Laplacian of divergence equals divergence of vector Laplacian

Question

Suppose $u$ is a vector including $(u_x, u_y, u_z)$. Is this correct?

$\nabla^2(\nabla\cdot u)=\nabla\cdot(\nabla^2 u)$

I think it is correct, but I don't know how to prove it.

Answer 1

They appear to be the same, expanding out in components. On either side, for example, $u_x$ has derivatives $xxx, xyy, xzz$ taken and summed.

Answer 2

According to Wikipedia, the vector Laplacian reads $$ \nabla^2 {\bf u} = \nabla (\nabla\cdot {\bf u}) - \nabla\times (\nabla \times {\bf u}) . $$ Taking the divergence, $$ \nabla\cdot (\nabla^2 {\bf u}) = \nabla\cdot \nabla (\nabla\cdot {\bf u}) = \nabla^2 (\nabla\cdot {\bf u}) $$ where we have used the fact that the curl of any vector field is divergence-free, and the definition of the Laplace operator.