I want to find the Richardson extrapolation of the 4th order Runga-Kutta method. I found a formula on Wikipedia for a general approximation and then I tried applying it to the specific method. I got this:
$A=A(h)+kh^4+O(h^{5})$
I know that the local truncation error is proportional to $h^5$ which is why I chose the last term to be of that order. For the other step size I get:
$A=A(h/2)+k(h/2)^4+O(h^{5})$
Then I can multiply the first equation by 16 and subtract the second equation and get:
$15A=16A-A=16A(h/2)-A(h)+15O(h^5)$
Dividing by 15, I finally get:
$A=\frac{16}{15}A(\frac{h}{2})-\frac{1}{15}A(h)+O(h^5)$
But is this the right way to do the extrapolation? I am not sure how to interpret the term $kh^4$. Does this term belong to the approximation or the error?
Best Answer
You compare 2 approximations for the exact value $y(t)=A=A(0)$ at some time $t=nh$ starting from the same point $y(0)=y_0$ (shift the actual times as necessary).
The multi-step approximation global error has indeed the formula $$ A=A(h)+kh^4+O(h^5) $$ and all your calculations follow. The two integrations together result in $3n$ RK4 steps. (Note that the Fehlberg method, as fixed-step method, gives order 5 in 6 steps, which is also the average step count for the RK4 extrapolation if seen as method with step size $h/2$.)
In the linked answer the case $n=1$ is considered, that is, $t=h$. The error coefficient $k=k(t)$ is initially mostly linear in $t$, $k=k_1t+O(t^2)$, so that the error formula gains another factor $h$. $$ A=A(h)+\underbrace{(k_1t)}_{=k_1h=constant}·h^4+O(h^6) $$