# A blow-up approach for singular elliptic problems with natural growth in the
gradient

###### Abstract.

We prove existence and nonexistence results concerning elliptic problems whose
basic model is

$\begin{cases}\displaystyle-\Delta u+\mu(x)\frac{|\nabla
u|^{2}}{(u+\delta)^{\gamma}}=\lambda u^{p},&x\in\Omega,\\\ u>0,&x\in\Omega,\\\
u=0,&x\in\partial\Omega,\end{cases}$

where $\Omega\subset\mathbb{R}^{N}(N\geq 3)$ is a bounded smooth domain,
$\lambda>0$, $p>1$, $\delta\geq 0$, $\gamma>0$ and $\mu\in
L^{\infty}(\Omega)$. The main achievement resides in handling a possibly
singular ($\delta=0$) first order term having a nonconstant coefficient $\mu$
in the presence of a superlinear zero order term. Our approach for the
existence results is based on fixed point theory. With the aim of applying it,
a previous analysis on a related non-homogeneous problem is carried out. The
required a priori estimates are proven via a blow-up method.

###### Key words and phrases:

Nonlinear elliptic equations, Singular gradient terms, Blow-up argument.
MSC: 35B44, 35B45, 35J25, 35J62, 35J75.
Research supported by PGC2018-096422-B-I00 (MCIU/AEI/FEDER, UE), Junta de
Andalucía FQM-116 and Programa de Contratos Predoctorales del Plan Propio de
la Universidad de Granada.

SALVADOR LÓPEZ-MARTÍNEZ
Inria, Univ. Lille, CNRS, UMR 8524 - Laboratoire Paul Painlevé, F-59000
Lille, France
Email adress<EMAIL_ADDRESS>
##  1\. Introduction

Let $\Omega\subset\mathbb{R}^{N}(N\geq 3)$ be a bounded domain of class
$\mathcal{C}^{2}$, $g:\Omega\times(0,+\infty)\to\mathbb{R}$ be a Carathéodory
function, and $f:[0,+\infty)\to[0,+\infty)$ be a continuous function. In this
work we will study the existence of solution to elliptic problems of the
following form:

($P_{\lambda}$) $\begin{cases}\displaystyle-\Delta u+g(x,u)|\nabla
u|^{2}=\lambda f(u),&x\in\Omega,\\\ u>0,&x\in\Omega,\\\
u=0,&x\in\partial\Omega,\end{cases}$

where $\lambda>0$ is a parameter. The precise conditions on functions $g,f$
and the statements of the main results regarding problem ($P_{\lambda}$) will
be shown in Section 2. For the sake of a clear presentation, we consider for
now a model problem:

(1.1) $\begin{cases}\displaystyle-\Delta u+\mu(x)\frac{|\nabla
u|^{2}}{(u+\delta)^{\gamma}}=\lambda u^{p},&x\in\Omega,\\\ u>0,&x\in\Omega,\\\
u=0,&x\in\partial\Omega,\end{cases}$

where $\delta\geq 0$, $\gamma>0$, $p>1$ and $\mu\in L^{\infty}(\Omega)$. Our
main goal is to allow $\mu$ to be nonconstant and even sign-changing, paying
special attention to the singular case $\delta=0$.

Problem (1.1) for $\mu\equiv 0$ becomes semilinear; as $p>1$, it is usually
referred to as superlinear. This superlinear case is classical and has been
extensively studied in the literature. Indeed, both variational (in [1]) and
topological (in [21, 25]) methods can be used to prove the existence of a
solution to (1.1) for all $\lambda>0$ provided $\mu\equiv 0$ and
$p\in(1,2^{*}-1)$, where $2^{*}=\frac{2N}{N-2}$. It is well-known that
Pohozaev’s identity (see [36]) implies that the restriction $p<2^{*}-1$ is
necessary for the existence of solution to the superlinear problem if the
domain is starshaped.

The study of problem (1.1) for a nontrivial $\mu$ was initiated in [35].
There, the authors consider $p\in(1,2^{*}-1)$, $\delta>0$ and
$\mu\equiv\text{constant}>0$. In this setting, they prove several results
which can be divided into two classes: on the one hand, those which lead to a
set of solutions similar to that for the classical semilinear problem (i.e.
existence for all $\lambda>0$) and, on the other hand, those which present
differences such as nonexistence of solution for $\lambda>0$ small. In fact,
the semilinear-like behavior is achieved provided $\gamma>1$, while the
differences appear if either $\gamma\in(0,1)$ or $\gamma=1$ and $\mu>p$. In
several proofs, the authors of [35] make use of the change of variable

(1.2)
$v=\psi(u)=\int_{0}^{u}e^{-\int_{0}^{s}\frac{\mu}{(t+\delta)^{\gamma}}dt}ds.$

It is easy to check formally that, if $\mu\equiv\text{constant}$ and
$\delta>0$, then the transformation (1.2) turns (1.1) into a semilinear
problem in the variable $v$. Thus, roughly speaking, the gradient term is
removed and semilinear techniques (such as variational methods) can be applied
in general. However, such a transformation can be performed only if $\mu$ is
constant.

We remark that problem (1.1) in the case $\gamma=1$ is specially interesting
because of the condition $\mu>p$ that appears in [35], which shows that the
interaction between the gradient term and the superlinear term in (1.1) plays
an important role. To this respect, some results concerning the case
$\gamma=1$ (always with $\delta>0$ and $\mu\equiv\text{constant}$) that
improve those in [35] in some directions can be found in [3, 29]. In the first
work, the authors prove nonexistence of solution for every $\lambda>0$ small
enough provided $\mu\geq p$, while in the second one the authors prove
existence of solution for every $\lambda>0$ provided
$\mu<\frac{2^{*}-1-p}{2^{*}-2}$. We point out that, in both mentioned works,
no restriction on $p$ from above is imposed. Nevertheless, if $p\geq 2^{*}-1$,
then the condition $\mu<\frac{2^{*}-1-p}{2^{*}-2}$ required by the existence
result in [29] forces $\mu$ to be negative. We also stress that a blow-up
argument is employed in [29] in order to obtain a priori estimates, even
though the change of unknown (1.2) is strongly used in order to get rid of a
quadratic gradient term from the general problem that the authors consider
and, in consequence, non-constant functions $\mu$ cannot be handled with their
approach.

Still focusing on problem (1.1) with $\gamma=1$, $\delta>0$ and
$\mu\equiv\text{constant}$, the range $\frac{2^{*}-1-p}{2^{*}-2}\leq\mu<p$ has
not been considered in the literature to our knowledge. However, in this
particular situation it is easy to see that the transformation (1.2) turns
(1.1) into a semilinear equation whose nonlinearity presents supercritical
growth at infinity and subcritical growth at zero. Therefore, [7, Theorem 8]
implies (after undoing the change of unknown) that there exists at least a
solution to (1.1) for every $\lambda>0$ large enough. Again, last (immediate)
result is based on the change of unknown, so does not cover problem (1.1) with
nonconstant $\mu$.

On the other hand, the singular case $\delta=0$ has been dealt with recently
in [18], one more time for $\mu\equiv\text{constant}>0$ (see also [19, 14] for
similar singular problems that involve a nonzero source term). The authors of
[18] show that, if $\gamma\in(0,1)$, then the situation is similar to the
nonsigular case $\delta>0$. Indeed, they prove a nonexistence result for
$\lambda>0$ small and an existence result for $\lambda>0$ large. On the
contrary, for $\gamma\geq 1$ they prove nonexistence results _for all_
$\lambda>0$. This fact exposes the remarkable influence of a strong
singularity in the equation. As far as we know, the
$\mu\not\equiv\text{constant}$ case for $\delta=0$ has not been studied in the
literature.

To sum up, in the present work we aim to develop an approach that permits to
deal with $x-$dependent $\mu$ in problem (1.1) and also with singular lower
order terms, i.e. $\delta=0$. In order to do so, we will employ topological
methods. More precisely, we will find solutions to (1.1) as fixed points of
certain compact operator that will be defined in Section 4. The well-
definition of such an operator will require the well-posedness of the
following problem:

(1.3) $\begin{cases}\displaystyle-\Delta u+\mu(x)\frac{|\nabla
u|^{2}}{u}=h(x),&x\in\Omega,\\\ u>0,&x\in\Omega,\\\
u=0,&x\in\partial\Omega,\end{cases}$

where $0\lneq h\in L^{q}(\Omega)$ for some $q>\frac{N}{2}$.

Singular problems of this kind have risen interest in the recent years. In
fact, the existence of solution with
$\|\mu\|_{L^{\infty}(\Omega)}<\frac{1}{2}$ has been proven in [11], and
extended to $\|\mu\|_{L^{\infty}(\Omega)}<1$ in [32]. As far as the uniqueness
of solution is concerned, some results are known for problems similar to
(1.3), even though they require either the singularity to be milder or $\mu$
to be constant (see [8, 4, 16]). We will prove that uniqueness for problem
(1.3) holds provided $\|\mu\|_{L^{\infty}(\Omega)}<1$; the proof is based on a
comparison principle that we state in Section 2 and prove in Section 4.
Furthermore, we will show that the condition $\|\mu\|_{L^{\infty}(\Omega)}<1$
is natural by proving a nonexistence result provided $\mu>1$ in a neighborhood
of $\partial\Omega$; the proof follows the ideas in [18, Lemma 2.5]. In next
statement we summarize the new uniqueness and nonexistence results that we
prove about problem (1.3) and we include the previously known ([11, 32])
existence part for completeness.

######  Theorem 1.1.

Let $0\lneq h\in L^{q}(\Omega)$ for some $q>\frac{N}{2}$ and let $0\lneq\mu\in
L^{\infty}(\Omega)$. The following statements hold true:

  1. (1)

If $\|\mu\|_{L^{\infty}(\Omega)}<1$, then there exists a unique finite energy
solution to problem (1.3).

  2. (2)

If there exist an open domain $\omega\subset\subset\Omega$ and a constant
$\tau>1$ such that $\mu(x)\geq\tau$ and $h(x)=0$, both for a.e.
$x\in\Omega\setminus\omega,$ then problem (1.3) admits no solution.

We point out that existence and uniqueness results for problem (1.3) are known
for general nonnegative $\mu\in L^{\infty}(\Omega)$ (i.e. without assuming
that $\|\mu\|_{L^{\infty}(\Omega)}<1$) provided $h$ is locally bounded away
from zero (see [2] for the existence and [16] for the uniqueness). Thus, we
clarify that the condition $h\equiv 0$ near the boundary in item (2) of
Theorem 1.1 is also natural for having nonexistence of solution.

Once we have shown that problem (1.3) is well-posed, we will be able to define
a compact operator whose fixed points are solutions to ($P_{\lambda}$) (see
Section 4). A version of a result in [27] (see [21]) will assure the existence
of a fixed point of the operator.

As it is mandatory for fixed point theorems, we will prove the existence of a
priori estimates on the solutions to a problem related to ($P_{\lambda}$). To
this task, we will adapt the blow-up method due to [25]. Roughly speaking,
this technique consists of assuming by contradiction that there exists a
sequence of solutions whose norms blow up as $n$ tends to infinity. The
conclusion follows by passing to the limit in a problem satisfied by a certain
normalized sequence. In fact, the limit function is a solution to a problem
which, however, does not admit any solution by virtue of some Liouville type
result. Therefore, one gets a contradiction, so any sequence of solutions must
be bounded. In this context, the difficulties that we find are twofold.
Firstly, the normalized sequence, say $\\{v_{n}\\}$, satisfies an equation
having a lower order term of type $\frac{|\nabla
v_{n}|^{2}}{v_{n}+\delta_{n}}$, where $0\leq\delta_{n}\to 0$ as $n\to+\infty$.
If we aim to pass to the limit, then we need to find positive lower bounds on
$\\{v_{n}\\}$, otherwise the lower order term may blow up as $n\to+\infty$.
And lastly, we arrive to a limit problem, having a quadratic gradient lower
order term of the form $\frac{|\nabla v|^{2}}{v}$, for which nonexistence
Liouville type results are not known in the literature (a non-exhaustive list
of references for Liouville type results about problems depending on the
gradient is [23, 24, 30, 34, 37, 38, 39]).

We overcome the first of the difficulties by proving Hölder estimates in spite
of the singular quadratic term. The proof follows the ideas of [28], which
have been widely used for singular problems (see [2, 6, 15, 17, 22, 31, 32],
among others). We will show that these estimates yield in turn positive lower
estimates from below and this will be enough to pass to the limit. Regarding
the second difficulty, we observe that the limit equation does admit a
convenient change of unknown which gets rid of the gradient term, so that we
may apply classical Liouville type results (see Section 3 below).

We state here the main existence result for problem (1.1) in the case
$\gamma=1$.

######  Theorem 1.2.

Let $p>1$, $\gamma=1$, $\delta\geq 0$ and $\mu\in L^{\infty}(\Omega)$. The
following statements hold true:

  1. (1)

If $\mu\in C(\overline{\Omega})$ and there exist two real numbers
$\tau,\sigma$ such that $2\sigma-1<\tau\leq\sigma<\frac{2^{*}-1-p}{2^{*}-2}$
and $\tau\leq\mu(x)\leq\sigma$ for all $x\in\overline{\Omega}$, then there
exists at least a solution to (1.1) for every $\lambda>0$. If, in addition,
either $\mu\geq 0$ or $\delta>0$, then there exists at least a finite energy
solution to (1.1) for every $\lambda>0$.

  2. (2)

If $\delta=0$ and there exist an open domain $\omega\subset\subset\Omega$ and
a constant $\tau>1$ such that $\mu(x)\geq\tau$ for a.e.
$x\in\Omega\setminus\omega$, then problem (1.1) admits no solution for any
$\lambda>0$.

Note that, in the first item of Theorem 1.2, $\mu$ is allowed to change sign
unless $p\geq 2^{*}-1$, in which case $\mu$ is necessarily negative. We also
point out that the smallness condition $\sigma<\frac{2^{*}-1-p}{2^{*}-2}$ in
Theorem 1.2 is natural since, in fact, problem (1.1) has no bounded solutions
provided $\gamma=1$, $\delta=0$,
$\mu\equiv\text{constant}\in\left[\frac{2^{*}-1-p}{2^{*}-2},1\right)$ and
$\Omega$ is starshaped (see Remark 2.5 below). Moreover, we will show later
that, strengthening the smallness condition conveniently (in terms of $p,N$),
one may assume $\mu$ to be either continuous only in a neighborhood of
$\partial\Omega$, or merely bounded in $\Omega$. Also about the existence part
of the theorem, it is worth to point out that we need to control $\mu$ from
below in order to prove the Hölder estimates that we mentioned. However, if
$\mu$ is constant, i.e. $\sigma=\tau$, then the condition $2\sigma-1<\tau$
becomes $\sigma=\tau<1$, which means no restriction since
$\sigma<\frac{2^{*}-1-p}{2^{*}-2}<1$. On the other hand, we stress that the
case $\mu\equiv 1$ and $\delta=0$ remains unsolved, i.e. there are neither
existence nor nonexistence results about problem (1.1) for
$\gamma=1,\delta=0,\mu\equiv 1,p>1$ in the literature.

Next result shows that our approach allows also to prove existence for
$\gamma>1$ and for all $\lambda>0$.

######  Theorem 1.3.

Let $p>1$, $\gamma>1$, $\delta\geq 0$ and $\mu\in L^{\infty}(\Omega)$. The
following statements hold true:

  1. (1)

If $\delta>0$ and $\mu$ satisfies

(1.4)
$2\|\mu^{+}\|_{L^{\infty}(\Omega)}+\|\mu^{-}\|_{L^{\infty}(\Omega)}<\delta^{\gamma-1},$

then there exists at least a finite energy solution to (1.1) for every
$\lambda>0$.

  2. (2)

If $\delta=0$ and there exist an open domain $\omega\subset\subset\Omega$ and
a constant $\tau>0$ such that $\mu(x)\geq\tau$ for a.e.
$x\in\Omega\setminus\omega$, then problem (1.1) admits no solution for any
$\lambda>0$.

Notice that, if $\mu\equiv\text{constant}>0$, then condition (1.4) becomes
$\mu<\frac{\delta^{\gamma-1}}{2}$. It is known that such a smallness condition
is not needed if $\mu$ is a positive constant (see [35, Theorem 1.2]).
Therefore, we presume that the assumption (1.4) is technical, even though we
cannot avoid it since it is used to prove the Hölder estimates mentioned
above. Concerning the nonexistence statement for the singular case $\delta=0$,
it is proven again following closely [18, Lemma 2.5]. We stress that the fact
that the singularity is strong, namely $\gamma>1$, allows to take $\mu>0$ near
$\partial\Omega$ (in contrast to the case $\gamma=1$, for which $\mu>1$ near
the boundary was needed).

We will be able to go beyond Theorem 1.2 and Theorem 1.3 and deal with any
$\gamma>0$. Indeed, for $\gamma\in(0,1)$ and $\delta\geq 0$, we will show that
existence of solution holds for every $\lambda>0$ large enough. We will also
deal with $\gamma\geq 1,\delta>0$ and $\mu$ a bounded function with arbitrary
size, i.e. we remove the size restrictions on $\mu$ in Theorem 1.2 and Theorem
1.3 as long as $\delta>0$, even though $\lambda$ must be large. The statement
of the result is the following.

######  Theorem 1.4.

Let $p>1$, $\gamma>0$, $\delta\geq 0$ and $\mu\in L^{\infty}(\Omega)$. If
$\gamma\geq 1$, assume in addition that $\delta>0$. Assume also one of the
following two conditions:

  1. (1)

$p<\frac{N+1}{N-1}$,

  2. (2)

$p<2^{*}-1$ and $0\leq\mu\in C(\overline{\Omega})$.

Then, there exists $\lambda_{0}>0$ such that there exists at least a solution
$u_{\lambda}$ to (1.1) for every $\lambda>\lambda_{0}$ satisfying
$\lim_{\lambda\to+\infty}\|u_{\lambda}\|_{L^{\infty}(\Omega)}=0.$ Moreover, if
either $\mu\geq 0$ or $\delta>0$, then $u_{\lambda}$ is a finite energy
solution.

If $\gamma\in(0,1)$, last result is consistent with the known results for
$\mu\equiv\text{ constant}>0$ which assure nonexistence for $\lambda>0$ small
(see [35] for $\delta>0$ and [18] for $\delta=0$). On the contrary, as we
pointed out above, for $\gamma>1$ we would expect an existence result for
every $\lambda>0$ (even for $\lambda$ small) without size restrictions on
$\mu$. As far as the case $\gamma=1$ and $\delta>0$ is concerned, we ignore
whether an existence result for $\lambda>0$ small and general $\mu$ should be
expected or not. Two exceptions are the ranges $\mu<\frac{2^{*}-1-p}{2^{*}-2}$
and $\mu\geq p$, for which existence (see Theorem 1.2 above and [29]) and
nonexistence (see [35, 4]) for $\lambda>0$ small are known respectively. In
other words, the existence of solution to (1.1) for $\gamma=1$, $\delta>0$,
$\frac{2^{*}-1-p}{2^{*}-2}\leq\mu<p$ and $\lambda>0$ small remains as an open
problem, even for $\mu\equiv\text{ constant}>0$.

We organize the paper as follows. In Section 2 we introduce the conditions on
$f,g$ and the statements of the general results about problem ($P_{\lambda}$).
Section 3 is devoted to proving some Liouville type results as well as several
propositions that provide the estimates via the blow-up method. In Section 4
we prove the results that we state in Section 2 and in the Introduction.
Finally, in the Appendix we gather some technical results that are required
throughout the paper.

Acknowledgments. The problems considered in this work have been proposed by T.
Leonori. The research was initiated with his collaboration during one week in
Granada and another week in Rome, when most of the ideas contained here
emerged. Moreover, for the conclusion of the paper, the interesting ideas,
comments and corrections by J. Carmona have also supposed a more than
remarkable contribution. In any case, their supervision and support have been
essential during the research period. This is why the author wants to warmly
thank both, collaborators and friends, for being examples of altruism to
follow in this competitive and sometimes hostile world of research. The author
wants to thank also D. Ruiz and C. De Coster for their kind and useful
suggestions and corrections.

##  2\. Hypotheses and main results

Let us consider a continuous function $f:[0,+\infty)\to[0,+\infty)$ and a
Carathéodory function $g:\Omega\times(0,+\infty)\to\mathbb{R}$. We will assume
throughout this paper that $g$ satisfies the following condition:

(2.1) $\exists g_{0}\in C((0,+\infty)):|g(x,s)|\leq g_{0}(s)\quad\text{a.e.
}x\in\Omega,\,\,\forall s>0.$

Last hypothesis is essentially the minimal condition that $g$ must satisfy so
that the weak formulation of ($P_{\lambda}$) is well defined:

######  Definition 2.1.

Let $f:[0,+\infty)\to[0,+\infty)$ be a continuous function and
$g:\Omega\times(0,+\infty)\to\mathbb{R}$ be a Carathéodory function satisfying
(2.1). A _solution_ to ($P_{\lambda}$) is a function $u\in H^{1}_{\mbox{\tiny
loc}}(\Omega)\cap L^{\infty}(\Omega)$ such that

$\begin{cases}&\displaystyle\forall\omega\subset\subset\Omega,\,\,\exists
c>0:\quad u(x)\geq c\text{ a.e. }x\in\omega,\\\ &\exists\gamma>0:\quad
u^{\gamma}\in H_{0}^{1}(\Omega),\text{ and }\\\
&\displaystyle\int_{\Omega}\nabla u\nabla\phi+\int_{\Omega}g(x,u)|\nabla
u|^{2}\phi=\int_{\Omega}f(u)\phi\quad\forall\phi\in
C_{c}^{1}(\Omega).\end{cases}$

Besides, we will say that $u$ is a _finite energy solution_ to ($P_{\lambda}$)
if it is a solution to ($P_{\lambda}$) with $\gamma=1$, i.e. if $u\in
H_{0}^{1}(\Omega)$.

