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\markboth{ Existence of many positive nonradial solutions }
{ Alfonso Castro \& Marcel B. Finan }
\begin{document}
\setcounter{page}{21}
\title{\vspace{-1in}\parbox{\linewidth}{\footnotesize\noindent
Nonlinear Differential Equations, \newline
Electron. J. Diff. Eqns., Conf. 05, 2000, pp. 21--31\newline
http://ejde.math.swt.edu or http://ejde.math.unt.edu
\newline ftp  ejde.math.swt.edu or ejde.math.unt.edu (login: ftp)}
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Existence of many positive nonradial solutions for a
superlinear Dirichlet problem on thin annuli
% 
\thanks{ {\em Mathematics Subject Classifications: 35J20, 35J25, 35J60}
 \hfil\break\indent 
{\em Key words: Superlinear Dirichlet problem, positive nonradial
solutions, \hfil\break\indent
variational methods} 
 \hfil\break\indent
\copyright 2000 Southwest Texas State University. 
\hfil\break\indent Published October 24, 2000. } } 

\date{}
\author{ Alfonso Castro \& Marcel B. Finan }  
\maketitle

\begin{abstract} 
We study the existence of many positive nonradial solutions of a
superlinear Dirichlet problem in an annulus in ${\mathbb R}^N$. Our 
strategy consists of finding the minimizer of the energy 
functional restricted to the Nehrai manifold of a subspace of 
functions with symmetries. The minimizer is a global critical point 
and therefore is a desired solution. Then we show that the minimal 
energy solutions in different symmetric classes have mutually 
different energies. The same approach has been used to prove the 
existence of many sign-changing nonradial solutions (see \cite{kn:0.1}). 
\end{abstract}

\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}

\section{Introduction}

In this article we study the existence of
many positive nonradial solutions of the equation
\begin{eqnarray}   
& \Delta u   +   f(u)   =   0  \quad\mbox{in } \Omega_{\epsilon}&\nonumber \\
&  u  >  0  \quad\mbox{in } \Omega_{\epsilon} &\label{P}\\
& u   =  0  \quad\mbox{on } \partial\Omega_{\epsilon}\,,&\nonumber
\end{eqnarray}
where 
\[ 
\Omega_{\epsilon} = \{x \in {\mathbb R}^3: 1-\epsilon < |x| < 1 \}.
\]
The non-linearity $f$ is of class $C^1({\mathbb R})$ and satisfies the 
following conditions:
\begin{description}

\item{\bf (C1)} $f(0) = 0$ and $f'(0) < \lambda_1$, where $\lambda_1$ is the smallest
eigenvalue of $-\Delta$ with zero Dirichlet boundary condition in 
$\Omega_{\epsilon}$.

\item{\bf (C2)}  $f'(u) > \frac{f(u)}{u}$ for all $u \neq 0$. 

\item{\bf (C3)} (Superlinearity)  
$\lim_{|u| \rightarrow \infty} \frac{f(u)}{u} = \infty$.

\item{\bf (C4)} (Subcritical growth) There exist  constants $C>0$ and 
$p \in (1, 5)$
such that
\[ 
|f'(u)| \leq C(|u|^{p-1} + 1), ~~~ \forall u \in {\mathbb R}.
\]

\item{\bf (C5)} There  exist constants $m \in (0,1)$ and $\rho$ such that
for $|u| > \rho$,
\[ 
uf(u) \geq \frac{2}{m} F(u) > 0,
\]
 where $F(u) = \int_{0}^{u} f(s)\,ds$.
\end{description}

\noindent This paper is motivated by the work of Coffman \cite {kn:1}, 
Li \cite{kn:2} and Lin \cite{kn:3}. In 1984, Coffman showed
that for $f(t) = -t + t^p$, where $p = 2N +1, N =2$, the above problem
has many positive nonradial solutions. His result was extended by Li
\cite{kn:2} to the N-dimensional case with $N \geq 4$ or $N = 2$
and for $p \in (1, \frac{N+2}{N-2})$, when the nonlinear term is
$\lambda t + t^p$ for $\lambda \leq 0$. Our main result (see Theorem
~\ref{main} below) concerns the case $N=3$. We will show that for $N=3$
our problem has many distinct nonradial solutions.  We
follow the strategy used in
\cite{kn:1,kn:2} and \cite{kn:3}. That is, we look for the
minimizer of
the energy functional restricted to the Nehari manifold of a subspace of
functions with symmetries. The minimizer is a global critical point and 
therefore is a solution
to (\ref{P}). Then we show that the minimal energy solutions
in different symmetric classes has mutually different energies. 

We would like to point out here that during the preparation of this
article we were unaware of the papers by Byeon \cite{kn:01} and  Catrina
and Wang \cite{kn:02}, where the above problem has been solved.
However, our approach is different from theirs.
 
Our main result is the following 

\begin{theorem} \label{main}
Let conditions (C1) - (C5) be satisfied. Then, for each positive
integer $k$ 
there exists $\epsilon_{1}(k) > 0$ such that if $0 < \epsilon < 
\epsilon_{1}(k)$ then (\ref{P}) has $k$ distinct positive nonradial
solutions.
\end{theorem}

We note that our argument works for $N \geq 2$ and not only for the
three dimensional case.

