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I am reading the following paper: http://ieeexplore.ieee.org/document/7798534/

For the equation (5):
$$\dot{a}=(1-a_i^2)a_i-\beta\sum_{k-i}(a_k\cos(\theta_k-\theta_i)-a_i)$$ $$\dot{\theta}_i=\beta\sum_{k-i}\frac{a_k}{a_i}\sin(\theta_i-\theta_k)$$

  1. Both are dynamics for amplitude $a$ and phase $\theta$.
  2. $k-i$ means there is an edge linking the node $k$ and $i$.
  3. $\beta$ is just a coefficient

My question is the following:

enter image description here

  1. Why the amplitude dynamics are invariant?

and the following:

enter image description here

  1. Why it is "global phase symmetry"
  2. Why the equilibria are semi-stable with respect to rotation?

Thanks!

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  1. If you add an $i$-independent time-dependent term $\delta\theta=\omega t+\gamma$ to $\theta_i$ then the difference $\theta_k-\theta_i$ that appears in the amplitude equation is unchanged. The phase equation is changed by a term $d\delta\theta/dt=\omega$.
  2. Global phase symmetry means that if $\psi_n=a_ne^{i\theta_n}$ is a solution, then also $\psi_n e^{i\phi t}$ with a constant phase $\phi$ is a solution. A local phase symmetry would have a $\phi_n$ that depends on $n$, which does not appply here.
  3. The limit cycle is semi-stable, rather than stable, because only amplitude perturbations decay in time, phase perturbations do not.
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  • $\begingroup$ To 1., can I use the same argument to say that the phase equation is unchanged and so phase dynamic is invariant? I am confused about 3., how to view it? Thanks! $\endgroup$ – sleeve chen Nov 3 '17 at 21:10
  • $\begingroup$ I clarified 1; concerning 3, the typical picture is the mexican hat --- amplitude perturbations are radial and damped, while phase fluctuations are azimuthal and undamped. $\endgroup$ – Carlo Beenakker Nov 3 '17 at 22:07
  • $\begingroup$ To 1., I am still confused that for amplitude eq it is $\theta_k-\theta_i$ and for the phase eq it is $\theta_i-\theta_k$. Why one is unchanged and the other one is changed by $\omega$? To 3., how to obtain the picture you mentioned? (By Taylor expansion which gives the linear equation for perturbation and then plot it?) Sorry for many questions. Thanks! $\endgroup$ – sleeve chen Nov 4 '17 at 2:07
  • $\begingroup$ it has nothing to do with the order $\theta_k-\theta_i$ or $\theta_i-\theta_k$; it just that $\delta\theta=\omega t+\gamma$ depends on time, so $(d/dt)(\theta_i+\delta\theta)=\dot{\theta}_i+\omega$; so the phase equation gets an offset $\omega$ which the amplitude equation does not. $\endgroup$ – Carlo Beenakker Nov 4 '17 at 10:02
  • $\begingroup$ But based on your argument, in the amplitude equation the value of the term $\cos(\theta_k-\theta_i-\delta \theta)$ will also be changed due to $\delta \theta$. $\endgroup$ – sleeve chen Nov 4 '17 at 10:09

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