar\Omega$ associated to pair
$(\lambda, \mu_{1}(\lambda))$.

Define the energy functionals on $W^{1,p}(\Omega)$ as
$$
\Phi_{\lambda}(u)=\frac{1}{p}\|u\|_{1,p}^p
 -\frac{\lambda}{p}\int_{\partial\Omega}\rho(x)|u|^pd\sigma=
\frac{1}{p}\|u\|_{1,p}^p+\Phi(u),\quad \lambda\in\mathbb{R}
$$
and
$$
\Psi(u)=\frac{1}{p}\int_{\partial\Omega}|u|^{p}d\sigma.
$$
It is clear that for any $\lambda\in\mathbb{R}$, solutions of
\eqref{e1.1}--\eqref{e1.2} are the critical points  of $\Phi_{\lambda}$
restricted to $\mathcal{M}$. We shall deal with operators $T$
acting from $W^{1,p}(\Omega)$ into $(W^{1,p}(\Omega))'$. $T$ is
said to belong to the class $(S_{+})$, if for any sequence $v_{n}$
weakly convergent to $v$ in $W^{1,p}(\Omega)$, and
$\limsup_{n\to +\infty}\langle Tv_{n}, v_{n}-v\rangle
\leq0$, it follows that $v_{n}\to v$ strongly in
$W^{1,p}(\Omega)$, where $(W^{1,p}(\Omega))'$ is the dual of
$W^{1,p}(\Omega)$ with respect to the pairing $\langle\cdot,\cdot\rangle$.

\section{Existence Results}

We will use Ljusternick-Schnirelmann theory on $C^1$-manifolds,
see \cite{s2}.
It is clear that for any $\lambda\in\mathbb{R}$, the functional
$\Phi_{\lambda}$ is even and bounded from below on $\mathcal{M}$.
Indeed, let $u\in\mathcal{M}$, then
$$
\Phi_{\lambda}(u)\geq
\frac{1}{p}(\|u\|_{1,p}^p-|\lambda|\|\rho\|_{\infty,\partial
\Omega}).
$$
So that
\begin{equation}
\Phi_{\lambda}(u)\geq
\frac{1}{p}(\lambda_{1}-|\lambda|\|\rho\|_{\infty,\partial \Omega})>-\infty,
\label{e3.1}
\end{equation}
where $\lambda_{1}=\mu_{1}(0)$ is the reciprocal of the optimal
constant in the Sobolev trace embedding
$W^{1,p}(\Omega)\hookrightarrow L^p(\partial \Omega)$.
By employing the Sobolev trace embedding, we deduce that
\begin{itemize}
    \item $\Psi$ and $\Phi$ are weakly continuous
    \item $\Psi'$ and $\Phi'$ are compact.
\end{itemize}

The following lemma is the key to show the existence of eigenvalues.

\begin{lemma} \label{lem3.1}
For each $\lambda\in\mathbb{R}$, we have
\begin{enumerate}
\item[(i)] $(\Phi_{\lambda})'$ maps the bounded sets in the
bounded sets;

\item[(ii)] if $u_{n}\rightharpoonup u$ (weakly)
in $W^{1,p}(\Omega)$ and $(\Phi_{\lambda})'(u_{n})$ converges
strongly in the space
 $(W^{1,p}(\Omega))'$,  then $u_n\to u$
(strongly) in $W^{1,p}(\Omega)$;

\item[(iii)] the functional
$\Phi_{\lambda}$ satisfies the Palais-Smale condition on
$\mathcal{M}$; i.e., for each sequence $(u_{n})_{n}\subset\mathcal{M}$,  if
$\Phi_{\lambda}(u_{n})$ is bounded and
\begin{equation}
(\Phi_{\lambda})'(u_{n})-c_{n}\Psi'(u_n)\to 0,
\label{e3.2}
\end{equation}
 with $c_n=\frac{\langle (\Phi_{\lambda})'(u_n),u_n\rangle}{\langle
\Psi'(u_n),u_n\rangle}$. Then, $(u_n)_n$ has a convergent
subsequence in $W^{1,p}(\Omega)$.
\end{enumerate}
\end{lemma}

\begin{proof} (i) Let $u,v \in W^{1,p}(\Omega).$ Thus
$$
\langle (\Phi_{\lambda})'(u),v\rangle =\int_{\Omega}|\nabla
u|^{p-2}\nabla u\nabla v dx+\int_{\Omega}|u|^{p-2}uv
dx+\int_{\partial\Omega}\rho(x)|u|^{p-2}uvd\sigma.
$$
By H\"older's inequality, we obtain
\begin{align*}
  |\langle(\Phi_{\lambda})'(u),v\rangle|
&\leq \Big(\int_{\Omega}|\nabla
u|^{(p-1)p'}dx\Big)^{1/p'} \|\nabla v\|_{p}
+\Big(\int_{\Omega}| u|^{(p-1)p'}dx\Big)^{1/p'}\|v\|_{p} \\
 &\quad + |\lambda|\|\rho\|_{\infty,\partial\Omega}
\Big(\int_{\partial\Omega}|
u|^{(p-1)p'}d\sigma\Big)^{1/p'}\|v\|_{p,\partial\Omega} \\
&= \|\nabla u\|_{p}^{p-1}\|\nabla
v\|_{p}+\|u\|_{p}^{p-1}\|v\|_{p}+
|\lambda|\|\rho\|_{\infty,\partial\Omega}
\|u\|_{p,\partial\Omega}^{p-1}\|v\|_{p,\partial\Omega}.
\end{align*}
Now, the trace Sobolev embedding
$W^{1,p}(\Omega)\hookrightarrow L^{p}(\partial\Omega)$ ensures
the existence of a constance $c>0$
such that
$$
\|w\|_{p,\partial\Omega}\leq c\|w\|_{1,p}, \quad \text{for each }
w\in W^{1,p}(\Omega).
$$
Therefore,
$$
\|(\Phi_{\lambda})'(u)\|\leq \|\nabla u\|_{p}^{p-1}\|\nabla
v\|_{p}+\|u\|_{p}^{p-1}\|v\|_{p}+
c^{p}|\lambda|\|\rho\|_{\infty,\partial\Omega})\|u\|_{1,p}^{p-1}\|v\|_{1,p}.
$$
It is clear that
$$
\|\nabla u\|_{p}^{p-1}\|\nabla
v\|_{p}+\|u\|_{p}^{p-1}\|v\|_{p}\leq \|u\|_{1,p}^{p-1}\|v\|_{1,p}.
$$
Combining the above inequalities, we conclude that
$$
|\langle(\Phi_{\lambda})'(u),v\langle|\leq
(1+c^{p}|\lambda|\|\rho\|_{\infty,\partial\Omega})\|u\|_{1,p}^{p-1}\|v\|_{1,p},
$$
for any $u,v\in W^{1,p}(\Omega)$. It follows that
$$
\|(\Phi_{\lambda})'(u)\|\leq
(1+c^{p}|\lambda|\|\rho\|_{\infty,\partial\Omega})\|u\|_{1,p}^{p-1},
$$
where $\|\cdot\|$ denotes the norm of  $(W^{1,p}(\Omega))'$.
This implies assertion (i).

