After integration by parts of the last two terms in the above inequality with respect to $x$, we obtain

$$
\begin{align*}
& \frac{\rho_f}{R} \int_0^t \int_{\Omega_0} |\partial_t v_\varepsilon|^2 dxdyds + \rho_s \int_0^t \int_0^L |\partial_{tt} \zeta_\varepsilon|^2 dxd s + \rho_s \int_0^t \int_0^L |\partial_{tt} \eta_\varepsilon|^2 dxd s + \gamma \int_0^t \int_0^L |\partial_{xxt} \eta_\varepsilon(t)|^2 dxd s \\
& \qquad + \frac{1}{2} \left( \frac{\mu\lambda(R)}{2} \int_{\Omega_0} |\nabla v_\varepsilon(t)|^2 dxdy + \varepsilon \int_0^L |\partial_{xxt} \zeta_\varepsilon(t)|^2 dx + \varepsilon \int_0^L |\partial_{xxt} \eta_\varepsilon(t)|^2 dx \right) \\
& \leq \beta \int_0^t \int_0^L |\partial_{xt} \zeta_\varepsilon|^2 dxd s + \alpha \int_0^t \int_0^L |\partial_{xxt} \eta_\varepsilon|^2 dxd s + \int_0^t \int_{\Omega_0} f \cdot \partial_t v_\varepsilon dxdyds \\
& \quad + \int_0^t \int_{\Omega_0} \operatorname{div} h \cdot \partial_t v_\varepsilon dxdyds - \int_0^t \int_0^L (h e_2)_{|y=1} \cdot (\partial_{tt} \zeta_\varepsilon, \partial_{tt} \eta_\varepsilon) dxd s + \int_0^t \int_0^L (\ell, m) \cdot (\partial_{tt} \zeta_\varepsilon, \partial_{tt} \eta_\varepsilon) dxd s \\
& \quad + \frac{1}{2} \left( \mu K(R) \int_{\Omega_0} |\nabla v_\varepsilon|^2 dxdy + \varepsilon \int_0^L |\partial_{xx} \zeta_\varepsilon|^2 dx + \varepsilon \int_0^L |\partial_{xx} \eta_\varepsilon|^2 dx \right) \\
& \quad - \beta \int_0^L \partial_x \zeta_\varepsilon(t) \partial_{xt} \zeta_\varepsilon(t) dx - \alpha \int_0^L \partial_{xx} \eta_\varepsilon(t) \partial_{xxt} \eta_\varepsilon(t) dx - \beta \int_0^L \partial_{xx} \zeta_0^\varepsilon \zeta_1^\varepsilon dx - \alpha \int_0^L \partial_{xxx} \eta_0^\varepsilon \partial_x \eta_1^\varepsilon dx.
\end{align*}
$$

Thanks to the way we have built $(v_0^\varepsilon, \zeta_0^\varepsilon, \zeta_1^\varepsilon, \eta_0^\varepsilon, \eta_1^\varepsilon)$, by Cauchy Schwartz, Young's and trace inequalities there exists $C_0$ which depends on the initial data and $R$ such that

$$
\begin{align*}
& \frac{\rho_f}{2R} \int_0^t \int_{\Omega_0} |\partial_t v_\varepsilon|^2 dxdyds + \frac{\rho_s}{2} \int_0^t \int_0^L |\partial_{tt} \zeta_\varepsilon|^2 dxd s + \frac{\rho_s}{2} \int_0^t \int_0^L |\partial_{tt} \eta_\varepsilon|^2 dxd s \\
& \qquad + \gamma \int_0^t \int_0^L |\partial_{xxt} \eta_\varepsilon(t)|^2 dxd s + \frac{\mu\lambda(R)}{4} \int_{\Omega_0} |\nabla v_\varepsilon(t)|^2 dxdy \\
& \qquad + \frac{\varepsilon}{2} \int_0^L |\partial_{xxt} \zeta_\varepsilon(t)|^2 dx + \frac{\varepsilon}{4} \int_0^L |\partial_{xxt} \eta_\varepsilon(t)|^2 dx \\
& \leq \beta \int_0^t \int_0^L |\partial_{xt} \zeta_\varepsilon|^2 dxd s + \alpha \int_0^t \int_0^L |\partial_{xxt} \eta_\varepsilon|^2 dxd s + \frac{\beta^2}{\varepsilon} \int_0^L |\partial_x \zeta_\varepsilon(t)|^2 dx \\
& \quad + \frac{\varepsilon}{4} \int_0^L |\partial_{xt} \zeta_\varepsilon(t)|^2 dx + \frac{\alpha^2}{\varepsilon} \int_0^L |\partial_{xx} \eta_\varepsilon(t)|^2 dx + C(R) \left( \|\operatorname{div} h\|_{L^2(0,t;L^2_\#(\Omega_0))}^2 + \|f\|_{L^2(0,t;L^2_\#(\Omega_0))}^2 \right) \\
& \quad + C(\rho_s, \Omega_0) \left( \|h\|_{L^2(0,t;H^1_\#(\Omega_0))}^2 + \|\ell\|_{L^2(0,t;L^2_\#(0,L))}^2 + \|m\|_{L^2(0,t;L^2_\#(0,L))}^2 \right) + C_0.
\end{align*}
$$

Thanks to the energy estimates (3.26), we obtain

$$
\begin{align}
& \frac{\rho_f}{2R} \|\partial_t v_\varepsilon\|_{L^2(0,t;L^2_\#(\Omega_0))}^2 + \frac{\rho_s}{2} \|\partial_{tt} \zeta_\varepsilon\|_{L^2(0,t;L^2_\#(\Omega_0))}^2 + \frac{\rho_s}{2} \|\partial_{tt} \eta_\varepsilon\|_{L^2(0,t;L^2_\#(\Omega_0))}^2 \nonumber \\
& \qquad + \gamma \|\partial_{xxt} \eta_\varepsilon(t)\|_{L^2(0,t;L^2_\#(\Omega_0))}^2 + \frac{\mu\lambda(R)}{4} \|\nabla v_\varepsilon(t)\|_{L^2_\#(\Omega_0)}^2 \nonumber \\
& \qquad + \frac{\varepsilon}{4} \|\partial_{xxt} \zeta_\varepsilon(t)\|_{L^2_\#(0,L)}^2 + \frac{\varepsilon}{4} \|\partial_{xxt} \eta_\varepsilon(t)\|_{L^2_\#(0,L)}^2 \le C(R) \left( \|\operatorname{div} h\|_{L^2(0,t;L^2_\#(\Omega_0))}^2 + \|f\|_{L^2(0,t;L^2_\#(\Omega_0))}^2 \right) \nonumber \\
& \qquad + C(\rho_s, \Omega_0) \left( \|h\|_{L^2(0,t;H^1_\#(\Omega_0))}^2 + \|\ell\|_{L^2(0,t;L^2_\#(0,L))}^2 + \|m\|_{L^2(0,t;L^2_\#(0,L))}^2 \right) + C_0 + C_\varepsilon, \tag{3.27}
\end{align}
$$

where $C_\varepsilon$ depends on initial data, source terms and $1/\varepsilon$.

Consequently by applying Gronwall Lemma, we obtain bounds on the solution in the following spaces:

• $v_\varepsilon$ in $H^1(0, T; L^2_\#(\Omega_0)) \cap L^\infty(0, T; H^1_\#(\Omega_0))$;

• $\zeta_\varepsilon$ in $H^2(0, T; L^2_\#(0, L))$, $\eta_\varepsilon$ in $H^2(0, T; H^1_\#(0, L))$;

• $\sqrt{\varepsilon}\zeta_\varepsilon$ in $W^{1,\infty}(0, T; H^2_\#(0, L))$, $\sqrt{\varepsilon}\eta_\varepsilon$ in $W^{1,\infty}(0, T; H^2_\#(0, L))$;
After integration by parts of the last two terms in the above inequality with respect to $x$, we obtain

$$
\begin{align*}
& \frac{\rho_f}{R} \int_0^t \int_{\Omega_0} |\partial_t v_\varepsilon|^2 dxdyds + \rho_s \int_0^t \int_0^L |\partial_{tt} \zeta_\varepsilon|^2 dxd s + \rho_s \int_0^t \int_0^L |\partial_{tt} \eta_\varepsilon|^2 dxd s + \gamma \int_0^t \int_0^L |\partial_{xxt} \eta_\varepsilon(t)|^2 dxd s \\
& \qquad + \frac{1}{2} \left( \frac{\mu\lambda(R)}{2} \int_{\Omega_0} |\nabla v_\varepsilon(t)|^2 dxdy + \varepsilon \int_0^L |\partial_{xxt} \zeta_\varepsilon(t)|^2 dx + \varepsilon \int_0^L |\partial_{xxt} \eta_\varepsilon(t)|^2 dx \right) \\
& \leq \beta \int_0^t \int_0^L |\partial_{xt} \zeta_\varepsilon|^2 dxd s + \alpha \int_0^t \int_0^L |\partial_{xxt} \eta_\varepsilon|^2 dxd s + \int_0^t \int_{\Omega_0} f \cdot \partial_t v_\varepsilon dxdyds \\
& \qquad + \int_0^t \int_{\Omega_0} \operatorname{div} h \cdot \partial_t v_\varepsilon dxdyds - \int_0^t \int_0^L (h e_2)_{|y=1} \cdot (\partial_{tt} \zeta_\varepsilon, \partial_{tt} \eta_\varepsilon) dxd s + \int_0^t \int_0^L (\ell, m) \cdot (\partial_{tt} \zeta_\varepsilon, \partial_{tt} \eta_\varepsilon) dxd s \\
& \qquad + \frac{1}{2} \left( \mu K(R) \int_{\Omega_0} |\nabla v_\varepsilon|^2 dxdy + \varepsilon \int_0^L |\partial_{xx} \zeta_\varepsilon|^2 dx + \varepsilon \int_0^L |\partial_{xx} \eta_\varepsilon|^2 dx \right) \\
& \quad - \beta \int_0^L \partial_x \zeta_\varepsilon(t) \partial_{xt} \zeta_\varepsilon(t) dx - \alpha \int_0^L \partial_{xx} \eta_\varepsilon(t) \partial_{xxt} \eta_\varepsilon(t) dx - \beta \int_0^L \partial_{xx} \zeta_0^\varepsilon \zeta_1^\varepsilon dx - \alpha \int_0^L \partial_{xxx} \eta_0^\varepsilon \partial_x \eta_1^\varepsilon dx.
\end{align*}
$$

Thanks to the way we have built ($v_0^\varepsilon, \zeta_0^\varepsilon, \zeta_1^\varepsilon, \eta_0^\varepsilon, \eta_1^\varepsilon$), by Cauchy Schwartz, Young's and trace inequalities there exists $C_0$ which depends on the initial data and $R$ such that

$$
\begin{align*}
& \frac{\rho_f}{2R} \int_0^t \int_{\Omega_0} |\partial_t v_\varepsilon|^2 dxdyds + \frac{\rho_s}{2} \int_0^t \int_0^L |\partial_{tt} \zeta_\varepsilon|^2 dxd s + \frac{\rho_s}{2} \int_0^t \int_0^L |\partial_{tt} \eta_\varepsilon|^2 dxd s \\
& \qquad + \gamma \int_0^t \int_0^L |\partial_{xxt} \eta_\varepsilon(t)|^2 dxd s + \frac{\mu\lambda(R)}{4} \int_{\Omega_0} |\nabla v_\varepsilon(t)|^2 dxdy \\
& \qquad + \frac{\varepsilon}{2} \int_0^L |\partial_{xxt} \zeta_\varepsilon(t)|^2 dx + \frac{\varepsilon}{4} \int_0^L |\partial_{xxt} \eta_\varepsilon(t)|^2 dx \\
& \leq \beta \int_0^t \int_0^L |\partial_{xt} \zeta_\varepsilon|^2 dxd s + \alpha \int_0^t \int_0^L |\partial_{xxt} \eta_\varepsilon|^2 dxd s + \frac{\beta^2}{\varepsilon} \int_0^L |\partial_x \zeta_\varepsilon(t)|^2 dx \\
& \qquad + \frac{\varepsilon}{4} \int_0^L |\partial_{xt} \zeta_\varepsilon(t)|^2 dx + \frac{\alpha^2}{\varepsilon} \int_0^L |\partial_{xx} \eta_\varepsilon(t)|^2 dx + C(R) \left( \|\operatorname{div} h\|_{L^2(0,t;L^2_\#(\Omega_0))}^2 + \|f\|_{L^2(0,t;L^2_\#(\Omega_0))}^2 \right) \\
& \qquad + C(\rho_s, \Omega_0) \left( \|h\|_{L^2(0,t;H^1_\#(\Omega_0))}^2 + \|\ell\|_{L^2(0,t;L^2_\#(0,L))}^2 + \|m\|_{L^2(0,t;L^2_\#(0,L))}^2 \right) + C_0.
\end{align*}
$$

Thanks to the energy estimates (3.26), we obtain

$$
\begin{align}
& \frac{\rho_f}{2R} \|\partial_t v_\varepsilon\|_{L^2(0,t;L^2_\#(\Omega_0))}^2 + \frac{\rho_s}{2} \|\partial_{tt} \zeta_\varepsilon\|_{L^2(0,t;L^2_\#(\Omega_0))}^2 + \frac{\rho_s}{2} \|\partial_{tt} \eta_\varepsilon\|_{L^2(0,t;L^2_\#(\Omega_0))}^2 \nonumber \\
& \qquad + \gamma \|\partial_{xxt} \eta_\varepsilon(t)\|_{L^2(0,t;L^2_\#(\Omega_0))}^2 + \frac{\mu\lambda(R)}{4} \|\nabla v_\varepsilon(t)\|_{L^2_\#(\Omega_0)}^2 \nonumber \\
& \qquad + \frac{\varepsilon}{4} \|\partial_{xxt} \zeta_\varepsilon(t)\|_{L^2_\#(0,L)}^2 + \frac{\varepsilon}{4} \|\partial_{xxt} \eta_\varepsilon(t)\|_{L^2_\#(0,L)}^2 \le C(R) \left( \|\operatorname{div} h\|_{L^2(0,t;L^2_\#(\Omega_0))}^2 + \|f\|_{L^2(0,t;L^2_\#(\Omega_0))}^2 \right) \nonumber \\
& \qquad + C(\rho_s, \Omega_0) \left( \|h\|_{L^2(0,t;H^1_\#(\Omega_0))}^2 + \|\ell\|_{L^2(0,t;L^2_\#(0,L))}^2 + \|m\|_{L^2(0,t;L^2_\#(0,L))}^2 \right) + C_0 + C_\varepsilon, \tag{3.27}
\end{align}
$$

where $C_\varepsilon$ depends on initial data, source terms and $1/\varepsilon$.

Consequently by applying Gronwall Lemma, we obtain bounds on the solution in the following spaces:

• $v_\varepsilon$ in $H^1(0, T; L^2_\#(\Omega_0)) \cap L^\infty(0, T; H^1_\#(\Omega_0))$;

• $\zeta_\varepsilon$ in $H^2(0, T; L^2_\#(0, L))$, $\eta_\varepsilon$ in $H^2(0, T; H^1_\#(0, L))$;

• $\sqrt{\varepsilon}\zeta_\varepsilon$ in $W^{1,\infty}(0, T; H^2_\#(0, L))$, $\sqrt{\varepsilon}\eta_\varepsilon$ in $W^{1,\infty}(0, T; H^2_\#(0, L))$;
These estimates furnish uniform bounds satisfied by approximate solutions that are sufficient to pass to the limit in the Galerkin approximated problems and obtain a solution of the weak formulation. We emphasize that (3.26) and (3.27) can be then obtained by passing to the limit in a Galerkin approximation. Note that the right hand side of (3.26) is independent of $\epsilon$, whereas the constant $C_\epsilon$ appearing in (3.27) depends on $1/\epsilon$. Thus, the energy estimate (3.26) is still valid when passing to the limit as $\epsilon \to 0$ whereas the bounds in (3.27) blow up when passing to the limit as $\epsilon \to 0$. However, we will see later that we can control the left hand side of (3.27) independently of $\epsilon$ thanks to additional estimates we derive in the last part of this subsection.

**Further regularity of the regularized solution** With the derived estimates we can further recover elliptic regularities and the strong formulations satisfied by the solution of the regularized problem. The key point here is that the equations of the longitudinal displacement $\zeta$ ant the transversal one $\eta$ are coupled through the constraint $\delta_{a,b}(\partial_t\zeta, \partial_t\eta) = 0$, and that further fixes the pressure average. Indeed we just obtained that there exists a triplet $(v_\epsilon, \zeta_\epsilon, \eta_\epsilon)$ such that

$$v_\epsilon \in H^1(0, T; L_\#^2(\Omega_0)) \cap L^\infty(0, T; H_\#^1(\Omega_0)), \quad (\zeta_\epsilon, \eta_\epsilon) \in (H^2(0, T; L_\#^2(0, L)) \cap W^{1,\infty}(0, T; H_\#^2(0, L)))^2;$$

satisfying (3.24) for any test function in $V$ and the initial conditions (3.25).

Then by considering $(\xi, \varphi) = (0,0)$ in (3.24), we obtain for any $w \in H_\#^1(\Omega_0)$ such that $\text{div}(B_{a,b}^T w) = 0$ and $w|_{y=1} = w|_{y=0} = 0$

$$\int_{\Omega_0} \rho_{f,a,b} \partial_t v_\epsilon \cdot w \, dxdy + \int_{\Omega_0} \left( \mu(A_{a,b}\nabla)v_\epsilon + \frac{\mu}{\det \nabla\chi_{a,b}} B_{a,b} \nabla v_\epsilon^T B_{a,b} \right) : \nabla w \, dxdy \\ = \int_{\Omega_0} f \cdot w \, dxdy - \int_{\Omega_0} h : \nabla w \, dxdy.$$

Since $B_{a,b}$ is invertible and belong to $H_\#^{1+s}(\Omega_0)$ which a multiplier space of $H_\#^1(\Omega_0)$ the standard de Rham Lemma can easily be adapted to the operators $(B_{a,b}\nabla)\cdot$ and $\text{div}(B_{a,b}^T\cdot)$ (see for instance [23], [25]), to get that there exists $s_{0,\epsilon} \in L_\#^2(\Omega_0)$ such that

$$(B_{a,b}\nabla)s_{0,\epsilon} = -\rho_{f,a,b}\partial_t v_\epsilon + \text{div}\left(\mu(A_{a,b}\nabla)v_\epsilon + \frac{\mu}{\det \nabla\chi_{a,b}} B_{a,b} \nabla v_\epsilon^T B_{a,b}\right) + f + \text{div} h. \quad (3.28)$$

As $f$ and $\text{div} h$ belong to $L^2(0, T; L_\#^2(\Omega_0))$ and by using the fact that $\partial_t v_\epsilon \in L^2(0, T; L_\#^2(\Omega_0))$, $(\partial_t \zeta_\epsilon, \partial_t \zeta_\epsilon) \in L^2(0, T; H_\#^2(\Omega_0))$, Proposition 2.3 implies

$$(v_\epsilon, s_{0,\epsilon}) \in L^2(0, T; H_\#^2(\Omega_0)) \times L^2(0, T; H_\#^1(\Omega_0) \cap L_\#^2(\Omega_0))$$

and consequently $(v_\epsilon, s_{0,\epsilon})$ satisfies (3.15) a.e on $(0, T) \times \Omega_0$. Hence, $(v_\epsilon, s_{0,\epsilon})$ satisfies (3.15)-(3.18) with periodic boundary conditions in $x$.

Next we recover the regularized beam equations. For any $(\xi, \varphi) \in H_\#^2(0, L)$, such that $\delta_{a,b}(\xi, \varphi) = 0$, where $\delta_{a,b}(\xi, \varphi)$ is defined by (3.10), we construct an admissible fluid test function $w = U(0, 0, \xi, \varphi) \in H_\#^2(\Omega_0)$ thanks again to Proposition 2.3. Thus, using $(w, \xi, \varphi)$ as a test function in (3.24), we obtain, after integration by parts in
These estimates furnish uniform bounds satisfied by approximate solutions that are sufficient to pass to the limit in the Galerkin approximated problems and obtain a solution of the weak formulation. We emphasize that (3.26) and (3.27) can be then obtained by passing to the limit in a Galerkin approximation. Note that the right hand side of (3.26) is independent of $\epsilon$, whereas the constant $C_\epsilon$ appearing in (3.27) depends on $1/\epsilon$. Thus, the energy estimate (3.26) is still valid when passing to the limit as $\epsilon \to 0$ whereas the bounds in (3.27) blow up when passing to the limit as $\epsilon \to 0$. However, we will see later that we can control the left hand side of (3.27) independently of $\epsilon$ thanks to additional estimates we derive in the last part of this subsection.

**Further regularity of the regularized solution** With the derived estimates we can further recover elliptic regularities and the strong formulations satisfied by the solution of the regularized problem. The key point here is that the equations of the longitudinal displacement $\zeta$ ant the transversal one $\eta$ are coupled through the constraint $\delta_{a,b}(\partial_t\zeta, \partial_t\eta) = 0$, and that further fixes the pressure average. Indeed we just obtained that there exists a triplet $(v_\epsilon, \zeta_\epsilon, \eta_\epsilon)$ such that

$$v_\epsilon \in H^1(0, T; L_\#^2(\Omega_0)) \cap L^\infty(0, T; H_\#^1(\Omega_0)), \quad (\zeta_\epsilon, \eta_\epsilon) \in (H^2(0, T; L_\#^2(0, L)) \cap W^{1,\infty}(0, T; H_\#^2(0, L)))^2;$$

satisfying (3.24) for any test function in $V$ and the initial conditions (3.25).

Then by considering $(\xi, \varphi) = (0,0)$ in (3.24), we obtain for any $w \in H_\#^1(\Omega_0)$ such that $\text{div}(B_{a,b}^T w) = 0$ and $w|_{y=1} = w|_{y=0} = 0$

$$\int_{\Omega_0} \rho_{f,a,b} \partial_t v_\epsilon \cdot w \, dxdy + \int_{\Omega_0} \left( \mu(A_{a,b} \nabla) v_\epsilon + \frac{\mu}{\det \nabla \chi_{a,b}} B_{a,b} \nabla v_\epsilon^T B_{a,b} \right) : \nabla w \, dxdy \\ = \int_{\Omega_0} f \cdot w \, dxdy - \int_{\Omega_0} h : \nabla w \, dxdy.$$

Since $B_{a,b}$ is invertible and belong to $H_\#^{1+s}(\Omega_0)$ which a multiplier space of $H_\#^1(\Omega_0)$ the standard de Rham Lemma can easily be adapted to the operators $(B_{a,b} \nabla) \cdot$ and $\text{div}(B_{a,b}^T \cdot)$ (see for instance [23], [25]), to get that there exists $s_{0,\epsilon} \in L_\#^2(\Omega_0)$ such that

$$(B_{a,b} \nabla) s_{0,\epsilon} = -\rho_{f,a,b} \partial_t v_\epsilon + \text{div} \left( \mu(A_{a,b} \nabla) v_\epsilon + \frac{\mu}{\det \nabla \chi_{a,b}} B_{a,b} \nabla v_\epsilon^T B_{a,b} \right) + f + \text{div} h. \quad (3.28)$$

As $f$ and $\text{div} h$ belong to $L^2(0, T; L_\#^2(\Omega_0))$ and by using the fact that $\partial_t v_\epsilon \in L^2(0, T; L_\#^2(\Omega_0))$, $(\partial_t \zeta_\epsilon, \partial_t \zeta_\epsilon) \in L^2(0, T; H_\#^2(\Omega_0))$, Proposition 2.3 implies

$$(v_\epsilon, s_{0,\epsilon}) \in L^2(0, T; H_\#^2(\Omega_0)) \times L^2(0, T; H_\#^1(\Omega_0) \cap L_\#^2(\Omega_0))$$

and consequently $(v_\epsilon, s_{0,\epsilon})$ satisfies (3.15) a.e on $(0, T) \times \Omega_0$. Hence, $(v_\epsilon, s_{0,\epsilon})$ satisfies (3.15)-(3.18) with periodic boundary conditions in $x$.

Next we recover the regularized beam equations. For any $(\xi, \varphi) \in H_\#^2(0, L)$, such that $\delta_{a,b}(\xi, \varphi) = 0$, where $\delta_{a,b}(\xi, \varphi)$ is defined by (3.10), we construct an admissible fluid test function $w = U(0, 0, \xi, \varphi) \in H_\#^2(\Omega_0)$ thanks again to Proposition 2.3. Thus, using $(w, \xi, \varphi)$ as a test function in (3.24), we obtain, after integration by parts in
space,

$$
\begin{align*}
& \int_{\Omega_0} (\rho_{f,a,b} \partial_t v_\varepsilon - \operatorname{div} \left( \mu(A_{a,b} \nabla) v_\varepsilon + \frac{\mu}{\operatorname{det} \nabla \chi_{a,b}} B_{a,b} \nabla v_\varepsilon^T B_{a,b} - \operatorname{div} h - f \right)) \cdot w \,dx dy \\
& \qquad + \int_0^L \left( \mu(A_{a,b} \nabla) v_\varepsilon + \frac{\mu}{\operatorname{det} \nabla \chi_{a,b}} B_{a,b} \nabla v_\varepsilon^T B_{a,b} \right)_{|y=1} e_2 \cdot (\xi e_1 + \varphi e_2) dx \\
& \qquad + \int_0^L (\rho_s \partial_{tt} \zeta_\varepsilon - \beta \partial_{xx} \zeta_\varepsilon + h_{|y=1} e_2 \cdot e_1 - \ell) \xi \,dx \\
& \qquad + \int_0^L (\rho_s \partial_{tt} \eta_\varepsilon + h_{|y=1} e_2 \cdot e_2 - m) \varphi \,dx + \gamma \int_0^L \partial_{ttx} \eta_\varepsilon \partial_x \varphi \,dx + \alpha \int_0^L \partial_{xx} \eta_\varepsilon \partial_{xx} \varphi \,dx \\
& \qquad + \varepsilon \int_0^L \partial_{xxt} \zeta_\varepsilon(x) \partial_{xx} \xi \,dx + \varepsilon \int_0^L \partial_{xxt} \eta_\varepsilon \partial_{xx} \varphi \,dx = 0.
\end{align*}
$$

This implies that in view of (3.28) that

$$
\begin{align*}
& \int_{\Omega_0} -(B_{a,b}\nabla)s_{0,\varepsilon} \cdot w \,dx dy + \int_0^L \left( \mu(A_{a,b}\nabla)v_\varepsilon + \frac{\mu}{\det\nabla\chi_{a,b}} B_{a,b}\nabla v_\varepsilon^T B_{a,b} \right)_{|y=1} e_2 \cdot (\xi e_1 + \varphi e_2) \,dx \\
& \qquad + \int_0^L (\rho_s \partial_{tt}\zeta_\varepsilon - \beta \partial_{xx}\zeta_\varepsilon + h_{|y=1} e_2 \cdot e_1 - \ell) \xi \,dx \\
& \qquad + \int_0^L (\rho_s \partial_{tt}\eta_\varepsilon + h_{|y=1} e_2 \cdot e_2 - m) \varphi \,dx + \gamma \int_0^L \partial_{ttx}\eta_\varepsilon \partial_x \varphi \,dx + \alpha \int_0^L \partial_{xx}\eta_\varepsilon \partial_{xx} \varphi \,dx \\
& \qquad + \varepsilon \int_0^L \partial_{xxt}\zeta_\varepsilon(x) \partial_{xx} \xi \,dx + \varepsilon \int_0^L \partial_{xxt}\eta_\varepsilon \partial_{xx} \varphi \,dx = 0.
\end{align*}
$$

Remembering that div($B_{a,b}^T w$) = 0, an integration by parts on the pressure term yields

$$
\int_0^L (\rho_s \partial_{tt} \zeta_\epsilon - \beta \partial_{xx} \zeta_\epsilon + h_{|y=1} e_2 \cdot e_1 + T_{a,b}[v_\epsilon, s_{0,\epsilon}] \cdot e_1 - \ell) \xi \,dx + \varepsilon \int_0^L \partial_{xxt} \zeta_\epsilon(x) \partial_{xx} \xi \,dx \\
+ \int_0^L (\rho_s \partial_{tt} \eta_\epsilon + h_{|y=1} e_2 \cdot e_2 + T_{a,b}[v_\epsilon, s_{0,\epsilon}] \cdot e_2 - m) \varphi \,dx + \gamma \int_0^L \partial_{ttx} \eta_\epsilon \partial_x \varphi \,dx + \alpha \int_0^L \partial_{xx} \eta_\epsilon \partial_{xx} \varphi \,dx + \varepsilon \int_0^L \partial_{xxt} \eta_\epsilon \partial_{xx} \varphi \,dx = 0.
$$

This identity is valid for any test function $(\xi, \varphi) \in H_{\#}^2(0, L)$, such that $\delta_{a,b}(\xi, \varphi) = 0$. Thus for any $(\xi, \varphi) \in H_{\#}^2(0, L)$, such that $\delta_{a,b}(\xi, \varphi) = 0$, we end up, a. e. in $t$, with an identity that writes

$$
\langle (F_1^\epsilon, F_2^\epsilon)(t), (\xi, \varphi) \rangle_{H_{\#}^{-2}(0,L), H_{\#}^2(0,L)} = 0.
$$

For any $(\xi, \bar{\varphi}) \in H_{\#}^2(0, L)$ we observe that $(\xi, \varphi) = (\xi, \bar{\varphi} - \frac{\delta_{a,b}(\xi, \bar{\varphi})}{L})$ satisfies the constraint $\delta_{a,b}(\xi, \varphi) = 0$ and we
can easily deduce that there exists a unique $k_{\varepsilon} : t \mapsto k_{\varepsilon}(t)$ such that, a. e. in $t$

$$
\langle (F_1^\varepsilon, F_2^\varepsilon)(t), (\xi, \bar{\varphi}) \rangle_{H_{\#}^{-2}(0,L), H_{\#}^2(0,L)} = k_\varepsilon(t) \delta_{a,b}(\xi, \bar{\varphi}), \quad \forall (\xi, \bar{\kappa}) \in H_{\#}^2(0,L).
$$

In a straightforward way $k_\epsilon(t) = \frac{1}{L} \int_0^L F_2^\epsilon(x,t)dx$. The fluid pressure $q_\epsilon$ is then given by $q_\epsilon(x,y,t) = s_{0,\epsilon}(x,y,t) + k_\epsilon(t)$ with $s_{0,\epsilon} \in L^2(0,T; H_\#^1(\Omega_0) \cap L_{\#0}^2(\Omega_0))$ and

$$
k_{\epsilon}(t) = \frac{1}{L} \int_{0}^{L} \left( \rho_{s} \partial_{tt} \eta_{\epsilon} + h_{|y=1} e_{2} \cdot e_{2} + T_{a,b}[v_{\epsilon}, s_{0,\epsilon}] \cdot e_{2} - m \right) dx. \quad (3.29)
$$

Hence, we have $k_\epsilon \in L^2(0, T)$ and $q_\epsilon \in L^2(0, T; H_\#^1(\Omega_0))$. This gives a meaning to $T_{a,b}[v_\epsilon, q_\epsilon]$ in $L^2(0, T; H_\#^{\frac{1}{2}}(0, L))$
space,

$$
\begin{align*}
& \int_{\Omega_0} (\rho_{f,a,b} \partial_t v_\varepsilon - \operatorname{div} \left( \mu(A_{a,b} \nabla) v_\varepsilon + \frac{\mu}{\operatorname{det} \nabla \chi_{a,b}} B_{a,b} \nabla v_\varepsilon^T B_{a,b} - \operatorname{div} h - f \right)) \cdot w \,dx dy \\
& \qquad + \int_0^L \left( \mu(A_{a,b} \nabla) v_\varepsilon + \frac{\mu}{\operatorname{det} \nabla \chi_{a,b}} B_{a,b} \nabla v_\varepsilon^T B_{a,b} \right)_{|y=1} e_2 \cdot (\xi e_1 + \varphi e_2) dx \\
& \qquad + \int_0^L (\rho_s \partial_{tt} \zeta_\varepsilon - \beta \partial_{xx} \zeta_\varepsilon + h_{|y=1} e_2 \cdot e_1 - \ell) \xi \,dx \\
& \qquad + \int_0^L (\rho_s \partial_{tt} \eta_\varepsilon + h_{|y=1} e_2 \cdot e_2 - m) \varphi \,dx + \gamma \int_0^L \partial_{ttx} \eta_\varepsilon \partial_x \varphi \,dx + \alpha \int_0^L \partial_{xx} \eta_\varepsilon \partial_{xx} \varphi \,dx \\
& \qquad + \varepsilon \int_0^L \partial_{xxt} \zeta_\varepsilon(x) \partial_{xx} \xi \,dx + \varepsilon \int_0^L \partial_{xxt} \eta_\varepsilon \partial_{xx} \varphi \,dx = 0.
\end{align*}
$$

