ExampleConstruction: Literature-sourcedVerification: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-26

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
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The Poisson integral of cos(theta) is r cos(theta)

Example

For the boundary datum $φ (e^{i θ}) = \cos θ$ , the Poisson integral is

$P [φ] (r e^{i ϕ}) = r \cos ϕ .$

Facts & Assumptions

Given: The boundary datum $φ (e^{i θ}) = \cos θ$ on $\partial D$ .

[L1]

The Poisson integral gives the unique continuous harmonic extension to the closed unit disc (The Poisson integral gives the unique continuous harmonic extension on the closed unit disc, The Poisson integral on the unit disc).

[L2]

The real part of a holomorphic polynomial is harmonic (Complex polynomials are entire with the power-rule derivative, and rational functions are holomorphic wherever their denominator is nonzero, The $C^{2}$ real and imaginary parts of a holomorphic function satisfy Laplace's equation and form a harmonic-conjugate pair).

Verification

technique · direct

1.1

The function $u (r e^{i ϕ}) = r \cos ϕ$ is the real part of $z$ , so [L2] makes it harmonic on $D$ .

2.1

On the boundary $∣ z ∣ = 1$ , the same formula gives $u (e^{i θ}) = \cos θ = φ (e^{i θ})$ . By [L1], the continuous harmonic extension of $φ$ is unique, so $P [φ] = u$ .

step 1.1L1∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

15 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Sigurdur Helgason, MIT 18.112 Lecture 16: Harmonic Functions (standard reference, not scraped)