\documentclass[12pt]{article}
\usepackage{latexsym}

\def\Def{Definition}
\def\a{for every }
\def\A{For every }
\def\e{there exists }
\def\E{There exists }
\def\pf{{\em Proof. }}
\def\eps{\varepsilon}
\def\bx{\bar{x}}
\def\by{\bar{y}}
\def\vphi{\varphi}

 \begin{document}

\newtheorem{dfn}{Definition}[section]
\newtheorem{eqn}{Equation}[section]
\newtheorem{lem}{Lemma}[section]
\newtheorem{thm}{Theorem}[section]
\newtheorem{cor}{Corollary}[section]

\title{A Brief Overview of Lyapunov Stability for Ordinary Differential Equations}
\author{David Glickenstein}
\date{May 2, 1996}
\maketitle

\abstract{Lyapunov stability is a way to characterize the qualitative
idea of stability in models of physical systems.  A stable solution to
a differential equation is one which does not change much when the
initial conditions are changed slightly.  We shall look at two methods
of checking the stability of solutions to ordinary differential
equations.  The first works by characterizing the solutions of linear
systems to study their stability and then looking at nonlinear systems
which are close to the linear case.  The second uses an auxilliary
function to find stability without having to characterize the
solutions.  Finally we look at some of the limitations of the Lyapunov
notion of stability.}

\section{Introduction}

We are going to look at general ordinary differential equations of the
following type:
\begin{equation}
	x'(t) = f(t,x(t))
	\label{gen}
\end{equation}
where $x: \Re \rightarrow \Re^n$ and $f:\Re \times \Re^n \rightarrow
\Re^n$ are functions and $t$, which usually represents time, is the
only independent variable.  The derivative $x'(t)$ is simply the
derivatives of each of the component functions, so if $x(t) =
(x_1(t),...,x_n(t))$ where $x_i(t)$ are real-valued functions on $\Re$
then
$$x'(t) = (x_1'(t),...,x_n'(t)).$$
In some cases we will be considering the autonomous case
\begin{equation}
	x'(t) = f(x(t))
	\label{aut}
\end{equation}
where $f:\Re^n \rightarrow \Re^n$.  We are going to assume the
existence of continuous solutions and uniqueness of a solution with
the initial condition $x(t_0) = x_0$ where $t_0$ is a constant and
$x_0$ is a constant n-vector.  In the simplest case, a solution will
tend to a constant as $t$ goes to infinity.  In the autonomous case,
we can think of the solution as a continuous curve through n-space
(called phase space), and when the solution tends to a constant, the
curve will converge to a point in phase space.  If a constant function
$c$ satisfies the differential equation, then the the derivative
$x'(t) = f = 0$ for $x_0 = c$ and $c$ is called an equilibrium point.
Periodic solutions satisfy $x(t) = x(t + n\tau)$ for some $\tau$ and
any integer $n$.  Other, more complicated solutions also exist which
may or may not be stable.  Nonautonomous systems have analogous types
of solutions.

We are going to be studying the stability of (\ref{gen}).  Stability
is a qualitative notion that if we perturb the initial condition
slightly, say $x(t_0) = x_0 + \eps$, then the new solution will be
close to the original system.  Stable differential equations are more
useful in practice since $x_0$ is often a  measured value and does
not have infinite precision.  A stable differential equation would
insure that our model would come up with an answer reasonably close to
the actual answer an exact initial condition.

