\documentclass[reqno]{amsart}
\usepackage{hyperref}
\AtBeginDocument{{\noindent\small \emph{Electronic Journal of
Differential Equations}, Vol. 2011(2011), No. 03, pp. 1--26.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or
http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2011 Texas State University - San Marcos.}
\vspace{9mm}}
\begin{document}
\title[\hfilneg EJDE-2011/03\hfil Renormalized solutions]
{Solvability of degenerated parabolic equations without sign
condition and three unbounded nonlinearities}
\author[Y. Akdim, J. Bennouna, M. Mekkour \hfil EJDE-2011/03\hfilneg]
{Youssef Akdim, Jaouad Bennouna, Mounir Mekkour} % in alphabetical order
\address{D\'epartement de Math\'ematiques,
Facult\'e des Sciences Dhar-Mahraz, F\`es, Morocco} \email[Y.
Akdim]{akdimyoussef@yahoo.fr} \email[J.
Bennouna]{jbennouna@hotmail.com} \email[M.
Mekkour]{mekkour.mounir@yahoo.fr}
\thanks{Submitted June 28, 2010. Published January 4, 2011.}
\subjclass[2000]{A7A15, A6A32, 47D20} \keywords{Weighted Sobolev
spaces; truncations; time-regularization;\hfill\break\indent
renormalized solutions}
\begin{abstract}
In this article, we study the problem
\begin{gather*}
\frac{\partial}{\partial t} b(x, u)-\operatorname{div}(a(x,t,u,D u))
+H(x,t,u,Du) = f\quad \text{in } \Omega\times ]0,T[,\\
b(x,u)(t=0)=b(x,u_0)\quad\text{in } \Omega,\\
u=0\quad\text{in } \partial\Omega\times ]0,T[
\end{gather*}
in the framework of weighted Sobolev spaces, with $b(x,u)$
unbounded function on $u$. The main contribution of our work is to
prove the existence of a renormalized solution without the sign
condition and the coercivity condition on $H(x,t,u,Du)$. The
critical growth condition on $H$ is with respect to
$Du$ and no growth condition with respect to $u$.
The second term $f$ belongs to $L^1(Q)$, and
$b(x,u_0)\in L^1(\Omega)$.
\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}
\allowdisplaybreaks
\newcommand{\norm}[1]{\|#1\|}
\section{Introduction}
Let $\Omega$ be a bounded open set of $\mathbb{R}^N$, $p$ be a real
number such that $2
\max (N,p)\quad\text{such that } w_i^{\frac{-r_0}{r_0-p}} \in
L^1_{\rm loc}(\Omega), \\
w_i \in L^1_{\rm loc}(\Omega),\label{2.1} \\
w_i^{\frac{-1}{p-1}} \in L^1_{\rm loc}(\Omega),\label{2.2}
\end{gather}
for any $0\leq i\leq N$. We denote by $W^{1,p}(\Omega,w)$ the space
of real-valued functions $u \in L^p(\Omega,w_0)$ such that the
derivatives in the sense of distributions fulfill
$$
\frac{\partial u}{\partial x_i} \in L^p(\Omega, w_i)\quad \text{for } i=1, \dots, N.
$$
Which is a Banach space under the norm
\begin{equation}
\|u\|_{1,p,w}=\Big[\int_{\Omega}|u(x)|^pw_0(x)\,dx +
\sum_{i=1}^N\int_\Omega |\frac{\partial u(x)}{\partial x_i}|^p
w_i(x)\,dx \Big]^{1/p}.\label{2.3}
\end{equation}
Condition \eqref{2.1} implies that $C_0^{\infty}(\Omega)$ is a
space of $W^{1,p}(\Omega,w)$ and consequently, we can introduce the
subspace $V=W_0^{1,p}(\Omega,w)$ of $W^{1,p}(\Omega,w)$ as the
closure of $C_0^{\infty}(\Omega)$ with respect to the norm
\eqref{2.3}. Moreover, condition \eqref{2.2} implies that
$W^{1,p}(\Omega,w)$ as well as $W_0^{1,p}(\Omega,w)$ are reflexive
Banach spaces.
We recall that the dual space of weighted Sobolev spaces
$W_0^{1,p}(\Omega,w)$ is equivalent to $W^{-1, p'}(\Omega, w^{*})$,
where $w^{*} = \{w_i^{*} = w_i^{1-p'}$, $i = 0, \dots, N \}$ and
where $p'$ is the conjugate of $p$; i.e., $p' = \frac{p}{p-1}$, (see
\cite{Kufner}).
\section{Basic assumptions}
\subsection*{Assumption (H1)} For $2\leq p<\infty$, we assume that
the expression
\begin{equation}
\||u|\|_{V}=\Big(\sum_{i=1}^N\int_\Omega |\frac{\partial u(x)}
{\partial x_i}|^pw_i(x)\,dx\Big)^{1/p}\label{h2.6}
\end{equation}
is a norm defined on $V$ which is equivalent to the norm
\eqref{2.3}, and there exists a weight function $\sigma$ on $\Omega$
such that,
$$
\sigma\in L^1(\Omega)\quad\text{and } \sigma^{-1}\in L^1(\Omega).
$$
We assume also the Hardy inequality,
\begin{equation}
\Big(\int_\Omega|u(x)|^p\sigma \,dx\Big)^{1/q} \leq
c\Big(\sum_{i=1}^N\int_\Omega |\frac{\partial u(x)} {\partial
x_i}|^pw_i(x)\,dx\Big)^{1/p},\label{h2.7}
\end{equation}
holds for every $u\in V$ with a constant $c>0$ independent of $u$,
and moreover, the imbedding
\begin{equation}
W^{1,p}(\Omega,w)\hookrightarrow\hookrightarrow L^p(\Omega, \sigma),
\label{h2.8}
\end{equation}
expressed by the inequality \eqref{h2.7} is compact. Notice that
$(V,\||\cdot|\|_V)$ is a uniformly convex (and thus reflexive)
Banach space.
\begin{remark} \label{rmk3.1} \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
\begin{equation}
w_i^{-\nu} \in L^1(\Omega)\quad\text{and}\quad w_i^{\frac{N}{N-1}}
\in L^1_{\rm loc}(\Omega)
\quad \text{for all } i=1,\dots,N. \label{2.4}
\end{equation}
Notice that the assumptions \eqref{2.1} and \eqref{2.4} imply
\begin{equation}
\||u\|| = \Big({ \sum_{i=1}^N}{\int_{\Omega}|\frac{\partial u}
{\partial x_i}|^p w_i(x)\,dx}\Big)^{1/p},\label{2.5}
\end{equation}
which is a norm defined on $W_0^{1,p}(\Omega,w)$ and its equivalent
to \eqref{2.3} and that, the imbedding
\begin{equation}
W_0^{1,p}(\Omega,w) \hookrightarrow L^p(\Omega), \label{2.6}
\end{equation}
is compact for all $1\leq q\leq p_1^*$ if
$p\nu0$, there exists
$\lambda_k>0$ and functions $A_k\in L^1(\Omega)$ and $B_k\in
L^p(\Omega)$ such that
\begin{equation}
\lambda_k\leq\frac{\partial b(x,s)}{\partial s}\leq A_k(x)\quad
\text{and}\quad \Big|D_x\Big(\frac{\partial b(x,s)}{\partial
s}\Big)\Big|\leq B_k(x) \label{condbb}
\end{equation}
for almost every $x\in \Omega$, for every $s$ such that $|s| \leq k$
, we denote by $D_x\big(\frac{\partial b(x,s)}{\partial s}\big)$ the
gradient of $\frac{\partial b(x,s)}{\partial s}$ defined in the
sense of distributions. For $i=1,\dots ,N$,
\begin{equation}
|a_i(x, t,s, \xi)|\leq \beta w_i^{1/p}(x) [ k(x,t) +
\sigma^{1/p'}|s|^{q/p'}+ { \sum_{j=1}^N}w_j^{1/p'}(x)|\xi_j|^{p-1}]
,\label{2.7}
\end{equation}
for a.e. $(x,t)\in Q$,all $(s,\xi)\in \mathbb{R}\times\mathbb{R}^N$,
some function $k(x,t)\in L^{p'}(Q)$ and $\beta>0$. Here $\sigma$ and
$q$ are as in (H1).
\begin{gather}
[a(x,t,s,\xi) -a(x,t,s,\eta)](\xi-\eta) > 0 \quad \text{for all }
(\xi,\eta) \in \mathbb{R}^N\times\mathbb{R}^N, \label{2.8} \\
a(x, t,s, \xi).\xi \geq \alpha {\sum_{i=1}^N}w_i |\xi_i|^p,
\label{2.9}
\end{gather}
Where $\alpha$ is a strictly positive constant.
\subsection*{Assumption (H3)}
Furthermore, let $H(x,t,s,\xi): \Omega\times
[0,T]\times\mathbb{R}\times\mathbb{R}^N \to \mathbb{R}$ be a
Carath\'eodory function such that for a.e $(x,t)\in Q$ and for all
$s\in \mathbb{R}, \xi\in \mathbb{R}^N$, the growth condition
\begin{equation}
|H(x, t,s, \xi)|\leq \gamma(x,t)+g(s) { \sum_{i=1}^N}w_i(x)|\xi_i|^p \label{H}
\end{equation}
is satisfied, where $g:\mathbb{R}\to \mathbb{R}^+$ is a continuous
positive positive function that belongs to $L^1(\mathbb{R})$, while
$\gamma(x,t)$ belongs to $L^1(Q)$.
