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<div class="section" id="equations-aux-derivees-partielles-utilisation-de-numpy">
<h1>4. Équations aux dérivées partielles : utilisation de NumPy<a class="headerlink" href="#equations-aux-derivees-partielles-utilisation-de-numpy" title="Lien permanent vers ce titre">¶</a></h1>
<p>On trouve dans le module <strong>NumPy</strong> les outils de manipulation des tableaux
pour le calcul numérique</p>
<blockquote>
<div><ul class="simple">
<li>Nombreuses fonctions de manipulation de tableaux</li>
<li>Bibliothèque mathématique importante</li>
</ul>
</div></blockquote>
<p>Il s’agit d’un
module stable, bien testé et relativement bien documenté.</p>
<p><a class="reference external" href="http://docs.scipy.org/doc/">http://docs.scipy.org/doc/</a>
<a class="reference external" href="http://docs.scipy.org/doc/numpy/reference/">http://docs.scipy.org/doc/numpy/reference/</a></p>
<p>Pour l’importer, on recommande d’utiliser</p>
<div class="highlight-python"><div class="highlight"><pre><span class="gp">>>> </span><span class="kn">import</span> <span class="nn">numpy</span> <span class="kn">as</span> <span class="nn">np</span>
<div class="newline"></div></pre></div>
</div>
<p>Toutes les fonctions NumPy seront alors préfixées par <em>np.</em></p>
<div class="section" id="introduction-rapide-a-numpy">
<h2>4.1. Introduction rapide à NumPy<a class="headerlink" href="#introduction-rapide-a-numpy" title="Lien permanent vers ce titre">¶</a></h2>
<p>Le module NumPy permet la manipulation simple et efficace des tableaux</p>
<div class="highlight-python"><div class="highlight"><pre><span class="gp">>>> </span><span class="n">x</span> <span class="o">=</span> <span class="n">np</span><span class="o">.</span><span class="n">arange</span><span class="p">(</span><span class="mi">0</span><span class="p">,</span><span class="mf">2.0</span><span class="p">,</span><span class="mf">0.1</span><span class="p">)</span> <span class="c"># De 0 (inclus) à 2 (exclus) par pas de 0.1</span>
<div class="newline"></div><span class="gp">>>> </span><span class="n">x</span>
<div class="newline"></div><span class="go">array([ 0. , 0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8, 0.9, 1. ,</span>
<div class="newline"></div><span class="go"> 1.1, 1.2, 1.3, 1.4, 1.5, 1.6, 1.7, 1.8, 1.9])</span>
<div class="newline"></div><span class="gp">>>> </span><span class="n">np</span><span class="o">.</span><span class="n">size</span><span class="p">(</span><span class="n">x</span><span class="p">)</span> <span class="c"># Sa taille</span>
<div class="newline"></div><span class="go">20</span>
<div class="newline"></div><span class="gp">>>> </span><span class="n">x</span><span class="p">[</span><span class="mi">0</span><span class="p">]</span> <span class="c"># Le premier élément</span>
<div class="newline"></div><span class="go">0.0</span>
<div class="newline"></div><span class="gp">>>> </span><span class="n">x</span><span class="p">[</span><span class="mi">1</span><span class="p">]</span> <span class="c"># Le deuxième élément</span>
<div class="newline"></div><span class="go">0.10000000000000001</span>
<div class="newline"></div><span class="gp">>>> </span><span class="n">x</span><span class="p">[</span><span class="mi">19</span><span class="p">]</span> <span class="c"># Le dernier élément</span>
<div class="newline"></div><span class="go">1.9000000000000001</span>
<div class="newline"></div><span class="gp">>>> </span><span class="n">x</span><span class="p">[</span><span class="mi">20</span><span class="p">]</span> <span class="c"># Pas un élément !</span>
<div class="newline"></div><span class="gt">Traceback (most recent call last):</span>
<div class="newline"></div> File <span class="nb">"<stdin>"</span>, line <span class="m">1</span>, in <span class="n"><module></span>
<div class="newline"></div><span class="gr">IndexError</span>: <span class="n">index 20 is out of bounds for axis 0 with size 20</span>
<div class="newline"></div><span class="gp">>>> </span><span class="n">a</span> <span class="o">=</span> <span class="n">np</span><span class="o">.</span><span class="n">array</span> <span class="p">([[</span><span class="mi">1</span><span class="p">,</span><span class="mi">2</span><span class="p">,</span><span class="mi">3</span><span class="p">],</span> <span class="p">[</span><span class="mi">4</span><span class="p">,</span><span class="mi">5</span><span class="p">,</span><span class="mi">6</span><span class="p">],</span> <span class="p">[</span><span class="mi">7</span><span class="p">,</span><span class="mi">8</span><span class="p">,</span><span class="mi">9</span><span class="p">]])</span>
<div class="newline"></div><span class="gp">>>> </span><span class="n">a</span>
<div class="newline"></div><span class="go">array([[1, 2, 3],</span>
<div class="newline"></div><span class="go"> [4, 5, 6],</span>
<div class="newline"></div><span class="go"> [7, 8, 9]])</span>
<div class="newline"></div><span class="gp">>>> </span><span class="n">b</span> <span class="o">=</span> <span class="mi">2</span> <span class="o">*</span> <span class="n">a</span> <span class="c"># Multiplication de chaque terme</span>
<div class="newline"></div><span class="gp">>>> </span><span class="n">c</span> <span class="o">=</span> <span class="n">a</span> <span class="o">+</span> <span class="n">b</span> <span class="c"># Sommation terme à terme</span>
<div class="newline"></div><span class="gp">>>> </span><span class="n">np</span><span class="o">.</span><span class="n">dot</span><span class="p">(</span><span class="n">a</span><span class="p">,</span> <span class="n">b</span><span class="p">)</span> <span class="c"># Produit de matrices</span>
<div class="newline"></div><span class="go">array([[ 60, 72, 84],</span>
