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Laplacian(Nabla x v)

  • Thread starter cscott
  • Start date
  • #1
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Homework Statement



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]


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:

Answers and Replies

  • #3
275
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Its a pain, but doable. You only have a theta direction, as a function of radius - so it'll simplify. Plug and chug.
http://en.wikipedia.org/wiki/Laplacian
 
  • #4
782
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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:
  • #5
782
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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
 
  • #6
275
2
Thats exactly what you were looking for right?
 
  • #7
782
1
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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