 Solution

A positive solution stays positive, sinceTherefore 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 $1/2$.