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\markboth{\hfil A Dirichlet problem in the strip \hfil EJDE--1996/10}%
{EJDE--1996/10\hfil Eugenio Montefusco \hfil}
\begin{document}
\title{\vspace{-1in}\parbox{\linewidth}{\footnotesize\noindent
{\sc  Electronic Journal of Differential Equations},
Vol.\ {\bf 1996}(1996), No.\ 10, pp. 1--9. \newline
ISSN: 1072-6691. URL: http://ejde.math.swt.edu or http://ejde.math.unt.edu
\newline ftp (login: ftp) 147.26.103.110 or 129.120.3.113}
 \vspace{\bigskipamount} \\
A Dirichlet problem in the strip
\thanks{ {\em 1991 Mathematics Subject Classifications:}
35J65, 35B50.\newline\indent
{\em Key words and phrases:} Maximum Principle, Sliding Method, 
\newline\indent Subsolution and Supersolution.
\newline\indent
\copyright 1996 Southwest Texas State University  and University of
North Texas.\newline\indent
Submitted June 8, 1996. Published October 26, 1996.} }
\date{}
\author{Eugenio Montefusco}
\maketitle

\begin{abstract}
In this paper we investigate a Dirichlet problem in a strip and, using 
the sliding method, we prove monotonicity for positive and bounded 
solutions. We obtain uniqueness of the solution and show that this 
solution is a function of only one variable.
From these qualitative properties we deduce existence of a classical
solution for this problem.
\end{abstract}

\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{proposition}[theorem]{Proposition}
\newtheorem{corollary}[theorem]{Corollary}

\section{Introduction}

In 1979 B. Gidas, W. M. Ni and L. Nirenberg studied the problem:
\begin{equation}
\label{Ginini}\left\{ 
\begin{array}{cl}
-\Delta u=f(u) & \mbox{in }B(0,r) \\ u\equiv 0 & \mbox{on }\partial B(0,r) 
\end{array}
\right. 
\end{equation}
where $f$ is a locally Lipschitz function. In \cite{GNN} they showed that
the solution of (\ref{Ginini}) is a radial function, therefore this solution
reflects the symmetry of the domain. The proof of this result is based on
the moving plane method and the maximum Principle.

In the last years the interest in qualitative properties of
solutions of nonlinear elliptic equations has increased. 
H. Berestycki and L. Nirenberg \cite{BN} have simplified the moving plane 
method and proved the symmetry of solutions of elliptic equations in 
nonsmooth domains. In the
same paper H. Berestycki and L. Nirenberg have also simplified the sliding
method, which is a technique for proving monotonicity of solutions of
nonlinear elliptic equations.

At the same time some mathematicians are interested in qualitative
properties of solutions of elliptic equations in unbounded domains. H.
Berestycki and L. Nirenberg have studied the flame propagation in
cylindrical domains \cite{BN2}. C. Li investigated elliptic equations
in various unbounded domains \cite{L}. H. Berestycki, M. Grossi and F.
Pacella showed (using the moving plane method) that an equation with
critical growth does not admit a solution in the half space \cite{BGP}.

In 1993, H. Berestycki, L. A. Caffarelli and L. Nirenberg considered a
Dirichlet problem in the half space, they showed that the solution is a
function of only one variable (under suitable hypotheses) using the sliding
method \cite{BCN}.

With the same technique we want to prove a similar result in the strip. In
fact, using the sliding method, we show that problem (\ref{striscia2}) has a
unique classical solution depending on one variable only. As a matter
of fact, the problem is reduced to an ODE.

This paper is organized as follows. In section 2, we study the 
qualitative properties of the solution to (\ref{striscia2}).
 In section 3 we show some simple corollaries to the qualitative study.

