Proof. Step 1. We recall the following classical result.

For any $\phi\in H^1(\widetilde {{\omega}}_{pq})$ and any $\psi\in H^1(\widetilde {\Omega}_{pq})$, we have (see Lemma 3.3 in [Reference Chakrabortty, Griso and Orlik5])

(4.9)\begin{equation} \begin{aligned} &\|\phi\|_{L^2(\widetilde {{\omega}}_{pq})}\leq C\big(\|\phi\|_{L^2({\omega}^H_{pq})}+\delta\|\nabla \phi\|_{L^2(\widetilde {{\omega}}_{pq})}\big),\\ &\|\psi\|_{L^2(\widetilde {\Omega}_{pq})}\leq C\big( \|\psi\|_{L^2(\Omega^H_{pq})}+\delta\|\nabla \psi\|_{L^2(\widetilde {\Omega}_{pq})}\big). \end{aligned} \end{equation}

The constant does not depend on $\varepsilon$ and $\delta$.

Step 2. We prove (4.7).

Consider the function ${\boldsymbol\phi}_{pq}=\mathfrak{R}-\textbf{R}_{pq}$ in $\widetilde {\omega}_{pq}$, for $(p,q)\in {\mathcal K}^*_\varepsilon$. First, observe that (4.6) yields ${\partial}_{\alpha}\mathfrak{V}=\textbf{R}_{pq}\textbf{e}_{\alpha}$ a.e. in ${\omega}^H_{pq}$. Therefore, this identity and (4.5) $_4$ give

\begin{equation*}\sum_{(p,q)\in {\mathcal K}^*_\varepsilon}\big\|(\mathfrak{R}-\textbf{R}_{pq})\textbf{e}_{\alpha}\big\|^2_{L^2({\omega}^H_{pq})}\leq {C\over \delta}\textbf{D}(v)^2,\end{equation*}

which, combined with (4.9) $_1$ and using (4.5) $_3$, yields

(4.10)\begin{equation} \sum_{(p,q)\in {\mathcal K}^*_\varepsilon}\|{\boldsymbol\phi}_{pq}\textbf{e}_{\alpha}\|^2_{L^2({\omega}^H_{pq})}=\sum_{(p,q)\in {\mathcal K}^*_\varepsilon}\|(\mathfrak{R}-\textbf{R}_{pq})\textbf{e}_{\alpha}\|^2_{L^2(\widetilde {\omega}_{pq})}\leq {C\over \delta}\textbf{D}(v)^2. \end{equation}

Since both $\mathfrak{R}$ and $\textbf{R}_{pq}$ belong to $\mathrm{SO}(3)$, we have $|\mathfrak{R}|_F=|\textbf{R}_{pq}|_F=\sqrt{3}$ and

\begin{equation*}{\boldsymbol\phi}_{pq}\textbf{e}_3=(\mathfrak{R}-\textbf{R}_{pq})\textbf{e}_3=(\mathfrak{R}-\textbf{R}_{pq})\textbf{e}_1\land\mathfrak{R}\textbf{e}_2+\textbf{R}_{pq}\textbf{e}_1\land(\mathfrak{R}-\textbf{R}_{pq})\textbf{e}_2,\end{equation*}

which, together with (4.10), gives (4.7) $_1$. Then, as a consequence of (4.7) $_1$ and (4.5) $_5$, we obtain (4.7) $_2$.

Similarly, we set $\phi_{pq}(x)=v(x)-a_{pq}-\textbf{R}_{pq}\overline{x}_{pq}$ in $\widetilde \Omega_{pq}$, for $(p,q)\in {\mathcal K}^*_\varepsilon$. Using the fact that $v(x)=\textbf{a}_{pq}+\textbf{R}_{pq}\overline{x}_{pq}$ a.e. in $\Omega^H_{pq}$, i.e., $\phi_{pq}=0$ a.e. in $\Omega^H_{pq}$, and estimate (4.7) $_2$ together with (4.9) $_2$, we obtain

\begin{equation*}\|\phi_{pq}\|_{L^2(\widetilde \Omega_{pq})}\leq C\delta\|\nabla v-\textbf{R}_{pq}\|_{L^2(\widetilde \Omega_{pq})}.\end{equation*}

Hence, after summation we obtain (4.7) $_3$.

Step 3. We prove (4.8).

Observe that, using (4.6) and (4.9) $_1$, we have

\begin{equation*}\int_{\widetilde {{\omega}}_{pq}}\big|\mathfrak{V}-a_{pq}-\textbf{R}_{pq}\overline{x}^{'}_{pq}\big|^2dx'\leq C\delta^2\sum_{{\alpha}=1}^2\|{\partial}_{\alpha}\mathfrak{V}-\textbf{R}_{pq}\textbf{e}_{\alpha}\|^2_{L^2(\widetilde {\omega}_{pq})},\end{equation*}

which, together with (4.5) $_3$ and (4.7) $_1$, yields (4.8). This completes the proof.

Proof. First, we set

(4.12)\begin{equation} \widetilde {\Omega}_{pq}\doteq \widetilde {{\omega}}_{pq} \times {\mathfrak{I}}_\delta,\quad\forall\, (p,q)\in {\mathcal K}^*_\varepsilon. \end{equation}

The proof is organized into the following steps.

Step 1. We recall the following classical result. For any $\Psi\in H^1(\widetilde {\Omega}_{pq})^3$, we have

(4.13)\begin{equation} \begin{aligned} &\|\Psi\|_{L^2(\widetilde {\Omega}_{pq})}\leq C\big(\|\Psi\|_{L^2(\Omega^H_{pq})}+\delta\|\mathrm{dist}(\nabla\Psi;\mathrm{SO}(3))\|_{L^2(\widetilde {\Omega}_{pq})}\big). \end{aligned} \end{equation}

The constant does not depend on $\varepsilon$ and $\delta$.

Below, we provide a brief proof of this inequality. Theorem II.1.1 in [Reference Blanchard and Griso1] yields a rigid deformation $\textbf{S}_{pq}={\mathcal{G}}_{pq}+\textbf{B}_{pq}\land \overline{x}_{pq}$, $({\mathcal{G}}_{pq},\textbf{B}_{pq})\in {\mathbb{R}}^3\times \mathrm{SO}(3)$ such that

\begin{equation*}\|\Psi-\textbf{S}_{pq}\|_{L^2(\widetilde {\Omega}_{pq})}\leq C\delta \|\mathrm{dist}(\nabla\Psi;\mathrm{SO}(3))\|_{L^2(\widetilde {\Omega}_{pq})}.\end{equation*}

Thus,

\begin{equation*}\begin{aligned} \|\Psi\|^2_{L^2(\widetilde {\Omega}_{pq})}\leq C\big(\|\textbf{S}_{pq}\|^2_{L^2(\widetilde {\Omega}_{pq})}+\delta^2\|\mathrm{dist}(\nabla\Psi;\mathrm{SO}(3))\|^2_{L^2(\widetilde {\Omega}_{pq})}\big)\\ \leq C\big((\varepsilon+\delta)^2|{\mathcal{G}}_{pq}|^2+(\varepsilon+\delta)^4|\textbf{B}_{pq}|^2 +\delta^2\|\mathrm{dist}(\nabla\Psi;\mathrm{SO}(3))\|^2_{L^2(\widetilde {\Omega}_{pq})}\big). \end{aligned} \end{equation*}

Besides, we have

\begin{equation*}(\varepsilon-\delta)^2|{\mathcal{G}}_{pq}|^2+(\varepsilon-\delta)^4|\textbf{B}_{pq}|^2\leq C \|\textbf{S}_{pq}\|^2_{L^2(\Omega^H_{pq})}\leq 2\big(\|\Psi-\textbf{S}_{pq}\|^2_{L^2(\Omega^H_{pq})}+\|\Psi\|^2_{L^2(\Omega^H_{pq})}\big).\end{equation*}

This completes the proof of (4.13).

Step 2. A preliminary result.

Let $v$ and $w$ be two rigid deformations

\begin{equation*} v(x)=\textbf{a}+\textbf{R} \left(\begin{matrix} x_1\\ x_2\\ x_3 \end{matrix}\right) \text{a.e. in } \Big(-{\delta\over 2},\varepsilon\Big)\times {\mathfrak{I}}_\varepsilon\times {\mathfrak{I}}_\delta,\quad (\textbf{a},\textbf{R})\in{\mathbb{R}}^3\times \mathrm{SO}(3) \end{equation*}

and

\begin{equation*} w(x)=\textbf{a}^{'}+\textbf{R}^{'}\left(\begin{matrix} x_1+\varepsilon\\ x_2\\ x_3 \end{matrix}\right) \text{a.e. in } \Big(-\varepsilon,{\delta\over 2}\Big)\times {\mathfrak{I}}_\varepsilon\times {\mathfrak{I}}_\delta,\quad (\textbf{a}^{'},\textbf{R}^{'})\in{\mathbb{R}}^3\times \mathrm{SO}(3). \end{equation*}

In $\displaystyle {\mathcal{O}}_{{\varepsilon\delta}}\doteq{\mathfrak{I}}_\delta\times {\mathfrak{I}}_\varepsilon\times {\mathfrak{I}}_\delta$, we have

\begin{equation*} \begin{aligned} \textbf{u}&=v-w=\textbf{a}-\textbf{a}^{'}-\varepsilon\textbf{R}^{'}\textbf{e}_1+(\textbf{R}-\textbf{R}^{'})x. \end{aligned} \qquad \hbox{in }\; {\mathcal{O}}_{{\varepsilon\delta}}. \end{equation*}

The $L^2$ norm of $v-w$ in ${\mathcal{O}}_{\varepsilon\delta}$ is given by

(4.14)\begin{equation} \begin{aligned} {1\over \varepsilon\delta^2}\|\textbf{u}\|^2_{L^2({\mathcal{O}}_{{\varepsilon\delta}})}&=\Big|\textbf{a}-\textbf{a}^{'}-\varepsilon\textbf{R}^{'}\textbf{e}_1\Big|^2 + {\delta^2\over 6}|(\textbf{R}-\textbf{R}^{'})\textbf{e}_1|^2\\ &\quad +{\delta^2\over 6}|(\textbf{R}-\textbf{R}^{'})\textbf{e}_3|^2+\Big({\varepsilon^2\over 12}+{\delta^2\over 12}\Big)|(\textbf{R}-\textbf{R}^{'})\textbf{e}_2|^2. \end{aligned} \end{equation}

Step 3. We prove (4.11).

We apply (4.13) with the deformations $v=u$ and $w_{pq}=\textbf{a}_{pq}+\textbf{R}_{pq}\overline{x}_{pq}$ in $\widetilde {\Omega}_{pq}$. We then apply it with $v=u$ and $w_{p-1q}=\textbf{a}_{p-1q}+\textbf{R}_{p-1q} \overline{x}_{p-1q}$ in $\widetilde {\Omega}_{p-1q}$. This allows us to estimate $w_{pq}-w_{p-1q}$ in $\widetilde {\Omega}_{pq}\cap \widetilde {\Omega}_{p-1q}$ using (4.14). We obtain (4.11) $_{1,3,7}$. Similarly, proceeding as above, we estimate $w_{pq}-w_{pq-1}$ in $\widetilde {\Omega}_{pq}\cap \widetilde {\Omega}_{pq-1}$, which, together with (4.14), yields (4.11) $_{2,4,8}$.

Proof. First, using the decomposition (4.35), we have

(4.63)\begin{equation} \begin{aligned} &\!\!\!\!\int_{\Omega_\delta}f_{\varepsilon\delta}\cdot (v-I_d)\,dx\\ &= \int_{\Omega_\delta}f_{\varepsilon\delta}\cdot\Big(({\mathcal{V}}-I_d)(x')+x_3(\textbf{R}-\textbf{I}_3)(x')\textbf{e}_3\Big)\,dx+\int_{\Omega^S_{\varepsilon\delta}}f_{\varepsilon\delta}\cdot\overline{v}\,dx,\\ &=\delta\int_{{\omega}_\delta}f_{\varepsilon\delta}(x')\cdot({\mathcal{V}}-I_d)(x')\,dx'+\int_{\Omega^S_{\varepsilon\delta}}f_{\varepsilon\delta}\cdot\overline{v}\,dx. \end{aligned} \end{equation}

First observe that the expression (4.60) and the estimate (4.38) $_3$ give

(4.64)\begin{equation} \left|\int_{\Omega^S_{\varepsilon\delta}}f_{\varepsilon\delta}\cdot \overline{v}\,dx\right|\leq \delta^{1/2}\|f_{\varepsilon\delta}\|_{L^2({\omega}_\delta)}\|\overline{v}\|_{L^2(\Omega_\delta)}\leq C\varepsilon\delta^{3/2}\|f\|_{L^2({\omega}_\delta)}\textbf{D}(v). \end{equation}

Again the estimates (4.56) $_{1,3}$ give

(4.65)\begin{equation} {\begin{aligned} \delta\left|\int_{{\omega}_{\delta}}f_{\varepsilon\delta}\cdot ({\mathcal{V}}-I_d)\,dx'\right| &\leq \delta\|f_{\varepsilon\delta,3}\|_{L^2({\omega}_\delta)}\|{\mathcal{U}}_3\|_{L^2({\omega}_\delta)}+\delta\sum_{\alpha=1}^2\|f_{\varepsilon\delta,\alpha}\|_{L^2({\omega}_\delta)}\|{\mathcal{U}}_\alpha\|_{L^2({\omega}_\delta)}\\ &\leq C\varepsilon^{1/2}\delta^{3/2}(\varepsilon^{1/2}+\delta^{1/2})\textbf{D}(v)\|f\|_{L^2({\omega}_\delta)}\\ &\quad +C_M\Big(\sum_{\alpha=1}^2\|f_\alpha\|_{L^2({\omega}_\delta)}\Big)\textbf{D}(v)^2. \end{aligned} } \end{equation}

