The Lumenary · Advanced undergraduate

Mathematical biology

Name: ______________________

Date: ______________________

You will learn to

Analyse the logistic equation and discrete-time population models, including fixed points, period doubling and the onset of chaos in the logistic map
Find the equilibria of predator–prey, competition and SIR models and classify their stability using the Jacobian and the trace–determinant test
Prove that the Lotka–Volterra model has a conserved quantity, so that its orbits are closed, and that its cycle averages equal the equilibrium
Derive the epidemic threshold $R_0 S(0)/N > 1$ , the final-size relation and the herd immunity threshold, and interpret $R_0$ and the assumptions behind it
Explain the limits of models, the difference between projections and forecasts, and why structurally unstable models mislead

1.
For the logistic map $x_{n+1} = 2.5\,x_n(1 - x_n)$ , what is the non-zero fixed point, in percent of the maximum population?
2.
For which growth rates $r$ is the non-zero fixed point $x^* = 1 - 1/r$ of the logistic map locally asymptotically stable? (Choose all that apply.)
- A. $r = 0.8$
- B. $r = 2$
- C. $r = 2.9$
- D. $r = 3.2$
3.
In the simple SIR model, an infection has $R_0 = 4$ . What percentage of the population must be immune for an introduced infection not to grow (the herd immunity threshold)?
4.
Which statement about the basic reproduction number $R_0 = \beta/\gamma$ is correct?
- A. $R_0$ is a fixed biological constant of each pathogen, like its genome length.
- B. $R_0$ depends on behaviour and environment as well as on the pathogen, and so can differ between populations.
- C. $R_0$ measures how deadly an infection is.
- D. $R_0$ is the average number of people each case infects at the current stage of an epidemic.
5.
At the coexistence equilibrium of the Lotka–Volterra predator–prey model the Jacobian has eigenvalues $\pm i\sqrt{\alpha\gamma}$ . What can we conclude from the linearisation alone?
- A.Nearby orbits are periodic, since the linearisation is a centre.
- B.Nothing about stability: the linearisation is inconclusive, and the conserved quantity $V$ is needed to show that orbits are closed.
- C.The equilibrium is unstable.
- D.The equilibrium is asymptotically stable.
6.
In the competition model $x' = x(1 - x - a_{12}y)$ , $y' = y(1 - y - a_{21}x)$ , which parameters give a stable coexistence equilibrium?
- A. $a_{12} = 0.5$ , $a_{21} = 0.6$
- B. $a_{12} = 1.5$ , $a_{21} = 1.6$
- C. $a_{12} = 0.5$ , $a_{21} = 1.6$
- D. $a_{12} = a_{21} = 1$
7.
A fish stock grows logistically with $r = 0.4$ per year and carrying capacity $K = 10\,000$ tonnes, and is harvested at a constant rate of $H$ tonnes per year. What is the largest $H$ for which a positive equilibrium exists (the maximum sustainable yield), in tonnes per year?
8.
In the SIR model with $R_0 > 1$ and almost everyone initially susceptible, when does the number of infected people reach its maximum?
- A.When half the population has been infected.
- B.When the susceptible fraction $S/N$ falls to $1/R_0$ .
- C.At the end of the epidemic, when $S$ stops changing.
- D.When the number of recoveries per day first exceeds the number of new infections, which can happen at any value of $S$ .
9.
Which Lean statement says that Volterra's quantity is conserved along a solution of the predator–prey equations at time t?
- A.(hx : 0 < x t) (hy : 0 < y t) (hx' : HasDerivAt x (α * x t - β * x t * y t) t) (hy' : HasDerivAt y (δ * x t * y t - γ * y t) t) : HasDerivAt (fun s => δ * x s - γ * log (x s) + β * y s - α * log (y s)) 0 t
- B.(hx' : HasDerivAt x (α * x t - β * x t * y t) t) (hy' : HasDerivAt y (δ * x t * y t - γ * y t) t) : δ * x t - γ * log (x t) + β * y t - α * log (y t) = 0
- C.(hx : 0 < x t) (hy : 0 < y t) : HasDerivAt (fun s => δ * x s - γ * log (x s) + β * y s - α * log (y s)) 0 t
- D.(hx' : HasDerivAt x (α * x t - β * x t * y t) t) (hy' : HasDerivAt y (δ * x t * y t - γ * y t) t) : HasDerivAt (fun s => x s + y s) 0 t
10.
Write final_size(R0) returning the fraction $1 - s$ of a population eventually infected in the SIR model, where $s \in (0, 1)$ solves $s = e^{-R_0(1 - s)}$ (return 0.0 when $R_0 \le 1$ ). Use bisection on $g(s) = s - e^{-R_0(1 - s)}$ . Since $g(1) = 0$ as well, bracket the non-trivial root in $[0, 1/R_0]$ , where $g(0) < 0 < g(1/R_0)$ .
Computational — Practice

