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@@ -6,7 +6,7 @@ Cours de contrôle optimal de l'ENSEEIHT pour le département Sciences du Numér
**CTD**
-* [Notes de cours - contrôle optimal](https://gitlab.irit.fr/toc/etu-n7/controle-optimal/-/raw/master/notes-co-2024.pdf)
+* [Notes de cours - contrôle optimal](https://gitlab.irit.fr/toc/etu-n7/controle-optimal/-/raw/master/notes-co-2025.pdf)
**TP**
@@ -24,10 +24,6 @@ Calcul différentiel et équation différentielle ordinaire
* [Sujet 4 : tp/derivative.ipynb](tp/derivative.ipynb) - Calcul numérique de dérivées
* [Sujet 5 : tp/runge-kutta.ipynb](tp/runge-kutta.ipynb) - Méthodes de Runge-Kutta
-Projet
-
-* [Projet : tp/transfert-orbital.ipynb](tp/transfert-orbital.ipynb) - Transfert orbital
-
**Notes de cours supplémentaires - ressources**
* [Automatique](https://gitlab.irit.fr/toc/etu-n7/controle-optimal/-/raw/master/ressources/notes-autom-2021.pdf)
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-{
- "cells": [
- {
- "attachments": {},
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "[<img src=\"https://gitlab.irit.fr/toc/etu-n7/controle-optimal/-/raw/master/ressources/Logo-toulouse-inp-N7.png\" alt=\"N7\" height=\"100\"/>](https://gitlab.irit.fr/toc/etu-n7/controle-optimal)"
- ]
- },
- {
- "attachments": {},
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<div style=\"width:90%;\n",
- " margin:10px;\n",
- " padding:8px;\n",
- " background-color:#FAA299;\n",
- " border:2px solid #BF381B;\n",
- " border-radius:20px;\n",
- " font-weight:bold;\n",
- " font-size:1em;\"> <!-- il faut laisser une ligne vide après celle-ci -->\n",
- "\n",
- "* Nom :\n",
- "* Prénom :\n",
- "</div>"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "# Transfert orbital\n",
- "\n",
- "Ce sujet (ou TP-projet) est à rendre (voir la date sur moodle et les modalités du rendu) et sera évalué pour faire partie de la note finale de la matière Contrôle Optimal."
- ]
- },
- {
- "attachments": {},
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<div style=\"width:90%;\n",
- " margin:10px;\n",
- " padding:8px;\n",
- " background-color:#afa;\n",
- " border:2px solid #bbffbb;\n",
- " border-radius:20px;\n",
- " font-weight:bold;\n",
- " font-size:1.5em;\n",
- " text-align:center;\">\n",
- "Transfert orbital à temps minimal\n",
- "</div>\n",
- "\n",
- "\n",
- "On considère le problème du transfert d'un satellite d'une orbite initiale à l'orbite géostationnaire à temps minimal. Ce problème s'écrit comme un problème de contrôle optimal sous la forme\n",
- "\n",
- "$$\n",
- "\\left\\lbrace\n",
- "\\begin{array}{l}\n",
- " \\min J(x, u, t_f) = t_f \\\\[1.0em]\n",
- " \\ \\ \\dot{x}_{1}(t) = ~ x_{3}(t) \\\\[0.5em]\n",
- " \\ \\ \\dot{x}_{2}(t) = ~ x_{4}(t) \\\\[0.5em]\n",
- " \\ \\ \\dot{x}_{3}(t) = -\\dfrac{\\mu\\, x_{1}(t)}{r^{3}(x(t))} + u_{1}(t) \\\\[1em]\n",
- " \\ \\ \\dot{x}_{4}(t) = -\\dfrac{\\mu\\, x_{2}(t)}{r^{3}(x(t))} + u_{2}(t), ~~ ||u(t)|| \\leq \\gamma_\\mathrm{max}, ~~ t \\in [0,t_f] ~~ \\text{p.p.}, ~~ u(t) = (u_1(t),u_2(t)), \\\\[1.5em]\n",
