CounterexampleConstruction: Literature-sourcedVerification: AI-adaptedPipeline-generatedjudge pass (gpt-6.1-sol)audited 2026-10-02

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Zero half-space trace does not ensure uniqueness without growth control

Statement refuted

For $n \geq 2$ the normal coordinate $u (x^{'}, t) = t$ is harmonic on $H = {t > 0}$ , continuous on its closure, and zero on the boundary plane, while $u$ is nonzero and unbounded. Hence the zero Dirichlet trace has at least the solutions $0$ and $t$ if boundedness or another valid growth condition is omitted.

Facts & Assumptions

Given: an integer $n \geq 2$ and the open upper half-space $H = {x = (x^{'}, t) \in R^{n - 1} \times R : t > 0}$ .

[F1]

For a $C^{2}$ function $f$ on an open set, $Δ f = \sum_{i < n} \partial_{i} \partial_{i} f$ , and $f$ is called harmonic when $Δ f = 0$ (The Laplacian of a $C^{2}$ function and of a $C^{2}$ vector field).

Counterexample

technique · direct

1.1givenF1algebra

Define $u : H \to R$ by $u (x^{'}, t) : = t$ , i.e. $u (x) = x_{n}$ with $x_{n}$ the last coordinate. Its first partial derivatives are $\partial_{i} u \equiv δ_{i n}$ and its second partial derivatives all vanish identically, so $Δ u = \sum_{i} \partial_{i} \partial_{i} u = 0$ on the open half-space $H$ ; hence $u$ is harmonic by [F1].

2.1step 1.1F1algebra

The same formula defines a continuous extension of $u$ to the closed half-space $\overline{H} = {x : x_{n} \geq 0}$ , and on the boundary plane $\partial H = {x_{n} = 0}$ this extension has the value $0$ . The zero function $0$ is harmonic on $H$ with the same zero boundary values.

3.1step 1.1step 2.1algebra

The two solutions differ and the second is unbounded: $u (e_{n}) = 1 \neq 0 = 0 (e_{n})$ at the point $e_{n} = (0, 1) \in H$ , and along the vertical ray ${s e_{n} : s > 0}$ the value $u (s e_{n}) = s$ tends to $+ \infty$ , so $\sup_{H} ∣ u ∣ = + \infty$ while $\sup_{H} ∣ 0 ∣ = 0$ . Thus $0$ and $u$ are two distinct solutions of the same zero Dirichlet problem on $H$ once boundedness --- or any other growth restriction excluding linear growth --- is dropped.

4.1step 1.1step 2.1step 3.1F1∎

Steps 1.1, 2.1 and 3.1 exhibit a nonzero unbounded harmonic function with the same continuous zero boundary trace as the zero function on the half-space; uniqueness of the half-space Dirichlet problem therefore requires a growth condition such as boundedness, and this witness is eliminated by it. The computation uses no choice principle.

Depends on

The Laplacian of a $C^2$ function and of a $C^2$ vector field

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

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Sources

Armin Schikorra, Partial Differential Equations I & II (2025) (standard reference, not scraped)