Dirichlet problem

For the Laplace or Poisson equation on a bounded connected domain, two solutions with the same Dirichlet boundary data are equal whenever the maximum principle or the corresponding energy argument applies.

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.

The Perron method constructs a harmonic solution with prescribed boundary data as the pointwise supremum of all subharmonic functions lying below those data. Harmonic lifting proves interior harmonicity; barriers determine whether the supremum attains the boundary values.

A continuous function $v$ is a Perron subfunction for $\Delta u=f$ when it lies below every classical solution of that Poisson equation on a relatively compact open ball whenever it lies below that solution on the ball's boundary. For $C^2$ functions, this is equivalent to $\Delta v\ge f$. A superfunction reverses the comparison and has $\Delta v\le f$ when $C^2$. Adding a boundary upper bound gives the family whose supremum defines the Perron method for the Dirichlet problem.

A barrier at $z\in \partial \Omega$ is a continuous superharmonic function that vanishes at $z$ and is strictly positive at every other point of the closure. It forces the Perron solution to converge to its prescribed boundary value at $z$.

A boundary point is regular for the Dirichlet problem when the Perron solution approaches the prescribed continuous boundary value there. The existence of a barrier implies regularity.

An open set satisfies an exterior cone condition at $z$ if a solid cone with vertex $z$ lies outside the set in some neighbourhood of $z$. The stronger global version requires the closures of the cone and the set to meet only at $z$. In two dimensions, a power barrier for an exterior cone makes such a point a regular boundary point for the Dirichlet problem. Local cone conditions suffice for local boundary regularity; a global cone is an additional geometric assumption.

For a planar set outside a global solid cone of half-angle $\alpha$, choose $0<{\alpha }_0<\min (\alpha ,\pi /2)$, $\beta =\pi /[2(\pi -{\alpha }_0)]$, and $0<\nu <\beta$. If $\psi$ is the angle from the opposite axis, the Laplacian in polar coordinates givesThe cosine has a positive uniform lower bound outside the original cone. On a bounded set, $\Delta g$ therefore has a positive uniform lower bound as well. The positive function $-g$ is a barrier for the Dirichlet problem. An unbounded set does not receive the same global lower bound from this construction.

A domain satisfies the exterior sphere condition at a boundary point when a ball contained in the complement is tangent to the boundary only at that point. Fundamental solutions of the Laplace equation centred at the ball's centre provide a harmonic barrier, so such a point is regular.

If $B(y,R)$ is an exterior ball tangent to a planar domain at $x_0$, thenis a harmonic function in the domain, vanishes at $x_0$, and is positive at every other point of its closure. It is therefore a barrier for the Dirichlet problem and proves that $x_0$ is a regular boundary point.