Problem 12 - Advanced Engineering Mathematics

Question

In Problems \(11-20\), classify (if possible) each critical point of the given plane autonomous system as a stable node, a stable spiral point, an unstable spiral point, an unstable node, or a saddle point. \(x^{\prime}=x^{2}-y^{2}-1\) \(y^{\prime}=2 y\)

Step-by-Step Solution

Answer

(1, 0) is an unstable node; (-1, 0) is a saddle point.

Step 1: Identify the Critical Points

To find the critical points, set the equations \( x' = x^2 - y^2 - 1 \) and \( y' = 2y \) equal to zero. Solving \( 2y = 0 \) gives \( y = 0 \). Substituting \( y = 0 \) in the first equation gives \( x^2 - 1 = 0 \), which results in \( x = \pm 1 \). Thus, the critical points are \( (1, 0) \) and \( (-1, 0) \).

Step 2: Linearize the System

The Jacobian matrix is used to analyze the stability by determining it at each critical point. For \( x' = x^2 - y^2 - 1 \), the partial derivatives are \( \frac{\partial f}{\partial x} = 2x \) and \( \frac{\partial f}{\partial y} = -2y \). For \( y' = 2y \), it's \( \frac{\partial g}{\partial x} = 0 \) and \( \frac{\partial g}{\partial y} = 2 \). So, the Jacobian matrix is: \[J = \begin{pmatrix} 2x & -2y \ 0 & 2 \end{pmatrix}\]

Step 3: Evaluate the Jacobian at the Critical Points

Evaluate the Jacobian at each critical point. For \( (1, 0) \), the Jacobian becomes: \[J(1,0) = \begin{pmatrix} 2 & 0 \ 0 & 2 \end{pmatrix}\]For \( (-1, 0) \), it is: \[J(-1,0) = \begin{pmatrix} -2 & 0 \ 0 & 2 \end{pmatrix}\]

Step 4: Determine the Eigenvalues

For \( (1,0) \), find the eigenvalues of \( J(1,0) \), which are solutions of \( \text{det}(J - \lambda I) = 0 \):\[\begin{vmatrix} 2-\lambda & 0 \ 0 & 2-\lambda \end{vmatrix} = 0\]This gives \( \lambda = 2 \), a repeated eigenvalue.For \( (-1,0) \), the eigenvalues for \( J(-1,0) \) are:\[\begin{vmatrix} -2-\lambda & 0 \ 0 & 2-\lambda \end{vmatrix} = 0\]This gives \( \lambda = -2 \) and \( \lambda = 2 \).

Step 5: Classify the Critical Points

The critical point \( (1, 0) \) has eigenvalues \( \lambda = 2 \), which are positive and identical, indicating it is an **unstable node**.For the critical point \( (-1, 0) \), the eigenvalues are \( \lambda = -2 \) and \( \lambda = 2 \), one positive and one negative, indicating it is a **saddle point**.

Key Concepts

Critical Points

In the study of plane autonomous systems, identifying critical points is a fundamental step. Critical points, also known as equilibrium points or stationary points, are where the system does not evolve over time; the derivatives of all involved functions equal zero. For the system given by

• \( x' = x^2 - y^2 - 1 \)
• \( y' = 2y \)

To find these points, we set both of these to zero. Solving \( y' = 2y = 0 \) gives \( y = 0 \). Substituting this into the first equation \( x' = x^2 - 1 = 0 \) yields \( x = \pm 1 \). Thus, critical points are \( (1, 0) \) and \( (-1, 0) \). These points are essential for analyzing the system's behavior near equilibrium.

Jacobian Matrix

Once the critical points are identified, the next step is to assess the system's stability using the Jacobian matrix. The Jacobian matrix provides a linear approximation of the autonomous system near each critical point by considering the partial derivatives of the system.
For our system, the Jacobian matrix \( J \) is composed of the first order partial derivatives of \( x' = x^2 - y^2 - 1 \) and \( y' = 2y \) with respect to \( x \) and \( y \):

• \( \frac{\partial f}{\partial x} = 2x \)
• \( \frac{\partial f}{\partial y} = -2y \)
• \( \frac{\partial g}{\partial x} = 0 \)
• \( \frac{\partial g}{\partial y} = 2 \)

Thus, \( J = \begin{pmatrix} 2x & -2y \ 0 & 2 \end{pmatrix} \). By evaluating this matrix at each critical point, we can predict the system's local stability, which brings us to the study of its eigenvalues.

Eigenvalues

Eigenvalues are crucial in determining the nature of critical points in a system. After obtaining the Jacobian matrix, we calculate its eigenvalues to understand the dynamics near each point. Specifically, we solve the characteristic equation \( \text{det}(J - \lambda I) = 0 \) for each critical point.
For the critical point \((1, 0)\), we find the eigenvalues are both \( \lambda = 2 \). Having both eigenvalues positive and identical indicates that this point is an **unstable node**.
For \((-1, 0)\), the eigenvalues \( \lambda = -2 \) and \( \lambda = 2 \) show one positive and one negative eigenvalue. This configuration signals that this point is a **saddle point**. These eigenvalues are key to understanding how solutions behave near the critical points of the system.

Stability Analysis

Stability analysis is used to predict the behavior of a system near its critical points. It lets us classify these points into different types based on the eigenvalues of the Jacobian matrix.
For a node, if both eigenvalues are positive, the system is an **unstable node**. Solutions move away from the critical point over time. If both eigenvalues are negative, it denotes a **stable node**, where solutions converge toward the point.
A saddle point, such as \((-1, 0)\) in our problem, has one positive and one negative eigenvalue. This mixed-sign configuration results in trajectories being drawn toward the point along one direction, but away in another, marking it as unstable. This dichotomy in behavior is fundamental to understanding complex dynamical systems. Properly classifying these points is essential for predicting long-term system behavior.