The consistent coupling boundary condition for the classical micromorphic model: existence, uniqueness and interpretation of parameters

Abstract

We consider the classical Mindlin–Eringen linear micromorphic model with a new strictly weaker set of displacement boundary conditions. The new consistent coupling condition aims at minimizing spurious influences from arbitrary boundary prescription for the additional microdistortion field $\varvec{P}$. In effect, $\varvec{P}$ is now only required to match the tangential derivative of the classical displacement $\varvec{u}$ which is known at the Dirichlet part of the boundary. We derive the full boundary condition, in adding the missing Neumann condition on the Dirichlet part. We show existence and uniqueness of the static problem for this weaker boundary condition. These results are based on new coercive inequalities for incompatible tensor fields with prescribed tangential part. Finally, we show that compared to classical Dirichlet conditions on $\varvec{u}$ and $\varvec{P}$, the new boundary condition modifies the interpretation of the constitutive parameters.

Appendix

1.1 The Lie-algebra $\mathfrak {so}(3)$ and the maps Anti and axl

Let us introduce the Lie-algebra

$$\begin{aligned} \mathfrak {s}\mathfrak {o}(3):=\left\{ \begin{pmatrix}0 &{} -v_{3} &{} v_{2}\\ v_{3} &{} 0 &{} -v_{1}\\ -v_{2} &{} v_{1} &{} 0 \end{pmatrix}\in \mathbb {R}^{3\times 3}\;\left. \right| \;\varvec{v}\in \mathbb {R}^3\right\} \end{aligned}$$

(108)

equipped with the matrix commutator bracket $\left[ A,B\right] =A\!\cdot \!\!B-B\!\cdot \!\!A.$ We identify it with the Lie-algebra $(\mathbb {R}^{3},\times )$ via the isomorphism

$$\begin{aligned} {{\,\mathrm{\text {axl}}\,}}:\mathfrak {s}\mathfrak {o}(3)\longrightarrow \mathbb {R}^{3},\qquad \begin{pmatrix}0 &{} -v_{3} &{} v_{2}\\ v_{3} &{} 0 &{} -v_{1}\\ -v_{2} &{} v_{1} &{} 0 \end{pmatrix}\mapsto \begin{pmatrix}v_{1}\\ v_{2}\\ v_{3} \end{pmatrix} \end{aligned}$$

(109)

whose inverse is

$$\begin{aligned} {{\,\mathrm{\text {Anti}}\,}}:\mathbb {R}^{3}\longrightarrow \mathfrak {s}\mathfrak {o}(3),\qquad \begin{pmatrix}v_{1}\\ v_{2}\\ v_{3} \end{pmatrix}\mapsto {{\,\mathrm{\text {Anti}}\,}}\begin{pmatrix}v_{1}\\ v_{2}\\ v_{3} \end{pmatrix}:=\begin{pmatrix}0 &{} -v_{3} &{} v_{2}\\ v_{3} &{} 0 &{} -v_{1}\\ -v_{2} &{} v_{1} &{} 0 \end{pmatrix}. \end{aligned}$$

(110)

These isomorphisms are constructed in a such a way that

$$\begin{aligned} \left( {{\,\mathrm{\text {Anti}}\,}}\varvec{u}\right) \!\cdot \! \varvec{v}=\varvec{u}\times \varvec{v}\qquad \forall (\varvec{u},\varvec{v})\in \mathbb {R}^{3}\times \mathbb {R}^{3}. \end{aligned}$$

(111)

Moreover,

$$\begin{aligned} ({{\,\mathrm{\text {Anti}}\,}}\varvec{u})\!\cdot \!\varvec{v}=\varvec{u}\times \varvec{v}=-\varvec{v}\times \varvec{u}=-({{\,\mathrm{\text {Anti}}\,}}\varvec{v})\!\cdot \!\varvec{u},\qquad \forall \varvec{u},\varvec{v}\in \mathbb {R}^{3}. \end{aligned}$$

(112)

We remember also the following useful relations: let $\varvec{A}\in \mathfrak {so}(3)$ and $\varvec{a}={{\,\mathrm{\text {axl}}\,}}\varvec{A}$, then inductively it can be proved that

$$\begin{aligned} \varvec{A}^{2n}=\left( -1\right) ^{n-1}\left\| \varvec{a}\right\| _{\mathbb {R}^{3}}^{2n-2}\varvec{A}^{2}\qquad \text {and}\qquad \varvec{A}^{2n-1}=\left( -1\right) ^{n-1}\left\| \varvec{a}\right\| _{\mathbb {R}^{3}}^{2n-2}\varvec{A}\qquad \forall n\in \mathbb {N}, \end{aligned}$$

(113)

i.e.,

$$\begin{aligned} \left( {{\,\mathrm{\text {Anti}}\,}}\varvec{a}\right) ^{2n}=\left( -1\right) ^{n-1}\left\| \varvec{a}\right\| _{\mathbb {R}^{3}}^{2n-2}\left( {{\,\mathrm{\text {Anti}}\,}}\varvec{a}\right) ^{2}\qquad \text {and}\qquad \left( {{\,\mathrm{\text {Anti}}\,}}\varvec{a}\right) ^{2n-1}=\left( -1\right) ^{n-1}\left\| \varvec{a}\right\| _{\mathbb {R}^{3}}^{2n-2}{{\,\mathrm{\text {Anti}}\,}}\varvec{a} \end{aligned}$$

(114)

for all $n\in \mathbb {N}$ and for all $\varvec{a}\in \mathbb {R}^{3}$. For $n=1$ we get

$$\begin{aligned} \varvec{A}^{2}\!\cdot \! \varvec{u}=\left( {{\,\mathrm{\text {Anti}}\,}}\varvec{a}\right) ^{2}\!\cdot \! \varvec{u}=\varvec{a}\times \left( \varvec{a}\times \varvec{u}\right) \overset{(120)}{=}\left\langle \varvec{a},\varvec{u}\right\rangle \varvec{a}-\left\| \varvec{a}\right\| ^{2}\varvec{u}=\left( \varvec{a}\otimes \varvec{a}-\left\| \varvec{a}\right\| ^{2}\mathbb {1}\right) \!\cdot \! \varvec{u} \end{aligned}$$

(115)

for all $(\varvec{A},\varvec{u})\in \mathfrak {so}(3)\times \mathbb {R}^3$, hence for $n=2$ we obtain

$$\begin{aligned} \varvec{A}^{3}\!\cdot \!\varvec{u}&=\varvec{A}\!\cdot \!\left( \varvec{A}^{2}\!\cdot \!\varvec{u}\right) \overset{(115)}{=}\varvec{A}\!\cdot \!\left( \left\langle \varvec{a},\varvec{u}\right\rangle \varvec{a}-\left\| \varvec{a}\right\| ^{2}\varvec{u}\right) =\left\langle \varvec{a},\varvec{u}\right\rangle \underbrace{\varvec{A}\!\cdot \!\varvec{a}}_{=\,0}-\left\| \varvec{a}\right\| ^{2}\varvec{A}\!\cdot \!\varvec{u} =-\left\| \varvec{a}\right\| ^{2}\varvec{A}\!\cdot \!\varvec{u}, \end{aligned}$$

(116)

because

$$\begin{aligned} \varvec{A}\!\cdot \!\varvec{a}=\left( {{\,\mathrm{\text {Anti}}\,}}\varvec{a}\right) \!\cdot \!\varvec{a}\overset{(111)}{=}\varvec{a}\times \varvec{a}=0\qquad \forall \varvec{a}\in \mathbb {R}^3. \end{aligned}$$

(117)

Moreover, considering a second-order tensor $\varvec{m}$ and a vector $\varvec{\nu }$, we have

$$\begin{aligned} \left( \varvec{m}\times \varvec{\nu }\right) _{ij}&=\left( \left( \varvec{m}\right) _{i}\times \varvec{\nu }\right) {}_{j}=\left( -\varvec{\nu }\times \left( \varvec{m}\right) _{i}\right) {}_{j}=\left( -({{\,\mathrm{\text {Anti}}\,}}\varvec{\nu }) \!\cdot \!\left( \varvec{m}\right) _{i}\right) {}_{j}=-\left( {{\,\mathrm{\text {Anti}}\,}}\varvec{\nu }\right) _{jk}\left( \left( \varvec{m}\right) _{i}\right) _{k}\nonumber \\&=-\left( {{\,\mathrm{\text {Anti}}\,}}\varvec{\nu }\right) _{jk}\varvec{m}_{ik}=-\varvec{m}_{ik}\left( {{\,\mathrm{\text {Anti}}\,}}\varvec{\nu }\right) _{jk}=\varvec{m}_{ik}\left( {{\,\mathrm{\text {Anti}}\,}}\varvec{\nu }\right) _{kj} \nonumber \\&=(\varvec{m}\!\cdot \!{{\,\mathrm{\text {Anti}}\,}}\varvec{\nu })_{ij} \qquad \forall i,j\in \left\{ 1,2,3\right\} . \end{aligned}$$

(118)

1.2 Derivation of consistent coupling mixed boundary conditions in the micromorphic model

In this section, for the convenience of the reader, we give the details of the first variation of the curvature part of the accounted action functional over the space ${\mathscr {H}}^{\,\sharp }(\Omega ):=\left\{ \varvec{P}\in H^{1}(\Omega ,\mathbb {R}^{3\times 3})\;\left. \right| \;\left. \varvec{P}\times \varvec{\nu }\right| _{\Gamma }=0\right\} ,$ i.e.,

