Is 2D rectangular billiard an integrable system? What's the form of explicit solution?

Question

Suppose the free particle moving inside 2D box. $$H=\frac{p_x^2}{2m}+\frac{p_y^2}{2m}+V$$ where the potential is zero inside the box and infinite outside the box.

It's clear that $p_x,p_y$ are not constant of motion but $p_x^2,p_y^2$ are. And these are in involution with each other so the system should be integrable.

How do you go about solving the problem? Can we write the explicit form of the solution? \begin{align*} \dot{x}&= \frac{p_x}{m}\ \ \&\ \ \dot{y}= \frac{p_y}{m}\\ \dot{p_x}&= -\partial_x V(x,y),\ \ \&\ \ \dot{p_y}= -\partial_yV(x,y) \end{align*} I don't How do you go about writing an explicit solution of these?

Answer 1

1. 2D rectangular billiard is Liouville integrable with 2 independent Poisson commuting integrals of motion $p_x^2$ and $p_y^2$. (They are preserved by the reflection conditions on the boundary. Moreover, $\vec{p}$ is preserved in the bulk.)

2. The solution is of the form $$ \begin{align} x(t)~=~& \frac{L_x}{2} \triangle \left(\frac{t-t_x}{T_x}\right) , \cr y(t)~=~& \frac{L_y}{2} \triangle \left(\frac{t-t_y}{T_y}\right) , \cr p_x(t)~=~& \frac{mL_x}{2T_x} \Box \left(\frac{t-t_x}{T_x}\right) , \cr p_y(t)~=~& \frac{mL_y}{2T_y} \Box \left(\frac{t-t_y}{T_y}\right) , \end{align} $$ where $\triangle$ and $\Box$ denote the triangle wave and square wave, respectively.

3. Interestingly, 2D rectangular billiard is never a superintegrable system, even if the side lengths $L_x/L_y\in\mathbb{Q}$ are commensurable.

Answer 2

I'm not sure I understand your argument for why the system should be integrable. I think that in general the system is not integrable. If you look at Wikipedia's article on "Dynamical billiards" and its references, you'll see that it depends on the shape of the box: some special cases are integrable but in general the system is not integrable and can be chaotic.