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I was surprised to learn from two Mathematica Demos by Enrique Zeleny that an elastic ball bouncing in a V or in a sinusoidal channel exhibits choatic behavior:
(The Poincaré map is shown above the red ball trajectories.) I wonder:

Q. Are there curves (other than a horizontal line!) for which such a ball drop (from some range of horizontal placements $x$ and vertical heights $y$) is not chaotic? In other words, are the V and sine wave special? Or (more likely) is it that non-chaotic curves are special?

(I cannot now access Zeleny's citation to Chaotic Dynamics: An Introduction Based on Classical Mechanics.)

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There appears to be some similarity between this and the logistic map x(1-x)alpha. Perhaps one could redraw the fixpoint line y=x as y=sin(x) or something similar and get a correspondence to your picture with the sinusoid. – The Masked Avenger Aug 16 '13 at 16:02
Wouldn't this be a reformulation of your "billiards under gravity" question ? Or do you imply that the curve is a function of x ? – Piyush Grover Aug 16 '13 at 21:39
@nonlinearism: First, there is no gravity in this elastic bounce (although one could add that). Second, I am seeking curves as a function of $x$ (without left and right box constraints)---as you suggest---that would not lead to chaotic bounces. – Joseph O'Rourke Aug 16 '13 at 23:35
If there is no gravity, what is pushing the ball down ? I see parabolas.. – Piyush Grover Aug 16 '13 at 23:38
A vertical line is a reflector, so a square well behaves like a horizontal line. But this could still be a very special case. – Noam D. Elkies Aug 17 '13 at 3:16
up vote 11 down vote accepted

At least one integrable case is known, the wedge billiard with an angle of $45^\circ$, see the first paper on this system:

H.E. Lehtieht and B.N. Miller, Numerical study of a billiard in a gravitational field, Physica D 21 93-104 (1986)

But, as usual in dynamical systems, completely regular cases are very special, so other examples are probably few/rare.

A link to the previous question: Billiard dynamics under gravity: a further beautiful example is given by Robert Israel.

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A possibly new example: a ball bouncing on a parabola ($=$ graph of a quadratic polynomial) is never chaotic. The associated dynamical system has an extra invariant that makes it "integrable" (if I have the terminology correctly). The general orbit is an elliptic curve $\cal E$ equipped with a point $P$ such that going from one bounce to the next corresponds to translation by $P$ on $\cal E$. The general orbit looks like one of these two pictures, depending on whether the curve has one or two real components:


In the two-component case, $P$ is on the non-identity component. So in either case the orbit is dense in $\cal E$, unless $P$ is torsion in which case the orbit is finite. The envelope curves are themselves parts of two parabolas, which meet at right angles in the one-component case. In the transition between the two cases the lower parabola narrows to a vertical line that meets the upper parabola at its center, as suggested by the next two pictures on either side of the transition:


At the transition point itself the ball quickly approaches a fixed point where it bounces vertically on the parabola's vertex.

There are two other limiting cases. In one direction, the ball bounces very close to the surface, and in the limit rolls on it:


Here $\cal E$ remains smooth but $P$ approaches $0$. In the other limiting case, $\cal E$ curve degenerates to two points related by translation by $P$; the second picture shows the one-parameter family of degenerate trajectories of this kind:


There's more to say here than is reasonable for a Mathoverflow answer, but let me at least get to the the derivation of the extra invariant and the elliptic curve.

The general setting can be described as follows. Choose units so that the gravitational constant is $1$, and assume that the ball bounces on the graph $(x, f(x)/2)$ of a differentiable function (the factor of $1/2$ will be convenient later). We further assume that $f$ is bounded below, or at least that $f(x) > -cx^2$ for all $c>0$, so that the ball always keeps hitting the graph. Then ball's path is completely described by a sequence of vectors $v_n = (x_n, y_n, z_n)$ where the $n$-th bounce occurs at $(x_n, f(x_n)/2)$ and shoots the ball with velocity $(y_n,z_n)$, with $z_n \geq f'(x_n) y_n/2$ so that $(y_n,z_n)$ is in the half-plane above the tangent to the graph at $(x_n, f(x_n)/2)$. This puts the ball on the path $(x_n + y_n t, \frac12 f(x_n) + z_n t - \frac12 t^2)$ with $t \geq 0$, so $x_{n+1}$ is the value of $x_n + y_n t$ at the smallest positive solution of $$ f(x_n) + 2 z_n t - t^2 = f(x_n + y_n t). $$ At that time the ball has velocity $(y_n, z_n - t)$. The new velocity vector $(y_{n+1}, z_{n+1})$ is then obtained by reflecting $(-y_n, t-z_n)$ about the normal to the graph at $(x_{n+1}, f(x_{n+1})/2)$, which is the line spanned by $(-\frac12 f'(x_{n+1}, 1)$ (orthogonal to the tangent $(1, \frac12 f'(x_{n+1})$).

