I am having a problem getting values to iterate with a while loop.

I need the values called u2(n) u3(n) u4(n) u5(n) to iterate with the four equations each time the loop runs. So it runs and say it gets u2=112 and u3=20, u4=20, u5=20 then I want it to use those values to run the equations again until the value of u5 is as close as possible to 250. The catch is that I need those values to start at 20 instead of nothing or zero which is causing me to get the error "Attempted to access u3(2); index out of bounds because nume1 (u3)=1." I know this has to be because I predefined those values as a scalar but I do not know any other way to do it. Thank you for your time.

clc; clear all

      n = 1;
      u1 = 500;
      r = .192;
      u6 = 20;
      h = 0;
      u5(n) = 20;
      u2(n) = 20;
      u3(n) = 20;
      u4(n) = 20;

while h < 250

      uc(n) = r*(u3(n) + u1) + (1 - 2*r)*(u2(n));
      uv(n) = r*(u4(n) + u2(n)) + (1 - 2*r)*(u3(n));
      ut(n) = r*(u5(n) + u3(n)) + (1 - 2*r)*(u4(n));
      ui(n) = r*(u6 + u4(n)) + (1 - 2*r)*(u5(n));

      u1 = 500;
      u2(n) = uc(n)
      u3(n) = uv(n)
      u4(n) = ut(n)
      u5(n) = ui(n)
      h = u5(n)
      u6 = 20;
      n = n+1;

end

EDIT:

I fixed the code with the help of the comments below and was able to get it to work, it is a little slow and could be sped up but it works none the less. Here it is for future reference, maybe it will help someone else.

clc; clear all

    n = 1;
    u1 = 500;
    r = .192;
    u6(n) = 20;
    u7(n) = 20;
    u8(n) = 20;
    u9 = 20;
    h = 0;
    u5(n) = 20;
    u2(n) = 20;
    u3(n) = 20;
    u4(n) = 20;
    g = 0;

while g < 250

    uc(n) = r*(u3(n) + u1) + abs(1 - 2*r)*(u2(n));
    uv(n) = r*(u4(n) + u2(n)) + abs(1 - 2*r)*(u3(n));
    ut(n) = r*(u5(n) + u3(n)) + abs(1 - 2*r)*(u4(n));
    ui(n) = r*(u6(n) + u4(n)) + abs(1 - 2*r)*(u5(n));
    uk(n) = r*(u7(n) + u5(n)) + abs(1 - 2*r)*(u6(n));
    ul(n) = r*(u8(n) + u6(n)) + abs(1 - 2*r)*(u7(n));
    uy(n) = r*(u9 + u7(n)) + abs(1 - 2*r)*(u8(n));

    u1 = 500;
    u2(n+1) = uc(n);
    u3(n+1) = uv(n);
    u4(n+1) = ut(n);
    u5(n+1) = ui(n);
    u6(n+1) = uk(n);
    u7(n+1) = ul(n);
    u8(n+1) = uy(n);
    u9 = 20;

    h = u5(n+1);
    n = n+1;
    g = h(end);

end

number_of_iterations = n-1;

t = (300/60)*number_of_iterations;

time = (0:5:575);

fprintf('Number of iterations to reach 250 degrees celcius is %.f\n',number_of_iterations) fprintf('Time it takes in minutes for x at 100 mm to reach 250 degrees celcius is %.f',t)

plot(time,u5)

xlabel('Time (Minutes)') ylabel('Iterations for U at 100 mm (Temperature)')

title('Temperature Vs. Time')

Answer 1

[Yao Li]

u2(n+1) = uc(n);
    u3(n+1) = uv(n);
    u4(n+1) = ut(n);
    u5(n+1) = ui(n);
    h = u5(n+1);

But it seems your algorithm needs improvement