Problem 39 - Advanced Engineering Mathematics

Question

(a) Letting \(x=\theta\) and \(y=x^{\prime}\) we obtain the system \(x^{\prime}=y\) and \(y^{\prime}=1 / 2-\sin x .\) since \(\sin \pi / 6=\sin 5 \pi / 6=1 / 2\) we see that \((\pi / 6,0)\) and \((5 \pi / 6,0)\) are critical points of the system. (b) The Jacobian matrix is $$\mathbf{g}^{\prime}(\mathbf{X})=\left(\begin{array}{cc} 0 & 1 \\ -\cos x & 0 \end{array}\right)$$ and so $$\mathbf{A}_{1}=\mathbf{g}^{\prime}=((\pi / 6,0))=\left(\begin{array}{cc} 0 & 1 \\ -\sqrt{3} / 2 & 0 \end{array}\right) \quad \text { and } \quad \mathbf{A}_{2}=\mathbf{g}^{\prime}=((5 \pi / 6,0))=\left(\begin{array}{cc} 0 & 1 \\ \sqrt{3} / 2 & 0 \end{array}\right)$$ since det \(\mathbf{A}_{1}>0\) and the trace of \(\mathbf{A}_{1}\) is \(0,\) no conclusion can be drawn regarding the critical point \((\pi / 6,0)\) since det \(\mathbf{A}_{2}<0,\) we see that \((5 \pi / 6,0)\) is a saddle point(c) From the system in part (a) we obtain the first-order differential equation $$\frac{d y}{d x}=\frac{1 / 2-\sin x}{y}$$ Separating variables and integrating we obtain $$\int y d y=\int\left(\frac{1}{2}-\sin x\right) d x$$ and $$\frac{1}{2} y^{2}=\frac{1}{2} x+\cos x+c_{1}$$ or $$y^{2}=x+2 \cos x+c_{2}$$ For \(x_{0}\) near \(\pi / 6,\) if \(\mathbf{X}(0)=\left(x_{0}, 0\right)\) then \(c_{2}=-x_{0}-2 \cos x_{0}\) and \(y^{2}=x+2 \cos x-x_{0}-2 \cos x_{0} .\) Thus, there are two values of \(y\) for each \(x\) in a sufficiently small interval around \(\pi / 6 .\) Therefore \((\pi / 6,0)\) is a center.

Step-by-Step Solution

Answer

\((\frac{5\pi}{6}, 0)\) is a saddle point; \((\frac{\pi}{6}, 0)\) is a center.

Step 1: Define the System of Equations

We start by defining the system \(x' = y\) and \(y' = \frac{1}{2} - \sin x\). By substituting \(x=\theta\), this system describes the behavior of the variables \(x\) and \(y\).

Step 2: Identify Critical Points

A critical point occurs where \(x' = 0\) and \(y' = 0\). We find that \(\sin x = \frac{1}{2}\) which leads to critical points at \((\frac{\pi}{6}, 0)\) and \((\frac{5\pi}{6}, 0)\).

Step 3: Compute the Jacobian Matrix

The Jacobian matrix \(\mathbf{g}'(\mathbf{X})\) is given as \(\begin{pmatrix} 0 & 1 \ -\cos x & 0 \end{pmatrix}\). This provides the linear approximation of the system near critical points.

Step 4: Evaluate Jacobian at Critical Points

At \((\frac{\pi}{6},0)\), the Jacobian \(\mathbf{A}_1 = \begin{pmatrix} 0 & 1 \ -\frac{\sqrt{3}}{2} & 0 \end{pmatrix}\). At \((\frac{5\pi}{6},0)\), it is \(\mathbf{A}_2 = \begin{pmatrix} 0 & 1 \ \frac{\sqrt{3}}{2} & 0 \end{pmatrix}\).

