Problem with RK4

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B.E. 2019-12-21

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https://ww2.mathworks.cn/matlabcentral/answers/497656-problem-with-rk4

评论： B.E. 2019-12-22

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

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Walter Roberson 2019-12-21

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.

B.E. 2019-12-22