Problem 28 - Advanced Engineering Mathematics

Question

Show that the dynamical system $$ \begin{aligned} &x^{\prime}=-\alpha x+x y \\ &y^{\prime}=1-\beta y-x^{2} \end{aligned} $$ has a unique critical point when \(\alpha \beta>1\) and that this critical point is stable when \(\beta>0\).

Step-by-Step Solution

Answer

The system has a unique critical point at \((0, \frac{1}{\beta})\); it is stable when \(\beta > 0\).

Step 1: Determine the Critical Points

To find the critical points of the system, we set \(x' = 0\) and \(y' = 0\).For \( x' = 0 \):\[-\alpha x + xy = 0\]\(x(-\alpha + y) = 0\). This gives \(x = 0\) or \(y = \alpha\).For \( y' = 0 \):\[ 1 - \beta y - x^2 = 0 \]Solving for \(y\), we get:\[ y = \frac{1 - x^2}{\beta} \]Substitute \(x = 0\) into the equation for \(y\), we get:\[ y = \frac{1}{\beta} \] Substitute \(y = \alpha\) into the equation for \(y\), we get:\[ 1 - \beta \alpha - x^2 = 0 \] Solving for \(x^2\), we find:\[ x^2 = 1 - \beta \alpha \] Since \(x^2 \geq 0\), \(1 - \beta \alpha\geq 0\) has to hold, but since \(\alpha\beta > 1\), no real solution for \(\beta\alpha\geq 0\) is possible with \(x eq 0\). Thus only critical point is when \((x, y) = (0, \frac{1}{\beta})\), assuming \(\beta > 0 \).

Step 2: Analyze Stability of the Critical Point

Now that we have the critical point at \((x, y) = (0, \frac{1}{\beta})\), we analyze its stability by considering the Jacobian matrix of the system.The Jacobian matrix \(J\) is given by:\[J = \begin{bmatrix}\frac{\partial x'}{\partial x} & \frac{\partial x'}{\partial y} \\frac{\partial y'}{\partial x} & \frac{\partial y'}{\partial y}\end{bmatrix}\]Calculating the partial derivatives, we have:\(\frac{\partial x'}{\partial x} = -\alpha + y, \; \frac{\partial x'}{\partial y} = x\) \(\frac{\partial y'}{\partial x} = -2x, \; \frac{\partial y'}{\partial y} = -\beta\)At the critical point \((x, y) = (0, \frac{1}{\beta})\), the Jacobian becomes:\[ J = \begin{bmatrix}\frac{1}{\beta} - \alpha & 0 \0 & -\beta\end{bmatrix}\]The eigenvalues of \(J\) are \(\lambda_1 = \frac{1}{\beta} - \alpha\) and \(\lambda_2 = -\beta\). Since we know \(\alpha \beta > 1\) implies \( \alpha > \frac{1}{\beta} \), it follows \(\lambda_1 < 0\) and \(\lambda_2 < 0\) for \(\beta > 0\). Therefore, the critical point is stable.

Key Concepts

Critical Points

In the world of dynamical systems, finding critical points is a crucial step when analyzing the behavior of a system. Critical points occur where the rates of change, denoted by derivatives, are equal to zero. By setting the derivatives of a system of equations to zero, you essentially identify where the system is momentarily at rest. For this dynamical system, we focus on the equations:

• \( x' = -\alpha x + xy \)
• \( y' = 1 - \beta y - x^2 \)

This results in the critical conditions: \( x(-\alpha + y) = 0 \) and \( 1 - \beta y - x^2 = 0 \). From these, we deduce possible solutions for \(x\) and \(y\), which helps us find points like \((x, y) = (0, \frac{1}{\beta})\). Identifying critical points is a foundational step to further analyze the system's behavior.

Stability Analysis

Once critical points are identified, the next step is to determine their stability. In stability analysis, we inspect if small perturbations or changes in the system's state will die out or grow. If they die out, the system returns to its steady state, indicating stability.

The critical point at \((x, y) = (0, \frac{1}{\beta})\) is analyzed for stability using matrix methods. Often, the classification of stability involves categorizing a point as stable, unstable, or a saddle point. Stability at a critical point is crucial because it helps predict the system's behavior near that point.

For our system, we determine this using the Jacobian matrix, which provides valuable information about dynamics near the critical point through its eigenvalues.

Jacobian Matrix

The Jacobian matrix is a powerful tool in examining local behavior around critical points of dynamical systems. It contains the first-order partial derivatives of your system's functions.
To construct the Jacobian matrix \( J \) for the system, calculate:

• \( \frac{\partial x'}{\partial x} = -\alpha + y \)
• \( \frac{\partial x'}{\partial y} = x \)
• \( \frac{\partial y'}{\partial x} = -2x \)
• \( \frac{\partial y'}{\partial y} = -\beta \)

Evaluating at the critical point \((x, y) = (0, \frac{1}{\beta})\) where \( x = 0 \) and \( y = \frac{1}{\beta} \), the Jacobian matrix simplifies to:

• \( J = \begin{bmatrix} \frac{1}{\beta} - \alpha & 0 \ 0 & -\beta \end{bmatrix} \)

The Jacobian matrix helps in determining the stability by examining its eigenvalues, which lead us to the next concept.

Eigenvalues

Eigenvalues play a critical role in the stability analysis of dynamical systems. By examining the eigenvalues of the Jacobian matrix, you can assess the stability of a critical point.

The eigenvalues are solutions to the characteristic equation, which for our Jacobian matrix \( J \) at the critical point \((0, \frac{1}{\beta})\), are:

• \( \lambda_1 = \frac{1}{\beta} - \alpha \)
• \( \lambda_2 = -\beta \)

Since \( \alpha\beta > 1 \), it implies \( \alpha > \frac{1}{\beta} \), rendering \( \lambda_1 < 0 \) and \( \lambda_2 < 0 \) when \( \beta > 0 \).
Negative eigenvalues indicate that perturbations to the system at this critical point will decay over time, leading to a stable critical point. Therefore, in this scenario, the critical point \((0, \frac{1}{\beta})\) is stable.