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A parabolic adjoint gives the gradient when the state map is differentiable

Statement

Assume Countable Choice and T>0, and use the real evolution triple (V,H,V∗) of Parameter-dependent parabolic variational states and objectives. On an open parameter set Θ⊆Rm, assume the state map θ↦uθ is Fr'echet differentiable into W(0,T;V), with derivative wη in direction η∈Rm. For each fixed θ and direction η, assume there is a null set Nθ,η⊂(0,T) such that the linearized equation below holds for every t∉Nθ,η and simultaneously every v∈V, using representatives of its data: ⟨wη′(t),v⟩+aθ(t;wη(t),v)=⟨fθ′[η](t),v⟩−aθ′[η](t;uθ(t),v),wη(0)=u0,θ′[η]. Let ϕ:H×Θ→R be C1, and let L:[0,T]×V×Θ→R be measurable in time and C1 in (u,θ) for almost every time, with a common integrable majorant that justifies differentiating its time integral. Assume Duϕ is represented by an element of H and DuL(t,uθ,θ)∈V∗ is square-integrable in time. Suppose the adjoint p∈W(0,T;V) satisfies the following equation simultaneously for every v∈V outside one null set Mθ⊂(0,T), and has the stated terminal trace: −⟨p′(t),v⟩+aθ(t;v,p(t))=DuL(t,uθ(t),θ)[v],p(T)=∇Hϕ(uθ(T),θ). Then for j(θ)=ϕ(uθ(T),θ)+∫0TL(t,uθ(t),θ) dt, Dj(θ)[η]=Dθϕ(uθ(T),θ)[η]+(p(0),u0,θ′[η])H+∫0T(DθL(t,uθ,θ)[η]+⟨fθ′[η](t),p(t)⟩−aθ′[η](t;uθ(t),p(t)))dt. The exceptional sets are independent of the test v; they need not be common to all parameters or directions. The forcing/form-derivative difference in the integrand is interpreted together. The hypotheses imply it is measurable and integrable: almost everywhere it equals ⟨wη′,p⟩+⟨p′,wη⟩+DuL[wη]. No global boundedness or coercivity of aθ, or separability of V, is required for this conditional identity.

This is conditional on the displayed differentiable state and adjoint; it does not prove either exists. The adjoint form is transposed as shown, so symmetry of aθ is not required.

Facts & Assumptions

Given: T>0, the real evolution triple, differentiable parameterized state and objective, its linearized equation, and a solution of the stated terminal adjoint equation, with both displayed variational equations holding simultaneously for all tests outside their specified null sets.

[F1]

The parameterized parabolic state and reduced objective use the variational equation on W(0,T;V) with H traces (Parameter-dependent parabolic variational states and objectives).

[F2]

For w,p∈W(0,T;V), the evolution-space pairing identity is (p(T),w(T))H−(p(0),w(0))H=∫0T(⟨p′,w⟩+⟨w′,p⟩)dt (The evolution-space energy identity and continuous H trace).

[F3]

Differentiable compositions and sums obey the Banach-space chain rule (Chain sum product and composition rules for Banach derivatives).

[A1]

Countable Choice (The Axiom of Countable Choice (ACω)) is required by the pairing supplier (The evolution-space energy identity and continuous H trace). No use is made of a parabolic well-posedness theorem to infer state or adjoint existence.

[A2]

For each fixed parameter and direction, the linearized and adjoint equations hold outside null sets independent of v. Their union, together with the null sets on which the W representatives are undefined, still has measure zero. This is an explicit hypothesis, not a consequence of the fixed-test state convention in [F1].

Proof

1.1F1F3given

Apply [F3] to the terminal term and the running integrand. The common integrable majorant permits differentiation under the integral, so Dj[η]=Dθϕ[η]+(∇Hϕ,wη(T))H+∫0T(DθL[η]+DuL[wη])dt. The terminal condition replaces ∇Hϕ by p(T).

1.2A1A2F2givenalgebra

Fix θ,η and discard the union of the null sets in [A2]. At every remaining time, wη(t),p(t)∈V and both equations hold for all tests, so the choices v=wη(t) in the adjoint equation and v=p(t) in the linearized equation are legitimate. Adding them and cancelling the two occurrences of aθ(t;wη(t),p(t)) gives ⟨fθ′[η],p⟩−aθ′[η](uθ,p)=⟨wη′,p⟩+⟨p′,wη⟩+DuL[wη]. Each term on the right belongs to L1(0,T) by the L2(V∗) and L2(V) hypotheses and Cauchy--Schwarz. Thus the difference on the left has a measurable integrable representative without assuming either summand separately integrable. Apply [F2] to wη,p; Countable Choice enters through this pairing supplier.

2.1F1step 1.1step 1.2algebra∎

The pairing identity now gives (p(T),wη(T))H=(p(0),u0,θ′[η])H+∫0T(−DuL[wη]+⟨fθ′[η],p⟩−aθ′[η](uθ,p))dt. Substitute this into step 1.1; the terms DuL[wη] cancel and yield the displayed formula.

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