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\topmatter
\title
THE SYSTEM OF TWO SPINNING\\
DISKS IN THE TORUS.
\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
November 4, 1993
\enddate
\abstract
\newpage
We study the system of two spinning disks in the torus. We show how to
reduce this nonhamiltonian system to a 3 dimensional measure
preserving reversible map. We establish that, in contrast to the
case of elastic collisions, this system may have periodic orbits with
all Floquet exponents on the unit circle.
\endabstract
\thanks
We would like to thank Eugene Gutkin and Dave Levermore
for helpful and enlightening discussions. The suggestions of
the referee are also gratefully acknowledged.
\endthanks
\endtopmatter
\par\newpage
\vskip.7cm
\subhead \S 0.Introduction \endsubhead
\vskip.4cm
The Boltzman-Sinai system of hard balls, or two dimensional disks,
moving in a torus and colliding elastically is nonuniformly hyperbolic
(\cite {S}, also \cite {S-Ch} and \cite {W1}). In particular, the
measure of the set of quasiperiodic motions is zero.
The elastic collision in such a system ignores the rotational degree
of freedom of the balls. Physically it describes the case of massive
particles with finite size but all the mass concentrated at the
center. Broomhead and Gutkin in a recent paper \cite {B-G} proposed
to study the dynamics of spinning disks. They introduced the collision
interaction which takes into account the rotation of the disks. It is
based on the assumptions of no-slip friction and energy preservation.
In the present paper we consider the system of two spinning disks in
the two dimensional torus. In the case of elastic collisions Sinai
\cite {S}, using the natural first integrals, reduced this Hamiltonian
system to the billiard system in the torus with a circular obstacle
(see also \cite {S-W}). The system of spinning disks is not
Hamiltonian but it can also be reduced to the system of a point
particle moving in the torus with a circular obstacle. The point
particle has additionally an intrinsic (angular)
velocity which affects the linear velocity after a collision with
the obstacle. The system has a natural invariant measure
and is reversible (in the sense that forward and backward dynamics are
conjugate by an involution).
We establish that in our system Floquet exponents for some periodic
orbits lie on the unit circle, if the disks and their moments of
inertia are sufficiently large. Thus in the reduced system we have
linearly stable orbits. As in the case of Hamiltonian systems the
linear stability does not prevent the actual orbits starting in the
neighborhood of our periodic solution from wandering away.
In Hamiltonian systems the information about the behavior of orbits in
the vicinity of an elliptic periodic orbit is furnished by the KAM theory.
Sevryuk \cite {Sev} developed systematically the KAM theory for
reversible systems. His results (Theorem 2.9) could be
applied to our case , but the calculation
of higher order terms and the checking of the nondegeneracy conditions
is cumbersome, and we did not do it. Taking for granted these
conditions, we would arrive at the conclusion that the union of
quasiperiodic orbits in the vicinity of our periodic orbit has
positive Lebesgue measure and thus our system is not ergodic.
The plan of the paper is as follows. In Section 1 we define the
no-slip collision of spinning disks.
In Section 2 we reduce the system of two
spinning disks in the torus to a four dimensional flow. We also
describe the simple periodic orbits in the system. In Section 3 we
calculate when the Floquet exponents for these periodic orbits lie on
the unit circle. We end with remarks about the impossibility of full
hyperbolicity in the system of two spinning disks.
\vskip.7cm
\subhead \S 1. Collisions of spinning disks \endsubhead
\vskip.4cm
Let us consider the system of two nonoverlapping disks in the plane
$\Bbb R^2$ or the two dimensional torus $\Bbb T^2 = \Bbb R^2/\Bbb Z^2$. Let
$q_i\in \Bbb R^2$ be the positions of the centers of the disks and
$v_i \in \Bbb R^2$ their linear velocities, $i=1,2$. By
$\omega_i,\ i=1,2,$ we denote the angular velocities of the
disks with respect to the centers. We assume that the two particles
are identical, in particular they have the same radius $r$, the same
mass $m$, and their moments of inertia $I$ are the same.
We require that the center of mass of the disk coincides with its center,
but we do not restrict the distribution of mass in the disk, so that
the moment of inertia $I$ may assume any value between $0$ (the mass
concentrated in the center) and $mr^2$ (the mass concentrated at the
perimeter). The uniform distribution of mass gives $I = \frac 14
mr^2$.
