LemmaStatement: AI-adaptedProof: AI-generatedPipeline-generated

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.

A tangent and adjoint pairing gives the ODE gradient boundary identity

Statement

Let $T > 0$ , let $S \in C^{1} ([0, T]; R^{d \times m})$ and $p \in C^{1} ([0, T]; R^{d})$ satisfy $S^{'} = A S + B$ and $p^{'} = - A^{T} p - g$ , where $A, B, g$ are continuous and have compatible dimensions. Then $p (T)^{T} S (T) - p (0)^{T} S (0) = \int_{0}^{T} (- g (t)^{T} S (t) + p (t)^{T} B (t)) d t .$

Facts & Assumptions

Given: $T > 0$ , the continuously differentiable matrix and vector functions and their differential equations in the Statement.

[F1]

The Euclidean pairing is the finite coordinate sum $x^{T} y = \sum_{i} x_{i} y_{i}$ (The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ ).

[F2]

The derivative of a product is $(f g)^{'} = f^{'} g + f g^{'}$ (Sums, scalar multiples, products and quotients: $(f + g)^{'} (c) = f^{'} (c) + g^{'} (c)$ , $(α f)^{'} (c) = α f^{'} (c)$ , $(f g)^{'} (c) = f^{'} (c) g (c) + f (c) g^{'} (c)$ , and $(f / g)^{'} (c) = (f^{'} (c) g (c) - f (c) g^{'} (c)) / g (c)^{2}$ when $g (c) \neq 0$ ).

[F3]

If $G^{'} = q$ and $q$ is integrable on $[0, T]$ , then $\int_{0}^{T} q = G (T) - G (0)$ (The second fundamental theorem: if $G$ is differentiable on $[a, b]$ with $G^{'} = f$ and $f$ is integrable, then $\int_{a}^{b} f = G (b) - G (a)$ ).

Proof

1.1givenF1F2algebra

For each column $j$ , write $(p^{T} S)_{j} = \sum_{i = 1}^{d} p_{i} S_{i j}$ . By [F2] coordinatewise, $((p^{T} S)_{j})^{'} = \sum_{i} p_{i}^{'} S_{i j} + p_{i} S_{i j}^{'}$ . Substituting the two equations and collecting the columns gives $(p^{T} S)^{'} = (- A^{T} p - g)^{T} S + p^{T} (A S + B) = - g^{T} S + p^{T} B$ , because $(A^{T} p)^{T} S = p^{T} A S$ .

2.1F3step 1.1∎

For each column $j$ , the scalar function $(p^{T} S)_{j}$ is $C^{1}$ , and the $j$ th component of the right-hand side in step 1.1 is continuous and hence integrable. Apply [F3] to each scalar component on $[0, T]$ ; collecting these identities gives the stated row-vector endpoint identity.

Depends on

Used by

Dependency tree · two levels

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Sources

William W. Hager, Runge-Kutta methods in optimal control and the transformed adjoint system, §3 (standard reference, not scraped)
Michael B. Giles and Niles A. Pierce, An Introduction to the Adjoint Approach to Design, §2.4 (standard reference, not scraped)