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Parabolic linear-quadratic control has an energy adjoint and reduced gradient

Statement

Assume Countable Choice. Let $T > 0$ and let $E \subset R^{d}$ be bounded and Lebesgue measurable. Let $V ↪ H ≅ H^{*} ↪ V^{*}$ be a real Gelfand triple with $V$ separable, and let $W = {w \in L^{2} (0, T; V) : w^{'} \in L^{2} (0, T; V^{*})}$ have the continuous $H$ -valued trace supplied by The evolution-space energy identity and continuous H trace. Let $a (t; \cdot, \cdot)$ be measurable, uniformly bounded and coercive as in Well-posedness for measurable coercive parabolic forms. Let $U = L^{2} (E \times (0, T))$ , let $B : U \to L^{2} (0, T; V^{*})$ be bounded and linear, and let $Q_{a d} = {q \in U : q_{a} \leq q \leq q_{b} a.e.}$ for $q_{a}, q_{b} \in U$ , $q_{a} \leq q_{b}$ . For fixed $u_{0} \in H$ , $f \in L^{2} (0, T; V^{*})$ , and $u_{d} \in L^{2} (0, T; H)$ , define the unique state by

u^{'} (t) + A (t) u (t) = f (t) + B q (t), u (0) = u_{0},

where $⟨ A (t) v, w ⟩ = a (t; v, w)$ and $⟨ A (t)^{*} p, v ⟩ = a (t; v, p)$ , and define

\hat{J} (q) = \frac{1}{2} ∥ u (q) - u_{d} ∥_{L^{2} (0, T; H)}^{2} + \frac{α}{2} ∥ q ∥_{U}^{2}, α > 0.

Then for every $q \in U$ the adjoint $p \in W$ uniquely solves

- p^{'} (t) + A (t)^{*} p (t) = u (q) (t) - u_{d} (t), p (T) = 0,

and $\hat{J}$ is Fr'echet differentiable with

\nabla \hat{J} (q) = B^{*} p + α q .

Define $B^{*} p \in U$ by $(B^{*} p, r)_{U} = \int_{0}^{T} ⟨ B r, p ⟩_{V^{*}, V} d t$ for every $r \in U$ ; boundedness of $B$ and Hilbert-space Riesz representation make it well-defined. The formula requires the terminal/initial traces and time-dependent form hypotheses above; it asserts no finite-element error rate. For the separate finite-element error splitting, assume additionally $u \in H^{1} (0, T; V)$ , $V_{h} \subset V$ is finite-dimensional, and $P_{h} \in C^{1} ([0, T]; L (V, V_{h}))$ in operator norm. Then its residual derivative is $ρ^{'} = P_{h}^{'} (t) u + P_{h} (t) u^{'} - u^{'}$ , including the $P_{h}^{'} (t) u$ term.

Facts & Assumptions

Given: Countable Choice; the real Gelfand triple with separable $V$ and its energy space; the stated bounded, coercive, measurable form; bounded linear $B$ ; fixed initial and desired data; and $α > 0$ .

[A1]

Countable Choice is the stated axiom for parabolic solvability and Hilbert-space Riesz representation (The Axiom of Countable Choice ( ${A C}_{ω}$ ), Riesz representation for Hilbert spaces).

[F1]

On the stated separable real evolution triple, for every right-hand side in $L^{2} (0, T; V^{*})$ and initial trace in $H$ , the parabolic equation has a unique energy solution in $W$ (Well-posedness for measurable coercive parabolic forms).

[F2]

Elements of $W$ have continuous $H$ traces, and for $v, w \in W$ the energy pairing obeys integration by parts, including the endpoint terms (Evolution triples and the parabolic energy space, The evolution-space energy identity and continuous H trace).

[F3]

Time reversal preserves measurability, boundedness, and coercivity for the transposed form; applying [F1] to the reversed equation gives the terminal adjoint.

