 Solution

For the dynamically regulated interpretation, take $X(0)=S$, $Y(0)=Z(0)=0$, and let X production respond to Y repression after the ligand appears. Define $h=K/S\in (0,1)$. Initially active X drives both Y and Z, giving $Y=Z=S(1-e^{-\alpha t})$. Y reaches $K$ at $t_1=-\ln (1-h)/\alpha$ and switches X production off. X then decays from $S$ to $K$, reaching it atUntil $t_2$, X remains active and Y keeps being produced. Thus $Y_2=S(1-h+h^2)$. At $t_2$, Y production stops, but Y still activates Z. Its subsequent decay crosses $K$ atImmediately afterwards both inputs to the Z OR gate are below threshold, so Z production stops and its derivative changes from positive to negative. Since Z has been continuously produced from zero until $t_3$,For $h=1/2$, $t_{\max 1}=\ln 6/\alpha$ and $Z_{\max 1}=5S/6$. Y repression is lifted at $t_3$, but X must first rebuild from a subthreshold value, so Z has a genuine first local maximum before any later response.

Wording qualification: if “X is always present at its steady state concentration” is read as an external clamp maintaining $X=S$ after ligand arrival, it overrides Y's repression and changes the problem. Then $Z=S(1-e^{-\alpha t})$ rises monotonically, with no finite first maximum. The finite-peak result uses the consistent initial-steady-state interpretation in which the stated negative feedback can act.