Pipeline-generated

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Direct transcription and numerical ode optimal control

1 · Prerequisites

2 · Summary

This page separates a measurable-control existence model from the pointwise smooth-data branch required to evaluate Runge–Kutta stages. It proves the bounded linear state map; under their stated Countable Choice hypotheses, it proves weak closedness of compact convex actuator bounds, weak-to-uniform state convergence, and strong recovery by cell averages. The LQR branch gives existence and uniqueness using the stated ultrafilter lemma, Dependent Choice, and Hahn–Banach assumptions, then derives the continuous adjoint, reduced gradient, and feasible-set variational inequality under their explicit choice hypotheses. The pointwise Hamiltonian condition is shown only for control-only constraints; a terminal-state example refutes the broader claim.

The transcription procedures cover single and multiple shooting, explicit stages, and implicit collocation with coupled stage residuals solved internally and their nonlinear-solve statuses returned. Declared path-constraint times are inserted as mesh nodes; sampled constraints do not certify continuous-path feasibility. Every equivalence-class datum used at a stage or quadrature time needs a chosen pointwise representative and sampling rule. Exact single/multiple shooting equivalence is finite-dimensional, and convex quadratic transcriptions reduce to QPs when sampled weights and constraints meet the stated assumptions. Mesh convergence is conditional on uniform objective consistency and feasible recovery. Strong convexity converts uniform state/quadrature and comparison-recovery bounds into an objective and control rate; an embedded RK difference alone is not a global error certificate.

The finite dispatcher reports decision variables, defects, constraints, KKT residuals, objective, and separate NLP, stage, IVP, quadrature, mesh, and budget statuses. Exact-Hessian SQP convergence is local for a fixed transcription under LICQ, strict complementarity, SOSC, and smoothness. A mesh-family KKT limit is also proved conditionally from residual consistency, uniformly bounded discrete KKT sensitivities, and reconstruction into common normed spaces; finite solver budgets may stop before convergence.

3 · Logical flowchart

4 · Definitions, theorems and proofs

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-6-luna)Open item page →

Finite-horizon bounded-control linear-quadratic ODE problem

Definition

Fix a finite horizon $T > 0$ , state and control dimensions $n, m \geq 1$ , and the real quotient space $L^{2} (0, T; R^{m})$ of measurable controls modulo equality almost everywhere (The space $L^{p} (μ)$ as the quotient by null functions). Let $A \in L^{\infty} (0, T; R^{n \times n})$ and $B \in L^{\infty} (0, T; R^{n \times m})$ be equivalence classes, let $x_{0}, x_{T} \in R^{n}$ , and let $x_{d} \in L^{2} (0, T; R^{n})$ . Let $U_{c} \subset R^{m}$ be nonempty, compact and convex. The pointwise control set is

U_{c} : = {u \in L^{2} (0, T; R^{m}) : u (t) \in U_{c} for almost every t} .

Let $Q \in L^{\infty} (0, T; R^{n \times n})$ and $R \in L^{\infty} (0, T; R^{m \times m})$ be symmetric almost everywhere, with $Q (t) ⪰ 0$ and $r I ⪯ R (t) ⪯ M I$ almost everywhere for constants $0 < r \leq M < \infty$ . Let $P \in R^{n \times n}$ be symmetric positive semidefinite. For each $u \in U_{c}$ , its state is the unique continuous solution of the Volterra equation

x_{u} (t) = x_{0} + \int_{0}^{t} (A (s) x_{u} (s) + B (s) u (s)) d s .

The same state equation is defined for every $u \in L^{2} (0, T; R^{m})$ , and the running and terminal cost on that ambient space is

J (u) = \frac{1}{2} \int_{0}^{T} ⁣ ((x_{u} (t) - x_{d} (t))^{T} Q (t) (x_{u} (t) - x_{d} (t)) + u (t)^{T} R (t) u (t)) d t + \frac{1}{2} (x_{u} (T) - x_{T})^{T} P (x_{u} (T) - x_{T}) .

The constrained problem minimizes this ambient functional over $F$ , where additional path or terminal constraints are represented by a nonempty convex weakly sequentially closed set $F \subseteq U_{c}$ ; with no additional state constraints, $F = U_{c}$ . “Weakly sequentially closed” means that every weakly convergent sequence in $F$ has its weak limit in $F$ (Weak convergence of nets and sequences). The definition asserts no optimizer or adjoint existence; the state-map well-definedness is supplied by The bounded measurable linear ODE has a unique affine state map ↗.

DefinitionDefinition: Literature-sourcedProof: Not applicableOpen item page →

Finite-horizon ODE optimal-control transcription data

Definition

For pointwise numerical transcription on $[0, T]$ , the data consist of integers $n, m \geq 1$ , open sets $Ω_{x} \subseteq R^{n}$ and $Ω_{u} \subseteq R^{m}$ , a nonempty compact convex control set $U_{c} \subset Ω_{u}$ , an initial state $x_{0} \in Ω_{x}$ , a pointwise-defined vector field $f : [0, T] \times Ω_{x} \times Ω_{u} \to R^{n}$ , a pointwise-defined running cost $L : [0, T] \times Ω_{x} \times Ω_{u} \to R$ , and a terminal cost $ϕ : Ω_{x} \to R$ . The admissible state solves $\dot{x} (t) = f (t, x (t), u (t))$ almost everywhere and the continuous objective is $\int_{0}^{T} L (t, x (t), u (t)) d t + ϕ (x (T))$ . Optional path and endpoint constraints are pointwise-defined functions evaluated at declared continuous times or transcription nodes. A compact tube is a compact subset of $[0, T] \times Ω_{x} \times Ω_{u}$ containing the state/control triples used by stages and trajectories. On such a tube, $f, L, ϕ$ and any constraint functions used in a local NLP theorem are continuous and have continuous state/control derivatives through order three (the $C^{3}$ convention of $C^{k}$ Euclidean maps and diffeomorphisms), with those derivatives bounded. An $L^{\infty}$ equivalence class alone has no specified value at a Runge–Kutta stage time, so it is not an admissible stage-evaluation input without an additional pointwise representative and sampling rule. This definition makes no existence, convergence, or continuous-path-feasibility claim.

LemmaStatement: Literature-sourcedProof: AI-generatedjudge pass (gpt-6-luna)Open item page →

The bounded measurable linear ODE has a unique affine state map

Statement

Let $T > 0$ , $A \in L^{\infty} (0, T; R^{n \times n})$ , $B \in L^{\infty} (0, T; R^{n \times m})$ , $u \in L^{2} (0, T; R^{m})$ , and $x_{0} \in R^{n}$ . Write $∥ A ∥_{\infty}$ and $∥ B ∥_{\infty}$ for the essential suprema of their Euclidean operator norms. There is a unique absolutely continuous $x : [0, T] \to R^{n}$ satisfying

x (t) = x_{0} + \int_{0}^{t} (A (s) x (s) + B (s) u (s)) d s .

Under Countable Choice it satisfies $\dot{x} = A x + B u$ almost everywhere, and it always satisfies

∥ x ∥_{\infty} \leq e^{∥ A ∥_{\infty} T} (∥ x_{0} ∥_{2} + ∥ B ∥_{\infty} \sqrt{T} ∥ u ∥_{2}) .

For fixed $A, B, x_{0}$ , the map $u \mapsto x_{u}$ is affine into $C ([0, T]; R^{n})$ , and its zero-initial-state linear part is bounded from $L^{2}$ to the supremum norm.

The almost-everywhere derivative assertion uses Countable Choice, as required by the cited $L^{1}$ fundamental theorem. The integral-equation existence, uniqueness, bound, and affine-map claims do not require that axiom.

Facts & Assumptions

Given: $T > 0$ , essentially bounded matrix classes $A, B$ , a control $u \in L^{2}$ , an initial state $x_{0}$ , and the standard Lebesgue integral.

[F1]

The scalar Lebesgue integral is linear and satisfies $∣ \int h ∣ \leq \int ∣ h ∣$ for $h \in L^{1}$ ; it is monotone on nonnegative integrable functions. Euclidean Cauchy--Schwarz gives $∣ q^{T} h ∣ \leq ∥ q ∥_{2} ∥ h ∥_{2}$ (The Lebesgue integral is linear on $L^{1} (μ)$ , The modulus of an integral is bounded by the integral of the modulus, Monotonicity and nonnegative homogeneity of the nonnegative integral, Cauchy-Schwarz $∣ ⟨ x, y ⟩ ∣ \leq ∥ x ∥_{2} ∥ y ∥_{2}$ with its equality case, the triangle inequality for $∥ \cdot ∥_{2}$ , the parallelogram law and polarisation).

[F2]

The indefinite integral of an $L^{1}$ function is absolutely continuous; under Countable Choice its derivative equals the integrand almost everywhere (The indefinite integral of an $L^{1}$ function is absolutely continuous, The indefinite integral of an $L^{1}$ function is differentiable almost everywhere).

[F3]

$C ([a, b]; R)$ is complete in the supremum metric; the finite product $C ([a, b]; R^{n})$ is therefore complete componentwise ( $C (K, R)$ is complete in the supremum metric for every nonempty compact metric space $K$ ).

[F4]

If $v (t) \leq c + M \int_{0}^{t} v (s) d s$ for continuous $v \geq 0$ and constants $c, M \geq 0$ , then $v (t) \leq c e^{M t}$ (Gronwall's integral inequality with variable and constant coefficients).

[F5]

$∥ u ∥_{L^{1} (0, T)} \leq \sqrt{T} ∥ u ∥_{L^{2} (0, T)}$ (Cauchy-Schwarz inequality for $L^{2}$ ).

[F6]

The Lebesgue integral is linear on $L^{1}$ ; changing an integrand on a null set does not change the integral (The Lebesgue integral is linear on $L^{1} (μ)$ ).

[F7]

For $0 \leq q < 1$ , the geometric series $\sum_{k = 0}^{\infty} q^{k}$ converges to $1 / (1 - q)$ (For $∣ r ∣ < 1$ , $\sum_{k \geq 0} r^{k} = 1 / (1 - r)$ , and for $∣ r ∣ \geq 1$ the series diverges).

[A1]

Countable Choice is assumed only for the almost-everywhere derivative conclusion (The Axiom of Countable Choice ( ${A C}_{ω}$ )).

Proof

1.1F1F2F5choosealgebra

For any Lebesgue-integrable vector function $h$ with $\int ∥ h ∥_{2} < \infty$ , write $I = \int h$ . If $I = 0$ , the norm inequality is immediate. Otherwise put $q = I / ∥ I ∥_{2}$ . Then $∥ I ∥_{2} = \int q^{T} h \leq \int ∣ q^{T} h ∣ \leq \int ∥ h ∥_{2}$ , by [F1], linearity componentwise, and $∥ q ∥_{2} = 1$ . Thus the vector integral estimate needed below holds for these Lebesgue integrands. Now put $g = B u \in L^{1} (0, T; R^{n})$ ; this follows from $B \in L^{\infty}$ , $u \in L^{2}$ , and [F5]. Choose a finite partition $0 = t_{0} < \dots < t_{N} = T$ with $∥ A ∥_{\infty} (t_{j + 1} - t_{j}) < 1$ on each subinterval when $∥ A ∥_{\infty} > 0$ ; if $∥ A ∥_{\infty} = 0$ , use $N = 1$ . Given the already constructed value $x (t_{j})$ , define $T z (t) = x (t_{j}) + \int_{t_{j}}^{t} (A (s) z (s) + g (s)) d s$ on $C ([t_{j}, t_{j + 1}]; R^{n})$ . By [F2], $T$ maps continuous functions to continuous functions, and the vector-integral estimate just proved gives $∥ T z - T w ∥_{\infty} \leq q_{j} ∥ z - w ∥_{\infty}$ with $q_{j} = ∥ A ∥_{\infty} (t_{j + 1} - t_{j}) < 1$ .

2.1F3F7step 1.1algebrainduction

Starting from the constant function $z_{0} = x (t_{j})$ , let $z_{k + 1} = T z_{k}$ . Then $∥ z_{k + 1} - z_{k} ∥_{\infty} \leq q_{j}^{k} ∥ z_{1} - z_{0} ∥_{\infty}$ . By [F7], the geometric majorant has finite sum, so the iterates are Cauchy in the supremum norm. Each coordinate converges uniformly by [F3], and the finite coordinate limit is a continuous vector function $z$ . The integral estimate in step 1.1 makes $T$ continuous in the supremum norm, hence $z = T z$ . If $z, w$ are two fixed points, $∥ z - w ∥_{\infty} \leq q_{j} ∥ z - w ∥_{\infty}$ , so $z = w$ . Solving the finitely many subintervals in order gives one continuous solution on $[0, T]$ , and the same uniqueness on each subinterval proves global uniqueness.

3.1A1F2F6step 2.1

The integrand $A x + B u$ lies in $L^{1}$ . Its indefinite integral is absolutely continuous by [F2], as required in the Statement. Under [A1], the $L^{1}$ fundamental theorem in [F2] differentiates the integral equation almost everywhere and yields $\dot{x} = A x + B u$ . The same equation is independent of the chosen representatives of $A, B, u$ by [F6].

3.2F4F5step 1.1step 2.1algebra

From the integral equation and step 1.1, $∥ x (t) ∥_{2} \leq ∥ x_{0} ∥_{2} + ∥ B ∥_{\infty} ∥ u ∥_{L^{1}} + ∥ A ∥_{\infty} \int_{0}^{t} ∥ x (s) ∥_{2} d s$ . Apply [F4] with $c = ∥ x_{0} ∥_{2} + ∥ B ∥_{\infty} ∥ u ∥_{L^{1}}$ and then [F5]; this gives the displayed supremum bound.

4.1F6step 2.1step 3.2algebra∎

Linearity of the integral [F6] shows that a linear combination of two zero-initial-state solutions is the solution for the same linear combination of controls; uniqueness from step 2.1 makes the zero-initial-state map linear. Step 3.2 with $x_{0} = 0$ bounds this linear map in the $L^{2}$ and supremum norms, while the fixed initial-state contribution is independent of $u$ . Hence the full map is affine and bounded as claimed.

LemmaStatement: AI-adaptedProof: AI-generatedOpen item page →

Under Countable Choice, pointwise compact convex control constraints are weakly closed in L2

Statement

Assume the Axiom of Countable Choice.

