I'm reading a paper https://doi.org/10.1063/1.2930766 about the Ott-Antonsen-Ansatz that is used to describe the dynamics of global coupled oscillator. There is a computational step from equation (4),(5) to (6),(7) that I don't understand.

I would really appreciate it if someone could explain to me what the author has done in these steps

  • 3
    $\begingroup$ Hi and welcome to physics.SE! It's currently unclear what exactly this question is asking without clicking on the link you provided. To make questions more accessible and guard against link rot, please include all relevant information, such as the equations you are talking about and the explanation of notation or specific terminology used, in your question. $\endgroup$
    – ACuriousMind
    Commented Apr 23, 2018 at 16:46
  • $\begingroup$ Just came across your question. Did you finally figure it out? From what I see the step you don't understand is simply replacing the $f$ in the Fourier time series with the constrain that all the coefficients $f_n$ are powers of the same quantity $\alpha$. This is just putting the the series of $f$ (with the constrain) inside the integral in (5) and you see that all the terms with $\theta$ vanish and only the $1$ survives… $\endgroup$
    – myradio
    Commented Jun 6, 2019 at 9:44

2 Answers 2



$$Z=Re^{i\Theta} = \int_{-\pi}^\pi \int_{-\infty}^\infty e^{i\theta} \rho(\theta,\omega,t)g(\omega) d\theta d\omega\tag{1}$$

be the complex meanfield of the Kuramoto model in the thermodynamic limes, where $\rho$ is the oscillator density and $g$ the distribution of the frequency.

The continuity equation for oscillator density will be $\frac{\partial \rho}{\partial \theta} + \frac{\partial }{\partial \theta} [\omega + KR\sin(\Theta - \theta)\rho]\tag{2}$

Make a Fourier series of $\rho$ in $\theta$

$$\rho(\theta,\omega,t)= \frac{g(\omega)}{2\pi} \sum_{l=-\infty}^{\infty} f_l(\omega,t) e^{il\theta}\qquad, f_l = f^*_{-l}, f_0=1$$

Assume that $f_l (\omega,t)$ is given by $f_l (\omega,t) = [\alpha (\omega,t)]]^n \qquad | \alpha(\omega,t)|\leq 1$

Now comes the part that I don't understand, the author writes: Substituting this series expansion into Eqs. (1) and (2), we find the remarkable result that this special form of f represents a solution to Eqs. (1) and (2) if

$$\frac{\partial \alpha}{\partial t} + \frac{K}{2}[Z\alpha^2 -Z^* ] + i\omega \alpha=0$$

$$Z^*=\int_{-\infty}^\infty d\omega \alpha (\omega,t) g(\omega) $$

  • $\begingroup$ If there is something not clear, feel free to ask. I would appriciate it, if someone could help me with this problem! $\endgroup$
    – Kreisel
    Commented Apr 24, 2018 at 7:59

Please try to compare the terms of $e^{in\theta}$ for specified $n$. You will find your answer. This question has been over two years. So I wish this answer should be useful for later ones.


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