我理解你试图解决系统of differential equation using ‘dde23’. To do this you need to do following steps.
- Define constant delays.
In this system of equation there are two lags i.e.
(t-1) and (t-0.2).
lags = [1 0.2];
2. Define Solution History.
It Defines the first solution from which the solver starts iterations.
functions = history(t)
s = ones(3,1);
end
3. Form the equation.
functiondydt = ddex1de(t,y,Z)
ylag1 = Z(:,1);
ylag2 = Z(:,2);
dydt = [ylag1(1);
ylag1(1)+ylag2(2);
y(2)];
end
Here In the call to ddex1de,
‘t’ is a scalar indicates the current ‘t’ in the equation
‘y’ is a column vector approximates y(t)
‘Z’ is a column vector approximates y(t –
α
j
) for delay
α
j
= lags(J).
In the following example Solution history is [1;1;1]. So Approximate values of lag ie ‘Z’ will be a matrix of size 3x2.In ‘Z’ Row defines the number of equations and column defines Number of lags.
4. Solve using ‘dde23’.
tspan = [0 5];
sol = dde23(@ddefun, lags, @history, tspan);
