로지스틱 방정식

(a) The equilibrium solutions for this logistic equation are $p = 0$ and $p = 1$. Since $p = 1$ is the carrying capacity and thus a stable equilibrium, $p(t) \to 1$ as $t \to \infty$ provided $p(0) > 0$. (b) We know that the general solution to the logistic equation $\frac{dP}{dt} = kP(N-P), \ P(0) = P_0$ is $$P(t) = \frac{N}{\left(\frac{N-P_0}{P_0}\right) e^{-kNt} + 1}\,.$$ In the context of this problem, we know that $k = 0.2$, $N = 1$, and $p(0) = 0.1$ and therefore the solution is $$p(t) = \frac{1}{\left(\frac{1-0.1}{0.1}\right) e^{-0.2t} + 1} = \frac{1}{9e^{-0.2t} + 1}\,.$$ (c) We know that $p$ is changing most rapidly when its derivative, $p'$, is largest. Recalling the original differential equation, we know that $$p' = 0.2p(1-p)$$ which shows that $p'$ is a quadratic function of $p$. That quadratic function has zeros at $p = 0$ and $p = 1$, so the vertex of this concave down quadratic function lies at $p = \frac{1}{2}$, and at this value of $p$ the rate of change of $p$ is largest. Thus, we solve the equation $p(t) = 0.5$ to find the time at which the population is increasing fastest. We see that $$0.5 = \frac{1}{9e^{-0.2t} + 1}$$ so $$9e^{-0.2t} + 1 = 2$$ and $e^{-0.2t} = 1/9$. It follows that $t = -5 \ln(1/9) \approx 10.986$ days. (d) Since $p(t)$ tells us the percentage of people who've heard the rumor, we need to solve the equation $p(t) = 0.8$. Thus we have $$0.8 = \frac{1}{9e^{-0.2t} + 1}$$ so $$9e^{-0.2t} + 1 = 1.25$$ and $e^{-0.2t} = 0.25/9$. It follows that $t = -5 \ln(0.25/9) \approx 17.19$ days.

(a) Let $y$ be the per capita growth rate as a function of $b$. Using the given data in the problem statement, we know that this linear function passes through the points $(6000,3)$ and $(9000,2)$. The equation of this line is thus $$y - 3 = -\frac{1}{3000}(b - 6000)$$ so $y = (3 + 2) - \frac{1}{3000}b = 5 - \frac{1}{3000}b$. Since the logistic equation comes from taking the product of the population at time $t$ with its per capita growth rate at that time, we find that $$\frac{db}{dt} = b \left( 5 - \frac{1}{3000}b \right) = \frac{1}{3000} b(15000 - b)$$ (b) From the form of the logistic equation, we see that the carrying capacity for the bacteria is $b = 15000$, which is a stable equilibrium for the population. (c) The bacteria is increasing most rapidly when $\frac{db}{dt}$ is greatest. Since $$\frac{db}{dt} = \frac{1}{3000} b(15000 - b)$$ is a quadratic function of $b$, $\frac{db}{dt}$ is maximized at the vertex of this concave down parabola, which occurs halfway between its zeros: $b = 0$ and $b = 15000$. Thus, when $b = 7500$, the population is growing fastest. To find the exact $t$-value when this occurs, we'd need to know an initial condition like in the next part of the problem. (d) Using $b(0) = 1000$, we solve the logistic equation IVP with $$\frac{db}{dt} = \frac{1}{3000} b(15000 - b)\,.$$ Using $N = 15000$, $k = \frac{1}{3000}$, and $b_0 = 1000$, we know from the standard form of the solution to the logistic equation that $$b(t) = \frac{15000}{\left(\frac{15000-1000}{1000}\right) e^{-5t} + 1} = \frac{15000}{14e^{-5t} + 1}\,.$$ To find how long it takes the bacteria to reach 80% of the carrying capacity, we solve the equation $$0.8 \cdot 15 = \frac{15000}{14e^{-5t} + 1}\,.$$ That equation implies that $$0.8 = \frac{1}{14e^{-5t} + 1}\,,$$ so $14e^{-5t} + 1 = 1.2$ and thus $e^{-5t} = 0.2/14 = 1/70$. It follows that $t = -\frac{1}{5} \ln(1/70) \approx 0.8497$ days.

(a) From the form of the given logistic equation, we can see the carrying capacity is 10000 fish. (b) In this situation, the initial condition is $P(0) = 10$, and by removing $20$% of the population per year, we subtract $0.2P$ from the rate that defines the differential equation, thus finding $$\frac{dP}{dt} = 0.1P(10 - P) - 0.2P\,.$$ (c) Expanding and rewriting the differential equation in (b), we see that $$\frac{dP}{dt} = P - 0.1P^2 - 0.2P = 0.8P - 0.1P^2 = 0.1P(8 - P)\,.$$ This, too, is a logistic equation, with carrying capacity $8000$ fish. (d) We can solve this logistic equation in the usual way (keeping in mind the initial condition $P(0) = 10$, and noting that the carrying capacity is $8$): $$P(t) = \frac{8}{\frac{8-10}{10}e^{-0.1 \cdot 8 t}+1} = \frac{8}{-0.2e^{-0.8t}+1}$$ After one year, the population of fish will thus be $$P(1) \approx 8.7899$$ thousand fish. (e) To be within 10% of the carrying capacity, we need $0.9(8) \le P(t) \le 1.1(8)$. Since the initial condition is larger than the carrying capacity, we want to know when $P(t) = 1.1(8) = 8.8$. Thus we solve the equation $$8.8 = \frac{8}{-0.2e^{-0.8t}+1}\,,$$ which implies $-0.2e^{-0.8t}+1 = 8/8.8 = 10/11$. Thus, $e^{-0.8t} = 5/11$, so $t = -1.25 \ln(5/11) \approx 0.986$ years.