Resolution of ODE Cauchy problem with delay

Question

We have the following ODE with delay $$ du(t)/dt= au(t-\tau) - bu(t), $$ $$ u(t)=u_0: -\tau < t < 0 $$ How we can calculate the solution of this problem? Please The delay make this problem so complicated for me

Thank you for the help

Answer 1

How to obtain an approximate solution for small $\tau$?.

Hint.

After transforming Laplace we have

$$ sU(s)-u(0) = a\int_0^{\infty}e^{-st}u(t-\tau)dt-bU(s) $$

but

$$ \int_0^{\infty}e^{-st}u(t-\tau)dt=e^{-\tau s}\int_{-\tau}^{\infty}e^{-s\xi}u(\xi)d\xi=e^{-\tau s}\left(\int_{-\tau}^{0}e^{-s\xi}u_0d\xi+U(s)\right)=e^{-\tau s}\left(\frac{u_0}{s}\left(e^{s\tau}-1\right)+U(s)\right) $$

so we have

$$ sU(s)= ae^{-\tau s}\left(\frac{u_0}{s}\left(e^{s\tau}-1\right)+U(s)\right)+bU(s)+u_0 $$

or

$$ U(s) = \frac{1}{s-b-a e^{-\tau s}}\left(ae^{\tau s}\left(\frac{u_0}{s}\left(e^{s\tau}-1\right)\right)+u_0\right) $$

now for small $\tau$ we can use the Padé approximation $e^{\tau s}\approx \frac{1+\frac{s \tau }{2}}{1-\frac{s \tau }{2}}$ and after that, we have

$$ U(s) = \frac{u_0 (2 a \tau +s \tau +2)}{s \tau (a-b+s)-2 (a+b-s)} $$

or

$$ U(s) = \frac{c_2}{s+r_1}+\frac{c_3}{s+r_2} $$

here $r_1,r_2$ are the roots of $s \tau (a-b+s)-2 (a+b-s)=0$

making the inversion, we have a quantitative-qualitative way to study it's behavior.

EDIT

Considering the denominator of $\frac{1}{s-b-a e^{-\tau s}}$ we can make considerations concerning the $u(t)$ behavior. The zeros of $s-b-a e^{-\tau s}$ will dictate if the solution is a stable solution or not. To have a stable solution is sufficient that the roots of $s-b-a e^{-\tau s}=0$ be inside the left half complex plane: then making $s=x+iy$ we have

$$ \cases{ x-b-a e^{-\tau x}\cos(\tau y) = 0\\ y+a e^{-\tau x}\sin(\tau y) = 0 }\ \ \ \ (1) $$

Supposing $a=1,b=5,\tau=10$ we have a root location map as follows. The intersections between the red and blue curves are the solutions for $s-b-ae^{-\tau s}=0$. In this case the solution $u(t)$ is limited for all $t\in \mathbb{R}$ because all poles are inside the left complex plane.

Comparing with $a=2,b=1,\tau = 2$ we can see that in this case the solution is unlimited because it has a pole into the right half complex plane.

EDIT-2

As a general rule, from $(1)$ we obtain

$$ \Phi(x,y) = \left(\frac{x-b}{a}\right)^2+\left(\frac{y}{a}\right)^2-e^{-2\tau x}=0 $$

so a sufficient condition for stability is that the curve $\Phi(x,y)$ were contained into the left half complex plane. In the previous plots, $\Phi(x,y)=0$ appears as a dashed black trace.