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The SIPDG broken energy expression is definite

Statement

Assume Countable Choice. Let ∥⋅∥DG be the energy expression from Symmetric interior-penalty DG form and energy seminorm on a finite matching simplicial mesh of a nonempty bounded polyhedral domain, with λ>0 and γ>0. Then ∥⋅∥DG is a norm on Vhbr.

Facts & Assumptions

Given: Countable Choice; a finite face-to-face triangulation of a bounded polyhedral open set, λ>0, γ>0, and the broken polynomial space and energy expression of Symmetric interior-penalty DG form and energy seminorm.

[F1]

The squared energy expression is the sum of nonnegative cell-gradient terms and positive-weight squared jumps on every interior and boundary facet. (Symmetric interior-penalty DG form and energy seminorm)

[F2]

A matching mesh is a finite union of nondegenerate simplices with pairwise-disjoint interiors; each interior facet is shared and each boundary facet is exterior. (Matching affine simplicial meshes and shape regularity)

[F3]

Products of L2 functions satisfy the integral Cauchy–Schwarz inequality. (Holder's inequality for integrals, including the endpoint cases)

[F5]

On each physical simplex, the Green identity applies to u=1 and the vector field G(x)=x, giving d∣K∣=∑F⊂∂K∫Fx⋅nK,F dS. (Sobolev traces and Gauss–Green formula on physical simplices)

[F6]

Countable Choice is the declared Sobolev/trace background of [F1] and the physical-simplex Green supplier [F5]. (The Axiom of Countable Choice (ACω))

Proof

1.1F1givenalgebra

Suppose ∥vh∥DG=0. Every cell-gradient term is zero, and positive definiteness of AK implies ∇(vh∣K)=0 almost everywhere. Since vh∣K is a polynomial on the connected simplex K, it is a constant cK.

2.1F1F2F5F6step 1.1algebra

Every interior-facet jump is zero by [F1], so adjacent cells have equal constants. Thus cK is constant on each connected component of the cell-facet adjacency graph. For any such component, sum [F5] over its finitely many cells. The cell volumes are positive, while the two oriented integrals on every shared facet cancel; if there were no unshared facet, this would give d∑K∣K∣=0, a contradiction. Every unshared facet is an exterior mesh facet by the matching property [F2]. On that facet the boundary jump is the cell trace itself, so [F1] forces the component's constant to be zero. Therefore vh=0 on every cell.

3.1F1F2F3F4step 2.1algebra∎

The expression is homogeneous under scalar multiplication. Define its associated pairing by bDG(vh,wh):=∑K∫K∇vhTAK∇wh+γ∑FhF−1∫F[vh]F[wh]F. For any vectors ξ,η, the quadratic (ξ+tη)TAK(ξ+tη) is nonnegative for all real t; if η≠0, its discriminant is nonpositive, giving ∣ξTAKη∣≤(ξTAKξ)1/2(ηTAKη)1/2, and the case η=0 is immediate. [F3] bounds the resulting cell integrals and each facet integral, while [F4] combines the finitely many cell and facet terms to give ∣bDG(vh,wh)∣≤∥vh∥DG∥wh∥DG. Expanding bDG(vh+wh,vh+wh) gives ∥vh+wh∥DG2≤(∥vh∥DG+∥wh∥DG)2. Thus the triangle inequality holds; together with step 2.1 this proves the norm axioms.

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