## Billiards

9. Victor Ivrii conjecture. Let $Q$ be a strictly convex domain, $F$ be the billiard map on the phase space $\Omega=\partial Q\times(0,\pi)$. Let $\omega$ be the Lebesgue measure of $\Omega$, and $\ell$ be the Lebesgue measure on $Q$.

Conjecture 1. $\omega(\text{Per}(F))=0$ for all $Q$ with $C^\infty$ boundaries.

Remark. This is about a general domain $Q$, not a generic domain.

Definition. A point $q\in\partial Q$ is said to be an absolute looping point, if $\omega_q(\bigcup_{n\neq0}F^n\Omega_q)>0$. Let $L(Q)$ be the set of absolute looping points.

Conjecture 2. $\ell(L(Q))=0$ for all $Q$.

Question: When $L(Q)=\emptyset$?

8. In Boltzmann gas model, the identical round molecules are confined by a box. Sinai has replaced the box by periodic boundary conditions so that the molecules move on a flat torus.

On circular and elliptic billiard tables, for all $p\ge 3$, the $(p,q)$-periodic orbits forms a continuous family and hence all the trajectories have the same length.

An invariant noncontractible topological annulus, $A\subset\Omega$, whose interior contains no invariant circles, is a Birkhoff instability region. The dynamics in an instability region
has positive topological entropy. Hence Birkhoff conjecture implies
that any non-elliptical billiard has positive topological entropy.

How to construct a strictly convex $C^1$-smooth billiard table with metric positive entropy? b) How to construct a convex $C^2$-smooth billiard table with positive metric entropy?

Recall Bunimovich stadium is not $C^2$, and not strictly convex.

A periodic orbit of period $q$ corresponds to an (oriented) closed polygon with $q$ sides, inscribed in $Q$, and satisfying the condition on the angles it makes with the boundary. Birkhoff called these the harmonic polygons.

Then the maximal circumference of 2-orbit yields the diameter of $Q$. The minimax circumference of 2-orbit corresponds
to the width of $Q$.

7. Sinai’s Fundamental Theorem of Dispersing Billiards. Suppose the orbit of $x$ never hit the singular set $\varphi=\pm\pi/2$. Then for any $\delta>0$, any $C>0$ there exists an open neighborhood $U_x$ such that

• for any unstable curve $W\subset U_x$: $m_W(y\in W:r^s(y)>C\cdot|W|)\ge(1-\delta)\cdot |W|$;
• for any stable curve $W\subset U_x$: $m_W(y\in W:r^u(y)>C\cdot|W|)\ge(1-\delta)\cdot |W|$;

In particular every points can be connected to invariant manifolds of larger size by $us$-paths.

Hopf argument. Consider the billiard map $(f,\mu)$ on a dispersing billiard table. A point $x$ is said to be Birkhoff-regular if the backward and forward Birkhoff averages exist at $x$ for every continuous function $\phi$. Denote by $B_f$ the set of such points. Then $\mu(B_f)=1$ by Birkhoff ergodic theorem.

A point $x$ is said to be good, if $x\in B_f$, the stable and unstable manifolds exist and are $B_f$-saturated. Denote by $G_f$ the set of such points. Then $\mu(G_f)=1$ by $C^2$-regularity and Pesin’s absolute continuity of these invariant foliations.

Property. For any given unstable manifold $W^u(p)$, $W^s(x)$ exists for $m_u$-a.e. $x\in W^u(p)$, and those points ‘whose $W^s$ are not $B_f$-saturated’ consist of $m_u$-null subset of $W^u(p)$ (by Pesin’s transverse absolute continuity of stable foliation).

In particular if $W^u(p)$ is $B_f$-saturated, then it is also $G_f$-saturated. Therefore $W^u(p)$ and $W^s(p)$ are $G_f$-saturated for each $p\in G_f$. So if we start with $A_1(p)=W^u(p)\cap G_f$, $A_2(p)=W^s(A_1(p))\cap G_f$, and build inductively $A_f(p)$, the accessibility class of $p$ cornering in $G_f$.

• $W^u(x)$ and $W^s(x)$ are $G_f$-saturated for every $x\in A_f(p)$.
• $\mu(A_f(p))\ge \mu(A_2(p))>0$ and $A_f(p)$ lies in one ergodic component of $\mu$.

Local ergodicty: For every $p\in G_f$, there exists an open neighborhood $U_p\subset A_f(p)$ (mod $\mu$).

