Minimum time orbit transfer
Introduction

Minimum time control of the Kepler equation (CNES / TAS / Inria / CNRS collaboration, check [1] and [2]):
\[\begin{cases} t_f \to \min,\\ \ddot{q} = -\mu\frac{q}{|q|^3} + \frac{u}{m}\,,\quad t \in [0,t_f],\\ \dot{m} = -\beta|u|,\quad |u| \leq T_{\mathrm{max}}. \end{cases}\]
Fixed initial and final Keplerian orbits (free final longitude).

using OptimalControl
using NLPModelsIpopt
using OrdinaryDiffEq
using Plots
using MINPACK
using ForwardDiff
using LinearAlgebra
Problem definition
Tmax = 60 # Maximum thrust in Newtons
cTmax = 3600^2 / 1e6; T = Tmax * cTmax # Conversion from Newtons to kg x Mm / h²
mass0 = 1500 # Initial mass of the spacecraft
β = 1.42e-02 # Engine specific impulsion
μ = 5165.8620912 # Earth gravitation constant
P0 = 11.625 # Initial semilatus rectum
ex0, ey0 = 0.75, 0 # Initial eccentricity
hx0, hy0 = 6.12e-2, 0 # Initial ascending node and inclination
L0 = π # Initial longitude
Pf = 42.165 # Final semilatus rectum
exf, eyf = 0, 0 # Final eccentricity
hxf, hyf = 0, 0 # Final ascending node and inclination
asqrt(x; ε=1e-9) = sqrt(sqrt(x^2+ε^2)) # Avoid issues with AD
function F0(x)
P, ex, ey, hx, hy, L = x
pdm = asqrt(P / μ)
cl = cos(L)
sl = sin(L)
w = 1 + ex * cl + ey * sl
F = zeros(eltype(x), 6) # Use eltype to allow overloading for AD
F[6] = w^2 / (P * pdm)
return F
end
function F1(x)
P, ex, ey, hx, hy, L = x
pdm = asqrt(P / μ)
cl = cos(L)
sl = sin(L)
F = zeros(eltype(x), 6)
F[2] = pdm * sl
F[3] = pdm * (-cl)
return F
end
function F2(x)
P, ex, ey, hx, hy, L = x
pdm = asqrt(P / μ)
cl = cos(L)
sl = sin(L)
w = 1 + ex * cl + ey * sl
F = zeros(eltype(x), 6)
F[1] = pdm * 2 * P / w
F[2] = pdm * (cl + (ex + cl) / w)
F[3] = pdm * (sl + (ey + sl) / w)
return F
end
function F3(x)
P, ex, ey, hx, hy, L = x
pdm = asqrt(P / μ)
cl = cos(L)
sl = sin(L)
w = 1 + ex * cl + ey * sl
pdmw = pdm / w
zz = hx * sl - hy * cl
uh = (1 + hx^2 + hy^2) / 2
F = zeros(eltype(x), 6)
F[2] = pdmw * (-zz * ey)
F[3] = pdmw * zz * ex
F[4] = pdmw * uh * cl
F[5] = pdmw * uh * sl
F[6] = pdmw * zz
return F
end
Direct solve
tf = 15 # Estimation of final time
Lf = 3π # Estimation of final longitude
x0 = [P0, ex0, ey0, hx0, hy0, L0] # Initial state
xf = [Pf, exf, eyf, hxf, hyf, Lf] # Final state
x(t) = x0 + (xf - x0) * t / tf # Linear interpolation
u = [0.1, 0.5, 0.] # Initial guess for the control
