Proving the equality in the attached Image

Question

I am trying to learn the mathematics behind Finite elements as an engineer. can anyone help me with the proof below ? I have almost figured out the solution. The problem is I get a multiplication of matrices that have unequal interior dimensions and I don't Know how to proceed from there. Note that the sigma is a 2*2 symmetric matrices.

Question: prove the equality

Here is what I have done so far:

my solution

In the first step, I have assumed that using the transpose of the divergence vector is possible. I don't know whether this assumption is valid though.

thanks.

Answer 1

Under convention that $\nabla$ is a column-vector like operator and $v$ is also a column vector: \begin{align} \nabla \cdot (\sigma v) &= \nabla (\sigma v)^T = \nabla (v^T \sigma) = \begin{pmatrix} \partial_x & \partial_y \end{pmatrix} \begin{pmatrix} v_1 \sigma_{11} + v_2 \sigma_{12} \\ v_1 \sigma_{21} + v_2 \sigma_{22} \end{pmatrix} \\ &= \partial_x(v_1 \sigma_{11} + v_2 \sigma_{12}) + \partial_y (v_1 \sigma_{21} + v_2 \sigma_{22}) \\ & = v_1 \partial_x \sigma_{11} + \sigma_{11} \partial_x v_1 + v_2 \partial_x \sigma_{21} + \sigma_{21} \partial_x v_2 \\ &+ v_1 \partial_y \sigma_{21} + \sigma_{21} \partial_y v_1 + v_2 \partial_y \sigma_{22} + \sigma_{22} \partial_y v_2 \\ & = \partial_x(\sigma_{11})v_1 + \partial_y(\sigma_{21})v_1 + \partial_x(\sigma_{12})v_2 + \partial_y(\sigma_{22})v_2 \\ & + \sigma_{11} \partial_x v_1 + \sigma_{12}\partial_x v_2 + \sigma_{21}\partial_y v_1 + \sigma_{22} \partial_y v_2 \\ &= \Big[\begin{pmatrix} \partial_x & \partial_y \end{pmatrix} \sigma \Big] \cdot v + \sigma \colon \Bigg[\begin{pmatrix} \partial_x \\ \partial_y \end{pmatrix} \begin{pmatrix}v_1 & v_2 \end{pmatrix} \Bigg] \\ & = \Big[\nabla^T \sigma \Big] \cdot v + \sigma \colon \nabla v^T \\ & \overset{\sigma = \sigma^T}{=} \Big[\nabla \cdot \sigma^T \Big] \cdot v + \sigma \colon \nabla v^T \end{align}

To generalize this to multiple dimensions (and simplify things considerably), typically index notation / einstein summation convention is employed: \begin{align} \partial_i (\sigma_{ij} v_j ) \overset{\text{Prod. rule of diff.}}{=} \partial_i (\sigma_{ij} ) v_j + \sigma_{ij} \partial_i v_j \overset{\sigma = \sigma^T}{=} \partial_i (\sigma_{ji} ) v_j + \sigma_{ij} \partial_i v_j = (\nabla \cdot \sigma^T \big) \cdot v_j + \sigma \colon \nabla v^T\end{align}