Proof. There exists $\epsilon^{s,u}_n(p)\to0$ and a rectangle $R_n(p)$ surrounding $p$ of stable size $\epsilon^s_n(p)$ and unstable size $\epsilon^u_n(p)$. Fix $R(p)=R_n(p)$ for some $n$ large. Although not every stable/unstable manifolds in $R_n$ can fully cross it, there are $us$-paths initializing from each point $x\in G_f\cap R(p)$, cornering in $G_f$ and fully crossing $R_n$, say $W^u(p_x)$. Given two points $x,y\in G_f\cap R(p)$, the two leaves $W^u(p_x)$ and $W^u(p_y)$, being $G_f$-saturated, admit a nontrivial stable holonomy $h^s: E\subset W^u(p_x)\to W^u(p_y)$. In particular $m_u(E\cap G_f)>0$ and $m_u(h(E\cap G_f))>0$: there exists $q\in W^u(p_x)\cap G_f$ such that $h^s(q)\in W^u(p_y)\cap G_f$. So the pair $(p_x, p_y)$ can be connected by a $us$-paths cornering in $G_f$, so is the pair $(x,y)$.

Let $E>0$ and consider the following Newton system: $\dot{\bf q}={\bf p}$, $\dot{\bf p}={\bf F}({\bf p})$ while ${\bf F}(u,v)=\langle E-u^2,-uv\rangle$. So the derivative $\frac{d}{dt}\frac{\|{\bf p}\|^2}{2}=p_x\dot{p}_x+p_y\dot{p}_y=p_x(E-p_x^2)-p_yp_xp_y=p_x(E-\|{\bf p}\|^2)$. In particular the constant kinetic surface $\|{\bf p}\|^2=E$ is a global attractor of the system.

6. A smooth curve $\Gamma$ is (relatively) focusing, if every incoming parallel wave-front will become focusing after the first reflection on $\Gamma$.

A focusing curve Γ is absolutely focusing, if every ray leave $\Gamma$ after finitely many reflections, and every incoming parallel wave-front become focusing after the last reflection in the series of consecutive reflections on $\Gamma$.

The above were introduced by Bunimovich in 1988 (under a different name: admissible). The local condition is studied by Donnay in 1991, where he prove the hyperbolicity of a class of billiards. Bunnimovich (1992) proved that weaker global condition implies the stronger local property.

Consider the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$. Without loss of generality we assume $a>1=b$.

The right half $x\ge0$ is absolutely focusing if and only if $a<\sqrt{2}$.

The larger half $x\ge-\delta$ is never absolutely focusing, since it traps rays of positive volume.

We parametrize the ellipse by ${\bf r}(\theta)$ for $\theta\in\mathbb{R}$.

Donnay (1991). The arc $a\le\theta\le b$ is absolutely focusing if the endpoints connection ${\bf r}(a){\bf r}(b)$ does not intersect the focal connection $F_1F_2$.

Proof. For such arc, only the winding orbits can hit it twice and parallel wave-fronts focus between each collision.

In particular the upper ellipse $y\ge0$ is absolutely focusing. Again the larger half $y\ge-\delta$ is not.

Dynamics on regular islands is characterized by a balance between focusing and defocusing, while defocusing dominates focusing on chaotic sea. Dispersing is a special case of defocusing, in the sense that the focusing time is negative.

Conjecture (Bunnimovich 1991): for every hyperbolic billiard, each focusing component of the boundary is absolutely focusing.

Bunnimovich (2003) and with Grigo (2010) proved that if a focusing arc $\Gamma$ is not absolutely focusing, then for any large $L>0$, $\Gamma$ can be included in some billiard table whose minimal free path leaving $\Gamma$ is larger than $L$, and has some elliptic periodic points on $\Gamma$. So the violation of the absolutely focusing cannot by compensated for by making the free path arbitrarily large.
and absolutely focusing is a necessary condition to construct universally chaotic billiard tables.

Note that if $\tau=\frac{d+d_1}{2}$, then the incoming parallel wave-front focus during the $\tau$-flight and leave $x_1$ as a parallel wave-front. Every arc of a round circle is absolutely focusing, so is the short enough focusing piece, since the curvature are almost constant and $\tau=\sim 2d\sim 2d_1>\frac{d+d_1}{2}$. Similarly any focusing boundary components with vanishing curvature are non-absolutely focusing, since $d_1\to\infty$. Bunimovich and Grigo introduced the following

A focusing curve $\Gamma$ is called non-absolutely focusing of minimal length, if every of its closed sub-arcs is absolutely focusing, but the curve itself is not absolutely focusing.