We're thus iterating a map $\beta: H \rightarrow H$ where $$ H = \{ (x_n, y_n, z_n) \in {\bf R}^3 : z_n \geq f'(x_n) y(n) / 2 \}, $$ and can factor $\beta$ as a product $\rho \sigma$ of involutions where $\sigma$ takes $(x_n, y_n, z_n)$ to $(x_{n+1}, -y_n, t-z_n)$ and $\rho$ fixes $x$ and reflects $(y,z)$ about $(-\frac12 f'(x),1)$. Time reversal corresponds to the inverse map $\beta^{-1} = \sigma \rho$. Here's a picture:


Black arrows are normal to the graph; blue tangent arrows $(y,z)$ coming from $(x,f(x)/2)$ represent the point $(x,y,z)$; the two arrows coming from the same point are related by the reflection $\rho$; and the two arrows at either end of a parabolic segment are related by $\sigma$.

By conservation of energy, $\sigma$ preserves $$ E(x,y,z) := f(x) + y^2 + z^2 $$ (that's why I wanted the factor of $1/2$ in $(x,f(x)/2)$); since the collisions are perfectly elastic, $\rho$ preserves $E$ as well. Moreover $\rho$ and $\sigma$, and thus also $\beta$, are volume-preserving bijections. So we have a non-dissipative dynamical system on each of the $2$-dimensional fibers of $E$.

Now suppose $f$ is quadratic. We translate and scale our coordinates (including the time coordinate $t$) to put the focus of the parabola at $(0,\frac12)$ while keeping the gravitational constant equal $1$. Then $f(x) = x^2$, and $E(x,y,z)$ is simply $x^2+y^2+z^2$; and $\frac12 f'(x)=x$, so $\beta$ moves $v = (x,y,z)$ around the part of the sphere of radius $E^{1/2}$ on which $z \geq xy$. The equation $f(x_n) + 2 z_n t - t^2 = f(x_n + y_n t)$ for the time between bounces $n$ and $n+1$ is now the quadratic equation $x_n^2 + 2 z_n t - t^2 = (x_n + y_n t)^2$. This equation has the known solution $t=0$, so the other solution must be a rational function of $x_n,y_n,z_n$. Thus $\sigma$, like $\rho$, is a rational map. We find that $t = 2(z_n - x_n y_n) / (1+y_n^2)$, and thus that $\sigma$ takes $(x,y,z)$ to $$ \left( \frac{(1-y^2)x + 2yz}{1+y^2}, \phantom. -y, \phantom. \frac{(1-y^2)z-2yx}{1+y^2} \right). $$ This resembles the formula for $\rho$ for good reason: while $\rho$ fixes $x$ and reflects $(y,z)$ about the line generated by $(-x,1)$, $\sigma$ fixes $y$ and reflects $(x,z)$ about the line generated by $(-y,1)$, then multiplies both $x$ and $y$ by $-1$.

It follows that both $\rho$ and $\sigma$ leave $z-xy$ invariant, as may be seen either by direct computation or by noting that $z-xy$ can be written as an inner product in two ways $$ z - xy = \langle (y,z), (-x,1) \rangle = \langle (x,z), (-y,1) \rangle $$ and that reflecting one vector about the span of another does not change their inner product. Therefore $\beta$ also preserves $z-yx$. It follows that the vectors $(x_n,y_n,z_n)$ are all on the intersection of a sphere $x^2+y^2+z^2 = E$ with another quadric $z-yx = F$, which for general $E,F$ (and thus for general initial conditions $(x_0,y_0,z_0)$) gives a genus-$1$ curve ${\cal E} = {\cal E}_{E,F}$. Finally, since $\rho,\sigma$ are rational involutions on ${\cal E}$ that are not translations, $\beta=\rho\sigma$ is translation on ${\cal E}$ by some point that depends on $E$ and $F$, and we're done.

