\documentclass[twoside]{article}
\usepackage{amssymb,amsmath} % font used for R in Real numbers
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\markboth{\hfil Unstable manifolds for delay differential equations
 \hfil EJDE--2002/32}
{EJDE--2002/32\hfil Hari P. Krishnan \hfil}

\begin{document}
\title{\vspace{-1in}\parbox{\linewidth}{\footnotesize\noindent
{\sc  Electronic Journal of Differential Equations},
Vol. {\bf 2002}(2002), No. 32, pp. 1--13. \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} \\
 %
  Existence of unstable manifolds for a certain class of delay 
  differential equations
 %
\thanks{ {\em Mathematics Subject Classifications:} 34K40.
\hfil\break\indent
{\em Key words:} invariant manifolds, functional differential equations.
\hfil\break\indent
\copyright 2002 Southwest Texas State University. \hfil\break\indent
Submitted December 1, 1999. Published April 1, 2002.} }
\date{}
%
\author{Hari P. Krishnan}
\maketitle

\begin{abstract}
  We prove a theorem for unstable manifolds in a differential 
  equation with a state-dependent delay.  Although the equation 
  cannot be formally linearized, we find an associated linear 
  delay equation whose dynamics are qualitatively similar near 
  the unstable manifold.  Our proof relies upon estimates of the 
  derivative of a trajectory on the unstable manifold near 
  equilibrium.
\end{abstract}

\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\allowdisplaybreaks

\section{Introduction}

Various authors have studied the delay equation $$
\dot{x}(t)=f(x(t),x(t-r)),\quad  r=r(x(t))\eqno(1.1) $$ over the
past few years. Mackey \cite{m1} proposed that a special case of
(1.1) could be used to model the spot price of an agricultural
commodity, such as corn. Here, $r(x(t))$ represents the time delay
between the production and delivery of the commodity and depends
upon the price $x(t)$. Cooke and Huang \cite{c1} have examined
some of the difficulties that arise when attempting to linearize
(1.1) and have developed decay estimates for solutions near the
origin. Mallet-Paret and Nussbaum \cite{m2} proved the existence
of a sawtooth-shaped, slowly oscillating periodic solution to the
equation $$ \varepsilon\dot{x}(t)=-x(t)+f(x(t-r)),\quad
r=r(x(t))\eqno(1.1) $$ in the singular limit as $\varepsilon$ goes
to 0. Kuang and Smith \cite{k1}, and Arino, Hadeler and Hbid
\cite{a1} have proved the existence of periodic solutions for
certain types of state-dependent delay equations, including (1.1).

When $r$ is a constant, solutions to (1.1) are typically embedded
in the space $C([-r,0])$, of continuous functions over the interval
$[-r,0]$. If $f$ is sufficiently well-behaved, it is possible to
prove the existence and uniqueness of solutions and also the
existence of a smooth unstable manifold relative to (1.1).

When $r=r(x(t))$ is non-constant, we run into difficulty since
solutions are not uniquely defined even locally in time.  Unless
we have a uniform bound for $|\dot{x}(t)|$, $x(t-r)$ and hence
$f(x(t-r))$ will not be a Lipschitz function in the argument
$x(t)$.  In this paper, we develop an alternative phase space
setting which allows us to control the derivative of $x(t)$; we
then prove existence, uniqueness and unstable manifold results.
Since the existence of an unstable manifold does not depend
on the choice of phase-space, our approach provides a consistent
way of looking at (1.1).

In section 2, we prove a straightforward existence result for
solutions to (1.1), relative to the phase space $W^{1,\infty}$. In
section 3, we show that (under appropriate technical conditions),
the unstable manifold is a smooth graph whenever $f$ is at least
$C^2$.  We will take advantage of the fact that trajectories on
the unstable manifold are defined from time $-\infty$ to the
present and are thus as smooth as possible.

\section{Technical Preliminaries}

Hale and Ladeira \cite{h1} have used the space $W^{1,\infty}$ to show
that solutions to the equation
$$\dot{x}(t)=f(x(t),x(t-\rho))\eqno(2.1)
$$
depend smoothly on the delay parameter $\rho$.
In \cite{h1}, $\rho$ is neither time- nor
state-dependent.  However, it turns out that the space
$W^{1,\infty}$ is useful in the case where the delay is
state-dependent.  We require the following definitions for which we use the
notation in \cite{h1}.

%\paragraph{Definition} % 2.1.}
 Suppose that $E$ is a linear
space equipped with the norms $|\cdot|$ and $N(\cdot)$; suppose
further that we define the set $B_{R,N}=\{x\in E : N(x)\leq R\}$
for any fixed $R>0$. ($B_{R,N}$ is a closed ball of radius $R$
centered at $0$ relative to the $N(\cdot)$ norm.)  Also suppose
that, for any fixed $R>0, (B_{R,N},|\cdot|)$ is a complete metric
space.  Then we refer to $E$ as a {\it quasi-Banach} space.

%\paragraph{Definition} % 2.2.}
Let $W^{1,\infty}([-r^\ast,0])$ be
the linear space of absolutely continuous functions $\phi :
[-r^\ast,0]\to \mathbb{R}$ whose derivatives are essentially
bounded.

Note that $W^{1,\infty}$ defines a quasi-Banach space when we set
$N(\phi)=\|\phi\|^\infty =|\phi(-r^\ast)|+
\:{\mathop{\rm ess\,sup}_{\:\:\theta \in
[-r^\ast,0]}} |\dot{\phi}(\theta)|$ and
$$|\phi|=\|\phi\|^1 =|\phi
(-r^\ast)|+\int^0_{-r^\ast} |\dot{\phi}(s)|ds.
$$
For technical reasons, we also require the following definitions.

Let $W^{1,\infty}_{\alpha,0}
([-r^\ast,\alpha])=\{\phi \in W^{1,\infty} ([-r^\ast,\alpha]) :
\phi (s)=0$ for $s\in [-r^\ast,0]\}$, where
$W^{1,\infty}_{\alpha,0}$ is equipped with the norms $\|\phi
\|^1_{\alpha}=\int^\alpha_0 |\dot{\phi}(s)|ds$ and $\|\phi
\|^\infty_\alpha =
{\mathop{\rm ess\,sup}_{s\in [0,\alpha]}}|\dot{\phi}(s)|$.