This implies that in view of (3.28) that

$$
\begin{align*}
& \int_{\Omega_0} -(B_{a,b}\nabla)s_{0,\varepsilon} \cdot w \,dx dy + \int_0^L \left( \mu(A_{a,b}\nabla)v_\varepsilon + \frac{\mu}{\det\nabla\chi_{a,b}} B_{a,b}\nabla v_\varepsilon^T B_{a,b} \right)_{|y=1} e_2 \cdot (\xi e_1 + \varphi e_2) \,dx \\
& \qquad + \int_0^L (\rho_s \partial_{tt}\zeta_\varepsilon - \beta \partial_{xx}\zeta_\varepsilon + h_{|y=1} e_2 \cdot e_1 - \ell) \xi \,dx \\
& \qquad + \int_0^L (\rho_s \partial_{tt}\eta_\varepsilon + h_{|y=1} e_2 \cdot e_2 - m) \varphi \,dx + \gamma \int_0^L \partial_{ttx}\eta_\varepsilon \partial_x \varphi \,dx + \alpha \int_0^L \partial_{xx}\eta_\varepsilon \partial_{xx} \varphi \,dx \\
& \qquad + \varepsilon \int_0^L \partial_{xxt}\zeta_\varepsilon(x) \partial_{xx} \xi \,dx + \varepsilon \int_0^L \partial_{xxt}\eta_\varepsilon \partial_{xx} \varphi \,dx = 0.
\end{align*}
$$

Remembering that div($B_{a,b}^T w$) = 0, an integration by parts on the pressure term yields

$$
\int_0^L (\rho_s \partial_{tt} \zeta_\epsilon - \beta \partial_{xx} \zeta_\epsilon + h_{|y=1} e_2 \cdot e_1 + T_{a,b}[v_\epsilon, s_{0,\epsilon}] \cdot e_1 - \ell) \xi \,dx + \varepsilon \int_0^L \partial_{xxt} \zeta_\epsilon(x) \partial_{xx} \xi \,dx \\
+ \int_0^L (\rho_s \partial_{tt} \eta_\epsilon + h_{|y=1} e_2 \cdot e_2 + T_{a,b}[v_\epsilon, s_{0,\epsilon}] \cdot e_2 - m) \varphi \,dx + \gamma \int_0^L \partial_{ttx} \eta_\epsilon \partial_x \varphi \,dx + \alpha \int_0^L \partial_{xx} \eta_\epsilon \partial_{xx} \varphi \,dx + \varepsilon \int_0^L \partial_{xxt} \eta_\epsilon \partial_{xx} \varphi \,dx = 0.
$$

This identity is valid for any test function $(\xi, \varphi) \in H_{\#}^2(0, L)$, such that $\delta_{a,b}(\xi, \varphi) = 0$. Thus for any $(\xi, \varphi) \in H_{\#}^2(0, L)$, such that $\delta_{a,b}(\xi, \varphi) = 0$, we end up, a. e. in $t$, with an identity that writes

$$
\langle (F_1^\varepsilon, F_2^\varepsilon)(t), (\xi, \varphi) \rangle_{H_{\#}^{-2}(0,L), H_{\#}^2(0,L)} = 0.
$$

For any $(\xi, \bar{\varphi}) \in H_{\#}^2(0, L)$ we observe that $(\xi, \varphi) = (\xi, \bar{\varphi} - \frac{\delta_{a,b}(\xi, \bar{\varphi})}{L})$ satisfies the constraint $\delta_{a,b}(\xi, \varphi) = 0$ and we
can easily deduce that there exists a unique $k_{\varepsilon} : t \mapsto k_{\varepsilon}(t)$ such that, a. e. in $t$

$$
\langle (F_1^\varepsilon, F_2^\varepsilon)(t), (\xi, \bar{\varphi}) \rangle_{H_{\#}^{-2}(0,L), H_{\#}^2(0,L)} = k_\varepsilon(t) \delta_{a,b}(\xi, \bar{\varphi}), \quad \forall (\xi, \bar{\kappa}) \in H_{\#}^2(0,L).
$$

In a straightforward way $k_\epsilon(t) = \frac{1}{L} \int_0^L F_2^\epsilon(x,t)dx$. The fluid pressure $q_\epsilon$ is then given by $q_\epsilon(x,y,t) = s_{0,\epsilon}(x,y,t) + k_\epsilon(t)$ with $s_{0,\epsilon} \in L^2(0,T; H_\#^1(\Omega_0) \cap L_{\#0}^2(\Omega_0))$ and

$$
k_{\epsilon}(t) = \frac{1}{L} \int_{0}^{L} \left( \rho_{s} \partial_{tt} \eta_{\epsilon} + h_{|y=1} e_{2} \cdot e_{2} + T_{a,b}[v_{\epsilon}, s_{0,\epsilon}] \cdot e_{2} - m \right) dx. \quad (3.29)
$$

Hence, we have $k_{\epsilon} \in L^2(0, T)$ and $q_{\epsilon} \in L^2(0, T; H_{\#}^1(\Omega_0))$. This gives a meaning to $T_{a,b}[v_{\epsilon}, q_{\epsilon}]$ in $L^2(0, T; H_{\#}^{\frac{1}{2}}(0, L))$
and we obtain the following weak formulation for any test function $(\xi, \varphi) \in H_{\#}^2(0, L)$

$$ \int_0^L (\rho_s \partial_{tt} \zeta_\epsilon - \beta \partial_{xx} \zeta_\epsilon + h|_{y=1} e_2 \cdot e_1 + T_{a,b}[v_\epsilon, q_\epsilon] \cdot e_1 - \ell) \xi \, dx + \epsilon \int_0^L \partial_{xxt} \zeta_\epsilon(x) \partial_{xx} \xi \, dx \\ + \int_0^L (\rho_s \partial_{tt} \eta_\epsilon + h|_{y=1} e_2 \cdot e_2 + T_{a,b}[v_\epsilon, q_\epsilon] \cdot e_2 - m) \varphi \, dx + \gamma \int_0^L \partial_{ttx} \eta_\epsilon \partial_x \varphi \, dx + \alpha \int_0^L \partial_{xxx} \eta_\epsilon \partial_{xx} \varphi \, dx + \epsilon \int_0^L \partial_{xxt} \eta_\epsilon \partial_{xx} \varphi \, dx = 0. \quad (3.30) $$

Thus we now can decouple the longitudinal and transversal equations of the beam. Finally by taking as test functions the projection of $\partial_{xxxxt}\zeta_\epsilon$ and $\partial_{xxxxt}\eta_\epsilon$ on the first Fourier modes on $(0, L)$, we obtain an identity which implies that $\zeta_\epsilon \in W^{1,\infty}(0, T; H_{\#}^2(0, L)) \cap H^1(0, T; H_{\#}^4(0, L))$ and $\eta_\epsilon \in W^{1,\infty}(0, T; H_{\#}^3(0, L)) \cap H^1(0, T; H_{\#}^4(0, L))$ (by letting the number of Fourier modes tends to infinity). Moreover we already know that $\zeta_\epsilon \in H^2(0, T; L_{\#}^2(0, L))$ and $\eta_\epsilon \in H^2(0, T; H_{\#}^1(0, L))$. It follows then, in view of (3.30), that equation (3.19) is satisfied a.e and (3.20) in $L^2(0, T; H_{\#}^{-1}(\Omega_0))$. From the regularities derived on $\eta_\epsilon$ we further obtain that $\partial_{tt}\eta_\epsilon \in L^2(0, T; H_{\#}^2(0, L))$ and consequently equation (3.20) is satisfied a.e.

**Remark 3.1** Note that, since $\delta_{ab}(\partial_t \zeta_\epsilon, \partial_t \eta_\epsilon) = 0$, by multiplying (3.19) by $-\partial_x b$ and (3.20) by $1 + \partial_x a$, $k_\epsilon$ is also given by

$$ k_\epsilon(t) = \frac{1}{\|(-\partial_x b, 1 + \partial_x a)\|_{L^2(0,L)}^2} \left( \beta \int_0^L \partial_x \zeta_\epsilon \partial_{xx} b \, dx - \alpha \int_0^L \partial_{xxx} \eta_\epsilon \partial_{xx} a \, dx + \gamma \int_0^L \partial_{xtt} \eta_\epsilon \partial_{xx} a \, dx \right. \\ \left. + \int_0^L T[v_\epsilon, s_{0,\epsilon}] \cdot (-\partial_x b e_1 + (1+\partial_x a)e_2) \, dx + \int_0^L h|_{y=1} e_2 \cdot (-\partial_x b e_1 + (1+\partial_x a)e_2) \, dx - \int_0^L (\ell, m) \cdot (-\partial_x b e_1 + (1+\partial_x a)e_2) \, dx \right). \quad (3.31) $$

### 3.2.4 Estimates independent of $\epsilon > 0$

Except for the energy estimate (3.26), the already obtained regularity estimates depend on the regularization parameter $\epsilon$. Thus in this subsection we derive regularity estimates independent of $\epsilon$. We multiply (3.15) by $\partial_t v_\epsilon$ and integrate the resulting equality over $\Omega_0 \times (0, t)$. If follows that

$$ \int_0^t \int_{\Omega_0} \rho_{f,a,b} |\partial_t v_\epsilon|^2 \, dxdyds - \int_0^t \int_{\Omega_0} \operatorname{div} \left( \mu(A_{a,b}\nabla)v_\epsilon + \frac{\mu}{\det \nabla \chi_{a,b}} B_{a,b} \nabla v_\epsilon^T B_{a,b} - q_\epsilon B_{a,b} \right) \cdot \partial_t v_\epsilon \, dxdyds \\ = \int_0^t \int_{\Omega_0} f \cdot \partial_t v_\epsilon \, dxdyds + \int_0^t \int_{\Omega_0} \operatorname{div} h \cdot \partial_t v_\epsilon \, dxdyds. \quad (3.32) $$

To integrate by parts the second term in the left hand side of the above equality, we need the following lemma as $\partial_t v_\epsilon \in L^2(0, T; L_{\#}^2(\Omega_0))$ only and it trace is not well defined on the fluid-structure interface.

**Lemma 3.1** For any $w \in L^2(0, T; H_{\#}^2(\Omega_0)) \cap H^1(0, T; L_{\#}^2(\Omega_0))$ such that $\det(B_{a,b}^T w) = 0$ on $\Omega_0$, $w(x, 1, t) = (\partial_t \zeta(x, t), \partial_t \eta(x, t))$ for some $(\zeta, \eta) \in H^2(0, T; L_{\#}^2(0, L))$ and $w|_{y=0} = 0$ and for any $q \in L^2(0, T; H_{\#}^1(\Omega_0))$, we have
$$ - \int_0^t \int_{\Omega_0} \operatorname{div} \left( \mu(A_{a,b}\nabla)w + \frac{\mu}{\det \nabla \chi_{a,b}} B_{a,b} \nabla w^T B_{a,b} \right) \cdot \partial_t w \, dxdyds + \int_0^t \int_{\Omega_0} (B_{a,b}\nabla)q \cdot \partial_t w \, dxdyds \\ = \frac{1}{4} \int_{\Omega_0} \frac{\mu}{\det \nabla \chi_{a,b}} \left| \left( (B_{a,b}\nabla)v_\epsilon(t) + ((B_{a,b}\nabla)v_\epsilon(t))^T \right) \right|^2 dxdy - \frac{1}{4} \int_{\Omega_0} \frac{\mu}{\det \nabla \chi_{a,b}} \left| \left( (B_{a,b}\nabla)v_0^\epsilon + ((B_{a,b}\nabla)v_0^\epsilon)^T \right) \right|^2 dxdy \\ - \int_0^t \int_0^L T_{a,b}[w, q] \cdot (\partial_{tt}\zeta e_1 + \partial_{tt}\eta e_2) \, dxds. \quad (3.33) $$
and we obtain the following weak formulation for any test function $(\xi, \varphi) \in H_{\#}^2(0, L)$

$$ 
\begin{aligned}
& \int_0^L (\rho_s \partial_{tt} \zeta_\epsilon - \beta \partial_{xx} \zeta_\epsilon + h_{|_{y=1}} e_2 \cdot e_1 + T_{a,b}[v_\epsilon, q_\epsilon] \cdot e_1 - \ell) \xi \, dx + \epsilon \int_0^L \partial_{xxt} \zeta_\epsilon(x) \partial_{xx} \xi \, dx \\
& + \int_0^L (\rho_s \partial_{tt} \eta_\epsilon + h_{|_{y=1}} e_2 \cdot e_2 + T_{a,b}[v_\epsilon, q_\epsilon] \cdot e_2 - m) \varphi \, dx + \gamma \int_0^L \partial_{txx} \eta_\epsilon \partial_x \varphi \, dx + \alpha \int_0^L \partial_{xx} \eta_\epsilon \partial_{xx} \varphi \, dx + \epsilon \int_0^L \partial_{xxt} \eta_\epsilon \partial_{xx} \varphi \, dx = 0.
\end{aligned}
\quad (3.30) $$

Thus we now can decouple the longitudinal and transversal equations of the beam. Finally by taking as test functions the projection of $\partial_{xxxxt}\zeta_\epsilon$ and $\partial_{xxxxt}\eta_\epsilon$ on the first Fourier modes on $(0, L)$, we obtain an identity which implies that $\zeta_\epsilon \in W^{1,\infty}(0, T; H_{\#}^2(0, L)) \cap H^1(0, T; H_{\#}^4(0, L))$ and $\eta_\epsilon \in W^{1,\infty}(0, T; H_{\#}^3(0, L)) \cap H^1(0, T; H_{\#}^4(0, L))$ (by letting the number of Fourier modes tends to infinity). Moreover we already know that $\zeta_\epsilon \in H^2(0, T; L_{\#}^2(0, L))$ and $\eta_\epsilon \in H^2(0, T; H_{\#}^1(0, L))$. It follows then, in view of (3.30), that equation (3.19) is satisfied a.e and (3.20) in $L^2(0, T; H_{\#}^{-1}(\Omega_0))$. From the regularities derived on $\eta_\epsilon$ we further obtain that $\partial_{tt}\eta_\epsilon \in L^2(0, T; H_{\#}^2(0, L))$ and consequently equation (3.20) is satisfied a.e.

**Remark 3.1** Note that, since $\delta_{ab}(\partial_t \zeta_\epsilon, \partial_t \eta_\epsilon) = 0$, by multiplying (3.19) by $-\partial_x b$ and (3.20) by $1 + \partial_x a$, $k_\epsilon$ is also given by

$$ 
\begin{aligned}
k_\epsilon(t) = & \frac{1}{\|(-\partial_x b, 1 + \partial_x a)\|_{L^2(0,L)}^2} \left( \beta \int_0^L \partial_x \zeta_\epsilon \partial_{xx} b \, dx - \alpha \int_0^L \partial_{xxx} \eta_\epsilon \partial_{xx} a \, dx + \gamma \int_0^L \partial_{xtt} \eta_\epsilon \partial_{xx} a \, dx \right. \\
& \left. + \int_0^L T[v_\epsilon, s_{0,\epsilon}] \cdot (-\partial_x b e_1 + (1+\partial_x a)e_2) \, dx + \int_0^L h_{|_{y=1}} e_2 \cdot (-\partial_x b e_1 + (1+\partial_x a)e_2) \, dx - \int_0^L (\ell, m) \cdot (-\partial_x b e_1 + (1+\partial_x a)e_2) \, dx \right).
\end{aligned}
\quad (3.31) $$

### 3.2.4 Estimates independent of $\epsilon > 0$

Except for the energy estimate (3.26), the already obtained regularity estimates depend on the regularization parameter $\epsilon$. Thus in this subsection we derive regularity estimates independent of $\epsilon$. We multiply (3.15) by $\partial_t v_\epsilon$ and integrate the resulting equality over $\Omega_0 \times (0, t)$. If follows that

$$ 
\begin{aligned}
& \int_0^t \int_{\Omega_0} \rho_{f,a,b} |\partial_t v_\epsilon|^2 \, dxdyds - \int_0^t \int_{\Omega_0} \operatorname{div} \left( \mu(A_{a,b}\nabla)v_\epsilon + \frac{\mu}{\det\nabla\chi_{a,b}} B_{a,b}\nabla v_\epsilon^T B_{a,b} - q_\epsilon B_{a,b} \right) \cdot \partial_t v_\epsilon \, dxdyds \\
& = \int_0^t \int_{\Omega_0} f \cdot \partial_t v_\epsilon \, dxdyds + \int_0^t \int_{\Omega_0} \operatorname{div} h \cdot \partial_t v_\epsilon \, dxdyds. \quad (3.32)
\end{aligned}
$$

To integrate by parts the second term in the left hand side of the above equality, we need the following lemma as $\partial_t v_\epsilon \in L^2(0, T; L_{\#}^2(\Omega_0))$ only and it trace is not well defined on the fluid-structure interface.

**Lemma 3.1** For any $w \in L^2(0, T; H_{\#}^2(\Omega_0)) \cap H^1(0, T; L_{\#}^2(\Omega_0))$ such that $\det(B_{a,b}^T w) = 0$ on $\Omega_0$, $w(x, 1, t) = (\partial_t \zeta(x, t), \partial_t \eta(x, t))$ for some $(\zeta, \eta) \in H^2(0, T; L_{\#}^2(0, L))$ and $w|_{|_{y=0}} = 0$ and for any $q \in L^2(0, T; H_{\#}^1(\Omega_0))$, we have

$$ 
\begin{aligned}
& - \int_0^t \int_{\Omega_0} \operatorname{div} \left( \mu(A_{a,b}\nabla)w + \frac{\mu}{\det\nabla\chi_{a,b}} B_{a,b}\nabla w^T B_{a,b} \right) \cdot \partial_t w \, dxdyds + \int_0^t \int_{\Omega_0} (B_{a,b}\nabla)q \cdot \partial_t w \, dxdyds \\
& = \frac{1}{4} \int_{\Omega_0} \frac{\mu}{\det\nabla\chi_{a,b}} \left| \left( (B_{a,b}\nabla)v_\epsilon(t) + ((B_{a,b}\nabla)v_\epsilon(t))^T \right) \right|^2 dxdy - \frac{1}{4} \int_{\Omega_0} \frac{\mu}{\det\nabla\chi_{a,b}} \left| \left( (B_{a,b}\nabla)v_0^\epsilon + ((B_{a,b}\nabla)v_0^\epsilon)^T \right) \right|^2 dxdy \\
& \qquad - \int_0^t \int_0^L T_{a,b}[w, q] \cdot (\partial_{tt}\zeta e_1 + \partial_{tt}\eta e_2) \, dxds. \quad (3.33)
\end{aligned}
$$
The proof of this lemma is done by an approximation process in time, the equality is first proved for smooth functions and obtained by passing to the limit.

By using the above Lemma, (3.32) becomes

$$ 
\begin{aligned}
& \int_0^t \int_{\Omega_0} \rho_{f,a,b} |\partial_t v_\epsilon|^2 dx dy ds + \frac{\mu}{4} \int_{\Omega_0} \frac{1}{\det \nabla \chi_{a,b}} |((B_{a,b} \nabla) v_\epsilon(t) + ((B_{a,b} \nabla) v_\epsilon(t))^T)|^2 dx dy ds \\
& = \int_0^t \int_{\Omega_0} f \cdot \partial_t v_\epsilon \, dx dy ds + \int_0^t \int_{\Omega_0} \operatorname{div} h \cdot \partial_t v_\epsilon \, dx dy ds \\
& + \frac{\mu}{4} \int_{\Omega_0} \frac{1}{\det \nabla \chi_{a,b}} |((B_{a,b} \nabla) v_\epsilon^T + ((B_{a,b} \nabla) v_\epsilon^T)^T)|^2 dx dy + \int_0^t \int_0^L T[v_\epsilon, q_\epsilon] \cdot (\partial_{tt} \zeta e_1 + \partial_{tt} \eta e_2) \, dx ds. \quad (3.34)
\end{aligned}
$$

Next we multiply (3.19) and (3.20) respectively by $\partial_{tt}\zeta_{\epsilon}$ and $\partial_{tt}\eta_{\epsilon}$ which belong respectively to $L^2(0, T; L_{\#}^2(0, L))$ and $L^2(0, T; H_{\#}^1(0, L))$ then we integrate the resulting equations over $(0, t) \times (0, L)$. This yields to:

$$ 
\begin{aligned}
& \rho_s \int_0^t \int_0^L |\partial_{tt}\zeta_{\epsilon}|^2 dx ds - \beta \int_0^t \int_0^L \partial_{xx}\zeta_{\epsilon} \partial_{tt}\zeta_{\epsilon} \, dx ds + \varepsilon \int_0^t \int_0^L \partial_{xxxxt}\zeta_{\epsilon} \partial_{tt}\zeta_{\epsilon} \, dx ds \\
& = - \int_0^t \int_0^L T_{a,b}[v_{\epsilon}, q_{\epsilon}] \cdot e_1 \partial_{tt}\zeta_{\epsilon} \, dx ds - \int_0^t \int_0^L h_{|_{y=1}} e_2 \cdot e_1 \partial_{tt}\zeta_{\epsilon} \, dx ds + \int_0^t \int_0^L \ell \partial_{tt}\zeta_{\epsilon} \, dx ds, \quad (3.35)
\end{aligned}
$$

$$ 
\begin{aligned}
& \rho_s \int_0^t \int_0^L |\partial_{tt}\eta_{\epsilon}|^2 dx ds + \gamma \int_0^t \int_0^L |\partial_{xxt}\eta_{\epsilon}|^2 dx ds + \alpha \int_0^t \int_0^L \partial_{xxxx}\eta_{\epsilon} \partial_{tt}\eta_{\epsilon} \, dx ds + \varepsilon \int_0^t \int_0^L \partial_{xxxxt}\eta_{\epsilon} \partial_{tt}\eta_{\epsilon} \, dx ds \\
& = - \int_0^t \int_0^L T_{a,b}[v_{\epsilon}, q_{\epsilon}] \cdot e_2 \partial_{tt}\eta_{\epsilon} \, dx ds - \int_0^t \int_0^L h_{|_{y=1}} e_2 \cdot e_2 \partial_{tt}\eta_{\epsilon} \, dx ds + \int_0^t \int_0^L m \partial_{tt}\eta_{\epsilon} \, dx ds. \quad (3.36)
\end{aligned}
$$

Since $\partial_{tt}\zeta_{\epsilon}$ and $\partial_{tt}\eta_{\epsilon}$ are not regular enough in space (only in $L^2(0, T; L_{\#}^2(0, L))$ and $L^2(0, T; H_{\#}^1(0, L))$ respectively), so integration by parts in space can not be performed directly. However, we can show by using a regularization argument that a.e. in $t$, the following equalities hold true

$$ \beta \int_0^t \int_0^L \partial_{xx}\zeta_{\epsilon} \partial_{tt}\zeta_{\epsilon} \, dx ds = \beta \int_0^t \int_0^L |\partial_{xt}\zeta_{\epsilon}|^2 \, dx ds + \beta \int_0^L \partial_{xx}\zeta_{\epsilon}(t) \partial_{t}\zeta_{\epsilon}(t) \, dx - \beta \int_0^L \partial_{xx}\zeta_0^{\epsilon} \zeta_1^{\epsilon} \, dx, $$

$$ \varepsilon \int_0^t \int_0^L \partial_{xxxxt}\zeta_{\epsilon} \partial_{tt}\zeta_{\epsilon} \, dx ds = \frac{\varepsilon}{2} \int_0^L |\partial_{xxt}\zeta_{\epsilon}(t)|^2 \, dx - \frac{\varepsilon}{2} \int_0^L |\partial_{xx}\zeta_1^{\epsilon}|^2 \, dx, $$

$$ \alpha \int_0^t \int_0^L \partial_{xxxx}\eta_{\epsilon} \partial_{tt}\eta_{\epsilon} \, dx ds = -\alpha \int_0^t \int_0^L |\partial_{txx}\eta_{\epsilon}|^2 \, dx ds - \alpha \int_0^L \partial_{xxx}\eta_{\epsilon}(t) \partial_{tx}\eta(t) \, dx + \alpha \int_0^L \partial_{xxx}\eta_0^{\epsilon} \partial_{x}\eta_1^{\epsilon} \, dx, $$

$$ \varepsilon \int_0^t \int_0^L \partial_{xxxxt}\eta_{\epsilon} \partial_{tt}\eta_{\epsilon} \, dx ds = \frac{\varepsilon}{2} \int_0^L |\partial_{xxt}\eta_{\epsilon}(t)|^2 \, dx - \frac{\varepsilon}{2} \int_0^L |\partial_{xx}\eta_1^{\epsilon}|^2 \, dx. $$

By combining the above equalities with (3.35) and (3.36) leads to

$$ 
\begin{aligned}
& \rho_s \int_0^t \int_0^L |\partial_{tt}\zeta_{\epsilon}|^2 dx ds + \frac{\varepsilon}{2} \int_0^L |\partial_{xxt}\zeta_{\epsilon}(t)|^2 dx ds = \beta \int_0^t \int_0^L |\partial_{xt}\zeta_{\epsilon}|^2 dx ds + \beta \int_0^L \partial_{xx}\zeta_{\epsilon}(t) \partial_{t}\zeta_{\epsilon}(t) \, dx - \beta \int_0^L \partial_{xx}\zeta_0^{\epsilon} \zeta_1^{\epsilon} \, dx \\
& - \int_0^t \int_0^L T_{a,b}[v_{\epsilon}, q_{\epsilon}] \cdot e_1 \partial_{tt}\zeta_{\epsilon} \, dx ds - \int_0^t \int_0^L h_{|_{y=1}} e_2 \cdot e_1 \partial_{tt}\zeta_{\epsilon} \, dx ds + \int_0^t \int_0^L \ell \partial_{tt}\zeta_{\epsilon} \, dx ds + \frac{\varepsilon}{2} \int_0^L |\partial_{xx}\zeta_1^{\epsilon}|^2 \, dx,
\end{aligned}
\quad (3.37)
$$
The proof of this lemma is done by an approximation process in time, the equality is first proved for smooth functions and obtained by passing to the limit.

By using the above Lemma, (3.32) becomes

$$ 
\begin{aligned}
& \int_0^t \int_{\Omega_0} \rho_{f,a,b} |\partial_t v_\epsilon|^2 dx dy ds + \frac{\mu}{4} \int_{\Omega_0} \frac{1}{\det \nabla \chi_{a,b}} |((B_{a,b} \nabla) v_\epsilon(t) + ((B_{a,b} \nabla) v_\epsilon(t))^T)|^2 dx dy ds \\
& = \int_0^t \int_{\Omega_0} f \cdot \partial_t v_\epsilon \, dx dy ds + \int_0^t \int_{\Omega_0} \operatorname{div} h \cdot \partial_t v_\epsilon \, dx dy ds \\
& + \frac{\mu}{4} \int_{\Omega_0} \frac{1}{\det \nabla \chi_{a,b}} |((B_{a,b} \nabla) v_\epsilon^T + ((B_{a,b} \nabla) v_\epsilon^T)^T)|^2 dx dy + \int_0^t \int_0^L T[v_\epsilon, q_\epsilon] \cdot (\partial_{tt} \zeta e_1 + \partial_{tt} \eta e_2) \, dx ds. \quad (3.34)
\end{aligned}
$$

Next we multiply (3.19) and (3.20) respectively by $\partial_{tt}\zeta_{\epsilon}$ and $\partial_{tt}\eta_{\epsilon}$ which belong respectively to $L^2(0, T; L^2_{\#}(0, L))$ and $L^2(0, T; H^1_{\#}(0, L))$ then we integrate the resulting equations over $(0, t) \times (0, L)$. This yields to:

$$ 
\begin{aligned}
& \rho_s \int_0^t \int_0^L |\partial_{tt}\zeta_{\epsilon}|^2 dx ds - \beta \int_0^t \int_0^L \partial_{xx}\zeta_{\epsilon} \partial_{tt}\zeta_{\epsilon} \, dx ds + \varepsilon \int_0^t \int_0^L \partial_{xxxxt}\zeta_{\epsilon} \partial_{tt}\zeta_{\epsilon} \, dx ds \\
& = - \int_0^t \int_0^L T_{a,b}[v_{\epsilon}, q_{\epsilon}] \cdot e_1 \partial_{tt}\zeta_{\epsilon} \, dx ds - \int_0^t \int_0^L h_{|_{y=1}} e_2 \cdot e_1 \partial_{tt}\zeta_{\epsilon} \, dx ds + \int_0^t \int_0^L \ell \partial_{tt}\zeta_{\epsilon} \, dx ds, \quad (3.35)
\end{aligned}
$$

$$ 
\begin{aligned}
& \rho_s \int_0^t \int_0^L |\partial_{tt}\eta_{\epsilon}|^2 dx ds + \gamma \int_0^t \int_0^L |\partial_{xxt}\eta_{\epsilon}|^2 dx ds + \alpha \int_0^t \int_0^L \partial_{xxxx}\eta_{\epsilon} \partial_{tt}\eta_{\epsilon} \, dx ds + \varepsilon \int_0^t \int_0^L \partial_{xxxxt}\eta_{\epsilon} \partial_{tt}\eta_{\epsilon} \, dx ds \\
& = - \int_0^t \int_0^L T_{a,b}[v_{\epsilon}, q_{\epsilon}] \cdot e_2 \partial_{tt}\eta_{\epsilon} \, dx ds - \int_0^t \int_0^L h_{|_{y=1}} e_2 \cdot e_2 \partial_{tt}\eta_{\epsilon} \, dx ds + \int_0^t \int_0^L m \partial_{tt}\eta_{\epsilon} \, dx ds. \quad (3.36)
\end{aligned}
$$

Since $\partial_{tt}\zeta_{\epsilon}$ and $\partial_{tt}\eta_{\epsilon}$ are not regular enough in space (only in $L^2(0, T; L^2_{\#}(0, L))$ and $L^2(0, T; H^1_{\#}(0, L))$ respectively), so integration by parts in space can not be performed directly. However, we can show by using a regularization argument that a.e. in $t$, the following equalities hold true

$$ \beta \int_0^t \int_0^L \partial_{xx}\zeta_{\epsilon} \partial_{tt}\zeta_{\epsilon} \, dx ds = \beta \int_0^t \int_0^L |\partial_{xt}\zeta_{\epsilon}|^2 \, dx ds + \beta \int_0^L \partial_{xx}\zeta_{\epsilon}(t) \partial_{t}\zeta_{\epsilon}(t) \, dx - \beta \int_0^L \partial_{xx}\zeta_0^{\epsilon} \zeta_1^{\epsilon} \, dx, $$

$$ \varepsilon \int_0^t \int_0^L \partial_{xxxxt}\zeta_{\epsilon} \partial_{tt}\zeta_{\epsilon} \, dx ds = \frac{\varepsilon}{2} \int_0^L |\partial_{xxt}\zeta_{\epsilon}(t)|^2 \, dx - \frac{\varepsilon}{2} \int_0^L |\partial_{xx}\zeta_1^{\epsilon}|^2 \, dx, $$

$$ \alpha \int_0^t \int_0^L \partial_{xxxx}\eta_{\epsilon} \partial_{tt}\eta_{\epsilon} \, dx ds = -\alpha \int_0^t \int_0^L |\partial_{txx}\eta_{\epsilon}|^2 \, dx ds - \alpha \int_0^L \partial_{xxx}\eta_{\epsilon}(t) \partial_{tx}\eta(t) \, dx + \alpha \int_0^L \partial_{xxx}\eta_0^{\epsilon} \partial_{x}\eta_1^{\epsilon} \, dx, $$

$$ \varepsilon \int_0^t \int_0^L \partial_{xxxxt}\eta_{\epsilon} \partial_{tt}\eta_{\epsilon} \, dx ds = \frac{\varepsilon}{2} \int_0^L |\partial_{xxt}\eta_{\epsilon}(t)|^2 \, dx - \frac{\varepsilon}{2} \int_0^L |\partial_{xx}\eta_1^{\epsilon}|^2 \, dx. $$

