The Laplacian of a Dyad in Cartesian using Indicial Notation

Question

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.

Answer 1

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 $(4)$.