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Imagine I have a point-like Brownian particle, with diffusion constant $D$, and I place it at some initial coordinate in a cage of known geometry. Assuming the volume $V$ of the cage is "everywhere" accessible, how long does it take the particle to lose its "memory" of the initialization coordinate and achieve a uniform probability distribution across the cage? How precisely can one measure this, and how important is the geometry of the cage?

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You should be able to study this by setting up the heat diffusion equation with appropriate reflecting boundary conditions. The geometry of the cage should be pretty important. Intuitively, if your cage looked like a "string of beads", it would be hard for your particle to pass from one bead to another. If your cage was convex, you would expect to see a uniform distribution much more quickly. See Gardiner, handbook of stochastic methods, chapter 5. – Simon Lyons Aug 10 '11 at 16:33

Let $\rho(t,x)$ be the probability that the particle is at location $x$ at time $t$. $\rho$ satisfies the equation $\rho_t = \rho_{xx}$, with $\rho(0,x)=\delta(x-x_0)$ where $x_0$ is the starting position. You have Neumann boundary conditions on the boundary of your domain. Suppose the eigenfunctions of the negative Laplacian of your domain with Neumann boundary conditions are $\phi_0, \phi_1, \ldots$ with corresponding eigenvalues $\lambda_0 \leq \lambda_1 \leq \lambda_2 \cdots$. $\phi_1$ is constant with $\lambda_0=0$. The solution to the equation is $$ \rho(t,x)= \sum_{j=0}^\infty c_j e^{- \lambda_j t} \phi_j (x) $$ where $c_j$ is the inner product of $\delta(x-x_0)$ with $\phi_j$ which gives $c_j=\phi_j(x_0)$. When $t$ goes to infinity you just get a constant for $\rho$. Your question is (I think) equivalent to asking how long does it take for all the $j>0$ terms to die out. Assuming $c_1 \neq 0$, this will be determined by $\lambda_1$; a bigger $\lambda_1$ means faster approach to uniform probability. There is not going to be an explicit formula for $\lambda_1$ for most domains. But you can get some intuition. As Alice says, if you have two equal volumes with a narrow connection, then $\lambda_1$ is going to be very small and it will take a long time for $\rho$ to be flat. (Unless you start exactly midway between the two volumes. This corresponds to $c_1=0$.)

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Seems like the timescale should depend on the number and sizes of spherical balls it takes to cover the entire space $V$. For example if your space has a narrow neck connecting two balloons then it would take a while for your particle to diffuse from the balloon where it started through the neck to the other side.

I would have entered as a comment but can't yet (sorry).

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@Alice, I'm assuming my particle is point-like, so I'm not sure if I understand the spherical ball suggestion. However, I think it's a good point that geometry of the cage matters - if you divide the cage into two equal volumes with a narrow connection, it seems likely to me that it might take longer for the probability distribution of the particle to become uniform throughout the space. – Rob Grey Aug 10 '11 at 15:29
@Rob The timescale to diffuse across a region is related to distance. In an infinite medium the distribution after time $t$ is a ball of radius $r(t) \sim \sqrt{Dt}$ (I am ignoring factors dependent on dimension). If you have a particle initially in the center of a balloon of radius $R$, for $t<<R^2/D$ then the expression for $r(t)$ is good. When $r(t)$ gets closer to $R$ then the boundary plays a role and the expression is no good. But $R^2/D$ gives you a handle on the time it takes to be smooth in your balloon. – Alice Aug 10 '11 at 16:56

For the initial distribution being delta function $\delta(y)$ the result will be the Green function of the Neumann problem for the heat equation. If the domain is regular enough the function can be written as $$ G(x,y,t)=\sum_{n=1}^\infty \varphi_n(x)\varphi_n(y)e^{-\lambda_n t} $$ where $\lambda_n$ and $\varphi_n$ are eigenvalues and normed in $L_2$ eigenfunctions of the corresponding elliptic Neumann problem. The first eigenvalue $\lambda_1=0$. So the rate of convergence to constant is exponential and determined by the next value $\lambda_2>0$.

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