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\topmatter
\title
TWO APPLICATIONS OF JACOBI FIELDS \\
TO THE BILLIARD BALL PROBLEM.
\endtitle
\author
Maciej P. Wojtkowski
\endauthor
\address
Maciej P. Wojtkowski,
Department of Mathematics,
University of Arizona,
Tucson, AZ 85721, USA.
\endaddress
\email
maciejw@math.arizona.edu
\endemail
\date
September 13, 93
\enddate
\abstract
\newpage
We present new proofs of two results on the billiard ball
problem by Rychlik \cite {R} and Bialy \cite {B}.
\endabstract
\thanks
We would like to thank Gil Bor, Jian Cheng, Victor Donnay
and Lenny Friedlander
for helpful and enlightening discussions. The suggestions of
the referee are also gratefully acknowledged.
\endthanks
\endtopmatter
\vskip.7cm
\subhead \S 0. INTRODUCTION \endsubhead
\vskip.4cm
We will give new proofs of two results on the billiard ball
problem by Rychlik \cite {R} and Bialy \cite {B}. The original proofs
were based on variational considerations. In our approach the variational
context is absent, the dynamical system takes the center stage.
We hope the simplifications provided by our method will make
possible some progress on the conjectures for which these results lend
partial support.
\vskip.7cm
\subhead \S 1.THE DYNAMICAL SYSTEM \endsubhead
\vskip.4cm
Let us consider a convex domain $Q$ in the plane. The billiard ball
system is the flow $\Phi^t$ on $Q\times\Bbb S^1$ defined by the free motion
of a point particle in $Q$, with elastic reflections at the boundary
$\partial Q$ (the angle of reflection equal to the angle of
incidence). The circle $\Bbb S^1$ represents unit velocities.
Strictly speaking, we need to identify the velocities at the boundary
according to the collision law. The flow $\Phi^t$ preserves the Liouville
measure $\nu$ equal to the product of the Lebesgue measures in $Q$
and $\Bbb S^1$.
Birkhoff \cite{Bir}
thought that this dynamical system is a very good model for
Hamiltonian dynamics. In the last thirty years his belief proved to be
strikingly accurate. We understand as much about the low dimensional
Hamiltonian dynamics, as we know about the billiard system.
The flow $\Phi^t$ has a natural section map $T : \Cal M \to \Cal M$,
where $\Cal M = \partial Q \times [0,\,\pi]$. The map $T$ describes
the dynamics ``from collision to collision''. The space $\Cal M$ is the set
of unit tangent vectors attached at the boundary and pointing inwards.
It can be coordinatized by $(s,\,\varphi)$, where $s$ is the arclength
parameter taken modulo $p$, the perimeter of $\partial Q$, and
$\varphi \in [0,\,\pi]$ is the angle that the unit velocity vector
makes with the tangent vector to the boundary.
The map $T$, the billiard ball map, preserves the measure $\mu = \sin
\varphi ds d\varphi$. This measure is obtained from $\nu$ by the
contraction (internal product) with the velocity vector field of the
flow $\Phi^t$. It follows that
$$
\int_{\Cal M} \tau(x) d\mu(x) = \nu (Q\times\Bbb S^1) = Area (Q) 2\pi,
\tag 1
$$
where $\tau(x)$ for $x\in \Cal M$ is the time to the next collision.
This formula is also well known in integral geometry (cf. \cite {S},
formula (3.6)).
\vskip.7cm
\subhead \S 2. JACOBI FIELDS \endsubhead
\vskip.4cm
Let us consider a family of billiard orbits (or straight lines
parametrized by the arc length)
$$
p(\epsilon) = p(\epsilon,\,t) = \gamma(\epsilon) + v(\epsilon)t,\ \
|\epsilon| < \epsilon_0,\ -\infty < t < +\infty.
$$
The Jacobi field $J(t)$ is defined as
$$
J(t) = \frac {\partial p}{\partial\epsilon}\vert_{\epsilon =0}.
