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\markboth{On the spectrum of the p-biharmonic operator}
{ Abdelouahed El Khalil,  Siham Kellati \&  Abdelfattah Touzani}

\begin{document}
\setcounter{page}{161}
\title{\vspace{-1in}\parbox{\linewidth}{\footnotesize\noindent
2002-Fez conference on Partial Differential Equations,\newline
Electronic Journal of Differential Equations,
Conference 09, 2002, pp 161--170. \newline
http://ejde.math.swt.edu or http://ejde.math.unt.edu
\newline ftp  ejde.math.swt.edu (login: ftp)}
\vspace{\bigskipamount} \\
%
  On the spectrum of the p-biharmonic operator
%
\thanks{ {\em Mathematics Subject Classifications:} 35P30, 34C23.
\hfil\break\indent
{\em Key words:}  p-biharmonic operator, Duality mapping,
     Palais-Smale condition, \hfil\break\indent
     unbounded weight.
\hfil\break\indent
\copyright 2002 Southwest Texas State University. \hfil\break\indent
Published December 28, 2002.} }

\date{}
\author{Abdelouahed El Khalil,  Siham Kellati \& Abdelfattah Touzani}
\maketitle

\begin{abstract}
  This work is devoted to the study of the spectrum for p-biharmonic
  operator with an indefinite weight in a bounded domain.
\end{abstract}

\numberwithin{equation}{section}
\newtheorem{thm}{Theorem}[section]
\newtheorem{lem}[thm]{Lemma}
\newtheorem{prop}[thm]{Proposition}
\newtheorem{rem}[thm]{Remark}
\newtheorem{cor}[thm]{Corrolary}


\section{Introduction}

Let $\Omega$ be a bounded domain in $\mathbb{R}^{N}$, $N\geq 1$, not
necessary
regular; $1<p<\infty $ and $\rho\in L^{r}(\Omega) $, $\rho\not= 0$, an
unbounded weight function which can change its sign, with
$r=r(N,p)$
satisfying the conditions
$$ %\label{C}
r\begin{cases} >\frac{N}{2p}& \mbox{for }\frac{N}{p}\geq 2\\
=1 &\mbox{for }\frac{N}{p}<2.
\end{cases}
$$
We assume that $|\Omega^{+}_{\rho}|\not=0$, where $\Omega^{+}_{\rho}=\{x\in
\Omega; \rho(x)>0\}$ and $\lambda\in \mathbb{R}$. We consider the
eigenvalue problem
$$\begin{gathered}
\Delta_p^{2}u=\lambda\rho(x)|u|^{p-2}u \quad\mbox{in } \Omega\\
u\in W_{0}^{2,p}(\Omega).
\end{gathered} \eqno{(1.1)}
$$
Here $\Delta_p^{2}:=\Delta(|\Delta u|^{p-2}\Delta u)$, the operator of
fourth order called the $p$-biharmonic operator. For $p=2$, the linear
operator
$\Delta_{2}^{2}=\Delta^{2}=\Delta.\Delta$ is the iterated Laplacian
that multiplied with positive constant appears often in  Navier-Stokes
equations as being a viscosity coefficient. Its reciprocal operator
denoted  $(\Delta^{2})^{-1}$ is the celebrated Green's operator \cite{Lio}.

It is important to indicate that here we don't suppose any
boundary conditions on the high order partial derivatives
$\frac{\partial^{2}u}{\partial x_{i}\partial x_{j}}$ on the boundary
set
$\partial\Omega$ of the domain $\Omega$. The particular case $\rho\equiv 1$
and $u=\Delta u=0$ on $\partial\Omega$ was considered recently by
Dr\'abek and \^Otani \cite{Dra}. There the authors proved the existence,
the simplicity, and the isolation of the first eigenvalue of (1.1) by
using a transformation of a problem to a known Poisson's problem,
and using the well-known advanced regularity of Agmon-Douglis-Niremberg
\cite{Gi-Tr}.
Note that this transformation processus is not applicable to our
situation because the quantity $\Delta u$ does not necessary vanished on
$\partial \Omega$ and the eventual regularity is not required in any bounded
domain.

The main objective of this work is to show that problem (1.1)
has at least one non-decreasing sequence of positive eigenvalues
$(\lambda_{k})_{k\geq 1}$, by using the
Ljusternich-schnirelmann theory on $C^{1}$ manifolds, see e.g. \cite{Szu}.
Our approach is based only on some properties of the considered operator.
So that we give a direct characterization of $\lambda_{k}$ involving a
minimax argument over sets of genus greater than $k$.

We set
$$\lambda_{1}=\inf\big\{\|\Delta v\|_p^{p}, v\in W_{0}^{2,p}(\Omega);
\int_{\Omega}\rho(x)|v|^{p}dx=1\big\},
$$
where $\|.\|_p$ denotes the
$L^{p}(\Omega)$-norm. It is not difficult to show that $\|\Delta u\|_p$
defines a norm in $W_{0}^{2,p}(\Omega)$ and
$W_{0}^{2,p}(\Omega)$ equipped with this norm is a uniformly convex
Banach space for $1<p<+\infty$. The norm $\|\Delta .\|_p$ is uniformly
equivalent on
$W_{0}^{2,p}(\Omega)$ to the usual norm of
$W_{0}^{2,p}(\Omega)$ \cite{Gi-Tr}.

This paper is organized as follows: In section 2, we establish
some definitions and show certain basic lemmas. In section 3, we use a
variational technique to prove the existence of a sequence of the
positive eigenvalues of p-biharmonic operator with any unbounded weight.

\section{Preliminaries}

Throughout this paper, all solutions are weak, i.e, $u\in
W_{0}^{2,p}(\Omega)$ is a solution of $(1.1)$, if for all $\varphi\in
C_{0}^{\infty}(\Omega)$,
$$
\int_{\Omega}|\Delta u|^{p-2}\Delta u\Delta\varphi\,
dx=\lambda\int_{\Omega}\rho(x)|u|^{p-2}u\varphi \,dx.
$$
If $u\in W_{0}^{2,p}(\Omega)\backslash \{0\}$, then $u$ shall be called an
eigenfunction of the $p$-biharmonic operator (or of (1.1)) associated to
the eigenvalue $\lambda$.
The following proposition states some fundamental properties of the
$p$-biharmonic operator.

\begin{prop} \label{prop2.1}
For any bounded domain $\Omega$ and $1<p<+\infty$, $\Delta_p^{2}$
satisfies the following: \begin{itemize}
\item[(i)] $\Delta_p^{2}$ is an hemicontinuous operator from
$W_{0}^{2,p}(\Omega)$ into $W^{-2,p'}(\Omega)$.

\item[(ii)] $\Delta_p^{2}$ is a bounded monotonous, and coercive operator.

\item[(iii)] $\Delta_p^{2}: W_{0}^{2,p}(\Omega)\to W^{-2,p'}(\Omega)$
is a bicontinuous operator. Here $p'=\frac{p}{p-1}$.
\end{itemize}
\end{prop}

\paragraph{Proof} (i) Define on $W_{0}^{2,p}(\Omega)$ the potential
functional
$$A(u)=\frac{1}{p}\|\Delta u\|_p^{p}.
$$
This functional is convex and of class $C^{1}$ on $W_{0}^{2,p}(\Omega)$.
Further its derivative operator is $A'=\Delta_p^{2}$. So this yields
the hemicontinuity.\\
(ii) By a simple calculation we can show  that
$\|\Delta_p^{2}u\|_{*}=\|\Delta u\|_p^{p-1}$, where $\|.\|_{*}$ is
the
dual norm associated to $\|\Delta.\|_p$. This implies that
$\Delta_p^{2}$ is bounded and is a monotonous operator. The
continuity and coercivity are obvious .\\
(iii)  The fact that, for any $u, v\in W_{0}^{2,p}(\Omega)$, $\|\Delta
u\|_p=\|\Delta v\|_p$ if $\Delta_p^{2}u=\Delta_p^{2}(v)$ and
($W_{0}^{2,p}(\Omega)$, $\|\Delta.\|_p)$ is a uniformly convex space
completes the proof.

\paragraph{Definition} %2.1
Let $X$ be a real reflexive Banach space and let $X^*$ stand for its
dual with respect to the pairing $\langle.,.\rangle$. $T$ a mapping acting
from $X$ into $X^*$. $T$ is said to belong to the class $(S^{+})$,
if for any sequence $\{u_{n}\}$ in $X$ with $u_{n}$ converges weakly
to $u\in X$ and
$\limsup_{n\to  +\infty}\langle Tu_{n},u_{n}-u \rangle \leq 0$.
It follows that $u_{n}$ converges strongly to $u$ in $X$. We write
$T\in (S^{+})$.

