Runge-Kutta fourth order (rk4)
Show older comments
When implementing the Runge-Kutta order 4 scheme(rk4) u1,...,un are computed using the algorithm:
for i = 0,1,...,n-1
k1 = f (ti, ui),
k2 = f (ti + h/2, ui + h/2*k1),
k3 = f (ti + h/2, ui + h/2*k2),
k4 = f (ti + h, ui + h*k3),
ui+1 = ui + h/6 (k1 + 2k2 + 2k3 + k4).
Would part of this code be written in matlab as the following:
for i = 1:n
k1 = h * feval ( f, u(:,i) );
k2 = h * feval ( f, u(:,i) + k1/2 );
k3 = h * feval ( f, u(:,i) + k2/2 );
k4 = h * feval ( f, u(:,i) + k3 );
u(:,i+1) = u(:,i) + h*( k1 + 2*k2 + 2*k3 + k4 ) / 6;
end;
I'm not too sure if the "h" in the last line (u(:,i+1)) should be there are not.
Any help would be very much appreciated.
Accepted Answer
More Answers (1)
Meysam Mahooti
on 5 May 2021
0 votes
https://www.mathworks.com/matlabcentral/fileexchange/61130-runge-kutta-fehlberg-rkf78?s_tid=srchtitle
Categories
Find more on Programming in Help Center and File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!