{
"cells": [
{
"cell_type": "code",
"execution_count": 1,
"metadata": {},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Python version 3.10.6 (main, Nov 14 2022, 16:10:14) [GCC 11.3.0]\n"
]
}
],
"source": [
"import sys #only needed to determine Python version number\n",
"print('Python version ' + sys.version) "
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"# M62 TP 3 - Implémentation de schémas\n",
"\n",
"Compléter le notebook en ajoutant l'implémentation des schémas indiqués et en vérifiant l'impléméntation sur l'exemple donné."
]
},
{
"cell_type": "markdown",
"metadata": {
"toc": true
},
"source": [
"Table of Contents
\n",
""
]
},
{
"cell_type": "markdown",
"metadata": {
"colab_type": "text",
"id": "WyG-bTRQE3f6",
"slideshow": {
"slide_type": "slide"
}
},
"source": [
"## Implémentation des schémas"
]
},
{
"cell_type": "markdown",
"metadata": {
"slideshow": {
"slide_type": "fragment"
}
},
"source": [
"On écrit les schémas numériques : \n",
"+ les $N+1$ nœuds d'intégration $[t_0,t_1,\\dots,t_{N}]$ sont contenus dans le vecteur `tt`\n",
"+ les $N+1$ valeurs $[u_0,u_1,\\dots,u_{N}]$ pour chaque méthode sont contenues dans le vecteur `uu`.\n",
"\n",
"Comme `len(tt)` $=N+1$ et `range(1,N)` produit 1,2,3,N-1 : \n",
"- pour un schéma à un pas $u_{n+1}=F(u_n)$ on initialise $u_0$ et on calcule $u_{n+1}$ pour $n$ de $0$ jusqu'à $N-1$, autrement dit `n in range(N)` soit encore `n in range(len(tt)-1)`\n",
"- pour un schéma à deux pas $u_{n+1}=F(u_n,u_{n-1})$ on initialise $u_0$ et $u_1$ et on calcule $u_{n+1}$ pour $n$ de $1$ jusqu'à $N-1$, autrement dit `n in range(1,N)` soit encore `n in range(1,len(tt)-1)`\n",
"- pour un schéma à trois pas $u_{n+1}=F(u_n,u_{n-1},u_{n-2})$ on initialise $u_0$, $u_1$ et $u_2$ et on calcule $u_{n+1}$ pour $n$ de $2$ jusqu'à $N-1$, autrement dit `n in range(2,N)` soit encore `n in range(2,len(tt)-1)`\n",
"- etc.\n"
]
},
{
"cell_type": "code",
"execution_count": 2,
"metadata": {},
"outputs": [
{
"data": {
"application/javascript": [
"IPython.notebook.set_autosave_interval(300000)"
]
},
"metadata": {},
"output_type": "display_data"
},
{
"name": "stdout",
"output_type": "stream",
"text": [
"Autosaving every 300 seconds\n"
]
}
],
"source": [
"%reset -f\n",
"%matplotlib inline\n",
"%autosave 300\n",
"\n",
"from matplotlib.pylab import *\n",
"from scipy.optimize import fsolve"
]
},
{
"cell_type": "markdown",
"metadata": {
"slideshow": {
"slide_type": "skip"
}
},
"source": [
"### Schémas explicites"