Let $U_{c} \subset R^{m}$ be nonempty, compact and convex, and let $U_{c}$ be the pointwise admissible set in Finite-horizon bounded-control linear-quadratic ODE problem. If $u_{k} \in U_{c}$ and $u_{k} ⇀ u$ weakly in real $L^{2} (0, T; R^{m})$ , then $u \in U_{c}$ . Thus $U_{c}$ is weakly sequentially closed.

Facts & Assumptions

Given: Countable Choice, a finite horizon $T > 0$ , a nonempty compact convex set $U_{c} \subset R^{m}$ , its pointwise admissible set $U_{c}$ , and a weakly convergent sequence in real $L^{2} (0, T; R^{m})$ .

[A1]

Countable Choice ensures the Lebesgue outer measure construction is a measure on its sigma-algebra (The Axiom of Countable Choice ( ${A C}_{ω}$ ), Assuming countable choice, $L (R^{n})$ is a sigma-algebra containing every elementary set and $λ_{n}$ is a complete measure extending elementary volume).

[F1]

$U_{c} = {u \in L^{2} : u (t) \in U_{c} for almost every t}$ (Finite-horizon bounded-control linear-quadratic ODE problem). The pairing $\int q^{T} w$ is well-defined for vector-valued $L^{2}$ classes and bounded by $∥ q ∥_{2} ∥ w ∥_{2}$ : apply Euclidean Cauchy--Schwarz pointwise and scalar $L^{2}$ Cauchy--Schwarz to the norm functions (Cauchy-Schwarz $∣ ⟨ x, y ⟩ ∣ \leq ∥ x ∥_{2} ∥ y ∥_{2}$ with its equality case, the triangle inequality for $∥ \cdot ∥_{2}$ , the parallelogram law and polarisation, Cauchy-Schwarz inequality for $L^{2}$ ).

[F2]

Weak convergence means convergence under every bounded linear functional; by [F1], $w \mapsto \int_{E} q^{T} w$ is bounded on $L^{2}$ whenever $E \subset [0, T]$ is measurable (Weak convergence of nets and sequences).

[F3]

A continuous real function on a nonempty compact metric space attains its minimum (A continuous real-valued function on a nonempty compact metric space is bounded and attains a greatest and a least value).

[F4]

The Lebesgue integral is linear on $L^{1}$ , and the nonnegative integral is monotone (The Lebesgue integral is linear on $L^{1} (μ)$ , Monotonicity and nonnegative homogeneity of the nonnegative integral).

[F5]

Rational open boxes form a countable basis of the product topology, and on $R^{m}$ that topology is the Euclidean metric topology ( $Q^{n}$ is a countable dense subset of $R^{n}$ , and rational open boxes form a countable basis, For $n \geq 1$ the product topology on $n$ copies of the usual topology of $R$ is the metric topology of $d_{\infty}$ on $R^{n}$ , and hence also of $d_{1}$ and $d_{2}$ , so $R^{n}$ as a product and $R^{n}$ as a metric space are one space).

[F6]

Under Countable Choice, Lebesgue measurable sets form a sigma-algebra and a countable union of measurable null sets is null, by countable subadditivity (Assuming countable choice, $L (R^{n})$ is a sigma-algebra containing every elementary set and $λ_{n}$ is a complete measure extending elementary volume, Finite and countable subadditivity of measures).

Given: The set and sequence in the Statement.

Proof

Given: The set and sequence in the Statement.

1.1F1F3F5choosealgebra

For $q \in R^{m}$ set $h (q) = \sup_{c \in U_{c}} q^{T} c$ , which is finite because $U_{c}$ is compact and hence bounded. If $z \notin U_{c}$ , [F3] gives a nearest point $y \in U_{c}$ to $z$ . For each $c \in U_{c}$ and $0 < θ \leq 1$ , convexity gives $y + θ (c - y) \in U_{c}$ ; minimality of $y$ implies $∥ z - y - θ (c - y) ∥_{2}^{2} \geq ∥ z - y ∥_{2}^{2}$ , and division by $θ$ followed by $θ ↓ 0$ gives $(z - y)^{T} (c - y) \leq 0$ . Thus $q_{0} = z - y$ satisfies $q_{0}^{T} z - h (q_{0}) \geq ∥ z - y ∥_{2}^{2} > 0$ . If $M_{c} = \sup_{c \in U_{c}} ∥ c ∥_{2}$ , then $∣ h (q) - h (q_{0}) ∣ \leq M_{c} ∥ q - q_{0} ∥_{2}$ and $∣ (q - q_{0})^{T} z ∣ \leq ∥ q - q_{0} ∥_{2} ∥ z ∥_{2}$ . By [F5], rational vectors are dense in Euclidean $R^{m}$ , so one can choose rational $q$ close enough to $q_{0}$ that $q^{T} z > h (q)$ . Therefore a point $z$ belongs to $U_{c}$ exactly when $q^{T} z \leq h (q)$ for every rational vector $q$ .

1.2A1F1F2F4givenalgebra

Fix a rational $q$ . Every $u_{k}$ is in $U_{c}$ , so $h (q) - q^{T} u_{k} (t) \geq 0$ almost everywhere. For any measurable $E \subseteq [0, T]$ , the test function $q 1_{E}$ lies in $L^{2}$ because $T < \infty$ ; weak convergence, [F2], and [F4] give $\int_{E} q^{T} u d t = \lim_{k} \int_{E} q^{T} u_{k} d t \leq h (q) ∣ E ∣$ . If $q^{T} u > h (q)$ on a set of positive measure, then for some integer $r \geq 1$ the measurable set $E_{r} = {t : q^{T} u (t) \geq h (q) + 1 / r}$ has positive measure, contradicting the integral inequality with $E = E_{r}$ . Hence $q^{T} u (t) \leq h (q)$ almost everywhere.

2.1A1F1F5F6step 1.1step 1.2algebra∎

The rational vectors form a countable set by [F5], so the exceptional null sets from step 1.2 have a null union by [A1, F6]. Outside that union all rational support inequalities hold simultaneously; step 1.1 then gives $u (t) \in U_{c}$ almost everywhere. This is exactly $u \in U_{c}$ .

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-6-luna)Open item page →

Time meshes and piecewise-constant control parameterization

Definition

For the horizon $[0, T]$ of Finite-horizon bounded-control linear-quadratic ODE problem, a time mesh is a finite partition $0 = t_{0} < t_{1} < \dots < t_{N} = T$ with $N \geq 1$ . Its cell widths are $h_{j} = t_{j + 1} - t_{j} > 0$ and its mesh size is $∣ Δ ∣ = \max_{0 \leq j < N} h_{j}$ . A piecewise-constant control vector is $(u_{0}, \dots, u_{N - 1})$ and represents the measurable control $u^{Δ} (t) = u_{j}$ for $t \in [t_{j}, t_{j + 1})$ ; its value at $T$ is immaterial in $L^{2}$ . If $u_{j} \in U_{c}$ for every $j$ , then $u^{Δ} \in U_{c}$ . A mesh family is refining when its mesh sizes tend to zero.

LemmaStatement: AI-adaptedProof: AI-generatedOpen item page →

Weakly convergent bounded controls have uniformly convergent states

Statement

For the data of Finite-horizon bounded-control linear-quadratic ODE problem, if $u_{k}, u \in U_{c}$ and $u_{k} ⇀ u$ weakly in $L^{2} (0, T; R^{m})$ , then $x_{u_{k}} \to x_{u}$ uniformly on $[0, T]$ .

Facts & Assumptions

Given: The bounded measurable linear state equation, the compact control set $U_{c}$ , and a sequence $u_{k} ⇀ u$ in the pointwise admissible set.

[F1]

The state map has a unique continuous solution and its zero-initial-state part is bounded and linear from $L^{2}$ to $C ([0, T]; R^{n})$ (The bounded measurable linear ODE has a unique affine state map).

[F2]

Every control in $U_{c}$ satisfies $∥ u_{k} (t) ∥_{2} \leq M_{c} : = \sup_{v \in U_{c}} ∥ v ∥_{2} < \infty$ almost everywhere (Finite-horizon bounded-control linear-quadratic ODE problem and compactness of $U_{c}$ ).

[F3]

Weak convergence means convergence under every bounded linear functional on $L^{2}$ (Weak convergence of nets and sequences).

[F4]

For a Lebesgue-integrable vector function $h$ , the componentwise integral obeys $∥ \int h ∥_{2} \leq \int ∥ h ∥_{2}$ : if $I = \int h \neq 0$ , test the integral against $I / ∥ I ∥_{2}$ and use componentwise linearity, Euclidean Cauchy--Schwarz, scalar integral monotonicity, and $∣ \int q^{T} h ∣ \leq \int ∣ q^{T} h ∣$ (The Lebesgue integral is linear on $L^{1} (μ)$ , Cauchy-Schwarz $∣ ⟨ x, y ⟩ ∣ \leq ∥ x ∥_{2} ∥ y ∥_{2}$ with its equality case, the triangle inequality for $∥ \cdot ∥_{2}$ , the parallelogram law and polarisation, The modulus of an integral is bounded by the integral of the modulus, Monotonicity and nonnegative homogeneity of the nonnegative integral); if $I = 0$ the bound is immediate.

Proof

1.1F1F2F4algebra

By [F1] and [F2], the states have a common bound $M_{x}$ in supremum norm. For $s, t \in [0, T]$ , the integral equation and [F4] give $∥ x_{u_{k}} (t) - x_{u_{k}} (s) ∥_{2} \leq L ∣ t - s ∣$ , where $L : = ∥ A ∥_{\infty} M_{x} + ∥ B ∥_{\infty} M_{c}$ . The same bound holds for $x_{u}$ .

1.2F1F3

For each fixed $t$ , evaluation at $t$ composed with the bounded linear zero-initial-state map in [F1] is a bounded linear functional on $L^{2}$ . Thus [F3] gives $x_{u_{k}} (t) \to x_{u} (t)$ coordinatewise.

2.1step 1.1step 1.2choosealgebra∎

If $L = 0$ , step 1.1 makes each state constant, and step 1.2 gives uniform convergence. If $L > 0$ , fix $ε > 0$ and choose a finite partition of $[0, T]$ with mesh smaller than $ε / (3 L)$ . By step 1.2 and finiteness of its nodes, there is an index after which $∥ x_{u_{k}} - x_{u} ∥_{2} < ε / 3$ at every node. Any $t$ is within the mesh of a node, so step 1.1 gives $∥ x_{u_{k}} (t) - x_{u} (t) ∥_{2} \leq L ∣ t - t_{j} ∣ + ∥ x_{u_{k}} (t_{j}) - x_{u} (t_{j}) ∥_{2} + L ∣ t - t_{j} ∣ < ε$ . Hence the convergence is uniform.

LemmaStatement: AI-adaptedProof: AI-generatedOpen item page →

Cell averages give a strong recovery sequence for compact convex controls

Statement

Assume the Axiom of Countable Choice.

Assume the constraint set in Finite-horizon bounded-control linear-quadratic ODE problem is only the pointwise box $F = U_{c}$ . For the uniform mesh $0 = t_{0} < \dots < t_{N} = T$ , $h = T / N$ , define $P_{N} u$ on $[t_{j}, t_{j + 1})$ by $h^{- 1} \int_{t_{j}}^{t_{j + 1}} u (s) d s$ . Then $P_{N} u \in U_{c}$ for every $u \in U_{c}$ , $P_{N} u$ is piecewise constant, and $∥ P_{N} u - u ∥_{L^{2}} \to 0$ as $N \to \infty$ . This recovery statement need not hold for additional state or terminal constraints unless they are separately shown to be preserved or approximated.

Facts & Assumptions

Given: Countable Choice, $T > 0$ , a nonempty compact convex $U_{c} \subset R^{m}$ , its admissible control set, and the uniform meshes in Time meshes and piecewise-constant control parameterization.

[F1]

Finite linear combinations of interval indicators are dense in scalar $L^{2} (R)$ under Countable Choice; applying this componentwise gives density of finite vector-valued interval step functions, and extending by zero applies it on $(0, T)$ (Finite linear combinations of box indicators are dense in $L^{p} (R^{n})$ for $1 \leq p < \infty$ , The Axiom of Countable Choice ( ${A C}_{ω}$ )).

[F2]

The componentwise Lebesgue vector integral is linear. For $h \in L^{2} (I; R^{m})$ on a cell $I$ of length $h_{I}$ , put $I_{h} = \int_{I} h$ . If $I_{h} \neq 0$ , then $q = I_{h} / ∥ I_{h} ∥_{2}$ satisfies $∥ I_{h} ∥_{2} = \int_{I} q^{T} h \leq \int_{I} ∥ h ∥_{2} \leq \sqrt{h_{I}} ∥ h ∥_{L^{2} (I)}$ , using pointwise Euclidean Cauchy--Schwarz, monotonicity, scalar Cauchy--Schwarz, and componentwise linearity; the zero case is immediate (The Lebesgue integral is linear on $L^{1} (μ)$ , Cauchy-Schwarz $∣ ⟨ x, y ⟩ ∣ \leq ∥ x ∥_{2} ∥ y ∥_{2}$ with its equality case, the triangle inequality for $∥ \cdot ∥_{2}$ , the parallelogram law and polarisation, Cauchy-Schwarz inequality for $L^{2}$ , The modulus of an integral is bounded by the integral of the modulus, Monotonicity and nonnegative homogeneity of the nonnegative integral).

[F3]

A continuous distance function attains its minimum on the nonempty compact set $U_{c}$ (A continuous real-valued function on a nonempty compact metric space is bounded and attains a greatest and a least value).

[F4]

For $v, w \in L^{2} (0, T; R^{m})$ , the pointwise Euclidean triangle inequality gives $∥ v + w ∥_{2} \leq ∥ v ∥_{2} + ∥ w ∥_{2}$ . Squaring, integrating by monotonicity, and applying scalar Minkowski to the scalar functions $∥ v (\cdot) ∥_{2}$ and $∥ w (\cdot) ∥_{2}$ yields $∥ v + w ∥_{L^{2}} \leq ∥ ∥ v (\cdot) ∥_{2} + ∥ w (\cdot) ∥_{2} ∥_{L^{2}} \leq ∥ v ∥_{L^{2}} + ∥ w ∥_{L^{2}}$ (Cauchy-Schwarz $∣ ⟨ x, y ⟩ ∣ \leq ∥ x ∥_{2} ∥ y ∥_{2}$ with its equality case, the triangle inequality for $∥ \cdot ∥_{2}$ , the parallelogram law and polarisation, Monotonicity and nonnegative homogeneity of the nonnegative integral, Minkowski's inequality for integrals, including $p = \infty$ ).