(ii) We use the condition $(S_+)$ as follows.
$(\Phi_{\lambda})'(u_n)$ being a convergent sequence strongly to
some $f\in (W^{1,p}(\Omega))'$. Thus, we have by calculation
\begin{equation}
\langle Au_{n},v\rangle =\langle
-\Delta_{p}u_{n},v\rangle+\int_{\Omega}|u_{n}|^{p-2}u_{n}vdx+
\int_{\partial\Omega}|\nabla u_{n}|^{p-2}\nabla u_{n}\nu
v\,d\sigma,\label{e3.3}
\end{equation}
for any $v\in W^{1,p}(\Omega)$, where $A$ is an operator defined
from $W^{1,p}(\Omega)$ into $(W^{1,p}(\Omega))'$ by
$$
\langle Au,v\rangle =\int_{\Omega}|\nabla u|^{p-2}\nabla u\nabla v
dx+\int_{\Omega}|u|^{p-2}uv\,dx.
$$
This operator  satisfies the condition $(S_{+})$ because
$-\Delta_{p}$ does it \cite{e1}.

If we take $v=u_{n}-u$ in \eqref{e3.3} we obtain
\begin{align*}
&\langle Au_{n},u_{n}-v\rangle\\
&=\langle -\Delta_{p}u_{n},u_{n}-v\rangle+\int_{\Omega}
|u_{n}|^{p-2}u_{n}(u_{n}-u)dx+\int_{\partial\Omega}|\nabla
u_{n}|^{p-2}\nabla u_{n}\nu (u_{n}-u)\, d\sigma.
\end{align*}
Introducing $(\Phi_{\lambda})'(u_{n})$, we deduce that
$$
\langle Au_{n},u_{n}-u\rangle=\langle
(\Phi_{\lambda})'(u_{n})-f,u_{n}-u\rangle +\langle
f,u_{n}-u\rangle -\langle(\Phi_{\lambda})'(u_n),u_n-u\rangle.
$$
Using the compactness of $\Phi'$, we find that as
$n\to \infty$,
$$
\limsup_{n\to +\infty}\langle Au_{n},u_{n}-u\rangle \geq 0.
$$
Hence $u_{n}\to u$ strongly in $W^{1,p}(\Omega)$, in virtue of the
condition $(S_{+})$.

(iii) From \eqref{e3.1} we deduce that $(u_{n})_{n}$ is bounded in
$W^{1,p}(\Omega)$. Thus, without loss of generality, we can assume
that $u_{n}\rightharpoonup u$ (weakly) in $W^{1,p}(\Omega)$ for
some function $u\in W^{1,p}(\Omega)$. It follows that
$\Psi'(u_{n})\to \Psi'(u)$ in $(W^{1,p}(\Omega))'$ and
$p\Psi(u)=1$, because $p\Psi(u_{n})=1, \forall n\in \mathbb{N}^*$.
Hence $u\in\mathcal{M}$. Since $(u_{n})_{n}$ is bounded, then
$(i)$ ensures that $\{(\Phi_{\lambda})'(u_{n})\}$ is bounded. By a
calculation we obtain via $\eqref{e3.2}$ that
$\{(\Phi_{\lambda})'(u_{n})\}$ converges strongly in
$(W^{1,p}(\Omega))'$.  Consequently, from (ii) we conclude that
$u_{n}\to u$ (strongly) in $W^{1,p}(\Omega)$. This achieves the
proof of Lemma.
\end{proof}

 Set $\Gamma_{k}=\left\{K\subset
\mathcal{M}: K\text{ symmetric, compact and }\  \gamma(K)=k
\right\}$, where $\gamma(K)=k$ is the genus of $K$; i.e., the
smallest integer $k$ such that there is an odd continuous map from
$K$ to $\mathbb{R}^k\backslash\{0\}$.

Next, we establish our existence result.

\begin{theorem} \label{thm3.1}
For each $\lambda\in \mathbb{R}$ and each integer $k\in \mathbb{N}^*$,
$$
\mu_{k}(\lambda):=\inf_{K\in \Gamma_{k}}\max_{u\in
K}\Phi_{\lambda}(u)
$$
is a critical value of $\Phi_{\lambda}$
restricted to $\mathcal{M}$. More precisely, there exists
$u_{k}(\lambda)\in\mathcal{M}$ such that
$$
\mu_{k}(\lambda)=p\Phi_{\lambda}(u_{k}(\lambda))=\max_{u\in
K}p\Phi_{\lambda}(u)
$$
and $(u_{k}(\lambda),\mu_{k}(\lambda))$ is
a solution of \eqref{e1.1}--\eqref{e1.2}. Moreover,
$\mu_{k}(\lambda)\to +\infty$,  as $k\to +\infty$.
\end{theorem}

\begin{proof}
In view of \cite{s2}, we need only to prove that for any $k\in
\mathbb{N}^*, \Gamma_{k}\neq\emptyset$ and the last assertion.
Indeed, since $W^{1,p}(\Omega)$ is separable, there exist
$(e_{i})_{i\geq 1}$ linearly dense in $W^{1,p}(\Omega)$ such that
$\mathop{\rm supp}e_{i}\cap \mathop{\rm supp}e_{j}=\empty$ if $i\neq j$,
where $\mathop{\rm supp}e_{i}$ denotes
the support of $e_{i}$.
We can suppose that
$e_{i}\in \mathcal{M}$ (if not we take $e_{i}'=\frac{e_{i}}{p\Psi(e_{i})}$).

