 Solution

ID: past-exam-of-the-topics-in-applied-mathematics-course-of-the-university-of-oxford/2007/a/2/iii/solution

Use

s = s_{0} u

c = e_{0} v

τ = k_{1} e_{0} t

. Dividing the equations by their respective concentration and time scales gives

ϵ = \frac{e _{0}}{s _{0}}, K = \frac{k _{- 1} + k _{2}}{k _{1} s _{0}}, λ = \frac{k _{2}}{k _{1} s _{0}}, η = \frac{k _{0}}{k _{1} e _{0}} .

Thus

u^{'} = η u - u + uv + (K - λ) v

and

ϵ v^{'} = u - uv - K v

. Notice that

K > λ > 0

and the physical enzyme fraction satisfies

0 \leq v \leq 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.

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