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\markboth{Nonlinear equations with natural growth terms and measure data }   
{ Alessio Porretta }   
   
\begin{document}
\setcounter{page}{183} 
\title{\vspace{-1in}\parbox{\linewidth}{\footnotesize\noindent   
2002-Fez conference on Partial Differential Equations,\newline   
Electronic Journal of Differential Equations,   
Conference 09, 2002, pp 183--202. \newline   
http://ejde.math.swt.edu or http://ejde.math.unt.edu   
\newline ftp  ejde.math.swt.edu (login: ftp)}   
 \vspace{\bigskipamount} \\   
%   
  Nonlinear equations with natural growth terms and measure data   
%   
\thanks{ {\em Mathematics Subject Classifications:} 35J60, 35J65, 35R05.   
\hfil\break\indent   
{\em Key words:} Nonlinear elliptic equations, natural growth terms, measure data.   
\hfil\break\indent   
\copyright 2002 Southwest Texas State University. \hfil\break\indent   
Published December 28, 2002. } }   
   
\date{}   
\author{Alessio  Porretta}   
\maketitle   
   
\begin{abstract}   
  We consider  a class of nonlinear elliptic equations containing
  a $p$-Laplacian type operator, lower order terms having natural 
  growth with respect to the gradient, and bounded measures as data. 
  The model example is the equation
  $$ -\Delta_p(u) + g(u)|\nabla u|^p=\mu  
  $$ 
  in a bounded set $\Omega\subset \mathbb{R}^N$, coupled with a  
  Dirichlet boundary condition. We provide a  review of the 
  results recently obtained in the absorption case (when $g(s)s\geq0$)
  and prove  a new  existence result  without any sign condition on $g$, 
  assuming only that $g\in L^1({\bf R})$. This latter assumption is 
  proved to be optimal for existence of solutions for any measure $\mu$.
\end{abstract}   
   
\numberwithin{equation}{section}   
\newtheorem{theorem}{Theorem}[section]   
\newtheorem{example}[theorem]{Example}   
\newtheorem{remark}[theorem]{Remark}   
   
\section{Introduction}   
   
In this work we focus our attention on nonlinear Dirichlet problems whose   
model is   
\begin{equation}\label{un}   
\begin{gathered}   
-\Delta_p(u) + g(u)|\nabla u|^p=\mu \quad\mbox{in }\Omega,\\   
u=0 \quad\mbox{on }\partial\Omega,   
\end{gathered}   
\end{equation}   
where $p>1$, $g:\mathbb{R}\to \mathbb{R}$ is a continuous function,   
 and $\mu$ is a bounded Radon measure on $\Omega$ which is a bounded subset   
of $\mathbb{R}^N$.   
   
Recently, many researchers have investigated the possibility to find solutions   
of \eqref{un} under the assumption that $g(s)s\geq 0$, in which case the term   
$g(u)|\nabla u|^p$ is said to be an absorption term. In this case   
 a detailed picture of what happens is now available, according to the growth   
at infinity of   
$g(s)$ and to whether   the measure $\mu$ charges or not sets   
of zero $p$-capacity (the capacity defined in $W^{1,p}_0(\Omega)$).   
In the next section, we try to give a quick review of these   
results and explain the main features of the problem in the absorption case,   
both for elliptic and for parabolic equations.   
   
No results for general measures $\mu$ are known to our knowledge if  the sign   
condition is not assumed to hold, possibly including the reaction case   
in which $g(s)s\leq 0$.  It is the purpose of the third section   
of this paper to give new results in this situation. Eventually, these new results seem to fit   
perfectly those proved in the absorption case, and we will prove (stated in more   
generality in Section 3) the following theorem, which extends that proved in \cite{Se} (under the same assumptions) for data   
$\mu\in L^1(\Omega)$.   
   
\begin{theorem}\label{tun}   
Let $\mu$ be  a nonnegative bounded  Radon measure on $\Omega$. Assume that   
$g\in  L^1(\mathbb{R})$. Then there exists a  distributional solution $u$   
of \eqref{un}.   
\end{theorem}   
   
Next we will give an  example which somehow expresses that the assumption   
$g\in L^1(\mathbb{R})$ in Theorem \ref{tun} is optimal;  if $\mu$ is   
the Dirac mass, we prove that no solution can be obtained by approximation. In particular,   
in the reaction case  ($g(s)s\leq 0$),   
if $\mu$ is approximated by  a  sequence of smooth functions, the sequence   
of approximating solutions converges to a solution of \eqref{un} if $g\in L^1(\mathbb{R})$,   
while it blows up everywhere in   
$\Omega$ if $g\not\in L^1(\mathbb{R})$. We recall that in \cite{MuPo} the absorption case   
$g(s)s\geq 0$ had already been studied; in that situation if the Dirac mass is approximated by   
smooth functions, the approximated solutions still converge to a solution  of the problem if   
$g\in L^1(\mathbb{R})$, while  they converge to zero if $g\not \in L^1(\mathbb{R})$. Thus, even if for different reasons,   
in both cases the assumption $g\in L^1(\mathbb{R})$ turns out to be optimal.   
   
   
\section{The absorption case:  a quick review}   
   
A  wide literature has dealt with elliptic and parabolic equations with measure   
data in the last decades. In particular, the techniques of a priori estimates and   
compactness of approximating solutions, firstly introduced in \cite{BG1}, have been   
proved to  work well enough for pseudomonotone operators of Leray-Lions type (\cite{LL}),   
providing several existence results in case of   
$L^1$ data.  The presence of absorbing lower order terms (i.e.   
satisfying a sign condition) often brings   in this kind of   
problems new features; for instance, as   in   
\cite{BGV}, \cite{BG2}, lower order terms may have a  regularizing effect on   
solutions of problems with $L^1$ data. The two main examples are the following   
problems:   
\begin{equation}\label{semil}   
\begin{gathered}   
 -\Delta_p u+|u|^{r-1}u=\mu \quad\mbox{in }\Omega,\\   
 u=0 \quad\mbox{on }\partial\Omega\,,   
\end{gathered}   
\end{equation}   
and   
\begin{equation}\label{ng}   
\begin{gathered}   
-\Delta_p u+u|\nabla u|^p=\mu \quad\mbox{in }\Omega,\\   
u=0\quad\mbox{on }\partial\Omega\,.   
\end{gathered}   
\end{equation}   
If $\mu\in L^1(\Omega)$, problem \eqref{semil} has a solution in $W^{1,q}_0(\Omega)$   
for any $q< {\frac{pr}{r+1}}$, while problem \eqref{ng} has a  finite energy solution $u$,  which   
belongs to $W^{1,p}_0(\Omega)$. In general, if the lower order term is absorbing, one can prove the   
existence of a solution with $L^1(\Omega)$ data; for instance,   the   
 problem:   
\begin{equation}\label{mp}   
\begin{gathered}   
-\Delta_p(u)+H(x,u,\nabla u)=f \quad\mbox{in }\Omega,\\   
u=0 \quad\mbox{on }\partial\Omega\,,   
\end{gathered}   
\end{equation}   
with $\xi\mapsto H(x,s,\xi)$ growing at most like $|\xi|^p$ (the   
so-called natural growth), always admits  a solution if $f\in L^1(\Omega)$ (see   
\cite{Po2}). In fact, dealing with the limit growth for $H(x,s,\xi)$ is not   
that easy and requires the strong compactness of truncations in   
the energy space; on the other hand, these   truncation methods  can be   
adapted to   several different contexts if  still dealing with $L^1(\Omega)$   
data, as obstacle problems or more general operators (see \cite{BeE}, \cite{BeEM}, \cite{EM}).   
   
When trying to extend the previous results to measure data, it   
turns out that precisely the regularizing effect mentioned above may be   
responsible for nonexistence of solutions. Actually, this fact was first observed   
in the pioneering works of H. Brezis (\cite{B,B1}) and in  a whole series of   
papers (see \cite{BaPi,BV,GM,Ve,VV,VV1} and the references therein) concerning problem   
\eqref{semil} in the linear case $p=2$.   
More recently, the  nonlinear case $p\neq 2$ has been dealt with in   
\cite{OP1,OP2,Bi-Ve}. Summing up these results, it is proved that   
 problem \eqref{semil} has a  solution   
for every given bounded measure $\mu$ only if $r<\frac{N(p-1)}{N-p}$, while if   
$r\geq \frac{N(p-1)}{N-p}$ then   no solution  exists if $\mu$ charges sets of zero   
$q$-capacity  with $\frac{q(p-1)}{q-p}<r$ (a necessary and sufficient condition in the linear case $p=2$ is given in   
\cite{GM}). As an example, if $\mu$ is the Dirac mass, then a solution of \eqref{semil}  exists if and   
only if $r<\frac{N(p-1)}{N-p}$.   
   
