demonstration of vector laplacian in cartesian coordinates

Question

I am stucked with the following demonstration. The vector laplacian formula is: $Δa = ∇(∇a) - ∇×(∇×a)$ , where $a$ is a vector field. I have to demonstrate that the vector laplacian in cartesian coordinates is: $Δa = (∇∇ax)ux +(∇∇ay)uy +(∇∇az)uz$ where: $ux$,$uy$ and $uz$ are the unit vectors, and $∇∇$ stands for nabla's operator to square. Thanks in advance.

Answer 1

\begin{eqnarray} \nabla^2 a &=& \nabla (\nabla \cdot a) - \nabla \times (\nabla \times a) \\ &=& \hat{\mathrm{e}}_i \left[ \partial_i \partial_j a_j- \varepsilon_{ijk} \varepsilon_{kbc} \partial_j \partial_b a_c\right] \\ &=& \hat{\mathrm{e}}_i \left[ \partial_i \partial_j a_j- (\delta_{ib} \delta_{jc}-\delta_{ic}\delta_{jb}) \partial_j \partial_b a_c\right]\\ &=& \hat{\mathrm{e}}_i \left[ \partial_i \partial_j a_j- \partial_j \partial_i a_j+ \partial_j \partial_j a_i\right] \\ &=& \hat{\mathrm{e}}_i \left[\partial_j \partial_j a_i\right] \, . \end{eqnarray}

Answer 2

I was also interested in this question so I am writing down what I came with. I've read on wiki that it can be seen as a particular case of the vector triple product formula (Lagrange's formula):

$\mathbf{u}\times (\mathbf{v}\times \mathbf{w}) = (\mathbf{u}\cdot\mathbf{w})\ \mathbf{v} - (\mathbf{u}\cdot\mathbf{v})\ \mathbf{w}$

The demonstration (also on wiki) for first coordinate is:

$\begin{align} (\mathbf{u} \times (\mathbf{v} \times \mathbf{w}))_x &= \mathbf{u}_y(\mathbf{v}_x\mathbf{w}_y - \mathbf{v}_y\mathbf{w}_x) - \mathbf{u}_z(\mathbf{v}_z\mathbf{w}_x - \mathbf{v}_x\mathbf{w}_z) \\ &= \mathbf{v}_x(\mathbf{u}_y\mathbf{w}_y + \mathbf{u}_z\mathbf{w}_z) - \mathbf{w}_x(\mathbf{u}_y\mathbf{v}_y + \mathbf{u}_z\mathbf{v}_z) \\ &= \mathbf{v}_x(\mathbf{u}_y\mathbf{w}_y + \mathbf{u}_z\mathbf{w}_z) - (\mathbf{u}_y\mathbf{v}_y + \mathbf{u}_z\mathbf{v}_z)\mathbf{w}_x + (\mathbf{u}_x\mathbf{v}_x\mathbf{w}_x - \mathbf{u}_x\mathbf{v}_x\mathbf{w}_x) \\ &= \mathbf{v}_x(\mathbf{u}_x\mathbf{w}_x + \mathbf{u}_y\mathbf{w}_y + \mathbf{u}_z\mathbf{w}_z) - (\mathbf{u}_x\mathbf{v}_x + \mathbf{u}_y\mathbf{v}_y + \mathbf{u}_z\mathbf{v}_z)\mathbf{w}_x \\ &= \mathbf{v}_x (\mathbf{u}\cdot\mathbf{w}) - (\mathbf{u}\cdot\mathbf{v})\mathbf{w}_x \end{align}$

(compared to initial demonstration, I've just made sure that the scalar is kept on the right when doing the substitution explained below to avoid any inconsistency)

now replacing $\mathbf{u}$ and $\mathbf{v}$ with $\nabla$ and $\mathbf{w}$ with $\mathbf{a}$, we get:

$ \begin{align} (\mathbf{\nabla} \times (\nabla \times \mathbf{a}))_x &= \left(\nabla (\nabla.\mathbf{a})\right)_x - (\nabla \cdot \nabla) a_x \\ &= \left(\nabla (\nabla.\mathbf{a})\right)_x - \Delta a_x \end{align} $