By combining the above equalities with (3.35) and (3.36) leads to

$$ 
\begin{aligned}
& \rho_s \int_0^t \int_0^L |\partial_{tt}\zeta_{\epsilon}|^2 dx ds + \frac{\varepsilon}{2} \int_0^L |\partial_{xxt}\zeta_{\epsilon}(t)|^2 dx ds = \beta \int_0^t \int_0^L |\partial_{xt}\zeta_{\epsilon}|^2 dx ds + \beta \int_0^L \partial_{xx}\zeta_{\epsilon}(t) \partial_{t}\zeta_{\epsilon}(t) \, dx - \beta \int_0^L \partial_{xx}\zeta_0^{\epsilon} \zeta_1^{\epsilon} \, dx \\
& - \int_0^t \int_0^L T_{a,b}[v_{\epsilon}, q_{\epsilon}] \cdot e_1 \partial_{tt}\zeta_{\epsilon} \, dx ds - \int_0^t \int_0^L h_{|_{y=1}} e_2 \cdot e_1 \partial_{tt}\zeta_{\epsilon} \, dx ds + \int_0^t \int_0^L \ell \partial_{tt}\zeta_{\epsilon} \, dx ds + \frac{\varepsilon}{2} \int_0^L |\partial_{xx}\zeta_1^{\epsilon}|^2 \, dx,
\end{aligned}
\quad (3.37)
$$
and for the vertical displacement

$$
\begin{align}
& \rho_s \int_0^t \int_0^L |\partial_{tt}\eta_\varepsilon|^2 dxds + \gamma \int_0^t \int_0^L |\partial_{xxt}\eta_\varepsilon|^2 dxds + \frac{\varepsilon}{2} \int_0^L |\partial_{xxt}\eta_\varepsilon(t)|^2 dx \nonumber \\
& = \alpha \int_0^t \int_0^L |\partial_{txx}\eta_\varepsilon|^2 dxds + \alpha \int_0^L \partial_{xxx}\eta_\varepsilon(t) \partial_{tx}\eta_\varepsilon(t) dx - \alpha \int_0^L \partial_{xxx}\eta_0^\varepsilon \partial_{x}\eta_1^\varepsilon dx \nonumber \\
& - \int_0^t \int_0^L T_{a,b}[v_\varepsilon, q_\varepsilon] \cdot e_2 \partial_{tt}\eta_\varepsilon dxds - \int_0^t \int_0^L h_{|y=1} e_2 \cdot e_2 \partial_{tt}\eta_\varepsilon dxds + \int_0^t \int_0^L m \partial_{tt}\eta_\varepsilon dxds + \frac{\varepsilon}{2} \int_0^L |\partial_{xx}\eta_1^\varepsilon|^2. \tag{3.38}
\end{align}
$$

Now by adding (3.34), (3.37) and (3.38), we obtain

$$
\begin{align*}
& \int_0^t \int_{\Omega_0} \rho_{f,a,b} |\partial_t v_\epsilon|^2 dxdyds + \rho_s \int_0^t \int_0^L |\partial_{tt}\zeta_\epsilon|^2 dxd s + \rho_s \int_0^t \int_0^L |\partial_{tt}\eta_\epsilon|^2 dxd s \\
& \qquad + \mu \int_{\Omega_0} \frac{1}{\det \nabla \chi_{a,b}} |((B_{a,b} \nabla) v_\epsilon)(t) + ((B_{a,b} \nabla) v_\epsilon)(t)^T)|^2 dxdy + \frac{\epsilon}{2} \left( \int_0^L |\partial_{xxt}\zeta_\epsilon(t)|^2 dx + \int_0^L |\partial_{xxt}\eta_\epsilon(t)|^2 dx \right) \\
& = \beta \int_0^t \int_0^L |\partial_{xt}\zeta_\epsilon|^2 dxd s + \alpha \int_0^t \int_0^L |\partial_{txx}\eta_\epsilon|^2 dxd s \\
& \qquad + \beta \int_0^L \partial_{xx}\zeta_\epsilon(t) \partial_t \zeta_\epsilon(t) dx + \alpha \int_0^L \partial_{xxx}\eta_\epsilon(t) \partial_{tx}\eta(t) dx - \beta \int_0^L \partial_{xx}\zeta_0^\epsilon \zeta_1^\epsilon dx - \alpha \int_0^L \partial_{xxx}\eta_0^\epsilon \partial_{x}\eta_1^\epsilon dx \\
& + \int_0^t \int_{\Omega_0} f \cdot \partial_t v_\epsilon dxdyds + \int_0^t \int_{\Omega_0} \operatorname{div} h \cdot \partial_t v_\epsilon dxdyds - \int_0^t \int_0^L h_{|y=1} e_2 \cdot (\partial_{tt}\zeta_\epsilon, \partial_{tt}\eta_\epsilon) dxd s + \int_0^t \int_0^L (\ell, m) \cdot (\partial_{tt}\zeta_\epsilon, \partial_{tt}\eta_\epsilon) dxd s \\
& \qquad + \mu \int_{\Omega_0} \frac{1}{\det \nabla \chi_{a,b}} |((B_{a,b} \nabla) v_0^\epsilon) + ((B_{a,b} \nabla) v_0^\epsilon)^T|^2 dxdy + \frac{\epsilon}{2} \left( \int_0^L |\partial_{xx}\zeta_1^\epsilon|^2 dx + \int_0^L |\partial_{xx}\eta_1^\epsilon|^2 dx \right).
\end{align*}
$$

Then by using uniform Korn's inequality (2.35) we have

$$
\begin{align}
& \frac{\rho_f}{2R} \|\partial_t v_\epsilon\|_{L^2(0,t;L^2_\#(\Omega_0))}^2 + \frac{\rho_s}{2} \|\partial_{tt}\zeta_\epsilon\|_{L^2(0,t;L^2_\#(0,L))}^2 + \frac{\rho_s}{2} \|\partial_{tt}\eta_\epsilon\|_{L^2(0,t;L^2_\#(0,L))}^2 \nonumber \\
& \qquad + \frac{\mu\lambda(R)}{4} \|\nabla v_\epsilon\|_{L^2(0,t;L^2_\#(\Omega_0))}^2 + \gamma \|\partial_{xxt}\eta_\epsilon\|_{L^2(0,t;L^2_\#(0,L))}^2 \nonumber \\
& \qquad + \frac{\epsilon}{2} \left( \|\partial_{xxt}\zeta_\epsilon(t)\|_{L^2_\#(0,L)}^2 + \|\partial_{xxt}\eta_\epsilon(t)\|_{L^2_\#(0,L)}^2 \right) \le C \left( \frac{R}{\rho_f} \right) \left( \|f\|_{L^2(0,t;L^2_\#(\Omega_0))}^2 + \|h\|_{L^2(0,t;H_\#^1(\Omega_0))}^2 \right) \nonumber \\
& \qquad + \frac{1}{2\rho_s} \left( C(\Omega_0) \|h\|_{L^2(0,t;H_\#^1(\Omega_0))}^2 + \|\ell\|_{L^2(0,t;L^2_\#(0,L))}^2 + \|m\|_{L^2(0,t;L^2_\#(0,L))}^2 \right) + C_0(R) \nonumber \\
& + \beta \|\partial_{xt}\zeta_\epsilon\|_{L^2(0,t;L^2_\#(\Omega_0))}^2 + \alpha \|\partial_{txx}\eta_\epsilon\|_{L^2(0,t;L^2_\#(\Omega_0))}^2 + \beta \int_0^L \partial_{xx}\zeta_\epsilon(t) \partial_t \zeta_\epsilon(t) dx + \alpha \int_0^L \partial_{xxx}\eta_\epsilon(t) \partial_{tx}\eta_\epsilon(t) dx, \tag{3.39}
\end{align}
$$

where $C_0(R)$ is a positive constant which depends on $R$ and on initial data and is independent of $\epsilon$.

It remains to bound the last four terms in the right hand side of the above inequality. To get a bound independent of $\epsilon$, we multiply the fluid equation (3.15) by $-\partial_{xx}v_\epsilon$, the longitudinal and transversal displacement of the structure equations (3.19) and (3.20) respectively by $-\partial_{xxt}\zeta_\epsilon$ and $-\partial_{xxt}\eta_\epsilon$. We underline that though $-\partial_{xx}v_\epsilon$ matches $(-\partial_{xxt}\zeta_\epsilon, -\partial_{xxt}\eta_\epsilon)$ at the fluid structure interface, but $(-\partial_{xx}v_\epsilon, -\partial_{xxt}\zeta_\epsilon, -\partial_{xxt}\eta_\epsilon)$ is not an admissible triplet of test functions as $\operatorname{div}(B_{a,b}^T\partial_{xx}v_\epsilon) \neq 0$. This will lead to the appearance of pressure terms that we have to carefully control, in particular boundary terms that do not appear when considering only the transversal case. At this step we need to assume that $(a, b) \in H_\#^2(0, L)$ to take care of these extra terms. By integrating the resulting equations
and for the vertical displacement

$$
\begin{align}
& \rho_s \int_0^t \int_0^L |\partial_{tt}\eta_\varepsilon|^2 dxds + \gamma \int_0^t \int_0^L |\partial_{xxt}\eta_\varepsilon|^2 dxds + \frac{\varepsilon}{2} \int_0^L |\partial_{xxt}\eta_\varepsilon(t)|^2 dx \nonumber \\
& = \alpha \int_0^t \int_0^L |\partial_{txx}\eta_\varepsilon|^2 dxds + \alpha \int_0^L \partial_{xxx}\eta_\varepsilon(t) \partial_{tx}\eta_\varepsilon(t) dx - \alpha \int_0^L \partial_{xxx}\eta_0^\varepsilon \partial_x \eta_1^\varepsilon dx \nonumber \\
& - \int_0^t \int_0^L T_{a,b}[v_\varepsilon, q_\varepsilon] \cdot e_2 \partial_{tt}\eta_\varepsilon dxds - \int_0^t \int_0^L h_{|y=1} e_2 \cdot e_2 \partial_{tt}\eta_\varepsilon dxds + \int_0^t \int_0^L m \partial_{tt}\eta_\varepsilon dxds + \frac{\varepsilon}{2} \int_0^L |\partial_{xx}\eta_1^\varepsilon|^2. \tag{3.38}
\end{align}
$$

Now by adding (3.34), (3.37) and (3.38), we obtain

$$
\begin{align*}
& \int_0^t \int_{\Omega_0} \rho_{f,a,b} |\partial_t v_\epsilon|^2 dxdyds + \rho_s \int_0^t \int_0^L |\partial_{tt}\zeta_\epsilon|^2 dxdss + \rho_s \int_0^t \int_0^L |\partial_{xxt}\eta_\epsilon|^2 dxdss + \gamma \int_0^t \int_0^L |\partial_{xxt}\eta_\epsilon|^2 dxdss \\
& \qquad + \mu \int_{\Omega_0} \frac{1}{\det \nabla \chi_{a,b}} |((B_{a,b} \nabla) v_\epsilon)(t) + ((B_{a,b} \nabla) v_\epsilon)(t))^T|^2 dxdy + \frac{\epsilon}{2} \left( \int_0^L |\partial_{xxt}\zeta_\epsilon(t)|^2 dx + \int_0^L |\partial_{xxt}\eta_\epsilon(t)|^2 dx \right) \\
& = \beta \int_0^t \int_0^L |\partial_{xt}\zeta_\epsilon|^2 dxdss + \alpha \int_0^t \int_0^L |\partial_{txx}\eta_\epsilon|^2 dxdss \\
& \qquad + \beta \int_0^L \partial_{xx}\zeta_\epsilon(t) \partial_t \zeta_\epsilon(t) dx + \alpha \int_0^L \partial_{xxx}\eta_\epsilon(t) \partial_{tx}\eta(t) dx - \beta \int_0^L \partial_{xx}\zeta_0^\epsilon \zeta_1^\epsilon dx - \alpha \int_0^L \partial_{xxx}\eta_0^\epsilon \partial_x \eta_1^\epsilon dx \\
& + \int_0^t \int_{\Omega_0} f \cdot \partial_t v_\epsilon dxdyds + \int_0^t \int_{\Omega_0} \operatorname{div} h \cdot \partial_t v_\epsilon dxdyds - \int_0^t \int_0^L h_{|y=1} e_2 \cdot (\partial_{tt}\zeta_\epsilon, \partial_{tt}\eta_\epsilon) dxdss + \int_0^t \int_0^L (\ell, m) \cdot (\partial_{tt}\zeta_\epsilon, \partial_{tt}\eta_\epsilon) dxdss \\
& \qquad + \mu \int_{\Omega_0} \frac{1}{\det \nabla \chi_{a,b}} |((B_{a,b} \nabla) v_0^\epsilon) + ((B_{a,b} \nabla) v_0^\epsilon)^T|^2 dxdy + \frac{\epsilon}{2} \left( \int_0^L |\partial_{xx}\zeta_1^\epsilon|^2 dx + \int_0^L |\partial_{xx}\eta_1^\epsilon|^2 dx \right).
\end{align*}
$$

Then by using uniform Korn's inequality (2.35) we have

$$
\begin{align}
& \frac{\rho_f}{2R} \|\partial_t v_\epsilon\|_{L^2(0,t;L^2_\#(\Omega_0))}^2 + \frac{\rho_s}{2} \|\partial_{tt}\zeta_\epsilon\|_{L^2(0,t;L^2_\#(0,L))}^2 + \frac{\rho_s}{2} \|\partial_{tt}\eta_\epsilon\|_{L^2(0,t;L^2_\#(0,L))}^2 \nonumber \\
& \qquad + \frac{\mu\lambda(R)}{4} \|\nabla v_\epsilon\|_{L^2(0,t;L^2_\#(\Omega_0))}^2 + \gamma \|\partial_{xxt}\eta_\epsilon\|_{L^2(0,t;L^2_\#(0,L))}^2 \nonumber \\
& \qquad + \frac{\epsilon}{2} \left( \|\partial_{xxt}\zeta_\epsilon(t)\|_{L^2_\#(0,L)}^2 + \|\partial_{xxt}\eta_\epsilon(t)\|_{L^2_\#(0,L)}^2 \right) \le C \left( \frac{R}{\rho_f} \right) \left( \|f\|_{L^2(0,t;L^2_\#(\Omega_0))}^2 + \|h\|_{L^2(0,t;H_\#^1(\Omega_0))}^2 \right) \nonumber \\
& \qquad + \frac{1}{2\rho_s} \left( C(\Omega_0) \|h\|_{L^2(0,t;H_\#^1(\Omega_0))}^2 + \|\ell\|_{L^2(0,t;L^2_\#(0,L))}^2 + \|m\|_{L^2(0,t;L^2_\#(0,L))}^2 \right) + C_0(R) \nonumber \\
& + \beta \|\partial_{xt}\zeta_\epsilon\|_{L^2(0,t;L^2_\#(\Omega_0))}^2 + \alpha \|\partial_{txx}\eta_\epsilon\|_{L^2(0,t;L^2_\#(\Omega_0))}^2 + \beta \int_0^L \partial_{xx}\zeta_\epsilon(t) \partial_t \zeta_\epsilon(t) dx + \alpha \int_0^L \partial_{xxx}\eta_\epsilon(t) \partial_{tx}\eta_\epsilon(t) dx, \tag{3.39}
\end{align}
$$

where $C_0(R)$ is a positive constant which depends on $R$ and on initial data and is independent of $\epsilon$.

It remains to bound the last four terms in the right hand side of the above inequality. To get a bound independent of $\epsilon$, we multiply the fluid equation (3.15) by $-\partial_{xx}v_\epsilon$, the longitudinal and transversal displacement of the structure equations (3.19) and (3.20) respectively by $-\partial_{xxt}\zeta_\epsilon$ and $-\partial_{xxt}\eta_\epsilon$. We underline that though $-\partial_{xx}v_\epsilon$ matches $(-\partial_{xxt}\zeta_\epsilon, -\partial_{xxt}\eta_\epsilon)$ at the fluid structure interface, but $(-\partial_{xx}v_\epsilon, -\partial_{xxt}\zeta_\epsilon, -\partial_{xxt}\eta_\epsilon)$ is not an admissible triplet of test functions as $\operatorname{div}(B_{a,b}^T \partial_{xx}v_\epsilon) \neq 0$. This will lead to the appearance of pressure terms that we have to carefully control, in particular boundary terms that do not appear when considering only the transversal case. At this step we need to assume that $(a, b) \in H_\#^2(0, L)$ to take care of these extra terms. By integrating the resulting equations
on $(0, t) \times (0, L)$ and after integration by parts in space, we obtain

$$
\begin{align}
& - \int_0^t \int_{\Omega_0} \rho_{f,a,b} \partial_t v_\epsilon \cdot \partial_{xx} v_\epsilon \,dx\,dy\,ds + \int_0^t \int_{\Omega_0} \operatorname{div} \left( \mu(A_{a,b} \nabla) v_\epsilon + \frac{\mu}{\det \nabla \chi_{a,b}} B_{a,b} \nabla v_\epsilon^T B_{a,b} - q_\epsilon B_{a,b} \right) \cdot \partial_{xx} v_\epsilon \,dx\,dy\,ds \nonumber \\
& + \frac{\rho_s}{2} \int_0^L |\partial_{tx} \zeta_\epsilon(t)|^2 dx + \frac{\rho_s}{2} \int_0^L |\partial_{tx} \eta_\epsilon(t)|^2 dx + \frac{\beta}{2} \int_0^L |\partial_{xxx} \zeta_\epsilon(t)|^2 dx + \frac{\alpha}{2} \int_0^L |\partial_{xxx} \eta_\epsilon(t)|^2 dx + \frac{\gamma}{2} \int_0^L |\partial_{txx} \eta_\epsilon(t)|^2 dx \nonumber \\
& + \varepsilon \left( \int_0^t \int_0^L |\partial_{xxxt} \zeta_\epsilon|^2 dx\,ds + \int_0^t \int_0^L |\partial_{xxxt} \eta_\epsilon|^2 dx\,ds \right) = - \int_0^t \int_{\Omega_0} f \cdot \partial_{xx} v_\epsilon \,dx\,dy\,ds - \int_0^t \int_{\Omega_0} \partial_{xh} : \nabla \partial_x v_\epsilon \,dx\,dy\,ds \nonumber \\
& \qquad + \int_0^t \int_0^L T[v_\epsilon, q_\epsilon] \cdot (\partial_{xxt} \zeta_\epsilon e_1 + \partial_{xxt} \eta_\epsilon e_2) \,dx\,ds - \int_0^t \int_0^L (\ell, m) \cdot (\partial_{xxt} \zeta_\epsilon e_1 + \partial_{xxt} \eta_\epsilon e_2) \,dx\,ds \nonumber \\
& \qquad + \frac{\rho_s}{2} \int_0^L |\partial_x \zeta_1^\epsilon|^2 dx + \frac{\rho_s}{2} \int_0^L |\partial_x \eta_1^\epsilon|^2 dx + \frac{\beta}{2} \int_0^L |\partial_{xxx} \zeta_0^\epsilon|^2 dx + \frac{\alpha}{2} \int_0^L |\partial_{xxx} \eta_0^\epsilon|^2 dx + \frac{\gamma}{2} \int_0^L |\partial_{xx} \eta_1^\epsilon|^2 dx. \tag{3.40}
\end{align}
$$

Since $\partial_{xx} v_\epsilon \in L^2(0, T; L_\#^2(\Omega_0))$, it is not regular enough to perform integration by parts in space directly. To proceed we state the following lemma and we refer the reader to [28] for a similar proof.

**Lemma 3.2** For $w \in L^2(0, T; H_\#^2(\Omega_0)) \cap H^1(0, T; L_\#^2(\Omega_0))$, such that $w|_{y=1} = (\partial_t\zeta, \partial_t\eta)$ with $(\zeta, \eta) \in H^1(0, T; H_\#^2(0, L))$ and $w|_{y=0} = 0$, and for $q \in L^2(0, T; H_\#^1(\Omega_0))$, we have

$$
\begin{align*}
& - \int_0^t \int_{\Omega_0} \rho_{f,a,b} \partial_t w \cdot \partial_{xx} w \,dx\,dy\,ds + \int_0^t \int_{\Omega_0} \operatorname{div} \left( \mu(A_{a,b} \nabla) w + \frac{\mu}{\det \nabla \chi_{a,b}} B_{a,b} \nabla w^T B_{a,b} - q B_{a,b} \right) \cdot \partial_{xx} w \,dx\,dy\,ds \\
&= \frac{1}{2} \int_{\Omega_0} \rho_{f,a,b} |\partial_x w(t)|^2 \,dx\,dy - \frac{1}{2} \int_{\Omega_0} \rho_{f,a,b} |\partial_x w(0)|^2 \,dx\,dy + \int_0^t \int_{\Omega_0} \partial_x \rho_{f,a,b} \partial_t w \cdot \partial_x w \,dx\,dy\,ds \\
&\quad + \mu \int_0^t \int_{\Omega_0} \left( (A_{a,b} \nabla) \partial_x w + \frac{1}{\det \nabla \chi_{a,b}} B_{a,b} \nabla \partial_x w^T B_{a,b} \right) : \nabla \partial_x w \,dx\,dy\,ds + \mu \int_0^t \int_{\Omega_0} (\partial_x A_{a,b} \nabla) w : \nabla \partial_x w \,dx\,dy\,ds \\
&\quad + \mu \int_0^t \int_{\Omega_0} \partial_x \left( \frac{1}{\det \nabla \chi_{a,b}} B_{a,b} \right) \nabla w^T B_{a,b} : \nabla \partial_x w \,dx\,dy\,ds + \mu \int_0^t \int_{\Omega_0} \frac{1}{\det \nabla \chi_{a,b}} B_{a,b} \nabla w^T \partial_x B_{a,b} : \nabla \partial_x w \,dx\,dy\,ds \\
&\quad - \int_0^t \int_{\Omega_0} \partial_x q B_{a,b} : \nabla \partial_x w \,dx\,dy\,ds - \int_0^t \int_{\Omega_0} q \partial_x B_{a,b} : \nabla \partial_x w \,dx\,dy\,ds \\
&\quad + \int_0^t \left\langle T_{a,b}[w, q] \cdot e_1, \partial_{xxt}\zeta \right\rangle_{H_\#^{\frac{1}{2}}(0,L), H_\#^{-\frac{1}{2}}(0,L)} \,ds + \int_0^t \left\langle T_{a,b}[w, q] \cdot e_2, \partial_{xxt}\eta \right\rangle_{H_\#^{\frac{1}{2}}(0,L), H_\#^{-\frac{1}{2}}(0,L)} \,ds
\end{align*}
$$

By applying the above lemma with $(w, q) = (v_\epsilon, q_\epsilon)$ and combining the resulting equation with (3.40), taking into account uniform Korn inequality (2.35), we obtain

$$
\begin{align}
& \frac{\rho_f}{2R} \int_{\Omega_0} |\partial_x v_\epsilon(t)|^2 dx dy + \frac{\rho_s}{2} \int_0^L |\partial_{tx} \zeta_\epsilon(t)|^2 dx + \frac{\rho_s}{2} \int_0^L |\partial_{tx} \eta_\epsilon(t)|^2 dx dy + \frac{\mu \lambda(R)}{2} \int_0^t \int_{\Omega_0} |\nabla \partial_x v_\epsilon|^2 dx dy ds \nonumber \\
& + \frac{\beta}{2} \int_0^L |\partial_{xxx} \zeta_\epsilon(t)|^2 dx + \frac{\alpha}{2} \int_0^L |\partial_{xxx} \eta_\epsilon(t)|^2 dx + \frac{\gamma}{2} \int_0^L |\partial_{txx} \eta_\epsilon(t)|^2 dx + \varepsilon \left( \int_0^t \int_0^L |\partial_{xxxt} \zeta_\epsilon|^2 dx ds + \int_0^t \int_0^L |\partial_{xxxt} \eta_\epsilon|^2 dx ds \right) \nonumber \\
& \leq - \int_0^t \int_{\Omega_0} f \cdot \partial_{xx} v_\epsilon \,dx\,dy\,ds - \int_0^t \int_{\Omega_0} \partial_{xh} : \nabla \partial_x v_\epsilon \,dx\,dy\,ds - \int_0^t \int_{\Omega_0} (\ell, m) \cdot (\partial_{xxt} \zeta_\epsilon, \partial_{xxt} \eta_\epsilon) \,dx\,ds - \int_0^t \int_{\Omega_0} \partial_x \rho_{f,a,b} \partial_t v_\epsilon \cdot \partial_x v_\epsilon \,dx\,dy\,ds \nonumber \\
& + \frac{1}{2} \int_{\Omega_0} \rho_{f,a,b} |\partial_x v_0^\epsilon|^2 dx dy + \frac{\rho_s}{2} \int_0^L |\partial_x \zeta_1^\epsilon|^2 dx + \frac{\rho_s}{2} \int_0^L |\partial_x \eta_1^\epsilon|^2 dx + \frac{\beta}{2} \int_0^L |\partial_{xxx} \zeta_0^\epsilon|^2 dx + \frac{\alpha}{2} \int_0^L |\partial_{xxx} \eta_0^\epsilon|^2 dx + \frac{\gamma}{2} \int_0^L |\partial_{xx} \eta_1^\epsilon|^2 dx \nonumber \\
& \quad - \mu \int_0^t \int_{\Omega_0} (\partial_x A_{a,b} \nabla) v_\epsilon : \nabla \partial_x v_\epsilon \,dx\,dy\,ds - \mu \int_0^t \int_{\Omega_0} \partial_x \left( \frac{1}{\det \nabla \chi_{a,b}} B_{a,b} \right) \nabla v_\epsilon^T B_{a,b} : \nabla \partial_x v_\epsilon \,dx\,dy\,ds \nonumber \\
& \quad - \mu \int_0^t \int_{\Omega_0} \frac{1}{\det \nabla \chi_{a,b}} B_{a,b} \nabla v_\epsilon^T \partial_x B_{a,b} : \nabla \partial_x v_\epsilon \,dx\,dy\,ds + \int_0^t \int_{\Omega_0} \partial_x q_\epsilon B_{a,b} : \nabla \partial_x v_\epsilon \,dx\,dy\,ds + \int_0^t \int_{\Omega_0} q_\epsilon \partial_x B_{a,b} : \nabla \partial_x v_\epsilon \,dx\,dy\,ds. \tag{3.41}
\end{align}
$$
on $(0, t) \times (0, L)$ and after integration by parts in space, we obtain

$$
\begin{align}
& - \int_0^t \int_{\Omega_0} \rho_{f,a,b} \partial_t v_\epsilon \cdot \partial_{xx} v_\epsilon \,dx\,dy\,ds + \int_0^t \int_{\Omega_0} \operatorname{div} \left( \mu(A_{a,b}\nabla)v_\epsilon + \frac{\mu}{\det \nabla\chi_{a,b}} B_{a,b} \nabla v_\epsilon^T B_{a,b} - q_\epsilon B_{a,b} \right) \cdot \partial_{xx} v_\epsilon \,dx\,dy\,ds \nonumber \\
& + \frac{\rho_s}{2} \int_0^L |\partial_{tx}\zeta_\epsilon(t)|^2 \,dx + \frac{\rho_s}{2} \int_0^L |\partial_{tx}\eta_\epsilon(t)|^2 \,dx + \frac{\beta}{2} \int_0^L |\partial_{xxx}\zeta_\epsilon(t)|^2 \,dx + \frac{\alpha}{2} \int_0^L |\partial_{xxx}\eta_\epsilon(t)|^2 \,dx + \frac{\gamma}{2} \int_0^L |\partial_{txx}\eta_\epsilon(t)|^2 \,dx \nonumber \\
& + \varepsilon \left( \int_0^t \int_0^L |\partial_{xxxt}\zeta_\epsilon|^2 \,dx\,ds + \int_0^t \int_0^L |\partial_{xxxt}\eta_\epsilon|^2 \,dx\,ds \right) = - \int_0^t \int_{\Omega_0} f \cdot \partial_{xx} v_\epsilon \,dx\,dy\,ds - \int_0^t \int_{\Omega_0} \partial_x h : \nabla \partial_x v_\epsilon \,dx\,dy\,ds \nonumber \\
& \qquad + \int_0^t \int_0^L T[v_\epsilon, q_\epsilon] \cdot (\partial_{xxt}\zeta_\epsilon e_1 + \partial_{xxt}\eta_\epsilon e_2) \,dx\,ds - \int_0^t \int_0^L (\ell, m) \cdot (\partial_{xxt}\zeta_\epsilon e_1 + \partial_{xxt}\eta_\epsilon e_2) \,dx\,ds \nonumber \\
& \qquad + \frac{\rho_s}{2} \int_0^L |\partial_x \zeta_1^\epsilon|^2 \,dx + \frac{\rho_s}{2} \int_0^L |\partial_x \eta_1^\epsilon|^2 \,dx + \frac{\beta}{2} \int_0^L |\partial_{xxx}\zeta_0^\epsilon|^2 \,dx + \frac{\alpha}{2} \int_0^L |\partial_{xxx}\eta_0^\epsilon|^2 \,dx + \frac{\gamma}{2} \int_0^L |\partial_{xx}\eta_1^\epsilon|^2 \,dx. \tag{3.40}
\end{align}
$$

Since $\partial_{xx} v_{\epsilon} \in L^2(0, T; L_{\#}^2(\Omega_0))$, it is not regular enough to perform integration by parts in space directly. To proceed we state the following lemma and we refer the reader to [28] for a similar proof.

**Lemma 3.2** For $w \in L^2(0, T; H_{\#}^2(\Omega_0)) \cap H^1(0, T; L_{\#}^2(\Omega_0))$, such that $w|_{y=1} = (\partial_t \zeta, \partial_t \eta)$ with $(\zeta, \eta) \in H^1(0, T; H_{\#}^2(0, L))$ and $w|_{y=0} = 0$, and for $q \in L^2(0, T; H_{\#}^1(\Omega_0))$, we have

$$
\begin{align*}
& - \int_0^t \int_{\Omega_0} \rho_{f,a,b} \partial_t w \cdot \partial_{xx} w \,dx\,dy\,ds + \int_0^t \int_{\Omega_0} \operatorname{div} \left( \mu(A_{a,b} \nabla) w + \frac{\mu}{\det \nabla \chi_{a,b}} B_{a,b} \nabla w^T B_{a,b} - q B_{a,b} \right) \cdot \partial_{xx} w \,dx\,dy\,ds \\
&= \frac{1}{2} \int_{\Omega_0} \rho_{f,a,b} |\partial_x w(t)|^2 \,dx\,dy - \frac{1}{2} \int_{\Omega_0} \rho_{f,a,b} |\partial_x w(0)|^2 \,dx\,dy + \int_0^t \int_{\Omega_0} \partial_x \rho_{f,a,b} \partial_t w \cdot \partial_x w \,dx\,dy\,ds \\
&\quad + \mu \int_0^t \int_{\Omega_0} \left( (A_{a,b} \nabla) \partial_x w + \frac{1}{\det \nabla \chi_{a,b}} B_{a,b} \nabla \partial_x w^T B_{a,b} \right) : \nabla \partial_x w \,dx\,dy\,ds + \mu \int_0^t \int_{\Omega_0} (\partial_x A_{a,b} \nabla) w : \nabla \partial_x w \,dx\,dy\,ds \\
&\quad + \mu \int_0^t \int_{\Omega_0} \partial_x \left( \frac{1}{\det \nabla \chi_{a,b}} B_{a,b} \right) \nabla w^T B_{a,b} : \nabla \partial_x w \,dx\,dy\,ds + \mu \int_0^t \int_{\Omega_0} \frac{1}{\det \nabla \chi_{a,b}} B_{a,b} \nabla w^T \partial_x B_{a,b} : \nabla \partial_x w \,dx\,dy\,ds \\
&\quad - \int_0^t \int_{\Omega_0} \partial_x q B_{a,b} : \nabla \partial_x w \,dx\,dy\,ds - \int_0^t \int_{\Omega_0} q \partial_x B_{a,b} : \nabla \partial_x w \,dx\,dy\,ds \\
&\quad + \int_0^t \left\langle T_{a,b}[w, q] \cdot e_1, \partial_{xxt}\zeta \right\rangle_{H_{\#}^{\frac{1}{2}}(0,L), H_{\#}^{-\frac{1}{2}}(0,L)} \,ds + \int_0^t \left\langle T_{a,b}[w, q] \cdot e_2, \partial_{xxt}\eta \right\rangle_{H_{\#}^{\frac{1}{2}}(0,L), H_{\#}^{-\frac{1}{2}}(0,L)} \,ds
\end{align*}
$$

By applying the above lemma with $(w, q) = (v_{\epsilon}, q_{\epsilon})$ and combining the resulting equation with (3.40), taking into account uniform Korn inequality (2.35), we obtain

$$
\begin{align}
& \frac{\rho_f}{2R} \int_{\Omega_0} |\partial_x v_\epsilon(t)|^2 dx dy + \frac{\rho_s}{2} \int_0^L |\partial_{tx} \zeta_\epsilon(t)|^2 dx + \frac{\rho_s}{2} \int_0^L |\partial_{tx} \eta_\epsilon(t)|^2 dx dy + \frac{\mu \lambda(R)}{2} \int_0^t \int_{\Omega_0} |\nabla \partial_x v_\epsilon|^2 dx dy ds \nonumber \\
& + \frac{\beta}{2} \int_0^L |\partial_{xxx} \zeta_\epsilon(t)|^2 dx + \frac{\alpha}{2} \int_0^L |\partial_{xxx} \eta_\epsilon(t)|^2 dx + \frac{\gamma}{2} \int_0^L |\partial_{txx} \eta_\epsilon(t)|^2 dx + \varepsilon \left( \int_0^t \int_0^L |\partial_{xxxt} \zeta_\epsilon|^2 dx ds + \int_0^t \int_0^L |\partial_{xxxt} \eta_\epsilon|^2 dx ds \right) \nonumber \\
& \le - \int_0^t \int_{\Omega_0} f \cdot \partial_{xx} v_\epsilon \,dx\,dy\,ds - \int_0^t \int_{\Omega_0} \partial_x h : \nabla \partial_x v_\epsilon \,dx\,dy\,ds - \int_0^t \int_{\Omega_0} (\ell, m) \cdot (\partial_{xxt} \zeta_\epsilon, \partial_{xxt} \eta_\epsilon) \,dx\,ds - \int_0^t \int_{\Omega_0} \partial_x \rho_{f,a,b} \partial_t v_\epsilon \cdot \partial_x v_\epsilon \,dx\,dy\,ds \nonumber \\
& + \frac{1}{2} \int_{\Omega_0} \rho_{f,a,b} |\partial_x v_0^\epsilon|^2 dx dy + \frac{\rho_s}{2} \int_0^L |\partial_x \zeta_1^\epsilon|^2 dx + \frac{\rho_s}{2} \int_0^L |\partial_x \eta_1^\epsilon|^2 dx + \frac{\beta}{2} \int_0^L |\partial_{xxx} \zeta_0^\epsilon|^2 dx + \frac{\alpha}{2} \int_0^L |\partial_{xxx} \eta_0^\epsilon|^2 dx + \frac{\gamma}{2} \int_0^L |\partial_{xx} \eta_1^\epsilon|^2 dx \nonumber \\
& \quad - \mu \int_0^t \int_{\Omega_0} (\partial_x A_{a,b} \nabla) v_\epsilon : \nabla \partial_x v_\epsilon \,dx\,dy\,ds - \mu \int_0^t \int_{\Omega_0} \partial_x \left( \frac{1}{\det \nabla \chi_{a,b}} B_{a,b} \right) \nabla v_\epsilon^T B_{a,b} : \nabla \partial_x v_\epsilon \,dx\,dy\,ds \nonumber \\
& \quad - \mu \int_0^t \int_{\Omega_0} \frac{1}{\det \nabla \chi_{a,b}} B_{a,b} \nabla v_\epsilon^T \partial_x B_{a,b} : \nabla \partial_x v_\epsilon \,dx\,dy\,ds + \int_0^t \int_{\Omega_0} \partial_x q_\epsilon B_{a,b} : \nabla \partial_x v_\epsilon \,dx\,dy\,ds + \int_0^t \int_{\Omega_0} q_\epsilon \partial_x B_{a,b} : \nabla \partial_x v_\epsilon \,dx\,dy\,ds. \tag{3.41}
\end{align}
$$
We start by bounding each term in the right hand side of the above inequality. By using Holder's and Young's inequality, we bound the terms including the external body forces:

$$ \left| \int_0^t \int_{\Omega_0} f \cdot \partial_{xx} v_{\varepsilon} \, dxdyds + \int_0^t \int_{\Omega_0} \partial_x h : \nabla \partial_x v_{\varepsilon} \, dxdyds \right| \le \int_0^t \left( \|f\|_{L^2(0,t;H^1_\frac{1}{2}(\Omega_0))} + \|h\|_{H^1(\Omega_0)} \right) \|\nabla \partial_x v_{\varepsilon}\|_{L^2(\Omega_0)} \\ \le C(R) \left( \|f\|_{L^2(0,t;L^2_\frac{1}{2}(\Omega_0))}^2 + \|h\|_{L^2(0,t;H^1_\frac{1}{2}(\Omega_0))}^2 \right) + \frac{\mu\lambda(R)}{N} \|\nabla \partial_x v_{\varepsilon}\|_{L^2(0,t;L^2_\frac{1}{2}(\Omega_0))}^2, \quad (3.42) $$

where N is a positive integer to be fixed later on.

Next, thanks to the equality of the fluid structure velocities at the interface (3.17) and by using Poincaré inequality for $\partial_x v_{\varepsilon}$ that is valid since $\partial_x v_{\varepsilon}|_{y=0} = 0$, we have

$$ \begin{aligned} \left| \int_0^t \int_0^L (\ell, m) \cdot (\partial_{xxt}\zeta_{\varepsilon}, \partial_{xxt}\eta_{\varepsilon}) \, dxds \right| & \le \|\partial_{xxt}\zeta_{\varepsilon}\|_{L^2(0,t;H^1_\frac{1}{2}(0,L))} \|\ell\|_{L^2(0,t;H^1_\frac{1}{2}(0,L))} \\ & \qquad + \|\partial_{tx}\eta_{\varepsilon}\|_{L^2(0,t;H^1_\frac{1}{2}(0,L))} \|m\|_{L^2(0,t;H^1_\frac{1}{2}(0,L))}, \\ & \le \|\partial_x v_{\varepsilon}\|_{L^2(0,t;H^1_\frac{1}{2}(0,L))} \|\ell\|_{L^2(0,t;H^1_\frac{1}{2}(0,L))} + \|\partial_x v_{\varepsilon}\|_{L^2(0,t;H^1_\frac{1}{2}(0,L))} \|m\|_{L^2(0,t;H^1_\frac{1}{2}(0,L))}, \\ & \le \frac{\mu\lambda(R)}{N} \|\partial_x \nabla v_{\varepsilon}\|_{L^2(0,t;L^2_\frac{1}{2}(\Omega_0))}^2 + C(R) \|\ell\|_{L^2(0,t;H^1_\frac{1}{2}(0,L))}^2 + C(R) \|m\|_{L^2(0,t;H^1_\frac{1}{2}(0,L))}^2. \end{aligned} \quad (3.43) $$

Next, since $\partial_x \rho_{f,a,b} = \rho_f \partial_x (\det \nabla \chi_{a,b}) \in H^s_\#(\Omega_0)$, then by using Holder's and Young's inequalities and Lemma 2.1, we obtain

$$ \left| \int_0^t \int_{\Omega_0} \partial_x \rho_{f,a,b} \partial_t v_{\varepsilon} \cdot \partial_x v_{\varepsilon} \, dxdyds \right| \le \int_0^t \|\partial_x \rho_{f,a,b} \partial_x v_{\varepsilon}\|_{L^2_\#(\Omega_0)} \|\partial_t v_{\varepsilon}\|_{L^2_\#(\Omega_0)} \, ds \\ \le K(R) \int_0^t \left( C_\delta \|\partial_x v_{\varepsilon}\|_{L^2_\#(\Omega_0)} + \delta \|\nabla \partial_x v_{\varepsilon}\|_{L^2_\#(\Omega_0)} \right) \|\partial_t v_{\varepsilon}\|_{L^2_\#(\Omega_0)} \, ds \\ \le K(R, \frac{1}{\kappa}) \int_0^t \left( C_\delta \|\partial_x v_{\varepsilon}\|_{L^2_\#(\Omega_0)} + \delta \|\nabla \partial_x v_{\varepsilon}\|_{L^2_\#(\Omega_0)} \right)^2 \, ds + \kappa \int_0^t \|\partial_t v_{\varepsilon}\|_{L^2_\#(\Omega_0)}^2 \, ds. $$

It yields

$$ \left| \int_0^t \int_{\Omega_0} \partial_x \rho_{f,a,b} \partial_t v_{\varepsilon} \cdot \partial_x v_{\varepsilon} \, dxdyds \right| \le K_\delta(R, \frac{1}{\kappa}) \|\partial_x v_{\varepsilon}\|_{L^2(0,t;L^2_\#(\Omega_0))}^2 \\ + \delta^2 K(R, \frac{1}{\kappa}) \|\nabla \partial_x v_{\varepsilon}\|_{L^2(0,t;L^2_\#(\Omega_0))}^2 + \kappa \|\partial_t v_{\varepsilon}\|_{L^2(0,t;L^2_\#(\Omega_0))}^2. \quad (3.44) $$

Thanks to the way we construct the approximated initial conditions, we have

$$ \frac{1}{2} \int_{\Omega_0} \rho_{f,a,b} |\partial_x v_0^\varepsilon|^2 + \frac{\rho_s}{2} \int_0^L |\partial_x \zeta_1^\varepsilon|^2 + \frac{\beta}{2} \int_0^L |\partial_{xx} \zeta_0^\varepsilon|^2 + \frac{\alpha}{2} \int_0^L |\partial_{xxx} \eta_0^\varepsilon|^2 + \frac{\gamma}{2} \int_0^L |\partial_{xxx} \eta_1^\varepsilon|^2 \, dx \le C_0(R), \quad (3.45) $$

where $C_0$ is a positive constant depending on initial data and R.

Again by using Holder's inequality, (2.64), Young's inequality and thanks to the energy estimate (3.26), we have:

$$ \begin{aligned} \left| \int_0^t \int_{\Omega_0} \mu(\partial_x A_{a,b} \nabla) v_{\varepsilon} : \nabla \partial_x v_{\varepsilon} \, dxdyds \right| & \le \mu \int_0^t \| (\partial_x A_{a,b} \nabla) v_{\varepsilon} \|_{L^2_\#(\Omega_0)} \| \nabla \partial_x v_{\varepsilon} \|_{L^2_\#(\Omega_0)} \, ds \\ & \le \mu K(R) \int_0^t \left( C_\delta \| \nabla v_{\varepsilon} \|_{L^2_\#(\Omega_0)} + \delta \| \nabla^2 v_{\varepsilon} \|_{L^2_\#(\Omega_0)} \right) \| \nabla \partial_x v_{\varepsilon} \|_{L^2_\#(\Omega_0)} \, ds \\ & \le K(R) \int_0^t \left( C_\delta \| \nabla v_{\varepsilon} \|_{L^2_\#(\Omega_0)} + \delta \| \nabla^2 v_{\varepsilon} \|_{L^2_\#(\Omega_0)} \right)^2 \, ds \\ & \qquad + \frac{\mu \lambda(R)}{N} \int_0^t \| \nabla \partial_x v_{\varepsilon} \|_{L^2_\#(\Omega_0)}^2 \, ds. \end{aligned} $$
We start by bounding each term in the right hand side of the above inequality. By using Holder's and Young's inequality, we bound the terms including the external body forces:

$$ \left| \int_0^t \int_{\Omega_0} f \cdot \partial_{xx} v_\varepsilon \, dxdyds + \int_0^t \int_{\Omega_0} \partial_x h : \nabla \partial_x v_\varepsilon \, dxdyds \right| \le \int_0^t \left( \|f\|_{L^2(0,t;H^1_\frac{1}{2}(\Omega_0))} + \|h\|_{H^1(\Omega_0)} \right) \|\nabla \partial_x v_\varepsilon\|_{L^2(\Omega_0)} \\ \le C(R) \left( \|f\|_{L^2(0,t;L^2_\frac{1}{2}(\Omega_0))}^2 + \|h\|_{L^2(0,t;H^1_\frac{1}{2}(\Omega_0))}^2 \right) + \frac{\mu\lambda(R)}{N} \|\nabla \partial_x v_\varepsilon\|_{L^2(0,t;L^2_\frac{1}{2}(\Omega_0))}^2, \quad (3.42) $$

where N is a positive integer to be fixed later on.

Next, thanks to the equality of the fluid structure velocities at the interface (3.17) and by using Poincaré inequality for $\partial_x v_\varepsilon$ that is valid since $\partial_x v_\varepsilon|_{y=0} = 0$, we have

$$ \left| \int_0^t \int_0^L (\ell, m) \cdot (\partial_{xxt}\zeta_\varepsilon, \partial_{xxt}\eta_\varepsilon) \, dxds \right| \le \|\partial_{xxt}\zeta_\varepsilon\|_{L^2(0,t;H^1_\frac{1}{2}(0,L))} \|\ell\|_{L^2(0,t;H^1_\frac{1}{2}(0,L))} \\ + \|\partial_{tx}\eta_\varepsilon\|_{L^2(0,t;H^1_\frac{1}{2}(0,L))} \|m\|_{L^2(0,t;H^1_\frac{1}{2}(0,L))}, \\ \le \|\partial_x v_\varepsilon\|_{L^2(0,t;H^1_\frac{1}{2}(0,L))} \|\ell\|_{L^2(0,t;H^1_\frac{1}{2}(0,L))} + \|\partial_x v_\varepsilon\|_{L^2(0,t;H^1_\frac{1}{2}(0,L))} \|m\|_{L^2(0,t;H^1_\frac{1}{2}(0,L))}, \\ \le \frac{\mu\lambda(R)}{N} \|\partial_x \nabla v_\varepsilon\|_{L^2(0,t;L^2_\frac{1}{2}(\Omega_0))}^2 + C(R) \|\ell\|_{L^2(0,t;H^1_\frac{1}{2}(0,L))}^2 + C(R) \|m\|_{L^2(0,t;H^1_\frac{1}{2}(0,L))}^2. \quad (3.43) $$

Next, since $\partial_x \rho_{f,a,b} = \rho_f \partial_x (\det \nabla \chi_{a,b}) \in H^s_\#(\Omega_0)$, then by using Holder's and Young's inequalities and Lemma 2.1, we obtain

$$ \left| \int_0^t \int_{\Omega_0} \partial_x \rho_{f,a,b} \partial_t v_\varepsilon \cdot \partial_x v_\varepsilon \, dxdyds \right| \le \int_0^t \|\partial_x \rho_{f,a,b} \partial_x v_\varepsilon\|_{L^2_\#(\Omega_0)} \|\partial_t v_\varepsilon\|_{L^2_\#(\Omega_0)} \, ds \\ \le K(R) \int_0^t \left( C_\delta \|\partial_x v_\varepsilon\|_{L^2_\#(\Omega_0)} + \delta \|\nabla \partial_x v_\varepsilon\|_{L^2_\#(\Omega_0)} \right) \|\partial_t v_\varepsilon\|_{L^2_\#(\Omega_0)} \, ds \\ \le K(R, \frac{1}{\kappa}) \int_0^t \left( C_\delta \|\partial_x v_\varepsilon\|_{L^2_\#(\Omega_0)} + \delta \|\nabla \partial_x v_\varepsilon\|_{L^2_\#(\Omega_0)} \right)^2 \, ds + \kappa \int_0^t \|\partial_t v_\varepsilon\|_{L^2_\#(\Omega_0)}^2 \, ds. $$

It yields

$$ \left| \int_0^t \int_{\Omega_0} \partial_x \rho_{f,a,b} \partial_t v_\varepsilon \cdot \partial_x v_\varepsilon \, dxdyds \right| \le K_\delta(R, \frac{1}{\kappa}) \|\partial_x v_\varepsilon\|_{L^2(0,t;L^2_\#(\Omega_0))}^2 \\ + \delta^2 K(R, \frac{1}{\kappa}) \|\nabla \partial_x v_\varepsilon\|_{L^2(0,t;L^2_\#(\Omega_0))}^2 + \kappa \|\partial_t v_\varepsilon\|_{L^2(0,t;L^2_\#(\Omega_0))}^2. \quad (3.44) $$

Thanks to the way we construct the approximated initial conditions, we have

$$ \frac{1}{2} \int_{\Omega_0} \rho_{f,a,b} |\partial_x v_0^\varepsilon|^2 + \frac{\rho_s}{2} \int_0^L |\partial_x \zeta_1^\varepsilon|^2 + \frac{\beta}{2} \int_0^L |\partial_{xx} \zeta_0^\varepsilon|^2 + \frac{\alpha}{2} \int_0^L |\partial_{xxx} \eta_0^\varepsilon|^2 + \frac{\gamma}{2} \int_0^L |\partial_{xxx} \eta_1^\varepsilon|^2 \, dx \le C_0(R), \quad (3.45) $$

where $C_0$ is a positive constant depending on initial data and R.

Again by using Holder's inequality, (2.64), Young's inequality and thanks to the energy estimate (3.26), we have:

$$ \left| \int_0^t \int_{\Omega_0} \mu(\partial_x A_{a,b} \nabla) v_\varepsilon : \nabla \partial_x v_\varepsilon \, dxdyds \right| \le \mu \int_0^t \| (\partial_x A_{a,b} \nabla) v_\varepsilon \|_{L^2_\#(\Omega_0)} \| \nabla \partial_x v_\varepsilon \|_{L^2_\#(\Omega_0)} \, ds \\ \le \mu K(R) \int_0^t \left( C_\delta \| \nabla v_\varepsilon \|_{L^2_\#(\Omega_0)} + \delta \| \nabla^2 v_\varepsilon \|_{L^2_\#(\Omega_0)} \right) \| \nabla \partial_x v_\varepsilon \|_{L^2_\#(\Omega_0)} \, ds \\ \le K(R) \int_0^t \left( C_\delta \| \nabla v_\varepsilon \|_{L^2_\#(\Omega_0)} + \delta \| \nabla^2 v_\varepsilon \|_{L^2_\#(\Omega_0)} \right)^2 \, ds \\ + \frac{\mu \lambda(R)}{N} \int_0^t \| \nabla \partial_x v_\varepsilon \|_{L^2_\#(\Omega_0)}^2 \, ds. $$
Remembering the energy estimate (3.26) it follows

$$ \int_0^t \int_{\Omega_0} \mu(\partial_x A_{a,b} \nabla) v_\epsilon : \nabla \partial_x v_\epsilon \, dxdyds \le C_0(R, \delta) + \left( \frac{\mu\lambda(R)}{N} + \delta^2 K(R) \right) \|\nabla \partial_x v_\epsilon\|_{L^2(0,t;L^2_\#(\Omega_0))}^2 \\ + \delta^2 K(R) \|\partial_{yy} v_\epsilon\|_{L^2(0,t;L^2_\#(\Omega_0))}^2, \quad (3.46) $$

where $C_0$ depends on $R$, on the data (physical parameters, initial data and applied forces) and on $\delta$. By using Proposition 2.3 and using (3.17), we have

$$ \begin{aligned} \|\partial_{yy} v_\epsilon\|_{L^2_\#(\Omega_0)} + \|\nabla s_{0,\epsilon}\|_{L^2_\#(\Omega_0)} &\le K(R) \left( \|f + \operatorname{div} h - \rho_{f,a,b} \partial_t v_\epsilon\|_{L^2_\#(\Omega_0)} + \|(\partial_t \zeta_\epsilon, \partial_t \eta_\epsilon)\|_{H^\frac{3}{2}_\#(0,L)} \right) \\ &\le K(R) \left( \|f\|_{L^2_\#(\Omega_0)} + \|h\|_{H^1_\#(\Omega_0)} + \|\partial_t v_\epsilon\|_{L^2_\#(\Omega_0)} \right. \\ &\qquad \left. + \|(\partial_t \zeta_\epsilon, \partial_t \eta_\epsilon)\|_{L^2_\#(0,L)} + \|(\partial_t x \zeta_\epsilon, \partial_t x \eta_\epsilon)\|_{H^\frac{1}{2}_\#(0,L)} \right) \\ &\le K(R) \left( \|f\|_{L^2_\#(\Omega_0)} + \|h\|_{H^1_\#(\Omega_0)} + \|\partial_t v_\epsilon\|_{L^2_\#(\Omega_0)} + \|(\partial_t \zeta_\epsilon, \partial_t \eta_\epsilon)\|_{L^2_\#(0,L)} + \|\partial_x v_\epsilon\|_{H^1_\#(\Omega_0)} \right). \end{aligned} $$

This implies

$$ \|\partial_{yy} v_\epsilon\|_{L^2_\#(\Omega_0)} + \|\nabla s_{0,\epsilon}\|_{L^2_\#(\Omega_0)} \le K(R) \left( \|f\|_{L^2_\#(\Omega_0)} + \|h\|_{H^1_\#(\Omega_0)} + \|\partial_t v_\epsilon\|_{L^2_\#(\Omega_0)} + \|(\partial_t \zeta_\epsilon, \partial_t \eta_\epsilon)\|_{L^2_\#(0,L)} \right. \\ \left. + \|\partial_x v_\epsilon\|_{L^2_\#(\Omega_0)} + \|\nabla \partial_x v_\epsilon\|_{L^2_\#(\Omega_0)} \right). \quad (3.47) $$

By combining the above inequality with (3.46) and the energy estimate (3.26), we obtain

$$ \int_0^t \int_{\Omega_0} \mu(\partial_x A_{a,b} \nabla) v_\epsilon : \nabla \partial_x v_\epsilon \, dxdyds \le C_0(R, \delta) + \left( \frac{\mu\lambda(R)}{N} + \delta^2 K(R, \mu) \right) \|\nabla \partial_x v_\epsilon\|_{L^2(0,t;L^2_\#(\Omega_0))}^2 \\ + \delta^2 K(R) \|\partial_t v_\epsilon\|_{L^2(0,T;L^2_\#(\Omega_0))}^2. \quad (3.48) $$

By using the same arguments we prove as above

$$ \left| \int_{\Omega_0} \partial_x \left( \frac{\mu}{\det \nabla \chi_{a,b}} B_{a,b} \right) \nabla v_\epsilon^T B_{a,b} : \nabla \partial_x v_\epsilon \, dxdy + \int_{\Omega_0} \frac{\mu}{\det \nabla \chi_{a,b}} B_{a,b} \nabla v_\epsilon^T \partial_x B_{a,b} : \nabla \partial_x v_\epsilon \, dxdy \right| \\ \le C_0(R, \delta) + \left( \frac{\mu\lambda(R)}{N} + \delta^2 K(R, \mu) \right) \|\nabla \partial_x v_\epsilon\|_{L^2(0,t;L^2_\#(\Omega_0))}^2 + \delta^2 K(R) \|\partial_{yy} v_\epsilon\|_{L^2(0,t;L^2_\#(\Omega_0))}^2. $$

By using (3.26) and (3.47), we obtain

$$ \left| \int_{\Omega_0} \partial_x \left( \frac{\mu}{\det \nabla \chi_{a,b}} B_{a,b} \right) \nabla v_\epsilon^T B_{a,b} : \nabla \partial_x v_\epsilon \, dxdy + \int_{\Omega_0} \frac{\mu}{\det \nabla \chi_{a,b}} B_{a,b} \nabla v_\epsilon^T \partial_x B_{a,b} : \nabla \partial_x v_\epsilon \, dxdy \right| \\ \le C_0(R, \delta) + \left( \frac{\mu\lambda(R)}{N} + \delta^2 K(R) \right) \|\nabla \partial_x v_\epsilon\|_{L^2(0,t;L^2_\#(\Omega_0))}^2 + \delta^2 K(R) \|\partial_t v_\epsilon\|_{L^2(0,t;L^2_\#(\Omega_0))}^2. \quad (3.49) $$

It remains to bound the pressure terms in (3.41). Since $\det(B_{a,b}^T v_\epsilon) = 0$, and thanks to the Piola identity,

$$ \begin{aligned} B_{a,b} : \nabla \partial_x v_\epsilon &= \partial_x (B_{a,b} : \nabla v_\epsilon) - \partial_x B_{a,b} : \nabla v_\epsilon \\ &= \partial_x \operatorname{div} (B_{a,b}^T v_\epsilon) - \partial_x B_{a,b} : \nabla v_\epsilon \\ &= -\partial_x B_{a,b} : \nabla v_\epsilon. \end{aligned} $$

Hence,

$$ \partial_x q_\epsilon B_{a,b} : \nabla \partial_x v_\epsilon = -\partial_x q_\epsilon \partial_x B_{a,b} : \nabla v_\epsilon. $$
Remembering the energy estimate (3.26) it follows

$$ \int_0^t \int_{\Omega_0} \mu(\partial_x A_{a,b} \nabla) v_\epsilon : \nabla \partial_x v_\epsilon \, dxdyds \le C_0(R, \delta) + \left( \frac{\mu\lambda(R)}{N} + \delta^2 K(R) \right) \|\nabla \partial_x v_\epsilon\|_{L^2(0,t;L^2_\#(\Omega_0))}^2 \\ + \delta^2 K(R) \|\partial_{yy} v_\epsilon\|_{L^2(0,t;L^2_\#(\Omega_0))}^2, \quad (3.46) $$

where $C_0$ depends on $R$, on the data (physical parameters, initial data and applied forces) and on $\delta$. By using Proposition 2.3 and using (3.17), we have

$$ \begin{aligned} \|\partial_{yy} v_\epsilon\|_{L^2_\#(\Omega_0)} + \|\nabla s_{0,\epsilon}\|_{L^2_\#(\Omega_0)} &\le K(R) \left( \|f + \operatorname{div} h - \rho_{f,a,b} \partial_t v_\epsilon\|_{L^2_\#(\Omega_0)} + \|(\partial_t \zeta_\epsilon, \partial_t \eta_\epsilon)\|_{H^\frac{3}{2}_\#(0,L)} \right) \\ &\le K(R) \left( \|f\|_{L^2_\#(\Omega_0)} + \|h\|_{H^1_\#(\Omega_0)} + \|\partial_t v_\epsilon\|_{L^2_\#(\Omega_0)} \right. \\ &\qquad \left. + \|(\partial_t \zeta_\epsilon, \partial_t \eta_\epsilon)\|_{L^2_\#(0,L)} + \|(\partial_t x \zeta_\epsilon, \partial_t x \eta_\epsilon)\|_{H^\frac{1}{2}_\#(0,L)} \right) \\ &\le K(R) \left( \|f\|_{L^2_\#(\Omega_0)} + \|h\|_{H^1_\#(\Omega_0)} + \|\partial_t v_\epsilon\|_{L^2_\#(\Omega_0)} + \|(\partial_t \zeta_\epsilon, \partial_t \eta_\epsilon)\|_{L^2_\#(0,L)} + \|\partial_x v_\epsilon\|_{H^1_\#(\Omega_0)} \right). \end{aligned} $$

This implies

$$ \|\partial_{yy} v_\epsilon\|_{L^2_\#(\Omega_0)} + \|\nabla s_{0,\epsilon}\|_{L^2_\#(\Omega_0)} \le K(R) \left( \|f\|_{L^2_\#(\Omega_0)} + \|h\|_{H^1_\#(\Omega_0)} + \|\partial_t v_\epsilon\|_{L^2_\#(\Omega_0)} + \|(\partial_t \zeta_\epsilon, \partial_t \eta_\epsilon)\|_{L^2_\#(0,L)} \right. \\ \left. + \|\partial_x v_\epsilon\|_{L^2_\#(\Omega_0)} + \|\nabla \partial_x v_\epsilon\|_{L^2_\#(\Omega_0)} \right). \quad (3.47) $$

By combining the above inequality with (3.46) and the energy estimate (3.26), we obtain

$$ \int_0^t \int_{\Omega_0} \mu(\partial_x A_{a,b} \nabla) v_\epsilon : \nabla \partial_x v_\epsilon \, dxdyds \le C_0(R, \delta) + \left( \frac{\mu\lambda(R)}{N} + \delta^2 K(R, \mu) \right) \|\nabla \partial_x v_\epsilon\|_{L^2(0,t;L^2_\#(\Omega_0))}^2 \\ + \delta^2 K(R) \|\partial_t v_\epsilon\|_{L^2(0,T;L^2_\#(\Omega_0))}^2. \quad (3.48) $$

By using the same arguments we prove as above

$$ \left| \int_{\Omega_0} \partial_x \left( \frac{\mu}{\det \nabla \chi_{a,b}} B_{a,b} \right) \nabla v_\epsilon^T B_{a,b} : \nabla \partial_x v_\epsilon \, dxdy + \int_{\Omega_0} \frac{\mu}{\det \nabla \chi_{a,b}} B_{a,b} \nabla v_\epsilon^T \partial_x B_{a,b} : \nabla \partial_x v_\epsilon \, dxdy \right| \\ \le C_0(R, \delta) + \left( \frac{\mu\lambda(R)}{N} + \delta^2 K(R, \mu) \right) \|\nabla \partial_x v_\epsilon\|_{L^2(0,t;L^2_\#(\Omega_0))}^2 + \delta^2 K(R) \|\partial_{yy} v_\epsilon\|_{L^2(0,t;L^2_\#(\Omega_0))}^2. $$

By using (3.26) and (3.47), we obtain

$$ \left| \int_{\Omega_0} \partial_x \left( \frac{\mu}{\det \nabla \chi_{a,b}} B_{a,b} \right) \nabla v_\epsilon^T B_{a,b} : \nabla \partial_x v_\epsilon \, dxdy + \int_{\Omega_0} \frac{\mu}{\det \nabla \chi_{a,b}} B_{a,b} \nabla v_\epsilon^T \partial_x B_{a,b} : \nabla \partial_x v_\epsilon \, dxdy \right| \\ \le C_0(R, \delta) + \left( \frac{\mu\lambda(R)}{N} + \delta^2 K(R) \right) \|\nabla \partial_x v_\epsilon\|_{L^2(0,t;L^2_\#(\Omega_0))}^2 + \delta^2 K(R) \|\partial_t v_\epsilon\|_{L^2(0,t;L^2_\#(\Omega_0))}^2. \quad (3.49) $$

It remains to bound the pressure terms in (3.41). Since $\det(B_{a,b}^T v_\epsilon) = 0$, and thanks to the Piola identity,

$$ \begin{aligned} B_{a,b} : \nabla \partial_x v_\epsilon &= \partial_x (B_{a,b} : \nabla v_\epsilon) - \partial_x B_{a,b} : \nabla v_\epsilon \\ &= \partial_x \operatorname{div} (B_{a,b}^T v_\epsilon) - \partial_x B_{a,b} : \nabla v_\epsilon \\ &= -\partial_x B_{a,b} : \nabla v_\epsilon. \end{aligned} $$

Hence,

$$ \partial_x q_\epsilon B_{a,b} : \nabla \partial_x v_\epsilon = -\partial_x q_\epsilon \partial_x B_{a,b} : \nabla v_\epsilon. $$
It follows from the above equality and (2.67) that

$$
\begin{aligned}
\left| \int_0^t \int_{\Omega_0} \partial_x q_\epsilon B_{a,b} : \nabla \partial_x v_\epsilon \, dxdyds \right| &= \left| \int_0^t \int_{\Omega_0} \partial_x q_\epsilon \partial_x B_{a,b} : \nabla v_\epsilon \, dxdyds \right| \\
&\le \int_0^t \| \partial_x q_\epsilon \|_{L^2(\Omega_0)} \| \partial_x B_{a,b} : \nabla v_\epsilon \|_{L^2(\Omega_0)} ds \\
&\le K(R) \int_0^t \| \partial_x q_\epsilon \|_{L^2(\Omega_0)} (C_\delta \| \nabla v_\epsilon \|_{L^2(\Omega_0)} + \delta \| \nabla^2 v_\epsilon \|_{L^2(\Omega_0)} ) ds.
\end{aligned}
$$

By using Young's inequality, it follows that

$$
\left| \int_0^t \int_{\Omega_0} \partial_x q_\epsilon B_{a,b} : \nabla \partial_x v_\epsilon \, dxdyds \right| \le \delta K(R) \int_0^t \left( \|\partial_x q_\epsilon\|_{L^2(\Omega_0)}^2 + \|\partial_{yy} v_\epsilon\|_{L^2(\Omega_0)}^2 + \|\nabla \partial_x v_\epsilon\|_{L^2(\Omega_0)}^2 \right) \\
\qquad + K(R, \delta) \|\nabla v_\epsilon\|_{L^2(0,t; L^2(\Omega_0))}^2.
$$

By using the energy estimate (3.26) and (3.47), and since $\partial_x q_\epsilon = \partial_x s_{0,\epsilon}$ we obtain

$$
\left| \int_0^t \int_{\Omega_0} \partial_x q_\epsilon B_{a,b} : \nabla \partial_x v_\epsilon \, dxdyds \right| \le C_0(R, \delta) + \delta K(R) \left( \|\partial_t v_\epsilon\|_{L^2(0,t;L^2(\Omega_0))}^2 + \|\nabla \partial_x v_\epsilon\|_{L^2(\Omega_0)}^2 \right). \quad (3.50)
$$

Finally, to bound the last term in the right hand side of (3.41) we integrate it by parts. Precisely, we have

$$
\int_0^t \int_{\Omega_0} q_\epsilon \partial_x B_{a,b} : \nabla \partial_x v_\epsilon \, dxdyds = - \int_0^t \int_{\Omega_0} \operatorname{div} (q_\epsilon \partial_x B_{a,b}) : \partial_x v_\epsilon \, dxdyds + \int_0^t \int_0^L (q_\epsilon \partial_x B_{a,b}^T \partial_x v_\epsilon)_{|_{y=1}} \cdot e_2 \, dx.
$$

By Piola identity, $\operatorname{div}(\partial_x B_{a,b}) = 0$. Hence,

$$
\int_0^t \int_{\Omega_0} q_\epsilon \partial_x B_{a,b} : \nabla \partial_x v_\epsilon \, dxdyds = - \int_0^t \int_{\Omega_0} \nabla q_\epsilon \partial_x B_{a,b}^T \partial_x v_\epsilon \, dxdyds + \int_0^t \int_0^L (q_\epsilon \partial_x B_{a,b}^T \partial_x v_\epsilon)_{|_{y=1}} \cdot e_2 \, dxdyds.
$$