Let $k\in \mathbb{N}^*$ and $\mathcal{F}_{k}=\mathop{\rm
span}\{e_{1},e_{2},\dots,e_{k}\}$. $\mathcal{F}_{k}$ is a vector
subspace and $\dim\mathcal{F}_{k}=k$. If $v\in \mathcal{F}_{k}$,
then there exist $\alpha_{1},\dots,\alpha_{k}  $ in $\mathbb{R}$
such that $v=\sum_{i=1}^{i=k}\alpha_{i}e_{i}$. Thus
\[
\Psi(v)=\sum_{i=1}^{i=k}|\alpha_{i}|^{p}\Psi(e_{i})
=\frac{1}{p}\sum_{i=1}^{i=k}|\alpha_{i}|^{p},
\]
because $\Psi(e_{i})=1$, for $ i=1,2,\dots,k$. It follows that the
map $v\to (p\Psi(v))^{1/p}$ is a norm on
$\mathcal{F}_{k}$. Hence, there is a constant $c>0$ so that
$$
c\|v\|_{1,p}\leq (p\Psi(v))^{1/p}\leq
\frac{1}{c}\|v\|_{1,p}, \quad \forall v\in \mathcal{F}_{k}.
$$
That is,
$$
c\|v\|_{1,p}\leq \Big(\int_{\partial\Omega}
|v|^pd\sigma\Big)^{1/p}\leq \frac{1}{c}\|v\|_{1,p},
\quad \forall v\in \mathcal{F}_{k}.
$$
This implies that the set
$$
\mathcal{V}=\mathcal{F}_{k}\cap
\left\{v\in W^{1,p}(\Omega): \|v\|_{p,\partial\Omega}\leq
1\right\}
$$
is bounded. Because
$\mathcal{V}\subset B(0,\frac{1}{c})$,
where
\[
B(0,\frac{1}{c})=\{v\in W^{1,p}: \|v\|_{1,p}\leq \frac{1}{c}\}.
\]
 Moreover $\mathcal{V}$ is a
symmetric bounded neighborhood of the origin $0$. Consequently,
from \cite[Proposition 2.3]{s2}, we deduce that
$\gamma(\mathcal{F}_{k}\cap \mathcal{M})=k$. Then
$\mathcal{F}_{k}\cap \mathcal{M}\in \Gamma_{k}$ (because
$\mathcal{F}_{k}\cap \mathcal{M}$ is compact, since it exactly
equals to the boundary of $\mathcal{V}$).

To complete the proof, it suffices to show that for any $\lambda
\in \mathbb{R}$, $\mu_{k}(\lambda)\to +\infty$, as $ k\to
+\infty$. Indeed, let $(e_{n},e_{j}^{*})_{n,j}$ be a biorthogonal
system such that $e_{n}\in W^{1,p}(\Omega)$, $e_{j}^{*}\in
(W^{1,p}(\Omega))'$,  the $(e_{n})_{n}$ are dense in
$W^{1,p}(\Omega)$; and the $(e_{j}^*)_{j}$ are total in
$(W^{1,p}(\Omega))'$. Set for any $k\in \mathbb{N}^*$
$$
\mathcal{F}_{k-1}^{\bot}=\overline{\mathop{\rm span}
(e_{k+1},e_{k+2},e_{k+3},\dots)}.
$$
Observe that for any for any
$K\in\Gamma_{k}$, $K\cap \mathcal{F}_{k-1}^{\bot}\neq \emptyset$
(by \cite[(g) of Proposition 2.3]{s2}).
Now, we claim that
$$
t_{k}:=\inf_{K\in\Gamma_{k}}\sup_{K\cap\mathcal{F}_{k-1}^{\bot}}p
\Phi_{\lambda}(u)\to +\infty, \quad \text{as } k\to +\infty.
$$
Indeed, to obtain the contradiction, assume for $k$ large enough
that there is $u_{k}\in \mathcal{F}_{k-1}^{\bot}$ with
$\int_{\partial\Omega}|u_{k}|^pd\sigma=1$ such that
$$
t_{k}\leq p\Phi_{\lambda}(u_{k})\leq M,
$$
for some $M>0$ independent of $k$. Therefore,
$$
\|u_{k}\|_{1,p}^p-\lambda\int_{\partial\Omega} \rho(x)|u_{k}|^pd\sigma\leq
M.
$$
Hence
$$
\|u_{k}\|_{1,p}^p\leq M+\lambda\|\rho\|_{\infty,\partial
\Omega}< \infty.
$$
This implies that $(u_{k})_{k}$ is bounded in
$W^{1,p}(\Omega)$. For a subsequence of $(u_{k})_{k}$ if
necessary, we can suppose that $(u_{k})$ converges weakly in
$W^{1,p}(\Omega)$ and strongly in $L^p(\partial\Omega)$. By our
choice of $\mathcal{F}_{k-1}^{\bot}$, we have
$u_{k}\rightharpoonup 0$ in $W^{1,p}(\Omega)$. Because
$\langle e_{n}^{*},e_{k}\rangle$, for all $k\geq n$.
This contradicts the fact that
$\int_{\partial\Omega}|u_{k}|^{p}d\sigma=1$, for all $k$
and the  the claim is proved.

Finally, since $\mu_{k}(\lambda)\geq t_{k}$
we conclude that
$\mu_{k}(\lambda)\to +\infty$, as $k\to+\infty$
and the proof is complete.
\end{proof}


\section{Simplicity and isolation of $\mu_{1}(\lambda)$}

\subsection{Simplicity}
First, observe that solutions of \eqref{e1.1}--\eqref{e1.2}, by
an well-known advanced regularity, belong to
$C^{1,\alpha}(\bar\Omega)$, see \cite{t1}.

\begin{lemma} \label{lem4.1}
Eigenfunctions associated to $\mu_{1}(\lambda)$ are either
positive or negative in $\Omega$. Moreover if $u\in
C^{1,\alpha}(\Omega)$ then $u$ has definite sign in
$\bar\Omega$.
\end{lemma}

\begin{proof}
Let $u$ be an eigenfunction associated to $\mu_{1}(\lambda)$,.
Since $\Phi_{\lambda}(|u|)\leq \Phi_{\lambda}(u)$ and
$\Psi(|u|)=\Psi(u)$, it follows from \eqref{e1.3} that $|u|$ is also
an eigenfunction associated to $\mu_{1}(\lambda)$.
Using Harnack's inequality, cf. \cite{g1}, we deduce that
$|u|>0$ in $\Omega$ and by continuity we conclude that
has definite sign in $\bar\Omega.$ In fact $|u|>0$ in $\bar\Omega$ because
  $\frac{\partial u}{\partial \nu}(x_{0})<0$ for any
$x_{0}\in\partial\Omega$ with
  $u(x_{0})=0$, by applying Hopf's Lemma, see \cite{v1}.
\end{proof}

\begin{theorem}[Uniqueness] \label{thm4.1}
For any $\lambda\in\mathbb{R}$, the eigenvalue $\mu_{1}(\lambda)$ defined
by $\eqref{e1.3}$ is a simple; i.e.,
the set of the eigenfunctions associated with $(\lambda,\mu_{1}(\lambda))$
is  $\{tu_{1}(\lambda): t\in \mathbb{R}\}$, where $u_{1}(\lambda)$
denotes the principal eigenfunction associated with
$(\lambda,\mu_{1}(\lambda))$.
\end{theorem}

\begin{proof}
By Theorem \ref{thm3.1} it is clear that $\mu_{1}(\lambda)$ is an
eigenvalue of the problem \eqref{e1.1}--\eqref{e1.2} for any
$\lambda\in\mathbb{R}$.
Let $u$ and $v$ be two eigenfunctions associated to
$(\lambda,\mu_{1}(\lambda))$, such that $u,v\in\mathcal{M}$. Thus
in virtue of Lemma \ref{lem4.1} we can assume that $u$ and $v$ are
positives.

