ExampleConstruction: AI-generatedVerification: AI-generatedPipeline-generatedprecheck pass

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.

Flux and expanded variable-coefficient stencils differ by a second-order term

Example

On $Ω = (0, 1)^{2}$ , take $a (x, y) = 1 + x^{2}$ and $u (x, y) = x^{2}$ . Then $\partial_{x} (a \partial_{x} u) = 2 + 6 x^{2}$ and the $y$ -flux is zero. Let $D_{h}^{f l u x}$ be the midpoint face-flux difference in the positive $x$ direction, and let $D_{h}^{e x p a n d e d} = a (x_{i}, y_{j}) δ_{x, h}^{2} u + a_{x} (x_{i}, y_{j}) D_{x, h}^{0} u$ , where $δ_{x, h}^{2}$ is the centered second difference and $D_{x, h}^{0}$ is the centered first difference. At each interior node,

D_{h}^{f l u x} u - D_{h}^{e x p a n d e d} u = h_{x}^{2} / 2.

Thus the two discrete operators are distinct even though both approximate the same continuous divergence.

Facts & Assumptions

Given: The midpoint face-flux convention of Rectangular grid and Dirichlet flux stencil, $a (x, y) = 1 + x^{2}$ , $u (x, y) = x^{2}$ , and a rectangular uniform grid on $(0, 1)^{2}$ .

[F1]

Rectangular grid and Dirichlet flux stencil: the positive $x$ -flux at a face is the midpoint value of $a$ times the forward or backward difference of $u$ across that face.

Verification

1.1givenF1algebra

At $x = x_{i}$ and $h = h_{x}$ , the two midpoint coefficients are $a_{+} = 1 + (x + h / 2)^{2}$ and $a_{-} = 1 + (x - h / 2)^{2}$ , while the one-sided differences of $x^{2}$ are $D_{+} u = 2 x + h$ and $D_{-} u = 2 x - h$ . Therefore $D_{h}^{f l u x} u = (a_{+} (2 x + h) - a_{-} (2 x - h)) / h = 2 + 6 x^{2} + h^{2} / 2$ .

2.1F1step 1.1algebra∎

The centered differences are $δ_{x, h}^{2} u = 2$ and $D_{x, h}^{0} u = 2 x$ , while $a (x, y) = 1 + x^{2}$ and $a_{x} (x, y) = 2 x$ . Hence $D_{h}^{e x p a n d e d} u = 2 (1 + x^{2}) + (2 x) (2 x) = 2 + 6 x^{2}$ , so subtracting from step 1.1 gives exactly $h_{x}^{2} / 2$ .

Depends on

Rectangular grid and Dirichlet flux stencil

Used by

Nothing in the library uses this result yet.

Dependency tree · one level

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Sources

Benjamin Seibold, MIT 18.336 Lecture 6: General Linear Second Order Equation (standard reference, not scraped)