I'm reading a paper about an oscillators model (Kuramoto model), and came across the following Fokker-Planck equation.
r(t) is defined as such:$re^{i\omega}=\int \int e^{i\theta}\rho(t,\theta.\omega )g(\omega)d\omega d\theta $ rho is a density (which had the constraint to integrate to unity) and is now perturbed in this way: $\rho (\omega,t,\theta) = \epsilon \eta(\omega,t,\theta) + 1/2\pi \ and \ \epsilon<<1$
now the Focker Planck equation reads : $ \varepsilon {\frac{\partial \eta }{\partial t}}=\varepsilon D{\frac{\partial ^{2}\eta }{\partial \theta ^{2}}}- {\frac{\partial }{\partial \theta}}\left [ \left ( \frac{1}{2 \pi}+\epsilon\eta \right )v(\omega,t,\theta) \right ] $
...then it says "Let's consider this equation at lowest order of $\eta$. To find the O($\epsilon$) contribution in the bracketed term, we observe that r(t) is O($\epsilon$). More specifically we find : $ r(t)=\epsilon r_1(t)+O(\epsilon^2) \\where:r_1e^{i\psi }=\int \int e^{i\theta}\eta(t,\theta.\omega )g(\omega)d\omega d\theta $
So Ok, they rewrite the first equation plugging in the modified density (eqn.2). Here is what I get:
$r(t)e^{i\psi}=\int\int e^{i\theta}\rho(\theta,t,\omega)g(\omega)d\omega d\theta\\=\int\int e^{i\theta}\left[\frac{1}{2\pi}+\epsilon\eta(\theta,t,\omega)\right]g(\omega)d\omega d\theta\\=\epsilon\int\int e^{i\theta}\eta(\theta,t,\omega)g(\omega)d\omega d\theta+\frac{1}{2\pi}\int\int e^{i\theta}g(\omega)d\omega d\theta$
$\Leftrightarrow r(t)=\epsilon e^{-i\psi}\int\int e^{i\theta}\eta(\theta,t,\omega)g(\omega)d\omega d\theta+\frac{e^{-i\psi}}{2\pi}\int_{-\infty}^{+\infty}\int_{0}^{2\pi}g(\omega)e^{i\theta}d\omega d\theta\\=\epsilon r_{1}(t)+\frac{e^{-i\psi}}{2\pi}\int_{-\infty}^{+\infty}g(\omega)d\omega\int_{0}^{2\pi}e^{i\theta}d\theta$ ...nothing is of order epsilon squared here ? Also I won't bet too much on it but it seems to me that the second integral vanishes because of the $[e^i\theta]_0^{2\pi}$...
Thanks.