Note that  $W^{1,p}(\Omega)\ni w\to\|\nabla w\|_{p}^{p}$; $w\to
\int_{\partial \Omega}|w|^pd\sigma$ and
$w\to\int_{\partial\Omega}\rho(x)|w|^pd\sigma$ are linear
functionals in $w^p$, for $w^p\geq 0$. Hence if we consider
$w=\left(\frac{u^p+v^p}{2}\right)^{1/p}$, then it belongs to
$W^{1,p}(\Omega)$ and $\int_{\partial
\Omega}|w|^pd\sigma=1$. Consequently, $w$ is admissible in the
definition of $\mu_{1}(\lambda)$. On the other hand, by the
convexity  of $\chi\to |\chi|^{p}$ we have by calculation  the
following inequalities
\begin{equation}
\begin{aligned}
\int_{\Omega}|\nabla w|^pdx
&= \frac{1}{2}\int_{\Omega}|u^{p-1}\nabla u+v^{p-1}\nabla
v|^{p}(u^{p}+v^{p})^{1-p}dx\\
&= \frac{1}{2}\int_{\Omega}\big|\frac{u^{p}}{u^{p}+v^{p}}\frac{\nabla
u}{u}+ \frac{v^{p}}{v^{p}+u^{p}}\frac{\nabla
v}{v}\big|^{p}(u^{p}+v^{p})^{1-p}dx \\
&\leq \frac{1}{2}\int_{\Omega}\Big(\frac{u^{p}}{u^{p}+v^{p}}
|\frac{\nabla u}{u}|^{p}+
\frac{v^{p}}{v^{p}+u^{p}} |\frac{\nabla v}{v}|^{p}\Big)dx  \\
&\leq \frac{1}{2}\int_{\Omega}(|\nabla u|^{p}+|\nabla v|^{p})dx.
\end{aligned}
\label{e4.1}
\end{equation}
By the choice of $u$ and $v$, we deduce that
\begin{equation}
\left| t\frac{\nabla u}{u}+ (1-t)\frac{\nabla v}{v}
\right|^{p}=t\left|\frac{\nabla
u}{u}\right|^{p}+(1-t)\left|\frac{\nabla
v}{v}\right|^{p},\label{e4.2}
\end{equation}
with $t=u^{p}/(u^{p}+v^{p})$.

Now, we claim that Now, we claim that $u=v \text{  a.e. on
}{\overline\Omega}.$ Indeed, consider the  auxiliary
function
$$
F({\chi}_{1},{\chi}_2)=\left|t{\chi}_{1}+ (1-t){\chi}_2\right|^{p}
-t\left|{\chi}_{1}\right|^{p}+(1-t)\left|{\chi}_2\right|^{p}.
$$
Since $t\neq 0$, critical points of $F$ are solutions of the
system
\begin{gather}
\frac{\partial F({\chi}_{1},{\chi}_2)}{\partial
{\chi}_{1}}= pt\left(\left|t{\chi}_{1}+
(1-t){\chi}_2\right|^{p-2}(t{\chi}_{1}-\left|{\chi}_{1}
\right|^{p-2}{\chi}_{1}\right)=0;\label{e4.3}\\
\frac{\partial F({\chi}_{1},{\chi}_{2})}{\partial
{\chi}_{2}}= p(t-1)\left(\left|t{\chi}_{1}+
(1-t){\chi}_{2}\right|^{p-2}(t{\chi}_{1}-\left|{\chi}_{2}
\right|^{p-2}{\chi}_{2}\right)=0.\label{e4.4}
\end{gather}
Thus \eqref{e4.2}, \eqref{e4.3} and \eqref{e4.4} imply that
$({\chi}_{1}=\frac{\nabla u}{u}, ({\chi}_{2}=\frac{\nabla v}{v})$
is a solution of the above system. Therefore,
$$
\left|\frac{\nabla u}{u}\right|^{p-2}\frac{\nabla
u}{u}=\left|\frac{\nabla v}{v}\right|^{p-2}\frac{\nabla v}{v}.
$$
Hence
$$
\frac{\nabla u}{u}=\frac{\nabla v}{v}\text{  a.e. in  }{\overline
\Omega}.
$$
This implies easily that $u=cv$ for some positive constant $c$. By
normalization we conclude that $c=1$. The proof is completed.
 \end{proof}

\begin{remark} \label{rmk4.1} \rm
Various proofs of the uniqueness result were given in Direchlet
$p$-Laplacian case by using $C^{1,\alpha}$-regularity and
$L^{\infty}$-estimation of the first eigenfunctions and by applying
either Picone's identity \cite{a1};
or Diaz-Sa\'{a}'s inequality  \cite{a2,c1,d2}, and or an abstract inequality
\cite{l1}.
\end{remark}

\subsection{Isolation}

\begin{proposition} \label{prop4.1}
For each $\lambda\in\mathbb{R}$, $\mu_{1}(\lambda)$ is the only eigenvalue
associated with $\lambda$, having an eigenfunction that does not change
sign on the boundary $\partial\Omega$.
\end{proposition}

