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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2012 (2012), No. 136, pp. 1--11.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2012 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2012/136\hfil Boundary behavior of large solutions ]
{Boundary behavior of large solutions for semilinear elliptic equations
 in borderline cases}

\author[Z. Zhang \hfil EJDE-2012/136\hfilneg]
{Zhijun Zhang}  % in alphabetical order

\address{Zhijun Zhang \newline
School of Mathematics and Information Science, Yantai University,
Yantai, Shandong, 264005, China}
\email{zhangzj@ytu.edu.cn, chinazjzhang@yahoo.com.cn}

\thanks{Submitted June 30, 2012. Published August 19, 2012.}
\thanks{Supported by grant 10671169 from NNSF of China
and  2009ZRB01795 from NNSF  of \hfill\break\indent Shandong Province}
\subjclass[2000]{35J55, 35J60, 35J65}
\keywords{Semilinear elliptic equations;  boundary blow-up;
 boundary  behavior;\hfill\break\indent  borderline cases}

\begin{abstract}
 In this article, we analyze  the boundary  behavior of solutions to
 the  boundary blow-up  elliptic problem
 $$
 \Delta u =b(x)f(u), \quad u\geq 0,\; x\in\Omega,\; u|_{\partial \Omega}=\infty,
 $$
 where $\Omega$ is  a bounded domain with smooth boundary in $\mathbb{R}^N$,
 $f(u)$   grows slower than any $u^p$ ($p > 1$) at infinity, and
 $b \in C^{\alpha}(\bar{\Omega})$ which is non-negative in $\Omega$
 and positive near $\partial\Omega$, may be vanishing on the boundary.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{proposition}[theorem]{Proposition}
\newtheorem{remark}[theorem]{Remark}
\newtheorem{definition}[theorem]{Definition}
\allowdisplaybreaks

\section{Introduction}

In this article, we consider the  boundary  behavior of solutions
 to the   boundary blow-up elliptic problem
 \begin{equation}\label{e1.1}
 \Delta u =b(x)f(u), \ u\geq 0,  \quad x\in \Omega,\quad  u|_{\partial \Omega}=\infty,
  \end{equation}
  where the last condition  means that $u(x) \to \infty$ as
$d(x)=\operatorname{dist}(x, \partial \Omega)  \to 0$, $\Omega$ is a
bounded domain with smooth boundary in $\mathbb{R}^N$, $f$ satisfies
\begin{itemize}
\item[(F1)]    $f\in C[0, \infty) \cap C^1(0, \infty)$,
 $f(0)=0$  and $f(s)$ is increasing on $(0, \infty)$;

\item[(F2)] the Keller-Osserman  (\cite{KE}, \cite{OS})
 condition
$$
   \Theta (r):= \int_r^\infty
\frac{ds}{\sqrt{2F(s)}}<\infty,  \quad \forall r>0,\quad
 F(s)=\int_0^s f(\tau) d\tau;
$$
\end{itemize}
the function $b$ satisfies
\begin{itemize}
\item[(B1)] $b \in C^{\alpha}(\bar{\Omega})$,
  is non-negative  in $\Omega$ and positive near $\partial\Omega$.
\end{itemize}
The model problem \eqref{e1.1}  arises from many branches of
mathematics and has generated a good deal of research, see, for
instance, \cite{AGQ}-\cite{BM}, \cite{CD}-\cite{DDGR},
\cite{KE}-\cite{LM},
\cite{OS}-\cite{Z}  and the references therein.

 When $b\equiv 1$ in $\Omega$ and $f$ satisfies (F1),
 it is well-known that \eqref{e1.1}
    has  one  solution $u\in C^2(\Omega)$ if and only if (F2) holds.
 Moreover, the blow-up rate of $u(x)$ near $\partial \Omega $ can be described
 by (see, e.g.,  \cite{BM} and \cite[Theorem 6.8]{DU})
\begin{equation}\label{e1.2}
\frac{\Theta (u(x))}{d(x)} \to 1 \quad\text{as }  d(x) \to 0.
\end{equation}
Moreover, if one assumes  that
\begin{equation}\label{e1.3}
\liminf_{r\to\infty} \frac {\Theta (\lambda r)}{\Theta (r)}> 1,\quad  
\forall  \lambda\in  (0, 1),
\end{equation}
then it holds (see  \cite{BM})
\begin{equation}\label{e1.4}
\frac{u(x)}{\phi(d(x))} \to 1 \quad\text{as }  d(x) \to 0,
\end{equation}
where  $\phi $ is the inverse of $\Theta$; i.e.,  $\phi$ satisfies
\begin{equation}\label{e1.5}
 \int_{\phi(t)}^\infty \frac{ds}{\sqrt{2F(s)}}=t, \quad \forall t>0.
\end{equation}
 However, there are  less results for the boundary behavior of
the solution to problem \eqref{e1.1} under the condition that
\begin{equation}\label{e1.6}
\lim_{r\to\infty} \frac {\Theta (\lambda r)}{\Theta (r)} =1,\quad
\forall  \lambda\in  (0, 1).
\end{equation}
When  $f$ satisfies
\begin{itemize}
\item[(A)]    $f$ is locally Lipschitz continuous and non-negative
on $[0, \infty)$,   and $f(s)/s$ is increasing on $(0,\ \infty)$;

\item[(B)] $f(s) = C_1^2s(\ln s)^{2\alpha}+
 C_2 s(\ln s)^{2\alpha-1}(1 + o(1))$  as
    $s \to \infty $
with $ C_1>0$, $\alpha>1$ and $C_2\in \mathbb{R}$,
\end{itemize}
C\^{i}rstea and Du \cite{CD} first showed that problem \eqref{e1.1}
has a unique  solution $u$ satisfying
\begin{equation}\label{e1.7}
\lim_{d(x)\to\infty} \frac {u(x)} {\exp \big((C_1(\alpha-1)
K(d(x)))^{-1/(\alpha-1)}\big)} =\exp (\xi_0),
\end{equation}
 where
 \begin{equation}\label{e1.8}
  \xi_0=\frac {1}{2}-\frac {C_2}{2\alpha
 C_1^2}.
 \end{equation}
 Then they  extended the above result
 to  weight $b$ which can be vanishing on the boundary.

