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

\AtBeginDocument{{\noindent\small
{\em Electronic Journal of Differential Equations},
Vol. 2006(2006), No. 64, pp. 1--9.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu  (login: ftp)}
\thanks{\copyright 2006 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2006/64\hfil A boundary blow-up]
{A boundary blow-up for sub-linear elliptic problems with a
nonlinear gradient term}
\author[Z. Zhang\hfil EJDE-2006/64\hfilneg]
{Zhijun Zhang}

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

\date{}
\thanks{Submitted February 5, 2006. Published May 20, 2006.}
\thanks{Supported  by grant 10071066 from the National Natural
Science Foundation of  China.}
\subjclass[2000]{35J60, 35B25, 35B50, 35R05}
\keywords{Semilinear elliptic equations; large solutions;
asymptotic behaviour}

\begin{abstract}
 By a perturbation method and constructing comparison functions,
 we show  the exact asymptotic behaviour of solutions to
 the semilinear elliptic problem
 $$
 \Delta u -|\nabla u|^q=b(x)g(u),\quad u>0 \quad\text{in }\Omega,
 \quad  u\big|_{\partial \Omega}=+\infty,
 $$
 where $\Omega$ is a bounded domain in $\mathbb{R}^N$ with smooth
 boundary, $q\in (1, 2]$,
 $g \in C[0,\infty)\cap C^1(0, \infty)$, $g(0)=0$,
 $g$ is increasing on $[0, \infty)$,
 and $b$ is non-negative non-trivial in $\Omega$,
 which may be singular or vanishing on the boundary.
\end{abstract}

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

\section{Introduction and statement of main results}

  The purpose of this paper is to investigate  the exact asymptotic
behaviour of  solutions near the boundary for the problem
\begin{equation}
\Delta u -|\nabla u|^q=b(x)g(u),  \quad u>0 \quad  \text{in } \Omega,\quad
u\big|_{\partial \Omega}=+\infty,
 \label{e1.1}
\end{equation}
where the last condition means that $u(x) \to +\infty$ as
$d(x)={\rm dist}(x, \partial \Omega)\to 0$, and the solution
is called ``a large solution'' or ``an explosive solution'',
$\Omega$ is a bounded domain with smooth boundary in
$\mathbb{R}^N$ $(N\geq 1)$,  $q \in (1, 2]$.
The functions  $g$ and $b$ satisfy
\begin{itemize}

\item[(G1)] $g\in C^1(0,\infty)\cap C[0, \infty)$, $g(0)=0$, $g$
is increasing on $[0, \infty).$

\item[(G2)]    $\int_t^\infty \frac{ds}{\sqrt{2G(s)}}=\infty$, for
all $t>0$, $ G(s)=\int_0^s g(z) dz.$

\item[(B1)]  $b \in C^\alpha({\Omega})$ for some $\alpha \in (0,1)$, is
non-negative  and non-trivial in $\Omega$.

\end{itemize}

The main feature of this paper is the presence of the three terms:
The  nonlinear term $g(u)$  which  is sub-linear at infinity,
the nonlinear gradient term $ |\nabla u|^q$, and the weight
 $b(x)$  which may be singular or vanishing on the boundary.

First, we  review  the  model
\begin{equation}
\Delta u=b(x)g(u)  \quad\text{in }  \Omega,\quad   u\big|_{\partial
\Omega}=+\infty.  \label{e1.2}
\end{equation}
For $g$ satisfying (G1) and the Keller-Osserman condition
\begin{itemize}
\item[(G3)] $\int_t^\infty \frac{ds}{\sqrt{2G(s)}}<\infty$,
\end{itemize}
problem \eqref{e1.2} arises in many branches of applied
mathematics and has been discussed by many authors; see for instance
\cite{b2,c1,c2,c3,d1,k1,k2,l1,l2,l4,l5,l6,m1,m2,m3,o1,t1,v1,v2,y1,z1,z2}.

