12-267/Numerical Methods

Summary of Numerical Methods

Based largely off of a note available here posted by Simon1 --Twine 20:55, 25 October 2012 (EDT)

Numerical methods: [math]\displaystyle{ \frac{dy}{dt} = f(t, y) }[/math] and [math]\displaystyle{ y(t_0) = y_0 }[/math], [math]\displaystyle{ y = \Phi(t) }[/math] is a solution.

1. Using the proof of Picard's Theorem:

[math]\displaystyle{ \Phi_0(x) = y_0 }[/math]

[math]\displaystyle{ \Phi_n(x) = y_0 + \int_{x_0}^x f(x, \Phi_{n-1}(x)) dx }[/math]

[math]\displaystyle{ \Phi_n(x) \rightarrow \Phi(x) }[/math]

2. The Euler Method:

[math]\displaystyle{ y_{n+1} = y_n + f(t_n, y_n)(t_{n+1} - t_n) = y_n + f_n h }[/math] if h is constant

Backward Euler formula: [math]\displaystyle{ y_{n+1} = y_n + h f(t_{n+1}, y_{n+1}) }[/math]

Local truncation error: [math]\displaystyle{ e_{n+1} = \frac{1}{2} \Phi''(t_n)h^2 \leq \frac{Mh^2}{2} }[/math] where [math]\displaystyle{ m = \mathrm{max} |\Phi''(t)| }[/math]

Local error is proportional to [math]\displaystyle{ h^2 }[/math].

Global error is proportional to h.

3. Improved Euler Formula (or Heun Formula):

[math]\displaystyle{ y_{n+1} = y_n + \frac{f_n + f(t_n + h, y_n + hf_n)}{2} h }[/math]

To determine local error we took the taylor expansions of [math]\displaystyle{ y }[/math] and [math]\displaystyle{ \Phi }[/math] and compared them.

[math]\displaystyle{ \Phi(x_1) = \Phi(x_0 + h) = \Phi(x_0) + h \Phi'(x_0) + \frac{h^2}{2} \Phi''(x_0) + O(h^3) }[/math]

[math]\displaystyle{ \Phi'(x) = f(x, \Phi(x)) \quad \Phi''(x) = \Phi'(x) = f_x(x, \Phi(x)) + f_y(x, \Phi(x))\Phi'(x) }[/math]

[math]\displaystyle{ \Phi(x_0 + h) = y_0 + h f(x_0, y_0) + \frac{h^2}{2}(f_x + f_y f) + O(h^3) }[/math]

We compare this to the computed value [math]\displaystyle{ y_1 }[/math]

[math]\displaystyle{ y_1 = y_0 +\frac{h}{2}(f(x_0, y_0) + f(x_0 + h, y_0 + hf(x_0, y_0))) }[/math]

Taking the taylor expansion we get

[math]\displaystyle{ y_1 = y_0 + h f(x_0, y_0) + \frac{h^2}{2}(f_x + f_y f) + O(h^3) }[/math]

So we have shown that the Improved Euler Formula is accurate up to an error term proportional to [math]\displaystyle{ h^3 }[/math]

Local truncation error is proportional to [math]\displaystyle{ h^3 }[/math]

Global truncation error is proportional to [math]\displaystyle{ h^2 }[/math]

4. The Runge-Kutta Method:

[math]\displaystyle{ y_{n+1} = y_n + \frac{k_{n1} + 2k_{n2} + 2k_{n3} + k_{n4}}{6} h }[/math]

where

[math]\displaystyle{ k_{n1} = f(t_n, y_n) \quad k_{n2} = f(t_n + \frac{1}{2} h, y_n + \frac{1}{2}hk_{n1}) }[/math]

[math]\displaystyle{ k_{n3} = f(t_n + \frac{1}{2} h, y_n + \frac{1}{2}hk_{n2}) \quad k_{n4} = f(t_n + h, y_n + hk_{n3}) }[/math]

Local truncation error is proportional to [math]\displaystyle{ h^5 }[/math].

Global truncation error is proportional to [math]\displaystyle{ h^4 }[/math].

Python Example of Euler's Method

In class on October 15th we discussed Euler's Method to numerically compute a solution to a differential equation. [math]\displaystyle{ x_0 }[/math] and [math]\displaystyle{ y_0 }[/math] are given as well as an increment amount [math]\displaystyle{ h }[/math], [math]\displaystyle{ x_{n+1} = x_n + h }[/math], and we use the guess [math]\displaystyle{ y_{n+1} = y_n + f(x_n, y_n)*h }[/math] where f computes the derivative as a function of x and y.

Here is an example of code (written in Python) which carries out Euler's Method for the example we discussed in class, [math]\displaystyle{ y' = -y }[/math]:

   def f(x, y):
       return -y
   
   def euler(x, y, f, h, x_max):
       """Take in coordinates x and y, a function f(x, y) which calculates
          dy/dx at (x, y), an increment h, and a maximum value of x.
          
          Return a list containing coordinates in the Euler's Method computation
          of the solution to Phi' = f(x, Phi(x)), Phi(x) = y, with the x
          values of those coordinates separated by h, and not exceeding x_max.
       """
       if x > x_max: # we have already calculated all our values
           return []
       x_next, y_next = (x + h, y + f(x, y)*h) # calculate the next x, y values
       # return the current coordinates, and every coordinates following it, in a list
       return [(x_next, y_next)] + euler(x_next, y_next, f, h, x_max) 
   
   if __name__ == '__main__':
       print euler(0, 1, f, 0.01, 1)[-1]