Navier Stokes Equation

[Brad_Ad23]

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

[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.

[Brad_Ad23]

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

[master_coda]

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

[Brad_Ad23]

So let me make sure I have this straight.

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

[master_coda]

I believe that is the correct interpretation.

[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

[lethe]

Originally posted by Brad_Ad23
So let me make sure I have this straight.

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

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