\documentclass[reqno]{amsart}
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\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2009(2009), No. 112, pp. 1--5.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2009 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2009/112\hfil Multiplicity of solutions]
{Multiplicity of solutions for some degenerate quasilinear
elliptic equations}

\author[V. Solferino \hfil EJDE-2009/112\hfilneg]
{Viviana Solferino}

\address{Viviana Solferino \newline
Department of Mathematics, University of Calabria, Via P. Bucci,
Arcavacata di Rende (CS), Italy} 
\email{solferino@mat.unical.it}

\thanks{Submitted June 1, 2009. Published September 10, 2009.}
\subjclass[2000]{35J20, 35J60, 35J70} 
\keywords{Quasilinear elliptic equations; variational methods; 
\hfill\break\indent
 noncoercive functionals;  nonsmooth critical point theory}

\begin{abstract}
 We show the existence of infinitely many solutions for
 a symmetric quasilinear problem whose principal part is degenerate.
\end{abstract}



\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{definition}[theorem]{Definition}
\newtheorem{proposition}[theorem]{Proposition }
\newtheorem{remark}[theorem]{Remark}

\section{Introduction and statement of main result}

Let $\Omega$ be a bounded open subset of $\mathbb{R}^n$ with $n\geq 2$.
We are interested in the solvability of the quasilinear
elliptic problem
\begin{equation}\label{eq:q}
\begin{gathered}
  -\mathop{\rm div}(j_\xi(x,u,\nabla u))+ j_s(x,u,\nabla u)=g(x,u)
  \quad \text{in }  \Omega \,,\\
u=0\quad \text{on }  \partial\Omega \,,
\end{gathered}
\end{equation}
where $j:\Omega \times\mathbb{R}\times \mathbb{R}^{n}
\to  \mathbb{R}$ is defined as
\[
j(x,s,\xi)=\frac{1}{2}\,a(x,s)|\xi|^2\,.
\]
A feature of quasilinear problems like \eqref{eq:q}
is that, for a general $u\in H^1_0(\Omega)$, the term
$j_s(x,u,\nabla u)$ belongs to $L^1(\Omega)$ under reasonable
assumptions, but not to $H^{-1}(\Omega)$.
As a consequence, the functional
\[
f(u) =  \int_\Omega j(x,u,\nabla u)-\int_\Omega G(x,u)\,,
\quad
G(x,s)=\int_0^s g(x,t)\, dt\,,
\]
whose Euler-Lagrange
equation is represented by \eqref{eq:q}, is continuous on
$H^1_0(\Omega)$, but not locally Lipschitz, in particular
not of class~$C^1$.

In spite of this fact, existence and multiplicity results have
been already obtained for this class of problems, also by means of
variational methods, starting from the case in which $a$ is
bounded and bounded away from zero (see \cite{Ar1, Ca}). Actually,
in \cite{Ar1} the nonsmoothness of the functional is overcome by a
suitable approximation procedure, while in \cite{Ca} a direct
approach, based on a critical point theory for continuous
functionals developed in \cite{ca-de,de-ma, ioffe_schwartzman1996,
katriel1994}, is used. However, it seems to be hard, in the
approach of \cite{Ar1}, to get multiplicity results when, e.g.,
$f$ is even.

Both approaches have been extended to the case in which
$a$ is still bounded away from zero, but possibly unbounded
(unbounded case), in \cite{Ar2, Pe-Sq}, respectively.
Again, multiplicity results when $f$ is even are proved
only in the second paper.

Finally, in \cite{ar-bo-or} the case in which $a$ is bounded,
but not bounded away from zero (degenerate case) is addressed
in the line of \cite{Ar1, Ar2}.
Also in this case, no multiplicity result is proved when $f$
is even.

