Double integrator: energy minimisation

Let us consider a wagon moving along a rail, whose acceleration can be controlled by a force $u$. We denote by $x = (x_1, x_2)$ the state of the wagon, where $x_1$ is the position and $x_2$ the velocity.

We assume that the mass is constant and equal to one, and that there is no friction. The dynamics are given by

\[ \dot x_1(t) = x_2(t), \quad \dot x_2(t) = u(t),\quad u(t) \in \R,\]

which is simply the double integrator system. Let us consider a transfer starting at time $t_0 = 0$ and ending at time $t_f = 1$, for which we want to minimise the transfer energy

\[ \frac{1}{2}\int_{0}^{1} u^2(t) \, \mathrm{d}t\]

starting from $x(0) = (-1, 0)$ and aiming to reach the target $x(1) = (0, 0)$.

First, we need to import the OptimalControl.jl package to define the optimal control problem, NLPModelsIpopt.jl to solve it, and Plots.jl to visualise the solution.

using OptimalControl
using NLPModelsIpopt
using Plots

Optimal control problem

Let us define the problem with the @def macro:

t0 = 0
tf = 1
x0 = [-1, 0]
xf = [0, 0]
ocp = @def begin
    t ∈ [t0, tf], time
    x ∈ R², state
    u ∈ R, control
    x(t0) == x0
    x(tf) == xf
    ẋ(t) == [x₂(t), u(t)]
    0.5∫( u(t)^2 ) → min
end

Solve and plot

Direct method

We can solve it simply with:

sol = solve(ocp)

▫ This is OptimalControl version v1.1.1 running with: direct, adnlp, ipopt.

▫ The optimal control problem is solved with CTDirect version v0.16.4.

   ┌─ The NLP is modelled with ADNLPModels and solved with NLPModelsIpopt.
   │
   ├─ Number of time steps⋅: 250
   └─ Discretisation scheme: trapeze

▫ This is Ipopt version 3.14.19, running with linear solver MUMPS 5.8.1.

Number of nonzeros in equality constraint Jacobian...:     3005
Number of nonzeros in inequality constraint Jacobian.:        0
Number of nonzeros in Lagrangian Hessian.............:      251

Total number of variables............................:     1004
                     variables with only lower bounds:        0
                variables with lower and upper bounds:        0
                     variables with only upper bounds:        0
Total number of equality constraints.................:      755
Total number of inequality constraints...............:        0
        inequality constraints with only lower bounds:        0
   inequality constraints with lower and upper bounds:        0
        inequality constraints with only upper bounds:        0

iter    objective    inf_pr   inf_du lg(mu)  ||d||  lg(rg) alpha_du alpha_pr  ls
   0  1.0000000e-01 1.10e+00 3.11e-14   0.0 0.00e+00    -  0.00e+00 0.00e+00   0
   1 -5.0000000e-03 7.36e-02 2.66e-15 -11.0 6.08e+00    -  1.00e+00 1.00e+00h  1
   2  6.0003829e+00 8.88e-16 1.78e-15 -11.0 6.01e+00    -  1.00e+00 1.00e+00h  1

Number of Iterations....: 2

                                   (scaled)                 (unscaled)
Objective...............:   6.0003828724303228e+00    6.0003828724303228e+00
Dual infeasibility......:   1.7763568394002505e-15    1.7763568394002505e-15
Constraint violation....:   8.8817841970012523e-16    8.8817841970012523e-16
Variable bound violation:   0.0000000000000000e+00    0.0000000000000000e+00
Complementarity.........:   0.0000000000000000e+00    0.0000000000000000e+00
Overall NLP error.......:   1.7763568394002505e-15    1.7763568394002505e-15


Number of objective function evaluations             = 3
Number of objective gradient evaluations             = 3
Number of equality constraint evaluations            = 3
Number of inequality constraint evaluations          = 0
Number of equality constraint Jacobian evaluations   = 3
Number of inequality constraint Jacobian evaluations = 0
Number of Lagrangian Hessian evaluations             = 2
Total seconds in IPOPT                               = 3.350

EXIT: Optimal Solution Found.

And plot the solution with:

plot(sol)

Indirect method

The first solution was obtained using the so-called direct method.^[1] Another approach is to use an indirect simple shooting method. We begin by importing the necessary packages.

using OrdinaryDiffEq # Ordinary Differential Equations (ODE) solver
using NonlinearSolve # Nonlinear Equations (NLE) solver

To define the shooting function, we must provide the maximising control in feedback form:

# maximising control, H(x, p, u) = p₁x₂ + p₂u - u²/2
u(x, p) = p[2]

# Hamiltonian flow
f = Flow(ocp, u)

# state projection, p being the costate
π((x, p)) = x

# shooting function
S(p0) = π( f(t0, x0, p0, tf) ) - xf

We are now ready to solve the shooting equations.

