\documentclass[twoside]{article}
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\markboth{Strongly nonlinear degenerated elliptic unilateral problems }
{ Y. Akdim, E. Azroul, \& A. Benkirane}
\begin{document}
\setcounter{page}{25}
\title{\vspace{-1in}\parbox{\linewidth}{\footnotesize\noindent
2002-Fez conference on Partial Differential Equations,\newline
Electronic Journal of Differential Equations,
Conference 09, 2002, pp 25--39. \newline
http://ejde.math.swt.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.swt.edu (login: ftp)}
\vspace{\bigskipamount} \\
%
Strongly nonlinear degenerated elliptic unilateral problems
via convergence of truncations
%
\thanks{ {\em Mathematics Subject Classifications:} 35J15, 35J70, 35J85.
\hfil\break\indent
{\em Key words:} Weighted Sobolev spaces, Hardy inequality,
variational ineqality, \hfil\break\indent
strongly nonlinear degenerated elliptic operators, truncations.
\hfil\break\indent
\copyright 2002 Southwest Texas State University. \hfil\break\indent
Published December 28, 2002. } }
\date{}
\author{Youssef Akdim, Elhoussine Azroul, \& Abdelmoujib Benkirane}
\maketitle
\begin{abstract}
We prove an existence theorem for a strongly nonlinear
degenerated elliptic inequalities involving nonlinear
operators of the form $Au+g(x,u,\nabla u)$.
Here $A$ is a Leray-Lions operator, $g(x,s,\xi)$ is a
lower order term satisfying some natural growth with respect
to $|\nabla u|$. There is no growth restrictions with respect to
$|u|$, only a sign condition. Under the assumption that the
second term belongs to $W^{-1,p'}(\Omega,w^*)$, we obtain
the main result via strong convergence of truncations.
\end{abstract}
\numberwithin{equation}{section}
\newtheorem{lem}{Lemma}[section]
\newtheorem{thm}{Theorem}[section]
\newtheorem{Def}{Definition}[section]
\newtheorem{rem}{Remark}[section]
\section{Introduction}
Let $\Omega$ be a bounded open set of $\mathbb{R}^N$ and $p$ a
real number such that $1
0$ independent of $u$.
Moreover, the imbedding
$$
X\hookrightarrow\hookrightarrow L^q(\Omega, \sigma), \eqno{(2.13)}
$$
expressed by the inequality (2.12) is compact.
Note that $(X,\||.|\|_X)$ is a uniformly convex (and thus reflexive)
Banach space.
\begin{rem} \label{rmk2.1} \rm
If we assume that $w_0(x)\equiv 1$ and in addition the integrability
condition: There exists $\nu\in ]\frac{N}{p},\infty[\cap
[\frac{1}{p-1},\infty[$ such that
$$ w_i^{-\nu}\in L^1(\Omega) $$
for all $i=1,\dots ,N$
(which is stronger than (2.2)). Then
$$
\||u|\|_X=\Big(\sum_{i=1}^N\int_\Omega |\frac{\partial
u(x)}{\partial x_i}|^p w_i(x)\,dx\Big)^{1/p}
$$
is a norm defined on $W_0^{1,p}(\Omega,w)$ and is equivalent to
(2.3). Moreover
$$
W_0^{1,p}(\Omega,w)\hookrightarrow\hookrightarrow L^q(\Omega),
$$
for all $1\leq q0,\ \mbox{ for all }
\xi\not=\eta\in \mathbb{R}^N,\tag{2.15}
\\
a(x,s, \xi).\xi\geq\alpha \sum_{i=1}^Nw_i|\xi_i|^{p},\tag{2.16}
\end{gather}
where $k(x)$ is a positive function in $L^{p'}(\Omega)$ and
$\alpha,\beta$ are strictly positive constants.
\paragraph{Assumption (H3)}
Let $g(x,s,\xi)$ be a Carath\'eodory function satisfying the following
assumptions:
\begin{gather}
g(x,s,\xi)s\geq 0\tag{2.17}\\
|g(x,s,\xi)|\leq b(|s|)\Big(\sum_{i=1}^Nw_i|\xi_i|^{p}+c(x)\Big),
\tag{2.18}
\end{gather}
where $b:\mathbb{R}^+\to \mathbb{R}^+$ is a continuous increasing function
and
$c(x)$ is a positive function which lies in $L^1(\Omega)$.
Now we recall some lemmas introduced in $\cite{akazbe}$ which will be
used later.
\begin{lem}[cf. \cite{akazbe}] \label{lem1}
Let $g\in L^r(\Omega,\gamma)$ and let $ g_n\in L^r(\Omega,\gamma)$,
with
$\|g_n\|_{r,\gamma} \leq c \ \ \ (10,\\
\mbox{for all }v\in W_0^{1,p}(\Omega,w) \quad v\geq\psi\mbox{ a.e.}\\
u\in W_0^{1,p}(\Omega,w)\quad u\geq\psi\mbox{ a.e.}\\
g(x,u,\nabla u)\in L^1(\Omega).
