Laplace Transform

Question

Let us assume that complex-valued differential equations as follows

$\dot{z}(t)=-Az(t)+Bz(t-\tau)$, $z\in \mathbb{C}$

How to find the solution of the above equation by using Laplace transform.

Answer 1

The LT of the solution is defined as

$$\int_0^{\infty} dt \, z(t) e^{-s t} = Z(s) $$

meaning that

$$\int_0^{\infty} dt \, z(t-\tau) e^{-s t} = e^{-s \tau} \int_{-\tau}^{\infty} dt \, z(t) \, e^{-s t}$$

If we assume as we usually do that $z(t) = 0$ when $t \lt 0$, we may then conclude that the LT of $z(t-\tau)$ is $e^{-s t} Z(s)$. Then the ODE is in $s$ space

$$s Z(s) - z(0) = \left ( -A + B e^{-s \tau} \right ) Z(s) $$

or

$$Z(s) = \frac{z(0)}{A+s-B e^{-s \tau}} $$

This is difficult to invert analytically, but all that's required once numerical values of $A$ and $B$ are given is to find the root(s) of the denominator numerically, say at $s=s_k$. The solution to the ODE is then

$$z(t) = z(0) \theta(t) \sum_k \frac{e^{s_k t}}{1+B \tau e^{-s_k \tau}} $$

where $\theta$ is the Heaviside step function.