added exercise 23, missing solutions

73fb02bb · Jochen Schulz · f66f1a08 · 73fb02bb · 73fb02bb
Commit 73fb02bb authored Sep 01, 2015 by Jochen Schulz
--- a/exercises/Exercises 23.ipynb
+++ b/exercises/Exercises 23.ipynb
+{
+  "nbformat": 4,
+  "metadata": {
+    "kernelspec": {
+      "display_name": "Haskell",
+      "language": "haskell",
+      "name": "haskell"
+    }
+  },
+  "cells": [
+    {
+      "metadata": {},
+      "source": [
+        "# Exercises 23\n",
+        "\n",
+        "## problem 1\n",
+        "\n",
+        "Consider the following system of ODE's\n",
+        "\n",
+        "$$y_1'(t) = - 0.04 y_1(t) + 10^4 y_2(t)y_3(t)$$\n",
+        "$$y_2'(t) = 0.04 y_1(t) - 10^4 y_2(t) y_3(t) - 3 \\cdot 10^7 y_2(t)^2$$\n",
+        "$$y_3'(t) = 3 \\cdot 10^7 y_2(t)^2$$\n",
+        "\n",
+        "Solve the system with initial value $(1,0,0)$ at $0$ with the time range$0 \\leq\n",
+        "t \\leq 3$. Use at least two different solvers and compare their results. Try to understand heuristicly why different solvers behave sometimes very differently."
+      ],
+      "cell_type": "markdown"
+    },
+    {
+      "metadata": {},
+      "source": [
+        "##problem 2\n",
+        "\n",
+        "Implement an explicit Euler method to solve ODE's. The explicit euler method has the parameters $c_i = 0$, $a_{ij} = 0$ and $b_1 = 0$ which then yields the update formula\n",
+        "\n",
+        "$$y_{n + 1} = y_n + h f ( t_n , y_ n )$$\n",
+        "\n",
+        "Choose a value h for the size of every step and set $t_n = t_0 + n h$ . \n"
+      ],
+      "cell_type": "markdown"
+    },
+    {
+      "metadata": {},
+      "source": [
+        "## problem 3\n",
+        "\n",
+        "*stiff* ODE's are somewhat *ill-conditioned* in the sense, that many algorithms for solving this type of ODE are then unstable. Consider the initial value problem: \n",
+        "\n",
+        "$$y^\\prime ( t ) = - 15 y ( t ) , t \\ge 0 , y ( 0 ) = 1$$\n",
+        "\n",
+        "and solve it with the `odeSolve` and the explicit euler. Look especially at $h= 1/4$ and $h = 1/8$ for the euler method and look what is happening.\n",
+        "\t"
+      ],
+      "cell_type": "markdown"
+    }
+  ],
+  "nbformat_minor": 0
+}
\ No newline at end of file
+%% Cell type:markdown id: tags:
+
+# Exercises 23
+
+## problem 1
+
+Consider the following system of ODE's
+
+$$y_1'(t) = - 0.04 y_1(t) + 10^4 y_2(t)y_3(t)$$
+$$y_2'(t) = 0.04 y_1(t) - 10^4 y_2(t) y_3(t) - 3 \cdot 10^7 y_2(t)^2$$
+$$y_3'(t) = 3 \cdot 10^7 y_2(t)^2$$
+
+Solve the system with initial value $(1,0,0)$ at $0$ with the time range$0 \leq
+t \leq 3$. Use at least two different solvers and compare their results. Try to understand heuristicly why different solvers behave sometimes very differently.
+
+%% Cell type:markdown id: tags:
+
+##problem 2
+
+Implement an explicit Euler method to solve ODE's. The explicit euler method has the parameters $c_i = 0$, $a_{ij} = 0$ and $b_1 = 0$ which then yields the update formula
+
+$$y_{n + 1} = y_n + h f ( t_n , y_ n )$$
+
+Choose a value h for the size of every step and set $t_n = t_0 + n h$ .
+
+%% Cell type:markdown id: tags:
+
+## problem 3
+
+*stiff* ODE's are somewhat *ill-conditioned* in the sense, that many algorithms for solving this type of ODE are then unstable. Consider the initial value problem:
+
+$$y^\prime ( t ) = - 15 y ( t ) , t \ge 0 , y ( 0 ) = 1$$
+
+and solve it with the `odeSolve` and the explicit euler. Look especially at $h= 1/4$ and $h = 1/8$ for the euler method and look what is happening.
+
--- a/solutions/Exercises 23_sol.ipynb
+++ b/solutions/Exercises 23_sol.ipynb
@@ -15,7 +15,7 @@
    "$$y_3'(t) = 3 \\cdot 10^7 y_2(t)^2$$\n",
    "\n",
    "Solve the system with initial value $(1,0,0)$ at $0$ with the time range$0 \\leq\n",
-    "t \\leq 3$. Use at least two different solvers and compare their results."
+    "t \\leq 3$. Use at least two different solvers and compare their results. Try to understand heuristicly why different solvers behave sometimes very differently."
   ]
  },
  {
@@ -131,11 +131,11 @@
   "source": [
    "## problem 3\n",
    "\n",
-    "*stiff* ODE's: Consider the initial value problem: \n",
+    "*stiff* ODE's are somewhat *ill-conditioned* in the sense, that many algorithms for solving this type of ODE are then unstable. Consider the initial value problem: \n",
    "\n",
    "$$y^\\prime ( t ) = - 15 y ( t ) , t \\ge 0 , y ( 0 ) = 1$$\n",
    "\n",
-    "and solve it with the odesolve\n",
+    "and solve it with the `odeSolve` and the explicit euler. Look especially at $h= 1/4$ and $h = 1/8$ for the euler method and look what is happening.\n",
    "\t"
   ]
  },

