Asymptotic Analysis for Domains Separated by a Thin Layer Made of Periodic Vertical Beams

Abstract

We consider a thin heterogeneous layer consisting of thin beams (of radius $r$) and study the limit behavior of this problem as the period $\varepsilon$, the thickness $\delta$ and the radius $r$ of the beams tend to zero. The decomposition of the displacement field into beams developed by Griso (J. Math. Pures Appl. 89:199–223, 2008) is used, which allows to obtain a priori estimates. Two types of unfolding operators are introduced to deal with different parts of the decomposition. In conclusion, we obtain the limit problem together with transmission conditions across the interface.

Appendices

Appendix A: Proofs of Result of Sect. 3

Proof of Lemma 3.1

Applying the 2D-Poincaré-Wirtinger’s inequality we obtain the following estimate:

$$ \| u - \mathcal {U} \|_{L^{2} (B'_{r,\varepsilon })} \leq Cr \| \nabla u \| _{L^{2} (B'_{r,\varepsilon })}, $$

(A.1)

where the constant does not depend on $r$ and $\varepsilon$.

Step 1. Estimate of $\mathcal {R}(0)$.

Recalling the definition of ℛ from (3.2) and using $\int_{D_{r}} x_{1} \,dx_{1} \,dx_{2} = \int_{D_{r}} x_{2} \,dx_{1} \,dx_{2} =0$, we can write

$$ \forall x_{3}\in[-\varepsilon ,0], \quad\mathcal {R}_{1} (x_{3}) = \frac {1}{I_{2} r^{4}} \int_{D_{r}} x_{2} \bigl(u_{3}(x) - \mathcal {U}_{3}(x_{3}) \bigr) \,dx_{1} \,dx_{2}. $$

By Cauchy’s inequality

$$ \begin{aligned} \forall x_{3}\in[-\varepsilon ,0], \quad \bigl|\mathcal {R}_{1}(x_{3})\bigr|^{2} &\leq \frac{1}{I_{2}^{2}r^{8}} \int_{D_{r}} x^{2} \,dx_{1}\,dx_{2} \times \int_{D_{r}} \bigl(u_{3}(x) - \mathcal {U}_{3}(x_{3})\bigr)^{2} \,dx_{1}\,dx_{2} \\ &\leq\frac{C}{r^{4}} \int_{D_{r}} \bigl(u_{3}(x) - \mathcal {U}_{3}(x_{3})\bigr)^{2} \,dx_{1} \,dx_{2}. \end{aligned} $$

Integrating with respect to $x_{3}$ gives

$$ \int_{-\varepsilon }^{0} \bigl|\mathcal {R}_{1}(x_{3})\bigr|^{2} \,dx_{3} \leq\frac{C}{r^{4}} \int_{B_{r,\varepsilon }} \bigl(u(x) - \mathcal {U}(x_{3}) \bigr)^{2} \,dx. $$

Using (A.1), we can write

$$ \| \mathcal {R}_{1} \|_{L^{2} (-\varepsilon ,0)} \leq \frac{C}{r} \| \nabla u \|_{L^{2} (B'_{r,\varepsilon })}. $$

(A.2)

The derivative of $\mathcal {R}_{1}$ is equal to

$$\frac{d \mathcal {R}_{1}}{dx_{3}} (x_{3}) = \frac{1}{I_{2} r^{4}} \int_{D_{r}} x_{2} \frac{\partial u_{3}(x)}{\partial x_{3}} \,dx_{1} \,dx_{2} $$

for a.e. $x_{3}\in(-\varepsilon ,0)$. Then proceeding as above, we obtain for a.e. $x_{3}\in(-\varepsilon ,0)$

$$\biggl\vert \frac{d \mathcal {R}_{1}}{dx_{3}} (x_{3}) \biggr\vert ^{2} \leq\frac {C}{r^{4}} \int_{D_{r}} \biggl\vert \frac{\partial u_{3}(x)}{\partial x_{3}} \biggr\vert ^{2} \,dx_{1} \,dx_{2}. $$

Hence

$$ \biggl\Vert \frac{d \mathcal {R}_{1}}{dx_{3}} \biggr\Vert _{L^{2}(-\varepsilon ,0)} \leq\frac{C}{r^{2}} \biggl\Vert \frac{\partial u_{3}}{\partial x_{3}} \biggr\Vert _{L^{2}(B'_{r,\varepsilon })}\leq\frac{C}{r^{2}} \| \nabla u \|_{L^{2} (B'_{r,\varepsilon })}. $$

(A.3)

We recall following classical estimates for $\phi\in H^{1}(-a,0)$, where $a>0$

$$ \begin{aligned} &\bigl|\phi(0)\bigr|^{2}\le \frac{2}{a}\|\phi\|^{2}_{L^{2}(-a,0)}+\frac{a}{2} \bigl\| \phi '\bigr\| ^{2}_{L^{2}(-a,0)}, \\ &\|\phi\|^{2}_{L^{2}(-a,0)}\le{2 a}\bigl|\phi(0)\bigr|^{2}+{a^{2}} \bigl\| \phi'\bigr\| ^{2}_{L^{2}(-a,0)}. \end{aligned} $$

(A.4)

Due to (A.2)–(A.3), (A.4)₁ with $a=r$ and $\varepsilon > r$, $\mathcal {R}_{1}(0)$ satisfies

$$\begin{gathered} \bigl| \mathcal {R}_{1} (0)\bigr|^{2} \leq\frac{C}{r^{3}} \| \nabla u \|^{2}_{L^{2} (B'_{r,\varepsilon })}. \end{gathered}$$

The estimates for $\mathcal {R}_{2}(0)$, $\mathcal {R}_{3}(0)$ are obtained in the same way. Hence we get (3.6)₁.

Step 2. Estimate of $\| \mathcal {R} \|_{L^{2} (0, \delta)}$.

Poincaré’s inequality yields

$$\bigl\| \mathcal {R} - \mathcal {R}(0)\bigr\| _{L^{2} (0, \delta)} \leq\delta\biggl\Vert \frac{d \mathcal {R}}{dx_{3}} \biggr\Vert _{L^{2}(0, \delta)}. $$

From (3.5)₃, (A.4)₂ and (3.6)₁ we get

$$ \| \mathcal {R} \|^{2}_{L^{2} (0, \delta)} \leq2\delta \bigl|\mathcal {R}(0) \bigr|^{2} + \delta^{2} \biggl\Vert \frac{d \mathcal {R}}{dx_{3}} \biggr\Vert ^{2}_{L^{2}(0, \delta)} \leq C \frac{\delta}{r^{3}} \| \nabla u \|^{2}_{L^{2} (B'_{r,\varepsilon })} + C \frac{\delta^{2}}{r^{4}} \bigl\| e(u) \bigr\| ^{2}_{L^{2} (B_{r, \delta})}. $$

(A.5)

Hence (3.6)₂ is proved.

Step 3. Estimate of $\mathcal {U}-\mathcal {U}(0)$.

Applying the inequality (3.5)₄ from Theorem 3.1 the following estimates on $\mathcal {U}$ hold:

$$\begin{gathered} \begin{aligned} \biggl\Vert \frac{d \mathcal {U}_{3}}{dx_{3}} \biggr\Vert _{L^{2} (0, \delta)} & \leq\frac{C}{r} \bigl\| e(u)\bigr\| _{L^{2}(B_{r, \delta})}, \\ \biggl\Vert \frac{d \mathcal {U}_{\alpha}}{dx_{3}} \biggr\Vert _{L^{2} (0, \delta )} & \leq\| \mathcal {R} \|_{L^{2} (0, \delta)} + \frac{C}{r} \bigl\| e(u) \bigr\| _{L^{2}(B_{r, \delta})}. \end{aligned} \end{gathered}$$

(A.6)

Combining (A.6)₂ with (A.5) gives

$$ \biggl\Vert \frac{d \mathcal {U}_{\alpha}}{dx_{3}} \biggr\Vert _{L^{2} (0, \delta )}^{2} \leq C \frac{\delta}{r^{3}} \| \nabla u \|^{2}_{L^{2} (B'_{r,\varepsilon })} + C \frac{\delta^{2}}{r^{4}} \bigl\| e(u) \bigr\| ^{2}_{L^{2} (B_{r, \delta})} + \frac{C}{r^{2}} \bigl\| e(u) \bigr\| ^{2}_{L^{2}(B_{r, \delta})}. $$

Taking into account assumption (2.7)₂, we obtain (3.6)₃. Then by (3.6)₃, (A.6)₁ and Poincaré’s inequality formulas (3.6)₄, (3.6)₅ follow.

Step 4. We prove the estimate (3.6)₆.

By Korn’s inequality, there exists a rigid displacement $\mathbf{r}$

$$\begin{gathered} \begin{aligned} &\mathbf{r}(x) = \mathbf{a} + \mathbf{b} \wedge \biggl( x + \frac {\varepsilon }{2} {\mathbf{e_{3}}} \biggr), \\ &{\mathbf{a} = \frac{1}{\varepsilon ^{3}} \int_{W_{\varepsilon }} u(x) \, dx}, \\ &\mathbf{b} = \frac{6}{\varepsilon ^{5}} \int_{W_{\varepsilon }} \biggl( x + \frac{\varepsilon }{2} {\mathbf{e_{3}}} \biggr) \wedge u(x) \, dx, \end{aligned} \end{gathered}$$

such that

$$\begin{gathered} \begin{aligned} \| u - \mathbf{r} \|_{L^{2} (W_{\varepsilon })} &\leq C \varepsilon \bigl\| e(u) \bigr\| _{L^{2} (W_{\varepsilon })}, \\ \bigl\| \nabla(u - \mathbf{r}) \bigr\| _{L^{2} (W_{\varepsilon })} &\leq C \bigl\| e(u) \bigr\| _{L^{2} (W_{\varepsilon })}. \end{aligned} \end{gathered}$$

(A.7)

Besides, by Poincaré-Wirtinger inequality we have

$$ \| u - \mathbf{a} \|_{L^{2} (W_{\varepsilon })} \leq C \varepsilon \| \nabla u \|_{L^{2} (W_{\varepsilon })}. $$

(A.8)

Using the following Sobolev embedding theorem

$$W^{s, p} \subset W^{t, q}, \quad\text{for } s, p, t, q: s \in \mathbb{R},\ s > t,\ p \in\mathbb{N} \text{ and } \frac{1}{q} = \frac{1}{p} - \frac{s - t}{n} $$

we get

$$ H^{1/2} \subset L^{4} \quad\text{or} \quad\| \varphi \|_{L^{4} (Y)} \leq C \| \varphi\|_{H^{1/2} (Y)}, \quad\forall\varphi\in H^{1/2} (Y). $$

Moreover, by definition of the $H^{1/2}$ norm for $W=Y\times(-1,0)$, we have

$$ \| \varphi\|_{H^{1/2} (Y)} \leq C \bigl( \| \varphi\|_{L^{2} (W)} + \| \nabla \varphi\|_{L^{2} (W)} \bigr), \quad\forall\varphi\in H^{1} (W). $$

Therefore,

$$ \| \varphi\|_{L^{4} (Y)} \leq C \bigl( \| \varphi\|_{L^{2} (W)} + \| \nabla \varphi\|_{L^{2} (W)} \bigr), \quad\forall\varphi\in H^{1} (W). $$

With the change of variables

$$y_{\varepsilon } = \varepsilon y, \quad\text{for } (y_{1}, y_{2}) \in Y,\ (y_{1}, y_{2}, y_{3}) \in W, $$

and defining

$$\varphi_{\varepsilon } (y_{\varepsilon }) = \varphi \biggl( \frac {y_{\varepsilon }}{\varepsilon } \biggr), $$

we obtain

$$ \frac{1}{\sqrt{\varepsilon}}\| \varphi_{\varepsilon } \|_{L^{4} (Y_{\varepsilon })} \leq C \biggl( \frac{1}{\varepsilon ^{3/2}}\| \varphi_{\varepsilon } \|_{L^{2} (W_{\varepsilon })} + \frac{1}{\sqrt {\varepsilon}}\| \nabla\varphi_{\varepsilon } \|_{L^{2} (W_{\varepsilon })} \biggr), \quad\forall\varphi_{\varepsilon } \in H^{1} (W_{\varepsilon }) $$

or

$$ \| \varphi\|_{L^{4} (Y_{\varepsilon })} \leq C \biggl( \frac {1}{\varepsilon }\| \varphi \|_{L^{2} (W_{\varepsilon })} + \| \nabla \varphi\|_{L^{2} (W_{\varepsilon })} \biggr), \quad\forall \varphi \in H^{1} (W_{\varepsilon }). $$

Therefore, (A.7) and the inequality above lead to

$$ \| u - \mathbf{r} \|_{L^{4} (Y_{\varepsilon })} \leq C \bigl\| e(u)\bigr\| _{L^{2}(W_{\varepsilon })}. $$

(A.9)

From the identity

$$ \frac{1}{\pi r^{2}} \int_{D_{r}} \bigl(u\bigl(x', 0\bigr) - \mathbf{r} \bigl(x', 0\bigr)\bigr) \,dx' = \mathcal {U} (0) - \mathbf{a} - \mathbf{b} \wedge\frac{\varepsilon }{2} e_{3}, $$

the estimate (A.9) and Hölder’s inequality we get

$$\begin{aligned} \biggl\vert \mathcal {U} (0) - \mathbf{a} - \mathbf{b} \wedge \frac{\varepsilon }{2} e_{3}\biggr\vert \leq& \frac{1}{\pi r^{2}} \biggl(\int _{D_{r}} 1^{4/3} \,dx' \biggr)^{3/4} \biggl( \int_{D_{r}} \bigl|u\bigl(x', 0\bigr) - \mathbf{r}\bigl(x', 0\bigr)\bigr|^{4} \,dx' \biggr)^{1/4} \\ \leq& \frac{C }{r^{1/2}} \bigl\| e(u) \bigr\| _{L^{2} (W_{\varepsilon })}. \end{aligned}$$

(A.10)

