/*! 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 21 Problems 20 through 22 deal with... [FREE SOLUTION] | 91Ó°ÊÓ

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Problems 20 through 22 deal with the case \(\epsilon=-1\), for which the system in (6) becomes $$ \frac{d x}{d t}=-3 x+x^{2}-x y, \quad \frac{d y}{d t}=-5 y+x y $$ and imply that the three critical points \((0,0),(3,0)\), and \((5,2)\) of \((8)\) are as shown in Fig. \(6.3 .17-\) with a nodal sink at the origin, a saddle point on the positive \(x\) -axis, and a spiral source at \((5,2) .\) In each problem use a graphing calculator of computer system to construct a phase plane portrait for the linearization at the indicated critical point. Do your local portraits look consistent with Fig. 6.3.17? Show that the linearization of the system in (8) at \((3,0)\) is \(u^{\prime}=3 u-3 v, v^{\prime}=-2 v\). Then show that the coefficient matrix of this linear system has the positive eigenvalue \(\lambda_{1}=3\) and the negative eigenvalue \(\lambda_{2}=-2\). Hence \((3,0)\) is a saddle point for \((8)\).

Short Answer

Expert verified
The system linearized at (3,0) is a saddle point, consistent with eigenvalues 3 and -2.

Step by step solution

01

Identify the critical point and system

We are tasked with linearizing the given system of differential equations at the critical point \((3,0)\). The system is: \( \frac{dx}{dt} = -3x + x^2 - xy \) and \( \frac{dy}{dt} = -5y + xy \).
02

Obtain Jacobian matrix

To linearize the system at \((3,0)\), compute the Jacobian matrix of the system. The Jacobian \(J\) is given by: \[ J = \begin{bmatrix} \frac{\partial}{\partial x}(-3x + x^2 - xy) & \frac{\partial}{\partial y}(-3x + x^2 - xy) \ \frac{\partial}{\partial x}(-5y + xy) & \frac{\partial}{\partial y}(-5y + xy) \end{bmatrix} \].
03

Compute partial derivatives

Calculate each element of the Jacobian at \((3,0)\). \[ \frac{\partial}{\partial x}(-3x + x^2 - xy) = -3 + 2x - y \] and at \((3,0)\), this becomes \(-3 + 6 = 3\). \[ \frac{\partial}{\partial y}(-3x + x^2 - xy) = -x \] and at \((3,0)\), this becomes \(-3\). \[ \frac{\partial}{\partial x}(-5y + xy) = y \] (evaluates to 0 at \((3,0)\)) and \[ \frac{\partial}{\partial y}(-5y + xy) = -5 + x \] and at \((3,0)\), this becomes \(-5 + 3 = -2\).
04

Construct the Jacobian matrix

Substitute the computed partial derivatives into the Jacobian matrix. The Jacobian matrix at \((3,0)\) is \[ J = \begin{bmatrix} 3 & -3 \ 0 & -2 \end{bmatrix} \].
05

Find eigenvalues of the Jacobian matrix

To find the eigenvalues, solve the characteristic equation \[ \det(J - \lambda I) = 0 \]. Substituting our matrix, this becomes \[ \det(\begin{bmatrix} 3 - \lambda & -3 \ 0 & -2 - \lambda \end{bmatrix}) = 0 \].
06

Compute the determinant

The determinant is calculated as\[ (3 - \lambda)(-2 - \lambda) = 0 \]. Expanding, we solve \[ \lambda^2 - \lambda - 6 = 0 \]. The roots are \(\lambda_1 = 3\) and \(\lambda_2 = -2\).
07

Interpret the eigenvalues

Since the eigenvalues at \((3,0)\) are \(\lambda_1 = 3\) (positive) and \(\lambda_2 = -2\) (negative), the critical point is a saddle point with trajectories moving away in one direction and towards in another.

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Key Concepts

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

Phase Plane Portrait
A phase plane portrait is a visual representation that displays the trajectory of a dynamical system in the plane, using a two-dimensional set of axes. This graphical depiction helps you to understand how the system behaves over time, given different initial conditions. In the context of our exercise, the system of differential equations provided generates a trajectory that can be plotted in the phase plane.
A phase plane portrait helps to identify critical points, which are points where the trajectory converges or diverges. These points are essential because different types of behavior such as nodes, saddle points, spirals, and more can be observed at these positions in the system. For example, the point \(3,0\) is identified as a saddle point in this exercise.
When constructing the phase plane portrait using tools such as graphing calculators or software, you look at local portraits around each critical point. This helps to verify the behavior predicted analytically (in our case at points like \(3,0\), \(0,0\) and \(5,2\)).
Linearization
Linearization is a method used to approximate non-linear systems near a critical point by a linear one. It involves calculating the Jacobian matrix at the critical point, which gives a linear system representing the behavior of the original system.
This technique is particularly useful because linear systems are easier to analyze than nonlinear systems. By studying the linearized system, you can infer the local behavior of the original nonlinear system around the critical point.
In our exercise, the system of differential equations is linearized at the critical point \(3,0\). The result is a simpler linear system \(u' = 3u - 3v\) and \(v' = -2v\), making it easier to compute and analyze the eigenvalues, and consequently the nature of \(3,0\) as a saddle point.
Jacobian Matrix
The Jacobian matrix is an essential tool in understanding the behavior of a dynamical system near a critical point. It contains partial derivatives of the system's equations and helps in linearization of the system.
In this exercise, the Jacobian matrix \(J\) is calculated by taking partial derivatives of the system equations with respect to both \(x\) and \(y\). For example: \(\frac{\partial}{\partial x}(-3x + x^2 - xy) = -3 + 2x - y\).
Evaluating these partial derivatives at the critical point \(3,0\), the Jacobian becomes \[J = \begin{bmatrix} 3 & -3 \ 0 & -2 \end{bmatrix},\]
which plays a crucial role in examining the nature of the critical point by allowing computation of eigenvalues.
Eigenvalues
Eigenvalues arise naturally when analyzing linearized systems via the Jacobian matrix. They are important because they determine the stability and type of behavior near critical points.
To find the eigenvalues of the Jacobian matrix, you solve the characteristic equation \(\det(J - \lambda I) = 0\). For our system, it becomes \(\lambda^2 + 2\lambda - 3 = 0\). Solving this gives eigenvalues \(\lambda_1 = 3\) and \(\lambda_2 = -2\).
These eigenvalues tell us that the critical point \(3,0\) behaves like a saddle point, since one eigenvalue is positive and the other is negative. Positive eigenvalues indicate instability and suggest that trajectories move away from the critical point, while negative ones indicate stability, guiding trajectories towards it.
Saddle Point
Understanding saddle points is key in analyzing the local dynamics near a critical point in a differential system. A saddle point is characterized by having both attracting and repelling directions due to eigenvalues of differing signs.
In our problem, the critical point \(3,0\) is identified as a saddle point. This finding comes from the eigenvalues of the Jacobian matrix: \(\lambda_1 = 3\) (positive) and \(\lambda_2 = -2\) (negative). One direction pushes trajectories away (due to the positive eigenvalue), while the other draws them in (due to the negative eigenvalue).
This results in a dynamic where trajectories near the saddle point move away in one direction and converge in another, leading to a distinctive, 'saddle-like' behavior in the phase plane portrait.

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Most popular questions from this chapter

In Problems 29 through 32, 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. $$ \frac{d x}{d t}=y^{2}-1, \quad \frac{d y}{d t}=x^{3}-y $$

Each of the systems in Problems 11 through 18 has a single critical point \(\left(x_{0}, y_{0}\right) .\) Apply Theorem 2 to classify this critical point as to type and stability. Verify your conclusion by using a computer system or graphing calculator to construct a phase portrait for the given system. $$ \frac{d x}{d t}=x-2 y-8, \quad \frac{d y}{d t}=x+4 y+10 $$

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. $$ \frac{d x}{d t}=4 y\left(1+x^{2}+y^{2}\right), \quad \frac{d y}{d t}=-x\left(1+x^{2}+y^{2}\right) $$

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 of 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 coefficient matrix of the linearization \(x^{\prime}=\) \(-2 x, y^{\prime}=-4 y\) of the system in (5) at \((0,0)\) has the negative eigenvalues \(\lambda_{1}=-2\) and \(\lambda_{2}=-4\). Hence \((0,0)\) is a nodal sink for (5).

In Problems, show that the given system is almost linear with \((0,0)\) as a critical point, and classify this critical point as to type and stability. Use a computer system or graphing calculator to construct a phase plane portrait that illustrates your conclusion. $$ \frac{d x}{d t}=\sin x \cos y-2 y, \frac{d y}{d t}=4 x-3 \cos x \sin y $$

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