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Bunimovich proved that the billiard-ball dynamics in the Bunimovich stadium is ergodic.
    Bunimovich
       (Image from Microwave_billiards_and_quantum_chaos.)

Q. Is it known that the billiard-ball dynamics in a rounded rectangle is ergodic?


    Billiard stadium
Here the corners are quarter circles; so can be viewed as a stadium with vertical segment inserts. Perhaps Bunimovich's proof also covers rounded rectangles? If anyone knows, I'd appreciate a pointer—Thanks!

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    $\begingroup$ One major difference between the Bunimovich stadium and the rounded rectangle is that there are one-parameter families of periodic orbits in the rounded rectangle: start with a centred diamond in the rounded rectangle stadium. This is a periodic orbit, but so are its friends if you start in the same direction with a slight perturbation of the initial point. This isn't enough to destroy ergodicity: the phase space of Billiards is considered to be two-dimensional - you consider returns to the boundary. These are parameterized by the point and the angle. The above is a 1D family of per. orbits. $\endgroup$ – Anthony Quas Feb 1 '14 at 17:38
  • $\begingroup$ So it has 0 Lebesgue measure. $\endgroup$ – Anthony Quas Feb 1 '14 at 17:39
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    $\begingroup$ @AnthonyQuas Yes, but the "bouncing ball" orbits in the stadium (period 2 orbits between the parallel sides) are also a 1-parameter family with similar properties. If you use the reflection trick on the rounded rectangle, you get a periodic lattice of "star" obstacles. Any family of lines that misses this gives periodic orbits in the original billiard. The smaller the rounded corners, the more such families. $\endgroup$ – user25199 Feb 5 '14 at 10:13
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I think the brief answer is Yes; the answer seems to be well-known to the grand master of billiards, Bunimovich himself. See http://www.scholarpedia.org/article/Dynamical_billiards and the references quoted therein.

The above example of rectangular billiard with round corners has several ergodic components in the invariant measure so it is not uniquely ergodic. But as long as there are round corners (focussing boundary components) there will be a continuous ergodic (w.r.t Lebesgue) component, while others, corresponding to periodic orbit (families), will be singular.

Other examples of this type of mixed invariant measure include the "mushroom" displayed by Bunimovich as Figure 3 in the above article–they are thus neither fully chaotic nor fully regular (the sharp rectangle would be non-hyperbolic). A billiard in such a mushroom has one integrable island formed by the trajectories that never leave the cap, and it is chaotic and ergodic on its complement. A mushroom becomes a semi-stadium when the width of the feet equals the width of the hat. Combining mushrooms together one gets examples of billiards with an arbitrary (finite or infinite) number of islands coexisting with an arbitrary (finite or infinite) number of chaotic components (see edit)

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  • $\begingroup$ Great article by Bunimovich---Thanks so much! And thanks for the summary explanation. $\endgroup$ – Joseph O'Rourke Feb 1 '14 at 21:06
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The relevant result is stated in Bunimovich, "Conditions of stochasticity of two-dimensional billiards" Chaos 1, 187-193 (1991): If the circles to which the arcs belong are completely contained in the billiard, the dynamics is ergodic. So

Yes, if the rounded arcs have equal radius.

If not, it is easy to construct rectangles for which the above condition fails.

Conjecture: There is an example with non-equal radii which is not ergodic.

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  • $\begingroup$ Nice conjecture! $\endgroup$ – Joseph O'Rourke Feb 5 '14 at 12:45

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