First, we will define our notion of stability.  What we shall refer to
as {\em stable} is more correctly termed {\em Lyapunov stable} since
there are many different mathematical notions of stability.

\begin{dfn}
  	A solution $x(t)$ is {\em stable} if \a $\eps >0$
	\e $\delta > 0$ such that if $\| x_0 - \bx_0 \| < \delta$ then
	$\| x(t) - \bx(t) \| < \eps$ \a $t \ge t_0$, where $\bx(t)$ is
	the solution to
	$$\bx'(t) = f(t, \bx(t))$$
with $x(t_0) = x_0$ and $\bx(t_0) = \bx_0$.
\end{dfn}

Sometimes our model works so well that if we change the initial
condition slightly, the solution will still converge to the exact
solution under the original initial conditions.  This means that when
we use a measured value, our approximate model (with an inexact
initial condition) not only stays close to the actual solution, but
gets closer and closer to the actual solution as $t$ increases.  This
notion is called {\em asymptotic stability}.

\begin{dfn}
	A solution $x(t)$ is {\em asymptotically stable} if it is stable
	and there is a $\rho > 0$ such that if $\| x_0 - \bx_0 \| < \rho$
	then 
$$\lim_{t \rightarrow \infty} \| x(t) - \bx(t) \| = 0.$$
\end{dfn}

In general, asymptotic stability is more desirable, since we know that
no matter where we place our initial condition, we will converge to
the actual answer if we are close enough to the actual initial
condition.  Unfortunately, asymptotic stability is not always
possible, as we shall see in the fourth section.

We will primarily use these two notions of stability.  There
are two main methods used to study Lyapunov stability, known
as Lyapunov's First and Second (or Direct) Methods.

\section {Lyapunov's First Method}

Lyapunov's First Method works by characterizing solutions to
differential equations of the type (\ref{gen}) and using those
characterizations to learn what we can about stability.  Thus our
approach will begin with simple linear systems and move on to certain
kinds of nonlinear systems which exhibit similar behavior to the
linear systems.

We will first look at linear differential equations 

\begin{equation}
	x'(t) = A(t)x(t)
    \label{linear}
\end{equation}

where $A: \Re ^n \rightarrow \Re ^n$ is linear and $x: \Re \rightarrow \Re^n.$

When working with linear systems of the form (\ref{linear}) we will
use the following theorem, without proof, to characterize solutions.
The theorem is presented as a matrix equation where the derivative
$C'(t)$ of a matrix function $C$ is simply defined as the derivative
of each of the entries of the matrix.

\begin{thm}
	If $\Phi '(t) = A(t) \Phi (t)$ with $\Phi$ an invertible $n \times
n$ matrix then all solutions of (\ref{linear}) are linear combinations
of the columns of $\Phi$ and, in particular, we can choose some $\Phi
(t)$ such that $\Phi (t_0) = I$ where $I$ is the identity matrix, so
that $x(t) = \Phi (t) x_0$ is a solution to (\ref{linear}).  \end{thm}

\pf Omitted.

\begin{dfn}
	$\Phi$ as described in the last theorem is called the {\em
fundamental matrix} of the differential equation.
\end{dfn}

The fundamental matrix depends only on the differential equation and
thus only on the matrix function $A(t)$.

We also will need to define the operator norm for a matrix A:

\begin{dfn}
	If $A$ is an $n \times n$ real matrix then
$$\| A \| = \sum^n_{i,j=1} |a_{ij}|$$
\end{dfn}

Note that $\| Ax \| \le \| A \| \  \|x\|$ where $\| A \|$ is the operator
norm and the other two norms are the usual Euclidean vector norm in $\Re^n$.

We are now ready to prove an important theorem.  It turns out that in
the linear case the property of solutions being stable is equivalent
to the property of solutions being bounded.

\begin{thm}
	All solutions of (\ref{linear}) are stable if and only if they are
bounded.
\end{thm}

% begin proof
\pf All solutions are of the form $\Phi (t) x_0$ where $\Phi$ is the
fundamental matrix such that $\Phi (t_0) = I$.  Let $x(t)$ and
$\bx(t)$ be solutions satisfying $x(t_0) = x_0$ and $\bx(t_0) =
\bar{x}_0$ respectively.
  