\tag 2
$$
The Jacobi field is the infinitesimal description of the family of
billiard orbits (lines) around the orbit $\epsilon =0$
$$
J(t) = \gamma'(0) + v'(0)t.
$$
The Jacobi fields satisfy the following differential equation
(the curvature of the plane is equal to zero)
$$
\frac {d^2}{dt^2}J = 0.
$$
At any point in the phase space the Jacobi fields form the tangent
subspace. They can be identified with pairs of vectors $(J(0),\,J'(0))
= (\gamma'(0),\, v'(0))$. Note that the prime denotes differentiation
with respect to $t$ or $\epsilon$, whichever applies. It is natural to
consider only the Jacobi fields for which $J(0) = \gamma'(0)$ is
orthogonal to the velocity vector $v(0)$. Note that $J'= v'(0)$ is
automatically orthogonal to $v(0)$. We will call such Jacobi fields
transversal. One can also think about the (two dimensional) space of
transversal Jacobi fields as the factor space by the (one dimensional)
subspace of Jacobi fields corresponding to variations of the initial
point along the orbit (time variations).
Transversal Jacobi fields can be considered as tangent
vectors to the section $\Cal M$. The connection is given by the
formulas
$$
\aligned
J = & \sin \varphi ds,\\
J'= & - k ds - d\varphi,
\endaligned
\tag {3}
$$
where $(J,\,J') \in \Bbb R^2$ are now linear coordinates after the appropriate
choice of a unit vector orthogonal to $v(0)$ was made, and $k= k(s)$ is the
curvature of the boundary $\partial Q$.
The evolution of $(J,\,J')$ between collisions is given by the matrix
$$
\left(\matrix
1 & \tau \\
0 & 1
\endmatrix\right).
\tag {4}
$$
The Jacobi fields can be naturally extended beyond
reflections in the boundary: with every family of lines including
a segment of a billiard orbit we can associate the family of reflected
lines which include the segment of the reflected orbit. The reflected
Jacobi field is now calculated by \thetag{2}. A
transversal Jacobi field $(J,\,J')$ is transformed at a reflection
by the linear map
$$
\left(\matrix
-1 & 0 \\
\frac 2{d} & -1
\endmatrix\right),\ \ \text{ where } \ \ d = \frac {\sin \varphi} {k(s)}.
\tag {5}
$$
\vskip.7cm
\subhead \S 3.THE THEOREM OF BIALY \endsubhead
\vskip.4cm
In the recent paper \cite {B} Bialy proved the following theorem
\proclaim
{Theorem 1(\cite{B}))}
If the phase space $\Cal M$ of the billiard ball map is foliated by
continuous closed invariant curves not null-homotopic in $\Cal M$,
then $Q$ is a disk.
\endproclaim
It is well known (\cite{Bir})that the billiard ball map in an ellipse is
integrable. No other examples of convex domains with integrable
billiard ball maps are known. It makes the following conjecture,
attributed to Birkhoff, plausible.
\proclaim{ Conjecture}
If a neighborhood of the boundary of $\Cal M$ is foliated by
continuous closed invariant curves not null-homotopic in $\Cal M$, then
$Q$ is an ellipse.
\endproclaim
We will first present the proof of a weaker result then Bialy's
Theorem, to make the line of thought perfectly clear. We will show
then that the same argument does apply in the general case.
\proclaim{Definition} A smooth closed curve contained
in the domain $Q$ is called a smooth caustic, if every billiard orbit with
one segment tangent to the curve has all its segments tangent to the
curve.
\endproclaim
The interior of a disk with the center removed is foliated by
smooth caustics, the concentric circles.
\proclaim {Theorem 2}
If the domain $Q$
is foliated by smooth caustics in such a way that almost every orbit
is tangent to a caustic, then $Q$ is a disk.
\endproclaim
\demo{Proof}
Let us fix $x\in M$ and the corresponding billiard orbit tangent to a
smooth caustic.
We denote by $l(x)$ the distance from the boundary to the
nearest point of tangency in the past with the smooth caustic. The
nearest point of tangency in the future is then equal to
$\tau(x) - l(Tx)$ (see Fig.1).
\topinsert
\vskip 4in
\hsize=4.5in
\raggedright
\noindent{\bf Figure 1} Billiard table and caustic.
\endinsert
The transversal Jacobi field defined
by the nearby
orbits tangent to the same smooth caustic vanishes at the points of
tangency. We get
from \thetag {3} and \thetag {4}
$$
\left(\matrix
1 & \tau(x)-l(Tx) \\
0 & 1
\endmatrix\right)
\left(\matrix
-1 & 0 \\
\frac 2{d(x)} & -1
\endmatrix\right)
\left(\matrix
1 & l(x) \\
0 & 1
\endmatrix\right)
\left(\matrix
0 \\
1
\endmatrix\right) =
\left(\matrix
0 \\
*
\endmatrix\right),
$$
where the $*$ indicates that the respective component is not
restricted. We obtain immediately
$$
d(x) = \frac {2l(x)(\tau(x)- l(Tx))}{l(x)+(\tau(x)- l(Tx))}.