(119)

1.2.1 Normal and tangential decomposition of $\varvec{{\mathfrak {m}}}\!\cdot \!\varvec{\nu }$

Starting from the classical triple product relation

$$\begin{aligned} \varvec{a}\times \left( \varvec{b}\times \varvec{c}\right) =\varvec{b}\left\langle \varvec{a},\varvec{c}\right\rangle -\varvec{c}\left\langle \varvec{a},\varvec{b}\right\rangle ,\qquad \text {for}\qquad \varvec{a},\varvec{b},\varvec{c}\in \mathbb {R}^{3}, \end{aligned}$$

(120)

let us consider two vectors $\varvec{u},\varvec{\nu }\in \mathbb {R}^{3}$ with $\left\| \varvec{\nu }\right\| =1$. Then,

$$\begin{aligned} \left\{ \begin{aligned}\varvec{u}&=\varvec{u}-\left\langle \varvec{u},\varvec{\nu }\right\rangle \varvec{\nu }+\left\langle \varvec{u},\varvec{\nu }\right\rangle \varvec{\nu }=\underbrace{\overbrace{\left\langle \varvec{\nu },\varvec{\nu }\right\rangle }^{=\,1}\varvec{u}-\left\langle \varvec{u},\varvec{\nu }\right\rangle \varvec{\nu }}_{=\,\varvec{\nu }\times \left( \varvec{u}\times \varvec{\nu }\right) }+\underbrace{\left\langle \varvec{u},\varvec{\nu }\right\rangle \varvec{\nu }}_{=\,\left( \varvec{\nu }\otimes \varvec{\nu }\right) \!\cdot \!\varvec{u}}\\ \varvec{u}&=\left( \mathbb {1}-\varvec{\nu }\otimes \varvec{\nu }\right) \!\cdot \!\varvec{u}+\left( \varvec{\nu }\otimes \varvec{\nu }\right) \!\cdot \!\varvec{u}\end{aligned} \right. \qquad \Longrightarrow \qquad \varvec{\nu }\times \left( \varvec{u}\times \varvec{\nu }\right) =\left( \mathbb {1}-\varvec{\nu }\otimes \varvec{\nu }\right) \!\cdot \!\varvec{u}.\nonumber \\ \end{aligned}$$

(121)

We want to give the equivalent formulation in terms of ${{\,\mathrm{\text {Anti}}\,}}\varvec{\nu }$ for $\varvec{\nu }\in \mathbb {R}^{3}$ with $\left\| \varvec{\nu }\right\| =1$:

$$\begin{aligned} \varvec{\nu }\times \left( \varvec{u}\times \varvec{\nu }\right)&=-\varvec{\nu }\times \left( \varvec{\nu }\times \varvec{u}\right) = -\varvec{\nu }\times \left( ({{\,\mathrm{\text {Anti}}\,}}\varvec{\nu })\!\cdot \!\varvec{u}\right) =-{{\,\mathrm{\text {Anti}}\,}}\varvec{\nu }\!\cdot \!\left( ({{\,\mathrm{\text {Anti}}\,}}\varvec{\nu })\!\cdot \!\varvec{u}\right) \nonumber \\&=\left( {{\,\mathrm{\text {Anti}}\,}}\varvec{\nu }\right) ^{T}\!\cdot \!\left( ({{\,\mathrm{\text {Anti}}\,}}\varvec{\nu })\!\cdot \!\varvec{u}\right) =\underbrace{\left( \left( {{\,\mathrm{\text {Anti}}\,}}\varvec{\nu }\right) ^{T}\!\cdot \!{{\,\mathrm{\text {Anti}}\,}}\varvec{\nu }\right) }_{\underset{\left\| \varvec{\nu }\right\| =1}{=}\,\mathbb {1}-\varvec{\nu }\otimes \varvec{\nu }}\!\cdot \!\,\varvec{u}=-\left( {{\,\mathrm{\text {Anti}}\,}}\varvec{\nu }\right) ^{2}\!\cdot \!\varvec{u}\,. \end{aligned}$$

(122)

We show now that

$$\begin{aligned} \left[ \varvec{\nu }\times \left( \varvec{u}\times \varvec{\nu }\right) \right] \times \varvec{\nu }=\varvec{u}\times \varvec{\nu } \end{aligned}$$

(123)

using the operator ${{\,\mathrm{\text {Anti}}\,}}$. Indeed

$$\begin{aligned} \left[ \varvec{\nu }\times \left( \varvec{u}\times \varvec{\nu }\right) \right] \times \varvec{\nu }&=-\varvec{\nu }\times \left[ \varvec{\nu }\times \left( \varvec{u}\times \varvec{\nu }\right) \right] \overset{(122)}{=}\varvec{\nu }\times \left[ \left( {{\,\mathrm{\text {Anti}}\,}}\varvec{\nu }\right) ^{2}\!\cdot \!\varvec{u}\right] = ({{\,\mathrm{\text {Anti}}\,}}\varvec{\nu })\!\cdot \!\left[ \left( {{\,\mathrm{\text {Anti}}\,}}\varvec{\nu }\right) ^{2}\!\cdot \!\varvec{u}\right] \nonumber \\&= \left( {{\,\mathrm{\text {Anti}}\,}}\varvec{\nu }\right) ^{3}\!\cdot \!\varvec{u}\overset{(114)}{=}-\left\| \varvec{\nu }\right\| ^2({{\,\mathrm{\text {Anti}}\,}}\varvec{\nu })\!\cdot \!\varvec{u}= -({{\,\mathrm{\text {Anti}}\,}}\varvec{\nu })\!\cdot \!\varvec{u}=-\varvec{\nu }\times \varvec{u}=\varvec{u}\times \varvec{\nu }\,. \end{aligned}$$

(124)

The identity (123) can be obtained also directly as follows

$$\begin{aligned} \left[ \varvec{\nu }\times \left( \varvec{u}\times \varvec{\nu }\right) \right] \times \varvec{\nu }&=\left[ \varvec{u}-\left\langle \varvec{u},\varvec{\nu }\right\rangle \varvec{\nu }\right] \times \varvec{\nu }=\varvec{u}\times \varvec{\nu }-\left\langle \varvec{u},\varvec{\nu }\right\rangle \underbrace{\varvec{\nu }\times \varvec{\nu }}_{=\,0}=\varvec{u}\times \varvec{\nu }. \end{aligned}$$

(125)

Thus, considering $\varvec{P}\in L^{2}(\Gamma ,\mathbb {R}^{3\times 3})$, since $\varvec{\nu }\in L^{\infty }(\Gamma ,\mathbb {R}^{3})$, we have the $L^2-$ decomposition

$$\begin{aligned} \varvec{P}&=\varvec{P}\!\cdot \!\left( \mathbb {1}-\varvec{\nu }\otimes \varvec{\nu }\right) +\varvec{P}\!\cdot \!\left( \varvec{\nu }\otimes \varvec{\nu }\right) \ =\varvec{P}\!\cdot \!\left( \left( {{\,\mathrm{\text {Anti}}\,}}\varvec{\nu }\right) ^{T}\!\cdot \!{{\,\mathrm{\text {Anti}}\,}}\varvec{\nu }\right) +\varvec{P}\!\cdot \!\left( \mathbb {1}-\left( {{\,\mathrm{\text {Anti}}\,}}\varvec{\nu }\right) ^{T}\!\cdot \!{{\,\mathrm{\text {Anti}}\,}}\varvec{\nu }\right) \nonumber \\&=-\,\varvec{P}\!\cdot \!\left( {{\,\mathrm{\text {Anti}}\,}}\varvec{\nu }\right) ^{2}+\varvec{P}\!\cdot \!\left( \mathbb {1}+\left( {{\,\mathrm{\text {Anti}}\,}}\varvec{\nu }\right) ^{2}\right) . \end{aligned}$$

(126)

1.2.2 Variation

We have

(127)

The product rule implies that

$$\begin{aligned}&\text {div}\left[ \left( \,\mu \,L_{\text {c}}^{2}\,a_{1}\, \text {D}\!\left( \text {dev}\,\text {sym}\,\varvec{P}\right) +a_{2}\, \text {D}\!\left( \text {skew}\,\varvec{P}\right) +\frac{2}{9}\,a_{3} \text {D}\!\left( \text {tr}\!\left( \varvec{P}\right) \mathbb {1}\right) \right) \! :\!\delta \varvec{P}\right] \nonumber \\&\quad =\,\mu \,L_{\text {c}}^{2}\,\langle \text {DIV}\left[ a_{1}\, \text {D}\left( \text {dev}\,\text {sym}\,\varvec{P}\right) +a_{2}\, \text {D}\!\left( \text {skew}\,\varvec{P}\right) +\frac{2}{9}\,a_{3} \text {D}\!\left( \text {tr}\!\left( \varvec{P}\right) \mathbb {1}\right) \right] ,\delta \varvec{P} \rangle _{\mathbb {R}^{3\times 3}}\nonumber \\&\qquad +\,\mu \,L_{\text {c}}^{2}\,\left( \langle a_{1}\, \text {D}\!\left( \text {dev}\,\text {sym}\,\varvec{P}\right) +a_{2}\, \text {D}\!\left( \text {skew}\,\varvec{P}\right) +\frac{2}{9}\, a_{3}\text {D}\!\left( \text {tr}\!\left( \varvec{P}\right) \mathbb {1}\right) , \text {D}\,\delta \varvec{P}\rangle _{\mathbb {R}^{3\times 3\times 3}}\right) \,. \end{aligned}$$