Between collisions the linear velocities of the centers of
the disks and their angular velocities stay constant.
The total energy of the system is equal to
$$
E = \frac 12m (v_1^2 + v_2^2) + \frac 12I (\omega_1^2+ \omega_2^2).
$$
At the collision of the disks, i.e., when
$$
q_2-q_1 = 2r e \ \ \text{ where } e = \frac{q_2-q_1}{\|q_2-q_1\|},
$$
(or simply $\|q_2-q_1\| = 2r$),
the velocities of the disks $(v_1,v_2,\omega_1,\omega_2)$ experience
instantenous change -- the collision interaction. Let $w \in \Bbb R^6$
denote the joint velocity vector of the two disks, $w =
(v_1,v_2,\omega_1,\omega_2)$. By $w^- =
(v_1^-,v_2^-,\omega_1^-,\omega_2^-)$ we denote the velocities
immediately before the collision and by $w^+ =
(v_1^+,v_2^+,\omega_1^+,\omega_2^+)$ the velocities immediately after
the collision.
The collision interaction of spinning disks was
recently studied by Broomhead and Gutkin in \cite {B-G}; also
Kozlov and Treschev discuss in \cite{K-T} the limit case of the
collision with a wall. We do not give here the derivation of the collision
rules from basic principles of mechanics. The reader is referred to
\cite{B-G} for such a derivation. Instead we list the many natural
properties which this interaction possess and finally present the formula.
\roster
\item
{\bf Linearity. }
There is a linear map $\Lambda$ independent of
the velocities such that
$$
w^+ = \Lambda w^-.
$$
\item
{\bf Reversibility.}
If $w^+ = \Lambda w^-$ then $ -w^- = \Lambda(- w^+) $. It is
equivalent to
$$
\Lambda^2 = Id = \text {identity matrix}.
$$
\item
{\bf Energy preservation.}
$$
E(w^+) = E(w^-).
$$
It is equivalent to the orthogonality of the matrix $\Lambda$ with
respect to the scalar product defined by the quadratic form $E$.
\item
{\bf Preservation of the linear momentum $U = mv_1 + mv_2$.}
$$
v_1^+ + v_2^+ = v_1^- + v_2^-.
$$
\item
{\bf Normal and tangential components interact separately.}
Let us consider the orthonormal frame $(e,f)$ at the point of contact.
We define as before
$$
e =\frac {q_2-q_1}{\|q_2-q_1\|}
$$
and $f$ is orthogonal to $e$ and such that $(e,f)$ has positive
(counterclockwise) orientation. We will call the $e$-components of
linear velocities normal and the $f$-components of the velocities
tangential. Denoting by $\langle\cdot,\,\cdot\rangle$ the standard
scalar product in tangent planes of $\Bbb T^2$, we have that, if $
\langle v_i^- ,\,e\rangle = 0,$ then $\langle v_i^+ ,\,e\rangle = 0, i
=1,2$ and if $ \langle v_i^- ,\,f\rangle = 0,$ then $\langle v_i^+
,\,f\rangle = 0, i =1,2$.
\item
{\bf Preservation of the angular momenta $A = \pm mr\langle
v,\,f\rangle + I\omega $.}
The angular momenta of the disks, $A_1 =-mr\langle v_1,\,f\rangle + I
\omega_1$ and $A_2 =mr\langle v_2,\,f\rangle + I \omega_2$,
calculated with respect to
the point of contact at the time of the collision are preserved. We
have
$$
\aligned
-mr\langle v_1^+,\,f\rangle + I \omega_1^+
&=- mr\langle v_1^-,\,f\rangle + I \omega_1^- \ \ \text{ and}\\
mr\langle v_2^+ ,\,f\rangle + I \omega_2^+
&= mr\langle v_2^- ,\,f\rangle + I \omega_2^-.
\endaligned
$$
Let us stress that the angular momentum is only a ``local'' integral;
it cannot be properly defined for the system of disks moving in a
torus.