######  Remark 2.2.

It can be proven by following [15, Appendix] that, in the previous definition,
one can take test functions $\phi\in H_{0}^{1}(\Omega)\cap L^{\infty}(\Omega)$
with compact support. Moreover, if $u$ is a finite energy solution, then one
can take any test function belonging to $H_{0}^{1}(\Omega)\cap
L^{\infty}(\Omega)$.

Let us fix $\sigma\in\mathbb{R}$. The following condition, stronger than
(2.1), means a more precise control on $g$ from above:

($g_{1}$) $\exists g_{0}\in C((0,+\infty)):\quad-g_{0}(s)\leq
g(x,s)\leq\frac{\sigma}{s}\quad\text{a.e. }x\in\Omega,\,\,\forall s>0.$

Let us consider also the monotonicity condition:

($g_{\nearrow}$) $s\mapsto sg(x,s)\,\,\text{ is nondecreasing for a.e.
}x\in\Omega.$

We state next a comparison principle that will be the key for proving the
uniqueness part of Theorem 1.1. The proof follows essentially the arguments in
[15], which in turn are inspired by [5].

######  Theorem 2.3.

Let $0\lneq h\in L^{1}_{\mbox{\tiny loc}}(\Omega)$ and let
$g:\Omega\times(0,+\infty)\to\mathbb{R}$ be a Carathéodory function satisfying
($g_{\nearrow}$) and ($g_{1}$) for some $\sigma\in(0,1)$. Let $u,v\in
C(\Omega)\cap W^{1,N}_{\mbox{\tiny loc}}(\Omega)$, with $u,v>0$ in $\Omega$,
be such that

(2.2) $\displaystyle\int_{\Omega}\nabla u\nabla\phi+\int_{\Omega}g(x,u)|\nabla
u|^{2}\phi\leq\int_{\Omega}h(x)\phi\quad\text{ and }$ (2.3)
$\displaystyle\int_{\Omega}\nabla v\nabla\phi+\int_{\Omega}g(x,v)|\nabla
v|^{2}\phi\geq\int_{\Omega}h(x)\phi$

for every $0\leq\phi\in H_{0}^{1}(\Omega)\cap L^{\infty}(\Omega)$ with compact
support. Suppose also that the following boundary condition holds:

(2.4) $\limsup_{x\to x_{0}}(u(x)^{1-\sigma}-v(x)^{1-\sigma})\leq 0\quad\forall
x_{0}\in\partial\Omega.$

Then, $u\leq v$ in $\Omega$.

We will present below four existence theorems that represent the core of this
work. The proofs are based on a fixed point result. The required a priori
estimates are obtained via a blow-up method (see Section 3).

Let us fix $p>1$ and $\delta\geq 0$. We establish now a growth condition and
two limit conditions on $f$ which make it behave like a superlinear power:

($f_{*}$) $\exists a\geq 1:\quad s^{p}\leq f(s)\leq
a(s+\delta)^{p}\quad\forall s\geq 0.$

($f_{\infty}$) $\exists
L\in(0,+\infty):\quad\lim_{s\to+\infty}\frac{f(s)}{s^{p}}=L.$

($f_{0}$) $\lim_{s\to 0}\frac{f(s)}{s}=0.$

Observe that ($f_{0}$) implies in particular that $f(0)=0$.

On the other hand, for fixed $\delta\geq 0$ and $\sigma,\tau\in\mathbb{R}$, we
set the following growth restriction on $g$ which will allow us to prove
certain Hölder estimates (see the Appendix below):

($g_{*}$) $\begin{cases}2\sigma-1<\tau\leq\sigma<1,\\\
\tau\leq(s+\delta)g(x,s)\leq\sigma\quad\text{a.e. }x\in\Omega,\,\,\forall
s>0.\end{cases}$

We will also need the following limit condition on $g$ at infinity for $p>1$:

($g_{\infty}$) $\begin{cases}&\exists\mu\in
C(\overline{\Omega}):\max_{x\in\overline{\Omega}}\mu(x)<\frac{2^{*}-1-p}{2^{*}-2},\\\
&\lim_{s\to+\infty}\|sg(\cdot,s)-\mu\|_{L^{\infty}(\Omega)}=0.\end{cases}$

A simple limit condition on $g$ at zero will be required too:

($g_{0}$) $\exists\lim_{s\to 0}(s+\delta)g(x,s)\quad\text{a.e. }x\in\Omega.$

Observe that, if $\delta>0$, then ($g_{0}$) is equivalent to $g(x,\cdot)$
being continuous at $s=0$ for a.e. $x\in\Omega$.

We are ready to state our first general existence result:

######  Theorem 2.4.

Let $f:[0,+\infty)\to[0,+\infty)$ be a continuous function and
$g:\Omega\times(0,+\infty)\to\mathbb{R}$ be a Carathéodory function. For
$\delta\geq 0$, $\sigma,\tau\in\mathbb{R}$ and $p>1$, assume that $f$
satisfies ($f_{*}$), ($f_{\infty}$) and ($f_{0}$), and that $g$ satisfies
($g_{*}$),($g_{\infty}$) and ($g_{0}$). Then, there exists at least a solution
to ($P_{\lambda}$) for every $\lambda>0$. If, in addition, either $g\geq 0$ or
$g(x,\cdot)$ is continuous at $s=0$ a.e. $x\in\Omega$, then there exists at
least a finite energy solution to ($P_{\lambda}$) for every $\lambda>0$.

######  Remark 2.5.

We point out that the smallness condition
$\max_{x\in\overline{\Omega}}\mu(x)<\frac{2^{*}-1-p}{2^{*}-2}$ in Theorem 2.4
coming from condition ($g_{\infty}$) is necessary for the existence of
solutions to ($P_{\lambda}$), at least in the model case of $\Omega$ a
satarshaped domain, $f(s)=\lambda s^{p}$ for some $\lambda>0$,
$p\in(1,2^{*}-1)$, and $sg(x,s)\equiv\mu$ for some constant $\mu<1$. Indeed,
assume by contradiction that $\mu\in\left[\frac{2^{*}-1-p}{2^{*}-2},1\right)$
and that $0<u\in H_{0}^{1}(\Omega)\cap L^{\infty}(\Omega)$ satisfies $-\Delta
u+\mu\frac{|\nabla u|^{2}}{u}=\lambda u^{p}$ in $\Omega$. Then, $v=cu^{1-\mu}$
satisfies $-\Delta v=v^{\frac{p-\mu}{1-\mu}}$ in $\Omega$ for some $c>0$.
Therefore, since $\frac{p-\mu}{1-\mu}\geq 2^{*}-1$ and $\Omega$ is starshaped,
the well-known Pohozaev’s identity (see [36]) yields a contradiction.

Let us fix again $p>1$. Next hypothesis is a limit condition on $g$ at
infinity weaker than ($g_{\infty}$) as it is only required to hold in a
neighborhood of $\partial\Omega$.

($\widetilde{g_{\infty}}$)
$\begin{cases}&\exists\omega\subset\subset\Omega,\,\,\mu\in
C(\overline{\Omega\setminus\omega}):\max_{x\in\overline{\Omega\setminus\omega}}\mu(x)<\frac{2^{*}-1-p}{2^{*}-2},\\\
&\lim_{s\to+\infty}\|sg(\cdot,s)-\mu\|_{L^{\infty}(\Omega\setminus\omega)}=0.\end{cases}$

The following result shows that, assuming a stronger control on $g$ from
above, one can relax the limit condition on $g$ at infinity to obtain a
solution to ($P_{\lambda}$).

######  Theorem 2.6.

Let $f:[0,+\infty)\to[0,+\infty)$ be a continuous function and
$g:\Omega\times(0,+\infty)\to\mathbb{R}$ be a Carathéodory function. For
$\delta\geq 0$, $\sigma,\tau\in\mathbb{R}$ and $p>1$, assume that $f$
satisfies ($f_{*}$), ($f_{\infty}$) and ($f_{0}$), and that $g$ satisfies
($g_{*}$), ($\widetilde{g_{\infty}}$) and ($g_{0}$). Assume in addition that
$\sigma\leq\frac{N-(N-2)p}{2}$. Then, there exists at least a solution to
($P_{\lambda}$) for every $\lambda>0$. If, in addition, either $g\geq 0$ or
$g(x,\cdot)$ is continuous at $s=0$ a.e. $x\in\Omega$, then there exists at
least a finite energy solution to ($P_{\lambda}$) for every $\lambda>0$.

Next theorem shows that the limit conditions at infinity are actually not
essential to obtain solutions to ($P_{\lambda}$). In return, one has to assume
an even stronger control on $g$ from above.

######  Theorem 2.7.

Let $f:[0,+\infty)\to[0,+\infty)$ be a continuous function and
$g:\Omega\times(0,+\infty)\to\mathbb{R}$ be a Carathéodory function. For
$\delta\geq 0$, $\sigma,\tau\in\mathbb{R}$ and $p>1$, assume that $f$
satisfies ($f_{*}$) and ($f_{0}$), and that $g$ satisfies ($g_{*}$) and
($g_{0}$). Assume in addition that $\sigma\leq\frac{N+1-(N-1)p}{2}$. Then,
there exists at least a solution to ($P_{\lambda}$) for every $\lambda>0$.
Moreover, the following holds.

  1. (1)

If either $g\geq 0$ or $g(x,\cdot)$ is continuous at $s=0$ a.e. $x\in\Omega$,
then there exists at least a finite energy solution to ($P_{\lambda}$) for
every $\lambda>0$.

  2. (2)

If $\delta=0$, then there exists $C>0$ such that, for every $\lambda>0$ and
for every solution $u$ to ($P_{\lambda}$), the estimate
$\lambda^{\frac{1}{p-1}}\|u\|_{L^{\infty}(\Omega)}\leq C$ holds.

Let $s_{0}\in(0,1)$, $\sigma,\tau\in\mathbb{R}$ and $p>1$. Consider the
following growth conditions near zero:

($\widetilde{f_{*}}$) $\exists a\geq 1:\quad s^{p}\leq f(s)\leq
as^{p}\quad\forall s\in(0,s_{0}).$

($\widetilde{g_{*}}$) $\begin{cases}2\sigma-1<\tau\leq\sigma<1,\\\ \tau\leq
sg(x,s)\leq\sigma\quad\text{a.e. }x\in\Omega,\,\,\forall
s\in(0,s_{0}).\end{cases}$

Notice that condition ($\widetilde{f_{*}}$) implies that $\lim_{s\to
0}\frac{f(s)}{s}=0$ and, in particular, $f(0)=0$. On the other hand, we point
out that, if $\tau>0$, then condition ($\widetilde{g_{*}}$) forces $g$ to be
singular at $s=0$. However, if ($\widetilde{g_{*}}$) holds for some $\tau<0$
and $\sigma>0$, then $g$ may be rather general at $s=0$ since it can be
continuous as well as unbounded from above and from below.

Our next existence theorem will require neither growth nor limit conditions at
infinity on $f$ and $g$. In exchange, the existence of solution holds only for
every $\lambda>0$ large enough.

######  Theorem 2.8.

Let $f:[0,+\infty)\to[0,+\infty)$ be a continuous function and
$g:\Omega\times(0,+\infty)\to\mathbb{R}$ be a Carathéodory function. For
$s_{0}\in(0,1)$, $\sigma,\tau\in\mathbb{R}$ and $p>1$, assume that $f$
satisfies ($\widetilde{f_{*}}$) and that $g$ satisfies ($\widetilde{g_{*}}$).
Assume in addition that there exists $\lim_{s\to 0}sg(x,s)$ for a.e.
$x\in\Omega$, and also that $\sigma\leq\frac{N+1-(N-1)p}{2}$. Then, there
exists $\lambda_{0}>0$ such that there exists at least a solution
$u_{\lambda}$ to ($P_{\lambda}$) for every $\lambda>\lambda_{0}$ satisfying
$\lim_{\lambda\to+\infty}\|u_{\lambda}\|_{L^{\infty}(\Omega)}=0$. If, in
addition, either $g\geq 0$ or $g(x,\cdot)$ is continuous at $s=0$ a.e.
$x\in\Omega$, then $u_{\lambda}$ is a finite energy solution.

Regarding Theorem 2.8, observe that, if $p>\frac{N+1}{N-1}$, then $\sigma<0$,
so $g(x,s)<0$ for $s$ near zero. Thus, for instance, the particular case
$g(x,s)=\frac{\mu}{s^{\gamma}}$ and $f(s)=s^{p}$ is not covered by Theorem 2.8
if $\gamma\in(0,1)$, $p\in\left[\frac{N+1}{N-1},2^{*}-1\right)$ and $\mu>0$ is
constant. The result that comes next does cover this particular case. In fact,
for $p\in(1,2^{*}-1)$, next theorem allows to consider general continuous
functions $g$ satisfying

(G) $\exists G\in L^{1}((0,s_{0})):\quad 0\leq g(x,s)\leq
G(s)\quad\forall(x,s)\in\Omega\times(0,s_{0}).$

($\widetilde{g_{0}}$) $\begin{cases}&\exists\mu_{0}\in
C(\overline{\Omega}):\max_{x\in\overline{\Omega}}\mu_{0}(x)<\frac{2^{*}-1-p}{2^{*}-2},\\\
&\lim_{s\to 0}\|sg(\cdot,s)-\mu_{0}\|_{L^{\infty}(\Omega)}=0.\end{cases}$

######  Remark 2.9.

Notice that (G) and ($\widetilde{g_{0}}$) imply that $\lim_{s\to
0}\|sg(\cdot,s)\|_{L^{\infty}(\Omega)}=0$, i.e. $\mu_{0}\equiv 0$ in condition
($\widetilde{g_{0}}$). Indeed, we first observe that the fact that $g\geq 0$
implies that

$\mu_{0}(x)\geq\mu_{0}(x)-sg(x,s)\geq-\|sg(\cdot,s)-\mu_{0}\|_{L^{\infty}(\Omega)}\quad\forall(x,s)\in\Omega\times(0,s_{0}).$

Passing to the limit as $s\to 0$ leads to $\mu_{0}\geq 0$. On the other hand,
we know that for every $\varepsilon>0$ there exists $s_{\varepsilon}\in(0,1)$
such that

$sg(x,s)\geq\mu_{0}(x)-\varepsilon\quad\forall(x,s)\in\Omega\times(0,s_{\varepsilon}).$

Let us assume by contradiction that there exists $x_{0}\in\Omega$ such that
$\mu_{0}(x_{0})>0$. Then we choose $\varepsilon=\frac{\mu_{0}(x_{0})}{2}$ and
we obtain

$G(s)\geq g(x_{0},s)\geq\frac{\varepsilon}{s}\quad\forall
s\in(0,s_{\varepsilon}).$

This contradicts hypothesis (G).

The result reads as follows:

######  Theorem 2.10.

Let $g:\overline{\Omega}\times(0,+\infty)\to\mathbb{R}$ be a continuous
function. For $s_{0}\in(0,1)$ and $p\in(1,2^{*}-1)$, assume that $g$ satisfies
(G) and ($\widetilde{g_{0}}$). Let us also consider the function
$f:[0,+\infty)\to[0+\infty)$ defined by $f(s)=s^{p}$ for all $s\geq 0$. Then,
there exists $\lambda_{0}>0$ such that there exists at least a finite energy
solution $u_{\lambda}$ to ($P_{\lambda}$) for every $\lambda>\lambda_{0}$
satisfying $\lim_{\lambda\to+\infty}\|u_{\lambda}\|_{L^{\infty}(\Omega)}=0$.

######  Remark 2.11.

We point out that Theorem 2.10 is valid for a very wide class of functions
$g$. For instance, if

$g(x,s)=\mu(x)h(s)\quad\forall(x,s)\in\Omega\times[0,+\infty),$

where $0\lneq\mu\in C(\overline{\Omega})$ and $h:[0,+\infty)\to[0,+\infty)$ is
continuous (also at $s=0$), then $g$ satisfies (G) and ($\widetilde{g_{0}}$).
On the other hand, a prototypical example of function $g$ singular at $s=0$
satisfying the conditions (G) and ($\widetilde{g_{0}}$) is

$g(x,s)=\frac{\mu(x)}{s^{\gamma}}\quad\forall(x,s)\in\Omega\times(0,+\infty)$

where $\gamma\in(0,1)$, $0\lneq\mu\in C(\overline{\Omega})$.

The last theorem of the section will be concerned with the following problem:

($H$) $\begin{cases}-\Delta u+g(x,u)|\nabla u|^{2}=h(x),&x\in\Omega,\\\
u>0,&x\in\Omega,\\\ u=0,&x\in\partial\Omega.\end{cases}$

The result provides a necessary condition for the existence of solution to
($H$). In order to prove it, we will assume, for fixed $\tau>0$, that $g(x,u)$
is sufficiently large near $\partial\Omega$ in the sense of the following
condition:

($g_{2}$) $\exists\omega\subset\subset\Omega,\,\,s_{0}\in(0,1):\quad
sg(x,s)\geq\tau\quad\text{a.e. }x\in\Omega\setminus\omega,\,\,\forall
s\in(0,s_{0}).$

The theorem states the following:

######  Theorem 2.12.

Let $h\in L^{1}(\Omega)$ and let $g:\Omega\times(0,+\infty)\to\mathbb{R}$ be a
Carathéodory function satisfying (2.1) and ($g_{2}$) for some $\tau>1$. Then,
every solution $u$ to ($H$) satisfies

$\int_{\Omega}\frac{|h(x)|}{u}=+\infty.$

##  3\. A priori estimates

We begin the section with two Liouville type results which will be the key
points for proving a priori estimates. The first of them is the following.

######  Lemma 3.1.

Let $p>1$ and let $h:(0,+\infty)\to[0,+\infty)$ be a continuous function
satisfying

$sh(s)\leq\sigma\quad\forall s>0$

for some $\sigma\in(0,1)$. Let us assume that there exists $\alpha>0$ such
that the function $\psi:[0,+\infty)\to[0,+\infty)$, defined by

$\psi(s)=\int_{0}^{s}e^{-\int_{\alpha}^{t}h(r)dr}dt\quad\forall s\geq 0,$

satisfies that

$s\mapsto\frac{\psi^{\prime}(s)s^{p}}{\psi(s)^{2^{*}-1}}\quad\text{ is
decreasing for all }s>0.$

Then, the following problem

(3.1) $\begin{cases}-\Delta u+h(u)|\nabla u|^{2}=u^{p},&x\in X,\\\ u>0,&x\in
X,\\\ u=0,&x\in\partial X,\end{cases}$

admits no solutions in $H^{1}_{\mbox{\tiny loc}}(X)\cap C(\overline{X})$,
where $X$ denotes either $\mathbb{R}^{N}$ or $\mathbb{R}^{N}_{+}$.

######  Remark 3.2.

We stress that Lemma 3.1 includes the particular case $h(s)=\frac{\sigma}{s}$
with $\sigma<\frac{2^{*}-1-p}{2^{*}-2}$. Indeed, in this case, it is easy to
check that
$\frac{\psi^{\prime}(s)s^{p}}{\psi(s)^{2^{*}-1}}=cs^{p-\sigma-(1-\sigma)(2^{*}-1)}$
for some constant $c>0$. Thus, it is decreasing if, and only if,
$\sigma<\frac{2^{*}-1-p}{2^{*}-2}$. Moreover, $\frac{2^{*}-1-p}{2^{*}-2}<1$,
so $sh(s)=\sigma<1$ for all $s>0$.

###### Proof of Lemma 3.1.

Arguing by contradiction, assume that there exists a solution $u\in
H^{1}_{\mbox{\tiny loc}}(X)\cap C(\overline{X})$ to (3.1). Straightforward
computations imply that $v=\psi(u)\in H^{1}_{\mbox{\tiny loc}}(X)\cap
C(\overline{X})$ satisfies

(3.2) $\begin{cases}-\Delta v=\varphi(v),&x\in X,\\\ v>0,&x\in X,\\\
v=0,&x\in\partial X,\end{cases}$

where $\varphi(t)=\psi^{\prime}(\psi^{-1}(t))\psi^{-1}(t)^{p}$ for all
$t\in\text{Im}(\psi)$. Moreover, classical elliptic regularity theory implies
that $v\in C^{2}(X)$. We claim now that, actually, problem (3.2) admits no
solutions $v\in C^{2}(X)\cap C(\overline{X})$.

