0 .
\end{equation}
It is also well-known that the Rayleigh quotient in \eqref{ray}
admits a minimizer which does not change sign in $\Omega $.
The Euler-Lagrange
equation associated with this minimization problem is
\begin{equation}\label{classiceq}
-\mathop{\rm div}(|\nabla u|^{p-2}\nabla u
)=\lambda \| u \|_{q}^{p-q} |u|^{q-2}u ,
\end{equation}
where $\| u\|_q$ denotes the norm of $u$ in $L^q(\Omega )$.
Usually in the literature, the function $u$ is normalized
in order to get rid of the apparently redundant factor
$\|u \|_{q}^{p-q}$. However, we prefer to keep it since it allows
to think of this problem as an eigenvalue problem. Indeed,
\eqref{classiceq} is homogeneous; i.e., if $u$ is a solution then
$ku $ is also a solution for all $k\in {\mathbb{R}}$, as one would
expect from an eigenvalue problem.
It turns out that $\lambda_1(p,q)$ is the smallest eigenvalue
of \eqref{classiceq} and we refer to it as the first eigenvalue.
The case $p=q$ has been largely investigated by many authors
and it has been often considered as a {\it typical eigenvalue
problem} (cf. e.g., Garc\'{\i}a Azorero and Peral Alonso~\cite{azal});
for extensive references on this subject we refer to
Lindqvist~\cite{lqv}. For the case $p\ne q$ we refer again
to \cite{azal}, \^Otani~\cite{ot88, ot84} and Dr\'{a}bek, Kufner
and Nicolosi~\cite{drabook} who consider an even more general
class of {\it nonhomogeneous eigenvalue problems}.
In this paper, we study several results and we indicate how to
adapt to the general case some proofs known in the case $p=q$.
First of all, we discuss the simplicity of $\lambda_1(p,q)$.
We recall that $\lambda_1(p,q)$ is simple if $q\le p$, as it is
proved in Idogawa and \^{O}tani~\cite{otani95}. If $q>p$ then
$\lambda_1(p,q)$ is not necessarily simple: for example,
simplicity does not hold if $\Omega $ is a sufficiently thin annulus,
see Kawohl~\cite{kawohl} and Nazarov~\cite{naza}.
However, if $\Omega $ is a ball then
the simplicity of $\lambda_1(p,q)$ is guaranteed also in the
case $q>p$: here we briefly describe the argument of Erbe and
Tang~\cite{erta}.
By adapting the argument
of Kawohl and Lindqvist~\cite{lqka}, we prove that if $q\le p$ then
the only eigenvalue admitting a non-negative eigenfunction is the
first one.
Moreover, by exploiting our point of view, we also give an alternative
proof of a uniqueness result of Dr\'{a}bek~\cite[Thm.~1.1]{dra} for
the equation $-\Delta_pu=|u|^{q-2}u$, see Theorem~\ref{drabrev}.
Finally, in the general case $1 0$ depending only on $N,p,q$
such that
\begin{equation} \label{poi}
\|u\|_{L^q(\Omega )} \le C |\Omega
|^{\frac{1}{q}-\frac{1}{p}+\frac{1}{N}}\| \nabla u\|_{L^p(\Omega)},
\end{equation}
for all $u\in W^{1,p}_0(\Omega )$. In particular it follows that
$\lambda_1(p,q)$ defined in \eqref{ray} is positive and satisfies
the inequality
\begin{equation}
\label{poibis}
\lambda_1(p,q)> \frac{1}{C^p|\Omega|^{p(\frac{1}{q}-\frac{1}{p}
+\frac{1}{N} )}}\,.
\end{equation}
Moreover, since the measure of $\Omega $ is finite, the
embedding $W^{1,p}_0(\Omega )\subset L^q(\Omega )$ is compact:
this combined with the reflexivity of $W^{1,p}_0(\Omega )$ guarantees
the existence of a minimizer in \eqref{ray}. As we mentioned in the
introduction, equation \eqref{classiceq} is the Euler-Lagrange
equation corresponding to the minimization problem \eqref{ray}.
