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Ideal-gas Euler states remain positive under Lax–Friedrichs flux splitting

Statement

Fix $γ > 1$ . For $U = (ρ, m, E) \in G_{γ}$ , let $e = E - m^{2} / (2 ρ)$ , $p = (γ - 1) e$ , $u = m / ρ$ , and $c = \sqrt{γ p / ρ}$ . If $a \geq ∣ u ∣ + c$ , then both split states $U_{δ} = U + δ F (U), δ \in {- 1 / a, 1 / a},$ belong to $G_{γ}$ , where $F (U) = (m, m^{2} / ρ + p, u (E + p))$ . More precisely, with $x = 1 + δ u$ their density and internal-energy density are $ρ_{δ} = ρ x, e_{δ} = e x - \frac{δ^{2} p^{2}}{2 ρ x} > 0.$

Facts & Assumptions

Given: An ideal-gas state $U \in G_{γ}$ , $γ > 1$ , and a speed $a \geq ∣ u ∣ + c$ . Prove positivity of both split states.

[F1]

Admissible states and pressure for one-dimensional ideal-gas Euler: $ρ > 0$ , $e = E - m^{2} / (2 ρ) > 0$ , $p = (γ - 1) e$ , $u = m / ρ$ , $c^{2} = γ p / ρ$ , and $F = (m, m^{2} / ρ + p, u (E + p))$ .

Proof

1.1givenF1algebra

Fix $δ \in {- 1 / a, 1 / a}$ . From [F1], direct substitution in $U + δ F (U)$ gives $ρ_{δ} = ρ (1 + δ u) = ρ x$ , $m_{δ} = ρ u (1 + δ u) + δ p$ , and $E_{δ} = E + δ u (E + p)$ . Expanding $E_{δ} - m_{δ}^{2} / (2 ρ_{δ})$ and using $E = e + ρ u^{2} / 2$ yields $e_{δ} = e x - δ^{2} p^{2} / (2 ρ x)$ .

2.1givenF1step 1.1algebra∎

Since $a \geq ∣ u ∣ + c$ and $c > 0$ , $x \geq 1 - ∣ u ∣ / a \geq c / a > 0$ , so $ρ_{δ} > 0$ . Also $∣ δ ∣ / x \leq 1 / c$ , hence $\frac{δ^{2} p^{2}}{2 ρ e x^{2}} \leq \frac{p^{2}}{2 ρ e c^{2}} = \frac{γ - 1}{2 γ} < 1$ . Therefore $e_{δ} = e x (1 - \frac{δ^{2} p^{2}}{2 ρ e x^{2}}) > 0$ , so $U_{δ} \in G_{γ}$ . The argument covers both signs of $δ$ and the equality case $a = ∣ u ∣ + c$ .

Depends on

Admissible states and pressure for one-dimensional ideal-gas Euler

Used by

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Sources

D. I. Ketcheson, R. J. LeVeque, and M. J. del Razo, Riemann Problems and Jupyter Solutions (standard reference, not scraped)
X. Zhang and C.-W. Shu, On Positivity-Preserving High Order Discontinuous Galerkin Schemes for Compressible Euler Equations on Rectangular Meshes (standard reference, not scraped)