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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2014 (2014), No. 09, pp. 1--20.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2014 Texas State University - San Marcos.}
\vspace{8mm}}

\begin{document}
\title[\hfilneg EJDE-2014/09\hfil Boundary blow-up solutions]
{Boundary blow-up solutions to semilinear elliptic equations with
nonlinear gradient terms}

\author[S. Liu, Y. Xu \hfil EJDE-2014/09\hfilneg]
{Shufang Liu, Yonglin Xu}  % in alphabetical order

\address{Shufang Liu \newline
Department of Mathematics, Gansu Normal University for Nationalities \\
Hezuo,  Gansu 747000, China}
\email{shuxueliushufang@163.com}

\address{Yonglin Xu (Corresponding author)\newline
School of Mathematics and Computer Science Institute,
Northwest University for Nationalities, Lanzhou, Gansu  730030, China}
\email{xuyonglin000@163.com}

\thanks{Submitted October 4, 2013. Published January 7, 2014.}
\subjclass[2000]{35J25, 35B50, 65J65}
\keywords{Boundary blow-up solutions; nonlinear gradient terms;
 \hfill\break\indent  Karamata regular variation}

\begin{abstract}
 In this article we study the blow-up rate of  solutions
 near the boundary for the semilinear elliptic  problem
 \begin{gather*}
 \Delta u\pm |\nabla u|^q=b(x)f(u), \quad  x\in\Omega,\\
 u(x)=\infty, \quad x\in\partial\Omega,
 \end{gather*}
 where $\Omega$ is a smooth bounded domain in $\mathbb{R}^N$, and $b(x)$ is a
 nonnegative weight function which may be bounded or singular on
 the boundary, and  $f$ is a regularly varying function at infinity.
 The results in this article emphasize the central role played by
 the nonlinear gradient term $|\nabla u|^q$ and the singular weight $b(x)$.
 Our main tools are the Karamata regular variation theory and the method of
 explosive upper and lower solutions.
\end{abstract}

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

\section{Introduction and statement of main results}

Let $\Omega\subset \mathbb{R}^N$ $(N\geq3)$ be a bounded domain with smooth
boundary. We are interested in the asymptotic behavior of  boundary blow-up
solutions to  the elliptic problem
\begin{equation} \label{Ppm}
\begin{gathered}
\Delta u\pm |\nabla u|^q=b(x)f(u), \quad x\in\Omega,\\
u(x)=\infty, \quad x\in\partial\Omega.
\end{gathered}
\end{equation}
For the functions $f(u)$ and $b(x)$, we assume the following hypotheses:
\begin{itemize}
\item[(F1)] $f\in C^1[0,\infty),f'(s)\geq 0$ for $s\geq 0$, $f(0)=0$ and
$f(s)>0$  for $s>0$.

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

\item[(B2)] $b$ has the property: if $x_0\in\Omega$ and $b(x_0)=0$,
then there exists a domain $\Omega_0$ such that $x_0\in\Omega_0
\subset\Omega$ and $b(x)>0$, for all $x\in\partial\Omega_0$.
\end{itemize}

The boundary condition $u(x)=\infty$, $x\in \partial\Omega$ is to be
understand as $u\to\infty$ when
$d(x)=\operatorname{dist}(x,\partial\Omega)\to 0+$. The solutions of
problem \eqref{Ppm} are called large solutions,  boundary blow-up
solutions or  explosive solutions; that is, the boundary blow-up solutions
provide uniform bounds for all other solutions to
$\Delta u\pm |\nabla u|^{q}=b(x)f(u)$ in $\Omega$,
regardless of the boundary data by the comparison principle.

The study of boundary blow-up solutions of $\Delta u=e^u$ in $\Omega$ was
initiated by Bieberbach \cite{B}, where $\Omega\subset \mathbb{R}^2$.
Problems of this type arise in Riemannian geometry, more precisely:
if a Riemannian metric of the form $|ds|^2=e^{2u(x)}|dx|^2$ has
constant Gaussian curvature $-b^2$, then $\Delta u=b^2e^{2u}$.
Rademacher \cite{RE} extended the results of Bieberbach to
$\Omega\subset \mathbb{R}^3$. Later, Lazer and McKenna \cite{LM1} generalized
the results of \cite{B,RE} to the case of bounded domains in $\mathbb{R}^N$ and
nonlinearities $b(x)e^u$.

Recently, C\^{i}rstea and R\v{a}dulescu\cite{C6,C7} opened a unified
new approach, the Karamata regular variation theory approach, to
study the uniqueness and asymptotic behavior of boundary blow-up solutions,
which enables us to obtain significant information about the
qualitative behavior of the boundary blow-up solutions in a general framework.
C\^{i}rstea \cite{C2}  obtained the asymptotic behavior of boundary blow-up
solutions to
\begin{equation}\label{1.1}
\Delta u+au=b(x)f(u),
\end{equation}
provided $f(x)$ and $b(x)$ satisfy
\begin{itemize}
\item[(F2)] $f\circ \mathcal {L}\in RV_\rho (\rho>0)$
(see Definition \ref{df1}) for some $\mathcal{L}\in C^2[A,\infty)$
satisfying $\lim_{u\to\infty}\mathcal {L}(u)=\infty$ and
$\mathcal{L}'\in NRV_{-1}$,

\item[(B3)] $\lim_{d(x)\to 0}\frac{b(x)}{k^2(d(x))}=1$,
$k(x)\in NRV_{\theta}(0+)$ (see Definition \ref{df2}) for some
$\theta\geq0$,  and $k$ is nondecreasing near the origin if
$\theta=0$.
\end{itemize}
They showed that the blowup rate of boundary blow-up solutions $u$ to problem
\eqref{1.1} can be expressed by
\begin{equation}\label{1.2}
\lim_{d(x)\to 0}\frac{u}{(\mathcal {L}\circ\Phi_1)(d(x))}=1,
\end{equation}
where the function $\Phi_1$ is defined as
\begin{equation}\label{1.3}
\int^\infty_{\Phi_1(t)}\frac{[\mathcal{L}'(y)]^{1/2}}
{y^\frac{\rho+1}{2}[L_f(y)]^{1/2}}dy 
=\int^t_0k(s)dt,\quad
\text{for all $x\in(0,\tau)$ with small } \tau>0.
\end{equation}
where $L_f$ is a normalised slowly varying function such that
\begin{equation}\label{1.4}
\lim_{u\to \infty}\frac{f(\mathcal {L}(u))}{u^\rho L_f(u)}=1.
\end{equation}

Elliptic boundary blow-up problems have been studied by a large
number of authors in the last century, see
\cite{CGR2005,CG2008,G2009,G2009-1,X2009} and references
therein.

For problem \eqref{Ppm}, with  $b\equiv 1$ in $\Omega$,  and$f(u)=u^p$, by
ordinary differential equation theory and  comparison principle,
Bandle and Giarrusso \cite{BG} showed the following results:

(1) If $p\geq 1$ and $q<\frac{2p}{p+1} (<2)$, then problem \eqref{Ppm}
possesses at least one solution. Every solution of \eqref{Ppm} satisfies
\begin{equation*}
\lim_{d(x)\to
0}\frac{u(x)}{(d(x))^{-2/(p-1)}}=\big[\sqrt{2(p+1)}/(p-1)\big]^{2/(p-1)}.
\end{equation*}

(2) The same statement for \eqref{Ppm} is true if
$\frac{2p}{p+1}<q<p$ except that in this case
\begin{equation*}
\lim_{d(x)\to
0}u(x)\Big(\frac{p-q}{q}d(x)\Big)^{q/(p-q)}=1.
\end{equation*}

(3) If $\max\{1,\frac{2p}{p+1}\}<q<2$, then \eqref{Ppm}, with the minus
sign,  possesses a solution. Each solution of \eqref{Ppm} with
the minus sign  satisfies
\begin{equation*}
\lim_{d(x)\to 0}u(x)(2-q)[(q-1)d(x)]^{\frac{2-q}{q-1}}=1.
\end{equation*}

(4) If $q=2$, \eqref{Ppm} with the minus sign  has a solution for
 all $p>0$ which satisfies
\begin{equation*}
\lim_{d(x)\to 0}u(x)/\ln d(x)=1.
\end{equation*}


Now we introduce the class of functions  $K_l$ consisting of
positive monotonic functions $k\in L^1(0,\vartheta)\cap
C^1(0,\vartheta)$ which satisfy
\begin{equation*}
\lim_{t\to0+}\frac{K(t)}{k(t)}=0,\quad
\lim_{t\to0+}\frac{d}{dt}\Big(\frac{K(t)}{k(t)}\Big)=l,
\quad\text{where } K(t)=\int^t_0k(s)ds.
\end{equation*}
We point out that for each $k\in\mathcal{K}_l$, $l\in[0,1]$ if $k$
is non-decreasing and $l\in[1,\infty)$ if $k$ is non-increasing. For
more propositions of $\mathcal{K}_l$, we refer reader to
\cite{C1,CGR2005}.

Some  examples of functions $k\in \mathcal {K}_l$ are:
\begin{enumerate}
\item $k(t)=t^q\in \mathcal {K}_l$ with  $l=1/(1+q)$;
\item $k(t)=(-\ln t)^q\in \mathcal {K}_l$ for $q<0$ with $l=1$;
\item $k(t)=-s/\ln t\in \mathcal {K}_l$ for $s>0$ with $l=1$;
\item $k(t)=t^s/\ln (1+t^{-1})\in \mathcal {K}_l$ for $s>0$ with
$l=1/(1+s)$.
\end{enumerate}

When $b$ satisfies (B1) and (B2), Zhang \cite{Z2} gave the
following results: Assume $f$ satisfies (F1), $f'(u)=u^\rho L(u)$,
$\rho>0$,  $L(u)$ is slowly varying at infinity,  $1<q<\rho+1$,
$b(x)$  satisfies (B1) with $b=0$ on $\partial\Omega$,
\begin{itemize}
\item[(B4)] $\lim_{d(x)\to 0}\frac{b(x)}{k^q(d(x))}=c_q$,
where $k(x)\in K_l$ for some $0<l\leq1$, 
\end{itemize}
$\varphi\in C^2(0,a)$ be uniquely determined by
\begin{equation*}
\int^\infty_{\varphi(t)}\frac{dt}{[f(y)]^{1/q}}=\int^t_0k(s)dt,\quad
\text{for all $x\in (0,\tau)$ with small } \tau>0.
\end{equation*}
(1) If $q=\frac{2(\rho+1)}{\rho+2}$ and
$\lim_{u\to\infty}L(u)=(1+\rho)\gamma\in(0,+\infty)$, then
every solution $u_+\in C^2(\Omega)$ to problem \eqref{Ppm}, with plus sign,
 satisfies
\begin{equation*}
\lim_{d(x)\to 0}\frac{u_+(x)}{\varphi(d(x))}=c_q^{-1/(\rho+1-q)},
\end{equation*}
where
\begin{equation*}
\varphi(t)=\Big(\frac{2-q}{\gamma^{1/q}(q-1)}\Big)^{(2-q)/(q-1)}
\Big(\int^t_0k(s)dt\Big)^{(q-2)/(q-1)},\quad t\in(0,a),
\end{equation*}

(2)  The same statement is true if
$\frac{2(\rho+1)}{\rho+2}<q\leq2$, where $\varphi\in
RVZ_{-q/l(\rho+1-q)}$ and there exists $H\in RVZ_{0}$ such that
$\varphi(t)=H(t)t^{-q/l(\rho+1-q)}$.


