I am trying to derive the critical coupling strength for synchronisation in a network of phase oscillators with noisy input.

I am following the steps outlined in Sakaguchi, Hidetsugu. "Cooperative phenomena in coupled oscillator systems under external fields." Progress of theoretical physics 79.1 (1988): 39-46.

However I am stuck in an expansion and I would be grateful if someone had an insight to help.

I have a problem to derive the expansion in Eq. 25 but i will also copy here the equations for convenience.

We consider phase oscillators with all-to-all coupling in the thermodynamic limit, and thereby have expressed the mean-field Fokker-Planck for the phase distribution

$$ \frac{\partial n(\psi ; \omega)}{\partial t}=-\frac{\partial}{\partial \psi}\{(\omega-K \sigma \sin \psi) n(\psi ; \omega)\}+D \frac{\partial^{2}}{\partial \psi^{2}} n(\psi ; \omega), $$

where $\psi$ denotes the phase and $\omega$ the natural frequency, $\sigma$ the synchronisation order parameter, and $K$ the coupling strength. We compute the stationary solution: $$ \begin{aligned} n(\psi ; \omega)=& \exp \left(\frac{-K \sigma+\omega \psi+K \sigma \cos \psi}{D}\right) n(0 ; \omega) \\ & \times\left\{1+\frac{\left(e^{-2 \pi \omega / D}-1\right) \int_{0}^{\phi} e^{(-\omega \phi-K \sigma \cos \phi) / D} d \phi}{\int_{0}^{2 \pi} e^{(-\omega \phi-K \sigma \cos \phi) / D} d \phi} \right\}, \end{aligned} $$

and we are about to write a self-consistent equation for the synchronisation order parameter $\sigma$. We have from previously (Eq.9) $$ \begin{aligned} \sigma \exp \left(i \phi_{0}\right) &=\int_{-\infty}^{\infty} d \omega g(\omega) \int_{0}^{2 \pi} d \psi n(\psi ; \omega) \exp (i \psi), \end{aligned} $$ where $g(\omega)$ denotes the natural frequency distribution, and $\phi_0$ stands for the mean phase.

We write the self consistent equation for the order parameter $$ \sigma=\int_{-\infty}^{\infty} d \omega g\left(\omega+\omega_{0}\right) \int_{0}^{2 \pi} d \psi n(\psi ; \omega) \exp (i \psi), $$ and to find the critical coupling strength we proceed to expand the expression for the order parameter in powers of $K \sigma/D$

$$ \begin{aligned} \sigma=& K \sigma\left[\frac{1}{2} \int_{-\infty}^{\infty} g\left(D \omega+\omega_{0}\right) \frac{d \omega}{\omega^{2}+1}-\frac{K^{2} \sigma^{2}}{4 D^{2}} \int_{-\infty}^{\infty} g\left(D \omega+\omega_{0}\right)\left\{\frac{1}{\omega^{2}+4}-\frac{\omega}{\left(\omega^{2}+1\right)^{2}}\right\} d \omega\right.\\ &\left.+O\left(\left(\frac{K \sigma}{D}\right)^{4}\right)\right]. \end{aligned} $$

So my question is how can I derive this expansion. The author mentions that they consider that $g(\omega+\omega_0)$ is symmetric about 0, and that the imaginary part of the self-consistent equation for the order parameter is zero. So the expansion is only for the real part.

I would be grateful for any insight!

  • 1
    $\begingroup$ with some effort I have arrived at the desired equation, with one difference, which I think is a typo in the cited paper: the integrand should have only even powers of $\omega$, while the formula as cited has a factor $\omega$ (numerator of second term in curly brackets); if you replace that by $\omega^2$ you find the result in the answer. $\endgroup$ Feb 1, 2023 at 18:41

1 Answer 1


To simplify the notation I denote $\tilde{K}=K\sigma/D$ and $\tilde{\omega}=\omega/D$.

First expand \begin{align} &n(\psi , \omega)= \exp \bigl(-\tilde{K} +\tilde{\omega} \psi+\tilde{K} \cos \psi\bigr) n(0 , \omega)\nonumber\\ &\times\biggl[1+\frac{\bigl(e^{-2 \pi \tilde{\omega}}-1\bigr) \int_{0}^{\phi} e^{(-\tilde{\omega} \phi-\tilde{K} \cos \phi)} d \phi}{\int_{0}^{2 \pi} e^{(-\tilde{\omega} \phi-\tilde{K} \cos \phi)} d \phi} \biggr] \end{align} to fourth order in $K$ and evaluate $$\int_0^{2\pi}n(\psi,\omega)\cos(\psi)\,d\psi=\pi n(0,\omega)\bigl[\frac{\tilde{K}}{1+\tilde{\omega}^2}-\frac{\tilde{K}^2}{(1+\tilde{\omega}^2)^2} +\frac{ \tilde{K}^3 \bigl(3 \tilde{\omega}^4+2 \tilde{\omega}^2+5\bigr)}{2 \bigl(\tilde{1+\omega}^2\bigr)^3 \bigl(4+\tilde{\omega}^2\bigr)}+{\cal O}(K^4)\bigr].$$

The function $n(0,\omega)$ is determined by the normalisation $$\int_0^{2\pi}n(\psi,\omega)\,d\psi=1,$$ which gives to fourth order in $K$ the equation $$n(0,\omega)=\frac{1}{2 \pi }+\frac{\tilde{K}}{2 \pi \bigl(\tilde{\omega}^2+1\bigr)}-\frac{\tilde{K}^2}{4\pi}\frac{\tilde{\omega}^2-2}{(\tilde{\omega}^2+1)(\tilde{\omega}^2+4)}-\frac{\tilde{K}^3}{4\pi}\frac{ \tilde{\omega}^4-17 \tilde{\omega}^2+6}{ \bigl(\tilde{\omega}^2+1\bigr)^2 \bigl(\tilde{\omega}^2+4\bigr) \bigl(\tilde{\omega}^2+9\bigr)}+{\cal O}(K^4).$$ We thus arrive at $$\int_0^{2\pi}n(\psi,\omega)\cos(\psi)\,d\psi=\tfrac{1}{2}\tilde{K}\frac{1}{1+\tilde{\omega}^2}+\tfrac{1}{4}\tilde{K}^3\frac{2\tilde{\omega}^2-1}{(1+\tilde{\omega}^2)^2(4+\tilde{\omega}^2)}+{\cal O}(K^4).$$ This then gives the equation for $\sigma$, $$ \sigma= \tfrac{1}{2} K \sigma\int_{-\infty}^{\infty} g\left(D \omega+\omega_{0}\right) \frac{d \omega}{\omega^{2}+1}$$ $$\qquad-\frac{K^{3} \sigma^{3}}{4 D^{3}} \int_{-\infty}^{\infty} g\left(D \omega+\omega_{0}\right)\left\{\frac{1}{\omega^{2}+4}-\frac{\color{red}{\omega^2}}{\left(\omega^{2}+1\right)^{2}}\right\} d \omega+{\cal O}(K^4). $$ The term of order $K$ agrees with the formula in the OP, the term of order $K^3$ is slightly different, the OP has $\omega$ instead of the red $\omega^2$. I think this is a typo (the integrand is an even function of $\omega$).


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