#  Large solutions of elliptic semilinear equations non-degenerate near the
boundary

G. Díaz Keywords: Large solutions, asymptotic behavior, singular boundary
conditions, degenerate problems. AMS Subject Classifications: 35B40, 35J25,
35J65.

###### Abstract

In this paper we study the so-called large solutions of elliptic semilinear
equations with non null sources term, thus solutions blowing up on the
boundary of the domain for which reason they are greater than any other
solution whenever Weak Maximum Principle holds. The main topic about large
solutions is uniqueness results and their behavior near the boundary. It is
much less than being simple. The structure of the semilinear equations
considered includes the well known Keller-Osserman integral and an assumption
on the ellipticity of the leading part of the differential operator. In our
study an uniform ellipticity near the boundary is required. We consider source
terms in the PDE whose boundary explosion is consistent with the Keller-
Osserman condition. Extra Keller-Osserman explosions on the source are also
studied, showing in particular that in some cases the PDE only admits large
solutions.

Dedicated to Ilde, brother and master

##  1 Introduction

This paper deals with solutions of semilinear elliptic equations with non null
source term

$-{\bf trace}~{}{\cal A}(\cdot){\rm D}^{2}u+\beta(u)=f\quad\hbox{in
$\Omega$},$ (1.1)

blowing up on the boundary, $u(x)=+\infty,~{}x\in\partial\Omega$. When a
Maximum Principle holds these solutions are called large solution because they
are bigger than any other bounded solution. An ingredient or the existence of
large solutions is the regularity on the domains compatible with a suitable
structure of the equation, preferably near the boundary. The main topic is
related with the uniquenees of large solutions. It is much less than being
simple. There is a wide bibliography on large solutions, mainly for the
homogeneous case $f\equiv 0$. A short list surely not the best would contain,
for instance, [2, 9, 10, 14, 19, 24, 27, 28, 29, 31, 34] and the references
therein.

The main tool near the boundary in this paper is the Lipschitz continuous
distance function ${\rm
d}(x)\doteq\hbox{dist}(x,\partial\Omega),~{}x\in\overline{\Omega},$ defined on
an open bounded set $\Omega\subset{\mathbb{R}}^{{\rm N}},~{}{\rm N}>1$. When
$\partial\Omega\in{\cal C}^{1}$ there exists a positive constant
$\delta_{\Omega}$ such that ${\rm d}\in{\cal C}^{1}$ in the subset
$\Omega_{\delta_{\Omega}}\doteq\\{x\in\Omega:~{}{\rm d}(x)<\delta_{\Omega}\\}$
as it was proved in [20] where also it is shown the eikonal equation

$|{\rm D}{\rm d}(x)|=1,\quad x\in\Omega_{\delta_{\Omega}},$ (1.2)

where we will denote ${\bf n}(x)=-{\rm D}{\rm d}(x)$ even if $x$ is not on the
boundary. If $x\in\partial\Omega$, ${\bf n}(x)$ is just the unit outward
normal to $\partial\Omega$ at $x$. Futhermore, when $\partial\Omega\in{\cal
C}^{m},~{}m\geq 2,$ one has ${\rm d}\in{\cal C}^{m}(\Omega_{\delta_{\Omega}})$
and

$\Delta{\rm d}(x)=-({\rm N}-1){\rm H}(z_{x})+o(1),\quad
x\in\Omega_{\delta_{\Omega}},$

where ${\rm H}(z_{x})$ is the mean curvature of $\partial\Omega$ at the
boundary point $z_{x}\in\partial\Omega$ such that ${\rm d}(x)=|x-z_{x}|$. So
that, we emphasize that all reasoning near the boundary $\partial\Omega$ will
be considered inside $\Omega_{\delta_{\Omega}}$.

###### Remark 1.1

In some papers the global regularity of $\partial\Omega\in{\cal C}^{2}$ can be
replaced by working on regular boundary points in the sense of an interior and
an exterior ball condition is fulfilled (see [32] of [1, Theorem 2]). From a
different approach the uniqueness of large solutions where the regularity of
$\partial\Omega$ is slightly lowered have been studies for several authors
(see, for instance, [26] or [31]).$\quad{}_{\Box}$

In this paper we will suppose that the leading term of the PDE is a Lipschitz
continuous function ${\cal A}:~{}\Omega\rightarrow{\cal S}^{{\rm
N}}_{+}\setminus\\{\text{\rm 0}\\}$ satisfying the general ellipticity
condition

$0\leq{\bf trace}~{}{\cal A}(x)\xi\otimes\xi\leq\Lambda,\quad
x\in\Omega,~{}\xi\in{\mathbb{R}}^{{\rm N}},~{}|\xi|=1$ (1.3)

that becomes uniform ellipticity near the boundary in the sense

$\lambda\leq{\bf trace}~{}{\cal A}(x){\bf n}(x)\otimes{\bf n}(x),\quad
x\in\Omega_{\delta_{\Omega}}$ (1.4)

for some positive constants $0<\lambda\leq\Lambda$. Certainly, (1.4) is a kind
of eikonal property (see (1.2)). So, that, we will assume in this paper that
the eikonal-like term ${\rm E}(x)\doteq{\cal A}(x){\bf n}(x)\otimes{\bf
n}(x),~{}x\in\Omega_{\delta_{\Omega}}$, takes values on the bounded interval
$[\lambda,\Lambda]$. As usual, ${\cal S}^{{\rm N}}$ is the symmetric ${\rm
N}\times{\rm N}$ real matrices set equipped with the ordering

${\rm Y}\in{\cal S}^{{\rm N}}_{+}\quad\Leftrightarrow\quad\hbox{the
eigenvalues of ${\rm Y}$ are nonnegative}.$

Since ${\cal A}\not\equiv{\bf 0}$ one satisfies

${\cal A}(x)\in{\cal S}^{{\rm N}}_{+},~{}x\in{\cal
A}\quad\Rightarrow\quad\Lambda_{{\cal A}}\geq\lambda_{{\cal A}}>0,$ (1.5)

where $\lambda_{{\cal A}}\doteq\inf_{x\in\Omega}{\bf trace}~{}{\cal A}(x){\rm
I}$ and $\Lambda_{{\cal A}}\doteq\sup_{x\in\Omega}{\bf trace}~{}{\cal
A}(x){\rm I}$. We also will use the notation ${\rm C}_{\partial\Omega,{\cal
A}}\doteq\displaystyle{\|{\bf trace}~{}{\cal A}(\cdot){\rm D}^{2}{\rm
d}\|}_{{\rm L}^{\infty}(\Omega_{\delta_{\Omega}})}<+\infty$.

###### Remark 1.2

Involving to stochastic process it is usual the case

${\cal A}(x)=\sigma(x)\sigma^{\tt t}(x),\quad x\in\Omega,$

where $\sigma:~{}\Omega\rightarrow{\mathbb{R}}^{{\rm
N}}\times{\mathbb{R}}^{p},~{}p\leq{\rm N}$, is a Lipschitz continuous
function. Clearly (1.3) holds by construction and then (1.4) only can be
available when $p={\rm N}$. If ${\cal A}\in{\rm W}^{2,\infty}(\Omega)$ then
$\sqrt{{\cal A}}$ is locally Lipschitz continuous in $\Omega$, as it was
proved in [18].$\quad{}_{\Box}$

In Section 7 we will prove a Strong Maximum Principle (see Theorem 7.2 below)
assuming that for any $\delta>0$ there exists $\lambda(\delta)>0$ such that

${\bf trace}~{}{\cal A}(x){\bf p}\otimes{\bf p}\geq\lambda(\delta),\quad
x\in\Omega,~{}\delta<|{\bf p}|<\delta^{-1}.$ (1.6)

as well as that for any ${\rm R}>0$ and $x\in\Omega$ there exist $\alpha>0$
and $\delta>0$ such that for all ${\bf p}(x)\in{\bf B}_{\delta}({\bf
0})\setminus\\{{\bf 0}\\}$ one satisfies

${\bf trace}~{}{\cal A}(x){\bf p}(x)\otimes{\bf
p}(x)\geq\dfrac{2\Lambda_{{\cal A}}}{\alpha{\rm R}^{2}}|{\bf p}(x)|^{2},\quad
x\in\Omega.$ (1.7)

###### Remark 1.3

Under an uniformly elliptic condition

${\bf trace}~{}{\cal A}(x){\bf p}\otimes{\bf p}\geq\lambda>0,\quad|{\bf p}|=1\
$

all properties (1.3), (1.4), (1.6) and (1.7) hold.$\quad{}_{\Box}$

We also consider zeroth order terms given by continuous and increasing
function $\beta:~{}{\mathbb{R}}_{+}\rightarrow{\mathbb{R}}_{+}$, with
$\beta(0)=0$ by simplicity. Certainly, some additional assumptions on the data
must be required in order to prove that the existence of classical large
solutions. An almost “unavoidable” hypothesis for these goal is the non-
degeneracy of the operator ${\cal A}$. In general, these condition will not be
required in this paper, as it ocurrs in some important examples of the
applications and consequently the ${\cal C}^{2}$ regularity even the
continuity of the solutions can not be guaranteed (see [18] or [33]). The
viscosity solution notion is adequate in order to remove non-degeneracy
hypothesis. We send to the monograph by M.G. Crandall, H. Ishii and P.L. Lions
[11] to understand how semi-continuous functions can solve (1.1) (see Section
4). The term viscosity solutions is an artifact motivated by their consistence
with the vanishing viscosity method. Since this viscosity notation is the
primary ones sometimes we drop the term viscosity and hereafter simply refer
to solution.

Some other structural assumptions are required when we consider the
differential operator

${\cal F}({\rm X},t,x)\doteq-{\bf trace}~{}\big{(}{\cal A}(x){\rm
X}\big{)}+\beta(t),\quad({\rm X},x,t)\in{\cal S}^{{\rm
N}}\times\Omega\times{\mathbb{R}}_{+}.$

When the boundary of $\Omega$ is smooth enough, the existence of large
solutions is equivalent to the so-called Keller-Osserman condition

$\int^{+\infty}\dfrac{ds}{\sqrt{{\rm G}(s)}}<+\infty\quad\hbox{where
$\displaystyle{\rm G}(s)=\int^{s}_{0}\beta(r)dr$}.$ (1.8)

We notice that under (1.8) we may define implicitly the decreasing function

$\delta=\Phi\big{(}\phi(\delta)\big{)}\quad\hbox{where
$\displaystyle\Phi(t)=\int^{+\infty}_{t}\dfrac{ds}{\sqrt{2{\rm G}(s)}},\quad
t>0$}$ (1.9)

that satisfies

$\phi(0)=+\infty\quad\hbox{and}\quad\phi^{\prime\prime}(\delta)=\beta\big{(}\phi(\delta)\big{)}\quad\hbox{for
small $\delta>0$.}$ (1.10)

Sometimes we say that $\phi$ provides the Keller-Osserman explosive profile.

###### Remark 1.4

In fact, by strightforward computations one proves that (1.8) implies

$\int^{+\infty}\dfrac{ds}{\sqrt{{\rm G}(s)-{\rm G}(a)}}<+\infty$

for any $a\geq 0$. Then one defines the function $\phi_{a}(\delta)$ by

$\delta=\Phi_{a}\big{(}\phi_{a}(\delta)\big{)}\quad\hbox{where
$\displaystyle\Phi_{a}(t)=\int^{+\infty}_{t}\dfrac{ds}{\sqrt{2\big{(}{\rm
G}(s)-{\rm G}(a)\big{)}}},\quad t>0$}$

that also satisfies (1.10). Moreover, one proves

$\lim_{\delta\searrow 0}\big{(}\phi_{a_{1}}(\delta)-\phi_{a_{2}}(\delta))=0$
(1.11)

(see [4, Theorem 1]).$\quad{}_{\Box}$

As it will be proved later, the function $\phi$ governs boundary profiles of
the large solutions independent on the primitive of $\beta$ (see Remark 1.11).

######  Remark 1.5 (Early examples)

In certain differential equations associated with the Riemannian metrics
appear large solutions relative to $\Lambda=\lambda=1,~{}f\equiv 0$ and
$\beta(t)=t^{\frac{{\rm N}+2}{{\rm N}-2}}$. It was studied in [30]. In
general, for the power like case $\beta_{m}(t)=t^{m}$ the condition (1.8)
becomes $m>1$ and

$\Phi_{m}(t)=\sqrt{\dfrac{m+1}{2}}\dfrac{2}{m-1}t^{-\frac{m-1}{2}}\quad\hbox{and}\quad\phi_{m}(\delta)=\left(\dfrac{(m-1)^{2}}{2(m+1)}\right)^{-\frac{1}{m-1}}\delta^{-\frac{2}{m-1}}.$

The study of large solutions trace back to [10] for
$\Lambda=\lambda=1,~{}f\equiv 0$ and $\beta(t)=e^{t}$ that satisfies (1.8) for
which

$\phi(\delta)=\log\left(\dfrac{2}{\delta^{2}}\right).$

The upcoming choice $\beta(t)=te^{2t}$ also satisfies (1.8). Then

$\phi(\delta)=\sqrt{2}\hbox{
erfc}^{-1}\left(\dfrac{\delta}{\sqrt{\pi}}\right),$

where
$\displaystyle\hbox{erfc}(\delta)=1-\hbox{erf}(\delta)=\dfrac{2}{\sqrt{\pi}}\int^{\infty}_{\delta}e^{-s^{2}}ds$.

###### Remark 1.6

From the inequality

$\widehat{\beta}(t)\geq\beta(t)\quad\hbox{for large $t$},$

it follows that if $\beta$ verifies (1.8) the same happens with
$\widehat{\beta}$. In particular, any function $\widehat{\beta}(t)\geq tq(t)$
such that

$\liminf_{t\rightarrow\infty}\dfrac{q(t)}{t^{\gamma}}\in(0,+\infty]\quad\hbox{for
some $\gamma>0$}$

satisfies (1.8). For instance, we may choose $q(t)\geq(\log
t)^{\gamma},~{}\gamma\geq 1,$ or
$q(t)\geq\log(\log(\cdots\log(t)\cdots)).\hfill\mbox{$\quad{}_{\Box}$}$

The plan in this paper is as follows. In Sections 2 and 3 we obtain a kind of
comparison of the boundary behavior by means of the property

$\limsup_{{\rm d}(x)\rightarrow 0}\dfrac{u^{*}(x)}{v_{*}(x)}\leq 1,$ (1.12)

for any subsolution $u$ and any nonnegative large supersolution $v$ of

$-{\bf trace}~{}{\cal A}(x){\rm D}^{2}u+\beta(u)=f\quad\hbox{in $\Omega$}.$

Since the large solutions are singular on the boundary, the condition (1.12)
is very useful when we try to transfer boundary comparison to the interior. We
note that (1.12) avoids working with the relaxed limit condition (see [11]).
First, in Theorems 2.1 and 2.2, in Section 2, the property (1.12) is obtained
under sources terms, $f$, with Keller-Osserman type growth on the boundary,
including explosive and bounded boundary behaviour of $f$.

In Section 3 we also obtain (1.12) now for any subsolution $u$ and any large
supersolution $v$ of

$-{\bf trace}~{}{\cal A}(x){\rm D}^{2}u+\beta(u)=f\quad\hbox{in $\Omega$},$

under sources terms, $f$, with extra Keller-Osserman type growth on the
boundary. Here, the boundary behaviour of $f$ only can be explosive. It is
collected in Theorem 3.2. It is about a genuine feature of the non homogeneous
equation

$-{\bf trace}~{}{\cal A}(\cdot){\rm D}^{2}u+\beta(u)=f\quad\hbox{in
$\Omega$}.$

The key idea of Theorem 3.2, announced in [14] and [1], is the scheme
$\beta(u)\approx f$ for large values of $u$ near the boundary while the
leading part of the equation grows up with a lower order. Certainly the source
term $f$ must be more explosive on the boundary than the studied in the
Keller-Osserman type growth case. Essentially we require a compatiblity
condition as

$f(x)=h\big{(}{\rm d}(x)\big{)}+o({\rm d}(x))\quad\hbox{with}\quad
h(\delta)>\beta\big{(}\phi(\delta)\big{)}\quad\hbox{for ${\rm d}(x)$ small}$
(1.13)

provided (1.8). We recall that, as it was proved in [24], the assumption (1.8)
is a necessary condition for the existence of large solutions. In Theorem 3.2
we will obtain the precise lower order estimate term

$u(x)=\beta^{-1}\big{(}h({\rm d}(x))\big{)}+o({\rm d}(x))$

near the boundary $\partial\Omega$. Here the boundedness of the eikonal like
term plays a secondary role. Since the proof of Theorem 3.2 is tedious, we
illustrate the reasoning for the power like case in Theorem 3.1 in which a
striking case appears by means of an argument of rescaling: the explositivity
of the sources is transferred to the supersolutions as it is proved in Theorem
3.3. Consequently, in this case the equation

$-{\bf trace}~{}{\cal A}(\cdot){\rm D}^{2}u+u^{m}=f\quad\hbox{in $\Omega$}$

only can admit large solutions.

We devote Section 4 to adequate versions of a Weak Maximum Principle. More
precisely, we obtain comparison results involved large solution as it is
proved in Theorem 4.1 under the useful estimate (1.12). So, by Theorem 4.1 we
transfer the comparison given by (1.12) to the interior

$u^{*}(x)\leq v_{*}(x),\quad x\in\Omega,$

provided

$\dfrac{\beta(t)}{t}\mbox{ is increasing for }t>0.$ (1.14)

Therefore

$-{\bf trace}~{}{\cal A}(\cdot){\rm D}^{2}u+\beta(u)=f\quad\hbox{in $\Omega$}$

admits at most a unique large solution $u$. Since it implies

$u^{*}(x)\leq u_{*}(x),\quad x\in\Omega$

the solution must be a continuous function. The proof adapts reasoning of the
[11, Theorem 3.3] of the Theory of Viscosity Solutions whose notions are
briefly recalled. We emphasize that no uniform ellipticity assumptions is
required in the proof of Theorem 4.1.

The existence of solutions is obtained in Theorems 4.4 and 4.5 by means of the
classical Perron’s Method (see [11] and the bibliography therein). As it is
well konwn the main ingredient of this method concerns to weak pass to the
limit in equations satisfying the Maximum Principle. In some sense, the
stability property of solutions can be seen as the analogue Minty’s device for
monotone operators. A main feature of the Viscosity Theory is the pass to
limit of semi-continuous solutions. Here we deal with it by appropriate
universal estimates obtained in Theorems 4.2 and 4.2. They are simple
consequence of the more general Theorem 6.1 proved in Section 6. We prove in
Theorems 4.4 and 4.5 that Perron function is in fact the unique solution.

Section 5 is devoted to the boundary explosive expansion of the large
solutions. Since it requires tedious and dark computations we illustrate by
considering only the second order term of the expansion. We emphasize with the
influence of the geometry provided of the secondary term that in some cases is
explosive, bounded or even vanishig on the boundary. While the case of Keller-
Osserman type growth on the boundary of the sources is studied in a general
way, for extra Keller-Osserman type growth we only consider the power like
case.

Two last sections are included in this article collecting some results
previously used. So, in Section 6 we take advantage of the Keller-Osserman
assumption (1.8) in order to obtain a universal upper bound very useful in
locally pass to the limits (see Theorem 6.1). On the other hand, some results
without coercivity term have been required. In particular, in the proof of
Theorem 2.2 we have used a Weak Maximum Principle without coercivity term (see
Theorem 7.1). Moreover, the uniqueness of Perron function as the unique large
solution, is proved by means of a Strong Maximum Principle without coercivity
term (see Theorem 7.2) under the assumptions (1.6) and (1.7). Here, the lack
of a coercive term is supplied with a kind of Kruzkov change of variable (see
[6] or [7]). Theorems 7.1 and 7.2 are been obtained in Section 7.

Some results of the paper admit extension when the function $\beta(r)$ is
replaced by suitable terms $\beta(x,r)$ as it is studied in [19] or [31].

##  2 Keller-Osserman type growth of the sources on the boundary

As it was pointed out, the Keller-Osserman condition (1.8)

$\int^{+\infty}\dfrac{ds}{\sqrt{{\rm G}(s)}}<+\infty\quad\hbox{where
$\displaystyle{\rm G}(s)=\int^{s}_{0}\beta(r)dr$}$

is equivalent to the existence of large solutions (see [24]). The uniqueness
of large solutions turns out to be delicate. In this section we study a kind
of uniqueness of the boundary behaviours of the large solutions. We begin with
some technicallities.

######  Lemma 2.1 (Lemma A.1 of [2])

Assumption (1.8) implies

$\lim_{t\rightarrow+\infty}\dfrac{t}{\sqrt{{\rm
G}(t)}}=0,\quad\lim_{t\rightarrow+\infty}\dfrac{t}{\beta(t)}=0$ (2.1)

and

$\lim_{t\rightarrow+\infty}\dfrac{\sqrt{{\rm G}(t)}}{\beta(t)}=0.$ (2.2)

In some cases we replace (1.8) by

$\dfrac{\beta(t)}{t^{\alpha}}\quad\hbox{is increasing for large $t$},$ (2.3)

where $\alpha>1$, slightly more restrictive.

###### Lemma 2.2

Under (2.3) one has the Keller-Osserman condition (1.8). Moreover, for $\nu>1$
and $\varepsilon>0$ the inequality

$\varepsilon+\nu\phi(\delta)\geq\phi\big{(}\nu^{-\frac{\alpha-1}{2}}\delta\big{)}\quad\hbox{for
$\delta$ small}$ (2.4)

holds.

Proof. Let us assume

$\dfrac{\beta(t)}{t^{\alpha}}\quad\hbox{is increasing for $t>a$}.$

Then from Remark 1.6 it follows that (2.3) leads to the Keller-Osserman
condition (1.8). Futhermore, it implies

${\rm G}(\nu t)-{\rm G}(a)=\int^{\nu
t}_{a}\beta(s)ds=\nu\int^{t}_{\frac{a}{\nu}}\beta(\nu
s)ds\geq\nu^{\alpha+1}\big{(}{\rm G}(t)-{\rm G}(a)\big{)}\quad\hbox{ for
$\nu>1$ },$

hence we deduce

$\Phi_{a}(\nu t)=\int^{+\infty}_{\nu t}\dfrac{d\tau}{\sqrt{2\big{(}{\rm
G}(\tau)-{\rm
G}(a)\big{)}}}=\nu\int^{+\infty}_{t}\dfrac{d\tau}{\sqrt{2\big{(}{\rm
G}(\nu\tau)-{\rm
G}(a)\big{)}}}\leq\nu^{\frac{1-\alpha}{2}}\Phi_{a}(t),\quad\hbox{for large
$t$},$

thus $\nu
t\geq\Phi_{a}^{-1}\big{(}\nu^{\frac{1-\alpha}{2}}\Phi_{a}(t)\big{)}$. The
change of variable $t=\phi_{a}(\delta)$ proves

$\nu\phi_{a}(\delta)\geq\phi_{a}\big{(}\nu^{-\frac{\alpha-1}{2}}\delta\big{)}\quad\hbox{for
$\delta$ small}.$

Finally, (2.4) follows from (1.11).$\quad{}_{\Box}$

###### Remark 2.1

The above reasonings prove for $\nu<1$ and $\varepsilon>0$ the inequality

$-\varepsilon+\nu\phi(\delta)\leq\phi\big{(}\nu^{-\frac{\alpha-1}{2}}\delta\big{)}\quad\hbox{for
$\delta$ small}.$ (2.5)

Clearly (2.3) is verified for the examples of Remark 1.5. $\quad{}_{\Box}$

The key idea of the boundary explosive estimate is the simple equality

$\phi^{\prime\prime}(\delta)=\beta\big{(}\phi(\delta)\big{)}=\mu\beta\big{(}\phi(\delta)\big{)}-(\mu-1)\beta\big{(}\phi(\delta)\big{)}$

(see (1.10)). The proof is based on following boundary behaviour

######  Proposition 2.1 (Boundary maximal universal upper bound)

Asume $\partial\Omega\in{\cal C}^{2}$. Suppose (1.3), (1.4) and (2.3). Then
any solution $u$

$-{\bf trace}~{}{\cal A}(x){\rm D}^{2}u+\beta(u)\leq f\quad\hbox{in $\Omega$}$

admits the maximal boundary behaviour given by

$\limsup_{{\rm d}(x)\searrow
0}\dfrac{u^{*}(x)}{\phi\left(\sqrt{\dfrac{1-\ell_{u}}{\Lambda}}{\rm
d}(x)\right)}\leq 1,$ (2.6)

provided

$\limsup_{{\rm d}(x)\searrow
0}\dfrac{f(x)}{\beta\left(\phi\left(\sqrt{\dfrac{1-\ell_{u}}{\Lambda}}{\rm
d}(x)\right)\right)}\leq\ell_{u}<1,$ (2.7)

for some positive constant $\ell_{u}$. Here $\phi$ is the function defined in
(1.9).

Proof. The idea is to construct a suitable classical large supersolution. Let
$\varepsilon>0$. So, for $0<\delta<\delta_{1}<\delta_{\Omega}$, we consider

${\rm
U}_{0}(x)=\phi_{a}\left(\sqrt{\dfrac{(1-\varepsilon)(1-\ell_{u})}{\Lambda}}({\rm
d}(x)-\delta)\right),\quad
x\in\Omega^{\delta}_{\delta_{1}}\doteq\\{x\in\Omega:~{}\delta<{\rm
d}(x)<\delta_{1}\\}.$

Here $\phi$ is the decreasing function defined implicitly in (1.9). In the
pioneer [30] this kind of method was used. By (1.10), straightforward
computations as in [2] or [14] prove

$\begin{array}[]{ll}{\bf trace}~{}{\cal A}(x){\rm D}^{2}{\rm
U}_{0}(x)&=\dfrac{(1-\varepsilon)(1-\ell_{u})}{\Lambda}\phi^{\prime\prime}(\sigma){\bf
trace}~{}{\cal A}(x){\rm D}{\rm d}(x)\otimes{\rm D}{\rm d}(x)\\\\[7.11317pt]
&+\sqrt{\dfrac{(1-\varepsilon)(1-\ell_{u})}{\Lambda}}\phi^{\prime}(\sigma){\bf
trace}~{}{\cal A}(x){\rm D}^{2}{\rm d}(x)\\\\[7.11317pt]
&\leq(1-\varepsilon)(1-\ell_{u})g\big{(}\phi(\sigma)\big{)}+{\rm
C}_{\partial\Omega,{\cal
A}}\sqrt{\dfrac{2(1-\varepsilon)(1-\ell_{u})}{\Lambda}{\rm
G}\big{(}\phi(\sigma)\big{)}},\quad
x\in\Omega^{\delta}_{\delta_{1}},\end{array}$

where we are denoting
$\sigma=\sqrt{\dfrac{(1-\varepsilon)(1-\ell_{u})}{\Lambda}}({\rm
d}(x)-\delta)$. Then

$\begin{array}[]{ll}-{\bf trace}~{}{\cal A}(x){\rm D}^{2}{\rm
U}_{0}(x)+\beta\big{(}{\rm
U}_{0}(x)\big{)}&\geq\beta\big{(}\phi(\sigma)\big{)}\bigg{[}\varepsilon(1-\ell_{u})+\ell_{u}\\\
&\left.-{\rm C}_{\partial\Omega,{\cal
A}}\sqrt{\dfrac{2(1-\varepsilon)(1-\ell_{u})}{\Lambda}}\dfrac{\sqrt{{\rm
G}\big{(}\phi(\sigma)\big{)}}}{\beta\big{(}\phi(\sigma)\big{)}}\right],\quad
x\in\Omega^{\delta}_{\delta_{1}}.\end{array}$

On the other hand, assumption (2.7) implies

$\ell_{u}\beta\big{(}\phi(\sigma)\big{)}\geq\ell_{u}\beta\left(\phi\left(\sqrt{\dfrac{1-\ell_{u}}{\Lambda}}{\rm
d}(x)\right)\right)\geq f(x),\quad x\in\Omega^{\delta}_{\delta_{1}},$

whence

$\begin{array}[]{ll}-{\bf trace}~{}{\cal A}(x){\rm D}^{2}{\rm
U}_{0}(x)+\beta\big{(}{\rm
U}_{0}(x)\big{)}-f(x)&\geq\beta\big{(}\phi(\sigma)\big{)}\bigg{[}\varepsilon(1-\ell_{u})\\\
&\left.-{\rm C}_{\partial\Omega,{\cal
A}}\sqrt{\dfrac{2(1-\varepsilon)(1-\ell_{u})}{\Lambda}}\dfrac{\sqrt{{\rm
G}\big{(}\phi(\sigma)\big{)}}}{\beta\big{(}\phi(\sigma)\big{)}}\right],\end{array}$
(2.8)

for $x\in\Omega^{\delta}_{\delta_{1}}$. Since
$0<\sigma<\sqrt{\dfrac{1-\ell_{u}}{\Lambda}}\delta_{1}$, taking account into
(2.2) we obtain

${\rm C}_{\partial\Omega,{\cal
A}}\sqrt{\dfrac{2(1-\varepsilon)(1-\ell_{u})}{\Lambda}}\dfrac{\sqrt{{\rm
G}\big{(}\phi(\sigma)\big{)}}}{\beta\big{(}\phi(\sigma)\big{)}}<\varepsilon(1-\ell_{u}),$

for some $\delta_{1}$ small enough. Because
$\beta\big{(}\phi(\sigma)\big{)}>0$ we deduce

$-{\bf trace}~{}{\cal A}(x){\rm D}^{2}{\rm U}_{0}(x)+\beta\big{(}{\rm
U}_{0}(x)\big{)}\geq f(x)\quad x\in\Omega^{\delta}_{\delta_{1}}.$

Clearly, ${\rm U}_{0}$ is a classical positive supersolution in
$\Omega^{\delta}_{\delta_{1}}$ such that

$u^{*}(x)<+\infty={\rm U}_{0}(x)\quad\hbox{if ${\rm d}(x)=\delta$}$

for any subsolution $u$. Moreover,

$\left\\{\begin{array}[]{l}-{\bf trace}~{}{\cal A}(x){\rm D}^{2}\big{(}{\rm
U}_{0}(x)+{\rm M}\big{)}+\beta\big{(}{\rm U}_{0}(x)+{\rm M}\big{)}\geq
f(x),\quad x\in\Omega^{\delta}_{\delta_{1}},\\\\[4.97931pt] {\rm
U}_{0}(x)+{\rm M}=+\infty>u^{*}(x)\quad\hbox{if ${\rm
d}(x)=\delta$},\end{array}\right.$

for any positive constant ${\rm M}$. In order to conclude (2.6) we need an
adequate inequality on the boundary ${\rm d}(x)=\delta_{1}$. In particular

${\rm U}_{0}(x)+{\rm M}\geq{\rm M}\geq\sup_{{\rm d}(x)=\delta_{1}}u^{*}\geq
u^{*}(x),\quad{\rm d}(x)=\delta_{1}$

for ${\rm M}\geq\displaystyle\sup_{{\rm d}(x)=\delta_{1}}u^{*}$ (we recall
that $u^{*}$ is an upper semi-continuous function). Remark 4.1 implies

$u^{*}(x)\leq{\rm U}_{0}(x)+{\rm M},\quad x\in\Omega_{\delta_{1}}^{\delta},$

for all $\delta<\delta_{\Omega}$, whence

$u^{*}(x)\leq\phi\left(\sqrt{\dfrac{(1-\varepsilon)(1-\ell_{u})}{\Lambda}}{\rm
d}(x)\right)+{\rm M},\quad x\in\Omega_{\delta_{1}}.$ (2.9)

This maximal bound (2.9) can be improved by means of (2.4). Indeed, let us
choose the constant $\nu=(1-\varepsilon)^{-\frac{1}{\alpha-1}}>1$ for which

$u^{*}(x)\leq\nu\phi\left(\sqrt{\dfrac{1-\ell_{u}}{\Lambda}}{\rm
d}(x)\right)+{\rm M}+\varepsilon,\quad x\in\Omega_{\delta_{1}}$

(see (2.4)) and

$\dfrac{u^{*}(x)}{\phi\left(\sqrt{\dfrac{1-\ell_{u}}{\Lambda}}{\rm
d}(x)\right)}\leq\nu+\dfrac{{\rm
M}+\varepsilon}{\phi\left(\sqrt{\dfrac{1-\ell_{u}}{\Lambda}}{\rm
d}(x)\right)}$

hold for $x\in\Omega_{\delta_{1}}$. Then

$\limsup_{{\rm d}(x)\searrow
0}\dfrac{u^{*}(x)}{\phi\left(\sqrt{\dfrac{1-\ell_{u}}{\Lambda}}{\rm
d}(x)\right)}\leq\nu.$

Taking $\varepsilon\searrow 0$ one concludes (2.6).$\quad{}_{\Box}$

We emphasize that in Theorem 2.2 we do not require any information of $u$ on
the $\partial\Omega$.

######  Proposition 2.2 (Boundary minimal universal lower bound)

Asume $\partial\Omega\in{\cal C}^{2}$. Suppose (1.3), (1.4) and (2.3). Then
any nonnegative large solution $v$ of

$-{\bf trace}~{}{\cal A}(x){\rm D}^{2}v+\beta(v)\geq f\quad\hbox{in $\Omega$}$

admits the minimal behaviour given by

$1\leq\liminf_{{\rm d}(x)\searrow
0}\dfrac{v_{*}(x)}{\phi\left(\sqrt{\dfrac{1-\ell_{d}}{\lambda}}{\rm
d}(x)\right)}$ (2.10)

provided

$\liminf_{{\rm d}(x)\searrow
0}\dfrac{f(x)}{\beta\left(\phi\left(\sqrt{\dfrac{1-\ell_{d}}{\lambda}}{\rm
d}(x)\right)\right)}\geq\ell_{d}\quad\hbox{with $0\leq\ell_{d}<1$},$ (2.11)

for some positive constant $\ell_{d}$. Here $\phi$ is the function defined in
(1.9).

Proof. For $\varepsilon>0,~{}\delta>0$ and $0<\delta_{0}<\delta_{\Omega}$, we
construct

${\rm
V}_{0}(x)=\phi\left(\sqrt{\dfrac{(1+\varepsilon)(1-\ell_{d})}{\lambda}}\big{(}{\rm
d}(x)+\delta\big{)}\right),\quad x\in\Omega_{\delta_{0}}.$

Again $\phi$ is the decreasing function defined implicitly in (1.9). Now,
straightforward computations prove

$\begin{array}[]{ll}{\bf trace}~{}{\cal A}(x){\rm D}^{2}{\rm
V}_{0}(x)&=\dfrac{(1+\varepsilon)(1-\ell_{d})}{\lambda}\phi^{\prime\prime}(\sigma){\bf
trace}~{}{\cal A}(x){\rm D}{\rm d}(x)\otimes{\rm D}{\rm d}(x)\\\\[7.11317pt]
&+\sqrt{\dfrac{(1+\varepsilon)(1-\ell_{d})}{\lambda}}\phi^{\prime}(\sigma){\bf
trace}~{}{\cal A}(x){\rm D}^{2}{\rm d}(x)\\\\[7.11317pt]
&\geq(1+\varepsilon)(1-\ell_{d})g\big{(}\phi(\sigma)\big{)}-{\rm
C}_{\partial\Omega,{\cal
A}}\sqrt{\dfrac{2(1+\varepsilon)(1-\ell_{d})}{\lambda}{\rm
G}\big{(}\phi(\sigma)\big{)}},\quad x\in\Omega_{\delta_{0}},\end{array}$

where we are denoting
$\sigma=\sqrt{\dfrac{(1+\varepsilon)(1-\ell_{d})}{\lambda}}\big{(}{\rm
d}(x)+\delta)$. Then

$\begin{array}[]{ll}-{\bf trace}~{}{\cal A}(x){\rm D}^{2}{\rm
V}_{0}(x)+\beta\big{(}{\rm
V}_{0}(x)\big{)}&\leq\beta\big{(}\phi(\sigma)\big{)}\bigg{[}-\varepsilon(1-\ell_{d})+\ell_{d}\\\
&\left.+{\rm C}_{\partial\Omega,{\cal
A}}\sqrt{\dfrac{2(1+\varepsilon)(1-\ell_{d})}{\lambda}}\dfrac{\sqrt{{\rm
G}\big{(}\phi(\sigma)\big{)}}}{\beta\big{(}\phi(\sigma)\big{)}}\right],\quad
x\in\Omega_{\delta_{0}}.\end{array}$

On the other hand, assumption (2.11) implies

$\ell_{d}\beta\big{(}\phi(\sigma)\big{)}\leq\ell_{d}\beta\left(\phi\left(\sqrt{\dfrac{(1+\varepsilon)(1-\ell_{d})}{\lambda}}{\rm
d}(x)\right)\right)\leq f(x),\quad x\in\Omega_{\delta_{0}}$

whence

$\begin{array}[]{ll}-{\bf trace}~{}{\cal A}(x){\rm D}^{2}{\rm
V}_{0}(x)+\beta\big{(}{\rm
V}_{0}(x)\big{)}-f(x)&\leq\beta\big{(}\phi(\sigma)\big{)}\bigg{[}-\varepsilon(1-\ell_{d})\\\
&\left.+{\rm C}_{\partial\Omega,{\cal
A}}\sqrt{\dfrac{2(1+\varepsilon)(1-\ell_{d})}{\lambda}}\dfrac{\sqrt{{\rm
G}\big{(}\phi(\sigma)\big{)}}}{\beta\big{(}\phi(\sigma)\big{)}}\right],\quad
x\in\Omega_{\delta_{0}}.\end{array}$

We also will use the property (2.2). So, fixed $\varepsilon_{0}>0$ and
$\varepsilon<\varepsilon_{0}$ we choose $0<\delta\leq\delta_{0}$ for which
$0<\sigma\leq
2\sqrt{\dfrac{(1+\varepsilon_{0})(1-\ell_{d})}{\lambda}}\delta_{0}$ and the
inequality

${\rm C}_{\partial\Omega,{\cal
A}}\sqrt{\dfrac{2(1+\varepsilon_{0})(1-\ell_{d})}{\lambda}}\dfrac{\sqrt{{\rm
G}\big{(}\phi(\sigma)\big{)}}}{\beta\big{(}\phi(\sigma)\big{)}}<\varepsilon$

holds for $\delta_{0}$ small enough. It implies that the function

${\rm
V}_{0}(x)=\phi\left(\sqrt{\dfrac{(1+\varepsilon)(1-\ell_{d})}{\lambda}}\big{(}{\rm
d}(x)+\delta\big{)}\right),\quad x\in\Omega_{\delta_{0}},$

satisfies

$-{\bf trace}~{}{\cal A}(x){\rm D}^{2}{\rm V}_{0}(x)+\beta\big{(}{\rm
V}_{0}(x)\big{)}\leq f(x)\quad x\in\Omega_{\delta_{0}}.$

Clearly, ${\rm V}_{0}$ is a classical bounded subsolution in
$\Omega_{\delta_{0}}$ such that

${\rm V}_{0}(x)<+\infty=v_{*}(x)\quad\hbox{if ${\rm d}(x)=0$}$

for any large supersolution $v$. So, for any nonnegative constant ${\rm M}$
one has

$\left\\{\begin{array}[]{l}-{\bf trace}~{}{\cal A}(x){\rm D}^{2}\big{(}{\rm
V}_{0}(x)-{\rm M}\big{)}+\beta\big{(}{\rm V}_{0}(x)-{\rm M}\big{)}\leq
f(x),\quad x\in\Omega_{\delta_{0}},\\\\[4.97931pt] {\rm V}_{0}(x)-{\rm
M}<+\infty=v_{*}(x)\quad\hbox{if ${\rm d}(x)=0$}.\end{array}\right.$

In order to conclude (2.10) we need an adequate inequality on the boundary
${\rm d}(x)=\delta_{0}$. In particular

${\rm V}_{0}(x)-{\rm
M}\leq\phi\left(2\delta_{0}\sqrt{\dfrac{(1+\varepsilon)(1-\ell_{d})}{\lambda}}\right)-{\rm
M}\leq 0\leq v_{*}(x),\quad{\rm d}(x)=\delta_{0}$

for ${\rm
M}\geq\phi\left(2\delta_{0}\sqrt{\dfrac{(1+\varepsilon)(1-\ell_{d})}{\lambda}}\right)$,
when we are taking $\delta=\delta_{0}$. Remark 4.2 implies

${\rm V}_{0}(x)-{\rm M}\leq v_{*}(x),\quad x\in\Omega_{\delta_{0}},$

thus

$\phi\left(\sqrt{\dfrac{(1+\varepsilon)(1-\ell_{d})}{\lambda}}\big{(}{\rm
d}(x)+\delta_{0}\big{)}\right)-{\rm M}\leq v_{*}(x),\quad
x\in\Omega_{\delta_{0}}$ (2.12)

holds. Again we may improve the minimal bound (2.12) by means of (2.5) here by
choosing $\nu=(1+\varepsilon)^{-\frac{1}{\alpha-1}}<1$. Indeed, (2.12) becomes

$\nu\phi\left(\sqrt{\dfrac{1-\ell_{d}}{\lambda}}\big{(}{\rm
d}(x)+\delta_{0}\big{)}\right)-({\rm M}+\varepsilon)\leq v_{*}(x),\quad
x\in\Omega_{\delta_{0}}.$

It leads to

$\nu-\dfrac{{\rm
M}+\varepsilon}{\phi\left(\sqrt{\dfrac{1-\ell_{d}}{\lambda}}{\rm
d}(x)\big{)}\right)}\leq\dfrac{v_{*}(x)}{\phi\left(\sqrt{\dfrac{1-\ell_{d}}{\lambda}}{\rm
d}(x)\big{)}\right)},\quad x\in\Omega_{\delta_{0}}$

and

$\nu\leq\liminf_{{\rm d}(x)\searrow
0}\dfrac{v_{*}(x)}{\phi\left(\sqrt{\dfrac{1-\ell_{d}}{\lambda}}{\rm
d}(x)\big{)}\right)}.$

Taking $\varepsilon\searrow 0$ one concludes (2.10).$\quad{}_{\Box}$

###### Remark 2.2

The property (2.4) avoids to consider the assumption as

$\limsup_{t\nearrow+\infty}\dfrac{\Phi(\eta t)}{\Phi(t)}<1\quad\hbox{for any
$\eta>1$}$

or the more sharp borderline case given by

$\limsup_{t\rightarrow\infty}\frac{\Phi(\eta_{0}t)}{\Phi(t)}=1\quad\mbox{for
some }\eta_{0}>1$

in the proofs of Propositions 2.1 and 2.2 as they were used in
[2].$\quad{}_{\Box}$

###### Remark 2.3

Obviously, any eventual nonnegative large solution of

$-{\bf trace}~{}{\cal A}(\cdot){\rm D}^{2}v+\beta(v)\geq f\quad\hbox{in
$\Omega$}$

admits a minimal behaviour given by (2.10), provided (1.3), (1.4), (2.3) and
$f\geq 0$, without the assumption (2.11).$\quad{}_{\Box}$

###### Remark 2.4

As it was pointed out, the assumption (2.3) is fulfilled for the choices of
Remark 1.5. Moreover, for the power like case $\beta_{m}(t)=t^{m}$ and $m>1$
the assumptions (2.7) and (2.11) correspond with

$\begin{array}[]{l}\displaystyle\left(\dfrac{2(m+1)}{(m-1)^{2}}\right)^{\frac{m}{m-1}}\left(\dfrac{\lambda}{1-\ell_{d}}\right)^{\frac{m}{m-1}}\ell_{d}\leq\liminf_{{\rm
d}(x)\searrow 0}f(x){\rm d}(x)^{\frac{2m}{m-1}}\\\\[5.69046pt] \hskip
71.13188pt\displaystyle\leq\limsup_{{\rm d}(x)\searrow 0}f(x){\rm
d}(x)^{\frac{2m}{m-1}}\leq\left(\dfrac{2(m+1)}{(m-1)^{2}}\right)^{\frac{m}{m-1}}\left(\dfrac{\Lambda}{1-\ell_{u}}\right)^{\frac{m}{m-1}}\ell_{u}.\end{array}$

For $\beta(t)=e^{t}$ the assumptions (2.7) and (2.11) correspond with

$\displaystyle 2\ell_{d}\dfrac{\lambda}{1-\ell_{d}}\leq\liminf_{{\rm
d}(x)\searrow 0}f(x){\rm d}(x)^{2}\leq\limsup_{{\rm d}(x)\searrow 0}f(x){\rm
d}(x)^{2}\leq 2\ell_{u}\dfrac{\Lambda}{1-\ell_{u}}.$

###### Remark 2.5

We note that under the assumptions of Propositions 2.1 and 2.2 we have

$\limsup_{{\rm d}(x)\searrow
0}\dfrac{u^{*}(x)}{\phi\left(\sqrt{\dfrac{1-\ell_{u}}{\Lambda}}{\rm
d}(x)\right)}\leq 1\leq\liminf_{{\rm d}(x)\searrow
0}\dfrac{v_{*}(x)}{\phi\left(\sqrt{\dfrac{1-\ell_{d}}{\lambda}}{\rm
d}(x)\right)}$

for any solution of

$-{\bf trace}~{}{\cal A}(\cdot){\rm D}^{2}u+\beta(u)\leq f\quad\hbox{in
$\Omega$},$

and any nonnegative large solution of

$-{\bf trace}~{}{\cal A}(\cdot){\rm D}^{2}v+\beta(v)\geq f\quad\hbox{in
$\Omega$}.$

From the above results we deduce

###### Theorem 2.1

Under the assumptions of Propositions 2.1 and 2.2 for any solution of

$-{\bf trace}~{}{\cal A}(\cdot){\rm D}^{2}u+\beta(u)\leq f\quad\hbox{in
$\Omega$}$

and any non negative large solution of

$-{\bf trace}~{}{\cal A}(\cdot){\rm D}^{2}v+\beta(v)\geq f\quad\hbox{in
$\Omega$}$

one has

$\limsup_{{\rm d}(x)\searrow 0}\dfrac{u^{*}(x)}{v_{*}(x)}\leq\liminf_{{\rm
d}(x)\searrow 0}\dfrac{\phi\left(\sqrt{\dfrac{1-\ell_{u}}{\Lambda}}{\rm
d}(x)\right)}{\phi\left(\sqrt{\dfrac{1-\ell_{d}}{\lambda}}{\rm d}(x)\right)}.$

Then (1.12) is fulfilled whenever

$\dfrac{1-\ell_{u}}{1-\ell_{d}}\geq\dfrac{\Lambda}{\lambda}$ (2.13)

holds. In particular if

$\sqrt{\dfrac{1-\ell_{u}}{\Lambda}}={\rm
C}=\sqrt{\dfrac{1-\ell_{d}}{\lambda}}\quad\Leftrightarrow\quad\dfrac{1-\ell_{u}}{1-\ell_{d}}=\dfrac{\Lambda}{\lambda},$
(2.14)

the precise boundary behaviour of any nonnegative large solution $u$ of

$-{\bf trace}~{}{\cal A}(\cdot){\rm D}^{2}u+\beta(u)=f\quad\hbox{in
$\Omega$},$

is given by

$\limsup_{{\rm d}(x)\searrow 0}\dfrac{u^{*}(x)}{\phi\big{(}{\rm C}{\rm
d}(x)\big{)}}\leq 1\leq\liminf_{{\rm d}(x)\searrow
0}\dfrac{u_{*}(x)}{\phi\big{(}{\rm C}{\rm d}(x)\big{)}},$

thus

$\lim_{{\rm d}(x)\searrow 0}\dfrac{u(x)}{\phi\big{(}{\rm C}{\rm
d}(x)\big{)}}=1.$

In the next result we complete the obtainement of the inequality (2.13) by
means of a sharp iterative device by M. Safonov (see [15] and [19]).

###### Theorem 2.2

Asume $\partial\Omega\in{\cal C}^{2}$. Let us suppose (1.3), (1.4), (2.3),
(4.12) and

$\dfrac{1-\ell_{u}}{1-\ell_{d}}<\dfrac{\Lambda}{\lambda}.$ (2.15)

Then any subsolution $u$ and any non negative large solution $v$ of (1.1)
satisfy the property (1.12) provided (2.7) and (2.11) with
$0\leq\ell_{u},\ell_{d}<1$.