Moreover, he also obtained some boundary blow-up rate of solutions to
problem \eqref{Ppm} if $0\leq q<2(1+l\rho)/(2+l\rho)$.  Zhang
\cite{Z1} considered problem \eqref{Ppm}  for a  weight $b$ that may be
singular on the boundary.

More recently, for $b$ satisfying (B1) and (B2), and $f$ satisfying
(F1) and (F2), Huang et al \cite{H} obtained the following:

(1) If $0\leq q<2$, and $b(x)$ satisfies (B3), then  \eqref{Ppm} has a large
solution $u_\pm$, which satisfy \eqref{1.2};

(2) If $q>2$, and $b(x)$ satisfy
\begin{itemize}
\item[(B5)] $\lim_{d(x)\to 0}\frac{b(x)}{k^q(d(x))}=1$,
where $k(x)\in NRV_{\theta}(0+)$ for some $\theta\geq0$,  and $k$ is
nondecreasing near the origin if $\theta=0$.
\end{itemize}
Then problem \eqref{Ppm}, with plus sign,  has a boundary blow-up
solution $u_+$ satisfying
\begin{equation}\label{1.5}
\lim_{d(x)\to 0}\frac{u_+}{(\mathcal
{L}\circ\Phi_2)(d(x))}=1,
\end{equation}
where $\Phi_2$ is given  by
\begin{equation}\label{1.6}
\int^\infty_{\Phi_2(t)}\frac{\mathcal
{L}'(y)}{y^{\rho/q}[L_f(y)]^{1/q}}dy=\int^t_0k(s)dt,\quad
\text{for all $x\in (0,\tau)$  with small } \tau>0.
\end{equation}

(3) If $q>2$, then $u_-=-\ln v$  is the unique solution to problem \eqref{Ppm},
with the minus sign, where $v$ is the unique solution to problem
$\Delta v=b(x)f(-\ln v)v$, $v>0$, $x\in\Omega$, $v|_{\partial\Omega}=0$.

(4) If $q=2$, and $b(x)$ satisfies (B3), then problem \eqref{Ppm}, with plus sign,
 has a unique solution $u_+$ satisfying
\begin{equation*}
u(x)\sim\frac{1}{\rho}\ln\Big(\frac{2+\rho(1+\theta)}{2}\Big)
+\ln\Psi(d(x))\quad\text{as }d(x)\to0,
\end{equation*}
where $\Psi(t)$ is given by
\[
\int^\infty_{\Psi(t)}\frac{dy}{y\sqrt{f(\ln y)}}=\int_0^tk(s)dt\quad
\text{for all $t\in(0,\tau)$, $\tau > 0$  small enough}.
\]

For more results of boundary blow-up problem with nonlinear gradient terms,
see \cite{GR2004,G2000,LW1999,CP2006,CGP2008,ML2013,CPW2013,AGQ2012,GS2013}.

We remark at this point that $\lim_{u\to\infty}\mathcal
{L}(u)=\infty$ with $\mathcal{L}'\in NRV_{-1}$ if and only if
\[
\mathcal{L}(u)=C\exp\Big\{\int^u_B\frac{s(t)}{t}dt\Big\},\quad\forall u>B>0,
\]
where $C>0$ is a constant and $s(t)$ is a normalised slowly varying
function satisfying
\[
\lim_{u\to\infty}s(u)=0,\quad
\lim_{u\to\infty}\int^u_B\frac{s(t)}{t}dt=\infty.
\]

Note that $f\circ \mathcal {L}\in RV_\rho (\rho>0)$ is equivalent to
the existence of $g\in RV_\rho$ so that
$f(u)=g(\mathcal {L}^\leftarrow (u))$ for $u$ large, where
$\mathcal {L}^\leftarrow$
denotes the inverse of $\mathcal {L}$, By Proposition \ref{prop5}, we know
that if $\mathcal{L}'\in NRV_{-1}$, then $\mathcal {L}^\leftarrow$
is rapidly varying with index $\infty$; i.e.,
\begin{equation*}
\lim_{u\to\infty}\frac{\mathcal {L}^\leftarrow(\lambda
u)}{\mathcal {L}^\leftarrow(u)}
=\begin{cases}
0, &\text{if }\lambda\in(0,1),\\
1, &\text{if } \lambda=1,\\
\infty, &\text{if } \lambda>1,
\end{cases}
\end{equation*}
Therefore, the nonlinear term $f(u)$ satisfies (F2), then it is
rapidly varying at infinity with index $\infty$, namely $f(u)\in RV_{\infty}$.

The main purpose of this article is to describe the asymptotic
behavior of the boundary blow-up solution to \eqref{Ppm}, when $f$ satisfies
\begin{itemize}
\item[(F3)] $f\circ \mathcal {L}\in RV_\rho$  $(\rho>0)$ for some
$\mathcal{L}\in C^2[A,\infty)$
satisfying $\mathcal{L}'\in NRV_{-r}$ with $0\leq r<1$.
\end{itemize}

Our main results are the following.

\begin{theorem}\label{thm1}
Let $f$ satisfy {\rm (F1), (F3)} with $q<\rho/(1-r)$,
$b(x)$ satisfies {\rm (B1), (B2)} and
\begin{itemize}
\item[(B6)] $\lim_{d(x)\to 0}\frac{b(x)}{k^2(d(x))}=c_0$,
where $k(x)\in K_l$ for some $0<l<\infty$.
\end{itemize}
(i) If
\begin{equation}\label{1.7}
0\leq q<\frac{2\rho}{\rho-r+1},
\end{equation}
then for any solution $u_\pm$ to problem \eqref{Ppm} satisfies
\begin{equation}\label{1.8}
\lim_{d(x)\to 0}\frac{u_\pm}{(\mathcal{L}\circ\Phi_1)(d(x))}=\xi_1,
\end{equation}
where
\begin{equation*}
\xi_1=\big[\frac{l(\rho+r-1)+2(1-r)}{2c_0}\big]^{\frac{1-r}{\rho+r-1}},
\end{equation*}
and $\Phi_1$ is defined by \eqref{1.3},  moreover,
\begin{equation*}
\Phi_1\in RV_{-\frac{2}{l(\rho+r-1)}}(0+).
\end{equation*}

(ii) The same statement is true if $q=\frac{2\rho}{\rho-r+1}$ and
$\lim_{u\to\infty}\mathcal {L}'(u)/u^{-r}=L_0$, moreover
\begin{equation}\label{1.9}
\lim_{d(x)\to0}u_\pm\Big(\int_0^tk(s)dt\Big)^{\frac{2(1-r)}{\rho+r-1}}
=L_0^{\frac{\rho}{\rho+r-1}}\Big(\frac{\rho+r-1}{2}
\Big)^{-\frac{2(1-r)}{\rho+r-1}}
\end{equation}
\end{theorem}

\begin{theorem}\label{thm2}
Let $b(x)$ satisfy {\rm (B1), (B2), (B4)} with $0<l<\infty$,
$f$ satisfy {\rm (F1)} and {\rm (F3)} with $q<\rho/(1-r)$. If
\begin{equation}\label{1.10}
\frac{2\rho}{\rho-r+1}<q<\frac{\rho}{1-r}.
\end{equation}
Then problem \eqref{Ppm}, with plus sign,  have a solution $u_+$,
 which satisfies
\begin{equation}\label{1.11}
\lim_{d(x)\to 0}\frac{u_+}{(\mathcal{L}\circ\Phi_2)(d(x))}=\xi_2,
\end{equation}
where
$\xi_2=c_q^{\frac{1-r}{q(1-r)-\rho}}$,
and $\Phi_2$ is defined by \eqref{1.5}; moreover,
\begin{equation*}
\Phi_2\in RV_{-\frac{q}{l(\rho+q(r-1))}}(0+).
\end{equation*}
\end{theorem}

\begin{theorem}\label{thm3}
Assume $f$ satisfies {\rm (F1)} and {\rm (F3)} with $q<\frac{\rho}{1-r}$, $b(x)$
satisfies {\rm (B1), (B2), (B6)}. If
\begin{equation}\label{1.12}
\max\big\{\frac{2\rho}{\rho+r-1},1\big\}<q<2,
\end{equation}
then  any solution $u_-$ of problem \eqref{Ppm}, with the minus sign,  satisfies
\begin{equation}\label{1.13}
\lim_{d(x)\to 0}\frac{u_-}{(\mathcal
{L}\circ\Phi_3)(d(x))}=\big[\frac{r-1}{q-2}\big]^{1/(q-1)},
\end{equation}
where $\Phi_3$ is given  by
\begin{equation}\label{1.14}
\int^\infty_{\Phi_3(t)}\frac{[\mathcal
{L}'(y)]^{\frac{1-q}{2-q}}}{y^\frac{1}{2-q}}dy=t, \forall x\in
(0,\tau)\quad \text{with small } \tau>0.
\end{equation}
and
$\Phi_3\in RV_{-\frac{q-2}{(q-1)(r-1)}}(0+)$.
\end{theorem}