\item
{\bf Friction at the point of contact}
If tangential components of the
linear velocities of the points in contact at the collision coincide,
then they are not changed (because there is no friction to change
them). More precisely, if $$
\langle v_1^-,\,f\rangle + r\omega_1^- = \langle v_2^-,\,f\rangle -
r\omega_2^-,
$$
then
$$
\langle v_i^+ - v_i^-,\,f\rangle = 0,\ \ \text { and }\ \
\omega_i^+ = \omega_i^-, i = 1,2.
$$
\item{\bf Exchange of velocities}
The disks exchange the linear velocities of their points in contact at
the collision, i.e.,
$$
v_1^+ + r\omega_1^+ f = v_2^- - r\omega_2^- f,
$$
and
$$
v_2^+ - r\omega_2^+ f = v_1^- + r\omega_1^- f.
$$
\endroster
Expanding the linear velocity vectors $v_1$ and $v_2$ in the
orthonormal basis $(e,\,f)$ let us consider the vector
$\left(\langle v_1,\,f\rangle,\,\langle v_2,\,f\rangle,\omega_1,\omega_2\right)
\in \Bbb R^4$. The operator $\Lambda$ projected to this $4$
dimensional subspace has the following form
$$
\Lambda = \left(
\matrix
\frac{1}{1+\xi ^2} & \frac{\xi ^2}{1+\xi ^2}&
\frac{-r\xi ^2}{1+\xi ^2} &\frac{-r\xi ^2}{1+\xi ^2} \\
\frac{\xi ^2}{1+\xi ^2} & \frac{1}{1+\xi ^2}&
\frac{r\xi ^2}{1+\xi ^2} &\frac{r\xi ^2}{1+\xi ^2} \\
\frac{-1}{r(1+\xi ^2)} & \frac{1}{r(1+\xi ^2)}&
\frac{\xi ^2}{1+\xi ^2} &\frac{-1}{1+\xi ^2} \\
\frac{-1}{r(1+\xi ^2)} & \frac{1}{r(1+\xi ^2)}&
\frac{-1}{1+\xi ^2} &\frac{\xi ^2}{1+\xi ^2}
\endmatrix\right)
$$
where $\xi = \sqrt{\frac I {mr^2}} \leq 1$. The description of
$\Lambda$ is completed by its action on the $2$-dimensional subspace
of vectors
$\left( \langle v_1,\,e\rangle,\, \langle v_2,\,e\rangle \right)$
$$
\Lambda = \left( \matrix
0 & 1 \\
1 & 0
\endmatrix\right).
$$
The Boltzman-Sinai system of hard disks is obtained in the limit as
$\xi$ goes to $0$ as a factor system. It occurs when the mass is
concentrated at the center. In this limit the tangential
components of the linear velocities are not affected by the angular
velocities. (It is interesting that in this limit the angular
velocities are changed depending on the linear velocities, but they
give no contribution to the energy.)
It is a consequence of the preservation of the linear and angular
momenta in the collision that
$\omega_1-\omega_2$ is preserved, i.e.,
$$
\omega_1^+-\omega_2^+ = \omega_1^--\omega_2^-.
$$
The operator $\Lambda$ has eigenvalues $1$ with multiplicity $4$ and
$-1$ with multiplicity $2$. The appearance of the double eigenvalue $-1$
is responsible for the {\bf non}preservation of the symplectic
structure.
\vskip.7cm
\subhead \S 2. Reduction of the system \endsubhead
\vskip.4cm
In the case of the elastic collisions Sinai \cite{S} (see also
\cite{S-W}) reduced the system of two disks to the billiard system in
the torus with a circular obstacle. It turns out that a similar
reduction is possible also in our case, notwithstanding the
nonhamiltonian character of the system.
We introduce the relative position of the disks $q = q_2 - q_1$ which
can be considered as a point in
$$
\Bbb T^2_r = \{q\in \Bbb T^2\ | \
\|q+k\| \geq 2r \ \ \text{for any integer vector}\ \ k\}.
$$
The relative velocity is $u = v_2 - v_1$. We have observed that
$\Omega = \omega_1 - \omega_2$ is the first integral in our system.
We introduce the following variable $z = r\xi(\omega_1+\omega_2)$.
