Poisson’s formula and boundary values

Try

Suppose the temperature prescribed around the unit circle is 1+2cos θ. The harmonic interior solution is 1+2r cos θ, where z=r exp(iθ). At the center, the angular variation disappears and the value is the boundary average one.

Poisson’s formula extends this construction from a few trigonometric modes to every continuous boundary function. Its kernel behaves like a weighted average concentrated near the nearest boundary angle as the interior point approaches the circle.

Predict

Predict whether the center value depends on a single boundary point. It does not; at r=0 the kernel is constant, so all angles receive equal weight.

Predict whether the kernel can exceed one near the boundary. It can become very large in a narrow angular region while its total integral remains fixed. The normalization controls total weight, not the pointwise height.

Experiment

Track single Fourier modes.

Boundary value g(θ)	Interior harmonic extension
1	1
cos θ	r cos θ
sin(2θ)	r²sin(2θ)
cos(3θ)	r³cos(3θ)

Higher frequencies decay more rapidly toward the center. These exact formulas describe harmonic smoothing, not an arbitrary numerical filter added to the boundary data.

Understand

For 0≤r<1 define P_r(t)=(1−r²)/(1−2r cos t+r²). It is positive. Expanding the real part of (1+r exp(it))/(1−r exp(it)) gives P_r(t)=1+2Σ rⁿcos(nt). For fixed r<1 this series converges uniformly, so integrating over a full period gives total mass 2π.

For real-valued, continuous 2π-periodic boundary data g, define u(r,θ)=(1/(2π))∫ from 0 to 2π of P_r(θ−t)g(t)dt. To see harmonicity directly, for each boundary point ζ=exp(it), the kernel is Re((ζ+z)/(ζ−z)). On any compact interior disk, the denominator and its derivatives are uniformly controlled, so the integral is the real part of a holomorphic function. Thus u is harmonic.

To prove boundary convergence, subtract g(θ) inside the integral, using normalization. Uniform continuity of g makes |g(θ−t)−g(θ)| small for small angular t. Positivity and total mass one control that near part. Outside a fixed small angular neighborhood, the denominator stays bounded below while 1−r² tends to zero. The far kernel mass therefore tends uniformly to zero. This proves u(r,θ)→g(θ) uniformly as r approaches one. Combined with continuity of g in angle, it gives continuous extension to the closed disk.

Uniqueness follows from the harmonic maximum principle. The difference of two continuous closed-disk solutions has zero boundary data; applying the maximum principle to it and its negative forces it to be zero everywhere.

Apply

Construct the solution for g(θ)=4+cos(2θ)−2sin θ. It is 4+r²cos(2θ)−2r sin θ. Verify it as the real part of 4+z²+2iz and check the boundary substitution.

Repair “The Poisson kernel converges pointwise to the boundary data.” The kernel and the data are different objects. Away from angle zero the kernel tends to zero, while near zero its height grows. Its integral against continuous data approaches that data because of concentration and normalization.

Continue

This solves the disk Dirichlet problem for continuous boundary values. Discontinuous data need a more careful statement about which boundary limits exist, and general domains require additional methods. Optional depth: transfer a disk solution through an explicit conformal map, checking how the prescribed boundary correspondence behaves at exceptional points such as infinity.