The first term on the right hand side can be estimated once again by applying (2.67) as follows

$$
\begin{aligned}
\left| \int_0^t \int_{\Omega_0} \nabla q_\epsilon \partial_x B_{a,b}^T \partial_x v_\epsilon \, dxdyds \right| &\le \int_0^t \|\nabla q_\epsilon\|_{L^2(\Omega_0)} \|\partial_x B_{a,b}^T \partial_x v_\epsilon\|_{L^2(\Omega_0)} \, ds \\
&\le K(R) \int_0^t \|\nabla q_\epsilon\|_{L^2(\Omega_0)} (C_\delta \|\partial_x v_\epsilon\|_{L^2(\Omega_0)} + \delta \|\nabla \partial_x v_\epsilon\|_{L^2(\Omega_0)} \, ds).
\end{aligned}
$$

This yields by Young's inequality to

$$
\left| \int_0^t \int_{\Omega_0} \nabla q_\epsilon \partial_x B_{a,b}^T \partial_x v_\epsilon \, dxdyds \right| \le \delta K(R) \int_0^t \left( \|\nabla q_\epsilon\|_{L^2(\Omega_0)}^2 + \|\nabla \partial_x v_\epsilon\|_{L^2(\Omega_0)}^2 \right) \, ds + K(R, \delta) \int_0^t \|\partial_x v_\epsilon\|_{L^2(\Omega_0)}^2 \, ds.
$$

Again (3.26) and (3.47) and $\nabla q_\epsilon = \nabla s_{0,\epsilon}$ implies

$$
\left| \int_0^t \int_{\Omega_0} \nabla q_\epsilon \partial_x B_{a,b}^T \partial_x v_\epsilon \, dxdyds \right| \le C_0(R, \delta) + \delta K(R) \left( \|\partial_t v_\epsilon\|_{L^2(0,t;L^2(\Omega_0))}^2 + \|\nabla \partial_x v_\epsilon\|_{L^2(\Omega_0)}^2 \right). \quad (3.51)
$$

Moreover, for the boundary term, taking into account the equality of the velocities at the interface (3.17), one can
easily verifies

$$
\int_0^t \int_0^L (q_\epsilon \partial_x B_{a,b}^T \partial_x v_\epsilon)_{|_{y=1}} \cdot e_2 \, dxdyds = - \int_0^t \int_0^L q_{\epsilon|_{y=1}} \partial_{xx} b \partial_{tx} \zeta_\epsilon \, dxdyds + \int_0^t \int_0^L q_{\epsilon|_{y=1}} \partial_{xx} a \partial_{tx} \eta_\epsilon \, dxdyds. \quad (3.52)
$$

Note that this term is equal to zero in the case where $\zeta_\epsilon = a = 0$, which is the case when the longitudinal displacement is neglected and one considers a beam in flexion only. We recall that $q_\epsilon$ writes $q_\epsilon(x, y, t) = s_{0,\epsilon}(x, y, t) + k_\epsilon(t)$ with $s_{0,\epsilon} \in L^2(0, T; H_\#^1(\Omega_0) \cap L_{\#0}^2(\Omega_0))$ and $k_\epsilon$ is given by (3.29) that writes

$$
k_{\epsilon}(t) = \frac{1}{L} \int_{0}^{L} \left( \rho_{s} \partial_{tt} \eta_{\epsilon} + h|_{y=1} e_{2} \cdot e_{2} + T_{a,b}[v_{\epsilon}, s_{0,\epsilon}] \cdot e_{2} - m \right) dx.
$$
It follows from the above equality and (2.67) that

$$
\begin{aligned}
\left| \int_0^t \int_{\Omega_0} \partial_x q_\epsilon B_{a,b} : \nabla \partial_x v_\epsilon \, dxdyds \right| &= \left| \int_0^t \int_{\Omega_0} \partial_x q_\epsilon \partial_x B_{a,b} : \nabla v_\epsilon \, dxdyds \right| \\
&\le \int_0^t \|\partial_x q_\epsilon\|_{L^2(\Omega_0)} \|\partial_x B_{a,b} : \nabla v_\epsilon\|_{L^2(\Omega_0)} ds \\
&\le K(R) \int_0^t \|\partial_x q_\epsilon\|_{L^2(\Omega_0)} \left( C_\delta \|\nabla v_\epsilon\|_{L^2(\Omega_0)} + \delta \|\nabla^2 v_\epsilon\|_{L^2(\Omega_0)} \right) ds.
\end{aligned}
$$

By using Young's inequality, it follows that

$$
\left| \int_0^t \int_{\Omega_0} \partial_x q_\epsilon B_{a,b} : \nabla \partial_x v_\epsilon \, dxdyds \right| \le \delta K(R) \int_0^t \left( \|\partial_x q_\epsilon\|_{L^2(\Omega_0)}^2 + \|\partial_{yy} v_\epsilon\|_{L^2(\Omega_0)}^2 + \|\nabla \partial_x v_\epsilon\|_{L^2(\Omega_0)}^2 \right) \\
\qquad + K(R, \delta) \|\nabla v_\epsilon\|_{L^2(0,t; L^2(\Omega_0))}^2.
$$

By using the energy estimate (3.26) and (3.47), and since $\partial_x q_\epsilon = \partial_x s_{0,\epsilon}$ we obtain

$$
\left| \int_0^t \int_{\Omega_0} \partial_x q_\epsilon B_{a,b} : \nabla \partial_x v_\epsilon \, dxdyds \right| \le C_0(R, \delta) + \delta K(R) \left( \|\partial_t v_\epsilon\|_{L^2(0,t;L^2(\Omega_0))}^2 + \|\nabla \partial_x v_\epsilon\|_{L^2(\Omega_0)}^2 \right). \quad (3.50)
$$

Finally, to bound the last term in the right hand side of (3.41) we integrate it by parts. Precisely, we have

$$
\int_0^t \int_{\Omega_0} q_\epsilon \partial_x B_{a,b} : \nabla \partial_x v_\epsilon \, dxdyds = - \int_0^t \int_{\Omega_0} \operatorname{div}(q_\epsilon \partial_x B_{a,b}) : \partial_x v_\epsilon \, dxdyds + \int_0^t \int_0^L (q_\epsilon \partial_x B_{a,b}^T \partial_x v_\epsilon)_{|_{y=1}} \cdot e_2 \, dx.
$$

By Piola identity, $\operatorname{div}(\partial_x B_{a,b}) = 0$. Hence,

$$
\int_0^t \int_{\Omega_0} q_\epsilon \partial_x B_{a,b} : \nabla \partial_x v_\epsilon \, dxdyds = - \int_0^t \int_{\Omega_0} \nabla q_\epsilon \partial_x B_{a,b}^T \partial_x v_\epsilon \, dxdyds + \int_0^t \int_0^L (q_\epsilon \partial_x B_{a,b}^T \partial_x v_\epsilon)_{|_{y=1}} \cdot e_2 \, dxdyds.
$$

The first term on the right hand side can be estimated once again by applying (2.67) as follows

$$
\begin{aligned}
\left| \int_0^t \int_{\Omega_0} \nabla q_\epsilon \partial_x B_{a,b}^T \partial_x v_\epsilon \, dxdyds \right| &\le \int_0^t \|\nabla q_\epsilon\|_{L^2(\Omega_0)} \|\partial_x B_{a,b}^T \partial_x v_\epsilon\|_{L^2(\Omega_0)} \, ds \\
&\le K(R) \int_0^t \|\nabla q_\epsilon\|_{L^2(\Omega_0)} \left( C_\delta \|\partial_x v_\epsilon\|_{L^2(\Omega_0)} + \delta \|\nabla \partial_x v_\epsilon\|_{L^2(\Omega_0)} \, ds \right).
\end{aligned}
$$

This yields by Young's inequality to

$$
\left| \int_0^t \int_{\Omega_0} \nabla q_\epsilon \partial_x B_{a,b}^T \partial_x v_\epsilon \, dxdyds \right| \le \delta K(R) \int_0^t \left( \|\nabla q_\epsilon\|_{L^2(\Omega_0)}^2 + \|\nabla \partial_x v_\epsilon\|_{L^2(\Omega_0)}^2 \right) \, ds + K(R, \delta) \int_0^t \|\partial_x v_\epsilon\|_{L^2(\Omega_0)}^2 \, ds.
$$

Again (3.26) and (3.47) and $\nabla q_\epsilon = \nabla s_{0,\epsilon}$ implies

$$
\left| \int_0^t \int_{\Omega_0} \nabla q_\epsilon \partial_x B_{a,b}^T \partial_x v_\epsilon \, dxdyds \right| \le C_0(R, \delta) + \delta K(R) \left( \|\partial_t v_\epsilon\|_{L^2(0,t;L^2(\Omega_0))}^2 + \|\nabla \partial_x v_\epsilon\|_{L^2(\Omega_0)}^2 \right). \quad (3.51)
$$

Moreover, for the boundary term, taking into account the equality of the velocities at the interface (3.17), one can
easily verifies

$$
\int_0^t \int_0^L (q_\epsilon \partial_x B_{a,b}^T \partial_x v_\epsilon)_{|_{y=1}} \cdot e_2 \, dxdyds = - \int_0^t \int_0^L q_{\epsilon|_{y=1}} \partial_{xx} b \partial_{tx} \zeta_\epsilon \, dxdyds + \int_0^t \int_0^L q_{\epsilon|_{y=1}} \partial_{xx} a \partial_{tx} \eta_\epsilon \, dxdyds. \quad (3.52)
$$

Note that this term is equal to zero in the case where $\zeta_\epsilon = a = 0$, which is the case when the longitudinal
displacement is neglected and one considers a beam in flexion only. We recall that $q_\epsilon$ writes $q_\epsilon(x, y, t) = s_{0,\epsilon}(x, y, t) +$
$k_\epsilon(t)$ with $s_{0,\epsilon} \in L^2(0, T; H_\#^1(\Omega_0) \cap L_{\#0}^2(\Omega_0))$ and $k_\epsilon$ is given by (3.29) that writes

$$
k_{\epsilon}(t) = \frac{1}{L} \int_{0}^{L} \left( \rho_{s} \partial_{tt} \eta_{\epsilon} + h_{|_{y=1}} e_{2} \cdot e_{2} + T_{a,b}[v_{\epsilon}, s_{0,\epsilon}] \cdot e_{2} - m \right) dx.
$$
We remark that for a. e. $t \in (0, T)$

$$|k_{\varepsilon}(t)| \le C \left( \|\partial_{tt}\eta_{\varepsilon}\|_{L_{\#}^2(0,L)} + \|(v_{\varepsilon}, s_{0,\varepsilon})\|_{H_{\#}^2(\Omega_0) \times H_{\#}^1(\Omega_0)} + \|h\|_{H_{\#}^1(\Omega_0)} + \|m\|_{L_{\#}^2(0,L)} \right).$$

Thanks to the energy estimate (3.26) and (3.47), since $s_{0,\varepsilon} \in L_{\#}^2(\Omega_0)$, we have

$$\|k_{\varepsilon}\|_{L^2(0,t)} \le C_0 + K(R) \left( \|\partial_{tt}\eta_{\varepsilon}\|_{L^2(0,t;L_{\#}^2(0,L))} + \|\partial_t v_{\varepsilon}\|_{L^2(0,t;L_{\#}^2(\Omega_0))} + \|\nabla \partial_x v_{\varepsilon}\|_{L^2(0,t;L_{\#}^2(\Omega_0))} \right) \quad (3.53)$$

By (3.52), the boundary term can be written as follows:

$$\int_0^t \int_0^L (q_\varepsilon \partial_x B_{a,b}^T \partial_x v_\varepsilon)_{|_{y=1}} \cdot e_2 \, dx \, ds = \int_0^t k_\varepsilon(s) \left( - \int_0^L \partial_{xx} b \partial_{tx} \zeta_\varepsilon + \int_0^L \partial_{xx} a \partial_{tx} \eta_\varepsilon \, dx \right) ds \\ - \int_0^t \int_0^L s_{0,\varepsilon|_{y=1}} \partial_{xx} b \partial_{tx} \zeta_\varepsilon \, dx \, ds + \int_0^t \int_0^L s_{0,\varepsilon|_{y=1}} \partial_{xx} a \partial_{tx} \eta_\varepsilon \, dx \, ds. \quad (3.54)$$

We start by bounding the first term on the right hand side of (3.54). By Holder's and Young's inequality, and taking into account (3.53), we have

$$\begin{aligned} & \left| \int_0^t k_\varepsilon(s) \left( - \int_0^L \partial_{xx} b \partial_{tx} \zeta_\varepsilon + \int_0^L \partial_{xx} a \partial_{tx} \eta_\varepsilon \, dx \right) ds \right| \\ & \le \int_0^t |k_\varepsilon(s)| \|(a,b)\|_{H_{\#}^2(0,L)} \left( \|\partial_{tx} \zeta_\varepsilon\|_{L_{\#}^2(0,L)} + \|\partial_{tx} \eta_\varepsilon\|_{L_{\#}^2(0,L)} \right) ds \\ & \le \kappa \|k_\varepsilon\|_{L^2(0,T)}^2 + \frac{R^2}{4\kappa} \left( \|\partial_{tx} \eta_\varepsilon\|_{L^2(0,t;L_{\#}^2(0,L))}^2 + \|\partial_{tx} \zeta_\varepsilon\|_{L^2(0,t;L_{\#}^2(0,L))}^2 \right) \end{aligned} \quad (3.55)$$

$$\begin{aligned} & \le C_0(R) + \kappa \left( \|\partial_{tt}\eta_{\varepsilon}\|_{L^2(0,t;L_{\#}^2(0,L))}^2 + \|\partial_t v_{\varepsilon}\|_{L^2(0,t;L_{\#}^2(\Omega_0))}^2 + \|\nabla \partial_x v_{\varepsilon}\|_{L^2(0,t;L_{\#}^2(\Omega_0))}^2 \right) \\ & + K(R, \frac{1}{\kappa}) \left( \|\partial_{tx}\eta_{\varepsilon}\|_{L^2(0,t;L_{\#}^2(0,L))}^2 + \|\partial_{tx}\zeta_{\varepsilon}\|_{L^2(0,t;L_{\#}^2(0,L))}^2 \right) \end{aligned} \quad (3.56)$$

Finally we have to estimate the second term on the right hand side of (3.54). Since $(a, b) \in H_{\#}^2(0, L)$, it follows by using Holder's and Young's inequality that

$$\left| \int_0^t \int_0^L s_{0,\varepsilon|_{y=1}} \partial_{xx} a \partial_{tx} \eta_{\varepsilon} ds \right| \le \int_0^t \|\partial_{xx} a\|_{L_{\#}^2(0,L)} \|s_{0,\varepsilon}\|_{L_{\#}^4(0,L)} \|\partial_{tx} \eta_{\varepsilon}\|_{L_{\#}^4(0,L)} ds.$$

Since $H_{\#}^l(0, L) \subset L_{\#}^4(0, L)$, for any $l \ge \frac{1}{4}$, using the interpolation inequality $\|\cdot\|_{H^{\frac{1}{4}}(0,L)} \le C \|\cdot\|_{L^2(0,L)}^{\frac{1}{2}} \|\cdot\|_{H^{\frac{1}{2}}(0,L)}^{\frac{1}{2}}$, and trace inequality and Poincaré inequality on $\partial_x v_\varepsilon$ (that is valid because $\partial_x v_{\varepsilon|_{y=0}} = 0$), we further obtain

$$\begin{aligned} \left| \int_0^t \int_0^L s_{0,\varepsilon|_{y=1}} \partial_{xx} a \partial_{tx} \eta_{\varepsilon} ds \right| & \le R \int_0^t \|s_{0,\varepsilon}\|_{H_{\#}^{\frac{1}{2}}(0,L)} \|\partial_{tx} \eta_{\varepsilon}\|_{L_{\#}^2(0,L)}^{\frac{1}{2}} \|\partial_{tx} \eta_{\varepsilon}\|_{H_{\#}^{\frac{1}{2}}(0,L)}^{\frac{1}{2}} ds \\ & \le C R \int_0^t \|s_{0,\varepsilon}\|_{H_{\#}^1(\Omega_0)} \|\partial_{tx} \eta_{\varepsilon}\|_{L_{\#}^2(0,L)}^{\frac{1}{2}} \|\nabla \partial_x v_{\varepsilon}\|_{L_{\#}^2(\Omega_0)}^{\frac{1}{2}} ds \\ & \le C R \int_0^t \|\nabla s_{0,\varepsilon}\|_{L_{\#}^2(\Omega_0)} \|\partial_{tx} \eta_{\varepsilon}\|_{L_{\#}^2(0,L)}^{\frac{1}{2}} \|\nabla \partial_x v_{\varepsilon}\|_{L_{\#}^2(\Omega_0)}^{\frac{1}{2}} ds. \end{aligned}$$
We remark that for a. e. $t \in (0, T)$

$$|k_{\varepsilon}(t)| \le C \left( \|\partial_{tt}\eta_{\varepsilon}\|_{L_{\#}^2(0,L)} + \|(v_{\varepsilon}, s_{0,\varepsilon})\|_{H_{\#}^2(\Omega_0) \times H_{\#}^1(\Omega_0)} + \|h\|_{H_{\#}^1(\Omega_0)} + \|m\|_{L_{\#}^2(0,L)} \right).$$

Thanks to the energy estimate (3.26) and (3.47), since $s_{0,\varepsilon} \in L_{\#}^2(\Omega_0)$, we have

$$\|k_{\varepsilon}\|_{L^2(0,t)} \le C_0 + K(R) \left( \|\partial_{tt}\eta_{\varepsilon}\|_{L^2(0,t;L_{\#}^2(0,L))} + \|\partial_t v_{\varepsilon}\|_{L^2(0,t;L_{\#}^2(\Omega_0))} + \|\nabla \partial_x v_{\varepsilon}\|_{L^2(0,t;L_{\#}^2(\Omega_0))} \right) \quad (3.53)$$

By (3.52), the boundary term can be written as follows:

$$\int_0^t \int_0^L (q_\varepsilon \partial_x B_{a,b}^T \partial_x v_\varepsilon)_{|_{y=1}} \cdot e_2 \, dx \, ds = \int_0^t k_\varepsilon(s) \left( - \int_0^L \partial_{xx} b \partial_{tx} \zeta_\varepsilon + \int_0^L \partial_{xx} a \partial_{tx} \eta_\varepsilon \, dx \right) ds \\ - \int_0^t \int_0^L s_{0,\varepsilon|_{y=1}} \partial_{xx} b \partial_{tx} \zeta_\varepsilon \, dx \, ds + \int_0^t \int_0^L s_{0,\varepsilon|_{y=1}} \partial_{xx} a \partial_{tx} \eta_\varepsilon \, dx \, ds. \quad (3.54)$$

We start by bounding the first term on the right hand side of (3.54). By Holder's and Young's inequality, and taking into account (3.53), we have

$$\begin{aligned} & \left| \int_0^t k_\varepsilon(s) \left( - \int_0^L \partial_{xx} b \partial_{tx} \zeta_\varepsilon + \int_0^L \partial_{xx} a \partial_{tx} \eta_\varepsilon \, dx \right) ds \right| \\ & \le \int_0^t |k_\varepsilon(s)| \|(a,b)\|_{H_{\#}^2(0,L)} \left( \|\partial_{tx} \zeta_\varepsilon\|_{L_{\#}^2(0,L)} + \|\partial_{tx} \eta_\varepsilon\|_{L_{\#}^2(0,L)} \right) ds \\ & \le \kappa \|k_\varepsilon\|_{L^2(0,T)}^2 + \frac{R^2}{4\kappa} \left( \|\partial_{tx} \eta_\varepsilon\|_{L^2(0,t;L_{\#}^2(0,L))}^2 + \|\partial_{tx} \zeta_\varepsilon\|_{L^2(0,t;L_{\#}^2(0,L))}^2 \right) \end{aligned} \quad (3.55)$$

$$\begin{aligned} & \le C_0(R) + \kappa \left( \|\partial_{tt}\eta_{\varepsilon}\|_{L^2(0,t;L_{\#}^2(0,L))}^2 + \|\partial_t v_{\varepsilon}\|_{L^2(0,t;L_{\#}^2(\Omega_0))}^2 + \|\nabla \partial_x v_{\varepsilon}\|_{L^2(0,t;L_{\#}^2(\Omega_0))}^2 \right) \\ & + K(R, \frac{1}{\kappa}) \left( \|\partial_{tx}\eta_{\varepsilon}\|_{L^2(0,t;L_{\#}^2(0,L))}^2 + \|\partial_{tx}\zeta_{\varepsilon}\|_{L^2(0,t;L_{\#}^2(0,L))}^2 \right) \end{aligned} \quad (3.56)$$

Finally we have to estimate the second term on the right hand side of (3.54). Since $(a, b) \in H_{\#}^2(0, L)$, it follows by using Holder's and Young's inequality that

$$\left| \int_0^t \int_0^L s_{0,\varepsilon|_{y=1}} \partial_{xx} a \partial_{tx} \eta_{\varepsilon} ds \right| \le \int_0^t \|\partial_{xx} a\|_{L_{\#}^2(0,L)} \|s_{0,\varepsilon}\|_{L_{\#}^4(0,L)} \|\partial_{tx} \eta_{\varepsilon}\|_{L_{\#}^4(0,L)} ds.$$

Since $H_{\#}^l(0, L) \subset L_{\#}^4(0, L)$, for any $l \ge \frac{1}{4}$, using the interpolation inequality $\|\cdot\|_{H^{\frac{1}{4}}(0,L)} \le C \|\cdot\|_{L^2(0,L)}^{\frac{1}{2}} \|\cdot\|_{H^{\frac{1}{2}}(0,L)}^{\frac{1}{2}}$, and trace inequality and Poincaré inequality on $\partial_x v_\varepsilon$ (that is valid because $\partial_x v_{\varepsilon|_{y=0}} = 0$), we further obtain

$$\begin{aligned} \left| \int_0^t \int_0^L s_{0,\varepsilon|_{y=1}} \partial_{xx} a \partial_{tx} \eta_{\varepsilon} ds \right| & \le R \int_0^t \|s_{0,\varepsilon}\|_{H_{\#}^{\frac{1}{2}}(0,L)} \|\partial_{tx} \eta_{\varepsilon}\|_{L_{\#}^2(0,L)}^{\frac{1}{2}} \|\partial_{tx} \eta_{\varepsilon}\|_{H_{\#}^{\frac{1}{2}}(0,L)}^{\frac{1}{2}} ds \\ & \le C R \int_0^t \|s_{0,\varepsilon}\|_{H_{\#}^1(\Omega_0)} \|\partial_{tx} \eta_{\varepsilon}\|_{L_{\#}^2(0,L)}^{\frac{1}{2}} \|\nabla \partial_x v_{\varepsilon}\|_{L_{\#}^2(\Omega_0)}^{\frac{1}{2}} ds \\ & \le C R \int_0^t \|\nabla s_{0,\varepsilon}\|_{L_{\#}^2(\Omega_0)} \|\partial_{tx} \eta_{\varepsilon}\|_{L_{\#}^2(0,L)}^{\frac{1}{2}} \|\nabla \partial_x v_{\varepsilon}\|_{L_{\#}^2(\Omega_0)}^{\frac{1}{2}} ds. \end{aligned}$$
The energy estimate (3.26) and the bound (3.47) on the pressure $s_{0,ε}$ imply

$$
\begin{aligned}
\left| \int_0^t \int_0^L s_{0,\epsilon}|_{y=1} \partial_{xx} a \partial_{tx} \eta_{\epsilon} ds \right| &\le C_0(R) + \kappa \|\partial_t v_{\epsilon}\|_{L^2(0,t;L_{\pi}^2(\Omega_0))}^2 + \kappa \|\nabla \partial_x v_{\epsilon}\|_{L^2(0,t;L_{\pi}^2(\Omega_0))}^2 \\
&\quad + K(R, \frac{1}{\kappa}) \int_0^t \|\partial_{tx} \eta_{\epsilon}\|_{L_{\pi}^2(0,L)} \|\nabla \partial_x v_{\epsilon}\|_{L_{\pi}^2(\Omega_0)} ds \\
&\le C_0(R) + \kappa \|\partial_t v_{\epsilon}\|_{L^2(0,t;L_{\pi}^2(\Omega_0))}^2 + 2\kappa \|\nabla \partial_x v_{\epsilon}\|_{L^2(0,t;L_{\pi}^2(\Omega_0))}^2 \\
&\quad + K(R, \frac{1}{\kappa}) \|\partial_{tx} \eta_{\epsilon}\|_{L^2(0,t;L_{\pi}^2(0,L))}^2.
\end{aligned}
\tag{3.57}
$$

In a similar way, one can show

$$
\begin{aligned}
\left| \int_0^t \int_0^L s_{0,\epsilon}|_{y=1} \partial_{xx} b \partial_{tx} \zeta_{\epsilon} ds \right| &\le C_0(R) + \kappa \|\partial_t v_{\epsilon}\|_{L^2(0,t;L_{\pi}^2(\Omega_0))}^2 + \kappa \|\nabla \partial_x v_{\epsilon}\|_{L^2(0,t;L_{\pi}^2(\Omega_0))}^2 \\
&\quad + K(R, \frac{1}{\kappa}) \int_0^t \|\partial_{tx} \eta_{\epsilon}\|_{L_{\pi}^2(0,L)} \|\nabla \partial_x v_{\epsilon}\|_{L_{\pi}^2(\Omega_0)} ds \\
&\le C_0(R) + \kappa \|\partial_t v_{\epsilon}\|_{L^2(0,t;L_{\pi}^2(\Omega_0))}^2 + 2\kappa \|\nabla \partial_x v_{\epsilon}\|_{L^2(0,t;L_{\pi}^2(\Omega_0))}^2 \\
&\quad + K(R, \frac{1}{\kappa}) \|\partial_{tx} \zeta_{\epsilon}\|_{L^2(0,t;L_{\pi}^2(0,L))}^2.
\end{aligned}
\tag{3.58}
$$

Now, by combining (3.56), (3.58) and (3.57) in (3.54), we obtain

$$
\left| \int_0^t \int_0^L (q_\epsilon \partial_x B_{\zeta_0, \eta_0}^T \partial_x v_\epsilon)|_{y=1} \cdot e_2 \, dx \right| \le C_0(R) + 3\kappa \|\partial_t v_\epsilon\|_{L^2(0,t;L_\pi^2(\Omega_0))}^2 + 5\kappa \|\nabla \partial_x v_\epsilon\|_{L^2(0,t;L_\pi^2(\Omega_0))}^2 \\
+ K(R, \frac{1}{\kappa}) \left( \|\partial_{tx} \zeta_\epsilon\|_{L^2(0,t;L_\pi^2(0,L))}^2 + \|\partial_{tx} \eta_\epsilon\|_{L^2(0,t;L_\pi^2(0,L))}^2 \right) \quad (3.59)
$$

We are now in a position to derive the desired estimate. By combining (3.42)-(3.45), (3.48)-(3.50), and (3.59) with
(3.41)

$$
\begin{aligned}
& \frac{\rho f}{2R} \|\partial_x v_\epsilon(t)\|_{L_\pi^2(\Omega_0)}^2 + \frac{\rho_s}{2} \|\partial_{tx} \zeta_\epsilon(t)\|_{L_\pi^2(0,L)}^2 + \frac{\rho_s}{2} \|\partial_{tx} \eta_\epsilon(t)\|_{L_\pi^2(0,L)}^2 + \frac{\mu \lambda(R)}{2} \|\nabla \partial_x v_\epsilon\|_{L^2(0,t;L_\pi^2(\Omega_0))}^2 \\
& \quad + \frac{\beta}{2} \|\partial_{xx} \zeta_\epsilon(t)\|_{L_\pi^2(0,L)}^2 + \frac{\alpha}{2} \|\partial_{xxx} \eta_\epsilon(t)\|_{L_\pi^2(0,L)}^2 + \frac{\gamma}{2} \|\partial_{txx} \eta_\epsilon(t)\|_{L_\pi^2(0,L)}^2 \\
& \quad + \epsilon \left( \|\partial_{xxxt} \zeta_\epsilon\|_{L^2(0,t;L_\pi^2(0,L))}^2 + \|\partial_{xxxt} \eta_\epsilon\|_{L^2(0,t;L_\pi^2(0,L))}^2 \right) \\
& \le C_0(R) + \left( \frac{4\mu\lambda(R)}{N} + 2\delta^2 K(R) + 2\delta K(R) + 5\kappa \right) \|\nabla \partial_x v_\epsilon\|_{L^2(0,t;L_\pi^2(\Omega_0))}^2 \\
& \quad + (2\delta^2 K(R) + 2\delta K(R) + 3\kappa) \|\partial_t v_\epsilon\|_{L^2(0,t;L_\pi^2(\Omega_0))}^2 \\
& \quad + K(R, \frac{1}{\kappa}) \left( \|\partial_{tx} \zeta_\epsilon\|_{L^2(0,t;L_\pi^2(0,L))}^2 + \|\partial_{tx} \eta_\epsilon\|_{L^2(0,t;L_\pi^2(0,L))}^2 \right).
\end{aligned}
$$

We choose $N$ big enough, $\delta$ and $\kappa$ small enough such that

$$
\frac{4\mu\lambda(R)}{N} + 2\delta^2 K(R) + 2\delta K(R) + 5\kappa < \frac{\mu\lambda(R)}{4}.
$$

With this choice we obtain

$$
\begin{aligned}
& \frac{\rho f}{2R} \|\partial_x v_\epsilon(t)\|_{L_\pi^2(\Omega_0)}^2 + \frac{\rho_s}{2} \|\partial_{tx}\zeta_\epsilon(t)\|_{L_\pi^2(0,L)}^2 + \frac{\rho_s}{2} \|\partial_{tx}\eta_\epsilon(t)\|_{L_\pi^2(0,L)}^2 + \frac{\mu\lambda(R)}{4} \|\nabla\partial_x v_\epsilon\|_{L^2(0,t;L_\pi^2(\Omega_0))}^2 \\
& \quad + \frac{\beta}{2} \|\partial_{xx}\zeta_\epsilon(t)\|_{L_\pi^2(0,L)}^2 + \frac{\alpha}{2} \|\partial_{xxx}\eta_\epsilon(t)\|_{L_\pi^2(0,L)}^2 + \frac{\gamma}{2} \|\partial_{txx}\eta_\epsilon(t)\|_{L_\pi^2(0,L)}^2 \\
& \quad + \epsilon \left( \|\partial_{xxxt}\zeta_\epsilon\|_{L^2(0,t;L_\pi^2(0,L))}^2 + \|\partial_{xxxt}\eta_\epsilon\|_{L^2(0,t;L_\pi^2(0,L))}^2 \right) \\
& \le C_0(R,T) + (2\delta^2 K(R) + 2\delta K(R) + 3\kappa) \|\partial_t v_\epsilon\|_{L^2(0,t;L_\pi^2(\Omega_0))}^2 + K(R, \frac{1}{\kappa}) \left( \|\partial_{tx}\zeta_\epsilon\|_{L^2(0,t;L_\pi^2(0,L))}^2 + \|\partial_{tx}\eta_\epsilon\|_{L^2(0,t;L_\pi^2(0,L))}^2 \right)
\end{aligned}
\tag{3.60}
$$
The energy estimate (3.26) and the bound (3.47) on the pressure $s_{0,ε}$ imply

$$
\begin{aligned}
\left| \int_0^t \int_0^L s_{0,\epsilon}|_{y=1} \partial_{xx} a \partial_{tx} \eta_{\epsilon} ds \right| &\le C_0(R) + \kappa \|\partial_t v_{\epsilon}\|_{L^2(0,t;L_{\pi}^2(\Omega_0))}^2 + \kappa \|\nabla \partial_x v_{\epsilon}\|_{L^2(0,t;L_{\pi}^2(\Omega_0))}^2 \\
&\quad + K(R, \frac{1}{\kappa}) \int_0^t \|\partial_{tx} \eta_{\epsilon}\|_{L_{\pi}^2(0,L)} \|\nabla \partial_x v_{\epsilon}\|_{L_{\pi}^2(\Omega_0)} ds \\
&\le C_0(R) + \kappa \|\partial_t v_{\epsilon}\|_{L^2(0,t;L_{\pi}^2(\Omega_0))}^2 + 2\kappa \|\nabla \partial_x v_{\epsilon}\|_{L^2(0,t;L_{\pi}^2(\Omega_0))}^2 \\
&\quad + K(R, \frac{1}{\kappa}) \|\partial_{tx} \eta_{\epsilon}\|_{L^2(0,t;L_{\pi}^2(0,L))}^2.
\end{aligned}
\quad (3.57)
$$