 %% Cell type:markdown id: tags:

 # Exercises 23

 ## problem 1

 Consider the following system of ODE's

 $$y_1'(t) = - 0.04 y_1(t) + 10^4 y_2(t)y_3(t)$$
 $$y_2'(t) = 0.04 y_1(t) - 10^4 y_2(t) y_3(t) - 3 \cdot 10^7 y_2(t)^2$$
 $$y_3'(t) = 3 \cdot 10^7 y_2(t)^2$$

 Solve the system with initial value $(1,0,0)$ at $0$ with the time range$0 \leq
-t \leq 3$. Use at least two different solvers and compare their results.
+t \leq 3$. Use at least two different solvers and compare their results. Try to understand heuristicly why different solvers behave sometimes very differently.

 %% Cell type:code id: tags:

 ``` haskell
 ```

 %% Cell type:markdown id: tags:

 ##problem 2

 Implement an explicit Euler method to solve ODE's. The explicit euler method has the parameters $c_i = 0$, $a_{ij} = 0$ and $b_1 = 0$ which then yields the update formula

 $$y_{n + 1} = y_n + h f ( t_n , y_ n )$$

 Choose a value h for the size of every step and set $t_n = t_0 + n h$ .

 %% Cell type:code id: tags:

 ``` haskell
 import Numeric.Container
 import Graphics.Rendering.Chart.Easy hiding (Matrix, Vector, scale)
 import Graphics.Rendering.Chart.Backend.Cairo

 harmonic w d t v = fromList [y2, -w^2*y1 -2*d*w*y2]
    where y1 = v @> 0
          y2 = v @> 1

 -- For stability: calculate time-step:
 --ht = h**2*h**2/( 2*a*(h**2+h**2) )

 e_euler :: (Vector Double -> Vector Double) -> Double -> Vector Double -> Vector Double
 e_euler f h y = y `add`  (h `scale` (f y))

 integr :: (Vector Double -> Vector Double) -> Vector Double -> Double -> Int -> [Vector Double]
 integr derivative initial_state h n = take n $ iterate (e_euler derivative h) initial_state
 ```

 %% Cell type:code id: tags:

 ``` haskell
 n = 4000
 h = 0.005

 expEq = integr (harmonic 1 0.1 0) (fromList [1::Double,0]) h n
 toRenderable $ do
    layout_title .= "Damped harmonic oscillator"
    setColors [opaque red]
    plot (line "damped (d= 0.1)" [zip (take n [0,0.2..] ) (map (head . toList) expEq)])
 ```

 %% Output



 %% Cell type:code id: tags:

 ``` haskell
 -- beware the step size
 n = 100
 h = 0.2

 expEq = integr (harmonic 1 0.1 0) (fromList [1::Double,0]) h n
 toRenderable $ do
    layout_title .= "Damped harmonic oscillator"
    setColors [opaque red]
    plot (line "damped (d= 0.1)" [zip (take n [0,0.2..] ) (map (head . toList) expEq)])
 ```

 %% Output



 %% Cell type:markdown id: tags:

 ## problem 3

-*stiff* ODE's: Consider the initial value problem:
+*stiff* ODE's are somewhat *ill-conditioned* in the sense, that many algorithms for solving this type of ODE are then unstable. Consider the initial value problem:

 $$y^\prime ( t ) = - 15 y ( t ) , t \ge 0 , y ( 0 ) = 1$$

-and solve it with the odesolve
+and solve it with the `odeSolve` and the explicit euler. Look especially at $h= 1/4$ and $h = 1/8$ for the euler method and look what is happening.


 %% Cell type:code id: tags:

 ``` haskell
 ```