Proof

1.1F2F3givenalgebra

On each interval $I_{j} = [t_{j}, t_{j + 1}]$ , let $m_{j} = h^{- 1} \int_{I_{j}} u (s) d s$ . If $m_{j} \notin U_{c}$ , compactness and [F3] give a nearest point $y \in U_{c}$ . For every $c \in U_{c}$ , the segment $y + θ (c - y)$ belongs to $U_{c}$ for $0 \leq θ \leq 1$ ; minimality of $y$ gives $∥ m_{j} - y - θ (c - y) ∥_{2}^{2} \geq ∥ m_{j} - y ∥_{2}^{2}$ . Divide by $θ > 0$ and let $θ ↓ 0$ to get $(m_{j} - y)^{T} (c - y) \leq 0$ . With $q_{0} = m_{j} - y \neq 0$ , this implies $q_{0}^{T} m_{j} > \sup_{c \in U_{c}} q_{0}^{T} c$ . But $q_{0}^{T} u (s) \leq \sup_{c \in U_{c}} q_{0}^{T} c$ almost everywhere, so integration contradicts the strict inequality. Thus every cell average lies in $U_{c}$ and $P_{N} u$ is pointwise admissible.

1.2F2F3F4algebra

For any $w \in L^{2}$ , [F2] on a cell $I_{j}$ gives $h ∥ h^{- 1} \int_{I_{j}} w ∥_{2}^{2} \leq \int_{I_{j}} ∥ w (s) ∥_{2}^{2} d s$ ; summing proves $∥ P_{N} w ∥_{2} \leq ∥ w ∥_{2}$ . For an interval indicator, $P_{N}$ agrees with it on cells not meeting an endpoint, and at most two endpoint cells contribute to the squared error, at most $2 h$ . By linearity of $P_{N}$ and the triangle inequality [F4], $P_{N} s \to s$ for every finite linear combination $s$ of interval indicators.

2.1F1F4step 1.2algebra∎

Fix $u \in L^{2}$ and $ε > 0$ . By [F1] choose such an interval step function $s$ with $∥ u - s ∥_{2} < ε$ . The contraction bound in step 1.2 and the triangle inequality [F4] give $∥ P_{N} u - u ∥_{2} \leq 2 ε + ∥ P_{N} s - s ∥_{2}$ . Taking $N \to \infty$ yields ${lim sup}_{N} ∥ P_{N} u - u ∥_{2} \leq 2 ε$ . Since $ε$ is arbitrary, $P_{N} u \to u$ strongly in $L^{2}$ .

DefinitionDefinition: Literature-sourcedProof: Not applicableOpen item page →

Runge–Kutta tableaux and stage collocation equations

Definition

Let $s \geq 1$ be a stage count, $A = (a_{i k}) \in R^{s \times s}$ , and $b, c \in R^{s}$ satisfy $c_{i} = \sum_{k = 1}^{s} a_{i k}$ for each $i$ . This is a (possibly implicit) Runge–Kutta tableau. On mesh cell $[t_{j}, t_{j + 1}]$ with width $h_{j}$ , a start state $X_{j, 0}$ and piecewise-constant control $u_{j}$ , define stage times $t_{j, i} = t_{j} + c_{i} h_{j}$ and the collocation stage equations $X_{j, i} = X_{j, 0} + h_{j} \sum_{k = 1}^{s} a_{i k} f (t_{j, k}, X_{j, k}, u_{j}), i = 1, \dots, s .$ The equations are admissibly defined only for $X_{j, 0} \in Ω_{x}$ , $u_{j} \in Ω_{u}$ , and stage inputs $(t_{j, i}, X_{j, i}, u_{j}) \in [0, T] \times Ω_{x} \times Ω_{u}$ for every $i$ , using the domains in Finite-horizon ODE optimal-control transcription data. A transcription procedure must check these memberships before each query of $f$ or $L$ , including trial stages of an implicit solve, and report a stage-time-domain failure for an invalid time or a stage-domain failure for an invalid state or control without making that query. The corresponding endpoint equation is $X_{j + 1, 0} = X_{j, 0} + h_{j} \sum_{i = 1}^{s} b_{i} f (t_{j, i}, X_{j, i}, u_{j})$ . In a finite transcription, an implicit tableau's coupled stage residuals are solved internally and the stage variables are eliminated from the outer NLP; the procedure returns the nonlinear stage-solve status and residuals. The quadrature contribution of the cell is $h_{j} \sum_{i = 1}^{s} b_{i} L (t_{j, i}, X_{j, i}, u_{j})$ . This definition asserts no stage-solution existence, uniqueness, order, stability, or error bound. If $A$ is strictly lower triangular, evaluate stages in increasing order $i = 1, \dots, s$ : form $X_{j, i} = X_{j, 0} + h_{j} \sum_{k < i} a_{i k} F_{j, k}$ , check the stated time-state-control memberships, and then evaluate $F_{j, i} = f (t_{j, i}, X_{j, i}, u_{j})$ , reporting an unavailable value as a stage-evaluation failure. This uses the explicit tableau coefficients and algebraic recursion of Explicit Runge–Kutta tableaus, stage maps, and failure status, with the finite-horizon domain checks specified here; it does not invoke that item's open-time-domain stage map or require an extension of $f$ beyond $[0, T]$ .

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-6-luna)Open item page →

Continuous adjoint equation for the linear-quadratic ODE

Definition

For the data of Finite-horizon bounded-control linear-quadratic ODE problem and an admissible control $u$ with state $x_{u}$ , a continuous LQR adjoint is a curve $p : [0, T] \to R^{n}$ whose coordinates are absolutely continuous in the scalar sense of Absolute continuity on a compact interval, satisfying, almost everywhere,

- \dot{p} (t) = A (t)^{T} p (t) + Q (t) (x_{u} (t) - x_{d} (t)), p (T) = P (x_{u} (T) - x_{T}) .

The equation defines the adjoint notion without asserting existence. Existence and uniqueness for each admissible control are established under Countable Choice by The backward LQR adjoint equation has a unique solution ↗. This sign convention is the one for the minimization Hamiltonian $H (t, x, p, u) = L (t, x, u) + p^{T} (A (t) x + B (t) u)$ .

LemmaStatement: AI-adaptedProof: AI-generatedOpen item page →

Under Countable Choice, the LQR objective is weakly lower semicontinuous

Statement

Assume the Axiom of Countable Choice.

For the data of Finite-horizon bounded-control linear-quadratic ODE problem, if $u_{k} \in U_{c}$ and $u_{k} ⇀ u$ weakly in $L^{2} (0, T; R^{m})$ , then $J (u) \leq {lim inf}_{k \to \infty} J (u_{k})$ .

Facts & Assumptions

Given: Countable Choice, the LQR data, the associated costs and states, and a weakly convergent sequence $u_{k} ⇀ u$ in real $L^{2}$ .

[A1]

Countable Choice is the hypothesis of the weak-closedness supplier (The Axiom of Countable Choice ( ${A C}_{ω}$ ), Under Countable Choice, pointwise compact convex control constraints are weakly closed in L2).

[F1]

The state and cost are those of Finite-horizon bounded-control linear-quadratic ODE problem; $Q (t) ⪰ 0$ , $P ⪰ 0$ , and $R (t) ⪰ r I$ almost everywhere.

[F2]

Weak convergence of controls in $U_{c}$ implies uniform convergence of their states (Weakly convergent bounded controls have uniformly convergent states).

[F3]

The pairing $⟨ v, w ⟩ = \int_{0}^{T} v (t)^{T} w (t) d t$ is well-defined on vector-valued $L^{2}$ classes. Pointwise Euclidean Cauchy--Schwarz followed by scalar $L^{2}$ Cauchy--Schwarz gives $∣ ⟨ v, w ⟩ ∣ \leq ∥ v ∥_{2} ∥ w ∥_{2}$ (Cauchy-Schwarz $∣ ⟨ x, y ⟩ ∣ \leq ∥ x ∥_{2} ∥ y ∥_{2}$ with its equality case, the triangle inequality for $∥ \cdot ∥_{2}$ , the parallelogram law and polarisation, Cauchy-Schwarz inequality for $L^{2}$ ).

[F4]

Under Countable Choice, the pointwise admissible set $U_{c}$ is weakly sequentially closed (Under Countable Choice, pointwise compact convex control constraints are weakly closed in L2).

Proof

1.1A1F1F2F4algebra

By [A1] and [F4], the weak limit $u$ belongs to $U_{c}$ , so [F2] applies. Put $e_{k} = x_{u_{k}} - x_{d}$ and $e = x_{u} - x_{d}$ ; their difference $e_{k} - e = x_{u_{k}} - x_{u}$ tends to zero uniformly. Since $Q$ is essentially bounded, the estimate $∣ a^{T} Q (t) a - b^{T} Q (t) b ∣ \leq ∥ Q (t) ∥_{2} (∥ a ∥_{2} + ∥ b ∥_{2}) ∥ a - b ∥_{2}$ shows that $\int_{0}^{T} e_{k}^{T} Q e_{k} d t \to \int_{0}^{T} e^{T} Q e d t$ . The terminal terms converge by the same estimate with $P$ and $x_{u_{k}} (T) \to x_{u} (T)$ .

1.2F1F3givenalgebra

Let $R v = R (\cdot) v (\cdot)$ . The bounds on $R$ imply $∥ R v ∥_{2} \leq M ∥ v ∥_{2}$ and $⟨ R v, v ⟩ \geq 0$ ; symmetry gives $⟨ R v, w ⟩ = ⟨ v, R w ⟩$ . For every fixed $w \in L^{2}$ , $⟨ R u_{k}, u_{k} ⟩ \geq 2 ⟨ R u_{k}, w ⟩ - ⟨ R w, w ⟩$ , because the difference is $⟨ R (u_{k} - w), u_{k} - w ⟩ \geq 0$ . The right-hand side converges to $2 ⟨ R u, w ⟩ - ⟨ R w, w ⟩$ by weak convergence, as [F3] makes $v \mapsto ⟨ R v, w ⟩ = ⟨ v, R w ⟩$ bounded. Taking the supremum over $w$ gives ${lim inf}_{k} ⟨ R u_{k}, u_{k} ⟩ \geq ⟨ R u, u ⟩$ , with equality in the supremum at $w = u$ .

2.1step 1.1step 1.2algebra∎

The state and terminal cost terms converge by step 1.1, while the control quadratic term is lower semicontinuous by step 1.2. Multiplying by $1 / 2$ and adding these three terms gives $J (u) \leq {lim inf}_{k} J (u_{k})$ .

LemmaStatement: AI-adaptedProof: AI-generatedOpen item page →

AC zero-derivative constancy and integrable-derivative reconstruction under Countable Choice

Statement

Assume Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )) and let $a \leq b$ .

If $F : [a, b] \to R$ is absolutely continuous and $F^{'} = 0$ almost everywhere on $(a, b)$ , then $F$ is constant.
Let $n \geq 1$ , let $y : [a, b] \to R^{n}$ have absolutely continuous coordinates, and let $h \in L^{1} ([a, b]; R^{n})$ . If $y_{j}^{'} = h_{j}$ almost everywhere for each coordinate $j$ , then, for every $t \in [a, b]$ , $y (t) = y (a) + \int_{a}^{t} h (s) d s .$ Here $h \in L^{1} ([a, b]; R^{n})$ means that each coordinate $h_{j}$ is Lebesgue integrable, and the integral is the componentwise Lebesgue integral. For $a = b$ it is zero.

Facts & Assumptions

Given: Countable Choice, a compact interval, and the scalar or vector data in the Statement.

[F1]

Absolute continuity controls sums of absolute endpoint increments over finite families with pairwise disjoint interiors and sufficiently small total length (Absolute continuity on a compact interval).

[F2]

Every gauge on a compact interval admits one finite fine tagged partition; every cell lies in its tag's open gauge neighborhood (Cousin's lemma: every gauge on a compact interval admits a fine tagged partition, Gauges and gauge-fine tagged partitions of a compact interval).

[F3]

Under Countable Choice, a null set has an open superset of arbitrarily small Lebesgue measure (Assuming countable choice, the Lebesgue outer measure of an arbitrary subset of $R^{n}$ is the infimum of the measures of the open sets containing it).

[F4]

Under Countable Choice, Lebesgue measure is complete, monotone and additive on disjoint measurable sets and agrees with interval length; finite sets are null and finite unions of null sets are null (Assuming countable choice, $L (R^{n})$ is a sigma-algebra containing every elementary set and $λ_{n}$ is a complete measure extending elementary volume, Assuming countable choice, Lebesgue outer measure is an outer measure that restricts to elementary volume).

[F5]

Every nonempty subset of the naturals has a least element, and $2^{- k} \to 0$ (The well-ordering principle, For $∣ r ∣ < 1$ the sequence $r^{k}$ is null, and for $∣ r ∣ > 1$ the sequence $∣ r ∣^{k}$ diverges to $+ \infty$ ).

[F6]

For $h_{j} \in L^{1}$ , its indefinite Lebesgue integral is absolutely continuous and, under Countable Choice, has derivative $h_{j}$ almost everywhere (The indefinite integral of an $L^{1}$ function is absolutely continuous, The indefinite integral of an $L^{1}$ function is differentiable almost everywhere).

[F7]

Derivatives use the difference-quotient limit and obey the sum and scalar rules (The derivative $f^{'} (c) = \lim_{x \to c} \frac{f (x) - f (c)}{x - c}$ of $f : A \to R$ at a point $c \in A$ that is a limit point of $A$ , and differentiability on a set, Sums, scalar multiples, products and quotients: $(f + g)^{'} (c) = f^{'} (c) + g^{'} (c)$ , $(α f)^{'} (c) = α f^{'} (c)$ , $(f g)^{'} (c) = f^{'} (c) g (c) + f (c) g^{'} (c)$ , and $(f / g)^{'} (c) = (f^{'} (c) g (c) - f (c) g^{'} (c)) / g (c)^{2}$ when $g (c) \neq 0$ ).

Proof

1.1givenF1F3F4choose

If $a = b$ , both conclusions hold directly. Otherwise fix $a \leq s < t \leq b$ and $ε > 0$ for claim 1. Let $E$ be a null exceptional set for $F^{'} = 0$ on $[s, t]$ , enlarged by $s, t$ ; it is null by [F4]. Restricting [F1] to $[s, t]$ , choose $d > 0$ so that total interval length below $d$ implies total absolute increment below $ε / 2$ . Put $η = ε / (2 (t - s + 1))$ . By [F3], choose one open $U \supseteq E$ with $λ (U) < d$ .

