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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2011 (2011), No. 51, pp. 1--19.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2011 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2011/51\hfil Second-order boundary estimates]
{Second-order boundary estimates for solutions to singular
elliptic equations in borderline cases}

\author[C. Anedda, G. Porru \hfil EJDE-2011/51\hfilneg]
{Claudia Anedda, Giovanni Porru} % in alphabetical order

\address{Claudia Anedda \newline
Dipartimento di Matematica e Informatica,
Universit\'a di Cagliari,
Via Ospedale 72, 09124 Cagliari, Italy}
\email{canedda@unica.it}

\address{Giovanni Porru \newline
Dipartimento di Matematica e Informatica,
Universit\'a di Cagliari,
Via Ospedale 72, 09124 Cagliari, Italy}
\email{porru@unica.it}

\thanks{Submitted January 10, 2011. Published April 13, 2011.}
\subjclass[2000]{35B40, 35J67}
\keywords{Elliptic problems; singular equations; \hfill\break\indent
second order boundary approximation}

\begin{abstract}
 Let $\Omega\subset R^N$ be a bounded smooth domain.
 We investigate the effect of the mean curvature of the boundary
 $\partial\Omega$ on the behaviour of the solution to the homogeneous
 Dirichlet boundary value problem for the equation $\Delta u+f(u)=0$.
 Under appropriate growth conditions on $f(t)$ as $t$ approaches zero,
 we find asymptotic expansions up to the second order of the solution
 in terms of the distance from  $x$ to the boundary $\partial\Omega$.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{corollary}[theorem]{Corollary}


\section{Introduction}

In this paper we study the Dirichlet problem
\begin{equation}\label{e1}
\begin{gathered}
\Delta u+f(u)=0 \quad\text{in }\Omega,\\
u=0\quad\text{on }\partial \Omega,
\end{gathered}
\end{equation}
where $\Omega$ is a bounded smooth domain in ${\mathbb R}^N$,
$N\ge 2$, and $f(t)$ is a decreasing and positive smooth function
in $(0,\infty)$, which approaches infinity as $t\to 0$.
Equation \eqref{e1} arises in problems of heat conduction and in fluid
mechanics.

Problems of this kind are discussed in many papers; see, for
instance, \cite{CGP,GR,Ka,LM1,ZC,ZY} and references therein.
For $f(t)=t^{-\gamma}$,
$\gamma>0$, in \cite{CRT} it is shown that there exists a positive
solution continuous up to the boundary $\partial\Omega$. For
$f(t)=t^{-\gamma}$, $\gamma>1$, in \cite{BCP} it is shown that there
exists a constant $B>0$ such that
$$
\big| u(x)-\Bigl(\frac{\gamma+1}
{\sqrt{2(\gamma-1)}}\delta\Bigr)^\frac{2}{1+\gamma}\big|< B\delta
^\frac{2\gamma}{\gamma+1},
$$
where $\delta=\delta(x)$ denotes the
distance from $x$ to the boundary $\partial \Omega$. For
$f(t)=t^{-\gamma}$, $\gamma>3$, in \cite{ACP} it is proved that
$$
u(x)=\Bigl(\frac{\gamma+1}
{\sqrt{2(\gamma-1)}}\delta\Bigr)^\frac{2}{1+\gamma}
\Bigl[1+\frac{1}{3-\gamma}H\delta+
o(\delta)\Bigr],
$$
where $H=H(x)$ is related with the mean curvature
of $\partial\Omega$ at the nearest point to $x$.


In  \cite{An}, more general nonlinearities are discussed.
More precisely, let
\begin{equation}\label{e2}
F(t)=\int_t^1f(\tau)d\tau,\quad
\lim_{t\to 0^+}F(t)=\infty,\quad
\frac{f'(t)F(t)}{(f(t))^2}=\frac{\gamma}{1-\gamma}+O(1)t^{\beta},
\end{equation}
where $\gamma\ge 3$, $\beta>0$ and $O(1)$ denotes a bounded quantity
as $t\to 0$. In addition, we suppose there is $M$ finite such that
for all $\theta\in (1/2,2)$ and for $t\in(0,1)$ we have
\begin{equation}\label{e3}
\frac{|f''(\theta t)|t^2}{f(t)}\le M.
\end{equation}
An example which satisfies these conditions is
$f(t)=t^{-\gamma}+t^{-\nu}$ with
$0<\nu<\gamma$; here $\beta=\min[\gamma-\nu,\gamma-1]$.

Let $\phi(\delta)$ be defined as
\begin{equation}\label{e4}
\int_0^{\phi(\delta)}\frac{1}{(2F(t))^{1/2}}dt=\delta.
\end{equation}
For $3<\gamma<\infty$, in \cite{An} it is proved that
\begin{equation}\label{e5}
u(x)=\phi(\delta)\Bigl[1+\frac{1}{3-\gamma}H\delta
+O(1)\delta^{\sigma+1}\Bigr],
\end{equation}
where $\sigma$ is any number such that
$0<\sigma<\min[\frac{\gamma-3}{\gamma+1},\frac{2\beta}{\gamma+1}]$.
Note that $\phi$ satisfies the one dimensional problem
$$
\phi''+f(\phi)=0,\quad \phi(0)=0.
$$
The estimate \eqref{e5} shows that the expansion of $u(x)$ in terms
of $\delta$ has the first part which is independent of the geometry
of the domain, and the second part which depends on the mean
curvature of the boundary as well as on $\gamma$.

In the present paper we investigate the borderline cases $\gamma=3$
and $\gamma=\infty$. In the case of $\gamma=3$ we find the expansion
\begin{equation}\label{e6}
u(x)=\phi(\delta)\Bigl[1+\frac{1}{4}H\delta\log \delta
+O(1)\delta(-\log \delta)^\sigma\Bigr],
\end{equation}
where $0<\sigma<1$ and $O(1)$ is bounded as $\delta\to 0$.

To discuss the case $\gamma=\infty$, we make the following
assumption
\begin{equation}\label{e7}
f(t)>0,\quad
\frac{f'(t)}{f(t)}=-\frac{\ell}{t^{\beta+1}}\bigl(1+O(1)t^\beta\bigr),
\end{equation}
with $\ell>0$ and $\beta>0$. Note that the above condition implies
\begin{equation}\label{e8}
\frac{F(t)}{f(t)}=\frac{t^{\beta+1}}{\ell}\bigl(1+O(1)t^\beta\bigr),
\quad
F(t)=\int_t^1f(\tau)d\tau.
\end{equation}
Furthermore, \eqref{e7}
together with \eqref{e8} imply \eqref{e2} with $\gamma=\infty$; that
is,
\begin{equation}\label{e9}
\frac{f'(t)F(t)}{(f(t))^2}=-1+O(1)t^{\beta}.
\end{equation}

Instead of \eqref{e3}, now we suppose that for some $m>2$ and some
$\epsilon\in (0,1)$, there is $M>0$ such that
\begin{equation}\label{e10}
\frac{|f''(\theta t)|t^{2}}{f(t)}\le M
\frac{1}{t^{2\beta}}(F(t))^{1/m},\quad \forall t\in (0,1/2),\;
\forall\theta\in (1-\epsilon,1+\epsilon).
\end{equation}
The function $f(t)=e^\frac{\ell}{\beta t^\beta}$ satisfies all
these conditions.

Under assumptions \eqref{e7} and \eqref{e10}, we find the estimate
$$
u(x)=\phi(\delta)\Bigr[1-\frac{1}{\ell}H\delta
\bigl(\phi(\delta)\bigr)^\beta+O(1)\delta
\bigl(\phi(\delta)\bigr)^{2\beta}\Bigr],
$$
where $\phi$ is defined as in \eqref{e4}.

Throughout this paper, the boundary $\partial\Omega$ is smooth in
the sense that it belongs to $C^4$.

