\documentclass[reqno]{amsart}
\AtBeginDocument{{\noindent\small
{\em Electronic Journal of Differential Equations},
Vol.~2000(2000), No.~62, pp.~1--6.\newline
ISSN: 1072-6691. URL: http://ejde.math.swt.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.swt.edu \quad ejde.math.unt.edu (login: ftp)}
\thanks{\copyright 2000 Southwest Texas State University.}
\vspace{1cm}}
\begin{document}
\title[\hfilneg EJDE--2000/62\hfil Analysis of a singular ODE]
{A geometric analysis of a singular ODE related to the study of a
quasilinear PDE}
\author[Jukka Tuomela \hfil EJDE--2000/62\hfilneg]
{ Jukka Tuomela }
\address{Jukka Tuomela \hfill\break\indent
Department of Mathematics,
University of Joensuu,
PL 111, 80101 Joensuu, Finland}
\email{jukka.tuomela@joensuu.fi}
\date{}
\thanks{Submitted April 14, 2000.
Revised September 12, 2000. Published October 11, 2000.}
\subjclass{35J60, 35B65, 34C10}
\keywords{singular ODE, quasilinear PDE}
\begin{abstract}
In this note we complement the analysis of a singular ODE given in
\cite{kopazu}. Using geometric arguments we are able to settle the structure
of the existence and non-existence regions in the parameter space and improve
the nonexistence bound.
\end{abstract}
\newtheorem{teo}{Theorem}
\newtheorem{lem}{Lemma}
\newtheorem{koro}{Corollary}
\maketitle
\section{Introduction}
In \cite{kopazu} existence and uniqueness questions of radial solutions of a
class of quasilinear elliptic PDEs in a ball, having strong dependence in the
gradient, are analysed in terms of the related singular ODE, see also
\cite{pasic}. The authors transform the ODE to an integral equation and look
for the fixed points of this operator equation. We complement their analysis
using only elementary geometric methods. We show that in addition to the
regular solution analysed in \cite{kopazu} there exist also infinitely many
singular solutions. It is not quite clear if these `new' solutions can be
used to construct new solutions to the original PDE, because at least they
are too singular to yield strong solutions. All solutions blow up in finite
time, and we give a nonexistence result which for all parameter values is
sharper than the one given in \cite{kopazu}. Finally we prove that the regions
of existence and nonexistence in the parameter space have a very simple
structure, answering rather completely the question raised in \cite{kopazu}.
\section{Results}
In \cite{kopazu} the following ODE is considered
\begin{equation}
t^\varepsilon \omega'-g_0\gamma t^{\gamma+\varepsilon-1}-f_0\omega^\delta=0
\label{alkuperyht}
\end{equation}
where $f_0$ and $g_0$ are assumed to be positive constants and other
parameters will be specified below. We are only interested in nonnegative
solutions which satisfy this equation for $t>0$ and which can be continuously
extended to $t=0$.
To simplify the study we first observe
\begin{lem} Let $u$ be a solution of
\begin{equation}
t^\varepsilon u'-\gamma t^{\gamma+\varepsilon-1}-u^\delta=0
\label{perusyht}
\end{equation}
and let us define $\omega(t)=c_1 u(c_2 t)$. Then $\omega$ is a solution of
\eqref{alkuperyht} if
\begin{align*}
c_2&=\big(f_0g_0^{\delta-1}\big)^{1/\beta}\\
c_1&=g_0\big(f_0g_0^{\delta-1}\big)^{-\gamma/\beta}=g_0c_2^{-\gamma}
\end{align*}
where $\beta=\gamma\delta-\gamma-\varepsilon+1$.
\label{skaalaus}
\end{lem}
\begin{proof} This is easy to check.
\end{proof}
Hence it is really sufficient to study \eqref{perusyht} because all solutions
of \eqref{alkuperyht} are obtained from the solutions of \eqref{perusyht} by
simple scaling. It is useful to make a further reduction.
\begin{lem} Let $x=t^{\varepsilon-1}$ and $y=u^{\delta-1}$ ($\varepsilon\ne 1$,
$\delta\ne 1$). The equation \eqref{perusyht} written in coordinates $(x,y)$
is
\begin{equation}
a x^2y'-y^2-\gamma x^ry^s=0
\label{muunyht}
\end{equation}
where $a=(\varepsilon-1)/(\delta-1)$, $r=1+\gamma/(\varepsilon-1)$ and
$s=(\delta-2)/(\delta-1)$.
\end{lem}
\begin{proof} This is also straightforward.