\tag 6
$$
The last formula is the classical ``mirror equation'' of the
geometric optics.
Since the harmonic mean of two positive numbers
does not exceed the arithmetic mean we
conclude that
$$
2d(x) \leq \tau(x)+ l(x) - l(Tx).
\tag 7
$$
Let us stress that to get \thetag 7 from \thetag 6, it is necessary to have
$$
l(x) > 0 \ \ \ \text{and} \ \ \ l(Tx) < \tau(x),
$$
(note that $d(x) > 0$).
We will now obtain the desired conclusion by integrating \thetag {7}
and comparing with the Isoperimetric Inequality, just as it was done
in \cite{B}. Integrating the right hand side of the inequality we
obtain from \thetag {1}
$$
\int_{\Cal M}\tau(x)+ l(x) - l(Tx) d\mu(x) =
\int_{\Cal M}\tau(x)d\mu(x) = 2\pi Area(Q).
$$
We have used the fact that $l(x)$ is bounded and defined almost
everywhere in $\Cal M$. The integration of the left hand side yields
$$
\int_{\Cal M}2 d(x) d\mu(x) =
\int_0^p\int_0^{\pi}\frac{2\sin^2\varphi}{k(s)} ds d\varphi =
\pi\int_0^p\frac1{k(s)} ds.
$$
By the Buniakovski - Cauchy - Schwartz inequality
$$
\int_0^p\frac1{k(s)} ds\int_0^p{k(s)} ds \geq p^2,
$$
with the equality only for the constant curvature, i.e. for the disk.
Since
$$
\int_0^p{k(s)} ds = 2\pi
\tag 8
$$ we conclude that
$$
2\pi Area(Q) \geq \frac {p^2}{2},
$$
which violates the Isoperimetric Inequality except when it is the
equality, i.e., when $Q$ is a disk.
\qed
\enddemo
Let us consider a measurable one dimensional subbundle $L$ of the tangent
bundle of $\Cal M$, i.e., for almost every $x\in \Cal M$ we have a one
dimensional subspace (a line) $L(x) \subset \Cal T_x\Cal M$ which depends
measurably on $x$. We assume that for almost every $x \in \Cal M$ the
subspace $L(x)$ is not vertical, i.e., it is not tangent to the
curve $\{s=const\}$. We choose an orientation of nonvertical
lines by the condition $ds > 0$. Let us call such an oriented
subbundle a monotone subbundle of the tangent bundle of $\Cal M$.
We call a monotone subbundle invariant ,if it is preserved, including
the orientation, under the
action of the derivative of $T$.
Let us recall that vectors tangent to $\Cal M$ can be naturally identified
with transversal Jacobi fields. If we have an invariant monotone
subbundle $L$, then Jacobi fields which correspond to nonzero tangent
vectors in the subbundle must vanish inside the billiard table, almost
everywhere in $\Cal M$. Indeed let $(J(x),\,J'(x)),\ x \in \Cal M$ be a
measurable family of nonzero Jacobi fields spanning the oriented
subbundle $L(x),\ x \in \Cal M$. We have $ J(x) > 0$ almost everywhere in
$\Cal M$ and there is a positive measurable function
$\alpha : \Cal M \to \Bbb R^+$ such that
$$
\left(\matrix
-1 & 0 \\
\frac 2{d(Tx)} & -1
\endmatrix\right)
\left(\matrix
1 & \tau(x) \\
0 & 1
\endmatrix\right)
\left(\matrix
J(x) \\
J'(x)
\endmatrix\right) =
\alpha(x)\left(\matrix
J(Tx) \\
J'(Tx)
\endmatrix\right).
$$
It follows that $J(x) + \tau(x)J'(x) = - \alpha(x)J(Tx)$ is almost
everywhere negative and hence almost everywhere the Jacobi field
vanishes inside the billiard table, i.e., there is a measurable
function $f(x),\, 0 < f(x) <
\tau(x)$ such that
$$
J(x) + f(x)J'(x) = 0.
$$
Let us now consider the measurable function
$$
l(x) = \tau(T^{-1}x) -
f(T^{-1}x).
$$
We have $ 0 < l(x) < \tau(T^{-1}x)$ almost everywhere. Moreover,
as a consequence of the invariance of the subbundle, the
function $l(x)$ satisfies \thetag 6.