\section{Main results}

We will use Ljusternick-Schnirelmann theory on $C^{1}$-manifolds
\cite{Szu}.
Consider the following two functionals defined on $W_{0}^{2,p}(\Omega)$:
$$
A(u)=\frac{1}{p}\|\Delta u\|_p^{p},\quad
B(u)=\frac{1}{p}\int_{\Omega}\rho(x)|u|^{p}dx.$$
We set ${\cal M}=\{ u\in W_{0}^{2,p}(\Omega); pB(u)=1\}$.

\begin{lem}
(i) $A$ and $B$ are even, and of class $C^{1}$ on
$W_{0}^{2,p}(\Omega)$.
(ii) ${\cal M}$ is a closed $C^{1}$-manifold.
\end{lem}

\paragraph{Proof}
(i) It is clear that $B$ is of class $C^{1}$ on $W_{0}^{2,p}(\Omega)$.
${\cal M}=B^{-1}\{\frac{1}{p}\}$ so $B$ is closed. Its derivative
operator
$B'$ satisfies $B'(u)\not=0\ \forall u\in{\cal M}$ (i.e., $B{'}(u)$
is
onto $\forall u\in{\cal M})$, so $B$ is a submersion, then ${\cal M}$
is a
$C^{1}$-manifold. \hfill$\square$

\begin{rem} \label{rmk3.1} \rm
Observe that $J: W_{0}^{2,p}(\Omega)\to  W^{-2,p'}(\Omega)$,
$$
J(u)=\begin{cases}
\|\Delta u\|_p^{2-p}\Delta_p^{2}u & \mbox {if } u\neq 0\\
0&\mbox{if }u=0 \end{cases}
$$
is the  duality mapping of ($W_{0}^{2,p}(\Omega)$, $\|\Delta.\|_p)$.
\end{rem}
The following lemma is the key to show existence.

\begin{lem} \label{lm3.2}
(i) $B': W_{0}^{2,p}(\Omega)\to  W^{-2,p'}(\Omega)$ is
completely continuous.\\
(ii) The functional $A$ satisfies the Palais-Smale condition on
${\cal M}$,
i.e., for $\{u_{n}\}\subset \cal{M}$, if $A(u_{n})$ is bounded and
$$
\epsilon_{n}:=A'(u_{n})-g_{n}B'(u_{n})\to  0\quad \mbox{as }n\to +\infty,
\eqno{(3.1)}
$$
where $g_{n}=\langle A'(u_{n}),u_{n}\rangle /
\langle B'(u_{n}),u_{n}\rangle$. Then $\{u_{n}\}_{n\geq 1}$ has a
convergent subsequence in $W_{0}^{2,p}(\Omega)$.
\end{lem}

\paragraph{Proof}
(i) Step 1: Definition of $B'$.\\
First case: $\frac{N}{p}>2$ and $r>\frac{N}{2p}$. Let $u,v \in
W_{0}^{2,p}(\Omega)$. By H\"older's inequality, we have
$$
\big|\int_{\Omega} \rho(x) |u(x)|^{p-2}u(x)v(x)dx\big|\leq \|
\rho\|_{r}\|u\|_{s}^{p-1}\| v\|_{p_{2}}
$$
where $\frac{1}{p_{2}}=\frac{1}{p}-\frac{2}{N}$ and $s$ is given by
$$
\frac{p-1}{s}+\frac{1}{p_{2}}+\frac{1}{r}=1.\eqno{(3.2)}
$$
Therefore,
$$\frac{p-1}{s}=1-\frac{1}{r}-\frac{1}{p_{2}}>
1-\frac{2p}{N}-\frac{1}{p_{2}}=\frac{p-1}{p_{2}}.
$$
Then it suffices to take
$$\max(1,p-1)<s<p_{2}\eqno{(3.3)}$$
so that $B'$ is well defined.\\
Second case: $\frac{N}{p}=2$ and $r>\frac{N}{2p}$.
In this case
$W_{0}^{2,p}(\Omega)\hookrightarrow L^{q}(\Omega)$,
for any $q\in[p,+\infty[$. There is $q\geq p$ such that
$\frac{1}{q}+\frac{1}{r}+\frac{p-1}{p}=\frac{1}{q}+\frac{1}{r}+\frac{1}{p'}=1$.\\
We obtain that
$$\frac{1}{q}=\frac{1}{p}-\frac{1}{r}\leq \frac{1}{p}.\eqno{(3.4)}$$
By H\"older's inequality, we arrive at
$$
\big|\int_{\Omega} \rho(x) |u(x)|^{p-2}u(x)v(x)dx\big|
\leq \|\rho\|_{r}\|u\|_p^{p-1}\| v\|_{q},
$$
for any $u,v\in W_{0}^{2,p}(\Omega)$. Then  $B'$ is also well
defined.\\
Third case:  $\frac{N}{p}<2$ and $r=1$. In this case
$W_{0}^{2,p}(\Omega)\hookrightarrow C(\overline{\Omega})\cap
L^{\infty}(\Omega)$.
Then for any $u,v \in W_{0}^{2,p}(\Omega)$, we have
$$
\big|\int_{\Omega} \rho(x) |u(x)|^{p-2}u(x)v(x)dx\big|<\infty,
$$
with $\rho\in L^{1}(\Omega)$, and $B'$ is well defined.