The approach used in proving Theorem \ref{main} is similar to the one
used in \cite {kn:2}, i.e., we consider the functionals
\[
J(u) = \int_{\Omega_{\epsilon}} \big\{ \frac{1}{2}|\nabla u|^2 - F(u)
\big\} dx
\]
and
\[
\gamma(u) = \int_{\Omega_{\epsilon}} \big(|\nabla u|^2 - uf(u)
\big)\,dx
\]
on $H_{0}^{1}(\Omega_{\epsilon})$, and for $k \geq 1$, we consider the
space of invariant functions
\begin{eqnarray*}
{H(\epsilon,\it k)} &:=& \mathop{\rm Fix} (G_{k} \times \{id_{{\mathbb R}}, - id_{{\mathbb R}}\})\\
 &=&
\big\{ v \in H_{0}^{1}(\Omega_{\epsilon}): 
v(g(x_1, x_2), Tx_3) = v(x_1, x_2, x_3), \\
&&\forall~(g,T) \in G_{k} \times \{id_{{\mathbb R}}, - id_{{\mathbb R}}\} \big\}\\[2pt]
& = & \big\{v \in H_{0}^{1}(\Omega_{\epsilon}): v(x_1, x_2, x_3) 
= u(x_1, x_2, |x_3|)\mbox{ for some }u\\
&&\mbox{satisfying } u(g(x_1, x_2), |x_3|) = u(x_1, x_2, |x_3|)\;
 \forall g \in G_k\big\}
\end{eqnarray*}
where $G_{k} = \{ g^{i}: 0 \leq i \leq k-1\}$ and 
$gz = e^{2\pi i/k} z$, $z \in {\mathbb C} \simeq {\mathbb R}^2$.
Also, we consider the Nehari manifold
\[ 
{S(\epsilon, \it k)} = \{ v \in {H(\epsilon, \it k)} \backslash
\{0\}: \gamma (v) = 0 \}.
\]
Similarly, we define the space of functions 
\begin{eqnarray*}
{H(\epsilon, \infty)} &:=& {\bf O}(2) \times \{id_{{\mathbb R}}, - id_{{\mathbb R}}\} \\[1pt]
& = & \big\{ v \in H_{0}^{1}(\Omega_{\epsilon}):
v(g(x_1, x_2), Tx_3) = v(x_1, x_2, x_3),\\
&& \forall (g,T) \in {\bf O}(2) \times \{id_{{\mathbb R}}, - id_{{\mathbb R}}\} 
\big\}\\[1pt]
&= &\big \{v \in H_{0}^{1}(\Omega_{\epsilon}): v(x_1, x_2, x_3) =
u(\sqrt{x_1^2 + x_2^2}, |x_3|),\mbox{ for some } u \big\},
\end{eqnarray*}
and the manifold
\[
{S(\epsilon, \infty)} = \{ v \in {H(\epsilon,\infty)}\backslash
\{0\}: \gamma (v) = 0\},
\]
where ${\bf O}(2)$ denotes the space of $2\times 2$ orthogonal matrices.
Note that if $u \in H(\epsilon,\infty)$ then $u$ is $\theta-$independent. 

Associated with the above sets, we consider the numbers
$$ j_{k}^{\epsilon} = \inf_{v \in S(\epsilon, \it k)} J(v)
\quad \mbox{and}\quad 
 j_{\infty}^{\epsilon} = \inf_{v \in S(\epsilon, \infty)} J(v).$$
We prove Theorem \ref{main} by establishing the following lemmas:

\begin{lemma} \label{lem1}
For $k = 1, 2, ..., \infty$, $j_{k}^{\epsilon}$ is achieved by some $u_{\epsilon,k} \in 
S(\epsilon, \it k)$ and $u_{\epsilon,k}$ is a critical
point of $J$ on ${H(\epsilon, \it k)}.$
\end{lemma}

\begin{lemma} \label{lem2}
Let $u_{\epsilon, k}$ be as in Lemma \ref{lem1}. Then $u_{\epsilon, k}$ is
a critical point of $J$ on $H_{0}^{1}(\Omega_{\epsilon}).$
\end{lemma}

\begin{lemma} \label{lem3}
Given a positive integer $k$, there exists $\epsilon_{1}(k)$ such that
for $0<\epsilon<\epsilon_{1}(k)$ we have
$$j_{k}^{\epsilon} < j_{\infty}^{\epsilon}.$$
\end{lemma}

\begin{lemma} \label{lem4}
For $n = 2,3,4,...$, $k = 1,2,3,...$, 
$j_{kn}^{\epsilon}
< j_{\infty}^{\epsilon}$ implies $j_{k}^{\epsilon} < j_{kn}^{\epsilon}.$
\end{lemma}

This article is organized as follows: In Section 2, we prove 
Lemmas \ref{lem1}--\ref{lem4}. In Section 3, we prove Theorem
\ref{main}.