[F4]

Apply Closed pointwise box of L2 controls to the bounded measurable space-time domain $E \times (0, T)$ . It supplies the pointwise box and the real Hilbert structure on $U = L^{2} (E \times (0, T))$ under Countable Choice, so Hilbert-space Riesz representation defines $B^{*} p$ .

[F5]

For the stated separable $V$ , if $u \in H^{1} (0, T; V)$ and $P_{h} \in C^{1} ([0, T]; L (V, V_{h}))$ in operator norm, then $(P_{h} u)^{'} = P_{h}^{'} u + P_{h} u^{'}$ and the parabolic residual contains $P_{h}^{'} u$ (Time derivatives of fixed and differentiable comparison projections, Semidiscrete Ritz error splitting and energy estimate).

Proof

1.1A1F1givenalgebra

For a direction $r \in U$ , the state difference $z = u (q + r) - u (q)$ satisfies $z^{'} + A (t) z = B r$ and $z (0) = 0$ . By [A1, F1] and linearity, $z = S r$ for a bounded linear solution map $S : U \to L^{2} (0, T; H)$ . Expanding the two squared norms gives $\hat{J} (q + r) - \hat{J} (q) = (u - u_{d}, z)_{L^{2} (0, T; H)} + α (q, r)_{U} + \frac{1}{2} ∥ z ∥_{L^{2} (0, T; H)}^{2} + \frac{α}{2} ∥ r ∥_{U}^{2} .$ The last two terms are quadratic in $r$ , so the first two terms give the Fr'echet derivative once the observation residual is paired with $z$ .

2.1A1F1F2F3step 1.1algebra

Set $g = u - u_{d}$ and reverse time: with $s = T - t$ , $\tilde{p} (s) = p (T - s)$ , $\tilde{g} (s) = g (T - s)$ , and ${\tilde{a}}^{*} (s; x, y) = a (T - s; y, x)$ , the backward equation becomes the forward equation $⟨ {\tilde{p}}^{'}, v ⟩ + {\tilde{a}}^{*} (s; \tilde{p}, v) = (\tilde{g}, v)_{H}$ with initial value $\tilde{p} (0) = 0$ . The transposed reversed form is measurable, bounded, and coercive with the same constants, so [A1, F1] supplies its unique energy solution; reflecting back proves [F3] and gives $p \in W$ . Its weak equation is $⟨ - p^{'}, v ⟩ + a (t; v, p) = (g, v)_{H}$ . By the energy integration-by-parts identity [F2], testing with the state variation $z$ and using $z (0) = 0$ and $p (T) = 0$ gives $\int_{0}^{T} (g, z)_{H} d t = \int_{0}^{T} (⟨ z^{'}, p ⟩ + a (t; z, p)) d t = \int_{0}^{T} ⟨ B r, p ⟩ d t = (B^{*} p, r)_{U} .$ Therefore the linear term in step 1.1 is $(B^{*} p, r)_{U}$ , and the derivative is $B^{*} p + α q$ .

3.1A1F4step 1.1step 2.1algebra

Boundedness of $S$ makes the remainder in step 1.1 bounded by $\frac{1}{2} ∥ S ∥^{2} ∥ r ∥_{U}^{2} + \frac{α}{2} ∥ r ∥_{U}^{2}$ , which is $o (∥ r ∥_{U})$ . Therefore the derivative is Fr'echet; Riesz representation in [A1] gives the stated gradient.

4.1F5givenalgebra∎

In the additional finite-element setting of [F5], differentiate $ρ (t) = P_{h} (t) u (t) - u (t)$ in the weak time sense. The product rule gives $ρ^{'} = P_{h}^{'} (t) u (t) + P_{h} (t) u^{'} (t) - u^{'} (t)$ . Thus the comparison residual contains the full displayed derivative, including $P_{h}^{'} u$ ; the reduced-gradient formula does not authorize dropping it.

Remarks

The proof uses the terminal condition on $p$ and the zero initial value of a state variation; omitting either endpoint term would invalidate the adjoint identity.

Depends on

Used by

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Sources

Christian Clason, Introduction to Finite Element Methods, lecture notes (standard reference, not scraped)