 It is  worthwhile to point out that   \eqref{e1.7}
depends not only on $C_1^2s(\ln s)^{2\alpha}$ but
also on the lower term $C_2 s(\ln s)^{2\alpha-1}$ in (B). This is
completely different from the case $f(s) = s^p[C_1 + o(1)]$
as $ s \to \infty $  for some $p > 1$, since problem
\eqref{e1.1} has a unique positive solution $u$ which satisfies
$$
\lim_{d(x)\to 0}u(x)(d(x))^{2/(p-1)}=
\Big(\frac {2(p+1)}{C_1(p-1)^2}\Big)^{1/(p-1)}
$$
in such a situation and $b\equiv 1$ in $\Omega$ (see  \cite{BM}).

On the other hand, when $b\equiv 1$ in $\Omega$, $f$ satisfies
(F1), (F2) and the conditions that
\begin{itemize}
\item[(F03)] there exists $\alpha>1$  such that
$$
\frac {2F(s) f'(s)}{f^2(s)}=1-(\alpha +o(1))(\ln s)^{-1}\quad\text{as }
  s\to \infty;
$$

\item[(F04)]  there exist $ \theta_0\in (0, 1)$ and $S_0>1$ such that
$$
\theta f(s)\geq f( \theta s),\quad \forall \theta
\in (\theta_0, 1),\; \forall s>S_0;
$$

\item[(F05)]  there exist $C_0>0$  and  $S_1\geq S_0$ such that
$$
\frac {s^2 |f''(\theta s)|}{f''(s)}\leq C_0 (\ln s)^{-1},\quad \forall s>S_1,\
 \forall\, \theta \in (1/2, 2),
$$
\end{itemize}
 Anedda and  Porru \cite{AP}
showed that for any $\varepsilon>0$ there is $C_\varepsilon > 0$
 such that the solution of problem \eqref{e1.1} satisfying
 \begin{align*}
& 1+ \frac {(\alpha- 1)(N-1)}{ 2(2\alpha-1)} K(x)d(x)-\varepsilon
d(x)- C_\varepsilon d^2(x) \\
&< \frac {u(x)}{ \phi(d(x))} \\
&< 1+ \frac {(\alpha- 1)(N-1)}{ 2(2\alpha-1)} K(x)d(x)+\varepsilon
d(x)+ C_\varepsilon d^2(x),
\end{align*}
where $ K(x)$ is the mean curvature of the surface
 $\{x\in \Omega: d(x) = {\rm constant} \}$.

 We also note that an example which satisfies the above requirements
is the following
$$
f(s) = 0,\quad  s\in [0, 1],\quad
f (s) = s (\ln s)^{2\alpha}, \quad s>1, \quad \alpha>1.
$$
Inspired by  the above works, in this article,  we analyze
the boundary behavior  of  solutions to problem \eqref{e1.1}
for more general $f$ which satisfies  the condition \eqref{e1.6}.
In particular,  we consider functions $f$ which satisfy  (F1),
(F2) and the following conditions  that
\begin{itemize}
\item[(F3)] there exist  two  functions
$f_1\in C^1[S_0,\ \infty)$ for some large $S_0>0$ and $f_2$  such
that
$$
f(s): =f_1(s)+f_2(s),\quad s\geq S_0;
$$

\item[(F4)]
 \begin{equation}\label{e1.9}
\frac {f_1'(s)s}{f_1(s)}:=1+g(s), \quad s\geq S_0,
\end{equation}
 with   $g\in C^1[S_0,\ \infty)$ satisfying
 \begin{gather}\label{e1.10}
g(s)>0,\quad s\geq S_0,\quad \lim_{s\to \infty}g(s)=0,\\
\label{e1.11}
\lim_{s\to \infty}\frac {sg'(s)}{g(s)}=0, \quad
\lim_{s\to \infty}\frac {sg'(s)}{g^2(s)}=C_g\in \mathbb{R}, \quad
\lim_{s\to \infty} \frac {\sqrt{\frac {s}{f_1(s)}}}{g(s)}=0;
 \end{gather}

\item[(F5)] either there exists a constant $E_1\neq 0$
  such that
 \begin{equation}\label{e1.12}
\lim_{s\to \infty}\frac {f_2(s)}{ g(s)f_1(s)}=E_1
\end{equation}
or
\begin{equation}\label{e1.13}
\lim_{s\to \infty}\frac {f_2(s)}{ g(s)f_1(s)}=0
\end{equation}
and there exists a constant   $\mu\leq 1$ such that
\begin{equation}\label{e1.14}
\lim_{s\to \infty}\frac{f_2(\xi s)}{f_2(s)}=\xi^\mu,\quad \forall \xi>0.
\end{equation}
\end{itemize}

