- Define constant delays.
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 年 5 月 4 日
I understand you are trying to solve system of differential equation using ‘dde23’. To do this you need to do following steps.
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);
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Guru Mohanty
2020 年 5 月 4 日
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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