Assume that all solutions to (\ref{linear}) are bounded.  Thus we can
find an $M$ such that $\| \Phi (t) \| < M$ for all $t > 0$.  Given
some $\eps > 0$, if $\| x_0 - \bx_0 \| < {\eps \over M}$ then
\begin{eqnarray*}
	\| x(t) - \bx(t) \| & = & \| \Phi (t)x_0 - \Phi (t) \bx_0 \| \\
	& \le & \| \Phi (t) \| \ \| x_0 - \bx_0 \| \\
	& < & \eps.
\end{eqnarray*}

Conversely, assume that all solutions are stable.  Thus the particular
solution $x(t) \equiv 0$ is stable, so \a $\eps > 0$ \e $\delta > 0$
such that if $\| \bx_0 \| < \delta$ then 
$$\| \bx (t) \| = \| \Phi (t) \bx_0\| < \eps.$$
In particular, let 
$$\bx_0^{(i)} = {\delta \over 2} e_i $$
where $e_i$ is the ith standard basis vector of $\Re^n$.  Then
$$\| \Phi (t) \bx_0^{(i)} \| = \| \phi_i(t) \| {\delta \over 2} <
\eps$$
where $\phi_i$ is the ith column of $\Phi$.  We can sum these over all
i and get:
$$\| \Phi (t) \| \le \sum^n_{i=1} \| \phi_i(t) \| \le {2n \eps \over
\delta} = k$$
so $$\| x(t) \| = \| \Phi (t) x_0 \| \le k \| x_0 \|$$
and $x(t)$ is bounded. $\Box$
%end proof

Note that boundedness is not necessarily true for stable systems in
general.  For instance, take the differential equation $x'(t) = 1$
whose solutions are all stable but not bounded.

Lyapunov's First Method consists mainly of showing that solutions of linear
systems, especially if $A(t)$ is a constant funtion, are bounded and
hence stable.  It then looks at functions which are nonlinear, but
simply small perturbations of linear functions.  We shall state
without proof several theorems.  For a more detailed explanation with
proofs, see [2].

Consider the constant linear equation:
\begin{equation}
	x'(t) = A \ x(t)
	\label{constlin}
\end{equation}
where $A$ is a constant matrix.  We can easily show the following.

\begin{thm}
	If all eigenvalues of A have negative real parts then every
solution to (\ref{constlin}) is asymptotically stable.
\end{thm}

\pf Omitted.

We can easily extend this to equations with slight perturbations.  So
we characterize the following nonlinear equation

\begin{equation}
	x'(t) = A \ x(t) + f(t,x(t))
	\label{nonlinear}
\end{equation}

with the following theorem.

\begin{thm}
	If f satisfies:

\begin{enumerate}
\item $f(t,x)$ is continuous for $\| x \| < a, \ 0 \le t <\infty$ and
\item $\lim_{\| x \| \rightarrow 0} {\| f(x,t) \| \over \| x \|} = 0$ 
\end{enumerate}
then the solution $x(t) \equiv 0$ is asymptotically stable.
\end{thm}

It turns out that to characterize the general linear case where $A$ is a
function of time, we need a slightly stronger notion of stability
called uniform stability.  The reader is again referred to [2].

\section{Lyapunov's Second Method}

Lyapunov's Second or Direct Method is unique in that it does not
require a characterization of the solutions to determine stability.
This method often allows us to determine whether a differential
equation is stable without knowing anything about what the solutions
look like, so it is ideal for dealing with nonlinear systems.