However, the statement of nonexistence of solutions needs to be suitably precised;   
how shall we express such a failure of existence? Three different ways have been   
suggested so far in  previous works: firstly,   nonexistence of solutions  for a   
general problem as \eqref{mp}   
may be deduced from removable singularity type results. This is a classical   
approach, and mostly used for linear operators;   
a set $K$ is removable if  any solution of   
\begin{gather*}   
-\Delta_p (u)+H(x,u,\nabla u)=f\quad\mbox{in}\quad \Omega\setminus K,\\   
u\in W^{1,p}(\Omega\setminus K)\,,   
\end{gather*}   
can be proved to be a solution in the whole of $\Omega$. If $K$ is removable,   
then we cannot have a solution of the equation with data concentrated on   
$K$.   
   
Alternatively, one studies the limit of   approximating equations:   
\begin{equation}\label{gpn}   
\begin{gathered}   
-\Delta_p (u_n)+H(x,u_n,\nabla u_n)=f_n \quad\mbox{in }\Omega\,,\\   
u_n=0\quad\mbox{on }\partial\Omega\,,   
\end{gathered}   
\end{equation}   
if $f_n$  converges to a measure $\mu$ in the so-called narrow topology,   
which means   
\begin{equation}\label{nar}   
\int_\Omega f_n \varphi dx \; \mathop{\to}^{n\to +\infty}\;   
\int_\Omega \varphi d\mu\,,   
 \end{equation}   
for any function $\varphi$ bounded and  continuous  on $\Omega$.   
This is the most natural way to approximate  a bounded Radon measure,   
so that, if a solution   
exists, we expect that  we can prove the convergence of $u_n$   
towards  a solution $u$ of   
\begin{equation}\label{mpmis}   
\begin{gathered}   
-\Delta_p(u)+H(x,u,\nabla u)=\mu \quad\mbox{in }\Omega\,\\   
u=0\quad\mbox{on }\partial\Omega\,,   
\end{gathered}   
\end{equation}   
like it happens if $f_n$ strongly converges in $L^1(\Omega)$. Thus   
studying the limit   
of $u_n$ is a constructive way to see whether and why existence may fail; thanks to \eqref{nar}   
$f_n$ is bounded in $L^1(\Omega)$ so that \lq\lq  a priori\rq\rq\ estimates are available, and usually compactness of $u_n$ can   
also be proved. The main task is to understand which is the limit of $u_n$ and what equation it   
satisfies.   
   
Finally,  a third approach  is in some sense the combination of the previous two.   
One studies   
\eqref{gpn} assuming only that $f_n$ converges in $L^1_{\rm loc}(\Omega\setminus K)$   
towards a function $f$, where $K$ is a compact subset of $\Omega$. Here no assumption is made on the behaviour of $f_n$ on   
$K$, so that no estimates on $K$ are obtained for $u_n$. If one proves that $u_n$   
still converges to a solution $u$ of \eqref{mp}, this means that   
perturbations of $f$ whatever singular, but localized on $K$, are not seen by the   
equation. As in the viewpoint of removable singularity, no solution can be expected   
for data concentrated on $K$.   
These three possible approaches were all investigated  as far as problem \eqref{semil} is concerned in some of the papers   
mentioned above.   
   
In a series of recent works, these questions have been studied for  problems with   
gradient dependent lower order terms. A particular case is given when the lower   
order term has   natural growth. When trying to find solutions for the model   
equation   
\begin{equation}\label{quad}   
\begin{gathered}   
-\Delta_p u+g(u)|\nabla u|^p=\mu\\   
u=0,   
\end{gathered}   
\end{equation}   
the growth at infinity of $g(s)$ and the regular or singular nature of $\mu$ play a  crucial role.   
Removable singularity   results were proved by H. Brezis and L. Nirenberg in   
\cite{BN} for $p=2$, showing that if $sg(s)\geq \frac\gamma{s^2}$ with $\gamma>1$,   
then any compact set of zero capacity (the standard Newtonnian capacity) is   
removable. In \cite{BGO2}, \cite{Po0}, \cite{MuPo}, the behaviour of  sequences of   
approximating solutions was studied if $\mu$ is approximated in the narrow topology,   
say by a  standard convolution. It is proved that if   
$g\in L^1(\mathbb{R})$, then  there exists a solution $u$ of \eqref{quad} for any measure   
$\mu$, while  if $g\not\in L^1(\mathbb{R})$ approximating solutions converge to a solution   
$u$ of the same problem but with datum $\mu_0$, the absolutely continuous part of   
$\mu$ with respect to $p$-capacity. Here $p$-capacity denotes the capacity defined in $W^{1,p}_0(\Omega)$ and we recall   
(see \cite{FST}) that any measure $\mu$ admits a unique decomposition as $\mu=\mu_0+\lambda$, where $\lambda$ is concentrated on  a set   
of zero $p$-capacity and $\mu_0(E)=0$ for any set $E$ of zero $p$-capacity. In   
other words, in the approximation method, one looses the singular part of the measure which is concentrated on sets of zero   
$p$-capacity;  if $\mu$ does not charge sets of zero $p$-capacity then existence is proved for any function $g(s)$.   
   
Removability properties in the stability approach are investigated in   
\cite{OPo}, where the approximating equations of \eqref{quad} are considered   
with data  $f_n$ only converging to a  function   
$f$ in $L^1_{\rm loc}(\Omega\setminus K)$, where $K$ has $p$-capacity zero. It is proved   
that, setting $ G(s)=\int_0^s g(t)dt$, if, roughly speaking,   
$\exp(-G(s)/(p-1))\in L^1(\mathbb{R})$ then $u_n$ still converges to  a  solution with datum   
$f$; thus,  whatever singular perturbations, provided they are  localized on sets of zero $p$-capacity, are   
not seen by the equation. This result somehow includes the removable   
singularity point of view, and  extends the result in   
\cite{BN} since the assumption that $\exp(-G(s)/(p-1))\in L^1(\mathbb{R})$, in the case $p=2$, is weaker than   
assuming  that $sg(s)\geq \frac\gamma{s^2}$ with $\gamma>1$.   
   
This kind of phenomena due to absorption terms has been investigated for   
parabolic equations as well. As it happens   
for the stationary case, the semilinear evolution problem   
\begin{equation}\label{psem}   
\begin{gathered}   
u_t-\Delta u+|u|^{r-1}u=\mu \quad\mbox{in }Q:=\Omega\times (0,T)\\   
u=0 \quad \mbox{on } \Sigma:=\partial \Omega\times (0,T)\\   
u(0)=u_0 \quad\mbox{in }\Omega,   
\end{gathered}   
\end{equation}   
does  not always have  a solution for any   
measure $\mu$ on $Q$ and any measure initial datum $u_0$. In   
\cite{BF}, the authors study the problem with $\mu=0$ concentrating the   
attention on the initial measure $u_0$. They point out that, ir $r$ is large enough,   
nonexistence phenomena may occurr, and can appear as  initial layer phenomena. In   
fact,    a singular measure as  initial condition may be lost   
while approximating the problem with smooth approximating problems.  Subsequently,   
in   
\cite{BaPi1}, necessary and sufficient conditions are given on the measures $\mu$   
and $u_0$ in order to have a  solution of \eqref{psem}; as expected, these conditions   
involve some notions of space-time dependent  capacity. Further results on nonlinear analogue of \eqref{psem}   
are proved in  \cite{And}, \cite{MV}, \cite{BCV} (see also the references in these papers).   
   
