# Area covered by Brownian motion of 2D disc

I would like to know the expected value for the area covered by a disc of radius $R$ whose center undergoes Brownian motion (diffusion).

Specifically, let $\mathbf{X}_t$ represent a two-dimensional Brownian motion, and define the covered set $C_t = \bigcup_t B(\mathbf{X}_t,R)$ to be the union of discs centered at every point along the realized trajectory. Define $A_t$ to be the area of the set $C_t$. What is the expected value $\mathbb{E}[A_t]$ as a function of $t$?

I am specifically interested in this for the mathematical modeling of a chemical reaction and would be happy with an approximate solution that works in the short-to-moderate time regime, where $|x_t|$ is at most (say) $5R$. I think that this time regime is too short to use the lattice random walk result for the number of distinct sites ($t / \log t$ from Dvoretzky and Erdös).

Thank you.

• There is a recent paper of Peres and Sousi on a related topic, showing that if a shape with area $A$ has its centre moved by a BM, then the shape that minimizes the total area covered is the disc. – Anthony Quas Jan 27 '14 at 22:57
• That is interesting to know, as the process I am modeling involves cells that are not spherical (but I am happy with a "spherical cow" approximation). – John Jumper Jan 27 '14 at 23:06

Let $u(r,t)$ solve the heat equation in polar coordinates in the complement of the disk: $$\partial_t u(r,t)= \frac{1}{2}\frac{1}{r}\partial_r r \partial_r u(r,t), \quad r>1$$ with initial condition $u(r,0)=0$ and boundary condition $u(1,t)=1$. I claim that the number you seek is $$A_t = \pi + 2\pi\int_1^\infty u(r,t)r dr.$$ (This is because $u(r,t)$ gives the probability that a Brownian particle released at radius $r$ has hit the disk by time $t$. The first term $\pi$ is just the area of the disk at time $0$. I have transferred the problem to one in which your particle remains fixed and potential target points move.)

There is an exact expression for $u$ (see Wendel, J. G. "Hitting spheres with Brownian motion". Ann. Prob. 8, 164 (1980)., which I learned of from the answer to this mathoverflow question, but ultimately traced back to a paper of Carslaw and Jaeger from 1940). It is most easily expressed in terms of the Laplace transform $F(r,\lambda)=\int_0^\infty u(r,t) e^{-\lambda t}$: $$F(r,\lambda) = \frac{K_0(\sqrt{2\lambda}r)}{\lambda K_0(\sqrt{2\lambda})}, \quad r\ge 1,$$ where $K_0$ is a modified Bessel function of the second kind and order $0$. In your case the integral over $r$ can be done exactly since $\int r K_0(r) dr = - K_1(r)$. So $$\int_0^\infty A_t e^{-\lambda t} d t \ = \ \frac{\pi}{\lambda} + \frac{2\pi K_1(\sqrt{2\lambda})}{\lambda (2\lambda)^{\frac{3}{2}}K_0(\sqrt{2\lambda})}.$$ You can recover $A_t$ by analytic continuation of the right hand side to complex $\lambda$ and Fourier inversion.

• It will take me more time to digest the solution, but I believe that transferring the problem to an absorbing disk at the origin is valid. Thank you. – John Jumper Jan 27 '14 at 23:04
• I should add that the contour integral you get for $A_t$ can be transformed into a form suitable for numerical integration. I wanted to compute $u(r,t)$ for various values of $r$ and $t$ a couple of years ago; I was able to do so easily in Matlab using off the shelf numerical integrators. – Jeff Schenker Jan 28 '14 at 14:57

Your process $C_t$ is better known by the amusing name of Wiener sausage, and its expected area $E[A_t]$ has been extensively studied (though more so for long times than short).

I found a 2010 paper by Yuji Hamana which gives small-time asymptotics for the area of the $d$-dimensional Wiener sausage; for $d=2$ it looks like we have, as $t \to 0$, $$E[A_t] \sim \pi R^2 + \sqrt{8 \pi t} R + o(\sqrt{t}).$$ There's also a 2013 paper by Kôhei Uchiyama that obtains a similar result.

MR2761916 Hamana, Yuji. On the expected volume of the Wiener sausage. J. Math. Soc. Japan 62 (2010), no. 4, 1113–1136. Open access at Project Euclid.

MR2988115 Uchiyama, Kôhei. The expected area of the Wiener sausage swept by a disc. Stochastic Process. Appl. 123 (2013), no. 1, 191–211.

• Thank you for the references. Unfortunately, I think I will have to use the inverse Laplace transform of the formula given above as I am in the regime where $\sqrt{t}$ is comparable to $R$. – John Jumper Jan 29 '14 at 0:34