# Steady-state of `degenerate' delayed differential equation

Consider the simple delayed differential equation:

$X'(t) = -a X(t) + a X(t - d)$

where $d$ and $a$ are positive constants. I'm interested in the possible steady-state (stationary) solutions of this and other similar equations as $t \rightarrow \infty$.

The steady-state seems to vary depending on the initial conditions, for example, I am interested in the case with initial data given by $X(t) = 0$ for $t \in [-d, 0)$ and $X(0) = 1$. Numerical integration with Matlab's dde23 routine gives the steady-state in the case at around 0.33. My question is whether there is another way (other than numerical integration for a long time period) to find such steady-states analogous to finding fixed points of ODEs. If you try and do that here, by assuming that the terms $X(t)$ and $X(t - d)$ become equal as $t \rightarrow \infty$, and that $X'(t) = 0$, you get the useless relation:

$0 = -a X^* + a X^*$

which, of course, tells you nothing about possible fixed points $X^*$.

Any ideas?

Thanks!

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Numerical integration is not the only way to tackle this specific problem. You can solve it successively on intervals $[0,d)$,$[d,2d)$,$[2d,3d)$, and so on. On each interval, you can see that the delayed term, $X(t-d)$, is known: it is the initial data on the first interval, and the solution on the previous interval for successive ones. You get a piecewise solution in this way, and for this problem, the solution on the $n$th interval is $e^{-at}P_{n-1}(t)$, where $P_n(t)$ is some polynomial of order $n$. However, studying the large $t$ behavior using such an approach may not be helpful. –  rajb245 Dec 16 '14 at 20:34
It may be possible to say something about the sequence of values $a_n = X(nd)$, and you might be able to prove that the sequence $a_n$ converges and find the value to which it converges using the "piecewise-solver" approach. –  rajb245 Dec 16 '14 at 20:55