- #1

Dafe

- 145

- 0

## Homework Statement

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.

## Homework Equations

[tex] \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}

[/tex]

The author turns the left side of the equation into:

[tex]-\nabla\cdot(p\vec{e} + \rho\vec{u}(\vec{u}\cdot\vec{e})) + \rho\vec{b}\cdot\vec{e} [/tex]

Just to be clear;

[tex]\vec{u}[/tex] is a vector valued function,

[tex]\vec{e}[/tex] is a fixed vector,

[tex]\rho, p[/tex] are scalar valued functions.

## The Attempt at a Solution

The first part is fine:

[tex]\nabla\cdot (p\vec{e}) = \nabla(p)\cdot\vec{e} + p(\nabla\cdot\vec{e})[/tex]

The divergence of a fixed vector is zero and so,

[tex]\nabla\cdot (p\vec{e}) = \nabla(p)\cdot\vec{e}[/tex]

Next I need to find

[tex] \nabla\cdot (\rho\vec{u}(\vec{u}\cdot\vec{e})) [/tex]

I am not sure what to do with is. [tex]\rho[/tex] is a scalar valued function, but so is I think [tex]\vec{u}\cdot\vec{e}[/tex]. 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.