# Existence and regularity results for nonlinear elliptic equations in Orlicz
spaces

Giuseppina Barletta
Dipartimento di Ingegneria Civile, dell’Energia, dell’Ambiente e dei Materiali
Università Mediterranea di Reggio Calabria
Via Zehender, 89122, Reggio Calabria (Italy)
e-mail<EMAIL_ADDRESS>
###### Abstract

We are concerned with the existence and regularity of the solutions to the
Dirichlet problem, for a class of quasilinear elliptic equations driven by a
general differential operator, depending on $(x,u,\nabla u)$, and with a
convective term $f$. The assumptions on the members of the equation are
formulated in terms of Young’s functions, therefore we work in the Orlicz-
Sobolev spaces. After establishing some auxiliary properties, that seem new in
our context, we present two existence and two regularity results. We conclude
with several examples.

00footnotetext:  Mathematics Subject Classification: 35J25, 35J60, 46E30,
47J05. Keywords: Nonlinear elliptic equations; Orlicz-Sobolev spaces; gradient
dependence; Leray-Lions operators; subsolution and supersolution. The author
is member of the Gruppo Nazionale per l’Analisi Matematica, la Probabilità e
le loro Applicazioni (GNAMPA) of the Istituto Nazionale di Alta Matematica
(INdAM).
This research is partially supported by the Ministry of Education, University
and Research of Italy, Prin 2017 Nonlinear Differential Problems via
Variational, Topological and Set-valued Methods (Project No. 2017AYM8XW).

##  1 Introduction

In the present paper we deal with the existence and regularity of solutions to
the the following nonlinear elliptic problem

(1.1) $\begin{cases}-{\rm div}(\mathcal{A}(x,u,\nabla u))=f(x,u,\nabla u)&{\rm
in}\,\,\,\Omega\\\ u=0&{\rm on}\,\,\,\partial\Omega\,.\end{cases}$

$\Omega$ is a subset of ${{\mathbb{R}}^{n}}$ having finite measure,
$\mathcal{A}:\Omega\times\mathbb{R}\times{{\mathbb{R}}^{n}}\to{{\mathbb{R}}^{n}}$
and $f:\Omega\times\mathbb{R}\times{{\mathbb{R}}^{n}}\to\mathbb{R}$ are
Carathéodory functions.
In spite of the many boundedness (see, for instance, [BaCiMa, Ci2, LU, MaWi]
and the references therein) or regularity (see [BaCoMi, BPS, DBe, Li, Li1,
Mi]) results, in which the existence of a solution is assumed a priori, there
are fewer results about the existence of solutions for (1.1), unless the
principal part is the $p-$Laplacian or growths like the $p-$Laplacian.
This is due to the fact that when the operator $\mathcal{A}$ in (1.1) depends
also from the unknown $u$, technical difficulties arise. First of all,
variational methods are not directly applicable (and this holds also for
problems with a convective term, namely with a nonlinearity $f$ depending also
on $\nabla u$), and the properties of $\mathcal{A}$ are actually well known
only in standard contexts (see [CLM, Tr]). This means, in particular, that in
most situations, the abstract framework for the study of (1.1) is the
classical Sobolev space $W_{0}^{1,p}(\Omega)$.
In this regard, we note that there are many papers dealing with the existence
of solutions to (1.1), whenever the principal part does not depends
explicitely on $u$, and the tecniques used are the most diverse. This also
testifies to the growing interest in recent years for the study of these
problems.
The two main motivations of the paper are to investigate problems whose
differential part depends explicitly also on the unknown, and to remove
restrictions of power type growth in the study of (1.1). To do this, we have
to establish some properties of $\mathcal{A}$, which are of independent
interest (see Proposition 2.17). Just to give an idea, in our contest we can
manage problems driven by an operator having a power-times-logarithmic type
growth, like (see Example 2.16)

(1.2) $\displaystyle\mathcal{A}(x,s,\xi)=$ $\displaystyle
a(x)|s|^{\beta}\lg^{\beta_{1}}(1+|s|)|\xi|^{p-2-\delta}\lg^{q(1-\frac{\delta}{p-1})}(1+|\xi|)\xi+|\xi|^{p-2}\lg^{q}(1+|\xi|)\xi\,.$

To the best of our knowledge no results are actually available for problems
driven by such operators. Conversely, many existence and regularity results
are available for (1.1), when $\mathcal{A}$ is the $p$-Laplacian, or the
$(p,q)$-Laplacian, namely when $\mathcal{A}(\nabla u)=|\nabla u|^{p-2}\nabla
u$ or $\mathcal{A}(\nabla u)=|\nabla u|^{p-2}\nabla u+|\nabla u|^{q-2}\nabla
u$ (see [AAP, AF, BF, FMP, LMZ, MW, NS, R, Z]). There is also an extensive
literature concerning problems in which the structure of $\mathcal{A}$ allows
to tackle (1.1) with the same techniques adopted for the $p-$Laplacian. We
refer to [FM, PZ] for problems with $\mathcal{A}(\nabla u)$, to [Boc, Boc1,
BNV, HT, MW, NS, T] for problems with $\mathcal{A}(x,\nabla u)$ and finally,
for problems with $\mathcal{A}(x,u,\nabla u)$ in $W_{0}^{1,p}(\Omega)$, we
cite [FMMT, G].
Much less is available regarding more general operators, built via a function
radial with respect to $\xi$, but not necesseraly a polinomyal. This new
situation requires the use of Young functions and Orlicz spaces. Existence
(and regularity) results for problems with $\mathcal{A}(\nabla u)=a(|\nabla
u|)\nabla u$ can be found in [BaTo1, BaTo2, CGSS, FMST]. In [DF] the authors
deal with an operator depending on the three variables via Young functions of
a real variable.
In this paper, the growth conditions on the terms appearing in (1.1) require
to replace the customary Sobolev space with an Orlicz space (see (1.2)). These
conditions cover several instances already studied in the papers cited above.
We stress that Young’s functions are also involved in the growth of the
convective term $f$. Similar hypotheses can be found in [BaTo1, BaTo2, DF].
Given the non-variational nature of the problem, we use the method of sub and
super solutions, togheter with truncation techniques and the theorem of
existence of zeros for monotone operators. For the regularity results, a main
tool is a Theorem due to Lieberman (see [Li1, Theorem 1.7] and Proposition
4.1).
The first step necessary to obtain our results consists in establishing some
properties of the operator $\mathcal{A}$ in Orlicz spaces. This properties, as
well as basic definitions and some other auxiliary results, are collected in
Section 2. The main existence theorems (and a general example of function
$\mathcal{A}$ satisfying our assumptions) are presented in Section 3: beside
to a general existence result, Theorem 3.4, in Theorem 3.8 we consider a
special instance of (1.1), where a subsolution and a supersolution are
obtained via variational methods. In Section 4 we prove our regularity
results. Theorems 3.8 and 4.4 have among the main hypothesis, the existence of
a subsolution and a supersolution. Starting from them, we construct a suitable
functional, in which appropriate truncations of $\mathcal{A}$ and of the
convective term $f$ are involved.
Finally, Section 5 is devoted to some examples where we highlight how the
method of sub and super solutions works well in all situations where the
convective term $f$ has two zeros of opposite sign, namely
$f(x,s_{1},0)=f(x,s_{2},0)=0$ for all $x\in\Omega$ and for some
$s_{1},s_{2}\in\mathbb{R}$, enjoying the condition $s_{1}\cdot s_{2}<0$.
In this regard, it should be noted that the use of the method of sub and super
solutions, combined with truncation techniques, is extremely valid when a mix
of conditions related to both $\mathcal{A}$ and $f$ occur. This is due to the
fact that once we have a sub and a super solution, we restrict our attention
to a suitable truncation of $\mathcal{A}(x,u,\nabla u)$. If the structure of
$f$ allow for sub and super solutions having good properties, such as, for
instance, boundedness, then the truncated operator satisfies the hypotheses of
the abstract result, even if the growth of $\mathcal{A}$ is more general than
one would expect (see Examples 5.1 and 5.3).

##  2 Preliminaries

In this Section we give the main definitions on Young functions and introduce
the Orlicz Sobolev spaces that we use in the sequel. We also collect some new
results, auxiliary for the proof of the main theorems. Classical results
concerning Young functions and Orlicz spaces can be found in [Chl, KrRu, RR1,
RR2].

###### Definition 2.1

A function $A:[0,\infty)\to[0,\infty]$ is called a Young function if it is
convex, vanishes at $0$, and is neither identically equal to $0$, nor to
infinity.

It is not restrictive to assume that any finite valued Young function is
continuous. For Young functions

(2.1) $A(\lambda t)\leq\lambda A(t)\quad\hbox{for $\lambda\leq 1$ and $t\geq
0$.}$

###### Definition 2.2

The Young conjugate of a Young function $A$ is the Young function
$\widetilde{A}$ defined as

$\widetilde{A}(s)=\sup\\{st-A(t):\ t\geq 0\\}\quad\hbox{for $s\geq 0$.}$

The following inequalities are a consequence of Definition 2.2

(2.2) $\frac{A(t)}{t}\leq\widetilde{A}^{-1}(A(t))\leq
2\frac{A(t)}{t}\quad\hbox{for}\ t>0\,,$ (2.3)
$st\leq\widetilde{A}(s)+A(t)\quad\hbox{for}\ s,t>0\,.$

The inverse function in (2.2) is the generalized right continuous inverse. In
general a Young function is left continuous (in effect it is continuous,
unless it takes the value $+\infty$), and when we deal with the inverse we
consider the right continuous one.

###### Definition 2.3

A Young function $A$ is said to dominate another Young function $B$ near
infinity, if there exist constants $k>0$ and $M\geq 0$ such that

(2.4) $B(t)\leq A(kt)\ \ \hbox{if }\,t\geq M\,.$

If (2.4) holds with $M=0$, then we say that $A$ dominates $B$ Two Young
functions $A$ and $B$ are called equivalent near infinity (globally) if they
dominate each other near infinity (globally) and w e briefly write $A\approx
B$ near infinity (globally).

###### Definition 2.4

Two functions $f,g:(0,\infty)\to[0,\infty)$, are equivalent near infinity
(briefly $f\approx g$ near infinity) if and only if there exist suitable
positive constants $c_{1}$, $c_{2}$ and $s_{0}$ such that

(2.5) $c_{1}g(c_{1}s)\leq f(s)\leq c_{2}g(c_{2}s)\quad\hbox{if}\ s>s_{0}\,.$

###### Definition 2.5

A Young function $B$ is said to increase essentially more slowly than $A$ near
infinity (briefly $B\ll A$), if $B$ is finite valued and

(2.6) $\lim_{t\to\infty}\frac{B(\lambda t)}{A(t)}=0\ \ \hbox{for
all}\;\lambda>0\,.$

Condition (2.6) is equivalent to

(2.7) $\lim_{t\to\infty}\frac{A^{-1}(t)}{B^{-1}(t)}=0\,,$

where $A^{-1}$ denotes the right continuous inverse of $A$. Note that if
$A(t)=+\infty$ for $t>t_{0}$, then $A^{-1}(s)$ is constant for $s$
sufficiently large, whatever inverse we take.

###### Proposition 2.6

Let $A$ and $B$ be two Young functions, such that $B\ll A$. Then there exists
a Young function $C$ such that $B\ll C\ll A$.

Proof.  Let us define $C^{-1}(t)=\sqrt{A^{-1}(t)B^{-1}(t)}$ for $t\geq 0$. One
can easily check that $C^{-1}$ is concave, strictly increasing (inasmuch $A$
and $B$ are strictly increasing), and

(2.8) $\displaystyle\lim_{t\to\infty}\frac{A^{-1}(t)}{C^{-1}(t)}=0,\ $
$\displaystyle\ \lim_{t\to\infty}\frac{C^{-1}(t)}{B^{-1}(t)}=0\,.$

Thus $C$ satisfies $B\ll C\ll A$.

###### Definition 2.7

A Young function $A$ is said to satisfy the $\Delta_{2}$-condition near
infinity (briefly $A\in\Delta_{2}$ near infinity) if it is finite valued and
there exist two constants $K\geq 2$ and $M\geq 0$ such that

(2.9) $A(2t)\leq KA(t)\quad\hbox{for }\ t\geq M\,.$

###### Definition 2.8

The function $A$ is said to satisfy the $\nabla_{2}$-condition near infinity
(briefly $A\in\nabla_{2}$ near infinity) if there exist two constants $K>2$
and $M\geq 0$ such that

(2.10) $A(2t)\geq KA(t)\quad\hbox{for }\ t\geq M\,.$

If (2.9) or (2.10) holds with $M=0$, then $A$ is said to satisfy the
$\Delta_{2}$-condition (globally), or the $\nabla_{2}$-condition (globally),
respectively. If (2.9) or (2.10) holds for $0\leq t\leq M$, then $A$ is said
to satisfy the $\Delta_{2}$-condition near zero, or the $\nabla_{2}$-condition
near zero, respectively. If $A\in\Delta_{2}\cap\nabla_{2}$ near infinity then
$\widetilde{A}\in\Delta_{2}\cap\nabla_{2}$ near infinity (see [BaCi1],
Proposition 6.6).

We give basic definitions and the main properties on the Orlicz spaces. Let
$\Omega$ be a measurable set in ${{\mathbb{R}}^{n}}$, with $n\geq 1$. Given a
Young function $A$, the Orlicz space $L^{A}(\Omega)$ is the set of all
measurable functions $u:\Omega\to\mathbb{R}$ such that the Luxemburg norm

$\|u\|_{L^{A}(\Omega)}=\inf\bigg{\\{}\lambda>0:\int_{\Omega}A\big{(}\tfrac{1}{\lambda}|u|\big{)}\,dx\leq
1\bigg{\\}}$

is finite. The functional $\|\cdot\|_{L^{A}(\Omega)}$ is a norm on
$L^{A}(\Omega)$, and it is a Banach space (see [Ad]). If $A$ and $B$ are Young
functions, $A\approx B$ near infinity and $|\Omega|<\infty$ then
$L^{A}(\Omega)=L^{B}(\Omega)$, and there exist
$k_{1}(,t_{0},|\Omega|),k_{2}(t_{0},|\Omega|)$ such that

(2.11) $\displaystyle k_{1}\|u\|_{L^{A}(\Omega)}\leq\|u\|_{L^{B}(\Omega)}\leq
k_{2}\|u\|_{L^{A}(\Omega)},\ \hbox{for all}\ u\in L^{A}(\Omega)\,,$

If $A$ is a Young function and $\widetilde{A}$ denotes its conjugate, then a
generalized Hölder inequality

(2.12) $\int_{\Omega}|uv|\,dx\leq
2\|u\|_{L^{A}(\Omega)}\|v\|_{L^{\widetilde{A}}(\Omega)}$

holds for every $u\in L^{A}(\Omega)$ and $v\in L^{\widetilde{A}}(\Omega)$.
If $A\in\Delta_{2}$ globally (or $A\in\Delta_{2}$ near infinity and $\Omega$
has finite measure) then

(2.13) $\int_{\Omega}A(k|u|)dx<+\infty\ \hbox{for all}\ u\in L^{A}(\Omega),\
\hbox{all}\ k\geq 0\,.$

Also, if $B\ll A$ near infinity and $\Omega$ has finite measure, then

(2.14) $\int_{\Omega}B(k|u|)dx<+\infty\ \hbox{for all}\ u\in L^{A}(\Omega),\
\hbox{all}\ k\geq 0\,.$

Given a Young function $A\in C^{1}([0,+\infty))$, define the quantities

(2.15) $\displaystyle i_{A}=\inf_{t>0}\frac{t\cdot
A^{\prime}(t)}{A(t)},\,s_{A}=\sup_{t>0}\frac{t\cdot A^{\prime}(t)}{A(t)},\ $
$\displaystyle\ i_{A}^{\infty}=\liminf_{t\to+\infty}\frac{t\cdot
A^{\prime}(t)}{A(t)},\,s_{A}^{\infty}=\limsup_{t\to+\infty}\frac{t\cdot
A^{\prime}(t)}{A(t)}\,.$

The conditions

$\displaystyle i_{A}>1\ \hbox{and}\ s_{A}<+\infty\ $ $\displaystyle\
(i_{A}^{\infty}>1\ \hbox{and}\ s_{A}^{\infty}<+\infty)$

are equivalent to the fact that $A\in\nabla_{2}\cap\Delta_{2}$ globally (near
infinity). The following result extends Lemma A.1 of [CM] to Young functions
$A\in\nabla_{2}$ near infinity.

