Solving system of ODEs

I have a system of DEs similar to the following:

If the variable Z is not present (i.e. eq (3) also), then I can solve it using ode45 command.

But the availability of Z is creating problem for me. To get the value of Z at time t, we have to solve the ODE (3) from time 0 to \tau, then we have to put that value of Z in equation (2), for evey instance 't'.

The main problem is, while solving DE using ode45, at any instance t, MATLAB uses the values of variables at that instance only. But here we require values of variables at time (t-\tau) to use at time t.

Please help me to solve the problem. Thanks in advance.

[darova]

Can you please attach your script?

[Ashok Das]

Main function is:

X_0 = 1; Y_0 = 1;  % Initial condition
t = linspace(0,10);  %TIme
[t, Var] = ode45(@myfun, t, [X_0 Y_0]);  % ODE solve
X= Var(:,1); 
Y= Var(:,2);

Create myfun :

function [dX dY] = myfun(t, IC)
X= IC(1);  Y=IC(2);
a=1; b=2; c=3; d=4;  % Values
dX = a*X + b*Y;
% Now here before this line I need to solve the DE 
% (3) upto time t= \tau to get the value of Z
%  (which I am not able to do).
dY = c*X+ d*Y + Z; 
end

Answer 1

[darova]

What if just solve 2 systems?

First one (t <= tau) with initial conditions: X(0) = X0, Y(0) = Y0

[t, Var] = ode45(...);  % ODE solve
X= Var(:,1); 
Y= Var(:,2);
x0 = X(end);        % initial conditions for 2d system
y0 = Y(end);

Second (t >= tau) with Z(tau) = Z0 (X and Y initial conditions give from the first solution)

What do you think about my idea?

[Ashok Das]

The problem is that, for the equation (3), the value of Z and its derivative is at time t' , but the variables X and Y are considered at time (t' +t - \tau).

[darova]

How t' depends on t? If t = tau+0.5, what will be t'?

Are you sure about correctness of (t' +t -tau)?

[Ashok Das]

In equation (3), t' is the only independent variable. t and \tau are constant regarding to equation (3). We have to solve (3) for t' =0 to t' = \tau, and fit that value in eqaution (2)

[Ashok Das]

If t = tau+0.5, then (t' +t -tau)= t' +0.5 .

The value of \tau is fixed throughout the simulation.

We have to find Z at t' = \tau for each value of t.

Answer 2

[Steven Lord]

I only skimmed your problem but it looks like when Z is not present, you have a set of ordinary differential equations. Once Z comes into the picture, you no longer have a set of ordinary differential equations. Your equations are now delay differential equations. Use the ODE solvers to solve the system until t reaches tau, then use the DDE solvers (like dde23 or ddesd) to solve the system using the results from the ODE solvers as the history.