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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2008(2008), No. 30, pp. 1--18.\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 2008 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2008/30\hfil Degenerate stationary problems]
{Degenerate stationary problems with homogeneous
boundary conditions}

\author[K. Ammar, H. Redwane\hfil EJDE-2008/30\hfilneg]
{Kaouther Ammar, Hicham Redwane}  % in alphabetical order

\address{Kaouther Ammar \newline
TU Berlin, Institut f\"ur Mathematik, MA 6-3 \\
Strasse des 17. Juni 136,  10623 Berlin, Germany}
\email{ammar@math.tu-berlin.de,  Fax:+4931421110, Tel: +4931429306}

\address{Hicham Redwane \newline
Facult\'e des sciences juridiques, Economiques et Sociales,
Universit\'e Hassan 1 \\
 B.P. 784,  Settat, Morocco}
\email{redwane\_hicham@yahoo.fr}


\thanks{Submitted January 8, 2008. Published February 28. 2008.}
\subjclass[2000]{35K65, 35F30, 35K35, 65M12}
\keywords{Degenerate; homogenous boundary conditions; diffusion;
\hfill\break\indent continuous flux}

\begin{abstract}
 We are interested in  the degenerate problem
 $$
 b(v)-\mathop{\rm div}a(v,\nabla g(v))=f
 $$
 with the homogeneous boundary condition
 $g(v)=0$  on some part of the boundary.
 The vector field $a$ is supposed to satisfy the
 Leray-Lions conditions and the functions $b,g$ to be continuous,
 nondecreasing and to verify the normalization condition
 $b(0)=g(0)=0$ and the range condition $R(b+g)=\mathbb{R}$.
 Using monotonicity methods,  we prove existence and
 comparison results for renormalized entropy solutions in
 the $L^1$ setting.
\end{abstract}

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

\section{Introduction}

 Let $\Omega$ be a  $C^{1,1}$ bounded open subset of  $\mathbb{R}^N$ with
 regular  boundary if $N >1$.
 We consider the  problem, ($P_{b,g}(f)$),
 \begin{equation} \label{Problemeprincipal}
\begin{gathered} b(v) -  \mathop{\rm div} a(v,\nabla g(v)) = f
\quad \text{in }\Omega \\
g(v)=0 \quad \text{on } \Gamma:= \partial \Omega,
\end{gathered}
\end{equation}
where $b,g:\mathbb{R}\to \mathbb{R}$ are  nondecreasing,
continuous such that  $b(0)=g(0)=0$, $R(b+g)=\mathbb{R}$ and
that  $f \in L^1(\Omega)$.

  The vector field $a:\mathbb{R}\times\mathbb{R}^N\to \mathbb{R}^N$
is supposed to be continuous,
to satisfy the growth condition
  \begin{equation} \label{growth}
|a(r,\xi)-a(r,0)|\leq C(|r|)|\xi|^{p-1}\quad\text{for all }
(r,\xi)\in\mathbb{R}\times\mathbb{R}^N
\end{equation}
  with $C:\mathbb{R}^+\to \mathbb{R}^+$ nondecreasing and
the weak coerciveness condition
\begin{equation} \label{coerciv}
(a(r,\xi)-a(r,0))\cdot\xi+{M}(|r|)\geq {\lambda}(|r|)|\xi|^p
\quad\text{for all }r\in\mathbb{R},\;\xi\in\mathbb{R}^N
\end{equation}
where ${M}:\mathbb{R}^+\to \mathbb{R}$,
${\lambda}:\mathbb{R}^+\to ]0,\infty[$ are continuous functions satisfying,
 for all $k>0$, $\lambda_k:=\inf_{\{r;\,|b(r)|\leq k\}}\lambda(r)>0$ and
$M_k:=\sup_{\{r;\,|b(r)|\leq k\}}M(r)<\infty$.

To prove  the uniqueness result, we assume that $a$ verifies  the
 additional condition
  \begin{equation} \label{additional}
(a(r,\xi)-a(s,\eta))\cdot (\xi-\eta)+B(r,s)(1+|\xi|^p+|\eta|^p)|r-s|
\geq \Gamma_1(r,s)\cdot\xi+{\Gamma}_2(r,s)\cdot\eta,
\end{equation}
  for all $r,s\in\mathbb{R}$, $\xi,\eta\in\mathbb{R}^N$,
for some continuous function $B:\mathbb{R}\times\mathbb{R}\to\mathbb{R}$
and continuous fields
$\Gamma_1,{\Gamma}_2:\mathbb{R}\times\mathbb{R}\to\mathbb{R}^N$.

It is well known that the above problem is ill-posed in the variational
setting. In the sense that there is no existence and uniqueness of a
weak solution in the distributional sense. In order to overcome this
difficulty, we use the notion of  entropy solution introduced
by Krushkov in \cite{Kr1} (see also \cite{Kr2}) and which coincides
which the ``physical'' solution. An other difficulty is related to the
irregularity of the data which is only supposed to be in $L^1(\Omega)$.
The suitable notion of solution which guarantees existence and uniqueness
results in this general frame-work is the so called renormalized entropy
solution (see \cite{BCW,CW,BK}).

The outline of the paper is as follows:
In Section 2, we define the renormalized entropy solution and present
our main results. Then, in section 3, we prove the comparison principle
for bounded solution. Finally, in Section 4, we prove the existence
 of a renormalized entropy solution, the comparison result in the
$L^1$-setting and give some possible extensions of our results.

\section{Definitions, notation  and main results}

\begin{definition}\label{weak} \rm
Let $f\in L^1(\Omega)$. A measurable function
$v:\Omega\to \mathbb{R}$ is said to be a weak solution of
 \eqref{Problemeprincipal}
 if $b(v)\in L^1(\Omega)$, $g(v)\in W^{1,p}(\Omega)$ and
$$
\int_\Omega b(v)\xi+\int_\Omega a(v,\nabla g(v))\cdot\nabla \xi=\int_\Omega f\xi$$
for all $\xi\in W_0^{1,p}(\Omega)\cap L^\infty(\Omega)$.
\end{definition}

\begin{definition}\label{vovelle} \rm
Let $f\in L^1(\Omega)$. A measurable function
 $v:\Omega\to \mathbb{R}$ is said to be a  renormalized entropy
solution of  \eqref{Problemeprincipal} if $b(v)\in L^1(\Omega)$,
$$
g(T_k v)\in W_0^{1,p}(\Omega),\quad \forall k>0
$$
and there exists some families  of non-negative bounded measures $\mu_l:= \mu_l(v)$ and $\nu_l=\nu_l(v)$ on $\overline{\Omega}$ such that
$$
\| \mu_l\|, \;\| \nu_{-l}\| \to 0,
\quad l\to\infty,
$$
and the  following entropy inequalities are satisfied:

 For all $k\in \mathbb{R}$, for all $\xi\in C_0^\infty( \mathbb{R}^N)$
such that $\xi\geq 0$ and sign$^+(-g(k))\xi=0$ a.e. on $\Gamma$,
for all $l\geq k$,
\begin{eqnarray} \label{locetrineq10}
 -\int_{\Omega}   b(v\wedge l) \chi_{\{v\wedge l>k\}}\xi
-\int_\Omega \chi_{\{v\wedge l>k\}} (a(v\wedge l, \nabla g(v\wedge l))-a(k,0)) \cdot
\nabla \xi \nonumber\\
\;\;\;\; +\int_\Omega \chi_{\{k>v\wedge l\}} f\xi
\geq -\langle \mu_l,\xi\rangle
\end{eqnarray}
and for all $k\in \mathbb{R}$, for all $\xi\in C_0^\infty( \mathbb{R}^N)$
such that $\xi\geq 0$ and sign$^+(g(k))\xi=0$ a.e. on $\Gamma$,
for $l\leq k$,
\begin{eqnarray} \label{locetrineq20}
 \int_{\Omega}   b(v\vee l)\chi_{\{k>v\vee l\}}\xi+\int_\Omega
 \chi_{\{k>v\vee l\}} (a(v\vee l, \nabla g(v\vee l))-a(k,0)) \cdot
 \nabla \xi\nonumber\\
\;\;\;\;- \int_\Omega \chi_{\{k>v\vee l\}} f\xi 
\geq -\langle \nu_l,\xi\rangle .
 \end{eqnarray}
\end{definition}

\begin{remark}\label{rema} \rm
(i) In the case where the data $f\in L^\infty(\Omega)$,
it is easily verified that a renormalized entropy solution $v$ of
\eqref{Problemeprincipal} is such that $b(v)\in L^1(\Omega)$,
 $g(v)\in W_0^{1,p}(\Omega)$ and $v$ satisfies the following entropy
inequalities:
For all $k\in \mathbb{R}$, for all $\xi\in C_0^\infty( \mathbb{R}^N)$ such that $\xi\geq 0$ and sign$^+(-g(k))\xi=0$ a.e. on $\Gamma$,
\begin{equation} \label{loclocetrineq1}
   -\int_\Omega \chi_{\{v>k\}} (a(v, \nabla g(v))-a(k,0))
\cdot \nabla \xi +\int_\Omega \chi_{\{v>k\}}f\xi\geq \int_{\Omega}
b(v)\chi_{\{v>k\}} \xi
\end{equation}
and for all $k\in \mathbb{R}$, for all $\xi\in C_0^\infty( \mathbb{R}^N)$
such that $\xi\geq 0$ and sign$^+(g(k))\xi=0$ a.e. on $\Gamma$,
\begin{equation} \label{loclocetrineq2}
\int_\Omega \chi_{\{k>v\}} (a(v, \nabla g(v))-a(k,0))\cdot
 \nabla \xi- \int_\Omega\chi_{\{k>v\}} f\xi \geq \int_{\Omega}
 -b(v)\chi_{\{k>v\}}\xi.
 \end{equation}
In this case, $v$ is called weak entropy solution of
 \eqref{Problemeprincipal}.

If moreover, the function $b$ is strictly increasing on $\mathbb{R}$
with $R(b)=\mathbb{R}$, then the weak entropy solution $v$ is also
in $L^\infty(\Omega)$.

 (ii) If $v$ is a renormalized entropy solution of
\eqref{Problemeprincipal}, then $-v$ is an entropy solution of
\begin{equation} \label{Problemeprincipall}
\begin{gathered}
\overline{b}(v) -  \mathop{\rm div} \overline{a}(v,\nabla \overline{g}(v))
= \tilde{f}\quad \text{in }\Omega \\
\overline{g}(v)=0 \quad \text{on } \Gamma:= \partial \Omega,
\end{gathered}
\end{equation}
where $\overline{b}(r)=-b(-r)$, $\overline{g}(r)=-g(-r)$ and
$\overline{a}(r,\xi)=-a(-r,\xi)$.
\end{remark}

 The main result of this paper is the following.

\begin{theorem}\label{main}
 For any $f \in L^1(\Omega)$,  there
exists  a unique pair  $(b(v),g(v))$ such that $v$ is a
 renormalized entropy solution of \eqref{Problemeprincipal}.
\end{theorem}

The uniqueness result  follows as a consequence of an   $L^1$-comparison
principle.