In view of the results mentioned above for elliptic equations, recent study has   
been devoted to   evolution problems as the following:   
\begin{equation}\label{parquad}   
\begin{gathered}   
u_t-\Delta_p u+g(u)|\nabla u|^p=0 \quad\mbox{in }Q:=\Omega\times (0,T)\\   
u=0 \quad \mbox{on } \Sigma:=\partial \Omega\times (0,T)\\   
u(0)=u_0 \quad\mbox{in }\Omega,   
\end{gathered}   
\end{equation}   
in case   $u_0$ is a bounded measure. The existence of a   
solution in case   
$u_0\in L^1(\Omega)$ is proved in \cite{Po1}. The   
possibility to extend this result to a  general measure initial datum is studied in   
\cite{BlPo}. Again, under the assumption that   
$g\in L^1(\mathbb{R})$, it is   proved the    existence of a solution for any measure   
$u_0$. On the other hand, if $g\not\in L^1(\mathbb{R})$, then initial layer phenomena   
occur;  in particular, if $u_{0n}$ is a convolution approximation of the measure   
$u_0$, the sequence of  approximating solutions   
$u_n$ of the same problem, with initial datum $u_{0n}$,  converges to  a solution   
$u$ of the problem having, as initial value, the absolutely continuous part of   
$u_0$ with respect to Lebesgue measure.  Sharp removable singularity type results,   
which in a  stronger way express the nonexistence of solution, still depend on the   
growth at infinity  of $g(s)$ and are obtained in \cite{Po3}.   
   
Eventually, one obtains for the evolution problem \eqref{parquad} the same type   
of results obtained for the elliptic problem \eqref{quad} replacing the role of   
$\mu$ with $u_0$ and the $p$-capacity (capacity in $W^{1,p}_0(\Omega)$) with the Lebesgue   
measure in   
$\Omega$. Are then these results consistent? The answer has to be found in the   
study of  the notion of capacity for parabolic equations. A  functional type   
presentation and construction of the parabolic $p$-capacity (capacity defined in   
the space $W=\{u\in L^p(0,T;W^{1,p}_0(\Omega))\,, u_t\in L^{p'}(0,T;W^{-1,p'}(\Omega))\}$) is given  in \cite{Pi}   
for $p=2$ and in \cite{DPP} for $p\neq 2$. In this last paper, it is proved  that   
given   
$B\subset \Omega$, the set   
$\{t=0\}\times B$ has zero parabolic capacity in $(0,T)\times \Omega$ if and only   
if  $B$ has zero Lebesgue measure. Thus,     if one looks at singularities at   
initial time as singularities on $Q$ concentrated at   
$t=0$, the results obtained on \eqref{parquad} reflect  perfectly those on   
\eqref{quad}.  Moreover, it becomes clear that in order to deal with problem   
\eqref{parquad} with interior space-time dependent measures as data, one has to   
follow the outlines of the stationary case and use a decomposition theorem for   
measures with respect to parabolic $p$-capacity. This latter result, which extends the   
stationary one given  in \cite{BGO1}, is proved in \cite{DPP} and states that   
any measure $\mu$ on $(0,T)\times \Omega$ which does not charge sets of zero   
parabolic $p$-capacity admits the decomposition (as a distribution)   
$$   
\mu=f+g_1-(g_2)_t   
$$   
with $f\in L^1(Q)$, $g_1\in L^{p'}(0,T;W^{-1,p'}(\Omega))$ and $g_2\in L^p(0,T;W^{1,p}_0(\Omega))$.   
   
 Finally, let us mention that, in the linear case ($p=2$),   
other existence and nonexistence results with gradient dependent   
lower order terms (absorbing or repulsive) and measure data are obtained in   
\cite{A,APi,BeLa,BSW} (see also the references cited therein).   
We point out that    the techniques used in these papers are  mainly based on   
 a linear operator and on the concept of   
distributional solution (with two integration by parts), or on semigroup theory   
 and the concept of integral solution.   
These approaches allow to have sharper nonexistence results especially for the case   
of subcritical growth, on the other hand  their study is mostly restricted to the case $g(u)\equiv 1$.   
   
   
   
\subsection{Natural growth reaction terms and measure data}   
   
As explained in the previous section, if the term $H(x,u,\nabla u)$ is an   
absorption term and has natural growth, the borderline case which allows to have   
solutions of \eqref{mpmis} for all measures $\mu$ is the case in which   
$$   
|H(x,u,\nabla u)|\leq g(u)|\nabla u|^p\,,\quad\mbox{with $g\in L^1(\mathbb{R})$.}   
$$   
Our aim is now to show that, somehow surprisingly, the same assumption is   
necessary and sufficient to have solutions for any measure even in the reaction   
case, that is without assuming any sign condition on $H(x,s,\xi)$. In particular,   
if we aim to have solutions of \eqref{quad} for {\emph {any}} given measure data, there is no   
difference between the reaction and the absorption case.   
   
Heuristically, this feature can be easily explained. In fact, the   
model equation   
\begin{equation}\label{mc}   
-\Delta u=g(u)|\nabla u|^2 +\mu\,,   
\end{equation}   
can be transformed, through a  change of unknown, into the equation   
\begin{equation}\label{eqmp}   
-\Delta v=\exp(G(u)) \mu\,,   
\end{equation}   
with $v=\int_0^{u}\exp(G(s))ds$ and   
$G(s)=\int_0^s g(r)dr$.   
   
In \cite{MuPo} we proved that equation \eqref{eqmp} has a solution if $\exp(G(u))$   
has a finite limit at infinity, which is the case whenever $g\in L^1(\mathbb{R})$, so that   
in this case \eqref{mc} is also expected to have  a solution. On the other hand, if   
$g\not\in L^1(\mathbb{R})$, then the right hand side of \eqref{eqmp} can be hardly handled   
since $\exp(G(u))$ is not bounded. We are going to provide an example where   
$\mu$ is the Dirac mass and no solution of   
\eqref{eqmp} can be found by approximation, precisely proving that approximated   
solutions of \eqref{mc} in this case blow up completely (i.e. at every point of   
$\Omega$).   
   
We will prove our result in a  more general situation. Assume that $a(x,s,\xi)$ and $H(x,s,\xi)$ are  Carath\'eodory
functions    satisfying, for almost every $x\in \Omega$, for every $s\in \mathbb{R}$, $\xi$, $\eta\in \mathbb{R}^N$
($\xi\neq \eta$):   
\begin{gather}   
\label{coerc}   
a(x,s,\xi) \cdot \xi \geq \alpha |\xi |^p\,,\quad   
\alpha>0\,,\; p>1\,,\\   
\label{gro}   
|a(x,s,\xi) |\leq \beta (k(x)+|s|^{p-1}+|\xi|^{p-1})\quad k(x)\in L^{p'}(\Omega),   
\;\beta>0\,, \\   
\label{mon}   
(a(x,s,\xi)-a(x,s,\eta))\cdot (\xi-\eta)>0\,,     
\end{gather}
and
\begin{equation}\label{creh}
\begin{gathered} 
|H(x,s,\xi)|\leq \gamma(x)+g(s)|\xi|^p\,,\;\gamma(x)\in L^1(\Omega)\\
\mbox{and }\quad g\,:{\bf R}\to {\bf R}^+\quad \mbox{continuous}\,,\quad g\geq 0\,,\quad g\in   
L^1(\mathbb{R})\,.   
\end{gathered}  
\end{equation} 
In the following we denote by $\mathop{\rm cap}_p(B)$ the $p$-capacity of a  borelian set   
$B\subset\Omega$, where the $p$-capacity is the standard notion of capacity   
defined in the Sobolev space $W^{1,p}_0(\Omega)$.   
Let us recall (see \cite{FST})  that any bounded Radon measure $\mu$ has  a unique   
decomposition as   
\begin{equation}\label{spl}   
\mu=\mu_0+\lambda\,,   
\end{equation}   
where $\mu_0$, $\lambda$ are bounded measures such that $\mu_0$ does not charge sets of   
zero $p$-capacity (i.e. $\mu_0(B)=0$ for every $B$ with $\mathop{\rm cap}_p(B)=0$) and $\lambda$ is   
concentrated on a set $E\subset   
\Omega$ such that $\mathop{\rm cap}_p(E)=0$. Moreover, if $\mu$ is nonnegative, then both $\mu_0$   
and $\lambda$ are nonnegative. For a presentation of the basic notions concerning   
measures and capacity the reader may refer to \cite{HKM}, \cite{DMOP}. We also have,   
from \cite{BGO1}, that $\mu_0$ furtherly  admits a decomposition (in distributional   
sense) as   
\begin{equation}\label{spli}   
\mu_0=f-\mathop{\rm div}(F)\,,\quad \mbox{$f\in L^1(\Omega)$, $F\in L^{p'}(\Omega)^N$.}   
\end{equation}   
Hereafter, let $\mu$ be a bounded nonnegative Radon measure on $\Omega$.   
Referring to the previous decomposition of $\mu$ and $\mu_0$ in \eqref{spl}, \eqref{spli},   
there exists a sequence $\mu_n$ of bounded functions such that   
\begin{equation}\label{mun}   
\begin{gathered}   
\mu_n=\mu_{0n}+\lambda_n\,,\quad \mbox{$\mu_{0n}\geq 0$, $\lambda_n\geq 0$,}\\   
 \mu_{0n}=f_n-\mathop{\rm div}(F_n)\,,\quad f_n\in L^\infty(\Omega)\,,\;   
F_n\in L^\infty(\Omega)^N\,,\\   
 f_n\to f\quad \mbox{strongly in $L^1(\Omega)$,}\\   
F_n \to F\quad \mbox{strongly in $L^{p'}(\Omega)^N$,}\\   
\int_\Omega\varphi \lambda_n dx\to \int_\Omega \varphi d\lambda\quad \forall  \varphi\in C_b(\Omega)\,,   
\end{gathered}   
\end{equation}   
where $C_b(\Omega)$ denotes the space of bounded continuous functions in $\Omega$.   
Such a  sequence $\mu_n$ can be constructed using convolution and a  suitable   
compactly supported approximation of $\mu$.   
   