We have
$$
E = \frac 1{4m} U^2 + \frac I{4} \Omega^2 +
\frac m4\left(u^2 + z^2\right).
$$
We can see that $u^2 + z^2$ has constant value. Between collisions
both $u$ and $z$ are constant. It turns out that at the collision
the relative velocities $u$ and $z$ change independently of the other
variables. Hence we obtain the following factor system. A point
particle moves in $\Bbb T^2_r$ with a constant velocity $u$. It also
carries an intrinsic scalar velocity $z$. When the point particle
reaches the circular obstacle both of these velocities are
instantaneously changed in the following way. Let $e$ be the unit
normal vector to the obstacle pointing outward, and let $f$ be the
unit vector orthogonal to $e$ (i.e., tangent to the obstacle) and
such that the frame $(e,\,f)$ is positively oriented.
Let further
$$
\aligned
u_n& = \langle u,\,e \rangle,\\
u_t& = \langle u,\,f \rangle
\endaligned
$$
be the normal and tangential components of the velocity $u$
respectively. At the collision the normal component is reversed
($u_n^+ = - u_n^-$) and
the tangential component interacts with the intrinsic velocity $z$ by
the formula
$$
\left(\matrix
u_t^+\\
z^+ \endmatrix \right) =
\left(\matrix
\frac{1 - \xi^2 }{1 + \xi^2 }& \frac{2\xi}{1 + \xi^2 } \\
\frac{2\xi}{1 + \xi^2 }& \frac{-1 + \xi^2}{1 + \xi^2}
\endmatrix \right)
\left(\matrix
u_t^-\\
z^- \endmatrix \right).
$$
Putting $\xi = \tan \alpha , \ 0 \leq \alpha \leq \frac \pi 2 $ we get
$$
\left(\matrix
u_t^+\\
z^+ \endmatrix \right) =
\left(\matrix
\cos 2\alpha& \sin 2\alpha \\
\sin 2\alpha&-\cos 2\alpha&
\endmatrix \right)
\left(\matrix
u_t^-\\
z^- \endmatrix \right).
$$
In this way we obtain the flow $\Phi^t:\Bbb T^2_r \times \Bbb S^2
\to \Bbb T^2_r \times \Bbb S^2$,$t\in \Bbb R$,
where $\Bbb S^2 = \{\ (u,\,z) \in \Bbb
R^2 \times \Bbb R\ | \ u^2 + z^2 =1 \ \}$ is the unit sphere of
velocities. At the
boundaries of $\Bbb T^2_r$ we identify the incoming velocities and the
outgoing velocities according to the reflection law above.
The flow preserves the measure $\mu$ equal to the product of the
standard Lebesgue measures in $\Bbb T^2_r$ and $\Bbb S^2$.
Moreover it
is a reversible system in the sense that, if we consider the map
$ R: \Bbb T^2_r \times \Bbb S^2 \to \Bbb T^2_r \times \Bbb S^2 $
$$
R(x,\,u,\,z) = (x,\,-u,\,-z),
$$
then $R^2$ is the identity map and
$\Phi^t\circ R = R \circ \Phi^{-t}$ for all times $t$.
This flow has many periodic orbits. Indeed, let us consider a segment
in $\Bbb T^2_r$ with both endpoints on the boundary of the obstacle
and perpendicular to the boundary, see Fig.1. This segment carries a
periodic orbit of our flow in the sense that, if the velocity $u$ is
parallel to the segment and $z=0$ then we have a periodic orbit. Also
every nearby parallel segment carries a periodic orbit when we choose
$z$ so that the velocity $u$ is reversed at the collisions.
\topinsert
\vskip 2.5in
\hsize=4.5in
\raggedright
\noindent{\bf Figure 1.} The periodic orbit.