(128)

Thus, remembering the definition of $\varvec{\mathfrak {m}}\in \mathbb {R}^{3\times 3\times 3}$ given in (23), we obtain

$$\begin{aligned} \delta \mathscr {F}_{\text {curv}}\left[ \varvec{P},\delta \varvec{P}\right]&=\int _{\Omega }\text {div}\left[ \varvec{\mathfrak {m}}\! : \!\delta \varvec{P}\right] \,\text {d}V-\int _{\Omega }\langle \text {DIV}\,\varvec{\mathfrak {m}},\delta \varvec{P}\rangle _{\mathbb {R}^{3\times 3}}\,\text {d}V. \end{aligned}$$

(129)

The theorem $\int _{\Omega }\langle \text {DIV}\,\varvec{\mathfrak {m}},\delta \varvec{P}\rangle \,\text {d}V$ will contribute a term in the bulk equilibrium equations, while thanks to the divergence theorem the term $\int _{\Omega }\text {div}\left[ \varvec{\mathfrak {m}}\! : \!\delta \varvec{P}\right] \,\text {d}V$ can be written as

$$\begin{aligned} \int _{\Omega }\text {div}\left[ \varvec{\mathfrak {m}}\! : \!\delta \varvec{P}\right] \,\text {d}V=\int _{\partial \Omega }\langle \varvec{\mathfrak {m}}\!\cdot \!\varvec{\nu },\delta \varvec{P}\rangle _{\mathbb {R}^{3\times 3}}\,\text {d}s. \end{aligned}$$

(130)

Recall that

$$\begin{aligned} \left. \varvec{P}\times \varvec{\nu }\right| _{\Gamma }=0\quad \Longrightarrow \quad \left. \delta \varvec{P}\times \varvec{\nu }\right| _{\Gamma }=0\quad \Longrightarrow \quad \left. \varvec{\nu }\times \left( \delta \varvec{P}\times \varvec{\nu }\right) \right| _{\Gamma }=\left. \delta \varvec{P}\!\cdot \!\left( \mathbb {1}-\varvec{\nu }\otimes \varvec{\nu }\right) \right| _{\Gamma }=0, \end{aligned}$$

(131)

so that it holds

$$\begin{aligned} \delta \varvec{P}=\delta \varvec{P}\!\cdot \! (\nu \otimes \nu ) \quad \text { along} \Gamma . \end{aligned}$$

(132)

Hence, for all tensor fields $\varvec{B}$ we have

$$\begin{aligned} \!\!\big \langle \varvec{B},\delta \varvec{P}\big \rangle _{\mathbb {R}^{3\times 3}}&=\big \langle \varvec{B},\delta \varvec{P}\!\cdot \! (\varvec{\nu }\otimes \varvec{\nu })\big \rangle _{\mathbb {R}^{3\times 3}}\!\!\!\overset{\Vert \varvec{\nu }\Vert =1}{=}\big \langle \varvec{B},\delta \varvec{P}\!\cdot \! (\varvec{\nu }\otimes \varvec{\nu })^2\big \rangle _{\mathbb {R}^{3\times 3}} \!\!= \big \langle \varvec{B}\!\cdot \!(\varvec{\nu }\otimes \varvec{\nu }),\delta \varvec{P}\!\cdot \! (\varvec{\nu }\otimes \varvec{\nu })\big \rangle _{\mathbb {R}^{3\times 3}} \end{aligned}$$

(133)

along $\Gamma$. Returning to the right-hand side of (130) we conclude:

$$\begin{aligned} \int _{\partial \Omega }\big \langle \varvec{{\mathfrak {m}}}\!\cdot \!\varvec{\nu },\delta \varvec{P}\big \rangle _{\mathbb {R}^{3\times 3}}\,\text {d}s&= \int _{\Gamma }\big \langle \varvec{{\mathfrak {m}}}\!\cdot \!\varvec{\nu },\delta \varvec{P}\big \rangle _{\mathbb {R}^{3\times 3}}\,\text {d}s + \int _{\partial \Omega \backslash {\overline{\Gamma }}}\big \langle \varvec{{\mathfrak {m}}}\!\cdot \!\varvec{\nu },\delta \varvec{P}\big \rangle _{\mathbb {R}^{3\times 3}}\,\text {d}s \nonumber \\&\!\!\!\overset{(133)}{=}\!\!\! \int _{\Gamma }\bigl \langle \left( \varvec{\mathfrak {m}}\!\cdot \!\varvec{\nu }\right) \!\cdot \! \left( \varvec{\nu }\otimes \varvec{\nu }\right) ,\underbrace{\delta \varvec{P}\!\cdot \!(\varvec{\nu }\otimes \varvec{\nu })}_{\text {free}} \bigr \rangle _{\mathbb {R}^{3\times 3}}\text {d}s+\int _{\partial \Omega \setminus {\overline{\Gamma }}} \langle \varvec{\mathfrak {m}}\!\cdot \!\varvec{\nu },\underbrace{\delta \varvec{P}}_{\text {free}}\rangle _{\mathbb {R}^{3\times 3}}\,\text {d}s. \end{aligned}$$

(134)

Since $\varvec{\nu }\otimes \varvec{\nu }=\left( \mathbb {1}-({{\,\mathrm{\text {Anti}}\,}}\varvec{\nu })^{T}\!\cdot \!{{\,\mathrm{\text {Anti}}\,}}\varvec{\nu }\right) =\left( \mathbb {1}+({{\,\mathrm{\text {Anti}}\,}}\varvec{\nu })^{2}\right)$ we can also rewrite the first term on the right-hand side of the last equation in terms of the ${{\,\mathrm{\text {Anti}}\,}}$ operator:

$$\begin{aligned} \int _{\Gamma }\bigl \langle \left( \varvec{\mathfrak {m}}\!\cdot \!\varvec{\nu }\right) \!\cdot \!\left( \varvec{\nu }\otimes \varvec{\nu }\right) ,\underbrace{\delta \varvec{P}\!\cdot \!(\varvec{\nu }\otimes \varvec{\nu })}_{\text {free}} \bigr \rangle _{\mathbb {R}^{3\times 3}}\text {d}s = \int _{\Gamma }\langle \left( \varvec{\mathfrak {m}}\!\cdot \!\varvec{\nu }\right) \!\cdot \!\left( \mathbb {1}+(\text {Anti}\, \varvec{\nu })^2\right) ,\underbrace{\delta \varvec{P}\!\cdot \!\left( \mathbb {1}+(\text {Anti}\,\varvec{\nu })^2\right) }_{\text {free}}\, \rangle _{\mathbb {R}^{3\times 3}}\,\text {d}s. \end{aligned}$$

(135)

Note that

$$\begin{aligned} \left( \mathbb {1}-\varvec{\nu }\otimes \varvec{\nu }\right) \!\cdot \!\left( \varvec{\nu }\otimes \varvec{\nu }\right)&=\varvec{0}\,,&\text {but}&\left( \mathbb {1}-\text {Anti}\,\varvec{\nu }\right) \!\cdot \!\text {Anti}\,\varvec{\nu }&\ne \varvec{0}. \end{aligned}$$

(136)

Thus

$$\begin{aligned} \delta \mathscr {F}_{\,\text {curv}}\left[ \varvec{P},\delta \varvec{P}\right]= & {} \int _{\Gamma } \bigl \langle \left( \varvec{\mathfrak {m}}\!\cdot \!\varvec{\nu }\right) \!\cdot \!\left( \varvec{\nu }\otimes \varvec{\nu }\right) , \underbrace{\delta \varvec{P}\!\cdot \!(\varvec{\nu }\otimes \varvec{\nu })}_{\text {free}}\bigr \rangle _{\mathbb {R}^{3\times 3}} \text {d}s+\int _{\partial \Omega \setminus {\overline{\Gamma }}}\langle \varvec{\mathfrak {m}}\!\cdot \!\varvec{\nu }, \underbrace{\delta \varvec{P}}_{\text {free}}\rangle _{\mathbb {R}^{3\times 3}}\,\text {d}s\nonumber \\&-\int _{\Omega }\langle \text {DIV}\,\varvec{\mathfrak {m}},\delta \varvec{P}\rangle _{\mathbb {R}^{3\times 3}}\,\text {d}V. \end{aligned}$$

(137)

The variations $\delta \varvec{P}$ in the bulk and at the boundary are independent, therefore $\delta \mathscr {F}_{\,\text {curv}}\left[ \varvec{P},\delta \varvec{P}\right] =0$ implies

$$\begin{aligned} \left. \left( \varvec{\mathfrak {m}}\!\cdot \!\varvec{\nu }\right) \!\cdot \!\left( \varvec{\nu }\otimes \varvec{\nu }\right) \,\right| _{\varvec{\Gamma }} =0\qquad \text {and}\qquad \left. \varvec{\mathfrak {m}}\!\cdot \!\varvec{\nu }\,\right| _{\partial \Omega \setminus {\overline{\Gamma }}}. \end{aligned}$$

Hence summarizing, the natural boundary conditions for the mixed problem for the classical micromorphic model are

1.3 Boundary conditions in the relaxed micromorphic model

For the convenience of the reader we repeat the above reasoning for the relaxed micromorphic model. We know that

$$\begin{aligned} \text {div}\,(\varvec{u}\times \varvec{v})=\left\langle \text {curl}\,\varvec{u},\varvec{v}\right\rangle -\left\langle \text {curl}\,\varvec{v},\varvec{u}\right\rangle \, , \end{aligned}$$

and with

$$\begin{aligned} \varvec{m}:=\mu \frac{L^{2}}{2}\,\mathbb {L}\,\text {Curl}\,\varvec{P}\in \mathbb {R}^{3\times 3} \, , \end{aligned}$$