\subsection*{Some notation}
Throughout this paper we use the operators
$$
H_\delta(s):=\min({{s^+}\over{\delta}},1), \quad
H_0(s)=\begin{cases}
1&\text{if }s>0\\
0&\text{if }s\leq 0
\end{cases}
$$
and we denote
\begin{equation} \label{E}
E:=\{r\in R(g)/(g^{-1})_0 \text{ is discontinuous at } r\}.
\end{equation}
For $k>0$, $T_k$ is  the truncation function defined on $\mathbb{R}$
by
$$
T_k(r)=\mathop{\rm sign}\nolimits^0(r)(|r|\wedge k)
$$
and for $r\in\mathbb{R}$, we define $r^+=r\vee 0$, $r^-=r\wedge 0$.

\section{Proofs of comparison and uniqueness results}

We first prove the comparison result in the $L^\infty$-setting.

\begin{theorem}\label{comp}
 For $i=1,2$, let $f_i\in L^\infty(\Omega) $ and
$v_i \in L^\infty(\Omega)$ be a weak  entropy solution of $P_{b,g}(f_i)$.
Then there exist $\kappa \in L^\infty (\Omega)$ with
$\kappa \in \mathop{\rm sign}\nolimits^+(v_1-v_2)$ a.e. in $\Omega$
such that, for any $\xi \in \mathcal{D}(\mathbb{R}^N)$, $\xi \geq 0$,
\begin{equation}  \label{L1comparison}
 \int_\Omega (b(v_1)-b(v_2))^+\xi
\leq \int_\Omega \kappa (f_1-f_2) \xi
- \int_\Omega\chi_{\{v_1 >v_2\}}(a(v_1,\nabla g(v_1))-a(v_2,
\nabla g(v_2))\cdot \nabla \xi.
\end{equation}
\end{theorem}

\begin{lemma}\label{Jose}
Let $f\in L^\infty(\Omega)$ and    $v$ be a weak  solution of
\eqref{Problemeprincipal}. Then
\begin{equation} \label{A}
\begin{aligned}
&\int_\Omega \chi_{\{v>k\}}((a(v,\nabla g(v)))-a(k,0))\cdot\nabla \xi
 +\int_\Omega \chi_{\{v>k\}}\{b(v)\xi-f\xi\}\,dx\, \\
&=-\lim_{\delta\to 0}\int_\Omega (a(v,\nabla g(v))-a(v,0))
\cdot \nabla g(v) H_\delta'(g(v)-g(k))\xi\,dx\,,
\end{aligned}
\end{equation}
for any $(k,\xi)\in \mathbb{R}\times \mathcal{D}^+({\Omega})$ such that
$g(k)\notin E$ and $(g(v)-g(k))^+\xi=0$ a.e. on $\Gamma$. Moreover,
\begin{equation} \label{B}
\begin{aligned}
&\int_\Omega \chi_{\{k>v\}}(a(v,\nabla g(v))-a(k,0))\cdot\nabla \xi+\int_\Omega \chi_{\{k>v\}}\{b(v)\xi-f\xi\}\,dx\, \\
&=\lim_{\delta\to 0}\int_\Omega (a(v,\nabla g(v))-a(v,0))
\cdot \nabla g(v)H_\delta'(g(k)-g(v))\xi\,dx\,,
\end{aligned}
\end{equation}
for any $(k,\xi)\in \mathbb{R}\times \mathcal{D}^+({\Omega})$ such
that $g(k)\notin E$ and $(g(k))^+\xi=0$ a.e. on $\Gamma$.
\end{lemma}
From now on, we denote $\tilde{a}(r,\xi)=a(r,\xi)-a(r,0),\;\;r\in{\mathbb R},\;\xi\in{\mathbb R}^N.$
 The proof of the above lemma follows the same lines as  the proof in
\cite[Lemma 2.5]{Ca1}.

\begin{proof}[Proof of Theorem \ref{comp}]
Let $(B_i)_{i=0 \dots m}$ be a covering of $\overline{\Omega}$
satisfying $B_0 \cap \partial \Omega= \emptyset$ and such that,
for each $i \geq 1$, $B_i$ is a ball  contained in some larger
ball ${B}_i$ with ${B}_i \cap \partial \Omega $ is part of the
graph of a Lipschitz function.
Let $(\wp_i)_{i=0\dots m}$ denote a partition of unity subordinate
to the covering $(B_i)_i$ and denote by  $\xi$ an arbitrary function
in $\mathcal{D}(\mathbb{R}^N)$, $\xi \geq 0$.

We use Kruzhkov's technique of doubling variables in order to prove
the comparison result ( see \cite{Kr1,Kr2,CW}, etc).
We choose two variables $x$ and $y$ and consider $v_1$ as function of $y$
and $v_2$ as function of $x\in \Omega$.
Define the test function
$\xi_{n}^i:(x,y)\mapsto \wp_i(x)\xi(x){\varrho}_n(x-y)$,
where $(\varrho_n)_n$ is a sequence of mollifiers in $\mathbb{R}^N$
such that $x\mapsto \varrho_n(x-y)\in \mathcal{D}(\Omega)$,
for all $y\in B_i$, $\sigma_n(x)=\int_\Omega \varrho_n(x-y)\,d\,y$
is an increasing sequence for all $x\in B_i$, and $\sigma_n(x)=1$ for
all $x\in B_i$ with $d(x,\mathbb{R}^N\setminus\Omega)>{{c}\over{N}}$
for some $c=c(i)$ depending on $B_i$. Then,  for $n$ sufficiently large,
\begin{gather*}
y \mapsto \xi_{n}^i(x,y) \in \mathcal{D}( \mathbb{R}^N), \quad
\text{for any } x \in \Omega,\\
 x \mapsto \xi_{n}^i(x,y) \in \mathcal{D}( \Omega), \quad
\text{ for any } y \in \Omega\\
\mathop{\rm supp}\nolimits_y(\xi_{n}^i(x,.))\subset B_i, \quad
\text{ for any }x\in \mathop{\rm supp}(\wp_i).
\end{gather*}
For convenience, we sometimes omit the index $i$  and  simply set
 $\wp=\wp_i$, $B=B_i$ and $\xi_{n}^i=\xi_{n}$.
Then $ \hat{\zeta}_n(x)  := \xi(x) \wp(x)\sigma_n({x})$
satisfies $\hat{\zeta}_{n} \in \mathcal{D}(\Omega)$,
 $0 \leq \hat{\zeta}_{n} \leq \xi$,   for all $n\in \mathbb{N}$.
 Let
\[
\Omega_1:=\{y\in \Omega/ v_1(y)\in E\},\quad
\Omega_2:=\{x\in \Omega/v_2(x)\in E\}.
\]
Then,  $\nabla_y g(v_1)=0$ a.e in $\Omega_1$ and $\nabla_xg(v_2)=0$
a.e in $\Omega_2$.
Moreover,
$ H_0(v_1-v_2)=H_0(g(v_1)-g(v_2))$ a.e in $(\Omega\setminus \Omega_1)\times \Omega\cup \Omega\times (\Omega\setminus \Omega_2)$.



\subsection*{First inequality}
 We first prove the following inequality:
\begin{equation} \label{firsthalfjdid}
\begin{aligned}
&\int_\Omega (b(v_1^+)-b(v_2^+)) \xi  \wp  \\
&\leq  \int_\Omega  \kappa_1 \chi_{\{ v_1 > 0\}}
(f_1-  \chi_{\{v_2 \geq 0\}}f_2 ) \xi  \wp \\
&\quad -\int_\Omega \chi_{\{v_1^+> v_2^+\}}(a(v_1^+,\nabla g(v_1^+))
-a(v_2^+,\nabla g(v_2^+))\cdot\nabla_x(\xi\wp)
+ \lim_{n \to \infty}\mathcal{L}(\xi \wp\sigma_n)
\end{aligned}
\end{equation}
where $\kappa_1\in L^\infty(\Omega),\,\kappa_1\in
\text{ sign}^+(v_1-v_2^+)$ and $\mathcal{L}$
is a linear operator which will be  defined later.