nlp_init = (state=x, control=u, variable=tf) # Initial guess for the NLP
ocp = @def begin
tf ∈ R, variable
t ∈ [0, tf], time
x = (P, ex, ey, hx, hy, L) ∈ R⁶, state
u ∈ R³, control
x(0) == x0
x[1:5](tf) == xf[1:5]
mass = mass0 - β * T * t
ẋ(t) == F0(x(t)) + T / mass * (u₁(t) * F1(x(t)) + u₂(t) * F2(x(t)) + u₃(t) * F3(x(t)))
u₁(t)^2 + u₂(t)^2 + u₃(t)^2 ≤ 1
tf → min
end
tf ∈ R, variable
t ∈ [0, tf], time
x = ((P, ex, ey, hx, hy, L) ∈ R⁶, state)
u ∈ R³, control
x(0) == x0
(x[1:5])(tf) == xf[1:5]
mass = mass0 - β * T * t
ẋ(t) == F0(x(t)) + (T / mass) * (u₁(t) * F1(x(t)) + u₂(t) * F2(x(t)) + u₃(t) * F3(x(t)))
u₁(t) ^ 2 + u₂(t) ^ 2 + u₃(t) ^ 2 ≤ 1
tf → min
The optimal control problem is of the form:
minimize J(x, u, tf) = g(x(0), x(tf), tf)
subject to
ẋ(t) = f(t, x(t), u(t), tf), t in [0, tf] a.e.,
ψ₋ ≤ ψ(t, x(t), u(t), tf) ≤ ψ₊,
ϕ₋ ≤ ϕ(x(0), x(tf), tf) ≤ ϕ₊,
where x(t) = (P(t), ex(t), ey(t), hx(t), hy(t), L(t)) ∈ R⁶, u(t) ∈ R³ and tf ∈ R.
Declarations (* required):
╭────────┬────────┬──────────┬──────────┬───────────┬────────────┬─────────────╮
│ times* │ state* │ control* │ variable │ dynamics* │ objective* │ constraints │
├────────┼────────┼──────────┼──────────┼───────────┼────────────┼─────────────┤
│ V │ V │ V │ V │ V │ V │ V │
╰────────┴────────┴──────────┴──────────┴───────────┴────────────┴─────────────╯
nlp_sol = OptimalControl.solve(ocp; init=nlp_init, grid_size=100)
This is Ipopt version 3.14.17, running with linear solver MUMPS 5.7.3.
Number of nonzeros in equality constraint Jacobian...: 11011
Number of nonzeros in inequality constraint Jacobian.: 303
Number of nonzeros in Lagrangian Hessian.............: 4445
Total number of variables............................: 910
variables with only lower bounds: 0
variables with lower and upper bounds: 0
variables with only upper bounds: 0
Total number of equality constraints.................: 611
Total number of inequality constraints...............: 101
inequality constraints with only lower bounds: 0
inequality constraints with lower and upper bounds: 0
inequality constraints with only upper bounds: 101
iter objective inf_pr inf_du lg(mu) ||d|| lg(rg) alpha_du alpha_pr ls
0 1.5000000e+01 2.14e-01 2.46e-01 0.0 0.00e+00 - 0.00e+00 0.00e+00 0
1 1.5632877e+01 1.73e-01 6.59e-01 -6.1 7.27e+00 - 5.95e-01 1.90e-01h 3
2 1.6096872e+01 1.51e-01 1.15e+00 -6.4 3.71e+00 0.0 4.78e-01 1.25e-01h 4
3 1.9018883e+01 6.12e-02 2.70e+00 -1.1 3.97e+00 -0.5 7.39e-01 6.03e-01H 1
