\documentclass[twoside]{article}
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\markboth{\hfil A Global Solution Curve \hfil}%
{\hfil Philip Korman \hfil}
\begin{document}
\setcounter{page}{119}
\title{\vspace{-1in}\parbox{\linewidth}{\footnotesize\noindent
{\sc  Differential Equations and Computational Simulations III}\newline
J. Graef, R. Shivaji, B. Soni J. \& Zhu (Editors)\newline
Electronic Journal of Differential Equations, Conference~01, 1997. \newline
ISSN: 1072-6691. URL: http://ejde.math.swt.edu or http://ejde.math.unt.edu
\newline ftp  147.26.103.110 or 129.120.3.113 (login: ftp)}
 \vspace{\bigskipamount} \\
 A Global Solution Curve for a Class of Semilinear Equations 
\thanks{ {\em 1991 Mathematics Subject Classifications:} 35J60.
\hfil\break\indent
{\em Key words and phrases:} Uniqueness of positive solution, Morse index.
\hfil\break\indent
\copyright 1998 Southwest Texas State University  and University of
North Texas. \hfil\break\indent
Published November 12, 1998.} }

\date{}
\author{Philip Korman}
\maketitle

\begin{abstract} 
We use bifurcation theory to give a simple proof of existence and
uniqueness of a positive solution for the problem
$$
\Delta u - \lambda u+u^p = 0 \quad
\mbox{for } |x| < 1, \quad u = 0 \quad \mbox{on } |x| = 1,
$$
where $x \in {\mathbb R}^n$, for any integer $n \geq 1$, and real 
$1<p < (n+2)/(n-2)$, $\lambda \geq 0$.
Moreover, we show that all solutions lie on a unique smooth curve of
solutions, and all solutions are non-singular. In the process we prove the
following assertion, which appears to be of independent interest: the Morse
index of the positive solution of
$$
\Delta u +u^p = 0 \quad
\mbox{for } |x| < 1, \quad u = 0 \quad \mbox{on } |x| = 1
$$
is one, for any $1<p < (n+2)/(n-2)$.
\end{abstract}

\newcommand{\ol}{\overline}
\newtheorem{thm}{Theorem}[section]
\newtheorem{lma}{Lemma}[section]

\section{Introduction}
Following the important work of B. Gidas, W.-M. Ni and L. Nirenberg \cite{GNN}
there has been a lot of interest in the radially symmetric 
solutions of the problem
\begin{equation}
\label{1.1}
\Delta u +  f(u) = 0 \quad
\mbox{for } |x| < 1, \quad u = 0 \quad \mbox{on } |x| = 1\,.
\end{equation}
Indeed they showed (under mild regularity assumption on $f(u)$) that all
positive solutions of (\ref{1.1}) are radially symmetric. (And the positive 
solutions are the only ones with a chance of being stable, a physically
significant property). Soon after publication of \cite{GNN} it was recognized
that uniqueness of radial solutions is very hard to prove, even for
relatively simple equations, see W.-M. Ni \cite{N}. 
One such equation is the focus of the present 
work:
\begin{equation}
\label{1.2}
\Delta u - \lambda u + u^p= 0 \quad
\mbox{for } |x| < 1, \quad u = 0 \quad \mbox{on } |x| = 1\,.
\end{equation}
Here $\lambda $ is a non-negative parameter, $p>1$ a real constant. M.K. Kwong
\cite{Kw} has shown that for $p$ subcritical, that is  
$p<(n+2)/(n-2)$ for $n>2$, the problem (\ref{1.2}) has a 
unique positive solution for any $\lambda \geq 0$, and in  
M.K. Kwong and Y. Li \cite{KL}, uniqueness is shown for any real $\lambda $,
and for more general equations; see also L. Zhang \cite{Z} and P.N. Srikanth 
\cite{S}. (In \cite{Kw} one has 
$\lambda =1$, but the ball has an arbitrary radius. By rescaling both $u$
and $x$, we see that the problems are equivalent.)
M.K. Kwong's result \cite{Kw} also covers
the cases of domain being an annulus and of whole of ${\mathbb R}^n$.
However, the proofs are rather involved (as the author himself mentions in
the introduction).  In this paper we use techniques from bifurcation theory
to give a short and simple proof of existence and uniqueness for a ball. 
 In addition to
simplicity, our proof provides some extra information on the solution set:
all solutions lie on one curve (which is significant for numerical 
computations), $u(0,\lambda)$ is increasing in $\lambda $, and it tends to infinity
with $\lambda $. (We denote the solution of (\ref{1.2}) by $u(r,\lambda )$.)

