Talk:Stokes drift

1D example of Falkovich
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(\tfrac12 k x - \tfrac12\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[

\frac12\, \left( \frac{k\hat{u}}{\omega} \right)^{\!2} + \frac18\, \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)        + \frac14\, \left( \frac{k\hat{u}}{\omega} \right)^{\!2}\, \sin(2k\xi_0-2\omega t)          + \frac12\, \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 - \frac12\, \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)