4 2.0358419e+01 1.56e-02 1.98e+00 -1.7 1.60e+00 -0.1 1.00e+00 8.38e-01h 1
5 2.0617233e+01 2.68e-04 4.98e-01 -3.6 2.59e-01 0.4 1.00e+00 1.00e+00h 1
6 2.0429778e+01 3.01e-04 1.53e-01 -3.7 1.87e-01 -0.1 1.00e+00 1.00e+00h 1
7 1.9857419e+01 2.52e-03 1.52e-01 -4.6 5.72e-01 -0.6 1.00e+00 1.00e+00f 1
8 1.9319499e+01 5.35e-03 1.22e-01 -2.7 1.03e+00 -1.1 1.00e+00 5.20e-01f 1
9 1.8399038e+01 4.64e-02 7.26e-02 -2.4 1.40e+00 -1.5 1.00e+00 6.59e-01f 1
iter objective inf_pr inf_du lg(mu) ||d|| lg(rg) alpha_du alpha_pr ls
10 1.7193105e+01 3.05e+00 2.62e-01 -2.1 3.01e+00 -2.0 1.00e+00 8.31e-01f 1
11 1.5106279e+01 1.10e+01 3.82e-01 -2.2 1.36e+01 -2.5 1.70e-01 3.23e-01h 1
12 1.5675085e+01 2.15e+00 2.08e-01 -2.2 1.62e+00 -1.2 7.98e-01 1.00e+00h 1
13 1.5767117e+01 1.59e-01 7.09e-02 -2.2 7.59e-01 -0.7 1.00e+00 1.00e+00h 1
14 1.5167619e+01 5.39e-01 1.06e-01 -1.9 6.62e+00 - 1.00e+00 4.46e-01h 1
15 1.5207976e+01 1.55e-01 2.60e-01 -2.2 1.87e+00 - 1.00e+00 7.88e-01h 1
16 1.4958081e+01 2.78e-02 2.02e-02 -4.0 4.59e-01 - 1.00e+00 9.93e-01h 1
17 1.5012097e+01 4.19e-01 7.47e-03 -2.5 6.61e-01 - 1.00e+00 9.34e-01h 1
18 1.4993570e+01 2.41e-04 5.76e-04 -2.7 3.51e-01 - 1.00e+00 1.00e+00h 1
19 1.4814517e+01 9.34e-04 3.26e-03 -8.7 2.05e-01 - 1.00e+00 1.00e+00h 1
iter objective inf_pr inf_du lg(mu) ||d|| lg(rg) alpha_du alpha_pr ls
20 1.4796807e+01 5.17e-04 5.95e-05 -5.1 6.84e-02 - 1.00e+00 1.00e+00h 1
21 1.4796405e+01 1.26e-07 1.37e-08 -11.0 1.93e-03 - 1.00e+00 1.00e+00h 1
Number of Iterations....: 21
(scaled) (unscaled)
Objective...............: 1.4796404863783373e+01 1.4796404863783373e+01
Dual infeasibility......: 1.3697506295873296e-08 1.3697506295873296e-08
Constraint violation....: 1.2616821831734626e-07 1.2616821831734626e-07
Variable bound violation: 0.0000000000000000e+00 0.0000000000000000e+00
Complementarity.........: 1.3417720767463476e-08 1.3417720767463476e-08
Overall NLP error.......: 1.2616821831734626e-07 1.2616821831734626e-07
Number of objective function evaluations = 33
Number of objective gradient evaluations = 22
Number of equality constraint evaluations = 33
Number of inequality constraint evaluations = 33
Number of equality constraint Jacobian evaluations = 22
Number of inequality constraint Jacobian evaluations = 22
Number of Lagrangian Hessian evaluations = 21
Total seconds in IPOPT = 6.251
EXIT: Optimal Solution Found.