Let %\paragraph{Definition} % 2.4.}
$$ \displaylines{
A (\alpha,\beta)=\{\phi \in
W^{1,\infty}_{\alpha,0} ([-r^\ast,\alpha]) :
\|\phi\|^\infty_\alpha \leq \beta \}\,,\cr
B (\alpha,\beta)=\{\phi \in
W^{1,\infty}_{\alpha,0} ([-r^\ast,\alpha]) : \|\phi\|^1_\alpha
\leq \beta\}\,.
}$$

When $\alpha \leq 1$, we note that  $A(\alpha,\beta)\subset
B(\alpha,\beta)$ since, for $\phi \in A(\alpha,\beta)$,
$\mathop{\rm ess\,sup}_{f\in[0,\alpha]}|\dot{\phi}(s)|\leq
\beta$ implies that ${\int^\alpha_0}
|\dot{\phi}(s)|ds \leq \beta {\int^\alpha_0} ds=\beta
\alpha \leq \beta$.  In general, we have the inclusion
$A(\alpha,\beta)\subset B(\alpha,\alpha \beta)$.  We may now
consider the initial value problem
$$\begin{gathered}
\dot{x}(t)=f(x(t),x(t-r))\,,\quad r=r(x(t))\,,\quad t>0\\
x(\theta)=\phi (\theta)\,,\quad  \theta \in [-r^\ast,0],
\phi (\cdot)\in W^{1,\infty}.
\end{gathered}\eqno(2.2)
$$

For the rest of the paper, we will assume that (2.2) satisfies
conditions (A1) and (A2) below.

\begin{enumerate}
\item[{(A1)}] $r:W^{1,\infty}([-r^\ast,0])\to \mathbb{R}$ is smooth,
 with ${\sup_{\phi \in W^{1,\infty}}} |D_\phi r|\leq c<\infty, D_\phi r$
 denoting the Fr\'{e}chet derivative of $r$ with respect to $\phi$.
 Also, \\
 $\alpha \leq {\inf_{\phi \in W^{1,\infty}}} r(\phi)\leq r^\ast <\infty$,
 for some $\alpha >0$ small.

\item[{(A2)}] $f:\mathbb{R}\times \mathbb{R} \to \mathbb{R}$
is a globally Lipschitz function.
In particular, for
$$(\xi_1,\eta_1), \xi_2,\eta_2)\in \mathbb{R}\times
\mathbb{R}, |f(\xi_1,\eta_1)-f(\xi_2,\eta_2)|\leq
L|\xi_1-\xi_2|+M|\eta_1-\eta_2|,
$$
which specifies the Lipschitz
constants $L$ and $M$.
\end{enumerate}
To give precise smoothness results, it will sometimes be
necessary to make the more stringent assumption
\begin{enumerate}

\item[{(A3)}] $f:\mathbb{R}\times \mathbb{R} \to \mathbb{R}$ is a
$C^k$ function, with $k\geq 2$, $f(0)=0$, and $f'(0)\neq 0$.
\end{enumerate}

We shall proceed by reformulating the initial-value problem (2.2)
as a fixed point equation in the phase space
$W^{1,\infty}([-r^\ast,\alpha])$.  First notice that $x(t)$
satisfies the system (2.2) over the interval $[-r^\ast,\alpha]$ if
and only if $x(t)=\phi_0(t)+z(t)$, $\phi_0(t)=\phi (t)$ for
$t\in [-r^\ast,0]$, $\phi_0(t) =\phi (0)$ for $t\in
[0,\alpha]$, and $z(t)$ satisfies
$$
z(t)=\left\{\begin{array}{ll}
0\,, & t\in [-r^\ast,0]\\[2pt]
{\int^t_0}f(\phi (0)+z(s)\,,\, \phi (s-r(\phi(0)+z(s))))ds
& t \in [0,\alpha].
\end{array}\right.\eqno(2.3)
$$
Here we have made the observation that, since
${\inf_\phi}\:( r(\phi))>\alpha , \phi_0
(s-r(x))=\phi (s-r(x))$ for any $x\in \mathbb{R}$ and
$z(s-r(\phi(0)+z(s)))=0$. We can now define the operator $T:
A(\alpha,\beta)\times B_{R,N}\times S \to
A(\alpha,\beta)$, with the function $r\in S$ if and only if
$r:W^{1,\infty} ([-r^\ast,0])\to \mathbb{R}^+$, and $r$
satisfies hypothesis (A1).

Solutions to (1.1) correspond to fixed points of the integral
equation
$$
T(z,\phi,r)(t)=\left\{\begin{array}{ll}
0\,, & t\in [-r^\ast,0]\\[3pt]
{\int^t_0}f(\phi (0)+z(s),\phi (s-r(\phi(0)+z(s))\\
+z(s-r(\phi (0)+z(s))))ds,& t \in [0,\alpha].
\end{array}\right.\eqno(2.4)
$$
defined on the interval $t\in [0,\alpha]$.  If we can show that
$z\in W^{1,\infty}_{\alpha,0}$ is a unique fixed point of the
equation $z=T(z,\phi,r)$, it follows from the relation
$x(t)=\phi_0(t)+z(t)$ that the map $x_t(\phi,\cdot)$ can be
continued from $t=0$ to $t=\alpha$.  We prove existence and
uniqueness using the contraction mapping principle, dividing our
proof into the following lemmas.

\begin{lemma} \label{lm2.1}
 Suppose that (A1) and (A2) are
satisfied.  Then for any fixed $R>0$ there exist $\alpha,\beta >0$
such that $T(\overline{B}(\alpha,\alpha \beta)\times B_R \times
S)\subset \overline{B}(\alpha,\alpha\beta)$ and also
$T(A(\alpha,\beta)\times B_R \times S)\subset A(\alpha,\beta)$.
Hence $T$ is a self-mapping with respect to the sets
$A(\alpha,\beta)$ and $\overline{B}(\alpha,\alpha \beta)$.
\end{lemma}

