In this review of the Kuramoto model, Eq. 14 is obtained by expanding the following integral in powers of $K r$,

$$ r = K r \int_{-\pi/2}^{\pi/2}\cos^2(\theta) g(K r \sin{\theta}) \mathrm{d}\theta $$ where $g(\omega)$ is some unknown function (though is later taken to be the p.d.f. of the Lorentz distribution $g(\omega) = (\gamma/\pi)/(\gamma^2 + \omega^2)$, for some parameter $\gamma$). The asymptotic expression they obtain via the expansion is $$ r \sim \sqrt{\frac{8(K-K_{c})}{-K_{c}^{3}g''(0)}} $$ where $K_c = 2/(\pi g(0))$. I don't see how you can expand this integral in powers of $Kr$ (I think this means a Taylor expansion in a new variable $z=Kr$ at $z=0$) without knowledge of $g$. Not much information is available, lots of other reviews appear to also omit this step. Is it a straightforward expansion? How is Eq. 14 obtained?


1 Answer 1


Presumably (because this is physics) $g$ is analytic in a neighborhood of the origin, so we can Taylor expand.

I think the really important part in deriving this equation for $r$ is the fact that the contribution from the first-order derivative term in $g$ is zero. Just underneath equation (10), they assume that the frequency distribution is an even function, i.e. $g(\omega) = g(-\omega)$. This implies that $g'(0) = 0$ and for that matter all the odd-order derivatives of $g$ are zero as well. You could also observe that $\cos^2\theta\cdot\sin\theta$ is an odd function, so even if $g'(0)$ were non-zero, that term would integrate to zero anyway.

If we Taylor expand to second order in $Kr$ we get:

$$\begin{align} 1 & = K\int_{-\pi/2}^{+\pi/2}\cos^2\theta\cdot g(Kr\sin\theta)d\theta \\ & = K\int_{-\pi/2}^{+\pi/2}\cos^2\theta\left(g(0) + \frac{g''(0)}{2}K^2r^2\sin^2\theta + \mathscr{O}(K^4r^4)\right)d\theta \end{align}$$

Do you see where to go from here?

  • 2
    $\begingroup$ Yes that's great, thanks. Maybe they should say find a power series for $g$ in $K r \sin(\theta)$. Then you can write $K_c = 2/(\pi g(0))$ and rearrange, having integrated term by term. $\endgroup$
    – apg
    Oct 21, 2022 at 17:00

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .