Problem 13 - 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. $$ \begin{aligned} &x^{\prime}=y-x^{2}+2 \\ &y^{\prime}=x^{2}-x y \end{aligned} $$

Step-by-Step Solution

Answer

The critical points are: \((0,-2)\) is a saddle point, \((2,2)\) is a stable spiral point, \((-1,-1)\) is an unstable spiral point.

Step 1: Identifying Critical Points

To find critical points, we set the system of equations equal to zero:\[ y - x^2 + 2 = 0 \] and \[ x^2 - xy = 0 \].From the second equation, we factor to get \( x(x-y) = 0 \). Thus, \( x = 0 \) or \( x = y \). Substituting \( x = 0 \) in the first equation gives \( y + 2 = 0 \), so \( y = -2 \). Thus, one critical point is \((0, -2)\).Substitute \( x = y \) in the first equation: \( y - y^2 + 2 = 0 \) gives \( y^2 - y - 2 = 0 \). Solving, \( (y-2)(y+1) = 0 \), so \( y = 2 \) or \( y = -1 \). Thus, additional critical points are \((2, 2)\) and \((-1, -1)\).

Step 2: Linearize the System

We linearize the system at each critical point by finding the Jacobian matrix:\[ J(x, y) = \begin{bmatrix} \frac{\partial x'}{\partial x} & \frac{\partial x'}{\partial y} \ \frac{\partial y'}{\partial x} & \frac{\partial y'}{\partial y} \end{bmatrix} = \begin{bmatrix} -2x & 1 \ 2x-y & -x \end{bmatrix} \]We will calculate this matrix for each critical point in subsequent steps.

Step 3: Compute Jacobian at (0, -2)

For the critical point \((0, -2)\), substitute into the Jacobian:\[ J(0, -2) = \begin{bmatrix} -2(0) & 1 \ 2(0) - (-2) & -(0) \end{bmatrix} = \begin{bmatrix} 0 & 1 \ 2 & 0 \end{bmatrix}. \]Now, we will compute eigenvalues to classify the type.

Step 4: Eigenvalues for (0, -2)

For the matrix \( \begin{bmatrix} 0 & 1 \ 2 & 0 \end{bmatrix} \), the characteristic equation is: \[ \lambda^2 - 2 = 0 \Rightarrow \lambda = \pm\sqrt{2}. \]The eigenvalues are real and of opposite signs, indicating a saddle point at \((0,-2)\).

Step 5: Compute Jacobian at (2, 2)

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

Step 6: Eigenvalues for (2, 2)

For the matrix \( \begin{bmatrix} -4 & 1 \ 2 & -2 \end{bmatrix} \), solve \( \det(J - \lambda I) = 0 \):\[ (\lambda + 4)(\lambda + 2) - 2 = 0 \Rightarrow \lambda^2 + 6\lambda + 6 = 0. \]The roots are \( \lambda = -3 \pm \sqrt{3}\), which are complex with negative real parts, indicating a stable spiral point at \((2, 2)\).

Step 7: Compute Jacobian at (-1, -1)

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

Step 8: Eigenvalues for (-1, -1)

For the matrix \( \begin{bmatrix} 2 & 1 \ -1 & 1 \end{bmatrix} \), solve \( \det(J - \lambda I) = 0 \):\[ (2-\lambda)(1-\lambda) + 1 = 0 \Rightarrow \lambda^2 - 3\lambda + 3 = 0. \]The roots \( \lambda = \frac{3 \pm i\sqrt{3}}{2} \) are complex with positive real parts, indicating an unstable spiral point at \((-1, -1)\).

Key Concepts

Critical Points

In the context of autonomous systems, critical points, also known as equilibrium points, occur where the system of equations is equal to zero. For a plane autonomous system, these points can be found by setting the derivatives of the system to zero. In our exercise:

• We have two equations: \( y - x^2 + 2 = 0 \) and \( x^2 - xy = 0 \).
• From these equations, solving for critical points involves simplifying and factoring terms.
• For example, the second equation simplifies to \( x(x-y) = 0 \), leading to solutions \( x = 0 \) or \( x = y \).

By substituting back into the first equation, we find the critical points: \((0, -2), (2, 2), (-1, -1)\). At these points, the system does not change, making them of particular interest in stability analysis.

Jacobian Matrix

The Jacobian matrix is a fundamental tool in analyzing the behavior of autonomous systems near their critical points. It provides a linear approximation of the system at these points through partial derivatives. In our exercise, the Jacobian matrix \( J(x, y) \) is given by:

• \[ J(x, y) = \begin{bmatrix} \frac{\partial x'}{\partial x} & \frac{\partial x'}{\partial y} \ \frac{\partial y'}{\partial x} & \frac{\partial y'}{\partial y} \end{bmatrix} = \begin{bmatrix} -2x & 1 \ 2x-y & -x \end{bmatrix} \]
• This matrix involves taking derivatives of the system's differential equations with respect to \( x \) and \( y \).

Evaluating this matrix at each critical point helps to determine the local behavior of the system. Each localized Jacobian corresponds to a linearized version that aids in understanding the dynamics around equilibrium points.

Eigenvalues

Understanding the local behavior of an autonomous system at critical points requires examining the eigenvalues of the Jacobian matrix. Eigenvalues provide insight into how solutions behave as they approach or move away from equilibrium points:

• For our exercise, the eigenvalues are derived from the characteristic equation of the Jacobian matrix: \( \text{det}(J - \lambda I) = 0 \).
• At each critical point, solving this equation yields different types of roots, either real or complex.
• These roots are crucial in classifying the type and stability of each critical point.

Real and differently signed eigenvalues suggest a saddle point, while complex roots with negative real parts indicate that the point is a stable spiral. Understanding how to calculate and interpret these eigenvalues helps in predicting system behavior.

Stability Analysis

Stability analysis in the context of autonomous systems focuses on determining the behavior around critical points—whether solutions converge to or diverge from these points. This analysis relies heavily on the computed eigenvalues of the Jacobian matrix:

• For a saddle point, as at \((0, -2)\), eigenvalues are real and of opposite signs, indicating instability in one direction.
• If eigenvalues are complex with negative real parts, as at \((2, 2)\), the system shows spiral behavior towards the point, thus demonstrating stability (a stable spiral point).
• Conversely, complex eigenvalues with positive real parts indicate an unstable spiral point, as seen at \((-1, -1)\).

Stability analysis provides vital predictions about the system's dynamic behavior, helping us understand how minor perturbations can affect the overall system evolution.