Problem 17 - Advanced Engineering Mathematics

Question

since \(x^{\prime}=-2 x y=0,\) either \(x=0\) or \(y=0 .\) If \(x=0, y\left(1-y^{2}\right)=0\) and \(s o(0,0),(0,1),\) and (0,-1) are critical points. The case \(y=0\) leads to \(x=0 .\) We next use the Jacobian matrix $$\mathbf{g}^{\prime}(\mathbf{X})=\left(\begin{array}{cc} -2 y & -2 x \\ -1+y & 1+x-3 y^{2} \end{array}\right)$$ to classify these three critical points. For \(\mathbf{X}=(0,0), \tau=1\) and \(\Delta=0\) and so the test is inconclusive. For \(\mathbf{X}=(0,1), \tau=-4, \Delta=4\) and so \(\tau^{2}-4 \Delta=0 .\) We can conclude that (0,1) is a stable critical point but we are unable to classify this critical point further in this borderline case. For \(\mathbf{X}=(0,-1), \Delta=-4<0\) and so (0,-1) is a saddle point.

Step-by-Step Solution

Answer

The critical points are \((0,0), (0,1), (0,-1)\); (0,1) is stable but not classified further, and (0,-1) is a saddle point.

Step 1: Identify Critical Points

Given the equation \(x' = -2xy = 0\), set either \(x = 0\) or \(y = 0\). If \(x = 0\), then we have another equation \(y(1 - y^2) = 0\). Solving this gives three solutions: \((0,0), (0,1),\) and \((0,-1)\). If \(y = 0\), substitute back to obtain \(x = 0\). Thus, the critical points are \((0,0), (0,1), (0,-1)\).

Step 2: Setup Jacobian Matrix

Use the given Jacobian matrix for analyzing the stability of these critical points: \[egin{bmatrix} -2y & -2x \ -1+y & 1+x-3y^2 d{bmatrix} \]

Step 3: Determine Stability at (0,0)

For the critical point \( (0,0) \), substitute into the Jacobian to find the trace \(\tau\) and determinant \(\Delta\). Calculation gives \(\tau = 1\) and \(\Delta = 0\). The result is inconclusive due to \(\Delta = 0\).

Step 4: Determine Stability at (0,1)

For the critical point \((0,1)\), substitute into the Jacobian to find \(\tau = -4\) and \(\Delta = 4\). Check \(\tau^2 - 4\Delta = 0\), indicating it's inconclusive about the type, but the point is stable.

Step 5: Determine Stability at (0,-1)

For the critical point \((0,-1)\), substitute into the Jacobian to find \(\Delta = -4\). Since \(\Delta < 0\), the point \((0,-1)\) is a saddle point, indicating instability.

Key Concepts

Jacobian Matrix

The Jacobian matrix is a fundamental tool in system stability analysis, especially for examining critical points in differential equations. It is essentially a matrix of all first-order partial derivatives of a vector-valued function. This matrix helps in understanding the local behavior of a system near equilibrium points.

In our exercise, the Jacobian matrix is given by: \[ \mathbf{g}^{\prime}(\mathbf{X})=\begin{bmatrix} -2y & -2x \ -1+y & 1+x-3y^2 \end{bmatrix} \]

To classify the critical points,

• We substitute each of the found critical points into this matrix.
• Calculate the trace \(\tau\) (sum of the diagonal elements).
• Calculate the determinant \(\Delta\) (product of the diagonal elements minus the product of the off-diagonal elements).

These calculations reveal the nature of the critical points, as different conditions on \(\tau\) and \(\Delta\) lead to varying classifications of stability.

Stability Analysis

Stability analysis involves evaluating whether a system will stay near its equilibrium state after a small perturbation. This is critical in understanding the long-term behavior of dynamic systems.

For each critical point, we consider the trace \(\tau\) and determinant \(\Delta\) of the Jacobian matrix:

• If \(\Delta > 0\), we have the potential for stable or unstable nodes or spirals.
• If \(\Delta < 0\), the critical point is a saddle point and unstable.
• If \(\Delta = 0\), the test is inconclusive, meaning further analysis is needed.

For instance, at the critical point \((0,0)\), we find \(\Delta = 0\), indicating an inconclusive test about stability.
For \((0,1)\), \(\tau = -4\) and \(\Delta = 4\) provide some stability insights, but the specific nature remains unclear. Lastly, at \((0,-1)\), \(\Delta = -4\) clearly identifies an unstable saddle point.

Saddle Point

A saddle point in the context of stability analysis is a type of critical point where the system's stability behavior changes direction. Typically, one direction is stable (attracting trajectories) and the other is unstable (repelling trajectories).

The hallmark of a saddle point is a Jacobian determinant \(\Delta < 0\). This negative value indicates that the critical point is inherently unstable.

In our example, the critical point \((0,-1)\) demonstrates \(\Delta = -4\), signaling it's a saddle point. Such points are important in dynamics as they represent areas where the system can diverge or change states profoundly when slightly perturbed. Understanding saddle points helps in predicting the paths and outcomes of dynamic systems.