###### Lemma 2.9

Let $\Omega$ be a subset of ${{\mathbb{R}}^{n}}$, with finite measure. Let $A$
be a Young function. Assume that $i_{A}^{\infty}>1$. Then there exists
$k_{3}(i_{A}^{\infty},\,|\Omega|)\geq 0$ such that

(2.16) $\int_{\Omega}A(|v|)dx\geq
k_{1}^{i^{\infty}_{A}}\|v\|_{L^{A}(\Omega)}^{i^{\infty}_{A}}-k_{3}\quad\hbox{for
every}\ v\in L^{A}(\Omega),\ \hbox{such that}\
\|v\|_{L^{A}(\Omega)}\geq\frac{1}{k_{1}}\,.$

Here $k_{1}$ is that of (2.11).

Proof. Since $i_{A}^{\infty}>1$, corresponding to
$\varepsilon\in(0,i_{A}^{\infty}-1)$ there exists $t_{0}>0$ such that
$\displaystyle{\frac{t\cdot A^{\prime}(t)}{A(t)}>i_{A}^{\infty}-\varepsilon}$
for $t\geq t_{0}$. Fix $\alpha>i_{A}^{\infty}$ and consider the Young function

(2.19) $\displaystyle
A_{1}(t)=\left\\{\begin{array}[]{ll}c_{1}t^{i_{A}^{\infty}}+c_{2}t^{\alpha}&\hbox{if}\
t\leq t_{0}\\\ A(t)&\hbox{if}\ t>t_{0}\end{array}\right.$

where $c_{1}$ and $c_{2}$ are chosen in such a way that $A_{1}\in
C^{1}([0,+\infty[)$. Then $A_{1}\approx A$ near infinity and
$i_{A_{1}}^{\infty}=i_{A}^{\infty}$. Let $v\in L^{A}(\Omega)$ be such that
$\|v\|_{L^{A}(\Omega)}\geq\frac{1}{k_{1}}.$

Using Lemma A.1 of [CM] and (2.11), with $B=A_{1}$

$\displaystyle\int_{\Omega}A(|v|)dx$
$\displaystyle=\int_{\Omega}A_{1}(|v|)dx+\int_{\\{|v|\leq
t_{0}\\}}(A(|v|)-A_{1}(|v|))dx$
$\displaystyle\geq\|v\|_{L^{A_{1}}(\Omega)}^{i_{A}^{\infty}}+m_{0}|\Omega|\geq
k_{1}^{i_{A}^{\infty}}\|v\|_{L^{A}(\Omega)}^{i_{A}^{\infty}}+m_{0}|\Omega|\,,$

where $m_{0}=\min_{0\leq t\leq t_{0}}(A(t)-A_{1}(t))\leq 0$. Thus (2.16)
follows with $k_{3}=-m_{0}|\Omega|\geq 0$.

The Lemma below is a consequence of the Theorems of Vitali and De la Vallée-
Poussin. It will be used several times in the paper.

###### Lemma 2.10

Let $\Omega\subseteq{{\mathbb{R}}^{n}}$ be such that $|\Omega|<\infty$ and let
$A$ be a Young function. If $\\{u_{k}\\}_{k\in N}\subseteq L^{A}(\Omega)$
satisfies

  * •

$u_{k}(x)\to u(x)$ a.e. in $\Omega$ ,

  * •

$\\{u_{k}\\}_{n\in N}$ is bounded in $L^{A}(\Omega)$ ,

then $u_{k}\to u$ in $L^{B}(\Omega)$ for all Young functions $B\ll A$ near
infinity.
If $L^{A}(\Omega)$ is reflexive, then $|u_{k}-u|\rightharpoonup 0$ in
$L^{A}(\Omega)$.

Proof. Let $B\ll A$ near infinity. Then $B$ is finite valued and we may
assume, without loss of generality, that $B$ is continuous. We must prove that
$\lim_{k\to+\infty}\int_{\Omega}B\left(\frac{|u_{k}-u|}{\lambda}\right)dx=0$,
for all $\lambda>0$.
If $A(t)\equiv+\infty$ for $t>t_{0}>0$, then
$L^{A}(\Omega)=L^{\infty}(\Omega)$ and the conclusion follows via the
dominated convergence theorem.
If $A$ is finite valued, then we may assume that it is continuous. Fix
$\lambda>0$ and let $M>0$ be such that $\|u_{k}\|_{L^{A}(\Omega)}\leq M$ for
all $n\in N$. Then, the continuity of $A$ and the Fatou’s lemma guarantee that
$u\in L^{A}(\Omega)$ and $\|u\|_{L^{A}(\Omega)}\leq M$. Let
$\Psi(t)=A\big{(}\frac{\lambda}{2M}B^{-1}(t)\big{)}$, for $t\geq 0$. Then
$\Psi$ is increasing and the condition $B\ll A$ guarantees that

(2.20) $\displaystyle\lim_{t\to+\infty}\frac{\Psi(t)}{t}=+\infty\,.$

Note that

(2.21)
$\displaystyle\left\\{B\left(\frac{|u_{k}-u|}{\lambda}\right)\right\\}_{k\in
N}\subseteq\left\\{v\in L^{1}(\Omega):\,\int_{\Omega}\Psi(|v|)dx\leq
1\right\\}\,.$

Thus the family
$\left\\{B\left(\frac{|u_{k}-u|}{\lambda}\right)\right\\}_{k\in N}$ is
equintegrable and we can apply the Vitali’s Theorem. This proves that
$u_{k}\to u$ in $L^{B}(\Omega)$.
For the last part of the proof we observe that $|u_{k}-u|\rightharpoonup v$ in
$L^{A}(\Omega)$, up to a subsequence. From the proof above we know that
$|u_{k}-u|\to 0$ in $L^{B}(\Omega)$. Thus $v\equiv 0$. This apply to all the
subsequences, thus it holds for the whole sequence.

From now and throughout the paper we assume that $\Omega$ is an open set in
${{\mathbb{R}}^{n}}$ with $|\Omega|<\infty$. The isotropic Orlicz-Sobolev
spaces $W^{1,A}(\Omega)$ and $W^{1,A}_{0}(\Omega)$ are defined as

(2.22) $\displaystyle W^{1,A}(\Omega)=\\{u:\Omega\to\mathbb{R}:$
$\displaystyle\,u\hbox{ is weakly differentiable in $\Omega$,\ $u$,\;$|\nabla
u|\in L^{A}(\Omega)$}\\}\,$

and

(2.23) $\displaystyle{W_{0}^{1,A}(\Omega)}=\\{u:\Omega\to\mathbb{R}:$ the
continuation of $u$ by $0$ outside $\Omega$ $\displaystyle\hbox{is weakly
differentiable in ${{\mathbb{R}}^{n}}$, \; u,\; $|\nabla u|\in
L^{A}(\Omega)$}\\}.$

The spaces $W^{1,A}(\Omega)$ and ${W_{0}^{1,A}(\Omega)}$ equipped with the
norms

(2.24) $\displaystyle\|u\|_{W^{1,A}(\Omega)}=\|u\|_{L^{A}(\Omega)}+\|\nabla
u\|_{L^{A}(\Omega)},\ \ \hbox{and}\ \ \|u\|_{{W_{0}^{1,A}(\Omega)}}=\|\nabla
u\|_{L^{A}(\Omega)},$

are Banach spaces. The norm on ${W_{0}^{1,A}(\Omega)}$ is equivalent to the
standard one

$\|u\|_{W_{0}^{1,A}(\Omega)}=\|u\|_{L^{A}(\Omega)}+\|\nabla
u\|_{L^{A}(\Omega)}\,.$

If $A\in\Delta_{2}\cap\nabla_{2}$ near infinity, then the Orlicz-Sobolev space
$W^{1,A}_{0}(\Omega)$ is reflexive ([BaCi1, Proposition 3.1]).

###### Definition 2.11

The optimal Sobolev conjugate of $A$ is defined by
$A_{n}:[0,\infty)\to[0,\infty]$

(2.25) $A_{n}(t)=A({H}^{-1}(t))\quad\hbox{for $t\geq 0$,}$

where ${H}:[0,\infty)\to[0,\infty)$ is given by

${H}(t)=\bigg{(}\int_{0}^{t}\bigg{(}\frac{\tau}{A(\tau)}\bigg{)}^{\frac{1}{n-1}}\,d\tau\bigg{)}^{\frac{n-1}{n}}\quad\hbox{for
$t\geq 0$,}$

provided that the integral is convergent. Here, ${H}^{-1}$ denotes the
generalized left-continuous inverse of ${H}$.

If

(2.26)
$\int_{0}\bigg{(}\frac{\tau}{A(\tau)}\bigg{)}^{\frac{1}{n-1}}\,d\tau<\infty\,,$

then (see [Ci3, Theorem 1])

(2.27) ${W_{0}^{1,A}(\Omega)}\to L^{A_{n}}(\Omega)\,.$

and there exists a constant $C=C(n)$ such that

(2.28) $\|u\|_{L^{A_{n}}(\Omega)}\leq
C\|u\|_{W_{0}^{1,A}(\Omega)}\quad\hbox{for every}\ u\in
W_{0}^{1,A}(\Omega)\,.$

Thus from [Ci1, Theorem 3] (or [HLZ, Theorem 3.1]), the embedding
${W_{0}^{1,A}(\Omega)}\to L^{E}(\Omega)$ is compact for every Young function
$E\ll A_{n}$ near infinity.
If

(2.29)
$\int^{\infty}\bigg{(}\frac{\tau}{A(\tau)}\bigg{)}^{\frac{1}{n-1}}\,d\tau=\infty$

then $A_{n}$ assumes only finite values, while when

(2.30)
$\int^{\infty}\bigg{(}\frac{\tau}{A(\tau)}\bigg{)}^{\frac{1}{n-1}}\,d\tau<\infty\,,$

then $A_{n}(t)=\infty$ for $t$ large and (2.28) yields

$\|u\|_{L^{\infty}(\Omega)}\leq C\|u\|_{W_{0}^{1,A}(\Omega)}\quad\hbox{for
every $u\in{W_{0}^{1,A}(\Omega)}$.}$

When (2.29) holds then (see [BaCi1, (3.13)])

(2.31) $\int_{\Omega}A_{n}(\lambda|u|)dx<\infty\ \hbox{for every}\ u\in
W_{0}^{1,A}(\Omega),\ \hbox{and every}\,\lambda>0.$

###### Example 2.12

A general Young function
Let $1<p<+\infty$, $q\in\mathbb{R}$. Consider a Young function
$B:[0,+\infty[\rightarrow[0,+\infty[$, such that

(2.32) $\displaystyle B^{\prime}(t)=t^{p-1}\lg^{q}(1+t)\quad\hbox{for}\
t>>1\,.$

Then

(2.33) $\displaystyle i_{B}^{\infty}=s_{B}^{\infty}=p>1\,,$

and $W_{0}^{1,B}(\Omega)$ is reflexive. Also

(2.34) $\displaystyle B(t)\approx t^{p}\lg^{q}(1+t)\quad\hbox{near infinity}.$

Thus, given any Young function $A$ such that

(2.35) $\displaystyle A(t)\approx t^{p}\lg^{q}(1+t)\ \hbox{near infinity}\,,$

and a set $\Omega$ with finite measure, it holds
${W_{0}^{1,A}(\Omega)}=W_{0}^{1,B}(\Omega)$, up to equivalent norms. The
conjugate $\widetilde{A}$ and the optimal Sobolev conjugate of $A$ satisfy

(2.36) $\displaystyle\widetilde{A}(t)\approx
t^{p^{\prime}}\lg^{-\frac{q}{p-1}}(1+t)\quad\hbox{near infinity},$

and

(2.41) $\displaystyle
A_{n}(t)\thickapprox\left\\{\begin{array}[]{lll}t^{\frac{np}{n-p}}\lg^{\frac{nq}{n-p}}(1+t)&\hbox{near
infinity}&\hbox{when}\ 1<p<n,\\\ e^{t^{\frac{n}{n-q-1}}}&\hbox{near
infinity}&\hbox{when}\ p=n,\ q<n-1,\\\ e^{e^{t^{\frac{n}{n-1}}}}&\hbox{near
infinity}&\hbox{when}\ p=n,\ q=n-1,\\\ +\infty&\hbox{near infinity}&\hbox{in
all the other cases}\,.\end{array}\right.$

If $p+q-1\geq 0$ then the function $B$ satisfying (2.32) for all $t>0$ is in
fact a Young function. When $p+q-1>0$ the function $B\in\nabla_{2}$ globally.
All the examples of the paper will involve functions $A$ complying with
(2.35).

We recall now some definitions and the theorem on pseudomonotone operators.
Then we introduce the conditions on the function $\mathcal{A}$ in (1.1).

###### Definition 2.13

Let $X$ be a real reflexive Banach space. A mapping ${\cal B}:X\to X^{*}$ is
called

  * (i)

coercive if $\lim_{\|u\|\to\infty}\frac{\langle{\cal
B}u,u\rangle}{\|u\|}=+\infty$;

  * (ii)

bounded if it maps bounded sets into bounded sets;

  * (iii)

pseudomonotone if $u_{n}\rightharpoonup u$ and
$\limsup_{k\to+\infty}\langle{\cal B}u_{k},u_{k}-u\rangle\leq 0$ imply that
${\cal B}u_{k}\rightharpoonup{\cal B}u$ and $\langle{\cal
B}u_{k},u_{k}\rangle\to\langle{\cal B}u,u\rangle$.

  * (iv)

satisfies the $(S)_{+}$ property if $u_{k}\rightharpoonup u$ and
$\limsup_{k\to+\infty}\langle{\cal B}u_{k},u_{k}-u\rangle\leq 0$ imply that
$u_{k}\to u$ (strongly) in $X$.

###### Theorem 2.14

(see [CLM, Theorem 2.99]) Let $X$ be a real reflexive Banach space and let
${\cal B}:X\to X^{*}$ be a bounded, coercive and pseudomonotone operator.
Then, for every $b\in X^{*}$ the equation ${\cal B}x=b$ has at least solution
$x\in X$.

Let us now introduce the hypotheses on the function $\mathcal{A}$ operator and
verify the properties of the integral operator associated with it, in a
standard way. We note that this is a Leray Lions type operator. The properties
of the auxiliary truncated operator will easily follow from those of the non-
truncated operator (see Proposition 2.17). We extend the results in [CLM, Tr]
in several directions: Orlicz spaces are considered, and even in the case of
Lebesgue spaces, the hypotheses are slightly more general.
Let $\Omega\subset{{\mathbb{R}}^{n}}$ be a set with finite measure and let $A$
be a Young function, $A\in\Delta_{2}\cap\nabla_{2}$ near infinity. Consider
the vector valued function
$\mathcal{A}:\Omega\times\mathbb{R}\times{{\mathbb{R}}^{n}}\to{{\mathbb{R}}^{n}}$,
$\mathcal{A}=(a_{1},\ldots a_{n})$, enjoying with the properties that each
$a_{i}(x,s,\xi)$ is a Carathéodory function, and

(2.42) $|\mathcal{A}(x,s,\xi)|\leq
q(x)+b\left[\widetilde{A}^{-1}(F(b|s|))+\widetilde{A}^{-1}(A(b|\xi|))\right]\
\hbox{for a.e.}\ x\in\Omega,\ \hbox{all}\
(s,\xi)\in\mathbb{R}\times{{\mathbb{R}}^{n}}\,.$

Here $q\in L^{\widetilde{A}}(\Omega)$, $b>0$, and $F$ is a Young function,
$F\ll A_{n}$ near infinity.

(2.43)
$\displaystyle\sum_{i=1}^{n}\left(a_{i}(x,s,\xi)-(a_{i}(x,s,\xi^{\prime})\right)\cdot(\xi_{i}-\xi_{i}^{\prime})>0\
$ $\displaystyle\ \hbox{for
a.e.}\,x\in\Omega,\,\hbox{all}\,s\in\mathbb{R},\,\hbox{all}\,\,\xi,\xi^{\prime}\in{{\mathbb{R}}^{n}},\,\xi\neq\xi^{\prime}\,.$
(2.44) $\displaystyle\sum_{i=1}^{n}a_{i}(x,s,\xi)\cdot\xi_{i}\geq
cA(c|\xi|)-dG(d|s|)-r(x)$ $\displaystyle\qquad\hbox{for a. e.}\ x\in\Omega,\
\hbox{all}\ (s,\xi)\in\mathbb{R}\times{{\mathbb{R}}^{n}}\,.$

Here $c,d>0$, $G$ is a Young function, $G\ll A_{n}$ near infinity, and $r\in
L^{1}(\Omega)$,.

