# A Simple Stochastic Dynamic Billiard

Consider the following stochastic dynamical system.

Fix $a > 0$, $b > 0$, and $v > 0$, and let $\mathbf{r}(t)=(x(t),y(t))$ be the position at time $t$ of a point which moves in the rectangle $R=\{ (x,y) \in \mathbb{R}^2: 0 \leq x \leq a, 0 \leq y \leq b \}$ with velocity of constant magnitude $v$ a according to the following rules:

(i) in the interior of $R$ the point is subject to no force, so that is moves with constant velocity $(v_x(t), v_y (t)) = \mathbf{v}(t)=\frac{d\mathbf{r}}{dt}(t)$;

(ii) when the point reaches the vertical side of $R$ corresponding to $x=0$ it is reflected diffusely, that is its velocity after the collision has always magnitude $v$, and the oriented angle $\theta$ that the versor $\mathbf{i}=(1,0)$ makes with the velocity $\mathbf{v}$ after the collision is uniformly distributed over $[-\pi/2,\pi/2]$;

(iii) if the point reaches any other side of $R$, then it is reflected elastically, that is in the case of the vertical side corresponding to $x=a$ the $y$-component of $\mathbf{v}$ is preserved, while the $x$-component of $\mathbf{v}$ changes sign, while in the case of either of the two horizontal sides the $x$-component of $\mathbf{v}$ is preserved, while the $y$-component of $\mathbf{v}$ changes sign;

(iv) finally, if the point reaches one of the vertices of $R$, then the reflection must be thought as the combination of two reflections by the the two edges involved; so e.g. if the point hits the vertex $(0,0)$ then it is reflected diffusely, that is its velocity after the collision has always magnitude $v$, and the oriented angle $\theta$ that the versor $\mathbf{i}=(1,0)$ makes with the velocity $\mathbf{v}$ after the collision is uniformly distributed over $[0,\pi/2]$; if instead the point hits e.g. the vertex $(a,0)$ then its velocity $\mathbf{v}$ before the collision is changed in $-\mathbf{v}$.

Consider a time $T> 0$, and let $N(T)$ be the number of time the point touches the vertical side of $R$ corresponding to $x=a$ (you can compute or not in $N(T)$ the times the point touches one of the vertices of this side: it should make no essential difference for what we want to prove, I think). Let $\theta_j$ be the angle that the velocity $\mathbf{v}$ just before the collision makes with $\mathbf{i}$ the $j$-th time the point touches the vertical sides of $R$ corresponding to $x=a$ (or one of its vertices, if you have considered also them in the computation of $N(t)$), and form the random sum $$\sum_{j=1}^{N(T)} cos \theta_j.$$ I would like to prove that for any initial conditions $(\mathbf{r}(0), \mathbf{v}(0))$, with $v_{x}(0) \neq 0$, the following equation holds $$\lim_{T \rightarrow \infty} \frac{1}{T} E \left[ \sum_{j=1}^{N(T)} cos \theta_j \right] = \frac{v}{4 a}.$$

Any help is welcome. For now, I have no idea about a possible proof.

NOTE. This problem has been suggested to me by an interesting proof of a physical law called Wien's displacement law given by Richtmyer, Kennard and Cooper in their book "Introduction to Modern Physics", Sixth Edition, Appendix of Chapter 5 (see in particular p. 145). I formulated a first model of dynamic stochastic billiard in my post Stochastic Dynamic Billiard, but it was considerably more complicated. Then I realized that the simplest model above could be a good formulation of the physical system as well. In the obvious three-dimensional version of this model, in which the point moves inside a parallelepiped, whose face corresponding to $x=0$ reflect diffusely (that is $v$ remains unchanged and $\mathbf{v}/v$ is uniformly distributed over a hemisphere) while all other faces (and all the edges and vertices) reflect elastically, we should get

$$\lim_{T \rightarrow \infty} \frac{1}{T} E \left[ \sum_{j=1}^{N(T)} cos \theta_j \right] = \frac{v}{6 a},$$

which has a deep thermodynamical meaning (see the reference above and also ter Haar and Wergeland, Elements of Thermodynamics, $\S 5.3.3$).

