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. $$ \begin{aligned} &x^{\prime}=-3 x-5 y \\ &y^{\prime}=2 x-y \end{aligned} $$

Short answer

Answer: The equilibrium point \(\mathbf{y}_e = \mathbf{0}\) is asymptotically stable.

Step-by-step solution

Identify matrix A

The given system of equations is $$ \begin{aligned} x' &= -3x - 5y, \\ y' &= 2x - y. \end{aligned} $$ Therefore, the matrix \(A\) is $$ A = \begin{bmatrix} -3 & -5 \\ 2 & -1 \end{bmatrix}. $$

Find the eigenvalues of matrix A

To find the eigenvalues, we need to solve the following characteristic equation: $$ \text{det}(A - \lambda I) = 0, $$ where \(I\) is the identity matrix. So, we have $$ \text{det} \left( \begin{bmatrix} -3 - \lambda & -5 \\ 2 & -1 - \lambda \end{bmatrix} \right) = 0. $$ Expanding the determinant, we get $$ (-3 - \lambda)(-1 - \lambda) - (-5)(2) = \lambda^2 + 4\lambda + 7 = 0. $$

Solve the characteristic equation for eigenvalues

To find the eigenvalues, we need to solve the characteristic equation: $$ \lambda^2 + 4\lambda + 7 = 0. $$ Since the discriminant, \(\Delta = 4^2 - 4 \cdot 7 < 0\), the quadratic equation has complex roots, which correspond to the eigenvalues. We find them using the quadratic formula: $$ \lambda_{1,2} = \frac{-4 \pm \sqrt{-8}}{2} = -2 \pm i\sqrt{2}. $$

Classify the equilibrium point based on the eigenvalues' real parts

Now we have the eigenvalues of matrix \(A\): \(\lambda_1 = -2 + i\sqrt{2}\) and \(\lambda_2 = -2 - i\sqrt{2}\). The real parts of both eigenvalues are negative (-2), so according to Theorem 6.3, we can conclude that the equilibrium point \(\mathbf{y}_e = \mathbf{0}\) is asymptotically stable.

Key concepts

Linear Systems of Differential Equations

Understanding linear systems of differential equations is fundamental to various applications in fields like engineering, physics, and economics. Such systems can be used to model real-world phenomena where variables change with respect to each other and time.

A linear system is typically written in the form \( \mathbf{y}' = A \mathbf{y} \), where \( \mathbf{y} \) is a vector of dependent variables, \( A \) is a matrix of constant coefficients, and primes denote derivatives with respect to an independent variable, often time. The solution to such systems provides insights into the behavior of dynamical models such as population growth, electrical circuits or mechanical systems.

In the context of our exercise, the matrix \( A \) is a 2x2 matrix with real, constant coefficients that describes how the system evolves over time. The equilibrium point, typically at \( \mathbf{y}_e = \mathbf{0} \), is a state where the system doesn't change, and this can lead to various types of stability, crucial for understanding long-term behavior of systems.

Eigenvalues and Eigenvectors

Moving from the general system to its underlying specifics, eigenvalues and eigenvectors are indispensable tools for analyzing linear systems. An eigenvector of a matrix \( A \) is a nonzero vector that, when multiplied with \( A \) yields a scalar multiple of itself. The corresponding scalar is known as the eigenvalue.

To find eigenvalues, we look for scalars \( \( \lambda \) \) such that \( A \mathbf{v} = \lambda \mathbf{v} \), where \( \mathbf{v} \) is the eigenvector. The importance of eigenvalues and eigenvectors lies in their ability to provide insights into the behavior of a system; for instance, they can reveal whether a system oscillates, grows, or decays over time.

In our exercise, the eigenvalues \( \lambda_1 \) and \( \lambda_2 \) were found using the characteristic equation of \( A \) and tell us much about the system's stability—dictating whether the system's state will converge to the equilibrium, oscillate, or diverge as time progresses.

Characteristic Equation

The characteristic equation is a pivotal concept when dealing with matrices and their eigenvalues. It's a polynomial whose roots are the eigenvalues of a matrix, revealing the intrinsic properties of the linear transformation that the matrix represents.

In mathematical terms, we derive the characteristic equation from \(\text{det}(A - \lambda I) = 0\), where \( A \) is our matrix, \( \lambda \) are the eigenvalues, and \( I \) is the identity matrix. Solving this determinant equation provides us with the eigenvalues.In our exercise, the resultant characteristic equation \( \lambda^2 + 4\lambda + 7 = 0 \) was key to determining the system’s eigenvalues which then gave us insight into the nature of the equilibrium point of the system.

Stability of Equilibrium

The concept of stability in equilibrium points is crucial to predict the response of a system to small disturbances. If after a disturbance the system returns to its equilibrium, it's considered stable; if it returns to equilibrium over time without perpetual oscillations, it's asymptotically stable; if the system deviates further from its equilibrium, it's deemed unstable.

For linear systems, the stability of an equilibrium point largely depends on the real parts of the eigenvalues of matrix \( A \). If all eigenvalues have negative real parts, the system is asymptotically stable and any small disturbance will dampen over time, as is the case in our example, with real parts of \( \lambda_1 \) and \( \lambda_2 \) both being -2.

Such information is tremendously useful. It enables us to understand phenomena such as why a bridge sways but doesn't collapse after a wind gust, or why an economy can adjust back to a steady-state after a shock. Thus, the stability of an equilibrium has practical implications in designing systems that need to return to a baseline after being pushed out of sync.