<div class="newline"></div><span class="go"> [132, 162, 192],</span>
<div class="newline"></div><span class="go"> [204, 252, 300]])</span>
<div class="newline"></div><span class="gp">>>> </span><span class="n">a</span> <span class="o">*</span> <span class="n">b</span> <span class="c"># Produit terme à terme</span>
<div class="newline"></div><span class="go">array([[ 2, 8, 18],</span>
<div class="newline"></div><span class="go"> [ 32, 50, 72],</span>
<div class="newline"></div><span class="go"> [ 98, 128, 162]])</span>
<div class="newline"></div></pre></div>
</div>
<p>On peut facilement effectuer des coupes dans un tableau numpy. Cette
fonctionnalité est particulièrement importante en calcul scientifique
(comme nous allons le voir) pour éviter l’utilisation de boucles.</p>
<div class="highlight-python"><div class="highlight"><pre><span class="gp">>>> </span><span class="n">t</span> <span class="o">=</span> <span class="n">np</span><span class="o">.</span><span class="n">array</span><span class="p">([</span><span class="mi">1</span><span class="p">,</span><span class="mi">2</span><span class="p">,</span><span class="mi">3</span><span class="p">,</span><span class="mi">4</span><span class="p">,</span><span class="mi">5</span><span class="p">,</span><span class="mi">6</span><span class="p">])</span>
<div class="newline"></div><span class="gp">>>> </span><span class="n">t</span><span class="p">[</span><span class="mi">1</span><span class="p">:</span><span class="mi">4</span><span class="p">]</span> <span class="c"># de l'indice 1 à l'indice 4 exclu !!!ATTENTION!!!</span>
<div class="newline"></div><span class="go">array([2, 3, 4])</span>
<div class="newline"></div><span class="gp">>>> </span><span class="n">t</span><span class="p">[:</span><span class="mi">4</span><span class="p">]</span> <span class="c"># du debut à l'indice 4 exclu</span>
<div class="newline"></div><span class="go">array([1, 2, 3, 4])</span>
<div class="newline"></div><span class="gp">>>> </span><span class="n">t</span><span class="p">[</span><span class="mi">4</span><span class="p">:]</span> <span class="c"># de l'indice 4 inclus à la fin</span>
<div class="newline"></div><span class="go">array([5, 6])</span>
<div class="newline"></div><span class="gp">>>> </span><span class="n">t</span><span class="p">[:</span><span class="o">-</span><span class="mi">1</span><span class="p">]</span> <span class="c"># excluant le dernier element</span>
<div class="newline"></div><span class="go">array([1, 2, 3, 4, 5])</span>
<div class="newline"></div><span class="gp">>>> </span><span class="n">t</span><span class="p">[</span><span class="mi">1</span><span class="p">:</span><span class="o">-</span><span class="mi">1</span><span class="p">]</span> <span class="c"># excluant le premier et le dernier</span>
<div class="newline"></div><span class="go">array([2, 3, 4, 5])</span>
<div class="newline"></div></pre></div>
</div>
<p>Pour extraire des sous-parties d’un tableau numpy, on a vu qu’on peut
faire de l’indexation simple <tt class="docutils literal"><span class="pre">t[0]</span></tt> et des coupes <tt class="docutils literal"><span class="pre">t[1:3]</span></tt>. Une autre
possibilité très pratique est de sélectionner certaines valeurs d’un
tableau grâce à un autre tableau de booléens (un “masque”), de taille
compatible avec le tableau d’intérêt. Cette opération s’appelle de
l’indexation par <strong>masque</strong></p>
<div class="highlight-python"><div class="highlight"><pre><span class="gp">>>> </span><span class="n">a</span> <span class="o">=</span> <span class="n">np</span><span class="o">.</span><span class="n">arange</span><span class="p">(</span><span class="mi">6</span><span class="p">)</span><span class="o">**</span><span class="mi">2</span>
<div class="newline"></div><span class="gp">>>> </span><span class="n">a</span>
<div class="newline"></div><span class="go">array([ 0, 1, 4, 9, 16, 25])</span>
<div class="newline"></div><span class="gp">>>> </span><span class="n">a</span> <span class="o">></span> <span class="mi">10</span> <span class="c"># le masque, tableau de booleens</span>
<div class="newline"></div><span class="go">array([False, False, False, False, True, True], dtype=bool)</span>
<div class="newline"></div><span class="gp">>>> </span><span class="n">a</span><span class="p">[</span><span class="n">a</span> <span class="o">></span> <span class="mi">10</span><span class="p">]</span> <span class="c"># une maniere compacte d'extraire les valeurs > 10</span>
<div class="newline"></div><span class="go">array([16, 25])</span>
<div class="newline"></div></pre></div>
</div>
<p><strong>Attention</strong> à la copie de tableau !</p>
<p>Pour un scalaire on a le comportement “intuitif” :</p>
<div class="highlight-python"><div class="highlight"><pre><span class="gp">>>> </span><span class="n">a</span> <span class="o">=</span> <span class="mf">1.0</span>
<div class="newline"></div><span class="gp">>>> </span><span class="n">b</span> <span class="o">=</span> <span class="n">a</span>
<div class="newline"></div><span class="gp">>>> </span><span class="n">b</span>
<div class="newline"></div><span class="go">1.0</span>
<div class="newline"></div><span class="gp">>>> </span><span class="n">a</span> <span class="o">=</span> <span class="mf">0.0</span>
<div class="newline"></div><span class="gp">>>> </span><span class="n">b</span>
<div class="newline"></div><span class="go">1.0</span>
<div class="newline"></div></pre></div>
</div>
<p>Pour un tableau NumPy, par defaut on ne copie que l’adresse du
tableau (pointeur) pas son contenu (les deux noms correspondent alors aux
mêmes adresses en mémoire).</p>
