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\documentclass[reqno]{amsart} 

\AtBeginDocument{{\noindent\small 
{\em Electronic Journal of Differential Equations},
Vol. 2004(2004), No. 12, pp. 1--9.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu  (login: ftp)}
\thanks{\copyright 2004 Texas State University - San Marcos.} 
\vspace{9mm}}

\begin{document} 

\title[\hfilneg EJDE--2004/12\hfil On the second eigenvalue]
{On the second eigenvalue of a Hardy-Sobolev operator} 

\author[K. Sreenadh\hfil EJDE--2004/12\hfilneg]
{K. Sreenadh}  

\address{Konijeti Sreenadh \hfill\break
T.I.F.R. Centre, Post Box No.1234,
Bangalore-560012, India}
\email{srinadh@math.tifrbng.res.in}


\date{}
\thanks{Submitted August 12, 2003. Published Janaury 22, 2004.}
\subjclass[2000]{35J20, 35J70, 35P05, 35P30}
\keywords{$p$-Laplacian, Hardy-Sobolev operator, unbounded domain}


\begin{abstract}
  In this note, we study the variational characterization and 
  some properties  of the second smallest eigenvalue of the
  Hardy-Sobolev operator
  $L_{\mu}:=-\Delta_{p}-\frac{\mu}{|x|^p}$ 
  with respect to an indefinite weight $V(x)$. 
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section] 
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{proposition}[theorem]{Propositioin}
\newtheorem{corollary}[theorem]{Corollary}


\section{ Introduction}

 Let $\Omega$ be a domain in $\mathbb{R}^N$  containing $0$.
We recall the classical Hardy-Sobolev inequality which states that, for $1<p<N$,
\begin{equation}
\label{eq:new1}
\int_{\Omega}|\nabla u|^pdx\ge \Big(\frac{N-p}{p}\Big)^p
\int_{\Omega}\frac{|u|^p}{{|x|}^p}dx,\quad \forall u\in C_{c}^{\infty}(\Omega).
\end{equation}
Let $D^{1,p}_0(\Omega)$ be the closure of $C^{\infty}_{c}(\Omega)$ with respect to
the norm $\|u\|_{1,p}=\|\nabla u\|_{L^p (\Omega)}$.
The Hardy-Sobolev operator $L_{\mu}$ on $D^{1, p}_0(\Omega)$ is defined as
\[
L_{\mu}u:=-\Delta_{p}u-\frac{\mu}{|x|^p}|u|^{p-2}u,\quad
0<\mu<\Big(\frac{N-p}{p}\Big)^p,
\]
where  $\Delta_{p}u=\mathop{\rm div}(|\nabla u|^{p-2}\nabla u)$, is the $p$-Laplacian.

 We are interested in the variational characterization and some properties 
 of the second smallest eigenvalue of the problem
\begin{equation}\label{eq:a1}
\begin{gathered}
L_{\mu}u= \lambda V(x)|u|^{p-2}u \quad \text{in } \Omega \\
 u = 0 \quad \text{on } \partial \Omega.
\end{gathered}
\end{equation}
On the weight on $V(x)$, we assume the following:
\begin{itemize}
\item[(H1)] $V\in L^{1}_{\rm loc}(\Omega)$,
$V^{+}=V_1+V_2  \not\equiv 0$ with $V_1\in L^{N/p}(\Omega)$
and $V_2$ is such that
$\lim_{x\to y,\; x\in \Omega}|x-y|^pV_2(x)=0$ for all
$y\in \overline{\Omega}$, $\lim_{|x|\to \infty, x\in \Omega}|x|^pV_2(x)=0$,
where $V^{+}(x)=\max \{V(x),0\}$.

\item[(H2)]  There exists $r>N/p$ and a closed subset $S$ of measure zero
 in $\mathbb{R}^N$ such that $\Omega\backslash S$ is connected and
 $V\in L^{r}_{\rm loc}(\Omega\backslash S)$.
\end{itemize}
Here we note that there is no global integrability condition assumed on $V^{-}$.

This work is motivated by the work in \cite{SW}. The eigenvalue problem with
indefinite weights has been studied for the
case $\mu=0$ by Szulkin-Willem \cite{SW}. However, some important properties, of
the smallest eigenvalue $\lambda_1$, such as  simplicity and being isolated were
shown only for $p=2$. Recently the author in  \cite{Sr2} proved the simplicity
of $\lambda_1$ and  sign changing nature of eigenfunctions corresponding
to other eigenvalues  when $\Omega$ is bounded. Infact in \cite{Sr2} the author studied these properties for $L_\mu$.
Following the same arguments, one can prove these results in the present case.
 However, showing that $\lambda_1$ is isolated and 
characterization of the second smallest eigenvalue,  were open questions.
To prove these properties, we follow the ideas in \cite{Sr1} and in \cite{CDG}.
Here we should mention that our results are new
even for the case $\mu=0$. We use the following results in later sections.