We recall that

(4.66)\begin{equation} \mathrm{dist}(F, \mathrm{SO}(3))\leq |F^TF-\textbf{I}_3| \quad \text{for all}\;\;F \in \textbf{M}_3\quad\hbox{such that}\;\; \mathrm{det}(F) \gt 0, \end{equation}

and due to the assumption (3.6), we have

(4.67)\begin{equation} c_0|F^{T}F-\textbf{I}_3|^2\leq \textbf{Q}_{\varepsilon\delta}\big(x , \textbf{E}(F)\big)\qquad \hbox{for a.e. } x\in \Omega^S_{\varepsilon\delta},\quad \forall\,F\in\textbf{M}_3. \end{equation}

Since we have $\textbf{J}_{\varepsilon\delta}(v)\leq \textbf{J}_{\varepsilon\delta}(I_d)=0$, then using the two inequalities (4.66)–(4.67), we obtain

\begin{align*}&c_0\textbf{D}(v)^2-\int_{\Omega_{\delta}}f_{\varepsilon\delta}\cdot(v-I_d)\,dx\leq c_0\|(\nabla v)^T(\nabla v)-\textbf{I}_3\|^2_{L^2(\Omega^S_{\varepsilon\delta})}\\ &\quad -\int_{\Omega_{\delta}}f_{\varepsilon\delta}\cdot(v-I_d)\,dx\leq \textbf{J}_{\varepsilon\delta}(v)\leq 0\end{align*}

which imply

(4.68)\begin{align} &c_0\|(\nabla v)^T(\nabla v)-\textbf{I}_3\|^2_{L^2(\Omega^S_{\varepsilon\delta})}\leq \left|\int_{\Omega_{\delta}}f_{\varepsilon\delta}\cdot (v-I_d)\,dx\right|, \nonumber\\ &c_0\textbf{D}(v)^2\leq 2 \left|\int_{\Omega_{\delta}}f_{\varepsilon\delta}\cdot (v-I_d)\,dx\right|. \end{align}

Therefore, the above inequalities with (4.64)–(4.65) give

\begin{equation*} c_0\textbf{D}(v)^2 \leq C\varepsilon\delta^{3/2}\textbf{D}(v)\|f\|_{L^2({\omega}_\delta)}+ 2C_M\Big(\sum_{\alpha=1}^2\|f_\alpha\|_{L^2({\omega}_\delta)}\Big)\textbf{D}(v)^2. \end{equation*}

So, with the assumption (4.61) and the inequalities (4.68), we obtain the estimates (4.62).

Proof. All the convergences (6.5) are direct consequences of the estimates (6.2).

From (6.2) $_5$, we obtain

\begin{equation*}{1\over (\varepsilon\delta)^{1/2}}\big((\textbf{R}_{\varepsilon\delta})^T+\textbf{R}_{\varepsilon\delta}-2\textbf{I}_3\big)\to 0\quad\text{strongly in}\;\;L^2({\omega})^{3\times 3}.\end{equation*}

Then, the convergence (6.5) $_2$ gives

\begin{equation*}{1\over ({\varepsilon\delta})^{1/2}}\big((\textbf{R}_{\varepsilon\delta})^T+\textbf{R}_{\varepsilon\delta}-2\textbf{I}_3\big)\rightharpoonup \textbf{A}+\textbf{A}^T\quad\text{weakly in}\;\;L^2({\omega})^{3\times 3}.\end{equation*}

This implies $\textbf{A}=-\textbf{A}^T$.

From the estimate (6.2) $_3$, we have

\begin{equation*}{1\over ({\varepsilon\delta})^{1/2}}(\partial_\alpha{\mathcal{U}}_{\varepsilon\delta}-(\textbf{R}_{\varepsilon\delta}-\textbf{I}_3)\textbf{e}_\alpha)\to 0\quad\text{strongly in}\ L^2({\omega})^3.\end{equation*}

Besides, the convergences (6.5) lead to

\begin{equation*}\begin{aligned} {1\over ({\varepsilon\delta})^{1/2}}(\partial_\alpha{\mathcal{U}}_{\varepsilon\delta}-(\textbf{R}_{\varepsilon\delta}-\textbf{I}_3)\textbf{e}_\alpha)\cdot\textbf{e}_\beta&\rightharpoonup -\textbf{A}\textbf{e}_\alpha\cdot\textbf{e}_\beta\quad&&\text{weakly in } L^2({\omega}),\\ {1\over ({\varepsilon\delta})^{1/2}}(\partial_\alpha{\mathcal{U}}_{\varepsilon\delta}-(\textbf{R}_{\varepsilon\delta}-\textbf{I}_3)\textbf{e}_\alpha)\cdot\textbf{e}_3&\rightharpoonup \partial_\alpha{\mathcal{U}}_3-\textbf{A}\textbf{e}_\alpha\cdot\textbf{e}_3\quad&&\text{weakly in } L^2({\omega}). \end{aligned} \end{equation*}

This gives (6.6) $_2$ and the expression of $\textbf{A}$ given by

(6.9)\begin{equation} \textbf{A}=\left(\begin{matrix} 0&0&-\partial_1{\mathcal{U}}_3\\ 0&0&-\partial_2{\mathcal{U}}_3\\ \partial_1{\mathcal{U}}_3 &\partial_2{\mathcal{U}}_3& 0 \end{matrix}\right). \end{equation}

Observe that due to the estimates (4.16), (4.37) $_4$ and inequality (6.1), we have

(6.10)\begin{align} &\|\textbf{R}^\diamond_{\varepsilon\delta}-\textbf{I}_3\|_{H^1({\omega}_\delta)}\leq C({\varepsilon\delta})^{1/2},\quad\|\textbf{R}_{\varepsilon\delta}-\textbf{R}^\diamond_{\varepsilon\delta}\|_{L^2({\omega}_\delta)}\leq C\varepsilon\sqrt{{\varepsilon\delta}}, \nonumber\\ &\|\partial_\alpha\textbf{R}_{\varepsilon\delta}^\diamond\textbf{e}_{3-\alpha}\|_{L^2({\omega}_\delta)}\leq C{\delta\over \varepsilon}\sqrt{{\varepsilon\delta}}, \end{align}

which imply

(6.11)\begin{equation} \begin{aligned} {1\over (\varepsilon\delta)^{1/2}}(\textbf{R}^\diamond_{\varepsilon\delta}-\textbf{I}_3)&\rightharpoonup \textbf{A}\quad&&\text{weakly in}\ H^1({\omega})^{3\times 3},\\ {1\over (\varepsilon\delta)^{1/2}}(\textbf{R}_{\varepsilon\delta}-\textbf{I}_3)&\to \textbf{A}\quad&&\text{strongly in}\ L^2({\omega})^{3\times 3} \end{aligned} \end{equation}

and $\textbf{A} \in H^1({\omega})^{3\times 3}$. So, with the expression (6.9) of $\textbf{A}$ using ${\mathcal{U}}_3$, we obtain ${\mathcal{U}}_3\in H^2({\omega})$. We also get $\partial_\alpha\textbf{A}\textbf{e}_{3-\alpha}=0$, which imply $\partial_{12}{\mathcal{U}}_3=0$. Due to the boundary condition satisfied by ${\mathcal{V}}_{\varepsilon\delta}$ and $\textbf{R}_{\varepsilon\delta}$, we obtained that ${\mathcal{U}}$ also satisfies the corresponding boundary conditions.

Using Lemma 4.8, we obtain

(6.12)\begin{equation} \begin{aligned} & \|(\textbf{R}^{\diamond}_{\varepsilon\delta}-\textbf{I}_3)\textbf{e}_1-d_1F_{\varepsilon\delta}^{\diamond}\|_{L^2({\omega}_\delta)}+\|(\textbf{R}^{\diamond}_{\varepsilon\delta}-\textbf{I}_3)\textbf{e}_2-d_2G_{\varepsilon\delta}^{\diamond}\|_{L^2({\omega}_\delta)}\leq C\sqrt{{\varepsilon\delta}}\Big({\delta\over \varepsilon}\Big),\\ &\|{\mathcal{U}}^{\diamond}_{{\varepsilon\delta}}-F^{\diamond}_{\varepsilon\delta}-G^{\diamond}_{\varepsilon\delta}\|_{H^1({\omega}_\delta)}\leq C\sqrt{{\varepsilon\delta}}\Big({\delta\over \varepsilon}\Big). \end{aligned} \end{equation}

The convergence (6.7) is direct consequence of (6.12). Observe that

(6.13)\begin{equation} \begin{aligned} \partial_1{\mathcal{U}}^{\diamond}_{\varepsilon\delta}-(\textbf{R}^\circ_{\varepsilon\delta}-\textbf{I}_3)\textbf{e}_1&=\partial_1{\mathcal{U}}^{\diamond}_{\varepsilon\delta}-d_1F^{\diamond}_{\varepsilon\delta}-(\textbf{R}^{\diamond}_{\varepsilon\delta}-\textbf{I}_3)\textbf{e}_1+d_1F^{\diamond}_{\varepsilon\delta}+(\textbf{R}_{\varepsilon\delta}^{\diamond}-\textbf{R}_{\varepsilon\delta}^\circ)\textbf{e}_1,\\ \partial_2{\mathcal{U}}^{\diamond}_{\varepsilon\delta}-(\textbf{R}^\circ_{\varepsilon\delta}-\textbf{I}_3)\textbf{e}_2&=\partial_2{\mathcal{U}}^{\diamond}_{\varepsilon\delta}-d_2G^{\diamond}_{\varepsilon\delta}-(\textbf{R}^{\diamond}_{\varepsilon\delta}-\textbf{I}_3)\textbf{e}_2+d_2G^{\diamond}_{\varepsilon\delta}+(\textbf{R}_{\varepsilon\delta}^{\diamond}-\textbf{R}_{\varepsilon\delta}^\circ)\textbf{e}_2, \end{aligned} \end{equation}

which together with the estimates (4.17) $_1$, (4.24) $_1$ and the convergences (6.7) give (6.8). Since, using (4.17) $_1$, (4.24) $_1$, and (6.1), we have

\begin{equation*}{\varepsilon\over \delta}{1\over \sqrt{{\varepsilon\delta}}}\|\partial_{\alpha}{\mathcal{U}}^{\diamond}_{\varepsilon\delta}-(\textbf{R}^\circ_{\varepsilon\delta}-\textbf{I}_3)\textbf{e}_{\alpha}\|_{L^2({\omega}_\delta)}\leq C\varepsilon,\quad {\varepsilon\over \delta}{1\over \sqrt{{\varepsilon\delta}}}\|\textbf{R}^{\diamond}_{\varepsilon\delta}-\textbf{R}^\circ_{\varepsilon\delta}\|_{L^2({\omega}_\delta)}\leq C{\varepsilon^2\over \delta}.\end{equation*}

The assumption (2.2) $_2$ is needed, so the last terms in (6.13) vanishes in the limit. This completes the proof.

Proof. Step 1: We prove (6.14).

Observe that the identity $\partial_\alpha {\mathcal{V}}_{\varepsilon\delta} - \textbf{R}_{\varepsilon\delta} \textbf{e}_\alpha = 0=\textbf{R}_{\varepsilon\delta}(\textbf{R}_{\varepsilon\delta})^T-\textbf{I}_3$ holds outside a set of measure at most $\displaystyle \textbf{O}\Big({\delta\over \varepsilon}\Big)$ (see (3.2)), then from the estimates (4.38) $_1$, (4.49), (6.1), we deduce the weak convergences

(6.17)\begin{equation} \begin{aligned} {1\over \varepsilon\delta} \big( \partial_\alpha {\mathcal{U}}_{\varepsilon\delta} - (\textbf{R}_{\varepsilon\delta} - \textbf{I}_3)\textbf{e}_\alpha \big) &\rightharpoonup 0 \quad &&\text{weakly in}\ L^2({\omega})^3,\\ {1\over \varepsilon\delta}\big(\textbf{R}_{\varepsilon\delta}(\textbf{R}_{\varepsilon\delta})^T-\textbf{I}_3\big)& \rightharpoonup 0 \quad&&\text{weakly in}\ L^2({\omega})^{3\times 3}. \end{aligned} \end{equation}

Combined with the identity

\begin{align*} e_{\alpha\beta}({\mathcal{U}}_{\varepsilon\delta}) - {1\over 2}(\textbf{R}_{\varepsilon\delta} + \textbf{R}_{\varepsilon\delta}^T - 2\textbf{I}_3)\textbf{e}_\alpha \cdot \textbf{e}_\beta &= {1\over 2}(\partial_\alpha {\mathcal{U}}_{\varepsilon\delta} - (\textbf{R}_{\varepsilon\delta} - \textbf{I}_3)\textbf{e}_\alpha) \cdot \textbf{e}_\beta \nonumber\\ &\quad + {1\over 2}(\partial_\beta {\mathcal{U}}_{\varepsilon\delta} - (\textbf{R}_{\varepsilon\delta} - \textbf{I}_3)\textbf{e}_\beta) \cdot \textbf{e}_\alpha. \end{align*}

This implies the weak convergence

(6.18)\begin{equation} {1\over {\varepsilon\delta}} \left[ e_{\alpha\beta}({\mathcal{U}}_{\varepsilon\delta}) - {1\over 2}(\textbf{R}_{\varepsilon\delta} + \textbf{R}_{\varepsilon\delta}^T - 2\textbf{I}_3)\textbf{e}_\alpha \cdot \textbf{e}_\beta \right] \rightharpoonup 0 \quad \text{weakly in}\ L^2({\omega}). \end{equation}

Using the identity (4.55) along with the convergence (6.17) $_2$ and (6.11) $_2$ gives

\begin{equation*}{1\over {\varepsilon\delta}}\big(\textbf{R}_{\varepsilon\delta}+(\textbf{R}_{\varepsilon\delta})^T-2\textbf{I}_3\big)\textbf{e}_\alpha\cdot\textbf{e}_\beta \rightharpoonup \textbf{A}^2\textbf{e}_\alpha\cdot\textbf{e}_\beta,\quad\text{weakly in}\ L^2({\omega}),\end{equation*}

which together with (6.5) $_{1}$ imply

\begin{align*} &{1\over {\varepsilon\delta}}\Big[e_{\alpha\beta}({\mathcal{U}}_{\varepsilon\delta})-{1\over 2}(\textbf{R}_{\varepsilon\delta}+(\textbf{R}_{\varepsilon\delta})^T-2\textbf{I}_3)\textbf{e}_\alpha\cdot\textbf{e}_\beta\Big]\rightharpoonup e_{\alpha\beta}({\mathcal{U}})-{1\over 2}\textbf{A}^2\textbf{e}_\alpha\cdot\textbf{e}_\beta,\\ &\quad \text{weakly in}\ L^2({\omega}).\end{align*}

So, from the expression (6.9) of $\textbf{A}$, we obtain $\textbf{A}^2\textbf{e}_\alpha\cdot\textbf{e}_\beta=-\partial_\alpha{\mathcal{U}}_3\partial_\beta{\mathcal{U}}_3$, which along with (6.18) imply (6.14).

