**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.



A toy model of heat death
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 $N$).


