Problem 13 - Advanced Engineering Mathematics

Question

Classify the critical point \((0,0)\) of the given linear system by computing the trace \(\tau\) and determinant \(\Delta\). (a) \(\begin{aligned} x^{\prime} &=-3 x+4 y \\ y^{\prime} &=-5 x+3 y \end{aligned}\) (b) \(\begin{aligned} x^{\prime} &=-3 x+2 y \\ y^{\prime} &=-2 x+y \end{aligned}\)

Step-by-Step Solution

Answer

(a) Center; (b) Stable node.

Step 1: Write the system in matrix form

For the given linear systems, we start by writing them in the form of a matrix. (a) The system is: \[ x' = -3x + 4y \] \[ y' = -5x + 3y \] This gives the matrix: \[ A = \begin{pmatrix} -3 & 4 \ -5 & 3 \end{pmatrix} \](b) The system is: \[ x' = -3x + 2y \] \[ y' = -2x + y \] This gives the matrix: \[ B = \begin{pmatrix} -3 & 2 \ -2 & 1 \end{pmatrix} \]

Step 2: Calculate the trace (\(\tau\))

The trace of a 2x2 matrix \( \begin{pmatrix} a & b \ c & d \end{pmatrix} \) is calculated as \( \tau = a + d \).(a) For matrix \( A \), the trace is: \[ \tau_A = -3 + 3 = 0 \](b) For matrix \( B \), the trace is: \[ \tau_B = -3 + 1 = -2 \]

Step 3: Calculate the determinant (\(\Delta\))

The determinant of a 2x2 matrix \( \begin{pmatrix} a & b \ c & d \end{pmatrix} \) is \( \Delta = ad - bc \).(a) For matrix \( A \), the determinant is: \[ \Delta_A = (-3)(3) - (4)(-5) = -9 + 20 = 11 \](b) For matrix \( B \), the determinant is: \[ \Delta_B = (-3)(1) - (2)(-2) = -3 + 4 = 1 \]

Step 4: Classify the critical point (\(0,0\))

Using the trace \( \tau \) and determinant \( \Delta \), we can classify the critical point based on the following conditions:- If \( \Delta > 0 \) and \( \tau^2 - 4\Delta > 0 \), the point is a node (stable if \( \tau < 0 \), unstable if \( \tau > 0 \)).- If \( \Delta > 0 \) and \( \tau^2 - 4\Delta < 0 \), the point is a spiral (stable if \( \tau < 0 \), unstable if \( \tau > 0 \)).- If \( \Delta < 0 \), the point is a saddle point.- If \( \Delta = 0 \), the point is a line of critical points.(a) For \( \Delta_A = 11 > 0 \) and \( \tau_A = 0 \): \( \tau_A^2 - 4\Delta_A = 0 - 44 = -44 < 0 \), the point is a center (since \( \Delta_A > 0 \) and \( \tau_A^2 - 4\Delta_A < 0 \)).(b) For \( \Delta_B = 1 > 0 \) and \( \tau_B = -2 \): \( \tau_B^2 - 4\Delta_B = 4 - 4 = 0 \), the point is a node (stable since \( \tau_B < 0 \)).

Key Concepts

Critical Points

In the context of linear systems of differential equations, critical points illustrate where a system's behavior changes. Consider them the 'resting points' in the system where there is no evolution of variables over time. In other words, all derivatives are zero at critical points.

For example, if you have a system:

• \(x' = -3x + 4y\)
• \(y' = -5x + 3y\)

Here, the critical point is where \(x' = 0\) and \(y' = 0\). This sums to a situation where the movement or flow stops, typically assessed as a coordinate like \((0,0)\) in our example. Understanding where and what kind of critical points exist helps us predict the system's behaviors and stability.

Matrix Trace

The trace of a matrix is a simple yet profound concept with vast implications in linear algebra. In a 2x2 matrix form like \( \begin{pmatrix} a & b \ c & d \end{pmatrix} \), the trace, denoted \( \tau \), is the sum of the diagonal elements. Simply put, calculate it as:

\[\tau = a + d\]

For example, consider a matrix \( A = \begin{pmatrix} -3 & 4 \ -5 & 3 \end{pmatrix} \). The trace is:

\[\tau_A = -3 + 3 = 0\]

The trace helps in determining certain properties of systems and matrix stability over time. This tool becomes crucial in systems classification, specifically when assessing structural stability and predicating dynamic behaviors.

Matrix Determinant

A matrix determinant, especially in a 2x2 matrix scenario, is central to understanding many properties of linear transformations and systems. The determinant offers insight into the system's geometry, such as area scaling factors.

The determinant of a matrix \( \begin{pmatrix} a & b \ c & d \end{pmatrix} \) is given as
\[\Delta = ad - bc\]

For instance, with matrix \( B = \begin{pmatrix} -3 & 2 \ -2 & 1 \end{pmatrix} \), the determinant is calculated as:

\[\Delta_B = (-3)(1) - (2)(-2) = -3 + 4 = 1\]

The determinant informs about the stability and nature of the critical points. For example, a positive determinant with a certain trace value and its properties guide us in classifying these points precisely.

System Classification

Through system classification, we determine how linear systems behave around critical points. This involves using both the trace and determinant calculated earlier.

Here are the general rules:

• If \( \Delta > 0 \) and \( \tau^2 - 4\Delta > 0 \), the point is a node. It is stable if \( \tau < 0 \), unstable if \( \tau > 0 \).
• If \( \Delta > 0 \) and \( \tau^2 - 4\Delta < 0 \), the point manifests as a spiral. It's stable if \( \tau < 0 \), and unstable if \( \tau > 0 \).
• If \( \Delta < 0 \), we have a saddle point, inherently unstable.
• If \( \Delta = 0 \), this point lies along a line of critical points.

Applying this knowledge, if we revisit the earlier matrix for \( A \), where \( \Delta_A = 11 > 0 \) and \( \tau_A = 0 \), we find a center since \( \tau_A^2 - 4\Delta_A < 0 \). Conversely, for matrix \( B \), where \( \Delta_B = 1 > 0 \) and \( \tau_B = -2 \), it's a stable node given \( \tau_B^2 - 4\Delta_B = 0 \). Understanding these classifications helps decipher the types of equilibria present and predict system dynamics.