Effective Model for Elastic Waves in a Substrate Supporting an Array of Plates/Beams with Flexural and Longitudinal Resonances

Abstract

In a previous study (Marigo et al. in J. Mech. Phys. Solids 143:104029, 2020) we have studied the effect of a periodic array of subwavelength plates or beams over a semi-infinite elastic ground on the propagation of waves hitting the interface. The study was restricted to the low frequency regime where only flexural resonances take place. Here, we present a generalization to higher frequencies which allows us to account for both flexural and longitudinal resonances and to evaluate their interplay. An effective model is obtained using asymptotic analysis and homogenization techniques, which can be expressed in terms of the ground alone with an effective dynamic (frequency-dependent) boundary conditions of the Robin’s type. For an in-plane wave at oblique incidence, the scattered displacement fields and the reflection coefficients are obtained in closed forms and their effectiveness to reproduce the actual scattering is inspected by comparison with direct numerics in a two-dimensional setting.

Appendices

Appendix A: Proofs of Some Properties

Throughout the appendix, we shall use that for a symmetric tensor $\tau _{ij}$ defined for $\hat{\mathbf{x}}' \in \hat{\mathsf{s}}_{\text{\tiny b}}$, and for a vector $a_{i}$ with antisymmetric tensor $\partial _{j} a_{i}$, we have

$$ \int _{\hat{\mathsf{s}}_{\text{\tiny b}}}a_{i}\partial _{j}\tau _{ij}\, \text{d}\hat{\mathbf{x}}'=\int _{\partial \hat{\mathsf{s}}_{\text{\tiny b}}} a_{i}\tau _{ij}n_{j}\,\text{d}\hat{\mathbf{x}}'- \int _{ \partial \hat{\mathsf{s}}_{\text{\tiny b}}} \partial _{j}a_{i}\tau _{ij} \,\text{d}\hat{\mathbf{x}}'=\int _{\partial \hat{\mathsf{s}}_{\text{\tiny b}}} a_{i}\tau _{ij}n_{j}\,\text{d}\hat{\mathbf{x}}', $$

(55)

since $\partial _{j}a_{i}\tau _{ij}=0$.

1.1 A.1 Determination of $({\mathbf{V}}^{0},{\mathbf{V}}^{1})$ and of $({\boldsymbol{\varSigma }}^{0},{\mathbf{V}}^{2})$

As previously said in §3.1.3, the dependence of ${\mathbf{V}}^{0}$ and ${\mathbf{V}}^{1}$ on $\hat{\mathbf{x}}'=(\hat{x},\hat{y})$ is obtained starting from $(\text{C'})^{-2}$ and $(\text{C'})^{-1}$ in (13) which tell us that there exists $({\mathbf{U}}^{0}({\mathbf{x}},\tilde{z}),\varOmega ^{0}({\mathbf{x}},\tilde{z}))$ and $({\mathbf{U}}^{1}({\mathbf{x}},\tilde{z}),\varOmega ^{1}({\mathbf{x}},\tilde{z}))$ such that

$$ \left \{ \textstyle\begin{array}{l@{\quad }l@{\quad }l} \displaystyle V_{x}^{0}=U_{x}^{0}({\mathbf{x}},\tilde{z})+\varOmega ^{0}({\mathbf{x}}, \tilde{z})\hat{y},\quad & \displaystyle V_{y}^{0}=U_{y}^{0}({\mathbf{x}}, \tilde{z})-\varOmega ^{0}({\mathbf{x}},\tilde{z})\hat{x},\quad & V_{z}^{0}= U_{z}^{0}({ \mathbf{x}}), \\ V_{x}^{1}=U_{x}^{1}({\mathbf{x}},\tilde{z})+\varOmega ^{1}({\mathbf{x}},\tilde{z}) \hat{y},& V_{y}^{1}=U_{y}^{1}({\mathbf{x}},\tilde{z})-\varOmega ^{1}({\mathbf{x}}, \tilde{z})\hat{x}.& \end{array}\displaystyle \right .$$

(56)

Next, as $\frac{\partial }{\partial \hat{x}}V_{z}^{1}=- \frac{\partial }{\partial \tilde{z}}V_{x}^{0}$, $\frac{\partial }{\partial \hat{y}}V_{z}^{1}=- \frac{\partial }{\partial \tilde{z}}V_{y}^{0}$ from $(\text{C'})^{-1}$, we have $\frac{\partial ^{2} }{\partial \hat{x}\partial \hat{y}}V_{z}^{1}=\pm \frac{\partial }{\partial \tilde{z}}\varOmega ^{0}=0$, hence

$$ \frac{\partial \varOmega ^{0}}{\partial \tilde{z}}=0, \qquad \displaystyle V_{z}^{1}= U_{z}^{1}({\mathbf{x}},\tilde{z})- \frac{\partial }{\partial \tilde{z}}\big(U_{x}^{0}\hat{x}+U_{y}^{0} \hat{y}\big). $$

(57)

The dependence of ${\mathbf{V}}^{2}$ with respect to $\hat{\mathbf{x}}'$ is more intricate as it requires to solve the following problem set in the section $\hat{\mathbf{x}}'\in \hat{\mathsf{s}}_{\text{\tiny b}}$ with $({\boldsymbol{\varSigma }}^{0},{\mathbf{V}}^{2})$ as unknowns:

$$ \left \{ \textstyle\begin{array}{l} \mathrm{div}_{{\hat{\mathbf{x}}}'}{\boldsymbol{\varSigma }^{0}}={\mathbf{0}}\quad \text{in } \hat{\mathsf{s}}_{\text{\tiny b}}, \qquad {\boldsymbol{\varSigma }}^{0}{ \mathbf{n}}={\mathbf{0}}\quad \text{on }\partial \hat{\mathsf{s}}_{\text{\tiny b}}, \\ \displaystyle \varepsilon ^{\hat{\mathbf{x}}'}\big({\mathbf{V}}^{2}\big)+ \varepsilon ^{\tilde{z}}\big({\mathbf{V}}^{1}\big)+\varepsilon ^{{\mathbf{x}}} \big({\mathbf{V}}^{0}\big)= \frac{1+\nu _{\text{\tiny b}}}{E_{\text{\tiny b}}}\,{\boldsymbol{\varSigma }}^{0} - \frac{\nu _{\text{\tiny b}}}{E_{\text{\tiny b}}}\,\text{tr}\big({ \boldsymbol{\varSigma }}^{0}\big)\,{\mathsf{\boldsymbol{I}}}, \\ \end{array}\displaystyle \right . $$

with the input ${\mathbf{V}}^{0}$ and ${\mathbf{V}}^{1}$ given by (56). For a circular section, this problem can be solved explicitly. Specifically there exists $({\mathbf{U}}^{2}({\mathbf{x}},\tilde{z}),\varOmega ^{2}({\mathbf{x}},\tilde{z}))$ such that the displacement ${\mathbf{V}}^{2}$ is given by

$$ \left \{ \textstyle\begin{array}{l@{\quad }l} V_{x}^{2}=&\displaystyle U_{x}^{2}({\mathbf{x}},\tilde{z})+\varOmega ^{2}({ \mathbf{x}},\tilde{z})\hat{y}-\frac{\partial U_{x}^{0}}{\partial x}\hat{x}- \left (\frac{\partial U_{y}^{0}}{\partial x}+ \frac{\partial U_{x}^{0}}{\partial y}\right )\frac{\hat{y}}{2} - \left ({\frac{\partial \varOmega ^{0}}{\partial x}\hat{x}+ \frac{\partial \varOmega ^{0}}{\partial y}\hat{y}}\right )\hat{y} \\ &\displaystyle -\nu _{\text{\tiny b}}\left ({ \frac{\partial U_{z}^{1}}{\partial \tilde{z}}}+ \frac{\partial U_{z}^{0}}{\partial z}\right )\hat{x}+\nu _{\text{\tiny b}}\frac{\partial ^{2} }{\partial \tilde{z}^{2}}\left (U_{x}^{0} \frac{\hat{x}^{2}-\hat{y}^{2}}{2}+ U_{y}^{0}\hat{x}\hat{y}\right ), \\ \\ V_{y}^{2}=&\displaystyle U_{y}^{2}({\mathbf{x}},\tilde{z})-\varOmega ^{2}({ \mathbf{x}},\tilde{z})\hat{x}-\frac{\partial U_{y}^{0}}{\partial y}\hat{y}- \left (\frac{\partial U_{y}^{0}}{\partial x}+ \frac{\partial U_{x}^{0}}{\partial y}\right )\frac{\hat{x}}{2}+\left ({ \frac{\partial \varOmega ^{0}}{\partial x}\hat{x}+ \frac{\partial \varOmega ^{0}}{\partial y}\hat{y}}\right )\hat{x} \\ &\displaystyle -\nu _{\text{\tiny b}}\left ({ \frac{\partial U_{z}^{1}}{\partial \tilde{z}}}+ \frac{\partial U_{z}^{0}}{\partial z}\right )\hat{y}+ \nu _{\text{\tiny b}}\frac{\partial ^{2} }{\partial \tilde{z}^{2}}\left (U_{x}^{0} \hat{x}\hat{y}- U_{y}^{0}\frac{\hat{x}^{2}-\hat{y}^{2}}{2}\right ), \\ \\ V_{z}^{2}=&\displaystyle U_{z}^{2}({\mathbf{x}},\tilde{z})-\left ( \frac{\partial U_{x}^{1}}{\partial \tilde{z}}+ \frac{\partial U_{z}^{0}}{\partial x}+{ \frac{\partial U_{x}^{0}}{\partial z}} \right )\hat{x}-\left ( \frac{\partial U_{y}^{1}}{\partial \tilde{z}}+ \frac{\partial U_{z}^{0}}{\partial y}+{ \frac{\partial U_{y}^{0}}{\partial z}} \right )\hat{y}, \end{array}\displaystyle \right .$$

