\documentclass[reqno]{amsart}
\usepackage{hyperref}

\AtBeginDocument{{\noindent\small
2005-Oujda International Conference on Nonlinear Analysis.
\newline {\em Electronic Journal of Differential Equations},
Conference 14, 2006, pp. 83--94.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or
http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu  (login: ftp)}
\thanks{\copyright 2006 Texas State University - San Marcos.}
\vspace{9mm}}
\setcounter{page}{83}

\begin{document}

\title[\hfilneg EJDE/Conf/14 \hfil Nonresonance problem]
{On a problem of lower limit in the study of nonresonance with
Leray-Lions operator}

\author[A. Anane, O. Chakrone, M. Chehabi \hfil EJDE/Conf/14 \hfilneg]
{Aomar Anane, Omar Chakrone, Mohammed Chehabi}  % in alphabetical order

\address{Aomar Anane \newline
D\'epartement de Math\'ematiques et Informatique,
 Facult\'e des Sciences,
 Universit\'e Mohammed 1er, Oujda,  Maroc}
\email{anane@sciences.univ-oujda.ac.ma}

\address{Omar Chakrone \newline
D\'epartement de Math\'ematiques et Informatique,
 Facult\'e des Sciences,
 Universit\'e Mohammed 1er, Oujda,  Maroc}
\email{chakrone@sciences.univ-oujda.ac.ma}

\address{Mohammed Chehabi \newline
D\'epartement de Math\'ematiques et Informatique,
 Facult\'e des Sciences,
 Universit\'e Mohammed 1er, Oujda,  Maroc}
\email{chehb\_md@yahoo.fr}

\date{}
\thanks{Published September 20, 2006.}
\subjclass[2000]{35J60, 35P30}
\keywords{Leray-Lions operator; Nonresonance; fist eigenvalue}

\begin{abstract}
 We prove the solvability of the Dirichlet problem
 \begin{gather*}
 Au  = f(u)+h \quad\text{in } \Omega , \\
 u  =   0 \quad\text{on }\partial \Omega
 \end{gather*}
 for a given $h$, under a condition involving only the
 asymptotic behaviour of the potential $F$ of $f$, where $A$ is a
 Leray-Lions operator.
\end{abstract}


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


\section{Introduction and statement of results}

This paper concerns the existence of solutions to the
problem
\begin{equation} \label{P}
\begin{gathered}
Au = f(u)+h \quad\text{in } \Omega , \\
u  = 0 \quad\text{on } \partial \Omega
\end{gathered}
\end{equation}
where $\Omega $ is a bounded domain of $\mathbb{R}^N$,
$N\geq 1$, $A$ is an operator of the form
$A(u)=-\sum_{i=1}^N \frac{\partial }{\partial x_{i}}A_{i}(\nabla u)$,
$f$ is a continuous function from $\mathbb{R}$ to $\mathbb{R}$ and
$h$ is a given function on $\Omega $.
Also we consider the problem
\begin{equation} \label{Pp}
\begin{gathered}
-\Delta _{p}u=f(u)+h\quad\text{in }\Omega \\
u =0 \quad\text{on }\partial \Omega
\end{gathered}
\end{equation}
where $\Delta _{p}$ denotes the $p$-Laplacian
$\Delta_{p}u=\mathop{\rm div}(|\nabla u|^{p-2}\nabla u)$, $1<p<\infty $.

A classical result, essentially due to Hammerstein \cite{H} asserts that if
$f$ satisfies a suitable polynomial growth restriction connected
with the Sobolev imbeddings and if
\begin{equation} \label{F1}
\limsup_{x\to \pm \infty}\frac{2F(s)}{|s|^2}<\lambda _{1},
\end{equation}
then problem \eqref{Pp} with $p=2$ is solvable for any
$h$. Here $F$ denotes the primitive $F(s)=\int_{0}^{s}f(t)dt$ and
$\lambda _{1}$ is the first eigenvalue of $-\Delta $ on
$H_{0}^{1}(\Omega )$. Several improvements of this result have
been considered in the recent years.

In $1989$, the case $N=1$ and $p=2$ was considered in \cite{Fe}.
It was shown there that \eqref{Pp} with $p=2$ is solvable
for any $h\in L^{\infty }(\Omega )$ if
\begin{equation} \label{F2}
\liminf_{x\to \pm \infty} \frac{2F(s)}{|s|^2}<\lambda _{1}.
\end{equation}
When $N\geq 1$ and $p=2$,  showed later in \cite{F} that
\eqref{Pp} is solvable for any $h\in L^{\infty }(\Omega )$ if
\begin{equation} \label{F3}
\liminf_{s\to \pm \infty} \frac{2F(s)}{|s|^2}
<\big( \frac{\pi }{2R(\Omega )}\big)^2,
\end{equation}
where $R(\Omega )$ denotes the radius of the smallest open ball
$B(\Omega )$ containing $\Omega $.
This result was extended to the $p$-Laplacian case in \cite{E}
where solvability of \eqref{Pp} was derived
under the condition
\begin{equation} \label{F4}
\liminf_{s\to \pm \infty} \frac{pF(s)}{|s|^p }
<(p-1)\left\{ \frac{1}{R(\Omega )}\int_{0}^{1}
\frac{dt}{(1-t^p )^{1/p}}\right\}^p .
\end{equation}
Note that this condition reduces to \eqref{F3} when $p=2$.

The question now naturally arises whether
$(p-1)\big\{ \frac{1}{R(\Omega )}\int_{0}^{1}\frac{dt}{(1-t^p )^{1/p}}\big\}^p$
can be replaced by $\lambda _{1}$ in \eqref{F4}, where $\lambda _{1}$
denotes the first eigenvalue of $-\Delta _{p}$ on
$W_{0}^{1,p}(\Omega )$ (cf\cite{A}).

