# A toy model of heat death

Motivation:

This is a toy model of how a closed system will always evolve towards the distribution of maximal entropy, where no further transfer of heat/energy is possible.

Problem set up:

Fix a positive integer $$N$$, and denote by $$[N]$$ the set $$\{1, \dots, N\}$$.

Let $$\mathcal L := [N] \times [N]$$ be a 2D lattice.

We model a flow of heat as follows - Initially at time $$0$$, all heat is concentrated at $$(1, 1)$$. At each time step $$t$$, for $$t \in \mathbb Z_+$$, an element of $$\mathcal L$$ is picked uniformly at random. At that stage, it and all its immediate neighbours to its east, west, north and south average their heat.

Formally, we have a sequence of iid uniformly distributed $$\mathcal L$$-valued random variables $$\varepsilon_n$$, for $$n \in \mathbb Z$$, and a sequence $$X_n$$ of functions $$\Omega \times \mathcal L \to [0, 1]$$, defined as follows:

$$X_0 = \mathbf 1_{(1, 1)}$$, almost surely.

Assume now $$X_0, \dots, X_n$$ have already been defined.

Write $$X_n = \sum_{z \in \mathcal L} \lambda_z \mathbf 1_{z}$$ for (random) $$\lambda_z \in [0, 1]$$ with $$\sum_{x \in \mathcal L} \lambda_z = 1$$, and $$\mathcal N(\varepsilon_n)$$ for the set consisting of $$\varepsilon_n$$ and its two to four, depending on the location of $$\varepsilon_n$$, neighbours.

Then define $$X_{n+1} := \left[\underset{x \in \mathcal N(\varepsilon_n)}{\sum} \underset{y \in \mathcal N(\varepsilon_n)}{\sum} \frac{\lambda_y}{|\mathcal N(\varepsilon_n)|} \mathbf 1_x \right]+ \underset{z \in \mathcal L \setminus \mathcal N(\varepsilon_n)}{\sum} \lambda_z \mathbf 1_z$$.

where $$|S|$$ denotes the cardinality of a finite set $$S$$.

Question: Let $$Y$$ be the “uniform distribution of heat” given by $$Y := \underset{z \in \mathcal L}{\sum} \frac{1}{|\mathcal L|} \mathbf 1_z$$. Is it true that almost surely, $$X_n \to Y$$ uniformly?

Thus the system evolves almost surely toward a distribution where no further transfer of heat is possible.

• Do you know the Ehrenfest dog-flea model (en.m.wikipedia.org/wiki/Ehrenfest_model)? This has a similar flavor. It’s in the Ehrenfests’ book on the Conceptual Foundations of the Statistical Approach in Mechanics, which is old but readable and interesting. Oct 15 at 12:33
• This does look very similar! Though I think my model is significantly simpler - no birth death processes involved. Oct 15 at 13:00

The Ehrenfest model (in discrete time, for simplicity) is just a Markov chain with the finite state space $$\{0,1,\dots, N\}$$ and the transition probabilities $$p(k,k-1)=k/N, \quad p(k,k+1)=1-k/N$$ described by a single transition (averaging) operator $$P$$. Its stationary distribution $$m$$ is the binomial one with the parameters $$\frac12,N$$, and $$\frac12 \theta (P^n+P^{n+1}) \to m$$ for any initial distribution $$\theta$$. [Since the operator $$P$$ has period 2, one has to take the average of $$\theta P^n$$ and $$\theta P^{n+1}$$.]

In your situation there is a family of averaging operators $$P_x$$ indexed by the points from the state space $$X$$ which have a unique common invariant measure $$m$$ (the uniform distribution on $$X$$). You take a sequence $$\boldsymbol x=(x_1,x_2,\dots)$$ of iid $$X$$-valued uniformly distributed random variables, and ask whether, given an initial distribution $$\theta$$ on $$X$$, the sequence $$\theta P_{x_1} P_{x_2} \dots P_{x_n}$$ converges to $$m$$ almost surely. Note that since we are talking about measures on a finite state space, all reasonable kinds of convergence are equivalent (in particular, the $$\ell^1$$ convergence in the total variation norm $$\|\cdot\|$$ and the $$\ell^\infty$$ "uniform" convergence).