For fixed $n\in \mathbb{N}$, since  $\mu_n\in L^\infty(\Omega)$,  under the previous   
assumptions it is proved in \cite{BST}  that there exists a weak solution $u_n\in   
W^{1,p}_0(\Omega)\cap L^\infty(\Omega)$   
 of the   
problem:   
\begin{equation}\label{appell}   
\begin{gathered}   
 -\mathop{\rm div}(a(x,u_n,\nabla u_n))=H(x, u_n,\nabla u_n)+\mu_n \quad   
 \mbox{in } \Omega\,,\\   
u_n=0 \quad\mbox{on } \partial \Omega \,.   
\end{gathered}   
\end{equation}   
Our main result is the following.   
   
\begin{theorem}\label{primo}   
Let $a(x,s,\xi)$ and $H(x,s,\xi)$ satisfy assumptions \eqref{coerc}--\eqref{creh}.   
Let $\mu$ be a nonnegative bounded Radon measure on $\Omega$. Then there exists a   
solution $u$ of the problem   
\begin{equation}\label{reac}   
\begin{gathered}   
-\mathop{\rm div}(a(x,u,\nabla u))=H(x, u,\nabla u)+\mu \quad\mbox{in } \Omega \,,\\   
u=0 \quad\mbox{on  }\partial \Omega\,.   
\end{gathered}   
\end{equation}   
\end{theorem}   
   
\paragraph{Proof.}   
 We essentially follow the method used   in \cite{Se}, which consists in   
multiplying the equation \eqref{appell} by $\exp(G(u_n))$ or by   
$\exp(-G(u_n))$,   
where   
$G(s)=\int_0^s g(t)/\alpha  dt$ (the function $g$ appears in \eqref{creh}). In   
other words this replaces the idea of the change of unknown which transforms   
the model problem   
\eqref{mc} into \eqref{eqmp}. After this multiplication, we will apply the   
techniques fully developed in \cite{Po0}, \cite{MuPo} to obtain the strong   
convergence of truncations.   
   