Observe that for $N>1$ and $p=2$,
$( \frac{\pi }{2R(\Omega)}) ^2<\lambda _{1}$, and a similar strict inequality
holds when $1<p<\infty $.
In \cite{AC}, it was showed that the constants in \eqref{F3} and
\eqref{F4} can be improved a little bit.

Denote by $l(\Omega )$ the length of the smallest edge of an
arbitrary parallelepiped containing $\Omega $.
If
\begin{equation} \label{F5}
\liminf_{s\to \pm \infty } \frac{pF(s)}{|s|^p }<C_{p}(l)
\end{equation}
where $C_{p}(l)=(p-1)\big\{ \frac{2}{l(\Omega )}\int_{0}^{1}\frac{dt}{%
(1-t^p )^{1/p}}\big\} ^p $ then for any
$h\in L^{\infty }(\Omega )$ the problem \eqref{Pp}
has a solution $u\in W_{0}^{1,p}(\Omega )\cap C^{1}(\Omega )$.

Observe that for $N=1$, $C_{p}=\lambda _{1}$ the  first eigenvalue
of $-\Delta $ on $\Omega =]0,l(\Omega )[$.

In particular, $C_{2}=\big( \frac{\pi }{l}\big) ^2$, and it
recovers the result of \cite{Fe}. It is clear that \eqref{F5} is a
weaker hypothesis than \eqref{F4}. The difference between \eqref{F5}
and \eqref{F4} is
particularly important when $\Omega $ is a rectangle or a triangle. However
$C_{p}(l)<\lambda _{1}$ when $N>1$, and the question raised above
remains open.

In this paper we investigate the question of replacing
$\Delta_{p}$ by the operator
 of the form
\[
A(u)=-\sum_{i=1}^N \frac{\partial }{\partial x_{i}}A_{i}(\nabla u).
\]
We assume the following hypotheses:
\begin{itemize}
\item[(A0)] For all $i\in \{1,2,\dots ,N\}$,
$A_{i}:\mathbb{R}^{\mathbb{N}}\to \mathbb{R}$ is continuous.

\item[(A1)] there exists $(c,k)\in (]0,+\infty [ )^2$ such that
$|A_{i}(\xi )| \leq c|\xi |^{p-1}+K$
for all $\xi \in \mathbb{R}^{\mathbb{N}}$,  and all $i\in \{1,2,\dots ,N\}$.

\item[(A2)]  \begin{itemize}
\item[(a)]$\sum_{i=1}^N (A_{i}(\xi)-A_{i}(\xi '))(\xi _{i}-\xi _{i}')>0$
for all $\xi \neq \xi '\in \mathbb{R}^N$;
\item[(b)]  for all $i\in \{1,2,\dots ,N\}$, the function
defined by \\
$r_{i}(s)=A_{i}(0,\dots,0,s,0,\dots ,0)$ for $s\in \mathbb{R}$
is odd;
\item[(c)] for each $i\in \{1,2,\dots ,N\}$, there exists
$a_{i}\in ]0,+\infty [$  such that \\
$\lim_{s\to +\infty} r_{i}(s)/ s^{p-1}=a_{i}$;
\item[(d)] for each $i\in \{1,2,\dots ,N\}$,
$r_{i}\in C^{1}(\mathbb{R}^{\ast })$
 and $\lim_{s\to 0} sr_{i}'(s)=0$;
\item[(e)] for all $i\in \{1,2,\dots ,N\}$, $A_{i}(\xi )=0$
for all $\xi \in \mathbb{R}^N$ such that $\xi _{i}=0$.
\end{itemize}
\end{itemize}

\begin{remark} \label{rmk1.1}
(1) The hypothesis (A2)(d) is in particular satisfied if we
suppose that for $i\in \{1,\dots ,N\}$, $r_{i}\in
C^{1}(\mathbb{R}^{\ast })$ and there exists $q_i$,
$1<q_{i}<\infty $,  there exists $\eta _{i}>0$, there exists
$(a,b)\in \mathbb{R}^2$, such that for all $|s|<\eta _{i}$,
$|r_{i}'(s)|\leq a|s|^{q_{i}-2}+b$.

\noindent (2) The assumption (A2)(d) is an hypothesis of
homogenization at  infinity for the operator $A$.
\end{remark}

\begin{definition} \label{def1.1}\rm
For  $i\in \{1,2,\dots ,N\}$, we define
\[
l_{i}(s)=\frac{1}{p-1}[sr_{i}(s)-\int_{0}^{s}r_{i}(t)dt]\quad
\forall s\in \mathbb{R}.
\]
\end{definition}
\begin{proposition} \label{prop1.2}
Assume (A0), (A1) and (A2). Then:
(1) The operator $A:W_{0}^{1,p}(\Omega )\to W^{-1,p'}(\Omega )$ is defined,
strictly monotone and
\[
\langle Au,v\rangle =\sum_{i=1}^N \int_{\Omega }A_{i}(\nabla u)
\frac{\partial v}{\partial x_{i}}dx\quad
\forall u,v\in W_{0}^{1,p}(\Omega ).
\]

\noindent (2) For each $i\in \{1,2,\dots ,N\}$, the function
$r_{i}:\mathbb{R}\to \mathbb{R}$ is continuous, strictly increasing and
$r_{i}(0)=0$.

