/*! 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 26 In Exercises \(21-30\), perform ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

In Exercises \(21-30\), perform steps i) and ii) for the linear system $$ \mathbf{x}^{\prime}=\mathbf{A x}, $$ where \(\mathbf{A}\) is the given matrix. i) Find the type of the equilibrium point at the origin. Do this without the aid of any technology or pplane6. You may, of course, check your answer using pplane. ii) Use pplane6 to plot several solution curves - enough to fully illustrate the behavior of the system. You will find it convenient to use the "linear system" choice from the Gallery menu. If the eigenvalues are real, use the "Keyboard input" option to include straight line solutions starting at \(\pm 10\) times the eigenvectors. If the equilibrium point is a saddle point, compute and plot the separatrices. $$ \left[\begin{array}{ll} 9 & 1 \\ 9 & 3 \end{array}\right] $$

Short Answer

Expert verified
The origin is an unstable node.

Step by step solution

01

Identify the Equilibrium Point

For the linear system \( \mathbf{x}' = \mathbf{A x} \), the equilibrium point is at the origin (0,0). This is because if \( \mathbf{x'} = \mathbf{0} \), then \( \mathbf{A x} = \mathbf{0} \) with the zero vector as a solution.
02

Calculate the Eigenvalues of A

The matrix given is \( \mathbf{A} = \begin{bmatrix} 9 & 1 \ 9 & 3 \end{bmatrix} \). To find the eigenvalues, solve the characteristic equation \( \det(\mathbf{A} - \lambda \mathbf{I}) = 0 \). The determinant is \( (9 - \lambda)(3 - \lambda) - 9 = \lambda^2 - 12\lambda + 18 = 0 \). Solving this quadratic equation gives the eigenvalues: \( \lambda_1 = 6 + \sqrt{18} \approx 10.24 \) and \( \lambda_2 = 6 - \sqrt{18} \approx 1.76 \).
03

Analyze the Eigenvalues

Both eigenvalues \( \lambda_1 \approx 10.24 \) and \( \lambda_2 \approx 1.76 \) are real and positive. When both eigenvalues are positive, the origin is an unstable equilibrium point known as a source or unstable node.
04

Plotting Using pplane6

Using pplane6, plot several solution curves. Since the eigenvalues are real and distinct, the system exhibits behavior typical of an unstable node. The trajectories will move away from the origin along paths approximately given by the eigenvectors.

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 Point
An equilibrium point in a linear system is a point where the system doesn't change; it is stationary or steady. In mathematical terms, for the system \( \mathbf{x}' = \mathbf{A x} \), the equilibrium point is at the origin \((0,0)\). This is because when there is no change in \( \mathbf{x} \) over time, \( \mathbf{x}' = \mathbf{0} \), leading to \( \mathbf{A x} = \mathbf{0} \). The origin becomes an equilibrium point when the matrix action results in a zero vector. This may sound fancy, but it's just saying that if we start at this point, our system stays right there and doesn't move anywhere. This characteristic makes the origin quite significant in studying the dynamics of linear systems. In our exercise, identifying the equilibrium point at \((0,0)\) is the first step in analyzing the behavior of the system.
Eigenvalues
Eigenvalues are fundamental in understanding the behavior of linear systems like \( \mathbf{x}' = \mathbf{A x} \). They, in combination with eigenvectors, determine how trajectories behave near an equilibrium point. To find the eigenvalues of a matrix, you solve the characteristic equation \( \det(\mathbf{A} - \lambda \mathbf{I}) = 0 \). For our matrix \( \mathbf{A} = \begin{bmatrix} 9 & 1 \ 9 & 3 \end{bmatrix} \), this results in \( \lambda^2 - 12\lambda + 18 = 0 \). Solving this gives you two eigenvalues, \( \lambda_1 \approx 10.24 \) and \( \lambda_2 \approx 1.76 \). These eigenvalues indicate the system's tendency to evolve over time. Knowing whether these values are real, complex, positive, or negative tells you whether points move toward or away from the equilibrium or spiral, oscillate, or stabilize.
Unstable Node
When exploring linear systems, existence of real and positive eigenvalues often indicates an unstable node at the equilibrium point. An unstable node is a type of equilibrium where trajectories move away from the critical point. In our system, both eigenvalues \( \lambda_1 \approx 10.24 \) and \( \lambda_2 \approx 1.76 \) are positive and real, suggesting the origin acts as an unstable equilibrium. Here's what happens:
  • The paths near the origin diverge away over time.
  • The direction and speed of movement away depend on the eigenvectors and eigenvalues, respectively.
The concept of an unstable node, simply put, is when you start close to the equilibrium, you don't remain there for long. Instead, the system "repels" itself outward, often resembling a source from which lines emanate.
pplane6
pplane6 is a powerful tool used to visualize the behavior of dynamical systems, especially linear ones like \( \mathbf{x}' = \mathbf{A x} \). It helps illustrate the concept of eigenvalues and equilibrium points by plotting solution curves. When using pplane6, you can:
  • Observe how trajectories behave around an equilibrium point.
  • Detect if the point is a stable node, unstable node, saddle point, or other types.
  • Use features like "linear system" choice to select system characteristics in the Gallery menu.
  • Employ "Keyboard input" for more precise analysis and representation.
The platform shows how solutions move—if they converge to or diverge from an equilibrium point and along which paths. This makes the abstract math much clearer and offers an intuitive grasp of the system's dynamics. For students, using pplane6 can transform linear algebra concepts into dynamic, understandable visuals.

