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\markboth{Marco Lenci}{Escape Orbits for Non-Compact Flat Billiards}
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\begin{document}
\title{Escape Orbits for Non-Compact Flat Billiards}
\author{
Marco Lenci\thanks{On leave from: Dipartimento di Matematica,
Universit\`a di Bologna, 40127 Bologna, Italy} \\
Mathematics Department \\
Princeton University \\
Princeton, NJ \ 08544 \\
U.S.A. \\
{\tt\small marco@math.princeton.edu}
}
\date{February 1996}
\maketitle
\begin{abstract}
It is proven that, under some conditions on $f$, the non-compact flat
billiard $\Omega = \{ (x,y) \in \R_0^{+} \times \R_0^{+};\ 0\le y \le
f(x) \}$ has no orbits going {\em directly} to $+\infty$. The
relevance of such sufficient conditions is discussed.
\end{abstract}
\section{Introduction}
Let $f$ be a smooth function from $\R_0^{+}$ to $\R^{+}$, bounded,
vanishing when $x\to +\infty$. No integrability assumptions are given.
Now construct a plane billiard table in the following way:
$\Omega = \{ (x,y) \in \R_0^{+} \times \R_0^{+};\ 0\le y \le f(x) \}$
as is shown in figure 1. Imagine to have a dimensionless
particle moving freely within $\Omega$ and reflecting elastically
at its boundary.
\par
Some interest seems to be focused recently on these (see
\cite{l,ss,k,ddl}) and other billiard systems over non-compact
manifolds (intriguing and related examples include \cite{b,gu,ga}).
In this contest an issue regards whether and how a trajectory may
leave any compact region of the configuration space. The study of
this is of theoretical interest in itself, as well serving as a possible
starting point for the investigation of the ergodic properties of
non-compact dynamical systems of which not much seems to be known.
Plus, and this is the main motivation for recent work, it has a direct
link with the search for {\em quantum chaos} in the quantized
version of these systems. Specifically,
unbounded orbits represent the other side of the coin compared to
periodic orbits. The distribution of the latter provides
information on the spectrum of the Laplace-Beltrami (i.e.
Hamiltonian) operator on the given manifold via the celebrated
Gutzwiller trace formula or similar asymptotic expansions \cite{g,st}.
Being able to ``count'' periodic orbits means gaining information on
the aysmptotics of the quantum energy levels. This has suggested the
idea of {\em coding} periodic orbits, i.e. associating an infinite
string of symbols to each periodic orbit, possibly in a bijective
way.
\footnote{There are quite a few works on that. \cite{gu} has good
analogies with the present paper.}
Understandably, unbounded trajectories represent a nuisance
for such a coding. Finally, it is known that the heat kernel
expansion is sensitive to finite or infinite cusps \cite{st}. One
would like to relate this fact to the presence of orbits eventually
falling into the cusp.
\par
The question we ask ourselves here is the following: are there any orbits
of our billiard in $\Omega$ which go {\em directly} to infinity, i.e.
maintaining for all times a positive $x$-component of their velocity?
To fix the notation let us call such trajectories {\em escape orbits}.
Of course there is always one of these: it is the orbit which lies on
the horizontal semi-axis. We call it the {\em trivial orbit}.
\par
We are going to show that no other orbits can share this property,
provided we require some conditions on $f$. The question was first
touched on by A.M.Leontovich
\footnote{To my knowledge.}
in 1962 (\cite{l}, Theorem 2) where -- even though he was searching
for oscillating unbounded orbits -- he obtained the above result for
{\em eventually convex} $f$'s, i.e. convex in a neighborhood of
$\infty$. This is the same result we find much more recently
in J.L.King's review \cite{k}. It can be explained easily, at least
for billiard tables of finite area. In that case the cusp has
asymptotically a vanishing measure. Now, if we have an escape orbit,
then, due to the hyperbolicity of the flow in that region of the
phase space, we can find a non-zero measure set of escape orbits,
that is which go into the cusp, with an obvious contradiction.
\footnote{This argument is described in \cite{k} as ``A gallon of water
won't fit inside a pint-sized cusp''.}
In fact we can always fix, as the initial point of our escape orbit,
a point $(x_0,y_0) = (x_0,f(x_0))$ on the upper boundary of $\Omega$,
far enough to lie in the region where $f$ in convex. The initial velocity
will have an $x$-component $v_x > 0$. Now it is easily seen that every
other set of initial conditions $x'_0 \ge x_0$ and $v'_x \ge v_x$ (provided
$v$ and $v'$ have the same modulus) would lead to a new escape orbit, due
to the dispersing effect of the convex upper wall.