The method uses a supplementary function called a Lyapunov function to
determine properties of the asymptotic behavior of solutions to a
differential equation of the general form (\ref{gen}).  We will need
the following definitions.

\begin{dfn}
	A function $V(t,x)$ is {\em positive [negative]
definite} if \e a real-valued function $\vphi (r)$ such that:

\begin{enumerate}
\item $\vphi (r)$ is strictly increasing on $0 \le r \le a$ and
$\vphi (0) = 0$ and
\item $V(t,x) \ge \vphi ( \| x \| ) \ [V(t,x) \le - \vphi ( \| x \| )]$
for all $(t,x) \in \{ (t,x): t_0 \le t < \infty, \| x \| \le b < a \}$.  
\end{enumerate}

\end{dfn}
	
\begin{dfn}
	$$V' = \sum^n_{i=1} {\partial V \over \partial x_i} f_i(t,x) +
{\partial V \over \partial t}$$
\end{dfn}
Note that $V'(t,x(t))$ is the time derivative of $V$.

\begin{dfn}
	A real function $V(t,x)$ is said to admit an {\em infinitesimal upper bound} if \e
$h > 0$ and a continuous, real-valued, strictly increasing function
$\psi$ with $\psi (0) = 0$ such that 
$$|V(t,x)| \le \psi( \| x \|) \ {\rm for} \ \| x \| < h \ {\rm and} \
t \ge t_0.$$
\end{dfn}

Using these definitions we shall construct a function $V$.  If we can
find such a $V$, then we can find stable solutions to the differential
equation.  Unfortunately, there is no general way to construct $V$
from the differential equation (\ref{gen}).  Note that we can define
$V$ in terms of $t$ and $x$ (not $x(t)$) and since our definition of $V'$
uses the function $f$, the existence of $V$ will have implications for
the differential equation.

\begin{thm}
	If a continuous real-valued function $V(t,x)$ exists such that:
\begin{enumerate}
\item $V(t,x)$ is positive definite and
\item $V'(t,x)$ is non-positive
\end{enumerate}
then $x(t) \equiv 0$ is a stable solution of (\ref{gen}).
\end{thm}

\pf We know that $V(t,x)$ is positive definite, so \e a strictly
increasing function $\vphi (r)$ such that 
$$0 < \vphi (\| x \| ) \le V(t,x)$$
for $0 < \| x \| < b$ and $t > t_0$.

Given $\eps > 0$ let 
$$m_{\eps} = \min_{\| x \| = \eps} \vphi (\| x \| ) = \vphi (\eps).$$ Notice that
$m_{\eps} > 0$.  Since $V$ is continuous and $V(t,0) = 0$ for all $t$
(by positive definiteness),
we can choose a $\delta > 0$ such that 
$$V(t_0, x_0) < m_{\eps}$$ 
if $\|x_0 \| < \delta$.  Now 
$$V'(t,x(t)) \le 0$$ 
for $t_1 \ge t_0$ and $\| x_0 \| < \delta$ implies
$$V(t_1, x(t_1)) \le V(t_0,x(t_0)) = V(t_0,x_0) < m_{\eps}$$

Now, suppose for some $t_1 \ge t_0$ that $\| x(t_1) \| \ge \eps$
whenever $\| x_0 \| < \delta$.  Then 
$$V(t_1,x(t_1)) \ge \vphi (\| x(t_1) \| ) \ge \vphi (\eps) =
m_{\eps}$$
which is a contradiction, so \a $\eps > 0$ \e $\delta > 0$ such that
$\| x(t) \| < \eps$ if $\| x_0 \| < \delta$ for $t \ge t_0$,
i.e. $x(t) \equiv 0$ is stable.  $\Box$

We call a function $V$ which satisfies the hypotheses of the theorem a
{\em Lyapunov function}.  In physical systems, the Lyapunov function
used is often an expression for energy, for instance stability for the
differential equation for a spring: 
\begin{eqnarray*}
x'(t) & = & y\\
y'(t) & = & -kx
\end{eqnarray*}
can be done with $V(x,y) = {1 \over 2} y^2 + {1 \over 2} k x^2$, which is the equation
for total energy of the system.