 As $v_1$ satisfies (\ref{A}) (with $v=v_1$,  $f=f_1$ ),
choosing $k = v_2^+(x)$ and $\xi(y)=\zeta_{n}(x,y)$ in
(\ref{loclocetrineq1}), integrating in  $x$ over $\Omega$,  we find
\begin{equation} \label{olfa}
\begin{aligned}
&\lim_{\delta\to 0}\int_{\{\Omega\setminus \Omega_1\}
 \times \{\Omega\setminus \Omega_2\}} \tilde{a}(v_1,\nabla_y g(v_1))\cdot
 \nabla_y g(v_1)H_\delta'(g(v_1)-g(v_2^+))\zeta_{n} \\
&=\lim_{\delta\to 0}\int_{\Omega\times \{\Omega\setminus \Omega_2\}}
 \tilde{a}(v_1,\nabla_y g(v_1))\cdot \nabla_y g(v_1)H_\delta'(g(v_1)
 -g(v_2^+))\zeta_{n} \\
&\leq  -\int_{\Omega\times \Omega}\chi_{\{v_1 > v_2^+\}}\{ b(v_1)
\zeta_{n}- f_1 \zeta_{n}+\tilde{a}(v_1^+,\nabla_y g(v_1^+ ))
 \cdot\nabla_y \zeta_{n}\\
&\quad +(a(v_1^+,0)-a(v_2^+,0))\cdot\nabla_y\zeta_n\}.
\end{aligned}
\end{equation}
Now, since
$x\mapsto \zeta_{n}(x,y)H_\delta(g(v_1^+ )-g(v_2^+))\in \mathcal{D}(\Omega)$
 for a.e. $y\in \Omega$,
we have
\begin{equation} \label{explication1}
\int_{\Omega\times \Omega} \tilde{a}(v_1^+ ,\nabla_y g(v_1^+ ))
\cdot\nabla_x(H_\delta(g(v_1^+ )-g(v_2^+))\zeta_{n})\,=0.
\end{equation}
 Therefore, going to the limit in $\delta$, we get
\begin{equation} \label{secret1}
\begin{aligned}
&\lim_{\delta\to 0}\int_{\{\Omega\setminus \Omega_1\}\times
\{\Omega\setminus \Omega_2\}}\tilde{a}(v_1^+ ,\nabla_y g(v_1^+ ))
\cdot \nabla_x g(v_2^+)
H_{\delta}' (g(v_1^+ )-g(v_2^+))\zeta_{n} \\
&=\int_{\Omega\times \Omega} H_0(g(v_1^+ )-g(v_2^+))\tilde{a}(v_1^+
 ,\nabla_y g(v_1^+ ))\cdot \nabla_x\zeta_{n}  \\
&=\int_{\Omega\times \Omega} H_0(v_1^+ -v_2^+) \tilde{a}(v_1,\nabla_y g(v_1^+ ))
 \cdot\nabla_x\zeta_{n}.
\end{aligned}
\end{equation}
Arguing as in \cite{Ca1}, inequality (\ref{olfa}) can  be written
as
\begin{equation} \label{najiha1}
\begin{aligned}
&\int_{\Omega\times \Omega}\{ -\tilde{a}(v_1^+ ,\nabla_y g(v_1^+ ))
\cdot \nabla_{x+y} \zeta_{n}-b(v_1^+ )
\zeta_{n}+f_1\zeta_{n}\\
&\;\;\;\;\;\;\;-(a(v_1^+,0)-a(v_2^+,0))\cdot\nabla_y\zeta_n\}H_0(v_1^+ -v_2^+) \\
&\geq \lim_{\delta\to 0}\int_{\{\Omega\setminus \Omega_1\}\times\{\Omega\setminus \Omega_2\}}\tilde{a}(v_1^+ ,\nabla_y g(v_1^+ ))
 \cdot \nabla_{x+y} (g(v_1^+ )- g(v_2^+))\\
 &\quad \times H_{\delta}'(g(v_1^+ )-g(v_2^+))\zeta_{n}
\end{aligned}
\end{equation}
with  $\nabla_{x+y}(\cdot):=\nabla_{x}(\cdot)+\nabla_{y}(\cdot)$.
Now, as  $v_2$ is a weak  entropy solution of \eqref{Problemeprincipal}
with $f_2$ instead of $f$,
choosing $k=v_1^+(y)$, $\xi(x)=\zeta_{n}(x,y)$ in \eqref{B}
(with $v=v_2$, $f=f_2$), integrating in  $y$ over  $ \Omega$,   we find
\begin{equation} \label{najwa}
\begin{aligned}
&-\lim_{\delta\to 0}\int_{\{\Omega\setminus \Omega_1\}\times \Omega}
 \tilde{a}(v_2,\nabla_x g(v_2))\cdot \nabla_x g(v_2)H_\delta'(g(v_1^+ )
 -g(v_2))\zeta_{n} \\
&=-\lim_{\delta\to 0}\int_{\{\Omega\setminus \Omega_1\}\times
 \{\Omega\setminus \Omega_2\}}\tilde{a}(v_2,\nabla_x g(v_2))\cdot
 \nabla_x g(v_2)  H_\delta'(g(v_1^+ )-g(v_2))\zeta_{n} \\
&\leq\int_{\Omega\times \Omega} \chi_{\{v_1^+  > v_2\}}\{
b(v_2) \zeta_{n}-f_2 \zeta_{n} + \tilde{a}(v_2,\nabla_x g(v_2))\cdot
\nabla_x \zeta_{n}+(a(v_1^+,0)\\
&\quad -a(v_2,0))\cdot \nabla_x\zeta_n\}
\end{aligned}
\end{equation}
It is easily verified that
\begin{equation} \label{21}
\begin{aligned}
&\int_{\{\Omega\setminus \Omega_1\}\times \{\Omega\setminus \Omega_2\}}
  \tilde{a}(v_2,\nabla_x g(v_2))\cdot \nabla_x g(v_2)H_\delta'(g(v_1^+ )
  -g(v_2))\zeta_{n} \\
&=\int_{\{\Omega\setminus \Omega_1\}\times \{\Omega\setminus \Omega_2\}}
\tilde{a}(v_2^+,\nabla_x g(v_2^+))\cdot \nabla_x g(v_2^+)H_\delta'
 (g(v_1^+ )-g(v_2^+))\zeta_{n} \\
&\quad +\int_{\{\Omega\setminus \Omega_1\}\times \{\Omega\setminus \Omega_2\}
}  \tilde{a}(v_2^-,\nabla_x g(v_2^-))\cdot \nabla_x g(v_2^-)
 H_\delta'(g(v_1^+ )-g(v_2^-))\zeta_{n}
\end{aligned}
\end{equation}
and that the second term in the right hand side of (\ref{21}) converges
to $0$ with $\delta\to 0$.
Moreover, the right hand side of (\ref{najwa}) is equal to
\begin{equation} \label{22}
\begin{aligned}
&\int_{\Omega\times \Omega} \chi_{\{v_1^+  > v_2^+\}}
\{b(v_2^+) \zeta_{n}-  \chi_{\{v_2 \geq 0\}}f_2\zeta_{n}
 +(\tilde{a}(v_2^+,\nabla_x g(v_2^+))-a(v_1^+,0) \\
 & +a(v_2^+,0))\cdot \nabla_x \zeta_{n}\} 
 + \int_{\Omega\times \Omega}\chi_{\{v_2<0\}}\{ b(v_2)\zeta_{n}
  - f_2\zeta_n- a(v_2,\nabla_xg(v_2))\cdot \nabla_x \zeta_{n}\}.
\end{aligned}
\end{equation}
Since
  $y\mapsto \zeta_{n}(x,y)H_\delta(g(v_1^+ )-g(v_2^+))\in
 \mathcal{D}(\Omega)$  for a.e. $(x)\in \Omega$, we have
\begin{equation} \label{explication2}
\int_{\Omega\times \Omega} \tilde{a}(v_2^+,\nabla_x g(v_2^+))
\cdot \nabla_y(H_\delta(g(v_1^+ )-g(v_2^+))\zeta_{n})=0.
\end{equation}
Therefore,
\begin{equation} \label{secret2}
\begin{aligned}
&-\lim_{\delta\to 0}\int_{\{\Omega\setminus \Omega_1\}\times
\{\Omega\setminus \Omega_2\}} \tilde{a}(v_2^+,\nabla_x g(v_2^+))\cdot
\nabla_y g(v_1^+ )H_{\delta}' (g(v_1^+ )-g(v_2^+))\zeta_{n} \\
&=\int_{\Omega\times \Omega} H_0(g(v_1^+ )-g(v_2^+))\tilde{a}(v_2^+,
\nabla_x g(v_2^+))\nabla_y\zeta_{n}.
\end{aligned}
\end{equation}
Then, the inequality (\ref{najwa}) can be equivalently written as
\begin{equation} \label{najwa'}
\begin{aligned}
&\int_{\Omega\times \Omega}
b(v_2^+)H_0(v_1^+ -v_2^+)\zeta_{n}-\int_{\Omega\times \Omega}
 \chi_{\{v_1^+ >v_2^+\}}\chi_{\{v_2\geq 0\}}f_2\zeta_{n} \\
& +\int_{\Omega\times \Omega} H_0(v_1^+ -v_2^+)\tilde{a}(v_2^+,
 \nabla_x g(v_2^+))\cdot (\nabla_y\zeta_{n}+\nabla_x\zeta_{n}) \\
&-\int_{\Omega\times \Omega} H_0(v_1^+ -v_2^+)(a(v_1^+,0)-a(v_2^+,0))\cdot\nabla_x\zeta_n\\
& +\int_{\Omega\times \Omega}\chi_{\{v_2<0\}}\{ b(v_2)\zeta_{n}
  - f_2\zeta_{n}- a(v_2,\nabla_xg(v_2))\cdot \nabla_x \zeta_{n}\} \\
&\geq \lim_{\delta\to 0}\int_{\{\Omega\setminus \Omega_1\}\times
  \{\Omega\setminus \Omega_2\}}
\tilde{a}(v_2^+,\nabla_x g(v_2^+))\cdot (\nabla_x g(v_2^+)-\nabla_y g(v_1^+ ))\\
&\quad\times H_\delta '(g(v_1^+ )-g(v_2^+))\zeta_{n}.
\end{aligned}
\end{equation}
Summing up inequalities (\ref{najiha1}) and (\ref{najwa'}),  we get
\begin{equation} \label{star}
\begin{aligned}
&\lim_{\delta\to 0}\int_{(\Omega\setminus \Omega_1)\times
(\Omega\setminus \Omega_2)}  (\tilde{a}(v_1^+ ,\nabla_y g(v_1^+ ))-\tilde{a}(v_2^+,
\nabla_x g(v_2^+)))\cdot
(\nabla_y g(v_1^+ )-\nabla_xg(v_2^+)) \\
&\times H_\delta'(g(v_1^+ )-g(v_2^+))\zeta_{n} \\
&\leq -\int_{\Omega\times \Omega} (b(v_1^+) -b(v_2^+))^+ \xi
  \wp {\varrho}_n - \int_{\Omega\times \Omega} b(v_2)\chi_{\{v_2<0\}}
 \zeta_{n}  \\
&\quad + \int_{\Omega\times \Omega}  \chi_{\{v_1^+  > v_2^+\}}
 \chi_{\{ v_1 > 0\}} (f_1 - \chi_{\{v_2 \geq 0\}}f_2 ) \zeta_{n}
 +\int_{\Omega\times \Omega} \chi_{\{v_2<0\}}f_2\zeta_{n}  \\
&\quad -\int_{\Omega\times \Omega} (a(v_1^+ ,\nabla_y g(v_1^+ ))-a(v_2^+,
 \nabla_x g(v_2^+)))\cdot(\nabla_{x+y}\zeta_{n})H_0(v_1^+ -v_2^+) \\
 &\quad - \int_{\Omega\times \Omega} \chi_{\{0 > v_2\}}a(v_2,
 \nabla_x g(v_2))\cdot\nabla_x\zeta_{n}.
\end{aligned}
\end{equation}
Denote the  integrals on the right hand side of (\ref{star}) by
 $I_1, \dots, I_6$ successively. Going to the limit with  $n$, one get
\begin{gather*}
\lim_{n\to \infty}I_1=-\int_\Omega (b(v_1^+ )-b(v_2^+))^+\xi\wp,\\
\limsup_{n \to \infty} I_3 \leq \int_\Omega \kappa_1 \chi_{\{ v_1 > 0\}}
(f_1-  \chi_{\{v_2 \geq 0\}}f_2) \xi \wp
\end{gather*}
 for some
\begin{equation}
\label{kappa}
\kappa_1 \in L^\infty(\Omega)\text{  with }
\kappa_1 \in \mathop{\rm sign}\nolimits^+ (v_1 - v_2^+)\quad
\text{a.e. in } \Omega,
\end{equation}
$$
\limsup_{m,n\to +\infty}I_5 =  - \int_\Omega H_0(v_1^+ -v_2^+)(a(v_1,
\nabla g(v_1^+ ))-a(v_2,\nabla g(v_2^+)))\cdot \nabla (\xi\wp),
$$
It remains to estimate
\[
I_2+I_4+I_6= \int_\Omega \chi_{\{v_2<0\}} \{b(v_2)\hat{\zeta}_n
 - f_2 \hat{\zeta}_n
+ a(v_2,\nabla_x g(v_2))\cdot\nabla_x \hat{\zeta}_n\}.
\]
Define the functional $\mathcal{L}$  on  $\mathcal{D}( \Omega)$  by
\begin{equation} \label{calL}
 \mathcal{L}(\zeta)=\int_\Omega b(v_2)\chi_{\{v_2<0\}} \zeta
 - \chi_{\{0 > v_2\}}  f_2 \zeta
+\int_\Omega \chi_{\{0 > v_2\}} a(v_2,\nabla_x g(v_2))\cdot\nabla_x\zeta.
\end{equation}
As $v_2$ is an entropy solution, we have $\mathcal{L}(\zeta) \geq 0$
for all $\zeta \in \mathcal{D}(\Omega)$, $\zeta \geq 0$, a.e.
$\mathcal{L}$ is a positive linear functional on $ \mathcal{D}(\Omega) $.
Since $(\hat{\zeta})_n=(\xi \sigma_n)_n$$\subset \mathcal{D}(\Omega)$
is an increasing sequence satisfying $0 \leq \xi \sigma_n \wp\leq \xi\wp$,
$\mathcal{L}(\hat{\zeta}_n) $ is a bounded and increasing sequence and
thus converges. As a consequence,
$I_2+I_4+I_6= \mathcal{L}(\xi \wp\sigma_n)$ converges as $n \to \infty$.

