Divergence of velocity of incompressible fluid in uniform gravity

Hi!

The velocity field as a function of poisition of an incompressible fluid in a uniform acceleration field, such as a waterfall accelerated by gravity can be found as follows:
The position is [itex]\vec{x}[/itex].
The velocity field is [itex] \vec{v} = \frac{d\vec{x}}{dt}[/itex].
The constant acceleration field is [itex]\vec{a}=\frac{d\vec{v}}{dt} =\frac{d\vec{v}}{dt}\frac{d\vec{x}}{d\vec{x}}= \frac{d\vec{v}}{d\vec{x}}\frac{d\vec{x}}{dt} = \frac{d\vec{v}}{d\vec{x}}\vec{v}[/itex].
Now we can find the velocity as a function of position by rearranging the above and integrating:
[itex]\int\vec{a}\cdot d\vec{x} = \int\vec{v}\cdot d\vec{v}[/itex]

[itex]\vec{a}\cdot \vec{x} = \frac{\vec{v}^2}{2}[/itex]

[itex]\vec{v} = (2\vec{a}\vec{x})^{\frac{1}{2}}[/itex]

The divergence of the velocity field is then

[itex]\vec\nabla\cdot\vec{v} = \vec\nabla\cdot(2\vec{a}\vec{x})^{\frac{1}{2}} = (2\vec{a})^{\frac{1}{2}}\vec\nabla\cdot \frac{\vec{x}^{-\frac{1}{2}}}{2} =\sqrt{\frac{\vec{a}}{2\vec{x}}}[/itex]

But shouldn't [itex]\vec\nabla\cdot\vec{v} = 0[/itex] in an incompressible fluid? The stramlines are all parallel to one another, as they follow the gravitational field, so they shouldn't diverge.

Where has my thinking gone wrong?