By strengthening the requirements on $V$ we can find conditions for
asymptotic stability.

\begin{thm}
	If a continuous function $V(t,x)$ exists satisfying:
\begin{enumerate}
\item $V(t,x)$ is positive definite,
\item $V(t,x)$ admits an infinitesimal upper bound, and
\item $V'(t,x)$ is negative definite
\end{enumerate}
then the solution $x(t) \equiv 0$ of (\ref{gen}) is asymptotically
stable.
\end{thm}

\pf  By the previous theorem, the solution must be stable.  Suppose
that $x(t)$ is not asymptotically stable, so \a $\eps >0$ there exist
$\delta > 0$ and $\lambda > 0$ such that any nonzero solution $x(t)$ where
$x(t_0) = x_0$ satisfies $\lambda \le \| x(t) \| < \eps$ if $t \ge
t_0$ and $\| x_0 \| < \delta$.

Since $V'(t,x)$ is negative definite, there is a strictly increasing function
$\vphi (r)$ vanishing at the origin such that $V'(t,x) \le -
\vphi(\| x \|)$.  We know that $\| x(t) \| \ge \lambda > 0$ for $t \ge
t_0$ so there is a $d > 0$ such that $V'(t,x(t)) \le -d$.  This
implies that 
\begin{eqnarray*}
	V(t,x(t)) & = & V(t_0,x_0) + \int^t_{t_0} V'(s,x(s)) ds \\
	& \le & V(t_0,x_0) - (t- t_0) d
\end{eqnarray*}
which is less less than zero for large $t$, contradicting the
assumption that $V(t,x)$ is positive definite.

Thus no such $\lambda$ exists and since $V(t,x(t))$ is positive definite and
decreasing with respect to $t$
$$\lim_{t \rightarrow \infty} V(t,x(t)) = 0$$
which implies that
$$\lim_{t \rightarrow \infty} \| x(t) \| = 0,$$
since $V(t,x(t)) \le \psi (\| x(t) \|)$ where $\psi (r)$ is strictly
increasing and vanishes at the origin (by the second hypothesis).
Thus as $V(t,x(t)) \rightarrow 0$, $\psi(\| x(t) \|) \rightarrow 0$,
and $\| x(t) \| \rightarrow 0$ (because $\psi$ is strictly increasing
and vanishes at 0).  Thus $x(t) \equiv 0$ is asymptotically stable. $\Box$

We can also use the direct method to determine instability in some cases.

\begin{thm}
	If a continuous real-valued function $V(t,x)$ exists on some set
$S = \{ (t,x) : t \ge t_1 \ and \ \| x \| < a \} $ satisfying:
\begin{enumerate}
\item $V(t,x)$ admits an infinitesimal upper bound,
\item $V'(t,x)$ is positive definite on S, and
\item \E a $T > t_1$ such that if $t_0 \ge T$ and $h > 0$ then \e $c
\in \Re^n$ such that $\| c \| < h$ and $V(t_0,c) > 0$
\end{enumerate}
then the solution $x(t) \equiv 0$ is not stable.
\end{thm}

\pf Suppose $x(t)$ is a solution not identically zero.  Then, by
uniqueness, $ \| x(t) \| \neq 0$ for all $t \ge t_1$.  If $x(t)$ is
defined at $t_0$ then for every $t \ge t_0$ for which $x(t)$ is
defined
$$V(t,x(t)) - V(t_0, x(t_0)) = \int^t_{t_0} V'(s,x(s)) ds > 0$$
since $V'(t,x(t))$ is positive definite.

