Consider the following logistic DE with a constant harvesting term:
$$\frac{dP}{dt}=rP(1-\frac{P}{b})-h,$$ where $r$ is the intrinsic growth rate of the population $P$, $b$ is the carrying capacity, and $h$ is the constant harvesting term. Suppose the units of time is in weeks. Now, we wish to consider the maximum harvesting rate every week, such that the population will never fall to $0$ at any point in time (i.e., the maximum sustainable yield).
Now, it is clear that $h_{\max}=\frac{rb}{4}$. To see this, write $rP(1-\frac{P}{b})-h$ as $-\frac{r}{b}P^2+rP-h$, and consider the discriminant, $\triangle =r^2-4\left(\frac{rh}{b}\right)$. If $h>\frac{rb}{4}$, $\triangle <0$, hence $\frac{dP}{dt}<0$ for all values of $P$, i.e. the population will always become extinct regardless of the value of the initial population.
If $h=\frac{rb}{4}$, then setting $\frac{dP}{dt}=0$ yields only $1$ equilibrium solution, which is $P=\frac{b}{2}$. A simple phase line diagram will tell us that if $P_0 \ge \frac{b}{2}$, where $P_0$ is our initial population size, the population will never go extinct (although $\frac{dP}{dt}< 0$, the population asymptotically tends towards the equilibrium solution $\frac{b}{2}$.) Of course, if $P_0 <\frac{b}{2}$, then the population will go extinct. But anyways, this justifies our claim that $h=\frac{rb}{4}$ is the maximum sustainable yield: the population will never go extinct as long as our initial population is large enough.
But let's consider the case where $h<\frac{rb}{4}$. Then, $\triangle >0$, hence yielding $2$ equilibrium solutions, namely: $P_1=\frac{b}{2}\left(1+\sqrt{1-\frac{4h}{rb}}\right)$, and $P_2=\frac{b}{2}\left(1-\sqrt{1-\frac{4h}{rb}}\right)$. Again, our phase line diagram will tell us that the population will not become extinct if and only if we have $P_0 \ge P_2$ ($P_1$ is a stable equilibrium while $P_2$ is an unstable equilibrium). But here comes the part that has been confusing me:
$$P_0 \ge \frac{b}{2}\left(1-\sqrt{1-\frac{4h}{rb}}\right) \iff h \le rP_0\left(1-\frac{P_0}{b}\right)$$
But how am I supposed to go about interpreting this last inequality, in conjunction to $h<\frac{rb}{4}$ ? Clearly, it does not make sense to speak of $rP_0\left(1-\frac{P_0}{b}\right)$ being larger than $\frac{rb}{4}$. But at the same time, if we have $h \le rP_0\left(1-\frac{P_0}{b}\right) < \frac{rb}{4} $, then since $h \le rP_0\left(1-\frac{P_0}{b}\right)$ is a necessary condition for our population to not drop to zero, doesn't this imply that there are some values of $h$ in between $rP_0\left(1-\frac{P_0}{b}\right)$ and $\frac{rb}{4}$ such that, no matter the value of $P_0$, our population always goes extinct? Intuitively this is contradictory – since there exists a range of values for $P_0$ that allows for sustainable growth at the maximum harvesting rate $\frac{rb}{4}$, then there should also be a range of values of $P_0$ that allows for sustainable growth at a harvesting rate less than the maximum harvesting rate?
Maybe this is a trivial issue and I am looking at things wrongly, but I can't resolve this apparent dilemma no matter how much I think about it.
Best Answer
Firstly, the transformation of the inequality $$P_0 \ge \frac{b}{2}\left(1-\sqrt{1-\frac{4h}{rb}}\right) \iff h \le rP_0\left(1-\frac{P_0}{b}\right)$$ is only partially correct. You have to be carefuly because taking the square is only an allowed operation if both sides are larger than zero. It should read: $$P_0 \ge \frac{b}{2}\left(1-\sqrt{1-\frac{4h}{rb}}\right) \iff \left(P_0\leq \frac{b}{2} \wedge h \le rP_0\left(1-\frac{P_0}{b}\right)\right)\vee P_0\geq\frac{b}{2}$$ From the second part we can directly deduce that the whole region $(h,P_0)\in[0,\frac{rb}{4}]\times[\frac{b}{2},\infty)$ is sustainable while the region $(h,P_0)\in(\frac{rb}{4},\infty)\times[0,\infty)$ is not. For the missing rectangle we indeed have the inequality $h \le rP_0\left(1-\frac{P_0}{b}\right)$, which together yields a region that somewhat looks like this: