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An active terminal constraint invalidates the pointwise control-only variational inequality

Statement refuted

Assume the Axioms of Countable Choice and Dependent Choice, which are the hypotheses used by the reduced-gradient supplier.

The pointwise control-only condition from LQR minimizers satisfy a feasible-set variational inequality holds for an LQR minimizer even when the feasible set includes arbitrary terminal constraints.

Facts & Assumptions

Given: Countable Choice and Dependent Choice. Take $T = 1$ , scalar state and control, $x (0) = 0$ , dynamics $\dot{x} = u$ , control set $U_{c} = [0, 2]$ , and terminal constraint $x (1) = 0$ . Let $Q = R = 1$ , $x_{d} (t) = 1$ , $P = 0$ , and $x_{T} = 0$ , so $F = {u \in L^{2} (0, 1) : 0 \leq u (t) \leq 2 a.e., \int_{0}^{1} u (t) d t = 0}$ . Use the adjoint and derivative conventions of Continuous adjoint equation for the linear-quadratic ODE and The LQR adjoint identity gives the reduced gradient.

[A1]

Countable Choice and Dependent Choice are assumed for the reduced-gradient 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).

[F1]

A nonnegative function with integral zero vanishes almost everywhere (A nonnegative measurable function has integral $0$ exactly when it vanishes almost everywhere).

[F2]

The adjoint satisfies $- \dot{p} = x - x_{d}$ and $p (1) = P (x (1) - x_{T})$ ; the reduced gradient is $p + u$ (Continuous adjoint equation for the linear-quadratic ODE, The LQR adjoint identity gives the reduced gradient).

Counterexample

The feasible set in the Given data is determined exactly by the zero terminal-state constraint.

1.1F1given

For every feasible $u$ , [F1] applied to $u \geq 0$ and $\int_{0}^{1} u = 0$ gives $u = 0$ almost everywhere. Thus $F = {0}$ and $u^{*} = 0$ is its unique feasible control and hence a minimizer.

2.1A1F2step 1.1algebra

The state is $x (t) = 0$ . The adjoint equation becomes $- \dot{p} = - 1$ with $p (1) = 0$ , whose unique solution is $p (t) = t - 1$ . The reduced gradient at $u^{*}$ is therefore $g (t) = p (t) = t - 1$ .

3.1step 1.1step 2.1algebra∎

For the pointwise control-only condition, choose the admissible point $c = 1$ . Then $g (t) (c - u^{*} (t)) = t - 1 < 0$ for every $t \in [0, 1)$ , a set of positive measure. Hence the asserted pointwise condition fails, although the global variational inequality over the singleton $F$ holds with equality.

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Sources

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