# Vector calculus identities mess

• Jan 11th 2012, 07:49 PM
Mollier
Vector calculus identities mess
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.
• Jan 11th 2012, 10:36 PM
FernandoRevilla
Re: Vector calculus identities mess
Quote:

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

We have $\displaystyle \nabla\cdot(f\vec{F})=\textrm{div}(f\vec{F})=f\; \textrm{div}\vec{F}+\vec{F}\cdot\nabla f=f(\nabla\cdot \vec{F})+\vec{F}\cdot \nabla f$ . In our case, $\displaystyle f=\rho (\vec{u}\cdot\vec{e})$ and $\displaystyle \vec{F}=\vec{u}$ .
• Jan 12th 2012, 08:51 PM
Mollier
Re: Vector calculus identities mess
Hi,

I tried that but ran into a few problems while trying to make the result look identical to the textbook.

$\displaystyle \nabla\cdot(\rho(\vec{u}\cdot\vec{e})\vec{u}) = \rho((\vec{u}\cdot\vec{e})(\nabla\cdot(\vec{u}))) + \vec{u}\cdot\nabla(\rho(\vec{u}\cdot\vec{e}))$

I play a bit with the last term and get,

$\displaystyle \vec{u}\cdot\nabla(\rho(\vec{u}\cdot\vec{e})) = \vec{u}\cdot [\rho\nabla(\vec{u}\cdot\vec{e})+(\vec{u}\cdot\vec{ e})\nabla(\rho)]$

by using the product rule for gradients. Now I know that there is an identity that tells us how to take the gradient of a vector dot product. I have here a dot product between a vector valued functon ($\displaystyle \vec{u}$) and a fixed vector ($\displaystyle \vec{e}$). If I treat both as vectors I get,

$\displaystyle \rho\nabla(\vec{u}\cdot\vec{e}) = \rho[(\vec{u}\cdot\nabla)\vec{e}+(\vec{e}\cdot\nabla ) \vec{u}+\vec{u}\times (\nabla\times\vec{e})+\vec{e}\times(\nabla \times \vec{u})]$

The second and third term are zero leaving me with

$\displaystyle \rho\nabla(\vec{u}\cdot\vec{e}) = \rho[(\vec{u}\cdot\nabla)\vec{e}+\vec{e}\times(\nabla \times \vec{u})]$

To get what I want, I need to let the last term be zero. Then,

$\displaystyle \rho\nabla(\vec{u}\cdot\vec{e})\cdot\vec{u} = \rho(\vec{u}\cdot\nabla)\vec{u}\cdot\vec{e}$

This looks like one of the terms in the textbook (see my first post).
I really hope I am missing something because this is a mess :) Thanks!