방향장과 정성적 해석

$\begin{array}{l | l} x_i & y_i \\ \hline 0.0 & 0.0000 \\ 0.5 & 0.5000 \\ 1.0 & 1.8591 \\ 1.5 & 10.5824 \\ 2.0 & 88378.1190 \\ \end{array}$

$\begin{array}{l | l | l | l} x_i & y(x) & h = 0.2 & h = 0.1 \\ \hline 0.0 & 0.5000 & 0.5000 & 0.5000 \\ 0.2 & 0.5412 & 0.5000 & 0.5201 \\ 0.4 & 0.6806 & 0.5816 & 0.6282 \\ 0.6 & 0.9747 & 0.7686 & 0.8622 \\ 0.8 & 1.5551 & 1.1250 & 1.3132 \\ 1.0 & 2.7183 & 1.7885 & 2.1788 \end{array}$

(a) Sketching the slopes at individual points by computing $\frac{dy}{dt}\vert_{(t,y)}$ at various values of $(t,y)$, we see the following slope field. (b) By following the slope field so that we pass through the points $(0,-4)$, $(0,-3)$, $\ldots$, $(0,4)$, we see the collection of curves in the figure above. (c) It looks like the solution that passes through $(0,-1)$ is a straight line with slope $m = 1$, and thus has the formula $y(t) = t-1$. To verify this function is a solution, we compute both $\frac{dy}{dt}$ and $t-y$. Doing so, we see $$\frac{dy}{dt} = 1$$ and $$t-y = t - (t-1) = 1\,.$$ Thus, $y(t) = t-1$ is indeed a solution to the differential equation. (d) From both the differential equation and the graphs we have sketched, it appears that $t$ and $y$ are equal at any point where a solution has a local minimum. This is because when $t = y$, $\frac{dy}{dt} = t - y = 0$, and having the derivative be zero is required for any local minimum on a differentiable function.

(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.

(a) If we let $y(t)$ be the population of the mice at year $t$ in thousands and it grows at a yearly rate that is twenty times the number of mice, the governing differential equation is $\frac{dy}{dt} = 20y$. (b) On their own, the mice add $20y$ thousand mice per year to their total population. With $C$ cats present, the cats now remove $M(y) = C\frac{y}{2+y}$ thousand mice per year. By taking both the rates at which mice are added and removed from the population into account, we find the updated differential equation $$\frac{dy}{dt} = 20y - C\frac{y}{2+y}$$ (c) For positive $y$ near $0$, $M(y) = \frac{y}{2+y} \approx 0$; for large values of $y$, $M(y) = \frac{y}{2+y} \approx 1$. This means that a single cat can remove about $1$ thousand mice per year from the population when the mice population is large. (d) Letting $C=1$, we are working with the differential equation $$\frac{dy}{dt} = 20y - \frac{y}{2+y}\,.$$ To find the equilibrium solutions, we first set $\frac{dy}{dt} = 0$. Doing so, we find that $20y = \frac{y}{2+y}$, so $40y + 20y^2 = y$, or $39y + 20y^2 = 0$. Factoring, we see $y(39y + 20) = 0$ and therefore $y = 0$ or $y = -\frac{20}{39}$. Since a negative equilibrium solution doesn't make sense in the context of the problem, the only equilibrium solution to study is $y = 0$. If we plot $dy/dt$ versus $y$ for $y > 0$, we see that $dy/dt$ is always positive, and thus the population will grow away from $y = 0$, making this equilibrium unstable. And this makes sense: the population of mice will grow without bound, because one cat is insufficient to keep up with a mice population growing at a rate of $20$ times its size per year. (e) Letting $C=60$, we are working with the differential equation $$\frac{dy}{dt} = 20y - \frac{60y}{2+y}\,.$$ Similar work as in (d) shows that the equilibrium solutions are $y = 0$ or $y = 1$, and if we plot $dy/dt$ versus $y$, we see that for $0 < y < 1$, $dy/dt < 0$, while for $y > 1$, $dt/dt > 0$. Thus, with $60$ cats present, as long as there are initially fewer than $1000$ mice, the cats will drive the mice population to $0$. If there are more than $1000$ mice initially present, the mice population will grow without bound. And if there are $1000$ mice initially present, this will be an (unstable) equilibrium. (f) In order to keep the mice population from growing arbitrarily large, we need to bring enough cats that for certain initial mice populations, we can make $dy/dt \le 0$. In other words, we need enough cats that there's a positive equilibrium solution. Considering the general differential equation, $$\frac{dy}{dt} = 20y - C\frac{y}{2+y}\,,$$ we thus set $dy/dt = 0$ and solve for $C$. Doing so, we find that $0 = 20y - C\frac{y}{2+y}$, so $20y(y+2) = Cy$ or $20y^2 + 40y - Cy = 0$. Factoring, we find that $$y(20y + (40-C)) = 0\,.$$ To have a positive equilibrium solution, we require that $40-C < 0$, and therefore the farmer must bring at least $41$ cats to the farm. Even in that circumstance, whether the cats can keep the number of mice from growing without bound will depend on the initial number of mice present.