Given a particle on the plane we can assume that it follows 2D Brownian motion. On the other hand if there is a lot of such a Brownian particles one can be interested in the evolution of the density of such particles.

How is the PDE for the density derived? I guess it should be similar to the heat equation and there should be connection with a gas dynamics. Could you refer me to the literature?


1 Answer 1


The equation resembles the heat equation very closely, so much so, that it is the heat equation. The literature on elementary things like this is uniformly terrible, so I cannot give a reference. To see why it is the heat equation, note that the law for the distribution function is the same as for a probability distribution, so it is linear equation, by the probability superposition principle.

If you evolve a little time in the future, the probability of being at any point gets incremented the average of the nearby neighbors. The average of the neighbors less the center point is the Laplacian, so the time derivtive is proportional to the Laplacian plus the center value, and since the integral over all space has to be unchanged with time, it must be just the Laplacian. Changing units of space, you can make the coefficient of the Laplacian 1:

$ {d\phi \over dt} = {1\over 2} \nabla^2 \phi $

But this still leaves a scaling invariance, if you rescale space by a certain amount, and time by the same amount squared, you get the same equation back. So the fundamental solution must be proportional to a scaling function: f({x^2\over t}) and you can solve for the solution, which is a spreading gaussian $\phi(x,t) = {1\over (2\pi t)^{d/2}} e^{-{x^2\over 2t}}$, where d is the dimension of space. (a product of one-dimensional Gaussians). This collection of ideas form a circle, because the same distribution is in the definition of Brownian motion, and comes from the central limit theorem.


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