Let $$f_n(\boldsymbol x) = \| \theta P_{x_1} P_{x_2} \dots P_{x_n} - m \| \;.$$ The sequence $$f_n$$ is non-increasing, and therefore convergent. By Kolmogorov's 0-1 law its limit $$f_\infty$$ is almost surely constant. Let $$k$$ be the minimal number such that for any $$x\in X$$ there is a sequence $$x_1,x_2,\dots, x_k\in X$$ with $$\text{supp}\,\delta_x P_{x_1} P_{x_2} \dots P_{x_k} = X \;.$$ Then there is $$\varepsilon > 0$$ such that $$\mathbf E [ f_{n+k} | f_n ] \le (1-\varepsilon) f_n \qquad \forall\,n\ge 0 \;,$$ whence $$f_\infty=0$$.

EDIT. The fact that the sequence $$f_n$$ is non-increasing is a consequence of the following inequality: $$\frac1d \sum_{i=1}^d |\theta_i - C| \ge \left| \frac1d \sum_i \theta_i - C \right| \;.$$ Here $$d$$ is the cardinality of the averaging set (i.e., between 3 and 5 in your example), and $$C=1/N^2$$ is the common value of the weights of the uniform distribution. After removing $$C$$ and the division by $$d$$ the above inequality amounts to the well-known $$\sum_i |\theta_i| \ge \left| \sum_i\theta_i \right| \;.$$ The expectation bound is just a constructive version of this inequality: if $$f_n(\boldsymbol x)=F>0$$, then there are two points $$z_1,z_2\in X$$ such that $$\theta P_{x_1} P_{x_2} \dots P_{x_n}(z_i) - m(z_i) \qquad i=1,2\;,$$ have absolute values comparable with $$F$$ and opposite signs. Therefore by the definition of $$k$$ there is at least one choice of $$x_{n+1},\dots, x_{n+k}$$ with \begin{aligned} &\|\theta P_{x_1} P_{x_2} \dots P_{x_n+k} - m \| \\ \\ &< (1-\epsilon) \cdot \|\theta P_{x_1} P_{x_2} \dots P_{x_n} - m \| \;, \end{aligned} where $$\epsilon$$ is an appropriate constant (which only depends on $$N$$).

• Very slick solution, but I had a hard time following some of the details - namely why does $f_n$ need to be non increasing? Oct 15 at 15:00
• As it is I think this is a little sketchy of a proof. Oct 15 at 15:01
• I have added more details
– R W
Oct 15 at 16:27


Indeed, let $$\ze_n:=\ep_n$$. Let $$L:=\mathcal L$$ and let $$H$$ be the Hilbert space of all functions $$f\colon L\to\R$$ with the norm $$\|f\|:=\sqrt{\sum_{z\in L}f(z)^2}$$, so that $$\dim H=|L|=M:=N^2$$. Then the $$X_n$$'s may be viewed as random vectors in $$H$$. Moreover, for all $$n=0,1,\dots$$, $$$$X_n=P_{\ze_n}\cdots P_{\ze_1}X_0,\tag{1}$$$$ where, for each $$z\in L$$, $$P_z$$ is the orthogonal projector (o.p.) of $$H$$ onto the space of all functions $$f\in H$$ that are harmonic at $$z$$. We say that a function $$f\in H$$ is harmonic at $$z$$ if $$f(z)$$ coincides with the average of the values of $$f$$ at all neighbors of $$z$$. Let us also say that a function $$f\in H$$ is harmonic if its harmonic at every $$z\in L$$.

Remark 1: Every harmonic function $$f\in H$$ is constant. Indeed, such a function $$f$$ attains its maximum $$M$$ at some point $$z\in L$$. Then $$f(y)=M$$ for all neighbors $$y$$ of $$z$$. Since any point of $$L$$ can be connected to any other point of $$L$$ by a chain linking neighbors, we conclude that $$f$$ is the constant $$M$$ on $$L$$. $$\quad\Box$$

It follows from (1) that $$\|X_0\|\ge\|X_1\|\ge\cdots$$ and hence there is a (random) limit $$$$l:=\lim_n\|X_n\|^2. \tag{2}$$$$

Next, for each real $$\ep>0$$ there is some real $$\de(\ep)>0$$ such that the following implication holds: $$$$\|f\|\le1\ \&\ \max_{z\in L}\|Q_z f\|\le\de(\ep)\implies \|Qf\|<\ep, \tag{3}$$$$ where $$Q_z:=I-P_z$$, $$Q:=I-P$$, $$I$$ is the identity operator on $$H$$, and $$P$$ is the o.p. of $$H$$ onto the space of all constant functions in $$H$$. Indeed, otherwise there would exist some real $$\ep>0$$ and a sequence $$(f_k)$$ in $$H$$ such that $$\|f_k\|\le1$$, $$\max_{z\in L}\|Q_z f_k\|\le1/k$$, and $$\|Qf_k\|\ge\ep$$ for all $$k$$. By compactness, without loss of generality $$f_k\to f$$ for some $$f\in H$$, so that $$\max_{z\in L}\|Q_z f\|=0$$ and $$\|Qf\|\ge\ep$$, so that $$f$$ is a non-constant harmonic function, which contradicts Remark 1.