\begin{proof}
Fix $\lambda\in\mathbb{R}$ and let $u_{1}(\lambda)$ be the
principal eigenfunction associated with
$(\lambda,\mu_{1}(\lambda))$. Suppose that there exists an
eigenfunction $v$ corresponding to a pair $(\lambda,\mu)$ with
$v\geq 0$ on $\partial\Omega$ and $v\in \mathcal{M}$. By the Maximum
Principle, $v>0$ on ${\overline\Omega}$. For simplify of
notation, set $u=u_{1}(\lambda)$. Let $\epsilon>0$ be small enough,
and write
\begin{gather}
u_{\epsilon}=u+\epsilon, \quad
v_{\epsilon}=v+\epsilon, \label{e4.5} \\
\phi(u_{\epsilon},v_{\epsilon})
=\frac{u_{\epsilon}^{p}-v_{\epsilon}^{p}}{u_{\epsilon}^{p-1}}.\label{e4.6}
\end{gather}
It is clear that
$\phi(u_{\epsilon},v_{\epsilon})\in W^{1,p}(\Omega)$ and it is an
 admissible test function in
\eqref{e1.1}--\eqref{e1.2}. Thus we obtain
\begin{equation}
\begin{aligned}
&\int_{\Omega}|\nabla u|^{p-2}\nabla u\nabla
\phi(u_{\epsilon},v_{\epsilon})dx+\int_{\Omega}
u^{p-1}\phi(u_{\epsilon},v_{\epsilon})dx\\
&=\int_{\partial\Omega}(\lambda \rho(x)+\mu_{1}(\lambda))
u^{p-1}\phi(u_{\epsilon},v_{\epsilon})d\sigma
\label{e4.7}
\end{aligned}
\end{equation}
and
\begin{equation}
\int_{\Omega}|\nabla v|^{p-2}\nabla v\nabla
\phi(u_{\epsilon},v_{\epsilon})dx+
\int_\Omega
v^{p-1}\phi(u_{\epsilon},v_{\epsilon})dx=\int_{\partial
\Omega}(\lambda
\rho(x)+\mu))v^{p-1}\phi(u_{\epsilon},v_{\epsilon})d\sigma.
\label{e4.8}
\end{equation}
 From \eqref{e4.7} and \eqref{e4.8},  we deduce by calculations
that
\begin{equation}
\begin{aligned}
&\int_{\Omega}|\nabla u|^{p-2}\nabla u\nabla
\phi(u_{\epsilon},v_{\epsilon})\, dx+
\int_{\Omega}|\nabla v|^{p-2}\nabla v\nabla
\phi(u_{\epsilon},v_{\epsilon})dx+
\int_{\Omega}|v|^{p-2}v\phi(u_{\epsilon},v_{\epsilon})\, dx\\
&=\int_{\partial \Omega} \lambda
\rho(x)\Big(\big(\frac{u}{u_{\epsilon}}\big)^{p-1}-
\big(\frac{v}{v_{\epsilon}}\big)^{p-1}\Big)
(u_{\epsilon}^p-v_{\epsilon}^p)d\sigma   \\
&\quad +\mu_{1}(\lambda)\int_{\partial
\Omega}u^{p-1}\big[u_{\epsilon}-\big(\frac{v_{\epsilon}}{u_{\epsilon}}
\big)^{p-1}v_{\epsilon}\big]d \sigma+\mu\int_{\partial \Omega}
u^{p-1}\big[v_{\epsilon}-\big(\frac{u_{\epsilon}}{v_{\epsilon}}
\big)^{p-1}u_{\epsilon}\big]d\sigma.
\end{aligned} \label{e4.9}
\end{equation}
On the other hand, by a long calculation again, we obtain
\begin{equation}
\nabla\phi(u_{\epsilon},v_{\epsilon})=
\big\{1+(p-1)\big(\frac{v_{\epsilon}}{u_{\epsilon}}\big)^p\big\}\nabla
u_{\epsilon}-
p\big(\frac{v_{\epsilon}}{u_{\epsilon}}\big)^{p-1}\nabla
v_{\epsilon} \label{e4.10}
\end{equation}
and
\begin{equation}
\int_{\Omega}
\left[u^{p-1}\phi(u_{\epsilon},v_{\epsilon})+v^{p-1}
\phi(u_{\epsilon},v_{\epsilon})\right]dx
=\int_{\Omega}\big[\big(\frac{u}{u_{\epsilon}}\big)^{p-1}
-\big(\frac {v}{v_{\epsilon}}\big)^{p-1}\big]
(u_{\epsilon}^p-v_{\epsilon}^p)dx.\label{e4.11}
\end{equation}
Therefore, \eqref{e4.9}, \eqref{e4.10} and \eqref{e4.11} yield
\begin{equation}
\begin{aligned}
&\int_{\Omega}\Big[
\big\{1+(p-1)\big(\frac{v_{\epsilon}}{u_{\epsilon}}\big)^p\big\}|\nabla
u_{\epsilon}|^p+
\big\{1+(p-1)\big(\frac{u_{\epsilon}}{v_{\epsilon}}\big)^p\big\}|\nabla
v_{\epsilon}|^p\Big]dx\\
&\quad +\int_{\Omega}
\Big[-p\big(\frac{v_{\epsilon}}{u_{\epsilon}}\big)^{p-1}|\nabla
v_{\epsilon}|^{p-2}\nabla u_{\epsilon}\nabla
v_{\epsilon}+p\big(\frac{u_{\epsilon}}{v_{\epsilon}}\big)^{p-1}|\nabla
u_{\epsilon}|^{p-2}\nabla u_{\epsilon}\nabla v_{\epsilon}\Big]dx\\
&= J_{\epsilon}+K_{\epsilon}-I_{\epsilon},
\end{aligned}
\label{e4.12}
\end{equation}
with
\begin{gather}
I_{\epsilon}=\int_{\Omega}
\Big(\big(\frac{u}{u_{\epsilon}}\big)^{p-1}
     -\big(\frac{v}{v_{\epsilon}}\big)^{p-1}\Big)
\left(u_{\epsilon}^p-v_{\epsilon}^p\right)dx\,,\label{e4.13}
\\
J_{\epsilon}=\lambda\int_{\partial \Omega}\rho(x)
\Big(\big(\frac{u}{u+\epsilon}\big)^{p-1}
-\big(\frac{v}{v+\epsilon}\big)^{p-1}\Big)
\left(u_{\epsilon}^p-v_{\epsilon}^p\right)d\sigma\,,\label{e4.14}
\\
K_{\epsilon}=\mu_{1}(\lambda)\int_{\partial \Omega} u^{p-1}
\Big[u_{\epsilon}-\big(\frac{v_{\epsilon}}{u_{\epsilon}}\big)^{p-1}
 v_{\epsilon}\Big]d \sigma
+\mu\int_{\partial \Omega}
u^{p-1}\Big[v_{\epsilon}-\big(\frac{u_{\epsilon}}{v_{\epsilon}}
 \big)^{p-1}u_{\epsilon}\Big]d \sigma.\label{e4.15}
\end{gather}
It is clear that $I_{\epsilon}\geq 0$. Now,
thanks to the inequalities of Lindqvist \cite{l1}, we can
distinguish tow cases according to $p$.