It is then natural to give the following definition where,
as usual, equation \eqref{classiceq} is interpreted in the weak sense.
\begin{definition}\label{weakdef} \rm
Let $\Omega$ be a domain in $\mathbb{R}^N$ with finite measure,
$1 0 $
is an eigenvalue of equation \eqref{classiceq} if there exists
$u\in W^{1,p}_0(\Omega )\setminus\{0\}$ such that
\begin{equation} \label{weakeq}
\int_{\Omega }|\nabla u |^{p-2}\nabla u\nabla \varphi dx
=\lambda \| u \|_{q}^{p-q} \int_{\Omega }|u|^{q-2}u\varphi dx,
\end{equation}
for all $\varphi \in W^{1,p}_0(\Omega )$. The eigenfunctions
corresponding to $\lambda $ are the solutions $u$ to \eqref{weakeq}.
\end{definition}
It is clear that all eigenvalues are positive and that
$\lambda_1(p,q)$ is the least eigenvalue. Moreover, the eigenfunctions
corresponding to $\lambda_1(p,q)$ are exactly the minimizers
in \eqref{ray}. We recall the following known result.
\begin{theorem}\label{bound}
Let $\Omega$ be a domain in $\mathbb{R}^N$ with finite measure,
$1 0 $ be an eigenvalue
of equation \eqref{weakeq} and $u\in W^{1,p}_0(\Omega )$ be a
corresponding eigenfunction. Then $u$ is bounded and its first
derivatives are locally H\"{o}lder continuous.
Moreover, if $u\geq 0$ in $\Omega $ then $u>0$ in $\Omega$.
\end{theorem}
As done in \cite[Lemma~5.2]{lqv} for the case $q=p$, the boundedness
of $u$ can be proved by using the method of \cite[Lemma~5.1]{ladur}.
The H\"older regularity of the first order derivatives follows
by Tolksdorf~\cite{tol}.
We note that the argument in \cite{lqv} allows to give a quantitative
bound for $u$. Namely, by a slight modification of \cite[Lemma~5.2]{lqv}
one can prove that there exists a constant $M>0$, depending only
on $p,q,N$, such that
\begin{equation}\label{qtbound}
\|u\|_{L^\infty(\Omega)} \le M \lambda^{\frac{1}{\delta p}}
\|u\|_{L^1(\Omega)},
\end{equation}
where $\delta = 1/N$ if $q\le p$ and $\delta = 1/q-1/p+1/N$ if $q>p$.
We refer to Franzina~\cite{fra} for details.
Finally, the fact that a non-negative eigenfunction does not vanish
in $\Omega$ can be deduced by the strong maximum principle in
Garc\'{i}a-Meli\`{a}n and Sabina de Lis~\cite[Theorem~1]{garmel}.
\begin{corollary}\label{signcor}
Let $\Omega$ be a domain in $\mathbb{R}^N$ with finite measure,
$1 0$ or $u<0$ in $\Omega $.
\end{corollary}
\begin{proof}
Clearly $u$ is a minimizer in \eqref{ray}.
Then also $|u|$ is a minimizer, hence a first eigenfunction.
Thus by Theorem~\ref{bound} $|u|$ cannot vanish in $\Omega $.
\end{proof}
\section{On the simplicity of $\lambda_1(p,q)$}
It is known that if $q\le p$ then $\lambda_1(p,q)$ is simple.
In fact we have the following theorem by Idogawa and
\^{O}tani~\cite[Theorem~4]{otani95} the proof of which works word
by word also when $\Omega$ is not bounded.
\begin{theorem}\label{uniray}
Let $\Omega$ be a domain in $\mathbb{R}^N$ with finite measure
and $1 0$ such that $\phi $ is twice
differentiable in $r$. Recall that by standard regularity theory
an eigenfunction $u$ is twice differentiable on the set
$\{x\in \Omega : \nabla u(x)\ne 0 \} $.
By writing \eqref{classiceq} in spherical coordinates and
multiplying both sides by $r^{N-1}$ it follows that if
$u=\phi (|x|)$ is a radial eigenfunction corresponding
to the eigenvalue $\lambda$ and $\|u\|_{L^q(\Omega )}=1$ then
\begin{equation}\label{radeq}
-( r^{N-1}|\phi '|^{p-2}\phi ' )'=\lambda r^{N-1} |\phi |^{q-2}\phi .
