10_Euler.ipynb

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   "source": [
    "import numpy as np\n",
    "import matplotlib.pyplot as plt\n",
    "from scipy.optimize import fsolve\n",
    "%matplotlib inline"
   ]
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    "# berechnet einen Schritt mit dem Expliziten Eulerverfahren\n",
    "def expl_euler(y, t, dt, f):\n",
    "    return y + dt * f(t, y)\n",
    "\n",
    "# berechnet einen Schritt mit dem Expliziten Eulerverfahren\n",
    "def impl_euler(y, t, dt, f):\n",
    "    # y_k = y\n",
    "    # y_{k+1} = x\n",
    "    g = lambda x: (x - y) / dt - f(t + dt, x)\n",
    "    y_new = fsolve(g, y)\n",
    "    return y_new\n",
    "    \n",
    "# berechnet die Nährerungslösung für y(t) mit einem beliebigen Einschrittverfahren\n",
    "def zeitintegration(y_0, t, f, verfahren):\n",
    "    n = t.shape[0]\n",
    "    y = np.zeros(n)\n",
    "    y[0] = y_0\n",
    "    for i in range(0, n-1):\n",
    "        dt = t[i+1] - t[i]\n",
    "        y[i+1] = verfahren(y[i], t[i], dt, f)\n",
    "    return y"
   ]
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   "source": [
    "# rechte Seite wie in Aufgabe 2\n",
    "def f(t, y):\n",
    "    return -12 * y**2\n",
    "\n",
    "# analytische Lösung für Aufgabe 2\n",
    "def y(t):\n",
    "    return 1.0 / (12*t - 11)"
   ]
  },
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   "cell_type": "code",
   "execution_count": null,
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    "t = np.linspace(1.0, 1.5, 8)\n",
    "y_ref = y(t)\n",
    "\n",
    "y_0 = y(1)\n",
    "y_expl = zeitintegration(y_0, t, f, expl_euler)\n",
    "y_impl = zeitintegration(y_0, t, f, impl_euler)\n",
    "    \n",
    "plt.plot(t, y_ref, label=\"Ref\")\n",
    "plt.plot(t, y_expl, label=\"Expliziter Euler\")\n",
    "plt.plot(t, y_impl, label=\"Impliziter Euler\")\n",
    "plt.legend()\n",
    "plt.show()"
   ]
  }
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