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Arthur DANJOU
f94ff07cab
- Standardized spacing around operators and function arguments in TP7_Kmeans.ipynb and neural_network.ipynb. - Enhanced the formatting of model building and training code in neural_network.ipynb for better clarity. - Updated the pyproject.toml to remove a specific TensorFlow version and added linting configuration for Ruff. - Improved comments and organization in the code to facilitate easier understanding and maintenance.
595 lines
20 KiB
Plaintext
595 lines
20 KiB
Plaintext
{
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"cells": [
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"# Computer session 2\n",
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"\n",
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"The goal of this computer class is to get a good feel of the Newton method and its\n",
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"variants. In a (maybe) surprising way, we actually start with the dichotomy method\n",
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"in the one-dimensional case.\n",
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"\n",
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"## The dichotomy method in the one-dimensional case\n",
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"\n",
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"When trying to solve the equation $\\phi(x) = 0$ in the one-dimensional case, the\n",
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"most naive method, which actually turns out to be quite efficient, is the dichotomy\n",
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"method. Namely, starting from an initial pair $(a_L , a_R ) \\in \\mathbb{R}^2$ with $a_L < a_R$ such\n",
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"that $\\phi(a_L)\\phi(a_R)<0$, we set $b ∶=\\frac{a_L+a_R}{2}$. If $\\phi(b) = 0$, the algorithm stops. If\n",
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"$\\phi(a_L)\\phi(b) < 0$ we set $a_L\\to a_L$ and $a_R \\to b$. In this way, we obtain a linearly\n",
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"converging algorithm. In particular, it is globally converging.\n",
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"\n",
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"\n",
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"Write a function `Dichotomy(phi,aL,aR,eps)` that take sas argument\n",
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"a function `phi`, an initial guess `aL,aR` and a tolerance `eps` and that runs the dichotomy\n",
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"algorithm. Your argument should check that the condition $\\phi(a_L)\\phi(a_R) < 0$ is satisfied,\n",
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"stop when the function `phi` reaches a value lower than `eps` and return the number\n",
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"of iteration. Run your algorithm on the function $f = tanh$ with initial guesses\n",
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"$a_L = −20$ , $a_R = 3$.\n",
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"\n"
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]
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},
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{
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"cell_type": "code",
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"execution_count": 1,
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"metadata": {
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"ExecuteTime": {
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"end_time": "2025-03-18T16:19:14.314484Z",
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"start_time": "2025-03-18T16:19:13.728014Z"
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}
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},
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"outputs": [
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{
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"name": "stdout",
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"output_type": "stream",
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"text": [
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"(-2.3283064365386963e-09, 31)\n"
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]
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}
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],
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"source": [
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"import numpy as np\n",
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"\n",
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"\n",
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"def dichotomy(phi, aL, aR, eps=1e-8):\n",
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" iter = 0\n",
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" b = (aL + aR) / 2\n",
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" while (aR - aL) / 2 > eps and phi(b) != 0:\n",
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" b = (aL + aR) / 2\n",
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" if phi(aL) * phi(b) < 0:\n",
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" aR = b\n",
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" else:\n",
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" aL = b\n",
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" iter += 1\n",
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" return b, iter\n",
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"\n",
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"\n",
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"f = lambda x: np.tanh(x)\n",
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"aL, aR = -20, 3\n",
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"print(dichotomy(f, aL, aR))"
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]
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},
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"## Solving one-dimensional equation with the Newton and the secant method\n",
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"\n",
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"We work again in the one-dimensional case with a function φ we want to find the\n",
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"zeros of.\n",
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"\n",
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"### Newton method\n",
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"\n",
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"Write a function `Newton(phi,dphi,x0,eps)` that takes, as arguments, a function\n",
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"`phi`, its derivative `dphi`, an initial guess `x0` and a tolerance `eps`\n",
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"and that returns an approximation of the solutions of the equation $\\phi(x) = 0$. The\n",
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"tolerance criterion should again be that $|\\phi| ≤\\text{\\texttt{eps}}$. Your\n",
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"algorithm should return an error message in the following cases:\n",
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"1. If the derivative is zero (look up the `try` and `except` commands in Python).\n",
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"2. If the method diverges.\n",
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"\n",
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"Apply this code to the minimisation of $x\\mapsto \\ln(e^x + e^{−x})$, with initial condition `x0=1.8`.\n",
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"Compare this with the results of Exercise 3.10."