\end{proof}
Note that identically zero function is a solution of \eqref{muunyht}
(if $s>0$), but not of \eqref{alkuperyht}. When in the following we say
`any solution' or `all solutions', it will always mean any or all solutions,
except identically zero solution. The notation $y\sim x^m$ will mean that
there exist positive constants $d_1$ and $d_2$ such that
$d_1 x^m\le y(x)\le d_2 x^m$ for small enough $x$.
Now let us list here for convenience the conditions the parameters are
supposed to satisfy in the rest of the paper.
\begin{align*}
\delta&>1,\ \varepsilon>1,\ \gamma>(\varepsilon-1)/(\delta-1)\\[1mm]
a&>0,\ \gamma>0,\ r>1,\ s<1,\ r+s>2,\ \beta>0
\end{align*}
Note that inequalities in the second line are consequences of the first line.
Let us also consider the following equations
\begin{align}
\label{ayht} & ax^2y'-y^2=0\\
\label{byht} & ax^2y'-\gamma x^ry^s=0
\end{align}
By elementary integration they have solutions
\begin{align}
\label{aratk} & y(x)=\frac{ax}{1+cax}\\[2mm]
\label{bratk} & y(x)=\Big(x^{\gamma/(\varepsilon-1)}+c\Big)^{\delta-1}
\end{align}
Hence \eqref{ayht} has an infinite number of (analytic) solutions with
$y(0)=0$ and $y'(0)=a$, but a unique solution can be specified by giving
the second derivative at origin (note that $y''(0)=-2a^2c$).
This phenomenon can be understood nicely in a geometric way with jet spaces,
see \cite{julk34} and references therein for more information.
These solutions can be used to study the solutions of \eqref{muunyht}.
In the following we will often specify some point $(x_0,y_0)$. It is always
assumed that $x_0$ and $y_0$ are positive.
\begin{lem} The equation \eqref{muunyht} does not have any solutions with
$\lim_{x\searrow 0}y(x)>0$.
\label{yakseli}
\end{lem}
\begin{proof} Let $y$ (resp. $z$) be a solution of \eqref{muunyht}
(resp. \eqref{ayht}), going through the point $(x_0,y_0)$. Comparing the
derivatives we see that $y(x)\le z(x)$ for all $0\le x\le x_0$ from which the
claim follows.
\end{proof}
We are interested in positive solutions of \eqref{muunyht} which can be
continuously extended to origin. We need the following simple result.
\begin{lem} Let $y$ be a solution of \eqref{muunyht} such that $y(x)>0$ for
$00$ for
$00$ for $00$. If $y$ is a solution of
\eqref{muunyht}, then on this line
\[
y'(x)=c^2/a+\gamma c^sx^{r+s-2}/a
\]
Because $r+s>2$ this implies that there exist $c$ and $d$ such that $y'(x)0$ for $x>0$. Then for any $b>0$ we have
\[
x^{\alpha}\le y(x)\le a(1+b)x
\]
where $\alpha=\gamma(\delta-1)/(\varepsilon-1)>1$. The first inequality is
valid for all $x$ and the second for sufficiently small $x$.
\end{lem}
\begin{proof} The second inequality follows from the proof of Lemma
\ref{yakseli}. To prove the other inequality let $(x_0,y_0)$ be a point on the
curve $y=x^\alpha$ or below it. Further let $y$ (resp. $z$) be a solution of
\eqref{muunyht} (resp. \eqref{byht}) going through $(x_0,y_0)$. We must show
that there exists $\tilde x>0$ such that $y(\tilde x)=0$. Now if
$y_00$.
If $y_0=x_0^\alpha$, essentially the same argument applies because the
solutions of \eqref{muunyht} intersect the curve $y=x^\alpha$ transversely.
\end{proof}
It is a straightforward task to check that no solution can approach the
origin as $y\sim x^m$ with $11$ and let $K_1$ (resp. $K_2$) be the curve $y=x^\alpha$
(resp. $y=cx^\alpha$). Let
\[
\Omega_d=\big\{ (x,y)\in\mathbb{R}^2\,|\, x^\alpha\le y\le cx^\alpha\ ,
\quad 0\le x\le d \big\}
\]
Then the boundary of $\Omega_d$ consists of the parts of the curves $K_1$ and
$K_2$ and the vertical part which will be called $\Gamma_d$. Then it is easy
to check that for $d$ small enough, the solutions go out of $\Omega_d$ only
through $\Gamma_d$. Now extend all solutions going through $\Gamma_d$
backwards in `time'. By continuity of the flow, there is at least one solution
$y$ such that $(x,y(x))\in\Omega_d$ for $0\le x\le d$.
It remains to prove the uniqueness. We will use some simple modifications of
the results in \cite{hubwes}. Suppose that there are two solutions $y_1$ and
$y_2$ such that $(x,y_i(x))\in\Omega_d$ for $0\le x\le d$, $i=1$, $2$. Denote
$v(x)=y_2(x)-y_1(x)$; without loss of generality we may assume that
$v(x)\ge 0$. The uniqueness will follow if we succeed in proving that $v(d)=0$.
