I am readintg a tutorial regarding fluid mechanics and I encountered this strange $\delta$ notation:$$ \begin{aligned} \frac{\partial^2\Psi}{\partial\mathbf{F}\partial\mathbf{F}}:\delta\mathbf{F}&=\delta\left(\frac{\partial\Psi}{\partial\mathbf{F}}\right)\\ &=......(\text{other calculations}) \end{aligned} $$ Here, $\mathbf{F}$ is $3\times3$ matrix, $\Psi$ is a scalar function of $\mathbf{F}$.
I just get totally confused by the notation $\delta$ and the colon (:) used here. I guess that here the $\delta$ notation refers to "differential", just like the $d$ notation in, for example, $\int f(x)dx$. But through this tutorial, there are other places using the $d$ notation, and why do the author use a different manner here?
Also, what exactly does the ":" notation mean here? I suppose this means that the derivative is taken with respect to $\mathbf{F}$? I can't find enough evidance to prove this.
Besides, the author suggests in the following context that, we have:$$ \delta[\det(\mathbf{F})]=(\det\mathbf{F})\mathbf{F}^{-T}:\delta\mathbf{F} \tag{1} $$ As far as I know, if you take a derivative over a determinant of a matrix, say $X$, you get (using notations that are more familiar to me):$$ d(\det(X))=\text{tr}(\det(X)X^{-1}dX) $$ It does not seem to like the form of equation (1) above. What is confusing to me is where the transpose of equation (1) come from, and, does equation (1) even mean taking the differentiate...?
Can somebody help me with that? Don't need to solve all the questions, just kindly drop a comment or a few hints and I would be greatly appreciate.
$ \def\p{\partial} \def\grad#1#2{\frac{\p #1}{\p #2}} \def\c#1{\color{red}{#1}} \def\H{{\cal H}} \def\C{{\cal C}} \def\A{{\cal A}} \def\B{{\cal B}} \def\qiq{\quad\implies\quad} $When applied to tensors, the double-contraction product sums over the last two indices of the first tensor and the first two indices of the second tensor.
For example, writing $\,\C = \A:\B\,$ in index notation yields $$\large{ \C_{ab\ldots il\ldots yz} = \sum_{j=1}^J\sum_{k=1}^K \,\A_{ab\ldots i\c{jk}}\,\B_{\c{jk}l\ldots yz} }$$ In your example, $\Psi$ is a scalar, $F$ is a matrix, the gradient is also a matrix, and the Hessian is a fourth-order tensor. Once again, index notation makes this clear $$\eqalign{ G_{ij} &= \grad{\Psi}{F_{ij}} \qiq \H_{ijkl} = \grad{G_{ij}}{F_{kl}} = \grad{^2\Psi}{F_{kl}\:\p F_{ij}} = \grad{^2\Psi}{F_{ij}\:\p F_{kl}} \\ }$$ The differential of the gradient can then be written as $$\eqalign{ dG &= \H:dF \qiq dG_{ij} &= \sum_{k=1}^3\sum_{l=1}^3\H_{ijkl}\:dF_{kl} \\ }$$ and the differential of the scalar function as $$\eqalign{ d\Psi &= G:dF\;=\; \sum_{i=1}^3\sum_{j=1}^3 G_{ij}\:dF_{ij} \\ }$$ You can also write the second order Taylor expansion as $$\eqalign{ \def\D{\Delta} \def\DF{\D F} \D\Psi &\,=\, G:\DF \;\;{\bf+}\;\; \tfrac12\DF:\H:\DF \\ }$$