\begin{proposition}[Boccardo-Murat \cite{BM}] \label{prop1.1}
Let $\Omega$ be a bounded domain in $\mathbb{R}^N$ and let  $u_{n} \in W^{1,p}(\Omega)$
satisfy
\[
-\Delta_{p} u_{n}=f_{n}+g_{n} \quad \mbox{in } \mathcal{D}'(\Omega)
\]
and \begin{itemize}
\item[(i)] $u_{n} \to u$ weakly in $W^{1,p}(\Omega)$
\item[(ii)] $u_{n} \to u$ in $L^p(\Omega)$
\item[(iii)] $f_{n} \to f$ in $W^{-1,p^{'}}$
\item[(iv)] $g_{n}$ is a bounded  sequence of Radon measures.
\end{itemize}
Then there exists a subsequence $ \{u_{n}\}$ of $\{u_{n}\}$
such that $\nabla u_{n} \to \nabla u$ a.e. in $\Omega $.
\end{proposition}

\begin{proposition}[Brezis-Lieb \cite{BL}] \label{prop1.2}
Let $f_{n}\to f $ a.e in $\Omega$ as $n\to \infty$ and $f_{n}$ be bounded in
$L^p(\Omega)$, for some $p>1$. Then
\[
\lim_{n\to \infty} \{\|f_{n}\|_{p}-\|f_{n}-f\|_{p}\}=\|f\|_{p}.
\]
\end{proposition}

Let $X$ be a Banach space and let $M=\{u\in X\;\vline \; g(u)=0\}$ with $g\in C^{1}$.
Also let $f:X \to \mathbb{R}$ be a $C^{1}$ functional and let $\tilde{f}$ be the
restriction of $f$ to $M$. Then we have the following form of the Mountain
pass Theorem \cite{St}.

\begin{proposition} \label{prop1.3}
Let $u,v\in M$ with $u\not\equiv v$ and suppose that
\[
c:=\inf_{h\in \Gamma}\max_{w\in h(t)} f(w)>\max\{f(u), f(v)\}
\]
where
\[
\Gamma:=\{h\in C([-1,+1],M)\;\vline h(-1)=u \quad \text{and}\quad h(1)=v\}\ne \emptyset
\]
Also suppose that $\tilde{f}$ satisfies Pailse-Smale (PS) condition on $M$.
Then $c$ is a critical value of $\tilde{f}$.
\end{proposition}

We define the norm
\[
\|\tilde{f}'\|_*=\inf \{\|f'(u)-tg'(u)\|_{X^{*}}: t\in \mathbb{R}\}.\]
 The variational characterization of the smallest eigenvalue is given by
\[
\lambda_1=\inf_{0\not\equiv u\in W^{1,p}_0(\Omega)} \frac{\int_{\Omega} |\nabla u|^pdx
-\int_{\Omega}\frac{|u|^p}{|x|^p}dx}{\int_{\Omega}|u|^p V(x) dx}
\]
and the corresponding eigenfunction is denoted by $\phi_1$, which is unique under
the condition  $\int_{\Omega}|\phi|^pV(x)dx=1$ (see \cite{Sr2}). We will prove the following property.

\begin{theorem} \label{thmI}
The eigenvalue $\lambda_1$ is isolated in the spectrum of $L_{\mu}$.
\end{theorem}

We will establish the following  variational characterisation of the second
smallest eigenvalue:
\[
\lambda_2=\inf_{\gamma\in \Gamma} \sup_{u\in \gamma} \frac{\int_{\Omega} |\nabla u|^pdx
-\int_{\Omega}\frac{|u|^p}{|x|^p}dx}{\int_{\Omega}|u|^p V(x) dx},
\]
where $\Gamma=\left\{\gamma\in C\left([-1,1]:M\right)\;\vline \gamma(-1)=-\phi_1, \gamma(1)
=\phi_1\right\}$ and $M$ is defined as in the next section.
We show also the following  property of $\lambda_2$.

\begin{theorem} \label{thmII}
If $V_a \le V_b$, then $\lambda_2(V_a)\ge \lambda_2(V_b)$.
\end{theorem}

\section{Proofs of results}

In this section we show that $\lambda_1$ is siolated and give a variational
 characterization for second smallest eigenvalue of $L_{\mu}$.

\begin{lemma} \label{lm2.1}
The mapping $u\longmapsto \int_{\Omega}V^{+}|u|^pdx $ is weakly continuous.
\end{lemma}

The proof of this lemma follows from (\ref{eq:new1}) and (H1). We refer the reader to \cite{SW} for more details.\\
Now, we consider the set
\[
M=\Big\{ u \in D^{1,p}_0(\Omega)\;\vline \;\int_{\Omega}|u|^pV(x)=1\Big\}.
\]
Since $M$ is not a manifold in $D^{1,p}_0(\Omega)$, we define 
$X=\{u\in D^{1,p}_0(\Omega)\;\vline\; \|u\|_X <\infty \}$, where
\[
\|u\|_{X}^p:=\int_{\Omega}|\nabla u|^p dx + \int_{\Omega} |u|^p V^{-} dx.
\]
Then $M$ is a $C^{1}$-manifold as a subset of the space $X$. On this space,
 we define the functional
\[
J_{\mu}(u)=\frac{\int_{\Omega}|\nabla u|^p\,dx
-\int_{\Omega}\frac{\mu}{|x|^p}|u|^p\,dx}{\int_{\Omega}|u|^p V\,dx}.
\]
Let $\tilde{J_\mu}$ denote the restriction of $J_\mu$ to $M$. and let
$\|u\|_{L^p(V)}^p=\int_{\Omega}|u|^pV(x) dx$.

\begin{lemma} \label{lm2.2}
The functional $\tilde{J_{\mu}}$ satisfies the Palais-Smale condition at any
positive level.
\end{lemma}

\begin{proof}
Let $\{u_{n}\}$ be a sequence in $M$ such that $J_{\mu}(u_{n})\to \lambda>0$ and
\begin{equation}
\label{eq:b1}
\langle J_{\mu}(u_{n}),\phi\rangle -J_{\mu}(u_{n})\int_{\Omega}|u_{n}|^{p-2}u_{n}
\phi Vdx =o(1).
\end{equation}
Using Hardy-Sobolev inequality and  $u_n\in M$, it follows that  $u_{n}$ is
bounded in $X$ which gives the existence of a subsequence $\{u_{n}\}$ of
$\{u_{n}\}$ and $u$ such that $u_{n}\to u$ weakly in $D^{1,p}_0(\Omega)$.
Since $\lambda>0$ we may assume that $J_{\mu}(u_{n})\ge 0$. Using Lemma
\ref{lm2.1} and (\ref{eq:b1}), we get
\[
\langle J_{\mu}' (u_{n})-J_{\mu}'(u), u_{n}-u\rangle +J_{\mu}(u_{n})
\int_{\Omega}\left[|u_{n}|^{p-2}u_{n}-|u|^{p-2}u\right](u_{n}-u)
V^{-}dx =o(1).
\]
By Fatou's Lemma,
\begin{align*}
0&=\int_{\Omega} \lim_{n\to \infty}\left[|u_{n}|^{p-2}u_{n}-|u|^{p-2}u\right]
[u_{n}-u]V^{-}\\
&\le \liminf_{n\to \infty}\int_{\Omega}\left[|u_{n}|^{p-2}u_{n}-|u|^{p-2}u\right]
[u_{n}-u]V^{-}dx.
\end{align*}
Also, $u_{n}$ satisfies
\[
-\Delta_{p}u_{n}-\frac{\mu}{|x|^p}|u_{n}|^{p-2}u_{n}-J_{\mu}(u_{n})|u_{n}|
^{p-2}u_{n} V(x) =o(1) \quad \text{in} \quad \mathcal{D}'(\Omega_{m}),
\]
where $\Omega_{m}$ is a bounded domain such that $\Omega=\cup_{m=1}^{\infty} \Omega_{m}$.
By Proposition \ref{prop1.1}, noting that
$\frac{\mu}{|x|^p} |u_{n}|^{p-2} u_{n}+ J_{\mu}(u_n) |u_n|^{p-2}u_n V^{-}$
is a bounded sequence of Radon measures,  there exists a subsequence
$\{u_{n}^{m}\}$ of $\{u_n\}$ an $u$ such that $\nabla u_{n}^{m}\to \nabla u$ a.e., in $\Omega_{m}$.
By the process of diagonalization we can choose a subsequence $\{u_{n}\}$ such
that $\nabla u_{n}\to \nabla u$ a.e. in $\Omega$.
By Proposition \ref{prop1.2}, we have
\begin{gather}
\label{eq:2new}
\|u_{n}-u\|_{1,p}^p=\|u_{n}\|_{1,p}^p-\|u\|_{1,p}^p+o(1)\\
\label{eq:2new2}
\|\frac{u_{n}-u}{|x|}\|_{L^p(1)}^p=\|\frac{u_{n}}{|x|}\|_{L^p(1)}^p-\|\frac{u}{|x|}\|_{L^p(1)}^p+o(1).
\end{gather}
We also have, by Fatau's lemma,
\begin{align*}
&\int_{\Omega} V^{-} ( |u_{n}|^p+|u|^p-|u_{n}|^{p-2} u_{n} u -|u|^{p-2} u u_{n} )dx \\
& \ge \int_{\Omega} V^{-} \left( |u_n|^p+|u|^p\right)
-\Big( \int_{\Omega} V^{-} |u_n|^p \Big)^{(p-1)/p}
 \Big(\int_{\Omega} V^{-}|u|^p\Big)^{1/p}\\
&\quad -\Big(\int_{\Omega} V^{-} |u|^p\Big)^{(p-1)/p}
 \Big(\int_{\Omega} V^{-}|u_n|^p\Big)^{1/p}\\
&=\Big[\Big(\int_{\Omega}V^- |u_n|^p\Big)^{(p-1)/p}
 - \Big( \int_{\Omega}V^- |u|^p \Big)^{(p-1)/p}\Big]\\
&\quad\times \Big[ \Big(\int_{\Omega} V^- |u_n|^p\Big)^{1/p}
 -\Big(\int_{\Omega} V^- |u|^p \Big) ^{\frac{1}{p}}\Big] \ge 0\,.
\end{align*}
Now using (\ref{eq:2new}) and (\ref{eq:2new2}),
\begin{align*}
o(1)=& \langle J_{\mu}(u_{n})-J_{\mu}(u), (u_{n}-u)\rangle 
 +J_{\mu}(u_{n})\int_{\Omega}[|u_{n}|^{p-2}u_{n}-|u|^{p-2}u](u_{n}-u) V^{-}dx\\
\ge& \int_{\Omega}|\nabla u_{n}-\nabla u|^p-\int_{\Omega}\frac{\mu}{|x|^p} |u_{n}-u|^p +o(1)\\
\ge& \big(1-\frac{\mu}{\lambda_{N}}\big) \|u_{n}-u\|_{1,p}+o(1).
\end{align*}
i.e., $u_{n} \to u$ in $D^{1,p}_0(\Omega)$.
 Notice that
\begin{align*}
o(1) = & \langle J_{\mu}(u_n)-J_{\mu}(u), u_{n}-u\rangle \\
=& \int_{\Omega} V^{-} \left( |u_n|^{p-2}u_n - |u|^{p-2}u\right) (u_n-u) dx+o(1)
\ge   0\,.
\end{align*}
Therefore, $\int_{\Omega} V^{-} |u_{n}|^p dx \to \int_{\Omega} V^{-} |u|^pdx$ and hence $\|u_n\|_{X}\to \|u\|_{X}$.
\end{proof}

