Brownian representation of the Dirichlet problem

Let $D$ be bounded, let $u$ be harmonic on $D$ and continuous on $\overline{D}$, and let ${\tau }_D$ be the first exit time of Brownian motion. The Itô formula makes $u(B_{t\wedge {\tau }_D})$ a bounded martingale, soIf $u=f$ on the boundary, this becomes $u(x)={\mathbb{E}}_x[f(B_{{\tau }_D})]$.

For a bounded domain whose boundary points are all regular boundary points, continuous Dirichlet boundary data $f$ have a unique solutionHere ${\tau }_D$ is the Brownian exit time. The Strong Markov property gives the mean value property for harmonic functions; a barrier for the Dirichlet problem gives continuity at the boundary. This is an existence theorem as well as the Brownian representation of the Dirichlet problem for an already known solution.

If $u$ is a bounded harmonic function on a domain $D$, continuous on its closure, and the Brownian exit time $T_D$ is almost surely finite from $z$, then $u(z)={\mathbb{E}}_z[u(B_{T_D})]$. The Itô formula and compact localization make $u(B_{t\wedge T_D})$ a bounded martingale. bounded convergence theorem at the exit time proves the representation even if the domain is unbounded. Removing boundedness of $u$ can invalidate this conclusion through loss of uniform integrability.

Suppose an unbounded domain contains the upper strip $(-1,1)\times (1,\infty )$ with its two vertical sides on the boundary. A bounded harmonic function with constant values $a$ on the right side and $b$ on the left side, and with valid bounded harmonic representation with almost sure Brownian exit, satisfies $u(x,y)\to a(1+x)/2+b(1-x)/2$ as $y\to \infty$. The horizontal Brownian exit time is finite; on that finite random interval the vertical Brownian path has a finite minimum. Consequently reaching the bottom before a side has probability tending to zero. The remaining two exit probabilities are the one-dimensional interval probabilities.

Almost-sure finiteness of the exit time does not replace boundedness of the domain. In the upper half-plane, $u(x,y)=y$ is harmonic with zero boundary values and Brownian motion reaches the boundary almost surely, but $u$ is positive inside. The stopped martingale fails to be uniformly integrable.