Hence we can repeat the proof of Theorem 2. We obtain the following
\proclaim {Theorem 3} If the billiard ball map $T$ has an invariant
monotone subbundle of the tangent bundle of $\Cal M$,
then the billiard table $Q$ must be a disk.
\endproclaim
Let us show how Bialy's Theorem 1 follows from Theorem 3.
The mapping $T: \Cal M \to \Cal M$ is a measure preserving twist map.
By the Birkhoff Theorem \cite{H}, any closed invariant curve not
null-homotopic in $\Cal M$ is the graph of a Lipschitz function
$\varphi = \varphi (s)$. Although a general Lipschitz function may be
nondifferentiable at some points, it must have finite upper and lower
derivatives
$$
\frac {d^{+}\varphi}{ds} = \limsup_{s_1,s_2 \to s }
\frac {\varphi(s_2) - \varphi(s_1)}{s_2 - s_1},\ \
\frac {d^{-}\varphi}{ds} = \liminf_{s_1,s_2 \to s }
\frac {\varphi(s_2) - \varphi(s_1)}{s_2 - s_1}.
$$
The upper and lower derivatives define respectively the upper and
lower tangent lines to the
graph. The bundle of upper (or lower)
tangent lines to the invariant curves is
defined everywhere and is clearly measurable and monotone (never
vertical). The Jacobi field which spans this oriented tangent line
can be chosen as (cf. \thetag 3) $ J = \sin \varphi(s),\, J' = -k(s) -
\frac {d^{+}\varphi}{ds}$. In turn, the positive function $l(x)$
(``the distance to the caustic'') is given by the formula
$$
l(s,\,\varphi) = \frac {\sin \varphi}{k(s) - \frac {d^{\pm}\varphi}{ds}}.
$$
It follows from the invariance of the curves that this monotone
subbundle is invariant. Indeed, the positive orientation of the curves
is preserved under $T$.
To summarize, the tangents to the invariant curves of the family, such as
described in the Bialy's Theorem 1, give us an invariant monotone
subbundle. So Theorem 3 implies Theorem 1.
\proclaim {Remark}
\endproclaim
Theorem 3 has an interesting connection to billiard tables with
hyperbolic billiard ball map. The stable and unstable subbundles are
invariant but by Theorem 3 they cannot be invariant as monotone
subbundles. Indeed, in the case of Bunimovich's stadium and its
generalizations by Donnay, \cite{D}, the positive orientation of
the (un)stable subbundle is reversed at some of the segments of the orbit
between the two flat pieces of the boundary (connecting the strictly
convex pieces).
In the case of billiard tables bounded by closed
convex scattering curves, introduced in \cite{W},
the situation is different. These are closed curves given by the natural
equation $k = k(s)$, where $k(s) > 0$ is a smooth function in the interval
$[0,\,p]$ such that
$$
\frac {d^2}{ds^2}\frac 1{k(s)} \leq 0
$$
with strict inequality holding almost everywhere in $s$. Hence the
closed curve is smooth except at one point. The examples include
perturbations of the circle and the cardioid.
For such billiard tables we have hyperbolicity in all of the phase
space $\Cal M$ and the positive orientation of
the (un)stable subbundle is preserved (cf.\cite{W}), so we do have two
invariant monotone subbundles. The apparent contradiction with
Theorem 3 is resolved
by the failure of the formula \thetag 8. Because of the singularity,
we have $\int_0^p{k(s)} ds > 2\pi$.
When we perturb the circle to a closed convex scattering curve, the
monotone invariant subbundle of the billiard ball map in the disk
splits into two subbundles (the stable and unstable ones). The
proof of Theorem 2 shows that it is only
the singularity of the boundary that provides room for such a splitting.
\vskip.7cm
\subhead \S 4. THE THEOREM OF RYCHLIK \endsubhead
\vskip.4cm
The problem of the measure of periodic orbits for the billiard ball
map arises in the spectral geometry \cite{M-M}. A version of Kupka-Smale
Theorem for billiard ball maps in convex domains was proven by Lazutkin
\cite {L}. It follows that for generic
convex domains the periodic orbits of bounded period are isolated,
and hence there are only countably many periodic orbits.
It is fairly clear that for convex domains with real analytic boundary
the measure of periodic orbits must be zero. At the same time it seems
plausible that by modifying the boundary in the $C^\infty$ category
one could produce a whole neighborhood filled with periodic orbits.
Rychlik proved in \cite{R} that it is not possible for orbits of period $3$.