Indeed, one can easily deduce that

$\psi(s)\geq\frac{\alpha^{\sigma}\left(s^{1-\sigma}-\alpha^{1-\sigma}\right)}{1-\sigma}\quad\forall
s>\alpha.$

Hence, $\lim_{s\to+\infty}\psi(s)=+\infty$. In consequence, the function
$\varphi$ is defined in $[0,+\infty)$. On the other hand, it is clear that the
function $t\mapsto\frac{\varphi(t)}{t^{2^{*}-1}}$ is decreasing for all $t>0$.
Therefore, in case $X=\mathbb{R}^{N}$, the claim follows from [9, Theorem 3].

Notice also that $\lim_{t\to 0}\frac{\varphi(t)}{t}=0$ and $\varphi(t)\geq
c\psi^{-1}(t)^{p-\sigma}$ for every $t>0$ and for some $c>0$. Hence, the claim
also holds true in case $X=\mathbb{R}^{N}_{+}$ by virtue of [20, Theorem 1.3].

In any case, we have arrived to a contradiction. The proof is now completed. ∎

We present here the second Liouville type result which is valid for
supersolutions.

######  Lemma 3.3.

Let $p>1$ and $\sigma\leq\frac{N-(N-2)p}{2}$. Then, the following problem

(3.3) $\begin{cases}\displaystyle-\Delta u+\sigma\frac{|\nabla
u|^{2}}{u}=u^{p},&x\in X,\\\ u>0,&x\in X,\\\ u=0,&x\in\partial X,\end{cases}$

admits no supersolutions in $H^{1}_{\mbox{\tiny loc}}(X)\cap C(\overline{X})$
provided $X=\mathbb{R}^{N}$ . On the other hand, if we assume that
$\sigma\leq\frac{N+1-(N-1)p}{2}$, then problem (3.3) with
$X=\mathbb{R}^{N}_{+}$ admits no supersolutions in $H^{1}_{\mbox{\tiny
loc}}(X)\cap C(\overline{X})$.

######  Remark 3.4.

Note that
$\frac{N+1-(N-1)p}{2}<\frac{N-(N-2)p}{2}<\frac{2^{*}-1-p}{2^{*}-2}<1$, so the
smallness conditions on $\sigma$ in Remark 3.2 and in Lemma 3.3 are gradually
more restrictive. We also stress that such conditions on $\sigma$ in Lemma 3.3
are sharp. Indeed, if $\sigma>\frac{N-(N-2)p}{2}$ (resp.
$\sigma>\frac{N+1-(N-1)p}{2}$), then one can find explicit supersolutions to
(3.3) for $X=\mathbb{R}^{N}$, see [33] (resp. $X=\mathbb{R}^{N}_{+}$, see
[10]).

###### Proof of Lemma 3.3.

Reasoning by contradiction, assume that there exists $u\in H^{1}_{\mbox{\tiny
loc}}(X)\cap C(\overline{X})$ a supersolution to (3.3). Then, there is a
constant $c>0$ such that $v=cu^{1-\sigma}\in H^{1}_{\mbox{\tiny loc}}(X)\cap
C(\overline{X})$ is a supersolution to

$\begin{cases}-\Delta v=v^{\frac{p-\sigma}{1-\sigma}},&x\in X,\\\ v>0,&x\in
X,\\\ v=0,&x\in\partial X.\end{cases}$

Hence, if $\sigma\leq\frac{N-(N-2)p}{2}$, then
$\frac{p-\sigma}{1-\sigma}\leq\frac{N}{N-2}$, so in case $X=\mathbb{R}^{N}$ we
arrived to a contraction with [33, Theorem 2.1]. On the other hand, if
$\sigma<\frac{N+1-(N-1)p}{2}$, then
$\frac{p-\sigma}{1-\sigma}<\frac{N+1}{N-1}$, so we have again a contradiction
with [10, Theorem 3.1]. ∎

Let $t\geq 0$, $\lambda>0$, $f:[0,+\infty)\to[0,+\infty)$ be a continuous
function satisfying ($f_{*}$) for some $p>1$ and
$g:\Omega\times(0,+\infty)\to[0,+\infty)$ be a Carathéodory function
satisfying ($g_{1}$) for some $\sigma\in(0,1)$. From now and up to the end of
the section, we will restrict ourselves to $g\geq 0$ and we will also impose
diverse upper bounds on $p$. However, we will show in Section 4 that, in most
cases, those restrictions can be relaxed for proving existence of solution.

Let us consider the following auxiliary problem:

($P^{t}$) $\begin{cases}\displaystyle-\Delta u+g(x,u)|\nabla u|^{2}=\lambda
f(u)+tu^{\sigma},&x\in\Omega,\\\ u>0,&x\in\Omega,\\\
u=0,&x\in\partial\Omega.\end{cases}$

Note that, in the term $tu^{\sigma}$, the exponent $\sigma$ is the same number
that appears in condition ($g_{1}$). We will derive a priori estimates on the
solutions to ($P^{t}$) which will provide the existence of solution to
($P_{\lambda}$).

Next proposition gives an a priori estimate on the parameter $t$ in problem
($P^{t}$).

######  Proposition 3.5.

Let $\lambda>0$, $f:[0,+\infty)\to[0,+\infty)$ be a continuous function and
$g:\Omega\times(0,+\infty)\to[0,+\infty)$ be a Carathéodory function. For
$p>1$ and $\sigma\in(0,1)$, assume that $f$ satisfies ($f_{*}$) and $g$
satisfies ($g_{1}$). Then, there exists $t_{0}>0$ such that problem ($P^{t}$)
admits no solution for any $t>t_{0}$.

###### Proof.

Let $u$ be a solution to ($P^{t}$) for some $t>0$. For a fixed smooth open set
$\omega\subset\subset\Omega$, let $\lambda_{1}$ be the principal eigenvalue to
the homogeneous Dirichlet eigenvalue problem in $\omega$, and let
$\varphi_{1}$ be any positive associated eigenfunction, i.e. $\lambda_{1}$ and
$\varphi_{1}$ satisfy

$\begin{cases}-\Delta\varphi_{1}=\lambda_{1}\varphi_{1},&x\in\omega,\\\
\varphi_{1}>0,&x\in\omega\\\ \varphi_{1}=0,&x\in\partial\omega.\end{cases}$

If we extend $\varphi_{1}\equiv 0$ in $\Omega\setminus\omega$, then the
function $\phi=\frac{\varphi_{1}}{u^{\sigma}}$ belongs to
$H_{0}^{1}(\Omega)\cap L^{\infty}(\Omega)$ and has compact support. Taking
$\phi$ as test function in ($P^{t}$) (which is allowed by virtue of Remark
2.2) we obtain

(3.4) $\int_{\Omega}\frac{\nabla
u\nabla\varphi_{1}}{u^{\sigma}}-\sigma\int_{\Omega}\frac{\varphi_{1}|\nabla
u|^{2}}{u^{\sigma+1}}+\int_{\Omega}g(x,u)\frac{\varphi_{1}|\nabla
u|^{2}}{u^{\sigma}}=\lambda\int_{\Omega}\frac{f(u)}{u^{\sigma}}\varphi_{1}+t\int_{\Omega}\varphi_{1}.$

On the one hand, it is clear by ($g_{1}$) that

$-\sigma\int_{\Omega}\frac{\varphi_{1}|\nabla
u|^{2}}{u^{\sigma+1}}+\int_{\Omega}g(x,u)\frac{\varphi_{1}|\nabla
u|^{2}}{u^{\sigma}}\leq 0.$

Thus, using also ($f_{*}$) we deduce from (3.4) that

(3.5)
$t\int_{\Omega}\varphi_{1}+\lambda\int_{\Omega}u^{p-\sigma}\varphi_{1}\leq\int_{\Omega}\frac{\nabla
u\nabla\varphi_{1}}{u^{\sigma}}.$

On the other hand, let us denote as $\nu$ the exterior normal unit vector to
$\partial\omega$. Then, Hopf’s lemma implies that $\nu\nabla\varphi_{1}<0$ on
$\partial\omega$. Hence, integration by parts in
$\int_{\omega}(-\Delta\varphi_{1})u^{1-\sigma}$ and Young’s inequality yield

$\int_{\Omega}\frac{\nabla
u\nabla\varphi_{1}}{u^{\sigma}}=\lambda_{1}\int_{\omega}\frac{\varphi_{1}u^{1-\sigma}}{1-\sigma}+\int_{\partial\omega}\frac{u^{1-\sigma}}{1-\sigma}\nu\nabla\varphi_{1}<\lambda_{1}\int_{\omega}\frac{\varphi_{1}u^{1-\sigma}}{1-\sigma}\leq\frac{\lambda}{2}\int_{\Omega}u^{p-\sigma}\varphi_{1}+C.$

In sum, from (3.5) we deduce that $t\leq t_{0}$ for some $t_{0}>0$, as we
wanted to prove. ∎

Next lemma provides a summability property that the solutions to ($P^{t}$)
satisfy. The interesting point is that such a property becomes better as the
singularity of $g$ becomes stronger.

######  Lemma 3.6.

Let $\lambda>0$, $\sigma\in(0,1)$, $f:[0,+\infty)\to[0,+\infty)$ be a
continuous function and $g:\Omega\times(0,+\infty)\to[0,+\infty)$ be a
Carathéodory function satisfying (2.1). For some $a>0$, $p\geq 1$ and
$\tau\in[0,1)$, assume that

$\displaystyle f(s)$ $\displaystyle\leq as^{p}\quad\forall s\geq 0,$
$\displaystyle sg(x,s)$ $\displaystyle\geq\tau\quad\text{a.e.
}x\in\Omega,\,\,\forall s>0.$

Let $u\in H^{1}_{\mbox{\tiny loc}}(\Omega)\cap L^{\infty}(\Omega)$ be a
solution to ($P^{t}$) for some $t\geq 0$. Then, $u^{\gamma}\in
H_{0}^{1}(\Omega)$ for every $\gamma>\frac{\max\\{1-\sigma,1-\tau\\}}{2}$.

###### Proof.

First of all, recall that Lemma 5.2 in the Appendix below implies that $u\in
C(\overline{\Omega})$. Now, for any $\alpha>0$, let us consider the function
$v=u^{\alpha}\in H^{1}_{\mbox{\tiny loc}}(\Omega)\cap C(\overline{\Omega})$.
It is easy to see that $v$ satisfies

$\displaystyle-\Delta v$
$\displaystyle=\frac{(1-\alpha-v^{\frac{1}{\alpha}}g(x,v^{\frac{1}{\alpha}}))|\nabla
v|^{2}}{\alpha v}+\alpha\lambda
v^{\frac{\alpha-1}{\alpha}}f(v^{\frac{1}{\alpha}})+\alpha
tv^{\frac{\alpha-1+\sigma}{\alpha}}$
$\displaystyle\leq\frac{(1-\alpha-\tau)|\nabla v|^{2}}{\alpha v}+\alpha\lambda
av^{\frac{\alpha-1+p}{\alpha}}+\alpha
tv^{\frac{\alpha-1+\sigma}{\alpha}},\quad x\in\Omega.$

Moreover, if we take $\alpha\geq\max\\{1-\sigma,1-\tau\\}$, then

(3.6) $-\Delta v\leq C,\quad x\in\Omega,$

for some constant $C>0$.

For every $\varepsilon,\beta>0$, let us now consider the function
$\phi=(v^{\beta}-\varepsilon)^{+}$. It is clear that $\phi\in
H^{1}_{\mbox{\tiny loc}}(\Omega)\cap C(\overline{\Omega})$. Furthermore, the
continuity of $v$ up to $\partial\Omega$ implies that $\phi$ has compact
support in $\Omega$. In sum, $\phi\in H_{0}^{1}(\Omega)\cap
C(\overline{\Omega})$ and has compact support. Therefore, even though $v$
might not belong to $H_{0}^{1}(\Omega)$, it follows from Remark 2.2 that one
can take $\phi$ as test function in (3.6). Thus, we obtain

$\int_{\Omega}\chi_{\\{v^{\beta}\geq\varepsilon\\}}\frac{|\nabla
v|^{2}}{v^{1-\beta}}\leq C,$

for another constant $C>0$ independent of $\varepsilon$. Now we let
$\varepsilon$ tend to zero and, by virtue of Fatou’s lemma, we deduce

$\int_{\Omega}\frac{|\nabla v|^{2}}{v^{1-\beta}}\leq C.$

Taking into account that

$\frac{|\nabla v|^{2}}{v^{1-\beta}}=C\left|\nabla
v^{\frac{1+\beta}{2}}\right|^{2}=C\left|\nabla
u^{\alpha\frac{1+\beta}{2}}\right|^{2},$

then we have proved that $u^{\alpha\frac{1+\beta}{2}}\in H_{0}^{1}(\Omega)$
for every $\alpha\geq\max\\{1-\sigma,1-\tau\\}$ and every $\beta>0$. This
proves the result. ∎

In the following result we prove a priori estimates on the solutions to
($P^{t}$) if $f$ satisfies ($f_{*}$) and ($f_{\infty}$) and $g$ satisfies
($g_{*}$) and ($g_{\infty}$). In the proof we adapt the blow-up method due to
[25].

######  Proposition 3.7.

Let $\lambda>0$, $f:[0,+\infty)\to[0,+\infty)$ be a continuous function and
$g:\Omega\times(0,+\infty)\to[0,+\infty)$ be a Carathédory function. For
$\delta,\tau\geq 0$, $\sigma>0$ and $p\in(1,2^{*}-1)$, assume that $f$
satisfies ($f_{*}$) and ($f_{\infty}$), and that $g$ satisfies ($g_{*}$) and
($g_{\infty}$). Then, there exists $C>0$ such that
$\|u\|_{L^{\infty}(\Omega)}\leq C$ for every solution $u$ to ($P^{t}$) for all
$t\in[0,t_{0}]$, where $t_{0}>0$ is given by Proposition 3.5.

###### Proof.

Reasoning by contradiction, we assume that there exist two sequences
$\\{t_{n}\\}\subset[0,t_{0}]$ and $\\{u_{n}\\}$ such that $u_{n}$ is a
solution to $(P^{t_{n}})$ for all $n$ and
$\|u_{n}\|_{L^{\infty}(\Omega)}\to+\infty$ as $n\to+\infty$. By virtue of
Lemma 5.4, we may consider a sequence $\\{x_{n}\\}\subset\Omega$ satisfying

$\|u_{n}\|_{L^{\infty}(\Omega)}=u_{n}(x_{n})\quad\forall n,\quad x_{n}\to
x_{0}\in\overline{\Omega}\text{, up to a subsequence}.$

We divide the rest of the proof into two parts. In the first of them we
consider the case $x_{0}\in\Omega$, while the second one is devoted to the
case $x_{0}\in\partial\Omega$. In turn, we divide each part into several
steps.

CASE 1) $x_{0}\in\Omega$.

Step 1.1) Scaling.

Denoting $d=\text{dist}(x_{0},\partial\Omega)/2>0$ and
$\eta_{n}=\|u_{n}\|_{L^{\infty}(\Omega)}^{-\frac{p-1}{2}}$, we define
$v_{n}:B_{d/\eta_{n}}(0)\to[0,+\infty)$ by

$v_{n}(y)=\eta_{n}^{\frac{2}{p-1}}u_{n}(x_{n}+\eta_{n}y)\quad\forall y\in
B_{d/\eta_{n}}(0).$

Therefore, $v_{n}\in H^{1}(B_{d/\eta_{n}}(0))\cap
L^{\infty}(B_{d/\eta_{n}}(0))$ and satisfies the equation

(3.7) $-\Delta v_{n}+u_{n}g_{n}(y,u_{n})\frac{|\nabla
v_{n}|^{2}}{v_{n}}=\lambda
v_{n}^{p}\frac{f(u_{n})}{u_{n}^{p}}+t_{n}\eta_{n}^{\frac{2(p-\sigma)}{p-1}}v_{n}^{\sigma},\quad
y\in B_{d/\eta_{n}}(0),$

where $g_{n}(y,s)=g(x_{n}+\eta_{n}y,s)$ for a.e. $y\in B_{d/\eta_{n}}(0)$ and
$s>0$. Moreover, $\|v_{n}\|_{L^{\infty}(B_{d/\eta_{n}}(0))}=v_{n}(0)=1.$ Our
aim now is to pass to the limit in (3.7). In the next step we will prove the a
priori estimates that will provide such a limit.

Step 1.2) A priori estimates.

Let us fix $R>0$ and denote $\omega=B_{R}(0)$. It is clear that $\omega\subset
B_{d/\eta_{n}}(0)$ for every $n$ sufficiently large, so $v_{n}$ satisfies the
equation (3.7) in $\omega$ and $\|v_{n}\|_{L^{\infty}(\omega)}=1$ for $n$
large. Of course, the same thing happens in $B_{2R}(0)$. Notice also that, if
$\delta=0$, then Lemma 3.6 implies that $u_{n}^{\gamma}\in H_{0}^{1}(\Omega)$
for all
$\gamma\in\left(\frac{\max\\{1-\sigma,1-\tau\\}}{2},1-\sigma\right]=\left(\frac{1-\tau}{2},1-\sigma\right]$,
where this last equality follows from $\tau<\sigma$ in condition ($g_{*}$).
Therefore, $v_{n}^{\gamma}\in H^{1}(B_{2R}(0))$ for all
$\gamma\in\left(\frac{1-\tau}{2},1-\sigma\right]$. This fact, together with
conditions ($g_{*}$) and ($f_{*}$), allow to apply Lemma 5.5 in the Appendix
below to deduce that there exist $C>0$, $\alpha\in(0,1)$ such that

$\|v_{n}\|_{C^{0,\alpha}(\overline{\omega})}\leq C$

for every $n$ large enough. As a consequence, there exists $v\in
C(\overline{\omega})$ such that, up to a subsequence,

$v_{n}\to v\text{ uniformly in }\overline{\omega}.$

Observe that $\|v\|_{L^{\infty}(\omega)}=1$ so, in particular, $v\not\equiv
0$.

On the other hand, let us consider a function $\phi\in C^{1}_{c}(B_{2R}(0))$
such that $0\leq\phi\leq 1$ in $B_{2R}(0)$ and $\phi\equiv 1$ in $\omega$. Now
we multiply both sides of (3.7) by $v_{n}\phi^{2}\in H^{1}_{0}(B_{2R}(0))\cap
L^{\infty}(B_{2R}(0))$ and integrate by parts, obtaining

$\int_{B_{2R}(0)}|\nabla v_{n}|^{2}\phi^{2}+2\int_{B_{2R}(0)}v_{n}\phi\nabla
v_{n}\nabla\phi\leq C,$

where we have used that $g\geq 0$, $\|v_{n}\|_{L^{\infty}(B_{2R}(0))}=1$ and,
one more time, condition ($f_{*}$). Hence, by Young’s inequality we easily
deduce that

$\int_{\omega}|\nabla v_{n}|^{2}\leq
C\left(\int_{B_{2R}(0)}|\nabla\phi|^{2}v_{n}^{2}+1\right)\leq C.$

That is to say, $\|v_{n}\|_{H^{1}(\omega)}\leq C$, and then, up to a
subsequence,

$v_{n}\rightharpoonup v\text{ weakly in }H^{1}(\omega).$

We will prove next that, for all $\omega_{0}\subset\subset\omega$, $v_{n}$ is
bounded from below in $\overline{\omega_{0}}$ by a positive constant
independent of $n$. The approach by comparison due to [32] is valid here.
Indeed, it is straightforward to see that the function
$w_{n}=\frac{v_{n}^{1-\sigma}}{1-\sigma}\in H^{1}(\omega)\cap
L^{\infty}(\omega)$ satisfies

$\int_{\omega}\nabla
w_{n}\nabla\phi=\lambda\int_{\omega}v_{n}^{p-\sigma}\frac{f(u_{n})}{u_{n}^{p}}\phi+\int_{\omega}(\sigma-
u_{n}g_{n}(y,u_{n}))\frac{|\nabla
v_{n}|^{2}\phi}{v_{n}^{\sigma+1}}+t_{n}\eta_{n}^{\frac{2(p-\sigma)}{p-1}}\int_{\omega}\phi$

for all $\phi\in C_{c}^{1}(\omega)$. Therefore, using conditions ($g_{*}$) and
($f_{*}$), we derive

$\int_{\omega}\nabla
w_{n}\nabla\phi\geq\lambda\int_{\omega}v_{n}^{p-\sigma}\phi$

for all $0\leq\phi\in C_{c}^{1}(\omega)$. On the other hand, let $z_{n}\in
H_{0}^{1}(\omega)\cap C(\overline{\omega})$ be the unique solution to

$\begin{cases}\displaystyle-\Delta z_{n}=\lambda
v_{n}^{p-\sigma},&y\in\omega,\\\ z_{n}=0,&y\in\partial\omega.\end{cases}$

It is clear that $z_{n}\to z$ uniformly in $\overline{\omega}$ and
$z_{n}\rightharpoonup z$ weakly in $H_{0}^{1}(\omega)$ for some $z\in
H_{0}^{1}(\omega)\cap C(\overline{\omega})$. In consequence, $z$ satisfies

$\begin{cases}\displaystyle-\Delta z=\lambda v^{p-\sigma},&y\in\omega,\\\
z=0,&y\in\partial\omega.\end{cases}$

Since $v\gneq 0$ in $\omega$, the strong maximum principle implies that, for
every $\omega_{0}\subset\subset\omega$, there exists $c>0$ such that $z\geq c$
in $\omega_{0}$. Besides, by comparison, $w_{n}\geq z_{n}$ in
$\overline{\omega}$. Hence, the uniform convergence of $\\{z_{n}\\}$ implies
that, for every $\varepsilon>0$, we may take $n$ large enough so that
$v_{n}^{1-\sigma}\geq(1-\sigma)(z-\varepsilon)$ in $\omega$. In sum, by
choosing $\varepsilon=\frac{c}{2}$ we conclude that

(3.8) $\forall\omega_{0}\subset\subset\omega,\,\,\exists
c_{\omega_{0}}>0:\quad v_{n}\geq c_{\omega_{0}},\quad
y\in\omega_{0},\,\,\forall n\text{ large.}$

From the previous estimates, it is straightforward to prove that $\\{\Delta
v_{n}\\}$ is bounded in $L^{1}_{\mbox{\tiny loc}}(\omega)$. Then, [12] implies
that, passing to a subsequence,

$\nabla v_{n}\to\nabla v,\quad y\in\omega.$

We are ready now to pass to the limit in (3.7).

