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Direct transcription and numerical ode optimal control — Examples

1 · Prerequisites

2 · Summary

The examples compute an exact scalar bounded-actuator LQR minimizer and its objective, compare an explicit-midpoint chemical-dosing trajectory with its exact ODE solution, and write the same two-interval control problem in single- and multiple-shooting variables. Each calculation identifies which values are exact and which belong only to a finite transcription.

3 · Logical flowchart

4 · Definitions, theorems and proofs

None yet.

5 · Examples, counterexamples and false statements

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

A bounded-actuator LQR control and its exact objective

Example

Take $T = 1$ , $\dot{x} = u$ , $x (0) = 0$ , $u (t) \in [- 1, 1]$ , running cost $\frac{1}{2} u (t)^{2}$ , and terminal cost $\frac{1}{2} (x (1) - 1)^{2}$ . The unique optimal actuator signal is the constant $u^{*} (t) = \frac{1}{2}$ , and its objective is $J (u^{*}) = \frac{1}{4}$ .

Facts & Assumptions

Given: The scalar problem above.

[F1]

The state is $x_{u} (t) = \int_{0}^{t} u (s) d s$ ; the objective is $J (u) = \frac{1}{2} ∥ u ∥_{2}^{2} + \frac{1}{2} (\int_{0}^{1} u - 1)^{2}$ (Finite-horizon bounded-control linear-quadratic ODE problem).

[F2]

Under Countable Choice, the backward adjoint exists uniquely and satisfies $- \dot{p} = 0$ and $p (1) = x (1) - 1$ (The backward LQR adjoint equation has a unique solution, Continuous adjoint equation for the linear-quadratic ODE).

[F3]

Under Countable Choice and Dependent Choice, the reduced gradient is $B^{T} p + R u$ (The LQR adjoint identity gives the reduced gradient).

[A1]

Countable Choice and Dependent Choice are assumed only for the reduced-gradient verification in step 3.1 (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).

[A2]

Exact real arithmetic is used.

Verification

1.1F1A2algebra

Put $m = \int_{0}^{1} u (t) d t$ . Since $\int_{0}^{1} (u - m)^{2} d t = \int_{0}^{1} u^{2} d t - m^{2}$ , direct expansion gives $J (u) = \frac{1}{2} ∥ u - m ∥_{2}^{2} + \frac{1}{2} [m^{2} + (m - 1)^{2}] = \frac{1}{2} ∥ u - m ∥_{2}^{2} + (m - \frac{1}{2})^{2} + \frac{1}{4}$ .

2.1step 1.1givenalgebra

Both squared terms are nonnegative, so $J (u) \geq \frac{1}{4}$ . Equality holds exactly when $u = m$ almost everywhere and $m = \frac{1}{2}$ , hence exactly for $u^{*} (t) = \frac{1}{2}$ almost everywhere. This control satisfies $[- 1, 1]$ and direct substitution gives $J (u^{*}) = \frac{1}{4}$ . Therefore it is the unique minimizer.

3.1A1F2F3step 2.1algebra∎

The state is $x (t) = t / 2$ . By [F2], the adjoint equation is $- \dot{p} = 0$ with terminal value $p (1) = x (1) - 1 = - 1 / 2$ . The constant curve $p (t) = - 1 / 2$ solves it, and uniqueness in [F2] makes this the adjoint. Under [A1], [F3] gives the reduced gradient $g = p + u^{*} = 0$ . Thus $g (t) (c - u^{*} (t)) = 0$ for every $c \in [- 1, 1]$ , so the control-only pointwise variational inequality holds with equality for this example. The optimizer and objective calculation in steps 1.1–2.1 used no choice axiom.

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

A chemical-dosing model under a constant admissible dose

Example

Let $x (t)$ be the concentration of a decaying reagent after dosing, with $\dot{x} = - x + u$ , $x (0) = 0$ , $0 \leq u \leq 1$ , and $0 \leq t \leq 1$ . Use zero running and terminal costs to focus on the state trajectory. For the admissible constant dose $u (t) = 1$ , the exact concentration is $x (t) = 1 - e^{- t}$ . A one-cell explicit-midpoint transcription from zero returns the stage value $X_{2} = 1 / 2$ and endpoint approximation $x_{1} = 1 / 2$ , whereas the exact endpoint is $1 - e^{- 1}$ . The controller's embedded difference alone is not an error certificate.

Facts & Assumptions

Given: The linear concentration model, constant input, one-cell mesh, and explicit midpoint tableau $a_{21} = 1 / 2$ , $b_{1} = 0$ , $b_{2} = 1$ , $c_{1} = 0$ , $c_{2} = 1 / 2$ .

[F1]

The unique state solves its Volterra integral equation for bounded measurable linear dynamics and an $L^{2}$ input (The bounded measurable linear ODE has a unique affine state map).