\section{Preliminary results}

\begin{lemma} \label{lem2.1}
Let $A(\rho,R)\subset \mathbb R^N$, $N\ge 2$, be the annulus
with radii $\rho$ and $R$ centered at the origin.
Let $f(t)>0$ smooth, decreasing for
$t>0$, and such that $\int_t^1(F(\tau))^{1/2}d\tau\to\infty$
as $t\to 0^+$, where $F(t)=\int_t^1f(\tau)d\tau$. We also suppose
that the function $s\mapsto (F(s))^{-1}\int_s^1(F(t))^{1/2}dt$
is increasing for $s$ close to $0$. If $u(x)$ is a solution to
problem \eqref{e1} in $\Omega=A(\rho,R)$ and $v(r)=u(x)$ for
$r=|x|$, then
\begin{equation}\label{e2.1}
v(r)>\phi(R-r)-C\frac{\int_v^1(F(t))^{1/2}dt}{(F(v))^{1/2}}(R-r),
\quad  \tilde r<r<R,
\end{equation}
and
\begin{equation}\label{e2.2}
v(r)<\phi(r-\rho)+C\phi'(r-\rho)\frac{\int_v^1(F(t))^{1/2}dt}{F(v)}(r-\rho),
\quad \rho<r<\overline r,
\end{equation}
where $\phi$ is defined as in \eqref{e4},
$\rho<\overline r\le \tilde r<R$ and $C$ is a
suitable positive constant.
\end{lemma}

\begin{proof}
 If $\Omega=A(\rho,R)$, the corresponding
solution $u(x)$ to problem \eqref{e1} is radially symmetric (by
uniqueness) and positive (by the maximum principle). With
$v(r)=u(x)$ for $r=|x|$ we have
\begin{equation}\label{e2.3}
v''+\frac{N-1}{r}v'+f(v)=0,\quad v(\rho)=v(R)=0.
\end{equation}
The latter equation can be rewritten as
$$
\bigl(r^{N-1}v'\bigr)'+r^{N-1}f(v)=0.
$$
Since $v(\rho)=v(R)$, we must have $v'(r_0)=0$ for some
$r_0\in(\rho,R)$. Integrating over $(r_0,r)$ we obtain
$$
r^{N-1}v'+\int_{r_0}^rt^{N-1}f(v)dt=0.
$$
Hence, $v(r)$ is increasing for $\rho<r<r_0$ and decreasing for
$r_0<r<R$. Multiplying \eqref{e2.3} by $v'$ and integrating over
$(r_0,r)$ we find
\begin{equation}\label{e2.4}
\frac{(v')^2}{2}+(N-1)\int_{r_0}^r\frac{(v')^2}{s}ds=F(v)-F(v_0),\quad
 v_0=v(r_0).
\end{equation}
Since $\int_t^1(F(\tau))^{1/2}d\tau\to\infty$ as $t\to 0$, we have
$F(t)\to\infty$ as $t\to 0$. Therefore, $F(v(r))\to\infty$ as $r\to
R$, and \eqref{e2.4} implies
\begin{equation}\label{e2.5}
|v'|<2(F(v))^{1/2},\quad  r\in(r_1,R),\quad r_0\le r_1<R.
\end{equation}
 As a consequence, with $v_1=v(r_1)$ we have
\begin{equation}\label{e2.6}
\int_{r_1}^r\frac{(v')^2}{s}ds
\le\frac{2}{r_1}\int_{r_1}^r(F(v))^{1/2}(-v')ds=
\frac{2}{r_1}\int_v^{v_1}(F(t))^{1/2}dt.
\end{equation}
Since
$$
\int_v^{v_1}(F(t))^{1/2}dt\le(F(v))^{1/2}v_1,
$$
using \eqref{e2.6} we find
\begin{equation}\label{e2.7}
 \lim_{r\to R}\frac{\int_{r_1}^r\frac{(v')^2}{s}ds}{F(v)}
=\lim_{r\to R}\frac{\int_v^{v_1}(F(t))^{1/2}dt}{F(v)}=0.
\end{equation}
Now, by \eqref{e2.4} we have
\begin{equation}\label{e2.8}
\frac{(v')^2}{2F(v)}=
1-\frac{(N-1)\int_{r_0}^r\frac{(v')^2}{s}ds+F(v_0)}{F(v)}.
\end{equation}
Note that, if $v_0>1$ then $F(v_0)<0$. We claim that
$$
(N-1)\int_{r_0}^r\frac{(v')^2}{s}ds+F(v_0)>0
$$
for $r$ close to $R$. Indeed, by \eqref{e2.7} and \eqref{e2.8} it
follows that $|v'|>(F(v))^{1/2}$ for $r\in(r_2,R)$. Hence,
$$
\int_{r_2}^r\frac{(v')^2}{s}ds>\frac{1}{R}\int_{r_2}^r(F(v))^{1/2}(-v')
ds=
\frac{1}{R}\int_{v(r)}^{v(r_2)}(F(\tau))^{1/2}d\tau.
$$
By using the assumption $\int_t^1(F(\tau))^{1/2}d\tau\to\infty$
as $t\to 0$, the latter inequality implies that
$\int_{r_2}^r\frac{(v')^2}{s}ds\to\infty$ as $r\to R$, and the claim
follows.

Equation \eqref{e2.8} yields
\begin{equation}\label{e2.9}
\frac{-v'}{(2F(v))^{1/2}}=1-\Gamma(r),
\end{equation}
where
$$
\Gamma(r)=1-\Bigl[1-\frac{(N-1)\int_{r_0}^r\frac{(v')^2}{s}ds
+F(v_0)}{F(v)}\Bigr]^{1/2}.
$$
Since
$$
1-[1-\epsilon]^{1/2}< \epsilon,\quad \forall\epsilon\in(0,1),
$$
using \eqref{e2.6} we find a constant $M$ such that, for $r$ close to
$R$,
\begin{equation}\label{e2.10}
0\le \Gamma(r)\le \frac{(N-1)\int_{r_0}^r\frac{(v')^2}{s}ds+F(v_0)}{F(v)}\le
M\frac{\int_v^{v_0}(F(t))^{1/2}dt}{F(v)}.
\end{equation}
Note that, by \eqref{e2.10} and \eqref{e2.7} we have
$\Gamma(r)\to 0$ as $r\to R$.

The inverse function of $\phi$ is
$$
\psi(s)=\int_0^s\frac{1}{(2F(t))^{1/2}}dt.
$$
Integration of \eqref{e2.9} over $(r,R)$ yields
$$
\psi(v)=R-r-\int_r^R\Gamma(s)ds,
$$
from which we find
\begin{equation}\label{e2.11}
v(r)=\phi\bigl(R-r-\int_r^R\Gamma(s)ds\bigr).
\end{equation}
By \eqref{e2.11}, we have
\begin{equation}\label{e2.12}
v(r)=\phi(R-r)-\phi'(\omega)\int_r^R\Gamma(s)ds,
\end{equation}
with
$$
R-r -\int_r^R\Gamma(s)ds<\omega<R-r.
$$
Since
$\phi'(\omega)=(2F(\phi(\omega)))^{1/2}$,
and since the function $t\to F(\phi(t))$ is decreasing we have
$$
\phi'(\omega)<\Bigl(2F\bigl(\phi(R-r-\int_r^R\Gamma(s)ds)
\bigr)\Bigr)^{1/2}
=(2F(v))^{1/2},
$$
where \eqref{e2.11} has been used in the last
step. Hence, by \eqref{e2.12} we have
$$
v(r)>\phi(R-r)-(2F(v))^{1/2}\int_r^R\Gamma(s)ds.
$$
Using \eqref{e2.10}, we find
\begin{equation}\label{e2.13}
v(r)>\phi(R-r)-(2F(v))^{1/2}M
\int_r^R\frac{\int_{v(s)}^{v_0}(F(\tau))^{1/2}d\tau}{F(v(s))}ds.
\end{equation}
Since $(F(t))^{-1}\int_t^1(F(\tau))^{1/2}d\tau$ is increasing
and since $v(s)$ is decreasing, for $s$ close to $R$ the function
$$
s\mapsto \frac{\int_{v(s)}^{v_0}(F(\tau))^{1/2}d\tau}{F(v(s))}
$$ is
decreasing. Using the monotonicity of this function, inequality
\eqref{e2.1} follows from \eqref{e2.13}.

