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Positivity, invariant domains and entropy-stable systems fluxes — Examples

1 · Prerequisites

2 · Summary

These worked cases exercise the exact assumptions of the companion page: near-vacuum Euler states on the symmetric HLL speed branch, a two-point Burgers entropy residual, a finite-node conservative positivity rescaling, and a coarse Burgers shock update that shows why semidiscrete entropy stability does not promise sharp shock resolution. The last example reports its forward-Euler state calculation without asserting a fully discrete entropy inequality.

3 · Logical flowchart

4 · Definitions, theorems and proofs

None yet.

5 · Examples, counterexamples and false statements

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Exact Tadmor entropy residual for a dissipative Burgers flux

Example

For Burgers' flux $f (u) = u^{2} / 2$ and entropy $η (u) = u^{2} / 2$ , the entropy flux is $q (u) = u^{3} / 3$ and the entropy potential is $ψ (u) = u^{3} / 6$ . The symmetric two-point flux $f^{e c} (u_{L}, u_{R}) = \frac{u_{L}^{2} + u_{L} u_{R} + u_{R}^{2}}{6}$ is entropy-conservative. Adding unit dissipation in entropy variables gives $f^{e s} = f^{e c} - (u_{R} - u_{L}) / 2$ ; at the jump $1 \to 0$ its Tadmor entropy residual is exactly $- 1 / 2$ .

Facts & Assumptions

Given: Burgers' flux, the quadratic entropy pair, and the two-point flux formula. Verify compatibility, the entropy-conservative identity, and the dissipative residual at $(u_{L}, u_{R}) = (1, 0)$ .

[F1]

Convex entropy pairs, entropy variables, and entropy potentials for systems: an entropy pair satisfies $q^{'} = η^{'} f^{'}$ , with $v = η^{'}$ and $ψ = v f - q$ .

[F2]

Tadmor entropy-conservative two-point flux: a continuous, consistent, symmetric two-point flux is entropy-conservative when $[v] f^{e c} = [ψ]$ for every pair of states, with right-minus-left jumps.

[F3]

Entropy-stable flux from positive-semidefinite entropy-variable dissipation: $f^{e s} = f^{e c} - D [v] / 2$ for symmetric positive-semidefinite $D$ .

Verification

1.1givenF1algebra

Here $η^{''} (u) = 1$ , $η^{'} (u) = u$ , and $q^{'} (u) = u^{2} = η^{'} (u) f^{'} (u)$ , so this is a convex entropy pair. Thus $v = u$ and $ψ = u f - q = u^{3} / 6$ by [F1].

2.1givenF1F2step 1.1algebra

The displayed $f^{e c}$ is polynomial, hence continuous, is symmetric, and satisfies $f^{e c} (u, u) = u^{2} / 2 = f (u)$ . Moreover, $(u_{R} - u_{L}) f^{e c} (u_{L}, u_{R}) = \frac{u_{R}^{3} - u_{L}^{3}}{6} = ψ (u_{R}) - ψ (u_{L})$ , so [F2] applies using $ψ = u^{3} / 6$ from step 1.1.

3.1givenF2F3step 2.1algebra∎

Taking $D = 1$ in [F3], at $(1, 0)$ we have $f^{e c} = 1 / 6$ , $[v] = - 1$ , and $[ψ] = - 1 / 6$ . Therefore $[v] f^{e s} - [ψ] = (- 1) (1 / 6 + 1 / 2) - (- 1 / 6) = - 1 / 2$ . In general the same substitution gives residual $- \frac{1}{2} (u_{R} - u_{L})^{2} \leq 0$ , using the formula in step 2.1.

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Two Euler quadrature nodes rescaled to a positive margin

Example

With weights $w_{1} = w_{2} = 1 / 2$ , cell mean $\overset{ˉ}{U} = (1, 0, 1)$ , and candidate conserved states $V_{1} = (1, 4, 1)$ and $V_{2} = (1, - 4, 1)$ , apply the limiter to $K_{7 / 5, 1 / 2}$ . It returns $θ_{*} = 1 / 4$ , giving states $(1, 1, 1)$ and $(1, - 1, 1)$ with positive pressure and the original cell mean.

Facts & Assumptions

Given: The weights, mean, nodes, and set $K_{7 / 5, 1 / 2}$ . Compute the maximal scaling and verify the mean and positive margin.

[F1]

Maximal convex scaling toward an admissible cell mean: the limiter chooses the largest $θ \in [0, 1]$ keeping each $\overset{ˉ}{U} + θ (V_{j} - \overset{ˉ}{U})$ in $K$ .

[F2]

Positive-margin admissible states for ideal-gas Euler: $K_{γ, ε}$ requires $ρ \geq ε$ and $E - m^{2} / (2 ρ) \geq ε$ .

[F3]

Admissible states and pressure for one-dimensional ideal-gas Euler: $p = (γ - 1) (E - m^{2} / (2 ρ))$ .

[F4]

Convex scaling preserves the cell mean and positive-margin node states: the maximal scaling exists, preserves the weighted mean, and returns nodes in the positive-margin set.

