\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 $$\Delta u -|\nabla u|^q=b(x)g(u), \quad u>0 \quad \text{in } \Omega,\quad u\big|_{\partial \Omega}=+\infty, \label{e1.1}$$ 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 $$\Delta u=b(x)g(u) \quad\text{in } \Omega,\quad u\big|_{\partial \Omega}=+\infty. \label{e1.2}$$ 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 $$\int_1^\infty\frac {ds}{g(s)}=\infty, \label{e1.3}$$ then \eqref{e1.2} has one solution if and only if $$\int_0^1(1-r)b(r)dr=\infty. \label{e1.4}$$ 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 $$b(x)\leq C(d(x))^{-2} (-\ln (d(x)))^{-p}, \label{e1.5}$$ 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 $10$, by the theory of ordinary differential equation and the comparison principle, Bandle and Giarrusso \cite{b3} showed that \\ (iii) if $10, \label{e1.9} then every solution to problem \eqref{e1.1} satisfies $$\lim_{d(x)\to 0}u(x)(d(x))^{(2-q)/(q-1)}=\xi_0, \label{e1.10}$$ 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 $$\frac{2-q}{q-1}=C_0\xi^{p-1}+(\frac {2-q}{q-1})^q \xi^{q-1}, \label{e1.11}$$ 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$00, \label{e1.13} then every solution $u$ to problem \eqref{e1.1} satisfies $$\lim_{d(x)\to 0}u(x)/{(-\ln (d(x)))}=\xi_0, \label{e1.14}$$ 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 $\varepsilon\in (0, \varepsilon_1)$ $$(1+\varepsilon x)^\alpha=1+\alpha\varepsilon x+o(\varepsilon^2). \label{e2.1}$$ \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 : 00 which only depends on \Omega such that $$d(x) \in C^2(\bar{\Omega}_{2\delta})\quad \text{and}\quad |\nabla d|\equiv 1\quad \text{on } \Omega_{2\delta}. \label{e2.2}$$ (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}. $$It follows by \eqref{e2.2} and \eqref{e1.8} that corresponding to \varepsilon\in (0,\varepsilon_1), there is \delta_\varepsilon\in (0, \delta) sufficiently small such that $$\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}$$ for all x\in \Omega_{2\delta_\varepsilon}, i=1, 2. \noindent (II) (i) When p=1, C_0\in \big(0, \frac{2-q}{(q-1)^2}\big). As the result of (I), we see that for \varepsilon\in (0, \varepsilon_1), $\left (\frac {2-q}{(q-1)^2}-C_0\right )\xi_{2\varepsilon}- \left(\frac{2-q}{q-1}\right)^q\xi_{2\varepsilon}^q =-(q-1)\left (\frac {2-q}{(q-1)^2}-C_0\right )\varepsilon+o(\varepsilon^2);$ and $\left (\frac {2-q}{(q-1)^2}-C_0\right )\xi_{1\varepsilon}- \left(\frac {2-q}{q-1}\right)^q\xi_{1\varepsilon}^q =(q-1)\left (\frac {2-q}{(q-1)^2}-C_0\right )\varepsilon+o(\varepsilon^2).$ (ii) When p\in (0, 1), C_0\in (0, \bar{C}). Since$$ \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 \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)))|\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 |\xi_{i\varepsilon}d(x)\Delta d(x)| +|b(x)(d(x))^2 g(-\xi_{i\varepsilon} \ln(d(x)))|