1
$\begingroup$

I am going through Nonlinear Dynamics and Chaos by Strogatz and I am getting stuck on a few questions:

To calculate the stability of a fixed point:

  1. Work on where it intersects the x axis ( fixed points )
  2. Take the derivative of the original function and then plug in the fixed points to determine stability.

Trying to apply this on a bifurcation equation in a nonuniform oscillator he gives the following:

Page 117. Question 4.3.3 Answers in the Answer manual Page 67

\begin{array}{l} \dot{\theta } =\mu \sin( \theta ) -\sin( 2\theta ) \end{array}

\begin{array}{l} \mu =-2\ sadle-node\ bifurcation\ at\ \theta =\pi \end{array}

\begin{array}{l} \mu =2\ sadle-node\ bifurcation\ at\ \theta =0 \end{array}

Solved the first one.

\begin{array}{l} replace\ \sin 2\theta \ with\ 2\sin \theta \ \cos \theta \ simplifies\ to\\ \\ \\ \mu =\ 2\ \cos( \theta )\\ \frac{d}{d\theta }( \mu ) =\frac{d}{d\theta } 2\ \cos( \theta )\\ 0=-2sin( \theta ) \ which\ is\ zero\ at\ 0\ and\ \pi \\ \end{array}

Additionally another question:

Page 117 Question 4.3.5 - Answers in the Answer manual Page 69

\begin{array}{l} \dot{\theta } =\mu +\cos( \theta ) +\cos( 2\theta ) \end{array}

\begin{array}{l} \mu =-2\ sadle-node\ bifurcation\ at\ \theta =0 \end{array}

\begin{array}{l} \mu =0\ sadle-node\ bifurcation\ at\ \theta =\pi \end{array}

\begin{array}{l} \mu =\frac{9}{8} \ @\ \theta =2\arctan\left(\sqrt{\frac{5}{3}}\right) \ and\ \theta =2\pi \ -2\arctan\left(\sqrt{\frac{5}{3}}\right) \end{array}

How is he analytically calculating these fixed points when mu is a variable? Is he looking at the phase plane? I have been over his Bifurcation chapter a number of times trying to piece it together and cant seem to figure it out.

I am basing these off of example 4.3.1 on page 99. Where he gives a general example of how to calculate fixed point stability with linear stability analysis.

Thanks

$\endgroup$
7
  • 1
    $\begingroup$ Could you perhaps be more specific about where in the book those examples are? And what do you mean by the “@” sign? (Also, don't forget the backslash in \sin, etc.) $\endgroup$
    – Hans Lundmark
    Commented May 16, 2023 at 8:28
  • 1
    $\begingroup$ @HansLundmark Okay thank you. Now updated $\endgroup$
    – SS1
    Commented May 16, 2023 at 8:36
  • 1
    $\begingroup$ There is no phase plane to look at – the phase space of $\dot \theta = f(\theta)$ is just a circle. $\endgroup$
    – Hans Lundmark
    Commented May 16, 2023 at 9:29
  • $\begingroup$ Anyway, you'll need to figure out what the function $f(\theta)$ looks like, depending on the parameter $\mu$. Maybe it's easier to start with Problem 4.3.5, since if you can just draw the graph of $g(\theta) = \cos \theta + \cos 2 \theta$, then you get the graph of $f(\theta) = \mu + \cos \theta + \cos 2 \theta$ by shifting the graph of $g$ up or down (according to the value of $\mu$). I don't know what they did to arrive at the expression $2 \arctan \sqrt{5/3}$ for one of the points where $g$ has its minimum; what comes out naturally for me is $\arccos(-1/4)$ (which is the same thing). $\endgroup$
    – Hans Lundmark
    Commented May 16, 2023 at 9:29
  • $\begingroup$ @HansLundmark Thank you. I graphed both of these and Desmos and I have a good understanding of how they change with the parameter \mu. My trouble is, how did they analytically calculate those fixed points? In the book he has a phase plane of Theta dot, vs Theta over the period 0-2pi. Any ideas how he worked out the fixed points? Because there are more created for different values of \mu. But he has only stated 2 for first one and 3 for second one. $\endgroup$
    – SS1
    Commented May 16, 2023 at 11:09

0

Reset to default

You must log in to answer this question.

Browse other questions tagged .