
According to Poincaré Birkhoff's theorem, there exists for each pair
(p,q) with p;SPMgt;1 and and 0;SPMlt;q/p;SPMlt;1 a periodic orbit of
period p which winds around the table q times. These periodic
orbits are called Birkhoff periodic orbits. In general, there exist
many more orbits of period p. It is an open question whether the
set of periodic orbits can form a set of positive Lebesgue measure in
the phase space. There exist some results for small periods
[20, 19]. We cannot
decide this question for our case.
Even if the set of periodic orbits forms a set of zero Lebesgue measure,
they still generally have a significant
influence on the global phase space picture. A stable
periodic orbit keeps a whole neighborhood nearby stable, which prevents
ergodicity. A first step in understanding the stability of the orbits
is finding out whether they are linear stable.
The stability of periodic orbits of small order can be decided
by hand. We will treat orbits
of period 2 and 4.
Proof. Look at an orbit of period 2 hitting the table at points with curvatures and and let l be the distance between the two points. A computation of the trace of the Jacobean of (see Proposition 4 of [22] or Proposition 3 of [23]) shows that the orbit is elliptic if and only if or .
Consider a 2orbit of length , that meets the billiard table at points with radii of curvature and , with . If one of the inequalities or is satisfied, then the orbit is hyperbolic, while if either or is satisfied, the orbit is elliptic. For our 2orbits, , so we get elliptic orbits when and hyperbolic orbits when .
We look first at the horizontal 2orbit . The vertical 2orbit is obviously identical to the horizontal one. The horizontal orbit has length 2. The curvature at the endpoints is , 1, and 0 for , respectively. This gives radii of curvature of 0, 1, and for , respectively. For and the orbit is hyperbolic. And for , and the orbit is elliptic. So the horizontal orbit is hyperbolic if p;SPMlt;2 and elliptic if p;SPMgt;2. The diagonal orbit has . The curvature at the endpoints is . The radius of curvature is then . The orbit is elliptic if , or
Multiplying both sides by , we get
, which means p;SPMlt;2. So the diagonal 2orbit is elliptic
if p;SPMlt;2. Similarly, the
diagonal 2orbit is hyperbolic if p;SPMgt;2.
End of the proof.
Proof. We calculate the Hessian for . Let be the central angle for the point of the orbit, be the angle between the path of the particle leaving the point and the directed tangent to the pcurve at that point. gives the distance of the straight line connecting the points given by and , and is the curvature at the point. The Hessian is then
with
In our case we have by symmetry and all of the . The determinant of the Hessian then simplifies to . Define . The orbit is hyperbolic if D;SPMgt;2 and and elliptic if D;SPMlt; 2 (see [22]). For the symmetric 4orbits we have
First look at the orbit. We get with , and , so that
We have already calculated at these points, so we get
So we get that the orbit is hyperbolic for p;SPMlt;2 and parabolic
for . (Numerical experimentation suggests that the orbit
is stable for p;SPMgt;2).
For the orbit , all four points have . We then calculate
and
The orbit is elliptic if
and .
So the orbit is elliptic if 1;SPMlt;p;SPMlt;2 and
hyperbolic for p;SPMgt;2.
End of the proof.
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