Problem 20 - 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. $$ \begin{aligned} &x^{\prime}=-2 x+y+10 \\ &y^{\prime}=2 x-y-15 \frac{y}{y+5} \end{aligned} $$

Step-by-Step Solution

Answer

The critical point \((5, 0)\) is a stable spiral point.

Step 1: Determine the Critical Points

To find the critical points of the system, we set the derivatives equal to zero: \(x' = -2x + y + 10 = 0\) and \(y' = 2x - y - 15\frac{y}{y+5} = 0\). By solving \(-2x + y + 10 = 0\), we get \(y = 2x - 10\). Substitute this into \(y'\) equation: \(2x - (2x - 10) - 15\frac{(2x-10)}{(2x-10)+5} = 0\). By solving these, we find the critical points: \((x, y) = (5, 0)\).

Step 2: Evaluate the Jacobean Matrix at Critical Point

The Jacobean matrix \( J \) of the system is given by the partial derivatives of \( x' \) and \( y' \): \[ J = \begin{bmatrix} \frac{\partial x'}{\partial x} & \frac{\partial x'}{\partial y} \ \frac{\partial y'}{\partial x} & \frac{\partial y'}{\partial y} \end{bmatrix} = \begin{bmatrix} -2 & 1 \ 2 - \frac{15}{y+5} & -1 + \frac{15y}{(y+5)^2} \end{bmatrix}.\] Substitute \((x, y) = (5, 0)\) into this matrix to get \[ J = \begin{bmatrix} -2 & 1 \ 0.5 & -2.5 \end{bmatrix}. \]

Step 3: Find the Eigenvalues of Jacobean Matrix

The eigenvalues of matrix \( J \) determine the type of critical point. The characteristic equation of \( J \) is \( \det(J - \lambda I) = 0 \), leading to \[ \det\begin{bmatrix} -2 - \lambda & 1 \ 0.5 & -2.5 - \lambda \end{bmatrix} = 0. \] Calculate \((-2-\lambda)(-2.5-\lambda) - 0.5 \cdot 1 = 0\), resulting in the characteristic polynomial \( \lambda^2 + 4.5\lambda + 4.5 = 0\).

Step 4: Solve Characteristic Polynomial for Eigenvalues

Solve the characteristic polynomial \( \lambda^2 + 4.5\lambda + 4.5 = 0 \) using the quadratic formula \( \lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). Here \(a = 1, b = 4.5, c = 4.5\). The roots are \( \lambda = \frac{-4.5 \pm \sqrt{4.5^2 - 4 \times 1 \times 4.5}}{2} = -2.25 \pm 1.299\).

Step 5: Classify the Critical Point

Assess the eigenvalues \( -2.25 \pm 1.299i \). Since they are complex with a negative real part, the critical point \((5, 0)\) is a stable spiral point.

Key Concepts

Critical Points

To find critical points in an autonomous system, we set the system's derivatives equal to zero. This step isolates the points in the plane where the system is at equilibrium—no net change occurs. For a system defined by its differential equations, such as - \(x' = -2x + y + 10\)- \(y' = 2x - y - 15\frac{y}{y + 5}\), we solve these equations simultaneously. In this example, setting \(x' = 0\) and \(y' = 0\) led to the critical point \((5, 0)\). This approach is essential because it lays the groundwork for analyzing the system's behavior near these points.

Jacobian Matrix

The Jacobian matrix is a powerful tool used to analyze the behavior of systems near critical points by considering the partial derivatives of the system. For a two-variable system, it has the form:\[ J = \begin{bmatrix} \frac{\partial x'}{\partial x} & \frac{\partial x'}{\partial y} \ \frac{\partial y'}{\partial x} & \frac{\partial y'}{\partial y} \end{bmatrix}. \]Evaluating this matrix at a critical point reveals how small perturbations in system variables impact the system's behavior. In our example, the matrix evaluated at \((5, 0)\) is given by:\[ J = \begin{bmatrix} -2 & 1 \ 0.5 & -2.5 \end{bmatrix}. \]This matrix helps us determine stability and system behavior near the critical point.

Eigenvalues

Eigenvalues, derived from the Jacobian matrix, provide insights into the nature of equilibrium points. They are solutions to the equation \( \det(J - \lambda I) = 0 \). In our system, solving\[ \det\begin{bmatrix} -2 - \lambda & 1 \ 0.5 & -2.5 - \lambda \end{bmatrix} = 0 \]results in the polynomial \( \lambda^2 + 4.5\lambda + 4.5 = 0 \). Using the quadratic formula, we get the eigenvalues, \( \lambda = -2.25 \pm 1.299i \). Eigenvalues with a negative real part indicate stability, whereas complex parts suggest rotational behaviors like spirals.

Stable Spiral Points

A Stable Spiral Point is characterized by eigenvalues that are complex with negative real parts. This combination signifies a system where trajectories spiral into the critical point over time. Such spirals indicate a returning motion to an equilibrium, albeit via oscillations. In our example, the eigenvalues were \( -2.25 \pm 1.299i \), confirming that the critical point \((5, 0)\) is a stable spiral point. Understanding stable spiral points is crucial in fields like control systems and ecology, where knowing how a system naturally stabilizes is beneficial.