Rene Nicholson

Answered

2022-10-20

When applying Richardson Extrapolation to Euler Method, when isn't the local truncation error of h^2 used to get (4*Euler(h/2) - Euler(h)) / 3 instead of using the global error of h to get 2*Euler(h/2) - Euler(h)?

Answer & Explanation

Szulikto

Expert

2022-10-21Added 22 answers

In general when you are applying Richardson extrapolation to a method $T(h)$, it should be an approximation of a fixed quantity $I$ which doesn't depend on $h$. Thus for instance for the Euler method, $T(h)$ is the result of running the Euler method with step size h for a fixed ODE up to a fixed finite time. So the error that matters is the global error.

rhenan5v

Expert

2022-10-22Added 3 answers

You want to compare two approximations at $t=h$. The error of two Euler steps at half step size is in the leading order

$2\cdot C{\left(\frac{h}{2}\right)}^{2}=\frac{1}{2}\cdot C{h}^{2}$

if the truncation error is locally $C{h}^{2}$ for a step of size $h$.

Thus you get, if $Euler(h)$ denotes the method application to get to a fixed time $t$ with step size $h$,

$Euler(h)=y+C{h}^{2}+O({h}^{3})$

and with two steps of half the step size

$Euler(h/2)=y+\frac{1}{2}C{h}^{2}+O({h}^{3}).$

Eliminating the term $C{h}^{2}$

$2\cdot Euler(h/2)-Euler(h)=y+O({h}^{3})$

This is incidentally the explicit midpoint method

$y(t+h)=y(t)+hf(t+{\textstyle \frac{h}{2}},\phantom{\rule{thinmathspace}{0ex}}y(t)+{\textstyle \frac{h}{2}}f(t,y(t)))+O({h}^{3})$

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