(58)

and the corresponding stress tensor ${\boldsymbol{\varSigma }}^{0}$ by

$$ \left \{ \textstyle\begin{array}{l} \displaystyle \varSigma _{xz}^{0}=\mu _{\text{\tiny b}}\left ( { \frac{\partial \varOmega ^{0}}{\partial z}+ \frac{\partial \varOmega ^{1}}{\partial \tilde{z}}}\right )\hat{y},\quad \varSigma _{yz}^{0}=-\mu _{\text{\tiny b}}\left ( { \frac{\partial \varOmega ^{0}}{\partial z}+ \frac{\partial \varOmega ^{1}}{\partial \tilde{z}}}\right )\hat{x}, \\ \\ \varSigma _{zz}^{0}=\displaystyle E_{\text{\tiny b}}\left ({ \frac{\partial U_{z}^{1}}{\partial \tilde{z}}}+ \frac{\partial U_{z}^{0}}{\partial z}- \frac{\partial ^{2} }{\partial \tilde{z}^{2}}\left (U_{x}^{0}\hat{x}+U_{y}^{0} \hat{y}\right )\right ),\quad \varSigma _{xx}^{0}=\varSigma _{xy}^{0}= \varSigma _{yy}^{0}=0. \end{array}\displaystyle \right .$$

(59)

1.2 A.2 Proof of $P_{1} :\frac{\partial U_{z}^{1}}{\partial \tilde{z}}=0$ in (15)

We use $(\text{E}_{z})^{-1}$ in (12), specifically

$$ \frac{\partial }{\partial \hat{x}}\varSigma _{xz}^{1}+ \frac{\partial }{\partial \hat{y}}\varSigma _{yz}^{1}+ \frac{\partial }{\partial \tilde{z}}\varSigma _{zz}^{0}=0, $$

that we integrate over $\hat{\mathbf{x}}'\in \hat{\mathsf{s}}_{\text{\tiny b}}$. Accounting for the boundary condition $\varSigma _{xz}^{1} n_{x}+\varSigma _{yz}^{1}n_{y}=0$ on $\partial \hat{\mathsf{s}}_{\text{\tiny b}}$, we obtain that $\frac{\partial }{\partial \tilde{z}}\int _{\hat{\mathsf{s}}_{\text{\tiny b}}}\varSigma _{zz}^{0}\,\text{d}\hat{\mathbf{x}}'=0$. Next, with the form of $\varSigma _{zz}^{0}$ given in (59), and accounting for the facts that (i) $U_{z}^{0}$ in (56) does not depend on $\tilde{z}$ and (ii) $\int _{\hat{\mathsf{s}}_{\text{\tiny b}}}\hat{x}\,\text{d}\hat{\mathbf{x}}'= \int _{\hat{\mathsf{s}}_{\text{\tiny b}}}\hat{y}\,\text{d}\hat{\mathbf{x}}'=0$, we obtain that $\frac{\partial ^{2} U_{z}^{1}}{\partial \tilde{z}^{2}}=0$. The field $U_{z}^{1}$ could be a linear function of $\tilde{z}$ but this is prevented by its $\tilde{z}$-periodicity hence

$$ \frac{\partial U_{z}^{1}}{\partial \tilde{z}}=0.$$

(60)

1.3 A.3 Proof of $P_{2} :\frac{\partial \varOmega ^{1}}{\partial \tilde{z}}=0$ in (15)

We start by introducing the torsion force ${\mathcal{T}}^{0}$

$$ {\mathcal{T}}^{0}=\int _{\hat{\mathsf{s}}_{\text{\tiny b}}}\left (\hat{y} \varSigma _{xz}^{0}-\hat{x}\varSigma _{yz}^{0}\right )\,\text{d}\hat{\mathbf{x}}'. $$

(61)

Using the forms of $\varSigma _{xz}^{0}$ and $\varSigma _{yz}^{0}$ in (59) we get that

$$ {\mathcal{T}}^{0}({\mathbf{x}},\tilde{z})=\mu _{\text{\tiny b}}J_{\text{\tiny b}}\left ( \frac{\partial \varOmega ^{0}}{\partial z}({\mathbf{x}})+ \frac{\partial \varOmega ^{1}}{\partial \tilde{z}}({\mathbf{x}},\tilde{z}) \right ),\qquad \text{with } J_{\text{\tiny b}}=\int _{\hat{\mathsf{s}}_{\text{\tiny b}}}(\hat{x}^{2}+\hat{y}^{2})\,\text{d}\hat{\mathbf{x}}'= \frac{\pi }{2}\hat{r}_{\text{\tiny b}}^{4}.$$

(62)

($J_{\text{\tiny b}}$ is the torsion constant for a circular cross-section.) Now, we use $\mathrm{div}_{{\hat{\mathbf{x}}}'}{\boldsymbol{\varSigma }^{1}}+\mathrm{div}_{ \tilde{z}}{\boldsymbol{\varSigma }^{0}}={\mathbf{0}}$, from $(\text{E}_{a})^{-1}$ and $(\text{E}_{z})^{-1}$ in (12), that we integrate over $\hat{\mathsf{s}}_{\text{\tiny b}}$ after multiplication by ${\boldsymbol{a}}=\hat{y}{\mathbf{e}}_{x}-\hat{x}{\mathbf{e}}_{y}$; we get

$$ 0=\int _{\hat{\mathsf{s}}_{\text{\tiny b}}}{\boldsymbol{a}}\cdot \mathrm{div}_{{ \hat{\mathbf{x}}}'}{\boldsymbol{\varSigma }^{1}}\,\text{d}\hat{\mathbf{x}}'+\int _{ \hat{\mathsf{s}}_{\text{\tiny b}}}{\boldsymbol{a}}\cdot \mathrm{div}_{\tilde{z}}{ \boldsymbol{\varSigma }^{0}}\,\text{d}\hat{\mathbf{x}}'=\int _{\partial \hat{\mathsf{s}}_{\text{\tiny b}}}{\boldsymbol{a}}\cdot ({\boldsymbol{\varSigma }^{1}}{\mathbf{n}}) \,\text{d}\hat{\mathbf{x}}'+\frac{\partial }{\partial \tilde{z}}\int _{ \hat{\mathsf{s}}_{\text{\tiny b}}}{\boldsymbol{a}}\cdot ({\boldsymbol{\varSigma }^{0}}{\mathbf{e}}_{z}) \,\text{d}\hat{\mathbf{x}}'. $$

(63)

In (63), we have applied (55) to the first integral involving ${\boldsymbol{\varSigma }^{1}}$. Indeed, it suffices to notice that $\mathrm{div}_{{\hat{\mathbf{x}}}'}{\boldsymbol{\varSigma }^{1}}= \mathrm{div}_{{ \hat{\mathbf{x}}}'}{\boldsymbol{\varSigma }^{1}}|_{\hat{\mathbf{x}}'}$ where ${{\boldsymbol{\varSigma }}^{1}}|_{\hat{\mathbf{x}}'}$ is the 2x2 symmetric restriction of ${\boldsymbol{\varSigma }}^{1}$ to the indices $(x,y)$ and that ${\boldsymbol{\nabla}}_{\hat{\mathbf{x}}'}{\boldsymbol{a}}$ is antisymmetric. Since ${{\boldsymbol{\varSigma }}^{1}}|_{\hat{\mathbf{x}}'}{\mathbf{n}}={{\boldsymbol{\varSigma }}^{1}}{\mathbf{n}}={ \mathbf{0}}$ over $\partial \hat{\mathsf{s}}_{\text{\tiny b}}$, we deduce that $\int _{\partial \hat{\mathsf{s}}_{\text{\tiny b}}}{\boldsymbol{a}}\cdot ({ \boldsymbol{\varSigma }^{1}}{\mathbf{n}})\,\text{d}\hat{\mathbf{x}}'=0$. Now, recalling the definition (61) of ${\mathcal{T}}^{0}$ as well as (62), we find

$$ 0=\frac{\partial }{\partial \tilde{z}}\int _{\hat{\mathsf{s}}_{\text{\tiny b}}}{\boldsymbol{a}}\cdot ({\boldsymbol{\varSigma }^{0}}{\mathbf{e}}_{z})\,\text{d} \hat{\mathbf{x}}'=\frac{\partial {\mathcal{T}}^{0}}{\partial \tilde{z}}= \mu _{\text{\tiny b}}J_{\text{\tiny b}} \frac{\partial ^{2} \varOmega ^{1}}{\partial \tilde{z}^{2}}. $$

(64)

Thus, $\varOmega ^{1}$ is independent of $\tilde{z}$ (a linear dependence w.r.t. $\tilde{z}$ is prevented by its $\tilde{z}$-periodicity), and ${\mathcal{T}}^{0}$ reduces to

$$ {\mathcal{T}}^{0}({\mathbf{x}})=\mu _{\text{\tiny b}}J_{\text{\tiny b}} \frac{\partial \varOmega ^{0}}{\partial z}({\mathbf{x}}), \quad \frac{\partial \varOmega ^{1}}{\partial \tilde{z}}=0. $$

(64)

1.4 A.4 Proof of $P_{3} : \varOmega ^{0}=0$ in (15)

Here, we shall see that the equation for torsion (64) can be complemented to get the wave equation for $\varOmega ^{0}$; in addition that the associated boundary conditions impose $\varOmega ^{0}=0$. (From (57) we already know that $\varOmega ^{0}({\mathbf{x}})$ does not depend of $\tilde{z}$.)

