Past exam of the short options course of the University of Oxford / 2021 / 9 / a / iv / Solution

For $r>\delta$, the sign of $f(N)$ is positive on $0<N<N^{\ast }$ and negative on $N>N^{\ast }$. Its graph passes through $0,N^{\ast }$, has positive initial slope $r-\delta$, and eventually behaves like $-\delta N$. The phase line therefore points toward $N^{\ast }$ from both sides.

The vector field is continuously differentiable, so solutions are unique. No positive solution crosses either equilibrium; one starting below $N^{\ast }$ increases and stays below it, and one starting above it decreases and stays above it. These bounded solutions exist for all forward time and have a limit. By continuity of $f$, any such limit must be a zero of $f$: otherwise the derivative would stay bounded away from zero near the limit. The increasing solution is bounded below by its strictly positive initial value, excluding zero; the decreasing one stays above $N^{\ast }$. ConsequentlyThe solution starting exactly at $N^{\ast }$ is constant. The zero initial condition remains at zero, which is why positivity is stipulated.