\end{equation}
If in addition $u$ is a first eigenfunction then $u$ does not change
sign in $\Omega $; thus, by integrating equation \eqref{radeq},
one can easily prove that $\phi '$ vanishes only at $r=0$.
Hence $\phi $ is twice differentiable for all $r>0$ and \eqref{radeq}
is satisfied in the classical sense for all $r>0$.
To prove the simplicity of $\lambda_1(p,q)$ we use the following Lemma.
The proof is more or less standard (further details can be
found in Franzina~\cite{fra}).
\begin{lemma}\label{uniqlem}
Let $1 0$. Then the Cauchy problem
\begin{equation}\label{cauchy}
\begin{gathered}
-( r^{N-1}|\phi '|^{p-2}\phi ' )'=\lambda r^{N-1} |\phi |^{q-2}\phi , \quad
r\in (0,R),\\
\phi (0)=c,\quad \phi'(0)=0,
\end{gathered}
\end{equation}
has at most one positive solution $\phi$ in $C^1[0,R]\cap C^2(0,R)$.
\end{lemma}
\begin{proof}
We consider the operator $T$ of $C[0,R]$ to $C[0,R]$ defined by
\begin{equation}
T (\phi) (r) = c - \int_0^r g^{-1}\Big( \frac{\lambda}{t^{N-1}}
\int_0^t s^{N-1}|\phi|^{q-2}\phi\,ds\Big)\,dt, \quad r\in [0,R],
\end{equation}
for all $\phi\in C[0,R]$, where $g(t)=|t|^{p-2}t$ if $t\ne 0$
and $g(0)=0$ and $g^{-1}$ denotes the inverse function of $g$.
It's easily seen that every positive solution to the Cauchy
problem \eqref{cauchy} is a fixed point of the operator $T$ of class
$C^1[0,R]\cap C^2(0,R)$.
Now let $\phi_1,\phi_2\in C^1[0,R]\cap C^2(0,R)$ be two positive
solutions to problem \eqref{cauchy}.
One can prove that there exists $\epsilon_1>0$ such that
\[
\|T(\phi_1)-T(\phi_2)\|_{C[0,\varepsilon]}
\leqslant C_1(\varepsilon) \|\phi_1 - \phi_2 \|_{C[0,\varepsilon]},
\]
for all $\varepsilon \in [0,\epsilon_1] $ where $C_1(\varepsilon)<1$.
It follows that $\phi_1=\phi_2$ in a neighborhood of zero.
Furthermore, let
$R_0 = \sup\{\varepsilon>0 : \phi_1=\phi_2\text{ on }[0,\varepsilon]\}$.
Arguing by contradiction, assume that
$R_0 0$.
Let $\phi_1,\phi_2\in C^1[0,R]\cap C^2(0,R)$ be two
positive solutions to the Cauchy problem \eqref{cauchy}
with $c=c_1, c_2$ respectively.
If $c_1\le c_2 $ then $\phi_1\le \phi_2 $.
\end{lemma}
By using Lemmas~\ref{uniqlem} and \ref{erbetanglem} we can
deduce the validity of the following result.
\begin{theorem}
Let $\Omega $ be a ball in ${\mathbb{R}}^{N}$, $1 1$ and that $\Omega $ is a ball of radius $R$ centered
at zero. Let $u_1, u_2$ be two nonzero eigenfunctions corresponding
to the first eigenvalue $\lambda_1(p,q)$. We have to prove that $u_1$
and $u_2$ are proportional. To do so we can directly assume that
$\| u_1\|_q=\| u_2\|_q=1$. Moreover, by Corollary \ref{signcor} we
can assume without loss of generality that $u_1,u_2>0$ on $\Omega$.
