\pdfoutput=1\relax\pdfpagewidth=8.26in\pdfpageheight=11.69in\pdfcompresslevel=9 \documentclass[twoside]{article} \usepackage{amsmath, amsfonts} \pagestyle{myheadings} \markboth{\hfil Boundary behavior of blow-up solutions \hfil EJDE--2002/78} {EJDE--2002/78\hfil Ahmed Mohammed \hfil} \begin{document} \title{\vspace{-1in}\parbox{\linewidth}{\footnotesize\noindent {\sc Electronic Journal of Differential Equations}, Vol. {\bf 2002}(2002), No. 78, pp. 1--15. \newline ISSN: 1072-6691. URL: http://ejde.math.swt.edu or http://ejde.math.unt.edu \newline ftp ejde.math.swt.edu (login: ftp)} \vspace{\bigskipamount} \\ % Boundary behavior of blow-up solutions to some weighted non-linear differential equations % \thanks{ {\em Mathematics Subject Classifications:} 49K20, 34B15, 34C11, 35J65. \hfil\break\indent {\em Key words:} Boundary behavior, blow-up solution, Keller-Osserman condition. \hfil\break\indent \copyright 2002 Southwest Texas State University. \hfil\break\indent Submitted March 21, 2002. Published September 19, 2002.} } \date{} % \author{Ahmed Mohammed} \maketitle \begin{abstract} We investigate, under appropriate conditions on the weight $g$ and the non-linearity $f$, the boundary behavior of solutions to $$\big(r^{\alpha}(u')^{p-1}\big)'=r^\alpha g(r)f(u),$$ $01$, $R>0$. For $00,\quad \int_0^1\bigl(\int_0^sf(t)dt\bigr)^{-1/p}ds=\infty, \\ f'(s)\ge 0\quad \mbox{for } s>0. \end{gathered} Solutions of (\ref{meqn}) are called blow-up solutions. Note that radial solutions of the$p$-Laplace equation $$\mathop{\rm div}(|\nabla u|^{p-2}\nabla u)=g(|x|)f(u),$$ in the ball$B:=B(0,R)\subseteq \mathbb{R}^N$satisfy the differential equation (\ref{meqn}) with$\alpha=N-1$. It will be convenient to rewrite (\ref{meqn}) as $$\label{meqnre} \bigl((u')^{p-1}\bigr)'+\frac{\alpha}{r}(u')^{p-1}=g(r)f(u), \quad u'(0)=0,\quad u(R)=\infty.$$ Throughout the paper we shall assume the following condition on the non-identically-zero weight function: $$\label{g0} g\geq 0,\quad g\in C([0,R)),\quad \mbox{and g is non-decreasing near R}.$$ Whenever the monotonicity condition on$g$is not required, we will state it explicitly. A necessary and sufficient condition for the existence of a solution to (\ref{meqn}), when$g(r)=C>0$, is the Keller-Osserman condition \cite{KEL,OSS}: $$\label{ok} \int_1^\infty\frac{ds}{F(s)^{1/p}} <\infty,\quad F(s)=\int_0^sf(t)dt.$$ If this condition holds for$f$, then the function $$\label{psidef} \psi(t):=\int_t^\infty\frac{1}{(qF(s))^{1/p}}\,ds,\quad \;t>0,$$ is well-defined, decreasing and convex. Here$q:=p/(p-1)$. Let$\phi$be the inverse function of$\psi$. Then$\lim_{s\to 0}\phi(s)=\infty$,$\lim_{s\to\infty}\phi(s)=0$, and$\phi$is known to satisfy the$1$-dimentional equation$(-(-\phi')^{p-1})'=f(\phi)$. The case$g=C>0$has been investigated