# Vector calculus, identities

1. Apr 28, 2010

### MaxManus

1. The problem statement, all variables and given/known data

Show that v$$\nabla$$v = $$\nabla$$xvxv
v · ∇v = ∇(0.5v2 + c × v

c=∇ × v

My attempt
∇(A · B)= B · ∇A + A · ∇B + B×(∇×A) + A×(∇×B)

Replace A and B with V

∇(v · v)= v · ∇v + v · ∇v + v×(∇×v) + v×(∇×v)

v · ∇v = ∇(0.5v2 - v×(∇×v)

Is v×(∇×v) = =∇ × v × v?
And am I on the right track?

I can't fint such a rule in my textbook

2. Apr 28, 2010

### lanedance

doesn't vxv = 0?

3. Apr 28, 2010

### gabbagabbahey

What you've written doesn't make much sense....Do you mean $v\matbf{\nabla}v=(\mathbf{\nabla}\times \textbf{v})\times\textbf{v}+(\textbf{v}\cdot\mathbf{\nabla})\textbf{v}$ ?

Well, $\textbf{a}\times\textbf{b}=-\textbf{b}\times\textbf{a}$, so you tell us whether or not $(\mathbf{\nabla}\times \textbf{v})\times\textbf{v}=\textbf{v}\times(\mathbf{\nabla}\times \textbf{v})$

Last edited: Apr 28, 2010
4. Apr 28, 2010

### MaxManus

Thanks to both of you.
The first line is wrong

I was suppose to show that
v · $$\nabla$$v = ∇$$\nabla$$(0.5v2) + c × v

c=∇$$\nabla$$× v.
I then usend
∇$$\nabla$$(A · B)=B · $$\nabla$$A + A · $$\nabla$$B + B×($$\nabla$$×A) + A×($$\nabla$$×B)

Replaced A and B with v

And got

v$$\nabla$$v = $$\nabla$$(0.5v2) - v×($$\nabla$$×v)

Using A×B = - B×A

- v×($$\nabla$$×v) = ($$\nabla$$×vv

And I have:

v · $$\nabla$$v = ∇$$\nabla$$(0.5v2) + c × v

c=∇$$\nabla$$× v.

Am I right?

Thanks again for all the help

5. Apr 28, 2010

### gabbagabbahey

Looks good to me

6. Apr 28, 2010

Thanks.