For $g(s)=s^p$, $p\in (0, 1]$, little is known.
Lair and Wood \cite{l3} showed  that if $b\in C(\bar{\Omega})$ then
\eqref{e1.2} has no solution. Then  Lair \cite{l2}  showed
 that if $g$ satisfies (G1), $b\in C(\bar{\Omega})$ is
non-negative  in $\Omega$ and is positive near the boundary then
\eqref{e1.2} has no  solution if and only if (G2) holds. Bachar
and Zeddini \cite[Theorem 3]{b1}  showed  that if $b\in
C(\bar{\Omega})$ and there exist positive constants $c_1, c_2$
such that $g(s)\leq c_1 s+c_2$, for all $s\geq 0$, then
\eqref{e1.2} has no solution.
 Chuaqui et al. \cite{c1} showed that when $\Omega=B$, $g(s)=s^p$,
$p\in (0, 1)$, and $b(|x|)=b(r)$ satisfies
\begin{itemize}
\item[(B2)] $\lim_{r\to 0^+}(1-r)^\gamma b(r)=c_0>0$ for some
$\gamma>0$,
\end{itemize}
then  \eqref{e1.2} has at least one solution if and only if
$\gamma\geq 2$. Moreover, if $\gamma> 2$, then,  for any solution
$u$, to problem \eqref{e1.2},
$$
\lim_{r\to 0^+}(1-r)^\beta u(r)=\Big(\frac
{c_0}{\beta(\beta+1)}\Big)^{1/{(1-p)}},
$$
 where $\beta=(\gamma-2)/{(1-p)}$. If $\gamma=2$, then,  for any
solution $u$ to problem \eqref{e1.2},
$$
\lim_{r\to 0^+}\frac {u(r)}
{(-\ln(1-r))^{1/{(1-p)}}}=\big(c_0(1-p)\big)^{1/{(1-p)}}.
$$
Yang \cite{y1}  showed that if $b\in C[0, 1)$ is non-negative
non-trivial in $[0, 1)$,  $g$ satisfies (G1) and
\begin{equation}
\int_1^\infty\frac {ds}{g(s)}=\infty,  \label{e1.3}
\end{equation}
then  \eqref{e1.2} has one solution if and only if
\begin{equation}
 \int_0^1(1-r)b(r)dr=\infty. \label{e1.4}
\end{equation}
Moreover, if $b(r)\sim (1-r)^{-\gamma}$ as $r\to 1$,
$\gamma\geq 2$, and $p\in (0, 1)$,  $g(s)\sim s(\ln s)^p$ as
$s\to\infty$,  then, for  any solution $u$ to problem
\eqref{e1.2},
$$
u(r)\sim \begin{cases}
(1-r)^{-(\gamma-2)/{(2-p)}}  &  \text{if } \gamma>2; \\
(-\ln(1-r))^{2/{(2-p)}} &  \text{if }   \gamma=2.
\end{cases}
$$
He also showed that \eqref{e1.2} has no solution provided
that  $\Omega$ is a bounded domain with smooth boundary in
$\mathbb{R}^N$  $(N\geq 1)$,  $g$ satisfies (G1) and \eqref{e1.3},
$b$ satisfies (B1) and
\begin{equation}
b(x)\leq C(d(x))^{-2} (-\ln (d(x)))^{-p},  \label{e1.5}
\end{equation}
where $p>1$  and $C>0$.

Let's return to problem \eqref{e1.1}.
When $b\equiv 1$  on $\Omega$: for $g(u)=u$,  Lasry and Lions \cite{l3}
established the model \eqref{e1.1} which  arises from the description of
the basic stochastic control problem, and showed by a perturbation
method and a sub-supersolutions method that if $q\in (1, 2]$ then
problem \eqref{e1.1}
 has a unique solution $u\in C^2(\Omega)$.  Moreover,
\\
(i)  when  $ 1<q<2$,
\begin{equation}
\lim_{d(x)\to 0}u(x)(d(x))^{(2-q)/(q-1)}
  =(2-q)^{-1}(q-1)^{-(2-q)/(q-1)}; \label{e1.6}
\end{equation}
(ii) when $q=2$,
\begin{equation}
\lim_{d(x)\to 0}u(x)/{(-\ln (d(x)))}=1. \label{e1.7}
\end{equation}
For $g(u)=u^p$, $p>0$, by the theory of ordinary differential equation
and the  comparison principle, Bandle and Giarrusso \cite{b3}
showed that
\\
(iii) if  $1<q\leq 2$, then problem \eqref{e1.1} has  one solution in
$C^2(\Omega)$;
\\
(iv)  if $\max \{2p/(p+1), 1\}<q<2$, then   every solution $u$ to
problem \eqref{e1.1}   satisfies \eqref{e1.6};
\\
(v) if $q=2$, then  every solution $u$ to problem \eqref{e1.1} satisfies
\eqref{e1.7}.