Step 2. $B'$ is completely continuous. Let $(u_{n}) \subset
W_{0}^{2,p}(\Omega)$ be a sequence such that $u_{n}\to  u$
weakly in
$W_{0}^{2,p}(\Omega)$. We have to show that $B'(u_{n})\to
B'(u)$
strongly in $W_{0}^{2,p}(\Omega)$, i.e.,\\
$$
\sup_{v\in W_{0}^{2,p}(\Omega)\,\, \|\Delta v\|_p\leq 1}
\Big|\int_{\Omega}\rho[|u_{n}|^{p-2}u_{n}-|u|^{p-2}u]v\,
dx\big|\to 0,\quad \mbox{as } n\to  +\infty.
$$
For this end, we distinguish three cases as in step 1 above.
For $\frac{N}{p}> 2$, and $r>\frac{N}{2p}$. Let  $s$ be as in (3.3).
Then
\begin{align*}
&\sup_{v\in W_{0}^{2,p}(\Omega),\, \|\Delta v\|_p\leq 1}
\Big|\int_{\Omega}\rho\big[|u_{n}|^{p-2}u_{n}-|u|^{p-2}u\big]v dx\Big|\\
&\leq \sup_{v\in W_{0}^{2,p}(\Omega),\, \|\Delta v\|_p\leq 1}
\big[\|\rho\|_{r} \big\| |u_{n}|^{p-2}u_{n}-|u|^{p-2}u\big\|_{\frac{s}{p-1}}
\|v\|_{p_{2}}\big]\\
&\leq c\|\rho\|_{r} \big\|
|u_{n}|^{p-2}u_{n}-|u|^{p-2}u\big\|_{\frac{s}{p-1}},
\end{align*}
where $c$ is the constant of Sobolev's embedding \cite{Ada}.
On the other hand, the Nemytskii's operator $u\mapsto |u|^{p-2}u$
is continuous from $L^{s}(\Omega)$ into $L^{\frac{s}{p-1}}(\Omega)$, and
$u_{n}\to  u$ weakly in $W_{0}^{2,p}(\Omega)$. So, we deduce
that $u_{n}\to  u$ strongly in $L^{s}(\Omega)$ because $s<p_{2}$.
Hence,
$$
\big\| |u_{n}|^{p-2}u_{n}-|u|^{p-2}u\big\|_{\frac{s}{p-1}}\to  0,\quad
\mbox{as } n\to  +\infty.
$$
This completes the proof of the claim.\\
If $\frac{N}{p}=2$ then
$$
\big|\int_{\Omega}\rho\big[|u_{n}|^{p-2}u_{n}-|u|^{p-2}u\big]v \,dx\big|
\leq \|\rho\|_{r}\big\| |u_{n}|^{p-2}u_{n}-|u|^{p-2}u
\big\|_p^{p-1}\|v\|_{q},
$$
where $q$ is given by (3.4). By Sobolev's embedding there exist $c>0$
such that
$$
\|v\|_{q}\leq c\|\Delta v\|_p,\quad \forall v\in W_{0}^{2,p}(\Omega).
$$
Thus
$$
\sup_{\stackrel{v\in W_{0}^{2,p}(\Omega)}{\|\Delta v\|_p\leq 1}}
\big|\int_{\Omega}\rho[|u_{n}|^{p-2}u_{n}-|u|^{p-2}u]v\,dx\big|
\leq c\|\rho\|_{r}\big\| |u_{n}|^{p-2}u_{n}-|u|^{p-2}u \big\|_p^{p-1}.
$$
From the continuity of $u\mapsto|u|^{p-1}u$ from $L^{p}(\Omega)$ into
$L^{p'}(\Omega)$, and from the compact embedding of
$W_{0}^{2,p}(\Omega)$
in $L^{p}(\Omega)$, we have the desired result.\\
If $\frac{N}{p}<2$ and $r=1,$ $W_{0}^{2,p}(\Omega)\subset
C(\overline{\Omega}),$
then we obtain
$$
\sup_{\stackrel{v\in W_{0}^{2,p}(\Omega)}{\|\Delta v\|_p\leq 1}}
\big|\int_{\Omega}\rho\big[|u_{n}|^{p-2}u_{n}-|u|^{p-2}u\big]v\,dx\big|
\leq c\|\rho\|_{1}\sup_{\overline{\Omega}}
\big| |u_{n}|^{p-2}u_{n}-|u|^{p-2}u\big|,
$$
where $c$ is the constant given by embedding of $W_{0}^{2,p}(\Omega)$
in $C(\overline{\Omega})\cap L^{\infty}(\Omega)$.
It is clear that
$$
\sup_{\overline{\Omega}}\big| |u_{n}|^{p-2}u_{n}-|u|^{p-2}u\big|\to
0,\quad \mbox{as } n\to +\infty.
$$
Hence $B'$ is completely continuous, also in this case.\\
(ii) $\{u_{n}\}$ is bounded in $W_{0}^{2,p}(\Omega)$. Hence without
loss of generality, we can assume that $u_{n}$ converges weakly in
$W_{0}^{2,p}(\Omega)$ for some function $u\in W_{0}^{2,p}(\Omega)$ and
$\|\Delta u_{n}\|_p\to  c$. For the rest we distinct two cases: \\
If $c=0$ then $u_{n}$ converges strongly to $0$ in
$W_{0}^{2,p}(\Omega)$.