Note that for all $$n=0,1,\dots$$ $$$$1\ge\|X_n\|^2\ge \|PX_n\|^2=1/M, \tag{4}$$$$ $$$$\|X_n\|^2-\|X_{n+1}\|^2=\|Q_{\ze_{n+1}}X_n\|^2. \tag{6}$$$$

On the event $$\{l>1/M\}$$, for all $$n$$ we have $$\|QX_n\|^2=\|X_n\|^2-\|PX_n\|^2=\|X_n\|^2-1/M\ge d:=l-1/M>0$$ and hence, by (3), $$$$\max_{z\in L}\|Q_z X_n\|>\de_*:=\de(\sqrt{d})>0. \tag{8}$$$$ So, there is a sequence $$(g_n)$$ of (deterministic) functions $$g_n\colon L^n\to L$$ such that on the event $$\{l>1/M\}$$
$$$$\|Q_{Z_n} X_n\|>\de_*>0 \tag{9}$$$$ for all $$n$$, where $$Z_n:=g_n(\ze_1,\dots,\ze_n)$$.

For natural numbers $$n,m$$, let
$$$$p_{n,m}:=P(\ze_{n+1}\ne Z_n,\dots,\ze_{n+m+1}\ne Z_{n+m}). \tag{10}$$$$ Then, conditioning on $$\ze_1,\dots,\ze_{n+m}$$, we get $$p_{n,m}=(1-1/M)p_{n,m-1}$$ if $$m\ge2$$ and hence $$p_{n,m}\le(1-1/M)^m\to0$$ as $$m\to\infty$$. It follows that the events $$\{\ze_{n+1}= Z_n\}$$ occur for all natural $$n$$ in a random set $$\nu$$ which is infinite almost surely (a.s.).

Thus, in view of (6) and (9), on the event $$\{l>1/M\}$$ we a.s. have
$$$$1=\|X_0\|^2\ge\sum_{n=0}^\infty(\|X_n\|^2-\|X_{n+1}\|^2) \\ \ge\sum_{n\in\nu}(\|X_n\|^2-\|X_{n+1}\|^2) \\ =\sum_{n\in\nu}\|Q_{\ze_{n+1}}X_n\|^2 \\ =\sum_{n\in\nu}\|Q_{Z_n}X_n\|^2 \\ \ge \sum_{n\in\nu}\de_*=\infty.$$$$ Thus, the event $$\{l>1/M\}$$ has probability $$0$$. That is, $$l=1/M$$ a.s. So, $$\|X_n\|^2\to1/M$$ a.s. So, $$\|X_n-PX_n\|^2=\|X_n\|^2-\|PX_n\|^2=\|X_n\|^2-1/M\to0$$ a.s. Since $$PX_n$$ is the constant function equal $$1/M$$, we have proved the desired result: $$X_n\to1/M$$ a.s.

• Wow this looks impressive.. I will go through this in detail tomorrow. Oct 15 at 17:01

After t steps, Let $$S_t=S(X_t) =\sum_{ z \in L} \lambda_z(t)^2] - 1/N^2=\sum_{ z \in L} [\lambda_z(t) -1/N^2]^2 \,.$$, Then $$S_t$$ is nonincreasing, so it must converge to some limiting random variable S*. Moreover, if $$S_t>0$$ then with positive probability $$S_{t+1}<(1-c)S_t$$ for some $$c=c(N)$$. This implies that that $$S_*=0$$ almost surely, whence $$\lambda_z(t)$$ must tend to $$1/N^2$$ almost surely for each $$z$$.

Details: If $$S_t=a^2$$, then there must exist vertices $$z,w$$ such that $$[\lambda_z(t)-\lambda_w(t)] \ge a/N$$, so there must exist two adjacent vertices $$u,v$$ such that $$[\lambda_u(t)-\lambda_v(t)] \ge a/(2N^2)$$. If $$u$$ or $$v$$ is the next vertex selected, then $$S_{t+1} \le S_t-a^2/(8N^4)$$. In other words, given the history up to time $$t$$, with probability at least $$2/N^2$$, we have $$S_{t+1} \le S_t [1-1/(8N^4)]$$. If $$S*>0$$ with positive probability, then on this event, find a random $$t$$ such that $$S_t and then wait until the first time one of $$u,v$$ is selected to get a contradiction.