For the other  results of large solutions to elliptic problems
with nonlinear gradient terms, see \cite{g1,g2,z3,z4,z5,z6} and the
references therein.
        In this note,  by a perturbation method and constructing
comparison functions, we show how the weight $b$   affects
the  exact asymptotic behaviour of solutions near the boundary, to
problems \eqref{e1.1}.

Our main results are state in the following theorems.

\begin{theorem} \label{thm1.1}
Let $1<q<2$, and assume (G1) and (B1).
\begin{itemize}
\item[(I)] If the following convergence is uniform for $\xi \in
[a,b]$ with $0<a<b$,
\begin{equation}
\lim_{d(x)\to 0}b(x)(d(x))^{\frac {q}{q-1}}g\big(\xi(d(x)
\big)^{-\frac {2-q}{q-1}})=0, \label{e1.8}
\end{equation}
then  every solution to problem \eqref{e1.1} satisfies \eqref{e1.6};

\item[(II)] if  $g(u)=u^p$, $p\in (0, 1]$ and
\begin{equation}
\lim_{d(x)\to 0}b(x)(d(x))^{\frac {q-p(2-q)}{q-1}}=C_0>0, \label{e1.9}
\end{equation}
 then every solution  to problem
\eqref{e1.1} satisfies
\begin{equation}
\lim_{d(x)\to 0}u(x)(d(x))^{(2-q)/(q-1)}=\xi_0, \label{e1.10}
\end{equation}
provided that
\\
{\rm (i)}  $p=1$, $C_0\in \big(0, \frac {2-q}{(q-1)^2}\big)$.
In this case,
$$
\xi_0=\big(\frac {q-1}{2-q}\big)^{q/(q-1)}
\big(\frac{2-q}{(q-1)^2}-C_0\big)^{1/(q-1)};
$$
{\rm (ii)}    $p\in (0, 1)$, $C_0\in (0, \bar{C})$   where
$$
\bar{C}=\Big(\frac {q(1-p)}{p(q-p)(q-1)}\Big)^{\frac{q-p}{q-1}}
\Big(\frac {p(1-p)}{q(q-1)}\Big)^{\frac{1-p}{q-1}}
\Big(\frac {2-q}{q-1}\Big)^{\frac{q(1-p)}{q-1}}.
$$
In this case, $\xi_0=\xi_2$, where the equation
\begin{equation}
\frac{2-q}{q-1}=C_0\xi^{p-1}+(\frac {2-q}{q-1})^q \xi^{q-1},
\label{e1.11}
\end{equation}
 has just two positive  solutions $\xi_1$   and  $\xi_2$  with
$$
0<\xi_1<\Big(\frac{C_0(1-p)}{q-1} \Big)^{1/(q-p)}
\Big(\frac{2-q}{q-1}\Big)^{q/(q-p)}<\xi_2.
$$
\end{itemize}
\end{theorem}

\begin{theorem} \label{thm1.2}
Let $q=2$, and assume (G1) and (B1).
\begin{itemize}
\item[(I)] If the following convergence is uniform for $\xi \in
[a,b]$ with $0<a<b$,
\begin{equation}
\lim_{d(x)\to 0}b(x)(d(x))^2 g(-\xi \ln(d(x)))=0, \label{e1.12}
\end{equation}
then  every solution to problem \eqref{e1.1} satisfies \eqref{e1.7};