extensively. For$p=2$,$\alpha=N-1$, asymptotic boundary estimates have been obtained in \cite{LMK, BAM2}. For$p>1$such estimates were obtained in \cite{GLP}. The question of existence of blow-up solutions when$g(r)$is bounded and vanishes on a set of positive measure has been considered in \cite{LAI}, when$p=2$and$\alpha=N-1$. Again for$p=2$and$\alpha=N-1$, the situation when$g(r)$is unbounded near$r=R$has recently been discussed in \cite{BAP}. In \cite{MPP}, equation (\ref{meqn}) was investigated in the general case when$g(r)$is unbounded near$R$. Our purpose in this paper is to study the boundary behavior of blow-up solutions of (\ref{meqn}) when$g$is unbounded near$R$and$f$satisfies the condition (\ref{ok}). For later reference, let us recall the following two results from \cite{RAW} and \cite{GLP}. The first is a comparison Lemma (see a proof in \cite{RAW}). For notational convenience in stating the Lemma, we let$L$denote the differential operator on the left hand side of equation (\ref{meqnre}) above. \begin{lemma}[Comparison] \label{comp} Let$0\leq a0$such that$u1, then $$\lim_{t\to\infty}\frac{t^p}{F(t)}=0.$$ \end{lemma} \section{Existence of blow-up solutions} Let us first make a few remarks about solutions of (\ref{meqnre}). Starting with the inequality (which follows from (\ref{meqnre})) ((u')^{p-1})'\leq g(r)f(u(r)),\quad 00 there is r_2 that depends on u and \epsilon such that \label{leq} (u')^{p-1}<\frac{r(1+\epsilon)}{\alpha+1}g(r)f(u),\quad r_2\frac{1-\alpha\epsilon}{\alpha+1}g(r)f(u(r)),\quad r_20\,. \end{gather} We recall the following two results from \cite{MPP}. \begin{theorem}\label{tg1}Assume that (\ref{g1}) holds. Then (\ref{meqn}) has a solution if and only if condition (\ref{ok}) holds. \end{theorem} \begin{theorem}\label{tg2} Assume that (\ref{g2}) holds. Then (\ref{meqn}) has a solution if and only if condition (\ref{ok}) fails to hold. \end{theorem} Let us give a proof of the sufficiency of the (\ref{ok}) condition for the existence of a blow-up solution in Theorem \ref{tg1}. For the proof of the rest of the assertions in Theorems \ref{tg1} and \ref{tg2}, we refer the reader to \cite{MPP}. \paragraph{Proof.} Let 0 w_k(0) then by the Comparison Lemma, we would have w_k(r)\leq w_{k+1}(r), 0 w_k(r) for some 0 w_k(r^*) and w'_{k+1}(r_*)=w'_k(r_*). But then by the Comparison Lemma again, we would have w_{k+1}>w_k on (r_*,R-1/k) which is obviously not possible. Therefore the claimed inequality holds. Using this and the fact that w_k and w_{k+1} satisfy equation (\ref{meqn}) we obtain \begin{align*}(w'_{k+1}(r))^{p-1}&=r^{-\alpha}\int_0^r s^\alpha g(s)f(w_{k+1}(s))\,ds \\&\leq r^{-\alpha}\int_0^r s^\alpha g(s)f(w_k(s))\,ds\\&=(w'_k(r))^{p-1}, \quad \;\;0k we have \begin{align*} |w_n(r)-w_n(t)|&=\big|\int_t^rw'_n(s)\,ds\big|\\ &\leq w'_n(\zeta)|r-t| \\ &\leq