We recall that, for
$k>1$ and $s$ in $\mathbb{R}$, the truncation is defined as
$$
T_k(s)=\begin{cases}
s & \text{if } |s| \leq k\\
k \frac{s}{|s|} & \text{if } |s|>k.
\end{cases}
$$
\section{Some technical results}
\subsection*{Characterization of the time mollification of a function
$u$} To deal with time derivative, we introduce a time mollification
of a function $u$ belonging to a some weighted Lebesgue space.
Thus we define for all $\mu\geq 0$ and all $(x,t)\in Q$,
$$
u_{\mu}=\mu \int_{\infty}^{t}\tilde{u}(x,s)\exp(\mu(s-t))ds
$$
where $\tilde{u}(x,s)=u(x,s)\chi _{(0,T)}(s)$.
\begin{proposition}[\cite{ahar2}] \label{mol}
(1) If $u\in L^p(Q,w_i)$ then $u_{\mu}$ is measurable in $Q$ and
$\frac{\partial u_{\mu}}{\partial t}=\mu(u-u_{\mu})$ and,
$$
\|u_{\mu}\|_{L^p(Q,w_i)}\leq \|u\|_{L^p(Q,w_i)}.
$$
(2) If $u\in W^{1,p}_{0}(Q,w)$, then $u_{\mu} \to u$ in
$W^{1,p}_{0}(Q,w)$ as $\mu\to \infty$.
(3) If $u_n \to u$ in $W^{1,p}_{0}(Q,w)$, then $(u_n)_{\mu} \to
u_{\mu}$ in $W^{1,p}_{0}(Q,w) $.
\end{proposition}
\subsection*{Some weighted embedding and compactness results}
In this section we establish some embedding and compactness results in
weighted Sobolev spaces, some trace results, Aubin's and Simon's
results \cite{simon}.
Let $V=W^{1, p}_0(\Omega,w)$,
$H=L^2(\Omega,\sigma)$ and let $V^*=W^{-1,p'}$,
with $(2\leq p< \infty)$.
Let $X=L^p(0,T;W^{1,p}_0(\Omega,w))$. The dual space of $X$ is
$X^*=L^{p'}(0,T,V^*)$ where $\frac{1}{p}+\frac{1}{p'}=1$ and
denoting the space $W^1_{p}(0,T,V,H)=\{ v\in X:v'\in X^*\}$ endowed
with the norm
$$
\|u\|_{W^1_{p}}=\|u\|_{X}+\|u'\|_{X^*},
$$
which is a Banach space. Here $u'$ stands for the generalized
derivative of $u$; i.e.,
$$
\int_0^Tu'(t)\varphi(t)dt=-\int_0^T u(t)\varphi'(t)dt \quad\text{for
all }\varphi\in C_0^{\infty}(0,T).
$$
\begin{lemma}[\cite{19}] \label{compact}
(1) The evolution triple $V\subseteq H\subseteq V^*$ is satisfied.
(2) The imbedding $W^1_{p}(0,T,V,H)\subseteq C(0,T,H)$ is
continuous.
(3) The imbedding $W^1_{p}(0,T,V,H)\subseteq L^p(Q,\sigma)$ is
compact.
\end{lemma}
\begin{lemma}[\cite{ahar2}] \label{gr}
Let $g\in L^r(Q,\gamma)$ and let $g_n\in L^r(Q,\gamma)$, with
$ \|g_n\|_{L^r(Q,\gamma)}\leq C$,
$10$, let us define the following approximation of $b, H, f$
and $u_0$;
\begin{equation}
b_n(x,r)=b(x,T_n(r))+\frac{1}{n}r\quad\text{for } n>0,\label{bn}
\end{equation}
In view of \eqref{bn}, $b_n$ is a Carath\'eodory function and
satisfies \eqref{condb}, there exist $\lambda_n>0$ and functions
$A_n\in L^1(\Omega)$ and $B_n\in L^p(\Omega)$ such that
$$
\lambda_n\leq \frac{\partial b_n(x,s)}{\partial s}\leq A_n(x)
\quad\text{and}\quad \big|D_x \Big(\frac{\partial b_n(x,s)}{\partial
s}\Big)\big|\leq B_n(x)
$$
a.e. in $\Omega$, $s\in\mathbb{R}$.
$$
H_n(x,t,s,\xi)=\frac{H(x,t,s,\xi)}{1+\frac{1}{n}
|H(x,t,s,\xi)|}\chi_{\Omega_n}.
$$
Note that $\Omega_n$ is a sequence of compacts covering the bounded
open set $\Omega$ and $\chi_{\Omega_n}$ is its characteristic
function.
\begin{gather}
f_n\in L^{p'}(Q), \quad\text{and}\quad f_n\to f\quad\text{a.e. in
$Q$ and strongly in $L^1(Q)$ as }
n\to +\infty,\label{fn} \\
u_{0n}\in D(\Omega),\quad \norm{b_n(x,u_{0n})}_{L^1}
\leq\norm{b(x,u_{0})}_{L^1}, \\
b_n(x,u_{0n})\to b(x,u_0)\quad\text{a.e. in $\Omega$ and strongly in
}L^1(\Omega). \label{uzn}
\end{gather}
Let us now consider the approximate problem:
\begin{equation} \label{Pn}
\begin{gathered}
\frac{\partial b_n(x,u_n)}{\partial t} - \operatorname{div}
(a(x,t,u_n,D u_n))+H_n(x,t,u_n,Du_n) =
f_n\quad\text{in } D'(Q),\\
u_n=0\quad\text{in }(0,T)\times \partial\Omega ,\\
b_n(x,u_n(t=0))=b_n(x,u_{0n}).
\end{gathered}
\end{equation}
Note that $H_n(x,t,s,\xi)$ satisfies the following conditions
$$
|H_n(x,t,s,\xi)|\leq H(x,t,s,\xi)\quad\text{and}\quad
|H_n(x,t,s,\xi)|\leq n.
$$
For all $u,v\in L^p(0,T;W^{1,p}_0(\Omega,w))$,
\begin{align*}
&\big|\int_Q H_n(x,t,u,Du)v\,dx\,dt\big|\\
&\leq \Big( \int_Q |H_n(x,t,u,Du)|^{q'}\sigma^{-\frac{q'}{q}}
\,dx\,dt\Big)^{1/q'}\Big( \int_Q|v|^q\sigma
\,dx\,dt\Big)^{1/q}\\
&\leq n\int_0^T\Big(\int_{\Omega_n}\sigma ^{1-q'}dx\Big)^{1/q'}dt
\|v\|_{L^q(Q,\sigma)}\\
&\leq C_n\norm{v}_{L^p(0,T;W^{1,p}_0(\Omega,w))}.
\end{align*}
Moreover, since $f_n\in L^{p'}(0,T;W^{-1,p'}(\Omega, w^*))$,
proving existence of a weak solution $u_n\in L^p(0,T;W^{1,p}_0(\Omega,w))$
of \eqref{Pn} is an easy task (see e.g. \cite{lions},\cite{ahar2}).
Let $\varphi\in L^p(0,T;W^{1,p}_0(\Omega,w))\cap L^{\infty}(Q)$ with
$\varphi >0$, choosing $v=\exp(G(u_n))\varphi$ as test function in
$\ref{Pn}$ where $G(s)=\int_0^s \frac{g(r)}{\alpha}dr$ (the function
$g$ appears in \eqref{H}). We have
\begin{align*}
&\int_Q \frac{\partial b_n(x,u_n)}{\partial t} \exp(G(u_n))\varphi
\,dx\,dt+\int_Q a(x,t,u_n,Du_n)D(\exp(G(u_n))\varphi)\,dx\,dt\\
&=\int_Q H_n(x,t,u_n,Du_n)\exp(G(u_n))\varphi \,dx\,dt +\int_Q f_n
\exp(G(u_n))\varphi \,dx\,dt.
\end{align*}
In view of \eqref{H}, we obtain
\begin{align*}
&\int_Q \frac{\partial b_n(x,u_n)}{\partial t} \exp(G(u_n))\varphi
\,dx\,dt\\
&+\int_Q
a(x,t,u_n,Du_n)Du_n\frac{g(u_n)}{\alpha}\exp(G(u_n))\varphi \,dx\,dt\\
&+\int_Q a(x,t,u_n,Du_n)\exp(G(u_n))D\varphi \,dx\,dt\\
&\leq \int_Q \gamma(x,t)\exp(G(u_n))\varphi \,dx\,dt +\int_Q
g(u_n)\sum_{i=1}^N
\big|\frac{\partial u_n}{\partial x_i}\big|w_i\exp(G(u_n))\varphi dx dt\\
&\quad +\int_Q f_n \exp(G(u_n))\varphi \,dx\,dt.
