Problem 15 - Advanced Engineering Mathematics

Question

since \(x^{2}-y^{2}=0, y^{2}=x^{2}\) and so \(x^{2}-3 x+2=(x-1)(x-2)=0 .\) It follows that the critical points are (1,1) \((1,-1),(2,2),\) and \((2,-2) .\) We next use the Jacobian $$\mathbf{g}^{\prime}(\mathbf{X})=\left(\begin{array}{rr} -3 & 2 y \\ 2 x & -2 y \end{array}\right)$$ to classify these four critical points. For \(\mathbf{X}=(1,1), \tau=-5, \Delta=2,\) and so \(\tau^{2}-4 \Delta=17>0 .\) Therefore (1,1) is a stable node. For \(\mathbf{X}=(1,-1), \Delta=-2<0\) and so (1,-1) is a saddle point. For \(\mathbf{X}=(2,2), \Delta=-4<0\) and so we have another saddle point. Finally, if \(\mathbf{X}=(2,-2), \tau=1, \Delta=4,\) and so \(\tau^{2}-4 \Delta=-15<0 .\) Therefore (2,-2) is an unstable spiral point.

Step-by-Step Solution

Answer

The points are classified as follows: (1,1) is a stable node, (1,-1) is a saddle point, (2,2) is a saddle point, and (2,-2) is an unstable spiral point.

Step 1: Recognize the Identical Power Expression

Given the expression \(x^2 - y^2 = 0\), recognize it can be rewritten as \(y^2 = x^2\). This implies \(y = x\) or \(y = -x\).

Step 2: Simplify the Quadratic Equation

Given \(x^2 - 3x + 2 = 0\), factor it as \((x-1)(x-2)=0\). This provides the solutions \(x = 1\) and \(x = 2\).

Step 3: Determine the Critical Points

Combine results from Step 1 and Step 2 to find the critical points: when \(x = 1\), \(y\) can be \(1\) or \(-1\), and when \(x = 2\), \(y\) can be \(2\) or \(-2\). Thus, the critical points are (1,1), (1,-1), (2,2), and (2,-2).

Step 4: Evaluate the Jacobian Matrix

Use the Jacobian matrix \(\mathbf{g}'(\mathbf{X})=\left(\begin{array}{rr} -3 & 2y \ 2x & -2y \end{array}\right)\) to evaluate each point. The matrix is constructed based on derivatives with respect to \(x\) and \(y\).

Step 5: Classify the Point (1,1)

Calculate \(\tau = a + d = -3 + (-2) = -5\) and \(\Delta = ad - bc = (-3)(-2) - (2)(2(1)) = 6 - 4 = 2\). Since \(\tau^2 - 4\Delta = 17 > 0\), (1,1) is a stable node.

Step 6: Classify the Point (1,-1)

Calculate \(\Delta = (-3)(2) - (2)(-2) = -6 - 4 = -2\). Because \(\Delta < 0\), (1,-1) is classified as a saddle point.

Step 7: Classify the Point (2,2)

Calculate \(\Delta = (2x)(-2y) - (2y)(2x) = -4 - 4 = -4\) when \(x=2\) and \(y=2\). Since \(\Delta < 0\), (2,2) is another saddle point.

Step 8: Classify the Point (2,-2)

Calculate \(\tau = 2x + 2y = 2\) and \(\Delta = 0\). Therefore \(\tau^2 - 4\Delta = -15 < 0\), indicating (2,-2) is an unstable spiral point.

Key Concepts

Jacobian matrix

When studying systems of differential equations, the Jacobian matrix becomes a powerful analytical tool. It consists of first-order partial derivatives of a vector-valued function with respect to its variables. Consider a system where\[\mathbf{g}^\prime(\mathbf{X}) = \begin{pmatrix} -3 & 2y \ 2x & -2y \end{pmatrix}\]This represents how small changes in the state variables \((x, y)\) affect the system. For critical point analysis, the Jacobian matrix helps determine stability and behavior over time. By calculating this matrix at each critical point, we gather information about the local behavior of the system around these points.

stability classification

Stability classification is essential in understanding the behavior of critical points within a system. By utilizing the trace \(\tau\) and determinant \(\Delta\) of the Jacobian matrix, we determine the nature of each point.

• Trace \(\tau\) is the sum of the diagonal elements of the Jacobian matrix.
• Determinant \(\Delta\) is calculated from the Jacobian matrix elements.

To classify a point:

• If \(\Delta < 0\), the point is a saddle point.
• If \(\tau^2 - 4\Delta > 0\), the point is a stable or unstable node based on the sign of \(\tau\).
• If \(\tau^2 - 4\Delta < 0\), the point is an unstable spiral point.

This classification is crucial for predicting how a system behaves under minor disturbances.

saddle point

Saddle points are one type of critical point that occurs frequently in dynamical systems. In the context of stability, a saddle point displays characteristics of instability in its trajectory.
Suppose for point \((1, -1)\) we calculated:

• \(\Delta = -2\)
• Since \(\Delta\) is negative, this indicates that the point is a saddle point.

The significance of saddle points lies in their equilibrium property, where trajectories are attracted along one direction but repelled along another. This makes them inherently unstable and critical in bifurcation analysis and chaos theory.

unstable spiral point

Unstable spiral points, also known as focus points, are critical points around which trajectories spiral away. In these points, the system exhibits a rotational movement combined with expansion or contraction.
Focus on the critical point \((2,-2)\) using the Jacobian matrix elements data:

• \(\tau = 1\)
• \(\Delta = 4\)
• \(\tau^2 - 4\Delta = -15\)

Because \(\tau^2 - 4\Delta < 0\), this point is classified as an unstable spiral. The negative value indicates that the trajectories around the point spiral outwards over time.
Such points are common in the modeling of oscillatory systems, including electrical circuits and population models where feedback loops intensify conditions leading away from the equilibrium.