= Solution
For <quasi-steady enzyme kinetics at low substrate saturation>, the <quasi-steady-state approximation> is controlled here by $\varepsilon=e_0/(s_0+K_m)$, rather than by $e_0/s_0$ alone. Even with $e_0\approx s_0$, taking $K_m\gg s_0$ makes $\varepsilon\ll1$ and $\sigma=s_0/K_m=O(\varepsilon)$.
After the initial fast layer $T=O(\varepsilon)$, the third equation gives
$$
y\sim\frac{\sigma x}{(1+\sigma)(1+\rho)\psi}=O(\varepsilon).
$$
Since $\phi=O(\varepsilon)$, the last equation gives $(e_i)_T=(\phi/\varepsilon)y=O(\varepsilon)$. Starting from $e_i(0)=0$, this implies $e_i=O(\varepsilon)$ on bounded intervals of the slow time. The quasi-steady second equation then gives $x=s+O(\varepsilon)$. In the first equation,
$$
s_T=-s+\frac\rho{1+\rho}s+O(\varepsilon)=-\frac{s}{1+\rho}+O(\varepsilon).
$$
Matching to the initial substrate value $s(0)=1$ therefore yields $\boxed{s(T)\sim e^{-T/(1+\rho)}}$. The relation $x\sim s$ is an outer approximation: it does not replace the initial layer needed to reconcile $x(0)=0$ with $s(0)=1$. Likewise the smallness of accumulated inactivated enzyme has been justified for bounded slow times, not arbitrarily long times.
Back to article page
