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Navier Stokes Equation

  1. Dec 28, 2003 #1
    I recently came across the vector version of the Navier Stokes equations for fluid flow.

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

    Ok, all is well until [tex]\bigtriangleup[/tex]. 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 [tex] \bigtriangledown[/tex] to the appropriate [tex]\bigtriangleup[/tex]
     
    Last edited: Dec 29, 2003
  2. jcsd
  3. Dec 29, 2003 #2
    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.
     
  4. Dec 29, 2003 #3
    Yes that is the Laplacian. Apparently they use that delta to represent it, it is also written as [tex]\bigtriangledown^2[/tex]
     
  5. Dec 29, 2003 #4
    [tex]\nabla^2\boldsymbol{v}=\nabla\left(\nabla\cdot\boldsymbol{v}\right)-\nabla\times\left(\nabla\times\boldsymbol{v}\right)[/tex]
     
  6. Dec 29, 2003 #5
    So let me make sure I have this straight.

    [tex]\bigtriangledown^2 \ v = grad \ div \ v - curl \ curl \ v[/tex]
     
  7. Dec 29, 2003 #6
    I believe that is the correct interpretation.
     
  8. Jan 4, 2004 #7
    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
     
  9. Jan 4, 2004 #8
    these operations are not associative, so you should not remove the parantheses.
     
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