In this paper we use frequently the following two theorems.

\begin{theorem}
Let $\Omega $ be an arbitrary bounded domain of $\Bbb R^N$ which is convex
in the $x_1$-direction. Let $u\in W_{loc}^{2,N}\left( \Omega \right) \cap
C\left( \overline{\Omega }\right) $ be a solution of:
\begin{equation}
\label{sliding}\left\{ 
\begin{array}{cl}
\Delta u+f(u)=0 & \mbox{in }\Omega \,, \\ u\equiv \varphi  & \mbox{on 
}\partial \Omega \,.
\end{array}
\right. 
\end{equation}
The function $f$ is assumed to be Lipschitz continuous. Here we assume that
for any three points $x'=(x'_1,y)\,$, $x=(x,y)\,$, $%
x'' =(x''_1,y)$ lying on a segment parallel to
the $x_1$-axis, $x'_1 <x_1 <x''_1\,$, with $%
x',\,x''\in \partial \Omega $, the following holds 
\begin{equation}
\label{monotonia}
\begin{array}{c}
\varphi (x')<u(x)<\varphi (x'')\quad 
\mbox{if }x\in \Omega , \\ \varphi (x')\le\varphi (x)\le\varphi
(x'')\quad \mbox{if }x\in \partial \Omega.
\end{array}
\end{equation}
Then $u$ is monotone with respect to $x_1$ in $\Omega $:
$$
u(x_1+\tau ,y)>u(x_1,y)\quad \mbox{for }(x_1,y)\,,\,(x_1+\tau ,y)\in \Omega
\;\mbox{and }\tau >0 . 
$$
Furthermore, if $f$ is differentiable, then $u_{x_1}>0$ in $\Omega $.
Finally, $u$ is the unique solution of (\ref{sliding}) in $%
W_{\small\mbox{loc}}^{2,N}(\Omega )\cap C\left( \overline{\Omega }
\right) $ satisfying (\ref{monotonia}).
\end{theorem}

\paragraph{Proof.} See Theorem 1.4 of \cite{BN}.

\begin{theorem}
Let $\Omega$ be a bounded domain and
suppose $u_1\in H^1\left( \Omega \right) $ is a subsolution while $u_2\in
H^1\left( \Omega \right) $ is a supersolution to problem (\ref{sliding}),
let be $f\in C( \Bbb R) $ and assume that with constants $%
c_1\,,\,c_2\in \Bbb R$ there holds $-\infty <c_1\leq u_1\leq u_2\leq
c_2<-\infty $, almost everywhere in $\Omega $. Then there exists a weak
solution $u\in H^1\left( \Omega \right) $ of (\ref{sliding}), satisfying the
condition $u_1\leq u\leq u_2$ almost everywhere in $\Omega $.
\end{theorem}

\paragraph{Proof.}See Theorem 2.4 in chapter I of \cite{Sw}.

\section{Statements and proofs}
\begin{theorem}
Let $u$ be a classical solution of 
\begin{equation}
\label{striscia2}\left\{ 
\begin{array}{cl}
-\Delta u=f(u) & \mbox{in }S \\ 0<u<M & \mbox{in }S \\ u\equiv 0 & \mbox{on }%
\{x_1=0\} \\ u\equiv M & \mbox{on }\{x_1=h\}
\end{array}
\right. 
\end{equation}
where $S := \{ x \in \Bbb R^N \quad \mbox{such that } 0<x_1<h \}$.
Let $f$ be a locally Lipschitz function with $f(0)\geq 0$ and $f(M)\leq 0$.
Then $u\equiv u(x_1)$ and $u_{x_1}>0$ in $S$. Moreover, $u$ is the unique
solution of (\ref{striscia2}).
\end{theorem}

The proof of Theorem 2.1 relies on the following propositions:

\begin{proposition}
There exists $w(t)$, a solution of 
\begin{equation}
\label{sopra}\left\{ 
\begin{array}{cc}
w''(t)+f(w(t))=0 & \mbox{in }(0,h) \\ 0<w<M & \mbox{in }(0,h)
\\ w(0)=0 &  \\ 
w(h)=M & 
\end{array}
\right. 
\end{equation}
such that $u(x)\leq w(x_1)$ in $S$.
\end{proposition}

\begin{proposition}
There exists $\rho (t)$, a solution of (\ref{sopra}), such that 
$u(x)\geq \rho(x_1)$ in $S$.
\end{proposition}

\paragraph{Proof of Theorem 2.1} By Theorem 1.1 problem (\ref{sopra}) has 
a unique solution such that $0<w<M$; indeed, we have that 
$\rho (t)\equiv w(t)$ in $(0,h)$.

