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Chapter 8: Problem 11

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).

Short Answer

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

Step by step solution

01

Define the function

First, define the function \( g(p) = 2p(1-p) - p \) as stated in the exercise. Simplify this equation: \( g(p) = 2p - 2p^2 - p = p - 2p^2. \)
02

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 \( p = 0.5 \).
03

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 \( p = 0.5 \).
04

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.
05

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.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

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Most popular questions from this chapter

For Problems \(57-66\) draw the vector field plot of the differential equation. Then, using the given initial conditions, sketch the solutions (i.e., draw a graph showing the dependent variable as a function of the independent variable). \(\frac{d y}{d t}=3 y-2\) (a) \(y(0)=2\), (b) \(y(0)=0\).

A cell constantly gains or loses small molecules to its environment because the small molecules are able to diffuse through the cell membrane. We will build a model for this process. Suppose a molecule is present in the cell at a concentration \(C(t)\), and present in its environment at a concentration \(C_{\infty}\) (you may assume \(C_{\infty}\) is a constant). One model for the diffusion of molecules across the cell membrane is that the rate at which molecules travel through the membrane is proportional to the difference in concentration between the cell and its surroundings. That is: Rate at which $$ \text { molecules flow out }=k\left(C-C_{\infty}\right) $$ of cell The constant \(k\) is known as the permeability of the membrane: \(k>0\), and \(k\) depends on the surface area of the cell and the chemistry of the membrane, as well as the type of molecule. (a) Starting with a word equation for the amount of small molecules in the cell, show, if the cell volume is \(V\), then: $$ \frac{d C}{d t}=-\frac{k}{V}\left(C-C_{\infty}\right) $$ (b) Find the equilibrium of \((8.53)\) and use a graphical analysis to determine whether it is stable or unstable. (c) Suppose that the molecule we are studying is produced within the cell. The cell produces the molecule at a rate \(r\); that is, a quantity \(r\) is produced (added to the cell) in unit time. Explain why the differential equation for the concentration of molecules in the cell should be modified to: $$ \frac{d C}{d t}=-\frac{k}{V}\left(C-C_{\infty}\right)+\frac{r}{V} $$ (d) Analyze Equation (8.54) to find the equilibrium value of the cell concentration. Is this equilibrium stable or unstable? You may use a graphical argument or calculate the eigenvalue to determine the equilibrium's stability.

You should treat \(h\) as a constant. For what values of \(h\) (if any) does each equation have equilibria? Use a graphical argument to show which of the equilibria (if any) are stable. $$ \frac{d y}{d t}=y-h $$

For make vector field plots of each of the differential equations. Find any equilibria of each differential equation and use your vector field plot to classify whether each equilibrium is stable or unstable. $$ \frac{d y}{d t}=2-y $$

Find the equilibria of the following differential equations. $$ \frac{d N}{d t}=\sin N $$
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