 Our main result is stated using the assumption
\begin{itemize}
  \item[(B2)]  There exist $k\in \Lambda$ and a positive
  constant $b_0$ such that
$$
 \lim_{d(x) \to 0 } \frac{b(x)}{(k(d(x)))^{2}}=b_0^2,
$$
where  $\Lambda$ denotes the set of all positive non-decreasing
functions in $C^1(0,\delta_0)$ ($\delta_0>0$) which satisfy
\begin{equation}\label{e1.15}
\lim_{t \to 0^+} \frac
{d}{dt}\left(\frac{K(t)}{k(t)}\right): = D_k\in [0, \infty),\ \
K(t)=\int_0^t k(s)ds,
\end{equation}
\end{itemize}


\begin{theorem}\label{thm1.1}
Let   $f$ satisfy {\rm (F1)--(F5)}.
If $b$  satisfies {\rm (B1)--(B2)},
then for any  solution $u$ of problem \eqref{e1.1},
\begin{equation}\label{e1.16}
    \lim_{d(x)\to 0}\frac {u(x)}{\psi(b_0K(d(x)))}=\exp (\xi_0),
     \end{equation}
where
    \begin{equation}\label{e1.17}
\begin{gathered}
\xi_0=\frac {1}{2}-E_2-(1-D_k)\big(\frac {1}{2}+C_g\big),\\
 E_2= \begin{cases}
E_1 &\text{if \eqref{e1.12} holds};\\
0, &\text{if \eqref{e1.13} and \eqref{e1.14} hold},
\end{cases}
\end{gathered}
\end{equation}
and $\psi$ is  the unique solution of the problem
\begin{equation}\label{e1.18}
 \int_{\psi(t)}^\infty \frac {ds} {\sqrt{sf_1(s)}}=t,  \quad \forall t>0.
\end{equation}
\end{theorem}

 \begin{remark}\label{rmk1.1} \rm
 (F3),  \eqref{e1.10}, and  \eqref{e1.12} or \eqref{e1.13} imply
$$
\lim_{s\to \infty} \frac {f_2(s)}{f(s)}=0,\quad
\lim_{s\to\infty}\frac {f_1(s)}{f(s)}=1.
$$
\end{remark}

 \begin{remark}\label{rmk1.2} \rm
Some basic examples which satisfy all our requirements are the
following:
\begin{itemize}
\item[(1)]  $f_1(s)=C_1^2s(\ln s)^{2\alpha}$ in (F3),
where  $\alpha>1$,
\begin{gather*}
g(s)=2\alpha (\ln s)^{-1}; \quad \lim_{s\to \infty}
\frac {\sqrt{\frac {s}{f_1(s)}}}{g(s)}=
 \frac {1}{2\alpha C_1}\lim_{s\to \infty}(\ln s)^{-(\alpha-1)}=0;
\\
 \frac {sg'(s)} {g^2(s)}\equiv C_g=-\frac {1}{2\alpha};\quad
 \lim_{s\to \infty}\frac {f_2(s)}{ g(s)f_1(s)}=\frac {1}{2\alpha C_1^2}\lim_{s\to
\infty}\frac {f_2(s)}{s(\ln s)^{2\alpha-1}}=E_2;
\\
\psi(t)=\exp\big(C_1(\alpha-1) t\big)^{-1/(\alpha-1)}.
\end{gather*}
In particular, when $f_2(s)=C_2 s^\mu (\ln s)^\beta$ with
$\beta\leq 2\alpha-1$,  $E_1=0$ for $\mu<1$ or $\mu=1$ and
$\beta< 2\alpha-1$, and $E_1=\frac {C_2}{2\alpha C_1^2}$ for $\mu=1$ and
$\beta= 2\alpha-1$.

\item[(2)]   $f_1(s)=C_1^2 s e^{(\ln s)^{q}}$
 in (F3), where $q\in (0, 1)$,
\begin{gather*}
g(s)=q(\ln s)^{-(1-q)};\quad
\lim_{s\to \infty} \frac {\sqrt{\frac {s}{f_1(s)}}}{g(s)}=
 \frac {1}{q C_1}\lim_{s\to \infty}
\frac { \exp (-\frac {1}{2}(\ln s)^q)}{(\ln s)^{-(1-q)}}=0;
\\
\lim_{s\to \infty}\frac {sg'(s)}{g^2(s)}=-\frac {1-q}{q} \lim_{s\to \infty}
 (\ln s)^{-q}=C_g=0;
\\
\lim_{s\to \infty}\frac {f_2(s)}{ g(s)f_1(s)}=\frac {1}{q C_1^2}\lim_{s\to
\infty}\frac {f_2(s)}{s(\ln s)^{-(1-q)}\exp ((\ln s)^q)}=E_2;
\\
\int_{\ln (\psi(t))}^\infty\exp(-s^q/2)ds=C_1 t.
\end{gather*}

\item[(3)] $f_1(s)=C_1^2 s(\ln s)^{2}(\ln (\ln s))^{2\alpha}$ in (F3),
 where  $\alpha>1$,
\begin{gather*}
g(s)=2(\ln s)^{-1}\big(1+\alpha (\ln (\ln s))^{-1}\big);
\\
\lim_{s\to \infty}  \frac {\sqrt{\frac {s}{f_1(s)}}}{g(s)}=
 \frac {1}{2 C_1}\lim_{s\to \infty}\frac {(\ln (\ln s))^{-\alpha}}{1+\alpha(\ln
(\ln s))^{-1}}=0;
\\
 \lim_{s\to \infty} \frac {sg'(s)}{g^2(s)}
=-\lim_{s\to \infty} \frac {1+\alpha(\ln (\ln s))^{-1}
+\alpha(\ln (\ln s ))^{-2}} {2(1+\alpha(\ln (\ln s ))^{-1})^2}=C_g=-\frac {1}{2};
\\
\lim_{s\to \infty}\frac {f_2(s)}{ g(s)f_1(s)}
=\frac {1}{2 C_1^2}\lim_{s\to \infty}\frac
{f_2(s)}{s \ln s (\ln (\ln s))^{2\alpha}(1+\alpha(\ln (\ln s
))^{-1})}=E_2;
\\
\psi(t)=\exp\big(\exp\big(C_1(\alpha-1)t\big)^{-1/(\alpha-1)}\big).
\end{gather*}
\end{itemize}
\end{remark}