1.4.1 A.4.1 The Wave Equation for Torsion

We start with the derivation of the equation satisfied by the macroscopic rotation $\varOmega ^{0}$. We use the relation $\mathrm{div}_{{\hat{\mathbf{x}}}'}{\boldsymbol{\varSigma }^{2}}+\mathrm{div}_{ \tilde{z}}{\boldsymbol{\varSigma }^{1}}+{\mathrm{div}}_{{\mathbf{x}}}{\boldsymbol{\varSigma }^{0}}+ \rho _{\text{\tiny b}}\omega ^{2}{\mathbf{V}}^{0}={\mathbf{0}}$ from $(\text{E}_{a})^{0}$ and $(\text{E}_{z})^{0}$ in (12), that we multiply by ${\boldsymbol{a}}=\hat{y}{\mathbf{e}}_{x}-\hat{x}{\mathbf{e}}_{y}$ and integrate over $(\hat{\mathbf{x}}',\tilde{z})\in \hat{\mathsf{s}}_{\text{\tiny b}}\times (0, \,\tilde{h}_{\text{\tiny F}})$. We get

$$ \int _{\hat{\mathsf{s}}_{\text{\tiny b}}\times (0,\tilde{h}_{\text{\tiny F}})}{\boldsymbol{a}}\cdot {\mathrm{div}}_{{\mathbf{x}}}{\boldsymbol{\varSigma }^{0}} \,\text{d}\hat{\mathbf{x}}'\,\text{d}\tilde{z}+\rho _{\text{\tiny b}}\omega ^{2} \int _{\hat{\mathsf{s}}_{\text{\tiny b}}\times (0,\tilde{h}_{\text{\tiny F}})}{\boldsymbol{a}}\cdot {\mathbf{V}}^{0}\,\text{d}\hat{\mathbf{x}}'\, \text{d}\tilde{z}=0, $$

where we have accounted for $\int _{\hat{\mathsf{s}}_{\text{\tiny b}}}{\boldsymbol{a}}\cdot \mathrm{div}_{{ \hat{\mathbf{x}}}'}{\boldsymbol{\varSigma }^{2}}\,\text{d}\hat{\mathbf{x}}'=0$ as in (63) and for the $\tilde{z}$-periodicity of ${\boldsymbol{\varSigma }}^{1}$. Next, owing to the form of ${\mathbf{V}}^{0}$ in (56) (with $\varOmega ^{0}=\varOmega ^{0}({\mathbf{x}})$ from (57)) and the form of ${\boldsymbol{\varSigma }}^{0}$ in (59) (with $\frac{\partial }{\partial \tilde{z}}\varOmega ^{1}=0$ from (64)), we obtain

$$ \frac{\partial {\mathcal{T}}^{0}}{\partial z}({\mathbf{x}})+\rho _{\text{\tiny b}}\omega ^{2}J_{\text{\tiny b}}\varOmega ^{0}({\mathbf{x}})=0, \quad \rightarrow \quad \mu _{\text{\tiny b}} \frac{\partial ^{2} \varOmega ^{0}}{\partial z ^{2}}({\mathbf{x}})+\rho _{\text{\tiny b}}\omega ^{2}\varOmega ^{0}({\mathbf{x}})=0, $$

(65)

with $J_{\text{\tiny b}}$ defined in (62).

1.4.2 A.4.2 The Corresponding Boundary Conditions

To conclude that $\varOmega ^{0}=0$, we have to specify the boundary conditions at $z=0$ and $h_{\text{\tiny b}}$ by analyzing the torsional behavior near the extremities.

At $z=h_{\text{\tiny b}}$

In the intermediate region located near $z=h_{\text{\tiny b}}$, we introduce the torsion force ${\widetilde{{\mathcal{T}}}^{0}}_{|\,h_{\text{\tiny b}}}({\mathbf{x}}', \tilde{z})$. Repeating the same procedure than in A.3 in the region of the beam, we find that

$$ \frac{\partial {\widetilde{{\mathcal{T}}}^{0}}_{|\,h_{\text{\tiny b}}}}{\partial \tilde{z}}=0, $$

(66)

hence ${\widetilde{{\mathcal{T}}}^{0}}_{|\,h_{\text{\tiny b}}}({\mathbf{x}}')$ does not depend on $\tilde{z}$. To determine its value, we go to the microscopic scale near $z=h_{\text{\tiny b}}$ with the expansions (27). Multiplying $\mathrm{div}_{\hat{\mathbf{x}}} \widehat{\boldsymbol{\varSigma }}^{0}|_{h_{\text{\tiny b}}}={\mathbf{0}}$ by ${\boldsymbol{a}}=\hat{y}{\mathbf{e}}_{x}-\hat{x}{\mathbf{e}}_{y}$, integrating over $\widehat{\mathsf{X}}_{h_{\text{\tiny b}}}=\left \{ \hat{\mathbf{x}}'\in \hat{\mathsf{s}}_{\text{\tiny b}},\hat{z}\in (-\infty ,0)\right \}$ and making use of the matching conditions and traction free boundary conditions, we deduce that

$$ {\widetilde{{\mathcal{T}}}^{0}}_{|\,h_{\text{\tiny b}}}({\mathbf{x}}')=\lim _{ \hat{z}\rightarrow -\infty } \int _{\hat{\mathsf{s}}_{\text{\tiny b}}} \left (\hat{y}\widehat{\varSigma _{xz}^{0}}_{|\,h_{\text{\tiny b}}}-\hat{x}{ \widehat{\varSigma _{yz}^{0}}}_{|\,h_{\text{\tiny b}}}\right )\,\text{d} \hat{\mathbf{x}}'=0. $$

(67)

Finally, using the matching condition between the intermediate region and the inner region of the beams we get

$$ {\mathcal{T}}^{0}({\mathbf{x}}',h_{\text{\tiny b}})=\lim _{\tilde{z} \rightarrow -\infty }{\widetilde{{\mathcal{T}}}^{0}}_{|\,h_{\text{\tiny b}}}({\mathbf{x}}')=0. $$

(68)

At $z=0$

In the intermediate region located near $z=0$, in virtue of (13) which holds also for the intermediate expansions (23), we have $\varepsilon ^{\hat{\mathbf{x}}'}\big ({\widetilde{{\mathbf{V}}}^{0}}_{|0}\big )={ \mathbf{0}}$, hence ${\widetilde{V_{x}^{0}}}_{|0}={\widetilde{U_{x}^{0}}}_{|0}({\mathbf{x}}', \tilde{z})+{\widetilde{\varOmega }^{0}}_{|0}({\mathbf{x}}')\hat{y}$. The goal is to show that ${\widetilde{\varOmega }^{0}}_{|0}({\mathbf{x}}')=0$. For that, we go at the microscopic scale near $z=0$, where we recall (see section (3.2.2)) that since $\varepsilon ^{\hat{\mathbf{x}}'}\big ({\widehat{\mathbf{V}}^{0}}_{|0})={\mathbf{0}}$, ${\widehat{\mathbf{V}}^{0}}_{|0}$ is a rigid body motion which can be reduced to a translation thanks to the periodic conditions acting at this scale. Therefore we get ${\widehat{V^{0}_{x}}}_{|0}={\widehat{U^{0}_{x}}}_{|0}({\mathbf{x}}')$. It now suffices to apply the micro-meso matching conditions, i.e., ${\widetilde{V^{0}_{x}}}_{|0}({\mathbf{x}}',0)=\lim _{\hat{z}\rightarrow + \infty } {\widehat{V^{0}_{x}}}_{|0}$, to show that ${\widetilde{\varOmega }^{0}}_{|0}({\mathbf{x}}')=0$. Finally, using again matching conditions between the intermediate region and the inner region of the beams we deduce that

$$ {\varOmega }^{0}({\mathbf{x}}',0)=\lim _{\tilde{z}\rightarrow +\infty }{ \widetilde{\varOmega }^{0}}_{|0}({\mathbf{x}}')=0. $$

(69)

Given the wave equation (65) and the corresponding boundary conditions (68)-(69), we deduce that outside the set of torsional resonant frequencies which correspond to $\omega _{n}=\sqrt{\frac{\mu _{\text{\tiny b}}}{\rho _{\text{\tiny b}}}} \frac{n\pi }{2h_{\text{\tiny b}}}$ with $n\in {Z}^{*}$, we have

$$ \varOmega ^{0}({\mathbf{x}})=0. $$

(70)

1.5 A.5 Proof of $P_{4} : \frac{\partial U_{x}^{0}}{\partial z}=\frac{\partial U_{y}^{0}}{\partial z}=0$ in (15)

1.5.1 A.5.1 Determination of ${\mathbf{V}}^{3}$ and ${\boldsymbol{\varSigma }}^{1}$

The first step is to solve the problem on $({\boldsymbol{\varSigma }}^{1},{\boldsymbol{V}}^{3})$ which reads, from $(\text{E}_{a})^{-1}$ and $(\text{E}_{z})^{-1}$ in (12) and from $(\text{C'})^{1}$ in (13),

$$ \left \{ \textstyle\begin{array}{l} \mathrm{div}_{{\hat{\mathbf{x}}}'}{\boldsymbol{\varSigma }^{1}}+\mathrm{div}_{ \tilde{z}}{\boldsymbol{\varSigma }^{0}}={\mathbf{0}}\quad \text{in } \hat{\mathsf{s}}_{\text{\tiny b}}, \qquad {\boldsymbol{\varSigma }}^{1}{\mathbf{n}}={\mathbf{0}}\quad \text{on } \partial \hat{\mathsf{s}}_{\text{\tiny b}}, \\ \displaystyle \varepsilon ^{\hat{\mathbf{x}}'}\big({\mathbf{V}}^{3}\big)+ \varepsilon ^{\tilde{z}}\big({\mathbf{V}}^{2}\big)+\varepsilon ^{{\mathbf{x}}} \big({\mathbf{V}}^{1}\big)= \frac{1+\nu _{\text{\tiny b}}}{E_{\text{\tiny b}}}\,{\boldsymbol{\varSigma }}^{1} - \frac{\nu _{\text{\tiny b}}}{E_{\text{\tiny b}}}\text{tr}\big({ \boldsymbol{\varSigma }}^{1}\big)\,{\mathsf{\boldsymbol{I}}}. \end{array}\displaystyle \right . $$

(71)