Step 1.3) Passing to the limit.

We already know that $v_{n}\geq c_{\omega_{0}}>0$ in
$\omega_{0}\subset\subset\omega$. Moreover,
$\|u_{n}\|_{L^{\infty}(\omega)}\to+\infty$. Therefore, $u_{n}\to+\infty$
locally uniformly in $\omega$. Then, by ($g_{\infty}$) we deduce that
$|u_{n}g_{n}(y,u_{n})-\mu(x_{n}+\eta_{n}y)|\to 0$ locally uniformly in
$\omega$. In consequence, the continuity of $\mu$ yields

$u_{n}g_{n}(y,u_{n})\to\mu(x_{0})\quad\text{ locally uniformly in }\omega.$

In sum, we have that

$0\leq u_{n}g_{n}(y,u_{n})\frac{|\nabla
v_{n}|^{2}}{v_{n}}\to\mu(x_{0})\frac{|\nabla v|^{2}}{v}\quad\text{ pointwise
in }\omega.$

Furthermore, it follows from conditions ($f_{\infty}$) and ($f_{*}$) and from
the Dominated Convergence Theorem that

$\int_{\omega}v_{n}^{p}\frac{f(u_{n})}{u_{n}^{p}}\phi\to\int_{\omega}v^{p}\phi\quad\forall\phi\in
C_{c}^{1}(\omega).$

Let us take $\phi\in C_{c}^{1}(\omega)$ such that $\phi\geq 0$ as test
function in the weak formulation of (3.7). By virtue of Fatou’s lemma and
using the convergences that we have proved, it is immediate to show that

$\int_{\omega}\nabla v\nabla\phi+\mu(x_{0})\int_{\omega}\frac{|\nabla
v|^{2}}{v}\phi\leq\lambda\int_{\omega}v^{p}\phi.$

If we take now $\frac{v\phi}{v_{n}}\in H_{0}^{1}(\omega)\cap
L^{\infty}(\omega)$ as test function in (3.7), we obtain

$\displaystyle\int_{\omega}\frac{\nabla v_{n}\nabla v}{v_{n}}\phi$
$\displaystyle-\int_{\omega}\frac{v|\nabla
v_{n}|^{2}}{v_{n}^{2}}\phi+\int_{\omega}\frac{v}{v_{n}}\nabla
v_{n}\nabla\phi+\int_{\omega}u_{n}g_{n}(y,u_{n})\frac{v|\nabla
v_{n}|^{2}}{v_{n}^{2}}\phi$
$\displaystyle=\lambda\int_{\omega}v_{n}^{p-1}v\frac{f(u_{n})}{u_{n}^{p}}\phi+t_{n}\eta_{n}^{\frac{2(p-\sigma)}{p-1}}\int_{\omega}\frac{v\phi}{v_{n}^{1-\sigma}}.$

Observe that

$0\leq(1-\sigma)\frac{v|\nabla
v_{n}|^{2}\phi}{v_{n}^{2}}\leq(1-u_{n}g_{n}(y,u_{n}))\frac{v|\nabla
v_{n}|^{2}\phi}{v_{n}^{2}}\to(1-\mu(x_{0}))\frac{|\nabla v|^{2}}{v}\phi.$

Then, again by Fatou’s lemma we derive

$\int_{\omega}\nabla v\nabla\phi+\mu(x_{0})\int_{\omega}\frac{|\nabla
v|^{2}}{v}\phi\geq\lambda\int_{\omega}v^{p}\phi.$

That is to say, $v\in H^{1}(\omega)\cap C(\overline{\omega})$ satisfies

$-\Delta v+\mu(x_{0})\frac{|\nabla v|^{2}}{v}=\lambda v^{p},\quad y\in\omega.$

Step 1.4) Conclusion.

Since $\omega=B_{R}(0)$ for arbitrary $R>0$, then a standard diagonal argument
(see [25]) implies that $v$ is well-defined in $\mathbb{R}^{N}$, it belongs to
$H^{1}_{\mbox{\tiny loc}}(\mathbb{R}^{N})\cap C(\mathbb{R}^{N})$ and it
satisfies

$-\Delta v+\mu(x_{0})\frac{|\nabla v|^{2}}{v}=\lambda v^{p},\quad
y\in\mathbb{R}^{N}.$

Furthermore, it is straightforward to check that the function
$w=\lambda^{\frac{1}{p-1}}v$ satisfies

$-\Delta w+\mu(x_{0})\frac{|\nabla w|^{2}}{w}=w^{p},\quad y\in\mathbb{R}^{N}.$

This is impossible by virtue of Lemma 3.1 (see also Remark 3.2).

CASE 2) $x_{0}\in\partial\Omega$.

Step 2.1) Scaling.

Recall that we are assuming that there exist sequences
$\\{t_{n}\\}\subset[0,t_{0}]$, $\\{u_{n}\\}$ and $\\{x_{n}\\}\subset\Omega$
such that $u_{n}$ is a solution to $(P^{t_{n}})$ for all $n$,
$\|u_{n}\|_{L^{\infty}(\Omega)}=u_{n}(x_{n})\to+\infty$ as $n\to+\infty$ and
$x_{n}\to x_{0}$ for some $x_{0}\in\partial\Omega$. Taking advantage of the
smoothness of $\partial\Omega$, we are allowed to perform a convenient change
of coordinates in such a way that $u_{n}$ is a solution to a similar problem
except that $\partial\Omega$ becomes flat near $x_{0}$ (see Lemma 5.1 in the
Appendix below for the detailed proof). In other words, we may assume without
loss of generality that $u_{n}$ is a solution to $(R^{t_{n}})$ for all $n$,
where

($R^{t}$) $\begin{cases}-\text{{\rm div}}(M(x)\nabla v)+b(x)\nabla
v+g(x,v)M(x)\nabla v\nabla v=\lambda f(v)+tv^{\sigma},\quad&x\in\Omega,\\\
v>0,\quad&x\in\Omega,\\\ v=0,\quad&x\in\Gamma,\end{cases}$

with $\Omega\subset\mathbb{R}^{N}_{+}$,
$\emptyset\not=\Gamma\subset\partial\Omega\cap\partial\mathbb{R}^{N}_{+}$
being open as a subset of $\partial\mathbb{R}^{N}_{+}$ and connected, $M\in
C^{1}(\overline{\Omega})^{N\times N}$ being uniformly elliptic, and $b\in
C(\overline{\Omega})^{N}$.

Since $\Gamma$ is open in $\partial\mathbb{R}^{N}_{+}$, then
$d_{n}=\text{dist}(x_{n},\partial\Omega)=\text{dist}(x_{n},\Gamma)=x_{n,N}$
for all $n$ large enough. Arguing as in the previous case, we define

$v_{n}(y)=\eta_{n}^{\frac{2}{p-1}}u_{n}(x_{n}+\eta_{n}y)\quad\forall
y\in\Omega_{n}\cup\Gamma_{n},$

where $\eta_{n}=\|u_{n}\|^{-\frac{p-1}{2}}_{L^{\infty}(\Omega)},$
$0\in\Omega_{n}=B_{d/\eta_{n}}(0)\cap\left\\{y_{N}>-d_{n}/\eta_{n}\right\\}$
and $\Gamma_{n}=B_{d/\eta_{n}}(0)\cap\left\\{y_{N}=-d_{n}/\eta_{n}\right\\}$
for some $d>0$. It is not difficult to see that $v_{n}$ is well-defined for
all $n$ large enough and it satisfies

$\begin{cases}\displaystyle-\text{{\rm div}}(M_{n}(y)\nabla
v_{n})+b_{n}(y)\nabla v_{n}+g_{n}(y,u_{n})u_{n}\frac{M_{n}(y)\nabla
v_{n}\nabla v_{n}}{v_{n}}&\\\ \displaystyle=\lambda
v_{n}^{p}\frac{f(u_{n})}{u_{n}^{p}}+\eta_{n}^{\frac{2(p-\sigma)}{p-1}}t_{n}v_{n}^{\sigma},&y\in\Omega_{n},\\\
v_{n}>0,&x\in\Omega_{n},\\\ v_{n}=0,&y\in\Gamma_{n},\end{cases}$

where $M_{n}(y)=M(x_{n}+\eta_{n}y)$, $b_{n}(y)=\eta_{n}b(x_{n}+\eta_{n}y)$ and
$g_{n}(y,\cdot)=g(x_{n}+\eta_{n}y,\cdot)$.

Now, if $\\{d_{n}/\eta_{n}\\}$ is unbounded, we can extract a subsequence such
that $d_{n}/\eta_{n}\to+\infty$ as $n\to+\infty$. In this case,
$\bigcup_{n\in\mathbb{N}}\Omega_{n}=\mathbb{R}^{N}$, so we can argue as in the
case $x_{0}\in\Omega$ without relevant changes and arrive to a contradiction.

Let us assume now that $\\{d_{n}/\eta_{n}\\}$ is bounded. Then, up to a (not
relabeled) subsequence, $d_{n}/\eta_{n}\to\kappa$ for some $\kappa\geq 0$.
Thus, $\bigcup_{n\in\mathbb{N}}\Omega_{n}=\\{y_{N}>\kappa\\}$. Let us prove
the estimates in this new situation.

Step 2.2) A priori estimates.

Let us denote

$T_{n}=(0,...,0,d_{n}/\eta_{n})\in\mathbb{R}^{N},\quad\Omega_{n}^{\prime}=B_{d/\eta_{n}}(T_{n})\cap\mathbb{R}^{N}_{+},\quad\Gamma_{n}^{\prime}=B_{d/\eta_{n}}(T_{n})\cap\partial\mathbb{R}^{N}_{+}.$

Observe that $\bigcup_{n\in\mathbb{N}}\Omega_{n}^{\prime}=\mathbb{R}^{N}_{+}$
and $\bigcup_{n\in\mathbb{N}}\Gamma_{n}^{\prime}=\partial\mathbb{R}^{N}_{+}$.
Let us define $w_{n}:\Omega_{n}^{\prime}\cup\Gamma_{n}^{\prime}\to[0,+\infty)$
by

$w_{n}(y)=v_{n}(y-T_{n})\quad\forall
y\in\Omega_{n}^{\prime}\cup\Gamma_{n}^{\prime}.$

Obviously, $w_{n}$ satisfies

(3.9) $\begin{cases}\displaystyle-\text{{\rm div}}(M_{n}^{\prime}(y)\nabla
w_{n})+b_{n}^{\prime}(y)\nabla
w_{n}+g^{\prime}_{n}\left(y,\frac{w_{n}}{\eta_{n}^{\frac{2}{p-1}}}\right)\frac{1}{\eta_{n}^{\frac{2}{p-1}}}M_{n}^{\prime}(y)\nabla
v_{n}\nabla v_{n}&\\\
\displaystyle=\lambda\eta_{n}^{\frac{2p}{p-1}}f\left(\frac{w_{n}}{\eta_{n}^{\frac{2}{p-1}}}\right)+\eta_{n}^{\frac{2(p-\sigma)}{p-1}}t_{n}w_{n}^{\sigma},&y\in\Omega^{\prime}_{n},\\\
w_{n}>0,&x\in\Omega^{\prime}_{n},\\\
w_{n}=0,&y\in\Gamma_{n}^{\prime},\end{cases}$

where $M_{n}^{\prime}(y)=M_{n}(y-T_{n})$, $b_{n}^{\prime}(y)=b_{n}(y-T_{n})$
and $g_{n}^{\prime}(y,\cdot)=g_{n}(y-T_{n},\cdot)$. We also have that
$\|w_{n}\|_{L^{\infty}(\Omega_{n}^{\prime})}=w_{n}(T_{n})=1$ for all $n$.

Let us denote $T_{\infty}=(0,...,0,\kappa)\in\mathbb{R}^{N}$ and, for
$R>\kappa$, let us consider the sets

$\omega=B_{R}(T_{\infty})\cap\mathbb{R}^{N}_{+},\quad\beta=B_{R}(T_{\infty})\cap\partial\mathbb{R}^{N}_{+}\quad\omega_{1}=B_{2R}(T_{\infty})\cap\mathbb{R}^{N}_{+},\quad\beta_{1}=B_{2R}(T_{\infty})\cap\partial\mathbb{R}^{N}_{+}.$

It is easy to see that, for all $n$ large enough,

$T_{n}\in\omega\subset\omega_{1}\subset\Omega_{n}^{\prime},\quad\quad
0\in\beta\subset\beta_{1}\subset\Gamma_{n}^{\prime},\quad\quad\overline{\omega}\subseteq\omega_{1}\cup\beta_{1}.$

In particular, $w_{n}$ satisfies problem (3.9) by changing
$\Omega_{n}^{\prime}$ with $\omega_{1}$ and $\Gamma_{n}^{\prime}$ with
$\beta_{1}$, and further, $\|w_{n}\|_{L^{\infty}(\omega_{1})}=1$ for all $n$
large enough. Thus, it follows again from Lemma 5.5 in the Appendix (thanks to
conditions ($g_{*}$) and ($f_{*}$) and to Lemma 3.6) that there exist
$C>0,\alpha\in(0,1)$ such that

$\|w_{n}\|_{C^{0,\alpha}(\overline{\omega})}\leq C$

for every $n$ large enough. As a consequence, there exists $v\in
C(\overline{\omega})$ such that, up to a subsequence, $w_{n}\to v$ uniformly
in $\overline{\omega}$. In particular, $v=0$ on $\beta$.

On the other hand, from the Hölder estimate one can also deduce that
$\kappa>0$. Indeed,

$1=|w_{n}(T_{n})-w_{n}(0)|\leq C|T_{n}|^{\alpha}=C(d_{n}/\eta_{n})^{\alpha}\to
C\kappa^{\alpha}.$

Next, an estimate on $\\{w_{n}\\}$ in $H^{1}_{\mbox{\tiny loc}}(\omega)$ can
be proven as in CASE 1, so that $w_{n}\rightharpoonup v$ weakly in
$H^{1}_{\mbox{\tiny loc}}(\omega)$, up to a subsequence. Moreover, the same
arguments are valid to prove that $\\{w_{n}\\}$ satisfies also (3.8) and,
furthermore, that $\nabla w_{n}\to\nabla v$ a.e. in $\omega$.

Step 2.3) Passing to the limit.

We can now pass to the limit as in CASE 1 and deduce that $v\in
H^{1}_{\mbox{\tiny loc}}(\omega)\cap C(\overline{\omega})$ is a solution to
the following problem:

$\begin{cases}\displaystyle-\text{{\rm div}}(M(x_{0})\nabla
v)+\mu(x_{0})\frac{M(x_{0})\nabla v\nabla v}{v}=\lambda
v^{p},\quad&y\in\omega,\\\ v>0,&y\in\omega,\\\
v=0,\quad&y\in\beta.\end{cases}$

Actually, as $R$ is arbitrary, the diagonal argument (see [25]) implies that
$v\in H^{1}_{\mbox{\tiny loc}}(\mathbb{R}^{N}_{+})\cap
C\left(\overline{\mathbb{R}^{N}_{+}}\right)$ and it satisfies

$\begin{cases}\displaystyle-\text{{\rm div}}(M(x_{0})\nabla
v)+\mu(x_{0})\frac{M(x_{0})\nabla v\nabla v}{v}=\lambda
v^{p},\quad&y\in\mathbb{R}^{N}_{+},\\\ v>0,&y\in\mathbb{R}^{N}_{+},\\\
v=0,\quad&y\in\partial\mathbb{R}^{N}_{+}.\end{cases}$

Step 2.4) Conclusion.

Observe that, if we denote $M(x_{0})=(m_{ij})$ for $i,j=1,...,N$, then the
previous equation may be written as

$\displaystyle\sum_{i,j=1}^{N}m_{ij}\frac{\partial^{2}v}{\partial
y_{i}\partial y_{j}}+\frac{\mu(x_{0})}{v}\sum_{i,j=1}^{N}m_{ij}\frac{\partial
v}{\partial y_{i}}\frac{\partial v}{\partial y_{j}}=\lambda v^{p},\quad
y\in\mathbb{R}^{N}_{+}.$

Since $i,j$ commute in both $\frac{\partial^{2}v}{\partial y_{i}\partial
y_{j}}$ and $\frac{\partial v}{\partial y_{i}}\frac{\partial v}{\partial
y_{j}}$, then a simple change of coordinates (see the conclusion of Case 1 in
Section 2 of [25] for the details) leads to finding a solution $w\in
H^{1}_{\mbox{\tiny loc}}(\mathbb{R}^{N}_{+})\cap
C\left(\overline{\mathbb{R}^{N}_{+}}\right)$ to

$\begin{cases}\displaystyle-\Delta w+\mu(y_{0})\frac{|\nabla
w|^{2}}{w}=w^{p},\quad&y\in\mathbb{R}^{N}_{+},\\\
w>0,&y\in\mathbb{R}^{N}_{+},\\\
w=0,\quad&y\in\partial\mathbb{R}^{N}_{+},\end{cases}$

for some $y_{0}\in\partial\Omega$. This contradicts Lemma 3.1 (see also Remark
3.2). The proof is concluded. ∎

The following result, as Proposition 3.7, provides a priori estimates on the
solutions to ($P^{t}$). The difference lies on the fact that we do not impose
the limit conditions at infinity ($g_{\infty}$) and ($f_{\infty}$) at the
expense of making a stronger restriction on $\sigma,p$.

######  Proposition 3.8.

Let $\lambda>0$, $f:[0,+\infty)\to[0,+\infty)$ be a continuous function and
$g:\Omega\times(0,+\infty)\to[0,+\infty)$ be a Carathéodory function. For
$\delta,\tau\geq 0$, $\sigma>0$ and $p\in\left(1,\frac{N+1}{N-1}\right)$,
assume that $f$ satisfies ($f_{*}$) and $g$ satisfies ($g_{*}$). Assume in
addition that $\sigma\leq\frac{N+1-(N-1)p}{2}$. Then, there exists $C>0$ such
that $\|u\|_{L^{\infty}(\Omega)}\leq C$ for every solution $u$ to ($P^{t}$)
for all $t\in[0,t_{0}]$, where $t_{0}>0$ is given by Proposition 3.5.

###### Proof.

The proof is very similar to that of Proposition 3.7. Here we give only a
sketch.

Arguing by contradiction, we assume that there exist two sequences
$\\{t_{n}\\}\subset[0,t_{0}]$ and $\\{u_{n}\\}$ such that $u_{n}$ is a
solution to $(P^{t_{n}})$ for all $n$ and
$\|u_{n}\|_{L^{\infty}(\Omega)}\to+\infty$ as $n\to+\infty$. We also consider
a sequence $\\{x_{n}\\}\subset\Omega$ satisfying

$\|u_{n}\|_{L^{\infty}(\Omega)}=u_{n}(x_{n})\quad\forall n,\quad x_{n}\to
x_{0}\in\overline{\Omega}\text{, up to a subsequence};$

this can be done by virtue of Lemma 5.4. We denote
$\eta_{n}=\|u_{n}\|_{L^{\infty}(\Omega)}^{-\frac{p-1}{2}}$. Let us assume that
$x_{0}\in\partial\Omega$ (we omit the simpler case $x_{0}\in\Omega$). Arguing
as in the proof of Proposition 3.7, we assume without loss of generality that
$u_{n}$ is a solution to $(R^{t_{n}})$ for all $n$, with
$\Omega=V\subset\mathbb{R}^{N}_{+}$ and
$\Gamma\subset\partial\Omega\cap\partial\mathbb{R}^{N}_{+}$, being $\Gamma$
open as a subset of $\partial\mathbb{R}^{N}_{+}$ and connected. Thus, there
exists a sequence of bounded domains $\\{\Omega_{n}\\}$ satisfying, for every
$n$, that $0\in\Omega_{n}$, $x_{n}+\eta_{n}y\in\Omega$ for all
$y\in\Omega_{n}$ and $\bigcup_{n\in\mathbb{N}}\Omega_{n}=X$, where $X$ may be
either $\mathbb{R}^{N}$ or $\\{y_{N}>\kappa\\}$ for some $\kappa\geq 0$.