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]
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},
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{
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"cell_type": "code",
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"execution_count": 2,
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"metadata": {
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"ExecuteTime": {
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"end_time": "2025-03-18T16:19:17.447647Z",
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"start_time": "2025-03-18T16:19:17.442560Z"
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}
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},
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"outputs": [
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{
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"name": "stdout",
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"output_type": "stream",
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"text": [
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"(inf, 'Method diverges')\n"
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]
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},
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{
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"name": "stderr",
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"output_type": "stream",
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"text": [
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"/var/folders/tp/_ld5_pzs6nx6mv1pbjhq1l740000gn/T/ipykernel_25957/3868809151.py:14: RuntimeWarning: overflow encountered in exp\n",
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" f = lambda x: np.log(np.exp(x) + np.exp(-x))\n"
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]
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}
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],
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"source": [
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"def Newton(phi, dphi, x0, eps=1e-10):\n",
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" iter = 0\n",
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" while abs(phi(x0)) > eps:\n",
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" try:\n",
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" x0 -= phi(x0) / dphi(x0)\n",
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" except ZeroDivisionError:\n",
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" return np.inf, \"Derivative is zero\"\n",
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" iter += 1\n",
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" if iter > 10000 or phi(x0) == np.inf:\n",
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" return np.inf, \"Method diverges\"\n",
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" return x0, iter\n",
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"\n",
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"\n",
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"f = lambda x: np.log(np.exp(x) + np.exp(-x))\n",
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"x0 = 1.8\n",
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"df = lambda x: (np.exp(x) - np.exp(-x)) / (np.exp(x) + np.exp(-x))\n",
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"print(Newton(f, df, x0))"
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]
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},
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"### Secant method\n",
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"\n",
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"Write a function `Secant(phi,x0,x1,eps)` that takes, as arguments, a function `phi`, two initial positions `x0`, `x1` and a tolerance\n",
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"`eps` and that returns an approximation of the solutions of the equation\n",
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"$\\phi(x) = 0$. The tolerance criterion should again be that $|\\phi| ≤\\text{\\texttt{eps}}$. Apply this code to the minimisation\n",
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"of $x\\mapsto \\ln(e^x + e^{−x})$, with initial conditions `x0=1`, `x1=1.9`, then\n",
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"`x0=1`, `x1=2.3` and\n",
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"`x0=1`, `x1=2.4`. Compare with the results of Exercise 3.10."
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]
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},
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{
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"cell_type": "code",
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"execution_count": 3,
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"metadata": {
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"ExecuteTime": {
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"end_time": "2025-03-18T16:19:19.649523Z",
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"start_time": "2025-03-18T16:19:19.456149Z"
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}
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},
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"outputs": [
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{
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"name": "stdout",
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"output_type": "stream",
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"text": [
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"(inf, 'Method diverges')\n",
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"(inf, 'Method diverges')\n",
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"(inf, 'Method diverges')\n"
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]
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}
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],
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"source": [
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"def Secant(phi, x0, x1, eps=1e-8):\n",
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" iter = 0\n",
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" while abs(phi(x1)) > eps:\n",
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" x1, x0 = x1 - phi(x1) * (x1 - x0) / (phi(x1) - phi(x0)), x0\n",
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" iter += 1\n",
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" if iter > 10000:\n",
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" return np.inf, \"Method diverges\"\n",
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" return x0, iter\n",
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"\n",
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"\n",
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"f = lambda x: np.log(np.exp(x) + np.exp(-x))\n",
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"xx = [(1, 1.9), (1, 2.3), (1, 2.4)]\n",
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"\n",
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"for x0, x1 in xx:\n",
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" print(Secant(f, x0, x1))"
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]
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},
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"## Combining dichotomy and the Newton method\n",
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"\n",
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"A possibility to leverage the advantages of dichotomy (the global convergence of\n",
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"the method) and of the Newton method (the quadratic convergence rate) is to\n",
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"combine both: start from an initial interval `[aL,aR]` of length `InitialLength`\n",
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"with $\\phi(a_L)\\phi(a_R)<0$ and fix a real\n",
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"number $s \\in [0; 1]$. Run the dichotomy algorithm until the new interval is of length\n",
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"`s*InitialLength`. From this point on, apply the Newton method.\n",
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"\n",
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"Implement this algorithm with `s = 0.1`. Include a possibility to switch\n",
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"back to the dichotomy method if, when switching to the Newton method, the new\n",
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"iterate falls outside of the computed interval `[aL,aR]`. Apply this to the minimisation\n",
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"of the function $f ∶ x\\mapsto \\ln(e^x + e^{−x})$ with an initial condition that made the Newton\n",
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"method diverge. What can you say about the number of iterations?"