\noindent (3) For each $i\in \{1,2,\dots ,N\}$, the function $l_{i}$ satisfies
\begin{itemize}
\item[(i)] $l_{i}$ is even, continuous and $l_{i}(0)=0$;

\item[(ii)]  $\lim_{s\to +\infty} \frac{l_{i}(s)}{s^p }=\frac{a_{i}}{p}$

\item[(iii)] $l_{i}\in C^{1}(\mathbb{R)}$ and
$l_{i}'(s)=\begin{cases} \frac{sr_{i}'(s)}{p-1} & \text{if }s\neq 0 \\
0 & \text{if } s=0.
\end{cases}$

\item[(iv)] $l_{i}$ is strictly increasing in $\mathbb{R}^{+}$.
\end{itemize}
\end{proposition}

\begin{proof}
(1) By (A0), (A1), it is clear that the operator $A$ is
defined from $W_{0}^{1,p}(\Omega )$ to $W^{-1,p'}(\Omega)$, we have
\[
\langle Au,v\rangle =\sum_{i=1}^N \int_{\Omega }A_{i}(\nabla u)\frac{%
\partial v}{\partial x_{i}}dx\quad \forall u,v\in W_{0}^{1,p}(\Omega )
\]
and by (A1)(a), we verify easily that $A$ is strictly
monotone.

(2) Let $i\in \{1,\dots ,N\}$. By (A0) and (A2)-(b),
$r_{i}$ is continuous and $r_{i}(0)=0$ , in the end $r_{i}$ is
strictly  increasing. Indeed, let $(s,s')\in \mathbb{R}^2$ such
that $s\neq s', $ we have
\[
(r_{i}(s)-r_{i}(s'))(s-s')=\sum_{i=1}^N (A_{i}(\xi )-A_{i}(\xi
'))(\xi _{i}-\xi _{i}')>0
\]
where $\xi =(0,\dots,s,\dots 0)$ and $\xi '=(0,\dots,s',\dots 0)$

(3)(i) By the foregoing, the function $l_{i}$ is even, continuous and
$l_{i}(0)=0$ for every $i\in \{1,\dots,N\}$

(3)(ii) We show  first that
\begin{equation} \label{ast}
\lim_{s\to +\infty } \frac{1}{s^p }
\int_{0}^{s}r_{i}(t)dt=\frac{a_{i}}{p}.
\end{equation}
Let $\varepsilon >0$, by (A2)(c), there exists
$\eta_{\varepsilon }=\eta $ such that
$|r_{i}(s)-a_{i}s^{p-1}| \leq \varepsilon s^{p-1}$ for all
$s\geq \eta $.

Integrating from $\eta $ to $s$, we obtain
\[
\big |\int_{0}^{s}r_{i}(t)dt-\int_{0}^{\eta }r_{i}(t)dt-\frac{a_{i}}{p}
[s^p -\eta ^p ]\big|\leq \frac{\varepsilon }{p}[s^p -\eta ^p ].
\]
Dividing by $s^p $ and letting $n\to +\infty $, we
obtain
\[
\lim_{s\to +\infty }\big|\frac{1}{s^p }
\int_{0}^{s}r_{i}(t)dt-\frac{a_{i}}{p}\big|=0
\]
i.e \eqref{ast} holds.
Writing
\[
\frac{l_{i}(s)}{s^p }=\frac{1}{p-1}\big\{ \frac{r_{i}(s)}{%
s^{p-1}}-\frac{1}{s^p }\int_{0}^{s}r_{i}(t)dt\big\}.
\]
By \eqref{ast} and (A2)(c), we have
$\lim_{s\to +\infty}\frac{l_{i}(s)}{s^p }=\frac{a_{i}}{p}$

(3)(iii) Since $r_{i}\in C^{1}(\mathbb{R}^{\ast })$, we have
$l_{i}\in C^{1}(\mathbb{R}^{\ast })$ and
$l_{i}'(s)=\frac{1}{p-1}sr_{i}'(s)$  for every $s\neq 0$.
On the other hand, for $s\neq 0$,  since $r_{i}$ is increasing
and odd, we have
\[
|\frac{l_{i}(s)}{s}|=\frac{1}{p-1}\big|
r_{i}(s)-\frac{1}{s}\big|\int_{0}^{s}r_{i}(t)dt\leq
\frac{2}{p-1}r_{i}(|s|).
\]
It results that $l_{i}'(0)$ exists and $l_{i}'(0)=0$. By
(A2)-(d) we obtain $\lim_{s\to 0}l_{i}'(s)=\lim_{s\to 0}sr_{i}'(s)$.
This proves that $l_{i}\in C^{1}(\mathbb{R})$.

(3)(iv) is a consequence of (3)(iii)
\end{proof}

\begin{example} \label{exa1.3} \rm
We give at first some examples for operators $A$ satisfying the hypothesis
(A0), (A1) and (A2). (1) Let
\[
Au=-\Delta _{p}u=-\sum_{i=1}^N \frac{\partial }{\partial x_{i}}
(|\nabla u|^{p-2}\frac{\partial u}{\partial x_{i}})
\]
Then we have
$A_{i}(\xi )=|\xi |^{p-2}\xi _{i}$  for every $\xi =(\xi _{i})\in \mathbb{R}^N$.
\\
$r(s)=r_{i}(s)=|s|^{p-2}s$  for every $s\in \mathbb{R}$ and every
$i\in \{1,\dots,N\}$.
\\
$l(s)=l_{i}(s)=\frac{1}{p}|s|^p $  for every $s\in\mathbb{R}$ and
every $i\in \{1,\dots,N\}$.

(2) Let
\[
Au=-\Delta _{p}u-\Delta _{q}u=-\sum_{i=1}^N
\frac{\partial }{\partial x_{i}}(|\nabla u|^{p-2}
\frac{\partial u}{\partial x_{i}}+|\nabla u|^{q-2}
\frac{\partial u}{\partial x_{i}})
\]
where $1<q<p<+\infty$.
Then we have
$A_{i}(\xi )=|\xi |^{p-2}\xi _{i}+|\xi |^{q-2}\xi
_{i}$ for every $\xi =(\xi _{i})\in \mathbb{R}^N$.
\\
$r(s)=r_{i}(s)=|s|^{p-2}s+|s|^{q-2}s$ \ for every
$s\in \mathbb{R}$ and every $i\in \{1,\dots,N\}$.
\\
$l(s)=l_{i}(s)=\frac{1}{p}|s|^p +\frac{q-1}{q(p-1)}|
s|^{q} $  for every $s\in \mathbb{R}$ and every
 $i\in \{1,\dots,N\}$.