\section{Proof of Lemmas \ref{lem1}--\ref{lem4}}

\noindent{\it Proof of Lemma \ref{lem1}:} The proof follows from
 \cite{kn:0} combined with the facts that the embeddings 
$H(\epsilon, k)\subset L^{p}(\Omega_{\epsilon})$ and 
$H(\epsilon, \infty) \subset L^{\infty}(\Omega_{\epsilon})$ are compact 
(See  \cite {kn:4}) \hfill$\diamondsuit$\smallskip

Lemma \ref{lem2} is a result of the symmetric criticality principle:
if $u_{\epsilon,k}$ is a critical point of $J$ on ${H(\epsilon, \it k)}$, 
then $u_{\epsilon,k}$ is a critical point of $J$ on $H_{0}^{1}(\Omega_{\epsilon})$
(See \cite {kn:4}).

To prove Lemma \ref{lem3} we need the following results.

\begin{lemma}[Poincar\'{e}'s inequality] \label{lem5}
Let $\Omega \subset {\mathbb R}^N$ be a smooth bounded domain with diameter $d.$
Then for $u \in
H_{0}^{1} (\Omega)$ we have
\[
\int_{\Omega} u^2dx \leq \frac{d^2}{N} \int_{\Omega} |\nabla u|^2 dx\,.
\]
\end{lemma}

\begin{lemma}[\cite{kn:0}] \label{lem6}
0 is a local minimum of $J$. If $u \in H(\epsilon,k) - \{0\}$, then there
exists a unique $\alpha = \alpha(u) \in (0,\infty)$ such that $\alpha u 
\in S(\epsilon,k)$. Moreover, $J(\alpha u) = \max_{\lambda > 0}{J(\lambda u)}
> 0$. 
\end{lemma}

\begin{lemma}[\cite{kn:0.2}] \label{lem7}
For $|v| > \rho$ and $s > 2$ we have
$$svf(sv) \geq Cs^{2/m} v f(v)$$
for some constant $C > 0$.     
\end{lemma}
\noindent
{\it Proof of Lemma \ref{lem3}:} Let $k \geq 1$
be an integer and $\epsilon > 0$ to be chosen below. According to Lemma
\ref{lem1}, there exist $u_{\epsilon,k} \in
S(\epsilon,k)$ and $u_{\epsilon,\infty} \in S(\epsilon,\infty)$ such that
$j_{k}^{\epsilon} = J(u_{\epsilon,k})$ and $j_{\infty}^{\epsilon} =
J(u_{\epsilon,\infty})$. By Lemma \ref{lem2}, $u_{\epsilon,k}$ and
$u_{\epsilon,\infty}$ are solutions to Problem \ref{P}. Let
$\Omega_{\epsilon}^{k}$ be the set of points $x=(x_1,x_2,x_3)
\equiv (r,\theta,x_3) \in \Omega_{\epsilon}$ such that $\theta \in
[0,2\pi/k]$.

