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SIPDG adjoint consistency for piecewise H2 dual solutions

Statement

Assume Countable Choice and the mesh and coefficient hypotheses of Symmetric interior-penalty DG form and energy seminorm. Let $e ∣_{K} \in H^{2} (K)$ on every cell, and let $z ∣_{K} \in H^{2} (K)$ be a dual solution with matching interior traces, zero boundary trace, matching normal coefficient fluxes, and $- \nabla \cdot (A_{K} \nabla z) = e$ on every cell. Then the SIPDG form, extended to these piecewise regular functions by its same formula, satisfies $a_{h} (e, z) = ∥ e ∥_{L^{2} (Ω)}^{2} .$

Facts & Assumptions

Given: Countable Choice; a finite matching affine simplicial mesh; piecewise constant symmetric positive-definite $A_{K}$ ; piecewise $H^{2}$ error $e$ ; and a piecewise $H^{2}$ dual solution $z$ with matching traces, zero boundary trace, matching normal fluxes, and $- \nabla \cdot (A_{K} \nabla z) = e$ cellwise.

[F1]

The SIPDG formula is symmetric in its two arguments, with face terms $- q_{F} (v) [w]_{F} - q_{F} (w) [v]_{F} + γ h_{F}^{- 1} [v]_{F} [w]_{F}$ . (Symmetric interior-penalty DG form and energy seminorm)

[F2]

For $v, G \in W^{1, 2}$ on a physical affine simplex, the trace Green identity is $\int_{K} v div G + \nabla v \cdot G = \sum_{F \subset \partial K} \int_{F} (γ_{F} v) (γ_{F} G) \cdot n_{K, F}$ . (Sobolev traces and Gauss–Green formula on physical simplices)

[F3]

Countable Choice is required by the imported reference-simplex Green identity [F2]. (The Axiom of Countable Choice ( ${A C}_{ω}$ ))

[F4]

$H^{2} (K)$ means $W^{2, 2} (K)$ , so $A_{K} \nabla z \in W^{1, 2} (K; R^{d})$ when $A_{K}$ is constant. (Integer-order Sobolev spaces and their norms)

Proof

1.1F1F2F3F4given

On each cell apply [F2] with scalar $v = e ∣_{K}$ and vector field $G = A_{K} \nabla z$ , which belongs to $W^{1, 2}$ by [F4]. Symmetry of $A_{K}$ and the dual equation give $\int_{K} A_{K} \nabla e \cdot \nabla z = \int_{K} e^{2} + \sum_{F \subset \partial K} \int_{F} e (A_{K} \nabla z \cdot n_{K, F})$ .

2.1step 1.1givenalgebra

Sum step 1.1 over all cells. Matching normal fluxes make each interior-face contribution $\int_{F} q_{F} (z) [e]_{F}$ , and the boundary contributions use $[e]_{F} = e$ . Thus $\sum_{K} \int_{K} A_{K} \nabla e \cdot \nabla z = ∥ e ∥_{L^{2} (Ω)}^{2} + \sum_{F} \int_{F} q_{F} (z) [e]_{F}$ .

3.1F1step 2.1givenalgebra∎

The traces of $z$ match on interior facets and vanish on boundary facets, so $[z]_{F} = 0$ for every $F$ . In $a_{h} (e, z)$ the term $- q_{F} (e) [z]_{F}$ and the penalty term therefore vanish; the remaining flux term is $- \sum_{F} \int_{F} q_{F} (z) [e]_{F}$ . Substituting step 2.1 proves the identity.

Depends on

Used by

SIPDG L2 error under adjoint consistency and dual approximation Theorem

Dependency tree · two levels

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Sources

Jean-Luc Guermond, MATH 661 Fall 2025, Part VIII, Chapter 38: Discontinuous Galerkin (standard reference, not scraped)
Douglas N. Arnold, Franco Brezzi, Bernardo Cockburn, and L. Donatella Marini, Unified Analysis of Discontinuous Galerkin Methods for Elliptic Problems, SIAM Journal on Numerical Analysis 39(5) (2002) (standard reference, not scraped)