(3) Let
\[
Au=-\Delta _{p,\varepsilon }u=-\sum_{i=1}^N
\frac{\partial }{\partial x_{i}}\big[ \varepsilon +|\nabla u|^2)
^{\frac{p-2}{2}}\frac{\partial u}{\partial x_{i}}\big]\,,
\]
where $\varepsilon >0$.
Then we have
$A_{i}(\xi )=(\varepsilon +|\xi |^2)^{\frac{p-2}{2}}\xi
_{i}$ for every $\xi =(\xi _{i})\in \mathbb{R}^N$.
\\
$r(s)=r_{i}(s)=(\varepsilon +|s|^2)^{\frac{p-2}{2}}s$  for every
$s\in \mathbb{R}$ and every $i\in \{1,\dots,N\}$.
\\
$l(s)=l_{i}(s)=(\varepsilon +|s|^2)^{\frac{p-2}{2}}
\big(\frac{s^2}{p}-\frac{\varepsilon }{p(p-1)}\big)
+\frac{1}{p(p-1)}\varepsilon ^{\frac{p}{2}}$
 for every $s\in \mathbb{R}$ and every $i\in \{1,\dots,N\}$.

\end{example}

\section{Proof of Main Theorem}

We consider the Dirichlet problem \eqref{P}
where $\Omega $ is a bounded domain of $\mathbb{R}^N$,
$N\geq 1$, $f$ is a continuous function from $\mathbb{R}$ to $\mathbb{R}$
and $h\in L^{\infty }(\Omega )$.

Denote by $[AB]$ the smallest edge of an arbitrary parallelepiped
containing $\Omega $. Making an orthogonal change of variables, we
can always suppose that $AB$ is parallel to one of the axis of
$\mathbb{R}^N$. So
$\Omega \subset P=\prod_{j=1}^N [a_{j},bj]$ with, for some $i$,
$|AB|=b_{i}-a_{i}=\min_{1\leq j\leq N} \{b_{j}-a_{j}\}$,
a quantity which we denote by $b-a$.

Denote by $l=l_{i}$, $r=r_{i}$, $F$ the primitive
$F(s)=\int_{0}^{s}f(t)dt$ and
\[
C_{p}=(p-1)\big\{ \frac{2}{b-a}\int_{0}^{1}\frac{dt}{(1-t^p )^{\frac{1}{p%
}}}\big\} ^p .
\]

\begin{theorem} \label{theor}
Assume
\begin{equation}\label{F}
\liminf_{s\to \pm \infty } \frac{F(s)}{l(s)}<C_{p}.
\end{equation}
Then for any $h\in L^{\infty }(\Omega )$, the problem \eqref{P}
has a solution $u\in W_{0}^{1,p}(\Omega )\cap L^{\infty }(\Omega
)$ in the weak sense; i.e
\[
\sum_{i=1}^N \int_{\Omega }A_{i}(\nabla u)
\frac{\partial \varphi }{\partial x_{i}}dx
=\int_{\Omega }f(u)\varphi +\int_{\Omega}h\varphi \quad
 \forall \varphi \in W_{0}^{1,p}(\Omega).
\]
\end{theorem}

\begin{definition} \label{def2.2}\rm
An upper solution for \eqref{P} is defined as a function
$\beta:\overline{\Omega}\to \mathbb{R}$ such that
\begin{itemize}
\item  $\beta \in C^{1}(\overline{\Omega })$

\item $A(\beta )\in C(\overline{\Omega })$

\item $A(\beta )(x)\geq f(\beta (x))+h(x)$  a e $x$ in $\Omega $.
\end{itemize}
A lower solution $\alpha $ is defined by reversing the
inequalities above.
\end{definition}

\begin{lemma} \label{lem2.1}
Assume that \eqref{P} admits an upper solution
$\beta $ and a lower solution $\alpha $ with
$\alpha (x)\leq \beta (x)$ in $\Omega $.
Then \eqref{P} admits a solution
$u\in W_{0}^{1,p}(\Omega )\cap C^{1}(\Omega )$, with
$\alpha (x)\leq u(x)\leq \beta (x)$ in $\Omega $.
\end{lemma}

\begin{proof} Let
\[
\widetilde{f}(x,s)=\begin{cases}
f(\beta (x))&\text{if } s\geq \beta (x), \\
f(s) & \text{if } \alpha (x)\leq s\leq \beta (x), \\
f(\alpha (x)) & \text{if } s\leq \alpha (x)
\end{cases}
\]
for every $(x,s)\in $ $\overline{\Omega }\times \mathbb{R}$ such that
$\widetilde{f}$ is bounded and continuous in
$\overline{\Omega }\times \mathbb{R}$, then the problem
\begin{equation} \label{Ptilde}
\begin{gathered}
Au = \widetilde{f}(x,u)+h(x)\quad \text{in } \Omega  \\
u = 0 \quad\text{on } \partial \Omega
\end{gathered}
\end{equation}
admits a solution $u\in W_{0}^{1,p}(\Omega )$ in the weak sense,
indeed the operator $A$ is strictly monotone, so we can use the
result of  Lions \cite{L} concerned the pseudomonotones
operators.

We claim that $\alpha (x)\leq u(x)\leq \beta (x)$ in $\Omega $,
which clearly implies the conclusion.