\begin{remark} \rm 
There are many  functions satisfying (F1) and (F3), for example:
\begin{itemize}
\item[(1)] $f(u)=u^{\frac{\rho}{1-r}}(\ln(u+1))^\alpha$,
for all $\alpha\geq0$.
\item[(2)] $f(u)=u^{\frac{\rho}{1-r}} \exp\{(\ln u)^\alpha_1 (\ln_2 u)^\alpha_2 \cdots (\ln_m
u)^\alpha_m\}$, where $\alpha_i\in(0,1)$ and \\
$\ln_m(\cdot)=\ln(\ln_{m-1}(\cdot))$.
\item[(3)] $f(u)=c_0u^{\frac{\rho}{1-r}}\exp\{\int_0^u\frac{s(t)}{t}dt\}$,
$u\geq0$, $s(t)\in C[0,+\infty)$ is nonnegative such that
$\lim_{t\to\infty}s(t)=0$ and
$\lim_{t\to\infty}s(t)/t\in[0,+\infty)$.
\end{itemize}
\end{remark}

\begin{remark} \rm
Define $\phi_1(K(dx))=\Phi_1(t)$, then $\phi_1$
satisfies
\begin{equation}\label{1.15}
\int^\infty_{\phi_1(t)}\frac{[\mathcal
{L}'(y)]^{1/2}}{y^\frac{\rho+1}{2}[L_f(y)]^{1/2}}dy=t.
\end{equation}
Define $\phi_2(K(dx))=\Phi_2(t)$, then $\phi_2$
satisfies
\begin{equation}\label{1.16}
\int^\infty_{\phi_2(t)}\frac{\mathcal
{L}'(y)}{y^{\rho/q}[L_f(y)]^\frac{1+-}{q}}dy=t.
\end{equation}
\end{remark}

\begin{remark} \rm
When $k\in \mathcal{K}_l$ with $0<l<\infty$,
instead of  $0<l\leq1$, then $b(x)$ may be singular near the
boundary, namely $\lim_{d(x)\to0+}b(x)=\infty$.
\end{remark}

\begin{remark} \rm
The existence of boundary blow-up solutions to \eqref{Ppm}
for $l\in(0,1]$, has been shown in \cite[Lemma 2.5]{Z2}.
The existence of boundary blow-up solutions to \eqref{Ppm}
for $l\in(1,\infty)$, has been shown in
 \cite[Theorem 1.4 and Remark 1.7]{Z1}.
\end{remark}

\begin{remark} \rm
Note that, the asymptotic behavior of the
boundary blow-up solutions to \eqref{Ppm} is independent on $|\nabla u|$ if
\eqref{1.7} holds.
The asymptotic behavior of the boundary blow-up solutions
to \eqref{Ppm}, with the minus sign,  is independent on the nonlinear
terms $b(x)f(u)$ if \eqref{1.12} holds.
\end{remark}

\begin{remark} \rm 
The above Theorems \ref{thm1} and \ref{thm2} are
independent of the choice of $L_f$. Indeed, if $\Phi_1(t)$ and
$\Phi_1'(t)$ are defined by \eqref{1.3} corresponding to $L_f$ and
$L_f'$, respectively, by \eqref{1.4} we infer that
$\lim_{u\to+\infty}L_f/L_f'=1$, in view of \cite[Lemma 2.4]{Z2},
we derive that
$\lim_{t\to0}\Phi_1(t)/\Phi_1'(t)=1$, this fact, combined
with
$\lim_{t\to0}\Phi_1(t)=\lim_{t\to0}\Phi_1'(t)=+\infty$,
shows that
\begin{equation*}
\lim_{t\to0}\frac{(\mathcal {L}\circ\Phi_1)(t)}{(\mathcal
{L}\circ\Phi_1')(t)}=1.
\end{equation*}
Subject to $\Phi_2(t)$, the same conclusion is holds.
\end{remark}


\begin{remark} \rm
Let
\begin{equation*}
\mathcal {J}_1(\xi)=\lim_{d(x)\to0}b(x)\frac{f(\xi\mathcal
{L}(\Phi_1(t)))}{\xi(\mathcal {L}\circ\Phi_1)''(t)}.
\end{equation*}
Then a direct computation shows that
\begin{align*}
\mathcal {J}_1(\xi)
&= \lim_{d(x)\to0}b(x)\frac{f(\xi\mathcal
{L}(\Phi_1(t)))}{\xi(\mathcal {L}\circ\Phi_1)''(t)}   \\
&= \xi^{\frac{\rho}{1-r}-1}\lim_{d(x)\to0}b(x)\frac{f(\mathcal
{L}(\Phi_1(t)))}{\mathcal {L}''(\Phi_1(t)(\Phi_1'(t))^2+\mathcal
{L}'(\Phi_1(t))(\Phi_1''(t)}\\
&= \xi^{\frac{\rho}{1-r}-1}\lim_{d(x)\to0}\frac{b(x)}{k^2(d(x))}\lim_{d(x)\to0}\frac{f(\mathcal
{L}(\Phi_1(t)))}{(\Phi_1(t))^\rho
L_f(\Phi_1(t))}\\
&\times\lim_{d(x)\to0}\frac{k^2(d(x))(\Phi_1(t))^{\rho}L_f(\Phi_1(t))}{\mathcal
{L}''(\Phi_1(t)(\Phi_1'(t))^2+\mathcal
{L}'(\Phi_1(t))(\Phi_1''(t)}\\
&= c\xi^{\frac{\rho}{1-r}-1}\lim_{d(x)\to0}\frac{(-\Phi_1(t))^2\mathcal
{L}'(\Phi_1(t))}{\Phi_1(t)(\mathcal
{L}''(\Phi_1(t)(\Phi_1'(t))^2+\mathcal
{L}'(\Phi_1(t))(\Phi_1''(t))}\\
&= c\xi^{\frac{\rho}{1-r}-1}\lim_{d(x)\to0}\frac{1}{\frac{\Phi_1(t)\mathcal
{L}''(\Phi_1(t))}{\mathcal
{L}'(\Phi_1(t))}+\frac{\Phi_1''(t)\Phi_1(t)}{(-\Phi_1'(t))^2}}\\
&= \frac{2c\xi^{\frac{\rho}{1-r}-1}}{l(\rho+r-1)+2(r-1)}.
\end{align*}
it can be easily seen that $\xi_1$ satisfies $\mathcal{J}_1(\xi_1)=1$.

In a similar way we can prove that $\xi_2$ satisfies $\mathcal
{J}_2(\xi_2)=1$, where
\begin{equation*}
\mathcal {J}_2(\xi)=\lim_{d(x)\to0}\xi^{q-1}\frac{[(\mathcal
{L}\circ\Phi_2)'(t)]^q}{(\mathcal {L}\circ\Phi_2)''(t)}.
\end{equation*}
\end{remark}

\begin{remark} \rm
It is important to notice that by
Proposition \ref{prop5}, $\mathcal{L}'\in NRV_{-r}$ with $0\leq r<1$ implies
that $\mathcal {L}^\leftarrow\in NRV_{1/(1-r)}$,
then $f(u)=g(\mathcal {L}^\leftarrow (u))\in RV_{\rho/(1-r)}$,
instead of $f(u)\in RV_{\infty}$ for $r=1$.
This fact will bring a significant change in the explosion speed of
the large solution of \eqref{Ppm}.
Firstly, by \cite[Lemmas 2.1 and 2.2]{H}, we know that
$\Phi_1\in NRV_{-\frac{2}{l\rho}}(0+)$,
which defined by \eqref{1.3} for $r=1$, and
$\Phi_2\in NRV_{-\frac{q}{l\rho}}(0+)$,
which defined by \eqref{1.6} for $r=1$, we conclude that
$\mathcal {L}\circ\Phi_1\in RV_ 0$, $\mathcal {L}\circ\Phi_2\in RV_0$.
By  \eqref{1.2} and \eqref{1.5}, we know that the solution
regularly varying at infinity with index 0, namely the solution to
problem \eqref{Ppm} is slowly varying functions if $r=1$.

 While  we replace $\mathcal{L}'\in NRV_{-1}$ by the
hypothesis $\mathcal{L}'\in NRV_{-r}$ with $0\leq r<1$, according to
Lemma \ref{lem1}, Lemma \ref{lem2} and Proposition \ref{prop5}
 (see below), we get that
\begin{equation*}
\mathcal {L}\circ\Phi_1\in RV_ {-\frac{2(1-r)}{l(\rho+r-1)}} ,\quad
\mathcal {L}\circ\Phi_2\in RV_ {-\frac{q(1-r)}{l(\rho+q(r-1))}}.
\end{equation*}
where $\Phi_1$ ($\Phi_2$) defined by \eqref{1.3}\eqref{1.6} )with
$0\leq r<1$, in this case, the solution regularly varying at
infinity with index
$$
-\frac{2(1-r)}{l(\rho+r-1)}\quad \big(-\frac{q(1-r)}{l(\rho+q(r-1))}\big).
$$

Secondly, for $r=1$,  we know that it is sufficient to know the bounds
of
\begin{equation}\label{1.17}
\lim_{d(x)\to0+}\frac{b(x)}{k^2(x)},
\end{equation}
we can obtain that \eqref{1.2} holds, namely,
\begin{equation*}
0<\liminf_{d(x)\to0+}\frac{b(x)}{k^2(x)}\quad  \text{and}\quad
\limsup_{d(x)\to0+}\frac{b(x)}{k^2(x)}<\infty,
\end{equation*}
implies \eqref{1.2} holds.

While for $0\leq r<1$, in order to get \eqref{1.8}, the weight
function $b(x)$ should satisfy (B6), that is we need to know
the exact value of \eqref{1.17}.
Indeed, if \eqref{1.7} holds and
\begin{equation*}
\liminf_{d(x)\to0+}\frac{b(x)}{k^2(x)}\geq c_*,
\end{equation*}
we can prove that
\begin{equation*}
\lim_{d(x)\to 0}\frac{u}{(\mathcal
{L}\circ\Phi_1)(d(x))}\leq\big[\frac{l(\rho+r-1)+2(1-r)}{2c_*}
\big]^{\frac{1-r}{\rho+r-1}},
\end{equation*}
and
\begin{equation*}
\liminf_{d(x)\to0+}\frac{b(x)}{k^2(x)}\leq c_*,
\end{equation*}
implies
\begin{equation*}
\lim_{d(x)\to 0}\frac{u}{(\mathcal
{L}\circ\Phi_1)(d(x))}\geq\big[\frac{l(\rho+r-1)+2(1-r)}{2c_*}
\big]^{\frac{1-r}{\rho+r-1}}.
\end{equation*}
\end{remark}

 The outline of the article is as follows. Section 2 gives some
notion and  results from regular variation theory. The main
Theorem will be proved in Section 3.