Step 2. We prove that there exists ${\mathcal{U}}^{(3-\alpha)}_3\in H_{(0,l)}^{2}\big((0,L)_{x_\alpha}\big)$ such that (6.15) holds.

First, since

\begin{equation*}{\mathcal{U}}_3 \in H^2_\gamma({\omega})\quad\text{and}\quad \partial_{12}{\mathcal{U}}_3=0\quad\text{a.e. in}\ {\omega},\end{equation*}

there exist ${\cal F} $ and ${\mathcal{G}}\in H^2(0,L)$ such that

\begin{equation*}{\mathcal{U}}_3(x_1,x_2)={\cal F}(x_1)+{\mathcal{G}}(x_2)\qquad \hbox{for a.e. } (x_1,x_2)\in {\omega}.\end{equation*}

Besides, we have

\begin{equation*} {\mathcal{U}}_3=0\quad\hbox{and}\quad \nabla {\mathcal{U}}_3=0\quad \text{a.e. on}\ \gamma.\end{equation*}

This implies

\begin{equation*} \begin{aligned} {\cal F}'(x_1)={\cal F}(x_1)+{\mathcal{G}}(0)=0\quad {\mathcal{G}}'(x_2)={\mathcal{G}}(x_2)+{\cal F}(0)=0\quad \text{for a.e.}\ (x_1,x_2)\in (0,l)^2. \end{aligned} \end{equation*}

So, we obtain (6.15) by setting

\begin{equation*}{\mathcal{U}}^{(2)}_3={\cal F}+{\mathcal{G}}(0),\qquad {\mathcal{U}}^{(1)}_3={\mathcal{G}}-{\mathcal{G}}(0).\end{equation*}

Step 3. We prove that

\begin{equation*} {\mathcal{U}}_3^{(1)}\equiv 0\qquad \hbox{or}\qquad {\mathcal{U}}_3^{(2)}\equiv 0. \end{equation*}

From (6.14), we have

\begin{equation*}e_{\alpha\beta}({\mathcal{U}}_m)+{1\over 2}\partial_\alpha{\mathcal{U}}_3\partial_\beta{\mathcal{U}}_3=0,\qquad (\alpha,\beta)\in\{1,2\}^2.\end{equation*}

We recall that for any membrane displacement belonging to $H^1({\omega})^2$, we have (It is called the Saint–Venant–Kirchhoff compatibility condition.)

\begin{equation*} \partial^2_{22}\big(e_{11}({\mathcal{U}}_m)\big)+\partial^2_{11}\big(e_{22}({\mathcal{U}}_m)\big)-2\partial^2_{12}\big(e_{12}({\mathcal{U}}_m)\big)=0 \qquad \hbox{in}\quad \mathcal{D}({\omega}), \end{equation*}

where $\mathcal{D}({\omega})$ denotes the space of compactly supported, infinitely differentiable functions on ${\omega}$. So

\begin{equation*}d^2_{11}{\mathcal{U}}^{(2)}_3(x_1)d^2_{22}{\mathcal{U}}^{(1)}_3(x_2)=0\qquad \hbox{for a.e. } (x_1,x_2)\in{\omega}\end{equation*}

which in turn implies the claim.

Step 4. Below, we determine the membrane displacement ${\mathcal{U}}_m$.

First, assume ${\mathcal{U}}^{(1)}_3\equiv 0$. Now, using the equation (6.14) $_1$, we have

\begin{equation*}\partial_1{\mathcal{U}}_1(x_1,x_2)=-{1\over 2}(d_1{\mathcal{U}}^{(2)}_3)^2(x_1),\quad \partial_2{\mathcal{U}}_2=0,\quad\text{and}\quad \partial_1{\mathcal{U}}_2=-\partial_2{\mathcal{U}}_1.\end{equation*}

The above conditions imply that there exists $a\in {\mathbb{R}}$ such that

\begin{equation*}{\mathcal{U}}_m(x_1,x_2)=\Big(-{1\over 2}\int_0^{x_1} \Big|{d\,{\mathcal{U}}^{(2)}_3\over dt}(t)\Big|^2dt+a x_2\Big)\textbf{e}_1-a x_1\textbf{e}_2\qquad \hbox{for a.e. } (x_1,x_2)\in {\omega}.\end{equation*}

The boundary conditions imply that $a=0$. This gives (6.16) in the case ${\mathcal{U}}^{(1)}_3\equiv 0$. Similarly, we prove the result in the case ${\mathcal{U}}^{(2)}_3\equiv 0$. This completes the proof.

Proof. Step 1: We prove (6.20).

Using the estimates (6.4) and the assumption (2.2), we obtain the convergences (6.20). Observe that (6.20) $_2$ is due to the scaling assumption (2.2) and fact that ((6.4) $_1$)

\begin{equation*}{1\over \varepsilon^2\sqrt{{\varepsilon\delta}}}\|\Pi^{({\alpha})}_{\varepsilon\delta}(\overline{v}_{\varepsilon\delta})\|_{L^2({\omega}\times {\cal Y}_{\alpha})}\leq C{\delta\over \varepsilon}.\end{equation*}

In addition, since $\overline{v}_{\varepsilon\delta}=0$ a.e. in $\partial{\omega}^H_{pq}\times {\mathfrak{I}}$ for all $(p,q)\in{\mathcal{K}}_\varepsilon^*$, passing to the limit gives (6.19).

Step 2: We prove (6.21) $_{2,3,4}$.

Observe that, using the estimates (5.5) and (6.2) $_{6,7}$, we have

(6.23)\begin{equation} \begin{aligned} &\big\|\mathcal{T}_{\varepsilon\delta}^{(\alpha)}\big((\textbf{R}_{\varepsilon\delta})^T\textbf{R}_{\varepsilon\delta}-\textbf{I}_3\big)\big\|_{L^2({\omega}\times Y_\alpha)}\leq C{\varepsilon\sqrt{{\varepsilon\delta}}},\qquad \big\|\mathcal{T}_{\varepsilon\delta}^{(\alpha)}\big(\nabla \textbf{R}\big)\big\|_{L^2({\omega}\times Y_\alpha)} \leq C{\varepsilon\over \delta}\sqrt{{\varepsilon\delta}}. \end{aligned} \end{equation}

We also have from (4.33) $_{3}$, for $(p,q)\in {\mathcal{K}}^{(2)}_\varepsilon$

(6.24)\begin{equation} \begin{aligned} & \partial_1\textbf{R}_{\varepsilon\delta}(p\varepsilon+z_1,q\varepsilon+z_2)={1\over \delta} \big(\textbf{R}_{{\varepsilon\delta},pq}-\textbf{R}_{{\varepsilon\delta},p-1q}\big),\\ & \partial_2\textbf{R}_{\varepsilon\delta}(p\varepsilon+z_1,q\varepsilon+z_2)=0 \end{aligned} \qquad\forall\,(z_1,z_2)\in{\mathfrak{I}}_\delta\times {\mathfrak{J}}_{\varepsilon\delta}. \end{equation}

First, due to the equality (6.24) $_2$, the estimates (4.11) $_2$, (4.45) $_3$, (5.5), and (6.1), we obtain

\begin{equation*}\big\|\mathcal{T}_{\varepsilon\delta}^{(2)}(\partial_2\textbf{R}_{\varepsilon\delta}) \big\|_{L^2({\omega}\times Y_2)}\leq C\varepsilon.\end{equation*}

Equality (6.24) $_2$ also implies that the support of the function $\mathcal{T}_{\varepsilon\delta}^{(2)}(\partial_2\textbf{R}_{\varepsilon\delta})$ has a measure of order $\displaystyle \textbf{O}\Big({\delta\over \varepsilon}\Big)$. As a consequence, we get (6.21) $_2$ for $\alpha=2$. In the same way, we show the convergence (6.21) $_2$ for $\alpha=1$.

Let $\phi$ be in $\mathcal{C}_c^\infty({\omega})^{3\times 3}$. The definition of $\mathcal{T}_{\varepsilon\delta}^{(2)}\big(\partial_1\textbf{R}_{\varepsilon\delta}\big)$ and the estimate (6.23) $_2$ lead to

(6.25)\begin{equation}\begin{aligned} \Big|\int_{{\omega}\times Y_2}{{\delta\over \varepsilon}{1 \over \sqrt{{\varepsilon\delta}}}}\mathcal{T}_{\varepsilon\delta}^{(2)}\big(\partial_1\textbf{R}_{\varepsilon\delta}\big):\phi\, dx'dy'-\int_{{\omega}\times Y^{(2)}_{\varepsilon\delta}}{\delta\over \varepsilon}{1 \over \sqrt{{\varepsilon\delta}}}\mathcal{T}_{\varepsilon\delta}^{(2)}\big(\partial_1\textbf{R}_{\varepsilon\delta}\big):\phi\, dx'dy'\Big|\\ \leq C{\delta\over \varepsilon}\|\phi\|_{L^\infty({\omega}\times Y_2)}, \end{aligned} \end{equation}

where $\displaystyle Y^{(2)}_{\varepsilon\delta}={\mathfrak{I}}\times {1\over \varepsilon}{\mathfrak{J}}_{\varepsilon\delta}=(-{1\over 2},{1\over 2})\times ({\delta\over 2\varepsilon},1-{\delta\over 2\varepsilon})$.

Then, for $\varepsilon$ small enough and due to (6.24) $_1$, we have

\begin{equation*} \begin{aligned} &\int_{{\omega}\times Y^{(2)}_{\varepsilon\delta}}{\displaystyle{\delta\over \varepsilon}{1 \over \sqrt{{\varepsilon\delta}}}}\mathcal{T}_{\varepsilon\delta}^{(2)}\big(\partial_1\textbf{R}_{\varepsilon\delta}\big):\phi\, dx'dy' \\ & \quad ={1\over \sqrt{{\varepsilon\delta}}}\Big(1-{\delta\over \varepsilon}\Big)\sum_{(p,q)\in {\mathcal{K}}^{(2)}_\varepsilon}\int_{{\omega}_{pq}} {\textbf{R}_{{\varepsilon\delta},pq}-\textbf{R}_{{\varepsilon\delta},p-1q}\over \varepsilon}:\phi\,dx'\\ &\quad ={1\over \sqrt{{\varepsilon\delta}}}\Big(1-{\delta\over \varepsilon}\Big)\sum_{(p,q)\in {\mathcal{K}}^{(2)}_\varepsilon}\int_{{\omega}_{pq}} \big(\textbf{R}_{{\varepsilon\delta},pq}-\textbf{I}_3\big):{\phi - \phi(\cdot+\varepsilon\textbf{e}_1)\over \varepsilon} dx'. \end{aligned} \end{equation*}

Now, thanks to (4.23), we replace $\textbf{R}_{{\varepsilon\delta},pq}$ by $\textbf{R}^\diamond_{\varepsilon\delta}$. Hence, the above equality and (6.25) yield

(6.26)\begin{equation}\begin{aligned} \Big|\int_{{\omega}\times Y_2}{\delta\over \varepsilon}{1 \over \sqrt{{\varepsilon\delta}}}\mathcal{T}_{\varepsilon\delta}^{(2)}\big(\partial_1\textbf{R}_{\varepsilon\delta}\big):\phi\, dx'dy'\\ \quad -\Big(1-{\delta\over \varepsilon}\Big) \int_{\omega} {1\over \sqrt{{\varepsilon\delta}}}\big(\textbf{R}^\diamond_{\varepsilon\delta}-\textbf{I}_3\big):{\phi - \phi(\cdot+\varepsilon\textbf{e}_1)\over \varepsilon} dx'\Big|\\ \leq C\Big(\varepsilon+{\delta\over \varepsilon}\Big)\|\phi\|_{L^\infty({\omega}\times Y_2)}. \end{aligned} \end{equation}

Convergence (6.11) $_1$ and the fact that

\begin{equation*}{\phi - \phi(\cdot+\varepsilon\textbf{e}_1)\over \varepsilon}\longrightarrow -\partial_1\phi\quad \hbox{strongly in }\; L^2({\omega})^{3\times 3}\end{equation*}

imply with (6.26)

\begin{align*}&\lim_{(\varepsilon,\delta)\to(0,0)}\int_{{\omega}\times Y_2}{\delta\over \varepsilon}{1 \over \sqrt{{\varepsilon\delta}}}\mathcal{T}_{\varepsilon\delta}^{(2)}\big(\partial_1\textbf{R}_{\varepsilon\delta}\big):\phi\, dx'dy'= -\int_{\omega} \textbf{A}:\partial_1\phi\,dx'\\ &=\int_{{\omega} \times Y_2} \partial_1\textbf{A} : \phi\,dx'dy'.\end{align*}

Note that, outside to a set of measure $\displaystyle \textbf{O}\Big({\delta\over \varepsilon}\Big)$, the function $\mathcal{T}_{\varepsilon\delta}^{(2)}\big(\partial_1\textbf{R}_{\varepsilon\delta}\big)$ does not depend on $y$. This ends the proof of the convergence (6.21) $_3$ for $\alpha=2$. In the same way, we show the convergence (6.21) $_4$ for $\alpha=1$.

Step 3: We prove (6.21) $_1$.

