Waves: Small Amplitude Approximation Why does the assumption of waves amplitude being small, compared to the wavelength, (as used for surface gravity waves in water) lead to the equivalent condition:
$$ {\partial v \over \partial t } >> { (v \nabla )v}$$
This seems to be one of the basic assumptions of certain classes of fluid mechanic problems.
I spend two hours in the library today reading the applicable chapters of about nine books and not in one the explanation was more detailed that to claim that $ (v \nabla )v$ must be on the order of $v^2<<1$.
 A: Background
The nonlinear steepening term, $\mathbf{v} \cdot \nabla \mathbf{v}$, is proportional to (unit-wise) a speed squared divided by a scale length, $L$.  It is more important for larger values of $v$ and/or smaller values of $L$.
To illustrate this, let's assume a linear approximation.  This allows us to use the following:
$$
\partial_{t} \rightarrow -i \ \omega \\
\nabla \rightarrow +i \ \mathbf{k}
$$
where $\omega$ is the angular frequency, $\mathbf{k}$ is the wavenumber, and we have assumed all quantities can be written as $Q \approx Q_{o} + \delta Q$, where:
$$
Q_{o} \equiv \text{ constant} \\
\delta Q \propto e^{i \left( \mathbf{k} \cdot \mathbf{x} - \omega \ t \right)}
$$
In other words, we assume that $\partial_{t} Q_{o} = \nabla Q_{o} = 0$.
Using these assumptions, we can rewrite the total derivative as:
$$
\frac{ d }{ dt } = \partial_{t} + \mathbf{v} \cdot \nabla \\
\approx -i \ \omega + i \ \mathbf{v} \cdot \mathbf{k}
$$
which shows that we have:
$$
\frac{ d \mathbf{v} }{ dt } \rightarrow \frac{ d \delta \mathbf{v} }{ dt } \approx -i \ \omega \ \delta \mathbf{v} + i \ \left(\delta \mathbf{v} \cdot \mathbf{k}\right) \delta \mathbf{v}
$$
where in the one-dimensional limit with $\delta \mathbf{v}$ entirely parallel to $\mathbf{k}$ reduces to:
$$
\frac{ d \delta v }{ dt } \sim -i \ k \left[ \frac{ \omega }{ k } \ \delta v + \left( \delta v \right)^{2} \right]
$$
Answer

Why does the assumption of waves amplitude being small, compared to the wavelength, (as used for surface gravity waves in water) lead to the equivalent condition: $$ \partial_{t} \mathbf{v} \gg \mathbf{v} \cdot \nabla \mathbf{v} $$

The idea is that the term $\delta v$ is the wave amplitude, namely that there is a velocity fluctuation about some mean value.  If the wavelength is large, that means $k$ is small which corresponds to a large phase velocity, $\omega/k$, assuming a constant frequency.
Thus, the small amplitude, $\delta v$, means that we have the following:
$$
\frac{ \omega }{ k } \ \delta v \gg \left( \delta v \right)^{2} \\
\text{or, in another form:} \\
\frac{ \omega }{ k } \gg \delta v
$$
It is just another way of stating that the velocity fluctuations are not large compared to the quasi-static terms.  One makes assumptions like this to avoid things like nonlinear wave steepening, dispersion, strong damping, etc.
Side Note
You will find this linearization method used in nearly all branches of physics for numerous topics for any system that can be decomposed into superpositions of multiple terms.  It is incredibly handy and often annoyingly accurate.
