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An entropy-stable Burgers flux still smears an under-resolved shock

Example

For the Burgers flux and quadratic entropy, the unit-dissipation entropy-stable flux has a negative Tadmor residual at the entropy shock $1 \to 0$ . On the four-cell periodic data $(1, 1, 0, 0)$ and with $ν = Δ t / Δ x = 1 / 2$ , one explicit forward-Euler update replaces the two states adjacent to that jump by $11 / 12$ and $1 / 3$ . Thus semidiscrete entropy stability coexists with visible shock smearing at this coarse resolution. This computation gives no fully-discrete entropy inequality for forward Euler.

Facts & Assumptions

Given: Burgers' flux $f (u) = u^{2} / 2$ , the entropy-stable flux from the preceding example, the four periodic cell states, and $ν = 1 / 2$ . Verify entropy admissibility of the exact shock, the semidiscrete flux residual, and the displayed finite update.

[F1]

Convex entropy pairs, entropy variables, and entropy potentials for systems: for $η = u^{2} / 2$ and $q = u^{3} / 3$ , $v = u$ and $ψ = u^{3} / 6$ .

[F2]

Tadmor entropy-conservative two-point flux: consistency and $[v] f^{e c} = [ψ]$ define entropy conservation.

[F3]

Entropy-stable flux from positive-semidefinite entropy-variable dissipation: with $D = 1$ , $f^{e s} = f^{e c} - [u] / 2$ .

[F4]

Semidiscrete cell entropy inequality for Tadmor-dissipative fluxes: this flux gives the semidiscrete cell entropy inequality, not a fully discrete time-step claim.

Verification

1.1givenF1algebra

The Rankine–Hugoniot speed from states $u_{L} = 1, u_{R} = 0$ is $(f (0) - f (1)) / (0 - 1) = 1 / 2$ . Since $f^{'} (0) = 0 < 1 / 2 < 1 = f^{'} (1)$ , it is a compressive shock. Its entropy production is $q (0) - q (1) - \frac{1}{2} (η (0) - η (1)) = - 1 / 3 + 1 / 4 = - 1 / 12 < 0$ , so it satisfies the quadratic entropy inequality.

2.1givenF1F2F3step 1.1algebra

The entropy-conservative Burgers flux is $f^{e c} = (u_{L}^{2} + u_{L} u_{R} + u_{R}^{2}) / 6$ : it is symmetric and diagonal-consistent, and $(u_{R} - u_{L}) f^{e c} = (u_{R}^{3} - u_{L}^{3}) / 6 = [ψ]$ , so [F2] applies. Hence $f^{e c} (1, 0) = 1 / 6$ and [F3] gives $f^{e s} (1, 0) = 2 / 3$ . With $[u] = - 1$ and $[ψ] = - 1 / 6$ from step 1.1, the Tadmor residual is $[u] f^{e s} - [ψ] = - 2 / 3 + 1 / 6 = - 1 / 2 < 0$ .

3.1givenF1F2F3F4step 2.1algebra∎

Equal-state faces have flux $f^{e s} (1, 1) = 1 / 2$ and $f^{e s} (0, 0) = 0$ by [F2]–[F3]. Thus the left state adjacent to the jump updates to $1 - \frac{1}{2} (2 / 3 - 1 / 2) = 11 / 12$ , and the right state updates to $0 - \frac{1}{2} (0 - 2 / 3) = 1 / 3$ . The jump is spread over two intermediate values. This is a finite computation only; [F4] does not certify the forward-Euler step's entropy inequality. The interface flux value was obtained in step 2.1.

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Sources

E. Tadmor, Entropy Stable Schemes for Hyperbolic Conservation Laws, Handbook of Numerical Methods for Hyperbolic Problems, Vol. 17 (standard reference, not scraped)
R. Bürger, P. E. Méndez, and C. Parés, On Entropy Stable Schemes for Degenerate Parabolic Multispecies Kinematic Flow Models (standard reference, not scraped)