I was trying to solve Lorenz System in matlab by using the method of RK2 and RK4. I had a script for both of the methods, the problem now is how can converge the following
y(1) = @(t,y) 10*(y(2)-y(1));
y(2) = @(t,y) y(1)*(28-y(3))-y(2);
y(3) = @(t,y) y(1)*y(2)-8*y(3)/3;
into simply an column vector of y.
here is what i was hoping for and it never worked:
y = zeros(3,1);
y(1) = @(t,y) 10*(y(2)-y(1));
y(2) = @(t,y) y(1)*(28-y(3))-y(2);
y(3) = @(t,y) y(1)*y(2)-8*y(3)/3;
and following is my RK2 function. my RK4 is similar to this RK2, manybe this can help you understand why I really need an vector of functions.
function y = RK2(fcn,lrange,urange,step,init)
%fcn = vector of functions
%lrange = lower bound
%urange = upper bound
%step = number of steps
%init = initial value
row = size(fcn,1);
stepsize = (urange-lrange)/step;
y = zeros(row,step);
%initializing vector of y
y(:,1) = init;
%initial condition
t = zeros(1,step+1);
%initializing vector of t
if row ~= size(init,1)
disp('number of functions and number of initial values do not match');
end
for n = 1:step
t(n) = (n-1)*stepsize;
t(step+1) = urange;
y1 = stepsize.*fcn(t(n),y(:,n));
y2 = stepsize.*fcn(t(n) + stepsize/2, y(:,n) + y1./2);
y(:,n+1) = y(:,n) + y2;
end
This should work, just make the function output a vector:
y = @(t,y) [10*(y(2)-y(1)), y(1)*(28-y(3))-y(2), y(1)*y(2)-8*y(3)/3];
y(1,[1;2;3])
size(y(1,[1;2;3]))