\end{gathered} \eqno(3.4) %tildeP
$$
Conversely, if $u$ is a solution of (3.4) then $u$ is a
solution of (3.3).
\end{lem}
The proof of this lemma is similar to the proof of
\cite[Remark 2.2]{beel} for the non weighted case.
\paragraph{Proof of theorem $\ref{thm1}$}
{\bf Step (1) The approximate problem and a priori estimate.}
Let $\Omega_\varepsilon$ be a sequence of compact subsets of $\Omega$
such
that $\Omega_\varepsilon$ increases to $\Omega$ as $\varepsilon
\to 0$.
We consider the sequence of approximate problems,
$$
\begin{gathered}
\langle Au_\varepsilon,v-u_\varepsilon\rangle +\int_\Omega
g_\varepsilon(x,u_\varepsilon,\nabla
u_\varepsilon)(v-u_\varepsilon)\,dx\geq
\langle f,v-u_\varepsilon\rangle \\
v\in W_0^{1,p}(\Omega,w)\quad v\geq\psi\mbox{ a.e.}\\
u_\varepsilon\in W_0^{1,p}(\Omega,w)\quad u_\varepsilon\geq\psi
\mbox{ a.e.}\\
\end{gathered} \eqno{(3.5)} %Pe
$$
where,
$$g_\varepsilon(x,s,\xi)=\frac{g(x,s,\xi)}{1+\varepsilon|g(x,s,\xi)|}
\chi_{\Omega_\varepsilon}(x),$$
and where $\chi_{\Omega_\varepsilon}$ is the characteristic function of
$\Omega_\varepsilon$.
Note that $g_\varepsilon(x,s,\xi)$ satisfies the following conditions,
$$
g_\varepsilon(x,s,\xi)s\geq 0,\quad |g_\varepsilon(x,s,\xi)|\leq
|g(x,s,\xi)|\quad \mbox{and}\quad | g_\varepsilon(x,s,\xi)|\leq
\frac{1}{\varepsilon}.
$$
We define the operator $G_\varepsilon:\ X\to X^*$ by,
$$\langle G_\varepsilon u,v\rangle =\int_\Omega
g_\varepsilon(x,u,\nabla
u)v\,dx.$$
Thanks to H\"older's inequality we have for all $u\in X$ and $v \in X$,
$$
\begin{aligned}
|\int_\Omega g_\varepsilon(x,u,\nabla u)v\,dx|
\leq&\Big(\int_\Omega |g_\varepsilon(x,u,\nabla
u)|^{q'}\sigma^{-\frac{q'}{q}}\,dx\Big)^{1/q'}
\Big(\int_\Omega |v|^{q}\sigma\,dx\Big)^{1/q}\\
\leq& \frac{1}{\varepsilon}
\Big(\int_{\Omega_\varepsilon}\sigma^{1-q'}
\,dx\Big)^{1/q'}\|v\|_{q,\sigma}
\leq c_\varepsilon\||v|\|.
\end{aligned}
\eqno{(3.6)}
$$
The last inequality is due to (2.11) and (2.13).
\begin{lem}\label{lemexsoap}
The operator $B_\varepsilon =A+G_\varepsilon$ from $X$ into its dual
$X^*$ is pseudo-monotone.
Moreover, $B_\varepsilon$ is coercive, in the sense that:
There exists $v_0\in K_\psi$ such that
$$
\frac{\langle B_\varepsilon v, v-v_0\rangle }{\||v|\|}\to
+\infty \quad \mbox{as } \||v|\|\to \infty,\quad v\in K_\psi.
$$
\end{lem}
The proof of this lemma will be presented below.
In view of lemma $\ref{lemexsoap}$,
(3.5) has a solution by the classical result (cf.
Theorem 8.1 and Theorem 8.2 chapter 2 $\cite{li}$).