First, we have

\begin{equation*} {1\over \varepsilon\sqrt{{\varepsilon\delta}}}\mathcal{T}_{\varepsilon\delta}^{(2)}\big(((\textbf{R}_{\varepsilon\delta})^T\textbf{R}_{\varepsilon\delta}-\textbf{I}_3)\textbf{1}_{Y_2\setminus Y^{(2)}_{\varepsilon\delta}}\big) \rightharpoonup 0 \qquad \text{weakly in } L^2({\omega}\times Y_2)^{3\times 3}.\end{equation*}

Now, we determine the weak limit of $\displaystyle {1\over \varepsilon\sqrt{{\varepsilon\delta}}}\mathcal{T}_{\varepsilon\delta}^{(2)}\big(((\textbf{R}_{\varepsilon\delta})^T\textbf{R}_{\varepsilon\delta}-\textbf{I}_3)\textbf{1}_{Y^{(2)}_{\varepsilon\delta}}\big)$. To do that, we consider the field $\textbf{R}^T_{\varepsilon\delta}\textbf{R}_{\varepsilon\delta}-\textbf{I}_3$ in $R^2_{pq}$ for $(p,q)\in{\mathcal{K}}_\varepsilon^{(2)}$. Using (4.33), we have for $(p,q)\in{\mathcal{K}}_\varepsilon^{(2)}$ and $(z_1,z_2)\in{\mathfrak{I}}_\delta\times {\mathfrak{J}}_{\varepsilon\delta}$

\begin{equation*}{\small \begin{aligned} 2\textbf{E}\big(\textbf{R}_{\varepsilon\delta}(p\varepsilon+z_1,q\varepsilon+z_2)\big)&={\delta^2-4x_1^2\over 4\delta^2}\big(\textbf{R}^T_{{\varepsilon\delta},pq}\textbf{R}_{\delta,p-1q}+\textbf{R}^T_{{\varepsilon\delta},p-1q}\textbf{R}_{{\varepsilon\delta},pq}-2\textbf{I}_3\big)\\ &={\delta^2-4x_1^2\over 4\delta^2}\big(\textbf{R}^T_{{\varepsilon\delta},pq}(\textbf{R}_{{\varepsilon\delta},p-1q}-\textbf{R}_{{\varepsilon\delta},pq})\\ &\quad +\textbf{R}^T_{{\varepsilon\delta},p-1q}(\textbf{R}_{{\varepsilon\delta},pq}-\textbf{R}_{{\varepsilon\delta},p-1q})\big). \end{aligned} }\end{equation*}

Let us set

\begin{equation*}\left.\begin{aligned} \textbf{A}_{\varepsilon\delta}(p\varepsilon+z_1,q\varepsilon+z_2)&=\textbf{R}_{{\varepsilon\delta},pq}-\textbf{R}_{{\varepsilon\delta},p-1q},\\ \textbf{B}_{\varepsilon\delta}(p\varepsilon+z_1,q\varepsilon+z_2)&=\textbf{R}^T_{{\varepsilon\delta},pq}, \end{aligned} \right\}\quad (p,q)\in{\mathcal{K}}_\varepsilon^{(2)}\ \text{and}\ (z_1,z_2)\in{\mathfrak{I}}_\delta\times {\mathfrak{J}}_{\varepsilon\delta}.\end{equation*}

Then, proceeding as in the previous step with the estimates (4.23), (6.1), and (6.10), we obtain

\begin{equation*} \begin{aligned} {1\over \varepsilon\sqrt{{\varepsilon\delta}}}\mathcal{T}_{\varepsilon\delta}^{(2)}(\textbf{A}_{\varepsilon\delta}\textbf{1}_{Y^{(2)}_{\varepsilon\delta}}) & \rightharpoonup \partial_1\textbf{A}\quad && \text{weakly in}\ L^2({\omega}\times Y_2)^{3\times 3},\\ \mathcal{T}_{\varepsilon\delta}^{(2)}(\textbf{B}_{\varepsilon\delta}\textbf{1}_{Y^{(2)}_{\varepsilon\delta}}) & \to \textbf{I}_3\quad && \text{strongly in}\ L^2({\omega}\times Y_2)^{3\times 3}. \end{aligned} \end{equation*}

Finally, we obtain

\begin{align*} &{1\over \varepsilon\sqrt{{\varepsilon\delta}}}\mathcal{T}_{\varepsilon\delta}^{(2)}\big(((\textbf{R}_{\varepsilon\delta})^T\textbf{R}_{\varepsilon\delta}-\textbf{I}_3)\textbf{1}_{Y^{(2)}_{\varepsilon\delta}}\big) \rightharpoonup \big(-\partial_1\textbf{A}+\partial_1\textbf{A}\big){1-4y^2_1\over 4}=0 \nonumber\\ &\quad \text{weakly in}\ L^1({\omega}\times Y_2)^{3\times 3}.\end{align*}

The estimate (6.23) $_1$ and the above convergence lead to (6.21) $_1$ for $\alpha=2$. The case $\alpha=1$ is similar.

Step 4: We prove the convergences (6.22).

From the estimate (6.2) $_3$ and the properties of $\mathcal{T}_{\varepsilon\delta}^{(\alpha)}$, we obtain

\begin{equation*} \|\mathcal{T}_{\varepsilon\delta}^{(\alpha)}\big(\partial_\beta{\mathcal{U}}_{\varepsilon\delta}-(\textbf{R}_{\varepsilon\delta}-\textbf{I}_3)\textbf{e}_\beta\big)\|_{L^2({\omega}\times Y_\alpha)}\leq C\varepsilon\sqrt{{\varepsilon\delta}}. \end{equation*}

As in the previous steps, we only have to consider $\partial_\beta{\mathcal{U}}_{\varepsilon\delta}-(\textbf{R}_{\varepsilon\delta}-\textbf{I}_3)\textbf{e}_\beta$ in the rectangles $R^2_{pq}$ for $(p,q)\in{\mathcal{K}}_\varepsilon^{(2)}$.

We recall from Step 3 of Lemma 4.9, for a.e. $(z_1,z_2)\in{\mathfrak{I}}_\delta\times {\mathfrak{J}}_{\varepsilon\delta}$ and $(p,q)\in {\mathcal{K}}^{(2)}_\varepsilon$

\begin{equation*} \begin{aligned} \big(\partial_2{\mathcal{V}}_{\varepsilon\delta}-\textbf{R}_{\varepsilon\delta}\textbf{e}_2\big)(p\varepsilon+z_1,q\varepsilon+z_2)&=0,\\ \big(\partial_1{\mathcal{V}}_{\varepsilon\delta}-\textbf{R}_{\varepsilon\delta}\textbf{e}_1\big)(p\varepsilon+z_1,q\varepsilon+z_2) &={1\over \delta}\big(\textbf{a}_{{\varepsilon\delta},pq}-\textbf{a}_{{\varepsilon\delta},p-1q}-\varepsilon\textbf{R}_{{\varepsilon\delta},p-1q}\textbf{e}_1\big)\\ &+{z_2\over \delta}(\textbf{R}_{{\varepsilon\delta},pq}-\textbf{R}_{{\varepsilon\delta},p-1q})\textbf{e}_2\\ &-{z_1\over \delta}(\textbf{R}_{{\varepsilon\delta},pq}-\textbf{R}_{{\varepsilon\delta},p-1q})\textbf{e}_1. \end{aligned} \end{equation*}

Proceeding as in the previous steps, the above expression give (6.22) $_1$ for $\alpha=2$. Then, using the estimates (4.11) $_{1,3,5}$ and (6.1), we obtain the convergence (6.22) $_2$ with ${\mathcal{Z}}^{(2)}\in L^2({\omega})^3$. The case $\alpha=1$ is similar, with ${\mathcal{Z}}^{(1)}\in L^2({\omega})^3$. Using the estimates (4.23), (4.29) with (6.1) and the convergences (6.7) $_{1,2}$ give

\begin{equation*}{\mathcal{Z}}^{(1)}=\partial_2{\mathcal{R}}^{(1)},\quad \text{in}\ {\omega},\quad {\mathcal{Z}}^{(2)}=\partial_1{\mathcal{R}}^{(2)},\quad \text{in}\ {\omega},\end{equation*}

which together with (6.8) imply ${\mathcal{Z}}^{(1)}={\mathcal{Z}}^{(2)}=\partial_{12}U$. Since $(\textbf{R}^\diamond_{\varepsilon\delta}-\textbf{I}_3)(-{\delta\over 2},\cdot)\textbf{e}_2$ is independent of $z_1$. Similarly $(\textbf{R}^\diamond_{\varepsilon\delta}-\textbf{I}_3)(\cdot,-{\delta\over 2})\textbf{e}_1$ is independent of $z_2$. This completes the proof.

Proof. As a consequence of the convergences (6.20) and (6.22)–(6.21), we obtain the following weak convergences in $L^2({\omega}\times {\mathcal{Y}}_2)^3$:

\begin{equation*} \begin{aligned} {1\over \varepsilon\sqrt{{\varepsilon\delta}}}\Pi_{\varepsilon\delta}^{(2)}\big(\nabla v_{\varepsilon\delta}-\textbf{R}_{\varepsilon\delta}\big)\textbf{e}_1\big)&\rightharpoonup Z^{(2)}+y_2{\mathcal{Z}}-y_1\partial_1\textbf{A}\textbf{e}_1+y_3\partial_1\textbf{A}\textbf{e}_3+\partial_{y_1}\overline{v}^{(2)},\\ {1\over \varepsilon\sqrt{{\varepsilon\delta}}}\Pi_{\varepsilon\delta}^{(2)}\big(\nabla v_{\varepsilon\delta}-\textbf{R}_{\varepsilon\delta}\big)\textbf{e}_2\big) &\rightharpoonup 0,\\ {1\over \varepsilon\sqrt{{\varepsilon\delta}}}\Pi_{\varepsilon\delta}^{(2)}\big(\nabla v_{\varepsilon\delta}-\textbf{R}_{\varepsilon\delta}\big)\textbf{e}_3\big)&\rightharpoonup \partial_{y_3}\overline{v}^{(2)}. \end{aligned} \end{equation*}

The above convergences and (6.21) $_1$ give the following

\begin{align*} {1\over \varepsilon\sqrt{{\varepsilon\delta}}}\Pi^{(2)}\big(\textbf{E}(v_{\varepsilon\delta})\big) & \rightharpoonup \left(\begin{matrix} Z^{(2)}_1+y_2{\mathcal{Z}}_1 & * & *\\ {1\over 2}(Z^{(2)}_2+y_2{\mathcal{Z}}_2)& 0& *\\ {1\over 2}(Z_3^{(2)}+y_2{\mathcal{Z}}_3)&0&0 \end{matrix}\right)+E^{(2)}({\cal U}_3^{(2)},\overline{v}^{(2)}), \nonumber\\ &\quad \text{weakly in}\ L^1({\omega}\times {\cal Y}_2)^{3\times 3}. \end{align*}

Then, setting as in (6.30) $_1$ gives (6.29) $_2$. Similarly, we show (6.29) $_1$.

Proof. Without loss of generality, we can assume that the sequence $\{v_{\varepsilon\delta}\}_{\varepsilon,\delta}$ satisfies $\textbf{J}_{\varepsilon\delta}(v_{\varepsilon\delta})\leq \textbf{J}_{\varepsilon\delta}(I_d)=0$. Then, due to (4.66)–(4.67), we have

\begin{align*}c_0\textbf{D}(v_{\varepsilon\delta})^2&-\int_{\Omega_{\delta}}f_{\varepsilon\delta}\cdot(v_{\varepsilon\delta}-I_d)\,dx\leq c_0\|(\nabla v_{\varepsilon\delta})^T(\nabla v_{\varepsilon\delta})-\textbf{I}_3\|^2_{L^2(\Omega^S_{\varepsilon\delta})}\\ &-\int_{\Omega_{\delta}}f_{\varepsilon\delta}\cdot(v_{\varepsilon\delta}-I_d)\,dx\leq \textbf{J}_{\varepsilon\delta}(v_{\varepsilon\delta})\leq 0.\end{align*}

Thus, from Lemma 4.14, we obtain

(7.7)\begin{equation} \textbf{D}(v_{\varepsilon\delta})+\|(\nabla v_{\varepsilon\delta})^T\nabla v_{\varepsilon\delta}-\textbf{I}_3\|_{L^2(\Omega^S_{\varepsilon\delta})}\leq C\varepsilon\delta^{3/2}. \end{equation}

The constant is independent of $\varepsilon$ and $\delta$.

Below, we apply the decomposition given in (4.35) for the deformation $v_{\varepsilon\delta}$. As in Section 6, we use the estimate (6.2) and the convergences stated in Lemma 6.1, Theorem 6.4, and Proposition 6.2, at least for a subsequence (still denoted by the same symbol).

Observe that, using the estimate (7.7) $_2$ and the convergence (6.29), and following the arguments of Theorem 6.4, there exist functions ${\mathcal{U}}_{0,3}^{(3-\alpha)}\in H_{(0,l)}^{2}\big((0,L)_{x_\alpha}\big)$ with ${\mathcal{U}}_{0,3}^{(1)}{\mathcal{U}}_{0,3}^{(2)}=0$ a.e. in $(0,L)$, $Z_0^{(\alpha)}\in L^2({\omega})^3$, ${\mathcal{Z}}_0\in L^2({\omega})^3$, and $\overline{v}^{(\alpha)}_0\in L^2({\omega}\times (0,1){y_\alpha};H^1({\mathfrak{I}}^2))^3$ satisfying (6.19), such that

(7.8)\begin{equation} \begin{aligned} {1\over 2\varepsilon\sqrt{{\varepsilon\delta}}}\Pi^{(1)}_{\varepsilon\delta}\big((\nabla v_{\varepsilon\delta})^T(\nabla v_{\varepsilon\delta})-\textbf{I}_3\big) &\rightharpoonup E^{(1)}({\mathcal{U}}_{0,3}^{(1)},\widehat v^{(1)}_0)\quad&&\text{weakly in }\; L^2({\omega}\times {\mathcal{Y}}_1)^{3\times 3},\\ {1\over 2\varepsilon\sqrt{{\varepsilon\delta}}}\Pi^{(2)}_{\varepsilon\delta}\big((\nabla v_{\varepsilon\delta})^T(\nabla v_{\varepsilon\delta})-\textbf{I}_3\big) &\rightharpoonup E^{(2)}({\mathcal{U}}_{0,3}^{(2)},\widehat v^{(2)}_0)\quad&&\text{weakly in }\; L^2({\omega}\times {\mathcal{Y}}_2)^{3\times 3}, \end{aligned} \end{equation}

where $\widehat v^{(\alpha)}_0$ is given by (7.6), and ${\mathcal{U}}_{0,3}^{(2)} + {\mathcal{U}}^{(1)}_{0,3} = {\mathcal{U}}_{0,3} \in {\mathbb{D}}_0$.