In any case, we define $v_{n}:\Omega_{n}\to\mathbb{R}$ by

$v_{n}(y)=\eta_{n}^{\frac{2}{p-1}}u_{n}(x_{n}+\eta_{n}y)\quad\forall
y\in\Omega_{n}.$

It is easy to check that $v_{n}$ satisfies the equation

$\displaystyle-\text{{\rm div}}(M_{n}(y)\nabla v_{n})$
$\displaystyle+b_{n}(y)\nabla v_{n}+u_{n}g_{n}(y,u_{n})\frac{M_{n}(y)\nabla
v_{n}\nabla v_{n}}{v_{n}}$ $\displaystyle=\lambda
v_{n}^{p}\frac{f(u_{n})}{u_{n}^{p}}+\eta_{n}^{\frac{2(p-\sigma)}{p-1}}t_{n}v_{n}^{\sigma},\quad
y\in\Omega_{n},$

where $M_{n}(y)=M(x_{n}+\eta_{n}y)$, $b_{n}(y)=\eta_{n}b(x_{n}+\eta_{n}y)$ and
$g_{n}(y,\cdot)=g(x_{n}+\eta_{n}y,\cdot)$. From the previous equation one
obtains the same local estimates as in Proposition 3.7. However, in this case
one cannot pass to the limit directly in the equation. We overcome this issue
by deducing from $sg_{n}(x,s)\leq\sigma$, $M_{n}(y)\nabla v_{n}\nabla
v_{n}\geq 0$ and $f(u_{n})\geq u_{n}^{p}$ the inequality

$-\text{{\rm div}}(M_{n}(y)\nabla v_{n})+b_{n}(y)\nabla
v_{n}+\sigma\frac{M_{n}(y)\nabla v_{n}\nabla v_{n}}{v_{n}}\geq\lambda
v_{n}^{p},\quad y\in\Omega_{n}.$

In fact, one can pass to the limit in the previous inequality as in
Proposition 3.7 (but using Fatou’s lemma only once). Applying after that a
convenient change of coordinates, we find a supersolution $v\in
H^{1}_{\mbox{\tiny loc}}(X)\cap C(\overline{X})$ to (3.3), where either
$X=\mathbb{R}^{N}$ or $X=\mathbb{R}^{N}_{+}$. This is a contradiction with
Lemma 3.3. ∎

Next proposition provides similar estimates as Proposition 3.7 and Proposition
3.8. The novelty is that $g$ satisfies now a limit condition at infinity only
in a neighborhood of $\partial\Omega$. This means that we need to impose
stronger restrictions on $\sigma,p$ than in Proposition 3.7, but milder than
in Proposition 3.8.

######  Proposition 3.9.

Let $\lambda>0$, $f:[0,+\infty)\to[0,+\infty)$ be a continuous function and
$g:\Omega\times(0,+\infty)\to[0,+\infty)$ be a Carathéodory function. For
$\delta,\tau\geq 0$, $\sigma>0$ and $p\in\left(1,\frac{N}{N-2}\right)$, assume
that $f$ satisfies ($f_{*}$) and ($f_{\infty}$), and that $g$ satisfies
($g_{*}$) and ($\widetilde{g_{\infty}}$). Assume in addition that
$\sigma\leq\frac{N-(N-2)p}{2}$. Then, there exists $C>0$ such that
$\|u\|_{L^{\infty}(\Omega)}\leq C$ for every solution $u$ to ($P^{t}$) for all
$t\in[0,t_{0}]$, where $t_{0}>0$ is given by Proposition 3.5.

###### Proof.

We argue similarly as for Proposition 3.7 and Proposition 3.8, so we provide
only the main ideas.

Assume by contradiction that there exist two sequences
$\\{t_{n}\\}\subset[0,t_{0}]$ and $\\{u_{n}\\}$ such that $u_{n}$ is a
solution to $(P^{t_{n}})$ for all $n$ and
$\|u_{n}\|_{L^{\infty}(\Omega)}\to+\infty$ as $n\to+\infty$. Thanks to Lemma
5.4, we may also consider a sequence $\\{x_{n}\\}\subset\Omega$ satisfying

$\|u_{n}\|_{L^{\infty}(\Omega)}=u_{n}(x_{n})\quad\forall n,\quad x_{n}\to
x_{0}\in\overline{\Omega}\text{ up to a subsequence}.$

Suppose that $x_{0}\in\Omega$. Since $x_{0}$ might belong to $\omega$ (where
we know nothing about the asymptotic behavior of $g$ at infinity), we cannot
proceed as in the proof of Proposition 3.7. Nevertheless, scaling $u_{n}$
conveniently and using ($g_{*}$) and ($f_{*}$) we may argue as in the proof of
Proposition 3.8 to find a supersolution $0<v\in H^{1}_{\mbox{\tiny
loc}}(\mathbb{R}^{N})\cap C(\mathbb{R}^{N})$ to

$\displaystyle-\Delta v+\sigma\frac{|\nabla v|^{2}}{v}=v^{p},\quad
y\in\mathbb{R}^{N}.$

This is a contradiction with Lemma 3.3.

On the other hand, if $x_{0}\in\partial\Omega$, then we may take advantage of
($\widetilde{g_{\infty}}$) to obtain, using also ($f_{\infty}$) and arguing as
in Proposition 3.7, a solution $v\in H^{1}_{\mbox{\tiny loc}}(X)\cap
C(\overline{X})$ to (3.3) for $\sigma=\mu(y_{0})$ and some
$y_{0}\in\overline{\Omega}$, and either $X=\mathbb{R}^{N}$ or
$X=\mathbb{R}^{N}_{+}$. This contradicts Lemma 3.1 (see also Remark 3.2). ∎

Next result provides an estimate for the solutions to problem ($P_{\lambda}$)
whose dependence on $\lambda$ is explicit. As a consequence, it is shown that
the norm of the solutions to problem ($P_{\lambda}$), if they exist, becomes
arbitrarily small as $\lambda$ tends to infinity.

######  Proposition 3.10.

Let $f:[0,+\infty)\to[0,+\infty)$ be a continuous function and
$g:\Omega\times(0,+\infty)\to[0,+\infty)$ be a Carathéodory function. For
$\delta=0$, $\tau\geq 0$, $\sigma>0$ and $p\in\left(1,\frac{N+1}{N-1}\right)$,
assume that $f$ satisfies ($f_{*}$) and $g$ satisfies ($g_{*}$). Assume in
addition that $\sigma\leq\frac{N+1-(N-1)p}{2}$. Then, there exists $C>0$ such
that

$\lambda^{\frac{1}{p-1}}\|u\|_{L^{\infty}(\Omega)}\leq C$

for every solution $u$ to ($P_{\lambda}$) for all $\lambda>0$.

###### Proof.

Arguing again as in the proof of Proposition 3.7, assume that there exist two
sequences $\\{\lambda_{n}\\}\subset[0,+\infty)$ and $\\{u_{n}\\}$ such that
$u_{n}$ is a solution to $(P_{\lambda_{n}})$ for all $n$ and
$\|z_{n}\|_{L^{\infty}(\Omega)}\to+\infty$ as $n\to+\infty$, where
$z_{n}=\lambda_{n}^{\frac{1}{p-1}}u_{n}$. It is easy to see that $z_{n}$
satisfies

$-\Delta
z_{n}+\lambda_{n}^{-\frac{1}{p-1}}g\left(x,\lambda_{n}^{-\frac{1}{p-1}}z_{n}\right)|\nabla
z_{n}|^{2}=\lambda_{n}^{\frac{p}{p-1}}f\left(\lambda_{n}^{-\frac{1}{p-1}}z_{n}\right),\quad
x\in\Omega.$

Since $z_{n}\in C(\overline{\Omega})$ by virtue of Lemma 5.4, then we may take
$\\{x_{n}\\}\subset\Omega$ such that
$z_{n}(x_{n})=\|z_{n}\|_{L^{\infty}(\Omega)}$ for all $n$. Let
$x_{0}\in\overline{\Omega}$ be such that, passing to a subsequence if
necessary, $x_{n}\to x_{0}\in\partial\Omega$ (the case $x_{0}\in\Omega$ is
analogous, so we omit it). Arguing as in the proof of Proposition 3.7 (Case
2), we may assume without loss of generality that $v_{n}\in
H^{1}(\Omega_{n})\cap L^{\infty}(\Omega_{n})$, defined by
$v_{n}(y)=\eta_{n}^{\frac{2}{p-1}}z_{n}(x_{n}+\eta_{n}y)$ for all
$y\in\Omega_{n}$, satisfies

(3.10) $-\text{{\rm div}}(M_{n}(y)\nabla v_{n})+b_{n}(y)\nabla
v_{n}+\frac{v_{n}}{L_{n}}g_{n}\left(y,\frac{v_{n}}{L_{n}}\right)\frac{M_{n}(y)\nabla
v_{n}\nabla v_{n}}{v_{n}}=L_{n}^{p}f\left(\frac{v_{n}}{L_{n}}\right),\quad
y\in\Omega_{n},$

where $\eta_{n}=\|z_{n}\|_{L^{\infty}(\Omega)}^{-\frac{p-1}{2}}$,
$L_{n}^{p-1}=\lambda_{n}\eta_{n}^{2}$ and
$\Omega_{n},\Gamma_{n},M_{n},b_{n},g_{n}$ are as in the proof of Proposition
3.7 (Case 2). From (3.10), and using conditions ($f_{*}$) and ($g_{*}$), one
can prove the same local estimates on $\\{v_{n}\\}$ as in the proof of
Proposition 3.7. Moreover, again by ($f_{*}$) and ($g_{*}$) we deduce that

$-\text{{\rm div}}(M_{n}(y)\nabla v_{n})+b_{n}(y)\nabla
v_{n}+\sigma\frac{M_{n}(y)\nabla v_{n}\nabla v_{n}}{v_{n}}\geq v_{n}^{p},\quad
y\in\Omega_{n}.$

Now we pass to the limit in the previous inequality by using Fatou’s lemma
once and obtain a supersolution $v\in H^{1}_{\mbox{\tiny loc}}(X)\cap
C(\overline{X})$ to (3.3), where either $X=\mathbb{R}^{N}$ or
$X=\mathbb{R}^{N}_{+}$. This is a contradiction by virtue of Lemma 3.3. ∎

In the last result of this section we impose stronger restrictions on $f$ and
$g$ to obtain an a priori estimate for the solutions to ($P_{\lambda}$)
similar to that in Proposition 3.10. The advantage is that $p$ can be chosen
near $2^{*}-1$.

######  Proposition 3.11.

Let $g:\overline{\Omega}\times(0,+\infty)\to[0,+\infty)$ be a continuous
function. For $\delta=0$, $\tau\geq 0$, $\sigma>0$ and
$p\in\left(1,2^{*}-1\right)$, assume that $g$ satisfies ($g_{*}$),
($g_{\infty}$) and ($\widetilde{g_{0}}$). Assume in addition that there exists
$\alpha>0$ such that, for each $x_{0}\in\overline{\Omega}$ and $L>0$, the
function $\psi:[0,+\infty)\to[0,+\infty)$, defined by

$\psi(s)=\int_{0}^{s}e^{-\frac{1}{L}\int_{\alpha
L}^{t}g\left(x_{0},\frac{r}{L}\right)dr}dt\quad\forall s\geq 0,$

satisfies that

(3.11) $s\mapsto\frac{\psi^{\prime}(s)s^{p}}{\psi(s)^{2^{*}-1}}\quad\text{ is
decreasing for all }s>0.$

Lastly, let us consider the function $f:[0,+\infty)\to[0,+\infty)$ defined by
$f(s)=s^{p}$ for all $s\geq 0$. Then, there exists $C>0$ such that

$\lambda^{\frac{1}{p-1}}\|u\|_{L^{\infty}(\Omega)}\leq C$

for every solution $u$ to ($P_{\lambda}$) for all $\lambda>0$.

###### Proof.

We may reproduce the proof of Proposition 3.10 up to (3.10), where
$\\{v_{n}\\}$ satisfies the same estimates as in the proof of Proposition 3.7.
Now, unlike in the proof of Proposition 3.10, we aim to pass to the limit
directly in (3.10). In order to do so, let us take a not relabeled subsequence
so that $L_{n}\to L$ for some $L\in[0,+\infty]$. Let us assume first that
$L=0$. In this case, we can pass to the limit in (3.10) as in the proof of
Proposition 3.7 and we obtain a solution $v\in H^{1}_{\mbox{\tiny loc}}(X)\cap
C(\overline{X})$ to (3.3) with $\sigma=\mu(y_{0})$ for some
$y_{0}\in\overline{\Omega}$, where either $X=\mathbb{R}^{N}$ or
$X=\mathbb{R}^{N}_{+}$. This is a contradiction by virtue of Lemma 3.1 (see
also Remark 3.2).

Therefore, necessarily $L>0$. Assume now that $L=+\infty$. Using
($\widetilde{g_{0}}$), we may pass to the limit again in (3.10) and we find a
solution $v\in H^{1}_{\mbox{\tiny loc}}(X)\cap C(\overline{X})$ to (3.3) with
$\sigma=\mu_{0}(y_{0})$ for some $y_{0}\in\overline{\Omega}$, where either
$X=\mathbb{R}^{N}$ or $X=\mathbb{R}^{N}_{+}$. One more time, this contradicts
Lemma 3.1 (see also Remark 3.2).

The only remaining possibility is $0<L<+\infty$. Thanks to the continuity of
$g$, we may pass to the limit in (3.10) once more to obtain a solution $v\in
H^{1}_{\mbox{\tiny loc}}(X)\cap C(\overline{X})$ to (3.1), where
$h(s)=\frac{1}{L}g\left(y_{0},\frac{s}{L}\right)$ for some
$y_{0}\in\overline{\Omega}$ and either $X=\mathbb{R}^{N}$ or
$X=\mathbb{R}^{N}_{+}$. Again, this is a contradiction with Lemma 3.1. ∎

##  4\. Proofs of the main results

We start by proving Theorem 2.3.

###### Proof of Theorem 2.3.

Let us first define the function
$\tilde{g}:\Omega\times(0,+\infty)\to[0,+\infty)$ by

$\tilde{g}(x,s)=\frac{\sigma-s^{\frac{1}{1-\sigma}}g\left(x,s^{\frac{1}{1-\sigma}}\right)}{(1-\sigma)s},\quad\text{a.e.
}x\in\Omega,\,\,\forall s>0.$

Thanks to ($g_{1}$) and ($g_{\nearrow}$), it is clear that $\tilde{g}$ is
nonincreasing in the $s$ variable. Moreover,

$|\tilde{g}(s,x)|\leq\frac{\sigma+s^{\frac{1}{1-\sigma}}g_{0}\left(s^{\frac{1}{1-\sigma}}\right)}{(1-\sigma)s},\quad\text{a.e.
}x\in\Omega,\,\,\forall s>0,$

so it is a bounded function in the $x$ variable.

For some $0\leq\phi\in H_{0}^{1}(\Omega)\cap L^{\infty}(\Omega)$ with compact
support, let us take $\frac{(1-\sigma)\phi}{u^{\sigma}}$ as test function in
(2.2). Then, if we denote $\tilde{u}=u^{1-\sigma}$, we deduce that

$\int_{\Omega}\nabla\tilde{u}\nabla\phi\leq\int_{\Omega}\tilde{g}(x,\tilde{u})|\nabla\tilde{u}|^{2}\phi+\int_{\Omega}\frac{(1-\sigma)h(x)}{\tilde{u}^{\frac{\sigma}{1-\sigma}}}\phi.$

Arguing similarly, $\tilde{v}=v^{1-\sigma}$ satisfies

$\int_{\Omega}\nabla\tilde{v}\nabla\phi\geq\int_{\Omega}\tilde{g}(x,\tilde{v})|\nabla\tilde{v}|^{2}\phi+\int_{\Omega}\frac{(1-\sigma)h(x)}{\tilde{v}^{\frac{\sigma}{1-\sigma}}}\phi.$

Moreover, $\tilde{u},\tilde{v}\in C(\Omega)\cap W^{1,N}_{\mbox{\tiny
loc}}(\Omega)$ and they satisfy $\limsup_{x\to
x_{0}}(\tilde{u}(x)-\tilde{v}(x))\leq 0$ for all $x_{0}\in\partial\Omega$.

At this point one can reproduce the proof of [31, Theorem 2.1] without
relevant changes and conclude that $\tilde{u}\leq\tilde{v}$ in $\Omega$.
Equivalently, $u\leq v$ in $\Omega$. ∎

Next we prove Theorem 2.12.

###### Proof of Theorem 2.12.

Let $v$ be a solution to ($H$). Recall that the definition of solution implies
that there exists $\gamma>0$ such that $v^{\gamma}\in H_{0}^{1}(\Omega)$. It
is easy to see that $u=v^{\gamma}$ satisfies the equation

(4.1) $-\Delta u+g_{\gamma}(x,u)|\nabla u|^{2}=\gamma v^{\gamma-1}h(x),\quad
x\in\Omega,$

where

$g_{\gamma}(x,s)=\frac{s^{\frac{1}{\gamma}}g\left(x,s^{\frac{1}{\gamma}}\right)+\gamma-1}{\gamma
s},\quad\text{a.e. }x\in\Omega,\,\,\forall s>0.$

It follows from ($g_{2}$) that

(4.2) $sg_{\gamma}(x,s)\geq\frac{\tau+\gamma-1}{\gamma}\quad\text{a.e.
}x\in\Omega\setminus\omega,\,\,\forall s\in(0,s_{0}^{\gamma}),$

where clearly $\frac{\tau+\gamma-1}{\gamma}>1$. From now we will denote
$\rho=\frac{\tau+\gamma-1}{\gamma}$ and $s_{1}=s_{0}^{\gamma}$.

Recall that there exists $c>0$ such that $v\geq c^{\frac{1}{\gamma}}$ in
$\omega$. For every $\varepsilon\in(0,\min\\{s_{1},c\\})$, let us define the
function $\varphi_{\varepsilon}:[0,+\infty)\to[0,+\infty)$ by

$\varphi_{\varepsilon}(s)=\begin{cases}\displaystyle\frac{1}{s}+\frac{s^{\rho-1}-\varepsilon^{\rho-1}}{(\rho-1)s^{\rho}},&s\geq\varepsilon,\\\
\displaystyle\frac{s}{\varepsilon^{2}},&0\leq s<\varepsilon.\end{cases}$

Clearly, $\varphi_{\varepsilon}(u)\in H_{0}^{1}(\Omega)\cap
L^{\infty}(\Omega)$. Hence, taking $\varphi_{\varepsilon}(u)$ as test function
in the weak formulation of (4.1) (this can be done thanks to Remark 2.2
because $u\in H_{0}^{1}(\Omega)$) we obtain

(4.3) $\int_{\Omega}\varphi_{\varepsilon}^{\prime}(u)|\nabla
u|^{2}+\int_{\Omega}g_{\gamma}(x,u)|\nabla
u|^{2}\varphi_{\varepsilon}(u)=\gamma\int_{\Omega}v^{\gamma-1}h(x)\varphi_{\varepsilon}(u).$

Observe that, from (4.2), it follows that

$\displaystyle\int_{\Omega}$ $\displaystyle g_{\gamma}(x,u)|\nabla
u|^{2}\varphi_{\varepsilon}(u)=\int_{\omega\cup\\{u\geq
s_{1}\\}}g_{\gamma}(x,u)|\nabla
u|^{2}\varphi_{\varepsilon}(u)+\int_{(\Omega\setminus\omega)\cap\\{u<s_{1}\\}}g_{\gamma}(x,u)|\nabla
u|^{2}\varphi_{\varepsilon}(u)$ (4.4)
$\displaystyle\geq-C+\int_{(\Omega\setminus\omega)\cap\\{\varepsilon<u<s_{1}\\}}\frac{\rho|\nabla
u|^{2}}{u}\left(\frac{1}{u}+\frac{u^{\rho-1}-\varepsilon^{\rho-1}}{(\rho-1)u^{\rho}}\right).$

On the other hand,

(4.5) $\int_{\Omega}\varphi_{\varepsilon}^{\prime}(u)|\nabla
u|^{2}=\int_{\\{u>\varepsilon\\}}\frac{\rho|\nabla
u|^{2}}{u^{2}}-\int_{\\{u>\varepsilon\\}}\frac{\rho|\nabla
u|^{2}}{u}\left(\frac{1}{u}+\frac{u^{\rho-1}-\varepsilon^{\rho-1}}{(\rho-1)u^{\rho}}\right)+\int_{\\{u\leq\varepsilon\\}}\frac{|\nabla
u|^{2}}{\varepsilon^{2}}.$

We will now absorb the negative term in (4.5) with the last term in (4).
Indeed,

$\displaystyle\int_{(\Omega\setminus\omega)\cap\\{\varepsilon<u<s_{1}\\}}$
$\displaystyle\frac{\rho|\nabla
u|^{2}}{u}\left(\frac{1}{u}+\frac{u^{\rho-1}-\varepsilon^{\rho-1}}{(\rho-1)u^{\rho}}\right)-\int_{\\{u>\varepsilon\\}}\frac{\rho|\nabla
u|^{2}}{u}\left(\frac{1}{u}+\frac{u^{\rho-1}-\varepsilon^{\rho-1}}{(\rho-1)u^{\rho}}\right)$
$\displaystyle=-\int_{[\\{\varepsilon<u<s_{1}\\}\cap\omega]\cup\\{u\geq
s_{1}\\}}\frac{\rho|\nabla
u|^{2}}{u}\left(\frac{1}{u}+\frac{u^{\rho-1}-\varepsilon^{\rho-1}}{(\rho-1)u^{\rho}}\right)$
$\displaystyle\geq-\int_{\\{u\geq\min\\{s_{1},c\\}\\}}\frac{\rho|\nabla
u|^{2}}{u}\left(\frac{1}{u}+\frac{u^{\rho-1}-\varepsilon^{\rho-1}}{(\rho-1)u^{\rho}}\right)\geq-C.$

In conclusion, from (4.3), (4.5) and from the previous discussion we deduce
that

$\int_{\\{u>\varepsilon\\}}\frac{\rho|\nabla
u|^{2}}{u^{2}}\leq\gamma\int_{\Omega}v^{\gamma-1}h(x)\varphi_{\varepsilon}(u)+C.$

Since $\varphi_{\varepsilon}(s)\leq\frac{\rho}{(\rho-1)s}$ for all $s>0$, we
finally deduce that

$\int_{\\{u>\varepsilon\\}}\frac{\rho|\nabla u|^{2}}{u^{2}}\leq
C\left(\int_{\Omega}\frac{v^{\gamma-1}|h(x)|}{u}+1\right)=C\left(\int_{\Omega}\frac{|h(x)|}{v}+1\right).$

Therefore, we let $\varepsilon$ tend to zero and by virtue of Fatou’s lemma we
obtain that

$\int_{\Omega}\frac{|\nabla u|^{2}}{u^{2}}\leq
C\left(\int_{\Omega}\frac{|h(x)|}{v}+1\right).$

Now, in [41] it is proven that $\int_{\Omega}\frac{|\nabla
u|^{2}}{u^{2}}=+\infty$. Therefore, $\int_{\Omega}\frac{|h(x)|}{v}=+\infty$
and the proof is finished. ∎

We are ready now to prove Theorem 1.1.

