 Solution

Put $h=K/S$, require $0<h<1$, and use the dimensionless time $s=\alpha t$. While X is produced, $X/S=1-e^{-s}$; afterwards $X/S=(1-e^{-\alpha T})e^{-\alpha (t-T)}$. X crosses the activation threshold on its rising branch atSymmetry keeps $Y=Z$. They initially receive production through X alone and need another interval $t_a$ of active input to reach $K$. If this happens while X is still above threshold, their mutual OR activation sustains production permanently. X remains above $K$ untilThe persistence correction to a gene-circuit pulse threshold uses the limiting latch condition is $t_b=2t_a$, yieldingFor $T>T_{\min }$, $Y$ and $Z$ cross their common threshold and approach $S$ even after X decays away. For $T<T_{\min }$ they either remain zero or make a subthreshold pulse and return to zero. Exact equality is sensitive to the convention for the ideal threshold, so robust switching requires a strict excess. At $h=1/2$, $T_{\min }=\ln 3/\alpha$, whereas $Y$ and $Z$ cross at $2\ln 2/\alpha$ in absolute time.

The shorter required production duration reflects X's persistence after production stops. Requiring $T\ge 2t_a$ would incorrectly discard that tail when all activating ligands remain present. If $K\ge S$, no finite pulse reaches the required feedback-active state. The second panel of the figure in solution contrasts pulses shorter and longer than $\ln 3/\alpha$.