First case: $p\geq 2$.
 From \eqref{e4.12} we have
\begin{equation}
J_{\epsilon}+K_{\epsilon}\geq \frac{1}{2^{p-2}-1}\int_{\Omega}
\left(\frac{1}{(u+1)^p}+\frac{1}{(v+1)^p}\right)|u\nabla v-v\nabla
u|^pdx\geq 0.\label{e4.16}
\end{equation}
Second case: $1<p<2$.
\begin{equation}
J_{\epsilon}+K_{\epsilon}\geq
c(p)\int_{\Omega}\frac{uv(u^p+v^p)}{\left(v|\nabla
u|+u|\nabla v|+1\right)^{2-p}}|u\nabla v-v\nabla u|^2dx\geq
0,\label{e4.17}
\end{equation}
where the constant $c(p)>0$ independent of $u,v,\lambda $ and
$\mu_{1}(\lambda)$.

The Dominated Convergence Theorem implies
$$
\lim_{\epsilon\to
0^{+}}J_{\epsilon}=\lim_{\epsilon\to
0^+}K_{\epsilon}=(\mu_{1}(\lambda)-\mu)\int_{\partial\Omega}
(u^p-v^p)d\sigma=0,
$$
because
\begin{equation}
\int_{\partial\Omega}u^pd\sigma=\int_{\partial\Omega}v^pd\sigma=1.
\label{e4.18}
\end{equation}
Now, letting $\epsilon\to 0^{+}$ in \eqref{e4.16} and \eqref{e4.17},
we arrive at
$u\nabla v=v\nabla u$  a.e. on $\Omega$.
Thus
$$
\nabla \left(\frac{u}{v}\right)=0\quad \text{a.e. on }\Omega.
$$
Hence, there exists $t>0$ such that $u=tv$ a.e. on $ \Omega$. By
continuity $u=v$ a.e. in $\overline\Omega$; and by the
normalization $\eqref{e4.18}$ we deduce that $t=1$ and $u=v$ a.e. on
$\partial\Omega$. This implies that $u=v$ a.e. on
$\overline\Omega$. Finally, we conclude that
$\mu=\mu_{1}(\lambda)$. Which completes the prof.
\end{proof}

\begin{remark} \label{rmk4.2} \rm
Proposition \ref{prop4.1} can also be shown by using {\it Picone's identity}.
A similar result was given by \cite{b4} in the restrictive case
$\lambda=0$.
\end{remark}

\begin{corollary} \label{coro4.1}
For each $\lambda\in\mathbb{R}$, if $u$ is an eigenfunction associated with
a pair $(\lambda, \mu)$ and $\mu\neq\mu_{1}(\lambda)$, then $u$
changes  sign on the boundary $\partial\Omega$. Moreover, we
have the  estimate
\begin{equation}
\min(|\partial\Omega^{-}|,|\partial\Omega^{+}|)\geq
c_{p^{*}}^{-N}(|\lambda|\|\rho\|_{\infty,\partial\Omega}
+|\mu|)^{-\eta},\label{e4.19}
\end{equation}
where
$$
\eta=\begin{cases} \frac{N}{p} &\text{if }1<p<N\\
2 &\text{if } p>N;
\end{cases}
$$
$c_{p^*}$ is the best constant in the Sobolev trace embedding
$W^{1,p}(\Omega)$ in $L^{p^*}(\partial\Omega)$; and
$|\partial \Omega^{\pm}|$ denotes the $(N-1)$-dimensional measure of
$\partial\Omega^{\pm}$. Here $p^{*}=\frac{p(N-1)}{N-p}$ is the
critical Sobolev exponent.
\end{corollary}

\begin{proof}
Set  $u^{+}=\max (u,0)$ and $u^{-}=\max(-u,0)$.
It follows from $\eqref{e2.1}$, where we put $v=u^{-}$,  that
$$
\int_{\Omega} |\nabla u^-|^pdx+\int_{\Omega}
|u^-|^pdx=\int_{\partial\Omega}(\lambda\rho(x)+\mu)|u^-|^pd\sigma.
$$
Thus
\begin{align*}
\|u^-\|_{1,p}&\leq \left(|\lambda|\|\rho\|_{\infty,\partial
\Omega}+|\mu|\right)\int_{\partial \Omega^-}|u^-|^pd\sigma\\
&\leq \left(|\lambda|\|\rho\|_{\infty,\partial
\Omega}+|\mu|\right)|\partial\Omega^-|^{p/N}
\Big(\int_{\partial \Omega}|u^-|^{p^*}\Big)^{p/p^*}.
\end{align*}
By the Sobolev embedding
$W^{1,p}(\partial\Omega)\hookrightarrow L^{p^*}(\partial \Omega)$, we deduce
that
$$
|\partial \Omega^-|\geq
c_{p^*}^{-N}\left(|\lambda|\|\rho\|_{\infty,\partial
\Omega}+|\mu|\right)^{-\eta}.
 $$
For $\partial\Omega^+$ the same estimate follows by taking $v=u^+$ in
\eqref{e2.1}. Hence  \eqref{e4.19} follows.
\end{proof}

\begin{remark} \label{rmk4.1b} \rm
(i) The right-hand side of \eqref{e4.19} is positive because
$\rho\not\equiv 0$ and if $\lambda=0$ then $\mu$ shall be an
eigenvalue of $p$-Laplacian related to trace embedding, so
$\mu-\lambda_{1}>0$, with $\lambda_{1}$ is the first eigenvalue of
\eqref{e1.1}--\eqref{e1.2} in the case $(\lambda=0)$.

(ii) As an easy consequence of Corollary \ref{coro4.1}, we get that the
number of the nodal components of each eigenfunction of
 \eqref{e1.1}--\eqref{e1.2} is finite.
\end{remark}

Using Proposition \ref{prop4.1} and Corollary \ref{coro4.1}, we can state the following
important result.

\begin{theorem} \label{thm4.2}
For each $\lambda\in\mathbb{R}$, $\mu_{1}(\lambda)$ is isolated.
\end{theorem}

\section{Variations of the weight}

Let $\mu_{1}(\lambda)=\mu_{1}(\rho)$ and
$u_{1}(\lambda)=u_{1}(\rho)$ (for indicating the dependance of the
weight $\rho$).