To prove \eqref{e2.2}, we observe that \eqref{e2.4} also
holds for $\rho<r<r_0$. Let us write equation \eqref{e2.4} as
\begin{equation}\label{e2.14}
\frac{(v')^2}{2}=F(v)-F(v_0)+(N-1)\int_r^{r_0}\frac{(v')^2}{s}ds,
\end{equation}
with $\rho<r<r_0$. By \eqref{e2.14}, $(v'(r))^2\to\infty$ as
$r\to\rho$. Moreover, since $v'(r)>0$ for $r\in(\rho,r_0)$, by
\eqref{e2.3} we have $v''(r)<0$. Hence, by \cite[Lemma 2.1]{LM2},
we have
$$
\lim_{r\to\rho}\frac{\int_r^{r_0}\frac{(v')^2}{t}dt}{(v'(r))^2}=0.
$$
Using this result and \eqref{e2.14} we find
$0<v'<2(F(v))^{1/2}$ for $r\in (\rho,r_3)$, $r_3\le r_0$. As a
consequence we have, with $v(r_3)=v_3$,
\begin{equation}\label{e2.15}
\int_r^{r_3}\frac{(v')^2}{s}ds\le\frac{2}{\rho}\int_r^{r_3}(F(v))^{1/2}v'ds=
\frac{2}{\rho}\int_v^{v_3}(F(t))^{1/2}dt.
\end{equation}
Since
$\int_v^{v_3}(F(t))^{1/2}dt\le(F(v))^{1/2}v_3$,
\eqref{e2.15} implies
\begin{equation}\label{e2.16}\lim_{r\to
\rho}\frac{\int_r^{r_0}\frac{(v')^2}{s}ds}{F(v)}=0.
\end{equation}
By  \eqref{e2.14}, we find
\begin{equation}\label{e2.17}
\frac{(v')^2}{2F(v)}=1+\frac{(N-1)\int_r^{r_0}\frac{(v')^2}{s}ds-F(v_0)}{F(v)}.\end{equation}
Using \eqref{e2.16} and \eqref{e2.17} and arguing as in the previous
case one finds that
$$
(N-1)\int_r^{r_0}\frac{(v')^2}{s}ds-F(v_0)>0
$$
for $r$ close to $\rho$.
Equation \eqref{e2.17} yields
\begin{equation}\label{e2.18}
\frac{v'}{(2F(v))^{1/2}}=1+\tilde\Gamma(r),\end{equation}
where
$$
\tilde\Gamma(r)=\Bigl[1+\frac{(N-1)\int_r^{r_0}
\frac{(v')^2}{s}ds-F(v_0)}{F(v)}\Bigr]^{1/2}-1.
$$
Since
$$
[1+\epsilon]^{1/2}-1<\epsilon,\quad \forall\epsilon>0,
$$
using \eqref{e2.15} one finds, for $r$ close to $\rho$,
\begin{equation}\label{e2.19}
0\le \tilde\Gamma(r)\le
\frac{(N-1)\int_r^{r_0}\frac{(v')^2}{s}ds-F(v_0)}{F(v)}\le \tilde
M\frac{\int_v^{v_0}(F(t))^{1/2}dt}{F(v)}.\end{equation}
Integration of \eqref{e2.18} over $(\rho,r)$ yields
$$
\psi(v)=r-\rho+\int_\rho^r\tilde\Gamma (s)ds,
$$
from which we find
\begin{equation}\label{e2.20}
v(r)=\phi(r-\rho)+\phi'(\omega_1)\int_\rho^r\tilde\Gamma(s)ds,
\end{equation}
with
$$
r-\rho<\omega_1<r-\rho+\int_\rho^r\tilde\Gamma(s)ds.
$$
Since $\phi'(s)$ is decreasing we have
$$
\phi'(\omega_1)< \phi'(r-\rho).
$$
The latter estimate, \eqref{e2.20} and \eqref{e2.19} imply
\begin{equation}\label{e2.21}
v(r)<\phi(r-\rho)+\phi'(r-\rho)\int_\rho^r\tilde
M\frac{\int_v^{v_0}(F(\tau))^{1/2}d\tau}{F(v)}ds.
\end{equation}
Since $v(s)$ is increasing for $s$ close to $\rho$, the function
$$
s\mapsto \frac{\int_{v(s)}^{v_0}(F(\tau))^{1/2}d\tau}{F(v(s))}
$$
is increasing. Hence, inequality \eqref{e2.2} follows from
\eqref{e2.21}. The lemma is proved.
\end{proof}

\begin{corollary} \label{coro2.2}
Assume the same notation and assumptions as in Lemma 2.1.
Given $\epsilon>0$ there are $r_\epsilon$ and $\tilde r_\epsilon$
such that
\begin{gather}\label{e2.22}
\phi(R-r)>v(r)>(1-\epsilon)\phi(R-r), \quad
r_\epsilon<r<R,\\
\label{e2.23}
\phi(r-\rho)<v(r)<(1+\epsilon)\phi(r-\rho), \ \ \ \rho<r<\tilde
r_\epsilon.
\end{gather}
\end{corollary}

\begin{proof} By \eqref{e2.9} we have
$$
\frac{-v'}{(2F(v))^{1/2}}<1.
$$
Integrating over $(r,R)$ we find $\psi(v)<R-r$, from which the left
hand side of \eqref{e2.22} follows. By \eqref{e2.1} we have
$$
v(r)>\Bigl[1-C\frac{\int_v^1(F(t))^{1/2}dt}{(F(v))^{1/2}}
\frac{R-r}{\phi(R-r)}\Bigr]\phi(R-r).
$$
Since $F(t)$ is decreasing we find
$$
\frac{\int_v^1(F(t))^{1/2}dt}{(F(v))^{1/2}}\le 1.
$$
Moreover, putting $R-r=\psi(s)$ we have
$$
0\le\lim_{r\to R}\frac{R-r}{\phi(R-r)}=\lim_{s\to 0}
\frac{\psi(s)}{s}\le \lim_{s\to 0}\frac{1}{(2F(s))^{1/2}}=0.
$$
The right hand side of \eqref{e2.22} follows from these estimates.

By \eqref{e2.18} we have
$$
\frac{v'}{(2F(v))^{1/2}}>1.
$$
Integrating over $(\rho,r)$, we find $\psi(v)>r-\rho$,
from which the
left hand side of \eqref{e2.23} follows.
By \eqref{e2.2} we have
$$
v(r)<\Bigl[1+C\phi'(r-\rho)\frac{\int_v^1(F(t))^{1/2}dt}{F(v)}
\frac{r-\rho}{\phi(r-\rho)}\Bigr]\phi(r-\rho).
$$
We find
$$
0\le\lim_{r\to \rho}\frac{\int_v^1(F(t))^{1/2}dt}{F(v)}\le
\lim_{r\to \rho}\frac{1}{(F(v))^{1/2}}=0.
$$
Moreover, putting $r-\rho=\psi(s)$, we have
$$
\frac{(r-\rho)\phi'(r-\rho)}{\phi(r-\rho)}
=\frac{\psi(s)(2F(s))^{1/2}}{s}\le 1.
$$
The right hand side of \eqref{e2.23} follows from these estimates.
The proof is complete.
\end{proof}

\section{The case $\gamma=3$}

Let $f(t)$ be a smooth, decreasing and positive function in
$(0,\infty)$. Assume \eqref{e2} with $\gamma=3$; that is,
\begin{equation}\label{e3.1}
F(t)=\int_t^1f(\tau)d\tau,\quad
\lim_{t\to 0^+}F(t)=\infty,\quad
\frac{f'(t)F(t)}{(f(t))^2}=-\frac{3}{2}+O(1)t^{\beta},
\end{equation}
where $\beta>0$ and $O(1)$ denotes a bounded quantity as $t\to 0$.
This condition implies, for $t$ small,
$$
-\frac{f'(t)}{f(t)}=\Bigl(\frac{3}{2}+O(1)t^\beta\Bigr)
\frac{f(t)}{F(t)}>\frac{5}{4}\frac{f(t)}{F(t)}.
$$
Integration over $(t,t_0)$, $t_0$ small, yields
$$
\log\frac{f(t)}{f(t_0)}>\frac{5}{4}\log\frac{F(t)}{F(t_0)},\quad
\frac{f(t)}{F(t)}>\frac{f(t_0)}{(F(t_0))^{5/4}}(F(t))^{1/4}.
$$
It follows that
\begin{equation}\label{e3.2}
\lim_{t\to 0}\frac{F(t)}{f(t)}= 0.
\end{equation}
Let us rewrite \eqref{e3.1} as
\begin{equation}\label{e3.3}
(F(t))^{-1/2}\Bigl(\frac{(F(t))
^{3/2}}{f(t)}\Bigr)'=O(1)t^{\beta}.\end{equation} Integrating
by parts over $(0,t)$ and using \eqref{e3.2} we find
\begin{equation}\label{e3.4}
\frac{F(t)}{t f(t)}=\frac{1}{2}+O(1)t^{\beta}.\end{equation} Using
the latter estimate and \eqref{e3.1} again we find
\begin{equation}\label{e3.5}
\frac{t f'(t)}{f(t)}=-3+O(1)t^{\beta}.\end{equation} Let us write
\eqref{e3.5} as
$$\frac{f'(t)}{f(t)}=-\frac{3}{t}+O(1)t^{\beta-1}.$$
Integration over $(t,1)$ yields
$$\log\frac{f(1)}{f(t)}=\log t^3+O(1).$$
Therefore, we can find two positive constants $C_1$, $C_2$ such that
\begin{equation}\label{e3.6}
C_1t^{-3}<f(t)<C_2t^{-3},\ \ \ \forall t\in (0,1).\end{equation}
Since $F(t)=\int_t^1f(\tau)d\tau$, using \eqref{e3.6} we find two
positive constants $C_3$, $C_4$ such that
\begin{equation}\label{e3.7}
C_3t^{-2}<F(t)<C_4t^{-2},\ \ \ \forall t\in (0,1/2).\end{equation}

\begin{lemma} \label{lem3.1}
If \eqref{e3.1} holds and if
$\phi(\delta)$ is defined as in \eqref{e4} then we have
\begin{equation}\label{e3.8}
\frac{\phi'(\delta)}{\delta f(\phi(\delta))}=2
+O(1)(\phi(\delta))^{\beta},\end{equation}
\begin{equation}\label{e3.9}
\frac{\phi(\delta)}{\delta\phi'(\delta)}=2+O(1)(\phi(\delta))^{\beta},\end{equation}
\begin{equation}\label{e3.10}
\frac{\phi(\delta)}{\delta^2f(\phi(\delta))}=
4+O(1)(\phi(\delta))^{\beta},\end{equation}
\begin{equation}\label{e3.11}
\phi(\delta)=O(1)\delta^{1/2}.
\end{equation}
\end{lemma}

For a proof of the above lemma, see \cite[Lemma 2.3]{An}.


\begin{lemma} \label{lem3.2}
Let $\Omega\subset \mathbb R^N$, $N\ge 2$, be a bounded smooth
domain, and let $f(t)>0$ be smooth, decreasing and satisfy
\eqref{e3.1} with $\beta>0$. If $u(x)$ is a solution to problem
\eqref{e1} then
\begin{equation}\label{e3.12}
\phi(\delta)\bigl[1-C\delta(-\log\delta)]<u(x)
<\phi(\delta)\bigl[1+C\delta(-\log\delta)\bigr],\end{equation} where
$\phi$ is defined as in \eqref{e4}, $\delta$ denotes the distance
from $x$ to $\partial\Omega$, and $C$ is a suitable constant.
\end{lemma}

\begin{proof}
If $P\in\partial\Omega$ we can consider a
suitable annulus of radii $\rho$ and $R$ contained in $\Omega$ and
such that its external boundary is tangent to $\partial\Omega$ in
$P$. If $v(x)$ is the solution of problem \eqref{e1} in this annulus,
by using the comparison principle for elliptic equations (\cite{GT},
Theorem 10.1) we have $u(x)\ge v(x)$ for $x$ belonging to the
annulus. Choose the origin in the center of the annulus and put
$v(x)=v(r)$ for $r=|x|$.

