This is pertaining to a concept that leads to a generalization of Hooke's Law in solid mechanics. From this point of view, stress is a symmetric, second order tensor. We denote this by $\sigma_{ij}$. Similarly, strain (more formally called Green strain) is also a second order tensor denoted by $\epsilon_{ij}$. By definition, $\epsilon_{ij} = \frac{1}{2}[u_{i, j} + u_{j,i}]$ where $u$ is the displacement vector. It is postulated that it is possible to define a strain energy density function $W$, for an elastic material. As an example, consider:
Let $W = \frac{1}{2}C_{ijkl}\epsilon_{ij}\epsilon_{kl}\\$ be the function where $C_{ijkl}\\$ are constants $\sigma_{ij}\\$ is the stress $\epsilon_{ij}$ is the strain $\sigma_{ij} = \frac{\partial{W}}{\partial\epsilon_{ij}}$ can be taken for granted. It can be shown that $\sigma_{ij} = C_{ijkl}\epsilon_{kl }$.
This what I have a problem with. How is possible to take a partial derivative with respect to $\epsilon_{ij}$? Any help is appreciated.
My working follows along the lines of a similar problem: Consider the function $f(x_1,x_2,...,x_n) = A_{ij}x_ix_j$ where the $A_{ij}$'s are constants. Calculate the partial derivatives $\frac{\partial{f}}{\partial{x_i}}\\$
Solution: Differentiate f , $\frac{\partial{f}}{\partial{x_i}} = A_{pq}[\frac{\partial{x_p}}{\partial{x_i}}x_q + x_p\frac{\partial{x_q}}{\partial{x_i}}]\\$ Since $x_i$'s are independent, $\frac{\partial{f}}{\partial{x_i}} = A_{pq}[\delta_{pi}x_q+x_p\delta_{qi}]\\$ We can use substitution rule to obtain: $\frac{\partial{f}}{\partial{x_i}} = (A_{ij} + A_{ji})x_j\\$ Extending this problem to the above one is not very obvious. Any help to connect the dots is appreciated. :)
We use the fact that ${\partial \varepsilon_{ij}}/{\partial \varepsilon_{k\ell}} = \delta_i^k\delta_j^\ell$, where $\delta_a^b$ denotes the Kronecker delta (cf. this, where Einstein's summation convention is used). Thus, \begin{aligned} \frac{\partial W}{\partial \varepsilon_{pq}} &= \frac{1}{2} C_{ijk\ell} \left( \frac{\partial \varepsilon_{ij}}{\partial \varepsilon_{pq}} \varepsilon_{k\ell} + \varepsilon_{ij} \frac{\partial \varepsilon_{k\ell}}{\partial \varepsilon_{pq}} \right)\\ & = \frac{1}{2} C_{ijk\ell} \left( \delta_i^p\delta_j^q \varepsilon_{k\ell} + \varepsilon_{ij} \delta_k^p\delta_\ell^q \right) \\ & = \frac{1}{2} C_{pqk\ell} \varepsilon_{k\ell} + \frac{1}{2} C_{ijpq} \varepsilon_{ij} \\ & = \frac{1}{2}\left( C_{pq ij} \varepsilon_{ij} + C_{ijpq} \varepsilon_{ij} \right) . \end{aligned} The requirement of symmetry for the stress tensor $\boldsymbol\sigma$ yields $C_{pqk\ell} = C_{k\ell pq}$ (cf. this article), and thus, $ {\partial W}/{\partial \varepsilon_{pq}} = C_{pq ij} \varepsilon_{ij} \, $.
One may prefer to work it out in intrinsic form with $W = \frac{1}{2} \boldsymbol\varepsilon : \boldsymbol C : \boldsymbol\varepsilon$: $$ \frac{\partial W}{\partial \boldsymbol\varepsilon} = \frac{1}{2} \left( \boldsymbol C : \boldsymbol\varepsilon + \boldsymbol \varepsilon :\boldsymbol C\right) , $$ where $:$ denotes the double-dot product. Similarly, we get Hooke's law $\boldsymbol\sigma = \boldsymbol C : \boldsymbol\varepsilon$, provided that $\boldsymbol C$ and $\boldsymbol\varepsilon$ commute. Lastly, the most convenient method consists in using the symmetry of $\boldsymbol\varepsilon$ by introducing the invariants $\varepsilon_\text{I}, \varepsilon_\text{II}, \varepsilon_\text{III}$ of $\boldsymbol\varepsilon$ and their derivatives: $$ \frac{\partial W}{\partial \boldsymbol\varepsilon} = \frac{\partial W}{\partial \varepsilon_\text{I}}\frac{\partial \varepsilon_\text{I}}{\partial \boldsymbol\varepsilon} + \frac{\partial W}{\partial \varepsilon_\text{II}}\frac{\partial \varepsilon_\text{II}}{\partial \boldsymbol\varepsilon} + \frac{\partial W}{\partial \varepsilon_\text{III}}\frac{\partial \varepsilon_\text{III}}{\partial \boldsymbol\varepsilon} \, . $$