Problem 11 - Calculus for Biology and Medicine

Question

Denote by \(p=p(t)\) the fraction of occupied sites in the patchy habitat model, and assume that $$ \frac{d p}{d t}=2 p(1-p)-p \quad \text { for } t \geq 0 $$ (a) Set \(g(p)=2 p(1-p)-p .\) Graph \(g(p)\) for \(p \in[0,1]\). (b) Find all equilibria of \((8.58)\) that are in \([0,1] .\) Use your graph from (a) to determine their stability. (c) Now use the eigenvalue approach to analyze the stability of the equilibria that you found in (b).

Step-by-Step Solution

Answer

Equilibria at \( p = 0 \) and \( p = 0.5 \). \( p = 0 \) is unstable, \( p = 0.5 \) is stable.

Step 1: Define the function

First, define the function \( g(p) = 2p(1-p) - p \) as stated in the exercise. Simplify this equation: \( g(p) = 2p - 2p^2 - p = p - 2p^2. \)

Step 2: Graph the function

Graph the function \( g(p) = p - 2p^2 \) over the interval \( p \in [0,1] \). You will notice that it is a downward-opening parabola with roots at \( p = 0 \) and \( p = 0.5 \).

Step 3: Find equilibria of the differential equation

Equilibria occur where \( \frac{dp}{dt} = g(p) = 0 \). Solving \( p - 2p^2 = 0 \), factor to find \( p(1 - 2p) = 0 \). Thus, the equilibria are \( p = 0 \) and \( p = 0.5 \).

Step 4: Assess stability using the graph

Use the graph of \( g(p) \) to assess stability. At \( p = 0 \), \( g'(p) < 0 \), indicating stability. At \( p = 0.5 \), \( g'(p) > 0 \), indicating instability.

Step 5: Analyze stability with the eigenvalue approach

Calculate the derivative \( g'(p) = 1 - 4p \) to analyze stability. Evaluate at equilibria: at \( p = 0 \), \( g'(0) = 1 \) (unstable because \( \frac{dg}{dp} > 0 \)); at \( p = 0.5 \), \( g'(0.5) = -1 \) (stable because \( \frac{dg}{dp} < 0 \)). Correct the earlier assessment by considering graphing errors.

Key Concepts

Equilibria

In differential equations, an equilibrium is a point where the rate of change of the system is zero. It represents a state of balance or rest for the system. For our problem, the differential equation is given as \( \frac{dp}{dt} = g(p) = p - 2p^2 \). Equilibria occur where the derivative is zero, so we set \( g(p) = 0 \). By solving \( p - 2p^2 = 0 \), we can factor it into \( p(1 - 2p) = 0 \). This gives us the equilibria at \( p = 0 \) and \( p = 0.5 \). These points are where the system does not change over time, making them the key points of interest for stability analysis.

Stability Analysis

Stability analysis goes a step further than finding equilibria by determining the behavior of the system around these points. In simpler terms, it tells us if an equilibrium is stable (where nearby points tend to move towards it) or unstable (where nearby points move away).

• If a small perturbation in the state returns to the equilibrium, it's stable.
• If it moves away from the equilibrium, it's unstable.

For this exercise, we graph the function \( g(p) = p - 2p^2 \) over the interval \([0, 1]\). The graph reveals a parabola opening downwards with roots at the equilibria, \( p = 0 \) and \( p = 0.5 \). Examining the slope around these points helps us understand their stability. For instance, at \( p = 0 \) if the derivative \( g'(0) < 0 \), it suggests that the equilibrium is stable. Contrarily, if \( g'(p) > 0 \) at \( p = 0.5 \), it indicates instability. Thus, graphical inspection allows us to make conclusions about the direction of movement close to the equilibria.

Eigenvalue Approach

The eigenvalue approach is a more technical method used to ascertain the stability of equilibria. While graphical techniques offer a visual insight, eigenvalues provide a precise algebraic assessment. In this context, we compute the derivative of the function, \( g'(p) = 1 - 4p \). Evaluating this derivative at the equilibria gives insight into their behavior.

• At \( p = 0 \), the eigenvalue \( g'(0) = 1 \), indicating instability because it is positive.
• Conversely, at \( p = 0.5 \), \( g'(0.5) = -1 \), signifying stability due to the negative value of the derivative.

The sign of these eigenvalues (or slopes) determines whether perturbations will die out (stable) or grow (unstable) at each equilibrium point. In summary, the eigenvalue approach provides a strong mathematical foundation for drawing conclusions about stability based on analyzing these derivatives directly.