\begin{theorem} \label{thm5.1}
For each $\lambda\in \mathbb{R}$, if $(\rho_{k})_{k}$ is a sequence of
functions in $L^{\infty}(\partial \Omega)$ that converges to $\rho$
and $\rho\not\equiv 0$,
then
\begin{gather}
\lim_{k\to \infty}\mu_{1}(\rho_{k})=\mu_{1}(\rho)\,,\label{e5.1} \\
\lim_{k\to \infty}\|u_{1}(\rho_{k})-u_{1}(\rho)\|_{1,p}^p=0\,.\label{e5.2}
\end{gather}
\end{theorem}

\begin{proof}
 If $\lambda =0$,  the result is evident because
$\mu_{1}(\rho_{k})=\mu_{1}(\rho)$, for all $k\in \mathbb{N}^*$.
If $\lambda\neq 0$, then for $v\in\mathcal{M}$,
$$
|\lambda\int_{\partial\Omega}(\rho_{k}-\rho)|v|^{p}d\sigma|
\leq |\lambda|\|\rho_{k}-\rho\|_{\infty,\partial\Omega}.
$$
Using the convergence of $\rho_{k}$ to $\rho$ in
$L^{\infty}(\partial \Omega)$,  for all $\epsilon>0$, there exists
$k_{\epsilon}\in \mathbb{N}$ such that for all $k\geq k_{\epsilon}$,
$$
|\lambda \int_{\partial\Omega}(|(\rho_{k}-\rho)|v|^{p}d\sigma|\leq
|\lambda|\frac{\epsilon}{|\lambda|}=\epsilon.
$$
This implies
\begin{gather}
\lambda\int_{\partial\Omega}\rho|v|^{p}d\sigma \leq
\epsilon+\lambda\int_{\partial\Omega}
\rho_{k}|v|^pd\sigma\,, \label{e5.3}\\
\lambda\int_{\partial \Omega }\rho_{k}|v|^pd\sigma \leq
\epsilon+\lambda \int_{\partial\Omega}\rho|v|^pd\sigma,
\label{e5.4}
\end{gather}
for  $v\in\mathcal{M}$, $\epsilon >0$  and $k\geq k_{\epsilon}$.

On the other hand, we have $\rho\not\equiv 0$. We take
$k_{\epsilon}$ large enough so that $\rho_{k}\not\equiv 0$. Thus
$$
\mu_{1}(\rho_{k})\leq \|v\|_{1,p}^p-\lambda \int_{\partial\Omega}
\rho_{k}|v|^pd\sigma.
$$
Combining \eqref{e5.3} and \eqref{e5.4}, we obtain
$$
\mu_{1}(\rho_{k})\leq \|v\|_{1,p}^p-\lambda\int_{\partial\Omega}\rho
|v|^pd\sigma +\epsilon.
$$
Passing to the infimum over $v\in\mathcal{M}$, we find
$$
\mu_{1}(\rho_{k})\leq\mu_{1}(\rho)+\epsilon, \quad
\mu_{1}(\rho)\leq \mu_{1}(\rho_{k})+\epsilon,\quad
\forall \epsilon>0,\; \forall k>k_{\epsilon}.
$$
Hence, we conclude the convergence \eqref{e5.1}.

For the strong convergence \eqref{e5.2} we argue as follows.
We have for $k$ large enough, $\rho_{k}\not\equiv 0$ and
\begin{equation}
\mu_{k}(\rho_{k})=\|u_{1}(\rho_{k})\|_{1,p}^p-\lambda\int_{\partial
\Omega}\rho_{k}(u_{1}(\rho_{k}))^pd\sigma.\label{e5.5}
\end{equation}
Thus
$$
\|u_{1}(\rho_{k})\|_{1,p}^p\leq
|\mu_{1}(\rho_{k})|+|\lambda|\|\rho_{k}\|_{\infty,\partial\Omega}.
$$
 From \eqref{e5.1} and the convergence of $\rho_{k}$ to $\rho$ in
$L^{\infty}(\partial\Omega)$, we deduce that
$(u_{1}(\rho_{k}))_{k}$ is a bounded sequence in
$W^{1,p}(\Omega)$. Since $W^{1,p}(\Omega)$ is reflexive and
compactly embedded in $L^p(\partial\Omega)$ we can extract a
subsequence of $(u_{1}(\rho_{k}))_{k}$ again labelled by $k$, such
that $u_{1}(\rho_{k})\rightharpoonup u$ (weakly) in
$W^{1,p}(\Omega)$ and $u_{1}(\rho_{k})\to u$ (strongly) in
$L^p(\partial\Omega)$,  as $k\to \infty$. We can also
suppose that $u_{1}(\rho_{k})\to u$ in $L^p(\Omega)$.
Passing to a subsequence if necessary, we can assume that
$u_{1}(\rho_{k})\to u$ a.e. in $\bar\Omega$. Thus $u\geq
0$ a.e. in $\overline\Omega$. We will prove that $u\equiv
u_{1}(\rho)$. To do this, using the Dominated Convergence Theorem
in $\partial\Omega$, we deduce that
$$
\int_{\partial\Omega}\rho_{k}(u_{1}(\rho_{k}))^pd\sigma\to
\int_{\partial\Omega}\rho u^p d\sigma,
$$
as $k\to \infty$. By \eqref{e5.5},  \eqref{e5.1} and the lower
weak semi-continuity of the norm we obtain that
\begin{equation}
\|u\|_{1,p}^p\leq\mu_{1}(\rho)+\lambda\int_{\partial \Omega}\rho
u^{p}d\sigma.\label{e5.6}
\end{equation}
The normalization $\int_{\partial\Omega} u^pd\sigma=1$ is proved.
Moreover, $u\geq 0$ a.e. in $\bar\Omega$, because $u_{1}(\rho_{k})>0$
in $\overline \Omega$ Thus $u$ is an admissible function in the variational
definition of $\mu_{1}(\lambda)$. So
$$
\mu_{1}(\lambda)\leq \|u\|_{1,p}^p-\lambda \int_{\partial \Omega}\rho
u^pd\sigma.
$$
This and \eqref{e5.6} yield
\begin{equation}
\mu_{1}(\rho)=\|u\|_{1,p}^p-\lambda\int_{\partial\Omega}\rho
u^pd\sigma.\label{e5.7}
\end{equation}
By the uniqueness of the principal
eigenfunction associated to $\mu_{1}(\lambda)$,  we must have
$u\equiv u_{1}(\rho)$. Consequently the limit function
$u_{1}(\rho)$ is independent of the choice of the (sub)sequence.
Hence, $u_{1}(\rho_{k})$ converges to $u_{1}(\rho)$ at least in
$L^p(\partial\Omega)$ and in $L^p(\Omega)$.
To complete the proof of $\eqref{e5.2}$, it suffices to use the Clarckson's
inequalities related to uniform convexity of $W^{1,p}(\Omega)$.
For this we distinguish two cases.