- " \\ \\ x_{1}(0)=x_{0,1}, ~~ x_{2}(0)=x_{0,2}, ~~ x_{3}(0)=x_{0,3}, ~~ x_{4}(0)=x_{0,4}, \\\\[1em]\n",
- " \\ \\ r(x(t_f)) = r_{f}, ~~ x_{3}(t_f)=-\\sqrt{\\dfrac{\\mu}{r_{f}^{3}}}x_{2}(t_f), ~~ x_{4} (t_f)= \\sqrt{\\dfrac{\\mu}{r_{f}^{3}}}x_{1}(t_f), \\\\\n",
- "\\end{array}\n",
- "\\right.\n",
- "$$\n",
- "\n",
- "avec $r(x)=\\sqrt{x_{1}^{2}+x_{2}^{2}}$.\n",
- "Les unités choisies sont le kilomètre pour les distances et l'heure pour les temps. On donne les paramètres suivants :\n",
- " \n",
- "$$\n",
- "\\mu=5.1658620912 \\times 10^{12} \\ \\mathrm{km}^{3}.\\mathrm{h}^{-2}, \\quad r_{f} = 42165 \\ \\mathrm{km}.\n",
- "$$\n",
- "\n",
- "Le paramètre $\\gamma_\\mathrm{max}$ dépend de la poussée maximale $F_\\mathrm{max}$ suivant la relation :\n",
- "\n",
- "$$\n",
- "\\gamma_\\mathrm{max} = \\frac{F_\\mathrm{max}\\times 3600^2}{m} \n",
- "$$\n",
- "\n",
- "où m est la masse du satellite qu'on fixe à $m=2000\\ \\mathrm{kg}$."
- ]
- },
- {
- "attachments": {},
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "### Résolution via du tir simple indirect"
- ]
- },
- {
- "cell_type": "code",
- "execution_count": null,
- "metadata": {},
- "outputs": [],
- "source": [
- "using DifferentialEquations, NLsolve, ForwardDiff, Plots, LinearAlgebra\n",
- "include(\"utils.jl\"); # fonctions utilitaires"
- ]
- },
- {
- "cell_type": "code",
- "execution_count": null,
- "metadata": {},
- "outputs": [],
- "source": [
- "# Les constantes du pb\n",
- "x0 = [-42272.67, 0, 0, -5796.72] # état initial\n",
- "μ = 5.1658620912*1e12\n",
- "rf = 42165\n",
- "F_max = 100\n",
- "γ_max = F_max*3600^2/(2000*10^3)\n",
- "t0 = 0\n",
- "rf3 = rf^3\n",
- "α = sqrt(μ/rf3);"
- ]
- },
- {
- "attachments": {},
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "✏️ **_Question 1:_**\n",
- " \n",
- "1. Donner le pseudo-hamiltonien associé au problème de contrôle optimal.\n",
- "2. Donner le pseudo système hamiltonien\n",
- "$$\n",
- " \\vec{H}(x, p, u) = \\left(\\frac{\\partial H}{\\partial p}(x, p, u), \n",
- " -\\frac{\\partial H}{\\partial x}(x, p, u) \\right).\n",
- "$$\n",
- "3. Calculer le contrôle maximisant. On supposera que $(p_3, p_4)\\neq (0,0)$."
- ]
- },
- {
- "attachments": {},
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "**Réponse** (double cliquer pour modifier le texte)\n",
- "\n",
- "<div style=\"width:90%;\n",
- " margin:10px;\n",
- " padding:20px;\n",
- " background-color:#CFF3F7;\n",
- " border:2px solid #063970;\n",
- " border-radius:20px;\n",
- " font-weight:bold;\n",
- " font-size:1em;\"> <!-- il faut laisser une ligne vide après celle-ci -->\n",
- "\n",
- "Ecrire la réponse ici\n",
- "</div>"
- ]
- },
- {
- "attachments": {},
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "✏️ **_Question 2:_** Compléter le code suivant contenant le pseudo-hamiltonien, le pseudo système hamiltonien et le contrôle maximisant."
