= Solution
Write $x=X/S$, $y=Y/S$, $z=Z/S$, $h=K/S$ and $s=\alpha t$. The <repressilator> in the stated <logic approximation for gene regulation>[logic approximation] is
$$
\boxed{\frac{dx}{ds}=H(h-z)-x,\quad
\frac{dy}{ds}=H(h-x)-y,\quad
\frac{dz}{ds}=H(h-y)-z,\quad(x,y,z)(0)=(0,1,0).}
$$
Assume $0<h<1$. 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 <past-exam-of-the-biological-physics-course-of-the-university-of-oxford/2019/3/a/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 $h\ge1$, 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.
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