###### Remark 2.15

Using the condition $A\in\Delta_{2}$ near infinity, and $|\Omega|<\infty$, it
can be shown that (2.44) is equivalent to

(2.45) $\sum_{i=1}^{n}a_{i}(x,s,\xi)\cdot\xi_{i}\geq
cA(|\xi|)-dG(d|s|)-r(x)\qquad\hbox{for a. e.}\ x\in\Omega,\ \hbox{all}\
(s,\xi)\in\mathbb{R}\times{{\mathbb{R}}^{n}}\,.$

We will use (2.45) in the sequel.

###### Example 2.16

We present some examples of functions $\mathcal{A}(x,s,\xi)$ satisfying
(2.42), (2.43) and (2.44).
Let $p,q\in\mathbb{R}$, satisfying

(2.46) $\displaystyle 1<p\leq n,\ $ $\displaystyle\ q\leq n-1,\ \ p-1+q>0\,.$

Consider also a Young function $D$, $D(t)\approx
t^{r}\lg^{-\frac{q}{p-1}}(1+t)$ near infinity, for some $r>1$.
When $p<n$ we take

(2.47) $\displaystyle\mathcal{A}(x,s,\xi)=$ $\displaystyle
a(x)|s|^{\beta}\lg^{\beta_{1}}(1+|s|)|\xi|^{p-2-\delta}\lg^{q(1-\frac{\delta}{p-1})}(1+|\xi|)\xi+|\xi|^{p-2}\lg^{q}(1+|\xi|)\xi\,,$
$\displaystyle\hbox{where}\ 0<\delta<p-1<n-1,\ 0<\beta<\frac{n\delta}{n-p},\
\beta+\beta_{1}>0,$ $\displaystyle a(x)>0,\ a\in L^{D}(\Omega),\ \hbox{and}\
r>\frac{np}{n\delta-\beta(n-p)}.$

When $p=n$, $q<n-1$, we define

(2.48) $\displaystyle\mathcal{A}(x,s,\xi)=$ $\displaystyle
a(x)e^{|s|^{\beta}}|\xi|^{n-2-\delta}\lg^{q(1-\frac{\delta}{n-1})}(1+|\xi|)\xi+|\xi|^{n-2}\lg^{q}(1+|\xi|)\xi\,,$
$\displaystyle\hbox{where}\ 0<\delta<n-1,\ 0<\beta<\frac{n}{n-q-1},$
$\displaystyle a(x)>0,\ a\in L^{D}(\Omega),\ \hbox{and}\ r>\frac{n}{\delta}.$

Finally, when $p=n$, $q=n-1$ we consider

(2.49) $\displaystyle\mathcal{A}(x,s,\xi)=$ $\displaystyle
a(x)e^{e^{|s|^{\beta}}}|\xi|^{n-2-\delta}\lg^{n-1-\delta}(1+|\xi|)\xi+|\xi|^{n-2}\lg^{n-1}(1+|\xi|)\xi\,,$
$\displaystyle\hbox{where}\ 0<\delta<n-1,\ 0<\beta<\frac{n}{n-1},$
$\displaystyle a(x)>0,\ a\in L^{D}(\Omega),\ \hbox{and}\ r>\frac{n}{\delta}.$

It holds

(2.50) $\displaystyle|\mathcal{A}(x,s,\xi)|\leq
a(x)|s|^{\beta}\lg^{\beta_{1}}(1+|s|)|\xi|^{p-1-\delta}\lg^{q(1-\frac{\delta}{p-1})}(1+|\xi|)+|\xi|^{p-1}\lg^{q}(1+|\xi|)\,,$
(2.51) $\displaystyle|\mathcal{A}(x,s,\xi)|\leq
a(x)e^{|s|^{\beta}}|\xi|^{n-1-\delta}\lg^{q(1-\frac{\delta}{n-1})}(1+|\xi|)+|\xi|^{n-1}\lg^{q}(1+|\xi|)\,,$

and

(2.52) $\displaystyle|\mathcal{A}(x,s,\xi)|\leq
a(x)e^{e^{|s|^{\beta}}}|\xi|^{n-1-\delta}\lg^{n-1-\delta}(1+|\xi|)+|\xi|^{n-1}\lg^{n-1}(1+|\xi|)\,,$

for all $(x,s,\xi)\in\Omega\times\mathbb{R}\times{{\mathbb{R}}^{n}}$,
respectively.
Now we use twice (2.3) in (2.50), (2.51) and (2.52): first with
$B(t)=t^{\frac{p-1}{\delta}}$ and then with $B(t)=t^{\frac{r\delta}{p}}$.

(2.53) $\displaystyle|\mathcal{A}(x,s,\xi)|$ $\displaystyle\leq
a(x)^{\frac{p-1}{\delta}}|s|^{\beta\frac{p-1}{\delta}}\lg^{\beta_{1}\frac{p-1}{\delta}}(1+|s|)+2|\xi|^{p-1}\lg^{q}(1+|\xi|)$
$\displaystyle\leq a(x)^{r\frac{p-1}{p}}+|s|^{\beta
r\frac{p-1}{r\delta-p}}\lg^{\beta_{1}r\frac{p-1}{r\delta-p}}(1+|s|)+2|\xi|^{p-1}\lg^{q}(1+|\xi|)\,.$
(2.54) $\displaystyle|\mathcal{A}(x,s,\xi)|$ $\displaystyle\leq
a(x)^{\frac{n-1}{\delta}}e^{\frac{n-1}{\delta}|s|^{\beta}}+2|\xi|^{n-1}\lg^{q}(1+|\xi|)$
$\displaystyle\leq
a(x)^{r\frac{n-1}{n}}+e^{\frac{r(n-1)}{r\delta-n}|s|^{\beta}}+2|\xi|^{n-1}\lg^{q}(1+|\xi|)\,.$
(2.55) $\displaystyle|\mathcal{A}(x,s,\xi)|$ $\displaystyle\leq
a(x)^{\frac{n-1}{\delta}}e^{\frac{n-1}{\delta}e^{|s|^{\beta}}}+2|\xi|^{n-1}\lg^{n-1}(1+|\xi|)$
$\displaystyle\leq
a(x)^{r\frac{n-1}{n}}+e^{\frac{(r)n-1}{r\delta-n\delta}e^{|s|^{\beta}}}+2|\xi|^{n-1}\lg^{n-1}(1+|\xi|)\,.$

Let $A(t)$ be defined as in (2.35). Taking into account the conditions in
(2.47) and in (2.48), for the function $a$ in (2.53) and (2.54) it holds
$a(x)^{r\frac{p-1}{\delta}}\in L^{\widetilde{A}}(\Omega)$. When (2.49) is in
force, then the function $a$ in (2.55) satisfies $a(x)^{r\frac{n-1}{n}}\in
L^{\widetilde{A}}(\Omega)$. Also,
$\widetilde{A}^{-1}(A(|\xi|))\approx|\xi|^{p-1}\lg^{q}(1+|\xi|)$ near
infinity.
When (2.47) is in force we consider the Young function
$F(s)\approx|s|^{\frac{\beta
rp}{r\delta-p}}\lg^{\frac{\beta_{1}rp}{r\delta-p}-\frac{q}{p-1}}(1+|s|)$ near
infinity. When (2.48) holds we choose the Young function $F(s)\approx
e^{s^{\beta}}$ near infinity. Similarly, when (2.49) is in effect, then we
choose $F(s)\approx e^{e^{s^{\beta}}}$ near infinity. In all the three cases
$F\ll A_{n}$ near infinity and $\widetilde{A}^{-1}(F(|s|))$ is equivalent,
near infinity, to the function in the $s$ variable, appearing in (2.53),
(2.54) and (2.55) respectively. Then (2.42) holds with $q(x)=a(x)+k$, for some
$k>0$.
Condition (2.43) is satisfied, because (2.46) guarantees that the functions
$B(t)=t^{p-1-\delta}\lg^{q(1-\frac{\delta}{p-1})}(1+t)$ and
$B(t)=t^{n-1-\delta}\lg^{n-1-\delta}(1+t)$ are increasing, for $t\geq 0$.
Condition (2.44) clearly holds true.
We now consider the borderline instances concernig $\beta$. When $\beta=0$,
then we can take $r=\frac{p}{\delta}$. When $\beta=\frac{n\delta}{n-p}$ in
(2.47), then in addition to $\beta_{1}>-\frac{n\delta}{n-p}$, it is necessary
$\beta_{1}<\frac{q\delta(n-1)}{(n-p)(p-1)}$ and $a\in L^{\infty}(\Omega)$.
When $q(x)\equiv 0$ then we have the classical $p$-Laplacian.

Define now the operator ${\cal
S}:{W_{0}^{1,A}(\Omega)}\to({W_{0}^{1,A}(\Omega)})^{*}$ as

(2.56) $\langle{\cal S}u,v\rangle=\int_{\Omega}\mathcal{A}(x,u,\nabla
u)\cdot\nabla v\,dx\quad\hbox{for $u,v\in{W_{0}^{1,A}(\Omega)}$.}$

The properties of ${\cal S}$ are listed in the next proposition.

###### Proposition 2.17

Let $A$ be a Young function, $A\in\nabla_{2}\cap\Delta_{2}$ near infinity.
Assume that $\mathcal{A}$ satisfies (2.42), (2.43), (2.44). Then the operator
${\cal S}$, introduced in (2.56), is well defined, bounded, continuous, and
satisfies the $(S_{+})$ property.

Proof. Let $u\in{W_{0}^{1,A}(\Omega)}$. From (2.42), (2.31) and Propositon 6.6
of [BaCi1] we can find $c_{1}>0$ such that

$\displaystyle\int_{\Omega}\widetilde{A}(|\mathcal{A}(x,u,\nabla u)|)dx$
$\displaystyle\leq\frac{1}{3}\int_{\Omega}\widetilde{A}(3q(x))dx+\frac{1}{3}\int_{\Omega}\widetilde{A}(3b\widetilde{A}^{-1}(F(b|u|)))dx+\frac{1}{3}\int_{\Omega}\widetilde{A}(3b\widetilde{A}^{-1}(A(b|\nabla
u|)))dx$ (2.57) $\displaystyle\leq
c_{1}\left[\int_{\Omega}\left(\widetilde{A}(3q(x))+F(b|u|)+A(b|\nabla
u|)\right)dx+1\right]=M<+\infty\,.$

Thus $|\mathcal{A}(x,u,\nabla u)|\in L^{\widetilde{A}}(\Omega)$ and
$\|\mathcal{A}(x,u,\nabla
u)\|_{L^{\widetilde{A}}(\Omega)}\leq\max\\{1,\,M\\}$. From (2.12)

$\displaystyle|\langle{\cal S}u,v\rangle|\leq 2\|\mathcal{A}(x,u,\nabla
u)\|_{L^{\widetilde{A}}(\Omega)}\|\nabla v\|_{L^{A}(\Omega)}\,.$

Similarly, if ${\cal C}\subseteq W^{1,A}_{0}(\Omega)$ is such that
$\|u\|_{W^{1,A}_{0}(\Omega)}\leq M_{0}$ for some $M_{0}>0$, then
$\|u\|_{L^{A_{n}}(\Omega)}\leq M_{1}$ for all $u\in{\cal C}$. From (2) and
Lemma 2.7 of [BaTo1], there exist $c_{1},L>0$ such that

$\int_{\Omega}\widetilde{A}\left(\mathcal{A}(x,u,\nabla u)\right)dx\leq
c_{1}\left(\int_{\Omega}\big{(}\widetilde{A}\left(3q(x)\right)+F\left(b|u|\right)+A\left(b|\nabla
u|\right)\big{)}\,dx+1\right)\leq L\ \hbox{for all}\ u\in{\cal C}.$

Then

(2.58) $\left\|\mathcal{A}(x,u,\nabla
u)\right\|_{L^{\widetilde{A}}(\Omega)}\leq\max\\{1,L\\}\ \hbox{for all}\
u\in{\cal C}\,,$

and

$\left\|{\cal S}(u)\right\|_{\left({W_{0}^{1,A}(\Omega)}\right)^{*}}\leq
2\max\\{1,L\\}\ \hbox{for all}\ u\in{\cal C}\,.$

Let now $\\{u_{k}\\}$ be a sequence in ${W_{0}^{1,A}(\Omega)}$ converging to
$u\in{W_{0}^{1,A}(\Omega)}$. Then, we can find a subsequence, say still
$\\{u_{k}\\}$, and two functions $g_{1}\in L^{A_{n}}(\Omega)$, $g_{2}\in
L^{A}(\Omega)$ such that $|u_{k}(x)|\leq g_{1}(x)$, $|u(x)|\leq g_{1}(x)$,
$|\nabla u_{k}(x)|\leq g_{2}(x)$ and $|\nabla u(x)|\leq g_{2}(x)$ for a.a.
$x\in\Omega$ and for all $k\in N$. Let $\lambda>0$. Using (2.42) and the
condition $\widetilde{A}\in\Delta_{2}$ near infinity we can find $c_{1}>0$
such that

(2.59) $\displaystyle\widetilde{A}\left(\frac{|\mathcal{A}(x,u_{k},\nabla
u_{k})-\mathcal{A}(x,u,\nabla u)|}{\lambda}\right)\leq$
$\displaystyle\frac{1}{5}\widetilde{A}\left(\frac{10q(x)}{\lambda}\right)$
$\displaystyle+\frac{1}{5}\bigg{[}\widetilde{A}\left(\frac{5b}{\lambda}\widetilde{A}^{-1}(F\left(b|u_{k}|\right))\right)+\widetilde{A}\left(\frac{5b}{\lambda}\widetilde{A}^{-1}(F\left(b|u|\right))\right)+$
$\displaystyle\widetilde{A}\left(\frac{5b}{\lambda}\widetilde{A}^{-1}(A\left(b|\nabla
u_{k}|\right))\right)+\widetilde{A}\left(\frac{5b}{\lambda}\widetilde{A}^{-1}(A\left(b|\nabla
u|\right))\right)\bigg{]}$
$\displaystyle\leq\frac{1}{5}\widetilde{A}\left(\frac{10q(x)}{\lambda}\right)+c_{1}(1+F\left(b|u_{k}|\right)+F\left(b|u|\right)+$
$\displaystyle A\left(b|\nabla u_{k}|\right)+A\left(b|\nabla u|\right))$
$\displaystyle\leq\frac{1}{5}\widetilde{A}\left(\frac{10q(x)}{\lambda}\right)+c_{1}(1+2F\left(bg_{1}(x)\right)+2A\left(bg_{2}(x)\right))$
$\displaystyle:=v_{\lambda}(x)\,.$

The function $v_{\lambda}\in L^{1}(\Omega)$, inasmuch both
$A,\widetilde{A}\in\Delta_{2}$ near inifnity and $F\ll A_{n}$ near infinity.
The continuity of $\mathcal{A}(x,\cdot,\cdot)$ guarantees that
$\lim_{k\to+\infty}\mathcal{A}(x,u_{k}(x),\nabla
u_{k}(x))=\mathcal{A}(x,u(x),\nabla u(x))$ a.e. in $\Omega$. We can thus apply
the Lebesgue theorem to obtain

(2.60)
$\displaystyle\lim_{k\to+\infty}\int_{\Omega}\widetilde{A}\left(\frac{|\mathcal{A}(x,u_{k},\nabla
u_{k})-\mathcal{A}(x,u,\nabla u)|}{\lambda}\right)dx=0\ \hbox{for all}\
\lambda>0\,.$

The arguments above apply to any subsequence of $\\{u_{k}\\}$. This means that
given any subsequence, we can extract a subsubsequence for which (2.60) holds.
Thus (2.60) holds for the whole sequence. Note that

$\displaystyle\left\|{\cal S}(u_{k})-{\cal
S}(u)\right\|_{\left({W_{0}^{1,A}(\Omega)}\right)^{*}}=\sup_{\|v\|_{{W_{0}^{1,A}(\Omega)}}\leq
1}|\langle{\cal S}(u_{k})-{\cal S}(u),v\rangle|\leq
2\|\mathcal{A}(x,u_{k},\nabla u_{k})-\mathcal{A}(x,u,\nabla
u)|\|_{L^{\widetilde{A}}(\Omega)}\,.$