With $v=\psi^+$ as test function in (3.5), we
deduce that
$\int_\Omega g_\varepsilon(x,u_\varepsilon,\nabla
u_\varepsilon)(u_\varepsilon-\psi^+)\geq 0$,
then, $\langle Au_\varepsilon,u_\varepsilon\rangle \leq \langle
f,u_\varepsilon-\psi^+\rangle +\langle Au_\varepsilon,\psi^+\rangle$,
i.e.,
$$\int_\Omega a(x,u_\varepsilon,\nabla u_\varepsilon)\nabla
u_\varepsilon\,dx\leq \langle f,u_\varepsilon-\psi^+\rangle
+\sum_{i=1}^N\int_\Omega a_i(x,u_\varepsilon,\nabla
u_\varepsilon)\frac{\partial \psi^+}{\partial x_i}\,dx,$$
then,
\begin{eqnarray*}
\lefteqn{\alpha\sum_{i=1}^N\int_\Omega w_i|\frac{\partial
u_\varepsilon}{\partial
x_i}|^p\,dx}\\
&=&\alpha\||u_\varepsilon|\|^p\\
&\leq& \|f\|_{X^*}(\||u_\varepsilon|\|+\||\psi^+|\|)+\\
&& +\sum_{i=1}^N\Big(\int_\Omega|a_i(x,u_\varepsilon,\nabla
u_\varepsilon)|^{p'}w_i^{1-p'}\,dx\Big)^{\frac{1}{p'}}
\Big(\int_\Omega |\frac{\partial \psi^+}{\partial x_i}|^pw_i\,dx
\Big)^{1/p}\\
&\leq&\|f\|_{X^*}(\||u_\varepsilon|\|+\||\psi^+|\|)+\\
&&+c\sum_{i=1}^N\Big(\int_\Omega
(k^{p'}+|u_\varepsilon|^q\sigma+\sum_{j=1}^{N}|\frac{\partial
u_\varepsilon}{\partial
x_j}|^pw_j)\,dx\Big)^{1/p'}\||\psi^+|\|.
\end{eqnarray*}
Using (2.13) the last inequality becomes,
$$\alpha \||u_\varepsilon|\|^p\leq c_1
\||u_\varepsilon|\|+c_2\||u_\varepsilon|\|^{\frac{q}{p'}}
+c_3\||u_\varepsilon|\|^{p-1}+c_4,
$$
where $c_i$ are various positive constants.
Then, thanks to (2.10) we can deduce that $u_\varepsilon$ remains
bounded in $W_0^{1,p}(\Omega,w)$, i.e.,
$$\||u_\varepsilon|\|\leq \beta_0,\eqno{(3.7)}
$$
where $\beta_0$ is some positive constant.
Extracting a subsequence (still denoted by $u_\varepsilon$) we get
$$
u_\varepsilon \rightharpoonup u\quad \mbox{weakly in $X$ and a.e. in }
\Omega.
$$
Note that $u\geq \psi$ a.e.
\noindent{\bf Step (2) Strong convergence of $T_k(u_\varepsilon)$.}
Thanks to (3.7) and (2.13) we can extract a subsequence still denoted
by $u_\varepsilon$ such that
$$
\begin{gathered}
u_\varepsilon\rightharpoonup u \quad \mbox{weakly in } W_0^{1,p}(\Omega,w,)\\
u_\varepsilon\to u\quad\mbox{a.e.\ in }\Omega.
\end{gathered}
\eqno{(3.8)}
$$
Let $k>0$ by lemma $\ref{lem4}$ we have
$$
T_k(u_\varepsilon) \rightharpoonup T_k(u)\quad \mbox{weakly in
}W_0^{1,p}(\Omega,w) \mbox{ as }\varepsilon\to 0.\eqno{(3.9)}
$$
Our objective is to prove that
$$T_k(u_\varepsilon)\to T_k(u)\quad
\mbox{strongly in }W_0^{1,p}(\Omega,w) \mbox{ as }\varepsilon\to
0.\eqno{(3.10)}
$$
Fix $k>\|\psi^+\|_\infty$, and use the notation
$z_\varepsilon=T_k(u_\varepsilon)-T_k(u)$.
We use, as a test function in (3.5),
$$
v_\varepsilon=u_\varepsilon-\eta\varphi_\lambda(z_\varepsilon)
\eqno{(3.11)}
$$
where $\varphi_\lambda(s)=se^{\lambda s^2}$ and
$\eta=e^{-4\lambda k^2}$.
Then we can check that $v_\varepsilon$ is admisible test function.