If $c\not =0$, then we argue as follows:
$$
\langle \Delta_p^{2}u_{n},u_{n}-u\rangle
=\|\Delta u_{n}\|_p^{p}-\langle \Delta_p^{2}(u_{n}),u\rangle .
$$
Applying $\epsilon_{n}$ of (3.1) to $u$, we deduce that
$$
\Theta_{n}:=\langle \Delta_p^{2}(u_{n}),u\rangle
-\|\Delta u\|_p^{p}\langle B'(u_{n}),u\rangle \to  0\quad \mbox{as }
n\to +\infty.\eqno{(3.5)}
$$
Thus
$$
\langle\Delta_p^{2}u_{n},u_{n}-u\rangle
=\|\Delta u_{n}\|_p^{p}-\Theta_{n}-\|\Delta u_{n}\|_p^{p}
\langle B'(u_{n}),u\rangle .
$$
That is,
$$\langle\Delta_p^{2}u_{n},u_{n}-u\rangle =\|\Delta
u_{n}\|_p^{p}(1-\langle B'(u_{n}),u\rangle )-\Theta_{n}.$$
Hence,
$$
\limsup_{n\to +\infty} \langle \Delta_p^{2}u_{n},u_{n}-u \rangle
\leq c^{p}{\limsup_{n\to +\infty}}(1-\langle B'(u_{n}),u\rangle ).
$$
On the other hand, from (i) $ B'(u_{n})\to B'(u)$ in
$W^{-2,p'}(\Omega)$ and $pB(u)=1$, because $pB(u_{n})=1$  for all
$n\in\mathbb{N}^*$. So $pB(u)= \langle B'(u),u\rangle =1$. This yields that
\begin{align*}
1-\langle B'(u_{n}),u \rangle =& \langle B'(u),u>-<B'(u_{n}),u\rangle
\leq &\|B'(u)-B'(u_{n})\|_{*}\|\Delta u\|_p,
\end{align*}
where $\|.\|_{*}$ is the dual norm associated with $\|\Delta.\|_p$.

From (i) again $ B'(u_{n})\to  B'(u)$ in
$W_{0}^{-2,p'}(\Omega)$, we deduce that
$$
\limsup_{n\to +\infty} \langle \Delta_p^{2}u_{n},u_{n}-u\rangle
\leq 0\eqno{(3.6)}
$$
We can write $\Delta_p^{2}u_{n}=\|\Delta u_{n}\|_p^{p-2}J(u_{n})$, since
$\|\Delta
u_{n}\|_p\not=0$ for $n$ large enough. Therefore,
$$
\limsup_{n\to +\infty} \langle \Delta_p^{2}u_{n},u_{n}-u\rangle
= c^{p-2} \limsup_{n\to  +\infty}\langle Ju_{n},u_{n}-u\rangle .
$$
According to (3.5) we conclude that
$$
\limsup_{n\to  +\infty} \langle Ju_{n},u_{n}-u \rangle \leq 0
$$
$J$ being a duality  mapping, thus it satisfies the condition $S^{+}$.
Therefore, $u_{n}\to  u$ strongly in $W_{0}^{2,p}(\Omega)$. This
achieves the proof of the lemma. \hfill$\square$

\begin{rem} \label{rmk3.2} \rm
$A'$ is continuous, odd, (p-1)-homogeneous, continuously invertible
and
$\|A'(u)\|_{*}=\|\Delta u\|_p^{p-1}$, $\forall u\in
W_{0}^{2,p}(\Omega)$.
\end{rem}

\begin{rem}\label{rmk3.3} \rm
We can give another method to prove  that the functional $A$ satisfies
the Palais-Smale condition on ${\cal M}$.
\end{rem}