<div class="highlight-python"><div class="highlight"><pre><span class="gp">>>> </span><span class="n">a</span> <span class="o">=</span> <span class="n">zeros</span><span class="p">((</span><span class="mi">2</span><span class="p">,</span> <span class="mi">2</span><span class="p">))</span>
<div class="newline"></div><span class="gp">>>> </span><span class="n">b</span> <span class="o">=</span> <span class="n">a</span>
<div class="newline"></div><span class="gp">>>> </span><span class="n">b</span>
<div class="newline"></div><span class="go">array([[ 0., 0.],</span>
<div class="newline"></div><span class="go"> [ 0., 0.]])</span>
<div class="newline"></div><span class="gp">>>> </span><span class="n">a</span><span class="p">[</span><span class="mi">1</span><span class="p">,</span> <span class="mi">1</span><span class="p">]</span> <span class="o">=</span> <span class="mi">10</span>
<div class="newline"></div><span class="gp">>>> </span><span class="n">b</span>
<div class="newline"></div><span class="go">array([[ 0., 0.],</span>
<div class="newline"></div><span class="go"> [ 0., 10.]])</span>
<div class="newline"></div></pre></div>
</div>
<p>Pour effectuer une copie des valeurs, il faut
utiliser <strong>.copy()</strong></p>
<div class="highlight-python"><div class="highlight"><pre><span class="gp">>>> </span><span class="n">c</span> <span class="o">=</span> <span class="n">b</span><span class="o">.</span><span class="n">copy</span><span class="p">()</span>
<div class="newline"></div><span class="gp">>>> </span><span class="n">c</span>
<div class="newline"></div><span class="go">array([[ 0., 0.],</span>
<div class="newline"></div><span class="go"> [ 0., 10.]])</span>
<div class="newline"></div><span class="gp">>>> </span><span class="n">b</span><span class="p">[</span><span class="mi">1</span><span class="p">,</span> <span class="mi">1</span><span class="p">]</span> <span class="o">=</span> <span class="mi">0</span>
<div class="newline"></div><span class="gp">>>> </span><span class="n">b</span>
<div class="newline"></div><span class="go">array([[ 0., 0.],</span>
<div class="newline"></div><span class="go"> [ 0., 0.]])</span>
<div class="newline"></div><span class="gp">>>> </span><span class="n">c</span>
<div class="newline"></div><span class="go">array([[ 0., 0.],</span>
<div class="newline"></div><span class="go"> [ 0., 10.]])</span>
<div class="newline"></div></pre></div>
</div>
<p><em>Remarque :</em> la même chose s’applique aux coupes :</p>
<div class="highlight-python"><div class="highlight"><pre><span class="gp">>>> </span><span class="n">a</span> <span class="o">=</span> <span class="n">np</span><span class="o">.</span><span class="n">arange</span><span class="p">(</span><span class="mi">10</span><span class="p">)</span>
<div class="newline"></div><span class="gp">>>> </span><span class="n">b</span> <span class="o">=</span> <span class="n">a</span><span class="p">[:</span><span class="mi">5</span><span class="p">]</span>
<div class="newline"></div><span class="gp">>>> </span><span class="n">a</span><span class="p">[</span><span class="mi">0</span><span class="p">]</span> <span class="o">=</span> <span class="mi">10</span>
<div class="newline"></div><span class="gp">>>> </span><span class="n">b</span>
<div class="newline"></div><span class="go">array([10, 1, 2, 3, 4])</span>
<div class="newline"></div></pre></div>
</div>
<p>Le module NumPy comporte beaucoup de fonctions qui permettent de créer
des tableaux spéciaux, manipuler des tableaux, de faire des opérations
sur ces tableaux, etc.</p>
<div class="highlight-python"><div class="highlight"><pre><span class="gp">>>> </span><span class="n">a</span> <span class="o">=</span> <span class="n">np</span><span class="o">.</span><span class="n">arange</span><span class="p">(</span><span class="mi">10</span><span class="p">)</span>
<div class="newline"></div><span class="gp">>>> </span><span class="n">np</span><span class="o">.</span><span class="n">sum</span><span class="p">(</span><span class="n">a</span><span class="p">)</span>
<div class="newline"></div><span class="go">45</span>
<div class="newline"></div><span class="gp">>>> </span><span class="n">np</span><span class="o">.</span><span class="n">mean</span><span class="p">(</span><span class="n">a</span><span class="p">)</span>
<div class="newline"></div><span class="go">4.5</span>
<div class="newline"></div></pre></div>
</div>
<div class="topic">
<p class="topic-title first">Application : calcul de pi</p>
<p>Nous reprenons ici un exemple de la section précédente, en utilisant
uniquement des tableaux et des fonctions de NumPy</p>
<div class="highlight-python"><div class="highlight"><pre><span class="gp">>>> </span><span class="n">x</span><span class="p">,</span> <span class="n">y</span> <span class="o">=</span> <span class="n">np</span><span class="o">.</span><span class="n">random</span><span class="o">.</span><span class="n">random</span><span class="p">((</span><span class="mi">2</span><span class="p">,</span> <span class="mi">100000</span><span class="p">))</span>
<div class="newline"></div><span class="gp">>>> </span><span class="n">x</span><span class="o">.</span><span class="n">shape</span><span class="p">,</span> <span class="n">y</span><span class="o">.</span><span class="n">shape</span>
<div class="newline"></div><span class="go">((100000,), (100000,))</span>
<div class="newline"></div><span class="gp">>>> </span><span class="n">mask</span> <span class="o">=</span> <span class="n">x</span><span class="o">**</span><span class="mi">2</span> <span class="o">+</span> <span class="n">y</span><span class="o">**</span><span class="mi">2</span> <span class="o"><</span> <span class="mi">1</span> <span class="c"># quart de disque</span>
