\documentclass[reqno]{amsart} \usepackage{hyperref} \usepackage{amssymb} \AtBeginDocument{{\noindent\small \emph{Electronic Journal of Differential Equations}, Vol. 2010(2010), No. 26, pp. 1--10.\newline ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu \newline ftp ejde.math.txstate.edu} \thanks{\copyright 2010 Texas State University - San Marcos.} \vspace{9mm}} \begin{document} \title[\hfilneg EJDE-2010/26\hfil Nonlinear eigenvalue problem] {Existence and uniqueness for a $p$-Laplacian nonlinear eigenvalue problem} \author[G. Franzina, P. D. Lamberti\hfil EJDE-2010/26\hfilneg] {Giovanni Franzina, Pier Domenico Lamberti} % in alphabetical order \address{Giovanni Franzina \newline Dipartimento di Matematica, University of Trento, Trento, Italy} \email{g.franzina@email.unitn.it} \address{Pier Domenico Lamberti \newline Dipartimento di Matematica Pura e Applicata, University of Padova, Padova, Italy} \email{lamberti@math.unipd.it} \thanks{Submitted January 14, 2010. Published February 16, 2010.} \subjclass[2000]{35P30} \keywords{$p$-laplacian; eigenvalues; existence; uniqueness results} \begin{abstract} We consider the Dirichlet eigenvalue problem $$ -\mathop{\rm div}(|\nabla u|^{p-2}\nabla u ) =\lambda \| u\|_q^{p-q}|u|^{q-2}u, $$ where the unknowns $u\in W^{1,p}_0(\Omega )$ (the eigenfunction) and $\lambda >0$ (the eigenvalue), $\Omega $ is an arbitrary domain in $\mathbb{R}^N$ with finite measure, $10 . \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 $1p$ 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 $10$ 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, $10 $ 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, $10 $ 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, $10$ 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 $1p$; 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, $1 0$ 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 $10$. 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 $R_01$ further technical work is required and the main step is the following Lemma for which we refer to Erbe and Tang~\cite[Lemma~3.1]{erta}. \begin{lemma}\label{erbetanglem} Let $N>1$, $10$. 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 ${\mathbb{R}}^{N}$, $11$ 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 $u_10$ in $\Omega$ and $\| u\|_q=1$; moreover, since $ku$ satisfies equation \eqref{eqnorm} then $k=\pm \lambda_{1}(p,q)^{\frac{1}{q-p}}$, hence $w$ is uniquely determined up to the sign. To conclude the proof it suffices to observe that if $v$ is an eigenfunction corresponding to $\lambda_1(p,q)$ and satisfies (\ref{jbis}) with $\lambda =\lambda_1(p,q)$ then $v$ minimizes $J_{|S_{pq}}$. \end{proof} By Lemma \ref{equiv} and Theorem~\ref{uniray} we deduce the following result of Dr\'{a}bek~\cite[Thm.~1.1]{dra} for the case $10$ 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) $. 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