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\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{}
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\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}
\end{equation}
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
\begin{equation} \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.
\end{equation}
Throughout the paper we shall assume the following condition on the
non-identically-zero weight function:
\begin{equation} \label{g0}
g\geq 0,\quad g\in C([0,R)),\quad
\mbox{and $g$ is non-decreasing near $R$}.
\end{equation}
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}:
\begin{equation} \label{ok}
\int_1^\infty\frac{ds}{F(s)^{1/p}} <\infty,\quad
F(s)=\int_0^sf(t)dt.
\end{equation}
If this condition holds for $f$, then the
function
\begin{equation} \label{psidef}
\psi(t):=\int_t^\infty\frac{1}{(qF(s))^{1/p}}\,ds,\quad \;t>0,
\end{equation}
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 $0\leq a**0$ 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}))
$$
((u')^{p-1})'\leq g(r)f(u(r)),\quad 00$ there is $r_2$ that depends on
$u$ and $\epsilon$ such that
\begin{equation} \label{leq}
(u')^{p-1}<\frac{r(1+\epsilon)}{\alpha+1}g(r)f(u),\quad r_2\frac{1-\alpha\epsilon}{\alpha+1}g(r)f(u(r)),\quad r_20\,.
\end{gather}
We recall the following two results from \cite{MPP}.
\begin{theorem}\label{tg1}Assume that (\ref{g1}) holds. Then
(\ref{meqn}) has a solution if and only if condition (\ref{ok}) holds.
\end{theorem}
\begin{theorem}\label{tg2}
Assume that (\ref{g2}) holds. Then
(\ref{meqn}) has a solution if and only if condition (\ref{ok}) fails to hold.
\end{theorem}
Let us give a proof of the sufficiency of the (\ref{ok}) condition for the
existence of a blow-up solution in Theorem \ref{tg1}.
For the proof of the rest of the assertions in Theorems \ref{tg1} and
\ref{tg2}, we refer the reader to \cite{MPP}.
\paragraph{Proof.}
Let $0 w_k(0)$ then by the
Comparison Lemma, we would have $w_k(r)\leq
w_{k+1}(r)$, $0 w_k(r)$ for some
$0 w_k(r^*)$ and
$w'_{k+1}(r_*)=w'_k(r_*)$. But then by the Comparison Lemma again,
we would have $w_{k+1}>w_k$ on $(r_*,R-1/k)$ which is obviously
not possible. Therefore the claimed inequality holds.
Using this and the fact that $w_k$ and $w_{k+1}$ satisfy equation
(\ref{meqn}) we obtain
\begin{align*}(w'_{k+1}(r))^{p-1}&=r^{-\alpha}\int_0^r s^\alpha g(s)f(w_{k+1}(s))\,ds
\\&\leq r^{-\alpha}\int_0^r s^\alpha g(s)f(w_k(s))\,ds\\&=(w'_k(r))^{p-1},
\quad \;\;0k$ we have
\begin{align*}
|w_n(r)-w_n(t)|&=\big|\int_t^rw'_n(s)\,ds\big|\\
&\leq w'_n(\zeta)|r-t| \\
&\leq w'_{k+1}(R-1/k)|r-t|,
\end{align*}
where $\zeta=\max\{r,t\}$. Thus $\{w_n\}_{n=k+1}^\infty$ is a
bounded equicontinuous family in $C([0,R-1/k])$, and hence has a
convergent subsequence. Let $u$ be the limit. We will show that
$u$ is the desired blow-up solution of (\ref{meqnre}) in $(0,R)$.
To see this, note that for $r\in[0,R-1/k]$ and $n>k$ the solution
$w_n$ satisfies the integral equation
$$
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.
$$
Letting $n\to\infty$
we see that $u$ satisfies the same integral equation. Since $k$ is
arbitrary we conclude that $u$ satisfies equation (\ref{meqnre})
in $(0,R)$. We now show that $u(r)\to\infty$ as $r\to R$. Let
$r_1$ such that (\ref{lesI}) holds for $w_m$ in $r_11$ and $01,\quad \mbox{for any }0<\beta<1.