###### Proof of Theorem 1.1.

The existence of a solution $u\in H_{0}^{1}(\Omega)$ (not necessarily bounded)
is a consequence of [32, Theorem 3.1]. Moreover, since $q>\frac{N}{2}$, the
well-known Stampacchia’s lemma (see [40]) implies that $u\in
L^{\infty}(\Omega)$.

On the other hand, Lemma 5.4 leads to $u\in C(\overline{\Omega})$.
Furthermore, $u\in W^{1,N}_{\mbox{\tiny loc}}(\Omega)$ by virtue of Lemma 5.2
and Remark 5.3 in the Appendix below. Therefore, the uniqueness of solution
follows directly from Theorem 2.3.

Finally, it is clear that, if $h\gneq 0$ has compact support, then
$\int_{\Omega}\frac{h(x)}{u}<+\infty$ for every solution $u$ to (1.3). Thus,
Theorem 2.12 implies the nonexistence part of the theorem. ∎

Now we prove Theorem 2.4. Since the proofs of Theorem 2.6 and Theorem 2.7 are
analogous, we will only make some comments below.

###### Proof of Theorem 2.4.

We divide the proof into several steps.

Step 1) An auxiliary problem

In addition to the hypotheses of Theorem 2.4, let us further assume for now
that $p<2^{*}-1$ and $g\gneq 0$. At the end of the proof of the theorem we
will make the extension to general $p,g$.

For every $0\lneq v\in C(\overline{\Omega})$, $t\geq 0$, $\lambda>0$, we
consider the following problem:

($Q^{t}$) $\begin{cases}\displaystyle-\Delta u+(v+\delta)g(x,v)\frac{|\nabla
u|^{2}}{u+\delta}=\lambda f(v)+tv^{\sigma},\quad&x\in\Omega,\\\
u>0,\quad&x\in\Omega,\\\ u=0,\quad&x\in\partial\Omega,\end{cases}$

where $\sigma$ is the same real number that appears in condition ($g_{*}$);
note that $\sigma\in(0,1)$ since $g\gneq 0$. It is clear that, if $u$ is a
solution to ($Q^{t}$) with $t=0$ and $v=u$, then it is actually a solution to
($P_{\lambda}$). In this step we will show that, if $v\gneq 0$, there exists a
unique finite energy solution to ($Q^{t}$).

Observe that, if $0\lneq v\in C(\overline{\Omega})$, then $0\lneq\lambda
f(v)+tv^{\sigma}\in C(\overline{\Omega})$ too. Therefore, if $\delta=0$, then
Theorem 1.1 implies directly that there exists a unique finite energy solution
to ($Q^{t}$), that we will denote as $u_{0}$. On the other hand, let us assume
that $\delta>0$. Then, $u_{0}$ is clearly a subsolution to ($Q^{t}$). Let us
consider the unique solution $w\in H_{0}^{1}(\Omega)\cap C(\overline{\Omega})$
to the linear problem

$\begin{cases}-\Delta w=\lambda f(v)+tv^{\sigma},&x\in\Omega,\\\
w>0,&x\in\Omega,\\\ w=0,&x\in\partial\Omega.\end{cases}$

Then, $w$ is a supersolution to ($Q^{t}$). Furthermore, $u_{0}\leq w$ in
$\Omega$ by comparison, so [13, Théorème 3.1] implies that there exists a
finite energy solution $u_{\delta}$ to ($Q^{t}$) such that $0<u_{0}\leq
u_{\delta}\leq w$ in $\Omega$. Moreover, every solution to ($Q^{t}$) belongs
to $C(\overline{\Omega})\cap W^{1,N}_{\mbox{\tiny loc}}(\Omega)$ by virtue of
Lemma 5.4, Lemma 5.2 and Remark 5.3. Also, the function
$(x,s)\mapsto\frac{(v(x)+\delta)g(x,v(x))}{s+\delta}$ satisfies ($g_{1}$) and
($g_{\nearrow}$). Therefore, Theorem 2.3 implies that $u_{\delta}$ is the
unique solution to ($Q^{t}$).

Step 2) Existence of a finite energy solution to ($P_{\lambda}$) if
$p<2^{*}-1$ and $g\gneq 0$.

Let $X=\\{w\in C(\overline{\Omega}):w(x)\geq 0\,\,\forall
x\in\overline{\Omega}\\}$. We have shown in the previous step that the
operator $K:X\times[0,+\infty)\to X$ given by

$\begin{cases}K(v,t)=u,\quad\forall v\in X\setminus\\{0\\},\,\,t\geq 0\text{,
where $u$ is the unique solution to \eqref{auxprob}},\\\ K(0,t)=0,\quad\forall
t\geq 0.\end{cases}$

is well-defined. We aim to prove that there exists $0\lneq u\in X$ such that
$K(u,0)=u$, which is equivalent to finding a solution to ($P_{\lambda}$).

We first prove that $K$ is continuous. Indeed, let $\\{v_{n}\\}\subset X$ and
$\\{t_{n}\\}\subset[0,+\infty)$ be such that $v_{n}\to v$ in
$C(\overline{\Omega})$ for some $v\in X$ and $t_{n}\to t$ for some $t\geq 0$.
Let us denote $u_{n}=K(v_{n},t_{n})$ and $h_{n}=\lambda
f(v_{n})+t_{n}v_{n}^{\sigma}$. Taking into account that $g\geq 0$, it is clear
that

(4.6) $-\Delta u_{n}\leq h_{n},\quad x\in\Omega,\,\,\forall n.$

We know that there exists $C>0$ such that $h_{n}\leq C$ for all $n$. Then,
(4.6) leads to $-\Delta u_{n}\leq C$ in $\Omega$. From this inequality, it is
straightforward to prove that $\\{u_{n}\\}$ is bounded in $H_{0}^{1}(\Omega)$.
Hence, up to a subsequence, $u_{n}\rightharpoonup u$ weakly in
$H_{0}^{1}(\Omega)$ and $u_{n}\to u$ strongly in $L^{m}(\Omega)$ for some
$0\leq u\in H_{0}^{1}(\Omega)$ and for all $m\in[1,2^{*})$. Furthermore, the
inequality (4.6) yields the following estimate by virtue of the well-known
Stampacchia’s lemma (see [40]):

(4.7) $\|u_{n}\|_{L^{\infty}(\Omega)}\leq
C\|h_{n}\|_{L^{\infty}(\Omega)},\quad\forall n.$

In particular, if $v\equiv 0$, (4.7) yields directly that $u_{n}\to 0$
strongly in $C(\overline{\Omega})$. That is to say, $K$ is continuous in
$\\{0\\}\times[0,+\infty)$.

Let us assume on the contrary that $v\gneq 0$. In this case, we will prove
following [32] that $\\{u_{n}\\}$ is locally bounded away from zero. Indeed,
first of all observe that, using that $h_{n}\leq C$, we deduce from (4.7) that
$\\{u_{n}\\}$ is bounded in $L^{\infty}(\Omega)$. Hence, using ($f_{*}$),
($g_{*}$) and the $L^{\infty}$ estimate on $u_{n}$, one can easily check that
the function $w_{n}=u_{n}^{1-\sigma}$ satisfies

$-\Delta w_{n}\geq\frac{(1-\sigma)\lambda v_{n}^{p}}{u_{n}^{\sigma}}\geq
Cv_{n}^{p},\quad x\in\Omega,$

for some constant $C>0$. On the other hand, let $z_{n}\in
H_{0}^{1}(\Omega)\cap C(\overline{\Omega})$ be the unique solution to the
linear problem

$\begin{cases}-\Delta z_{n}=Cv_{n}^{p},&x\in\Omega,\\\
z_{n}=0,&x\in\partial\Omega.\end{cases}$

By comparison, $w_{n}\geq z_{n}$ for all $n$. Moreover, it is clear that
$\\{z_{n}\\}$ is bounded in $H_{0}^{1}(\Omega)$ and, following [32] (which in
turn is based on [28]), we also have that $\\{z_{n}\\}$ is bounded in
$C^{0,\alpha}(\overline{\Omega})$ for some $\alpha\in(0,1)$. Hence, passing to
a subsequence, $z_{n}$ converges weakly in $H_{0}^{1}(\Omega)$ and strongly in
$C(\overline{\Omega})$ to the unique solution $z\in H_{0}^{1}(\Omega)\cap
C(\overline{\Omega})$ to

$\begin{cases}-\Delta z=Cv^{p},&x\in\Omega,\\\
z=0,&x\in\partial\Omega.\end{cases}$

Since $v\gneq 0$, the strong maximum principle implies that, for all
$\omega\subset\subset\Omega$, there exists $c_{\omega}>0$ such that $z\geq
2c_{\omega}$ in $\omega$. Furthermore, since $z_{n}\to z$ uniformly in
$\overline{\Omega}$, it follows that $z_{n}\geq z-c_{\omega}$ in
$\overline{\Omega}$ for all $n$ large enough. In sum,

$u_{n}^{1-\sigma}=w_{n}\geq z_{n}\geq c_{\omega},\quad x\in\omega$

for all $n$ large enough, as we claimed.

It is clear now that $\\{\Delta u_{n}\\}$ is bounded in $L^{1}_{\mbox{\tiny
loc}}(\Omega)$. Hence, from [12] we deduce that $\nabla u_{n}\to\nabla u$ a.e.
in $\Omega$. We may now pass to the limit using Fatou’s lemma twice, similarly
as in the proof of Proposition 3.7 (Case 1, Step 1.3), so that $u$ is the
unique solution to ($Q^{t}$) i.e. $u=K(v,t)$.

It remains to prove that $u_{n}\to u$ strongly in $C(\overline{\Omega})$.
Indeed, since $\\{u_{n}\\}$ is locally bounded away from zero, Lemma 5.2
implies that, for every $\omega\subset\subset\Omega$, one may take a
subsequence such that $u_{n}\to u$ strongly in $C(\overline{\omega})$.
Actually, the uniqueness of $u$ assures that the original sequence
$\\{u_{n}\\}$, and not merely a subsequence, converges itself to $u$ strongly
in $C(\overline{\omega})$. On the other hand, let $\zeta_{n}\in
H_{0}^{1}(\Omega)\cap C(\overline{\Omega})$ be the unique solution to

$\begin{cases}-\Delta\zeta_{n}=h_{n},&x\in\Omega,\\\
\zeta_{n}=0,&x\in\partial\Omega.\end{cases}$

By comparison, $u_{n}\leq\zeta_{n}$ in $\Omega$ for all $n$. Moreover,
$\zeta_{n}\to\zeta$ strongly in $C(\overline{\Omega})$, where $\zeta\in
H_{0}^{1}(\Omega)\cap C(\overline{\Omega})$ is the unique solution to

$\begin{cases}-\Delta\zeta=\lambda f(v)+tv^{\sigma},&x\in\Omega,\\\
\zeta=0,&x\in\partial\Omega.\end{cases}$

In particular, for fixed $\varepsilon>0$, we deduce that

$|u_{n}(x)-u(x)|\leq\zeta_{n}(x)+u(x)<\zeta(x)+u(x)+\frac{\varepsilon}{2}\quad\forall
x\in\overline{\Omega},\,\,\forall n\text{ large enough.}$

Therefore, since $u,\zeta\in C(\overline{\Omega})$ with $u=\zeta=0$ on
$\partial\Omega$, there exits $\omega\subset\subset\Omega$ such that

$|u_{n}(x)-u(x)|<\varepsilon\quad\forall
x\in\overline{\Omega}\setminus\omega,\,\,\forall n\text{ large enough.}$

Since $u_{n}\to u$ strongly in $C(\overline{\omega})$, we conclude that

$|u_{n}(x)-u(x)|<\varepsilon\quad\forall x\in\overline{\Omega},\,\,\forall
n\text{ large enough.}$

In other words, $u_{n}\to u$ strongly in $C(\overline{\Omega})$, so $K$ is
continuous.

We prove now that $K$ is compact, i.e. it maps bounded sets to relatively
compact sets. Indeed, let $\\{v_{n}\\}\subset X$ and
$\\{t_{n}\\}\subset[0,+\infty)$ be bounded sequences. Taking subsequences,
$v_{n}\to v$ in the weak-$\star$ topology of $L^{\infty}(\Omega)$ for some
$v\in L^{\infty}(\Omega)$, while $t_{n}\to t$ for some $t\geq 0$. This is
enough to pass to the limit in the equations as above. We conclude that, up to
subsequences, $K(v_{n},t_{n})\to K(v,t)$ strongly in $C(\overline{\Omega})$.
This proves that $K$ is compact.

We will prove next that there exist $0<r<R$ and $t_{1}\geq 0$ such that

  1. (1)

$u\not=sK(u,0)\,\,\forall s\in[0,1],\,\,\forall u\in X$ with
$\|u\|_{L^{\infty}(\Omega)}=r$,

  2. (2)

$u\not=K(u,t)\,\,\forall t\geq 0,\,\,\forall u\in X$ with
$\|u\|_{L^{\infty}(\Omega)}=R$,

  3. (3)

$u\not=K(u,t)\,\,\forall t\geq t_{1},\,\,\forall u\in X$ with
$\|u\|_{L^{\infty}(\Omega)}\leq R$.

In order to prove item (1), let us assume by contradiction that, for all
$r>0$, there exist $s\in[0,1]$ and $u\in X$ with
$\|u\|_{L^{\infty}(\Omega)}=r$ such that $u=sK(u,0)$. In particular, $u>0$ in
$\Omega$ and

$\begin{cases}-\Delta u\leq s\lambda f(u),&x\in\Omega,\\\ u=0,\
&x\in\partial\Omega.\end{cases}$

Since ($f_{0}$) holds, we can choose $r>0$ such that $\lambda
f(t)\leq\frac{\lambda_{1}}{2}t$ for all $t\in[0,r]$, where $\lambda_{1}$
stands for the principal eigenvalue of $-\Delta$ in $\Omega$ with zero
Dirichlet boundary conditions. Furthermore, $u\leq r$ in $\Omega$, so we have
that

$\begin{cases}\displaystyle-\Delta u\leq\frac{\lambda_{1}}{2}u,&x\in\Omega,\\\
u>0,&x\in\Omega,\\\ u=0,&x\in\partial\Omega.\end{cases}$

This is a clear contradiction with the definition of $\lambda_{1}$.

On the other hand, if we take $R>C$, where $\|u\|_{L^{\infty}(\Omega)}\leq C$
(see Proposition 3.7) then it is clear that (2) holds. Moreover, if we take
$t_{1}>t_{0}$, where $t_{0}>0$ is given by Proposition 3.5, then (3) also
holds.

In conclusion, [21, Proposition 2.1 and Remark 2.1] can be applied and in
consequence we obtain a positive fixed point of $K_{0}$, i.e. a finite energy
solution to ($P_{\lambda}$).

Step 3) Extension to the general case

From this point we assume only the hypotheses of Theorem 2.4, i.e. $g$ need
not be non-negative and $p$ might be greater than or equal to $2^{*}-1$.

Fix $\lambda>0$. For some $\gamma>1$ that will be chosen later, let us
consider the following problem:

(4.8) $\begin{cases}\displaystyle-\Delta v+g_{\gamma}(x,v)|\nabla
v|^{2}=\lambda f_{\gamma}(v),&x\in\Omega,\\\ v>0,&x\in\Omega,\\\
v=0,&x\in\partial\Omega,\end{cases}$

where

$\displaystyle f_{\gamma}(s)$
$\displaystyle=b(s+\delta^{\gamma})^{\frac{\gamma-1}{\gamma}}f\left((s+\delta^{\gamma})^{\frac{1}{\gamma}}-\delta\right),\quad\forall
s\geq 0,$ $\displaystyle g_{\gamma}(x,s)$
$\displaystyle=\frac{(s+\delta^{\gamma})^{\frac{1}{\gamma}}g\left(x,(s+\delta^{\gamma})^{\frac{1}{\gamma}}-\delta\right)+\gamma-1}{\gamma(s+\delta^{\gamma})},\quad\text{a.e.
}x\in\Omega,\,\,\forall s>0,$

and $b>0$ is a constant, which depends only on $p,\delta,\gamma$, that will be
chosen below too. It is easy to see that, if $v$ is a solution to (4.8), then
$u=(v+\delta^{\gamma})^{\frac{1}{\gamma}}-\delta$ is a solution to
($P_{\lambda}$). Moreover, if $v\in H_{0}^{1}(\Omega)$ and $\delta>0$, then
$u\in H_{0}^{1}(\Omega)$ too, so it has finite energy. We will now choose
$\gamma>1$ so that problem (4.8) admits a finite energy solution.

Indeed, let us first denote $p_{\gamma}=\frac{\gamma-1+p}{\gamma}$. Obviously,
$p_{\gamma}>1$. It is easy to see that the function

$s\mapsto\frac{s^{p_{\gamma}}}{\left(s+\delta^{\gamma}\right)^{\frac{\gamma-1}{\gamma}}\left[\left(s+\delta^{\gamma}\right)^{\frac{1}{\gamma}}-\delta\right]^{p}}$

is bounded in $[0,+\infty)$. Then, using that $f$ satisfies ($f_{*}$), $b>0$
can be chosen (depending only on $p,\delta,\gamma$) such that

$f_{\gamma}(s)\geq s^{p_{\gamma}}\quad\forall s\geq 0.$

Moreover, since $f$ satisfies ($f_{*}$), it is clear that

$f_{\gamma}(s)\leq ab\gamma(s+\delta^{\gamma})^{p_{\gamma}}\quad\forall s\geq
0.$

We have proved that $f_{\gamma}$ satisfies ($f_{*}$). Taking into account that
$f$ satisfies ($f_{\infty}$) and ($f_{0}$), it is straightforward to check
that also $f_{\gamma}$ satisfies ($f_{\infty}$) and ($f_{0}$) respectively.

On the other hand, since $g$ satisfies ($g_{\infty}$), then

$\lim_{s\to+\infty}\left\|sg_{\gamma}(\cdot,s)-\frac{\mu+\gamma-1}{\gamma}\right\|_{L^{\infty}(\Omega)}=0.$

Using that $\max_{x\in\overline{\Omega}}\mu(x)<\frac{2^{*}-1-p}{2^{*}-2}$, it
is easy to see that

$\max_{x\in\overline{\Omega}}\frac{\mu(x)+\gamma-1}{\gamma}<\frac{2^{*}-1-p_{\gamma}}{2^{*}-2}.$

Thus, $g_{\gamma}$ satisfies ($g_{\infty}$). It is also immediate to check
that $g_{\gamma}$ satisfies ($g_{0}$) using that $g$ does too. Notice finally
that, since $g$ satisfies ($g_{*}$), then

$\frac{\tau+\gamma-1}{\gamma}\leq\left(s+\delta^{\gamma}\right)g_{\gamma}(s,x)\leq\frac{\sigma+\gamma-1}{\gamma}\quad\text{a.e.
}x\in\Omega,\,\,\forall s>0.$

It is clear that

$2\frac{\sigma+\gamma-1}{\gamma}-1<\frac{\tau+\gamma-1}{\gamma}\leq\frac{\sigma+\gamma-1}{\gamma}<1.$

Therefore, $g_{\gamma}$ satisfies ($g_{*}$).