Thus

(2.61) $\displaystyle\lim_{k\to+\infty}\left\|{\cal S}(u_{k})-{\cal
S}(u)\right\|_{\left({W_{0}^{1,A}(\Omega)}\right)^{*}}=0\,.$

This proves the continuity af ${\cal S}$.
Let us now demonstrate the $(S)_{+}$ property. Let $\\{u_{k}\\}$ be a sequence
in ${W_{0}^{1,A}(\Omega)}$, $u_{k}\rightharpoonup u$ and

(2.62) $\limsup_{k\to+\infty}\langle{\cal S}u_{k},u_{k}-u\rangle\leq 0\,.$

We divide the proof in four steps.
Step 1 $\nabla u_{k}\to\nabla u$ a.e.
Let $F$ be as in (2.42). From Proposition 2.6 there exists a Young function
$F_{1}$ such that $F\ll F_{1}\ll A_{n}$ near infinity. Thus $\\{u_{k}\\}$
strongly converges to $u$ in $L^{F_{1}}(\Omega)$, and we can find a function
$g\in L^{F_{1}}(\Omega)$ and a subsequence of $\\{u_{k}\\}$, say still
$\\{u_{k}\\}$, such that $|u_{k}(x)|,|u(x)|\leq g(x)$ a.e. in $\Omega$. Thus

(2.63) $\displaystyle\widetilde{A}\left(\frac{|\mathcal{A}(x,u_{k},\nabla
u)-\mathcal{A}(x,u,\nabla u)|}{\lambda}\right)\leq$
$\displaystyle\frac{1}{5}\widetilde{A}\left(\frac{10q(x)}{\lambda}\right)$
$\displaystyle+\frac{1}{5}\bigg{[}\widetilde{A}\left(\frac{5b}{\lambda}\widetilde{A}^{-1}(F\left(b|u_{k}|\right))\right)+\widetilde{A}\left(\frac{5b}{\lambda}\widetilde{A}^{-1}(F\left(b|u|\right))\right)+$
$\displaystyle
2\widetilde{A}\left(\frac{5b}{\lambda}\widetilde{A}^{-1}(A\left(b|\nabla
u|\right))\right)\bigg{]}$
$\displaystyle\leq\frac{1}{5}\widetilde{A}\left(\frac{10q(x)}{\lambda}\right)+c_{1}(1+2F\left(bg_{1}(x)\right)+2A\left(b|\nabla
u|\right))$ $\displaystyle:=v_{\lambda}(x)\,.$

The function $v_{\lambda}\in L^{1}(\Omega)$ because of (2.14) and standard
arguments used several times in this proof. Thus, arguing as for (2.60),
$\mathcal{A}(x,u_{k},\nabla u)-\mathcal{A}(x,u,\nabla u)\to 0$ in
$L^{\widetilde{A}}(\Omega)$ and

(2.64) $\lim_{k\to+\infty}\int_{\Omega}(\mathcal{A}(x,u_{k},\nabla
u)-\mathcal{A}(x,u,\nabla u))(\nabla u_{k}-\nabla u)dx=0\,.$

By definition of weak convergence

(2.65) $\lim_{k\to+\infty}\int_{\Omega}\mathcal{A}(x,u,\nabla u)(\nabla
u_{k}-\nabla u)dx=0\,.$

From (2.43)

(2.66) $\displaystyle 0\leq$
$\displaystyle\int_{\Omega}(\mathcal{A}(x,u_{k},\nabla
u_{k})-\mathcal{A}(x,u_{k},\nabla u))(\nabla u_{k}-\nabla u)dx$
$\displaystyle=\int_{\Omega}(\mathcal{A}(x,u_{k},\nabla
u_{k})-\mathcal{A}(x,u,\nabla u))(\nabla u_{k}-\nabla u)dx$
$\displaystyle+\int_{\Omega}(\mathcal{A}(x,u,\nabla
u)-\mathcal{A}(x,u_{k},\nabla u))(\nabla u_{k}-\nabla u)dx\,.$

Passing to the limit in (2.66) and using (2.62), (2.65) and (2.64)

(2.67) $\lim_{k\to+\infty}\int_{\Omega}(\mathcal{A}(x,u_{k},\nabla
u_{k})-\mathcal{A}(x,u_{k},\nabla u))(\nabla u_{k}-\nabla u)dx=0\,.$

Thus the sequence $(\mathcal{A}(x,u_{k},\nabla
u_{k})-\mathcal{A}(x,u_{k},\nabla u))(\nabla u_{k}-\nabla u)\to 0$ in
$L^{1}(\Omega)$, and we can find a subsequence, say still $\\{u_{k}\\}$, and a
set $\Omega_{0}\subset\Omega$, such that $|\Omega_{0}|=0$ and (recall that
$u_{k}\to u$ in $L^{F_{1}}(\Omega)$)

(2.68) $\displaystyle\mathcal{A}(x,u_{k},\nabla
u_{k})-\mathcal{A}(x,u_{k},\nabla u))(\nabla u_{k}-\nabla u)\to 0,\
u_{k}(x)\to u(x)\ \hbox{in}\ \Omega\setminus\Omega_{0}\,.$

We prove that for every $x\in\Omega\setminus\Omega_{0}$ there exists $M>0$
such that $|\nabla u_{k}(x)|\leq M$ for all $k\in N$. We argue by
contradiction. Assume that there exists $x\in\Omega\setminus\Omega_{0}$ such
that for every $h>|\nabla u(x)|+1$ there exists $k_{h}\in N$ such that
$|\nabla u_{k_{h}}(x)|>h$. In particular

(2.69) $\displaystyle|\nabla u_{k_{h}}(x)-\nabla u(x)|>1\,.$

The sequence $\left\\{\frac{\nabla u_{k_{h}}(x)-\nabla u(x)}{|\nabla
u_{k_{h}}(x)-\nabla u(x)|}\right\\}$ converges to $\xi\in{{\mathbb{R}}^{n}}$,
up to a subsequence. We keep the same notation as above for the subsequence
and use (2.43) and (4.14)

(2.70) $\displaystyle 0\leq\left(\mathcal{A}\left(x,u_{k_{h}}(x),\nabla
u(x)+\frac{\nabla u_{k_{h}}(x)-\nabla u(x)}{|\nabla u_{k_{h}}(x)-\nabla
u(x)|}\right)-\mathcal{A}(x,u_{k_{h}}(x),\nabla u(x))\right)\frac{\nabla
u_{k_{h}}(x)-\nabla u(x)}{|\nabla u_{k_{h}}(x)-\nabla u(x)|}=$
$\displaystyle\left(\mathcal{A}\left(x,u_{k_{h}}(x),\nabla u(x)+\frac{\nabla
u_{k_{h}}(x)-\nabla u(x)}{|\nabla u_{k_{h}}(x)-\nabla
u(x)|}\right)-\mathcal{A}(x,u_{k_{h}}(x),\nabla
u_{k_{h}}(x))\right)\frac{\nabla u_{k_{h}}(x)-\nabla u(x)}{|\nabla
u_{k_{h}}(x)-\nabla u(x)|}$
$\displaystyle+\left(\mathcal{A}(x,u_{k_{h}}(x),\nabla
u_{k_{h}}(x))-\mathcal{A}(x,u_{k_{h}}(x),\nabla u(x))\right)\frac{\nabla
u_{k_{h}}(x)-\nabla u(x)}{|\nabla u_{k_{h}}(x)-\nabla u(x)|}\leq$
$\displaystyle\left(\mathcal{A}(x,u_{k}(x),\nabla
u_{k}(x))-\mathcal{A}(x,u_{k}(x),\nabla u(x))\right)(\nabla
u_{k_{h}}(x)-\nabla u(x))\,.$

From (2.70) and (2.68)

(2.71) $\displaystyle\lim_{h\to+\infty}\left(\mathcal{A}(x,u_{k_{h}},\nabla
u+\frac{\nabla u_{k_{h}}(x)-\nabla u(x)}{|\nabla u_{k_{h}}(x)-\nabla
u(x)|})-\mathcal{A}(x,u_{k_{h}},\nabla u)\right)\frac{\nabla
u_{k_{h}}(x)-\nabla u(x)}{|\nabla u_{k_{h}}(x)-\nabla u(x)|}=0$

and this leads to $\left(\mathcal{A}(x,u(x),\nabla
u(x)+\xi)-\mathcal{A}(x,u(x),\nabla u(x))\right)\xi=0$, namely $\xi=0$. This
contradicts $|\xi|=1$, thus $|\nabla u_{k_{h}}(x)|\leq M$, for some $M>0$ and
we can find a subsequence, say still $\\{\nabla u_{k_{h}}\\}$, converging to
$\eta\in{{\mathbb{R}}^{n}}$. Then from (2.68) and the convergence of
$\\{\nabla u_{k_{h}}\\}$ to $\eta\in{{\mathbb{R}}^{n}}$

(2.72) $\displaystyle
0=\lim_{h\to+\infty}\left(\mathcal{A}(x,u_{k_{h}}(x),\nabla
u_{k_{h}}(x))-\mathcal{A}(x,u_{k_{h}}(x),\nabla u(x))\right)(\nabla
u_{k_{h}}(x)-\nabla u(x))$
$\displaystyle=\left(\mathcal{A}(x,u(x),\eta)-\mathcal{A}(x,u(x),\nabla
u(x))\right)(\eta-\nabla u(x))\,.$

From (2.43) we deduce $\eta=\nabla u(x)$. We have so proved that every
subsequence of $\\{\nabla u_{k}\\}$ has a subsequence converging to $\nabla
u(x)$. Thus the whole sequence converges to $\nabla u(x)$ in
$\Omega\setminus\Omega_{0}$.
Step 2 $|a_{i}(x,u_{k},\nabla u_{k})-a_{i}(x,u,\nabla u)|\rightharpoonup 0$ in
$L^{\widetilde{A}}(\Omega)$.
Due to step 1 and to the continuity of $\mathcal{A}(x,\cdot,\cdot)$, we have
$\mathcal{A}(x,u_{k},\nabla u_{k})\to\mathcal{A}(x,u,\nabla u)$ a.e. in
$\Omega$. $\\{u_{k}\\}_{k\in N}$ is bounded in ${W_{0}^{1,A}(\Omega)}$ and,
from (2.58), $\\{|\mathcal{A}(x,u_{k},\nabla u_{k})|\\}_{k\in N}$ is bounded
in $L^{\widetilde{A}}(\Omega)$. Since $\widetilde{A}\in\nabla_{2}$ near
infinity we can apply Corollary 2.10 to obtain

$|a_{i}(x,u_{k},\nabla u_{k})-a_{i}(x,u,\nabla u)|\rightharpoonup 0\
\hbox{in}\ L^{\widetilde{A}}(\Omega)\,.$

Step 3. $\mathcal{A}(x,u_{k},\nabla u_{k})\cdot\nabla
u_{k}\to\mathcal{A}(x,u,\nabla u)\cdot\nabla u$ in $L^{1}(\Omega)$.
It holds

(2.73) $\displaystyle|\mathcal{A}(x,u,\nabla u)\cdot\nabla
u-\mathcal{A}(x,u_{k},\nabla u_{k})\cdot\nabla u_{k}|=$ $\displaystyle
2(\mathcal{A}(x,u,\nabla u)\cdot\nabla u-\mathcal{A}(x,u_{k},\nabla
u_{k})\cdot\nabla u_{k})^{+}$ $\displaystyle-\mathcal{A}(x,u,\nabla
u)\cdot\nabla u+\mathcal{A}(x,u_{k},\nabla u_{k})\cdot\nabla u_{k}\,.$

Put $\sigma_{k}(x)=\mathcal{A}(x,u,\nabla u)\cdot\nabla
u-\mathcal{A}(x,u_{k},\nabla u_{k})\cdot\nabla u_{k}$. One has
$\sigma_{k}^{+}\leq\mathcal{A}(x,u,\nabla u)\cdot\nabla u$ and $\sigma_{k}\to
0$ a.e in $\Omega$. Then

(2.74) $\displaystyle\sigma_{k}(x)^{+}\to 0\ $ $\displaystyle\ \hbox{in}\
L^{1}(\Omega).$

From (2.73)

$\displaystyle 0\leq\int_{\Omega}|\mathcal{A}(x,u_{k},\nabla u_{k})\cdot\nabla
u_{k}-\mathcal{A}(x,u,\nabla u)\cdot\nabla u|dx=$ $\displaystyle
2\int_{\Omega}\sigma_{k}^{+}(x)dx-\int_{\Omega}\mathcal{A}(x,u,\nabla
u)\cdot\nabla udx$ $\displaystyle+\int_{\Omega}\mathcal{A}(x,u_{k},\nabla
u_{k})\left(\nabla u_{k}-\nabla u\right)dx+$
$\displaystyle\int_{\Omega}\mathcal{A}(x,u_{k},\nabla u_{k})\nabla udx\,.$

So, using (2.74), (2.62) and Step 2

(2.75) $\lim_{k\to+\infty}\int_{\Omega}|\mathcal{A}(x,u_{k},\nabla
u_{k})\cdot\nabla u_{k}-\mathcal{A}(x,u,\nabla u)\cdot\nabla u|dx=0\,.$

Step 4 $\nabla u_{k}\to\nabla u$ in $L^{A}(\Omega)$.
From (2.45), $cA(|\nabla u_{k}(x)|)\leq\mathcal{A}(x,u_{k},\nabla
u_{k})\cdot\nabla u_{k}+dG(d|u_{k}|)+r(x)$ for all $x\in\Omega$, all $k\in N$.
From Step 3 and the assumption $G\ll A_{n}$, we know that there exists a
subsequence of $\\{u_{k}\\}$, say still $\\{u_{k}\\}$ and a function $g\in
L^{1}(\Omega)$, such that $\mathcal{A}(x,u_{k},\nabla u_{k})\cdot\nabla
u_{k}+dG(d|u_{k}|)\leq g(x)$ a.e. in $\Omega$. Taking into account that
$A(c|\nabla u_{k}(x)|)\to A(c|\nabla u(x)|)$ a.e. in $\Omega$, we deduce that

(2.76) $\lim_{k\to+\infty}\int_{\Omega}A(c|\nabla
u_{k}(x)|)dx=\int_{\Omega}A(c|\nabla u(x)|)dx\,.$

A standard and repeatedly used argument shows that it holds for the whole
sequence. Equation (2.76) and the $\Delta_{2}$ condition on $A$, guarantees
that $\nabla u_{k}\to\nabla u$ in $L^{A}(\Omega)$.

We now construct the truncation of ${\cal S}$ that we use in the proof of our
results. For every $r\in\mathbb{R}$, we set $r^{+}=\max\\{r,0\\}$,
$r^{-}=\max\\{-r,0\\}$.
Let $\underline{u},\,\overline{u}\in W^{1,A}(\Omega)$ be such that
$(\overline{u})^{-},\,\underline{u}^{+}\in{W_{0}^{1,A}(\Omega)}$, and
$\underline{u}\leq\overline{u}$ a.e. in $\Omega$. The truncation operator
$T:{W_{0}^{1,A}(\Omega)}\to{W_{0}^{1,A}(\Omega)}$ is defined by

(2.77) $T(u)=\left\\{\begin{array}[]{cc}\overline{u}&\hbox{if}\
u>\overline{u}\\\ u&\hbox{if}\ \underline{u}\leq u\leq\overline{u}\\\
\underline{u}&\hbox{if}\ u<\underline{u}\end{array}\right.$

The properties of $\underline{u},\,\overline{u}$ guarantee that
$T(u),\,(u-\overline{u})^{+},\,(u-\underline{u})^{-}\in{W_{0}^{1,A}(\Omega)}$.
In particular $T$ is well defined.
It is known (see [H], p.20) that

(2.78) $\nabla
T(u)=\left\\{\begin{array}[]{cc}\nabla\overline{u}(x)&\hbox{a.e. on the set}\
\\{u>\overline{u}\\}\\\ \nabla u(x)&\hbox{a.e. on the set}\
\\{\underline{u}\leq u\leq\overline{u}\\}\\\ \nabla\underline{u}(x)&\hbox{a.e.
on the set}\ \\{u<\underline{u}\\}\end{array}\right.$

Given the functions $\underline{u},\,\overline{u}\in W^{1,A}(\Omega)$ as
above, and such that $\underline{u},\,\overline{u}\in L^{A_{n}}(\Omega)$, let
us define the operator ${\cal
S}_{T}:W_{0}^{1,A}(\Omega)\to(W_{0}^{1,A}(\Omega))^{*}$, as

(2.79) $\langle{\cal S}_{T}u,v\rangle=\int_{\Omega}\mathcal{A}(x,Tu,\nabla
u)\cdot\nabla v\,dx\,.$

###### Corollary 2.18

Assume that $A:[0,+\infty[\to 0,+\infty[$ is a Young function,
$A\in\Delta_{2}\cap\nabla_{2}$ near infinity, and that $\mathcal{A}$ satisfies
(2.42), (2.43) and (2.44). Then the operator ${\cal S}_{T}$, introduced in
(2.79) is well defined, bounded, continuous and has the $(S_{+})$ property.

