ExampleConstruction: AI-generatedVerification: AI-generatedPipeline-generatedprecheck passjudge pass (gpt-6-luna)

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

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Variable-foot pointwise interpolation can change the grid sum

Example

Use the period- $3$ one-dimensional grid with nodes $0, 1, 2$ , $h = 1$ , and one time step $Δ t = 1$ . Set $b (x) = \frac{5}{12} - \frac{1}{6} \cos ⁣ (\frac{2 π x}{3}) - \frac{1}{2 \sqrt{3}} \sin ⁣ (\frac{2 π x}{3}) .$ This is a smooth periodic velocity. Start with grid values $(U_{0}, U_{1}, U_{2}) = (0, 0, 1)$ , which are samples of the smooth nonnegative period- $3$ function $u_{0} (x) = \frac{16}{9} \sin^{2} ⁣ (\frac{π x}{3}) \sin^{2} ⁣ (\frac{π (x - 1)}{3}) .$

Facts & Assumptions

Given: The grid, time step, velocity, data, and explicit Euler/multilinear update in the Example.

[F1]

Periodic tensor grids and multilinear interpolation: a foot at fraction $θ$ between consecutive nodes has weights $1 - θ, θ$ , with node indices wrapped modulo $3$ .

[F2]

Explicit Euler characteristic backtracking with multilinear interpolation: at arrival node $x_{i}$ the lifted foot is $x_{i} - Δ t b (x_{i})$ and the new value is the previous grid data interpolated there.

Proof

1.1givenF1F2algebra

The squared-sine factors are smooth and period $3$ ; the value is zero at nodes $0$ and $1$ and equals $1$ at node $2$ . Direct substitution into $b$ gives $(b (0), b (1), b (2)) = (1 / 4, 1 / 4, 3 / 4)$ . Therefore the three lifted feet are $- 1 / 4, 3 / 4, 5 / 4$ . Their interpolation rows, in node order $0, 1, 2$ , are $(3 / 4, 0, 1 / 4)$ , $(1 / 4, 3 / 4, 0)$ , and $(0, 3 / 4, 1 / 4)$ .

2.1givenstep 1.1algebra∎

Applying those rows to $(0, 0, 1)$ gives the new grid values $(1 / 4, 0, 1 / 4)$ . The grid sum changes from $1$ to $1 / 2$ . This is a pointwise interpolation update, so the grid sum is a diagnostic only; the result shows why pointwise interpolation is not itself a conservative remap.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

4 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

Lukas Einkemmer and Alexander Ostermann, On the error propagation of semi-Lagrange and Fourier methods for advection problems, Computers & Mathematics with Applications 69 (2015), 170–179 (standard reference, not scraped)