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Chapter 6: Problem 19

In Problems, investigate the type of the critical point \((0,0)\) of the given almost linear system. Verify your conclusion by using a computer system or graphing calculator to construct a phase portrait. Also, describe the approximate locations and apparent types of any other critical points that are visible in your figure. Feel free to investigate these additional critical points; you can use the computational methods discussed in the application material for this section. $$ \frac{d x}{d t}=x-3 y+2 x y, \quad \frac{d y}{d t}=4 x-6 y-x y $$

Short Answer

Expert verified
The critical point \((0,0)\) is a stable node due to the negative real eigenvalues.

Step by step solution

01

Identify the Linear System

First, we need to identify the linear part of the system by ignoring the nonlinear terms. The given system is \( \frac{dx}{dt} = x - 3y + 2xy \) and \( \frac{dy}{dt} = 4x - 6y - xy \). The linear system is then \( \frac{dx}{dt} = x - 3y \) and \( \frac{dy}{dt} = 4x - 6y \).
02

Determine the Jacobian Matrix

Next, calculate the Jacobian matrix of the linear part of the system at \((0,0)\). The Jacobian is \[ J = \begin{bmatrix} \frac{\partial}{\partial x}(x - 3y) & \frac{\partial}{\partial y}(x - 3y) \ \frac{\partial}{\partial x}(4x - 6y) & \frac{\partial}{\partial y}(4x - 6y) \end{bmatrix} = \begin{bmatrix} 1 & -3 \ 4 & -6 \end{bmatrix} \].
03

Analyze the Eigenvalues

To determine the type of critical point, find the eigenvalues of the Jacobian matrix. The eigenvalues \(\lambda\) are found by solving the characteristic equation \(\det(J - \lambda I) = 0\). This results in:\[ \det \begin{bmatrix} 1-\lambda & -3 \ 4 & -6-\lambda \end{bmatrix} = 0 \].Calculate the determinant and solve for \(\lambda\):\((1-\lambda)(-6-\lambda) - (-3)(4) = \lambda^2 + 5\lambda + 6 = 0\), giving eigenvalues \(\lambda = -2\) and \(\lambda = -3\).
04

Type of Critical Point

Since both eigenvalues are real and negative, \((0,0)\) is a stable node based on the signs and nature of the eigenvalues.
05

Sketch the Phase Portrait

Use computer software or a graphing calculator to sketch the phase portrait of the system. Observe the behavior around the critical point \((0,0)\), and identify any other visible critical points. The system should converge to the stable node \((0,0)\).
06

Identify Other Critical Points

To find other critical points, set both \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \) to zero, and solve the equations:1. \( x - 3y + 2xy = 0 \)2. \( 4x - 6y - xy = 0 \).Solving these will reveal approximate locations of other critical points visible in the figure.

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Key Concepts

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

Phase Portrait
A phase portrait is a graphical representation that provides a visual overview of the trajectories of a dynamical system in a plane. It is used to depict the behavior of the system's solutions over time. Particularly, the phase portrait helps us understand the tendencies of the system near critical points, like the one at \((0,0)\).

When constructing a phase portrait, the arrows in the graph represent the direction of movement within the system regarding its variables. Areas where arrows converge indicate stable regions, whereas areas where arrows diverge prompt instability.

In the given exercise, creating a phase portrait allows us to observe that the arrows converge towards the point \((0,0)\), confirming it as a stable node. Phase portraits are crucial in analyzing the behavior of complex systems and validating analytical findings visually.
Stable Node
A stable node is a type of critical point found in dynamical systems where trajectories surrounding it move towards it over time. This kind of behavior usually indicates stability within the system.

In the current exercise, \((0,0)\) is identified as a stable node. The fact that both eigenvalues are real and negative, which are \(-2\) and \(-3\), confirms this stability. Stable nodes are known for their settling behavior, where solutions to the system tend to reach a steady state, moving towards the stable node as time evolves.