So that
$$-\langle Au_\varepsilon,\eta \varphi_\lambda z_\varepsilon \rangle
-\int_\Omega g_\varepsilon(x,u_\varepsilon,\nabla
u_\varepsilon)\eta\varphi_\lambda(z_\varepsilon)\,dx\geq -\langle
f,\eta\varphi_\lambda(z_\varepsilon)\rangle
$$
which implies that
$$\langle Au_\varepsilon,\varphi_\lambda (z_\varepsilon) \rangle
+\int_\Omega g_\varepsilon(x,u_\varepsilon,\nabla
u_\varepsilon)\varphi_\lambda(z_\varepsilon)\,dx\leq \langle
f,\varphi_\lambda(z_\varepsilon)\rangle.\eqno{(3.12)}$$
Since $\varphi_\lambda(z_\varepsilon)$ is bounded in $X$ and converges
a.e.\ in $\Omega$ to zero and using (2.13), we have
$\varphi_\lambda(z_\varepsilon) \rightharpoonup 0 $ weakly in $X$ as
$\varepsilon\to 0$. Then
$$\eta_1(\varepsilon)=\langle
f,\varphi_\lambda(z_\varepsilon)\rangle\to
0,\eqno{(3.13)}$$
and since $g_\varepsilon(x,u_\varepsilon,\nabla
u_\varepsilon)\varphi_\lambda(z_\varepsilon) \geq 0$ in the subset
$\{x\in \Omega : |u_\varepsilon(x)| \geq k\}$ hence (3.12) and
(3.13) yield
$$\langle
Au_\varepsilon,\varphi_\lambda(z_\varepsilon)\rangle
+\int_{\{|u_\varepsilon|\leq
k\}}g_\varepsilon(x,u_\varepsilon,\nabla
u_\varepsilon)\varphi_\lambda(z_\varepsilon)\,dx \leq
\eta_1(\varepsilon).\eqno{(3.14)}
$$
We study each term in the left hand side of (3.14). We have,
$$
\begin{aligned}
\langle Au_\varepsilon,
\varphi_\lambda(z_\varepsilon)\rangle
=& \int_\Omega a(x,u_\varepsilon,\nabla
u_\varepsilon)\nabla(T_k(u_\varepsilon)-T_k(u))
\varphi_\lambda'(z_\varepsilon)\,dx\\
=& \int_\Omega a(x,T_k(u_\varepsilon),\nabla
T_k(u_\varepsilon))\nabla(T_k(u_\varepsilon)-T_k(u))\varphi_\lambda'(
z_\varepsilon)\, dx\\
&-\int_{\{|u_\varepsilon|>k\}}
a(x,u_\varepsilon,\nabla
u_\varepsilon)\nabla T_k(u)\varphi_\lambda'(z_\varepsilon)\,dx\\
=& \int_\Omega \left(a(x,T_k(u_\varepsilon),\nabla
T_k(u_\varepsilon))-a(x,T_k(u_\varepsilon),\nabla
T_k(u))\right)\nabla(T_k(u_\varepsilon)\\
&-T_k(u))\varphi_\lambda'(z_\varepsilon)\,dx
+\eta_2(\varepsilon),
\end{aligned}\eqno{(3.15)}
$$
where,
$$
\begin{aligned}
\eta_2(\varepsilon)=&\int_\Omega a(x,T_k(u_\varepsilon),\nabla
T_k(u))\nabla(T_k(u_\varepsilon)-T_k(u))\varphi_\lambda'(
z_\varepsilon)\, dx\\
&-\int_{\{|u_\varepsilon|>k\}}a(x,u_\varepsilon,\nabla
u_\varepsilon)\nabla T_k(u)\varphi_\lambda'(z_\varepsilon)\,dx,
\end{aligned}
$$
which converges to 0 as $\varepsilon\to 0$.
On the other hand,
$$
\begin{aligned}
|&\int_{\{|u_\varepsilon|\leq k\}}
g_\varepsilon (x,u_\varepsilon,\nabla
u_\varepsilon)\varphi_\lambda(z_\varepsilon)\,dx|\\
\leq& \int_{\{|u_\varepsilon|\leq
k\}}b(k)[c(x)+\sum_{i=1}^N|\frac{\partial u_\varepsilon}{\partial
x_i}|^pw_i]|\varphi_\lambda(z_\varepsilon)|\,dx\\
\leq& b(k)\int_{\{|u_\varepsilon|\leq k\}} c(x)
|\varphi_\lambda(z_\varepsilon)|\;
dx+\frac{b(k)}{\alpha}\int_{\{|u_\varepsilon|\leq k\}}
a(x,u_\varepsilon,\nabla u_\varepsilon)\nabla u_\varepsilon
|\varphi_\lambda(z_\varepsilon)|\,dx\\
=& \eta_3(\varepsilon) +\frac{b(k)}{\alpha}\int_\Omega
a(x,T_k(u_\varepsilon),\nabla T_k(u_\varepsilon))\nabla
T_k(u_\varepsilon) |\varphi_\lambda(z_\varepsilon)|\,dx\\
=& \frac{b(k)}{\alpha}\int_\Omega
\left(a(x,T_k(u_\varepsilon),\nabla
T_k(u_\varepsilon))-a(x,T_k(u_\varepsilon),\nabla T_k(u))\right)\nabla
(T_k(u_\varepsilon)\\
&-T_k(u)) |\varphi_\lambda(z_\varepsilon)|\,dx
+\eta_4(\varepsilon)
\end{aligned}\eqno{(3.16)}
$$
where
$$\eta_3(\varepsilon)= b(k)\int_{\{|u_\varepsilon|\leq k\}}
c(x) |\varphi_\lambda(z_\varepsilon)|\,dx\to 0\mbox{ as
}\varepsilon \to 0$$
and
\begin{eqnarray*}
\eta_4(\varepsilon)&=&\eta_3(\varepsilon)+\frac{b(k)}{\alpha}\int_\Omega
a(x,T_k(u_\varepsilon),\nabla T_k(u))\nabla (T_k(u_\varepsilon)-T_k(u))
|\varphi_\lambda(z_\varepsilon)|\,dx\\
& &+\frac{b(k)}{\alpha}\int_\Omega a(x,T_k(u_\varepsilon),\nabla
T_k(u_\varepsilon))\nabla T_k(u) |\varphi_\lambda(z_\varepsilon)|\;
dx\to 0 \quad\mbox{as }\varepsilon \to 0.
