 Solution

Put $\theta =x-t$. Taking the gradient of the velocity potential givesThe motion is not generally incompressible; a velocity potential ensures irrotationality, not zero divergence. Both components have zero time average at each fixed point because their sine and cosine averages vanish.

The fluid acceleration is the material derivative $D\mathbf{u}/Dt={\partial }_t\mathbf{u}+(\mathbf{u}\cdot \nabla )\mathbf{u}$. Direct differentiation givesAveraging over one period at fixed $(x,y)$ therefore yieldsThis need not be the derivative of the mean velocity: the nonlinear advective term remains after averaging.

The dyed particle obeys the Lagrangian trajectory equationsExpand $x=ϵx_1+{ϵ}^2{x}_2+O({ϵ}^3)$ and $y=ϵy_1+{ϵ}^2{y}_2+O({ϵ}^3)$, for bounded times. The first-order equations give $x_1=0$, ${\dot{y}}_1=\cos t$, so $y_1=\sin t$. At second order,Integrating with zero initial values verifiesThe velocity averaged along this particle over $0\le t\le 2\pi$ is its displacement divided by $2\pi$. HenceThis Stokes drift is a Lagrangian mean; it does not contradict the zero Eulerian mean velocity at a fixed point. The secular term is a small-parameter expansion, not an assertion of uniform accuracy as $t\to \infty$.