 \begin{remark}\label{rmk1.3} \rm
 When  $f$ further satisfies the condition $f(s)/s$ being increasing on
 $(0, \infty)$, in a  similar  proof in
\cite{CD}, problem \eqref{e1.1}  has a unique solution.
\end{remark}

\begin{remark}\label{rmk1.4} \rm
 For the existence of the minimal solution to problem \eqref{e1.1}, see \cite{LA}.
\end{remark}

\begin{remark}\label{rmk1.5}  \rm
 For each $k\in \Lambda$,  $D_k\in [0, 1]$ and
\begin{equation}\label{e1.19}
\lim_{t \to 0^+} \frac{K(t)}{k(t)}=0,\quad
\lim_{t \to 0^+} \frac{K(t)k'(t)}{k^2(t)}=1-\lim_{t
\to 0^+} \frac {d}{dt}\Big(\frac{K(t)}{k(t)}\Big)=1-D_k.
\end{equation}
\end{remark}

\section{Preliminaries}

Our approach relies on Karamata regular variation theory established
by Karamata in 1930 which is a basic tool in stochastic process
(see,  for instance,   Bingham,  Goldie and Teugels \cite{BGT},
 Maric \cite{MA} and the references therein.),
and has been applied to study the asymptotic
behavior of solutions to differential equations and problem
\eqref{e1.1} (see Maric \cite{MA}, C\^{i}rstea and R\v{a}dulescu
\cite{CR}, R\v{a}dulescu \cite{RAD}, C\^{i}rstea and Du \cite
{CD}, the authors \cite{Z} and the references therein.).
In this section, we present some bases of Karamata regular variation
theory.

\begin{definition}\label{def2.1}\rm
A positive measurable function $f$ defined on $[a,\infty)$, for some $a>0$,
is called \emph{regularly varying at infinity} with index $\rho$,
 written $f \in RV_\rho$, if for each $\xi>0$ and some $\rho \in \mathbb{R}$,
\begin{equation}\label{e2.1}
\lim_{t \to \infty} \frac{f(\xi t)}{f(t)}= \xi^\rho.
\end{equation}
\end{definition}

In particular, when $\rho=0$, $f$ is called
\emph{slowly varying at infinity}.
Clearly, if $f\in RV_\rho$, then $L(t):\ =f(t)/{t^\rho}$ is slowly
varying at infinity.

Some basic  examples of slowly varying functions at infinity are
\begin{itemize}
\item[(i)]  every measurable function on $[a, \infty)$ which has a
positive limit at infinity;

\item[(ii)]  $(\ln t)^q$ and $\big(\ln (\ln t)\big)^q$,   $q\in \mathbb{R}$;
\item[(iii)]  $ e^{(\ln t)^q}$, $0<q<1$.
\end{itemize}
We also say that a positive measurable function $g$ defined on
$(0,a)$  for some $a>0$,  is \emph{regularly varying  at zero} with
index $\rho$ (and denoted by  $g \in RVZ_\rho$) if
$t\to g(1/t)$ belongs to $RV_{-\rho}$.

\begin{proposition}[Uniform convergence theorem]\label{pro2.1}
If $f\in RV_\rho$, then   \eqref{e2.1}   holds uniformly for
$\xi \in [c_1, c_2]$ with $0<c_1<c_2$. Moreover, if $\rho<0$, then uniform
convergence holds on intervals of the form $(a_1, \infty)$ with $a_1>0$;
if $\rho>0$, then uniform convergence holds on intervals $(0, a_1]$ provided $f$
is bounded on $(0, a_1]$ for all $a_1>0$.
\end{proposition}

\begin{proposition}[Representation theorem]\label{prop2.2}
 A function $L$ is slowly varying at infinity if and only if it may be written
in the form
\begin{equation}\label{e2.2}
L(t)=\varphi(t)  \exp \Big( \int_{a_1}^t \frac {y(\tau)}{\tau}
d\tau \Big), \quad  t \geq a_1,
\end{equation}
for some $a_1\geq a$, where the functions $\varphi$ and $y$ are
measurable and for $t \to \infty$, $y(t)\to 0$ and
$\varphi(t)\to c_0$, with $c_0>0$.
\end{proposition}

We say that
\begin{equation}\label{e2.3}
 \hat{L}(t)=c_0  \exp \Big( \int_{a_1}^t\frac {y(\tau)}{\tau} d\tau \Big), \quad
 t \geq a_1,
 \end{equation}
is \emph{normalized} slowly varying  at infinity and
 \begin{equation}\label{e2.4}
  f(t)=t^\rho\hat{L}(t), \quad  t \geq a_1,
\end{equation}
is \emph{normalized} regularly varying at infinity with
 index $\rho$  (and written $f\in NRV_\rho$).

 A function $f\in RV_\rho$ belongs to $NRV_\rho$ if and only  if
\begin{equation}\label{e2.5}
 f\in C^1[a_1, \infty),\text{ for  some $a_1>0$  and } \lim_{t \to \infty}
  \frac{tf'(t)}{f(t)}=\rho.
  \end{equation}
Then, we see that $f_1\in NRV_1$, $f_2\in RV_\mu$, $f\in RV_1$ and
$g$ is normalized slowly varying  at infinity   in (F3)-(F5).