Then Theorem 2.1 is proved, since we have that $\rho (x_1)\leq u(x)\leq
w(x_1)\equiv \rho (x_1)$, so $u$ is a function of only $x_1$; furthermore ,
Theorem 1.1 shows that $u_{x_1}>0$ in $S$ and $u$ is the unique solution of (%
\ref{striscia2}).

Now we must prove Propositions 2.2 and 2.3, which are more difficult than
Theorem 2.1.

\subsection*{Proof of Proposition 2.2}
Since the function $u$ is a classical solution of (\ref{striscia2}),
 by the Schauder estimates we can say that $|\nabla u|\leq K$ in the 
strip, with $K$
depending only on $\max _{[0,h]}f(s)$, (see Theorem 8.33 in \cite{GT}).

For $\varepsilon \in \left( 0,\min \left[1/K,h/M\right]\right)$,
 consider the function on $\Bbb R^{+}$:
$$
\sigma _\varepsilon (t)=\left\{ 
\begin{array}{cc}
\frac t\varepsilon & \mbox{in }[0,\varepsilon M] \\ M & \mbox{in }[%
\varepsilon M,h]. 
\end{array}
\right. 
$$
Set $\Omega _R:=\left\{ x\in S:\left( x_2^2+\dots +x_N^2\right) ^{1/2}
<R\right\} $, the cylinder in $S$ of radius $R$.

We note that there is a unique function $w_\varepsilon $ such that:
$$
\left\{ 
\begin{array}{cl}
-\Delta w_\varepsilon =f(w_\varepsilon ) & \mbox{in }\Omega _R \\ 
u<w_\varepsilon <M & \mbox{in }\Omega _R \\ w_\varepsilon \equiv \sigma
_\varepsilon (x_1) & \mbox{on }\partial \Omega _R\,.
\end{array}
\right.  
$$
In fact, $u$ is a subsolution of this problem (on the boundary $u$ takes
values less than $w$ because $u$ cannot increases faster, since $|\nabla
u|\leq K$), and the constant function $M$ is a supersolution. By
the theorem of sub and supersolution (see Theorem 1.2), there exists a
solution.

This solution is smooth by the classical regularity results (recall that 
$f$ is locally Lipschitz, and we can apply Lemma B.3 of \cite{Sw}); 
therefore, by Theorem 1.1 the function $w_\varepsilon $ is unique and 
strictly increasing in $x_1$ (note that $u<w_\varepsilon<M$ holds by 
the strong maximum Principle).

If we slide $u$ vertically (i.e. without changing $x_1$), we find that
similarly to the proof of the first Lemma 6 in \cite{BCN}:
$$
u(x_1,x^{\prime }+a)<w_\varepsilon \mbox{\quad in }\Omega _R\mbox{, }\forall
a\in \Bbb R^{N-1}. 
$$
Now let $\varepsilon \rightarrow 0$ through a sequence. The sequence 
$w_\varepsilon $ is bounded in $C^2(\Omega _R)$ by Theorem 8.33 in 
\cite{GT},
indeed it is compact in $C^1(A)$, where $A$ is a compact subset of $\Omega
_R\setminus \{(0,x^{\prime }):|x^{\prime }|=R\}$.