This understanding helps in predicting long-term behaviors of systems, essential for fields such as engineering and physics, where real-world phenomena often need to be stable and predictable.
Jacobian Matrix
The Jacobian matrix provides vital information about how a system behaves near its critical points. It's a matrix of first-order partial derivatives that represents the linear part of a non-linear system.

In the given problem, the Jacobian matrix at critical point \((0,0)\) is \[ J = \begin{bmatrix} 1 & -3 \ 4 & -6 \end{bmatrix} \]. This matrix enables us to assess the local behavior and stability of the system. By examining how small changes affect the system variables, the Jacobian gives us insights on linear approximations around critical points.

Understanding the Jacobian matrix is crucial because it forms the basis for determining the nature of critical points like nodes or saddles, ultimately aiding the analysis of system stability.
Eigenvalues
Eigenvalues reveal the nature and stability of a critical point in the system. They are solutions to the characteristic equation derived from the Jacobian matrix.

In this exercise, the characteristic equation turns out to be \( \lambda^2 + 5\lambda + 6 = 0 \), with solutions \( \lambda = -2 \) and \( \lambda = -3 \). The eigenvalues, since real and negative, indicate that the critical point \((0,0)\) is a stable node.

Analyzing eigenvalues is an essential aspect of understanding the behavior of dynamical systems. Not only do they indicate stability, but they also determine the rate at which the system approaches or diverges from equilibrium. This information helps in the design and control of systems to ensure desired behaviors.

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Most popular questions from this chapter

Solve each of the linear systems to determine whether the critical point \((0,0)\) is stable, asymptotically stable, or unstable. Use a computer system or graphing calculator to construct a phase portrait and direction field for the given system. Thereby ascertain the stability or instability of each critical point, and identify it visually as a node, a saddle point, a center; or a spiral point. \(\frac{d x}{d t}=y, \quad \frac{d y}{d t}=-5 x-4 y\)

The term bifurcation generally refers to something "splitting apart." With regard to differential equations or systems involving a parameter, it refers to abrupt changes in the character of the solutions as the parameter is changed contimously. Problems illustrate sensitive cases in which small perturbations in the coefficients of a linear or almost linear system can change the type or stability (or both) of a critical point. $$ \begin{aligned} &\text { This problem deals with the almost linear system }\\\ &\frac{d x}{d t}=y+h x\left(x^{2}+y^{2}\right), \quad \frac{d y}{d t}=-x+h y\left(x^{2}+y^{2}\right) \end{aligned} $$ in illustration of the sensitive case of Theorem 2, in which the theorem provides no information about the stability of the critical point \((0,0)\). (a) Show that \((0,0)\) is a center of the linear system obtained by setting \(h=0 .\) (b) Suppose that \(h \neq 0\). Let \(r^{2}=x^{2}+y^{2}\), then apply the fact that $$ x \frac{d x}{d t}+y \frac{d y}{d t}=r \frac{d r}{d t} $$ to show that \(d r / d t=h r^{3} . \quad\) (c) Suppose that \(h=-1 .\) Integrate the differential equation in (b); then show that \(r \rightarrow 0\) as \(t \rightarrow+\infty\). Thus \((0,0)\) is an asymptotically stable critical point of the almost linear system in this case. (d) Suppose that \(h=+1 .\) Show that \(r \rightarrow+\infty\) as \(t\) increases, so \((0,0)\) is an unstable critical point in this case.

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. $$ \frac{d x}{d t}=2 \sin x+\sin y, \frac{d y}{d t}=\sin x+2 \sin y $$

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

In Problems, apply Theorem I to determine the type of the critical point \((0,0)\) and whether it is asymptotically stable, stable, or unstable. Verify your conclusion by using a computer system or graphing calculator to construct a phase portrait for the given linear system. $$ \frac{d x}{d t}=4 x-y, \quad \frac{d y}{d t}=2 x+y $$
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