]
},
{
"cell_type": "markdown",
"metadata": {
"colab_type": "text",
"id": "y69SGZjfIDo9",
"slideshow": {
"slide_type": "slide"
}
},
"source": [
"#### Schéma d'Euler progressif = de Adam-Bashforth à 1 pas\n",
"$$\n",
"\\begin{cases}\n",
"u_0=y_0,\\\\\n",
"u_{n+1}=u_n+h\\varphi(t_n,u_n)& n=0,1,2,\\dots N-1\n",
"\\end{cases}\n",
"$$"
]
},
{
"cell_type": "code",
"execution_count": 3,
"metadata": {
"colab": {
"autoexec": {
"startup": false,
"wait_interval": 0
}
},
"colab_type": "code",
"id": "_Bgo6mNyIQgu",
"scrolled": true,
"slideshow": {
"slide_type": "fragment"
}
},
"outputs": [],
"source": [
"def EE(phi,tt,y0):\n",
" h = tt[1]-tt[0]\n",
" uu = [y0]\n",
" for i in range(len(tt)-1):\n",
" uu.append(uu[i]+h*phi(tt[i],uu[i]))\n",
" return uu"
]
},
{
"cell_type": "markdown",
"metadata": {
"colab_type": "text",
"id": "bJ2pbhejIQM2",
"slideshow": {
"slide_type": "slide"
}
},
"source": [
"#### Schéma de Adam-Bashforth à 2 pas\n",
"$$\n",
"\\begin{cases}\n",
"u_0=y_0,\\\\\n",
"u_{1}=u_0+h\\varphi(t_0,u_0),\\\\\n",
"u_{n+1}=u_n+\\frac{h}{2}\\Bigl(3\\varphi(t_n,u_n)-\\varphi(t_{n-1},u_{n-1})\\Bigr)& n=1,2,3,4,5,\\dots N-1\n",
"\\end{cases}\n",
"$$"
]
},
{
"cell_type": "markdown",
"metadata": {
"colab_type": "text",
"id": "nI8swyc6RxIR",
"slideshow": {
"slide_type": "slide"
}
},
"source": [
"#### Schéma de Adam-Bashforth à 3 pas\n",
"$$\n",
"\\begin{cases}\n",
"u_0=y_0,\\\\\n",
"u_{1}=u_0+h\\varphi(t_0,u_0),\\\\\n",
"u_{2}=u_1+\\frac{h}{2}\\Bigl(3\\varphi(t_1,u_1)-\\varphi(t_{0},u_{0})\\Bigr),\\\\\n",
"u_{n+1}=u_n+\\frac{h}{12}\\Bigl(23\\varphi(t_n,u_n)-16\\varphi(t_{n-1},u_{n-1})+5\\varphi(t_{n-2},u_{n-2})\\Bigr)& n=2,3,4,5,\\dots N-1\n",
"\\end{cases}\n",
"$$"
]
},
{
"cell_type": "markdown",
"metadata": {
"colab_type": "text",
"id": "awcWzBp7SXvQ",
"slideshow": {
"slide_type": "slide"
}
},
"source": [
"#### Schéma de Adam-Bashforth à 4 pas\n",
"$$\n",
"\\begin{cases}\n",
"u_0=y_0,\\\\\n",
"u_{1}=u_0+h\\varphi(t_0,u_0),\\\\\n",
"u_{2}=u_1+\\frac{h}{2}\\Bigl(3\\varphi(t_1,u_1)-\\varphi(t_{0},u_{0})\\Bigr),\\\\\n",
"u_{3}=u_2+\\frac{h}{12}\\Bigl(23\\varphi(t_2,u_2)-16\\varphi(t_{1},u_{1})+5\\varphi(t_{0},u_{0})\\Bigr),\\\\\n",
"u_{n+1}=u_n+\\frac{h}{24}\\Bigl(55\\varphi(t_n,u_n)-59\\varphi(t_{n-1},u_{n-1})+37\\varphi(t_{n-2},u_{n-2})-9\\varphi(t_{n-3},u_{n-3})\\Bigr)& n=3,4,5,\\dots N-1\n",
"\\end{cases}\n",
"$$"
]
},
{
"cell_type": "markdown",
"metadata": {
"colab_type": "text",
"id": "heLmvMe_S0y6",
"slideshow": {