(138)

we have

$$\begin{aligned} \!\!\!\!\mu \frac{L^{2}}{2}\sum _{i=1}^{3}\left\langle \left( \mathbb {L}\,\text {Curl}\,\varvec{P}\right) _{i},\left( \text {Curl}\,\delta \varvec{P}\right) _{i}\right\rangle _{\mathbb {R}^{3}}&=\sum _{i=1}^{3}\left\langle \left( \varvec{m}\right) _{i},\text {curl}\,\!\left( \delta \varvec{P}\right) _{i}\right\rangle _{\mathbb {R}^{3}}\nonumber \\&=\sum _{i=1}^{3}\left( -\,\text {div}\,[\left( \varvec{m}\right) _{i}\times \left( \delta \varvec{P}\right) _{i}]+\left\langle \text {curl}\,\!\left( \varvec{m}\right) _{i},\left( \delta \varvec{P}\right) _{i}\right\rangle _{\mathbb {R}^{3}}\right) . \end{aligned}$$

(139)

Thus,

$$\begin{aligned} \delta \mathscr {F}_{\,\text {curv}}^{\text {relax}}\left[ \varvec{P},\delta \varvec{P}\right]&=\mu \frac{L^{2}}{2}\int _\Omega \left\langle \mathbb {L}\,\text {Curl}\,\varvec{P},\text {Curl}\,\delta \varvec{P}\right\rangle _{\mathbb {R}^{3\times 3}} \,\text {d}V = \mu \frac{L^{2}}{2}\sum _{i=1}^{3}\int _\Omega \left\langle \left( \mathbb {L}\,\text {Curl}\,\varvec{P}\right) _{i},\left( \text {Curl}\,\delta \varvec{P}\right) _{i}\right\rangle _{\mathbb {R}^{3}} \,\text {d}V \nonumber \\&=\sum _{i=1}^{3} \left( \int _{\Omega }-\,\text {div}\,[\left( \varvec{m}\right) _{i}\times \left( \delta \varvec{P}\right) _{i}]\,\text {d}V+\int _{\Omega }\left\langle \text {curl}\,\!\left( \varvec{m}\right) _{i},\left( \delta \varvec{P}\right) _{i}\right\rangle _{\mathbb {R}^{3}}\text {d}V\right) \nonumber \\&=\sum _{i=1}^{3}\int _{\Omega }-\,\text {div}\,[\left( \varvec{m}\right) _{i}\times \left( \delta \varvec{P}\right) _{i}]\,\text {d}V+\int _{\Omega }\left\langle \text {Curl}\,\varvec{m},\delta \varvec{P}\right\rangle _{\mathbb {R}^{3\times 3}}\text {d}V \end{aligned}$$

(140)

and

(141)

where in the last step we have again used that $(-{{\,\mathrm{\text {Anti}}\,}}\varvec{\nu })^2=\mathbb {1}-\varvec{\nu }\otimes \varvec{\nu }$. Thus

$$\begin{aligned} \delta \mathscr {F}_{\,\text {curv}}^{\text {relax}}\left[ \varvec{P},\delta \varvec{P}\right] =\int _{\partial \Omega \setminus {\overline{\Gamma }}}\left\langle \varvec{m}\!\cdot \!{{\,\mathrm{\text {Anti}}\,}}\varvec{\nu },\delta \varvec{P}\!\cdot \!\left( \mathbb {1}-\varvec{\nu }\otimes \varvec{\nu }\right) \right\rangle _{\mathbb {R}^{3\times 3}}\text {d}s +\int _{\Omega }\left\langle \text {Curl}\,\varvec{m},\delta \varvec{P}\right\rangle _{\mathbb {R}^{3\times 3}}\text {d}V. \end{aligned}$$

(142)

The variations $\delta \varvec{P}$ in the bulk and at the boundary are independent, therefore $\delta \mathscr {F}_{\,\text {curv}}^{\text {relax}}\left[ \varvec{P},\delta \varvec{P}\right] =0$ implies

$$\begin{aligned} \left. \varvec{m}\!\cdot \!{{\,\mathrm{\text {Anti}}\,}}\varvec{\nu }\right| _{\partial \Omega \setminus {\overline{\Gamma }}}\overset{(150)}{=}\left. \varvec{m}\times \varvec{\nu }\right| _{\partial \Omega \setminus {\overline{\Gamma }}}=0. \end{aligned}$$

(143)

Hence summarizing, the natural boundary conditions for the mixed problem for the relaxed micromorphic model are

$$\begin{aligned} \begin{array}{cccccccc} \text {Dirichlet} &{} &{} &{} &{} &{} &{} &{} \text {Neumann}\\ \\ \left. \varvec{P}\times \varvec{\nu }\right| _{\Gamma }=0 &{} &{} &{} &{} &{} &{} &{} \left. \varvec{m}\times \varvec{\nu }\right| _{\partial \Omega \setminus {\overline{\Gamma }}}=0 \end{array} \end{aligned}$$

1.4 Derivation of mixed boundary conditions for a linear second gradient material

The second gradient linearized elasticity model (without mixed terms) reads (see for example [13,14,15,16,17,18, 27, 41, 51, 66])

$$\begin{aligned} \mathscr {F}:\mathbb {H}&\rightarrow \mathbb {R}^{+},&\varvec{u}\mapsto \mathscr {F}\left[ \varvec{u}\right]&=\underbrace{\int _{\Omega }\frac{1}{2}\left\langle \mathbb {C}\,\text {sym}\,\text {D}\varvec{u},\text {sym}\,\text {D}\varvec{u}\right\rangle \text {d}V}_{(\text {I})}+\underbrace{\int _{\Omega }\frac{1}{2}\left\langle \mathbb {G}\,\text {D}^{2}\varvec{u},\text {D}^{2}\varvec{u}\right\rangle \text {d}V}_{(\text {II})}, \end{aligned}$$

(144)

where

$$\begin{aligned} \mathbb {H}:=\left\{ u\in H_{0,\Gamma }^{1}\left( \Omega ,\mathbb {R}^{3}\right) \cap H^{\,2}\!\left( \Omega ,\mathbb {R}^{3}\right) \;\left. \right| \;\mathbb {G}\,\text {D}^{2}\varvec{u}\in H\!\left( \text {Div}\,\!\left( \text {DIV}\,\right) ;\Omega ,\mathbb {R}^{3}\otimes \text {Sym}\,\!(3)\right) \right\} \end{aligned}$$

and

$$\begin{aligned} H\!\left( \text {Div}\,(\text {DIV}\,\!);\Omega ,\mathbb {R}^{3}\otimes \text {Sym}\,\!(3)\right)&:=\left\{ \varvec{U}\in H\!\left( \text {DIV}\,;\Omega ,\mathbb {R}^{3}\otimes \text {Sym}\,\!(3)\right) \;\left. \right| \;\text {Div}\,\text {DIV}\,\varvec{U}\in L^{2}\left( \Omega ,\mathbb {R}^{3}\right) \right\} \nonumber \\&\;=\left\{ \varvec{U}\in H\!\left( \text {DIV}\,;\Omega ,\mathbb {R}^{3}\otimes \text {Sym}\,\!(3)\right) \;\left. \right| \;\text {DIV}\,\varvec{U}\in H\!\left( \text {Div}\,;\Omega ,\mathbb {R}^{3\times 3}\right) \right\} . \end{aligned}$$

(145)

In order to evaluate the variational derivative $\delta {\mathcal {I}}\left[ \varvec{u},\delta \varvec{u}\right]$, with $\delta \varvec{u}\in \mathbb {H}$ we study the two terms $(\text {I})$ and $(\text {II})$ separately. To make the calculations easier, we do them assuming the boundary and the involved fields regular. Concerning part $\left( \text {I}\right)$ we have

$$\begin{aligned} \delta \int _{\Omega }\frac{1}{2}&\left\langle \mathbb {C}\,\text {sym}\,\text {D}\varvec{u},\text {sym}\,\text {D}\varvec{u}\right\rangle \text {d}V=-\int _{\Omega }\left\langle \text {Div}\,\!\left[ \mathbb {C}\,\text {sym}\,\text {D}\varvec{u}\right] ,\delta \varvec{u}\right\rangle \text {d}V+\underbrace{\int _{\partial \Omega \setminus {\overline{\Gamma }}}\left\langle \left( \mathbb {C}\,\text {sym}\,\text {D}\varvec{u}\right) \!\cdot \!\varvec{\nu },\delta \varvec{u}\right\rangle _{\mathbb {R}^{3}}\text {d}s}_{(a)} \end{aligned}$$

and we obtain the normal stress on $\partial \Omega \setminus {\overline{\Gamma }}$

$$\begin{aligned} \left. \varvec{\sigma }\!\cdot \!\varvec{\nu }\,\right| _{\partial \Omega \setminus {\overline{\Gamma }}}=\left. (\mathbb {C}\,\text {sym}\,\text {D}\varvec{u})\!\cdot \!\varvec{\nu }\,\right| _{\partial \Omega \setminus {\overline{\Gamma }}}. \end{aligned}$$

(146)