Observe that $\tilde{J_\mu}(u)\ge \lambda_1$ and $\tilde{J_\mu}(\pm \phi_1)=\lambda_1$.
So $+ \phi_1$ and $-\phi_1$ are two global minima of $\tilde{J_\mu}$.
Now consider
\[\Gamma=\{ \gamma\in C([-1,1];M) \;\vline \;\gamma(-1)=-\phi_1,
\gamma(1)=\phi_1\}. 
\]
By Proposition \ref{prop1.3}, there exists $u\in X$ such that
$\tilde{J}_{\mu}'(u) =0$ and
$J_{\mu}(u)={\mathcal{C}}$, where
\begin{equation} \label{Cvalue}
\mathcal{C} =\inf_{\gamma\in\Gamma} \sup_{u\in\gamma} \tilde{J}_{\mu}(u).
\end{equation}

\begin{lemma} \label{lm2.3}
\begin{itemize}
\item[(i)] $M$ is locally arc wise connected  
\item[(ii)] Any connected open subset $B$ of $M$ is arcwise connected 
\item[(iii)] if $B'$ is a component of an open set $A$, then 
$\partial B' \cap B$ is empty.
\end{itemize}
\end{lemma}

The proof of this lemma follows from the fact that $M$ is a Banach Manifold. For a proof we refer the reader to \cite{CDG}.
 Define $\mathcal{O}=\{u\in M \;\vline \; \tilde{J_{\mu}}(u)< r \}$

\begin{lemma} \label{lm2.4}
 Each component of $\mathcal{O}$ contains a critical point of $\tilde{J_{\mu}}$.
\end{lemma}

\begin{proof} Let $\mathcal{O}_1$ be a component of $\mathcal{O}$ and let
$d=\inf \{\tilde{J_\mu}(u), u\in \mathcal{O}_1\}$, where
$\overline{\mathcal{O}_1}$ is $X$-closure of $\mathcal{O}$.
Suppose this infimum is achieved by $v \in \overline{\mathcal{O}_1}$.
Then by Lemma \ref{lm2.3} this cannot be in $\partial \mathcal{O}_1$ and hence
$v$ is in $\mathcal{O}_1$ and is a critical point of $\tilde{J_\mu}$.

