 Solution

ID: past-exam-of-the-topics-in-applied-mathematics-course-of-the-university-of-oxford/2008/a/2/b/ii/solution

In the first-order Taylor expansion correction it suffices to use the zeroth-order derivatives. They give

\dot{I} S + I \dot{S} = (r I S - a I) S + I (- r I S) = I S {r (S - I) - a} .

The delayed product is consequently

I (t - τ) S (t - τ) = I S + τ I S {a + r (I - S)} + O (τ^{2}) .

Multiplying by

r

and substituting into the two transfer equations produces

f (I, S, r, a) = r {a + r (I - S)}, g (I, S, r, a) = - r {a + r (I - S)} .

The opposite signs of

f

and

g

are also required by conservation law of

S + I + R

: this correction changes the timing of the transfer between

S

and

I

, while leaving the instantaneous death term unchanged.

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