\paragraph{Proof.}
Since the proofs are similar, we show only
that $T(\overline{B}(\alpha,\alpha \beta)\times B_R \times
S)\subset \overline{B}(\alpha,\alpha\beta)$.  From (A2),
\begin{align*}
\|T(z,\phi,r)(t)\|^1_\alpha =& \int^\alpha_0 |f(\phi
(0)+z(s)\,,\quad  \phi (s-r(z(s)+\phi (0))))|ds \\
\leq& \alpha
\{\sup_{S\in [0,\alpha]} M |\phi (0)+z(s)| +\sup_{S\in [0,\alpha]}
N |\phi (s-r(z(s)+\phi (0)))|\} \\
\leq &\alpha \{ M(R+\alpha \beta)+NR\};
\end{align*}
thus, for $\alpha \leq \frac{1}{\beta}$
and $\beta \geq R(M+N)+1$,
$\alpha\{M(R+\alpha\beta)+NR\}\leq \alpha \beta$, and the proof is
complete. \hfill$\Box$\smallskip

Note that $\alpha >0$ can be made as small as necessary by choosing $\beta$
large.
Thus the assumption that $0\leq \alpha < {\inf_{x\in \mathbb{R}}} r(x)$ is
not overly restrictive.

\begin{lemma} \label{lm2.2}
  $T$ is a uniform contraction, with
respect to the norms $\|\cdot\|^1_\alpha$ and
$\|\cdot\|^\infty_\alpha$, over $\overline{B}(\alpha,\alpha
\beta)$.
\end{lemma}

\paragraph{Proof.}  We need to show that, for $\alpha >0$
sufficiently small and for any pair $z,w \in \overline{B}
(\alpha,\alpha \beta)$, there exists an element $0<c_0 <1$ with
$c_0=c_0(\alpha)$ such that
$\|T(z,\phi,r(z))-T(w,\phi,r(w))\|^1_\alpha \leq
c_0\|z-w\|^1_\alpha$ for any fixed $\phi \in B_R$.  To do this, we
rewrite the quantity $\|T(z,\phi,r(z))-T(w,\phi,r(w))\|^1_\alpha$
explicitly.  We then obtain
\begin{align*}
\|T(z,&\phi,r(z))-T(w,\phi,r(w))\|^1_\alpha \\
=&{\int^\alpha_0} |f(\phi(0)+z(s), \phi
(s-r(\phi(0)+z(s))))\\
& -f(\phi (0)+w(s), \phi (s-r(\phi (0)+w(s))))|ds \\
\leq &\int^\alpha_0 M|z(s)-w(s)|+N|\phi(s-r(\phi (s-r(\phi
(0)+w(s)))|ds\,.
\end{align*}
This last inequality follows from (A2). Now, since
$z(0)=0=w(0)$, it follows that ${\int^\alpha_0}
|z(s)-w(s)|ds\leq \alpha {\int^\alpha_0}
|\dot{z}(s)-\dot{w}(s)|ds$ so it remains to estimate
${\int^\alpha_0}|\phi (s-r(\phi (0)+rz(s)))-\phi
(s-r(\phi(0)+w(s)))|ds$.
 However, since $\phi \in B_R$, it follows that
\begin{align*}
|\phi& (s-r(\phi(0)+z(s)))-\phi(s-r(\phi(0)+w(s)))| \\
&\leq R|r(\phi(0)+z(s))-r(\phi(0)+w(s))|\\
&\leq Rc|z(s)-w(s)|\,,
\end{align*}
where above inequality follows Assumption (A1). Then we obtain
\begin{multline*}
\int^\alpha_0
|\phi (s-r(\phi (0)+z(s)))-\phi(s-r(\phi(0)+w(s))|ds \\
 \leq Rc {\int^\alpha_0}|z(s)-w(s)|ds
 \leq Rc \alpha {\int^\alpha_0}|\dot{z}(s)-\dot{w}(s)|ds.
\end{multline*}
Combining the above inequalities gives
\begin{multline*}
\|T(z,\phi,r(z))-T(w,\phi,r(w))\|^1_\alpha \\
\leq c \int^\alpha_0|\dot{z}(s)-\dot{w}(s)|ds+\alpha Rc \int^\alpha_0
|\dot{z}(s)-\dot{w}(s)|ds=\alpha(1+Rc)\|z-w\|^1_\alpha.
\end{multline*}
 Suppose
that we fix any $\alpha <{\frac{1}{1+Rc}}$; it
follows that, with respect to the constant $c_0=\alpha
(1+Rc)<1$, $T$ is a contraction map on $\overline{B}(\alpha,
\alpha \beta)$. The proof for the $\|\cdot \|^\infty_\alpha$ norm
is similar and therefore omitted.

\begin{lemma} \label{lm2.3}
  Consider $T : B(\alpha,\alpha
\beta)\times B_R \times S \to B(\alpha,\alpha \beta)$ with
$r\in S$ fixed.  Then $T=T(r,z,\phi)$ is a Lipschitz continuous
function in $z$ and $\phi$.
\end{lemma}

