/*! 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 34 First describe the type of \(x\)... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

First describe the type of \(x\) - and \(y\) -populations involved \((e x\) ponential or logistic) and the nature of their interactioncompetition, cooperation, or predation. Then find and characterize the system's critical points (as to type and stability). Determine what nonzero \(x-\) and \(y\) -populations can coexist. Finally, construct a phase plane portrait that enables you to describe the long-term behavior of the two populations in terms of their initial populations \(x(0)\) and \(y(0)\). $$ \frac{d x}{d t}=30 x-2 x^{2}-x y, \quad \frac{d y}{d t}=20 y-4 y^{2}+2 x y $$

Short Answer

Expert verified
The system exhibits competitive logistic growth; the critical point \((10,5)\) is stable, indicating possible coexistence.

Step by step solution

01

Identify the Type of Populations and Interaction

The provided equations \( \frac{dx}{dt} = 30x - 2x^2 - xy \) and \( \frac{dy}{dt} = 20y - 4y^2 + 2xy \) suggest that both the \(x\) and \(y\) populations exhibit logistic growth, as evidenced by the terms \(-2x^2\) and \(-4y^2\) which represent self-limiting factors. The interaction term \(-xy\) and \(+2xy\) indicates that a competition interaction exists between the two populations for resources.
02

Finding Critical Points

Set \( \frac{dx}{dt} = 0 \) and \( \frac{dy}{dt} = 0 \) to find the critical points:1. \( 30x - 2x^2 - xy = 0 \)2. \( 20y - 4y^2 + 2xy = 0 \)From the first equation, factor to get \( x(30 - 2x - y) = 0 \), giving \( x = 0 \) or \( y = 30 - 2x \).From the second equation, factor to get \( y(20 - 4y + 2x) = 0 \), giving \( y = 0 \) or \( x = 2y - 10 \).Solve these together to find critical points:- \((0,0)\)- \((0,5)\)- \((15,0)\)- \((10,5)\).
03

Analyzing Stability of Critical Points

Use the Jacobian matrix \( J = \begin{bmatrix} \frac{\partial}{\partial x}(30x - 2x^2 - xy) & \frac{\partial}{\partial y}(30x - 2x^2 - xy) \ \frac{\partial}{\partial x}(20y - 4y^2 + 2xy) & \frac{\partial}{\partial y}(20y - 4y^2 + 2xy) \end{bmatrix} \) to find the nature and stability of the critical points:Calculate the eigenvalues for \( J \) at each critical point to determine stability:- \((0,0)\) : Saddle Point- \((0,5)\) : Saddle Point- \((15,0)\): Saddle Point- \((10,5)\): Stable Node or Spiral, check specific eigenvalues for true nature.
04

Determine Nonzero Coexistence

Evaluate the possibility of coexistence without extinction; the nonzero critical point \((10,5)\) represents a potential state where both populations \(x\) and \(y\) can coexist without one pushing the other to extinction, assuming it is stable.
05

Constructing the Phase Plane Portrait

Draw a phase plane with \(x\)-axis representing the \(x\) population and \(y\)-axis representing the \(y\) population. Mark the critical points: \((0,0)\), \((0,5)\), \((15,0)\), and \((10,5)\).Sketch trajectories based on analysis:- Arrows from saddle points \( (0,0) \), \( (0,5) \), and \( (15,0) \) indicate population either dies out or diverges to \((10,5)\).- Use the Jacobian-derived stability to indicate attractive near the coexistent point \((10,5)\).Assimilate trajectories to predict long-term behaviors from different initial populations \(x(0)\) and \(y(0)\).

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.

Population Dynamics
Population dynamics is the study of how and why populations change over time. In our exercise, we examine two interacting populations, represented by the variables \(x\) and \(y\). Both populations exhibit logistic growth. This is evident from the self-limiting terms \(-2x^2\) and \(-4y^2\), which imply that the populations grow rapidly at first but slow as they approach a carrying capacity.

Logistic growth is common in nature, as resources are limited, causing population growth to taper off when those limits are reached. Additionally, the interaction terms \(-xy\) and \(+2xy\) show a competitive interaction between the populations. This means both populations vie for the same resources, impacting each other's growth.

Understanding how these populations interact can help predict future population dynamics and provide insights into managing ecosystems, pest control, or conservation efforts.
Critical Points
Critical points are values of \(x\) and \(y\) where the rate of change (i.e., the derivatives) of the populations is zero. These points help us understand potential equilibrium states—values where the populations might stabilize if left undisturbed.

From our step-by-step solution, we identify critical points by setting \(\frac{dx}{dt} = 0\) and \(\frac{dy}{dt} = 0\). The four critical points found are \((0,0)\), \((0,5)\), \((15,0)\), and \((10,5)\).

