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Zero mean of the divergence under zero normal trace

Statement

Assume Countable Choice. Let D⊂Rd be bounded and open. If v∈H0(div⁡;D), then div⁡v∈L1(D) and ∫Ddiv⁡v dx=0.

Facts & Assumptions

Given: Countable Choice, a bounded open set D, and v with zero weak normal trace.

[A1]

Countable Choice is inherited from the Sobolev and trace-quotient interfaces. (The Axiom of Countable Choice (ACω))

[F1]

H0(div⁡;D) is the kernel of the weak normal trace. (Zero-normal-trace H(div) space)

[F2]

Green duality defines ⟨γnv,[w]⟩=∫D(v⋅∇w+(div⁡v)w). (Weak normal trace by Green duality)

[F3]

Bounded sets have finite Lebesgue measure, so the constant function 1 belongs to L2(D) and has zero weak gradient in H1(D). (Lebesgue measure is sigma-finite, and every metrically bounded subset of Rn has finite outer measure, Integer-order Sobolev spaces and their norms)

[F4]

Holder's inequality implies ∥g∥L1(D)≤∣D∣1/2∥g∥L2(D) for g∈L2(D). (Holder's inequality for integrals, including the endpoint cases)

[F5]

The divergence of an H(div) field belongs to L2(D). (The real H(div) space)

Proof

1.1A1F3F4F5construct

Since div⁡v∈L2(D) by [F5], and ∣D∣<∞ by [F3], [F4] shows div⁡v∈L1(D). The constant class 1 is an admissible element of H1(D) by [F3] and the definition of the weak Sobolev norm, so its trace class [1] lies in TD.

2.1F2step 1.1algebra

Apply [F2] with w=1. Since ∇1=0, Green duality gives ⟨γnv,[1]⟩=∫Ddiv⁡v dx.

3.1F1step 2.1∎

By [F1], γnv=0 as an element of TD′ acting on TD, so its value at [1]∈TD is zero. The identity in step 2.1 therefore gives ∫Ddiv⁡v=0.

Depends on

Used by

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Sources