Asymptotics of a delay differential equation 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.
 A: I can’t give a closed form expression for the constant $c$, but I believe there is a plausible argument for a lower bound that is strictly greater than $\frac{1}{4}$.
Suppose we define the solution piecewise, with $u_n(t)$ the solution for $n\leq t \lt n+1$, where $n\in \mathbb{N}$.  We have:
$$u_0(t) = t$$
$$u_n(t) = \frac{2}{t-1}\left( I_{n-1} + \int_{n-1}^{t-1}u_{n-1}(s) \space\mathrm{d}s\right)$$
where
$$I_0=0$$
$$I_n = I_{n-1} + \int_{n-1}^{n}u_{n-1}(s)\space\mathrm{d}s$$
We can find closed-form solutions up to $u_4$:
$$u(t)=
\begin{array}{cc}
  & 
\begin{array}{cc}
 t & 0\leq t<1 \\
 -1+t & 1\leq t<2 \\
 -3+\frac{2}{-1+t}+t & 2\leq t<3 \\
 \frac{17+(-8+t) t+4 \log (-2+t)}{-1+t} & 3\leq t<4 \\
 \frac{2 \left(\frac{49}{2}+\frac{\pi ^2}{3}+\frac{1}{2} (-16+t) t-8 \log (2)+2 (5+2 \log (-3+t))
   \log (-2+t)+4 \text{Li}_2(3-t)\right)}{-1+t} & 4\leq t<5 \\
\end{array}
 \\
\end{array}
$$
The plot below shows the closed-form solutions for $u$, in blue, along with the line $c_4(t+1)$, in yellow, for a value $c_4 = 0.2522189$ that we will explain shortly.

Intuitively at least, it is apparent that $u_4(t)$ is well-approximated by the line $c_4(t+1)$ here.
Without making that notion precise, suppose that $u_n(t)$ can be approximated by $c_{n}(t+1)$, for all $n \ge 4$.  We then have:
$$\begin{array}{rcl}
c_n(t+1) & \approx & \frac{2}{t-1}\left( I_{n-1} + \int_{n-1}^{t-1}c_{n-1}(s+1) \space\mathrm{d}s\right) \\
c_n & \approx & \frac{1}{t^2-1}\left(2 I_{n-1} + c_{n-1} (t^2 - n^2)\right)
\end{array}
$$
Putting $t=n+1$:
$$\begin{array}{rcl}
c_n & \approx & \frac{(2n+1)c_{n-1} + 2I_{n-1}}{n(n+2)}
\end{array}
$$
We also have:
$$\begin{array}{rcl}
I_n & \approx & I_{n-1} + \int_{n-1}^n c_{n-1}(s+1)\space\mathrm{d}s \\
  & \approx & I_{n-1} + (n+1/2) c_{n-1}
\end{array}
$$
Note that the condition for $c_{n} = c_{n-1}$ is that $I_{n-1} = \frac{c_{n-1} (n^2-1)}{2}$, and the sequence will be increasing or decreasing at this step according to whether $I_{n-1}$ is greater than or less than this value.  And the relationship between $c_n$ and $I_n$, when they are computed according to the formulas above, is such that all subsequent values of $c_n$ will be the same.
Now, the value we have chosen for $c_4 = 0.2522189$ is that which yields the correct value of $I_5$ from $I_4$, where the values obtained from our closed-form solutions are:
$$I_4 \approx 3.02983$$
$$I_5 \approx 4.41703$$
This is in the region where the sequence is increasing, and we find:
$$c_5 \approx 0.2524018 $$
Carrying forward the calculations for $c_n$ and $I_n$ together, with the formulas we are using $c_n$ will remain at this value.
Now, to produce a rigorous lower bound on $c$ would require putting intervals on all the approximations.  But I suspect that both the separation of $c_4$ above $\frac{1}{4}$ and the relationship with $I_4$ that implies a subsequent increase in $c_n$ are robust enough to guarantee a lower bound above $\frac{1}{4}$.
A: 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*}