2.1F2F5F7step 1.1construct

For $x \notin E$ , define $k (x)$ as the least $k \in N_{0}$ such that $∣ F (y) - F (x) ∣ \leq η ∣ y - x ∣$ whenever $y \in [s, t]$ and $∣ y - x ∣ < 2^{- k}$ . This set is nonempty by $F^{'} (x) = 0$ , [F7] and [F5], including $y = x$ by equality. For $x \in E$ , instead take the least $k$ with $(x - 2^{- k}, x + 2^{- k}) \subseteq U$ ; openness and [F5] give existence. Set $γ (x) = 2^{- k (x)}$ . These uniquely specified least integers define one positive gauge; no arbitrary pointwise radius selections are needed. Obtain one finite $γ$ -fine partition ${([x_{i - 1}, x_{i}], ξ_{i})}_{i = 1}^{m}$ from [F2].

3.1F1F4step 1.1step 2.1algebra

Cells tagged in $E$ lie in $U$ . Their interiors are disjoint, and their total length is the measure of their finite union, since shared endpoints form a finite null set. By [F4] this sum is at most $λ (U) < d$ , so [F1] bounds their total absolute $F$ -increment by $ε / 2$ ; for an empty family the sum is zero.

4.1F2step 1.1step 2.1step 3.1algebra

For every other cell, strict fineness and step 2.1 give $∣ F (x_{i}) - F (x_{i - 1}) ∣ \leq ∣ F (x_{i}) - F (ξ_{i}) ∣ + ∣ F (ξ_{i}) - F (x_{i - 1}) ∣ \leq η (x_{i} - x_{i - 1})$ , because the tag lies in the cell. The sum over these cells is at most $η (t - s) < ε / 2$ . Telescoping all cells and applying step 3.1 yields $∣ F (t) - F (s) ∣ < ε$ . Arbitrary positive $ε$ gives equality, and arbitrary $s, t$ prove claim 1.

5.1givenF1F4F6F7step 4.1algebra∎

For claim 2 with $a < b$ , fix a coordinate $j$ and put $I_{j} (t) = \int_{a}^{t} h_{j} (s) d s$ . By [F6], $I_{j}$ is absolutely continuous and $I_{j}^{'} = h_{j}$ almost everywhere. The function $H_{j} = y_{j} - y_{j} (a) - I_{j}$ is absolutely continuous: for each positive tolerance use the [F1] moduli for $y_{j}, I_{j}$ at half that tolerance and take their minimum. Outside the union of the two exceptional null sets, [F7] gives $H_{j}^{'} = 0$ . Claim 1 makes $H_{j}$ constant, and $H_{j} (a) = 0$ , so $y_{j} (t) = y_{j} (a) + I_{j} (t)$ for all $t$ . Applying this argument to the finitely many coordinates proves the displayed vector identity.

LemmaStatement: AI-adaptedProof: AI-generatedOpen item page →

The backward LQR adjoint equation has a unique solution

Statement

Assume the Axiom of Countable Choice.

For every LQR data set in Finite-horizon bounded-control linear-quadratic ODE problem and every admissible control $u$ , the adjoint in Continuous adjoint equation for the linear-quadratic ODE exists and is unique.

Facts & Assumptions

Given: Countable Choice; the bounded measurable matrices $A, Q$ , state $x_{u}$ , target $x_{d} \in L^{2}$ , and terminal value $P (x_{u} (T) - x_{T})$ .

[F1]

The linear state lemma supplies a unique absolutely continuous solution of its Volterra integral equation for $C, D \in L^{\infty}$ and $v \in L^{2}$ ; under Countable Choice this constructed solution also satisfies $\dot{y} = C (t) y + D (t) v (t)$ almost everywhere (The bounded measurable linear ODE has a unique affine state map, The Axiom of Countable Choice ( ${A C}_{ω}$ )).

[F2]

The adjoint is defined by $- \dot{p} = A^{T} p + Q (x_{u} - x_{d})$ and $p (T) = P (x_{u} (T) - x_{T})$ (Continuous adjoint equation for the linear-quadratic ODE).

[F3]

Under Countable Choice, an absolutely continuous vector curve with known $L^{1}$ derivative equals its initial value plus the componentwise Lebesgue integral of that derivative (AC zero-derivative constancy and integrable-derivative reconstruction under Countable Choice).

[F4]

An absolutely continuous scalar function is continuous, and a continuous scalar function on a compact interval is bounded; these apply to each of the finitely many coordinates ( $C^{1}$ implies Lipschitz, Lipschitz implies absolutely continuous, and absolutely continuous implies continuous and bounded variation, A continuous real function on a compact subset of $R$ is bounded).

[F5]

On the finite horizon, $∥ g ∥_{L^{1}} \leq \sqrt{T} ∥ g ∥_{L^{2}}$ componentwise (Cauchy-Schwarz inequality for $L^{2}$ ).

[F6]

The reflection $τ (s) = T - s$ is a $C^{1}$ diffeomorphism of $(0, T)$ onto itself with $∣ τ^{'} (s) ∣ = 1$ . The $L^{1}$ change-of-variables formula therefore preserves integrals and null sets under $τ$ , so reflection preserves the $L^{2}$ and $L^{\infty}$ classes used below (A C^1 diffeomorphism satisfies the change-of-variables formula for L^1 functions).

Proof

1.1F1F2F4F6givenconstruct

The forcing $g (s) : = Q (T - s) (x_{u} (T - s) - x_{d} (T - s))$ lies in $L^{2} (0, T; R^{n})$ by [F6]: $Q$ is essentially bounded, $x_{u}$ is continuous on a compact interval and hence bounded, and $x_{d} \in L^{2}$ . Also $C (s) : = A (T - s)^{T} \in L^{\infty}$ by [F6]. Apply [F1] with $D (s) = I$ and initial value $q (0) = P (x_{u} (T) - x_{T})$ to obtain an absolutely continuous $q$ satisfying the forward Volterra integral equation and $\dot{q} = C q + g$ almost everywhere.

2.1F6step 1.1F2algebra

Set $p (t) = q (T - t)$ . Each coordinate remains absolutely continuous because reflection preserves interval lengths and disjoint interiors in the defining finite-family condition of Absolute continuity on a compact interval. At points where the coordinate derivatives exist, the difference-quotient definition and the chain rule for $t \mapsto T - t$ give $\dot{p} (t) = - \dot{q} (T - t)$ ; [F6] carries the exceptional null set to a null set. The forward equation for $q$ therefore gives $- \dot{p} (t) = A (t)^{T} p (t) + Q (t) (x_{u} (t) - x_{d} (t))$ almost everywhere, and $p (T) = q (0) = P (x_{u} (T) - x_{T})$ . Thus $p$ satisfies [F2].

3.1F1F3F4F5F6step 1.1step 2.1algebra∎

If $\tilde{p}$ is another adjoint solution, reflection gives a coordinatewise absolutely continuous $\tilde{q} (s) = \tilde{p} (T - s)$ as in step 2.1, with $\tilde{q} (0) = q (0)$ and $\dot{\tilde{q}} = C \tilde{q} + g$ almost everywhere by [F6]. By [F4], $\tilde{q}$ is continuous and bounded componentwise. Since $C \in L^{\infty}$ , $g \in L^{2}$ and [F5] gives $g \in L^{1}$ , the function $h = C \tilde{q} + g$ belongs to $L^{1}$ componentwise. Apply [F3] to obtain $\tilde{q} (s) = q (0) + \int_{0}^{s} (C (r) \tilde{q} (r) + g (r)) d r$ for every $s$ . This is exactly the forward Volterra equation used to construct $q$ in step 1.1. Its actual integral-equation uniqueness in [F1] gives $\tilde{q} = q$ , hence $\tilde{p} = p$ .

TheoremStatement: AI-adaptedProof: AI-generatedOpen item page →

A bounded-control linear-quadratic ODE has a unique minimizer

Statement

Assume the ultrafilter lemma, Dependent Choice, and Hahn–Banach. For the LQR data of Finite-horizon bounded-control linear-quadratic ODE problem, if the feasible-control set $F$ is nonempty, convex, and weakly sequentially closed, then $J$ attains its minimum on $F$ at a unique control $u^{*}$ . The direct-method branch uses exactly these three principles; Dependent Choice also supplies Countable Choice for the $L^{2}$ Hilbert-space/reflexivity suppliers. No use of the full Axiom of Choice is asserted.

Facts & Assumptions

Given: The LQR data, a nonempty convex weakly sequentially closed $F \subseteq U_{c}$ , the ultrafilter lemma, Dependent Choice, and Hahn–Banach.

[A1]

The ultrafilter lemma, Dependent Choice and Hahn–Banach are the exact axiom assumptions of the direct-method supplier; the uses below are weak-subsequence extraction and the stated reflexivity route (The ultrafilter extension principle (UL/BPI), The axiom of dependent choice: a relation in which every element is related to something admits an $N$ -indexed chain, The real dominated-extension principle as an additional hypothesis over ZF).

[F1]

Under Countable Choice, real $L^{2} (0, T; R^{m})$ with its integral pairing is a Hilbert space ( $L^{2}$ with the integral pairing is a Hilbert space).

[F2]

A real Hilbert space is reflexive under Countable Choice (Hilbert spaces are reflexive).

[F3]

Dependent Choice implies Countable Choice (Dependent choice implies countable choice).

[F4]

Under the ultrafilter lemma, Dependent Choice and Hahn–Banach, a proper coercive weakly sequentially lower-semicontinuous functional on a nonempty weakly sequentially closed subset of a real reflexive Banach space attains its infimum (The direct method in a reflexive Banach space).

[F5]

Under Countable Choice (supplied by Dependent Choice through [F3]), the pointwise control set is weakly sequentially closed; additional path/terminal constraints are included only through the assumed weakly sequentially closed set $F$ (Under Countable Choice, pointwise compact convex control constraints are weakly closed in L2, Finite-horizon bounded-control linear-quadratic ODE problem).

[F6]

The quadratic objective $J$ is weakly lower semicontinuous (Under Countable Choice, the LQR objective is weakly lower semicontinuous).

[F7]

The state map is unique, affine, and bounded from $L^{2}$ to continuous states (The bounded measurable linear ODE has a unique affine state map).

Proof

1.1A1F1F2F3

By [A1] and [F3], Countable Choice is available. Hence [F1] makes $L^{2} (0, T; R^{m})$ a real Hilbert space and [F2] makes it reflexive.

1.2F5F7givenalgebra

The set $F$ is nonempty by hypothesis and weakly sequentially closed by [F5]. For $u \in F$ , every state and terminal quadratic term in $J$ is nonnegative, while $R (t) ⪰ r I$ gives $J (u) \geq \frac{r}{2} ∥ u ∥_{2}^{2}$ . The cost is finite on $F$ by [F7], so $J$ is proper and coercive there.

2.1A1F1F2F4F6step 1.1step 1.2

By [F6], $J$ is weakly sequentially lower semicontinuous on $F$ . The axioms in [A1] and the reflexivity and admissibility checks in steps 1.1–1.2 verify every hypothesis of [F4]; therefore there exists $u^{*} \in F$ with $J (u^{*}) = \inf_{u \in F} J (u)$ .

3.1F1F5F7step 1.2step 2.1algebra∎

The state map is affine by [F7]. For $0 < θ < 1$ , each positive-semidefinite quadratic state or terminal form satisfies $q (θ a + (1 - θ) b) = θ q (a) + (1 - θ) q (b) - θ (1 - θ) q (a - b) \leq θ q (a) + (1 - θ) q (b)$ . The control form satisfies $\frac{1}{2} ⟨ R (θ u + (1 - θ) v), θ u + (1 - θ) v ⟩ = \frac{θ}{2} ⟨ R u, u ⟩ + \frac{1 - θ}{2} ⟨ R v, v ⟩ - \frac{θ (1 - θ)}{2} ⟨ R (u - v), u - v ⟩$ . If $u \neq v$ , then $⟨ R (u - v), u - v ⟩ \geq r ∥ u - v ∥_{2}^{2} > 0$ , so $J$ is strictly convex on the convex set $F$ . Two distinct minimizers would have a feasible strict convex combination with smaller cost, contradicting the minimum value established in step 2.1. Thus $u^{*}$ is unique.

Remarks

The source notes' direct-method outline is not used as a substitute for lower semicontinuity: Under Countable Choice, the LQR objective is weakly lower semicontinuous proves that requirement for this objective. The proof does not claim an axiom-minimal version without the three principles in [A1].

AlgorithmAlgorithm: AI-adaptedProof: Not applicableOpen item page →

Finite-budget indirect shooting for an interior-control ODE optimality system

Definition

This procedure is restricted to a smooth, unconstrained-interior control branch. Let the smooth data be those of Finite-horizon ODE optimal-control transcription data, with no active path or endpoint constraints apart from the prescribed initial state. Any optional inequality constraints must remain strictly inactive throughout the branch; additional endpoint equalities or active endpoint inequalities require an augmented boundary residual and are outside this procedure. Let $H (t, x, p, u) = L (t, x, u) + p^{T} f (t, x, u)$ . Supply a pointwise-defined twice continuously differentiable feedback $ψ (t, x, p)$ on an open tube such that $H_{u} (t, x, p, ψ (t, x, p)) = 0$ and $ψ (t, x, p) \in int U_{c}$ throughout that tube; differentiability uses the convention of $C^{k}$ Euclidean maps and diffeomorphisms. Every successful IVP evaluation must keep its trajectory in this tube. If the evaluator cannot certify that condition for a queried trajectory, return inadmissible-interior-branch with no shooting result. Define $H_{x}^{f i x} (t, x, p, u) : = D_{x} [L (t, x, u) + p^{T} f (t, x, u)]$ , with the derivative taken holding $u$ fixed, and only then substitute $u = ψ (t, x, p)$ . Define $z = (x, p) \in R^{2 n}$ and $G (t, z) = (\begin{array}{c} f (t, x, ψ (t, x, p)) \\ - H_{x}^{f i x} (t, x, p, ψ (t, x, p)) \end{array}), B (z_{i n i t}, z_{t e r m}) = (\begin{array}{c} x_{i n i t} - x_{0} \\ p_{t e r m} - ϕ_{x} (x_{t e r m}) \end{array}),$ where $z_{i n i t} = (x_{i n i t}, p_{i n i t})$ and $z_{t e r m} = (x_{t e r m}, p_{t e r m})$ . The indirect shooting procedure takes a certified differentiable IVP evaluator for $z^{'} = G (t, z)$ , an initial trial $z_{0}$ , residual tolerance $τ \geq 0$ , and correction cap $M \in N_{0}$ , and calls Finite-budget single shooting for a nonlinear two-point BVP on the square BVP $(G, B)$ . It returns that algorithm's trajectory and status without changing the root-tolerance meaning. In particular, root-tolerance-met certifies only the computed boundary residual for the numerical IVP map; it is not an exact BVP, optimality, global-optimality, or discretization-error certificate. If the control is on a boundary or no such smooth feedback is supplied, this procedure is inapplicable.

