 Solution

Write $c=c_1$, $\beta =(k_{-1}+k_1^{+})/(k_1{S}_0)$ and $\kappa =k_1^{+}/(k_1{S}_0)$. Since $c_0=1-c$, the exact equations areAt the prescribed initial data $s=1,c=0$, $r_1=1$, rather than zero. The quasi-steady-state approximation would instead demand $c=s/(s+\beta )$, giving $c(0^{+})=1/(1+\beta )$. This mismatch signals an initial layer.

Set $\sigma =\tau /ϵ$ and expand $s=s_0+ϵs_1+\cdots$, $c=c_0+\cdots$. At leading order $s_0^{\prime }=0$, so $s_0=1$, while $c_0^{\prime }=1-(1+\beta )c_0$ and $c_0(0)=0$. ThusFor a more informative substrate approximation, integration of $s_1^{\prime }=-1+(1+\beta -\kappa )c_0$ with $s_1(0)=0$ givesOn the slow scale $c_{\mathrm{out}}=s_{\mathrm{out}}/(s_{\mathrm{out}}+\beta )$ and $s_{\mathrm{out},\tau }=-\kappa s_{\mathrm{out}}/(s_{\mathrm{out}}+\beta )$, with matched leading value $s_{\mathrm{out}}(0)=1$. Integrating gives $s_{\mathrm{out}}+\beta \log s_{\mathrm{out}}=1-\kappa \tau$. In the overlap $1\ll \sigma \ll 1/ϵ$, the inner complex tends to $1/(1+\beta )$ and its substrate slope matches the expansion of this outer solution. A leading composite asymptotic expansion isIt satisfies both initial values at leading order and smoothly joins the initial complex build-up to substrate depletion. The displayed $O(ϵ)$ inner substrate term records the small binding-induced initial slip if greater accuracy is wanted.