To estimate the first term in the left hand side of (\ref{star}),
we use the additional hypothesis (\ref{additional}) on the vector
field $a$:
\begin{equation} \label{khorma}
\begin{aligned}
&\int_{(\Omega\setminus \Omega_1)\times (\Omega\setminus \Omega_2)}
(a(v_1^+ ,\nabla_y g(v_1^+ )-a(v_2^+,\nabla_x g(v_2^+)))
\cdot (\nabla_y g(v_1^+ )-\nabla_xg(v_2^+)) \\
&\times H_\delta'(g(v_1^+ )-g(v_2^+))\zeta_{n} \\
&\geq -{{1}\over{\delta}}\int_{(\Omega\setminus\Omega_1)\times
(\Omega\setminus\Omega_2)} \zeta_{n} B(v_1^+ , v_2^+)
\times (1+|\nabla_y g(v_1^+ ))|^p+|\nabla_x g(v_2^+)|^p)|v_1^+ -v_2^+| \\
&\quad\times \chi_{\{0\leq g(v_1^+ )-g(v_2^+)\leq \delta\}} \\
&\quad +{{1}\over{\delta}}\int_{(\Omega\setminus\Omega_1)\times
 (\Omega\setminus\Omega_2)}
\zeta_{n}\Gamma_1(v_1^+ ,v_2^+)\cdot \nabla_y g(v_1^+ )
 \chi_{\{0\leq g(v_1^+ )-g(v_2^+)\leq \delta\}} \\
&\quad +{{1}\over{\delta}}\int_{(\Omega\setminus\Omega_1)\times
 (\Omega\setminus\Omega_2)}
\zeta_{n}{\Gamma}_2(v_1^+ ,v_2^+)\cdot \nabla_x g(v_2^+)
 \chi_{\{0\leq g(v_1^+ )-g(v_2^+)\leq \delta\}}.
\end{aligned}
\end{equation}
The two last terms in the right hand side of (\ref{khorma}) can be
estimated as follows
\begin{align*}
&{{1}\over{\delta}}\int_{(\Omega\setminus\Omega_1)\times
(\Omega\setminus\Omega_2)}\zeta_{n}\Gamma_1(v_1^+ ,v_2^+)
\cdot \nabla_y g(v_1^+ )\chi_{\{0\leq g(v_1^+ )-g(v_2^+)\leq \delta\}} \\
&=\int_{(\Omega\setminus \Omega_1)\times (\Omega\setminus \Omega_2)}
\Big(\int_0^{\gamma(v_1,v_2)}
\Gamma_1((g^{-1})_0(g(v_2^+)+\delta r),(g^{-1})_0((g(v_2^+))\,dr\Big)
\nabla_y\zeta_{n}
\end{align*}
and
\begin{align*}
&{{1}\over{\delta}}\int_{(\Omega\setminus\Omega_1)\times
(\Omega\setminus\Omega_2)}
 \zeta_{n}{\Gamma}_2(v_1^+ ,v_2^+)\cdot \nabla_x g(v_2^+)
 \chi_{\{0\leq g(v_1^+ )-g(v_2^+)\leq \delta\}} \\
&=\int_{(\Omega\setminus \Omega_1)\times (\Omega\setminus \Omega_2)}
\Big(\int_0^{\gamma(v_1,v_2)}
\Gamma_2((g^{-1})_0(g(v_1^+ )),(g^{-1})_0(g(v_1^+ )-\delta r))\,dr\Big)
\nabla_x\zeta_{n},
\end{align*}
where
$$
\gamma(v_1,v_2):=\inf  (g(v_1^+ )-g(v_2^+))^+/\delta,1).
$$
Due to the continuity of $\Gamma((g^{-1})_0(r),\xi)$ in
$r, g(r)\notin E$, it follows that
the two terms converge to 0 with $\delta$.
In order to estimate the remaining term, we use the estimation
$$
|r-s|=|(g^{-1})_0(g(r))-(g^{-1})_0(g(r))|
 \leq C |g(r )-g(s)|, \quad g(r)\notin E,\,g(s)\not\in E
$$
where $C$ is the Lipschitz constant of $(g^{-1})_0$ on
$\{r\in\mathbb{R}, b(r)\notin E, |r|\leq |g(v_1)+g(v_2)|\}$.
Then, we have
\begin{align*}
&-\lim_{\delta\to 0}{{1}\over{\delta}}\int_{(\Omega\setminus\Omega_1)
 \times (\Omega\setminus\Omega_2)}
 \zeta_{n} B(v_1^+ , v_2^+)\times (1+|\nabla_y g(v_1^+ )|^p
 +|\nabla_x g(v_2^+)|^p)|v_1^+ -v_2^+| \\
&\times \chi_{\{0\leq g(v_1^+ )-g(v_2^+))\}} \\
&\geq -C \lim_{\delta\to 0}\int_{(\Omega\setminus\Omega_1)\times
 (\Omega\setminus\Omega_2)}
\zeta_{n} B(v_1^+ , v_2^+)\times (1+|\nabla_y g(v_1^+ )|^p
 +|\nabla_x g(v_2^+)|^p) \\
&\quad\times \chi_{\{0\leq g(v_1^+ )-g(v_2^+)\}}
=0.
\end{align*}
Using similar arguments, we prove that
\begin{align*}
 &-\lim_{\delta\to 0}\int_{(\Omega\setminus \Omega_1)\times
(\Omega\setminus \Omega_2)}  (a(v_1^+ ,0)-a(v_2^+,0))\cdot
(\nabla_y g(v_1^+ )-\nabla_xg(v_2^+)) \\
&\times H_\delta'(g(v_1^+ )-g(v_2^+))\zeta_{n}=0. 
\end{align*}

Combining the  estimates of $I_1 , \dots, I_6$, we get
\begin{equation} \label{firsthalf}
\begin{aligned}
&\int_\Omega (b(v_1^+) -b(v_2^+))^+ \xi  \wp \\
&\leq  \int_\Omega  \kappa_1 \chi_{\{ v_1 > 0\}}
 (f_1-  \chi_{\{v_2 \geq 0\}}f_2 ) \xi  \wp
 + \lim_{n \to \infty}\mathcal{L}(\xi \wp\sigma_n) \\
&\quad -\int_\Omega (a(v_1^+ ,\nabla_xg(v_1^+))-a(v_2^+,
\nabla g(v_2^+)))\cdot\nabla_x(\xi\wp)\chi_{\{v_1^+ > v_2^+ \}}.
\end{aligned}
\end{equation}
 This is ``half'' of the inequality to be proved.

\subsection*{Second inequality:}
In view of Remark \ref{rema}, inequality (\ref{firsthalf})
is still true when $v_1$ is replaced by $-v_2$,  $v_2$ is replaced
 by $-v_1$, $f_1$ by $-f_2$, $f_2$ by $-f_1$, $b$ by $\overline{b}$,
$g$ by $\overline{g}$ and $a$ by $\overline{a}$. Then we have
\begin{equation}  \label{secondhalf}
\begin{aligned}
&\int_\Omega (b(v_1^-) -b(v_2^-))^+ \xi \wp_i\\
&\leq  \int_\Omega  \kappa_2 \chi_{\{ v_2 <  0\}} (\chi_{\{v_1 \leq 0\}}
f_1-  f_2 ) \xi \wp_i \\
&\quad -\int_\Omega \!\chi_{\{v_1^- \geq v_2^- \}} (a(v_1^-,
\nabla g(v_1^-))-a(v_2^-,\nabla g( v_2^-))) \cdot
\nabla_x (\xi \wp_i)+ \lim_{n \to \infty}\mathcal{L}(\xi \sigma_n\wp_i) ,
\end{aligned}
\end{equation}
where
\[
\mathcal{L}(\xi):= \int_\Omega (b(v_1))^+\zeta
+ \int_\Omega\chi_{\{v_1 >0\}} \{a(v_1,\nabla g(v_1))\cdot
\nabla_y \zeta
+f_1\zeta\}.
\]
Using the same arguments as above, we can prove that
$(\mathcal{L}(\xi \sigma_n\wp_i))$ converges
(as $\mathcal{L}(\xi\sigma_n\wp_i)$)  with $n$.

