Brownian motion and hitting a Quadrilateral

Question

I want to compute the hitting probability of a bounded plane by a Brownian motion starting at the origin. In other words, given the coordinates of a quadrilateral A , can we compute $P(T_{A}<\infty)$? How can I go about it?

First I try the specific case where A is centered at the x-axis and is parallel to the zx plane. Also, let A be a rectangle.

Then, $\{T_{A} <\infty\}=\{B_{1}(t)=a, |B_{2}(t)|\leq b,|B_{3}(t)|\leq c$ for some $t>0\}$. Here by a,b,c I mean the distance from origin, length and width of the rectangle $A$.

So I have to compute: $P_{0}\{(B_{1}(t)=a)\cap (|B_{2}(t)|\leq b)\cap (|B_{3}(t)|\leq c)$ for some $t>0\}$

Can I use the independence of the coordinates of a Brownian motion? I claim no, because the above events require a common t.

Update

Can I someone solve for the above A. My friend advised me to use Fourier transforms.

$$\nabla^2\Phi(\vec{r})=-\delta(\vec{r}),\;\;\Phi(\vec{r})=0\;\;\text{for}\;\;\vec{r}\in A,$$

$$P_A=\int_A \frac{\partial\Phi}{\partial\vec{r}}\cdot\hat{n}\;dS,$$

with $\hat{n}$ a unit vector normal to $A$ and pointing outward.

Answer 1

The probability $P_A$ to eventually reach the surface $A$ by Brownian motion, starting from the origin, is equivalent to an electrostatic problem: Integrate the electric field on the grounded surface $A$ induced by a point charge at the origin. The corresponding equations are

$$\nabla^2\Phi(\vec{r})=-\delta(\vec{r}),\;\;\Phi(\vec{r})=0\;\;\text{for}\;\;\vec{r}\in A,$$

$$P_A=\int_A \frac{\partial\Phi}{\partial\vec{r}}\cdot\hat{n}\;dS,$$

with $\hat{n}$ a unit vector normal to $A$ and pointing outward.

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First example: $A$ is the infinite plane $z=0$ at a distance $d$ from the point charge, then $$\Phi(\vec{r})=\frac{1}{4\pi}\left([x^2+y^2+(z-d)^2]^{-1/2}-[x^2+y^2+(z+d)^2]^{-1/2}\right)$$ and one finds $P_A=1$.

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Second example: $A$ is a sphere of radius $R$ and the point charge is at a distance $D>R$ from its center, then $$\Phi(\vec{r})=\frac{1}{4π}\left([r^2+D^2-2Dr\cos\theta]^{-1/2}-[(rD/R)^2+R^2-2Dr\cos\theta]^{-1/2}\right)$$ and one finds $P_A=R/D$.

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In both these examples the potential can be found using the method of image charges. There is no general closed-form solution for the potential for abitrary $A$. Using a numerical Poisson solver to find $\Phi$ seems the simplest way to make progress in your case.

Answer 2

$\nabla^2\Phi(\vec{r})=-\delta(\vec{r})\Rightarrow [(2\pi \xi_{1})^2+(2\pi \xi_{2})^2+(2\pi \xi_{3})^2]\hat{\Phi(\xi)}=(2\pi)^2|\xi|^{2}\hat{\Phi(\xi)}=-1\Rightarrow \Phi(x)=c_{2}\frac{1}{4\pi^{2}} | x|^{-1}$

Then, $P_{A}=c_{2}\frac{1}{4\pi^{2}}\int_{A}\frac{\partial|x|^{-1}}{dx}\cdot \vec{n} dS...$ still working on it.