Proof. We claim that in this case the inequality

$\limsup_{{\rm d}(x)\rightarrow 0}\dfrac{u^{*}(x)}{v_{*}(x)}\leq 1$

also holds. Indeed, (2.15) implies that for $\varepsilon>0$ the inequality

$\varepsilon+\nu\phi\left(\sqrt{\dfrac{1-\ell_{d}}{\lambda}}{\rm
d}(x)\big{)}\right)\geq\phi\left(\sqrt{\dfrac{1-\ell_{u}}{\Lambda}}{\rm
d}(x)\big{)}\right)\big{)}\quad\hbox{for ${\rm d}(x)$ small}$

(see (2.4)) holds, where

$\nu=\left(\dfrac{(1-\ell_{d})\Lambda}{(1-\ell_{u})\lambda}\right)^{\frac{1}{\alpha-1}}>1.$

It leads to

$0<\limsup_{{\rm d}(x)\searrow 0}\dfrac{u^{*}(x)}{v_{*}(x)}={\rm
L}\leq\left(\dfrac{(1-\ell_{d})\Lambda}{(1-\ell_{u})\lambda}\right)^{\frac{1}{\alpha-1}}.$

Assume ${\rm L}>1$, otherwise the proof ends. Then for every constant $k$ near
${\rm L}$ such that $1<k<{\rm L}$ there exists
$x_{0}\in\Omega_{\delta_{\Omega}}$ satisfying

$u^{*}(x_{0})>kv_{*}(x_{0})\quad\hbox{for ${\rm d}(x_{0})$ small.}$

We will use the notation $[u^{*}>kv_{*}]\doteq\big{\\{}x\in\Omega_{{\rm
d}(x_{0})}:u^{*}(x)>kv_{*}(x)\big{\\}}$. It is clear, that in the open set
$[u^{*}>kv_{*}]$ the assumption (2.3) implies

$\beta(u^{*})>\beta(kv_{*})>k^{\alpha}\beta(v_{*}).$

Let us consider $r=\dfrac{{\rm d}(x_{0})}{2}$, for which ${\rm
B}_{r}(x_{0})\subset\Omega_{\delta_{\Omega}}$, and define the smooth function

$w(x)={\rm C}\big{(}r^{2}-|x-x_{0}|^{2}\big{)},\quad x\in{\rm B}_{r}(x_{0})$

that satisfies

${\bf trace}~{}{\cal A}(x){\rm D}^{2}w(x)\geq-2{\rm C}\Lambda,\quad x\in{\rm
B}_{r}(x_{0}).$

Then

$\left\\{\begin{array}[]{l}-{\bf trace}~{}{\cal A}(\cdot){\rm D}^{2}u^{*}\leq
f-k^{\alpha}\beta(v_{*}),\\\\[4.26773pt] -{\bf trace}~{}{\cal A}(\cdot){\rm
D}^{2}\big{(}kv_{*}-w\big{)}\geq k\big{(}f-\beta(v_{*})\big{)}-2{\rm
C}\Lambda\end{array}\right.\quad\hbox{in $[u^{*}>kv_{*}]\cap{\rm
B}_{r}(x_{0})$ }$

in viscosity sense. Since $r<{\rm d}(x)<3r$, one proves

$\phi\left(\sqrt{\dfrac{1-\ell_{d}}{\lambda}}{\rm
d}(x)\big{)}\right)\geq\phi\left(\sqrt{\dfrac{1-\ell_{d}}{\lambda}}3r\big{)}\right),$

hence Proposition 2.2 implies

$k(k^{\alpha-1}-1)\beta\big{(}v_{*}(x)\big{)}\geq
k(k_{0}^{\alpha}-1)\beta\left((1+\varepsilon)\phi\left(\sqrt{\dfrac{1-\ell_{d}}{\lambda}}3r\big{)}\right)\right)$

for $\varepsilon>0$ and ${\rm d}(x_{0})$ small, where $k\geq k_{0}>1$. So
that,

$f-k^{\alpha}\beta(v_{*})-k\big{(}f-\beta(v_{*})\big{)}+2{\rm C}\Lambda\leq
0\quad\hbox{in $[u^{*}>kv_{*}]\cap{\rm B}_{r}(x_{0})$},$

provided

${\rm
C}=\dfrac{k(k_{0}^{\alpha}-1)}{2\Lambda}\beta\left((1+\varepsilon)\phi\left(\sqrt{\dfrac{1-\ell_{d}}{\lambda}}3r\big{)}\right)\right).$

By means of the Weak Maximum Principle without coercivity term (see Theorem
7.1) one proves that there exists
$x_{1}\in\partial\big{(}[u^{*}>kv_{*}]\cap{\rm B}_{r}(x_{0})\big{)}$ such that

$u^{*}(x)-kv_{*}(x)+w(x)\leq u^{*}(x_{1})-kv_{*}(x_{1})+w(x_{1}),\quad
x\in[u^{*}>kv_{*}]\cap{\rm B}_{r}(x_{0}).$ (2.16)

In fact, $x_{1}\in\partial{\rm B}_{r}(x_{0})$ because $x_{1}\in{\rm
B}_{r}(x_{0})$ implies $x_{1}\in\partial[u^{*}>kv]$, thus
$u^{*}(x_{1})=kv(x_{1})$ whence the contradiction

${\rm C}r^{2}=w(x_{0})\leq w(x_{1})<{\rm C}r^{2}$

follows. So that, $x_{1}\in\partial{\rm B}_{r}(x_{0})$ and $w(x_{1})=0$ as
well as

$u^{*}(x_{1})-kv_{*}(x_{1})=u^{*}(x_{1})-kv_{*}(x_{1})+w(x_{1})\geq
u^{*}(x_{0})-kv_{*}(x_{0})+w(x_{0})\geq w(x_{0})={\rm C}r^{2}.$

Thus

$u^{*}(x_{1})-kv_{*}(x_{1})\geq
k\dfrac{\big{(}k_{0}^{\alpha-1}-1\big{)}r^{2}}{2\Lambda}\beta\left((1+\varepsilon)\phi\left(\sqrt{\dfrac{1-\ell_{d}}{\lambda}}3r\right)\right).$

Now the property (2.1) implies

$u^{*}(x_{1})-kv_{*}(x_{1})\geq
k\dfrac{(1+\varepsilon)\big{(}k_{0}^{\alpha-1}-1\big{)}r^{2}}{2\Lambda\varepsilon_{0}}\phi\left(\sqrt{\dfrac{1-\ell_{d}}{\lambda}}3r\right),$

for $0<\varepsilon_{0}$ and ${\rm d}(x_{0})$ small. On the other hand, from
(2.4) we deduce

$\varepsilon+\nu\phi\left(\sqrt{\dfrac{1-\ell_{d}}{\lambda}}3r\right)\geq\phi\left(\sqrt{\dfrac{1-\ell_{u}}{\Lambda}}r\right)$

where

$\nu\doteq\left(9\dfrac{(1-\ell_{d})\Lambda}{(1-\ell_{u})\lambda}\right)^{\frac{1}{\alpha-1}}>1$

(see (2.15)). Since $v$ is a solution of (1.1) we lead to

$\phi\left(\sqrt{\dfrac{1-\ell_{d}}{\lambda}}3r\right)\geq\dfrac{1}{\nu}\phi\left(\sqrt{\dfrac{1-\ell_{u}}{\Lambda}}r\right)-\dfrac{\varepsilon}{\nu}\geq\dfrac{1}{\nu}v^{*}(x_{1})-\dfrac{\varepsilon}{\nu}$

(see Proposition 2.1). So that,

$u^{*}(x_{1})\geq
k\left(1+\dfrac{(1+\varepsilon)\big{(}k_{0}^{\alpha-1}-1\big{)}r^{2}}{2\Lambda\nu\varepsilon_{0}(1-\varepsilon)}\right)v_{*}(x_{1})-\dfrac{(1+\varepsilon)\varepsilon\big{(}k_{0}^{\alpha-1}-1\big{)}r^{2}}{2\Lambda\nu\varepsilon_{0}}.$

Therefore, from

$0<\limsup_{{\rm d}(x)\searrow 0}\dfrac{u^{*}(x)}{v_{*}(x)}={\rm L}<+\infty$

we deduce

${\rm L}+\varepsilon\geq
k\left(1+\dfrac{(1+\varepsilon)\big{(}k_{0}^{\alpha-1}-1\big{)}r^{2}}{2\Lambda\nu\varepsilon_{0}}\right).$

So, the choice $k={\rm L}-\varepsilon>k_{0}>1$, thus $\varepsilon<{\rm
L}-k_{0}$, implies

${\rm L}+\varepsilon\geq({\rm
L}-\varepsilon)\left(1+\dfrac{(1+\varepsilon)\big{(}k_{0}^{\alpha-1}-1\big{)}r^{2}}{2\Lambda\nu\varepsilon_{0}(1-\varepsilon)}\right).$

Therefore one arrives to a contradiction by letting $\varepsilon\searrow
0.\hfill\mbox{$\quad{}_{\Box}$}$

###### Remark 2.6

We also may obtain (2.16) by using Theorem 7.3 provided (1.6) and
(1.7).$\quad{}_{\Box}$

###### Remark 2.7

Property (2.14) requires $\ell_{d}\leq\ell_{u}$ and

$\ell_{d}\leq\liminf_{{\rm d}(x)\searrow
0}\dfrac{f(x)}{\beta\left(\phi\big{(}{\rm C}{\rm
d}(x)\big{)}\right)}\leq\limsup_{{\rm d}(x)\searrow
0}\dfrac{f(x)}{\beta\left(\phi\big{(}{\rm C}{\rm
d}(x)\big{)}\right)}\leq\ell_{u}<1.$

When $\Lambda=\lambda$ equality (2.14) follows if $0\leq\ell_{u}=\ell_{d}<1$.
It extends the result obtained in [14] for the unit operator ${\cal
A}(\cdot)\equiv{\rm I}.\hfill\mbox{$\quad{}_{\Box}$}$

##  3 Extra Keller-Osserman type growth of the sources on the boundary

As it was pointed out, first we illustrate the power like case corresponding
to

$-{\bf trace}~{}{\cal A}(\cdot){\rm D}^{2}u+u^{m}=d^{-q}\quad\hbox{in
$\Omega$},$

where we require

$q>\dfrac{2m}{m-1},~{}m>1,$

in order to consider an extra Keller-Osserman type growth at the boundary of
solutions for which we want to obtain the property (1.12)

$\limsup_{{\rm d}(x)\rightarrow 0}\dfrac{u^{*}(x)}{v_{*}(x)}\leq 1.$

######  Theorem 3.1 (Comparison on the boundary (power-like case))

Asume $\partial\Omega\in{\cal C}^{2}$. Suppose (1.3), (1.4) and

$\limsup_{{\rm d}(x)\searrow 0}f(x){\rm
d}(x)^{q}\leq\ell_{u}\quad(\ell_{u}>0),$ (3.1)

for $q>\dfrac{2m}{m-1},~{}m>1$. Then any solution $u$ of

$-{\bf trace}~{}{\cal A}(\cdot){\rm D}^{2}u+u^{m}\leq f\quad\hbox{in
$\Omega$}$

admits a maximal behaviour given by

$\limsup_{{\rm d}(x)\searrow 0}u^{*}(x){\rm
d}(x)^{\frac{q}{m}}\leq\ell_{u}^{\frac{1}{m}}.$ (3.2)

On the other hand, under

$\liminf_{{\rm d}(x)\searrow 0}f(x){\rm d}(x)^{q}\geq\ell_{d}>0,$ (3.3)

any large non negative supersolution $v$ of

$-{\bf trace}~{}{\cal A}(\cdot){\rm D}^{2}v+v^{m}\geq f\quad\hbox{in
$\Omega$}$

admits the behaviour given by

$\liminf_{{\rm d}(x)\searrow 0}v_{*}(x){\rm
d}(x)^{\frac{q}{m}}\geq\ell_{d}^{\frac{1}{m}}.$ (3.4)

Therefore, one satisfies (1.12).

Proof. Once more the idea is to construct suitable classical large
supersolutions. So, for $0<\delta<\delta_{\Omega}$, we consider

${\rm U}_{0}(x)={\rm C}_{0}\sigma^{-\frac{q}{m}},\quad
x\in\Omega^{\delta}_{\delta_{\Omega}}\doteq\\{x\in\Omega:~{}\delta<{\rm
d}(x)<\delta_{\Omega}\\},$

where ${\rm C}_{0}>0$ is independent on $\delta$. Here $\sigma\doteq{\rm
d}(x)-\delta$. Straightforward computations show

$-{\bf trace}~{}{\cal A}(x){\rm D}^{2}{\rm U}_{0}(x)={\rm
C}_{0}\dfrac{q}{m}\sigma^{-\frac{q+2m}{m}}\bigg{(}-\frac{q+m}{m}{\rm
E}(x)+\sigma{\bf trace}~{}{\cal A}(x){\rm D}^{2}{\rm d}(x)\bigg{)},\quad
x\in\Omega^{\delta}_{\delta_{1}},$

where $0<\delta_{1}<\delta_{\Omega}$. On the other hand, let
$0<\varepsilon<{\rm C}_{0}^{m}-\ell_{u}$, assumed ${\rm C}_{0}^{m}>\ell_{u}$.
Then, the condition (3.1) implies

$-f(x)\geq-(\ell_{u}+\varepsilon){\rm
d}(x)^{-q}\geq-(\ell_{u}+\varepsilon)\sigma^{-q},\quad
x\in\Omega^{\delta}_{\delta_{1}}.$

So that

$\begin{array}[]{ll}-{\bf trace}~{}{\cal A}(\cdot){\rm D}^{2}{\rm U}_{0}+{\rm
U}_{0}^{m}-f&\geq\sigma^{-q}\bigg{[}-{\rm
C}_{0}\dfrac{q}{m}\sigma^{\frac{q(m-1)-2m}{m}}\bigg{(}\dfrac{q+m}{m}{\rm
E}(x)+\sigma{\bf trace}~{}{\cal A}(x){\rm D}^{2}{\rm d}(x)\bigg{)}\\\ &+{\rm
C}_{0}^{m}-(\ell_{u}+\varepsilon)\bigg{]},\end{array}$

for $x\in\Omega^{\delta}_{\delta_{1}}$. We note the inequality

$\dfrac{q(m-1)-m}{m}>\dfrac{q(m-1)-2m}{m}>0,$

whence

$-{\bf trace}~{}{\cal A}{\rm D}^{2}{\rm U}_{0}+{\rm
U}_{0}^{m}-f\geq\sigma^{-q}\left[-{\rm
C}_{0}\dfrac{q}{m}\dfrac{q+m}{m}\Lambda\delta_{1}^{\frac{q(m-1)-2m}{m}}-{\rm
C}_{0}\dfrac{q}{m}{\rm C}_{\partial\Omega,{\cal
A}}\delta_{1}^{\frac{q(m-1)-m}{m}}+{\rm
C}_{0}^{m}-(\ell_{u}+\varepsilon)\right],$

holds in $\Omega^{\delta}_{\delta_{1}}$. Since

${\rm
C}_{0}\left[\dfrac{q}{m}\delta_{1}^{\frac{q(m-1)-2m}{m}}\left(\frac{q+m}{m}\Lambda-{\rm
C}_{\partial\Omega,{\cal A}}\delta_{1}\right)\right]\leq{\rm
C}_{0}^{m}-(\ell_{u}+\varepsilon)$

is fulfilled for $\delta_{1}$ small enough uniformly on ${\rm C}_{0}$, we
obtain

$-{\bf trace}~{}{\cal A}(x){\rm D}^{2}{\rm U}_{0}(x)+({\rm U}_{0}(x))^{m}\geq
f(x),\quad x\in\Omega^{\delta}_{\delta_{1}}.$

By means of arguments used in Proposition 2.1 one deduces

$\limsup_{{\rm d}(x)\nearrow 0}u^{*}(x){\rm d}(x)^{\frac{q}{m}}\leq{\rm
C}_{0}$

and then (3.2) holds by letting ${\rm C}_{0}\searrow\ell_{u}^{\frac{1}{m}}$.
Next for $0<\delta$ and $0<\delta_{1}\leq\delta_{1}$, we construct

${\rm V}_{0}(x)={\rm C}_{0}\sigma^{-\frac{q}{m}},\quad
x\in\Omega_{\delta_{1}}.$

for $\sigma={\rm d}(x)+\delta$. As above, straightforward computations show

$-{\bf trace}~{}{\cal A}(x){\rm D}^{2}{\rm V}_{0}(x)={\rm
C}_{0}\dfrac{q}{m}\sigma^{-\frac{q+2m}{m}}\bigg{(}-\frac{q+m}{m}{\rm
E}(x)+\sigma{\bf trace}~{}{\cal A}(x){\rm D}^{2}{\rm d}(x)\bigg{)},\quad
x\in\Omega_{\delta_{1}}.$

On the other hand, let $0<\varepsilon<\ell_{d}-{\rm C}_{0}^{m}$, assumed
$\ell_{d}>{\rm C}_{0}^{m}$. Then, the condition (3.3) implies

$-f(x)\leq-(\ell_{d}-\varepsilon){\rm
d}(x)^{-q}\geq(\ell_{d}-\varepsilon)\big{(}{\rm d}(x)+\delta\big{)}^{-q},\quad
x\in\Omega_{\delta_{1}}.$

So that

$\begin{array}[]{ll}-{\bf trace}~{}{\cal A}{\rm D}^{2}{\rm
V}_{0}(x)+\big{(}{\rm V}_{0}(x)\big{)}^{m}-f(x)&\leq\sigma^{-q}\bigg{[}-{\rm
C}_{0}\dfrac{q}{m}\dfrac{q+m}{m}{\rm
E}(x)\sigma^{\frac{q(m-1)-2m}{m}}\\\\[5.69046pt] &+{\rm
C}_{0}\dfrac{q}{m}\sigma^{\frac{q(m-1)-m}{m}}{\bf trace}~{}{\cal A}(x){\rm
D}^{2}{\rm d}(x)+{\rm C}_{0}^{m}-(\ell_{d}-\varepsilon)\bigg{]}\\\\[5.69046pt]
&\leq\sigma^{-q}\bigg{[}{\rm C}_{0}\dfrac{q}{m}{\rm C}_{\partial\Omega,{\cal
A}}\sigma^{\frac{q(m-1)-m}{m}}+{\rm
C}_{0}^{m}-(\ell_{d}-\varepsilon)\bigg{]},\end{array}$

for $x\in\Omega_{\delta_{1}}$. We note $\ell_{d}-{\rm
C}_{0}^{m}-\varepsilon>0$. Again we use the inequality

$\dfrac{q(m-1)-m}{m}>\dfrac{q(m-1)-2m}{m}\geq 0.$

Therefore, taking $\delta=\delta_{1}<\delta_{\Omega}$

${\rm C}_{0}\dfrac{q}{m}(2\delta_{1})^{\frac{q(m-1)-m}{m}}{\rm
C}_{\partial\Omega,{\cal A}}\leq(\ell_{d}-\varepsilon)-{\rm C}_{0}^{m}$

holds for some $\delta_{1}$ small enough uniformly on ${\rm C}_{0}$. Thus

$-{\bf trace}~{}{\cal A}(x){\rm D}^{2}{\rm V}_{0}(x)+\big{(}{\rm
V}_{0}(x)\big{)}^{m}\leq f(x),\quad x\in\Omega_{\delta_{1}}.$

So, by means of arguments used in Proposition 2.2 we deduce

$\liminf_{{\rm d}(x)\nearrow 0}v_{*}(x){\rm d}(x)^{\frac{q}{m}}\geq{\rm
C}_{0}$

and then (3.4) holds by letting ${\rm C}_{0}\nearrow\ell_{d}^{\frac{1}{m}}$.
Since $\ell_{d}\leq\ell_{u}$ the property (1.12) hods.$\quad{}_{\Box}$

###### Remark 3.1

Since the diffusive term of the PDE is negligible with respect to the other
one the ellipticity constans $\lambda$ and $\Lambda$ do not appear in the
estimations (3.2) and (3.4). In Theorem 5.2 we obtain a more general
result.$\quad{}_{\Box}$

With the above simple ideas we get the general result

######  Theorem 3.2 (Comparison on the boundary (general case))

Asume $\partial\Omega\in{\cal C}^{2}$ and $\beta\in{\cal C}^{2}$. Suppose
(1.3), (1.4) and

$\liminf_{{\rm d}(x)\searrow 0}\dfrac{f(x)}{h\big{(}{\rm d}(x)\big{)}}\leq 1$
(3.5)

for some ${\cal C}^{2}$ decreasing function
$h:~{}{\mathbb{R}}_{+}\rightarrow{\mathbb{R}}_{+}$ such that $h(0)=\infty$ and
$h^{\prime}(0)=0$ satisfying the compatiblity condition (1.13). Futhermore, we
assume that

$\left\\{\begin{array}[]{l}{\rm
H}_{1}(\sigma)\doteq\dfrac{h(\sigma)}{\beta^{\prime}\big{(}\beta^{-1}\big{(}h(\sigma)\big{)}}\left(h^{\prime\prime}(\sigma)-\dfrac{\big{(}h^{\prime}(\sigma)\big{)}^{2}\beta^{\prime\prime}\big{(}\beta^{-1}\big{(}h(\sigma)\big{)}\big{)}}{\big{(}\beta^{\prime}\big{(}\beta^{-1}\big{(}h(\sigma)\big{)}\big{)}^{2}}\right),\\\\[5.69046pt]
{\rm
H}_{2}(\delta)\doteq-\dfrac{h^{\prime}(\sigma)}{h(\sigma)\beta^{\prime}\big{(}\beta^{-1}\big{(}h(\sigma)\big{)}\big{)}}\end{array}\right.$
(3.6)

are increasing functions vanishing at the origin. Then any subsolution $u$ of

$-{\bf trace}~{}{\cal A}(\cdot){\rm D}^{2}u+\beta(u)=f\quad\hbox{in $\Omega$}$

admits a maximal behaviour given by

$\limsup_{{\rm d}(x)\searrow 0}\dfrac{u^{*}(x)}{\beta^{-1}\big{(}h({\rm
d}(x))\big{)}}\leq 1.$ (3.7)

On the other hand, under

$\liminf_{{\rm d}(x)\searrow 0}\dfrac{f(x)}{h\big{(}{\rm d}(x)\big{)}}\geq 1$
(3.8)

any large nononegative supersolution $v$ admits the minimal behaviour given by

$\liminf_{{\rm d}(x)\searrow 0}\dfrac{v_{*}(x)}{\beta^{-1}\big{(}h({\rm
d}(x))\big{)}}\geq 1.$ (3.9)

Hence assuming

$\lim_{{\rm d}(x)\searrow 0}\dfrac{f(x)}{h\big{(}{\rm d}(x)\big{)}}=1,$ (3.10)

one has (1.12).

Proof. The idea is to construct a classical large subsolution. Let

${\rm U}_{0}(x)=\varrho\beta^{-1}\big{(}h({\rm d}(x)-\delta)\big{)},\quad
x\in\Omega^{\delta}_{\delta_{1}},$

where $0<\delta<\delta_{1}<\delta_{\Omega}$ and $\varrho>1$. Denoting
$\sigma={\rm d}(x)-\delta$, straightforward computations show

$\begin{array}[]{ll}-{\bf trace}~{}{\cal A}(x){\rm D}^{2}{\rm
U}_{0}(x)&\geq\Lambda\left(-\dfrac{h^{\prime\prime}(\sigma)}{\beta^{\prime}\big{(}\beta^{-1}\big{(}h(\sigma)\big{)}\big{)}}+\dfrac{\big{(}h^{\prime}(\sigma)\big{)}^{2}g\beta^{\prime\prime}\big{(}\beta^{-1}\big{(}h(\sigma)\big{)}\big{)}}{\big{(}\beta^{\prime}\big{(}\beta^{-1}\big{(}h(\sigma)\big{)}\big{)}^{3}}\right)\\\
&+\dfrac{h^{\prime}(\sigma)}{\beta^{\prime}\big{(}\beta^{-1}\big{(}h(\sigma)\big{)}\big{)}}{\cal
C}_{\partial\Omega,{\cal A}},\quad x\in\Omega^{\delta}_{\delta_{1}},\quad
x\in\Omega^{\delta}_{\delta_{1}}\end{array}$

(see (3.6)). On the other hand, let $0<\varepsilon<\varrho-1$. Then, the
condition (3.5) implies

$-f(x)\geq-(1+\varepsilon)h({\rm d}(x))\geq-(1+\varepsilon)h({\rm
d}(x)-\delta),\quad x\in\Omega^{\delta}_{\delta_{1}}.$

Consequently

$-{\bf trace}~{}{\cal A}(x){\rm D}^{2}{\rm U}_{0}(x)+\beta\big{(}{\rm
U}_{0}(x)\big{)}-f(x)\geq h(\delta_{1})\left[\varrho\bigg{(}-\Lambda{\rm
H}_{1}(\delta_{1})-{\rm H}_{2}(\delta_{1}){\cal C}_{\partial\Omega,{\cal
A}}\bigg{)}+\varrho-(1+\varepsilon)\right],$

for $x\in\Omega^{\delta}_{\delta_{1}}$. Since

$\Lambda{\rm H}_{1}(\delta_{1})+{\rm
H}_{2}(\delta_{1})\leq\dfrac{\varrho-(1+\varepsilon)}{\varrho}$

is fulfilled for $\delta_{1}$ small enough (see (3.6)), we obtain

$-{\bf trace}~{}{\cal A}(x){\rm D}^{2}{\rm U}_{0}(x)+\beta\big{(}{\rm
U}_{0}(x)\big{)}\geq f(x)\quad\hbox{for $x\in\Omega^{\delta}_{\delta_{1}}$}.$

Finally one deduces

$\limsup_{{\rm d}(x)\nearrow 0}\dfrac{u^{*}(x)}{\beta^{-1}\big{(}h({\rm
d}(x))\big{)}}\leq\varrho$

and then (3.7) holds by letting $\varrho\searrow 1$.

