Problem with RK4
2 visualizaciones (últimos 30 días)
Mostrar comentarios más antiguos
I have a problem with this code when I choose Nbite equal to 1
function y=RK4(Myfun,s1,s2,Nbite,y0)
h=(s2-s1)/(Nbite-1);
tt=s1:h:s2;
y=zeros(1,Nbite);
y(1)=y0;
for i=1:Nbite-1
k1 = h*Myfun(tt(i), y(i));
k2 = h*Myfun(tt(i) + h/2, y(i) + k1/2 );
k3 = h*Myfun(tt(i) + h/2, y(i) + k2/2 );
k4 = h*Myfun(tt(i+1), y(i) + k3);
y(i+1) = y(i) + (k1+2*k2+2*k3+k4)/6;
end
G=@(t,x) x.*log(1./x)-(1+tanh(100*(x-10))).*cos(t).*sin(2*t)
z=RK4(G,0.0.1,1,1)
ans z= 1????????????????
There is a problem because the time step in the code is worth 0
4 comentarios
Walter Roberson
el 21 de Dic. de 2019
B.E. is partly right. When Nbite = 1, h does go to infinity, but s1:inf:s2 is defined and is s1 .
But you have y(1)=y0 and your y0 is 1, so you initialize y(1) to 1. Then you loop over i=1:Nbite-1 which is i=1:0 so you do not execute the loop body so you do not change y at all. The answer is just going to be the y0 that you started with.
Respuestas (0)
Ver también
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!