# PDE for the probability of Brownian motion staying in an area (reference request)

I am looking for a (preferably some monograph) reference on the following fact: $$u ( t, x ) = \mathbb{P} \{ x + B_s \in A \ \text{for all} \ s \leq t \}$$ satisfies the heat equation $$\frac{\partial u}{\partial t} = \frac{1}{2} \Delta u \quad \text{in} \quad \mathbb{R}_+ \times A$$ and the following initial-boundary value conditions: $$u \ |_{t = 0} = 1 \quad \text{and} \quad u \ |_{\partial A} = 0.$$

I'm struggling to find this particular fact among many facts of the same type. Correct me if I'm wrong, but it doesn't seem to follow (at least directly) from Feynman-Kac formula, nor does it follow from the other BM-PDE links I keep finding.

Additionally to the reference request, I'd like to know what's the most straightforward way to see this fact?

P.S. It seems to formally follow from Feynman-Kac, which says that $$u ( t, x ) = \mathbb{E} \left\{ f ( B_t ) \exp \left( -\int_0^t v(B_s) \, ds \right) \right\}$$ solves $$\frac{\partial u}{\partial t} = \frac{1}{2} \Delta u - vu$$ with $$u \ |_{t=0} = f$$. Letting $$f=1$$ and formally $$v ( x ) = \begin{cases} 0, & x \in A, \\ \infty, & x \not\in A \end{cases}$$ seems to give the result... Maybe it's possible to do it more rigorously by taking $$v_n(x) = n$$ on $$x \not\in A$$ and then letting $$n \to \infty$$.

• What do you assume from $A$?
– m7e
Commented Mar 31 at 16:10
• @m7e, I would actually like to know what should I assume of it... Being compact isn't necessary, but I have no idea how far the result stretches... Commented Mar 31 at 16:21
• S. Port and C. Stone, Brownian Motion and Classical Potential Theory,
– mike
Commented Apr 1 at 5:03

The easiest way to show this is to check that if $$\hat{u}$$ is a bounded solution to your boundary value problem, then $$\hat{u}(t-s,B_{s\wedge\tau}+x)$$ is a martingale, where $$\tau$$ is the minimum of the exit time and $$t$$. This can be done e.g. using Itô calculus. But by Markov property of the Brownian motion, we have $$\mathbb{P}(\tau= t|\sigma(B_{[0,s]})=u(t-s,B_{s\wedge\tau}+x)$$, hence this is also a (Doob/Lévy) martingale. So, there difference is a bounded martingale, and we have by optional stopping theorem, $$0=\mathbb{E}(u(t-\tau,B_{\tau}+x)-\hat{u}(t-\tau,B_{\tau}+x))=u(t,x)-\hat{u}(t,x),$$ where the expression in the first expectation is zero almost surely since $$u$$ and $$\hat{u}$$ have the same boundary and initial conditions.

• I don't quite see how this proof works. First, showing that if $\hat{u}$ is a solution of heat equation $\implies \hat{u} ( t - s, B_{t \wedge \tau} + x )$ is a martingale is straightforward, I agree. But how do we show that $u(t-s, B_{s \wedge \tau} + x )$ with $u$ from my post is a martingale? Second, how do boundary conditions come into play here? And third, it seems that your proof requires $\tau$ to be bounded, but does it also work if $A$ is not compact? Commented Mar 31 at 11:14
• @tsnao, I added some details, hope that helps! In the case $A$ is not compact, the solution to the parabolic problem is not unique, so you need to pick a bounded one (then the above argument shows that a bounded solution is unique). Commented Mar 31 at 11:55
• I still cannot reproduce your argument for why $u(t-s, B_{s \wedge \tau} + x)$ is a martingale. I'm probably doing something stupid, but what you wrote doesn't seem enough... Commented Mar 31 at 23:10

I think your idea ("PS") works fine, at least when $$A$$ has finite measure and is a moderately reasonable set (let's say open) and probably in general with more effort. It does seem to get a bit technical though. Here's a sketch:

We can also take $$f=\chi_A$$, $$v_n=n\chi_{A^c}$$, and then define $$u_n(x,t) = E\left( f(B_t)e^{-\int_0^t v(B_s)\, ds}\right) ,$$ as you suggested. Then $$0\le u_n\le 1$$, $$u_n$$ decreases to $$u$$ and $$u_n(t)=e^{-tH_n}f$$ in $$L^2(\mathbb R^d)$$ with $$H_n=-\Delta/2+v_n$$. We have $$(H_n-i)^{-1}\to (H_A-i)^{-1}\oplus 0$$ strongly, with $$H_A$$ denoting minus one half times the Dirichlet Laplacian on $$L^2(A)$$.

This is not quite strong resolvent convergence since $$0$$ is not the resolvent of an operator, but it is similar in spirit and leads to similar conclusions. In particular, we will again obtain $$e^{-tH_n}g\to e^{-tH_A} g$$ for $$g\in L^2(A)$$ and since also $$\chi_A u_n\to u$$ in $$L^2(A)$$ by monotone or dominated convergence, this shows that $$u=e^{-tH_A} 1$$, as desired.

• Nice argument. Does $(H_A - i)^{-1} \oplus 0$ means that it sends functions with support outside $A$ to zero? How do I formally see this convergence? Commented Mar 31 at 20:16
• In one space dimension, we can solve the ODE $H_ny=iy$ explicitly (at least when $A, A^c$ are not too complicated) and build the integral kernel of the resolvent from these solutions. I'm not sure off the top of my head what a good general approach would be, but the fact that an infinite potential barrier imposes a Dirichlet boundary condition is "well known" (certainly to physicists). Commented Mar 31 at 21:24
• Infinite potential barrier is exactly what I had in mind! Maybe I should look for a reference on how do people in PDE/spectral theory treat this subject... Is there a chance that it also works for unbounded domains? Commented Mar 31 at 22:00
• Yes, I don't think any of the above assumptions are necessarily essential, I was just very lazy with the details. For example, $A=(0,\infty)\subseteq \mathbb R$ will certainly work. Commented Mar 31 at 22:03