 Similarly, $g$ is called   \emph{normalized} regularly
varying at zero with index $\rho$, and denoted by
$g \in NRVZ_\rho$, if $t\to g(1/t)$ belongs to $NRV_{-\rho}$.

\begin{proposition}\label{prop2.3}
If  functions $L, L_1$ are slowly varying at infinity, then
\begin{itemize}
 \item[(i)]   $L^\rho$ (for every $\rho\in \mathbb{R}$),
   $L\circ L_1$
 (if $L_1(t)\to \infty$ as $t\to \infty$) , are also slowly varying at infinity.

\item[(ii)]    For every $\rho >0$ and $t\to \infty$,
$$
t^{\rho} L(t)\to \infty, \quad t^{-\rho} L(t)\to 0.
$$

 \item[(iii)]   For $\rho\in\mathbb{R}$ and $t\to \infty$,
$\ln (L(t))/{\ln t}\to 0$ and $\ln (t^\rho L(t))/{\ln t}\to \rho$.
\end{itemize}
\end{proposition}

Our  results in the section are summarized as follows.

\begin{lemma}[{\cite[Lemma 2.1]{Z}}] \label{lem2.1}
 Let $k\in \Lambda$.
\begin{itemize}
\item[(i)]  When $D_k\in (0, 1)$,
$k$ is \emph{normalized} regularly varying at zero with  index $(1-D_k)/{D_k}$;

\item[(ii)] when $D_k=1$, $k$  is   normalised slowly varying  at  zero;

\item[(iii)]  when $D_k=0$, $k$   grows faster  than any $t^p$ ($p > 1$)  near  zero.
\end{itemize}
\end{lemma}

Denote
\begin{equation}\label{e2.6}
\Theta(r)=\int_r^\infty\frac {ds}{\sqrt{2F(s)}},\quad
\Theta_1(r)=\int_r^\infty\frac {ds}{\sqrt{s f_1(s)}}, \quad r>0.
\end{equation}
Then
\begin{equation}\label{e2.7}
 \Theta'(r)=-\frac {1}{\sqrt{2F(r)}},\quad
  \ \Theta_1'(r)=-\frac {1}{\sqrt{r f_1(r)}},\quad r>0.
\end{equation}

\begin{lemma}\label{lem2.2}
 Under the hypotheses in Theorem \ref{thm1.1}:
\begin{itemize}
\item[(i)]
$$
\lim_{r\to\infty} \frac {\Theta (\lambda r)}{\Theta ( r)} =
\lim_{r\to\infty} \frac {\Theta_1 (\lambda r)}{\Theta_1 (
r)}=1,\quad \forall  \lambda\in  (0, 1);
$$

\item[(ii)]
$$
\lim_{r\to \infty}  \frac {(r/ f_1(r))^{1/2} }
 {\Theta_1(r)g(r)}=\frac {1}{2}+C_g;
$$

\item[(iii)]
$$
\lim_{r\to \infty} \frac {\frac {f_1(\xi r)} {\xi f_1(r)}-1}{g(r)}= \ln \xi
$$
uniformly for $\xi \in [c_1, c_2]$ with $0<c_1<c_2$;

\item[(iv)]
 $$ \lim_{r\to \infty}\frac {f_2(\xi r)}{\xi g(r)f_1(r)}=E_2
$$
uniformly for $\xi \in [c_1, c_2]$ with
$0<c_1<c_2$.
\end{itemize}
\end{lemma}

\begin{proof}
 (i)   By  $f, f_1\in RV_1$ and the l'Hospital's rule, we
have
\begin{gather*}
\lim_{r\to\infty} \frac {F (\lambda r)}{F (r)}
=\lambda\lim_{r\to\infty} \frac {f (\lambda r)}{f (r)}=\lambda^2,
\\
 \lim_{r\to\infty} \frac {\Theta (\lambda r)}{\Theta (r)}
=\lambda\lim_{r\to\infty} \frac {\Theta' (\lambda r)}{\Theta' (r)}
=\lambda\lim_{r\to\infty} \Big (\frac {F (\lambda r)}{F (r)}\Big)^{-1/2}=1;
\\
 \lim_{r\to\infty} \frac {\Theta_1 (\lambda r)}{\Theta_1 (r)}
=\lambda \lim_{r\to\infty} \frac {\Theta_1' (\lambda r)}{\Theta_1' (r)}
=\lambda\lim_{r\to\infty} \Big (\frac
{\lambda f_1 (\lambda r)}{f_1 ( r)}\Big)^{-1/2}=1.
\end{gather*}

(ii)   By   \eqref{e1.11} and the l'Hospital's rule, we obtain
\begin{align*}
& \lim_{r\to \infty}
 \frac {\big(\frac {r}{f_1(r)}\big)^{1/2} } {\Theta_1(r)g(r)}\\
&=\lim_{r\to \infty} \frac {(g(r))^{-1}
\big(\frac {r}{f_1(r)}\big)^{1/2} } {\Theta_1(r)}\\
&= \lim_{r\to \infty}  \frac {-(g(r))^{-2}g'(r)
\big(\frac {r}{f_1(r)}\big)^{1/2}+\frac {1}{2}(g(r))^{-1}
  \big(\frac {r}{f_1(r)}\big)^{-1/2}
 \frac {f_1(r)-r f_1'(r)}{f_1^2(r)}} {-(r f_1(r))^{-1/2}}
\\
&= \lim_{r\to \infty}\Big( \frac {1}{2g(r)}
 \frac {r f_1'(r)-f_1(r)}{f_1(r)}+\frac {r g'(r)}{g^2(r)}
 \Big)=\frac {1}{2}+C_g.
\end{align*}