\endinsert
\vskip.7cm
\subhead \S 3. Linear stability of the periodic orbit \endsubhead
\vskip.4cm
Let us to introduce a section of the flow. As it is done in the case
of the Sinai billiards, we consider the dynamics from collision to
collision. Our reduced phase space is the solid torus $\Cal M = \Bbb
S^1 \times \Bbb S^2_+$, where $\Bbb S^1$ is the boundary of the
obstacle and $\Bbb S^2_+$ is the semisphere of velocities with the
$u$-component pointing outwards. We obtain the map $\Psi : \Cal M \to
\Cal M$ which is a local diffeomorphism with discontinuities produced
by the orbits tangent to the obstacle. We introduce coordinates
$(s,\,u_t,\,z)$ into $\Cal M$, where $s$ is the arc length along the
boundary of the obstacle and $u_t$ is the tangential component of the
velocity $u$, so that $u_t^2 + z^2 \leq 1$. The measure $ds\wedge
du_t\wedge dz $ is invariant under the map $\Psi$. We introduce
the involution
$\Upsilon : \Cal M \to \Cal M$,
$\Upsilon(s,\,u_t,\,z) = (s,\,-\cos 2\alpha u_t - \sin 2\alpha z,\,
-\sin 2\alpha u_t + \cos 2\alpha z)$. The map $\Psi$ is reversible
with respect to this involution, i.e., $ \Upsilon \circ \Psi=
\Psi^{-1} \circ \Upsilon.$
In the calculation of the derivative of the Poincar\'e map for our
periodic orbit we will use the ideas developed for other ``flows with
collisions'' \cite {W2}. It is also advantageous to extend the
dynamics to the large ambient space $\Bbb T^2 \times \Bbb R^2 \times
\Bbb R$. Let
$$
\widetilde{\Phi}^t: \Bbb T^2 \times \Bbb R^2 \times \Bbb R \to
\Bbb T^2 \times \Bbb R^2 \times \Bbb R
$$
be the flow describing the free motion of a point particle in
$\Bbb T^2 $, i.e.,
$$
\widetilde{\Phi}^t(q,\,u,\,z) = (q+tu,\,u,\,z),
$$
where $q\in \Bbb T^2,\, u\in \Bbb R^2,\, z\in \Bbb R$.
Let further
$$
\Gamma: \Bbb S^1 \times \Bbb R^2 \times \Bbb R \to
\Bbb S^1 \times \Bbb R^2 \times \Bbb R
$$
be the following ``collision map'' $ \Gamma(s^-,\,u^-,\,z^-) = (s^+,\,
u^+,\,z^+)$, where
$$
\aligned
s^+ &= s^-,\\
u^+ &= -\langle u^-,\,e\rangle e + \left(\cos 2\alpha \langle u^-,\,f\rangle+
\sin 2\alpha z^-\right)f,\\
z^+ &= \sin 2\alpha \langle u^-,\,f\rangle-\cos 2\alpha z^-,
\endaligned
$$
$s$ is the arc length parameter
and $(e,\,f) =(e(s),\,f(s)) $ is the positively oriented orthonormal
frame at the boundary $\Bbb S^1$ of the obstacle, $e$ is the unit
normal vector pointing outward and $f$ is the unit tangent vector.
Let us denote by $\tau : \Bbb S^1 \times \Bbb S^2_+ \to \Bbb R$ the
return time to the section $\Cal M = \Bbb S^1 \times \Bbb S^2_+.$
Now the section map $\Psi : \Cal M \to \Cal M $ can be described as
the following composition
$$
\Psi = \Gamma \circ \widetilde{\Phi}^\tau_{\vert\Cal M}.
$$
This composition corresponds to the natural separation of the dynamics
into the free motion followed by the collision with the obstacle.
We will be able to find the derivative of $\Psi$, if we learn how to
differentiate $\widetilde{\Phi}^\tau$ and $\Gamma$.
We have
$$
D\widetilde{\Phi}^t = \left(dq+ tdu,\,du,\,dz\right).
\tag 1
$$
The derivative of $\widetilde{\Phi}^\tau$ is obtained by composing
\thetag {1} with the projection on $\Cal M$ in the direction of the
velocity vector of the flow (equal to $(u,\,0,\,0)$).
In the case when the orbit is orthogonal to the
obstacle ($u$ parallel to $e$) we get
$$
D\widetilde{\Phi}^\tau_{\vert\Cal M} = \left(dq+ \tau\langle
du,\,f\rangle f,\,du,\,dz\right).
\tag 2
$$
We proceed with the differentiation of $\Gamma$. Let us stress that
$\Gamma$ is linear in the {\bf variable} frame. We have
$$
\aligned
de &= \kappa ds f,\\
df &= -\kappa ds e
\endaligned
\tag {3}
$$
where $\kappa = (2r)^{-1}$ is the curvature of the obstacle $\Bbb
S^1$.
