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Euler positive states are convex and positive-margin sets are closed convex

Statement

Fix $γ > 1$ and $ε > 0$ . The ideal-gas admissible set $G_{γ} = {(ρ, m, E) : ρ > 0, E - \frac{m^{2}}{2 ρ} > 0}$ and the positive-margin set $K_{γ, ε}$ from Positive-margin admissible states for ideal-gas Euler are convex, and $K_{γ, ε}$ is closed in $R^{3}$ .

Facts & Assumptions

Given: $γ > 1$ , $ε > 0$ , and the two state sets just defined. Prove their convexity and the closedness of the margin set.

[F1]

Admissible states and pressure for one-dimensional ideal-gas Euler: $G_{γ}$ is given by $ρ > 0$ and $E - m^{2} / (2 ρ) > 0$ .

[F2]

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

Proof

1.1givenF1algebra

Let $U_{k} = (ρ_{k}, m_{k}, E_{k}) \in G_{γ}$ for $k = 1, 2$ and $0 \leq θ \leq 1$ . Put $ρ_{θ} = θ ρ_{1} + (1 - θ) ρ_{2}$ and similarly define $m_{θ}, E_{θ}$ . Weighted Cauchy–Schwarz applied to $m_{θ} = \sqrt{θ ρ_{1}} (\sqrt{θ} m_{1} / \sqrt{ρ_{1}}) + \sqrt{(1 - θ) ρ_{2}} (\sqrt{1 - θ} m_{2} / \sqrt{ρ_{2}})$ gives $m_{θ}^{2} \leq ρ_{θ} (θ m_{1}^{2} / ρ_{1} + (1 - θ) m_{2}^{2} / ρ_{2})$ . Thus $ρ_{θ} > 0$ and $E_{θ} - m_{θ}^{2} / (2 ρ_{θ}) \geq θ (E_{1} - m_{1}^{2} / (2 ρ_{1})) + (1 - θ) (E_{2} - m_{2}^{2} / (2 ρ_{2})) > 0$ . Therefore $G_{γ}$ is convex.

2.1givenF2step 1.1algebra

For $U_{1}, U_{2} \in K_{γ, ε}$ , the weighted Cauchy–Schwarz inequality in step 1.1 gives $ρ_{θ} \geq ε$ and $E_{θ} - m_{θ}^{2} / (2 ρ_{θ}) \geq θ ε + (1 - θ) ε = ε$ . Hence $K_{γ, ε}$ is convex.

3.1givenF2algebra∎

If $U_{n} = (ρ_{n}, m_{n}, E_{n}) \in K_{γ, ε}$ converges to $U = (ρ, m, E)$ , then $ρ \geq ε > 0$ and continuity of $(ρ, m, E) \mapsto E - m^{2} / (2 ρ)$ on $ρ \geq ε$ gives $E - m^{2} / (2 ρ) \geq ε$ . Thus $U \in K_{γ, ε}$ , proving closedness.

Depends on

Used by

Cited to discharge well-definedness by Positive-margin admissible states for ideal-gas Euler.

Dependency tree · two levels

3 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

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)