The main purpose of this paper is to prove that, in the model case
mentioned in \cite{ar-bo-or} and with a symmetry assumption,
problem \eqref{eq:q} possesses infinitely many solutions. We also
show that it is not necessary to restart all the machinery of the
previous cases to get the result. By a change of variable, the
degenerate case can be reduced to that of \cite{Ca}. Apart from
the model case, it would be interesting to check if, up to a
change of variable, the degenerate case can be reduced to a
suitable form of the unbounded case and viceversa.

To state our result, consider the model case
\[
a(x,s) = \frac{1}{{(b(x)+s^2)}^\alpha}\,,
\]
where $b$ is a measurable function satisfying
$0<\beta_1\leq b(x)\leq \beta_2$ a.e. in $\Omega$
and $\alpha \in [0,\frac{n}{2n-2})$.

Let also $g:\Omega \times  \mathbb{R}\to  \mathbb{R}$
be a  Carath\'{e}odory function  satisfying the following assumptions:
\begin{itemize}
    \item  there exists $b,d >0$ and $2<p<2^*(1-\alpha)$ such that
           \begin{equation}\label{quattro}
            |g(x,s)|\leq b|s|^{p-1}+d
            \end{equation}
            for almost every $x \in \Omega $ and every $s \in \mathbb{R}$
\item for almost every $x \in \Omega $ and every $s \in \mathbb{R}$,
 \begin{equation}\label{alfa}
 g(x,-s)=-g(x,s).
\end{equation}

\end{itemize}
We set $G(x,s)=\int_0^s g(x,t)\, dt$ and we assume
that there exist $\nu>2,R>0$ such that
\begin{equation}\label{cinque}
  0<\nu G(x,s)\leq s g(x,s).
\end{equation}
for almost every  $x\in \Omega
    $ and all $s \in \mathbb{R}$ with $|s|\geq R$.

It easily follows that the function $a$ satisfies the
following conditions:
\begin{itemize}
    \item  there exist   $c_1,c_2>0$ such that
              \begin{equation}\label{uno}
            \frac{c_1}{{(1+|s|)}^{2\alpha}}\leq a(x,s)\leq c_2
             \end{equation}
              for almost every $x \in \Omega$, for every
              $s \in \mathbb{R}$,

\item for almost every $x$ in $\Omega$ the function $a(x,.)$ is
          differentiable on $\mathbb{R}$ and there exist $c_3>0$
          such that, for almost every $x \in \Omega$,
its derivative
$$
a_s(x,s)\equiv \frac{\partial a}{\partial
          s}(x,s)=\frac{-2\alpha\,s\,a(x,s)}{b(x)+s^2}
$$
satisfies
\begin{equation}\label{due}
    |a_s(x,s)|\leq c_3 \quad \forall s \in     \mathbb{R}
\end{equation}

 \item for almost every $x \in \Omega$ and all $s \in \mathbb{R}$
          \begin{equation}\label{quattro1}
                     a(x,s)=a(x,-s).
             \end{equation}
 \end{itemize}


\begin{definition} \label{def1.1}
\rm We say that $u$ is a weak solution of \eqref{eq:q} if $u \in
H_0^1(\Omega)$ and
$$
\int_\Omega j_\xi(x,u,\nabla u)\nabla v+j_s(x,u,\nabla u)v
=\int_\Omega g(x,u)\,v
$$
for every $v \in C_0^\infty(\Omega)$.
\end{definition}

 We are now able to state our main result.

\begin{theorem}\label{undicio}
Assume that conditions \eqref{quattro}, \eqref{alfa} and
\eqref{cinque} hold. Then there exists a sequence
$\{u_h\}\subset  H_0^1(\Omega)\cap L^\infty(\Omega)$
of weak solutions of (\ref{eq:q}) such that
$$
 \int_\Omega j(x,u_h,\nabla u_h)-\int_\Omega G(x,u_h)
$$
approaches $+\infty$ as $h\to +\infty$.
\end{theorem}