Next we recall a bifurcation theorem of M.G. Crandall and P.H. Rabinowitz,
which will be used to continue the solution curve, and to describe the
structure of the solution set near any possible turning point.

\begin{thm}
\cite{CR}  Let $X$ and $Y$ be Banach spaces.  
Let $(\ol{\lambda}, \ol{x}) \in {\mathbb R} \times X$ and let $F$
be a continuously differentiable mapping of an open neighborhood of
$(\ol{\lambda}, \ol{x})$ into $Y$.
Let the null-space $N(F_x(\ol{\lambda}, \ol{x}))= \mbox{span}
\; x_0$ be one-dimensional
and $\mbox{codim} \; R(F_x(\ol{\lambda}, \ol{x})) = 1$.  
Let $F_\lambda (\ol{\lambda}, \ol{x}) \not\in R(F_x(\ol{\lambda}, \ol{x}))$.  
If $Z$ is a complement of $\mbox{span} \; x_0$ in $X$, then the solutions of 
$F(\lambda,x) = F(\ol{\lambda}, \ol{x})$ near $(\ol{\lambda}, \ol{x})$ form a curve $(\lambda(s), x(s)) = 
(\ol{\lambda} + \tau(s), \ol{x} + sx_0 + z(s))$, where $s \rightarrow (\tau(s), 
z(s)) \in {\mathbb R} \times Z$ is a continuously differentiable function near 
$s=0$ and $\tau(0) = \tau'(0) = 0$, $z(0) = z'(0) = 0$.
\end{thm}


\section{A global solution curve}
\setcounter{equation}{0}
We study positive solutions of the Dirichlet problem for the semilinear elliptic equation on 
the unit ball 
\begin{equation}
\label{2.0}
\Delta u + \lambda f(u) = 0 \quad
\mbox{for } |x| < 1, \quad u = 0 \quad \mbox{on } |x| = 1,
\end{equation}
i.e, a Dirichlet problem on a ball in ${\mathbb R}^n$, depending on a positive parameter
$\lambda $. 
By the classical theorem of B. Gidas, W.-M. Ni and L. Nirenberg 
\cite{GNN} positive solutions of (\ref{2.1}) 
are radially symmetric, which reduces (\ref{2.0}) to 
\begin{equation}
\label{2.1}
u'' + \frac{n-1}{r} u' + \lambda f(u) = 0 \quad \mbox{for } 0 < r < 1,
\quad u'(0) = u(1) = 0\,.
\end{equation}
We shall also need the corresponding linearized equation 
\begin{equation}
\label{2.2}
w'' + \frac{n-1}{r} w' + \lambda f'(u) w = 0 \quad \mbox{for } 0 < r < 1,
\quad w'(0) = w(1) = 0\,.
\end{equation}
For multiplicity and uniqueness results it is 
important to know if $w(r)$ is of one sign on $(0,1)$. This is part of
the strategy we developed jointly with Y. Li and T. Ouyang; see \cite{KLO} 
which also has references to our earlier papers. However, positivity of
$w(r)$ is usually very hard to prove. Recently we noticed in \cite{K3} that
it sometimes suffices to prove that $w(r)$ cannot change sign exactly once.
This observation is rooted in the important work of M.K. Kwong and L. Zhang
\cite{KZ}, and it will be crucial in the present paper. 
We begin with some some preliminary results. 

The following lemma was proved in \cite{K}. We present its proof for
completeness.

\begin{lma}   Assume that the function $f(u) \in C^2(\bar {\mathbb R}_+)$, and the
problem \mbox{\rm (\ref{2.2})} has a nontrivial solution $w$ at some $\lambda $. Then
\begin{equation}
\label{l}
\int_0^1 f(u)wr^{n-1} \, dr =\frac{1}{2\lambda } u'(1)w'(1). 
\end{equation}
\end{lma}
\paragraph{Proof.} The function $v=ru_r-u_r(1)$ satisfies
\begin{eqnarray}
\label{l1}
& \Delta v+\lambda f'(u)v=-2\lambda f(u)-
\lambda f'(u)u'(1) \quad \mbox{for $|x|<1$,  } \\ 
& v=0 \quad \mbox{for $|x|=1$}.     \nonumber
\end{eqnarray}
By the Fredholm alternative the right hand side of (\ref{l1}) is orthogonal
to $w$, i.e.
\begin{equation}
\label{l2}
\int_0^1 f(u)wr^{n-1} \, dr =-\frac{1}{2} u'(1)
        \int_0^1 f'(u)wr^{n-1} \, dr\,.
        \end{equation}
Integrating the equation (\ref{2.2}), we obtain
\[
\lambda \int_0^1 f'(u)wr^{n-1} \, dr =-w'(1).
\]
Using this in (\ref{l2}), we conclude the lemma. \hfill$\diamondsuit$\medskip