w'_{k+1}(R-1/k)|r-t|, \end{align*} where \zeta=\max\{r,t\}. Thus \{w_n\}_{n=k+1}^\infty is a bounded equicontinuous family in C([0,R-1/k]), and hence has a convergent subsequence. Let u be the limit. We will show that u is the desired blow-up solution of (\ref{meqnre}) in (0,R). To see this, note that for r\in[0,R-1/k] and n>k the solution w_n satisfies the integral equation w_n(r)=w_n(0)+\int_0^r\left(\int_0^t\left(\frac{s}{t}\right)^\alpha g(s)f(w_n(s))\,ds\right)^{\frac{1}{p-1}}\,dt. $$Letting n\to\infty we see that u satisfies the same integral equation. Since k is arbitrary we conclude that u satisfies equation (\ref{meqnre}) in (0,R). We now show that u(r)\to\infty as r\to R. Let r_1 such that (\ref{lesI}) holds for w_m in r_11 and 01,\quad \mbox{for any }0<\beta<1. This condition implies the following Lemma given in \cite{BAM3} without proof. Since subsequent results rely on this Lemma, we shall include the short proof for the readers' convenience as well as for completeness. \begin{lemma}\label{bm} Let \psi\in C[t_0,\infty). Suppose that \psi is strictly monotone decreasing and satisfies (\ref{psi}). Let \phi:=\psi^{-1}. Given a positive number \gamma there exist positive numbers \eta_\gamma, \delta_\gamma such that the following hold: \begin{enumerate} \item If \gamma >1, then \phi((1-\eta)\delta)\leq \gamma\phi(\delta) for all \eta\in[0,\eta_\gamma], \delta\in[0,\delta_\gamma] \item If \gamma <1, then \phi((1+\eta)\delta)\geq \gamma\phi(\delta) for all \eta\in[0,\eta_\gamma], \delta\in[0,\delta_\gamma]. \end{enumerate} \end{lemma} \paragraph{Proof.} We prove (1) only, as (2) is an easy consequence of (1). Let \gamma>1 be given. By hypothesis,$$ \liminf_{t\to\infty}\frac{\psi(\gamma^{-1}t)}{\psi(t)}=\alpha,\quad \mbox{for some } \alpha=\alpha(\gamma)>1. $$Let us fix 1<\theta<\alpha. Then for some t_\gamma we have $$\label{q} \psi(\gamma^{-1}t)\geq\theta\psi(t),\quad \mbox{for all }t\geq t_\gamma.$$ Now let \delta_\gamma:=\theta\psi(t_\gamma), and \eta_\gamma:=(\theta-1)/\theta. If 0<\delta\leq\delta_\gamma, then \phi(\delta/\theta)\geq t_\gamma, and hence by (\ref{q}), we obtain \phi(\delta/\theta)\leq \gamma\phi(\delta). Thus if 0<\eta\leq \eta_\gamma so that (1-\eta)\delta\geq \delta/\theta then we conclude that \phi((1-\eta)\delta)\leq \phi(\delta/\theta)\leq \gamma\phi(\delta) as desired. \hfill\square We will also need the following consequence of the above Lemma. \begin{lemma}\label{teclemma} Let \psi and \phi be as in Lemma \ref{bm}, and let \gamma>0, C>0 be given. Then there is a positive constant \delta_0=\delta_0(C,\gamma) and a positive integer m=m(C,\gamma) such that the following hold: \begin{enumerate} \item If \gamma>1, then \phi(C\delta)\leq \gamma^m\phi(\delta) for all 0<\delta<\delta_0 \item If \gamma<1, then \phi(C\delta)\geq \gamma^m\phi(\delta) for all 0<\delta<\delta_0. \end{enumerate} \end{lemma} \paragraph{Proof.