Then, using the assumption on the forces (4.60), together with the expression (4.63), and applying the convergences (6.5) $_{1,2}$ and the estimate (6.4) $_1$, we obtain

(7.9)\begin{equation} \begin{aligned} &\lim_{(\varepsilon,\delta)\to(0,0)}{1\over \varepsilon^2\delta^3}\int_{\Omega_\delta}f_{\varepsilon\delta}\cdot (v_{\varepsilon\delta}-I_d)\,dx\\ &\hskip 5mm= \lim_{(\varepsilon,\delta)\to(0,0)}\big({1\over \varepsilon^2\delta^2}\int_{{\omega}_\delta}\hskip -3mmf_{\varepsilon\delta}\cdot({\mathcal{V}}_{\varepsilon\delta}-I_d)\,dx'+{1\over \varepsilon^3\delta}\int_{{\omega}\times {\mathcal{Y}}_\alpha}\hskip -3mm\Pi_{\varepsilon\delta}^{(\alpha)}\big(f_{\varepsilon\delta}\cdot \overline{v}_{\varepsilon\delta}\big)\,dx'dy\big)\\ &\hskip 5mm=\int_{{\omega}} f\cdot{\mathcal{U}}_0\,dx'. \end{aligned} \end{equation}

We claim that

(7.10)\begin{equation} \begin{aligned} \textbf{1}_{({\omega}\setminus{\omega}_{\varepsilon\delta}^J)\times {\mathcal{Y}}_1}\big({1\over \varepsilon\sqrt{{\varepsilon\delta}}}\Pi_{\varepsilon\delta}^{(1)}\big(\textbf{E}(\nabla v_{\varepsilon\delta})\big)\big)\rightharpoonup E^{(1)}({\mathcal{U}}_{0,3}^{(1)},\widehat v_0^{(1)})\quad \text{weakly in}\ L^2({\omega}\times {\mathcal{Y}}_1)^{3\times 3},\\ \textbf{1}_{({\omega}\setminus{\omega}_{\varepsilon\delta}^J)\times {\mathcal{Y}}_2}\big({1\over \varepsilon\sqrt{{\varepsilon\delta}}}\Pi_{\varepsilon\delta}^{(2)}\big(\textbf{E}(\nabla v_{\varepsilon\delta})\big)\big)\rightharpoonup E^{(2)}({\mathcal{U}}_{0,3}^{(2)},\widehat v^{(2)}_0)\quad \text{weakly in}\ L^2({\omega}\times {\mathcal{Y}}_2)^{3\times 3}, \end{aligned} \end{equation}

where $\textbf{1}_{({\omega}\setminus{\omega}_{\varepsilon\delta}^J)\times {\mathcal{Y}}_\alpha}$ is the characteristic function for ${({\omega}\setminus{\omega}_{\varepsilon\delta}^J)\times {\mathcal{Y}}_\alpha}$ where we put

(7.11)\begin{equation} \Omega^J_{\varepsilon\delta}=\Omega_{\varepsilon\delta}^{(1)}\cap\Omega_{\varepsilon\delta}^{(2)}\cap \Omega_\delta\quad \hbox{and}\quad {\omega}^J_{\varepsilon\delta}=\Omega^J_{\varepsilon\delta}\cap {\omega}_\delta. \end{equation}

We have $\displaystyle \textbf{1}_{({\omega}\setminus{\omega}_{\varepsilon\delta}^J)\times {\mathcal{Y}}_\alpha}\in L^\infty({\omega}\times {\mathcal{Y}}_\alpha)$. Then, since $|{\omega}\times {\mathcal{Y}}_\alpha|=L^2$ and $\displaystyle |{\omega}_{\varepsilon\delta}^J\times {\mathcal{Y}}_\alpha|=4\delta^2|{\mathcal{K}}_\varepsilon|^2=\textbf{O}\Big({\delta^2\over \varepsilon^2}\Big)$, so $|{\omega}^J_{\varepsilon\delta}\times {\mathcal{Y}}_\alpha|\to0$. Hence

(7.12)\begin{equation} \textbf{1}_{({\omega}\setminus{\omega}_{\varepsilon\delta}^J)\times {\mathcal{Y}}_\alpha}(x',y)\to \textbf{1}_{{\omega}\times {\mathcal{Y}}_\alpha}(x',y) \quad \hbox{for a.e. } (x',y)\in {\omega}\times {\mathcal{Y}}_\alpha. \end{equation}

Using the estimate (7.7) $_2$ and properties of re-scaling operator (5.2), we have

\begin{align*} &\left\|\Pi_{\varepsilon\delta}^{(\alpha)}\big(\textbf{E}(\nabla v_{\varepsilon\delta})\big)\right\|_{L^2({\omega}\times {\mathcal{Y}}_\alpha)}\leq C\varepsilon\sqrt{{\varepsilon\delta}}\\ &\quad \implies \Big\|\textbf{1}_{({\omega}\setminus{\omega}_{\varepsilon\delta}^J)\times {\mathcal{Y}}_\alpha}{1\over \varepsilon\sqrt{{\varepsilon\delta}}}\Pi_{\varepsilon\delta}^{(\alpha)}\big(\textbf{E}(\nabla v_{\varepsilon\delta})\big)\Big\|_{L^2({\omega}\times {\mathcal{Y}}_\alpha)}\leq C.\end{align*}

Together with (7.12) and the weak convergence (7.8) gives (7.10).

Using the unfolding operator, we transform $\textbf{W}_{\varepsilon\delta}$ (see (3.3) for the definition) to get,

\begin{equation*} \begin{aligned} &{1\over \varepsilon^2\delta^3}\int_{\Omega^S_{\varepsilon\delta}}\textbf{W}_{\varepsilon\delta}(x,\nabla v_{\varepsilon\delta})\,dx \geq \sum_{\alpha=1}^2{1\over \varepsilon^2\delta^3}\int_{\Omega^{(\alpha)}_{\varepsilon\delta}\setminus \Omega^J_{\varepsilon\delta}}\textbf{W}_{\varepsilon\delta}(x,\nabla v_{\varepsilon\delta})\,dx\\ &\geq \sum_{\alpha=1}^2\int_{{\omega}\times {\mathcal{Y}}_\alpha} \left\langle \textbf{1}_{({\omega}\setminus{\omega}_{\varepsilon\delta}^J)\times {\mathcal{Y}}_\alpha}\big({1\over \varepsilon\sqrt{{\varepsilon\delta}}}\Pi_{\varepsilon\delta}^{(\alpha)}\big(\textbf{E}(\nabla v_{\varepsilon\delta})\big)\big)\Pi^{(\alpha)}_{\varepsilon\delta}({\mathbb{A}}_{{\varepsilon\delta}}), \right.\\ &\left.\quad {1\over \varepsilon\sqrt{{\varepsilon\delta}}}\Pi_{\varepsilon\delta}^{(\alpha)}\big(\textbf{E}(\nabla v_{\varepsilon\delta})\big)\right\rangle\,dx'dy. \end{aligned} \end{equation*}

So, the above inequality along with convergences (7.9) and (7.10) together with the weak lower semi-continuity of the function $\textbf{J}_{\varepsilon\delta}$ give

\begin{equation*}\liminf_{(\varepsilon,\delta)\to(0,0)}{\textbf{J}_{\varepsilon\delta}(v_{\varepsilon\delta})\over \varepsilon^2\delta^3}\geq\textbf{J}\big({\mathcal{U}}_{0,3},\widehat v_0^{(1)},\widehat v_0^{(2)}\big).\end{equation*}

Finally, using the equation (7.5), we obtain

\begin{equation*}\liminf_{(\varepsilon,\delta)\to(0,0)}{\textbf{m}_{\varepsilon\delta}\over \varepsilon^2\delta^3}=\liminf_{(\varepsilon,\delta)\to(0,0)} {\textbf{J}_{\varepsilon\delta}(v_{\varepsilon\delta})\over \varepsilon^2\delta^3}\geq\textbf{J}\big({\mathcal{U}}_{0,3},\widehat v_0^{(1)},\widehat v_0^{(2)}\big).\end{equation*}

This completes the proof.

Proof. See Appendix A.

Proof. See Appendix A.

Proof. Observe that

\begin{equation*}\textbf{m}\leq \textbf{J}(0,0,0)=0,\quad \text{since}\ (0,0,0)\in {\mathbb{D}},\end{equation*}

which imply $\textbf{m}\in [-\infty,0]$.

Step 1. We show that $\textbf{m}\in(-\infty,0]$. Let $({\mathcal{U}}_{3},\widehat v^{(1)},\widehat v^{(2)})\in{\mathbb{D}}$ be such that

(7.15)\begin{equation} \textbf{J}({\mathcal{U}}_3,\widehat v^{(1)},\widehat v^{(2)})\leq \textbf{J}(0,0,0)=0. \end{equation}

Since, $\widehat v^{(\alpha)}\in L^2\big({\omega}; \textbf{L}^2\big((0,1)_{y_\alpha};H^1\big({\mathfrak{I}}^2\big)\big)^3 \big)$, there exist $\overline{v}^{(\alpha)}$, $Z^{(\alpha)}$, and ${\mathcal{Z}}$ such that $\overline{v}^{(\alpha)}$ satisfies (6.19) and $\widehat v^{(\alpha)}$ is given by (6.30). We have ${\mathcal{U}}_3^{(3-\alpha)}\in H_{(0,l)}^{2}\big((0,L)_{x_\alpha}\big)$ such that ${\mathcal{U}}_3^{(1)}{\mathcal{U}}_3^{(2)}=0$ a.e. in $(0,L)$. So, using the Coercivity (7.4) of $\textbf{Q}^{(\alpha)}$ and (7.14), there exists $0 \lt C\in{\mathbb{R}}$ such that

(7.16)\begin{equation} \begin{aligned} \|d^2_{22}{\mathcal{U}}_3^{(1)}\|^2_{L^2({\omega})}+\|Z^{(1)}\|^2_{L^2({\omega})}&+\|{\mathcal{Z}}\|^2_{L^2({\omega})}+\|\widehat v^{(1)}\|^2_{L^2({\omega}\times (0,1)_{y_1};H^1({\mathfrak{I}}^2))}\\ & \leq C\|E^{(1)}({\mathcal{U}}_3^{(1)},\widehat v^{(1)})\|^2_{L^2({\omega}\times {\mathcal{Y}}_1)}\\ &\leq C\int_{{\omega}\times {\mathcal{Y}}_1}\textbf{Q}^{(1)}(y,E^{(1)}({\mathcal{U}}_3^{(1)},\widehat v^{(1)}))\,dydx',\\ \|d^2_{11}{\mathcal{U}}_3^{(2)}\|^2_{L^2({\omega})}+\|Z^{(2)}\|^2_{L^2({\omega})}&+\|{\mathcal{Z}}\|^2_{L^2({\omega})}+\|\widehat v^{(2)}\|^2_{L^2({\omega}\times (0,1)_{y_2};H^1({\mathfrak{I}}^2))}\\ &\leq C\|E^{(2)}({\mathcal{U}}_3^{(2)},\widehat v^{(2)})\|^2_{L^2({\omega}\times {\mathcal{Y}}_2)}\\ &\leq C\int_{{\omega}\times {\mathcal{Y}}_2}\textbf{Q}^{(2)}(y,E^{(2)}({\mathcal{U}}_3^{(2)},\widehat v^{(2)}))\,dydx'. \end{aligned} \end{equation}

The above inequalities along with the boundary condition satisfied by ${\mathcal{U}}_3^{(\alpha)}$, and using Poincaré inequality we obtain

(7.17)\begin{align} \|{\mathcal{U}}_3^{(2)}\|^2_{H^2(0,L)}+\|{\mathcal{U}}_3^{(1)}\|^2_{H^2(0,L)}&\leq C\big(\|d^2_{11}{\mathcal{U}}_3^{(2)}\|^2_{L^2(0,L)}+\|d^2_{22}{\mathcal{U}}_3^{(1)}\|^2_{L^2(0,L)}) \nonumber\\ &\leq C\textbf{W}({\mathcal{U}}_3,\widehat v^{(1)},\widehat v^{(2)}). \end{align}

Since, ${\mathcal{U}}_{\alpha}\in H_{(0,l)}^{1}\big((0,L)_{x_\alpha}\big)$ and $d_\alpha{\mathcal{U}}_{\alpha}=-{1\over 2}|d_\alpha{\mathcal{U}}_3^{(3-\alpha)}|^2$ a.e. in $(0,L)$, we get

\begin{equation*}\|d_\alpha{\mathcal{U}}_{\alpha}\|^2_{L^2(0,L)}\leq C\|d_\alpha{\mathcal{U}}_3^{(3-\alpha)}\|^4_{L^4(0,L)}.\end{equation*}

Using Poincaré inequality, (7.17) and the fact that $H^2(0,L)$ is continuously embedded in $W^{1,4}(0,L)$, we get

(7.18)\begin{equation} \sum_{{\alpha}=1}^2\|{\mathcal{U}}_{\alpha}\|^2_{L^2(0,L)}\leq C\sum_{{\alpha}=1}^2\|{\mathcal{U}}_3^{(3-\alpha)}\|^4_{H^2(0,L)}\leq C\left[\textbf{W}({\mathcal{U}}_3,\widehat v^{(1)},\widehat v^{(2)})\right]^2. \end{equation}

