Chapter 11: Problem 16
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. $$ \begin{aligned} &x^{\prime}=x y-3 y-4 \\ &y^{\prime}=y^{2}-x^{2} \end{aligned} $$
Short Answer
Expert verified
(2,2) is an unstable spiral; (-2,-2) is a stable node.
Step by step solution
01
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 \((-2, -2)\).
02
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:\[J = \begin{bmatrix} y & x - 3 \ -2x & 2y\end{bmatrix}\]
03
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 \((-2, -2)\):\[J(-2, -2) = \begin{bmatrix} -2 & -5 \ 4 & -4\end{bmatrix}\]
04
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 \( \lambda_1 = -3 \), \( \lambda_2 = -3 \).
05
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.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
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 \( y \).
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 \( (-2, -2) \).
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.