<div class="newline"></div><span class="gp">>>> </span><span class="n">np</span><span class="o">.</span><span class="n">mean</span><span class="p">(</span><span class="n">mask</span><span class="p">)</span> <span class="c"># fraction des points dans le disque</span>
<div class="newline"></div><span class="go">0.78474999999999995</span>
<div class="newline"></div><span class="gp">>>> </span><span class="n">np</span><span class="o">.</span><span class="n">pi</span> <span class="o">/</span> <span class="mi">4</span>
<div class="newline"></div><span class="go">0.7853981633974483</span>
<div class="newline"></div></pre></div>
</div>
</div>
</div>
<div class="section" id="equation-de-la-chaleur-1d">
<h2>4.2. Équation de la chaleur 1D<a class="headerlink" href="#equation-de-la-chaleur-1d" title="Lien permanent vers ce titre">¶</a></h2>
<div class="section" id="discretisation-en-differences-finies">
<h3>4.2.1. Discrétisation en différences finies<a class="headerlink" href="#discretisation-en-differences-finies" title="Lien permanent vers ce titre">¶</a></h3>
<p>On va s’intéresser dans un premier temps à l’équation de la chaleur
(diffusion thermique) en une dimension d’espace</p>
<div class="math">
<p><img src="_images/math/636a9aa84deeb492aaaade73721b56b7748e3519.png" alt="\frac{\partial T}{\partial t} = \kappa \, \frac{\partial^2 T}{\partial
x^2} \, ,"/></p>
</div><p>on considèrera les conditions aux limites suivantes</p>
<div class="math">
<p><img src="_images/math/0a3a3e58890b2e782079b14e646b79c0d4824fbc.png" alt="\forall t \qquad T=0 \, ,\qquad \text{en} \,\, x=0 \,\, \text{et} \,\, x=1 \, ,\\[3mm]
T=\sin(2\pi\,x)\, , \qquad \text{en $t=0$}\, ."/></p>
</div><p>On va chercher à discrétiser ce problème pour en chercher une solution
approchée.</p>
<p>La discrétisation la plus simple que l’on puisse envisager (aux différences
finies s’écrit)</p>
<div class="math">
<p><img src="_images/math/6cd5717e08d1c54fdd92ea7a9580d07a2af7ada0.png" alt="\frac{T_{j}^{n+1}-T_{j}^{n}}{\Delta t} =
\kappa \,
\frac{\frac{T_{j+1}^n-T_{j}^{n}}{\Delta
x}-\frac{T_{j}^n-T_{j-1}^{n}}{\Delta x}}{\Delta x} \, ,"/></p>
</div><p>que l’on peut re-écrire</p>
<div class="math">
<p><img src="_images/math/006d58851eb4425c8f2ee83871f67bcbd5e24075.png" alt="T_{j}^{n+1} = T_{j}^{n} + c \, (T_{j-1}^{n}-2\, T_{j}^{n}+T_{j+1}^{n}) \, ,
\qquad \text{avec}\quad
c\equiv \frac{{\Delta t}\, \kappa}{\Delta x^2} \, ."/></p>
</div><p>Cela s’implémente très simplement en Python, par exemple sous la forme</p>
<div class="highlight-python"><div class="highlight"><pre><span class="k">for</span> <span class="n">j</span> <span class="ow">in</span> <span class="nb">range</span><span class="p">(</span><span class="mi">1</span><span class="p">,</span> <span class="n">NX</span> <span class="o">-</span> <span class="mi">1</span><span class="p">):</span>
<div class="newline"></div> <span class="n">RHS</span><span class="p">[</span><span class="n">j</span><span class="p">]</span> <span class="o">=</span> <span class="n">dt</span> <span class="o">*</span> <span class="n">K</span> <span class="o">*</span> <span class="p">(</span><span class="n">T</span><span class="p">[</span><span class="n">j</span> <span class="o">-</span> <span class="mi">1</span><span class="p">]</span> <span class="o">-</span> <span class="mi">2</span> <span class="o">*</span> <span class="n">T</span><span class="p">[</span><span class="n">j</span><span class="p">]</span> <span class="o">+</span> <span class="n">T</span><span class="p">[</span><span class="n">j</span> <span class="o">+</span> <span class="mi">1</span><span class="p">])</span> <span class="o">/</span> <span class="p">(</span><span class="n">dx</span><span class="o">**</span><span class="mi">2</span><span class="p">)</span>
<div class="newline"></div>
<div class="newline"></div><span class="k">for</span> <span class="n">j</span> <span class="ow">in</span> <span class="nb">range</span> <span class="p">(</span><span class="mi">1</span><span class="p">,</span> <span class="n">NX</span> <span class="o">-</span> <span class="mi">1</span><span class="p">):</span>
<div class="newline"></div> <span class="n">T</span><span class="p">[</span><span class="n">j</span><span class="p">]</span> <span class="o">+=</span> <span class="n">RHS</span><span class="p">[</span><span class="n">j</span><span class="p">]</span>
<div class="newline"></div></pre></div>
</div>
<div class="figure align-center">
<a class="reference external image-reference" href="auto_examples/edp1_1D_heat_loops.html"><img alt="_images/plot_edp1_1D_heat_loops_1.png" src="_images/plot_edp1_1D_heat_loops_1.png" style="width: 640.0px; height: 480.0px;" /></a>
</div>
<p>[<a class="reference internal" href="auto_examples/edp1_1D_heat_loops.html#example-edp1-1d-heat-loops-py"><em>Python source code</em></a>]</p>
</div>
<div class="section" id="convergence">
<h3>4.2.2. Convergence<a class="headerlink" href="#convergence" title="Lien permanent vers ce titre">¶</a></h3>
<p>En introduisant un développement de Taylor, on peut estimer la qualité de
l’approximation numérique (évolution de l’erreur en fonction de
<img class="math" src="_images/math/1eb29f9de3753a59530941141fcb5c7aa3fa2e38.png" alt="\Delta x" style="vertical-align: 0px"/> et <img class="math" src="_images/math/a1ffc0a012620941fe660cedabff822ce7162eca.png" alt="\Delta t" style="vertical-align: 0px"/>).</p>
<p>En écrivant</p>
<div class="math">
<p><img src="_images/math/a2c4eccdf7047450106142df985c5c7d5d58064c.png" alt="T_{j+\alpha}^n = T_{j}^n
+ \alpha \, \Delta x \left(\frac{\partial T}{\partial x}\right)_{j}^n
+ \alpha^2 \, \frac{\Delta x^2}{2} \left(\frac{\partial^2 T}{\partial x^2}\right)_{j}^n
+ \alpha^3 \, \frac{\Delta x^3}{3!} \left(\frac{\partial^3 T}{\partial
x^3}\right)_{j}^n"/></p>
</div><div class="math">
