I am attempting to work through a common problem entailing variable harvesting of a population growing logistically. I am struggling to determine all the properties for stable harvesting. I will detail my work to show where I need help
The given ODE is, $\frac{dN}{dt}=RN(1-\frac{N}{K})-HN$
Where,
- $N$ is the population
- $R$ is the instinsic growth rate of the population
- $K$ is the carrying capacity
- $H$ is relative portion of the population to be harvested
The goal is to determine properties for $H$ that allow for equilibrium of the population. The given ODE has no method for finding a general solution as it's nonlinear. Instead, I choose a more pratical method by setting $K=1$ to represent the maximum relative carry capacity. This would in turn mean that any choices for $R$ and the resulting $N$ would be in decimal form to be fractional pieces of $K=1$. This provides,
$\frac{dN}{dt}=RN(1-\frac{N}{1})-HN=\frac{dN}{dt}=RN(1-N)-HN$
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Next I determine equilibrium $N_e$ through finding solutions to $\frac{dN_e}{dt}=0$
$\frac{dN_e}{dt}=RN_e(1-N_e)-HN_e=RN_e-RN_e^2-HN_e=N_e(R-RN_e-H)=0$.
Two solutions exist, $N_{e_1}=0$ and $N_{e_2}=\frac{R-H}{R}=1-\frac{H}{R}$. An equilibrium point for a population must be greater than zero, otherwise the population is tending towards extinction. Thus, $N_{e_1}=0$ is an invalid solution and we must have $N_{e_2}=1-\frac{H}{R}>0$. By solving inequality of $N_{e_2}$ for $H$, we discover $H<R$. Which is a logical property. If the relative portion of the population harvested $H$ was greater than the growth rate $R$, the population would reach extinction.
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The next step is to determine the maximum harvest which sustains $N_{e_2}$. To do so, we employ the second derivative test.
$\frac{d^2N}{dt^2}=\frac{d}{dt}[RN_{e_2}(1-N_{e_2})-HN_{e_2}]=R(1-2N_{e_2})-H=R[1-2(1-\frac{H}{R})]-H$
By the properties of the second derivative test, for a local maximum to exist, we must have $\frac{d^2N}{dt^2}<0$. Thus,
$R[1-2(1-\frac{H}{R})]-H<0$
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This is where I fail to continue the problem. When solving the above inequality for $H$ I always find $H<R$, which is the exact same property discovered earlier in the solution. There should be a unique maximum for $H$. I'm unsure of where I made error in my logic or what other methods to employ. Help would be greatly appreciated. Thank you.
Best Answer
What you should be maximizing is not $N$ or even $H$, but instead $HN$, which gives you the total amount harvested at a point in time. With $N=1-\frac{H}{R}$, this is $R(1-N)N$. which is maximized at $N=1/2$, corresponding to $H=R/2$ (i.e. half the growth rate).
This makes sense from a physical standpoint - the best time to harvest is when the population is growing the fastest, which is when the current population is half the carrying capacity ($N=K/2$). At that point, you should harvest half the growth rate ($H=R/2$) in order to maintain a steady population.