PropositionStatement: AI-adaptedProof: AI-adaptedPipeline-generatedprecheck pending

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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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The two-stage Rosenbrock update reuses one shifted factorization for two solves

Statement

On every successful call of Two-stage second-order Rosenbrock W-method with shared factorization, the procedure forms one matrix $W = I - α h \tilde{J}$ , factorizes it once, and uses that stored factor object for exactly two stage right-hand sides, one for $k_{1}$ and one for $q_{2}$ . Thus it uses one shifted-matrix factorization and two triangular solves. This is an operation-count statement, not a claim that either solve succeeds.

Facts & Assumptions

Given: A successful execution of the two-stage method with a finite-dimensional shifted matrix.

[F1]

Two-stage second-order Rosenbrock W-method with shared factorization: the procedure factors $W$ once, solves the first stage, then calls the same factor object's solve operation for $W q_{2} = F_{1} - k_{1}$ .

[F2]

Reusable Householder QR solve for a shifted matrix with residual status: one factor object stores a single QR factorization and each call to its solve operation applies that factorization and performs one triangular substitution.

Proof

Given: The successful execution in the Statement.

1.1givenF1F2

The first instruction of [F1] constructs $W$ and makes one factorization call, producing the factor object specified by [F2].

2.1F1F2step 1.1

The algorithm calls that object's solve once for the $k_{1}$ right-hand side and once for the $q_{2}$ right-hand side; neither call constructs a new factor object. By [F2], each such call performs one triangular solve.

3.1step 1.1step 2.1algebra∎

Counting the calls in step 1.1 and step 2.1 gives exactly one factorization and exactly two triangular solves on a successful execution, as asserted.

Depends on

Used by

Nothing in the library uses this result yet.

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Sources

Daniel Kressner, Advanced Numerical Analysis, §3.5 Rosenbrock methods (standard reference, not scraped)