Solving Vector Calculus: (a+2b)∇(∇⋅u) - b∇x∇xu - (3a+2b)c∇T(r)=0

[mcfc]

I'm unsure how to do this problem:

(a + 2b)\nabla(\nabla \cdot \vec u) - b \nabla \times \nabla \times \vec u - (3a + 2b)c\nabla T(r)= \vec 0
\hat u = U_r \hat r + u_\theta \hat \theta +u_z \hat z
a,b,c constants
how would I solve this for u?

 

[Atropos]

i know that \nabla(\nabla\bullet\vec{u}) - \nabla\times\nabla\times\vec{u} = \nabla^{2}\vec{u}

i dunno, what about this?..

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