My code is giving me zeros everywhere for my solution vectors but I don't know why. I've decomposed a coupled second order ODE into 4 1st order ODE's.
I have my function defined as xp.m
function zprime = f(t,z)
a = 1;
b = 1;
c = 1.5;
zprime = zeros(4,1);
zprime(1) = z(2);
zprime(2) = -a*z(1) + b*(z(3) - z(1));
zprime(3) = z(4);
zprime(4) = -c*(z(3) - z(1));
end
I run it in matlab with the following command:
[t,z] = ode45('xp',[1,100],[0 0 0 0]);
since my initial conditions are all 0. Is it that my initial conditions give the 0 solution or something else? When I change the ic's, the solutions change, as expected.
Thanks
Your actual question has more to do with mathematics than Matlab or programming. If you plug in zeros to your function f, you'll see immediately that it can give no other answer besides zero. You should look up equilibrium points or fixed points. Even if an equilibrium is unstable (imagine the top of a hill), a state exactly at that point with no disturbances (external inputs, numerical error) will remain balanced forever. If you're going to be working with differential equations, it's good to know how to find equilibrium points and then how to calculate the Jacobian matrix evaluated at these points in order to determine system properties. If you have further questions in this area, I'd suggest going to math.stackexchange.com.
Also, you're using an old scheme for calling your integration function. You can also pass in your parameters. Define your integration function as a sub-function in your main function or as a separate function file (same file name as function name – you'll want something other than f)
function zprime = f(t,z,a,b,c)
zprime(1,1) = z(2);
zprime(2,1) = -a*z(1) + b*(z(3) - z(1));
zprime(3,1) = z(4);
zprime(4,1) = -c*(z(3) - z(1));
Then, in your main function, call
a = 1;
b = 1;
c = 1.5;
[t,z] = ode45(@(t,z)f(t,z,a,b,c),[1 100],[0 0 0 0]);