\end{align*}
By \eqref{2.9}, we obtain
\begin{equation}
\begin{aligned}
& \int_Q \frac{\partial b_n(x,u_n)}{\partial t} \exp(G(u_n))\varphi
\,dx\,dt +\int_Q
a(x,t,u_n,Du_n)\exp(G(u_n))D\varphi \,dx\,dt\\
&\leq \int_Q \gamma(x,t)\exp(G(u_n))\varphi \,dx\,dt+ \int_Q f_n
\exp(G(u_n))\varphi \,dx\,dt,
\end{aligned} \label{positif}
\end{equation}
for all $\varphi \in L^p(0,T;W^{1,p}_0(\Omega,w)) \cap L^{\infty
}(Q), \varphi >0$. On the other hand, taking
$v=\exp(-G(u_n))\varphi$ as test function in \eqref{Pn} we deduce,
as in \eqref{positif}, that
\begin{align}
&\int_Q \frac{\partial b_n(x,u_n)}{\partial t} \exp(-G(u_n))\varphi
\,dx\,dt
+\int_Q a(x,t,u_n,Du_n)\exp(-G(u_n))D\varphi \,dx\,dt \notag \\
&+\int_Q \gamma(x,t)\exp(-G(u_n))\varphi \,dx\,dt \notag\\
&\geq \int_Q f_n \exp(-G(u_n))\varphi \,dx\,dt, \label{negatif}
\end{align}
for all $\varphi \in L^p(0,T;W^{1,p}_0(\Omega,w)) \cap L^{\infty
}(Q), \varphi >0$. Let $\varphi=T_k(u_n)^+\chi_{(0,\tau)}$, for
every $\tau\in [0,T]$, in \eqref{positif} we have,
\begin{equation} \label{599}
\begin{aligned}
&\int_{\Omega} B _k^n(x,u_n(\tau)) \exp(G(u_n)) dx +\int_{Q_{\tau}}
a(x,t,u_n,Du_n)\exp(G(u_n))DT_k(u_n)^+ \,dx\,dt\\
& \leq \int_{Q_{\tau}} \gamma(x,t)\exp(G(u_n))T_k(u_n)^+
\,dx\,dt+ \int_{Q_{\tau}}f_n \exp(G(u_n))T_k(u_n)^+ \,dx\,dt\\
&\quad +\int_{\Omega} B _k^n(x,u_{0n})dx,
\end{aligned}
\end{equation}
where $B_k^n(x,r)=\int_0^rT_k(s)^+\frac{\partial b_n(x,s)}{\partial
s}ds$. Due to this definition, we have
\begin{equation}
0\leq\int_{\Omega} B _k^n(x,u_{0n})dx\leq k\int_{\Omega}
|b_n(x,u_{0n})|dx\leq k\norm{b(x,u_0)}_{L^1(\Omega)}. \label{511}
\end{equation}
Using this inequality, $B_k^n(x,u_n)\geq 0$ and
$G(u_n)\leq \frac{\norm{g}_{L^1(\mathbb{R})}}{\alpha}$,
we deduce
\begin{align*}
& \int_{Q_{\tau}}
a(x,t,u_n,DT_k(u_n)^+)DT_k(u_n)^+ \exp(G(u_n)) \,dx\,dt
\\
&\leq k\exp\Big( \frac{\|g\|_{L^1(\mathbb{R})}} {\alpha}\Big)
\Big(\norm{u_{0n}}_{L^1(\Omega)}
+\norm{f_n}_{L^1(Q)} +\norm{\gamma}_{L^1(Q)}
+\norm{b_n(x,u_{0n})}_{L^1(\Omega)}\Big)
\\
&\leq c_1 k.
\end{align*}
Thanks to \eqref{2.9}, we have
\begin{equation}
\alpha \int_{Q_{\tau}}\sum_{i=1}^Nw_i(x)\big|
\frac{\partial T_k(u_n)^+}{\partial x_i}\big|^p
\exp(G(u_n))\,dx\,dt\leq c_1k. \label{eqtk}
\end{equation}
We deduce that
\begin{equation}
\alpha \int_Q\sum_{i=1}^Nw_i(x)\big| \frac{\partial
T_k(u_n)^+}{\partial x_i}\big|^p\,dx\,dt\leq c_1k .\label{tk1}
\end{equation}
Similarly to \eqref{tk1}, we take
$\varphi=T_k(u_n)^-\chi_{(0,\tau)}$ in \eqref{negatif} we deduce
that
\begin{equation}
\alpha \int_Q\sum_{i=1}^Nw_i(x)\big| \frac{\partial
T_k(u_n)^-}{\partial x_i}\big|^p\,dx\,dt\leq c_2k \label{tk2}
\end{equation}
where $c_2$ is a positive constant. Combining \eqref{tk1} and
\eqref{tk2} we conclude that
\begin{equation}
\norm{T_k(u_n)}^p_{L^p(0,T;W^{1,p}_0(\Omega,w))}\leq ck. \label{tk}
\end{equation}
We deduce from the above inequality, \eqref{599} and \eqref{511},
that
\begin{equation}
\int_{\Omega} B _k^n(x,u_n)dx\leq k(\norm{f}_{L^1(Q)}+
\norm{b(x,u_{0})}_{L^1(\Omega)})\equiv Ck .\label{514}
\end{equation}
Then, $T_k(u_n)$ is bounded in $L^p(0,T;W^{1,p}_0(\Omega,w))$, and
$T_k(u_n)\rightharpoonup v_k$ in the space
$L^p(0,T;W^{1,p}_0(\Omega,w))$, and by the compact imbedding
\eqref{2.6} gives
$$
T_k(u_n)\to v_k\quad\text{strongly in $L^p(Q,\sigma)$ and a.e. in
}Q.
$$
Let $k>0$ be large enough and $B_R$ be a ball of $\Omega$, we have
\begin{align*}
& k\operatorname{meas}(\{ |u_n|>k\}\cap B_R\times[0,T])\\
&=\int_{0}^{T}\int_{\{ |u_n|>k\}\cap B_R}|T_k(u_n)|\,dx\,dt\\
&\leq \int_{0}^{T}\int_{B_R}|T_k(u_n)|\,dx\,dt\\
&\leq \Big( \int_Q|T_k(u_n)|^p \sigma \,dx\,dt\Big)^{1/p}
\Big(\int_{0}^{T}\int_{B_R} \sigma^{1-p'} \,dx\,dt \Big)^{1/p'}\\
&\leq T c_R\Big(\int_{Q}\sum_{i=1}^Nw_i(x)
\big|\frac{\partial T_k(u_n)}{\partial x_i}\Big|^p\,dx\,dt\Big)^{1/p} \\
&\leq c k^{1/p},
\end{align*}
which implies
$$
\operatorname{meas}(\{ |u_n|>k\}\cap B_R\times[0,T])\leq
\frac{c_1}{k^{1-\frac{1}{p}}},\quad \forall k\geq 1.
$$
So, we have
$$
\lim_{k\to +\infty }(\operatorname{meas}(\{ |u_n|>k\} \cap
B_R\times[0,T]))=0.
$$
Now we turn to prove the almost every convergence of $u_n$ and
$b_n(x,u_n)$. Consider now a function non decreasing $g_k\in
C^2(\mathbb{R})$ such that $g_k(s)=s$ for $|s|\leq \frac{k}{2}$ and
$g_k(s)=k$ for $|s|\geq k $. Multiplying the approximate equation by
$g'_k(b_n(x,u_n))$, we obtain
\begin{equation}
\begin{aligned}
&\frac{\partial g_k(b_n(x,u_n))}{\partial t}-\operatorname{div}
(a(x,t,u_n,Du_n)g'_k(b_n(x,u_n)))\\
&+a(x,t,u_n,Du_n)g''_k(b_n(x,u_n))D_x\Big(\frac{\partial
b_n(x,u_n)}{\partial s}\Big)Du_n\\
&+H_n(x,t,u_n,Du_n)g'_k(b_n(x,u_n))\\
&=f_n g'_k(b_n(x,u_n))
\end{aligned}\label{518}
\end{equation}
in the sense of distributions, which implies that
\begin{gather}
g_k(b_n(x,u_n)) \text{ is bounded in }L^p(0,T;W^{1,p}_0(\Omega,w)), \label{519}\\
\frac{\partial g_k(b_n(x,u_n))}{\partial t} \text{ is bounded in
}X^*+L^1(Q), \label{520}
\end{gather}
independent of $n$ as long as $k0$,)
\begin{align*}
&\operatorname{meas}(\{ \big|b_n(x,u_n)-b_m(x,u_m)\big|>\lambda\}
\cap B_R\times[0,T])\\
&\leq \operatorname{meas}(\{ |b_n(x,u_n)|>k\}\cap B_R\times[0,T])
+\operatorname{meas}(\{ |b_m(x,u_m)|>k\}\cap B_R\times[0,T])\\
&\quad +\operatorname{meas}(\{ \big|g_k(b_n(x,u_n))
-g_k(b_m(x,u_m))\big|>\lambda\}).
\end{align*}
Let $\varepsilon >0$, then there exist $k(\varepsilon)>0$ such that
$$
\operatorname{meas}(\{ \big|b_n(x,u_n)-b_m(x,u_m)\big|
>\lambda\}\cap B_R\times[0,T])
\leq \varepsilon
$$
for all $n,m\geq n_0(k(\varepsilon),\lambda,R)$. This proves that
$(b_n(x,u_n))$ is a Cauchy sequence in measure in $B_R\times[0,T]$,
thus converges almost everywhere to some measurable function $v$.
