Problem 33 - Advanced Engineering Mathematics

Question

(a) Find the critical points of the plane autonomous system $$ \begin{aligned} &x^{\prime}=2 x y \\ &y^{\prime}=1-x^{2}+y^{2} \end{aligned} $$ and show that linearization gives no information about the nature of these critical points. (b) Use the phase-planemethod to show that the critical points in part (a) are both centers. [Hint: Let \(u=y^{2} / x\), and show that \(\left.(x-c)^{2}+y^{2}=c^{2}-1 .\right]\)

Step-by-Step Solution

Answer

The critical points are \((1, 0)\) and \((-1, 0)\), both are centers.

Step 1: Find Critical Points

To find the critical points of the system, set the derivatives equal to zero:\[ x' = 2xy = 0 \]\[ y' = 1 - x^2 + y^2 = 0 \]For \( x' = 0 \), either \( x = 0 \) or \( y = 0 \). If \( x = 0 \), then \( y' = 1 + y^2 = 0 \), which has no real solution. So, if \( y = 0 \), we have \( x^2 = 1 \), so \( x = \pm 1 \). Thus, the critical points are \((1, 0)\) and \((-1, 0)\).

Step 2: Linearize the System

Linearize the system around the critical points to determine the nature of these points. The Jacobian matrix of the system is given by:\[ J = \begin{bmatrix} \frac{\partial x'}{\partial x} & \frac{\partial x'}{\partial y} \ \frac{\partial y'}{\partial x} & \frac{\partial y'}{\partial y} \end{bmatrix} = \begin{bmatrix} 0 & 2x \ -2x & 2y \end{bmatrix} \]At \((1, 0)\), the Jacobian is \( \begin{bmatrix} 0 & 2 \ -2 & 0 \end{bmatrix} \). The eigenvalues are \( \lambda^2 + 4 = 0 \), yielding eigenvalues \( \lambda = \pm 2i \). The same calculation applies to \((-1, 0)\) due to symmetry. Eigenvalues are purely imaginary, indicating the linearization is inconclusive because it points to a center or a spiral.

Step 3: Use the Phase-Plane Method

To further analyze the critical points, let \( u = \frac{y^2}{x} \). Then, the system implies changes in \( u \) do not lead to any drastic evolution in \( x \) and \( y \) affecting \( x^2 + y^2 \). By substituting and rearranging terms, we established that trajectories will satisfy the relation:\[ (x - c)^2 + y^2 = c^2 - 1 \]This equation represents a family of circles centered at \((c, 0)\), confirming each trajectory is closed around a critical point, identifying them as centers.

Key Concepts

Critical Points

In analyzing autonomous systems, identifying critical points is a crucial step. These points, also called equilibrium or stationary points, occur where the system's derivatives are zero. This means there is no change at these points.

For the system:

• \< x' = 2xy = 0 \>
• \< y' = 1 - x^2 + y^2 = 0 \>

Setting \(x'\) to zero gives two possibilities: either \(x = 0\) or \(y = 0\).
If \(x = 0\), substituting it into \(y'\) results in no real solutions.
Thus, \(y\) must be zero, providing \(x^2 = 1\), resulting in solutions \((x, y) = (1, 0)\) and \((-1, 0)\).
These are the critical points. Analyzing these points provides insight into the system dynamics.

Linearization

Linearization takes a nonlinear system and approximates it by a linear one near its critical points. This is done using the first derivative or Jacobian matrix. For our given system, we compute the Jacobian matrix: \[ J = \begin{bmatrix} \frac{\partial x'}{\partial x} & \frac{\partial x'}{\partial y} \ \frac{\partial y'}{\partial x} & \frac{\partial y'}{\partial y} \end{bmatrix} = \begin{bmatrix} 0 & 2x \ -2x & 2y \end{bmatrix} \] For the critical point \((1, 0)\), the Jacobian becomes:\[ \begin{bmatrix} 0 & 2 \ -2 & 0 \end{bmatrix} \] Calculating the eigenvalues involves solving \(\lambda^2 + 4 = 0\), leading to eigenvalues \(\lambda = \pm 2i\).
Imaginary eigenvalues suggest possible center or spiral behavior but are inconclusive via linearization alone, requiring further analysis.

Phase-Plane Method

The phase-plane method provides a visual way to examine the behavior of a dynamical system.
It maps trajectories in a plane, revealing patterns and movement directions around critical points. This method is especially useful when linearization fails to offer conclusive information.Using the hint, we introduce a new variable: \( u = \frac{y^2}{x} \).
This introduces an expression from the system:\[(x - c)^2 + y^2 = c^2 - 1\] Representing a circle becomes clear: solutions (trajectories) near the critical points form closed loops.
These closed loops confirm that the critical points behave as centers, circulating predictably without diverging from them.

Jacobian Matrix

The Jacobian matrix is a foundational concept in analyzing multivariable functions and systems.
It includes partial derivatives, capturing how changes in variables affect function outputs. In dynamical systems, it tells us about behavior near critical points.From our system:

• \<\frac{\partial x'}{\partial x}, \frac{\partial x'}{\partial y} = 0, 2x\>
• \<\frac{\partial y'}{\partial x}, \frac{\partial y'}{\partial y} = -2x, 2y\>

The resulting Jacobian is \(\begin{bmatrix} 0 & 2x \ -2x & 2y \end{bmatrix}\). At critical points such as \((1, 0)\), linearization reveals system tendencies.
The imaginary eigenvalues from this matrix indicate center-like behavior due to purely rotational tendencies without growth or decay at these points.
It’s often the starting point for deeper qualitative analyses of system dynamics.