(iii)   When $\xi = 1$, the result is obvious. Let $\xi \neq 1$.
By $f_1\in RV_1$, one can see that
$$
\frac {f_1(\xi r)}{\xi f_1(r)}-1 =
\exp \big(\int_r^{\xi r}\frac {g(\tau)}{\tau}d\tau \big)-1.
$$
 It follows by $g\in NRV_0$ and Proposition  \ref{prop2.2} that
$$
\lim_{r\to\infty}\frac {g(r \nu)}{\nu}=0,\quad
\lim_{r\to\infty} \frac {g(r \nu)}{g(r)}=1
$$
uniformly with respect to $\nu\in [c_1, c_2]$.
So
\begin{gather*}
\lim_{r\to\infty}\int_r^{\xi r}\frac {g(\tau)}{\tau}d\tau =
\lim_{r\to\infty} \int_1^{\xi}\frac {g(r \nu)}{\nu} d\nu=0,
\\
\lim_{r\to\infty} \int_1^{\xi}\frac {g(r \nu)}{g(r)\nu} d\nu=
\int_1^{\xi}\nu^{-1}d\nu=\ln \xi.
\end{gather*}
Since
$e^s-1\cong s$ as $s\to 0$, this leads to
$$
\frac {f_1(\xi r)}{\xi f_1(r)}-1\cong g(r) \ln \xi \quad\text{as }
r\to \infty
$$
uniformly for $\xi \in [c_1, c_2]$ with $0<c_1<c_2$ by
 Proposition \ref{prop2.2}.

(iv) Note that
$$
\lim_{r\to \infty}\frac {f_2(\xi r)}{\xi
g(r)f_1(r)}=\lim_{r\to \infty}\frac {f_2(\xi r)}{\xi
f_2(r)}\lim_{r\to \infty}\frac {f_2(r)}{ g(r) f_1(r)}.
$$
When \eqref{e1.13} and \eqref{e1.14} hold,
$$
\lim_{r\to \infty}\frac {f_2(\xi r)}{\xi g(r)f_1(r)}=0.
$$
 When \eqref{e1.12} holds. Let
$$
\frac {f_2(s)}{ g(s)f_1(s)}-E_1=h(s)\quad \text{with }
\lim_{s\to \infty}h(s)=0.
$$
It follows by $g\in NRV_0$  and  $f_1\in NRV_1$ that
$$
\lim_{r\to \infty}\frac {f_2(\xi r)}{
f_2(r)}=\lim_{r\to \infty}\frac {f_1(\xi r)}{ f_1(r)}\frac
{g(\xi r)}{g(r)} \frac {E_1+h(\xi s)}{E_1+h(s)} =\xi;
$$
thus
$$
\lim_{r\to \infty}\frac {f_2(\xi r)}{\xi g(r)f_1(r)}=E_1.
$$
\end{proof}

\begin{lemma}\label{lem2.3} 
Under the hypotheses of Theorem \ref{thm1.1}, let
 $\psi$ be  the solution to the problem
$$
\int_{\psi(t)}^\infty \frac {ds}{\sqrt{s f_1(s)}}=t,\quad \forall  t>0.
$$
Then
\begin{itemize}
\item[(i)] $-\psi'(t)=\sqrt{\psi(t)f_1(\psi(t))}$,
 $\psi(t)>0$, $t>0$, $\psi(0):=\lim_{t\to 0^+}\psi(t)=\infty$,
$\psi''(t)=\frac {1}{2}\big(f_1(\psi(t))+\psi(t)f_1'(\psi(t))\big)$,
$t>0$;

\item[(ii)] 
$$\lim _{t\to 0}\big(g(\psi(t))\big)^{-1}\Big(\frac {1}{2}\big( 1+\frac
{\psi(t)f_1'(\psi(t))}{f_1(\psi(t))}\big)-\frac {f_1(\xi
\psi(t))}{\xi f_1(\psi(t))}\Big)=\frac {1}{2}-\ln \xi;
$$

\item[(iii)]  
$$\lim _{t\to 0} \frac  {\sqrt{\psi(t)f_1(\psi(t))}}
 {t g(\psi(t))f_1(\psi(t))}=\frac {1}{2}+C_g;
$$

\item[(iv)]  
$$
\lim _{t\to 0} \frac {f_2(\xi\psi(t))}{\xi g(\psi(t))f_1(\psi(t))}=E_2
$$
 uniformly for $\xi \in [c_1, c_2]$ with $0<c_1<c_2$.
\end{itemize}
\end{lemma}

\begin{proof}
 By  the definition of $\psi$ and a
direct  calculation, we can show (i).
Statements (ii)--(iv)  follow by Lemma \ref{lem2.2}, letting  $u=\psi(t)$.  
\end{proof}


\section {Proof of Theorem \ref{thm1.1}}

 First, by the same proof of \cite[Lemma 2.4]{DKS},  we have the 
following result.

\begin{lemma}[{Comparison principle \cite[Lemma 2.1]{DKS}}] \label{lem3.1}  
 Let $\Omega\subset \mathbb{R}^N$ be a bounded domain,  and
{\rm (F1), (B1)} be satisfied.
 Assume that $u_1, u_2\in C^2(\Omega)$ satisfy
$\Delta u_1 \geq b(x) f(u_1)$ and $\Delta u_2 \leq b(x)f(u_2)$ in $\Omega$. 
If  $\liminf_{x\to \partial\Omega} (u_2-u_1)(x)\geq 0$,  then $u_2\geq u_1 $ in
$\Omega$.
\end{lemma}

