My question is related to continuum mechanics, taking partial derivative of tensor with respect to tensor.
$$\mathbf{\sigma} = \lambda \hspace{1pt} tr(\mathbf{\epsilon})+ 2\mu\mathbf{\epsilon}$$
Where, $\mathbf{\sigma,\epsilon}$ are second order tensors, $tr(\mathbf{\epsilon})$ is trace of the tensor.
I want to find
$$\frac{\partial \mathbf{\sigma}}{\partial\mathbf{\epsilon}}$$
I start like this:
$$ \mathbb{C_{ijmn}}=\frac{\partial {\sigma_{ij}}}{\partial{\epsilon_{mn}}} = \frac{\partial}{\partial \epsilon_{mn}} \big(\lambda\delta_{ij}\epsilon_{kk} + 2\mu\epsilon_{ij} \big)$$
$$=\lambda\delta_{ij}\frac{\partial}{\partial \epsilon_{mn}}\epsilon_{kk} + 2\mu\frac{\partial}{\partial \epsilon_{mn}}\epsilon_{ij}$$
$$=\lambda\delta_{ij}\delta_{km}\delta_{kn} + 2\mu\delta_{im}\delta_{jn}$$
$$=\lambda\delta_{ij}\delta_{mn}+2\mu\delta_{im}\delta_{jn}$$
Is this correct?
Best Answer
Actually, I think the result is wrong. To show it recall that the stiffness tensor has minor symmetries due the symmetry of both stress and deformation tensors (see e.g. [1]). According to your results, e.g. $C_{1212}\neq C_{1221}$ which shows that the tensor $C$ is not symmetric.
To derive the correct result we have to take into account that:
$$ \frac{\partial \epsilon_{ij}}{\partial \epsilon_{kl}}=\frac{1}{2}\left(\delta_{ik}\delta_{jl}+\delta_{il}\delta_{jk}\right)=I_{ijkl} $$
where $I_{ijkl}$ is known as the fourth-order symmetric identity tensor (see e.g. [2]).
P.S. Don't forget to multiply the first component of the rhs of the first equation by the second-order identity tensor, otherwise you are adding a scalar with a second-order tensor.