Deriving the Diffusion Equation from a Random Walk

The diffusion equation $$ \frac{\partial \rho}{\partial t} = D\nabla^2 \rho $$ appears in a remarkable number of places. I've previously covered the diffusion equation in one guise as the heat equation (and the cool patterns produced by reaction-diffusion equations)

Here we'll derive the diffusion equation starting from a random walk.

Imagine at the initial moment, \(t_0=0\), we observe the position of a random-walker. This walker will move from its initial position \(x_0\) to position \(x_1\) after \(t_1\) seconds with probability \(\chi(x_1 - x_0)\). This density \(\chi(\ell)\) describes the probability of the random-walker taking a step of length \(\ell\). The density \(\rho(x,t)\) gives the probability of finding the walker at location \(x\) at time \(t\).

We want to describe the evolution of this density \(\rho\) through time. The probability density at \((x,t + \Delta t)\) should be linked to the density at \((x, t)\) in some way. It seems reasonable that, assuming the \(\Delta t\) is small enough, the density should be a small evolution of the the values at the prior time.

If we know that at \(t+\Delta t\) that the walker is at \(x^\prime\), we can determine the density by thinking about all the ways to end up at \(x^\prime\). For instance we could end up at \(x^\prime\) by starting at \(x_0\) and taking a step of length \(x^\prime - x_0\). The probability to start at \(x_0\) at time \(t\) is given by \(\rho(x_0, t)\) while the probability to take that step is \(\chi(x^\prime - x_0)\). The walker could start anywhere that the density \(\rho(x,t)\) is defined so we have to consider contributions from the whole domain. This leads to the integral $$ \rho(x^\prime, t+\Delta t) = \int_{-\infty}^{\infty}\chi(x^\prime - x)\rho(x, t)dx $$

Carefully, while minding the various negative signs, use the subtitution \(u=x^\prime - x\) in this integral to get $$ \rho(x^\prime, t+\Delta t) = \int_{-\infty}^{\infty}\chi(u)\rho(x-u, t)du $$

Now for a decisive approximation! Taylor expand the density in the integrand around x, truncating at the second-order term. This gives us $$ \rho(x-u, t) \approx \rho(x,t) + u\frac{\partial \rho}{\partial x} + \frac{u^2}{2}\frac{\partial^2 \rho}{\partial x^2} $$ for sufficiently small \(u\). Use this to approximate the integral obtained earlier. $$ \rho(x^\prime, t+\Delta t) \approx \int_{-\infty}^{\infty}\chi(u)\left(\rho(x,t) + u\frac{\partial \rho}{\partial x} + \frac{u^2}{2}\frac{\partial^2 \rho}{\partial x^2}\right)du $$

Let's halt here and consider whether this procedure makes sense or not. After all, the Taylor expansion is valid only in small neighborhoods of \(x\), yet in the integral its being computed over the entirety of the real number line \(\mathbb{R}\). What gives?

The idea (I think) is that over very small displacements of time \(\Delta t\), under reasonable assumptions about the motion of our random-walker, like that it moves continuously (no teleporting, etc.) at some reasonable velocity, then we should expect the step probability \(\chi(\ell)\) to vanish over most of \(\mathbb{R}\).

We can write then $$ \rho(x^\prime, t+\Delta t) = \int_{\alpha}^{\beta}\chi(u)\rho(x-u, t)du $$ as the integral over the region where the step probability is non-vanishing.

On a subset of this region we should expect our Taylor series approximation to hold quite well: $$ \rho(x^\prime, t+\Delta t) \approx \int_{\alpha_1}^{\beta_1}\chi(u)\left(\rho(x,t) + u\frac{\partial \rho}{\partial x} + \frac{u^2}{2}\frac{\partial^2 \rho}{\partial x^2}\right)du $$ The main idea though is the step probabilities control the integrand far away from the prior position, assuming that the particle moves reasonably in infinitesimally small time slices.

Now with our handy approximation, we can compute the integral: $$ \int_{\alpha_1}^{\beta_1}\chi(u)\left(\rho(x,t) + u\frac{\partial \rho}{\partial x} + \frac{u^2}{2}\frac{\partial^2 \rho}{\partial x^2}\right)du = \rho(x,t)\int_{-\infty}^{\infty}\chi(u)du + \frac{\partial\rho}{\partial x}\int_{-\infty}^{\infty}u\chi(u)du + \frac{1}{2}\frac{\partial^2 \rho}{\partial x^2}\int_{-\infty}^{\infty}u^2\chi(u)du $$

Observe that each of the remaining three integrals on the right-hand side are easily determined. The first integral $$ \int_{-\infty}^{\infty}\chi(u)du = 1 $$ by definition of a probability distribution.

The second integral is the first moment of the step distribution $$ \int_{-\infty}^{\infty}u\chi(u)du = \mu $$

While the final integral is its second moment $$ \int_{-\infty}^{\infty}u^2\chi(u)du = \sigma^2 $$

For this problem we'll conveniently set the mean \(\mu = 0\).

We're now left with $$ \rho(x^\prime, t+\Delta t) \approx \rho(x,t) + \frac{\sigma^2}{2}\frac{\partial^2 \rho}{\partial x^2} $$

We assumed previously that \(\Delta t\) was small. Particularly, let's assume that its small enough so that we have $$ \rho(x, t+\Delta t) - \rho(x, t) \approx \frac{\partial \rho}{\partial t}\Delta t $$

Plugging that into the result of the integral approximation gives us $$ \frac{\partial \rho}{\partial t} \approx \frac{\sigma^2}{2\Delta t}\frac{\partial^2 \rho}{\partial x^2} $$

This result holds for the multidimensional case, giving us the diffusion equation $$ \frac{\partial \rho}{\partial t} = D \nabla^2 \rho $$ where \(D={\sigma^2}/{2\Delta t}\) is the diffusion constant.