Past exam of the mathematics course of the University of Cambridge / 2019 / ii / Paper 4 / 14C (Mathematical Biology)

Write $N=S+I$. Inthe term $N$ represents production of healthy new plants by both healthy and diseased plants at unit per-capita rate. The term $-NS$ is density-dependent loss of healthy plants through competition, and $-\beta IS$ is loss through mass-action infection. Inthe term $-NI$ is the same density-dependent loss for diseased plants, while $+\beta IS$ is the matching gain when healthy plants become diseased. Thus $\beta$ is the transmission-rate coefficient per healthy-diseased encounter.

Adding the equations cancels the infection terms and gives the logistic differential equationThe disease therefore changes how the population is divided between healthy and diseased plants but has no effect on the total-population dynamics: infected plants reproduce at the same rate, all offspring enter the healthy class, and infection causes no additional death in this model.

The fixed-point conditions areOn $I=0$ they give $(S,I)=(0,0)$ and the disease-free equilibrium $(1,0)$. An interior fixed point requires $I=(\beta -1)S$, so it exists only for $\beta >1$; substitution in the first equation gives the endemic equilibrium

The Jacobian matrix isAt $(1,0)$ its eigenvalues are $-1$ and $\beta -1$. Hence $(1,0)$ is asymptotically stable for $\beta <1$ and a saddle point for $\beta >1$. At the endemic equilibrium, the equivalent $(N,\theta )$ coordinates give eigenvalues $-1$ and $1-\beta$, so it is a stable node whenever it exists. The origin is unstable because every solution with $N(0)>0$ has $N(t)\to 1$.

For the phase portrait, the line $I=0$ is invariant and points toward $(1,0)$. Alsoso trajectories move toward the line $S+I=1$. If $\beta <1$, then $\dot{I}<0$ throughout the interior and every nonzero solution approaches $(1,0)$. If $\beta >1$, the additional infective nullcline is the ray $I=(\beta -1)S$; $I$ increases below this ray and decreases above it, and every trajectory with $I(0)>0$ approaches $(1/\beta ,1-1/\beta )$. At $\beta =1$ the disease-free and endemic branches exchange stability in a transcritical bifurcation.

Defineso $S=N(1-\theta )$ and $I=N\theta$. Part (a) already givesUsing the quotient rule,and hence the reduced Plant SI model with logistic total population is

For every $N(0)>0$, logistic growth gives $N(t)\to 1$. The asymptotic prevalence equation is thereforeIf $\beta <1$, its only nonnegative equilibrium is $\theta =0$, which attracts all prevalences. If $\beta >1$, $heta=0$ is unstable and the stable value is ${\theta }_{\ast }=1-1/\beta$. Combining these values with $N_{\ast }=1$ recovers exactly the equilibria and outcomes in part (b).

Uniform per-capita culling subtracts $kS$ and $kI$ from the two equations:Adding and again differentiating $\theta =I/N$ gives the Uniform per-capita culling in the plant SI model equationsThe direct culling terms cancel from the prevalence equation because both classes are culled equally, but culling lowers the eventual total population. For $0\le k<1$, $N(t)\to 1-k$, and a rare infection grows precisely whenTherefore, when $\beta >1$, disease is eliminated while a positive plant population remains exactly forFor $\beta \le 1$ the disease already dies out without culling. If $k\ge 1$, culling drives the entire plant population to extinction, which removes the disease only by removing its host.