Talk:Stokes drift

1D example of Falkovich

Animation over time $t,$ of the Lagrangian parcel position $\xi (\xi _{0},t)$ as a function of $\xi _{0}.$ The red line is the exact solution and the blue line is the 2nd-order perturbation-series solution. The parameters are $\omega =k=1$ and ${\hat {u}}=0.3.$ Note the slightly larger Stokes drift velocity ${\overline {u}}_{S}$ and lower phase speed $\sigma /k$ of the exact solution.
The red dots show Lagrangian drifter positions $x=\xi (\xi _{0},t)$ for equidistant labels $\xi _{0}$ evolving with time.

Note that the Stokes drift in Falkovich' example of 1D flow has an exact solution. In this case, the Eulerian velocity is taken as $u={\hat {u}}\,\cos(kx-\omega t)$ – where instead of the sine as used by Falkovich, the cosine is used because of symmetry conditions of $\xi (\xi _{0},t)$ at $\xi _{0}=0$ and $t=0.$ Now the Lagrangian parcel position is denoted as $x=\xi (\xi _{0},t),$ with the position label $\xi _{0}$ taken equal to $\xi _{0}=x.$ The position $\xi (\xi _{0},t)$ is the solution of:

${\frac {\partial \xi }{\partial t}}=u(\xi ,t)={\hat {u}}\,\cos(k\xi -\omega t).$

The additional condition on $\xi (\xi _{0},t)$ is that at $t=0$ the Stokes drift is equal to zero, i.e. that the spatial mean value of the oscillation $\xi (\xi _{0},t)-\xi _{0}$ is zero: $\left\langle \xi (\xi _{0},0)-\xi _{0}\right\rangle =0.$ Then the progressive wave solution is:

$\xi ={\frac {2}{k}}\arctan \,\left({\frac {\sigma }{\omega +k{\hat {u}}}}\,\tan({\tfrac {1}{2}}kx-{\tfrac {1}{2}}\sigma t)\right)+{\frac {\omega t+n\,2\pi }{k}},$

where

$\sigma =\omega \,{\sqrt {1-\left({\frac {k{\hat {u}}}{\omega }}\right)^{2}}}\qquad {\text{and}}\qquad n=\operatorname {round} \left({\frac {kx-\sigma t}{2\pi }}\right),$

with the round function denoting rounding to the nearest integer.

It can directly be observed that the Lagrangian moving parcel experiences a different (lower) frequency $\sigma$ than the Eulerian velocity frequency $\omega .$ The Stokes drift velocity ${\overline {u}}_{S}$ is simply the difference in positions after one Lagrangian wave period $2\pi /\sigma$ has passed, divided by the Lagrangian wave period. So the exact expression for the Stokes drift velocity is:

${\overline {u}}_{S}={\frac {\sigma }{2\pi }}\left[\xi (0,2\pi /\sigma )-\xi (0,0)\right]={\frac {\omega -\sigma }{k}}={\frac {\omega }{k}}\left(1-{\sqrt {1-\left({\frac {k{\hat {u}}}{\omega }}\right)^{\!2}}}\right).$

It has the Taylor expansion:

${\overline {u}}_{S}={\frac {\omega }{k}}\left[{\frac {1}{2}}\,\left({\frac {k{\hat {u}}}{\omega }}\right)^{\!2}+{\frac {1}{8}}\,\left({\frac {k{\hat {u}}}{\omega }}\right)^{\!4}+{\mathcal {O}}\left(\left({\frac {k{\hat {u}}}{\omega }}\right)^{\!6}\right)\right],$

in agreement with Falkovich' perturbation solution. Which is in this case – with a cosine for the velocity field, $u={\hat {u}}\,\cos(kx-\omega t)$:

$\xi =\xi _{0}+{\frac {1}{k}}\,\left[-{\frac {k{\hat {u}}}{\omega }}\,\sin(k\xi _{0}-\omega t)+{\frac {1}{4}}\,\left({\frac {k{\hat {u}}}{\omega }}\right)^{\!2}\,\sin(2k\xi _{0}-2\omega t)+{\frac {1}{2}}\,\left({\frac {k{\hat {u}}}{\omega }}\right)^{\!2}\,\omega t+{\mathcal {O}}\left(\left({\frac {k{\hat {u}}}{\omega }}\right)^{\!3}\right)\right],\quad {\text{and}}$

$\sigma =\omega \,\left[1-{\frac {1}{2}}\,\left({\frac {k{\hat {u}}}{\omega }}\right)^{2}\right]+{\mathcal {O}}\left(\left({\frac {k{\hat {u}}}{\omega }}\right)^{\!4}\right).$

-- Crowsnest (talk) 15:29, 6 March 2017 (UTC)