= 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 at
$$
t_a=\frac{-\ln(1-h)}\alpha.
$$
Symmetry 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 $K$ until
$$
t_b=T+\frac1\alpha\ln\frac{1-e^{-\alpha T}}h
=\frac1\alpha\ln\frac{e^{\alpha T}-1}h.
$$
The <persistence correction to a gene-circuit pulse threshold> uses the limiting latch condition is $t_b=2t_a$, yielding
$$
\boxed{T_{\min}=\frac1\alpha\ln\left(1+\frac{h}{(1-h)^2}\right).}
$$
For $T>T_{\min}$, $Y$ 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 $h=1/2$, $T_{\min}=\ln3/\alpha$, whereas $Y$ and $Z$ cross at $2\ln2/\alpha$ in absolute time.
The shorter required \i[production duration] reflects X's persistence after production stops. Requiring $T\ge2t_a$ 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 <past-exam-of-the-biological-physics-course-of-the-university-of-oxford/2019/3/a/solution> contrasts pulses shorter and longer than $\ln3/\alpha$.
Back to article page