Next for $0<\delta$ and $0<\delta_{1}\leq\delta_{\Omega}$, we construct

${\rm V}_{0}(x)=\varrho\beta^{-1}\big{(}h({\rm d}(x)+\delta)\big{)},\quad
x\in\Omega_{\delta_{1}},$

where $0<\varrho<1$. Let $0<\varepsilon<1-\varrho$. Then, the condition (3.8)
implies

$-f(x)\leq-(1-\varepsilon)h({\rm d}(x))\leq-(1-\varepsilon)h({\rm
d}(x)+\delta),\quad x\in\Omega_{\delta_{1}}.$

As above, by straightforward computations we have

$-{\bf trace}~{}{\cal A}(x){\rm D}^{2}{\rm V}_{0}(x)+\beta\big{(}{\rm
V}_{0}(x)\big{)}-f(x)\leq h(\delta_{1})\left[\varrho\bigg{(}-\lambda{\rm
H}_{1}(2\delta_{\Omega})+{\rm H}_{2}(2\delta_{1}){\cal
C}_{\partial\Omega,{\cal A}}\bigg{)}+\varrho-(1-\varepsilon)\right],$

for $x\in\Omega_{\delta_{1}}$ (see (3.6)). Here by simplicity we choose
$\delta=\delta_{1}<\delta_{\Omega}$. Since

${\rm H}_{2}(2\delta_{1})\leq\dfrac{(1-\varepsilon)-\varrho}{\varrho}$

is fulfilled for $\delta_{1}$ small enough, and

$-{\bf trace}~{}{\cal A}(x){\rm D}^{2}{\rm V}_{0}(x)+\beta\big{(}{\rm
V}_{0}(x)\big{)}\leq f(x)\quad\hbox{for $x\in\Omega^{\delta}_{\delta_{1}}$}$

holds. Finally one deduces

$\liminf_{{\rm d}(x)\nearrow 0}\dfrac{v_{*}(x)}{\beta^{-1}\big{(}h({\rm
d}(x))\big{)}}\geq\varrho.$

Now (3.9) follows by letting $\varrho\nearrow 1.\hfill\mbox{$\quad{}_{\Box}$}$

###### Remark 3.2

For the power-like case $\beta_{m}(t)=t^{m},~{}m>1$, and
$h_{q}(\delta)=\delta^{-q}$ one has

$\left\\{\begin{array}[]{l}{\rm
H}_{1}(\sigma)=\dfrac{q}{m}\dfrac{q+m}{m}\sigma^{-\frac{q(m-1)-2m}{m}},\\\\[7.82433pt]
{\rm
H}_{2}(\sigma)=\dfrac{q}{m}\sigma^{\frac{q(m-1)-m}{m}}.\end{array}\right.$

Certainly, both functions are increasing functions vanishing at the origin
provided $q>\dfrac{2m}{m-1}.\hfill\mbox{$\quad{}_{\Box}$}$

######  Remark 3.3 (Bieberbach choice)

For the exponential case $\beta(t)=e^{t}$ the Keller-Osserman boundary profile
is governed by

$\phi(\delta)=\log\left(\dfrac{2}{\delta^{2}}\right)$

(see Remark 1.5). Here, available extra Keller-Osserman boundary profiles are
governed by functions $h(\delta)$ satisfying

$h(\delta)>\dfrac{2}{\delta^{2}}\quad\hbox{for $\delta$ small}$

(see (1.13)) such that

$\left\\{\begin{array}[]{l}{\rm
H}_{1}(\sigma)=h^{\prime\prime}(\sigma)-\dfrac{\big{(}h^{\prime}(\sigma)\big{)}^{2}}{h(\sigma)},\\\\[7.82433pt]
{\rm
H}_{2}(\sigma)=-\dfrac{h^{\prime}(\sigma)}{\big{(}h(\sigma)\big{)}^{2}}\end{array}\right.$

are increasing functions vanishing at the origin. In order to avoid tedious
computations we only give the simple example
$h(\delta)=\dfrac{\mu}{\delta^{2}}$ for $\mu>2.\hfill\mbox{$\quad{}_{\Box}$}$

In this case of extra Keller-Osserman type growth by means of an argument of
rescaling the explositivity of the sources is transferred to the
supersolutions.

######  Theorem 3.3 (Transferring the explosivity)

Suppose $q\geq\dfrac{2m}{m-1}$. Let $f$ be a source for which

$\liminf_{{\rm d}(x)\searrow 0}f(x){\rm d}(x)^{q}=\ell_{d}>0.$ (3.11)

The any nononegative supersolution $v$ of

$-{\bf trace}~{}{\cal A}(\cdot){\rm D}^{2}v+v^{m}=f\quad\hbox{in $\Omega$}$

is a large supersolution.

Proof. Let $x_{0}\in\Omega_{\delta_{\Omega}}$ and $2r={\rm d}(x_{0})$ with
$r<1$. Then the supersolution $v_{*}$ solves, in the viscosity sense,

$\left\\{\begin{array}[]{ll}-{\bf trace}~{}{\cal A}(\cdot){\rm
D}^{2}v_{*}+v_{*}^{m}\geq\widehat{{\rm C}}r^{-q},&\quad x\in{\bf
B}_{r}(x_{0}),\\\\[4.26773pt] v_{*}(x)\geq 0,&\quad x\in\partial{\bf
B}_{r}(x_{0}),\end{array}\right.$

for $\widehat{{\rm C}}=\ell_{d}2^{-q}$. Since ${\cal A}$ is uniform elliptic
in $\Omega_{\delta_{\Omega}}$ there exists a classical solution ${\rm W}$ of

$\left\\{\begin{array}[]{ll}-{\bf trace}~{}{\cal A}(x){\rm D}^{2}{\rm
W}(x)+{\rm W}(x)^{m}=\widehat{{\rm C}}r^{-q},&\quad x\in{\bf
B}_{r}(x_{0}),\\\\[4.26773pt] {\rm W}(x)=0,&\quad x\in\partial{\bf
B}_{r}(x_{0}).\end{array}\right.$

Then $v_{*}(x)\geq{\rm W}(x),~{}x\in\overline{{\bf B}}_{r}(x_{0})$. Next, we
construct $\widehat{{\rm W}}(x)={\rm W}(x-x_{0})$ that solves

$\left\\{\begin{array}[]{ll}-{\bf trace}~{}{\cal A}(x-x_{0}){\rm
D}^{2}\widehat{{\rm W}}(x)+\widehat{{\rm W}}(x)^{m}=\widehat{{\rm
C}}r^{-q},&\quad x\in{\bf B}_{r}(0),\\\\[4.26773pt] \widehat{{\rm
W}}(x)=0,&\quad x\in\partial{\bf B}_{r}(0),\end{array}\right.$

hence

$-{\bf trace}~{}{\cal A}(x-x_{0}){\rm D}^{2}\widehat{{\rm W}}(x)+\widehat{{\rm
W}}(x)^{m}\geq\widehat{{\rm C}}r^{-\frac{2m}{m-1}},\quad x\in{\bf B}_{r}(0)$

or

$r^{\frac{2m}{m-1}}{\bf trace}~{}{\cal A}(x-x_{0}){\rm D}^{2}\widehat{{\rm
W}}(x)+r^{\frac{2m}{m-1}}\widehat{{\rm W}}(x)^{m}\geq\widehat{{\rm C}},\quad
x\in{\bf B}_{r}(0)$

for $r$ small. Since $\dfrac{2}{m-1}+2=\dfrac{2m}{m-1}$, the function ${\rm
W}_{r}(x)=r^{\frac{2}{m-1}}\widehat{{\rm W}}(rx)$ satisfies

$-{\bf trace}~{}{\cal A}(x-x_{0}){\rm D}^{2}{\rm W}_{r}(x)+{\rm
W}_{r}(x)^{m}\geq\widehat{{\rm C}},\quad x\in{\bf B}_{1}(0),$

and we get the problem

$\left\\{\begin{array}[]{ll}-{\bf trace}~{}{\cal A}(x-x_{0}){\rm D}^{2}{\rm
V}(x)+{\rm V}(x)^{m}=\widehat{{\rm C}},&\quad x\in{\bf
B}_{1}(0),\\\\[4.26773pt] {\rm V}(x)=0,&\quad x\in\partial{\bf
B}_{1}(0).\end{array}\right.$

A simple adaptation of Strong Maximum Principle (see [37]) shows ${\rm V}>0$
in ${\bf B}_{1}(0)$ whence

$0<{\rm V}(0)\leq\left(\dfrac{{\rm
d}(x_{0})}{2}\right)^{\frac{2}{m-1}}v_{*}(x_{0})$

concludes the result. $\quad{}_{\Box}$

##  4 Comparison, uniqueness and existence of solutions

As it was pointed out in Section 1 the eventual degeneracy inside the domain
of the leading part of the equation prevents, in general, to consider ${\cal
C}^{2}$ solutions. They are replaced by viscosity solutions that solve the PDE
more intrinsically. We send to the mongraphy [11] for a full introduction of
the Viscosity Solution Theory. In short, we recall that for any function
$u:{\cal O}\rightarrow{\mathbb{R}},~{}{\cal O}\subseteq{\mathbb{R}}^{{\rm
N}}$, locally bounded from above, one defines the upper semi-continuous
envelope

$u^{*}(x)=\limsup_{y\rightarrow x\atop y\in{\cal O}}u(y),\quad x\in{\cal O}$

and the superjets

${\cal J}^{2,+}_{{\cal O}}u^{*}(x)=\big{\\{}({\bf p},{\rm
X})\in{\mathbb{R}}^{{\rm N}}\times{\cal S}^{{\rm N}}:~{}u^{*}(y)\leq
u^{*}(x)+\displaystyle{\langle{\bf
p},y-x\rangle}+\dfrac{1}{2}\displaystyle{\langle{\rm
X}(y-x),y-x\rangle}+o(|y-x|^{2})\big{\\}}.$

In particular, a locally bounded from above function
$u:\Omega\rightarrow{\mathbb{R}}$ is a (viscosity) subsolution of

$-{\bf trace}~{}{\cal A}(\cdot){\rm D}^{2}u+\beta(u)=f\quad\hbox{in $\Omega$}$

if

$-{\bf trace}~{}{\cal A}(x){\rm X}+\beta\big{(}u^{*}(x)\big{)}\leq f(x)$

for all $({\bf p},{\rm X})\in{\cal J}^{2,+}_{\Omega}u^{*}(x),~{}x\in\Omega$.
By simplicity, sometimes we say that $u$ is a solution of

$-{\bf trace}~{}{\cal A}(\cdot){\rm D}^{2}u+\beta(u)\leq f\quad\hbox{in
$\Omega$}.$

###### Remark 4.1

From the above definition we deduce a property near the well known
verification results of the Optimal Control Problems. More precisely any
(viscosity) subsolution of

$-{\bf trace}~{}{\cal A}(\cdot){\rm D}^{2}u+\beta(u)=f\quad\hbox{in $\Omega$}$

satisfies

$u^{*}(x)\leq\inf_{{\rm V}\in{\cal V}_{+}^{u}}{\rm V}(x):,\quad
x\in\overline{\Omega},$ (4.1)

where

${\cal V}_{+}^{u}=\sup\big{\\{}{\rm V}\in{\cal C}^{2}\hbox{ supersolution in
$\Omega$ such that $u^{*}\leq{\rm V}$ on $\partial\Omega$}\big{\\}}.$

Indeed, given a such function ${\rm V}$, if there exists a positive interior
maximum point of $u^{*}-{\rm V}$ at some $x_{0}\in\Omega$ one has

$u^{*}(x)-u^{*}(x_{0})\leq{\rm V}(x)-{\rm V}(x_{0})=\displaystyle{\langle{\rm
D}{\rm V}(x_{0}),x-x_{0}\rangle}+\dfrac{1}{2}\displaystyle{\langle{\rm
D}^{2}{\rm V}(x_{0}),x-x_{0}\rangle}+o(|x-x_{0}|^{2})^{,}$

thus $\big{(}{\rm D}{\rm V}(x_{0}),{\rm D}^{2}{\rm V}(x_{0})\big{)}\in{\cal
J}^{2,+}_{\Omega}u^{*}(x_{0})$ for which we derive the contradiction

$-{\bf trace}~{}{\cal A}(x_{0}){\rm D}^{2}{\rm
V}(x_{0})+\beta\big{(}u^{*}(x_{0})\big{)}\leq-{\bf trace}~{}{\cal
A}(x_{0}){\rm D}^{2}{\rm V}(x_{0})+\beta\big{(}{\rm V}(x_{0})\big{)}.$

Therefore

$u^{*}(x)-{\rm V}(x)\leq u^{*}(x_{0})-{\rm V}(x_{0})\leq 0,\quad
x\in\overline{\Omega}$

and (4.1) holds.$\quad{}_{\Box}$

Analogously, for $v:{\cal O}\rightarrow{\mathbb{R}},~{}{\cal
O}\subseteq{\mathbb{R}}^{{\rm N}}$, locally bounded from below, one defines
the lower semi-continuous envelope

$v^{*}(x)=\limsup_{y\rightarrow x\atop y\in{\cal O}}v(y),\quad x\in{\cal O}$

and the subjets

${\cal J}^{2,-}_{{\cal O}}u^{*}(x)=\big{\\{}({\bf p},{\rm
X})\in{\mathbb{R}}^{{\rm N}}\times{\cal S}^{{\rm N}}:~{}v_{*}(y)\geq
v_{*}(x)+\displaystyle{\langle{\bf
p},y-x\rangle}+\dfrac{1}{2}\displaystyle{\langle{\rm
X}(y-x),y-x\rangle}+o(|y-x|^{2})\big{\\}}.$

In particular, a locally bounded from below function
$v:\Omega\rightarrow{\mathbb{R}}$ is a (viscosity) supersolution of

$-{\bf trace}~{}{\cal A}(\cdot){\rm D}^{2}v+\beta(v)=f\quad\hbox{in $\Omega$}$

if

$-{\bf trace}~{}{\cal A}(x){\rm X}+\beta\big{(}v_{*}(x)\big{)}\geq f(x)$

for all $({\bf p},{\rm X})\in{\cal J}^{2,-}_{\Omega}v_{*}(x),~{}x\in\Omega$.
Sometimes we say that $v$ is a solution of

$-{\bf trace}~{}{\cal A}(\cdot){\rm D}^{2}v+\beta(u)\geq f\quad\hbox{in
$\Omega$}.$

###### Remark 4.2

Arguing as in Remark 4.2 one proves that any (viscosity) supersolution of

$-{\bf trace}~{}{\cal A}(\cdot){\rm D}^{2}v+\beta(v)=f\quad\hbox{in $\Omega$}$

satisfies

$\sup_{{\rm U}\in{\cal V}_{-}^{v}}{\rm U}(x)\leq v_{*}(x),\quad
x\in\overline{\Omega},$ (4.2)

where

${\cal V}_{-}^{v}=\big{\\{}{\rm U}\in{\cal C}^{2}\hbox{ is a subsolution in
$\Omega$ such that ${\rm U}\leq v_{*}$ on $\partial\Omega$}\big{\\}}.$

Clearly, a locally bounded function $u:~{}\Omega\rightarrow{\mathbb{R}}$ is a
(viscosity) solution if its upper semi-continuous envelope $u^{*}$ and its
lower semi-continuous envelope $u_{*}$ are sub and supersolution,
respectively. We send to the monography [11] for the theory of (viscosity)
solutions.

Next, we study a kind of superlinearity property

$q(t+s)\geq q(s)+q(t),\quad r,s\geq 0,$ (4.3)

which can be transferred under a suitable condition

######  Lemma 4.1 (Superlinearity transferred)

Let $q$ and $\beta$ two nonnegative functions on ${\mathbb{R}}_{+}$ such that

$\dfrac{\beta(t)}{q(t)}\quad\hbox{is nondecreasing}.$

Then if $q(t)$ verifies (4.3) the same property holds for $\beta(t)$.

Proof. It follows from

$\beta(t)+\beta(s)=q(t)\dfrac{\beta(t)}{q(t)}+q(s)\dfrac{\beta(s)}{q(s)}\leq\big{(}q(t)+q(s)\big{)}\dfrac{\beta(t+s)}{q(t+s)}\leq\beta(t+s).$

###### Remark 4.3

Since the identity is a function satisfying (4.3) and $t^{\alpha-1}$ is a
nondecreasing function if $\alpha\geq 1$ we deduce that
$q(t)=t^{\alpha},~{}\alpha\geq 1$ satisfies (4.3).$\quad{}_{\Box}$

We give some details on the result of uniqueness. First we obtain the
uniqueness of eventual classical solutions of (1.1) in a direct way when

$\limsup_{{\rm d}(x)\rightarrow 0}\dfrac{u(x)}{v(x)}\leq 1$

is assumed (see (1.12)). Since we only intend to give an idea of the reasoning
we omit additional asumptions on the ellipticity of the ${\cal A}(x)$ for
which (1.1) admits classical solutions in this motivation (see [17] or [35]).

######  Proposition 4.1 (Classical Maximum Principle)

Suppose (1.14). Let $f$ and $g$ be two continuous functions. If
$u,v\in\mathcal{C}^{2}(\Omega)$ are two nonnegative functions verifying

$\left\\{\begin{array}[]{l}-{\bf trace}~{}{\cal A}(x){\rm
D}^{2}u(x)+\beta\big{(}u(x)\big{)}\leq f(x)\\\\[5.69046pt] -{\bf
trace}~{}{\cal A}(x){\rm D}^{2}v(x)+\beta\big{(}v(x)\big{)}\geq
g(x),\end{array}\quad x\in\Omega,\right.$

and

$u(x)\leq v(x),\quad x\in\partial\Omega$ (4.4)

then

$u(x)\leq
v(x)+\beta^{-1}\left(\displaystyle{\|(f-g)_{+}\|}_{\infty}\right),\quad
x\in\Omega.$ (4.5)

When we replace (4.4) by

$\limsup_{x\rightarrow\partial\Omega}\frac{u(x)}{v(x)}\leq 1\quad\hbox{and
$v>0$ near $\partial\Omega$}$ (4.6)

the inequality (4.5) also holds. Here $r_{+}=\max\\{r,0\\}$.

Proof. If the maximum of the continuous function $u-v$ is attained on
$\partial\Omega$ or it is a nonegative value the inequality (4.5) follows. On
the contrary, let us assume that $u-v$ admits some point $x_{0}\in\Omega$ such
that

$(u-v)(x_{0})=\max_{\Omega}~{}(u-v)>0.$

Since

$0\geq{\bf trace}~{}{\cal A}(x_{0}){\rm D}^{2}(u-v)(x_{0})$ (4.7)

we deduce the inequality

$(f-g)(x_{0})+{\bf trace}~{}{\cal A}(x_{0}){\rm
D}^{2}(u-v)(x_{0})\geq\beta\big{(}u(x_{0})\big{)}-\beta\big{(}v(x_{0})\big{)}>0.$
(4.8)

Then from Lemma 4.1 we get

$\beta\big{(}(u-v)(x_{0})\big{)}\leq(f-g)_{+}(x_{0})$

and (4.5) follows. On the other hand, it is clear that

$\limsup_{x\rightarrow\partial\Omega}\frac{u(x)}{v(x)}<1\quad\hbox{and $v>0$
near $\partial\Omega$}$ (4.9)

lead to $u\leq v$ near $\partial\Omega$ and the above reasoning is true. The
more general assumption (4.6) implies

$\limsup_{x\rightarrow\partial\Omega}\frac{u(x)}{(1+\varepsilon)v(x)}<1,$

for all $\varepsilon>0$ and then $u\leq(1+\varepsilon)v$ near
$\partial\Omega$. Moreover, from (1.14) we have

$-(1+\varepsilon){\bf trace}~{}{\cal A}(\cdot){\rm
D}^{2}v+\beta\big{(}(1+\varepsilon)v\big{)}\geq(1+\varepsilon)\left(-{\bf
trace}~{}{\cal A}(\cdot){\rm
D}^{2}v+\beta(v)\right)\geq(1+\varepsilon)g\quad\hbox{in $\Omega$}.$

The above reasoning leads to

$\beta(u-(1+\varepsilon)v)(x_{0})\leq(f-g)(x_{0})-\varepsilon
g(x_{0})\leq\displaystyle{\|(f-g)_{+}\|}_{\infty}+\varepsilon\displaystyle{\|g\|}_{\infty}$

thus

$u(x)-(1+\varepsilon)v(x)\leq\beta^{-1}\left(\displaystyle{\|(f-g)_{+}\|}_{\infty}+\varepsilon\displaystyle{\|g\|}_{\infty}\right),\quad
x\in\Omega,$

and the result follows by letting $\varepsilon\rightarrow
0.\hfill\mbox{$\quad{}_{\Box}$}$

###### Remark 4.4

For a direct Comparison Principle as

$u\leq v$ on $\partial\Omega$ and $f\leq g$ in $\Omega\quad\Rightarrow\quad
u\leq v$ in $\Omega$

the assumption (1.14) is not required (see (4.7) and (7.10)).$\quad{}_{\Box}$

A simple consequence proves that the condition

$\displaystyle\limsup_{x\rightarrow\partial\Omega}\frac{u(x)}{v(x)}=1$ (4.10)

implies uniqueness.

