Alphabeta Math
LemmaStatement: AI-adaptedProof: AI-adaptedPipeline-generatedprecheck passverified 2026-10-22 (gpt-6.1-sol)
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.

The finite-element discrete adjoint gives the reduced gradient

Statement

Let O⊆Rm be open and V=Rn have its standard Euclidean inner product. Let R:O×V→V and G:O×V→Y be continuously differentiable, where Y is a finite-dimensional real Euclidean space. Suppose S:O→V is continuously differentiable, R(q,S(q))=0 for every q∈O, and the matrix DuR(q,S(q)) is invertible at a fixed q∈O. Let y∈Y, q0∈Rm, and α>0. Define

J(q,u):=12∥G(q,u)−y∥Y2+α2∥q−q0∥22,Φ(q):=J(q,S(q)).

Let p∈V be the unique solution of

(DuR(q,S(q)))Tp=∇uJ(q,S(q)),

where ∇qJ and ∇uJ are the Euclidean representatives of the two partial derivative functionals. Then for every h∈Rm,

DΦ(q)[h]=⟨∇qJ(q,S(q))−(DqR(q,S(q)))Tp, h⟩Rm.

Consequently ∇Φ(q)=∇qJ(q,S(q))−(DqR(q,S(q)))Tp. This is the exact derivative of the specified finite-dimensional model; it asserts no convergence of an iteration that changes q.

Facts & Assumptions

Given: The open set, Euclidean spaces, continuously differentiable residual, observation, and state maps, state identity, invertible state Jacobian, data, reference parameter, positive penalty, and adjoint equation in the Statement.

[F1]

The hypotheses give the state identity R(q,S(q))=0 on O.

[F2]

The vector p is defined in the Statement by (DuR(q,S(q)))Tp=∇uJ(q,S(q)). The assumed invertibility of DuR(q,S(q)) makes this solution unique.

[F3]

For continuously differentiable maps between finite-dimensional real Euclidean spaces, the derivative of a composition is the composition of the derivatives (The chain rule for total derivatives: D(g∘f)(a)=Dg(f(a))∘Df(a)).

[F4]

Every linear functional on a finite-dimensional real inner-product space has a unique representing vector (Finite-dimensional Riesz representation: every functional is uniquely v↦⟨v,w⟩).

[F5]

The Euclidean pairing is bilinear and satisfies ⟨ATp,v⟩=⟨p,Av⟩ for every real matrix A of compatible dimensions (The Euclidean inner product ⟨x,y⟩=∑k<nxkyk on Rn).

Proof

1.1givenF1F3algebra

Fix an arbitrary direction h∈Rm. Differentiate [F1] at q using [F3]. The chain rule gives DqR(q,S(q))h+DuR(q,S(q))DS(q)[h]=0. Since DuR(q,S(q)) is invertible by the Statement, this identity determines DS(q)[h].

1.2givenF3F5algebra

Apply [F3] to Φ(q)=J(q,S(q)). For Euclidean vectors z,w, bilinearity [F5] gives 12∥z+tw∥22−12∥z∥22=t⟨z,w⟩+t22∥w∥22, so the derivative of a squared-norm term is its residual paired with the derivative of its argument. Applying the chain rule to the data and penalty terms in J gives DΦ(q)[h]=⟨∇qJ(q,S(q)),h⟩Rm+⟨∇uJ(q,S(q)),DS(q)[h]⟩V.

2.1F2F5step 1.1step 1.2algebra

Substitute [F2] in the second term of step 1.2. The transpose identity [F5] gives ⟨(DuR)Tp,DS[h]⟩V=⟨p,DuR DS[h]⟩V. By step 1.1, DuR DS[h]=−DqR h, so DΦ(q)[h]=⟨∇qJ(q,S(q))−(DqR(q,S(q)))Tp, h⟩Rm.

3.1F4step 2.1∎

The expression in step 2.1 represents the derivative functional on every h. By [F4] its representing vector is unique, so that vector is ∇Φ(q) and has the stated formula.

Remarks

The transpose convention is the Euclidean coefficient-space convention fixed in the definition. A continuous PDE adjoint or an adjoint for another coefficient inner product must be derived using that model's own operator pairing.

Depends on

Used by

Dependency tree · two levels

30 results within two dependency steps 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