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]
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},
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{
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"cell_type": "code",
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"execution_count": 22,
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"metadata": {
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"ExecuteTime": {
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"end_time": "2025-03-18T16:44:41.592150Z",
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"start_time": "2025-03-18T16:44:41.584318Z"
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}
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},
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"outputs": [
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{
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"name": "stdout",
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"output_type": "stream",
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"text": [
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"(inf, 'Method diverges')\n"
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]
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},
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{
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"name": "stderr",
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"output_type": "stream",
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"text": [
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"/var/folders/tp/_ld5_pzs6nx6mv1pbjhq1l740000gn/T/ipykernel_25957/1578277506.py:23: RuntimeWarning: overflow encountered in exp\n",
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" f = lambda x: np.log(np.exp(x) + np.exp(-x))\n"
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]
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}
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],
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"source": [
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"def DichotomyNewton(phi, dphi, aL, aR, s=0.1, eps=1e-10):\n",
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" iter = 0\n",
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" inital_length = aR - aL\n",
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" while (aR - aL) >= s * inital_length:\n",
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" b = (aL + aR) / 2\n",
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" if phi(aL) * phi(b) < 0:\n",
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" aR = b\n",
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" else:\n",
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" aL = b\n",
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" iter += 1\n",
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" x0 = (aL + aR) / 2\n",
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" while abs(phi(x0)) > eps:\n",
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" try:\n",
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" x0 -= phi(x0) / dphi(x0)\n",
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" except ZeroDivisionError:\n",
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" return np.inf, \"Derivative is zero\"\n",
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" iter += 1\n",
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" if iter > 10000 or phi(x0) == np.inf:\n",
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" return np.inf, \"Method diverges\"\n",
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" return x0, iter\n",
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"\n",
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"\n",
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"f = lambda x: np.log(np.exp(x) + np.exp(-x))\n",
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"df = lambda x: (np.exp(x) - np.exp(-x)) / (np.exp(x) + np.exp(-x))\n",
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"print(DichotomyNewton(f, df, -20, 3))"
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]
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},
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"## Solving an optimisation problem using the Newton method\n",
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"\n",
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"An island (denoted by a point $I$ below) is situated 2 kilometers from the shore (its projection on the shore\n",
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"is a point $P$). A guest staying at a nearby hotel $H$ wants to go from the hotel to the\n",
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"island and decides that he will run at 8km/hr for a distance $x$, before swimming at\n",
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"speed 3km/hr to reach the island.\n",
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"\n",
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"\n",
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"\n",
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"Taking into account the fact that there are 6 kilometers between the hotel $H$ and $P$,\n",
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"how far should the visitor run before swimming?\n",
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"\n",
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"Model the situation as a minimisation problem, and solve it numerically.\n",
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"Compare the efficiency of the dichotomy method and of the Newton algorithm."
