/*! 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 38 We will analyze the population d... [FREE SOLUTION] | 91影视

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Chapter 8: Problem 38

We will analyze the population dynamics that are predicted by (8.69), for different values of b and \(c\). Suppose \(b=c\), then \(\frac{d x}{d t}=\frac{k n b}{2} x(1-x)^{2}\). Assuming \(k=1\), \(n=1\), and \(b=1\), sketch the vector field plot for \(x\).

Short Answer

Expert verified
The vector field shows arrows pointing toward \( x=0.5 \), with equilibrium at \( x=0 \) and \( x=1 \).

Step by step solution

01

Write down the given differential equation

We are given the differential equation \( \frac{dx}{dt} = \frac{k n b}{2} x (1-x)^2 \). By substituting \( k = 1 \), \( n = 1 \), and \( b = 1 \) into the equation, it simplifies to \( \frac{dx}{dt} = \frac{1 \cdot 1 \cdot 1}{2} x (1-x)^2 = \frac{1}{2} x (1-x)^2 \).
02

Analyze the differential equation

The equation \( \frac{dx}{dt} = \frac{1}{2} x (1-x)^2 \) is a nonlinear first-order differential equation where \( x \) is the variable dependent on time \( t \). This equation describes the rate of change of \( x \) over time. The term \( x (1-x)^2 \) implies that \( x = 0 \) and \( x = 1 \) are equilibrium points.
03

Identify equilibrium points

Set the right-hand side of the equation \( \frac{1}{2} x (1-x)^2 = 0 \) to determine the equilibrium points. The solutions are \( x = 0 \) and \( x = 1 \) since these make the equation equal to zero. These points are where the rate of change \( \frac{dx}{dt} \) is zero.
04

Sketch the vector field plot

To sketch the vector field, evaluate \( \frac{1}{2} x (1-x)^2 \) at several points within the interval \( x = [0, 1] \). At \( x = 0 \), the derivative is zero. As \( x \) increases toward 0.5, \( \frac{dx}{dt} \) is positive, indicating \( x \) increases. Beyond \( x = 0.5 \), \( \frac{dx}{dt} \) decreases but remains positive until \( x \) approaches 1, where the derivative again becomes zero. The direction of the arrows at each point should reflect these changes, with arrows pointing towards \( x = 0.5 \).

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Key Concepts

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

Differential Equations in Population Dynamics
Differential equations are a powerful tool for modeling how populations change over time. In this context, they are equations involving derivatives, which represent rates of change. The differential equation given in our exercise is a first-order nonlinear equation: \[ \frac{dx}{dt} = \frac{1}{2} x (1-x)^2 \] This equation consists of:
  • An expression for the rate of change \( \frac{dx}{dt} \).
  • A continuous variable \( x \), representing the population size at any given time.
  • A function of \( x \) on the right-hand side that dictates how this change occurs over time.
Differential equations can describe processes that change smoothly in time, making them ideal for understanding population dynamics. Here, the term \( x(1-x)^2 \) suggests the growth rate depends non-linearly on the current size of the population, showing more complex dynamics than simple exponential growth or decay.
As \( x \) approaches zero or one, the rate of change also approaches zero, hinting at equilibrium points where the population stabilizes.
Equilibrium Points in Population Models
Equilibrium points in a population model are values of \( x \) where the rate of change is zero. In our context, this occurs when \( \frac{dx}{dt} = 0 \). Identifying these points helps predict stable states of the population. By setting the right-hand side of the equation equal to zero: \[ \frac{1}{2} x (1-x)^2 = 0 \] We find the solutions at \( x = 0 \) and \( x = 1 \).
These solutions are significant because:
  • At \( x = 0 \), the population remains extinct unless initially disturbed.
  • At \( x = 1 \), the population has reached its carrying capacity, or the maximum sustainable population size.
Knowing equilibrium points allows us to understand under which conditions the population will stop changing and what stable states are possible.
This can inform management strategies or predict long-term outcomes in ecological or biological systems.
Understanding Vector Field Plots
A vector field plot is a graphical representation that describes how a population's growth rate changes across different population sizes, particularly over a specified interval. It provides a visual understanding of the dynamics described by the differential equation. To construct a vector field plot for our equation, we evaluate \( \frac{1}{2} x (1-x)^2 \) at various points on the interval from \( x = 0 \) to \( x = 1 \).
Here's how to interpret these plots:
  • Arrows are drawn to represent the direction and magnitude of the population's rate of change at each point.
  • Where arrows converge or diverge indicates attraction to or from equilibrium points.
  • At \( x = 0.5 \), the derivative is maximized, suggesting the fastest growth rate.
The vector field thus shows that the population grows towards \( x = 0.5 \) where it experiences its fastest increase, but eventually stabilizes as it approaches the equilibrium points \( x = 0 \) and \( x = 1 \).
This visual tool is invaluable for understanding non-linear dynamics and predicting how population sizes will change over time.

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Most popular questions from this chapter

Find the equilibria of the following differential equations. $$ \frac{d x}{d t}=x^{2}-3 x+2 $$

To study the effects of habitat destruction on a single species, we modify the Levins model in the following way: We assume that a fraction \(D\) of patches is permanently destroyed. Consequently, only patches that are vacant and undestroyed can be successfully colonized. A fraction \(1-p(t)-D\) of patches is both vacant and undestroyed where \(p(t)\) is the fraction of occupied patches. Then: $$ \frac{d p}{d t}=c p(1-p-D)-m p $$ (a) Explain in words the meaning of the different terms in \((8.62)\) (b) Assume that \(m=0.2, c=2\), and \(D=0.2\). Show that \((8.62)\) predicts a nontrivial equilibrium value for \(p(t)\) and that this equilibrium is stable.

Insulin pumps treat patients with type I diabetes by releasing insulin continuously into the fat in the patient's stomach or thigh. We will develop a model for the transport of insulin from the site where it is released by the pump, by treating the fat as a compartment in a single-compartment model. Let's suppose that the pump releases insulin at a constant rate, \(r(r\) is the amount added in one unit of time). (a) Explain why, if insulin is not transported from the site of release, the amount of insulin at the site of release, \(a(t)\), will obey a differential equation: $$ \frac{d a}{d t}=r $$ (b) From the fat, the insulin enters the patient's bloodstream. Suppose that a fraction \(p\) of the insulin present in the patient's fat enters the blood in unit time. Explain why: $$ \frac{d a}{d t}=r-p a $$ (c) Find the equilibrium from the differential equation in part (b) and determine whether this equilibrium is stable or unstable.

To derive the model for the growth or decline of the population of cooperators interacting in a snowdrift game, we modeled the proportion of cooperators using a model. $$ \frac{d x}{d t}=k n x(1-x)(-(b-c / 2) x+(b-c)) $$ where \(b>0\) represents the benefit of interaction if one player is a cooperator and \(c>0\) is the cost of cooperation. In Section 8.3.4 we analyzed Equation \((8.68)\) if \(b>c / 2 .\) Determine the equilibria and what their stability is if \(b=c / 2\).

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).
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