<p><img src="_images/math/c9982c97aeb6ddaf45d8a556748b6745b280930a.png" alt="+ \alpha^4 \, \frac{\Delta x^4}{4!} \left(\frac{\partial^4 T}{\partial x^4}\right)_{j}^n
+ \alpha^5 \, \frac{\Delta x^5}{5!} \left(\frac{\partial^5 T}{\partial x^5}\right)_{j}^n
+ {\cal O}(\Delta x^6) \, ,"/></p>
</div><p>et en sommant les expressions pour <img class="math" src="_images/math/997a209dcb90543b0b7472e90e7cf855091b8b00.png" alt="\alpha=-1" style="vertical-align: -1px"/> et <img class="math" src="_images/math/5d739b3faa1ca596d500cae6ad311fd86a8b6937.png" alt="\alpha=1" style="vertical-align: -1px"/>,
on a</p>
<div class="math">
<p><img src="_images/math/6092d70c08bb1b555f02616d4dfb359a24aa33d5.png" alt="T_{j-1} + T_{j+1} = 2 T_{j} + \Delta x^2 \left.\frac{\partial^2
T}{\partial x^2}\right|_{j}^n + \frac{\Delta
x^4}{12}\left.\frac{\partial^4 T}{\partial x^4}\right|_{j}^n + \mathcal{O}(\Delta
x^6) \, ,"/></p>
</div><p>donc</p>
<div class="math">
<p><img src="_images/math/c2e2f196a1283e51d0136d2d44c3416655d7c84e.png" alt="\left.\frac{\partial ^2 T}{\partial x ^2} \right|_j^n =
\frac{T_{j-1}^n-2T_j^n+T_{j+1^n}}{\Delta x ^2} - \frac{\Delta
x^2}{12}\left.\frac{\partial^4T}{\partial x^4}\right|_j^n + \mathcal{O}(\Delta x^4)
\, ."/></p>
</div><p>Un calcul similaire en temps permet d’estimer l’erreur “de troncature”
associée à notre schéma discret</p>
<div class="math">
<p><img src="_images/math/cfb4205a530b2bd937c25288c8b8ec72a9c3b9d9.png" alt="R(T)=
\frac{\Delta t}{2}\left.\frac{\partial^2 T}{\partial t^2}\right|_j^n
- \kappa\frac{\Delta x^2}{12}\left.\frac{\partial^4 T}{\partial x^4}\right|_j^n
+ \mathcal{O}(\Delta t^2)+\mathcal{O}(\Delta x^4) \, ."/></p>
</div><p>On peut essayer de vérifier numériquement que le schéma utilisé est bien
d’ordre deux en espace.</p>
<p>Pour cela on va effectuer une boucle extérieure sur la résolution et mesurer
une norme de l’erreur entre la solution calculée et la solution analytique</p>
<div class="highlight-python"><div class="highlight"><pre><span class="n">scale</span> <span class="o">=</span> <span class="n">np</span><span class="o">.</span><span class="n">exp</span><span class="p">(</span><span class="o">-</span><span class="mi">4</span><span class="o">*</span><span class="p">(</span><span class="n">np</span><span class="o">.</span><span class="n">pi</span><span class="o">**</span><span class="mi">2</span><span class="p">)</span><span class="o">*</span><span class="n">K</span><span class="o">*</span><span class="n">Time</span><span class="p">)</span>
<div class="newline"></div><span class="n">TO</span> <span class="o">=</span> <span class="n">np</span><span class="o">.</span><span class="n">sin</span><span class="p">(</span><span class="mi">2</span><span class="o">*</span><span class="n">np</span><span class="o">.</span><span class="n">pi</span><span class="o">*</span><span class="n">x</span><span class="p">)</span>
<div class="newline"></div><span class="n">DDX</span><span class="p">[</span><span class="n">k</span><span class="p">]</span> <span class="o">=</span> <span class="n">dx</span>
<div class="newline"></div><span class="n">ERR</span><span class="p">[</span><span class="n">k</span><span class="p">]</span> <span class="o">=</span> <span class="nb">max</span><span class="p">(</span><span class="nb">abs</span><span class="p">(</span><span class="n">T</span><span class="o">-</span><span class="n">TO</span><span class="o">*</span><span class="n">scale</span><span class="p">))</span>
<div class="newline"></div></pre></div>
</div>
<div class="figure align-center">
<a class="reference external image-reference" href="auto_examples/edp2_1D_heat_loops_conv.html"><img alt="_images/plot_edp2_1D_heat_loops_conv_1.png" src="_images/plot_edp2_1D_heat_loops_conv_1.png" style="width: 640.0px; height: 480.0px;" /></a>
</div>
<p>[<a class="reference internal" href="auto_examples/edp2_1D_heat_loops_conv.html#example-edp2-1d-heat-loops-conv-py"><em>Python source code</em></a>]</p>
<p>On constate que le schéma semble bien être d’ordre 2 en espace, mais que le
calcul devient insupportablement long.</p>
<p>C’est qu’en fait ce code est mal écrit car il ne tire pas profit des
possibilités de calcul vectoriel offertes par NumPy.</p>
<p>Pour cela il faut remplacer les lignes</p>
<div class="highlight-python"><div class="highlight"><pre><span class="k">for</span> <span class="n">j</span> <span class="ow">in</span> <span class="nb">range</span><span class="p">(</span><span class="mi">1</span><span class="p">,</span> <span class="n">NX</span> <span class="o">-</span> <span class="mi">1</span><span class="p">):</span>
<div class="newline"></div> <span class="n">RHS</span><span class="p">[</span><span class="n">j</span><span class="p">]</span> <span class="o">=</span> <span class="n">dt</span> <span class="o">*</span> <span class="n">K</span> <span class="o">*</span> <span class="p">(</span><span class="n">T</span><span class="p">[</span><span class="n">j</span> <span class="o">-</span> <span class="mi">1</span><span class="p">]</span> <span class="o">-</span> <span class="mi">2</span> <span class="o">*</span> <span class="n">T</span><span class="p">[</span><span class="n">j</span><span class="p">]</span> <span class="o">+</span> <span class="n">T</span><span class="p">[</span><span class="n">j</span> <span class="o">+</span> <span class="mi">1</span><span class="p">])</span> <span class="o">/</span> <span class="p">(</span><span class="n">dx</span><span class="o">**</span><span class="mi">2</span><span class="p">)</span>
<div class="newline"></div>
<div class="newline"></div><span class="k">for</span> <span class="n">j</span> <span class="ow">in</span> <span class="nb">range</span> <span class="p">(</span><span class="mi">1</span><span class="p">,</span> <span class="n">NX</span> <span class="o">-</span> <span class="mi">1</span><span class="p">):</span>
