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Well-balanced balance laws and wet/dry fronts — Examples

1 · Prerequisites

Well-balanced balance laws and wet/dry fronts

2 · Summary

Examples

The finite calculations verify a rest lake with a dry crest, a first HLL step that wets a neighboring dry cell while preserving water, and a concrete central-flux/centered-source update that creates a spurious momentum from a discrete lake. The counterexample distinguishes exact well-balancedness of the specified hydrostatic reconstruction from any general claim about centered source approximations.

3 · Logical flowchart

4 · Definitions, theorems and proofs

None yet.

5 · Examples, counterexamples and false statements

CounterexampleConstruction: AI-generatedVerification: AI-generatedprecheck passjudge pass (gpt-6-luna)Open item page →

A centered pressure flux and centered bed source move a discrete lake

Statement refuted

On a periodic three-cell grid, the central physical-flux rule ${\hat{F}}_{i + 1 / 2} = (F (U_{i}) + F (U_{i + 1})) / 2$ with centered bed source $(0, - g h_{i} (b_{i + 1} - b_{i - 1}) / (2 Δ x))$ preserves every lake-at-rest state under the update $U_{i}^{+} = U_{i} - \frac{Δ t}{Δ x} ({\hat{F}}_{i + 1 / 2} - {\hat{F}}_{i - 1 / 2}) + Δ t (0, - \frac{g h_{i} (b_{i + 1} - b_{i - 1})}{2 Δ x}) .$

Facts & Assumptions

Given: The periodic central-flux/centered-source update in the refuted statement, $Δ x = g = 1$ , any $Δ t > 0$ , bed values $(b_{0}, b_{1}, b_{2}) = (0, 1 / 2, 0)$ , and states $m_{i} = 0$ , $h_{i} = 1 - b_{i} = (1, 1 / 2, 1)$ . These are admissible lake-at-rest data with level $η_{0} = 1$ under One-dimensional shallow-water balance law with a dry-state convention. Show one updated momentum is nonzero.

[F1]

One-dimensional shallow-water balance law with a dry-state convention: for a wet rest state, $m_{i} = 0$ , $h_{i} + b_{i} = η_{0}$ , and the momentum flux is $g h_{i}^{2} / 2$ .

Counterexample

1.1givenF1algebra

The pressure values are $p_{0} = 1 / 2$ , $p_{1} = 1 / 8$ , and $p_{2} = 1 / 2$ by [F1]. Thus the central right and left momentum fluxes at cell $0$ are ${\hat{F}}_{1 / 2}^{m} = (p_{0} + p_{1}) / 2 = 5 / 16$ and ${\hat{F}}_{- 1 / 2}^{m} = (p_{2} + p_{0}) / 2 = 1 / 2$ .

2.1givenstep 1.1algebra∎

The central-flux contribution to $m_{0}^{+}$ is $- Δ t (5 / 16 - 1 / 2) = 3 Δ t / 16$ . The centered source contributes $- Δ t h_{0} (b_{1} - b_{2}) / 2 = - Δ t / 4 = - 4 Δ t / 16$ . Therefore $m_{0}^{+} = - Δ t / 16 \neq 0$ , so the stated update moves this lake and the universal preservation claim is false.

ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passjudge pass (gpt-6-luna)Open item page →

A lake at rest over a bump with a dry crest

Example

Take five periodic cells, $g = 1$ , bed values $(b_{0}, b_{1}, b_{2}, b_{3}, b_{4}) = (0, 1, 2, 1, 0)$ , and free-surface level $η_{0} = 2$ . Set $m_{i} = 0$ and $h_{i} = (2 - b_{i})_{+}$ , so $(h_{i}) = (2, 1, 0, 1, 2)$ ; the crest cell is dry.

Facts & Assumptions

Given: The periodic five-cell bed and states above. Verify the exact one-step hydrostatic HLL update.

[F1]

One-dimensional shallow-water balance law with a dry-state convention: on wet rest cells, $m_{i} = 0$ and $h_{i} + b_{i} = η_{0}$ ; a dry cell has $(h_{i}, m_{i}) = (0, 0)$ with $b_{i} \geq η_{0}$ .

[F2]

Hydrostatic reconstruction with a dry-safe HLL flux and one-sided pressure corrections: each side corrected flux is the common HLL flux plus its cell pressure minus reconstructed pressure.

[F3]

The hydrostatic HLL update preserves lakes at rest, including dry cells: if the reconstructed states at a face agree at rest, the cell-side corrections leave equal hydrostatic pressures on the two faces of each cell.

Verification

1.1givenF1F2algebra

The data are admissible by [F1]. The four face maxima, including the periodic face between cells $4$ and $0$ , are $(1, 2, 2, 1, 0)$ , so the paired reconstructed depths are respectively $(1, 1)$ , $(0, 0)$ , $(0, 0)$ , $(1, 1)$ , and $(2, 2)$ . All reconstructed velocities are zero.

2.1F2step 1.1algebra

At these equal states the common HLL fluxes are $(0, 1 / 2)$ , $(0, 0)$ , $(0, 0)$ , $(0, 1 / 2)$ and $(0, 2)$ . Adding the one-sided pressure corrections in [F2] makes the two corrected fluxes adjacent to each cell both equal $(0, h_{i}^{2} / 2)$ : the five cell pressures are $2, 1 / 2, 0, 1 / 2, 2$ .

3.1F2F3step 2.1algebra∎

Each cell therefore has zero corrected-flux difference in the update, so every state is unchanged, including the dry crest. This verifies the exact lake-at-rest preservation in [F3] for this finite wet/dry example.

ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passOpen item page →

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.

Well-balanced balance laws and wet/dry fronts — Examples

1 · Prerequisites

2 · Summary

Examples

3 · Logical flowchart

4 · Definitions, theorems and proofs

5 · Examples, counterexamples and false statements

A centered pressure flux and centered bed source move a discrete lake

Statement refuted

Facts & Assumptions

Counterexample

A lake at rest over a bump with a dry crest

Example

Facts & Assumptions

Verification

A first HLL step wets a dry cell without negative depth

Example

Facts & Assumptions

Verification

Sources