From Cauchy–Schwarz’s inequality and taking into account (A.8), we derive

$$\begin{aligned} |{\mathbf{b}}| \leq& \frac{C}{\varepsilon ^{5}} \biggl( \int _{W_{\varepsilon }} \biggl\vert x + \frac{\varepsilon }{2} {\mathbf {e_{3}}} \biggr\vert ^{2} \,dx \biggr)^{1/2} \biggl( \int_{W_{\varepsilon }} \bigl| u(x) - \mathbf{a} \bigr|^{2} \,dx \biggr)^{1/2} \\ \leq&\frac{C}{\varepsilon ^{5}} \cdot\varepsilon \cdot\varepsilon ^{3/2} \| u - \mathbf{a} \|_{L^{2} (W_{\varepsilon })} \leq\frac {C}{\varepsilon ^{5/2}} \varepsilon \| \nabla u \|_{L^{2} (W_{\varepsilon })} \leq\frac{C}{\varepsilon ^{3/2}} \| \nabla u \| _{L^{2} (W_{\varepsilon })}. \end{aligned}$$

(A.11)

Using (A.10) and (A.11), we obtain

$$\begin{aligned} \bigl\vert \mathcal {U}(0) - \mathbf{a} \bigr\vert \leq&\biggl\vert \mathcal {U} (0) - \mathbf{a} - \mathbf{b} \wedge\frac{\varepsilon }{2} { \mathbf{e_{3}}} \biggr\vert + \biggl\vert {\mathbf{b}} \wedge \frac {\varepsilon }{2} {\mathbf{e_{3}}} \biggr\vert \\ \leq& \frac{C }{r^{1/2}} \bigl\| e(u) \bigr\| _{L^{2} (W_{\varepsilon })} + \frac{C}{\varepsilon ^{1/2}} \| \nabla u \|_{L^{2} (W_{\varepsilon })}. \end{aligned}$$

(A.12)

The estimates (A.4) and (A.8) yield

$$ \bigl\| u(\cdot, \cdot, 0) - \mathbf{a} \bigr\| ^{2}_{L^{2}(Y_{\varepsilon })} \leq C \varepsilon \| \nabla u \|^{2}_{L^{2} (W_{\varepsilon })}. $$

(A.13)

Combining (A.12) and (A.13) gives

$$ \begin{aligned} \bigl\| u(\cdot, \cdot, 0) - \mathcal {U}(0) \bigr\| ^{2}_{L^{2} (Y_{\varepsilon })} &\leq C \bigl(\bigl\| u (\cdot, \cdot, 0) - \mathbf{a} \bigr\| ^{2}_{L^{2} (Y_{\varepsilon })} + \bigl\| \mathcal {U}(0) - \mathbf{a}\bigr\| ^{2}_{L^{2} (Y_{\varepsilon })}\bigr) \\ &\leq C \varepsilon \| \nabla u \|^{2}_{L^{2} (W_{\varepsilon })} + C \frac{\varepsilon ^{2}}{r} \bigl\| e(u) \bigr\| ^{2}_{L^{2} (W_{\varepsilon })} + C\varepsilon \|\nabla u \|^{2}_{L^{2} (W_{\varepsilon })} \\ &\leq C \varepsilon \| \nabla u \|^{2}_{L^{2} (W_{\varepsilon })} + C \frac{\varepsilon ^{2}}{r} \bigl\| e(u) \bigr\| ^{2}_{L^{2} (W_{\varepsilon })}. \end{aligned} $$

Hence we get (3.6)₆. □

Proof of Lemma 3.3

Using (3.6)₆ and then summing over all of the periodicity cells gives

$$ \bigl\| u(\cdot, \cdot, 0) - \widetilde{\mathcal {U}}(\cdot, \cdot, 0) \bigr\| ^{2}_{L^{2} (\widehat{\omega}_{\varepsilon })} \leq C \varepsilon \| \nabla u \|^{2}_{L^{2} (\varOmega^{-})} + C \frac{\varepsilon ^{2}}{r} \| u \|^{2}_{V}. $$

(A.14)

In the same way, the following estimate is derived:

$$ \bigl\| u(\cdot, \cdot, \delta) - \widetilde{\mathcal {U}}(\cdot, \cdot, \delta) \bigr\| ^{2}_{L^{2} (\widehat{\omega}_{\varepsilon })} \leq C \varepsilon \| \nabla u \|^{2}_{L^{2} (\varOmega^{+}_{\delta})} + C \frac {\varepsilon ^{2}}{r} \| u \|^{2}_{V}. $$

Applying (3.7)₂ we can write

$$ \bigl\| \widetilde{\mathcal {U}}_{3}(\cdot, \cdot, \delta) - \widetilde {\mathcal {U}}_{3} (\cdot, \cdot, 0)\bigr\| _{L^{2} (\widehat{\omega }_{\varepsilon })}^{2} \leq\delta\biggl\Vert \frac{\partial\widetilde {\mathcal {U}}_{3}}{\partial x_{3}} \biggr\Vert ^{2}_{L^{2} (\widehat{\omega }_{\varepsilon } \times(0, \delta))} \leq C \frac{\varepsilon ^{2} \delta}{r^{2}} \| u \|^{2}_{V}. $$

(A.15)

From (3.7)₆ we have

$$ \bigl\| \widetilde{\mathcal {U}}_{\alpha} (\cdot, \cdot, \delta) - \widetilde{\mathcal {U}}_{\alpha}(\cdot, \cdot, 0)\bigr\| _{L^{2} (\widehat {\omega}_{\varepsilon })}^{2} \leq C \frac{\varepsilon ^{2} \delta ^{2}}{r^{3}} \| u \|^{2}_{V} + C \frac{\varepsilon ^{2} \delta^{3}}{r^{4}} \| u \| ^{2}_{V} \leq C \frac{\varepsilon ^{2} \delta^{3}}{r^{4}} \| u \|^{2}_{V}. $$

(A.16)

Using (A.16) and the estimates above we obtain (3.11), (3.12). □

Proof of Lemma 3.4

From Korn’s inequality and the trace theorem we derive

$$\begin{gathered} \begin{aligned} \| u \|_{L^{2} (\varSigma)} &\leq C \| u \|_{H^{1} (\varOmega^{-})} \leq C_{1} \bigl\| e(u) \bigr\| _{L^{2} (\varOmega^{-})},\\ \| u \|_{H^{1} (\varOmega_{\delta}^{+})} &\leq C \bigl(\bigl\| e(u) \bigr\| _{L^{2}(\varOmega_{\delta}^{+})} + \| u \|_{L^{2} (\varSigma^{+}_{\delta})} \bigr). \end{aligned} \end{gathered}$$

(A.17)

We know that there exists a rigid displacement $\mathbf{r}$

$$\forall x\in{\mathbb{R}}^{3},\quad\mathbf{r}(x)=\mathbf{a}+\mathbf {b}\land(x-\delta{\mathbf{e_{3}}}),\quad\mathbf{a},\mathbf{b}\in {\mathbb{R}}^{3}, $$

such that

$$ \| u - \mathbf{r} \|_{L^{2} (\varSigma_{\delta}^{+})} \leq C \|u-\mathbf {r} \|_{H^{1}(\varOmega_{\delta}^{+})}\leq C\bigl\| e(u) \bigr\| _{L^{2}(\varOmega_{\delta}^{+})}, $$

(A.18)

where the constant does not depend on $\delta$ (since $|\varSigma _{\delta}^{+}|$ is independent of $\delta$ and $\varOmega_{\delta }^{+}\subset\varOmega^{+}$ and $|\varOmega^{+}\setminus\varOmega_{\delta }^{+}|=|\omega|\delta$). Then, we get

$$ \bigl\| (u-\mathbf{r}) (\cdot, \cdot, \delta)\bigr\| _{L^{2}(\widehat{\omega }_{\varepsilon })} \leq\| u - \mathbf{r} \|_{L^{2} (\varSigma_{\delta }^{+})} \leq C\bigl\| e(u) \bigr\| _{L^{2}(\varOmega_{\delta}^{+})}. $$

(A.19)

Using

$$ \bigl\| u(\cdot, \cdot, 0) \bigr\| _{L^{2} (\widehat{\omega}_{\varepsilon })} \leq\| u \|_{L^{2} (\varSigma)}, $$

(A.20)

from (3.11) and (3.12) we obtain

$$\begin{gathered} \begin{aligned} \bigl\| u_{\alpha}(\cdot, \cdot, \delta) \bigr\| _{L^{2} (\widehat{\omega }_{\varepsilon })} &\leq C \varepsilon ^{1/2} \| \nabla u \|_{L^{2} (\varOmega^{+}_{\delta})} + C \frac{\varepsilon \delta^{3/2}}{r^{2}} \| u \|_{V} + C\| u \|_{V}, \\ \bigl\| u_{3}(\cdot, \cdot, \delta) \bigr\| _{L^{2} (\widehat{\omega }_{\varepsilon })} &\leq C \varepsilon ^{1/2} \| \nabla u \|_{L^{2} (\varOmega^{+}_{\delta})} + C \frac{\varepsilon ^{3/2} \delta^{1/2}}{r} \| u \|_{V} + C\| u \|_{V}. \end{aligned} \end{gathered}$$

(A.21)

Combining this with (A.19) gives

$$\begin{gathered} \begin{aligned} \bigl\| {\mathbf{r}}_{\alpha} (\cdot, \cdot, \delta) \bigr\| _{L^{2}(\widehat {\omega}_{\varepsilon })} &\leq C \varepsilon ^{1/2} \| \nabla u \| _{L^{2} (\varOmega^{+}_{\delta})} + C \frac{\varepsilon \delta ^{3/2}}{r^{2}} \| u \|_{V} + C\| u \|_{V}, \\ \bigl\| {\mathbf{r}}_{3} (\cdot, \cdot, \delta) \bigr\| _{L^{2}(\widehat{\omega }_{\varepsilon })} &\leq C \varepsilon ^{1/2} \| \nabla u \|_{L^{2} (\varOmega^{+}_{\delta})} + C \frac{\varepsilon \delta^{1/2}}{r} \| u \| _{V} + C\| u \|_{V}. \end{aligned} \end{gathered}$$

(A.22)

Therefore,

$$\begin{gathered} |{\mathbf{a_{1}}}| + |{\mathbf{a_{2}}}| + |{ \mathbf{b_{3}}}| \leq C \varepsilon ^{1/2} \| \nabla u \|_{L^{2} (\varOmega^{+}_{\delta})} + C \frac{\varepsilon \delta^{3/2}}{r^{2}} \| u \|_{V} + C\| u \|_{V}, \\ |{\mathbf{a_{3}}}| + |{\mathbf{b_{1}}}| + |{ \mathbf{b_{2}}}| \leq C \varepsilon ^{1/2} \| \nabla u \|_{L^{2} (\varOmega^{+}_{\delta})} + C \frac{\varepsilon \delta^{1/2}}{r} \| u \|_{V} + C\| u \|_{V}. \end{gathered}$$

These estimates together with (A.18) allow us to obtain estimates on $u_{1}, u_{2}, u_{3}$. This yields

$$\begin{gathered} \begin{aligned} \| u_{\alpha} \|_{H^{1} (\varOmega_{\delta}^{+})} &\leq C \varepsilon ^{1/2} \| \nabla u \|_{L^{2} (\varOmega^{+}_{\delta})} + C \frac {\varepsilon \delta^{3/2}}{r^{2}} \| u \|_{V} + C\| u \|_{V}, \\ \| u_{3} \|_{H^{1} (\varOmega_{\delta}^{+})} &\leq C \varepsilon ^{1/2} \| \nabla u \|_{L^{2} (\varOmega^{+}_{\delta})} + C \frac{\varepsilon \delta ^{1/2}}{r} \| u \|_{V} + C\| u \|_{V}. \end{aligned} \end{gathered}$$

(A.23)

Therefore,

$$ \| \nabla u \|_{L^{2} (\varOmega_{\delta}^{+})} \leq C \varepsilon ^{1/2} \| \nabla u \|_{L^{2} (\varOmega^{+}_{\delta})} + C \frac{\varepsilon \delta^{3/2}}{r^{2}} \| u \|_{V} + C\| u \|_{V}. $$

For $\varepsilon$ small enough the following holds true:

$$ \| \nabla u \|_{L^{2} (\varOmega_{\delta}^{+})} \leq C \frac{\varepsilon \delta^{3/2}}{r^{2}} \| u \|_{V} + C\| u \|_{V}. $$

Inserting this in (A.23) we derive (3.13)–(3.14). □

Proof of Lemma 3.5

From the estimates in Theorem 3.1, (3.6)₂ and (3.6)₃ and after summation over all the beams, we get (making use of the assumption (2.7)₂)

$$ \|\nabla u\|^{2}_{L^{2} (\varOmega_{r, \varepsilon , \delta}^{i})}\le C \biggl( \frac{\delta}{r} \| \nabla u \|^{2}_{L^{2} (\varOmega^{-})} +\frac{\delta ^{2}}{r^{2}} \bigl\| e(u) \bigr\| ^{2}_{L^{2} (\varOmega_{r, \varepsilon , \delta }^{i})} \biggr) \leq C \frac{\delta^{2}}{r^{2}} \| u \|^{2}_{V}. $$

(A.24)

From (A.14) and (A.17)₁, it follows that

$$\begin{aligned} \sum_{\xi\in\varXi_{\varepsilon }} \varepsilon ^{2} \bigl| \mathcal {U}_{\xi} (0) \bigr|^{2} =& \bigl\| \widetilde{\mathcal {U}} (\cdot, \cdot, 0) \bigr\| ^{2}_{L^{2} (\widehat{\omega}_{\varepsilon })} \leq C \frac{\varepsilon ^{2}}{r}\| u \|^{2}_{V} + C\| u \|^{2}_{V}, \\ \sum_{\xi\in\varXi_{\varepsilon }} \bigl| \mathcal {U}_{\xi} (0)\bigr|^{2} \leq& C \biggl(\frac{1}{r} + \frac{1}{\varepsilon ^{2}} \biggr)\| u \|^{2}_{V}. \end{aligned}$$

