# Laplacian(Nabla x v)

## 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.

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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

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?

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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

Thats exactly what you were looking for right?

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...