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]
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},
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{
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"cell_type": "code",
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"execution_count": 38,
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"metadata": {
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"ExecuteTime": {
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"end_time": "2025-03-18T17:43:43.061916Z",
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"start_time": "2025-03-18T17:43:43.042625Z"
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}
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},
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"outputs": [
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{
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"name": "stdout",
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"output_type": "stream",
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"text": [
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"Optimal point (Newton): 0.9299901531755377\n",
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"Objective function value at optimal point (Newton): 1.9329918821224974\n",
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"Number of iterations (Newton): 100\n",
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"Optimal point (Dichotomy): inf\n",
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"Objective function value at optimal point (Dichotomy): inf\n",
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"Number of iterations (Dichotomy): Method diverges\n"
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]
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}
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],
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"source": [
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"\n",
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"u = lambda x: np.sqrt((6 - x) ** 2 + 4)\n",
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"\n",
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"\n",
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"def objective_function(x):\n",
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" return x / 8 + u(x) / 3\n",
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"\n",
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"\n",
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"def gradient(x):\n",
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" return 1 / 8 + (6 - x) / (3 * u(x))\n",
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"\n",
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"\n",
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"def hessian(x):\n",
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" return (12 * u(x) - (2 * x - 12) ** 2 / 12 * u(x)) / 36 * u(x) ** 2\n",
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"\n",
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"\n",
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"# Newton's method for optimization\n",
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"def newton_method(initial_guess, tolerance=1e-6, max_iterations=100):\n",
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" x = initial_guess\n",
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" iterations = 0\n",
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" while iterations < max_iterations:\n",
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" grad = gradient(x)\n",
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" hess = hessian(x)\n",
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" if np.abs(grad) < tolerance:\n",
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" break\n",
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" x -= grad / hess\n",
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" iterations += 1\n",
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" return x, iterations\n",
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"\n",
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"\n",
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"# Dichotomy method for optimization\n",
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"def dichotomy_method(aL, aR, eps=1e-6, max_iterations=1000):\n",
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" iterations = 0\n",
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" x0 = (aL + aR) / 2\n",
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" grad = gradient(x0)\n",
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" hess = hessian(x0)\n",
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" while abs(grad) > eps:\n",
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" try:\n",
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" x0 -= grad / hess\n",
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" except ZeroDivisionError:\n",
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" return np.inf, \"Derivative is zero\"\n",
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" iterations += 1\n",
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" if iterations > max_iterations or grad == np.inf:\n",
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" return np.inf, \"Method diverges\"\n",
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" grad = gradient(x0)\n",
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" hess = hessian(x0)\n",
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" return x0, iterations\n",