tf = variable(nlp_sol)
p = costate(nlp_sol)
p0 = p(0)
plot(nlp_sol)
Shooting (1/2), Tmax = 60 Newtons
function ur(x, p, tf) # Regular maximising control
H1 = p' * F1(x)
H2 = p' * F2(x)
H3 = p' * F3(x)
u = [H1, H2, H3]
u = u / sqrt(u[1]^2 + u[2]^2 + u[3]^2)
return u
end
fr = Flow(ocp, ur) # Regular flow (first version)
function shoot(ξ::Vector{T}) where T
tf, p0 = ξ[1], ξ[2:end]
xf_, pf = fr(0, x0, p0, tf)
s = zeros(T, 7)
s[1:5] = xf_[1:5] - xf[1:5]
s[6] = pf[6]
s[7] = p0[1]^2 + p0[2]^2 + p0[3]^2 + p0[4]^2 + p0[5]^2 + p0[6]^2 - 1
return s
end
p0 = p0 / norm(p0) # Normalization |p0|=1 for free final time
ξ = [tf; p0]; # Initial guess
jshoot(ξ) = ForwardDiff.jacobian(shoot, ξ)
shoot!(s, ξ) = (s[:] = shoot(ξ); nothing)
jshoot!(js, ξ) = (js[:] = jshoot(ξ); nothing)
bvp_sol = fsolve(shoot!, jshoot!, ξ; show_trace=true); println(bvp_sol)
Iter f(x) inf-norm Step 2-norm Step time
------ -------------- -------------- --------------
1 1.441506e+00 0.000000e+00 7.911273
2 7.828904e-02 5.822092e-05 14.433610
3 1.805789e-02 1.793255e-05 0.006112
4 4.039623e-03 6.129907e-07 0.005667
5 1.138619e-04 1.107407e-08 0.015535
6 6.538025e-05 1.359255e-10 0.005620
7 6.244845e-06 4.489006e-12 0.009841
8 3.534559e-08 5.342350e-14 0.005779
9 3.606218e-10 1.853859e-18 0.008501
Results of Nonlinear Solver Algorithm
* Algorithm: Modified Powell (User Jac, Expert)
* Starting Point: [14.796404863783373, -0.016164433702769165, -0.9050962222165153, -0.3306787679250943, -0.09648293225364692, 0.035148118586239945, 0.24626559970809675]
* Zero: [14.800364354664758, -0.016189714921626915, -0.9084889547974425, -0.3252598589830436, -0.09447217088704539, 0.03432296283542438, 0.24184432891069754]
* Inf-norm of residuals: 0.000000
* Convergence: true
* Message: algorithm estimates that the relative error between x and the solution is at most tol
* Total time: 22.402010 seconds
* Function Calls: 9
* Jacobian Calls (df/dx): 1
Shooting (2/2), Tmax = 0.7 Newtons
hr = (t, x, p) -> begin # Regular maximised Hamiltonian (more efficient)
H0 = p' * F0(x)
H1 = p' * F1(x)
H2 = p' * F2(x)
H3 = p' * F3(x)
mass = mass0 - β*T*t
h = H0 + T / mass * sqrt(H1^2 + H2^2 + H3^2)
return h
end
hr = Hamiltonian(hr; autonomous=false)
fr = Flow(hr) # Regular flow (again)
Tmax = 0.7 # Maximum thrust (Newtons)
cTmax = 3600^2 / 1e6; T = Tmax * cTmax # Conversion from Newtons to kg x Mm / h²
tf = 1.210e3; p0 =-[-2.215319700438820e+01, -4.347109477345140e+01, 9.613188807286992e-01, 3.181800985503019e+02, -2.307236094862410e+00, -5.797863110671591e-01] # Tmax = 0.7 Newtons
p0 = p0 / norm(p0) # Normalization |p0|=1 for free final time
ξ = [tf; p0]; # Initial guess
bvp_sol = fsolve(shoot!, jshoot!, ξ; show_trace=true); println(bvp_sol)
Iter f(x) inf-norm Step 2-norm Step time
------ -------------- -------------- --------------
1 1.073118e+00 0.000000e+00 4.371657
2 1.431619e-01 1.222060e+01 11.056950
3 1.090263e-01 1.104500e+01 0.289040
4 1.847681e-02 2.971575e+00 0.317280
5 1.996342e-02 4.468500e-01 0.262949
6 8.607891e-03 2.181855e-02 0.293898
7 2.826201e-03 2.618597e-03 0.267660
8 1.241794e-03 7.993872e-05 0.293262
9 1.555157e-03 6.602970e-05 0.289012
10 7.599961e-05 2.809120e-05 0.290977
11 4.206481e-05 1.773353e-07 0.297937
