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Chapter 4: Problem 4

For each of the following questions, draw the phase portrait as function of the control parameter \(\mu\). Classify the bifurcations that occur as \(\mu\) varies, and find all the bifurcation values of \(\mu\). $$ \dot{\theta}=\frac{\sin \theta}{\mu+\cos \theta} $$

Short Answer

Expert verified
The fixed points of the differential equation \(\dot{\theta}=\frac{\sin \theta}{\mu+\cos \theta}\) are \(\theta = 0, \pi\). The stability of these points changes depending on the parameter \(\mu\). There are two saddle-node bifurcations at \(\mu = -1\) for \(\theta = 0\) and \(\mu = 0\) for \(\theta = \pi\).

Step by step solution

01

Identify the Fixed Points

To find the fixed points, we must find the values of \(\theta\) for which \(\dot{\theta}=0\). From the given equation: $$ \frac{\sin \theta}{\mu+\cos \theta}=0 $$ The fixed points are obtained when \(\sin \theta=0\). We have: $$ \theta = 0, \pi $$
02

Determine the Stability of Fixed Points

To evaluate the stability of the fixed points, we need to study the sign of \(\dot{\theta}\). We have: $$ \dot{\theta}=\frac{\sin \theta}{\mu+\cos \theta} $$ Now we analyze the stability for both fixed points separately: 1. For \(\theta=0\): If \(\mu > -1\), then \(\mu + \cos\theta > 0\), and \(\sin\theta = 0\). So the fixed point is stable. If \(\mu < -1\), then \(\mu + \cos\theta < 0\), and \(\sin\theta = 0\). So the fixed point is unstable. 2. For \(\theta=\pi\): If \(\mu > 0\), then \(\mu + \cos\theta > 0\), and \(\sin\theta = 0\). So the fixed point is stable. If \(\mu < 0\), then \(\mu + \cos\theta < 0\), and \(\sin\theta = 0\). So the fixed point is unstable.
03

Analyze the Bifurcations

From the stability analysis, we can see that the bifurcations occur at \(\mu=-1\) for \(\theta=0\) and \(\mu=0\) for \(\theta=\pi\). Since the stability changes as \(\mu\) crosses these values, these bifurcations are considered as saddle-node bifurcations.
04

Draw the Phase Portrait

To draw the phase portrait, we plot the fixed points and their stability regions on the \(\mu\)-\(\theta\) plane. For \(\theta = 0\) and \(\mu > -1\), the fixed point is stable, and for \(\mu < -1\), it is unstable. For \(\theta = \pi\) and \(\mu > 0\), the fixed point is stable, and for \(\mu < 0\), it is unstable. By plotting these regions and the fixed points, we can obtain the phase portrait.
05

Find the Bifurcation Values of \(\mu\)

Based on the analysis, we found two bifurcations (saddle-node bifurcations) at: 1. \(\mu = -1\) for \(\theta = 0\) 2. \(\mu = 0\) for \(\theta = \pi\) These are the bifurcation values of \(\mu\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Phase Portrait
A phase portrait provides a visual representation of the trajectory of states through phase space in a dynamical system. Essentially, it's a graphical way to understand how the system evolves over time based on its starting conditions.

To plot a phase portrait, we usually seek the fixed points first, which are akin to landmarks on our map. We then draw curves that represent the trajectories of the system as it moves through different states, governed by its differential equations. These trajectories are like paths that show where the system will go, depending on where it starts.

For the given system \( \dot{\theta}=\frac{\sin \theta}{\mu+\cos \theta} \), the phase portrait changes as the control parameter \(\mu\) varies, showing different behaviors of the system, including stability and potential for bifurcations, as certain parameter values are crossed.
Fixed Points
Fixed points are the 'crossroads' of a dynamical system. They occur where there is no change in state over time (hence, 'fixed'). Mathematically, we locate these points by setting the time derivative of the system variable(s) to zero and solving the resultant equations.

In our exercise, \(\dot{\theta}=0\) leads to solving \(\sin \theta=0\), which yields fixed points at \(\theta = 0, \pi\). Identifying these points is critical, as they often underpin the analysis of system stability and bifurcations, paving the way to understanding the system's overall behavior.
Stability of Dynamical Systems
The concept of stability tells us whether small perturbations or changes to the system's state will dissipate over time or grow, potentially leading to a vastly different outcome. In layman's terms, it's what makes a system predictable (stable) or wildly sensitive to small changes (unstable).