Using (3.5)₄, (3.6)₃, (A.4), we obtain

$$\begin{gathered} \begin{aligned} \sum_{\xi\in\varXi_{\varepsilon }} \| \mathcal {U}_{\xi, 3} \|^{2}_{L^{2} (0, \delta)} &\leq C \biggl( \frac{\delta }{\varepsilon ^{2}} + \frac {\delta^{2}}{r^{2}} \biggr) \| u \|^{2}_{V}, \\ \sum_{\xi\in\varXi_{\varepsilon }}\| \mathcal {U}_{\xi, \alpha} \| ^{2}_{L^{2} (0, \delta)} &\leq C \biggl( \frac{\delta }{\varepsilon ^{2}} + \frac{\delta^{4}}{r^{4}} \biggr) \| u \|^{2}_{V}. \end{aligned} \end{gathered}$$

(A.25)

Additionally,

$$ \sum_{\xi\in\varXi_{\varepsilon }} \| \bar{u}_{\xi} \|^{2}_{L^{2} (B_{r, \delta})} \leq C r^{2} \bigl\| e(u)\bigr\| ^{2}_{L^{2} (\varOmega_{r, \varepsilon , \delta}^{i})} \leq C r^{2} \| u \|^{2}_{V}. $$

(A.26)

Then (3.6)₂, (A.25) and (A.26) give

$$\begin{aligned} \sum_{\xi\in\varXi_{\varepsilon }} \| u_{\xi, \alpha} \|^{2}_{L^{2} (B_{r, \delta})} \le& C \biggl( \frac{r^{2}\delta }{\varepsilon ^{2}} + \frac{\delta^{4}}{r^{2}} + \delta^{2} + r^{2} \biggr) \| u \|^{2}_{V} \leq C \frac{r^{2}\delta }{\varepsilon ^{2}} \biggl(1+ \frac{\varepsilon ^{2}\delta^{3}}{r^{4}} \biggr) \| u \|^{2}_{V}, \\ \sum_{\xi\in\varXi_{\varepsilon }} \| u_{\xi, 3} \|^{2}_{L^{2} (B_{r, \delta})} \le& C \biggl( \frac{r^{2}\delta }{\varepsilon ^{2}} + \delta ^{2} + r^{2} \biggr) \| u \|^{2}_{V} \leq C \frac{r^{2}\delta }{\varepsilon ^{2}} \biggl( 1 + \frac{\varepsilon ^{2}\delta}{r^{2}} \biggr) \| u \|^{2}_{V}. \end{aligned}$$

From the last inequalities, we derive (3.17)₂ and (3.17)₃. □

Appendix B: Properties of the Periodic Unfolding Operators. Estimates for the Compactness

Lemma B.1

(Properties of the operators $\mathcal {T}_{\varepsilon }$, $\mathcal {T}_{\varepsilon }^{'}$)

• 1. $\forall v,w \in L^{2}(\omega\times(0, \delta))$

  $$\mathcal {T}_{\varepsilon }(vw) = \mathcal {T}_{\varepsilon }(v) \mathcal {T}_{\varepsilon }(w), $$

  $\forall v,w \in L^{2}(\omega\times B_{r, \delta})$

  $$\mathcal {T}_{\varepsilon }^{'}(vw) = \mathcal {T}_{\varepsilon }^{'}(v) \mathcal {T}_{\varepsilon }^{'}(w). $$

• 2. $\forall u \in L^{1}(\omega\times(0, \delta))$

  $$\delta \int_{\omega\times(0, 1)} \mathcal {T}_{\varepsilon }(u) \, ds_{1} \,ds_{2}\,dX_{3} = \int_{\widehat{\omega}_{\varepsilon } \times(0, \delta)} u\,ds_{1}\,ds_{2} \,dx_{3}, $$

  $\forall u \in L^{1}(\omega\times B_{r, \delta})$

  $$r^{2} \delta \int_{\omega\times B_{1}} \mathcal {T}_{\varepsilon }^{'}(u)\, ds_{1}\,ds_{2}\,dX_{1}\,dX_{2} \,dX_{3} = \int_{\widehat{\omega}_{\varepsilon } \times B_{r, \delta}} u\,ds_{1}\,ds_{2} \,dx_{1}\,dx_{2}\,dx_{3}. $$

• 3. $\forall u \in L^{2}(\omega\times(0, \delta))$

  $$\bigl\| \mathcal {T}_{\varepsilon }(u) \bigr\| _{L^{2} (\omega\times(0, 1))} \leq \frac{1}{\sqrt{\delta}} \| u \|_{L^{2}(\omega\times(0, \delta))}, $$

  $\forall u \in L^{2}(\omega\times B_{r, \delta})$

  $$\bigl\| \mathcal {T}_{\varepsilon }^{'}(u) \bigr\| _{L^{2} (\omega\times B_{1})} \leq \frac{1}{r \sqrt{\delta}} \| u\|_{L^{2}(\omega\times B_{r, \delta})}. $$

• 4. Let $u$ be in $L^{2}(\omega, H^{1}(0, \delta))$. Then we have

  $$\delta\mathcal {T}_{\varepsilon }(\nabla_{x_{3}} u) = \nabla_{X_{3}} \mathcal {T}_{\varepsilon } (u). $$

  Let $u$ be in $L^{2}(\omega, H^{1}(B_{r, \delta}))$. Then we have

  $$r \mathcal {T}_{\varepsilon }^{'}(\nabla_{x_{\alpha}} u) = \nabla _{X_{\alpha}} \mathcal {T}_{\varepsilon }^{'} (u), \qquad\delta \mathcal {T}_{\varepsilon }^{'}(\nabla_{x_{3}} u) = \nabla_{X_{3}} \mathcal {T}_{\varepsilon }^{'} (u),\quad \textit{where } \alpha= 1,2. $$

Proof

Properties 1–3 are obtained similarly to the proof of Lemma 5.1 of [2].

Property 4 is the direct consequence of the chain rule:

$$\begin{gathered} \frac{\partial(\mathcal {T}_{\varepsilon }^{'} (u))}{\partial X_{\alpha}} = r \mathcal {T}_{\varepsilon }^{'} \biggl( \frac{\partial u}{\partial x_{\alpha}} \biggr), \quad\alpha= 1, 2, \\ \frac{\partial(\mathcal {T}_{\varepsilon } (u))}{\partial X_{3}} = \delta\mathcal {T}_{\varepsilon } \biggl( \frac{\partial u}{\partial x_{3}} \biggr), \qquad\frac{\partial(\mathcal {T}_{\varepsilon }^{'} (u))}{\partial X_{3}} = \delta\mathcal {T}_{\varepsilon }^{'} \biggl( \frac{\partial u}{\partial x_{3}} \biggr). \end{gathered}$$

 □

From Lemmas 3.2 and B.1 we obtain the following result.

Lemma B.2

There exists a constant $C$, independent of $\varepsilon$, $\delta$ and $r$, such that

$$\begin{gathered} \bigl\| \mathcal {T}_{\varepsilon }(\widetilde{\mathcal {U}}_{\varepsilon }) \bigr\| _{L^{2}(\omega,H^{1}(0, 1))} \leq C , \end{gathered}$$

(B.1)

$$\begin{gathered} \bigl\| \mathcal {T}_{\varepsilon }(\widetilde{\mathcal {U}}_{\varepsilon ,3}) - \widetilde{\mathcal {U}}_{\varepsilon ,3 }(\cdot, \cdot, 0)\bigr\| _{L^{2}(\omega, H^{1}(0, 1))} \leq C \frac{r}{\delta} , \end{gathered}$$

(B.2)

$$\begin{gathered} \bigl\| \mathcal {T}_{\varepsilon }(\widetilde{\mathcal {R}}_{\varepsilon }) \bigr\| _{L^{2}(\omega, H^{1} (0, 1))} \leq\frac{C}{\delta} , \end{gathered}$$

(B.3)

$$\begin{gathered} \biggl\| \frac{\partial\mathcal {T}_{\varepsilon }(\widetilde{\mathcal {U}}_{\varepsilon })}{\partial X_{3}}- \delta\mathcal {T}_{\varepsilon }(\widetilde{\mathcal {R}}_{\varepsilon })\land{\mathbf{e_{3}}} \biggr\| _{L^{2} (\omega\times(0, 1))} \leq C \frac{r}{\delta} , \end{gathered}$$

(B.4)

$$\begin{gathered} \bigl\| \mathcal {T}_{\varepsilon }^{'}(\widetilde{{u}}_{\varepsilon }) \bigr\| _{L^{2}(\omega\times(0,1), H^{1}( D_{1}))} \leq C \frac{r^{2}}{\delta^{2}}, \end{gathered}$$

(B.5)

$$\begin{gathered} \biggl\Vert \frac{\partial\mathcal {T}_{\varepsilon }^{'}(\widetilde {{u}}_{\varepsilon })}{\partial X_{3}}\biggr\Vert _{L^{2} (\omega\times B_{1})} \leq C \frac{r}{\delta}. \end{gathered}$$

(B.6)

Proof of Proposition 4.1

The convergences in (4.1)–(4.3), (4.5), (4.6), (4.8), (4.12) and (4.14) follow from estimate (3.23) and those in Lemma B.2.

The equalities in (4.4) are consequences of (3.8)₁–(3.8)₂. To obtain (4.11), take into account that from (4.14) we have

$$ \frac{\partial\widetilde{\mathcal {U}}}{\partial X_{3}} - \widetilde {\mathcal {R}} \wedge{\mathbf{e_{3}}} = \left ( \textstyle\begin{array}{c} \displaystyle\frac{\partial\widetilde{\mathcal {U}}_{1}}{\partial X_{3}} - \widetilde{\mathcal {R}}_{2}\\ \displaystyle\frac{\partial\widetilde{\mathcal {U}}_{2}}{\partial X_{3}} + \widetilde{\mathcal {R}}_{1}\\ \displaystyle\frac{\partial\widetilde{\mathcal {U}}_{3}}{\partial X_{3}} \end{array}\displaystyle \right )=0. $$

Then (4.4) yields (4.10). Equalities in (4.7) are the consequences of $\frac{\partial\widetilde{\mathcal {U}}_{3}}{\partial X_{3}}=0$ and estimates (3.9), (3.10). Again due to (3.9), (3.10), we obtain

$$\begin{gathered} \widetilde{\mathcal {U}}_{\alpha} \bigl(x', 0\bigr) = {u_{\alpha}^{-}}_{|\varSigma }\bigl(x'\bigr),\qquad \widetilde{\mathcal {U}}_{\alpha} \bigl(x', 1\bigr) = {u_{\alpha}^{+}}_{|\varSigma }\bigl(x'\bigr),\quad\text{for a.e. } x' \in\omega. \end{gathered}$$

From Lemma B.2 we have $\| \mathcal {T}_{\varepsilon }^{'}(\widetilde{{u}}_{\varepsilon }) \|_{L^{2}(\omega, H^{1}( B_{1}))} \leq C \frac{r}{\delta}$. From this and (4.12) we deduce (4.13). □

The strain tensor of the displacement $u_{\varepsilon}$ is

$$\begin{aligned} \mathcal {T}_{\varepsilon }^{'} \bigl(\widetilde{e(u_{\varepsilon })} \bigr)_{ij} =& \mathcal {T}_{\varepsilon }^{'} \bigl(\widetilde {e(u_{\varepsilon })} \bigr)_{ij}, \quad i, j = 1, 2, \\ \mathcal {T}_{\varepsilon }^{'} \bigl(\widetilde{e(u_{\varepsilon })} \bigr)_{13} =& \frac{1}{2} \biggl( \biggl(\frac{1}{\delta } \frac{\partial\mathcal {T}_{\varepsilon }(\widetilde{\mathcal {U}}_{\varepsilon ,1})}{\partial X_{3}} - \mathcal {T}_{\varepsilon }(\widetilde{\mathcal {R}}_{\varepsilon ,2}) \biggr) - \frac{r}{\delta } \frac{\partial\mathcal {T}_{\varepsilon }(\widetilde{\mathcal {R}}_{\varepsilon ,3})}{\partial X_{3}} X_{2} \biggr) + \mathcal {T}_{\varepsilon }^{'} \bigl( \widetilde{e(u_{\varepsilon })} \bigr)_{13}, \\ \mathcal {T}_{\varepsilon }^{'} \bigl(\widetilde{e(u_{\varepsilon })} \bigr)_{23} =& \frac{1}{2} \biggl( \biggl(\frac{1}{\delta } \frac{\partial\mathcal {T}_{\varepsilon } (\widetilde{\mathcal {U}}_{\varepsilon ,2})}{\partial X_{3}} + \mathcal {T}_{\varepsilon }(\widetilde{\mathcal {R}}_{\varepsilon ,1}) \biggr) + \frac{r}{\delta }\frac{\partial\mathcal {T}_{\varepsilon }(\widetilde{\mathcal {R}}_{\varepsilon ,3})}{\partial X_{3}} X_{1} \biggr) + \mathcal {T}_{\varepsilon }^{'} \bigl( \widetilde{e(u_{\varepsilon })} \bigr)_{23}, \\ \mathcal {T}_{\varepsilon }^{'} \bigl(\widetilde{e(u_{\varepsilon })} \bigr)_{33} =& \frac{1}{\delta}\frac{\partial\mathcal {T}_{\varepsilon }(\widetilde{\mathcal {U}}_{\varepsilon ,3})}{\partial X_{3}} + \frac{r}{\delta} \frac{\partial\mathcal {T}_{\varepsilon }(\widetilde{\mathcal {R}}_{\varepsilon ,1})}{\partial X_{3}} X_{2} - \frac {r}{\delta} \frac{\partial\mathcal {T}_{\varepsilon }(\widetilde {\mathcal {R}}_{\varepsilon ,2})}{\partial X_{3}} X_{1} + \mathcal {T}_{\varepsilon }^{'} \bigl(\widetilde{e(u_{\varepsilon })} \bigr)_{33}. \end{aligned}$$

Define the field $\widetilde{u'}\in L^{2}(\omega\times(0,1), H^{1}(D_{1},{\mathbb{R}}^{3}))$ by

$$\begin{gathered} \widetilde{u'}_{\alpha} = \widetilde{{u}}_{\alpha}, \qquad \widetilde{u'}_{3} = \widetilde{{u}}_{3} + X_{1} Z_{1} + X_{2} Z_{2}. \end{gathered}$$

Then

$$\begin{gathered} \frac{\partial\widetilde{u'}_{3}}{\partial X_{1}} = \frac{\partial \widetilde{{u}}_{3}}{\partial X_{1}} + Z_{1},\qquad \frac{\partial\widetilde{u'}_{3}}{\partial X_{2}} = \frac{\partial \widetilde{{u}}_{3}}{\partial X_{2}} + Z_{2}. \end{gathered}$$

Appendix C: Derivation of the Limit Problem

3.1 C.1 Equations for the Domain $\varOmega_{\varepsilon }^{i}$

Proof of Proposition 5.1

Step 1. Obtain the limit equations in $\varOmega_{\varepsilon }^{i}$.