We short-circuit the proof by assuming, or anticipating, that

$$ \varSigma _{xx}^{1}=\varSigma _{xy}^{1}=\varSigma _{yy}^{1}=0,$$

(72)

which can be verified a posteriori. As a result, from (71) and with $\text{tr}\big ({\boldsymbol{\varSigma }}^{1}\big )=\varSigma _{zz}^{1}$, we have $\varSigma _{zz}^{1}=E_{\text{\tiny b}}\left ( \frac{\partial V_{z}^{2}}{\partial \tilde{z}}+ \frac{\partial V_{z}^{1}}{\partial z}\right )$. Next, using $V_{z}^{2}$ in (58), $V_{z}^{1}$ in (57) and with $V_{z}^{0}= U_{z}^{0}({\mathbf{x}})$ independent of $\tilde{z}$ from (56), we obtain

$$ \varSigma _{zz}^{1}=E_{\text{\tiny b}}\left ( \frac{\partial U_{z}^{2}}{\partial \tilde{z}}+ \frac{\partial U_{z}^{1}}{\partial z}- \frac{\partial ^{2} }{\partial \tilde{z}^{2}}\left (U_{x}^{1}\hat{x}+U_{y}^{1} \hat{y}\right ) -2\frac{\partial ^{2} }{\partial z \partial \tilde{z}} \left (U_{x}^{0}\hat{x}+U_{y}^{0}\hat{y}\right ) \right ). $$

(73)

Let us consider now in (71) (second line) the strains associated to the 2D vanishing stress components $(\varSigma _{xx}^{1},\varSigma _{xy}^{1},\varSigma _{yy}^{1})$. We obtain

$$ \displaystyle E_{\text{\tiny b}}\left ( \frac{\partial V_{x}^{3}}{\partial \hat{x}} + \frac{\partial V_{x}^{1}}{\partial x}\right )=-\nu _{\text{\tiny b}} \varSigma _{zz}^{1},\quad \displaystyle E_{\text{\tiny b}}\left ( \frac{\partial V_{y}^{3}}{\partial \hat{y}} + \frac{\partial V_{y}^{1}}{\partial y}\right )=-\nu _{\text{\tiny b}} \varSigma _{zz}^{1}, $$

(74)

and

$$ \displaystyle \frac{\partial V_{y}^{3}}{\partial \hat{x}}+ \frac{\partial V_{x}^{3}}{\partial \hat{y}}+ \frac{\partial V_{y}^{1}}{\partial x}+ \frac{\partial V_{x}^{1}}{\partial y}=0. $$

We find that there exists $(U_{x}^{3}({\mathbf{x}},\tilde{z}),U_{y}^{3}({\mathbf{x}},\tilde{z}),\varOmega ^{3}({ \mathbf{x}},\tilde{z}))$ such that the displacements $( V_{x}^{3}, V_{y}^{3})$ read

$$ \left \{ \textstyle\begin{array}{l@{\quad }l} \displaystyle V_{x}^{3} =&\displaystyle U_{x}^{3}({\mathbf{x}},\tilde{z})+ \varOmega ^{3}({\mathbf{x}},\tilde{z})\hat{y}- \frac{\partial \varOmega ^{1}}{\partial x}\hat{x}\hat{y}- \frac{\partial \varOmega ^{1}}{\partial y}\hat{y}^{2}- \frac{\partial U_{x}^{1}}{\partial x}\hat{x}-\left ( \frac{\partial U_{y}^{1}}{\partial x}+ \frac{\partial U_{x}^{1}}{\partial y}\right )\frac{\hat{y}}{2} \\ & \displaystyle -\nu _{\text{\tiny b}}\left ( \frac{\partial U_{z}^{2}}{\partial \tilde{z}}+ \frac{\partial U_{z}^{1}}{\partial z}\right )\hat{x}+\nu _{\text{\tiny b}}\frac{\partial }{\partial \tilde{z}}\left (\left ( \frac{\partial U_{x}^{1}}{\partial \tilde{z}}+2 \frac{\partial U_{x}^{0}}{\partial z}\right ) \frac{\hat{x}^{2}-\hat{y}^{2}}{2}\right. \\ &\left.{}+\left ( \frac{\partial U_{y}^{1}}{\partial \tilde{z}}+2 \frac{\partial U_{y}^{0}}{\partial z} \right )\hat{x}\hat{y}\right ), \\ \\ \displaystyle V_{y}^{3} =&\displaystyle U_{y}^{3}({\mathbf{x}},\tilde{z})- \varOmega ^{3}({\mathbf{x}},\tilde{z})\hat{x}+ \frac{\partial \varOmega ^{1}}{\partial y}\hat{x}\hat{y}+ \frac{\partial \varOmega ^{1}}{\partial x}\hat{x}^{2}- \frac{\partial U_{y}^{1}}{\partial y}\hat{y}-\left ( \frac{\partial U_{y}^{1}}{\partial x}+ \frac{\partial U_{x}^{1}}{\partial y}\right )\frac{\hat{x}}{2} \\ & \displaystyle -\nu _{\text{\tiny b}}\left ( \frac{\partial U_{z}^{2}}{\partial \tilde{z}}+ \frac{\partial U_{z}^{1}}{\partial z}\right )\hat{y}+\nu _{\text{\tiny b}}\frac{\partial }{\partial \tilde{z}}\left (\left ( \frac{\partial U_{x}^{1}}{\partial \tilde{z}}+2 \frac{\partial U_{x}^{0}}{\partial z}\right )\hat{x}\hat{y}\right.\\ &\left.{}+\left ( \frac{\partial U_{y}^{1}}{\partial \tilde{z}}+2 \frac{\partial U_{z}^{0}}{\partial z} \right ) \frac{\hat{x}^{2}-\hat{y}^{2}}{2}\right ). \end{array}\displaystyle \right .$$

(74)

The remaining part of the solution, $(\varSigma _{xz}^{1},\varSigma _{yz}^{1})$ and $V_{z}^{3}$, is more demanding. From (71) with $\varSigma _{zz}^{0}$ in (59) and using that $\frac{\partial }{\partial \tilde{z}}U_{z}^{0}= \frac{\partial }{\partial \tilde{z}}U_{z}^{1}=0$ from (56) and (60), these fields satisfy

$$ \left \{ \textstyle\begin{array}{l} \displaystyle \frac{\partial \varSigma _{xz}^{1}}{\partial \hat{x}}+ \frac{\partial \varSigma _{yz}^{1}}{\partial \hat{y}}=- \frac{\partial \varSigma _{zz}^{0}}{\partial \tilde{z}}=E_{\text{\tiny b}}\frac{\partial ^{3} }{\partial \tilde{z}^{3} }\left (U_{x}^{0} \hat{x}+U_{y}^{0}\hat{y}\right ) \quad \text{in } \hat{\mathsf{s}}_{\text{\tiny b}}, \\ \displaystyle \varSigma _{xz}^{1}=\mu _{\text{\tiny b}}\left ( \frac{\partial V_{z}^{3}}{\partial \hat{x}}+ \frac{\partial V_{x}^{2}}{\partial \tilde{z}}+ \frac{\partial V_{z}^{1}}{\partial x}+ \frac{\partial V_{x}^{1}}{\partial z}\right ), \quad \varSigma _{yz}^{1}= \mu _{\text{\tiny b}}\left (\frac{\partial V_{z}^{3}}{\partial \hat{y}}+ \frac{\partial V_{y}^{2}}{\partial \tilde{z}}+ \frac{\partial V_{z}^{1}}{\partial y}+ \frac{\partial V_{y}^{1}}{\partial z}\right ), \\ \varSigma _{xz}^{1}n_{x}+\varSigma _{yz}^{1} n_{y}=0\quad \text{on } \partial \hat{\mathsf{s}}_{\text{\tiny b}}. \end{array}\displaystyle \right .$$

(74)

(We have used the constitutive equation $(\text{C})^{1}$ instead of $(\text{C'})^{1}$ in the second line.) Now, we rearrange the constitutive behavior using (i) $(V_{x}^{2},V_{y}^{2})$ in (58) (with $\varOmega ^{0}=0$ from (70)), (ii) $(V_{x}^{1},V_{y}^{1},V_{z}^{1})$ in (56)-(57) and (iii) $\frac{\partial }{\partial \tilde{z}}U_{z}^{0}=0$ from 56, resulting in

$$ \frac{1}{\mu _{\text{\tiny b}}} \underbrace{\left (\textstyle\begin{array}{c} \varSigma _{xz}^{1} -\mu _{\text{\tiny b}}S \hat{y}\\ \varSigma _{yz}^{1} +\mu _{\text{\tiny b}}S \hat{x}\end{array}\displaystyle \right )}_{{ \boldsymbol{\varSigma }}'} =\displaystyle {\boldsymbol{\nabla}}_{\hat{\mathbf{x}}'} \underbrace{\left (V_{z}^{3}+V_{z}'\right )}_{V'} +\nu _{\text{\tiny b}}\frac{\partial ^{3} U_{x}^{0}}{\partial \tilde{z}^{3} } \left ( \textstyle\begin{array}{c} \frac{\hat{x}^{2}-\hat{y}^{2}}{2} \\ \hat{x}\hat{y}\end{array}\displaystyle \right ) +\nu _{\text{\tiny b}} \frac{\partial ^{3} U_{y}^{0}}{\partial \tilde{z}^{3} }\left ( \textstyle\begin{array}{c} \hat{x}\hat{y} \\ \frac{-\hat{x}^{2}+\hat{y}^{2}}{2} \end{array}\displaystyle \right ), $$