In a similar way, one can show

$$
\begin{aligned}
\left| \int_0^t \int_0^L s_{0,\epsilon}|_{y=1} \partial_{xx} b \partial_{tx} \zeta_{\epsilon} ds \right| &\le C_0(R) + \kappa \|\partial_t v_{\epsilon}\|_{L^2(0,t;L_{\pi}^2(\Omega_0))}^2 + \kappa \|\nabla \partial_x v_{\epsilon}\|_{L^2(0,t;L_{\pi}^2(\Omega_0))}^2 \\
&\quad + K(R, \frac{1}{\kappa}) \int_0^t \|\partial_{tx} \eta_{\epsilon}\|_{L_{\pi}^2(0,L)} \|\nabla \partial_x v_{\epsilon}\|_{L_{\pi}^2(\Omega_0)} ds \\
&\le C_0(R) + \kappa \|\partial_t v_{\epsilon}\|_{L^2(0,t;L_{\pi}^2(\Omega_0))}^2 + 2\kappa \|\nabla \partial_x v_{\epsilon}\|_{L^2(0,t;L_{\pi}^2(\Omega_0))}^2 \\
&\quad + K(R, \frac{1}{\kappa}) \|\partial_{tx} \zeta_{\epsilon}\|_{L^2(0,t;L_{\pi}^2(0,L))}^2.
\end{aligned}
\quad (3.58)
$$

Now, by combining (3.56), (3.58) and (3.57) in (3.54), we obtain

$$
\left| \int_0^t \int_0^L (q_\epsilon \partial_x B_{\zeta_0, \eta_0}^T \partial_x v_\epsilon)|_{y=1} \cdot e_2 \, dx \right| \le C_0(R) + 3\kappa \|\partial_t v_\epsilon\|_{L^2(0,t;L_\pi^2(\Omega_0))}^2 + 5\kappa \|\nabla \partial_x v_\epsilon\|_{L^2(0,t;L_\pi^2(\Omega_0))}^2 \\
+ K(R, \frac{1}{\kappa}) \left( \|\partial_{tx} \zeta_\epsilon\|_{L^2(0,t;L_\pi^2(0,L))}^2 + \|\partial_{tx} \eta_\epsilon\|_{L^2(0,t;L_\pi^2(0,L))}^2 \right) \quad (3.59)
$$

We are now in a position to derive the desired estimate. By combining (3.42)-(3.45), (3.48)-(3.50), and (3.59) with
(3.41)

$$
\begin{aligned}
& \frac{\rho_f}{2R} \|\partial_x v_\epsilon(t)\|_{L_\pi^2(\Omega_0)}^2 + \frac{\rho_s}{2} \|\partial_{tx} \zeta_\epsilon(t)\|_{L_\pi^2(0,L)}^2 + \frac{\rho_s}{2} \|\partial_{tx} \eta_\epsilon(t)\|_{L_\pi^2(0,L)}^2 + \frac{\mu\lambda(R)}{2} \|\nabla \partial_x v_\epsilon\|_{L^2(0,t;L_\pi^2(\Omega_0))}^2 \\
& + \frac{\beta}{2} \|\partial_{xx} \zeta_\epsilon(t)\|_{L_\pi^2(0,L)}^2 + \frac{\alpha}{2} \|\partial_{xxx} \eta_\epsilon(t)\|_{L_\pi^2(0,L)}^2 + \frac{\gamma}{2} \|\partial_{txx} \eta_\epsilon(t)\|_{L_\pi^2(0,L)}^2 \\
& \qquad + \epsilon \left( \|\partial_{xxxt} \zeta_\epsilon\|_{L^2(0,t;L_\pi^2(0,L))}^2 + \|\partial_{xxxt} \eta_\epsilon\|_{L^2(0,t;L_\pi^2(0,L))}^2 \right) \\
\le & C_0(R) + \left( \frac{4\mu\lambda(R)}{N} + 2\delta^2 K(R) + 2\delta K(R) + 5\kappa \right) \|\nabla \partial_x v_\epsilon\|_{L^2(0,t;L_\pi^2(\Omega_0))}^2 \\
& + (2\delta^2 K(R) + 2\delta K(R) + 3\kappa) \|\partial_t v_\epsilon\|_{L^2(0,t;L_\pi^2(\Omega_0))}^2 \\
& + K(R, \frac{1}{\kappa}) \left( \|\partial_{tx} \zeta_\epsilon\|_{L^2(0,t;L_\pi^2(0,L))}^2 + \|\partial_{tx} \eta_\epsilon\|_{L^2(0,t;L_\pi^2(0,L))}^2 \right).
\end{aligned}
$$

We choose $N$ big enough, $\delta$ and $\kappa$ small enough such that

$$
\frac{4\mu\lambda(R)}{N} + 2\delta^2 K(R) + 2\delta K(R) + 5\kappa < \frac{\mu\lambda(R)}{4}.
$$

With this choice we obtain

$$
\begin{aligned}
& \frac{\rho_f}{2R} \|\partial_x v_\epsilon(t)\|_{L_\pi^2(\Omega_0)}^2 + \frac{\rho_s}{2} \|\partial_{tx}\zeta_\epsilon(t)\|_{L_\pi^2(0,L)}^2 + \frac{\rho_s}{2} \|\partial_{tx}\eta_\epsilon(t)\|_{L_\pi^2(0,L)}^2 + \frac{\mu\lambda(R)}{4} \|\nabla\partial_x v_\epsilon\|_{L^2(0,t;L_\pi^2(\Omega_0))}^2 \\
& + \frac{\beta}{2} \|\partial_{xx}\zeta_\epsilon(t)\|_{L_\pi^2(0,L)}^2 + \frac{\alpha}{2} \|\partial_{xxx}\eta_\epsilon(t)\|_{L_\pi^2(0,L)}^2 + \frac{\gamma}{2} \|\partial_{txx}\eta_\epsilon(t)\|_{L_\pi^2(0,L)}^2 \\
& \qquad + \epsilon \left( \|\partial_{xxxt}\zeta_\epsilon\|_{L^2(0,t;L_\pi^2(0,L))}^2 + \|\partial_{xxxt}\eta_\epsilon\|_{L^2(0,t;L_\pi^2(0,L))}^2 \right) \\
\le & C_0(R,T) + (2\delta^2 K(R) + 2\delta K(R) + 3\kappa) \|\partial_t v_\epsilon\|_{L^2(0,t;L_\pi^2(\Omega_0))}^2 \\
& + K(R, \frac{1}{\kappa}) \left( \|\partial_{tx}\zeta_\epsilon\|_{L^2(0,t;L_\pi^2(0,L))}^2 + \|\partial_{tx}\eta_\epsilon\|_{L^2(0,t;L_\pi^2(0,L))}^2 \right)
\end{aligned}
\quad (3.60)
$$
**Final regularity estimates:** We are now in a position to obtain a regularity estimate independent of the parameter of regularization $\epsilon$. To this end, we consider a linear combination of the previous estimates, namely $(E)_\Lambda = \Lambda(3.39) + (3.60)$, with a parameter $\Lambda > 0$ that will be chosen afterwards. Then we obtain

$$
\begin{aligned}
& \frac{\Lambda \rho_f}{2R} \|\partial_t v_\epsilon\|_{L^2(0,t;L^2_\#(\Omega_0))}^2 + \frac{\Lambda \rho_s}{2} \|\partial_{tt}\zeta_\epsilon\|_{L^2(0,t;L^2_\#(0,L))}^2 + \frac{\Lambda \rho_s}{2} \|\partial_{tt}\eta_\epsilon\|_{L^2(0,t;L^2_\#(0,L))}^2 + \frac{\Lambda \mu \lambda(R)}{4} \|\nabla v_\epsilon\|_{L^2(0,t;L^2_\#(\Omega_0))}^2 \\
& \qquad + \gamma \Lambda \|\partial_{xtt}\eta_\epsilon\|_{L^2(0,t;L^2_\#(0,L))}^2 + \frac{\Lambda \epsilon}{2} \left( \|\partial_{xxt}\zeta_\epsilon(t)\|_{L^2_\#(0,L)}^2 + \|\partial_{xxt}\eta_\epsilon(t)\|_{L^2_\#(0,L)}^2 \right) \\
& \qquad + \frac{\rho_f}{2R} \|\partial_x v_\epsilon(t)\|_{L^2_\#(\Omega_0)}^2 + \frac{\rho_s}{2} \|\partial_{tx}\zeta_\epsilon(t)\|_{L^2_\#(0,L)}^2 + \frac{\rho_s}{2} \|\partial_{tx}\eta_\epsilon(t)\|_{L^2_\#(0,L)}^2 + \frac{\mu \lambda(R)}{4} \|\nabla \partial_x v_\epsilon\|_{L^2(0,t;L^2_\#(\Omega_0))}^2 \\
& \qquad + \frac{\beta}{2} \|\partial_{xx}\zeta_\epsilon(t)\|_{L^2_\#(0,L)}^2 + \frac{\alpha}{2} \|\partial_{xxx}\eta_\epsilon(t)\|_{L^2_\#(0,L)}^2 + \frac{\gamma}{2} \|\partial_{txx}\eta_\epsilon(t)\|_{L^2_\#(0,L)}^2 + \epsilon \left( \|\partial_{xxt}\zeta_\epsilon\|_{L^2(0,t;L^2_\#(0,L))}^2 + \|\partial_{xxt}\eta_\epsilon\|_{L^2(0,t;L^2_\#(0,L))}^2 \right) \\
& \le C_0(R) + (2\delta^2 K(R) + 2\delta K(R) + 3\kappa) \|\partial_t v_\epsilon\|_{L^2(0,t;L^2_\#(\Omega_0))}^2 + K(R, \frac{1}{\kappa}) \left( \|\partial_{tx}\zeta_\epsilon\|_{L^2(0,t;L^2_\#(0,L))}^2 + \|\partial_{tx}\eta_\epsilon\|_{L^2(0,t;L^2_\#(0,L))}^2 \right) \\
& \qquad + \Lambda \beta \|\partial_{xt}\zeta_\epsilon\|_{L^2(0,t;L^2_\#(\Omega_0))}^2 + \Lambda \alpha \|\partial_{txx}\eta_\epsilon\|_{L^2(0,t;L^2_\#(\Omega_0))}^2 + \Lambda \beta \int_0^L \partial_{xx}\zeta_\epsilon(t) \partial_t \zeta_\epsilon(t) dx + \Lambda \alpha \int_0^L \partial_{xxx}\eta_\epsilon(t) \partial_{tx}\eta_\epsilon(t) dx,
\end{aligned}
$$

It follows by choosing $\Lambda = \min\left(\frac{\rho_s}{2\beta}, \frac{\gamma}{2\alpha}, \frac{1}{2}, \frac{\rho_s}{2\alpha}\right)$ and by using Young's inequality that

$$
\begin{aligned}
& \frac{\Lambda \rho_f}{2R} \|\partial_t v_\epsilon\|_{L^2(0,t;L^2_\#(\Omega_0))}^2 + \frac{\Lambda \rho_s}{2} \|\partial_{tt}\zeta_\epsilon\|_{L^2(0,t;L^2_\#(0,L))}^2 + \frac{\Lambda \rho_s}{2} \|\partial_{tt}\eta_\epsilon\|_{L^2(0,t;L^2_\#(0,L))}^2 + \frac{\Lambda \mu \lambda(R)}{4} \|\nabla v_\epsilon\|_{L^2(0,t;L^2_\#(\Omega_0))}^2 \\
& \qquad + \gamma \Lambda \|\partial_{xtt}\eta_\epsilon\|_{L^2(0,t;L^2_\#(0,L))}^2 + \frac{\Lambda \epsilon}{2} \left( \|\partial_{xxt}\zeta_\epsilon(t)\|_{L^2_\#(0,L)}^2 + \|\partial_{xxt}\eta_\epsilon(t)\|_{L^2_\#(0,L)}^2 \right) \\
& \qquad + \frac{\rho_f}{2R} \|\partial_x v_\epsilon(t)\|_{L^2_\#(\Omega_0)}^2 + \frac{\rho_s}{2} \|\partial_{tx}\zeta_\epsilon(t)\|_{L^2_\#(0,L)}^2 + \frac{\rho_s}{2} \|\partial_{tx}\eta_\epsilon(t)\|_{L^2_\#(0,L)}^2 + \frac{\mu \lambda(R)}{4} \|\nabla \partial_x v_\epsilon\|_{L^2(0,t;L^2_\#(\Omega_0))}^2 \\
& \qquad + \frac{\beta}{2} \|\partial_{xx}\zeta_\epsilon(t)\|_{L^2_\#(0,L)}^2 + \frac{\alpha}{2} \|\partial_{xxx}\eta_\epsilon(t)\|_{L^2_\#(0,L)}^2 + \frac{\gamma}{2} \|\partial_{txx}\eta_\epsilon(t)\|_{L^2_\#(0,L)}^2 + \epsilon \left( \|\partial_{xxt}\zeta_\epsilon\|_{L^2(0,t;L^2_\#(0,L))}^2 + \|\partial_{xxt}\eta_\epsilon\|_{L^2(0,t;L^2_\#(0,L))}^2 \right) \\
& \le C_0(R) + (2\delta^2 K(R) + 2\delta K(R) + 3\kappa) \|\partial_t v_\epsilon\|_{L^2(0,t;L^2_\#(\Omega_0))}^2 + K(R, \frac{1}{\kappa}) \left( \|\partial_{tx}\zeta_\epsilon\|_{L^2(0,t;L^2_\#(0,L))}^2 + \|\partial_{tx}\eta_\epsilon\|_{L^2(0,t;L^2_\#(0,L))}^2 \right) \\
& \qquad + \frac{\rho_s}{2} \|\partial_{xt}\zeta_\epsilon\|_{L^2(0,t;L^2_\#(\Omega_0))}^2 + \frac{\gamma}{2} \|\partial_{txx}\eta_\epsilon\|_{L^2(0,t;L^2_\#(\Omega_0))}^2 + \frac{\beta}{4} \|\partial_{xx}\zeta_\epsilon(t)\|_{L^2_\#(0,L)}^2 + \frac{\rho_s}{4} \|\partial_t \zeta_\epsilon(t)\|_{L^2_\#(0,L)}^2 \\
& \qquad + \frac{\alpha}{4} \|\partial_{xxx}\eta_\epsilon(t)\|_{L^2_\#(0,L)}^2 + \frac{\rho_s}{4} \|\partial_{tx}\eta_\epsilon(t)\|_{L^2_\#(0,L)}^2
\end{aligned}
$$

Hence, by choosing $\kappa$ and $\delta$ small enough such that $2\delta^2 K(R) + 2\delta K(R) + 3\kappa < \frac{\Lambda \rho_f}{4R}$ remembering the energy estimate (3.26) it follows that

$$
\begin{aligned}
& \frac{\Lambda \rho_f}{4R} \|\partial_t v_\epsilon\|_{L^2(0,t;L^2_\#(\Omega_0))}^2 + \frac{\Lambda \rho_s}{2} \|\partial_{tt}\zeta_\epsilon\|_{L^2(0,t;L^2_\#(0,L))}^2 + \frac{\Lambda \rho_s}{2} \|\partial_{tt}\eta_\epsilon\|_{L^2(0,t;L^2_\#(0,L))}^2 + \frac{\Lambda \mu \lambda(R)}{4} \|\nabla v_\epsilon\|_{L^2(0,t;L^2_\#(\Omega_0))}^2 \\
& \qquad + \gamma \Lambda \|\partial_{xtt}\eta_\epsilon\|_{L^2(0,t;L^2_\#(0,L))}^2 + \frac{\Lambda \epsilon}{2} \left( \|\partial_{xxt}\zeta_\epsilon(t)\|_{L^2_\#(0,L)}^2 + \|\partial_{xxt}\eta_\epsilon(t)\|_{L^2_\#(0,L)}^2 \right) \\
& \qquad + \frac{\rho_f}{2R} \|\partial_x v_\epsilon(t)\|_{L^2_\#(\Omega_0)}^2 + \frac{\rho_s}{2} \|\partial_{tx}\zeta_\epsilon(t)\|_{L^2_\#(0,L)}^2 + \frac{\rho_s}{4} \|\partial_{tx}\eta_\epsilon(t)\|_{L^2_\#(0,L)}^2 + \frac{\mu \lambda(R)}{4} \|\nabla \partial_x v_\epsilon\|_{L^2(0,t;L^2_\#(\Omega_0))}^2 \\
& \qquad + \frac{\beta}{4} \|\partial_{xx}\zeta_\epsilon(t)\|_{L^2_\#(0,L)}^2 + \frac{\alpha}{4} \|\partial_{xxx}\eta_\epsilon(t)\|_{L^2_\#(0,L)}^2 + \frac{\gamma}{2} \|\partial_{txx}\eta_\epsilon(t)\|_{L^2_\#(0,L)}^2 + \epsilon \left( \|\partial_{xxt}\zeta_\epsilon\|_{L^2(0,t;L^2_\#(0,L))}^2 + \|\partial_{xxt}\eta_\epsilon\|_{L^2(0,t;L^2_\#(0,L))}^2 \right) \\
& \le C_0(R) + \left( K(R, \frac{1}{\kappa}) + \frac{\rho_s}{2} \right) \|\partial_{tx}\zeta_\epsilon\|_{L^2(0,t;L^2_\#(0,L))}^2 + \frac{\gamma}{2} \|\partial_{txx}\eta_\epsilon\|_{L^2(0,t;L^2_\#(\Omega_0))}^2.
\end{aligned}
\quad (3.61)
$$

From the above inequality we deduce

$$
\int_0^L |\partial_{xt}\zeta_\epsilon|^2 dxds + \int_0^L |\partial_{xxt}\eta_\epsilon|^2 dxds \le \frac{2}{\min(\rho_s, \gamma)} \left( C_0(R) + M \left( \int_0^t \int_0^L |\partial_{xt}\zeta_\epsilon|^2 dxds + \int_0^t \int_0^L |\partial_{xxt}\eta_\epsilon|^2 dxds \right) \right),
$$
**Final regularity estimates:** We are now in a position to obtain a regularity estimate independent of the parameter of regularization $\epsilon$. To this end, we consider a linear combination of the previous estimates, namely $(E)_\Lambda = \Lambda(3.39) + (3.60)$, with a parameter $\Lambda > 0$ that will be chosen afterwards. Then we obtain

$$
\begin{align*}
& \frac{\Lambda \rho_f}{2R} \|\partial_t v_\epsilon\|_{L^2(0,t;L^2_\#(\Omega_0))}^2 + \frac{\Lambda \rho_s}{2} \|\partial_{tt}\zeta_\epsilon\|_{L^2(0,t;L^2_\#(0,L))}^2 + \frac{\Lambda \rho_s}{2} \|\partial_{tt}\eta_\epsilon\|_{L^2(0,t;L^2_\#(0,L))}^2 + \frac{\Lambda \mu \lambda(R)}{4} \|\nabla v_\epsilon\|_{L^2(0,t;L^2_\#(\Omega_0))}^2 \\
& \qquad + \gamma \Lambda \|\partial_{xtt}\eta_\epsilon\|_{L^2(0,t;L^2_\#(0,L))}^2 + \frac{\Lambda \epsilon}{2} \left( \|\partial_{xxt}\zeta_\epsilon(t)\|_{L^2_\#(0,L)}^2 + \|\partial_{xxt}\eta_\epsilon(t)\|_{L^2_\#(0,L)}^2 \right) \\
& \qquad + \frac{\rho_f}{2R} \|\partial_x v_\epsilon(t)\|_{L^2_\#(\Omega_0)}^2 + \frac{\rho_s}{2} \|\partial_{tx}\zeta_\epsilon(t)\|_{L^2_\#(0,L)}^2 + \frac{\rho_s}{2} \|\partial_{tx}\eta_\epsilon(t)\|_{L^2_\#(0,L)}^2 + \frac{\mu \lambda(R)}{4} \|\nabla \partial_x v_\epsilon\|_{L^2(0,t;L^2_\#(\Omega_0))}^2 \\
& \qquad + \frac{\beta}{2} \|\partial_{xx}\zeta_\epsilon(t)\|_{L^2_\#(0,L)}^2 + \frac{\alpha}{2} \|\partial_{xxx}\eta_\epsilon(t)\|_{L^2_\#(0,L)}^2 + \frac{\gamma}{2} \|\partial_{txx}\eta_\epsilon(t)\|_{L^2_\#(0,L)}^2 + \epsilon \left( \|\partial_{xxt}\zeta_\epsilon\|_{L^2(0,t;L^2_\#(0,L))}^2 + \|\partial_{xxt}\eta_\epsilon\|_{L^2(0,t;L^2_\#(0,L))}^2 \right) \\
& \le C_0(R) + (2\delta^2 K(R) + 2\delta K(R) + 3\kappa) \|\partial_t v_\epsilon\|_{L^2(0,t;L^2_\#(\Omega_0))}^2 + K(R, \frac{1}{\kappa}) \left( \|\partial_{tx}\zeta_\epsilon\|_{L^2(0,t;L^2_\#(0,L))}^2 + \|\partial_{tx}\eta_\epsilon\|_{L^2(0,t;L^2_\#(0,L))}^2 \right) \\
& \qquad + \Lambda \beta \|\partial_{xt}\zeta_\epsilon\|_{L^2(0,t;L^2_\#(\Omega_0))}^2 + \Lambda \alpha \|\partial_{txx}\eta_\epsilon\|_{L^2(0,t;L^2_\#(\Omega_0))}^2 + \Lambda \beta \int_0^L \partial_{xx}\zeta_\epsilon(t) \partial_t \zeta_\epsilon(t) dx + \Lambda \alpha \int_0^L \partial_{xxx}\eta_\epsilon(t) \partial_{tx}\eta_\epsilon(t) dx,
\end{align*}
$$

It follows by choosing $\Lambda = \min\left(\frac{\rho_s}{2\beta}, \frac{\gamma}{2\alpha}, \frac{1}{2}, \frac{\rho_s}{2\alpha}\right)$ and by using Young's inequality that

$$
\begin{align*}
& \frac{\Lambda \rho_f}{2R} \|\partial_t v_\epsilon\|_{L^2(0,t;L^2_\#(\Omega_0))}^2 + \frac{\Lambda \rho_s}{2} \|\partial_{tt}\zeta_\epsilon\|_{L^2(0,t;L^2_\#(0,L))}^2 + \frac{\Lambda \rho_s}{2} \|\partial_{tt}\eta_\epsilon\|_{L^2(0,t;L^2_\#(0,L))}^2 + \frac{\Lambda \mu \lambda(R)}{4} \|\nabla v_\epsilon\|_{L^2(0,t;L^2_\#(\Omega_0))}^2 \\
& \qquad + \gamma \Lambda \|\partial_{xtt}\eta_\epsilon\|_{L^2(0,t;L^2_\#(0,L))}^2 + \frac{\Lambda \epsilon}{2} \left( \|\partial_{xxt}\zeta_\epsilon(t)\|_{L^2_\#(0,L)}^2 + \|\partial_{xxt}\eta_\epsilon(t)\|_{L^2_\#(0,L)}^2 \right) \\
& \qquad + \frac{\rho_f}{2R} \|\partial_x v_\epsilon(t)\|_{L^2_\#(\Omega_0)}^2 + \frac{\rho_s}{2} \|\partial_{tx}\zeta_\epsilon(t)\|_{L^2_\#(0,L)}^2 + \frac{\rho_s}{2} \|\partial_{tx}\eta_\epsilon(t)\|_{L^2_\#(0,L)}^2 + \frac{\mu \lambda(R)}{4} \|\nabla \partial_x v_\epsilon\|_{L^2(0,t;L^2_\#(\Omega_0))}^2 \\
& \qquad + \frac{\beta}{2} \|\partial_{xx}\zeta_\epsilon(t)\|_{L^2_\#(0,L)}^2 + \frac{\alpha}{2} \|\partial_{xxx}\eta_\epsilon(t)\|_{L^2_\#(0,L)}^2 + \frac{\gamma}{2} \|\partial_{txx}\eta_\epsilon(t)\|_{L^2_\#(0,L)}^2 + \epsilon \left( \|\partial_{xxt}\zeta_\epsilon\|_{L^2(0,t;L^2_\#(0,L))}^2 + \|\partial_{xxt}\eta_\epsilon\|_{L^2(0,t;L^2_\#(0,L))}^2 \right) \\
& \le C_0(R) + (2\delta^2 K(R) + 2\delta K(R) + 3\kappa) \|\partial_t v_\epsilon\|_{L^2(0,t;L^2_\#(\Omega_0))}^2 + K(R, \frac{1}{\kappa}) \left( \|\partial_{tx}\zeta_\epsilon\|_{L^2(0,t;L^2_\#(0,L))}^2 + \|\partial_{tx}\eta_\epsilon\|_{L^2(0,t;L^2_\#(0,L))}^2 \right) \\
& \qquad + \frac{\rho_s}{2} \|\partial_{xt}\zeta_\epsilon\|_{L^2(0,t;L^2_\#(\Omega_0))}^2 + \frac{\gamma}{2} \|\partial_{txx}\eta_\epsilon\|_{L^2(0,t;L^2_\#(\Omega_0))}^2 + \frac{\beta}{4} \|\partial_{xx}\zeta_\epsilon(t)\|_{L^2_\#(0,L)}^2 + \frac{\rho_s}{4} \|\partial_t \zeta_\epsilon(t)\|_{L^2_\#(0,L)}^2 \\
& \qquad + \frac{\alpha}{4} \|\partial_{xxx}\eta_\epsilon(t)\|_{L^2_\#(0,L)}^2 + \frac{\rho_s}{4} \|\partial_{tx}\eta_\epsilon(t)\|_{L^2_\#(0,L)}^2
\end{align*}
$$

Hence, by choosing $\kappa$ and $\delta$ small enough such that $2\delta^2 K(R) + 2\delta K(R) + 3\kappa < \frac{\Lambda \rho_f}{4R}$ remembering the energy estimate (3.26) it follows that

$$
\begin{align*}
& \frac{\Lambda \rho_f}{4R} \|\partial_t v_\epsilon\|_{L^2(0,t;L^2_\#(\Omega_0))}^2 + \frac{\Lambda \rho_s}{2} \|\partial_{tt}\zeta_\epsilon\|_{L^2(0,t;L^2_\#(0,L))}^2 + \frac{\Lambda \rho_s}{2} \|\partial_{tt}\eta_\epsilon\|_{L^2(0,t;L^2_\#(0,L))}^2 + \frac{\Lambda \mu \lambda(R)}{4} \|\nabla v_\epsilon\|_{L^2(0,t;L^2_\#(\Omega_0))}^2 \\
& \qquad + \gamma \Lambda \|\partial_{xtt}\eta_\epsilon\|_{L^2(0,t;L^2_\#(0,L))}^2 + \frac{\Lambda \epsilon}{2} \left( \|\partial_{xxt}\zeta_\epsilon(t)\|_{L^2_\#(0,L)}^2 + \|\partial_{xxt}\eta_\epsilon(t)\|_{L^2_\#(0,L)}^2 \right) \\
& \qquad + \frac{\rho_f}{2R} \|\partial_x v_\epsilon(t)\|_{L^2_\#(\Omega_0)}^2 + \frac{\rho_s}{2} \|\partial_{tx}\zeta_\epsilon(t)\|_{L^2_\#(0,L)}^2 + \frac{\rho_s}{4} \|\partial_{tx}\eta_\epsilon(t)\|_{L^2_\#(0,L)}^2 + \frac{\mu \lambda(R)}{4} \|\nabla \partial_x v_\epsilon\|_{L^2(0,t;L^2_\#(\Omega_0))}^2 \\
& \qquad + \frac{\beta}{4} \|\partial_{xx}\zeta_\epsilon(t)\|_{L^2_\#(0,L)}^2 + \frac{\alpha}{4} \|\partial_{xxx}\eta_\epsilon(t)\|_{L^2_\#(0,L)}^2 + \frac{\gamma}{2} \|\partial_{txx}\eta_\epsilon(t)\|_{L^2_\#(0,L)}^2 + \epsilon \left( \|\partial_{xxt}\zeta_\epsilon\|_{L^2(0,t;L^2_\#(0,L))}^2 + \|\partial_{xxt}\eta_\epsilon\|_{L^2(0,t;L^2_\#(0,L))}^2 \right) \\
& \le C_0(R) + \left( K(R, \frac{1}{\kappa}) + \frac{\rho_s}{2} \right) \|\partial_{tx}\zeta_\epsilon\|_{L^2(0,t;L^2_\#(0,L))}^2 + \frac{\gamma}{2} \|\partial_{txx}\eta_\epsilon\|_{L^2(0,t;L^2_\#(\Omega_0))}^2. \quad (3.61)
\end{align*}
$$

From the above inequality we deduce

$$
\int_0^L |\partial_{xt}\zeta_\epsilon|^2 dxds + \int_0^L |\partial_{xxt}\eta_\epsilon|^2 dxds \le \frac{2}{\min(\rho_s, \gamma)} \left( C_0(R) + M \left( \int_0^t \int_0^L |\partial_{xt}\zeta_\epsilon|^2 dxds + \int_0^t \int_0^L |\partial_{xxt}\eta_\epsilon|^2 dxds \right) \right),
$$
where $M = \max(K(R, \frac{1}{\kappa}) + \frac{\rho_s}{2}, \frac{\gamma}{2})$.

Gronwall's Lemma implies

$$ \int_{0}^{L} |\partial_{xt}\zeta_{\epsilon}|^{2} dxds + \int_{0}^{L} |\partial_{xxt}\eta_{\epsilon}|^{2} dxds \leq \frac{2C_{0}(R)}{\min(\rho_{s}, \gamma)} \exp\left(\frac{2MT}{\min(\rho_{s}, \gamma)}\right). $$

Combining (3.61) with the above inequality implies for $T \leq T^*$

$$ \begin{aligned} & \frac{\Lambda \rho_f}{4R} \|\partial_t v_\epsilon\|_{L^2(0,t;L^2_\#(\Omega_0))}^2 + \frac{\Lambda \rho_s}{2} \|\partial_{tt}\zeta_\epsilon\|_{L^2(0,t;L^2_\#(0,L))}^2 + \frac{\Lambda \rho_s}{2} \|\partial_{tt}\eta_\epsilon\|_{L^2(0,t;L^2_\#(0,L))}^2 + \gamma \Lambda \|\partial_{xxt}\eta_\epsilon\|_{L^2(0,t;L^2_\#(0,L))}^2 \\ & + \frac{\rho_f}{2R} \|\partial_x v_\epsilon(t)\|_{L^2_\#(\Omega_0)}^2 + \frac{\rho_s}{2} \|\partial_{tx}\zeta_\epsilon(t)\|_{L^2_\#(0,L)}^2 + \frac{\rho_s}{4} \|\partial_{tx}\eta_\epsilon(t)\|_{L^2_\#(0,L)}^2 + \frac{\mu\lambda(R)}{4} \|\nabla\partial_x v_\epsilon\|_{L^2(0,t;L^2_\#(\Omega_0))}^2 \\ & + \frac{\beta}{4} \|\partial_{xxx}\zeta_\epsilon(t)\|_{L^2_\#(0,L)}^2 + \frac{\alpha}{4} \|\partial_{xxx}\eta_\epsilon(t)\|_{L^2_\#(0,L)}^2 + \frac{\gamma}{2} \|\partial_{txx}\eta_\epsilon(t)\|_{L^2_\#(0,L)}^2 \leq C_{0,T^*}(R), \end{aligned} \quad (3.62) $$

with $C_{0,T^*}(R)$ non decreasing with respect to $R$. The elliptic estimate (3.47) further implies

$$ \|\partial_{yy} v_{\epsilon}\|_{L^2(0,t;L^2_\#(\Omega_0))} + \|\nabla q_{\epsilon}\|_{L^2(0,t;L^2_\#(\Omega_0))} \leq C_{0,T^*}(R). \quad (3.63) $$

where $C_0$ depends again only on initial data, source terms, $R$ and the different positive physical parameters.

**Limit as $\epsilon$ goes to zero and uniqueness.** By standard compactness arguments, we may extract a weakly converging subsequence in the family of $\epsilon$- solutions $(v_\epsilon, q_\epsilon, \zeta_\epsilon, \eta_\epsilon)$ satisfying (3.26), (3.62) and (3.63). Then applying classical arguments, we obtain that the weak limit $(v, q, \zeta, \eta)$ is a solution to our linear problem with the expected regularity and satisfying (3.14). We left to the reader all these computations as they are completely standard. Finally, suppose that we have two solutions to our linear system with the same data, then the difference would satisfy (3.26) with vanishing data. We deduce the uniqueness of the solution to the homogeneous linear problem (3.1). This ends the proof of Proposition 3.1.