###### Corollary 4.1

Suppose (1.14). If $u,v\in\mathcal{C}^{2}(\Omega)$ are two non negative
functions for which

$-{\bf trace}~{}{\cal A}(x){\rm
D}^{2}u(x)+\beta\big{(}u(x)\big{)}=f(x)\quad\hbox{in $\Omega$}$

and (4.10) hold, then $u=v$ on $\Omega$, provided that $f$ is a non negative
continuous function.$\quad{}_{\Box}$

So that, a uniqueness condition on the boundary as

$\lim_{x\rightarrow\partial\Omega}\dfrac{u(x)}{v(x)}=1$

plays an important role. Next, we extend Proposition 4.1 in the Viscosity
Solution Theory. Hark back to the above reasoning. Let us assume that the
upper semi continuous function $u^{*}-v_{*}$ admits some interior point
$x_{0}\in\Omega$ such that

$(u^{*}-v_{*})(x_{0})=\max_{\overline{\Omega}}~{}(u^{*}-v_{*})>0.$

As in the proof of Proposition 4.1, the aim is to work by means of an
inequality as (4.7). Unfortunately $u^{*}$ and $v_{*}$ are not ${\cal C}^{2}$
functions. If we use semijets $(p,{\rm X})\in{\cal J}^{2,+}u^{*}(x_{0})$ and
$(q,{\rm Y})\in{\cal J}^{2,-}v_{*}(x_{0})$ the device goes if

${\bf trace}~{}{\cal A}(x_{0})({\rm X}-{\rm Y})\leq 0$

holds for adequate ${\rm X},{\rm Y}\in{\cal S}^{{\rm N}}$ but strong
dificulties appear due to the semijets ${\cal J}^{2,+}u^{*}(x_{0})$ and ${\cal
J}^{2,-}v_{*}(x_{0})$ are set valued functions. The reasoning of the Jensen
Maximum Principle is very advantageous (see [23], [22] or [11] for details).
Essentially, it is about working close $x_{0}$ by means of a device that
doubles the number of variables penalizating this doubling (see the works by
N.S. Kruzkov, for instance [25]). Indeed, for $\alpha>0$ the upper semi-
continuous function

$\Psi(x,y)\doteq
u^{*}(x)-v_{*}(y)-\dfrac{\alpha}{2}|x-y|^{2},\quad(x,y)\in\Omega\times\Omega$

has a finite supremum equals to ${\rm M}_{\alpha}$ and admits
$(x_{\alpha},y_{\alpha})$ such that

$\lim_{\alpha\nearrow\infty}\left({\rm
M}_{\alpha}-\Psi(x_{\alpha},y_{\alpha})\right)=0$

and

$\left\\{\begin{array}[]{l}\displaystyle\lim_{\alpha\rightarrow\infty}\alpha|x_{\alpha}-y_{\alpha}|^{2}=0,\\\\[4.26773pt]
\displaystyle\lim_{\alpha\rightarrow\infty}{\rm
M}_{\alpha}=(u^{*}-v_{*})(x_{0})\end{array}\right.$

(see [11, Lemma 3.1]). Moreover, there exist ${\rm X},{\rm Y}\in{\cal S}^{{\rm
N}}$ such that

$\big{(}\alpha(x_{\alpha}-y_{\alpha},{\rm X}\big{)}\in\overline{{\cal
J}}^{2,+}u^{*}(x_{\alpha})\quad\hbox{and}\quad\big{(}\alpha(y_{\alpha}-x_{\alpha},{\rm
Y}\big{)}\in\overline{{\cal J}}^{2,-}v_{*}(y_{\alpha})$

and

$-3\alpha\left(\begin{array}[]{ll}\text{\rm I}&{\bf 0}\\\ {\bf 0}&\text{\rm
I}\end{array}\right)\leq\left(\begin{array}[]{lr}{\rm X}&{\bf 0}\\\ {\bf
0}&-{\rm Y}\end{array}\right)\leq 3\alpha\left(\begin{array}[]{lr}\text{\rm
I}&-\text{\rm I}\\\ \text{\rm I}&\text{\rm I}\end{array}\right)$ (4.11)

In fact, (4.11) implies ${\rm X}\leq{\rm Y}$ (see [11, Theorem 3.2]). So that,
some suitable adptations near $x_{0}$ can be available in order that the aim
can go. For large $\alpha$ one has

$(u^{*}-v_{*})(x_{0})\leq{\rm M}_{\alpha}\leq
u^{*}(x_{\alpha})-v_{*}(y_{\alpha})$

and

$\beta\big{(}u^{*}(x_{\alpha})\big{)}-\beta\big{(}v_{*}(y_{\alpha})\big{)}\leq{\bf
trace}~{}{\cal A}(x_{\alpha}){\rm X}-{\bf trace}~{}{\cal A}(y_{\alpha}){\rm
Y}\leq\omega\big{(}\alpha|x_{\alpha}-y_{\alpha}|^{2}\big{)}$

assuming that there exists a kind of modulus of continuity
$\omega:{\mathbb{R}}_{+}\rightarrow{\mathbb{R}}_{+}$, with $\omega(0^{+})=0$
such that

${\bf trace}~{}{\cal A}(x){\rm X}-{\bf trace}~{}{\cal A}(y){\rm
Y}\leq\omega\big{(}\alpha|x-y|^{2}\big{)}$ (4.12)

whenever $x,y\in\Omega,~{}\alpha>0,$ and ${\rm X},{\rm Y}\in{\cal S}^{{\rm
N}}$ for which (4.11) holds. Again, Lemma 4.1 implies

$\begin{array}[]{ll}\beta\big{(}(u^{*}-v_{*})(x_{0})\big{)}&\leq\beta\big{(}u^{*}(x_{\alpha})-v_{*}(y_{\alpha})\big{)}\leq\beta\big{(}u^{*}(x_{\alpha})\big{)}-\beta\big{(}v_{*}(y_{\alpha})\big{)}\\\\[4.26773pt]
&\leq\omega\big{(}\alpha|x_{\alpha}-y_{\alpha}|^{2}\big{)}+f(x_{\alpha})-g(y_{\alpha})\\\\[4.26773pt]
&\leq\omega\big{(}\alpha|x_{\alpha}-y_{\alpha}|^{2}\big{)}+\omega_{f}\big{(}|x_{\alpha}-y_{\alpha}|\big{)}+\displaystyle{\|(f-g)_{+}\|}_{\infty}\end{array}$
(4.13)

where $\omega_{f}$ is the modulus of continuity of $f$. Finally, the reasoning
of Proposition 4.1 ends by letting $\alpha\nearrow\infty$. What we have tried
is

######  Theorem 4.1 (Weak Maximum Principle)

Assume (1.14) and (4.12). Let $u$ be a discontinuous subsolution of

$-{\bf trace}~{}{\cal A}(\cdot){\rm D}^{2}u+\beta(u)=f,\quad x\in\Omega$

where $f$ is a uniformly continuous function and $v$ be a discontinuous
supersolution of

$-{\bf trace}~{}{\cal A}(\cdot){\rm D}^{2}v+\beta(v)=g,\quad x\in\Omega$

where $g$ is a continuous function. Then

$\limsup_{x\rightarrow\partial\Omega}\frac{u^{*}(x)}{v_{*}(x)}\leq 1$ (4.14)

implies

$u^{*}(x)\leq
v_{*}(x)+\beta^{-1}\left(\displaystyle{\|(f-g)_{+}\|}_{\infty}\right),\quad
x\in\Omega.$

Moreover, if $u$ is a solution of

$-{\bf trace}~{}{\cal A}(\cdot){\rm D}^{2}u+\beta(u)=f\quad x\in\Omega$

one has

$u^{*}(x)\leq u_{*}(x),\quad x\in\Omega,$

whence the function $u$ is continuous in
$\Omega.\hfill\mbox{$\quad{}_{\Box}$}$

###### Remark 4.5

We send to Theorems 2.1 and 2.2 in order to obtain (4.14).$\quad{}_{\Box}$

###### Remark 4.6

In the proof of Theorem 4.1 the assumption of degenerate ellipticity is not
assumed explicitly because (4.12) implies that the matrix function ${\cal A}$
is degenerate elliptic. Indeed, given ${\rm Y}\in{\cal S}_{+}^{{\rm N}}$
Cauchy inequality implies

$-\displaystyle{\langle{\rm
Y}\eta,\eta\rangle}\leq\varepsilon|\eta|^{2}+\dfrac{1}{\varepsilon}\displaystyle{\|{\rm
Y}\|}^{2}\displaystyle{\|\eta\|}^{2},\quad\varepsilon>0,~{}\eta\in{\mathbb{R}}^{{\rm
N}},$

where $\displaystyle{\|{\rm Y}\|}=\sup_{\displaystyle{\|\eta\|}\leq
1}|\displaystyle{\langle{\rm Y}\eta,\eta\rangle}|$. Clearly, we also have the
inequalty

$-\displaystyle{\langle{\rm
Y}\eta,\eta\rangle}\leq\varepsilon|\eta|^{2}+\left(1+\dfrac{\displaystyle{\|{\rm
Y}\|}}{\varepsilon}\right)\displaystyle{\|{\rm
Y}\|}\displaystyle{\|\eta\|}^{2}$

that may be rewiten as

$\left(\begin{array}[]{lc}{\bf 0}&{\bf 0}\\\ {\bf 0}&-({\rm Y}+\varepsilon{\rm
I})\end{array}\right)\leq\left(1+\dfrac{\displaystyle{\|{\rm
Y}\|}}{\varepsilon}\right)\displaystyle{\|{\rm
Y}\|}\left(\begin{array}[]{lr}{\rm I}&-{\rm I}\\\ {\rm I}&{\rm
I}\end{array}\right).$

By choosing $\alpha>\displaystyle{\|{\rm
Y}\|}\max\left\\{1,1+\dfrac{\displaystyle{\|{\rm Y}\|}}{\varepsilon}\right\\}$
and $\varepsilon$ small one obtains (4.11) for the matrix ${\rm X}={\bf 0}$
and ${\rm Y}+\varepsilon{\rm I}$, whence (4.12) implies

$-{\cal A}\left(y-\frac{z}{\alpha}\right)({\rm Y}+\varepsilon{\rm I})={\cal
A}(y){\bf 0}-{\cal A}\left(y-\frac{z}{\alpha}\right)({\rm Y}+\varepsilon{\rm
I})\leq\omega\left(\dfrac{\displaystyle{\|z\|}^{2}}{\alpha}\right)$

for any $z\in{\mathbb{R}}^{{\rm N}}\setminus\\{0\\}$. Then letting
$\alpha\rightarrow\infty$ and $\varepsilon\searrow 0$ concludes

$0\leq{\cal A}(y)\xi\otimes\xi,\quad y\in\Omega,~{}\xi\in{\mathbb{R}}^{{\rm
N}},$

provided ${\rm Y}=\xi\otimes\xi\in{\cal S}_{+}^{{\rm
N}},~{}\xi\in{\mathbb{R}}^{{\rm N}}$. See [11, Remark 3.4] for a more general
proof. When

${\cal A}(x)=\sigma(x)\sigma(x)^{\tt t}$

where $\sigma:~{}\Omega\rightarrow{\mathbb{R}}^{{\rm
N}}\times{\mathbb{R}}^{N}$is a Lipschitz continuous function with constant
${\rm L}$ the assumption (4.12) becomes

${\bf trace}~{}{\cal A}(x){\rm X}-{\bf trace}~{}{\cal A}(y){\rm Y}\leq 3{\rm
L}^{2}\alpha|x-y|^{2},$ (4.15)

thus $\omega(r)=3{\rm L}^{2}r$ (see [11, Example 3.6]).$\quad{}_{\Box}$

###### Remark 4.7

Remark 4.4 can be extended to Theorem 4.1. Indeed, a direct comparison is
deduced from (4.13) that becomes

$\beta\big{(}u^{*}(x_{\alpha})\big{)}-\beta\big{(}v_{*}(y_{\alpha})\big{)}\leq\omega\big{(}\alpha|x_{\alpha}-y_{\alpha}|^{2}\big{)}+\omega_{f}\big{(}|x_{\alpha}-y_{\alpha}|\big{)}$

if we assume $f\leq g$ in $\Omega$. Therefore an inequality as

$(u^{*}-v_{*})(x_{0})=\max_{\overline{\Omega}}~{}(u^{*}-v_{*})>0$

implies the contradiction

$0<\beta\big{(}u^{*}(x_{0})\big{)}-\beta\big{(}v_{*}(x_{0})\big{)}\leq 0.$

Consequently, we have proved a direct Comparison Principle as

$\hbox{$u^{*}\leq v_{*}$ on $\partial\Omega$ and $f\leq g$ in
$\Omega\quad\Rightarrow\quad u^{*}\leq v_{*}$ in $\Omega$},$

under (4.12) and the uniformly continuous of the function $f$. We emphasize
that the assumption (1.14) is not required.$\quad{}_{\Box}$

The existence of solution follows from the well known Perron method as, for
instance [21] or [11]. For several reasons first we sketch an adaptation of
the plan of [14] on the elliptic regularization

$\left\\{\begin{array}[]{ll}-\varepsilon\Delta u_{\varepsilon,n}-{\bf
trace}~{}{\cal A}(\cdot){\rm
D}^{2}u_{\varepsilon,n}+\beta(u_{\varepsilon,n})=\min\\{f,n\\}&\quad\hbox{in
$\Omega$},\\\\[4.26773pt] u_{\varepsilon,n}\equiv n&\quad\hbox{in
$\partial\Omega$},\\\ \end{array}\right.$ (4.16)

where $f$ is a nonnegative uniformly continuous function. From classical
argument, there exists $u_{\varepsilon,n}\in{\cal C}^{2}(\Omega)$ solving
(4.16) (see [17], [14] or [35]). In fact, $\\{u_{\varepsilon,n}\\}_{n}$ is an
increasing sequence. The consideration of the assumption (1.8) enables us to
obtain universal bounds as it is proved in Section 6 below. More precisely
under (1.8) we define

$\Psi_{\varepsilon}(\zeta)\doteq\dfrac{1}{{\rm A}}\Phi^{-1}\left(\dfrac{{\rm
B}}{\sqrt{\Lambda+\varepsilon}}\zeta\right),\quad 0\leq\zeta<{\rm R},$ (4.17)

with ${\rm R}<\dfrac{\sqrt{\Lambda+\varepsilon}}{{\rm
B}}\Phi(0^{+})\leq+\infty$, where

$\Phi(t)=\int^{+\infty}_{t}\dfrac{ds}{\sqrt{2{\rm G}(s)}}.$

was defined in Introduction (see (1.9)).

###### Remark 4.8

For the power like case $\beta_{m}(t)=t^{m},~{}m>1,$ one satisfies
$\Phi_{m}(0^{+})=+\infty$. Also for $\beta(t)=e^{t}$ and for which
$\beta(t)=te^{2t}$ one has $\Phi(0^{+})=+\infty.\hfill\mbox{$\quad{}_{\Box}$}$

A simple consequence of the proof of Theorem 6.1 below in Section 6 we prove

###### Theorem 4.2

Assume (1.3), (1.8) as well as (4.12). Then there exists two positive
constants ${\rm A}$ and ${\rm B}$, independent on $n$, for which

$u_{\varepsilon,n}(x)\leq\sum_{i=1}^{{\rm N}}\Psi_{\varepsilon}\big{(}{\rm
R}^{2}-|x_{i}|^{2}\big{)}+\beta^{-1}\big{(}\displaystyle{\|f\|}_{{\cal
Q}_{{\rm R}}(x_{0})}\big{)},\quad x\in{\cal Q}_{{\rm R}}(x_{0})$ (4.18)

holds in any cube ${\cal Q}_{{\rm R}}(x_{0})\doteq\\{x\in{\mathbb{R}}^{{\rm
N}}:~{}|x_{i}-x_{0,i}|<{\rm
R}\\}\subset\subset\Omega.\hfill\mbox{$\quad{}_{\Box}$}$

So that, the elliptic regularization implies

$u_{\varepsilon}(x)=\sup_{n}u_{\varepsilon,n}(x)<+\infty,\quad x\in\Omega,$
(4.19)

whence one proves that $u_{\varepsilon}$ is a ${\cal C}^{2}$ large solution of

$-\varepsilon\Delta u_{\varepsilon}-{\bf trace}~{}{\cal A}(\cdot){\rm
D}^{2}u_{\varepsilon}+\beta(u_{\varepsilon})=f\quad\hbox{in $\Omega$}.$ (4.20)

We send to [14] for details. We note by construction that the function defined
in (4.19) is the minimal large solution of (4.20). A refinement in the proof
of Theorem 6.1 leads to a simple consequence

###### Theorem 4.3

Assume (1.3), (1.8) as well as (4.12). Then there exists two positive
constants ${\rm A}$ and ${\rm B}$, independent on $n$ for which

$u_{\varepsilon}(x)\leq\sum_{i=1}^{{\rm N}}\Psi_{1}\big{(}{\rm
R}^{2}-|x_{i}|^{2}\big{)}+\beta^{-1}\big{(}\displaystyle{\|f\|}_{{\cal
Q}_{{\rm R}}(x_{0})}\big{)},\quad x\in{\cal Q}_{{\rm R}}(x_{0})$ (4.21)

holds in any cube ${\cal Q}_{{\rm R}}(x_{0})\subset\subset\Omega$, where we
are assuming $0<\varepsilon<1.\hfill\mbox{$\quad{}_{\Box}$}$

Now the estimate (4.21) enables us to define the functions

$\overline{u}(x)=\limsup_{y\mapsto x\atop\varepsilon\searrow
0}u_{\varepsilon}(y)$

and

$\underline{u}(x)=\liminf_{y\mapsto x\atop\varepsilon\searrow
0}u_{\varepsilon}(y).$

Moreover, the stablity results of the Viscosity Solution Theory (see [11])
show that $\overline{u}(x)$ and $\underline{u}(x)$ are respectively viscosity
sub and supersolution of

$-{\bf trace}~{}{\cal A}(\cdot){\rm D}^{2}u+\beta(u)=f\quad\hbox{in $\Omega$}$

(see also [8], [5]). Both functions blow up on the boundary. Defining the set

${\cal F}_{-}=\big{\\{}u\hbox{ is a (viscosity) subsolution in
$\Omega$}\big{\\}}.$

one has $\overline{u}\in{\cal F}_{-}$. Therefore the Perron function

${\cal U}(x)\doteq\sup_{u\in{\cal F}_{-}}u^{*}(x),\quad x\in\Omega$ (4.22)

is well defined under the assumptions of Theorem 6.1 and verifies

$\overline{u}(x)\leq{\cal U}(x)\leq+\infty,\quad x\in\Omega$

Futhermore, Perron function, ${\cal U}$, blows up on the boundary. In fact,
adapting the reasonings of [11, Theorem 4.1] (see also [13, Theorem 4]) the
Perron function solves

$-{\bf trace}~{}{\cal A}(\cdot){\rm D}^{2}{\cal U}+\beta({\cal
U})=f\quad\hbox{in $\Omega$}.$

A suitable adaptation of the reasoning of the proof of Theorem 4.1 gives

$u_{\varepsilon,n}(x)\leq v_{*}(x),\quad x\in\Omega$

for any large supersolution of (1.1), the Perron solution is the minimum large
solution of (1.1) in the sense

${\cal U}^{*}(x)\leq v_{*}(x),\quad x\in\Omega.$ (4.23)

We collect the result.

###### Theorem 4.4

Let us assume $\Omega$ is a bounded set in ${\mathbb{R}}^{{\rm N}},~{}{\rm
N}>1,~{}\partial\Omega\in{\cal C}^{2}$, (1.3), (1.4), (1.6), (1.7), (2.3) as
well as (4.12). Let $f$ be a uniformly continuous function satisfying

$\limsup_{{\rm d}(x)\searrow
0}\dfrac{f(x)}{\beta\big{(}\phi\left(\sqrt{\dfrac{1-\ell}{\Lambda}}{\rm
d}(x)\right)}\leq\ell\leq\liminf_{{\rm d}(x)\searrow
0}\dfrac{f(x)}{\beta\big{(}\phi\left(\sqrt{\dfrac{1-\ell}{\lambda}}{\rm
d}(x)\right)}$

with $0\leq\ell<1$. Then ${\cal U}\in{\cal C}(\Omega)$ is the unique large
solution of the equation (1.1) whose boundary behaviour satisfies the
inequality

$\limsup_{{\rm d}(x)\searrow 0}\dfrac{{\cal
U}(x)}{\phi\left(\sqrt{\dfrac{1-\ell}{\Lambda}}{\rm d}(x)\right)}\leq
1\leq\liminf_{{\rm d}(x)\searrow 0}\dfrac{{\cal
U}(x)}{\phi\left(\sqrt{\dfrac{1-\ell}{\lambda}}{\rm d}(x)\right)}$

(see Theorem 2.2 ). Here $\phi$ is the function defined in (1.9).

Proof. Notice that we do not consider the assumption (1.14). Let $v$ be an
arbitrary continuous large solution of (1.1). Then

${\cal U}^{*}(x)\leq v(x),\quad x\in\Omega$

holds (see (4.23)). We claim that in fact ${\cal U}_{*}\equiv v$. Indeed,
assume that there exists $x_{0}\in\Omega$ such that ${\cal U}_{*}(x_{0})\leq
v(x_{0})$. Since $\Omega$ is bounded we deduce that

${\cal O}^{\varepsilon}=\\{x\in\Omega:~{}(1+\varepsilon){\cal
U}_{*}(x)<v(x)\\}\subset\subset\Omega$

is a non empty open subset, provided $\varepsilon>0$ small. The assumption
(2.3) implies the property

$\dfrac{\beta(t)}{t}\quad\hbox{is increasing for $t$ greater than some large
$t_{0}$}$

and choose $\mu>0$ so small that ${\cal U}_{*}(x)\geq t_{0}$ holds in
$\Omega_{\mu}=\\{x\in\Omega:~{}{\rm d}(x)<\mu\\}$. As $\varepsilon$ is small
${\cal O}^{\varepsilon}_{\mu}={\cal O}^{\varepsilon}\cap\Omega_{\mu}$ is a non
empty open subset. Moreover

$-{\bf trace}~{}{\cal A}(\cdot){\rm D}^{2}(1+\varepsilon){\cal
U}+\beta\big{(}(1+\varepsilon){\cal U}\big{)}\geq(1+\varepsilon)\big{(}-{\bf
trace}~{}{\cal A}(\cdot){\rm D}^{2}{\cal U}+\beta({\cal U})\big{)}\geq 0,\quad
x\in{\cal O}^{\varepsilon}_{\mu}$

in the viscosity sense. Then adapting Remark 4.7 we obtain

$v(x)-(1+\varepsilon){\cal U}_{*}(x)\leq\sup_{\partial{\cal
O}^{\varepsilon}_{\mu}}\big{(}v-(1+\varepsilon){\cal U}_{*}\big{)},\quad
x\in{\cal O}^{\varepsilon}_{\mu}.$

By construction the maximum of $v-(1+\varepsilon){\cal U}_{*}$ can not be
achieved in $\partial{\cal O}^{\varepsilon}$, whence

$v(x)-(1+\varepsilon){\cal U}_{*}(x)\leq\sup_{\\{y\in\partial{\cal
O}^{\varepsilon},~{}{\rm d}(y)=\mu\\}}\big{(}v(y)-{\cal U}_{*}(y)\big{)},\quad
x\in{\cal O}^{\varepsilon}_{\mu}$

and

$v(x)-{\cal U}_{*}(x)\leq\sup_{y\in\partial{\cal O}^{\varepsilon},~{}{\rm
d}(y)=\mu\\}}\big{(}v(y)-{\cal U}_{*}(y)\big{)}\equiv{\rm C}_{\mu},\quad
x\in{\cal O}^{\varepsilon}_{\mu}$

hold. On the other hand, by mean of Theorem 7.3 the inequality $\beta({\cal
U}^{*})\leq\beta(v)$ in $\Omega$ gives

$v(x)-{\cal U}_{*}(x)\leq{\rm C}_{\mu},\quad x\in\Omega\setminus\Omega_{\mu},$

thus

$v(x)-{\cal U}_{*}(x)\leq{\rm C}_{\mu},\quad x\in\Omega.$

Since ${\rm C}_{\mu}=v(z_{0})-{\cal U}_{*}(z_{0})\geq 0$ for some
$z_{0}\in\partial{\cal O}^{\varepsilon}$ and ${\rm d}(z_{0})=\mu>0$, the upper
semi-continuous function $u-{\cal U}_{*}$ attains a maximim value in $\Omega$
at some interior point. Since ${\cal U}\leq v$ we may apply The Strong Maximum
Theorem without coercive term (Theorem 7.2) and deduce

$v(x)-{\cal U}_{*}(x)\equiv{\rm C}_{\mu},\quad x\in\Omega.$

The relative version of (4.13) for $v$ and ${\cal U}_{*}$ at $z_{0}\in\Omega$
becomes

$\beta\big{(}v(x_{\alpha})\big{)}-\beta\big{(}{\cal
U}_{*}(y_{\alpha})\big{)}\leq\omega\big{(}\alpha|x_{\alpha}-y_{\alpha}|^{2}\big{)}+\omega_{f}\big{(}|x_{\alpha}-y_{\alpha}|\big{)}$

whence letting $\alpha\nearrow\infty$ we get

$\beta\big{(}{\cal U}_{*}(z_{0})+{\rm
C}_{\mu}\big{)}=\beta\big{(}v(z_{0})\big{)}\leq\beta\big{(}{\cal
U}_{*}(z_{0})\big{)}$

implies ${\rm C}_{\mu}=0$. Notice that the inequality

$v={\cal U}_{*}\leq{\cal U}^{*}\leq v$

concludes the continuity of ${\cal U}.\hfill\mbox{$\quad{}_{\Box}$}$

A similar program can be developed when the source term has the extra Keller-
Osserman type growth studied in Section 3. Then one deduces

###### Theorem 4.5

Let us assume $\Omega\in{\cal C}^{2},~{}\beta\in{\cal C}^{2}$, (1.3), (1.4),
(1.6), (1.7), (2.3) as well as (4.12). Let $f$ be a uniformly continuous
function satisfying

$\lim_{{\rm d}(x)\searrow 0}\dfrac{f(x)}{h\big{(}{\rm d}(x)\big{)}}=1,$

for some ${\cal C}^{2}$ decreasing function
$h:~{}{\mathbb{R}}_{+}\rightarrow{\mathbb{R}}_{+}$ such that $h(0)=\infty$ and
$h^{\prime}(0)=0$ satisfying the compatiblity condition (1.13) as well as

$\left\\{\begin{array}[]{l}{\rm
H}_{1}(\sigma)\doteq\dfrac{h(\sigma)}{\beta^{\prime}\big{(}\beta^{-1}\big{(}h(\sigma)\big{)}}\left(h^{\prime\prime}(\sigma)-\dfrac{\big{(}h^{\prime}(\sigma)\big{)}^{2}\beta^{\prime\prime}\big{(}\beta^{-1}\big{(}h(\sigma)\big{)}\big{)}}{\big{(}\beta^{\prime}\big{(}\beta^{-1}\big{(}h(\sigma)\big{)}\big{)}^{2}}\right),\\\\[5.69046pt]
{\rm
H}_{2}(\delta)\doteq-\dfrac{h^{\prime}(\sigma)}{h(\sigma)\beta^{\prime}\big{(}\beta^{-1}\big{(}h(\sigma)\big{)}\big{)}}\end{array}\right.$

are increasing, vanishing at the origin (see (3.6)). Then the equation (1.1)
has a unique large continuous solution ${\cal U}\in{\cal C}(\Omega)$ whose
boundary behaviour satisfies

$\limsup_{{\rm d}(x)\searrow 0}\dfrac{{\cal U}(x)}{\beta^{-1}\big{(}h({\rm
d}(x))\big{)}}=1.$

##  5 Second order term in the boundary assymptotic expansion

As it was pointed out in Introduction, in Sections 2 and 3 we only have
obtained the leading term of the boundary assymptotic expansion of the large
solutions. Here we study the second order term of this expansion. Relative to
Keller-Osserman type growth of the sources on the boundary (see (2.7)) we
replace the classical large subsolution near the boundary ${\rm U}_{0}$, used
in the proof of Proposition 2.1, by