Let $t_0 \ge T$ and $\eps > 0$.  There exists a $c$ such that $\| c \|
< \min(a,\eps)$ and $V(t_0,c) > 0$ by the third hypothesis.  Since $V$
admits an infinitesimal upper bound and is continuous, there is a
$\lambda \in (0,a)$ such that $\| x \| < \lambda$ for $t \ge t_1$.
Therefore,
$$| V(t,x) | < V(t_0, c).$$

Let $x(t)$ satisfy the initial condition $x(t_0) = c$.  Then $V(t,x(t)
> V(t_0, x(t_0)) > 0$ (from the positive definiteness argument), so
$\| x(t) \| \ge \lambda$.  Also by the second hypothesis we know that
there is a strictly increasing function $\vphi(r)$ such that $V'(t,x)
\ge \vphi(\| x \|)$.  Let
$$ \mu = \min_{\| x \| \in [\lambda, a]} \vphi(\| x \|) = \vphi (\lambda).$$
then
$$V'(t,x(t)) \ge \vphi(\| x(t) \|) \ge \mu.$$
Using the integral inequality, we get
$$V(t,x(t)) \ge V(t_0, x(t_0)) + (t - t_0) \mu.$$ Thus we can make
$V(t, x(t))$ arbitrarily large.  Since V(t,x) admits an infinitesimal
upper bound, there exists a strictly increasing function $\psi (r)$
such that $|V(t,x(t))| < \psi (\| x \|)$ and $\psi (0) = 0$.  So since
$V(t,x(t))$ can be arbitrarily large, $\psi(\| x(t) \|)$ can also be
arbitrarily large, and thus $\| x(t) \|$ gets arbitrarily large and the
solution is not stable. $\Box$

\section {Limitations of Lyapunov Stability}

The Lyapunov notion of stability has some problems, some of which we
shall look into here.  Asymptotic stability as we have defined it does
not always work well for periodic systems, so often other notions such
as orbital stability are used when studying differential equations
which have periodic solutions.  In addition, certain classes of
systems such as Hamiltonian Systems do not react well to a Lyapunov
analysis.  Finally, there is the question of whether Lyapunov
stability actually reflects our qualitative idea of stability.

We will first look at how Lyapunov stability works with periodic
solutions to the autonomous system (\ref{aut}).  It turns out that our
notion of asymptotic stability is incompatible with such solutions,
since the points will not get closer together, even thought two
solutions may get closer and closer to the same final state (a
periodic cycle).  To prove this we will first need a Lemma that
solutions to autonomous systems are invariant under translation.

\begin{lem}
	If $x(t)$ is a solution to (\ref{aut}) then $x(t+h)$ is also a
solution for any real number $h$.
\end{lem}

\pf Let $\bx (t) = x(t+h)$ and $s = t+h$.  Then
\begin{eqnarray*}
	\bx ' (t) & = & {dx \over ds}{ds \over dt} \\
	& = & x'(s)\\
	& = & f(x(s))\\
	& = & f(x(t+h))\\
	& = & f(\bx(t))\\
\end{eqnarray*}
so $\bx(t) = x(t+h)$ is a solution to (\ref{aut}). $\Box$

Now we can use the fact that translations of solutions are also
solutions to show that nontrivial periodic solutions cannot be
asymptotically stable.  By nontrivial periodic we mean that the
solutions do not converge to a constant vector, i.e., $x(t) \neq x(t +
\eps)$ for some small $\eps > 0$.

\begin{thm}
	If $x(t)$ is a nontrivial periodic solution of an autonomous
	system (\ref{aut}) then $x(t)$ is not asymptotically stable.
\end{thm}

\pf Suppose $x(t)$ is a nontrivial periodic solution.  Thus \e $t_0$
such that $f(x(t_0)) \neq 0$ because if not, $x'(t) = 0$ for all $t$
and the solution is constant (trivially periodic).  Suppose also that
$x(t)$ is asymptotically stable.

The stability assumption says that given $\eps > 0$ \e $\hat{\delta}
> 0$ such that $\| x(t) - \bx(t) \| < \eps$ if $ \| x_0 - \bx_0 \| <
\hat{\delta}$ where $x(t)$ satisfies $x(t_0) = x_0$ and $\bx(t)$ is a
solution to the same equation satisfying the initial condition
$\bx(t_0) = \bx_0$.