In the following, we omit for shortness the dependence on $x$ in the integrals,   
and we denote by $c$ any positive constant   
independent on $n$. Let $\varphi\in W^{1,p}_0(\Omega)\cap L^\infty(\Omega)$;  choosing   
$\exp(G(u_n)) \varphi$ as test function in   
\eqref{appell} we have   
\begin{align*}   
 \int_\Omega &\exp(G(u_n))a(u_n,\nabla u_n)\nabla \varphi+   
\int_\Omega \frac{g(u_n)}{\alpha}\exp(G(u_n))a(u_n,\nabla u_n)\nabla u_n \varphi\\   
&= \int_\Omega H(u_n,\nabla u_n) \exp(G(u_n)) \varphi + \int_\Omega \varphi\exp(G(u_n)) \mu_n\,.   
\end{align*}   
For any $\varphi\geq 0$, thanks to \eqref{coerc} and \eqref{creh} we obtain   
\begin{equation}\label{poseq}   
\begin{gathered}   
 \int_\Omega \exp(G(u_n))a(u_n,\nabla u_n)\nabla \varphi\leq   \int_\Omega   
\gamma(x) \varphi\exp(G(u_n))+ \int_\Omega   
\varphi\exp(G(u_n)) \mu_n \\   
\forall \varphi\in W^{1,p}_0(\Omega)\cap L^\infty(\Omega)\,,\varphi\geq 0\,.   
\end{gathered}   
\end{equation}   
Similarly, taking $\exp(-G(u_n)) \varphi$ as test function in   
\eqref{appell}   we obtain   
\begin{multline}\label{negeq}   
 \int_\Omega \exp(-G(u_n))a(u_n,\nabla u_n)\nabla \varphi   
 +  \int_\Omega \gamma(x) \varphi\exp(-G(u_n)) \\   
 \geq \int_\Omega \varphi\exp(-G(u_n)) \mu_n   
  \quad \forall \varphi\in W^{1,p}_0(\Omega)\cap   
L^\infty(\Omega)\,,\varphi\geq 0\,.   
\end{multline}   
Let $\varphi=T_k(u_n)^+$ in \eqref{poseq} and   
$\varphi=T_k(u_n)^-$ in \eqref{negeq}.   
Also let $G(\pm \infty)=\frac1\alpha\int_0^{\pm\infty}g(s)ds$ which are well   
defined since $g\in L^1(\mathbb{R})$. Since   
$ \exp(G(-\infty))\leq \exp(G(s))\leq   
\exp(G(+\infty))$ and $\exp(|G(\pm \infty)|)\leq \exp   
(\|g\|_{L^1(\mathbb{R})}/\alpha)$,  using   
\eqref{coerc}, we obtain   
\begin{equation}\label{stitr}   
\|T_k(u_n)\|_{W^{1,p}_0(\Omega)}^p\leq   
\frac1\alpha \exp\big(\frac{\|g\|_{L^1(\mathbb{R})}}\alpha\big) k(\|\gamma\|_{L^1(\Omega)}+   
\|\mu_n\|_{L^1(\Omega)})\leq c k\,.   
\end{equation}   
Standard estimates (see \cite{BBGGPV}) imply that $u_n$ is bounded in   
the Marcinkiewicz space $M^{\frac{N(p-1)}{N-p}}(\Omega)$ and $|\nabla u_n|$   
is bounded in   
the Marcinkiewicz space $M^{\frac{N(p-1)}{N-1}}(\Omega)$. In particular we have   
from \eqref{gro} that $a(x,u_n,\nabla u_n)$ is bounded in $L^q(\Omega)^N$ for any   
$q<\frac N{N-1}$. Furthermore, there exist a function $u$ and a  subsequence such   
that   
\begin{align*}   
& u_n\to u\quad \mbox{a.e. in $\Omega$,}\\   
&T_k(u_n)\to T_k(u)\quad \mbox{weakly in $W^{1,p}_0(\Omega)$ and a.e. in $\Omega$ for any   
$k>0$.}   
\end{align*}   
Let us take $\varphi=T_1(u_n-T_j(u_n))^-$ in \eqref{negeq}; we obtain   
\begin{multline}\label{entry}   
 \int_{\{-(j+1)\leq u_n\leq -j\}}   
 a(u_n,\nabla u_n)\nabla u_n+   
\int_\Omega \exp(-G(u_n)) T_1(u_n-T_j(u_n))^-  \mu_n\\   
\leq \gamma \int_\Omega \exp(-G(u_n))T_1(u_n-T_j(u_n))^-  \,.   
\end{multline}   
The term with $\mu_n$ can be neglected since it is   
nonnegative. In the right hand side we can pass to the limit in $n$ and in $j$ by   
Lebesgue's theorem, using that $G$ is bounded; indeed, since   
$$   
\exp(-G(u))T_1(u -T_j(u ))^- \leq \exp\big(\frac{\|g\|_{L^1(\mathbb{R})}}\alpha\big)   
\chi_{\{u <-j\}}   
$$   
we have   
$$   
\int_\Omega \exp(-G(u_n))T_1(u_n-T_j(u_n))^- \;   
\mathop{\to}\limits^{n\to \infty} \;   
\int_\Omega \exp(-G(u ))T_1(u -T_j(u ))^-\;   
\mathop{\to}\limits^{j\to \infty} \;  0\,,   
$$   
so that we deduce from \eqref{entry}   
\begin{equation}\label{dmop}   
\lim_{j\to \infty}\limsup_{n\to \infty}\quad   
\int_{\{-(j+1)\leq u_n\leq -j\}}  a(u_n,\nabla u_n)\nabla u_n=0\,.   
\end{equation}   
We are going now to prove that the truncations strongly converge in $W^{1,p}_0(\Omega)$.   
Following the idea introduced in \cite{DMOP}, this is done by using a   
suitable sequence of cut-off functions. Indeed, let $\delta>0$;   
since $\lambda$ is a regular measure concentrated on $E$ and since $E$ has zero   
$p$-capacity, there exist a compact set   
$K_\delta\subset E$ and a sequence $\{\psi_\delta\}$ of functions   
in $C^{\infty}_c(\Omega)$ with the properties  that   
\begin{equation}\label{pside}   
\begin{gathered}   
 \lambda(E\setminus K_\delta)<\delta\,,\quad   
 0\leq \psi_\delta\leq 1\,,\\   
 \mbox{$\psi_\delta\equiv 1$ on an open neighbourhood $A_\delta$ of $K_\delta$}\\   
\psi_\delta\mathop{\to}\limits^{\delta\to 0}\; 0 \quad \mbox{strongly in $W^{1,p}_0(\Omega)$.}   
\end{gathered}   
\end{equation}   
Take now   
$\varphi=(k-T_k(u_n)) (1-|T_1(u_n-T_j(u_n)|) \psi_\delta$ in   
\eqref{negeq}, with $j>k$. Observe that $\varphi=(k-u_n)\psi_\delta$ if $|u_n|<k$ and $\varphi=0$ if $u_n>k$.   
Thus we get, using also that $\exp(-G(u_n))\leq   
\exp\big(\frac{\|g\|_{L^1(\mathbb{R})}}\alpha\big)$ and $\psi_\delta\leq 1$,   
\begin{align}\label{near}   
& \int_\Omega \exp(-G(u_n))a(u_n,\nabla u_n)\nabla   
 \psi_\delta (k-T_k(u_n)) (1-|T_1(u_n-T_j(u_n)|) \nonumber\\   
&+ 2k\exp\big(\frac{\|g\|_{L^1(\mathbb{R})}}\alpha\big)   
\int_{\{-(j+1)\leq u_n\leq -j\}} \hskip -5mm   
 a(u_n,\nabla u_n)\nabla u_n  +2k\exp\big(\frac{\|g\|_{L^1(\mathbb{R})}}   
 \alpha\big)\int_\Omega  \gamma \psi_\delta \nonumber\\   
&\geq \exp\big(-\frac{\|g\|_{L^1(\mathbb{R})}}\alpha\big)   
\int_\Omega a(T_k(u_n),\nabla T_k(u_n))\nabla T_k(u_n) \psi_\delta \\   
&\quad +\int_\Omega   
(k-T_k(u_n))\exp(-G(u_n))  (1-|T_1(u_n-T_h(u_n)|) \psi_\delta  \mu_n\,.   
\nonumber   
\end{align}   
Since   
$$   
|a(u_n,\nabla u_n) (1-|T_1(u_n-T_j(u_n)|)|\leq   
|a(T_{j+1}(u_n),\nabla T_{j+1}(u_n))|\,,   
$$   
and since last term is bounded in $L^{p'}(\Omega)$ and $G$ is bounded, we have   
that there exists $\Lambda_j\in L^{p'}(\Omega)^N$ such that   
$$   
\exp(-G(u_n))a(u_n,\nabla u_n)(k-T_k(u_n)) (1-|T_1(u_n-T_j(u_n)|)\to   
\Lambda_j   
$$   
weakly in $L^{p'}(\Omega)^N$.   
Thus we get   
\begin{multline*}   
\lim_{n\to \infty}   
\int_\Omega \exp(-G(u_n))a(u_n,\nabla u_n)\nabla   
 \psi_\delta  (k-T_k(u_n)) (1-|T_1(u_n-T_j(u_n)|)\\   
 = \int_\Omega \Lambda_j\nabla \psi_\delta \,,   
\end{multline*}   
and then, as $\delta$ tends to zero, thanks to \eqref{pside}  we have   
$$   
\lim_{\delta\to 0}\lim_{n\to \infty}   
\int_\Omega \exp(-G(u_n))a(u_n,\nabla u_n)\nabla   
 \psi_\delta  (k-T_k(u_n)) (1-|T_1(u_n-T_j(u_n)|)=0\,.   
$$   
The third integral in \eqref{near}    easily goes to zero since $\psi_\delta$ converges to zero   
and $\gamma\in L^1(\Omega)$.   
Furthermore, the term with $\mu_n$ can again be neglected since it is   
nonnegative. Therefore, passing to the limit first in   
$n$, then in   
$\delta$  we obtain from   
\eqref{near}   
\begin{align*}   
& \lim_{\delta\to 0}\limsup_{n\to \infty}   
\int_\Omega a(T_k(u_n),\nabla T_k(u_n))\nabla T_k(u_n) \psi_\delta\\   
&\quad \leq \limsup_{n\to \infty}   
2k\exp\big(\frac{2\|g\|_{L^1(\mathbb{R})}}\alpha\big)\int_{\{-(j+1)\leq u_n\leq   
-j\}} a(u_n,\nabla u_n)\nabla u_n\,.   
\end{align*}   
Then, as $j$ goes to infinity, using \eqref{dmop} and since   
 $a(x,s,\xi)\cdot \xi\geq 0$, we get   
\begin{equation}\label{half}   
\lim_{\delta\to 0}\limsup_{n\to \infty}   
\int_\Omega a(T_k(u_n),\nabla T_k(u_n))\nabla T_k(u_n) \psi_\delta = 0\,.   
\end{equation}   
Let  now $w_n=T_{2k}(u_n-T_h(u_n)+T_k(u_n)-T_k(u))$, we take   
$\varphi=w_n^+(1-\psi_\delta)$ in \eqref{poseq} and $\varphi=w_n^-(1-\psi_\delta)$ in   
\eqref{negeq} to obtain   
\begin{align*}   
& \int_{\{w_n\geq 0\}} \exp(G(u_n))a(u_n,\nabla u_n)\nabla w_n   
(1-\psi_\delta)\\   
&\leq \int_\Omega \gamma w_n^+\exp(G(u_n)) (1-\psi_\delta)   
+ \int_\Omega w_n^+\exp(G(u_n)) (1-\psi_\delta)\mu_n \\   
&\quad +\int_\Omega \exp(G(u_n))a(u_n,\nabla u_n)\nabla \psi_\delta  w_n^+   
\end{align*}   
and   
\begin{align*}   
& \int_{\{w_n\leq 0\}} \exp(-G(u_n))a(u_n,\nabla u_n)\nabla w_n   
(1-\psi_\delta) \\   
&\leq   \int_\Omega\gamma w_n^-\exp(-G(u_n)) (1-\psi_\delta)   
 - \int_\Omega w_n^-\exp(-G(u_n)) (1-\psi_\delta)\mu_n  \\   
&\quad - \int_\Omega \exp(-G(u_n))a(u_n,\nabla u_n)\nabla \psi_\delta  w_n^-\,.   
\end{align*}   
Setting $M=h+4k$ and using  $a(x,s,\xi)\cdot\xi\geq 0$, we have   
\begin{align*}   
a(u_n,\nabla u_n)\nabla w_n\geq&   
a(T_k(u_n),\nabla T_k(u_n))\nabla (T_k(u_n)-T_k(u))\\   
&-|a(x,{T_M(u_n)},{\nabla T_M(u_n)})| |\nabla T_k(u)| \chi_{\{|u_n|>k\}}\,.  
\end{align*}   
Then   
\begin{equation}\label{una} 
\begin{aligned}   
& \int_{\{w_n\geq 0\}} \exp(G(u_n))a(T_k(u_n),\nabla T_k(u_n))\nabla   
(T_k(u_n)-T_k(u)) (1-\psi_\delta)\\   
&\leq  \int_\Omega \gamma w_n^+\exp(G(u_n)) (1-\psi_\delta)   
+\int_\Omega w_n^+\exp(G(u_n))  (1-\psi_\delta)\mu_n  \\   
&\quad +   
\int_\Omega \exp(G(u_n))a(u_n,\nabla u_n)\nabla \psi_\delta  w_n^+\cr &\quad +   
\int_\Omega \exp(G(u_n))|a(x,{T_M(u_n)},{\nabla T_M(u_n)})| |\nabla   
T_k(u)| \chi_{\{|u_n|>k\}} (1-\psi_\delta)   
\end{aligned}   
\end{equation}   
and   
\begin{equation}\label{altra}   
\begin{aligned}   
&\int_{\{w_n\leq 0\}}   
\exp(-G(u_n))a(T_k(u_n),\nabla T_k(u_n))\nabla (T_k(u_n)-T_k(u)) (1-\psi_\delta)\\   