It seems reasonable to guess that one can prove, for any smooth $f$ with a local minimum, that the bouncing-ball dynamical system will be non-chaotic for $E$ close enough to the minimum, by deforming the second invariant $z-yx$ of the approximating parabola; but I'm too far from my expertise here to carry out such a proof.

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I should have Googled the wedge solution first: seems that the parabolic example was described in a paper A new integrable gravitational billiard by H.J.Horsch and J.Lang, J. Phys. A 24 (1990), p.45 = . Some of the geometry I described here might still be new, though. – Noam D. Elkies Aug 25 '13 at 2:23
This seems very closely related to Poncelet's porism: do you know if there is a real connection? – M P Aug 25 '13 at 5:22
Wow! Beautiful analysis, decorated with graceful, gorgeous figures! Could almost be published as-is. – Joseph O'Rourke Aug 25 '13 at 13:23
@M P: I was wondering the same thing. The algebraic structure is there but I didn't see a natural translation that explains the coincidence. $$ $$ @Joseph O'Rourke: Thanks! I hope there will be a venue for publishing about this even though the system's integrability has been shown already; as I wrote, there are a few more things of interest here that I felt would be too much for a MO answer. I'll mention one or two of them without proof in the next edit, though. – Noam D. Elkies Aug 25 '13 at 14:59
@John It turns out that with Newtonian gravity other conics work - see my second answer. – Carl Aug 30 '13 at 11:13

Just to verify Carl Dettmann's claim (in his answer) concerning the $45^\circ$ "wedge billiard," here is an example using that same Mathematica code I cited, showing a structured Poincaré map when the wedge sides have slope $1$:
   45 deg

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This is really a collection of comments to Noam's wonderful parabola example, but a bit long...

The set of known integrable ordinary (non-gravitational) billiards consists of pieces of confocal conics, or their limits (straight lines). The internal angles are all 90 degrees apart from some exceptions I will ignore henceforth: The 60-60-60 and 30-60-90 triangles and circular wedges. The most common examples are the circle and ellipse, but also shapes such an elliptical and a hyperbolic arc, such that the two have the same foci.

In the case of a single conic, the equation to obtain the location of the next collision is linear: The intersection of a straight line and a conic is a quadratic equation, but we can divide by the known solution given by the initial point.

Remark: This feature also holds for similarly oriented parabolas, and hence also the parabola gravitational billiard.

Question: Is a gravitational billiard consisting of two confocal parabolic arcs also integrable?

Noam also conjectures that gravitational billiards with local minima cannot be chaotic: I have a recent preprint including some examples (circle, ellipse and oval gravitational billiards): Indeed, the low energy orbits appear regular. There is also interesting higher energy dynamics: We also ask (with some numerical justification)

Conjecture: Are there any energies for which these gravitational billiards are ergodic?

Addendum: A generalisation of Noam's example is that of Newtonian gravity (ie inverse square force toward a centre), with the parabola replaced with a conic with a focus at the centre.

Newtonian billiard

Here the gravitational attraction is to the black dot, billiard boundary is one branch of a hyperbola with this as focus, and the caustics are arcs of an ellipse and a hyperbola confocal with the boundary hyperbola. As with the parabola, the caustics have a nice corner despite the smooth force field and billiard boundary.

Publishable? Alas it is a trivial consequence of the conclusion of Horsch and Lang. The second constant of motion for the two centre problem is preserved by billiard collisions with a confocal boundary. So, I suspect not at present, though would love to be contradicted...

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Both Joseph O'Rourke's 45° wedge and Noam Elkies parabola are limiting cases of conic sections with Newtownian gravity. The wedge is a degenerate conic. Constant gravity is the limit as the focus moves far away. The foliations and algebraic invariants are still of interest. – john mangual Aug 30 '13 at 14:23
@John Thanks. I agree about the parabola. The wedge is the limit of one branch of a hyperbola, but with both foci limiting to the corner, not to infinity. So I can't see how it follows as a limit of a conic with Newtonian gravity. – Carl Aug 30 '13 at 16:19
A further reference on integrable quadric billiards with potentials: V. Dragovic and B. Jovanovic, J. Math. Phys. 38, 3063-3068 (1997). – Carl Sep 4 '13 at 14:39

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