\section{Proof of the main result}

Let $\varphi \in C^2(\mathbb{R})$ be
 defined as
$$
\varphi(s)=s\,{(1+s^2)}^{\frac{\alpha}{2(1-\alpha)}}.
$$

\begin{remark}\label{nota4}
\rm We observe that $\varphi$ is odd and that there exists
$\gamma>0$ such that
  \begin{equation}\label{one}
    \varphi'(s)\geq \gamma(1+|\varphi(s)|)^\alpha.
\end{equation}
Moreover we have
\begin{equation}\label{ai}
 \lim_{s\to \pm \infty}
\frac{s\,\varphi'(s)}{\varphi(s)}= \lim_{s\to \pm \infty}
\Big(1+s\frac{\varphi''(s)}{\varphi'(s)}\Big)=\frac{1}{1-\alpha}.
\end{equation}
\end{remark}

Let us consider the change of variable
$u=\varphi(v)$. We can define on $H_0^1(\Omega)$ the functional
$$
\tilde{f}(v)=\frac{1}{2}\int_\Omega A(x,v)|\nabla v|^2-
\int_\Omega \widetilde{G}(x,v),
$$
where
$$
A(x,s)=a(x,\varphi(s))\cdot (\varphi'(s))^2, \quad
\widetilde{G}(x,s)=G(x,\varphi(s))=\int_0^s \tilde{g}(x,t)dt,
$$
 with $\tilde{g}(x,s)=g(x,\varphi(s))\cdot \varphi'(s)$.

 Now let us consider the integrand $\tilde{j}: \Omega\times
 \mathbb{R}\times {\mathbb{R}}^n\to\mathbb{R}$
 defined by
$$
\tilde{j}(x,s,\xi)=\frac{1}{2}A(x,s)|\xi|^2.
$$

\begin{remark}\label{notaz2}
\rm It is readily seen that  \eqref{one} and the left inequality
of (\ref{uno}) imply that for almost every $x \in \Omega$ and for
every $(s,\xi)\in \mathbb{R}\times {\mathbb{R}}^n$, there holds
$$
\tilde{j}(x,s,\xi)\geq \alpha_0|\xi|^2,
$$
where $\alpha_0=\frac{c_1 \,\gamma^2}{2}$.
\end{remark}

\begin{remark}\label{notazz} \rm
By Remark \ref{nota4} there exists $\Lambda>0$  such
that, for a.e. $x\in \Omega$ and for every $s
\in \mathbb{R}$, we have
\begin{gather}\label{del}
    A(x,s)\leq \Lambda, \\
\label{del1}
   |A_s(x,s)|\leq \Lambda.
\end{gather}
\end{remark}

\begin{proposition}\label{mill1}
Assume condition \eqref{cinque}. Then
\begin{equation}\label{cedo}
\frac{\nu}{1-\alpha}\,\widetilde{G}(x,s)\leq s\tilde{g}(x,s)
\end{equation}
for every $s\in \mathbb{R}$ with $|s|\geq R$.
\end{proposition}

\begin{proof}
Condition (\ref{cinque}) implies
$$
\nu \widetilde{G}(x,s)\,\varphi'(s)\leq \varphi(s)\,
\widetilde{G}_s(x,s)
$$
hence
 $$
\nu \frac{s \varphi'(s)}{\varphi(s)}\widetilde{G}(x,s)\leq
s \widetilde{G}_s(x,s)
$$
and taking into account Remark
\ref{nota4}, we get the thesis.
\end{proof}

\begin{proposition}\label{notaz4}
There exists $\mu< \frac{\nu}{1-\alpha}-2$ such that,
for almost every  $x \in \Omega$, for every
$\xi \in {\mathbb{R}}^n$, for every $s \in \mathbb{R}$ with
$|s|\geq R$,
we have
\begin{equation}\label{cici}
0\leq s\tilde{j}_s(x,s,\xi)\leq \mu\, \tilde{j}(x,s,\xi).
\end{equation}
\end{proposition}