Proof. The inequality $|T(u(x))|\leq|u(x)|$ for all $x\in\Omega$ guarantees
that the operator ${\cal S}_{T}$ is well defined and bounded. Thanks to Lemma
4.1 in [BaTo1], the arguments used in Proposition 2.17 work also for ${\cal
S}_{T}$. Thus ${\cal S}_{T}$ is continuous and has the $(S)_{+}$ property.

##  3 Main results

In this Section we state two of the main results of the paper (Theorems 3.4
and 3.8).
First we give the fundamental definitions of weak solution, subsolution and
supersolution to (1.1).

###### Definition 3.1

A function $u\in{W_{0}^{1,A}(\Omega)}$ is a weak solution to problem (1.1) if

$\int_{\Omega}\mathcal{A}(x,u,\nabla u)\cdot\nabla
vdx=\int_{\Omega}f(x,u,\nabla u)vdx\ \hbox{for all
$v\in{W_{0}^{1,A}(\Omega)}$},$

and $\int_{\Omega}\mathcal{A}(x,u,\nabla u)\cdot\nabla vdx\in\mathbb{R}$ for
all $v\in{W_{0}^{1,A}(\Omega)}$.

###### Definition 3.2

We say that $\overline{u}\in W^{1,A}(\Omega)$ is a supersolution to (1.1) if
$(\overline{u})^{-}\in{W_{0}^{1,A}(\Omega)}$,

$+\infty>\int_{\Omega}\mathcal{A}(x,\overline{u},\nabla\overline{u})\cdot\nabla
vdx\geq\int_{\Omega}f(x,\overline{u},\nabla\overline{u})vdx>-\infty$

for all $v\in{W_{0}^{1,A}(\Omega)}$, $v\geq 0$ a.e. in $\Omega$.

###### Definition 3.3

We say that $\underline{u}\in W^{1,A}(\Omega)$ is a subsolution to (1.1) if
$\underline{u}^{+}\in{W_{0}^{1,A}(\Omega)}$ and

$-\infty<\int_{\Omega}\mathcal{A}(x,\underline{u},\nabla\underline{u})\cdot\nabla
vdx\leq\int_{\Omega}f(x,\underline{u},\nabla\underline{u})vdx<+\infty$

for all $v\in{W_{0}^{1,A}(\Omega)}$, $v\geq 0$ a.e. in $\Omega$.

###### Theorem 3.4

Let $\Omega$ be an open set in ${{\mathbb{R}}^{n}}$, with $n\geq 2$, such that
$|\Omega|<\infty$. Let $A\in C^{1}([0,+\infty))$ be a Young function,
$A\in\Delta_{2}\cap\nabla_{2}$ near infinity. Assume also that $A$ satisfies
(2.26) and (2.29), or (2.30). Let $\underline{u}$ and $\overline{u}$ be a
subsolution and a supersolution of problem (1.1), respectively, with
$\underline{u}\leq\overline{u}$ a.e. in $\Omega$, and
$\underline{u},\overline{u}\in L^{A_{n}}(\Omega)$. Assume that the function
$\mathcal{A}$ satisfies (2.42), (2.43), (2.45). Let
$f:\Omega\times\mathbb{R}\times\mathbb{R}^{n}\to\mathbb{R}$ be a Carathéodory
function fulfilling

(3.1)
$|f(x,s,\xi)|\leq\sigma(x)+\overline{\gamma}\widetilde{E}^{-1}(A(|\xi|))\
\hbox{ for a.e.}\ x\in\Omega,\ \hbox{all}\
s\in[\underline{u}(x),\overline{u}(x)],\ \hbox{all}\
\xi\in{{\mathbb{R}}^{n}}\,,$

where $\sigma\in L^{\widetilde{A}_{n}}(\Omega)$, $\overline{\gamma}>0$ and
$E:[0,+\infty[\to[0,+\infty[$ is a Young function, $E\ll A_{n}$ near infinity.
Then problem $(P)$ has a solution $u\in{W_{0}^{1,A}(\Omega)}$ such that
$\underline{u}\leq u\leq\overline{u}$ a.e. in $\Omega$.

To prove Theorem 3.4, we perturb problem (1.1) and formulate an auxiliary one.
Let $\Pi:W_{0}^{1,A}(\Omega)\to(W_{0}^{1,A}(\Omega))^{*}$, given by

(3.2) $\Pi(u)(v)=\int_{\Omega}\pi(x,u(x))v(x)dx\,,\ \hbox{for}\
u,v\in{W_{0}^{1,A}(\Omega)},$

where

$\pi(x,s)=\left\\{\begin{array}[]{cc}\widetilde{E}^{-1}(E(s-\overline{u}(x))&\hbox{if}\
s>\overline{u}(x)\\\ 0&\hbox{if}\ \underline{u}(x)\leq s\leq\overline{u}(x)\\\
-\widetilde{E}^{-1}(E(\underline{u}(x)-s))&\hbox{if}\
s<\underline{u}(x).\end{array}\right.$

Let ${\cal N}_{f}\circ T:{W_{0}^{1,A}(\Omega)}\to({W_{0}^{1,A}(\Omega)})^{*}$
be the operator defined as

$\langle{\cal N}_{f}\circ T(u),v\rangle=\int_{\Omega}f(x,Tu,\nabla
Tu)v(x)dx\,,\ \hbox{for}\ u,v\in{W_{0}^{1,A}(\Omega)},$

Given $\mu>0$, we consider the following problem

(3.3) $\begin{cases}-{\rm div}(\mathcal{A}(x,Tu,\nabla
u))+\mu\Pi(u)=N_{f}(Tu)&\hbox{in}\;\Omega,\\\
u=0&\hbox{on}\;\partial\Omega.\end{cases}$

The result below guarantees that problem (3.3) has a solution, provided the
parameter $\mu>0$ is sufficiently large.

###### Theorem 3.5

Under the same assumtions of Theorem 3.4, there exists $\mu_{0}>0$ such that
(3.3) admits a solution whenever $\mu\geq\mu_{0}$.

Proof. For all $\mu>0$ consider the operator ${\cal
A}_{\mu}:W_{0}^{1,A}(\Omega)\to(W_{0}^{1,A}(\Omega))^{*}$, defined by

$\displaystyle\langle{\cal A}_{\mu}(u),v\rangle=$
$\displaystyle\int_{\Omega}\mathcal{A}(x,Tu,\nabla u)\cdot\nabla
v\,dx+\mu\int_{\Omega}\pi(x,u)v\,dx-\int_{\Omega}f(x,Tu,\nabla Tu)v\,dx=$
$\displaystyle\langle{\cal S}_{T}u+\mu\Pi u-{\cal N}_{f}\circ
T(u),v\rangle\qquad\qquad\hbox{for }\ u,v\in W_{0}^{1,A}(\Omega)\,.$

We prove that ${\cal A}_{\mu}$ is well defined, bounded, pseudomonotone and
there is $\mu_{0}>0$ such that ${\cal A}_{\mu}$ is coercive for all
$\mu>\mu_{0}$.
Due to Corollary 2.18, Propositions 4.3 and 4.5 of [BaTo1], ${\cal A}_{\mu}$
is well defined, bounded and continuous. To prove that it is pseudomonotone,
we take $u\in{W_{0}^{1,A}(\Omega)}$, and a sequence
$\\{u_{k}\\}\subset{W_{0}^{1,A}(\Omega)}$ such that

$u_{k}\rightharpoonup u\quad\hbox{in ${W_{0}^{1,A}(\Omega)}$},\ \hbox{and}\
\limsup_{k\to\infty}\,\langle{\cal A}_{\mu}(u_{k}),u_{k}-u\rangle\leq 0\,.$

Equations (4.6) and (4.17) of [BaTo1] allow to write

$\limsup_{k\to\infty}\,\langle{\cal S}_{T}(u_{k}),u_{k}-u\rangle\leq 0\,.$

Thus $u_{k}\to u$ in ${W_{0}^{1,A}(\Omega)}$ (see Corollary 2.18), and
consequently

$\lim_{k\to\infty}\|{\cal A}_{\mu}(u_{k})-{\cal
A}_{\mu}(u)\|_{({W_{0}^{1,A}(\Omega)})^{*}}=0\,,$

so $\langle{\cal A}_{\mu}(u_{k}),u_{k}\rangle\to\langle{\cal
A}_{\mu}(u),u\rangle$, $\langle{\cal A}_{\mu}(u_{k}),v\rangle\to\langle{\cal
A}_{\mu}(u),v\rangle$ for all $v\in{W_{0}^{1,A}(\Omega)}$ and ${\cal A}_{\mu}$
is a pseudomonotone operator.
It remains to prove that ${\cal A}_{\mu}$ is coercive for some $\mu>0$.
Arguing like for equation (5.3) of [BaTo1] we can find a constant $c_{1}>0$
such that

(3.4) $\overline{\gamma}\int_{\Omega}\widetilde{E}^{-1}(A(|\nabla
Tu|))|u|dx\leq
c_{1}+2\overline{\gamma}\int_{\Omega}E\left(\frac{|u|}{2}\right)dx+\frac{c}{2}\int_{\Omega}A(|\nabla
u|)dx\ \hbox{for all}\ u\in{W_{0}^{1,A}(\Omega)}\,.$

Here $c$ is that of (2.45). From (3.1), (2.12), (2.28), and (3.4)

$\displaystyle\left|\int_{\Omega}f(x,Tu,\nabla
Tu)udx\right|\leq\left|\int_{\Omega}\sigma(x)u(x)dx\right|+\overline{\gamma}\left|\int_{\Omega}\widetilde{E}^{-1}(A(|\nabla
Tu|))udx\right|\leq$ (3.5) $\displaystyle\leq
2C\|\sigma\|_{L^{\widetilde{B}}(\Omega)}\|u\|_{W_{0}^{1,A}(\Omega)}+\frac{c}{2}\int_{\Omega}A(|\nabla
u|)dx+2\overline{\gamma}\int_{\Omega}E\left(\frac{|u|}{2}\right)dx+c_{1}\,.$

Let $v(x)=\max\\{|\underline{u}(x)|,|\overline{u}(x)|\\}$ Then $|T(u)(x)|\leq
v(x)$ for all $x\in\Omega$, and $\int_{\Omega}G(dv(x))dx<+\infty$. From (2.45)

(3.6) $\displaystyle\int_{\Omega}\mathcal{A}(x,Tu,\nabla u)\cdot\nabla u\,dx$
$\displaystyle\geq c\int_{\Omega}A(|\nabla
u|)dx-d\int_{\Omega}G(d|T(u)|)dx-\int_{\Omega}r(x)dx$ $\displaystyle\geq
c\int_{\Omega}A(|\nabla u|)dx-c_{2}\ \ \hbox{for any}\
u\in{W_{0}^{1,A}(\Omega)}\,.$

So, choosing $\mu>a$, $u\in{W_{0}^{1,A}(\Omega)}$ with
$\|u\|_{{W_{0}^{1,A}(\Omega)}}>>1$, and using (3), (3.6), Lemma 4.6 of
[BaTo1], and (2.16)

$\displaystyle\frac{\langle{\cal
A}_{\mu}(u),u\rangle}{\|u\|_{{W_{0}^{1,A}(\Omega)}}}\geq$
$\displaystyle\frac{\frac{c}{2}\int_{\Omega}A(|\nabla
u|)dx-2C\|\sigma\|_{L^{\widetilde{B}}(\Omega)}\|u\|_{{W_{0}^{1,A}(\Omega)}}+2(\mu-\overline{\gamma})\int_{\Omega}E\left(\frac{|u|}{2}\right)dx-
c_{3}}{\|u\|_{{W_{0}^{1,A}(\Omega)}}}$
$\displaystyle\geq\frac{\frac{ck_{1}^{i^{\infty}_{A}}}{2}\|u\|_{{W_{0}^{1,A}(\Omega)}}^{i_{A}^{\infty}}-2C\|\sigma\|_{L^{\widetilde{B}}(\Omega)}\|u\|_{{W_{0}^{1,A}(\Omega)}}+2(\mu-\overline{\gamma})\int_{\Omega}E\left(\frac{|u|}{2}\right)dx-
c_{4}}{\|u\|_{{W_{0}^{1,A}(\Omega)}}}\,.$

Thus

$\lim_{\|u\|\to+\infty}\frac{\langle{\cal
A}_{\mu}(u),u\rangle}{\|u\|_{{W_{0}^{1,A}(\Omega)}}}=+\infty\,.$

Theorem 2.14 guarantees that there exists $u\in W^{1,A}_{0}(\Omega)$ such that
${\cal A}_{\mu}(u)\equiv 0$. Thus

(3.7) $\int_{\Omega}\mathcal{A}(x,Tu,\nabla u)\cdot\nabla
vdx+\mu\int_{\Omega}\pi(x,u(x))v(x)dx-\int_{\Omega}f(x,Tu,\nabla Tu)vdx=0\,.$

for all $v\in W_{0}^{1,A}(\Omega)$.

###### Remark 3.6

The proof above works also if we weaken (2.42)$\ldots$(2.44), requiring them
to hold for $s\in[\underline{u}(x),\overline{u}(x)]$ rather than for all
$s\in\mathbb{R}$. This is because in the proof we consider only the truncated
function $\mathcal{A}(x,Tu,\nabla u)$.

Prooof of Theorem 3.4. By Theorem 3.5 there exists a solution
$u\in{W_{0}^{1,A}(\Omega)}$ of the truncated auxiliary problem (3.3) provided
$\mu>0$ is sufficiently large. Let us fix such a $\mu$ and $u$.
Via the same comparison arguments of the proof of Theorem 3.6 of [BaTo1] we
can prove that the solution of (3.3) has the enclosure property
$u\in[\underline{u},\overline{u}]$. Thus, it follows from (2.77) and (3.2)
that $Tu=u$ and $\Pi(u)=0$. Consequently, $u$ is a solution of(1.1).

The proof of the Corollary below follows the same lines as that of Corollary
5.2 of [BaTo1].

###### Corollary 3.7

Let $\Omega$ be an open set in ${{\mathbb{R}}^{n}}$, with $n\geq 2$, such that
$|\Omega|<\infty$. Let $A\in C^{1}([0,+\infty))$ be a Young function,
$A\in\Delta_{2}\cap\nabla_{2}$ near infinity. Assume also that $A$ satisfies
(2.26) and (2.29), or (2.30). Let $\underline{u}$ and $\overline{u}$ be a
subsolution and a supersolution of problem (1.1), respectively, with
$\underline{u}\leq\overline{u}$ a.e. in $\Omega$,
$\underline{u},\overline{u}\in{W_{0}^{1,A}(\Omega)}$, and such that the
Carathéodory function
$f:\Omega\times\mathbb{R}\times\mathbb{R}^{n}\to\mathbb{R}$ fulfills

(3.8)
$|f(x,s,\xi)|\leq\rho(x)+g(|s|)+\overline{\gamma}\widetilde{E}^{-1}(A(|\xi|))\
\hbox{a.e.}\ x\in\Omega,\ \hbox{all}\ s\in[\underline{u}(x),\overline{u}(x)],\
\xi\in{{\mathbb{R}}^{n}}\,,$

where $\rho\in L^{\widetilde{A}_{n}}(\Omega)$, $\overline{\gamma},\,E,$ are as
in Theorem 3.4, and $g:[0,+\infty[\to[0,+\infty[$ is a nondecreasing function
such that $g(0)=0$ and there exist $s_{0},\,h>0$ such that $g(|s|)|s|\leq
A_{n}(h|s|)$ for all $|s|\geq s_{0}$.
Then problem $(P)$ possesses a nontrivial solution
$u\in{W_{0}^{1,A}(\Omega)}$.