From (7.15) along with (7.17)–(7.18), we get

\begin{equation*} \begin{aligned} \textbf{W}\big({\mathcal{U}}_{3},\widehat v^{(1)},\widehat v^{(2)}\big) &\leq \left|\int_{\omega} f\cdot {\mathcal{U}}\,dx'\right|\\ & \leq L \sum_{\alpha=1}^2\big(\|f_\alpha\|_{L^2({\omega})}\|{\mathcal{U}}_\alpha\|_{L^2(0,L)}+\|f_3\|_{L^2({\omega})}\|{\mathcal{U}}_3^{(\alpha)}\|_{L^2(0,L)}\big)\\ &\leq C\left(\sum_{{\alpha}=1}^2\|f_{\alpha}\|_{L^2({\omega})}\right)\left[\textbf{W}({\mathcal{U}}_3,\widehat v^{(1)},\widehat v^{(2)})\right]\\ &\quad +C\|f_3\|_{L^2({\omega})}\left[\sqrt{\textbf{W}({\mathcal{U}}_3,\widehat v^{(1)},\widehat v^{(2)})}\right]. \end{aligned} \end{equation*}

So, from the above inequality along with the assumption of forces (4.61) ( $C\|f_{\alpha}\|_{L^2({\omega})}\leq {1\over 2}$), we get

\begin{equation*} \textbf{W}\big({\mathcal{U}}_{3},\widehat v^{(1)},\widehat v^{(2)}\big) \leq C\quad\text{and}\quad \left|\int_{\omega} f\cdot {\mathcal{U}}\,dx'\right|\leq C. \end{equation*}

Finally, if $({\mathcal{U}}_{3},\widehat v^{(1)},\widehat v^{(2)})\in{\mathbb{D}}$ is such that it satisfies (7.15), then using the above inequality along with (7.16)–(7.18) we obtain for $\alpha=1,2$

(7.19)\begin{equation} \|{\mathcal{U}}_\alpha\|^2_{L^2(0,L)}+\|{\mathcal{U}}_3^{(\alpha)}\|^2_{H^2(0,L)}+\|Z^{(\alpha)}\|^2_{L^2({\omega})}+\|{\mathcal{Z}}\|^2_{L^2({\omega})}+\|\overline{v}^{(\alpha)}\|^2_{L^2({\omega}\times (0,1)_{y_\alpha};H^1({\mathfrak{I}}^2))}\leq C, \end{equation}

and

\begin{equation*}-C\leq -\int_{\omega} f\cdot {\mathcal{U}},\,dx'\leq \textbf{W}({\mathcal{U}}_3,\widehat v^{(1)},\widehat v^{(2)})-\int_{\omega} f\cdot{\mathcal{U}}\,dx'\leq 0,\end{equation*}

which implies $\textbf{J}({\mathcal{U}}_3,\widehat v^{(1)},\widehat v^{(2)})\in [-C,0]$. As a consequence, we obtain $\textbf{m}\in(-\infty,0]$.

Step 2: We prove the main statement of the theorem.

Since $\textbf{m}\in (-\infty,0]$, there exists a minimizing sequence $\{({\mathcal{U}}_{n,3},\widehat v^{(1)}_{n},\widehat v^{(2)}_n)\}_n\subset {\mathbb{D}}$ satisfying (7.15) and

\begin{equation*} \textbf{m}=\inf_{({\mathcal{U}}_{3},\widehat v^{(1)},\widehat v^{(2)})\in{\mathbb{D}}}\textbf{J}({\mathcal{U}}_3,\widehat v^{(1)},\widehat v^{(2)})=\liminf_{n\to+\infty}\textbf{J}({\mathcal{U}}_{n,3},\widehat v^{(1)}_n,\widehat v^{(2)}_n), \end{equation*}

where

\begin{align*} &{\mathcal{U}}_n(x')={\mathcal{U}}_{n,1}(x_1)\textbf{e}_1+{\mathcal{U}}_{n,2}(x_2)\textbf{e}_2+({\mathcal{U}}_{n,3}^{(2)}(x_1)+{\mathcal{U}}_{n,3}^{(1)}(x_2))\textbf{e}_3,\\ & {\mathcal{U}}_{n,3}^{(1)} {\mathcal{U}}_{n,3}^{(2)}=0,\quad\text{a.e. in}\ (0,L),\end{align*}

with ${\mathcal{U}}_{n,\alpha}\in H_{(0,l)}^{1}\big((0,L)_{x_\alpha}\big)$ given by

\begin{equation*}d_\alpha{\mathcal{U}}_{n,\alpha}=-{1\over 2}|d_\alpha{\mathcal{U}}_{n,3}^{(3-\alpha)}|^2\,\text{a.e. in}\ (0,L),\end{equation*}

and

\begin{equation*}\widehat v^{(1)}_n=\overline{v}^{(1)}_n+y_2 Z_n^{(1)}+y_1y_2{\mathcal{Z}}_n,\quad\widehat v^{(2)}_n=\overline{v}^{(2)}_n+y_1Z^{(2)}_n+y_1y_2{\mathcal{Z}}_n.\end{equation*}

We have at least for a subsequence (denoted by the same subscript)

\begin{equation*} \begin{aligned} {\mathcal{U}}_{n,3}^{(3-\alpha)}&\rightharpoonup {\mathcal{W}}_3^{(3-\alpha)}\quad &&\text{weakly in}\ H_{(0,l)}^{2}\big((0,L)_{x_\alpha}\big),\\ {\mathcal{U}}_{n,\alpha} &\rightharpoonup {\mathcal{W}}_{\alpha}\quad &&\text{weakly in}\ H_{(0,l)}^{1}\big((0,L)_{x_\alpha}\big),\\ Z^{(\alpha)}_n&\rightharpoonup Z_0^{(\alpha)}\quad &&\text{weakly in}\ L^2({\omega})^3,\quad {\mathcal{Z}}_n\rightharpoonup {\mathcal{Z}}_0\quad \text{weakly in}\ L^2({\omega})^3,\\ \overline{v}^{(\alpha)}_n &\rightharpoonup \overline{w}^{(\alpha)}\quad &&\text{weakly in}\ L^2({\omega}\times (0,1)_{y_\alpha};H^1({\mathfrak{I}}^2))^3. \end{aligned} \end{equation*}

Then, since $\textbf{W}$ is weak lower semi-continuous, we obtain

\begin{equation*} \textbf{W}({\mathcal{W}}_3,\widehat w^{(1)},\widehat w^{(2)})-\int_{{\omega}} f\cdot{\mathcal{W}}\,dx' \leq\liminf_{n\to+\infty}\big(\textbf{W}\big({\mathcal{U}}_{n,3},\widehat v_n^{(1)},\widehat v_n^{(2)}\big)-\int_{{\omega}} f\cdot{\mathcal{U}}_n\,dx'\big)=\textbf{m}. \end{equation*}

So, we have the existence of minimizer $({\mathcal{W}}_{3},\widehat w^{(1)},\widehat w^{(2)})\in{\mathbb{D}}$ and the fact that $\textbf{m}$ is a minimum by sequential criteria. This completes the proof.

Proof. Since $\widehat \chi^{(1)}\in \textbf{L}^2\big((0,1)_{y_1};H^1\big({\mathfrak{I}}^2\big)\big)^3$, there exist $\overline{\chi}^{(1)}\in L^2\big((0,1)_{y_1};H^1\big({\mathfrak{I}}^2\big)\big)^3$ satisfying (7.2) $_1$ and $\textbf{a},\textbf{b}\in{\mathbb{R}}^3$ such that

\begin{equation*}\widehat \chi^{(1)}(y)=\overline{\chi}^{(1)}(y)+\textbf{a} y_2+\textbf{b} y_1 y_2,\quad \forall\,y\in {\mathcal{Y}}_1.\end{equation*}

Let us set (see (A.8))

\begin{equation*}\widetilde {\chi}^{(1)}(y)= \overline{\chi}^{(1)}(y)+\textbf{a} y_2+\textbf{b} y_1 y_2-y_2 y_3\textbf{e}_3,\quad \forall\,y\in {\mathcal{Y}}_1,\end{equation*}

then, using (7.4) and (7.13), we get

\begin{equation*}\begin{aligned} {\mathcal{C}}^{(1)}_{hom}\geq c_0\|E^{(1)}(0,\widehat \chi^{(1)})-\textbf{M}^{(1)}\|^2_{L^2({\mathcal{Y}}_1)}= c_0\|E^{(1)}(0,\widetilde \chi^{(1)})\|^2_{L^2({\mathcal{Y}}_1)}\\ \geq c_0\big(1+|\textbf{a}|^2+|\textbf{b}|^2+\|\overline{\chi}^{(1)}\|^2_{L^2((0,1)_{y_1};H^1({\mathfrak{I}}^2)}\big)=C_1 \gt 0. \end{aligned} \end{equation*}

Similarly, we get $\mathcal{C}^{(2)}_{hom}\geq C_2 \gt 0$. Taking $C_0=\min\{C_1,C_2\}$ gives $\mathcal{C}^{(\alpha)}_{hom}\geq C_0 \gt 0$.

First observe that, proceeding as in the proof of Proposition 7.4, there exists a minimizer of $\textbf{J}^{(\alpha)}$ in ${\mathbb{D}}^{(\alpha)}$ and $\textbf{m}^{(\alpha)}$ is in fact a minimum by the sequential criterion.

From Proposition 7.4, we have (since ${\mathcal{W}}_3\in {\mathbb{D}}_0$, we have ${\mathcal{W}}_3={\mathcal{W}}_3^{(2)}+{\mathcal{W}}_3^{(1)}$ and ${\mathcal{W}}_3^{(1)}{\mathcal{W}}_3^{(2)}\equiv0$ a.e. in $(0,L)$ with ${\mathcal{W}}_3^{(3-\alpha)}\in H_{(0,l)}^{2}((0,L)_{x_\alpha})$)

\begin{equation*}\textbf{J}({\mathcal{W}}_3,\widehat w^{(1)},\widehat w^{(2)})= \textbf{m} = \inf_{({\mathcal{U}}_{3},\widehat v^{(1)},\widehat v^{(2)})\in {\mathbb{D}}}\textbf{J}({\mathcal{U}}_3,\widehat v^{(1)},\widehat v^{(2)}).\end{equation*}

Without loss of generality, let us assume ${\mathcal{W}}_3^{(1)}\equiv0$, which using (7.21) $_1$ gives ${\mathcal{Z}}=0$, a.e. in ${\omega}$. Then, using (7.21), we have

\begin{align*}\textbf{m}&=\textbf{J}({\mathcal{W}}^{(2)}_3,0,\widehat w^{(2)})=\textbf{W}\big({\mathcal{W}}^{(2)}_3,0,\widehat w^{(2)}\big)\\ &\quad -\int_{{\omega}} f\cdot\big({\mathcal{W}}_1\textbf{e}_1+{\mathcal{W}}^{(2)}_3\textbf{e}_3\big)\,dx'=\textbf{J}^{(2)}({\mathcal{W}}_3^{(2)},\widehat w^{(2)}) ,\end{align*}

where $d_1{\mathcal{W}}_1=-{1\over 2}|d_1{\mathcal{W}}_3^{(2)}|^2$ a.e. in $(0,L)$. So, we have $\textbf{m}^{(2)}\leq \textbf{m}$, since $({\mathcal{W}}_3^{(2)},\widehat w^{(2)})\in{\mathbb{D}}^{(2)}$.

On the other hand, since $({\mathcal{W}}_3^{(2)},0,\widehat w^{(2)})\in {\mathbb{D}}$, we have $\textbf{m}\leq \textbf{m}^{(2)}$. Finally, we have $\textbf{m}=\textbf{m}^{(2)}$.

Similarly, if we assume ${\mathcal{W}}_3^{(2)}\equiv0$, which using (7.21) $_2$ gives ${\mathcal{Z}}=0$ a.e. in ${\omega}$. Then analogously we obtain

\begin{align*}\textbf{m}&=\textbf{J}({\mathcal{W}}_3^{(1)},\widehat w^{(1)},0)=\textbf{W}\big({\mathcal{W}}^{(1)}_3,\widehat w^{(1)},0\big)-\int_{{\omega}} f\cdot\big({\mathcal{W}}_2\textbf{e}_2+{\mathcal{W}}^{(1)}_3\textbf{e}_3\big)\,dx'\\ &=\textbf{m}^{(1)}=\textbf{J}^{(1)}({\mathcal{W}}_3^{(1)},\widehat w^{(1)}).\end{align*}

Consequently,

\begin{equation*}\textbf{m}=\textbf{m}^{(1)} \quad\text{or}\quad \textbf{m}=\textbf{m}^{(2)}.\end{equation*}

Using the homogenized coefficients (7.23), we have

\begin{equation*}\begin{aligned} \textbf{m}^{(1)}&=\textbf{J}^{(1)}({\mathcal{W}}_3^{(1)},\widehat w^{(1)})=\int_{(0,L)}{\mathcal{C}}^{(1)}_{hom}\,[d^2_{22}{\mathcal{W}}_3^{(1)}]^2\,dx_2\\ &\quad -\int_{{\omega}} f\cdot\big({\mathcal{W}}_2\textbf{e}_2+{\mathcal{W}}^{(1)}_3\textbf{e}_3\big)\,dx'=\textbf{J}^{(1)}_{hom}({\mathcal{W}}_3^{(1)}),\\ \textbf{m}^{(2)}&=\textbf{J}^{(2)}({\mathcal{W}}_3^{(2)},\widehat w^{(2)})=\int_{(0,L)}{\mathcal{C}}^{(2)}_{hom}\,[d^2_{11}{\mathcal{W}}_3^{(2)}]^2\,dx_1\\ &\quad -\int_{{\omega}} f\cdot\big({\mathcal{W}}_1\textbf{e}_1+{\mathcal{W}}^{(1)}_3\textbf{e}_3\big)\,dx'=\textbf{J}^{(2)}_{hom}({\mathcal{W}}_3^{(2)}), \end{aligned} \end{equation*}

so, we have

\begin{equation*}\textbf{m}=\min\{\textbf{m}^{(1)},\textbf{m}^{(2)}\}.\end{equation*}

This completes the proof.

Proof. Below, we assume that $\textbf{m}=\textbf{m}^{(2)}$, and we show that

\begin{equation*}\limsup_{(\varepsilon,\delta)\to(0,0)}{\textbf{m}_{\varepsilon\delta}\over \varepsilon^2\delta^3}\leq \textbf{m}^{(2)}.\end{equation*}

The proof in the case $\textbf{m}= \textbf{m}^{(1)}$ is similar. In Proposition 7.5, we obtained that $\textbf{m}^{(2)}$ is the minimum of the functional $\textbf{J}^{(2)}$ over the space ${\mathbb{D}}^{(2)}\subset {\mathbb{D}}$.

