LemmaStatement: AI-adaptedProof: AI-adaptedPipeline-generatedprecheck passjudge pass (gpt-6-luna)

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
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AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

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Variable-step Adams weights are integrals of the actual-node Lagrange bases

Statement

Let $k \geq 1$ and let $t_{n + 1} > t_{n} > \dots > t_{n - k + 1}$ be distinct time nodes. For $j = 0, \dots, k - 1$ , let

$L_{j}^{P} (s) = \prod_{\begin{array}{c} 0 \leq ℓ < k \\ ℓ \neq j \end{array}} \frac{s - t_{n - ℓ}}{t_{n - j} - t_{n - ℓ}}, b_{j}^{n} = \int_{t_{n}}^{t_{n + 1}} L_{j}^{P} (s) d s .$

These coefficients give the explicit Adams predictor

$Y_{n + 1}^{P} = Y_{n} + \sum_{j = 0}^{k - 1} b_{j}^{n} f (t_{n - j}, Y_{n - j})$

obtained by integrating the degree- $(k - 1)$ interpolant through the known derivative values $f (t_{n - j}, Y_{n - j})$ .

For the corrector, let ${\hat{L}}_{j} (s)$ , $j = 0, \dots, k$ , be the Lagrange basis on the $k + 1$ distinct nodes $t_{n + 1}, t_{n}, \dots, t_{n + 1 - k}$ and define ${\hat{b}}_{j}^{n} = \int_{t_{n}}^{t_{n + 1}} {\hat{L}}_{j} (s) d s$ . These are the coefficients of the integrated degree- $k$ interpolant, with $j = 0$ the endpoint derivative at $t_{n + 1}$ . The fully implicit corrector is

$Y_{n + 1} = Y_{n} + \sum_{j = 0}^{k} {\hat{b}}_{j}^{n} f (t_{n + 1 - j}, Y_{n + 1 - j}),$

where the $j = 0$ term uses the unknown new endpoint value.

Facts & Assumptions

Given: Strictly increasing time nodes, derivative samples at the listed nodes, and the Lagrange basis polynomials defined in the statement.

[A1]

The integer $k$ is at least one and all listed nodes are distinct; a product over an empty index set, which occurs in the predictor when $k = 1$ , equals $1$ .

Proof

1.1givenA1algebra

Every denominator $t_{n - j} - t_{n - ℓ}$ in $L_{j}^{P}$ is nonzero because the nodes are distinct. Direct substitution gives $L_{j}^{P} (t_{n - r}) = 1$ if $j = r$ and $0$ otherwise; for $k = 1$ the empty product is $1$ , so this also covers the one-node predictor.

2.1step 1.1algebra

Therefore $P (s) = \sum_{j = 0}^{k - 1} L_{j}^{P} (s) f (t_{n - j}, Y_{n - j})$ is the unique polynomial of degree at most $k - 1$ interpolating the known derivative samples: it has the prescribed value at each node, and the difference of two such polynomials would have $k$ distinct roots while having degree at most $k - 1$ .

3.1step 2.1algebra

Integrating the identity defining $P$ from $t_{n}$ to $t_{n + 1}$ gives $\int_{t_{n}}^{t_{n + 1}} P (s) d s = \sum_{j = 0}^{k - 1} b_{j}^{n} f (t_{n - j}, Y_{n - j})$ . The Adams relation $Y_{n + 1}^{P} = Y_{n} + \int P$ yields exactly the predictor coefficients stated.

4.1givenalgebra∎

The same cardinal-value argument on the $k + 1$ corrector nodes gives the unique degree-at-most- $k$ interpolant; integrating it gives $\sum_{j = 0}^{k} {\hat{b}}_{j}^{n} f (t_{n + 1 - j}, Y_{n + 1 - j})$ , with the $j = 0$ term evaluated at the new endpoint for the fully implicit formula.

Used by

Dependency tree · 0 levels

Nothing. This result depends on no other item in the library.

Sources

Ernst Hairer, Syvert P. Nørsett, and Gerhard Wanner, Solving Ordinary Differential Equations I: Nonstiff Problems, 2nd ed., Chapter III.5, pp. 397–400 (standard reference, not scraped)
Tobin A. Driscoll and Richard J. Braun, Fundamentals of Numerical Computation, §6.6 (standard reference, not scraped)