Therefore, summation of  (\ref{firsthalf}) and (\ref{secondhalf}) yields
\begin{equation}  \label{inequi}
\begin{aligned}
 &\int_\Omega (b(v_1) -b(v_2))^+ \xi \wp_i\\
& \leq  \int_\Omega  \kappa (f_1-  f_2 ) \xi \wp_i
  -\int_\Omega \chi_{\{v_1 \geq v_2 \}} (a(v_1,
 \nabla g(v_1))-a(v_2,\nabla g(v_2) )) \cdot
\nabla_x (\xi \wp_i)  \\
&\quad + \lim_{n \to \infty} \mathcal{L}(\xi\wp_i \sigma_n)
+ \lim_{n \to \infty}\mathcal{L}(\xi \wp_i\sigma_n),
\end{aligned}
\end{equation}
for any $\xi \in \mathcal{D}( \mathbb{R}^N)$,
$\xi \geq 0$, for all $ i \in \{1, \dots, m\}$.

\begin{remark} \label{rmk3.3} \rm
 The method of doubling variables allows to prove the following local
 comparison result: for all $\xi\in \mathcal{D} ( \Omega)$,
here exists $\kappa \in L^\infty(\Omega)$ with
 $\kappa \in \mathop{\rm sign}\nolimits^+(v_1-v_2)$ a.e. in
$\Omega$ such that,  for any $\zeta \in \mathcal{D}(\Omega)$,
 $\zeta \geq 0$,
\begin{equation}  \label{local}
\begin{aligned}
&\int_\Omega (b(v_1)-b(v_2))^+\zeta +\int_\Omega  \chi_{\{v_1 >v_2\}}
(a(v_1,\nabla g(v_1))-a(v_2,\nabla g(v_2))\cdot \nabla\zeta \\
&\leq \int_\Omega \kappa (f_1-f_2) \zeta.
\end{aligned}
\end{equation}
The proof in this case is easier as the global comparison result.
Indeed, as $\xi=0$ on $\Gamma$, we can choose $k=v_2(x)$
(resp $k=v_1(s,x)$) in (\ref{loclocetrineq1})
(resp in \ref{loclocetrineq2}) and we have  only to add the obtained
inequalities, then to go to the limit on $n$ in order to get
(\ref{local}).
\end{remark}

As $\xi =\xi(1-\sigma_m) + \xi \sigma_m$ and $\xi \sigma_m \in
\mathcal{D}( \Omega)$ for $m$ sufficiently large, applying the local
comparison principle (\ref{local}) with $\zeta = \xi \sigma_m$,
the global estimate (\ref{inequi}) with $\xi(1-\sigma_m)$, we obtain
\begin{align*}
&-\int_\Omega (b(v_1) -b(v_2) )^+ \xi \wp_i
 - \chi_{\{v_1 \geq v_2 \}} (a(v_1,\nabla g(v_1))-a(v_2,\nabla g(v_2) ))
 \cdot \nabla_x (\xi \wp_i)  \\
&\geq  \int_\Omega (b(v_1) -b(v_2) )^+ (\xi(1-\sigma_m) )\wp_i
 + \int_\Omega  \kappa (f_1-  f_2 ) \xi (1-\sigma_m) \wp_i  \\
&\quad -\int_\Omega \chi_{\{v_1 \geq v_2 \}} (a(v_1,\nabla g(v_1))
 -a(v_2,\nabla g(v_2) )) \cdot
\nabla_x (\xi (1-\sigma_m)\wp_i)  \\
&\geq   - \lim_{n \to \infty} \mathcal{L}(\xi \wp_i (1-\sigma_m) \sigma_n)
 - \lim_{n \to \infty} \mathcal{L}(\xi \wp_i (1-\sigma_m) \sigma_n) \\
&=  - \lim_{n \to \infty} \mathcal{L}(\xi \wp_i (\sigma_n
 -\sigma_m \sigma_n))- \lim_{n \to \infty} \mathcal{L}(\xi \wp_i
 (\sigma_n -\sigma_m \sigma_n)) .
\end{align*}
Note that $\wp_i \sigma_n\sigma_m =\wp_i \sigma_m$ for $n$ sufficiently
large. Therefore,
\[
\lim_{m \to \infty} \lim_{n \to \infty} \mathcal{L} (\xi \wp_i
(\sigma_n -\sigma_m \sigma_n))=\lim_{m \to \infty}
 \lim_{n \to \infty} \mathcal{L}(\xi \wp_i (\sigma_n
-\sigma_m \sigma_n))=0,
\]
 and thus, passing to the limit with
$m \to \infty$ in the preceding inequality yields
\begin{align*}
& \int_\Omega (b(v_1) -b(v_2) )^+ \xi  \wp_i + \chi_{\{v_1 \geq v_2 \}} (a(v_1,\nabla g(v_1))-a(v_2,\nabla g(v_2) )) \cdot
\nabla_x (\xi \wp_i)  \\
& \leq \int_\Omega  \kappa (f_1-  f_2 ) \xi \wp_i
\end{align*}
 After summation over $i$, we deduce (\ref{L1comparison}).
\end{proof}

\section{Existence of entropy solution}
The proof of the existence result consists of two steps.
In a first step,  we prove existence of a bounded entropy solution
 of the  problem
\begin{equation} \label{pbalphagf}
\begin{gathered}
b_\alpha(v)-\mathop{\rm div} a(v, \nabla g(v))=f \quad\text{ in }{\Omega}\\
g(v)=0\quad \text{on }{\Gamma},
\end{gathered}
\end{equation}
where $f\in L^1(\Omega)$ and $b_\alpha$ is an increasing Lipschitz
continuous function on $\mathbb{R}$ such that $b_\alpha(0)=0$ and
$\lim_{\alpha\to 0}b_\alpha(r)= b(r)$, for all $r\in\mathbb{R}$.

This is done via approximation with the elliptic-parabolic problems
with homogeneous boundary conditions:
%P_{b_\alpha,g_\varepsilon}(f)\left\{\begin{array}{ll}
\begin{equation} \label{pbalphagepsilonf}
\begin{gathered}
b_\alpha(v)-\mathop{\rm div} a(v, \nabla g_\varepsilon(v))=f \quad
\text{in }{\Omega}\\
v=0\quad \text{on  }{\Gamma},
\end{gathered}
\end{equation}
where $g_\varepsilon(r)=g(r)+\varepsilon r$.
In the second step,  we pass to the limit with  $\alpha$ to 0 and
prove the existence result for $L^1$-data.

\subsection{First step}

\begin{proposition} \label{smoothexist}
 For all  $\varepsilon>0$ and  $f \in L^\infty(\Omega)$,  there exists
 a unique  $v \in L^\infty(\Omega)$  entropy solution of
\eqref{pbalphagepsilonf} i.e.
  $v\in W_0^{1,p}({\Omega})$ and $v$  satisfies the following
entropy inequalities:
For all $k\in \mathbb{R}$, for all $\xi\in C_0^\infty( \mathbb{R}^N)$
such that $\xi\geq 0$ and sign$^+(-k)\xi=0$ a.e. on $\Gamma$,
\begin{equation} \label{loclocetrineqq1}
  \int_{\Omega}   b_\alpha(v) \chi_{\{v>k\}}\xi \leq \int_\Omega
\chi_{\{v>k\}} (f\xi -(a (v,\nabla g_\varepsilon(v))-a(k,0))\cdot \nabla \xi)
\end{equation}
and for all $k\in \mathbb{R}$, for all $\xi\in C_0^\infty( \mathbb{R}^N)$
such that $\xi\geq 0$ and sign$^+(k)\xi=0$ a.e. on $\Gamma$,
\begin{equation} \label{loclocetrineqq2}
\int_{\Omega}   -b_\alpha(v)\chi_{\{k>v\}}\xi \leq
- \int_\Omega \chi_{\{k>v\}} (f\xi -(a(v,\nabla g_\varepsilon(v))-a(k,0))\cdot
\nabla \xi ).
 \end{equation}
\end{proposition}

\begin{proof}
 The existence of a unique  weak solution $v$   of
\eqref{pbalphagepsilonf} is already proved in \cite{KP1}.
Indeed the Problem can be equivalently formulated as follows:
 %(E)(f,\varepsilon)
\begin{equation} \label{Efepsilon}
\begin{gathered}
(b_\alpha\circ g_\varepsilon^{-1})(v)-\mathop{\rm div}
a(g_\varepsilon^{-1}(v), \nabla v)=f \quad \text{in }\Omega\\
v=0\quad \text{on }{\Gamma}.
\end{gathered}
\end{equation}
As $(r,\xi)\mapsto a(g_\varepsilon^{-1}(v),\xi),\;r\in\mathbb{R},
\,\xi\in \mathbb{R}^N$ satisfies the same hypothesis as the vector
field $a$ thanks to the strict monotonicity of $g_\varepsilon$,
it is sufficient to apply the results of \cite{Lions}.
In order to prove that the week solution satisfies the entropy
inequalities, we proceed as in \cite{Ca1}.
\end{proof}

\begin{proposition}\label{limitsmoothexist}
For all  $f \in L^\infty({\Omega})$,  there exists  a unique
$v \in L^\infty({\Omega})$   weak (and entropy ) solution of
\eqref{pbalphagf} i.e.  $g(v)\in W_0^{1,p}({\Omega})$ and $v$
satisfies the following entropy inequalities:
 For all $k\in \mathbb{R}$, for all $\xi\in C_0^\infty( \mathbb{R}^N)$
such that $\xi\geq 0$ and sign$^+(-g(k))\xi=0$ a.e. on ${\Gamma}$,
\begin{equation} \label{alpha}
 \int_{{\Omega}}   b_\alpha(v) \chi_{\{v>k\}}\xi \leq \int_{{\Omega}}
 \chi_{\{v>k\}} (f\xi -(a (v,\nabla g(v))-a(k,0))\cdot \nabla \xi)
\end{equation}
and for all $k\in \mathbb{R}$, for all $\xi\in C_0^\infty( \mathbb{R}^N)$
such that $\xi\geq 0$ and sign$^+(g(k))\xi=0$ a.e. on ${\Gamma}$,
\begin{equation} \label{beta}
\int_{{\Omega}}   -b_\alpha(v)\chi_{\{k>v\}}\xi
\leq - \int_{{\Omega}} \chi_{\{k>v\}} (f\xi -(a(v,\nabla g(v))-a(k,0))\cdot
\nabla \xi ).
 \end{equation}
\end{proposition}

\begin{proof}
  According to Proposition \ref{smoothexist}, for
${f}\in L^\infty({\Omega})$, there exists a unique
$v_\varepsilon\in L^\infty({\Omega})$  entropy solution of
\eqref{pbalphagepsilonf}. i.e.
$v_\varepsilon\in L^\infty({\Omega})$,
$g_\varepsilon(v_\varepsilon)\in W_0^{1,p}({\Omega}))$ and $v_\varepsilon$
 satisfies the entropy inequalities (\ref{loclocetrineqq1})
and (\ref{loclocetrineqq2}):

