Problem 12 - Calculus for Biology and Medicine

Question

Denote by \(p=p(t)\) the fraction of occupied sites in the patchy habitat 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 \((8.59)\) 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

Answer

The equilibrium at \(p=0\) is stable based on both the graph and eigenvalue analysis.

Step 1: Define the function g(p)

The problem defines the function as \(g(p) = 0.5p(1-p) - 1.5p\). Simplifying the expression gives \(g(p) = 0.5p - 0.5p^2 - 1.5p\), which further simplifies to \(g(p) = -0.5p^2 - 1p\).

Step 2: Graph g(p) for p ∈ [0,1]

To graph \(g(p)\) over the interval \([0,1]\), note that \(g(p) = -0.5p^2 - 1p\) is a downward-opening quadratic, intersecting the y-axis at \(g(0) = 0\) and \(g(1) = -0.5 - 1 = -1.5\). The graph of \(g(p)\) shows the curve dropping from 0 at \(p=0\) to \(-1.5\) at \(p=1\).

Step 3: Find Equilibria

Equilibria occur when \(\frac{dp}{dt} = g(p) = 0\). This implies \(-0.5p^2 - 1p = 0\). Factoring out \(p\) gives \(p(-0.5p - 1) = 0\), providing the solutions \(p = 0\) and \(p = -2\). Only \(p = 0\) is within the interval \([0,1]\).

Step 4: Determine Stability Using the Graph

For \(p=0\), since the derivative \(g'(p) = -p - 1 < 0\) for \(p > 0\), the graph slopes downward at \(p=0\). This indicates that \(p=0\) is stable as any small perturbation away from 0 will result in \(p\) returning to 0.

Step 5: Analyze Stability Using Eigenvalues

To find stability using eigenvalues, calculate \(g'(p) = -p - 1\). At \(p = 0\), we find \(g'(0) = -0 - 1 = -1\). Since the eigenvalue \(-1 < 0\), the equilibrium at \(p=0\) is stable.

Key Concepts

Equilibrium Analysis

In differential equations, equilibrium points are crucial because they help us understand where a system might naturally settle.
For the patchy habitat model given by the equation \( \frac{d p}{d t}=0.5 p(1-p)-1.5 p \), an equilibrium is found where the rate of change over time is zero.
This means we set the equation equal to zero and solve for \( p \):

• \( 0.5p(1-p) - 1.5p = 0 \)
• Simplies to \( p(-0.5p - 1) = 0 \)
• These conditions provide candidates for equilibrium: \( p = 0 \) and \( p = -2 \).

When focusing on \( p \) in the interval \([0,1]\), only \( p = 0 \) is relevant.
This is because it is the only point that lies within the possible range for a fraction of occupied sites in a patchy habitat model.
Thus, \( p = 0 \) serves as a potential equilibrium point.

Stability Analysis

To determine the stability of an equilibrium point, graphing \( g(p) \) can be very insightful.
From the given exercise, \( g(p) = -0.5p^2 - p \) represents a downward-opening parabola.
This means, visually, the curve will decrease at an increasing rate.
For stability, we observe the slope of the graph near our equilibrium point:

• If the slope at the equilibrium is negative, \( p \) will typically return to the equilibrium when disturbed, indicating a stable point.
• If positive, disturbing the equilibrium means moving further away, leading to instability.

For the point \( p = 0 \), examining \( g'(p) = -p - 1 \), gives us \( g'(0) = -1 \).
Since \( -1 < 0 \), the point is stable, which implies any slight deviation in \( p \) quickly returns to this equilibrium.

Eigenvalues

In stability analysis, eigenvalues offer a powerful mathematical tool to verify findings.
An eigenvalue simplifies understanding whether a system naturally stabilizes at an equilibrium or diverges.
In the given problem, calculating the derivative of \( g(p) = -0.5p^2 - p \) gives:

• \( g'(p) = -p - 1 \)

Evaluating this at \( p = 0 \) results in \( g'(0) = -1 \).
The eigenvalue, \(-1\), reveals further details about stability:

• If \( ext{eigenvalue} < 0 \): The equilibrium is stable, as disturbances return to equilibrium.
• If \( ext{eigenvalue} > 0 \): The system is unstable, implying deviations cause the system to diverge further from equilibrium.

Here, \( g'(0) = -1 \) confirms \( p = 0 \) is stable, validating our initial stability analysis.

Patchy Habitat Model

A patchy habitat model is a framework to understand species occupancy across fragmented environments.
Often represented in a differential equation, it evaluates the fraction of occupied patches over time.
The model simplifies complex ecological dynamics into a mathematical context:

• \( p = p(t) \): The time-dependent fraction of occupied sites.
• Rate of change \( \frac{dp}{dt} \) reflects growth minus loss.

In this exercise, the model (\( 0.5 p(1-p) - 1.5p \)) considers the dynamic interaction of occupancy with factors:

• "Growth" modeled by \( 0.5p(1-p) \), indicating how local reproduction and migration might fill patches.
• "Loss" represented by \( -1.5p \) could involve mortality or migration outflows reducing occupancy.

Understanding these dynamics is vital in conservational ecology, providing insights into species' survival and habitat connectivity.
It helps policymakers decide on interventions necessary to preserve biodiversity in fragmented landscapes.