} We prove (1) only, as (2) is an immediate consequence of (1). If C\geq 1, then there is nothing to prove. So assume that 01, let \eta_\gamma and \delta_\gamma be the positive constants given in Lemma \ref{bm}. Choose 0<\eta<\eta_\gamma such that 0m we have$$ ((w'_j)^{p-1})'+\frac{\alpha}{r}(w'_j)^{p-1}\geq (\alpha+1) g(M_m)f(w_j),\quad \;\mbox{on} \quad (0,R-1/m), by the Comparison Lemma we conclude that w_j(0)\leq w_m(0). In particular we have w_j(0)\leq w_k(0) for all j>k. On using Remark \ref{ginc} we get \begin{align*} w_j'(r)&\geq (qg(M_j))^{1/p}(F(w_j(r))-F(w_j(0)))^{1/p} \\&=(g(M_j))^{1/p}(qF(w_j(r)))^{1/p}\Big(1-\frac{F(w_j(0))} {F(w_j(r))}\Big)^{1/p}. \end{align*} Rewriting this expression, and using the inequality w_j(0)\leq w_k(0) for all j>k we find that \frac{w_j'(r)}{(qF(w_j(r)))^{1/p}}\geq (g(M_j))^{1/p} \Big(1-\frac{F(w_k(0))}{F(w_j(r))}\Big)^{1/p}. Integrating this on (r,R-1/j) we obtain \begin{align*} &\int_{w_j(r)}^\infty\frac{1}{(qF(t))^{1/p}}\,dt\\ &\geq (g(M_j))^{1/p}(R-1/j-r) \Big[\frac{1}{R-1/j-r} \int_r^{R-1/j}\Big(1-\frac{F(w_k(0))}{F(w_j(s))}\Big)^{1/p}\,ds\Big]. \end{align*} Note that the expression in the bracket on the right tends to one as r approaches R-1/j. Thus we have obtained \psi(w_j(r))\geq (g(M_j))^{1/p}(R-1/j-r)(1-\eta_j(r)), $$where$$ \eta_j(r)=1-\Big[\frac{1}{R-1/j-r}\int_r^{R-1/j}\big(1-\frac{F(w_k(0))} {F(w_j(s))}\big)^{1/p}\,ds\Big], so that \eta_j(r)\to 0 as r\to R-1/j. Hence we obtain, by Lemma \ref{bm}, that for any \epsilon>0 there is r_0(\epsilon,j) such that (recall that g(M_j)\geq g(r), 0\leq r\leq R-1/j) \begin{align*} w_j(r)&\leq \phi\bigl((g(M_j))^{1/p}(R-1/j-r)(1-\eta_j(r)\bigr)\\ &\leq (1+\epsilon)\phi\left(g^{1/p}(r)(R-1/j-r)\right), \quad r_0(\epsilon,j)r_0, \begin{align*} \int_{u(r)}^\infty\frac{1}{(qF(t))^{1/p}}\,dt&\leq \Big(1+\frac{u'(r_0)^p}{qg(r)F(u(r))}\Big)^{1/p}\int_r^Rg(t)^{1/p}\,dt\\ &= (1+\vartheta(r))\int_r^Rg(t)^{1/p}\,dt, \end{align*} where \vartheta(r)\to 0 as r\to R. Therefore, u(r)\geq\phi\Big( (1+\vartheta(r))\int_r^Rg(t)^{1/p}\,dt \Big). $$By Lemma \ref{bm}, we see that for any \epsilon>0 we can find r_1(\epsilon)>r_0 such that$$ u(r)>(1-\epsilon)\phi\Big(\int_r^Rg(t)^{1/p}\,dt\Big),\quad r_1(\epsilon)0 be given. Then by Lemma \ref{bm} we pick $\eta_\epsilon$ and $\delta_\epsilon$ such that $$\label{ep} \phi((1-\eta)\delta)\leq (1+\epsilon)\phi(\delta),\quad \mbox{and}\quad \phi((1+\eta)\delta)\geq (1-\epsilon)\phi(\delta),$$ for all $\eta\in[0,\eta_\epsilon],\;\delta\in[0,\delta_\epsilon]$. By (\ref{g3}), we choose $r_\epsilon>0$ such that for all $r_\epsilon1$. Let $\delta>0$ such that $\lambda<1-\delta$ or $1+\delta<\Lambda$. Then we pick some sequence $r_m$ that converges to $R$ so that $$(1+\delta)g^{1/p}(r_m)(R-r_m)<\int_{r_m}^Rg^{1/p}(s)\,ds,\quad\mbox{if }\Lambda>1$$ or $$(1-\delta)g^{1/p}(r_m)(R-r_m)>\int_{r_m}^Rg^{1/p}(s)\,ds,\quad\mbox{if }\lambda<1.