${\rm U}(x)={\rm U}_{0}(x)+{\rm U}_{1}(x)\doteq\phi(\sigma)+\widehat{\rm
C}(x,\varepsilon,\delta)\Theta\big{(}\phi(\sigma)\big{)},\quad
x\in\Omega^{\delta}_{\delta_{1}}\quad(0<\delta<\delta_{1}<\delta_{\Omega})$

where $\sigma=\widehat{\sigma}_{\varepsilon}\big{(}{\rm d}(x)-\delta)$ with
$\widehat{\sigma}_{\varepsilon}=\sqrt{\dfrac{(1-\varepsilon)(1-\ell_{u})}{\Lambda}}$.
The functions $\widehat{\rm
C}(\cdot,\varepsilon,\delta):\Omega_{\delta_{1}}\rightarrow{\mathbb{R}}$ and
$\Theta:{\mathbb{R}}_{+}\rightarrow{\mathbb{R}}_{+}$ will be determined from
the reasoning that follows. In particular, we will suppose $\widehat{\rm
C}(\cdot,\varepsilon,\delta)\in{\rm
W}^{2,\infty}\big{(}\Omega^{\delta}_{\delta_{1}}\big{)}$. On the other hand,
let $\Theta(t)$ be a function satisfying

$\Theta^{\prime}(t)=-t\Phi^{\prime}(t)\quad\hbox{for large $t$}.$ (5.1)

See Remark 5.3 below for suitable choices of the function $\Theta$. Clearly,
if $\Theta\big{(}\phi(0)\big{)}=\Theta(\infty)<\infty$ one has

$\lim_{\sigma\rightarrow
0}\dfrac{\Theta\big{(}\phi(\sigma)\big{)}}{\phi(\sigma)}=\lim_{t\rightarrow\infty}\dfrac{\Theta(t)}{t}=0.$
(5.2)

it implies that $\widehat{\rm
C}(x,\varepsilon,\delta)\Theta\big{(}\phi(\sigma)\big{)}$ is a bounded second
order term in the explosive boundary expansion. When
$\Theta\big{(}\phi(0)\big{)}=\Theta(\infty)=\infty$ the property (5.1) implies
that $\widehat{\rm C}(x,\varepsilon,\delta)\Theta\big{(}\phi(\sigma)\big{)}$
is an explosive second order term in the expansion. Indeed, by using the
L’Hôpital rule and (2.1) one deduces

$\lim_{\sigma\rightarrow
0}\dfrac{\Theta\big{(}\phi(\sigma)\big{)}}{\phi(\sigma)}=\lim_{t\rightarrow\infty}\dfrac{\Theta(t)}{t}=\lim_{t\rightarrow\infty}\Theta^{\prime}(t)=\lim_{t\rightarrow\infty}\dfrac{t}{\sqrt{2{\rm
G}(t)}}=0.$ (5.3)

Let $0<\varepsilon<1$ and $0\leq\ell_{u}<1$. Denoting $\widehat{\rm C}(x)={\rm
C}(x,\varepsilon,\delta)$, one has

$\begin{array}[]{ll}{\rm D}_{ij}{\rm
U}_{1}(x)&=\Theta\big{(}\phi(\sigma)\big{)}{\rm D}_{ij}\widehat{\rm
C}(x)+\sigma_{\varepsilon}\Theta^{\prime}\big{(}\phi(\sigma)\big{)}\phi^{\prime}(\sigma)\big{(}{\rm
D}_{i}\widehat{\rm C}(x){\rm D}_{j}{\rm d}(x)+{\rm D}_{j}\widehat{\rm
C}(x){\rm D}_{i}{\rm d}(x)\big{)}\\\\[4.26773pt]
&+\widehat{\sigma}_{\varepsilon}^{2}\widehat{\rm
C}(x)\big{(}\Theta^{\prime\prime}\big{(}\phi(\sigma)\big{)}\big{(}\phi^{\prime}(\sigma)\big{)}^{2}+\Theta^{\prime}\big{(}\phi(\sigma)\big{)}\phi^{\prime\prime}(\sigma)\big{)}{\rm
D}_{i}{\rm d}(x){\rm D}_{j}{\rm d}(x)\\\\[4.26773pt]
&+\widehat{\sigma}_{\varepsilon}\widehat{\rm
C}(x)\Theta^{\prime}\big{(}\phi(\sigma)\big{)}\phi^{\prime}(\sigma){\rm
D}_{ij}{\rm d}(x),\quad x\in\Omega^{\delta}_{\delta_{1}},\end{array}$

whence

$\begin{array}[]{ll}{\bf trace}~{}{\cal A}(x){\rm
U}_{1}(x)&=\Theta\big{(}\phi(\sigma)\big{)}{\bf trace}~{}{\cal A}(x){\rm
D}^{2}\widehat{\rm C}(x)\\\\[4.26773pt]
&+\sigma_{\varepsilon}\Theta^{\prime}\big{(}\phi(\sigma)\big{)}\phi^{\prime}(\sigma){\bf
trace}~{}{\cal A}(x)\big{(}{\rm D}\widehat{\rm C}(x)\otimes{\rm D}{\rm
d}(x)+{\rm D}{\rm d}(x)\otimes{\rm D}\widehat{\rm C}(x)\big{)}\\\\[4.26773pt]
&+\widehat{\sigma}_{\varepsilon}^{2}\widehat{\rm
C}(x)\big{(}\Theta^{\prime\prime}\big{(}\phi(\sigma)\big{)}\big{(}\phi^{\prime}(\sigma)\big{)}^{2}+\Theta^{\prime}\big{(}\phi(\sigma)\big{)}\phi^{\prime\prime}(\sigma)\big{)}{\rm
E}(x)\\\\[4.26773pt] &+\widehat{\sigma}_{\varepsilon}\widehat{\rm
C}(x)\Theta^{\prime}\big{(}\phi(\sigma)\big{)}\phi^{\prime}(\sigma){\bf
trace}~{}{\cal A}(x){\rm D}^{2}{\rm d}(x),\quad
x\in\Omega^{\delta}_{\delta_{1}},\end{array}$

where ${\rm E}(x)\doteq{\bf trace}~{}{\cal A}(x){\rm D}{\rm d}(x)\otimes{\rm
D}{\rm d}(x)\in[\lambda,\Lambda]$ is an eikonal-like term. On the other hand

$\begin{array}[]{ll}\beta({\rm U}(x))&=\beta({\rm U}_{0}(x))+{\rm
U}_{1}(x)\beta^{\prime}({\rm U}_{0}(x))+\dfrac{{\rm
U}_{1}(x)^{2}}{2}\beta^{\prime\prime}(\xi)\\\\[4.26773pt]
&=\beta\big{(}\phi(\sigma)\big{)}+\widehat{\rm
C}(x)\Theta\big{(}\phi(\sigma)\big{)}\beta^{\prime}\big{(}\phi(\sigma)\big{)}+\dfrac{\big{(}\widehat{\rm
C}(x)\Theta\big{(}\phi(\sigma)\big{)}\big{)}^{2}}{2}\beta^{\prime\prime}\big{(}\xi(\sigma)\big{)},\end{array}$

for some $\xi(\sigma)$ between $\phi(\sigma)$ and $\phi(\sigma)+\widehat{\rm
C}(x)\Theta\big{(}\phi(\sigma)\big{)}$. We are interested in to prove the
inequality

$-{\bf trace}~{}{\cal A}(x){\rm D}^{2}{\rm U}(x)+\beta\big{(}{\rm
U}(x)\big{)}-f(x)\geq 0,\quad x\in\Omega^{\delta}_{\delta_{1}}$

for $\delta_{1}$ small enough. The computations are very tedious. So that, we
consider suitable groups of terms. Moreover, we will use the change of
variable $t=\phi(\sigma)$. As in the proof of Proposition 2.1 the first group
is

${\cal G}_{1}\doteq-{\bf trace}~{}{\cal A}(x){\rm D}^{2}{\rm
U}_{0}(x)+\beta\big{(}{\rm
U}_{0}(x)\big{)}-f(x)+\sigma_{0}\phi^{\prime}(\sigma){\bf trace}~{}{\cal
A}{\rm D}^{2}{\rm d}\geq\beta(t)\big{(}\varepsilon(1-\ell_{u})\big{)}>0$

provided (2.7) (see (2.8)). The second group is

$\begin{array}[]{ll}{\cal
G}_{2}&\doteq-\widehat{\sigma}_{\varepsilon}\phi^{\prime}(\sigma){\bf
trace}~{}{\cal A}{\rm D}^{2}{\rm d}(x)\\\\[4.26773pt] &-\widehat{\rm
C}(x)\big{[}\widehat{\sigma}_{\varepsilon}^{2}\big{(}\Theta^{\prime\prime}\big{(}\phi(\sigma)\big{)}\big{(}\phi^{\prime}(\sigma)\big{)}^{2}+\Theta^{\prime}\big{(}\phi(\sigma)\big{)}\phi^{\prime\prime}(\sigma)\big{)}\big{)}{\rm
E}(x)-\Theta\big{(}\phi(\sigma)\big{)}\beta^{\prime}\big{(}\phi(\sigma)\big{)}\big{]}\end{array}$

for the particular case $t=\phi(\sigma)$ Then

$\begin{array}[]{ll}{\cal G}_{2}&=\widehat{\sigma}_{\varepsilon}{\bf
trace}~{}{\cal A}{\rm D}^{2}{\rm d}\sqrt{2{\rm G}(t)}-\widehat{\rm
C}(x)\big{[}\widehat{\sigma}_{\varepsilon}^{2}\big{(}2\Theta^{\prime\prime}(t){\rm
G}(t)+\Theta^{\prime}(t)\beta(t)\big{)}{\rm
E}(x)-\Theta(t)\beta^{\prime}(t)\big{]}.\end{array}$

From (5.1) we have

$\Theta^{\prime}(t)=\dfrac{t}{\sqrt{2{\rm G}(t)}}$

and

$2\Theta^{\prime\prime}(t){\rm G}(t)+\Theta^{\prime}(t)\beta(t)=\sqrt{2{\rm
G}(t)}.$

So, the choice

$\widehat{\rm C}(x)=\widehat{\sigma}_{\varepsilon}\dfrac{\sqrt{2{\rm
G}\big{(}\phi(\sigma)\big{)}}}{\widehat{\sigma}_{\varepsilon}^{2}\sqrt{2{\rm
G}\big{(}\phi(\sigma)\big{)}}{\rm
E}(x)-\Theta\big{(}\phi(\sigma)\big{)}\beta^{\prime}\big{(}\phi(\sigma)\big{)}}{\bf
trace}~{}{\cal A}(x){\rm D}^{2}{\rm d}(x),$ (5.4)

for $x\in\Omega^{\delta}_{\delta_{1}}$, leads ${\cal G}_{2}=0$.

###### Remark 5.1

We may write (5.4) as

$\begin{array}[]{ll}\widehat{\rm
C}(x)&=\dfrac{\widehat{\sigma}_{\varepsilon}}{\widehat{\sigma}_{\varepsilon}^{2}{\rm
E}(x)-\dfrac{\Theta(t)\beta^{\prime}(t)}{\sqrt{2{\rm G}(t)}}}{\bf
trace}~{}{\cal A}(x){\rm D}^{2}{\rm d}(x)\\\\[28.45274pt]
&=-\widehat{\sigma}_{\varepsilon}\dfrac{\sqrt{2{\rm
G}(t)}}{\Theta(t)\beta^{\prime}(t)}\dfrac{1}{1-\widehat{\sigma}_{\varepsilon}^{2}{\rm
E}(x)\dfrac{\sqrt{2{\rm G}(t)}}{\Theta(t)\beta^{\prime}(t)}}{\bf
trace}~{}{\cal A}(x){\rm D}^{2}{\rm d}(x)\\\\[28.45274pt]
&=-\widehat{\sigma}_{\varepsilon}\dfrac{\sqrt{2{\rm
G}(t)}}{\Theta(t)\beta^{\prime}(t)}\displaystyle\sum_{n\geq}\left(\widehat{\sigma}_{\varepsilon}^{2}{\rm
E}(x)\dfrac{\sqrt{2{\rm G}(t)}}{\Theta(t)\beta^{\prime}(t)}\right)^{n}{\bf
trace}~{}{\cal A}(x){\rm D}^{2}{\rm d}(x)\\\\[28.45274pt]
&=-\widehat{\sigma}_{\varepsilon}\dfrac{\sqrt{2{\rm
G}\big{(}\phi(\sigma)\big{)}}}{\Theta\big{(}\phi(\sigma)\big{)}\beta^{\prime}\big{(}\phi(\sigma)\big{)}}{\bf
trace}~{}{\cal A}(x){\rm D}^{2}{\rm d}(x)\big{(}1+o(1)\big{)}{\bf
trace}~{}{\cal A}(x){\rm D}^{2}{\rm d}(x),\end{array}$ (5.5)

provided

$\widehat{\sigma}_{\varepsilon}^{2}{\rm E}(x)\dfrac{\sqrt{2{\rm
G}\big{(}\phi(\sigma)\big{)}}}{\beta^{\prime}\big{(}\phi(\sigma)\big{)}}\dfrac{1}{\Theta\big{(}\phi(\sigma)\big{)}\big{)}}<1.$
(5.6)

Analogously

$\begin{array}[]{ll}\widehat{\rm
C}(x)\Theta\big{(}\phi(\sigma)\big{)}&=\dfrac{\widehat{\sigma}_{\varepsilon}\Theta(t)}{\widehat{\sigma}_{\varepsilon}^{2}{\rm
E}(x)-\dfrac{\Theta(t)\beta^{\prime}(t)}{\sqrt{2{\rm G}(t)}}}{\bf
trace}~{}{\cal A}(x){\rm D}^{2}{\rm d}(x)\\\\[28.45274pt]
&=-\dfrac{\widehat{\sigma}_{\varepsilon}\sqrt{2{\rm
G}(t)}}{\beta^{\prime}(t)}\dfrac{1}{1-\widehat{\sigma}_{\varepsilon}^{2}{\rm
E}(x)\dfrac{\beta^{\prime}(t)}{\sqrt{2{\rm
G}(t)}}\left(\Theta\big{(}\phi(\sigma)\big{)}\right)^{-1}}{\bf trace}~{}{\cal
A}(x){\rm D}^{2}{\rm d}(x)\\\\[28.45274pt]
&\displaystyle=-\dfrac{\widehat{\sigma}_{\varepsilon}\sqrt{2{\rm
G}(t)}}{\beta^{\prime}(t)}\sum_{n\geq
0}\bigg{(}\widehat{\sigma}_{\varepsilon}^{2}{\rm
E}(x)\dfrac{\beta^{\prime}(t)}{\sqrt{2{\rm
G}(t)}}\left(\Theta(t)^{-1}\right)^{n}{\bf trace}~{}{\cal A}(x){\rm D}^{2}{\rm
d}(x)\\\\[21.33955pt] &=-\dfrac{\widehat{\sigma}_{\varepsilon}\sqrt{2{\rm
G}\big{(}\phi(\sigma)\big{)}}}{\beta^{\prime}\big{(}\phi(\sigma)\big{)}}{\bf
trace}~{}{\cal A}(x){\rm D}^{2}{\rm d}(x)\big{(}1+o(1)\big{)},\end{array}$
(5.7)

provided

$\widehat{\sigma}_{\varepsilon}^{2}{\rm
E}(x)\dfrac{\beta^{\prime}\big{(}\phi(\sigma)\big{)}}{\sqrt{2{\rm
G}\big{(}\phi(\sigma)\big{)}}}\dfrac{1}{\Theta\big{(}\phi(\sigma)\big{)}\big{)}}<1.$
(5.8)

We note that the claim implies that if such secondary term exists it is unique
term influenced by the geometry by means of ${\bf trace}~{}{\cal A}(x){\rm
D}^{2}{\rm d}(x)$ whenever $\Lambda=\lambda=1.\hfill\mbox{$\quad{}_{\Box}$}$

The third group is

${\cal G}_{3}\doteq-\Theta\big{(}\phi(\sigma)\big{)}{\bf trace}~{}{\cal
A}(x){\rm D}^{2}\widehat{\rm C}(x)=-\Theta(t){\bf trace}~{}{\cal A}(x){\rm
D}^{2}\widehat{\rm C}(x)$

for which

$\lim_{t\nearrow\infty}\dfrac{\Theta(t)}{\beta(t)}=\lim_{t\nearrow\infty}\dfrac{\Theta(t)}{t}\frac{t}{\beta(t)}=0$

(see (2.1) and (5.2) or (5.3)). The next group to be considered is

${\cal G}_{4}\doteq\dfrac{1}{2}\big{(}\widehat{\rm
C}(x)\Theta\big{(}\phi(\sigma)\big{)}\big{)}^{2}\beta^{\prime\prime}\big{(}\xi(\sigma)\big{)}.$

Since $\xi(\sigma)\rightarrow\infty$ as $\sigma\rightarrow 0$ and
$\displaystyle\lim_{t\nearrow\infty}\dfrac{\sqrt{2{\rm G}(t)}}{\beta(t)}=0$
(see (2.2)), we will use the property

$\lim_{\sigma\searrow 0}\dfrac{\big{(}\widehat{\rm
C}(x)\Theta\big{(}\phi(\sigma)\big{)}\big{)}^{2}\beta^{\prime\prime}\big{(}\phi(\sigma)\big{)}}{\sqrt{2{\rm
G}\big{(}\phi(\sigma)\big{)}}}\dfrac{\sqrt{2{\rm
G}\big{(}\phi(\sigma)\big{)}}}{\beta\big{(}\phi(\sigma)\big{)}}=0$ (5.9)

deduced from the assumption

$\lim_{\sigma\searrow 0}\dfrac{\big{(}\widehat{\rm
C}(x)\Theta\big{(}\phi(\sigma)\big{)}\big{)}^{2}\beta^{\prime\prime}\big{(}\phi(\sigma)\big{)}}{\sqrt{2{\rm
G}\big{(}\phi(\sigma)\big{)}}}=0.$ (5.10)

The last group is

$\begin{array}[]{ll}{\cal
G}_{5}&\doteq-\sigma_{\varepsilon}\Theta^{\prime}\big{(}\phi(\sigma)\big{)}\phi^{\prime}(\sigma)\left[{\bf
trace}~{}{\cal A}(x)\big{(}{\rm D}\widehat{\rm C}(x)\otimes{\rm D}{\rm
d}(x)+{\rm D}{\rm d}(x)\otimes{\rm D}\widehat{\rm
C}(x)\big{)}\right.\\\\[4.26773pt] &\quad\left.+\widehat{\rm C}(x){\bf
trace}~{}{\cal A}(x){\rm D}^{2}{\rm d}(x)\right]\\\\[4.26773pt]
&=\sigma_{\varepsilon}\Theta^{\prime}(t)\sqrt{2{\rm G}(t)}\left[{\bf
trace}~{}{\cal A}(x)\big{(}{\rm D}\widehat{\rm C}(x)\otimes{\rm D}{\rm
d}(x)+{\rm D}{\rm d}(x)\otimes{\rm D}\widehat{\rm C}(x)\big{)}+\widehat{\rm
C}(x){\bf trace}~{}{\cal A}(x){\rm D}^{2}{\rm d}(x)\right]\\\\[4.26773pt]
&=-\sigma_{\varepsilon}t\left[{\bf trace}~{}{\cal A}(x)\big{(}{\rm
D}\widehat{\rm C}(x)\otimes{\rm D}{\rm d}(x)+{\rm D}{\rm d}(x)\otimes{\rm
D}\widehat{\rm C}(x)\big{)}+\widehat{\rm C}(x){\bf trace}~{}{\cal A}(x){\rm
D}^{2}{\rm d}(x)\right]\end{array}$

for which we recall again the property

$\lim_{t\nearrow\infty}\dfrac{t}{\beta(t)}=0$

(see (2.1)). With the above reasonings we have

######  Theorem 5.1 (Second order term)

Let us assume $\partial\Omega\in{\cal C}^{4},~{}\beta\in{\cal C}^{2}$ and
${\cal A}(\cdot)$ adequately smooth. Under the assumptions of Proposition 2.1
for any solution of

$-{\bf trace}~{}{\cal A}(\cdot){\rm D}^{2}u+\beta(u)\leq f\quad\hbox{in
$\Omega$}$

we have

$\limsup_{{\rm d}(x)\searrow
0}\dfrac{u^{*}(x)}{\phi\big{(}\widehat{\sigma}_{0}{\rm d}(x)\big{)}+{\rm
C}_{u}(x)\Theta\left(\phi\big{(}\widehat{\sigma}_{0}{\rm
d}(x)\big{)}\right)}\leq 1$ (5.11)

provided (2.7) and (5.10), where
$\widehat{\sigma}_{0}=\sqrt{\dfrac{1-\ell}{\Lambda}}$ and the function
$\Theta$ is given by (5.1) and

${\rm C}_{u}(x)=\dfrac{\Lambda\widehat{\sigma}_{0}}{(1-\ell_{u}){\rm
E}(x)-\Lambda\dfrac{\Theta\left(\phi\big{(}\widehat{\sigma}_{0}{\rm
d}(x)\big{)}\right)}{\sqrt{2{\rm G}\left(\phi\big{(}\widehat{\sigma}_{0}{\rm
d}(x)\big{)}\right)}}\beta^{\prime}\big{(}\phi\big{(}\widehat{\sigma}_{0}{\rm
d}(x)\big{)}}{\bf trace}~{}{\cal A}(x){\rm D}^{2}{\rm d}(x),$

for ${\rm d}(x)$ small (see (5.7)), is assumed in ${\rm W}^{2,\infty}$ and
${\rm E}(x)\doteq{\bf trace}~{}{\cal A}(x){\rm D}{\rm d}(x)\otimes{\rm D}{\rm
d}(x)$. Moreover

${\rm C}_{u}(x)\Theta\big{(}\phi(\widehat{\sigma}_{0}{\rm
d}(x))\big{)}=-\dfrac{\widehat{\sigma}_{0}\sqrt{2{\rm
G}\left(\phi\big{(}\widehat{\sigma}_{0}{\rm
d}(x)\big{)}\right)}}{\beta^{\prime}\big{(}\phi\big{(}\widehat{\sigma}_{0}{\rm
d}(x)\big{)}}{\bf trace}~{}{\cal A}(x){\rm D}^{2}{\rm
d}(x)\big{(}1+o(1)\big{)},$ (5.12)

for ${\rm d}(x)$ small (see (5.7)). Analogously, under the assumptions of
Proposition 2.2 for any nonnegative large solution of

$-{\bf trace}~{}{\cal A}(\cdot){\rm D}^{2}v+\beta(v)\geq f\quad\hbox{in
$\Omega$}$

we have

$1\leq\liminf_{{\rm d}(x)\searrow
0}\dfrac{v_{*}(x)}{\phi\big{(}\widetilde{\sigma}_{0}{\rm d}(x)\big{)}+{\rm
C}_{d}(x)\Theta\left(\phi\big{(}\widetilde{\sigma}_{0}{\rm
d}(x)\big{)}\right)}$ (5.13)

provided (2.11) and (5.10), where
$\widetilde{\sigma}_{0}=\sqrt{\dfrac{1-\ell}{\lambda}}$ the function $\Theta$
is given by (5.1) and

${\rm C}_{d}(x)=\dfrac{\lambda\widetilde{\sigma}_{0}}{(1-\ell_{d}){\rm
E}(x)-\lambda\dfrac{\Theta\left(\phi\big{(}\widetilde{\sigma}_{0}{\rm
d}(x)\big{)}\right)}{\sqrt{2{\rm G}\left(\phi\big{(}\widetilde{\sigma}_{0}{\rm
d}(x)\big{)}\right)}}\beta^{\prime}\big{(}\phi\big{(}\widetilde{\sigma}_{0}{\rm
d}(x)\big{)}}{\bf trace}~{}{\cal A}(x){\rm D}^{2}{\rm d}(x),$

for ${\rm d}(x)$ small, assumed in ${\rm W}^{2,\infty}$. Moreover

${\rm C}_{d}(x)\Theta\big{(}\phi(\widetilde{\sigma}_{0}{\rm
d}(x))\big{)}=-\dfrac{\widehat{\sigma}_{0}\sqrt{2{\rm
G}\left(\phi\big{(}\widetilde{\sigma}_{0}{\rm
d}(x)\big{)}\right)}}{\beta^{\prime}\big{(}\phi\big{(}\widetilde{\sigma}_{0}{\rm
d}(x)\big{)}}{\bf trace}~{}{\cal A}(x){\rm D}^{2}{\rm
d}(x)\big{(}1+o(1)\big{)},$ (5.14)

for ${\rm d}(x)$ small (see (5.7)).

