# Brownian motion in a box

It is often said that the Brownian motion for a particle in a box, thus a finite domain, is described by a uniform probability distribution in the longtime limit. One may easily imagine this maybe intuitively, but is there actually an easy way of showing this? How is it that even though there are boundaries, no bias is introduced in the longtime limit of where the particle may be in the box?

If one assumes the discrete case, so the box is populated with only a finite number of points the Brownian particle can occupy, does the above statement still hold? Please feel free to give references that you see fit for such questions, I imagine these are all solved problems and that's why most people often state them using the adverb trivially.

• I assume it's just noticing that since we have about $10^{23}$ particles at hand, the number of particles hitting the wall is about $(10^{23})^{2/3}$, i.e. only a fraction of order $10^{-8}$ of all particles, which is probably a much smaller effect than those introduced by other assumptions of the theory (e.g. no rotational degrees of freedom etc.). – Lukas Berns Jun 13 '17 at 11:51
• nature.com/articles/ncomms9558 – user126422 Jun 13 '17 at 12:15
• I assume that you are discussing the Wiener process and the simple random walk, not more realistic models of (real) Brownian motion. If yes, then relevant keywords are: reversible Markov chain, invariant measure, random walk on a graph (for the discrete case); Ito diffusion, invariant measure, reversibility, gradient flow (for the continuous case). – Yvan Velenik Jun 13 '17 at 13:52
• @YvanVelenik Exactly. Do you personally happen to know of relevant references tackling such questions? – user929304 Jun 13 '17 at 14:40
• It very much depends on your background (basically any probability textbook dealing with Markov chains and/or diffusions will do). A simple reference for the Markov chain part would be Finite Markov Chains and Algorithmic Applications (see Example 6.1 therein). The continuous setting (diffusions) is technically more demanding, so it depends on what you know/want to know. – Yvan Velenik Jun 13 '17 at 15:16

I decided to run a simple simulation of a one-dimensional random walk on lattice with closed boundary condition. To implement the boundary conditions, I basically forced to "turn back" the particle every time it hits the boundaries of the segment.

I took every lattice site to be an integer, and the boundaries to be at $[-L,L]$. This is the result for $L=50$ and $10^8$ steps: So, it looks like in this simple lattice model the distribution is uniform everywhere except from the very boundaries of the segment, i.e. it is uniform in $(-L,L)$.

I don't know if this still holds in a continuous model, also because it becomes more tricky to do the simulation in this way, because the particle can never really hit the boundary, so it is possible that the result depends on how the boundary conditions are implemented exactly.

Relevant literature

This problem is commonly known as reflected brownian motion (RBM). Many articles can be found regarding RBM in open regions (like $[0,\infty)$), but we are more interested in RBM in closed regions.

The Stationary Distribution of Reflected Brownian Motion in a Planar Region - . M. Harrison, H. J. Landau, L. A. Shepp

It gives an explicit -and rather complicated- expression for the stationary probability distribution. The article is quite technical, but from what I understand the stationary distribution is not uniform.

• Thanks for your efforts again. I have accepted your answer as it addresses the question directly, by numerical means and provides interesting literature references. By the way, eager to hear your take of alarge's comment above. – user929304 Jun 28 '17 at 12:26

I suspect there is a natural way to treat the general case, but I will only deal with $$1$$ boundary condition. Say you are in a the one dimensional box $$[-L,L]$$, (we will see in a moment that the same argument works in the $$n$$ dimensional box). Brownian motion has infinitesimal generator equal to $$1/2$$ times the Laplacian. This is equivalent to saying that if we begin our Brownian motion with probability distribution $$\mu$$, then for any bounded continuous function $$f$$, the function $$u_t(x)=E_{\mu}[f(B_t)]=\int f(x) P(B_t=x),$$ solves the equation $$\partial_t u=\frac{1}{2} \Delta u.$$ The subscript $$\mu$$ on the expectation means the Brownian motion $$B_t$$ at time zero has probability distribution $$\mu$$, it may be simplest to think of $$\mu$$ being a delta function.

