Question

In each exercise, consider the linear system \(\mathbf{y}^{\prime}=A \mathbf{y}\). Since \(A\) is a constant invertible \((2 \times 2)\) matrix, \(\mathbf{y}=\mathbf{0}\) is the unique (isolated) equilibrium point. (a) Determine the eigenvalues of the coefficient matrix \(A\). (b) Use Table \(6.2\) to classify the type and stability characteristics of the equilibrium point at the phase-plane origin. If the equilibrium point is a node, designate it as either a proper node or an improper node. $$ \mathbf{y}^{\prime}=\left[\begin{array}{rr} 7 & -24 \\ 2 & -7 \end{array}\right] \mathbf{y} $$

Short answer

2. What is the classification and stability of the equilibrium point at the phase-plane origin?

Step-by-step solution

Identify the matrix A

We are provided the matrix A in the exercise: $$ A=\left[\begin{array}{rr} 7 & -24 \\ 2 & -7 \end{array}\right] $$

Calculate the Eigenvalues of A

To find the eigenvalues, first form the characteristic equation by subtracting \(\lambda I\) from the matrix A, and find the determinant of the resulting matrix. Set this determinant equal to 0 and solve for \(\lambda\): $$ \operatorname{det}(A-\lambda I)=\operatorname{det}\left(\left[\begin{array}{cc} 7-\lambda & -24 \\ 2 & -7-\lambda \end{array}\right]\right)=(7-\lambda)(-7-\lambda)-(-24)(2)=0 $$ Now we need to solve this quadratic equation for \(\lambda\): $$ \lambda^2-7\lambda\cdot(-1)+1\cdot(-24\cdot2+7\cdot7)=\lambda^2+7\lambda-65=0 $$ Solving this equation, we obtain the eigenvalues: $$ \lambda_{1,2} = \frac{-7 \pm \sqrt{7^2 - 4(1)(-65)}}{2(1)} = \frac{-7\pm 8}{2} = \{-\frac{1}{2}, -15\} $$ So the eigenvalues of matrix A are \(\lambda_1 = -\frac{1}{2}\) and \(\lambda_2 = -15\).

Classify the Equilibrium Point

Let's recall the stability classification criteria using Table 6.2. Since both eigenvalues are real and negative, the equilibrium point at the phase-plane origin is a stable node. Now we need to determine if it is proper or improper. The given matrix A is diagonalizable since the algebraic multiplicity and geometric multiplicity of both eigenvalues is 1. Since A is diagonalizable, the equilibrium point is a proper node.

Write the Final Answer

We have determined the following: 1. Eigenvalues of matrix A: \(\lambda_1 = -\frac{1}{2}\) and \(\lambda_2 = -15\). 2. The equilibrium point at the phase-plane origin is a stable node, specifically a proper node.

Key concepts

Eigenvalues

When studying a linear system like \( \mathbf{y}^{\prime}=A \mathbf{y} \), eigenvalues of the matrix \( A \) play a crucial role in understanding the system's behavior. Eigenvalues are essentially special numbers that tell us how a matrix can be transformed or stretched in different directions.

For our specific problem, the matrix given is \[A=\left[\begin{array}{rr}7 & -24 \2 & -7\end{array}\right]\]Finding eigenvalues involves solving the characteristic equation, which is derived from \( \text{det}(A - \lambda I) = 0 \), where \( I \) is the identity matrix and \( \lambda \) represents the eigenvalue.

This leads to a quadratic equation whose solutions give us the eigenvalues. In our case, the solutions are \( \lambda_1 = -\frac{1}{2} \) and \( \lambda_2 = -15 \).

• An eigenvalue indicates the factor by which a transformation matrix scales a vector.
• Real negative eigenvalues suggest a change in direction and decrease in magnitude over time.
• The presence of distinct eigenvalues hints that the matrix might be diagonalizable.

Stability Analysis

Stability analysis involves assessing how systems react to small disturbances, which is particularly important in systems modeled by differential equations.

For linear systems of differential equations, eigenvalues are key to this analysis.

In our scenario, since \( \lambda_1 = -\frac{1}{2} \) and \( \lambda_2 = -15 \) are both real and negative, the corresponding eigenvectors lead to solutions that decay over time to the equilibrium point at the origin.

• Real negative eigenvalues mean absolute stability; deviations from equilibrium die out over time.
• The type of node (proper or improper) further influences stability characteristics.
• Since both eigenvalues here are distinct, and the matrix is diagonalizable, we confirm a proper node and thus classical stability.

Understanding stability helps predict both short-term behavior and long-term trends of the system with remarkable precision.

Equilibrium Points

An equilibrium point is where a system remains if unperturbed. It represents a state where no change occurs if the system begins at this point.

In our linear system, \( \mathbf{y} = \mathbf{0} \) is the isolated equilibrium point since the matrix \( A \) is constant and invertible, meaning it provides a unique solution.

Equilibrium points are often analyzed to determine the local behavior of solutions around them.

• If all nearby solutions converge towards the equilibrium point, it is stable.
• In this exercise, with both eigenvalues being negative, the equilibrium point \( \mathbf{y} = \mathbf{0} \) is confirmed to be stable, assuring predictable behavior close to this point.
• The nature of surrounding trajectories (approaching or diverging) gives insights into overall system dynamics.

Understanding equilibrium points sets the stage for more detailed dynamics analysis, ensuring comprehensive understanding and prediction of real-world systems.