\item[(II)] if  $g(u)=u^p$, $p\in (0, 1]$ and
\begin{equation}
\lim_{d(x)\to 0}b(x)(d(x))^2(-\ln(d(x)))^p=C_0>0, \label{e1.13}
\end{equation}
then  every solution $u$ to problem \eqref{e1.1} satisfies
\begin{equation}
\lim_{d(x)\to 0}u(x)/{(-\ln (d(x)))}=\xi_0, \label{e1.14}
\end{equation}
provided that
\\
{\rm (i)} $p=1$, $C_0\in (0, 1)$, $\xi_0=1-C_0$;
\\
{\rm (ii)} $p\in (0, 1)$, $C_0=2^p/4$, $\xi_0=1/2$;
\\
{\rm (iii)}  $p\in (0, 1)$,  $C_0\in (0, 2^p/4)$,
  $\xi_0=\xi_2$, where  the equation
$$
\xi-\xi^2=C_0\xi^p,
$$ %\label{e1.15}
has just two positive  solutions $\xi_1$   and  $\xi_2$  with
$0<\xi_1<1/2<\xi_2<1$.
\end{itemize}
\end{theorem}

 \section{Proof of theorems}

\begin{lemma}[{The comparison principle, \cite[Theorem 10.1]{g3}}]
\label{lem2.1}
Let $\Psi(x,s,\xi)$ satisfy the following two conditions
\begin{itemize}
\item[(D1)]  $\Psi$ is non-increasing in $s$ for each $(x,\xi)\in
(\Omega \times \mathbb{R}^N)$;

\item[(D2)]   $\Psi$  is continuously differentiable with respect to
the variable  $\xi$  in $\Omega \times (0,\infty)\times \mathbb{R}^N$.
\end{itemize}
If $u,v\in C(\bar{\Omega})\cap C^2(\Omega)$ satisfy
$\Delta u+\Psi(x,u,\nabla u)\geq \Delta v+\Psi(x,v,\nabla v)$ in
$\Omega$ and $u\leq v$ on $\partial \Omega$, then $u\leq v$ in $\Omega$.
\end{lemma}

\begin{lemma}[Taylor's formula]  \label{lem2.2}
Let $\alpha\in \mathbb{R}$, $x\in [-x_0, x_0]$ with $x_0\in (0, 1)$.
Then there exists $\varepsilon_1>0$ small enough such that for
$\varepsilon\in (0, \varepsilon_1)$
\begin{equation}
(1+\varepsilon x)^\alpha=1+\alpha\varepsilon x+o(\varepsilon^2).
\label{e2.1}
\end{equation}
\end{lemma}

