Laplacian of Nabla x v in Cylindrical Coordinates

[cscott]

Homework Statement

In cylindrical:

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

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

where \vec{v} = v(\rho)\hat{z}

The Attempt at a Solution

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

But how do I apply the Laplacian? I can't even get that out of Maple.

 

[cscott]

I think I got it with this: http://mathworld.wolfram.com/VectorLaplacian.html

EDIT: maybe not

 

[lzkelley]

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

 

[cscott]

So I should be applying \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} (all should be partials)

to simply f = -\frac{dv}{d\rho}\hat{\theta}?

and use the derivatives of the unit vectors here http://mathworld.wolfram.com/CylindricalCoordinates.html?

 

[cscott]

I got \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

This making sense anyone? Thanks

 

[lzkelley]

Thats exactly what you were looking for right?

 

[cscott]

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...