With a particular choice of test functions and thanks to the strict
monotonicity of $b_\alpha$, one can prove  that
$(v_\varepsilon)_\varepsilon$ and
$(|\nabla  g_\varepsilon(v_\varepsilon)|)_\varepsilon$
are  uniformly bounded in $L^\infty({\Omega})$ and
$L^p({\Omega})$ respectively. Thanks to the growth condition
(\ref{growth}) on $a$, it follows that
$(a(v_\varepsilon,\nabla g_\varepsilon(v_\varepsilon)))_\varepsilon$
is bounded in $L^{p'}({\Omega})^N$ as well.  Following classical
arguments, extracting a subsequence if necessary,
 we can prove that as $\varepsilon\to 0$,
$$
g(v_\varepsilon)\text{ converges  to some }w\in L^\infty({\Omega})\cap W_0^{1,p}({\Omega})$$
 weakly in $W_0^{1,p}({\Omega})$ and strongly in $L^p(\Omega)$. Moreover,
$$a(v_\varepsilon, \nabla g_\varepsilon(v_\varepsilon))\text{ converges weakly in } L^{p'}({\Omega})^N \text{ to some }\chi \in L^{p'}({\Omega})^N.$$
In order  to prove the strong convergence of $v_\varepsilon$
(in $L^1_{Loc}$ for example) to some $v$,  we can use the method of
compensated compactness ( see \cite{AlKar} and  \cite{KarTow})
but this requires some additional conditions on the flux function $\Phi$.
An other approach consists in using the $L^\infty$ uniform bound
on $(v_\varepsilon)$ in order to deduce the weak-$*$ convergence
of $(v_\varepsilon)$ to a function $v$. Then, going to the limit in
the approximate entropy inequalities, we prove that $v$ is  an entropy
process solution of \eqref{pbalphagf}
(see Definition \ref{Process} below). Finally using a ``stronger''
principle of uniqueness, we show that $v$ is the entropy solution
of \eqref{pbalphagf} and that the convergence holds strongly in
$L^1(\Omega)$.
\end{proof}

\begin{definition}\label{defweakconv} \rm
 Let $\Omega$ be an open subset of $\mathbb{R}^N$ ($N\geq 1$), $(u_n)$
be a bounded sequence of $L^\infty(\Omega)$ and
$u\in L^\infty(\Omega\times (0,1))$. The sequence $(u_n)$ converges
towards $u$ in the ``nonlinear weak-$*$ sense'' if
\begin{equation}
\int_\Omega g(u_n(x))\psi(x)\,dx\to \int_0^1\int_\Omega g(u(x,\mu))\psi(x)
\,dx\,d\mu,\quad\text{as }n\to \infty,
\end{equation}
for all $\psi\in L^1(\Omega)$, for all
$g\in \mathcal{C}(\mathbb{R},\mathbb{R})$.
\end{definition}

\begin{lemma}\label{weakconv}
  Let $\Omega$ be an open subset of $\mathbb{R}^N$ ($N\geq 1$)
and $(u_n)$ be a bounded sequence of $L^\infty(\Omega)$.
Then $(u_n)$ admits a subsequence converging in the nonlinear weak-$*$
sense.
\end{lemma}

For the proof of the above lemma see \cite{AlEy,Dip}.
According to Lemma \ref{weakconv}, the sequence $(v_\varepsilon)$
is convergent in the nonlinear weak-$*$ sense to some
$v\in L^\infty(\Omega\times (0,1))$.
 We will prove that $v$ is a weak entropy process solution of
\eqref{pbalphagf} in the following sense.

\begin{definition}\label{Process} \rm
Let $u \in L^\infty((0,1)\times \Omega)$ with $g(u)\in W_0^{1,p}(\Omega)$.
The function $u$ is a weak entropy process solution of
\eqref{pbalphagf} if
for all $k\in \mathbb{R}$, for all $\xi\in C_0^\infty(\mathbb{R}^N)$
such that $\xi\geq 0$ and sign$^+(-g(k))\xi=0$ a.e. on $\partial\Omega$,
\begin{equation} \label{lloclocetrineqqq1b}
 \int_0^1\int_{\Omega}   b_\alpha(u)\chi_{\{u>k\}}\xi\,d\mu
\leq \int_0^1\int_{\Omega}  \chi_{\{u>k\}} (f\xi -(a (u,\nabla g(u))-a(k,0))
\cdot \nabla \xi )\,d\mu
\end{equation}
and for all $k\in \mathbb{R}$, for all $\xi\in C_0^\infty(\mathbb{R}^N)$
such that $\xi\geq 0$ and sign$^+(g(k))\xi=0$ a.e. on $\tilde{\Sigma}$,
\begin{equation} \label{lloclocetrineqqq2}
-\int_0^1\int_{\Omega}   b_\alpha(u)\chi_{\{k>u\}}\xi\,d\mu
\leq -\int_0^1\int_{\Omega} \chi_{\{k>u\}} (f\xi -(a(u,\nabla g(u))-a(k,0))\cdot
\nabla \xi)\,d\mu ).
 \end{equation}
\end{definition}

Taking into account the above estimates, it follows that
\begin{equation} \label{convg(v)2}
 g(v_\varepsilon)\text{ converges  to  }
g(v)\in L^\infty(\Omega)\cap W_0^{1,p}(\Omega)
\end{equation}
 strongly in $L^p(\Omega)$ and weakly in $W^{1,p}(\Omega)$.
In particular, it follows that $g({v})$ is independent of $\mu$.

To pass to the limit in (\ref{loclocetrineqq1}) and
(\ref{loclocetrineqq2}),  it remains to prove that
\begin{equation} \label{passagealalimite}
 \int_\Omega a(v_\varepsilon,\nabla g_\varepsilon(v_\varepsilon))
\cdot \nabla \xi\to \int_0^1(\int_\Omega a(v,\nabla g(v))\cdot
\nabla \xi)\,d\mu
\end{equation}
 By the Minty Browder argument, we have only to prove that
$$
\lim_{\varepsilon\to 0}\int_\Omega a(v_\varepsilon,
\nabla g(v_\varepsilon))\cdot \nabla (g(v_\varepsilon)-g(v))=0.
$$
As $v_\varepsilon$ is also a weak solution of \eqref{pbalphagepsilonf},
 we have
\begin{align*}
&\lim_{\varepsilon\to 0}\int_\Omega a(v_\varepsilon,
 \nabla g(v_\varepsilon))\cdot \nabla (g(v_\varepsilon)-g(v)) \\
&=-\lim_{\varepsilon\to 0}\Big[\int_\Omega b(v_\varepsilon)
 (g(v_\varepsilon)-g(v))+\int_\Omega f (g(v_\varepsilon)-g(v))\Big]
=0
\end{align*}
where the last equality follows by the strong convergence in
$L^p(\Omega)$ of $g(v_\varepsilon)$   to $g(v)$ and the weak$*$-convergence
of $v_\varepsilon $ to $v$.
By the standard pseudo-monotonicity argument it follows that
\begin{equation}  \label{degidentify}
\int_\Omega \chi\cdot\nabla\xi
= \int_0^1\int_\Omega a(v,\nabla g (v))\cdot\nabla\xi \quad
 \text{for all  }  \xi\in \mathcal{D}({\Omega}).
\end{equation}
 Indeed, for $\xi \in \mathcal{D}({\Omega})$, $\xi \geq 0$,
 $\alpha \in \mathbb{R}$, we have
\begin{align*}
\alpha \int_{{\Omega}} \chi \nabla\xi
&= \lim_{\varepsilon \to 0}\int_{{\Omega}} \alpha
 a(v_\varepsilon,\nabla g_\varepsilon (v_\varepsilon))\cdot \nabla\xi\\
 &\geq
\limsup_{\varepsilon \to 0}
 \int_{{\Omega}}   a(v_\varepsilon,\nabla g_\varepsilon (v_\varepsilon))\cdot
 \nabla(g_\varepsilon(v_\varepsilon) -g (v) + \alpha \xi )\\
&\geq
\limsup_{\varepsilon \to 0}
 \int_{{\Omega}}  a(v_\varepsilon,\nabla (g (v) - \alpha \xi))\cdot\nabla (g_\varepsilon (v_\varepsilon) -g (v) +\alpha \xi )\\
&\geq
 \int_{{\Omega}} \alpha  a(v,\nabla(g (v) - \alpha \xi))\cdot \nabla\xi.
\end{align*}
 Dividing by $\alpha >0$ (resp. $\alpha<0$), passing to the limit with
$\alpha  \to 0$,  we obtain (\ref{degidentify}).
We can now pass to the limit in  (\ref{loclocetrineqq1}) and
(\ref{loclocetrineqq2}) to get   for all $k\in \mathbb{R}$,
for all $\xi\in C_0^\infty( \mathbb{R}^N)$ such that $\xi\geq 0$ and
sign$^+(-g(k))\xi=0$ a.e. on ${\Gamma}$,
\begin{equation} \label{lloclocetrineqqq1}
  \int_0^1\int_{{\Omega}}   b_\alpha(v)\chi_{\{v>k\}}\xi
 \leq \int_0^1\int_{{\Omega}}  \chi_{\{v>k\}} (f\xi -(a (v,\nabla g(v))-a(k,0))
\cdot \nabla \xi)
\end{equation}
and for all $k\in \mathbb{R}$, for all $\xi\in C_0^\infty( \mathbb{R}^N)$
such that $\xi\geq 0$ and sign$^+(g(k))\xi=0$ a.e. on ${\Gamma}$,
\begin{equation} \label{lloclocetrineqqq2b}
\int_0^1\int_{{\Omega}}   -b_\alpha(v)\chi_{\{k>v\}}\xi
\leq  - \int_0^1\int_{{\Omega}} \chi_{\{k>v\}}( f\xi
+(a(v,\nabla g(v))-a(k,0))\cdot \nabla \xi ).
 \end{equation}
  Hence we have shown that $v$
  is a weak entropy process solution of \eqref{pbalphagf}.
Now,  to prove that $v$  is the week  entropy solution of
 \eqref{pbalphagf}, we use the following ``reinforced''
comparison principle.

\begin{proposition}\label{compforce}
 Let $f_i\in L^\infty(\Omega)$ and $v_i\in L^\infty(\Omega\times (0,1)$
be a weak entropy process of $P_{b_\alpha,g}(f_i)$ $i=1,2$. Then
there exists $\kappa\in L^\infty(\Omega\times (0,1))$ with $\kappa\in$
sign$^+(v_1-v_2)$ a.e. in $\Omega\times (0,1)$ such that
\[
  \int_0^1\int_{{\Omega}}(b_\alpha(v_1(x,\alpha))-b_\alpha(v_2(x,\mu)))^+
\xi\,dx\,d\alpha\,d\mu
\leq\int_0^1\int_{\Omega}\kappa(f_1-f_2)\xi\,dx.
\]
\end{proposition}

In particular, when $f_1=f_2$, we have
$$
v_1(x,\alpha)=v_2(x,\mu)\quad \text{for a.e. }
(x,\alpha,\mu)\in \Omega\times (0,1)\times (0,1).
$$
Defining the function $w(x)=\int_0^1 v_1(x,\alpha)\,d\alpha$,
we deduce that $w(x)=v_1(x,\alpha)=v_2(x,\beta)$ for a.e.
$(x,\alpha,\beta)\in \Omega\times (0,1)\times (0,1)$.