Let  $v_0 \in C^{2+\alpha} (\Omega)\cap C^1 (\bar{\Omega}) $ be the
unique solution of the problem
\begin{equation}\label{e3.1}
-\Delta v=1,\quad  v>0, \quad  x\in\Omega,\quad  v|_{\partial\Omega}=0.
\end{equation}
By the H\"{o}pf  maximum  principle \cite[Lemma 3.4]{GT}, we
see that
\begin{equation}\label{e3.2}
\nabla v_0 (x) \neq 0, \;  \forall x \in
\partial \Omega \text{ and }  c_1 d(x)\leq v_0(x) \leq c_2 d(x),\;  \forall
  x \in \Omega, 
\end{equation}
 where $c_1$, $c_2$ are positive
constants.

  Denote   $\varsigma_0=\exp (\xi_0)$, where
 $\xi_0$ is given in \eqref{e1.17}, 
$$
\varsigma_2=\varsigma_0+\varepsilon,\quad
\varsigma_1=\varsigma_0-\varepsilon,\quad
\varepsilon\in (0,\min\{\varsigma_0,\ b_0^2\}/2).
$$ 
It follows that
$$\varsigma_0/2<\varsigma_1<\varsigma_2<2\varsigma_0, \quad
\lim_{\varepsilon\to 0}\varsigma_1
=\lim_{\varepsilon\to 0}\varsigma_2=\varsigma_0.
$$ 
Since
$\ln(1+s)\cong s$  as $s\to 0^+$, we can choose
$\varepsilon$ sufficiently small such that
\begin{gather}\label{e3.3}
\ln (\varsigma_0)-\ln (\varsigma_2)=\ln \big(1- \frac
{\varepsilon}{\varsigma_0+\varepsilon}\big)
< -\frac {1}{4\varsigma_0}\varepsilon;
\\ \label{e3.4}
 \ln (\varsigma_0)-\ln (\varsigma_1)=\ln \big(1+ \frac
{\varepsilon}{\varsigma_0-\varepsilon}\big)> \frac {1}{4
\varsigma_0}\varepsilon.
\end{gather}
  Fix the above $\varepsilon$.  For any $\delta>0$, we define 
$\Omega_\delta =\{x\in\Omega: 0<d(x)<\delta \}$. Since $\Omega$ is $C^2$-smooth, 
choose $\delta_1\in (0, \delta_0)$ such that 
(see,  14.6. Appendix: Boundary Curvatures and the Distance Function in \cite{GT})
\begin{equation}\label{e3.5}
 d\in C^{2}(\Omega_{\delta_1}), \quad
|\nabla d(x)|= 1,\quad
 \Delta d(x)  =-(N-1)H(\bar{x})+o(1), \quad \forall x\in \Omega_{\delta_1}.
 \end{equation}

\begin{proof}[Proof of Theorem \ref{thm1.1}]
 By  Lemma \ref{lem2.3},   \eqref{e1.19},   \eqref{e3.3},    \eqref{e3.5} and
 $K\in C[0, \delta_0)$ with $K(0)=0$,    we see that there
are $\delta_{1\varepsilon}$, $\delta_{2\varepsilon} \in \big(0,
\min\{1, {\delta_1}/2\}\big)$ (which are corresponding to
$\varepsilon$) sufficiently small such that
\begin{itemize}
\item[(i)]  $(b_0^2-\varepsilon)k^2(d(x)-\sigma)\leq
(b_0^2-\varepsilon)k^2(d(x))< b(x)$, 
$x\in D_\sigma^-=\Omega_{2\delta_{1\varepsilon}}/{\bar{\Omega}_\sigma}$;
$b(x)<(b_0^2+\varepsilon)k^2(d(x))\leq
(b_0^2+\varepsilon)k^2(d(x)+\sigma)$,
$x\in D_\sigma^+=\Omega_{2\delta_{1 \varepsilon}-\sigma}$,  where
 $\sigma \in (0,\delta_{1\varepsilon})$;

\item[(ii)] $b_0K(d(x))\leq \delta_{2\varepsilon}$,
$x\in \Omega_{2\delta_{1\varepsilon}}$;

\item[(iii)]  for all $(x,t)\in
\Omega_{2\delta_{1\varepsilon}}\times(0, 2\delta_{2\varepsilon})$,
\begin{align*}
& (g(\psi(t))^{-1}\Big(\frac {1}{2}\big( 1+\frac
{\psi(t)f_1'(\psi(t))}{f_1(\psi(t))}\big)  -\frac
{f_1(\varsigma_2\psi(t))}{\varsigma_2f_1(\psi(t))}\Big)-\frac
{f_2(\varsigma_2\psi(t))} {\varsigma_2g(\psi(t))f_1(\psi(t))} \\
&-\frac  {\sqrt{\psi(t)f_1(\psi(t))}}
 {t g(\psi(t))f_1(\psi(t))}\frac
 {K(d(x))k'(d(x))}{k^2(d(x))}\leq - \frac
 {1}{4\varsigma_0}\varepsilon;
\end{align*}

