Problem 90 - Calculus for Biology and Medicine

Question

Suppose that a fish population evolves according to the logistic equation and that a fixed number of fish per unit time are removed. That is, $$ \frac{d N}{d t}=r N\left(1-\frac{N}{K}\right)-H $$ Assume that \(r=2\) and \(K=1000\). (a) Find possible equilibria, and discuss their stability when \(H=100\) (b) What is the maximal harvesting rate that maintains a positive population size?

Step-by-Step Solution

Answer

Equilibria are at \(N_1=947.2\) (stable) and \(N_2=52.8\) (unstable). Maximal harvesting rate is \(H=500\).

Step 1: Determine Equilibrium Points

To find the equilibrium points, set the differential equation to zero: \[ rN\left(1-\frac{N}{K}\right) - H = 0 \] Plug in the given values for \(r\) and \(K\), which are 2 and 1000, respectively.\[ 2N\left(1-\frac{N}{1000}\right) = H \]Therefore, \[ 2N - \frac{2N^2}{1000} = H \]Simplify and rearrange the equation:\[ 2N^2 - 2000N + 1000H = 0 \]This is a quadratic equation in terms of \(N\).

Step 2: Solve Quadratic Equation for Equilibrium

To solve the quadratic equation from Step 1, use the quadratic formula:\[ N = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]For our equation \(2N^2 - 2000N + 1000H = 0\), the coefficients are \(a=2\), \(b=-2000\), \(c=1000H\).Substitute \(H=100\):\[ N = \frac{2000 \pm \sqrt{(-2000)^2 - 4 \cdot 2 \cdot 1000 \cdot 100}}{2 \cdot 2} \]Calculate the discriminant and solve for \(N\):\[ N = \frac{2000 \pm \sqrt{4000000 - 800000}}{4} = \frac{2000 \pm \sqrt{3200000}}{4} \]\(\sqrt{3200000} = 1788.85\), thus:\[ N = \frac{2000 \pm 1788.85}{4} \]This yields two solutions: \(N_1 = 947.2\) and \(N_2 = 52.8\).

Step 3: Determine Stability of Equilibrium Points

To assess stability, we consider the sign of the derivative of the growth factor \[ g(N) = rN\left(1-\frac{N}{K}\right) - H \] evaluated at the equilibrium points. The derivative is \[ g'(N) = 2 - \frac{4N}{1000} = 2 - \frac{N}{250} \]Evaluating at \(N_1 = 947.2\),\[ g'(947.2) = 2 - \frac{947.2}{250} = 2 - 3.7888 = -1.7888 \] (Negative, Stable)Evaluating at \(N_2 = 52.8\),\[ g'(52.8) = 2 - \frac{52.8}{250} = 2 - 0.2112 = 1.7888 \] (Positive, Unstable).Thus, \(N_1\) is a stable equilibrium and \(N_2\) is an unstable equilibrium.

Step 4: Find the Maximum Harvest Rate for Positive Population

For a positive equilibrium, the discriminant of the quadratic \(2N^2 - 2000N + 1000H = 0\) must be non-negative:\[ 4000000 - 800000H \geq 0 \]Solving for \(H\):\[ 800000H \leq 4000000 \]\[ H \leq 5 \]Thus, the maximum rate \(H\) that maintains a positive population is 500.

Key Concepts

Equilibrium Points

In population dynamics, especially within the logistic equation model, an equilibrium point is where the population doesn't change over time. To find these points, we first set the rate of change of the population to zero. This transforms the equation\[ rN\left(1 - \frac{N}{K}\right) - H = 0 \]into a quadratic form to solve.
Using specific values (like \( r = 2 \) and \( K = 1000 \)), and solving the quadratic,\[ 2N^2 - 2000N + 1000H = 0 \]yields two potential values for the population, \( N_1 \) and \( N_2 \), known as equilibrium points.
Plugging in \( H = 100 \), we discerned that the equilibria are approximately \( N_1 = 947.2 \) and \( N_2 = 52.8 \). But what do these numbers mean?

• \( N_1 = 947.2 \) represents a state where the population stays stable if initially close to that number.
• \( N_2 = 52.8 \) shows a critical yet unstable point, a threshold below which the population will decrease further if disrupted.

Understanding these points is fundamental in predicting future changes or potential risks in managing a population like fish.

Stability Analysis

Determining the stability of equilibrium points is crucial for sustainable population management. A stable equilibrium keeps the population in balance, whereas an unstable one means even small changes can cause significant shifts.
To perform this analysis, we look at the sign of the derivative of the growth function, \( g(N) \), at these points. The function is derived as:\[ g'(N) = 2 - \frac{N}{250} \]Evaluating this derivative helps us understand how the population changes near each equilibrium:

• **At \( N_1 = 947.2 \):** \( g'(947.2) = -1.7888 \), indicating that any deviation will naturally push the population back to this point, confirming its stability.
• **At \( N_2 = 52.8 \):** \( g'(52.8) = 1.7888 \) implies that small changes can destabilize the population, making this point unstable.

The sign of the derivative informs us if the equilibrium attracts the population toward it (negative, stable) or repels it (positive, unstable). With these insights, decisions on how to maintain or adjust harvesting rates can be made more effectively.

Harvesting Rate

An important aspect of managing a population is determining the maximum rate of removal, or harvesting rate, that can keep the population from declining to zero. This involves ensuring that the discriminant of the equilibrium equation remains non-negative to allow a sustainable population size.
Given the equation:\[ 2N^2 - 2000N + 1000H = 0 \]the condition to maintain a viable population is that the discriminant\[ 4000000 - 800000H \geq 0 \]must be satisfied.

• Solving this inequality, we find \( H \leq 500 \).
• This means that the maximum sustainable harvesting rate is 500 fish per unit time.
• Exceeding this rate risks depleting the population faster than it can recover.

Understanding and maintaining correct harvesting rates ensure that the population remains stable and does not fall below sustainable levels. This knowledge aids in preventing over-exploitation and promoting long-term ecological balance.