LemmaStatement: AI-adaptedProof: AI-generatedOpen item page →

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.

TheoremStatement: AI-adaptedProof: AI-generatedOpen item page →

LQR minimizers satisfy a feasible-set variational inequality

Statement

Assume the Axioms of Countable Choice and Dependent Choice.

For the LQR data of Finite-horizon bounded-control linear-quadratic ODE problem, let $F$ be nonempty and convex and let $u \in F$ . Put $g (t) = B (t)^{T} p (t) + R (t) u (t)$ , with the adjoint from Continuous adjoint equation for the linear-quadratic ODE. If $u$ minimizes $J$ on $F$ , then $\int_{0}^{T} g (t)^{T} (v (t) - u (t)) d t \geq 0 (v \in F) .$ Conversely, if this variational inequality holds at $u$ , then $u$ is the unique minimizer on $F$ .

If $F = U_{c}$ , the variational inequality is equivalent (under Dependent Choice, hence Countable Choice) to the existence of one null set $N$ such that $g (t)^{T} (c - u (t)) \geq 0$ for every $t \notin N$ and every $c \in U_{c}$ . The pointwise condition is not asserted for additional state or terminal constraints.

Facts & Assumptions

Given: The LQR data, a nonempty convex feasible set $F$ , a control $u \in F$ , its continuous adjoint, and the variational-inequality hypothesis when proving sufficiency or the reverse pointwise implication.

[F1]

The objective has bounded first-order map $D J (u) [w] = \int_{0}^{T} (B^{T} p + R u)^{T} w d t$ , with remainder $O (∥ w ∥_{2}^{2})$ (The LQR adjoint identity gives the reduced gradient).

[F2]

The state map is affine, and for every $u, v \in L^{2}$ the quadratic expansion has the exact remainder $J (v) - J (u) - D J (u) [v - u] = \frac{1}{2} \int_{0}^{T} (δ x^{T} Q δ x + (v - u)^{T} R (v - u)) d t + \frac{1}{2} δ x (T)^{T} P δ x (T)$ , where $δ x = x_{v} - x_{u}$ (Finite-horizon bounded-control linear-quadratic ODE problem, The bounded measurable linear ODE has a unique affine state map).

[F3]

Cauchy–Schwarz makes the integral pairing finite for $L^{2}$ factors (Cauchy-Schwarz inequality for $L^{2}$ ).

[F4]

[F5]

Under Countable Choice, Lebesgue measurable sets form a sigma-algebra and countable unions of measurable null sets are null by countable subadditivity (Assuming countable choice, $L (R^{n})$ is a sigma-algebra containing every elementary set and $λ_{n}$ is a complete measure extending elementary volume, Finite and countable subadditivity of measures).

[F6]

[A1]

Dependent Choice and Countable Choice are available for the countable dense subset of $U_{c}$ used in the pointwise equivalence (The axiom of dependent choice: a relation in which every element is related to something admits an $N$ -indexed chain, The Axiom of Countable Choice ( ${A C}_{ω}$ ), Dependent choice implies countable choice).

Proof

1.1F1givenchoosealgebra

Suppose $u$ minimizes $J$ on $F$ and fix $v \in F$ . Convexity gives $u_{θ} = u + θ (v - u) \in F$ for $0 \leq θ \leq 1$ . Hence $(J (u_{θ}) - J (u)) / θ \geq 0$ for $0 < θ \leq 1$ . The first-order expansion [F1] with increment $θ (v - u)$ has quotient $D J (u) [v - u] + O (θ ∥ v - u ∥_{2}^{2})$ , so its limit is $D J (u) [v - u]$ . This proves the displayed variational inequality.

1.2F3givenalgebra

Suppose the pointwise inequalities hold outside one null set and $F = U_{c}$ . For any $v \in U_{c}$ , substitute $c = v (t)$ outside that null set and the null set on which $v (t) \notin U_{c}$ . The vector $g = B^{T} p + R u$ lies in $L^{2}$ because $B, R$ are essentially bounded, $p$ is continuous on the compact horizon, and $u \in L^{2}$ ; [F3] makes $g^{T} (v - u)$ integrable. Integrating proves the global variational inequality.

1.3F2givenalgebra

Assume the variational inequality at $u$ . For any $v \in F$ , [F2] and $Q ⪰ 0$ , $R ⪰ r I$ give $J (v) - J (u) \geq D J (u) [v - u] + \frac{r}{2} ∥ v - u ∥_{2}^{2}$ . The linear term is nonnegative by the assumed inequality; hence $J (v) \geq J (u)$ , with strict inequality if $v \neq u$ . Thus $u$ is the unique minimizer.

2.1A1F3F4F5F6givenchoosealgebra∎

Assume the global variational inequality and suppose no common null set makes all pointwise inequalities true. By [F4] and [A1], choose one point from each rational open box meeting $U_{c}$ ; the resulting countable set $D \subseteq U_{c}$ is dense. Define $M = ⋃_{d \in D, k \geq 1} E_{d, k}$ , where $E_{d, k} = {t : g (t)^{T} (d - u (t)) \leq - 1 / k}$ . These sets are measurable, and $M$ is measurable by [F5]. Continuity in $c$ and density show that $M$ is exactly the set where some $c \in U_{c}$ violates the pointwise inequality, so the absence of a common null set means $∣ M ∣ > 0$ . By [F5], some $E_{d, k}$ has positive measure. Put $v (t) = d$ on that set and $v (t) = u (t)$ elsewhere. This measurable control lies in $U_{c}$ . On $E_{d, k}$ , $- g (t)^{T} (d - u (t)) \geq 1 / k$ ; [F6] and monotonicity give $\int_{0}^{T} g^{T} (v - u) d t \leq - ∣ E_{d, k} ∣ / k < 0$ , contradicting the global variational inequality. Therefore one null set suffices for every $c \in U_{c}$ .

Remarks

For arbitrary active state or terminal constraints the valid first-order condition is the integral inequality over the feasible set. The multiplier $p$ in the unconstrained adjoint equation alone does not encode normal multipliers for those additional constraints.

CounterexampleConstruction: AI-generatedVerification: AI-generatedOpen item page →

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.

AlgorithmAlgorithm: AI-adaptedProof: Not applicableOpen item page →

Finite-horizon shooting and Runge–Kutta collocation transcription

Definition

Inputs. Supply pointwise smooth data on $[0, T]$ from Finite-horizon ODE optimal-control transcription data, an original mesh, a piecewise-constant control parameterization, an explicit tableau from Explicit Runge–Kutta tableaus, stage maps, and failure status or a possibly implicit tableau from Runge–Kutta tableaux and stage collocation equations, one of single-shooting, multiple-shooting, or collocation mode, a finite smooth representation of the control constraints when a solver-ready NLP is requested, finite sets of declared path-constraint times, endpoint constraints, the tableau quadrature rule, an initial decision vector $z_{0}$ and multiplier vector $μ_{0}$ , a stage residual tolerance $τ_{s} \geq 0$ , a quadrature error tolerance $τ_{q} \geq 0$ , a stage Newton correction cap $M_{s} \in N_{0}$ , and a maximum number of mesh nodes. For the direct linear-stage branch, optionally supply exact pointwise coefficient maps $A : [0, T] \to R^{n \times n}$ and $B : [0, T] \to R^{n \times m}$ satisfying $f (t, x, u) = A (t) x + B (t) u$ at every queried stage input; only this explicit representation enables that branch. Supply a deterministic derivative evaluator for the actual returned stage maps and resulting finite NLP on an open neighborhood of the queried candidate. On a locally fixed strict stopping branch of the finite stage-iteration procedure, differentiate that finite graph, including when the residual happens to be exactly zero at the queried point. Implicit differentiation as in Implicit differentiation and adjoint formulas under a nonsingular residual Jacobian is permitted only when the selected returned stage map is certified to agree on that neighborhood with the exact local solution branch and its residual Jacobian is nonsingular there; an exact residual at the queried point alone does not establish this agreement. Return derivative-failure when no derivative of the actual returned map is available. Supply a quadrature error evaluator which returns a certified upper bound for the objective quadrature discrepancy or unavailable. Each supplied evaluator returns a value or failure in finite time. An optional trajectory-estimation request supplies the inputs of Finite-budget adaptive Runge–Kutta controller. $X_{j, i} = X_{j, 0} + h_{j} \sum_{k < i} a_{i k} F_{j, k}, t_{j, i} = t_{j} + c_{i} h_{j}, F_{j, i} = f (t_{j, i}, X_{j, i}, u_{j}), Φ_{j} (X_{j, 0}, u_{j}) = X_{j, 0} + h_{j} \sum_{i} b_{i} F_{j, i} .$

Mesh and constraints. Check that the original mesh satisfies $0 = t_{0} < \dots < t_{N} = T$ and that all declared constraint and quadrature times lie in $[0, T]$ ; if either check fails, return mesh-domain-failure without evaluation. If the original node count exceeds the supplied cap, return mesh-budget-exhausted. Insert every declared path-constraint time not already a node, unless this exceeds the node cap, in which case return mesh-budget-exhausted. Tie each inserted subcell's control to its original containing cell, so the parameterization is unchanged. On the resulting mesh, check all stage times $t_{j} + c_{i} h_{j}$ lie in $[0, T]$ before querying $f$ or $L$ ; an invalid time returns mesh-domain-failure with no evaluation. Each tableau stage and its quadrature contribution use the control $u_{j}$ of that cell, as in the stage equations below. At a declared path- or endpoint-constraint time that is a mesh node before $T$ , use the right-hand original-cell value from the half-open representative in Time meshes and piecewise-constant control parameterization; at $T$ , use the last original-cell control value unless a separate endpoint-control variable is supplied. Use the same convention at separately declared non-stage quadrature times. An endpoint constraint is evaluated at its declared endpoint. This is finite sampling and does not assert continuous-path feasibility.

Stage evaluation. On each cell write $h_{j} = t_{j + 1} - t_{j}$ and $u_{j}$ for its tied constant control. Before forming any stages, check $X_{j, 0} \in Ω_{x}$ and $u_{j} \in Ω_{u}$ ; a failed check returns stage-domain-failure with the cell and no objective candidate. For an explicit tableau, form the stages recursively by the displayed equations. Before each query of $f$ or $L$ , check that the time is in $[0, T]$ , the state lies in $Ω_{x}$ , and the control lies in $Ω_{u}$ ; a failed membership check returns stage-domain-failure, and an unavailable function value returns stage-evaluation-failure with the cell and stage index. Apply the same checks to every declared path-constraint evaluation.

For an implicit tableau, do not leave the collocation stages as outer NLP decision variables. Form the coupled slope residual $R_{j} (K; X_{j, 0}, u_{j})_{i} = K_{i} - f ⁣ (t_{j} + c_{i} h_{j}, X_{j, 0} + h_{j} \sum_{k} a_{i k} K_{k}, u_{j}), i = 1, \dots, s .$ With the stage times fixed and already checked in $[0, T]$ , its open evaluation domain consists of stage vectors for which every induced stage state lies in $Ω_{x}$ and the cell control lies in $Ω_{u}$ ; require $R_{j}$ to be $C^{1}$ on this domain. A compact tube used for bounds does not replace this open evaluation domain. If exact coefficient maps $A, B$ were supplied and satisfy $f (t, x, u) = A (t) x + B (t) u$ at every queried stage input, form the finite matrix $M_{j}$ with blocks $(M_{j})_{i k} = δ_{i k} I - h_{j} a_{i k} A (t_{j, i})$ and right-hand side $r_{i} = A (t_{j, i}) X_{j, 0} + B (t_{j, i}) u_{j}$ . Use exact Gaussian elimination on the stored sample coefficients: scan columns left to right, pivot on the first nonzero entry at or below the current row, swap, eliminate below, and back-substitute. If no pivot is available, return singular-stage-matrix; otherwise form every induced stage state and check every stage time-state-control input belongs to the smooth-data tube. A failed domain check returns stage-domain-failure with the failing cell and stage index, the partial result record, and no objective candidate. If all checks pass, the exact solution of $M_{j} K = r$ is the stage vector and the nonlinear-solve status is exact-linear-stage-solve. In all other smooth cases, form the deterministic initial guess $K_{i}^{(0)} = f (t_{j}, X_{j, 0}, u_{j})$ after checking that this query is defined. Before calling Newton, check that the entire seed $K^{(0)}$ belongs to the declared open domain of $R_{j}$ , equivalently that every induced stage state lies in $Ω_{x}$ and every required control lies in $Ω_{u}$ . If this check fails, return stage-domain-failure with the failing cell and stage index, the partial result record, and no objective candidate. Otherwise call Finite-budget Newton for a square vector residual on $R_{j}$ with that open evaluation domain, tolerance $τ_{s}$ and cap $M_{s}$ . Record its iterate, residual vector, and exact status for each cell. Only root supplies a stage candidate; it means $∥ R_{j} ∥_{2} \leq τ_{s}$ , not an exact root unless $τ_{s} = 0$ . Any other status returns the partial result record with nlp_status=stage-solve-failure, the failing cell, and no objective candidate. On a returned stage vector use $X_{j, i} = X_{j, 0} + h_{j} \sum_{k} a_{i k} K_{k}$ and $Φ_{j} = X_{j, 0} + h_{j} \sum_{i} b_{i} K_{i}$ . A singular stage Jacobian or failed stage query is reported, not interpreted as lack of an exact solution.