Indeed, $\{u_{n}\}$ is bounded in $W_{0}^{2,p}(\Omega)$, we can assume
for a subsequence if necessary  that $u_{n}$ converges weakly in
$W_{0}^{2,p}(\Omega)$. The claim is to prove that $u_{n}$ is of Cauchy in
$W_{0}^{2,p}(\Omega)$.
Set
$$
G(u_{n},u_{m})=\int_{\Omega}(|\Delta u_{n}|^{p-2}\Delta u_{n}-|\Delta
u_{m}|^{p-2}\Delta u_{m}).\Delta(u_{n}-u_{m}).
$$
From (ii) of the proposition \ref{prop2.1}, $\Delta_p^{2}$ is a monotonous
operator on $W_{0}^{2,p}(\Omega)$. So that
\begin{align*}
0\leq G(u_{n},u_{m})
=&\langle \Delta_p^{2}u_{n}-\Delta_p^{2}u_{m},u_{n}-u_{m}\rangle\\
=&\langle \epsilon_{n}-\epsilon_{m},u_{n}-u_{m}\rangle+\langle
h_{n}-h_{m},u_{n}-u_{m}\rangle ,
\end{align*}
with $\epsilon_{n}$ defined as in (3.1) and
$h_{n}=\|\Delta u_{n}\|_p^{p}B'(u_{n})$.
$$
G(u_{n},u_{m})\leq \|\epsilon_{n}-\epsilon_{m}\|_{*}
\|\Delta u_{n}-\Delta u_{m}\|_p+\|h_{n}-h_{m}\|_{*}\|\Delta u_{n}
-\Delta u_{m}\|_p.
$$
Or $h_{n}$ converges for a subsequence if necessary in
$W_{0}^{2,p}(\Omega)$. Indeed, from (i) of Lemma \ref{lm3.2}
$B': W_{0}^{2,p}(\Omega)\to  W^{-2,p'}(\Omega)$ is completely
continuous. On the other hand, for a subsequence if necessary
$\|\Delta u_{n}\|\to  c\geq 0$. It follows that $(h_{n})_{n\geq 0}$ is
convergent in $W^{-2,p'}(\Omega)$. Then
$$
G(u_{n},u_{m})\to  0,\quad \mbox{as }n\to +\infty.\eqno{(3.7)}
$$
From \cite{Lin}, we have the following inequality
$$
|t_{1}-t_{2}|^{p}\leq c \{(|t_{1}|^{p-2}t_{1}-|t_{2}|^{p-2}t_{2}).
(t_{1}-t_{2})\}^{\frac{\gamma}{2}}(|t_{1}|^{p}
+|t_{2}|^{p})^{1-\frac{\gamma}{2}},
$$
for any $t_{1},t_{2}\in\mathbb{R}$, with $\gamma=p$ if $1<p<2$ and
$\gamma=2$
if
$p\geq 2$.
By applying H\"older's inequality, we deduce that
$$\|\Delta u_{n}-\Delta u_{m}\|_p^{p}\leq c
\{G(u_{n},u_{m})\}^{\frac{\gamma}{2}}(\|\Delta u_{n}\|_p^{p}+\|\Delta
u_{m}\|_p^{p})^{1-\frac{\gamma}{2}}\eqno{(3.8)}$$
for some  positive constante $c$ independent of $n$ and $m$.
According to (3.7), (3.8) shows that $(u_{n})_{n}$ is a Cauchy's
sequence in
$W_{0}^{2,p}(\Omega)$. This proves the claim.
\hfill$\square$

Set
$$\Gamma_{k}=\{K\subset \cal {M}: \mbox{ K is symmetric,compact and }
\gamma(K)\geq k \},
$$
where $\gamma(K)=k$ is the genus of $k$, i.e., the smallest integer $k$
such that there exists an odd continuous map from $K$ to
$\mathbb{R}^{k}-\{0\}$.