\item[(iv)]
$$
\frac  {\sqrt{\psi(t)f_1(\psi(t))}}
 {t g(\psi(t))f_1(\psi(t))} \frac {K(d(x))}
 {k(d(x))} |\Delta d(x)|\leq \frac {1}{8\varsigma_0}\varepsilon,
\quad \forall  (x,t)\in \Omega_{2\delta_{1\varepsilon}}\times(0,
2\delta_{2\varepsilon}).
$$
\end{itemize}
Now we define
\begin{gather}\label{e3.6}
d_1(x)=d(x)-\sigma,\quad d_2(x)=d(x)+\sigma; \\
\label{e3.7}
\bar{u}_\varepsilon=\varsigma_2\psi(\sqrt{b_0^2-\varepsilon}
K(d_1(x))),\quad x\in D_\sigma^-; \\
\label{e3.8}
\underline{u}_\varepsilon=\varsigma_1\psi(\sqrt{b_0^2+\varepsilon}
K(d_2(x))),\quad x\in D_\sigma^+.
\end{gather}
 Then, by (i)--(iv), \eqref{e3.5} and a direct calculation, we see that 
for $x\in D_\sigma^-$
and  $r=\sqrt{b_0^2-\varepsilon} K(d_1(x))$,
\begin{align*}
& \Delta \bar{u}_\varepsilon(x)-b(x)
f(\bar{u}_\varepsilon(x))\\
&=\varsigma_2 (b_0^2-\varepsilon) k^2(d_1(x))\psi''(r)
 +\varsigma_2 \sqrt{b_0^2-\varepsilon}\psi'(r)\big(k'(d_1(x))
 +k(d_1(x))\Delta d(x)\big)\\
&\quad -b(x) ( f_1(\varsigma_2\psi(r))+f_2(\varsigma_2\psi(r))
\\
&\leq  \varsigma_2
(b_0^2-\varepsilon)f_1(\psi(r))g(\psi(r))k^2(d_1(x))
\Big[(g(\psi(r))^{-1}\Big(\frac {1}{2}\big( 1+\frac
{\psi(r)f_1'(\psi(r))}{f_1(\psi(r))}\big) \\
&\quad -\frac{f_1(\varsigma_2\psi(r))}{\varsigma_2f_1(\psi(r))}\Big)-\frac
{f_2(\varsigma_2\psi(r))} {\varsigma_2g(\psi(r))f_1(\psi(r))} \\
&\quad -\frac  {\sqrt{\psi(r)f_1(\psi(r))}}
 {r g(\psi(r))f_1(\psi(r))}\Big(\frac {K(d_1(x))k'(d_1(x))}{k^2(d_1(x))}
 + \frac {K(d_1(x))}{k(d_1(x))} \Delta d(x)\Big)\Big]
\leq  0;
\end{align*}
i.e., $\bar{u}_\varepsilon$ is a supersolution of
 \eqref{e1.1} in $D_\sigma^-$.

In a similar way, we can show that $
\underline{u}_\varepsilon=\varsigma_1\psi(\sqrt{b_0^2+\varepsilon}K(d_2(x)))$
 is a subsolution of   \eqref{e1.1} in  $ D_\sigma^+$.
 Now let  $u$ be an arbitrary solution to
problem \eqref{e1.1}, we can  choose  a large $M$  such that
\begin{equation}\label{e3.9}
u\leq \bar{u}_\varepsilon+M v_0 \quad \text{on }
\partial D_\sigma^-,\quad \underline{u}_\varepsilon
 \leq u+M v_0   \quad\text{on } \partial D_\sigma^+,
\end{equation}
where $v_0$ is the solution of  \eqref{e3.1}. 

Also by (F1), we  that $u+M v_0$ and $\bar{u}_\varepsilon+M v_0$ are two
 supersolutions of  equation \eqref{e1.1} in $\Omega$ and in $D_\sigma^-$.
  Since $u <\infty$
on $d = \sigma$; $\bar{u}_\varepsilon (x)=\infty$ on $d = \sigma$;
$u=\infty$ on $\partial\Omega$, it follows by (F1)  and 
 Lemma \ref{lem3.1} that
\begin{equation}\label{e3.10}
u(x)\leq M v_0(x)+\bar{u}_\varepsilon(x), \quad x\in D_\sigma^-;\quad
\underline{u}_\varepsilon (x)\leq u(x)+M v_0(x), \quad x\in D_\sigma^+.
\end{equation}
Hence, letting $\sigma \to 0$, we have for $x\in \Omega_{2\delta_{1\varepsilon}}$,
$$
1-\frac{M v_0(x)}{\varsigma_{1}\psi(\sqrt{b_0^2+\varepsilon}K(d(x)))}\leq
\frac{u(x)}{\varsigma_{1} \psi(\sqrt{b_0^2+\varepsilon}K(d(x)))};
$$
and
$$
 \frac{u(x)}{\varsigma_{2}\psi(\sqrt{b_0^2-\varepsilon}K(d(x)))} \leq 1+
\frac{M v_0(x)}{\varsigma_{2}
\psi(\sqrt{b_0^2-\varepsilon}K(d(x)))}.
$$
Consequently, by  $K(0)=0$ and $\psi(0)=\infty$,
\begin{gather*}
1\leq \lim_{d(x) \to 0 } \inf \frac{u(x)}{\varsigma_{1}
\psi(\sqrt{b_0^2+\varepsilon}K(d(x)))},\\
\lim_{d(x) \to 0 } \sup \frac{u(x)}{\varsigma_{2}
\psi(\sqrt{b_0^2-\varepsilon}K(d(x)))} \leq 1.
\end{gather*}
Thus letting $\varepsilon\to 0$, we obtain
$$
\lim_{d(x) \to 0 }\frac {u(x)} {\psi(b_0K(d(x)))}=\varsigma_0.
$$
This completes the proof.
\end{proof}

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\end{document}