Dynamical interpretation of reflecting boundary conditions in the Fokker-Planck equation

Question

Background:

For a particle driven by the dynamical equation $$ \dot{x}(t) = a(x,t) + b(x)\xi(t),$$ where $\xi(t)$ is a Gaussian white noise, the probability distribution of position $x$ is governed by the Fokker-Planck equation $$ \frac{\partial}{\partial t}P(x,t) = -\frac{\partial}{\partial x}J(x,t)$$ where $$J(x,t) = a(x,t)P(x,t) - \frac{1}{2}\frac{\partial}{\partial x}\Big[b(x,t)^2P(x,t)\Big]$$ is the probability current. One can solve the Fokker-Planck equation for different boundary conditions. One possibility is "reflecting" boundary conditions when the probability current vanishes at some reflection point $x=a$, which can be written $$J(a,t)=0. $$

Question:

I am wondering how to understand what the individual particles are actually doing in a Fokker-Planck equation with reflecting boundary conditions. How are individual particle trajectories modified by reflecting boundary conditions in the Fokker-Planck equation? Is it as if they are undergoing elastic collisions with the barrier? Could one interpret reflecting boundary conditions as a potential term in the original equation of motion?

Answer 1

It means that the current in $x=a$ is 0, but the density can be whatever.

You can indeed imagine it as particles bouncing off a wall: in a small line element $[a-dx, a]$ you have $N$ particles, but $N/2$ are going towards left and $N/2$ towards the right, so the net current is $0$. The ones going "on the right" will then bounce off the wall and from then on going towards left, while the same number of particles will come from the left from the area element before ($[a-2dx, a-dx]$) to compensate. Again, the net current is 0. And so on. Of course this is true on average not for every particle.

Consider for example $N$ molecules moving in a closed box. Of course they can not escape the box, so the current $J(\text{box side}, t)=0$. But their concentration is constant everywhere ($N/\text{volume of the box}$). In terms of potential, it would mean that the energy on the sides of the box is infinite and particles can not escape it.

Answer 2

Consider the special case where $a\equiv 0$ and $b\equiv 1\,.$ Then $x(t)$ is a standard Brownian motion. With reflection condition at zero this process has the transition density $$ p(t,x,y)=\frac{1}{\sqrt{2\pi t}}\left\{\exp\left(-\frac{(x-y)^2}{2t}\right)+\exp\left(-\frac{(x+y)^2}{2t}\right)\right\}\, $$ which solves the heat equation $$ \frac{\partial p}{\partial t}=\frac{1}{2}\frac{\partial^2 p}{\partial x^2}\,. $$ According to exercise 2.8.4 of [1], this is the transition density of the modulus of Brownian motion $|x(t)|\,.$

In other words: the trajectory of the reflected particle is $|x(t)|\,.$

The picture in [2] is also very enlightning.

[1] I. Karatzas, S. Shreve, Brownian Motion and Stochastic Calculus. Springer Graduate Texts in Mathematics. 1991.

[2] Wikipedia Reflection principle (Wiener process).