(75)

where

$$ \left \{ \textstyle\begin{array}{l@{\quad }l} V_{z}'= &\displaystyle \left ( \frac{\partial U_{x}^{2}}{\partial \tilde{z}}+ \frac{\partial U_{z}^{1}}{\partial x}+ \frac{\partial U_{x}^{1}}{\partial z}\right )\hat{x}+ \left ( \frac{\partial U_{y}^{2}}{\partial \tilde{z}} + \frac{\partial U_{z}^{1}}{\partial y}+ \frac{\partial U_{y}^{1}}{\partial z}\right )\hat{y} \\ &\displaystyle - \frac{\partial }{\partial \tilde{z}}\left ( \frac{\partial U_{x}^{0}}{\partial x}\hat{x}^{2}+ \frac{\partial U_{y}^{0}}{\partial y}\hat{y}^{2} +\left ( \frac{\partial U_{y}^{0}}{\partial x}+ \frac{\partial U_{x}^{0}}{\partial y}\right )\hat{x}\hat{y}\right ), \\ \displaystyle S=&\displaystyle \frac{\partial \varOmega ^{2}}{\partial \tilde{z}}+ \frac{\partial \varOmega ^{1}}{\partial z}-\frac{1}{2} \frac{\partial }{\partial \tilde{z}}\left ( \frac{\partial U_{y}^{0}}{\partial x}- \frac{\partial U_{x}^{0}}{\partial y}\right ). \end{array}\displaystyle \right .$$

(76)

Basically, the idea is to work on the auxiliary displacement variable $V'$ which has absorbed $V_{z}'$ associated to a gradient. Thus, it is now sufficient to solve the problem on $({\boldsymbol{\varSigma }}', V')$ which satisfy (75) along with $\mathrm{div}_{{\hat{\mathbf{x}}}'}{\boldsymbol{\varSigma }}'=E_{\text{\tiny b}} \frac{\partial ^{3} }{\partial \tilde{z}^{3} }\left (U_{x}^{0}\hat{x}+U_{y}^{0} \hat{y}\right )$ and ${\boldsymbol{\varSigma }}'{\boldsymbol{n}}=0$; by linearity, we can set

$$ \left \{ \textstyle\begin{array}{l} \displaystyle {\boldsymbol{\varSigma }}'({\mathbf{x}},\tilde{z},\hat{\mathbf{x}}')= \frac{\partial ^{3} }{\partial \tilde{z}^{3} }U_{x}^{0}({\mathbf{x}}, \tilde{z})\;{\boldsymbol{\sigma }}^{x}(\hat{\mathbf{x}}')+ \frac{\partial ^{3} }{\partial \tilde{z}^{3} }U_{y}^{0}({\mathbf{x}}, \tilde{z})\;{\boldsymbol{\sigma }}^{y}(\hat{\mathbf{x}}'), \\ \displaystyle V'({\mathbf{x}},\tilde{z},\hat{\mathbf{x}}')= \frac{\partial ^{3} }{\partial \tilde{z}^{3} }U_{x}^{0}({\mathbf{x}}, \tilde{z})\; v^{x}(\hat{\mathbf{x}}')+ \frac{\partial ^{3} }{\partial \tilde{z}^{3} }U_{y}^{0}({\mathbf{x}}, \tilde{z}) \;v^{y}(\hat{\mathbf{x}}'), \end{array}\displaystyle \right . $$

(77)

and solve

$$ \left \{ \textstyle\begin{array}{l} \displaystyle {\mathrm{div}}{\boldsymbol{\sigma }}^{x}=E_{\text{\tiny b}}\hat{x} \quad \text{in } \hat{\mathsf{s}}_{\text{\tiny b}}, \\ {\boldsymbol{\sigma }}^{x}\,{\boldsymbol{n}}=0 \quad \text{ on } \partial \hat{\mathsf{s}}_{\text{\tiny b}}, \\ \displaystyle {\boldsymbol{\sigma }}^{x}=\mu _{\text{\tiny b}}{\boldsymbol{\nabla}}v^{x}+ \mu _{\text{\tiny b}}\nu _{\text{\tiny b}}\left ( \textstyle\begin{array}{c} \frac{\hat{x}^{2}-\hat{y}^{2}}{2} \\ \hat{x}\hat{y}\end{array}\displaystyle \right ), \end{array}\displaystyle \right .\qquad \left \{ \textstyle\begin{array}{l} \displaystyle {\mathrm{div}}{\boldsymbol{\sigma }}^{y}=E_{\text{\tiny b}}\hat{y} \quad \text{in } \hat{\mathsf{s}}_{\text{\tiny b}}, \\ {\boldsymbol{\sigma }}^{y}\,{\boldsymbol{n}}=0\quad \text{ on } \partial \hat{\mathsf{s}}_{\text{\tiny b}}, \\ \displaystyle {\boldsymbol{\sigma }}^{y}=\mu _{\text{\tiny b}}{\boldsymbol{\nabla}}v^{y}+ \mu _{\text{\tiny b}}\nu _{\text{\tiny b}}\left ( \textstyle\begin{array}{c} \hat{x}\hat{y} \\ -\frac{\hat{x}^{2}-\hat{y}^{2}}{2}\end{array}\displaystyle \right ). \end{array}\displaystyle \right . $$

The above problems can be solved in polar coordinates with $\hat{x}=\hat{r}\cos \theta$, $\hat{y}=\hat{r}\sin \theta$ by looking for a solution of the form $v^{x}= g(\hat{r})\cos \theta$ and $v^{y}= g(\hat{r})\sin \theta$, with $g$ satisfying $g''+\frac{g'}{\hat{r}}-\frac{g}{\hat{r}^{2}}=2\hat{r}$, along with the boundary condition $(2 g'(\hat{r}_{\text{\tiny b}})+\nu _{\text{\tiny b}}\hat{r}_{\text{\tiny b}}^{2})=0$. It follows that

$$ g(\hat{r})=\frac{\hat{r}^{3}}{4}-\left (\frac{3}{4}+ \frac{\nu _{\text{\tiny b}}}{2}\right )\hat{r}_{\text{\tiny b}}^{2}\hat{r}.$$

Eventually the solution reads

$$\begin{aligned} \left \{ \textstyle\begin{array}{l} \displaystyle V_{z}^{3}=-V'_{z}+ \frac{\partial ^{3} }{\partial \tilde{z}^{3} }\left (\left (U_{x}^{0} \hat{x}+U_{y}^{0}\hat{y}\right )\frac{g}{\hat{r}}\right ), \\ \displaystyle \frac{1}{\mu _{\text{\tiny b}}}\varSigma _{xz}^{1}=S\hat{y}+ \frac{\partial ^{3} }{\partial \tilde{z}^{3} }\left [U_{x}^{0}\left ( \left (\frac{g'}{\hat{r}^{2}}+\frac{\nu _{\text{\tiny b}}}{2}\right ) \hat{x}^{2}+ \left (\frac{g}{\hat{r}^{3}}- \frac{\nu _{\text{\tiny b}}}{2}\right )\hat{y}^{2}\right )+U_{y}^{0} \left (\frac{g'}{\hat{r}^{2}}-\frac{g}{\hat{r}^{3}}+\nu _{\text{\tiny b}}\right )\hat{x}\hat{y}\right ], \\ \displaystyle \frac{1}{\mu _{\text{\tiny b}}}\varSigma _{yz}^{1}=-S\hat{x}+ \frac{\partial ^{3} }{\partial \tilde{z}^{3} }\!\left [ U_{x}^{0}\left ( \frac{g'}{\hat{r}^{2}}-\frac{g}{\hat{r}^{3}}+\nu _{\text{\tiny b}}\!\right ) \hat{x}\hat{y}+U_{y}^{0}\left (\!\left (\frac{g'}{\hat{r}^{2}}+ \frac{\nu _{\text{\tiny b}}}{2}\!\right )\hat{y}^{2}+\left ( \frac{g}{\hat{r}^{3}}-\frac{\nu _{\text{\tiny b}}}{2}\right )\hat{x}^{2} \! \right ) \!\right ], \end{array}\displaystyle \right . \end{aligned}$$

(77)

with $\hat{r}=\sqrt{\hat{x}^{2}+\hat{y}^{2}}$ and with $(V'_{z},S)$ defined in (76).

1.5.2 A.5.2 Macroscopic Equation for the Flexural Motions $U_{a}^{1}$

Now, we want to derive the relations equivalent to (19), (20) and (21). To do so, we iterate the procedure conducted in §3.1.5. First, we obtain $M^{1}_{a}$, $a=x,y$, by integrating $\varSigma _{zz}^{1}$ (whose form is in (73)) over $\hat{\mathbf{x}}'\in \hat{\mathsf{s}}_{\text{\tiny b}}$ after multiplication by $\hat{x}$ then by $\hat{y}$. Next, the equilibrium is obtained starting with $(\text{E}_{a})^{1}$ in (12): $\frac{\partial }{\partial \hat{x}}\varSigma _{xa}^{3}+ \frac{\partial }{\partial \hat{y}}\varSigma _{ya}^{3}+ \frac{\partial }{\partial \tilde{z}}\varSigma _{az}^{2}+ \frac{\partial }{\partial z}\varSigma _{az}^{1}+\rho _{\text{\tiny b}} \omega ^{2} V_{a}^{1}=0$, $a=x,y$ (since $\varSigma ^{1}_{xa}=\varSigma ^{1}_{ya}=0$ from (72)), that we integrate over $\hat{\mathbf{x}}'\in \hat{\mathsf{s}}_{\text{\tiny b}}$; then we use $\varSigma _{az}^{1}$ in (77) and $V^{1}_{a}$ in (56). Eventually, the relations between bending moments and shear forces are written from $(\text{E}_{z})^{0}$ in (12): $\frac{\partial }{\partial \hat{x}}\varSigma _{xz}^{2}+ \frac{\partial }{\partial \hat{y}}\varSigma _{yz}^{2}+ \frac{\partial }{\partial \tilde{z}}\varSigma _{zz}^{1}+ \frac{\partial }{\partial z}\varSigma _{zz}^{0}+\rho _{\text{\tiny b}} \omega ^{2} V_{z}^{0}=0$ (since $\varSigma _{xz}^{0}=\varSigma _{yz}^{0}=0$ from (16)), that we integrate over $\hat{\mathbf{x}}'\in \hat{\mathsf{s}}_{\text{\tiny b}}$ after multiplication by $\hat{a}=\hat{x},\hat{y}$; then, we use $\varSigma _{zz}^{0}$ in (16) and $V_{z}^{0}$ in (56). We obtain

