LemmaStatement: AI-adaptedProof: 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.

Second-order consistency of the variable-coefficient flux stencil

Statement

Let $Ω = (0, L_{x}) \times (0, L_{y})$ , $a \in C^{3} (\overline{Ω})$ with $a (x, y) \geq a_{0} > 0$ , and $u \in C^{4} (\overline{Ω})$ . Set $f = - \partial_{x} (a \partial_{x} u) - \partial_{y} (a \partial_{y} u)$ . For $r = 0, 1, 2, 3$ and $s = 1, 2, 3, 4$ , define $A_{r}^{x} = ∥ \partial_{x}^{r} a ∥_{L^{\infty} (Ω)}$ and $U_{s}^{x} = ∥ \partial_{x}^{s} u ∥_{L^{\infty} (Ω)}$ , where $\partial_{x}^{0} a = a$ . Define

C_{x} = \frac{A_{0}^{x} U_{4}^{x}}{24} + \frac{A_{1}^{x} U_{3}^{x}}{12} + \frac{A_{2}^{x} U_{2}^{x}}{16} + \frac{A_{3}^{x} U_{1}^{x}}{48} + L_{x} (\frac{A_{1}^{x} U_{4}^{x}}{48} + \frac{A_{2}^{x} U_{3}^{x}}{48} + \frac{A_{3}^{x} U_{2}^{x}}{96}) + L_{x}^{2} (\frac{A_{2}^{x} U_{4}^{x}}{192} + \frac{A_{3}^{x} U_{3}^{x}}{288}) + L_{x}^{3} \frac{A_{3}^{x} U_{4}^{x}}{1152} .

For $r = 0, 1, 2, 3$ and $s = 1, 2, 3, 4$ , also set $A_{r}^{y} = ∥ \partial_{y}^{r} a ∥_{L^{\infty} (Ω)}$ and $U_{s}^{y} = ∥ \partial_{y}^{s} u ∥_{L^{\infty} (Ω)}$ ; define $C_{y}$ by the same displayed expression with $x, L_{x}, A_{r}^{x}, U_{s}^{x}$ replaced by $y, L_{y}, A_{r}^{y}, U_{s}^{y}$ . If $u_{i j}^{*} = u (x_{i}, y_{j})$ and $f_{h, i j} = f (x_{i}, y_{j})$ , then for every interior node,

∣ (A_{h}^{a} u^{*})_{i j} - f_{h, i j} ∣ \leq 2 C_{x} h_{x}^{2} + 2 C_{y} h_{y}^{2},

where $A_{h}^{a}$ is the exact-midpoint flux stencil of Rectangular grid and Dirichlet flux stencil. The estimate includes every interior row next to the boundary; the Dirichlet boundary samples themselves are imposed exactly.

Facts & Assumptions

Given: The rectangle $Ω$ , $a \in C^{3} (\overline{Ω})$ with $a \geq a_{0} > 0$ , $u \in C^{4} (\overline{Ω})$ , the pointwise equation $f = - \partial_{x} (a \partial_{x} u) - \partial_{y} (a \partial_{y} u)$ , and the rectangular grids of Rectangular grid and Dirichlet flux stencil.

[F1]

Rectangular grid and Dirichlet flux stencil: $A_{h}^{a}$ is the negative difference of adjacent midpoint face fluxes in each coordinate direction, divided by that coordinate's mesh spacing; its face values are sampled at the exact face midpoints.

[A1]

The coordinate derivatives $\partial_{x}^{r} a$ for $0 \leq r \leq 3$ and $\partial_{x}^{s} u$ for $1 \leq s \leq 4$ are continuous and bounded on $\overline{Ω}$ ; the analogous $y$ derivatives are also continuous and bounded.