We recall that solution of (\ref{2.1}) is called singular 
provided the corresponding linearized problem (\ref{2.2}) has a 
nontrivial solution.
The following lemma follows immediately from the equations (\ref{2.1})
and (\ref{2.2}). We assume that $f(u) \in C^2({\mathbb R}_+)$ for the rest
of the paper.
\begin{lma}
Let $(\lambda ,u)$ be a singular solution of \mbox{\rm (\ref{2.1})}. Then
\begin{equation}
\label{i1}
\int _0^1 \left(f(u)-f'(u)u \right) \,wr^{n-1} \, dr=0\,.
\end{equation}
\end{lma}

The following lemma is a consequence of the first two.
\begin{lma}
Let $(\lambda ,u)$ be a 
singular solution of \mbox{\rm (\ref{2.1})}. Then for any real $\gamma$
\begin{equation}
\label{i2}
\int _0^1 \left(\gamma f(u)-f'(u)u \right) \,wr^{n-1} \, dr=
\frac{\gamma-1}{2\lambda } u'(1)w'(1).
\end{equation}
\end{lma}

\paragraph{Proof.} Multiplying (\ref{l}) by $\gamma -1$, and adding
(\ref{i1}), we obtain (\ref{i2}). \medskip

The following lemma is known, see e.g. E.N. Dancer \cite{D}. We present its
proof for completeness.
\begin{lma}
Positive 
solutions of the problem \mbox{\rm (\ref{2.1})} are globally parameterized by their
maximum values $u(0,\lambda )$. I.e., for every $p>0$ there is at most one
$\lambda >0$, for which $u(0,\lambda )=p$.
\end{lma}

\paragraph{Proof.} If $u(r,\lambda )$ is a solution of (\ref{2.1}) with
$u(0,\lambda )=p$, then $v \equiv u(\frac{1}{\sqrt{\lambda }}r)$ solves
\begin{equation}
\label{++}
v''+\frac{n-1}{r}v'+f(v)=0, \quad \mbox{$v(0)=p$, $v'(0)=0$}.
\end{equation}
If $u(0,\mu)=p$ for some $\mu \ne \lambda $, then $u(\frac{1}{\sqrt{\mu }}r)$
is another solution of the same problem. This is a contradiction in view of
the uniqueness of solutions for initial value problems of the type (\ref{++}),
see \cite{PS}. \hfill$\diamondsuit$\medskip

We shall need the eigenvalue problem corresponding to an arbitrary solution
of (\ref{2.1})
\begin{eqnarray}
\label{i3}
& \phi'' + \frac{n-1}{r} \phi' + \lambda f'(u) \phi +\mu \phi =0
\quad \mbox{for }0 < r < 1, \\ 
& \phi'(0) = \phi(1) = 0.            \nonumber
\end{eqnarray}
We  define Morse index of any 
solution of (\ref{2.1}) to be the number of negative eigenvalues of 
(\ref{i3}).  \medskip

We shall work with a standard test function $v(r)=ru_r+\mu u$,
where $\mu$ is a positive constant, to be specified. One verifies
that $v$ satisfies
\begin{equation}
\label{2.3}
v'' + \frac{n-1}{r} v' + \lambda f'(u) v = \lambda \left[ 
\mu uf_u-(\mu +2)f \right] \quad \mbox{for } 0 < r < 1\,.
\end{equation}
We assume that there is a number $u_0>0$, such that $f(u)<0$ on
$(0,u_0)$, while $f(u)>0$ on $(u_0,\infty)$.
We define a function $G(u) \equiv \frac{uf'(u)}{f(u)}$. The following lemma
follows the idea of M.K. Kwong and L. Zhang \cite[Lemma 8]{KZ}.