We consider now a special instance of (1.1), in which $\mathcal{A}$ does not
depend on $s$ and has a potential with respect to $\xi$. So, let
$\Omega\subset{{\mathbb{R}}^{n}}$ be a set of finite measure and let $A,B$ be
two Young functions such that $A\in\Delta_{2}\cap\nabla_{2}$ near infinity,
$B\in\nabla_{2}$ near zero, and $A\circ B^{-1}$ is a Young function too. We
assume that $\mathcal{A}:\Omega\times{{\mathbb{R}}^{n}}\to{{\mathbb{R}}^{n}}$,
$\mathcal{A}=(a_{1},\ldots a_{n})$, is such that each $a_{i}(x,\xi)$ is a
Carathéodory function, and

(3.9) $\displaystyle|\mathcal{A}(x,\xi)|\leq
q(x)\widetilde{B}^{-1}(B(b|\xi|))+b\widetilde{A}^{-1}(A(b|\xi|))$
$\displaystyle\hbox{for some}\ q\in L^{\widetilde{A\circ B^{-1}}}(\Omega),\
\hbox{some}\ b>0,\,\hbox{for a.e.}\,x\in\Omega,\,\hbox{all}\
\xi\in{{\mathbb{R}}^{n}}\,,$ (3.10)
$\displaystyle\sum_{i=1}^{n}\left(a_{i}(x,\xi)-a_{i}(x,\xi^{\prime})\right)\cdot(\xi_{i}-\xi_{i}^{\prime})>0\quad\hbox{for
a. e.}\ x\in\Omega,\ \hbox{all}\ \xi,\xi^{\prime}\in{{\mathbb{R}}^{n}}\,,\
\xi\neq\xi^{\prime}\,,$ (3.11)
$\displaystyle\sum_{i=1}^{n}a_{i}(x,\xi)\cdot\xi_{i}\geq
cA(c|\xi|)\quad\hbox{for some}\ c>0,\ \hbox{for a.e.}\ x\in\Omega,\
\hbox{all}\ \xi\in{{\mathbb{R}}^{n}}\,.$

Furthermore, we assume that there exists a measurable function
$\Phi:\Omega\times{{\mathbb{R}}^{n}}\to\mathbb{R}$, even with respect to
$\xi\in{{\mathbb{R}}^{n}}$ and such that

(3.12) $\displaystyle\Phi_{\xi}(x,\xi)=\mathcal{A}(x,\xi)\ \hbox{for all}\
(x,\xi)\in\Omega\times{{\mathbb{R}}^{n}},$ $\displaystyle\ \Phi(x,0)=0\
\hbox{for all}\ x\in\Omega.$

Since $A\circ B^{-1}$ is a Young function, it follows that $A$ dominates $B$
near infinity and $B$ dominates $A$ near zero. Thus

(3.13) $\displaystyle A(t)\leq B(\overline{k}t)\ \hbox{for}\ 0\leq
t\leq\overline{t}\quad\hbox{and}$ $\displaystyle\quad B(t)\leq A(\tilde{k}t)\
\hbox{for}\ t\geq\tilde{t}>0\,.$

Also, if $u\in L^{A}(\Omega)$ then $B(k|u|)\in L^{A\circ B^{-1}}(\Omega)$ for
all $k>0$.
Condition (3.10) ensures that $\Phi(x,\cdot)$ is convex for every
$x\in\Omega$. From (3.9) and (3.11) there exist $k_{4},k_{5}>0$ such that

(3.14) $k_{4}A\left(k_{4}|\xi|\right)\leq\Phi(x,\xi)\leq
2q(x)B(b|\xi|)+k_{5}A(k_{5}|\xi|)\ \hbox{for all}\
(x,\xi)\in{{\mathbb{R}}^{n}}\,.$

Now, problem (1.1) reads as

(3.15) $\begin{cases}-{\rm div}(\Phi_{\xi}(x,\nabla u))=f(x,u,\nabla u)&{\rm
in}\,\,\,\Omega\\\ u=0&{\rm on}\,\,\,\partial\Omega\,.\end{cases}$

For functions $f$ satisfying suitable growth conditions we can construct a sub
or a supersolution for problem (3.15), via variational methods. The
$\Delta_{2}$ and $\nabla_{2}$ conditions play a crucial role here.

###### Theorem 3.8

Let $\Omega$ be an open set in ${{\mathbb{R}}^{n}}$, with $n\geq 2$, such that
$|\Omega|<\infty$. Let $A\in C^{1}([0,+\infty))$ be a Young function,
$A\in\Delta_{2}\cap\nabla_{2}$ at infinity. Assume also that $A$ satisfies
(2.26) and (2.29), or (2.30). Let
$\mathcal{A}:\Omega\times{{\mathbb{R}}^{n}}\to{{\mathbb{R}}^{n}}$ and
$\Phi:\Omega\times{{\mathbb{R}}^{n}}\to\mathbb{R}$ be two Carathéodory
functions satisfying (3.9)$\ldots$(3.12). Let
$f:\Omega\times\mathbb{R}\times\mathbb{R}^{n}\to\mathbb{R}$ be a Carathéodory
function fulfilling

(3.16) $\displaystyle-\rho_{1}(x)-g_{1}(|s|)\leq f(x,s,\xi)\leq$
$\displaystyle\rho_{2}(x)+g_{2}(|s|)+\overline{\gamma}\widetilde{E}^{-1}(A(|\xi|))\
\hbox{for a.e.}\,x\in\Omega,\ \hbox{all}\ s\leq 0,$ $\displaystyle\
\hbox{all}\ \xi\in{{\mathbb{R}}^{n}},\ f(x,0,0)\leq 0\ \hbox{in}\ \Omega$
$\displaystyle\ \hbox{and}\ f(x,0,0)<0\ \hbox{on a set of positive measure},$

or

(3.17)
$\displaystyle-\rho_{2}(x)-g_{2}(|s|)-\overline{\gamma}\widetilde{E}^{-1}(A(|\xi|))$
$\displaystyle\leq f(x,s,\xi)\leq\rho_{1}(x)+g_{1}(|s|)\ \hbox{for a.e.}\
x\in\Omega,\ \hbox{all}\ s\geq 0,$ $\displaystyle\ \hbox{all}\
\xi\in{{\mathbb{R}}^{n}},\ f(x,0,0)\geq 0\ \hbox{in}\ \Omega,$ $\displaystyle\
\hbox{and}\ f(x,0,0)>0\ \hbox{on a set of positive measure},$

where $\overline{\gamma}>0$, $E$ is a Young function, $E\ll A_{n}$ near
infinity, $\rho_{1},\rho_{2}:\Omega\to[0,+\infty[$ are two measurable
functions, $\rho_{i}\in L^{\widetilde{A}_{n}}(\Omega),\ i=1,2,$,
$g_{1},g_{2}:[0,+\infty[\to[0,+\infty[$ are two non decreasing functions such
that $g_{1}(0)=g_{2}(0)=0$ and there exist
$s_{0}>0,\,h_{0}\in\left]0,\tau\omega_{n}^{\frac{1}{n}}|\Omega|^{-\frac{1}{n}}\right[$,
$h_{1}>0$ such that

(3.18) $\displaystyle g_{1}(|s|)|s|\leq A(h_{0}|s|)\ \hbox{and}\
g_{2}(|s|)|s|\leq A_{n}(h_{1}|s|)\ \hbox{for all}\ |s|\geq s_{0}\,.$

Here $\omega_{n}$ is the measure of the unit ball in ${{\mathbb{R}}^{n}}$,
$\tau=\min\\{1,k_{4}^{2}\\}$ where $k_{4}$ is that of (3.14).
Then problem (3.15) possesses a nontrivial constant sign solution
$u\in{W_{0}^{1,A}(\Omega)}$.

Proof. Suppose that (3.16) is in force. We construct a subsolution
$\underline{u}\leq 0$ a.e., $\underline{u}\not\equiv 0$, and show that
$\overline{u}\equiv 0$ is a supersolution but not a solution to (3.15). Then,
we show that $f$ satisfies (3.8).
Put $G_{1}(t)=\int_{0}^{t}g_{1}(\tau)d\tau,\ t\geq 0$ and consider the
functional $J:{W_{0}^{1,A}(\Omega)}\to\mathbb{R}$, defined as

$J(u)=\int_{\Omega}\left(\Phi(x,\nabla
u)+\rho_{1}(x)u-G_{1}(|u|)\right)dx\quad\hbox{for}\
u\in{W_{0}^{1,A}(\Omega)}.$

We prove that $J$ is well defined, weakly lower semicontinuous, coercive and

(3.19) $J^{\prime}(u)v=\int_{\Omega}\Phi_{\xi}(x,\nabla u)\nabla
vdx+\int_{\Omega}\rho_{1}(x)v(x)dx-\int_{\Omega}g_{1}(|u|)sign\,u\,v(x)dx$

for all $u,v\in{W_{0}^{1,A}(\Omega)}$. We examine separately the three
integrals.
Due to (3.14), the fact that $A\in\Delta_{2}$ at infinity, and the convexity
of $\Phi(x,\cdot)$, for all $x\in\Omega$, the functional
$u\mapsto\int_{\Omega}\Phi(x,\nabla u)dx$ is well defined in
${W_{0}^{1,A}(\Omega)}$, convex. We briefly sketch the proof of its
regularity, because it makes use of standard arguments like the Lebesgue
Theorem, and the properties of Young’s functions. Let
$u,v\in{W_{0}^{1,A}(\Omega)}$. For all $x\in\Omega$, all $t>0$, $t<<1$, there
exists $\mu_{t,x}\in(0,1)$ such that

(3.20) $\displaystyle\left|\frac{\Phi(x,\nabla u+t\nabla v)-\Phi(x,\nabla
u)}{t}\right|=$ $\displaystyle\left|\Phi_{\xi}(x,\nabla u+\mu_{t,x}t\nabla
v)\nabla v\right|$ $\displaystyle\leq q(x)\widetilde{B}^{-1}(B(b|\nabla
u+\mu_{t,x}t\nabla v|))|\nabla v|+$ $\displaystyle
b\widetilde{A}^{-1}(A(b|\nabla u+\mu_{t,x}t\nabla v|))|\nabla v|$
$\displaystyle\leq 2q(x)\frac{B(b(|\nabla u|+|\nabla v|))}{b(|\nabla
u|+|\nabla v|)}|\nabla v|+$ $\displaystyle 2\frac{A(b(|\nabla u|+|\nabla
v|))}{|\nabla u|+|\nabla v|}|\nabla v|$
$\displaystyle\leq\frac{2q(x)}{b}B(b(|\nabla u|+|\nabla v|))+$ $\displaystyle
2bA(b(|\nabla u|+|\nabla v|))\,.$

We used (3.9), the monotonicity of $\widetilde{A}^{-1}\circ A$ and
$\widetilde{B}^{-1}\circ B$, and (2.2). Now, the condition $A\in\Delta_{2}$
near infinity, (3.13) and the Lebesgue Theorem, allow to prove that the
functional $u\mapsto\int_{\Omega}\Phi(x,\nabla u)dx$ is $C^{1}$.
Thus the weak lower semicontinuity of $J$ and equation (3.19) follow.
To prove the coercivity of $J$ we need the following inequality, that can be
found in [BaCi1, Proposition 3.2],

$\int_{\Omega}A(|u|)dx\leq\int_{\Omega}A(\omega_{n}^{-\frac{1}{n}}|\Omega|^{\frac{1}{n}}|\nabla
u|)dx\ \hbox{for all}\ u\in{W_{0}^{1,A}(\Omega)}\,.$

Let $\varepsilon>0$ be such that
$h_{0}\omega_{n}^{-\frac{1}{n}}|\Omega|^{\frac{1}{n}}<\tau-\varepsilon$. Using
(3.18) and the inequality above

(3.21) $\displaystyle\int_{\Omega}G_{1}(|u|)dx\leq$
$\displaystyle\int_{\\{|u|\leq
s_{0}\\}}G_{1}(s_{0})dx+\int_{\\{|u|>s_{0}\\}}A(h_{0}|u|)dx$
$\displaystyle\leq
G_{1}(s_{0})|\Omega|+\int_{\Omega}A((\tau-\varepsilon)|\nabla u|)dx\leq$
$\displaystyle
G_{1}(s_{0})|\Omega|+\sqrt{\tau-\varepsilon}\int_{\Omega}A(k_{4}|\nabla u|)dx\
\hbox{for all}\ u\in{W_{0}^{1,A}(\Omega)}\,.$

Take now $u\in{W_{0}^{1,A}(\Omega)}$, $\|u\|_{{W_{0}^{1,A}(\Omega)}}>1$ and
use (3.14), (2.12), (3.21) and (2.16)

$\displaystyle\frac{J(u)}{\|u\|_{{W_{0}^{1,A}(\Omega)}}}$
$\displaystyle\geq\frac{\left(k_{4}-\sqrt{\tau-\varepsilon}\right)\int_{\Omega}A(k_{4}|\nabla
u|)dx-
c_{4}\|u\|_{{W_{0}^{1,A}(\Omega)}}-G_{1}(s_{0})|\Omega|}{\|u\|_{{W_{0}^{1,A}(\Omega)}}}$
$\displaystyle\geq\left(k_{4}-\sqrt{\tau-\varepsilon}\right)(k_{4})^{i^{\infty}_{A}}\|u\|_{{W_{0}^{1,A}(\Omega)}}^{i_{A}^{\infty}-1}-c_{4}-\frac{k_{3}+G_{1}(s_{0})|\Omega|}{\|u\|_{{W_{0}^{1,A}(\Omega)}}}\,.$

This proves that $J$ is coercive. Thus it has a global minimum. Let
$\underline{u}$ be a global minimum point for $J$. We prove that
$\underline{u}\not\equiv 0$. To this end consider a function $v\in
C_{0}^{1}(\Omega)$, such that $b|\nabla v(x)|\leq\overline{t}$ and
$k_{5}\overline{k}|\nabla v(x)|\leq\overline{t}$ for all $x\in\Omega$. Also,
$v\leq 0$ and $\rho_{1}(x)v(x)\not\equiv 0$ in $\Omega$. The inequality
$\frac{B(t_{1})}{B(t_{0})}>\left(\frac{t_{1}}{t_{0}}\right)^{k_{B}}$ holds for
$0<t_{0}<t_{1}<\overline{t}$, and some $k_{B}>1$, by virtue of the
$\nabla_{2}$ condition near zero. Then, choosing once $t_{1}=b|\nabla v|$,
$t_{0}=bt|\nabla v|$, and secondly $t_{1}=k_{5}\overline{k}|\nabla v|$,
$t_{0}=tt_{1}$, with $t<1$, and taking into account (3.14)

$\displaystyle J(tv)\leq 2t^{k_{B}}\int_{\Omega}q(x)B(b|\nabla
v|)dx+k_{5}t^{k_{B}}\int_{\Omega}B(k_{5}\overline{k}|\nabla
v|)dx+t\int_{\Omega}\rho_{1}(x)vdx<0\ \ \hbox{for}\ t<<1\,,$

and this proves that $J(\underline{u})<0$. Using $J(-|\underline{u}|)\geq
J(\underline{u})$ and the fact that $\Phi(x,\cdot)$ is even, we obtain
$\underline{u}\leq 0$ a.e. in $\Omega$.
Now we prove that $\underline{u}$ is a subsolution and $u\equiv 0$ is a
supersolution but not a solution to (1.1). Note that

(3.22)
$J^{\prime}(\underline{u})(v)=\int_{\Omega}\Phi_{\xi}(x,\nabla\underline{u})\nabla
vdx+\int_{\Omega}(\rho_{1}(x)+g_{1}(|\underline{u}(x)|))vdx\equiv 0,\
\hbox{for all}\ v\in{W_{0}^{1,A}(\Omega)}\,.$

Acting with any $v\in{W_{0}^{1,A}(\Omega)}$, $v\geq 0$, in (3.22) and using
(3.16)

$\int_{\Omega}\Phi_{\xi}(x,\nabla\underline{u})\nabla
vdx-\int_{\Omega}f(x,\underline{u},\nabla\underline{u})vdx\leq 0\,,$

that is $\underline{u}$ is a subsolution to (3.15). Using (3.16) and choosing
$v\in{W_{0}^{1,A}(\Omega)}$, $v\geq 0$

$0-\int_{\Omega}f(x,0,0)vdx\geq 0\,,$

thus $u\equiv 0$ is a supersolution to (1.1) and the assumption on $f(x,0,0)$
guarantees that it is not a solution.
We put $\rho(x)=\max\\{\rho_{i}(x),\ i=1,2\\}$, $g(|s|)=\max\\{g_{i}(|s|),\
i=1,2\\}$ and use (3.16)

$|f(x,s,\xi)|\leq\rho(x)+g(|s|)+\overline{\gamma}\widetilde{E}^{-1}(A(|\xi|))\
\ \hbox{for}\ x\in\Omega,\ s\in[\underline{u}(x),0],\
\xi\in{{\mathbb{R}}^{n}}\,.$

Then $f$ satisfies (3.8) and from Corollary 3.7 problem (3.15) has a
nontrivial solution $u\in{W_{0}^{1,A}(\Omega)}$ and $u\in[\underline{u},0]$.
When (3.17) is in force we consider $f_{1}(x,s,\xi)=-f(x,-s,-\xi)$. Then, by
virtue of the proof above, problem

(3.23) $\begin{cases}-{\rm div}\left(\Phi_{\xi}(x,\nabla
v)\right)=f_{1}(x,v,\nabla v)&{\rm in}\,\,\,\Omega\\\ v=0&{\rm
on}\,\,\,\partial\Omega\,,\end{cases}$

has a nontrivial solution $v\in{W_{0}^{1,A}(\Omega)}$, $v\leq 0$ a.e. in
$\Omega$. Then the function $u=-v$ is a nontrivial solution to (1.1) and
$u\geq 0$ a.e. in $\Omega$.

