LemmaStatement: AI-adaptedProof: AI-generatedPipeline-generatedprecheck passjudge pass (gpt-6-luna)

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A dry HLL face sends water only into the dry side

Statement

At any face in Hydrostatic reconstruction with a dry-safe HLL flux and one-sided pressure corrections, if the reconstructed right state is dry, then ${\hat{F}}^{h} \geq 0$ , and $\hat{F} = (0, 0)$ whenever ${\hat{F}}^{h} = 0$ . If the reconstructed left state is dry, then ${\hat{F}}^{h} \leq 0$ , and again ${\hat{F}}^{h} = 0$ implies $\hat{F} = (0, 0)$ .

Facts & Assumptions

Given: The HLL branches, shallow-water flux and speed bounds from Hydrostatic reconstruction with a dry-safe HLL flux and one-sided pressure corrections and HLL depth flux is a bounded outflow/inflow combination. Check the two orientations of a dry face, including both-zero speeds.

[F1]

Hydrostatic reconstruction with a dry-safe HLL flux and one-sided pressure corrections: a dry reconstructed state is $(h^{*}, m^{*}) = (0, 0)$ with $u^{*} = c^{*} = 0$ and $F (0, 0) = (0, 0)$ .

[F2]

HLL depth flux is a bounded outflow/inflow combination: in the middle branch, ${\hat{F}}^{h} = p h_{L}^{*} - q h_{R}^{*}$ , with $p, q \geq 0$ , and $p > 0$ when $h_{L}^{*} > 0$ and $α, β > 0$ , while $q > 0$ when $h_{R}^{*} > 0$ and $α, β > 0$ .

Proof

1.1F1F2algebra

Suppose $h_{R}^{*} = 0$ . Then $u_{R}^{*} = c_{R}^{*} = 0$ . If both speeds are zero, [F1] makes the full flux zero. If $a_{L} < 0 = a_{R}$ , the second branch is $F_{R}^{*} = (0, 0)$ . If $a_{L} = 0 < a_{R}$ , the speed definition forces $u_{L}^{*} \geq c_{L}^{*}$ . When $h_{L}^{*} > 0$ , this gives ${\hat{F}}^{h} = h_{L}^{*} u_{L}^{*} > 0$ ; when $h_{L}^{*} = 0$ , both states and the flux are dry and zero. In the middle branch, [F2] gives ${\hat{F}}^{h} = p h_{L}^{*} \geq 0$ ; it is strictly positive for a wet left state and otherwise both states are dry and the full flux is zero.

1.2F1F2algebra

Suppose instead $h_{L}^{*} = 0$ . Then $u_{L}^{*} = c_{L}^{*} = 0$ . If both speeds vanish, the full flux is zero. If $a_{L} = 0 < a_{R}$ , the first branch is $F_{L}^{*} = (0, 0)$ . If $a_{L} < 0 = a_{R}$ , the speed definition forces $u_{R}^{*} \leq - c_{R}^{*}$ . For $h_{R}^{*} > 0$ this makes ${\hat{F}}^{h} = h_{R}^{*} u_{R}^{*} < 0$ ; if $h_{R}^{*} = 0$ , the full flux is zero. In the middle branch, [F2] gives ${\hat{F}}^{h} = - q h_{R}^{*} \leq 0$ , strictly negative for a wet right state and otherwise zero with both states dry.

2.1step 1.1step 1.2F1∎

The two orientations show that zero mass flux at a dry face occurs only when the selected outer state is dry or both reconstructed states are dry; in each case [F1] gives zero momentum flux as well.

Depends on

Used by

The hydrostatic HLL step conserves water and preserves dry admissibility Theorem

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

E. Audusse, F. Bouchut, M.-O. Bristeau, R. Klein, and B. Perthame, A Fast and Stable Well-Balanced Scheme with Hydrostatic Reconstruction for Shallow Water Flows (standard reference, not scraped)
D. I. Ketcheson, R. J. LeVeque, and M. J. del Razo, Riemann Problems and Jupyter Solutions, Chapter 12 (standard reference, not scraped)