Let $Q$ be a simply connected billiard table, whose boundary $\Gamma=\partial Q$ consists finitely many piece-wisely smooth components. The defocusing mechanism applies to absolutely focusing components.

Although compositions of twist maps, and iterates of a twist map, are in general no longer twist maps, Donnay observed that, then the iterates of the billiard map will still be twist maps (when restricting to absolutely focusing boundary components).

5. Reuleaux triangle: take an equilateral triangle with vertices A, B, C. Draw the arc BC on the circle centered at A, the arc CA on the circle centered at B, and the arc AB on the circle centered at C. The resulting figure is of constant width.

More generally, we start with a triangle ABC (say $AB$ is the longest edge) and put a wood stick over $AB$ with length $d>|AB|$. We fix the stick at $A$ and rotate it countclockwise until it reach the point $C$. Then we fix the stick at $C$ and start a new rotation to hit $B$. Finally we fix the stick at $B$ and start a new rotation to hit $A$. The resulting domain is of prescribed width $d$. Moreover, it is easy to see that the boundary consists of six arcs of different radii and is of length $d\pi$. Similar construction can be carried out over any convex polygon with an odd number of sides.

Click or go to here for the following gif:

The evolute $\gamma$ of a curve $\Gamma$ is the locus of all its centers of curvature. Equivalently, an evolute is the envelope of the normals to a curve.

A vertex of a smooth curve $\Gamma$ is a critical point of the curvature. Equivalently, a vertex is a point at which the osculating circle has the third order tangency with the curve.

This is typically a local maximum or minimum of curvature, and is the definition used by some author (For example, a round circle has constant curvature, then every point would be a vertex). According to the four-vertex theorem, every closed curve must have at least four vertices.

At a vertex of $\Gamma$, the evolute $\gamma$ has a stationary point, generically, a cusp. A generic cusp is semi-cubic: in appropriate local coordinates, it is given by the equation $y^2=x^3$.

4. Let $E$ be the ellipse given by $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, or ${\bf p}\cdot E{\bf p}=1$, where ${\bf p}=(x,y)$ and $E=\text{diag}(a^{-2},b^{-2})$. Taking derivatives, we get ${\bf \tau}\cdot E{\bf p}=0$, where $\tau$ is a vector tangent to $E$. Suppose a particle at ${\bf p}\in E$ moving inward the interior of the ellipse, say its direction ${\bf v}$. After hitting $E$ again elastically, we record the new position $p_1\in E$ and direction ${\bf v}_1$.

Theorem 4.4 in Geometry and Billiards (link).
(a) Suppose the trajectory between one collision does not intersect $F_1F_2$, so is the trajectory between next collision. Moreover they are tangent to a confocal ellipse.
(b). The quantity ${\bf v}\cdot E{\bf p}$ is preserved. That is, the ${\bf v}_1\cdot E{\bf p}_1={\bf v}\cdot E{\bf p}$.

Proof. (a). We reflect $F_1$ along the trajectory, say $F_1'$, and connect $F_1'F_2$ and intersect with the trajectory at $B$. Then $2\hat a=|BF_1|+|BF_2|$ stays the same and specifies the longer axis of the confocal ellipse.

(b). Note that $({\bf p}_1-{\bf p})\cdot E({\bf p}_1+{\bf p})={\bf p}_1\cdot E{\bf p}_1+{\bf p}_1\cdot E {\bf p}-{\bf p}\cdot E{\bf p}_1-{\bf p}\cdot E{\bf p}=0$, since $E$ is symmetric. So ${\bf v}\cdot E({\bf p}_1+{\bf p})=0$ since ${\bf v}$ is parallel to ${\bf p}_1-{\bf p}$. Therefore ${\bf v}\cdot E{\bf p}_1=-{\bf v}\cdot E{\bf p}$.

Note that ${\bf \tau}_1\cdot E{\bf p}_1=0$, and ${\bf v}+{\bf v}_1$ is parallel to ${\bf \tau}_1$, we see $({\bf v}+{\bf v}_1)\cdot E{\bf p}_1=0$, or equivalently, ${\bf v}_1\cdot E{\bf p}_1=-{\bf v}\cdot E{\bf p}_1$. Putting together, we get that ${\bf v}_1\cdot E{\bf p}_1={\bf v}\cdot E{\bf p}$. QED

Once again let $E$ ($b=1) be an ellipse parametrized by $0\le \phi\le 2\pi$, the inner unit velocity parametrized by the angle with the tangent direction, $0\le \theta\le\pi$. Another integral of the billiard map is $F(\phi,\theta)=\frac{\sin^2\theta}{1-e^2\cos^2\phi}$, where $e=\sqrt{a^2-1}$ is the eccentricity of $E$. What is the relation with $G({\bf p},{\bf v})={\bf v}\cdot E{\bf p}$?