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"\n",
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"\n",
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"# Initial guess for Newton's method\n",
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"initial_guess_newton = 4\n",
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"\n",
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"# Run Newton's method\n",
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"optimal_point_newton, iterations_newton = newton_method(initial_guess_newton)\n",
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"print(f\"Optimal point (Newton): {optimal_point_newton}\")\n",
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"print(\n",
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" f\"Objective function value at optimal point (Newton): {objective_function(optimal_point_newton)}\"\n",
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")\n",
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"print(f\"Number of iterations (Newton): {iterations_newton}\")\n",
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"\n",
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"# Initial interval for dichotomy method\n",
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"aL, aR = 0, 6\n",
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"\n",
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"# Run dichotomy method\n",
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"optimal_point_dichotomy, iterations_dichotomy = dichotomy_method(aL, aR)\n",
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"print(f\"Optimal point (Dichotomy): {optimal_point_dichotomy}\")\n",
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"print(\n",
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" f\"Objective function value at optimal point (Dichotomy): {objective_function(optimal_point_dichotomy)}\"\n",
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")\n",
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"print(f\"Number of iterations (Dichotomy): {iterations_dichotomy}\")"
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]
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},
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"## The Newton method to solve boundary value problems\n",
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"\n",
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"### Shooting method\n",
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"\n",
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"We consider the following non-linear ODE\n",
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"\n",
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"\\begin{equation}\n",
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"y''= f(x,y,y'),\\quad x\\in[a,b],\\quad y(a)=\\alpha, y(b)=\\beta.\n",
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"\\end{equation}\n",
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"\n",
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"\n",
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"To classically integrate such an ODE, we usually don’t have endpoints for $y$,\n",
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"but initial values for $y$ and $y'$. So, we cannot start at $x=a $ and integrate\n",
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"up to $x=b$. This is a _boundary value problem_.\n",
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"\n",
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"One approach is to approximate $y$ by somme finite difference and then arrive at\n",
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"a system for the discrete values $y(x_i)$ and finally solve large linear\n",
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"systems.\n",
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"\n",
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"Here, we will see how we can formulate the problem as a _shooting_ method, and\n",
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"use Newton method so solve it.\n",
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"\n",
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"The idea is to use a _guess_ for the initial value of $y'$. Let $s$ be a\n",
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"parameter for a fonction $y(\\;\\cdot\\;;s)$ solution of (1) such that\n",
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"$$y(a;s)=\\alpha,\\text{ and }y'(a;s)=s.$$\n",
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"\n",
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"There is no chance that $y(b;s)=y(b)=\\beta$ but we can adjust the value of $s$,\n",
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"refining the guess until it is (nearly equal to) the right value.\n",
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"\n",
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"This method is known as _shooting_ method in analogy to shooting a ball at a\n",
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"goal, determining the unknown correct velocity by throwing it too fast/too\n",
|
||
"slow until it hits the goal exactly.\n",
|
||
"\n",
|
||
"#### In Practice\n",
|
||
"\n",
|
||
"For the parameter $s$, we integrate the following ODE:\n",
|
||
"\n",
|
||
"\\begin{equation}\n",
|
||
"y''= f(x,y,y'),\\quad x\\in[a,b],\\quad y(a)=\\alpha, y'(a)=s.\n",
|
||
"\\end{equation}\n",
|
||
"\n",
|
||
"We denote $y(\\;\\cdot\\;;s)$ solution of (2).\n",
|
||
"\n",
|
||
"Let us now define the _goal function_. Here, we want that $y(b;s)=\\beta$, hence,\n",