The same argument may be used to deduce that an infinite cusp on the
Poincar\'e's disc does not allow non-trivial trajectories to collapse
into it, which is implicitly stated in \cite{gu}.
\par
We are going to relax the hypothesis for the non-existence result to
hold: asymptotic hyperbolicity is not a necessary condition at all.
We may allow $f$ to have flex points and abrupt negative variations
(see, for instance, figure 2c). The proofs are presented in the
next section, while examples of non-convex $f$'s fulfilling our
hypotheses are discussed in the last section in order to understand what
the new assumptions actually mean and how far they can be pushed.
\section{The result}
\begin{theorem}
Consider the plane billiard table generated as above by the function
$f$ defined on $\R_0^{+})$, twice differentiable, positive, bounded,
such that
$$
f(x) \searrow 0 \mbox{ as } x\to +\infty.
\eqno{{\em (H1)}}
$$
Also, for $x$ sufficiently large,
$$
f'(x) < 0,
\eqno{{\em (H2)}}
$$
Then, under either one of these assumptions:
$$
\limsup \frac{f'(x)}{f(x)} <0;
\eqno{{\em (A1)}}
$$
or
$$
\lim_{x\to +\infty} f'(x) =0 \mbox{ and } \limsup_{x\to +\infty}
\frac{f'' f}{f'}(x) < +\infty;
\eqno{{\em (A2)}}
$$
no orbits but the trivial one are escape orbits.
\end{theorem}
\skippar
{\sc Remark.} The assumption about the convexity of $f$ is contained in
(A2): in fact if $f'' \geq 0$, then necessarily $f' \nearrow 0$ and
$f'' f/f' \leq 0$.
\skippar
{\sc Proof.} Suppose, contrary to our goal, we have a non-trivial escape
orbit: let us fix without loss of generality an initial point in a
neighborhood of $+\infty$ where all the asymptotic hypotheses hold.
For instance, (H2) would do, and (A2), if this is the case, would be
read as
\begin{equation}
f'(x) \geq -\eps \mbox{ and } \frac{f'' f}{f'}(x) < k_1;
\label{newA2}
\end{equation}
for some $\eps>0$. Also, for some consistency of notation let us
suppose the initial point is a bouncing point on the upper wall, i.e.
$(x_0,y_0) = (x_0,f(x_0))$.
\skippar
Using the notations of figure 1 we have the fundamental relation:
\begin{equation}
\tan\theta_{n+1} (x_{n+1}-x_n) = f(x_n) + f(x_{n+1}).
\label{fundamental}
\end{equation}
With a bit of geometry, looking at the same picture, we get
\begin{displaymath}
\theta_{n+1} = \theta_n + \pi - 2\alpha_n = \theta_n + 2\delta_n,
\end{displaymath}
calling $\delta_n = - \arctan(f'(x_n)) > 0$. This summarizes to
\begin{displaymath}
\theta_n = \theta_1 + 2\ \sum_{k=1}^{n-1} \delta_k.
\end{displaymath}
Thus $\{ \theta_n \}_{n \ge 1}$ is an increasing sequence. Since we
have assumed the particle never go backwards, then $\theta_n <
\pi/2$ for all $n \ge 1$, so $\theta_n \nearrow
\theta_\infty \in [\theta_1,\pi/2]$. Hence $\tan\theta_n \ge
\tan\theta_1 =: k_2 > 0$. From this, the monotonicity on $f$, and
(\ref{fundamental}) we have
\begin{equation}
x_{n+1} - x_n \leq \frac{2}{k_2} f(x_n).
\label{estimate1}
\end{equation}
What stated above implies that $\sum_k \delta_k < +\infty$. Therefore
$\delta_k \to 0$. As a consequence, we see that $\exists k_3\in ]0,1[$
such that $\delta_k = \geq k_3 \, \tan\delta_k = k_3 |f'(x_k)|$.
If we are able to prove that
\begin{equation}
-\sum_{k=0}^{\infty} f'(x_k) = +\infty,
\label{to-violate}
\end{equation}
that inequality implies that $\sum_k \delta_k$ cannot converge,
creating a contradiction which finishes the proof.
\skippar
Defining $g:= -f'/f >0$ will greatly simplify our notation.
>From (\ref{estimate1}), we obtain, for some constant $k_4$,
\begin{equation}
-\sum_n f'(x_n) \geq k_4 \sum_n g(x_n) (x_{n+1} - x_n).
