# Consider this equation at lowest order in epsilon , what does it mean?

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.

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Perhaps it'd be clearer what that means if one looked at what happened immediately after in the paper? –  anon May 5 '12 at 6:35
I edited a bit. Thanks. –  zebullon May 5 '12 at 7:19
I don't see how we could possibly have anything to say about this from what you've written, since $\rho$ doesn't occur anywhere outside the last line. I think you'll have to provide more context, or ideally link to the paper that you're reading. –  joriki May 5 '12 at 10:21
I added some details, and this is from "stability incoherence in a population of coupled oscillators" p.7. –  zebullon May 5 '12 at 11:15

## 1 Answer

In perturbation theory you have an exact solution for a problem that is close to your real problem. You might have a potential that is close to a harmonic oscillator, which you express as $V(x)=kx^2+\epsilon W(x)$. You know an exact solution in the limit $\epsilon =0$ and are interested in an approximate solution when $\epsilon \ne 0$. If $\epsilon$ is small, you hope the process will converge and the first or first few terms will be a good approximation. Often you can plug your altered potential into the equations and follow along the derivation of the exact solution. $\epsilon$ keeps track of the order of the terms. You often find the solution comes out as a power series in $\epsilon$. The first order term comes from the original solution reacting to the changed potential. The second order term comes from the perturbation to the original solution reacting to the changed potential and so on. You keep going until the soltution is as accurate as you want or you get tired.

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Thanks for your answer, this sounds like what I could understand from the wiki article that i checked before asking, though in that precise case : If I plug the new density (2) into the double integral (1) I get : $$r(t)=\varepsilon { r }_{ 1 }(t)+\frac { e^{-i\psi} }{ 2\pi } \iint { { e }^{ i(\theta ) }g(\omega ) }$$. ...And here I see no epsilon, or epsilon squared in the second r.h.s term ?! –  zebullon May 6 '12 at 11:28
@zebullon: the second line will multiply it by epsilon, as the integral has $\eta$ while the second line has $\epsilon \eta$ –  Ross Millikan May 6 '12 at 15:17
I edited my question to include how I derive from the plug-in step so that you can tell me where I'm wrong. Thanks. –  zebullon May 7 '12 at 3:15