Talk:Predictor–corrector method

Proposed example

Latest comment: 16 years ago1 comment1 person in discussion

Example of a trapezoidal predictor-corrector method

${\displaystyle y'=f(t,y),\quad y(t_{0})=y_{0}.}$ 

first calculate an initial guess value ${\displaystyle {\tilde {y}}_{g}}$ 

${\displaystyle {\tilde {y}}_{g}=y_{i}+hf(t_{i},y_{i})}$ 

next calculate successive guesses

${\displaystyle {\tilde {y}}_{g+1}=y_{i}+{\frac {h}{2}}(f(t_{i},y_{i})+f(t_{i+1},{\tilde {y}}_{g})).}$ 

${\displaystyle {\tilde {y}}_{g+2}=y_{i}+{\frac {h}{2}}(f(t_{i},y_{i})+f(t_{i+1},{\tilde {y}}_{g+1})).}$ 

...

${\displaystyle {\tilde {y}}_{g+n}=y_{i}+{\frac {h}{2}}(f(t_{i},y_{i})+f(t_{i+1},{\tilde {y}}_{g+n-1})).}$ 

until the guesses converge to within some error tolerance e:

${\displaystyle |{\tilde {y}}_{g+n}-{\tilde {y}}_{g+n-1}|<=e}$ 

Once convergence is reached, then use the final guess as the next step:

${\displaystyle y_{i+1}={\tilde {y}}_{g+n}.}$ 

If the guesses don't converge within some number of steps, such as ${\displaystyle n=16}$  reduce h and repeat the step. To optimize this, if the steps converge too soon, such as 4 steps, then increase h. If I remember correctly, the iterative process converges quadratically.

Jeffareid (talk) 04:09, 21 July 2009 (UTC)Reply