\begin{proof}[Proof of Theorem \ref{thm1.1}]
Given an arbitrary $\varepsilon\in (0, \xi_0/2)$, let
$\xi_{2\varepsilon}=\xi_0+\varepsilon$,
$\xi_{1\varepsilon}=\xi_0-\varepsilon$.
It follows that
$$
\frac {1}{2}\xi_0 <\xi_{1\varepsilon}<\xi_{2\varepsilon}<2\xi_0.
$$
For $\delta>0$, we define
$$
\Omega_\delta=\{x \in \Omega  : 0<d(x)<\delta \}.
$$
Since $\partial\Omega\in C^2$, there exists a constant $\delta>0$ which
only depends on $\Omega$ such that
\begin{equation}
d(x) \in C^2(\bar{\Omega}_{2\delta})\quad \text{and}\quad
  |\nabla d|\equiv 1\quad  \text{on }  \Omega_{2\delta}.
\label{e2.2}
\end{equation}
(I)  When \eqref{e1.8} holds, $\xi_0=(2-q)^{-1}(q-1)^{-(2-q)/(q-1)}$.
It follows from Lemma \ref{lem2.2}  that there exists $\varepsilon_1>0$
small enough such that for $\varepsilon\in (0, \varepsilon_1)$
\begin{align*}
\frac {2-q}{(q-1)^2}\xi_{2\varepsilon}-   (\frac
{2-q}{q-1})^q\xi_{2\varepsilon}^q
&=\frac {2-q}{(q-1)^2}(\xi_0+\varepsilon)-   (\frac
{2-q}{q-1})^q(\xi_0+\varepsilon)^q\\
&=\frac {2-q}{(q-1)^2}\varepsilon-(\frac
{2-q}{q-1})^q\xi_0^q\big((1+\frac
{\varepsilon}{\xi_0})^q-1\big)\\
&=-\frac {(q-1)(2-q)}{(q-1)^2}\varepsilon+o(\varepsilon^2);
\end{align*}
and
\begin{align*}
\frac {2-q}{(q-1)^2}\xi_{1\varepsilon}-   (\frac
{2-q}{q-1})^q\xi_{1\varepsilon}^q
&=\frac {2-q}{(q-1)^2}(\xi_0-\varepsilon)-   (\frac
{2-q}{q-1})^q(\xi_0-\varepsilon)^q\\
&=\frac {(q-1)(2-q)}{(q-1)^2}\varepsilon+o(\varepsilon^2).
\end{align*}
Denote
$$
c_1=\frac {(q-1)(2-q)}{(q-1)^2}.
$$
It follows by \eqref{e2.2} and \eqref{e1.8} that corresponding
to $\varepsilon\in (0,\varepsilon_1)$, there is
$\delta_\varepsilon\in (0, \delta)$
sufficiently small such that
\begin{equation}
\frac {2-q}{q-1}|\xi_{i\varepsilon}d(x)\Delta d(x)|
+|b(x)(d(x))^2 g\left(-\xi_{i\varepsilon} \ln(d(x))\right)|<\frac
{c_1}{2}\varepsilon, \label{e2.3}
\end{equation}
 for all $x\in \Omega_{2\delta_\varepsilon}$, $i=1, 2$.


