Question

Each exercise lists a linear system \(\mathbf{y}^{\prime}=A \mathbf{y}\), where \(A\) is a real constant invertible \((2 \times 2)\) matrix. Use Theorem \(6.3\) to determine whether the equilibrium point \(\mathbf{y}_{e}=\mathbf{0}\) is asymptotically stable, stable but not asymptotically stable, or unstable. $$ \mathbf{y}^{\prime}=\left[\begin{array}{rr} 0 & -2 \\ 2 & 0 \end{array}\right] \mathbf{y} $$

Short answer

Answer: The equilibrium point y_e=0 is stable but not asymptotically stable.

Step-by-step solution

Write down the matrix A

The given matrix A is $$ A = \left[\begin{array}{rr} 0 & -2 \\ 2 & 0 \end{array}\right] $$

Find the eigenvalues of matrix A

To find the eigenvalues, we need to solve the characteristic equation \(\det(A - \lambda I) = 0\), where I is the identity matrix and \(\lambda\) is the eigenvalue. $$ \det\left(\left[\begin{array}{cc} 0-\lambda & -2 \\ 2 & 0-\lambda \end{array}\right]\right) = (0-\lambda)^2 - (-2)(2) = \lambda^2 + 4 = 0 $$ Solve the equation \(\lambda^2 + 4 = 0\) for eigenvalues: $$ \lambda_{1,2} = \pm 2i $$

Analyze the eigenvalues and determine the stability

The eigenvalues of the given matrix A are \(\lambda_1 = 2i\) and \(\lambda_2 = -2i\). Both eigenvalues have real parts equal to 0. According to Theorem 6.3, if the eigenvalues have positive real parts, the equilibrium point is unstable. If they have negative real parts, the equilibrium point is asymptotically stable. If the eigenvalues have non-positive real parts and at least one real part is 0, the equilibrium point is stable but not asymptotically stable. Since the real parts of both eigenvalues are 0, the equilibrium point \(\mathbf{y}_e=\mathbf{0}\) is stable but not asymptotically stable.

Key concepts

Linear Systems

In mathematics, a linear system, often encountered in algebra and calculus, is a collection of linear equations relating to the same set of variables.
For example, in a differential equation telling how a system evolves over time, the linear system can be expressed as:

• \(\mathbf{y}^{\prime} = A \mathbf{y}\) where \(\mathbf{y}\) is a vector of variables and \(A\) is a matrix of coefficients.

Linear systems are foundational to understanding more complex systems and have vast applications from engineering to finance. In the context of dynamics, they help model systems which can be analyzed using techniques from linear algebra, such as eigenvectors and eigenvalues. Understanding whether a system is stable or not often starts with inspecting its linearized form.

Eigenvalues

Eigenvalues represent the scalar values in a matrix equation and are fundamental when analyzing the behavior of a linear system.
Mathematically, when a system is represented by a matrix \(A\), the eigenvalues \(\lambda\) can be found using the characteristic equation:

• \(\det(A - \lambda I) = 0\)

Here, \(I\) is the identity matrix.
The solutions to this characteristic equation are the eigenvalues. They tell us important information about a system's behavior, particularly regarding stability.
For instance, in our matrix \(A = \left[\begin{array}{rr} 0 & -2 \ 2 & 0 \end{array}\right]\), the discovered eigenvalues are purely imaginary (\(\pm 2i\)), which directly inform our understanding of the system's stability.

Asymptotic Stability

Asymptotic stability is a concept used to describe the long-term behavior of an equilibrium point in a dynamical system.
An equilibrium point is asymptotically stable if, after small perturbations, the system tends to return to this point over time. Mathematically, this is identifiable when all eigenvalues of the system's matrix

• have strictly negative real parts.

In the context of our exercise, since the eigenvalues are \(\pm 2i\) with zero real parts, the equilibrium point is not asymptotically stable. This means that although solutions may hover around the equilibrium, they do not necessarily converge to it. Recognizing such characteristics helps in predicting whether a real-world system will naturally settle down after disturbance or not.

Differential Equations

Differential equations are equations that relate a function with its derivatives.
They play a crucial role in modeling phenomena where change is constant, such as in physics, biology, and economics. A linear differential equation is an equation that can be expressed in the form:

• \(\mathbf{y}^{\prime} = A \mathbf{y}\)

This specific structure allows us to apply various mathematical techniques, like matrix analysis, to deduce important properties of the system.
By solving these equations, we understand how a system evolves and responds over time. The particular setup with matrix \(A\) allows us to predict stability based on eigenvalues, as demonstrated in our exercise.

Matrix Analysis

Matrix analysis is the area of mathematics dedicated to the study of matrices and their algebraic properties. This discipline provides tools necessary for solving systems of linear equations.
The matrix \(A\) in our differential equation serves as the pivotal object of study.
By analyzing matrices, we can:

• Determine eigenvalues and eigenvectors.
• Study stability properties of the system.
• Simplify complex systems using diagonalization.

In our example, matrix analysis was key to identifying the eigenvalues \(\pm 2i\), which told us about the stability of the equilibrium point.
Thorough understanding of matrices not only aids in solving abstract mathematical problems but also has practical applications in fields using complex data systems.