Then for a subsequence denoted again $u_n$,
\begin{gather}
u_n\to u\quad\text{a.e. in }Q, \label{523}\\
b_n(x,u_n)\to b(x,u)\quad\text{a.e. in } Q.\label{524}
\end{gather}
We can deduce from \eqref{tk} that
\begin{equation}
T_k(u_n)\rightharpoonup T_k(u)\quad\text{weakly in
}L^p(0,T;W^{1,p}_0(\Omega,w)) \label{faible}
\end{equation}
and then, the compact imbedding \eqref{h2.8} gives
$$
T_k(u_n)\to T_k(u)\quad\text{strongly in $L^q(Q,\sigma)$ and a.e. in
}Q.
$$
Which implies, by using \eqref{2.7}, for all $k>0$ that there exists
a function $h_k\in \prod_{i=1}^N L^{p'}(Q,w_i^*)$, such that
\begin{equation}
a(x,t,T_k(u_n),DT_k(u_n))\rightharpoonup h_k \quad \text{weakly in }
\prod_{i=1}^N L^{p'}(Q,w_i^*). \label{hk}
\end{equation}
We now establish that $b(x,u)$ belongs to
$L^{\infty}(0,T;L^1(\Omega))$. Using \eqref{523} and passing to the
limit-inf in \eqref{514} as $n$ tends to $+\infty$, we obtain that
$$
\frac{1}{k}\int_{\Omega}B_k(x,u)(\tau)dx \leq
[\norm{f}_{L^1(Q)}+\norm{u_0}_{L^1(\Omega)}]\equiv C,
$$
for almost any $\tau$ in $(0,T)$. Due to the definition of
$B_k(x,s)$ and the fact that $\frac{1}{k}B_k(x,u)$ converges
pointwise to $b(x,u)$, as $k$ tends to $+\infty$, shows that
$b(x,u)$ belong to $L^{\infty}(0,T;L^1(\Omega))$.
\begin{lemma} \label{lem5.4}
Let $u_n$ be a solution of the approximate problem \eqref{Pn}. Then
\begin{equation}\lim _{m\to
\infty}\limsup_{n\to \infty}\int_{\{m\leq|u_n|\leq m+1\}}
a(x,t,u_n,Du_n)Du_n\,dx\,dt=0 \label{an}
\end{equation}
\end{lemma}
\begin{proof}
Considering the function
$\varphi=T_1(u_n-T_m(u_n))^-:=\alpha_m(u_n)$ in \eqref{negatif} this
function is admissible since $\varphi\in
L^p(0,T;W^{1,p}_0(\Omega,w))$ and $\varphi\geq 0$. Then, we have
\begin{align*}
& \int_Q \frac{\partial b_n(x,u_n)}{\partial t} \alpha_m(u_n)
\,dx\,dt +\int_{\{-(m+1)\leq u_n\leq -m\}} a(x,t,u_n,Du_n)Du_n
\alpha'_m(u_n)
\,dx\,dt \\
&\quad + \int_Q f_n \exp(-G(u_n))\alpha_m(u_n) \,dx\,dt \\
&\leq \int_Q \gamma(x,t)\exp(-G(u_n))\alpha_m(u_n) \,dx\,dt.
\end{align*}
Which, by setting $B_n^{m}(x,r)=\int_0^r \frac{\partial
b_n(x,s)}{\partial s}\alpha_m(s) ds$, gives
\begin{align*}
&\int_{\Omega}B_n^{m}(x,u_n)(T)dx +\int_{\{-(m+1)\leq u_n\leq
-m\}} a(x,t,u_n,Du_n)Du_n \alpha'_m(u_n) \,dx\,dt \\
&+ \int_Q f_n \exp(-G(u_n))\alpha_m(u_n) \,dx\,dt \\
&\leq \int_Q \gamma(x,t)\exp(-G(u_n))\alpha_m(u_n) \,dx\,dt+
\int_{\Omega}B_n^{m}(x,u_{0n})dx.
\end{align*}
Since $B_n^{m}(x,u_n)(T)\geq 0$ and by Lebesgue's theorem, we have
\begin{equation}\lim _{m\to
\infty}\lim_{n\to \infty}\int_Q f_n \exp(-G(u_n))\alpha_m(u_n)
\,dx\,dt =0.\label{fexp}
\end{equation}
Similarly, since $\gamma \in L^1(\Omega)$, we obtain
\begin{equation}\lim _{m\to
\infty}\lim_{n\to \infty}\int_Q \gamma \exp(-G(u_n))\alpha_m(u_n)
\,dx\,dt =0.\label{gexp}
\end{equation}
We conclude that
\begin{equation}\lim _{m\to
\infty}\limsup_{n\to \infty}\int_{\{-(m+1)\leq u_n\leq -m\}}
a(x,t,u_n,Du_n)Du_n\,dx\,dt=0 .\label{an1}
\end{equation}
On the other hand, let $\varphi=T_1(u_n-T_m(u_n))^+$ as test
function in \eqref{positif} and reasoning as in the proof of
\eqref{an1} we deduce that
\begin{equation}
\lim _{m\to \infty}\limsup_{n\to \infty}\int_{\{m)\leq u_n\leq
m+1\}} a(x,t,u_n,Du_n)Du_n\,dx\,dt=0 .\label{an2}
\end{equation}
Thus \eqref{an} follows from \eqref{an1} and \eqref{an2}.
\end{proof}
\subsection*{Step 2: Almost everywhere convergence of the gradients.}
This step is devoted to introduce for $k\geq 0$ fixed a time
regularization of the function $T_k(u)$ in order to perform the
monotonicity method. This kind of regularization has been first
introduced by R. Landes (see Lemma 6 and proposition 3, p.230, and
proposition 4, p.231, in\cite{landes}).
Let $\psi_i\in D(\Omega)$ be a sequence which converge strongly to
$u_0$ in $L^1(\Omega)$. Set $w_{\mu}^i=(T_k(u))_{\mu}+e^{-\mu
t}T_k(\psi_i)$ where $(T_k(u))_{\mu}$ is the mollification with
respect to time of $T_k(u)$. Note that $w_{\mu}^i$ is a smooth
function having the following properties:
\begin{gather}
\frac{\partial w_{\mu}^i}{\partial t}=\mu(T_k(u)-w_{\mu}^i),\quad
w_{\mu}^i(0)=T_k(\psi_i),\quad \big|w_{\mu}^i\big|\leq k,\\
w_{\mu}^i\to T_k(u) \quad\text{in }L^p(0,T;W^{1,p}_0(\Omega,w)),
\end{gather}
as $\mu\to \infty$.
We introduce the following function of one real:
$$
h_m(s)=\begin{cases}
1 & \text{if } |s| \leq m\\
0 & \text{if } |s|\geq m+1\\
m+1- s & \text{if } m\leq s\leq m+1\\
m+1+s & \text{if } -(m+1)\leq s\leq -m
\end{cases}
$$
where $m>k$.
Let $\varphi=(T_k(u_n)-w_{\mu}^i)^+h_m(u_n)\in
L^p(0,T;W^{1,p}_0(\Omega,w))\cap L^{\infty}(Q)$ and $\varphi\geq 0$,
then we take this function in \eqref{positif}, we obtain
\begin{equation} \label{test}
\begin{aligned}
&\int_{\{T_k(u_n)-w_{\mu}^i\geq 0\}} \frac{\partial
b_n(x,u_n)}{\partial t}
\exp(G(u_n))(T_k(u_n)-w_{\mu}^i)h_m(u_n) \,dx\,dt\\
&+\int_{\{T_k(u_n)-w_{\mu}^i\geq 0\}}
a(x,t,u_n,Du_n)D(T_k(u_n)-w_{\mu}^i)h_m(u_n) \,dx\,dt\\
& - \int_{\{m\leq u_n\leq m+1\}}
\exp(G(u_n))a(x,t,u_n,Du_n)Du_n(T_k(u_n)-w_{\mu}^i)^+ \,dx\,dt\\
&\leq \int_Q \gamma(x,t)\exp(G(u_n))(T_k(u_n)-w_{\mu}^i)^+h_m(u_n)
\,dx\,dt\\
&\quad +\int_Q f_n \exp(G(u_n))(T_k(u_n)-w_{\mu}^i)^+h_m(u_n)
\,dx\,dt.
\end{aligned}
\end{equation}
Observe that
\begin{align*}
&\int_{\{m\leq u_n\leq m+1\}}
\exp(G(u_n))a(x,t,u_n,Du_n)Du_n(T_k(u_n)-w_{\mu}^i)^+ \,dx\,dt\\
&\leq 2k \int_{\{m\leq u_n\leq m+1\}} a(x,t,u_n,Du_n)Du_n\,dx\,dt.
\end{align*}
Thanks to \eqref{an} the third integral tend to zero as $n$ and $m$
tend to infinity, and by Lebesgue's theorem, we deduce that the
right hand side converge to zero as $n$, $m$ and $\mu$ tend to
infinity. Since
\begin{gather*}
(T_k(u_n)-w_{\mu}^i)^+h_m(u_n)\rightharpoonup
(T_k(u)-w_{\mu}^i)^+h_m(u) \quad\text{weakly* in $L^{\infty}(Q)$,
as $n\to \infty$},\\
\text{and } (T_k(u)-w_{\mu}^i)^+h_m(u)\rightharpoonup 0 \quad
\text{weakly* in $L^{\infty}(Q)$ as $\mu\to \infty$}.
\end{gather*}
Let $\varepsilon_l(n,m,\mu,i)$ $l=1,\dots ,n$ various functions tend
to zero as $n$, $m$, $i$ and $\mu$ tend to infinity.