The differentiation of $\Gamma$ yields $ds^+ = ds^-$, and
$$
\aligned
du^+ =& -\langle du^-,\,e\rangle e -\langle u^-,\,de\rangle e -\langle
u^-,\,e\rangle de + \\
&\left(\cos 2\alpha \langle du^-,\,f\rangle
+\cos 2\alpha \langle u^-,\,df\rangle +
\sin 2\alpha dz^-\right)f +\left(\cos 2\alpha \langle u^-,\,f\rangle+
\sin 2\alpha z^-\right)df,\\
dz^+ =& \sin 2\alpha \langle du^-,\,f\rangle +\sin 2\alpha \langle
u^-,\,df\rangle
-\cos 2\alpha dz^-,
\endaligned
$$
We restrict our attention to orbits perpendicular to the obstacle,
which allows us to use $\langle
u^-,\,f\rangle = 0$ and $z^- = 0$. Taking this and \thetag {3}
into account we get
$$
\aligned
du^+ &= -\langle du^-,\,e\rangle e +\left (- \langle
u^-,\,e\rangle \kappa ds + \cos 2\alpha \langle du^-,\,f\rangle -\cos
2\alpha \langle u^-,\,e\rangle\kappa ds + \sin 2\alpha dz^-\right)f
,\\
dz^+ &= \sin 2\alpha \langle du^-,\,f\rangle-\sin 2\alpha \langle
u^-,\,e\rangle
\kappa ds-\cos 2\alpha dz^-.
\endaligned
\tag {4}
$$
Let us consider the periodic orbit carried by the segment
perpendicular to the obstacle at its both endpoints, as described in
the previous section (Fig.1). The Poincar\'e section map for this
periodic orbit is $\Psi^2$. On this orbit $\langle u,f\rangle = 0$, $z
= 0$ and $\tau\langle u,e\rangle = l$, the length of the
orbit. Using the formulas \thetag {2} and \thetag {4} we can obtain
the derivative of $\Psi$ on this orbit. In the coordinates
$(ds,\,\langle du,e \rangle,\, \langle du,f \rangle ,\, dz) \in \Bbb R^4$ we get the product of the following
$4\times 4$ matrices
$$
\left(\matrix 1 & 0 & 0 & 0 \\
0 & -1 & 0& 0 \\
c(1+\cos 2\alpha) & 0 & \cos 2\alpha & \sin 2\alpha \\
c\sin 2\alpha & 0 & \sin 2\alpha & -\cos 2\alpha
\endmatrix\right)
\left(\matrix -1 & 0 & -\tau & 0 \\
0 & -1 & 0 & 0 \\
0 & 0 & -1 & 0 \\
0 & 0 & 0& 1 \endmatrix\right),
$$
where $c = \kappa l\tau^{-1}$.
Dropping the redundant component $\langle du,\,e \rangle$ we get the
following product of $3\times 3$ matrices
$$
D\Psi =
\left(\matrix 1 & 0 & 0 \\
c(1+\cos 2\alpha) & \cos 2\alpha & \sin 2\alpha \\
c\sin 2\alpha & \sin 2\alpha & - \cos 2\alpha \endmatrix\right)
\left(\matrix -1 & -\tau & 0 \\
0 & -1 & 0 \\
0 & 0 & 1 \endmatrix\right).
\tag 5
$$
To obtain the eigenvalues of this matrix, let us observe that it is
reversible (as expected in view of the reversibility of $\Psi$). The
determinant of the matrix is $-1$. Hence one of the eigenvalues is
$-1$ and the product of the other two $\lambda_1,\lambda_2$ is equal to
$1$. It follows that $\lambda_1$ and $\lambda_2$ are on the unit
circle if and only if the trace of the matrix is between $-3$ and $1$.
Calculating the trace of the matrix we get
$$
-2 \leq -(1+\cos 2\alpha)\kappa l -2\cos 2\alpha \leq 2.
\tag 6
$$
Since in our situation the curvature $\kappa$ is positive,
it follows that $D\Psi$ and $D\Psi^2$ have eigenvalues on the unit
circle if and only if
$$
\kappa l \leq 2\xi^2
\tag 7
$$
With these formulas we can address the question of linear stability in
the system of a spinning disk inside a planar domain. It was
established in \cite {B-G} and \cite {K-T} that the collisions of a
spinning disk with a wall are the same as in the reduced system above.
Let us consider a periodic orbit carried by the segment perpendicular
to the boundary of the domain at both endpoints.