\section{Preliminaries}

In this section, we collect some notions and properties of regularly
varying functions. For more details, we refer the reader to \cite{BT,R,S}.

\begin{definition}\label{df1}\rm
A positive measurable function $f$ defined on $[D,\infty)$ for some
$D>0$, is called regularly varying (at infinity) with index $q\in R$
(written $f\in RV_q$) if for all $\xi>0$
$$
\lim_{u\to\infty}\frac{f(\xi u)}{f(u)}=\xi^q.
$$
\end{definition}

When the index of regular variation $q$ is zero, we say that the
function is slowly varying. We say that $f(u)$ is regularly varying
(on the right) at the origin with index $q\in \mathbb{R}$
(in short $f\in RV_q(0+)$) provided $f(1/u)\in RV_{-q}$. The transformation
$f(u)=u^qL(u)$ reduces regular variation to slow variation. Some
typical example of slowly varying functions are given by:
(1) Every measurable function on $[A,\infty)$ which has a positive
limit at $\infty$.
(2) The logarithm $\log u$, its iterates $\log_m u$ and
powers of $\log_mu$.
(3)  $L(u)=\exp\{(\log u)^{1/3}\cos((\log u)^{1/3})\}$,
exhibits infinite oscillation in the sense that
\begin{equation*}
\lim _{u\to \infty}\inf L(u)=0 \quad\text{and} \quad
\lim_{u \to \infty}\sup L(u)=\infty.
\end{equation*}
This shows that the behavior at infinity for a slowly varying
function cannot be predicted.
Next we state a uniform convergence theorem,

\begin{proposition}\label{prop1}
The convergence $L(\xi
u)/L(u)\to1$ as $u\to\infty$ holds uniformly on each
compact $\varepsilon-$set in $(0,\infty)$.
\end{proposition}

 Now, we have some elementary properties of slowly varying functions.

\begin{proposition} \label{prop2}
If $L$ is slowly varying, then
\begin{itemize}
\item[(1)] For any $\alpha>0$, $u^\alpha L(u)\to\infty$,
$u^{-\alpha}L(u)\to0$  as $u\to\infty$;
\item[(2)] $(L(u))^\alpha$ varies slowly for every $\alpha\in \mathbb{R}$;
\item[(3)] If $L_1$ varies slowly, so do $L(u)L_1(u)$ and $L(u)+
L_1(u)$.
\end{itemize}
\end{proposition}

\begin{proposition}[Representation Theorem]\label{prop3}
The function $L(u)$ is slowly
varying if and only if it can be written in the form
\begin{equation}\label{2.1}
L(u)=M(u)\exp \big\{\int^u_B \frac{y(t)}{t}dt \big\}\quad (u\geq B)
\end{equation}
for some $B>0$, where $y\in C[B,\infty)$ satisfies
$\lim_{u\to\infty}y(u)=0$ and $M(u)$ is measurable on
$[B,\infty)$ such that
$lim_{u\to\infty}M(u)=M\in(0,\infty)$.
\end{proposition}

If $M(u)$ is replaced by $\hat{M}$ in \eqref{2.1}, we get a normalised
regularly varying function.

\begin{definition}\label{df2}\rm
A function $f(u)$ defined for $u>B$ is called a normalised regularly
varying function of index $q$ (in short $f\in NRV_q$) if it is $C^1$
and satisfies
\begin{equation}\label{2.2}
\lim_{u\to\infty}\frac{uf'(u)}{f(u)}=q.
\end{equation}
\end{definition}

Note that $f\in NRV_{q+1}$ if and only if $f$ is $C^1$ and $f'\in
RV_q$. And  $NRV_q(0+)$ (resp.,$ NRV_q$) denote the set of all
normalised regularly varying functions at 0 (resp.,$\infty$) of
index $q$. A typify example function $f(u)=u^{q+1}+\sin(u^{q+2})$
(defined for large $u$) belongs to $RV_{q+1}$ but not $NRV_{q+1}$.

Next we presente[Karamata's Theorem, direct half.

\begin{proposition} \label{prop4}
Let $f\in RV_q$ be locally bounded in $[A,\infty)$. Then
 (1) For any $j\geq-(q+1)$,
\begin{equation}\label{2.3}
\lim_{u\to\infty}\frac{u^{j+1}f(u)}{\int^u_Ax^jf(x)dx}=j+q+1.
\end{equation}
(2)  For any $j<-(q+1)$, (and for $j=-(q+1)$ if
$\int^\infty x^{-(q+1)}f(x) dx <\infty)$
\begin{equation}\label{2.4}
\lim_{u\to\infty}\frac{u^{j+1}f(u)}{\int^\infty_ux^jf(x)dx}=-(j+q+1).
\end{equation}
\end{proposition}

\begin{definition}\label{df3}\rm
A non-decreasing function $f$ defined on $(A,\infty)$ is
$\Gamma-$varying at $\infty$ (written $f\in \Gamma$) if
$\lim_{u\to \infty}f(u)=\infty$ and there exists $\chi:
(A,\infty)\to (0,\infty)$ such that
\begin{equation*}
\lim_{u\to\infty}\frac{f(u+\lambda\chi(u))}{f(u)}=
e^\lambda,\quad \text{for all } \lambda\in \mathbb{R}.
\end{equation*}
\end{definition}

The function $\chi$ is called an auxiliary function and is unique up
to asymptotic equivalence. The following functions $f$  with the
specified auxiliary functions $\chi$.
\begin{itemize}
\item[(1)] $f(x)=\exp(x^p)$  for $p>0$ with
\begin{equation*}
\chi=\begin{cases}
1, &\text{for }x\leq0,\\
p^{-1}x^{1-p}, &\text{for } x>0.
\end{cases}
\end{equation*}

\item[(2)] $f(x)=\exp(x\ln_+x)$ with
\begin{equation*}
\chi=\begin{cases}
1, &\text{for } x\leq1, \\
(\ln x)^{-1}, &\text{for }  x>1.
\end{cases}
\end{equation*}

\item[(3)] $f(x)=\exp(e^x)$ with $\chi=e^{-x}$.
\end{itemize}

For a non-decreasing function $H$ on R, we define the (left
continuous) inverse of $H$ by
\begin{equation*}
H^\leftarrow (y)=\inf \{s: H(s)\geq y\}.
\end{equation*}

\begin{proposition} \label{prop5}
We have
\begin{itemize}
\item[(i)] If $f(u)\in RV_q$, then $\lim_{u\to
\infty}\ln f(u)/\ln u=q$.
\item[(ii)] If $f_1(u)\in RV_q$ and $f_2(u)\in RV_s$ with $\lim_{u\to
\infty}f_2(u)=\infty$, then $f_1\circ f_2\in RV_{qs}$.
\item[(iii)] Suppose $f(u)$ is non-decreasing, $\lim_{u\to
\infty}f(u)=\infty$ and $f(u)\in RV_q$, $0<q<\infty$, then
$f^\leftarrow \in RV_{1/q}$.
\end{itemize}
\end{proposition}

Now we a Characterization of $\Phi_1$.
\begin{lemma} \label{lem1}
Suppose that $f$ satisfies {\rm (F3)}. Then
\begin{itemize}
\item[(i)] The function $\Phi_1$ given by \eqref{1.2} is well defined.
Moreover, $\Phi_1\in C^2(0,\tau)$ satisfies $\lim_{t\to 0+}
\Phi(t)=\infty$;
\item[(ii)] $\Phi_1\in NRV_{-\frac{2}{l(r+\rho-1)}}(0+)$ satisfies
\begin{equation*}
\lim_{t\to 0+}\frac{\ln_m\Phi_1(t)}{\ln_mt}=
\begin{cases} -\frac{2}{l(r+\rho-1)},  & m=1,\\
 -1,   & m\geq 2,
\end{cases}
\end{equation*}
where we set  $\ln_{m+1}(\cdot)=\ln(\ln_m(\cdot))$, $m\geq1$.
\item[(iii)] $\lim_{t\to 0+}\frac{\Phi_1(t)}{\Phi_1'(t)}
=\lim_{t\to0+}\frac{\Phi_1'(t)}{\Phi_1''(t)}
=\lim_{t\to0+}\frac{\Phi_1(t)}{\Phi_1''(t)}=0$.

\item[(iv)] $\lim_{t\to0+}\frac{\Phi_1''(t)\Phi_1(t)}{|\Phi_1'(t)|^2}
=1+\frac{l(r+\rho-1)}{2}$.

\item[(v)] If \eqref{1.7} holds, $\lim_{t\to0+}
\frac{(-\Phi_1'(t))^{2-q}}{(\Phi_1(t))^{r(q-1)-1}}=0$.
\end{itemize}
\end{lemma}

\begin{proof}
In a similar way as \cite[Lemma 3.4]{C2}, we can prove
(i)-(iv). Here we only prove (v).
We differentiate \eqref{1.3} to obtain
\begin{equation*}
(-\Phi'_1(t))^2\mathcal {L}'(\Phi_1(t))=(\Phi_1(t))^{\rho+1}
L_f(\Phi_1(t))k^2(t),
\end{equation*}
then we have
\begin{align*}
&\lim_{t\to0+}\frac{(-\Phi_1'(t))^{2-q}}{(\Phi_1(t))^{r(q-1)-1}}\\
&=\lim_{t\to0+}L_f^{\frac{q-2}{2}}(\Phi_1(t))k^{q-2}(t))
\left((\Phi_1(t))^r\mathcal {L}'(\Phi_1(t))\right)^{\frac{q-2}{2}}
(\Phi_1(t))^{\frac{q-2}{2}(\rho+1-r)-(r-1)}.
\end{align*}
Recalling that $L_f \in NRV_0$, $\mathcal {L}\in NRV_{-r}$ and
\eqref{1.7}, by Proposition \ref{prop2}, we get that  (v)  holds.
\end{proof}

\begin{corollary} \label{coro2.2}
The function $\phi_1$ given by \eqref{1.15} is well
defined and satisfies
\begin{itemize}
\item[(i)] $\phi_1\in C^2(0,\tau)$ and $\lim_{t\to 0+}\Phi(t)=\infty$;
\item[(ii)] $\phi_1\in NRV_{-\frac{2}{r+\rho-1}}(0+)$ satisfies
\begin{equation*}
\lim_{t\to 0+}\frac{\ln_m\phi_1(t)}{\ln_mt}=
\begin{cases} -\frac{2}{r+\rho-1},  & m=1,\\
 -1,  & m\geq 2.
\end{cases}
\end{equation*}
\item[(iii)] $\lim_{t\to 0+}\frac{\phi_1(t)}{\phi_1'(t)}
=\lim_{t\to0+}\frac{\phi_1'(t)}{\phi_1''(t)}
=\lim_{t\to0+}\frac{\phi_1(t)}{\phi_1''(t)}=0$.
\item[(iv)] $\lim_{t\to0+}\frac{\phi_1''(t)\phi_1(t)}{|\phi_1'(t)|^2}
 =1+\frac{2}{r+\rho-1}$.
\item[(v)] $\lim_{t\to0+}\frac{(-\phi_1'(t))^{2-q}}{(\phi_1(t))^{r(q-1)-1}}=0$.
\end{itemize}
\end{corollary}
 Next we Characterize $\Phi_2$.

