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Laplacian(Nabla x v)

  1. May 17, 2008 #1
    1. The problem statement, all variables and given/known data

    In cylindrical:

    Get [tex]\frac{1}{\rho} \frac{d}{dp} \left( \rho\frac{d^2 v}{d\rho^2}\right) - \frac{1}{\rho^2}\frac{dv}{dp} = 0[/tex]

    Out of [tex]\nabla^2\left(\nabla \times \vec{v}\right) = 0[/tex]

    where [tex]\vec{v} = v(\rho)\hat{z}[/tex]

    3. The attempt at a solution

    I get [tex]\nabla \times \vec{v} = -\frac{dv}{dp}\hat{\theta}[/tex]

    But how do I apply the Laplacian? I can't even get that out of Maple.
    Last edited: May 17, 2008
  2. jcsd
  3. May 17, 2008 #2
    Last edited: May 17, 2008
  4. May 17, 2008 #3
    Its a pain, but doable. You only have a theta direction, as a function of radius - so it'll simplify. Plug and chug.
  5. May 19, 2008 #4
    So I should be applying [tex]\nable^2 f = \frac{1}{\rho} \frac{d}{d\rho}\left(\rho\frac{df}{d\rho}\right) + \frac{1}{\rho^2}\frac{d^2f}{d\theta^2} + \frac{d^2f}{dz^2}[/tex] (all should be partials)

    to simply [tex]f = -\frac{dv}{d\rho}\hat{\theta}[/tex]?

    and use the derivatives of the unit vectors here http://mathworld.wolfram.com/CylindricalCoordinates.html?
    Last edited: May 19, 2008
  6. May 19, 2008 #5
    I got [tex]\left[ \frac{1}{\rho^2}\frac{dv}{dp} - \frac{1}{\rho} \frac{d}{dp} \left( \rho\frac{d^2 v}{d\rho^2}\right) \right] \hat{\theta} = 0[/tex]

    This making sense anyone? Thanks
  7. May 19, 2008 #6
    Thats exactly what you were looking for right?
  8. May 19, 2008 #7
    Yeah. I guess applying the Laplacian to a vector messed me up a bit.

    I also wonder why they don't give the answer in vector form...
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