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

Question

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?

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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 $n \to \infty$.

Answer 1

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.

Answer 2

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.