Derivative of inverse of second-order tensor with respect to tensor using indices

Question

my question is related to evaluating the derivatives of products of tensors using indices. I have gone through the questions already asked on math.stackexchange, (search on tensor derivative yields 902 results, only this one is closer but still unanswered) but after going through them for one week, I couldn't find the answer. So unfortunately here comes another question related to tensor derivatives.

I have a tensor ${\bf F}$ (deformation gradient in solid mechanics) and I have dozens of relations where I need derivatives of tensorial functions $$\mathbb G(\bf F) = \frac{\partial {\boldsymbol \sigma}(\bf F)}{\partial \bf F}$$
I know that derivative of $\bf F$ with itself is the fourth rank identity tensor $\mathbb I$: $$ \frac{\partial \bf F}{\partial \bf F} = {\bf I}\overline {\otimes} {\bf I} = \mathbb I $$ or in indical notation: $$ \frac{\partial F_{ij}}{\partial F_{pq}} = \delta_{ip} \delta_{jq} $$ given that components of $\bf F$ are not inter-related through further dependency.

The problem comes in when evaluating product of two tensors, example: $$ \frac{\partial (\bf F^{-1}\cdot \bf F)}{\partial \bf F} = \frac{\partial (F^{-1}_{ij}\cdot F_{jk})}{\partial F_{pq}} {\bf e_i}\otimes{\bf e_k}\otimes{\bf e_p}\otimes{\bf e_q} = {\bf 0} $$ Question: is it correct to write above equation, using chain rule as: $$ \begin{split} \frac{\partial (\bf F^{-1}\cdot \bf F)}{\partial \bf F} &= \frac {\partial\bf F^{-1}}{\partial \bf F} \cdot {\bf F} + {\bf F^{-1}} \cdot \frac {\partial\bf F}{\partial \bf F} \\ &= \left[ \frac{\partial F^{-1}_{ij}}{\partial F_{pq}}\,F_{jk} + F^{-1}_{ij}\,\frac{\partial F_{jk}}{\partial F_{pq}} \right] {\bf e_i}\otimes{\bf e_k}\otimes{\bf e_p}\otimes{\bf e_q} = {\bf 0} \end{split} $$ The terms in square brackets now become: $$ \begin{split} \, & \frac{\partial F^{-1}_{ij}}{\partial F_{pq}}\,F_{jk} + F^{-1}_{ij}\,\frac{\partial F_{jk}}{\partial F_{pq}} = 0 \\ \Longrightarrow & \frac{\partial F^{-1}_{ij}}{\partial F_{pq}} = - F^{-1}_{ij}\,\frac{\partial F_{jk}}{\partial F_{pq}} F^{-1}_{jk} \\ &\quad \qquad = - F^{-1}_{ij}\,\delta_{jp} \delta_{kq} F^{-1}_{jk} \end{split} $$ So the final result is: $$ \frac{\partial F^{-1}_{ij}}{\partial F_{pq}} = - F^{-1}_{ip}\, F^{-1}_{jq} $$ However, in Matrix Cook Book (page 10 eq. 60) (MCB), and this Wikipedia article, the result is: $$ \frac{\partial F^{-1}_{ij}}{\partial F_{pq}} = - F^{-1}_{ip}\, F^{-1}_{qj} $$ The interchanged indices mean that my formulation has transposed the "actual" result (as per MCB and Wikipedia).

What is wrong with the derivation method using indices, which I have used.

Thanks.

Note: please note that my question is not about the derivative of inverse of a tensor, but the method itself, i.e. derivative of an expressions consisting of products of tensors, with respect to a tensor!

Answer 1

Here, the Euclidean space $\Bbb R^3$ is described by an orthonormal basis. The notation with indices is incorrect, since the index $j$ appears three times in one term (see e.g. this article about the Einstein notation). Moreover, the dot-product of two second-order tensors $\boldsymbol{F}$ and $\boldsymbol{F}^{-1}$ satisfies $$ F_{\ell k} F^{-1}_{kj} = \delta_{\ell j} \, . $$ Since $j$ is repeated twice in one term, it is correct to rename it $\ell$: $$ \frac{\partial F_{ij}^{-1}}{\partial F_{pq}} F_{jk} + F_{ij}^{-1} \frac{\partial F_{jk}}{\partial F_{pq}} = \frac{\partial F_{i\ell}^{-1}}{\partial F_{pq}} F_{\ell k} + F_{i\ell}^{-1} \frac{\partial F_{\ell k}}{\partial F_{pq}} \, . $$ Hence, multiplying by $F^{-1}_{kj}$ on the right, one has \begin{aligned} \frac{\partial F_{ij}^{-1}}{\partial F_{pq}} &= - F_{i\ell}^{-1} \frac{\partial F_{\ell k}}{\partial F_{pq}} F_{kj}^{-1} \\ &= -F_{i\ell}^{-1} \delta_{\ell p} \delta_{k q} F_{kj}^{-1} \\ &= -F_{i p}^{-1} F_{qj}^{-1} \, . \end{aligned}