We note that our assumptions imply those of Lemma 2.1. Indeed, the
condition $\int_t^1(F(\tau))^{1/2}d\tau\to\infty$ as $t\to 0$,
follows from \eqref{e3.7}. Furthermore, using \eqref{e3.7} again and
\eqref{e3.6}, for $s$ close to $0$ we have
$$
\frac{d}{ds}\Bigl[(F(s))^{-1}\int_s^1(F(t))^{1/2}dt\Bigr]=
(F(s))^{-1/2}\Bigl[\frac{f(s)\int_s^1(F(\tau))^{1/2}d\tau}
{(F(s))^{3/2}}-1\Bigr]>0.
$$
Therefore, we can use Lemma 2.1
and Corollary 2.2. By \eqref{e2.1}, we have
\begin{equation}\label{e3.13}
v(r)>\phi(R-r)-C_1\frac{\int_v^1(F(t))^{1/2}dt}
{(F(v))^{1/2}}(R-r),\ \ \ \tilde r<r<R.
\end{equation}
By using \eqref{e3.7} we find that
$$
\lim_{r\to R}\int_{v(r)}^1(F(t))^{1/2}dt=\infty=\lim_{r\to R}
v(r)\bigl(F(v(r))\bigr)^{1/2}\log(R-r)^{-1}.
$$
Using de l'H\^opital rule and \eqref{e3.4} we find
\begin{align*}
&\lim_{r\to
R}\frac{\int_v^1(F(t))^{1/2}dt}{v(F(v))^{1/2}\log(R-r)^{-1}}\\
&=\lim_{r\to
R}\frac{-(F(v))^{1/2}v'}{v'\Bigl((F(v))^{1/2}-\frac{v
f(v)}
{2(F(v))^{1/2}}\Bigr)\log(R-r)^{-1}+\frac{v(F(v))^{1/2}}{R-r}}\\
&=\lim_{r\to R}\frac{1}{\Bigl(-1+\frac{v f(v)}
{2F(v)}\Bigr)\log(R-r)^{-1}-\frac{v}{v'(R-r)}}\\
&=\lim_{r\to R}\frac{1}{O(1)v^\beta
\log(R-r)^{-1}-\frac{v}{v'(R-r)}}.
\end{align*}
By \eqref{e2.22} we have $v(r)<\phi(R-r)$. Using this inequality and
\eqref{e3.11} with $\delta=R-r$ we obtain
$$
\lim_{r\to R}v^\beta\log(R-r)^{-1}=0.
$$
Moreover, using \eqref{e2.9}, de l'H\^opital rule and \eqref{e3.4} we
find
\begin{align*}
&\lim_{r\to R}\frac{v}{-v'(R-r)}= \lim_{r\to
R}\frac{v(2F(v))^{-1/2}}{R-r}\\
&=\lim_{r\to
R}(-v')\Bigl((2F(v))^{-1/2}+v(2F(v))^{-\frac{3}{2}}f(v)\Bigr)\\
&=\lim_{r\to R}\Bigl(1+\frac{vf(v)}{2F(v)}\Bigr)=2.
\end{align*}
Hence,
\begin{equation}\label{e3.14}
\lim_{r\to R}\frac{\int_v^1(F(\tau))^{1/2}d\tau}{v(F(v))^{1/2}
\log(R-r)^{-1}}=\frac{1}{2}.
\end{equation}
 From \eqref{e3.13} and \eqref{e3.14} we find
$$
v(r)>\phi(R-r)-C_2v(r)(R-r)\log(R-r)^{-1}.
$$
By \eqref{e2.22}, $v(r)<\phi(R-r)$, hence
\begin{equation}\label{e3.15}
v(r)>\phi(R-r)\bigl(1-C_2(R-r)\log(R-r)^{-1}\bigr).
\end{equation}
For $x$ near to $P$ we have $\delta=R-r$; therefore, \eqref{e3.15}
and the inequality $u(x)\ge v(x)$ yield the left hand side of
\eqref{e3.12}.

Consider a new annulus of radii $\rho$ and $R$ containing $\Omega$
and such that its internal boundary is tangent to $\partial\Omega$
in $P$. If $w(x)$ is the solution of problem \eqref{e1} in this
annulus, by using the comparison principle for elliptic equations we
have $u(x)\le w(x)$ for $x$ belonging to $\Omega$. Choose the origin
in the center of the annulus and put $w(x)=w(r)$ for $r=|x|$. By
\eqref{e2.2} of Lemma 2.1 (with $w$ in place of $v$) we have
\begin{equation}\label{e3.16}
w(r)<\phi(r-\rho)+C_3(r-\rho)\phi'(r-\rho)\frac{\int_w^1
(F(t)^{1/2}dt}{F(w)}, \quad  \rho<r<\overline
r.
\end{equation}
The same proof used to get \eqref{e3.14} yields
$$
\lim_{r\to \rho}\frac{\int_w^1(F(t))^{1/2}dt}
{w(F(w))^{1/2}\log(r-\rho)^{-1}}=\frac{1}{2}.
$$
Hence, for $r$ near $\rho$,
\begin{equation}\label{e3.17}
\frac{\int_w^1(F(t))^{1/2}dt}{F(w)}\le
C_4(F(w))^{-1/2}w\log(r-\rho)^{-1}.
\end{equation}
Since
$\phi'=(2F(\phi))^{1/2}$, \eqref{e3.16} and \eqref{e3.17} imply
$$
w(r)<\phi(r-\rho)+C_5(r-\rho)\Bigl(\frac{F(\phi)}{F(w)}
\Bigr)^{1/2}w\log(r-\rho)^{-1}.
$$
By \eqref{e3.7} and \eqref{e2.23} (with $w$ instead of $v$) we have
$$
\Bigl(\frac{F(\phi)}{F(w)}\Bigr)^{1/2}w\le C_6\phi.
$$
Hence,
$$
w(r)<\phi(r-\rho)\bigl(1+C_7(r-\rho)\log(r-\rho)^{-1}\bigr).
$$
For $x$ near to $P$, this estimate and the inequality $u(x)\le w(x)$
yield the right hand side of \eqref{e3.12}. The lemma is proved.
\end{proof}

To state the next theorem, we define
\begin{equation}\label{e18}
H(x)=\sum_{i=1}^{N-1}\frac{-k_i}{1-k_i\delta},
\end{equation}
where $\delta=\delta(x)$ denotes the distance from $x$ to $\partial
\Omega$, and $k_i=k_i(\overline x)$ denote the principal curvatures
of $\partial\Omega$ at $\overline x$, the nearest point to $x$. We
note that in several papers, instead of $H(x)$, the function
$\frac{1}{N-1}H(x)$ is considered.

\begin{theorem} \label{thm3.3}
 Let $\Omega\subset \mathbb R^N$, $N\ge 2$, be a bounded smooth
domain, and let $f(t)>0$ be smooth, decreasing and satisfy
\eqref{e3.1}, as well as \eqref{e3}. If $u(x)$ is a solution to
problem \eqref{e1}, then
$$
\phi(\delta)\Bigl[1+\frac{1}{4}H\delta\log \delta
-C\delta(-\log
\delta)^\sigma\Bigr]<u(x)<\phi(\delta)\Bigl[1+\frac{1}{4}H\delta\log
\delta +C\delta(-\log \delta)^\sigma\Bigr],
$$
where $\phi$ is defined as in \eqref{e4}, $H=H(x)$ is defined
as in \eqref{e18},
$0<\sigma<1$ and $C$ is a suitable constant.
\end{theorem}

