5
$\begingroup$

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.

Bounded Plane 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.

$\endgroup$
4
$\begingroup$

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.


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 $P_A=1$.


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 $P_A=R/D$.


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.

$\endgroup$
  • $\begingroup$ For A symmetrically arranged to the origin, can you expand/give advice on how to compute that integral? $\endgroup$ – TKM Sep 8 '14 at 19:42
  • $\begingroup$ Would I be using a Fourier Transform to find $\Phi$? $\endgroup$ – TKM Sep 8 '14 at 20:16
  • $\begingroup$ can you please give reference for the "image chargers" $\endgroup$ – TKM Sep 8 '14 at 20:21
  • 2
    $\begingroup$ en.wikipedia.org/wiki/Method_of_image_charges $\endgroup$ – Carlo Beenakker Sep 8 '14 at 20:22
0
$\begingroup$

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

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.

$\endgroup$
  • $\begingroup$ this $\Phi$ doesn't satisfy the boundary condition on A; I'm sorry, but since the boundary condition is in real space, Fourier transformation to reciprocal space is unlikely to be helpful. $\endgroup$ – Carlo Beenakker Sep 9 '14 at 6:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.