Problem with RK4
1 visualización (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
Categorías
Más información sobre Loops and Conditional Statements en Help Center y File Exchange.
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!