Now, by the Ljusternick-Schnirelmann theory, see e.g. \cite{Szu}, we
have our main result formulated as follows.

\begin{thm} \label{thm3.1}
For any integer $k\in \mathbb{N}^*$,
$$
\lambda_{k}:=\inf_{K\in \Gamma_{k}}\max_{u\in K}pA(u)
$$
is a critical value of A restricted on $\cal{M}$. More precisely, there
exists $u_{k}\in K_{k}\in\Gamma_{k}$ such that
$$
\lambda_{k}=pA(u_{k})=\sup_{u\in K_{k}}pA(u)
$$
and $(\lambda_{k},u_{k})$ is a solution of (1.1) associated with the
positive eigenvalue $\lambda_{k}$. Moreover,
$$
\lambda_{k}\to +\infty,\quad \mbox{as }k\to +\infty.
$$
\end{thm}

\paragraph{Proof}
We need only to prove that for any $k\in\mathbb{N}^*$,
$\Gamma_{k}\not= \emptyset$ and the least assertion.
Indeed, since $W_{0}^{2,p}(\Omega)$ is separable, there exist
$(e_{i})_{i\geq 1}$ linearly dense in $W_{0}^{2,p}(\Omega)$ such that
$\mathop{\rm supp}e_{i}\cap \mathop{\rm supp} e_{j}=\emptyset$ if
$i\not=$j. We can assume that $e_{i}\in\cal{M}$.
Let $k\in\mathbb{N}^*$, denote $F_{k}=\mathop{\rm
span}\{e_{1},e_{2},\dots,e_{k}\}$.
$F_{k}$ is a vectorial subspace and $\dim F_{k}=k$.
If $v\in F_{k}$, then there exist $\alpha_{1},\dots\alpha_{k}$ in
$\mathbb{R}$
such that $v=\sum_{i=1}^{k}\alpha_{i}e_{i}$. Thus
$B(v)=\sum_{i=1}^{k}|\alpha_{i}|^{p}B(e_{i})=\frac{1}{p}
\sum_{i=1}^{k}|\alpha_{i}|^{p}$.
It follows that the map $v\mapsto (pB(v))^{1/p}:=\|v\|$
defines a norm on $F_{k}$.
Consequently, there is a constant $c>0$ such that
$$
c\|\Delta u\|_p\leq\|v\|\leq \frac{1}{c}\|\Delta u\|_p.
$$
This implies that the set
$$
V=F_{k}\cap\{v\in W_{0}^{2,p}(\Omega):B(v)\leq \frac{1}{p} \}
$$
is bounded. Thus $V$ is a symmetric bounded neighbourhood of
$0 \in F_{k}$. By (f) in \cite[Prop. 2.3]{Szu},
$\gamma(F_{k}\cap\cal{M})= k. $
Because $F_{k}\cap\cal{M}$ is compact and $\Gamma_{k}\not=\emptyset$.
Now, we claim that $\lambda_{k}\to +\infty$, as
$k\to +\infty$. Indeed let be $(e_{n},e_{j}^*)_{n,j}$ a
bi-orthogonal system such that $e_{n}\in W_{0}^{2,p}(\Omega)$ and
$e_{j}^*\in W^{-2,p'}(\Omega)$, the $e_{n}$ are linearly dense in
$W_{0}^{2,p}(\Omega)$; and the $e_{j}^*$ are total for
$W^{-2,p'}(\Omega)$, see e.g. \cite{Szu}. For $k\in\mathbb{N}^*$, set
$$
F_{k}=\mathop{\rm span}\{e_{1},\dots,e_{k}\},\quad
F_{k}^{\bot}=\mathop{\rm span}\{e_{k+1,e_{k+2},\dots}\}.
$$
By (g) of Proposition 2.3 in \cite{Szu}, we have for any A$\in \Gamma_{k},$
$A\cap F_{k-1}^{\bot}\not=\emptyset$. Thus
$$
t_{k}:=\inf_{A\in\Gamma_{k}}\sup_{u\in A\cap F_{k-1}^{\bot}}pA(u)\to
+\infty.
$$
Indeed, if not, for $k$ is large, there exists $u_{k}\in
F_{k-1}^{\bot}$
with $\|u_{k}\|_p=1$ such that
$$t_{k}\leq pA(u_{k})\leq M,
$$
for some $M>0$ independent of $k$. Thus $\|\Delta u_{k}\|_p\leq M$.
This implies that $(u_{k})_{k}$ is bounded in $W_{0}^{2,p}(\Omega)$.
For a subsequence of $\{u_{k}\}$ if necessary, we can assume that
$\{u_{k}\}$
converge weakly in $W_{0}^{2,p}(\Omega)$ and strongly in
$L^{p}(\Omega)$.
By our choice of $F_{k-1}^{\bot}$ , we have $u_{k}\hookrightarrow 0$
weakly
in $W_{0}^{2,p}(\Omega)$. Because $\langle e_{n}^*,e_{k}\rangle=0$,
$\forall k\geq n$.
This contradicts the fact that $\|u_{k}\|_p=1\, \forall k$. Since
$\lambda_{k}\geq t_{k}$, the claim is proved.
This completes the proof. \hfill$\square$