$$ \left \{ \textstyle\begin{array}{l} \displaystyle M^{1}_{a}=E_{\text{\tiny b}}I_{\text{\tiny b}} \frac{\partial ^{2} U_{a}^{1}}{\partial \tilde{z}^{2}}+2E_{\text{\tiny b}}I_{\text{\tiny b}} \frac{\partial ^{2} U^{0}_{a}}{\partial z \partial \tilde{z}},\quad \displaystyle \frac{\partial T_{a}^{2}}{\partial \tilde{z}}+\rho _{\text{\tiny b}}\omega ^{2}\varphi U_{a}^{1}+\mu _{\text{\tiny b}}A_{\text{\tiny b}} \frac{\partial ^{4} U^{0}_{a}}{\partial z \partial \tilde{z}^{3}}=0, \\ \displaystyle T^{2}_{a}+\frac{\partial M_{a}^{1}}{\partial \tilde{z}}+E_{\text{\tiny b}}I_{\text{\tiny b}} \frac{\partial ^{3} U_{a}^{0}}{\partial z \partial \tilde{z}^{2}}=0, \end{array}\displaystyle \right . $$

with $A_{\text{\tiny b}}=-\frac{1}{2}(1+\nu _{\text{\tiny b}})\pi r_{\text{\tiny b}}^{4}$. The above system leads to

$$ \frac{\partial ^{4} U^{1}_{a}}{\partial \tilde{z}^{4} }-\kappa ^{4} U_{a}^{1}=B_{\text{\tiny b}} \frac{\partial ^{4} U_{a}^{0}}{\partial z \partial \tilde{z}^{3}}, \quad a=x,y, $$

(78)

with $\kappa ^{4}= \frac{\rho _{\text{\tiny b}}\varphi }{E_{\text{\tiny b}}I_{\text{\tiny b}}} \omega ^{2}$ and $B_{\text{\tiny b}}= \frac{\mu _{\text{\tiny b}}A_{\text{\tiny b}}}{E_{\text{\tiny b}}I_{\text{\tiny b}}}-3=-4$. If $U^{0}_{a}$ depends on $z$ (and we want to show that it is not the case), then $U^{0}_{a}$ solution of (22) reads $U^{0}_{a}({\mathbf{x}},\tilde{z})=A_{a}({\mathbf{x}})\sin \kappa \tilde{z}+B_{a}({ \mathbf{x}})\cos \kappa \tilde{z}$. Integrating (89) over $\tilde{z}\in (0,2\pi /\kappa )$ after multiplication by $\cos \kappa \tilde{z}$ (resp. by $\sin \kappa \tilde{z}$) and integrating by parts four times provides $\frac{\partial }{\partial z}A_{a}=0$ (resp. $\frac{\partial }{\partial z}B_{a}=0$), which allows us to conclude that

$$ \frac{\partial U_{x}^{0}}{\partial z}= \frac{\partial U_{y}^{0}}{\partial z}=0. $$

Appendix B: The Case of a Two-Dimensional Array of Plates

We provide in this section the effective model for the alternative situation of a periodic array of identical plates supported by a soil substrate. We transpose the notations of the beams case, assuming that the plates have a height $h_{\text{\tiny b}}$, a thickness $2r_{\text{\tiny b}}$ and that the periodicity of the array is $\ell$. We denote $(\lambda _{\text{\tiny b}},\mu _{\text{\tiny b}},\rho _{\text{\tiny b}})$ the material parameters of the plates and $(\lambda _{\text{\tiny s}},\mu _{\text{\tiny s}},\rho _{\text{\tiny s}})$ those of the substrate. The same scaling as for the beams case is used, see (7). For in-plane incident wave, the resulting displacement fields remain in-plane with

$$ U_{y}=0,\qquad \frac{\partial U_{x}}{\partial y}= \frac{\partial U_{z}}{\partial y}=0,\qquad \sigma _{xy}=\sigma _{yz}=0. $$

Following the same asymptotic procedure as for the beams, we derive an effective model for the array of plates. We provide below the main results of the derivation. With ${\mathbf{x}}=(x,z)$, the equivalent of (14) in the region of the plates and far from their extremities reads

$$ \left \{ \textstyle\begin{array}{l@{\quad }l} \displaystyle V_{x}^{0}=U_{x}^{0}(x,\tilde{z}), \ & V_{z}^{0}= U_{z}^{0}({\mathbf{x}}), \\ V_{x}^{1}=U_{x}^{1}({\mathbf{x}},\tilde{z}),& \displaystyle V_{z}^{1}= U_{z}^{1}({ \mathbf{x}})-\frac{\partial }{\partial \tilde{z}}U_{x}^{0}\,\hat{x}, \\ (U_{x}^{0},U_{x}^{1})\quad \tilde{z}-\text{periodic}, \end{array}\displaystyle \right .$$

(79)

see also forthcoming Remark (87). From $(\text{E}_{a})^{-2}$, $(\text{E}_{z})^{-2}$ and from $(\text{C})^{0}$ in (12), we have $\frac{\partial }{\partial \hat{x}}\varSigma _{xx}^{0}= \frac{\partial }{\partial \hat{x}}\varSigma _{xz}^{0}=0$ with ${\boldsymbol{\varSigma }}^{0}={\mathsf{A}}\,{\boldsymbol{\varepsilon }}^{\hat{x}}\big ({\mathbf{V}}^{2} \big )+{\mathsf{A}}\,{\boldsymbol{\varepsilon }}^{\tilde{z}}\big ({\mathbf{V}}^{1}\big )+{ \mathsf{A}}\,{\boldsymbol{\varepsilon }}^{{\mathbf{x}}}\big ({\mathbf{V}}^{0}\big )$, and the system on $({\boldsymbol{\varSigma }}^{0},{\mathbf{V}}^{2})$ has again explicit solutions which reads

$$ \left \{ \textstyle\begin{array}{l} V_{x}^{2}=\displaystyle U_{x}^{2}({\mathbf{x}},\tilde{z})- \left ( \frac{\partial U_{x}^{0}}{\partial x} + \frac{\lambda _{\text{\tiny b}}}{\lambda _{\text{\tiny b}}+2\mu _{\text{\tiny b}}} \frac{\partial U_{z}^{0}}{\partial z}\right )\hat{x}+ \frac{\lambda _{\text{\tiny b}}}{\lambda _{\text{\tiny b}}+2\mu _{\text{\tiny b}}} \frac{\partial ^{2} U_{x}^{0}}{\partial \tilde{z}^{2}} \frac{\hat{x}^{2}}{2}, \\ V_{z}^{2}=\displaystyle U_{z}^{2}({\mathbf{x}},\tilde{z})-\left ( \frac{\partial U_{x}^{1}}{\partial \tilde{z}}+ \frac{\partial U_{z}^{0}}{\partial x}+{ \frac{\partial U_{x}^{0}}{\partial z}} \right )\hat{x}, \end{array}\displaystyle \right .$$

and the corresponding stress tensor ${\boldsymbol{\varSigma }}^{0}$ by

$$ \displaystyle \varSigma _{xx}^{0}=\varSigma _{xz}^{0}=0,\quad \varSigma _{zz}^{0}= \displaystyle E_{\text{\tiny b}}^{*}\left ( \frac{\partial U_{z}^{0}}{\partial z}- \frac{\partial ^{2} U_{x}^{0}}{\partial \tilde{z}^{2}}\hat{x}\right ), \qquad E_{\text{\tiny b}}^{*}= \frac{4\mu _{\text{\tiny b}}(\lambda _{\text{\tiny b}}+\mu _{\text{\tiny b}})}{\lambda _{\text{\tiny b}}+2\mu _{\text{\tiny b}}}= \frac{E_{\text{\tiny b}}}{1-\nu _{\text{\tiny b}}^{2}}, $$

(80)

which is the two-dimensional version of (58)-(59). The axial stress is $N_{zz}^{0}= \int _{\hat{\mathsf{s}}_{\text{\tiny b}}}\varSigma _{zz}^{0} \text{d}\hat{x}$ with now $\hat{\mathsf{s}}_{\text{\tiny b}}=\{\hat{x}\in \left (-\hat{r}_{\text{\tiny b}},\hat{r}_{\text{\tiny b}}\right )\}$ and the equilibrium is deduced from the two-dimensional version of $(\text{E}_{z})^{0}$ in (12) which reads $\frac{\partial }{\partial \hat{x}}\varSigma _{xz}^{2}+ \frac{\partial }{\partial \tilde{z}}\varSigma _{zz}^{1}+ \frac{\partial }{\partial z}\varSigma _{zz}^{0}+\rho _{\text{\tiny b}} \omega ^{2}U_{z}^{0}=0$

$$ N_{zz}^{0}({\mathbf{x}})=\displaystyle 2 E_{\text{\tiny b}}^{*}\hat{r}_{\text{\tiny b}}\,\frac{\partial U_{z}^{0}}{\partial z}({\mathbf{x}}), \qquad \frac{\partial N_{zz}^{0}}{\partial z}({\mathbf{x}})+2\rho _{\text{\tiny b}}\omega ^{2}\hat{r}_{\text{\tiny b}}U_{z}^{0}({\mathbf{x}})=0, $$