\begin{lemma} \label{lem2}
Suppose that $f$ satisfies {\rm (F3)}. Then
\begin{itemize}
\item[(i)] The function $\Phi_2$ given by \eqref{1.6} is well defined.
Moreover, $\Phi_2\in C^2(0,\tau)$ satisfies $\lim_{t\to 0^+}
\Phi_2(t)=\infty$;
\item[(ii)]  $\Phi_2\in NRV_{-\frac{q}{l(\rho+q(r-1))}}(0+)$ satisfies
\begin{equation}\label{2.5}
\lim_{t\to 0+}\frac{\ln_m\Phi_2(t)}{\ln_mt}=
\begin{cases}
-\frac{q}{l(\rho+q(r-1))},  & m=1,\\
-1,  & m\geq 2,
\end{cases}
\end{equation}
where we set  $\ln_{m+1}(\cdot)=\ln(\ln_m(\cdot)), m\geq1$;

\item[(iii)] $\lim_{t\to 0+}\frac{\Phi_2(t)}{\Phi_2'(t)}
=\lim_{t\to 0+}\frac{\Phi_2'(t)}{\Phi_2''(t)}
=\lim_{t\to 0+}\frac{\Phi_2(t)}{\Phi_2''(t)}=0$;

\item[(iv)]
$\lim_{t\to0+}\frac{\Phi_2''(t)\Phi_2(t)}{|\Phi_2'(t)|^2}
=1+\frac{l(\rho+q(r-1))}{q}$;

\item[(v)]  If \eqref{1.9} holds,
$\lim_{t\to0+}\frac{(\Phi_2'(t))^{2-q}}{(\Phi_2(t))^{r(1-q)+1}}=0$.
\end{itemize}
\end{lemma}

\begin{proof}
(i) Let $b>0$ such that $\mathcal {L}'( t)$,
 $L_f(t)$ are positive on $(b,\infty)$. Since $\mathcal {L}'\in
RV_{-r}$ and $L_f\in RV_0$, by Proposition \ref{prop2}, we have
\begin{equation*}
\lim_{t\to \infty} \frac{\mathcal {L}'(t)}
{t^{\rho/q}[L_f(t)]^{1/q}}t^{1+\tau}=\lim_{t\to
\infty}\frac{t^r\mathcal{L}'(t)}{[L_f(t)]^{1/q}}
t^{1+\tau-\frac{\rho}{q}-r}=0, \text{ for~ some }~ \tau\in
(0,\frac{\rho}{q}+r-1).
\end{equation*}
This shows that, for some $D>0$,
\begin{equation*}
h(x)=\int^\infty_x\frac{\mathcal {L}'(t)}
{t^{\rho/q}[L_f(t)]^{1/q}}dt<\infty, \quad \text{for all }x>D.
\end{equation*}
So, $\Phi_2$ is well defined on $(0,\tau)$ for small enough $\tau$.

We easily see that $h:(D,\infty)\to (0, h(D))$ is bijective
and $\lim_{t\to 0}\int^t_0k(s)ds=0$, $\Psi= h^{-1}(\int
^t_0k(s)ds)$ for $t\in(0,\tau)$, $\tau $ is small enough. Then
$\lim_{t\to0}\Phi_2(t)=\infty$. Moreover, by direct
differentiating, we have $\Phi_2\in C^2$.

(ii), Note that, $k(t)\in NRV_\theta(0+)$ with
$\theta=1/l-1$, then by Definition \ref{df2} and Proposition
\ref{prop4}, it follows that
\begin{equation} \label{2.6}
\lim_{t\to 0}\frac{tk'(t)}{k(t)}=\theta,~~
\lim_{t\to0}\frac{\int^t_0k(s)ds}{tk(t)}= l,
\end{equation}
on the other hand, by \eqref{1.6}, we have
 \begin{equation} \label{2.7}
\frac{-\Phi_2'(t)\mathcal {L}'(\Phi_2(t))}
{\Phi_2(t)^{\rho/q}[L_f(\Phi_2(t))]^{1/q}}=k(t),~~
\forall t\in(0,\tau),
\end{equation}
thanks to Proposition \ref{prop4}, we obtain
\begin{equation*}
\lim_{t\to \infty}\frac{\mathcal
{L}'(t)}{t^{\frac{\rho}{q}-1}[L_f(t)]^{1/q}h(t)}
=-(1-\frac{\rho}{q}-r)=\frac{\rho}{q}+r-1,
\end{equation*}
hence, in view of \eqref{1.6},
\begin{equation} \label{2.8}
\lim_{t\to 0+}\frac{\mathcal {L}'(\Phi_2(t))}
{\Phi_2(t)^{\frac{\rho}{q}-1}[L_f(\Phi_2(t))]^{1/q}
\int^t_0k(s)dt}=\frac{\rho}{q}+r-1,
\end{equation}
which, together with \eqref{2.7}, yields,
\begin{equation} \label{2.9}
\lim_{t\to0+}\frac{\Phi_2'(t)\int^t_0k(s)ds}{\Phi_2(t)k(t)}
=-\frac{q}{\rho+q(r-1)},
\end{equation}
by \eqref{2.6} and \eqref{2.9},
\begin{equation} \label{2.10}
\lim_{t\to0+}\frac{t\Phi_2'(t)}{\Phi_2(t)}
=\lim_{t\to0+}\frac{\Phi_2'(t)\int^t_0k(s)ds}{\Phi_2(t)k(t)}\times
\frac{tk(t)}{\int^t_0k(s)ds}=-\frac{q}{l(\rho+q(r-1))},
\end{equation}
this implies
$$
\Phi_2\in NRV_{-\frac{q}{l(\rho+q(r-1))}}(0+).
$$
By \eqref{2.10} and L'Hospital's rule, we obtain
\begin{equation*}
\lim_{t\to0+}\frac{\ln\Phi_2(t)}{\ln t}
=\lim_{t\to0+}\frac{t\Phi_2'(t)}{\Phi_2(t)}=-\frac{q}{l(\rho+q(r-1))},
\end{equation*}
and
\begin{equation*}
\lim_{t\to0+}\frac{\ln(\ln\Phi_2(t))}{\ln(\ln t)}
=\lim_{t\to0+}\frac{t \Phi_2'(t)}{\Phi_2(t)}\cdot\frac{\ln
t}{\ln \Psi} =1,
\end{equation*}
we now prove \eqref{2.5} by induction, Let $m=n(n\geq2)$, we have
\begin{equation*}
\lim_{t\to0+}\frac{\ln_n\Phi_2(t)}{\ln_nt}=1.
\end{equation*}
Then, if $m=n+1$, we obtain
\begin{equation*}
\lim_{t\to0+}\frac{\ln_{n+1}\Phi_2(t)}{\ln_{n+1}t}
=\lim_{t\to0+}\frac{\ln(\ln_n\Phi_2(t))}{\ln(\ln_nt)}
=\lim_{t\to0+}\frac{\ln_nt}{\ln_n\Phi_2(t)}=1,
\end{equation*}
this prove \eqref{2.5}.

(iii) Following from (ii), $\Phi_2\in
NRV_{-\frac{q}{l(\rho+q(r-1))}}(0+)$, then the claim of (iii) is clear.

(iv) Differentiating \eqref{2.7}, we deduce that
\begin{equation}\label{2.11}
\begin{aligned}
\Phi_2''(t)
&= -\frac{\Phi_2'(t)k(t)\Phi_2(t)^{\frac{\rho}
{q}-1}[L_f(\Phi_2(t))]^{1/q}}{\mathcal{L}'(\Phi_2(t))} \\
&\quad \Big[\frac{\rho}{q}+\frac{k'(t)\Phi_2(t)}{k(t)\Phi_2'(t)}
+\frac{L'_f(\Phi_2(t))\Phi_2(t)}{qL_f(\Phi_2(t))}-\frac{\Phi_2(t)\mathcal
{L}''(\Phi_2(t))}{\mathcal {L}'(\Phi_2(t))} \Big],
\end{aligned}
\end{equation}
since $L_f\in NRV_0$ and $\mathcal {L}'\in NRV_{-r}$, by Definition
\ref{df2}, we have
\begin{eqnarray}\label{2.12}
\lim_{t\to0+}\frac{\Phi_2(t)L_f'(\Phi_2(t))}{L_f(\Phi_2(t))}=0,\quad
\lim_{t\to0+}\frac{\Phi_2(t)\mathcal {L}''(\Phi_2(t))}{\mathcal {L}'(\Phi_2(t))}
=-r,
\end{eqnarray}
which combined \eqref{2.7} with \eqref{2.11}, leads to
\[
\lim_{t\to0+}\frac{\Phi_2''(t)\mathcal{L}'(\Phi_2(t))}
{\Phi_2'(t)k(t)\Phi_2(t)^{\frac{\rho}{q}-1}[L_f(\Phi_2(t))]^{1/q}}
=-(r+\frac{l(\rho+q(r-1))}{q}),
\]
then, thanks to \eqref{2.7}, we have
\[
\lim_{t\to0+}\frac{\Phi_2''(t)\Phi_2(t)}{|\Phi_2'(t)|^2}
=r+\frac{l(\rho+q(r-1))}{q}.
\]