\noindent (II) (i)  When  $p=1$, $C_0\in \big(0, \frac{2-q}{(q-1)^2}\big)$.
As the result of (I), we see that for
$\varepsilon\in (0, \varepsilon_1)$,
\[
\left (\frac {2-q}{(q-1)^2}-C_0\right )\xi_{2\varepsilon}-
\left(\frac{2-q}{q-1}\right)^q\xi_{2\varepsilon}^q
 =-(q-1)\left (\frac {2-q}{(q-1)^2}-C_0\right
)\varepsilon+o(\varepsilon^2);
\]
and
\[
\left (\frac {2-q}{(q-1)^2}-C_0\right )\xi_{1\varepsilon}-
\left(\frac
{2-q}{q-1}\right)^q\xi_{1\varepsilon}^q
=(q-1)\left (\frac {2-q}{(q-1)^2}-C_0\right
)\varepsilon+o(\varepsilon^2).
\]
 (ii) When $p\in (0, 1)$, $C_0\in (0, \bar{C})$. Since
$$
\left (\frac{C_0(1-p)}{q-1}(\frac {q-1}{2-q})^q \right )^{1/(q-p)}<\xi_0,
$$
it follows that
$$
(q-1) (\frac {2-q}{q-1})^q \xi_0^{q-p}-C_0 (1-p)>0.
$$
Then by  Lemma \ref{lem2.2}, there exists $\varepsilon_1>0$
small enough such that for $\varepsilon\in (0, \varepsilon_1)$
\begin{align*}
&\frac {2-q}{(q-1)^2}\xi_{2\varepsilon}-   \left(\frac
{2-q}{q-1}\right)^q\xi_{2\varepsilon}^q-C_0\xi_{2\varepsilon}^p\\
&=\frac {2-q}{(q-1)^2}(\xi_0+\varepsilon)-   \left(\frac
{2-q}{q-1}\right)^q(\xi_0+\varepsilon)^q-C_0(\xi_0+\varepsilon)^p\\
&=\frac {2-q}{(q-1)^2}\varepsilon -C_0\xi_0^p\left ((1+\frac
{\varepsilon}{\xi_0})^p-1\right)-\left (\frac
{2-q}{q-1}\right)^q\xi_0^q\left ((1+\frac
{\varepsilon}{\xi_0})^q-1\right)\\
&=-\xi_0^{-1}\left( q\left(\frac
{2-q}{q-1}\right)^q\xi_0^q+pC_0\xi_0^p-\frac
{2-q}{(q-1)^2}\xi_0\right)\varepsilon+o(\varepsilon^2)\\
&=-\xi_0^{-(2-p)}\left((q-1)\left (\frac {2-q}{q-1} \right )^q
\xi_0^{(q-p)}-C_0 (1-p)\right)\varepsilon+o(\varepsilon^2);
\end{align*}
and
\begin{align*}
&\frac {2-q}{(q-1)^2}\xi_{1\varepsilon}-   \left(\frac
{2-q}{q-1}\right)^q\xi_{1\varepsilon}^q-C_0\xi_{1\varepsilon}^p\\
&=\frac {2-q}{(q-1)^2}(\xi_0-\varepsilon)-   \left(\frac
{2-q}{q-1}\right)^q(\xi_0-\varepsilon)^q-C_0(\xi_0-\varepsilon)^p\\
&=\xi_0^{-(2-p)}\left((q-1)\left (\frac {2-q}{q-1} \right )^q
\xi_0^{(q-p)}-C_0 (1-p)\right)\varepsilon+o(\varepsilon^2).
\end{align*}
Denote
$$
c_2=\xi_0^{-(2-p)}\left((q-1)\left (\frac {2-q}{q-1}
\right)^q \xi_0^{(q-p)}-C_0 (1-p)\right).
$$
We see by \eqref{e2.2} and \eqref{e1.9} that corresponding to
$\varepsilon\in (0, \varepsilon_1)$,
there is $\delta_\varepsilon\in (0, \delta)$ sufficiently small
such that
\begin{equation}
\frac {2-q}{q-1}|\xi_{i\varepsilon}d(x)\Delta d(x)|
+|b(x)(d(x))^2 g(-\xi_{i\varepsilon}
\ln(d(x)))|<C_0\xi_{i\varepsilon}^p+\frac {c_2}{2}\varepsilon,
\label{e2.4}
\end{equation}
for all $x\in \Omega_{2\delta_\varepsilon}$, $i=1, 2$.
Let $\beta \in (0,\delta_\varepsilon)$ be arbitrary, we define
\begin{gather*}
\overline{u}_\beta=\xi_{2\varepsilon}(d(x)-\beta)^{-(2-q)/(q-1)},