We will use the following test function:

$$\begin{aligned} &v_{\varepsilon } (x) = \frac{r}{\delta}\psi(\varepsilon\xi)\left ( \textstyle\begin{array}{c} \displaystyle \frac{\delta}{r}\varphi_{1} \biggl( \frac{x_{3}}{\delta} \biggr) - \frac{x_{2} - \varepsilon \xi_{2}}{r} \varphi_{4} \biggl( \frac {x_{3}}{\delta} \biggr)\\ \displaystyle \frac{\delta}{r}\varphi_{2} \biggl( \frac{x_{3}}{\delta} \biggr) + \frac{x_{1} - \varepsilon \xi_{1}}{r} \varphi_{4} \biggl( \frac {x_{3}}{\delta} \biggr)\\ \displaystyle \varphi_{3} \biggl(\frac{x_{3}}{\delta} \biggr) - \frac{x_{1} - \varepsilon \xi_{1}}{r} \frac{d \varphi_{1}}{d X_{3}} \biggl( \frac {x_{3}}{\delta} \biggr) - \frac{x_{2} - \varepsilon \xi_{2}}{r} \frac{d \varphi_{2}}{d X_{3}} \biggl(\frac{x_{3}}{\delta} \biggr) \end{array}\displaystyle \right ),\\ &\quad\xi= \biggl[\frac{x'}{\varepsilon } \biggr]_{Y}, \end{aligned}$$

where $\psi\in C^{\infty}_{c} (\omega)$, $\varphi_{3}$ and $\varphi_{4} \in H_{0}^{1} (0, 1)$, $\varphi_{1}$ and $\varphi_{2} \in H_{0}^{2} (0, 1)$. Computation of the symmetric strain tensor gives

$$ e(v_{\varepsilon }) = \frac{r}{\delta^{2}}\psi(\varepsilon \xi )\left ( \textstyle\begin{array}{c@{\quad}c@{\quad}c} 0 & 0 & \displaystyle -\frac{1}{2} \frac{x_{2} - \varepsilon \xi_{2}}{r } \frac {d \varphi_{4}}{d X_{3}}\\ \ldots& 0 & \displaystyle \frac{1}{2}\frac{x_{1} - \varepsilon \xi_{1}}{r } \frac{d \varphi_{4}}{d X_{3}} \\ \ldots& \ldots& \displaystyle \biggl( \frac{d \varphi_{3}}{d X_{3}} - \frac{x_{1} - \varepsilon \xi_{1}}{r} \frac{d^{2} \varphi_{1}}{d X_{3}^{2}} - \frac{x_{2} - \varepsilon \xi_{2}}{r} \frac{d^{2} \varphi_{2}}{d X_{3}^{2}} \biggr)\\ \end{array}\displaystyle \right )\quad\hbox{in } \varepsilon \xi+B_{1}. $$

Then

$$\begin{aligned} &\frac{\delta^{2}}{r} \mathcal {T}_{\varepsilon }^{'} \bigl(\widetilde {e(v_{\varepsilon })} \bigr) \rightarrow\psi\bigl(x'\bigr)\left ( \textstyle\begin{array}{c@{\quad}c@{\quad}c} 0 & 0 & \displaystyle -\frac{1}{2} X_{2} \frac{d \varphi_{4}}{d X_{3}}\\ \ldots& 0 & \displaystyle \frac{1}{2}X_{1} \frac{d \varphi_{4}}{d X_{3}} \\ \ldots& \ldots& \displaystyle \frac{d \varphi_{3}}{d X_{3}} - X_{1} \frac{d^{2} \varphi_{1}}{d X_{3}^{2}} - X_{2} \frac{d^{2} \varphi_{2}}{d X_{3}^{2}}\\ \end{array}\displaystyle \right ) = V\bigl(x', X\bigr) \\ &\quad\text{strongly in } L^{2} (\omega\times B_{1}). \end{aligned}$$

Moreover,

$$ \mathcal {T}_{\varepsilon }^{'} (\widetilde{v}_{\varepsilon }) \rightarrow\psi\bigl(x'\bigr)\left ( \textstyle\begin{array}{c} \varphi_{1} (X_{3})\\ \varphi_{2} (X_{3})\\ 0 \end{array}\displaystyle \right ) \quad\text{strongly in } L^{2} (\omega\times B_{1}). $$

Unfolding the integral over $\varOmega_{\varepsilon }^{i}$, we obtain

$$ \begin{aligned} \int_{\varOmega^{i}_{\varepsilon }} \sigma_{\varepsilon } : e(v_{\varepsilon }) \,dx &= \sum_{\xi\in\varXi_{\varepsilon }} \int _{\varepsilon \xi+ B_{r, \delta}} \sigma_{\varepsilon } : \widetilde{e(v_{\varepsilon })} \,dx \\ &= r^{2} \delta\sum_{\xi\in\varXi_{\varepsilon }} \int_{B_{1}} \mathcal {T}_{\varepsilon }^{'} ( \sigma_{\varepsilon }) : \mathcal {T}_{\varepsilon }^{'} \bigl( \widetilde{e(v_{\varepsilon })} \bigr) \,dx' \, dX_{1} \, dX_{2} \, dX_{3} \\ &= \frac{r^{2} \delta}{\varepsilon ^{2}} \int_{\omega\times B_{1}} \mathcal {T}_{\varepsilon }^{'} ( \sigma_{\varepsilon }) : \mathcal {T}_{\varepsilon }^{'} \bigl( \widetilde{e(v_{\varepsilon })} \bigr) \,dx' \, dX_{1} \, dX_{2} \, dX_{3}. \end{aligned} $$

In the same way, integrating the forces we get

$$ \int_{\varOmega^{i}_{\varepsilon }} f_{\varepsilon }\cdot v_{\varepsilon}dx = \frac{r^{2} \delta}{\varepsilon ^{2}} \int_{\omega\times B_{1}} \mathcal {T}_{\varepsilon }^{'} (f_{\varepsilon })\cdot\mathcal {T}_{\varepsilon }^{'}( \widetilde{v}_{\varepsilon }) \,dx' \, dX_{1} \, dX_{2} \, dX_{3}. $$

Taking the limit gives

$$ \frac{\kappa_{0}^{4}}{\kappa_{1}^{3}} \int_{\omega\times B_{1}} \varTheta: V \,dx' \, dX = \sum _{\alpha=1}^{2} \int_{\omega\times B_{1}} F^{m}_{\alpha}\bigl(x', X\bigr) \psi\bigl(x'\bigr) \varphi_{\alpha}( X) \,dx' \, dX. $$

(C.1)

We can localize the equation above. This gives

$$\begin{aligned} &\frac{\pi\kappa_{0}^{4}}{4 \kappa_{1}^{3}}\mu^{m} \int_{\omega\times(0, 1)} \frac{\partial\widetilde{\mathcal {R}}_{3}}{\partial X_{3}} \frac{d \varphi_{4}}{d X_{3}} \psi \, dx' \, dX_{3} \\ &\qquad{}+ \frac{\pi\kappa_{0}^{4}}{4 \kappa_{1}^{3}} E^{m} \int_{\omega\times(0, 1)} \biggl( 4 \frac{\partial \widetilde{\mathcal {U}}^{'}_{3}}{\partial X_{3}} \frac{d \varphi_{3}}{d X_{3}} + \frac{\partial^{2} \widetilde{\mathcal {U}}_{1}}{\partial X_{3}^{2}} \frac{d^{2} \varphi_{1}}{d X_{3}^{2}} + \frac{\partial^{2} \widetilde {\mathcal {U}}_{2}}{\partial X_{3}^{2}} \frac{d^{2} \varphi_{2}}{d X_{3}^{2}} \biggr) \psi\, dx' \, dX_{3} \\ &\quad{}=\int_{\omega\times(0, 1)} \bigl(\widetilde{F}^{m}_{1} \varphi_{1} + \widetilde{F}^{m}_{2} \varphi_{2} \bigr) \psi \,dx' \, dX_{3}. \end{aligned}$$

(C.2)

The density of the tensor product $\mathcal{C}^{\infty}_{c}(\omega )\otimes H^{1}_{0}(0,1)$ (resp. $\mathcal{C}^{\infty}_{c}(\omega)\otimes H^{2}_{0}(0,1)$) in $L^{2}(\omega ; H^{1}_{0} (0,1))$ (resp. $L^{2}(\omega ; H^{2}_{0} (0,1))$) implies

$$\begin{aligned} &\frac{\pi\kappa_{0}^{4}}{4 \kappa_{1}^{3}}\mu^{m} \int_{\omega\times(0, 1)} \frac{\partial\widetilde{\mathcal {R}}_{3}}{\partial X_{3}} \frac {\partial\varPhi_{4}}{\partial X_{3}} \, dx' \, dX_{3} \\ &\qquad{} + \frac{\pi\kappa _{0}^{4}}{4 \kappa_{1}^{3}} E^{m} \int_{\omega\times(0, 1)} \biggl( 4 \frac {\partial\widetilde{\mathcal {U}}^{'}_{3}}{\partial X_{3}} \frac{\partial \varPhi_{3}}{\partial X_{3}} + \frac{\partial^{2} \widetilde{\mathcal {U}}_{1}}{\partial X_{3}^{2}} \frac{\partial^{2} \varPhi_{1}}{\partial X_{3}^{2}} + \frac{\partial^{2} \widetilde{\mathcal {U}}_{2}}{\partial X_{3}^{2}} \frac {\partial^{2} \varPhi_{2}}{\partial X_{3}^{2}} \biggr) \, dx' \, dX_{3} \\ &\quad{}= \int_{\omega\times(0, 1)} \bigl(\widetilde{F}^{m}_{1} \varPhi_{1} + \widetilde{F}^{m}_{2} \varPhi_{2} \bigr) \,dx' \, dX_{3} \\ &\quad\forall \varPhi_{3}, \varPhi_{4}\in L^{2}\bigl(\omega; H^{1}_{0}(0,1)\bigr),\ \forall\varPhi_{1},\varPhi _{2}\in L^{2}\bigl(\omega; H^{2}_{0}(0,1) \bigr). \end{aligned}$$

(C.3)

Step 2. Obtain $\widetilde{\mathcal {R}}_{3}$, $\widetilde{\mathcal {U}}^{'}_{3}$.

Since $\varphi_{3}\in H_{0}^{1} (0, 1)$ is not in the right-hand side of equation (C.2), we obtain

$$ E^{m} \int_{0}^{1} \frac{\partial\widetilde{\mathcal {U}}'_{3}}{\partial X_{3}} \frac{d \varphi_{3}}{d X_{3}} \,dX_{3} = 0\quad\Rightarrow\quad\frac {\partial^{2} \widetilde{\mathcal {U}}'_{3}}{\partial X_{3}^{2}} = 0 \quad \text{a.e. in } \omega\times(0,1). $$

(C.4)

Moreover, we have $\widetilde{\mathcal {U}}'_{3}(x', 0)=0$ for a.e. $x' \in\omega$. Therefore, there exists $a\in L^{2}(\omega)$ such that

$$ \widetilde{\mathcal {U}}'_{3} \bigl(x', X_{3}\bigr)= X_{3} a\bigl(x'\bigr), \quad \hbox{for a.e. } \bigl(x', X_{3}\bigr)\in\omega \times(0,1). $$

Similarly, recalling $\varphi_{4}\in H_{0}^{1} (0, 1)$ and taking $\varphi _{1} = \varphi_{2} = \varphi_{3} = 0$ in (C.2), leads to

$$ \mu^{m} \int_{0}^{1} \frac{\partial\widetilde{\mathcal {R}}_{3}}{\partial X_{3}} \frac{d \varphi_{4}}{d X_{3}} \, dX_{3} = 0\quad\Rightarrow\quad \frac{\partial^{2} \widetilde{\mathcal {R}}_{3}}{\partial X_{3}^{2}} = 0 \quad \text{ a.e. in } \omega\times(0,1). $$

This together with the boundary conditions (4.4) from Proposition 4.1 gives $\widetilde{\mathcal {R}}_{3} = 0$. □

The variational problem (5.1) and the boundary conditions (4.9)–(4.10) allow us to determine ${\mathcal {U}}_{\alpha}$ ($\alpha=1,2$) in terms of the applied forces $\widetilde {F}^{m}_{\alpha}$ and the traces $u^{\pm}_{\alpha|\varSigma}$.