(v) In a similar way to Lemma \ref{lem1} (v), we can prove that (v)
holds, here we omit its proof.
\end{proof}

\begin{corollary} \label{coro2.3}
The function $\phi_2$ given by \eqref{1.16} is well
defined and
\begin{itemize}
\item[(i)]  $\phi_2\in C^2(0,\tau)$ and $\lim_{t\to 0^+}
\phi_2(t)=\infty$;
\item[(ii)]  $\phi_2\in NRV_{-\frac{q}{\rho+q(r-1)}}(0+)$ satisfies
\begin{equation*}
\lim_{t\to 0+}\frac{\ln_m\phi_2(t)}{\ln_mt}=
\begin{cases} -\frac{q}{\rho+q(r-1)},
& m=1, \\
-1,   & m\geq 2.
\end{cases}
\end{equation*}
\item[(iii)]  $\lim_{t\to 0+}\frac{\phi_2(t)}{\phi_2'(t)}
=\lim_{t\to 0+}\frac{\phi_2'(t)}{\phi_2''(t)}
=\lim_{t\to 0+}\frac{\phi_2(t)}{\phi_2''(t)}=0$.
\item[(iv)]
$\lim_{t\to0+}\frac{\phi_2''(t)\phi_2(t)}{|\phi_2'(t)|^2}=1+\frac{\rho+q(r-1)}{q}$.
\item[(v)]
$\lim_{t\to0+}\frac{(\phi_2'(t))^{2-q}}{(\phi_2(t))^{r(1-q)+1}}=0$.
\end{itemize}
\end{corollary}

Next we Characterize of $\Phi_3$.

\begin{lemma} \label{lem3}
Suppose that $f$ satisfies {\rm (F3)}. Then
\begin{itemize}
\item[(i)] $\Phi_3\in C^2(0,\tau)$ and $\lim_{t\to 0^+}
\Phi_3(t)=\infty$;

\item[(ii)] $\Phi_3\in NRV_{-\frac{q-2}{(q-1)(r-1)}}(0+)$ satisfies
\begin{equation*}
\lim_{t\to 0+}\frac{\ln_m\Phi_3(t)}{\ln_mt}=
\begin{cases} -\frac{q-2}{(q-1)(r-1)}, & m=1,\\
-1,  & m\geq 2;
\end{cases}
\end{equation*}
\item[(ii)]  $\lim_{t\to 0+}\frac{\Phi_3(t)}{\Phi_3'(t)}
=\lim_{t\to 0+}\frac{\Phi_3'(t)}{\Phi_3''(t)}
=\lim_{t\to 0+}\frac{\Phi_3(t)}{\Phi_3''(t)}=0$;
\item[(iii)]
$\lim_{t\to0+}\frac{\Phi_3''(t)\phi_2(t)}{|\Phi_3'(t)|^2}
=1+\frac{(q-1)(r-1)}{q-2}$;
\item[(iv)]
$\lim_{t\to0+}\frac{(\Phi_3'(t))^{2-q}}{(\Phi_3(t))^{r(1-q)+1}}=0$.
\end{itemize}
\end{lemma}

The Proof of the above Lemma is similarly to the previous lemmas, here we
omit it.

\begin{proposition}\label{prop6}
Let $\Psi(x, s, \xi )$ satisfy the following two conditions
\begin{itemize}
\item[(i)]  $\Psi$ is non-increasing in $s$ for each $(x, \xi )\in\Omega
\times \mathbb{R}^N$.
\item[(ii)] $\Psi$ is continuously differentiable with respect to the $\xi$
variable 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{proposition}

\section{Proof of main results}

In this section we  prove Theorems \ref{thm1}-\ref{thm3}. The proof of
each theorem will be split in two cases according to the values of
$l$.
Given $\delta>0$, for $\forall\beta\in(0,\delta)$, denote
\begin{gather*}
\Omega_\delta=\{x\in\Omega, 0<d(x)< \delta\},\quad
\partial\Omega_\delta=\{x\in\Omega, d(x)=\delta\},\\
\Omega^-_\beta=\Omega_{2\delta}\backslash\bar{\Omega}_\beta,\quad
\Omega^+_\beta=\Omega_{2\delta-\beta},
\end{gather*}


\begin{proof}[Proof of Theorem \ref{thm1}]
\textbf{Case 1}: (i) $l\in(0,1]$.  Set
\begin{eqnarray}\label{3.1}
\xi^\pm=\Big(\frac{2-2r+l(\rho+r-1)}{2(c_0\pm\varepsilon)}\Big)^\frac{1-r}{\rho+r-1},
\end{eqnarray}
where $\varepsilon\in(0, c_0)$ is arbitrary. We now diminish
$\delta\in(0, \beta/2)$, such that
\begin{itemize}
\item[(i)] $d(x)$ is a $C^2$-function on the set $\{x\in R^N:d(x)<2\delta\}$;
\item[(ii)] $k(x)$ is non-decreasing on $(0, 2\delta)$;
\item[(iii)] $ c_0k^2(d(x)-\beta)<b(x)<c_0k^2(d(x)+\beta)$;
for all $x\in\{x\in R^N:d(x, \partial\Omega)<2\delta\}$.
\end{itemize}

Let $\beta\in(0,\delta)$ be arbitrary, we define
\begin{gather*}
u_\beta^+=\xi^+\mathcal{L}(\Phi_1(d(x)+\beta)),
\quad x\in\Omega^+_\beta,\\
u_\beta^-=\xi^-\mathcal {L}(\Phi_1(d(x)-\beta)), \quad
x\in\Omega^-_\beta,
\end{gather*}
by the definition of $u_\beta^\pm$ we derive
\[
\nabla u_\beta^\pm=\xi^\pm\mathcal {L}'(\Phi_1(d(x)\pm\beta))
\Phi_1'(d(x)\pm\beta)\nabla d(x);
\]
since $|\nabla d(x)|=1$, it follows that
\begin{align*}
\Delta u_\beta^\pm
&= \xi^\pm\mathcal
{L}''(\Phi_1(d(x)\pm\beta))[\Phi_1'(d(x)\pm\beta)]^2
+\xi^\pm\mathcal{L}'(\Phi_1(d(x)\pm\beta))\Phi_1''(d(x)\pm\beta)\\
&\quad +\xi^\pm\mathcal{L}'(\Phi_1(d(x)\pm\beta))\Phi_1'(d(x)\pm\beta)\Delta
d(x).
\end{align*}
Then
\begin{align*}
& \Delta u_\beta^+\pm|\nabla u_\beta^+(x)|^q-b(x)f(u_\beta^+)\\
&\geq k^2(d(x)+\beta)f(u_\beta^+)(A_1^+(d(x)+\beta)+
A_2^+(d(x)+\beta)\\
&\quad +A_3^+(d(x)+\beta)\pm A_4^+(d(x)+\beta)-c_0),
\end{align*}
and
\begin{align*}
& \Delta u_\beta^-\pm|\nabla u_\beta^-(x)|^q-b(x)f(u_\beta^-)\\
&\leq k^2(d(x)-\beta)f(u_\beta^-)(A_1^-(d(x)-\beta)+
A_2^-(d(x)-\beta)\\
&\quad +A_3^-(d(x)-\beta)\pm A_4^-(d(x)-\beta)-c_0),
\end{align*}
where we denote
\begin{gather*}
A_1^\pm(t)= \frac{\xi^\pm\mathcal
{L}''(\Phi_1(t))[\Phi_1'(t)]^2}{k^2(t)f(\xi^\pm\mathcal{L}(\Phi_1(t)))},\quad
A_2^\pm(t)=\frac{\xi^\pm\mathcal{L}'(\Phi_1(t))\Phi_1''(t)}{k^2(t)
 f(\xi^\pm\mathcal{L}(\Phi_1(t)))},\\
A_3^\pm(t)= \frac{\xi^\pm\mathcal{L}'(\Phi_1(t))\Phi_1'(t)\Delta
d(x)}{k^2(t)f(\xi^\pm\mathcal{L}(\Phi_1(t)))},\quad
A_4^\pm(t)=\frac{(\xi^\pm)^q(\mathcal {L}'(\Phi_1(t)))^q
(-\Phi_1'(t))^q}{k^2(t)f(\xi^\pm\mathcal{L}(\Phi_1(t)))}.
\end{gather*}
According to $\mathcal{L}'\in NRV_{-r}$ and \eqref{1.4}, it is clear
that
\begin{gather*}
\lim_{t\to0}A_1^\pm=(\xi^\pm)^{1-\frac{\rho}{1-r}}\lim_{t\to0}\frac{\mathcal
{L}''(\Phi_1(t))[\Phi_1'(t)]^2}{k^2(t)f(\mathcal{L}(\Phi_1(t)))},\\
\lim_{t\to0}A_2^\pm=(\xi^\pm)^{1-\frac{\rho}{1-r}}\lim_{t\to0}\frac{\mathcal{L}'(\Phi_1(t))\Phi_1''(t)}{k^2(t)f(\mathcal{L}(\Phi_1(t)))},\\
\lim_{t\to0}A_3^\pm=(\xi^\pm)^{1-\frac{\rho}{1-r}}\lim_{t\to0}\frac{
\mathcal{L}'( \Phi_1(t))\Phi_1'(t)\Delta d(x)}{k^2(t)f(\mathcal{L}(
\Phi_1(t)))},\\
\lim_{t\to0}A_4^\pm=(\xi^\pm)^{q-\frac{\rho}{1-r}}\lim_{t\to0}\frac{(\mathcal
{L}'(\Phi_1(t)))^q
(-\Phi_1'(t))^q}{k^2(t)f(\mathcal{L}(\Phi_1(t)))}.
\end{gather*}
Thanks to \eqref{1.3} and \eqref{1.4}, we have
\begin{align*}
\lim_{t\to0}\frac{(\Phi_1(t))^\rho
L_f(\Phi_1(t))}{f(\mathcal{L}(\Phi_1(t)))}=1,\quad
\lim_{t\to0}\frac{(-\Phi_1(t))^2\mathcal{L}'(\Phi_1(t))}{k^2(t)
(\Phi_1(t))^{\rho+1}L_f(\Phi_1(t))}=1.
\end{align*}
By $\mathcal{L}'\in NRV_{-r}$, we obtain
\begin{align*}
\lim_{t\to0}\frac{\Phi_1(t)\mathcal{L}''(\Phi_1(t))}{\mathcal{L}'(\Phi_1(t))}=-r.
\end{align*}
Then
\begin{align*}
&\lim_{t\to0}\frac{\mathcal
{L}''(\Phi_1(t))[\Phi_1'(t)]^2}{k^2(t)f(\mathcal{L}(\Phi_1(t)))}\\
&= \lim_{t\to0}\frac{(\Phi_1(t))^\rho
L_f(\Phi_1(t))}{f(\mathcal{L}(\Phi_1(t)))}
\lim_{t\to0}\frac{(-\Phi_1(t))^2\mathcal{L}'(\Phi_1(t))}
{k^2(t)(\Phi_1(t))^{\rho+1}L_f(\Phi_1(t))}
\lim_{t\to0}\frac{\Phi_1(t)\mathcal{L}''(\Phi_1(t))}{\mathcal{L}'(\Phi_1(t))}
= -r,
\end{align*}
which implies
\[
\lim_{t\to0}A_1^\pm=\frac{-r}{(\xi^\pm)^{\frac{\rho}{1-r}-1}}.
\]
In view of Lemma \ref{lem1}, we have
\begin{align*}
&\lim_{t\to0}\frac{\mathcal{L}'(\Phi_1(t))\Phi_1''(t)}{k^2(t)f(\mathcal{L}
(\Phi_1(t)))}\\
&= \lim_{t\to0}\frac{(\Phi_1(t))^\rho
L_f(\Phi_1(t))}{f(\mathcal{L}(\Phi_1(t)))}
\lim_{t\to0}\frac{(-\Phi_1(t))^2\mathcal{L}'(\Phi_1(t))}
{k^2(t)(\Phi_1(t))^{\rho+1}L_f(\Phi_1(t))}
\lim_{t\to0}\frac{\Phi_1(t)\Phi_1''(t)}{(-\Phi_1(t))^2}\\
&= 1+\frac{l(r+\rho-1)}{2},
\end{align*}
which yields
\[
\lim_{t\to0}A_2^\pm=\frac{1+\frac{l(r+\rho-1)}{2}}{(\xi^\pm)
^{\frac{\rho}{1-r}-1}}.
\]
We notice that
\begin{align*}
A_3^\pm=A_2^\pm\frac{\Phi_1'(t)}{\Phi_1''(t)},
\end{align*}
and we infer that $\lim_{t\to0}A_3^\pm=0$.