Let us first write \eqref{muunyht} as
\[
y'=f(x,y)=y^2/(ax^2)+\alpha x^{r-2}y^s
\]
Then we can express $v$ using
\[
v'(x)=y_2'(x)-y_1'(x)=f(x,y_2)-f(x,y_1)=\int_{y_1(x)}^{y_2(x)}
\frac{\partial f}{\partial y}(x,s) ds
\]
Hence we need an estimate for $\partial f/\partial y$. A simple computation
shows that in $\Omega_d$
\begin{equation}
\frac{\partial f}{\partial y}=\frac{2y}{ax^2}+s\alpha x^{r-2}y^{s-1}\le
\begin{cases}
\tfrac{2c}{a}\,x^{\alpha-2}+c^{s-1}\alpha x^{-1}&,\ 0~~-1$ this implies that $w(0)=c_2v(d)$ for some $c_2>0$.
On the other hand $v(x)\ge w(x)$ if $x\le d$. Combined with
$\lim_{x \searrow 0}v(x)=0$, this is possible only if $v(d)=0$. This is
essentially Theorem 4.7.5 (or rather its proof) in \cite[p. 188]{hubwes}.
Case 2: $0~~~~1$.
This is essentially the extension of Theorem 4.7.5 in \cite[p. 188]{hubwes},
outlined in Exercise 4.7\#3 \cite[p. 196]{hubwes}
\end{proof}
Hence there is one solution behaving like $x^\alpha$ (let us call this the
regular solution) and infinitely many solutions which behave like $x$ near
the origin (called singular solutions). Translating the above results back to
the original coordinates we get
\begin{koro}
If $T$ is small enough, equation \eqref{alkuperyht} has
\begin{itemize}
\item infinitely many solutions in $V_m$ if $m\le (\varepsilon-1)/(\delta-1)$
\item a unique solution in $V_m$ if $(\varepsilon-1)/(\delta-1)\gamma$
\end{itemize}
where
\[
V_m=\big\{u\in C[0,T]\,|\, 0\le u(t)\le M_u t^m\big\}
\]
and $M_u$ may depend on $u$.
\label{lause}
\end{koro}
Let us then take a look at the `lifespan' of these solutions.
\begin{teo} All solutions of \eqref{muunyht} blow up in a finite time.
\end{teo}
\begin{proof} If $p=(x_0,y_0)$ is above the line $y=ax$, then the solution of
\eqref{muunyht} through $p$ blows up by comparing it to the corresponding
solution of \eqref{ayht}. Further, solutions of \eqref{muunyht} intersect
$y=ax$ transversely. It remains to show that any solution starting below this
line will eventually meet it. Let $p=(x_0,y_0)$ be below the line $y=ax$; let
$y$ be a solution of \eqref{muunyht} through $p$ and $z$ a solution of
\eqref{byht} through $p$. Then $y(x)\ge z(x)$ for $x\ge x_0$. On the other
hand $z$ will meet $y=ax$ because it grows superlinearly.
\end{proof}
Finally let us consider the problem raised in \cite{kopazu}. What is the
structure of the set of parameter values $(f_0,g_0)$ for which the problem
\eqref{alkuperyht} has (resp. does not have) a solution on a given interval
$[0,T]$ ? Using Lemma \ref{skaalaus} this question has a very simple answer.
\begin{teo} There exists $c>0$ (depending on $T$, $\gamma$, $\varepsilon$
and $\delta$) such that the problem \eqref{alkuperyht}
\begin{itemize}
\item has a solution if $f_0g_0^{\delta-1}(\varepsilon-1)/(\delta-1)$),
and later the solutions cannot cross by the uniqueness of the solutions.
Hence it follows that \eqref{perusyht} has (infinitely many) solutions on the
interval $[0,T]$ if $T1$ and let $y$ (resp. $z$) be a solution of
\eqref{muunyht} (resp. \eqref{ayht}) through $p$. Then $y$ blows up before
\[
x=\frac{d}{d-1}\,(da)^{(\varepsilon-1)/\beta}
\]
because $z$ blows up there. Getting back to $(t,u)$ coordinates this means
that
\[
T_b\le \left(\frac{ad^\alpha}{(d-1)^{\alpha-1}}\right)^{1/\beta}
\]
The right hand side is minimised if $d=\alpha$ which gives the result.
\end{proof}
Note that our estimate is better than \eqref{vanha} for all parameter values
because
\[
\left(\frac{\alpha}{\alpha-1}\right)^{\alpha-1}~~