The definition of the sequence $w_{\mu}^i$ makes it possible to
establish the following lemma, which will be proved in the Appendix.
\begin{lemma}\cite{R1}\label{znn}
For $k\geq 0$ we have
\begin{equation}
\int_{\{T_k(u_n)-w_{\mu}^i\geq 0\}} \frac{\partial
b_n(x,u_n)}{\partial t} \exp(G(u_n))(T_k(u_n)-w_{\mu}^i)h_m(u_n)
\,dx\,dt\geq \varepsilon(n,m,\mu,i)\label{zn}
\end{equation}
\end{lemma}
On the other hand, the second term of left hand side of \eqref{test}
reads as follows
\begin{align*}
&\int_{\{T_k(u_n)-w_{\mu}^i\geq 0\}}
a(x,t,u_n,Du_n)D(T_k(u_n)-w_{\mu}^i)h_m(u_n) \,dx\,dt\\
&=\int_{\{T_k(u_n)-w_{\mu}^i\geq 0,|u_n|\leq k\}}
a(x,t,T_k(u_n),DT_k(u_n))D(T_k(u_n)-w_{\mu}^i)h_m(u_n) \,dx\,dt \\
&\quad - \int_{\{T_k(u_n)-w_{\mu}^i\geq 0,|u_n|\geq k\}}
a(x,t,u_n,Du_n)Dw_{\mu}^ih_m(u_n) \,dx\,dt.
\end{align*}
Since $m>k$, $ h_m(u_n)=0$ on $\{|u_n|\geq m+1\}$, One has
\begin{equation}
\begin{aligned}
&\int_{\{T_k(u_n)-w_{\mu}^i\geq 0\}}
a(x,t,u_n,Du_n)D(T_k(u_n)-w_{\mu}^i)h_m(u_n) \,dx\,dt\\
&=\int_{\{T_k(u_n)-w_{\mu}^i\geq 0\}}
a(x,t,T_k(u_n),DT_k(u_n))D(T_k(u_n)-w_{\mu}^i)h_m(u_n) \,dx\,dt \\
&\quad - \int_{\{T_k(u_n)-w_{\mu}^i\geq 0,|u_n|\geq k\}}
a(x,t,T_{m+1}(u_n),DT_{m+1}(u_n))Dw_{\mu}^ih_m(u_n)
\,dx\,dt\\
&=J_1+J_2
\end{aligned}\label{tm}
\end{equation}
In the following we pass to the limit in \eqref{tm}: first we let
$n$ tend to $+\infty$, then $\mu$ and finally $m$, tend to
$+\infty$. Since $a(x,t,T_{m+1}(u_n),DT_{m+1}(u_n))$ is bounded in
$\prod_{i=1}^N L^{p'}(Q,w_i^*)$, we have that
$$
a(x,t,T_{m+1}(u_n),DT_{m+1}(u_n))h_m(u_n) \chi_{\{|u_n|>k\}}\to
h_mh_m(u) \chi_{\{|u|>k\}}
$$
strongly in $\prod_{i=1}^N L^{p'}(Q,w_i^*)$ as $n$ tends to
infinity, it follows that
\begin{align*}
J_2&=\int_{\{T_k(u_n)-w_{\mu}^i\geq
0 \}}h_mDw_{\mu}^ih_m(u)\chi_{\{|u|>k\}}\,dx\,dt+\varepsilon(n)\\
&=\int_{\{T_k(u_n)-w_{\mu}^i\geq 0 \}}h_m(DT_k(u)_{\mu}-e^{-\mu
t}DT_k(\psi_i))h_m(u)\chi_{\{|u|>k\}}\,dx\,dt+\varepsilon(n).
\end{align*}
By letting $\mu\to +\infty$,
$$
J_2=\int_{\{T_k(u_n)-w_{\mu}^i\geq 0
\}}h_mDT_k(u)\,dx\,dt+\varepsilon(n,\mu).
$$
Using now the term $J_1$ of \eqref{tm} one can easily show that
\begin{equation} \label{i1}
\begin{aligned}
&\int_{\{T_k(u_n)-w_{\mu}^i\geq 0\}}
a(x,t,T_k(u_n),DT_k(u_n))D(T_k(u_n)-w_{\mu}^i)h_m(u_n) \,dx\,dt\\
&=\int_{\{T_k(u_n)-w_{\mu}^i\geq 0\}}
\left[a(x,t,T_k(u_n),DT_k(u_n))-a(x,t,T_k(u_n),DT_k(u))\right]\\
&\quad\times\left[DT_k(u_n)-DT_k(u)\right]h_m(u_n) \,dx\,dt\\
&\quad +\int_{\{T_k(u_n)-w_{\mu}^i\geq 0\}}
a(x,t,T_k(u_n),DT_k(u))(DT_k(u_n)-DT_k(u))h_m(u_n) \,dx\,dt\\
&\quad +\int_{\{T_k(u_n)-w_{\mu}^i\geq 0\}}
a(x,t,T_k(u_n),DT_k(u_n))DT_k(u)h_m(u_n) \,dx\,dt\\
&\quad -\int_{\{T_k(u_n)-w_{\mu}^i\geq 0\}}
a(x,t,T_k(u_n),DT_k(u_n))Dw_{\mu}^ih_m(u_n)
\,dx\,dt\\
&=K_1+K_2+K_3+K_4.
\end{aligned}
\end{equation}
We shall go to the limit as $n$ and $\mu\to +\infty$ in the three
integrals of the right-hand side. Starting with $K_2$, we have by
letting $n\to +\infty$,
\begin{equation}
K_2=\varepsilon(n).\label{k2}
\end{equation}
About $K_3$, we have by letting $n\to +\infty$ and using \eqref{hk},
$$
K_3=\int_{\{T_k(u_n)-w_{\mu}^i\geq 0
\}}h_kDT_k(u)h_m(u)\chi_{\{|u|>k\}}\,dx\,dt+\varepsilon(n)
$$
By letting $\mu\to +\infty$,
\begin{equation}
K_3=\int_{\{T_k(u_n)-w_{\mu}^i\geq 0
\}}h_kDT_k(u)\,dx\,dt+\varepsilon(n,\mu).\label{k3}
\end{equation}
For $K_4$ we can write
$$
K_4=-\int_{\{T_k(u_n)-w_{\mu}^i\geq 0
\}}h_kDw_{\mu}^ih_m(u)\,dx\,dt+\varepsilon(n),
$$
By letting $\mu\to +\infty$,
\begin{equation}
K_4=-\int_{\{T_k(u_n)-w_{\mu}^i\geq 0
\}}h_kDT_k(u)\,dx\,dt+\varepsilon(n,\mu).\label{k4}
\end{equation}
We then conclude that
\begin{align*}
&\int_{\{T_k(u_n)-w_{\mu}^i\geq 0\}}
a(x,t,T_k(u_n),DT_k(u_n))D(T_k(u_n)-w_{\mu}^i)h_m(u_n) \,dx\,dt \\
&=\int_{\{T_k(u_n)-w_{\mu}^i\geq 0\}}
\left[a(x,t,T_k(u_n),DT_k(u_n))-a(x,t,T_k(u_n),DT_k(u))\right]\\
&\quad \times\left[DT_k(u_n)-DT_k(u)\right]h_m(u_n) \,dx\,dt
+\varepsilon(n,\mu).
\end{align*}
On the other hand, we have
\begin{equation} \label{545}
\begin{aligned}
&\int_{\{T_k(u_n)-w_{\mu}^i\geq 0\}}
\left[a(x,t,T_k(u_n),DT_k(u_n))-a(x,t,T_k(u_n),DT_k(u))\right]\\
&\quad \times\left[DT_k(u_n)-DT_k(u)\right] \,dx\,dt\\
&=\int_{\{T_k(u_n)-w_{\mu}^i\geq 0\}}
\left[a(x,t,T_k(u_n),DT_k(u_n))-a(x,t,T_k(u_n),DT_k(u))\right]\\
&\quad \times\left[DT_k(u_n)-DT_k(u)\right]h_m(u_n) \,dx\,dt\\
&\quad +\int_{\{T_k(u_n)-w_{\mu}^i\geq 0\}}
a(x,t,T_k(u_n),DT_k(u_n))(DT_k(u_n)-DT_k(u))\\
&\quad\times (1-h_m(u_n)) \,dx\,dt\\
&\quad -\int_{\{T_k(u_n)-w_{\mu}^i\geq 0\}}
a(x,t,T_k(u_n),DT_k(u))(DT_k(u_n)-DT_k(u))\\
&\quad\times (1-h_m(u_n)) \,dx\,dt.
\end{aligned}
\end{equation}
Since $h_m(u_n)=1\quad\text{in }\{|u_n|\leq m\}$ and $ \{|u_n|\leq
k\}\subset \{|u_n|\leq m\}$ for $m$ large enough, we deduce from
\eqref{545} that
\begin{align*}
&\int_{\{T_k(u_n)-w_{\mu}^i\geq 0\}}
\left[a(x,t,T_k(u_n),DT_k(u_n))-a(x,t,T_k(u_n),DT_k(u))\right]\\
&\times\left[DT_k(u_n)-DT_k(u)\right] \,dx\,dt\\
&=\int_{\{T_k(u_n)-w_{\mu}^i\geq 0\}}
\left[a(x,t,T_k(u_n),DT_k(u_n))-a(x,t,T_k(u_n),DT_k(u))\right]\\
&\quad \times\left[DT_k(u_n)-DT_k(u)\right]h_m(u_n) \,dx\,dt\\
&\quad +\int_{\{T_k(u_n)-w_{\mu}^i\geq 0,|u_n|>k\}}
a(x,t,T_k(u_n),DT_k(u))DT_k(u)(1-h_m(u_n)) \,dx\,dt.