Then $w_\varepsilon $ converges to a function $w_R$ satisfying:
\begin{equation}
\label{cilstrip}\left\{ 
\begin{array}{cl}
-\Delta w_R=f(w_R) & \mbox{in }\Omega _R \\ w(0,x^{\prime })\equiv 0 & \mbox{%
for }|x^{\prime }|<R \\ w\equiv M & \mbox{on }\partial \Omega _R\setminus
\{x_1=0\} \\ w\mbox{ is increasing in }x_1 &  \\ 
u(x_1,x^{\prime }+a)<w_R<M & \mbox{in }\Omega _R\mbox{, }\forall a\in \Bbb R%
^{N-1}. 
\end{array}
\right. 
\end{equation}

Now we want to prove that $w_R$ is the unique solution of (\ref{cilstrip}).
In order to prove the uniqueness we cannot use Theorem 1.1, since $w_R$ is
not continuous on $\overline{\Omega }_R$ ,but in \cite{BCN} (section 2, p.
34-35) this fact is proved, using the sliding method and a little bit of
care.

Now we can consider $R^{\prime }>R$. For $a\in \Bbb R^{N-1}$, such that $%
|a|<R^{\prime }-R$, and for $\delta \in (0,h)$, we slide $\Omega _R$ so that
its center is at $\left( a,\delta -\frac h2\right) $. For $\delta $ small,
by continuity, the translated $w_R$ is greater than $w_{R^{\prime }}$ in the
overlapped region with $\Omega _{R^{\prime }}$. Moving the displaced $\Omega
_R$ up, and using the sliding method as above, we conclude that
\begin{equation}
\label{disug}w_R(x)>w_{R^{\prime }}(x_1,x^{\prime }+a)\quad \forall a\in 
\Bbb R^{N-1}\mbox{, with }|a|<R^{\prime }-R. 
\end{equation}
It follows that $w_R$ is a decreasing sequence in $R$. We let $R\rightarrow
\infty $, through a sequence, and we find that the sequence converges to a
function $w$ which satisfies problem (\ref{striscia2}).

From (\ref{disug}) we find that
$$
w(x)\geq w(x_1,x^{\prime }+a)\quad \forall a\in \Bbb R^{N-1}\,. 
$$
Then it follows that $w$ is independent of $x^{\prime }$, since $a$ is an
arbitrary vector in $\Bbb R^{N-1}$. We also have 
$$
u(x)\le w(x_1)\le M, 
$$
(it follows from the formulation of problem (\ref{cilstrip}) and letting $%
R\rightarrow \infty $).

Then the function $w$ satisfies problem (\ref{sopra}) and is greater than 
$u$%; thus, Proposition 2.2 is proved.

\subsection*{Proof of Proposition 2.3}
We wish to prove Proposition 2.3 following the proof of Proposition 2.2;
this is possible because we are in the strip and we can build a suitable
boundary condition for the problem in the cylinder.
In \cite{BCN} this is not possible, 
since the only condition at the boundary that
we can consider is the constant $0$, and with this
condition we cannot apply the Maximum Principle.

For $\varepsilon \in \left( 0,\min \left[1/K,h/M\right]\right)$,
define the function
$$
\gamma (t)=\left\{ 
\begin{array}{cl}
0 & \mbox{in }[0,h-\varepsilon M] \\ 
(t-h)/\varepsilon +M & \mbox{in }[%
h-\varepsilon M,h]. 
\end{array}
\right. 
$$
As above, there exists a unique function $\rho _\varepsilon $ such that:
$$
\left\{ 
\begin{array}{cl}
-\Delta \rho _\varepsilon =f(\rho _\varepsilon ) & \mbox{in }\Omega _R \\ 
0<\rho _\varepsilon <u & \mbox{in }\Omega _R \\ \rho _\varepsilon \equiv
\gamma (x_1) & \mbox{on }\partial \Omega _R\;. 
\end{array}
\right.  
$$
In this case, $u$ is a supersolution of the problem (on the
boundary $u$ takes values greater than $\rho _\varepsilon $), and the
constant $0$ is a subsolution (as $f(0)\geq 0$). By the theorem of
sub and supersolution, there is a solution of the equation.

