# Derivative of the inverse of the Right Cauchy-Green Deformation Tensor wrt itself

In continuum mechanics, we define the Right-Cauchy-Green Deformation Tensor as

$\boldsymbol{C}=\boldsymbol{F}^T\boldsymbol{F}$

I want to compute $\frac{\partial \boldsymbol{C}^{-1}}{\partial \boldsymbol{C}}$.

In the Wikipedia Article covering Tensor Derivatives, it says that in index form $\left(\frac{\partial \boldsymbol{C}^{-1}}{\partial \boldsymbol{C}}\right)_{IJKL} = -C^{-1}_{IK}C^{-1}_{LJ}$. (Correct?)

What I'm confused is this part here found in a paper regarding continuum mechanics (but also appears in many other places) :

Note that I am trying to figure out the step where we go from the Second-Piola-Kirchhoff to the Material Tangent Tensor. The Second Piola-Kirchhoff Stress $\boldsymbol{S}$ is related to the Material Tangent Tensor $\mathbb{C}$ by

$\mathbb{C} = 2\frac{\partial \boldsymbol{S}}{\partial \boldsymbol{C}}$.

and we should be able to use the tensor derivative given in the wikipedia article to get what we need.

What I don't get is how the derivative $\left(\frac{\partial \boldsymbol{C}^{-1}}{\partial \boldsymbol{C}}\right)_{IJKL}$ got splitted into two parts, $(IKLJ)$ and $(ILJK)$. There is only one term in the Wikipedia article.

Can somebody correct my misunderstanding?

• I was the original author of the wikipedia article and recall that I had mentioned that for symmetric 2-tensors you can write the expression for the derivative of the inverse as a sum of two equivalent terms. Not sure whether I provided a proof. My suggestion is to actually plug in numbers for $C$ to see what those two terms mean. – Biswajit Banerjee Dec 13 '16 at 21:40

Let's say you want to compute the derivative of any matrix function $X=X(C)$ with regard to entry $C_{ij}$: $$\frac{\partial X}{\partial C_{ij}}.$$ In other words, you ask how the matrix $X$ changes as you change $C_{ij}$ a bit. This can easily be answered by considering that $XX^{-1}=I$, independent of any of the elements of $C$. In other words, $$\frac{\partial XX^{-1}}{\partial C_{ij}}=0$$ because $XX^{-1}=I$ does not change if you modify $C_{ij}$. But by the product rule, you have $$\frac{\partial XX^{-1}}{\partial C_{ij}}= X\frac{\partial X^{-1}}{\partial C_{ij}} + \frac{\partial X}{\partial C_{ij}}X^{-1} = 0$$ and consequently $$X\frac{\partial X^{-1}}{\partial C_{ij}} = - \frac{\partial X}{\partial C_{ij}}X^{-1}$$ or equivalently, assuming that the matrix $X$ is invertible: $$\frac{\partial X^{-1}}{\partial C_{ij}} = - X^{-1} \frac{\partial X}{\partial C_{ij}}X^{-1}.$$ This is true for any function $X=X(C)$. In particular, it is true if you consider $X(C)=C$, in which case you get $$\frac{\partial C^{-1}}{\partial C_{ij}} = - C^{-1} \frac{\partial C}{\partial C_{ij}}C^{-1}.$$ In component notation, this can be written as $$\frac{\partial (C^{-1})_{kl}}{\partial C_{ij}} = - C^{-1}_{km} \frac{\partial C_{mn}}{\partial C_{ij}}C^{-1}_{nl}.$$ Now, the derivative of $C_{mn}$ with regard to $C_{ij}$ is only nonzero if $m=i,n=j$, and in that case it is in fact one. So $$\frac{\partial C_{mn}}{\partial C_{ij}} = \delta_{mi}\delta_{nj},$$ with the Kronecker delta symbol, which then finally yields this: $$\frac{\partial (C^{-1})_{kl}}{\partial C_{ij}} = - C^{-1}_{ki} C^{-1}_{jl}.$$ This also explains what exactly the various indices in your formula correspond to.
We know that $$C C^{-1} = I$$ If we perturb C a little bit, by a matrix $X$, and recompute the inverse, the result should stay roughly the same. In symbols: $$(C + X) \left(C^{-1} + \underbrace{\frac{\partial C^{-1}}{\partial C}\cdot X}_{\text{matrix}}\right) \approx I,$$ where "$\frac{\partial C^{-1}}{\partial C}\cdot X$" is the first order correction to $C^{-1}$ due to the perturbation by $X$ (we want to determine this).
Multiply things out: $$I + C \frac{\partial C^{-1}}{\partial C}\cdot X + X C^{-1} + \underbrace{X \frac{\partial C^{-1}}{\partial C}\cdot X}_{O(X^2) \approx 0} = I.$$ Then drop the term that is quadratic in $X$ (since tends to zero much faster than terms that are linear in $X$), and solve the resulting equation for $\frac{\partial C^{-1}}{\partial C}\cdot X$. You get the following formula: $$\frac{\partial C^{-1}}{\partial C}\cdot X = -C^{-1} X C^{-1}.$$ This completely describes how to compute the directional derivative of of the map $C \mapsto C^{-1}$ in direction $X$.
Viewing $X$ as a free variable, this formula describes a 4'th order multilinear map. When written out in components, it becomes: \begin{align} \overbrace{\frac{\partial C^{-1}}{\partial C}}^{4'\text{th order tensor}} : (u,X ,v) \mapsto & -u^T C^{-1} X C^{-1} v \\ =& -\sum_{IJKL} \left( C^{-1} \right)_{IK} \left( C^{-1} \right)_{LJ} u_I X_{KL} v_J, \end{align} and hence the tensor $-\left( C^{-1} \right)_{IK} \left( C^{-1} \right)_{LJ}$ represents $\frac{\partial C^{-1}}{\partial C}$ when all quantities are expressed in a given basis.