Answer key — Mathematical biology

1.
$r(1 - x) = 1$ gives $x^* = 1 - 1/r = 1 - 0.4 = 0.6$ , i.e. $60\%$ . Its multiplier is $f'(x^*) = 2 - r = -0.5$ , with $|{-0.5}| < 1$ , so it is stable; orbits approach it with alternating overshoots, since the multiplier is negative.
2.
$f_r'(x^*) = 2 - r$ , so the fixed point is stable exactly for $1 < r < 3$ : here $r = 2$ and $r = 2.9$ . At $r = 3$ the multiplier reaches $-1$ and a period-doubling bifurcation produces a 2-cycle, stable for $3 < r < 1 + \sqrt 6$ .
3.
Growth requires $R_0(1 - p) > 1$ , so the threshold is $p_c = 1 - 1/R_0 = 1 - \tfrac14 = 75\%$ . An uncontrolled epidemic overshoots it: the final-size relation $s = e^{-4(1 - s)}$ gives $s \approx 0.02$ , so about $98\%$ would be infected.
4.
$R_0 = \beta \cdot \frac1\gamma$ is a contact-and-transmission rate times an infectious period. Both factors depend on the setting — crowding, contact patterns, ventilation, vectors — so $R_0$ is not a biological constant. During an outbreak the relevant quantity is $R_t \approx R_0 S/N$ , further reduced by interventions.
5.
The eigenvalues have zero real part, so linearisation is inconclusive. The exact conserved quantity $V = \delta x - \gamma\ln x + \beta y - \alpha\ln y$ shows that orbits are closed curves, so the equilibrium is stable but not asymptotically stable. Adding prey self-limitation, for instance, turns it into a stable spiral.
6.
With trace $-(x^* + y^*) < 0$ , the coexistence equilibrium is stable iff $\det J = x^*y^*(1 - a_{12}a_{21}) > 0$ ; with $x^*, y^* > 0$ this means $a_{12}, a_{21} < 1$ . So only $(0.5, 0.6)$ gives stable coexistence: intraspecific competition must outweigh interspecific competition.
7.
A positive equilibrium exists iff $H \le \max_N rN(1 - N/K) = rK/4 = 0.4 \times 10\,000/4 = 1000$ tonnes per year. At $H = rK/4$ the two equilibria merge (saddle-node bifurcation); any larger harvest drives the stock to extinction — and near the maximum, the stable equilibrium is close to the unstable one, so small shocks can tip the stock into collapse.
8.
Since $I' = \gamma I(R_0S/N - 1)$ and $S$ decreases, $I$ increases while $S/N > 1/R_0$ and decreases afterwards: the peak is at $S/N = 1/R_0$ , which is also the herd immunity threshold. The invariant $S + I - \frac{N}{R_0}\ln S$ then gives the peak value $I_{\max}/N = 1 - (1 + \ln R_0)/R_0$ when $S_0 \approx N$ .
9.
Statement (a). (b) confuses "constant" with "zero", (c) drops the differential equations, and (d) names the wrong quantity. Integrating (a) over time (by the mean value theorem) shows that $V$ is constant along a solution.
10.
```
import math

def final_size(R0):
    if R0 <= 1:
        return 0.0
    g = lambda s: s - math.exp(-R0 * (1 - s))
    lo, hi = 0.0, 1.0 / R0
    for _ in range(200):
        mid = (lo + hi) / 2
        if g(mid) < 0:
            lo = mid
        else:
            hi = mid
    return 1 - (lo + hi) / 2
```
Why is the root below $1/R_0$ ? At $s = 1/R_0$ , $e^{-R_0(1 - s)} = e^{-(R_0 - 1)} < \frac{1}{1 + (R_0 - 1)} = \frac{1}{R_0}$ , so $g(1/R_0) > 0$ , while $g(0) < 0$ . The fraction escaping is below $1/R_0$ — the epidemic overshoots the herd immunity threshold.