# auxiliary in-place NLE function
nle!(s, p0, λ) = s[:] = S(p0)

# initial guess for the Newton solver
p0_guess = [1, 1]

# NLE problem with initial guess
prob = NonlinearProblem(nle!, p0_guess)

# resolution of S(p0) = 0
sol = solve(prob; show_trace=Val(true))
p0_sol = sol.u # costate solution

# print the costate solution and the shooting function evaluation
println("\ncostate: p0 = ", p0_sol)
println("shoot: S(p0) = ", S(p0_sol), "\n")

Algorithm: NewtonRaphson(
    descent = NewtonDescent(),
    autodiff = AutoForwardDiff(),
    vjp_autodiff = AutoReverseDiff(
        compile = false
    ),
    jvp_autodiff = AutoForwardDiff(),
    concrete_jac = Val{false}()
)

----    	-------------       	-----------
Iter    	f(u) inf-norm       	Step 2-norm
----    	-------------       	-----------
0       	6.66666667e-01      	0.00000000e+00
1       	7.06715337e-15      	1.20830460e+01
Final   	7.06715337e-15
----------------------

costate: p0 = [12.000000000000014, 5.999999999999988]
shoot: S(p0) = [-7.067153371923864e-15, 4.159252963096603e-15]

To plot the solution obtained by the indirect method, we need to build the solution of the optimal control problem. This is done using the costate solution and the flow function.

sol = f((t0, tf), x0, p0_sol; saveat=range(t0, tf, 100))
plot(sol)

State constraint

Direct method

We add the path constraint

\[x_2(t) \le 1.2.\]

Let us model, solve and plot the optimal control problem with this constraint.

# the upper bound for x₂
a = 1.2

# the optimal control problem
ocp = @def begin
    t ∈ [t0, tf], time
    x ∈ R², state
    u ∈ R, control
    x₂(t) ≤ a
    x(t0) == x0
    x(tf) == xf
    ẋ(t) == [x₂(t), u(t)]
    0.5∫( u(t)^2 ) → min
end

# solve with a direct method using default settings
sol = solve(ocp)

# plot the solution
plt = plot(sol; label="Direct", size=(800, 600))

Indirect method

The pseudo-Hamiltonian is (considering the normal case):

\[H(x, p, u, \mu) = p_1 x_2 + p_2 u - \frac{u^2}{2} + \mu\, c(x),\]

with $c(x) = x_2 - a$. Along a boundary arc we have $c(x(t)) = 0$. Differentiating, we obtain:

\[ \frac{\mathrm{d}}{\mathrm{d}t}c(x(t)) = \dot{x}_2(t) = u(t) = 0.\]

The zero control is maximising; hence, $p_2(t) = 0$ along the boundary arc.

\[ \dot{p}_2(t) = -p_1(t) - \mu(t) \quad \Rightarrow \mu(t) = -p_1(t).\]

Since the adjoint vector is continuous at the entry time $t_1$ and the exit time $t_2$, we have four unknowns: the initial costate $p_0 \in \mathbb{R}^2$ and the times $t_1$ and $t_2$. We need four equations: the target condition provides two, reaching the constraint at time $t_1$ gives $c(x(t_1)) = 0$, and finally $p_2(t_1) = 0$.

# flow for unconstrained extremals
f = Flow(ocp, (x, p) -> p[2])

ub = 0          # boundary control
c(x) = x[2]-a   # constraint: c(x) ≥ 0
μ(p) = -p[1]    # dual variable

# flow for boundary extremals
g = Flow(ocp, (x, p) -> ub, (x, u) -> c(x), (x, p) -> μ(p))

# shooting function
function shoot!(s, p0, t1, t2)
    x_t0, p_t0 = x0, p0
    x_t1, p_t1 = f(t0, x_t0, p_t0, t1)
    x_t2, p_t2 = g(t1, x_t1, p_t1, t2)
    x_tf, p_tf = f(t2, x_t2, p_t2, tf)
    s[1:2] = x_tf - xf
    s[3] = c(x_t1)
    s[4] = p_t1[2]
end

We are now ready to solve the shooting equations.

# auxiliary in-place NLE function
nle!(s, ξ, λ) = shoot!(s, ξ[1:2], ξ[3], ξ[4])

# initial guess for the Newton solver
ξ_guess = [40, 10, 0.25, 0.75]

# NLE problem with initial guess
prob = NonlinearProblem(nle!, ξ_guess)

# resolution of the shooting equations
sol = solve(prob; show_trace=Val(true))
p0, t1, t2 = sol.u[1:2], sol.u[3], sol.u[4]

# print the costate solution and the entry and exit times
println("\np0 = ", p0, "\nt1 = ", t1, "\nt2 = ", t2)

Algorithm: NewtonRaphson(
    descent = NewtonDescent(),
    autodiff = AutoForwardDiff(),
    vjp_autodiff = AutoReverseDiff(
        compile = false
    ),
    jvp_autodiff = AutoForwardDiff(),
    concrete_jac = Val{false}()
)

----    	-------------       	-----------
Iter    	f(u) inf-norm       	Step 2-norm
----    	-------------       	-----------
0       	5.00000000e-02      	0.00000000e+00
1       	1.21434328e-14      	1.64924225e+00
Final   	1.21434328e-14
----------------------

p0 = [38.400000000000226, 9.600000000000033]
t1 = 0.24999999999999944
t2 = 0.7500000000000008

To reconstruct the trajectory obtained with the state constraint, we concatenate the flows: one unconstrained arc up to the entry time $t_1$, a boundary arc between $t_1$ and $t_2$, and finally another unconstrained arc up to $t_f$. This concatenation allows us to compute the complete solution — state, costate, and control — which we can then plot together with the direct solution for comparison.

# concatenation of the flows
φ = f * (t1, g) * (t2, f)

# compute the solution: state, costate, control...
flow_sol = φ((t0, tf), x0, p0; saveat=range(t0, tf, 100))

# plot the solution on the previous plot
plot!(plt, flow_sol; label="Indirect", color=2, linestyle=:dash)