CounterexampleConstruction: AI-generatedVerification: AI-generatedPipeline-generatedprecheck passjudge pass (gpt-6-luna)

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Underintegrated Burgers collocation can increase quadratic entropy

Statement refuted

The SBP matrix identity alone guarantees that a naively collocated strong-form discretization of a nonlinear conservation law cannot increase its discrete quadratic entropy.

Witness Data

Use one reference interval $[- 1, 1]$ with three Legendre–Gauss–Lobatto nodes $(- 1, 0, 1)$ , weights $H = diag (1 / 3, 4 / 3, 1 / 3)$ , derivative matrix $D = (\begin{array}{ccc} - 3 / 2 & 2 & - 1 / 2 \\ - 1 / 2 & 0 & 1 / 2 \\ 1 / 2 & - 2 & 3 / 2 \end{array}),$ and boundary matrix $B = diag (- 1, 0, 1)$ . For periodic Burgers flux $f (u) = u^{2} / 2$ , define the naive collocated strong form $\dot{U} + D f (U) + H^{- 1} B ({\hat{f}}_{\partial} - f (U_{\partial})) = 0,$ where ${\hat{f}}_{\partial} = (f (U_{0}) + f (U_{2})) / 2$ is used at both periodic endpoints. Take $U = (- 3, - 1, - 2)^{T}$ .

Facts & Assumptions

Given: The three-node nodal vector, positive diagonal quadrature norm, exact SBP identity, pointwise Burgers flux, central periodic boundary flux, and quadratic entropy $E (U) = \frac{1}{2} U^{T} H U$ .

[F1]

Diagonal-norm summation-by-parts element data: SBP data satisfy $D 1 = 0$ and $H D + D^{T} H = B$ with positive diagonal $H$ .

Verification

1.1givenF1algebra

Direct multiplication gives $D 1 = 0$ and $H D + D^{T} H = diag (- 1, 0, 1) = B$ , so the derivative and norm satisfy the stated SBP identity. The three-point Gauss–Lobatto rule is exact through degree three, while both the quadratic entropy integrand $u_{h}^{2}$ and cubic entropy flux $u_{h}^{3}$ exceed that degree for a quadratic polynomial; the pointwise collocation therefore underintegrates these nonlinear products.

2.1givenstep 1.1algebra

At the chosen state, $f (U) = (9 / 2, 1 / 2, 2)^{T}$ and ${\hat{f}}_{\partial} = 13 / 4$ . Hence $D f (U) = (- 27 / 4, - 5 / 4, 17 / 4)^{T}$ and $H^{- 1} B ({\hat{f}}_{\partial} - f (U_{\partial})) = (15 / 4, 0, 15 / 4)^{T}$ . The semidiscrete residual is therefore $(- 3, - 5 / 4, 8)^{T}$ , so $\dot{U} = (3, 5 / 4, - 8)^{T}$ .

3.1givenstep 2.1algebra∎

The discrete entropy rate is $\dot{E} = U^{T} H \dot{U} = (- 3) (1 / 3) (3) + (- 1) (4 / 3) (5 / 4) + (- 2) (1 / 3) (- 8) = 2 / 3 > 0$ . This exact periodic state contradicts the asserted universal nonincrease, despite the SBP identity; it does not contradict the entropy-stable flux-differencing theorem, which uses a different volume flux.

Depends on

Diagonal-norm summation-by-parts element data

Used by

Nothing in the library uses this result yet.

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