\documentclass[reqno]{amsart} \usepackage{hyperref} \AtBeginDocument{{\noindent\small \emph{Electronic Journal of Differential Equations}, Vol. 2009(2009), No. 90, pp. 1--15.\newline ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu \newline ftp ejde.math.txstate.edu} \thanks{\copyright 2009 Texas State University - San Marcos.} \vspace{9mm}} \begin{document} \title[\hfilneg EJDE-2009/90\hfil Second-order boundary estimates] {Second-order boundary estimates for solutions to singular elliptic equations} \author[C. Anedda\hfil EJDE-2009/90\hfilneg] {Claudia Anedda} \address{Claudia Anedda \newline Dipartimento di Matematica e Informatica, Universit\'a di Cagliari, Via Ospedale 72, 09124 Cagliari, Italy} \email{canedda@unica.it} \thanks{Submitted June 17, 2009. Published July 30, 2009.} \subjclass[2000]{35B40, 35B05, 35J25} \keywords{Elliptic problems; singular equations; \hfill\break\indent second order boundary approximation} \begin{abstract} Let $\Omega\subset R^N$ be a bounded smooth domain. We investigate the effect of the mean curvature of the boundary $\partial\Omega$ in the behaviour of the solution to the homogeneous Dirichlet boundary value problem for the singular semilinear equation $\Delta u+f(u)=0$. Under appropriate growth conditions on $f(t)$ as $t$ approaches zero, we find an asymptotic expansion up to the second order of the solution in terms of the distance from $x$ to the boundary $\partial\Omega$. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{corollary}[theorem]{Corollary} \section{Introduction} In this article, we study the Dirichlet problem \begin{equation}\label{eq1} \begin{gathered} \Delta u+f(u)=0 \quad \text{in }\Omega\\ u=0\quad \text{on } \partial \Omega, \end{gathered} \end{equation} where $\Omega$ is a bounded smooth domain in $\mathbb{R}^N$, $N\ge 2$, and $f(t)$ is a decreasing and positive smooth function in $(0,\infty)$, which approaches infinity as $t\to 0$. Equation \eqref{eq1} arises in problems of heat conduction and in fluid mechanics. Problems with singular data are discussed in many papers; see, for instance, \cite{GP,GR2, GR, Ka,ZC,ZY} and references therein. Let $f(t)=t^{-\gamma}$. For $\gamma>0$, in \cite{CRT} it is shown that there exists a positive solution continuous up to the boundary $\partial\Omega$. For $\gamma>1$, in \cite{LM1} it is shown that the solution $u$ satisfies $$ 00$ is discussed. For equation \eqref{eq1} in case $1<\gamma<3$, in \cite{BCP} it is shown that there exists a constant $B>0$ such that $$ \Big| u(x)-\Bigl(\frac{\gamma+1} {\sqrt{2(\gamma-1)}}\delta\Bigr)^\frac{2}{1+\gamma}\Big|< B\delta ^\frac{2\gamma}{\gamma+1}. $$ For $\gamma>3$, in \cite{MR} it is proved that $$ \Big| u(x)-\Bigl(\frac{\gamma+1} {\sqrt{2(\gamma-1)}}\delta\Bigr)^\frac{2}{1+\gamma}\Big|< B\delta ^\frac{\gamma+3}{\gamma+1}. $$ In \cite{ACP}, for $\gamma>3$, it is proved that $$ u(x)=\Bigl(\frac{\gamma+1} {\sqrt{2(\gamma-1)}}\delta\Bigr)^\frac{2}{1+\gamma} \Bigl[1+\frac{N-1}{3-\gamma}K\delta+ o(\delta)\Bigr], $$ where $K=K(x)$ stands for the mean curvature of the surface $\{x\in\Omega:\delta(x)=\textrm{constant}\}$. In this article we extend the latter estimate to more general nonlinearities. More precisely, assume \begin{equation}\label{eq2} \frac{f'(t)F(t)}{f^2(t)}=\frac{\gamma}{1-\gamma}+O(1)t^{\beta},\quad F(t)=\int_t^1f(\tau)d\tau \end{equation} where $\gamma > 3$, $\beta>0$ and $O(1)$ denotes a bounded quantity as $t\to 0$. In addition, we suppose there is $M$ finite such that for all $\theta\in (1/2,2)$ and for $t$ small we have \begin{equation}\label{eq3} \frac{|f''(\theta t)|t^2}{f(t)}\le M. \end{equation} An example which satisfies these conditions is $f(t)=t^{-\gamma}+t^{-\nu}$ with $0<\nu<\gamma$. Here $\beta=\min[\gamma-\nu,\gamma-1]$. When $\phi(\delta)$ is defined as \begin{equation}\label{eq4} \int_0^{\phi(\delta)}(2F(t))^{-1/2}dt=\delta \end{equation} and $\gamma>3$, we prove that \begin{equation}\label{5} u(x)=\phi(\delta)\Bigl[1+\frac{N-1}{3-\gamma}K\delta +O(1)\delta^{\sigma+1}\Bigr], \end{equation} where $\sigma$ is any number such that $0<\sigma<\min[\frac{\gamma-3}{\gamma+1},\frac{2\beta}{\gamma+1}]$. Note that $\phi$ is a solution to the one dimensional problem $$ \phi''+f(\phi)=0,\quad \phi(0)=0. $$ The estimate (\ref{5}) shows that the expansion of $u(x)$ in terms of $\delta$ has the first part which is independent of the geometry of the domain, and the second part which depends on the mean curvature of the boundary as well as on $\gamma$. For $1<\gamma<3$, the first part of the expansion is still independent of the geometry of the domain, but it is not possible to find the second part in terms of the mean curvature even in the special case $f(t)=t^{-\gamma}$, see \cite{ACP} for details. We observe that similar results are known for boundary blow-up problems, see \cite{AP2,Ba,BM}. In \cite{AP2} the problem $$ \Delta u=f(u),\quad u\to\infty \quad\text{as } x\to\partial \Omega $$ is discussed under the assumption that $f(t)$ is positive, increasing and satisfying $$ \frac{f'(t)\mathcal{F}(t)}{f^2(t)}=\frac{p}{1+p}+O(1)t^{-\beta},\quad \mathcal{F}(t)=\int_0^tf(\tau)d\tau, $$ where $p>1$, $\beta>0$ and $O(1)$ is a bounded quantity as $t\to\infty$. Under some additional conditions for $f$, in \cite{AP2} it is shown that \begin{equation}\label{eq15} u(x)=\Phi(\delta)\Bigl[1+\frac{N-1}{p+3}K\delta+ o(\delta)\Bigr], \end{equation} where $\Phi$ is defined as $$ \int_{\Phi(\delta)}^\infty(2F(t))^{-1/2}dt=\delta. $$ Note that $\Phi$ satisfies $$ \Phi''=f(\Phi),\quad \Phi(0)=\infty. $$ We underline that \eqref{eq15} holds for $p>1$, in contrast to (\ref{5}) which holds when $\gamma>3$. Moreover, when $f(t)=t^p$ with $p$ close to one, it is possible to find other terms in the expansion \eqref{eq15}. For example, the third term depends on the mean curvature $K$ as well as on its gradient $\nabla K$ (see \cite{AP1}). Instead, the estimate (\ref{5}) cannot be improved for any value of $\gamma$ because $0<\sigma<1$. \section{Preliminary results} Let $f(t)$ be a decreasing and positive smooth function in $(0,\infty)$, which approaches infinity as $t\to 0$. Assume condition \eqref{eq2} with $\gamma>1$ and $\beta>0$. Let us rewrite \eqref{eq2} as \begin{equation}\label{2.1} (F(t))^\frac{1}{1-\gamma}\Bigl(\frac{(F(t)) ^\frac{\gamma}{\gamma-1}}{f(t)}\Bigr)'=O(1)t^{\beta}. \end{equation} Since by \cite[Lemma 2.1]{LM2}, \begin{equation}\label{2.2} \lim_{t\to 0}\frac{F(t)}{f(t)}= 0, \end{equation} integration by parts on $(0,t)$ of (\ref{2.1}) yields \begin{equation}\label{2.3} \frac{F(t)}{t f(t)}=\frac{1}{\gamma-1}+O(1)t^{\beta}. \end{equation} Using the latter estimate and \eqref{eq2} again we find \begin{equation}\label{2.4} \frac{t f'(t)}{f(t)}=-\gamma+O(1)t^{\beta}. \end{equation} Let us write (\ref{2.4}) as $$ \frac{f'(t)}{f(t)}=-\frac{\gamma}{t}+O(1)t^{\beta-1}. $$ Integration over $(t,1)$ yields $$ \log\frac{f(1)}{f(t)}=\log t^\gamma+O(1). $$ Therefore, we can find two positive constants $C_1$, $C_2$ such that \begin{equation}\label{2.5} C_1t^{-\gamma}0$ smooth, decreasing for $t>0$ and such that $f(t)\to\infty$ as $t\to 0$. Assume condition \eqref{eq2} with $\gamma>3$. If $u(x)$ is a solution to problem \eqref{eq1} in $\Omega=A(\rho,R)$ and $v(r)=u(x)$ for $r=|x|$, then \begin{equation}\label{2.7} v(r)>\phi(R-r)-C\frac{\int_v^1(F(t))^{1/2}dt} {(F(v))^{1/2}}(R-r), \quad\tilde rc(F(v))^{1/2}$ for some positive constant $c$. Hence, $$ \int_{r_0}^r\frac{(v')^2}{s}ds >\frac{c}{r_0}\int_{r_0}^r(F(v))^{1/2}(-v')ds= \frac{c}{r_0}\int_{v}^{v_0}(F(t))^{1/2}dt. $$ By using the estimate (\ref{2.6}) for $F(t)$, the latter inequality implies that $\int_{r_0}^r\frac{(v')^2}{s}ds\to\infty$ as $r\to R$. As a consequence, we have $$ (N-1)\int_{r_0}^r\frac{(v')^2}{s}ds+F(v_0)>0 $$ for $r$ close to $R$. Since for $0< \epsilon< 1$, we have $1-[1-\epsilon]^{1/2}< \epsilon$. Using (\ref{2.12}) we find a constant $M$ such that \begin{equation}\label{2.15} 0\le \Gamma(r)\le \frac{\frac{2(N-1)}{r_0}\int_v^{v_0}(F(t))^{1/2}dt+F(v_0)}{F(v)}\le M\frac{\int_v^{v_0}(F(t))^{1/2}dt}{F(v)}. \end{equation} The inverse function of $\phi$ is $$ \psi(s)=\int_0^s\frac{1}{(2F(t))^{1/2}}dt. $$ Integration of (\ref{2.14}) over $(r,R)$ yields $$ \psi(v)=R-r-\int_r^R\Gamma(s)ds, $$ from which we find \begin{equation}\label{2.16} v(r)=\phi\bigl(R-r-\int_r^R\Gamma(s)ds\bigr). \end{equation} By (\ref{2.16}) we have \begin{equation}\label{2.17} v(r)=\phi(R-r)-\phi'(\omega)\int_r^R\Gamma(s)ds,\end{equation} with $$ R-r>\omega>R-r-\int_r^R\Gamma(s)ds. $$ Since $$ \phi'(\omega)=(2F(\phi(\omega)))^{1/2}, $$ and since the function $t\to F(\phi(t))$ is decreasing we have $$ \phi'(\omega)<\Bigl(2F\bigl(\phi(R-r-\int_r^R\Gamma(s)ds)\bigr)\Bigr)^{1/2} =(2F(v))^{1/2}, $$ where (\ref{2.16}) has been used in the last step. Hence, by (\ref{2.17}) we find $$ v(r)>\phi(R-r)-(2F(v))^{1/2}\int_r^R\Gamma(s)ds. $$ Using (\ref{2.15}) we also have \begin{equation}\label{2.18} v(r)>\phi(R-r)-(2F(v))^{1/2}M\int_r^R \frac{\int_v^{v_0}(F(t))^{1/2}dt}{F(v)}ds. \end{equation} We claim that the function $$ r\to \frac{\int_{v(r)}^{v_0}(F(t))^{1/2}dt}{F(v(r))} $$ is decreasing for $r$ close to $R$. This is equivalent to show that the function $$ s\to (F(s))^{-1}\int_{s}^{v_0}(F(t))^{1/2}dt $$ is increasing for $s$ close to $0$. We have \begin{align*} \frac{d}{ds}\Bigl[(F(s))^{-1}\int_{s}^{v_0}(F(t))^{1/2}dt\Bigr] &= (F(s))^{-2}f(s)\int_{s}^{v_0}(F(t))^{1/2}dt-(F(s))^{-1/2}\\ &=(F(s))^{-1/2}\Bigl[\frac{\int_{s}^{v_0}(F(t))^{1/2}dt}{(F(s)) ^\frac{3}{2}(f(s))^{-1}}-1\Bigr]. \end{align*} By (\ref{2.5}) and (\ref{2.6}) it follows that $\int_{s}^{v_0}(F(t))^{1/2}dt$ and $(F(s))^\frac{3}{2}(f(s))^{-1}$ tend to $\infty$ as $s\to 0$. Therefore, we can use de l'H\^opital rule and condition \eqref{eq2} to find $$ \lim_{s\to 0}\frac{\int_{s}^{v_0}(F(t))^{1/2}dt} {(F(s))^\frac{3}{2}(f(s))^{-1}}-1 =\lim_{s\to 0}\frac{2}{3+2(F(s))(f(s))^{-2}f'(s)}-1 =\frac{\gamma+1}{\gamma-3}. $$ Hence, since $\gamma>3$, the function $(F(s))^{-1/2}\int_{s}^{v_0}(F(t))^{1/2}dt$ is increasing for $s$ close to $0$, and the claim follows. Using this fact, inequality (\ref{2.7}) follows from (\ref{2.18}). To prove (\ref{2.8}), let us write equation (\ref{2.10}) as \begin{equation}\label{2.19} \frac{(v')^2}{2}=F(v)-F(v_0)+(N-1)\int_r^{r_0}\frac{(v')^2}{s}ds, \end{equation} with $\rho0$, we have \cite[Lemma 2.1]{LM2} $$ \lim_{r\to\rho}\frac{\int_r^{r_0}\frac{(v')^2}{t}dt}{(v')^2}=0. $$ Hence, (\ref{2.19}) implies $00 $$ for $r$ close to $\rho$. Since for $\epsilon>0$ we have $[1+\epsilon]^{1/2}-1<\epsilon$, the function $\Gamma(r)$ satisfies (\ref{2.15}) (possibly with a different constant $M$). Integration of (\ref{2.20}) over $(\rho,r)$ yields $$ \psi(v)=r-\rho+\int_\rho^r\Gamma (s)ds, $$ from which we find \begin{equation}\label{2.21} v(r)=\phi(r-\rho)+\phi'(\omega_1)\int_\rho^r\Gamma(s)ds, \end{equation} with $$ r-\rho<\omega_10$ there are $r_1$ and $r_2$ such that \begin{gather}\label{2.23} \phi(R-r)>v(r)>(1-\epsilon)\phi(R-r), \quad r_1\Bigl[1-C\frac{\int_v^1(F(t))^{1/2}dt}{(F(v))^{1/2}} \frac{R-r}{\phi(R-r)}\Bigr]\phi(R-r). $$ Since $F(t)$ is decreasing we have $$ \frac{\int_v^1(F(t))^{1/2}dt}{(F(v))^{1/2}}\le 1. $$ Moreover, putting $R-r=\psi(s)$ we have $$ \lim_{r\to R}\frac{R-r}{\phi(R-r)}=\lim_{s\to 0}\frac{\psi(s)}{s} \le \lim_{s\to 0}(2F(s))^{-1/2}=0. $$ The right hand side of (\ref{2.23}) follows from these estimates. By (\ref{2.20}) we have $$ \frac{v'}{(2F(v))^{1/2}}>1. $$ Integrating over $(\rho,r)$ we find $\psi(v)>r-\rho$, from which the left hand side of (\ref{2.24}) follows. By (\ref{2.8}) we have $$ v(r)<\Bigl[1+C\phi'(r-\rho)\frac{\int_v^1(F(t))^{1/2}dt}{F(v)} \frac{r-\rho}{\phi(r-\rho)}\Bigr]\phi(r-\rho). $$ We find $$ \lim_{r\to \rho}\frac{\int_v^1(F(t))^{1/2}dt}{F(v)}\le \lim_{r\to \rho}\frac{1}{(F(v))^{1/2}}=0. $$ Moreover, putting $r-\rho=\psi(s)$ we have $$ \frac{(r-\rho)\phi'(r-\rho)}{\phi(r-\rho)}=\frac{\psi(s)(2F(s))^{1/2}}{s} \le 1. $$ The right hand side of (\ref{2.24}) follows from these estimates. The corollary is proved. \end{proof} \begin{lemma} \label{lem2.3} If \eqref{eq2} holds with $\gamma>1$ and if $\phi(\delta)$ is defined as in \eqref{eq4}, then \begin{gather}\label{2.25} \frac{\phi'(\delta)}{\delta f(\phi(\delta))}=\frac{\gamma+1}{\gamma-1} +O(1)(\phi(\delta))^{\beta},\\ \label{2.26} \frac{\phi(\delta)}{\delta\phi'(\delta)}=\frac{\gamma+1}{2} +O(1)(\phi(\delta))^{\beta},\\ \label{2.27} \frac{\phi(\delta)}{\delta^2f(\phi(\delta))}= \frac{(\gamma+1)^2}{2(\gamma-1)}+O(1)(\phi(\delta))^{\beta},\\ \label{2.28} \phi(\delta)=O(1)\delta^\frac{2}{\gamma+1}, \end{gather} where $O(1)$ is a bounded quantity as $\delta\to 0$. \end{lemma} \begin{proof} By the relation $$ -1-2\Bigl(\frac{\gamma}{1-\gamma}+O(1)t^\beta\Bigr) =\frac{\gamma+1}{\gamma-1}+O(1)t^\beta, $$ using \eqref{eq2} we have $$ -1-2F(t)f'(t)(f(t))^{-2}=\frac{\gamma+1}{\gamma-1}+O(1)t^\beta, $$ where $O(1)$ is bounded as $t\to 0$. Multiplying by $(2F(t))^{-1/2}$ we find $$ -(2F(t))^{-1/2}-(2F(t))^{1/2}f'(t)(f(t))^{-2}= \frac{\gamma+1}{\gamma-1}(2F(t))^{-1/2}+O(1)t^\beta(2F(t))^{-1/2}, $$ and \begin{equation}\label{2.29} \bigl((2F(t))^{1/2}(f(t))^{-1}\bigr)'= \frac{\gamma+1}{\gamma-1}(2F(t))^{-1/2}+O(1)t^\beta(2F(t))^{-1/2}. \end{equation} Using (\ref{2.5}) and (\ref{2.6}) we find that $(2F(t))^{1/2}(f(t))^{-1}\to 0$ as $t\to 0$. Hence, integrating (\ref{2.29}) on $(0,s)$ we obtain \begin{equation}\label{2.30} (2F(s))^{1/2}(f(s))^{-1} =\frac{\gamma+1}{\gamma-1}\int_0^s\bigl(2F(t)\bigr)^{-1/2}dt +O(1)\int_0^{s}t^\beta(2F(t))^{-1/2}dt, \end{equation} where $O(1)$ is bounded as $s\to 0$. Since $$ \int_0^{s}t^\beta(2F(t))^{-1/2}dt\le s^\beta\int_0^{s}(2F(t))^{-1/2}dt, $$ equation (\ref{2.30}) implies $$ \frac{(2F(s))^{1/2}}{f(s)}= \frac{\gamma+1}{\gamma-1}\int_0^s\bigl(2F(t)\bigr)^{-1/2}dt +O(1)s^\beta\int_0^s (2F(t))^{-1/2}dt. $$ Putting $s=\phi(\delta)$ and recalling that $\phi'(\delta)=(2F(\phi(\delta)))^{1/2}$, estimate (\ref{2.25}) follows. Recall that \eqref{eq2} implies (\ref{2.3}). Hence, we have \begin{gather*} \frac{tf(t)}{2F(t)}=\frac{\gamma-1}{2}+O(1)t^\beta,\\ \frac{2F(t)+tf(t)}{2F(t)}=\frac{\gamma+1}{2}+O(1)t^\beta,\\ \frac{(2F(t))^{-1/2}+tf(t)(2F(t))^{-3/2}}{(2F(t)) ^{-1/2}}=\frac{\gamma+1}{2}+O(1)t^\beta,\\ \bigl(t(2F(t))^{-1/2}\bigr)'=\frac{\gamma+1}{2}(2F(t))^{-1/2}+O(1) t^\beta(2F(t))^{-1/2}. \end{gather*} By \eqref{2.6}, $t(2F(t))^{-1/2}\to 0$ as $t\to 0$. Hence, integrating over $(0,s)$ we find $$ s(2F(s))^{-1/2}=\frac{\gamma+1}{2}\psi(s)+O(1)\int_0^st^\beta(2F(t))^{-1/2}dt =\frac{\gamma+1}{2}\psi(s)+O(1)s^\beta\psi(s),$$ where $$ \psi(s)=\int_0^s(2F(t))^{-1/2}dt. $$ Since $\psi'(s)=(2F(s))^{-1/2}$ we find $$ \frac{s\psi'(s)}{\psi(s)}=\frac{\gamma+1}{2}+O(1)s^\beta. $$ Putting $s=\phi(\delta)$ and noting that $\phi$ is the inverse function of $\psi$ we get (\ref{2.26}). By (\ref{2.25}) we have $$ \frac{\phi(\delta)}{\delta^2f(\phi(\delta))}= \frac{\phi(\delta)\Bigl(\frac{\gamma+1}{\gamma-1}+ O(1)(\phi(\delta))^{\beta}\Bigr)}{\delta\phi'(\delta)}. $$ Using the latter estimate and (\ref{2.26}) we get $$ \frac{\phi(\delta)}{\delta^2f(\phi(\delta))} =\Bigl(\frac{\gamma+1}{2}+O(1)(\phi(\delta))^{\beta}\Bigr) \Bigl(\frac{\gamma+1}{\gamma-1}+ O(1)(\phi(\delta))^{\beta}\Bigr), $$ from which (\ref{2.27}) follows. Estimate (\ref{2.28}) follows immediately from (\ref{2.6}). The lemma is proved. \end{proof} \section{Main result} \begin{lemma} \label{lem3.1} Let $\Omega\subset \mathbb{R}^N$, $N\ge 2$ be a bounded smooth domain, and let $f(t)>0$ smooth, decreasing for $t>0$ and such that $f(t)\to\infty$ as $t\to 0$. Assume condition \eqref{eq2} with $\gamma>3$ and $\beta>0$. If $u(x)$ is a solution to problem \eqref{eq1}, then \begin{equation}\label{3.1} \phi(\delta)\bigl(1-C\delta\bigr)\phi(R-r)-C_1\frac{\int_v^1(F(t))^{1/2}dt}{(F(v))^{1/2}}(R-r), \quad \tilde r3$, by (\ref{2.6}) we find that $\int_t^1(F(\tau))^{1/2}d\tau$ and $t(F(t))^{1/2}$ approach infinite as $t$ approaches zero. Using de l'H\^opital rule and (\ref{2.3}) (which follows from \eqref{eq2}) we find \begin{equation}\label{3.3} \lim_{t\to 0}\frac{\int_t^1(F(\tau))^{1/2}d\tau}{t(F(t))^{1/2}}= \lim_{t\to 0}\frac{-(F(t))^{1/2}}{(F(t))^{1/2}-\frac{t f(t)}{2(F(t))^{1/2}}} =\lim_{t\to 0}\frac{1}{-1+\frac{tf(t)}{2F(t)}}=\frac{2}{\gamma-3}. \end{equation} Therefore, (\ref{3.2}) implies $$ v(r)>\phi(R-r)-C_2v(r)(R-r). $$ The latter estimate and the left hand side of (\ref{2.23}) yield $$ v(r)>\phi(R-r)\bigl(1-C_2(R-r)\bigr). $$ For $x$ near $\partial\Omega$ we have $\delta=R-r$; therefore, the latter estimate and the inequality $u(x)\ge v(x)$ yield the left hand side of (\ref{3.1}). Consider a new annulus of radii $\rho$ and $R$ containing $\Omega$ and such that its internal boundary is tangent to $\partial\Omega$ in $P$. If $w(x)$ is the solution of problem \eqref{eq1} in this annulus, by using the comparison principle for elliptic equations we have $u(x)\le w(x)$ for $x$ belonging to $\Omega$. Choose the origin in the center of the annulus and put $w(x)=w(r)$ for $r=|x|$. By Lemma \ref{lem2.1} with $w$ in place of $v$ we have \begin{equation}\label{3.4} w(r)<\phi(r-\rho)+C_1(r-\rho)\phi'(r-\rho)\frac{\int_w^1 (F(t))^{1/2}dt}{F(w)}, \quad \ \rho0$ smooth, decreasing for $t>0$ and such that $f(t)\to\infty$ as $t\to 0$. Assume condition \eqref{eq2} with $\gamma>3$ and $\beta>0$. If $u(x)$ is a solution to \eqref{eq1}, then \begin{equation}\label{3.7} \phi(\delta)\Bigl[1+\frac{N-1}{3-\gamma}K\delta -C\delta^{1+\sigma}\Bigr]f(\phi)\Bigl[-1+\gamma A\delta+\alpha\gamma \delta^{1+\sigma}-C_5\phi^{\beta}\delta-C_6\delta^2-C_7\alpha \phi^\beta\delta^{1+\sigma} \\ &\quad -C_8(\alpha\delta^{1+\sigma})^2\Bigr]. \end{aligned} \end{equation} Since by (\ref{3.8}), $$ -\Bigl(A+\frac{\gamma+1}{\gamma-1}(H-2A)\Bigr)=\gamma A, $$ by (\ref{3.11}) and (\ref{3.15}), we have \begin{equation}\label{3.16} \Delta w<-f(w)\end{equation} provided that \begin{align*} &C_1\phi^{\beta}\delta+C_2\delta^2 +\alpha\delta^{1+\sigma}\Bigl(-1+2(1+\sigma) \frac{\gamma+1}{\gamma-1}+\sigma(1+\sigma) \frac{(\gamma+1)^2}{2(\gamma-1)}+C_3\phi^{\beta}+C_4\delta\Bigr)\\ &<\alpha\gamma \delta^{1+\sigma}-C_5\phi^{\beta}\delta-C_6\delta^2- C_7\alpha \phi^\beta\delta^{1+\sigma}-C_8(\alpha\delta^{1+\sigma})^2. \end{align*} Rearranging terms, \begin{equation}\label{3.17} \begin{aligned} &(C_1+C_5)\phi^{\beta}\delta^{-\sigma}+(C_2+C_6)\delta^{1-\sigma}\\ &<\alpha\Bigl(\gamma+1-2(1+\sigma)\frac{\gamma+1}{\gamma-1} -\sigma(1+\sigma)\frac{(\gamma+1)^2}{2(\gamma-1)}-(C_3+C_7) \phi^{\beta}\\ &\quad -C_4\delta-C_8\alpha\delta^{1+\sigma}\Bigr). \end{aligned} \end{equation} Using (\ref{2.28}) we find $$ \phi^{\beta}\delta^{-\sigma}0. $$ Hence, we can take $\alpha_0$ large and $\delta_0$ small so that (\ref{3.13}) and (\ref{3.17}) hold for $\alpha\ge \alpha_0$, $\delta\le\delta_0$ with $\alpha\delta^{1+\sigma}\le\alpha_0\delta_0^{1+\sigma}$. Let us show now that we can choose $\delta_1$ so that $u(x)\le w(x)$ for $\delta(x)=\delta_1$. Using Lemma \ref{lem3.1} we have $$ w(x)-u(x)\ge\phi(\delta)\Bigl[\frac{(N-1)K}{3-\gamma}\delta +\alpha\delta^{1+\sigma}-C\delta\Bigr]. $$ Take $\alpha_0$ and $\delta_0$ so that (\ref{3.13}) and (\ref{3.17}) hold, and put $q=\alpha_0\delta_0^{1+\sigma}$. Decrease $\delta$ and increasing $\alpha$ according to $\alpha\delta^{1+\sigma}=q$ until $$ \frac{(N-1)K}{3-\gamma}\delta+q-C\delta>0 $$ for $\delta(x)=\delta_1$. Then $w(x)> u(x)$ for $\delta(x)=\delta_1$. By (\ref{3.16}) and the comparison principle \cite[Theorem 10.1]{GT}, it follows that $u(x)\le w(x)$ in $\{x\in\Omega:\delta(x)<\delta_1\}$. We look for a sub solutions of the form $$ v(x)=\phi(\delta)\bigl(1+A\delta-\alpha\delta^{1+\sigma}\bigr), $$ where $A$ and $\sigma$ are the same as before and $\alpha$ is a positive constant to be determined. Instead of (\ref{3.10}) now we have \begin{equation}\label{3.18} \begin{aligned} \Delta v&=f(\phi)\Bigl[-1-A\delta+\alpha\delta^{1+\sigma} -\Bigl(\frac{\gamma+1}{\gamma-1}+O(1)\phi^\beta\Bigr)\delta H \bigl(1+A\delta-\alpha\delta^{1+\sigma}\bigr)\\ &\quad +2\Bigl(\frac{\gamma+1}{\gamma-1}+O(1)\phi^\beta\Bigr)\delta \bigl(\nabla A\cdot\nabla\delta\delta+A-\alpha(1+\sigma) \delta^{\sigma}\bigr)\\ &\quad +\Bigl(\frac{(\gamma+1)^2}{2(\gamma-1)}+O(1)\phi^\beta\Bigr)\delta^2 \bigl(\Delta A \delta+2\nabla A\cdot\nabla\delta -A H-\alpha\sigma(1+\sigma)\delta^{\sigma-1}\\ &\quad + \alpha(1+\sigma)\delta^{\sigma}H\bigr)\Bigr]. \end{aligned} \end{equation} This means that we can find suitable constants $C_i$ (not necessarily the same as before) such that \begin{equation}\label{3.19} \begin{aligned} \Delta v&>f(\phi)\Bigl[-1-\Bigl(A+\frac{\gamma+1}{\gamma-1}(H-2A)\Bigr) \delta-C_1\phi^{\beta}\delta-C_2\delta^2\\ &\quad -\alpha\delta^{1+\sigma}\Bigl(-1+2(1+\sigma) \frac{\gamma+1}{\gamma-1}+\sigma(1+\sigma) \frac{(\gamma+1)^2}{2(\gamma-1)}+C_3\phi^{\beta} +C_4\delta\Bigr)\Bigr]. \end{aligned} \end{equation} After $\alpha$ is fixed, we consider only points $x\in\Omega$ such that \begin{equation}\label{3.20} -\frac{1}{2}-f(v) \end{equation} when \begin{align*} &-C_1\phi^{\beta}\delta-C_2\delta^2\\ &-\alpha\delta^{1+\sigma}\Bigl(-1+2(1+\sigma) \frac{\gamma+1}{\gamma-1}+\sigma(1+\sigma) \frac{(\gamma+1)^2}{2(\gamma-1)}+C_3\phi^\beta+C_4\delta\Bigr)\\ &>-\alpha\gamma \delta^{1+\sigma}+C_5\phi^{\beta}\delta+C_6\delta^2+C_7\alpha \phi^\beta\delta^{1+\sigma} +C_8(\alpha\delta^{1+\sigma})^2. \end{align*} Rearranging terms, \begin{equation}\label{3.23} \begin{aligned} &(C_1+C_5)\phi^{\beta}\delta^{-\sigma}+(C_2+C_6)\delta^{1-\sigma}\\ &<\alpha\Bigl(\gamma+1-2(1+\sigma)\frac{\gamma+1}{\gamma-1} -\sigma(1+\sigma)\frac{(\gamma+1)^2}{2(\gamma-1)}-(C_3+C_7) \phi^{\beta}\\ &\quad - C_4\delta-C_8\alpha\delta^{1+\sigma}\Bigr), \end{aligned} \end{equation} which is the same as (\ref{3.17}) (possibly with different constants). Therefore, we can take $\delta$ small and $\alpha$ large in order to satisfy this inequality. Take $\alpha_0$ and $\delta_0$ so that (\ref{3.20}) and (\ref{3.23}) hold, and put $q=\alpha_0\delta_0^{1+\sigma}$. Using Lemma \ref{lem3.1}, $$ v(x)-u(x)\le\phi(\delta)\Bigl[\frac{(N-1)K}{3-\gamma}\delta -\alpha\delta^{1+\sigma}+C\delta\Bigr]. $$ Decrease $\delta$ and increasing $\alpha$ according to $\alpha\delta^{1+\sigma}=q$ until $$ \frac{(N-1)K}{3-\gamma}\delta-q+C\delta<0 $$ for $\delta(x)=\delta_2$. Then $v(x)