Let $j$ be a positive integer to be chosen independent of $k$ and 
$\epsilon$.  Define
\[
\omega (r,\theta,x_3) = \left\{ \begin{array}{ll}
 u_{\epsilon,\infty}(r,|x_3|)\sin{(jk\theta)} 
 & \mbox{for $\theta \in [0, \pi/(jk)]$}\\[2pt]
 0 & \mbox{for $\theta \in [\pi/(jk), 2\pi/k]$}\,,
                          \end{array} \right.
\]
where $r = (x_1^2 + x_2^2)^{1/2}$. Extend $\omega$ periodically 
to all of $\Omega_{\epsilon}$ in the $\theta$ direction.
 Let $z \in H(\epsilon,k)$ be the resulting extension. 
Since $u_{\epsilon,\infty}>0$ then $z>0$. By the chain
rule we have
\begin{eqnarray*}
|\nabla \omega(r, \theta, x_3)|^2 
& = & (u_{\epsilon,\infty})_{r}^2\sin^2{(jk\theta)} + \frac{1}{r^2}
(u_{\epsilon,\infty})^2
(jk)^2\cos^2{(jk\theta)}\\
& &+  |\nabla_{x_3}u_{\epsilon,\infty}|^2\sin^2{(jk\theta)}
\end{eqnarray*}
if $\theta \in [0, \pi/(jk)]$, otherwise $\nabla \omega = 0.$
Thus, 
$$|\nabla \omega (r, \theta,x_3)|^2 \leq (u_{\epsilon,\infty})_r^2 +
 \frac{1}{r^2}
(jk)^2(u_{\epsilon,\infty})^2 + |\nabla_{x_3} u_{\epsilon,\infty}|^2.$$
By Lemma \ref{lem5} we have
\begin{eqnarray*}
\int_{\Omega_{\epsilon}^{2jk}} u_{\epsilon,\infty}^2\,dx_1\,x_2\,dx_3 \leq 
\frac{(\mathop{\rm diam}(\Omega_{\epsilon}^{2jk}))^2}{3} 
\int_{\Omega_{\epsilon}^{2jk}} |\nabla u_{\epsilon,\infty}|^2 \,dx_1\,x_2\,dx_3.
\end{eqnarray*}
\noindent 
Thus,
\begin{eqnarray}
\lefteqn{\int_{\Omega_{\epsilon}} |\nabla z|^2 \,dx_1\,x_2\,dx_3 }\nonumber\\
 & = & k\int_{\Omega_{\epsilon}^
{2jk}} |\nabla \omega|^2 \,dx_1\,x_2\,dx_3 \nonumber \\
& \leq & k \int_{\Omega_{\epsilon}^{2jk}} [(1 + \frac{16(jk)^2}{27}\epsilon^2)
(u_{\epsilon,\infty})_{r}^2 + |\nabla_{x_3} u_{\epsilon,\infty}|^2]\,dx_1\,x_2\,dx_3 \nonumber \\
& \leq & 2k \int_{\Omega_{\epsilon}^{2jk}} ( (u_{\epsilon,\infty})_r^2 +
 |\nabla_{x_3}
u_{\epsilon,\infty}|^2)\,dx_1\,x_2\,dx_3 \label{eq10} \\
& = & 2k\int_{\Omega_{\epsilon}^{2jk}}  |\nabla u_{\epsilon,\infty}|^2 
\,dx_1\,x_2\,dx_3 \nonumber
\end{eqnarray}
provided that $\epsilon < 3\sqrt{3}/(4jk)$. 
By Lemma \ref{lem6}  we can find $\alpha > 0$ such that
$\gamma (\alpha z)= 0$. Let
$D = \{(r,\theta,x_3): u_{\epsilon,\infty}(r,|x_3|) > \rho, \theta \in
[\frac{\pi}{4jk},\frac
{\pi}{2jk}] \}$. Suppose that $\alpha > \sqrt{2}$. Then for
$(r,\theta,x_3)
\in D$ we have $\alpha \sin{(jk\theta)} > 2$. This, the fact that $tf(t)$ is
bounded from below, say by $E$, and Lemma \ref{lem7} imply
\begin{eqnarray}
\lefteqn{\int_{\Omega_{\epsilon}} \alpha zf(\alpha z)\,dx_1\,x_2\,dx_3 
}\nonumber\\
& = & k
\int_{\Omega_{\epsilon}^{2jk}} \alpha \sin{(jk\theta)}u_{\epsilon,\infty}
f(\alpha \sin{(jk\theta)}u_{\epsilon,\infty})\,dx_1\,x_2\,dx_3 \nonumber \\   
& \geq & k\left( E |\Omega_{\epsilon}^{2jk}| + C\alpha^{2/m}\int_{D}
u_{\epsilon,\infty}f(u_{\epsilon,\infty})\,dx_1\,x_2\,dx_3 \right) \label{eq7} \\  
& = & k \left( E |\Omega_{\epsilon}^{2jk}| + C\alpha^{2/m}
(\int_{u_{\epsilon,\infty}> \rho} u_{\epsilon,\infty}f(u_{\epsilon,\infty})r\,dr\,dx_3) 
( \int_{\pi/(4jk)}^{\pi/(2jk)} 
d\theta) \right) \nonumber \\
& = & k E|\Omega_{\epsilon}^{2jk}| +k\frac{C}{4}\alpha^{2/m}
(\int_{u_{\epsilon,\infty} > \rho} u_{\epsilon,\infty}f(u_{\epsilon,\infty})r\,dr\,dx_3) 
( \int_{0}^{\pi/(jk)} d\theta)\,, \nonumber
\end{eqnarray}
where $|\Omega_{\epsilon}|$ denotes the volume of $\Omega_{\epsilon}$.