Let ${\bf p}=(x,y)$ be a point on the ellipse, ${\bf n}=(x,y)=E{\bf p}$ be the out-normal direction at ${\bf p}$, ${\bf v}=(u,v)$ be an inward unit vector. Then the orbit along $({\bf p},{\bf v})$ is tangent to a caustic $\displaystyle C_\lambda: \frac{x^2}{a^2-\lambda^2}+\frac{y^2}{y^2-\lambda^2}=1$, where $\displaystyle \lambda=-ab\cdot (\frac{xu}{a^2}+\frac{yv}{b^2})=-ab{\bf v}\cdot E{\bf p}$.

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Let $A=\begin{pmatrix}a & b \\ c & d \end{pmatrix}\in\mathrm{SL}(2,\mathbb{R})$ preserving the diagonal $x=y$. So $ad-bc=1$ and $a+b=c+d$. Substitute $d$: $1=a(a+b-c)-bc=(a-c)(a+b)$. Solve $c$: $c=a-\frac{1}{a+b}$ and hence $A=\begin{pmatrix}a & b \\ a-\frac{1}{a+b} & b+\frac{1}{a+b}\end{pmatrix}$ and $\mathrm{Tr}(A)=a+b+\frac{1}{a+b}$ is either parabolic (if $a+b=\pm1$) or hyperbolic (generic case).

(Not that meaningful if all of them are not convex….)

3. Let’s consider the inscribed square in the unit circle and squeeze the two side arcs. Consider the periodic 2 orbit. We have $d=d_1$ increase (from $1$ to $+\infty$), and $\tau$ decrease from $2$ to $\sqrt{2}$. In particular $\mathrm{Tr}(D_pT^2)=4(\frac{\tau}{d}-1)^2-2$ varies

In particular the orbit starts as a parabolic one, turns into an elliptic one and then parabolic again. Moreover there are elliptic island surrounding that orbit developing and then shrinking.

The case $\tau=d$. In this case $D_pT^2=\begin{pmatrix}-1 & 0 \\ 0 & -1\end{pmatrix}$. Although it is a rotation, a small perturbation will make it hyperbolic. So it is reasonable to put it in the parabolic category.

Suppose we put a twist at the collision: $(r,\phi)\to(r,\phi_1)$ where $\phi_1=g(\phi)$. Then the time-reversibility is equivalent to say that $g(-\phi_1)=-\phi$. This is equivalent to say that the graph $G(g)$ is symmetric with respect to the line $x+y=0$: the symmetric point of $(x,y)$ with respect to the line $x+y=0$ is $(-y,-x)$.

Let $x=(p,\theta)\in\Omega$ and $X=aX_r+bX_\theta\in T_x\Omega$ be a tangent vector. There is a smooth curve $c(t)\in \Omega$ such that $c(0)=x$ and $\dot{c}(0)=X$. For each $c(t)$, consider the ray on the table and the intersection point $D(t)$ with the ray of $x$ (may be at infinite). The limiting point $D(X)=\lim_{t\to0} D(t)$ is the forward focus point of the tangent vector $X\in T_x\Omega$ and the distance $d(X)=\pm|D(X)-p|$ is the forward focus distance of $X\in T_x\Omega$. In particular we let $d(x)=d(X_r)$. Similarly we define the backward focus distance $d^-(X)=d(D\sigma X)$, where $\sigma$ be the time-reverse on $\Omega$

Let $\nu(X)=\frac{b}{a}$ be the slope of $X$, $y=Tx$ and $Y\in T_y\Omega$.

Donnay: If $\tau(x)> d(X)+d^-(Y)$, then $\nu(DT X)> \nu(Y)$.

2. Note that $h(s_0,s_1)=|\Gamma(s_1)-\Gamma(s_0)|$ is a generating function of $T$: $\displaystyle (s_{i+1},\theta_{i+1})=T(s_i,\theta_i)$ if $h_1(s_i,s_{i+1})+h_2(s_{i-1},s_i)=0$. It is easy to see $h_1(s_i,s_{i+1})=-\cos\theta_i=-h_2(s_{i-1},s_i)$. So we might say that $h$ is the generating function for a more proper coordinate $(s,r)$ with $r=\cos\theta$. But sometime we won’t distinguish them.