Transcription modes. In single shooting, the decision variables are the original-cell controls; set $X_{0, 0} = x_{0}$ and recursively compute $X_{j + 1, 0} = Φ_{j} (X_{j, 0}, u_{j})$ . In multiple shooting, the decision variables are node states and controls; fix $X_{0, 0} = x_{0}$ and impose the continuity equalities $d_{j} = X_{j + 1, 0} - Φ_{j} (X_{j, 0}, u_{j}) = 0$ for $j = 0, \dots, N - 1$ . In collocation mode, use the same node/control variables, impose the same continuity equalities $d_{j} = 0$ , and use solved internal stage states; the collocation equations are solved internally by the implicit-stage procedure above. Explicit stages are likewise internal deterministic evaluations. After every interval map in every mode, check $Φ_{j} \in Ω_{x}$ before using it as a later start or evaluating a path constraint, terminal cost, or endpoint constraint; otherwise return stage-domain-failure with no objective candidate. Independently of the computed endpoints and defect residuals, check $X_{N, 0} \in Ω_{x}$ before evaluating $ϕ (X_{N, 0})$ or any terminal constraint; in multiple shooting and collocation this node is an independent decision variable even when $Φ_{N - 1} \in Ω_{x}$ . Before any endpoint-constraint query, also check every actual state and control argument, including a separately supplied endpoint-control variable, belongs to that function's declared domain. Any failed membership check returns stage-domain-failure with no objective candidate. In every mode the finite feasible set includes $u_{j} \in U_{c}$ for each original cell. The finite objective is $\sum_{j} h_{j} \sum_{i} b_{i} L (t_{j, i}, X_{j, i}, u_{j}) + ϕ (X_{N, 0})$ , and the finite constraints are exactly the fixed initial-state condition, the continuity equalities $d_{j} = 0$ in multiple shooting and collocation, the declared node/time constraints, the represented control constraints, and endpoint constraints. Include every continuity defect in the equality residual $h$ and its corresponding multiplier in $ν$ ; in single shooting the initial state and continuity are enforced by recursive elimination. If $U_{c}$ is available only as a compact set-membership constraint and has no finite smooth representation, return the finite set-constrained model with nlp_status=not-smooth-nlp; the dispatcher must not treat it as an SQP-ready NLP. A QP solver route additionally requires the sampled control constraints to be affine.

Output. Evaluate the supplied initial candidate and return a record containing its ordered decision variables, defect residuals, per-cell implicit-stage residuals, path/endpoint constraint residuals, KKT residuals, and objective value, when available. For equality residual $h$ , inequality residual $g \leq 0$ , and multiplier pair $(ν, λ)$ , report $r_{p} = (h, [g]_{+})$ , $r_{d} = \nabla J_{N} + D h^{T} ν + D g^{T} λ$ , $r_{λ} = [- λ]_{+}$ , and $r_{c} = λ ⊙ g$ . Include a model that can reevaluate these quantities at a later solver candidate. Return separate nlp_status, per-cell implicit_stage_status, optional ivp_status, quadrature_status, mesh_status, and budget_status fields. A quadrature bound at most $τ_{q}$ is bound-met; a larger bound is bound-not-met; an unavailable bound is unavailable. The build-stage budget field is not-run; the solver dispatcher fills its final value. A failed derivative evaluation leaves KKT residuals unavailable with its exact status. An unevaluated optional IVP estimate is not-requested; if requested, report the adaptive controller's exact finite status, not an error certificate.

This direct transcription samples pointwise data; an $L^{\infty}$ equivalence class without a chosen representative and sampling rule is not stage-evaluable. For a single-shooting reduced objective represented as the finite state-step graph of Finite parameterized state-step graphs and objectives, its exact real-arithmetic reduced gradient is given by The reverse recursion is the exact gradient of a finite state-step objective. Agreement with the continuous ODE adjoint gradient is only conditional on the weighted discrete-gradient representation and the separate state, adjoint, derivative-data, endpoint, and quadrature estimates in Discrete and continuous ODE gradients agree to order p under augmented error bounds. The optional adaptive RK estimate is separate from the fixed NLP and its embedded difference is not a certified global error bound. If all construction, stage, sample-domain and objective checks succeed, return nlp_status=constructed for a solver-ready NLP, or nlp_status=not-smooth-nlp for a finite set-constrained model without a smooth control representation; set mesh_status=valid and retain budget_status=not-run. The algorithm makes no continuous-path feasibility, convergence-rate, global-optimality, or unconditioned discrete-adjoint accuracy claim.

LemmaStatement: AI-adaptedProof: AI-generatedOpen item page →

Exact single and multiple shooting have the same reduced feasible controls

Statement

Fix a mesh, controls $u_{0}, \dots, u_{N - 1}$ , and a successful deterministic Runge–Kutta interval map $Φ_{j}$ from Finite-horizon shooting and Runge–Kutta collocation transcription. In the single-shooting NLP, define states recursively from $X_{0, 0} = x_{0}$ and $X_{j + 1, 0} = Φ_{j} (X_{j, 0}, u_{j})$ . In the multiple-shooting NLP, impose the same initial state and continuity equations. Then projection from the multiple-shooting feasible set onto its control coordinates is a bijection onto the single-shooting feasible controls. Any objective and node constraints that use the same recursively determined states have identical reduced values and feasibility. This is an algebraic equivalence of the discrete NLPs; it is not an equivalence to the continuous ODE problem.

Facts & Assumptions

Given: The fixed mesh, the deterministic interval maps, and controls for which every interval map succeeds.

[F1]

An interval map is available only after its explicit stages or implicit stage residual solve return a successful status; it assigns exactly one deterministic endpoint to that successful start/control pair, and single shooting applies it recursively (Finite-horizon shooting and Runge–Kutta collocation transcription).

[F2]

The multiple-shooting variables satisfy $X_{0, 0} = x_{0}$ and $X_{j + 1, 0} = Φ_{j} (X_{j, 0}, u_{j})$ (Finite-horizon shooting and Runge–Kutta collocation transcription).

Proof

1.1F1givenconstruct

Given any control tuple feasible in single shooting, calculate $X_{1, 0} = Φ_{0} (x_{0}, u_{0})$ , then each later $X_{j + 1, 0} = Φ_{j} (X_{j, 0}, u_{j})$ . Uniqueness of each deterministic map makes this recursively defined state tuple well-defined; by construction it satisfies every multiple-shooting continuity equation.

2.1F1F2step 1.1induction

Conversely, a multiple-shooting feasible tuple has fixed $X_{0, 0} = x_{0}$ and satisfies each continuity equation. Induction on $j$ forces its $X_{j, 0}$ to equal the recursive single-shooting value for the same controls. Hence there is exactly one multiple-shooting state tuple for each feasible control tuple and the projection is bijective.

3.1step 1.1step 2.1givenalgebra∎

Under this bijection the states at all shared nodes agree. Therefore every node constraint has the same value and every objective formed from the same node/stage values is identical. Thus reduced feasibility and objective values coincide.

LemmaStatement: AI-adaptedProof: AI-generatedOpen item page →

A linear-quadratic transcription with affine constraints is a convex QP

Statement

Consider a finite mesh and an LQR model evaluated by a fixed Runge–Kutta tableau, explicit or implicit, whose stage quadrature weights are nonnegative and whose sampled running and terminal quadratic weights are positive semidefinite. If $A, B, Q, R, x_{d}$ are supplied as almost-everywhere equivalence classes, choose pointwise representatives and state the sampling rule at every stage and quadrature time; the finite QP claim concerns only that resulting sampled system, since an equivalence class alone has no stage value. For an implicit tableau, require that each cell's sampled linear stage-residual matrix is nonsingular and use its unique solution. Suppose every control, path, terminal, initial, and shooting/collocation constraint is specified by finitely many affine equalities or inequalities in the transcription variables. Then the finite transcription of Finite-horizon shooting and Runge–Kutta collocation transcription is a convex quadratic program (QP): its objective is a convex quadratic polynomial and its equality and inequality constraints are affine. This is the finite NLP/QP branch used with the convex models in Conic, LP, and convex QP primal-dual models.

This algebraic claim does not make an $L^{\infty}$ equivalence class a valid input to the pointwise smooth-data procedure. If that procedure is called, its separate pointwise-domain and smoothness hypotheses must also hold; otherwise this lemma concerns only the specified finite sampled equations.

Facts & Assumptions

Given: A finite mesh, chosen pointwise representatives and a sampling rule for any equivalence-class data, linear dynamics, a fixed explicit or implicit Runge–Kutta tableau, nonnegative stage weights, positive-semidefinite sampled quadratic cost weights, nonsingular sampled stage matrices when implicit stages are eliminated, and affine constraints.

[F1]

An explicit RK stage is a finite affine recursion in prior stages. For an implicit linear cell, the slope equations have matrix $M_{i k} = δ_{i k} I - h a_{i k} A (t_{i})$ and right-hand side $A (t_{i}) X_{0} + B (t_{i}) u$ ; if $M$ is nonsingular, their unique solution and the endpoint are affine in $(X_{0}, u)$ (Explicit Runge–Kutta tableaus, stage maps, and failure status, Runge–Kutta tableaux and stage collocation equations).

[F2]

The transcription uses the explicit recursions or solved implicit interval maps for the node/defect equations and uses the displayed quadrature objective at the chosen sample values (Finite-horizon shooting and Runge–Kutta collocation transcription).

[F3]

A convex QP has a positive-semidefinite quadratic objective Hessian and affine equalities/inequalities (Conic, LP, and convex QP primal-dual models, Inequality constrained programs, tangent cones, constraint qualifications, and KKT).

Proof

1.1F1F2giveninductionalgebra

On each interval with linear dynamics, an explicit tableau has affine stages by recursion: the first stage is affine in the start and control, and if previous stages are affine then $X_{i} = X_{0} + h \sum_{k < i} a_{i k} F_{k}$ and $F_{i} = A (t_{i}) X_{i} + B (t_{i}) u$ are affine. For an implicit tableau, the sampled slope equations form the affine system $M K = A X_{0} + B u$ described in [F1]. Nonsingularity makes every stage and endpoint an affine function of $(X_{0}, u)$ ; the algorithm's exact linear stage-solve branch reports failure if this matrix is singular. Thus the endpoint/defect equations and any retained stage equations are affine.

2.1F2F3givenstep 1.1algebra∎

Each quadrature term is a nonnegative scalar weight times a positive-semidefinite quadratic form applied to an affine state/control map, hence has positive-semidefinite Hessian; the terminal form has the same property. A finite sum of such Hessians is positive semidefinite, so the objective is convex quadratic. The remaining constraints are affine by hypothesis; thus [F3] identifies the transcription as a convex QP.

TheoremStatement: AI-adaptedProof: AI-generatedOpen item page →

Mesh-consistent LQR minimizers converge to the unique continuous minimizer

Statement

Assume the axiom branch of A bounded-control linear-quadratic ODE has a unique minimizer, and let $u^{*}$ be the unique minimizer of the LQR objective $J$ on a nonempty convex weakly sequentially closed feasible set $F \subseteq U_{c}$ . For each mesh index $N$ , let $F_{N} \subseteq F$ be nonempty, and let $J_{N}$ be the discrete transcription objective. Assume there are $ε_{N} \to 0$ such that $∣ J_{N} (v) - J (v) ∣ \leq ε_{N}$ for every $v \in F_{N}$ . Assume also a recovery sequence $v_{N} \in F_{N}$ with $∥ v_{N} - u^{*} ∥_{L^{2}} \to 0$ . If $u_{N}$ is a global minimizer of $J_{N}$ on $F_{N}$ , then $u_{N} ⇀ u^{*}$ in $L^{2}$ and $x_{u_{N}} \to x_{u^{*}}$ uniformly.

For the control-only case $F = U_{c}$ and piecewise-constant mesh controls, the cell-average lemma supplies the strong recovery sequence. If active path or terminal constraints are present, preservation or recovery of those constraints is an additional hypothesis; node sampling alone does not establish continuous-path feasibility. This theorem is conditional on a uniform objective consistency bound; an embedded adaptive RK difference alone does not provide that bound.

Facts & Assumptions

Given: The LQR data, its unique minimizer, feasible discrete sets, uniform consistency, recovery controls, and global discrete minimizers.

[F1]

The continuous LQR minimizer exists uniquely under the stated closedness and choice-axiom hypotheses (A bounded-control linear-quadratic ODE has a unique minimizer).

[F2]

The control set $U_{c}$ is bounded in $L^{2}$ because $U_{c}$ is compact, and is weakly sequentially closed under Countable Choice (supplied by the theorem's axiom branch) (Finite-horizon bounded-control linear-quadratic ODE problem, Under Countable Choice, pointwise compact convex control constraints are weakly closed in L2).

[F3]

A bounded sequence in a real reflexive Banach space has a weakly convergent subsequence under the ultrafilter lemma, Dependent Choice, and Hahn–Banach (A bounded sequence in a reflexive Banach space has a weakly convergent subsequence).

[F4]

The LQR objective is weakly sequentially lower semicontinuous on $U_{c}$ (Under Countable Choice, the LQR objective is weakly lower semicontinuous).

[F5]

Weakly convergent controls in $U_{c}$ have uniformly convergent states (Weakly convergent bounded controls have uniformly convergent states).

[F6]

The state map is affine and bounded from $L^{2}$ to the supremum norm, so the LQR objective is continuous under strong $L^{2}$ convergence (The bounded measurable linear ODE has a unique affine state map, Finite-horizon bounded-control linear-quadratic ODE problem).

[A1]

The ultrafilter lemma, Dependent Choice, and Hahn–Banach are assumed for [F3] and real Hilbert-space reflexivity (The ultrafilter extension principle (UL/BPI), The axiom of dependent choice: a relation in which every element is related to something admits an $N$ -indexed chain, The real dominated-extension principle as an additional hypothesis over ZF, $L^{2}$ with the integral pairing is a Hilbert space, Hilbert spaces are reflexive).

Proof

1.1F2F3A1givenalgebra

Since $u_{N} \in F_{N} \subseteq U_{c}$ and $U_{c}$ is compact, $\sup_{N} ∥ u_{N} ∥_{2} \leq \sqrt{T} \sup_{c \in U_{c}} ∥ c ∥_{2} < \infty$ . Real $L^{2}$ is reflexive by [A1], so [F3] gives a weakly convergent subsequence from every subsequence of $(u_{N})$ .