Proof. We only show the result relative to $u^{*}$. The proof for $v_{*}$ is
analogous (see Proposition 2.2). Following the above reasonings the function

${\rm U}(x)={\rm U}_{0}(x)+{\rm U}_{1}(x)\doteq\phi(\sigma)+\widehat{\rm
C}(x)\Theta\big{(}\phi(\sigma)\big{)},\quad
x\in\Omega^{\delta}_{\delta_{1}}\quad(0<\delta<\delta_{1}<\delta_{\Omega})$

satisfies

$\begin{array}[]{ll}-{\bf trace}~{}{\cal A}(x){\rm D}^{2}{\rm
U}(x)+\beta\big{(}{\rm U}(x)\big{)}-f(x)&\geq{\cal G}_{1}+{\cal G}_{2}+{\cal
G}_{3}+{\cal G}_{4}+{\cal G}_{5}\\\
&\geq\beta(t)\left(\varepsilon(1-\ell_{u})+\dfrac{{\cal G}_{3}+{\cal
G}_{4}+{\cal G}_{5}}{\beta(t)}\right),\quad
x\in\Omega^{\delta}_{\delta_{1}},\end{array}$

for $\Theta(t)$ and $\widehat{\rm C}(x)$ are defined in (5.1) and (5.4)
respectively. Since the assumptions imply

$\lim_{t\nearrow\infty}\dfrac{{\cal G}_{3}+{\cal G}_{4}+{\cal
G}_{5}}{\beta(t)}=0$

and $0<\sigma<\widehat{\sigma}_{0}\delta_{1}$ we deduce

$\varepsilon(1-\ell_{u})+\dfrac{{\cal G}_{3}+{\cal G}_{4}+{\cal
G}_{5}}{\beta\big{(}\phi(\sigma)\big{)}}>0$

for $\delta_{1}$ small enough. Therefore

$-{\bf trace}~{}{\cal A}(x){\rm D}^{2}{\rm U}(x)+\beta\big{(}{\rm
U}(x)\big{)}\geq f(x)\quad x\in\Omega^{\delta}_{\delta_{1}}$

for $\delta_{1}$ small enough. As in the proof of Proposition 2.1 we obtain

$u^{*}(x)\leq\phi\left(\sqrt{\dfrac{1-\varepsilon)(1-\ell_{u})}{\Lambda}}{\rm
d}(x)\right)+\widehat{\rm
C}(x)\Theta\left(\phi\left(\sqrt{\dfrac{1-\varepsilon)(1-\ell_{u})}{\Lambda}}{\rm
d}(x)\right)\right)+{\rm M},\quad x\in\Omega_{\delta_{1}}.$

where ${\rm M}\geq\displaystyle\sup_{{\rm d}(x)=\delta_{1}}u^{*}$. Then the
estimate (5.11) follows.$\quad{}_{\Box}$

###### Remark 5.2

In this paper we omit general assumptions for which ${\rm C}_{u}(x),{\rm
C}_{d}(x)\in{\rm W}^{2,\infty}$. The above proof adapts and simplifies results
obtained in [2] whenever $f\equiv 0$ and $\Lambda=\lambda=1$. See [3] for
other reasonings.$\quad{}_{\Box}$

######  Remark 5.3 (Second order boundary assymtotic)

From (5.7) we have

${\rm
C}_{u}(x)\Theta\big{(}\phi(\sigma)\big{)}=-\widehat{\sigma}_{0}\dfrac{\sqrt{2{\rm
G}\big{(}\phi(\sigma))}}{\beta^{\prime}\big{(}\phi(\sigma)\big{)}}{\bf
trace}~{}{\cal A}(x){\rm D}^{2}{\rm d}(x)\big{(}1+o(1)\big{)}$

for $\sigma=\widehat{\sigma}_{0}{\rm d}(x)$, with
$\widehat{\sigma}_{0}=\sqrt{\dfrac{1-\ell_{u}}{\Lambda}}$, provided (5.8).
Then, denoting

$\lim_{t\nearrow\infty}\dfrac{\sqrt{2{\rm G}(t)}}{\beta^{\prime}(t)}={\rm L}$
(5.15)

we deduce:

i) If ${\rm L}=\infty$ the second order term ${\rm
C}_{u}(x)\Theta\big{(}\phi(\sigma)\big{)}$ is explosive on the boundary where
the influence of the geometry appears,

ii) If $0<{\rm L}<\infty$ the second order term ${\rm
C}_{u}(x)\Theta\big{(}\phi(\sigma)\big{)}$ is bounded and the influence of the
geometry appears on the boundary,

iii) If ${\rm L}=0$ the second order term ${\rm
C}_{u}(x)\Theta\big{(}\phi(\sigma)\big{)}$ vanishes on the boundary and the
influence of the geometry is null on the boundary.

Let us come back to the choice of function $\Theta$ satisfying (5.1):

a) Under

$\int^{+\infty}\dfrac{sds}{\sqrt{2{\rm G}(s)}}=\infty$ (5.16)

we will consider the function

$\Theta(t)=\int^{t}_{t_{0}}\dfrac{sds}{\sqrt{2{\rm G}(s)}},\quad\hbox{for $t$
large}$ (5.17)

where $t_{0}<\infty$ is arbitrary. We note that $\Theta(\infty)=\infty$ and
$0\leq{\rm L}\leq\infty$. It is preferable $t_{0}=0$ if

$\int_{0^{+}}\dfrac{sds}{\sqrt{2{\rm G}(s)}}<+\infty$ (5.18)

holds.

b) Under

$\int^{+\infty}\dfrac{sds}{\sqrt{2{\rm G}(s)}}<\infty$ (5.19)

and

$\int_{0^{+}}\dfrac{sds}{\sqrt{2{\rm G}(s)}}=+\infty$ (5.20)

we define

$\Theta(t)=-\int^{\infty}_{t}\dfrac{sds}{\sqrt{2{\rm G}(s)}},\quad\hbox{for
$t$ large}.$ (5.21)

We note that $\Theta(\infty)=0$ and ${\rm L}=0$.

c) Under

$\int^{+\infty}\dfrac{sds}{\sqrt{2{\rm G}(s)}}<\infty$ (5.22)

and

$\int_{0^{+}}\dfrac{sds}{\sqrt{2{\rm G}(s)}}<+\infty$ (5.23)

we define

$\Theta(t)=\int^{t}_{0}\dfrac{sds}{\sqrt{2{\rm G}(s)}},\quad\hbox{for $t$
large}.$ (5.24)

We note that
$\displaystyle\Theta(\infty)=\int^{\infty}_{0^{+}}\dfrac{sds}{\sqrt{2{\rm
G}(s)}}<+\infty$ and $0<{\rm L}<\infty$.

We send to [2] for some details on the explosive boundary expansion of
solutions including non explosive second order terms for homogeneous sources
$f\equiv 0$ and $\Lambda=\lambda=1.\hfill\mbox{$\quad{}_{\Box}$}$

######  Example 5.1 (Power like case)

For $\beta_{m}(t)=t^{m},~{}m>1,$ one has

$\sqrt{2{\rm G}_{m}(t)}=\sqrt{\frac{2}{m+1}}t^{\frac{m+1}{2}},\hskip
5.69046pt\Phi_{m}(t)=\dfrac{\sqrt{2(m+1)}}{m-1}t^{-\frac{m-1}{2}}\hskip
5.69046pt\hbox{and}\hskip
5.69046pt\phi_{m}(\delta)=\left(\dfrac{\sqrt{2(m+1)}}{m-1}\right)^{\frac{2}{m-1}}\delta^{-\frac{2}{m-1}}$

(see Remark 1.5). Let us recall

$\Theta_{m}^{\prime}(t)=-t\Phi_{m}^{\prime}(t)=\dfrac{t}{\sqrt{2{\rm
G}_{m}(t))}}=\sqrt{\frac{m+1}{2}}t^{-\frac{m-1}{2}}$

(see (5.1)) and

$\int^{t_{2}}_{t_{1}}\dfrac{sds}{\sqrt{2{\rm
G}_{m}(s)}}=\sqrt{\dfrac{m+1}{3}}\dfrac{2}{3-m}\left(t_{2}^{\frac{3-m}{2}}-t_{1}^{\frac{3-m}{2}}\right).$
(5.25)

a) If $1<m<3$ the properties

$\int^{\infty}\dfrac{sds}{\sqrt{2{\rm
G}_{m}(s)}}=\infty\quad\hbox{and}\quad\int_{0^{+}}\dfrac{sds}{\sqrt{2{\rm
G}_{m}(s)}}<\infty$

hold. Then

$\dfrac{\sqrt{2{\rm
G}_{m}\big{(}\phi_{m}(\sigma))}}{\beta_{m}^{\prime}\big{(}\phi_{m}(\sigma)\big{)}}=\sqrt{2}\left(\dfrac{(m+1)^{2-m}}{(m-1)^{3-m}}\right)^{\frac{1}{m-1}}\sigma^{-\frac{3-m}{m-1}}.$

On the other hand, we define

$\Theta_{m}(t)=\int^{t}_{0}\dfrac{sds}{\sqrt{2{\rm
G}_{m}(s)}}=\sqrt{\dfrac{m+1}{2}}\dfrac{2}{3-m}t^{\frac{3-m}{2}}$ (5.26)

that satisfies $\Theta_{m}(\infty)=\infty$. Since (5.7) and (5.8) hold, we
deduce that ${\rm C}_{u,m}(x)\Theta\big{(}\phi_{m}(\widehat{\sigma}_{0}{\rm
d}(x))\big{)}$ is an explosive second order term and the influence of the
geometry appears on the boundary (see Remak 5.3). More precisely

${\rm
C}_{u,m}(x)\Theta\big{(}\phi_{m}(\sigma)\big{)}=-\widehat{\sigma}_{0}^{\frac{m-2}{m-1}}\sqrt{2}\left(\dfrac{(m+1)^{2-m}}{(m-1)^{3-m}}\right)^{\frac{1}{m-1}}{\rm
d}(x)^{-\frac{3-m}{m-1}}{\bf trace}~{}{\cal A}(x){\rm D}^{2}{\rm
d}(x)\big{(}1+o(1)\big{)}$ (5.27)

for $\sigma=\widehat{\sigma}_{0}{\rm d}(x)$, with
$\widehat{\sigma}_{0}=\sqrt{\dfrac{1-\ell_{u}}{\Lambda}}$. We may detail the
explosive boundary expansion. Indeed, we note that

$\Theta_{m}\big{(}\phi_{m}(\sigma)\big{)}=\sqrt{\dfrac{m+1}{2}}\dfrac{2}{3-m}\left(\dfrac{\sqrt{2(m+1)}}{m-1}\right)^{-1+\frac{2}{m-1}}\sigma^{1-\frac{2}{m-1}},$
(5.28)

whence

$\begin{array}[]{l}\phi_{m}\big{(}\sigma)+{\rm
C}_{u,m}(x)\Theta_{m}\big{(}\phi_{m}(\sigma)\big{)}=\phi_{m}\big{(}\sigma)\bigg{(}1+\\\
\hskip 85.35826pt\left.+{\rm
C}_{u,m}(x)\sqrt{\dfrac{m+1}{2}}\dfrac{2}{3-m}\left(\dfrac{\sqrt{2(m+1)}}{m-1}\right)^{-1}\sigma{\bf
trace}~{}{\cal A}(x){\rm D}^{2}{\rm d}(x)\right).\end{array}$

Moreover, one has

$\begin{array}[]{ll}{\rm
C}_{u,m}(x)&=\dfrac{\widehat{\sigma}_{0}}{\widehat{\sigma}_{0}^{2}{\rm
E}(x)-\dfrac{\Theta_{m}\big{(}\phi_{m}(\sigma)\big{)}}{\sqrt{2{\rm
G}_{m}\big{(}\phi_{m}(\sigma)\big{)}}}\beta_{m}^{\prime}\left(\big{(}\phi_{m}(\sigma)\big{)}\right)}{\bf
trace}~{}{\cal A}(x){\rm D}^{2}{\rm d}(x)\\\\[22.76228pt]
&=\dfrac{\widehat{\sigma}_{0}(m-3)}{m^{2}+\widehat{\sigma}_{0}^{2}{\rm
E}(x)+1\big{)}m-3\widehat{\sigma}_{0}^{2}{\rm E}(x)}{\bf trace}~{}{\cal
A}(x){\rm D}^{2}{\rm d}(x).\end{array}$ (5.29)

One proves the property

$m^{2}+\widehat{\sigma}_{0}^{2}{\rm
E}(x)+1\big{)}m-3\widehat{\sigma}_{0}^{2}{\rm E}(x)\neq 0$

whence ${\rm C}_{u,m}\in{\rm W}^{2,\infty}$, provided $\partial\Omega\in{\cal
C}^{4}$ and ${\cal A}(\cdot)$ is adequately smooth.

The computation for $\sigma=\widetilde{\sigma}_{0}{\rm d}(x)$, with
$\widetilde{\sigma}_{0}=\sqrt{\dfrac{1-\ell_{d}}{\lambda}}$ are analogous.
Finally, one proves that (5.10) holds and Theorem 5.1 applies. We note that
the Lowener-Nirenberg choice (see [30]) $\beta(t)=t^{\frac{{\rm N}+2}{{\rm
N}-2}}$ belongs this case if ${\rm N}>4$.

b) When $m>3$ the properties

$\int^{\infty}\dfrac{sds}{\sqrt{2{\rm
G}_{m}(s)}}<\infty\quad\hbox{and}\quad\int_{0^{+}}\dfrac{sds}{\sqrt{2{\rm
G}_{m}(s)}}=\infty$

hold (see (5.25). Here one defines the function

$\Theta_{m}(t)=-\int^{\infty}_{t}\dfrac{sds}{\sqrt{2{\rm
G}_{m}(s)}}=\sqrt{\dfrac{m+1}{2}}\dfrac{2}{3-m}t^{\frac{3-m}{2}}$

that coincides with (5.26), but now $m>3$ where the formulae (5.28) and (5.29)
are the same. Therefore, we deduce that the second order ${\rm
C}_{u,m}(x)\Theta_{m}\big{(}\phi_{m}(\widehat{\sigma}_{0}{\rm d}(x))\big{)}$
vanishes on the boundary, hence the influence of the geometry is null on the
boundary. The Lowener-Nirenberg choice belongs this case if ${\rm N}=3$.

c) If $m=3$ the properties

$\int^{\infty}\dfrac{sds}{\sqrt{2{\rm
G}_{3}(s)}}<\infty\quad\hbox{and}\quad\int_{0^{+}}\dfrac{sds}{\sqrt{2{\rm
G}_{3}(s)}}=\infty$

hold. Then one define

$\Theta_{3}(t)=\int^{t}_{t_{0}}\dfrac{sds}{\sqrt{2{\rm
G}_{3}(s)}}=\sqrt{2}\log\bigg{(}\dfrac{t}{t_{0}}\bigg{)},\quad t>t_{0}$

where $t_{0}<\infty$ is arbitrary. Here $\Theta_{3}(\infty)=\infty$, thus
$\Theta_{3}\big{(}\phi_{3}(0)\big{)}=+\infty$. Take $t_{0}=1$ by simplicity.
In this case (5.27) implies that the second order ${\rm
A}_{u,3}(x)\Theta_{3}\big{(}\phi_{3}(\widehat{\sigma}_{0}{\rm d}(x))\big{)}$
is bounded where the influence of the geometry on the boundary appears (see
Remak 5.3). Then

$\Theta_{3}\big{(}\phi_{3}(\sigma)\big{)}=\sqrt{2}\log\bigg{(}\dfrac{\sqrt{2}}{\sigma}\bigg{)}.$

and

$\begin{array}[]{ll}{\rm
C}_{u,3}(x)&=\dfrac{\widehat{\sigma}_{0}}{\widehat{\sigma}_{0}^{2}{\rm
E}(x)-\dfrac{\Theta_{3}\big{(}\phi_{3}(\sigma)\big{)}}{\sqrt{2{\rm
G}_{3}\big{(}\phi_{m}(\sigma)\big{)}}}\beta_{3}^{\prime}\bigg{(}\big{(}\phi_{3}(\sigma)\big{)}\bigg{)}}{\bf
trace}~{}{\cal A}(x){\rm D}^{2}{\rm d}(x)\\\\[34.14322pt]
&=\dfrac{\widehat{\sigma}_{0}}{\widehat{\sigma}_{0}^{2}{\rm
E}(x)-6\log\left(\dfrac{\sqrt{2}}{\sigma}\right)}{\bf trace}~{}{\cal A}(x){\rm
D}^{2}{\rm d}(x).\end{array}$

Since

$\widehat{\sigma}_{0}^{2}{\rm
E}(x)-6\log\bigg{(}\dfrac{\sqrt{2}}{\sigma}\bigg{)}<0\quad\hbox{for $\sigma$
small enough}$

one deduces ${\rm C}_{u,3}\in{\rm W}^{2,\infty}$, provided
$\partial\Omega\in{\cal C}^{4}$ and ${\cal A}(\cdot)$ is adequately smooth.
Moreover,

$\begin{array}[]{ll}{\rm
C}_{u,3}(x)\Theta_{3}\big{(}\phi_{3}(\sigma)\big{)}&=\dfrac{\widehat{\sigma}_{0}\sqrt{2}\log\bigg{(}\dfrac{\sqrt{2}}{\sigma}\bigg{)}}{\widehat{\sigma}_{0}^{2}{\rm
E}(x)-6\log\bigg{(}\dfrac{\sqrt{2}}{\sigma}\bigg{)}}{\bf trace}~{}{\cal
A}(x){\rm D}^{2}{\rm d}(x)\\\\[28.45274pt]
&=-\dfrac{\widehat{\sigma}_{0}\sqrt{2}}{6}\dfrac{1}{1-\dfrac{\widehat{\sigma}_{0}^{2}{\rm
E}(x)}{6}\left(\log\bigg{(}\dfrac{\sqrt{2}}{\sigma}\bigg{)}\right)^{-1}}{\bf
trace}~{}{\cal A}(x){\rm D}^{2}{\rm d}(x)\\\\[28.45274pt] \end{array}$

(see (5.7)) whence the full expansion is

$\phi_{3}\big{(}\widehat{\sigma}_{0}{\rm
d}(x)\big{)}-\dfrac{\widehat{\sigma}_{0}\sqrt{2}}{6}{\bf trace}~{}{\cal
A}(x){\rm D}^{2}{\rm d}(x)+o(1).$

Here the assumption (5.10) also holds. The Lowener-Nirenberg choice belongs
this case if ${\rm N}=4$.

For $f\equiv 0,~{}\ell_{u}=0$ and $\Lambda=\lambda=1$ the relative result was
obtained by first time in [12]. The case $0\leq\ell_{u}<1$ and
$\Lambda=\lambda=1$ was obtained in [1].$\quad{}_{\Box}$

######  Example 5.2 (Bieberbach choices)

One proves that for $\beta(t)=e^{t}$ the conditions (5.22) and (5.23) hold.
Now

$\lim_{t\nearrow\infty}\dfrac{\sqrt{2{\rm
G}(t)}}{\beta^{\prime}(t)}=\lim_{t\nearrow\infty}\sqrt{2}\big{(}e^{-t}-e^{-2t}\big{)}=0.$

Moreover, one proves

$\Theta(t)=\sqrt{2}\left(2-(t+2)e^{-\frac{t}{2}}\right).$

Then the second order term ${\rm C}_{u}(x)\Theta\big{(}\phi(\sigma)\big{)}$
vanishes on the boundary and the influence of the geometry is null on the
boundary (see Remak 5.3). The full explosive boundary expasion is

$\phi(\sigma)+o(\sigma).$

The conditions (5.22) and (5.23) also hold for the choice
$\beta(t)=te^{t^{2}}$. Here

$\lim_{t\nearrow\infty}\dfrac{\sqrt{2{\rm
G}(t)}}{\beta^{\prime}(t)}=\lim_{t\nearrow\infty}\dfrac{1-e^{-t^{2}}}{1-2t^{2}}=0.$

Moreover, one proves

$\Theta(t)=\sqrt{2}\left(12-e^{-\frac{t^{2}}{2}}\right).$

Therefore the second order term ${\rm
C}_{u}(x)\Theta\big{(}\phi(\sigma)\big{)}$ vanishes on the boundary and the
influence of the geometry is also null on the boundary (see Remak 5.3). The
full explosive boundary expasion is

$\phi(\sigma)+o(\sigma).$

For general nonlinear function $\beta(t)$ the consideration of the upper terms
in the explosive boundary expansion of solution under extra Keller-Osserman
type growth of the sources on the boundary (see (3.10)) is very tangled. The
study for the power like choice $\beta_{m}=t^{m},~{}m>1$, was detailed in [1]
when $\lambda=\Lambda=1$. We end this paper by extending the results. By
simplicity in the exposition we only detail the reasonings on the equation

$-{\bf trace}~{}{\cal A}(\cdot){\rm D}^{2}u^{*}+\big{(}u^{*}\big{)}^{m}\leq
f\quad\hbox{in $\Omega$}$

under

$\limsup_{{\rm d}(x)\searrow 0}f(x){\rm
d}(x)^{q}\leq\ell_{u}\quad(\ell_{u}>0),$

for $q>\dfrac{2m}{m-1},~{}m>1$ (see (3.1)). Here, we replace ${\rm U}_{0}$,
used in the proof of Theorem 3.1, by

${\rm U}(x)\doteq{\rm C}_{0}\sigma^{-\frac{q}{m}}\bigg{(}(1+\varepsilon)+{\rm
C}_{1}(x,\varepsilon,\delta)\sigma^{\gamma}\bigg{)},\quad
x\in\Omega^{\delta}_{\delta_{1}}\quad(0<\delta<\delta_{1}<\delta_{\Omega}),$

where $\sigma\doteq{\rm d}(x)-\delta$ and $\widehat{\rm
C}_{1}(\cdot)\doteq{\rm
C}_{1}(\cdot,\varepsilon,\delta):\Omega_{\delta_{1}}\rightarrow{\mathbb{R}}$
are real functions yet to be determined as well as the constant ${\rm C}_{0}$.
We will be assumed that $\widehat{\rm C}_{1}\in{\rm
W}^{2,\infty}\big{(}\Omega_{\delta_{1}}\big{)}$. Straightforward computations
yield

$\begin{array}[]{ll}{\bf trace}~{}{\cal A}(x){\rm
D}^{2}\sigma^{-\frac{q}{m}}&=\sigma^{-q}\left[\dfrac{(q+m)q}{m^{2}}{\rm
E}(x)\sigma^{\frac{q(m-1)-2m}{m}}-\dfrac{q}{m}{\bf trace}~{}{\cal A}(x){\rm
D}^{2}{\rm d}(x)\sigma^{\frac{q(m-1)-m}{m}}\right],\\\\[7.82433pt] {\bf
trace}~{}{\cal A}(x){\rm D}^{2}\widehat{\rm
C}_{1}(x)\sigma^{-\frac{q}{m}+\gamma}&=\sigma^{-q}\left[\left(\dfrac{m\gamma-q}{m}\right)\left(\dfrac{m(\gamma-1)-q}{m}\right){\rm
E}(x)\widehat{\rm
C}_{1}(x)\sigma^{\frac{q(m-1)+m(\gamma-2)}{m}}\right.\\\\[7.82433pt]
&+\dfrac{m\gamma-q}{m}\widehat{\rm
F}_{1}(x)\sigma^{\frac{q(m-1)+m(\gamma-1)}{m}}+{\bf trace}~{}{\cal A}(x){\rm
D}^{2}\widehat{\rm
C}_{1}(x)\sigma^{\frac{q(m-1)+m\gamma}{m}}\bigg{]},\end{array}$

where $\widehat{\rm F}_{1}(x)={\bf trace}~{}{\cal A}(x)\big{(}{\rm D}{\rm
d}(x)\otimes{\rm D}\widehat{\rm C}_{1}(x)+{\rm D}\widehat{\rm
C}_{1}(x)\otimes{\rm D}{\rm d}(x)\big{)}+\widehat{\rm C}_{1}(x){\bf
trace}~{}{\cal A}(x){\rm D}^{2}{\rm d}(x)$ is a bounded function. On the other
hand, we may write

$\big{(}{\rm U}(x))^{m}={\rm
C}_{0}^{m}\sigma^{-q}\left((1+\varepsilon)^{m}+m\widehat{\rm
C}_{1}(x)\sigma^{\gamma}+\dfrac{m(m-1)}{2}\widehat{\rm
C}_{1}(x)^{2}\sigma^{2\gamma}\xi(\sigma)^{m-2}\right)$

for some $\xi(\sigma)$ in between $1$ and $1+{\rm C}_{1}(x)\sigma^{\gamma}$,
determinating a bounded interval. Therefore,