###### Remark 3.9

From [BaCiMa, Theorem 3.1], when $\rho_{1},\ \rho_{2}\in
L^{M,\infty}(\Omega)$, for a suitable Young function $M$ (see equation $(3.9)$
in [BaCiMa]), then the solution $u$ is essentially bounded.

##  4 Regularity results

In this section we give some existence and regularity results, Theorems 4.4
and 4.5. We strenghten the hypotheses on $\Omega$ and on $A$ (see (4.1)), in
order to apply regularity theory (see Proposition 4.1).
The proof of the existence is based on sub and supersolution methods, while
the main tool for the regularity is Theorem 1.7 of [Li1] (see also the remark
after that result and [Li]), that we recall below.

###### Proposition 4.1

(see [Li1, Theorem 1.7]) Let $\Omega$ be a bounded domain in $R^{n}$ with a
$C^{1,\alpha}$ boundary, for some $0<\alpha\leq 1$. Let $A$ be a Young
function satisfying

(4.1) $\displaystyle A\in C^{2}(]0,+\infty[),$ $\displaystyle\ \hbox{and there
exist two positive constants}\ \delta,\,g_{0}>0$ such that $\displaystyle\
\delta\leq\frac{tA^{\prime\prime}(t)}{A^{\prime}(t)}\leq g_{0}\quad\hbox{for}\
t>0\,.$

Let
$\mathcal{A}:\Omega\times\mathbb{R}\times{{\mathbb{R}}^{n}}\to{{\mathbb{R}}^{n}}$
be a vector valued function, with Carathéodory components, $a_{i}$,
$i=1,\ldots,n$. Consider the problem

$-div(\mathcal{A}(x,u,\nabla u))=f(x,u,\nabla u)\quad\hbox{in}\ \Omega\,.$

Suppose $\mathcal{A}$ and $f$ satisfy the structure conditions (here
$a_{ij}(x,s,\eta)=\frac{\partial a_{i}}{\partial\eta_{j}}$)

(4.2)
$\displaystyle\sum_{i,j=1}^{n}a_{ij}(x,s,\eta)\xi_{i}\xi_{j}\geq\frac{A^{\prime}(|\eta|)}{|\eta|}|\xi|^{2}$
(4.3)
$\displaystyle\sum_{i,j=1}^{n}|a_{ij}(x,s,\xi)|\leq\Lambda\frac{A^{\prime}(|\xi|)}{|\xi|}$

(4.4)
$\displaystyle|\mathcal{A}(x,s,\xi)-\mathcal{A}(y,w,\xi)|\leq\Lambda_{1}(1+A^{\prime}(|\xi|)(|x-y|^{\alpha}+|s-w|^{\alpha})$
(4.5) $\displaystyle|f(x,s,\xi)|\leq\Lambda_{1}(1+A^{\prime}(|\xi|)|\xi|),$

for some positive constants $\Lambda$, $\Lambda_{1}$, $M_{0}$, for all $x$ and
$y\in\Omega$, for all $s,w\in[-M_{0},M_{0}]$ and for all
$\xi\in\mathbb{R}^{n}$. Then, any solution $u\in W^{1,A}(\Omega)$, with
$|u|\leq M_{0}$ in $\Omega$, is $C^{1,\beta}(\overline{\Omega})$ for some
positive $\beta$.

We point out that (4.1) guarantees that $A^{\prime}(0)=0$ and
$A\in\nabla_{2}\cap\Delta_{2}$ globally.

###### Lemma 4.2

Let $A$ be a Young function satisfying (4.1). If
$\mathcal{A}:\Omega\times\mathbb{R}\times{{\mathbb{R}}^{n}}\to{{\mathbb{R}}^{n}}$
has Carathéodory components, $a_{i}(x,s,0)\equiv 0$, for a.e. $x\in\Omega$,
all $s\in\mathbb{R}$, all $i=1,\ldots,n$, and satisfies (4.2) and (4.3) for
all $s\in\mathbb{R}$, then (2.42), (2.43) and (2.44) hold with $q(x)\equiv
r(x)\equiv 0$, $F\equiv 0$, and $d=0$.

Proof. Let $H(t)=a_{i}(x,s,t\xi),\ t>0$. Using (4.3), $a_{i}(x,s,0)\equiv 0$,
and (4.1)

(4.6) $|H(1)|\leq\int_{0}^{1}|\nabla
a_{i}(x,s,t\xi)||\xi|dt\leq\Lambda\int_{0}^{1}\frac{A^{\prime}(t|\xi|)}{t|\xi|}|\xi|dt\leq\frac{\Lambda}{\delta}A^{\prime}(|\xi|).$

Then, thanks to equation (6.23) in [BaCi1]

(4.7)
$|\mathcal{A}(x,s,\xi)|\leq\frac{\Lambda\sqrt{n}}{\delta}A^{\prime}(|\xi|)\leq\frac{\Lambda\sqrt{n}}{\delta}\widetilde{A}^{-1}(A(2|\xi|)\leq
b\widetilde{A}^{-1}(A(b|\xi|),\ \hbox{for}\
b=\max\left\\{\frac{\Lambda\sqrt{n}}{\delta},2\right\\}\,,$

and (2.42) is proved.
For any $\xi,\xi^{\prime}\in{{\mathbb{R}}^{n}}$, write
$\xi=\xi^{\prime}+\eta$, $\eta\neq 0_{{{\mathbb{R}}^{n}}}$. Define
$H(t)=\sum_{i=1}^{n}\left(a_{i}(x,s,\xi^{\prime}+t\eta)-a_{i}(x,s,\xi^{\prime})\right)\eta_{i}$,
for $t\in[0,1]$. From (4.2),
$H^{\prime}(t)\geq\frac{A^{\prime}(|\xi^{\prime}+t\eta|)}{|\xi^{\prime}+t\eta|}\cdot|\eta|^{2}$
and this guarantees that $H(1)>H(0)=0$, namely (2.43).
Choose now $\xi^{\prime}=0$ in the function $H$ just used. For $\eta\neq
0_{{{\mathbb{R}}^{n}}}$, (4.2) and (4.1) give

$H(1)\geq\int_{0}^{1}\frac{A^{\prime}(t|\eta|)}{t|\eta|}|\eta|^{2}dt\geq\int_{0}^{1}\frac{t^{g_{0}}A^{\prime}(|\eta|)|\eta|}{t}dt\geq\frac{1}{g_{0}}A\left(|\eta|\right)\,.$

If $\eta=0_{{{\mathbb{R}}^{n}}}$ then $H(1)=0$, and (2.44) is proved.

Following the spirit of Example 5.1, we present a function $\mathcal{A}$
satisfying (4.2), (4.3) and (4.4). This general function will be used in
Example 5.3, but we prefere to introduce this function here, to emphasize its
general structure.

###### Example 4.3

Let $p>1$, $q\in\mathbb{R}$ and $p+q-1>0$. Consider the Young function
$A:[0,+\infty[\to[0,+\infty[$ complying with

(4.8) $\displaystyle A^{\prime}(t)=t^{p-1}\lg^{q}(1+t)\quad\ \hbox{for}\ t\geq
0\,.$

Then $A$ satisfies (4.1) because

(4.9) $\displaystyle
0<\min\\{p+q-1,p-1\\}\leq\frac{tA^{\prime\prime}(t)}{A^{\prime}(t)}$
$\displaystyle\leq\max\\{p+q-1,p-1\\}<\infty\ \hbox{for}\ t>0\,.$

Let us define

(4.10)
$\displaystyle\mathcal{A}(x,s,\xi)=(\|x\|^{\gamma}|s|^{\delta}+1)|\xi|^{p-2}\lg^{q}(1+|\xi|)\xi\
$ $\displaystyle\ \hbox{for}\
(x,s,\xi)\in\Omega\times\mathbb{R}\times{{\mathbb{R}}^{n}}\,.$

Here $\Omega$ is a bounded domain with a $C^{1,\alpha}$ boundary,
$\beta,\delta\geq\alpha$ and $p,q$ are like above. We show that $\mathcal{A}$
satisfies (4.2), (4.3) and (4.4). It holds

(4.11)
$\displaystyle\sum_{i,j=1}^{n}\partial_{j}a_{1}(x,s,\eta)\xi_{i}\xi_{j}=$
$\displaystyle(\|x\|^{\gamma}|s|^{\delta}+1)|\eta|^{p-4}\lg^{q-1}(1+|\eta|)$
$\displaystyle\left[\left((p-2)|\lg(1+|\eta|)+q\frac{|\eta|}{1+|\eta|}\right)\langle\xi,\eta\rangle^{2}+|\eta|^{2}\lg(1+|\eta|)|\xi|^{2}\right]$

Let now $\mu=\min\\{1,p+q-1,p-1\\}$. Then

$\displaystyle\left((p-2)|\lg(1+|\eta|)+q\frac{|\eta|}{1+|\eta|}\right)\langle\xi,\eta\rangle^{2}+|\eta|^{2}\lg(1+|\eta|)|\xi|^{2}\geq\mu|\eta|^{2}|\xi|^{2}\lg(1+|\eta|)\
\hbox{for}\ \eta,\xi\in{{\mathbb{R}}^{n}}\,.$

This guarantees that

$\displaystyle\sum_{i,j=1}^{n}\partial_{j}a_{i}(x,s,\eta)\xi_{i}\xi_{j}\geq\mu\frac{A^{\prime}(|\eta|)}{|\eta|}|\xi|^{2}\
$ $\displaystyle\ \hbox{for}\ \eta,\xi\in{{\mathbb{R}}^{n}}\,,$

and (4.2) holds. A simple calculation shows that (4.3) holds too. For what
concerns (4.4), let $M>0$ and take $x,y\in\Omega,\ s,w\in[-M,M],\
\xi\in{{\mathbb{R}}^{n}}$. Then

(4.12) $\displaystyle|\mathcal{A}(x,s,\xi)-\mathcal{A}(y,w,\xi)|$
$\displaystyle\leq|\|x\|^{\gamma}|s|^{\delta}-\|y\|^{\gamma}|w|^{\delta}|A^{\prime}(|\xi|)$
$\displaystyle\leq((|\|x\|^{\gamma}-\|y\|^{\gamma}|)|s|^{\delta}+\|y\|^{\gamma}(||s|^{\delta}-|w|^{\delta}|))A^{\prime}(|\xi|)$
$\displaystyle\leq C(\|x-y\|^{\alpha}+|s-w|^{\alpha})A^{\prime}(|\xi|)\
\hbox{for some}\ C>0\,.$

For the last inequality in (4.12) we used (4.15) and (4.16) further down. They
are obtained via the inequalities below.
If $\rho\leq 1$ then there exists $c>0$ such that

(4.13) $\displaystyle|t^{\rho}-z^{\rho}|\leq
c\frac{|t-z|}{t^{1-\rho}+z^{1-\rho}}\ $ $\displaystyle\ \hbox{for all}\
t,z>0,$

If $\rho>1$ then there exists $c>0$ such that

(4.14) $\displaystyle|t^{\rho}-z^{\rho}|\leq c|t-z||(t^{\rho-1}+z^{\rho-1})\ $
$\displaystyle\ \hbox{for all }\ t,z>0,$

Take $s,w\in[-M,M]$, for $M>0$. If $0<\delta\leq 1$, from
$(|s|^{1-\delta}+|w|^{1-\delta})^{\frac{1}{1-\delta}}\geq|s|+|w|$ and (4.13)

(4.15) $\displaystyle||s|^{\delta}-|w|^{\delta}|$ $\displaystyle\leq
c\frac{||s|-|w||}{|s|^{1-\delta}+|w|^{1-\delta}}\leq
c\frac{||s|-|w||^{1-\alpha+\alpha}}{(|s|+|w|)^{1-\delta}}$ $\displaystyle\leq
c||s|-|w||^{\alpha}(|s|+|w|)^{\beta-\alpha}\leq
c(2M)^{\delta-\alpha}|s-w|^{\alpha}\,.$

A similar argument, when $1\leq\delta\leq\alpha$, via (4.14), leads to

(4.16) $\displaystyle||s|^{\delta}-|w|^{\delta}|$ $\displaystyle\leq c\leq
c2^{2-\alpha}M^{\delta-\alpha}|s-w|^{\alpha}\,.$

The same holds for $|\|x\|^{\gamma}-\|y\|^{\gamma}|,\ x,y\in\Omega$. Thus
(2.45) remains true.

For the first existence and regularity Theorem we assume that problem (1.1)
admits a subsolution and a supersolution $\underline{u}$, $\overline{u}\in
W^{1,\infty}(\Omega)$.

###### Theorem 4.4

Let $\Omega$ be a bounded domain in ${{\mathbb{R}}^{n}}$ with a $C^{1,\alpha}$
boundary. Let the functions $A$ and $\mathcal{A}$ be as in Proposition 4.1.
Assume further that (4.2), (4.3) and (4.4) hold for all $s\in\mathbb{R}$. Let
$\underline{u}$, $\overline{u}\in W^{1,\infty}(\Omega)$ be a subsolution and a
supersolution for problem (1.1), with $\underline{u}(x)<\overline{u}(x)$ a.e.
$x\in\Omega$. Let
$f:\Omega\times\mathbb{R}\times{{\mathbb{R}}^{n}}\to\mathbb{R}$ be a
Carathéodory function satisfying

(4.17)
$\displaystyle|f(x,s,\xi)|\leq\sigma(x)+\overline{\gamma}(s)A^{\prime}(|\xi|)|\xi|\quad\hbox{for
a.e.}\ x\in\Omega,\ \hbox{all}\ s\in[\underline{u}(x),\overline{u}(x)],\
\hbox{all}\ \xi\in{{\mathbb{R}}^{n}}\,,$

where $\sigma\in L^{\infty}(\Omega)$ and
$\overline{\gamma}:[0,+\infty[\to[0,+\infty[$ is locally essentially bounded.
Then problem (1.1) admits at least a solution $u\in
C_{0}^{1,\beta}(\overline{\Omega})$. Moreover $\underline{u}(x)\leq
u(x)\leq\overline{u}(x)$ a.e in $\Omega$.