By the asymptotic stability assumption we can find a $\rho > 0$ such
that 
$$\lim_{t \rightarrow \infty} \| x(t) - \bx(t) \| = 0$$
if $\| x_0 - \bx_0 \| < \rho$.

Let $\delta = \min (\hat{\delta}, \rho)$.  By the continuity of $x(t)$
we can fix some small $\bar{t}$ so that 
$$\| x(t_0) - x(t_0 + \bar{t}) \| < \delta.$$
By the Lemma, $x(t + \bar{t})$ is a solution to the differential
equation, so by our asymptotic stability assumption
$$\lim_{t \rightarrow \infty} \| x(t) - x(t + \bar{t})\|  = 0.$$

We know that $x'(t_0) \neq 0$ so we can find $r > 0$ so that $\|x(t_0)
- x(t_0 + \bar{t}) \| > r$.  Also $x(t)$ is periodic, so assume it has
period $\tau$.  Then
$$\| x(t_0 + n \tau) - x(t_0 + \bar{t} + n \tau) \| > r >0$$
for all $n$, so the limit cannot go to zero, which contradicts our
assumption of asymptotic stability.  Thus $x(t)$ cannot be
asymptotically stable. $\Box$ 

Another class of problems which for which asymptotic stability does
not work is Time-Independent Hamiltonian systems.  They are defined as follows.

\begin{dfn}
	A {\em time-independent Hamiltontian system (TIHS)} is the 2n
dimensional system
\begin{eqnarray*}
	x'(t) & = & H_y(x(t), y(t)) \\
	y'(t) & = & -H_x(x(t), y(t))
\end{eqnarray*}
where $x(t)$ and $y(t)$ are both functions from $\Re$ to $\Re^n$ and
$H(x,y)$ is a real valued function with continuous partial
derivatives.
\end{dfn}

Hamiltonian Systems are used to model a large class of physical systems
and have many applications.  It turns out that, like nontrivial
periodic solutions to autonomous systems, no solutions to a TIHS can be
asymptotically stable.

\begin{thm}
	If $(x(t), y(t))$ is a solution to a TIHS then it is not
asymptotically stable.
\end{thm}

\pf Suppose $(x(t),y(t))$ is an asymptotically stable solution.  Then
we can find $t_0$ and $\delta > 0$ such that if $(\bx(t), \by(t))$ is
a solution and if $\| (x(t_0),y(t_0)) - (\bx(t_0), \by(t_0)) \| < \delta$ then
$$\lim_{t \rightarrow \infty} \| (x(t),y(t)) - (\bx(t), \by(t)) \| =
0.$$

But also, 
$${d \over dt} H(x,y) = {\partial H \over \partial x}{dx \over dt} +
{\partial H \over \partial y}{dy \over dt} = H_x (H_y) + H_y (-H_x) =
0$$
so $H(x,y)$ is constant on any solution $(x(t),y(t))$.  By the continuity of $H$,
it must be constant on some $\delta$-neighborhood N of
$(x(t_0),y(t_0))$, so every point in N is an equilibrium point.  Thus
$(x(t),y(t))$ cannot be asymptotically stable (for initial condition
$(\bx(t_0),\by(t_0)) \in {\rm N}$, the solution
$(\bx(t),\by(t))$ will not converge to $(x(t),y(t))$. $\Box$

It turns out that although Lyapunov stability is extremely useful in
mathematically defining our qualitative view of stability, it is
neither necessary nor sufficient for something to appear stable on a
physical level.  Some examples can be found at the end of Chapter 4 of [1].

\begin{thebibliography}{99}
\bibitem{Cronan} J. Cronin. {\em Differential Equations: Introduction
and Qualitative Theory.} Marcel Dekker, New York, 1994.
\bibitem{Sanchez} D.A. Sanchez. {\em Ordinary Differential Equations
and Stability Theory: An Introduction.}  Dover Publications, New York,
1979.
\end{thebibliography}

\end{document}