- ]
- },
- {
- "cell_type": "code",
- "execution_count": null,
- "metadata": {},
- "outputs": [],
- "source": [
- "#####\n",
- "##### A COMPLETER\n",
- "\n",
- "# pseudo-Hamiltonien\n",
- "function H(x, p, u)\n",
- " h = 0\n",
- " return h\n",
- "end\n",
- "\n",
- "# pseudo système hamiltonien\n",
- "function Hv(x, p, u)\n",
- " n = size(x, 1)\n",
- " dx = zeros(eltype(x), n)\n",
- " dp = zeros(eltype(x), n)\n",
- " return dx, dp\n",
- "end\n",
- "\n",
- "# Contrôle maximisant\n",
- "function control(p)\n",
- " u = zeros(eltype(p),2)\n",
- " return u\n",
- "end\n",
- "\n",
- "#####\n",
- "##### FIN A COMPLETER\n",
- "\n",
- "# flot hamiltonien pour le calcul des extrémales\n",
- "f = Flow((x, p) -> Hv(x, p, control(p)));"
- ]
- },
- {
- "attachments": {},
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "On note \n",
- "$\n",
- " \\alpha := \\sqrt{\\frac{\\mu}{r_f^3}}.\n",
- "$\n",
- "La condition terminale peut se mettre sous la forme $c(x(t_f)) = 0$, avec $c \\colon \\mathbb{R}^4 \\to \\mathbb{R}^3$.\n",
- "\n",
- "✏️ **_Question 3:_** Donner l'expression de $c(x)$."
- ]
- },
- {
- "attachments": {},
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "**Réponse** (double cliquer pour modifier le texte)\n",
- "\n",
- "<div style=\"width:90%;\n",
- " margin:10px;\n",
- " padding:20px;\n",
- " background-color:#CFF3F7;\n",
- " border:2px solid #063970;\n",
- " border-radius:20px;\n",
- " font-weight:bold;\n",
- " font-size:1em;\"> <!-- il faut laisser une ligne vide après celle-ci -->\n",
- "\n",
- "Ecrire la réponse ici\n",
- "</div>"
- ]
- },
- {
- "attachments": {},
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "Le temps final étant libre, on a la condition au temps final \n",
- "\n",
- "$$\n",
- " H(x(t_f), p(t_f), u(t_f)) = -p^0 = 1. \\quad \\text{(on se place dans le cas normal)}\n",
- "$$\n",
- "\n",
- "De plus, la condition de transversalité \n",
- "\n",
- "$$\n",
- "p(t_f) = c'(x(t_f))^T \\lambda, ~~ \\lambda \\in \\mathbb{R}^3,\n",
- "$$\n",
- "\n",
- "conduit à la relation suivante (où $\\lambda$ n'apparaît plus)\n",
- "\n",
- "$$\n",
- "\\Phi(x(t_f), p(t_f)) := x_2(t_f) \\Big( p_1(t_f) + \\alpha\\, p_4(t_f) \\Big) - x_1(t_f) \\Big( p_2(t_f) - \\alpha\\, p_3(t_f) \\Big) = 0.\n",
- "$$\n",
- "\n",
- "En considérant la condition aux limites, la condition finale sur le pseudo-hamiltonien et la condition de transversalité, la fonction de tir simple est donnée par \n",
- "\n",
- "\\begin{equation*}\n",
- " \\begin{array}{rlll}\n",
- " S \\colon & \\mathbb{R}^5 & \\longrightarrow & \\mathbb{R}^5 \\\\\n",
- " & (p_0, t_f) & \\longmapsto &\n",
- " S(p_0, t_f) := \\begin{pmatrix}\n",
- " c(x(t_f, x_0, p_0)) \\\\[0.5em]\n",
- " \\Phi(z(t_f, x_0, p_0)) \\\\[0.5em]\n",
- " H(z(t_f, x_0, p_0), u(z(t_f, x_0, p_0))) - 1\n",
- " \\end{pmatrix}\n",
- " \\end{array}\n",
- "\\end{equation*}\n",
- "\n",
- "où $z(t_f, x_0, p_0)$ est la solution au temps $t_f$ du pseudo système hamiltonien bouclé par\n",
- "le contrôle maximisant, partant au temps $t_0=0$ du point $(x_0, p_0)$. On rappelle que l'on note\n",
- "$z=(x, p)$."
