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.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

The SBP inflow SAT produces the boundary energy term

Example

On the interval $[0, 1]$ , use two endpoint nodes, $H = \frac{1}{2} I$ , $D = (\begin{array}{cc} - 1 & 1 \\ - 1 & 1 \end{array})$ , left and right selectors $R_{L} = (1, 0)$ and $R_{R} = (0, 1)$ , and $B = diag (- 1, 1)$ . For $u_{t} + a u_{x} = 0$ with $a > 0$ , inflow value $g (t)$ at $x = 0$ , and upwind outflow at $x = 1$ , the SBP-SAT ODE is $H \dot{U} + a H D U + a e_{L} (U_{L} - g) = 0.$

Facts & Assumptions

Given: The two-node SBP data, the upwind inflow SAT sign, and the discrete energy $E = \frac{1}{2} U^{T} H U$ .

[F1]

Upwind SBP-SAT discretization of constant transport: the inflow SAT in the strong form is $- a e_{L} (g - U_{L}) = a e_{L} (U_{L} - g)$ and the outflow SAT is zero.

[F2]

Upwind SBP-SAT transport satisfies the semidiscrete L2 energy identity: the SBP identity converts the volume derivative term into the signed endpoint quadratic boundary term.

Verification

1.1givenF2algebra

The matrices satisfy $H D = (\begin{array}{cc} - 1 / 2 & 1 / 2 \\ - 1 / 2 & 1 / 2 \end{array})$ and $H D + D^{T} H = B$ , so $U^{T} H D U = \frac{1}{2} (U_{R}^{2} - U_{L}^{2})$ .

2.1givenF1step 1.1algebra∎

Multiply the ODE in [F1] by $U^{T}$ . The inflow penalty contributes $a U_{L} (U_{L} - g)$ , and step 1.1 gives $\dot{E} + \frac{a}{2} U_{R}^{2} + \frac{a}{2} (U_{L} - g)^{2} = \frac{a}{2} g^{2}$ . Thus the penalty has the sign required by the upwind energy identity.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

5 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.