We can formally write a solution to the differential equation which $$u$$ satisfies as $$u_t(x)=e^{\frac{t}{2} \Delta} u_0(x),$$ where we are interpreting $$e^{\frac{t}{2} \Delta}$$ as the power series expansion $$e^{\frac{t}{2} \Delta}=\sum_{k=0}^{\infty} \frac{(t \Delta/2)^k}{k!}.$$ We expect this to work because formally differentiating $$\partial_t e^{\frac{t}{2} \Delta}$$ gives $$\frac{1}{2} \Delta$$. If you are rightfully suspicious of this formal differentiation you can check directly that $$\partial_t \sum_{k=0}^{\infty} \frac{(t \Delta(u)/2)^k}{k!}=\frac{1}{2}\Delta u.$$ Dealing with this exponential seems tricky on first sight so lets diagonalize the Laplacian keeping in mind that we can exponentiate a diagonal matrix by just exponentiating each entry. Lets consider the boundary conditions such that $$\partial_x u(L)=\partial_x u(-L)=0$$ i.e. force the derivative to be $$0$$ on the boundary. Then the Laplacian has eigenfunctions $$\psi_{2n}(x)=\cos\left(\frac{(2n)\pi}{2 L} x\right)$$ with eigenvalues $$\lambda_{2n}=-\left(\frac{(2n)\pi}{L}\right)^2,$$ and $$\psi_{2n+1}(x)=\sin\left(\frac{(2n+1) \pi}{2L}x\right)$$ with eigenvalues $$\lambda_{2n+1}=-\left( \frac{(2n+1) \pi}{L} \right)^2$$ for each $$n \geq 0$$. Note that every one of these eigenvalues is negative except for the eigenvalue of the constant function $$\psi_0(x)$$, which is $$0$$.

Fourier analysis tells us that we can write any reasonable function $$u_0(x)$$ on the interval $$[-L,L]$$ whose derivative at the boundary is $$0$$ as a sum $$u_0(x)=\sum_{n \geq 0} a_n \psi_{n}(x)$$. In particular this works for any bounded continuous function. Now we have $$u_t(x)=e^{\frac{t}{2} \Delta} \sum_{n \geq 0} a_n \psi_{n}(x)=\sum_{n=0}^{\infty} e^{\frac{t \lambda_n}{2}} a_n \psi_n(x).$$ Because all eigenvalues are negative except for $$\lambda_0$$, as $$t \to \infty$$, $$u_t(x)$$ approaches the constant function $$a_0$$. In fact all other terms have exponential decay in $$t$$, so $$u_t(x)$$ converges to a constant quite quickly.

Now intuitively we would like to set $$f(x)=\delta(y-x)$$ so that $$u_t(x)=E_{\mu}[\delta(y-x)]=P(B_t=y),$$ is approaching a constant quickly. This would allow us to conclude that the distribution of a brownian motion approaches the uniform distribution regardless of its initial distribution $$\mu$$. Unfortunately a delta function is not bounded and continuous, so we can't do this directly but we can approximate the delta function arbitrarily well by a bounded continuous function and thus get arbitrarily good approximations of $$P(B_t=y)$$ which converge to a constant independent of $$y$$.

In a higher dimensional box $$\prod_{i=1}^n [-L_i, L_i]$$ we can make a similar argument, where the eigenfunctions of the $$n$$ dimensional Laplacian can be written in terms of our $$1$$ dimensional eigenfunctions as $$\prod_{i=1}^n \psi_n(x)$$ and similarly the eigenvaules as $$\sum_{i=1}^n \lambda_n$$. The important observation is that the constant function still has eigenvalue $$0$$ and every other eigenvalue is still negative.

The argument for periodic boundary conditions will be almost identical because the eigenfunctions of the Laplacian can still be written in terms of sin and cosine which have negative eigenvalues unless they are constant. I have not thought much about eigenfunctions of the Laplacian with sticky boundary conditions.

Let's examine the one dimensional case to understand this. We will set up a probability equation for where the particle is, and then take the continuum limit in both time and space. The result will be a familiar friend.

Consider a box of length $L$ divided into $N$ bins. Assume that a particle starts off close to the center of the box, and in each timestep has an equal probability of jumping to the right or to the left, $\alpha\Delta t\le .5$. Note that this is proportional to $\Delta t$, so $\alpha$ is a transition rate. The probability that it stays in the same bin is then $1-2\alpha\Delta t$.

