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.



Area covered by Brownian motion of 2D disc
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).](http://projecteuclid.org/DPubS?service=UI&version=1.0&verb=Display&handle=euclid.aop/1176994833), which I learned of from the answer to this [mathoverflow question](https://mathoverflow.net/questions/96703/finite-time-hitting-probabilities-for-brownian-motion-in-the-plane), but ultimately traced back to a [paper of Carslaw and Jaeger](http://plms.oxfordjournals.org/content/s2-46/1/361.full.pdf) 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. 


