Prove Laplacian of Dyad: Indicial Notation

[ganondorf29]

Homework Statement

[/B]
Given the dyad formed by two arbitrary position vector fields, u and v, use indicial notation in Cartesian coordinates to prove:

$$\nabla^2 ({\vec u \vec v}) = \vec v \nabla^2 {\vec u} + \vec u \nabla^2 {\vec v} + 2\nabla {\vec u} \cdot {(\nabla \vec v)}^T
$$

Homework Equations

[/B]
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

[/B]
$$
\nabla^2 {\mathbf {uv}} = (u_{i}v_{j})_{,kk} = u_{i,kk}v_{j} + u_{i}v_{j,kk} \\
u_{i,kk}v_{j} + u_{i}v_{j,kk} = \vec v \nabla^2 {\vec u} + \vec u \nabla^2 {\vec v}
$$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.

 

[fresh_42]

Homework Statement

[/B]
Given the dyad formed by two arbitrary position vector fields, u and v, use indicial notation in Cartesian coordinates to prove:

$$\nabla^2 ({\vec u \vec v}) = \vec v \nabla^2 {\vec u} + \vec u \nabla^2 {\vec v} + 2\nabla {\vec u} \cdot {(\nabla \vec v)}^T
$$

Homework Equations

[/B]
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

[/B]
$$
\nabla^2 {\mathbf {uv}} = (u_{i}v_{j})_{,kk} = u_{i,kk}v_{j} + u_{i}v_{j,kk} \\
u_{i,kk}v_{j} + u_{i}v_{j,kk} = \vec v \nabla^2 {\vec u} + \vec u \nabla^2 {\vec v}
$$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.

I'm not good with coordinates, so I leave this up to you.
But $\nabla (\mathbf{u}\mathbf{v}) = \mathbf{u}(\nabla \mathbf{v}) + (\nabla \mathbf{u})\mathbf{v}$ and the next differentiation gives you the third term (twice).