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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?

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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 $$ K=\sqrt{\frac{8}{\pi}}\sigma $$

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$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}% \newcommand{\dd}{{\rm d}}% \newcommand{\down}{\downarrow}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\half}{{1 \over 2}}% \newcommand{\ic}{{\rm i}}% \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\ol}[1]{\overline{#1}}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ \begin{align} 1 &= K \int_{-\pi/2}^{\pi/2}\cos^{2}\pars{\theta} {\rm g}\pars{KR\sin\pars{\theta}}\,\dd\theta\ \\[3mm]&\stackrel{R \sim 0}{\sim}\ K \int_{-\pi/2}^{\pi/2}\cos^{2}\pars{\theta} \braces{{\rm g}\pars{0} + {\rm g}'\pars{0}KR\sin\pars{\theta} + \half\,{\rm g}''\pars{0}\bracks{KR\sin\pars{\theta}}^{2}}\,\dd\theta \\[3mm]&=K{\rm g}\pars{0} \underbrace{\int_{-\pi/2}^{\pi/2}\cos^{2}\pars{\theta}\,\dd\theta} _{\ds{=\ {\pi \over 2}}} + {1 \over 8}\,K^{3}{\rm g}''\pars{0}R^{2} \underbrace{\int_{-\pi/2}^{\pi/2}\sin^{2}\pars{2\theta}\,\dd\theta}_{\ds{=\ {\pi \over 2}}} \end{align} $\color{#00f}{\large\ds{K = {2 \over \pi{\rm g}\pars{0}}}}$

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    $\begingroup$ Would you mind to add some remark on how this gives something new to the previous answer? $\endgroup$
    – Dirk
    Commented Jan 19, 2014 at 10:33
  • $\begingroup$ @Dirk It did not assume any particular form of ${\rm g}\pars{x}$ besides the information provided by the OP. Thanks. $\endgroup$
    – Felix Marin
    Commented Jan 19, 2014 at 10:34
  • $\begingroup$ Yes, and I thought it would be good to have some text along with the answer. $\endgroup$
    – Dirk
    Commented Jan 19, 2014 at 10:38
  • $\begingroup$ This answer is absolutely impossible to read with mathjax turned off. $\endgroup$
    – Ricardo Andrade
    Commented Jan 19, 2014 at 19:42

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