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:
- Work on where it intersects the x axis ( fixed points )
- 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