Question

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. $$ \left[\begin{array}{ll} 2 & -2 \\ 4 & -4 \end{array}\right] $$

Short answer

The origin is a stable node due to eigenvalues \(\lambda_1 = 0\) and \(\lambda_2 = -2\).

Step-by-step solution

Write the System Equation

The linear system of differential equations is given by \(\mathbf{x}^{\prime}=\mathbf{A x}\), where \(\mathbf{A}\) is the matrix \(\begin{bmatrix} 2 & -2 \ 4 & -4 \end{bmatrix}\). We will first determine the type of equilibrium point at the origin.

Find the Eigenvalues of the Matrix

To find the eigenvalues, solve the characteristic equation \(\det(\mathbf{A} - \lambda \mathbf{I}) = 0\), where \(\mathbf{I}\) is the identity matrix. 1. Calculate \(\mathbf{A} - \lambda \mathbf{I} = \begin{bmatrix} 2-\lambda & -2 \ 4 & -4-\lambda \end{bmatrix}\).2. Find the determinant: \[ (2-\lambda)(-4-\lambda) - (-2)(4) = \lambda^2 + 2\lambda = 0 \].3. Solve for \(\lambda\): \(\lambda(\lambda+2) = 0\), giving eigenvalues \(\lambda_1 = 0\) and \(\lambda_2 = -2\).

Determine the Equilibrium Type

With eigenvalues \(\lambda_1 = 0\) and \(\lambda_2 = -2\), the system is a linearly stable node due to one negative eigenvalue and one zero eigenvalue. Generally, the equilibrium point may be a non-isolated (line of equilibria) or unstable node, depending on the eigenvectors.

Use pplane6 for Visualization

Open pplane6 and load the linear system using the Gallery option.1. Input the matrix \(\mathbf{A} = \begin{bmatrix} 2 & -2 \ 4 & -4 \end{bmatrix}\).2. Plot several trajectories to observe the behavior.3. For real eigenvalues, include straight line solutions through eigenvectors using 'Keyboard input'. Since we have zero eigenvalue, focus on any other linear solutions or separatrices.

Key concepts

Equilibrium Point Analysis

In the analysis of linear systems of differential equations, determining the nature of equilibrium points is crucial. An equilibrium point of the system \( \mathbf{x}' = \mathbf{Ax} \) is where the system's behavior rests (i.e., where \( \mathbf{x}' = 0 \)). For this specific system, since we are given the matrix \( \mathbf{A} = \begin{bmatrix} 2 & -2 \ 4 & -4 \end{bmatrix} \), the steps to analyze this involve finding the eigenvalues. These provide insights into the nature of these equilibrium points.

In equilibrium point analysis without technology, we first find the eigenvalues of the matrix using the characteristic equation \( \det(\mathbf{A} - \lambda \mathbf{I}) = 0 \). This leads to the solutions \( \lambda_1 = 0 \) and \( \lambda_2 = -2 \), indicating a linearly stable node due to the presence of a zero and a negative eigenvalue. Such a node implies that the equilibrium point isn’t isolated, often forming a line of equilibria parallel to the corresponding eigenvector of the zero eigenvalue. Unlike saddle points or spirals, nodes can indicate stability but require detailed examination.

Nodes, especially those with zero eigenvalues, often demand additional insights from their corresponding eigenvectors to understand the direction and nature of any flow around the node. Analyzing these aspects provides a deeper appreciation of the system’s dynamics.

Eigenvalues and Eigenvectors

Eigenvalues and eigenvectors are vital tools in the study of linear systems of differential equations. Let's dive deeper into their applications. Eigenvalues \( \lambda \) are solutions to the characteristic equation \( \det(\mathbf{A} - \lambda \mathbf{I}) = 0 \). These values help identify the behavior of system solutions as they relate to equilibrium points. In our current context, we found \( \lambda_1 = 0 \) and \( \lambda_2 = -2 \), indicating at least one direction of stability.

After identifying the eigenvalues, the next step is to find the corresponding eigenvectors. Eigenvectors determine the direction in which the system evolves over time. Solving \((\mathbf{A} - \lambda \mathbf{I})\mathbf{v} = \mathbf{0} \) for each eigenvalue will reveal these vectors.

The zero eigenvalue generally points to a direction of no exponential change in the system's trajectory; however, its corresponding eigenvector might indicate directional flow related to the zero eigenvalue within the phase portrait. In this exercise, finding the eigenvectors helps to visualize the phase space and structural pattern of the system behavior near the equilibrium point.

Phase Portraits with pplane6

Phase portraits are visual representations of the trajectories of a dynamical system over time in the phase plane. They provide an intuitive way to understand the qualitative behavior of the system by showing the paths that solutions take over time. In this exercise, using pplane6 is suggested to visualize the system's behavior given by the matrix \( \mathbf{A} = \begin{bmatrix} 2 & -2 \ 4 & -4 \end{bmatrix} \).

Here's how to conduct this with the software:

• Load the matrix using the Gallery option in pplane6.
• Plot multiple trajectories to capture the variety of solutions that the system might take, especially the straight-line solutions through the eigenvectors, which represent the direction of eigenvalues.
• If you have real eigenvalues, their own trajectories emerge more distinctly, and for the zero eigenvalue, notice the line of equilibria.

These portraits vividly show converging and diverging patterns, especially how trajectories approach or move away from equilibrium points. This exercise in using pplane6 is beneficial because visual insights aid in comprehending the system dynamics beyond simple numerical or algebraic solutions.