The Laplacian of a Dyad in Cartesian using Indicial Notation 
*

*The problem statement, all variables and given/known data


Given the dyad formed by two arbitrary position vector fields, u and v, use indicial notation in Cartesian coordinates to prove:
$$\nabla^2 ({\vec u \vec v}) = \vec v \nabla^2 {\vec u} + \vec u \nabla^2 {\vec v} + 2\nabla {\vec u} \cdot {(\nabla \vec v)}^T
$$


*Relevant equations


Per my professor's notes, the Laplacian of a dyad (also a tensor) is given as:
$$
\nabla^2 {\mathbf {S}} = \nabla \cdot {S_{ij,k} \mathbf{e_{i}e_{j}e_{k}}} = S_{ij,kk} \mathbf{e_{i}e_{j}}
$$


*The attempt at a solution


$$
\nabla^2 {\mathbf {uv}} = (u_{i}v_{j})_{,kk} = u_{i,kk}v_{j} + u_{i}v_{j,kk} \\
u_{i,kk}v_{j} + u_{i}v_{j,kk} = \vec v \nabla^2 {\vec u} + \vec u \nabla^2 {\vec v}
$$
I don't know where the following terms come from:
$$
2\nabla {\vec u} \cdot {(\nabla \vec v)}^T
$$
Does anyone have any suggestions? I feel that I am missing a step or something.
 A: The "missing term" in the OP comes from applying the product rule for differentiating 

$$\nabla (\phi \psi)=\phi \nabla(\psi)+\nabla(\phi)\psi \tag 1$$

and 

$$\nabla \cdot(\phi \vec A)=\phi \nabla\cdot(\vec A)+\nabla(\phi)\cdot \vec A \tag 2$$

Note that we can write 
$$\begin{align}
\nabla^2(\vec u\vec v)&=\hat x_i\hat x_j\nabla^2 (u_iv_j)\\\\
&=\hat x_i\hat x_j\nabla \cdot \nabla(u_iv_j)\\\\
&=\hat x_i\hat x_j\nabla \cdot (u_i\nabla (v_j) + \nabla(u_i)v_j )\tag 3\\\\
&=\hat x_i\hat x_j \left(u_i\nabla^2(v_j)+2\nabla (u_i)\cdot \nabla (v_j)+v_j\nabla ^2(u_i)\right)\tag 4\\\\
&=\vec u\nabla^2(\vec v)+2\nabla(\vec u)\cdot \nabla(\vec v)+\vec v\nabla ^2(\vec u)
\end{align}$$
as was to be shown!  
Note that we applied $(1)$ with $\phi =u_i$ and $\psi=v_j$ to arrive at $(3)$ and applied $(2)$ with $\phi= u_i$ ($\phi= v_j$)and $\vec A=\nabla v_j$ ($\vec A=\nabla u_i$) to arrive at $(4)$.
A: We need to evaluate $\nabla^2 \{ \mathbf{a} \mathbf{b}\}$, where $\mathbf{a}$ and $\mathbf{b}$ are vectors
Let's write $\nabla^2$ as $\nabla_x^2 + \nabla_y^2 + \nabla_z^2$
First, let's evaluate $\nabla_x^2 \{\mathbf{a} \mathbf{b} \}$
$$
\begin{align}
\nabla_x^2 \{\mathbf{a} \mathbf{b}\} &= \nabla_x \nabla_x \{\mathbf{a} \mathbf{b}\} \\\\
 &= \nabla_x \{   \left(  \nabla_x \mathbf{a}     \right) \mathbf{b} +  \mathbf{a} \nabla_x \mathbf{b}         \} \\\\
 &= \left(\nabla_x^2 \mathbf{a} \right) \mathbf{b} + 2 \left( \nabla_x \mathbf{a} \right) \left( \nabla_x \mathbf{b} \right) + \mathbf{a} \nabla_x^2 \mathbf{b}
\end{align} \tag{1}$$
Similarly we evaluate the y and z components of the laplacian as under
$$
\begin{align}
\nabla_y^2 \{\mathbf{a} \mathbf{b}\} &= \nabla_y \nabla_y \{\mathbf{a} \mathbf{b}\} \\\\
&= \nabla_y \{   \left(  \nabla_y \mathbf{a}     \right) \mathbf{b} +  \mathbf{a} \nabla_y \mathbf{b}         \} \\\\
&= \left(\nabla_y^2 \mathbf{a} \right) \mathbf{b} + 2 \left( \nabla_y \mathbf{a} \right) \left( \nabla_y \mathbf{b} \right) + \mathbf{a} \nabla_y^2 \mathbf{b}
\end{align} \tag{2}$$
$$
\begin{align}
\nabla_z^2 \{\mathbf{a} \mathbf{b}\} &= \nabla_z \nabla_z \{\mathbf{a} \mathbf{b}\} \\\\
&= \nabla_z \{   \left(  \nabla_z \mathbf{a}     \right) \mathbf{b}  +  \mathbf{a} \nabla_z \mathbf{b}         \} \\\\
&= \left(\nabla_z^2 \mathbf{a} \right) \mathbf{b} + 2 \left( \nabla_z \mathbf{a} \right) \left( \nabla_z \mathbf{b} \right) + \mathbf{a} \nabla_z^2 \mathbf{b}
\end{align} \tag{3} $$
Now, combining the above equations and noting that
$$
\begin{align}
\left( \nabla_x \mathbf{a} \right) \left( \nabla_x \mathbf{b} \right) + \left( \nabla_y \mathbf{a} \right) \left( \nabla_y \mathbf{b} \right) +  \left( \nabla_z \mathbf{a} \right) \left( \nabla_z \mathbf{b} \right) =   \left(\vec{\nabla} \mathbf{a} \right) \cdot \left( \vec{\nabla} \mathbf{b} \right)
\end{align} \tag{4}$$
, we get
$$
\begin{align}
    \nabla^2 \{ \mathbf{a} \mathbf{b} \} &= \left(\nabla^2 \mathbf{a} \right) \mathbf{b} + 2 \left(\vec{\nabla} \mathbf{a} \right) \cdot \left( \vec{\nabla} \mathbf{b} \right) + \mathbf{a} \nabla^2 \mathbf{b}
\end{align} \tag{5}$$
Remarks :-

*

*Note that $\nabla_x, \nabla_y$ and $\nabla_z$ are scalars and hence we could write $\nabla_x \{ \mathbf{a} \mathbf{b}    \} = \left( \nabla_x \mathbf{a} \right) \mathbf{b} + \mathbf{a} \{ \nabla_x \mathbf{b}\}$ and similarly for the y and z components.


*Since $\mathbf{a} \mathbf{b}$ is a dyadic product, we made sure that we don't change this order throughout the calculation.


*My final result differs from that by Mark Viola because in evaluating the term $\hat x_i\hat x_j v_j\nabla ^2(u_i)$ eq(4) of his answer, he has changed the order of the dyadic product $\hat x_i\hat x_j$, which is not allowed. According to me the said term should be evaluated as
$$ 
\hat x_i\hat x_j v_j\nabla ^2(u_i) = \nabla ^2(u_i) v_j \hat x_i\hat x_j = \nabla ^2(u_i) \hat x_i v_j \hat{x_j} = \left( \nabla^2  \vec{u} \right) \vec{v} \tag{6}
$$
Note that since $v_j$ and $\nabla^2 \left( u_i \right)$ are scalars and hence their orders could be swapped but the order of the dyadic product $\hat{x_i} \hat{x_j} $ couldn't be and it is because of this that my final result differs from Mark and the one in the question itself.
