Question

The equation $$ u^{\prime \prime}-\mu\left(1-\frac{1}{3} u^{\prime 2}\right) u^{\prime}+u=0 $$ is often called the Rayleigh equation. (a) Write the Rayleigh equation as a system of two first order equations. (b) Show that the origin is the only critical point of this system. Determine its type and whether it is stable or unstable. (c) Let \(\mu=1 .\) Choose initial conditions and compute the corresponding solution of the system on an interval such as \(0 \leq t \leq 20\) or longer. Plot \(u\) versus \(t\) and also plot the trajectory in the phase plane. Observe that the trajectory approaches a closed curve (limit cycle). Estimate the amplitude \(A\) and the period \(T\) of the limit cycle. (d) Repeat part (c) for other values of \(\mu,\) such as \(\mu=0.2,0.5,2,\) and \(5 .\) In each case estimate the amplitude \(A\) and the period \(T\). (e) Describe how the limit cycle changes as \(\mu\) increases. For example, make a table of values and/or plot \(A\) and \(T\) as functions of \(\mu .\)

Short answer

Question: Convert the given Rayleigh equation into a system of first-order differential equations, analyze the critical points and their stability, and investigate the behavior of the system for different values of the parameter \(\mu\). Briefly describe the changes of limit cycle as a function of \(\mu\). Answer: The Rayleigh equation can be converted into a system of first-order differential equations as follows: \begin{cases} u'(t) = v(t) \\ v'(t) = \mu(1-\frac{1}{3}v^2(t))v(t) - u(t) \end{cases} The origin \((0,0)\) is the only critical point, and it is a center. Based on linearized analysis, the stability of the origin is inconclusive. By solving the system numerically for different values of \(\mu\), we observe that the limit cycle changes as \(\mu\) increases. In general, the amplitude and the period of the limit cycle depend on \(\mu\), and understanding this dependency further deepens our insights into the behavior and stability of the Rayleigh equation.

Step-by-step solution

Convert to first-order system

Introduce new variables \(v(t) = u'(t)\) and \(w(t) = u''(t)\) to convert the second-order Rayleigh equation into a system of first-order equations. Then, the given equation can be written as: \begin{cases} u'(t) = v(t) \\ v'(t) = \mu(1-\frac{1}{3}v^2(t))v(t) - u(t) \end{cases} Notice that now we have a system of two first-order equations. Step 2: Identify the critical points and determine their type and stability

Find critical points

A critical point (equilibrium) occurs when \(u'(t) = v'(t) = 0\). From our first-order system, we have: \begin{cases} 0 = v(t) \\ 0 = \mu(1-\frac{1}{3}v^2(t))v(t) - u(t) \end{cases} Since \(v(t) = 0\), the second equation becomes \(u(t) = 0\). Hence, the origin \((0,0)\) is the only critical point.

Determine stability and type

To determine the type and stability of the origin, we linearize the system and find the Jacobian matrix: $$ J(u,v) = \begin{pmatrix} \frac{\partial{u'}}{\partial{u}} & \frac{\partial{u'}}{\partial{v}} \\ \frac{\partial{v'}}{\partial{u}} & \frac{\partial{v'}}{\partial{v}} \end{pmatrix} = \begin{pmatrix} 0 & 1 \\ -1 & -\mu v \end{pmatrix} $$ Now, evaluate the Jacobian at the origin, \(J(0,0) = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}\). The eigenvalues of this matrix are \(\pm i\), which have zero real part. Therefore, the origin is a center, and its stability is inconclusive based on linearized analysis alone. Step 3: Solve the system for \(\mu = 1\)

Solve and plot the solution

For \(\mu = 1\), choose initial conditions such as \(u(0) = 0\) and \(v(0) = 1\). Numerically solve the first-order system in the given interval \(0 \leq t \leq 20\), using software such as MATLAB, Python, or a numerical solver. Plot \(u\) vs. \(t\) and the phase plane trajectory. Depending on the chosen initial conditions, observe that the trajectory approaches a closed curve (limit cycle), and estimate the amplitude \(A\) and the period \(T\). Step 4: Repeat step 3 for other values of \(\mu\)

Solve and analyze for different values of \(\mu\)

Repeat step 3 for different values of \(\mu\), such as \(\mu = 0.2, 0.5, 2,\) and \(5\). In each case, estimate the amplitude \(A\) and the period \(T\). Compare the results to observe how the system's behavior changes for varying values of the parameter \(\mu\). Step 5: Describe the limit cycle changes as a function of \(\mu\)

Discuss the dependency on \(\mu\)

Using the results from Step 4, describe how the limit cycle changes as \(\mu\) increases. Consider making a table of values and/or plot \(A\) and \(T\) as functions of \(\mu\). This will help to understand how the system evolves depending on the parameter \(\mu\) and gain further insights into the behavior and stability of the Rayleigh equation.

Key concepts

First-order System

When solving differential equations, transforming them into a system of first-order equations is a common technique. The Rayleigh equation given in the exercise is a second-order non-linear differential equation. We want to convert this into a system of first-order equations. To do this, we introduce new variables. Let's define \( v(t) = u'(t) \) to represent the first derivative of \( u(t) \) and express the second-order differential in terms of the first-order derivatives. Therefore, we form the system:

• \( u'(t) = v(t) \)
• \( v'(t) = \, \mu \left(1-\frac{1}{3}v^2(t)\right)v(t) - u(t) \)

This system allows us to analyze the behavior of the solution in a more manageable form, establishing a foundation for further analysis of the Rayleigh equation.

Limit Cycle

A limit cycle is an isolated closed trajectory in the phase plane. Such cycles are significant in systems as they indicate periodic solutions, a common feature in oscillatory systems like the Rayleigh equation. When solving the Rayleigh equation for specific parameters, we are expected to observe that the trajectory approaches a closed loop in the phase plane. With various values of \( \mu \), you might observe that the solution starts from an initial point and settles into a repetitive path, which doesn't converge to an equilibrium but rather repeats in a cycle. This cycle's characteristics—its amplitude and period—can depend significantly on the parameter \( \mu \). Observing how these features change as we vary \( \mu \) gives us insight into the non-linear dynamics of the system.

Jacobians and Stability

In analyzing the stability of a system, finding critical points and the Jacobian matrix is crucial. For a system converted into first-order equations, the critical points occur where all derivatives equate to zero. In this exercise, substituting these conditions into our equations shows that the origin \((0,0)\) is the only critical point.
The Jacobian matrix is used to determine the stability and type of these points. For our system, the Jacobian is:\[J(u,v) = \begin{pmatrix}0 & 1 \-1 & -\mu v\end{pmatrix}\]Evaluating this at the origin yields \(J(0,0)\) with eigenvalues \(\pm i\). This indicates that the critical point is a center, which is typically stable but with the caveat of being neutrally stable in a linearized analysis. The eigenvalues' zero real components make it tricky to determine stability from the Jacobian alone, hence why further analysis, such as a non-linear examination, may be necessary for a complete understanding.

Phase Plane Analysis

Phase plane analysis provides a graphical method to study the behavior of dynamic systems by plotting trajectories. In the case of the Rayleigh equation, representing the first-order system on a phase plane enables us to visualize how the solutions evolve over time.Start by plotting \( u(t) \) versus \( v(t) \), creating a visual map of the system's possible states and how they change. By doing so in our Rayleigh system for different initial conditions and values of \( \mu \), you'll observe various trajectories.
Some will quickly approach a closed path representing the limit cycle. This cycle's appearance indicates how energy is conserved over time, instead of simply diminishing as in some dissipative systems. It also allows us to see whether the system is conservative or if energy persists, shaping our understanding of the system's stability and long-term behavior.