&\leq  \int_\Omega \gamma w_n^-\exp(-G(u_n)) (1-\psi_\delta)   
 - \int_\Omega w_n^-\exp(-G(u_n))  (1-\psi_\delta)\mu_n \\   
&\quad -\int_\Omega \exp(-G(u_n))a(u_n,\nabla u_n)\nabla \psi_\delta  w_n^-\\   
&\quad +\int_\Omega\exp(-G(u_n)) |a(x,{T_M(u_n)},{\nabla T_M(u_n)})|   
|\nabla T_k(u)| \chi_{\{|u_n|>k\}}(1-\psi_\delta)\,.   
\end{aligned}   
\end{equation}   
Since $a(u_n,\nabla u_n)$ is bounded in $L^q(\Omega)^N$ for any   
$q<\frac N{N-1}$, there exists $\nu\in L^q(\Omega)^N$ such that   
$a(u_n,\nabla u_n)$ weakly converges to   
$\nu$ in $L^q(\Omega)^N$. Since $\psi_\delta\in W^{1,\infty}_0(\Omega)$, and $G$ is bounded,   
we get   
\begin{equation}\label{p}   
\begin{aligned}   
&\int_\Omega \exp(G(u_n))a(u_n,\nabla u_n)\nabla \psi_\delta  w_n^+\\   
&\mathop{\to}\limits^{n\to \infty}   
\; \int_\Omega \exp(G(u))\nu\nabla \psi_\delta  T_{2k}(u-T_h(u))   
\; \mathop{\to}\limits^{h\to \infty}\; 0\,.   
\end{aligned}   
\end{equation}   
Using that $|\nabla T_k(u)| \chi_{\{|u_n|>k\}}$ strongly converges to zero in   
$L^p(\Omega)$ and that $\nabla T_M(u_n)$ is bounded in $L^{p'}(\Omega)^N$ we also have that   
\begin{equation}\label{o}   
\int_\Omega \exp(G(u_n))|a(x,{T_M(u_n)},{\nabla T_M(u_n)})| |\nabla   
T_k(u)| \chi_{\{|u_n|>k\}} (1-\psi_\delta)\; \mathop{\to}\limits^{n\to \infty}   
\; 0\,.   
\end{equation}   
 Similarly, using   
the weak convergence of $T_k(u_n)$ to $T_k(u)$ in $W^{1,p}_0(\Omega)$,   
we have   
\begin{equation}\label{r}   
\int_\Omega \exp(-G(u_n))a(T_k(u_n),\nabla T_k(u))\nabla (T_k(u_n)-T_k(u))   
(1-\psi_\delta)\; \mathop{\to}\limits^{n\to \infty}   
\; 0\,,   
\end{equation}   
and, since $\gamma\in L^1(\Omega)$,   
\begin{equation}\label{e}   
\begin{aligned}   
&\int_\Omega   
\gamma w_n^+\exp(G(u_n)) (1-\psi_\delta)\\   
&\mathop{\to}\limits^{n\to \infty}   
\; \int_\Omega \gamma\exp(G(u))(1- \psi_\delta ) T_{2k}(u-T_h(u))^+   
\;\mathop{\to}\limits^{h\to \infty} \; 0\,.   
\end{aligned}   
\end{equation}   
Moreover, we have, using the decomposition of $\mu_n$ in \eqref{mun},   
\begin{align*}   
&\int_\Omega w_n^+\exp(G(u_n)) (1-\psi_\delta)\mu_n \\   
&=\int_\Omega w_n^+\exp(G(u_n))(1-\psi_\delta) d \mu_{0n}+ \int_\Omega w_n^+\exp(G(u_n))   
(1-\psi_\delta)\lambda_n \\   
&\leq \exp\big(\frac{\|g\|_{L^1(\mathbb{R})}}\alpha\big)   
\int_\Omega w_n^+(1-\psi_\delta) d \mu_{0n}+   
2k\exp\big(\frac{\|g\|_{L^1(\mathbb{R})}}\alpha\big) \int_\Omega (1-\psi_\delta)\lambda_n\,.   
\end{align*}   
Since $w_n^+$  converges to $T_{2k}(u-T_h(u))^+$ weakly-$*$ in $L^\infty(\Omega)$   
and weakly in   
$W^{1,p}_0(\Omega)$, using the convergence of $ \mu_{0n}$ (which is strong in   
$L^1(\Omega)+W^{-1,p'}(\Omega)$) and $\lambda_n$ we obtain   
\begin{equation}\label{majo}   
\begin{aligned}   
&\limsup_{n\to \infty}\int_\Omega w_n^+\exp(G(u_n)) (1-\psi_\delta)\mu_n \\   
&\leq \exp\big(\frac{\|g\|_{L^1(\mathbb{R})}}\alpha\big)   
\int_\Omega T_{2k}(u-T_h(u))^+(1-\psi_\delta) d\mu_{0}\\   
&\quad + 2k\exp\big(\frac{\|g\|_{L^1(\mathbb{R})}}\alpha\big)   
\int_\Omega(1-\psi_\delta) \,d\lambda\,.   
\end{aligned}   
\end{equation}   
Since  $T_k(u)\in W^{1,p}_0(\Omega)$ for any $k>0$ and   
\eqref{stitr} holds true, we have (see e.g. Remark 2.11 in \cite{DMOP}) that $u$ has   
a cap-quasi continuous representative which is cap-quasi everywhere finite, that   
is  there exists  a function $\tilde u$ such that $\tilde u=u$ almost everywhere   
and  ${\rm cap} \{|\tilde u|=+\infty\}=0$. In particular, since $\mu_0$ does not charge sets of zero capacity, we have that   
$\tilde u$ is finite $\mu_0$-quasi everywhere, hence $T_{2k}(\tilde u-T_h(\tilde u))$   
converges to zero $\mu_0$-quasi everywhere. Letting $h$ go to infinity we deduce that   
$$   
\lim_{h\to \infty}\int_\Omega T_{2k}(u-T_h(u))^+(1-\psi_\delta)\,d\mu_{0}=0\,,   
$$   
 so that \eqref{majo} implies   
\begin{equation}\label{t}   
\lim_{h\to \infty} \limsup_{n\to \infty}\int_\Omega w_n^+\exp(G(u_n))   
(1-\psi_\delta)\mu_n \leq   
2k\exp\big(\frac{\|g\|_{L^1(\mathbb{R})}}\alpha\big) \int_\Omega(1-\psi_\delta)\, d\lambda   
\end{equation}   
Then, as $n$ and then $h$ go to infinity, using \eqref{p}, \eqref{o},   
\eqref{r}, \eqref{e}, \eqref{t}, we obtain from \eqref{una},   
\begin{align*}   
&\limsup_{h\to \infty}\limsup_{n\to \infty}   
 \int_{\{w_n\geq 0\}}   
\exp(G(u_n))\big[a(T_k(u_n),\nabla T_k(u_n))\\   
&-a(T_k(u_n),\nabla T_k(u))\big]\nabla(T_k(u_n)-T_k(u)) (1-\psi_\delta)\\   
&\quad \leq 2k\exp(\frac{\|g\|_{L^1(\mathbb{R})}}\alpha) \int_\Omega   
(1-\psi_\delta)d \lambda  \leq   
2k\exp\big(\frac{\|g\|_{L^1(\mathbb{R})}}\alpha\big)   
\lambda(\Omega\setminus K_\delta)\,.   
\end{align*}   
By means of \eqref{mon} and recalling \eqref{pside} we deduce   
\begin{multline*}   
\limsup_{\delta \to 0}\limsup_{h\to \infty}\limsup_{n\to \infty}   
\int_{\{w_n\geq 0\}} \big[a(T_k(u_n),\nabla T_k(u_n))\\   
-a(T_k(u_n),\nabla T_k(u))\big]\nabla(T_k(u_n)-T_k(u)) (1-\psi_\delta)\leq 0.   
\end{multline*}   
In the same way we work on \eqref{altra}, obtaining   
\begin{multline*}   
\limsup_{\delta \to 0}\limsup_{h\to \infty}\limsup_{n\to \infty}   
 \int_{\{w_n\leq 0\}} \big[a(T_k(u_n),\nabla T_k(u_n))\\   
 - a(T_k(u_n),\nabla T_k(u))\big]\nabla (T_k(u_n)-T_k(u)) (1-\psi_\delta)\leq 0.   
\end{multline*}   
Adding the two inequalities  we conclude   
\begin{multline}\label{shalf}   
\limsup_{\delta\to 0}\limsup_{n\to \infty}   
\int_\Omega \big[a(T_k(u_n),\nabla T_k(u_n))\\   
-a(T_k(u_n),\nabla T_k(u))\big]\nabla (T_k(u_n)-T_k(u)) (1-\psi_\delta)=0\,.   
\end{multline}   
Now, we have   
\begin{align*}   
&  \int_\Omega [a(T_k(u_n),\nabla T_k(u_n))-   
a(T_k(u_n),\nabla T_k(u))]\nabla (T_k(u_n)-T_k(u))\\   
&= \int_\Omega [a(T_k(u_n),\nabla T_k(u_n))-   
a(T_k(u_n),\nabla T_k(u))]\nabla (T_k(u_n)-T_k(u))(1-\psi_\delta)\\   
&\quad + \int_\Omega a(T_k(u_n),\nabla T_k(u_n))\nabla T_k(u_n) \psi_\delta   
-\int_\Omega a(T_k(u_n),\nabla T_k(u_n))\nabla T_k(u) \psi_\delta\\   
&\quad -\int_\Omega a(T_k(u_n),\nabla T_k(u))\nabla (T_k(u_n)-T_k(u)) \psi_\delta\,.   
\end{align*}   
Using the weak convergence of $T_k(u_n)$ to $T_k(u)$ last term converges to   
zero as $n$ goes to infinity. Similarly, we have that   
$a(T_k(u_n),\nabla T_k(u_n))$ is bounded in $L^{p'}(\Omega)^N$   
uniformly on $n$ while $\nabla T_k(u)\psi_\delta$ converges to zero in   
$L^p(\Omega)^N$ as $\delta$ tends to zero. Using also   
\eqref{shalf} and \eqref{half}   we finally get, letting first $n$ go   
to infinity and then $\delta$ to zero,   
$$   
\lim_{n\to \infty}   
\int_\Omega [a(T_k(u_n),\nabla T_k(u_n))-   
a(T_k(u_n),\nabla T_k(u))]\nabla   
(T_k(u_n)-T_k(u))=0.   
$$   
Under assumptions \eqref{coerc}--\eqref{mon}, it is well known that this   
implies   
\begin{equation}\label{sconv}   
T_k(u_n)\to T_k(u)\quad \mbox{strongly in $W^{1,p}_0(\Omega)$ for any $k>0$.}   
\end{equation}   
Moreover, using that $\mathop{\rm meas}\{|u_n|>k\}$ goes to zero as $k$ goes to infinity   
uniformly on $n$, as  a consequence of \eqref{sconv} we also have that, up to   
subsequences, $\nabla u_n$ almost everywhere converges to $\nabla u$ in $\Omega$.   
In turns, this implies that   
\begin{equation}\label{aun}   
a(x,{u_n},{\nabla u_n})\to a(x,u,\nabla u)\quad   
\mbox{strongly in $L^q(\Omega)^N$ for any $q<\frac N{N-1}$.}   
\end{equation}   
Let $\varphi=\int_0^{u_n} g(s)\chi_{\{s>h\}}ds$ in   
\eqref{poseq}; since $|\varphi|\leq \int_h^\infty g(s)ds$   we have   
\begin{align*}   
&\int_\Omega a(u_n,\nabla u_n)\nabla u_n g(u_n)\chi_{\{u_n>h\}}\\   
&\leq \exp\big(\frac{\|g\|_{L^1(\mathbb{R})}}\alpha\big)   
\Big(\int_h^\infty g(s)ds\Big) (\|\gamma\|_{L^1(\Omega)}+ \|\mu_n\|_{L^1(\Omega)})\,.   
\end{align*}   
Using \eqref{coerc} and  the fact that $\mu_n$ is bounded in $L^1(\Omega)$ gives   
$$   
\alpha \int_{\{u_n>h\}}g(u_n)|\nabla u_n|^p\leq c\Big(\int_h^\infty   
g(s)ds\Big)\,,   
$$   
and then since $g\in L^1(\mathbb{R})$ we obtain   
$$   
\lim_{h\to \infty}\quad \sup_{n\in \mathbb{N}}   
\int_{\{u_n>h\}}g(u_n)|\nabla u_n|^p=0\,.   
$$   
Similarly, taking $\varphi=\int_{u_n}^0 g(s)\chi_{\{s<-h\}}ds$ in \eqref{negeq}   
we obtain the corresponding result on the set $\{u_n<-h\}$, hence   
\begin{equation}\label{equ}   
\lim_{h\to \infty}\quad \sup\limits_{n\in \mathbb{N}}   
\int_{\{|u_n|>h\}}g(u_n)|\nabla u_n|^p=0\,.   
\end{equation}   
 A standard argument   
allows to conclude from   
\eqref{sconv} and   
\eqref{equ} that $g(u_n)|\nabla u_n|^p$ strongly converges in $L^1(\Omega)$ to   
$g(u)|\nabla u|^p$. Then from \eqref{creh}, the almost everywhere convergence of   
$u_n$ and $\nabla u_n$  and Lebesgue's theorem we conclude that   
\begin{equation}\label{hn}   
H(x,u_n,\nabla u_n)\to H(x,u,\nabla u)\quad \mbox{strongly in $L^1(\Omega)$.}   
\end{equation}   
Thanks to \eqref{aun} and \eqref{hn} we can pass to the limit  in \eqref{appell}   
and we obtain that $u$ is a distributional solution of \eqref{reac}.   
\hfill$\square$   
   