Concerning part $\left( \text {II}\right)$ we obtain

$$\begin{aligned}&\delta \int _{\Omega }\frac{1}{2} \left\langle \mathbb {G}\,\text {D}^{2}\varvec{u},\text {D}^{2}\varvec{u}\right\rangle \text {d}V\nonumber \\&\quad =\int _{\Omega }\left\langle \mathbb {G}\,\text {D}^{2}\varvec{u},\text {D}^{2}\delta \varvec{u}\right\rangle \text {d}V=\int _{\Omega }\left\langle \varvec{\mathfrak {m}},\text {D}^{2}\delta \varvec{u}\right\rangle \text {d}V =-\int _{\Omega }\left\langle \text {DIV}\,\varvec{\mathfrak {m}},\text {D}\delta \varvec{u}\right\rangle \text {d}V+\int _{\partial \Omega }\left\langle \varvec{\mathfrak {m}}\!\cdot \!\varvec{\nu },\text {D}\delta \varvec{u}\right\rangle _{\mathbb {R}^{3}}\,\text {d}s\nonumber \\&\quad =\int _{\Omega }\left\langle \text {Div}\,\text {DIV}\,\varvec{\mathfrak {m}},\delta \varvec{u}\right\rangle \text {d}V -\underbrace{\int _{\partial \Omega \setminus {\overline{\Gamma }}}\left\langle \left( \text {DIV}\,\varvec{\mathfrak {m}}\right) \!\cdot \!\varvec{\nu },\delta \varvec{u}\right\rangle _{\mathbb {R}^{3}}\text {d}s}_{(b)} +\int _{\partial \Omega }\left\langle \varvec{\mathfrak {m}}\!\cdot \!\varvec{\nu },\text {D}\delta \varvec{u}\right\rangle _{\mathbb {R}^{3}}\text {d}s. \end{aligned}$$

(147)

Using the surface differential operators [41], we can develop further the term in duality with $\left. \text {D}\delta \varvec{u}\right| _{\partial \Omega }$. Indeed, remarking that

$$\begin{aligned} \left. \text {D}\delta \varvec{u}\right| _{\partial \Omega }=\left( \left( \mathbb {1}-\varvec{\nu }\otimes \varvec{\nu }\right) +\left( \varvec{\nu }\otimes \varvec{\nu }\right) \right) \left. \text {D}\delta \varvec{u}\right| _{\partial \Omega }, \end{aligned}$$

(148)

we have

$$\begin{aligned} \int _{\partial \Omega }\left\langle \varvec{\mathfrak {m}}\!\cdot \!\varvec{\nu },\text {D}\delta \varvec{u}\right\rangle _{\mathbb {R}^{3\times 3}}\text {d}s&=\int _{\partial \Omega }\left\langle \varvec{\mathfrak {m}}\!\cdot \!\varvec{\nu },\text {D}\delta \varvec{u}\!\cdot \!\left( \left( \mathbb {1}-\varvec{\nu }\otimes \varvec{\nu }\right) +\left( \varvec{\nu }\otimes \varvec{\nu }\right) \right) \right\rangle _{\mathbb {R}^{3\times 3}}\text {d}s\nonumber \\&=\int _{\partial \Omega }\left\langle \varvec{\mathfrak {m}}\!\cdot \!\varvec{\nu },\text {D}\delta \varvec{u}\!\cdot \!\left( \mathbb {1}-\varvec{\nu }\otimes \varvec{\nu }\right) \right\rangle _{\mathbb {R}^{3\times 3}}\text {d}s+\int _{\partial \Omega }\left\langle \varvec{\mathfrak {m}}\!\cdot \!\varvec{\nu },\text {D}\delta \varvec{u}\!\cdot \!\left( \varvec{\nu }\otimes \varvec{\nu }\right) \right\rangle _{\mathbb {R}^{3\times 3}}\text {d}s\nonumber \\&=\int _{\partial \Omega }\left\langle -\text {Div}\,_{\partial \Omega ,\top }\left[ \varvec{\mathfrak {m}}\!\cdot \!\varvec{\nu }\right] ,\delta \varvec{u}\right\rangle _{\mathbb {R}^{3}}\text {d}s+\int _{\partial \Omega }\left\langle \varvec{\mathfrak {m}}\!\cdot \!\varvec{\nu },\text {D}\delta \varvec{u}\!\cdot \!\left( \varvec{\nu }\otimes \varvec{\nu }\right) \right\rangle _{\mathbb {R}^{3\times 3}}\text {d}s\nonumber \\&=\underbrace{\int _{\partial \Omega \setminus {\overline{\Gamma }}}\left\langle -\text {Div}\,_{\partial \Omega ,\top }\left[ \varvec{\mathfrak {m}}\!\cdot \!\varvec{\nu }\right] ,\delta \varvec{u}\right\rangle _{\mathbb {R}^{3}}\text {d}s}_{(c)} +\int _{\partial \Omega }\left\langle \varvec{\mathfrak {m}}\!\cdot \!\varvec{\nu },\text {D}\delta \varvec{u}\!\cdot \!\left( \varvec{\nu }\otimes \varvec{\nu }\right) \right\rangle _{\mathbb {R}^{3\times 3}}\text {d}s. \end{aligned}$$

(149)

Now, we develop further the term $\int _{\partial \Omega }\left\langle \varvec{\mathfrak {m}}\!\cdot \!\varvec{\nu },\text {D}\delta \varvec{u}\!\cdot \!\left( \varvec{\nu }\otimes \varvec{\nu }\right) \right\rangle _{\mathbb {R}^{3\times 3}}\text {d}s$. We have,

(150)

Summing up all contributions (a), (b) and (c), we obtain

$$\begin{aligned} \int _{\partial \Omega \setminus {\overline{\Gamma }}}\left\langle \left( \mathbb {C}\,\text {sym}\,\text {D}\varvec{u}\right) \!\cdot \!\varvec{\nu }-\left( \text {DIV}\,\varvec{\mathfrak {m}}\right) \!\cdot \!\varvec{\nu }-\text {Div}\,_{\partial \Omega ,\top }\left[ \varvec{\mathfrak {m}}\!\cdot \!\varvec{\nu }\right] ,\delta \varvec{u}\right\rangle _{\mathbb {R}^{3}}\text {d}s. \end{aligned}$$

(151)

Thus

$$\begin{aligned} \delta \mathscr {F}\!\left[ \varvec{u},\delta \varvec{u}\right]&= \int _{\partial \Omega \setminus {\overline{\Gamma }}}\left\langle \left( \mathbb {C}\,\text {sym}\,\text {D}\varvec{u}\right) \!\cdot \!\varvec{\nu }-\left( \text {DIV}\,\varvec{\mathfrak {m}}\right) \!\cdot \!\varvec{\nu }-\text {Div}\,_{\partial \Omega ,\top }\left[ \varvec{\mathfrak {m}}\!\cdot \!\varvec{\nu }\right] ,\delta \varvec{u}\right\rangle _{\mathbb {R}^{3}}\text {d}s \nonumber \\&\quad -\int _{\Omega }\left\langle \text {Div}\,\!\left[ \mathbb {C}\,\text {sym}\,\text {D}\varvec{u}\right] ,\delta \varvec{u}\right\rangle \text {d}V+\int _{\Omega }\left\langle \text {Div}\,\text {DIV}\,\varvec{\mathfrak {m}},\delta \varvec{u}\right\rangle \text {d}V \nonumber \\&\quad +\int _{\partial \Omega }\left\langle \left( \varvec{\mathfrak {m}}\!\cdot \!\varvec{\nu }\right) \!\cdot \!\varvec{\nu },\text {D}\delta \varvec{u}\!\cdot \!\varvec{\nu }\right\rangle _{\mathbb {R}^{3}}\text {d}s. \end{aligned}$$

(152)

The variations $\delta \varvec{u}$ in the bulk and at the boundary are independent, therefore $\delta \mathscr {F}\!\left[ \varvec{u},\delta \varvec{u}\right] =0$ implies

(153a)

(153b)

1.5 Relaxed micromorphic model as a particular case of the general micromorphic model

The relaxed micromorphic model can also be written formally as the classical micromorphic model since $\text {Curl}\,\varvec{P}$ is related to $\text {D}\varvec{P}$ via a linear map. This means, in particular, that there exists a linear map

$$\begin{aligned} {\mathcal {L}}:\mathbb {R}^{3\times 3\times 3}\rightarrow \mathbb {R}^{3\times 3}\qquad \text {such that}\qquad {\mathcal {L}}\,\text {D}\varvec{P}=\text {Curl}\,\varvec{P}, \end{aligned}$$

(154)

allowing us to rewrite the curvature term in the energy density of the relaxed micromorphic model as follows

$$\begin{aligned} \left\langle \mathbb {L}\,\text {Curl}\,\varvec{P},\text {Curl}\,\varvec{P}\right\rangle =\left\langle \mathbb {L}\,{\mathcal {L}}\,\text {D}\varvec{P},{\mathcal {L}}\text {D}\varvec{P}\right\rangle =\bigl \langle \underbrace{{\mathcal {L}}^{T}\,\mathbb {L}\,{\mathcal {L}}}_{=:\,{\widehat{\mathbb {L}}}} \,\text {D}\varvec{P},\text {D}\varvec{P}\bigr \rangle . \end{aligned}$$

(155)

Such a map ${\mathcal {L}}$ can be given explicitly. Indeed, introducing the Levi-Civita third-order tensor $\varvec{\epsilon }$, the classical $\text {curl}\,$ operator can be written in terms of the Jacobian as follows

$$\begin{aligned} \text {curl}\,\varvec{u}=\varvec{\epsilon }:\text {D}\varvec{u}^T \end{aligned}$$

