Chain rule for scalar-valued function with tensor argument

Question

my question comes from continuum Mechanics, where we have the so-called deformation gradient $F$, which essentially is the Jacobian map of a transformation $x$. Then define $C=F^TF$ to be the right Cauchy-Green strain tensor. And finally we have a function $\sigma(F^TF)$ which is scalar valued.

Question: I need to take the derivative w.r.t. $F$ of $\sigma(F^TF)$, so of course I need to apply the chain rule: $$[D_F \sigma(F^TF)]_{ij} = \frac{\partial \sigma(C)}{\partial C_{kl}} \frac{\partial C_{kl}}{\partial F_{ij}}$$

and then I have that $\frac{\partial C_{kl}}{\partial F_{ij}}=\frac{\partial F_{mk} F_{ml}}{\partial F_{ij}} = \delta_{ij} \delta_{jk}F_{ml} + F_{mk} \delta_{im}\delta_{jl}$

However, I think that the result of the whole operation should be something like $D_C \sigma(C) F^T$, but I am really stuck! Any help is highly appreciated!

Answer 1

In coordinates, we have : $$\frac{\partial \sigma}{\partial F_{ij}}= \frac{ \partial\sigma}{\partial C_{kl}} \cdot \left( \delta_{jk}F_{il} + F_{ik}\delta_{jl}\right) = F_{il} \frac{ \partial\sigma}{\partial C_{jl}} + F_{ik}\frac{ \partial\sigma}{\partial C_{kj}} $$

The synthetic notation for this is : $$D_F\sigma = F \cdot (D_C\sigma)^T + F\cdot D_C\sigma$$

Answer 2

$ \def\s{\sigma}\def\p{\partial} \def\L{\left}\def\R{\right}\def\LR#1{\L(#1\R)} \def\trace#1{\operatorname{Tr}\LR{#1}} \def\grad#1#2{\frac{\p #1}{\p #2}} $Given the variable definitions $$C = F^TF,\quad\quad \s = \s(C),\quad\quad G=\grad{\s}{C}$$ Note that since $C$ is symmetric, the gradient is too: $\;G=G^T$

Use the known gradient to write the differential. Then perform a change of variable from $C\to F.\,$ Then recover the new gradient. $$\eqalign{ d\s &= G:dC \\&= G:(F^TdF+dF^TF) \\&= (G^T+G):F^TdF \\&= 2FG:dF \\ \grad{\s}{F} &= 2FG \\ \\ }$$

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In the steps above, a colon is used to denote the matrix inner product, which is a convenient notation for the trace $$\eqalign{ A:B &= \sum_{i=1}^m\sum_{j=1}^n A_{ij}B_{ij} \;=\; \trace{AB^T} \\ A:A &= \big\|A\big\|^2_F \\ }$$ The properties of the underlying trace function allow the terms in such a product to be rearranged in many different ways, e.g. $$\eqalign{ A:B &= B:A \\ A:B &= A^T:B^T \\ C:AB &= A^TC:B = CB^T:A \\ }$$

Notice that higher-order tensors like $\LR{\grad{C_{k\ell}}{F_{ij}}}$ were not needed. This is true for most problems in Matrix Calculus.