"slide_type": "slide"
}
},
"source": [
"#### Schéma de Adam-Bashforth à 5 pas\n",
"$$\n",
"\\begin{cases}\n",
"u_0=y_0,\\\\\n",
"u_{1}=u_0+h\\varphi(t_0,u_0),\\\\\n",
"u_{2}=u_1+\\frac{h}{2}\\Bigl(3\\varphi(t_1,u_1)-\\varphi(t_{0},u_{0})\\Bigr),\\\\\n",
"u_{3}=u_2+\\frac{h}{12}\\Bigl(23\\varphi(t_2,u_2)-16\\varphi(t_{1},u_{1})+5\\varphi(t_{0},u_{0})\\Bigr),\\\\\n",
"u_{4}=u_3+\\frac{h}{24}\\Bigl(55\\varphi(t_3,u_3)-59\\varphi(t_{2},u_{2})+37\\varphi(t_{1},u_{1})-9\\varphi(t_{0},u_{0})\\Bigr),\\\\\n",
"u_{n+1}=u_n+\\frac{h}{720}\\Bigl(1901\\varphi(t_n,u_n)-2774\\varphi(t_{n-1},u_{n-1})+2616\\varphi(t_{n-2},u_{n-2})-1274\\varphi(t_{n-3},u_{n-3})+251\\varphi(t_{n-4},u_{n-4})\\Bigr)& n=4,5,\\dots N-1\n",
"\\end{cases}\n",
"$$"
]
},
{
"cell_type": "markdown",
"metadata": {
"colab_type": "text",
"id": "OldEmxFfTJfq",
"slideshow": {
"slide_type": "slide"
}
},
"source": [
"#### Schéma de Nylström à 2 pas\n",
"$$\n",
"\\begin{cases}\n",
"u_0=y_0,\\\\\n",
"u_{1}=u_0+h\\varphi(t_0,u_0),\\\\\n",
"u_{n+1}=u_{n-1}+2h\\varphi(t_{n},u_{n})& n=1,2,3,4,5,\\dots N-1\n",
"\\end{cases}\n",
"$$"
]
},
{
"cell_type": "markdown",
"metadata": {
"colab_type": "text",
"id": "dADQEhyYTVQz",
"slideshow": {
"slide_type": "slide"
}
},
"source": [
"#### Schéma de Nylström à 3 pas\n",
"$$\n",
"\\begin{cases}\n",
"u_0=y_0,\\\\\n",
"u_{1}=u_0+h\\varphi(t_0,u_0),\\\\\n",
"u_{2}=u_{0}+2h\\varphi(t_{1},u_{1}),\\\\\n",
"u_{n+1}=u_{n-1}+\\frac{h}{3}\\Bigl(7\\varphi(t_{n},u_{n})-2\\varphi(t_{n-1},u_{n-1})+\\varphi(t_{n-2},u_{n-2})\\Bigr)& n=2,3,4,5,\\dots N-1\n",
"\\end{cases}\n",
"$$"
]
},
{
"cell_type": "markdown",
"metadata": {
"colab_type": "text",
"id": "LsgdqQnfTf66",
"slideshow": {
"slide_type": "slide"
}
},
"source": [
"#### Schéma de Nylström à 4 pas\n",
"$$\n",
"\\begin{cases}\n",
"u_0=y_0,\\\\\n",
"u_{1}=u_0+h\\varphi(t_0,u_0),\\\\\n",
"u_{2}=u_{0}+2h\\varphi(t_{1},u_{1}),\\\\\n",
"u_{3}=u_{1}+\\frac{h}{3}\\Bigl(7\\varphi(t_{2},u_{2})-2\\varphi(t_{1},u_{1})+\\varphi(t_{0},u_{0})\\Bigr),\\\\\n",
"u_{n+1}=u_{n-1}+\\frac{h}{3}\\Bigl(8\\varphi(t_{n},u_{n})-5\\varphi(t_{n-1},u_{n-1})+4\\varphi(t_{n-2},u_{n-2})-\\varphi(t_{n-3},u_{n-3})\\Bigr)& n=3,4,5,\\dots N-1\n",
"\\end{cases}\n",
"$$"
]
},
{
"cell_type": "markdown",
"metadata": {
"colab_type": "text",
"id": "p4f0txAsIwNG",
"slideshow": {
"slide_type": "slide"
}
},
"source": [
"#### Schéma d'Euler modifié\n",
"$$\n",
"\\begin{cases}\n",
"u_0=y_0,\\\\\n",
"\\tilde u = u_n+\\frac{h}{2}\\varphi(t_n,u_n),\\\\\n",
