Applying Richardson Extrapolation to Euler Method

Question

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 1

You want to compare two approximations at $t=h$. The error of two Euler steps at half step size is in the leading order $$ 2\cdot C\left(\frac h2\right)^2=\frac12\cdot Ch^2 $$ if the truncation error is locally $Ch^2$ 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 $h$, $$ Euler(h)=y+Ch^2 +O(h^3) $$ and with two steps of half the step size $$ Euler(h/2) =y+\frac12Ch^2+O(h^3). $$ Eliminating the term $Ch^2$ $$ 2\cdot Euler(h/2)-Euler(h)=y+O(h^3) $$ This is incidentally the explicit midpoint method $$ y(t+h)=y(t)+hf(t+\tfrac h2,\,y(t)+\tfrac h2f(t,y(t)))+O(h^3) $$

Answer 2

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.