Now, let $\epsilon < 1/4$. Then $K = \inf_{u \in S(1/4,
k)} J(u) \leq  \inf_{u \in S(\epsilon, k)} J(u)$. 
Let $u_{\epsilon,\infty}$ also
denote the zero extension of $u_{\epsilon,\infty}$ to all of
$\Omega_{1/4}$. Then
$u_{\epsilon,\infty} \in S(1/4, k)$. Thus, $J(u_{\epsilon,\infty})
\geq K$
and consequently
\begin{eqnarray} 
\int_{\Omega_{\epsilon}} |\nabla u_{\epsilon,\infty}|^2 \,dx_1\,x_2\,dx_3 &
\geq & 2K + 2M|\Omega_{\epsilon}| \label{e3}
\end{eqnarray}
\noindent
where $M = \inf{\{F(t): t \in {\mathbb R}\}}.$\\
If we choose  $\epsilon$ so that $|\Omega_{\epsilon}| <K/(-2M)$ then 
(\ref{e3}) implies
\begin{eqnarray}  \label{eq11}
\int_{\Omega_{\epsilon}} |\nabla u_{\epsilon,\infty}|^2 dx_1\,dx_2\,dx_3
  \geq K 
\end{eqnarray}
\noindent
and this leads to
\begin{eqnarray*}
\lefteqn{\int_{u_{\epsilon,\infty} > \rho}
u_{\epsilon,\infty}f(u_{\epsilon,\infty})r\,dr\,dx_3}\\
& = & \int_{u_{\epsilon,\infty} \geq 0}u_{\epsilon,\infty}
f(u_{\epsilon,\infty})r\,dr\,dx_3 
- \int_{u_{\epsilon,\infty}\leq \rho} u_{\epsilon,\infty}f(u_{\epsilon,\infty})r\,dr\,dx_3 \\ \\
& = &\big[ 1 - \frac{\int_{u_{\epsilon,\infty} \leq \rho} 
u_{\epsilon,\infty}f(u_{\epsilon,\infty})r\,dr\,dx_3}{\int_{u_{\epsilon,\infty} \geq
0} u_{\epsilon,\infty}f(u_{\epsilon,\infty})r\,dr\,dx_3} \big] \int_{u_{\epsilon,\infty} \geq 0} 
u_{\epsilon,\infty}f(u_{\epsilon,\infty})r\,dr\,dx_3 \\ \\
& = & \big[ 1 - \frac{2\pi \int_{u_{\epsilon,\infty} \leq \rho}
u_{\epsilon,\infty}f(u_{\epsilon,\infty})r\,dr\,dx_3}{\int_{\Omega_{\epsilon}}
|\nabla u_{\epsilon,\infty}|^2 \,dx_1\,x_2\,dx_3} \big] \int_{u_{\epsilon,\infty}
\geq 0} u_{\epsilon,\infty}f(u_{\epsilon,\infty})r\,dr\,dx_3 \\ \\
& \geq & (1 - C\epsilon) \int_{u_{\epsilon,\infty} \geq 0}
u_{\epsilon,\infty}f(u_{\epsilon,\infty})r\,dr\,dx_3
\end{eqnarray*} 
where the constant $C$ is independent of $(\epsilon, j, k)$. By choosing 
$\epsilon$ in such a way that $1 - C\epsilon > 1/2$ we
conclude
\[ 
\int_{u_{\epsilon,\infty} > \rho}
u_{\epsilon,\infty}f(u_{\epsilon,\infty})r\,dr\,dx_3 > \frac{1}{2} 
\int_{\Omega_{\epsilon}} u_{\epsilon,\infty}f(u_{\epsilon,\infty})
r\,dr\,dx_3.
\]
This reduces (\ref{eq7}) to
\begin{equation}  \label{eq8}
\int_{\Omega_{\epsilon}} \alpha zf(\alpha z)\,dx_1\,dx_2\,dx_3 
\geq kE|\Omega_{\epsilon}^{2jk}| + kC\alpha^{2/m}
 \int_{\Omega_{\epsilon}^{2jk}} |\nabla
u_{\epsilon,\infty}|^2\,dx_1\,dx_2\,dx_3.
\end{equation}
On the other hand,  using (\ref{eq10}) we have
\begin{eqnarray} \label{eq9}
\int_{\Omega_{\epsilon}} \alpha zf(\alpha z)\,dx_1\,x_2\,dx_3 & = & \alpha^2
\int_{\Omega_{\epsilon}} |\nabla z|^2 \,dx_1\,x_2\,dx_3 \nonumber \\
& \leq & 2k\alpha^2 \int_{\Omega_{\epsilon}^{2jk}} |\nabla
u_{\epsilon,\infty}|^2 \,dx_1\,x_2\,dx_3.
\end{eqnarray}
Combining (\ref{eq8}) and (\ref{eq9}) to obtain
\[
(C\alpha^{2/m} - 2\alpha^2)\int_{\Omega_{\epsilon}^{2jk}}
 |\nabla u_{\epsilon,\infty}|^2
\,dx_1\,x_2\,dx_3 \leq \frac{C\epsilon}{jk}
\]
Hence,
\begin{eqnarray*}
C\alpha^{2/m} - 2\alpha^2 & \leq & \frac{C\epsilon/(jk)}{
\int_{\Omega_{\epsilon}^{2jk}} |\nabla u_{\epsilon,\infty}|^2\,dx_1\,x_2\,dx_3}
\\ 
& \leq & \frac{C\epsilon}{\int_{\Omega_{\epsilon}} |\nabla
u_{\epsilon,\infty}|^2 \,dx_1\,x_2\,dx_3} \leq M_{1}\,,
\end{eqnarray*}
where we have used (\ref{eq11}). But the function $g(\alpha) =
 C\alpha^{2/m} - 2 \alpha^2$ satisfies $g(0) = 0$ and 
$\lim_{\alpha \rightarrow\infty} g(\alpha) = +\infty$. Thus, if 
$g(\alpha) \leq M_{1}$ then there
