matlabdelayodedde

Solving DDE with small time steps in for loop

I want to solve following DDE using a for loop in MATLAB:

xdot(t) = Ax(t) + BKx(t-h)

in which:

A = [0 1 ; -1 0.1];
B = [0 ; 1];
h = 0.2;
K = [-0.0469 -1.7663];
t = [0 5]

Solving this with conventional procedure is simple and the results are acceptable.

sol = dde23(ddefun,lags,history,tspan,options,varargin)

However, when I try to solve it using for loop, the results are wrong. Here is a simple code for my for loop.

time = 0:0.001:5;
for i = 2:5001
x(:,1) = [1 -1];
history(:,1) = [1 -1];
[t h] = ode23(@(t,h)histExam1(t,h,A,B,K),[time(i-1)    time(i)],history(:,i-1));
history(:,i)= h(end,:);
sol = dde23(((@(t,y,z)ddefun(t,y,z,A,B,K))),0.2,history(:,i),[time(i-1) time(i)]);
x(:,i)=sol.y(:,end);
end

I think, the only problem in this code is my time steps and delay input. I use same dde function for both codes so It cannot be a problem.The reason I want to solve DDE in a for loop is "BK" value which is state dependent (not in this simple example) and in each time step I need to update "BK".
enter image description here

The correct answer plotted above with conventional method. enter image description here And the wrong answer I get using "for loop" is plotted above. It is interesting to mention that correct answer is sensitive to delay. But delay doesn’t affect the answer from 2nd method.

Solution

  • Ok. After weeks of thinking, finally found a solution. Simply:

    sol = dde23(((@(t,y,z)ddefun(t,y,z,A,B,K))),0.2,[1;-1],[0 time(i)]);

    and let the magics happen. This code helps you with updating states at each time step. I hope it will help you in the future.

    All the Best,

    Sina