Problem 14 - 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 \(11.2 .12 .\) $$ \begin{aligned} &x^{\prime}=\frac{3}{2} x+\frac{1}{4} y \\ &y^{\prime}=-x+\frac{1}{2} y \end{aligned} $$

Step-by-Step Solution

Answer

The critical point \((0, 0)\) is a center node.

Step 1: Write the System as a Matrix

The given system is \(x' = \frac{3}{2}x + \frac{1}{4}y\) and \(y' = -x + \frac{1}{2}y\). This can be written as a matrix equation: \[\begin{pmatrix} x' \ y' \end{pmatrix} = \begin{pmatrix} \frac{3}{2} & \frac{1}{4} \ -1 & \frac{1}{2} \end{pmatrix} \begin{pmatrix} x \ y \end{pmatrix}.\] The matrix \(A\) is \(\begin{pmatrix} \frac{3}{2} & \frac{1}{4} \ -1 & \frac{1}{2} \end{pmatrix}\).

Step 2: Compute the Trace \(\tau\)

The trace of a matrix \(A\) is the sum of its diagonal elements. For the matrix \(A = \begin{pmatrix} \frac{3}{2} & \frac{1}{4} \ -1 & \frac{1}{2} \end{pmatrix}\), the trace \(\tau\) is calculated as follows: \[ \tau = \frac{3}{2} + \frac{1}{2} = 2. \]

Step 3: Compute the Determinant \(\Delta\)

The determinant of a 2x2 matrix \(\begin{pmatrix} a & b \ c & d \end{pmatrix}\) is given by \(ad - bc\). For our matrix \(A = \begin{pmatrix} \frac{3}{2} & \frac{1}{4} \ -1 & \frac{1}{2} \end{pmatrix}\), the determinant is calculated as follows: \[ \Delta = \left(\frac{3}{2}\right)\left(\frac{1}{2}\right) - \left(-1\right)\left(\frac{1}{4}\right) = \frac{3}{4} + \frac{1}{4} = 1. \]

Step 4: Classify Using Trace and Determinant

Using Figure 11.2.12, we classify the critical point \((0,0)\) based on the trace \(\tau = 2\) and the determinant \(\Delta = 1\). Since \(\tau^2 - 4\Delta = 4 - 4 = 0\), the system has a double root and it is a center or a degenerate point, typically corresponding to a stable center node.

Key Concepts

Trace of a Matrix

The trace of a matrix is a simple yet important concept in linear algebra. For any square matrix, the trace is the sum of the elements on its main diagonal.
In simpler terms, the main diagonal consists of elements that run from the top left to the bottom right.
This is a useful property because it remains invariant under changes of coordinates or similarity transformations.The mathematical notation for the trace of a matrix \( A \) is expressed as \( \text{tr}(A) \). Consider a 2x2 matrix:\[A = \begin{pmatrix} a & b \ c & d \end{pmatrix}\]The trace of this matrix, \( \tau \), can be calculated as:\[\text{tr}(A) = a + d\]

• For our specific matrix, \( A = \begin{pmatrix} \frac{3}{2} & \frac{1}{4} \ -1 & \frac{1}{2} \end{pmatrix} \), the trace \( \tau \) is \( \frac{3}{2} + \frac{1}{2} = 2 \).

Determinant of a Matrix

The determinant of a matrix is a scalar value that can be computed from the elements of a square matrix.
For a 2x2 matrix, the determinant helps in determining the unique properties of the matrix, such as invertibility and eigenvalues.For a matrix \( A = \begin{pmatrix} a & b \ c & d \end{pmatrix} \), the determinant \( \Delta \) is given by:\[\Delta = ad - bc\]This formula arises from the need to scale, rotate, or reflect transformations described by the matrix. For the matrix:\[A = \begin{pmatrix} \frac{3}{2} & \frac{1}{4} \ -1 & \frac{1}{2} \end{pmatrix}\]We calculate the determinant as:\[\Delta = \left(\frac{3}{2}\right)\left(\frac{1}{2}\right) - \left(-1\right)\left(\frac{1}{4}\right) = \frac{3}{4} + \frac{1}{4} = 1\]

• Determinant value indicates properties like stability and eigenvalues; in our case, \( \Delta = 1 \).

Stability Analysis

Stability analysis in the context of linear systems is critical for understanding the behavior of solutions near equilibrium points.
This involves examining the trace and determinant of the matrix derived from the system's equations.The stability of the system can be analyzed using the characteristic equation, which includes the terms \( \tau \) and \( \Delta \):\[\lambda^2 - \tau \lambda + \Delta = 0\]

• The nature of the roots of this equation (real or complex) directly informs us about the stability.
• For our system, we analyze \( \tau^2 - 4\Delta \).
• If \( \tau^2 - 4\Delta > 0 \), the system has real roots indicating saddle points or nodes (stable/unstable).
• If \( \tau^2 - 4\Delta = 0 \), we have a double root, pointing towards a center or degenerate case, which might be stable.

In the exercise, given \( \tau = 2 \) and \( \Delta = 1 \), \( \tau^2 - 4\Delta = 0 \) implies the critical point at \((0,0)\) is a center node or potentially stable.

Eigenvalues and Eigenvectors

Eigenvalues and eigenvectors play a key role in determining the dynamics of a system described by a matrix.
They provide insight into the system's behavior, particularly around equilibrium points.For a matrix \( A \), eigenvalues \( \lambda \) are solutions to the characteristic polynomial:\[\det(A - \lambda I) = 0\]

• The corresponding eigenvectors are non-zero vectors that, when multiplied by the matrix \( A \), do not change direction.
• The linear transformation defined by \( A \) scales the eigenvectors by their eigenvalues.

In our exercise, we already classify the point using the trace and determinant, which indirectly provide the nature of eigenvalues.
If \( \tau^2 - 4\Delta = 0 \), the matrix might have repeated eigenvalues pointing toward special cases like a center. Thus, understanding eigenvalues and eigenvectors helps in bifurcation analysis and control theory.