is a constant $K'$ such that $\alpha \leq K'$. By letting
\begin{equation}
\alpha \leq \max{\big\{\sqrt{2}, K'\big\}} \equiv K'' \label{e4}
\end{equation}
we conclude that $\alpha$ is bounded. 
Let
$w = \alpha z \in S(\epsilon, k).$
\noindent
Since $F(t)$ is bounded from below, say by $M$, then (\ref{e4}) and
(\ref{eq10}) imply
\begin{eqnarray}
J(w) & = & \int_{\Omega_{\epsilon}} \left( \frac{|\nabla w|^2}{2} - F(w)
\right) \,dx_1\,x_2\,dx_3    \nonumber \\ 
& = & \int_{\Omega_{\epsilon}} \frac{\alpha^2 |\nabla z|^2 }{2} \,dx_1\,x_2\,dx_3
 - \int_{\Omega_{\epsilon}} F(w)\,dx_1\,x_2\,dx_3 
 \nonumber \\
& \leq & 2(K'')^2 k \int_{\Omega_{\epsilon}^{2jk}} \frac{|\nabla u_{\epsilon,\infty}
|^2}{2} \,dx_1\,x_2\,dx_3 - kM|\Omega_{\epsilon}^{2jk}| \nonumber \\ 
& \leq &   2\frac{(K'')^2}{j} \int_{\Omega_{\epsilon}} \frac{|u_{\epsilon,\infty}|^2}{2}
\,dx_1\,x_2\,dx_3 + \frac{kC\epsilon}{jk} \nonumber \\
& \leq & \frac{(K'')^2}{j} \int_{\Omega_{\epsilon}} \frac{|\nabla u_{\epsilon,\infty}|^2}{2}
\,dx_1\,x_2\,dx_3
+ \frac{1}{j} \int_{\Omega_{\epsilon}} \frac{|\nabla u_{\epsilon,\infty}|^2}{2}
\,dx_1\,x_2\,dx_3
\nonumber \\ 
& \leq & \frac{C}{j}\int_{\Omega_{\epsilon}} \frac{|\nabla u_{\epsilon,\infty}|^2}{2}
\,dx_1\,x_2\,dx_3
\label{e6}          
\end{eqnarray} 
where we have used (\ref{e3}) with $\epsilon$ chosen in such a way that
$C\epsilon< K$ and $|\Omega_{\epsilon}| < K/(-2M)$.
We claim that there exists a constant $C$ such that 
$$ \int_{\Omega_{\epsilon}} \frac{|\nabla
u_{\epsilon,\infty}|^2}{2}\,dx_1\,x_2\,dx_3 \leq CJ(u_{\epsilon,\infty}).
$$
Indeed, since $\gamma (u_{\epsilon,\infty}) = 0$  and by (C5) we have
\begin{eqnarray*}
\lefteqn{\int_{\Omega_{\epsilon}} u_{\epsilon,\infty} 
f(u_{\epsilon,\infty})\,dx_1\,x_2\,dx_3 }\\
& = & \int_{\Omega_{\epsilon}} |\nabla u_{\epsilon,\infty}|^2\,dx_1\,x_2\,dx_3\\
& = & \int_{\Omega_{\epsilon}} ( |\nabla u_{\epsilon,\infty}|^2 - 2F(u_{\epsilon,
\infty}))\,dx_1\,x_2\,dx_3 + 2\int_{\Omega_{\epsilon}} F(u_{\epsilon,\infty})\,dx_1\,x_2\,dx_3\\  
&\leq & 2 (J(u_{\epsilon,\infty}) + \frac{m}{2}\int_{\Omega_{\epsilon}}
u_{\epsilon,\infty}f(u_{\epsilon,\infty}) \,dx_1\,x_2\,dx_3 + 
C|\Omega_{\epsilon}| ).
\end{eqnarray*}
It follows that
\begin{equation} \label{e5}
(1-m) \int_{\Omega_{\epsilon}} u_{\epsilon,\infty}f(u_{\epsilon,\infty})
\,dx_1\,x_2\,dx_3 \leq 2(J(u_{\epsilon,\infty}) +C|\Omega_{\epsilon}|).
\end{equation}
On the other hand, using (\ref{e3}) and (C5) we have
\begin{eqnarray*}
J(u_{\epsilon,\infty}) & = & \int_{\Omega_{\epsilon}} (\frac{1}{2}|\nabla
u_{\epsilon,\infty}|^2 - F(u_{\epsilon,\infty})) \,dx_1\,x_2\,dx_3 \\
& \geq & (\frac{1-m}{2})\int_{\Omega_{\epsilon}} u_{\epsilon,\infty}f(u_{\epsilon,\infty})
\,dx_1\,x_2\,dx_3 - C'|\Omega_{\epsilon}| \\ 
&\geq & (\frac{1-m}{2})K - C'|\Omega_{\epsilon}|.
\end{eqnarray*}
By choosing $\epsilon$ in such a way that 
$C'|\Omega_{\epsilon}| <\frac{1}{4}(1-m)K$,
we see that $J(u_{\epsilon,\infty}) > (\frac{1-m}{4})K.$
Now, we choose $\epsilon$ such that $C|\Omega_{\epsilon}| < (\frac{1-m}{4})K.$
Using this in (\ref{e5}) we obtain
\[
\int_{\Omega_{\epsilon}} u_{\epsilon,\infty}f(u_{\epsilon,\infty})
\,dx_1\,x_2\,dx_3 \leq CJ(u_{\epsilon,\infty}).
\]
Also, using this in (\ref{e6}) we obtain
$$ 
J(w) \leq \frac{C}{j} J(u_{\epsilon,\infty}).
$$ 
Choosing $j$ such that $j >2C$ we obtain $J(w) < J(u_{\epsilon,\infty})$ 
and this concludes the present proof. 
\hfill$\diamondsuit$\smallskip