\proclaim{Theorem \cite {R}}
The set of periodic orbits of period $3$ of the billiard ball map
is nowhere dense.
\endproclaim
Rychlik shows that actually the set must have measure zero.
His proof is based on the study of the length functional, and it
required computer assistance. Recently Stojanov \cite {St} simplified
the calculations, so that there is no need for the use of symbolic
computation systems. We propose a more dynamical proof.
\demo{Proof}
Let $\{x_0,\,x_1,\,x_2\}$ be a periodic orbit of period $3$ with a
neighborhood filled by periodic orbits of period 3.
\topinsert
\vskip 4in
\hsize=4.5in
\raggedright
\noindent{\bf Figure 2} Periodic orbit with period 3.
\endinsert
We have that $T^3$ and its derivative $DT^3$ is equal to identity in
the neighborhood of $x_0$. The last property can be formulated in the
language of Jacobi fields. In particular we get from \thetag {4} and
\thetag {5}, see Fig. 2,
$$
\aligned
&\left(\matrix
1 & \tau(x_0) \\
0 & 1
\endmatrix\right)
\left(\matrix
-1 & 0 \\
\frac 2{d(x_0)} & -1
\endmatrix\right)
\left(\matrix
1 & \tau(x_2) \\
0 & 1
\endmatrix\right)
\left(\matrix
0 \\
1
\endmatrix\right) =\\
&\left(\matrix
-1 & 0 \\
-\frac 2{d(x_1)} & -1
\endmatrix\right)
\left(\matrix
1 & -\tau(x_1) \\
0 & 1
\endmatrix\right)
\left(\matrix
-1 & 0 \\
-\frac 2{d(x_2)} & -1
\endmatrix\right)
\left(\matrix
0 \\
1
\endmatrix\right).
\endaligned
$$
The equality of the first components of the two vectors yields
$$
\tau(x_0) + \tau(x_2) - \tau(x_1) = \frac
{2\tau(x_0)\tau(x_2)}{d(x_0)}.
\tag 8
$$
The total length of the periodic orbits in our neighborhood must be
constant. Let us denote it by $L$,
$$
L = \tau(x_0) + \tau(x_1) + \tau(x_2).
$$
>From the Cosine Formula we get
$$
\tau(x_0) + \tau(x_2) - \tau(x_1) = \frac
{4\tau(x_0)\tau(x_2)\sin^2\varphi_0}{L}
\tag 9
$$
where $x_0 = (s_0,\,\varphi_0)$. Combining \thetag {8} and \thetag {9}
we obtain
$$
k(s_0) = \frac {2\sin^3\varphi_0}{L}.
\tag 10
$$
Such a relation has to hold also for all nearby orbits. In particular for
all nearby orbits starting at the same point of the boundary with different
velocity vectors, i.e., at $x(\varphi) = (s_0,\, \varphi)$ with
$\varphi$ close to $\varphi_0$. But this is impossible because the
function on the right hand side of \thetag {10} is not constant in any
interval.
\qed
\enddemo
In the same fashion as it was done in \cite {R}, one can now argue
that the set of periodic orbits of period $3$ must have measure zero.
Indeed, let us assume that the set of periodic points of period $3$
has positive measure. If $x_0$ is the Lebesgue density point of the
set of periodic points with period $3$, then although $T^3$ may be
different from identity in the neighborhood of $x_0$, $DT^3$ must be equal
to identity at $x_0$. Hence the formulas \thetag {8} and \thetag {10}
must hold for such a point. We obtain that for any Lebesgue
density point $x= (s,\,\varphi)$
$$
L(s,\,\varphi)k(s) - 2\sin^3\varphi = 0,
\tag 11
$$
where $L(s,\,\varphi)$ is defined as the perimeter of the triangle
with the vertices at the foot points of $T^{-1}x,x$ and $Tx$. Every
periodic point with period $3$ is the critical point of the function
$L(s,\,\varphi)$ defined in this way.
It follows from \thetag {11} that at every Lebesgue density point of
the set of periodic points of period $3$ we must have also
$$
\frac {\partial}{\partial \varphi}\left(L(s,\,\varphi)k(s) -
2\sin^3\varphi\right) = 0.
\tag 12
$$
In view of the criticality of $L$ at the periodic points, the partial
derivative \thetag {12} is zero only if $\varphi$ assumes values
$0,\frac \pi 2, \pi$. But this is impossible for an orbit of period
$3$. Hence the set of density points is empty, contrary to our
assumption.
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\end