Problem 24

Question

Denote the size of a population at time $t$ by $N(t)$, and assume that $$\frac{d N}{d t}=2 N(N-10)\left(1-\frac{N}{100}\right) \quad \text { for } t \geq 0$$ (a) Find all equilibria of $(8.72)$. (b) Use the eigenvalue approach to determine the stability of the equilibria you found in (a). (c) Set $$g(N)=2 N(N-10)\left(1-\frac{N}{100}\right)$$ for $N \geq 0$, and graph $g(N) .$ Identify the equilibria of $(8.72)$ 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

Verified

Answer

Equilibria are at $N = 0$, $N = 10$, $N = 100$; $N = 0$ and $N = 10$ are unstable, $N = 100$ is stable.

1Step 1: Identifying Equilibrium Points

Equilibrium points occur where $ \frac{dN}{dt} = 0 $. Thus, we need to solve the equation, $ 2N(N-10)\left(1 - \frac{N}{100}\right) = 0 $. This can be broken down into the factors: $ N = 0 $, $ N = 10 $, and $ 1 - \frac{N}{100} = 0 $, which simplifies to $ N = 100 $. Therefore, the equilibria are $ N = 0 $, $ N = 10 $, and $ N = 100 $.

2Step 2: Analyzing Stability with Eigenvalues

To determine stability, we calculate the derivative $ g'(N) = \frac{d}{dN} \left[ 2N(N-10)(1-\frac{N}{100}) \right] $. After simplification, we find $ g'(N) = 2(3N^2 - 220N + 1000) $. Plugging in each equilibrium point: - For $ N = 0 $, find $ g'(0) = 2 \times 1000 = 2000 $, indicating instability. - For $ N = 10 $, find $ g'(10) = 2 \times 300 = 600 $, indicating instability. - For $ N = 100 $, find $ g'(100) = -2000 $, indicating stability.

3Step 3: Graphing g(N) and Interpreting Results

Graph $ g(N) = 2N(N-10)(1-\frac{N}{100}) $ for $ N \geq 0 $. The graph will show zeros at $ N = 0 $, $ N = 10 $, and $ N = 100 $, corresponding to the equilibria. - The graph decreases through $ N = 0 $ and $ N = 10 $, confirming instability as determined through eigenvalues. At $ N = 100 $, $ g(N) $ changes sign from positive to negative, confirming stability. The graphical interpretation confirms the eigenvalue analysis, indicating $ N = 0 $ and $ N = 10 $ are unstable, while $ N = 100 $ is stable.

Key Concepts

Equilibrium Points in Population DynamicsStability Analysis Through EigenvaluesUnderstanding Eigenvalues in the Context of Population Dynamics

Equilibrium Points in Population Dynamics

Equilibrium points in population dynamics are specific population sizes at which the population does not change over time. To find these points, we set the population growth rate equation equal to zero. In our exercise, the equation describing the population dynamics is $ \frac{d N}{d t}=2 N(N-10)\left(1-\frac{N}{100}\right) $.
By setting $ \frac{dN}{dt} = 0 $, we solve for the values of $ N $ where this condition holds true. Breaking down the equation into its factors, we find:

$ N = 0 $
$ N = 10 $
$ N = 100 $

These points, $ N = 0, 10, 100 $, are where the population size remains constant, making them our equilibrium points. Understanding these equilibria helps in managing populations and predicting long-term outcomes.

Stability Analysis Through Eigenvalues

Stability analysis involves determining whether small changes in the population size will return to an equilibrium point or deviate further from it. This is crucial when managing ecosystems or any system influenced by population changes.
One method of stability analysis involves using eigenvalues. Once we find the equilibrium points, we evaluate the first derivative of the growth function $ g(N) = 2 N(N-10)\left(1-\frac{N}{100}\right) $.
The derivative, $ g'(N) = 2(3N^2 - 220N + 1000) $, is evaluated at each equilibrium point:

For $ N = 0 $, $ g'(0) = 2000 $, indicating instability.
For $ N = 10 $, $ g'(10) = 600 $, indicating instability.
For $ N = 100 $, $ g'(100) = -2000 $, showing stability.

A positive derivative at an equilibrium suggests instability, as small changes lead the population away from this state. Conversely, a negative derivative indicates stability, as perturbations will dampen, returning the system to equilibrium.

Understanding Eigenvalues in the Context of Population Dynamics

Eigenvalues play a significant role in understanding the stability of equilibrium points in population dynamics. In simple terms, eigenvalues help determine how a system responds to small changes or disturbances. When looking at an equilibrium, eigenvalues provide a mathematical indication of the direction and magnitude of the system's movement when slightly altered.
If the eigenvalue associated with an equilibrium point is positive, it means that the equilibrium is unstable. Small deviations from the equilibrium will grow over time. Conversely, if the eigenvalue is negative, the equilibrium is stable, and any small disturbances will diminish, keeping the system close to equilibrium.
It's essential in population dynamics models as it can help predict long-term behavior of populations. For instance, understanding stability via eigenvalues can aid in forming strategies for conservation or ensuring the sustainability of a species.

Problem 23

Problem 24

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