\end{eqnarray*}
Note that, when $\lambda\geq\left(\frac{b(k)}{2\alpha}\right)^2$
we have $$\varphi_\lambda'(s)-\frac{b(k)}{\alpha}|\varphi(s)|\geq
\frac{1}{2}.$$
Which combining with (3.14),(3.15) and (3.16) one obtains
\begin{multline*}
\int_\Omega\big(a(x,T_k(u_\varepsilon),\nabla
T_k(u_\varepsilon))-
a(x,T_k(u_\varepsilon),\nabla T_k(u))\big)\nabla
(T_k(u_\varepsilon)-T_k(u)) \,dx\\
\leq \eta_5(\varepsilon)=2(\eta_1(\varepsilon)-\eta_2(\varepsilon)
+\eta_4(\varepsilon))\to 0\quad \mbox{as }\varepsilon \to 0.
\end{multline*}
Finally lemma $\ref{lem5}$ implies (3.10) for any fixed $k\geq
\|\psi\|_\infty$.
\noindent {\bf Step (3) Passage to the limit.}
In view of (3.10) we have for a subsequence,
$$
\nabla u_\varepsilon\to \nabla u\quad\mbox{a.e.\ in }
\Omega,\eqno{(3.17)}
$$
which with (3.8) imply,
$$
\begin{gathered}
a(x,u_\varepsilon,\nabla u_\varepsilon)\to a(x,u,\nabla u)
\quad\mbox{a.e.\ in }\Omega,\\
g_\varepsilon(x,u_\varepsilon,\nabla u_\varepsilon)\to
g(x,u,\nabla u)\mbox{a.e.\ in }\Omega,\\
g_\varepsilon(x,u_\varepsilon,\nabla
u_\varepsilon)u_\varepsilon\to g(x,u,\nabla u)u\quad
\mbox{a.e.\ in }\Omega.
\end{gathered}
\eqno{(3.18)}
$$
On the other hand, thanks to (2.14) and (3.7) we have
$a(x,u_\varepsilon,\nabla u_\varepsilon)$ is bounded in
$\prod_{i=1}^NL^{p'}(\Omega,w_i^*)$
then by lemma \ref{lem1} we obtain
$$
a(x,u_\varepsilon,\nabla u_\varepsilon)\rightharpoonup a(x,u,\nabla
u)\quad
\mbox{weakly in }\prod_{i=1}^NL^{p'}(\Omega,w_i^*).\eqno{(3.19)}
$$
We shall prove that,
$$ g_\varepsilon(x,u_\varepsilon,\nabla u_\varepsilon)\to
g(x,u,\nabla u)\quad\mbox{ strongly in }\ L^1(\Omega).\eqno{(3.20)}
$$
By (3.18), to apply Vitali's theorem it suffices to prove
that
$g_\varepsilon(x,u_\varepsilon,\nabla u_\varepsilon)$ is uniformly
equi-integrable. Indeed,
thanks to (2.17), (3.6) and (3.7) we obtain,
$$0\leq \int_\Omega g_\varepsilon(x,u_\varepsilon,\nabla
u_\varepsilon)u_\varepsilon\,dx\leq c_0,\eqno{(3.21)}$$
where $c_0$ is some positive constant.