<div class="newline"></div> <span class="n">T</span><span class="p">[</span><span class="n">j</span><span class="p">]</span> <span class="o">+=</span> <span class="n">RHS</span><span class="p">[</span><span class="n">j</span><span class="p">]</span>
<div class="newline"></div></pre></div>
</div>
<p>par des instructions vectorielles (les “boucles” sont alors gérées par du
code compilé et non par du code interpreté)</p>
<div class="highlight-python"><div class="highlight"><pre><span class="n">RHS</span><span class="p">[</span><span class="mi">1</span><span class="p">:</span><span class="o">-</span><span class="mi">1</span><span class="p">]</span> <span class="o">=</span> <span class="n">dt</span> <span class="o">*</span> <span class="n">K</span> <span class="o">*</span> <span class="p">(</span><span class="n">T</span><span class="p">[:</span><span class="o">-</span><span class="mi">2</span><span class="p">]</span> <span class="o">-</span> <span class="mi">2</span> <span class="o">*</span> <span class="n">T</span><span class="p">[</span><span class="mi">1</span><span class="p">:</span><span class="o">-</span><span class="mi">1</span><span class="p">]</span> <span class="o">+</span> <span class="n">T</span><span class="p">[</span><span class="mi">2</span><span class="p">:])</span> <span class="o">/</span> <span class="p">(</span><span class="n">dx</span><span class="o">**</span><span class="mi">2</span><span class="p">)</span>
<div class="newline"></div><span class="n">T</span> <span class="o">+=</span> <span class="n">RHS</span>
<div class="newline"></div></pre></div>
</div>
<p>On constate que l’execution est alors quasi-instantanée.</p>
<p>[<a class="reference internal" href="auto_examples/edp3_1D_heat_vect_conv.html#example-edp3-1d-heat-vect-conv-py"><em>Python source code</em></a>]</p>
</div>
<div class="section" id="formulation-matricielle">
<h3>4.2.3. Formulation matricielle<a class="headerlink" href="#formulation-matricielle" title="Lien permanent vers ce titre">¶</a></h3>
<p>Que se passe t’il si on pousse l’analyse vers de plus petits pas d’espace ???</p>
<p>On est de fait limité par un critère de stabilité.</p>
<p>Pour une résolution spatiale fixée, celui-ci nous impose donc un nombre
minimum d’iterations pour atteindre un temps donné.</p>
<p>On peut cependant chercher à obtenir directement la solution du problème
stationnaire.</p>
<p>Considérons le système modifié avec terme source (pour éviter une solution
stationnaire triviale)</p>
<div class="math">
<p><img src="_images/math/cb6b4d4805983815927a9d6d5f42eca0273aceba.png" alt="\frac{\partial T}{\partial t} = \kappa \, \frac{\partial^2 T}{\partial x^2} + S \, ,"/></p>
</div><p>On a alors la solution stationnaire en résolvant</p>
<div class="math">
<p><img src="_images/math/45bd0ca900377f586c8d8bc0097ad04bc80ebf38.png" alt="\kappa \, \frac{\partial^2 T}{\partial x^2} = - S \, ,"/></p>
</div><p>Pour cela il faut donc résoudre un système linéaire</p>
<div class="math">
<p><img src="_images/math/36de7c74deb83c2f57c34827dab88f2a1e4c3cd5.png" alt="\kappa (T_{j-1}^{n}-2\, T_{j}^{n}+T_{j+1}^{n}) = -S \, \Delta x^2 \, ."/></p>
</div><p>qui peut s’écrire, avec nos conditions aux limites (<img class="math" src="_images/math/caa786a1615e0e7e6b6770c530359a8b4863822a.png" alt="T=0" style="vertical-align: -1px"/> en
<img class="math" src="_images/math/34ea5b6f6de584d56d19a89cc4923a9d9a5cfa41.png" alt="x=0" style="vertical-align: -1px"/> et <img class="math" src="_images/math/4c52e02a35db152495396584bfeae1d0599cd753.png" alt="x=1" style="vertical-align: -1px"/>) sous forme matricielle (avec la convention de
Python pour les indices, i.e. de 0 à N-1) :</p>
<div class="math">
<p><img src="_images/math/4877061abd65b08a3210acf82d77d2004ede63f2.png" alt="\kappa \, \left(
\begin{array}{ccccc}
-2 & 1 & 0 & \cdots & 0 \\
1 & -2 & 1 & &\vdots\\
0 &\ddots&\ddots&\ddots& 0\\
\vdots & & 1 & -2 & 1\\
0 & \cdots & 0 & 1 & -2
\end{array}
\right)
\left(
\begin{array}{c}
T_1\\
T_2\\
\vdots\\
T_{N-3}\\
T_{N-2}
\end{array}
\right)
=
-S \, \Delta x^2 \,
\left(
\begin{array}{c}
1\\
1\\
\vdots\\
1\\
1
\end{array}
\right)"/></p>
</div><p>Pour résoudre ce problème en Python, on peut définir une matrice creuse (tridiagonale).</p>
<dl class="docutils">
<dt>Pour cela on utilise le format de matrices creuses de SciPy ::</dt>
<dd>import scipy.sparse as sp</dd>
</dl>
<p>Ainsi que le solveur associé:</p>
<div class="highlight-python"><div class="highlight"><pre><span class="kn">from</span> <span class="nn">scipy.sparse.linalg.dsolve</span> <span class="kn">import</span> <span class="n">spsolve</span>
<div class="newline"></div></pre></div>
</div>
<p>On peut alors définir le problème aux différences finies:</p>
<div class="highlight-python"><div class="highlight"><pre><span class="n">data</span> <span class="o">=</span> <span class="p">[</span><span class="n">np</span><span class="o">.</span><span class="n">ones</span><span class="p">(</span><span class="n">N</span><span class="p">),</span> <span class="o">-</span><span class="mi">2</span><span class="o">*</span><span class="n">np</span><span class="o">.</span><span class="n">ones</span><span class="p">(</span><span class="n">N</span><span class="p">),</span> <span class="n">np</span><span class="o">.</span><span class="n">ones</span><span class="p">(</span><span class="n">N</span><span class="p">)]</span> <span class="c"># Diagonal terms</span>
<div class="newline"></div><span class="n">offsets</span> <span class="o">=</span> <span class="n">np</span><span class="o">.</span><span class="n">array</span><span class="p">([</span><span class="o">-</span><span class="mi">1</span><span class="p">,</span> <span class="mi">0</span><span class="p">,</span> <span class="mi">1</span><span class="p">])</span> <span class="c"># Their positions</span>