  • At \((0,0)\), both populations are absent, an equilibrium from which neither population can grow.
  • The point \((0,5)\) shows the \(y\)-population existing without the \(x\) population.
  • Conversely, \((15,0)\) indicates the \(x\)-population alone.
  • Finally, \((10,5)\) is a point of interest where both populations coexist simultaneously.
Finding and analyzing these critical points allows for predicting how the populations might evolve or interact over time. Such insights are vital for ecological policy-making or understanding competitive species interactions.
Phase Plane Analysis
The phase plane is a graphical representation of all the possible states of a dynamical system, showing the relationships and trajectories of the populations over time. In phase plane analysis, the \(x\)-axis usually represents one population while the \(y\)-axis represents the other. This exercise guides us to plot the trajectories based on initial conditions \(x(0)\) and \(y(0)\).

By marking the critical points \((0,0)\), \((0,5)\), \((15,0)\), and \((10,5)\) on the phase plane, we can visualize the flow of system states. The trajectories around these points can tell us a lot about the system's dynamics:

  • Saddle points like \((0,0)\), \((0,5)\), and \((15,0)\) suggest instability; the populations tend to diverge from these points.
  • The phase plane thus uses the trajectories to depict how the population might increase, decrease, or stabilize.
Understanding the phase plane of a system provides crucial information on how perturbations or changes in the system will affect population dynamics, facilitating data-driven decisions in ecology.
Stability of Systems
The stability of a system relates to how it reacts to small disturbances or changes from equilibrium. In the context of population dynamics, it describes whether populations return to previous states or move to a new state after disturbances.

In our exercise, we assess stability by examining the Jacobian matrix at each critical point. This matrix helps determine which critical points are stable or unstable.

Calculating the eigenvalues from the Jacobian reveals:
  • All points except \((10,5)\) are saddle points, indicating instability—small changes can lead to significantly different outcomes.
  • The \((10,5)\) critical point, however, is a stable node or spiral, suggesting resilience—populations tend to return here even after disturbances.
Realizing which systems are stable helps in predicting long-term behaviors and making informed decisions, such as which species conservation efforts are more likely to succeed or where control measures might not yield expected outcomes.

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

Find and classify each of the critical points of the almost linear systems. Use a computer system or graphing calculator to construct a phase plane portrait that illustrates your findings. $$ \frac{d x}{d t}=-x+\sin y, \frac{d y}{d t}=2 x $$

A system \(d x / d t=F(x, y), d y / d t=\) \(G(x, y)\) is given. Solve the equation $$ \frac{d y}{d x}=\frac{G(x, y)}{F(x, y)} $$ to find the trajectories of the given system. Use a computer system or graphing calculator to construct a phase portrait and direction field for the system, and thereby identify visually the apparent character and stability of the critical point \((0,0)\) of the given system. \(\frac{d x}{d t}=y^{3} e^{x+y}, \quad \frac{d y}{d t}=-x^{3} e^{x+y}\)

First describe the type of \(x\) - and \(y\) -populations involved \((e x\) ponential or logistic) and the nature of their interactioncompetition, cooperation, or predation. Then find and characterize the system's critical points (as to type and stability). Determine what nonzero \(x-\) and \(y\) -populations can coexist. Finally, construct a phase plane portrait that enables you to describe the long-term behavior of the two populations in terms of their initial populations \(x(0)\) and \(y(0)\). $$ \frac{d x}{d t}=3 x-x^{2}-\frac{1}{4} x y, \quad \frac{d y}{d t}=x y-2 y $$

In Problems, find all critical points of the given system, and investigate the type and stability of each. Verify your conclusions by means of a phase portrait constructed using a computer system or graphing calculator. $$ \frac{d x}{d t}=x y-2, \quad \frac{d y}{d t}=x-2 y $$

Problems 14 through 17 deal with the predator-prey system $$ \begin{aligned} &\frac{d x}{d t}=x^{2}-2 x-x y \\ &\frac{d y}{d t}=y^{2}-4 y+x y \end{aligned} $$ Here each population-the prey population \(x(t)\) and the predator population \(y(t)\) -is an unsophisticated population (like the alligators of Section \(2.1)\) for which the only alternatives (in the absence of the other population) are doomsday and extinction. Problems 14 through 17 imply that the four critical points \((0,0),(0,4),(2,0)\), and \((3,1)\) of the system in (5) are as shown in Fig. \(6.3 .15-a\) nodal sink at the origin, \(a\) saddle point on each coordinate axis, and a spiral source interior to the first quadrant. This is a two-dimensional version of "doomsday versus extinction." If the initial point \(\left(x_{0}, y_{0}\right)\) lies in Region \(I\), then both populations increase without bound (until doomsday), whereas if it lies in Region II, then both populations decrease to zero (and thus both become extinct). In each of these problems use a graphing calculator or computer system to construct a phase plane portrait for the linearization at the indicated critical point. Do your local portraits look consistent with Fig. 6.3.15? Show that the linearization of (5) at \((0,4)\) is \(u^{\prime}=-6 u\), \(v^{\prime}=4 u+4 v\). Then show that the coefficient matrix of this linear system has the negative eigenvalue \(\lambda_{1}=-6\) and the positive eigenvalue \(\lambda_{2}=4\). Hence \((0,4)\) is a saddle point for the system in (5).

See all solutions

Recommended explanations on Math 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.