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Vector calculus

  1. Dec 2, 2009 #1
    I'm unsure how to do this problem:

    [itex](a + 2b)\nabla(\nabla \cdot \vec u) - b \nabla \times \nabla \times \vec u - (3a + 2b)c\nabla T(r)= \vec 0[/itex]
    [itex]\hat u = U_r \hat r + u_\theta \hat \theta +u_z \hat z[/itex]
    a,b,c constants
    how would I solve this for u?
     
  2. jcsd
  3. Dec 2, 2009 #2
    i know that [tex]\nabla[/tex]([tex]\nabla\bullet\vec{u}[/tex]) - [tex]\nabla\times\nabla\times\vec{u}[/tex] = [tex]\nabla^{2}\vec{u}[/tex]

    i dunno, what about this?..

    Solve this equation for [tex]\nabla\times\nabla\times\vec{u}[/tex], then substitute into your equation, and simplify the resulting equation. The first term of your equation should reduce to (a+b)[tex]\nabla[/tex]([tex]\nabla\bullet\vec{u}[/tex]) +b[tex]\nabla^{2}\vec{u}[/tex] - (3a+2b)c[tex]\nabla[/tex]T(r) = 0.

    You should be able to write out the laplacian explicitly and simplify all the individual components. If you can find some way to turn the value (3a+2b) into (3a+3b) you could probably simplify things a lot.
     
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