 \  x\in D_\beta^-
=\Omega_{2\delta_\varepsilon}/{\bar{\Omega}_\beta};\\
%
\underline{u}_\beta=\xi_{1\varepsilon}
(d(x)+\beta)^{-(2-q)/(q-1)},\  x\in
D_\beta^+=\Omega_{2\delta_\varepsilon-\beta}.
\end{gather*}
It follows that for $(x, \beta)\in  D_\beta^-\times(0,\delta_\varepsilon)$,
 \begin{align*}
&\Delta \overline{u}_\beta(x)-|\nabla \overline{u}_\beta(x)|^q -b(x)g(\overline{u}_\beta(x))\\
&=  (d(x)-\beta)^{-q/(q-1)}\Big(\frac
{\xi_{2\varepsilon}(2-q)}{(q-1)^2}- \frac
{\xi_{2\varepsilon}(2-q)}{q-1}(d(x)-\beta)\Delta d(x)
- \xi_{2\varepsilon}^q(\frac {2-q}{q-1})^q \\
&\quad  -b(x)g\left(\xi_{2\varepsilon}(d(x)-\beta)^{-(2-q)/(q-1)}\right)
(d(x)-\beta)^{q/(q-1)}\Big)\\
&\leq 0;
\end{align*}
and   for $(x, \beta)\in  D_\beta^+\times(0, \delta_\varepsilon)$
\begin{align*}
&\Delta \underline{u}_\beta(x)-|\nabla \underline{u}_\beta(x)|^q -b(x)g(\underline{u}_\beta(x))\\
&=  (d(x)+\beta)^{-q/(q-1)}\Big(\frac
{\xi_{1\varepsilon}(2-q)}{(q-1)^2}- \frac
{\xi_{1\varepsilon}(2-q)}{q-1}(d(x)+\beta)\Delta d(x)
- \xi_{1\varepsilon}^q(\frac {2-q}{q-1})^q \\
&\quad -b(x)g\left(\xi_{2\varepsilon}(d(x)+\beta)^{-(2-q)/(q-1)}
\right)(d(x)+\beta)^{q/(q-1)}\Big)\\
&\geq  0.
\end{align*}
Now let  $u$ be an arbitrary solution of problem \eqref{e1.1} and
$M_{u}(2\delta_\varepsilon)=\max_{d(x)\geq 2\delta_\varepsilon}u(x)$.
We see that
$$
u\leq M_u(2\delta_\varepsilon)+\overline{u}_\beta\quad\text{on }
  \partial D_\beta^-.
$$
Since  $\underline{u}_\beta =\xi_{1\varepsilon}
(2\delta_\varepsilon)^{-(2-q)/(q-1)}=M_{\underline{u}}(2\delta_\varepsilon)$
whenever $ d(x)= 2\delta_{2\varepsilon}-\beta$, we see that
$$
\underline{u}_\beta \leq
u+M_{\underline{u}}(2\delta_\varepsilon)\quad\text{on }
\partial D_\beta^+.
$$
It follows by (G1) and  Lemma \ref{lem2.1} that
\begin{gather*}
u\leq M_u(2\delta_\varepsilon)+\overline{u}_\beta, \quad x\in  D_\beta^-;\\
\underline{u}_\beta \leq u+M_{\underline{u}}(2\delta_\varepsilon),
\quad  x\in  D_\beta^+.
\end{gather*}
Hence, for $x\in D_\beta^-\cap D_\beta^+$,
and letting $\beta \to 0$, we have
$$
\xi_{1\varepsilon}-
\frac{M_{\underline{u}}(2\delta_\varepsilon)}{(d(x))^{-(2-q)/(q-1)}}\leq
\frac{u(x)}{(d(x))^{-(2-q)/(q-1)}}
\leq  \xi_{2\varepsilon}+\frac
{M_{u}(2\delta_\varepsilon)}{(d(x))^{-(2-q)/(q-1)}};
$$
i.e.,
$$
\xi_{1\varepsilon}\leq
\lim_{d(x) \to 0 } \inf \frac{u(x)}{(d(x))^{-(2-q)/(q-1)}}
\leq \lim_{d(x) \to 0 } \sup
\frac{u(x)}{(d(x))^{-(2-q)/(q-1)}} \leq \xi_{2\varepsilon}.
$$
Letting $\epsilon \to 0$, and using the definitions of
$\xi_{1\varepsilon}$ and $\xi_{2\varepsilon}$, we have
$$
\lim_{d(x) \to 0 }
\frac{u(x)}{(d(x))^{-(2-q)/(q-1)}}=\xi_0.
$$
\end{proof}