The proof of Proposition \ref{compforce} follows the same lines as
those of Theorem \ref{comp} and is omitted. The reader is referred
among others to \cite{vovelle} and  \cite{Dip} in order to verify the
technical tools which are necessary  to deal with measure-valued
functions. The result of Proposition \ref{compforce} implies that $v$
is the unique weak entropy solution of \eqref{pbalphagf} and
 the first step of the proof is complete.


\subsection{Second step}
The comparison principle is again the main tool in this last step:
Let $f\in L^1(\Omega)$. For $m,n\in \mathbb{N}$, let
$f_{m,n}=f\wedge m\vee (-n)$
and define $b_{m,n}:r\mapsto b(r)+{{1}\over{m}}r^+-{{1}\over{n}}r^-$.
Denote by $v_{m,n}$ the unique weak entropy solution of
$P_{b_{m,n},g}(f_{m,n})$ (which exists by the result of the first step).
Then,
 \begin{equation} \label{one}
 0 \leq \int_\Omega -\chi_{\{v_{m,n}>k\}} \{(a(v_{m,n},
\nabla g(v_{m,n}))-a(k,0))\cdot \nabla \xi
+  f_{m,n}\xi -  b_{m,n}(v_{m,n})\xi\}
\end{equation}
 for any $\xi \in \mathcal{D}(  \mathbb{R}^N)$, $\xi \geq 0$, for all
 $k \in \mathbb{R}$ such that sign$^+(-g(k))\xi=0$ on $\Gamma$,
\begin{equation} \label{two}
0\leq \int_\Omega   \chi_{\{k>v_{m,n}\}} \{(a(v_{m,n},
\nabla g(v_{m,n}))-a(k,0)) \cdot\nabla \xi
-  f_{m,n}\xi + b_{m,n}(v_{m,n})\xi \}
\end{equation}
for any $\xi \in \mathcal{D}(\mathbb{R}^N)$, $\xi \geq 0$, for all
 $k \in \mathbb{R}$ such that sign$^+(g(k))\xi=0$ on $\Gamma$.

By Theorem \ref{comp},  there exists
${\kappa_{m_1,m_2}} \in L^\infty(\Omega)$ and ${\kappa}_{n_1,n_2}
 \in L^\infty(\Omega)$
with ${\kappa_{m_1,m_2}} \in \mathop{\rm sign}\nolimits^+(v_{m_1,n}
-v_{m_2,n})$, ${\kappa}_{n_1,n_2} \in \mathop{\rm sign}\nolimits^+
(v_{m,n_1}-v_{m,n_2})$  such that,
for all $\xi \in \mathcal{D}^+(\mathbb{R}^N)$, $\xi \geq 0$,
\begin{equation} \label{comparison11}
\begin{aligned}
&\int_\Omega({{1}\over{m_2}}(v_{m_1,n}^+)-{{1}\over{m_2}}
 (v_{m_2,n}^+))^+\xi+{{1}\over{n}}(-v_{m_1,n}^-+v_{m_2,n}^-)^+\xi \\
&\leq -\int_\Omega (b(v_{m_1,n})-b(v_{m_2,n}))^+\xi
 + \int_\Omega \kappa_{m_1,m_2} ({{1}\over{m_2}}-{{1}\over{m_1}})
 v_{m_1,n}^+\xi  \\
&-\int_\Omega \chi_{\{v_{m_1,n} >v_{m_2,n}\}}
(a(v_{m_1,n},\nabla g(v_{m_1,n}))-a(v_{m_2,n},\nabla g(v_{m_2,n})))
\cdot \nabla \xi.
\end{aligned}
\end{equation}
and
\begin{equation} \label{comparison12}
\begin{aligned}
&\int_\Omega({{1}\over{n_2}}v_{m,n_1}^--{{1}\over{n_2}}v_{m,n_2}^-)^+
\xi+{{1}\over{m}}(v_{m,n_1}^+-v_{m,n_2}^+)^+\xi \\
&\leq \int_\Omega -(b(v_{m,n_1})-b(v_{m,n_2}))^+\xi
- \int_\Omega {\kappa}_{n_1,n_2} ({{1}\over{n_2}}
-{{1}\over{n_1}})v_{m,n_1}^- \xi \\
&\quad +\int_\Omega \chi_{\{v_{m,n_1}>v_{m,n_2}\}}
(a(v_{m,n_1},\nabla g(v_{m,n_1}))-a(v_{m,n_2},\nabla
g(v_{m,n_2})))\cdot \nabla \xi.
\end{aligned}
\end{equation}
This yields that $v_{m_1,n}\leq v_{m_2,n}$ for $m_1\leq m_2$ and
$v_{m,n_1}\leq v_{m,n_2}$ for $n_1\geq n_2$.
Therefore, $v_{m,n} \uparrow _m  v_n$
 a.e. on $\Omega$ where $v_n: \Omega \to \overline{\mathbb{R}}$ is a measurable
function.
 Here, we use the notation $\uparrow_n$ resp. $\downarrow_n$
 to denote convergence of a sequence which is monotone increasing, resp.
 decreasing in $n$. Moreover, from (\ref{comparison11}) and
(\ref{comparison12}), it follows that
\begin{equation} \label{key}
 b(v_{m,n})_m\to b(v_n) \quad \text{in } L^1(\Omega).
\end{equation}
Applying a diagonal argument, we may assume that
 for some subsequence $(m(n))_n$ we have
\begin{equation}
 v_{m(n),n} \to v_n \quad \text{a.e. in } \Omega,\quad
 b(v_{m(n),n}) \to b(v_n) \quad \text{in } L^1(\Omega).
\end{equation}
 where  $v_n$ is the weak entropy solution of
 $P_{b_n,g}(f_n)$ with  $b_n:= b_{m(n),n}$, $f_n=f_{m(n),n}$.
Next, we prove that $v_n$ is finite a.e. in $\Omega$:
Suppose first that $b(+\infty):=\lim_{r\to+\infty}b(r)<\infty$.
 Then, by the Range condition, it follows that
$\lim_{r\to+\infty}g(r)=\infty$.

As $v_{m,n}$ is a week solution of $P_{b_{m,n},g}(f_{m,n})$,
choosing $g(T_k(v_{m,n}^+))$ as test function, taking into account
the growth condition on $a$, we find
$$
\lambda_{b(+\infty)}\int_\Omega |\nabla g(T_k v_{m,n}^+)|^p
\leq M_{b(+\infty)}+g(k)\int_\Omega |f_{m,n}|
$$
(see condition (\ref{growth}) on $a$ ).
Hence, by Poincar\'e's inequality,
$$
|\{v_{m,n}^+\geq k\}|\leq {{C(1+g(k))}\over{g(k)^p}}$$ for some constant
$C$ independent of $m,n$ and $k$. Passing the limit with $m\to \infty$
and then with $k\to \infty$ in the above inequality, we find that
$v_n$ is finite a.e. on $\Omega$. In the case where $b(+\infty)=+\infty$,
the last assertion follows from (\ref{key}).
Using $T_kg(v_{m,n})$ as test function in the weak formulation, by
the coerciveness assumption on $a$, we obtain
$$
\int_\Omega |\nabla T_k g(v_{m,n})|^p\leq C(k),
$$
for a constant $C(k) $ depending only on $k$.
Therefore, we can assume that the sequence
$(T_kg(v_{m(n),n}))_n$ converges weakly in $W_0^{1,p}(\Omega))$ to
$T_kg(v_n)$.
Going to the limit with $n\to \infty$, proceeding as above,
we can extract a subsequence still denoted $(v_n)_n$ such that
$$
v_n\to v\quad \text{a.e.  in }\Omega,\quad
b(v_n)\to b(v)\quad \text{in }L^1(\Omega)
$$
where $v$ is finite a.e. in $\Omega$. Moreover,
$T_kg(v)\in W_0^{1,p}(\Omega)$ and
$(T_kg(v_{n}))_n$ converges weakly in $W_0^{1,p}(\Omega))$ to $T_kg(v)$.
Applying again the argument of Minty Browder, we can  prove for
our diagonal sequence that $a(T_kv_n,\nabla g(T_kv_n))$ $\to$ $ a(T_kv,\nabla g(T_kv))$ weakly in $(L^{p'}(\Omega))^N$.
It remains only to prove the inequalities (\ref{locetrineq10})
and (\ref{locetrineq20}).
 To this end, let us first verify that $v_n$ satisfies
(\ref{locetrineq10}) and (\ref{locetrineq20}) for all $n\in\mathbb{N}$:
 For all $k\in \mathbb{R}$, for all $l\geq k$, for any
$\xi \in \mathcal{D}( \mathbb{R}^N)$, $\xi \geq 0$, we have
\begin{align*}
&\int_\Omega   -b_n(v_n\wedge l)\chi_{\{v_n\wedge l>k\}} \xi + \chi_{\{v_n\wedge l>k\}} f_n\xi  \\
& - \chi_{\{v_n\wedge l>k\}} (a(v_n\wedge l,\nabla g(v_n\wedge l))-a(k,0))
\cdot \nabla \xi\\
&=\int_\Omega  \chi_{\{v_n>k\}}\{- (b_n(v_n)-f_n) \xi -  (a(v_n,\nabla g(v_n))-a(k,0)
\cdot \nabla \xi \}\\
&\quad  +\int_\Omega \chi_{\{v_n>l\}} (b_n(v_n)-b_n(l)-f_n) \xi+ (a(v_n,\nabla g(v_n))-a(l,0))
\cdot \nabla \xi \}+f_n^-\\
&\geq  \int_\Omega \chi_{\{v_n>l\}}\{ (b_n(v_n)-b_n(l)+f_n)\xi
+ (a(v_n,\nabla g(v_n))-a(l,0))
\cdot \nabla \xi   -f_n^-\xi\},
\end{align*}
Let
\[
\langle \mu^n_l,\xi\rangle :=
-\int_\Omega \chi_{\{v_n>l\}} \{ (b_n(v_n)-b_n(l)) \xi
+ f_n\xi +( a(v_n,\nabla g(v_n))-a(l,0)) \cdot \nabla \xi -f_n^-\xi\}.
\]
Then, $\mu^n_l$ is a non-negative measure on $\overline{\Omega}$ and
$\mu^n_l\equiv 0$ for $l\geq \| v_n\|_{L^\infty(\Omega)}$.
Moreover,
$$
\|\mu^n_l\|\leq \int_\Omega |f_n|\chi_{\{v_n>l\}}.
$$
Working on  the second entropy inequality, we construct a family of
bounded non-negative measures $(\nu^n_l)_l$ on $\overline{\Omega}$
\[
\langle \nu^n_l,\xi\rangle
:=-\int_\Omega \chi_{\{l>v_n\}}\{ (b_n(l)- (b_n(v_n))) \xi-f_n^+\xi
- f_n\xi +(a(l,0)-a(v_n,\nabla g(v_n))) \cdot \nabla \xi \}
\]
such that
\[
\int_\Omega  \chi_{\{k>v_n\vee l\}}\{b_n(v_n\vee l) \xi
-  f_n\xi+ a(v_n\vee l,\nabla g(v_n\vee l)) \cdot \nabla \xi\}
\geq -\langle \nu^n_l,\xi\rangle
\]
for all $\xi\in \mathcal{D}^+(\Omega)$ and $k\in\mathbb{R}$ with
$(g( k))^+\xi=0$ on $\Gamma$
and  $\|\nu^n_l\|\leq \int_\Omega |f_n|\chi_{\{v_n<l\}}$.
It follows that $(\mu^n_l)_n$ and $(\nu^n_l)_n$ are uniformly bounded
with respect to $n$. Therefore, we can extract two subsequences still
denoted by $(\mu^n_l)_n$ and $(\nu^n_l)_n$ which are  convergent with
 respect to the weak$-*$ topology on $C(\overline{\Omega})$ to $\mu_l$
and $\nu_l$ respectively.
Now, combining all the estimates on the sequence $(v_n)_n$, we can
pass the limit in the above inequalities to (\ref{locetrineq10})
and (\ref{locetrineq20}).
 The measures $\mu_l$ and $\nu_l$ are defined as follows:
\begin{gather*}
\langle \mu_l,\xi\rangle :=-\int_\Omega \chi_{\{v>l\}}\{(b(v)-b(l))\xi
-f\xi+ (a(v,D g(v))-a(l,0)  \cdot\nabla\xi-f^-\xi\},
\\
\langle \nu_l,\xi\rangle :=-\int_\Omega \chi_{\{l>v\}}\{(b(l)-b(v))\xi
+(a(l,0)-  a(v,D g(v)))\cdot\nabla\xi+f\xi-f^+\xi\}.
\end{gather*}
 Here, $Dg(v)$ is defined by $\chi_{\{-k<v< k\}}
Dg(v)=\nabla g(T_k v)$ for all $k>0$.
%\end{proof}