\end{align*}
It is easy to see that the last terms of the last equality tend to
zero as $n\to +\infty$, which implies
\begin{align*}
&\int_{\{T_k(u_n)-w_{\mu}^i\geq 0\}}
\left[a(x,t,T_k(u_n),DT_k(u_n))-a(x,t,T_k(u_n),DT_k(u))\right]\\
&\times\left[DT_k(u_n)-DT_k(u)\right] \,dx\,dt\\
&=\int_{\{T_k(u_n)-w_{\mu}^i\geq 0\}}
\left[a(x,t,T_k(u_n),DT_k(u_n))-a(x,t,T_k(u_n),DT_k(u))\right]\\
&\quad \times\left[DT_k(u_n)-DT_k(u)\right]h_m(u_n) \,dx\,dt
+\varepsilon(n)
\end{align*}
Combining \eqref{zn}, \eqref{i1}, \eqref{k2}, \eqref{k3}, \eqref{k4}
and \eqref{545}, we obtain
\begin{equation} \label{qa}
\begin{aligned}
&\int_{\{T_k(u_n)-w_{\mu}^i\geq 0\}}
\left[a(x,t,T_k(u_n),DT_k(u_n))-a(x,t,T_k(u_n),DT_k(u))\right]\\
&\times\left[DT_k(u_n)-DT_k(u)\right] \,dx\,dt\leq
\varepsilon(n,\mu,m)
\end{aligned}
\end{equation}
Passing to the limit in \eqref{qa} as $n$ and $m$ tend to infinity,
we obtain
\begin{equation} \label{plus}
\begin{aligned}
&\lim_{n\to \infty}\int_{\{T_k(u_n)-w_{\mu}^i\geq 0\}}
\left[a(x,t,T_k(u_n),DT_k(u_n))-a(x,t,T_k(u_n),DT_k(u))\right]\\
&\times\left[DT_k(u_n)-DT_k(u)\right] \,dx\,dt=0.
\end{aligned}
\end{equation}
On the other hand, taking $\varphi=(T_k(u_n)-w_{\mu}^i)^-h_m(u_n)$
in \eqref{negatif}, we deduce as in \eqref{plus} that
\begin{equation} \label{moin}
\begin{aligned}
&\lim_{n\to \infty}\int_{\{T_k(u_n)-w_{\mu}^i\leq 0\}}
\left[a(x,t,T_k(u_n),DT_k(u_n))-a(x,t,T_k(u_n),DT_k(u))\right]\\
&\times\left[DT_k(u_n)-DT_k(u)\right] \,dx\,dt=0.
\end{aligned}
\end{equation}
Combining \eqref{plus} and \eqref{moin}, we conclude
\begin{equation}
\begin{aligned}
&\lim_{n\to \infty}\int_Q
\left[a(x,t,T_k(u_n),DT_k(u_n))-a(x,t,T_k(u_n),DT_k(u))\right]\\
&\times\left[DT_k(u_n)-DT_k(u)\right] \,dx\,dt=0.
\end{aligned}\label{ll}
\end{equation}
Which, by lemma \eqref{lem4.5}, implies
\begin{equation}
T_k(u_n)\to T_k(u)\quad\text{strongly in
$L^p(0,T;W^{1,p}_0(\Omega,w))$ for all }k. \label{tfort}
\end{equation}
Now, observe that for every $\sigma>0$,
\begin{align*}
&\operatorname{meas}\{(x,t)\in\Omega\times [0,T]:
|Du_n-Du|>\sigma\}\\
&\leq \operatorname{meas}\{(x,t)\in\Omega\times [0,T]:|Du_n|>k\}\\
&\quad +\operatorname{meas}\{(x,t)\in\Omega\times [0,T]:|u|>k\}\\
&\quad +\operatorname{meas}\{(x,t)\in\Omega\times [0,T]:
\big|DT_k(u_n)-DT_k(u)\big|>\sigma\}
\end{align*}
then as a consequence of \eqref{tfort} we have that $Du_n$ converges
to $Du$ in measure and therefore, always reasoning for a
subsequence,
\begin{equation}
Du_n\to Du \quad\text{a. e. in }Q .\label{grad}
\end{equation}
Which implies
\begin{equation}
a(x,t,T_k(u_n),DT_k(u_n))\rightharpoonup a(x,t,T_k(u),DT_k(u))
\quad\text{in } \prod_{i=1}^NL^{p'}(Q,w_i^*). \label{atk}
\end{equation}
\subsection*{Step 3: Equi-integrability of the nonlinearity sequence}
We shall now prove that $H_n(x,t,u_n,Du_n)\to H(x,t,u,Du)$ strongly
in $L^1(Q)$ by using Vitali's theorem. Since $H_n(x,t,u_n,Du_n)\to
H(x,t,u,Du)$ a.e. in $Q$, Consider a function $\rho_h(s)=\int_0^s
g(\nu)\chi_{\{\nu>h\}}d\nu$, take
$\varphi=\rho_h(u_n)=\int_0^{u_n}g(s)\chi_{\{s>h\}}ds$ as test
function in \eqref{positif}, we obtain
\begin{align*}
&\Big[\int_{\Omega} B_h^n(x,u_n) dx\Big]_0^T
+\int_Q a(x,t,u_n,Du_n)Du_ng(u_n)\chi_{\{u_n>h\}} \,dx\,dt \\
&\leq \Big( \int_h^{\infty}g(s)\chi_{\{s>h\}}ds\Big) \exp\Big(
\frac{\norm{g}_{L^1(\mathbb{R})}}{\alpha}\Big)
\Big(\norm{\gamma}_{L^1(Q)}+\norm{f_n}_{L^1(Q)}\Big),
\end{align*}
where $B_h^n(x,r)=\int_0^r\frac{\partial b_n(x,s)}{\partial s}
\rho_h(s)ds$, which implies, since $B_h^n(x,r)\geq 0$,
\begin{align*}
&\int_Q a(x,t,u_n,Du_n)Du_ng(u_n)\chi_{\{u_n>h\}} \,dx\,dt \\
&\leq \Big( \int_h^{\infty}g(s)ds\Big)
\exp\Big(\frac{\norm{g}_{L^1(\mathbb{R})}}{\alpha}\Big)
\left(\norm{\gamma}_{L^1(Q)} +
\norm{f_n}_{L^1(Q)}\right)+\int_{\Omega} B_h^n(x,u_{0n}) dx.
\end{align*}
Using \eqref{2.9}, we have
$$
\int_{\{u_n>h\}}g(u_n)\sum_{i=1}^N w_i \big|\frac{\partial
u_n}{\partial x_i}\big|^p\,dx\,dt \leq C\int_h^{\infty}g(s)\,ds.
$$
Since $g\in L^1(\mathbb{R})$, we have
$$
\lim_{h\to \infty}\sup_{n\in
\mathbb{N}}\int_{\{u_n>h\}}g(u_n)\sum_{i=1}^Nw_i \big|\frac{\partial
u_n}{\partial x_i}\big|^p\,dx\,dt=0.
$$
Similarly, let
$\varphi=\int_{u_n}^0g(s)\chi_{\{s<-h\}}ds$ as a test function in
\eqref{negatif}, we conclude that
$$
\lim_{h\to \infty}\sup_{n\in
\mathbb{N}}\int_{\{u_n<-h\}}g(u_n)\sum_{i=1}^Nw_i
\big|\frac{\partial u_n}{\partial x_i}\big|^p\,dx\,dt=0.
$$
Consequently,
$$
\lim_{h\to +\infty}\sup_{n\in
\mathbb{N}}\int_{\{|u_n|>h\}}g(u_n)\sum_{i=1}^Nw_i
\big|\frac{\partial u_n}{\partial x_i}\big|^p\,dx\,dt=0,
$$
which, for $h$ large enough, implies
\begin{align*}
\int_Q g(u_n)\sum_{i=1}^Nw_i\big|\frac{\partial u_n}{\partial
x_i}\big|^p\,dx\,dt &\leq \int_{\{|u_n|0$,
there exist $\lambda_k>0$ and functions $A_k\in L^1(\Omega)$ and
$B_k\in L^p(\Omega)$ such that
\begin{gather}
\lambda_k\leq\frac{\partial b(x,s)}{\partial s}\leq A_k(x),\quad
\big|D_x\Big(\frac{\partial b(x,s)}{\partial s}\Big)\big|
\leq B_k(x), \label{condb}\\
H(x,t,s,\xi)=\rho \sin
(s)\exp(s^{-2}) \sum_{i=1}^N w_i(x)|\xi_i|^p,\quad \rho\in\mathbb{R},\\
a_i(x,t,s,d)=w_i(x)|d_i|^{p-1}\operatorname{sgn}(d_i),\quad
i=1,\dots ,N,
\end{gather}
with $w_i(x)$, ($i=1,\dots ,N$), a weight function strictly
positive, $x\in Q$. Then, we can consider the Hardy inequality in
the form
$$
\Big(\int_{\Omega}|u(x)|^p\sigma(x)dx\Big)^{1/p} \leq c
\Big(\int_{\Omega}|Du(x)|^pw(x)dx\Big)^{1/p}.
$$
It is easy to show that the $a_i(t,x,s,d)$ are Caratheodory functions
satisfying the growth condition \eqref{2.7} and the coercivity
\eqref{2.9}. On the order hand the monotonicity condition is
verified. In fact,
\begin{align*}
&\sum_{i=1}^N\left( a_i(x,t,d)-a(x,t,d')\right)(d_i-d'_i)\\
&= w(x)\sum_{i=1}^{N-1}\left( |d_i|^{p-1}\operatorname{sgn}(d_i)-
|d'_i|^{p-1}\operatorname{sgn}(d'_i)\right)(d_i-d'_i)> 0,
\end{align*}
for almost all $x\in \Omega$ and for all $d, d'\in\mathbb{R}^N$.
