Problem 21

Question

Denote by $p=p(t)$ the fraction of occupied patches in a metapopulation model, and assume that $$\frac{d p}{d t}=0.5 p(1-p)-1.5 p \quad \text { for } t \geq 0$$ (a) Set $g(p)=0.5 p(1-p)-1.5 p .$ Graph $g(p)$ for $p \in[0,1]$. (b) Find all equilibria of $(8.69)$ that are in $[0,1] .$ Use your graph in (a) to determine their stability. (c) Use the eigenvalue approach to analyze the stability of the equilibria that you found in (b).

Step-by-Step Solution

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Answer

Equilibrium at $ p=0 $ is stable.

1Step 1: Define g(p)

We are given the differential equation $ \frac{dp}{dt} = 0.5p(1-p) - 1.5p $. We need to set $ g(p) = 0.5p(1-p) - 1.5p $ which simplifies to $ g(p) = -p^2 - p $.

2Step 2: Graph g(p) over [0,1]

To graph $ g(p) = -p^2 - p $, evaluate the function at the endpoints of the interval [0,1] and key points within. The function is quadratic and opens downwards. It has zeros at $ p = 0 $ and $ p = -1 $. The vertex is at $ p = -\frac{b}{2a} = -\frac{-1}{2(-1)} = \frac{1}{2} $, giving $ g(0.5) = -0.5^2 - 0.5 = -0.75 $. The function decreases from 0, reaches a minimum at 0.5 and continues to decrease away from 0.5.

3Step 3: Determine Equilibria

Equilibria occur where $ g(p) = 0 $. For $ g(p) = -p^2 - p = 0 $, we factor to find $ p(p+1)=0 $. The solutions are $ p = 0 $ and $ p = -1 $. However, because $ p $ represents a fraction and must be within [0,1], the only valid equilibrium is $ p = 0 $.

4Step 4: Analyze Graph for Stability

From the graph of $ g(p) = -p^2 - p $, we observe that around $ p = 0 $, the graph slopes downwards. Since $ g(p) > 0 $ for $ p < 0 $ and $ g(p) < 0 $ for all $ p > 0 $, $ p = 0 $ is a stable equilibrium.

5Step 5: Use Eigenvalue Approach for Stability

To apply the eigenvalue approach, evaluate the derivative $ g'(p) = -2p - 1 $ at the equilibrium $ p = 0 $. We find $ g'(0) = -1 $. The negative sign of $ g'(0) $ indicates stable equilibrium, confirming that $ p = 0 $ is stable.

Key Concepts

Equilibrium AnalysisStability GraphEigenvalue Approach

Equilibrium Analysis

Equilibrium in a metapopulation model refers to a state where the fraction of occupied patches, denoted as $ p $, is constant over time. In the given equation $ \frac{d p}{d t}=0.5 p(1-p)-1.5 p $, setting $ g(p) = 0.5p(1-p) - 1.5p $ allows us to find equilibrium points. At equilibrium, the rate of change is zero: $ \frac{dp}{dt} = 0 $.
To find these points:

We set $ g(p) = 0 $, which simplifies to $ -p^2 - p = 0 $.
Factoring gives $ p(p+1) = 0 $.
This yields potential equilibria at $ p = 0 $ and $ p = -1 $.

However, since $ p $ represents a fraction, it must be between 0 and 1, meaning $ p = -1 $ is not valid.
Consequently, the only equilibrium within the valid range is $ p = 0 $. This indicates that, initially, all patches become uninhabited, unless other factors change $ p $.
Understanding the concept of equilibrium in this model helps predict the state of a population over time.

Stability Graph

The stability of the equilibrium point $ p = 0 $ can be understood by examining the graph of the function $ g(p) = -p^2 - p $. A graph provides a visual interpretation,
conveying how $ p $ behaves near an equilibrium point:

The graph of $ g(p) $ is a downward opening parabola, visualized by solving for key points.
Calculate the vertex, $ p = \frac{1}{2} $, and its value is $ g(0.5) = -0.75 $.
Equilibrium occurs where the curve crosses the horizontal axis, here at $ p = 0 $ and $ p = -1 $.

For stability analysis, observe how the curve behaves around these points. Near $ p = 0 $, if $ g(p) > 0 $ for $ p < 0 $ and $ g(p) < 0 $ for $ p > 0 $, it shows stability.
This happens because slight perturbations in $ p $ will still lead back to the equilibrium at $ p = 0 $, showing the system's tendency to return to this state. The graph illustrates that as $ p $ increases or decreases slightly, $ g(p) $ remains negative, pulling $ p $ back to zero, confirming stability.

Eigenvalue Approach

The eigenvalue approach provides a mathematical technique for analyzing the stability of the equilibria by examining the derivative of the function $ g(p) $.
This approach focuses on calculating the eigenvalue of a linear approximation near the equilibrium point.

First, calculate the derivative $ g'(p) = -2p - 1 $.
Evaluate $ g'(p) $ at the equilibrium point $ p = 0 $.
This gives us $ g'(0) = -1 $.

In stability analysis, the sign of the eigenvalue (in this case $ g'(0) $) is crucial. A negative eigenvalue indicates that small deviations from equilibrium will diminish over time.
For $ p = 0 $, since $ g'(0) = -1 $, the negative value confirms the stability of this point.
This analysis aligns with our earlier interpretation, reinforcing that the equilibrium at $ p = 0 $ is indeed stable, and perturbations will not lead away from this state. The eigenvalue provides a concise numerical validation of the graphically observed stability.

Problem 20

Problem 21

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