\begin{proof}
We look for a super-solutions of the kind
$$
w(x)=\phi(\delta)\Bigl[1+A\delta\log\delta
+\alpha\delta(-\log \delta)^\sigma\Bigr],\quad
 A=\frac{H}{4},$$
where $\alpha$ is a positive constant to be determined. We have
\begin{align*}
w_{x_i}&=\phi'\delta_{x_i}\Bigl[1+A\delta\log\delta+\alpha\delta(-\log
\delta)^\sigma\Bigr]+\phi\Bigl[A_{x_i}\delta\log\delta\\
&\quad +A\log(e\delta)\delta_{x_i}+\alpha\delta_{x_i}(-\log\delta)^\sigma
-\alpha\sigma\delta_{x_i}(-\log\delta)^{\sigma-1}\Bigr].
\end{align*}
We know that (see for example \cite[page 355]{GT})
\begin{equation}\label{e3.19}
\sum_{i=1}^N \delta_{x_i}\delta_{x_i}=1,\quad
\sum_{i=1}^N \delta_{x_i x_i}=-H.
\end{equation}
Using \eqref{e3.19} we find
\begin{align*}
\Delta w&=\phi''\Bigl[1+A\delta\log\delta+\alpha\delta(-\log \delta)^\sigma\Bigr]
-\phi' H\Bigl[1+A\delta\log\delta+\alpha\delta(-\log
\delta)^\sigma\Bigr]\\
&\quad +2\phi'\Bigl[\nabla A\cdot\nabla \delta\;
\delta\log\delta+A+A\log\delta+\alpha(-\log\delta)^\sigma-
\alpha\sigma(-\log\delta)^{\sigma-1}\Bigr]\\
&\quad +\phi\Bigl[\Delta A\; \delta\log\delta+2\nabla
A\cdot\nabla\delta\log(e\delta)+A\delta^{-1} -AH\log(e\delta)
-\alpha H(-\log\delta)^{\sigma}\\
&\quad -\alpha \sigma(-\log\delta)^{\sigma-1}\delta^{-1} +\alpha\sigma
H(-\log\delta)^{\sigma-1}+
\alpha\sigma(\sigma-1)(-\log\delta)^{\sigma-2}\delta^{-1}\Bigr].
\end{align*}
By using the equation $\phi''=-f(\phi)$, as well as \eqref{e3.8} and
\eqref{e3.10}, we find
\begin{align*}
\Delta w&=f(\phi)\Bigl\{-1-A\delta\log\delta-\alpha\delta(-\log
\delta)^\sigma -\bigl(2+O(1)\phi^\beta\bigr)\delta H
\Bigl[1+A\delta\log\delta\\
&\quad +\alpha\delta(-\log \delta)^\sigma\Bigr]
+2\bigl(2+O(1)\phi^\beta\bigr)\delta \Bigl[\nabla
A\cdot\nabla\delta\; \delta\log\delta+A
+A\log\delta\\
&\quad +\alpha(-\log\delta)^\sigma-
\alpha\sigma(-\log\delta)^{\sigma-1}\Bigr]
+\bigl(4+O(1)\phi^\beta\bigr)\delta^2 \Bigl[\Delta A\;
\delta\log\delta +A\delta^{-1}\\
&\quad +2\nabla A\cdot\nabla\delta\log(e\delta) -AH\log(e\delta) -\alpha
H(-\log\delta)^{\sigma}-\alpha
\sigma(-\log\delta)^{\sigma-1}\delta^{-1}\\
&\quad +\alpha\sigma H(-\log\delta)^{\sigma-1}+
\alpha\sigma(\sigma-1)(-\log\delta)^{\sigma-2}\delta^{-1}\Bigr]\Bigr\}.
\end{align*}
After some simplification,
\begin{align*}
\Delta w&=f(\phi)\Bigl\{-1+3A\delta\log\delta+3\alpha\delta(-\log
\delta)^\sigma -2H\delta+O(1)\delta^2\log\delta+O(1)\phi^\beta\delta\log\delta\\
&\quad +8A\delta-8 \alpha\sigma\delta(-\log \delta)^{\sigma-1}+\alpha
O(1)\phi^\beta\delta(-\log\delta)^\sigma + \alpha
O(1)\delta(-\log\delta)^{\sigma-2}\Bigr\}.
\end{align*}
Hence, since $-2H+8A=0$, for some positive constants $C_1$, $C_2$
and $C_3$ we have
\begin{equation}\label{e3.20}
\begin{split}\Delta
w&<f(\phi)\Bigl\{-1+3A\delta\log\delta
+C_1\delta^2(-\log\delta)+C_2\phi^\beta\delta(-\log\delta)\\
&\quad +\alpha\delta(-\log\delta)^{\sigma}\Bigl[3-8\sigma(-\log\delta)^{-1}+C_3(-\log\delta)^{-2}
\Bigr]\Bigr\}.\end{split}
\end{equation}
Note that \eqref{e3.11} has
been used to compare $\phi^\beta\delta(-\log\delta)^\sigma$ with
$\delta(-\log\delta)^{\sigma-2}$.

On the other hand, using Taylor's expansion we have
\begin{equation}\label{e3.21}
\begin{split}
f(w)&=f(\phi)\Bigl\{1+\phi\frac{f'(\phi)}{f(\phi)}\bigl[A\delta\log\delta+\alpha\delta(-\log\delta)^\sigma\bigr]\\
&\quad + \phi^2\frac{f''(\overline\phi)}{2f(\phi)}
\bigl[A\delta\log\delta+\alpha\delta(-\log\delta)^\sigma\bigr]^2
\Bigr\},\end{split}
\end{equation}
with $\overline\phi$ between $\phi$ and
$\phi\bigl(1+A\delta\log\delta+\alpha\delta(-\log\delta)^\sigma\bigr)$.
We consider points $x\in\Omega$ such that
\begin{equation}\label{e3.22}
-\frac{1}{2}<A\delta\log\delta+\alpha\delta(-\log\delta)^\sigma<1.
\end{equation}
This means that
$1/2<1+A\delta\log\delta+\alpha\delta(-\log\delta)^\sigma<2$;
therefore, the term $\overline\phi$ which appears in \eqref{e3.21}
satisfies $\overline\phi=\theta\phi$, with $1/2<\theta<2$. Using the
estimates \eqref{e3.5} and \eqref{e3}, by \eqref{e3.21} we find
\begin{equation}\label{e3.23}
\begin{split}
f(w)&=f(\phi)\Bigl\{1+\bigl(-3+O(1)\phi^{\beta}\bigr)
A\delta\log\delta+O(1)(\delta\log\delta)^2\\
&\quad +\alpha\delta(-\log\delta)^\sigma\Bigl[-3+O(1)\phi^{\beta}
+O(1)\alpha\delta(-\log\delta)^\sigma\Bigr]\Bigr\}.
\end{split}\end{equation}
By \eqref{e3.23}, we can take suitable positive constants $C_4$,
$C_5$, $C_6$ and $C_7$ such that
\begin{equation}\label{e3.24}
\begin{split}f(w)&< f(\phi)\Bigl\{1-3
A\delta\log\delta+C_4\phi^\beta\delta(-\log\delta)+C_5(\delta\log\delta)^{2}\\
&\quad +\alpha\delta(-\log\delta)^{\sigma}\Bigl[-3 +C_6\phi^{\beta}
+C_7\alpha\delta(-\log\delta)^\sigma\Bigr]\Bigr\}.
\end{split}
\end{equation}
By \eqref{e3.20} and \eqref{e3.24} we have
\begin{equation}\label{e3.25}
\Delta w+f(w)<0
\end{equation}
whenever
\begin{align*}
&C_1\delta^2(-\log\delta)+C_2\phi^\beta\delta(-\log\delta)
+\alpha\delta(-\log\delta)^{\sigma}\Bigl[-8\sigma(-\log\delta)^{-1}+C_3(-\log\delta)^{-2}
\Bigr]\\
&+C_4\phi^\beta\delta(-\log\delta)+C_5(\delta\log\delta)^{2}
+\alpha\delta(-\log\delta)^{\sigma}\Bigl[C_6\phi^{\beta}
+C_7\alpha\delta(-\log\delta)^\sigma\Bigr]<0.
\end{align*}
Rearranging we find
\begin{equation}\label{e3.26}
\begin{split}
&C_1\delta(-\log\delta)^{2-\sigma}+(C_2+C_4)\phi^\beta(-\log\delta)^{2-\sigma}+
C_5\delta(-\log\delta)^{3-\sigma}\\
&<\alpha\Bigl[8\sigma-C_3(-\log\delta)^{-1}-C_6\phi^\beta(-\log\delta)-
C_7\alpha\delta(-\log\delta)^{1+\sigma}\Bigr].
\end{split}
\end{equation}
Since, by \eqref{e3.11}, $\phi^\beta\le C\delta^\frac{\beta}{2}$, and
since $\sigma>0$, \eqref{e3.26} holds for $\alpha$ fixed and $\delta$
small enough.