 Now we show that $d$ is achieved. Let $u_n\in \mathcal{O}_1$ be a minimizing
sequence with $\tilde{J_\mu}(u_n)\le d+\frac{1}{n^2}$.
By Ekeland Variational Principle, we get $v_n \in \mathcal{O}_1$ such that
\begin{gather}
\label{eq:Lem2.41}
\tilde{J_\mu}(v_n)\le \tilde{J_\mu}(u_n), \\
\|v_n -u_n \|_X \le \frac{1}{n}, \\
\tilde{J_\mu}(v_n)\le \tilde{J_\mu}(v)+\frac{1}{n}\|v_n -v\|_X,
\quad \forall v\in \mathcal{O}_1\,. \label{eq:Lem2.42}
\end{gather}
 From (\ref{eq:Lem2.41}) it follows that $\tilde{J_\mu}(v_n)$ is bounded.
Now we claim that $\|\tilde{J_\mu}'(v_n)\|_{*}\to 0$.
We  fix $n$ and choose $w\in X$ tangent to $M$ at $v_n$, i.e., $\int_\Omega 
|v_n|^{p-2}v_n w V=0$.  Now we consider the path
\[
u_t =\frac{v_n +t w}{\|v_n+tw\|_{L^p(V)}}.
\]
Since $\tilde{J_\mu}(v_n)\le d+\frac{1}{n}<r $ for $n$ large, we have $v_n\in 
\overline{\mathcal{O}}_1$ and by  Lemma \ref{lm2.3} (iii),  $v_n\notin \partial \mathcal{O}_1$. So $u_t \in {\mathcal{O}_1}$ for $|t|$ small.
 Taking  $v=u_t$ in \eqref{eq:Lem2.42} we obtain
\begin{equation}\label{eq:2new4}
\begin{aligned}
&\frac{\tilde{J_\mu}(v_n)-\tilde{J_\mu}(v_n+t w)}{t} \\
&\le \frac{1}{n t}\|v_n(\frac{1}{r(t)}-1)\|_X
+\frac{1}{n}\|w\|+
\frac{1}{t}\big(\frac{1}{r(t)^p}-1\big) \tilde{J}_\mu (v_n+tw),
\end{aligned}
\end{equation}
where $r(t)=\|v_n+tw\|_{L^p(V)}$.
The last term in (\ref{eq:2new4}) involves $\frac{r(t)^p -1}{t}$ which can be
calculated as
\[
\frac{d}{dt} r(s)^p \big|_{s=0}=\lim_{t\to 0} \frac{r(t)^p -1}{t}.
\]
On the other hand since $w$ is tangent to $M$ at $v_n$,
\[
\frac{d}{dt} r(s)^p \big|_{s=0}= p\int_{\Omega}|v_n|^{p-2}v_n w V(x)dx=0.
\]
Therefore, we have $\frac{r(t)^p -1}{t} \to 0$ as $t\to 0$ and that the second term goes to $0$. Similarly, the first term also goes to zero
as $t\to 0$. Taking limit $t\to 0$ in (\ref{eq:2new4}) we get
\[
\langle J_{\mu}' (v_n), w\rangle \le \frac{1}{n} \|w\|_X, \quad
\text{for all $ w \in X$ tangent to $M$ at $v_n$}.
\]
Now if $w$ is arbitrary in $X$. We choose $\alpha_n$ so that $(w-\alpha_n v_n)$
is tangent to $M$ at $v_n$. i.e., $\alpha_n =\int_{\Omega} |v_n|^{p-2}v_n w V(x) dx$.
So (\ref{eq:2new4}) gives,
\[
|\langle J_{\mu}'(v_n), w \rangle- \langle J_{\mu}'(v_n), v_n\rangle
\int_{\Omega} |v_n|^{p-2} v_n w| \le \frac{1}{n} \|w-\alpha_n v_n\|_X
\]
Since $\|\alpha_n v_n \|_X \le C. \|w\|_X$, we have
\[
|\langle J_\mu '(v_n), w\rangle - t_n \int_\Omega |v_n|^{p-2}v_n w V(x) dx |
\le \epsilon_n \|w\|_X
\]
where $t_n=\langle J_\mu '(v_n), v_n\rangle$ and $\epsilon_n \to 0$.
Therefore, $\|\tilde{J_\mu}'(v_n)\|_* \to 0$ and $v_n$ is a Palais-Smale 
sequence. Hence by Lemma \ref{lm2.2}, $\{v_n\}$ has a convergent subsequence 
with limit, say, $v$. Then $d$ is achieved at $v$.
\end{proof}

\begin{lemma} \label{lm2.5}
The number $\mathcal{C}$ defined by \eqref{Cvalue} is the second smallest
eigenvalue of $L_{\mu}$
\end{lemma}