3.2 C.2 Equations for the Macroscopic Domain

3.2.1 C.2.1 Determination of $\widetilde{\mathcal {U}}^{'}_{3}$

Proof of Lemma 5.1

For the sake of simplicity we extend $\phi$ to a function belonging to $W^{1,\infty}({\mathbb{R}}^{2})$ which is still denoted by $\phi$. We take

$$\widetilde{\varXi}_{\varepsilon}= \bigl\{ \xi\in{\mathbb{Z}}^{2}\ (\varepsilon \xi+\varepsilon Y)\cap\omega\neq\emptyset \bigr\} . $$

Observe that $\varXi_{\varepsilon}\subset\widetilde{\varXi}_{\varepsilon}$. Consider the following estimate:

$$\begin{aligned} \| \phi_{\varepsilon ,r} - \phi \|_{L^{\infty}(\omega)} =& \biggl\Vert \chi \biggl(\frac{\varepsilon}{r} \biggl\{ \frac{\cdot}{\varepsilon } \biggr\} _{Y} \biggr) \biggl( \phi \biggl( \varepsilon \biggl[\frac{\cdot}{\varepsilon } \biggr]_{Y} \biggr) - \phi \biggr) \biggr\Vert _{L^{\infty}(\omega)} \\ \leq& \sup_{\xi\in\widetilde{\varXi}_{\varepsilon}} \biggl\Vert \chi \biggl(\frac{\cdot}{r} \biggr) \bigl( \phi(\varepsilon \xi) - \phi (\varepsilon \xi+ \cdot)\bigr) \biggr\Vert _{L^{\infty}(Y_{\varepsilon })} \\ =& \sup_{\xi\in\widetilde{\varXi}_{\varepsilon}} \biggl\Vert \chi \biggl( \frac{\varepsilon}{r} \cdot \biggr) \bigl( \phi(\varepsilon \xi) - \phi ( \varepsilon \xi+ \varepsilon \cdot)\bigr) \biggr\Vert _{L^{\infty}(Y)} \\ \leq& \varepsilon \|\chi\|_{L^{\infty}({\mathbb{R}}^{2})}\|\nabla\phi \|_{L^{\infty}({\mathbb{R}}^{2})}. \end{aligned}$$

(C.5)

The partial derivative of $\phi_{\varepsilon ,r} - \phi$ with respect to $x_{\alpha}$ is

$$\begin{aligned} &\frac{\partial(\phi_{\varepsilon ,r} - \phi)}{\partial x_{\alpha }} \bigl(x'\bigr) = \frac{1}{r} \frac{\partial\chi}{\partial X_{\alpha}} \biggl(\frac{\varepsilon}{r} \biggl\{ \frac{x'}{\varepsilon } \biggr\} _{Y} \biggr) \biggl( \phi \biggl( \varepsilon \biggl[\frac{x'}{\varepsilon } \biggr]_{Y} \biggr) - \phi \bigl(x'\bigr) \biggr) - \chi \biggl(\frac{\varepsilon}{r} \biggl\{ \frac{x'}{\varepsilon } \biggr\} _{Y} \biggr) \frac{\partial\phi}{\partial x_{\alpha}} \bigl(x'\bigr),\\ &\quad\hbox{for a.e. } x'\in\omega,\\ &\frac{\partial(\phi_{\varepsilon ,r} - \phi)}{\partial x_{\alpha }} \bigl(\varepsilon \xi+\varepsilon y'\bigr) = \frac{1}{r} \frac{\partial\chi }{\partial X_{\alpha}} \biggl(\frac{\varepsilon}{r} y' \biggr) \bigl( \phi(\varepsilon \xi) - \phi\bigl(\varepsilon \xi+\varepsilon y'\bigr) \bigr) - \chi \biggl(\frac{\varepsilon}{r}y' \biggr) \frac{\partial \phi}{\partial x_{\alpha}}\bigl(\varepsilon \xi+\varepsilon y' \bigr),\\ &\quad \xi\in\widetilde{\varXi}_{\varepsilon}, \hbox{ for a.e. } y'\in Y. \end{aligned}$$

Since $\chi$ has compact support in ${\mathbb{R}}^{2}$, there exists $R>0$ such that $\operatorname{supp}(\chi)\subset D_{R}$. Thus, the support of the function $y'\longmapsto \chi (\frac{\varepsilon}{r}y' )$ is included in the disc $D_{rR/\varepsilon }$. As a consequence we get for a.e. $y'\in D_{rR/\varepsilon }$

$$\bigl\vert \phi(\varepsilon \xi) - \phi\bigl(\varepsilon \xi+\varepsilon y'\bigr)\bigr\vert \le r R\|\nabla\phi\|_{L^{\infty}({\mathbb{R}}^{2})}. $$

Using the above estimate, the norms of the derivatives satisfy

$$\begin{aligned} \biggl\Vert \frac{\partial(\phi_{\varepsilon ,r} - \phi)}{\partial x_{\alpha}} \biggr\Vert ^{p}_{L^{p} (\varepsilon \xi+\varepsilon Y)}\! =&\varepsilon ^{2} \biggl\Vert \frac{\partial\chi}{\partial X_{\alpha }} \biggl(\frac{\varepsilon}{r}\cdot \biggr) \frac{\phi(\varepsilon \xi) - \phi(\varepsilon \xi+\varepsilon \cdot)}{r} - \chi \biggl(\frac{\varepsilon}{r}\cdot \biggr) \frac{\partial\phi}{\partial x_{\alpha}} (\varepsilon \xi+\varepsilon \cdot) \biggr\Vert ^{p}_{L^{p} (Y)} \\ \leq& Cr^{2} \|\nabla\chi\|^{p}_{L^{\infty}({\mathbb{R}}^{2})}\|\nabla \phi\|^{p}_{L^{\infty}({\mathbb{R}}^{2})}. \end{aligned}$$

The constant $C$ does not depend on $\varepsilon$ and $r$. Combining the above estimates for $\xi\in\widetilde{\varXi}_{\varepsilon}$, that gives

$$ \bigl\Vert \nabla(\phi_{\varepsilon ,r} - \phi) \bigr\Vert _{L^{p} (\omega)} \leq C \biggl(\frac{r}{\varepsilon } \biggr)^{2/p}\| \nabla \chi\|_{L^{\infty}({\mathbb{R}}^{2})}\|\nabla\phi\|_{L^{\infty}({\mathbb{R}}^{2})}. $$

(C.6)

The constant does not depend on $r$ and $\varepsilon$. Hence, estimates (C.5) and (C.6) imply that $\phi _{\varepsilon}$ strongly converges toward $\phi$ in $W^{1,p}(\omega)$. □

Proof of Lemma 5.2

For any $\psi_{3} \in{\mathcal{C}}^{1} (\overline{\omega} \times[0, 1])$ satisfying $\psi_{3}(x',0)=0$ for every $x'\in\omega$, we consider the following test function:

$$\begin{aligned} &v_{\varepsilon ,\alpha}(x)= 0\quad\hbox{for a.e. } x\in \varOmega _{\varepsilon},\ \alpha=1,2, \\ &v_{\varepsilon }(x) = 0 \quad\hbox{for a.e. } x\in\varOmega ^{-} , \\ &v_{\varepsilon ,3}(x) = \frac{r}{\delta} \biggl[\psi_{3} \bigl(x',1\bigr) \biggl(1 -\chi \biggl( \frac{\varepsilon }{r} \biggl\{ \frac{x' }{\varepsilon } \biggr\} _{Y} \biggr) \biggr) + \psi_{3} \biggl(\varepsilon \biggl[\frac{x' }{\varepsilon } \biggr]_{Y}, 1 \biggr) \chi \biggl( \frac{\varepsilon }{r} \biggl\{ \frac{x' }{\varepsilon } \biggr\} _{Y} \biggr) \biggr],\\ &\quad \hbox{for a.e. } x\in \varOmega ^{+}_{\varepsilon},\\ &v_{\varepsilon ,3}(x) = \frac{r}{\delta} \psi_{3} \biggl( \varepsilon \biggl[\frac{x' }{\varepsilon } \biggr]_{Y}, \frac {x_{3}}{\delta} \biggr), \quad\hbox{for a.e. } x\in \varOmega ^{i}_{\varepsilon}. \end{aligned}$$

If $\frac{r}{\varepsilon }$ is small enough, $v_{\varepsilon}$ is an admissible test function. The symmetric strain tensor in $\varOmega ^{i}_{\varepsilon }$ is given by

$$ e(v_{\varepsilon }) = \frac{r}{\delta^{2}}\left ( \textstyle\begin{array}{c@{\quad}c@{\quad}c} 0 & 0 & 0 \\ \ldots& 0 & 0 \\ \ldots& \ldots& \displaystyle \frac{\partial\psi_{3}}{\partial X_{3}} (\varepsilon \xi, \frac{x_{3}}{\delta} )\\ \end{array}\displaystyle \right ) \quad\hbox{a.e. in } \varepsilon \xi+B_{r,\delta }. $$

Then

$$ \frac{\delta^{2}}{r} \mathcal {T}_{\varepsilon }^{'} \bigl(\widetilde {e(v_{\varepsilon })} \bigr) \rightarrow \left ( \textstyle\begin{array}{c@{\quad}c@{\quad}c} 0 & 0 &0 \\ \ldots& 0 & 0 \\ \ldots& \ldots& \displaystyle \frac{\partial\psi_{3}}{\partial X_{3}} (x',X_{3})\\ \end{array}\displaystyle \right ) = V\bigl(x', X\bigr) \quad\text{strongly in } L^{2} (\omega\times B_{1}). $$

The elements of the symmetric strain tensor in $\varOmega_{\varepsilon }^{+}$ are written as follows:

$$\begin{aligned} e_{11}(v_{\varepsilon }) =& e_{22}(v_{\varepsilon }) = e_{12}(v_{\varepsilon }) = e_{33}(v_{\varepsilon }) =0, \\ e_{\alpha3}(v_{\varepsilon }) =& e_{3\alpha}(v_{\varepsilon }) = \frac{1}{2}\frac{r}{\delta} \frac{\partial\psi_{3}}{\partial x_{\alpha}} \bigl(x',1 \bigr) \bigl(1 - \chi(y)\bigr)\\ &{}+ \frac{1}{2\delta} \frac{\partial \chi}{\partial y_{\alpha}}(y) \biggl(\psi_{3}\bigl(x',1\bigr) - \psi_{3} \biggl(\varepsilon \biggl[\frac{x'}{\varepsilon } \biggr]_{Y}, 1 \biggr) \biggr), \end{aligned}$$

where $y = \frac{\varepsilon }{r} \{\frac{x'}{\varepsilon } \}_{Y}$.

Using Lemma 5.1 and taking into account $\frac{r}{\delta} \rightarrow0$, the following convergences hold:

$$ \begin{aligned} &v_{\varepsilon }(\cdot+ \delta e_{3}) \longrightarrow0 \quad\text{strongly in } H^{1}\bigl(\varOmega^{+};{ \mathbb{R}}^{3}\bigr), \\ &e(v_{\varepsilon }) \longrightarrow0 \quad\text{strongly in } L^{2} \bigl(\varOmega^{+} ; {\mathbb{R}}^{9}\bigr). \end{aligned} $$

Moreover,

$$\begin{gathered} \mathcal {T}_{\varepsilon }' (v_{\varepsilon }) \longrightarrow0 \quad \text{strongly in } H^{1} \bigl(\omega\times B_{1} ; { \mathbb{R}}^{3}\bigr). \end{gathered}$$

Using $v_{\varepsilon }$ as a test function in (2.9) and passing to the limit in the unfolded formulation, gives

$$ \int_{\omega\times(0, 1)} \frac{\partial\widetilde{\mathcal {U}}'_{3}}{\partial X_{3}}\bigl(x',X_{3} \bigr) \frac{\partial\psi_{3}}{\partial X_{3}}\bigl(x', X_{3}\bigr) \, dx' \, dX = \int_{\omega\times(0, 1)} a\bigl(x'\bigr) \frac{\partial \psi_{3}}{\partial X_{3}} \bigl(x', X_{3}\bigr) \, dx' \, dX = 0. $$

Hence $a = 0$. Since the test functions are dense in

$${\mathbb{V}}_{s}= \bigl\{ \varPsi\in L^{2}\bigl(\omega; H^{1}(0,1)\bigr)\bigm| \varPsi \bigl(x',0\bigr)=0 \text{ a.e. in } \omega \bigr\} $$

we obtain

$$ \int_{\omega\times(0, 1)} \frac{\partial\widetilde{\mathcal {U}}'_{3}}{\partial X_{3}}\bigl(x',X_{3} \bigr) \frac{\partial\varPsi}{\partial X_{3}}\bigl(x', X_{3}\bigr) \, dx' \, dX = 0\quad\forall\varPsi\in{\mathbb{V}}_{s}. $$

(C.7)

 □

3.2.2 C.2.2 Determination of $u^{\pm}_{\alpha}$ and $u_{3}$

Proof of Theorem 5.1

For any $v\in{\mathbb{V}}$ such that $v_{|\varOmega^{-}}\in W^{1,\infty }(\varOmega ^{-} , {\mathbb{R}}^{3})$ and $v_{|\varOmega^{+}}\in W^{1,\infty }(\varOmega ^{+} , {\mathbb{R}}^{3})$, we first define the displacement $v_{\varepsilon ,r}$ in the following way:

$$\begin{aligned} &v_{\varepsilon ,r} (x) = v(x) \biggl(1 -\chi \biggl( \frac{\varepsilon }{r} \biggl\{ \frac{x' }{\varepsilon } \biggr\} _{Y} \biggr) \biggr) + v \biggl(\varepsilon \biggl[\frac{x'}{\varepsilon } \biggr]_{Y}, x_{3} \biggr)\chi \biggl( \frac{\varepsilon }{r} \biggl\{ \frac{x' }{\varepsilon } \biggr\} _{Y} \biggr), \\ &\quad\hbox{for a.e. } x \in \varOmega ^{-}\cup\varOmega ^{+}. \end{aligned}$$

(C.8)

Then let $h$ denote the following function belonging to $W^{1,\infty} (-L,L)$:

$$ h (x_{3}) = \left\{ \textstyle\begin{array}{l@{\quad}l} \frac{x_{3}+L}{L}, & x_{3} \in[- L, 0],\\ 1, & x_{3}\geq0. \end{array}\displaystyle \right. $$

(C.9)

Now consider the test displacement

$$\begin{aligned} v'_{\varepsilon }(x) =& v(x) \bigl(1-h(x_{3}) \bigr)+ v_{\varepsilon ,r} (x) h(x_{3}), \quad\hbox{for a.e. } x\in\varOmega ^{-} , \\ v'_{\varepsilon }(x) =& v_{\varepsilon ,r} \bigl(x', x_{3}-\delta \bigr), \quad \hbox{for a.e. } x \in\varOmega ^{+}_{\varepsilon}, \\ v'_{\varepsilon }(x) =& \left ( \textstyle\begin{array}{c} \displaystyle \psi_{1} \biggl( \varepsilon \biggl[\frac{x' }{\varepsilon } \biggr]_{Y}, \frac{x_{3}}{\delta} \biggr)\\ \displaystyle \psi_{2} \biggl( \varepsilon \biggl[\frac{x' }{\varepsilon } \biggr]_{Y}, \frac{x_{3}}{\delta} \biggr)\\ \displaystyle v_{3} \biggl(\varepsilon \biggl[\frac{x' }{\varepsilon } \biggr]_{Y}, 0 \biggr)-\frac{\varepsilon}{\delta } \biggl\{ \frac{x' }{\varepsilon } \biggr\} _{Y}\cdot\frac{\partial\psi}{\partial X_{3}} \biggl( \varepsilon \biggl[\frac{x' }{\varepsilon } \biggr]_{Y}, \frac{x_{3}}{\delta} \biggr) \end{array}\displaystyle \right ) \quad\hbox{for a.e. } x\in \varOmega ^{i}_{\varepsilon}, \end{aligned}$$

where $\psi_{\alpha} \in{\mathcal{C}}^{1} (\overline{\omega} ; \mathcal{C}^{3}([0,1])), \alpha= 1,2$, satisfies

$$ \psi_{\alpha} \bigl(x', 0\bigr) = v_{\alpha|\varOmega^{-}} \bigl(x', 0\bigr), \qquad\psi _{\alpha} \bigl(x', 1\bigr) = v_{\alpha|\varOmega^{+}} \bigl(x', 0\bigr)\quad\text{for every } x'\in\omega. $$

If $\frac{r}{\varepsilon }$ is small enough, $v'_{\varepsilon}$ is an admissible test displacement.