\begin{cor} \label{cor3.1}
(i)  $\lambda_{1}=\inf\{\|\Delta v\|_p^{p}, v\in W_{0}^{2,p}(\Omega);
\int_{\Omega}\rho(x)|v|^{p}dx=1\}$.\\
(ii)
$0<\lambda_{1}\leq\lambda_{2}\leq\dots\leq\lambda_{n}\to +\infty$.
\end{cor}

\paragraph{Proof}
(i) For $u\in \cal{M}$, we put $K_{1}=\{u,-u\}$. It is clear that
$\gamma (K_{1})=1 $, that $A$ is even and that
$$
pA(u)=\max_{K_{1}}pA\geq\inf_{K\in \Gamma_{1}}\max_{K}pA.
$$
Hence
$$
\inf_{u\in\cal{M}}pA(u)\geq \inf_{K\in \Gamma_{1}}\max_{K}pA=\lambda_{1}.
$$
On the other hand, $\forall K\in\Gamma_{1},\ \forall u\in K$,
$$
\sup_{K}pA\geq pA(u)\geq \inf_{u\in\cal{M}}pA(u).
$$
So
$$ \inf_{K\in \Gamma_{1}}\max_{K}pA
=\lambda_{1}\geq\inf_{u\in\cal{M}}pA(u).
$$
Thus
$$\lambda_{1}=\inf_{u\in\cal{M}}pA(u)=\inf\{\|\Delta
v\|_p^{p}, v\in W_{0}^{2,p}(\Omega):\int_{\Omega}\rho(x)|v|^{p}dx=1\}.
$$
(ii) For all $i\geq j$, $\Gamma_{i}\subset\Gamma_{j}$. From the
definition of
$\lambda_{i},i\in\mathbb{N}^*$, we have $\lambda_{i}\geq\lambda_{j}$.
$\lambda_{n}\to +\infty$ is already proved in Theorem \ref{thm3.1}.
Which completes the proof. \hfill$\square$

\begin{cor}\label{coro3.2}
Assume that $|\Omega^{-}_{\rho}|\not=0$ with
$\Omega^{-}_{\rho}=\{x\in\Omega: \rho(x)<0\}$. Then $\Delta_p^{2}$
has a decreasing sequence of the negative eigenvalues
$(\lambda_{-n})(\rho)_{n\geq 0}$, such that
$\lim_{n\to +\infty}\lambda_{-n}=-\infty$.
\end{cor}

\paragraph{Proof}
First, remark that $\Omega^{-}_{\rho}=\Omega^{+}_{(-\rho)},$ so
$|\Omega^{+}_{(-\rho)}|=|\Omega^{-}_{\rho}|\not=0$. From Theorem
\ref{thm3.1},
$\Delta_p^{2}$ has an increasing sequence of the positive eigenvalues
$\lambda_{n}(-\rho)$, such that
$\lim_{n\to+\infty}\lambda_{n}(-\rho)=+\infty$. Note that
$\lambda_{n}(-\rho)$ satisfies
$$
\Delta_p^{2}u=\lambda_{n}(-\rho)(-\rho)|u|^{p-2}u
=-\lambda_{n}(-\rho)\rho|u|^{p-2}u,
$$ for $u\in W_{0}^{2,p}(\Omega)$.
Put $ \lambda_{-n}(\rho):=-\lambda_{n}(-\rho)$
then $\lambda_{n}(-\rho)_{n\geq 0}$ is an increasing positive sequence so
$(\lambda_{-n})(\rho)_{n\geq 0}$ is a negative decreasing sequence.
On the other hand, $\lim_{n\to  +\infty}\lambda_{n}(-\rho)=+\infty$.
So
$$\lim_{n\to  +\infty}\lambda_{-n}(\rho)=-\infty.$$

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\end{thebibliography}

\noindent\textsc{Abdelouahed El Khalil} (e-mail: lkhlil@hotmail.com)  \\
\textsc{Siham Kellati} (e-mail: siham360@caramail.com)\\
\textsc{Abdelfattah Touzani} (e-mail: atouzani@iam.net.ma)\\[2pt]
Department of Mathematics and Informatic,\\
Faculty of Sciences Dhar-Mahraz, \\
P.O. Box 1796 Atlas-Fez, Morocco

\end{document}