Problem 11 - 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. $$ \begin{aligned} &x^{\prime}=1-2 x y \\ &y^{\prime}=2 x y-y \end{aligned} $$

Step-by-Step Solution

Answer

Critical points: (0,0) is a stable saddle-type; (1,0) is a source (unstable point).

Step 1: Identify the Critical Points

To find the critical points, solve the system by setting \( x' = 0 \) and \( y' = 0 \). This gives us the equations: \( 1 - 2xy = 0 \) and \( 2xy - y = 0 \). Solving these simultaneously, we find the critical points are \( (0, 0) \) and \( (1, 0) \).

Step 2: Find the Jacobian Matrix

The Jacobian matrix \( J \) of the system at a point \( (x, y) \) is given by evaluating the partial derivatives: \[ J = \begin{bmatrix} \frac{\partial x'}{\partial x} & \frac{\partial x'}{\partial y} \ \frac{\partial y'}{\partial x} & \frac{\partial y'}{\partial y} \end{bmatrix} = \begin{bmatrix} 0 & -2x \ 2y & 2x-1 \end{bmatrix} \].

Step 3: Evaluate Jacobian at Critical Point (0,0)

Substitute \( (0, 0) \) into the Jacobian matrix.\[ J(0, 0) = \begin{bmatrix} 0 & 0 \ 0 & -1 \end{bmatrix} \].\ The eigenvalues of this matrix are \( \lambda_1 = 0 \) and \( \lambda_2 = -1 \). Since one eigenvalue is zero and the other is negative, this suggests a non-isolated critical point, potentially a line of equilibria, but more analysis could suggest a stability akin to stable saddle.

Step 4: Evaluate Jacobian at Critical Point (1,0)

Substitute \( (1, 0) \) into the Jacobian matrix.\[ J(1, 0) = \begin{bmatrix} 0 & -2 \ 0 & 1 \end{bmatrix} \].\ The eigenvalues of this matrix are \( \lambda_1 = 1 \) and \( \lambda_2 = 0 \). The positive eigenvalue indicates that this critical point is unstable, suggesting a direction of flow from the point.

Key Concepts

Critical Points

Critical points of a system occur where the system experiences no change, meaning both derivatives (in our case, both \( x' \) and \( y' \)) are zero. These points can inform us about the nature of the system’s behavior at specific locations and are important for determining stability.

To find the critical points of a two-equation system, set each derivative equation to zero and solve. In our example:

• For \( x' = 1 - 2xy \), set \( 1 - 2xy = 0 \). For \( y' = 2xy - y \), set \( 2xy - y = 0 \).

The solutions to these equations give critical points. For the provided autonomous system, these are found to be \((0, 0)\) and \((1, 0)\). Each critical point must then be analyzed further to determine its nature by examining how other solutions behave around these points.

Jacobian Matrix

The Jacobian Matrix is a crucial component in analyzing the behavior around critical points in a dynamic system. It is composed of the first order partial derivatives of the system's functions. Essentially, it helps reveal how changes in the system’s state variables impact its future behavior.

For the given autonomous system:

• The Jacobian matrix \( J \) is expressed as follows: \[ J = \begin{bmatrix} \frac{\partial x'}{\partial x} & \frac{\partial x'}{\partial y} \ \frac{\partial y'}{\partial x} & \frac{\partial y'}{\partial y} \end{bmatrix} = \begin{bmatrix} 0 & -2x \ 2y & 2x - 1 \end{bmatrix} \]

The partial derivatives in the matrix measure the sensitivity of \( x' \) and \( y' \) to changes in \( x \) and \( y \). Substitute the critical points into this matrix to understand the local dynamics.

Stability Analysis

Stability analysis uses the Jacobian matrix evaluated at the critical points to decipher the type and stability of those points. The eigenvalues of the Jacobian tell us if a point is stable, unstable, or something more complex like a saddle point.

Let’s discuss our critical points:

• For \((0, 0)\), substituting into the Jacobian yields \( J(0, 0) = \begin{bmatrix} 0 & 0 \ 0 & -1 \end{bmatrix} \). The eigenvalues are \( \lambda_1 = 0 \) and \( \lambda_2 = -1 \). A zero eigenvalue suggests non-isolated stability possibly indicating a continuum of rest points or potentially stable saddle point behavior.
• For \((1, 0)\), the matrix is \( J(1, 0) = \begin{bmatrix} 0 & -2 \ 0 & 1 \end{bmatrix} \), with eigenvalues \( \lambda_1 = 1 \) and \( \lambda_2 = 0 \). A positive eigenvalue indicates instability, showing a direction of flow away from the point.

Through these eigenvalues, stability analysis helps predict whether solutions will tend toward or away from a critical point. This can further classify critical points as stable or unstable and gives insight into the overall behavior of the dynamic system.