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The hydrostatic HLL step conserves water and preserves dry admissibility

Statement

Let $N \geq 1$ cells form a finite chain, with two admissible ghost states used only to define the boundary faces, or let the cells be periodic. Apply one step of Hydrostatic reconstruction with a dry-safe HLL flux and one-sided pressure corrections to states in $K$ from One-dimensional shallow-water balance law with a dry-state convention. If $λ (a_{R, i + 1 / 2} - a_{L, i - 1 / 2}) \leq 1 (1 \leq i \leq N),$ then each updated state remains in $K$ . The depth is nonnegative; if it is zero then the momentum is zero. On a finite chain the total water changes by the net boundary discharge, and on a periodic mesh total water is conserved.

Facts & Assumptions

Given: Admissible cell and ghost states, the hydrostatic update, the stated cellwise CFL, and the HLL coefficient and dry-face lemmas. Prove one-step admissibility and water balance.

[F1]

Hydrostatic reconstruction with a dry-safe HLL flux and one-sided pressure corrections: one-sided corrections change only momentum; the first component uses one shared face mass flux, and each reconstructed depth is no larger than its original cell depth.

[F2]

HLL depth flux is a bounded outflow/inflow combination: at a face, ${\hat{F}}^{h} = p h_{L}^{*} - q h_{R}^{*}$ with $0 \leq p \leq a_{R}$ and $0 \leq q \leq - a_{L}$ ; the corresponding coefficient is strictly below its speed bound when its reconstructed side is wet.

[F3]

A dry HLL face sends water only into the dry side: at a dry reconstructed side the mass flux points into the dry side, and zero mass flux implies the full vector flux is zero.

[F4]

One-dimensional shallow-water balance law with a dry-state convention: every state with positive depth lies in $K$ , while a state with zero depth lies in $K$ exactly when its momentum is zero.

Proof

1.1givenF1F2algebra

Write $p_{R}$ for the left-state coefficient at face $i + 1 / 2$ , $q_{L}$ for the right-state coefficient at face $i - 1 / 2$ , and use analogous notation for the two inflow coefficients. Expanding the first component of [F1] with [F2] gives $h_{i}^{n + 1} = h_{i}^{n} - λ (p_{R} h_{i, +}^{*} + q_{L} h_{i, -}^{*}) + λ (q_{R} h_{i + 1, -}^{*} + p_{L} h_{i - 1, +}^{*})$ . The inflow terms are nonnegative and the total outflow is at most $(a_{R, i + 1 / 2} - a_{L, i - 1 / 2}) h_{i}^{n}$ , so the CFL gives $h_{i}^{n + 1} \geq 0$ .

2.1F1F2F4step 1.1algebra

If $h_{i}^{n} > 0$ and at least one outgoing speed is positive, the corresponding outflow term is strictly below its speed bound times $h_{i}^{n}$ : this follows from [F2] if the reconstructed depth is positive, and is immediate if that reconstructed depth is zero. By step 1.1 and the CFL, $h_{i}^{n + 1} > 0$ . If both outgoing speeds are zero, the outflow is zero and step 1.1 gives $h_{i}^{n + 1} \geq h_{i}^{n} > 0$ . Thus a wet cell remains wet and is in $K$ by [F4].

2.2F1F3step 1.1algebra

If $h_{i}^{n} = 0$ , then $m_{i}^{n} = 0$ and [F1] gives $h_{i, +}^{*} = h_{i, -}^{*} = 0$ , so both corrections on this cell vanish. Its left face has dry right state and its right face has dry left state. By [F3], the left mass flux is nonnegative and the right mass flux nonpositive; if the updated depth is zero, nonnegativity forces both mass fluxes to vanish.

3.1F1F3F4step 2.1step 2.2algebra

In the zero-depth case of step 2.2, [F3] makes both full HLL face fluxes zero. The cell corrections also vanish, so its updated momentum is zero and its updated state lies in $K$ by [F4]. If instead that initially dry cell has positive updated depth, its updated state lies in $K$ by [F4] as well. Together with step 2.1, this proves that every updated state lies in $K$ under the stated CFL condition.

4.1F1step 2.1step 3.1algebra∎

Multiplying the first component of each update by $Δ x$ and summing over the finite chain cancels every interior shared mass flux, because the one-sided corrections have zero first component. This gives $Δ x \sum_{i = 1}^{N} (h_{i}^{n + 1} - h_{i}^{n}) = - Δ t ({\hat{F}}_{N + 1 / 2}^{h} - {\hat{F}}_{1 / 2}^{h})$ . Thus only boundary discharge changes finite-domain water; periodic identification makes the right-hand side zero. Combining this identity with steps 2.1 and 3.1 proves the theorem.

Depends on

Used by

A first HLL step wets a dry cell without negative depth 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)
D. I. Ketcheson, R. J. LeVeque, and M. J. del Razo, Riemann Problems and Jupyter Solutions, Chapter 12 (standard reference, not scraped)