(156)

where the contraction operator  :  is defined component-wise in the following manner:

$$\begin{aligned} \left( \text {curl}\,\varvec{u}\right) _{k}=\epsilon _{kij}(\text {D}\varvec{u}^T)_{ij}=\epsilon _{kij}u_{j,i}\qquad \forall k\in \left\{ 1,2,3\right\} . \end{aligned}$$

Hence, from the definition (4) we obtain

$$\begin{aligned} \text {Curl}\,\varvec{P}&=\begin{pmatrix}\text {curl}\,\left( \varvec{P}\right) _{1}\\ \text {curl}\,\left( \varvec{P}\right) _{2}\\ \text {curl}\,\left( \varvec{P}\right) _{3} \end{pmatrix}=\begin{pmatrix}\varvec{\epsilon }:\left( \text {D}\left( \varvec{P}\right) _{1}\right) ^{T}\\ \varvec{\epsilon }:\left( \text {D}\left( \varvec{P}\right) _{2}\right) ^{T}\\ \varvec{\epsilon }:\left( \text {D}\left( \varvec{P}\right) _{3}\right) ^{T} \end{pmatrix} \end{aligned}$$

(157)

giving component-wise

$$\begin{aligned} \left( \text {Curl}\,\varvec{P}\right) _{\alpha \beta }=\epsilon _{\beta ij}P_{\alpha j,i}. \end{aligned}$$

Remark 1

The positive-definiteness of the bilinear form $\left\langle \mathbb {L}\,\text {Curl}\,\varvec{P},\text {Curl}\,\varvec{P}\right\rangle$ in terms of $\text {Curl}\,\varvec{P}$ does not imply the positive definiteness of $\bigl \langle {\widehat{\mathbb {L}}}\,\text {D}\varvec{P},\text {D}\varvec{P}\bigr \rangle$ in terms of $\text {D}\varvec{P}$ but only its positive semi-definiteness. Thus, proving an existence result using the formulation of the relaxed micromorphic model as a classical one is not straightforward but needs new function spaces and new coercive inequalities [37, 39, 40, 61, 63].

Denoting by $\varvec{\mathfrak {m}}:={\widehat{\mathbb {L}}}\,\text {D}\varvec{P}$ the relative third-order stress tensor and remembering that $\varvec{m}=\mathbb {L}\,\text {Curl}\,\varvec{P}$, we would like to understand if the following relation holds

$$\begin{aligned} \varvec{\mathfrak {m}}\!\cdot \!\varvec{\nu }=\varvec{m}\times \varvec{\nu }\in \mathbb {R}^{3\times 3}, \qquad \varvec{\mathfrak {m}}\in \mathbb {R}^{3\times 3\times 3},\quad \varvec{m}\in \mathbb {R}^{3\times 3}. \end{aligned}$$

(158)

Considering for simplicity the constitutive tensor $\mathbb {L}$ trivial in the component-wise calculation, we obtain

$$\begin{aligned}&\left\langle \smash {\overbrace{{\mathcal {L}}^{T}\left( {\mathcal {L}}\,\text {D}\varvec{P}\right) }^{=\,\varvec{\mathfrak {m}}}},\text {D}\varvec{P}\right\rangle \nonumber \\&\quad =\left\langle {\mathcal {L}}\,\text {D}\varvec{P},{\mathcal {L}}\,\text {D}\varvec{P}\right\rangle =\left\langle \text {Curl}\,\varvec{P},\text {Curl}\,\varvec{P}\right\rangle =P_{\alpha j,i}\epsilon _{\beta ij}P_{\alpha n,m}\epsilon _{\beta mn} component-wise calculation\nonumber \\&\quad =P_{\alpha j,i}\epsilon _{ij\beta }P_{\alpha n,m}\epsilon _{\beta mn}=\left( P_{\alpha j,i}\epsilon _{ij\beta }\epsilon _{\beta mn}\right) P_{\alpha n,m} =\underbrace{\left( P_{\alpha j,i}\epsilon _{ij\beta }\epsilon _{\beta mn}\right) }_{=\,\varvec{\mathfrak {m}}\,=\,\varvec{m}\!\cdot \!\epsilon }P_{\alpha n,m}, \end{aligned}$$

(159)

hence

$$\begin{aligned} \left( \varvec{\mathfrak {m}}\!\cdot \!\varvec{\nu }\right) _{\alpha \beta }=M_{\alpha \beta \gamma }\nu _{\gamma }=\left( P_{\alpha j,i}\epsilon _{\beta ij}\epsilon _{\beta q\gamma }\right) \nu _{\gamma }=m_{\alpha \beta }\epsilon _{\beta q\gamma }v_{\gamma }=\left( \varvec{m}\times \varvec{\nu }\right) _{\alpha \beta }, \end{aligned}$$

(160)

i.e.,

$$\begin{aligned} \varvec{\mathfrak {m}}\!\cdot \!\varvec{\nu }&=\left( {\widehat{\mathbb {L}}}\,\text {D}\varvec{P}\right) \!\cdot \!\varvec{\nu }=\left( ({\mathcal {L}}^{T}\, \mathbb {L}\,{\mathcal {L}})\,\text {D}\varvec{P}\right) \!\cdot \!\varvec{\nu }=\left( ({\mathcal {L}}^{T}\,\mathbb {L})\,({\mathcal {L}} \,\text {D}\varvec{P})\right) \!\cdot \!\varvec{\nu }\nonumber \\&=\big ({\mathcal {L}}^{T}\,\smash {\underbrace{\mathbb {L}\,\text {Curl}\,\varvec{P}}_{=\,\varvec{m}}}\big )\!\cdot \! \varvec{\nu }=\left( {\mathcal {L}}^{T}\varvec{m}\right) \!\cdot \!\varvec{\nu }=\varvec{m}\times \varvec{\nu },\qquad \text {as claimed.} \end{aligned}$$

(161)

We finally also point out that

$$\begin{aligned} \text {DIV}\,\left( {\mathcal {L}}^{T}\left( {\mathcal {L}}\,\text {D}\varvec{P}\right) \right)&=\left( {\mathcal {L}}^{T}\left( {\mathcal {L}}\,\text {D}\varvec{P}\right) \right) _{\alpha mn,n}=\varvec{\mathfrak {m}}_{\alpha mn,n}=\left( \varvec{m}\!\cdot \!\epsilon \right) _{\alpha mn,n}=\left( P_{\alpha j,in}\epsilon _{ij\beta }\epsilon _{\beta mn}\right) _{\alpha m}=\\&=\left( -\,P_{\alpha j,in}\epsilon _{ij\beta }\epsilon _{\beta nm}\right) _{\alpha m}=-\,\text {Curl}\,\text {Curl}\,\varvec{P}. \end{aligned}$$

since in three dimensions the Levi-Civita tensors is symmetric for a cyclical permutation of the indexes while antisymmetric for an anticyclic permutation.

1.6 Operatorial structure of the anisotropic micromorphic model with consistent coupling boundary conditions

In Romano et al. [73] the authors present the classical linear micromorphic model in the operatorial form which is very useful when we deal with the full anisotropic micromorphic model. In this section, we want to present the anisotropic micromorphic model with consistent coupling boundary condition in this operatorial form. Setting $\mathbb {H}_{1}:=H^{1}(\Omega ,\mathbb {R}^{3})\times H^{1}(\Omega ,\mathbb {R}^{3\times 3})$, let us introduce the formal operator

$$\begin{aligned} \begin{array}{rcl} \mathscr {A}_{1}:\mathbb {H}_{1} &{} -\!\!-\!\!\!\longrightarrow &{} \mathbb {H}_{2}\\ \\ \begin{pmatrix}\varvec{u}\\ \varvec{P}\end{pmatrix} &{} \longmapsto &{} \begin{pmatrix}\text {sym}\,\text {D} &{} -\text {sym}\,\\ \text {skew}\,\text {D} &{} -\text {skew}\,\\ 0 &{} \text {sym}\,\\ 0 &{} \text {D} \end{pmatrix}\!\cdot \!\begin{pmatrix}\varvec{u}\\ \varvec{P}\end{pmatrix}=\begin{pmatrix}\text {sym}\,\!\left( \text {D}\varvec{u}-\varvec{P}\right) \\ \text {skew}\,\!\left( \text {D}\varvec{u}-\varvec{P}\right) \\ \text {sym}\,\varvec{P}\\ \text {D}\varvec{P}\end{pmatrix} \end{array} \end{aligned}$$

(162)

where

$$\begin{aligned} \mathbb {H}_{2}:=L^{2}(\Omega ,\text {Sym}\,(3))\times L^{2}(\Omega ,\mathfrak {so}(3))\times H^{1}(\Omega ,\text {Sym}\,(3))\times L^{2}(\Omega ,\mathbb {R}^{3\times 3\times 3}). \end{aligned}$$

(163)