\begin{remark} \rm   
The assumption that $\mu$ is nonnegative is not essential in Theorem \ref{primo}.   
In order to deal with changing sign measures it is enough to follow the same lines   
of the previous proof with suitable modifications while proving the strong   
convergence of truncations similar to those developed in \cite{DMOP}.   
\end{remark}   
   
\begin{example} \rm   
Let $\mu=\delta_{0}$ be the Dirac mass at the origin and let $\Omega=B(0,1)$ be the unit   
ball in $\mathbb{R}^N$, with $N\geq 3$. Let $\mu_n=n^N\chi_{B(0,\frac1n)}$; clearly $\mu_n$ converges, in the narrow topology, to   
$\lambda\delta_0$ for some constant $ \lambda>0$. Note that in particular $\mu_n$ satisfies \eqref{mun}   
(with $f_n=F_n=0$).   
Let  $u_n$ be any sequence of solutions of   
\begin{equation}\label{apex}   
\begin{gathered}   
-\Delta u_n=g(u_n)|\nabla u_n|^2+\mu_n\quad \mbox{in }\Omega\,,\\   
u_n=0 \quad \mbox{on }\partial\Omega.   
\end{gathered}   
\end{equation}   
We claim that if the following assumption holds:   
\begin{equation}\label{assex}   
\begin{gathered}   
\mbox{$\exists h\in C(\mathbb{R},\mathbb{R}^+)$: $g(s)\geq h(s)$ for every $s\in \mathbb{R}^+$,}\\   
\mbox{$h$ is nonincreasing, $\lim_{s\to +\infty}h(s)=0$ and   
$h\not\in L^1(\mathbb{R}^+)$,}   
\end{gathered}   
\end{equation}   
then the sequence $u_n$ blows up completely, namely $u_n(x)\to +\infty$ for   
every $x\in \Omega$.   
   
