Lemma 4.1 extends obviously to arbitrary area-preserving twist maps, where the action functional replaces the length. Invariant circles with rational rotation number happen for a thin set of twist maps.
Deformations of minimal periodic orbits.
A strictly convex piecewise curve , parametrized by the arc length is completely determined by its radius of curvature . We will assume that is piecewise continuous. At each of the finite number of points of discontinuity as well as at the end points of the curve, we assume the existence of the one-sided limits .
A closed convex, piecewise curve is, by
our convention, a finite union of strictly convex arcs and straight line
segments called flat arcs. Let be the subset of
If belongs to a flat arc of or if
belongs to a strictly convex arc and on at least one side,
we say that A is a flat point of . When
belongs to a strictly convex arc, and , we say that A is a
vertex or a corner point.
We will freely identify periodic orbits of the billiard map with polygons, inscribed in T. A periodic orbit, P is said to admit a deformation, if there exists a continuous deformation of periodic orbits P. More precisely if are the arclength coordinates, we assume that are continuous. We will also consider one-sided deformations P(t), , or , assuming the same conditions, except the derivatives at t=0 are one-sided.
1) Let be a billiard map, and let be an invariant circle. Let . If , then contains minimal periodic orbits of type (p,q). Periodic points, , satisfy . Typically, they are isolated.
2) If , for a convex caustic , then the one-sided derivatives, , exist, and are given by eq. 5. The jumps of the derivative are due to the flat arcs of . If is strictly convex, then f is , and is given by Eq. 5.
3) With the same assumptions as in the preceeding remark, let be a periodic point, and let P be the corresponding inscribed polygon (Figures 8,10). Let be strictly convex at the supporting points of P. If , then the periodic orbit P consists of hyperbolic points. If , then the points are parabolic. By Lemma 4.3, the condition that P be parabolic is necessary for the existence of periodic deformations of P. A stronger assertion holds: if P is not an isolated periodic orbit, then P is parabolic.
If P is a hyperbolic periodic orbit, then it is a repelling periodic orbit for the circle map if , and an attracting periodic orbit for the circle map if (see Figure 11). If P is parabolic, and isolated, P may be repelling or attracting on each side independently. It is also possible that contains isolated periodic points, and nontrivial intervals of periodic points as well (see Figure 11 b)). There is a similar analysis of the one-sided derivatives of , and a natural extension of eq. 5.
Next: Aubry-Mather sets and the Up: Billiards that share a Previous: Devil's staircaseOliver Knill