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Planar rotation carries H(div) normal traces to tangential traces

Statement

Assume Countable Choice. For a nondegenerate oriented triangle K⊂R2, set R(a,b)=(−b,a), let n be cell-outward and t=Rn. If σ∈H(div⁡;K) is real, then Rσ∈H(curl⁡;K), curl⁡(Rσ)=div⁡σ distributionally, and ∫K(Rσ⋅R∇ϕ+curl⁡(Rσ)ϕ)=⟨γnσ,[ϕ]⟩ for all ϕ∈H1(K). Thus weak tangential trace equals weak normal trace under t=Rn.

Here H(curl⁡;K):={w∈L2(K;R2):∂1w2−∂2w1∈L2(K)} with its graph norm. The weak tangential trace is the functional on H1(K)/H01(K) defined by the integral in the Statement; the computation below proves it is the rotation of the PE08 weak normal trace.

Facts & Assumptions

Given: Countable Choice and real σ∈H(div⁡;K).

[A1]

Countable Choice is declared for trace interfaces. (The Axiom of Countable Choice (ACω))

[F1]

H(div) consists of real L2 vectors with L2 distributional divergence. (The real H(div) space)

[F2]

The weak normal trace acts by ∫K(σ⋅∇ϕ+div⁡σ ϕ). (Weak normal trace by Green duality)

Proof

1.1givenF1algebra

Writing Rσ=(−σ2,σ1) gives curl⁡(Rσ)=∂1σ1+∂2σ2=div⁡σ as distributions; [F1] places it in L2.

2.1A1F2step 1.1algebra∎

Orthogonality gives Rσ⋅R∇ϕ=σ⋅∇ϕ, so step 1.1 makes the trace integral exactly [F2]. Also Rσ⋅t=Rσ⋅Rn=σ⋅n.

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Sources