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Kernel projection gives dense conforming gauge approximants

Statement

Assume Countable Choice. Let Ω be a bounded connected Lipschitz polyhedron and let {Th} be a shape-regular family of matching affine tetrahedral meshes with maximum cell diameter h→0, as in Matching affine simplicial meshes and shape regularity and the mesh hypothesis of Smooth fields are approximated by the lowest-order edge family. For each mesh let Kh=ker⁡(curl⁡∣Vh) and Xh=Vh∩Kh⊥, with the L2 inner product. Then for every u∈X and every sequence of mesh sizes hn→0, there are uhn∈Xhn such that ∥uhn−u∥H(curl⁡;Ω)⟶0.

Facts & Assumptions

Given: Countable Choice, the constant ϵ>0 from the Maxwell definition, u∈X, the finite-dimensional PEC edge spaces Vh, and the shape-regular refining mesh family from Smooth fields are approximated by the lowest-order edge family.

[A1]

Countable Choice is inherited from the smooth edge-approximation supplier. (The Axiom of Countable Choice (ACω))

[F1]

H0(curl⁡) is the graph-norm closure of Cc∞(Ω;R3). Since Vh⊂V and every vector in Kh has zero distributional curl, Kh⊂K. (Zero tangential trace H(curl) space, PEC Maxwell cavity eigenproblem and full-kernel gauge)

[F2]

For every compactly supported smooth field w, the canonical edge interpolants Ihw are in Vh for sufficiently small h and converge to w in the H(curl⁡) norm. (Smooth fields are approximated by the lowest-order edge family)

[F3]

Every finite-dimensional subspace of the real L2 Hilbert space has an orthonormal basis; a closed subspace has an orthogonal projection. (Every finite-dimensional real or complex inner product space has an orthonormal basis, Orthogonal decomposition by a closed subspace)

[F4]

Since ϵ is a positive constant scalar, the weighted and ordinary L2 orthogonal complements of K and Kh coincide. (PEC Maxwell cavity eigenproblem and full-kernel gauge)

Proof

Given: Countable Choice, u∈X, and an arbitrary mesh sequence hn→0.

1.1A1F1F2givenchoose

By [F1], for every j≥1 there is φj∈Cc∞(Ω;R3) with ∥u−φj∥H(curl⁡)<1/j. Countable Choice selects one such field for each j. By [F2], for each fixed j there is an index Nj such that for every n≥Nj, Ihnφj∈Vhn and ∥Ihnφj−φj∥H(curl⁡)<1/j; choose the least such Nj, and replace the sequence by increasing thresholds Mj:=max⁡{j,N1,…,Nj,Mj−1+1}, with M0=0.

1.2A1F3givenconstruct

Each Khn is a finite-dimensional subspace of the real L2 Hilbert space. By Countable Choice [A1], choose an orthonormal basis for each Khn using [F3]. If a sequence in Khn converges in L2, its coordinates against this finite basis converge, and the limit is the corresponding finite linear combination; thus Khn is closed. By [F3] it has an orthogonal projection Pn satisfying ∥Pny∥2≤∥y∥2, including Pn=0 when Khn={0}.

2.1step 1.1algebra

For each n with {j≤n:Mj≤n}≠∅, let j(n) be its largest member and set wn:=Ihnφj(n); for the finitely many initial indices where the set is empty put wn=0. Since Mj≥j and is strictly increasing, j(n)→∞. For all later indices, ∥wn−u∥H(curl⁡)<2/j(n), so wn→u in graph norm.

3.1F1F3F4step 2.1step 1.2

Since Khn⊂K by [F1] and u∈K⊥ by [F4], Pnu=0. Thus ∥Pnwn∥2=∥Pn(wn−u)∥2≤∥wn−u∥2. Set uhn:=wn−Pnwn. Both terms lie in Vhn and the projection residual is orthogonal to Khn, so uhn∈Xhn.

4.1step 2.1step 3.1algebra∎

Because curl⁡(Pnwn)=0, ∥uhn−u∥2≤2∥wn−u∥2 and ∥curl⁡(uhn−u)∥2=∥curl⁡(wn−u)∥2. Consequently ∥uhn−u∥H(curl⁡)≤3∥wn−u∥H(curl⁡)→0 by step 2.1.

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