convolution integral with dde

8 Ansichten (letzte 30 Tage)
Angie
Angie am 10 Jun. 2019
Kommentiert: Satheesh oe am 20 Dez. 2019
Hi guys, I am trying to solve this differential equation using dde. I have a problem with the integral term. In the code shown below, tau is the delay but I cannot just specify a constant value because it is also in the integral which goes from 0 to t. Does anywone know how to deal with it? Thank you!
%%
function sol = exer_3
sol = dde23(@exer3f,tau,[0; 0],[0, 10]);
figure
plot(sol.x,sol.y)
function v = exer3f(t,y,Z)
k = 125; m = 5; F = 1; w = 8;
c=@(t)exp(-t^2);
ylag = Z(:,1);
v = zeros(2,1);
v(1)=y(2);
v(2) = -(k*y(1) - F*cos(w*t) + integral(@(tau)c(tau).*ylag(1), 0, t,'ArrayValued',true))./m;
  4 Kommentare
Torsten
Torsten am 12 Jun. 2019
Bearbeitet: Torsten am 12 Jun. 2019
If you substitute
y = u'
you arrive at the integro-differential equation
y'(t) = F/m * cos(w*t) - k/m*u(0) - 1/m * integral_{0}^{t} y(s)*(c(t-s)+k) ds
which has the required form.
Satheesh oe
Satheesh oe am 20 Dez. 2019
Hi,
can i know why there is (+k) tem in the integral "(c(t-s)+k)".
Regards,
satheesh

Melden Sie sich an, um zu kommentieren.

Antworten (0)

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!

Translated by