## 3.3 Non homogeneous divergence constraint.

In this subsection, we study the linear problem (3.1)-(3.7) in case when $g$ is not equal to zero. More precisely, we establish the following proposition:

**Proposition 3.2** Let $(a, b) \in H^2_\#(0, L)$ such that $\chi_{a,b}$ is a injective on $\bar{\mathcal{F}}_0$. Given $T > 0$, initial data $(v_0, \zeta_0, \zeta_1, \eta_0, \eta_1) \in X^0$ and a source term $(f, h, g, 0, 0) \in S_T$. Then there exists a unique solution $(v, q, \zeta, \eta) \in X_{f,T} \times X_{s,T}$ to Problem (3.1)-(3.7) and a non decreasing function $K : [0, \infty) \to [0, \infty)$ such that assuming $\|(a, b)\|_{H^2_\#(0,L)} + \|(\det \nabla \chi_{a,b})^{-1}\|_{C_\#(\bar{\Omega}_0)} \leq R$, the solution $(v, q, \zeta, \eta)$ satisfies

$$ \|(v, q)\|_{X_{f,T}} + \|(\zeta, \eta)\|_{X_{s,T}} \leq K(R)(\|(v_0, \zeta_0, \zeta_1, \eta_0, \eta_1)\|_{X^0} + \|(f, h, g, 0, 0)\|_{S_T}) \quad (3.64) $$

Proof. We look for $(v, q, \zeta, \eta)$ solution to (3.1)-(3.7) with initial data $(v_0, \zeta_0, \zeta_1, \eta_0, \eta_1) \in X^0$ and source term $(f, h, g, 0, 0) \in S_T$. We are going to lift the right hand side of the divergence in order to apply the previous result obtained in the case of divergence free constraint. To do so we need to modify the fluid velocity unknown but also the elastic displacement one. This is due to the fact that the average of $g$ is not equal to zero. It is a key difference compared to the case where the longitudinal displacement is neglected. Thus let us set $\delta_g = \int_{\Omega_0} gdx$. As
where $M = \max(K(R, \frac{1}{\kappa}) + \frac{\rho_s}{2}, \frac{\gamma}{2})$.

Gronwall's Lemma implies

$$ \int_{0}^{L} |\partial_{xt}\zeta_{\epsilon}|^{2} dx ds + \int_{0}^{L} |\partial_{xxt}\eta_{\epsilon}|^{2} dx ds \leq \frac{2C_{0}(R)}{\min(\rho_{s}, \gamma)} \exp \left( \frac{2MT}{\min(\rho_{s}, \gamma)} \right). $$

Combining (3.61) with the above inequality implies for $T \leq T^*$

$$ \frac{\Lambda \rho_f}{4R} \|\partial_t v_\epsilon\|_{L^2(0,t;L^2_\#(\Omega_0))}^2 + \frac{\Lambda \rho_s}{2} \|\partial_{tt}\zeta_\epsilon\|_{L^2(0,t;L^2_\#(0,L))}^2 + \frac{\Lambda \rho_s}{2} \|\partial_{tt}\eta_\epsilon\|_{L^2(0,t;L^2_\#(0,L))}^2 + \gamma \Lambda \|\partial_{xxt}\eta_\epsilon\|_{L^2(0,t;L^2_\#(0,L))}^2 \\ + \frac{\rho_f}{2R} \|\partial_x v_\epsilon(t)\|_{L^2_\#(\Omega_0)}^2 + \frac{\rho_s}{2} \|\partial_{tx}\zeta_\epsilon(t)\|_{L^2_\#(0,L)}^2 + \frac{\rho_s}{4} \|\partial_{tx}\eta_\epsilon(t)\|_{L^2_\#(0,L)}^2 + \frac{\mu\lambda(R)}{4} \|\nabla\partial_x v_\epsilon\|_{L^2(0,t;L^2_\#(\Omega_0))}^2 \\ + \frac{\beta}{4} \|\partial_{xx}\zeta_\epsilon(t)\|_{L^2_\#(0,L)}^2 + \frac{\alpha}{4} \|\partial_{xxx}\eta_\epsilon(t)\|_{L^2_\#(0,L)}^2 + \frac{\gamma}{2} \|\partial_{txx}\eta_\epsilon(t)\|_{L^2_\#(0,L)}^2 \leq C_{0,T^*}(R), \quad (3.62) $$

with $C_{0,T^*}(R)$ non decreasing with respect to $R$. The elliptic estimate (3.47) further implies

$$ \|\partial_{yy} v_{\epsilon}\|_{L^2(0,t;L^2_\#(\Omega_0))} + \|\nabla q_{\epsilon}\|_{L^2(0,t;L^2_\#(\Omega_0))} \leq C_{0,T^*}(R). \quad (3.63) $$

where $C_0$ depends again only on initial data, source terms, $R$ and the different positive physical parameters.

**Limit as $\epsilon$ goes to zero and uniqueness.** By standard compactness arguments, we may extract a weakly converging subsequence in the family of $\epsilon$- solutions $(v_\epsilon, q_\epsilon, \zeta_\epsilon, \eta_\epsilon)$ satisfying (3.26), (3.62) and (3.63). Then applying classical arguments, we obtain that the weak limit $(v, q, \zeta, \eta)$ is a solution to our linear problem with the expected regularity and satisfying (3.14). We left to the reader all these computations as they are completely standard. Finally, suppose that we have two solutions to our linear system with the same data, then the difference would satisfy (3.26) with vanishing data. We deduce the uniqueness of the solution to the homogeneous linear problem (3.1). This ends the proof of Proposition 3.1.

## 3.3 Non homogeneous divergence constraint.

In this subsection, we study the linear problem (3.1)-(3.7) in case when $g$ is not equal to zero. More precisely, we establish the following proposition:

**Proposition 3.2** Let $(a, b) \in H^2_\#(0, L)$ such that $\chi_{a,b}$ is a injective on $\bar{\mathcal{F}}_0$. Given $T > 0$, initial data $(v_0, \zeta_0, \zeta_1, \eta_0, \eta_1) \in X^0$ and a source term $(f, h, g, 0, 0) \in S_T$. Then there exists a unique solution $(v, q, \zeta, \eta) \in X_{f,T} \times X_{s,T}$ to Problem (3.1)-(3.7) and a non decreasing function $K : [0, \infty) \to [0, \infty)$ such that assuming $\|(a, b)\|_{H^2_\#(0,L)} + \|(\det \nabla \chi_{a,b})^{-1}\|_{C_\#(\bar{\Omega}_0)} \leq R$, the solution $(v, q, \zeta, \eta)$ satisfies

$$ \|(v, q)\|_{X_{f,T}} + \|(\zeta, \eta)\|_{X_{s,T}} \leq K(R)(\|(v_0, \zeta_0, \zeta_1, \eta_0, \eta_1)\|_{X^0} + \|(f, h, g, 0, 0)\|_{S_T}) \quad (3.64) $$

Proof. We look for $(v, q, \zeta, \eta)$ solution to (3.1)-(3.7) with initial data $(v_0, \zeta_0, \zeta_1, \eta_0, \eta_1) \in X^0$ and source term $(f, h, g, 0, 0) \in S_T$. We are going to lift the right hand side of the divergence in order to apply the previous result obtained in the case of divergence free constraint. To do so we need to modify the fluid velocity unknown but also the elastic displacement one. This is due to the fact that the average of $g$ is not equal to zero. It is a key difference compared to the case where the longitudinal displacement is neglected. Thus let us set $\delta_g = \int_{\Omega_0} g dx$. As
$g \in L^2(0, T; H_{\#}^1(\Omega_0)) \cap H^1(0, T; (H_{\#}^1(\Omega_0))')$, then $\delta_g = \langle g, 1 \rangle_{(H_{\#}^1(\Omega_0))', H_{\#}^1(\Omega_0)} \in H^1(0, T)$, and by Proposition 2.3 there exists unique couple $(w, q_1) \in L^2(0, T; H_{\#}^2(\Omega_0)) \times L^2(0, T; H_{\#}^1(\Omega_0) \cap L_{\#0}^2(\Omega_0))$ such that

$$ -\mu \operatorname{div} \left( (A_{a,b} \nabla)w + \frac{1}{\det \nabla \chi_{a,b}} B_{a,b} \nabla w^T B_{a,b} \right) + (B_{a,b} \nabla)q_1 = 0, \quad \text{on } \Omega_0, \qquad (3.65) $$

$$ \operatorname{div}(B_{a,b}^T w) = g, \quad \text{in } \Omega_0, \qquad (3.66) $$

$$ w|_{y=1} = \left(0, \frac{\delta_g}{L}\right), \qquad (3.67) $$

$$ w|_{y=0} = 0. \qquad (3.68) $$

that further satisfies

$$ \|w\|_{L^2(0,T;H_{\#}^2(\Omega_0))} + \|q_1\|_{L^2(0,T;H_{\#}^1(\Omega_0))} \le K(R) \left( \|g\|_{L^2(0,T;H_{\#}^1(\Omega_0))} + \|\delta_g\|_{L^2(0,T)} \right) \le K(R) \|g\|_{L^2(0,T;H_{\#}^1(\Omega_0))}. \qquad (3.69) $$

Now, we are in position to derive an estimate of $\partial_t w$ in $L^2(0, T; L^2(\Omega_0))$. Let $\varphi \in D(\Omega_0 \times (0, T))$, then by integration by parts we have

$$ \int_0^T \int_{\Omega_0} \partial_t w \cdot \varphi \, dxdydt = - \int_0^T \int_{\Omega_0} w \cdot \partial_t \varphi \, dxdydt. \qquad (3.70) $$

Again by using Proposition 2.3 there exists unique couple $(z_\varphi, q_\varphi) \in C_c^1([0, T]; H_{\#}^2(\Omega_0)) \times C_c^1([0, T]; H_{\#}^1(\Omega_0) \cap L_{\#0}^2(\Omega_0))$, such that

$$ -\mu \operatorname{div} \left( (A_{a,b} \nabla)z_\varphi + \frac{1}{\det \nabla \chi_{a,b}} B_{a,b} \nabla z_\varphi^T B_{a,b} \right) + (B_{a,b} \nabla)q_\varphi = \varphi, \quad \text{in } \Omega_0, $$

$$ \operatorname{div}(B_{a,b}^T z_\varphi) = 0, \quad \text{in } \Omega_0, $$

$$ z|_{y=1} = 0, \quad z|_{y=0} = 0, $$

and satisfying

$$ \|z_\varphi\|_{L^2(0,T;H_{\#}^2(\Omega_0))} + \|q_\varphi\|_{L^2(0,T;H_{\#}^1(\Omega_0))} \le K(R) \|\varphi\|_{L^2(0,T;L_{\#}^2(\Omega_0))}. \qquad (3.71) $$

It follows that

$$ \int_0^T \int_{\Omega_0} \partial_t w \cdot \varphi \, dxdydt = - \int_0^T \int_{\Omega_0} w \cdot \left( -\mu \operatorname{div} \left( (A_{a,b} \nabla) \partial_t z_\varphi + \frac{1}{\det \nabla \chi_{a,b}} B_{a,b} \nabla \partial_t z_\varphi^T B_{a,b} \right) + (B_{a,b} \nabla) \partial_t q_\varphi \right) \, dxdydt. \qquad (3.72) $$

By integration by parts and since $\operatorname{div}(B_{a,b}^T w) = g$, $w|_{y=0} = 0$ and $w|_{y=1} = (0, \frac{\delta_g}{L})$, we obtain

$$ - \int_{\Omega_0} (B_{a,b} \nabla) \partial_t q_\varphi \cdot w \, dxdy = \int_{\Omega_0} g \partial_t q_\varphi \, dxdy - \frac{\delta_g}{L} \int_0^L \partial_t q_{\varphi|_{y=1}} (1 + \partial_x a) \, dx \qquad (3.73) $$

On the other hand, by integration by parts and using again that $w|_{y=0} = 0$ and $w|_{y=1} = (0, \frac{\delta_g}{L})$, we get

$$ \int_{\Omega_0} \operatorname{div} \left( (A_{a,b} \nabla) \partial_t z_\varphi + \frac{1}{\det \nabla \chi_{a,b}} B_{a,b} \nabla \partial_t z_\varphi^T B_{a,b} \right) \cdot w \, dxdy \\ = - \int_{\Omega_0} \left( (A_{a,b} \nabla) \partial_t z_\varphi + \frac{1}{\det \nabla \chi_{a,b}} B_{a,b} \nabla \partial_t z_\varphi^T B_{a,b} \right) : \nabla w \, dxdy \\ + \frac{\delta_g}{L} \int_0^L \left( (A_{a,b} \nabla) \partial_t z_\varphi + \frac{1}{\det \nabla \chi_{a,b}} B_{a,b} \nabla \partial_t z_\varphi^T B_{a,b} \right) \Big|_{y=1} e_2 \cdot e_2 \, dx. $$
$g \in L^2(0, T; H_{\#}^1(\Omega_0)) \cap H^1(0, T; (H_{\#}^1(\Omega_0))')$, then $\delta_g = \langle g, 1 \rangle_{(H_{\#}^1(\Omega_0))', H_{\#}^1(\Omega_0)} \in H^1(0, T)$, and by Proposition 2.3 there exists unique couple $(w, q_1) \in L^2(0, T; H_{\#}^2(\Omega_0)) \times L^2(0, T; H_{\#}^1(\Omega_0) \cap L_{\#0}^2(\Omega_0))$ such that

$$ -\mu \operatorname{div} \left( (A_{a,b} \nabla)w + \frac{1}{\det \nabla \chi_{a,b}} B_{a,b} \nabla w^T B_{a,b} \right) + (B_{a,b} \nabla)q_1 = 0, \quad \text{on } \Omega_0, \qquad (3.65) $$

$$ \operatorname{div}(B_{a,b}^T w) = g, \quad \text{in } \Omega_0, \qquad (3.66) $$

$$ w|_{y=1} = \left(0, \frac{\delta_g}{L}\right), \qquad (3.67) $$

$$ w|_{y=0} = 0. \qquad (3.68) $$

that further satisfies

$$ \|w\|_{L^2(0,T;H_{\#}^2(\Omega_0))} + \|q_1\|_{L^2(0,T;H_{\#}^1(\Omega_0))} \le K(R) \left( \|g\|_{L^2(0,T;H_{\#}^1(\Omega_0))} + \|\delta_g\|_{L^2(0,T)} \right) \le K(R) \|g\|_{L^2(0,T;H_{\#}^1(\Omega_0))}. \qquad (3.69) $$

Now, we are in position to derive an estimate of $\partial_t w$ in $L^2(0, T; L^2(\Omega_0))$. Let $\varphi \in \mathcal{D}(\Omega_0 \times (0, T))$, then by integration by parts we have

$$ \int_0^T \int_{\Omega_0} \partial_t w \cdot \varphi \, dxdydt = - \int_0^T \int_{\Omega_0} w \cdot \partial_t \varphi \, dxdydt. \qquad (3.70) $$

Again by using Proposition 2.3 there exists unique couple $(z_\varphi, q_\varphi) \in C_c^1([0, T]; H_{\#}^2(\Omega_0)) \times C_c^1([0, T]; H_{\#}^1(\Omega_0) \cap L_{\#0}^2(\Omega_0))$, such that

$$ -\mu \operatorname{div} \left( (A_{a,b} \nabla)z_\varphi + \frac{1}{\det \nabla \chi_{a,b}} B_{a,b} \nabla z_\varphi^T B_{a,b} \right) + (B_{a,b} \nabla)q_\varphi = \varphi, \quad \text{in } \Omega_0, $$

$$ \operatorname{div}(B_{a,b}^T z_\varphi) = 0, \quad \text{in } \Omega_0, $$

$$ z|_{y=1} = 0, \quad z|_{y=0} = 0, $$

and satisfying

$$ \|z_\varphi\|_{L^2(0,T;H_{\#}^2(\Omega_0))} + \|q_\varphi\|_{L^2(0,T;H_{\#}^1(\Omega_0))} \le K(R) \|\varphi\|_{L^2(0,T;L_{\#}^2(\Omega_0))}. \qquad (3.71) $$

It follows that

$$ \int_0^T \int_{\Omega_0} \partial_t w \cdot \varphi \, dxdydt = - \int_0^T \int_{\Omega_0} w \cdot \left( -\mu \operatorname{div} \left( (A_{a,b} \nabla) \partial_t z_\varphi + \frac{1}{\det \nabla \chi_{a,b}} B_{a,b} \nabla \partial_t z_\varphi^T B_{a,b} \right) + (B_{a,b} \nabla) \partial_t q_\varphi \right) \, dxdydt. \qquad (3.72) $$

By integration by parts and since $\operatorname{div}(B_{a,b}^T w) = g$, $w|_{y=0} = 0$ and $w|_{y=1} = (0, \frac{\delta_g}{L})$, we obtain

$$ - \int_{\Omega_0} (B_{a,b} \nabla) \partial_t q_\varphi \cdot w \, dxdy = \int_{\Omega_0} g \partial_t q_\varphi \, dxdy - \frac{\delta_g}{L} \int_0^L \partial_t q_{\varphi|_{y=1}} (1 + \partial_x a) \, dx \qquad (3.73) $$

On the other hand, by integration by parts and using again that $w|_{y=0} = 0$ and $w|_{y=1} = (0, \frac{\delta_g}{L})$, we get

$$ \int_{\Omega_0} \operatorname{div} \left( (A_{a,b} \nabla) \partial_t z_\varphi + \frac{1}{\det \nabla \chi_{a,b}} B_{a,b} \nabla \partial_t z_\varphi^T B_{a,b} \right) \cdot w \, dxdy \\ = - \int_{\Omega_0} \left( (A_{a,b} \nabla) \partial_t z_\varphi + \frac{1}{\det \nabla \chi_{a,b}} B_{a,b} \nabla \partial_t z_\varphi^T B_{a,b} \right) : \nabla w \, dxdy \\ + \frac{\delta_g}{L} \int_0^L \left( (A_{a,b} \nabla) \partial_t z_\varphi + \frac{1}{\det \nabla \chi_{a,b}} B_{a,b} \nabla \partial_t z_\varphi^T B_{a,b} \right) \Big|_{y=1} e_2 \cdot e_2 \, dx. $$
Since $A_{a,b}$ is symmetric, we get

$$ 
\begin{aligned}
\int_{\Omega_0} \operatorname{div} \left( (A_{a,b} \nabla) \partial_t z_\varphi + \frac{1}{\det \nabla \chi_{a,b}} B_{a,b} \nabla \partial_t z_\varphi^T B_{a,b} \right) \cdot w \, dxdy \\
&= - \int_{\Omega_0} \left( (A_{a,b} \nabla)w + \frac{1}{\det \nabla \chi_{a,b}} B_{a,b} \nabla w^T B_{a,b} \right) : \nabla \partial_t z_\varphi \, dxdy \\
&\quad + \frac{\delta_g}{L} \int_0^L \left( (A_{a,b} \nabla) \partial_t z_\varphi + \frac{1}{\det \nabla \chi_{a,b}} B_{a,b} \nabla \partial_t z_\varphi^T B_{a,b} \right) \Big|_{y=1} e_2 \cdot e_2 \, dx.
\end{aligned}
$$

Again by integration by parts and using that $z_{\varphi|_{y=0}} = z_{\varphi|_{y=0}} = 0$, we have

$$ 
\begin{aligned}
\int_{\Omega_0} \operatorname{div} \left( (A_{a,b} \nabla) \partial_t z_\varphi + \frac{1}{\det \nabla \chi_{a,b}} B_{a,b} \nabla \partial_t z_\varphi^T B_{a,b} \right) \cdot w \, dxdy \\
&= \int_{\Omega_0} \operatorname{div} \left( (A_{a,b} \nabla)w + \frac{1}{\det \nabla \chi_{a,b}} B_{\zeta_0, \eta_0} \nabla w^T B_{a,b} \right) \cdot \partial_t z_\varphi \, dxdy \\
&\quad + \frac{\delta_g}{L} \int_0^L \left( (A_{a,b} \nabla) \partial_t z_\varphi + \frac{1}{\det \nabla \chi_{a,b}} B_{a,b} \nabla \partial_t z_\varphi^T B_{a,b} \right) \Big|_{y=1} e_2 \cdot e_2 \, dx.
\end{aligned}
$$

It follows thanks to (3.65) that

$$ 
\begin{aligned}
\int_{\Omega_0} \operatorname{div} \left( (A_{a,b} \nabla) \partial_t z_\varphi + \frac{1}{\det \nabla \chi_{a,b}} B_{a,b} \nabla \partial_t z_\varphi^T B_{a,b} \right) \cdot w \, dxdy \\
&= - \int_{\Omega_0} \left( (B_{a,b} \nabla)q_1 \right) \cdot \partial_t z_\varphi \, dxdy \\
&\quad + \frac{\delta_g}{L} \int_0^L \left( (A_{a,b} \nabla) \partial_t z_\varphi + \frac{1}{\det \nabla \chi_{a,b}} B_{a,b} \nabla \partial_t z_\varphi^T B_{a,b} \right) \Big|_{y=1} e_2 \cdot e_2 \, dx.
\end{aligned}
$$

Consequently, performing an integration by parts in the first term in the right hand side and by noting that $\operatorname{div}(B_{a,b}^T \partial_t z_\varphi) = 0$ and $z_{\varphi|_{y=0}} = z_{\varphi|_{y=1}} = 0$, we obtain

$$ 
\begin{aligned}
\int_{\Omega_0} \operatorname{div} \left( (A_{a,b} \nabla) \partial_t z_\varphi + \frac{1}{\det \nabla \chi_{a,b}} B_{a,b} \nabla \partial_t z_\varphi^T B_{a,b} \right) \cdot w \, dxdy \\
&= \frac{\delta_g}{L} \int_0^L \left( (A_{a,b} \nabla) \partial_t z_\varphi + \frac{1}{\det \nabla \chi_{a,b}} B_{a,b} \nabla \partial_t z_\varphi^T B_{a,b} \right) \Big|_{y=1} e_2 \cdot e_2 \, dx.
\end{aligned}
$$

Combining the above equality with (3.72) in view of (3.73), we get for any $\varphi \in \mathcal{D}(\Omega_0 \times (0, T))$

$$ 
\begin{aligned}
\int_0^T \int_{\Omega_0} \partial_t w \cdot \varphi \, dx \, dy \, dt = \int_0^T \int_{\Omega_0} g \partial_t q_\varphi \, dx \, dy \, dt &+ \int_0^T \frac{\delta_g}{L} \left( - \int_0^L \partial_t q_{\varphi|_{y=1}} (1 + \partial_x a) \, dx \right. \\
& \left. + \int_0^L \left( (A_{a,b} \nabla) \partial_t z_\varphi + \frac{1}{\det \nabla \chi_{a,b}} B_{a,b} \nabla \partial_t z_\varphi^T B_{a,b} \right) \Big|_{y=1} e_2 \cdot e_2 \, dx \right) \, dt.
\end{aligned}
\quad (3.74) $$

By integrating the right hand side with respect to time, we obtain for any $\varphi \in \mathcal{D}(\Omega_0 \times (0, T))$

$$ 
\begin{aligned}
\int_0^T \int_{\Omega_0} \partial_t w \cdot \varphi \, dx \, dy \, dt = - \int_0^T \langle \partial_t g, q_\varphi \rangle_{(H_\sharp^1(\Omega_0))', H^1(\Omega_0)} \, dt &- \int_0^T \frac{\dot{\delta}_g}{L} \left( - \int_0^L q_{\varphi|_{y=1}} (1 + \partial_x a) \, dx \right. \\
& \left. + \int_0^L \left( (A_{a,b} \nabla) z_\varphi + \frac{1}{\det \nabla \chi_{a,b}} B_{a,b} \nabla z_\varphi^T B_{a,b} \right) \Big|_{y=1} e_2 \cdot e_2 \, dx \right) \, dt.
\end{aligned}
\quad (3.75) $$

It follows

$$ 
\begin{aligned}
\left| \int_0^T \int_{\Omega_0} \partial_t w \cdot \varphi \, dx \, dy \, dt \right| \le K(R) \left( \|\partial_t g\|_{L^2(0,T;H_\sharp^1(\Omega_0))'} \|q_\varphi\|_{L^2(0,T;H_\sharp^1(\Omega_0))} + \|\dot{\delta}_g\|_{L^2(0,T)} \|q_\varphi\|_{L^2(0,T;H_\sharp^1(\Omega_0))} \right. \\
\left. + \|\dot{\delta}_g\|_{L^2(0,T)} \|z_\varphi\|_{L^2(0,T;H_\sharp^2(\Omega_0))} \right).
\end{aligned}
$$
Since $A_{a,b}$ is symmetric, we get

$$ 
\begin{aligned}
\int_{\Omega_0} \operatorname{div} \left( (A_{a,b} \nabla) \partial_t z_\varphi + \frac{1}{\det \nabla \chi_{a,b}} B_{a,b} \nabla \partial_t z_\varphi^T B_{a,b} \right) \cdot w \, dxdy \\
&= - \int_{\Omega_0} \left( (A_{a,b} \nabla)w + \frac{1}{\det \nabla \chi_{a,b}} B_{a,b} \nabla w^T B_{a,b} \right) : \nabla \partial_t z_\varphi \, dxdy \\
&\quad + \frac{\delta_g}{L} \int_0^L \left( (A_{a,b} \nabla) \partial_t z_\varphi + \frac{1}{\det \nabla \chi_{a,b}} B_{a,b} \nabla \partial_t z_\varphi^T B_{a,b} \right) \Big|_{y=1} e_2 \cdot e_2 \, dx.
\end{aligned}
$$

Again by integration by parts and using that $z_{\varphi|_{y=0}} = z_{\varphi|_{y=0}} = 0$, we have

$$ 
\begin{aligned}
\int_{\Omega_0} \operatorname{div} \left( (A_{a,b} \nabla) \partial_t z_\varphi + \frac{1}{\det \nabla \chi_{a,b}} B_{a,b} \nabla \partial_t z_\varphi^T B_{a,b} \right) \cdot w \, dxdy \\
&= \int_{\Omega_0} \operatorname{div} \left( (A_{a,b} \nabla)w + \frac{1}{\det \nabla \chi_{a,b}} B_{\zeta_0, \eta_0} \nabla w^T B_{a,b} \right) \cdot \partial_t z_\varphi \, dxdy \\
&\quad + \frac{\delta_g}{L} \int_0^L \left( (A_{a,b} \nabla) \partial_t z_\varphi + \frac{1}{\det \nabla \chi_{a,b}} B_{a,b} \nabla \partial_t z_\varphi^T B_{a,b} \right) \Big|_{y=1} e_2 \cdot e_2 \, dx.
\end{aligned}
$$

It follows thanks to (3.65) that

$$ 
\begin{aligned}
\int_{\Omega_0} \operatorname{div} \left( (A_{a,b} \nabla) \partial_t z_\varphi + \frac{1}{\det \nabla \chi_{a,b}} B_{a,b} \nabla \partial_t z_\varphi^T B_{a,b} \right) \cdot w \, dxdy \\
&= - \int_{\Omega_0} \left( (B_{a,b} \nabla)q_1 \right) \cdot \partial_t z_\varphi \, dxdy \\
&\quad + \frac{\delta_g}{L} \int_0^L \left( (A_{a,b} \nabla) \partial_t z_\varphi + \frac{1}{\det \nabla \chi_{a,b}} B_{a,b} \nabla \partial_t z_\varphi^T B_{a,b} \right) \Big|_{y=1} e_2 \cdot e_2 \, dx.
\end{aligned}
$$

Consequently, performing an integration by parts in the first term in the right hand side and by noting that $\operatorname{div}(B_{a,b}^T \partial_t z_\varphi) = 0$ and $z_{\varphi|_{y=0}} = z_{\varphi|_{y=1}} = 0$, we obtain

$$ 
\begin{aligned}
\int_{\Omega_0} \operatorname{div} \left( (A_{a,b} \nabla) \partial_t z_\varphi + \frac{1}{\det \nabla \chi_{a,b}} B_{a,b} \nabla \partial_t z_\varphi^T B_{a,b} \right) \cdot w \, dxdy \\
&= \frac{\delta_g}{L} \int_0^L \left( (A_{a,b} \nabla) \partial_t z_\varphi + \frac{1}{\det \nabla \chi_{a,b}} B_{a,b} \nabla \partial_t z_\varphi^T B_{a,b} \right) \Big|_{y=1} e_2 \cdot e_2 \, dx.
\end{aligned}
$$

Combining the above equality with (3.72) in view of (3.73), we get for any $\varphi \in \mathcal{D}(\Omega_0 \times (0, T))$

$$ 
\begin{aligned}
\int_0^T \int_{\Omega_0} \partial_t w \cdot \varphi \, dx \, dy \, dt = \int_0^T \int_{\Omega_0} g \, \partial_t q_\varphi \, dx \, dy \, dt + \int_0^T \frac{\delta_g}{L} \left( - \int_0^L \partial_t q_{\varphi|_{y=1}} (1 + \partial_x a) \, dx \right. \\
\left. + \int_0^L \left( (A_{a,b} \nabla) \partial_t z_\varphi + \frac{1}{\det \nabla \chi_{a,b}} B_{a,b} \nabla \partial_t z_\varphi^T B_{a,b} \right) \Big|_{y=1} e_2 \cdot e_2 \, dx \right) \, dt.
\end{aligned}
\quad (3.74) $$

By integrating the right hand side with respect to time, we obtain for any $\varphi \in \mathcal{D}(\Omega_0 \times (0, T))$

$$ 
\begin{aligned}
\int_0^T \int_{\Omega_0} \partial_t w \cdot \varphi \, dx \, dy \, dt = - \int_0^T \langle \partial_t g, q_\varphi \rangle_{(H_\sharp^1(\Omega_0))', H^1(\Omega_0)} \, dt - \int_0^T \frac{\dot{\delta}_g}{L} \left( - \int_0^L q_{\varphi|_{y=1}} (1 + \partial_x a) \, dx \right. \\
\left. + \int_0^L \left( (A_{a,b} \nabla) z_\varphi + \frac{1}{\det \nabla \chi_{a,b}} B_{a,b} \nabla z_\varphi^T B_{a,b} \right) \Big|_{y=1} e_2 \cdot e_2 \, dx \right) \, dt.
\end{aligned}
\quad (3.75) $$

It follows

$$ 
\begin{aligned}
\left| \int_0^T \int_{\Omega_0} \partial_t w \cdot \varphi \, dx \, dy \, dt \right| \le K(R) \left( \|\partial_t g\|_{L^2(0,T;H_\sharp^1(\Omega_0))'} \|q_\varphi\|_{L^2(0,T;H_\sharp^1(\Omega_0))} + \|\dot{\delta}_g\|_{L^2(0,T)} \|q_\varphi\|_{L^2(0,T;H_\sharp^1(\Omega_0))} \right. \\
\left. + \|\dot{\delta}_g\|_{L^2(0,T)} \|z_\varphi\|_{L^2(0,T;H_\sharp^2(\Omega_0))} \right).
\end{aligned}
$$
In view of (3.71) this implies

$$ \left| \int_0^T \int_{\Omega_0} \partial_t w \cdot \varphi \, dx \, dy \, dt \right| \le K(R) \|\partial_t g\|_{L^2(0,T;H_\sharp^1(\Omega_0))'} \|\varphi\|_{L^2(0,T;L_\sharp^2(\Omega_0))}. $$

Hence,

$$ \|\partial_t w\|_{L^2(0,T,L_\sharp^2(\Omega_0))} \le K(R) \|\partial_t g\|_{L^2(0,T;H_\sharp^1(\Omega_0))'} \quad (3.76) $$

Let us set $\tilde{v} = v - w$ and $\tilde{q} = q - q_1$, then $(\tilde{v}, \tilde{q})$ satisfies in $\Omega_0 \times (0, T)$

$$ \rho_{f,a,b} \partial_t \tilde{v} - \operatorname{div} \left( \mu(A_{a,b} \nabla) \tilde{v} + \frac{\mu}{\operatorname{det} \nabla \chi_{a,b}} B_{a,b} \nabla \tilde{v}^T B_{a,b} - \tilde{q} B_{a,b} \right) = f + \operatorname{div} h - \rho_{f,a,b} \partial_t w, \quad (3.77) $$

$$ \operatorname{div}(B_{a,b}^T \tilde{v}) = 0, \quad (3.78) $$

and the boundary conditions

$$ \tilde{v}_{|_{y=1}} = (\partial_t \zeta, \partial_t \eta - \frac{\delta_g}{L}), \quad (3.79) $$

$$ \tilde{v}_{|_{y=0}} = 0. \quad (3.80) $$

Due to (3.79) we need also to change of unknown for the structure displacement. Let us denote

$$ (\zeta, \tilde{\eta}) = \left( \zeta, \eta - \int_{0}^{t} \frac{\delta_g(s)}{L} ds \right), \quad (3.81) $$

we end up with the following coupled system

$$ \rho_{f,a,b} \partial_t \tilde{v} - \operatorname{div} \left( \mu(A_{a,b} \nabla) \tilde{v} + \frac{\mu}{\operatorname{det} \nabla \chi_{a,b}} B_{a,b} \nabla \tilde{v}^T B_{a,b} - \tilde{q} B_{a,b} \right) = f + \operatorname{div} h - \rho_{f,a,b} \partial_t w, \quad \text{in } \Omega_0 \times (0, T), \quad (3.82) $$

$$ \operatorname{div}(B_{a,b}^T \tilde{v}) = 0, \quad \text{in } \Omega_0 \times (0, T), \quad (3.83) $$

$$ \tilde{v}_{|_{y=1}} = (\partial_t \tilde{\zeta}, \partial_t \tilde{\eta}), \quad (3.84) $$

$$ \tilde{v}_{|_{y=0}} = 0, \quad (3.85) $$

$$ \rho_s \partial_{tt} \tilde{\zeta} - \beta \partial_{xx} \tilde{\zeta} = -T_{a,b}[\tilde{v}, \tilde{q}] \cdot e_1 - h_{|_{y=1}} e_2 \cdot e_1 + \ell, \quad \text{in } (0, L), \quad (3.86) $$

$$ \rho_s \partial_{tt} \tilde{\eta} + \alpha \partial_{xxxx} \tilde{\eta} - \gamma \partial_{txx} \tilde{\eta} = -T_{a,b}[\tilde{v}, \tilde{q}] \cdot e_2 - h_{|_{y=1}} e_2 \cdot e_2 + m, \quad \text{in } (0, L). \quad (3.87) $$

with

$$ \ell = -T_{a,b}[w, q_1] \cdot e_1, \quad (3.88) $$

$$ m = -T_{a,b}[w, q_1] \cdot e_2 - \rho_s \frac{\dot{\delta}_g}{L}. \quad (3.89) $$

