Problem 37 - Advanced Engineering Mathematics

Question

The corresponding plane autonomous system is $$x^{\prime}=y, \quad y^{\prime}=-\frac{\alpha}{L} x-\frac{\beta}{L} x^{3}-\frac{R}{L} y$$ where \(x=q\) and \(y=q^{\prime} .\) If \(\mathbf{X}=(x, y)\) is a critical point, \(y=0\) and \(-\alpha x-\beta x^{3}=-x\left(\alpha+\beta x^{2}\right)=0 .\) If \(\beta>0\) \(\alpha+\beta x^{2}=0\) has no real solutions and so (0,0) is the only critical point. since $$\mathbf{g}^{\prime}(\mathbf{X})=\left(\begin{array}{cr} 0 & 1 \\ \frac{-\alpha-3 \beta x^{2}}{L} & -\frac{R}{L} \end{array}\right)$$ \(\tau=-R / L<0\) and \(\Delta=\alpha / L>0 .\) Therefore (0,0) is a stable critical point. If \(\beta<0,(0,0)\) and \((\pm \hat{x}, 0),\) where \(\hat{x}^{2}=-\alpha / \beta\) are critical points. At \(\mathbf{X}(\pm \hat{x}, 0), \tau=-R / L<0\) and \(\Delta=-2 \alpha / L<0 .\) Therefore both critical points are saddles.

Step-by-Step Solution

Answer

If \(\beta>0\), (0,0) is the only stable critical point. If \(\beta<0\), (0,0) and \((\pm \hat{x}, 0)\) are critical points; the latter are saddles.

Step 1: Define the System and Identify Variables

The given plane autonomous system is \[x^{\prime}=y, \quad y^{\prime}=-\frac{\alpha}{L} x-\frac{\beta}{L} x^{3}-\frac{R}{L} y\]here, the variables are defined as follows:- \(x = q\)- \(y = q^{\prime}\).

Step 2: Determine Critical Points

A critical point occurs when both derivatives are zero, i.e., \((x', y') = (0, 0)\). Thus, we have:1. For \(y = 0\) in \(x^{'} = y\), we need: \[-\alpha x - \beta x^3 = 0.\]2. Factoring gives: \[x(-\alpha - \beta x^2) = 0.\]This implies \(x = 0\) or \(x^2 = -\frac{\alpha}{\beta}\).

Step 3: Check the Case When \(\beta > 0\)

If \(\beta > 0\), then \(-\frac{\alpha}{\beta} < 0\), making \(x^2 = -\frac{\alpha}{\beta}\) impossible since the square of a real number can't be negative. Therefore, the only critical point is \((0, 0)\).

Step 4: Check the Case When \(\beta < 0\)

If \(\beta < 0\), then \(-\frac{\alpha}{\beta} > 0\), meaning \(x^2 = -\frac{\alpha}{\beta}\) has real solutions. These are \((\pm \hat{x}, 0)\) where \(\hat{x}^2 = -\frac{\alpha}{\beta}\). The critical points are \((0,0)\) and \((\pm \hat{x}, 0)\).

Step 5: Evaluate Stability for \((0, 0)\)

Using the Jacobian matrix \[\mathbf{g}^{\prime}(\mathbf{X})=\begin{pmatrix} 0 & 1 \ \frac{-\alpha-3 \beta x^{2}}{L} & -\frac{R}{L} \end{pmatrix}\] at \((0, 0)\), the entries evaluate to:- \(\tau = \text{trace} = -\frac{R}{L} < 0\)- \(\Delta = \text{determinant} = \frac{\alpha}{L} > 0\)Thus, \((0, 0)\) is a stable node because \(\tau < 0\) and \(\Delta > 0\).

Step 6: Evaluate Stability for \((\pm \hat{x}, 0)\)

For the points \((\pm \hat{x}, 0)\):- \(\tau = -\frac{R}{L} < 0\)- \(\Delta = -\frac{2\alpha}{L} < 0\)This indicates that these points are saddle points since \(\Delta < 0\).

Key Concepts

Critical Points

In an autonomous system, critical points are crucial because they represent the states where the system is in equilibrium. Here, the system's derivatives are zero, meaning that there is no change occurring in the state variables. For the autonomous system in question, these points occur when both \(x'\) and \(y'\) become zero simultaneously. This leads to the equations \(x' = y = 0\) and \(y' = -\frac{\alpha}{L} x - \frac{\beta}{L} x^3 - \frac{R}{L} y = 0\). Solving these conditions, we find that either \(x = 0\) or \(x^2 = -\frac{\alpha}{\beta}\), depending on the value of \(\beta\).
If \(\beta > 0\), the equation \(x^2 = -\frac{\alpha}{\beta}\) has no solution since a square cannot be negative, leaving \((0,0)\) as the sole critical point. However, if \(\beta < 0\), \(x^2 = -\frac{\alpha}{\beta}\) becomes possible and yields additional critical points: \((\pm \hat{x}, 0)\). These critical points are significant in determining the behavior and stability of the system, as each point indicates an equilibrium condition.

Jacobian Matrix

The Jacobian matrix is a mathematical tool used to analyze the behavior of dynamical systems near their critical points. For the given autonomous system, the Jacobian matrix \(\mathbf{g}'(\mathbf{X})\) is crucial because it contains partial derivatives that provide insights into how the system behaves around a critical point. The matrix is given by:

• \(a = 0, b = 1\)
• \(c = \frac{-\alpha - 3\beta x^2}{L}, d = -\frac{R}{L}\)

The entries \(a\), \(b\), \(c\), and \(d\) determine the nature of the critical points by influencing the trace (\(\tau\)) and determinant (\(\Delta\)) of the matrix. Calculating these gives us the conditions necessary for analyzing stability.
In our example, the trace and determinant at \((0,0)\) can be computed directly as \(\tau = -\frac{R}{L}\) and \(\Delta = \frac{\alpha}{L}\), providing a basis for further stability analysis. The analysis is extended to other critical points by plugging in the respective values of \(x\), which differ based on the sign of \(\beta\).

Stability Analysis

Stability analysis involves examining the nature of each critical point. It's vital to understand whether small perturbations around these points will dampen out, indicating stability, or grow over time, indicating instability.
For the point \((0,0)\), with the trace \(\tau = -\frac{R}{L} < 0\) and determinant \(\Delta = \frac{\alpha}{L} > 0\), the system is stable. The negative trace signifies that any trajectories around this point naturally return to it, and the positive determinant ensures that perturbations die out.
Contrastingly, for points \((\pm \hat{x}, 0)\), both \(\tau\) and \(\Delta\) tell a different story. Here, \(\tau = -\frac{R}{L} < 0\) remains negative, but \(\Delta = -\frac{2\alpha}{L} < 0\) is also negative, indicating saddle points. In these cases, small disturbances will not return to the critical point but will move away along certain paths, signifying instability.
This analysis is pivotal for scientists and engineers because it guides them in predicting the behavior of dynamic systems in real-world applications, helping them design systems that either harness or mitigate such stability properties.