As far as assumption \eqref{assex} is concerned,   
observe that if $g$ is nonincreasing, converges to zero at infinity  and $g\not \in   
L^1(\mathbb{R})$, we can clearly take   
$h=g$ in \eqref{assex}; this includes the main examples of $g$ around the borderline   
case $g\in L^1(\mathbb{R})$, as $g(s)=1/(|s|+1)$ or   
$g(s)=1/((1+|s|)\log(1+|s|))$.   
Anyway, assumption   
\eqref{assex} is stated in this generality to include most examples of $g$; in   
particular, note that the it requires $g$ to be {\bf larger} than  a   
nonincreasing function which is not integrable, so that $g$ itself may also   
be unbounded.   
   
In order to prove our claim, we adapt an  idea    
used in a   context of sublinear equations by  L. Orsina (\cite{Oper}).   
Let us set $H(s)=\int_0^s h(\xi)d\xi$, $\psi(s)=\int_0^s \exp(H(\xi))d\xi$   
and define   
$v_n :=\psi(u_n)$ (the function $h$ is defined in \eqref{assex}).   
Observe that $\psi$ is an increasing unbounded function, so that $v_n$   
goes to infinity if and only if $u_n$ goes to infinity.   
Since $g(u_n)\geq h(u_n)$, $v_n$ satisfies   
\begin{equation}\label{eqv}   
\begin{gathered}   
-\Delta v_n \geq \exp(H(u_n)) \mu_n \quad \mbox{in } \Omega\,,\\   
v_n=0\quad \mbox{on }\partial \Omega.   
\end{gathered}   
\end{equation}   
In particular, by definition of $\mu_n$, we have that $v_n$ is a supersolution   
of the problem   
\begin{equation}\label{auco}   
\begin{gathered}   
-\Delta z=\exp(H(\psi^{-1}(z))) n^N \quad\mbox{in } B(0,\frac1n),\\   
z=0\quad \mbox{on } \partial B(0,\frac1n)\,.   
\end{gathered}   
\end{equation}   
Let $\varphi_{1,n}$ be the first eigenfunction of the Laplacian on $B(0,\frac1n)$,   
normalized so  that  $\|\varphi_{1,n}\|_{L^\infty(\Omega)}=1$, and let $\lambda_{1,n}$ be the   
first eigenvalue. Let us set   
$$   
B(s) :=\frac{\exp(H(\psi^{-1}(s)))}{s}.   
$$   
Since $h$ is nonincreasing we have   
\begin{align*}   
\frac{d}{dr}\Big(\frac{\exp(H(r))}{\psi(r)}\Big)=&   
\frac{\exp(H(r))}{\psi(r)^2}\Big( h(r)\int_0^r   
\exp(H(\xi))d\xi-\exp(H(r))\Big)\\   
\leq& \frac{\exp(H(r))}{\psi(r)^2}\Big(  \int_0^r   
\exp(H(\xi)) h(\xi)d\xi-\exp(H(r))\Big)<0\,,   
\end{align*}   
so that $B(\psi(s))$ is decreasing. Since $\psi$ is increasing, we deduce that   
$B$ is a decreasing function. Let us set $T_n=B^{-1}(\frac{\lambda_{1,n}}{n^N})$.   
Since $B$ is decreasing, we deduce that   
$$   
\frac{\lambda_{1,n}}{n^N}=B(T_n)=B(T_n\|\varphi_{1,n}\|_{L^\infty(\Omega)})\leq   
B(T_n \varphi_{1,n}(x))\quad   
\forall x\in B(0,\frac1n)\,,   
$$   
which implies, by definition of $B$,   
$$   
\lambda_{1,n} T_n \varphi_{1,n}(x)\leq \exp(H(\psi^{-1}(T_n \varphi_{1,n}(x)))) n^N  \quad   
\forall x\in B(0,\frac1n)\,.   
$$   
Since $\lambda_{1,n} T_n \varphi_{1,n}=-\Delta (T_n \varphi_{1,n})$ we conclude that   
$T_n \varphi_{1,n}$ is a subsolution of \eqref{auco}. Since   
$\exp\big(H(\psi^{-1}(z))\big)/z=B(z)$ is decreasing,  a well-known   
comparison principle holds   
for positive sub-super solutions of \eqref{auco} (see for example \cite{BK}),   
so that we get $v_n\geq T_n \varphi_{1,n}$ in $B(0,\frac1n)$.   
By scaling arguments we know that   
$$   
\varphi_{1,n}(x)=\varphi_{1,1}(nx)\,,\quad \lambda_{1,n}=\lambda_{1,1} n^2\,,   
$$   
hence we obtain   
$$   
\forall x\in B(0,\frac1{2n}) : \quad v_n(x)\geq   
B^{-1}(\frac{\lambda_{1,1}}{n^{N-2}})\min_{B(0,\frac1{2})}\varphi_{1,1}\,.   
$$   
Since $\varphi_{1,1}$ is radial, we have   
$\min_{B(0,\frac1{2})}\varphi_{1,1}=\varphi_{1,1}(\frac12)$, so that   
\begin{equation}\label{uf}   
\min_{B(0,\frac1{2n})} v_n\geq   
B^{-1}(\frac{\lambda_{1,1}}{n^{N-2}}) \varphi_{1,1}(\frac12)\,.   
\end{equation}   
Now observe that, using De L'Hospital's theorem and the fact that $h(s)$ goes to   
zero at infinity, we have   
$\lim_{s\to +\infty} B(s)=0$.   
Since $\lambda_{1,1}/n^{N-2}$ converges to zero as $n$ tends to infinity,   
we end up with   
$$   
\lim_{n\to +\infty}B^{-1}(\frac{\lambda_{1,1}}{n^{N-2}})=+\infty\,,   
$$   
and then from \eqref{uf}   
\begin{equation}\label{minex}   
\lim_{n\to +\infty} \min_{B(0,\frac1{2n})} v_n=+\infty\,.   
\end{equation}   
Let  now $G(x,y)$ be the kernel of the Laplacian with zero boundary condition;   
we have from \eqref{eqv}   
\begin{equation}\label{pref}   
\begin{aligned}   
v_n(x)&\geq \int_\Omega G(x,y)\exp(H(u_n))(y) \mu_n(y) dy\\   
&\geq \min_{B(0,\frac1{2n})}\left( \exp(H(\psi^{-1}(v_n)))\right)   
\int_{B(0,\frac1{2n})} G(x,y) n^N\, dy\,.   
\end{aligned}   
\end{equation}   
Since there exists a constant $c>0$ such that   
$$   
\int_{B(0,\frac1{2n})} G(x,y) n^N dy\to c \int_\Omega G(x,y)d\delta_0(y)>0\,,   
$$   
and since both $\psi^{-1}$ and  $H$ go to infinity at infinity (because $h\not\in   
L^1(\mathbb{R}^+)$), we deduce using \eqref{minex} that the right hand side of \eqref{pref}   
goes to infinity as $n$ goes to infinity. We then conclude   
$$   
\lim_{n\to +\infty} v_n(x)=+\infty\quad \forall x\in \Omega\,.   
$$   
Since $\psi$ is unbounded and $u_n=\psi^{-1}(v_n)$, we have proved that the   
solutions $u_n$ of \eqref{apex} blow up completely in $\Omega$.   
This is in sharp contrast with what proved in Theorem \ref{primo} when $g\in   
L^1(\mathbb{R})$, so that this assumption is optimal in the existence result above.   
\end{example}   
   
   
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\noindent\textsc{Alessio Porretta}\\   
Dipartimento di Matematica,   
Universit\`a di Roma \lq\lq Tor Vergata\rq\rq,\\   
Via della Ricerca Scientifica 1, 00133, Roma, Italia.\\   
email: porretta@mat.uniroma2.it   
\end{document}   