(81)

being the equivalent of (17). For the flexural motions, (i) we obtain the bending moment $M_{x}^{0}$ by integrating $\varSigma _{zz}^{0}$, in (80), over $\hat{x}\in \hat{\mathsf{s}}_{\text{\tiny b}}$ after multiplication by $\hat{x}$, (ii) next, from (80), we use that $\varSigma _{xz}^{0}=0$, (iii) eventually we integrate $\frac{\partial }{\partial \hat{x}}\varSigma _{xx}^{2}+ \frac{\partial }{\partial \tilde{z}}\varSigma _{xz}^{1}+\rho _{\text{\tiny b}}\omega ^{2} V_{x}^{0}=0$ over $\hat{x}\in \hat{\mathsf{s}}_{\text{\tiny b}}$. We get

$$ M_{x}^{0}(x,\tilde{z})=E_{\text{\tiny b}}^{*} \frac{2\hat{r}_{\text{\tiny b}}^{3}}{3} \frac{\partial ^{2} U_{x}^{0}}{\partial \tilde{z}^{2}}(x,\tilde{z}), \displaystyle \quad T_{x}^{0}=0, \quad \frac{\partial T_{x}^{1}}{\partial \tilde{z}}(x,\tilde{z})+2\rho _{\text{\tiny b}}\omega ^{2}\hat{r}_{\text{\tiny b}}\; U_{x}^{0}(x,\tilde{z})=0. $$

(82)

The relation between the bending moment and the shear force is obtained from the two-dimensional version of $(\text{E}_{z})^{-1}$ in (12), which reads $\frac{\partial }{\partial \hat{x}}\varSigma _{xz}^{1}+ \frac{\partial }{\partial \tilde{z}}\varSigma _{zz}^{0}=0$ that we integrate over $\hat{x}\in \hat{\mathsf{s}}_{\text{\tiny b}}$ after multiplication by $\hat{x}$. It results that

$$ \frac{\partial M_{x}^{0}}{\partial \tilde{z}}(x,\tilde{z})+T_{x}^{1}(x, \tilde{z})=0,$$

(83)

as in (21), which along with the above relation provides the wave equation for the flexural motion $U_{x}$ in the form

$$ \frac{\partial ^{4} U_{x}^{0}}{\partial \tilde{z}^{4} }(x,\tilde{z}) - \tilde{\kappa }^{4}\, U_{x}^{0}(x,\tilde{z})=0, \quad \text{with } \tilde{\kappa }=\left ( \frac{3\rho _{\text{\tiny b}}\omega ^{2}}{E_{\text{\tiny b}}^{*}\hat{r}_{\text{\tiny b}}^{2}} \right )^{1/4}. $$

(84)

At the intermediate scale near the ends, the displacement $\widetilde{U}{_{x}^{0}}(x,\tilde{z})$ is not $\tilde{z}$-periodic anymore. We obtain the same flexural equation as in (82)-(84), namely

$$ \left \{ \textstyle\begin{array}{l} \displaystyle \widetilde{M}{_{x}^{0}}_{|\,z^{*}}(x,\tilde{z})=E_{\text{\tiny b}}^{*}\frac{2\hat{r}_{\text{\tiny b}}^{3}}{3} \frac{\partial ^{2} \widetilde{U}{_{x}^{0}}_{|\,z^{*}}}{\partial \tilde{z}^{2}}(x, \tilde{z}),\quad \frac{\partial \widetilde{M}{_{x}^{0}}_{|\,z^{*}}}{\partial \tilde{z}}(x, \tilde{z})+\widetilde{T}{_{x}^{1}}_{|\,z^{*}}(x,\tilde{z})=0, \\ \displaystyle \frac{\partial \widetilde{T}{_{x}^{1}}_{|\,z^{*}}}{\partial \tilde{z}}(x, \tilde{z})+2\rho _{\text{\tiny b}}\omega ^{2}\hat{r}_{\text{\tiny b}}\; \widetilde{U}{_{x}^{0}}_{|\,z^{*}}(x,\tilde{z})=0, \end{array}\displaystyle \right .$$

with $\tilde{z}\in (0,+\infty )$ if $z^{*}=0$ and $\tilde{z}\in (-\infty ,0)$ if $z^{*}=h_{\text{\tiny b}}$. When $|\tilde{z}|\rightarrow \infty$, the fields $( \widetilde{U}{_{x}^{0}}_{|\,z^{*}},\widetilde{M}{_{x}^{0}}_{|\,z^{*}}, \widetilde{T}{_{x}^{1}}_{|\,z^{*}})$ match their $\tilde{z}$-periodic counterparts $( {U}{_{x}^{0}},{M}{_{x}^{0}}, {T}{_{x}^{1}})$ defined inside the plates. Near $\tilde{z}=0$, the boundary conditions are obtained in the same way as the matchings developed in §3.2 for beams. Doing so, we obtain

$$ \left \{ \textstyle\begin{array}{l} u^{0}_{x}(x,0)=\widetilde{U}{_{x}^{0}}_{|0}(x,0),\quad \displaystyle \frac{\partial \widetilde{U}{_{x}^{0}}_{|0}}{\partial \tilde{z}}(x,0)=0, \\ \widetilde{T}{_{x}^{1}}_{|\,h_{\text{\tiny b}}}(x,0)=0, \quad \widetilde{M}{_{x}^{0}}_{|\,h_{\text{\tiny b}}}(x,0)=0, \\ \sigma ^{0}_{xz}(x,0)=\hat{\ell }^{-1}\widetilde{T}{_{x}^{0}}_{|0}(x,0)=0, \quad \displaystyle \sigma ^{1}_{xz}(x,0)=\hat{\ell }^{-1}\widetilde{T}{_{x}^{1}}_{|0}(x,0). \end{array}\displaystyle \right . $$

Similarly, we get the boundary conditions for the longitudinal wave equation (81)

$$ u^{0}_{z}(x,0)={U}{_{z}^{0}}(x,0), \quad \sigma ^{0}_{zz}(x,0)= \hat{\ell }^{-1}N{^{0}_{zz}}(x,0),\quad N_{zz}^{0}(x,h_{\text{\tiny b}})=0. $$

As in §3.3, we construct the problem at the dominant order for the displacement $U_{z}({\mathbf{x}})=U_{z}^{0}({\mathbf{x}})$ and $U_{x}({\mathbf{x}})=\left (\widetilde{U}{_{x}^{0}}_{|0}(x,\tilde{z}),U_{x}(x, \tilde{z}),\widetilde{U}{_{x}^{0}}_{|\,h_{\text{\tiny b}}}(x,\tilde{z}) \right )$ accounting for $\tilde{z}=z/\eta$ and $r_{\text{\tiny b}}=\eta ^{2}\hat{r}_{\text{\tiny b}}$. Gathering the results in the inner region of the plates and the intermediate region near the extremities, we get the final set of homogenized equations for $z\in (0,h_{\text{\tiny b}})$ as follows. In the substrate, the displacement field is governed by the classical balance of linear momentum equation and constitutive equation of the substrate. The array of the plates is replaced by frequency dependent boundary conditions which encapsulate both the longitudinal and flexural resonant motions of the plates

$$ \left \{ \textstyle\begin{array}{l} {\mathrm{div}}{\boldsymbol{\sigma }}+\rho _{\text{\tiny s}}\omega ^{2} {\mathbf{u}}={ \mathbf{0}}, \quad {\boldsymbol{\sigma }}=2\mu _{\text{\tiny s}}{\boldsymbol{\varepsilon }}+ \lambda _{\text{\tiny s}}\text{tr}({\boldsymbol{\varepsilon }}) {\mathsf{\boldsymbol{I}}}, \quad \text{for } z\in (-\infty ,0), \\ \displaystyle \sigma _{xz}(x,0)=\mu _{\text{\tiny s}}k_{\text{\tiny T}}\,f_{\text{\tiny F}}(\omega ,h_{\text{\tiny b}})\,u_{x}(x,0),\qquad \displaystyle \sigma _{zz}(x,0)=\mu _{\text{\tiny s}}k_{\text{\tiny T}}\,f_{\text{\tiny L}}(\omega ,h_{\text{\tiny b}})\,u_{z}(x,0), \\ \end{array}\displaystyle \right .$$

(85)

and

$$ \left \{ \textstyle\begin{array}{l@{\quad }l} \displaystyle f_{\text{\tiny F}}(\omega ,h_{\text{\tiny b}})=\varphi \; \frac{\rho _{\text{\tiny b}}}{\rho _{\text{\tiny s}}} \frac{k_{\text{\tiny T}}}{\kappa }\; \frac{\text{sh}\kappa h_{\text{\tiny b}}\cos \kappa h_{\text{\tiny b}}+\text{ch}\kappa h_{\text{\tiny b}}\sin \kappa h_{\text{\tiny b}}}{1+\cos \kappa h_{\text{\tiny b}}\text{ch}\kappa h_{\text{\tiny b}}}, & \displaystyle \kappa =\left ( \frac{3\rho _{\text{\tiny b}}\omega ^{2}}{E_{\text{\tiny b}}^{*}r_{\text{\tiny b}}^{2}} \right )^{1/4}, \\ \displaystyle f_{\text{\tiny L}}(\omega ,h_{\text{\tiny b}})=\varphi \; \frac{\rho _{\text{\tiny b}}}{\rho _{\text{\tiny s}}} \frac{k_{\text{\tiny T}}}{K} \;\tan Kh_{\text{\tiny b}}, &\displaystyle K= \sqrt{\frac{\rho _{\text{\tiny b}}}{E_{\text{\tiny b}}^{*}}}\,\omega , \end{array}\displaystyle \right .$$

with

$$ \varphi =\frac{2r_{\text{\tiny b}}}{\ell },\quad E_{\text{\tiny b}}^{*}= \frac{4\mu _{\text{\tiny b}}(\lambda _{\text{\tiny b}}+\mu _{\text{\tiny b}})}{\lambda _{\text{\tiny b}}+2\mu _{\text{\tiny b}}}= \frac{E_{\text{\tiny b}}}{1-\nu _{\text{\tiny b}}^{2}}. $$

(86)