This last inequality can not be strict, since for $d\neq d'$ , since
$w>0$ a.e. in $\Omega$.
While the Carath\'eodory function $H(x,t,s,\xi)$ satisfies the
condition \eqref{H} indeed
$$
|H(x,t,s,\xi)|\leq |\rho|\exp(s^{-2}) \sum_{i=1}^N
w_i(x)|\xi_i|^p=g(s)\sum_{i=1}^N w_i(x)|\xi_i|^p
$$
where $g(s)=|\rho|\exp(s^{-2}$ is a function positive continuous
which belongs to $L^1(\mathbb{R})$. Note that $H(x,t,s,\xi)$ does
not satisfy the sign condition \eqref{sign} and the coercivity
condition \eqref{coer}.
In particular, let us use special weight function, $w$, expressed
in terms of the distance to the bounded $\partial \Omega$. Denote
$d(x)=\operatorname{dist}(x,\partial\Omega)$ and set
$w(x)=d^{\lambda}(x)$, $\sigma(x)=d^{\mu}(x)$.
Finally, the hypotheses of Theorem \ref{thm1} are satisfied.
Therefore, for all $f\in L^1(Q)$, the problem
\begin{gather*}
b(x,u)\in L^{\infty}([0,T];L^1(\Omega));\quad T_k(u)\in L^p(0,T;W^{1,p}_0(\Omega,w)), \\
\lim_{m\to +\infty} \int_{\{m\leq |u|\leq
m+1\}}d^{\lambda}(x)\sum_{i=1}^N \big|\frac{\partial u}{\partial
x_i}\big|^{p-1}\operatorname{sgn} (\frac{\partial u}{\partial
x_i})\frac{\partial u }{\partial x_i}
\,dx\,dt=0;\\
B_S(x,r)=\int_0^r \frac{\partial b(x,\sigma)}{\partial \sigma}
S'(\sigma)d\sigma,\\
\begin{aligned}
&\int_{\Omega}B_S(x,u(T))\varphi(T)dx
-\int_Q B_S(x,u)\frac{\partial \varphi}{\partial t}\,dx\,dt\\
&+\int_Q S'(u)d^{\lambda}(x)\sum_{i=1}^N \big|\frac{\partial
u}{\partial x_i}\big|^{p-1}\operatorname{sgn} (\frac{\partial
u}{\partial x_i})
\frac{\partial\varphi }{\partial x_i}\,dx\,dt\\
&+\int_Q S''(u) d^{\lambda}(x)\sum_{i=1}^N \big|\frac{\partial
u}{\partial x_i}\big|^{p-1}\operatorname{sgn} (\frac{\partial
u}{\partial x_i})
\frac{\partial u }{\partial x_i}\varphi \,dx\,dt\\
&+\int_Q \rho S'(u) \sin (u)\exp(u^{-2}) \sum_{i=1}^N
w_i\big|\frac{\partial u}{\partial x_i}\big|^{p-1} \varphi \,dx\,dt\\
&=\int_Q fS'(u)\varphi \,dx\,dt+\int_{\Omega}B_S(x,u_0)\varphi(0)dx,
\end{aligned} \\
B_S(x,u)(t=0)=B_S(x,u_0)\quad\text{in }\Omega,
\end{gather*}
for all $\varphi \in C^{\infty}_0(Q)$ and $S\in W^{1,
\infty}(\mathbb{R})$ with $S'\in C^{\infty}_0(\mathbb{R})$, has at
least one renormalised solution.
\section{Appendix}
\begin{proof}[Proof of Lemma \ref{znn}] (see also \cite{R2})
Integration by parts and the use of the properties
of $(w)_{\mu}^i$ yield
\begin{equation}
\begin{aligned}
&\int_{0}^{T}\int_{\{x\in \Omega; T_k(u_n)-w_{\mu}^i\geq 0\}}
\frac{\partial b_n(x,u_n)}{\partial t}
h_m(u_n)\exp(G(u_n))(T_k(u_n)-w_{\mu}^i)
\,dx\,dt\\
&=\int_{0}^{T}\int_{\{x\in \Omega; T_k(u_n)-w_{\mu}^i\geq 0\}}
\frac{\partial b_n(x,u_n)}{\partial t} h_m(u_n)T_k(u_n)\exp(G(u_n)),dx\,dt\\
&-\int_{0}^{T}\int_{\{x\in \Omega; T_k(u_n)-w_{\mu}^i\geq 0\}}
\frac{\partial b_n(x,u_n)}{\partial t}
h_m(u_n)\exp(G(u_n))w_{\mu}^i dx dt\\
&=I_1^n+I_2^{n,\mu}.
\end{aligned}\label{611}
\end{equation}
We denote
\begin{gather*}
B_{m,k}^n(x,r)=\int_0^r\frac{\partial b_n(x,s)} {\partial
s}h_m(s)T_k(s)\exp(G(s))ds,\\
B_m^n(x,r)=\int_0^r\frac{\partial b_n(x,s)}{\partial
s}h_m(s)\exp(G(s))ds.
\end{gather*}
By a standard argument we can write the first term on the right-hand
side of \eqref{611} as
\begin{equation}\begin{aligned}
I_1^n&=\Big[\int_{\{x\in\Omega;\ T_k(u_n)-w_{\mu}^i\geq
0\}}B_{m,k}^n(x,u_n)dx
\Big]_0^T\\
&=\int_{\{x\in \Omega; \ T_k(u_n)(T)-w_{\mu}^i(T)\geq 0\}}
B_{m,k}^n(x,T_m(u_n)(T))dx\\
&-\int_{\{x\in \Omega;\ T_k(u_n)(0)-w_{\mu}^i(0)\geq
0\}}B_{m,k}^n(x,T_m(u_n)(0))dx.
\end{aligned} \label{622}
\end{equation}
We observe that
$$
\frac{\partial b_n(x,T_m(u_n))}{\partial
s}h_m(u_n)=\left(\frac{\partial b_n(x,T_m(u_n))}{\partial
s}+\frac{1}{n}\right)h_m(u_n)
$$
for $n>m$ with $\operatorname{supp}h_m\subset [-m; m]$. Passing to
the limit in \eqref{622} as $n\to +\infty$, we deduce that
\begin{equation}
\begin{aligned}
I_1^n &=\int_{\{x\in \Omega;\ T_k(u)(T)-w_{\mu}^i(T)\geq
0\}}B_{m,k}(x,T_m(u(T)))dx\\
&-\int_{\{x\in \Omega;\ T_k(u)(0)-w_{\mu}^i(0)\geq
0\}}B_{m,k}(x,T_m(u_{0}))dx+\varepsilon(n).
\end{aligned} \label{6333}
\end{equation}
where $B_{m,k}(x,r)=\int_0^r\frac{\partial b(x,s)}{\partial
s}h_m(s)T_k(s)\exp(G(s))ds$. Passing to the limit in \eqref{6333} as
$i\to +\infty$ and $\mu\to +\infty$, we have
\begin{equation}
\begin{aligned}
I_1^n
=\int_{\Omega}[B_{m,k}(x,u(T))-B_{m,k}(x,u_{0})]dx+\varepsilon(n,\mu,i).
\end{aligned} \label{633}
\end{equation}
The second term on the right-hand side
of \eqref{611} can be written as
\begin{equation}
\begin{aligned}
I_2^{n,\mu} &=-\int_{0}^{T}\int_{\{x\in \Omega;/
T_k(u_n)-w_{\mu}^i\geq 0\}}\frac{\partial
b_n(x,u_n)}{\partial t} h_m(u_n)\exp(G(u_n))w_{\mu}^i dx dt\\
&=-\left[\int_{\{x\in \Omega;\ T_k(u_n)-w_{\mu}^i\geq
0\}}B_m^n(x,u_n)w_{\mu}^idx\right]_0^T\\
&\int_0^T\quad \int_{\{x\in \Omega;\ T_k(u_n)-w_{\mu}^i\geq 0\}}
B_m^n(x,u_n)\frac{\partial w_{\mu}^i}{\partial t}\,dx\,dt\\
&=-\int_{\{x\in \Omega;\ T_k(u_n)(T)-w_{\mu}^i(T)\geq
0\}}B_m^n(x,T_m(u_n(T)))w_{\mu}^i(T)dx\\
&+\int_{\{x\in \Omega;\ T_k(u_n)(0)-w_{\mu}^i(0)\geq
0\}}B_m^n(x,u_{0n})w_{\mu}^i(0)dx\\
&\quad +\mu\int_0^T\int_{\{x\in \Omega;\
T_k(u_n)-w_{\mu}^i\geq 0\}}B_m^n(x,u_n)(T_k(u)-w_{\mu}^i)\,dx\,dt.
