convolution integral with dde

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Angie
Angie 2019 年 6 月 10 日
コメント済み: Satheesh oe 2019 年 12 月 20 日
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 件のコメント
Torsten
Torsten 2019 年 6 月 12 日
編集済み: Torsten 2019 年 6 月 12 日
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 2019 年 12 月 20 日
Hi,
can i know why there is (+k) tem in the integral "(c(t-s)+k)".
Regards,
satheesh

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