If the curvature $\kappa$ of the boundary at these endpoints is the
same, then the necessary and sufficient condition for the Floquet exponents
to be on the unit circle is given by the condition \thetag 6. If the
boundary is locally concave at the collision points (nonnegative
$\kappa$) then this condition is equivalent to \thetag 7.
In the case of convex
boundary (negative $\kappa$) we get from \thetag 6 the condition
$$ -\kappa l \leq 2,
\tag 8
$$ which coincides with the condition of linear stability of an
ordinary billiard trajectory (it has to be the case since now the
parameter $\xi$ is absent).
Let us note that when $\kappa = 0$ we cover the case of
the system, considered by Broomhead and Gutkin \cite {B-G}, of the
disk between two parallel lines. They established the (nonlinear)
stability of the periodic orbit perpendicular to the lines. Our
conditions \thetag {7} and \thetag 8 show how much we can curve the
boundaries without making the orbit linearly unstable.
For the system of a spinning disk inside a planar circular domain, we
obtain that the periodic orbits passing through the center of the disk
have all Floquet exponents equal to one ($\kappa l= -2$) but the orbit
is unstable. Indeed, the system is completely integrable and
we can describe the orbits in the following way.
Because of the symmetry of the domain, for a
segment of the orbit between collisions, the tangential component of
the velocity is the same at the departure from the boundary and at the
next arrival. The normal component is constant in absolute value. Hence,
although after the reflection the tangent component gets changed, it
will assume the original value after the next reflection. Consequently
every second segment is tangent to the same concentric circle. We
obtain a ``double caustic'' and all motions are quasiperiodic, Fig.2.
\topinsert
\vskip 3in
\hsize=4.5in
\raggedright
\noindent{\bf Figure 2.} The double caustic.
\endinsert
Let us finally consider the case of the periodic orbit perpendicular
to the boundary with different curvatures $\kappa_1,\kappa_2$ at the
two endpoints. To get a hold of the eigenvalues of $D\Psi^2$ we use
again reversibility of the product of the two matrices \thetag 5 with
different values of $c$. Since the determinant of this product is equal
to $1$ we can conclude that one eigenvalue is equal to $1$ and the
other two eigenvalues are on the unit circle if and only if the trace
of the matrix is between $-1$ and $3$. After a lengthy but
straightforward calculation we obtain that this condition is equivalent to
$$
0 \leq (\kappa_1 l +1 -\xi^2)(\kappa_2 l +1 -\xi^2) \leq (1+\xi^2)^2.
\tag 9
$$
When $\xi = 0$ the condition \thetag 9 coincides with the condition of
linear stability of the ordinary billiard orbit (cf. \cite {W3},
Proposition 3).
\vskip.7cm
\subhead \S 4. Concluding remarks \endsubhead
\vskip.4cm
By the condition \thetag {7} we have linear stability of the periodic
orbit when $\xi$ is sufficiently close to $1$ (away from the elastic
collisions) and the length of the orbit is sufficiently
short (the obstacle, or the disks are large). The orbit seems to be
degenerate (it has a one parameter family of periodic orbits in its
vicinity), but it is only the minimal degeneration caused by the
reversibility. One would like to apply the KAM theory to the
neighborhood of the periodic orbit. The KAM theory for reversible maps
was developed systematically by Sevryuk \cite{Sev}. We found the task
of calculating the higher order terms in the expansion of our map
prohibitively cumbersome and so we were unable to establish for what
values of the parameters the nondegeneracy conditions of Theorem 2.9
of \cite{Sev} hold. If and when this is the case, then most of the
nearby orbits are quasi periodic and will stay close forever. In
particular we would establish that the system is not ergodic.
Our dynamical system $\Psi : \Cal M \to \Cal M$ cannot have all
Lyapunov exponents different from zero almost everywhere and be
ergodic. Indeed the number of positive (or negative) exponents is an
invariant function. If the system is ergodic the number of positive
and the number of negative exponents must be constant almost
everywhere. These numbers are also equal because of reversibility.
Hence if the system is ergodic, there may be at most one positive and
one negative Lyapunov exponent and at least one equal to zero.
Because of this observation we do not expect to find systems of this
type with hyperbolicity in all of the phase space.
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\enddocument