"u_{n+1}=u_n+h\\varphi\\left(t_n+\\frac{h}{2},\\tilde u\\right)& n=0,1,2,\\dots N-1\n",
"\\end{cases}\n",
"$$"
]
},
{
"cell_type": "markdown",
"metadata": {
"colab_type": "text",
"id": "_50Xo95mT9tC",
"slideshow": {
"slide_type": "slide"
}
},
"source": [
"#### Schéma de Runke-Kutta RK4-1\n",
"$$\\begin{cases}\n",
"u_0\t = y_0 \\\\\n",
"K_1 = \\varphi\\left(t_n,u_n\\right)\\\\\n",
"K_2 = \\varphi\\left(t_n+\\frac{h}{2},u_n+\\frac{h}{2} K_1)\\right)\\\\\n",
"K_3 = \\varphi\\left(t_n+\\frac{h}{2},u_n+\\frac{h}{2}K_2\\right)\\\\\n",
"K_4 = \\varphi\\left(t_{n+1},u_n+h K_3\\right)\\\\\n",
"u_{n+1} = u_n + \\frac{h}{6}\\left(K_1+2K_2+2K_3+K_4\\right) & n=0,1,\\dots N-1\n",
"\\end{cases}\n",
"$$"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"### Schémas implicites"
]
},
{
"cell_type": "markdown",
"metadata": {
"colab_type": "text",
"id": "y69SGZjfIDo9",
"slideshow": {
"slide_type": "slide"
}
},
"source": [
"#### Schéma d'Euler régressif\n",
"$$\n",
"\\begin{cases}\n",
"u_0=y_0,\\\\\n",
"u_{n+1}=u_n+h\\varphi(t_{n+1},u_{n+1})& n=0,1,2,\\dots N-1\n",
"\\end{cases}\n",
"$$\n",
"avec $u_{n+1}$ zéro de la fonction $$x\\mapsto -x+u_n+h\\varphi(t_{n+1},x)$$"
]
},
{
"cell_type": "code",
"execution_count": 4,
"metadata": {},
"outputs": [],
"source": [
"def EI(phi,tt,y0):\n",
" h = tt[1]-tt[0]\n",
" uu = [y0]\n",
" for i in range(len(tt)-1):\n",
" temp = fsolve(lambda x: -x+uu[i]+h*phi(tt[i+1],x), uu[i])\n",
" uu.append(temp[0])\n",
" return uu"
]
},
{
"cell_type": "markdown",
"metadata": {
"colab_type": "text",
"id": "23PyYGzhQwuo",
"slideshow": {
"slide_type": "slide"
}
},
"source": [
"#### Schéma de Crank-Nicolson (AM-1)\n",
"$$\n",
"\\begin{cases}\n",
"u_0=y_0,\\\\\n",
"u_{n+1}=u_n+\\frac{h}{2}\\Bigl(\\varphi(t_n,u_n)+\\varphi(t_{n+1},u_{n+1})\\Bigr)& n=0,1,2,\\dots N-1\n",
"\\end{cases}\n",
"$$\n",
"avec $u_{n+1}$zéro de la fonction $$x\\mapsto -x+u_n+\\frac{h}{2}(\\varphi(t_n,u_n)+\\varphi(t_{n+1},x))$$"
]
},
{
"cell_type": "markdown",
"metadata": {
"colab_type": "text",
"id": "23PyYGzhQwuo",
"slideshow": {
"slide_type": "slide"
}
},
"source": [
"#### Schéma de AM-2\n",
"$$\n",
"\\begin{cases}\n",
"u_0=y_0,\\\\\n",
"u_1=u_0+\\frac{h}{2}\\Bigl(\\varphi(t_1,u_1)+\\varphi(t_{0},u_{0})\\Bigr),\\\\\n",
"u_{n+1}=u_n+\\frac{h}{12}\\Bigl(5\\varphi(t_{n+1},u_{n+1})+8\\varphi(t_n,u_n)-\\varphi(t_{n-1},u_{n-1})\\Bigr)& n=1,2,\\dots N-1\n",
"\\end{cases}\n",
"$$"
]
},
{
"cell_type": "markdown",
"metadata": {
"colab_type": "text",