\begin{proof} We follow the proof in \cite{CDG}.
Assume by contradiction that there exists an eigenvalue $\delta$ such that
$\lambda_1 <\delta<{\mathcal{C}}$.  In other words, $\tilde{J_{\mu}}$ has
a critical value $\delta$ with $\lambda_1<\delta< {\mathcal{C}}$.
We will construct a path in $\Gamma$ on which $\tilde{J_{\mu}}$ remains
$\le \delta$, which yields a contradiction with the definition of $\mathcal{C}$.
Let $u\in$ M satisfies the equation
\[
-\Delta_{p}u-\frac{\mu}{|x|^p}|u|^{p-2}u=\delta V(x)|u|^{p-2}u
\quad\text{in }  \mathcal{D}'(\Omega),
\]
and $u$ changes sign in $\Omega$. Taking $u^{+}$ and $u^{-}$ as test function
we get
\begin{gather*}
\int_{\Omega} |\nabla u^{+}|^p\,dx -\int_{\Omega}
\frac{\mu}{|x|^p}|u^{+}|^p\,dx=\delta \int_{\Omega} (u^{+})^p V(x)\,dx\,\\
\int_{\Omega} |\nabla u^{-}|^p\,dx -\int_{\Omega} \frac{\mu}{|x|^p}(u^{-})^p
\,dx=\delta \int_{\Omega} (u^{-})^pV(x)\,dx\,.
\end{gather*}
Consequently
\[
\tilde{J_{\mu}}(u)= \tilde{J_{\mu}}(\frac{u^{+}}{\|u^{+}\|_{L^p(V)}})=\tilde{J_{\mu}}(\frac{-u^{-}}{\|u^{-}\|_{L^p(V)}})=
\tilde{J_{\mu}}(\frac{u^{-}}{\|u_{-}\|_{L^p(V)}})=\delta.
\]
We will consider the following three paths in $M$, which go respectively
from $u$ to $\frac{u^{+}}{\|u^{+}\|_{L^p(V)}}$, from
$\frac{u^{+}}{\|u^{+}\|_{L^p(V)}}$ to $\frac{u^{-}}{\|u^{-}\|_{L^p(V)}}$
and $\frac{-u^{-}}{\|u^{-}\|_{L^p(V)}}$ to $u$:
\begin{gather*}
u_1(t)=\frac{tu+(1-t) u^{+}}{\|tu+(1-t)u^{+}\|_{L^p(V)}},\\
u_2(t)=\frac{tu^{+}+(1-t) u^{-}}{\|tu+(1-t)u^{-}\|_{L^p(V)}},\\
u_{3}(t)=\frac{-tu^{-}+(1-t) u}{\|-tu^{-}+(1-t)u\|_{L^p(V)}}.
\end{gather*}
Also we have
\[
\tilde{J_{\mu}}(u_1(t)) = \tilde{J_{\mu}}(u_2(t))= \tilde{J_{\mu}}(u_{3}(t))
=\delta.
\]
By joining the paths $u_1(t)$ and $u_2(t)$ we get a new path which connects
$u$ and $\frac{u^{-}}{\|u^{-}\|_{L^{p}(V)}}$ and stays at levels $\le \delta$.
Call this path as $u_{4}(t)$.
 Now we define $O=\{ v \in M\;\vline\; \tilde{J_{\mu}}(v)<\delta \}$. Clearly
$\phi_1, -\phi_1\in O$.
Since $\frac{u^{-}}{\|u^{-}\|_{L^{p}(V)}}$ does not change sign and vanishes on a set
of positive measure it is not a critical point of $\tilde{J_{\mu}}$.
So $\frac{u^{-}}{\|u^{-}\|_{L^{p}(V)}}$ is a regular value of $\tilde{J_{\mu}}$,
and consequently there exists a $C^{1}$ path $\eta:[-\epsilon, \epsilon]\to M$ with
$\eta(0)=\frac{u^{-}}{\|u^{-}\|_{L^p(V)}}$ and $\frac{d}{dt} (\tilde{J_{\mu}}(\eta(t))
\big|_{t=0}\ne 0$. choose a point $v\in O$ on this path (this is possible because
$\tilde{J_{\mu}}'(\eta(t))\big|_{t=0}\ne 0$)
 we can thus move from $\frac{u^{-}}{\|u^{-}\|_{L^p(V)}}$ to $v$ through this path
which lies at levels $<\delta $. Taking the component of $O$
which contains $v$ and applying Lemma \ref{lm2.3}  together with Lemma
\ref{lm2.4}, we can connect $v$ to $+\phi_1$ (or to $-\phi_1$) with a path in $M$
at levels $ <\delta $. Let us assume that this is $+\phi_1$ which is reached
in this way. Now call this path connecting $\frac{u^{-}}{\|u^{-}\|_{L^p(V)}}$
and $\phi_1$ as $u_{5}(t)$, and consider the
 symmetric path $-u_{5}(t)$, which goes from $-\frac{u^{-}}{\|u^{-}\|_{L^p(V)}}$
 to $-\phi_1$. We evaluate the functional $\tilde{J_{\mu}}$ along $-u_{5}(t)$.
Since $\tilde{J_\mu}$ is even,
\[
\tilde{J_{\mu}}(-u_{5}(t))= \tilde{J_{\mu}}(u_{5}(t)) \le \delta.
\]
 Finally with $u_{3}(t)$ we can connect
$-\frac{u^{-}}{\|u^{-}\|_{L^p(V)}}$ with $u$ by a path which stays at level
$\delta$ . Putting every
thing together we get a path connecting $-\phi_1$ and $\phi_1$ staying at
levels $\le \delta$. This concludes the proof.
\end{proof}

Note that Theorem \ref{thmI} is an immediate consequence of Lemma \ref{lm2.5}.
So we have the following characterization of $\lambda_2$, the second smallest
eigenvalue of $L_{\mu}$,
\[
\lambda_2=\inf_{\gamma\in\Gamma}\sup_{u\in \gamma}\int_{\Omega}|\nabla u|^pdx
-\int_{\Omega}\frac{|u|^p}{|x|^p}dx.
\]
 Now let  $\mu_{k}$ be the sequence of eigenvalues obtained in \cite{Sr2} which
are characterized as 
\[
\mu_{k}=\inf_{{\mathcal{A}}\in {\mathcal{F}}} \sup_{{u\in\mathcal{A}}}
\int_{\Omega}|\nabla u|^p-\int_{\Omega}\frac{\mu}{|x|^p}|u|^p\,,
\]
where ${\mathcal{F}} =\{\mathcal{A}\subset M \;\vline\; \text{ the genus of }
\mathcal{A} \ge k \}$.