- ]
- },
- {
- "attachments": {},
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "✏️ **_Question 4:_** Compléter le code suivant de la fonction de tir."
- ]
- },
- {
- "cell_type": "code",
- "execution_count": null,
- "metadata": {},
- "outputs": [],
- "source": [
- "#####\n",
- "##### A COMPLETER\n",
- "\n",
- "# Fonction de tir\n",
- "function shoot(p0, tf)\n",
- " \n",
- " s = zeros(eltype(p0), 5)\n",
- "\n",
- " ...\n",
- " \n",
- " return s\n",
- "\n",
- "end;"
- ]
- },
- {
- "attachments": {},
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "✏️ **_Question 5:_** Expliquer simplement ce que fait le code suivant."
- ]
- },
- {
- "attachments": {},
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "**Réponse** (double cliquer pour modifier le texte)\n",
- "\n",
- "<div style=\"width:90%;\n",
- " margin:10px;\n",
- " padding:20px;\n",
- " background-color:#CFF3F7;\n",
- " border:2px solid #063970;\n",
- " border-radius:20px;\n",
- " font-weight:bold;\n",
- " font-size:1em;\"> <!-- il faut laisser une ligne vide après celle-ci -->\n",
- "\n",
- "Ecrire la réponse ici\n",
- "</div>"
- ]
- },
- {
- "cell_type": "code",
- "execution_count": null,
- "metadata": {},
- "outputs": [],
- "source": [
- "# Itéré initial\n",
- "y_guess = [1.0323e-4, 4.915e-5, 3.568e-4, -1.554e-4, 13.4] # pour F_max = 100N\n",
- "\n",
- "# Jacobienne de la fonction de tir\n",
- "foo(y) = shoot(y[1:4], y[5])\n",
- "jfoo(y) = ForwardDiff.jacobian(foo, y)\n",
- "\n",
- "# Résolution de shoot(p0, tf) = 0\n",
- "nl_sol = nlsolve(foo, jfoo, y_guess; xtol=1e-8, method=:trust_region, show_trace=true);\n",
- "\n",
- "# On récupère la solution si convergence\n",
- "if converged(nl_sol)\n",
- " p0_sol_100 = nl_sol.zero[1:4]\n",
- " tf_sol_100 = nl_sol.zero[5]\n",
- " println(\"\\nFinal solution:\\n\", nl_sol.zero)\n",
- "else\n",
- " error(\"Not converged\")\n",
- "end"
- ]
- },
- {
- "cell_type": "code",
- "execution_count": null,
- "metadata": {},
- "outputs": [],
- "source": [
- "# Fonction d'affichage d'une solution\n",
- "function plot_solution(p0, tf)\n",
- "\n",
- " # On trace l'orbite de départ et d'arrivée\n",
- " gr(dpi=300, size=(500,400), thickness_scaling=1)\n",
- " r0 = norm(x0[1:2])\n",
- " v0 = norm(x0[3:4])\n",
- " a = 1.0/(2.0/r0-v0*v0/μ)\n",
- " t1 = r0*v0*v0/μ - 1.0;\n",
- " t2 = (x0[1:2]'*x0[3:4])/sqrt(a*μ);\n",
- " e_ellipse = norm([t1 t2])\n",
- " p_orb = a*(1-e_ellipse^2);\n",
- " n_theta = 101\n",
- " Theta = range(0, stop=2*pi, length=n_theta)\n",
- " X1_orb_init = zeros(n_theta)\n",
- " X2_orb_init = zeros(n_theta)\n",
- " X1_orb_arr = zeros(n_theta)\n",
- " X2_orb_arr = zeros(n_theta)\n",
- "\n",
- " # Orbite initiale\n",
- " for i in 1:n_theta\n",
- " theta = Theta[i]\n",
- " r_orb = p_orb/(1+e_ellipse*cos(theta));\n",
- " X1_orb_init[i] = r_orb*cos(theta);\n",
- " X2_orb_init[i] = r_orb*sin(theta);\n",
- " end\n",
- "\n",
