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

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HLL depth flux is a bounded outflow/inflow combination

Statement

At a face of the scheme in Hydrostatic reconstruction with a dry-safe HLL flux and one-sided pressure corrections, write $α = a_{R}$ and $β = - a_{L}$ . There are coefficients $p, q \geq 0$ such that ${\hat{F}}^{h} = p h_{L}^{*} - q h_{R}^{*}, p \leq α, q \leq β .$ If $α > 0$ and $h_{L}^{*} > 0$ , then $p < α$ ; if $β > 0$ and $h_{R}^{*} > 0$ , then $q < β$ . These statements include the branches where one speed is zero and the all-dry branch.

Facts & Assumptions

Given: The hydrostatic reconstructed states, physical flux and HLL branch definitions. Derive the mass-flux coefficients and their bounds without assuming exact Riemann speeds.

[F1]

Hydrostatic reconstruction with a dry-safe HLL flux and one-sided pressure corrections: $α = \max (0, u_{L}^{*} + c_{L}^{*}, u_{R}^{*} + c_{R}^{*})$ , $β = \max (0, c_{L}^{*} - u_{L}^{*}, c_{R}^{*} - u_{R}^{*})$ , and the middle mass flux is the first component of the displayed HLL formula.

[F2]

Hydrostatic reconstruction with a dry-safe HLL flux and one-sided pressure corrections: $F^{h} (U_{s}^{*}) = h_{s}^{*} u_{s}^{*}$ , and $c_{s}^{*} = \sqrt{g h_{s}^{*}}$ ; when $h_{s}^{*} = 0$ , both $u_{s}^{*}$ and $F (U_{s}^{*})$ are zero.

Proof

1.1givenF1F2algebra

In the middle branch $α, β > 0$ . Define $p = α (u_{L}^{*} + β) / (α + β)$ and $q = β (α - u_{R}^{*}) / (α + β)$ . Substituting $a_{L} = - β$ and $a_{R} = α$ into [F1], then using [F2], gives ${\hat{F}}^{h} = (α h_{L}^{*} u_{L}^{*} + β h_{R}^{*} u_{R}^{*} - α β (h_{R}^{*} - h_{L}^{*})) / (α + β) = p h_{L}^{*} - q h_{R}^{*}$ .

2.1F1F2step 1.1algebra

The speed definitions give $u_{L}^{*} + β \geq c_{L}^{*} \geq 0$ , $α - u_{L}^{*} \geq c_{L}^{*} \geq 0$ , $α - u_{R}^{*} \geq c_{R}^{*} \geq 0$ , and $β + u_{R}^{*} \geq c_{R}^{*} \geq 0$ . Hence the coefficients in step 1.1 satisfy $p, q \geq 0$ , $p \leq α$ , and $q \leq β$ . When $h_{L}^{*} > 0$ , $c_{L}^{*} > 0$ and $α - u_{L}^{*} > 0$ , so $p < α$ ; when $h_{R}^{*} > 0$ , $c_{R}^{*} > 0$ and $β + u_{R}^{*} > 0$ , so $q < β$ .

3.1F1F2step 2.1algebra∎

If $a_{L} = 0 < a_{R}$ , the first HLL branch gives ${\hat{F}}^{h} = h_{L}^{*} u_{L}^{*}$ ; the speed definitions imply $u_{L}^{*} \geq c_{L}^{*} \geq 0$ and $α \geq u_{L}^{*} + c_{L}^{*}$ , so take $p = u_{L}^{*}$ and $q = 0$ . If $a_{L} < 0 = a_{R}$ , the second branch gives ${\hat{F}}^{h} = h_{R}^{*} u_{R}^{*}$ , with $u_{R}^{*} \leq - c_{R}^{*} \leq 0$ and $β \geq c_{R}^{*} - u_{R}^{*}$ , so take $p = 0$ and $q = - u_{R}^{*}$ . If both speeds vanish, their definitions force $u_{s}^{*} - c_{s}^{*} \geq 0$ and $u_{s}^{*} + c_{s}^{*} \leq 0$ on each side, hence $c_{s}^{*} = u_{s}^{*} = h_{s}^{*} = 0$ and the first branch returns zero. These choices satisfy the bounds in step 2.1 and cover every equality-speed branch.

Depends on

Hydrostatic reconstruction with a dry-safe HLL flux and one-sided pressure corrections

Used by

Dependency tree · two levels

2 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)