A fairly easy vector calculus identity question?

[quarky2001]

I'm working on simplifying a big physical expression (I don't like the Navier-Stokes equations at all anymore), and I'm curious how to simplify the following term:

$$\vec{v}\cdot (\vec{v}\cdot\nabla )\vec{v}$$

where v is a fluid velocity - i.e. definitely spatially varying.

I'm just not sure about the order of operations here.

 

[dextercioby]

The operator between the brackets is a differential operator which acts to the right

$$\vec{v}\cdot\nabla = v_i \partial _i$$

So your whole expression is nothing but

$$v_j v_i\partial_i v_j$$

a scalar.

 

[quarky2001]

Sorry - I'm not sure what you mean by that subscript notation.

I know v dotted with nabla is an operator on the v on the far right, and I know the result will be a scalar. I just don't know how I could re-write that expression in a simpler fashion.

 

[dextercioby]

Well, it's the handy tensor notation using Einstein's summing notation. Nothing more.

You can't write that in a simpler fashion, unless v is subject to some conditions (divergence-less for example).

 

[quarky2001]

Actually, I finally did manage to rewrite it. (although I'm not familiar with Einstein's tensor notation)

Assuming my math is correct, the following equality should hold:

$$\vec{v}\cdot (\vec{v}\cdot\nabla )\vec{v} = \frac{1}{2}(\vec{v}\cdot\nabla )v^2$$

 

[dextercioby]

Sure, you're right. But is that really simpler ? I can say it's shorter, it takes less space on a piece of paper.