<div class="newline"></div><span class="n">LAP</span> <span class="o">=</span> <span class="n">sp</span><span class="o">.</span><span class="n">dia_matrix</span><span class="p">((</span><span class="n">data</span><span class="p">,</span> <span class="n">offsets</span><span class="p">),</span> <span class="n">shape</span><span class="o">=</span><span class="p">(</span><span class="n">N</span><span class="p">,</span> <span class="n">N</span><span class="p">))</span>
<div class="newline"></div></pre></div>
</div>
<p>et utiliser le
solver inclus dans SciPy :</p>
<div class="highlight-python"><div class="highlight"><pre><span class="n">f</span> <span class="o">=</span> <span class="o">-</span><span class="n">np</span><span class="o">.</span><span class="n">ones</span><span class="p">(</span><span class="n">N</span><span class="p">)</span> <span class="o">*</span> <span class="n">dx</span><span class="o">**</span><span class="mi">2</span>
<div class="newline"></div><span class="n">T</span> <span class="o">=</span> <span class="n">spsolve</span><span class="p">(</span><span class="n">LAP</span><span class="p">,</span> <span class="n">f</span><span class="p">)</span>
<div class="newline"></div></pre></div>
</div>
<p>[<a class="reference internal" href="auto_examples/edp4_1D_heat_solve.html#example-edp4-1d-heat-solve-py"><em>Python source code</em></a>]</p>
<p><em>Remarque :</em> la même approche pourrait être utilisée pour l’équation
d’évolution en temps en utilisant le schéma implicite</p>
<div class="math">
<p><img src="_images/math/7aa1d9a3922969087537c1eb21cd3e42d26de0b4.png" alt="T_{j}^{n+1} = T_{j}^{n} + c \, (T_{j-1}^{n+1}-2\, T_{j}^{n+1}+T_{j+1}^{n+1}) \, ,
\qquad \text{avec}\quad
c\equiv \frac{{\Delta t}\, \kappa}{\Delta x^2} \, ."/></p>
</div></div>
</div>
<div class="section" id="equation-de-la-chaleur-2d">
<h2>4.3. Équation de la chaleur 2D<a class="headerlink" href="#equation-de-la-chaleur-2d" title="Lien permanent vers ce titre">¶</a></h2>
<p>On peut traiter le problème équivalent en deux dimensions d’espace</p>
<div class="math">
<p><img src="_images/math/e77c2d37ad9bbd6757dfcb7c6b5da1095420d6eb.png" alt="\frac{\partial T}{\partial t} = \kappa \, \Delta T + S\, ,"/></p>
</div><p>de la même manière, avec un schéma explicite en temps</p>
<div class="math">
<p><img src="_images/math/14c10a9bca7da27d3311194cfbbe19e11bb6b40b.png" alt="T_{i,j}^{n+1} = T_{i,j}^{n} + {\Delta t}\, \kappa \, \left[
(T_{i-1,j}^{n} - 2\, T_{i,j}^{n} + T_{i+1,j}^{n})/{\Delta x^2}\right."/></p>
</div><div class="math">
<p><img src="_images/math/fba90f7de856e16f4848cc33b18e0812f21a89d6.png" alt="\left. +
(T_{i,j-1}^{n} - 2\, T_{i,j}^{n} + T_{i,j+1}^{n})/{\Delta y^2}
\right] \, ."/></p>
</div><p>Ce qui devient en Python:</p>
<div class="highlight-python"><div class="highlight"><pre><span class="k">for</span> <span class="n">n</span> <span class="ow">in</span> <span class="nb">range</span><span class="p">(</span><span class="mi">0</span><span class="p">,</span> <span class="n">NT</span><span class="p">):</span>
<div class="newline"></div> <span class="n">RHS</span><span class="p">[</span><span class="mi">1</span><span class="p">:</span><span class="o">-</span><span class="mi">1</span><span class="p">,</span> <span class="mi">1</span><span class="p">:</span><span class="o">-</span><span class="mi">1</span><span class="p">]</span> <span class="o">=</span> <span class="n">dt</span> <span class="o">*</span> <span class="n">K</span> <span class="o">*</span> <span class="p">(</span> <span class="p">(</span><span class="n">T</span><span class="p">[:</span><span class="o">-</span><span class="mi">2</span><span class="p">,</span> <span class="mi">1</span><span class="p">:</span><span class="o">-</span><span class="mi">1</span><span class="p">]</span> <span class="o">-</span> <span class="mi">2</span> <span class="o">*</span> <span class="n">T</span><span class="p">[</span><span class="mi">1</span><span class="p">:</span><span class="o">-</span><span class="mi">1</span><span class="p">,</span> <span class="mi">1</span><span class="p">:</span><span class="o">-</span><span class="mi">1</span><span class="p">]</span> <span class="o">+</span> <span class="n">T</span><span class="p">[</span><span class="mi">2</span><span class="p">:,</span> <span class="mi">1</span><span class="p">:</span><span class="o">-</span><span class="mi">1</span><span class="p">])</span> <span class="o">/</span> <span class="p">(</span><span class="n">dx</span><span class="o">**</span><span class="mi">2</span><span class="p">)</span> \
<div class="newline"></div> <span class="o">+</span> <span class="p">(</span><span class="n">T</span><span class="p">[</span><span class="mi">1</span><span class="p">:</span><span class="o">-</span><span class="mi">1</span><span class="p">,</span> <span class="p">:</span><span class="o">-</span><span class="mi">2</span><span class="p">]</span> <span class="o">-</span> <span class="mi">2</span> <span class="o">*</span> <span class="n">T</span><span class="p">[</span><span class="mi">1</span><span class="p">:</span><span class="o">-</span><span class="mi">1</span><span class="p">,</span> <span class="mi">1</span><span class="p">:</span><span class="o">-</span><span class="mi">1</span><span class="p">]</span> <span class="o">+</span> <span class="n">T</span><span class="p">[</span><span class="mi">1</span><span class="p">:</span><span class="o">-</span><span class="mi">1</span><span class="p">,</span><span class="mi">2</span><span class="p">:])</span> <span class="o">/</span> <span class="p">(</span><span class="n">dy</span><span class="o">**</span><span class="mi">2</span><span class="p">))</span>