To prove the first inequality, one multiplies \eqref{Ptilde}
 by $w=u-u_{\alpha }$, where $u_{\alpha }(x)=\max(u(x),\alpha (x))$,
integrates by parts and uses the fact that
$\alpha $ is a lower solution we obtain
$\langle A(u)-A(u-w),w\rangle \leq 0$,
which implies $w=0$ (since $A$ is strictly monotone).
\end{proof}

\begin{lemma} \label{lem2.2} Let $a<b$ and $M>0$, and assume
\begin{equation} \label{F+}
\liminf_{s\to +\infty }\frac{F(s)}{l(s)}<C_{p}.
\end{equation}
then there exists $\beta _{1}\in C^{1}(I)$ such that
$(r(\beta_{1}'(t)))'\in C(I)$ and
\begin{gather*}
-(r(\beta _{1}'(t)))'  \geq f(\beta_{1}(t))+M \quad \forall t\in I, \\
\beta _{1}(t) \geq 0 \quad \forall t\in I
\end{gather*}
where $I=[a,b]$.
\end{lemma}

\begin{lemma} \label{lem2.3}
Assume
\begin{equation} \label{F-}
\liminf_{s\to -\infty }\frac{F(s)}{l(s)}<C_{p}.
\end{equation}
then there exists $\alpha _{1}\in C^{1}(I)$ such that
$(r(\alpha_{1}'(t)))'\in C(I)$ and
\begin{gather*}
-(r(\alpha _{1}'(t)))'\leq f(\alpha_{1}(t))-M \quad \forall t\in I \\
\alpha _{1}(t) \leq 0 \quad \forall t\in I
\end{gather*}
where $I=[a,b]$.
\end{lemma}

Accepting for a moment the conclusion of these two lemmas, let us
turn to the Proof of Theorem \ref{theor}. By Lemma \ref{lem2.1} it
suffices to show the existence of an upper solution and a lower
solution for \eqref{P}. Let us describe the construction of
the upper solution (that of the lower solution is similar).

Let $M>\| h\| _{\infty }$ and $i\in \{1,2,\dots,N\}$ such that
$b=b_{i}$, $a=a_{i}$.
By Lemma \ref{lem2.2} there exists $\beta _{1}:I\to
\mathbb{R}$ such that $\beta _{1}\in C^{1}(I)$,
$(r(\beta_{1}'(t)))'\in C(I)$ and
\begin{gather*}
-(r(\beta _{1}'(t)))'\geq f(\beta_{1}(t))+M \quad \forall t\in I, \\
 \beta _{1}(t)\geq 0\quad \forall t\in I.
\end{gather*}

Writing $\beta (x)=\beta _{1}(x_{i})$ for all
$x=(x_{i})\in \overline{\Omega }$, it is clear that
$\beta \in C^{1}(\overline{\Omega })$,
$A(\beta (x))=A(\beta _{1}(x_{i}))\in C(\overline{\Omega })$, and
we have by (A2)(e):
\begin{align*}
A(\beta (x)) &=-\sum_{j=1}^n \frac{\partial}{\partial x_{j}}A_{j}
(\nabla \beta (x)) \\
&=-\frac{\partial }{\partial x_{i}}(r_{i}(\beta _{1}'(x_{i}))) \\
&=-(r(\beta _{1}'(x_{i})))' \\
&\geq f(\beta _{1}(x_{i}))+M \\
&=f(\beta (x))+M \\
&\geq f(\beta (x))+h(x)\text{ \ \ a.e.}x\in \Omega
\end{align*}
The proof of Theorem \ref{theor} is thus complete.

Next, we present the proof of Lemma \ref{lem2.2}.
The proof of Lemma \ref{lem2.3} follows similarly.

\textbf{First case.}
Suppose $\inf_{s\geq 0} f(s)=-\infty $. Then there exists
$\beta \in \mathbb{R}^{\ast }\mathbb{+}$ such that $f(\beta )<-M$,
and the constant function $\beta $ provides a solution to the
problem in Lemma \ref{lem2.2}.

\noindent\textbf{Second case.}
Suppose now $\inf_{s\geq 0} f(s)>-\infty $. Let $K>M$ such that
$\inf_{s\geq 0} f(s)>-K+1$.
Thus $f(s)+K\geq 1$ for all $s\geq 0$. Define
$g:\mathbb{R}\to \mathbb{R}$ by
\[
g(s)=\begin{cases}
f(s)+K & \text{if } s\geq 0 \\
f(0)+K & \text{if } s<0
\end{cases}
\]
and denote $G(s)=\int_{0}^{s}g(t)dt$ for all $s$ in $\mathbb{R}$.
It is easy to see that $g(s)\geq 1$ for all $s$ in $\mathbb{R}$ and
that
\[
0\leq \liminf_{s\to +\infty } \frac{G(s)}{l(s)}=
\liminf_{s\to +\infty } \frac{F(s)}{l(s)}<C_{p}.
\]
Now it is clearly sufficient to prove the existence of a function
$\beta_{1}:I\to \mathbb{R}$ such that $\beta _{1}\in C^{1}(I)$,
$(r(\beta _{1}'(t)))'\in C(I)$ and
\begin{gather*}
-(r(\beta _{1}'(t)))'=g(\beta _{1}(t)) \quad \forall t\in I \\
 \beta _{1}(t) \geq 0\quad \forall t\in I
\end{gather*}
For this purpose we will need the following four Lemmas.