Proof

1.1givenA1F1algebra

Fix an interior node $(x_{i}, y_{j})$ and write $x = x_{i}$ , $h = h_{x}$ , $A_{r} = A_{r}^{x}$ , and $U_{s} = U_{s}^{x}$ . Taylor's theorem with bounded third derivative for $a$ and bounded fourth derivative for $u$ gives $a (x \pm h / 2, y_{j}) = a \pm h a_{x} / 2 + h^{2} a_{x x} / 8 + R_{a}^{\pm}$ with $∣ R_{a}^{\pm} ∣ \leq A_{3} h^{3} / 48$ , and $D_{+} u = u_{x} + h u_{x x} / 2 + h^{2} u_{x x x} / 6 + R_{u}^{+}$ , $D_{-} u = u_{x} - h u_{x x} / 2 + h^{2} u_{x x x} / 6 + R_{u}^{-}$ with $∣ R_{u}^{\pm} ∣ \leq U_{4} h^{3} / 24$ , where $D_{+} u = (u (x + h, y_{j}) - u (x, y_{j})) / h$ and $D_{-} u = (u (x, y_{j}) - u (x - h, y_{j})) / h$ .

2.1givenA1step 1.1algebra

Multiplying the coefficient and difference expansions yields, with $F_{\pm} = a (x \pm h / 2, y_{j}) D_{\pm} u$ , $F_{\pm} = a u_{x} \pm h (a u_{x x} + a_{x} u_{x}) / 2 + h^{2} (a u_{x x x} / 6 + a_{x} u_{x x} / 4 + a_{x x} u_{x} / 8) + R_{F}^{\pm}$ . The absolute values of the discarded products are bounded by $h^{3} (A_{0} U_{4} / 24 + A_{1} U_{3} / 12 + A_{2} U_{2} / 16 + A_{3} U_{1} / 48) + h^{4} (A_{1} U_{4} / 48 + A_{2} U_{3} / 48 + A_{3} U_{2} / 96) + h^{5} (A_{2} U_{4} / 192 + A_{3} U_{3} / 288) + h^{6} A_{3} U_{4} / 1152$ . Since $h \leq L_{x}$ , this is at most $C_{x} h^{3}$ for the stated $C_{x}$ .

3.1givenstep 2.1algebra

The coefficient of $h^{2}$ in step 2.1 is the same for $F_{+}$ and $F_{-}$ , so it cancels on subtraction. Thus $∣ (F_{+} - F_{-}) / h - \partial_{x} (a \partial_{x} u) (x_{i}, y_{j}) ∣ \leq (∣ R_{F}^{+} ∣ + ∣ R_{F}^{-} ∣) / h \leq 2 C_{x} h_{x}^{2}$ , because $\partial_{x} (a \partial_{x} u) = a_{x} u_{x} + a u_{x x}$ . Interchanging $x$ and $y$ proves the corresponding vertical-flux bound $2 C_{y} h_{y}^{2}$ .

4.1F1givenstep 3.1algebra∎

By [F1], $A_{h}^{a} u^{*}$ is the negative sum of the two discrete flux divergences. The continuous right-hand side $f$ is the negative sum of the matching derivatives. Subtracting these expressions and applying the horizontal and vertical bounds from step 3.1 gives $∣ (A_{h}^{a} u^{*})_{i j} - f_{h, i j} ∣ \leq 2 C_{x} h_{x}^{2} + 2 C_{y} h_{y}^{2}$ at every interior node, including nodes whose stencil touches a boundary value.

Depends on

Rectangular grid and Dirichlet flux stencil

Used by

Second-order grid L2 error for rectangular Dirichlet flux differences Theorem

Dependency tree · one level

1 result within one dependency step 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

Endre Süli, Numerical Solution of PDEs, complete lecture notes (standard reference, not scraped)
Benjamin Seibold, MIT 18.336 Lecture 5: Finite Difference Approximation (standard reference, not scraped)
Benjamin Seibold, MIT 18.336 Lecture 6: General Linear Second Order Equation (standard reference, not scraped)