We introduce also

$$\begin{aligned} \begin{array}{rcl} \mathscr {A}_{2}:\mathbb {H}_{2} &{} -\!\!-\!\!\!\longrightarrow &{} \mathbb {H}_{2}\\ \\ \begin{pmatrix}\text {sym}\,\!\left( \text {D}\varvec{u}-\varvec{P}\right) \\ \text {skew}\,\!\left( \text {D}\varvec{u}-\varvec{P}\right) \\ \text {sym}\,\varvec{P}\\ \text {D}\varvec{P}\end{pmatrix} &{} \longmapsto &{} {\displaystyle \frac{1}{2}}\begin{pmatrix}\mathbb {C}_{\text {e}} &{} 0 &{} 0 &{} 0\\ 0 &{} \mathbb {C}_{\text {c}} &{} 0 &{} 0\\ 0 &{} 0 &{} \mathbb {C}_{\text {micro}} &{} 0\\ 0 &{} 0 &{} 0 &{} \mu L_{\text {c}}^{2}{\widehat{\mathbb {L}}} \end{pmatrix}\!\cdot \!\begin{pmatrix}\text {sym}\,\!\left( \text {D}\varvec{u}-\varvec{P}\right) \\ \text {skew}\,\!\left( \text {D}\varvec{u}-\varvec{P}\right) \\ \text {sym}\,\varvec{P}\\ \text {D}\varvec{P}\end{pmatrix}={\displaystyle \frac{1}{2}}\begin{pmatrix}\mathbb {C}_{\text {e}}\,\text {sym}\,\!\left( \text {D}\varvec{u}-\varvec{P}\right) \\ \mathbb {C}_{\text {c}}\,\text {skew}\,\!\left( \text {D}\varvec{u}-\varvec{P}\right) \\ \mathbb {C}_{\text {micro}}\,\text {sym}\,\varvec{P}\\ \mu L_{\text {c}}^{2}{\widehat{\mathbb {L}}}\,\text {D}\varvec{P}\end{pmatrix}. \end{array} \end{aligned}$$

The anisotropic potential energy density for the micromorphic model without mixed terms can be rewritten as follows

$$\begin{aligned} W\left( \text {D}\varvec{u},\varvec{P},\text {D}\varvec{P}\right) =\left\langle \frac{1}{2}\begin{pmatrix}\mathbb {C}_{\text {e}} &{} 0 &{} 0 &{} 0\\ 0 &{} \mathbb {C}_{\text {c}} &{} 0 &{} 0\\ 0 &{} 0 &{} \mathbb {C}_{\text {micro}} &{} 0\\ 0 &{} 0 &{} 0 &{} \mu L_{\text {c}}^{2}{\widehat{\mathbb {L}}} \end{pmatrix}\!\cdot \!\begin{pmatrix}\text {sym}\,\!\left( \text {D}\varvec{u}-\varvec{P}\right) \\ \text {skew}\,\!\left( \text {D}\varvec{u}-\varvec{P}\right) \\ \text {sym}\,\varvec{P}\\ \text {D}\varvec{P}\end{pmatrix},\begin{pmatrix}\text {sym}\,\!\left( \text {D}\varvec{u}-\varvec{P}\right) \\ \text {skew}\,\!\left( \text {D}\varvec{u}-\varvec{P}\right) \\ \text {sym}\,\varvec{P}\\ \text {D}\varvec{P}\end{pmatrix}\right\rangle , \end{aligned}$$

where

Since

$$\begin{aligned} L^{2}\!\left( \Omega ,\text {Sym}\,(3)\right) \times L^{2}\!\left( \Omega ,\mathfrak {so}(3)\right) \simeq L^{2}\!\left( \Omega ,\text {Sym}\,(3)\oplus \mathfrak {so}(3)\right) =L^{2}\!\left( \Omega ,\mathbb {R}^{3\times 3}\right) , \end{aligned}$$

(164)

we can identify

$$\begin{aligned} \begin{pmatrix} \mathbb {C}_{\text {e}}\,\text {sym}\,\!\left( \text {D}\varvec{u}-\varvec{P}\right) \\ \mathbb {C}_{\text {c}}\,\text {skew}\,\!\left( \text {D}\varvec{u}-\varvec{P}\right) \\ \mathbb {C}_{\text {micro}}\,\text {sym}\,\varvec{P}\\ \mu L_{\text {c}}^{2}\,{\widehat{\mathbb {L}}}\,\text {D}\varvec{P}\end{pmatrix} \longmapsto \begin{pmatrix} \mathbb {C}_{\text {e}}\,\text {sym}\,\!\left( \text {D}\varvec{u}-\varvec{P}\right) +\mathbb {C}_{\text {c}}\,\text {skew}\,\!\left( \text {D}\varvec{u}-\varvec{P}\right) \\ \mathbb {C}_{\text {micro}}\,\text {sym}\,\varvec{P}\\ \mu \,L_{\text {c}}^{2}\,{\widehat{\mathbb {L}}}\,\text {D}\varvec{P}\end{pmatrix} = \begin{pmatrix} \varvec{\sigma } \\ \mathbb {C}_{\text {micro}}\,\text {sym}\,\varvec{P}\\ \varvec{\mathfrak {m}}\end{pmatrix} \end{aligned}$$

(165)

and improving the regularity, i.e., asking for

$$\begin{aligned} \begin{pmatrix}\varvec{\sigma }\\ \mathbb {C}_{\text {micro}}\,\text {sym}\,\varvec{P}\\ \varvec{\mathfrak {m}}\end{pmatrix}&\in H\!\left( \text {Div}\,;\Omega ,\mathbb {R}^{3\times 3}\right) \times H^{1}\!\left( \Omega ,\text {Sym}\,\left( 3\right) \right) \times H\!\left( \text {DIV}\,;\Omega ,\mathbb {R}^{3\times 3\times 3}\right) =:\mathbb {H}_{3} \end{aligned}$$

(166)

we can write also the ensuing PDE system in strong form, via the equilibrium operator

$$\begin{aligned} \begin{array}{rcl} \mathscr {A}_{3}:\mathbb {H}_{3} &{} -\!\!-\!\!\!\longrightarrow &{} L^2(\Omega ,\mathbb {R}^3)\times L^2(\Omega ,\mathbb {R}^{3\times 3})\\ \\ \begin{pmatrix}\varvec{\sigma }\\ \mathbb {C}_{\text {micro}}\,\text {sym}\,\varvec{P}\\ \varvec{\mathfrak {m}}\end{pmatrix} &{} \longmapsto &{} \begin{pmatrix}\text {Div}\,&{} 0 &{} 0\\ \mathbb {1}&{} -\mathbb {1}&{} \text {DIV}\,\end{pmatrix}\!\cdot \!\begin{pmatrix}\varvec{\sigma }\\ \mathbb {C}_{\text {micro}}\,\text {sym}\,\varvec{P}\\ \varvec{\mathfrak {m}}\end{pmatrix}=\begin{pmatrix}\text {Div}\,\varvec{\sigma }\\ \\ \varvec{\sigma }-\mathbb {C}_{\text {micro}}\,\text {sym}\,\varvec{P}+\text {DIV}\,\varvec{\mathfrak {m}}\end{pmatrix}. \end{array} \end{aligned}$$

Assuming the partition $\Gamma \cup (\partial \Omega \setminus {\overline{\Gamma }})\subseteq \partial \Omega$, the boundary condition operator for regular fields is

1.6.1 Generalized consistent coupling boundary conditions

Recently, generalizations of theorem 1 have been proved in [37]. Let us consider the larger Sobolev spaces

$$\begin{aligned} H(\text {sym}\,\text {Curl}\,;\Omega ,\mathbb {R}^{3\times 3})&=\left\{ \varvec{P}\in L^{2}(\Omega ,\mathbb {R}^{3\times 3})\;\left. \right| \;\text {sym}\,\text {Curl}\,\varvec{P}\in L^{2}(\Omega ,\mathbb {R}^{3\times 3})\right\} ,\\ H(\text {dev}\,\text {sym}\,\text {Curl}\,;\Omega ,\mathbb {R}^{3\times 3})&=\left\{ \varvec{P}\in L^{2}(\Omega ,\mathbb {R}^{3\times 3})\;\left. \right| \;\text {dev}\,\text {sym}\,\text {Curl}\,\varvec{P}\in L^{2}(\Omega ,\mathbb {R}^{3\times 3})\right\} , \end{aligned}$$

equipped, respectively, with the norms

$$\begin{aligned} \left\| \varvec{P}\right\| _{H\left( \text {sym}\,\text {Curl}\,\!\right) }^{2}&=\left\| \varvec{P}\right\| _{L^{2}}^{2}+\left\| \text {sym}\,\text {Curl}\,\varvec{P}\right\| _{L^{2}}^{2},\\ \left\| \varvec{P}\right\| _{H\left( \text {dev}\,\text {sym}\,\text {Curl}\,\!\right) }^{2}&=\left\| \varvec{P}\right\| _{L^{2}}^{2}+\left\| \text {dev}\,\text {sym}\,\text {Curl}\,\varvec{P}\right\| _{L^{2}}^{2}. \end{aligned}$$

