# Finding Kuramoto Model coupling strength with limits?

The following is an equation describing the coupled phases of N oscillators according to the Kuramoto model:

$$1 = K \int_{-\pi/2}^{\pi/2}\cos^{2}\left(\theta\right)\, {\rm g}\left(KR\sin\left(\theta\right)\right)\,{\rm d}\theta$$

We assume that g is a symmetric distribution with zero mean, since the equation is written from the point of view of a reference frame rotating together with the oscillator.

I need to take the $\lim_{R \rightarrow0^+}$ to solve for the coupling strength Kc. This can't be solved with Wolfram Alpha since g is not read as a gaussian distribution but instead as a constant.

How else can you solve this limit?

-

This problem can be approached by a series in $KR$ and assuming for $g$ a Gaussian distribution. So, we have to manage $$1=K\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\cos^2\theta\frac{1}{\sqrt{2\pi}\sigma} e^{-\frac{K^2R^2\sin^2\theta}{2\sigma^2}}d\theta.$$ The limit $R\rightarrow 0^+$ can be taken under the integrale but we prefer a series in $KR$ that yields $$1=K\frac{1}{\sqrt{2\pi}\sigma}\left(\frac{\pi}{2}-\frac{\pi}{2^4 \sigma ^2} K^2 R^2+\frac{\pi}{2^7 \sigma ^4}K^4 R^4+O(K^6R^6)\right).$$ The required limit provides $$K=\sqrt{\frac{8}{\pi}}\sigma$$
@Dirk It did not assume any particular form of ${\rm g}\pars{x}$ besides the information provided by the OP. Thanks. –  Felix Marin Jan 19 '14 at 10:34