A couple of weeks ago my professor gave me a paper on Stokes' law and how to derive it from the equations of motion and continuity. (http://www.ux.uis.no/~finjord/pdf/flu/stokes.pdf)
In one of the parts (the one to derive eq 4.18) I found the following step, where $u_0$ is a constant vector and $f$ only depends on $r$: $\nabla^2($grad div $ - \nabla^2)(\nabla f \times u_0) = - (\nabla^2)^2(\nabla f \times u_0)$
However I don't see why the first term in the LHS disappears. I did check it on WolframAlpha, but this only tells me it's zero and not why.
Can anybody explain why it is zero?
Thanks in advance.
This is really just straightforward calculation, assuming that $f$ has at least continuous second partial derivatives (Schwarz theorem): $$\nabla \times u_0 = \partial_i f \phantom{0} u_{0j}\phantom{0} \varepsilon_{ijk} \vec{e}_k$$ therefore $$div (\nabla \times u_0) = \partial_{ik} f \phantom{0} u_{0j}\phantom{0} \varepsilon_{ijk} = 0$$ If it is not clear why this term is zero, I suggest to think about symmetry properties.