"id": "23PyYGzhQwuo",
"slideshow": {
"slide_type": "slide"
}
},
"source": [
"#### Schéma de AM-3\n",
"$$\n",
"\\begin{cases}\n",
"u_0=y_0,\\\\\n",
"u_1=u_0+\\frac{h}{2}\\Bigl(\\varphi(t_1,u_1)+\\varphi(t_{0},u_{0})\\Bigr),\\\\\n",
"u_{2}=u_1+\\frac{h}{12}\\Bigl(5\\varphi(t_{2},u_{2})+8\\varphi(t_1,u_1)-\\varphi(t_{0},u_{0})\\Bigr),\\\\\n",
"u_{n+1}=u_n+\\frac{h}{24}\\Bigl(9\\varphi(t_{n+1},u_{n+1})+19\\varphi(t_n,u_n)-5\\varphi(t_{n-1},u_{n-1})+\\varphi(t_{n-2},u_{n-2})\\Bigr)& n=2,3,\\dots N-1\n",
"\\end{cases}\n",
"$$"
]
},
{
"cell_type": "markdown",
"metadata": {
"colab_type": "text",
"id": "23PyYGzhQwuo",
"slideshow": {
"slide_type": "slide"
}
},
"source": [
"#### Schéma de AM-4\n",
"$$\n",
"\\begin{cases}\n",
"u_0=y_0,\\\\\n",
"u_1=u_0+\\frac{h}{2}\\Bigl(\\varphi(t_1,u_1)+\\varphi(t_{0},u_{0})\\Bigr),\\\\\n",
"u_{2}=u_1+\\frac{h}{12}\\Bigl(5\\varphi(t_{2},u_{2})+8\\varphi(t_1,u_1)-\\varphi(t_{0},u_{0})\\Bigr),\\\\\n",
"u_{3}=u_2+\\frac{h}{24}\\Bigl(9\\varphi(t_{3},u_{3})+19\\varphi(t_2,u_2)-5\\varphi(t_{1},u_{1})+\\varphi(t_{0},u_{0})\\Bigr),\\\\\n",
"u_{n+1}=u_n+\\frac{h}{720}\\Bigl(251\\varphi(t_{n+1},u_{n+1})+646\\varphi(t_n,u_n)-264\\varphi(t_{n-1},u_{n-1})+106\\varphi(t_{n-2},u_{n-2})-19\\varphi(t_{n-3},u_{n-3})\\Bigr)& n=3,4,\\dots N-1\n",
"\\end{cases}\n",
"$$"
]
},
{
"cell_type": "markdown",
"metadata": {
"colab_type": "text",
"id": "23PyYGzhQwuo",
"slideshow": {
"slide_type": "slide"
}
},
"source": [
"#### Schéma BDF2\n",
"$$\n",
"\\begin{cases}\n",
"u_0=y_0,\\\\\n",
"u_1=u_0+h\\varphi(t_1,u_1),\\\\\n",
"u_{n+1}=\\frac{4}{3}u_n-\\frac{1}{3}u_{n-1}+\\frac{2}{3}h\\varphi(t_{n+1},u_{n+1})& n=1,2,\\dots N-1\n",
"\\end{cases}\n",
"$$"
]
},
{
"cell_type": "markdown",
"metadata": {
"colab_type": "text",
"id": "23PyYGzhQwuo",
"slideshow": {
"slide_type": "slide"
}
},
"source": [
"#### Schéma BDF3\n",
"$$\n",
"\\begin{cases}\n",
"u_0=y_0,\\\\\n",
"u_1=u_0+h\\varphi(t_1,u_1),\\\\\n",
"u_{2}=\\frac{4}{3}u_1-\\frac{1}{3}u_{0}+\\frac{2}{3}h\\varphi(t_{2},u_{2}),\\\\\n",
"u_{n+1}=\\frac{18}{11}u_n-\\frac{9}{11}u_{n-1}+\\frac{2}{11}u_{n-2}+\\frac{6}{11}h\\varphi(t_{n+1},u_{n+1})& n=2,3,\\dots N-1\n",
"\\end{cases}\n",
"$$"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"#### Schéma RK1_M\n",
"$$\n",
"\\begin{cases}\n",
"u_0=y_0,\\\\\n",
"u_{n+1}=u_n+h\\varphi\\left(\\frac{t_n+t_{n+1}}{2},\\frac{u_n+u_{n+1}}{2}\\right)& n=0,1,2,\\dots N-1\n",
"\\end{cases}\n",
"$$"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"### Schémas predicteur-correcteur"
