I'm reading in a fluid dynamics book and in it the author shortens an equation using identities my rusty vector calculus brain cannot reproduce.

$\displaystyle \vec{e} \cdot \frac{\partial}{\partial t}(\rho \vec{u}) = -\nabla\cdot (\rho\vec{u})\cdot\vec{e} - \rho(\vec{u}\cdot\nabla)\vec{u}\cdot\vec{e} - (\nabla p)\cdot\vec{e} + \rho\vec{b}\cdot\vec{e} $

The author turns the left side of the equation into: $\displaystyle -\nabla\cdot(p\vec{e} + \rho\vec{u}(\vec{u}\cdot\vec{e})) + \rho\vec{b}\cdot\vec{e} $

Just to be clear; $\displaystyle \vec{u}$ is a vector valued function,

$\displaystyle \vec{e}$ is a fixed vector,

$\displaystyle \rho, p$ are scalar valued functions.

The first part is fine: $\displaystyle \nabla\cdot (p\vec{e}) = \nabla(p)\cdot\vec{e} + p(\nabla\cdot\vec{e})$

The divergence of a fixed vector is zero and so, $\displaystyle \nabla\cdot (p\vec{e}) = \nabla(p)\cdot\vec{e}$

Next I need to find $\displaystyle \nabla\cdot (\rho\vec{u}(\vec{u}\cdot\vec{e})) $

I am not sure what to do with is. $\displaystyle \rho$ is a scalar valued function, but so is I think $\displaystyle \vec{u}\cdot\vec{e}$. I know of the product rule between a scalar and a vector valued function, but what happens when there are two scalar valued functions?

Any suggestions are welcome, thanks.