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Symmetric interior-penalty discontinuous Galerkin — Examples

1 · Prerequisites

2 · Summary

The interval calculations use a self-contained one-dimensional extension of the jump, average, and penalty formulas, with counting measure on point faces and $h_{F}$ equal to the common interval-cell length. They compute an exact interior consistency contribution and show that a too-small penalty makes the symmetric form negative on a nonzero test function. A separate broken polynomial example computes its distributional derivative and verifies why a nonzero facet jump is incompatible with global $H^{1}$ regularity.

3 · Logical flowchart

4 · Definitions, theorems and proofs

None yet.

5 · Examples, counterexamples and false statements

ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passjudge pass (gpt-6-luna)Open item page →

A broken polynomial with a nonzero jump is not globally H1

Example

On the interval partition $K^{-} = (0, 1 / 2)$ , $K^{+} = (1 / 2, 1)$ , let $v = 1$ on $K^{-}$ and $v = 0$ on $K^{+}$ . This is a broken piecewise-constant polynomial with jump $[v] = 1$ at the interior face. Its distributional derivative is a point mass, so $v \notin W^{1, 2} (0, 1)$ .

Facts & Assumptions

Given: Countable Choice; the interval partition; the piecewise-constant function above; and the weak-derivative definition of $W^{1, 2}$ .

[F1]

A weak first derivative $g \in L^{2} (0, 1)$ must satisfy $\int_{0}^{1} v φ^{'} = - \int_{0}^{1} g φ$ for every $φ \in C_{c}^{\infty} (0, 1)$ . (Integer-order Sobolev spaces and their norms)

[F2]

The interior scalar jump is the left trace minus the right trace. (Jump and average conventions for symmetric interior-penalty DG)

[F3]

The Sobolev-space definition assumes Countable Choice. (The Axiom of Countable Choice ( ${A C}_{ω}$ ))

[F4]

The $L^{2}$ case of Holder's inequality gives $∣ \int f g ∣ \leq ∥ f ∥_{2} ∥ g ∥_{2}$ . (Holder's inequality for integrals, including the endpoint cases)

Verification

1.1F2F3givenalgebra

The traces at $x = 1 / 2$ are $v^{-} = 1$ and $v^{+} = 0$ , so [F2] gives $[v] = 1$ . For every smooth compactly supported $φ$ , direct integration on the two intervals gives $\int_{0}^{1} v φ^{'} = \int_{0}^{1 / 2} φ^{'} = φ (1 / 2)$ .

2.1F1F4step 1.1givenconstruct∎

If an $L^{2}$ weak derivative $g$ existed, [F1] would give $φ (1 / 2) = - \int_{0}^{1} g φ$ . Use the explicit bump $η (t) = \exp (1 - (1 - 4 t^{2})^{- 1})$ for $∣ t ∣ < 1 / 2$ and $η (t) = 0$ otherwise; it is smooth because the exponential is flat at the endpoints, is supported in $[- 1 / 2, 1 / 2] \subset (- 1, 1)$ , and satisfies $η (0) = 1$ . Let $φ_{ε} (x) = η ((x - 1 / 2) / ε)$ for $0 < ε < 1 / 2$ . Then $φ_{ε} \in C_{c}^{\infty} (0, 1)$ , $φ_{ε} (1 / 2) = 1$ , and $∥ φ_{ε} ∥_{L^{2}} = \sqrt{ε} ∥ η ∥_{L^{2} (- 1, 1)} \to 0$ . The integral Cauchy–Schwarz inequality [F4] forces $∣ \int g φ_{ε} ∣ \leq ∥ g ∥_{2} ∥ φ_{ε} ∥_{2} \to 0$ , contradicting step 1.1. Thus no such $g$ exists and $v \notin W^{1, 2} (0, 1)$ .

ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passjudge pass (gpt-6-luna)Open item page →

One-dimensional SIPDG jump and penalty calculation

Example

This is a self-contained one-dimensional interval specialization of the jump convention in Jump and average conventions for symmetric interior-penalty DG. Let $K^{-} = (0, 1 / 2)$ and $K^{+} = (1 / 2, 1)$ , take unit coefficient, and set $h_{F} = 1 / 2$ at the interior point face. Define $v (x) = x$ on $K^{-}$ and $v (x) = x - 1$ on $K^{+}$ . Then $v (0) = v (1) = 0$ , while the interior jump at $x = 1 / 2$ is $1$ and the average derivative is $1$ . The face contributions are $- {v^{'}} [v] - {v^{'}} [v] + γ h_{F}^{- 1} [v]^{2}$ . With penalty parameter $γ = 1$ , the bilinear form has value $a_{h} (v, v) = 1$ .