Eventually, as in the three-dimensional case, once the effective problem (85) in the substrate has been resolved, the solution in the region of the plates can be post-processed, as $U_{x}$ and $U_{z}$ in the region of the plates are solutions to

$$ \left \{ \textstyle\begin{array}{l} \displaystyle \frac{\partial ^{4} U_{x}}{\partial z^{4} } -\kappa ^{4} \, U_{x}=0,\quad \displaystyle U_{x}(x,0)=u_{x}(x,0), \quad \frac{\partial U_{x}}{\partial z}(x,0)= \frac{\partial ^{2} U_{x}}{\partial z ^{2}}(x,h_{\text{\tiny b}})= \frac{\partial ^{3} U_{x}}{\partial z^{3} }(x,h_{\text{\tiny b}})=0. \\ \displaystyle \frac{\partial ^{2} U_{z}}{\partial z ^{2}}+K^{2} U_{z}=0, \quad \displaystyle U_{z}(x,0)=u_{z}(x,0),\quad \frac{\partial U_{z}}{\partial z}(x,h_{\text{\tiny b}})=0. \end{array}\displaystyle \right .$$

Solving these two problems for which the inputs are the prescribed displacements ${\mathbf{u}}(x,0)$ at the junction with the substrate gives

$$ \displaystyle U_{x}({\mathbf{x}})=u_{x}(x,0)V_{\text{\tiny F}}(z), \qquad U_{z}({\mathbf{x}})=u_{z}(x,0)V_{\text{\tiny L}}(z), \qquad \displaystyle \text{with } \displaystyle V_{\text{\tiny L}}(z), V_{\text{\tiny F}}(z) \text{ defined in (6).} $$

Remark 1

The expressions of the displacements in (79) have been obtained using

$$ p_{1} : \frac{\partial U_{z}^{1}}{\partial \tilde{z}}=0, \quad p_{4} : \frac{\partial U_{x}^{0}}{\partial z}=0, $$

(87)

being the equivalent of (15). A priori we have $V_{x}^{0}=U_{x}^{0}({\mathbf{x}},\tilde{z})$, $V_{z}^{0}= U_{z}^{0}({\mathbf{x}})$ and $V_{x}^{1}=U_{x}^{1}({\mathbf{x}},\tilde{z})$, $V_{z}^{1}= U_{z}^{1}({\mathbf{x}},\tilde{z})- \frac{\partial }{\partial \tilde{z}}U_{x}^{0}\,\hat{x}$, eventually $\varSigma _{zz}^{0}= E_{\text{\tiny b}}^{*}\left ({ \frac{\partial U_{z}^{1}}{\partial \tilde{z}}}+ \frac{\partial U_{z}^{0}}{\partial z}- \frac{\partial ^{2} U_{x}^{0}}{\partial \tilde{z}^{2}}\hat{x}\right )$. The proof of $p_{1}$ is similar to that of $P_{1}$ provided in A.2 using the two-dimensional version of $(\text{E}_{z})^{-1}$ in (12), $\frac{\partial }{\partial \hat{x}}\varSigma _{xz}^{1}+ \frac{\partial }{\partial \tilde{z}}\varSigma _{zz}^{0}=0$, that we integrate over $\hat{x}\in \hat{\mathsf{s}}_{\text{\tiny b}}$ and we get $\frac{\partial ^{2} U_{z}^{1}}{\partial \tilde{z}^{2}}=0$ hence $p_{1}$ as a linear dependence of $U_{z}^{1}$ on $\tilde{z}$ is prevented by its $\tilde{z}$-periodicity. The proof of $p_{4}$ – We start with the two-dimensional version of (71) (hence $\mathrm{div}_{{\hat{\mathbf{x}}}'}\to {\mathrm{div}}_{\hat{x}}$ and ($\varSigma ^{1}_{xx},\varSigma _{xz}^{1},\varSigma _{zz}^{1}$) given by $(\text{C})^{1}$ in (12)). The problem can be solved on $(\varSigma _{xx}^{1},\varSigma _{zz}^{1},V^{3}_{x})$ then on $(\varSigma ^{1}_{xz},V_{z}^{3})$ resulting in

$$ \left \{ \textstyle\begin{array}{l} \displaystyle \varSigma _{xx}^{1}=0,\qquad \varSigma _{xz}^{1}= \displaystyle \frac{E_{\text{\tiny b}}^{*}}{2} \frac{\partial ^{3} U_{x}^{0}}{\partial \tilde{z}^{3} }\left (\hat{x}^{2}- \hat{r}_{\text{\tiny b}}^{2}\right ), \\ \displaystyle \varSigma _{zz}^{1}=\displaystyle E_{\text{\tiny b}}^{*} \left (\frac{\partial U_{z}^{2}}{\partial \tilde{z}}+ \frac{\partial U_{z}^{1}}{\partial z}- \frac{\partial }{\partial \tilde{z}}\left ( \frac{\partial U_{x}^{1}}{\partial \tilde{z}}+2 \frac{\partial U_{x}^{0}}{\partial z}\right )\hat{x}\right ), \end{array}\displaystyle \right .$$

and

$$ \left \{ \textstyle\begin{array}{l@{\quad }l} V_{x}^{3}=&\displaystyle U_{x}^{3}({\mathbf{x}},\tilde{z})- \\ &\displaystyle \left (\frac{\partial U_{x}^{1}}{\partial x} + \frac{\lambda _{\text{\tiny b}}}{\lambda _{\text{\tiny b}}+2\mu _{\text{\tiny b}}} \left (\frac{\partial U_{z}^{2}}{\partial \tilde{z}}+ \frac{\partial U_{z}^{1}}{\partial z}\right )\right )\hat{x}+ \frac{\lambda _{\text{\tiny b}}}{\lambda _{\text{\tiny b}}+2\mu _{\text{\tiny b}}} \frac{\partial }{\partial \tilde{z}}\left ( \frac{\partial U_{x}^{1}}{\partial \tilde{z}}+2 \frac{\partial U_{x}^{0}}{\partial z}\right )\frac{\hat{x}^{2}}{2}, \\ V_{z}^{3}=& \displaystyle U_{z}^{3}({\mathbf{x}},\tilde{z})-\left ( \frac{\partial U_{x}^{2}}{\partial \tilde{z}}+ \frac{\partial U_{z}^{1}}{\partial x}+{ \frac{\partial U_{x}^{1}}{\partial z}} \right )\hat{x}+ \frac{\partial }{\partial \tilde{z}}\left ( \frac{\partial U^{0}_{x}}{\partial x}\hat{x}^{2}+ \frac{\partial ^{2} U_{x}^{0}}{\partial \tilde{z}^{2}}\left (\alpha \hat{x}^{3}-\beta \hat{r}_{\text{\tiny b}}^{2}\hat{x}\right ) \right ), \end{array}\displaystyle \right .$$

where $\alpha = \frac{3\lambda _{\text{\tiny b}}+4\mu _{\text{\tiny b}}}{6(\lambda _{\text{\tiny b}}+2\mu _{\text{\tiny b}})}$ and $\beta = \frac{2(\lambda _{\text{\tiny b}}+\mu _{\text{\tiny b}})}{\lambda _{\text{\tiny b}}+2\mu _{\text{\tiny b}}}$. To conclude the proof of $P_{4}$, we use the procedure conducted in A.5.2. First, we obtain $M^{1}_{x}$ by integrating $\hat{x}\varSigma _{zz}^{1}$ over $\hat{x}\in \hat{\mathsf{s}}_{\text{\tiny b}}$. Next, the equilibrium is obtained integrating $\frac{\partial }{\partial \hat{x}}\varSigma _{xx}^{3}+ \frac{\partial }{\partial \tilde{z}}\varSigma _{xz}^{2}+ \frac{\partial }{\partial z}\varSigma _{xz}^{1}+\rho _{\text{\tiny b}} \omega ^{2} V_{x}^{1}=0$ over $\hat{x}\in \hat{\mathsf{s}}_{\text{\tiny b}}$. Eventually, we use $\frac{\partial }{\partial \hat{x}}\varSigma _{xz}^{2}+ \frac{\partial }{\partial \tilde{z}}\varSigma _{zz}^{1}+ \frac{\partial }{\partial z}\varSigma _{zz}^{0}+\rho _{\text{\tiny b}} \omega ^{2} V_{z}^{0}=0$ that we integrate over $\hat{x}\in \hat{\mathsf{s}}_{\text{\tiny b}}$ after multiplication by $\hat{x}$. We obtain

$$ \left \{ \textstyle\begin{array}{l} \displaystyle M^{1}_{x}=\frac{2}{3}E_{\text{\tiny b}}^{*}\hat{r}_{\text{\tiny b}}^{3}\frac{\partial }{\partial \tilde{z}}\left ( \frac{\partial U_{x}^{1}}{\partial \tilde{z}}+2 \frac{\partial U^{0}_{x}}{\partial z}\right ),\quad \displaystyle \frac{\partial T_{x}^{2}}{\partial \tilde{z}}+2\rho _{\text{\tiny b}} \omega ^{2}\hat{r}_{\text{\tiny b}}U_{x}^{1}- \frac{2}{3}E_{\text{\tiny b}}^{*}\hat{r}_{\text{\tiny b}}^{3} \frac{\partial ^{4} U^{0}_{x}}{\partial z \partial \tilde{z}^{3}}=0, \\ \displaystyle T^{2}_{x}+\frac{\partial M_{x}^{1}}{\partial \tilde{z}}+ \frac{2}{3}E_{\text{\tiny b}}^{*}\hat{r}_{\text{\tiny b}}^{3} \frac{\partial ^{3} U_{x}^{0}}{\partial z \partial \tilde{z}^{2}}=0, \end{array}\displaystyle \right .$$

(88)

which leads to

$$ \frac{\partial ^{4} U^{1}_{x}}{\partial \tilde{z}^{4} }-\kappa ^{4} U_{x}^{1}=-4 \frac{\partial ^{4} U_{x}^{0}}{\partial z \partial \tilde{z}^{3}}, $$

(89)

as in (89). The conclusion is identical: $\frac{\partial U_{x}^{0}}{\partial z}=0$.