If we now choose $\gamma>1$ large enough so that
$\frac{\tau+\gamma-1}{\gamma}>0$ and $p_{\gamma}<2^{*}-1$, then we may apply
the previous step in order to obtain a finite energy solution to (4.8). The
proof is now finished. ∎

###### Proofs of Theorem 2.6 and Theorem 2.7.

The proofs are analogous to that for Theorem 2.4. The only essential
difference lies in how the estimates on the solutions are obtained. In fact,
instead of Proposition 3.7, we have to apply Proposition 3.9 for Theorem 2.6
and both Proposition 3.8 and Proposition 3.10 for Theorem 2.7. ∎

Here we carry out the proof of Theorem 2.8.

###### Proof of Theorem 2.8.

Let us define the Carathéodory function
$\bar{g}:\Omega\times(0,+\infty)\to\mathbb{R}$ by

$\bar{g}(x,s)=\begin{cases}g(x,s)\quad&\text{if
}x\in\Omega,\,\,s\in(0,s_{0}),\\\
\displaystyle\frac{s_{0}g(x,s_{0})}{s}\quad&\text{if }x\in\Omega,\,\,s\geq
s_{0}.\end{cases}$

Let us also consider the continuous function
$\bar{f}:[0,+\infty)\to[0,+\infty)$ defined by

$\bar{f}(s)=\begin{cases}f(s)\quad&\text{if }s\in(0,s_{0}),\\\
\displaystyle\frac{f(s_{0})}{s_{0}^{p}}s^{p}&\text{if }s\geq
s_{0}.\end{cases}$

It is clear that $\bar{f}$ and $\bar{g}$ satisfy the hypotheses of Theorem
2.7. Thus, if we write $(\bar{P}_{\lambda})$ for denoting problem
($P_{\lambda}$) with $\bar{g}$ and $\bar{f}$ instead of $g$ and $f$, then
Theorem 2.7 implies that there exists a solution $u_{\lambda}$ to
$(\bar{P}_{\lambda})$ for all $\lambda>0$ that satisfies

$\lambda^{\frac{1}{p-1}}\|u_{\lambda}\|_{L^{\infty}(\Omega)}\leq
C\quad\forall\lambda>0.$

In particular, there exists $\lambda_{0}>0$ such that
$\|u_{\lambda}\|_{L^{\infty}(\Omega)}<s_{0}$ for every
$\lambda\geq\lambda_{0}$. Hence, $u_{\lambda}$ is, in fact, a solution to
($P_{\lambda}$) for every $\lambda\geq\lambda_{0}$. The proof is finished. ∎

It is left to prove Theorem 2.10.

###### Proof of Theorem 2.10.

Let
$\sigma\in\left(0,\min\left\\{\frac{1}{2},\frac{2^{*}-1-p}{2^{*}-2}\right\\}\right)$.
By virtue of Remark 2.9, we may take $\delta\in(0,1)$ such that

$sg(x,s)\leq\sigma\quad\forall(x,s)\in\overline{\Omega}\times(0,\delta].$

Notice that $\delta$ can be chosen as small as necessary. In fact, it will be
taken small enough several times throughout the proof, depending its size only
on $N,p,G$.

Let us define the function
$\bar{g}:\overline{\Omega}\times(0,+\infty)\to[0,+\infty)$ by

$\bar{g}(x,s)=\begin{cases}g(x,s)\quad&\text{if
}(x,s)\in\overline{\Omega}\times(0,\delta],\\\ \displaystyle\frac{\delta
g(x,\delta)}{s}\quad&\text{if
}(x,s)\in\overline{\Omega}\times(\delta,+\infty).\end{cases}$

It is clear that $\bar{g}$ satisfies ($g_{*}$), ($g_{\infty}$) and ($g_{0}$).
Thus, if we write $(\bar{P}_{\lambda})$ for denoting problem ($P_{\lambda}$)
with $\bar{g}$ instead of $g$, then Theorem 2.4 implies that there exists a
solution to $(\bar{P}_{\lambda})$ for all $\lambda>0$. We aim now to prove
that the hypotheses of Proposition 3.11 hold true, so that there exists $C>0$
such that

(4.9) $\lambda^{\frac{1}{p-1}}\|u\|_{L^{\infty}(\Omega)}\leq C\quad\forall
u\text{ solution to }(\bar{P}_{\lambda})\text{ with }\lambda>0.$

Observe that, if this is true, in particular there exists $\lambda_{0}>0$ such
that $\|u\|_{L^{\infty}(\Omega)}\leq\delta$ for every solution $u$ to
$(\bar{P}_{\lambda})$ with $\lambda\geq\lambda_{0}$. Hence, the solutions to
$(\bar{P}_{\lambda})$ with $\lambda\geq\lambda_{0}$ are, in fact, solution to
($P_{\lambda}$). Thus, the proof will finish as soon as we prove (4.9).

We will now prove that (3.11) in Proposition 3.11 holds. Indeed, for any
$L>0$, $x_{0}\in\overline{\Omega}$, let us consider the function
$\psi:[0,+\infty)\to[0,+\infty)$ given by

$\psi(s)=\int_{0}^{s}e^{-\frac{1}{L}\int_{\delta
L}^{t}\bar{g}\left(x_{0},\frac{r}{L}\right)dr}dt\quad\forall s\geq 0.$

It is straightforward to check that

$\psi(s)=L\int_{0}^{\frac{s}{L}}e^{-\int_{\delta}^{t}\bar{g}\left(x_{0},r\right)dr}dt\quad\forall
s\geq 0.$

We will prove next that
$s\mapsto\frac{\psi^{\prime}(s)s^{p}}{\psi(s)^{2^{*}-1}}$ is decreasing for
all $s>0$, or equivalently, that

(4.10)
$(p-s\bar{g}(x_{0},s))\frac{\psi(sL)}{L}<(2^{*}-1)s\psi^{\prime}(sL)\quad\forall
s>0.$

Notice that neither $\frac{\psi(sL)}{L}$ nor $\psi^{\prime}(sL)$ depends on
$L$.

In order to prove (4.10), we distinguish two possibilities: either
$0<s\leq\delta$ or $s>\delta$. Assume first that $0<s\leq\delta$. Then,

$\displaystyle(p-s\bar{g}(x_{0},s))\frac{\psi(sL)}{L}-(2^{*}-1)s\psi^{\prime}(sL)$
$\displaystyle=$
$\displaystyle(p-sg(x_{0},s))\int_{0}^{s}e^{-\int_{\delta}^{t}g(x_{0},r)dr}dt-(2^{*}-1)se^{-\int_{\delta}^{s}g(x_{0},r)dr}.$

It follows from the mean value theorem that there exists $\theta\in(0,1)$ such
that

$\displaystyle(p-sg(x_{0},s))\int_{0}^{s}e^{-\int_{\delta}^{t}g(x_{0},r)dr}dt-(2^{*}-1)se^{-\int_{\delta}^{s}g(x_{0},r)dr}$
$\displaystyle=$ $\displaystyle
se^{\int_{s}^{\delta}g(x_{0},r)dr}\left[(p-sg(x_{0},s))e^{\int_{\theta
s}^{s}g(x_{0},r)dr}-(2^{*}-1)\right].$

Thanks to (G) we derive

$(p-sg(x_{0},s))e^{\int_{\theta s}^{s}g(x_{0},r)dr}-(2^{*}-1)\leq
pe^{\int_{0}^{\delta}G(r)dr}-(2^{*}-1).$

Using that $\lim_{\delta\to 0}\int_{0}^{\delta}G(r)dr=0$ and $p<2^{*}-1$, we
are allowed to choose $\delta$ small enough so that

$pe^{\int_{0}^{\delta}G(r)dr}<2^{*}-1.$

Hence, (4.10) holds for $0<s\leq\delta$.

We assume now that $s>\delta$. In this case,

$\psi(s)=AL+\delta^{\tau}L\frac{\left(\frac{s}{L}\right)^{1-\tau}-\delta^{1-\tau}}{1-\tau},$

where $\tau=\delta g(x_{0},\delta)$ and

$A=\frac{\psi(\delta
L)}{L}=\int_{0}^{\delta}e^{\int_{t}^{\delta}g(x_{0},r)dr}dt.$

Recall that $\tau\leq\sigma<\frac{2^{*}-1-p}{2^{*}-2}$. Then,

$\frac{p-\tau}{1-\tau}\leq\frac{p-\sigma}{1-\sigma}<2^{*}-1.$

Let us denote $\gamma=2^{*}-1-\frac{p-\sigma}{1-\sigma}$. We deduce that

$\displaystyle(p$
$\displaystyle-s\bar{g}(x_{0},s))\frac{\psi(sL)}{L}-(2^{*}-1)s\psi^{\prime}(sL)$
$\displaystyle=(p-\tau)\left(A+\delta^{\tau}\frac{s^{1-\tau}-\delta^{1-\tau}}{1-\tau}\right)-(2^{*}-1)\delta^{\tau}s^{1-\tau}$
$\displaystyle\leq(2^{*}-1)\left((1-\tau)A+\delta^{\tau}(s^{1-\tau}-\delta^{1-\tau})-\delta^{\tau}s^{1-\tau}\right)-\gamma\left(A+\delta^{\tau}\frac{s^{1-\tau}-\delta^{1-\tau}}{1-\tau}\right)$
$\displaystyle\leq(2^{*}-1)(A-\delta)-\gamma
A=A\left((2^{*}-1)\left(1-\frac{\delta}{A}\right)-\gamma\right)$
$\displaystyle\leq
A\left((2^{*}-1)\left[1-\delta\left(\int_{0}^{\delta}e^{\int_{t}^{\delta}G(r)dr}dt\right)^{-1}\right]-\gamma\right).$

It is clear that

$\lim_{\delta\to
0}\frac{\int_{0}^{\delta}e^{\int_{t}^{\delta}G(r)dr}dt}{\delta}=\lim_{\delta\to
0}\frac{e^{\int_{0}^{\delta}G(r)dr}\int_{0}^{\delta}e^{-\int_{0}^{t}G(r)dr}dt}{\delta}=\lim_{\delta\to
0}e^{-\int_{0}^{\delta}G(r)dr}=1.$

Then, we can take $\delta$ small enough so that

$(2^{*}-1)\left[1-\delta\left(\int_{0}^{\delta}e^{\int_{t}^{\delta}G(r)dr}dt\right)^{-1}\right]<\gamma.$

In conclusion, (4.10) holds for $s>\delta$, and thus, for all $s>0$.

In sum, it follows that the hypotheses of Proposition 3.11 are fulfilled for
problem $(\bar{P}_{\lambda})$ so our claim was true. The proof is now
finished. ∎

We conclude the section by proving Theorem 1.2, Theorem 1.3 and Theorem 1.4 as
consequences of the general results in Section 2.

###### Proof of Theorem 1.2.

The first item of the theorem is a direct consequence of Theorem 2.4, while
the second one follows from Theorem 2.12. ∎

###### Proof of Theorem 1.3.

Let us denote $g(x,s)=\frac{\mu(x)}{(s+\delta)^{\gamma}}$. In order to prove
the first item, we assume that $\delta>0$ and that $\mu$ satisfies (1.4). It
is easy to see that

$\inf_{(x,s)\in\Omega\times[0,+\infty)}(s+\delta)g(x,s)=-\frac{\|\mu^{-}\|_{L^{\infty}(\Omega)}}{\delta^{\gamma-1}},\quad\sup_{(x,s)\in\Omega\times[0,+\infty)}(s+\delta)g(x,s)=\frac{\|\mu^{+}\|_{L^{\infty}(\Omega)}}{\delta^{\gamma-1}}.$

Therefore, $\tau\leq(s+\delta)g(x,s)\leq\sigma$ for
$\tau=-\frac{\|\mu^{-}\|_{L^{\infty}(\Omega)}}{\delta^{\gamma-1}}$ and
$\sigma=\frac{\|\mu^{+}\|_{L^{\infty}(\Omega)}}{\delta^{\gamma-1}}$. Hence,
(1.4) implies that $g$ satisfies ($g_{*}$). On the other hand, it is clear
that $\|(s+\delta)g(\cdot,s)\|_{L^{\infty}(\Omega)}\to 0$ as $s\to+\infty$.
Thus, $g$ also satisfies ($g_{\infty}$). In conclusion, Theorem 2.4 implies
that there exists a finite energy solution to (1.1) for any $\lambda>0$.

On the contrary, let us assume now that $\delta=0$ and $\mu(x)\geq\tau>0$ for
a.e. $x\in\Omega\setminus\omega$. Then, ($g_{2}$) holds. Thus, Theorem 2.12
implies that problem (1.1) admits no solution for any $\lambda>0$. ∎

###### Proof of Theorem 1.4.

Let us denote $g(x,s)=\frac{\mu(x)}{(s+\delta)^{\gamma}}$. We know that, for
every $\varepsilon>0$, there exists $s_{0}>0$ such that

(4.11)
$s|g(x,s)|\leq\|\mu\|_{L^{\infty}(\Omega)}\frac{s}{(s+\delta)^{\gamma}}<\varepsilon\quad\forall
s\in(0,s_{0}),\,\,\text{a.e. }x\in\Omega.$

Therefore, ($\widetilde{g_{0}}$) holds (in particular, $\lim_{s\to 0}sg(x,s)$
obviously exists for a.e. $x\in\Omega$). If we assume that $0\lneq\mu\in
C(\overline{\Omega})$ and $p<2^{*}-1$, then the hypotheses of Theorem 2.10 are
fulfilled, so we conclude.

Let us assume on the contrary that $p<\frac{N+1}{N-1}$. Then, in (4.11) we may
take
$\sigma=-\tau=\varepsilon\in\left(0,\min\left\\{\frac{1}{3},\frac{N+1-(N-1)p}{2}\right\\}\right)$
in such a way that $g$ satisfies ($\widetilde{g_{*}}$). Thus, the result
follows after applying Theorem 2.8.

The proof is finished. ∎

##  5\. Appendix

###  5.1. Technical lemma

In this subsection we prove a technical result that is required by the blow-up
method.

######  Lemma 5.1.

Let $\Omega\subset\mathbb{R}^{N}(N\geq 3)$ be a bounded domain with boundary
of class $\mathcal{C}^{2}$, $\lambda>0$,
$g:\Omega\times(0,+\infty)\to\mathbb{R}$ be a Carathéodory function, and
$f:[0,+\infty)\to\mathbb{R}$ be a continuous function. Then, for every
$x_{0}\in\partial\Omega$, there exist $U\subset\mathbb{R}^{N}$ a neighborhood
of $x_{0}$ and an injective and $C^{2}$ map $y:U\to\mathbb{R}^{N}$, with
$C^{2}$ inverse, such that $V=y(U\cap\Omega)\subset\mathbb{R}^{N}_{+}$,
$\Gamma=y(U\cap\partial\Omega)\subset\partial V\cap\partial\mathbb{R}^{N}_{+}$
and, if $u$ is a solution to ($P_{\lambda}$), then the function $v=u\circ
y^{-1}:V\to(0,+\infty)$ is a solution to

(5.1) $\begin{cases}-\text{{\rm div}}(M(y)\nabla v)+b(y)\nabla
v+j(y,v)M(y)\nabla v\nabla v=f(v),\quad&y\in V,\\\ v>0,\quad&y\in V,\\\
v=0,\quad&y\in\Gamma,\end{cases}$

where $M\in C^{1}(\overline{V})^{N\times N}$ is uniformly elliptic, $b\in
C(\overline{V})^{N}$ and $j(\cdot,\cdot)=g(y^{-1}(\cdot),\cdot)$, being $M$
and $b$ independent of $u$.

###### Proof.

Since $\partial\Omega$ is of class $\mathcal{C}^{2}$, there exist
$U\subset\mathbb{R}^{N}$ a neighborhood of $x_{0}$ and a $C^{2}$ function
$\psi:U^{\prime}\to\mathbb{R}$, where
$U^{\prime}=\\{x^{\prime}\in\mathbb{R}^{N-1}:\exists
x_{N}\in\mathbb{R},(x^{\prime},x_{N})\in U\\}$, such that

$\displaystyle\psi(x^{\prime})<x_{N}\quad$
$\displaystyle\forall(x^{\prime},x_{N})\in U\cap\Omega,$
$\displaystyle\psi(x^{\prime})=x_{N}\quad$
$\displaystyle\forall(x^{\prime},x_{N})\in U\cap\partial\Omega.$

Let us define the change of variables $y:U\to\mathbb{R}^{N}$ by

$y(x)=(x^{\prime},x_{N}-\psi(x^{\prime}))\quad\forall x\in U.$

It is clear that

$\displaystyle y(x)\in\mathbb{R}^{N}_{+}\quad$
$\displaystyle\forall(x^{\prime},x_{N})\in U\cap\Omega,$ $\displaystyle
y(x)\in\partial\mathbb{R}^{N}_{+}\quad$
$\displaystyle\forall(x^{\prime},x_{N})\in U\cap\partial\Omega.$

This proves that $V\subset\mathbb{R}^{N}_{+}$ and
$\Gamma\subset\partial\mathbb{R}^{N}_{+}$. Moreover, it is a simple exercise
to prove that $\Gamma\subset\partial V$. It is also easy to see that the
function $y^{-1}:y(U)\to U$ given by

$y^{-1}(z)=(z^{\prime},z_{N}+\psi(z^{\prime}))\quad\forall z\in y(U)$

is the inverse function of $y$. Note that $y^{-1}$ is well defined since
$z^{\prime}\in U^{\prime}$ for every $z\in y(U)$.

Let us now define $v:V\cup\Gamma\to\mathbb{R}$ by

$v(z)=u(y^{-1}(z))\quad\forall z\in V\cup\Gamma.$

Observe that $v=0$ on $\Gamma$ and $u(x)=v(y(x))$ for all $x\in U\cap\Omega$.
We will show next that $v$ satisfies an equation in $V$.

Now we compute the derivatives that we need. We emphasize that such
derivatives can be understood in a pointwise sense due to Remark 5.3 and to
the $C^{2}$ regularity of $\psi$. We stress also that, as $\psi$ does not
depend on $x_{N}$, it will be understood that $\frac{\partial\psi}{\partial
x_{N}}(x)=0$ for $x\in U$.

  1. a)

$\displaystyle Dy(x)=\left(\begin{array}[]{cccc|c}1&0&\cdots&0&0\\\
0&1&&\vdots&\vdots\\\ \vdots&&\ddots&0&\vdots\\\ 0&\cdots&0&1&0\\\
\hline\cr-\frac{\partial\psi}{\partial
x_{1}}(x^{\prime})&\cdots&\cdots&-\frac{\partial\psi}{\partial
x_{N-1}}(x^{\prime})&1\end{array}\right),$

  2. b)

$\begin{cases}\displaystyle\frac{\partial u}{\partial x_{i}}(x)=\frac{\partial
v}{\partial z_{i}}(y(x))-\frac{\partial v}{\partial
z_{N}}(y(x))\frac{\partial\psi}{\partial x_{i}}(x^{\prime}),\quad
i=1,\cdots,N-1,\\\ \\\ \displaystyle\frac{\partial u}{\partial
x_{N}}(x)=\frac{\partial v}{\partial z_{N}}(y(x)),\end{cases}$

  3. c)

$|\nabla u(x)|^{2}=|\nabla v(y(x))|^{2}+\left(\frac{\partial v}{\partial
z_{N}}(y(x))\right)^{2}|\nabla\psi(x^{\prime})|^{2}-2\frac{\partial
v}{\partial z_{N}}(y(x))\nabla v(y(x))\nabla\psi(x^{\prime}),$

  4. d)

$\displaystyle\frac{\partial^{2}u}{\partial x_{i}^{2}}(x)$
$\displaystyle=\frac{\partial^{2}v}{\partial z_{1}\partial
z_{i}}(y(x))\frac{\partial y_{1}}{\partial
x_{i}}(x)+\cdots+\frac{\partial^{2}v}{\partial z_{N}\partial
z_{i}}(y(x))\frac{\partial y_{N}}{\partial x_{i}}(x)-\frac{\partial
v}{\partial z_{N}}(y(x))\frac{\partial^{2}\psi}{\partial
x_{i}^{2}}(x^{\prime})$ $\displaystyle-\left[\frac{\partial^{2}v}{\partial
z_{1}\partial z_{N}}(y(x))\frac{\partial y_{1}}{\partial
x_{i}}(x)+\cdots+\frac{\partial^{2}v}{\partial z_{N}^{2}}(y(x))\frac{\partial
y_{N}}{\partial x_{i}}(x)\right]\frac{\partial\psi}{\partial
x_{i}}(x^{\prime})$ $\displaystyle=\frac{\partial^{2}v}{\partial
z_{i}^{2}}(y(x))-2\frac{\partial^{2}v}{\partial z_{i}\partial
z_{N}}(y(x))\frac{\partial\psi}{\partial
x_{i}}(x^{\prime})+\frac{\partial^{2}v}{\partial
z_{N}^{2}}(y(x))\left(\frac{\partial\psi}{\partial
x_{i}}(x^{\prime})\right)^{2}$ $\displaystyle-\frac{\partial v}{\partial
z_{N}}(y(x))\frac{\partial^{2}\psi}{\partial x_{i}^{2}}(x^{\prime}),\quad
i=1,\cdots,N-1,$

  5. e)

$\displaystyle\Delta u=\Delta v-2\nabla\frac{\partial v}{\partial
z_{N}}\nabla\psi+\frac{\partial^{2}v}{\partial
z_{N}^{2}}|\nabla\psi|^{2}-\frac{\partial v}{\partial z_{N}}\Delta\psi.$

Let us denote $j(z,s)=g(y^{-1}(z),s)$ for a.e. $z\in V$ and for all $s>0$.
Thus, $v=v(y)$ (from this point $y$ will simply denote variable in $V$)
satisfies the equation

$\displaystyle-\Delta v$ $\displaystyle-\frac{\partial^{2}v}{\partial
y_{N}^{2}}|\nabla\psi|^{2}+2\nabla\frac{\partial v}{\partial
y_{N}}\nabla\psi+\frac{\partial v}{\partial y_{N}}\Delta\psi$ (5.2)
$\displaystyle=f(v)-j(y,v)\left(|\nabla v|^{2}+\left(\frac{\partial
v}{\partial y_{N}}\right)^{2}|\nabla\psi|^{2}-2\frac{\partial v}{\partial
y_{N}}\nabla v\nabla\psi\right),\quad y\in V.$

Let us define the matrix

$M(y)=\left(\begin{array}[]{cccc|c}1&0&\cdots&0&-2\frac{\partial\psi}{\partial
x_{1}}(y^{\prime})\\\ 0&1&&\vdots&\vdots\\\ \vdots&&\ddots&0&\vdots\\\
0&\cdots&0&1&-2\frac{\partial\psi}{\partial x_{N-1}}(y^{\prime})\\\ \hline\cr
0&\cdots&\cdots&0&1+|\nabla\psi(y^{\prime})|^{2}\end{array}\right),\quad y\in
V,$

and also the vector $b(y)=(0,\cdots,0,-\Delta\psi(y^{\prime})),\,y\in V$.
Then, one can check that $v$ is a solution to (5.1).