To prove Lemma \ref{lem4} we need

\begin{lemma} \label{lem0}
Let $v \in H_{0}^{1}(\Omega_{\epsilon})$. Then the function 
$$P_{v}(\lambda) = \frac{\lambda vf(\lambda v)}{2} - F(\lambda v)$$ is increasing
on $(0,\infty).$
\end{lemma}
\noindent
{\it Proof.} Differentiating $P-{v}$ with respect to $\lambda$ we find
$$P_{v}'(\lambda) = \frac{\lambda v^2}{2} \left( f'(\lambda v) - 
\frac{f(\lambda v)}{\lambda v}\right) > 0\,,$$
where we have used (C2). This completes the proof \hfill$\diamondsuit$
\smallskip

\noindent {\it Proof of Lemma \ref{lem4}:}
Fix $k$ and $n$. Let $0 < \epsilon < \epsilon_{1}(kn)$. Lemma \ref{lem1}
guarantees the existence
of a minimizer $u_{\epsilon,kn}$ of $J$ on $S(\epsilon,kn)$. 
 From Lemma \ref{lem2} we see that $u_{\epsilon,kn}$ is a solution to
(\ref{P}).
Furthermore, from Lemma \ref{lem3} we know that $u_{\epsilon,kn}$ is
nonradial.
Now, by the regularity theory of elliptic equations we know 
that $u_{\epsilon,kn}$ is a $C^2$ function. Let $x = (r, \theta)$ be the
polar coordinates 
of $x \in {\mathbb R}^2$ and write $u = u_{\epsilon,kn}(r, \theta, |x_3|)$. Then
$$ \int_{\Omega_{\epsilon}} |\nabla u|^2 \,dx_1\,x_2\,dx_3 = 
\int_{(r,|x_3|)} \int_{0}^{2\pi}
(u_{r}^2 + \frac{1}{r^2}u_{\theta}^2 + |\nabla_{x_3} u|^2)r\,dr\,d\theta \,dx_3$$
and
$$ \int_{\Omega_{\epsilon}} F(u)\,dx_1\,x_2\,dx_3 = \int_{(r,|x_3|)} 
\int_{0}^{2\pi} F(u)r\,dr\,d\theta \,dx_3\,.$$
Define the function
$$ v(r, \theta, |x_3|) = u(r, \frac{\theta}{n}, |x_3|),\quad 0 \leq \theta \leq 2\pi.$$
Then
\begin{eqnarray*}
v(r, \theta + \frac{2\pi}{k}, |x_3|) & = & u(r, \frac{\theta}{n} + 
\frac{2\pi}{kn}, |x_3|) \\
& = & u(r, \frac{\theta}{n}, |x_3|) \\
& = & v(r, \theta, |x_3|).
\end{eqnarray*}
It follows that $v \in H(\epsilon,k)$. On the other hand,
it is easy to check that the following equalities hold
$$\displaylines{
 v_{r} (r, \theta, |x_3|) = u_{r} (r, \frac{\theta}{n}, |x_3|)\,, \cr
 v_{\theta} (r, \theta, |x_3|) = \frac{1}{n} u_{\theta} 
(r, \frac{\theta}{n}, |x_3|) \,,\cr
 \nabla_{x_3} v(r, \theta, |x_3|) = \nabla_{x_3} u(r, \frac{\theta}{n},
|x_3|). 
}$$
Therefore, 
\begin{eqnarray*}
\int_{\Omega_{\epsilon}} |\nabla v|^2\,dx_1\,x_2\,dx_3 
&=& k \int_{(r,|x_3|)} 
\int_{0}^{2\pi/k} (u_{r}^2(r,\frac{\theta}{n},|x_3|) 
 +  \frac{1}{r^2n^2}u_{\theta}^2(r, \frac{\theta}{n},|x_3|) \\
 &&+  |\nabla_{x_3} u(r,\frac{\theta}{n},|x_3|)|^2) r\,dr\,d\theta \,dx_3 \\
& = & kn \int_{0}^{\frac{2\pi}{nk}} (u_{r}^2(r,\theta,|x_3|) + 
\frac{1}{r^2n^4}u_{\theta}^2(r,\theta,|x_3|) \\
&&+ |\nabla_{x_3} u(r,\theta,|x_3|)|^2)r\,dr\,d\theta \,dx_3 \\
& = & \int_{(r,|x_3|)} \int_{0}^{2\pi} (u_{r}^2(r,\theta,|x_3|) + 
\frac{1}{r^2n^4}u_{\theta}^2(r,\theta,|x_3|) \\
&&+ |\nabla_{x_3} u(r,\theta,|x_3|)|^2)r\,dr\,d\theta \,dx_3. 
\end{eqnarray*}
Also
\begin{eqnarray*}
\int_{\Omega_{\epsilon}} F(v)\,dx_1\,x_2\,dx_3 &=& k \int_{(r,|x_3|)}
 \int_{0}^{2\pi/k}
F(u(r,\frac{\theta}{n},|x_3|))r\,dr\,d\theta \,dx_3 \\
& = & \int_{(r,|x_3|)} \int_{0}^{2\pi} F(u(r,\theta,|x_3|))r\,dr\,d\theta
\,dx_3\,.
\end{eqnarray*}
Since $u$ does not belong to $S(\epsilon, \infty)$ we have
\begin{eqnarray*}
\int_{(r,|x_3|)} \int_{0}^{2\pi} u_{\theta}^2(r,\theta,|x_3|)
r\,dr\,d\theta dx_3> 0\,.
\end{eqnarray*}
It follows that
\begin{eqnarray*}
\gamma (v) & = & \int_{\Omega_{\epsilon}} (|\nabla v|^2
 - vf(v))\,dx_1\,x_2\,dx_3 \\
& = &  \int_{(r,|x_3|)} \int_{0}^{2\pi} (u_{r}^2(r,\theta,|x_3|) + 
\frac{1}{r^2n^4}u_{\theta}^2(r,\theta,|x_3|) \\
&&+|\nabla_{x_3} u(r,\theta,|x_3|)|^2 - uf(u))r\,dr\,d\theta \,dx_3\\
& < & \int_{(r,|x_3|)} \int_{0}^{2\pi} (u_{r}^2(r,\theta,|x_3|) + 
\frac{1}{r^2}u_{\theta}^2(r,\theta,|x_3|) \\
&&+ |\nabla_{x_3} u(r,\theta,|x_3|)|^2 - uf(u))r\,dr\,d\theta \,dx_3 \\
& = & \int_{\Omega_{\epsilon}} (|\nabla u|^2 - uf(u))\,dx_1\,x_2\,dx_3 = 0.
\end{eqnarray*}
This yields
\begin{eqnarray} \label{v}
\int_{\Omega_{\epsilon}} |\nabla v|^2 \,dx_1\,x_2\,dx_3 < \int_{\Omega_{\epsilon}} 
vf(v)\,dx_1\,x_2\,dx_3\,.
\end{eqnarray}
Now, by Lemma  \ref{lem6} we can find $0< \alpha < 1$  such that 
$\alpha v \in  S(\epsilon, k)$.  Let $w = \alpha v \in
S(\epsilon, k)$. Using Lemma \ref{lem0} and the definition 
of $j_{k}^{\epsilon}$ we have
\begin{eqnarray*}
j_{k}^{\epsilon} & \leq & J(w) = J(\alpha v) 
                 = P_{v}(\alpha)  \\
                 & < & P_{v}(1)  \\
                 & = & \int_{\Omega_{\epsilon}} (\frac{1}{2}vf(v) - F(v))\,dx_1\,x_2\,dx_3 \\
                 & = & \int_{\Omega_{\epsilon}} (\frac{1}{2}uf(u) - F(u))\,dx_1\,x_2\,dx_3\\
                 & = & J(u) = j_{kn}^{\epsilon}.
\end{eqnarray*}
Putting together all the arguments above we conclude a proof of the 
lemma \hfill$\diamondsuit$

