How to Extract Delayed State Terms in a Model with Distributed Delay?
3 Ansichten (letzte 30 Tage)
Ältere Kommentare anzeigen
Muhammad
am 22 Aug. 2024
Bearbeitet: Torsten
am 22 Aug. 2024
I'm working on a model with distributed delays, and I'm using the ddesd function to solve the delay differential equations. My setup involves a distributed delay, defined by:
tau = 1;
gamma = 0.5;
number_of_delays = 11; % should be odd
lags = linspace(tau-gamma,tau+gamma,number_of_delays);
tspan=[0 600];
sol=ddesd(@(t,y,Z)ddefunc(t,y,Z,lags),lags,[0.2; 0.08],tspan);
p=plot(sol.x,sol.y);
set(p,{'LineWidth'},{2;2})
title('y(t)')
xlabel('Time(days)'), ylabel('populations')
legend('x','y')
function yp = ddefunc(~,y,Z,lags)
a=0.1;
b=0.05;
c=0.08;
d=0.02;
yl1 = trapz(lags,Z(2,:));
yp = [a*y(1)-b*y(1)*yl1;
c*y(1)*y(2)-d*y(2)];
end
In the case of discrete delays, we can easily extract the delayed state terms using the deval or interp1 commands. However, I'm unsure how to proceed with extracting or examining the delayed state terms for distributed delay case.
0 Kommentare
Akzeptierte Antwort
Torsten
am 22 Aug. 2024
Bearbeitet: Torsten
am 22 Aug. 2024
So in the distributed case you want to extract
y2_delayed(t) = integral_{tau = t-1.5]^{tau = t-0.5} y2(tau) dtau
thus the term yl1 in ddefunc ?
I plotted it as "Lag term" below.
tau = 1;
gamma = 0.5;
lags = [tau-gamma;tau+gamma];
tspan = [0 600];
sol = dde23(@ddefunc,lags,[0.2; 0.08; 0.08*2*gamma],tspan,ddeset('RelTol',1e-6,'AbsTol',1e-6));
figure(1)
p = plot(sol.x,sol.y(1:2,:));
set(p,{'LineWidth'},{2;2})
title('y(t)')
xlabel('Time(days)'), ylabel('populations')
legend('x','y')
figure(2)
p = plot(sol.x,sol.y(3,:));
set(p,{'LineWidth'},{2})
title('Lag(t)')
xlabel('Time(days)'), ylabel('Lag term')
legend('Lag')
function yp = ddefunc(~,y,Z)
a=0.1;
b=0.05;
c=0.08;
d=0.02;
yp = [a*y(1)-b*y(1)*y(3);
c*y(1)*y(2)-d*y(2);
Z(2,1)-Z(2,2)];
end
Weitere Antworten (1)
Torsten
am 22 Aug. 2024
Verschoben: Torsten
am 22 Aug. 2024
What exactly do you want to extract ? y2, evaluated at (t-lags) ?
And you know that your equation can be solved using dde23 because you can treat
integral_{tau = t-1.5]^{tau = t-0.5} y2(tau) dtau
in the same way as I suggested here:
?
In this case, you get an equation with discrete delays of length 1.5 and 0.5.
Hint:
d/dt (integral_{tau = t-1.5]^{tau = t-0.5} y2(tau) dtau) = y2(t-0.5) - y2(t-1.5)
2 Kommentare
Siehe auch
Kategorien
Mehr zu Numerical Integration and Differential Equations finden Sie in Help Center und File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!