For any measurable subset $E$ of $\Omega$ and any $m>0$ we have,
$$\int_E |g_\varepsilon(x,u_\varepsilon,\nabla
u_\varepsilon)|\,dx=\int_{E\cap
X_m^\varepsilon}|g_\varepsilon(x,u_\varepsilon,\nabla
u_\varepsilon)|\,dx+\int_{E\cap Y_m^\varepsilon}
|g_\varepsilon(x,u_\varepsilon,\nabla u_\varepsilon)|\,dx$$
where,
$$
X_m^\varepsilon=\{x\in \Omega,\ |u_\varepsilon(x)|\leq m\},\quad
Y_m^\varepsilon=\{x\in \Omega,\ |u_\varepsilon(x)|> m\}.
$$
From these expressions, (2.18), and (3.21), we have
$$
\begin{aligned}
\int_E& |g_\varepsilon(x,u_\varepsilon,\nabla
u_\varepsilon)|\,dx\\
=& \int_{E\cap X_m^\varepsilon}|g_\varepsilon(x,u_\varepsilon,\nabla T_m
(u_\varepsilon))|\,dx+\int_{E\cap Y_m^\varepsilon}
|g_\varepsilon(x,u_\varepsilon,\nabla u_\varepsilon)|\,dx\\
\leq& \int_{E\cap
X_m^\varepsilon}|g_\varepsilon(x,u_\varepsilon,\nabla T_m
(u_\varepsilon))|\,dx+\frac{1}{m}\int_\Omega
g_\varepsilon(x,u_\varepsilon,\nabla u_\varepsilon)u_\varepsilon\,dx\\
\leq & b(m)\int_E (\sum_{i=1}^{N}w_i|\frac{\partial
T_m(u_\varepsilon)}{\partial
x_i}|^p+c(x))+\frac{c_0}{m}.
\end{aligned}\eqno{(3.22)}
$$
Since the sequence ($\nabla T_m(u_\varepsilon)$) strongly converges in
$\prod_{i=1}^NL^p(\Omega,w_i),$ then (3.22) implies the
equi-integrability
of $g_\varepsilon(x,u_\varepsilon,\nabla u_\varepsilon)$.
Moreover, since $g_\varepsilon(x,u_\varepsilon,\nabla
u_\varepsilon)u_\varepsilon\geq 0$ a.e.\ in $\Omega$, then by
(3.18), (3.21) and Fatou's lemma, we have
$g(x,u,\nabla u)u\in L^1(\Omega)$.
On the other hand, for $v\in L^\infty(\Omega)$, set $h=k+\|v\|_\infty$,
then
\begin{eqnarray*}
|\frac{\partial T_k(v-u_\varepsilon)}{\partial x_i}|w_i^{1/p}
&=& \chi_{\{|v-u_\varepsilon|\leq k\}}|\frac{\partial v}{\partial
x_i}-\frac{\partial u_\varepsilon}{\partial x_i}|w_i^{1/p}\\
&\leq& \chi_{\{|u_\varepsilon|\leq h\}}|\frac{\partial v}{\partial
x_i}-\frac{\partial u_\varepsilon}{\partial x_i}|w_i^{1/p}\\
&\leq& |\frac{\partial v}{\partial
x_i}|w_i^{1/p}+|\frac{\partial
T_h(u_\varepsilon)}{\partial x_i}|w_i^{1/p}
\end{eqnarray*}
which implies, using Vitali's theorem with (3.10) and (3.17) that
$$ \nabla T_k(v-u_\varepsilon)\to \nabla T_k(v-u)\quad
\mbox{strongly in } \prod_{i=1}^NL^p(\Omega,w_i)\eqno{(3.23)}
$$
for any $v\in W_0^{1,p}(\Omega,w)\cap L^\infty(\Omega)$.
Thanks to lemma $\ref{lemeqso}$ and from (3.19), (3.20) and (3.23) we
can
pass to the limit in
$$\langle Au_\varepsilon,T_k(v-u_\varepsilon)\rangle+\int_\Omega
g_\varepsilon (x,u_\varepsilon,\nabla
u_\varepsilon)T_k(v-u_\varepsilon)\geq\langle
f,T_k(v-u_\varepsilon)\rangle$$
and we obtain,
$$
\langle Au,T_k(v-u)\rangle+\int_\Omega g(x,u,\nabla
u)T_k(v-u)\geq\langle f,T_k(v-u)\rangle \eqno{(3.24)}
$$
for any $v\in W_0^{1,p}(\Omega,w)\cap L^\infty(\Omega)$
and for all $k>0$.
Taking for any $v\in W_0^{1,p}(\Omega,w)$ and $v\geq \psi$ the test
function $T_m(v)$ which belongs to
$W_0^{1,p}(\Omega,w)\cap L^\infty(\Omega) $
for $m\geq \|\psi^+\|_\infty$ and passing to the limit in (3.24) as
$m\to \infty$, then $u$ is a solution of (3.4).
Using again lemma \ref{lemeqso} we obtain the desired result, i.e.,
$u$ is a solution of (3.3).