Proof. Let
$M=\max\\{\|\overline{u}\|_{\infty},\|\underline{u}\|_{\infty}\\}+1>0$,
$R>\max\\{\|\nabla\overline{u}\|_{\infty},\|\nabla\underline{u}\|_{\infty}\\}$.
Let $I=[-M,M]$ and $\overline{\gamma}=\|\overline{\gamma}\|_{L^{\infty}(I)}$.
Then $\overline{\gamma}<+\infty$ and
$\overline{\gamma}(s)\leq\overline{\gamma}$ for a.e. $s\in I$. Let
$I_{0}\subset I$ be a set of null measure, such that
$\overline{\gamma}(s)>\overline{\gamma}$ for all $s\in I_{0}$. For $s\in
I_{0}$ it holds

$\displaystyle|f(x,s,\xi)|=\lim_{t\to s}|f(x,t,\xi)|=\liminf_{t\to
s}|f(x,t,\xi)|$ $\displaystyle\leq\sigma(x)+\liminf_{t\to
s}\overline{\gamma}(t)A^{\prime}(|\xi|)|\xi|$
$\displaystyle\leq\sigma(x)+\overline{\gamma}A^{\prime}(|\xi|)|\xi|$
$\displaystyle\quad\hbox{for a.e.}\;x\in\Omega,\;\hbox{all}\
\xi\in{{\mathbb{R}}^{n}}\,.$

Let the truncated functions $a_{i}^{M}$ and $f_{R}$ be defined as

$a_{i}^{M}(x,s,\xi)=\left\\{\begin{array}[]{cc}a_{i}(x,-M,\xi)&\hbox{if}\
s\leq-M,\\\ a_{i}(x,s,\xi)&\hbox{if}\ -M<s<M\,,\\\ a_{i}(x,M,\xi)&\hbox{if}\
s\geq M\,,\end{array}\right.$

and

$f_{R}(x,s,\xi)=\left\\{\begin{array}[]{cc}f(x,s,\xi)&\hbox{if}\ |\xi|\leq
R,\\\ f(x,s,\xi)\cdot\frac{A^{\prime}(R)R}{A^{\prime}(|\xi|)|\xi|}&\hbox{if}\
|\xi|>R\,.\end{array}\right.$

Consider the problem

(4.18) $\begin{cases}-div(\mathcal{A}^{M}(x,u,\nabla u))=f_{R}(x,u,\nabla
u)&{\rm in}\,\,\,\Omega\\\ u=0&{\rm on}\,\,\,\partial\Omega\,.\end{cases}$

In view of the choice of $R$, $\underline{u}$ and $\overline{u}$ are a
subsolution and a supersolution to (4.18) respectively. The function $f$
satisfies satisfies the hypotheses of Proposition 4.1. In particular, (4.5)
holds ith $\Lambda_{1}=\max\\{\|\sigma\|_{\infty},\overline{\gamma}\\}$. Since
$|f_{R}|\leq|f|$ the same holds for $f_{R}$ whatever $R$ is. Also
$\mathcal{A}^{M}$ satisfies (4.2), (4.3) and (4.4). Due to Proposition 4.1
there exist two positive constants $0<\beta\leq 1$ and $C$, independent from
$R$, such that any solution to (4.18) belongs in
$C_{0}^{1,\beta}(\overline{\Omega})$ and
$\|u\|_{C_{0}^{1,\beta}(\overline{\Omega})}\leq C$. Choosing $R>C$ we deduce
that $u$ is a solution to (1.1).

The next Theorem is related to Theorem 3.8, because it deals with problem
(3.15). Now, we are concerned with the existence of a regular solution to
(3.15).

###### Theorem 4.5

Let $\Omega$ be a bounded domain in ${{\mathbb{R}}^{n}}$ with $C^{1,\alpha}$
boundary. Let $A$ be a Young function satisfying (4.1), and let
$\mathcal{A}:\Omega\times{{\mathbb{R}}^{n}}\to{{\mathbb{R}}^{n}}$ be a vector
valued function satisfying (4.2) and (4.3) for a.e. $x\in\Omega$, all
$\xi\in{{\mathbb{R}}^{n}}$, and such that $a_{i}(x,0)\equiv 0$ for a.e.
$x\in\Omega$, all $i=1,\ldots,n$. We further assume that (3.12) holds. Let
$f:\Omega\times\mathbb{R}\times\mathbb{R}^{n}\to\mathbb{R}$ be a Carathéodory
function fulfilling

(4.19)
$\displaystyle-\rho_{2}(x)-g_{2}(s)-\overline{\gamma}(s)A^{\prime}(|\xi|)|\xi|$
$\displaystyle\leq f(x,s,\xi)\leq\rho_{1}(x)+g_{1}(s)\ \hbox{for a.e.}\
x\in\Omega,\ \hbox{all}\ s\geq 0,$ $\displaystyle\ \hbox{all}\
\xi\in{{\mathbb{R}}^{n}},\ f(x,0,0)\geq 0$ $\displaystyle\ \hbox{in}\ \Omega,\
\hbox{and}\ f(x,0,0)>0\ \hbox{on a set of positive measure},$

or

(4.20) $\displaystyle-\rho_{1}(x)-g_{1}(|s|)\leq f(x,s,\xi)$
$\displaystyle\leq\rho_{2}(x)+g_{2}(|s|)+\overline{\gamma}(s)A^{\prime}(|\xi|)|\xi|\
\hbox{for a.e.}\,x\in\Omega,\ \hbox{all}\ s\leq 0,$ $\displaystyle\
\hbox{all}\ \xi\in{{\mathbb{R}}^{n}},\ f(x,0,0)\leq 0$ $\displaystyle\
\hbox{in}\ \Omega,\hbox{and}\ f(x,0,0)<0\ \hbox{on a set of positive
measure},$

Here $\rho_{1},\rho_{2}:\Omega\to[0,+\infty[$ are two measurable functions,
$\rho_{1},\,\rho_{2}\in L^{\infty}(\Omega)$; $g_{1}$ is like in Theorem 3.8,
$g_{2}:[0,+\infty[\to[0,+\infty[$ is a non-decreasing function such that
$g_{2}(0)=0$ and $\overline{\gamma}(s)$ is a locally essentially bounded
function.
Then problem $(P)$ has a nontrivial, solution $u\in
C_{0}^{1,\beta}(\overline{\Omega})$.
If (4.19) jolds, then $u\geq 0$ in $\Omega$. In the other case $u\leq 0$ in
$\Omega$.

Proof. From the proof of Theorem 3.8 we know that, when (4.19) is in force,
there exists a nontrivial solution $\overline{u}\geq 0$, to problem (3.15).
Remark 3.9 guarantees that $\overline{u}$ is bounded. From Proposition 4.1, we
have that $\overline{u}\in C_{0}^{1,\beta}(\overline{\Omega})$. The
inequalities in (4.19) show that $\overline{u}$ is a supersolution to problem
(1.1) and $\underline{u}=0$ is a subsolution to problem (1.1). The assumptions
on $\rho_{2}$ guarantee that $\underline{u}=0$ is not a solution. If we put
$M=\|\overline{u}\|_{\infty}$,
$\overline{\gamma}=\|\overline{\gamma}\|_{L^{\infty}(0,M)}$,
$\sigma(x)=\max\\{\rho_{1}(x)+g_{1}(M),\;\rho_{2}(x)+g_{2}(M)\\}$ for a.e.
$x\in\Omega$, then (4.19) leads to

(4.21)
$\displaystyle|f(x,s,\xi)|\leq\sigma(x)+\overline{\gamma}A^{\prime}(|\xi|)|\xi|$
$\displaystyle\quad\hbox{for
a.e.}\;x\in\Omega,\;s\in[0,\overline{u}(x)],\;\xi\in\mathbb{R}^{n}\,.$

So, from Theorem 4.4, problem (3.15) admits at least a nontrivial solution
$u\in C_{0}^{1,\beta}(\overline{\Omega})$ such that $0\leq u\leq\overline{u}$.
For the other case, it is enough to put $f_{1}(x,s,\xi)=-f(x,-s,-\xi)$ and to
use the first part of the proof.

##  5 Examples

In this Section we present some applications of Theorem 3.4 (Subsection 5.1)
where the structure of $f$ and the condition $\mathcal{A}(x,s,0)\equiv
0_{{{\mathbb{R}}^{n}}}$ guarantee the existence of constant sub and
supersolutions for (1.1). This situation is really interesting and meaningful
because hilights how in this setting the growth of $\mathcal{A}$ with respect
to $s$ can be whatever we want. Then we present an application of Theorem 3.8
(Subsection 5.2) and finally some applications of Theorem 4.5 (Subsection
5.3). In the latter case the growth of $\mathcal{A}$ with respect to $s$ comes
once again into play (see equation (4.4)).
In all the examples $\Omega$ is an open subset in ${{\mathbb{R}}^{n}}$ with
finite measure.

###  5.1 Applications of Theorem 3.4

In this example we do not impose growth conditions with respect to $s$ for
$\mathcal{A}$, but only with respect to $x$ and $\xi$. The structure of the
convection term guarantees that (1.1) has a pair of constant sub and
supersolutions.
Let $0<\delta<p-1$, $p+q>1$, $a:\Omega\to[0,+\infty[$ be a measurable
function, $a\in L^{\frac{p}{\delta}}(\Omega)$, and let
$b:[0,+\infty[\to[0,+\infty[$ be a continuous function.
Consider the problem

(5.1) $\begin{cases}-{\rm div}\left(\left(a(x)b(|u|)|\nabla
u|^{p-2-\delta}\lg^{q(1-\frac{\delta}{p-1})}(1+|\nabla u|)+|\nabla
u|^{p-2}\lg^{q}(1+|\nabla u|)\right)\nabla u\right)=f(x,u,\nabla u)&{\rm
in}\,\,\Omega\\\ u=0&{\rm on}\,\partial\Omega\,,\end{cases}$

The Young function $A$ governing the differential operator $\mathcal{A}$ obeys
(2.35).
The convection term
$f:\Omega\times\mathbb{R}\times{{\mathbb{R}}^{n}}\to\mathbb{R}$ is defined as

(5.2) $\displaystyle f(x,s,\xi)=(h(x)+k(|\xi|))g(s)\quad\hbox{for}\
(x,s,\xi)\in\Omega\times\mathbb{R}\times{{\mathbb{R}}^{n}}\,.$

Here $h:\Omega\to[0,+\infty[$ is a measurable function, $h\in
L^{\widetilde{A}_{n}}(\Omega)$, $k:[0,+\infty]\to\mathbb{R}$ is a continuous
function, $k(0)>0$, and $k$ has the following behavior near infinity

(5.7) $\displaystyle\left\\{\begin{array}[]{ll}|k(s)|\approx
s^{\frac{p}{(p^{*})^{\prime}}}\lg^{r}(s)&\ \hbox{for some}\
r<\frac{(n+1)q}{n},\ \hbox{when}\ p<n\,,\\\ |k(s)|\approx s^{n}\lg^{r}(s)&\
\hbox{for some},\ r<q-1+\frac{q+1}{n},\ \hbox{when}\ p=n,\ q<n-1\,,\\\
|k(s)|\approx s^{n}\lg^{n-1}(s)\lg^{-\frac{1}{r}}(\lg s)&\ \hbox{for some},\
r<\frac{n-1}{n},\ \hbox{when}\ p=n,\ q=n-1\,,\\\ |k(s)|\approx
s^{p}\lg^{r}(s)&\ \hbox{for some},\ r<q,\ \hbox{when}\ p>n\ \hbox{or}\
p=n,q>n-1\,.\end{array}\right.$

The function $g:\mathbb{R}\to\mathbb{R}$ is continuous, $g(s)>0$ for
$s\in[0,\overline{s})$ and $g(\overline{s})=0$. First of all we note that
$u_{1}=0$ and $u_{2}=\overline{s}$ are a subsolution and a supersolution to
(5.1) and $u\equiv 0$ is not a solution. The growth of the functions $h$ and
$k$ quarantee that $f$ satisfies (3.1).
Due to the continuity of $b$, and taking into account Example 2.16 (with
$\beta=\beta_{1}=0$), we see that conditions (2.42), (2.43) and (2.44) hold,
for a.e. $x\in\Omega$, all $s\in[0,\overline{s}]$, all
$\xi\in{{\mathbb{R}}^{n}}$. Thus, by Theorem 3.4, problem (5.1) has a
nontrivial solution $u\in[u_{1},u_{2}]$.
The same arguments work for different choices of $h$ and $g$. In particular we
see that Theorem 3.4 works well with all nonlinearities having two zeros
$s_{1}$ and $s_{2}$, with $s_{1}<0<s_{2}$, ($f(x,s_{1},0)=f(x,s_{2},0)=0$ for
all $x\in\Omega$) and $f(x,s,0)$ has constant sign for $s\in]s_{1},s_{2}[$, or
for which $f(x,0,0)$ has constant sign and there exists $s\in\mathbb{R}$ such
that $f(x,s,0)\equiv 0$ and $s\cdot f(x,0,0)>0$.

###  5.2 Applications of Theorem 3.8

Let $p,q,r\in\mathbb{R}$ be such that $1<r<p<n$, $1<q<p$, and let $m<0$.
Let $a,\rho:\Omega\to[0,+\infty[$ be two measurable functions, $a\in
L^{\frac{p}{p-r}}(\Omega)$ and $\rho>0$ on a subset of $\Omega$ having
positive measure.
We show that problem

(5.8) $\begin{cases}-{\rm div}\left(\left(a(x)|\nabla u|^{r-2}+|\nabla
u|^{p-2}\right)\nabla u\right)=\frac{\rho(x)+|u|^{q-1}}{1+|\nabla u|}-|\nabla
u|^{\frac{p}{{p^{*}}^{\prime}}}|\lg(|\nabla u|)|^{\frac{m}{p}^{*}}&{\rm
in}\,\,\Omega\\\ u=0&{\rm on}\,\partial\Omega\,,\end{cases}$

has a nontrivial solution $u\in W_{0}^{1,p}(\Omega)$, $u\geq 0$.
The functions $\Phi(x,\xi)=a(x)\frac{|\xi|^{r}}{r}+\frac{|\xi|^{p}}{p}$, and
$\mathcal{A}(x,\xi)=\left(a(x)|\xi|^{r-2}+|\xi|^{p-2}\right)\xi$ satisfy
(3.9), (3.10), (3.11) and (3.12). Condition (3.17) holds too, with
$\rho_{1}(x)\equiv\rho(x)$, $g_{1}(|s|)=|s|^{q-1}$,
$\rho_{2}(x)=-M^{\frac{p}{{p^{*}}^{\prime}}}|\lg(|M|)|^{\frac{m}{p}^{*}}$, for
a suitable constant $M>0$, $g_{2}(|s|)\equiv 0$ and $E(t)\approx
t^{p^{*}}\lg^{m}(t)$, for $t>>1$.
Thus from Theorem 3.8 problem (5.8) has a nontrivial solution $u\in
W_{0}^{1,p}(\Omega)$, $u\geq 0$.

###  5.3 Applications of Theorem 4.4

Let $\Omega$ be a bounded domain with a $C^{1,\alpha}$ boundary. Let
$\gamma\geq\alpha$, $p>1$ and $q\in\mathbb{R}$, satisfying $p+q-1>0$. Let
$h:\Omega\to[0,+\infty[$ be a measurable function, $h\in L^{\infty}(\Omega)$,
$h\neq 0$ on a subset of $\Omega$ having finite measure, and let
$g:\mathbb{R}\to\mathbb{R}$ be a continuous function such that $g(s)>0$ for
all $s\in[0,\overline{s}[$ and $g(\overline{s})\equiv 0$. We consider the
problem

(5.9) $\begin{cases}-{\rm div}\left(\|x\|^{\gamma}e^{|u|}|\nabla
u|^{p-2}\lg^{q}(1+|u|)\nabla u\right)=\left(h(x)+|\nabla
u|^{p}\lg^{q}(1+|\nabla u|)\right)g(u)&{\rm in}\,\,\Omega\\\ u=0&{\rm
on}\,\partial\Omega\,,\end{cases}$

and prove that it has a nontrivial solution $u\in W_{0}^{1,A}(\Omega)$, $u\geq
0$. The Young function $A$ is defined via (4.8).
Using the Mac Laurin expansion of $k(t)=e^{t}$ we see that
$|e^{|s|}-e^{|w|}|\leq\frac{e^{M}(e^{M}-1)}{M^{\alpha}}|s-w|^{\alpha}$ for all
$s,w\in[-M,M]$. Thus, following the arguments used in the Example 4.3, we can
prove that the operator $\mathcal{A}$ and the function $f$, defined
respectively as
$\mathcal{A}(x,s,\xi)=\|x\|^{\gamma}e^{|s|}|\xi|^{p-2}\lg^{q}(1+|\xi|)\xi$ and
$f(x,s,\xi)=\left(h(x)+|\xi|^{p}\lg^{q}(1+|\xi|)\right)g(s)$ satisfy the
hypotheses of Theorem 4.4. Furthermore, $\underline{u}\equiv 0$ and
$\overline{u}=\overline{s}$ are a subsolution and a supersolution to (5.9),
and $\underline{u}\equiv 0$ is not a solution. Thus, from Theorem 4.4, the
problem which we are dealing with, has a regular solution
$u\in[0,\overline{s}]$.

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