Say we have the line segment $L(t) = [0,t]$, and randomly remove open intervals of length $1$ from $L(t)$ until no more open intervals of length $1$ remain. Define $u(t)$ as the expected measure of what remains.


Note $u(t) = t$ for $0 \le t < 1$ and $u(t) = t-1$ for $1 \le t < 2$. In general, we have the integral equation $$u(t) = \frac{2}{t-1} \int\_0^{t-1} u(s)\ ds.$$ Rearranging and differentiating, we get the delay differential equation $$u'(t) = \frac{-1}{t-1} u(t) + \frac{2}{t-1} u(t-1).$$ Note the equilibrium solution is $u(t) = c(t+1)$ for some constant $c$.


Main question - what's the value of $c$? My guess is $c = \frac{1}{4}$, but I have no proof. More generally, what are the asymptotics?


One additional observation. Using the transformation $u(t) = \frac{v(t)}{t-1}$, we can write our delay differential equation in the simplified form $$v'(t) = \frac{2}{t-2} v(t-1).$$


Here's the motivation for this question. Say we have a very large auditorium, and pairs of people fill it up, who sit together in adjacent seats. When no more pairs can be seated, about what fraction of the seats remain unoccupied? The answer is about $\frac{1}{e^2}$. Now instead of pairs, we seat groups of $n$ people, all of whom sit adjacently. As $n$ gets very large, about what fraction of seats are expected to remain unoccupied when the auditorium is filled? The answer should be the same as for my question.



Asymptotics of a delay differential equation
Let us show that
\begin{equation\*}
 u(t)\ge c(1+t)\text{ for some real $c>1/4$ and all $t\ge4$. }\tag{1}
\end{equation\*}
Let
\begin{equation\*}
 u\_n(z):=u(n+z),\quad w\_n(z):=u\_n(z)-\tfrac14\,(1+n+z); 
\end{equation\*}
here and in what follows, $z\in[0,1]$ and $n=0,1,\dots$.
Then for $n\ge1$
\begin{equation\*}
 w\_n(z)=\frac2{n-1+z}\,\Big(\sum\_{k=0}^{n-2}J\_k+\int\_0^z w\_{n-1}(x)\,dx\Big), 
\end{equation\*}
where
\begin{equation\*}
 J\_k:=\int\_0^1 w\_k(x)\,dx. 
\end{equation\*}
Next, $J\_0=1/8=-J\_1$, and hence
\begin{equation\*}
 w\_n(z)=\frac2{n-1+z}\,\Big(\sum\_{k=2}^{n-2}J\_k+\int\_0^z w\_{n-1}(x)\,dx\Big) \tag{2}
\end{equation\*}
for $n\ge4$. Moreover, $J\_2=0.01\ldots>0$ and
\begin{equation\*}
 w\_3(x)=\frac{x (3 x-14)+16 \ln (x+1)}{4 (x+2)}>0
\end{equation\*}
for $x\in(0,1]$, so that $J\_3>0$. It follows now from (2) by induction that $w\_n>0$ on $(0,1]$ and $J\_n>0$ for all $n\ge3$.


Now it follows from (2) that
\begin{equation\*}
 J\_n\ge\frac2n\,\sum\_{k=2}^{n-2}J\_k \tag{3}
\end{equation\*}
for all $n\ge4$. Recall that $J\_n>0$ for all $n\ge2$. So,
\begin{equation\*}
 b:=\min\_{2\le k\le10}\frac{J\_k}{6+k}>0, 
\end{equation\*}
and
\begin{equation\*}
 J\_n\ge b(6+n) \tag{4}
\end{equation\*}
for $n=2,\dots,10$.
Also,
\begin{equation\*}
 \frac2n\,\sum\_{k=2}^{n-2}(6+k)=9 - \frac{36}n + n\ge6+n
\end{equation\*}
for $n\ge12$. So, by induction, it follows from (3) that (4) holds for all $n\ge2$. So, in view of (2),
\begin{equation\*}
 w\_n(z)\ge\frac2n\,\sum\_{k=2}^{n-2}J\_k
 \ge b(6+n) \tag{5}
\end{equation\*}
for $n\ge4$. So,
\begin{equation\*}
 u(n+z)=u\_n(z)=w\_n(z)+\tfrac14\,(1+n+z)\ge b(6+n)+\tfrac14\,(1+n+z)
 \ge c(1+n+z)
\end{equation\*}
for $n\ge4$, where $c:=b+\tfrac14>\tfrac14$. Thus, (1) is proved.




---


It is much easier to show that
\begin{equation\*}
 u(t)\le C(1+t)\text{ for some real $C>0$ and all $t\ge0$. }
\end{equation\*}


