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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 $\ln 3/\alpha$.

Write $x=X/S$, $y=Y/S$, $z=Z/S$, $h=K/S$ and $s=\alpha t$. The repressilator in the stated logic approximation isAssume $0<h<1$. Initially low Z allows X production; high Y blocks Z; and Y is initially at its unrepressed steady level. X first rises through $h$ at $s_1=a=-\ln (1-h)$, turning Y production off. Y then decays through $h$ at $s_2=a+b$, where $b=-\ln h$, allowing Z to rise. Z reaches $h$ at $s_3=2a+b$ and switches X production off. At this point $x_3=1-h(1-h{)}^2$, so X falls through $h$ at $s_4=s_3+\ln (x_3/h)$ and releases Y production.

Y's rise then switches Z off, Z's fall releases X again, and the sequence repeats. Every segment is exponential relaxation towards one if its production gate is on, or towards zero if off. For a rising segment starting at $v_0<h$, the next threshold time is $\ln [(1-v_0)/(1-h)]$; for a falling segment starting at $v_0>h$, it is $\ln (v_0/h)$. These rules determine the whole trace without pretending the concentrations are square waves. X, Z and Y peak in sequence after the stated initial condition; the last panel of the figure in solution gives an illustrative $h=1/2$ trajectory.

The delayed sequence of three inhibitory links supplies negative feedback with finite response times. Equal rates make the cycling symmetric after the initial transient, not simultaneous. If $h\ge 1$, repression is never reliably engaged and the proteins approach their unrepressed levels instead; oscillation is not implied for arbitrary thresholds or for every smooth regulatory law replacing the ideal relay.