Due to Lemma 5.1, the following convergences hold:

$$\begin{gathered} \begin{aligned} &v'_{\varepsilon }(\cdot+ \delta{ \mathbf{e_{3}}}) \longrightarrow v \quad\text{strongly in } H^{1}\bigl(\varOmega^{+};{\mathbb{R}}^{3}\bigr), \\ &v'_{\varepsilon } \longrightarrow v \quad\text{strongly in } H^{1}\bigl(\varOmega^{-};{\mathbb{R}}^{3}\bigr), \\ &e\bigl(v'_{\varepsilon }\bigr) \longrightarrow e(v) \quad\text{strongly in } L^{2} \bigl(\varOmega^{+} \cup\varOmega^{-} ; { \mathbb{R}}^{9}\bigr). \end{aligned} \end{gathered}$$

Computing the strain tensor in $\varOmega_{\varepsilon }^{i}$ gives

$$ \begin{aligned} e_{ij}\bigl(v'_{\varepsilon } \bigr) &=0\quad(i,j)\neq(3,3), \\ e_{33}\bigl(v'_{\varepsilon }\bigr) &= - \frac{r}{\delta^{2}} \biggl( X_{1} \frac {\partial^{2} \psi_{1}}{\partial X_{3}^{2}} \biggl( \varepsilon \biggl[\frac {x' }{\varepsilon } \biggr]_{Y}, X_{3} \biggr) + X_{2} \frac{\partial^{2} \psi _{2}}{\partial X_{3}^{2}} \biggl( \varepsilon \biggl[ \frac{x' }{\varepsilon } \biggr]_{Y}, X_{3} \biggr) \biggr). \end{aligned} $$

Therefore,

$$\begin{gathered} \mathcal {T}_{\varepsilon }' \bigl(v'_{\varepsilon } \bigr)\longrightarrow \left ( \textstyle\begin{array}{c} \psi_{1} (x', X_{3})\\ \psi_{2} (x', X_{3})\\ v_{3} (x', 0) \end{array}\displaystyle \right ) \quad \text{strongly in } L^{2}\bigl(\omega\times B_{1}; {\mathbb {R}}^{3}\bigr), \\ \frac{\delta^{2}}{r} \mathcal {T}_{\varepsilon }' \bigl(e \bigl(v'_{\varepsilon }\bigr)^{33} \bigr) \longrightarrow- \biggl( X_{1} \frac {\partial^{2} \psi_{1}}{\partial X_{3}^{2}} \bigl(x', X_{3}\bigr) + X_{2} \frac{\partial ^{2} \psi_{2}}{\partial X_{3}^{2}} \bigl(x', X_{3}\bigr) \biggr) \quad\text{strongly in } L^{2}(\omega \times B_{1}). \end{gathered}$$

Unfolding and taking the limit in (2.9) gives

$$\begin{aligned} &\int_{\varOmega^{\pm}} \sigma^{\pm} : e(v) \, dx - \frac{\kappa _{0}^{4}}{\kappa_{1}^{3}} \int_{\omega\times B_{1}} \varTheta: \biggl( X_{1} \frac {\partial^{2} \psi_{1}}{\partial X_{3}^{2}} + X_{2} \frac{\partial^{2} \psi _{2}}{\partial X_{3}^{2}} \biggr) \, dx' \, dX \\ &\quad{}= \int_{\varOmega^{\pm}} F v \, dx + \int_{\omega\times B_{1}} \bigl(F^{m}_{1} \psi_{1} + F^{m}_{2} \psi_{2} + F_{3} v_{3}\bigr)\, dx' \, dX. \end{aligned}$$

Since the space $W^{1,\infty}(\varOmega ^{+} ; {\mathbb{R}}^{3})$ is dense in $H^{1}(\varOmega ^{+};{\mathbb{R}}^{3})$, the space of these functions in $W^{1,\infty}(\varOmega ^{-} , {\mathbb{R}}^{3})$ vanishing on $\varGamma$ is dense in $H^{1}(\varOmega ^{-};{\mathbb{R}}^{3})$ and since the space $\mathcal{C}^{1} (\overline{\omega} ; \mathcal{C}^{3}([0,1]))$ is dense in $L^{2}(\omega; H^{1}(0,1))$, the equality above holds for every $v$ in ${\mathbb{V}}$ and every $\psi_{1}$, $\psi_{2}$ in $L^{2}(\omega ; H^{1}(0,1))$ satisfying

$$\psi_{\alpha} \bigl(x', 0\bigr) = v_{\alpha|\varOmega^{-}} \bigl(x', 0\bigr), \qquad\psi _{\alpha} \bigl(x', 1\bigr) = v_{\alpha|\varOmega^{+}} \bigl(x', 0\bigr)\quad\text{for a.e. } x'\in\omega. $$

Finally, integrating over $D_{1}$ and making use of (C.3), (C.7) and (5.2) yields the result. □

3.2.3 C.2.3 Case (i)

We introduce the classical unfolding operator.

Definition C.1

For a Lebesgue-measurable function $\varphi$ on $\omega$, the unfolding operator $\mathcal {T}''_{\varepsilon }$ is defined as follows:

$$ \mathcal {T}''_{\varepsilon } (\varphi) (s, y) = \left \{ \textstyle\begin{array}{l@{\quad}l} \displaystyle \varphi \biggl(\varepsilon \biggl[\frac{s}{\varepsilon} \biggr]_{Y}+\varepsilon y \biggr), & \text{for a.e. } (s, y) \in\widehat{\omega}_{\varepsilon } \times Y,\\ 0, & \text{for a.e. } (s, y) \in\varLambda_{\varepsilon} \times Y. \end{array}\displaystyle \right . $$

Recall that (see [6]) the following lemma holds true.

Lemma C.1

Let $\phi$ be in $W^{1,\infty}(\omega)$ and let $\phi_{\varepsilon}$ be defined by

$$\phi_{\varepsilon}\bigl(x'\bigr)=\chi \biggl( \biggl\{ \frac{x'}{\varepsilon } \biggr\} _{Y} \biggr)\phi \biggl(\varepsilon \biggl[\frac{x'}{\varepsilon } \biggr]_{Y} \biggr)+ \biggl[1-\chi \biggl( \biggl\{ \frac{x'}{\varepsilon } \biggr\} _{Y} \biggr) \biggr] \phi\bigl(x'\bigr)\quad\textit{for a.e. } x'\in\omega. $$

Then we have

$$\begin{aligned} &\mathcal {T}''_{\varepsilon } (\phi_{\varepsilon})\longrightarrow\phi \quad\textit{strongly in } L^{2}\bigl(\omega; H^{1}(Y)\bigr), \\ &\mathcal {T}''_{\varepsilon } (\nabla \phi_{\varepsilon})\longrightarrow\nabla\phi\quad\textit{strongly in } L^{2}(\omega \times Y). \end{aligned} $$

Theorem C.1

The variational formulation for the problem (2.9) in the case $(i)$ is given by

$$\begin{aligned} &\int_{\varOmega^{+} \cup\varOmega^{-}} \sigma^{\pm} : e(v) \, dx + \frac {\pi\kappa_{0}^{4}}{4 \kappa_{1}^{3}} E^{m} \int_{\omega\times(0, 1)} \sum_{\alpha= 1}^{2} \frac{\partial^{2} \psi_{\alpha}}{\partial X_{3}^{2}} \frac{\partial^{2} \widetilde{\mathcal {U}}_{\alpha}}{\partial X_{3}^{2}} \, dx'\, dX_{3} \\ &\qquad{}+\frac{\pi\kappa_{0}^{4}}{4 \kappa_{1}^{3}}\mu^{m} \int_{\omega\times(0, 1)} \frac{\partial\widetilde{\mathcal {R}}_{3}}{\partial X_{3}} \frac {\partial\psi_{4}}{\partial X_{3}} \, dx' \, dX_{3} + \frac{\pi\kappa _{0}^{4}}{\kappa_{1}^{3}} E^{m} \int_{\omega\times(0, 1)} \frac{\partial \widetilde{\mathcal {U}}^{'}_{3}}{\partial X_{3}} \frac{\partial\varPhi _{3}}{\partial X_{3}} \, dx' \, dX_{3} \\ &\quad{}= \int_{\varOmega^{+} \cup\varOmega^{-}} F v \, dx + \int_{\omega\times (0, 1)} \sum_{\alpha= 1}^{2} \widetilde{F}_{\alpha}^{m} \psi_{\alpha}\, dx' \, dX_{3} + \int_{\omega} \overline{F}_{3}^{m} v_{3} \,dx', \\ &\quad{}\forall(v, \psi_{1}, \psi_{2}, \psi_{3}, \psi_{4})\in{\mathbb {V}}_{T}, \ \forall \varPhi_{3} \in L^{2}\bigl(\omega; H^{1}_{0}(0,1)\bigr). \end{aligned}$$

(C.10)

Proof

Step 1. Take the limit in the weak formulation.

In addition to (4.1) and (4.2) we have

$$\begin{gathered} \mathcal {T}''_{\varepsilon } (u_{\varepsilon }) \rightharpoonup u^{-} \quad\text{weakly in } L^{2} \bigl(\varOmega^{-}; H^{1}(Y)\bigr), \end{gathered}$$

(C.11)

$$\begin{gathered} \mathcal {T}''_{\varepsilon } (\nabla u_{\varepsilon }) \rightharpoonup \nabla u^{-} +\nabla_{y} \widehat{u}^{-} \quad\text{ weakly in } L^{2}\bigl(\varOmega^{-}\times Y\bigr), \end{gathered}$$

(C.12)

$$\begin{gathered} \mathcal {T}''_{\varepsilon } (u_{\varepsilon }) (\cdot+ \delta {\mathbf{e_{3}}},\cdot\cdot) \rightharpoonup u^{+} \quad\text{weakly in } L^{2}\bigl(\varOmega^{+}; H^{1}(Y)\bigr), \end{gathered}$$

(C.13)

$$\begin{gathered} \mathcal {T}''_{\varepsilon } (\nabla u_{\varepsilon }) (\cdot+ \delta {\mathbf{e_{3}}},\cdot\cdot) \rightharpoonup\nabla u^{+} +\nabla_{y} \widehat{u}^{+} \quad\text{weakly in } L^{2}\bigl(\varOmega^{+}\times Y\bigr), \end{gathered}$$

(C.14)

where $\widehat{u}^{-}$ belongs to $L^{2}(\varOmega^{-}; H^{1}_{per}(Y;{\mathbb {R}}^{3}))$ and $\widehat{u}^{+}$ belongs to $L^{2}(\varOmega^{+}; H^{1}_{per}(Y;{\mathbb{R}}^{3}))$.

Remark C.1

Here the third variable of $u_{\varepsilon }$ is considered as a parameter, on which the unfolding operator $\mathcal {T}''_{\varepsilon }$ does not have any effect.

Step 2. Determination of $\mathcal{U}'_{3}$.

To determine the function $a$ introduced in Proposition 5.1, take $\psi_{3} \in{\mathcal{C}}^{1} (\overline{\omega} \times[0, 1])$, satisfying $\psi_{3}(x',0)=0$ for every $x'\in\omega$, and consider the following test function:

$$\begin{aligned} &v_{\varepsilon ,\alpha}(x) =0\quad\hbox{for a.e. } x\in \varOmega _{\varepsilon},\ \alpha=1,2, \\ &v_{\varepsilon }(x) = 0 \quad\hbox{for a.e. } x\in\varOmega ^{-} , \\ &v_{\varepsilon ,3}(x) = \varepsilon ^{1/3} \biggl[\psi_{3} \bigl(x',1\bigr) \biggl(1 -\chi \biggl( \biggl\{ \frac{x' }{\varepsilon } \biggr\} _{Y} \biggr) \biggr) + \psi_{3} \biggl(\varepsilon \biggl[\frac{x'}{\varepsilon } \biggr]_{Y}, 1 \biggr)\chi \biggl( \biggl\{ \frac{x' }{\varepsilon } \biggr\} _{Y} \biggr) \biggr]\\ &\quad\hbox{for a.e. } x\in \varOmega ^{+}_{\varepsilon}, \\ &v_{\varepsilon , 3}(x) = \varepsilon ^{1/3} \psi_{3} \biggl( \varepsilon \biggl[\frac{x'}{\varepsilon } \biggr]_{Y}, \frac {x_{3}}{\varepsilon ^{2/3}} \biggr) \chi \biggl( \biggl\{ \frac {x'}{\varepsilon } \biggr\} _{Y} \biggr) \quad\hbox{for a.e. } x\in \varOmega ^{i}_{\varepsilon}. \end{aligned}$$

We obtain the following convergences:

$$\begin{aligned} &v_{\varepsilon }(\cdot+ \delta{\mathbf{e_{3}}}) \longrightarrow0 \quad\text{strongly in } H^{1}\bigl(\varOmega^{+} \cup \varOmega^{-};{\mathbb {R}}^{3}\bigr), \\ &e(v_{\varepsilon }) \longrightarrow0 \quad\text{strongly in } L^{2} \bigl(\varOmega^{+} \cup\varOmega^{-}; {\mathbb{R}}^{9}\bigr),\\ &\mathcal {T}_{\varepsilon }' (v_{\varepsilon }) \longrightarrow0 \quad \text{ strongly in } H^{1} \bigl(\omega\times B_{1} ; { \mathbb{R}}^{3}\bigr). \end{aligned}$$

Unfolding and taking the limit as in the Sect. 5.2.1 we obtain that $a = 0$.