$$ If the limits in (\ref{limits}) hold, then clearly we have the limit $$\lim_{m\to \infty}\frac{\phi\left( g^{1/p}(r_m)(R-r_m)\right)}{\phi \big(\int_{r_m}^Rg^{1/p}(s)\,ds\big)}=1.$$ By the mean value theorem, and using that $-\phi'$ is decreasing, we have \label{dec} \begin{aligned} \Big|&\frac{\phi\left( g^{1/p}(r_m)(R-r_m)\right)}{\phi\big(\int_{r_m}^Rg^ {\frac1p}(s)\,ds\big)}-1\Big| \\ &=\frac{-\phi'(\vartheta(r_m))\big|\int_{r_m}^Rg^ {\frac1p}(s)\,ds-g^{1/p}(r_m)(R-r_m)\big|}{\phi\big(\int_{r_m}^Rg^ {\frac1p}(s)\,ds\big)} \\ &\geq \frac{-\phi'\big(\int_{r_m}^Rg^{1/p}(s)\,ds\big)\cdot \big(\int_{r_m}^Rg^{1/p}(s)\,ds\big)}{\phi\big(\int_{r_m}^Rg^{1/p}(s) \,ds\big)}\,\Big|1-\frac{ g^{1/p}(r_m)(R-r_m)}{\int_{r_m}^Rg^{1/p}(s) \,ds}\Big| \end{aligned} From (\ref{dec}) and the above limit, we conclude that $$\lim_{m\to \infty}\frac{-\phi'\big(\int_{r_m}^Rg^{1/p}(s)\,ds\big)\cdot \big(\int_{r_m}^Rg^{1/p}(s)\,ds\big)}{\phi\big(\int_{r_m}^Rg^{1/p} (s)\,ds\big)}\Big|1-\frac{g^{1/p}(r_m)(R-r_m)}{\int_{r_m}^Rg^{1/p}(s)\,ds} \Big|=0.$$ Since $$\Big|1-\frac{g^{1/p}(r_m)(R-r_m)}{\int_{r_m}^Rg^{1/p}(s)\,ds}\Big|\geq \frac{\delta}{1+\delta}>0,$$ we conclude that the limit of the other factor must be zero. Recalling that $$-\phi'(t)=(qF(\phi(t)))^{1/p},$$ and hence letting $s=\phi(t)$ so that $s\to\infty$ if and only if $t\to0$, we have the following chain of equalities. \begin{align*} \lim_{t\to0}\frac{-\phi'(t)t}{\phi(t)}&=\lim_{t\to0}\frac{(qF(\phi(t)))^ {1/p}t} {\phi(t)}\\&=\lim_{s\to\infty}\frac{\psi(s)}{s/(qF(s))^{1/p}} \\&=\lim_{s\to\infty}\frac{-1}{1-(sf(s))/(pF(s))}=1/(E/p-1). \end{align*} In computing the above limit, we have used L'H\^{o}pital's rule which is justified by Lemma \ref{AF}. Going back to (\ref{dec}), since the first term on the right side of this inequality tends to zero as $m\to\infty$ we thus conclude that the limit in (\ref{f3}) is $E=\infty$. \hfill$\square$ \begin{theorem} Suppose $f$ satisfies \eqref{f1}, \eqref{ok}, \eqref{psi}, \eqref{f3} and $g$ satisfies \eqref{g1}. If for any solution $u$ of \eqref{meqnre}, we have either $$\limsup_{r\to R}\frac{u(r)}{\phi(g^{1/p}(r)(R-r))}=0,\quad \mbox{or} \quad \;\liminf_{r\to R}\frac{u(r)}{\phi(\int_r^Rg^{1/p}(s)\,ds)}=\infty,$$ then either $\lambda=0$ or $\Lambda=\infty$. \end{theorem} \paragraph{Proof.} We will prove the case when the limit superior is zero, the other case being similar. Suppose contrary to our conclusion we have $\lambda>0$ and $\Lambda<\infty$. Let us fix $\lambda_0$ and $\Lambda_0$ with $0<\lambda_0<\lambda$ and $\Lambda_0>\Lambda$. Then for some $r_0$ we have  \int_r^Rg^{1/p}(s)\,ds>\lambda_0g^{1/p} (r)(R-r), \quad \;r_0