Problem 12 - Advanced Engineering Mathematics

Question

Classify the critical point \((0,0)\) of the given linear system by computing the trace \(\tau\) and determinant \(\Delta\) and using Figure \(11.2 .12 .\) $$ \begin{aligned} &x^{\prime}=-5 x+3 y \\ &y^{\prime}=-7 x+4 y \end{aligned} $$

Step-by-Step Solution

Answer

The critical point (0,0) is a stable spiral.

Step 1: Write the System in Matrix Form

Represent the system as a matrix equation. The system is: \[ \begin{aligned} & x' = -5x + 3y \ & y' = -7x + 4y \end{aligned} \] This can be written in matrix form as \[ \begin{bmatrix} x' \ y' \end{bmatrix} = \begin{bmatrix} -5 & 3 \ -7 & 4 \end{bmatrix} \begin{bmatrix} x \ y \end{bmatrix}. \]

Step 2: Analyze Matrix to Find Trace and Determinant

The matrix A is \( \begin{bmatrix} -5 & 3 \ -7 & 4 \end{bmatrix} \). The trace \( \tau \) is the sum of the diagonal elements: \( \tau = -5 + 4 = -1 \). The determinant \( \Delta \) is calculated as: \[ \Delta = (-5)(4) - (3)(-7) = -20 + 21 = 1. \]

Step 3: Classify the Critical Point (0,0)

Using the computed trace \( \tau = -1 \) and determinant \( \Delta = 1 \), and referring to the classification criteria, we analyze stability. Since \( \tau^2 - 4\Delta = (-1)^2 - 4(1) = 1 - 4 = -3 < 0 \), the eigenvalues are complex with a negative real part. Therefore, the critical point \((0,0)\) is a stable spiral.

Key Concepts

Critical Points

The concept of critical points is essential when analyzing linear systems of differential equations. A critical point is a point in the phase space where the system does not change. In simpler terms, it's a steady state or equilibrium point around which the system's behavior can be assessed. For the given linear system, the critical point is \((0,0)\). This means we want to determine how the system behaves around this point.

• Understanding critical points helps in predicting long-term system dynamics.
• They are particularly important for assessing the stability of the system.
• Finding the nature of the equilibrium point helps in visualizing how trajectories behave near it.

Just identifying the critical points is not enough; we need further analysis to classify them, which involves further mathematical tools like matrices and eigenvalues.

Matrix Form

Converting a system of differential equations into matrix form is a fundamental step in analyzing linear systems. For the given problem, each equation was represented in a compact form using matrices. The system: \[ \begin{aligned} & x' = -5x + 3y & y' = -7x + 4y \end{aligned}\] agrees with the matrix representation as follows: \[ \begin{bmatrix} x' \ y' \end{bmatrix} = \begin{bmatrix} -5 & 3 \ -7 & 4 \end{bmatrix} \begin{bmatrix} x \ y \end{bmatrix}. \]

• Matrix form simplifies the handling and solving of systems.
• Allows the use of linear algebraic methods for analysis.
• Reveals patterns and properties of the system more clearly.

This step not only simplifies the calculations but also lays the groundwork for finding other properties like trace, determinant, and eigenvalues without manually solving the differential equations.

Trace and Determinant

Computing the trace and determinant of a system's matrix are crucial steps in understanding the system's behavior. The trace, \( \tau \), of the matrix is the sum of its diagonal entries, while the determinant, \( \Delta \), comes from specific arithmetic based on its elements.
For our system, the matrix is: \[ \begin{bmatrix} -5 & 3 \ -7 & 4 \end{bmatrix}. \]
- **Trace** \( \tau = -5 + 4 = -1 \)- **Determinant** \( \Delta = (-5)(4) - (3)(-7) = -20 + 21 = 1 \)- **Why it matters**:

• The trace and determinant help determine the nature of eigenvalues.
• They aid in defining the system's stability.

These values are pivotal as they determine whether eigenvalues are real or complex, affecting the system's stability at the critical point.

Eigenvalues

Eigenvalues are numbers that give insight into the dynamics around critical points of the system. After obtaining the trace and determinant, we can deduce the eigenvalues using the characteristic equation of the matrix. They are determined by:
\[ \lambda^2 - \tau \lambda + \Delta = 0 \]
For this exercise:- \( \lambda^2 + \lambda + 1 = 0 \) given \( \tau = -1 \) and \( \Delta = 1 \).

• This results in complex eigenvalues \( \lambda = -\frac{1}{2} \pm \frac{\sqrt{3}}{2}i \).

- **Why eigenvalues are important**:

• They tell us about the type and stability of equilibrium points.
• Complex eigenvalues often indicate oscillatory behavior near critical points.
• Real parts of eigenvalues indicate whether the solutions grow, shrink or maintain stability over time.

Thus, understanding eigenvalues helps predict how the system behaves as time progresses.

Stability Analysis

Stability analysis of a linear system assesses whether a system returns to equilibrium after small disturbances. Using the computed trace and determinant, we make deductions regarding the stability near the critical point.
For the given linear system, with trace \( \tau = -1 \) and determinant \( \Delta = 1 \), further calculations show:
- \( \tau^2 - 4\Delta = (-1)^2 - 4(1) = -3 < 0 \), indicating complex eigenvalues.- **Result**: Eigenvalues with negative real parts mean the system is a stable spiral, as any perturbations will lead back to the equilibrium point.

• Stability types include stable/unstable nodes, spirals, and centers.
• Complex eigenvalues often imply spirals, with their real parts dictating stability.
• Real eigenvalues determine if nodes are stable or unstable.

In conclusion, the critical point \((0,0)\) in this system is a stable spiral, meaning trajectories cycle inward towards the equilibrium over time.