Facts & Assumptions

Given: The two-cell interval partition, unit coefficient, counting measure on all point faces, $h_{F} = 1 / 2$ at each face (the common interval-cell length), the piecewise affine function $v$ , and $γ = 1$ . For functions $v, w$ on these cells, define $a_{h} (v, w) : = \sum_{K \in {K^{-}, K^{+}}} \int_{K} v^{'} w^{'} d x - \sum_{F \in {0, 1 / 2, 1}} (q_{F} (v) [w]_{F} + q_{F} (w) [v]_{F}) + γ \sum_{F \in {0, 1 / 2, 1}} h_{F}^{- 1} [v]_{F} [w]_{F},$ where $q_{1 / 2} (v) = {v^{'}}_{1 / 2}$ , $q_{0} (v) = - v^{'} (0 +)$ , and $q_{1} (v) = v^{'} (1 -)$ ; the endpoint jumps are the one-sided value traces and all face integrals use counting measure.

[F1]

On an interior face the scalar jump is the minus trace minus the plus trace, and the average is their arithmetic mean. (Jump and average conventions for symmetric interior-penalty DG)

Verification

1.1givenF1algebra

The derivative is $1$ on both cells, so the broken volume term is $\int_{0}^{1 / 2} 1 d x + \int_{1 / 2}^{1} 1 d x = 1$ . At the interior point $x = 1 / 2$ , [F1] gives $[v] = 1 / 2 - (- 1 / 2) = 1$ and ${v^{'}} = 1$ , while both boundary traces vanish.

2.1givenstep 1.1algebra∎

The interior consistency terms contribute $- 2 {v^{'}} [v] = - 2$ , and the penalty contributes $γ h_{F}^{- 1} [v]^{2} = 1 \cdot 2 \cdot 1 = 2$ . Adding these to the volume term from step 1.1 gives $a_{h} (v, v) = 1 - 2 + 2 = 1$ .

ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passjudge pass (gpt-6-luna)Open item page →

A too-small SIPDG penalty destroys coercivity

Example

Use the two-cell interval specialization and piecewise affine function described in Given below. At this mesh and with penalty $γ = 1 / 4$ , the function satisfies $a_{h} (v, v) = - 1 / 2$ . Hence the form is not coercive in this case, even though its energy expression on $v$ is positive.

Facts & Assumptions

Given: $K^{-} = (0, 1 / 2)$ , $K^{+} = (1 / 2, 1)$ , unit coefficient, counting measure on all point faces, $h_{F} = 1 / 2$ at each face (the common interval-cell length), the piecewise affine function $v (x) = x$ on $K^{-}$ and $v (x) = x - 1$ on $K^{+}$ , zero endpoint values, and penalty $γ = 1 / 4$ . For functions $v, w$ on these cells, define $a_{h} (v, w) : = \sum_{K \in {K^{-}, K^{+}}} \int_{K} v^{'} w^{'} d x - \sum_{F \in {0, 1 / 2, 1}} (q_{F} (v) [w]_{F} + q_{F} (w) [v]_{F}) + γ \sum_{F \in {0, 1 / 2, 1}} h_{F}^{- 1} [v]_{F} [w]_{F},$ where $q_{1 / 2} (v) = {v^{'}}_{1 / 2}$ , $q_{0} (v) = - v^{'} (0 +)$ , and $q_{1} (v) = v^{'} (1 -)$ ; the endpoint jumps are the one-sided value traces and all face integrals use counting measure. Define the energy expression by $∥ v ∥_{D G}^{2} : = \sum_{K \in {K^{-}, K^{+}}} \int_{K} ∣ v^{'} ∣^{2} d x + γ \sum_{F \in {0, 1 / 2, 1}} h_{F}^{- 1} [v]_{F}^{2} .$

[F1]

The one-dimensional jump is $v^{-} - v^{+}$ and the interior average is the arithmetic mean of the two traces. (Jump and average conventions for symmetric interior-penalty DG)

Verification

1.1givenF1algebra

Both cell derivatives equal $1$ , so the volume term is $1$ . The endpoint terms vanish because $v (0) = v (1) = 0$ . At the interior point, [F1] gives $[v] = 1$ and ${v^{'}} = 1$ . The energy expression is $∥ v ∥_{D G}^{2} = 1 + (1 / 4) (1 / (1 / 2)) \cdot 1^{2} = 3 / 2 > 0$ .

2.1givenstep 1.1algebra∎

The two consistency terms sum to $- 2$ , while the penalty term is $(1 / 4) (1 / (1 / 2)) \cdot 1^{2} = 1 / 2$ . Therefore $a_{h} (v, v) = 1 - 2 + 1 / 2 = - 1 / 2 < 0$ , which contradicts coercivity with any positive constant in a positive energy norm.