2.1F1F4F6givenstep 1.1algebra

Let a subsequence $u_{N_{k}}$ converge weakly to $\overset{ˉ}{u}$ . Because each $u_{N_{k}} \in F$ and $F$ is weakly sequentially closed, $\overset{ˉ}{u} \in F$ . By [F4] and consistency, $J (\overset{ˉ}{u}) \leq {lim inf}_{k} J (u_{N_{k}}) \leq {lim sup}_{k} (J_{N_{k}} (u_{N_{k}}) + ε_{N_{k}})$ . Global discrete minimality gives $J_{N_{k}} (u_{N_{k}}) \leq J_{N_{k}} (v_{N_{k}}) \leq J (v_{N_{k}}) + ε_{N_{k}}$ . To see $J (v_{N_{k}}) \to J (u^{*})$ , [F6] gives $∥ x_{v_{N_{k}}} - x_{u^{*}} ∥_{\infty} \leq C ∥ v_{N_{k}} - u^{*} ∥_{2} \to 0$ ; the control quadratic difference is bounded by $\frac{M}{2} (∥ v_{N_{k}} ∥_{2} + ∥ u^{*} ∥_{2}) ∥ v_{N_{k}} - u^{*} ∥_{2}$ , and state and terminal quadratic differences tend to zero by the corresponding bounded-quadratic estimate and uniform state convergence. Hence $J (v_{N_{k}}) \to J (u^{*})$ , so $J (\overset{ˉ}{u}) \leq J (u^{*})$ .

3.1F1step 1.1step 2.1algebra

Since $\overset{ˉ}{u} \in F$ and $u^{*}$ is the unique minimizer there, step 2.1 forces $\overset{ˉ}{u} = u^{*}$ . Every subsequence therefore has a weakly convergent subsubsequence whose limit is $u^{*}$ . If the full sequence failed to converge weakly to $u^{*}$ , some bounded linear functional and some positive tolerance would witness failure along a subsequence; a weakly convergent subsubsequence from step 1.1 would contradict that witness. Thus $u_{N} ⇀ u^{*}$ .

4.1F5step 3.1∎

Apply [F5] to the weakly convergent admissible controls from step 3.1 to obtain $x_{u_{N}} \to x_{u^{*}}$ uniformly on $[0, T]$ .

TheoremStatement: AI-adaptedProof: AI-generatedOpen item page →

Strong convexity converts transcription objective error into a control bound

Statement

Let $u^{*}$ minimize the LQR objective $J$ on a convex feasible set $F$ , with control weight $R (t) ⪰ r I$ for $r > 0$ . Let $u_{N} \in F$ and let $v_{N} \in F$ be a recovered comparison control. Suppose the computed objective $J_{N}$ satisfies $∣ J_{N} (w) - J (w) ∣ \leq δ_{N}$ for $w = u_{N}, v_{N}$ , and $J_{N} (u_{N}) \leq J_{N} (v_{N}) + η_{N}$ . If $J (v_{N}) \leq J (u^{*}) + ρ_{N}$ , then $J (v) - J (u^{*}) \geq \frac{r}{2} ∥ v - u^{*} ∥_{2}^{2}$ for every $v \in F$ , and $0 \leq J (u_{N}) - J (u^{*}) \leq 2 δ_{N} + η_{N} + ρ_{N}, ∥ u_{N} - u^{*} ∥_{L^{2}} \leq \sqrt{\frac{2 (2 δ_{N} + η_{N} + ρ_{N})}{r}} .$ The conclusion is conditional on the stated objective consistency and comparison bounds. A raw embedded RK error estimate is not a certified value of $δ_{N}$ .

Here are sufficient bounds that derive the two error terms from transcription errors and recovery. Set $C_{s} = e^{∥ A ∥_{\infty} T} ∥ B ∥_{\infty} \sqrt{T}$ , $M_{c} = \sup_{c \in U_{c}} ∥ c ∥_{2}$ , $M_{x} = e^{∥ A ∥_{\infty} T} (∥ x_{0} ∥_{2} + ∥ B ∥_{\infty} T M_{c})$ , $E = \sqrt{T} M_{x} + ∥ x_{d} ∥_{2}$ , and $M_{T} = M_{x} + ∥ x_{T} ∥_{2}$ . Suppose, for $w = u_{N}, v_{N}$ , a pointwise representative and sampling rule are fixed, $J_{N} (w)$ has the form $J_{N} (w) = \sum_{k} ω_{N k} {\tilde{ℓ}}_{N, w} (τ_{N k}) + \frac{1}{2} ({\tilde{x}}_{N} [w] (T) - x_{T})^{T} P ({\tilde{x}}_{N} [w] (T) - x_{T}),$ where ${\tilde{ℓ}}_{N, w} (t) = \frac{1}{2} (({\tilde{x}}_{N} [w] (t) - x_{d} (t))^{T} Q (t) ({\tilde{x}}_{N} [w] (t) - x_{d} (t)) + w (t)^{T} R (t) w (t))$ , $ω_{N k} \geq 0$ , $\sum_{k} ω_{N k} = T$ , and the chosen samples satisfy $∥ Q (τ_{N k}) ∥_{2} \leq ∥ Q ∥_{\infty}$ . Suppose $∥ {\tilde{x}}_{N} [w] - x_{w} ∥_{\infty} \leq e_{N}$ , $∥ x_{w} (τ_{N k}) - x_{d} (τ_{N k}) ∥_{2} \leq E_{N}$ , $∥ x_{w} (T) - x_{T} ∥_{2} \leq M_{T}$ , and $∣ \sum_{k} ω_{N k} ℓ_{w} (τ_{N k}) - \int_{0}^{T} ℓ_{w} (t) d t ∣ \leq q_{N},$ where $ℓ_{w} (t) = \frac{1}{2} ((x_{w} (t) - x_{d} (t))^{T} Q (t) (x_{w} (t) - x_{d} (t)) + w (t)^{T} R (t) w (t))$ . Then one may take $δ_{N} = q_{N} + \frac{T}{2} ∥ Q ∥_{\infty} (2 E_{N} + e_{N}) e_{N} + \frac{1}{2} ∥ P ∥ (2 M_{T} + e_{N}) e_{N} .$ If a feasible recovery sequence also satisfies $∥ v_{N} - u^{*} ∥_{2} \leq σ_{N}$ , then one may take $ρ_{N} = C_{J} σ_{N}$ , where $C_{J} = ∥ Q ∥_{\infty} E \sqrt{T} C_{s} + ∥ P ∥ M_{T} C_{s} + M \sqrt{T} M_{c},$ with $M$ any essential upper bound for $∥ R (t) ∥_{2}$ . In the control-only case, under Countable Choice the cell-average lemma supplies $v_{N} = P_{N} u^{*}$ and $σ_{N} \to 0$ . If $u^{*}$ is Lipschitz with constant $L_{u}$ , then $σ_{N} = \sqrt{T} L_{u} h_{N}$ is a valid rate for mesh size $h_{N}$ . For a general $L^{2}$ minimizer, no rate for $σ_{N}$ is asserted.

Facts & Assumptions

Given: The LQR model, $r > 0$ , a continuous minimizer, two feasible controls, and the finite objective inequalities in the Statement. For the explicit numerical-error estimate, also assume the stated pointwise sampling, state, and quadrature bounds.

[F1]

The state map is affine and bounded from $L^{2}$ to the supremum norm (The bounded measurable linear ODE has a unique affine state map).

[F2]

Expanding the LQR quadratic forms for $d = v - u$ gives $J (v) - J (u) = G_{u} [d] + \frac{1}{2} \int_{0}^{T} (δ x^{T} Q δ x + d^{T} R d) d t + \frac{1}{2} δ x (T)^{T} P δ x (T),$ where $δ x = x_{v} - x_{u}$ and $G_{u} [d] = \int_{0}^{T} ((x_{u} - x_{d})^{T} Q δ x + u^{T} R d) d t + (x_{u} (T) - x_{T})^{T} P δ x (T)$ ; this follows from [F1] and the definition of $J$ (Finite-horizon bounded-control linear-quadratic ODE problem, The bounded measurable linear ODE has a unique affine state map).

[F3]

For admissible controls, $∥ u ∥_{2} \leq \sqrt{T} M_{c}$ and $∥ x_{u} ∥_{\infty} \leq M_{x}$ by compactness of $U_{c}$ and the state-map bound (Finite-horizon bounded-control linear-quadratic ODE problem, The bounded measurable linear ODE has a unique affine state map).

[A1]

Countable Choice is assumed only for the control-only cell-average recovery specialization (The Axiom of Countable Choice ( ${A C}_{ω}$ ), Cell averages give a strong recovery sequence for compact convex controls).

Proof

1.1F2givenalgebra

Fix $v \in F$ and put $d = v - u^{*}$ . Convexity gives $u^{*} + θ d \in F$ for $0 \leq θ \leq 1$ . By [F2], $J (u^{*} + θ d) - J (u^{*}) = θ G_{u^{*}} [d] + θ^{2} R_{d}$ , where $R_{d} = \frac{1}{2} \int_{0}^{T} (δ x_{d}^{T} Q δ x_{d} + d^{T} R d) d t + \frac{1}{2} δ x_{d} (T)^{T} P δ x_{d} (T)$ . Minimality makes this expression nonnegative for $0 < θ \leq 1$ ; divide by $θ$ and let $θ ↓ 0$ to get $G_{u^{*}} [d] \geq 0$ . Also $R_{d} \geq \frac{r}{2} ∥ d ∥_{2}^{2}$ since $Q, P ⪰ 0$ and $R ⪰ r I$ . Taking $θ = 1$ proves $J (v) - J (u^{*}) \geq \frac{r}{2} ∥ v - u^{*} ∥_{2}^{2}$ .

1.2givenF3algebra

For the stated sampled objective, subtract the quadrature of $ℓ_{w}$ evaluated at the exact state. At every quadrature node, the state-cost difference is bounded by $\frac{1}{2} ∥ Q ∥_{\infty} (2 E_{N} + e_{N}) e_{N}$ ; the control-cost summand is unchanged. Nonnegative weights with sum $T$ therefore bound the accumulated state perturbation by $\frac{T}{2} ∥ Q ∥_{\infty} (2 E_{N} + e_{N}) e_{N}$ . The exact-state quadrature discrepancy is at most $q_{N}$ . At $T$ , the terminal quadratic difference is at most $\frac{1}{2} ∥ P ∥ (2 M_{T} + e_{N}) e_{N}$ . Thus $∣ J_{N} (w) - J (w) ∣ \leq δ_{N}$ for $w = u_{N}, v_{N}$ with the displayed $δ_{N}$ .

2.1givenstep 1.1algebra

The assumed inequalities and consistency give $J (u_{N}) \leq J_{N} (u_{N}) + δ_{N} \leq J_{N} (v_{N}) + η_{N} + δ_{N} \leq J (v_{N}) + η_{N} + 2 δ_{N} \leq J (u^{*}) + ρ_{N} + η_{N} + 2 δ_{N}$ . Subtract $J (u^{*})$ and combine with step 1.1 to get the objective-gap bound.

3.1step 1.1step 2.1algebra

Apply the growth inequality of step 1.1 at $v = u_{N}$ . Substitute the upper objective-gap bound from step 2.1, divide by $r / 2 > 0$ , and take nonnegative square roots to obtain the control-error estimate.

4.1A1F1F2F3givenalgebra∎

For any feasible $v$ with $∥ v - u^{*} ∥_{2} \leq σ$ , [F2] and [F3] bound the absolute difference of the state running terms by $∥ Q ∥_{\infty} E \sqrt{T} C_{s} σ$ , the terminal terms by $∥ P ∥ M_{T} C_{s} σ$ , and the control terms by $M \sqrt{T} M_{c} σ$ . Hence $J (v) - J (u^{*}) \leq C_{J} σ$ . Under [A1] the cell-average lemma gives strong recovery for the control-only case. If $u^{*}$ is $L_{u}$ -Lipschitz, for each $t$ in a cell its distance from the cell average is at most $L_{u} h_{N}$ , so $∥ P_{N} u^{*} - u^{*} ∥_{2} \leq \sqrt{T} L_{u} h_{N}$ . Substitution gives the stated $ρ_{N}$ rate and completes the sufficient-estimate branch.

AlgorithmAlgorithm: AI-adaptedProof: Not applicableOpen item page →

Finite-budget barrier or SQP solve for a transcription NLP

Definition

Inputs are a transcription record from Finite-horizon shooting and Runge–Kutta collocation transcription, a primal initial iterate $z_{0}$ and full multiplier iterate $μ_{0} = (ν_{0}, λ_{0})$ in the fixed equality/inequality constraint ordering, an optional canonical QP start $(x^{0}, y^{0}, s^{0})$ with nonnegative primal variables $x^{0}$ , free equality multipliers $y^{0}$ , and nonnegative dual slacks $s^{0}$ , an SQP model builder, and finite nonnegative tolerances $τ_{p}, τ_{d}, τ_{c}, τ_{s t a g e}, τ_{q u a d}$ . The stage and quadrature tolerances must match those bound into the transcription record. The KKT tolerances $τ_{p}, τ_{d}, τ_{c}$ bound primal feasibility, stationarity, and multiplier-feasibility/complementarity; in the QP route $τ_{c}$ also bounds the primal-dual gap. Also supply finite integer SQP and QP update caps $K, N \in N_{0}$ , positive finite integer penalty/backtracking caps $R, J_{m a x} \in N$ with $R, J_{m a x} \geq 1$ , merit parameters $ρ_{0} > 0$ , $γ > 1$ , $0 < β, σ < 1$ , and the finite-budget exact-Hessian branch selector and identified active set when that branch is requested. The IVP estimator uses its own declared controller tolerances; its status is not treated as an error certificate.

The SQP model builder is a deterministic finite procedure which, at each queried full state $(z_{k}, ν_{k}, λ_{k})$ , returns either a failure status or the finite NLP data $f_{k}, h_{k}, g_{k}, D h_{k}, D g_{k}$ , its objective gradient, a symmetric positive-definite SQP model matrix $W_{k}$ when the merit branch is selected, and the exact Lagrangian Hessian when the exact-Hessian branch is selected. The model matrix may depend on the supplied current multiplier vectors. For the exact-Hessian branch with identified active set $I$ , evaluate that Hessian with equality multipliers $ν_{k}$ , active inequality multipliers $(λ_{k})_{I}$ , and zero inactive multipliers; its required positivity is the separate reduced-curvature condition of the exact step, not full positive definiteness. It must use $g_{i} \leq 0$ and nonnegative inequality multipliers. In the merit branch, the associated quadratic subproblem is exactly $\min_{p} \nabla f_{k}^{T} p + \frac{1}{2} p^{T} W_{k} p$ subject to $h_{k} + D h_{k} p = 0$ and $g_{k} + D g_{k} p \leq 0$ . For that branch, the builder also supplies the active equality/inequality row data required by L1 merit globalization with finite penalty and backtracking budgets. Solver tolerances are used as follows: $τ_{p}$ bounds the norm of equality and positive inequality violations, $τ_{d}$ bounds the stationarity residual, and $τ_{c}$ bounds multiplier-feasibility and complementarity residuals. The stage and quadrature tolerances are passed to the transcription's corresponding residual and certified-error checks. The IVP estimator uses its own declared controller tolerances and has a separate status; its embedded difference is not a certified error bound. Any called builder, evaluator, canonicalizer, or solver returns a value or failure status in finite time.

