Let $\Omega$ be a domain in $\mathbb{R}^2$ with smooth boundary. A billiard trajectory is a continuous curve $c: \mathbb{R}\supseteq I \longrightarrow \overline{\Omega}$ such that
- $c(t) \in \partial \Omega$ only for a discrete set of times $t$
- if $c(t) \notin \partial \Omega$, then $c|_{[t-\varepsilon, t+\varepsilon]}$ is a straight line for some $\varepsilon>0$.
- If $c(t) \in \partial \Omega$, then both one-sided derivatives exist and we have $\dot{c}(t+) = \dot{c}(t-) - 2\langle \dot{c}(t-), \mathbf{n}\rangle \mathbf{n}$ where $\mathbf{n}$ is the (exterior or interior) unit normal to $\Omega$.
That is, billiard particles move in straight lines and reflect making the angle of incidence equal to the angle of reflection.
In [1], the following is proven: If $\Omega$ is convex (and has smooth boundary as always assumed here), then for each $x \in \Omega$ and each velocity vector $v \in \mathbb{R}^2$, there exists a unique billiard trajectory $c$ with $c(0) = x$ and $\dot{c}(0) = v$. Conversely, it is shown that if the boundary is merely $C^2$, then there exists a billiard trajectory that hits the boundary infinitely often in finite time.
It is also standard in the theory of billiards that for almost all $(x, v)$ there exists a unique billiard trajectory starting at $(x, v)$.
Question: What about the smooth, but non-convex case?
It is clear that if a billiard trajectory is "stuck", that is, its boundary-hitting times $t_n$ have a finite limit $t_0$ and hence $c(t_n)$ converges to a point $x \in \partial \Omega$, then the curvature of the boundary must vanish at $x$.
But is there an example for a "stuck" billiard trajectory on a domain with smooth (but non-convex) boundary?
[1] B. Halpern, Strange Billiard Tables, Transactions of the AMS, 1977
