 Solution

ID: past-exam-of-the-mathematical-and-computational-biology-course-of-the-university-of-cambridge/2025/paper-1/5/d/solution

H = rH e^{- a P}

and

P = b H (1 - e^{- a P})

. Taking

H = 0

P = 0

H \neq = 0

, the first equation gives

e^{- a P} = 1/ r

and hence

P_{*} = \frac{lo g r}{a}, H_{*} = \frac{r lo g r}{ab ( r - 1 )} (r \neq = 1) .

These are the only isolated candidates obtained from the equations. There is an exceptional parameter case: at

r = 1

P = 0

and every

H \geq 0

is an equilibrium. Thus the assertion of exactly two equilibria needs

r \neq = 1

, and only the positive candidate belongs to the biological state space when

r > 1

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