Considering $A_4^\pm$, since
\begin{align*}
&\frac{(\mathcal {L}'(\Phi_1(t)))^q
(-\Phi_1'(t))^q}{k^2(t)f(\mathcal{L}(\Phi_1(t)))}\\
&= \frac{(\Phi_1(t))^\rho L_f(\Phi_1(t))}{f(\mathcal{L}(\Phi_1(t)))}
\frac{(-\Phi_1(t))^2\mathcal{L}'(\Phi_1(t))}
{k^2(t)(\Phi_1(t))^{\rho+1}L_f(\Phi_1(t))}
((\Phi_1(t))^r\mathcal{L}'(\Phi_1(t)))^{q-1}\\
&\quad \times\frac{(-\Phi_1'(t))^{2-q}}{(\Phi_1(t))^{r(q-1)-1}}
\end{align*}
we use Lemma \ref{lem1} (v) to obtain $\lim_{t\to0}A_4^\pm=0$.
Then we have
\begin{align*}
&\lim_{d(x)+\beta\to0}\left(A_1^+(d(x)+\beta)+
A_2^+(d(x)+\beta) +A_3^+(d(x)+\beta)\pm A_4^+(d(x)+\beta)-c_0\right)\\
& =+\varepsilon,
 \\
&\lim_{d(x)\to\beta}\left(A_1^-(d(x)-\beta)+
A_2^-(d(x)-\beta) +A_3^-(d(x)-\beta)\pm
A_4^-(d(x)-\beta)-c_0\right)\\
&=-\varepsilon,
\end{align*}
Now we can choose $\delta>0$ small enough so that
\begin{gather*}
\Delta u_\beta^+\pm|\nabla
d(x)|^q-b(x)f(u_\beta^+)\leq0, \quad x\in\Omega^+_\beta,\\
\Delta u_\beta^-\pm|\nabla d(x)|^q-b(x)f(u_\beta^-)\geq0, \quad
x\in\Omega^-_\beta,
\end{gather*}

Let $u(x)$ be a non-negative solution of  \eqref{Ppm} and
$M(2\delta)=\max_{d(x)\geq2\delta}u(x)$, 
$N(2\delta)=\mathcal {L}(\xi^-B(2\delta))$, it follows that
\begin{gather*}
u(x)\leq M(2\delta)+u_\beta^-,\quad x\in\partial \Omega^-_\beta,\\
u_\beta^+\leq N(2\delta)+u(x),\quad  x\in\partial\Omega^+_\beta,
\end{gather*}
This, combined with Proposition \ref{prop6}, yields
\begin{gather*}
u(x)\leq M(2\delta)+u_\beta^-, \quad x\in \Omega^-_\beta,\\
u_\beta^+\leq N(2\delta)+u(x),~ x\in\Omega^+_\beta,
\end{gather*}
for each $x\in\Omega^-_\beta\cap\Omega^+_\beta$, we have
$$
u_\beta^+-N(2\delta)\leq u\leq M(2\delta)+u_\beta^-,
$$
we arrive at
\begin{align*}
\frac{u_\beta^+}{(\mathcal
{L}\circ\Phi_1)(d(x))}-\frac{N(2\delta)}{(\mathcal
{L}\circ\Phi_1)(d(x))}
&\leq\frac{u(x)}{(\mathcal{L}\circ\Phi_1)(d(x))} \\
&\leq\frac{u_\beta^-}{(\mathcal
{L}\circ\Phi_1)(d(x))}+\frac{M(2\delta)}{(\mathcal
{L}\circ\Phi_1)(d(x))},
\end{align*}
we note that \eqref{1.8} and Proposition \ref{prop5} leads to
\[
\mathcal {L}\circ\Phi_1\in RV_{\frac{2(r-1)}{l(\rho+r-1)}}(0+)
\]
thus, we deduce that $\lim_{d(x)\to0}\mathcal{L}\circ\Phi_1(d(x)=\infty$.
Then letting $d(x)\to0$, we conclude \eqref{1.2}.

\noindent\textbf{Case 2:} $l\in(1,+\infty)$.
We now diminish $\delta\in(0, \beta/2)$, such that
\begin{itemize}
\item[(i)] $d(x)$ is a $C^2$-function for all $x\in\Omega_{2\delta}$;
\item[(ii)] $k(x)$ is non-increasing on $(0, 2\delta)$;
\item[(iii)] $c_0k^{2}(d(x))<b(x)<c_0k^2(d(x))$
for all $x\in\Omega_{2\delta}$.
\end{itemize}
Let $\beta\in(0,\delta)$ be arbitrary, we define
\begin{gather*}
u_\beta^+=\xi^+\mathcal{L}(\phi_1(d(x)+\beta)),
\quad x\in\Omega^+_\beta,\\
u_\beta^-=\xi^-\mathcal {L}(\phi_1(d(x)-\beta)), \quad
x\in\Omega^-_\beta,
\end{gather*}
where $\xi^\pm$ defined by \eqref{3.1} and $\phi_1$ defined by
\eqref{1.15}.
We infer that
\begin{align*}
\Delta u_\beta^\pm
&= \xi^\pm\mathcal
{L}''(\xi^\pm\phi_1(d(x)\pm\beta))[\phi_1'(d(x)\mp\beta)]^2k^2(d(x))\\
&\quad+\xi^\pm\mathcal{L}'(\xi^\pm\phi_1(d(x)\pm\beta))\phi_1''(d(x)
 \pm\beta)k^2(d(x))\\
&\quad+ \xi^\pm\mathcal{L}'(\xi^\pm\phi_1(d(x)\pm\beta))\phi_1'(d(x)
 \pm\beta)k'(d(x))\\
&\quad+ \xi^\pm\mathcal{L}'(\xi^\pm\phi_1(d(x)\pm\beta))\phi_1'(d(x)
 \pm\beta)k(d(x))\Delta d(x).
\end{align*}
Then we obtain
\begin{align*}
& \Delta u_\beta^+\pm|\nabla u_\beta^+(x)|^q-b(x)f(u_\beta^+)\\
&\geq k^2(d(x)+\beta)f(u_\beta^+)(A_1^+(d(x)+\beta)+
A_2^+(d(x)+\beta)\\
&\quad + A_3^+(d(x)+\beta)+A_4^+(d(x)+\beta)\pm A_5^+(d(x)+\beta)-c_0),
\end{align*}
and
\begin{align*}
& \Delta u_\beta^-\pm|\nabla u_\beta^-(x)|^q-b(x)f(u_\beta^-)\\
&\leq k^2(d(x)-\beta)f(u_\beta^-)(A_1^-(d(x)-\beta)+
A_2^-(d(x)-\beta)\\
&\quad + A_3^-(d(x)-\beta)+A_4^-(d(x)-\beta)\pm A_5^-(d(x)-\beta)-c_0),
\end{align*}
where we denote
\begin{gather*}
A_1^\pm(t)= \frac{(\xi^\pm)^2\mathcal
{L}''(\xi^\pm\phi_1(t))[\phi_1'(t)]^2}{f(\mathcal{L}(\xi^\pm\phi_1(t)))},\quad
A_2^\pm(t)=\frac{\xi^\pm\mathcal{L}'(\xi^\pm\phi_1(t))\phi_1''(t)}{f(\mathcal{L}(\xi^\pm\phi_1(t)))},\\
A_3^\pm(t)= \frac{\xi^\pm\mathcal{L}'(\xi^\pm\phi_1(t))
 \phi_1'(t)k'(d(x))}{k^2(d(x))f(\mathcal{L}(\xi^\pm\phi_1(t)))},\\
A_4^\pm(t)=\frac{\xi^\pm\mathcal{L}'(\xi^\pm\phi_1(t))\Phi_1'(t)k'(d(x))\Delta
d(x)}{k^2(d(x))f(\mathcal{L}(\xi^\pm\phi_1(t)))},\\
A_5^\pm(t)= \frac{(\xi^\pm)^q(\mathcal {L}'(\xi^\pm\phi_1(t)))^q
(-\phi_1'(t))^q}{k^{2-q}(d(x))f(\mathcal{L}(\xi^\pm\phi_1(t)))}.
\end{gather*}
Following the  same arguments as above we obtain
\begin{gather*}
\lim_{t\to0}A_1^\pm(t)=\frac{-r}{(\xi^\pm)^{\rho+r-1}},\quad
\lim_{t\to0}A_2^\pm(t)=\frac{1+\frac{r+\rho-1}{2}}{(\xi^\pm)^{\rho+r-1}},\\
\lim_{t\to0}A_3^\pm(t)=\frac{\frac{r+\rho-1}{2}(l-1)}{(\xi^\pm)^{\rho+r-1}},\quad
\lim_{t\to0}A_4^\pm(t)=0,\quad \lim_{t\to0}A_5^\pm(t)=0.
\end{gather*}
Then we obtain
\begin{align*}
&\lim_{d(x)+\beta\to0}(A_1^+(d(x)+\beta)+ A_2^+(d(x)+\beta)\\
&+ A_3^+(d(x)+\beta)+A_4^+(d(x)+\beta)\pm A_5^+(d(x)+\beta)-c_0)=+\varepsilon,
\\
&\lim_{d(x)\to\beta}(A_1^-(d(x)-\beta)+ A_2^-(d(x)-\beta)\\
&+ A_3^-(d(x)-\beta)+A_4^-(d(x)-\beta)\pm
A_5^-(d(x)-\beta)-c_0)=-\varepsilon,
\end{align*}
The remaining arguments in  case 1 also apply here, so that
case 2 is proved.

