= Solution
The four nonnegative <equilibria> are $0,a_1,a_2,a_3$. Since all factors are positive below $a_1$, the sign of $h(N)$ on consecutive intervals is
$$
(0,a_1):+,qquad(a_1,a_2):- ,qquad(a_2,a_3):+,qquad(a_3,\infty):-.
$$
The <phase line> therefore points away from zero, towards $a_1$, away from $a_2$, and towards $a_3$. Equivalently $h'(0)=a_1a_2a_3>0$, and at each simple positive root $a_i$, $h'(a_i)=-a_i\prod_{j\ne i}(a_j-a_i)$. Their signs alternate negative, positive, negative.
Hence \b[extinction and $a_2$ are unstable, while $a_1$ and $a_3$ are <asymptotically stable>]. Initial positive populations below $a_2$ tend to $a_1$; those above $a_2$ tend to $a_3$; an initial population exactly at $a_2$ remains there. This is <bistability> with a threshold separating the two basins.
Back to article page