\begin{lma} Assume that on $(u_0,\infty)$ the function  $G(u)$ is 
strictly decreasing and $\lim_{u \rightarrow \infty} G(u) >1$, while $G(u)<1$
on $(0,u_0)$. Then any non-trivial solution of \mbox{\rm (\ref{2.2})} $w(r)$ cannot
have exactly one zero on $(0,1)$.
\end{lma}

\paragraph{Proof.} Since $w(0) \neq 0$, see \cite{PS} for the appropriate
uniqueness result (if $w(0)=w'(0)=0$ then $w \equiv 0$), we may assume that
$w(0)>0$. Assume that on the contrary $w(r)$ has exactly one root at some
$r=r_0$, i.e.
\begin{equation}
\label{2.4}
w(r)>0 \quad \mbox{on $(0,r_0)$}, \quad w(r)<0 \quad \mbox{on $(r_0,1)$}.
\end{equation}
We claim that $f(u(r_0))>0$, so that $G(u(r_0))>1$ by our conditions. 
Indeed, assuming otherwise, we would have that
\begin{equation}
\label{44}
f(u)<0 \quad \mbox{on $(r_0,1)$}.
\end{equation}
We now set $\mu=0$ in (\ref{2.3}), then multiply (\ref{2.3}) by $w$,
subtract from that equation (\ref{2.2}) multiplied by $v$, and then
integrate over $(r_0,1)$. Obtain
\begin{equation}
\label{45}
r_0^{n-1}v(r_0)w'(r_0)=-2\lambda \int_{r_0}^1fw r^{n-1} \, dr\,.
\end{equation}
By (\ref{44}) the right hand side of (\ref{45}) is negative, while the left
side is positive, a contradiction, proving the claim.

Setting $\gamma=G(u(r_0))$, we see that the horizontal line $y=\gamma>1$
intersects the graph of $y=G(u)$ exactly once, and
\begin{equation}
\label{2.5}
 \gamma f(u) - uf'(u) 
\left\{ \begin{array}{lll} < 0 & \mbox{for all} & u < u(r_0) \\
> 0 & \mbox{for all} & u > u(r_0). \end{array}\right. 
\end{equation}
  Since $\gamma>1$,
we obtain by Lemma 2.3 (notice that $w'(1)>0$ by (\ref{2.4}))
\begin{equation}
\label{2.6}
\int_0^1 \left[  \gamma f(u) - uf'(u) \right]w(r)r^{n-1}\, dr<0\,.
\end{equation}
In view of (\ref{2.4}) and (\ref{2.5}) the quantity on the left is
positive, and we have a contradiction in (\ref{2.6}). \hfill$\diamondsuit$
\medskip

We now restrict our attention  to the problem
\begin{equation}
\label{2.7}
u'' + \frac{n-1}{r} u' - \lambda u+u^p = 0 \quad \mbox{for } 0 < r < 1,
\quad u'(0) = u(1)= 0\,,
\end{equation}
where $p>1$ is a constant. One checks 
that the Lemma 2.5 applies for $f(u)=-\lambda u+u^p$ at
any $\lambda \geq 0$. Indeed, $\displaystyle 
G(u)=1+(p-1)\frac{u^p}{-\lambda u+u^p}<1$, where
$f(u)<0$, and $G'(u)<0$ for all $u>0$. \medskip

We now consider the problem
\begin{equation}
\label{2.13}
u'' + \frac{n-1}{r} u' + u^p = 0 \quad \mbox{for } 0 < r < 1,
 \quad u'(0) = u(1)= 0\,.
\end{equation}
For $1<p<(n+2)/(n-2)$ the problem is known to have a unique positive
solution, see \cite{GNN}.

\begin{lma}
Assume that $1<p \leq n/(n-2)$ in case $n>2$. (We allow any 
$p>1$ for $n=1,2$.) Then
the Morse index of the positive solution of \mbox{\rm (\ref{2.13})} is one.
Moreover, this solution is non-singular (i.e. the second eigenvalue of
\mbox{\rm (\ref{2.14})} is positive).
\end{lma}

\paragraph{Proof.} We claim that the Morse index cannot be zero. Indeed, assuming
that Morse index is zero, let $\mu \geq 0$ and $\phi(r)>0$ be solutions
of (\ref{i3}) with $f(u)=u^p$. We now multiply (\ref{i3}) by $u$, and subtract
the equation (\ref{2.1}) multiplied by $\phi$, then integrate over $(0,1)$.
Obtain
\[
\int_0^1 \left[(p-1)u^p+\mu u \right]\phi \, dr=0,
\]
which is impossible, since the quantity on the left is positive.

