/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 22 Denote by \(p=p(t)\) the fractio... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Denote by \(p=p(t)\) the fraction of occupied patches in a metapopulation model, and assume that $$\frac{d p}{d t}=c p(1-p)-p^{2} \quad \text { for } t \geq 0$$ where \(c>0 .\) The term \(p^{2}\) describes the density-dependent extinction of patches; that is, the per-patch extinction rate is \(p\), and a fraction \(p\) of patches are occupied, resulting in an extinction rate of \(p^{2}\). The colonization of vacant patches is the same as in the Levins model. (a) Set \(g(p)=c p(1-p)-p^{2}\) and sketch the graph of \(g(p)\). (b) Find all equilibria of \((8.70)\) in \([0,1]\), and determine their stability. (c) Is there a nontrivial equilibrium when \(c>0 ?\) Contrast your findings with the corresponding results in the Levins model.

Short Answer

Expert verified
Equilibria are at \( p = 0 \) and \( p = \frac{c-1}{c} \). Stable is \( p = \frac{c-1}{c} \) when \( c > 1 \); \( p = 0 \) is unstable for all \( c > 0 \). A nontrivial equilibrium exists for \( c > 1 \).

Step by step solution

01

Understand the Function g(p)

First, we need to rewrite the function given by the differential equation. The function is defined as \( g(p) = c p(1-p) - p^2 \). This function represents the rate of change in the fraction of occupied patches in the population model.
02

Analyze the Function g(p) for Equilibrium

Find the equilibria by setting \( \frac{d p}{d t} = g(p) = 0 \). This gives the equation \( c p(1-p) - p^2 = 0 \). We will solve for \( p \).
03

Solve the Equation for p

Set the quadratic equation \( cp(1-p) - p^2 = 0 \) as \( cp - cp^2 - p^2 = 0 \), which simplifies to \( p(cp - c - p) = 0 \). From this, the solutions are \( p = 0 \) or \( cp - c - p = 0 \).
04

Solve the Linear Equation for p

Solving \( cp - c - p = 0 \) gives \( (c-1)p = c \). Thus, \( p = \frac{c}{c-1} \). Since \( p \) must be between [0, 1], this solution is invalid if \( c < 1 \).
05

Determine Valid Equilibria in [0,1]

The valid equilibria are \( p = 0 \) when \( c \geq 0 \) and possibly \( p = \frac{c-1}{c} \) for \( c > 1 \). When \( c > 1 \), \( \frac{c-1}{c} \) is a valid nontrivial equilibrium in the range.
06

Determine Stability of Equilibria

Evaluate the derivative \( g'(p) = c(1-2p) - 2p \). For \( p = 0 \), \( g'(0) = c > 0 \), suggesting instability. For \( p = \frac{c-1}{c} \) if it exists, \( g' \) is negative because \( g'(\frac{c-1}{c}) = c(1-2\frac{c-1}{c}) - 2\frac{c-1}{c} < 0 \), indicating stability.
07

Compare with Levins Model

The Levins model typically implies \( p_{eq} = 1 - \frac{e}{c} \), but this model indicates possible stability at non-zero \( p \), beyond \( c = 1 \), which was not noted prior.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Equilibrium Analysis
In the context of metapopulation models, equilibrium analysis is key to understanding the behavior of populations through differential equations. An equilibrium point occurs where the rate of change in the system is zero. In this exercise, we identify these points by setting the equation for the rate of change \[ g(p) = c p(1-p) - p^2 = 0 \]to zero. This implies that the population neither grows nor declines at these points.

For this particular model, we seek the values of \(p\) between 0 and 1 that satisfy this equation. By simplifying and solving the quadratic equation derived from setting \(g(p) = 0\), we determine the equilibria to be \(p = 0\) and \(p = \frac{c-1}{c}\) for \(c > 1\). The equilibrium point \(p = 0\) indicates no occupied patches, suggesting extinction. Conversely, \(p = \frac{c-1}{c}\) represents a nontrivial state of occupancy when \(c > 1\), where a proportion of patches remain occupied in equilibrium.

Identifying these equilibria helps analyze the long-term outcomes of populations, especially under varied environmental conditions.
Stability of Equilibria
Stability of an equilibrium in a metapopulation model helps predict how a system will respond to small changes. If an equilibrium is stable, the population will return to this state after small perturbations. Conversely, if unstable, the population will diverge, moving away from this point.

For assessing stability, we look at the derivative of \(g(p)\), the function modeling population change, which is given as\[ g'(p) = c(1-2p) - 2p. \]

Evaluating this for the equilibrium points, we find:
  • At \(p = 0\), \(g'(0) = c > 0\), indicating that this point is unstable. A small increase in \(p\) leads to a further increase away from zero, suggesting extinction cannot be maintained indefinitely.
  • At \(p = \frac{c-1}{c}\), if this is within [0,1], \(g'\) evaluates to a negative value, showing the equilibrium is stable. This results because any small deviation from \(p\) will result in dynamics pushing the population back to this state.
Understanding stability is crucial for predicting how real-world ecosystems might behave in response to changes like habitat loss or restoration.
Differential Equations
Differential equations are mathematical tools used to describe how quantities change with respect to another, often over time. In the setting of a metapopulation model, these equations help capture the dynamic processes influencing population sizes.

The given differential equation in our exercise is\[ \frac{d p}{d t} = c p(1-p) - p^2, \]where \( p = p(t) \) represents the fraction of occupied patches over time. The equation includes terms for colonization \((c p(1-p))\) and extinction \((- p^2)\), helping predict how the system evolves.

Crucially, differential equations in ecological models allow scientists to simulate scenarios and explore potential alterations to biodiversity. By analyzing solutions to these equations, like we did for equilibrium and stability, researchers can forecast how real-world populations might shift in response to variables such as climate change or human activities. Thus, differential equations enable a robust framework for understanding and managing ecological dynamics.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

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).

Solve each pure-time differential equation. \(\frac{d x}{d t}=\frac{1}{1-t}\), where \(x(0)=2\)

Assume that the size of a population, denoted by \(N(t)\), evolves according to the logistic equation. Find the intrinsic rate of growth if the carrying capacity is \(100, N(0)=10\), and \(N(1)=20\).

Suppose that \(N(t)\) denotes the size of a population at time \(t .\) The population evolves according to the logistic equation, but, in addition, predation reduces the size of the population so that the rate of change is given by $$\frac{d N}{d t}=N\left(1-\frac{N}{50}\right)-\frac{9 N}{5+N}$$ The first term on the right-hand side describes the logistic growth; the second term describes the effect of predation. (a) Set $$g(N)=N\left(1-\frac{N}{50}\right)-\frac{9 N}{5+N}$$ and graph \(g(N)\). (b) Find all equilibria of \((8.65)\). (c) Use your graph in (a) to determine the stability of the equilibria you found in (b). (d) Use the method of eigenvalues to determine the stability of the equilibria you found in (b).

Suppose that a population, whose size at time \(t\) is given by \(N(t)\), grows according to $$\frac{d N}{d t}=\frac{1}{100} N^{2}, \quad \text { with } N(0)=10$$ (a) Solve \((8.43)\). (b) Graph \(N(t)\) as a function of \(t\) for \(0 \leq t<10\). What happens as \(t \rightarrow 10 ?\) Explain in words what this means.

See all solutions

Recommended explanations on Biology Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.