In theorem 3.3 [37] it is proved that

$$\begin{aligned} H_{0,\Gamma }(\text {sym}\,\text {Curl}\,;\Omega ,\mathbb {R}^{3\times 3})&:=\left\{ \varvec{P}\in H(\text {sym}\,\text {Curl}\,;\Omega ,\mathbb {R}^{3\times 3})\;\left. \right| \;\text {sym}\,\!(\!\left. \varvec{P}\times \varvec{\nu }\,\right| _{\Gamma })=0\right\} ,\\ H_{0,\Gamma }(\text {dev}\,\text {sym}\,\text {Curl}\,;\Omega ,\mathbb {R}^{3\times 3})&:=\left\{ \varvec{P}\in H(\text {dev}\,\text {sym}\,\text {Curl}\,;\Omega ,\mathbb {R}^{3\times 3})\;\left. \right| \;\text {dev}\,\text {sym}\,\!(\!\left. \varvec{P}\times \varvec{\nu }\,\right| _{\Gamma })=0\right\} , \end{aligned}$$

are closed subspaces of $H(\text {sym}\,\text {Curl}\,;\Omega ,\mathbb {R}^{3\times 3})$ and $H(\text {dev}\,\text {sym}\,\text {Curl}\,;\Omega ,\mathbb {R}^{3\times 3})$, respectively, and there exist $c_{1},c_{2}>0$ such that the following inequalities hold

$$\begin{aligned} \left\| \varvec{P}\right\| _{L^{2}}^{2}&\leqslant c_{1}\left( \left\| \text {sym}\,\varvec{P}\right\| _{L^{2}}^{2}+\left\| \text {sym}\,\text {Curl}\,\varvec{P}\right\| _{L^{2}}^{2}\right)&\forall \varvec{P}\in H_{0,\Gamma }(\text {sym}\,\text {Curl}\,;\Omega ,\mathbb {R}^{3\times 3}),\\ \left\| \varvec{P}\right\| _{L^{2}}^{2}&\leqslant c_{2}\left( \left\| \text {sym}\,\varvec{P}\right\| _{L^{2}}^{2}+\left\| \text {dev}\,\text {sym}\,\text {Curl}\,\varvec{P}\right\| _{L^{2}}^{2}\right)&\forall \varvec{P}\in H_{0,\Gamma }(\text {dev}\,\text {sym}\,\text {Curl}\,;\Omega ,\mathbb {R}^{3\times 3}). \end{aligned}$$

Moreover, the authors proved that the accounted boundary conditions allow to control $\text {skew}\,\varvec{P}$ with $\text {sym}\,\text {Curl}\,\varvec{P}$ and $\text {dev}\,\text {sym}\,\text {Curl}\,\varvec{P}$, respectively, in the introduced closed subspaces. The proposed boundary conditions were derived as usual via integration by parts. We can use these results to further weaken the consistent coupling boundary condition. Indeed, it is straightforward to prove that the two following boundary-value problems are well posed:

Problem 1: find $(\varvec{u},\varvec{P})\in H^{1}(\Omega ,\mathbb {R}^{3})\times H^{1}(\Omega ,\mathbb {R}^{3\times 3})$ such that

$$\begin{aligned} \left. \begin{aligned} \text {Div}\,\varvec{\sigma }&=0&\text {in}&\;\Omega \\ \quad \varvec{\sigma }-2\mu _{\text {micro}}\,\text {sym}\,\varvec{P}-\lambda _{\text {micro}}\text {tr}(\varvec{P})\mathbb {1}+\,\text {DIV}\,\varvec{\mathfrak {m}}&=0&\text {in}&\;\Omega \\ \\ \left. \varvec{u}\right| _{\Gamma }&={\widehat{\varvec{u}}}&\text {on}&\;\Gamma \\ \text {sym}\,\!(\!\left. \varvec{P}\times \varvec{\nu }\right| _{\Gamma })&=\text {sym}\,\!(\!\left. \text {D}\varvec{u}\times \varvec{\nu }\right| _{\Gamma })&\text {on}&\;\Gamma \\ \\ {\left. \varvec{\sigma }\!\cdot \!\varvec{\nu }\,\right| _{\partial \Omega \setminus {\overline{\Gamma }}}}\,&{=0}&\text {on}&\;\partial \Omega \setminus {\overline{\Gamma }}\\ {\left. \varvec{\mathfrak {m}}\!\cdot \!\varvec{\nu }\,\right| _{\partial \Omega \setminus {\overline{\Gamma }}}}\,&{=0}&\text {on}&\;\partial \Omega \setminus {\overline{\Gamma }}\\ {\left. \left( \varvec{\mathfrak {m}}\!\cdot \!\varvec{\nu }\right) \!\cdot \!\left( \varvec{\nu }\otimes \varvec{\nu }\right) \,\right| _{{{\varvec{\Gamma }}}}}\,&{=0}&\text {on}&\;\Gamma \\ {\text {skew}\,\!\left[ \!\left. \left( \varvec{\mathfrak {m}}\!\cdot \!\varvec{\nu }\right) \!\cdot \!\left( \mathbb {1}-\varvec{\nu }\otimes \varvec{\nu }\right) \, \right| _{{{\varvec{\Gamma }}}}\right] }\,&{=0}&\text {on}&\;\Gamma \end{aligned} \right\} \end{aligned}$$

where $\varvec{\sigma }$ and $\varvec{\mathfrak {m}}$ are as in (22) and (23),

Problem 2: find $(\varvec{u},\varvec{P})\in H^{1}(\Omega ,\mathbb {R}^{3})\times H^{1}(\Omega ,\mathbb {R}^{3\times 3})$ such that

$$\begin{aligned} \left. \begin{aligned} \text {Div}\,\varvec{\sigma }&=0&\text {in}&\;\Omega \\ \quad \varvec{\sigma }-2\mu _{\text {micro}}\,\text {sym}\,\varvec{P}-\lambda _{\text {micro}}\text {tr}(\varvec{P})\mathbb {1}+\,\text {DIV}\,\varvec{\mathfrak {m}}&=0&\text {in}&\;\Omega \\ \\ \left. \varvec{u}\right| _{\Gamma }&={\widehat{\varvec{u}}}&\text {on}&\;\Gamma \\ \text {dev}\,\text {sym}\,\!(\!\left. \varvec{P}\times \varvec{\nu }\right| _{\Gamma })&=\text {dev}\,\text {sym}\,\!(\!\left. \text {D}\varvec{u}\times \varvec{\nu }\right| _{\Gamma })&\text {on}&\;\Gamma \\ \\ {\left. \varvec{\sigma }\!\cdot \!\varvec{\nu }\,\right| _{\partial \Omega \setminus {\overline{\Gamma }}}}\,&{=0}&\text {on}&\;\partial \Omega \setminus {\overline{\Gamma }}\\ {\left. \varvec{\mathfrak {m}}\!\cdot \!\varvec{\nu }\,\right| _{\partial \Omega \setminus {\overline{\Gamma }}}}\,&{=0}&\text {on}&\;\partial \Omega \setminus {\overline{\Gamma }}\\ {\left. \left( \varvec{\mathfrak {m}}\!\cdot \!\varvec{\nu }\right) \!\cdot \!\left( \varvec{\nu }\otimes \varvec{\nu }\right) \,\right| _{{{\varvec{\Gamma }}}}}\,&{=0}&\text {on}&\;\Gamma \\ {\text {skew}\,\!\left[ \!\left. \left( \varvec{\mathfrak {m}}\!\cdot \!\varvec{\nu }\right) \!\cdot \!\left( \mathbb {1}-\varvec{\nu }\otimes \varvec{\nu }\right) \,\right| _{{{\varvec{\Gamma }}}}\right] }\,&{=0}&\text {on}&\;\Gamma \\ {\text {tr}\,\!\left[ \!\left. \left( \varvec{\mathfrak {m}}\!\cdot \!\varvec{\nu }\right) \!\cdot \!\left( \mathbb {1}-\varvec{\nu }\otimes \varvec{\nu }\right) \,\right| _{{{\varvec{\Gamma }}}}\right] }\,&{=0}&\text {on}&\;\Gamma \end{aligned} \right\} \end{aligned}$$

Exactly as in (51), we can prove that the following spaces

$$\begin{aligned} \mathscr {H}_{\text {sym}\,}^{\sharp }\!(\Omega )&:=H^{1}(\Omega ,\mathbb {R}^{3\times 3})\cap H_{0,\Gamma }(\text {sym}\,\text {Curl}\,;\Omega ,\mathbb {R}^{3\times 3}),\\ \mathscr {H}_{\text {dev}\,\text {sym}\,}^{\sharp }\!(\Omega )&:=H^{1}(\Omega ,\mathbb {R}^{3\times 3})\cap H_{0,\Gamma }(\text {dev}\,\text {sym}\,\text {Curl}\,;\Omega ,\mathbb {R}^{3\times 3}), \end{aligned}$$

equipped with the norm $\left\| \varvec{P}\right\| _{\sharp }^{2}=\left\| \text {sym}\,\varvec{P}\right\| _{L^2}^{2}+\left\| {\text {D}}\!\varvec{P}\right\| _{L^2}^{2}$ are Hilbert spaces (i.e., we cannot have generalized rigid body motion $\varvec{A}\in \mathfrak {so}(3)$) and that $\left\| \cdot \right\| _{\sharp }^{2}$ is equivalent to $\left\| \cdot \right\| _{1,2,\Omega }^{2}$. The well-posedness of the associated homogeneous problems

$$\begin{aligned} \int _{\Omega }W(\text {D}\varvec{u},\varvec{P},\text {D}\varvec{P})\,\text {d}V\longrightarrow \min ,\qquad \left( \varvec{u},\varvec{P}\right) \in H_{0,\Gamma }^{1}(\Omega ,\mathbb {R}^{3})\times \mathscr {H}_{\text {sym}\,}^{\sharp }\!(\Omega ) \end{aligned}$$

(167)

and

$$\begin{aligned} \int _{\Omega }W(\text {D}\varvec{u},\varvec{P},\text {D}\varvec{P})\,\text {d}V\longrightarrow \min ,\qquad \left( \varvec{u},\varvec{P}\right) \in H_{0,\Gamma }^{1}(\Omega ,\mathbb {R}^{3})\times \mathscr {H}_{\text {dev}\,\text {sym}\,}^{\sharp }\!(\Omega ), \end{aligned}$$

(168)

respectively, follows as before.