Assume now that Morse index is at least two. Then
the second eigenvalue of the corresponding linearized operator is negative. 
That is, we can find a $\mu<0$ and a function $\phi(r)$, which changes
sign exactly once on $(0,1)$, such that ($\phi(r)$ is of course the second
eigenfunction)
\begin{eqnarray}
\label{2.14}
& \phi'' + \frac{n-1}{r} \phi' + pu^{p-1}\phi + \mu \phi= 0 \quad 
\mbox{for }0 < r < 1, \\ 
& \phi'(0) = \phi(1) = 0\,. \nonumber
\end{eqnarray}

We shall use a test function of M. Ramaswamy and P.N. Srikanth \cite{RS},
$z=ru'+\frac{2}{p-1}u$. From (\ref{2.3}) we notice that $z(r)$ satisfies
\begin{equation}
\label{2.15}
z'' + \frac{n-1}{r} z' + pu^{p-1}z = 0 \quad 
\mbox{for }0 < r < 1, \quad z(1)=u'(1). 
\end{equation}
We claim that $z(r)$ is decreasing on $(0,1)$. Indeed,
\[
z'=ru''+\frac{p+1}{p-1}u' \leq ru''+(n-1)u'=-ru^p<0\,, 
\]
since our condition $p \leq \frac{n}{n-2}$ is equivalent to 
$\frac{p+1}{p-1} \geq n-1$. Since $z(0)>0$, while $z(1)<0$, it follows
that $z(r)$ changes sign exactly once on $(0,1)$, say at some $
\alpha \in (0,1)$. Assume for definiteness that $\phi(0)>0$, and denote
by $\xi$ the point where $\phi(r)$ changes sign.
From the equations (\ref{2.14}) and (\ref{2.15}) we conclude
\begin{equation}
\label{2.16}
\left[ r^{n-1} \left(\phi'z-\phi z'\right) \right]'+\mu \phi z r^{n-1}=0.
\end{equation}

We consider two possibilities.

\noindent {\bf Case 1.} Assume that $\xi \leq \alpha$. Integrating
(\ref{2.16}) over $(0,\xi)$, we have
\[
\xi^{n-1}\phi '(\xi)z(\xi)+\mu \int_0^\xi \phi z r^{n-1} \, dr=0\,.
\]
Since the first term on the left is non-positive, while the second one is
negative, we have a contradiction.

\noindent
{\bf Case 2.} Assume that $\alpha<\xi$. Since $z$ changes sign exactly once,
it follows that $z(r)<0$ on $(\xi,1]$. Integrating (\ref{2.16}) over
$(\xi,1)$, we obtain
\[
\phi'(1)z(1)-\xi^{n-1}\phi'(\xi)z(\xi)+\mu \int_\xi^1 \phi z
r^{n-1} \, dr=0\,.
\]
Since all terms on the left are negative, we again have a contradiction,
completing the proof. \hfill$\diamondsuit$\medskip

That solution is non-singular was proved first in \cite{RS}. This also follows
from the above argument. Indeed, if solution is singular then $\mu=0$ is an
eigenvalue of (\ref{2.14}), and it has to be the second eigenvalue, since we
already proved that the Morse index is one. Then we get the same
contradiction as above, except if $\xi=\alpha$. But then $\phi(r)$ and $z(r)$ are 
solutions of the same linear equation, having the same root. Hence
they have to be constant multiples of each other, which is impossible
since only one of them vanishes at $r=1$.

\begin{thm}
Assume that $1<p <(n+2)/(n-2)$ in case $n>2$. (We allow any 
$p>1$ for $n=1,2$.) Then
the Morse index of the positive solution of \mbox{\rm (\ref{2.13})} is one.
\end{thm}

\paragraph{Proof.} Examining the proof of Lemma 2.6, where a special case of this
theorem was proved, we see that the same proof works, provided the function
$z(r)$ changes sign exactly once on $(0,1)$. This was shown to be true for
$p \leq n/(n-2)$. Let us now vary $p$ from $n/(n-2)$ to
$(n+2)/(n-2)$. Since by M. Ramaswamy and P.N. Srikanth \cite{RS} the 
positive solution of (\ref{2.13}) is non-degenerate, it follows by
the implicit function theorem that positive solution of (\ref{2.13}) varies
continuously with $p$ in $C^1$ norm, and hence $z(r)$ varies continuously 
with $p$. But this implies that $z(r)$ cannot have more than one zero
on $(0,1)$. Indeed, assuming otherwise, we can find a $p \in 
(\frac{n}{n-2},\frac{n+2}{n-2})$ and a $\xi \in (0,1)$ such that
$z(\xi)=z'(\xi)=0$. By the uniqueness theorem for ODE, $z(r) \equiv 0$,
which contradicts $z(0)>0$. \hfill$\diamondsuit$\smallskip

