For loop dies in the middle of solving

So the function works until a point in the for loop. I debugged and figured out the exact spot is in the for loop when i=.6 if you step until the line that says" u(:,i)=uvnp1(1:2,1)" the function basically gives up. And it gives me this error message "??? Subscript indices must either be real positive integers or logicals." It doesnt make sense to me because the i IS a real positive integer. Here is the function run it using the given numbers and you will see what I am talking about

 function oscillation(M,C,delta_t,tf,u0,v0)
% M is the masses matrix
% C is the spring constants matrix
% delta_t is the time step
% tf is the total time
% u0 is the initial positions
% v0 is the initial velocities

 A=[1 0;-1 1];
K=A'*C*A;
GT1=eye(4);
GT1(3:4,1:2)=-delta_t*eye(2)/2;
GT1(1:2,3:4)=((delta_t)*(M\K))/2;
GT2=eye(4);
GT2(3:4,1:2)=delta_t*eye(2)/2;
GT2(1:2,3:4)=-((delta_t)*(M\K))/2;
GT=GT1\GT2;
time=0:delta_t:tf;
u=zeros(2,(tf/delta_t)+1);
v=zeros(2,(tf/delta_t)+1);
u(:,1)=u0;
v(:,1)=v0;
uvn=[u0;v0];
for i=(delta_t:delta_t:tf)
    uvnp1=GT*uvn;
    u(:,(i/delta_t)+1)=uvnp1(1:2,1)
    v(:,(i/delta_t)+1)=uvnp1(3:4,1)
    uvn=uvnp1;
end
end
% oscillation([2 0;0 1],[1 0;0 2],0.1,1,[0;0],[1;-1])

Answer 1

[Image Analyst]

Since when is 0.6 an integer? Try this. I don't know if it's right, but at least it produces the arrays without crashing:

function test()
  oscillation([2 0;0 1],[1 0;0 2],0.1,1,[0;0],[1;-1])
end

function oscillation(M,C,delta_t,tf,u0,v0)
  % M is the masses matrix
  % C is the spring constants matrix
  % delta_t is the time step
  % tf is the total time
  % u0 is the initial positions
  % v0 is the initial velocities

  A=[1 0;-1 1];
  K=A'*C*A;
  GT1=eye(4);
  GT1(3:4,1:2)=-delta_t*eye(2)/2;
  GT1(1:2,3:4)=((delta_t)*(M\K))/2;
  GT2=eye(4);
  GT2(3:4,1:2)=delta_t*eye(2)/2;
  GT2(1:2,3:4)=-((delta_t)*(M\K))/2;
  GT=GT1\GT2;
  time=0:delta_t:tf;
  u=zeros(2,(tf/delta_t)+1);
  v=zeros(2,(tf/delta_t)+1);
  u(:,1)=u0;
  v(:,1)=v0;
  uvn=[u0;v0];
  counter = 1;
  for i=(delta_t:delta_t:tf)
    uvnp1=GT*uvn;
    u(:,counter)=uvnp1(1:2,1)
    v(:,counter)=uvnp1(3:4,1)
    uvn=uvnp1;
    counter = counter + 1;
  end
end

[Kyle]

Well the i=0.6 was divided by delta_t=0.1 which made it six. But this solved my problem, thanks.

[Walter Roberson]

It only made it approximately 6.

http://matlab.wikia.com/wiki/FAQ#Why_is_0.3_-_0.2_-_0.1_.28or_similar.29_not_equal_to_zero.3F

Answer 2

[Azzi Abdelmalek]

The problem can be caused by

u(:,(i/delta_t)+1)

Are you sur that i/delta_t is real positive integer or logical