Past exam of the short options course of the University of Oxford / 2023 / 9 / b

Abbreviate $A={\alpha }_{12}$, $B={\alpha }_{21}$. The coordinate factors immediately give the three boundary equilibria $S_1,S_2,S_3$. The Jacobian matrix isAt $S_1$, the eigenvalues are $1,\rho \gamma >0$, so it is a repelling equilibrium. At the other two points, triangularity givesTherefore $S_2$ is asymptotically stable for $\gamma <B$ and a saddle point for $\gamma >B$; $S_3$ is asymptotically stable for $\gamma >1/A$ and a saddle point for $\gamma <1/A$. Equality gives a zero eigenvalue, so linearization alone is inconclusive there.

For later use, the coexistence point is physically positive exactly when $B<\gamma <1/A$ if $AB<1$, or $1/A<\gamma <B$ if $AB>1$. Its stability is as supplied in the question. If $AB=1$ the displayed formula for $S_4$ does not apply: when also $\gamma =B=1/A$, the coincident non-axis nullclines give a whole segment of equilibria.

The non-axis nullclines are $u_1+A{u}_2=1$ and $B{u}_1+u_2=1$. For $A>1$, $B<1$, the first has intercepts $(1,0),(0,1/A)$ and the second has $(1/B,0),(0,1)$. Their axis intercept order rules out a positive intersection. Below the first line $u_1$ increases and above it decreases; below the second $u_2$ increases and above it decreases. Thus $S_2$ is a saddle point and $S_3=(0,1)$ is stable.

For strictly positive initial populations, both coordinates stay positive and are bounded above by their individual logistic equations. There are no interior periodic orbits: the Dulac function $1/(u_1{u}_2)$ makes the divergence of the multiplied vector field $-1/u_2-\rho /u_1<0$, so the Bendixson-Dulac theorem applies. The stable manifold of $S_2$ is the invariant red-only axis, and $S_1$ is repelling. The phase portrait, or the Poincaré-Bendixson theorem with these boundary directions, therefore givesThe qualification $u_2(0)>0$ matters: a grey population identically absent cannot drive extinction. In dimensional variables, positive grey presence leads to $N_1\to 0$, $N_2\to K_2$ in this no-hunting case.

Under $A>1$, $B<1$, $AB>1$, the thresholds are ordered $0<1/A<B<1$. As $\gamma$ decreases from one, $S_2$ first becomes stable atFor $1/A<\gamma <B$, both boundary equilibria are stable, and the positive coexistence point $S_4$ is a saddle point. Its stable manifold is the separatrix between their basins of attraction. For every fixed $u_1(0)>0$, a sufficiently small positive $u_2(0)$ lies in the red-survival basin. To justify the quantifier beyond a neighborhood of $S_2$, start on the red-only axis: its logistic equation approaches $S_2$. Continuous dependence over the finite time needed to enter a stable neighborhood shows that sufficiently small positive initial grey populations do so as well.

At the second threshold,$S_4$ meets $S_3$. For $0<\gamma <1/A$, $S_3$ is a saddle point, its stable manifold is the grey-only axis, and there is no positive coexistence equilibrium. Boundedness and the same Bendixson-Dulac theorem argument rule out an interior cycle. Consequently every initial state with $u_1(0)>0$ and $u_2(0)\ge 0$ approaches $S_2=(1,0)$. This is the removal regime that eliminates grey squirrels regardless of their initial abundance. At threshold equality linearization is nonhyperbolic; the strict parameter intervals above avoid inferring stability from a zero eigenvalue.