The uniqueness result in the $L^1$ setting follows from the following
proposition.

\begin{proposition}\label{comparison2}
 Let $f_1\in L^\infty(\Omega)$, $f_2\in L^1(\Omega)$ and
$v_1$, $v_2$ be an  entropy solution and a  renormalized entropy solution
of  \eqref{Problemeprincipal} with $f_1$ instead of $f$, and
\eqref{Problemeprincipal} with $f_2$ in stead of $f$, respectively.
Then, there exists ${\kappa} \in L^\infty(\Omega)$ with
${\kappa} \in \mathop{\rm sign}\nolimits^+(v_1-v_2\vee l)$ a.e.
in $\Omega$ such that,  for any $\zeta \in \mathcal{D}^+( \mathbb{R}^N)$,
\begin{equation} \label{globalnew}
\begin{aligned}
-\langle \nu_l,\zeta \rangle
&\leq  -\int_\Omega\chi_{\{v_1 >v_2\vee l\}}
(a(v_1,\nabla g(v_1))-a(v_2\vee l,\nabla g(v_2\vee l )))\cdot
\nabla \zeta  \\
&\quad -\int_\Omega (b(v_1)-b(v_2\vee l))^+\zeta + \int_\Omega \kappa
(f_1-f_2) \zeta.
\end{aligned}
\end{equation}
\end{proposition}

For the proof of the above proposition can be found in \cite{KJP}.
Let us show how to deduce uniqueness of the renormalized entropy solution:
Let $v$ be a renormalized entropy solution of \eqref{Problemeprincipal} and $v_n$ be
the  entropy solution of $P_{b,g}(f_n)$ constructed above.
Then by Proposition \ref{comparison2}, there exists
${\kappa_n} \in L^\infty(\Omega)$ with
${\kappa_n} \in \mathop{\rm sign}\nolimits^+(v_n-v\vee l_n)$ a.e. in
$\Omega$ such that,  for any $\zeta \in \mathcal{D}( \mathbb{R}^N)$,
$\zeta \geq 0$, for any $l_n\geq n$,
\begin{equation}  \label{globall}
\begin{aligned}
-\langle \nu_{-l_n},\zeta \rangle
&\leq -\int_\Omega\chi_{\{v_n>v\vee (-l_n)\}}
(a(v_n,\nabla g(v_n))-a(v\vee (-l_n),\nabla g(v\vee (-l_n))))\cdot
 \nabla \zeta\\
&\quad   \int_\Omega -(b_n(v_n)-b_n(v\vee (-l_n)))^+\zeta
+ \int_\Omega \kappa_n (f_n-f) \zeta.
\end{aligned}
\end{equation}
Similarly, we prove that  there exists
$\tilde{\kappa_n} \in L^\infty(\Omega)$ with
$\tilde{\kappa_n} \in \mathop{\rm sign}\nolimits^+(v\wedge l_n-v_1)$
a.e. in $\Omega$ such that,  for any
$\zeta \in \mathcal{D}( \mathbb{R}^N)$, $\zeta \geq 0$, for any
$l_n\geq n$,
\begin{equation}  \label{global2new}
\begin{aligned}
-\langle \mu_{l_n},\zeta \rangle
&\leq \int_\Omega-\chi_{\{v_n<v\wedge l_n\}}
(a(v\wedge l_n,\nabla g(v\wedge l_n))-a(v_n,\nabla g(v_n)))\cdot
\nabla \zeta  \\
&\quad -\int_\Omega (b_n(v\wedge l_n)-b_n(v_n))^+\zeta
+ \int_\Omega \tilde{\kappa}_n (f-f_n) \zeta.
\end{aligned}
\end{equation}
Summing up (\ref{globall}) and  (\ref{global2new}),
letting $n\to +\infty$,  we get $b(v)=\lim_{n\to +\infty }b(v_n)$.

Let us define  the operator
$A_{b,g}$ in
$L^\infty(\Omega)\times L^\infty(\Omega)\subset L^1(\Omega)
\times L^1(\Omega)$ by $(u,f)\in A_{b,g}$ if and only if there exists $v$
measurable such that $b(v)=u$ and $v$ is an entropy solution of
$P_{b,g}(f+u)$.

\begin{proposition}\label{end}
Let $b$ be strictly increasing with $b(0)=0$. Then
\begin{itemize}
\item[(i)] The operator $A_{b,g}$ is $T-accretive$ in $L^1(\Omega)$; i.e.,
for all $(u_i,f_i)\in A_{b,g}$,
\[
\int_\Omega \kappa (f_1-f_2)\geq 0\quad \text{for some }\kappa\in
 \mathop{\rm sign}(v_1-v_2).
\]

\item[(ii)] For any $\alpha>0$, $R(I+\alpha A_{b,g})=L^\infty(\Omega)$,

\item[(iii)] $\overline{D(A_{b,g}})^{L^1(\Omega)}
=\{u\in L^1(\Omega),\;u(x)\in \overline{R(b)}\,a.e. x\in \Omega\}$
\end{itemize}
\end{proposition}

\begin{proof}
(i) and (ii) are direct consequences of Theorem \ref{comp} and
 the existence result.
To prove (iii), let $f\in L^\infty(\Omega)$ be such that
$f\pm \epsilon\in R(b)$ and let $v_h$ be an entropy solution of
\begin{gather*}
b(v)-h\mathop{\rm div}a(v,\nabla g(v))=f\quad \text{in }\Omega\\
g(v)=0 \quad \text{on }\Gamma
\end{gather*}
with $h>0$. Then
\begin{equation}\label{ki}
\| b(v_h)\|_{L^q(\Omega)} \leq \| f\|_{L^q(\Omega)}
\end{equation}
 for every $1\leq q\leq +\infty$. In particular,
$\| v_h\|_{L^\infty(\Omega)}\leq C(f)$ and by the growth condition,
it follows that $h\mathop{\rm div}a(v_h,\nabla g(v_h))\to 0$
in $\mathcal{D}'(\Omega)$. Therefore, $b(v_h)\to f$ in
$\mathcal{D}'(\Omega)$ and weakly in $L^p(\Omega)$. Whence,
$\liminf_{h\to 0}\| b(v_h)\|_{L^p(\Omega)}\geq \| f \|_{L^p(\Omega)}$.
Taking into account (\ref{ki}), we deduce that $b(v_h)\to f$
strongly in $L^1(\Omega)$. The proof is complete.
\end{proof}

\begin{remark} \label{rmk4.9}\rm
 Proposition \ref{end}  allows to study the Cauchy problem associated
to \eqref{Problemeprincipal} from the point of view of semi-groups theory.
 The elliptic parabolic problem
$$
b(v)_t-\mathop{\rm div} a(v,\nabla g(v))=f\in (0,T)\times\Omega
$$
with initial condition and general boundary condition will be
 treated by the first author in a forthcoming paper.
\end{remark}

\begin{corollary} \label{coro4.10}
 For every $f\in L^1((0,T)\times\Omega)$ and every
$v_0\in \overline{D(A_{b,g})}$, there exists a unique integral
solution of
\begin{equation} \label{pv0fAbg} %P(v_0,f)(A_{b,g})
\begin{gathered}
  u_t+A_{b,g}(u)\ni f\\
v(0)=v_0,
 \end{gathered}
\end{equation}
with $u$ in $C([0,T),L^1(\Omega))$.
Moreover, a comparison principle holds.
\end{corollary}

\begin{thebibliography}{00}

\bibitem{AL} H. W. Alt,  S. Luckhaus,
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