- " # Orbite d'arrivée\n",
- " for i in 1:n_theta\n",
- " theta = Theta[i]\n",
- " X1_orb_arr[i] = rf*cos(theta) ;\n",
- " X2_orb_arr[i] = rf*sin(theta);\n",
- " end;\n",
- "\n",
- " # Calcul de la trajectoire\n",
- " ode_sol = f((t0, tf), x0, p0)\n",
- " t = ode_sol.t\n",
- " x1 = [ode_sol[1, j] for j in 1:size(t, 1) ]\n",
- " x2 = [ode_sol[2, j] for j in 1:size(t, 1) ]\n",
- " u = zeros(2, length(t))\n",
- "\n",
- " for j in 1:size(t, 1)\n",
- " u[:,j] = control(ode_sol[5:8, j])\n",
- " end\n",
- "\n",
- " px = plot(x1, x2, color=:red, legend=:best, label=\"Trajectoire\",\n",
- " border=:none, size = (800,400), aspect_ratio=:equal)\n",
- " plot!(px, X1_orb_init, X2_orb_init, color=:green, label=\"Orbite initiale\")\n",
- " plot!(px, X1_orb_arr, X2_orb_arr, color=:blue, label=\"Orbite d'arrivée\")\n",
- " plot!(px, [x0[1]], [x0[2]], seriestype=:scatter, color=:green, markersize = 5, label=\"Départ\")\n",
- " xf = ode_sol[1:2, end]\n",
- " plot!(px, [xf[1]], [xf[2]], seriestype=:scatter, color=:red, markersize = 5, label=\"Arrivée\")\n",
- " plot!(px, [0.0], [0.0], color = :blue, seriestype=:scatter, markersize = 25, label=\"Terre\")\n",
- "\n",
- " pu1 = plot(t, u[1,:], color=:red, label=\"u₁(t)\", legend=:best)\n",
- " pu2 = plot(t, u[2,:], color=:red, label=\"u₂(t)\", legend=:best)\n",
- "\n",
- " display(plot(pu1, pu2, layout = (1,2), size = (800,400)))\n",
- " display(px)\n",
- " \n",
- "end;"
- ]
- },
- {
- "cell_type": "code",
- "execution_count": null,
- "metadata": {},
- "outputs": [],
- "source": [
- "# On affiche la solution pour Fmax = 100\n",
- "plot_solution(p0_sol_100, tf_sol_100)"
- ]
- },
- {
- "attachments": {},
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "✏️ **_Question 6:_** Que vaut le temps final $t_f$ (à 5 digits près) ? Combien de révolutions complètes autour de la Terre a réalisé le satellite."
- ]
- },
- {
- "attachments": {},
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "**Réponse** (double cliquer pour modifier le texte)\n",
- "\n",
- "<div style=\"width:90%;\n",
- " margin:10px;\n",
- " padding:20px;\n",
- " background-color:#CFF3F7;\n",
- " border:2px solid #063970;\n",
- " border-radius:20px;\n",
- " font-weight:bold;\n",
- " font-size:1em;\"> <!-- il faut laisser une ligne vide après celle-ci -->\n",
- "\n",
- "Ecrire la réponse ici\n",
- "</div>"
- ]
- },
- {
- "attachments": {},
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "### Etude du temps de transfert en fonction de la poussée maximale\n",
- "\n",
- "✏️ **_Question 7:_** \n",
- "\n",
- "* A l'aide de ce que vous avez fait précédemment, déterminer $t_f$ (attention, penser à bien itinialiser la\n",
- "méthode de tir) pour\n",
- " \n",
- "$$\n",
- " F_\\mathrm{max} \\in \\{100, 90, 80, 70, 60, 50, 40, 30, 20 \\}.\n",
- "$$\n",
- " \n",
- "* Tracer ensuite $t_f$ en fonction de $F_\\mathrm{max}$ et commenter le résultat."