|
||
"we define:\n",
|
||
"$$g:s\\mapsto \\left.y(x;s)\\right|_{x=b}-\\beta$$\n",
|
||
"\n",
|
||
"We seek $s^*$ such that $g(s^*)=0$.\n",
|
||
"\n",
|
||
"Note that computing $g(s)$ involves the integration of an ODE, so each\n",
|
||
"evaluation of $g$ is expensive. Newton’s method seems then to be a good\n",
|
||
"way due to its fast convergence.\n",
|
||
"\n",
|
||
"To be able to code a Newton’s method, we need to compute the derivative of $g$.\n",
|
||
"For this purpose, let define\n",
|
||
"$$z(x;s)=\\frac{\\partial y(x;s)}{\\partial s}.$$\n",
|
||
"\n",
|
||
"Then by differentiating (2) with respect to $s$, we get\n",
|
||
"$$z''=\\frac{\\partial f}{\\partial y}z+\\frac{\\partial f}{\\partial y'}z',\\quad\n",
|
||
"z(a;s)=0,\\text{ and }z'(a;s)=1.$$\n",
|
||
"\n",
|
||
"The derivative of $g$ can now be expressed in term of $z$:\n",
|
||
"$$g'(z)=z(b;s).$$\n",
|
||
"\n",
|
||
"Putting this together, we can code the Newton's method:\n",
|
||
"$$s_{n+1}=s_n-\\frac{g(s_n)}{g'(s_n)}.$$\n",
|
||
"\n",
|
||
"To sum up, a shooting method requires an ODE solver and a Newton solver.\n",
|
||
"\n",
|
||
"#### Example\n",
|
||
"\n",
|
||
"Apply this method to\n",
|
||
"$$y''=2y^3-6y-2x^3,\\quad y(1)=2,y(2)=5/2,$$\n",
|
||
"with standard library for integration, and your own Newton implementation.\n",
|
||
"\n",
|
||
"Note that you may want to express this with one order ODE. Moreover, it may be\n",
|
||
"simpler to solve only one ODE for both $g$ and $g'$.\n",
|
||
"\n",
|
||
"With python, you can use `scipy.integrate.solve_ivp` function:\n",
|
||
"https://docs.scipy.org/doc/scipy/reference/generated/scipy.integrate.solve_ivp.html#scipy.integrate.solve_ivp\n",
|
||
"\n",
|
||
"Plot the solution $y$.\n",
|
||
"\n",
|
||
"For numerical parameters, compute the solution up to a precision of $10^{-8}$\n",
|
||
"and get the function on a grid of 1000 points.\n",
|
||
"\n",
|
||
"## Finite differences\n",
|
||
"\n",
|
||
"Here, we are going to use a different approach to solve the boundary value\n",
|
||
"problem:\n",
|
||
"\n",
|
||
"\\begin{equation}\n",
|
||
"y''= f(x,y,y'),\\quad x\\in[a,b],\\quad y(a)=\\alpha, y(b)=\\beta.\n",
|
||
"\\end{equation}\n",
|
||
"\n",
|
||
"This problem can be solved by the following direct process:\n",
|
||
"\n",
|
||
"1. We discretize the domain choosing an integer $N$, grid points $\\{x_n\\}_{n=0,\\dots,N}$ and\n",
|
||
" we define the discrete solution $\\{y_n\\}_{n=0,\\dots,N}$.\n",
|
||
"1. We discretize the ODE using derivative approximation with finite differences\n",
|
||
" in the interior of the domain.\n",
|
||
"1. We inject the boundary conditions (here $y_0=\\alpha$ and $y_N=\\beta$) in the\n",
|
||
" discretized ODE.\n",
|
||
"1. Solve the system of equation for the unknows $\\{y_n\\}_{n=1,\\dots,N-1}$.\n",
|
||
"\n",
|
||
"We use here a uniform grid :\n",
|
||
"$$h:=(b-a)/N, \\quad \\forall n=0,\\dots, N\\quad x_n=hn.$$\n",
|
||
"If we use a centered difference formula for $y''$ and $y'$, we obtain:\n",
|
||
"$$\\forall n=1,\\dots,N-1,\\quad \\frac{y_{n+1}-2y_n+y_{n-1}}{h^2}=f\\left(x_n,y_n,\\frac{y_{n+1}-y_{n-1}}{2h}\\right).$$"
|
||
]
|
||
},
|
||
{
|
||
"cell_type": "markdown",
|
||
"metadata": {},
|
||
"source": [
|
||
"The result is a system of equations for $\\mathbf{y}=(y_1,\\dots,y_{N-1})$ :\n",
|
||
"$$G(\\mathbf{y})=0,\\quad G:\\mathbb{R}^{N-1}\\to\\mathbb{R}^{N-1}.$$"
|
||
]
|
||
},
|
||
{
|
||
"cell_type": "markdown",
|
||
"metadata": {},
|
||
"source": [
|
||
"This system can be solved using Newton's method. Note that the Jacobian\n",
|
||
"$\\partial G/\\partial \\mathbb{y}$ is _tridiagonal_.\n",
|
||
"\n",
|
||
"Of course, here, we are in the multidimensional context, so you will have to\n",
|
||
"code a Newton algorithm well suited.\n",
|
||
"\n",
|
||
"#### Example\n",
|
||
"\n",
|
||
"Apply this method to\n",
|
||
"$$y''=2y^3-6y-2x^3,\\quad y(1)=2,y(2)=5/2.$$\n",
|
||
"\n",
|
||
"Plot the solution $y$.\n",
|
||
"\n",
|
||
"For numerical parameters, compute the solution up to a precision of $10^{-8}$\n",
|
||
"and get the function on a grid of 1000 points."
|
||
]
|
||
},
|
||
{
|
||
"cell_type": "markdown",
|
||
"metadata": {},
|
||
"source": [
|
||
"**Remark:** In the context of numerical optimal control, these two numerical\n",
|
||
"methods are often called _indirect method_ (for the shooting method) and _direct\n",
|
||
"method_ (for the finite difference method)."
|
||
]
|
||
},
|
||
{
|
||
"cell_type": "markdown",
|
||
"metadata": {},
|
||
"source": [
|
||
"## Who’s the best?\n",
|
||
"\n",
|
||
"Compare the two methods playing with parameters (grid discretization, precision,\n",
|
||
"initialization, etc.). Mesure the time computation."
|
||
]
|
||
},
|
||
{
|
||
"cell_type": "markdown",
|
||
"metadata": {},
|
||
"source": []
|
||
},
|
||
{
|
||
"cell_type": "code",
|
||
"execution_count": null,
|
||
"metadata": {},
|
||
"outputs": [],
|
||
"source": []
|
||
},
|
||
{
|
||
"cell_type": "code",
|
||
"execution_count": null,
|
||
"metadata": {},
|
||
"outputs": [],
|
||
"source": []
|
||
},
|
||
{
|
||
"cell_type": "code",
|
||
"execution_count": null,
|
||
"metadata": {},
|
||
"outputs": [],
|
||
"source": []
|
||
}
|
||
],
|
||
"metadata": {
|
||
"kernelspec": {
|
||
"display_name": "base",
|
||
"language": "python",
|
||
"name": "python3"
|
||
},
|
||
"language_info": {
|
||
"codemirror_mode": {
|
||
"name": "ipython",
|
||
"version": 3
|
||
},
|
||
"file_extension": ".py",
|
||
"mimetype": "text/x-python",
|
||
"name": "python",
|
||
"nbconvert_exporter": "python",
|
||
"pygments_lexer": "ipython3",
|
||
"version": "3.12.2"
|
||
}
|
||
},
|
||
"nbformat": 4,
|
||
"nbformat_minor": 4
|
||
}
|