Problem 20 - Advanced Engineering Mathematics

Question

Let \(\mathbf{X}=\mathbf{X}(t)\) be the response of the linear dynamical system $$ \begin{aligned} &x^{\prime}=\alpha x-\beta y \\ &y^{\prime}=\beta x+\alpha y \end{aligned} $$ that satisfies the initial condition \(\mathbf{X}(0)=\mathbf{X}_{0} .\) Determine conditions on the real constants \(\alpha\) and \(\beta\) that will ensure \(\lim _{t \rightarrow \infty} \mathbf{X}(t)=(0,0) . \operatorname{Can}(0,0)\) be a node or saddle point?

Step-by-Step Solution

Answer

The origin \((0,0)\) is stable if \( \alpha < 0 \), but cannot be a node or saddle point.

Step 1: Rewrite as Matrix System

The given system of equations can be represented in matrix form as \( \mathbf{X}' = A\mathbf{X} \), where \( A = \begin{pmatrix} \alpha & -\beta \ \beta & \alpha \end{pmatrix} \) and \( \mathbf{X}(t) = \begin{pmatrix} x(t) \ y(t) \end{pmatrix} \).

Step 2: Find Eigenvalues of Matrix A

To determine the stability of the system, calculate the eigenvalues of the matrix \( A \). The characteristic equation is given by \( \text{det}(A - \lambda I) = 0 \), leading to \( (\alpha - \lambda)^2 + \beta^2 = 0 \).

Step 3: Solve Characteristic Equation

Solving \((\alpha - \lambda)^2 + \beta^2 = 0\), we find the eigenvalues \( \lambda_1, \lambda_2 = \alpha \pm i\beta \). Since these are complex with non-zero imaginary parts, they indicate a center under certain conditions on \( \alpha \) and \( \beta \).

Step 4: Analyze Conditions for Stability

The system is stable if the real part of the eigenvalues is negative. Here, since \( \lambda_1, \lambda_2 \) have real part \( \alpha \), they will decay to zero at \( t \rightarrow \infty \) if \( \alpha < 0 \).

Step 5: Determine Type of Equilibrium

Since the eigenvalues \( \lambda_1, \lambda_2 = \alpha \pm i\beta \) are complex with \( \alpha = 0 \), they do not represent a node or saddle point due to the imaginary part, but a center. Therefore, \( (0,0) \) cannot be a node or saddle point without modifying \( \alpha \) and \( \beta \).

Key Concepts

Eigenvalues

Understanding eigenvalues is crucial in analyzing the behavior of linear dynamical systems. Eigenvalues, often denoted by \( \lambda \), are special numbers that arise from solving the equation \( \text{det}(A - \lambda I) = 0 \). Here, \( A \) is the system matrix, and \( I \) is the identity matrix. The result of this determinant gives us a polynomial, whose roots are the eigenvalues.

These eigenvalues provide insights into the system's dynamics. For instance:

• If all eigenvalues have negative real parts, the system tends to reach a stable equilibrium over time.
• If any eigenvalue has a positive real part, the system may exhibit instability and potentially diverge.
• Complex eigenvalues often indicate oscillatory behavior, with the real part affecting whether these oscillations grow, decay, or maintain constant amplitude.

In our exercise, the eigenvalues \( \lambda_1 \) and \( \lambda_2 = \alpha \pm i\beta \) are complex, suggesting oscillatory dynamics due to their imaginary components.

Stability Analysis

Stability analysis determines whether the trajectory of a dynamical system remains close to an equilibrium point over time. In our situation, stability hinges on the real part of the eigenvalues derived from the system matrix \( A \).

When the real part of all eigenvalues is negative, solutions of the system will decay exponentially to the equilibrium point, indicating a stable system. Conversely, a positive real part suggests that solutions could grow without bound, leading to instability.

In the given system:\[ \lambda_1, \lambda_2 = \alpha \pm i\beta \]The real part is \( \alpha \). Thus, for stability, we need \( \alpha < 0 \). When \( \alpha = 0 \), the system exhibits neutral stability and behaves as a center, meaning solutions will neither converge to nor diverge from the equilibrium point. Instead, they maintain a constant amplitude of oscillation.

Equilibrium Points

Equilibrium points are crucial in understanding the behavior of dynamical systems. These are points where the system does not change over time, corresponding to where the derivative \( x', y' \) becomes zero for continuous systems.

In our matrix form \( \mathbf{X}' = A\mathbf{X} \), the equilibrium point is \( (x, y) = (0, 0) \), meaning no changes occur in the system at this point. However, the nature of this equilibrium can vary:

• A **node** is a stable equilibrium where trajectories move directly towards the equilibrium point.
• A **saddle point** is an unstable equilibrium; some trajectories approach, while others diverge.
• A **center** oscillates around the equilibrium, like in our system due to complex eigenvalues.

In this case, because the eigenvalues are complex with a zero real part and the same magnitude, \( (0, 0) \) cannot be a node or saddle point. Instead, it exhibits the characteristics of a center, maintaining consistent oscillatory patterns.