\begin{corollary} \label{cocro2.6}
With the notation above, $\mu_2=\lambda_2$.
\end{corollary}

\begin{proof}
Let $\gamma$ be a curve in $\Gamma$. By joining this with its symmetric path
$-\gamma(t) $ we can get a set of genus $\ge 2$ where $J_{\mu}$ does not
increase its values. Therefore,
$\lambda_2 \ge \mu_2$. But by Theorem \ref{thmI}, there is no eigenvalue between
$\lambda_1$ and $\lambda_2$. Hence $\lambda_2=\mu_2$.
\end{proof}

\begin{lemma} \label{lm2.7}
Let $u\in X$ be a solution of (\ref{eq:a1}) and let $\mathcal{O}$ be a
component of $\{x\in \Omega \;\vline \; u(x)>0\}$.
Then $u \big|_{\mathcal{O}} \in D^{1,p}_0(\mathcal{O})$
\end{lemma}

\begin{proof}
Let $u_{n} \in C_{c}(\Omega)\cap D^{1,p}_0(\Omega)$ such that $u_{n}\to u$
in $D^{1,p}_0(\Omega)$. Then $u_{n}^{+}\to u^{+}$ in $D^{1,p}_0(\Omega)$.
Let $v_{n}=\min(u_{n},u)$ and let $\phi:\mathbb{R}\to \mathbb{R}$ be a $C^{1}$ function
such that
$$
\phi(t)= \begin{cases}
0&\text{for }  t\le 1/2\\
1&\text{for } t\ge 1 \end{cases}
$$
and $|\phi'|\le 1$. Let $\psi_{r}(x)=\phi\big(d(x,S)/r\big)$ where
$d(x,S)=\mathop{\rm dist}(x,S)$. Then
$$
\psi_{r}(x)\begin{cases}
0 &\text{for } d(x,S)\le r/2 \\
1 &\text{for } d(x,S)\ge r \end{cases}
$$
and $|\nabla \psi_{r}(x)| \le C/r$  for some constant $C$.
Now we define  $w_{n,r}(x)=\psi_{r}v_{n}(x)\big|_\mathcal{O}$. Since
$\psi_{r}v_{n}\in C(\overline{\Omega})$, we have
$w_{n,r} \in C(\overline{\mathcal{O}})$ and vanishes on the boundary
$\partial \mathcal{O}$. Indeed for $x\in \partial \mathcal{O}\cap S$
then $\psi_{r}(x)=0$ and so $w_{n,r}(x)=0$. If $x\in \partial \mathcal{O} \cap \Omega$
and $x\notin S$ then $u(x)=0$(since $u$ is continuous except at 0) and so
$v_{n}(x)=0$ . If $x\in \partial \Omega$ then $u_{n}(x)=0$ and hence
$v_{n}(x)=0$. So in all the cases $w_{n,r}(x)=0$ for
$x \in \partial \mathcal{O}$. Therefore,
$w_{n,r}\in D^{1,p}_0(\mathcal{O})$.
\begin{align*}
\int_{\Omega}|\nabla(w_{n,r})-\nabla(\psi_{r}u)|^p
&=\int_{\mathcal{O}}|(\nabla \psi_{r}) v_{n}+\psi_{r} \nabla v_{n}
  -(\nabla \psi_{r}) u -\psi_{r} \nabla u|^p\, dx\\
&\le \|\nabla \psi_{r}v_{n}-\nabla \psi_{r}u\|_{L^p(\mathcal{O})}^p
+\|\psi_{r}\nabla v_{n}- \psi_{r} \nabla u\|_{L^p(\mathcal{O})}^p
\end{align*}
which goes to $0$ as $n\to \infty$.
i.e., $w_{n,r}\to \psi_{r}u\big|_{\mathcal{O}}$ in $D^{1,p}_0(\mathcal{O})$.
Now
\[
\int_{\mathcal{O}}|\nabla \psi_{r}u +\psi_{r} \nabla u -u|^p \le
\int_{\mathcal{O}}|\psi_{r} \nabla u -\nabla u|^p
+\int_{\mathcal{O}\cap \{r/2<|x|<r\}}|\nabla \psi_{r}|^p u 
\]
$\to 0$ as $r\to 0$ by (\ref{eq:new1}). Therefore, $u\big|_{\mathcal{O}}
\in D^{1,p}_0(\mathcal{O})$.
\end{proof}