Verification

1.1givenF2algebra

The weighted mean is $(V_{1} + V_{2}) / 2 = (1, 0, 1)$ . Its density is $1$ and its internal-energy density is $1$ , so it belongs to $K_{7 / 5, 1 / 2}$ by [F2]. The scaled states are $(1, 4 θ, 1)$ and $(1, - 4 θ, 1)$ , each with internal energy $1 - 8 θ^{2}$ . Hence they lie in $K_{7 / 5, 1 / 2}$ exactly when $1 - 8 θ^{2} \geq 1 / 2$ , or $0 \leq θ \leq 1 / 4$ .

2.1givenF1F2F3F4algebra∎

The largest feasible parameter is $θ_{*} = 1 / 4$ by [F1]. The outputs are $(1, 1, 1)$ and $(1, - 1, 1)$ ; each has internal-energy density $1 / 2$ and pressure $(2 / 5) (1 / 2) = 1 / 5 > 0$ by [F3]. Their weighted mean is $(1, 0, 1)$ , also as guaranteed by [F4].

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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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A near-vacuum Euler state remains positive under symmetric HLL speeds

Example

Take $γ = 7 / 5$ , density $ρ = 10^{- 8}$ , pressure $p = 10^{- 12}$ , and velocities $u_{L} = - 1 / 100$ , $u_{R} = 1 / 100$ . The conserved states have $m_{L, R} = \mp 10^{- 10}$ and $E_{L, R} = 3 \times 10^{- 12}$ . Choose the symmetric HLL speed $a = 11 / 500$ at every face and Courant ratio $ν = Δ t / h = 125 / 11$ . For the four-cell periodic data $(U_{L}, U_{L}, U_{R}, U_{R})$ , one HLL forward-Euler update preserves positive density and pressure despite the very small initial values. The calculation illustrates a first-order positivity branch; it does not claim accuracy of the shock profile.

Facts & Assumptions

Given: The ideal-gas Euler flux and HLL update, with the parameters above. Verify the states, speed branch, CFL condition, and positivity after one update.

[F1]

Admissible states and pressure for one-dimensional ideal-gas Euler: $p = (γ - 1) (E - m^{2} / (2 ρ))$ and $c^{2} = γ p / ρ$ .

[F2]

Ideal-gas Euler states remain positive under Lax–Friedrichs flux splitting: if $a \geq ∣ u ∣ + c$ , then $U \pm F (U) / a \in G_{γ}$ .

[F3]

Euler positive states are convex and positive-margin sets are closed convex: $G_{γ}$ is convex.

[F4]

First-order HLL updates are averages of adjacent half-cell fan averages: under $2 ν a \leq 1$ , the HLL update is the mean of adjacent half-cell fan averages.

[F5]

Symmetric HLL speeds preserve ideal-gas density and internal-energy positivity: the symmetric HLL branch with $a \geq ∣ u ∣ + c$ and $2 ν a \leq 1$ preserves $G_{γ}$ .

[F6]

First-order finite-volume update using supplied HLL speeds: $U_{i}^{+} = U_{i} - ν ({\hat{F}}_{i + 1 / 2} - {\hat{F}}_{i - 1 / 2})$ .

[F7]

HLL two-wave approximate flux: the symmetric middle-branch HLL flux is used at every interface, with equal-state data substituted into its displayed formula.

Verification

1.1givenF1algebra

The kinetic energy density is $ρ u^{2} / 2 = 5 \times 10^{- 13}$ and $p / (γ - 1) = 5 / 2 \times 10^{- 12}$ , so $E = 3 \times 10^{- 12}$ . Hence [F1] gives the specified pressure and positive internal energy, so both states lie in $G_{γ}$ .

2.1givenF1F2F5step 1.1algebra

Their sound speed satisfies $c^{2} = 7 / 50000 < (3 / 250)^{2}$ , hence $∣ u ∣ + c < 1 / 100 + 3 / 250 = 11 / 500 = a$ . Also $ν a = 1 / 4$ , so $2 ν a = 1 / 2 \leq 1$ . The symmetric-speed hypotheses in [F2]–[F5] hold, with the positive admissibility verified in step 1.1.

3.1givenF1F2F3F4F6F7step 2.1algebra∎

At the discontinuity, substitution of $S_{L} = - a, S_{R} = a$ into [F7] and the HLL intermediate-state formula of [F4] gives $U_{*} = (U_{L} + F_{L} / a + U_{R} - F_{R} / a) / 2$ . Both split states are in $G_{γ}$ by [F2], so $U_{*} \in G_{γ}$ by [F3]. At equal-state faces, the same middle-state formula gives $U_{*} = U_{L}$ or $U_{R}$ . Substituting the shared face fluxes in [F6], or equivalently using [F4], yields $U_{i}^{+} = (3 / 4) U_{L} + (1 / 4) U_{*}$ and $U_{i + 1}^{+} = (3 / 4) U_{R} + (1 / 4) U_{*}$ . Convexity keeps both states in $G_{γ}$ , and [F1] gives positive density and pressure, using the CFL verification in step 2.1.

Positivity, invariant domains and entropy-stable systems fluxes — Examples

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

5 · Examples, counterexamples and false statements

Exact Tadmor entropy residual for a dissipative Burgers flux

Example

Facts & Assumptions

Verification

Two Euler quadrature nodes rescaled to a positive margin

Example

Facts & Assumptions

Verification

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

Example

Facts & Assumptions

Verification

A near-vacuum Euler state remains positive under symmetric HLL speeds

Example

Facts & Assumptions

Verification

Sources