Problem 10 - Calculus for Biology and Medicine

Question

Suppose that a fish population evolves according to a logistic equation and that fish are harvested at a rate proportional to the population size. If \(N(t)\) denotes the population size at time \(t\), then $$\frac{d N}{d t}=r N\left(1-\frac{N}{K}\right)-h N$$ Assume that \(r=2\) and \(K=1000\). (a) Find possible equilibria, use the graphical approach to discuss their stability when \(h=0.1\), and find the maximal harvesting rate that maintains a positive population size. (b) Show that if \(h

Step-by-Step Solution

Answer

Equilibria: \( N=0 \) (unstable), \( N=950 \) (stable for \( h=0.1 \)). Maximal \( h=2 \).

Step 1: Write the Logistic Equation

The logistic equation with harvesting is given by \( \frac{d N}{d t} = r N \left( 1 - \frac{N}{K} \right) - h N \). Substitute the given values \( r = 2 \) and \( K = 1000 \), so the equation becomes \( \frac{d N}{d t} = 2N \left( 1 - \frac{N}{1000} \right) - h N \).

Step 2: Find Equilibrium Points

Equilibrium points occur when \( \frac{d N}{d t} = 0 \). Set \( 2N \left( 1 - \frac{N}{1000} \right) - h N = 0 \), which simplifies to \( 2N - \frac{2N^2}{1000} - hN = 0 \) or \( N(2 - \frac{2N}{1000} - h) = 0 \). Solving gives \( N = 0 \) or \( 2 - h = \frac{2N}{1000} \).

Step 3: Solve for Non-zero Equilibrium

From \( 2 - h = \frac{2N}{1000} \), solve for \( N \) to get \( N = 1000(1 - \frac{h}{2}) \). This is the non-zero equilibrium point.

Step 4: Graphical Analysis of Stability for Part (a)

For \( h = 0.1 \), find the equilibria: \( N = 0 \) and \( N = 1000 \times (1 - \frac{0.1}{2}) = 950 \). The equilibrium at \( N = 0 \) is unstable, and \( N = 950 \) is stable because the function \( f(N) = 2N - \frac{2N^2}{1000} - 0.1N \) decreases to zero as \( N \to 950^- \), and increases from zero as \( N \to 950^+ \).

Step 5: Maximum Harvesting Rate for Positive Population

Solve \( 1000(1 - \frac{h}{2}) = 0 \) for \( h \) to prevent \( N \to 0 \). Thus \( 1 - \frac{h}{2} = 0 \) gives \( h = 2 \). So, the maximum allowable \( h \) is just less than 2 for some positive \( N \).

Step 6: Show Nonzero Equilibrium for \( h < 2 \) (Part b)

As shown earlier, for \( h < 2 \), the equilibrium \( N = 1000(1 - \frac{h}{2}) \) is positive. Therefore, there is a nontrivial equilibrium.

Step 7: Analyze Equilibrium Stability Using Eigenvalue Approach (Part c - i)

Linearize the system around equilibrium \( N = 1000(1-\frac{h}{2}) \) by finding the derivative \( \frac{d^2 N}{d t dN} \) at \( N = 950 \). Calculate and find that if the derivative's slope is negative, \( N = 950 \) is stable. Eigenvalue \( \lambda \) being negative indicates stability.

Step 8: Graphical Stability Analysis (Part c - ii)

Use the graphical profile of \( f(N) = 2N(1 - \frac{N}{1000}) - hN \) with \( h = 0.1 \). The graph indicates stability at \( N = 950 \) because the slope at this equilibrium is negative at \( f'(950) < 0 \), showing a return to equilibrium after small disruptions.

Key Concepts

Equilibrium Stability

Analyzing equilibrium stability in a logistic model involves examining how a population behaves around equilibrium points. Equilibrium points occur where the rate of change in population size is zero. In the given logistic equation with harvesting, \[\frac{d N}{d t} = r N \left(1 - \frac{N}{K}\right) - hN,\]setting the derivative to zero identifies these points. For this model, with the parameters \(r = 2\) and \(K = 1000\), equilibria are found from:

• \(N = 0\), which represents the extinction point.
• \(N = 1000(1 - \frac{h}{2})\), a non-zero equilibrium when \(h < 2\).

For \(h = 0.1\), the non-zero equilibrium \(N = 950\) is stable. Stability implies that small disturbances to the population will result in it returning to \(N = 950\) rather than moving away. The zero equilibrium is unstable, as any increase in population will lead to further increases, moving the system away from extinction.

Eigenvalue Analysis

Eigenvalue analysis helps in determining stability by analyzing linearized approximations of the differential equations. To perform this analysis, consider small perturbations around the equilibrium.Let's linearize the system:1. Identify the equilibrium point \(N = 1000(1-\frac{h}{2})\).2. Calculate the derivative, \(\frac{d^2N}{dtdN}\), at the equilibrium for the linearization.3. Use this derivative as the slope in the differential equation approximation.The eigenvalue, \(\lambda\), derived from this linear approximation, indicates stability:

• If \(\lambda < 0\), disturbances decay over time, and the system returns to equilibrium, indicating stability.
• If \(\lambda > 0\), disturbances grow, leading to instability.

In our model, for the equilibrium \(N = 950\) with \(h = 0.1\), the author calculated that the slope (and hence \(\lambda\)) is negative, confirming that this point is stable.

Harvesting Rate

The concept of the harvesting rate in logistic equations is crucial for understanding how external influences affect population sustainability. The harvesting rate, denoted by \(h\), is proportional to the population size—meaning the amount harvested increases with a larger population.For a positive, sustainable population size, the harvesting rate must be carefully managed:

• Maximum harvest sustainable: For the nontrivial equilibrium to exist, \(h\) must satisfy \(h < 2\) considering \(r = 2\). This ensures a positive population remains possible.
• Calculated equilibrium point: If \(h < 2\), the equilibrium is \(N = 1000(1-\frac{h}{2})\). For example, with \(h = 0.1\), the population stabilizes at \(N = 950\).
• If \(h\) approaches or exceeds \(2\), the population may collapse, as the maximum allowable \(h\) was calculated to prevent extinction.

Understanding and applying these principles ensure that management practices keep the population viable under harvesting pressure.