Moreover $\displaystyle \frac{\partial h}{\partial s_1}=\cos\theta_1$ and $\displaystyle \frac{\partial h}{\partial s_0}=-\cos\theta_0$. So we have the total differential $dh=-\cos\theta_0 ds_0+\cos\theta_1 ds_1$. Taking exterior differential, we get $0=d(dh)=\sin\theta_0 d\theta_0\wedge ds_0-\sin\theta_1 d\theta_1\wedge ds_1$. In particular $\omega=\sin\theta_0 d\theta_0\wedge ds_0=\sin\theta_1 d\theta_1\wedge ds_1=\omega_1$ is invariant.

Given $s\in\Gamma$, consider the normal line $I(s)$ based on $\Gamma(s)$ and let $L(s)$ be the length of $I(s)\cap Q$. Let the $x$-axis be the tangent direction at $\Gamma(s)$. Then locally $y=ax^2+o(x)$ around $\Gamma(s)$ and $\hat y=l+k\hat x+o(\hat x)$ around the other endpoint. Then $L^2(s)=(x-\hat x)^2+(y-\hat y)^2=l^2+2lk(1-2al)x+O(x^2)$. Suppose $\bar s$ is a critical point of $L(s)$. Then $k(1-2al)=0$. Two possible choices:

– either $k=0$, then $I(\bar s)$ is orthogonal to $\Gamma$ at the other endpoint and hence a periodic 2 orbit;

– or $l=\frac{1}{2a}=r(s)$ (the radius of curvature). In general we can pre-exclude this case.

Let’s consider an optical wave-front. For small times, such a wave front is diffeomorphic to a sphere. After a reflection, there will be (in general) points where the front ceases to be an immersed surface, for example points where different parts of the front self-intersect. The set of all points where the immersion fails is called the caustic.

Caustics also can also be observed as the optical wave-front pass through the homogeneous or non-flat medium, since these spheres become more and more deformed as time increase until the front ceases to be an immersed surface. For example after the light bundle passes a glass of water.

1. Let $Q$ be a convex domain with $C^2$ boundary, $\Omega$ be the (open) phase space and $T:\Omega\to\Omega$ be the billiard map on $\Omega$. A point $p=(s,\theta)$ is said to be glancing if its orbit gets arbitrarily close to the upper and lower boundaries $\{\theta=0,\pi\}$ of $\Omega$. It is clear that the existence of an invariant curve implies no glancing point. The converse is also true:

Birkhoff: Assume no glancing point/orbit of $(\Omega,T)$, then there exists an invariant curve.

Sketch of Proof. For any $\epsilon>0$, there exists $\delta>0$ such that points from $U_\delta=\{p=(s,\theta):\theta<\delta\}$ never visit $V_\epsilon=\{p=(s,\theta):\theta>\pi-\epsilon\}$. So the union $\bigcup_{\mathbb{Z}}T^nU_\delta$ is an open subset of $\{p=(s,\theta):\theta\le\pi-\epsilon\}$, whose boundary should be an invariant curve.

Further analysis shows that such a curve must be the graph of a Lipshitz continuous function $\theta=\theta(s)$. In fact Birkhoff proved that, for any invariant open set $U\subset \Omega\backslash\{p:\theta=\pi\}$ homeomorphic to $\Gamma\times[0,\pi)$, its boundary is a Lipschitz graph.

Mather: If the curvature of $\Gamma=\partial\Omega$ vanishes at some point, then for every $\epsilon>0$, there exists a trajectory visiting $U_\epsilon$ and $V_\epsilon$. For example the table enclosed by $x^4+y^4=1$.

Lazutkin introduced a special coordinate near the boundary: $\displaystyle (x,y)=(C^{-1}\cdot\int_0^s k(r)^{-2/3}dr,4C^{-1}k(s)^{1/3}\sin\theta/2)$, where $\displaystyle C=\int_0^l k(s)^{-2/3}ds$ is a normalizing constant. Under this new coordinate, the map near the boundary can be represented by $T:(x,y)\mapsto (x+y+O(y^3),y+O(y^4))$, as the perturbation of $(x,y)\mapsto(x+y,y)$.

Lazutkin: Consider a smooth billiard table with strictly positive curvature. Then there are invariant curves of positive volume near the boundary $\{\pi=0,\pi\}$. In other words, there are caustics of positive volume near the boundary $\partial\Gamma$.