\begin{proof}[Proof of Theorem \ref{thmII}]
We denote $\tilde{J}_\mu$ corresponding to $V_b$ with  $\tilde{J}_{\mu,b}$.
Let $u_{a}$ be a solution to
\[
-\Delta_{p}u-\frac{\mu}{|x|^p}|u|^{p-2}u=\lambda_2V_{a}(x)|u|^{p-2}u
\quad \text{in } \mathcal{D}'(\Omega).
\]
Assuming that the claim below is true, we have
\[
\tilde{J}_{\mu,b}\left(\frac{v^+}{\|v^+\|_{L^p (V_b)}}\right) < \lambda_2 (V_a),
\tilde{J}_{\mu,b}\left(\frac{v^-}{\|v^-\|_{L^p (V_b)}}\right) < \lambda_2 (V_a),
\tilde{J}_{\mu,b}\left(v\right) < \lambda_2 (V_a).
\]
Define $\mathcal{O}_b =\{u\in X, \int_\Omega |u|^p V_b=1, \tilde{J}_{\mu,b}\left(v\right) < \lambda_2 (V_a).\}$ Now we proceed as in Lemma \ref{lm2.5}, to define the paths $v_i (t),i=1,...,5$ on which $\tilde{J}_{\mu,b} < \lambda_2(V_a)$. We join these
paths in a way described in Lemma \ref{lm2.5} to obtain a path $\gamma(t)$ in 
$\Gamma_b$ (the family of paths corresponds to $V_b$) such that $\tilde{J}_{\mu,b}
(\gamma(t))<\lambda_2 (V_a)$. This completes  the proof.\\
\noindent {\bf Claim:} There exists $v\in X$, which changes sign and
\begin{equation}\label{eq:b2}
\begin{gathered}
\frac{\int_{\Omega}|\nabla v^{+}|^p\,dx-\int_{\Omega}
\frac{\mu}{|x|^p}|v^{+}|^p\,dx}{\int_{\Omega} (v^{+})^pV_{b}\,dx}<\lambda_2(V_{a}),\\
\frac{\int_{\Omega}|\nabla v^{-}|^p\,dx-\frac{\mu}{|x|^p}|v^{-}|^p\,dx}{\int_{\Omega}
 (v^{-})^pV_{b}\,dx}<\lambda_2(V_{a}).
\end{gathered}
\end{equation}
{\it Proof of Claim:} Since $u_a$ is an eigenfunction corresponding to $\lambda_2> \lambda_1$, it has to change sign in $\Omega$ (see \cite{Sr2}). Let  $O_1$ and $O_2$ be positive and negative nodal 
domains of $u_a$ respectively  such that
\[
\int_{O_1}V_{a}\; (u_{a}^{+})^p\,dx<\int_{O_1}V_{b}\; (u_{a}^{+})^p\,dx \quad \text{and }
\int_{O_2}V_{a}\;(u_{a}^{-})^p\,dx \le \int_{O_2}V_{b}\; (u_{a}^{-})^p\,dx.
\]
By Lemma \ref{lm2.7}, $u_a \big|_{O_1}\in D^{1,p}_0(O_1)$ and also in $L^p(O_1, V^-)$.
We have
\[
\lambda_1(O_1,V_{b})\le\frac{\int_{O_1} |\nabla u_a|^p 
-\frac{\mu}{|x|^p}|u_a|^p}{\int_{O_1}|u_a|^pV_{b}}<\lambda_2(V_{a}).
\]
Therefore, $\lambda_1(O_1,V_{b})<\lambda_2(V_{a})$. Simillarily 
$\lambda_1(O_2,V_{b})\le \lambda_2(V_{a})$.
Now we modify $O_1$ and $O_2$ to get $\tilde{O_1}$ and $\tilde{O_2}$ with
empty intersection and $\lambda_1(\tilde{O_1},V_b)<\lambda_2(V_{a})$ and 
$\lambda_1(\tilde{O_2},V_b)<\lambda_2 $.
For $\eta>0$, let $O_1(\eta) = \{x\in O_1 \;\vline \;dist(x,O_1^{c})>\eta\}$. 
Then $\lambda_1(O_1(\eta),V_b)\ge \lambda_1(O_1,V_b) $ and 
$ \lambda_1(O_1(\eta),V_b)\to \lambda_1(O_1,V_b)$  as $\eta \to 0$.
Therefore, there exists $\eta_0 >0$ such that 
$\lambda_1(O_1(\eta),V_b)<\lambda_2(V_{a})$ for $0<\eta<\eta_0$. 
Let $x\in \partial O_2\cap \Omega$ and 
$0<\eta <\min\{\eta_0, \mathop{\rm dist}(x_0, \Omega^{c})\}$.
 Now define $\tilde{O_2}=O_2\cup B(x_0, \eta/2)$.
 Then $\tilde{O_2}\cap O_1(\eta) =\emptyset$, $\lambda_1
(\tilde{O_2},V_b)<\lambda_1(O_2, V_b)<\lambda_2(V_{a})$. Now we consider the function
$v=v_1-v_2$, where $v_i$ are the extensions by zero outside
$\tilde{O_i}$ of the eigenfunctions associated to 
$\lambda_1 (\tilde{O_i}, V_b)$. Then $v$ satisfies (\ref{eq:b2}).
\end{proof}

\subsection*{Acknowledgements} 
The Author was supported by grant 40/1/2002-R\&D-II/165 from the
National Board for Higher Mathematics(NBHM), DAE, Govt. of India.


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\end{document}