/*! 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 9 Find all equilibria of each syst... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 11: Problem 9

Find all equilibria of each system of differential equations and use the analytical approach to determine the stability of each equilibrium. $$ \begin{array}{l} \frac{d x_{1}}{d t}=4 x_{1}\left(1-x_{1}\right)-2 x_{1} x_{2} \\ \frac{d x_{2}}{d t}=x_{2}\left(2-x_{2}\right)-x_{2} \end{array} $$

Short Answer

Expert verified
Equilibria are \((0,0), (1,0), (0,1), (1,1)\); stability requires Jacobian evaluation.

Step by step solution

01

Identifying Equilibrium Conditions

For an equilibrium point, the derivatives \( \frac{dx_1}{dt} \) and \( \frac{dx_2}{dt} \) must be equal to zero. This yields the equilibrium conditions: \[ 4x_1(1-x_1) - 2x_1x_2 = 0 \]\[ x_2(2 - x_2) - x_2 = 0 \].
02

Solve for \(x_2\) Equilibrium Points

Starting with the second equation: \[ x_2(2-x_2) - x_2 = 0 \] simplifies to \[ x_2(1-x_2) = 0 \] which gives \( x_2 = 0 \) or \( x_2 = 1 \). These are the equilibrium values for \( x_2 \).
03

Solve for \(x_1\) with \(x_2 = 0\)

Substitute \( x_2 = 0 \) into the first equation:\[ 4x_1(1-x_1) = 0 \]which factors to \( x_1(1-x_1) = 0 \), giving \( x_1 = 0 \) or \( x_1 = 1 \). So, the equilibria are \((0,0)\) and \((1,0)\).
04

Solve for \(x_1\) with \(x_2 = 1\)

Substitute \( x_2 = 1 \) into the first equation:\[ 4x_1(1-x_1) - 2x_1 = 0 \]This simplifies to \[ 2x_1^2 - 2x_1 = 0 \]factoring gives \( x_1(x_1 - 1) = 0 \), thereby \( x_1 = 0 \) or \( x_1 = 1 \). Thus, additional equilibria are \((0,1)\) and \((1,1)\).
05

Stability Analysis of Each Equilibrium Point

The stability of an equilibrium can be determined by evaluating the Jacobian matrix at each equilibrium point. The Jacobian for the system is: \[ J = \begin{bmatrix} \frac{\partial}{\partial x_1} (4x_1(1-x_1) - 2x_1x_2) & \frac{\partial}{\partial x_2} (4x_1(1-x_1) - 2x_1x_2) \ \frac{\partial}{\partial x_1} (x_2(2-x_2) - x_2) & \frac{\partial}{\partial x_2} (x_2(2-x_2) - x_2) \end{bmatrix} \] Calculate the Jacobian and evaluate it at each equilibrium to determine stability based on the eigenvalues.
06

Evaluate Jacobian at \((0,0)\)

The Jacobian at \((0,0)\) is:\[ J(0,0) = \begin{bmatrix} 4 & 0 \ 0 & 1 \end{bmatrix} \]Eigenvalues are \( \lambda_1 = 4 \) and \( \lambda_2 = 1 \), both positive, indicating \((0,0)\) is an unstable equilibrium.
07

Evaluate Jacobian at \((1,0)\)

The Jacobian at \((1,0)\) is:\[ J(1,0) = \begin{bmatrix} -4 & -2 \ 0 & 1 \end{bmatrix} \]Eigenvalues are \( \lambda_1 = -4 \) and \( \lambda_2 = 1 \). One positive and one negative eigenvalue indicate \((1,0)\) is a saddle point, thus unstable.
08

Evaluate Jacobian at \((0,1)\)

The Jacobian at \((0,1)\) is:\[ J(0,1) = \begin{bmatrix} 4 & 0 \ 0 & 0 \end{bmatrix} \]Eigenvalues are \( \lambda_1 = 4 \) and \( \lambda_2 = 0 \), indicating a non-hyperbolic equilibrium that requires further investigation for stability, potentially unstable here.
09

Evaluate Jacobian at \((1,1)\)

The Jacobian at \((1,1)\) is:\[ J(1,1) = \begin{bmatrix} -2 & -2 \ 0 & 0 \end{bmatrix} \]Eigenvalues are \( \lambda_1 = -2 \) and \( \lambda_2 = 0 \), indicating a non-hyperbolic point. With \( \lambda_1 < 0 \), it suggests local stability but ultimately requires further nonlinear analysis.

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 91影视!