By Theorem~\ref{rad} $u_1,u_2$ are radial functions hence they can
be written as $u_1=\phi_1(|x|)$,
$u_2=\phi_2(|x|)$ for suitable positive functions
$\phi_1,\phi_2\in C^1[0,R]\cap C^2(0,R)$ satisfying condition
$\phi_1'(0)=\phi_2'(0)=0$ and equation \eqref{radeq} with
$\lambda =\lambda_1(p,q)$. If $\phi_1(0)\ne \phi_2(0)$, say
$\phi_1(0)<\phi_2(0)$, then by Lemma~\ref{erbetanglem}
$\phi_1\le \phi _2 $ in $\Omega $ hence $\| u_1\|_q< \| u_2\|_q$,
since by continuity $u_1p$ are treated
separately; instead, here we adopt a unified approach.
We point out that in this paper we do not assume that $\Omega $
is bounded as largely done in the literature, but only that its
measure is finite.
\section{The eigenvalue problem}
Let $\Omega $ be a domain in $\mathbb{R}^N$ with finite measure
and $1

p$;
see \cite{kawohl} and \cite{naza} where the case of a sufficiently
thin annulus is considered. However, as one may expect,
if $\Omega$ is a ball then $\lambda_1(p,q)$ is simple. Basically,
this depends on the following theorem, cf. \cite{kawohl}.
\begin{theorem}\label{rad}
Let $\Omega $ be a ball in ${\mathbb{R}}^{N}$ centered at zero,
$1

0$ there
exists $n\in {\mathbb{N}}$ such that $\sup_{u\in {\mathcal{A}}}E(u) >L$
for all symmetric subsets ${\mathcal{A}}$ of $M$ such that
${\mathcal{A}}$ is compact in $W^{1,p}_0(\Omega )$ and
$\gamma ({\mathcal{A}})\geq n$. Assume to the contrary that there
exists $L>0$ such that this is not the case and set
$B_L=\{u\in M:\ E(u)\le L \}$. By means of a simple contradiction
argument one can prove that there exists $n_L\in {\mathbb{N}}$
such that $\inf_{u\in B_L}\| P_{n_L}u\|_{W^{1,p}_0(\Omega )}>0 $
hence $P_{n_L}u\ne 0$ for all $u\in B_L$.
Let $k_L=\dim X_{n_L}+1$. By assumption there exists a symmetric
subset ${\mathcal{A}}$ of $M$ such that ${\mathcal{A}}$ is compact,
$\gamma ({\mathcal{A}})\geq k_L$ and $\sup_{u\in {\mathcal{A}}}E(u)\le L$. Since ${\mathcal{A}}\subset B_L$ then $P_{n_L}u\ne 0$ for all $u\in {\mathcal{A}}$
hence $\gamma (P_{n_L}({\mathcal{A}}))\geq k_L$; on the other
hand $P_{n_L}({\mathcal{A}})\subset X_{n_L} $ hence
$\gamma (P_{n_L}({\mathcal{A}}))\le \dim X_{n_L}= k_L -1$,
a contradiction.
\end{proof}
We remark that, despite the results of
Binding and Rynne~\cite{bin} who have recently provided examples
of nonlinear eigenvalue problems for which not all eigenvalues are
variational, it is not clear yet whether for our problem the
variational eigenvalues exhaust the spectrum if $N>1$, not even in
the classical case $p=q$. However, a complete description of
$\sigma (p,q)$ is available for $N=1$, see \^Otani~\cite{ot84}
and Dr\'{a}bek and Man\'{a}sevich~\cite{draman}.
The following theorem is a restatement of
\cite[Theorems~3.1, 4.1]{draman}. We include a detailed proof
of \eqref{onedimeq} for the convenience of the reader.
Recall that the function defined by
$$
\arcsin_{pq}(t)= \frac{q}{2}\int ^{\frac{2t}{q}}_0
\frac{ds}{(1-s^q)^{\frac{1}{p}}}\, ,
$$
for all $t\in [0,q/2] $, is a strictly increasing
function of $[0,q/2]$ onto $[0, \pi_{pq}/2]$ where
$\pi_{pq}= 2\arcsin _{pq} (q/2)=B(1/q, 1-1/p)$ and $B$
denotes the Euler Beta function. The inverse function
of $\arcsin _{pq} $, which is denoted by $\sin_{pq}$,
is extended to $[-\pi_{pq}, \pi_{pq}]$ by setting
$\sin_{pq}(\theta )=\sin_{pq}(\pi_{pq}-\theta )$ for all
$\theta \in ]\pi_{pq}/2, \pi_{pq}] $,
$\sin_{pq}(\theta )=-\sin_{pq}(-\theta ) $ for all
$\theta \in [-\pi_{pq},0[$, and then it is extended by periodicity
to the whole of ${\mathbb{R}}$.