Here smoothness is determined by KAM theorem. Douady showed that $C^6$ is sufficient.

Mather: If there is a point of zero curvature on the boundary a convex table, then there is no caustics passing that point.

Proof. Suppose $K(s)=0$ and pick a point $x=(s,\theta)$. According to the formula $\frac{1}{f_-(x)}+\frac{1}{f_+(x)}=\frac{2K(s)}{\sin\theta}=0$, we see $f_+(x)=-f_-(x)$. So the backward and forward focusing lie on the two sides of the boundary.

Andrea Hubacher: a discontinuity in the curvature of $\Gamma=\partial Q$ does not allow caustics near the boundary. For example, tables obtained by the string construction around a triangle.

Knill noted that Hubacher obtained this result when she was an undergraduate student at ETH.

Proof. Suppose $s=0$ is the jumping point with radii $r and there are caustics approaching the boundary of phase space, say $\theta=0$. On each caustics we can find a trajectory orthogonal to the normal line of $\Gamma$ at $s=0$, say $(s_0,\theta_0)\to(s_1,\theta_1)$. Then he proved that $\theta_1<\rho\cdot \theta_0$, where $\rho=\sqrt{r/R}<1$.

Quantitative version: estimate the size of the region free of caustics near the boundary. In particular for a typical table by the string construction of a polygon, the free of caustics region is the annulus between table and the polygon.

Now consider a trajectory starting at $\hat s_0=0$, say $(0,\hat \theta_0)$, following the same caustics. According to Mather, these two trajectories can’t cross. To avoid backward crossing, we need $\hat \theta_0\sim \theta_0$. To avoid forward crossing, we need $\hat \theta_0\sim \theta_1$. Recall that $\theta_1<\rho\cdot \theta_0$. So we can’t avoid both crossings: the trajectories of $(s_0,\theta_0)$, $(0,\hat \theta_0)$ will cross and can’t be on some caustics. So the caustics, (if exist) will detour when coming near the jumping point and accumulate to a bump over $\theta=0$, which violates the twisting+area-preserving property.

Open questions:

1. Are periodic orbits dense in the annulus for a smooth Birkhoff billiard? Are the set of periodic $n$ orbits not dense?

2. Does there exist a smooth convex billiards with positive Lyapunov exponents on a set of positive measure?

3. Is there a Birkhoff billiard with a caustic which is a fractal: a set with Hausdorff dimension between 1 and 2?

For example, the Bunimovich stadium are hyperbolic and ergodic whose underline table is $C^1$.

Birkhoff-Poritski conjecture: Let $Q$ be a convex billiard with $C^k$ boundary. If the billiard dynamics on $Q$ is integrable, then $Q$ is an ellipse.

This is a bundle of many problems, because it depends on the notion of integrability.

A special case has been treated by Misha Bialy, Convex billiards and a theorem by E. Hopf, Math. Z, 124 (1): 147–154, 1993:

Theorem: If $\Omega$ is continuously foliated by invariant curves which are not null-homotopic, then $Q$ is a round disk.

Recently a conditional version has been proved by Kaloshin and Sorrentino: On conjugacy of convex billiards

Theorem. If an integrable billiard is $C^2$ conjugate to an ellipse (resp. a circle) in a neighborhood of the boundary, then it is an ellipse (resp. a circle).

Guillemin problem: if the dynamics of two Birkhoff billiards are topologically conjugate, then their tables are similar.

Consider the trajectory $p(a)$ along the minor axis of the ellipse $\frac{x^2}{a^2}+y^2=1$ ($a\ge1$). Then

1) $p(1)$ is parabolic (not Lyapunov stable),

2) $p(a)$ is elliptic for $a\in(1,\sqrt{2})$,

3) $p(\sqrt{2})$ is parabolic (but Lyapunov stable),

4) $p(a)$ goes back to elliptic for $a>\sqrt{2}$.

Now let’s choose $b\in(0,1)$ as the parameter $x^2+\frac{y^2}{b^2}=1$. We can paste two different half-ellipses: say the upper/lower with $0. Then that orbit will be hyperbolic if $\displaystyle \frac{1}{b_1}>b_1+b_2>\frac{1}{b_2}$, which is true for all $b_1$ small and $b_2$ close to 1. Interesting domain is around the point $(b_1,b_2)=(1/\sqrt{2},1/\sqrt{2})$. In particular hyperbolicity holds for all $b_1<1/\sqrt{2} with $b_1+b_2=\sqrt{2}$.