By regularity theory, the function $\rho _\varepsilon $ is smooth;
indeed, by the sliding method (Theorem 1.14), $\rho _\varepsilon $ is 
unique and strictly increasing in $x_1$ (note that $0<\rho _\varepsilon <u$ is
true by the strong maximum Principle). As for the function $w_\varepsilon $
we find, by the sliding method, that
$$
0<\rho _\varepsilon <u(x_1,x^{\prime }+a)\quad \mbox{in }\Omega _R\mbox{, }%
\forall a\in \Bbb R^{N-1}. 
$$

Now let $\varepsilon \rightarrow 0$ through a sequence. The sequence $\rho
_\varepsilon $ is bounded in $C^2(\Omega _R)$ by the Schauder theory,
then by the Ascoli-Arzel\`a Theorem, it is compact in $C^1(A)$, where 
$A$ is a compact subset of $\Omega _R\setminus \{(h,x^{\prime }):|x^{\prime
}|=R\}$. Then $\rho _\varepsilon $ converges to a function $\rho _R$
satisfying:
\begin{equation}
\label{cilstrip2}\left\{ 
\begin{array}{cl}
-\Delta \rho _R=f(\rho _R) & \mbox{in }\Omega _R \\ \rho _R\equiv 0 & \mbox{%
on }\partial \Omega _R\setminus \{x_1=h\} \\ \rho _R(h,x^{\prime }) & \mbox{%
for }|x^{\prime }|<R \\ \rho _R\mbox{ is increasing in }x_1 &  \\ 
0<\rho _R<u(x_1,x^{\prime }+a) & \mbox{in }\Omega _R\mbox{, }\forall a\in 
\Bbb R^{N-1}. 
\end{array}
\right. 
\end{equation}

To show that the function $\rho _R$ is the unique solution of (\ref
{cilstrip2}), we proceed as in \cite{BCN}. Let $\zeta $ be another
solution of (\ref{cilstrip2}). Let $\Sigma _\delta $ denote the intersection 
$\Omega _R\cap \{\Omega _R-(h-\delta )e_1\}$, and let $\rho ^{\prime }$ be
the shifted $\rho _R$.
Then the function $\rho ^{\prime }$ satisfies:
$$
-\Delta \rho ^{\prime }=f(\rho ^{\prime })\quad \mbox{in }\Sigma _\delta 
$$
and
$$
\lim \inf _{x\rightarrow \partial \Sigma _\delta }(\rho ^{\prime }-\zeta
)\geq 0. 
$$
Since the function $f$ is locally Lipschitz, by the mean value Theorem, $%
\rho _\delta :=(\rho ^{\prime }-\zeta )$ satisfies:
\begin{equation}
\label{eqcap}-\Delta \rho _\delta =c(x)\rho _\delta \quad \mbox{in }\Sigma
_\delta \,, 
\end{equation}
where $|c(x)|$ is a function bounded by the Lipschitz constant of $f$.

For $\delta $ small we may apply the maximum Principle in narrow domains
(see \cite{GNN} Corollary p. 213), as $\Sigma _\delta $ is narrow in the $%
x_1 $ direction, and conclude that $\rho _\delta >0$ (using also the strong
maximum Principle).

We want to slide the translated $\Omega _R$ by increasing $\delta $, and use
the maximum Principle to show that $\rho _\delta >0$ in $\Sigma _\delta $
for every positive $\delta <h$. Suppose that $\rho _\delta >0$ in $\Sigma
_\delta $ for a maximal open interval $(0,\mu )$, with $\mu \le \delta $.

We want to show that $\mu =h$ by contradiction. Assume that $\mu <h$.

By continuity, $\rho _\mu \geq 0$ in $\Sigma _\mu $ and satisfies (\ref
{eqcap}), since $\rho _\mu >0$ on $\{x_1=0\}$ we have that $\rho _\mu \not
\equiv 0$; therefore, by the maximum Principle, we can say that $\rho _\mu
>0 $ in $\Sigma _\mu $.

We choose a small positive real number $\alpha $ such that $\alpha <\min
[(h-\mu ),R]$, and consider the subset $A:=\{(x,x^{\prime })\in \Omega
_R:x_1<(\mu -\alpha )\,,\,|x^{\prime }|<(R-\alpha )\}$.

