\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 
$0<x\le x^*$. Then the (right hand) limit $\lim_{x\searrow 0}y(x)$ exists.
\label{raja}
\end{lem}
\begin{proof} Writing \eqref{muunyht} as
\[
             y'=y^2/(ax^2)+\gamma x^{r-2}y^s/a
\]
we see that $y'$ is positive (in the relevant region), and hence $y$ is 
monotonically increasing. The result follows because monotonic functions have 
right (and left) hand limits.
\end{proof}
\begin{koro} Let $y$ be a solution of \eqref{muunyht} such that $y(x)>0$ for 
$0<x\le x^*$. Let us set $y(0)=0$. Then $y\,:\,[0,x^*]\rightarrow \mathbb{R}$ 
is continuous.
\label{jatkuvuus}
\end{koro}
\begin{proof} By Lemma \ref{raja} the limit $\lim_{x\searrow 0}y(x)$ exists. 
By Lemma \ref{yakseli} this limit must be zero.
\end{proof}
Hence positive solutions can be continuously extended to zero, if they exist 
on $(0,x^*]$. Our next task is then to prove existence.
\begin{lem} For $x^*$ small enough, equation \eqref{muunyht} has an infinite 
number of solutions with $y(x)>0$ for $0<x\le x^*$ and 
$\lim_{x\searrow 0}y(x)=0$.
\label{origo}
\end{lem}
\begin{proof} Let us define 
\begin{align*}
\Omega_d^c&=\big\{(x,y)\in \mathbb{R}^2\,|\,0\le x\le d,\ y\ge cx\big\}\\
\Gamma_d^c&=\big\{(x,y)\in \mathbb{R}^2\,|\,0< x\le d,\ y= cx\big\}\cup 
\big\{(x,y)\in \mathbb{R}^2\,|\, x=d,\ y\ge cd\big\}
\end{align*} 
Consider the line $y=cx$ with some $c>0$. 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 $y'(x)<c$ 
for all $0< x\le d$. Hence no solution of \eqref{muunyht} enters $\Omega_d^c$ 
through $\Gamma_d^c$. Hence by Corollary \ref{jatkuvuus} we can continuously 
extend the solutions of  \eqref{muunyht}  which meet $\Gamma_d^c$ backwards 
in `time' to origin.
\end{proof}

\begin{lem} Let $y$ be a solution of \eqref{muunyht} with 
$\lim_{x\searrow 0}y(x)=0$ and $y(x)>0$ 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 
$y_0<x_0^\alpha$, $y(x)\le z(x)=(x^{\gamma/(\varepsilon-1)}+c)^{\delta-1}$ for
 some $c<0$ which shows that $\tilde x\ge (-c)^{(\varepsilon-1)/\gamma}>0$. 
 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 $1<m<\alpha$. 
\begin{teo} There is precisely one solution of \eqref{muunyht} with  
$y\sim x^\alpha$ for small $x$.
\end{teo}
\begin{proof} Let $c>1$ 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<s<1\\
              \tfrac{2c}{a}\,x^{\alpha-2}&,\ s\le 0
           \end{cases}
\label{arvio}
\end{equation}
Case 1: $s\le 0$. Now $v'(x)\le c_1 x^{\alpha-2} v(x)$ where $c_1=2c/a$. 
Consider the differential equation $w'(x)= c_1 x^{\alpha-2} w(x)$ with initial 
condition $w(d)=v(d)$. The solution is given by
\[
       w(x)=v(d)\exp(-\int_x^d c_1 s^{\alpha-2}ds)
\]
and because $\alpha-2>-1$ 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: $0<s<1$. Let us write the inequality in \eqref{arvio} a bit differently:
\[
      \frac{\partial f}{\partial y}\le
       \Big(  \frac{2c x^{\alpha-1}}{a}+\frac{\gamma}{c^{1-s}a}\Big) x^{-1}\le
       \Big(  \frac{2c d^{\alpha-1}}{a}+\frac{\gamma}{c^{1-s}a}\Big) x^{-1}
\]
Now  we can first choose $c$ sufficiently big so that $\gamma/(c^{1-s}a)\le 1/2$
 and then $d$ sufficiently small so that $2cd^{\alpha-1}/a\le 1/2$. Hence for 
 suitable $c$ and $d$ we have $\partial f/\partial y\le 1/x$ in $\Omega_d$.

