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A degree-independent SIPDG penalty loses coercivity

Statement refuted

For the symmetric interior penalty discontinuous Galerkin (SIPDG) form, choosing a penalty coefficient $η$ independent of polynomial degree always makes the form coercive on discontinuous piecewise polynomials.

Facts & Assumptions

Given: Integer $p \geq 2$ , fixed $η > 0$ , the two unit intervals, and the stated interior jump, average, and bilinear form.

[F1]

Weighted Jacobi endpoint kernel on the reference interval for $d = 1$ gives shifted orthogonal polynomials $ϕ_{n}$ on $[0, 1]$ with $∥ ϕ_{n} ∥_{2}^{2} = 1 / (2 n + 1)$ .

Counterexample

Use the two-cell mesh $(- 1, 0) \cup (0, 1)$ with cell size one, natural exterior boundary, degree $p$ on each cell, jump $[v] = v (0^{-}) - v (0^{+})$ , average derivative ${v^{'}} = (v^{'} (0^{-}) + v^{'} (0^{+})) / 2$ , and symmetric interior penalty form

a_{h} (v, v) = \sum_{K} \int_{K} ∣ v^{'} ∣^{2} - 2 {v^{'}} [v] + η [v]^{2},

where $η > 0$ is fixed independent of $p$ .

1.1F1givenalgebra

For $d = 1$ , the Rodrigues polynomial in [F1] is $ϕ_{n} (t) = (- 1)^{n} D^{n} (t^{n} (1 - t)^{n}) / n!$ . Its value at $t = 1$ is $1$ : writing $s = 1 - t$ , the first nonzero term of $t^{n} (1 - t)^{n} = (1 - s)^{n} s^{n}$ is $s^{n}$ , and $D_{t}^{n} = (- 1)^{n} D_{s}^{n}$ . Define $K_{p - 1} (t) = \sum_{n = 0}^{p - 1} (2 n + 1) ϕ_{n} (t)$ . Orthogonality, the norm in [F1], and $ϕ_{n} (1) = 1$ give $K_{p - 1} (1) = \sum_{n = 0}^{p - 1} (2 n + 1) = p^{2}$ , $\int_{0}^{1} K_{p - 1} = 1$ , and $∥ K_{p - 1} - 1 ∥_{2}^{2} = p^{2} - 1$ .

2.1F1step 1.1givenalgebra∎

Put $q_{p} = (K_{p - 1} - 1) / 2$ . Then $\deg q_{p} \leq p - 1$ , $\int_{0}^{1} q_{p} = 0$ , $q_{p} (1) = (p^{2} - 1) / 2$ , and $\int_{0}^{1} q_{p}^{2} = (p^{2} - 1) / 4$ . Define $v_{p} (x) = 1 + \int_{0}^{x + 1} q_{p} (t) d t$ on $(- 1, 0)$ and $v_{p} (x) = 0$ on $(0, 1)$ . Its cell restrictions have degree at most $p$ , its jump is $1$ , its left derivative at the interface is $q_{p} (1)$ , and its right derivative is $0$ . Therefore the broken derivative energy is $(p^{2} - 1) / 4$ and the interface flux term is $- 2 (q_{p} (1) / 2) (1) = - (p^{2} - 1) / 2$ . Substituting into the stated form yields $a_{h} (v_{p}, v_{p}) = η - (p^{2} - 1) / 4$ , which is negative whenever $p^{2} > 4 η + 1$ . Meanwhile $\sum_{K} ∥ v_{p}^{'} ∥_{2}^{2} + [v_{p}]^{2} = (p^{2} - 1) / 4 + 1 > 0$ . This disproves coercivity for every fixed $η$ once $p$ is sufficiently large.

Depends on

Weighted Jacobi endpoint kernel on the reference interval

Used by

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Sources

Tim Warburton and Jan S. Hesthaven, On the Constants in hp-Finite Element Trace Inverse Inequalities (standard reference, not scraped)
Leszek F. Demkowicz, Mathematical Theory of Finite Elements (standard reference, not scraped)