12-267/Numerical Methods

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]

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].

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