most recent 30 from http://mathoverflow.net 2013-05-22T10:11:39Z http://mathoverflow.net/feeds/question/115139 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/115139/finding-kuramoto-model-coupling-strength-with-limits Mr Robert 2012-12-02T02:44:57Z 2012-12-02T15:07:53Z

<p>The following is an equation describing the coupled phases of N oscillators according to the Kuramoto model:</p> <p>$ 1 = K \int_{-\pi/2}^{\pi/2}cos^2(\theta)g(KRsin(\theta))d\theta$</p> <p>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.</p> <p>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.</p> <p>How else can you solve this limit?</p>

http://mathoverflow.net/questions/115139/finding-kuramoto-model-coupling-strength-with-limits/115181#115181 Jon 2012-12-02T15:07:53Z 2012-12-02T15:07:53Z

<p>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 $$ K=\sqrt{\frac{8}{\pi}}\sigma $$</p>