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The LQR adjoint identity gives the reduced gradient

Statement

Assume the Axioms of Countable Choice and Dependent Choice.

For the LQR problem of Finite-horizon bounded-control linear-quadratic ODE problem, let $u \in U_{c}$ , let $x = x_{u}$ , and let $p$ be its adjoint from Continuous adjoint equation for the linear-quadratic ODE. Define the bounded linear map $D J (u) : L^{2} (0, T; R^{m}) \to R$ by $D J (u) [w] = \int_{0}^{T} (B (t)^{T} p (t) + R (t) u (t))^{T} w (t) d t .$ It is the first-order norm derivative of $J$ at $u$ in the precise sense that $J (u + w) - J (u) - D J (u) [w] = O (∥ w ∥_{2}^{2})$ as $∥ w ∥_{2} \to 0$ .

Facts & Assumptions

Given: The LQR data, $u, w \in L^{2}$ , the state variation $δ x = x_{u + w} - x_{u}$ , and the adjoint $p$ .

[F1]

The state map is affine; its difference $δ x$ satisfies $\dot{δ x} = A δ x + B w$ and $δ x (0) = 0$ (The bounded measurable linear ODE has a unique affine state map).

[F2]

The adjoint is absolutely continuous and satisfies $- \dot{p} = A^{T} p + Q (x - x_{d})$ and $p (T) = P (x (T) - x_{T})$ (Continuous adjoint equation for the linear-quadratic ODE, The backward LQR adjoint equation has a unique solution).

[F3]

For absolutely continuous scalar functions, integration by parts gives $\int_{0}^{T} p^{T} \dot{δ x} + {\dot{p}}^{T} δ x = p (T)^{T} δ x (T) - p (0)^{T} δ x (0)$ componentwise (Integration by parts for absolutely continuous functions).

[F4]

The integral pairing is bounded by the product of the $L^{2}$ norms (Cauchy-Schwarz inequality for $L^{2}$ ).

[A1]

Countable Choice and Dependent Choice are assumed for [F3], exactly as required by that supplier (The Axiom of Countable Choice ( ${A C}_{ω}$ ), The axiom of dependent choice: a relation in which every element is related to something admits an $N$ -indexed chain).

Proof

1.1F1givenalgebra

By [F1], the state variation $δ x = x_{u + w} - x$ is linear in $w$ and satisfies $∥ δ x ∥_{\infty} \leq C_{0} ∥ w ∥_{2}$ for a fixed constant $C_{0}$ . Expanding the running and terminal quadratic forms gives the linear term $\int_{0}^{T} (x - x_{d})^{T} Q δ x d t + \int_{0}^{T} u^{T} R w d t + (x (T) - x_{T})^{T} P δ x (T)$ and a remainder $\frac{1}{2} \int_{0}^{T} (δ x^{T} Q δ x + w^{T} R w) d t + \frac{1}{2} δ x (T)^{T} P δ x (T)$ . Boundedness of $Q, R, P$ and the state-map estimate bound this remainder by $C ∥ w ∥_{2}^{2}$ . Thus the displayed linear term is the first-order norm derivative before rewriting its state contributions through the adjoint.

1.2F1F2F3A1algebra

Apply [F3] to $p$ and $δ x$ , whose absolute continuity follows from [F1] and [F2]. Using their differential equations gives $p (T)^{T} δ x (T) - p (0)^{T} δ x (0) = - \int_{0}^{T} (x - x_{d})^{T} Q δ x d t + \int_{0}^{T} p^{T} B w d t$ . Since $δ x (0) = 0$ and $p (T) = P (x (T) - x_{T})$ , rearranging yields $\int_{0}^{T} (x - x_{d})^{T} Q δ x d t + (x (T) - x_{T})^{T} P δ x (T) = \int_{0}^{T} p^{T} B w d t$ .

2.1F1F2F4step 1.1step 1.2algebra∎

Substitute the identity from step 1.2 into the derivative coefficient of step 1.1 and use $(p^{T} B w) = (B^{T} p)^{T} w$ and $(u^{T} R w) = (R u)^{T} w$ because $R = R^{T}$ almost everywhere. The resulting coefficient is $D J (u) [w]$ . Moreover, $B^{T} p + R u \in L^{2}$ since $B, R$ are essentially bounded, $p$ is continuous, and $u \in L^{2}$ ; [F4] therefore makes $D J (u)$ a bounded linear map. The quadratic remainder bound in step 1.1 proves the stated first-order norm expansion.

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Sources

J. Monnier, Optimal Control of ODEs: The Linear-Quadratic (LQ) Case, course notes, Chapter 2 §2.5.2 (standard reference, not scraped)
W. W. Hager, Runge-Kutta methods in optimal control and the transformed adjoint system, Numerische Mathematik 87 (2000) (standard reference, not scraped)