First case: $p\geq 2$.
We have
\begin{align*}
&  \int_{\Omega} \Big|\frac{\nabla
u_{1}(\rho_{k})-\nabla u_{1}(\rho)}{2}\Big|^pdx +
\int_{\Omega} \left|\frac{\nabla
u_{1}(\rho_{k})+\nabla u_{1}(\rho)}{2}\right|^pdx  \\
& \leq \frac{1}{2}\int_{\Omega} |\nabla
u_{1}(\rho_{k})|^pdx+ \frac{1}{2}\int_{\Omega}|\nabla
u_{1}(\rho)|^pdx
\end{align*}
and
\begin{align*}
&\mu_{1}(\rho_{k})\int_{\partial
\Omega}\Big(\frac{u_{1}(\rho_{k})+ u_{1}(\rho)}{2}\Big)^pd\sigma \\
&\leq \int_{\Omega}
\Big|\frac{\nabla u_{1}(\rho_{k}) + \nabla u_{1}(\rho)}{2}\Big|^pdx
- \lambda\int_{\partial
\Omega}\rho_{k}\Big(\frac{u_{1}(\rho_{k})+u_{1}(\rho)}{2}\Big)^pd\sigma.
\end{align*}
Moreover,
$$
\int_{\Omega}
\Big|\frac{u_{1}(\rho_{k})-u_{1}(\rho)}{2}\Big|^pdx\leq
\int_{\Omega}\Big|\frac{u_{1}(\rho_{k})+u_{1}(\rho)}{2}\Big|^pdx+
\frac{1}{2}\|u_{1}(\rho_{k})\|_p^p+\frac{1}{2}\|u_{1}(\rho)\|_{p}^{p}.
$$
Hence
\begin{align*}
&\|u_{1}(\rho_{k})-u_{1}(\rho)\|_{1,p}^p \\
&\leq -\mu_{1}(\rho_{k})\int_{\partial
\Omega}\Big(\frac{u_{1}(\rho_{k})+u_{1}(\rho)}{2}\Big)^pd\sigma-
\lambda\int_{\partial
\Omega}\rho_{k}\Big(\frac{u_{1}(\rho_{k})+u_{1}(\rho)}{2}\Big)^pd\sigma\\
&\quad + \frac{1}{2}\Big(\mu_{1}(\rho_{k})- \lambda
\int_{\partial\Omega}\rho_{k}(x)u_{1}(\rho_{k})d\sigma\Big)
+ \frac{1}{2}\Big(\mu_{1}(\rho)-\lambda\int_{\partial\Omega}\rho
u_{1}^pd\sigma\Big).
\end{align*}
Then, by using the Dominated Convergence Theorem we deduce that
$$
\limsup_{k\to +\infty}\|u_{1}(\rho_{k})-u_{1}(\rho)\|_{1,p}^{p}=0.
$$

Second case: $1<p<2$. In this case, we have
\begin{align*}
&\Big\{\int_{\Omega} \Big|\frac{\nabla u_{1}(\rho_{k})
- \nabla u_{1}(\rho)}{2}\Big|^pdx\Big\}^{\frac{1}{p-1}}
+ \Big\{\int_{\Omega} \Big|\frac{\nabla u_{1}(\rho_{k})+
\nabla u_{1}(\rho)}{2}\Big|^pdx\Big\}^{\frac{1}{p-1}}
\\
&\leq \Big\{ \frac{1}{2}\int_{\Omega} |\nabla
u_{1}(\rho_{k})|^pdx+
\frac{1}{2}\int_{\Omega} |\nabla u_{1}(\rho)|^pdx\Big\}^{\frac{1}{p-1}}
\end{align*}
and
\begin{align*}
&\mu_{1}(\rho_{k})\int_{\partial
\Omega}\Big(\frac{u_{1}(\rho_{k})+ u_{1}(\rho)}{2}\Big)^{p}d\sigma \\
&\leq \int_{\Omega}
\Big|\frac{\nabla u_{1}(\rho_{k})+ \nabla u_{1}(\rho)}{2}\Big|^p
-\lambda\int_{\partial
\Omega}\rho_{k}\Big(\frac{u_{1}(\rho_{k})+u_{1}(\rho)}{2}\Big)^pd\sigma.
\end{align*}
Hence, by  the definitions of $\mu_{1}(\rho_{k})$ and
$\mu_{1}(\rho)$; and the second Clarckson's inequality we obtain
the convergence \eqref{e5.2}.
\end{proof}

\begin{corollary} \label{coro5.1}
For any bounded domain $\Omega$,  the function $\lambda
\to \mu_{1}(\lambda)$ is differentiable on $\mathbb{R}$ and the
function $\lambda\to u(\lambda)$ is continuous from $\mathbb{R}$
into $W^{1,p}(\Omega)$. More precisely
$$
\mu_{1}'(\lambda_{0})=
-\int_{\partial\Omega}\rho(x)(u_{1}(\lambda_{0}))^pd\sigma,\quad
\forall \lambda_{0} \in \mathbb{R}.
$$
\end{corollary}

\begin{proof}
 Denote by $\mu_{1}(\lambda,\rho)$ the principal eigenvalue
associated with $\lambda$ and the weight $\rho$ and by
$u_{1}(\lambda,\rho)$ the principal eigenfunction corresponding.
Suppose that $\lambda_{k}\to\lambda_{0}$ in $\mathbb{R}$, then
$h_{k}=\lambda_{k}\rho\to \lambda_{0}\rho=h$ in
$L^{\infty}(\partial\Omega)$. From Theorem \ref{thm5.1} we deduce that
\begin{gather*}
\mu_{1}(\lambda_{k})=\mu_{1}(1,h_{k})\to
\mu_{1}(1,h)=\mu_{1}(\lambda_{0}),\\
u_{1}(\lambda_{k})=u_{1}(1,h_{k})\to
u_{1}(1,h)=u_{1}(\lambda_{0}) \text{ in } W^{1,p}(\Omega).
\end{gather*}
For the differentiability, it suffices to use the variational
characterization of $\mu_{1}(\lambda)$ and of
$\mu_{1}(\lambda_{0})$, so that
$$
(\lambda-\lambda_{0})\int_{\partial\Omega}\rho(x)(u_{1}(\lambda))^pd\sigma\leq
\mu_{1}(\lambda)-\mu_{1}(\lambda_{0})\leq
(\lambda_{0}-\lambda)\int_{\partial\Omega}(u_{1}(\lambda_{0}))^pd\sigma,$$
for all $\lambda,\lambda_{0}\in \mathbb{R}$.
This completes the proof.
\end{proof}

\paragraph{\textbf{Acknowledgement}}
The first author is  supported by grant number 22/12 from the
Al-Imam Muhammad Ibn Saud Islamic University, Riyadh, KSA.

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\end{document}