Step 5: Determine Nature of Critical Points

For \((\frac{\pi}{6},0)\), the determinant is positive and trace is zero, indicating this point is indecisive as a center or a spiral. For \((\frac{5\pi}{6},0)\), the determinant is negative, indicating it is a saddle point.

Step 6: Derive First-order Differential Equation

From \(y' = \frac{1}{2} - \sin x\), we derive \(\frac{dy}{dx} = \frac{1/2 - \sin x}{y}\), a separable differential equation.

Step 7: Integration and Solution

Separating variables and integrating, we find \(\int y \, dy = \int (\frac{1}{2} - \sin x) \, dx\). The integrated result is \(\frac{1}{2} y^2 = \frac{1}{2} x + \cos x + c_1\), leading to \(y^2 = x + 2\cos x + c_2\).

Step 8: Analyze Around Critical Point

For \(x_0\) near \(\frac{\pi}{6}\), set initial condition \(\mathbf{X}(0) = (x_0, 0)\). Solving gives \(c_2 = -x_0 - 2 \cos x_0\), rendering \(y^2 = x + 2\cos x - x_0 - 2\cos x_0\). Therefore, around \(\frac{\pi}{6}\), \((\frac{\pi}{6}, 0)\) is a center.

Key Concepts

Jacobian Matrix

The Jacobian matrix is a powerful tool used in analyzing systems of differential equations. It represents the linearization of a system of differential equations at a given point and is structured as a matrix of first-order partial derivatives. This matrix helps us study the behavior of a system near its critical points.
For our current system, the Jacobian matrix is:

• \[\mathbf{g}^{\prime}(\mathbf{X})=\begin{pmatrix} 0 & 1 \ -\cos x & 0 \end{pmatrix}\]

The Jacobian matrix provides insights into the dynamics of the system around specific points, such as whether these points are stable or unstable. By evaluating the Jacobian at critical points, which are locations where the system's behavior changes, we can determine the type of stability present at those points.
The determinant and trace of the Jacobian matrix are particularly useful in this analysis, as they allow us to classify the nature of critical points in the system.

System of Differential Equations

A system of differential equations consists of multiple equations that describe how variables change in relation to each other over time. These systems are typically used to model dynamic processes that depend on more than one changing factor.
Let's consider the system defined in this exercise:

• \(x' = y\)
• \[y' = \frac{1}{2} - \sin x\]

This particular system illustrates how the rate of change of \(x\) is equal to \(y\), while the rate of change for \(y\) depends on \(\frac{1}{2} - \sin x\). Solving these systems often involves identifying critical points, analyzing stability, and understanding the interactions between the variables. Systems like these allow us to model real-world phenomena by understanding how different variables influence each other.

Saddle Point Analysis

Saddle points are critical points where the system shows characteristics of both stability and instability, depending on the direction of approach. In mathematical terms, a saddle point is characterized by an equilibrium that is stable in some directions and unstable in others.
In our exercise, we found:

• The point \((5\pi/6,0)\) was identified as a saddle point.
• It was discovered by noting that the determinant of the Jacobian at this point was negative.

This negative determinant is a key indicator of a saddle point because it implies that the critical point is hyperbolic and hence an unstable equilibrium. Such points are critical in systems analysis as they indicate regions where small perturbations can cause the system to shift drastically.

Separable Differential Equations

Separable differential equations are a subset of differential equations that can be solved by separating the variables into two distinct sides of an equation. This method facilitates easier integration and solution finding.
In our exercise, the differential equation derived was:

• \[\frac{dy}{dx} = \frac{\frac{1}{2} - \sin x}{y}\]

To solve this, we separated the variables to isolate all terms involving \(y\) on one side and all terms involving \(x\) on the other:

• \[\int y \, dy = \int \left(\frac{1}{2} - \sin x\right) \, dx\]

Integrating both sides gave a solution involving \(y^2\), \(x\), and \(\cos x\). This process highlights how separating and integrating can provide clear solutions to problems, making it a preferred technique for dealing with specific types of differential equations.