\paragraph{Proof of lemma $\ref{lemexsoap}$}
By proposition 2.6 chapter 2 $\cite{li}$, it is sufficient to show
that
$B_\varepsilon$ is of the calculus of variations type in the sense of
definition $\ref{defcava}$. Indeed put,
$$
b_1(u,v,\tilde w)=\sum_{i=1}^N\int_\Omega a_i(x,u,\nabla v)\nabla
\tilde w\,dx,\quad
b_2(u,\tilde w)=\int_\Omega g_\varepsilon(x,u,\nabla u)\tilde
w\,dx.
$$
Then the mapping $\tilde w\mapsto b_1(u,v,\tilde w)+b_2(u,\tilde w)$ is
continuous in $X$. Then
$$ b_1(u,v,\tilde w)+b_2(u,\tilde w)=b(u,v,\tilde
w)=\langle B_\varepsilon(u,v),\tilde w\rangle ,\quad
B_\varepsilon(u,v)\in W^{-1,p'}(\Omega,w^*)
$$
and we have
$$B_\varepsilon(u,u)=B_\varepsilon u.
$$
Using (2.14) and H\"older's inequality we can show that $A$ is bounded
as in \cite{drkumu}, and thanks to (3.6)
$B_\varepsilon$ is bounded. Then, it is sufficient to check
(2.6)-(2.9).
Next we show that (2.6) and (2.7) are true.
By (2.15) we have,
$$ (B_\varepsilon(u,u)-B_\varepsilon(u,v),
u-v)=b_1(u,u,u-v)-b_1(u,v,u-v)\geq 0.
$$
The operator $v\to B_\varepsilon(u,v)$ is bounded
hemicontinuous. Indeed, we have
$$ a_i(x,u,\nabla(v_1+\lambda v_2))\to a_i(x,u,\nabla
v_1)\quad \mbox{
strongly in }L^{p'}(\Omega,w_i^*)\mbox{ as }\lambda\to
0.\eqno{(3.25)}
$$
On the other hand, $\left(g_\varepsilon(x,u_1+\lambda
u_2,\nabla(u_1+\lambda u_2))\right)_\lambda$ is bounded in
$L^{q'}(\Omega,\sigma^{1-q'})$
and $g_\varepsilon(x,u_1+\lambda u_2,\nabla(u_1+\lambda
u_2))\to
g_\varepsilon(x,u_1,\nabla u_1)\ \ \ a.e.$ in $\Omega$, hence lemma
\ref{lem1} gives
$$\begin{gathered}
g_\varepsilon(x,u_1+\lambda u_2,\nabla(u_1+\lambda
u_2))\rightharpoonup
g_\varepsilon(x,u_1,\nabla u_1)\\
\mbox{ weakly in }L^{q'}(\Omega,\sigma^{1-q'})\mbox{ as }\lambda \to 0.
\end{gathered}\eqno{(3.26)}
$$
Using (3.25) and (3.26) we can write
$$b(u,v_1+\lambda v_2,\tilde w)\to b(u,v_1,\tilde w)\ \
\mbox{
as }\lambda \to 0 \ \ \forall u,v_i,\tilde w\in X.$$
Similarly we can prove (2.7).
\noindent Proof of assertion (2.8).
Assume that $u_n\rightharpoonup u$ weakly in $X$ and
$(B(u_n,u_n)-B(u_n,u),u_n-u)\to 0$.
We have,
\begin{multline*}
(B(u_n,u_n)-B(u_n,u),u_n-u)\\
= \sum_{i=1}^N\int_\Omega \left(a_i(x,u_n,\nabla u_n)-a_i(x,u_n,\nabla
u)\right)\nabla(u_n-u)\,dx\to 0,
\end{multline*}
then, by lemma $\ref{lem5}$,
$u_n\to u$ strongly in $X$,
which gives
$$ b(u_n,v,\tilde w)\to b(u,v,\tilde w)\quad \forall \tilde
w\in X,
$$
i.e.,
$B_\varepsilon(u_n,v)\rightharpoonup B_\varepsilon(u,v)
\quad \mbox{weakly in }X^*$.
It remains to prove (2.9).