One can then use Bayes' rule to calculate the probability that the particle is in the $n^{th}$ bin at time $t+\Delta t$:

\begin{align} p(n,t+\Delta t) &= \sum_{n'}p(n,t+\Delta t|n',t)p(n',t)\\ &=p(n+1,t)\alpha\Delta t+(1-2\alpha\Delta t)p(n,t)+p(n-1,t)\alpha\Delta t \end{align} Rearranging things and taking the limit as $\Delta t\rightarrow 0$ we find the Master equation $$\frac{\partial p}{\partial t}(n,t) = \alpha\left[p(n+1,t)-2p(n,t)+p(n-1,t)\right]$$ Now let $x = n\Delta x$ and $\Delta x = L/N$. Define the probability density that a particle is in the $n^{th}$ bin via $p(n,t) = \rho(x,t)\Delta x$. We can recast the above into the form $$\frac{\partial \rho}{\partial t}=\alpha[\rho(x+\Delta x,t)-2\rho(x,t)+\rho(x-\Delta x,t)]$$ We can then define the diffusivity as $D=\alpha\Delta x^2$, and take the continuum limit $\Delta x\rightarrow 0$. This is done by expanding the density in Taylor series. Note that the first order terms will vanish, and we will be left with $$\frac{\partial \rho}{\partial t}=D\frac{\partial^2 \rho}{\partial x^2}$$ This is nothing more than our old friend the diffusion equation. One can then solve this with no flux boundary conditions. A numerical procedure can be found here. The result is that no matter what initial distribution you start with, if you wait long enough (a timescale larger than $L^2/D$), the distribution will approach uniformity.

This procedure is the same for higher dimensions. Things get complicated if you choose to include interactions between particles. That, however, isa story for another day.

• This is just essentially a (nonrigorous) derivation of the Fokker-Planck equation (or the Feynman-Kac relation), and the answer as is basically entirely skips the part from where it is trivial to see that the PDE results in a uniform distribution (i.e. the actual question), just essentially stating the fact ("The result is that no matter what initial distribution you start with, if you wait long enough (a timescale larger than $L^2/D$), the distribution will approach uniformity"). – alarge Jun 23 '17 at 2:15
• I'm sure you've solved the diffusion equation in the past, and have some physical intuition for it. If not, the mathematics stack exchange has several posts on the topic. Good luck! – Damian Sowinski Jun 23 '17 at 3:19
• That's an odd comment. Why not just say that surely the OP has solved the simple SDE already and should anyway have physical intuition as to how Brownian processes work, and thus needs not be asking the question. Or if they do, they ought to go to maths stack exchange for SDEs and stochastic processes have been discussed at length there? – alarge Jun 23 '17 at 3:39
• I made a decision based on the tension between a rigorous answer and reaching a broader audience. From a pedagogic standpoint, the diffusion equation is first encountered during undergrad training. SDE's are not part of any, to my knowledge, physics curricula at the undergrad level. For that matter, unless your graduate work is in statmech, it's unlikely that a course on stochastic calculus is ever encountered. E&M, QM, and even fluid mechanics courses are, and physical intuition is built in those courses for PDEs. Of course you're free to make a different decision. – Damian Sowinski Jun 23 '17 at 3:53

I think the distribution remains flat for diffusion in a box, unless the boundary conditions are really strange or unphysical. Let me try with this not-so-accurate argument. Free diffusion gives a probability distribution which starts at $t=0$, for a single particle, as a delta function centered at the initial position, and evolves as a Gaussian $G_{free}$ with a variance that grows linearly in time.

Now, if you have a box around the particle, I'd say that the probability distribution $G_{box}$ at a time $t$ will be some sort of "folded" version of the Gaussian $G_{free}$ (with the tails of $G_{free}$ i)folded back for reflective boundary conditions, or ii) copied and pasted on the on the side for periodic boundary conditions, or iii) just removed for sticky boundaries iv)... ). This can create a strange $G_{box}$ at finite times, but eventually in the long run everything will be smoothed out and $G_{box}$ will become flat.

(On the other hand, if your box is not a rigid box but a harmonic potential well (the particle feels a force linear with displacement from the center), then $G_{box}$ will remain always the same Gaussian.)