방향장과 정성적 해석

(a) A graph of $f$ against $P$ is given in blue in the figure below. This graph and the differential equation show that the equilibrium solutions occur when $f(P)=0$, or when $P=0$ or $P=6$. Since $P' < 0$ for $P$ slightly less than 0, the solutions with $P < 0$ decrease away from the equilibrium $P=0$. Also, $P'>0$ for $P$ slightly larger than 0 means that the solutions for $P >$ increase away from the equilibrium $P=0$. So $P=0$ is an unstable equilibrium. If $P$ is slightly less than $6$ we see that $P > 0$, so these solutions increase toward the equilibrium $P=0$. When $P$ is slightly larger than $6$, we have $P' < 0$ and so these solutions decrease toward the equilibrium $P = 6$. We conclude that the equilibrium solution $P=6$ is stable. (b) Since $P$ will decrease by $1$ thousand fish every month through harvesting, the new differential equation is $$\frac{dP}{dt} = g(P) = P(6-P)-1\,.$$ A sketch of $g$ against $P$ is given in green in the figure in (a). The equilibrium solutions now occur when $g(P) = P(6-P)-1 = 0$. This happens when $P^2-6P+1 = 0$. The quadratic formula tells us that the solutions to this quadratic equation are $$P = \frac{6+\sqrt{32}}{2} \approx 5.83 \ \ \ \text{ and } \ \ \ P = \frac{6-\sqrt{32}}{2} \approx 0.172\,.$$ The same argument as in (a) shows that the equilibrium at $P\approx 0.172$ is unstable and the equilibrium at $P \approx 5.83$ is stable. (c) Assuming that there are fish in the lake, so that $P > 0$, there are two cases to consider. If $P < \frac{6-\sqrt{32}}{2}$, then $P' < 0$ and the fish population will die out. If $\frac{6-\sqrt{32}}{2} < P < \frac{6+\sqrt{32}}{2}$, then $P' > 0$ and the fish population will increase and approach a stable equilibrium population of $\frac{6+\sqrt{32}}{2}$ thousand fish. If $P > \frac{6+\sqrt{32}}{2}$, then $P' < 0$ and the fish population will decrease and approach a stable equilibrium population of $\frac{6+\sqrt{32}}{2}$ thousand fish. (d) In this case the differential equation becomes $$\frac{dP}{dt} = g(P) = P(6-P)-h\,.$$ We will have equilibrium solutions when $P^2 - 6P + h = 0$ or when $$P = \frac{6+\sqrt{36-4h}}{2} \ \ \ \text{ and } \ \ \ P = \frac{6-\sqrt{36-4h}}{2}\,.$$ (e) We need a positive stable equilibrium. This will will happen as long as $36-4h \ge 0$ or when $h \le 9$. So the maximum harvest is $9000$ fish. Note that $P'$ is negative, so we need to start with more than $3000$ fish. Since $P(6-P)$ has a maximum value of $9$, the fish will reproduce quickly enough so that harvesting $9000$ fish will maintain a stable population.