Stability is examined around fixed points. If a system returns to a fixed point after a small disturbance, that point is considered stable; if it moves away, it is unstable. Our system's stability depends on the relationship between \(\mu\) and \(\cos\theta\), as seen in the provided solution steps. Changes in \(\mu\) can switch a system from stable to unstable at the fixed points, revealing the critical nature of system parameters in influencing dynamical behavior.
Saddle-Node Bifurcation
A saddle-node bifurcation occurs when a pair of fixed points, one stable and one unstable, collide and annihilate each other as a parameter is varied. Imagine two opposing traffic flows coming together and disappearing at a particular point鈥攖his conceptualizes the 'bifurcation' effect.

In the provided exercise, \(\mu=-1\) and \(\mu=0\) are the bifurcation points for \(\theta=0\) and \(\theta=\pi\), respectively. At these values, the behavior of the system changes fundamentally, representing a shift in stability. This type of bifurcation is fundamental to understanding many aspects of dynamical systems, including transitions to chaos and the onset of certain types of oscillations in physical and biological systems.

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Most popular questions from this chapter

(Understanding \(\sin \theta(t))\) By imagining the rotational motion of an overdamped pendulum, sketch \(\sin \theta(t)\) vs, \(t\) for a typical solution of \(\theta^{\prime}=\gamma-\sin \theta .\) How does the shape of the waveform depend on \(\gamma\) ? Make a series of graphs for different \(\gamma\), including the limiting cases \(\gamma=1\) and \(\gamma \gg 1\). For the pendulum, what physical quantity is proportional to \(\sin \theta(t) ?\)

The oscillation period for the nonuniform oscillator is given by the integral \(T=\int_{\pi}^{k} \frac{d \theta}{\omega-a \sin \theta}\), where \(\omega>a>0\). Evaluate this integral as follows. a) Let \(u=\tan \frac{\theta}{2} .\) Solve for \(\theta\) and then express \(d \theta\) in terms of \(u\) and \(d u\). b) Show that \(\sin \theta=2 u /\left(I+u^{2}\right)\). (Hint: Draw a right triangle with base 1 and height \(u .\) Then \(\frac{\theta}{2}\) is the angle opposite the side of length \(u\), since \(u=\tan \frac{g}{2}\) by definition. Finally, invoke the half-angle formula \(\sin \theta=2 \sin \frac{f}{2} \cos \frac{f}{2}\).) c) Show that \(u \rightarrow \pm \infty\) as \(\theta \rightarrow \pm \pi\), and use that fact to rewrite the limits of integration. d) Express \(T\) as an integral with respect to \(u\). e) Finally, complete the square in the denominator of the integrand of (d), and reduce the integral to the one studied in Exercise \(4.3 .1\), for a suitable choice of \(x\) and \(r\).

For each of the following vector fields, find and classify all the fixed points, and sketch the phase portrait on the circle. $$ \dot{\theta}=\sin \theta+\cos \theta $$

(Alternative derivation of scaling law) For systems close to a saddle-node bifurcation, the scaling law \(T_{\text {tweleneck }}=O\left(r^{-1 / 2}\right)\) can also be derived as follows. a) Suppose that \(x\) has a characteristic scale \(O\left(r^{n}\right)\), where \(a\) is unknown for now. Then \(x=r^{\prime \prime} u\), where \(u-O(1)\). Similarly, suppose \(t=r^{6} \tau\), with \(\tau \sim O(1)\). Show that \(\dot{x}=r+x^{2}\) is thereby transformed to \(r^{o-b} \frac{d u}{d \tau}=r+r^{2 i} u^{2}\) b) Assume that all terms in the equation have the same order with respect to \(r\). and thereby derive \(a=\frac{1}{2}, b=-\frac{1}{2}\).

For each of the following vector fields, find and classify all the fixed points, and sketch the phase portrait on the circle. $$ \dot{\theta}=\sin k \theta \text { where } k \text { is a positive integer. } $$
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