A Gradient of higher rank tensor

[chowdhury]

@anuttarasammyak : yes in most cases,

Query 1.)

here is what I try to follow
$$\nabla \cdot (\bf{\epsilon}^{S} \cdot \nabla \phi) = \nabla \cdot (\bf{e} : \nabla_{s} \bf{u}) $$
Now
$$\nabla_{i} \epsilon_{ij}^{S} \nabla_{j} \phi = \nabla_{i} \epsilon_{ij}^{S} \phi_{,j} = \epsilon_{ij}^{S}\nabla_{i} \phi_{,j} = \epsilon_{ij}^{S}\phi_{,ij}$$

I don't know how to full the full 3-index notation for e. It was immensely helpful by @Orodruin : With his insight, it turned out to be,

$$ \nabla \cdot (\bf{e} : \nabla_{s} \bf{u}) = \nabla_{i} e_{ijk} \nabla_{j} u_{k} = e_{ijk} u_{k, ji}$$

Hence
$$\epsilon_{ij}^{S}\phi_{,ij} = e_{ijk} u_{k, ji}$$

I do not understand what was the basis for the book to say symmetry condition for the e-coeficient in their index?
View attachment 297832

Query 2.)
$$ \bf{c^{eff}} = \bf{c^{E}} + e^{transpose} \cdot (\bf{(\epsilon^{S}})^{-1} \bf{e})$$

The notation my be incorrect, but the true nature is CORRECT in the above expression, then when I wanted to express in index notation, I get into trouble,

$$ c^{eff}_{IJ} = c^{E}_{IJ} + e^{transpose}_{iI} (\epsilon^{S}_{??})^{-1} e_{??})
$$

I cannot write in single index notation, meaning not combined capital I,J etc.

Query 3.) For this equation,

$$\nabla \cdot (\bf{c}^{E} : \nabla_{s}\bf{u}) -\rho \frac{\partial^2 \bf{u}}{\partial t^2} = - \nabla \cdot (\bf{e}^{transpose} \cdot \nabla \phi)$$

I derived
$$(c^{E}_{ijkl} u_{k,l})_{,j} - \rho u_{i,tt} = - (e^{transpose}_{ijk} \phi_{,k})_{,j} =- (\color{red}{e}_{kji} \phi_{,jk}) $$
But the book derived as in the picture above. There might be difference in the indices, I cannot reconcile between mine and the book.

Does anybody know how to derivemy above three queries? Thanks.