]
},
{
"cell_type": "markdown",
"metadata": {
"colab_type": "text",
"id": "23PyYGzhQwuo",
"slideshow": {
"slide_type": "slide"
}
},
"source": [
"#### Schéma de Heun\n",
"$$\n",
"\\begin{cases}\n",
"u_0=y_0,\\\\\n",
"\\tilde u = u_n+h\\varphi(t_n,u_n)\\\\\n",
"u_{n+1}=u_n+\\frac{h}{2}\\Bigl(\\varphi(t_n,u_n)+\\varphi(t_{n+1},\\tilde u)\\Bigr)& n=0,1,2,\\dots N-1\n",
"\\end{cases}\n",
"$$"
]
},
{
"cell_type": "markdown",
"metadata": {
"colab_type": "text",
"id": "23PyYGzhQwuo",
"slideshow": {
"slide_type": "slide"
}
},
"source": [
"#### Schéma AM-2 AB-1\n",
"$$\n",
"\\begin{cases}\n",
"u_0=y_0,\\\\\n",
"u_1=u_0+h\\varphi(t_0,u_0),\\\\\n",
"\\tilde u=u_n+h\\varphi(t_n,u_n),\\\\\n",
"u_{n+1}=u_n+\\frac{h}{12}\\Bigl(5\\varphi(t_{n+1},\\tilde u)+8\\varphi(t_n,u_n)-\\varphi(t_{n-1},u_{n-1})\\Bigr)& n=1,2,\\dots N-1\n",
"\\end{cases}\n",
"$$"
]
},
{
"cell_type": "markdown",
"metadata": {
"colab_type": "text",
"id": "23PyYGzhQwuo",
"slideshow": {
"slide_type": "slide"
}
},
"source": [
"#### Schéma AM-3 AB-2\n",
"$$\n",
"\\begin{cases}\n",
"u_0=y_0,\\\\\n",
"u_1=u_0+h\\varphi(t_0,u_0),\\\\\n",
"u_2=u_0+\\frac{h}{2}(3\\varphi(t_1,u_1)-\\varphi(t_{0},u_{0})),\\\\\n",
"\\tilde u=u_n+\\frac{h}{2}(3\\varphi(t_n,u_n)-\\varphi(t_{n-1},u_{n-1})),\\\\\n",
"u_{n+1}=u_n+\\frac{h}{24}\\Bigl(9\\varphi(t_{n+1},\\tilde u)+19\\varphi(t_n,u_n)-5\\varphi(t_{n-1},u_{n-1})+\\varphi(t_{n-2},u_{n-2})\\Bigr)& n=2,3,\\dots N-1\n",
"\\end{cases}\n",
"$$"
]
},
{
"cell_type": "markdown",
"metadata": {
"slideshow": {
"slide_type": "slide"
}
},
"source": [
"## Comparaison sur un exemple\n",
"Considérons le problème de Cauchy\n",
">trouver la fonction $y \\colon I\\subset \\mathbb{R} \\to \\mathbb{R}$ définie sur l'intervalle $I=[0,1]$ telle que\n",
"$$\\begin{cases}\n",
"y'(t) = \\sin(t)+y(t), &\\forall t \\in I=[0,1],\\\\\n",
"y(0) = 0\n",
"\\end{cases}$$\n",
"\n",
"1. Calculer la solution exacte en utilisant le module `sympy`.\n",
"1. Calculer la solution approchée obtenue avec la méthode d'Euler explicite avec $h=1/N$ et $N=8$ (pour bien visualiser les erreurs);\n",
"2. même exercice pour les méthodes *multipas explicites*;\n",
"3. même exercice pour les méthodes *multipas implicites*;\n",
"4. même exercice pour les méthodes *predictor-corrector*.\n",
"5. Pour chaque méthode, afficher solution exacte *vs* solution approchée ainsi que le maximum de la valeur absolue de l'erreur."
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