\paragraph{Proof.}  Fix $r$ and consider $(z,\phi)$,
$(w,\psi) \in B(\alpha,\alpha \beta)\times B_R$; we need to show
that $\|T(z,\phi)=T(w,\psi)\|^1_\alpha \leq c\{\|z-w\|^1_\alpha
+\|\phi -\psi\|^1\}$ for some constant $0\leq c=c(\alpha)<\infty$.
Because of the triangle inequality, it
suffices to show that $\|T(z,\phi)-T(w,\phi)\|^1_\alpha
+\|T(w,\phi)-T(w,\psi)|^1_\alpha \leq c\{\|z-w\|^1_\alpha
+\|\phi-\psi\|^1\}$.  From the proof of Lemma \ref{lm2.2}, we know that
$\|T(z\phi)-T(w,\phi)\|^1_\alpha \leq c\|z-w\|^1_\alpha$; thus, it
remains to show that $\|T(w,\phi)-T(w,\psi)\|^1_\alpha \leq
c\{\|z-w\|^1_\alpha +\|\phi-\psi\|^1\}$ for some $0\leq c<\infty$.
We write
\begin{align*}
\|T(w,\phi)-T(w,\psi)\|^1_\alpha
=&{\int^\alpha_0}\Big| f(\phi (0)+w(s)\,,\,
\phi (s-r(\phi(0)+w(s))))\\
&-f(\psi (0)+w(s)\,,\,  \psi (s-r(\psi(0)+ w(s))))\Big| ds\\
\leq& M {\int^\alpha_0} \big|\phi (0)-\psi (0)|ds
+N {\int^\alpha_0} \Big|\phi(s-r(\phi (0)+w(s)))\\
&-\psi (s-r(\psi(0)+w(s)))\Big|ds.
\end{align*}
We know that
\begin{align*}
 M {\int^\alpha_0}|\phi (0)-\psi (0)|ds
 =& M \alpha |\alpha (0)-\psi (0)|\\
 =&M \alpha |\phi (-r^\ast)+{\int^0_{-r^\ast}}
\dot{\phi} (\theta)d\theta -\psi
(-r^\ast)-{\int^0_{-r^\ast}} \dot{\psi}
(\theta)d\theta |\\
\leq &M\alpha \left\{|\phi(-r^\ast)-\psi (-r^\ast)|+{\int^0_{-r^\ast}}|
\dot{\phi}(\theta)-\dot{\psi}(\theta)|d\theta \right\}\\
=&M\alpha \|\phi-\psi\|^1_\alpha
\end{align*}
Also we know that
\begin{align*}
N &{\int^\alpha_0} | \phi(s-r(\phi(0)+w(s)))-\psi (s-r(\psi(0)+w(s)))|ds\\
\leq &N \Big\{{\int^\alpha_0} |\phi (s-r(\phi (0)+w(s)))-\phi (s-r(\psi (0)
+w(s)))|ds\\
&+{\int^\alpha_0} |\phi (s-r(\psi (0)+w(s)))-\psi(s-r(\psi(0)+w(s)))|
ds\Big\}\\
\leq &N R\alpha c |\psi (0)-\psi (0)| +N {\int^0_{-r^\ast}}
|\phi (\theta)-\psi (\theta)|d\theta\\
\leq& N R\alpha c|\phi (0)-\psi (0)|+N {\int^0_{-r^\ast}}
[|\phi (-r^\ast)-\psi(-r^\ast)|+{\int^\theta_{-r^\ast}} |\dot{\phi}(\sigma)
-\dot{\psi}(\sigma)|d\sigma]d\theta\\
\leq& NR\alpha c |\phi(0)-\psi (0)|+Nr^\ast\|\phi-\psi\|^1\\
\leq& N(R\alpha^2 cM+r^\ast)\|\phi-\psi\|^1.
\end{align*}
The above estimates complete the proof. \hfill$\Box$\smallskip

Using Lemmas \ref{lm2.1}, \ref{lm2.2}, and \ref{lm2.3}, with
$\alpha >0$ and sufficiently small, the contraction mapping theorem
gives the following result.

\begin{theorem} \label{thm2.1}  Consider the initial-value problem
(2.2) subject to assumptions (A1) and (A2), with $\phi (\cdot)\in
B_R$ and $R>0$ fixed.  Then there exists a real number $\alpha
=\alpha (R)>0$, independent of $\phi$, such that $x(t,\phi)$
exists and is unique on $[0,\alpha]$.
\end{theorem}

It is also possible to establish a simple sufficient condition for
global existence (i.e., existence on the interval $t\in
[0,\infty)$), using the fact that equation (1.1) is autonomous.
The theorem that follows implies that the only way in which a
solution $x(t,\phi)$ can fail to exist is if it blows up in finite
time.

\begin{theorem} \label{thm2.2}
 Consider the initial-value problem (2.2) subject to assumptions (A1)
and (A2), with $\phi (\cdot)\in
B_R$ and $R>0$ fixed.  Further, suppose that the solution
$x(t,\phi)$ is non-continuable on the interval $[0,b)$ for some
$0<b<\infty$.  Then, for any fixed $c>0$, there exists a $t\in
[0,b)$ such that $|x(t,\phi)|>c$.
\end{theorem}

\paragraph{Proof.}
We apply a contradiction argument.  Suppose that for some fixed $\mu>0$,
 ${\sup_{t\in [0,b)}} {\sup_{\theta \in [-r^\ast ,0]}} |x(t+\theta)|<\mu$
and that $x(t,\phi)$ is not continuable on $[0,b)$.  From assumption (A2),
 we know that, for $|\xi|, |\eta|<\mu$, ${\sup_{\xi,\eta}} |f(\xi,\eta)|\leq (M+N)\mu <\infty$.
 Now we have assumed that $\|\phi (\cdot)\|^\infty <R<\infty$; hence, by definition,
 ${\mathop{\rm ess\,sup}_{\theta \in [-r^\ast,0]}}\:\: |\dot{\phi}(\theta)|\leq R$ and
 $\|x_t(\phi,\cdot)\|^\infty \leq \mu + \max (\mu,R)$ and for all
 $t\in [0,b)$, $x_t(\phi,\cdot)\in B_{\mu + \max (\mu,R)}$.
 Setting $R_c=\mu + \max (\mu, R)$, we know that, for all $t\in [0,b)$, there exists an $\alpha_0=\alpha (R_0)$
 such that $x(t,\phi)$ is continuable from $t$ to $t+\alpha_0$.  Thus we gain a contradiction
 if we choose $t=b-{\frac{\alpha_0}{2}}$, since $x(t,\phi)$ can now be continued beyond $b$.

\section{An Unstable Manifold Theorem}

In this section, we prove a theorem on local unstable manifolds for
equation (1.1).  Significantly, it will turn out that the unstable
manifold forms a smooth graph, even though the vector field
defined by (2.2) is not Fr\'{e}chet differentiable in the initial
conditions.  We set up our problem in the following way.

Consider the following two equations, $
\dot{x}(t)=f(x(t),x(t-r)),\quad  r=r(x(t))$ and  $
\dot{x}(t)=f(x(t),x(t-r(0)))$ under the assumption that $f(0)=0$.
We do not claim that the second equation is a formal linearization
of the first; however, we will show that it gives a good
description of the local dynamics near the unstable manifold of
$
\dot{x}(t)=f(x(t),x(t-r)),\quad  r=r(x(t))$.

