/*! 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 2 The point \((0,0)\) is always an... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

The point \((0,0)\) is always an equilibrium. Use the analytical approach to investigate its stability \(\frac{d x_{1}}{d t}=-x_{1}-x_{2}+x_{1}^{2}\) \(\frac{d x_{2}}{d t}=x_{2}-x_{1}^{2}\)

Short Answer

Expert verified
Eigenvalues of \(J\) are \(-1, 1\); \((0,0)\) is a saddle point, hence unstable.

Step by step solution

01

Identify Equilibrium Points

The given system of differential equations is \( \frac{d x_1}{d t}=-x_1-x_2+x_1^2 \) and \( \frac{d x_2}{d t}=x_2-x_1^2 \). The point \((0,0)\) is given as an equilibrium. To confirm, set both equations to zero and solve: \(-x_1-x_2+x_1^2=0\) and \(x_2-x_1^2=0\). Solving these, \(x_1=0\) and \(x_2=0\) satisfy both, confirming \((0,0)\) as an equilibrium point.
02

Linearization Around Equilibrium

Linearize the system around the equilibrium point \((0,0)\) by computing the Jacobian matrix. First, calculate partial derivatives: the Jacobian matrix \(J=\begin{bmatrix} \frac{\partial f_1}{\partial x_1} & \frac{\partial f_1}{\partial x_2} \ \frac{\partial f_2}{\partial x_1} & \frac{\partial f_2}{\partial x_2} \end{bmatrix}\) evaluated at \((0,0)\). For \(f_1 = -x_1 - x_2 + x_1^2\), partials are \(-1\) and \(-1\) respectively. For \(f_2 = x_2 - x_1^2\), partials are \(-2x_1\) evaluated to \(0\) and \(1\). Thus, \(J=\begin{bmatrix} -1 & -1 \ 0 & 1 \end{bmatrix}.\)

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.

Jacobian matrix
The Jacobian matrix is a useful tool in the stability analysis of differential equations. When analyzing a system of differential equations, the Jacobian matrix helps to approximate the system near an equilibrium point. Simply put, it involves taking the partial derivatives of each function in a system relative to each variable and organizing these derivatives in a matrix format.

For the system given in the exercise, the Jacobian matrix was formed by determining the partial derivatives of the functions with respect to each variable, then evaluating them at the equilibrium point \((0,0)\). This resulted in a matrix \(J=\begin{bmatrix} -1 & -1 \ 0 & 1 \end{bmatrix}.\)

The importance of the Jacobian lies in its influence on subsequent steps in stability analysis, particularly linearization and obtaining eigenvalues, which indicate whether the equilibrium is stable or not.
Linearization
Linearization is a technique used to simplify the analysis of a system of differential equations by approximating it near an equilibrium point using a linear model. The idea is to use the Jacobian matrix to write a linear system that describes the behavior of the original, possibly nonlinear, system close to the equilibrium.

In this exercise, the system of equations was linearized around the point \((0, 0)\) using the previously calculated Jacobian matrix:
  • The matrix \(\begin{bmatrix} -1 & -1 \ 0 & 1 \end{bmatrix}\) represents the best linear approximation of the system near the equilibrium point.
  • Linearization views small deviations from the equilibrium and predicts the system’s steady behavior based on these small deviations.
By transforming the complex differential equations into a linear form, it becomes much easier to analyze and predict the stability around the equilibrium point.
Differential equations
Differential equations are mathematical models used to describe how a particular quantity changes over time, often representing physical processes like motion, growth, or decay. They involve derivatives and are central to modeling continuous change.

In the given exercise, the system is defined by two differential equations:
  • \(\frac{dx_1}{dt} = -x_1 - x_2 + x_1^2\)
  • \(\frac{dx_2}{dt} = x_2 - x_1^2\)
These equations describe how \(x_1\) and \(x_2\) change with respect to time. To solve such systems, it's common to look for equilibrium points like (0,0) where the system remains unchanged over time. Solving differential equations, especially nonlinear ones, often requires qualitative approaches like stability analysis rather than solving them directly by finding exact solutions.
Analytical approach
An analytical approach in stability analysis involves using mathematical techniques to explore the properties of a system of equations. It often includes evaluating equilibrium points, finding the Jacobian matrix, and performing linear stability analysis.

In this scenario, the analytical approach was used to examine the stability of the equilibrium point \((0, 0)\). Steps included finding the Jacobian matrix and determining the system's behavior near this point. Here’s a brief breakdown:
  • First, identify equilibrium points in the system where all derivatives equal zero.
  • Then, derive the Jacobian matrix to linearize the system around those points.
  • Analyzing the eigenvalues of the Jacobian matrix at these points enables predictions about stability.
The system's stability can be deduced depending on whether the eigenvalues are negative, positive, or complex. This structured analytical strategy provides clarity and insight into the complex dynamics of differential systems.

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. $$ A=\left[\begin{array}{ll} 0 & 2 \\ 3 & 7 \end{array}\right] $$

Assume that \(N(t)\) denotes the density of an insect species at time \(t\) and \(P(t)\) denotes the density of its predator at time \(t\). The insect species is an agricultural pest, and its predator is used as a biological control agent. Their dynamics are given by the system of differential equations $$ \begin{array}{l} \frac{d N}{d t}=5 N-3 P N \\ \frac{d P}{d t}=2 P N-P \end{array} $$ (a) Explain why $$ \frac{d N}{d t}=5 N $$ describes the dynamics of the insect in the absence of the predator. Solve (11.85). Describe what happens to the insect population in the absence of the predator. (b) Explain why introducing the insect predator into the system can help to control the density of the insect. (c) Assume that at the beginning of the growing season the insect density is \(0.5\) and the predator density is \(2 .\) You decide to control the insects by using an insecticide in addition to the predator. You are careful and choose an insecticide that does not harm the predator. After you spray, the insect density drops to \(0.01\) and the predator density remains at \(2 .\) Use a graphing calculator to investigate the long-term implications of your decision to spray the field. In particular, investigate what would have happened to the insect densities if you had decided not to spray the field, and compare your results with the insect density over time that results from your application of the insecticide.

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 complex conjugates. Analyze the stability of the equilibrium \((0,0)\), and classify the equilibrium according to whether it is a stable spiral, an unstable spiral, or a center. $$ A=\left[\begin{array}{ll} 1 & 3 \\ -2 & 1 \end{array}\right] $$

Analyze the stability of the equilibrium \((0,0)\), and classify the equilibrium. $$ A=\left[\begin{array}{rr} -2 & 3 \\ 1 & -4 \end{array}\right] $$

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 complex conjugates. Analyze the stability of the equilibrium \((0,0)\), and classify the equilibrium according to whether it is a stable spiral, an unstable spiral, or a center. $$ A=\left[\begin{array}{rr} 1 & 2 \\ -5 & -3 \end{array}\right] $$

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.