\begin{lemma}\label{lem2.4}
Let $0<c<\infty $ and $t\in ]0,1[$, then
\[
\lim_{\alpha \to +\infty} \frac{\alpha }{l^{-1}(c(l(\alpha )
-l(\alpha t)))}=\frac{1}{c^{1/p}(1-t^p )^{\frac{1}{p}}}
\]
In particular by, Fatou Lemma,
\[
\frac{1}{c^{1/p}}\int_{0}^{1}\frac{dt}{(1-t^p )^{1/p}}\leq
\liminf_{\alpha \to +\infty} \int_{0}^{1}\frac{\alpha dt}{l^{-1}
(c(l(\alpha )-l(\alpha t)))}
\]
\end{lemma}

\begin{proof}
Denote $s(\alpha )=\frac{\alpha }{l^{-1}(c(l(\alpha)-l(\alpha t)))}$
and $\frac{a_{i}}{p}=d$.
By Proposition \ref{prop1.2} (3)(ii), we have
\[
\lim_{s\to +\infty} \frac{s^{1/p}}{l^{-1}(s)}=d^{1/p}.
\]
On the other hand,
\[
\lim_{\alpha \to +\infty} [c(l(\alpha)-l(\alpha t))]=+\infty ,
\]
 and more generally,
\[
\lim_{\alpha \to +\infty} \frac{l(\alpha )-l(\alpha t)}{\alpha ^p }
=d(1-t^p )>0\,.
\]
Writing
\[
s(\alpha )=\frac{1}{[\frac{c(l(\alpha )-l(\alpha t))}{\alpha ^p }]
^{1/p}}\frac{[c(l(\alpha )-l(\alpha t))]^{1/p}}{l^{-1}(c(l(\alpha )
-l(\alpha t)))}
\]
Letting $n\to +\infty $ and by the three limits above, we have
\[
\lim_{\alpha \to +\infty} s(\alpha )=\frac{1}{c^{1/p}(1-t^p )^{1/p}}
\]
\end{proof}

\begin{lemma} \label{lem2.5}
For $d>0$, define
\[
\tau _{G}(d)=\int_{0}^{d}\frac{ds}{l^{-1}[\frac{G(d)-G(s)}{p-1}]}\,.
\]
 Then
\[
\limsup_{d\to +\infty} \tau _{G}(d)\geq
\Big(\int_{0}^{1}\frac{dt}{(1-t^p )^{1/p}}\Big)
\Big(\frac{1}{p-1}\liminf_{s\to +\infty } \frac{G(s)}{l(s)}\Big)^{1/p}.
\]
In particular \eqref{F+} implies
$\limsup_{d\to +\infty} \tau _{G}(d)>(b-a)/2$.
\end{lemma}

\begin{proof}
Let $\rho $ be a positive number such that $\liminf_{s\to +\infty
} \frac{G(s)}{l(s)}<\rho <C_{l}$. Then $\underset{s\to +\infty
}{\lim \sup }[\rho l(s)-G(s)]=+\infty $. Let $w_{n}$ be the
smallest number in $[0,n]$ such that $\max_{0\leq s\leq n}
K(s)=K(w_{n})$ where $K(s)=\rho l(s)-G(s);$ it is easily seen that
$(w_{n})$ is increasing with respect to $n$. Since $\rho
l(s)-G(s)<\rho l(w_{n})-G(w_{n})$ for all $s\in [0,w_{n}[$, we
have $\frac{G(w_{n})-G(s)}{p-1} <\frac{\rho }{p-1}(l(w_{n})-l(s))$
for all $s\in [0,w_{n}[$, since $l:[0,+\infty [\to [0,+\infty [$
is an increasing \emph{homeomorphism}, we have
\[
\frac{1}{l^{-1}\big[ \frac{\rho }{p-1}(l(w_{n})-l(s))\big] }
<\frac{1}{l^{-1}\big[ \frac{1}{p-1}(G(w_{n})-G(s))\big] }.
\]
Integrating from $0$ to $w_{n}$ and changing variable $s=uw_{n}$
in the first member of inequality, we obtain
\[
\int_{0}^{1}\frac{w_{n}}{l^{-1}[ \frac{\rho }{p-1}(l(w_{n})-l(w_{n}s))] }
ds\leq \tau _{G}(w_{n}).
\]
Letting $n\to +\infty $, we obtain
\[
\liminf_{n\to +\infty} \int_{0}^{1}\frac{w_{n}}{l^{-1}
\big[ \frac{\rho }{p-1}(l(w_{n})-l(w_{n}s))\big] }ds\leq
 \limsup_{n\to +\infty}\tau _{G}(w_{n}).
\]
By Lemma \ref{lem2.4}, it results
\[
\limsup_{d\to +\infty} \tau _{G}(d)\geq
\Big[\int_{0}^{1}\frac{dt}{(1-t^p )^{1/p}}\Big]
\Big[\frac{\rho }{p-1}\Big] ^{\frac{-1}{p}}.
\]
Letting $\rho \to \liminf_{s\to +\infty } \frac{G(s)}{l(s)}$,  the
Lemma is proved.
\end{proof}

\begin{lemma} \label{lem2.6}
Let $d>0$ and consider the mapping $T_{d}$ defined by
\[
T_{d}(u)=d-\int_{a}^{t}r^{-1}\Big( \Big[ \int_{a}^{\tau
}g(u(s))ds\Big] ^{1/(p-1)}\Big) d\tau
\]
in the Banach space $C(I)$. Then $T_{d}$ has a fixed point.
\end{lemma}

\begin{proof}
Clearly by Ascoli's theorem $T_{d}$ is compact. The proof of Lemma
\ref {lem2.6} uses an homotopy argument based on the Leray
Schauder topological degree. So $T_{d}$ will have a fixed point if
the following condition holds:

There exists $\rho >0$ such that $(I-\lambda T_{d})(u)\neq 0$ for all
$u\in \partial B(0,\rho )$ for all $\lambda \in [0,1]$, where
$\partial B(0,\rho )=\{u\in C(I);\| u\| _{\infty }=\rho \}$.

