CorollaryStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck 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 $\overline{Ω}$ and harmonic on $Ω$ . If $u = v$ on $\partial Ω$ , then $u = v$ on $\overline{Ω}$ .

Facts & Assumptions

Given: A bounded complex domain $Ω$ , continuous functions $u, v$ on $\overline{Ω}$ , both harmonic on $Ω$ , and equality $u = v$ on $\partial Ω$ .

[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

1.1

Let $w : = u - v$ . Then $w$ is continuous on $\overline{Ω}$ , harmonic on $Ω$ , and satisfies $w = 0$ on $\partial Ω$ .

givenalgebra

2.1

Applying [L1] to $w$ gives $\sup_{\overline{Ω}} w = \sup_{\partial Ω} w = 0$ , so $w \leq 0$ on $\overline{Ω}$ ; applying [L1] to $- w$ gives $\sup_{\overline{Ω}} (- w) = 0$ , so $w \geq 0$ on $\overline{Ω}$ . Hence $w = 0$ everywhere.

step 1.1L1algebra

3.1

Therefore $u = v$ on $\overline{Ω}$ .

step 2.1∎

Depends on

Maximum and minimum principles for plane harmonic functions

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

Jeremy Orloff, MIT 18.04 Topic 5: Introduction to Harmonic Functions (standard reference, not scraped)