Using Lemma 3.2 we find
$$
w(x)-u(x)\ge \phi(\delta)\bigl(-\log \delta \bigr)^{-1}\bigl[-A\delta(\log\delta)^2
+\alpha\delta(-\log \delta)^{1+\sigma} -C\delta(\log
\delta)^2\bigr].
$$
If $\alpha$ and $\delta$ are such that
\eqref{e3.22} and \eqref{e3.26} hold, define
$q=\alpha\delta(-\log\delta)^{1+\sigma}$ and decrease $\delta$
(increasing $\alpha$) so that
$\alpha\delta(-\log\delta)^{1+\sigma}=q$ until
$$
-A\delta(\log\delta)^2+q-C\delta(\log \delta)^2>0
$$
for $\delta(x)=\delta_1$. Then, applying the comparison principle to
\eqref{e3.25} and \eqref{e1} we find
$$
w(x)\ge u(x),\quad  x\in\Omega:\delta(x)<\delta_1.
$$
By a similar argument one finds a sub-solution of the kind
$$
w(x)=\phi(\delta)\Bigl(1+A\delta\log\delta
-\alpha\delta(-\log \delta)^\sigma\Bigr),
$$
where  $A$ and $\sigma$ are the same as before and $\alpha$ is a
suitable positive constant. The theorem follows.
\end{proof}

\section{The case $\gamma=\infty$}

Let $f(t)$ be a smooth, decreasing and positive function in
$(0,\infty)$. In this section we assume conditions \eqref{e7} and
\eqref{e10}. By \eqref{e7} one finds positive constants $c_1$, $c_2$,
$\ell_1$ and $\ell_2$ such that
\begin{equation}\label{e4.1}
c_1e^{\ell_1/t^\beta}<f(t)<c_2e^{\ell_2/t^\beta},\quad t>0.
\end{equation}
Similarly, by \eqref{e8} (which follows from
\eqref{e7}), one finds
\begin{equation}\label{e4.2}
c_3e^{\ell_1/t^\beta}<F(t)<c_4e^{\ell_2/t^\beta},\quad
t\in\bigl(0,\frac{1}{2}\bigr).
\end{equation}
By \eqref{e4.2}, for $m>\ell_2 2^{\beta+1}/\ell_1$, we find
\begin{equation}\label{e4.3}
\sup_{0<t< 1/2}\frac{(F(t))^\frac{2}{m}}{F(2t)}<\infty.
\end{equation}

\begin{lemma} \label{lem4.1}
If \eqref{e7} holds, we have
\begin{equation}\label{e4.4}
\frac{\phi'(\delta)}{f(\phi(\delta))}=\delta +O(1)\delta
(\phi(\delta))^\beta,
\end{equation}
where $\phi(\delta)$ is defined as in \eqref{e4}.
\end{lemma}

\begin{proof} Recall that \eqref{e7} implies \eqref{e9}. Using
\eqref{e9} and the relation
$$
-1-2\bigl[-1+O(1)t^\beta\bigr]=1+O(1)t^\beta,
$$
we have
$$
-1-2F(t)f'(t)(f(t))^{-2}=1+O(1)t^\beta.
$$
Multiplying by
$(2F(t))^{-1/2}$ we find
$$
-(2F(t))^{-1/2}-(2F(t))^{1/2}f'(t)(f(t))^{-2}=
(2F(t))^{-1/2}+O(1)t^\beta(2F(t))^{-1/2},
$$
and
\begin{equation}\label{e4.5}
\bigl((2F(t))^{1/2}(f(t))^{-1}\bigr)'=
(2F(t))^{-1/2}+O(1)t^\beta(2F(t))^{-1/2}.
\end{equation}
By \eqref{e8} we have
$$
\frac{(F(t))^{1/2}}{f(t)}=\frac{1}{(F(t))^{1/2}}
\frac{F(t)}{f(t)}=
\frac{1}{(F(t))^{1/2}}\frac{t^{\beta+1}}{\ell}\bigl(1+O(1)t^\beta\bigr).
$$
The latter estimate yields
$$
\lim_{t\to 0}(F(t))^{1/2}(f(t))^{-1}= 0.
$$
Hence,
integrating \eqref{e4.5} on $(0,s)$ we obtain
\begin{equation}\label{e4.6}
(2F(s))^{1/2}(f(s))^{-1}=\int_0^s\bigl(2F(t)\bigr)^{-1/2}dt
+O(1)\int_0^{s}t^\beta(2F(t))^{-1/2}dt.
\end{equation}
Since $t^\beta$ is increasing we have
$$
0\le \int_0^{s}t^\beta (2F(t))^{-1/2}dt
\le s^\beta\int_0^{s}(2F(t))^{-1/2}dt,
$$
and equation \eqref{e4.6} implies
$$
\frac{(2F(s))^{1/2}}{f(s)}=
\int_0^s\bigl(2F(t)\bigr)^{-1/2}dt +O(1)s^\beta\int_0^s
(2F(t))^{-1/2}dt.
$$
Putting $s=\phi(\delta)$ and recalling
that $\phi'(\delta)=(2F(\phi(\delta)))^{1/2}$, \eqref{e4.4}
follows and the lemma is proved.
\end{proof}

\begin{lemma} \label{lem4.2}
Let $\Omega\subset \mathbb R^N$, $N\ge 2$, be a bounded smooth
domain, let $f(t)>0$ be smooth, decreasing and satisfying
\eqref{e7}$. If $u(x)$ is a solution to problem \eqref{e1}$ then
\begin{equation}\label{e4.7}
\phi\bigl[1-C\delta \phi^\beta\bigr]<u(x)<\phi
\Bigl[1+C\delta\Bigl(\frac{F(\phi)}{F(2\phi)}\Bigr)^{1/2}
\phi^\beta\Bigr],
\end{equation}
where $\phi=\phi(\delta)$ is defined
as in \eqref{e4}, $C$ is a suitable constant and $\delta=\delta(x)$
denotes the distance from $x$ to $\partial\Omega$.
\end{lemma}

\begin{proof}
We proceed as in the proof of Lemma 3.2 using
the same notation. We prove first that our assumptions imply those
of Lemma 2.1. Indeed, estimate \eqref{e4.2} implies
$$
\lim_{t\to 0}\int_t^1(F(\tau))^{1/2}d\tau=\infty.
$$
To prove
the monotonicity of the function
$s\mapsto(F(s))^{-1}\int_s^1(F(t))^{1/2}dt$ for $s$ close to
$0$, we claim that
$$
\frac{d}{ds}\Bigl[(F(s))^{-1}\int_s^1(F(t))^{1/2}dt\Bigr]=
(F(s))^{-1/2}\Bigl[\frac{\int_s^1(F(\tau))^{1/2}d\tau}
{(F(s))^{3/2}(f(s))^{-1}}-1\Bigr]>0.
$$
Indeed, using \eqref{e9}, for $s$ close to $0$ we have
\begin{align*}
(F(s))^{3/2}(f(s))^{-1}
&=-\int_s^1\Bigl((F(t))^{3/2}(f(t))^{-1}\Bigr)'dt\\
&=\int_s^1(F(t))^{1/2}\Bigl(\frac{3}{2}+F(t)f'(t)(f(t))^{-2}\Bigr)dt\\
&>\frac{1}{4}\int_s^1(F(t))^{1/2}dt.
\end{align*}
The above estimate and \eqref{e4.2} yield
$$
\lim_{s\to 0}(F(s))^{3/2}(f(s))^{-1}=+\infty.
$$
Using de l'H\^opital rule and \eqref{e9} we find
$$
\lim_{s\to 0}\frac{\int_s^1(F(\tau))^{1/2}d\tau}
{(F(s))^{3/2}(f(s))^{-1}}=\lim_{s\to 0}\frac{1}
{\frac{3}{2}+F(s)(f(s))^{-2}f'(s)}=2.
$$
It follows that
$$
\frac{d}{ds}\Bigl[(F(s))^{-1}\int_s^1(F(t))^{1/2}dt\Bigr]>0,
$$
as claimed.

Now we can use Lemma 2.1 and its Corollary. By  \eqref{e2.1},
\begin{equation}\label{e4.8}
v(r)>\phi(R-r)-C\frac{\int_v^1(F(t))^{1/2}dt}
{(F(v))^{1/2}}(R-r),\quad \tilde r<r<R.
\end{equation}
By \eqref{e4.2} we have
$$
\lim_{t\to 0}t^{\beta+1}(F(t))^{1/2}=+\infty.
$$
Using de l'H\^opital rule and \eqref{e8} we find
\begin{equation}\label{e4.9}
\lim_{t\to
0}\frac{\int_t^1(F(\tau))^{1/2}d\tau}{t^{\beta+1}(F(t))^{1/2}}=
\lim_{t\to
0}\frac{1}{-(\beta+1)t^\beta+\frac{t^{\beta+1}f(t)}{2F(t)}}
=\frac{2}{\ell}.
\end{equation}
Equations \eqref{e4.8} and \eqref{e4.9} imply
$$
v(r)>\phi(R-r)-C_1(v(r))^{\beta+1}(R-r).
$$
By \eqref{e2.22}, $v(r)<\phi(R-r)$. Hence,
\begin{equation}\label{e4.10}
v(r)>\phi(R-r)\bigl[1-C_1(\phi(R-r))^\beta(R-r)\bigr].
\end{equation}
Arguing as in the proof of Lemma 3.2, one proves that \eqref{e4.10}
implies the left hand side of \eqref{e4.7}.

