TheoremStatement: Literature-sourcedProof: AI-generatedPipeline-generatedprecheck passjudge pass (gpt-6-luna)

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The hydrostatic HLL update preserves lakes at rest, including dry cells

Statement

For any discrete bed values, the update of Hydrostatic reconstruction with a dry-safe HLL flux and one-sided pressure corrections leaves every lake-at-rest profile of One-dimensional shallow-water balance law with a dry-state convention unchanged, including a wet/dry shoreline with dry cells above the free-surface level. On a finite chain, the supplied ghost states and bed values must extend the same lake-at-rest profile.

Facts & Assumptions

Given: $g > 0$ , arbitrary cell bed values $b_{i}$ , a level $η_{0}$ , and cell states with $m_{i} = 0$ , $h_{i} + b_{i} = η_{0}$ when wet, and $h_{i} = 0$ , $b_{i} \geq η_{0}$ when dry. On a finite chain these conditions also hold for the supplied ghost data. Verify every cell update.

[F1]

One-dimensional shallow-water balance law with a dry-state convention: a dry cell has state $(0, 0)$ and the physical flux there is $(0, 0)$ ; a wet lake-at-rest cell has $m = 0$ and $h + b = η_{0}$ .

[F2]

Hydrostatic reconstruction with a dry-safe HLL flux and one-sided pressure corrections: the face bed is the maximum of the two adjacent beds, reconstructed depths are positive parts of free surface minus that bed, and each corrected side flux adds its cell pressure minus reconstructed pressure.

Proof

1.1givenF1F2algebra

Fix a face and set $b^{*} = \max (b_{L}, b_{R})$ . If a side cell is wet, its free surface is $η_{0}$ , so its reconstructed depth is $(η_{0} - b^{*})_{+}$ . If a side cell is dry, then $b_{s} \geq η_{0}$ and $b^{*} \geq b_{s}$ , so its reconstructed depth is $(b_{s} - b^{*})_{+} = 0 = (η_{0} - b^{*})_{+}$ . Thus both reconstructed depths equal $H = (η_{0} - b^{*})_{+}$ , and both reconstructed velocities are zero.

2.1F1F2step 1.1algebra

If $H > 0$ , the states on the face are equal to $(H, 0)$ and the middle HLL formula has speeds $- \sqrt{g H}$ and $\sqrt{g H}$ , so its flux is $F (H, 0) = (0, g H^{2} / 2)$ . If $H = 0$ , both states and speeds are zero and the dry branch returns $(0, 0)$ . Thus in either case the common flux is $(0, g H^{2} / 2)$ .

3.1F2step 2.1algebra∎

On the side belonging to a cell of depth $h_{s}$ , the correction in [F2] changes this common flux to $(0, g h_{s}^{2} / 2)$ . Therefore the right corrected flux of cell $i$ and the left corrected flux of that same cell are both $(0, g h_{i}^{2} / 2)$ . Their difference is zero, so the update leaves every $U_{i}$ unchanged.

Depends on

Used by

A lake at rest over a bump with a dry crest Example

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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)
F. Bouchut, III. Shallow water problem – Hydrostatic reconstruction scheme (standard reference, not scraped)