As $\rho _\mu >0$ in $\Sigma _\mu $, there exists some constant $\varepsilon
>0$ such that $\rho _\mu \ge \varepsilon $ in $\overline{A}$. Thus for $\mu
^{\prime }>\mu $, with $(\mu ^{\prime }-\mu )$ sufficiently small, we obtain
that $\rho _{\mu ^{\prime }}>0$ in $\overline{A}$.

To conclude that $\rho _{\mu ^{\prime }}>0$ in $\Sigma _{\mu ^{\prime }}$ we
use the maximum Principle again. In $\Sigma _{\mu ^{\prime }}\setminus 
\overline{A}$ the function $\rho _{\mu ^{\prime }}$ verifies $-\Delta \rho
_{\mu ^{\prime }}=c(x)\rho _{\mu ^{\prime }}$, and since $\rho _{\mu
^{\prime }}>0$ in $\overline{A}$, we also have $\lim \inf _{x\rightarrow
\partial (\Sigma _{\mu ^{\prime }}\setminus \overline{A})}\rho _{\mu
^{\prime }}(x)\ge 0$. Since $\left( \Sigma _{\mu ^{\prime }}\setminus 
\overline{A}\right) $ has small measure for $(\mu ^{\prime }-\mu )$ small,
the maximum principle holds in $\left( \Sigma _{\mu ^{\prime }}\setminus 
\overline{A}\right) $ (see Proposition 1.1 in \cite{BN}), and we conclude
that $\rho _{\mu ^{\prime }}>0$ in $\left( \Sigma _{\mu ^{\prime }}\setminus 
\overline{A}\right) $, and hence in all of $\Sigma _{\mu ^{\prime }}$. This
is impossible for the maximality of $(0,\mu )$, therefore we have proved
that $\rho _\delta >0$ in $\Sigma _\delta ,\;\forall \delta <h$.

Now let $\delta \rightarrow h$; by continuity, it follows that $\rho _R\ge
\zeta $ in $\Omega _R$. We may interchange the roles of $\rho _R$ and $\zeta 
$ in the proof, and state that $\rho _R\equiv \zeta $.

Now we can consider $R^{\prime }>R$. For $a\in \Bbb R^{N-1}$ such that $%
|a|<\left( R^{^{\prime }}-R\right) $, and for $\delta \in (0,h)$, we slide $%
\Omega _R$ so that its center is at $\left( a,\delta -\frac h2\right) $. For 
$\delta $ small, by continuity, the translated $\rho _R$ is less than $\rho
_{R^{\prime }}$ in the overlapped region with $\Omega _{R^{\prime }}$.
Moving the displaced $\Omega _R$, and using the sliding method, we conclude
that
\begin{equation}
\label{disug2}\rho _{R^{\prime }}\left( x\right) >\rho _R(x_1,x^{\prime
}+a)\quad \forall a\in \Bbb R^{N-1}\mbox{, with }|a|<R^{\prime }-R. 
\end{equation}

It follows that $\rho _R$ is an increasing sequence in $R$. We let $%
R\rightarrow \infty $, through a sequence, and find that the sequence
converges to a function $\rho $, which satisfies equation (\ref{sopra}).
From (\ref{disug2}) we find that
$$
\rho (x)\ge \rho (x_1,x^{\prime }+a)\quad \forall a\in \Bbb R^{N-1}, 
$$
then it follows that $\rho $ is independent of $x^{\prime }$, since $a$ is
arbitrary. Furthermore, we also have that

$$
0\le \rho (x_1)\le u(x), 
$$
(it follows from the formulation of problem (\ref{sopra}) and letting $%
R\rightarrow \infty $).
This concludes the proof of Proposition 2.3.