Step 3.

For any $v\in{\mathbb{V}}$ such that $v_{|\varOmega^{-}}\in W^{1,\infty }(\varOmega ^{-}; {\mathbb{R}}^{3})$ and $v_{|\varOmega^{+}}\in W^{1,\infty }(\varOmega ^{+}; {\mathbb{R}}^{3})$, we define the displacement $v_{\varepsilon}$ in the following way:

$$\begin{aligned} &v_{\varepsilon } (x) = v(x) \biggl(1 -\chi \biggl( \biggl\{ \frac{x' }{\varepsilon } \biggr\} _{Y} \biggr) \biggr) + v \biggl(\varepsilon \biggl[\frac {x' }{\varepsilon } \biggr]_{Y}, x_{3} \biggr)\chi \biggl( \biggl\{ \frac{x' }{\varepsilon } \biggr\} _{Y} \biggr), \\ &\quad{} \hbox{for a.e. } x \in\varOmega ^{-}\cup\varOmega ^{+}. \end{aligned}$$

(C.15)

Consider the following test displacement:

$$\begin{aligned} &v'_{\varepsilon }(x) = v(x) \bigl(1-h(x_{3}) \bigr)+v_{\varepsilon }(x) h(x_{3}) + \varepsilon \varPsi^{(-)}\bigl(x',x_{3}\bigr) \widehat{v} \biggl( \biggl\{ \frac{x'}{\varepsilon } \biggr\} _{Y} \biggr) , \quad\hbox{for a.e. } x\in\varOmega ^{-} , \\ &v'_{\varepsilon }(x) = v_{\varepsilon }\bigl(x', x_{3}-\delta \bigr) + \varepsilon \varPsi^{(+)} \bigl(x',x_{3}-\delta \bigr)\widehat{v} \biggl( \biggl\{ \frac{x'}{\varepsilon } \biggr\} _{Y} \biggr), \quad\hbox{for a.e. } x \in \varOmega ^{+}_{\varepsilon}, \\ &v'_{\varepsilon }(x) = \left ( \textstyle\begin{array}{c} \displaystyle \psi_{1} \biggl( \varepsilon \biggl[\frac{x' }{\varepsilon } \biggr]_{Y}, \frac{x_{3}}{\delta} \biggr)\\ \displaystyle \psi_{2} \biggl( \varepsilon \biggl[\frac{x' }{\varepsilon } \biggr]_{Y}, \frac{x_{3}}{\delta} \biggr)\\ \displaystyle v_{3} \biggl(\varepsilon \biggl[\frac{x' }{\varepsilon } \biggr]_{Y}, 0 \biggr)- \frac{\varepsilon}{\delta } \biggl\{ \frac{x' }{\varepsilon } \biggr\} _{Y}\cdot\frac{\partial\psi}{\partial X_{3}} \biggl( \varepsilon \biggl[\frac{x' }{\varepsilon } \biggr]_{Y}, \frac{x_{3}}{\delta} \biggr) \end{array}\displaystyle \right )\quad\hbox{for a.e. } x\in \varOmega ^{i}_{\varepsilon}, \end{aligned}$$

where

• $\widehat{v} \in H^{1}_{per} (Y; {\mathbb{R}}^{3})$,

• $\psi_{\alpha} \in{\mathcal{C}}^{1} (\overline{\omega} ; \mathcal{C}^{3}([0,1])), \alpha= 1,2$, satisfies

  $$ \psi_{\alpha} \bigl(x', 0\bigr) = v_{\alpha|\varOmega^{-}} \bigl(x', 0\bigr), \qquad\psi _{\alpha} \bigl(x', 1\bigr) = v_{\alpha|\varOmega^{+}} \bigl(x', 0\bigr)\quad\text{for every } x'\in\omega, $$

• $\varPsi^{(-)}\in W^{1,\infty}(\varOmega^{-})$, $\varPsi^{(+)}\in W^{1,\infty}(\varOmega^{+})$ satisfies

  $$\varPsi^{(\pm)}\bigl(x', 0\bigr)=0,\quad\hbox{a.e. in } \omega,\qquad \varPsi ^{(-)}=0\quad\hbox{on } \varGamma, $$

• $h(x_{3})$ is defined as in (C.9).

Using (5.2), we obtain the following convergences:

$$\begin{gathered} \begin{aligned} &\mathcal {T}''_{\varepsilon } \bigl(v'_{\varepsilon } (\cdot, \cdot\cdot )\bigr) \longrightarrow v \quad\hbox{strongly in } L^{2}\bigl(\varOmega^{-}; H^{1}(Y)\bigr), \\ &\mathcal {T}''_{\varepsilon } \bigl(\nabla v'_{\varepsilon } (\cdot, \cdot \cdot)\bigr) \longrightarrow\nabla v + \varPsi^{(-)}\nabla_{y} \widehat{v} \quad\hbox{strongly in } L^{2}\bigl(\varOmega^{-}\times Y\bigr), \\ &\mathcal {T}''_{\varepsilon } \bigl(v'_{\varepsilon } (\cdot+\delta {\mathbf{e_{3}}}, \cdot\cdot)\bigr) \longrightarrow v \quad\hbox{strongly in } L^{2}\bigl(\varOmega^{+}; H^{1}(Y)\bigr), \\ &\mathcal {T}''_{\varepsilon } \bigl(\nabla v'_{\varepsilon } (\cdot +\delta {\mathbf{e_{3}}}, \cdot \cdot)\bigr) \longrightarrow\nabla v + \varPsi^{(+)}\nabla_{y} \widehat{v} \quad\hbox{strongly in } L^{2}\bigl(\varOmega^{+} \times Y\bigr). \end{aligned} \end{gathered}$$

Moreover,

$$\begin{aligned} &\mathcal {T}_{\varepsilon } \bigl(\mathcal {T}''_{\varepsilon } \bigl(v'_{\varepsilon }\bigr)\bigr) \longrightarrow \left ( \textstyle\begin{array}{c} \displaystyle \psi_{1}(x', X_{3})\\ \displaystyle \psi_{2}(x', X_{3})\\ \displaystyle v_{3} (x', 0) \end{array}\displaystyle \right ) \quad\hbox{strongly in } L^{2}\bigl(\omega; H^{1}(Y \times B_{1})\bigr), \\ &\frac{\delta^{2}}{r}\mathcal {T}_{\varepsilon } \bigl(\mathcal {T}''_{\varepsilon } \bigl( e_{33}\bigl(v'_{\varepsilon }\bigr) \bigr) \bigr) \longrightarrow- X_{1} \frac{\partial^{2} \psi_{1}}{\partial X_{3}^{2}} \bigl(x', X_{3}\bigr) - X_{2} \frac{\partial^{2} \psi_{2}}{\partial X_{3}^{2}} \bigl(x', X_{3}\bigr)\\ &\quad\hbox{strongly in } L^{2}(\omega\times Y \times B_{1}). \end{aligned}$$

Unfolding and taking the limit, we obtain

$$\begin{aligned} &\int_{\varOmega^{\pm} \times Y} \bigl(\sigma^{\pm} + \widehat{\sigma }^{\pm}\bigr) : \bigl(e(v) + \varPsi^{(\pm)} e_{y}( \widehat{v}) \bigr) \, dx\, dy \\ &\qquad{}- \frac{\kappa_{0}^{4}}{\kappa_{1}^{3}} \int_{\omega\times B_{1}} \varTheta: \biggl( X_{1} \frac{\partial^{2} \psi_{1}}{\partial X_{3}^{2}} + X_{2} \frac {\partial^{2} \psi_{2}}{\partial X_{3}^{2}} \biggr) \,dx' \, dX \\ &\quad{}= \int_{\varOmega^{\pm}} F v \, dx + \int_{\omega\times B_{1}} \bigl(F^{m}_{1} \psi_{1} + F^{m}_{2} \psi_{2} + F_{3} v_{3}\bigr)\, dx' \, dX. \end{aligned}$$

(C.16)

Since $\sigma^{\pm}$ and $e(v)$ do not depend on $y$ and due to the periodicity of the fields $\widehat{v}$ and $\widehat{u}^{\pm}$, the equality above reads

$$\begin{aligned} &\int_{\varOmega^{\pm} } \sigma^{\pm} : e(v) \, dx + \int_{\varOmega ^{\pm} \times Y} \widehat{\sigma}^{\pm} : \varPsi^{(\pm)} e_{y}(\widehat{v}) \, dx\, dy\\ &\qquad{}- \frac{\kappa_{0}^{4}}{\kappa_{1}^{3}} \int_{\omega \times B_{1}} \varTheta: \biggl( X_{1} \frac{\partial^{2} \psi_{1}}{\partial X_{3}^{2}} + X_{2} \frac{\partial^{2} \psi_{2}}{\partial X_{3}^{2}} \biggr) \,dx' \, dX \\ &\quad{}=\int_{\varOmega^{\pm}} F v \, dx + \int_{\omega\times B_{1}} \bigl(F^{m}_{1} \psi_{1} + F^{m}_{2} \psi_{2} + F_{3} v_{3}\bigr)\, dx' \, dX. \end{aligned}$$

Step 3. To determine $\widehat{\sigma}$ we first take $v=0$. We then obtain

$$\begin{aligned} &\int_{\varOmega^{\pm} \times Y} \widehat{\sigma}^{\pm} : \varPsi^{(\pm )} e_{y}(\widehat{v}) \, dx\, dy- \frac{\kappa_{0}^{4}}{\kappa_{1}^{3}} \int _{\omega\times B_{1}} \varTheta: \biggl( X_{1} \frac{\partial^{2} \psi _{1}}{\partial X_{3}^{2}} + X_{2} \frac{\partial^{2} \psi_{2}}{\partial X_{3}^{2}} \biggr) \,dx' \, dX\\ &\quad{}= \int_{\omega\times B_{1}} \bigl(F^{m}_{1} \psi_{1} + F^{m}_{2} \psi_{2}\bigr)\, dx' \, dX. \end{aligned}$$

Since the right-hand side does not contain $\widehat{v}$,

$$\int_{\varOmega^{\pm} \times Y} \widehat{\sigma}^{\pm} : \varPsi^{(\pm )} e_{y}(\widehat{v}) \, dx \,dy = 0, $$

which corresponds to the strong formulation

$$ \left\{ \textstyle\begin{array}{l@{\quad}l} \sum_{j = 1}^{3} \frac{\partial\widehat{\sigma}^{\pm}_{ij}}{\partial y_{j}} = 0, &\text{in } \varOmega^{\pm} \times Y, \\ \sum_{j = 1}^{3} \widehat{\sigma}^{\pm}_{ij} = 0, &\text{on } \partial\bigl(\varOmega^{\pm} \times Y\bigr), \end{array}\displaystyle \right. $$

for $i = 1, 2, 3$. Therefore, $\widehat{\sigma}^{\pm} = 0$, and (C.16) is rewritten as

$$\begin{aligned} &\int_{\varOmega^{\pm} } \sigma^{\pm} : e(v) \, dx - \frac{\kappa _{0}^{4}}{\kappa_{1}^{3}} \int_{\omega\times B_{1}} \varTheta: \biggl( X_{1} \frac {\partial^{2} \psi_{1}}{\partial X_{3}^{2}} + X_{2} \frac{\partial^{2} \psi _{2}}{\partial X_{3}^{2}} \biggr) \,dx' \, dX \\ &\quad{}= \int_{\varOmega^{\pm}} F v \, dx + \int_{\omega\times B_{1}} \bigl(F^{m}_{1} \psi_{1} + F^{m}_{2} \psi_{2} + F_{3} v_{3}\bigr)\, dx' \, dX. \end{aligned}$$

(C.17)

Since the space $W^{1,\infty}(\varOmega ^{+} ; {\mathbb{R}}^{3})$ is dense in $H^{1}(\varOmega ^{+};{\mathbb{R}}^{3})$, since the space of functions in $W^{1,\infty}(\varOmega ^{-} , {\mathbb{R}}^{3})$ vanishing on $\varGamma$ is dense in $H^{1}(\varOmega ^{-};{\mathbb{R}}^{3})$ and since the space $\mathcal{C}^{1} (\overline{\omega} ; \mathcal{C}^{3}([0,1]))$ is dense in $L^{2}(\omega; H^{1}(0,1))$, the equality above holds for every $v$ in ${\mathbb{V}}$ and every $\psi_{1}$, $\psi_{2}$ in $L^{2}(\omega ; H^{1}(0,1))$ satisfying

$$\psi_{\alpha} \bigl(x', 0\bigr) = v_{\alpha|\varOmega^{-}} \bigl(x', 0\bigr), \qquad\psi _{\alpha} \bigl(x', 1\bigr) = v_{\alpha|\varOmega^{+}} \bigl(x', 0\bigr)\quad\text{for a.e. } x'\in\omega. $$

Finally, integrating over $D_{1}$ and using (5.2), we obtain the result.  □

3.3 C.3 Convergences

Theorem C.2

Under the assumptions (3.22) on the applied forces, we first have (convergence of the stress energy)

$$ \begin{aligned} \lim_{\varepsilon \to0}{ \mathcal{E}}(u_{\varepsilon}) = & \int_{\varOmega^{+} \cup\varOmega^{-}} \sigma^{\pm} : e(u) \, dx + \frac{\pi\kappa_{0}^{4}}{4 \kappa_{1}^{3}} E^{m} \int_{\omega\times(0, 1)} \sum_{\alpha= 1}^{2} \biggl|\frac{\partial^{2} \widetilde{\mathcal {U}}_{\alpha}}{\partial X_{3}^{2}} \biggr|^{2} \, dx' \, dX_{3}. \end{aligned} $$

(C.18)

The sequence $(u_{\varepsilon }, \sigma_{\varepsilon })$ shows the following convergence behavior:

• $u_{\varepsilon } \rightarrow u^{-}$ strongly in $H^{1}(\varOmega ^{-})$, $u_{\varepsilon }(\cdot+ \delta{\mathbf{e_{3}}}) \rightarrow u^{+}$ strongly in $H^{1}(\varOmega^{+})$,