\label{estimate2}
\end{equation}
\skippar
{\sc Case} (A1). Obviously (A1) reads $g \geq k_5 > 0$. Hence, since
by hypothesis $x_n \to\infty$, (\ref{estimate2}) gives
(\ref{to-violate}). It may worth remind that (A1) means we have
exponential decay for $f$. In fact, after an integration, we see
that $\forall x_2 > x_1 \geq 0$,
\begin{displaymath}
f(x_2) \leq f(x_1)\: e^{-k_5(x_2 - x_1)}.
\end{displaymath}
\skippar
{\sc Case} (A2). The proof is little more involved here. We use our
assumption on the limit of $f'$ to prove the following
\begin{lemma}
There exists a constant $k_6$ such that $\forall n\in\N$
\begin{displaymath}
\frac{f(x_n)}{f(x_{n+1})} \leq k_6.
\end{displaymath}
\end{lemma}
{\sc Proof.} Let us call $\bar x_n$ the point in $[x_n,x_{n+1}]$ provided
by the Lagrange mean value theorem. We can write
\begin{displaymath}
\frac{f(x_n)}{f(x_{n+1})} = 1 - \frac{f'(\bar x_n) (x_{n+1}-x_n)}
{f(x_{n+1})}.
\end{displaymath}
Using (\ref{estimate1}), this is turned into
\begin{displaymath}
\frac{f(x_n)}{f(x_{n+1})} \left( 1 + \frac{2 f'(\bar x_n)}{k_2} \right)
\leq 1,
\end{displaymath}
which yields the lemma, since the term in the brackets is positive
for $n$ large enough, because of the assumption about the vanishing of
$f'$.
\skippar
This enables us to prove
\begin{lemma}
There exists a constant $k_7 \in ]0,1[$ such that, for $n$
sufficiently large, $g(x_n) \geq k_7 \max_{[x_n,x_{n+1}]} g$.
\end{lemma}
{\sc Proof.} Proving the statement is equivalent to proving that we can
find a $k_8 > 0$ for which
\begin{displaymath}
\log g(\tilde x_n) - \log g(x_n) \leq k_8,
\end{displaymath}
where $\tilde x_n$ maximizes $g$ in $[x_n,x_{n+1}]$. Using again the
Lagrange mean value theorem, the fact that $\tilde x_n-x_n \le (2/k_2)
f(x_n)$ (a consequence of (\ref{estimate1})), and the previous lemma, we
obtain
\begin{eqnarray*}
\log g(\tilde x_n) - \log g(x_n) & = & \frac{g'}{g}(\hat x_n)
(\tilde x_n-x_n) \\
& \leq & \frac{2}{k_2} \left( \frac{f''}{f'} - \frac{f'}{f}
\right) (\hat x_n)\ f(x_n) \\
& \leq & k_9 \left( \frac{f''}{f'} - \frac{f'}{f} \right) (\hat x_n)
\ f(x_{n+1}) \\
& \leq & k_9 \left( \frac{f''f}{f'} - f' \right) (\hat x_n) \leq k_9
(k_1 + \eps),
\end{eqnarray*}
having used (\ref{newA2}) in the last step.
\skippar
We are now prompted to get (\ref{to-violate}) in this case, too.
Looking at (\ref{estimate2}) we can write:
\begin{eqnarray*}
\sum_n g(x_n) (x_{n+1} - x_n) & \ge & k_7 \sum_n
(\max_{[x_n,x_{n+1}]} g) (x_{n+1} - x_n) \\
& \ge & k_7 \int_{x_0}^{+\infty} g(x)dx = +\infty,
\end{eqnarray*}
since $-\int^\infty (f'/f) = -\lim_{x\to\infty} (\log f(x) + const) =
+\infty$. This ends the proof of the theorem.
\section{Discussion}
The obvious news the theorem says, compared to the mentioned condition
$f''>0$, is the possibility for $f'$ to oscillate,
to a certain extent. Dynamically speaking, the change in direction
our particle gets every time it bounces against the upper wall
($\delta_k = -\arctan(f'(x_k))\,$) need not be a monotone sequence. As
a matter of fact, (A2) precisely controls the amount of such an
oscillation. An example will illustrate the case: for $\alpha >1,
\beta >0, c > 1$ define $f'_1(x) := -x^{-\alpha} (\sin(x^{\beta}) + c)
< 0$. This means that we define $f_1(x) := -\int_{x}^{\infty} f'_1
(z)dz$, which makes sense as a convergent integral. Therefore
$f''_1(x) = -\beta x^{-\alpha+\beta-1} \cos(x^{\beta}) +
O(x^{-\alpha-1})$, showing that $f_1$ is not convex. Now, the
asymptotic behavior of $f_1$ and its derivatives is easily extracted
to yield
\begin{displaymath}
\frac{f''_1 f_1}{f'_1}(x) \asymp x^{-\alpha+\beta}.