12 7.672444e-06 3.852977e-07 0.275044
13 5.441236e-07 7.260827e-09 0.265315
14 5.909087e-08 4.510506e-11 0.380645
15 2.813787e-08 2.110683e-13 0.664074
16 4.440955e-09 3.591076e-17 0.345580
Results of Nonlinear Solver Algorithm
* Algorithm: Modified Powell (User Jac, Expert)
* Starting Point: [1210.0, 0.06881811251625697, 0.13504139789537645, -0.0029863026489984875, -0.9884150724670144, 0.007167346236588738, 0.001801085395597836]
* Zero: [1214.5922195529836, 0.06933409656211512, 0.5161919854246422, -0.0002316690086500129, -0.8536506970530086, 0.004363238004614628, -9.327820837705058e-5]
* Inf-norm of residuals: 0.000000
* Convergence: true
* Message: algorithm estimates that the relative error between x and the solution is at most tol
* Total time: 19.961381 seconds
* Function Calls: 16
* Jacobian Calls (df/dx): 1
Plots
tf = bvp_sol.x[1]
p0 = bvp_sol.x[2:end]
ode_sol = fr((0, tf), x0, p0)
t = ode_sol.t; N = size(t, 1)
P = ode_sol[1, :]
ex = ode_sol[2, :]
ey = ode_sol[3, :]
hx = ode_sol[4, :]
hy = ode_sol[5, :]
L = ode_sol[6, :]
cL = cos.(L)
sL = sin.(L)
w = @. 1 + ex * cL + ey * sL
Z = @. hx * sL - hy * cL
C = @. 1 + hx^2 + hy^2
q1 = @. P *((1 + hx^2 - hy^2) * cL + 2 * hx * hy * sL) / (C * w)
q2 = @. P *((1 - hx^2 + hy^2) * sL + 2 * hx * hy * cL) / (C * w)
q3 = @. 2 * P * Z / (C * w)
plt1 = plot3d(1; xlim = (-60, 60), ylim = (-60, 60), zlim = (-5, 5), title = "Orbit transfer", legend=false)
@gif for i = 1:N
push!(plt1, q1[i], q2[i], q3[i])
end every N ÷ min(N, 100)

Reproducibility
The documentation of this package was built using these direct dependencies,
Status `~/work/Kepler.jl/Kepler.jl/docs/Project.toml`
[e30172f5] Documenter v1.10.1
⌅ [f6369f11] ForwardDiff v0.10.38
[459d104a] Kepler v0.2.0 `~/work/Kepler.jl/Kepler.jl`
[4854310b] MINPACK v1.3.0
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[1dea7af3] OrdinaryDiffEq v6.93.0
[91a5bcdd] Plots v1.40.13
[37e2e46d] LinearAlgebra v1.11.0
Info Packages marked with ⌅ have new versions available but compatibility constraints restrict them from upgrading. To see why use `status --outdated`
and using this machine and Julia version.
Julia Version 1.11.5
Commit 760b2e5b739 (2025-04-14 06:53 UTC)
Build Info:
Official https://julialang.org/ release
Platform Info:
OS: Linux (x86_64-linux-gnu)
CPU: 4 × AMD EPYC 7763 64-Core Processor
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A more complete overview of all dependencies and their versions is also provided.
Status `~/work/Kepler.jl/Kepler.jl/docs/Manifest.toml`
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Info Packages marked with ⌃ and ⌅ have new versions available. Those with ⌃ may be upgradable, but those with ⌅ are restricted by compatibility constraints from upgrading. To see why use `status --outdated -m`
You can also download the manifest file and the project file.
References
- 1Bonnard, B.; Caillau, J.-B.; Trélat, E. Geometric optimal control of elliptic Keplerian orbits. Discrete Contin. Dyn. Syst. Ser. B 5 (2005), no. 4, 929-956.
- 2Caillau, J.-B.; Gergaud, J.; Noailles, J. 3D Geosynchronous Transfer of a Satellite: continuation on the Thrust. J. Optim. Theory Appl. 118 (2003), no. 3, 541-565.