• $\sigma_{\varepsilon } \rightarrow\sigma^{-}$ strongly in $L^{2}(\varOmega^{-})$, $\sigma_{\varepsilon }(\cdot+ \delta{\mathbf{e_{3}}}) \rightarrow \sigma^{+}$ strongly in $L^{2}(\varOmega^{+})$,

• $\frac{\delta ^{2}}{r} \mathcal {T}_{\varepsilon }' (\sigma _{\varepsilon }) \rightarrow\varTheta$ strongly in $L^{2}(\omega\times B_{1})$, where

  $$\varTheta_{ij} = \left\{ \textstyle\begin{array}{l@{\quad}l} -E^{m}\biggl(X_{1}\frac{\partial^{2} \widetilde{\mathcal {U}}_{1}}{\partial X_{3}^{2}} + X_{2} \frac{\partial^{2} \widetilde{\mathcal {U}}_{2}}{\partial X_{3}^{2}} \biggr), & (i, j) = (3, 3), \\ 0, &\textit{otherwise}, \end{array}\displaystyle \right. $$

• $\delta\mathcal {T}_{\varepsilon }(\widetilde{\mathcal {R}}_{\varepsilon , \alpha}) \rightarrow\widetilde{\mathcal {R}}_{\alpha}, \quad\alpha= 1,2$ strongly in $L^{2} (\omega; H^{1} (0, 1))$,

• $\mathcal {T}_{\varepsilon } (\widetilde{\mathcal {U}}_{\varepsilon ,\alpha}) \rightarrow\widetilde{\mathcal {U}}_{\alpha }, \quad\alpha= 1,2$ strongly in $L^{2}(\omega; H^{1} (0, 1))$, $\mathcal {T}_{\varepsilon } (\widetilde{\mathcal {U}}_{\varepsilon ,3}) \rightarrow u_{3}^{\pm} (\cdot, 0)$ strongly in $L^{2}(\omega; H^{1}(0, 1))$, $\frac{\delta}{r}\mathcal {T}_{\varepsilon }(\widetilde{\mathcal {U}}_{\varepsilon ,3} - \widetilde{\mathcal {U}}_{\varepsilon ,3} (\cdot , \cdot, 0)) \rightarrow0$ strongly in $L^{2} (\omega; H^{1} (0, 1))$.

Proof

Step 1. We prove (C.18).

We first recall a classical identity: if $T$ is a symmetric $3\times3$ matrix we have

$$\begin{aligned} &\lambda^{m} \operatorname{Tr}(T) \operatorname{Tr}(T) + \sum_{i,j=1}^{3} 2\mu^{m} T_{ij} T_{ij} \\ &\quad{}= E^{m} T_{33}^{2} + \frac{E^{m}}{(1+\nu^{m})(1-2\nu^{m})}\bigl(T_{11}+T_{22}+2\nu^{m} T_{33}\bigr)^{2} \\ &\qquad{}+\frac{E^{m}}{2(1+\nu^{m})}\bigl[(T_{11}-T_{22})^{2} + 4\bigl(T_{12}^{2}+T_{13}^{2}+T_{23}^{2} \bigr)\bigr]. \end{aligned}$$

(C.19)

Now, we consider the total elastic energy of the displacement $u_{\varepsilon}$, given by (2.10):

$$ {\mathcal{E}}(u_{\varepsilon})= \int_{\varOmega_{\varepsilon }} \sigma _{\varepsilon } : e(u_{\varepsilon}) \, dx = \int_{\varOmega _{\varepsilon }} f_{\varepsilon }\cdot u_{\varepsilon}\,dx. $$

(C.20)

The left-hand side of (C.20) is

$$\begin{aligned} {\mathcal{E}}(u_{\varepsilon}) =& \int_{\varOmega_{\varepsilon }} \sigma _{\varepsilon } : e(u_{\varepsilon})\, dx \\ =& \int_{\varOmega^{-}} \sigma_{\varepsilon } : e(u_{\varepsilon}) \,dx + \int_{\varOmega^{i}_{\varepsilon }} \sigma_{\varepsilon } : e(u_{\varepsilon}) \,dx + \int_{\varOmega^{+}_{\varepsilon }} \sigma _{\varepsilon } : e(u_{\varepsilon}) \,dx \\ =& \int_{\varOmega^{-}} \sigma_{\varepsilon } : e(u_{\varepsilon}) \,dx + \frac{r^{2}\delta}{\varepsilon ^{2}} \int_{\omega\times B_{1}} {\mathcal {T}}'_{\varepsilon}( \sigma_{\varepsilon }) : \mathcal {T}'_{\varepsilon}\bigl(e(u_{\varepsilon}) \bigr) \,dx' \,dX \\ &{}+ \int_{\varOmega ^{+}_{\varepsilon }} \sigma_{\varepsilon } : e(u_{\varepsilon}) \,dx. \end{aligned}$$

(C.21)

The second term of the right-hand side in the equation above is transformed using identity (C.19):

$$\begin{aligned} &\int_{\omega\times B_{1}} {\mathcal{T}}'_{\varepsilon}(\sigma _{\varepsilon }) : \mathcal{T}'_{\varepsilon}\bigl(e(u_{\varepsilon}) \bigr) \,dx' \,dX \\ &\quad{}= \int_{\omega\times B_{1}} \biggl( E^{m} e_{33}^{2}(u_{\varepsilon}) + \frac{E^{m}}{(1+\nu^{m})(1-2\nu^{m})} \bigl(e_{11}(u_{\varepsilon})+ e_{22}(u_{\varepsilon})+2\nu^{m} e_{33}(u_{\varepsilon}) \bigr)^{2} \\ &\qquad{}+\frac{E^{m}}{2(1+\nu^{m})} \bigl[ \bigl(e_{11}(u_{\varepsilon}) - e_{22}(u_{\varepsilon}) \bigr)^{2}+ 4 \bigl( \bigl(e_{12}(u_{\varepsilon})\bigr)^{2} \\ &\qquad{}+ \bigl(e_{13}(u_{\varepsilon})\bigr)^{2} + \bigl(e_{23}(u_{\varepsilon})\bigr)^{2} \bigr) \bigr] \biggr) \,dx'\,dX. \end{aligned}$$

Then by standard weak lower semi-continuity, Lemma 4.1 and (4.16) give (we recall that $\widetilde{\mathcal{R}}_{3}=0$)

$$ \int_{\varOmega^{+} \cup\varOmega^{-}} \sigma^{\pm} : e(u) \, dx + \frac {\kappa_{0}^{4}}{\kappa_{1}^{3}} \int_{\omega\times B_{1}} E^{m} e_{33}^{2}(u) \, dx'\,dX \leq\liminf_{\varepsilon \to0} {\mathcal {E}}(u_{\varepsilon}). $$

(C.22)

Besides, the convergences in Proposition 4.1 lead to

$$\begin{aligned} \limsup_{\varepsilon \to0}{ \mathcal{E}}(u_{\varepsilon})&=\limsup_{\varepsilon \to0} \int_{\varOmega_{\varepsilon }} f_{\varepsilon }\cdot u_{\varepsilon}\,dx=\lim _{\varepsilon \to0} \int_{\varOmega _{\varepsilon }} f_{\varepsilon }\cdot u_{\varepsilon}\,dx \\ &= \int_{\varOmega^{+} \cup\varOmega^{-}} F \cdot u \,dx + \int_{\omega \times(0, 1)} \sum_{\alpha= 1}^{2} \widetilde{F}_{\alpha}^{m} \widetilde{\mathcal{U}}_{\alpha} \,dx' \, dX_{3} + \int_{\omega} \overline{F}_{3}^{m} u_{3} \,dx'. \end{aligned} $$

Hence

$$\limsup_{\varepsilon \to0}{\mathcal{E}}(u_{\varepsilon}) \leq \liminf _{\varepsilon \to0} {\mathcal{E}}(u_{\varepsilon}). $$

Therefore (we recall that $\widetilde{\mathcal{U}}'_{3}=0$),

$$ \lim_{\varepsilon \to0}{\mathcal{E}}(u_{\varepsilon}) = \int _{\varOmega^{+} \cup\varOmega^{-}} \sigma^{\pm} : e(u) \, dx + \frac{\pi \kappa_{0}^{4}}{4 \kappa_{1}^{3}} E^{m} \int_{\omega\times(0, 1)} \sum_{\alpha= 1}^{2} \biggl|\frac{\partial^{2} \widetilde{\mathcal {U}}_{\alpha}}{\partial X_{3}^{2}} \biggr|^{2} \, dx' \, dX_{3}. $$

(C.23)

Step 2. As an immediate consequence of the convergence (C.23) above, we have

$$\begin{gathered} \begin{aligned} &\sigma_{\varepsilon } \rightarrow \sigma^{-} \quad\text{strongly in }L^{2}\bigl(\varOmega^{-}\bigr), \\ &\sigma_{\varepsilon }(\cdot+ \delta{\mathbf{e_{3}}}) \rightarrow \sigma^{+} \quad\text{strongly in }L^{2}\bigl(\varOmega^{+}\bigr), \\ &\frac{\delta^{2}}{r} \mathcal {T}_{\varepsilon }^{'} (\sigma _{\varepsilon }) \rightarrow\varTheta\quad\text{strongly in } L^{2} ( \omega\times B_{1}). \end{aligned} \end{gathered}$$

(C.24)

Hence

$$\begin{gathered} \begin{aligned} &e(u_{\varepsilon }) \rightarrow e \bigl(u^{-}\bigr) \quad\text{strongly in }L^{2}\bigl(\varOmega^{-}\bigr), \\ &e\bigl(u_{\varepsilon } (\cdot+ \delta{\mathbf{e_{3}}})\bigr) \rightarrow e\bigl(u^{+}\bigr) \quad\text{strongly in }L^{2}\bigl( \varOmega^{+}\bigr), \\ &\frac{\delta^{2}}{r} \mathcal {T}_{\varepsilon }^{'}\bigl(\widetilde {e(u_{\varepsilon })}\bigr) \rightarrow X \quad\text{strongly in } L^{2} (\omega\times B_{1}). \end{aligned} \end{gathered}$$

(C.25)

From (C.25)₁ and Korn’s inequality (2.11) in $\varOmega ^{-}$, we deduce that

$$u_{\varepsilon } \rightarrow u^{-} \quad \hbox{strongly in } H^{1} \bigl(\varOmega^{-}\bigr). $$

The strong convergence above and the estimates (3.9)–(3.10) yield

$$ u_{\varepsilon }(\cdot, 0)1_{\widehat{\omega}_{\varepsilon}}\rightarrow u^{-}_{|\varSigma}=\widetilde{\mathcal{U}}(\cdot, 0)\quad \hbox{strongly in } L^{2}(\varSigma). $$

(C.26)

From (C.25)₃ and (4.15)₄ we derive that

$$\begin{aligned} &\frac{\delta}{r}\frac{\partial\mathcal {T}_{\varepsilon }(\widetilde {\mathcal {U}}_{\varepsilon ,3})}{\partial X_{3}} + \delta\frac{\partial \mathcal {T}_{\varepsilon }(\widetilde{\mathcal {R}}_{\varepsilon ,1})}{\partial X_{3}} X_{2} - \delta\frac{\partial\mathcal {T}_{\varepsilon }(\widetilde{\mathcal {R}}_{\varepsilon ,2})}{\partial X_{3}} X_{1} + \frac{\delta^{2}}{r} \mathcal {T}_{\varepsilon }^{'}\bigl(\widetilde{e(u_{\varepsilon })} \bigr)_{33} \rightarrow \frac {\partial\widetilde{\mathcal {R}}_{1}}{\partial X_{3}} X_{2} - \frac {\partial\widetilde{\mathcal {R}}_{2}}{\partial X_{3}} X_{1} \\ &\quad{}\text{strongly in } L^{2}(\omega\times B_{1}). \end{aligned}$$

Hence, the equalities (3.4)₁–(3.4)₃ with the estimates (3.8)₁–(3.11) and the convergence in (C.26) lead to

$$\begin{gathered} \delta\mathcal {T}_{\varepsilon }(\widetilde{\mathcal {R}}_{\varepsilon , \alpha}) \rightarrow\widetilde{\mathcal {R}}_{\alpha}, \quad\alpha = 1,2 \text{ strongly in } L^{2} \bigl(\omega; H^{1} (0, 1)\bigr), \end{gathered}$$

(C.27)

$$\begin{gathered} \frac{\delta}{r}\mathcal {T}_{\varepsilon }\bigl(\widetilde{\mathcal {U}}_{\varepsilon ,3} - \widetilde{\mathcal {U}}_{\varepsilon ,3} (\cdot , \cdot, 0) \bigr) \rightarrow0 \quad\text{strongly in } L^{2} \bigl(\omega; H^{1} (0, 1)\bigr), \end{gathered}$$

(C.28)

$$\begin{gathered} \mathcal {T}_{\varepsilon }(\widetilde{\mathcal {U}}_{\varepsilon ,3}) \rightarrow u^{-}_{3|\varSigma} =u^{+}_{3|\varSigma} \quad\text{strongly in } L^{2} \bigl(\omega, H^{1}(0, 1)\bigr). \end{gathered}$$

(C.29)

The fourth estimate in Lemma B.2, the convergences in (C.26)–(C.27) and the equalities (4.11) imply that

$$ \mathcal {T}_{\varepsilon } (\widetilde{\mathcal {U}}_{\varepsilon ,\alpha}) \rightarrow\widetilde{\mathcal {U}}_{\alpha}, \quad\alpha = 1,2 \quad\text{strongly in } L^{2}\bigl(\omega; H^{1} (0, 1)\bigr). $$

(C.30)

From the convergences in (C.29)–(C.30) and the estimates (3.11)–(3.12) we obtain

$$ u_{\varepsilon }(\cdot, \delta)1_{\widehat{\omega}_{\varepsilon}}\rightarrow u^{+}_{|\varSigma}=\widetilde{\mathcal{U}}(\cdot, 1)\quad \hbox{strongly in } L^{2}(\varSigma). $$

(C.31)

Finally, due to the strong convergence above together with (C.25)₂ and (A.17)₂, we get $u_{\varepsilon}(\cdot+ \delta{\mathbf{e_{3}}}) \rightarrow u^{+}$ strongly in $H^{1}(\varOmega^{+})$. □