First audit the transcription dimensions, sampled domains, and data status fields. If stage construction failed, return the partial record with its stage and NLP statuses and do not call a solver. If the model satisfies A linear-quadratic transcription with affine constraints is a convex QP, map its finite quadratic objective and affine constraints to canonical LP/QP form: replace each free variable $z_{i}$ by $z_{i}^{+} - z_{i}^{-}$ with $z_{i}^{+}, z_{i}^{-} \geq 0$ ; for each affine inequality $a_{i}^{T} z + d_{i} \leq 0$ , add a primal slack $ξ_{i} \geq 0$ and equality $a_{i}^{T} z + ξ_{i} = - d_{i}$ ; retain existing equalities; and transform the quadratic objective by substitution. The transcription record supplies finite stored coefficient arrays. Canonicalize using exact rational arithmetic on their dyadic values, then round each resulting coefficient once to the nearest normal representable value under Operationwise real and complex floating point arithmetic with rounded square roots and fused multiply-add, preserving exact zeros; any range failure makes the QP route inapplicable. Check the resulting equality matrix by exact rational row elimination; if it is rank deficient or the conversion is otherwise inadmissible, mark the QP route inapplicable and proceed to SQP. The optional start $(x^{0}, y^{0}, s^{0})$ is in these converted coordinates and is checked constructively: require every primal coordinate, including each added $ξ_{i}$ , and every dual slack coordinate $s^{0}$ to be strictly positive, and require $y^{0}$ to be finite with the correct equality dimension. If it is absent or fails this check, record strict-start-unavailable and proceed to SQP; do not infer infeasibility or decide whether some other strict start exists. If it passes, call Finite-budget primal-dual Newton methods for LP, convex QP, and symmetric cones once with centering and fraction-to-boundary parameters both $1 / 2$ , $ϵ_{p} = τ_{p}$ , $ϵ_{d} = \min (τ_{d}, τ_{c})$ , $ϵ_{g} = τ_{c}$ , and cap $N$ . The minimum in $ϵ_{d}$ ensures the supplier's single tolerance for both its dual and complementarity residuals meets each requested bound; $ϵ_{g}$ separately bounds the certified primal-dual gap. If it returns invalid-input or strict-start-unavailable, record that QP status and proceed to SQP. Return every other status and audited residuals. certified-gap certifies the resulting stored canonical floating-point QP, interpreted as exact dyadic rationals, under Fraction-to-boundary steps preserve positivity and exact primal-dual feasibility certifies a gap; it is not a certificate for the unsampled continuous problem. A residual-candidate remains a candidate. No other QP status is reinterpreted as success.

If the QP route is inapplicable or lacks a supplied strict start, run the SQP branch only when the transcription status is solver-ready constructed; otherwise return its not-smooth-nlp or model-construction status without an SQP call. Initialize $(z, ν, λ) = (z_{0}, ν_{0}, λ_{0})$ and verify the full multiplier dimensions. An invalid state returns model-build-failure with its last available record. For $k = 0, 1, \dots, K$ , in order, query the model builder at the entire current state $(z_{k}, ν_{k}, λ_{k})$ , propagating any builder failure without changing that state, and compute $r_{p} = (h_{k}, [g_{k}]_{+}), r_{d} = \nabla f_{k} + D h_{k}^{T} ν_{k} + D g_{k}^{T} λ_{k}, r_{λ} = [- λ_{k}]_{+}, r_{c} = (λ_{k, i} g_{k, i})_{i},$ where positive parts are componentwise. If $∥ r_{p} ∥ \leq τ_{p}$ , $∥ r_{d} ∥ \leq τ_{d}$ , $∥ r_{λ} ∥ \leq τ_{c}$ , and $∥ r_{c} ∥ \leq τ_{c}$ , return kkt-residual-met with the current decision and residual record; this is an approximate KKT candidate unless all residuals are exactly zero. If $k = K$ , return budget-exhausted. Otherwise, if the exact-Hessian branch is selected and an active set $I$ has been correctly identified, call Full-step exact-Hessian SQP after active-set identification at the current pair $(z_{k}, (ν_{k}, (λ_{k})_{I}))$ with inactive multipliers set to zero, using the exact Hessian evaluated at that same pair. On its successful full Newton-pair return $(z^{+}, μ^{+})$ , use $z^{+}$ , place the equality and active-inequality components of $μ^{+}$ into the full vectors $(ν, λ)$ , and set inactive inequality components to zero; any failure stops with that status without changing the current state. In the merit branch, call L1 merit globalization with finite penalty and backtracking budgets with the current model $W_{k}$ , $ρ_{0}, γ, β, σ, R, J_{m a x}$ . On accepted or accepted-soc, use its actual returned accepted point $z^{+}$ , including the SOC displacement when present, and full returned QP multipliers $(ν^{Q P}, λ^{Q P})$ to set $(z, ν, λ) = (z^{+}, ν^{Q P}, λ^{Q P})$ for the next model-builder query. Retain the complete supplier record, including its retained equality basis and selected inequality support. Do not replace the full vectors by reduced support multipliers or infer the true nonlinear active set from QP support. Any other supplier status stops the loop without changing the current state. The exact-Hessian local branch has only the conditional convergence guarantee in Exact-Hessian SQP converges quadratically on the identified active KKT system.

Every return is a result record with the last available decision variables, the full current SQP multiplier vectors when that branch has been reached, its last supplier record when one exists, defect residuals, path/endpoint constraint residuals, KKT residuals, objective value, and separate nlp_status, implicit_stage_status, ivp_status, quadrature_status, mesh_status, and budget_status fields. Fields not reached are marked not-run; failed evaluations carry their failure status and do not supply a candidate value. The finite caps bound the procedure. It does not guarantee a start, solver success, a global solution of a nonconvex problem, continuous-path feasibility from sampled constraints, or a certified approximation error for the continuous ODE problem.

TheoremStatement: AI-adaptedProof: AI-generatedOpen item page →

Exact-Hessian SQP converges locally to a regular transcription KKT point

Statement

Fix a mesh and a finite transcription generated from the pointwise smooth data in Finite-horizon ODE optimal-control transcription data. Suppose its NLP functions have continuous derivatives through order three on an open neighborhood of a KKT point $(z_{*}, μ_{*})$ , LICQ holds there, every active inequality multiplier is strictly positive, and the second-order sufficient condition holds on the critical cone. Then, after the correct active set is identified, the exact-Hessian full-step branch using Full-step exact-Hessian SQP after active-set identification is well-defined from a sufficiently close primal-dual start and converges quadratically to $(z_{*}, μ_{*})$ . The finite-budget dispatch of Finite-budget barrier or SQP solve for a transcription NLP may terminate before this local convergence is realized.

This also has the following conditional mesh-family conclusion. For each mesh index $N$ , let $K_{N} : Z_{N} \to Y_{N}$ be the square active-set KKT residual map of a finite transcription, with finite-dimensional normed spaces $Z_{N}, Y_{N}$ , and let $z_{N}^{0} \in Z_{N}$ be a reference primal-dual pair. Let $Z$ be a common Banach space containing the continuous primal and multiplier fields, with bounded linear reconstruction maps $E_{N} : Z_{N} \to Z$ that use the same fixed norms and satisfy $\sup_{N} ∥ E_{N} ∥ \leq C_{E}$ . Suppose a continuous KKT pair $z^{†} \in Z$ and mesh-consistent constraints satisfy $∥ E_{N} z_{N}^{0} - z^{†} ∥_{Z} \to 0$ and $∥ K_{N} (z_{N}^{0}) ∥_{Y_{N}} \leq ε_{N} \to 0$ . Suppose the discrete sensitivity maps $D K_{N} (z_{N}^{0}) : Z_{N} \to Y_{N}$ are invertible with $∥ D K_{N} (z_{N}^{0})^{- 1} ∥ \leq β$ uniformly, and $D K_{N}$ is $L$ -Lipschitz on a common radius $r$ about each $z_{N}^{0}$ , with constants independent of $N$ . For all sufficiently large $N$ , require $2 β ε_{N} \leq r$ and $2 β^{2} L ε_{N} \leq \frac{1}{2}$ . Then there is a unique active-set KKT root $z_{N}^{*}$ in the radius- $2 β ε_{N}$ ball, and $∥ E_{N} z_{N}^{*} - z^{†} ∥_{Z} \leq ∥ E_{N} z_{N}^{0} - z^{†} ∥_{Z} + 2 C_{E} β ε_{N} \to 0$ . If these roots satisfy LICQ, strict complementarity, SOSC, and a common local full-step basin for exact-Hessian SQP, then the exact-Hessian iterates from starts in that basin converge quadratically to $z_{N}^{*}$ for each such mesh. The varying systems are compared through the reconstruction maps and their fixed norms; this conclusion makes no global optimizer, path-feasibility, or finite-budget convergence claim.

Facts & Assumptions

Given: A fixed smooth finite transcription, a KKT pair, LICQ, strict complementarity, the second-order sufficient condition, a correct identified active set, and an initial primal-dual point sufficiently close to the KKT pair.

[F1]

The transcription is a finite NLP with the objective and constraints returned by Finite-horizon shooting and Runge–Kutta collocation transcription. The C3 regularity of these actual reduced functions on the stated neighborhood is an explicit hypothesis; smooth data alone do not certify a differentiable implicit-stage solve branch (Finite-horizon ODE optimal-control transcription data).

[F2]

The KKT convention is $g_{i} \leq 0$ with nonnegative active multipliers (Inequality constrained programs, tangent cones, constraint qualifications, and KKT).

[F3]

The exact-Hessian local SQP theorem gives quadratic convergence for a $C^{2}$ NLP with LICQ, strict complementarity, SOSC, and locally Lipschitz active-KKT Jacobian (Exact-Hessian SQP converges quadratically on the identified active KKT system).

[F4]

The strict second-order sufficient condition used here is positivity of the Lagrangian Hessian on every nonzero critical-cone direction (Strict second-order positivity on the critical cone gives a strict local minimum).

[F5]

The actual NLP functions are $C^{3}$ by [F1]. On a closed ball compactly contained in their open $C^{3}$ neighborhood, their third derivatives are bounded, so their Hessians and the active KKT Jacobian are locally Lipschitz.

[F6]

The active-set KKT Jacobian and full-step exact-Hessian SQP branch have the local properties stated in the fixed-system theorem (Exact-Hessian SQP converges quadratically on the identified active KKT system).

[F7]

Each finite-dimensional normed space $Z_{N}$ is complete (Every finite-dimensional normed space is Banach). A closed ball in $Z_{N}$ is complete: every Cauchy sequence in the ball converges in $Z_{N}$ , and continuity of the norm keeps its limit in the ball. A contraction of a nonempty closed ball into itself therefore has a unique fixed point by A contraction of a nonempty complete metric space into itself has exactly one fixed point, the limit of the iterates from any starting point.

[A1]

For the mesh-family conclusion, the reconstruction, residual-consistency, uniform inverse, uniform Jacobian-Lipschitz, active-set, and common-basin assumptions in the Statement hold.

Proof

1.1F1given

A fixed transcription has finitely many decision variables. By the regularity hypothesis in the Statement and [F1], its actual objective and constraint functions have continuous derivatives through order three on a neighborhood of the KKT point. For an implicit-stage transcription this hypothesis includes the selected local stage-solve branch; no differentiability of that branch is inferred from the smooth-data definition alone.

1.2A1F7algebra

For the mesh-family branch, set $A_{N} = D K_{N} (z_{N}^{0})$ and $T_{N} (z) = z - A_{N}^{- 1} K_{N} (z)$ . If $ε_{N} = 0$ , $z_{N}^{0}$ is already a root. Otherwise take the closed ball of radius $R_{N} = 2 β ε_{N}$ . The residual bound gives $∥ T_{N} (z_{N}^{0}) - z_{N}^{0} ∥ \leq β ε_{N} = R_{N} / 2$ . For $z$ in that ball, the uniform Lipschitz bound gives $∥ D T_{N} (z) ∥ = ∥ I - A_{N}^{- 1} D K_{N} (z) ∥ \leq β L R_{N} \leq 1 / 2$ . Thus $T_{N}$ maps the ball into itself and is a contraction. Completeness [F7] gives its unique fixed point $z_{N}^{*}$ , which is exactly the unique root of $K_{N}$ there and satisfies $∥ z_{N}^{*} - z_{N}^{0} ∥ \leq R_{N}$ .

2.1F2F3F4F5givenstep 1.1

The KKT map for the identified active set is built from the first derivatives of those objective and constraint functions. Its Jacobian is locally Lipschitz by continuity and local boundedness of their third derivatives in [F5]. The assumed LICQ, strict complementarity, and SOSC from [F4] are exactly the remaining hypotheses of [F3].

3.1F3givenstep 2.1algebra

Apply [F3] to the active KKT system. It provides a neighborhood in which exact-Hessian full Newton steps are defined and converge quadratically from a sufficiently close start, which is the asserted fixed-mesh branch. The dispatcher has a finite outer cap, so it can stop before the local sequence converges.

4.1A1F6step 1.2algebra∎

By boundedness of $E_{N}$ , $∥ E_{N} z_{N}^{*} - z^{†} ∥_{Z} \leq C_{E} ∥ z_{N}^{*} - z_{N}^{0} ∥ + ∥ E_{N} z_{N}^{0} - z^{†} ∥_{Z} \leq 2 C_{E} β ε_{N} + ∥ E_{N} z_{N}^{0} - z^{†} ∥_{Z}$ , which tends to zero. The assumed LICQ, strict complementarity, SOSC, local Lipschitz conditions, and common basin let [F6] apply at each $z_{N}^{*}$ ; hence each exact-Hessian full-step sequence from that basin converges quadratically.

5 · Examples, counterexamples and false statements

None yet.