Suppose that $0$ is a hyperbolic equilibrium point of $
\dot{x}(t)=f(x(t),x(t-r(0)))$; it follows that the set
$\Lambda=\{\lambda \in \mathbb{C} : \lambda =-\alpha
+f'(0)e^{-r(0)\lambda}, Re \lambda >0\}$ is finite and that the
space $W^{1,\infty}([-r^\ast,0])$ can be decomposed as
$W^{1,\infty}=U\oplus S$, as in Hale and Lunel \cite{h2}.  Here we
define $U=\{\phi \in W^{1,=\infty}: x_t(\phi,\cdot)$ exists and
remains bounded (in the $\|\cdot \|^1$-norm) for all $t\leq 0\}$
and set $\phi \in U$ equal to $\phi^U$.  Similarly, we define
$S=\{\phi \in W^{1,\infty} : x_t (\phi,\cdot)$ exists and remains
bounded for all $t\geq 0\}$, where $x_t(\phi,\cdot)$ is a solution
to the initial-value problem (2.2). The dimension of $U$ is equal
to the elements (including multiplicities) in $\Lambda$, and is
finite.  Associated with the sets $U$ and $S$ are the projections
$\pi_U : W^{1,\infty}\to U$ and $\pi_S : W^{1,\infty}\to S$, with
$\pi_U U=U$ and $\pi_S S=(I-\pi_U)S=S$.

We now examine at the dynamics of the solution map $L:U \to U$
defined by $ \dot{x}(t)=f(x(t),x(t-r(0)))$, where $U$ is a
finite-dimensional subspace of $W^{1,\infty}$.  Since the origin
is a hyperbolic equilibrium point, there exist constants $M_0,
\beta_0>0$, independent of $\phi^U \in U$, such that, for $t\leq
0, {\sup_{\theta \in [-r^\ast,0]}} |T(t)\phi^U (\theta)|\leq M_0
e^{\beta_0 t} {\sup_{\theta \in [-r^\ast,0]}} |\phi^U(\theta)|$.
Since all norms are equivalent in finite dimensions, we know that
there exist constants $M,\beta>0$, independent of $\phi^U$, such
that $\|T(t)\phi^U\|^1\leq M e^{\beta t} \|\phi^U\|^1$.  If we
consider the restriction of $g$ to $U$, the following smoothness
result can be proved.

\begin{lemma} \label{lm3.1}
 Suppose that, in (1.1)
$\dot{x}(t)=-\alpha x(t)+f(x(t-r)), r=r(x(t)), r:
\mathbb{R}\to \mathbb{R}^+$ and $f: \mathbb{R} \to \mathbb{R}$ are
$C^2$ functions.  Then there exists a neighbourhood $N_\delta
(0)\subset U$ such that the mapping $g:U\to \mathbb{R}$ is
Fr\'{e}chet differentiable at all points $x_t(\cdot)\in N_\delta
(0)$.
\end{lemma}

\paragraph{Proof.}  Suppose that we fix a point $x_t
(\cdot)\in N_\delta (0)$.  Since $f\in C^2$, from the chain rule
it is sufficient to show that the function $g_1:U\to
\mathbb{R}$, defined by $g_1(x_t(\cdot))=x(t-r(x(t)))$, is Fr\'{e}chet
differentiable.  We now prove that, for $h_t(\cdot)\in U, D_\phi
g_1(x_t(\cdot)) h_t=-\frac{dr}{dx} \dot{x} (t-r(x(t)))
h(t)+h(t-r(x(t)))$, where $D_\phi g_1$ is a linear operator since
$x_t (\cdot)$ is fixed. We can write
\begin{align*}
\big|g_1(x_t&+h_t)-g_1(x_t)-D_\phi g_1(x_t)h_t\big|\\
=&\big|x(t-r(x(t)+h(t)))+h(t-r(x(t)+h(t-r(x(t)+h(t)))\\
 &-x(t-r(x(t)))+\frac{dr}{dx} \dot{x} (t-r(x(t)))h(t)-h(t-r(x(t)))\big|\\
=&\big|x(t-r(x(t))-\frac{dr}{dx} h(t)+o(h))+h(t-r(x(t))-\frac{dr}{dx}
  h(t)+o(h))\\
 &-x(t-r(x(t)))+\frac{dr}{dx} \dot{x}(t-r(x(t)))h(t)-h(t-r(x(t)))\big|\\
=&\big|\dot{x}(t-r(x(t))\big(-\frac{dr}{dx} h(t)+o(h)\big)+\dot{h}(t-r(x(t)))
 \big(-\frac{dr}{dx}h(t)+o(h)\big)\\
 &+x(t-r(x(t)))+h(t-r(x(t)))-x(t-r(x(t)))\\
 &+\frac{dr}{dx} \dot{x}(t-r(x(t)))h(t)-h(t-r(x(t)))\big|\\
=&|-\frac{dr}{dx} \dot{h}(t-r(x(t)))h(t) +o(h)|.
\end{align*}
The proof will be complete if we can show that
$-\frac{dr}{dx} \dot{h}(t-r(x(t)))h(t)$ is $o(h)$.  In
particular, we need to show that, for any sequence $\{h_n\}$, with
each $h_n\in U, {\lim_{n\to \infty}} \|h_n\|^1=0$
implies that ${\lim_{n\to \infty}} {\sup_{\theta
\in [-r^\ast,0]}} |\dot{h}_n(\theta)|=0$.  We proceed by
contradiction and suppose that there exists an $\varepsilon >0$
such that, for all $N\in Z\!\!\!Z^+$ and some $n\geq N,
{\sup_{\theta \in [-r^\ast,0]}}
|\dot{h}_n(\theta)|>\varepsilon$.  From the fundamental theorem of
calculus, it follows that ${\sup_{\theta \in
[-r^\ast,o]}}|h_n(\theta)|\leq \|h_n\|^1$. Also, it is obvious
that ${\lim_{n\to \infty}} {\sup_{\theta \in
[-r^\ast,0]}} |h_n (\theta -r^\ast)|=0$ whenever
${\lim_{n\to \infty}} {\sup_{\theta \in
[-r^\ast,0]}} |h_n(\theta)|$, since $h_n(\cdot)\in U$.  Next, we
substitute $h_n(\cdot-r^\ast)$ into $\dot{x}(t)=x(t-r(0))$, thus
defining $\dot{h}_n(\cdot)$ over the interval $[-r^\ast,0]$.  We
now find that, for all $\varepsilon >0$, there exists a
$\delta=\delta (\varepsilon)>0$ such that $\sup
|\dot{h}_n(\cdot)|<\varepsilon$ whenever $\|h_n\|^1<\delta$. Since
this property holds for all $n\geq N=N(\varepsilon)$, we arrive at
a contradiction and the proof is complete. \hfill$\Box$