\end{equation}
This condition implies the following Lemma given in \cite{BAM3}
without proof. Since subsequent results rely on this Lemma, we
shall include the short proof for the readers' convenience as well
as for completeness.
\begin{lemma}\label{bm}
Let $\psi\in C[t_0,\infty)$. Suppose that $\psi$ is strictly
monotone decreasing and satisfies (\ref{psi}). Let
$\phi:=\psi^{-1}$. Given a positive number $\gamma$ there exist
positive numbers $\eta_\gamma, \delta_\gamma$ such that the
following hold:
\begin{enumerate}
\item If $\gamma >1$, then
$\phi((1-\eta)\delta)\leq \gamma\phi(\delta)$ for all
$\eta\in[0,\eta_\gamma]$, $\delta\in[0,\delta_\gamma]$
\item If $\gamma <1$, then $\phi((1+\eta)\delta)\geq \gamma\phi(\delta)$
for all $\eta\in[0,\eta_\gamma]$, $\delta\in[0,\delta_\gamma]$.
\end{enumerate}
\end{lemma}
\paragraph{Proof.}
We prove (1) only, as (2) is an easy consequence of (1). Let
$\gamma>1$ be given. By hypothesis,
$$
\liminf_{t\to\infty}\frac{\psi(\gamma^{-1}t)}{\psi(t)}=\alpha,\quad
\mbox{for some } \alpha=\alpha(\gamma)>1.
$$
Let us fix $1<\theta<\alpha$. Then for
some $t_\gamma$ we have
\begin{equation} \label{q}
\psi(\gamma^{-1}t)\geq\theta\psi(t),\quad \mbox{for all }t\geq t_\gamma.
\end{equation}
Now let $\delta_\gamma:=\theta\psi(t_\gamma)$, and $\eta_\gamma:=(\theta-1)/\theta$.
If $0<\delta\leq\delta_\gamma$, then $\phi(\delta/\theta)\geq t_\gamma$, and hence by (\ref{q}),
we obtain $\phi(\delta/\theta)\leq \gamma\phi(\delta)$. Thus if $0<\eta\leq \eta_\gamma$ so that
$(1-\eta)\delta\geq \delta/\theta$ then we conclude that
$\phi((1-\eta)\delta)\leq \phi(\delta/\theta)\leq \gamma\phi(\delta)$ as desired.
\hfill$\square$
We will also need the following consequence of the above Lemma.
\begin{lemma}\label{teclemma}
Let $\psi$ and $\phi$ be as in Lemma \ref{bm}, and let $\gamma>0$, $C>0$
be given. Then there is a positive constant $\delta_0=\delta_0(C,\gamma)$
and a positive integer $m=m(C,\gamma)$ such that the following hold:
\begin{enumerate}
\item If $\gamma>1$, then $\phi(C\delta)\leq \gamma^m\phi(\delta)$
for all $0<\delta<\delta_0$
\item If $\gamma<1$, then $\phi(C\delta)\geq \gamma^m\phi(\delta)$
for all $0<\delta<\delta_0$.
\end{enumerate}
\end{lemma}
\paragraph{Proof.}
We prove (1) only, as (2) is an immediate consequence of (1). If
$C\geq 1$, then there is nothing to prove. So assume that $01$, let $\eta_\gamma$ and $\delta_\gamma$ be the
positive constants given in Lemma \ref{bm}. Choose
$0<\eta<\eta_\gamma$ such that $0m$ we have
$$
((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
\begin{equation} \label{ep} \phi((1-\eta)\delta)\leq
(1+\epsilon)\phi(\delta),\quad \mbox{and}\quad
\phi((1+\eta)\delta)\geq (1-\epsilon)\phi(\delta),
\end{equation}
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
\begin{equation} \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}
\end{equation}
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**