\begin{proof}
 Indeed
\begin{equation} \label{uu}
\begin{aligned}
\tilde{j}_s(x,s,\xi)
&=\frac{1}{2}A_s(x,s)|\xi|^2\\
&= \frac{1}{2}[a_s(x,\varphi(s))\cdot
(\varphi'(s))^3+2\varphi'(s)\cdot\varphi''(s)\cdot
a(x,\varphi(s))]|\xi|^2\\
&=a(x,\varphi(s))\cdot \varphi'(s)\Big[\frac{-\alpha
\,\varphi(s)\,
(\varphi'(s))^2}{b(x)+(\varphi(s))^2}+\varphi''(s)\Big]|\xi|^2.
\end{aligned}
\end{equation}
Let $s>0$.
Then recalling that $a(x,\varphi(s))$ and
$\varphi'(s)$ are positive functions, it suffices to prove that
the square bracket is non negative. Note that the expression is equal
to
$$
\frac{\alpha
s{(1+s^2)}^{\frac{\alpha}{2(1-\alpha)}}}{(1-\alpha){(1+s^2)}^2}
\Big[\frac{-(s^2+1-\alpha){(1+s^2)}^{\frac{\alpha}{(1-\alpha)}}}{b(x)
+{(1+s^2)}^{\frac{\alpha}{(1-\alpha)}}\,s^2}+
\frac{(s^2+1-\alpha)\,b(x)}{(1-\alpha)(b(x)+{(1+s^2)}
^{\frac{\alpha}{1-\alpha}}\,s^2)}+2\Big].
$$
Observing that the second term in square bracket is positive and
the sum of the first and third is equal to
$$
\frac{{(1+s^2)}^{\frac{\alpha}{1-\alpha}}(s^2-(1-\alpha))
+2\,b(x)}{b(x)+{(1+s^2)}^{\frac{\alpha}{1-\alpha}}\,s^2},
$$
the assertion follows if we assume $R=\sqrt{1-\alpha}$. On the other
hand if $s<0$, taking into account that $\varphi''(s)$ is an odd
function, we deduce that the square bracket in (\ref{uu}) is
negative.  Now we prove the right inequality. Since
$\varphi'(s)\geq 0$ and $a_s(x,\varphi(s))\leq 0$, we have
$$
s\,\tilde{j}_s(x,s,\xi)\leq
2\,\tilde{j}(x,s,\xi)\Big(\frac{\varphi''(s)\,s}{\varphi'(s)}\Big)
$$
and  by Remark \ref{nota4} it follows the assertion with $R$ large
enough and  $\mu= \frac{2\,\alpha}{1-\alpha}$.
\end{proof}

We now are able to prove the main result of the paper.

\begin{proof}[Proof of Theorem \ref{undicio}]
By Remark \ref{notaz2} and \ref{notazz} and Proposition
\ref{mill1} and \ref{notaz4} we are able to apply Theorem 2.6 in
\cite{Ca}. So  obtaining a sequence of  weak solutions $\{v_h\}
\subset H_0^1(\Omega)$ of the problem
$$
\int_\Omega A(x,v)\nabla v \nabla w+
\frac{1}{2}\int_\Omega A_s(x,v)|\nabla u|^2\,w=
\int_\Omega \tilde{g}(x,v)w
$$
for every $w \in C_0^\infty(\Omega)$
with $f(v_h)\to\infty$. By Theorem 7.1 in \cite{Pe-Sq}
these solutions belong to  $L^\infty(\Omega)$. If we set
$u_h=\varphi(v_h)$, it is clear that $u_h\in H_0^1(\Omega)\cap
L^\infty(\Omega)$ and an easy calculation shows that each $u_h$ is
a weak solution of (\ref{eq:q}) with
$$
\int_\Omega j(x,u_h,\nabla u_h)-\int_\Omega G(x,u_h)
\to +\infty
$$
as $h\to +\infty$.
 \end{proof}

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\end{thebibliography}
\end{document}
