On the wikipedia article for stokes drift (https://en.wikipedia.org/wiki/Stokes_drift) they show that for:
$\dot{\zeta} = u\sin(k\zeta - wt)$ has approximate solution (by perturbation theory):
$\zeta(\zeta_0,t) = \zeta_0+\frac{u}{w}\cos(k\zeta_0-wt)-\frac{ku^2}{4w^2}\sin(2(k\zeta_0-wt))+\frac{ku^2t}{2w}$.
Could someone spell out how this was obtained? If not by perturbation theory then perhaps some other way.
A similar, but not identical, expression can be obtained using perturbation theory. (I suspect there are missing terms in the given approximation, as it does not satisfy the initial condition $\zeta(\zeta_0,0)=\zeta_0$.)
In order to derive the perturbative solution, let's write $\zeta(t)$ as a power series in $u$: $$ \zeta(t)=f_0(t)+f_1(t)u+f_2(t)u^2+\ldots \tag{1} $$ Plugging $(1)$ into the ODE $\dot{\zeta}=u\sin(k\zeta-wt)$, and expanding both sides in powers of $u$, we obtain \begin{align} \dot{f_0}+\dot{f_1}u+\dot{f_2}u^2+\ldots &=u\sin(k(f_0+f_1u+f_2u^2+\ldots)-wt) \\ &=u\sin(kf_0-wt)+ku^2\cos(kf_0-wt)f_1+O(u^3). \tag{2} \end{align} Equating terms multiplying equal powers of $u$, we obtain the ODEs \begin{align} \dot{f_0}&=0, \tag{3} \\ \dot{f_1}&=\sin(kf_0-wt), \tag{4} \\ \dot{f_2}&=k\cos(kf_0-wt)f_1. \tag{5} \\ \end{align} Equation $(3)$ implies $f_0(t)=C$. Since $\zeta(0)=\zeta_0$ does not depend on $u$, we must have $C=\zeta_0$. Plugging this result into $(4)$ and integrating, we find $$ f_1(t)=\int_0^t\sin(k\zeta_0-wt')\,dt'=\frac{1}{w}\left[\cos(k\zeta_0-wt)-\cos(k\zeta_0)\right]. \tag{6} $$ Equation $(5)$ then yields \begin{align} f_2(t)&=\frac{k}{w}\int_0^t\cos(k\zeta_0-wt')\left[\cos(k\zeta_0-wt')-\cos(k\zeta_0)\right]dt' \\ &=\frac{kt}{2w}-\frac{k}{4w^2}\left[\sin(2(k\zeta_0-wt))-\sin(2k\zeta_0)\right] \\ &\quad\,+\frac{k\cos(k\zeta_0)}{w^2}\left[\sin(k\zeta_0-wt)-\sin(k\zeta_0)\right]. \tag{7} \end{align} Collecting all pieces together, we finally obtain \begin{align} \zeta(\zeta_0,t)&=\zeta_0+\frac{u}{w}\left[\cos(k\zeta_0-wt)\color{red}{-\cos(k\zeta_0)}\right] \\ &\quad-\frac{ku^2}{4w^2}\left[\sin(2(k\zeta_0-wt))\color{red}{-\sin(2k\zeta_0)}\right] \\ &\quad\, \color{red}{+\,\frac{ku^2\cos(k\zeta_0)}{w^2}\left[\sin(k\zeta_0-wt)-\sin(k\zeta_0)\right]}+\frac{ku^2t}{2w}, \tag{8} \end{align} where the terms highlighted in red are the ones missing in the approximate solution that you have quoted.