Dirichlet problem

In $\mathrm{Re}z>0$ outside the circle $|z-d|=R$ with $d>R>0$, the displayed bounded harmonic function is zero on the imaginary axis and $-V_0$ on the circle. Put ${\tau }_0=\mathrm{arcosh}(d/R)$. The circle is the Apollonius circle $|(z-a)/(z+a)|=e^{-{\tau }_0}$; its centre and radius are $a\coth {\tau }_0=d$ and $a/\sinh {\tau }_0=R$. The point $a$ lies inside the excluded disk and $-a$ outside the half-plane, so the logarithmic modulus is harmonic throughout the physical domain. Local branches of the complex logarithm suffice; its real part is globally single-valued even though a global complex logarithm need not exist. The potential tends to zero at infinity. A boundedness or far-field condition is needed to infer uniqueness from the finite boundary data. This is also the geometry of cylinder-plane image charges.

A harmonic extension of boundary data $f$ is a harmonic function satisfying those Dirichlet boundary data in a specified solution class. Existence and uniqueness depend on the domain, the regularity of the data and, on unbounded domains, growth conditions. A Poisson kernel constructs such an extension on standard domains.

For bounded piecewise continuous Dirichlet boundary data $f$ on a circle of radius $R$, write its Fourier series as $a_0+{\sum }_{n\ge 1}[a_n\cos (n\theta )+b_n\sin (n\theta )]$. The displayed series gives the bounded exterior harmonic extension for $r>R$. The radial inversion in a circle $r\mapsto R^2/r$ converts it into the ordinary interior Poisson kernel solution. Values at discontinuities are understood through the usual boundary trace, with midpoint radial limits at jumps; changing the assigned values at finitely many boundary points does not change the extension.

The limit at infinity is forced to equal the angular mean $a_0=(2\pi {)}^{-1}{\int }_{-\pi }^{\pi }f(\theta )d\theta$. Indeed the angular mean of any exterior harmonic function solves $(r{\overline{u}}^{\prime }{)}^{\prime }=0$, and boundedness excludes its logarithmic term. Equivalently, inversion makes infinity an isolated removable singularity of a bounded planar harmonic function. Uniqueness follows from interior uniqueness after inversion, including the logarithmic cutoff proof of harmonic uniqueness with finitely many boundary exceptions when needed.

The bounded harmonic function on the open unit disk with boundary values one on its upper semicircle and zero on its lower semicircle is the displayed function. The Möbius transformation $w=i(1-z)/(1+z)$ has real part $2y/|1+z{|}^2$ and imaginary part $(1-|z{|}^2)/|1+z{|}^2$. Pulling back the upper-half-plane step-data solution proves the formula by conformal invariance of harmonicity. The discontinuity points $z=\pm 1$ have no path-independent boundary limit.

For $a>0$, a harmonic function on $x\ge 0$, $0\le y\le a$ with zero data on the horizontal sides, sine data $\sin (n\pi y/a)$ at the end, and decay at infinity is given by the displayed expression for every integer $n$. The second derivatives cancel since the decay exponent squared equals the transverse wavenumber squared. The absolute value is necessary for negative integers; $n=0$ gives the zero solution. The solution is bounded and decays uniformly in $y$.

For a bounded regular region and a sufficiently smooth real harmonic function, apply the divergence theorem to $u\nabla u$. Its divergence is $|\nabla u{|}^2+u\Delta u$. The boundary term vanishes, hence the integral of the nonnegative gradient square is zero. The gradient vanishes and the boundary values fix each component's constant as zero. Taking the difference of two solutions gives uniqueness for the Poisson equation with Dirichlet boundary data. Unbounded regions require an additional condition at infinity.

Let $w$ be a bounded harmonic function in a bounded connected smooth planar region, with zero continuous boundary limits except at finitely many points. Then $w=0$. This extends zero-boundary harmonic uniqueness to bounded solutions with isolated boundary exceptions, without assuming their initial Dirichlet energy is finite.

Choose ${\eta }_{ϵ}=0$ within distance ${ϵ}^2$ of each exceptional point, ${\eta }_{ϵ}=1$ beyond distance $ϵ$, and interpolate linearly in the logarithm of distance. Each annular transition has $\int |\nabla {\eta }_{ϵ}{|}^2=O(1/|\log ϵ|)$. Apply the divergence theorem to the equation multiplied by ${\eta }_{ϵ}^{2}w$. The ordinary boundary terms vanish, and the Cauchy-Schwarz inequality givesEvery fixed compact subset eventually has ${\eta }_{ϵ}=1$, so the gradient vanishes everywhere. The nonexceptional boundary data then fix the constant to zero. Differences of bounded harmonic extensions with identical piecewise continuous data satisfy this hypothesis. The individual extensions can have infinite Dirichlet energy at a jump; the cutoff argument avoids applying an unjustified finite-energy identity to them.

Unbounded-domain Dirichlet problems can need a condition at infinity. On all of ${\mathbb{R}}^n$, constant functions have identical empty boundary data and are distinct finite-energy stationary points of $\frac{1}{2}\int |Du{|}^2$. With nonempty boundary, on the exterior of the unit ball in dimension at least three, $1-|x{|}^{2-n}$ is a nonzero harmonic function vanishing on the sphere, with finite Dirichlet energy and a nonzero limit at infinity. It is stationary for compactly supported variations. The permitted class of variations and the behaviour at infinity must be specified for a global variational uniqueness claim.

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 nonempty collection of real continuous subharmonic functions on a connected Riemann surface is a Perron family if it is closed under pairwise maxima and harmonic lifting of a continuous subharmonic function on every relatively compact coordinate disk. Its pointwise supremum is either everywhere infinite or a harmonic function on a Riemann surface. At a point where the supremum is finite, take increasing disk lifts approaching that value. The Harnack inequality for harmonic functions gives a finite harmonic limit on the disk. Lifting their maxima with any other member gives a larger harmonic limit with the same value at the chosen point, hence the same limit by the strong maximum principle for harmonic functions. The supremum is therefore harmonic on that disk; connectedness propagates finiteness to the whole surface.

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.