What is the vector Laplacian of a vector $AS$ I am trying to better understand vector Laplacians and, in an effort to do that, I am trying to calculate the Laplacian of a known vector $A$ by hand.
I am wondering if anyone can verify the following:
Given $A = [2xy^2z]\;\hat{\imath}\;+\;[0.5y^2z]\;\hat{\jmath}\;+\;[2z^2x^{1/2}]\;\hat{k}$
Find $\nabla^2A$.

I have proceeded solving this using the identity $\nabla^2A=\nabla(\nabla \cdot A)-\nabla \times(\nabla \times A)$
My answer is:
$\nabla ^2 A = [4zx^{1/2}-4xz]\;\hat{\imath}\;+\;[z]\;\hat{\jmath}\;+\;[4y^2+4x^{1/2}+0.5z^2x^{-3/2}]\;\hat{k}$
See below for the full work:




$\nabla\cdot A = 2y^2z+yz+4zx^{1/2} = \alpha $
$\nabla\alpha = [2zx^{-1/2}]\hat{\imath}+[4yz+z]\hat{\jmath}+[2y^2+y+4x^{-1/2}]\hat{k}$
Therefore, we have solved for $\nabla (\nabla\cdot A)$
Next step is to find $\nabla \times (\nabla \times A)$:
$\nabla \times A = [-0.5y^2]\hat{\imath}-[z^2x^{-1/2}-2xy^2]\hat{\jmath}-[4xyz]\hat{k} = B$
$\nabla \times B = [-4xz-2zx^{-1/2}]\hat{\imath}-[-4yz]\hat{\jmath}+[-0.5z^2x^{-3/2}-2y^2+y]\hat{k}$
Therefore, we have solved for $\nabla\times (\nabla\times A)$
Putting it all together:
$\nabla^2 A=\nabla\alpha -\nabla \times B = [4zx^{1/2}-4xz]\;\hat{\imath}\;+\;[z]\;\hat{\jmath}\;+\;[4y^2+4x^{1/2}+0.5z^2x^{-3/2}]\;\hat{k}$

If you find any errors, please point them out! Also, if there is an easier way to solve this by hand, I would also appreciate the info!
Thanks

Your $\nabla \cdot A$ is wrong because $\frac{\partial}{\partial x}(2xy^2)=2y^2$ (without $z$). So it is difficult to control the other steps.
Anyway, in Cartesian coordinates, the vector laplacian of a vector field $A=(A_x,A_y,A_z)$ is:
$$
\nabla^2 A=(\nabla^2 A_x,\nabla^2 A_y,\nabla^2 A_z)
$$
that is more simple to evaluate
