CounterexampleConstruction: AI-adaptedVerification: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-09

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The wrong normal gives the wrong sign

Statement refuted

The claim that the divergence formula remains valid with the inward unit normal is false. A witness is $F (x) = x$ on $B_{R} (0) \subset R^{n}$ , $n \geq 2$ , $R > 0$ : the inward flux is $- n ∣ B_{R} ∣$ but the divergence integral is $n ∣ B_{R} ∣ > 0$ . Use ${A C}_{ω}$ for the integration convention.

Facts & Assumptions

Given: Assume ${A C}_{ω}$ , $n \geq 2$ , $R > 0$ . Take the ball $B_{R} (0)$ , field $F (x) = x$ , and the inward normal as the proposed witness.

[F1]

For F=x the outward ball flux is n times its positive volume. (Flux and scaling on balls).

Counterexample

1.1

The ball is bounded with smooth boundary, and F is a polynomial C1 field on its closure. Directly $div F = \sum_{i = 1}^{n} 1 = n$ , so its integral is $n ∣ B_{R} ∣$ . This is positive: $(- R / (2 \sqrt{n}), R / (2 \sqrt{n}))^{n} \subset B_{R}$ has positive product volume.

givenalgebra

2.1

The inward unit normal is $- x / R$ on $∣ x ∣ = R$ . Consequently $F \cdot ν_{i n} = - ∣ x ∣^{2} / R = - R$ and its flux is $- R ∣ \partial B_{R} ∣ = - n ∣ B_{R} ∣$ by F1. Step 1.1 proves this differs from the positive divergence integral, although every domain and field regularity hypothesis holds. It is exactly the orientation hypothesis that fails.

step 1.1F1algebra∎

Source notes

Hunter §1.12 Theorem 1.46, printed p. 17 (PDF p. 23), explicitly requires the outward normal. This sign counterexample is its ball specialization.

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Sources

Hunter, Notes on Partial Differential Equations (standard reference, not scraped)