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A parabolic adjoint gives the gradient when the state map is differentiable
Statement
Assume Countable Choice and , and use the real evolution triple of Parameter-dependent parabolic variational states and objectives. On an open parameter set , assume the state map is Fr'echet differentiable into , with derivative in direction . For each fixed and direction , assume there is a null set such that the linearized equation below holds for every and simultaneously every , using representatives of its data: Let be , and let be measurable in time and in for almost every time, with a common integrable majorant that justifies differentiating its time integral. Assume is represented by an element of and is square-integrable in time. Suppose the adjoint satisfies the following equation simultaneously for every outside one null set , and has the stated terminal trace: Then for , The exceptional sets are independent of the test ; 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 . No global boundedness or coercivity of , or separability of , 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 is not required.
Facts & Assumptions
Given: , 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.
The parameterized parabolic state and reduced objective use the variational equation on with traces (Parameter-dependent parabolic variational states and objectives).
For , the evolution-space pairing identity is (The evolution-space energy identity and continuous H trace).
Differentiable compositions and sums obey the Banach-space chain rule (Chain sum product and composition rules for Banach derivatives).
Countable Choice (The Axiom of Countable Choice ()) 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.
For each fixed parameter and direction, the linearized and adjoint equations hold outside null sets independent of . Their union, together with the null sets on which the representatives are undefined, still has measure zero. This is an explicit hypothesis, not a consequence of the fixed-test state convention in [F1].
Proof
Apply [F3] to the terminal term and the running integrand. The common integrable majorant permits differentiation under the integral, so The terminal condition replaces by .
Fix and discard the union of the null sets in [A2]. At every remaining time, and both equations hold for all tests, so the choices in the adjoint equation and in the linearized equation are legitimate. Adding them and cancelling the two occurrences of gives Each term on the right belongs to by the and hypotheses and Cauchy--Schwarz. Thus the difference on the left has a measurable integrable representative without assuming either summand separately integrable. Apply [F2] to ; Countable Choice enters through this pairing supplier.
The pairing identity now gives . Substitute this into step 1.1; the terms cancel and yield the displayed formula.
Depends on
Used by
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Sources
- Christian Clason, Introduction to Finite Element Methods, complete lecture notes, §11.1 (standard reference, not scraped)
- Hannes Meinlschmidt, Optimization with Partial Differential Equations, complete lecture notes, §§1.1–1.2 and Chapter 3 (standard reference, not scraped)