[F2]

The one-step stage and endpoint formulas for an explicit RK tableau are those of Explicit Runge–Kutta tableaus, stage maps, and failure status.

[F3]

Piecewise-constant controls are defined interval by interval on the mesh (Time meshes and piecewise-constant control parameterization).

[F4]

An embedded adaptive RK estimate is a local pair difference; controller termination does not certify a global exact error (Finite-budget adaptive Runge–Kutta controller, The bounded adaptive Runge–Kutta controller terminates).

[A1]

Exact real arithmetic is used for the displayed values.

Verification

1.1F1givenalgebra

The exact integral equation is $x (t) = \int_{0}^{t} (- x (s) + 1) d s$ . The function $x (t) = 1 - e^{- t}$ has $x (0) = 0$ and derivative $e^{- t} = - (1 - e^{- t}) + 1$ , so it solves the equation; uniqueness follows from [F1].

2.1F2F3A1step 1.1algebra

On the one-cell mesh, the explicit midpoint first slope is $K_{1} = - x_{0} + u = 1$ . The second stage is $X_{2} = x_{0} + \frac{1}{2} K_{1} = \frac{1}{2}$ , so $K_{2} = - X_{2} + u = \frac{1}{2}$ . The output is $x_{1} = x_{0} + K_{2} = \frac{1}{2}$ by [F2]; the exact endpoint is $1 - e^{- 1}$ by step 1.1.

3.1F4step 2.1algebra∎

The exponential series gives $e = 1 + 1 + \sum_{k \geq 2} 1 / k! > 2$ , so $1 - e^{- 1} > 1 / 2$ and the midpoint endpoint differs from the exact endpoint. The procedure returns the stated value, not a bound on the true error, by [F4].

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

Two-interval single and multiple shooting for x-prime equals u

Example

For $\dot{x} = u$ , $x (0) = 0$ , and mesh $0, 1 / 2, 1$ , take the compact control set $U_{c} = [- 1, 1]$ and let the two constant controls satisfy $v_{0}, v_{1} \in U_{c}$ . Since the dynamics are constant on each interval, every consistent explicit RK tableau with $\sum_{i} b_{i} = 1$ and stage abscissae $c_{i} \in [0, 1]$ gives the exact interval updates on this mesh. Single shooting yields $x_{1} = v_{0} / 2$ and $x_{2} = (v_{0} + v_{1}) / 2$ . Multiple shooting retains $x_{1}, x_{2}$ and imposes $x_{1} - v_{0} / 2 = 0$ , $x_{2} - x_{1} - v_{1} / 2 = 0$ . The shared objective $\frac{1}{4} (v_{0}^{2} + v_{1}^{2}) + \frac{1}{2} (x_{2} - 1)^{2}$ therefore has the same reduced value in both formulations.

Facts & Assumptions

Given: The two-cell mesh, compact control set $U_{c} = [- 1, 1]$ , constant controls $v_{0}, v_{1} \in U_{c}$ , exact real arithmetic, and a consistent explicit tableau with $\sum_{i} b_{i} = 1$ and $c_{i} \in [0, 1]$ .

[F1]

Piecewise-constant controls assign one control value per mesh cell (Time meshes and piecewise-constant control parameterization).

[F2]

The single and multiple shooting recursions use the same deterministic interval maps (Finite-horizon shooting and Runge–Kutta collocation transcription).

[F3]

Their feasible control sets and reduced objectives agree by projection (Exact single and multiple shooting have the same reduced feasible controls).

[A1]

The explicit tableau is consistent, meaning $\sum_{i} b_{i} = 1$ ; all arithmetic shown is exact.

Verification

1.1F1A1givenalgebra

On the first cell every stage slope equals $v_{0}$ , so its endpoint increment is $\frac{1}{2} v_{0} \sum_{i} b_{i} = \frac{1}{2} v_{0}$ . The second cell adds $\frac{1}{2} v_{1} \sum_{i} b_{i} = \frac{1}{2} v_{1}$ . Thus single shooting gives the stated $x_{1}$ and $x_{2}$ .

2.1F2F3step 1.1algebra∎

The multiple-shooting continuity equations uniquely impose $x_{1} = \frac{1}{2} v_{0}$ and $x_{2} = x_{1} + \frac{1}{2} v_{1} = \frac{1}{2} (v_{0} + v_{1})$ . Substitution into its objective gives exactly the single-shooting objective. This is the concrete bijection and reduced-objective identity stated in [F3].

Direct transcription and numerical ode optimal control — Examples

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

5 · Examples, counterexamples and false statements

A bounded-actuator LQR control and its exact objective

Example

Facts & Assumptions

Verification

A chemical-dosing model under a constant admissible dose

Example

Facts & Assumptions

Verification

Two-interval single and multiple shooting for x-prime equals u

Example

Facts & Assumptions

Verification

Sources