# Solving complex integro-differential equation

조회 수: 6 (최근 30일)
Jose Aroca 2020년 11월 13일
편집: Bruno Luong 2020년 11월 13일
I want to solve the following integro-differential equation: , with the conditon c(0)=1, and plot its real part, that should look like a decaying exponential. I want to be able to choose the value of Omega. This is what I have tried so far but Matlab says "Warning: Unable to find symbolic solution". The line c1(t) = subs(c1(t),t,t/om) is for the x axis to be in dimensionless units (Omega*t)
clearvars
close all
omega = 0.3;
syms t om tau c1(t)
f(t) = exp(1i*om*(t-tau));
Fx = -int(f,tau,[-inf,inf]);
ode = diff(c1,t) == c1(t)/2*Fx;
cond = c1(0) == 1;
c1(t) = dsolve (ode);
c1(t) = subs(c1(t),t,t/om);
c1(t) = subs(c1(t),om,omega);
fplot ((real(c1(t))).^2,[0,10])
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Walter Roberson 2020년 11월 13일
If it is a convolution there should be an f(tau)*f(t-tau) and that would make a big difference in the integral. You accidentally rewrote an integral that just might be convergent into an one that is not for real-valued omega.
Jose Aroca 2020년 11월 13일
My original problem is , where is as described above. I have already solved this previously using a Laplace transform, but for this special case c is now only dependent on t, not tau (c(t)) so I am supposed to be able to plug it out of the integral and then .

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### 채택된 답변

Bruno Luong 2020년 11월 13일
편집: Bruno Luong 2020년 11월 13일
For omega with imag(omega) < 0, the solution of the integro-differential eqt
(dc/dt)(t) = -c(t)/2 * integral_0^inf exp(i*omega*(t-tau)) dtau
has analytic form and is
c(t) = c0 * exp( exp(i*omega*t) / (2*omega^2) )
where c0 is an arbitrary constant.

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