\paragraph{Remark.} % 3.1.}
We may regard Lemma \ref{lm3.1} from the
following perspective. Although, relative to the background space
$W^{1,\infty}([-r^\ast,0])$, the $\|\cdot\|^\infty$-norm is not
equivalent to the $\|\cdot\|^1$-norm, relative to $U$, the
$\|\cdot\|^\infty$- and $\|\cdot \|^1$-norms are equivalent.  This
property guarantees the Fr\'{e}chet differentiability of $g$ when
restricted to $U$.

Trajectories on the unstable manifold appear as solutions to an
integral equation associated with $ \dot{x}(t)=f(x(t),x(t-r(0)))$.
In order to specify the integral equation, we require some
additional notation. Suppose that $\Phi$ is a basis for $U$ and
$\Psi$ is a basis for $U^T$, with $(\Phi,\Psi)=1$, so that $\pi_U
\phi=\Phi(\Psi,\phi)$. We then define $X(\cdot)$ to be the
fundamental matrix solution of $ \dot{x}(t)=f(x(t),x(t-r(0)))$,
$$K(t,s)(\theta)=\int^s_0 X(t+\theta-\tau)d\tau\,,\quad X^u_0=\Phi
\Psi (0)\,,\quad K(t,s)^u=\int^s_0 T(t-\tau)X^u_0 d\tau, $$ and
$$K(t,s)^s=\pi_s K(t,s)=K(t,s)-\Phi (\Psi,K(t,s)).$$ A function
$x^\ast_t (\phi,\cdot)\in W^{1,\infty}([-r^\ast,0])$ satisfying
equation $ \dot{x}(t)=f(x(t),x(t-r(0)))$ must also satisfy the
variation-of-constants formula (3.1) $x_t (\phi,\cdot)=T(t)\phi
+\int^t_0 d[K(t,\tau)]g(x_\tau)$ (for a full treatment of (3.1),
see \cite{h2}).  The following lemma, which is proved in
\cite{h2}, allows us to characterize the unstable set of $
\dot{x}(t)=f(x(t),x(t-r(0)))$. For convenience, we shall set
$W^u(0)=\{\phi \in W^{1,\infty} : x_t(\cdot,\phi)$ exists for all
$t\leq 0$ and ${\lim_{t\to -\infty}} x_t (\cdot,\phi)=0\}$, and
$W^s(0)=\{\phi \in W^{1,\infty} : x_t (\cdot,\phi)$ exists for all
$t\geq 0$ and ${\lim_{t\to +\infty}} x_t (\cdot,\phi)=0\}$.

\begin{lemma} \label{lm3.2}
Suppose that $x^\ast(t,\phi)$ is a solution of (1.1) that is defined and
bounded for all $t\leq 0$.
Then $x^\ast_t (\cdot,\phi)\in W^{1,\infty}([-r^\ast,0])$
satisfies the integral equation $$x_t(\cdot)=T(t)\phi^u + \int^t_0
T(t-\tau)X^u_0 f(x_\tau) d\tau +\int^t_{-\infty} d[K(t,\tau)^s]
f(x_\tau).\eqno(3.2)
$$
\end{lemma}

The proof of this theorem can be found in \cite{h2}.
We can now prove our main theorem. We remark that if $f\in C^k$ in
the theorem below, $k\geq 2$, then $W^u_{{\rm{loc}}}(0)$ will be a
$C^{k-1}$-manifold.

\begin{theorem} \label{thm3.1}  Suppose that $0$ is a hyperbolic
equilibrium point of equation (1.1) and $f\in C^2$.  Then there
exists a neighborhood $N_\delta (0)$ of $0$ in $W^{1,\infty}
([-r^\ast,0])$and a map $\pi:U \cap N_\delta(0) \longrightarrow
W^u(0)$ such that $(\cdot,\pi(\cdot))$ defines a smooth graph.

\end{theorem}

\paragraph{Proof.}  We start by proving the existence of the
unstable manifold $W^u(0)$. In Lemma \ref{lm3.1}, we proved that
$g:W^{1,\infty}([-r^\ast,0])\to \mathbb{R}$ is Lipschitz
continuous.  Also, after writing
\begin{align*}
\dot{x}(t) =&-\alpha x(t)+f(x(t-r))\\
=&-\alpha x(t)+f'(0)x(t-r(0)) +(f(x(t-r))-f'(0)x(t-r(0)))\\
=&-\alpha x(t)+f'(0)x(t-r(0)) + f_1(x(t-r)),
\end{align*}
with $r=r(x(t))$, we know that there exists a monotone increasing,
continuous function $\eta (r) : [0,\infty)\to [0,\infty)$ with
$\eta (0)=0$ such that, for any pair $\phi, \psi \in
T(r^\ast)W^{1,\infty}([-r^\ast,0])$ with $\|\phi\|^1, \|\psi
\|^1\leq \sigma, |f_1(\phi)-f_1(\psi)|\leq \eta
(\sigma)\|\phi-\psi\|^1$.  We now apply the contraction mapping
principle to equation (3.2) for $\|\phi\|^1$ sufficiently small.
Since $\|T(t)\phi^u\|^1 \leq Me^{\beta t} \|\phi^u\|^1$ for $t\leq
0$, we know that there exists a constant $C_0>0$ such that $$
\|T|_{u}x_t \|^1 \leq C_0 (e^{\beta t}\|\phi^u\|^1 +\eta
(\delta)\int^0_t e^{\beta (t-\tau)}\|x_\tau\|^1 d\tau+\eta
(\delta)\int^t_{-\infty} e^{-\beta (t-\tau)} \|x_t \|^1 d\tau), $$
where $$T|_u x_t =T(t)\phi^u + \int^t_0 T(t-\tau) X^u_0 f(x_\tau)
d\tau +\int^t_{-\infty} d[K(t,\tau)^s] f(x_t)$$ and
$\|\phi\|^1<\frac{\delta}{2C_0} <\!< 1$.  We may view $T|_{U}$ as
a self-mapping over the set $$
S(\phi,\delta)=\big\{x_t:(-\infty,0]\to
W^{1,\infty}([-r^\ast,0]),\, \pi_u x_0=\phi^u \in U,\, \sup_{t\in
[-\infty,0)} \|x_t\|^1 \leq \delta\big\}. $$ It follows that
$S(\phi,\delta)$ is closed and bounded with respect to the norm
$\|x_t\|={\sup_{t\in (-\infty,0]}} \|x_t\|^1$. Now, if $\delta$ is
chosen so that $\eta(\delta)<\frac{\beta}{4C_0}$, then an
application of Gronwall's inequality gives the estimate $\|T|_U
x_t\|^1<\delta (\frac{1}{2}+\frac{2C_0}{\beta} \eta
(\delta))<\delta$, and $T|_U=T(\phi^u)$ defines a contraction map.
Thus, $T|_U$ has a unique fixed point, which we call $x^\ast_t
(\phi^U,\cdot)$. $x^\ast_t (\phi^U,\cdot)$ is an absolutely
continuous function.

