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For smooth planar domains and a conformal map $\Phi :D\to \Omega$ extending as a $C^1$ boundary diffeomorphism with nonzero derivative, the Poisson kernel as a density against boundary arclength satisfiesThe derivative is evaluated at the boundary point, not at the interior observation point.

To derive it, use conformal invariance of harmonic measure and change boundary variables: $d{s}_{\Phi (\zeta )}=|{\Phi }^{\prime }(\zeta )|d{s}_{\zeta }$. The two Poisson integrals must give the same harmonic extension for every continuous boundary datum, identifying the densities. Kernels defined against an angular measure instead require the corresponding measure conversion.

For $\Phi (z)=Rz$ from the unit disk to the radius-$R$ disk, the rule gives $P_{R\mathbb{D}}(Rz,R\zeta )=P_{\mathbb{D}}(z,\zeta )/R$. At the center this is $1/(2\pi R)$, whose integral around circumference $2\pi R$ is one. This normalization check detects a misplaced derivative factor.

For a finite undirected graph with unit edge conductances, define the Discrete Dirichlet energy of a real-valued function $f$ byEach undirected edge is counted once. More generally, multiply each term by its edge conductance. The energy is nonnegative and vanishes exactly when $f$ is constant on every connected component. Large differences across strongly conducting edges are expensive; this is why Discrete Dirichlet energy appears in harmonic extension and electrical-network principles.

Normalization depends on the chain. For simple random walk on a finite connected unweighted graph with $m$ edges, the Stationary distribution is $\pi (x)=\mathrm{deg}(x)/(2m)$, so its Dirichlet form of a Markov chain is ${\mathcal{E}}_P(f,f)={\mathcal{E}}_G(f,f)/(2m)$. On the three-vertex path with values $0,1,0$, the graph energy is $2$, whereas the simple-random-walk energy is $1/2$. Stating which normalization is used matters when comparing spectral gaps or effective resistances.

On a disk of radius $a$, prescribe $V_1$ on the upper open semicircle and $V_2$ on the lower one. The bounded harmonic extension has the formula in the title for $r<a$. The boundary data have only a constant term and odd sine Fourier coefficients, giving $u=(V_1+V_2)/2+[2(V_1-V_2)/\pi ]{\sum }_{m\ge 0}(r/a{)}^{2m+1}\sin ((2m+1)\theta )/(2m+1)$. The complex series is half of $\log [(1+w)/(1-w)]$, where $w=(r/a)e^{i\theta }$. Its quotient has positive real part, so taking the imaginary part yields the displayed principal arctangent without an argument ambiguity. At the two boundary jumps, the radial limit is the mean; the harmonic function has no continuous extension matching both adjacent constant arcs there.

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.

A real harmonic function on the unit disk has the displayed expansion, convergent locally uniformly with its derivatives in the interior. If its continuous boundary data have Fourier coefficients $A_0,A_n,B_n$ in the convention $f(\theta )=A_0/2+{\sum }_{n\ge 1}(A_n\cos (n\theta )+B_n\sin (n\theta ))$, its harmonic extension has $a_0=A_0/2$, $a_n=A_n$, $b_n=B_n$.

To obtain the radial factors, separation of variables in the Laplace equation in polar coordinates gives $r^2{R}^{\prime \prime }+r{R}^{\prime }-\lambda R=0$ and ${\Theta }^{\prime \prime }+\lambda \Theta =0$. Angular periodicity restricts nonconstant modes to $\lambda =n^2$ for positive integers $n$. The radial factors are $r^n,r^{-n}$, with $1,\log r$ for the constant angular mode. Boundedness at the origin removes $r^{-n}$ and $\log r$. Each surviving factor $r^n\cos (n\theta )$ or $r^n\sin (n\theta )$ is a harmonic polynomial, the real or imaginary part of $(x+iy{)}^n$. The expansion can also be derived from the Poisson integral on the unit disk for the disk.

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.