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

In Exercises \(21-30\), perform steps i) and ii) for the linear system $$ \mathbf{x}^{\prime}=\mathbf{A x}, $$ where \(\mathbf{A}\) is the given matrix. i) Find the type of the equilibrium point at the origin. Do this without the aid of any technology or pplane6. You may, of course, check your answer using pplane. ii) Use pplane6 to plot several solution curves - enough to fully illustrate the behavior of the system. You will find it convenient to use the "linear system" choice from the Gallery menu. If the eigenvalues are real, use the "Keyboard input" option to include straight line solutions starting at \(\pm 10\) times the eigenvectors. If the equilibrium point is a saddle point, compute and plot the separatrices. $$ \left[\begin{array}{rr} 1 & 1 \\ -18 & 10 \end{array}\right] $$

Consider the nonlinear system $$ \begin{aligned} &x^{\prime}=x(1-x)-x y, \\ &y^{\prime}=y(2-y)+2 x y . \end{aligned} $$ Show that \((-1 / 3,4 / 3)\) is an equilibrium point of the system. a) Without the use of technology, calculate the Jacobian of the system at the equilibrium point \((-1 / 3\), \(4 / 3)\). What is the equation of the linearization at this equilibrium point? Use [v,e]=eig( \(\mathrm{J})\) to find the eigenvalues and eigenvectors of this Jacobian. b) Enter the system in pplane6. Find the equilibrium point at \((-1 / 3,4 / 3)\). Does the data in the Equilibrium point data window agree with your findings in part (a)? Note: The eigenvalues of the Jacobian predict classification of the equilibrium point. In this case, the point \((-1 / 3,4 / 3)\) is a saddle because the eigenvalues are real and opposite in sign. c) Display the linearization. Does the equation of the linearization agree with your findings in part (a)?

In Exercises \(1-4\), without the aid of technology, using only your algebra skills, sketch the nullclines and find the equilibrium point(s) of the assigned system. Indicate the flow of the the vector field along each nullcline, similar to that shown in Figure 13.1. Check your result with pplane6. If the Symbolic Toolbox is available, use the solve command to find the equilibrium point(s). $$ \begin{aligned} &x^{\prime}=2 x-y+3 \\ &y^{\prime}=y-x^{2} \end{aligned} $$

If a system has two distinct negative eigenvalues, then both straight line solutions will decay to the origin with the passage of time. Consequently, all solutions will decay to the origin. Enter the system, \(x^{\prime}=-4 x+y, y^{\prime}=-2 x-y\), in pplane6 and plot the straight line solutions. Plot several more solutions and note that they also decay to the origin. Select Solutions \(\rightarrow\) Find an equilibrium point, find the equilibrium point at the origin, then read its classification from the PPLANE6 Equilibrium point data window.

In Exercises \(1-4\), without the aid of technology, using only your algebra skills, sketch the nullclines and find the equilibrium point(s) of the assigned system. Indicate the flow of the the vector field along each nullcline, similar to that shown in Figure 13.1. Check your result with pplane6. If the Symbolic Toolbox is available, use the solve command to find the equilibrium point(s). $$ \begin{aligned} &x^{\prime}=x+2 y-4 \\ &y^{\prime}=2 x-y-2 \end{aligned} $$

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.