Key Concepts

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

Equilibrium Points in Differential Equations
Equilibrium points are crucial in understanding the dynamics of differential equations. To find them, you set the derivatives of the system equal to zero. In our case, we need to solve the equations \( \frac{dx_1}{dt} = 0 \) and \( \frac{dx_2}{dt} = 0 \). This is done by solving the algebraic equations derived from the system:
  • \( 4x_1(1-x_1) - 2x_1x_2 = 0 \)
  • \( x_2(1-x_2) = 0 \)
For \( x_2 \), the solutions are straightforward: \( x_2 = 0 \) or \( x_2 = 1 \). By substituting these values into the first equation, we find the corresponding values for \( x_1 \) which are also 0 or 1. Thus, the equilibrium points for this system are \((0,0)\), \((1,0)\), \((0,1)\), and \((1,1)\). These points represent conditions where the system is in balance, meaning no change over time.
Stability Analysis of Equilibrium Points
Once we identify the equilibrium points, the next step is to determine their stability. Stability analysis helps us understand how the system behaves near these points. Are they stable like a pendulum at rest, or unstable like a pencil balanced on its tip? To analyze stability, we consider the small perturbations around these equilibrium points. If a small disturbance causes the system to return to equilibrium, it's stable. If the disturbance grows, it's unstable.
Jacobian Matrix in Differential Systems
The Jacobian matrix is an essential tool in analyzing stability. It helps us understand how functions change around the equilibrium points in a multi-dimensional system. For a system with two variables like ours, the Jacobian matrix \( J \) is constructed by taking partial derivatives of the system equations:
  • The first row of \( J \) contains the partial derivatives of the first equation with respect to \( x_1 \) and \( x_2 \).
  • The second row contains the derivatives of the second equation in the same manner.
This matrix helps us identify how minor changes can shift the system dynamics, hinting at the nature of equilibrium stability.
Eigenvalues and Stability Conclusions
Eigenvalues derived from the Jacobian matrix are key to determining stability. Here's how they function:
  • If all eigenvalues have negative real parts, the equilibrium is stable (attracting).
  • If at least one eigenvalue has a positive real part, the equilibrium is unstable (repelling).
  • Zero or complex eigenvalues need more sophisticated assessment, as they could imply stability depending on further criteria.
For our equilibrium points:
  • \((0,0)\): With eigenvalues \( \lambda_1 = 4 \) and \( \lambda_2 = 1 \), both positive, indicating instability.
  • \((1,0)\): Eigenvalues \( \lambda_1 = -4 \) and \( \lambda_2 = 1 \) feature one of each, a classic sign of a saddle point, thus unstable.
  • \((0,1)\): Eigenvalues \( \lambda_1 = 4 \) and \( \lambda_2 = 0 \) require further analysis as the zero raises questions about local stability.
  • \((1,1)\): With \( \lambda_1 = -2 \) and \( \lambda_2 = 0 \), it tends towards local stability but isn't fully certain without additional analysis.
Understanding eigenvalues offers a quick glimpse into whether equilibrium points can retain their balance in real-world scenarios.

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

We consider differential equations of the form $$ \frac{d \mathbf{x}}{d t}=A \mathbf{x}(t) $$ where $$ A=\left[\begin{array}{ll} a_{11} & a_{12} \\ a_{21} & a_{22} \end{array}\right] $$ The eigenvalues of A will be real, distinct, and nonzero. Analyze the stability of the equilibrium \((0,0)\), and classify the equilibrium according to whether it is a sink, a source, or a saddle point. $$ A=\left[\begin{array}{rr} -2 & -3 \\ 1 & 3 \end{array}\right] $$

Write each system of differential equations in matrix form. $$ \begin{array}{l} \frac{d x_{1}}{d t}=x_{3}-2 x_{1} \\ \frac{d x_{2}}{d t}=-x_{1} \end{array} $$

Find the general solution of each given system of differential equations and sketch the lines in the direction of the eigenvectors. Indicate on each line the direction in which the solution would move if it starts on that line. $$ \left[\begin{array}{l} \frac{d x_{1}}{d t} \\ \frac{d x_{2}}{d t} \end{array}\right]=\left[\begin{array}{rr} -3 & 3 \\ 6 & 4 \end{array}\right]\left[\begin{array}{l} x_{1}(t) \\ x_{2}(t) \end{array}\right] $$

Consider communities composed of two species. The abundance of species 1 at time \(t\) is given by \(N_{1}(t)\), the abundance of species 2 at time \(t\) by \(N_{2}(t) .\) Their dynamics are described by $$ \begin{array}{l} \frac{d N_{1}}{d t}=f_{1}\left(N_{1}, N_{2}\right) \\ \frac{d N_{2}}{d t}=f_{2}\left(N_{1}, N_{2}\right) \end{array} $$ Assume that when both species are at low abundances their abundances increase and that \(f_{1}\) and \(f_{2}\) change sign when crossing their zero isoclines. In each problem, determine the sign structure of the community matrix at the nontrivial equilibrium (indicated by a dot) on the basis of the graph of the zero isoclines. Determine the stability of the equilibria if possible.

An unrealistic feature of the Lotka-Volterra model is that the prey exhibits unlimited growth in the absence of the predator. The model described by the following system remedies this shortcoming (in the model, we assume that the prey evolves according to logistic growth in the absence of the predator; the other features of the model are retained): $$ \begin{array}{l} \frac{d N}{d t}=N\left(1-\frac{N}{K}\right)-4 P N \\ \frac{d P}{d t}=P N-5 P \end{array} $$ Here, \(K>0\) denotes the carrying capacity of the prey in the absence of the predator. In what follows, we will investigate how the carrying capacity affects the outcome of this predator-prey interaction. (a) Draw the zero isoclines of \((11.88)\) for (i) \(K=10\) and (ii) \(K=\) \(3 .\) (b) When \(K=10\), the zero isoclines intersect, indicating the existence of a nontrivial equilibrium. Analyze the stability of this nontrivial equilibrium. (c) Is there a minimum carrying capacity required in order to have a nontrivial equilibrium? If yes, find it and explain what happens when the carrying capacity is below this minimum and what happens when the carrying capacity is above this minimum.
See all solutions

Recommended explanations on Biology Textbooks

Biological Molecules

Read Explanation

Biology Experiments

Read Explanation

Energy Transfers

Read Explanation

Organ Systems

Read Explanation

Responding to Change

Read Explanation

Plant Biology

Read Explanation
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.