I'm trying to find an approximate solution to a nonlinear differential equation. It involves something to the effect of $\frac{d\Psi}{ds} = \sin{\Psi} + \dots$ , where $\Psi$ is a small variable.
If I were to expand $\Psi\sim\Psi_0+\epsilon\Psi_1+\dots$ , I have a few options for how to proceed:
- $\sin{\Psi}\approx\Psi+\dots\approx\Psi_0+\epsilon\Psi_1+\dots$, then take the O(1) equation
- $\sin{\Psi}\approx[\Psi_0+\epsilon\Psi_1]-\frac{1}{6}[\Psi_0+\epsilon\Psi_1]^3+\dots = \sin{\Psi_0}+O(\epsilon)$
- $\sin{\Psi}\approx\Psi+\frac{1}{6}\Psi^3+\dots\approx\Psi_0+\frac{1}{6}\Psi_0^3+O(\epsilon)$
etc.
Is there a best way of doing this? How should one expand a small variable within a function that can itself be expanded? I'm leaning towards the first option because it makes my work easier, but I'm less certain that this is the correct one when finding the O($\epsilon$) solution.
To perform an asympotoic expansion it helps make things clear if the only 'small' parameter is $\epsilon$. You say that you want to consider small $\Psi$, so we expand as $$ \Psi(s) = \epsilon \Psi_1(s)+\epsilon^2 \Psi_2(s) + \ldots $$ Substituting this into $\sin$ yields $$ \sin(\Psi) = (\epsilon \Psi_1(t)+\epsilon^2 \Psi_2(t)+\ldots) - \frac{1} {6} (\epsilon \Psi_1(t)+\epsilon^2 \Psi_2(t)+\dots)^3+\ldots = \epsilon \Psi_1(t)+\epsilon^2 \Psi_2(t) + O(\epsilon^3) $$ Note that the third order term will include contributions from both $\Psi_3$ and $\Psi_1^3$.