$\lim_{\sigma\searrow 0}\sigma^{2\gamma}\xi(\sigma)^{m-2}=0.$

Let us choose

$\gamma=\dfrac{q(m-1)-2m}{m}>0\quad\hbox{and}\quad\dfrac{(q+m)q}{m^{2}}{\rm
E}(x)=m{\rm C}_{0}^{m-1}\widehat{\rm C}_{1}(x).$ (5.30)

So that, the assumption (3.1) on the source $f$ implies

$-f(x)\geq-{\rm
d}(x)^{-q}(\ell_{u}+\varepsilon)\geq(\ell_{u}+\varepsilon)\sigma^{-q},$

hence

$-{\bf trace}~{}{\cal A}(\cdot){\rm D}^{2}{\rm U}(x)+\big{(}{\rm
U}(x)\big{)}^{m}-f(x)\geq\sigma^{-q}\bigg{(}(1+\varepsilon)^{m}{\rm
C}_{0}^{m}-(\ell_{u}+\varepsilon)+{\cal Q}_{1}(x,\sigma)\bigg{)},\quad
x\in\Omega^{\delta}_{\delta_{1}},$

where

$\begin{array}[]{ll}{\cal Q}_{1}(x,\sigma)&\doteq-\dfrac{q}{m}{\bf
trace}~{}{\cal A}(x){\rm D}^{2}{\rm d}(x){\rm
C}_{0}\sigma^{\frac{q(m-1)-m}{m}}\\\\[4.26773pt]
&-\left(\dfrac{m\gamma-q}{m}\right)\left(\dfrac{m(\gamma-1)-q}{m}\right){\rm
E}(x){\rm C}_{0}\widehat{\rm
C}_{1}(x)\sigma^{\frac{q(m-1)+m(\gamma-2)}{m}}\\\\[7.82433pt]
&-\dfrac{m\gamma-q}{m}{\rm C}_{0}\widehat{\rm
F}_{1}(x)\sigma^{\frac{q(m-1)+m(\gamma-1)}{m}}-{\rm C}_{0}{\bf trace}~{}{\cal
A}(x){\rm D}^{2}\widehat{\rm
C}_{1}(x)\sigma^{\frac{q(m-1)+m\gamma}{m}}\\\\[5.69046pt]
&+\dfrac{m(m-1)}{2}\widehat{\rm
C}_{1}(x)^{2}\sigma^{\frac{2\gamma+q(m-2)}{m}}\xi(\sigma)^{m-2},\quad
x\in\Omega^{\delta}_{\delta_{1}}.\end{array}$ (5.31)

The above reasonings lead to

######  Theorem 5.2 (Second order term)

Let us assume $\partial\Omega\in{\cal C}^{4}$ and ${\cal A}(\cdot)$ adequately
smooth. Suppose (1.3) and (1.4). Let

$q>\dfrac{2m}{m-1}\quad\hbox{and}\quad\gamma=\dfrac{q(m-1)-2m}{m}>0,\quad
m>1.$ (5.32)

For any solution of

$-{\bf trace}~{}{\cal A}(\cdot){\rm D}^{2}u+u^{m}\leq f\quad\hbox{in
$\Omega$}$

we have

$\limsup_{{\rm d}(x)\searrow 0}\dfrac{u^{*}(x)}{{\rm
d}(x)^{-\frac{q}{m}}\big{(}1+{\rm C}_{1,u}(x){\rm
d}(x)^{\gamma}\big{)}}\leq\ell_{u}^{\frac{1}{m}},$ (5.33)

provided

$\limsup_{{\rm d}(x)\searrow 0}f(x){\rm
d}(x)^{q}\leq\ell_{u}\quad(\ell_{u}>0),$

where

${\rm C}_{1,u}(x)=\dfrac{(q+m)q}{m^{3}}\ell_{u}^{\frac{m}{m-1}}{\rm E}(x).$
(5.34)

Analogously, for any nonnegative large solution of

$-{\bf trace}~{}{\cal A}(\cdot){\rm D}^{2}v+v^{m}\geq f\quad\hbox{in
$\Omega$}$

we have

$\liminf_{{\rm d}(x)\searrow 0}\dfrac{v_{*}(x)}{{\rm
d}(x)^{-\frac{q}{m}}\big{(}1+{\rm C}_{1,d}(x){\rm
d}(x)^{\gamma}\big{)}}\geq\ell_{d}^{\frac{1}{m}},$ (5.35)

provided

$\limsup_{{\rm d}(x)\searrow 0}f(x){\rm d}(x)^{q}\geq\ell_{d}>0$ (5.36)

and

${\rm C}_{1,d}(x)=\dfrac{(q+m)q}{m^{3}}\ell_{d}^{\frac{m}{m-1}}{\rm E}(x).$

Proof. We only show the result relative to $u^{*}$. The proof for $v_{*}$ is
analogous (see Theorem 3.1). With the above notation one proves that the
function ${\cal Q}_{1}(x,\sigma)$ (see (5.31)) satisfies

$\lim_{\sigma\searrow 0}{\cal Q}_{1}(x,\sigma)=0$

uniformly in $\Omega^{\delta}_{\delta_{1}}$, for $\delta_{1}$ small enough.
Then by choosing

${\rm C}_{0}>\dfrac{(\ell_{u}+\varepsilon)^{\frac{1}{m}}}{1+\varepsilon}$

we deduce

$-{\bf trace}~{}{\cal A}(\cdot){\rm D}^{2}{\rm U}(x)+\big{(}{\rm
U}(x)\big{)}^{m}\geq f(x),\quad x\in\Omega^{\delta}_{\delta_{1}},$

for $\delta_{1}$ small. As in the proof of Theorem 3.1 we deduce

$\limsup_{{\rm d}(x)\nearrow 0}\dfrac{u^{*}(x)}{{\rm
d}(x)^{-\frac{q}{m}}\big{(}1+\widehat{\rm C}_{1}(x){\rm
d}(x)^{\gamma}\big{)}}\leq{\rm C}_{0},$

whence letting ${\rm C}_{0}\searrow\ell_{u}^{\frac{1}{m}}$ one concludes the
proof.$\quad{}_{\Box}$

###### Remark 5.4

From Theorem 5.2 one deduces that for extra Keller-Osserman type growth of the
sources on the boundary the second order in the explosive boundary expansion
of large solutions is independent on the geometry whenever
$\lambda=\Lambda=1$, as it was obtained in [1].$\quad{}_{\Box}$

###### Remark 5.5

Since

$-\dfrac{q}{m}+\gamma=\dfrac{q(m-2)-2m}{m}$

we deduce that if $\dfrac{2m}{m-1}<q<\dfrac{2m}{m-2}$ the second term of the
expansion is explosive on the boundary. If $q=\dfrac{2m}{m-2}>0$ the second
term is bounded on the boundary while $0<\dfrac{2m}{m-2}<q$ implies that these
second term of the expansion vanishes on the boundary.$\quad{}_{\Box}$

##  6 Universal bounds

Here we adapt the proof of Lemma 1 and Theorem 1 of [13] (see also the pioneer
work [36]) for the solution of the equation

$-{\bf trace}~{}{\cal A}(x){\rm D}^{2}u+\beta(u)\leq f\quad\hbox{in
$\Omega\subseteq{\mathbb{R}}^{{\rm N}}$}$

where $\beta$ is a function satisfying (1.8) and (1.3) holds. Let us consider
the decreasing function

$\Phi(t)=\int^{+\infty}_{t}\dfrac{ds}{\sqrt{2{\rm G}(s)}}\quad\hbox{where
${\rm G}^{\prime}(s)=\beta(s)$}$

defined in Introduction (see (1.9)). We note that $\Phi(a)<+\infty$. For each
couple of positive constants ${\rm A}$ and ${\rm B}$ we consider the ${\cal
C}^{2}$ convex function

$\Psi(\zeta)\doteq\dfrac{1}{{\rm A}}\Phi^{-1}\left(\dfrac{{\rm
B}}{\sqrt{\Lambda}}\zeta\right),\quad 0\leq\zeta<{\rm R},$

where ${\rm R}<\dfrac{\sqrt{\Lambda}}{{\rm B}}\Phi(a)$ (see (4.17)).
Straightforward computations imply

$\Psi^{\prime}(\zeta)=-\dfrac{{\rm B}}{{\rm A}\sqrt{\Lambda}}\sqrt{2{\rm
G}\big{(}{\rm
A}\Psi(\zeta)\big{)}}\quad\hbox{and}\quad\Psi^{\prime\prime}(\zeta)=\dfrac{{\rm
B}^{2}}{{\rm A}\Lambda}\beta\big{(}{\rm A}\Psi(\zeta)\big{)}.$

Next, we define the ${\cal C}^{2}$ convex real function

${\rm W}(x)=\Psi\big{(}\zeta(x)\big{)},\quad\zeta(x)={\rm
R}^{2}-|x|^{2},\quad|x|<{\rm R},$

for which

$\begin{array}[]{ll}{\rm
W}^{\prime\prime}(x)&=\Psi_{\zeta\zeta}\big{(}\zeta(x)\big{)}\big{(}\zeta^{\prime}(x)\big{)}^{2}+\Psi_{\zeta}\big{(}\zeta(x)\big{)}\zeta^{\prime\prime}(x)\\\\[5.69046pt]
&=4x^{2}\Psi_{\zeta\zeta}\big{(}\zeta(x)\big{)}+2\dfrac{{\rm B}}{{\rm
A}\sqrt{\Lambda}}\sqrt{2{\rm G}\big{(}{\rm
A}\Psi(\zeta(x))\big{)}}\\\\[5.69046pt] &\leq 4{\rm R}^{2}\dfrac{{\rm
B}^{2}}{{\rm A}\Lambda}\beta\big{(}\Psi(\zeta(x))\big{)}+2\dfrac{{\rm B}}{{\rm
A}\sqrt{\Lambda}}\sqrt{2{\rm
A}\Psi\big{(}\zeta(x)\big{)}\beta\big{(}\Psi\big{(}\zeta(x)\big{)}\big{)}},\quad|x|<{\rm
R}.\end{array}$

On the other hand, if $0<{\rm A}<1$ the inequality

$\dfrac{{\rm B}}{\sqrt{\Lambda}}{\rm R}^{2}\geq\dfrac{{\rm
B}}{\sqrt{\Lambda}}\zeta(x)\geq\int^{\Psi(\zeta(x))}_{{\rm
A}\Psi(\zeta(x))}\dfrac{ds}{\sqrt{2{\rm G}(s)}}\geq\dfrac{(1-{\rm
A})\Psi\big{(}\zeta(x)\big{)}}{\sqrt{2{\rm
G}\big{(}\Psi\big{(}\zeta(x)\big{)}\big{)}}}\geq\dfrac{(1-{\rm
A})\Psi\big{(}\zeta(x)\big{)}}{\sqrt{2\Psi\big{(}\zeta(x)\big{)}\beta\big{(}\Psi\big{(}\zeta(x)\big{)}\big{)}}}$

implies

$\sqrt{2{\rm A}\Psi\big{(}\zeta(x)\big{)}\beta\big{(}{\rm
A}\Psi\big{(}\zeta(x)\big{)}\big{)}}\leq\dfrac{2{\rm A}{\rm B}{\rm
R}^{2}}{(1-{\rm A})}\sqrt{\dfrac{{\rm A}}{\Lambda}}\beta\big{(}{\rm
A}\Psi(\zeta(x))\big{)},$

whence

$\Lambda{\rm W}^{\prime\prime}(x)\leq 4{\rm B}^{2}{\rm
R}^{2}\left(\dfrac{1}{{\rm A}}+\dfrac{1}{(1-{\rm A})\sqrt{{\rm
A}}}\right)\beta\big{(}{\rm W}(x)\big{)},\quad|x|<{\rm R}.$

Denote

$c({\rm A})\doteq\dfrac{1}{\sqrt{\dfrac{1}{{\rm A}}+\dfrac{1}{(1-{\rm
A})\sqrt{{\rm A}}}}},\quad 0<{\rm A}<1.$

Since $c(0)=c(1)=0$ the continuous function $c({\rm A})$ attains a positive
maximum at some ${\rm A}_{\infty}$ in the interval $(0,1)$. Then for

${\rm B}=\dfrac{c({\rm A}_{\infty})}{2{\rm R}^{2}}$

one has

$\Lambda{\rm W}^{\prime\prime}(x)\leq\left(\dfrac{2{\rm B}{\rm R}}{c({\rm
A}_{\infty})}\right)^{2}\beta\big{(}{\rm W}(x)\big{)},\quad|x|<{\rm R}.$ (6.1)

###### Theorem 6.1

Assume (1.3) and (1.8). Then there exists two positive constants ${\rm
A}_{\infty}$ and ${\rm B}$, satisfying (6.4) below, for which

$u^{*}(x)\leq\sum_{i=1}^{{\rm N}}\Psi\big{(}{\rm
R}^{2}-|x_{i}|^{2}\big{)}+\beta^{-1}\big{(}\displaystyle{\|f\|}_{{\cal
Q}_{{\rm R}}(x_{0})}\big{)},\quad x\in{\cal Q}_{{\rm R}}(x_{0})$ (6.2)

holds in any cube ${\cal Q}_{{\rm R}}(x_{0})\subset\subset\Omega$, for any
subsolution $u$ of

$-{\bf trace}~{}{\cal A}(\cdot){\rm D}^{2}u+\beta(u)\leq f\quad\hbox{in
$\Omega$}.$

Proof. Assume with no loss of generality $x_{0}=0$. The above reasoning proves
that the function

${\cal W}(x)=\sum_{i=1}^{{\rm
N}}\Psi\big{(}\zeta(x_{i})\big{)},\quad\zeta(x_{i})={\rm
R}^{2}-|x_{i}|^{2},\quad|x_{i}|<{\rm R},$

satisfies in the cube ${\cal Q}_{{\rm R}}(0)$ the inequality

${\bf trace}~{}{\cal A}(x){\rm D}^{2}{\cal W}(x)\leq\Lambda\sum_{i=1}^{{\rm
N}}{\rm W}^{\prime\prime}(x_{i})\leq\left(\dfrac{2{\rm B}{\rm R}}{c({\rm
A}_{\infty})}\right)^{2}\sum_{i=1}^{{\rm
N}}\beta\big{(}\Psi(\zeta(x_{i}))\big{)}\leq{\rm N}\left(\dfrac{2{\rm B}{\rm
R}}{c({\rm A}_{\infty})}\right)^{2}\beta\big{(}{\cal W}(x)\big{)},$

thus

$-{\bf trace}~{}{\cal A}(x){\rm D}^{2}{\cal W}(x)+\beta\big{(}{\cal
W}(x)\big{)}\geq 0,\quad x\in{\cal Q}_{{\rm R}}(0),$ (6.3)

provided

${\rm N}\left(\dfrac{2{\rm B}{\rm R}}{c({\rm A}_{\infty})}\right)^{2}=1.$
(6.4)

Since ${\cal W}(x)=+\infty$, Remark 4.1 concludes the
estimate.$\quad{}_{\Box}$

###### Remark 6.1

We emphasize that no uniform ellipticity assumption is required in the above
proof. On the other hand, the choice (6.4) is independent on the constant
$\Lambda.\hfill\mbox{$\quad{}_{\Box}$}$

###### Remark 6.2

Relative to the equation

$-\varepsilon\Delta u_{\varepsilon}-{\bf trace}~{}{\cal A}(\cdot){\rm
D}^{2}u_{\varepsilon}+\beta(u_{\varepsilon})=f\quad\hbox{in $\Omega$}$

with $0<\varepsilon<1$, the estimate (4.18) follows from (6.2) by replacing
$\Lambda$ by $\Lambda+\varepsilon$ in the definition (4.17) of the function
$\Phi(\zeta)$. The estimate (4.21) also follows from (6.2) by replacing
$\Lambda$ by $\Lambda+1$. Indeed (6.3) becomes

$(\varepsilon+\Lambda){\rm W}^{\prime\prime}(x)\leq(1+\Lambda){\rm
W}^{\prime\prime}(x)\leq\left(\dfrac{2{\rm B}{\rm R}}{c({\rm
A}_{\infty})}\right)^{2}\beta\big{(}{\rm W}(x)\big{)},\quad|x|<{\rm R},$

whence one deduces

$-\varepsilon\Delta{\cal W}(x)-{\bf trace}~{}{\cal A}(x){\rm D}^{2}{\cal
W}(x)+\beta\big{(}{\cal W}(x)\big{)}\geq 0,\quad x\in{\cal Q}_{{\rm R}}(0).$

##  7 Strong Maximum Principle without coercivity term

Here we group some results used in above reasoning. So, in the proof of
Theorem 2.2 we require a version of Weak Maximum Principle without coercivity
term.

######  Theorem 7.1 (Weak Maximum Principle without coercivity term (I))

Assume (4.12). Let $u$ be a discontinuous subsolution of

$-{\bf trace}~{}{\cal A}(\cdot){\rm D}^{2}u=f,\quad x\in\Omega$

where $f$ is a uniformly continuous function and $v$ be a discontinuous
supersolution of

$-{\bf trace}~{}{\cal A}(\cdot){\rm D}^{2}v=g,\quad x\in\Omega$

where $g$ is a continuous function. Then

$u^{*}(x)-v_{*}(x)\leq\sup_{\partial\Omega}~{}(u^{*}-v_{*})+\dfrac{{\rm
C}_{\Omega}}{2\lambda_{{\cal A}}}\displaystyle{\|(f-g)_{+}\|}_{{\rm
L}_{\infty}},\quad x\in\Omega,$

where $\displaystyle{\rm C}_{\Omega}=\max_{x\in\Omega}|x|^{2}$. See the
Introduction for the positive constant $\lambda_{{\cal A}}$.

Proof. It is clear that the smooth function ${\rm
W}(x)=-|x|^{2},~{}x\in\Omega$, satisfies

$-{\bf trace}~{}{\cal A}(x){\rm D}^{2}{\rm W}(x)\geq 2\lambda_{{\cal A}},\quad
x\in\Omega.$

So, for any positive constant $a$ we have

$-{\bf trace}~{}{\cal A}(\cdot){\rm D}^{2}(v_{*}+a{\rm W})\geq
g+2a\lambda_{{\cal A}},\quad x\in\Omega.$

Arguing on an eventual interior maximum of $u^{*}-(v_{*}+a{\rm W})$ at some
$x_{0}\in\Omega$, as in the Theorem 4.1, we get

$0\leq\omega\big{(}\alpha|x_{\alpha}-y_{\alpha}|^{2}\big{)}+\omega_{f}\big{(}|x_{\alpha}-y_{\alpha}|\big{)}+f(y_{\alpha})-g(y_{\alpha})-2a\lambda_{{\cal
A}}$

(see (4.13)). Then by letting $\alpha\nearrow\infty$ one derives the
contradiction

$0\leq f(x_{0})-g(x_{0})-2a\lambda_{{\cal A}}<0$

provided $a=\dfrac{\displaystyle{\|(f-g)_{+}\|}_{\infty}}{2\lambda_{{\cal
A}}}+\varepsilon$ with $\varepsilon>0$. So that, we have obtained the
inequality

$\big{(}u^{*}-v_{*}\big{)}(x)\leq\big{(}u^{*}-v_{*}-a{\rm
W}\big{)}(x)\leq\sup_{\partial\Omega}~{}\big{(}u^{*}-v_{*}-a{\rm
W}\big{)}\leq\sup_{\partial\Omega}~{}(u^{*}-v_{*})+\left(\dfrac{\displaystyle{\|(f-g)_{+}\|}_{\infty}}{2\lambda_{{\cal
A}}}+\varepsilon\right){\rm C}_{\Omega},$

for $x\in\Omega$. The result follows sending $\varepsilon\searrow
0.\hfill\mbox{$\quad{}_{\Box}$}$

Next we detail the application of certain results and reasonings of [6] in
what follows. The classical device by E. Hopf (see [20]) enables us to obtain
a Strong Maximum Principle as follows. It was used in the proof of Theorem
4.4.

######  Lemma 7.1 (Hopf Lemma)

Assume (4.12) and (1.7). Let $u$ be a discontinuous subsolution of

$-{\bf trace}~{}{\cal A}(\cdot){\rm D}^{2}u=f,\quad x\in\Omega$

and $v$ be a discontinuous supersolution of

$-{\bf trace}~{}{\cal A}(\cdot){\rm D}^{2}v=f,\quad x\in\Omega$

where $f$ is a uniformly continuous function. Suppose that $\Omega$ is an open
set such that there exists $x_{0}\in\partial\Omega$ for which

${\bf B}_{{\rm R}}(y)\subset\Omega\quad\hbox{and}\quad\partial{\bf B}_{{\rm
R}}(y)\cap\partial\Omega=\\{x_{0}\\}\quad\hbox{{\rm(}interior sphere condition
at $x_{0}\in\partial\Omega${\rm)}}$

and

$\big{(}u^{*}-v_{*}\big{)}(x)\leq\big{(}u^{*}-v_{*}\big{)}(x_{0}),\quad
x\in{\bf B}_{{\rm R}}(y)$

hold. Then

$\liminf_{x\rightarrow
x_{0}}\dfrac{\big{(}u^{*}-v_{*}\big{)}(x_{0})-\big{(}u^{*}-v_{*}\big{)}(x)}{|x-x_{0}|}>0.$

Proof. The smooth and positive function

${\rm W}(x)=e^{-\alpha|x-y|^{2}}-e^{-\alpha{\rm R}^{2}},\quad x\in{\bf
B}_{{\rm R}}(y)\subset\Omega$

satisfies ${\rm D}{\rm W}(x)=-2\alpha e^{-\alpha|x-y|^{2}}(x-y)$ whence

$\left\\{\begin{array}[]{l}\alpha{\rm R}e^{-\alpha{\rm R}^{2}}\leq|{\rm D}{\rm
W}(x)|=2\alpha e^{-\alpha|x-y|^{2}}|x-y|\leq 2{\rm R}\alpha
e^{-\alpha\frac{{\rm R}^{2}}{4}},\\\\[4.26773pt] \begin{array}[]{ll}{\rm
D}^{2}{\rm W}(x)&=-2\alpha e^{-\alpha|x-y|^{2}}{\rm
I}+4\alpha^{2}e^{-\alpha|x-y|^{2}}\big{(}x-y\big{)}\otimes\big{(}x-y\big{)}\\\\[4.26773pt]
&\geq-\dfrac{2}{{\rm R}}|{\rm D}{\rm W}(x)|{\rm I}+\alpha{\rm R}\dfrac{{\rm
D}{\rm W}(x)\otimes{\rm D}{\rm W}(x)}{|{\rm D}{\rm
W}(x)|}\end{array}\end{array}\right.\qquad x\in{\bf B}_{{\rm
R}}(y)\setminus{\bf B}_{\frac{{\rm R}}{2}}(y).$

Then from (1.7) we have

${\bf trace}~{}{\cal A}(x){\rm D}^{2}{\rm W}(x)\geq 0,\quad x\in{\bf B}_{{\rm
R}}(y)\setminus{\bf B}_{\frac{{\rm R}}{2}}(y),$

for some provided $\alpha$. Therefore

$-{\bf trace}~{}{\cal A}(\cdot){\rm D}^{2}(u^{*}+\varepsilon{\rm W})\leq
f\quad\hbox{in ${\bf B}_{{\rm R}}(y)\setminus{\bf B}_{\frac{{\rm
R}}{2}}(y)$},$

for any $\varepsilon>0$. Moreover

$(u^{*}-v_{*}+\varepsilon{\rm W})(x)\leq(u^{*}-v_{*})(x_{0}),\quad
x\in\partial{\bf B}_{\frac{{\rm R}}{2}}(y)$

if

$\varepsilon=\dfrac{(u^{*}-v_{*})(x_{0})-\sup_{x\in\partial{\bf B}_{\frac{{\rm
R}}{2}}(y)}(u^{*}-v_{*})(x)}{e^{-\alpha\frac{{\rm R}^{2}}{4}}-e^{-\alpha{\rm
R}^{2}}}>0.$

So that, the Weak Maximum Principle (Theorem 7.1) applied to the set ${\bf
B}_{{\rm R}}(y)\setminus\overline{{\bf B}}_{\frac{{\rm R}}{2}}(y)$ leads to

$(u^{*}-v_{*})(x_{0})-(u^{*}-v_{*})(x)\geq{\rm W}(x),\quad{\bf B}_{{\rm
R}}(y)\setminus\overline{{\bf B}}_{\frac{{\rm R}}{2}}(y).$

Sinces ${\rm W}(x_{0})=0$ one concludes

$\liminf_{x\rightarrow
x_{0}}\dfrac{(u^{*}-v_{*})(x_{0})-(u^{*}-v_{*})(x)}{|x-x_{0}|}\geq\varepsilon\displaystyle{\langle{\rm