\end{aligned} \label{644}
\end{equation}
By passing to the limit as $n$ tends to infinity in \eqref{644},
we obtain
\begin{align*}
I_2^{n,\mu}
&=-\int_{\{x\in \Omega;\ T_k(u)-w_{\mu}^i\geq 0\}}[B_m(x,u(T))
w_{\mu}^i(T)-B_m(x,u_{0})w_{\mu}^i(0)dx\\
&\quad +\mu\int_{\{x\in \Omega;\ T_k(u)-w_{\mu}^i\geq
0\}}\int_0^TB_m(x,u)(T_k(u)-w_{\mu}^i)\,dx\,dt+\varepsilon(n),
\end{align*}
where $B_m(x,r)=\int_0^r\frac{\partial b(x,s)}{\partial
s}h_m(s)\exp(G(s))ds$. Therefore, passing to the limit, in $i$ and
$\mu$ , in the first terms on the right-hand side of the last
equality, we deduce that
\begin{equation}\begin{aligned}
&\int_{\{x\in \Omega;\ T_k(u)-w_{\mu}^i\geq
0\}}[B_m(x,u(T))w_{\mu}^i(T)-B_m(x,u_{0})w_{\mu}^i(0)dx\\
&=\int_{\Omega}[B_m(x,u(T))(T_k(u(T))-B_m(x,u_{0})T_k(u_0))dx+\varepsilon(n,\mu,i).
\end{aligned}\label{666}
\end{equation}
The second term on the right-hand side of \eqref{644} can be
rewritten as
\begin{equation}\begin{aligned}
&\mu\int_0^T\int_{\{x\in \Omega;\ T_k(u)-w_{\mu}^i\geq
0\}}B_m(x,u)(T_k(u)-w_{\mu}^i)\,dx\,dt\\
&=\mu\int_0^T\int_{\{x\in \Omega;\ T_k(u)-w_{\mu}^i\geq
0\}}(B_m(x,u)-B_m(x,T_k(u)))(T_k(u)-w_{\mu}^i)\,dx\,dt\\
&+\mu\int_0^T\int_{\{x\in \Omega;\ T_k(u)-w_{\mu}^i\geq
0\}}(B_m(x,T_k(u))-B_m(x,w_{\mu}^i)(T_k(u)-w_{\mu}^i)\,dx\,dt\\
&+\mu\int_0^T\int_{\{x\in \Omega;\ T_k(u)-w_{\mu}^i\geq
0\}}B_m(x,w_{\mu}^i)(T_k(u)-w_{\mu}^i)\,dx\,dt\\
&=J_1+J_2+J_3,
\end{aligned}\label{677}
\end{equation}
where
\begin{equation}\begin{aligned}
J_1&=\mu\int_0^T\int_{\{x\in \Omega;\ T_k(u)-w_{\mu}^i\geq
0;u>k\}}(B_m(x,u)-B_m(x,k))(k-w_{\mu}^i)\,dx\,dt\\
&\quad +\mu\int_0^T\int_{\{x\in \Omega;\ T_k(u)-w_{\mu}^i\geq
0;u<-k\}}(B_m(x,u)-B_m(x,-k))(-k-w_{\mu}^i)\,dx\,dt\\
&\geq 0.
\end{aligned}\label{688}
\end{equation}
As $B_m(x,z)$ is non-decreasing for $z$ and $-k\leq w_{\mu}^i\leq
k$, it follows that
\begin{equation}
J_2\geq 0.\label{699}
\end{equation}
Moreover,
\begin{equation}\begin{aligned}
J_3&=\mu\int_0^T\int_{\{x\in \Omega;\ T_k(u)-w_{\mu}^i\geq
0\}}B_m(x,w_{\mu}^i)(T_k(u)-w_{\mu}^i)\,dx\,dt\\
&=\int_0^T\int_{\{x\in \Omega;\ T_k(u)-w_{\mu}^i\geq
0\}}B_m(x,w_{\mu}^i)\frac{\partial(w)_{\mu}^i}{\partial t}dx
dt\\
&=\int_{\{x\in \Omega;\ T_k(u)(T)-w_{\mu}^i(T)\geq
0\}}\overline{B}(x,w_{\mu}^i(T))dx\\
&-\int_{\{x\in \Omega;\ T_k(u)(0)-w_{\mu}^i(0)\geq
0\}}\overline{B}((x,w_{\mu}^i(0))dx,
\end{aligned}\label{610}
\end{equation}
where $\overline{B}(x,z)=\int_0^zB_m(x,r)dr$. Also $w_{\mu}^i\to
T_k(u)$ a.e. in $Q$ as $i$ and $\mu$ tends to $+\infty$ and
$|w_{\mu}^i|\leq k$. Then Lebegue's convergence theorem shows that
\begin{equation}
J_3=\int_{\Omega}(\overline{B}(x,T_k(u(T)))-\overline{B}(x,T_k(u_0)))dx
+\varepsilon(n,\mu,i).\label{6111}
\end{equation}
In view of \eqref{666}-\eqref{6111}, one has
\begin{equation}
\begin{aligned}
I_2^{n,\mu}
&\geq -\int_{\Omega}[B_m(x,u(T))T_k(u(T))-B_m(x,u_0)T_k(u_0)]dx\\
&\quad +\int_{\Omega}(\overline{B}(x,T_k(u(T)))-\overline{B}(x,T_k(u_0)))dx
+\varepsilon(n,\mu,i).
\end{aligned}\label{6122}
\end{equation}
As a consequence of \eqref{611}, \eqref{633} and \eqref{6122}, we
deduce that
\begin{equation}
\begin{aligned}
&\int_{\{(x,t)\in\Omega\times(0,T);\ \ T_k(u)-w_{\mu}^i\geq
0\}}\frac{\partial b_n(x,u_n)}{\partial
t} h_m(u_n)\exp(G(u_n))(T_k(u_n)-w_{\mu}^i)dxdt\geq\\
&\geq\int_{\Omega}[B_{m,k}(x,u(T))-B_{m,k}(x,u_{0})]dx\\
&\quad -\int_{\Omega}[B_m(x,u(T))T_k(u(T))-B_m(x,u_0)T_k(u_0)]dx\\
&\quad
+\int_{\Omega}(\overline{B}(x,T_k(u(T)))-\overline{B}(x,T_k(u_0)))dx
+\varepsilon(n,\mu,i).
\end{aligned}\label{6133}
\end{equation}
Observe that for any $z\in \mathbb{R}$ and for almost every $x\in\Omega$,
we have
$$
\overline{B}(x,T_k(z))=B_m(x,z)T_k(z)-B_{m,k}(x,z).
$$
Indeed,
\begin{equation}
\begin{aligned}
\overline{B}(x,T_k(z))
&=\int_0^{T_k(z)}B_m(x,r)dr\\
&=\Big[r\int_0^r \frac{\partial b(x,\sigma)} {\partial
\sigma}h_m(\sigma)\exp(G(\sigma))d\sigma\Big]_0^{T_k(z)}\\
&\quad -\int_0^{T_k(z)}r \frac{\partial b(x,r)}{\partial r}h_m(r)\exp(G(r))dr\\
&=T_k(z)\int_0^{T_k(z)} \frac{\partial b(x,r)}{\partial
r}h_m(r)\exp(G(r))dr\\
&\quad -\int_0^{T_k(z)}T_k(r) \frac{\partial b(x,r)}{\partial r}h_m(r)\exp(G(r))dr\\
&=T_k(z)B_m(x,T_k(z))-B_{m,k}(x,T_k(z)).
\end{aligned}\label{6144}
\end{equation}
This is due to the fact that for $|r|k$ we have
\begin{align*}
&B_{m,k}(x,r)\\
&=\int_0^k\frac{\partial b(x,\sigma)}{\partial \sigma}
h_m(\sigma)\sigma \exp(G(\sigma)) d\sigma+ k\int_k^r\frac{\partial
b(x,\sigma)}{\partial \sigma}h_m
(\sigma)\exp(G(\sigma))d\sigma,
\end{align*}
\begin{align*}
&-T_k(r)B_m(x,r)\\
&=-k\int_0^k\frac{\partial b(x,\sigma)} {\partial
\sigma}h_m(\sigma)\exp(G(\sigma))d\sigma
- k\int_k^r\frac{\partial b(x,\sigma)}{\partial
\sigma}h_m(\sigma)\exp(G(\sigma))d\sigma,
\end{align*}
and
\[
\overline{B}(x,k)=k\int_0^k\frac{\partial b(x,\sigma)} {\partial
\sigma}h_m(\sigma)\exp(G(\sigma))d\sigma- k\int_0^k\frac{\partial
b(x,\sigma)}{\partial \sigma}h_m(\sigma)\exp(G(\sigma))\sigma\,d\sigma.
\]
The case $r<-k$ is similar to the previous one. This conclude the
proof.
\end{proof}
\subsection*{Acknowledgements}
The authors are grateful to Professor H. Redwane for his comments
and suggestions. His article \cite{R2} was the motivation for writing
this article.
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\end{document}