Derivative of vector-tensor-vector product wrt tensor

Question

In the context of continuum mechanics I need to compute the derivative

$$ \frac{\partial}{\partial C}\left( \vec{e}_1\cdot\bar C \vec{e}_1\right)$$

with $\vec{e}_1$ the first cartesian unit vector, $C$ a symmetric rank 2 tensor and $\bar C = J^{-\frac{2}{3}} C$, $J = \det(C)^\frac{1}{2}$.

Question: Is the following statement true? $$ \frac{\partial}{\partial C} \left(\vec{e}_1\cdot\bar C \vec{e}_1\right) = \vec{e}_1\cdot\frac{\partial\bar C}{\partial C} \vec{e}_1 $$

Attempt to prove:

• text book results: $$ \frac{\partial\bar C}{\partial C} = J^{-\frac{2}{3}}(\mathbb{I} - \frac{1}{3} C\otimes C^{-1}) $$ ($\mathbb{I}$: rank 4 identity) $$ \frac{\partial J^{-\frac{2}{3}}}{\partial C} = -\frac{1}{3}J^{-\frac{2}{3}} C^{-1} $$

• 1. alternative:

$$ \frac{\partial}{\partial C} \left(\vec{e}_1\cdot\bar C \vec{e}_1\right) = \frac{\partial J^{-\frac{2}{3}}C_{11}}{\partial C} = -\frac{1}{3}C_{11}J^{- \frac{2}{3}} C^{-1} + J^{-\frac{2}{3}}\frac{\partial C_{11}}{\partial C} $$ where the last derivative is $\frac{\partial C_{11}}{\partial C_{ij}}\vec e_i\otimes\vec e_j = \vec e_1\otimes\vec e_1$, hence

$$ \frac{\partial}{\partial C} \left(\vec{e}_1\cdot\bar C \vec{e}_1\right) =J^{- \frac{2}{3}}(\vec e_1\otimes\vec e_1 -\frac{1}{3}C_{11} C^{-1}) $$

• 2. alternative:

$$ \vec{e}_1\cdot\frac{\partial\bar C}{\partial C} \vec{e}_1 = \vec e_1\cdot\left(J^{-\frac{2}{3}}(\mathbb{I} - \frac{1}{3} C\otimes C^{-1})\right)\vec e_1 = J^{- \frac{2}{3}}(\vec e_1\otimes\vec e_1 - \frac{1}{3}\vec e_1\cdot(C\otimes C^{-1})\vec e_1) $$

The last expression can be written in tensor notation as $$ C_{ij}C^{-1}_{pq}\vec e_1\cdot(\vec e_i \otimes\vec e_j\otimes\vec e_p\otimes\vec e_q)\vec e_1 = C_{ij}C^{-1}_{pq}\delta_{1i} (\vec e_j\otimes\vec e_p)\delta_{q1} = C_{1j}C^{-1}_{1p} \vec e_j\otimes\vec e_p $$

which differs from $C_{11}C^{-1}_{ij}$.

Maybe this proves that $ \frac{\partial}{\partial C}\left( \vec{e}_1\cdot\bar C \vec{e}_1\right) \neq \vec{e}_1\cdot\frac{\partial\bar C}{\partial C} \vec{e}_1 $ but I don't quite trust my analysis. Thank you for any comments!

Answer 1

I personally prefer "bashing" such expressions. What I mean by that is to reduce everything to einstein summation convention, and then compute the derivative using regular calculus. I'm also going to call $\vec e_1$ as just $e$ for simplicity.

$$ J = \sqrt{det(C)} = \sqrt{ \epsilon_{ijk} C_{1i} C_{2j} C_{3k}} \\ \overline{C}_{ab} = (J^{2/3}C)_{ab} = (\epsilon_{ijk} C_{1i} C_{2j} C_{3k})^{1/3}C_{ab} \\ (\overline Ce)_p = \overline C_{ap} {e}_p \\ $$

Now that we have all the expressions, we can compute the expression:

\begin{align*} e \cdot \overline C e &= e_p (\overline C e)_p \\ &= e_p (\overline C_{ap} {e}_p) \\ &= \overline C_{ap} e_p^2 \\ &= [(\epsilon_{ijk} C_{1i} C_{2j} C_{3k})^{1/3}C_{ap}] e_p^2 \end{align*}

from where we can mechanically now find the derivative $\frac{\partial(e \cdot \overline C e)}{\partial C_{xy}}$ using the chain rule and that $\frac{\partial C_{ab}}{\partial C_{xy}}= \delta_{ab} \delta{xy}$