- ]
- },
- {
- "attachments": {},
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "**Réponse** (double cliquer pour modifier le texte et donner votre commentaire)\n",
- "\n",
- "<div style=\"width:90%;\n",
- " margin:10px;\n",
- " padding:20px;\n",
- " background-color:#CFF3F7;\n",
- " border:2px solid #063970;\n",
- " border-radius:20px;\n",
- " font-weight:bold;\n",
- " font-size:1em;\"> <!-- il faut laisser une ligne vide après celle-ci -->\n",
- "\n",
- "Ecrire la réponse ici\n",
- "</div>"
- ]
- },
- {
- "cell_type": "code",
- "execution_count": null,
- "metadata": {},
- "outputs": [],
- "source": [
- "# Les différentes valeurs de poussées\n",
- "F_max_span = range(100, stop=20, length=11);\n",
- "γ_max_span = [F_max_span[j]*3600^2/(2000*10^3) for j in 1:size(F_max_span,1)];"
- ]
- },
- {
- "cell_type": "code",
- "execution_count": null,
- "metadata": {},
- "outputs": [],
- "source": [
- "# Solution calculée précédemment\n",
- "y_guess = [p0_sol_100; [tf_sol_100]]\n",
- "\n",
- "# Pour le stockage des solutions\n",
- "tf_sols = zeros(length(γ_max_span)) # vecteur des temps de transfert\n",
- "p0_sols = zeros(4, length(γ_max_span)) # matrice des co-états initiaux\n",
- "\n",
- "#####\n",
- "##### A COMPLETER\n",
- "\n",
- "...\n"
- ]
- },
- {
- "cell_type": "code",
- "execution_count": null,
- "metadata": {},
- "outputs": [],
- "source": [
- "# Affichage de tf en fonction de la poussée maximale\n",
- "plot(F_max_span, tf_sols, aspect_ratio=:equal, legend=:best, label=\"tf(Fmax)\")"
- ]
- },
- {
- "cell_type": "code",
- "execution_count": null,
- "metadata": {},
- "outputs": [],
- "source": [
- "# Plots sol pour F_max = 20N\n",
- "i = 11\n",
- "γ_max = γ_max_span[i]\n",
- "plot_solution(p0_sols[:, i], tf_sols[i])"
- ]
- },
- {
- "attachments": {},
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "### Animation\n",
- "\n",
- "Juste pour s'amuser"
- ]
- },
- {
- "cell_type": "code",
- "execution_count": null,
- "metadata": {},
- "outputs": [],
- "source": [
- "include(\"space.jl\"); using .Space"
- ]
- },
- {
- "cell_type": "code",
- "execution_count": null,
- "metadata": {},
- "outputs": [],
- "source": [
- "# solution choice\n",
- "i = 11\n",
- "γ_max = γ_max_span[i] # γ_max must be updated for the use of the flow\n",
- "\n",
- "# animation\n",
- "# nFrame = 100; d = 30; fps = floor(Int, nFrame/d) => 2 minutes and 30 seconds of computation\n",
- "nFrame = 100; d = 30; fps = floor(Int, nFrame/d)\n",
- "Space.animation(p0_sols[:, i], tf_sols[i], f, γ_max, nFrame=nFrame, fps=fps)"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<div style=\"width:90%;\n",
- " margin:10px;\n",
- " padding:8px;\n",
- " background-color:#afa;\n",
- " border:2px solid #bbffbb;\n",
- " border-radius:20px;\n",
- " font-weight:bold;\n",
- " font-size:1.5em;\n",
- " text-align:center;\">\n",
- "Transfert orbital à consommation minimale\n",