To prove that this condition holds, suppose by contradiction that
for all $n=1,2,\dots$ there exists $u_{n}\in \partial B(0,n)$, $
\lambda _{n}\in [0,1]$ such that: $u_{n}=\lambda
_{n}T_{d}(u_{n})$. The latter relation implies
\begin{equation} \label{e1}
u_{n}=\lambda _{n}d-\lambda _{n}\int_{a}^{t}r^{-1}
\Big(\Big[ \int_{a}^{\tau }g(u(s))ds\Big] ^{\frac{1}{p-1}}\Big) d\tau
\end{equation}
Therefore, $u_{n}\in C^{1}(I)$ and we have successively
\begin{equation} \label{e2}
\begin{gathered}
u_{n}'(t)=-\lambda _{n}r^{-1}\Big( \Big[\int_{a}^{\tau }g(u(s))ds\Big]
^{\frac{1}{p-1}}\Big) <0 \quad \forall t\in ]a,b], \\
u_{n}'(a) = 0,
\end{gathered}
\end{equation}
$\big( r[ \frac{u_{n}'(t)}{\lambda _{n}}]\big) '\in C(I)$
and
\begin{equation} \label{e3}
-\Big( r\Big( \frac{u_{n}'(t)}{\lambda _{n}}
\Big) \Big) '=g(u_{n}(t))\quad \forall t\in I.
\end{equation}
Note that by \eqref{e2}, $u_{n}'(t)<0$ in $]a,b]$, so that
$u_{n}$ is decreasing.
Hence, for $n>d$, $u_{n}(b)=-n$. Multiplying the equation \eqref{e3} by
$u_{n}'(t)$, we obtain
\begin{equation} \label{e4}
-\lambda _{n}\Big( l\Big( \frac{u_{n}'(t)}{\lambda _{n}}\Big) \Big) '
=\frac{1}{p-1}\frac{d}{dt}G(u_{n}(t)).
\end{equation}
Indeed
\begin{align*}
\Big( l\Big( \frac{u_{n}'(t)}{\lambda _{n}}\Big)\Big)'
&= \Big[ l\Big( r^{-1}\Big(
r\big( \frac{u_{n}'(t)}{\lambda _{n}}\big) \Big) \Big) \Big] ' \\
&= \big(l\circ r^{-1}\big)'
\Big( r\big( \frac{u_{n}'(t)}{\lambda _{n}}\big) \Big)
\Big( r\big( \frac{u_{n}'(t)}{\lambda _{n}}\big) \Big) ' \\
&= \frac{1}{p-1}\frac{u_{n}'(t)}{\lambda _{n}}
\Big( r\big( \frac{u_{n}'(t)}{\lambda _{n}}\big) \Big) '
\end{align*}
By \eqref{e4}, we have
\[
\lambda _{n}\Big( l\big( \frac{u_{n}'(t)}{\lambda
_{n}}\big) \Big) =\frac{1}{p-1}(G(\lambda _{n}d)-G(u_{n}(t))
\]
and
\[
-\frac{u_{n}'(t)}{\lambda _{n}l^{-1}\big[
\frac{G(\lambda _{n}d)-G(u_{n}(t))}{(p-1)\lambda _{n}}\big]}=1.
\]
Integrating from $a$ to $b$ and changing variable $s=u_{n}(t)$ \
$(u_{n}(a)=\lambda _{n}d$ and $u_{n}(b)=-n)$, we obtain
\[
\int_{-n}^{\lambda _{n}d}\frac{ds}{\lambda _{n}l^{-1}\big[
\frac{G(\lambda _{n}d)-G(s)}{(p-1)\lambda _{n}}\big] }=b-a
\]
i.e.
\[
\int_{0}^{\lambda _{n}d}\frac{ds}{\lambda _{n}l^{-1}\big[
\frac{G(\lambda_{n}d)-G(s)}{(p-1)\lambda _{n}}\big] }
= b-a+\int_{0}^{-n}\frac{ds}{\lambda _{n}l^{-1}
\big[ \frac{G(\lambda _{n}d)-G(s)}{(p-1)\lambda _{n}}\big] } \\
\geq 0
\]
Since $G(s)=sg(0)$ for $s\leq 0$ and changing variable $s=-u$, we
obtain
\begin{equation} \label{e5}
 0\leq (b-a)-\int_{0}^{n}\frac{ds}{\lambda
_{n}l^{-1}\big[ \frac{G(\lambda _{n}d)-sg(0)}{(p-1)\lambda
_{n}}\big] }
\end{equation}
Denote by $l(u)=\frac{G(\lambda _{n}d)-G(s)}{(p-1)\lambda _{n}}$ such that
$l'(u)du=\frac{g(0)}{(p-1)\lambda _{n}}ds$ and
$ds=\frac{\lambda _{n}}{g(0)}r'(u)udu$ for $u\neq 0$
and denote
$\alpha _{n}=l^{-1}\big[ \frac{G(\lambda _{n}d)}{(p-1)\lambda _{n}}\big] $ and
$\beta _{n}=l^{-1}\big[\frac{(G(\lambda _{n}d)+ng(0))}{(p-1)\lambda _{n}}\big]$.
By \eqref{e5}, we obtain
\begin{align*}
0 &\leq  (b-a)-\int_{\alpha _{n}}^{\beta _{n}}\frac{r'(u)}{g(0)}du \\
&= (b-a)-\frac{1}{g(0)}r\big\{ l^{-1}\big[
 \frac{G(\lambda _{n}d)-ng(0)}{(p-1)\lambda _{n}}\big] \big\}
+\frac{1}{g(0)}r\big\{l^{-1}\big[ \frac{G(\lambda _{n}d)}{(p-1)\lambda _{n}}
\big]\big\} .
\end{align*}
Since
\[
\frac{G(\lambda _{n}d)-ng(0)}{(p-1)\lambda _{n}}
\geq \frac{ng(0)}{(p-1)}, \quad
\frac{G(\lambda _{n}d)}{(p-1)\lambda _{n}}
\leq \frac{d}{p-1}\max_{0\leq s\leq d}|g(s)|
\]
and $r\circ l^{-1}$ is increasing, it results that
\[
0\leq (b-a)-\frac{1}{g(0)}r\big\{ l^{-1}\big[
\frac{ng(0)}{(p-1)\lambda_{n}}\big] \big\}
+\frac{1}{g(0)}r\big\{ l^{-1}\big[ \frac{d}{p-1}
\underset{0\leq s\leq d}{\max }|g(s)|\big] \big\} .
\]
Letting $n\to +\infty $, we get a contradiction.
Let us denote by $u_{d}\in C(I)$ a fixed point of the mapping
$T_{d}$ of Lemma \ref{lem2.6}
\end{proof}

\begin{lemma} \label{lem2.7} There exists $d>0$ such that
$u_{d}(t)\geq 0$ for all $t\in [a,\frac{a+b}{2}[$.
\end{lemma}

\begin{proof}
We know that $u_{d}$ is decreasing and that $u_{d}(a)=d$ for all
$d>0$. Let us distinguish two cases.

