# Navier Stokes Equation

1. Dec 28, 2003

I recently came across the vector version of the Navier Stokes equations for fluid flow.

$$\displaystyle{\frac{\partial \mathbf{u}}{\partial \mathbf{t}}} + ( \mathbf{u} \cdot \bigtriangledown) \mathbf{u} = v \bigtriangleup \mathbf{u} - grad \ p$$

Ok, all is well until $$\bigtriangleup$$. I know this represents the laplacian. What is the formulation of the Laplacian for this since it is a vector? Is it just simply the second partials dot product with the respective terms of the vector? Or is it something else?

edit: changed text where I say problem is $$\bigtriangledown$$ to the appropriate $$\bigtriangleup$$

Last edited: Dec 29, 2003
2. Dec 29, 2003

### pnaj

For whatever reason, I can't seem to use laTex ... have to do a bit more reading first.

But, are you sure that the delta you've picked out is the Laplacian ... looks like grad to me.

3. Dec 29, 2003

Yes that is the Laplacian. Apparently they use that delta to represent it, it is also written as $$\bigtriangledown^2$$

4. Dec 29, 2003

### master_coda

$$\nabla^2\boldsymbol{v}=\nabla\left(\nabla\cdot\boldsymbol{v}\right)-\nabla\times\left(\nabla\times\boldsymbol{v}\right)$$

5. Dec 29, 2003

So let me make sure I have this straight.

$$\bigtriangledown^2 \ v = grad \ div \ v - curl \ curl \ v$$

6. Dec 29, 2003

### master_coda

I believe that is the correct interpretation.

7. Jan 4, 2004

### dhris

Well, that is an identity for the operators. But why don't you like the idea of the usual Laplacian acting on a vector? It's just a derivative operator, which is allowable on vectors as long as you remember that the basis vectors also have to be differentiated.

dhris

8. Jan 4, 2004

### lethe

these operations are not associative, so you should not remove the parantheses.