Past exam of the topics in applied mathematics course of the University of Oxford / 2007 / A / 2 / iii

Use $s=s_{0}u$, $c=e_{0}v$ and slow time $\boxed{\tau =k_1{e}_{0}t}$. Dividing the equations by their respective concentration and time scales givesThus $u^{\prime }=\eta u-u+uv+(K-\lambda )v$ and $ϵv^{\prime }=u-uv-Kv$. Notice that $K>\lambda >0$ and the physical enzyme fraction satisfies $0\le v\le 1$. The initial values are $u(0)=1$, $v(0)=0$ for this particular normalization; the threshold analysis compares different initial substrate abundances using a fixed reference scale.