understanding DDE23 function format
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I look at this example https://www.mathworks.com/help/matlab/math/dde-with-constant-delays.html
DDE23 function has a format of
ddex1de(t,y,Z)
How should one understand Z?
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Guru Mohanty
2020-5-4
I understand you are trying to solve system 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.
function s = history(t)
s = ones(3,1);
end
3. Form the equation.
function dydt = 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 – αj) for delay αj= 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);
2 个评论
Guru Mohanty
2020-5-4
As it is a System of 3 Equations, 'y', 'dydt', 'ylag1' 'ylag2' have dimension 3x1. There are two constant Delay present, So The dimention of 'Z' is 3x2.
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