Problem 25 - Calculus for Biology and Medicine

Question

Denote the size of a population at time \(t\) by \(N(t)\), and assume that $$\frac{d N}{d t}=0.3 N(N-17)\left(1-\frac{N}{200}\right) \quad \text { for } t \geq 0$$ (a) Find all equilibria of \((8.73)\). (b) Use the eigenvalue approach to determine the stability of the equilibria you found in (a). (c) Set $$g(N)=0.3 N(N-17)\left(1-\frac{N}{200}\right)$$ for \(N \geq 0\), and graph \(g(N)\). Identify the equilibria of \((8.73)\) on your graph, and use the graph to determine the stability of the equilibria. Compare your results with your findings in (b). Use your graph to give a graphical interpretation of the eigenvalues associated with the equilibria.

Step-by-Step Solution

Answer

Equilibria: \( N = 0 \) (stable), \( N = 17 \) (unstable), \( N = 200 \) (potentially stable). Stability is consistent across methods.

Step 1: Find Equilibria

The equilibria occur where the rate of change of the population is zero, i.e., \( \frac{dN}{dt} = 0 \). Hence, we solve: \[ 0.3 N(N - 17)\left(1 - \frac{N}{200}\right) = 0 \] This equation has three possible solutions: 1. \( N = 0 \) 2. \( N = 17 \) 3. \( 1 - \frac{N}{200} = 0 \Rightarrow N = 200 \) Thus, the equilibria are \( N = 0, 17, \) and \( 200 \).

Step 2: Determine Stability Using Eigenvalues

To determine stability, we need the derivative of \( g(N) \):\[ g'(N) = 0.3\left( (N-17)\left(1-\frac{N}{200}\right) + N\left(1-\frac{N}{200}\right) - \frac{N(N-17)}{200} \right) \]Evaluate this at each equilibrium:- **For \( N = 0 \):** \\( g'(0) = 0.3(-17)(1) = -5.1 \) (stable)- **For \( N = 17 \):** \\( g'(17) = 0.3 \times 17 \times \left(0.91 - 0.085\right) = 7.128 \) (unstable)- **For \( N = 200 \):** \\( g'(200) = 0.3 \times 183 \times (0) = 0 \) (stability cannot be determined via linearization) The eigenvalue method indicates: \( N = 0 \) is stable; \( N = 17 \) is unstable; \( N = 200 \) requires further investigation.

Step 3: Graphical Representation of g(N)

Graphs provide a visual understanding. Graph \( g(N) = 0.3N(N-17)(1-\frac{N}{200}) \). - **Identify equilibrium points**: These are where the graph crosses the x-axis: at \( N = 0, 17, 200 \).- **Stability:** - \( N = 0 \) (crosses from above): Stable. - \( N = 17 \) (crosses from below): Unstable. - \( N = 200 \) (a tangent point): Potentially stable or neutral (depends on higher-order terms). This graphical analysis coincides with the eigenvalue analysis results.

Step 4: Compare Results and Interpret Stability Graphically

Comparing the eigenvalue and graphical results: - Both indicate \( N = 0 \) is stable and \( N = 17 \) is unstable.- Both suggest that \( N = 200 \) requires more information to confirm stability.- Graphically, \( N = 200 \) is where the curve touches the axis and turns, typically suggesting a semi-stable or neutral equilibrium, but further investigation of dynamics around \( N > 200 \) is generally needed to fully classify."

Key Concepts

Equilibrium Points

In population dynamics, equilibrium points represent the population sizes where the growth or decline rate is zero. This concept is crucial in understanding how populations stabilize or change over time. To find equilibrium points, we set the rate of change of the population, given by \( \frac{dN}{dt} \), equal to zero.

For this specific problem, the differential equation is given as:

• \( \frac{dN}{dt}=0.3 N(N-17)\left(1-\frac{N}{200}\right) \)

Solving this equation leads to three possible equilibrium points: \( N = 0, 17, \) and \( 200 \). Each of these points corresponds to a situation where the population is not changing, meaning the birth and death rates, or any other inflows and outflows affecting the population size, are balanced.

The equilibrium \( N = 0 \) suggests extinction, \( N = 17 \) could indicate a small persistent population, and \( N = 200 \) suggests a potentially larger, sustainable population under ideal conditions.

Stability Analysis

Stability analysis determines whether a population can stay at an equilibrium point over time. Essentially, it helps us understand if small changes will cause the population to return to an equilibrium or deviate from it.

In this exercise, the stability around each equilibrium point is analyzed using the derivative of the function \( g(N) \). The derivative, \( g'(N) \), helps in determining the nature of each equilibrium:

• **Stable**: A slight disturbance causes the system to return to equilibrium.
• **Unstable**: A disturbance leads the system away from equilibrium.
• **Neutral or Semi-stable**: The stability can't be clearly determined just by small disturbances.

For the given function, evaluating the derivative \( g'(N) \) at the equilibrium points provides insights:

• At \( N = 0 \), \( g'(0) = -5.1 \), indicating stable equilibrium since the derivative is negative.
• At \( N = 17 \), \( g'(17) \approx 7.128 \), showing an unstable equilibrium due to the positive derivative.
• At \( N = 200 \), \( g'(200) = 0 \), and further investigation is required since linear stability analysis is inconclusive here.

Eigenvalues

Eigenvalues are mathematical tools used to analyze the stability of equilibrium points in dynamic systems, including population models. They provide valuable insights into how populations respond to disturbances when at equilibrium.

In stability analysis, eigenvalues are calculated from the derivative of the function that represents population change. The sign of the eigenvalue determines stability:

• A **negative eigenvalue** typically implies stability, meaning any small change will die out over time.
• A **positive eigenvalue** suggests instability, where small changes grow and move the system away from equilibrium.
• **Zero eigenvalue** calls for further investigation as it might indicate neutral stability or transitions requiring a second look at higher-order terms or nonlinearities.

Examining the derivative \( g'(N) \) at equilibrium points gives the needed eigenvalues:

• \( N = 0 \) results in a negative value \( -5.1 \), making this equilibrium stable due to the returning tendency of the population.
• \( N = 17 \) yields a positive eigenvalue \( 7.128 \), indicating potential growth away from the equilibrium.
• \( N = 200 \) produces an eigenvalue of zero, highlighting the need for deeper investigation as it could possibly be a neutral equilibrium point.

Understanding these eigenvalues helps researchers predict the long-term behavior of populations under different starting conditions.