\section{Proof of Main Theorem}

For any integer $k \geq 1$, according to Lemma \ref{lem3} there exists $\epsilon_1
(2^k)$ such that if $0 < \epsilon < \epsilon_{1}(2^k)$ then 
$$ j_{2^k}^{\epsilon} < j_{\infty}^{\epsilon}.$$
\noindent
It follows from Lemma \ref{lem4} that
\begin{equation} \label{2}
j_{2}^{\epsilon} < j_{2^2}^{\epsilon} < ... < j_{2^k}^{\epsilon} <
j_{\infty}^{\epsilon}.
\end{equation}
\noindent
According to Lemma \ref{lem1}, there exists $u_{\epsilon,i} \in S(\epsilon, i), i = 1,
..., k$, such that
$$ j_{2^i}^{\epsilon} = J(u_{\epsilon,i}).$$
\noindent
Moreover, by Lemma \ref{lem2} $u_{\epsilon,i}$ is a solution of Problem
\ref{P}, $i = 1,\dots ,k$. By (\ref{2}), $\{u_{\epsilon,i} \}_{i=1}^{k}$, 
are nonrotationally equivalent and nonradial. This completes
the proof of the theorem. \hfill$\diamondsuit$ 

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}\end{thebibliography}\medskip

\noindent{\sc Alfonso Castro}\\
 Division of Mathematics and Statistics\\
 University of Texas at San Antonio\\
 San Antonio, TX 78249 USA.\\
e-mail: castro@math.utsa.edu \medskip

\noindent {\sc Marcel B. Finan}\\
Department of Mathematics \\
The University of Texas at Austin\\
Austin, TX 78712 USA. \\
e-mail: mbfinan@math.utexas.edu

\end{document}