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A first HLL step wets a dry cell without negative depth

Example

On the full-line unit grid, set $g = 1$ , $b_{i} = 0$ , $U_{i}^{0} = (1, 0)$ for $i \leq 0$ , and $U_{i}^{0} = (0, 0)$ for $i \geq 1$ . Use $λ = Δ t / Δ x = 1 / 2$ for one hydrostatic HLL step.

Facts & Assumptions

Given: The step data and time step above. Compute the wet/dry interface fluxes, verify the cellwise CFL and update the two cells adjacent to the initial front.

[F1]

Hydrostatic reconstruction with a dry-safe HLL flux and one-sided pressure corrections: with flat bed the hydrostatic states equal the cell states; the HLL flux uses the specified wave-speed extrema and the finite-volume flux-difference update.

[F2]

The hydrostatic HLL step conserves water and preserves dry admissibility: on a finite chain with admissible ghost states, including a single updated cell, the face-specific CFL $λ (a_{R} at right - a_{L} at left) \leq 1$ preserves admissible depth and dry momentum.

Verification

1.1F1givenalgebra

At every wet/wet face the states agree at $(1, 0)$ , the speeds are $(- 1, 1)$ , and the HLL flux is $(0, 1 / 2)$ . At every dry/dry face the speeds and flux are zero. At the unique wet/dry face, $F_{L} = (0, 1 / 2)$ , $F_{R} = (0, 0)$ , $U_{R} - U_{L} = (- 1, 0)$ and $a_{L} a_{R} = - 1$ , so the middle formula in [F1] gives $\hat{F} = (1 / 2, 1 / 4)$ .

2.1F1F2step 1.1algebra

For a wet cell adjacent to the front the outgoing-speed sum is $1 + 1 = 2$ , so $λ$ times that sum is $1$ ; for the first dry cell it is $1 + 0 = 1$ , so the product is $1 / 2$ ; all other cells have either the wet/wet value $1$ or the dry/dry value $0$ on each side, and satisfy the same bound. For each $i$ , apply [F2] to the single-cell chain consisting of cell $i$ , with its actual initial neighbors $U_{i - 1}^{0}, U_{i + 1}^{0} \in K$ as ghost states and their bed values zero. Its two face fluxes and update coincide with those on the full-line grid by [F1]. Thus every updated cell is admissible.

3.1F1F2step 1.1step 2.1algebra∎

The wet cell $i = 0$ has left flux $(0, 1 / 2)$ and right flux $(1 / 2, 1 / 4)$ , hence $U_{0}^{1} = (1, 0) - \frac{1}{2} ((1 / 2, 1 / 4) - (0, 1 / 2)) = (3 / 4, 1 / 8)$ . The dry cell $i = 1$ has left flux $(1 / 2, 1 / 4)$ and right flux $(0, 0)$ , hence $U_{1}^{1} = (0, 0) - \frac{1}{2} ((0, 0) - (1 / 2, 1 / 4)) = (1 / 4, 1 / 8)$ . Both depths are positive, and their sum remains $1$ , so the front wets the adjacent dry cell without losing water.

Depends on

Used by

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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)