Control-free problems

Control-free problems are optimal control problems without a control variable. They are used for optimizing constant parameters in dynamical systems, such as:

• Identifying unknown parameters from observed data (parameter estimation)
• Finding optimal parameters for a given performance criterion

This page demonstrates two simple examples with known analytical solutions.

First, we import the necessary packages:

using OptimalControl
using NLPModelsIpopt
using Plots

Example 1: Exponential growth rate estimation

Consider a system with exponential growth:

\[\dot{x}(t) = p \cdot x(t), \quad x(0) = 2\]

where $p$ is an unknown growth rate parameter. We have observed data following $x_{\text{obs}}(t) = 2 e^{0.5 t}$ and want to estimate $p$ by minimizing the squared error:

\[\min_{p} \int_0^{10} (x(t) - x_{\text{obs}}(t))^2 \, dt\]

The analytical solution is $p = 0.5$.

Problem definition

# observed data (analytical solution)
data(t) = 2.0 * exp(0.5 * t)

# optimal control problem (parameter estimation)
ocp1 = @def begin
    p ∈ R, variable              # growth rate to estimate
    t ∈ [0, 10], time
    x ∈ R, state

    x(0) == 2.0
    ẋ(t) == p * x(t)

    ∫((x(t) - data(t))^2) → min  # fit to observed data
end

Solution

sol1 = solve(ocp1; display=false)
println("Estimated growth rate: p = ", variable(sol1))
println("Objective value: ", objective(sol1))
println("Expected: p = 0.5, objective ≈ 0.0")

Estimated growth rate: p = 0.49997784682846513
Objective value: 2.673091409826263e-6
Expected: p = 0.5, objective ≈ 0.0

plot(sol1; size=(800, 400))

The estimated parameter should be very close to $p = 0.5$, and the objective (squared error) should be nearly zero since we're fitting to the exact analytical solution.

Example 2: Harmonic oscillator pulsation optimization

Consider a harmonic oscillator:

\[\ddot{q}(t) = -\omega^2 q(t)\]

with initial conditions $q(0) = 1$, $\dot{q}(0) = 0$ and final condition $q(1) = 0$. We want to find the minimal pulsation $\omega$ satisfying these constraints:

\[ \begin{aligned} & \text{Minimise} && \omega^2 \\ & \text{subject to} \\ & && \ddot{q}(t) = -\omega^2 q(t), \\[1.0em] & && q(0) = 1, \quad \dot{q}(0) = 0, \\[0.5em] & && q(1) = 0. \end{aligned}\]

The analytical solution is $\omega = \pi/2 \approx 1.5708$, giving $q(t) = \cos(\pi t / 2)$.

Problem definition

# optimal control problem (pulsation optimization)
ocp2 = @def begin
    ω ∈ R, variable              # pulsation to optimize
    t ∈ [0, 1], time
    x = (q, v) ∈ R², state

    q(0) == 1.0
    v(0) == 0.0
    q(1) == 0.0                  # final condition

    ẋ(t) == [v(t), -ω^2 * q(t)]

    ω^2 → min   # minimize pulsation
end

Solution

sol2 = solve(ocp2; display=false)
println("Optimal pulsation: ω = ", variable(sol2))
println("Objective value: ω² = ", objective(sol2))
println("Expected: ω = π/2 ≈ 1.5708, ω² ≈ 2.4674")

Optimal pulsation: ω = 1.570801494616592
Objective value: ω² = 2.4674173354897193
Expected: ω = π/2 ≈ 1.5708, ω² ≈ 2.4674

plot(sol2; size=(800, 400))

The optimal pulsation should be close to $\omega = \pi/2 \approx 1.5708$, and the objective $\omega^2 \approx 2.4674$.

Comparison with analytical solutions

For the harmonic oscillator, we can compare the numerical solution with the analytical one:

# analytical solution
t_analytical = range(0, 1, 100)
q_analytical = cos.(π * t_analytical / 2)
v_analytical = -(π/2) * sin.(π * t_analytical / 2)

# plot comparison
plt = plot(sol2; size=(800, 600))
plot!(plt, t_analytical, q_analytical;
      label="q (analytical)", linestyle=:dash, linewidth=2, subplot=1)
plot!(plt, t_analytical, v_analytical;
      label="v (analytical)", linestyle=:dash, linewidth=2, subplot=2)

The numerical and analytical solutions should match very closely.