Power series expansion of the order parameter in the Kuramoto model

Question

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?

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?