12-267/Numerical Methods

Numerical methods: ${\frac {dy}{dt}}=f(t,y)$ and $y(t_{0})=y_{0}$, $y=\Phi (t)$ is a solution.

1. Using the proof of Picard's Theorem:

$\Phi _{0}(x)=y_{0}$

$\Phi _{n}(x)=y_{0}+\int _{x_{0}}^{x}f(x,\Phi _{n-1}(x))dx$

$\Phi _{n}(x)\rightarrow \Phi (x)$

2. The Euler Method:

$y_{n+1}=y_{n}+f(t_{n},y_{n})(t_{n+1}-t_{n})=y_{n}+f_{n}h$ if h is constant

Backward Euler formula: $y_{n+1}=y_{n}+hf(t_{n+1},y_{n+1})$

Local truncation error: $e_{n+1}={\frac {1}{2}}\Phi ''(t_{n})h^{2}\leq {\frac {Mh^{2}}{2}}$ where $m=\mathrm {max} |\Phi ''(t)|$

Local error is proportional to $h^{2}$.

Global error is proportional to h.

3. Improved Euler Formula (or Heun Formula):

$y_{n+1}=y_{n}+{\frac {f_{n}+f(t_{n}+h,y_{n}+hf_{n})}{2}}h$

Local truncation error is proportional to $h^{3}$

Global truncation error is proportional to $h^{2}$

4. The Runge-Kutta Method:

$y_{n+1}=y_{n}+{\frac {k_{n1}+2k_{n2}+2k_{n3}+k_{n4}}{6}}h$

where

$k_{n1}=f(t_{n},y_{n})\quad k_{n2}=f(t_{n}+{\frac {1}{2}}h,y_{n}+{\frac {1}{2}}hk_{n1})$

$k_{n3}=f(t_{n}+{\frac {1}{2}}h,y_{n}+{\frac {1}{2}}hk_{n2})\quad k_{n4}=f(t_{n}+h,y_{n}+hk_{n3})$

Local truncation error is proportional to $h^{5}$.

Global truncation error is proportional to $h^{4}$.

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