Shallow water approximation, justification of $\frac{Dv}{Dt}\sim 0$? In the shallow water approximation (a.k.a. long wave approximation) it is often taken that the material derivative:
$$\frac{Dv}{Dt}\sim 0$$
Where $v$ is the vertical component of the velocity. 
I have found a justification of this a posterior here but was wondering if there where any justifications we could make for this to be the case a priori?
 A: The material derivative of the vertical (or $\hat{\mathbf{z}}$ direction) fluid velocity, $\mathbf{w}$, is given by:
$$
\frac{ D \ \mathbf{w} }{ D t } = \partial_{t} \ \mathbf{w} + \mathbf{u} \cdot \nabla \mathbf{w} \tag{1}
$$
where $\mathbf{u}$ is the bulk flow velocity orthogonal to $\mathbf{w}$ and along the wave propagation direction (i.e., the direction of the phase velocity and/or group velocity, or the $\hat{\mathbf{x}}$ direction) and $\nabla$ is the gradient operator.  If we reduce this to a 2D problem where the slope of the sea floor varies only along, say, the $\hat{\mathbf{x}}$ direction, then Equation 1 reduces to:
$$
\frac{ D \ w }{ D t } = \partial_{t} \ w + u \ \partial_{x} w \tag{2}
$$
The trick here is to assume that $w \propto e^{i \left( k \ x - \omega \ t \right)}$ (called linearization or making a linear approximation), thus Equation 2 goes to:
$$
\partial_{t} \ w + u \ \partial_{x} w \rightarrow -i \ \omega \ w + u \ \left( i \ k \ w \right) \tag{3}
$$
If $u$ represents then phase velocity, then the second term on the right-hand side goes to $i \ \omega/k \ \left( k \ w \right) = i \ \omega \ w$.  Thus, the two terms cancel.