(ii) The proof this case  follows from \cite[Lemma 2.4]{Z2}.
\end{proof}

\begin{proof}[Proof of Theorem \ref{thm2}]
\textbf{Case 1:} $l\in(0,1]$. Set
\begin{align*}
\xi^\pm=\Big(\frac{1}{c_q\pm\varepsilon}\Big)^{\frac{1}{\rho-q(1-r)}},
\end{align*}
where $\varepsilon \in (0, c_q)$ is arbitrary. We now diminish
$\delta>0$, such that
\begin{itemize}
\item[(i)] $d(x)$ is a $C^2-$function on the set $\{x\in R^N:
d(x)<2\delta\}$;
\item[(ii)] $k(x)$ is non-decreasing on $(0, 2\delta)$;
\item[(iii)] $c_qk^q(d(x)-\beta)<b(x)<c_qk^q(d(x)+\beta)$,
for all $x \in\{ x \in R^N:d(x, \partial\Omega)<2\delta\}$.
\end{itemize}
Let $\beta\in(0,\delta)$ be arbitrary, we define
\begin{gather*}
u_\beta^+=\mathcal {L}(\xi^+\Phi_2(d(x)+\beta)), \quad x\in\Omega^+_\beta,\\
u_\beta^-=\mathcal {L}(\xi^-\Phi_2(d(x)-\beta)), \quad x\in\Omega^-_\beta.
\end{gather*}
By the definition of $u_\beta^\pm$ we have
\begin{align*}
\nabla u_\beta^\pm=\xi^\pm\mathcal {L}'(\xi^\pm\Phi_2(d(x)\pm\beta))
\Phi_2'(d(x)\pm\beta)\nabla d(x).
\end{align*}
Since $|\nabla d(x)|=1$ it follows that
\begin{align*}
\Delta u_\beta^\pm
&= (\xi^\pm)^2\mathcal
{L}''(\xi^\pm(d(x)\pm\beta))[\Phi_2'(d(x)\pm\beta)]^2\\
&\quad +\xi^\pm\mathcal {L}'(\xi^\pm\Phi_2(d(x)\pm\beta))
\Phi_2''(d(x)\pm\beta)\\
&\quad+ \xi^\pm\mathcal
{L}'(\xi^\pm\Phi_2(d(x)\pm\beta)) \Phi_2'(d(x)\pm\beta)\Delta d(x)\,.
\end{align*}
Then 
\begin{align*}
& \Delta u_\beta^++|\nabla u_\beta^+(x)|^q-b(x)f(u_\beta^+)\\
&\geq k^q(d(x)+\beta)f(u_\beta^+)[B_1^+(d(x)+\beta)+
B_2^+(d(x)+\beta)\\
&\quad + B_3^+(d(x)+\beta)+B_4^+(d(x)+\beta)-c_q],
\end{align*}
and
\begin{align*}
& \Delta u_\beta^-+|\nabla u_\beta^-(x)|^q-b(x)f(u_\beta^-)\\
&\leq k^q(d(x)-\beta)f(u_\beta^-)[B_1^-(d(x)-\beta)+
B_2^-(d(x)-\beta)\\
&\quad+ B_3^-(d(x)-\beta)+B_4^-(d(x)-c_q],
\end{align*}
where we denote
\begin{gather*}
B_1^\pm(t)= \frac{(\xi^\pm)^2\mathcal
{L}''(\xi^\pm\Phi_2(t))[\Phi_2'(t)]^2}{k^q(t)f(\mathcal{L}(\xi^\pm\Phi_2(t)))},
\quad 
B_2^\pm(t)=\frac{\xi^\pm\mathcal{L}'(\xi^\pm\Phi_2(t))\Phi_2''(t)}{k^q(t)
 f(\mathcal{L}(\xi^\pm\Phi_2(t)))},\\
B_3^\pm(t)= \frac{\xi^\pm\mathcal{L}'(\xi^\pm\Phi_2(t))\Phi_2'(t)\Delta
d(x)}{k^q(t)f(\mathcal{L}(\xi^\pm\Phi_2(t)))},\quad
B_4^\pm(t)=\frac{(\xi^\pm)^q(\mathcal {L}'(\xi^\pm\Phi_2(t)))^q
(-\Phi_2'(t))^q}{k^q(t)f(\mathcal{L}(\xi^\pm\Phi_2(t)))}.
\end{gather*}
A direct computation shows that
\[
\lim_{t\to0}B_1^\pm(t)=0,\quad
\lim_{t\to0}B_2^\pm(t)=0,~~\lim_{t\to0}B_3^\pm(t)=0,\quad
\lim_{t\to0}B_4^\pm(t)=\frac{1}{(\xi^\pm)^{q(r-1)+\rho}}.
\]
Thus
\begin{align*}
&\lim_{d(x)+\beta\to0}\left(B_1^+(d(x)+\beta)+
B_2^+(d(x)+\beta) +B_3^+(d(x)+\beta)+ B_4^+(d(x)+\beta)-c_q\right)\\
&=+\varepsilon,\\
&\lim_{d(x)\to\beta}\left(B_1^-(d(x)-\beta)+
B_2^-(d(x)-\beta) +B_3^-(d(x)-\beta)+
B_4^-(d(x)-\beta)-c_q\right)\\
&=-\varepsilon,
\end{align*}
Similar arguments show that \eqref{1.11} holds.

\textbf{Case 2}: $l\in(1,\infty)$.
This case is similarly, here we omit it.
\end{proof}

\begin{proof}[Proof of Theorem \ref{thm3}]
The main idea is  the same as in the proof of Theorems \ref{thm1} and 
\ref{thm2}.
We consider two cases, and give the proof of the 
case $l\in(0,1]$, the other case is omitted.
Set
\begin{align*}
\xi^\pm=\Big(\frac{r-1}{(q-2)(1\pm\varepsilon)}\Big)^{\frac{1}{(q-1)(1-r)}}.
\end{align*}
Define
\[
u_\beta^\pm=\xi^\pm\mathcal{L}(\Phi_3(d(x)\pm\beta)), \quad
x\in\Omega^\pm_\beta.
\]
We infer that
\begin{align*}
& \Delta u_\beta^++|\nabla u_\beta^+(x)|^q-b(x)f(u_\beta^+)\\
&\geq (\xi^\pm)^q
(\mathcal{L}'(\Phi_3(d(x)+\beta)))^q(\Phi_3'(d(x)+\beta))^q[C_1^+(d(x)+\beta)+
C_2^+(d(x)+\beta)\\
&\quad + C_3^+(d(x)+\beta)-1-C_4^+(d(x)+\beta)],
\end{align*}
and
\begin{align*}
& \Delta u_\beta^-+|\nabla u_\beta^-(x)|^q-b(x)f(u_\beta^-)\\
&\leq (\xi^\pm)^q
(\mathcal{L}'(\Phi_3(d(x)-\beta)))^q(\Phi_3'(d(x)-\beta))^q[C_1^-(d(x)-\beta)+
C_2^-(d(x)-\beta)\\
&\quad + C_3^-(d(x)-\beta)-1+C_4^-(d(x)],
\end{align*}
where
\begin{gather*}
C_1^\pm(t)= \frac{\xi^\pm\mathcal
{L}''(\Phi_1(t))[\Phi_1'(t)]^2}{(\xi^\pm)^q
(\mathcal{L}'(\Phi_3(t)))^q(\Phi_3'(t))^q},\quad 
C_2^\pm(t)=\frac{\xi^\pm\mathcal{L}'(\Phi_1(t))\Phi_1''(t)}{(\xi^\pm)^q
(\mathcal{L}'(\Phi_3(t)))^q(\Phi_3'(t))^q},\\
C_3^\pm(t)= \frac{\xi^\pm\mathcal{L}'(\Phi_1(t))\Phi_1'(t)\Delta
d(x)}{(\xi^\pm)^q (\mathcal{L}'(\Phi_3(t)))^q(\Phi_3'(t))^q},\quad
C_4^\pm(t)=\frac{b(x)f(u_\beta^\pm)}{(\xi^\pm)^q
(\mathcal{L}'(\Phi_3(t)))^q(\Phi_3'(t))^q}.
\end{gather*}
A direct computation shows that
\begin{gather*}
\lim_{t\to0}C_1^\pm(t)=\frac{-r}{(\xi^\pm)^{(q-1)(1-r)}},\quad
\lim_{t\to0}C_2^\pm(t)=\frac{1+\frac{(q-1)(r-1)}{q-2}}{(\xi^\pm)^{(q-1)(1-r)}},
\\
\lim_{t\to0}C_3^\pm(t)=\lim_{t\to0}C_4^\pm(t)=0.
\end{gather*}
then we can choose $\delta>0$ small enough so that
\begin{gather*}
\Delta u_\beta^+\pm|\nabla
d(x)|^q-b(x)f(u_\beta^+)\leq0, \quad x\in\Omega^+_\beta,\\
\Delta u_\beta^-\pm|\nabla d(x)|^q-b(x)f(u_\beta^-)\geq0, \quad
x\in\Omega^-_\beta,
\end{gather*}
In a similar way we can prove that that \eqref{1.13} holds.
\end{proof}

\subsection*{Acknowledgments}
Shufang Liu  was supported by the Presidential Foundation of Gansu
Normal University  for Nationalities (No. 12-15).
Yonglin Xu was supported by the Fundamental Research Funds for the Central
Universities (No. 31920130006)
and by the Middle-Younger Scientific Research Fund (No. 12XB39).

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\end{document}