Question

Use the information given about the nature of the equilibrium point at the origin to determine the value or range of permissible values for the unspecified entry in the coefficient matrix. The origin is an asymptotically stable proper node of \(\mathbf{y}^{\prime}=\left[\begin{array}{rr}-2 & 0 \\ \alpha & -2\end{array}\right] \mathbf{y} ;\) determine the value(s) of \(\alpha\).

Short answer

Answer: The range of permissible values for α is any real number, as it doesn't affect the eigenvalues of the coefficient matrix, which already have negative real parts. So, α ∈ ℝ.

Step-by-step solution

Write down the coefficient matrix

The coefficient matrix is given as: $$ A = \begin{bmatrix} -2 & 0 \\ \alpha & -2 \end{bmatrix} $$ where α is an unspecified entry.

Find the eigenvalues of the coefficient matrix

To find the eigenvalues, we need to first calculate the characteristic equation. The characteristic equation is given by: $$ \det(A - \lambda I) = 0 $$ where λ is an eigenvalue, and I is the identity matrix. Applying this to our coefficient matrix, we have: $$ \det\begin{bmatrix} -2 - \lambda & 0 \\ \alpha & -2 - \lambda \end{bmatrix} = 0 $$

Solve for the eigenvalues

To find the eigenvalues, we compute the determinant of the matrix above: $$ (-2 - \lambda)(-2 - \lambda) - 0*\alpha = 0 \\ \lambda^2 + 4\lambda + 4 = 0 $$ This is a quadratic equation with two eigenvalues λ₁ and λ₂. We notice that α is not involved in this equation. Let's find the eigenvalues and check their real parts.

Calculate the eigenvalues

To find the eigenvalues, we can use the quadratic formula: $$ \lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $$ where a = 1, b = 4, and c = 4. Applying the formula, we get: $$ \lambda_{1,2} = \frac{-4 \pm \sqrt{16 - 16}}{2} \\ \lambda_{1,2} = -2 $$

Check the real parts of the eigenvalues

Since both eigenvalues are real, we check their real parts: $$ \operatorname{Re}(\lambda_1) = -2 \\ \operatorname{Re}(\lambda_2) = -2 $$ Both real parts are negative, which implies that the origin is an asymptotically stable proper node regardless of the value of α.

Determine the permissible values for α

Since both eigenvalues have negative real parts and no relation to the value of α, any value of α will maintain the asymptotically stable proper node property at the origin. Thus, α can have any real value: $$ \alpha \in \mathbb{R} $$

Key concepts

Asymptotically Stable

An equilibrium point is said to be **asymptotically stable** if, after a small disturbance, the system returns to that point over time. In simpler terms, if you slightly budge the system away from equilibrium, it will naturally settle back down. This concept is crucial in evaluating whether solutions to differential equations will remain bounded and nicely behaved over time. In the context of our mathematical problem, the equilibrium point at the origin is said to be asymptotically stable.
- This means any solution starting close to the origin will tend to approach and stay at that point as time goes on. - For linear systems, asymptotic stability is often concerned with the eigenvalues of the system's matrix.
Real eigenvalues with negative real parts ensure that perturbations to initial conditions will decay, leading the solutions back towards the equilibrium point. Thus, once it is confirmed that all eigenvalues have negative real parts, it solidifies the system's asymptotic stability.

Coefficient Matrix

The **coefficient matrix**, denoted as matrix \( A \), is fundamental for describing linear systems of differential equations. It dictates the system's behavior by governing how state variables interact and change over time. In our problem, the matrix is given as \( A = \begin{bmatrix} -2 & 0 \ \alpha & -2 \end{bmatrix} \).
- This matrix decides how each variable in vector \( \mathbf{y} \) contributes to their derivatives. - Each entry in the matrix corresponds to how strongly one variable influences another.
The *unspecified entry* \( \alpha \) introduces a degree of freedom, allowing us to analyze how varying this parameter can affect the system's stability properties. In this scenario, since the eigenvalues were rendered independent of \( \alpha \), the coefficient matrix stays a consistent descriptor of behavior, confirming the stability without specific bounds on \( \alpha \).

Stability Analysis

**Stability analysis** is the process of determining whether a system's equilibrium points are stable, unstable, or asymptotically stable. The analysis primarily involves assessing the eigenvalues of the system's coefficient matrix. For our differential equation system, this involves computing the eigenvalues of matrix \( A \).
- The first step is forming the characteristic equation by setting the determinant of \( A - \lambda I \) to zero. From there, solving for \( \lambda \) gives us the eigenvalues. - Analyzing these eigenvalues helps us understand how perturbations will evolve over time.
In our particular exercise, the computed eigenvalues were \( -2 \), both real and negative, which signifies asymptotic stability around the equilibrium point. This indicates any perturbation will decrease exponentially, allowing the system to settle back at the origin over time. The takeaway from the analysis is that the matrix structure ensures stability regardless of \( \alpha \)'s value, highlighting the robustness of the system's design.