MHB Index Notation Proof: Proving $\nabla\cdot\left(\phi\textbf{u}\right)$

[SamJohannes]

Hi Everyone!

I'm looking to prove $\nabla\cdot\left(\phi\textbf{u}\right)=\phi\nabla\cdot\textbf{u} + \textbf{u}\cdot\nabla\phi$ in index notation where u is a vector and phi is a scalar field.

I'm unsure how to represent phi in index notation. For instance, is the first line like
${\partial}_{i}\phi{u}_{i}$ with phi represented without an index?

I've sort of been put in the deep end within my course and any guidance would be greatly appreciated.

 

[Opalg]

Hi Everyone!

I'm looking to prove $\nabla\cdot\left(\phi\textbf{u}\right)=\phi\nabla\cdot\textbf{u} + \textbf{u}\cdot\nabla\phi$ in index notation where u is a vector and phi is a scalar field.

I'm unsure how to represent phi in index notation. For instance, is the first line like
${\partial}_{i}\phi{u}_{i}$ with phi represented without an index?

Yes. More exactly, you should write it as $\partial_{i}(\phi{u}_{i})$, and then use the product rule for differentiation to say that this is equal to $\phi(\partial_{i}{u}_{i}) + (\partial_{i}\phi){u}_{i}.$

 

[SamJohannes]

Ah, I see.

So I got:

$\nabla\cdot\left(\phi\textbf{u}\right)=\partial_{i}\left(\phi{u}_{i}\right)=\phi\left(\partial_{i}{u}_{i}\right)+{u}_{i}\left(\partial_{i}\phi\right)=\phi\left(\nabla\cdot\textbf{u}\right)+\textbf{u}\cdot\left(\nabla\phi\right)$

 

[SamJohannes]

So I've done a heap more. But now I've come up against some more tricky ones. They are
i)$\nabla(u\cdot v)=(u\cdot\nabla)v+(v\cdot\nabla)u+u\times(\nabla\times v)+v\times (\nabla \times u)$
and
ii)$u\times (\nabla\times u) = \frac{1}{2}\nabla (u\cdot u) - (u\cdot\nabla )u$

For number one I really have no idea.

For number two I have
$u\times (\nabla\times u) =\varepsilon_{ijk}\varepsilon_{kmn}{u}_{j}\partial_{m}{u}_{n}=\varepsilon_{kij}\varepsilon_{kmn}{u}_{j}\partial_{m}{u}_{n}=(\delta_{im}\delta_{jn}-\delta_{in}\delta_{jm}){u}_{j}\partial_{m}{u}_{n}$
$=\delta_{im}\delta_{jn}{u}_{j}\partial_{m}{u}_{n}-\delta_{in}\delta_{jm}{u}_{j}\partial_{m}{u}_{n}={u}_{n}\partial_{i}{u}_{n}-{u}_{m}\partial_{m}{u}_{i}$
which seems close but I'm not certain...