<div class="newline"></div> <span class="n">T</span><span class="p">[</span><span class="mi">1</span><span class="p">:</span><span class="o">-</span><span class="mi">1</span><span class="p">,</span><span class="mi">1</span><span class="p">:</span><span class="o">-</span><span class="mi">1</span><span class="p">]</span> <span class="o">+=</span> <span class="p">(</span><span class="n">RHS</span><span class="p">[</span><span class="mi">1</span><span class="p">:</span><span class="o">-</span><span class="mi">1</span><span class="p">,</span> <span class="mi">1</span><span class="p">:</span><span class="o">-</span><span class="mi">1</span><span class="p">]</span> <span class="o">+</span> <span class="n">dt</span> <span class="o">*</span> <span class="n">S</span><span class="p">)</span>
<div class="newline"></div></pre></div>
</div>
<div class="figure align-center">
<a class="reference external image-reference" href="auto_examples/edp5_2D_heat_vect.html"><img alt="_images/plot_edp5_2D_heat_vect_1.png" src="_images/plot_edp5_2D_heat_vect_1.png" style="width: 640.0px; height: 480.0px;" /></a>
</div>
<p>[<a class="reference internal" href="auto_examples/edp5_2D_heat_vect.html#example-edp5-2d-heat-vect-py"><em>Python source code</em></a>]</p>
<p>Pour résoudre directement la solution stationnaire en 2D, en revanche le
système linéaire est plus difficile à formuler.</p>
<p>La température dépend à présent de deux indices <img class="math" src="_images/math/34857b3ba74ce5cd8607f3ebd23e9015908ada71.png" alt="i" style="vertical-align: 0px"/> et <img class="math" src="_images/math/8122aa89ea6e80784c6513d22787ad86e36ad0cc.png" alt="j" style="vertical-align: -6px"/>.</p>
<p>Pour formuler le problème sous la forme</p>
<div class="math">
<p><img src="_images/math/fca7af9be1bf892d9c408b9fff28a64e457c9ace.png" alt="\left[ A\right]
\left(T\right)
=
-\left(S\right)"/></p>
</div><p>il faut numéroter les <img class="math" src="_images/math/c3668f65cba99ec9907b51acefb909a2ff285b1c.png" alt="T_{i,j}" style="vertical-align: -8px"/> sous la forme d’une grand vecteur et
utiliser le produit de Kronecker</p>
<div class="highlight-python"><div class="highlight"><pre><span class="n">LAP2</span> <span class="o">=</span> <span class="n">sp</span><span class="o">.</span><span class="n">kron</span><span class="p">(</span><span class="n">LAP</span><span class="p">,</span> <span class="n">I1D</span><span class="p">)</span> <span class="o">+</span> <span class="n">sp</span><span class="o">.</span><span class="n">kron</span><span class="p">(</span><span class="n">I1D</span><span class="p">,</span><span class="n">LAP</span><span class="p">)</span>
<div class="newline"></div></pre></div>
</div>
<p>il ne reste alors qu’à résoudre le système linéaire</p>
<div class="highlight-python"><div class="highlight"><pre><span class="n">T</span> <span class="o">=</span> <span class="n">spsolve</span><span class="p">(</span><span class="n">LAP2</span><span class="p">,</span><span class="n">f2</span><span class="p">)</span>
<div class="newline"></div></pre></div>
</div>
<p>et à transformer le résultat (qui est un vecteur de taille NxN) sous la forme d’une
matrice de taille (N,N)</p>
<div class="highlight-python"><div class="highlight"><pre><span class="n">T</span><span class="o">.</span><span class="n">reshape</span><span class="p">(</span><span class="n">N</span><span class="p">,</span><span class="n">N</span><span class="p">)</span>
<div class="newline"></div></pre></div>
</div>
<dl class="docutils">
<dt>Le code complet est disponible ci-dessous :</dt>
<dd>[<a class="reference internal" href="auto_examples/edp6_2D_heat_solve.html#example-edp6-2d-heat-solve-py"><em>Python source code</em></a>]</dd>
</dl>
<div class="green topic">
<p class="topic-title first"><strong>Exercice</strong>: Une équation d’onde en deux dimensions</p>
<p>Modifiez l’exemple d’intégration en temps explicite de
l’équation de la chaleur en deux dimensions (vu ci-dessus)</p>
<p>[<a class="reference internal" href="auto_examples/edp5_2D_heat_vect.html#example-edp5-2d-heat-vect-py"><em>Python source code</em></a>]</p>
<p>pour traiter (toujours en deux dimensions) l’équation d’ondes suivante</p>
<div class="math">
<p><img src="_images/math/4df1a3a24308ff237af5cdab2831dd28c325cc99.png" alt="\frac{\partial ^2 u}{\partial t ^2} = c^2 \, \Delta u \, ."/></p>
</div><p>de la même manière, on peut écrire un schéma explicite en temps</p>
<div class="math">
<p><img src="_images/math/fd90916f9459eca1d7af2b3d650dc1c958aa9405.png" alt="u_{i,j}^{n+1} = 2 \, u_{i,j}^{n} - u_{i,j}^{n-1}
+ {\Delta t ^2}\, c^2 \, \left[
(u_{i-1,j}^{n} - 2\, u_{i,j}^{n} + u_{i+1,j}^{n})/{\Delta x^2}\right."/></p>
</div><div class="math">
<p><img src="_images/math/9e71a80394bdc5d6356cc5529d3733292d315826.png" alt="+\left.
(u_{i,j-1}^{n} - 2\, u_{i,j}^{n} + u_{i,j+1}^{n})/{\Delta y^2}
\right] \, ."/></p>
</div><p>On pourra considérer une condition initiale
<img class="math" src="_images/math/10d6a4b8107292c46cfc32be4dd12f42cf6bea11.png" alt="u|_0" style="vertical-align: -7px"/> sous la forme d’une gaussienne
et <img class="math" src="_images/math/53e04ab1b42dd4034b00b804304c5c83e887ab75.png" alt="\partial_t u|_0 = 0" style="vertical-align: -7px"/>.</p>
</div>
<div class="figure align-center">
<a class="reference external image-reference" href="auto_examples/edp7_waves.html"><img alt="_images/plot_edp7_waves_1.png" src="_images/plot_edp7_waves_1.png" style="width: 640.0px; height: 480.0px;" /></a>
</div>
<div class="topic">
<p class="topic-title first">Correction...</p>
<p>Le code complet est disponible ci-dessous :
[<a class="reference internal" href="auto_examples/edp7_waves.html#example-edp7-waves-py"><em>La solution...</em></a>]</p>
</div>
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