\begin{theorem}
If $N=1$ and $\Omega =(0, a)$ with $a>0$ then
\begin{equation}\label{onedimeq}
\lambda_1(p,q)=q^{\frac{p(1-q)}{q}}
\Big(\frac{2\pi_{pq}}{a^{\frac{1}{q}-\frac{1}{p}+1}}\Big) ^p
\Big(1-\frac{1}{p}\Big)\Big(\frac{1}{q}-\frac{1}{p}+1 \Big)^{\frac{p-q}{q}},
\end{equation}
and $\lambda _n(p,q)= n^p\lambda_1(p,q)$ for all $n\in {\mathbb{N}}$.
Moreover, $\sigma (p,q)=\{\lambda_n(p,q):\ n\in {\mathbb{N}}\}$,
$\lambda_n(p,q) $ is simple for all $n\in {\mathbb{N}}$ and the
corresponding eigenspace is spanned by the function
$u_n(x)=\sin_{pq}\left(\frac{n\pi_{pq}}{a}x \right)$, $x\in (0,a)$.
\end{theorem}
\begin{proof}
Let $u$ be an eigenfunction corresponding to $\lambda_1(p,q)$ with
$u>0$ on $(0,a)$ and $\|u\|_{L^q(0,a)}=1$.
By \cite[Lemma~2.5]{ot84} it follows that
\begin{equation}\label{ident}
\frac{p-1}{p}|u'(x)|^p+\frac{\lambda_1(p,q) }{q}|u(x)|^q=\frac{p-1}{p}|u'(0)|^p,
\end{equation}
for all $x\in (0,a)$.
Recall that $u'(x) >0 $ for all $x\in (0,a/2)$ and $u'(a/2)=0$. Thus, by setting $y=u(x)/u(a/2)$ and by means of a change if variables in integrals it follows by \eqref{ident} that
\begin{equation}\label{ident1}
\begin{aligned}
a&=2\int_0^{a/2}dx\\
&= 2\Big(\frac{q(p-1)}{\lambda_1(p,q)p}
\Big)^{1/q}|u'(0)|^{p/q-1}\int_0^1(1-y^q)^{-1/p}dy \\
& = \frac{2}{q}\Big(\frac{q(p-1)}{\lambda_1(p,q)p}
\Big)^{1/q}|u'(0)|^{p/q-1}\pi_{pq}.
\end{aligned}
\end{equation}
By integrating \eqref{ident} and recalling that
$\lambda_1(p,q)=\| u'\|^p_{L^p(0,a)} $ it follows that
\begin{equation}
\label{ident2}
\Big( \frac{p-1}{p}+\frac{1}{q} \Big)\lambda_1(p,q)
=\frac{p-1}{p}|u'(0)|^pa.
\end{equation}
By combining \eqref{ident1} and \eqref{ident2} we deduce \eqref{onedimeq}.
By \cite[Proposition~4.2, Theorem~II]{ot84}, one can easily deduce
that $\sigma (p,q)=\{ n^p\lambda_1(p,q):\ n\in {\mathbb{N}} \}$
which, combined with the argument used in
\cite[Proposition~4.6]{cue} for the case $p=q$, allows to conclude
that $ \lambda _n(p,q)= n^p\lambda_1(p,q) $. For the proof of the
last part of the statement we refer to \cite{draman}.
\end{proof}
\subsection*{Acknowledgements}
The authors are thankful to Professors B. Kawohl and
P. Lindqvist for useful discussions and references.
Detailed proofs of most of the results presented in this paper
were discussed by G. Franzina in a part of his dissertation~\cite{fra}
under the guidance of P.D. Lamberti.
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\end{document}