By \eqref{e2.2} of Lemma 2.1 (with $w$ in place of $v$) we have
\begin{equation}\label{e4.11}
w(r)<\phi(r-\rho)+C\phi'(r-\rho)\frac{\int_w^1
(F(t))^{1/2}dt}{F(w)}(r-\rho), \quad \rho<r<\tilde
r.
\end{equation}
By \eqref{e4.9} we can find a constant $C_2$ such
that
$$
\frac{\int_w^1(F(t))^{1/2}dt}{F(w)}
\le C_2\frac{1}{(F(w))^{1/2}}w^{\beta+1}.
$$
By using this estimate and the equation
$\phi'=(2F(\phi))^{1/2}$, from \eqref{e4.11} we find
\begin{equation}\label{e4.12}
w(r)<\phi+C_3(r-\rho)\Bigl(\frac{F(\phi)}{F(w)}\Bigr)^{1/2}w^{\beta+1}.
\end{equation}
By \eqref{e2.23} (with $w$ in place of $v$ and with $\epsilon=1$),
for $r$ close to $\rho$ we have $w(r)<2\phi(r-\rho)$. Hence, from
\eqref{e4.12} we find
$$
w(r)<\phi\Bigl[1+C_4(r-\rho)\Bigl(\frac{F(\phi)}{F(2\phi)}\Bigr)^{1/2}
\phi^\beta \Bigr].
$$
Proceeding as in the proof of Lemma 3.2, we obtain the right
hand side of \eqref{e4.7}. The proof is complete.
\end{proof}

\begin{theorem} \label{thm4.3}
Let $\Omega\subset \mathbb R^N$, $N\ge 2$, be a bounded smooth
domain, let $f(t)$ be smooth, decreasing and satisfying \eqref{e7}
and \eqref{e10}. If $u(x)$ is a solution to problem \eqref{e1}
then
$$
\phi\Bigl[1-\frac{1}{\ell}H\delta \phi^{\beta}
-C\delta \phi^{2\beta}\Bigr]\le
u(x)\le \phi\Bigl[1-\frac{1}{\ell}H\delta \phi^{\beta}+C\delta
\phi^{2\beta}\Bigr],
$$
where $\phi=\phi(\delta)$ is defined as in
\eqref{e4}, $H=H(x)$ is defined as in \eqref{e18}, and $C$ is a
suitable positive constant.
\end{theorem}

\begin{proof}
We look for a super-solution of the form
$$
w(x)=\phi(\delta)-A\delta \phi^{\beta+1}+\alpha\delta
\phi^{2\beta+1},\quad A=\frac{1}{\ell}H,
$$
where $\alpha$ is a positive constant to be determined. We have
$$
w_{x_i}=\phi'\delta_{x_i}-A_{x_i}\delta \phi^{\beta+1}-
A\bigl[\phi^{\beta+1}+(\beta+1)\delta\phi^{\beta}
\phi'\bigr]\delta_{x_i}
+\alpha\bigl[\phi^{2\beta+1}+(2\beta+1)\delta\phi^{2\beta}
\phi'\bigr]\delta_{x_i}.
$$
Recalling \eqref{e3.19} we find
\begin{equation}\label{e4.13}
\begin{split}
\Delta w&=\phi''-\phi' H -\Delta A\delta \phi^{\beta+1} -2\nabla
A\cdot\nabla \delta\bigl(\phi^{\beta+1}+(\beta+1)\delta \phi^{\beta}
\phi'\bigr)\\
&\quad - A\bigl[2(\beta+1)\phi^\beta\phi'+(\beta+1)\beta \delta
\phi^{\beta-1} (\phi')^2+(\beta+1)\delta
\phi^\beta\phi''\bigr]\\
&\quad +AH\bigl[\phi^{\beta+1}+(\beta+1)\delta\phi^{\beta} \phi'\bigr ]
+\alpha\bigl[2(2\beta+1)\phi^{2\beta}\phi'+(2\beta+1)2\beta\delta\\
&\quad -\phi^{2\beta-1}(\phi')^2 +(2\beta+1) \delta \phi^{2\beta}\phi''
-\bigl(\phi^{2\beta+1}+(2\beta+1)\delta\phi^{2\beta}
\phi'\bigr)H\bigr].
\end{split}
\end{equation}
Equation \eqref{e4.4} yields
\begin{equation}\label{e4.14}
\phi'=\bigl[1+O(1)\phi^{\beta}\bigr]\delta f(\phi).
\end{equation}
Since $\phi''=-f(\phi)$, by \eqref{e4.13} and \eqref{e4.14} we find
\begin{equation}\label{e4.15}
\begin{split}
\Delta w&=f(\phi)\Bigl[-1-H\delta+O(1)\delta
\phi^\beta+O(1)\frac{\phi^{\beta+1}}{f(\phi)}+O(1)\delta^3\phi^{\beta-1}f(\phi)\\
&\quad +\alpha O(1)\delta \phi^{2\beta}+\alpha
O(1)\frac{\phi^{2\beta+1}}{f(\phi)}+\alpha O(1)\delta^3
\phi^{2\beta-1}f(\phi)\Bigr].
\end{split}
\end{equation}
We claim that, for $\delta$ small,
\begin{equation}\label{e4.16}
\frac{\phi^{\beta+1}}{f(\phi)}\le \delta \phi^\beta.
\end{equation}
Rewrite \eqref{e4.16} as
$$
\frac{\phi}{\delta f(\phi)}\le 1.
$$
The latter inequality follows by the  statement
\begin{align*}
\lim_{\delta\to 0}\frac{\phi}{\delta f(\phi)}
&=\lim_{t\to 0}\frac{t(f(t))^{-1}}{\psi(t)}= \lim_{t\to
0}\frac{(f(t))^{-1}-t(f(t))^{-2}f'(t)}{(2F(t))^{-1/2}}\\
&=\lim_{t\to
0}\Bigl[\Bigl(\frac{2F(t)}{f(t)}\Bigr)^{1/2}\frac{1}{(f(t))^{1/2}}
-\frac{t}{(2F(t))^{1/2}}\frac{2F(t)f'(t)}{(f(t))^2}\Bigr]=0.
\end{align*}
In the last step we have used \eqref{e8}, \eqref{e9}, \eqref{e4.1} and
\eqref{e4.2}.

Now we claim that, for $\delta$ small,
\begin{equation}\label{e4.17}
\delta^3\phi^{\beta-1}f(\phi)\le \delta \phi^\beta.
\end{equation}
Rewrite \eqref{e4.17} as
$$
\frac{\delta^2 f(\phi)}{\phi}\le 1.
$$
The latter inequality follows by the statement
\begin{align*}
\lim_{\delta\to 0}\frac{\delta}{\phi^{1/2}(f(\phi))^{-1/2}}
&=\lim_{t\to 0}
\frac{\psi(t)}{t^{1/2}(f(t))^{-1/2}}\\
&=\lim_{t\to 0} \frac{2(2F(t))^{-1/2}}{(t
f(t))^{-1/2}-t^{1/2}(f(t))^{-\frac{3}{2}}f'(t)}\\
&=\lim_{t\to 0} \frac{\sqrt
2\Bigl(\frac{F(t)}{tf(t)}\Bigr)^{1/2} }{\frac{F(t)}{t f(t)}-
\frac{F(t)f'(t)}{(f(t))^2}}=0,
\end{align*}
where
\eqref{e8} and \eqref{e9} have been used.

Let us consider now the terms containing $\alpha$. By \eqref{e4.16},
for $\delta$ small we have
\begin{equation}\label{e4.18}
\frac{\phi^{2\beta+1}}{f(\phi)}\le  \delta
\phi^{2\beta}.\end{equation}

Finally, by \eqref{e4.17} we find
\begin{equation} \label{e4.19}
\delta^3 \phi^{2\beta-1}f(\phi)\le \delta
\phi^{2\beta}.
\end{equation}
Therefore, by \eqref{e4.15} and estimates
\eqref{e4.16}-\eqref{e4.19}, we find suitable positive constants
$M_1$, $M_2$, such that
\begin{equation}\label{e4.20}
\Delta w<f(\phi)\bigl[-1-H\delta+M_1\delta \phi^\beta+\alpha
M_2\delta \phi^{2\beta}\bigr].
\end{equation}

On the other hand, by Taylor's formula we have
\begin{equation}\label{e4.21}
f(t+\omega t)=f(t)\Bigl[1+\frac{t
f'(t)}{f(t)}\omega+\frac{1}{2}\frac{t^2 f''(\theta t)}
{f(t)}\omega^2\Bigr],
\end{equation}
 where $\theta$ is between $1$
and $1+\omega$. If $-\epsilon<\omega<\epsilon$ we can use
\eqref{e10}; using also \eqref{e7}, from \eqref{e4.21} we find
$$
f(t+\omega
t)=f(t)\Bigr[1-\frac{\ell}{t^\beta}\bigl(1+O(1)t^\beta\bigr)\omega+
O(1)\frac{1}{t^{2\beta}}(F(t))^{1/m}\omega^2\Bigr].
$$
Here $m$ is so large that \eqref{e10} and \eqref{e4.3} hold. Let
$$
\omega= -A\delta \phi^\beta+\alpha\delta\phi^{2\beta},
$$
and take $\alpha$ and $\delta_0$ so that, for
$\{x\in\Omega:\delta(x)<\delta_0\}$
\begin{equation}\label{e4.22}
-\epsilon< -A\delta
\phi^\beta+\alpha\delta\phi^{2\beta}<\epsilon.\end{equation} With
$t=\phi(\delta)$ we have $t+t\omega=w$, and
\begin{align*}
f(w)&=f(\phi)\Bigr[1-\ell\bigl(1+O(1)\phi^\beta\bigr)
\bigl(-A\delta+\alpha\delta \phi^\beta\bigr)+
O(1)\Bigl(-A\delta+\alpha\delta\phi^\beta
\Bigr)^2(F(\phi))^{1/m}\Bigr]\\
&=f(\phi)\Bigr[1+\ell A\delta-\alpha\ell\delta
\phi^{\beta}+O(1)\delta \phi^\beta +\alpha O(1)\delta \phi^{2\beta}+
O(1)\delta^2(F(\phi))^{1/m}\\
&\quad  +\alpha^2
O(1)\delta^2\phi^{2\beta}(F(\phi))^{1/m}\Bigr].
\end{align*}
Note that, using \eqref{e8}, \eqref{e4.2}, and recalling
that $m>2$ we find
\begin{align*}
0\le \lim_{\delta\to 0}\frac{\delta^2(F(\phi))^{1/m}}
{\delta \phi^\beta}=& \lim_{\delta\to 0}\frac{\delta}{\phi^\beta
(F(\phi))^{-1/m}}=\lim_{t\to 0}\frac{\psi(t)}{t^\beta
(F(t))^{-1/m}}\\
&=\lim_{t\to 0}\frac{(2F(t))^{-1/2}}{\beta
t^{\beta-1}(F(t))^{-1/m}+\frac{1}{m}t^\beta
(F(t))^{-\frac{1}{m}-1}f(t)}\\
& \le \frac{m}{\sqrt 2}\lim_{t\to
0}\frac{F(t)}{f(t)}\frac{1}{t^\beta
(F(t))^{\frac{1}{2}-\frac{1}{m}}}=0.
\end{align*}
Hence, we can find positive constants $M_3$, $M_4$, $M_5$ such that
$$
f(w)<f(\phi)\bigr[1+\ell A\delta-\alpha\ell\delta \phi^\beta
+M_3\delta \phi^\beta
+\alpha M_4\delta \phi^{2\beta}+\alpha^2 M_5\delta^2
\phi^{2\beta}(F(\phi))^{1/m}\bigr].
$$
Recalling that $H=\ell A$, by \eqref{e4.20} and the latter
inequality we have
\begin{equation}\label{e4.23}
\Delta w+f(w)<0\end{equation}
provided
$$
M_1\delta \phi^\beta+\alpha
M_2\delta \phi^{2\beta}-\alpha\ell\delta \phi^\beta+M_3\delta
\phi^\beta +\alpha M_4\delta \phi^{2\beta}+\alpha^2 M_5\delta^2
\phi^{2\beta}(F(\phi))^{1/m}<0.
$$
Rearranging we find
\begin{equation}\label{e4.24}
M_1+M_3<\alpha\bigl [\ell-(M_2+M_4)\phi^\beta-\alpha M_5\delta
\phi^\beta (F(\phi))^{1/m}\bigr].
\end{equation}
Since
$$
\lim_{\delta\to 0}\delta(F(\phi))^{1/m}
=\lim_{t\to 0}\psi(t)(F(t))^{1/m}\le
\lim_{t\to 0}t(F(t))^{\frac{1}{m}-\frac{1}{2}}=0,
$$
it follows that \eqref{e4.24}
holds for $\delta$ small and $\alpha$ large.

Using the right hand side of \eqref{e4.7} we have
$$
w-u>\phi^{\beta+1}(F(\phi))^{-1/m}\Bigl[-A\delta(F(\phi))^{1/m}
+\alpha\delta \phi^{\beta}(F(\phi))^{1/m}-
C\delta\frac{(F(\phi))^{\frac{1}{2}+\frac{1}{m}}}
{(F(2\phi))^{1/2}} \Bigr].
$$
Take $\alpha_1$ large and
$\delta_1$ small so that \eqref{e4.22} and \eqref{e4.24} hold for
$\{x\in\Omega:\delta(x)<\delta_1\}$, and define
$$
q=\alpha_1\delta_1 \phi^{\beta}(F(\phi))^{1/m}.
$$
Let us show that we can decrease $\delta$ increasing $\alpha$
according to $\alpha\delta
\phi^\beta(F(\phi))^{1/m}=q$ until
\begin{equation}\label{e4.25}
-A\delta(F(\phi))^{1/m}
+q- C\delta\frac{(F(\phi))^{\frac{1}{2}+\frac{1}{m}}}
{(F(2\phi))^{1/2}} >0
\end{equation}
for $\{x\in\Omega: \delta(x)=\delta_2\}$. Indeed, we have
$$
0\le \lim_{\delta\to 0}\delta(F(\phi))^{1/m}=\lim_{t\to
0}\psi(t)(F(t))^{1/m}\le \lim_{t\to
0}(F(t))^{-\frac{1}{2}+\frac{1}{m}}=0.
$$
Furthermore, using \eqref{e4.3} we find
$$
0\le\lim_{\delta\to 0}\delta\frac{(F(\phi))^{\frac{1}{2}
+\frac{1}{m}}}{(F(2\phi))^{1/2}}= \lim_{t\to
0}\frac{\psi(t)(F(t))^{\frac{1}{2} +\frac{1}{m}}}
{(F(2t))^{1/2}}\le \lim_{t\to 0}\frac{t(F(t))^{1/m}}
{(F(2t))^{1/2}}=0.
$$


If \eqref{e4.25} holds, then $w-u>0$ for $\delta(x)=\delta_2$. Since
$w-u=0$ on $\partial\Omega$, by \eqref{e4.23} and \eqref{e1} we have
$w-u\ge 0$ on $\{x\in\Omega: \delta(x)<\delta_2\}$. We have proved
that, for $C$ large,
$$
u(x)<\phi\Bigl[1-\frac{1}{\ell}H\delta \phi^{\beta}+C\delta
\phi^{2\beta}\Bigr].
$$
In a very similar manner, using the left hand side of \eqref{e4.7},
one finds that
$$
v=\phi-\frac{1}{\ell}H\delta \phi^{\beta+1}-\alpha \delta
\phi^{2\beta+1},
$$
satisfies $v-u\le 0$ in a neighborhood of
$\partial\Omega$ provided $\alpha$ is large enough.
The proof is complete.
\end{proof}



\begin{thebibliography}{00}

\bibitem{An} C. Anedda;
\emph{Second order boundary estimates for solutions to singular
elliptic equations}, Electronic Journal of Differential Equations,
Vol 2011 (2011), no. 90, 1-15.

\bibitem{ACP} C. Anedda, F. Cuccu and G. Porru;
\emph{Boundary estimates for solutions to singular elliptic
equations}, Le Matematiche, LX (2005), 339-352.

\bibitem{BCP}
S. Berhanu, F. Cuccu and G. Porru; \emph{On the boundary behaviour,
including second order effects, of solutions to singular elliptic
problems}, Acta Math. Sin. (Engl. Ser.), 23 (2007), 479-486.

\bibitem{CRT} M. G. Crandall, P. H. Rabinowitz and L. Tartar;
\emph{On a Dirichlet problem with a singular nonlinearity}, Comm.
Part. Diff. Eq., 2 (1997), 193-222.

\bibitem{CGP} F. Cuccu, E. Giarrusso and G. Porru;
 \emph{Boundary behaviour for solutions of elliptic singular equations
with a gradient term}, Nonlinear Analysis 69 (2008), 4550-4566.

\bibitem{GR} M. Ghergu and V. R\u{a}dulescu;
\emph{On a class of sublinear singular elliptic problems with
convection term}, J. Math. Anal. Appl., 311 (2005), 635-646.

\bibitem{GT} D. Gilbarg and N. S. Trudinger;
\emph{Elliptic Partial Differential Equations of Second Order},
 Springer Verlag, Berlin (1977).

\bibitem{Ka} B. Kawohl;
\emph{On a class of singular elliptic equations}, Progress in PDE:
elliptic and parabolic problems, Pitman Res. Notes Math. Series,
266, Longman (1992), 156-163.

\bibitem{LM1} A. Lazer and P. J. McKenna;
\emph{On a singular nonlinear elliptic boundary value problem},
Proc. American. Math. Soc., 111 (1991), 721-730.

\bibitem{LM2}
A. Lazer and P. J. McKenna; \emph{Asymptotic behaviour of solutions
of boundary blow-up problems}, Differential and Integral Equations,
7 (1994), 1001-1019.

\bibitem{ZC} Z. Zhang and J. Cheng;
\emph{Existence and optimal estimates of solutions for singular
nonlinear Dirichlet problems}, Nonlinear Analysis, 57 (2004),
473-484.

\bibitem{ZY} Z. Zhang and J. Yu;
\emph{On a singular nonlinear Dirichlet problem with a convection
term}, SIAM J. Math. Anal. 32 (2000), 916-927.

\end{thebibliography}


\end{document}