In the following, for the sake of simplicity, we omit the superscript $\phantom{A}^{(2)}$ in the test functions but use the superscript for the unfolding operators and spaces.

That is why we start with $({\mathcal{W}}_3,\widehat w)$ an element in ${\mathbb{D}}^{(2)}\cap\big( \mathcal{C}^2([0,L]_{x_1})\times \mathcal{C}^1([0,L]_{x_1}\times \overline{{\mathcal{Y}}_2})^3\big)$, $\widehat w$ with compact support with respect to $x_1$.

We define ${\mathcal{W}}_1\in H_{(0,l)}^{1}\big((0,L)_{x_1}\big)\cap \mathcal{C}^1([0,L])$ as

\begin{equation*}d_1{\mathcal{W}}_1=-{1\over 2}\big|d_1{\mathcal{W}}_3\big|^2,\quad\text{in}\ (0,L),\end{equation*}

and

\begin{align*} &{\mathcal{W}}(x')={\mathcal{W}}_{1}(x_1)\textbf{e}_1+{\mathcal{W}}_{3}(x_1)\textbf{e}_3,\\ &\textbf{A}(x_1)=\left(\begin{matrix} 0&0&-d_1{\mathcal{W}}_3(x_1)\\ 0&0&0\\ d_1{\mathcal{W}}_3(x_1)&0&0 \end{matrix}\right),\quad\text{for}\ x'\in{\omega}.\end{align*}

We also have

(7.25)\begin{equation} \widehat w=\overline{w}+y_1 Z\in {\mathfrak{W}}((0,L)_{x_1}\times {\mathcal{Y}}_2) \cap \mathcal{C}^1([0,L]_{x_1}\times \overline{{\mathcal{Y}}_2})^3. \end{equation}

Due to (7.21), we can take $Z=0$ a.e. in $(0,l)$.

Step 1. In this step, to any function $\psi\in H^1\big((0,L)_{x_1})$, we associate $\displaystyle \psi^{\diamond}_{\varepsilon\delta}\in H^1\Big(\Big(-{\delta\over 2},L+{\delta\over 2}\Big)_{x_1}\Big)$.

First we extend $\psi$ by setting (the extension of $\psi$ is still denoted $\psi$)

\begin{equation*} \psi(x_1)=\psi(0)\qquad \forall x_1\in \Big[-{\delta\over 2},0\Big],\qquad \psi(x_1)=\psi(L)\qquad \forall x_1\in \Big[L,L+{\delta\over 2}\Big]. \end{equation*}

Then, we define the function $\psi^{\diamond}_{\varepsilon\delta}$ by

(7.26)\begin{equation} \psi^{\diamond}_{\varepsilon\delta}(x_1)=\left\{\begin{aligned} &\psi\Big(p\varepsilon+{\varepsilon\over \delta}z_1\Big),\quad&& x_1=p\varepsilon+z_1,\quad z_1\in{\mathfrak{I}}_\delta,\quad p\in\{0,\ldots,N_\varepsilon\},\\ &\psi\Big(p\varepsilon+{\varepsilon\over 2}\Big) ,\quad && x_1\in(p\varepsilon+{\delta\over 2},p\varepsilon+\varepsilon-{\delta\over 2}),\quad p\in\{0,\ldots,N_\varepsilon-1\}. \end{aligned} \right. \end{equation}

Note that

(7.27)\begin{align} &d_1\psi^{\diamond}_{\varepsilon\delta}(x_1) \nonumber\\ &=\left\{\begin{aligned} &{\varepsilon\over \delta}d_1\psi\Big(p\varepsilon+{\varepsilon\over \delta}z_1\Big),\quad&& x_1=p\varepsilon+z_1,\quad z_1\in{\mathfrak{I}}_\delta,\quad p\in\{0,\ldots,N_\varepsilon\},\\ &0 ,\quad && x_1\in(p\varepsilon+{\delta\over 2},p\varepsilon+\varepsilon-{\delta\over 2}),\quad p\in\{0,\ldots,N_\varepsilon-1\}. \end{aligned} \right. \end{align}

Step 2. In this step, we construct the sequence of admissible deformations $\{w_{{\varepsilon\delta}}\}_{\varepsilon,\delta}\subset \textbf{V}_{\varepsilon\delta}$.

The sequence of test deformations $w_{\varepsilon\delta}$ is

\begin{equation*}w_{{\varepsilon\delta}}=I_d+U_{e,{\varepsilon\delta}}+\overline{w}_{\varepsilon\delta}\qquad \hbox{in}\;\; \Omega_\delta\end{equation*}

where

\begin{align*} &\overline{w}_{{\varepsilon\delta}}(p\varepsilon+\delta y_1,q\varepsilon+\varepsilon y_2,\delta y_3)= ({\varepsilon\delta})^{3/2}\overline{w}(p\varepsilon,q\varepsilon, y)\\ &\quad \text{for a.e.}\ y\in {\mathcal{Y}}_2\ \text{and for all}\ (p,q)\in {\mathcal{K}}_\varepsilon \end{align*}

and $\overline{w}_{{\varepsilon\delta}}=0$ a.e. in $\Omega_\delta\setminus{\Omega^{(2)}_{\varepsilon\delta}}$.

The field $U_{e,{\varepsilon\delta}}$ is defined as follows: In $\Omega_\delta$ by

(7.28)\begin{equation} \begin{aligned} U_{e,{\varepsilon\delta}}(x)={\mathcal{W}}_{{\varepsilon\delta}}(x_1)+\int_{-{\delta\over 2}}^{x_1}(\textbf{R}^{\diamond}_{{\varepsilon\delta}}(t)-\textbf{I}_3)\textbf{e}_1\,dt +(\textbf{R}^{\diamond}_{{\varepsilon\delta}}(x_1)-\textbf{I}_3)x_3\textbf{e}_3, \end{aligned} \end{equation}

where the field of matrices $\textbf{R}_{\varepsilon\delta}^{\diamond} \in \displaystyle W^{1,\infty}\Big(\Big(-{\delta\over 2},L+{\delta\over 2}\Big)_{x_1}; \mathrm{SO}(3)\Big)$ is constructed using Step 1 by transforming $\textbf{R}_{\varepsilon\delta}\in \mathcal{C}^1([0,L]_{x_1};\mathrm{SO}(3))$, defined by

\begin{equation*}\textbf{R}_{{\varepsilon\delta}}(x_1)= \exp\big(\sqrt{{\varepsilon\delta}}\textbf{A}(x_1)\big),\qquad x_1\in(0,L).\end{equation*}

Observe that $(\textbf{R}_{\varepsilon\delta}-\textbf{I}_3)\textbf{e}_2=0$ in $\left(-{\delta\over 2},L+{\delta\over 2}\right)$. The field $\displaystyle {\mathcal{W}}_{\varepsilon\delta}\in W^{1,\infty}\Big(\Big(-{\delta\over 2},L+{\delta\over 2}\Big)_{x_1}\Big)^3$ is defined by

(7.29)\begin{equation} {\mathcal{W}}_{\varepsilon\delta}={\varepsilon\delta} W_{\varepsilon\delta}+\delta\sqrt{{\varepsilon\delta}}\textbf{Z}_{\varepsilon\delta}^{\diamond}\qquad \hbox{a.e. in }\; (-{\delta\over2},L+{\delta\over2}) \end{equation}

where $\textbf{Z}^{\diamond}_{\varepsilon\delta}$ is constructed as in Step 1 by transforming

(7.30)\begin{equation} \textbf{Z}(x_1)=\int_0^{x_1}Z(t)\,dt,\quad x_1\in[0,L], \end{equation}

$ W_{\varepsilon\delta}$ is given by

(7.31)\begin{equation} \begin{aligned} {W}_{\varepsilon\delta} (x_1)&={4(x_1-p\varepsilon)^2-\delta^2\over 8\delta^2}d_1\textbf{R}_{\varepsilon\delta}^{\diamond}(p\varepsilon)\textbf{e}_1\qquad x_1\in p\varepsilon+{\mathfrak{I}}_\delta,\qquad p\in \big\{0,\ldots,N_\varepsilon\big\},\\ {W}_{\varepsilon\delta} (x_1)&=0\hskip 30mm x_1\in \Big(p\varepsilon+{\delta\over 2},p\varepsilon+\varepsilon-{\delta\over 2}\Big)\qquad p\in \big\{0,\ldots,N_\varepsilon-1\big\}. \end{aligned} \end{equation}

So, by construction, we have $w_{{\varepsilon\delta}}\in H^1(\Omega_\delta)^3$ and $w_{{\varepsilon\delta}}=I_d$ a.e. on $\Gamma_\delta$ due to the boundary condition satisfied by ${\mathcal{W}}_3$ since $\textbf{R}_{\varepsilon\delta}^{\diamond}=\textbf{I}_3$ in $(-\delta/2,l)$ (due the boundary condition satisfied by $\textbf{A}$) which in turn give $U_{e,{\varepsilon\delta}}=0$ on $\Gamma_\delta$.

Due to (7.28), we have the following:

\begin{equation*} \nabla U_{e,{\varepsilon\delta}}-(\textbf{R}_{\varepsilon\delta}^{\diamond}-\textbf{I}_3)=0, \quad\text{a.e. in } \Omega^H_{pq},\quad (p,q)\in{\mathcal{K}}_\varepsilon^* \end{equation*}

since (see (7.29)–(7.30) and (7.27))

\begin{equation*} \begin{aligned} d_1{\mathcal{W}}_{\varepsilon\delta}(x_1)&=0,\quad x_1\in \Big(p\varepsilon+{\delta\over 2},p\varepsilon+\varepsilon-{\delta\over 2}\Big),\, p\in \big\{0,\ldots,N_\varepsilon-1\big\},\\ d_1{\mathcal{W}}_{\varepsilon\delta}(x_1)&={\varepsilon\delta} d_1 W_{\varepsilon\delta}+\delta\sqrt{{\varepsilon\delta}}\,d_1\textbf{Z}^{\diamond}_{\varepsilon\delta}, \qquad x_1\in p\varepsilon+{\mathfrak{I}}_\delta,\quad p\in \big\{0,\ldots,N_\varepsilon\big\}. \end{aligned} \end{equation*}

So, we have $\nabla w_{\varepsilon\delta}\in \mathrm{SO}(3)$ a.e. in $\Omega^H_{\varepsilon\delta}$. This implies $w_{\varepsilon\delta}\in \textbf{V}_{\varepsilon\delta}$.

Step 3. In this step, we give convergence of the fields.

Observe that

(7.32)\begin{equation} {1\over ({\varepsilon\delta})^{3/2}}\Pi_{\varepsilon\delta}(\overline{w}_{\varepsilon\delta}) \to \overline{w}\quad \text{strongly in}\ L^2\big({\omega}\times (0,1)_{y_2};H^1({\mathfrak{I}}^2)\big)^3. \end{equation}

Recalling $\textbf{A}$ belongs to ${\mathcal{C}}^1\big([0,L]_{x_1}\big)^{3\times 3}$ and is uniformly bounded in this space. Since ${\varepsilon\delta}\leq 1$, the classical properties of the exponential of a matrix yield

(7.33)\begin{equation} \begin{aligned} &\big\|\textbf{R}_{\varepsilon\delta}-\textbf{I}_3\big\|_{L^\infty(0,L)}\leq \sqrt{{\varepsilon\delta}}\|\textbf{A}\|_{L^\infty(0,L)}\exp\big(\|\textbf{A}\|_{L^\infty(0,L)}\big)\leq C\sqrt{{\varepsilon\delta}},\\ &\big\|\textbf{R}_{\varepsilon\delta}-\textbf{I}_3-\sqrt{{\varepsilon\delta}}\textbf{A}\big\|_{L^\infty(0,L)}\leq {\varepsilon\delta}\|\textbf{A}\|^2_{L^\infty(0,L)}\exp\big(\|\textbf{A}\|_{L^\infty(0,L)}\big)\leq C {\varepsilon\delta},\\ & \Big\|\textbf{R}_{\varepsilon\delta}-\textbf{I}_3-\sqrt{{\varepsilon\delta}}\textbf{A}-{{\varepsilon\delta}\over 2} \textbf{A}^2\Big\|_{L^\infty(0,L)}\leq ({\varepsilon\delta})^{3/2}\|\textbf{A}\|^3_{L^\infty(0,L)}\exp\big(\|\textbf{A}\|_{L^\infty(0,L)}\big)\\ &\leq C ({\varepsilon\delta})^{3/2},\\ &\big\|d_1\textbf{R}_{\varepsilon\delta}-\sqrt{{\varepsilon\delta}}\,d_1\textbf{A}\big\|_{L^\infty(0,L)}\leq {\varepsilon\delta}\|d_1\textbf{A}\|_{L^\infty(0,L)}\|\textbf{A}\|_{L^\infty(0,L)} \exp\big(\|\textbf{A}\|_{L^\infty(0,L)}\big)\leq C{\varepsilon\delta}. \end{aligned} \end{equation}

The above estimates along with the fact that (recall that $\displaystyle d_1{\mathcal{W}}_1=-{1\over 2}\big|d_1{\mathcal{W}}_3\big|^2$)

\begin{equation*} \begin{aligned} & \textbf{A}\textbf{e}_1\cdot\textbf{e}_1=0,\qquad \textbf{A}\textbf{e}_1\cdot\textbf{e}_3=d_1{\mathcal{W}}_3,\qquad \textbf{A}^2\textbf{e}_1\cdot\textbf{e}_1=2d_1{\mathcal{W}}_1 \end{aligned} \end{equation*}

give the following convergences:

(7.34)\begin{equation} \begin{aligned} \textbf{R}_{\varepsilon\delta} & \to \textbf{I}_3\quad &&\text{strongly in}\ H^1(0,L)^{3\times 3},\\ {1\over \sqrt{{\varepsilon\delta}}}(\textbf{R}_{\varepsilon\delta}-\textbf{I}_3) & \to \textbf{A}\quad &&\text{strongly in}\ H^1(0,L)^{3\times 3},\\ {1\over \sqrt{{\varepsilon\delta}}}(\textbf{R}_{\varepsilon\delta}-\textbf{I}_3)\textbf{e}_1\cdot\textbf{e}_3&\to d_1 {\mathcal{W}}_3\quad &&\text{strongly in}\ L^2(0,L),\\ {1\over {\varepsilon\delta}}(\textbf{R}_{\varepsilon\delta}-\textbf{I}_3)\textbf{e}_1\cdot\textbf{e}_1&\to d_1 {\mathcal{W}}_1\quad &&\text{strongly in}\ L^2(0,L). \end{aligned} \end{equation}

Due to the definitions of $\mathcal{T}_{\varepsilon\delta}^{(2)}$ (see (5.4)) and (7.26), we have

\begin{equation*} \begin{aligned} \mathcal{T}_{\varepsilon\delta}^{(2)}(\textbf{R}_{\varepsilon\delta}^{\diamond})(x_1,x_2,y_1,y_2) &=\textbf{R}_{\varepsilon\delta}\big(p\varepsilon+\delta y_1\big)\quad (x_1, y_1)\in [p\varepsilon,p\varepsilon+\varepsilon)\times {\mathfrak{I}},\\ & p\in\{0,\ldots,N_\varepsilon\}, \quad (x_2,y_2)\in (0,L)\times (0,1). \end{aligned} \end{equation*}

This leads to using (7.27) and (7.34) $_{1,2}$

(7.35)\begin{equation} \begin{aligned} \textbf{R}^{\diamond}_{\varepsilon\delta}&\to \textbf{I}_3\quad&&\text{strongly in } L^2(0,L)^{3\times 3},\\ {1\over \sqrt{\varepsilon\delta}}\big(\textbf{R}^{\diamond}_{\varepsilon\delta}- \textbf{I}_3\big)& \to \textbf{A} \quad&&\text{strongly in } L^2(0,L)^{3\times 3},\\ {{\delta\over \varepsilon}{1\over \sqrt{\varepsilon\delta}}}\mathcal{T}_{\varepsilon\delta}^{(2)}(d_1\textbf{R}_{\varepsilon\delta}^{\diamond})& \to d_1\textbf{A}\quad&&\text{strongly in}\ L^2({\omega}\times Y_2)^{3\times 3}. \end{aligned} \end{equation}

We also obtain using the definition (7.29)–(7.31) of ${\mathcal{W}}_{\varepsilon\delta}$

(7.36)\begin{equation} \begin{aligned} {1\over \varepsilon\sqrt{{\varepsilon\delta}}}\mathcal{T}_{\varepsilon\delta}^{(2)}( d_1{\mathcal{W}}_{\varepsilon\delta})&\to Z+y_1 d_1\textbf{A}\textbf{e}_1\quad &&\text{strongly in}\ L^2({\omega}\times Y_2)^3,\\ {1\over {{\varepsilon\delta}}}{\mathcal{W}}_{\varepsilon\delta}&\to 0\quad &&\text{strongly in}\ L^2((0,L))^3. \end{aligned} \end{equation}

Step 4. In this step, we present the convergence of the Green–St. Venant’s strain tensor.

Observe that

(7.37)\begin{equation} \begin{aligned} \nabla w_{\varepsilon\delta}-\textbf{R}^{\diamond}_{\varepsilon\delta}&=\nabla U_{e,{\varepsilon\delta}}-(\textbf{R}^{\diamond}_{\varepsilon\delta}-\textbf{I}_3)+\nabla \overline{w}_{\varepsilon\delta}\quad &&\text{in}\ \Omega^{(2)}_{\varepsilon\delta},\\ \nabla w_{\varepsilon\delta}-\textbf{R}^{\diamond}_{\varepsilon\delta}&=\nabla U_{e,{\varepsilon\delta}}-(\textbf{R}^{\diamond}_{\varepsilon\delta}-\textbf{I}_3)\quad &&\text{in}\ \Omega_\delta\setminus\overline{\Omega_{\varepsilon\delta}^{(2)}}, \end{aligned} \end{equation}

and

\begin{equation*}\begin{aligned} (\nabla w_{\varepsilon\delta})^T(\nabla w_{\varepsilon\delta})-\textbf{I}_3 & =(\nabla w_{\varepsilon\delta}-\textbf{R}^{\diamond}_{{\varepsilon\delta}})^T(\nabla w_{\varepsilon\delta})+(\textbf{R}^{\diamond}_{{\varepsilon\delta}})^T(\nabla w_{\varepsilon\delta}-\textbf{R}^{\diamond}_{{\varepsilon\delta}}). \end{aligned} \end{equation*}

We have in $\displaystyle\Big(-{\delta\over 2},L+{\delta\over 2}\Big)^2\times {\mathfrak{I}}_\delta$, (using (7.28) and (7.29)–(7.30))

(7.38)\begin{equation} \begin{aligned} (\textbf{I}_3+\nabla U_{e,{\varepsilon\delta}}-\textbf{R}_{\varepsilon\delta}^{\diamond})\textbf{e}_2&=0,\quad (\textbf{I}_3+\nabla U_{e,{\varepsilon\delta}}-\textbf{R}_{\varepsilon\delta}^{\diamond})\textbf{e}_3=0,\\ (\textbf{I}_3+\nabla U_{e,{\varepsilon\delta}}-\textbf{R}_{\varepsilon\delta}^{\diamond})\textbf{e}_1&=d_1{\mathcal{W}}_{\varepsilon\delta} +d_1\textbf{R}^{\diamond}_{\varepsilon\delta} x_3\textbf{e}_3. \end{aligned} \end{equation}

Then, using (5.1), (7.35), and (7.36), we obtain

\begin{equation*} {1\over \varepsilon\sqrt{{\varepsilon\delta}}}\Pi_{\varepsilon\delta}^{(2)}\big((\nabla U_{e,{\varepsilon\delta}}-\textbf{R}_{\varepsilon\delta}^{\diamond}+\textbf{I}_3)\big)\textbf{e}_1\to Z+{y_1}d_1\textbf{A}\textbf{e}_1+{y_3}d_1\textbf{A}\textbf{e}_3 \;\text{strongly in}\ L^2({\omega}\times {\mathcal{Y}}_2)^3. \end{equation*}

Using identities (7.37)–(7.38) and the above convergence and (7.32) yield the following strong convergence:

(7.39)\begin{equation} {1\over 2\varepsilon\sqrt{{\varepsilon\delta}}}\Pi_{\varepsilon\delta}^{(2)}\big(((\nabla w_{\varepsilon\delta})^T(\nabla w_{\varepsilon\delta})-\textbf{I}_3)\big) \to E({\mathcal{W}}_{3},\widehat w)\quad \text{strongly in}\ L^2({\omega}\times {\mathcal{Y}}_2)^{3\times 3} \end{equation}

by proceeding as in Theorem 6.4.

Step 5. In this step, we provide the convergence of the right-hand side forces.

With definition of forces (4.60) and $U_{e,{\varepsilon\delta}}$ (7.28), we get

\begin{equation*} \int_{\Omega_\delta}f_{\varepsilon\delta}\cdot U_{e,{\varepsilon\delta}}\,dx=\delta\int_{{\omega}_\delta} f_{\varepsilon\delta}\cdot {\mathcal{W}}_{{\varepsilon\delta}}\,dx' +\delta\int_{{\omega}_\delta} f_{\varepsilon\delta}\cdot \big(\int_{-{\delta\over 2}}^{x_1}(\textbf{R}_{\varepsilon\delta}^{\diamond}-\textbf{I}_3)\textbf{e}_1\,dt\big)dx' \end{equation*}

and the scaling of $\overline{w}_{\varepsilon\delta}$, ${\varepsilon\delta} W_{\varepsilon\delta}$, $\delta\sqrt{{\varepsilon\delta}}\textbf{Z}^{\diamond}_{\varepsilon\delta}$ give the following convergences of the forces with the convergences (7.32), (7.36) $_2$

(7.40)\begin{equation} \begin{aligned} &\lim_{(\varepsilon,\delta)\to(0,0)}{1\over \varepsilon^2\delta^2}\big(\int_{{\omega}_\delta} f_{\varepsilon\delta}\cdot{\mathcal{W}}_{\varepsilon\delta}\,dx'\big)=0= \lim_{(\varepsilon,\delta)\to(0,0)}{1\over \varepsilon^2\delta^3}\big(\int_{{\omega}_\delta} f_{\varepsilon\delta}\cdot\overline{w}_{\varepsilon\delta}\,dx'\big),\\ &\lim_{(\varepsilon,\delta)\to(0,0)}{1\over \varepsilon^2\delta^2}\int_{{\omega}_\delta} f_{1,{\varepsilon\delta}} \big(\int_{-{\delta\over 2}}^{x_1}(\textbf{R}_{\varepsilon\delta}^{\diamond}-\textbf{I}_3)\textbf{e}_1\cdot\textbf{e}_1\,dt\big)\,dx' =\int_{\omega} f_1{\mathcal{W}}_1\,dx',\\ &\lim_{(\varepsilon,\delta)\to(0,0)}{1\over \varepsilon^2\delta^2}\int_{{\omega}_\delta}f_{3,{\varepsilon\delta}} \big(\int_{-{\delta\over 2}}^{x_1}(\textbf{R}_{\varepsilon\delta}^{\diamond}-\textbf{I}_3)\textbf{e}_1\cdot\textbf{e}_3\,dt\big)\,dx'=\int_{\omega} f_3{\mathcal{W}}_3\,dx'. \end{aligned} \end{equation}

Step 6. We present the main result of the theorem.

From expressions (7.28), along with the estimates (7.33) and the definitions of the field ${\mathcal{W}}_{\varepsilon\delta}$ and $\textbf{R}_{\varepsilon\delta}^\diamond$, we have

\begin{equation*}\begin{aligned} &\|\nabla w_{\varepsilon\delta}-\textbf{R}^\diamond_{\varepsilon\delta}\|_{L^\infty(\Omega_\delta)}\leq C\varepsilon\sqrt{{\varepsilon\delta}}. \end{aligned} \end{equation*}

The above estimate give

\begin{equation*} \|\nabla w_{\varepsilon\delta}-\textbf{I}_3\|_{L^\infty(\Omega_\delta)}\leq \|\nabla w_{{\varepsilon\delta}}-\textbf{R}_{\varepsilon\delta}^{\diamond}\|_{L^\infty(\Omega_\delta)}+C\|\textbf{R}_{\varepsilon\delta}^{\diamond}-\textbf{I}_3\|_{L^\infty(-{\delta\over 2},L+{\delta\over2})}\leq C\sqrt{{\varepsilon\delta}}, \end{equation*}

where the constant $C$ does not depend on $\varepsilon$ and $\delta$. Then, we obtain for $\alpha=1,2$

(7.41)\begin{equation} \begin{aligned} \Pi_{\varepsilon\delta}^{(\alpha)}(\nabla w_{\varepsilon\delta})\to \textbf{I}_3,\quad \text{strongly in}\ L^\infty({\omega}\times {\mathcal{Y}}_\alpha)^{3\times 3},\\ \text{and}\quad \mathrm{det}(\nabla w_{\varepsilon\delta}) \gt 0,\quad\text{a.e. in}\ \Omega_\delta. \end{aligned} \end{equation}

So, the test function sequence $\{w_{\varepsilon\delta}\}_{\varepsilon,\delta}\subset \textbf{V}_{\varepsilon\delta}$ satisfies $\textbf{W}_{\varepsilon\delta}(x,\nabla w_{\varepsilon\delta})=\textbf{Q}_{\varepsilon\delta}(x,\textbf{E}(\nabla w_{\varepsilon\delta}))$.

Due to (7.41) and $\textbf{m}_{\varepsilon\delta}\leq \textbf{J}_{\varepsilon\delta}(w_{\varepsilon\delta})$, we have after passing $(\varepsilon,\delta)\to(0,0)$,

\begin{align*}\limsup_{(\varepsilon,\delta)\to(0,0)}{\textbf{m}_{\varepsilon\delta}\over \varepsilon^2\delta^3} &\leq \lim_{(\varepsilon,\delta)\to(0,0)}{\textbf{J}_{\varepsilon\delta}(w_{\varepsilon\delta})\over \varepsilon^2\delta^3}=\textbf{W}\big({\mathcal{W}}_{3},0,\widehat {w}\big)\\ &\quad -\int_{{\omega}} f\cdot{\mathcal{W}}\,dx'= \textbf{J}({\mathcal{W}}_3,0,\widehat {w})=\textbf{J}^{(2)}({\mathcal{W}}_3,\widehat w).\end{align*}

Since ${\mathbb{D}}^{(2)}\cap\big( \mathcal{C}^2([0,L]_{x_1})\times \mathcal{C}^1([0,L]_{x_1}\times \overline{{\mathcal{Y}}_2})^3\big)$ is dense in ${\mathbb{D}}^{(2)}$ and $({\mathcal{W}}_3,\widehat w)\in {\mathbb{D}}^{(2)}\cap\big( \mathcal{C}^2([0,L]_{x_1})\times \mathcal{C}^1([0,L]_{x_1}\times \overline{{\mathcal{Y}}_2})^3\big)$ is arbitrary, we get

\begin{equation*} \limsup_{(\varepsilon,\delta)\to(0,0)}{\textbf{m}_{\varepsilon\delta}\over \varepsilon^2\delta^3}\leq\textbf{J}^{(2)}({\mathcal{W}}_3,\widehat w),\quad \forall\,({\mathcal{W}}_3,\widehat w)\in{\mathbb{D}}^{(2)}. \end{equation*}

Finally, using the pair $({\mathcal{W}}_3,\widehat w)\in {\mathbb{D}}^{(2)}$ which minimizes the functional $\textbf{J}^{(2)}$ over ${\mathbb{D}}^{(2)}$, we obtain

\begin{equation*}\limsup_{(\varepsilon,\delta)\to(0,0)}{\textbf{m}_{\varepsilon\delta}\over \varepsilon^2\delta^3}\leq\textbf{m}^{(2)}.\end{equation*}

This completes the proof.

Proof. This theorem is an immediate consequence of Propositions 7.1, 7.4, 7.5, and 7.6.