Moreover, let $a>1$ be such that $(a-1)\|\nabla\psi\|_{L^{\infty}(V)}^{2}<1$.
Then, by Cauchy-Schwarz’s and Young’s inequalities, the following holds for
every $\xi\in\mathbb{R}^{N}$:

$\displaystyle M(y)\xi\xi$
$\displaystyle=|\xi|^{2}-2\xi_{N}\nabla\psi\xi^{\prime}+\xi_{N}^{2}|\nabla\psi|^{2}$
$\displaystyle\geq|\xi|^{2}-(a-1)\xi_{N}^{2}|\nabla\psi|^{2}-\frac{1}{a}|\xi^{\prime}|^{2}$
$\displaystyle=\left(1-\frac{1}{a}\right)|\xi^{\prime}|^{2}+(1-(a-1)|\nabla\psi|^{2})\xi_{N}^{2}$
$\displaystyle\geq\left(1-\frac{1}{a}\right)|\xi^{\prime}|^{2}+(1-(a-1)\|\nabla\psi\|_{L^{\infty}(V)}^{2})\xi_{N}^{2}$
$\displaystyle\geq\min\left\\{1-\frac{1}{a},1-(a-1)\|\nabla\psi\|_{L^{\infty}(V)}^{2}\right\\}|\xi|^{2}.$

Then, $M$ is uniformly elliptic. The proof is finished. ∎

###  5.2. Local Hölder estimate and global continuity

This subsection is devoted to proving a local Hölder condition on the
solutions to the problems that concern this manuscript. In particular, the
solutions are continuous in the interior of $\Omega$. We will show at the end
of the section that the solutions are, in fact, continuous up to the boundary.

######  Lemma 5.2.

Let $g:\Omega\times(0,+\infty)\to\mathbb{R}$ be a Carathéodory function
satisfying (2.1) and let $h\in L^{q}(\Omega)$ for some $q>\frac{N}{2}$. Then,
for every $M>c>0$ and every $\omega\subset\subset\Omega$, there exist
$\alpha\in(0,1)$, $L>0$ such that every solution $u\in H^{1}_{\mbox{\tiny
loc}}(\Omega)\cap L^{\infty}(\Omega)$ to

(5.3) $\begin{cases}-\Delta u+g(x,u)|\nabla u|^{2}=h(x),&x\in\Omega,\\\ c\leq
u\leq M,&x\in\Omega,\end{cases}$

satisfies

(5.4) $\|u\|_{C^{0,\alpha}(\overline{\omega})}\leq L.$

###### Proof.

Let us consider $m_{0}=\max_{s\in[c,M]}g_{0}(s)$, where
$g_{0}:(0,+\infty)\to[0,+\infty)$ is the continuous function given by (2.1),
and let us take $\delta\in\left(0,\frac{1}{2m_{0}}\right)$. With the notation
of [28, Chapter 2, Section 6, p. 81], we will show that $u$ belongs to the
class $\mathcal{B}_{2}\left(\Omega,M,\gamma,\delta,\frac{1}{2q}\right)$ for
some $\gamma>0$ to be determined later. This implies (5.4) by virtue of [28,
Theorem 6.1, p.90].

Indeed, let us fix $\rho>0$ and let $B_{\rho}$ be an open ball of radius
$\rho$ such that $B_{\rho}\subset\Omega$. Let us fix also $\sigma\in(0,1)$,
and consider a function $\zeta\in C_{c}^{\infty}(B_{\rho})$ satisfying that
$0\leq\zeta\leq 1$ in $B_{\rho}$, $\zeta\equiv 1$ in the concentric open ball
$B_{\rho-\sigma\rho}$, and $|\nabla\zeta|<\frac{b}{\sigma\rho}$ in $B_{\rho}$
for some constant $b>0$ independent of $\rho,\sigma$.

Let us denote $k_{\rho}=\sup_{x\in B_{\rho}}u(x)-\delta$, and take $k\geq
k_{\rho}$. We also consider the set

$A_{k,\rho}=\\{x\in B_{\rho}:u(x)\geq k\\}.$

Clearly, $(u-k)^{+}\zeta^{2}\in H_{0}^{1}(\Omega)\cap L^{\infty}(\Omega)$ and
has compact support in $\Omega$, so it may be taken as test function in (5.3).
Thus we obtain

$2\int_{A_{k,\rho}}\zeta(u-k)\nabla\zeta\nabla u+\int_{A_{k,\rho}}|\nabla
u|^{2}\zeta^{2}+\int_{A_{k,\rho}}g(x,u)|\nabla
u|^{2}(u-k)\zeta^{2}=\int_{A_{k,\rho}}h(x)(u-k)\zeta^{2}.$

On the one hand, by (2.1) and by the definitions of $k_{\rho},m_{0}$ we deduce
that

$-\int_{A_{k,\rho}}g(x,u)|\nabla u|^{2}(u-k)\zeta^{2}\leq\delta
m_{0}\int_{A_{k,\rho}}|\nabla u|^{2}\zeta^{2}.$

On the other hand, Young’s inequality yields

$-2\int_{A_{k,\rho}}\zeta(u-k)\nabla\zeta\nabla u\leq
2\int_{A_{k,\rho}}(u-k)^{2}|\nabla\zeta|^{2}+\frac{1}{2}\int_{A_{k,\rho}}|\nabla
u|^{2}\zeta^{2}.$

Therefore, we derive

$\left(\frac{1}{2}-\delta m_{0}\right)\int_{A_{k,\rho}}|\nabla
u|^{2}\zeta^{2}\leq
2\int_{A_{k,\rho}}(u-k)^{2}|\nabla\zeta|^{2}+\int_{A_{k,\rho}}h(x)(u-k)\zeta^{2}.$

Next, it follows from Hölder’s inequality that

$\int_{A_{k,\rho}}h(x)(u-k)\zeta^{2}\leq\delta\|h\|_{L^{q}(\Omega)}|A_{k,\rho}|^{\frac{1}{q^{\prime}}}$

and

$\displaystyle 2\int_{A_{k,\rho}}(u-k)^{2}\zeta^{2}$
$\displaystyle\leq\frac{2b}{\rho^{2}\sigma^{2}}\int_{A_{k,\rho}}(u-k)^{2}\leq\frac{2b}{\rho^{2}\sigma^{2}}\|(u-k)^{2}\|_{L^{\infty}(A_{k,\rho})}|A_{k,\rho}|$
$\displaystyle\leq\frac{2bC(N)}{\rho^{2-\frac{N}{q}}\sigma^{2}}\|(u-k)^{2}\|_{L^{\infty}(A_{k,\rho})}|A_{k,\rho}|^{\frac{1}{q^{\prime}}},$

where $|B_{\rho}|=C(N)^{q}\rho^{N}$. In sum,

(5.5) $\int_{A_{k,\rho-\sigma\rho}}|\nabla
u|^{2}\leq\gamma\left(\frac{1}{\rho^{2-\frac{N}{q}}\sigma^{2}}\|(u-k)^{2}\|_{L^{\infty}(A_{k,\rho})}+1\right)|A_{k,\rho}|^{\frac{1}{q^{\prime}}},$

where

$\gamma=\frac{\max\left\\{2bC(N),\delta\|h\|_{L^{q}(\Omega)}\right\\}}{\frac{1}{2}-\delta
m_{0}}.$

Furthermore, notice that $v=-u$ satisfies

$-\Delta v-g(x,-v)|\nabla v|^{2}=-h(x),\quad x\in\Omega.$

Therefore, it is straightforward to check that the inequality (5.5) also holds
by substituting $u$ with $-u$. In conclusion,
$u\in\mathcal{B}_{2}\left(\Omega,M,\gamma,\delta,\frac{1}{2q}\right)$ and the
proof is finished. ∎

######  Remark 5.3.

Let $g:\Omega\times(0,+\infty)\to\mathbb{R}$ be a Carathéodory function
satisfying (2.1), $h\in L^{q}(\Omega)$ for some $q>\frac{N}{2}$ and $u\in
H^{1}_{\mbox{\tiny loc}}(\Omega)\cap L^{\infty}(\Omega)$. Using the local
Hölder regularity given by Lemma 5.2 one can prove that, if $u\in
H^{1}_{\mbox{\tiny loc}}(\Omega)\cap L^{\infty}(\Omega)$ is a solution to
(5.6) that is locally bounded away from zero, then $u\in W^{1,2q}_{\mbox{\tiny
loc}}(\Omega)$. The proof is based on a standard bootstrap argument, see [15,
Appendix] for further details. Moreover, combining this local summability of
the gradients with the classical Calderon-Zygmund regularity theory one can
easily prove that, if $h\in L^{\infty}(\Omega)$, then $u\in
W^{2,q}_{\mbox{\tiny loc}}(\Omega)$ for every $q<\infty$.

We will show next that every solution to ($H$) is continuous up to
$\partial\Omega$.

######  Lemma 5.4.

Let $g:\Omega\times(0,+\infty)\to[0,+\infty)$ be a Carathédory function
satisfying (2.1) and let $h\in L^{q}(\Omega)$ for some $q>\frac{N}{2}$. Then,
every solution to ($H$) belongs to $C(\overline{\Omega})$.

###### Proof.

Let $u\in H^{1}_{\mbox{\tiny loc}}(\Omega)\cap L^{\infty}(\Omega)$ be a
solution to ($H$). Clearly, Lemma 5.2 implies that $u\in C(\Omega)$. It is
left to prove the continuity on the boundary. In order to do so, recall that
there exists $\gamma>0$ such that $u^{\gamma}\in H_{0}^{1}(\Omega)$. We may
assume without loss of generality that $\gamma>1$. Let us denote
$v=u^{\gamma}$. Using that $g\geq 0$, we deduce that

$-\Delta v=-\gamma(\gamma-1)u^{\gamma-2}|\nabla u|^{2}+\gamma
u^{\gamma-1}(-\Delta u)\leq\gamma u^{\gamma-1}h(x)\leq C|h(x)|,\quad
x\in\Omega,$

for some constant $C>0$. On the other hand, let us consider the unique
solution $w\in H_{0}^{1}(\Omega)\cap C(\overline{\Omega})$ to the following
problem:

$\begin{cases}-\Delta w=|h(x)|,&x\in\Omega,\\\
w=0,&x\in\partial\Omega.\end{cases}$

By standard comparison, $0<v\leq w$ in $\Omega$. Therefore, $\lim_{x\to
y}v(x)=0$ for every $y\in\partial\Omega$ and, in particular, $u\in
C(\overline{\Omega})$. ∎

###  5.3. Global Hölder estimates

In the previous subsection we have shown that the solutions to the problems in
this paper are locally Hölder continuous. In the proof, we have strongly used
the fact that the solutions are locally bounded away from zero in order to
avoid the possible singularity of $g(x,u(x))$ as $x$ approaches
$\partial\Omega$. In the present subsection we aim to prove a _global_ Hölder
estimate, i.e. the singularity cannot be avoided since we need a Hölder
estimate to hold even near $\partial\Omega$. Therefore, we perform a different
proof that requires $g$ to satisfy the stronger condition ($g_{*}$).

######  Lemma 5.5.

Let $\Omega\subset\mathbb{R}^{N}$ be a smooth bounded domain. Let
$\\{M_{n}\\}\subset L^{\infty}(\Omega)^{N\times N}$ and $\\{b_{n}\\}\subset
L^{\infty}(\Omega)^{N}$ be bounded sequences in the norm of their respective
spaces. Assume that there exists $\eta>0$ such that

$M_{n}(y)\xi\xi\geq\eta|\xi|^{2}\quad\text{a.e.
}y\in\Omega,\,\,\forall\xi\in\mathbb{R}^{N},\,\,\forall n.$

For some $p>1$, $\delta,\tau\geq 0$, $\sigma\in(0,1)$ independent of $n$, let
$f:[0,+\infty)\to[0,+\infty)$ be a continuous function satisfying ($f_{*}$)
and, for all $n$, let $g_{n}:\Omega\times(0,+\infty)\to[0,+\infty)$ be a
Carathédory function satisfying ($g_{*}$). Let $\\{v_{n}\\}\subset
H^{1}(\Omega)\cap C(\overline{\Omega})$ be such that $0<v_{n}\leq C$ in
$\Omega$ for all $n$ and for some $C>0$. For some sequences
$\\{L_{n}\\}\subset(0,+\infty)$, $\\{\epsilon_{n}\\}\subset[0,1]$, assume that
$v_{n}$ satisfies

$-\text{{\rm div}}(M_{n}(y)\nabla v_{n})+b_{n}(y)\nabla
v_{n}+\frac{1}{L_{n}}g_{n}\left(y,\frac{v_{n}}{L_{n}}\right)M_{n}(y)\nabla
v_{n}\nabla
v_{n}=L_{n}^{p}f\left(\frac{v_{n}}{L_{n}}\right)+\epsilon_{n}v_{n}^{\sigma},\quad
y\in\Omega,\,\,\forall n.$

If $\delta=0$, assume in addition that there exists
$\gamma\in\left(\frac{1-\tau}{2},1-\sigma\right]$ such that $v_{n}^{\gamma}\in
H^{1}(\Omega)$ for all $n$. For some (possibly empty) set
$\Gamma\subset\partial\Omega$ that is open in the topology of $\partial\Omega$
and connected, let us suppose also that

$v_{n}=0,\quad y\in\Gamma.$

Furthermore, if $\\{L_{n}\\}$ diverges, assume additionally that $\delta=0$
and $\epsilon_{n}=0$ for all $n$. Then, for every smooth bounded domain
$\omega$ satisfying that $\overline{\omega}\subseteq\Omega\cup\Gamma$, there
exist $\alpha\in(0,1)$, $C>0$ such that, taking a (not relabeled) subsequence
if necessary,

$\|v_{n}\|_{C^{0,\alpha}(\overline{\omega})}\leq C\quad\forall n.$

###### Proof.

Let us denote $\delta_{n}=L_{n}\delta$ and consider the function
$u_{n}=(v_{n}+\delta_{n})^{\gamma}-\delta_{n}^{\gamma}$ for
$\gamma\in\left(\frac{1-\tau}{2},1-\sigma\right]$; if $\delta=0$, then
$\gamma$ must be chosen so that $u_{n}\in H^{1}(\Omega)$ for all $n$. It is
easy to see that $u_{n}\in C(\overline{\Omega})$, $0<u_{n}\leq C$ in $\Omega$,
$u_{n}=0$ on $\Gamma$, and it satisfies

(5.6) $-\text{{\rm div}}(M_{n}(y)\nabla u_{n})+b_{n}(y)\nabla
u_{n}=\tilde{g}_{n}(y,u_{n})M_{n}(y)\nabla u_{n}\nabla
u_{n}+\tilde{f}_{n}(u_{n}),\quad y\in\Omega,$

where

$\displaystyle\tilde{g}_{n}(y,s)$
$\displaystyle=\frac{1-\gamma-\frac{(s+\delta_{n}^{\gamma})^{\frac{1}{\gamma}}}{L_{n}}g_{n}\left(y,\frac{(s+\delta_{n}^{\gamma})^{\frac{1}{\gamma}}}{L_{n}}-\delta\right)}{\gamma(s+\delta_{n}^{\gamma})}\quad\text{a.e.
}y\in\Omega,\,\,\forall s>0,\,\,\forall n,$ $\displaystyle\tilde{f}_{n}(s)$
$\displaystyle=\gamma(s+\delta_{n}^{\gamma})^{\frac{\gamma-1}{\gamma}}\left(L_{n}^{p}f\left(\frac{(s+\delta_{n}^{\gamma})^{\frac{1}{\gamma}}}{L_{n}}-\delta\right)+\epsilon_{n}\left((s+\delta_{n}^{\gamma})^{\frac{1}{\gamma}}-\delta_{n}\right)^{\sigma}\right)\quad\text{a.e.
}y\in\Omega,\,\,\forall n.$

We claim that $\\{u_{n}\tilde{f}_{n}(u_{n})\\}$ admits a subsequence bounded
in $L^{\infty}(\Omega)$ and, furthermore,

(5.7) $0\leq s\tilde{g}_{n}(y,s)\leq c\quad\text{a.e. }y\in\Omega,\,\,\forall
s>0,\,\,\forall n,$

for some $c\in(0,1)$. In order to prove the claim, we first observe that
($f_{*}$) implies

$\tilde{f}_{n}(s)\leq
a(s+\delta_{n}^{\gamma})^{\frac{p-1+\gamma}{\gamma}}+\epsilon_{n}(s+\delta_{n}^{\gamma})^{\frac{\sigma-1+\gamma}{\gamma}}.$

On the one hand, let us assume that $\\{L_{n}\\}$ is not divergent. Then, it
admits a (not relabeled) bounded subsequence. Hence, taking into account that
$p>1$, it clearly holds that

$au_{n}(u_{n}+\delta_{n}^{\gamma})^{\frac{p-1+\gamma}{\gamma}}\leq
C\quad\forall n.$

Besides, since $\gamma\geq\frac{1-\sigma}{2}$, then $\sigma-1+2\gamma\geq 0$,
so

$\gamma\epsilon_{n}u_{n}(u_{n}+\delta_{n}^{\gamma})^{\frac{\sigma-1+\gamma}{\gamma}}\leq(u_{n}+\delta_{n}^{\gamma})^{\frac{\sigma-1+2\gamma}{\gamma}}\leq
C\quad\forall n.$

Thus, $\\{u_{n}\tilde{f}_{n}(u_{n})\\}$ is bounded in $L^{\infty}(\Omega)$. On
the other hand, assume that $\\{L_{n}\\}$ diverges. Then,
$\delta_{n}=\epsilon_{n}=0$ for all $n$, so $\tilde{f}_{n}(u_{n})\leq
au_{n}^{\frac{p-1+\gamma}{\gamma}}\leq C$ for all $n$ and, again,
$\\{u_{n}\tilde{f}_{n}(u_{n})\\}$ is bounded in $L^{\infty}(\Omega)$.

Let us now verify that (5.7) holds. Indeed, from ($g_{*}$) and
$\gamma\in\left(\frac{1-\tau}{2},1-\sigma\right]$, it follows

$0\leq\frac{1-\gamma-\sigma}{\gamma}\leq(s+\delta_{n}^{\gamma})\tilde{g}_{n}(y,s)\leq\frac{1-\gamma-\tau}{\gamma}<1.$

Thus, our claim holds.

In sum, we may apply the arguments in [15, Appendix] without relevant changes
to prove that $\|u_{n}\|_{C^{0,\alpha}(\overline{\omega})}\leq C$ for some
$C>0,\alpha\in(0,1)$. In particular, using that the function $s\mapsto
s^{\frac{1}{\gamma}}$ is locally Lipschitz for $s\geq 0$, we have that

$\displaystyle|v_{n}(x)-v_{n}(y)|$
$\displaystyle=|v_{n}(x)+\delta_{n}-v_{n}(y)-\delta_{n}|$ $\displaystyle\leq
C|(v_{n}(x)+\delta_{n})^{\gamma}-(v_{n}(y)+\delta_{n})^{\gamma}|$
$\displaystyle=C|u_{n}(x)-u_{n}(y)|\leq C|x-y|^{\alpha}\quad\forall
x,y\in\omega.$

In conclusion, $\|v_{n}\|_{C^{0,\alpha}(\overline{\omega})}\leq C$, as we
wanted to prove. ∎

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