If $S$ is a symmetric matrix, then rewriting $\int{S:\nabla \phi} dx$

Question

I am trying to prove that:

If $S$ is a symmetric matrix, then one can rewrite $\int{S:\nabla \phi} \text{dx}$ as $\int{S:D(\phi)} \text{dx}$, where $D(\phi)$ is the symmetric gradient of $\phi$.

Any kind of hint will be helpful! (Question was to only rewrite the expression, however after looking through some other notes, I believe that putting a condition of $S$ being symmetric should be sufficient?)

Answer 1

I don't think this has anything to do with the integral. If $S=S^T$ then \begin{align} S : \nabla \phi = \sum_{i,j} S_{ij} \partial_j \phi_i &= \frac12\left(\sum_{i,j} S_{ij} \partial_j \phi_i + \sum_{i,j} S_{ij} \partial_j \phi_i\right) \\ &= \frac12 \left(\sum_{i,j} S_{ij} \partial_j \phi_i + \sum_{i,j} S_{\color{red}{ji}} \partial_j \phi_i\right)\\ &= \frac12 \left(\sum_{i,j} S_{ij} \partial_j \phi_i + \sum_{j',i'} S_{i',j'} \partial_{i'} \phi_{j'}\right) \\ &= \sum_{i,j}S_{ij} \frac12 \left( \partial_j \phi_i + \partial_i \phi_j \right)\\ &= S : D(\phi). \end{align} Or, if you prefer to see less indices, since $A^T:B = A:B^T$ \begin{align} S: \nabla \phi &= \frac12(S + S^T ): \nabla \phi \\ &= \frac12\left(S:\nabla \phi + S^T: \nabla \phi\right) \\ &= \frac12\left(S:\nabla \phi + S: \nabla \phi^T\right) \\ &= S : \frac12(\nabla \phi + \nabla\phi^T)\\ & = S:D(\phi). \end{align}