Use separation of variables, $\Phi =F(r)H(\theta )$, in the Laplace equation in polar coordinates. Dividing by $FH/r^2$ givesSingle-valued $2\pi$-periodic angular factors have integer $n\ge 0$. For $n\ge 1$, they are $\cos n\theta ,\sin n\theta$, and the radial equation $r^2{F}^{\prime \prime }+r{F}^{\prime }-n^{2}F=0$ has solutions $r^n,r^{-n}$. The zero angular mode gives $F=a_0+b_0\log r$. These separated products give the requested solutions; arbitrary sums may use independent coefficients for each radial-angular combination.

Boundedness at the origin removes the logarithm and all negative radial powers. The boundary function is even, so its sine Fourier coefficients vanish. With the convention $f(\theta )=a_0+{\sum }_{n\ge 1}{a}_n\cos n\theta$, its Fourier coefficients areThus the bounded interior harmonic extension isFor $r<R$ the series and its derivatives converge locally uniformly, so it is harmonic. It is a rotated and scaled instance of the harmonic measure of a semicircle in a disk. For example, summing the power series for the complex arctangent, or differentiating the following expression, gives the equivalent real formThis also proves boundedness directly.

At the two jumps, the boundary interpretation matters. At every continuity point the radial limit equals the prescribed value. At $\theta =\pm \pi /2$, the radial limit is $C/2$, the midpoint of the two one-sided values. Consequently for $C\ne 0$ there is no harmonic solution continuous on the closed disk with exactly the printed boundary values. The standard bounded Dirichlet problem imposes the trace away from the jumps, or almost everywhere; assigning zero separately at those two boundary points does not alter the interior solution.

For uniqueness let $w=\tilde{\Phi }-\Phi$. Formally the divergence theorem applied to $w\nabla w$ givesFor continuous zero boundary data, this immediately forces $w=0$. With the actual discontinuous data, one should justify the two exceptional points rather than assume finite Dirichlet energy there. Let $|w|\le M$ and choose a cutoff ${\eta }_{ϵ}$ which is zero within ${ϵ}^2$ of those points, one beyond $ϵ$, and linear in logarithmic distance between. Its gradient satisfies $\int |\nabla {\eta }_{ϵ}{|}^2=O(1/|\log ϵ|)$. Applying the divergence theorem to ${\eta }_{ϵ}^{2}w\nabla w$ givesThe resulting energy tends to zero. Hence $\nabla w=0$ on every compact subset and $w$ is constant; its zero boundary trace on the open arcs makes it zero. This is the logarithmic cutoff proof of harmonic uniqueness with finitely many boundary exceptions.

For the exterior harmonic extension of disk boundary data, boundedness instead excludes $r^n$ and the logarithm. The same boundary coefficients giveThis is ${\Phi }_{\mathrm{in}}(R^2/r,\theta )$, using inversion in a circle. Its limit at infinity is $C/2$. To see why that value is forced independently of this construction, average any bounded exterior harmonic function over a circle. The mean satisfies $(r{\overline{\Phi }}^{\prime }{)}^{\prime }=0$, hence is $A+B\log r$. Boundedness forces $B=0$, and the boundary mean fixes $A=C/2$. A constant limit at infinity must therefore be $C/2$.

Finally suppose two bounded exterior solutions have this limit. Their difference $w$ tends to zero. Under planar inversion in a circle, it is bounded and harmonic near the origin, so the isolated point there is removable. Its local expansion gives $w=O(r^{-1})$, ${\partial }_{r}w=O(r^{-2})$ in the original exterior. Apply the preceding divergence theorem energy argument on $R<r<L$, with logarithmic cutoffs at the two boundary jumps. After removing those cutoffs, the inner boundary term vanishes; the outer one is ${\int }_{r=L}w{\partial }_{r}wds=O(L^{-2})$. Letting $L\to \infty$ forces $\nabla w=0$, and the zero limit gives $w=0$. Thus the bounded exterior solution is unique as well.

A Poisson kernel represents the harmonic extension of Dirichlet boundary data by an integral over the boundary. When the Dirichlet Green function is normalized by $-\Delta G_{\Omega }=\delta$, its negative outward normal derivative gives the displayed kernel. The Poisson kernel for the upper half-space and the Poisson kernel for the upper half-plane give explicit examples with different boundary dimensions. Boundedness or other stated growth conditions are needed when selecting a solution in an unbounded domain.