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SIPDG L2 error under adjoint consistency and dual approximation

Statement

Assume Countable Choice and the SIPDG setting of Symmetric interior-penalty DG form and energy seminorm. Let u satisfy all hypotheses of Exact consistency of the symmetric interior-penalty DG form, and let uh∈Vhbr satisfy ah(uh,vh)=ℓh(vh) for all vh∈Vhbr. Set e:=u−uh, let V:=H01(Ω;R)=W01,2(Ω;R) as in Zero-boundary Sobolev space as a norm closure, and let h:=max⁡K∈ThK, with 0<h≤ℓΩ:=diam⁡(Ω). Suppose s∈(1/2,1] and a dual solution ze∈V exists satisfying ∑K∈T∫KAK∇v⋅∇ze dx=∫Ωve dx(v∈V), with the following properties:

  1. Its one-sided normal coefficient-flux traces are in L2(F) in the weak Green sense defining Dh. In the branch s=1, assume also ze∣K∈H2(K) on every cell, with matching interior value traces, zero boundary value trace, and matching interior normal coefficient fluxes; then SIPDG adjoint consistency for piecewise H2 dual solutions gives ah(e,ze)=∥e∥L2(Ω)2. In the branch s<1, assume this adjoint-consistency identity as an additional hypothesis. The weak dual equation gives div⁡(AK∇ze)∣K=−e∣K∈L2(K), so the stated flux and trace conditions place ze in the form domain Dh of Symmetric interior-penalty DG form and energy seminorm.
  2. The specified normed dual-regularity space Rs is a subspace of V∩Dh, and ∥ze∥Rs≤Creg∥e∥L2(Ω) for some Creg≥0.
  3. There is zh∈Vhbr such that ∥ze−zh∥DG,∗≤CapphsℓΩ1−s∥ze∥Rs. Here Capp≥0.

Then ∥e∥L2(Ω)≤(2+2γ−1/2)CappCreghsℓΩ1−s∥e∥DG,∗. The theorem retains the dual regularity exponent and the half-order face-flux contribution in ∥⋅∥DG,∗. It does not infer the dual approximation estimate or adjoint-consistency identity from elliptic regularity alone.

Facts & Assumptions

Given: Countable Choice; the exact solution u satisfying Exact consistency of the symmetric interior-penalty DG form and the discrete solution uh satisfying the equation in the Statement; the dual solution ze∈H01(Ω;R); an exponent s∈(1/2,1]; the branch-specific adjoint-consistency hypotheses in the Statement; the dual regularity and approximation inequalities in the Statement; and the augmented SIPDG expression.

[F1]

Exact consistency and the discrete equation imply ah(e,vh)=0 for every vh∈Vhbr. (Exact consistency of the symmetric interior-penalty DG form)

[F2]

On its form domain, the SIPDG bilinear form satisfies ∣ah(v,w)∣≤CB∥v∥DG,∗∥w∥DG,∗ with CB=2+2γ−1/2. (Boundedness of the SIPDG form in the augmented norm)

[F3]

In the s=1 branch, piecewise H2 dual solutions with matching traces and normal flux satisfy ah(e,ze)=∥e∥L2(Ω)2. (SIPDG adjoint consistency for piecewise H2 dual solutions)

[F4]

Countable Choice is the stated background for the exact-consistency, adjoint-consistency, physical-trace, and SIPDG form-domain suppliers used here. (The Axiom of Countable Choice (ACω))

[F5]

The zero-boundary space V is the norm closure of Cc∞(Ω) in W1,2(Ω), so a smooth test compactly supported inside a cell belongs to V. (Zero-boundary Sobolev space as a norm closure)

[F6]

The reference trace maps transfer to bounded value traces on each physical simplex facet for every cellwise W1,2 function. (Sobolev traces and Gauss–Green formula on physical simplices)

[F7]

The exact solution is cellwise H2 by the hypotheses of exact consistency, and the SIPDG definition places cellwise H2 functions and broken polynomials in Dh. (Exact consistency of the symmetric interior-penalty DG form, Symmetric interior-penalty DG form and energy seminorm)

[A1]

For s<1, the dual solution is assumed to have the stated weak L2 normal-flux traces and adjoint-consistency identity; this carrier does not prove the fractional trace theorem that would discharge those assumptions.

Proof

1.1F1F3F4A1givenchoose

The result is immediate if ∥e∥L2=0. Otherwise choose zh from the stated approximation hypothesis. For s=1, [F3] gives ah(e,ze)=∥e∥L22 under the stated piecewise-H2 and trace-matching hypotheses; for s<1 this identity is [A1]. Exact consistency plus the discrete equation give ah(e,zh)=0 by [F1]. Hence ∥e∥L22=ah(e,ze−zh).

1.2F2F4F5F6F7given

By [F7], u is cellwise H2 and uh,zh are broken polynomials, so e=u−uh and zh belong to Dh. For each cell K, a test in Cc∞(K) extended by zero belongs to V by [F5]; the weak dual equation then gives div⁡(AK∇ze)∣K=−e∣K∈L2(K) in distributions. Item 1 supplies the one-sided L2 normal flux traces, and ze∈V supplies its cellwise H1 restrictions and value traces by [F6]. Thus ze∈Dh and [F2] gives ∣ah(e,ze−zh)∣≤CB∥e∥DG,∗∥ze−zh∥DG,∗ with CB=2+2γ−1/2.

2.1step 1.1step 1.2givenalgebra∎

Apply step 1.2 to step 1.1 and then use the dual approximation and regularity assumptions. This gives ∥e∥L22≤CBCappCreghsℓΩ1−s∥e∥DG,∗∥e∥L2. Divide by the positive ∥e∥L2 to obtain the asserted estimate.

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