 Solution

ID: past-exam-of-the-mathematics-and-the-environment-and-mathematical-physiology-course-of-the-university-of-oxford/2008/b/5/ii/solution

\dot{S} = - k_{1} SE + k_{- 1} C, \dot{E} = - k_{1} SE + (k_{- 1} + k_{2}) C, \dot{C} = k_{1} SE - (k_{- 1} + k_{2}) C, \dot{P} = k_{2} C .

Thus

E + C = E_{0}

and

S + C + P = S_{0}

are conserved. Eliminating free enzyme yields the stated two-equation system.

With

s = S / S_{0}

c = C / E_{0}

and

τ = t / T

, substitution gives

α = k_{1} E_{0} T, β = \frac{k _{- 1} E _{0} T}{S _{0}}, γ = k_{1} S_{0} T, δ = (k_{- 1} + k_{2}) T .

The dimensionless initial data are

s (0) = 1, c (0) = 0

. The parameter named

δ

here is an enzyme kinetic coefficient and is unrelated to the loss parameter used in the later blood-cell question.

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