use a strengthened set of requirements which implies (\[DefPos\]), namely $$\mu_{e}>0,\qquad\mu_{c}>0,\qquad2\lambda_{e}+\mu_{e}>0,\qquad\mu_{h}>0,\qquad2\lambda_{h}+\mu_{h}>0,\qquad\alpha_{c}>0.\label{DefPos-1}$$

### Comparison with Mindlin and Eringen models

It can be checked that the proposed strain energy density (\[KinPot\]) represents a particular case of the strain energy density proposed by Mindlin (cf. Eq. (5.5) of [@Mindlin]). Indeed, considering that our micro-strain tensor is the transposed of the one introduced by Mindlin ($P_{ij}=\psi_{ji}$) and using the substitutions (cf. also [@Neff_Forest_jel05]): $$\begin{gathered}
\mu=\mu_{h},\quad\lambda=\lambda_{h},\quad b_{1}=\lambda_{e}+\lambda_{h},\quad b_{2}=\mu_{e}+\mu_{h}+\mu_{c},\quad b_{3}=\mu_{e}+\mu_{h}-\mu_{c},\quad g_{1}=-\lambda_{h},\quad g_{2}=-\mu_{h},\label{Identification1}\\
\nonumber \\
a_{10}=\alpha_{c},\quad a_{14}=-\alpha_{c},\qquad a_{1}=a_{2}=a_{3}=a_{4}=a_{5}=a_{8}=a_{11}=a_{13}=a_{15}=0,\label{Identification2}\end{gathered}$$ we are able to recover that our relaxed model can be obtained as a particular case of Mindlin’s one. The relaxed strain energy density (\[KinPot\]) is not positive definite in the sense of Mindlin and Eringen, but it gives rise to a well posed model (see [@NeffRelaxed]). It is evident that the representation of the strain energy density (\[KinPot\]) is more suitable for applications than Mindlin’s one due to the reduced number of parameters. Indeed, if we consider all the terms with the space derivatives of $\mathbf{P}$ to be vanishing ($\alpha_{c}=0$ in our model and $a_{i}=0$ in Mindlin) we have complete equivalence of the two models when using the parameters identification (\[Identification1\]). We can hence start noticing that we 