Assume that
$$u_n\rightharpoonup u\quad\mbox{ weakly in }X\eqno{(3.27)}
$$
and that
$$B(u_n,v)\rightharpoonup \psi\quad\mbox{weakly in }X^*.\eqno{(3.28)}
$$
Thanks to (2.13), (2.14) and (3.27) we obtain,
$$a_i(x,u_n,\nabla v)\to a_i(x,u,\nabla v)\quad
\mbox{in }L^{p'}(\Omega,w_i^*)\mbox{ as } n\to \infty,
$$
then,
$$b_1(u_n,v,u_n)\to b_1(u,v,u).\eqno{(3.29)}
$$
On the other hand, by H\"older's inequality,
\begin{align*}
|b_2(u_n,u_n-u)|\leq &\left(\int_\Omega|g_\varepsilon(x,u_n,\nabla
u_n)|^{q'}\sigma^{\frac{-q'}{q}}\;
dx\right)^{1/q'}\left(\int_\Omega|u_n-u|^q\sigma\,dx
\right)^{1/q}\\
\leq &\frac{1}{\varepsilon}\left(\int_{\Omega_\varepsilon}\sigma^{\frac{-q'}{q}}\;
dx\right)^{1/q'}\|u_n-u\|_{L^q(\Omega,\sigma)}\to
0\quad \mbox{as }n\to \infty,
\end{align*}
i.e.,
$$b_2(u_n,u_n-u)\to 0\mbox{ as }n\to
\infty,\eqno{(3.30)}
$$
but in view of (3.28) and (3.29) we obtain
$$b_2(u_n,u)=(B_\varepsilon(u_n,v),u)-b_1(u_n,v,u)\to
(\psi,u)-b_1(u,v,u)
$$
and from (3.30) we have
$b_2(u_n,u_n)\to (\psi,u)-b_1(u,v,u)$.
Then,
$$(B_\varepsilon(u_n,v),u_n)=b_1(u_n,v,u_n)+b_2(u_n,u_n)\to
(\psi,u).
$$
Now show that $B_\varepsilon$ is coercive. Let $v_0\in K_\psi$.
From H\"older's inequality, the growth condition (2.14) and the compact
imbedding (2.13) we have
\begin{eqnarray*}
\langle Av,v_0\rangle&=&\sum_{i=1}^N \int_\Omega a_i(x,v,\nabla
v)\frac{\partial v_0}{\partial x_i}\,dx\\
&\leq& \sum_{i=1}^N \Big(\int_\Omega |a_i(x,v,\nabla
v)|^{p'}w_i^{\frac{-p'}{p}}\,dx\Big)^{\frac{1}{p'}}
\Big(\int_\Omega
|\frac{\partial v_0}{\partial x_i}|^pw_i\,dx\Big)^{1/p}\\
&\leq & c_1\||v_0|\|\Big( \int_\Omega
k(x)^{p'}+|v|^q\sigma+\sum_{j=1}^N
|\frac{\partial v}{\partial x_j}|^pw_j \,dx\Big)^{\frac{1}{p'}}\\
&\leq & c_2(c_3+\||v|\|^{\frac{q}{p'}}+\||v|\|^{p-1}),
\end{eqnarray*}
where $c_i$ are various constants.
Thanks to (2.16), we obtain
$$ \frac{\langle Av,v\rangle}{\||v|\|}-\frac{\langle
Av,v_0\rangle}{\||v|\|}
\geq \alpha \||v|\|^{p-1}-\||v|\|^{p-2}-
\||v|\|^{\frac{q}{p'}-1}-\frac{c}{\||v|\|}.
$$
In view of (2.10) we have $p-1>\frac{q}{p'}-1$.
Then,
$$\frac{\langle Av,v-v_0\rangle}{\||v|\|}\to \infty
\quad \mbox{as }\||v|\|\to \infty.
$$
Since $\langle G_\varepsilon v,v \rangle \geq 0$ and $\langle
G_\varepsilon v,v_0 \rangle$ is bounded, we have
$$
\frac{\langle B_\varepsilon v,v-v_0\rangle}{\||v|\|}
\geq \frac{\langle Av,v-v_0\rangle}{\||v|\|}-\frac{\langle
G_\varepsilon
v,v_0\rangle}{\||v|\|}\to \infty \quad\mbox{as }\||v|\|\to \infty.
$$
\begin{rem} \label{rmk3.2} \rm
Assumption (2.10) appears to be necessary to prove the
boundedness of $(u_\varepsilon)_\varepsilon$ in $W_0^{1,p}(\Omega,w)$
and the coercivity of the operator $B_\varepsilon$. While Assumption
(2.11) is necessary to prove the boundedness of $G_\varepsilon$ in
$W_0^{1,p}(\Omega,w)$. Thus, when $g\equiv 0$, we don't need to
assume (2.11).
\end{rem}
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\noindent\textsc{Youssef Akdim} (e-mail: y.akdim1@caramail.com)\\
\textsc{Elhoussine Azroul} (e-mail: elazroul@caramail.com)\\
\textsc{Abdelmoujib Benkirane} (e-mail: abenkirane@fsdmfes.ac.m)\\[2pt]
D\'epartement de Math\'ematiques et Informatique\\
Facult\'e des Sciences Dhar-Mahraz\\
B.P 1796 Atlas F\`es, Maroc
\end{document}