Problem 16 - 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. $$ \begin{aligned} &x^{\prime}=x y-3 y-4 \\ &y^{\prime}=y^{2}-x^{2} \end{aligned} $$

Step-by-Step Solution

Answer

(2,2) is an unstable spiral; (-2,-2) is a stable node.

Step 1: Find the Critical Points

To find the critical points, set \( x' = 0 \) and \( y' = 0 \). From \( x' = xy - 3y - 4 = 0 \), factor to get \( y(x - 3) = 4 \).From \( y' = y^2 - x^2 = 0 \), we have \( y^2 = x^2 \), giving \( y = \pm x \).Solving both equations, we find two critical points: \((2, 2)\) and \((-2, -2)\).

Step 2: Determine Jacobian Matrix

Calculate the Jacobian matrix by finding the partial derivatives of \( x' \) and \( y' \). \[J = \begin{bmatrix} \frac{\partial}{\partial x}(xy - 3y - 4) & \frac{\partial}{\partial y}(xy - 3y - 4) \ \frac{\partial}{\partial x}(y^2 - x^2) & \frac{\partial}{\partial y}(y^2 - x^2)\end{bmatrix}\]Calculating, we get:\[J = \begin{bmatrix} y & x - 3 \ -2x & 2y\end{bmatrix}\]

Step 3: Evaluate Jacobian at Each Critical Point

Substitute each critical point into the Jacobian matrix.For \((2, 2)\):\[J(2, 2) = \begin{bmatrix} 2 & -1 \ -4 & 4\end{bmatrix}\]For \((-2, -2)\):\[J(-2, -2) = \begin{bmatrix} -2 & -5 \ 4 & -4\end{bmatrix}\]

Step 4: Compute Eigenvalues of the Jacobian

Compute the eigenvalues for each Jacobian using the formula \( \det(J - \lambda I) = 0 \).For \( (2, 2) \): Determine \( \lambda \) by solving\[\det \begin{bmatrix} 2-\lambda & -1 \ -4 & 4-\lambda \end{bmatrix} = 0\]This simplifies to \( (2-\lambda)(4-\lambda) + 4 = 0 \), giving roots \( \lambda_1 = 3 - i \), \( \lambda_2 = 3 + i \).For \((-2, -2)\): Compute \( \lambda \) by solving\[\det \begin{bmatrix} -2-\lambda & -5 \ 4 & -4-\lambda \end{bmatrix} = 0\]This results in \( \lambda_1 = -3 \), \( \lambda_2 = -3 \).

Step 5: Classify the Critical Points

For \( (2, 2) \), the eigenvalues \( 3 \pm i \) are complex with positive real parts, indicating an unstable spiral.For \((-2, -2)\), the eigenvalues are both \( -3 \), indicating a stable node.

Key Concepts

Jacobian Matrix

The Jacobian Matrix is a crucial concept when analyzing plane autonomous systems. It provides a linear approximation of a system around a critical point. To construct the Jacobian for a system, you need to compute partial derivatives of each equation in the system with respect to each variable involved. Take, for instance, a system where \( x' \) and \( y' \) denote the derivatives of \( x \) and \( y \), respectively. The Jacobian \( J \) is set up as follows:

• First row: Derivatives of \( x' \) with respect to \( x \) and \( y \).
• Second row: Derivatives of \( y' \) with respect to \( x \) and \( y \).

For the given exercise, the Jacobian is computed from the system \( x' = xy - 3y - 4 \) and \( y' = y^2 - x^2 \). After calculating the necessary partial derivatives, the matrix becomes:\[J = \begin{bmatrix} y & x - 3 \ -2x & 2y \end{bmatrix}\] Substituting critical points into this matrix will allow us to further analyze the system's dynamics.

Critical Points Classification

Critical points in a dynamical system are where the system doesn't change, meaning both \( x' \) and \( y' \) are zero. For our exercise, setting \( x' = xy - 3y - 4 = 0 \) and \( y' = y^2 - x^2 = 0 \) provides the critical points.

• Simplifying gives \( y(x - 3) = 4 \) from the first equation, and \( y^2 - x^2 = 0 \) implies \( y = \pm x \).
• These lead to the critical points: \( (2, 2) \) and \( (-2, -2) \).

Finding these points is essential as they indicate states where the system equilibrates or changes stability. Analyzing around these points with tools like the Jacobian matrix reveals more about the behavior of the system.

Eigenvalues of Matrices

Eigenvalues are vital in understanding the behavior near a critical point. They are calculated from the Jacobian matrix by solving the characteristic equation, \( \det(J - \lambda I) = 0 \), where \( J \) is the Jacobian matrix and \( I \) is the identity matrix. For the critical point \( (2, 2) \), substituting it into the Jacobian gives \[J(2, 2) = \begin{bmatrix} 2 & -1 \ -4 & 4 \end{bmatrix}\]The determinant of \( J - \lambda I \) yields a quadratic in \( \lambda \), leading to eigenvalues \( \lambda_1 = 3 + i \) and \( \lambda_2 = 3 - i \), showing complex roots.Similarly, for \( (-2, -2) \), the matrix becomes \[J(-2, -2) = \begin{bmatrix} -2 & -5 \ 4 & -4 \end{bmatrix}\],resulting in both eigenvalues being real and negative: \( \lambda_1 = -3 \) and \( \lambda_2 = -3 \). Understanding the nature of these eigenvalues helps in classifying the critical points.

Stability Analysis

Stability analysis determines how a system behaves near its critical points. The nature of the eigenvalues calculated from the Jacobian matrix guides this classification:

• If eigenvalues have negative real parts, the system's critical point is stable (attracting trajectory).
• If eigenvalues have positive real parts, the critical point is unstable (repelling trajectory).
• If the eigenvalues are complex, their real parts dictate spiral behavior: positive indicates an unstable spiral, while negative suggests a stable spiral.

For the critical point \( (2, 2) \), complex eigenvalues with positive real parts (\( 3 \pm i \)) indicate an unstable spiral. In contrast, \( (-2, -2) \) with real negative eigenvalues (\( -3, -3 \)) is classified as a stable node. This method allows us to anticipate how a plane autonomous system will behave over time, helping predict system stability and response to small perturbations.