\end{displaymath}
Thus, (A2) holds if, and only if, $\alpha \ge \beta$, meaning that the
faster $f_1$ vanishes the more violent the oscillation of $f'_1$ is
allowed to be.
\par
Another example may be interesting to present, to show that there are
cases where (A1) holds but (A2) does not. Pick up a $\phi \in
C^{\infty}(\R)$ supported in $]-1/2,1/2[$ with $\int \phi = 1-e^{-1}$.
Call, for $k \in\N$,
\begin{displaymath}
\phi_k(x) := \phi((x-k-1/2)\, e^k),
\end{displaymath}
supported in $]k+1/2-e^{-k}/2,k+1/2+e^{-k}/2[$. Let us now define
$h(x) := \sum_{k=0}^{\infty} \phi_k(x)$. The result is shown in
figure 2a. We see that
\begin{displaymath}
\int_k^{k+1} h(x)dx = \int_k^{k+1} \phi_k(x)dx = (1-e^{-1}) e^{-k}.
\end{displaymath}
Also denote by $H(x) := \int_{x}^{\infty} h(z)dz$. Finally, let us
introduce $f'_2(x) := -e^{-x} -h(x)$, corresponding to $f_2(x) =
e^{-x} + H(x)$. Their graphs are displayed in figure 2b and 2c,
respectively. Certainly $f'_2 \not\to 0$ and (A2) is not verified.
Now, from above we can estimate the value of $f_2$. In fact
$e^{-[x]-1} \leq H(x) \leq e^{-[x]}$ giving $H(x) \leq e^{-x+1}$.
Therefore
\begin{displaymath}
\frac{|f'_2|}{f_2}(x) = \frac{e^{-x}+h(x)}{e^{-x}+H(x)} \geq
\frac{e^{-x}}{e^{-x}+e^{-x+1}} = \frac{1}{1+e} \geq 0.
\end{displaymath}
That is: (A1) holds as well as the result, in this case.
\skippar
Is it difficult to say to what extent our theorem is inclusive of the
general case or how it can be refined. The point here is that finding
a sufficient condition for the non-existence
of an escape orbit is much more direct than
finding a necessary condition. The shape of $f$ can be {\em pathological}
enough but not in a suitable way that allows a trajectory to go directly
to infinity. One thing can be said, though: hypotheses (H1) and (H2) do
not suffice and one needs some extra assumption to control a possible
wild behaviour of $f'$. As a matter of fact we may now sketch the
construction of a billiard table verifying those hypotheses and having one
escape orbit. We will start by first drawing the orbit and then a compatible
$f$.
\par
Consider the polyline shown in figure 3a with $\theta_1 \in ]0,\pi/2[$ and
$\{y_n \}$ any non-integrable sequence such that $y_n \searrow 0$. If this
were an escape orbit then we would have $f(x_n) = y_n,\, f'(x_n) = 0$ and
$\theta_n = \theta_1\ \forall n$. Furthermore $x_{n+1} - x_n =
(y_{n+1} + y_n)/\tan\theta_1$ so that $\lim_{n\to\infty} x_n = \infty$,
because of our assumption on $\{y_n \}$. Of course any $f$ giving rise
to such an orbit cannot satisfy (H2), because of the flat tangent at
the bouncing points, but we can slightly modify
our picture in order to fit it. Take an
integrable sequence $\{\delta_n \},\: \delta_n>0$ such that
$\theta_\infty := \theta_1 + 2 \sum_n \delta_n < \pi/2$.
Now modify the trajectory in figure 3a, ``shrinking''
it in order to have $\theta_n := \theta_1 + 2 \sum_{k=1}^{n-1} \delta_k$;
keep $y_n$ fixed. The result is drawn in figure 3b. This is again an
escape orbit since, due to our choice of $\theta_\infty$, the contraction
of the little triangles has a lower bound, i.e. $x_{n+1} - x_n \geq
(y_{n+1} + y_n)/\tan\theta_\infty$. One can now very easily construct
an $f$ which satisfies (H1) and (H2) and whose graph is an upper wall
for this trajectory.
\par
This proves our remark.
\section{Acknowledgments}
I wish to thank Ya.G.Sinai, C.Liverani, E.Gutkin and N.Chernov for
their kindness in discussing with me about this subject. I also
express my gratitude to A.Parmeggiani for his support, during
the preparation of this paper.
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\newblock {\it Escape Orbits for a Class of Infinite Step Billiards}
\newblock In preparation
\end{thebibliography}
\end{document}
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