The Laplacian of a Dyad in Cartesian using Indicial Notation 1. 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: $$\nabla^2 ({\vec u \vec v}) = \vec v \nabla^2 {\vec u} + \vec u \nabla^2 {\vec v} + 2\nabla {\vec u} \cdot {(\nabla \vec v)}^T $$ 2. Relevant equations Per my professor's notes, the Laplacian of a dyad (also a tensor) is given as: $$ \nabla^2 {\mathbf {S}} = \nabla \cdot {S_{ij,k} \mathbf{e_{i}e_{j}e_{k}}} = S_{ij,kk} \mathbf{e_{i}e_{j}} $$ 3. 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.

The "missing term" in the OP comes from applying the product rule for differentiating

> $$\
abla (\phi \psi)=\phi \
abla(\psi)+\
abla(\phi)\psi \tag 1$$

and

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

Note that we can write

$$\begin{align} \
abla^2(\vec u\vec v)&=\hat x_i\hat x_j\
abla^2 (u_iv_j)\\\\\\\ &=\hat x_i\hat x_j\
abla \cdot \
abla(u_iv_j)\\\\\\\ &=\hat x_i\hat x_j\
abla \cdot (u_i\
abla (v_j) + \
abla(u_i)v_j )\tag 3\\\\\\\ &=\hat x_i\hat x_j \left(u_i\
abla^2(v_j)+2\
abla (u_i)\cdot \
abla (v_j)+v_j\
abla ^2(u_i)\right)\tag 4\\\\\\\ &=\vec u\
abla^2(\vec v)+2\
abla(\vec u)\cdot \
abla(\vec v)+\vec v\
abla ^2(\vec u) \end{align}$$

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 $\vec A=\
abla v_j$ ($\vec A=\
abla u_i$) to arrive at $(4)$.