= Solution
The non-axis <nullclines> are $u_1+Au_2=1$ and $Bu_1+u_2=1$. 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.
\Image[/past-exam-of-the-short-options-course-of-the-university-of-oxford/2023-squirrel-phase-planes.png]
{title=Competition nullclines and trajectories as grey-squirrel removal increases}
{description=Original dimensionless model plots with $A=2$, $B=0.75$, and $\rho=1$. The three panels illustrate grey dominance, bistability, and red dominance. Solid lines are the red and grey non-axis <nullclines>; arrows and trajectories show the direction of motion.}
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> $1/(u_1u_2)$ 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 gives
$$
\boxed{(u_1,u_2)\longrightarrow(0,1)}.
$$
The qualification $u_2(0)>0$ matters: a grey population identically absent cannot drive extinction. In dimensional variables, positive grey presence leads to $N_1\to0$, $N_2\to K_2$ in this no-hunting case.
Back to article page
