= Solution
{c}
Write $f(N)=N[r/(1+bN^2)-\delta]$. A <steady state> is a nonnegative zero of $f$. If $r<\delta$, the bracket is strictly negative for every $N\ge0$, so only $N=0$ exists. If $r>\delta$, there are exactly two nonnegative <steady states>,
$$
\boxed{N=0,\qquad N^*=\sqrt{\frac{r-\delta}{b\delta}}.}
$$
For an autonomous scalar <ordinary differential equation>, a <steady state> with $f'<0$ is locally asymptotically stable, while one with $f'>0$ is unstable. Here
$$
f'(N)=\frac{r(1-bN^2)}{(1+bN^2)^2}-\delta,\qquad f'(0)=r-\delta.
$$
At the positive <steady state>, use $1+b(N^*)^2=r/\delta$ to obtain
$$
\boxed{f'(N^*)=-\frac{2\delta(r-\delta)}r<0.}
$$
Therefore zero is \b[stable for $r<\delta$ and unstable for $r>\delta$], whereas the positive <steady state> is \b[stable whenever it exists]. At the omitted equality $r=\delta$, $f(N)=-\delta bN^3/(1+bN^2)$; zero is nonhyperbolic but still attracts every nonnegative initial state.
Back to article page