\begin{proof}[Proof of Theorem \ref{thm1.2}]
We proceed as in the proof of Theorem \ref{thm1.1}.
Given an arbitrary $\varepsilon\in (0, \xi_0/2)$,
let
$$
 \xi_{2\varepsilon}=\xi_0+\varepsilon,\quad
\xi_{1\varepsilon}=\xi_0-\varepsilon.
$$
Note that
$$
\frac {1}{2}\xi_0 <\xi_{1\varepsilon}<\xi_{2\varepsilon}<2\xi_0.
$$
When $p=1$, $C_0\in (0, 1)$, $\xi_0=1-C_0$, we see that
$$
(1-C_0)\xi_{2\varepsilon}-\xi_{2\varepsilon}^2=-\varepsilon\xi_0
-o(\varepsilon^2)\quad\text{and}\quad
 (1-C_0)\xi_{1\varepsilon}-\xi_{1\varepsilon}^2
 =\varepsilon\xi_0-o(\varepsilon^2).
$$
 It follows  by \eqref{e2.2} and \eqref{e1.12} that  there is
$\delta_\varepsilon\in (0, \delta)$ sufficiently small such that
\begin{equation}
|\xi_{i\varepsilon}d(x)\Delta d(x)|
+|b(x)(d(x))^2 g(-\xi_{i\varepsilon} \ln(d(x)))|<\frac
{\xi_0}{2}\varepsilon,  \label{e2.5}
\end{equation}
for all $x\in \Omega_{2\delta_\varepsilon}$, $i=1, 2$.
When  $p\in (0, 1)$ and $\xi_0 \geq\frac {1}{2}$, we see that
$\xi_0>\frac {1-p}{2-p}$ and for $\varepsilon\in (0,\varepsilon_1)$,
\begin{align*}
\xi_{2\varepsilon}-
 \xi_{2\varepsilon}^2-C_0\xi_{2\varepsilon}^p
 &=\xi_0+\varepsilon-(\xi_0+\varepsilon)^2-C_0(\xi_0+\varepsilon)^p\\
 &=-(2-p)\left(\xi_0-\frac {1-p}{2-p}\right)\varepsilon+o(\varepsilon^2);
\end{align*}
and
\begin{align*}
\xi_{1\varepsilon}- \xi_{1\varepsilon}^2-C_0\xi_{1\varepsilon}^p
&=\xi_0-\varepsilon-(\xi_0-\varepsilon)^2-C_0(\xi_0-\varepsilon)^p\\
&=(2-p)\left(\xi_0-\frac {1-p}{2-p}\right)\varepsilon+o(\varepsilon^2).
\end{align*}
Denote
$$
c_3=(2-p)\Big(\xi_0-\frac {1-p}{2-p}\Big).
$$
 It follows  by \eqref{e2.2} and \eqref{e1.13} that  there is
$\delta_\varepsilon\in (0, \delta)$ sufficiently small such that
\begin{equation}
|\xi_{i\varepsilon}d(x)\Delta d(x)|
+|b(x)(d(x))^2 g(-\xi_{i\varepsilon}
\ln(d(x)))|<C_0\xi_{i\varepsilon}^p+\frac {c_3}{2}\varepsilon, \label{e2.6}
\end{equation}
for all $x\in \Omega_{2\delta_\varepsilon}$, $i=1, 2$.
Let $\beta \in (0,\delta_\varepsilon)$ be arbitrary, we define
\begin{gather*}
\overline{u}_\beta=-\xi_{2\varepsilon}\ln(d(x)-\beta),
 \quad  x\in D_\beta^-
=\Omega_{2\delta_\varepsilon}/{\bar{\Omega}_\beta};
\\
 \underline{u}_\beta=-\xi_{1\varepsilon}\ln (d(x)+\beta),\quad
  x\in D_\beta^+=\Omega_{2\delta_\varepsilon-\beta}.
\end{gather*}
It follows that for $(x, \beta)\in  D_\beta^-\times(0,\delta_\varepsilon)$,
 \begin{align*}
\Delta \overline{u}_\beta(x)-|\nabla \overline{u}_\beta(x)|^2
-b(x)g(\overline{u}_\beta(x)) &=
(d(x)-\beta)^2\Big(\xi_{2\varepsilon}-\xi_{2\varepsilon}(d(x)-\beta)
  \Delta d(x)- \xi_{2\varepsilon}^2 \\
&\quad -b(x)(d(x)-\beta)^2g(\xi_{2\varepsilon} \ln (d(x)-\beta))\Big)\\
&\leq 0;
\end{align*}
and  for $(x, \beta)\in  D_\beta^+\times(0, \delta_\varepsilon)$,
\begin{align*}
\Delta \underline{u}_\beta(x)-|\nabla \underline{u}_\beta(x)|^2
-b(x)g(\underline{u}_\beta(x)) &=   (d(x)+\beta)^2
\Big(\xi_{2\varepsilon}-\xi_{2\varepsilon}(d(x)+\beta)\Delta
d(x)- \xi_{2\varepsilon}^2\\
&\quad -b(x)(d(x)+\beta)^2g(\xi_{2\varepsilon}
\ln (d(x)+\beta))\Big)\\
&\geq  0.
\end{align*}
The rest of the proof is the same as in Theorem \ref{thm1.1}, we omit it.
\end{proof}

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\end{document}