We are ready to state our main result. 
\begin{thm}
Assume that $1<p< (n+2)/(n-2)$ in case $n>2$, and $p>1$ in dimensions
one and two. Then for any $\lambda
\geq 0$ there exists a unique positive solution of 
\mbox{\rm (\ref{2.7})}, which is moreover 
non-singular. In addition, $u(0,\lambda )$ is increasing in $\lambda $, and
$\lim_{\lambda \rightarrow \infty} u(0,\lambda )=\infty$.
\end{thm}

\paragraph{Proof.} 
We begin with the unique positive solution at $\lambda=0$. Since this solution 
is non-singular, we can continue it for small $\lambda>0$ using the implicit
function theorem. By scaling $u=\lambda ^{1/(p-1)}v$, 
we can put the problem 
(\ref{2.7}) in the form (\ref{2.1}) with $f(u)=-u+u^p$.
Hence we can use Lemma 2.1 to continue the solution curve globally in 
$\lambda $ (i.e. at each point either implicit function theorem or
the Crandall-Rabinowitz theorem 1.1 applies, see e.g. \cite{KLO} for more 
details). It is easy to check that rescaling does not affect the Morse index
of solution. We claim that solutions on this curve stay non-singular.
Indeed, by Theorem 2.1 at $\lambda =0$ zero lies between the first and second
eigenvalues of the corresponding linearized problem. If a singular solution
appears on the curve, it means that zero is now an eigenvalue, and 
it is either
the first or the second one. It cannot be the second one, since then 
the corresponding eigenfunction satisfies the linearized equation (\ref{2.2}),
and it vanishes exactly once on $(0,1)$, which is excluded by Lemma 2.5 . If
on the other hand, zero is the first or principal eigenvalue, then
the nontrivial solution of (\ref{2.1}) is of one sign, and we may assume
that in fact $w(r)>0$ on $(0,1)$. But this leads to a contradiction by
Lemma 2.2, since $f(u)-f'(u)u<0$ for all $r \in (0,1)$.

Since solutions on this curve
cannot go to infinity at a finite $\lambda $ by the result of B. Gidas and
J. Spruck \cite{GS}, it follows that this curve continues without turns
for all $\lambda \geq 0$. We claim that there are no other solution curves.
Indeed, if there were other solutions, not lying on
the above curve, they would have to lie on another curve of solutions, 
for which there
would be no place to go for decreasing $\lambda $. Hence solution curve is
unique, and so for each $\lambda \geq
0$ there exists a unique solution.

Finally, we turn to the properties of $u(0,\lambda )$. By Lemma 2.4 we know
that $u(0,\lambda )$ is either strictly increasing or decreasing. Assume 
contrary to what we want to prove that it is decreasing. Then it tends to 
a non-negative limit as $\lambda \rightarrow \infty$. If the limit is positive, we
see from (\ref{2.7}) that $u''(0)$ would have to become positive for large
$\lambda $, which is impossible by \cite{GNN}. If the limit is zero, then
multiplying the equation (\ref{2.7}) and integrating, we see that the 
quantity on the left is negative for large $\lambda $, a contradiction.
Hence $u(0,\lambda )$ is increasing in $\lambda $. If $u(0,\lambda )$ failed
to go to infinity, it would have to tend to a finite limit, which was 
already shown to be impossible. \hfill$\diamondsuit$

\paragraph{Remark.} It follows from the work of P.N. Srikanth \cite{S} that
our solution curve continues for negative $\lambda $.
In fact, combining both results, we see that a curve
of positive non-singular solutions of (\ref{2.7}) bifurcates off $\lambda =
-\lambda _1$, and continues for all $\lambda >-\lambda _1$. Here $\lambda _1$
is the principal eigenvalue of the $-\Delta $ on the unit ball.

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\end{thebibliography}
\bigskip

{\sc Philip Korman} \\ 
Institute for Dynamics and
Department of Mathematical Sciences \\ 
University of Cincinnati \\ 
Cincinnati Ohio 45221-0025  USA\\
E-mail address: kormanp@math.uc.edu
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