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TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-26
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A harmonic function is recovered from its values on any containing circle by the Poisson formula

Statement

Let u be harmonic on an open set containing the closed disc D(a,R)‾, and let z=a+ρeiϕ with 0≤ρ<R. Then

u(z)=12π∫02πR2−ρ2R2−2Rρcos⁡(ϕ−t)+ρ2 u(a+Reit) dt.

So a harmonic function on a disc is recovered from its boundary values on any larger concentric circle lying inside its domain.

Facts & Assumptions

Given: A harmonic function u on a neighbourhood of D(a,R)‾.

[L1]

The Poisson integral gives the unique continuous harmonic extension from the unit-circle boundary to the closed unit disc (The Poisson integral gives the unique continuous harmonic extension on the closed unit disc).

[L2]

Two continuous harmonic functions on the same bounded domain with the same boundary values agree (The bounded plane Dirichlet problem has at most one continuous harmonic solution).

Proof

technique · direct
1.1givenalgebra

Define v(w):=u(a+Rw) on D‾. Then v is continuous on D‾, harmonic on D, and has boundary values v(eit)=u(a+Reit).

2.1step 1.1L1L2

By [L1], the Poisson integral of the boundary function t↦u(a+Reit) is a continuous harmonic function on D‾ with the same boundary values as v; [L2] therefore makes it equal to v throughout D‾.

3.1step 2.1algebra∎

Evaluating step 2.1 at w=(z−a)/R=(ρ/R)eiϕ gives exactly the displayed formula, because the unit-disc kernel there is 1−(ρ/R)21−2(ρ/R)cos⁡(ϕ−t)+(ρ/R)2=R2−ρ2R2−2Rρcos⁡(ϕ−t)+ρ2.

Depends on

Used by

Dependency tree · two levels

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Sources