• Your random reflection law is somewhat odd. It is most usual in this type of problems (especially with a physics motivation) to use the Knudsen law (replace uniform out angle by the law with density proportional to the cosine of the angle with the normal). The reason is that Knudsen law makes the Liouville measure (i.e. uniform measure on pairs (position, direction) stationary (as does deterministic elastic reflection). There are many works on such random billiard. – Benoît Kloeckner Nov 10 '17 at 21:32
• This choice of reflection law seems to be the problem causing Iosif Pinelis negative answer to your question. By putting more weight toward $-\pi/2$ and $\pi/2$ than a typical gas inside your domain would do, you are creating many very steep trajectories, which take a very long time to hit the $a$-wall. By the way, having had a quick look at your other question, the Liouville measure ($dx dy d\theta$ up to normalization) is certainly not invariant (let alone ergodic) if you use a uniform reflection angle. – Benoît Kloeckner Nov 10 '17 at 21:40
• Last remark: you can reduce to a band $\{0\le x \le a\}$ up to some harmless change of sign in half the angles, by unfolding the rectangle by the symetries with respect to the horizontal sides. This simplifies a bit the model but has no effect on the limit you are interested in. – Benoît Kloeckner Nov 10 '17 at 21:43
• @BenoîtKloeckner You are absolutely right: the problem is the law of reflection. I have no experience in dynamic billiards (this is not my field of research): I would be very grateful to you if you could give me some reference on the stochastic models with Knudsen law you quoted. In any case thank you very very much for your insightful comments on this problem. – Maurizio Barbato Nov 11 '17 at 9:21
• @BenoîtKloeckner Dear Benoît, I have emended my model according to your suggestions and I have created a new post A Really Simple Stochastic Dynamic Billiard in order to clarify definitively the matter. Iosif Pinelis's argument below shows that my conjecture holds in this new model. I kindly invite you and Iosif Pinelis to answer my new post, since you deserve the merit of the right formulation of the model. – Maurizio Barbato Nov 11 '17 at 11:02

To avoid trivialities, assume that the initial position of the particle is in the interior of the rectangle. Then the limit is actually $0$. This is trivial in the case when the $x$-coordinate of the initial velocity is $0$.
Otherwise, in a nonrandom finite time, depending only on the initial position and velocity, the particle will hit the side $x=0$. So, without loss of generality let us assume the movement begins right after that hitting time, with the initial velocity of magnitude $v$ at a random angle $\theta_1\in[-\pi/2,\pi/2]$ with the $x$-axis. Note that the horizontal coordinate of the velocity may change in magnitude only right after hitting the side $x=0$; except for this, it may only change in sign right after hitting the side $x=a$. Also, the cosine of the angle between the velocity and the $x$-axis does not vary while the particle is traveling from the side $x=0$ toward the side $x=a$. Therefore, introducing $Y_j:=\cos\theta_j$, $X_i:=\frac{a}{vY_i}$ -- the $i$th traveling time from the side $x=0$ to the side $x=a$, $T_j:=\sum_1^j X_i$, $N_t:=\sum_{j=1}^\infty I\{T_j\le t\}$ (with $I\{\cdot\}$ denoting the indicator function), we see that $N(t)\le N_t$ for real $t>0$; here, we understand $N(t)$ as the number of times the particle hits the vertical side $x=a$ before or at time $t$. So,
$$\sum_{j=1}^{N(t)}\cos\theta_j\le R_t:=\sum_{j=1}^{N_t}Y_j.$$ By the Renewal Reward Theorem, we have $$\frac1t\,ER_t\underset{t\to\infty}\longrightarrow\frac{EY_1}{EX_1}=\frac{2/\pi}\infty=0.$$ Thus, $$0\le\frac1t\,E\sum_{j=1}^{N(t)}\cos\theta_j\le \frac1t\,ER_t\underset{t\to\infty}\longrightarrow0.$$
• Indeed. The situation gets a bit better (and more trivial) if you sum $\frac 1\cos\theta_j$ (in that case you just need to show is that it cannot happen that a single round takes a noticeable portion of the total elapsed time infinitely often), but I have no idea where $3$ in the denominator for the 3D box can possibly come from... – fedja Nov 10 '17 at 20:41
• As I had had no experience working on such problems, the clarity came to me only in the hindsight, alas: that, because $EX_1=\infty$, hits of $x=a$ will be just too infrequent for a positive reward limit. – Iosif Pinelis Nov 10 '17 at 21:23