= Solution
A positive solution stays positive, since
$$
u(t)=u(0)\exp\left(\int_0^t[1-u(s-\tau)]\,ds\right).
$$
Therefore the sign of $\dot u$ is exactly the sign of $1-u(t-\tau)$. Initially the given history below one makes $\dot u>0$. If $u$ crosses one upwards at time $t_0$, it continues growing until $t_0+\tau$, because the feedback still sees earlier values below one. It reaches a maximum when the delayed value crosses one. Similarly, after a downward crossing at $t_1$, it continues decreasing until $t_1+\tau$ and reaches a minimum then.
This delayed reversal makes overshoots and undershoots possible. In a derivative sketch, positive lobes persist until one delay after upward crossings, followed by negative lobes persisting until one delay after downward crossings. Sustained oscillation depends on the delay; a short delay can instead give convergence to one. The example shown has $\tau=2>\pi/2$ and a history identically $1/2$.
\Image[/past-exam-of-the-mathematical-physiology-course-of-the-university-of-oxford/2011-delay-feedback.png]
{title=Overshoot, delayed derivative sign changes, and the absence of a nonnegative real characteristic root}
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