We now show that $x^\ast_t (\phi^u,\cdot)$ is Lipschitz in
$\phi^u$. In particular, for any pair $\phi^u, \psi^u \in U$ with
$\|\phi^u\|^1, \|\psi^u\|^1$ sufficiently small, we find a
constant $C_1>0$, independent of $\phi^u, \psi^u$, such that
$\|x^\ast_t(\phi^u,\cdot)-x^\ast_t (\psi^u,\cdot)\|^1 \leq C_1
\|\phi^u-\psi^u\|^1$ for $t\leq 0$.  But now
\begin{align*}
x_t^\ast(\phi^u,&\cdot)-x^\ast_t(\psi^u,\cdot)\\
=&T(t)(\phi^u-\psi^u)+\int^t_0 T(t-\tau)X^u_0(f(x^\ast_\tau(\phi^u,\cdot))
  -f(x^\ast_\tau(\psi^u,\cdot))d\tau\\
 &+\int^t_{-\infty}d[K(t,\tau)^s](f(x^\ast_\tau (\phi^u,\cdot))
 -f(x^\ast_\tau(\psi^u,\cdot))
\end{align*}
 and hence
\begin{align*}
\|x^\ast_t (\phi^u,&\cdot)-x^\ast_t(\psi^u,\cdot)\|^1\\
\leq &C_0\Big(e^{\beta t} \|\phi^u-\psi^u\|^1+\eta (\delta)
\int^0_t e^{\beta (t-\tau)}\|x^\ast_\tau (\phi^u,\cdot)
-x^\ast_\tau (\psi^u,\cdot)\|^1d\tau\\
&+\eta (\delta)\int^t_{-\infty} e^{-\beta
(t-\tau)} \|x^\ast_\tau (\phi^u,\cdot) -x^\ast_\tau
(\psi^u,\cdot)\|d\tau \Big).
\end{align*}
Our proof relies upon the
estimate $$\|x^\ast_t (\phi^u,\cdot)-x^\ast_t (\psi^u,\cdot)\|^1
\leq C e^{\beta_0t} \|\phi^u-\psi^u\|^1,$$ which is valid for some
$\beta_0, C>0$ and for all $t\leq 0$ and $\|\phi^u\|^1,
\|\psi^u\|^1\leq \frac{\delta}{2C_0}$.  In this case,
$x^\ast_t(\phi^u,\cdot)$ must be Lipschitz in $\phi^u$ with
respect to the constant $C$.  We verify the estimate by first
defining the weighted norm
$\|x_t(\phi^u,\cdot)\|^{1,\beta_0}={\sup_{-\infty<t\leq 0}}
e^{-\beta_0t} \|x_t (\phi^u,\cdot)\|^1$ on $S(\phi,\delta)$.  If
$\beta_0 \in (\beta-1,\beta), \beta_0>0$, it immediately follows
that $\|x_t (\phi^u,\cdot)\|^{1,\beta_0} \leq \delta$.  Also,
\begin{align*}
e&^{-\beta_0t} \|x_t(\phi^u,\cdot)-x_t(\psi^u,\cdot)\|^1\\
\leq &C_0 \Big[e^{(\beta-\beta_0)t}\|\phi^u-\psi^u\|^1
+\eta(\delta)\int^0_t e^{(\beta-\beta_0)(t-\tau)}
e^{-\beta_0\tau} \|x^\ast_\tau (\phi^u,\cdot)-x^\ast_\tau
(\psi^u,\cdot)\|^1d\tau\\
&+ \eta (\delta) \int^t_{-\infty}
e^{-(\beta+\beta_0)(t-\tau)} \|x^\ast_\tau
(\phi^u,\cdot)-x^\ast_\tau (\psi^u,\cdot)\|^1 d\tau\Big]\\
\leq &C_0 \Big[e^{(\beta-\beta_0)t} \|\phi^u-\psi^u\|^1 +\eta
(\delta)\|x^\ast_t (\phi^u,\cdot)-x^\ast_t
(\phi^u,\cdot)\|^{1,\beta_0} \int^0_t e^{(\beta-\beta_0)(t-\tau)}
d\tau \\
&+\eta (\delta)\|x^\ast_t
(\phi^u,\cdot)-x^\ast_t (\psi^u,\cdot)\|^{1,\beta_0}
\int^t_{-\infty} e^{-(\beta+\beta_0)(t-\tau)} d\tau\Big]\\
\leq &C_0 \Big[ e^{(\beta-\beta_0)t} \|\phi^u-\psi^u\|^1
+\eta(\delta) \big(\frac{1}{\beta -\beta_0} -1\big) \|x^\ast_t
(\phi^u,\cdot)-x^\ast_t (\psi^u,\cdot)\|^{1,\beta_0}\\
& +\eta (\delta) \frac{1}{\beta +\beta_0} \|x^\ast_t
(\phi^u,\cdot)-x^\ast_t (\psi^u,\cdot)\|^{1,\beta_0}\Big]\\
\leq& C_0 \|\phi^u-\psi^u\|^1+C_0 \eta (\delta)
\big[\frac{1}{\beta-\beta_0}-1+\frac{1}{\beta+\beta_0}\big]
\|x^\ast_t (\phi^u,\cdot)-x^\ast_t (\psi^u,\cdot)\|^{1,\beta_0}.
\end{align*}
But now $\|x_t(\phi_t (\phi^u,\cdot)-x_t(\phi^u,\cdot)\|^1 \leq C
e^{\beta_0t}\|\phi^u-\psi^u\|^1$, where
$$C=\frac{C_0}{1-C_0 \eta
(\delta)\big[\frac{1}{\beta-\beta_0}-1+\frac{1}{\beta
+\beta_0}\big]}\,,$$
and thus $W^u (0,N_\delta (0))$ defines a Lipschitz graph.