- "</div>"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "On considère maintenant le problème du transfert d'un satellite d'une orbite initiale à une orbite géostationnaire à temps final fixé en cherchant à minimiser la consommation de carburant. Ce problème s'écrit comme un problème de contrôle optimal sous la forme\n",
- "\n",
- "$$\n",
- "\\left\\lbrace\n",
- "\\begin{array}{l}\n",
- " \\displaystyle \\min J(x, u) = \\int_{0}^{t_f} \\Vert u(t)\\Vert \\mathrm{d}t \\\\[1.0em]\n",
- " \\ \\ \\dot{x}_{1}(t) = ~ x_{3}(t) \\\\[0.5em]\n",
- " \\ \\ \\dot{x}_{2}(t) = ~ x_{4}(t) \\\\[0.5em]\n",
- " \\ \\ \\dot{x}_{3}(t) = -\\dfrac{\\mu\\, x_{1}(t)}{r^{3}(x(t))} + u_{1}(t) \\\\[1em]\n",
- " \\ \\ \\dot{x}_{4}(t) = -\\dfrac{\\mu\\, x_{2}(t)}{r^{3}(x(t))} + u_{2}(t), ~~ ||u(t)|| \\leq \\gamma_\\mathrm{max}, ~~ t \\in [0,t_f] ~~ \\text{p.p.}, ~~ u(t) = (u_1(t),u_2(t)), \\\\[1.5em]\n",
- " \\ \\ x_{1}(0)=x_{0,1}, ~~ x_{2}(0)=x_{0,2}, ~~ x_{3}(0)=x_{0,3}, ~~ x_{4}(0)=x_{0,4}, \\\\[1em]\n",
- " \\ \\ r(x(t_f)) = r_{f}, ~~ x_{3}(t_f)=-\\sqrt{\\dfrac{\\mu}{r_{f}^{3}}}x_{2}(t_f), ~~ x_{4} (t_f)= \\sqrt{\\dfrac{\\mu}{r_{f}^{3}}}x_{1}(t_f). \\\\\n",
- "\\end{array}\n",
- "\\right.\n",
- "$$\n",
- "\n",
- "Toutes les constantes sont identiques au problème précédent. On considéra le problème avec $F_\\mathrm{max} = 100$ et on prendre comme temps final : $t_f = 1.5 T^{100}_{\\min}$, où $T^{100}_{\\min}$ correspond au temps minimal solution du problème précédent pour $F_\\mathrm{max} = 100$.\n",
- "\n",
- "**Remarques importantes.** \n",
- "\n",
- "* On ne considèrera que des extrémales normales, c'est-à-dire $p^0 = -1$.\n",
- "* Le problème a de la structure, le contrôle optimal est composé d'arcs où la poussée est nulle $u(t) = 0$ et où la poussée est maximale $\\Vert u(t)\\Vert = \\gamma_\\mathrm{max}$.\n",
- "* Vous devrez dans un premier temps résoudre le problème avec le coût suivant ce qui permet de le régulariser et ainsi d'utiliser une méthode de tir simple :\n",
- "\n",
- "$$\n",
- "\\int_{0}^{t_f} \\left( \\Vert u(t)\\Vert - \\varepsilon \\left( \\ln(\\Vert u(t)\\Vert) + \\ln(\\gamma_\\mathrm{max}-\\Vert u(t)\\Vert) \\right) \\right) \\mathrm{d}t.\n",
- "$$"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "✏️ **_Question 8:_** Résoudre via du tir simple le problème régularisé pour différentes valeurs de $\\varepsilon$ suffisamment petites pour déterminer la structure optimale, en commençant avec $\\varepsilon$ suffisamment grand pour faire converger la méthode de tir plus facilement."
- ]
- },
- {
- "cell_type": "code",
- "execution_count": null,
- "metadata": {},
- "outputs": [],
- "source": []
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "✏️ **_Question 9:_** Résoudre via du tir multiple le problème de transfert orbital à consommation minimale. Vous utilisez la structure obtenue précédemment et vous donnerez au solveur une bonne condition initiale à l'aide de la question précédente."
- ]
- },
- {
- "cell_type": "code",
- "execution_count": null,
- "metadata": {},
- "outputs": [],
- "source": []
- }
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