Problem 18 - Advanced Engineering Mathematics

Question

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. $$ \begin{aligned} &x^{\prime}=x\left(1-x^{2}-3 y^{2}\right) \\ &y^{\prime}=y\left(3-x^{2}-3 y^{2}\right) \end{aligned} $$

Step-by-Step Solution

Answer

All critical points are saddle points.

Step 1: Identify Critical Points

To find the critical points, we set \(x' = 0\) and \(y' = 0\). Solving, we have the equations \(x(1-x^2-3y^2)=0\) and \(y(3-x^2-3y^2)=0\). From \(x(1-x^2-3y^2)=0\), either \(x = 0\) or \(1-x^2-3y^2 = 0\). From \(y(3-x^2-3y^2)=0\), either \(y = 0\) or \(3-x^2-3y^2 = 0\). We solve these to find the critical points: \((0,0), (1,0), (0,1), (-1,0), (0,-1), (\sqrt{3}, \frac{1}{\sqrt{3}}), (-\sqrt{3}, \frac{1}{\sqrt{3}}), (\sqrt{3}, -\frac{1}{\sqrt{3}}), (-\sqrt{3}, -\frac{1}{\sqrt{3}})\).

Step 2: Calculate the Jacobian Matrix

The Jacobian matrix \(J\) is calculated using partial derivatives. For the system, it is: \[ J = \begin{bmatrix} \frac{\partial x'}{\partial x} & \frac{\partial x'}{\partial y} \ \frac{\partial y'}{\partial x} & \frac{\partial y'}{\partial y} \end{bmatrix} = \begin{bmatrix} 1 - 3x^2 - 3y^2 & -6xy \ -2xy & 3 - x^2 - 9y^2 \end{bmatrix} \] Calculate \(J\) at each critical point to determine the type of equilibrium.

Step 3: Evaluate Jacobian at Critical Points

Evaluate the Jacobian matrix at each critical point. For example, at \((0,0)\), \(J = \begin{bmatrix} 1 & 0 \ 0 & 3 \end{bmatrix}\), and at \((1,0)\), \(J = \begin{bmatrix} -2 & 0 \ 0 & 2 \end{bmatrix}\). Repeat this process for each critical point.

Step 4: Analyze Eigenvalues of Jacobian

Determine the eigenvalues of \(J\) for each critical point. The eigenvalues (\(\lambda_1\) and \(\lambda_2\)) will help classify the critical point. Eigenvalues that are both negative indicate a stable node, both positive indicate an unstable node, a mix of positive and negative indicates a saddle point, and complex eigenvalues indicate spiral points.

Step 5: Classify Each Critical Point

Using the signs of the eigenvalues or their real parts, classify each critical point: 1. \((0,0)\): Saddle point (eigenvalues 1 and 3).2. \((1,0), (-1,0)\): Saddle points (eigenvalues -2, 2).3. \((0,1), (0,-1)\): Saddle points (eigenvalues -8, 3).4. \((\pm\sqrt{3}, \pm\frac{1}{\sqrt{3}})\): Saddle points due to mixed signs in eigenvalues.

Key Concepts

Critical Points

In autonomous systems, critical points are the coordinates where the system's rate of change is zero. These points are found by setting the derivatives equal to zero. For our exercise, we analyze the given equations:

• \(x' = x(1-x^2-3y^2) = 0\)
• \(y' = y(3-x^2-3y^2) = 0\)

We solve these equations to find where the system's behavior is stationary. Critical points represent potential states of equilibrium where the system might settle or change its dynamics.
The critical points found for our exercise are:

• \((0,0), (1,0), (0,1), (-1,0), (0,-1)\)
• \((\sqrt{3}, \frac{1}{\sqrt{3}}), (-\sqrt{3}, \frac{1}{\sqrt{3}})\)
• \((\sqrt{3}, -\frac{1}{\sqrt{3}}), (-\sqrt{3}, -\frac{1}{\sqrt{3}})\)

Each of these points is a candidate for further stability analysis.

Jacobian Matrix

The Jacobian matrix is a crucial tool for assessing the stability of critical points in a dynamical system. It is a matrix of first-order partial derivatives of a vector-valued function. In our exercise, the Jacobian matrix \(J\) is formed from partial derivatives of the given system:

• \(J = \begin{bmatrix} \frac{\partial x'}{\partial x} & \frac{\partial x'}{\partial y} \ \frac{\partial y'}{\partial x} & \frac{\partial y'}{\partial y} \end{bmatrix}\)
• \(J = \begin{bmatrix} 1 - 3x^2 - 3y^2 & -6xy \ -2xy & 3 - x^2 - 9y^2 \end{bmatrix}\)

At each critical point, the Jacobian is evaluated to provide a linear approximation of the system near those points. This helps in determining the local behavior of the system.
By plugging in the critical points into the Jacobian matrix, we form specific matrices used for calculating eigenvalues, which are essential for classification.

Eigenvalues

Eigenvalues arise from solving the characteristic equation of the Jacobian matrix at each critical point. They are fundamental in classifying the type of equilibrium. Different combinations of eigenvalues provide insights into the stability properties of each critical point:

• If both eigenvalues are negative, the point is a stable node.
• If both are positive, it is an unstable node.
• Mixed signs suggest a saddle point.
• Complex eigenvalues indicate a spiral point, with stability determined by the sign of the real part.

For instance, by evaluating the Jacobian matrix at \((0,0)\), we found eigenvalues \(1\) and \(3\), indicating it is a saddle point. Each critical point's unique set of eigenvalues helps determine whether it's stable, unstable, or has a special behavior like spirals.

Stability Analysis

Stability analysis involves evaluating the nature of critical points by examining the eigenvalues derived from the Jacobian matrix. This process helps predict the system’s responses from equilibrium points to small disturbances:

• Stable points attract nearby trajectories.
• Unstable points repel them.
• Saddle points have one attracting and one repelling direction.

In our exercise, by assessing the eigenvalues associated with each critical point, we determine their stability type. For example:

• \((0,0)\): Saddle point, due to eigenvalues \(1\) and \(3\).
• \((1,0)\) & \((-1,0)\): Saddle points, given eigenvalues \(-2\) and \(2\).
• Other points also classify based on similar eigenvalue logic.

Stable spiral points or nodes indicate potential equilibrium, while unstable configurations suggest divergence or complex behavior. This type of analysis is crucial for understanding the dynamic characteristics of a system.