\section{Remarks and corollaries}
\begin{proposition}
Theorem 2.1 is true even if the function $f$ depends on $x_1$ provided the
function $f(t,s)$ is increasing in the variable $t$.
\end{proposition}

\paragraph{Proof} Indeed if $f(s,u)$ is increasing in the first variable,
we obtain
\begin{eqnarray*}
-\Delta \rho _\delta &=& f(x_1+(\delta -h),\rho )-f(x_1,\zeta ) \\ 
&\leq &f(x_1,\rho )-f(x_1,\zeta )\\
&=&c(x)(\rho -\zeta )\;. \end{eqnarray*}
This inequality is sufficient  for the proofs of the above Propositions.
In fact the maximum Principle holds with inequality, and we don't need the
equality.

\begin{corollary}
There exists a unique classical solution to problem (\ref{striscia2}) which
satisfies the claim of Theorem 2.1, supposing that $f$ is a locally
Lipschitz function such that $f(0)\ge 0$ and $f(M)\le 0$.
\end{corollary}

\paragraph{Proof} We have proved that every solution of the problem is a
function depending only on $x_1$, thus problem (\ref{striscia2}) has a
solution if and only if problem (\ref{sopra}) has a solution. But the
one-dimensional problem admits a weak solution $u$ by the Theorem of sub and
supersolution (see theorem 1.2), since the constants $0$ and $M$ are
functions of $H^1(0,h)$ and they are sub and supersolution of (\ref{sopra}),
respectively.

Since $f$ is a locally Lipshitz function this weak solution $u$ is a
classical solution (see appendix B in \cite{Sw}) and is unique by the
Theorem 1.1. The Corollary is proved.

This Corollary is a simple generalization of Theorem 1.2; in fact our domain
is not bounded, but the domain $S$ has a particular geometry and this is
crucial for the proof of Theorem 2.3 and Corollary 3.2.

\begin{corollary}
Consider the problem: 
\begin{equation}
\label{cilindro}\left\{ 
\begin{array}{cl}
-\Delta u=f(u) & \mbox{in }(0,h)\times \Omega  \\ 0<u<M & \mbox{in }%
(0,h)\times \Omega  \\ u\equiv 0 & \mbox{on }\{0\}\times \Omega  \\ u\equiv
M & \mbox{on }\{h\}\times \Omega  \\ u_\nu \equiv 0 & \mbox{on }(0,h)\times
\partial \Omega \;,
\end{array}
\right. 
\end{equation}
where $\Omega$ is a Domain in $\Bbb R^{N-1}$ and $u_\nu$ is the exterior 
normal derivative.
Let $f$ be a locally Lipschitz function with $f(0)\ge 0$ and $f(M)\le 0$.

Then there exists at least a classical solution of (\ref{cilindro})
depending only on the variable $x_1$ and strictly increasing in the $x_1$
direction.
\end{corollary}

\paragraph{Proof.} Under these hypotheses we know that if there is a unique
classical solution to this problem with $\Omega =\Bbb R^{N-1}$, this
solution satisfies problem (\ref{cilindro}). In fact the function $u$ is a
function of $x_1$ only, thus the derivative $u_\nu $ is zero on $(0,h)\times
\partial \Omega $ since the normal vector $\nu $ is orthogonal to the $x_1-$%
direction on this part of the boundary. Then the solution of (\ref{striscia2}%
) is a solution of (\ref{cilindro}), and this concludes the proof of the
statement.

\paragraph{Remark.} In Corollary 3.3 we don't require all of the 
hypotheses on the domain $\Omega \subset \Bbb R^{N-1}$. In fact, this 
domain can be unbounded, and the argument still holds.

\paragraph{Acknowledgments.} I wish to express my deep gratitude to
professor Filomena Pacella for having suggested the topic, and for her
inspiring guidance. Moreover, I wish to thank professor Louis Nirenberg for
his encouragement and concern, and to professor  Luigi Orsina for his
useful  discussions.


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\bibitem[S]{Sw}  M. Struwe, {\sl Variational Methods}, Springer-Verlag,
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\end{thebibliography}

\medskip\noindent 
{\sc Eugenio Montefusco\\
 Scuola Normale Superiore\\
 Piazza dei Cavalieri 7\\
 56126 Pisa}\\
 E-mail address: montefus@cibs.sns.it

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