Rene Nicholson

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)?

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Szulikto

Expert

In general when you are applying Richardson extrapolation to a method $T\left(h\right)$, it should be an approximation of a fixed quantity $I$ which doesn't depend on $h$. Thus for instance for the Euler method, $T\left(h\right)$ 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.

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rhenan5v

Expert

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 $Euler\left(h\right)$ denotes the method application to get to a fixed time $t$ with step size $h$,
$Euler\left(h\right)=y+C{h}^{2}+O\left({h}^{3}\right)$
and with two steps of half the step size
$Euler\left(h/2\right)=y+\frac{1}{2}C{h}^{2}+O\left({h}^{3}\right).$
Eliminating the term $C{h}^{2}$
$2\cdot Euler\left(h/2\right)-Euler\left(h\right)=y+O\left({h}^{3}\right)$
This is incidentally the explicit midpoint method
$y\left(t+h\right)=y\left(t\right)+hf\left(t+\frac{h}{2},\phantom{\rule{thinmathspace}{0ex}}y\left(t\right)+\frac{h}{2}f\left(t,y\left(t\right)\right)\right)+O\left({h}^{3}\right)$

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