Now proceed as before: $v'(x)\le x^{-1} v(x)$ and let $w$ be a solution of 
$w'(x)= x^{-1} w(x)$ with $w(d)=v(d)$. By elementary integration $w(x)=v(d)x/d$
 and again $v(x)\ge w(x)$ if $x\le d$. Now $v(x)\le (c-1)x^\alpha$ which 
implies that
\[
          v(d)x/d\le (c-1)x^\alpha
\]
for all $0\le x\le d$. This is possible only if $v(d)=0$ because $\alpha>1$. 
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 $(\varepsilon-1)/(\delta-1)<m \le\gamma$
\item no solutions in $V_m$ if $m>\gamma$
\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 $f_0g_0^{\delta-1}<c$ 
\item does not have a solution if $f_0g_0^{\delta-1}\ge c$
\end{itemize} 
\end{teo}
\begin{proof} Let $u$ be the regular solution of \eqref{perusyht}. Then there 
exists a well defined number $T_b$ such that 
$\lim_{t\nearrow T_b}u(t)=\infty$ and no $\tilde T<T_b$ has this property. 
Now the regular solution is the last one to blow up, because initially the 
regular solution ($u\sim t^\gamma$) must be `below' the singular solutions 
($u_s\sim t^{(\varepsilon-1)/(\delta-1)}$, $\gamma>(\varepsilon-1)/(\delta-1)$),
 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 $T<T_b$ and does not have any solutions on $[0,T]$ if 
$T\ge T_b$. To get the general result we scale the `time' variable by   
$c_2=\big(f_0g_0^{\delta-1}\big)^{1/\beta}$ like in Lemma \ref{skaalaus}. 
Then it is easily seen that $c=(T_b/T)^\beta$.
\end{proof}  
So the whole problem is essentially characterized by one parameter, namely 
$T_b$. It remains to estimate $T_b$. The results in \cite{kopazu} give
\begin{equation}
    \left(\frac{(\gamma\delta-\varepsilon+1)(\delta-1)^{\delta-1}}
    {\delta^\delta}\right)^{1/\beta} < T_b\le  
    \left(\gamma\delta^{\delta/(\delta-1)}\right)^{1/\beta}
\label{vanha}
\end{equation}
We can improve the upper bound.
\begin{lem} 
\[
         T_b\le \left(\gamma \left(\frac{\alpha}{\alpha-1}\right)^{\alpha-1}
         \right)^{1/\beta}
\]
\end{lem}
\begin{proof} Consider the problem \eqref{muunyht}. All solutions remain above 
the curve $y=x^\alpha$. This curve meets the line $y=dax$ at the point 
$(x_0,dax_0)$ where
\[
    x_0=(da)^{(\varepsilon-1)/\beta}
\]
Let $p=(x_0,dax_0)$ where $d>1$ 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
\[
   \left(\frac{\alpha}{\alpha-1}\right)^{\alpha-1}<e<\delta^{\delta/(\delta-1)}
\]
For example if one takes $\gamma=1$, $\delta=8$ and $\varepsilon=7$, then our 
result gives
\[
       T_b\le 7^{1/6}\approx 1.38
\]
while \eqref{vanha} gives
\[
       T_b\le 8^{8/7} \approx 10.8 
\]

\providecommand{\bysame}{\leavevmode\hbox to3em{\hrulefill}\thinspace}
\begin{thebibliography}{1}

\bibitem{hubwes}
J.~H. Hubbard and B.~H. West, \emph{Differential equations. {A} dynamical
  systems approach, {P}art 1}, Texts in applied mathematics, vol.~5, Springer,
  1991.

\bibitem{kopazu}
L.~Korkut, M.~Pa\v{s}i\'{c}, and D.~\v{Z}ubrini\'{c}, \emph{A singular {ODE}
  related to quasilinear elliptic equations}, Elect. J. Diff. Eq. \textbf{2000}
  (2000), no.~12, 1--37.

\bibitem{pasic}
M.~Pa\v{s}i\'{c}, \emph{Nonexistence of spherically symmetric solutions for
  $p$--{L}aplacian in the ball}, C. R. Math. Rep. Acad. Sci. Canada \textbf{21}
  (1999), 16--22.

\bibitem{julk34}
J.~Tuomela, \emph{On the resolution of singularities of ordinary differential
  systems}, Num. Alg. \textbf{19} (1998), 247--259.

\end{thebibliography}


\end{document}