It remains to prove that $W^u(0,N_\delta (0))$ defines a smooth
graph over the domain $U\cap N_\delta (0)$.  From Lemma
\ref{lm3.1}, we know that $D_\phi g$ is continuous, so that the
proof follows from the estimate
\begin{align*}
\|x^\ast_t &(\psi^u+h,\cdot)-x^\ast_t
(\psi^u,\cdot)-x^\ast_t(\phi^u+h,\cdot)+x^\ast_t(\phi^u,\cdot)\|^1\\
=&\|T(t)(\psi^u+h-\psi^u-\phi^u-h+\phi^u)\\ &+\int^0_t
T(t-\tau)X^u_0\left[f((x^\ast_\tau(\psi^u+h,\cdot))-
 f(x^\ast_\tau(\psi^u,\cdot))\right]d\tau\\
&- \int^0_t T(t-\tau)X^u_0 \left[f(x^\ast_\tau
 (\phi^u+h,\cdot))-f(x^\ast_\tau (\psi^u,\cdot)\right]d\tau\\
&+\int^t_{-\infty} d[K(t,\tau)^s]
 [f(x^\ast_\tau(\psi^u+h,\cdot))-f(x^\ast_\tau(\psi^u,\cdot))]\\
&-\int^t_{-\infty}d[K(t,\tau)^s][f(x^\ast_\tau(\phi^u+h,\cdot))
 -f(x^\ast_\tau(\phi^u,\cdot))]\|^1\\
=&\|\int^0_tT(t-\tau)X^u_0 D_\phi f(x^\ast_\tau(\psi^u,\cdot))h_\tau
 d\tau-\int^0_t T(t-\tau) X^u_0 D_\phi f(x^\ast_\tau
 (\phi^u,\cdot))h_\tau d\tau\\
\leq& \|\int^0_t T(t-\tau) X^u_0
(D_\phi f(x^\ast_\tau (\phi^u,\cdot))-D_\phi f(x^\ast_\tau
(\phi^u,\cdot))) h_\tau d\tau\|^1\\
&+\|\int^t_{-\infty}\frac{d[K(t,\tau)^s]}{d\tau} (D_\phi f(x^\ast_\tau
(\psi^u,\cdot))-D_\phi f(x^\ast_\tau (\phi^u,\cdot))) h_\tau d\tau
\|^1+o(h)\\
\leq& C_1\| \int^0_t e^{\beta(t-\tau)} (D_\phi
f(x^\ast_\tau (\psi^u,\cdot))-D_\phi
f(x^\ast_\tau(\phi^u,\cdot)))h_\tau d\tau\|^1\\
&+C_2 \|\int^t_{-\infty} e^{-\beta (t-\tau)} (D_\phi
f(x^\ast_\tau(\psi^u,\cdot))-D_\phi
f(x^\ast_\tau(\phi^u,\cdot)))h_\tau d\tau\|^1+o(h)\\
\leq&C_1\|\sup_{t\leq 0} \|(D_\phi f(x^\ast_t (\psi^u,\cdot))-D_\phi
f(x^\ast_t(\phi^u,\cdot))h_t \|^1 \int^0_t e^{\beta(t-\tau)}d\tau\|^1\\
&+C_2 \|\sup_{t\leq 0}\|(D_\phi f(x^\ast_t
(\psi^u,\cdot))-D_\phi
f(x^\ast_t(\phi^u,\cdot)))h_t\|^1\int^t_{-\infty} e^{-\beta
(t-\tau)} d\tau\|^1+o(h)\\
=&o(h),
\end{align*}
since the Fr\'{e}chet derivative $D_\phi f$ is continuous.  Thus
$x^\ast_t(\phi^u,\cdot)$ varies smoothly in $\phi^u$, and
$W^u(0,N_\delta(0))$ defines a smooth graph.  The proof of Theorem
\ref{thm3.1} is now complete. \hfill$\Box$\smallskip

 The set $W^u(0,N_\delta(0))$ defines a smooth graph over
$U$. Thus, a typical solution $x_t(\phi,\cdot)$ will have a saddle
structure near the origin.  More precisely, $x_t (\phi,\cdot)$
will approach $0$ along a path nearby $W^s(0,N_\delta (0))$ before
asymptotically tending to the smooth set $W^u(0,N_\delta (0))$. In
this sense any solution which does not decay to $0$ becomes more
well-behaved as $t$ increases.  Here we have defined
$W^s(0)=\{\phi \in W^{1,\infty}([-r^\ast,0]) :
{\lim_{t\to \infty}} x_t(\phi,\cdot)=0\}$, but have
made no claims about the structure or smoothness properties of
$W^s(0)$.  In general, it is to be expected that the dynamics on
$W^s(0)$ will be considerably more complicated than the dynamics
on $W^u(0)$.

\paragraph{Acknowledgements}
This paper forms part of the author's Ph.D thesis (applied math,
Brown University, 1998).  The author would like to thank the
reviewer for his helpful comments, his advisor John Mallet-Paret,
and Jan Garbarek.

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\end{thebibliography}

\bigbreak

\noindent\textsc{Hari P. Krishnan}\\ 
Vice President,\\
Morgan Stanley Corporation \\
Penthouse, 77 West Wacker Street \\
Chicago, IL 60601, USA\\ 
e-mail: Hari.Krishnan@morganstanley.com


\end{document}