Alphabeta Math
CorollaryStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-26
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The bounded plane Dirichlet problem has at most one continuous harmonic solution

Statement

Let Ω be a bounded complex domain, and let u,v be continuous on Ω‾ and harmonic on Ω. If u=v on ∂Ω, then u=v on Ω‾.

Facts & Assumptions

Given: A bounded complex domain Ω, continuous functions u,v on Ω‾, both harmonic on Ω, and equality u=v on ∂Ω.

[L1]

For a bounded domain, a continuous harmonic function attains its maximum and minimum on the boundary unless it is constant (Maximum and minimum principles for plane harmonic functions).

Proof

technique · direct
1.1givenalgebra

Let w:=u−v. Then w is continuous on Ω‾, harmonic on Ω, and satisfies w=0 on ∂Ω.

2.1step 1.1L1algebra

Applying [L1] to w gives sup⁡Ω‾w=sup⁡∂Ωw=0, so w≤0 on Ω‾; applying [L1] to −w gives sup⁡Ω‾(−w)=0, so w≥0 on Ω‾. Hence w=0 everywhere.

3.1step 2.1∎

Therefore u=v on Ω‾.

Depends on

Used by

Dependency tree · two levels

7 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