First if there exists $d>0$ such that $u_{d}(b)\geq 0$, then the
conclusion of Lemma \ref{lem2.7} clearly follows.
So we can assume that $u_{d}(b)<0$ for every $d>0$. Since
$u_{d}(a)=d>0$, there exists $\delta _{d}\in ]a,b[$ such that
$u_{d}(\delta _{d})=0$. It is clear that
$u_{d}(t)\geq 0$ for all $t\in [a,\delta _{d}[$, and so it
is sufficient to show that $\limsup_{d\to +\infty}\delta _{d}>\frac{a+b}{2}$.
Processing as in the proof of Lemma \ref{lem2.6}, we obtain
\[
-u_{d}'(t)\big\{ l^{-1}\big(
\frac{G(d)-G(u_{d}(t))}{p-1}\big) \big\} ^{-1}=1.
\]
Integrating from $a$ to $\delta _{d}$ and changing variable
$s=u_{d}(t)$, one gets
\[
\tau _{G}(d)=\int_{0}^{d}\frac{ds}{l^{-1}\big[ \frac{G(d)-G(s)}{p-1}%
\big] }=\delta _{d}-a,
\]
consequently
\[
\limsup_{d\to +\infty} \delta _{d}>a+\frac{b-a}{2}=\frac{a+b}{2}
\]
\end{proof}

\begin{proof}[Proof of Lemma \ref{lem2.2} continued].
Denoting $u_{d}(t)$ by $u(t)$, we have $u\in C^{1}(I)$,
$(r(u'))'\in C(I)$ and
\begin{gather*}
-(r(u'))' = g(u(s)) \quad \forall t\in I, \\
u(t) \geq 0 \quad \forall t\in [a,\frac{a+b}{2}[, \\
u'(a)=0.
\end{gather*}
Define a function $\beta _{1}$ from $[a,b]$ to $\mathbb{R}$ by
\[
\beta _{1}(t)=\begin{cases}
u(\frac{3a+b}{2}-t)& \text{if } t\in [a,\frac{a+b}{2}], \\
u(t-\frac{b-a}{2}) & \text{if } t\in [\frac{a+b}{2},b].
\end{cases}
\]
We will show that this function $\beta $ fulfills the conditions
of Lemma \ref{lem2.2}. To see this it is sufficient to show that
\begin{itemize}
\item[(a)] $\beta _{1}$ is nonnegative in $[a,b]$,

\item[(b)] $\beta _{1}\in C^{1}([a,b])$,

\item[(c)] $(r(\beta _{1}'(t)))'\in C([a,b])$ and
$-(r(\beta _{1}'(t)))'=g(\beta _{1}(t))$ for all $t\in [a,b]$.
\end{itemize}
Proof of (a). If $a\leq t\leq \frac{a+b}{2}$, then
$a\leq \frac{3a+b}{2}-t\leq \frac{a+b}{2}$, and if
$\frac{a+b}{2}\leq t\leq b$, then $a\leq t-\frac{b-a}{2}\leq \frac{a+b}{2}$,
so that the conclusion follows
from the sign of $u$ on $[a,\frac{a+b}{2}]$.

Proof of (b). $\beta _{1}\in C^{1}([a,\frac{a+b}{2}[)$,
$\beta_{1}\in C^{1}(]\frac{a+b}{2},b])$, and moreover
$\frac{d}{dt^{+}}\beta _{1}(\frac{a+b}{2})=u'(a)=0$ and
$\frac{d}{dt^{-}}\beta _{1}(\frac{a+b}{2})=u'(a)=0$.

Proof of (c). We know that $-(r(u'(t)))'=g(u(t))$ for
$t\in [a,b]$, therefore
\[
-(r(u'(t))=\int_{a}^{t}g(u(s))ds.
\]
If $a\leq t\leq \frac{a+b}{2}$ then
$a\leq \frac{3a+b}{2}-t\leq \frac{a+b}{2}$, which gives
\[
\beta _{1}(t)=u\big( \frac{3a+b}{2}-t\big) \quad \text{and}\quad
\beta_{1}'(t)=-u'\big( \frac{3a+b}{2}-t\big).
\]
We obtain
\[
-(r(u'(\frac{a+b}{2}-t))=r(\beta _{1}'(t)).
\]
The change of variable $u=\frac{3a+b}{2}-s$ yields
\[
\int_{a}^{\frac{3a+b}{2}-t}g(u(s))ds=\int_{t}^{\frac{a+b}{2}}
g(u(\frac{3a+b}{2}-s))ds,
\]
hence
\[
r(\beta _{1}'(t))=\int_{t}^{\frac{a+b}{2}}g(\beta
_{1}(s))ds\quad \forall t\in [a,\frac{a+b}{2}]
\]
and
\[
-(r(\beta _{1}'(t)))'=g(\beta _{1}(t))\quad \forall t\in [a,\frac{a+b}{2}]
\]
The proof is similar for all $t\in [\frac{a+b}{2},b]$.
\end{proof}

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\end{document}