By Proposition 3.1, there exists $(\tilde{v}, \tilde{q}, \tilde{\zeta}, \tilde{\eta}) \in X_{f,T} \times X_{s,T}$ such that

$$ \| (\tilde{v}, \tilde{q}) \|_{X_{f,T}} + \|(\tilde{\zeta}, \tilde{\eta})\|_{X_{s,T}} \le K(R) \left( \| (v_0, \zeta_0, \zeta_1, \eta_0, \eta_1) \|_{X^0} + \|(f, h, 0, 0, 0)\|_{S_T} \right. \\ \left. + \| \partial_t w \|_{L^2(0,T;L_\sharp^2(\Omega_0))} + \| w \|_{L^2(0,T;H_\sharp^2(\Omega_0))} + \| \dot{\delta}_g \|_{L^2(0,T)} \right). \quad (3.90) $$
In view of (3.71) this implies

$$ \left| \int_0^T \int_{\Omega_0} \partial_t w \cdot \varphi \, dx \, dy \, dt \right| \le K(R) \|\partial_t g\|_{L^2(0,T;H_\sharp^1(\Omega_0))'} \|\varphi\|_{L^2(0,T;L_\sharp^2(\Omega_0))}. $$

Hence,

$$ \|\partial_t w\|_{L^2(0,T,L_\sharp^2(\Omega_0))} \le K(R) \|\partial_t g\|_{L^2(0,T;H_\sharp^1(\Omega_0))'} \quad (3.76) $$

Let us set $\tilde{v} = v - w$ and $\tilde{q} = q - q_1$, then $(\tilde{v}, \tilde{q})$ satisfies in $\Omega_0 \times (0, T)$

$$ \rho_{f,a,b} \partial_t \tilde{v} - \operatorname{div} \left( \mu(A_{a,b} \nabla) \tilde{v} + \frac{\mu}{\det \nabla \chi_{a,b}} B_{a,b} \nabla \tilde{v}^T B_{a,b} - \tilde{q} B_{a,b} \right) = f + \operatorname{div} h - \rho_{f,a,b} \partial_t w, \quad (3.77) $$

$$ \operatorname{div}(B_{a,b}^T \tilde{v}) = 0, \quad (3.78) $$

and the boundary conditions

$$ \tilde{v}_{|_{y=1}} = (\partial_t \zeta, \partial_t \eta - \frac{\delta_g}{L}), \quad (3.79) $$

$$ \tilde{v}_{|_{y=0}} = 0. \quad (3.80) $$

Due to (3.79) we need also to change of unknown for the structure displacement. Let us denote

$$ (\zeta, \tilde{\eta}) = \left( \zeta, \eta - \int_{0}^{t} \frac{\delta_g(s)}{L} ds \right), \quad (3.81) $$

we end up with the following coupled system

$$ \rho_{f,a,b} \partial_t \tilde{v} - \operatorname{div} \left( \mu(A_{a,b} \nabla) \tilde{v} + \frac{\mu}{\det \nabla \chi_{a,b}} B_{a,b} \nabla \tilde{v}^T B_{a,b} - \tilde{q} B_{a,b} \right) = f + \operatorname{div} h - \rho_{f,a,b} \partial_t w, \quad \text{in } \Omega_0 \times (0, T), \quad (3.82) $$

$$ \operatorname{div}(B_{a,b}^T \tilde{v}) = 0, \quad \text{in } \Omega_0 \times (0, T), \quad (3.83) $$

$$ \tilde{v}_{|_{y=1}} = (\partial_t \tilde{\zeta}, \partial_t \tilde{\eta}), \quad (3.84) $$

$$ \tilde{v}_{|_{y=0}} = 0, \quad (3.85) $$

$$ \rho_s \partial_{tt} \tilde{\zeta} - \beta \partial_{xx} \tilde{\zeta} = -T_{a,b}[\tilde{v}, \tilde{q}] \cdot e_1 - h_{|_{y=1}} e_2 \cdot e_1 + \ell, \quad \text{in } (0, L), \quad (3.86) $$

$$ \rho_s \partial_{tt} \tilde{\eta} + \alpha \partial_{xxxx} \tilde{\eta} - \gamma \partial_{txx} \tilde{\eta} = -T_{a,b}[\tilde{v}, \tilde{q}] \cdot e_2 - h_{|_{y=1}} e_2 \cdot e_2 + m, \quad \text{in } (0, L). \quad (3.87) $$

with

$$ \ell = -T_{a,b}[w, q_1] \cdot e_1, \quad (3.88) $$

$$ m = -T_{a,b}[w, q_1] \cdot e_2 - \rho_s \frac{\dot{\delta}_g}{L}. \quad (3.89) $$

By Proposition 3.1, there exists $(\tilde{v}, \tilde{q}, \tilde{\zeta}, \tilde{\eta}) \in X_{f,T} \times X_{s,T}$ such that

$$ \| (\tilde{v}, \tilde{q}) \|_{X_{f,T}} + \|(\tilde{\zeta}, \tilde{\eta})\|_{X_{s,T}} \le K(R) \left( \| (v_0, \zeta_0, \zeta_1, \eta_0, \eta_1) \|_{X^0} + \|(f, h, 0, 0, 0)\|_{S_T} \right. \\ \left. + \| \partial_t w \|_{L^2(0,T;L_\sharp^2(\Omega_0))} + \| w \|_{L^2(0,T;H_\sharp^2(\Omega_0))} + \| \dot{\delta}_g \|_{L^2(0,T)} \right). \quad (3.90) $$
Combining the above inequality with estimates (3.69)-(3.76) implies, for $T \le T^*$

$$ \|(\tilde{v}, \tilde{q})\|_{X_{f,T}} + \|(\tilde{\zeta}, \tilde{\eta})\|_{X_{s,T}} \le K_{T^*}(R) \left( \|(v_0, \zeta_0, \zeta_1, \eta_0, \eta_1)\|_{X^0} + \|(f, h, 0, 0)\|_{S_T} \right. \\ \left. + \|\partial_t g\|_{L^2(0,T;H^1(\Omega_0))'} + \|g\|_{L^2(0,T;H^1(\Omega_0))} \right). $$

Hence,

$$ \|(\tilde{v}, \tilde{q})\|_{X_{f,T}} + \|(\tilde{\zeta}, \tilde{\eta})\|_{X_{s,T}} \le K_{T^*}(R) \left( \|(v_0, \zeta_0, \zeta_1, \eta_0, \eta_1)\|_{X^0} + \|(f, h, g, 0, 0)\|_{S_T} \right). $$

By noting that $v = \tilde{v} + w$, $q = \tilde{q} + q_1$, $\zeta = \tilde{\zeta}$, $\eta = \tilde{\eta} + 1/L \int_0^t \delta_g(s)ds$ and using (3.69)-(3.76), we finally obtain the existence and uniqueness of a solution $(v, q, \zeta, \eta) \in X_{f,T} \times X_{s,T}$ of Problem (3.1)-(3.7) satisfying the following estimates

$$ \|(v, q)\|_{X_{f,T}} + \|(\zeta, \eta)\|_{X_{s,T}} \le K_{T^*}(R) \left( \|(v_0, \zeta_0, \zeta_1, \eta_0, \eta_1)\|_{X^0} + \|(f, h, g, 0, 0)\|_{S_T} \right). \quad (3.91) $$

This concludes the study of the linearized non homogeneous problem.

# 4 Proof of Theorem 2.1

Consider $(v_0, \zeta_0, \zeta_1, \eta_0, \eta_1) \in H^1_\#(\mathcal{F}(0)) \times H^2_\#(0, L) \times H^1_\#(0, L) \times H^3_\#(0, L) \times H^1_\#(0, L)$ such that (2.3) is satisfied and such that there exists a mapping $\chi_{\zeta_0, \eta_0}$ satisfying $\det\nabla\chi_{\zeta_0, \eta_0} > 0$ on $\bar{\Omega}_0$ and $\chi_{\zeta_0, \eta_0}$ is injective on $\bar{\mathcal{F}}_0$. We denote by

$$ R := \|u_0\|_{H^1(\mathcal{F}(0))} + \|\zeta_0\|_{H^2_\#(0,L)} + \|\zeta_1\|_{H^1_\#(0,L)} + \|\eta_0\|_{H^3_\#(0,L)} + \|\eta_1\|_{H^1_\#(0,L)} + \|(\det \nabla \chi_{\zeta_0, \eta_0})^{-1}\|_{C_\#^2(\bar{\Omega}_0)}. $$

For $T > 0$ and $M > 0$, we define the set

$$ \mathcal{K}_T(M) := \{(z, r, \bar{\zeta}, \bar{\eta}) \in X_{f,T} \times E_{s,T}^0, z|_{y=0} = 0, z|_{y=1} = \partial_t \zeta e_1 + \partial_t \eta e_2, \|(z, r, \bar{\zeta}, \bar{\eta})\|_{X_{f,T} \times X_{s,T}} \le M\}, \quad (4.1) $$

where

$$ E_{s,T}^0 := \{(\bar{\zeta}, \bar{\eta}) \in X_{s,T} : \bar{\zeta}|_{t=0} = \bar{\eta}|_{t=0} = 0\}. $$

We define the mapping $\mathcal{L}_T^0$ from the set $\mathcal{K}_T(M)$ into $X_{f,T} \times E_{s,T}^0$ as follows:

$$ \mathcal{L}_T^0(z, r, \bar{\zeta}, \bar{\eta}) = (v, q, \bar{\mathcal{X}}, \bar{\mathcal{Y}}), $$

with $(v, q, \bar{\mathcal{X}}, \bar{\mathcal{Y}}) = (v, q, \mathcal{X} - \zeta_0, \mathcal{Y} - \eta_0)$ and $(v, q, \mathcal{X}, \mathcal{Y})$ is a solution of Problem (3.1)-(3.6) for $(a, b) = (\zeta_0, \eta_0)$ and source term equals to $(f, h, g, 0, 0)$, where $f, g, h$ are defined as follows:

$$ f[z, \zeta, \eta] = \rho_f(\det \nabla \chi_{\zeta_0, \eta_0} - \det \nabla \chi_{\zeta, \eta}) \partial_t z - \rho_f \left( (z - \partial_t \chi_{\zeta, \eta}) \cdot (B_{\zeta, \eta} \nabla) \right) z, \quad (4.2) $$

$$ h[z, r, \zeta, \eta] = -\mu[(A_{\zeta_0, \eta_0} - A_{\zeta, \eta})\nabla]z - \mu\left(\frac{1}{\det \nabla \chi_{\zeta_0, \eta_0}} B_{\zeta_0, \eta_0} \nabla z^T B_{\zeta_0, \eta_0} - \frac{1}{\det \nabla \chi_{\zeta, \eta}} B_{\zeta, \eta} \nabla z^T B_{\zeta, \eta}\right) + r(B_{\zeta_0, \eta_0} - B_{\zeta, \eta}). \quad (4.3) $$

$$ g[z, \zeta, \eta] = \operatorname{div}((B_{\zeta_0, \eta_0}^T - B_{\zeta, \eta}^T) z) = (B_{\zeta_0, \eta_0}^T - B_{\zeta, \eta}^T) : \nabla z, \quad (4.4) $$
Combining the above inequality with estimates (3.69)-(3.76) implies, for $T \le T^*$

$$ \| (\tilde{v}, \tilde{q}) \|_{X_{f,T}} + \| (\tilde{\zeta}, \tilde{\eta}) \|_{X_{s,T}} \le K_{T^*}(R) \left( \| (v_0, \zeta_0, \zeta_1, \eta_0, \eta_1) \|_{X^0} + \| (f, h, 0, 0) \|_{S_T} \right. \\ \left. + \| \partial_t g \|_{L^2(0,T; H^1_\#(\Omega_0))'} + \| g \|_{L^2(0,T; H^1_\#(\Omega_0))} \right). $$

Hence,

$$ \| (\tilde{v}, \tilde{q}) \|_{X_{f,T}} + \| (\tilde{\zeta}, \tilde{\eta}) \|_{X_{s,T}} \le K_{T^*}(R) \left( \| (v_0, \zeta_0, \zeta_1, \eta_0, \eta_1) \|_{X^0} + \| (f, h, g, 0, 0) \|_{S_T} \right). $$

By noting that $v = \tilde{v} + w$, $q = \tilde{q} + q_1$, $\zeta = \tilde{\zeta}$, $\eta = \tilde{\eta} + 1/L \int_0^t \delta_g(s)ds$ and using (3.69)-(3.76), we finally obtain the existence and uniqueness of a solution $(v, q, \zeta, \eta) \in X_{f,T} \times X_{s,T}$ of Problem (3.1)-(3.7) satisfying the following estimates

$$ \| (v, q) \|_{X_{f,T}} + \| (\zeta, \eta) \|_{X_{s,T}} \le K_{T^*}(R) \left( \| (v_0, \zeta_0, \zeta_1, \eta_0, \eta_1) \|_{X^0} + \| (f, h, g, 0, 0) \|_{S_T} \right). \quad (3.91) $$

This concludes the study of the linearized non homogeneous problem.

# 4 Proof of Theorem 2.1

Consider $(v_0, \zeta_0, \zeta_1, \eta_0, \eta_1) \in H^1_\#(\mathcal{F}(0)) \times H^2_\#(0, L) \times H^1_\#(0, L) \times H^3_\#(0, L) \times H^1_\#(0, L)$ such that (2.3) is satisfied and such that there exists a mapping $\chi_{\zeta_0, \eta_0}$ satisfying $\det \nabla \chi_{\zeta_0, \eta_0} > 0$ on $\bar{\Omega}_0$ and $\chi_{\zeta_0, \eta_0}$ is injective on $\bar{\mathcal{F}}_0$. We denote by

$$ R := \|u_0\|_{H^1(\mathcal{F}(0))} + \|\zeta_0\|_{H^2_\#(0,L)} + \|\zeta_1\|_{H^1_\#(0,L)} + \|\eta_0\|_{H^3_\#(0,L)} + \|\eta_1\|_{H^1_\#(0,L)} + \|(\det \nabla \chi_{\zeta_0, \eta_0})^{-1}\|_{C^2(\bar{\Omega}_0)}. $$

For $T > 0$ and $M > 0$, we define the set

$$ \mathcal{K}_T(M) := \{(z, r, \bar{\zeta}, \bar{\eta}) \in X_{f,T} \times E_{s,T}^0, z|_{y=0} = 0, z|_{y=1} = \partial_t \zeta e_1 + \partial_t \eta e_2, \|(z, r, \bar{\zeta}, \bar{\eta})\|_{X_{f,T} \times X_{s,T}} \le M\}, \quad (4.1) $$

where

$$ E_{s,T}^0 := \{(\bar{\zeta}, \bar{\eta}) \in X_{s,T} : \bar{\zeta}|_{t=0} = \bar{\eta}|_{t=0} = 0\}. $$

We define the mapping $\mathcal{L}_T^0$ from the set $\mathcal{K}_T(M)$ into $X_{f,T} \times E_{s,T}^0$ as follows:

$$ \mathcal{L}_T^0(z, r, \bar{\zeta}, \bar{\eta}) = (v, q, \bar{\mathcal{X}}, \bar{\mathcal{Y}}), $$

with $(v, q, \bar{\mathcal{X}}, \bar{\mathcal{Y}}) = (v, q, \mathcal{X} - \zeta_0, \mathcal{Y} - \eta_0)$ and $(v, q, \mathcal{X}, \mathcal{Y})$ is a solution of Problem (3.1)-(3.6) for $(a, b) = (\zeta_0, \eta_0)$ and source term equals to $(f, h, g, 0, 0)$, where $f, g, h$ are defined as follows:

$$ f[z, \zeta, \eta] = \rho_f (\det \nabla \chi_{\zeta_0, \eta_0} - \det \nabla \chi_{\zeta, \eta}) \partial_t z - \rho_f ((z - \partial_t \chi_{\zeta, \eta}) \cdot (B_{\zeta, \eta} \nabla)) z, \quad (4.2) $$

$$ h[z, r, \zeta, \eta] = -\mu[(A_{\zeta_0, \eta_0} - A_{\zeta, \eta})\nabla]z - \mu\left(\frac{1}{\det \nabla \chi_{\zeta_0, \eta_0}} B_{\zeta_0, \eta_0} \nabla z^T B_{\zeta_0, \eta_0} - \frac{1}{\det \nabla \chi_{\zeta, \eta}} B_{\zeta, \eta} \nabla z^T B_{\zeta, \eta}\right) + r(B_{\zeta_0, \eta_0} - B_{\zeta, \eta}). \quad (4.3) $$

$$ g[z, \zeta, \eta] = \operatorname{div}((B_{\zeta_0, \eta_0}^T - B_{\zeta, \eta}^T) z) = (B_{\zeta_0, \eta_0}^T - B_{\zeta, \eta}^T) : \nabla z, \quad (4.4) $$
where $(\zeta, \eta) = (\bar{\zeta} + \zeta_0, \bar{\eta} + \eta_0)$. We will prove that the source terms $f, g, h$ defined through (4.2)-(4.4) are in the good spaces so that $\mathcal{L}_T^0$ is well defined. We underline that $(v, q, \bar{\zeta}, \bar{\eta})$ is a fixed point of the mapping $\mathcal{L}_T^0$ if and only if $(v, q, \zeta, \eta)$ is a solution of Problem (2.19) with initial data (3.2)-(3.6) on $\Omega_0 \times (0, T_0)$. Moreover, if the mapping $\mathcal{L}_T^0$ admits a unique fixed point in $\mathcal{K}_T(M)$ for any $T \le T_0$, then Problem (2.19) with initial data (3.2)-(3.6) has a unique strong solution on $\Omega_0 \times (0, T_0)$. Hence, to prove Theorem 2.1 we will establish that for a well chosen $M \in (0, \infty)$ and $T$ small enough, depending on $R$ and $M$:

i. $\mathcal{L}_T^0(\mathcal{K}_T(M)) \subset \mathcal{K}_T(M);$

ii. the mapping $\mathcal{L}_T^0: \mathcal{K}_T(M) \to \mathcal{K}_T(M)$ is a contraction.

We start to establish the first assertion i.e. for $T$ small enough and $M \ge M_0$, the mapping $\mathcal{L}_T^0$ maps $\mathcal{K}_T(M)$ into itself. Let $(z, r, \bar{\zeta}, \bar{\eta}) \in \mathcal{K}_T(M)$ and denote by $(v, q, \bar{\mathcal{X}}, \bar{\mathcal{Y}}) = \mathcal{L}_T^0(z, r, \bar{\zeta}, \bar{\eta})$. We recall that from the definition of $\mathcal{L}_T^0$, we have $(\bar{\zeta}, \bar{\eta}) = (\zeta - \zeta_0, \eta - \eta_0)$ with $(v, q, \mathcal{X}, \mathcal{Y})$ is a solution of Problem (3.1)-(3.6) for $(a, b) = (\zeta_0, \eta_0)$ and source term $(f, h, g, 0, 0)$ defined through (4.2)-(4.4). If the source term $(f, h, g, 0, 0) \in S_T$, then by applying Proposition 3.2 for $(a, b) = (\zeta_0, \eta_0)$ we get that there exists a unique solution $(v, q, \mathcal{X}, \mathcal{Y}) \in X_{f,T} \times X_{s,T}$ of Problem (3.1)-(3.7) satisfying

$$ \|(v, q)\|_{X_{f,T}} + \|(\mathcal{X}, \mathcal{Y})\|_{X_{s,T}} \le C(R)(\|(v_0, \zeta_0, \zeta_1, \eta_0, \eta_1)\|_{X^0} + \|(f, h, g, 0, 0)\|_{S_T}). $$

It follows that

$$ \|\mathcal{L}_T^0(z, r, \bar{\zeta}, \bar{\eta})\|_{X_{f,T} \times X_{s,T}} \le \|(\zeta_0, \eta_0)\|_{H_\#^2(0,L) \times H_\#^3(0,L)} + C(R)(\|(v_0, \zeta_0, \zeta_1, \eta_0, \eta_1)\|_{X^0} + \|(f, h, g, 0, 0)\|_{S_T}). \quad (4.5) $$

Taking $M \ge M_0 = 2(\|(\zeta_0, \eta_0)\|_{H_\#^2(0,L) \times H_\#^3(0,L)} + C(R)(\|(v_0, \zeta_0, \zeta_1, \eta_0, \eta_1)\|_{X^0}))$, we get

$$ \|\mathcal{L}_T^0(z, r, \bar{\zeta}, \bar{\eta})\|_{X_{f,T} \times X_{s,T}} \le \frac{M}{2} + C(R)\|(f, h, g, 0, 0)\|_{S_T}. \quad (4.6) $$

To proceed we need to bound the source terms $f, h, g$ defined through (4.2)-(4.4) respectively in $L^2(0, T; L_\#^2(\Omega_0))$, $L^2(0, T; H_\#^1(\Omega_0))$, $L^2(0, T; H_\#^1(\Omega_0)) \cap H_{00}^1(0, T; (H_\#^1(\Omega_0))')$ in the right hand side of (4.6). In order to establish such bounds, we gather in the following lemma some properties of the matrix $B_{\zeta,\eta}$ and $A_{\zeta,\eta}$ that are defined by (2.11) and (2.12) respectively.

**Lemma 4.1** Assuming that $(\zeta, \eta) \in L^\infty(0, T; H_\#^2(0, L)) \cap W^{1,\infty}(0, T; H_\#^1(0, L))$ and $(\zeta, \eta)|_{t=0} = (\zeta_0, \eta_0) \in H_\#^2(0, L)$, then, for $0 < s < \frac{1}{2}$

$$ \begin{aligned} \|B_{\zeta, \eta} - B_{\zeta_0, \eta_0}\|_{L^\infty(0,T;H_\#^{1+s}(\Omega_0))} &\le C T^{\frac{1}{2}-s} \|(\bar{\zeta}, \bar{\eta})\|_{L^\infty(0,T;H_\#^2(0,L))}^{\frac{1}{2}+s} \|(\bar{\zeta}, \bar{\eta})\|_{W^{1,\infty}(0,T;H_\#^1(0,L))}^{\frac{1}{2}-s} \\ \|\partial_t B_{\zeta, \eta}\|_{L^\infty(0,T;L_\#^4(\Omega_0))} &\le C \|(\bar{\zeta}, \bar{\eta})\|_{L^\infty(0,T;H_\#^2(0,L)) \cap W^{1,\infty}(0,T;H_\#^1(0,L))} \end{aligned} $$

where $(\bar{\zeta}, \bar{\eta}) = (\zeta - \zeta_0, \eta - \eta_0)$.

Furthermore, if we assume that

• $\|(\zeta_0, \eta_0)\|_{H_\#^2(0,L)} + \|(\det \nabla \chi_{\zeta_0, \eta_0})^{-1}\|_{C_\#(\bar{\Omega}_0)} \le R,$

• $\|(\zeta, \eta)\|_{L^\infty(0,T;H_\#^2(0,L)) \cap W^{1,\infty}(0,T;H_\#^1(0,L))} \le M,$
where $(\zeta, \eta) = (\bar{\zeta} + \zeta_0, \bar{\eta} + \eta_0)$. We will prove that the source terms $f, g, h$ defined through (4.2)-(4.4) are in the good spaces so that $\mathcal{L}_T^0$ is well defined. We underline that $(v, q, \bar{\zeta}, \bar{\eta})$ is a fixed point of the mapping $\mathcal{L}_T^0$ if and only if $(v, q, \zeta, \eta)$ is a solution of Problem (2.19) with initial data (3.2)-(3.6) on $\Omega_0 \times (0, T_0)$. Moreover, if the mapping $\mathcal{L}_T^0$ admits a unique fixed point in $\mathcal{K}_T(M)$ for any $T \le T_0$, then Problem (2.19) with initial data (3.2)-(3.6) has a unique strong solution on $\Omega_0 \times (0, T_0)$. Hence, to prove Theorem 2.1 we will establish that for a well chosen $M \in (0, \infty)$ and $T$ small enough, depending on $R$ and $M$:

i. $\mathcal{L}_T^0(\mathcal{K}_T(M)) \subset \mathcal{K}_T(M);$

ii. the mapping $\mathcal{L}_T^0: \mathcal{K}_T(M) \to \mathcal{K}_T(M)$ is a contraction.

We start to establish the first assertion i.e. for $T$ small enough and $M \ge M_0$, the mapping $\mathcal{L}_T^0$ maps $\mathcal{K}_T(M)$ into itself. Let $(z, r, \bar{\zeta}, \bar{\eta}) \in \mathcal{K}_T(M)$ and denote by $(v, q, \bar{\mathcal{X}}, \bar{\mathcal{Y}}) = \mathcal{L}_T^0(z, r, \bar{\zeta}, \bar{\eta})$. We recall that from the definition of $\mathcal{L}_T^0$, we have $(\bar{\zeta}, \bar{\eta}) = (\zeta - \zeta_0, \eta - \eta_0)$ with $(v, q, \mathcal{X}, \mathcal{Y})$ is a solution of Problem (3.1)-(3.6) for $(a, b) = (\zeta_0, \eta_0)$ and source term $(f, h, g, 0, 0)$ defined through (4.2)-(4.4). If the source term $(f, h, g, 0, 0) \in S_T$, then by applying Proposition 3.2 for $(a, b) = (\zeta_0, \eta_0)$ we get that there exists a unique solution $(v, q, \mathcal{X}, \mathcal{Y}) \in X_{f,T} \times X_{s,T}$ of Problem (3.1)-(3.7) satisfying

$$ \|(v, q)\|_{X_{f,T}} + \|(\mathcal{X}, \mathcal{Y})\|_{X_{s,T}} \le C(R)(\|(v_0, \zeta_0, \zeta_1, \eta_0, \eta_1)\|_{X^0} + \|(f, h, g, 0, 0)\|_{S_T}). $$

It follows that

$$ \|\mathcal{L}_T^0(z, r, \bar{\zeta}, \bar{\eta})\|_{X_{f,T} \times X_{s,T}} \le \|(\zeta_0, \eta_0)\|_{H_\#^2(0,L) \times H_\#^3(0,L)} + C(R)(\|(v_0, \zeta_0, \zeta_1, \eta_0, \eta_1)\|_{X^0} + \|(f, h, g, 0, 0)\|_{S_T}). \quad (4.5) $$

Taking $M \ge M_0 = 2(\|(\zeta_0, \eta_0)\|_{H_\#^2(0,L) \times H_\#^3(0,L)} + C(R)(\|(v_0, \zeta_0, \zeta_1, \eta_0, \eta_1)\|_{X^0}))$, we get

$$ \|\mathcal{L}_T^0(z, r, \bar{\zeta}, \bar{\eta})\|_{X_{f,T} \times X_{s,T}} \le \frac{M}{2} + C(R)\|(f, h, g, 0, 0)\|_{S_T}. \quad (4.6) $$

To proceed we need to bound the source terms $f, h, g$ defined through (4.2)-(4.4) respectively in $L^2(0, T; L_\#^2(\Omega_0))$, $L^2(0, T; H_\#^1(\Omega_0))$, $L^2(0, T; H_\#^1(\Omega_0)) \cap H_{00}^1(0, T; (H_\#^1(\Omega_0))')$ in the right hand side of (4.6). In order to establish such bounds, we gather in the following lemma some properties of the matrix $B_{\zeta,\eta}$ and $A_{\zeta,\eta}$ that are defined by (2.11) and (2.12) respectively.

**Lemma 4.1** Assuming that $(\zeta, \eta) \in L^\infty(0, T; H_\#^2(0, L)) \cap W^{1,\infty}(0, T; H_\#^1(0, L))$ and $(\zeta, \eta)|_{t=0} = (\zeta_0, \eta_0) \in H_\#^2(0, L)$, then, for $0 < s < \frac{1}{2}$

$$ \begin{aligned} \|B_{\zeta, \eta} - B_{\zeta_0, \eta_0}\|_{L^{\infty}(0,T;H_{\#}^{1+s}(\Omega_0))} &\le C T^{\frac{1}{2}-s} \|(\bar{\zeta}, \bar{\eta})\|_{L^{\infty}(0,T;H_{\#}^2(0,L))}^{\frac{1}{2}+s} \|(\bar{\zeta}, \bar{\eta})\|_{W^{1,\infty}(0,T;H_{\#}^1(0,L))}^{\frac{1}{2}-s} \\ \|\partial_t B_{\zeta, \eta}\|_{L^{\infty}(0,T;L_{\#}^4(\Omega_0))} &\le C \|(\bar{\zeta}, \bar{\eta})\|_{L^{\infty}(0,T;H_{\#}^2(0,L)) \cap W^{1,\infty}(0,T;H_{\#}^1(0,L))} \end{aligned} $$

where $(\bar{\zeta}, \bar{\eta}) = (\zeta - \zeta_0, \eta - \eta_0)$.

Furthermore, if we assume that

* $\|(\zeta_0, \eta_0)\|_{H_\#^2(0,L)} + \|(\det \nabla \chi_{\zeta_0, \eta_0})^{-1}\|_{C_\#(\bar{\Omega}_0)} \le R,$

* $\|(\zeta, \eta)\|_{L^\infty(0,T;H_\#^2(0,L)) \cap W^{1,\infty}(0,T;H_\#^1(0,L))} \le M,$
we have

$$ \|\det \nabla \chi_{\zeta, \eta} - \det \nabla \chi_{\zeta_0, \eta_0}\|_{L^{\infty}(0,T; H_{\#}^{1+s}(\Omega_0))} \le C(R, M)T^{\frac{1}{2}-s} $$

and there exists a time $T(R, M)$ depending on $R$ and $M$ such that for $T \le T(R, M)$, $\det(\nabla\chi_{\zeta,\eta})^{-1}$ and $A_{\zeta,\eta}$ are well defined in $C([0,T]; H_{\#}^{1+s}(\Omega_0))$, for $0 < s < \frac{1}{2}$, and

$$ \begin{aligned} & \|(\det \nabla \chi_{\zeta, \eta})^{-1} - (\det \nabla \chi_{\zeta_0, \eta_0})^{-1}\|_{L^{\infty}(0,T; H_{\#}^{1+s}(\Omega_0))} \le C(R, M)T^{\frac{1}{2}-s} \\ & \|A_{\zeta, \eta} - A_{\zeta_0, \eta_0}\|_{L^{\infty}(0,T; H_{\#}^{1+s}(\Omega_0))} \le C(R, M)T^{\frac{1}{2}-s}, \\ & \|A_{\zeta, \eta} - A_{\zeta_0, \eta_0}\|_{L^{\infty}(0,T; H_{\#}^{1+s}(\Omega_0))} \le C(R, M)T^{\frac{1}{2}-s}, \end{aligned} $$

The proof of the above lemma relies on classical arguments and in particular the fact that $L^{\infty}(0,T; H_{\#}^2(0,L)) \cap W^{1,\infty}(0,T; H_{\#}^1(0,L))$ continuously embeds in $C^{0,1/2-s}([0,T]; H_{\#}^{1+s}(\Omega_0)) \subset C([0,T]; C_{\#}^1(\bar{\Omega}_0))$, $\det(\nabla\chi_{\zeta,\eta})^{-1}$ and $A_{\zeta,\eta}$ are well defined in $C([0,T]; H_{\#}^{1+s}(\Omega_0))$, for $0 < s < \frac{1}{2}$, and that the following inequality is satisfied, for $0 < s < \frac{1}{2}$, for $\bar{\phi} \in L^{\infty}(0,T; H_{\#}^2(0,L)) \cap W^{1,\infty}(0,T; H_{\#}^1(0,L))$, such that $\bar{\phi}|_{t=0} = 0$,

$$ \|\bar{\phi}\|_{L^{\infty}(0,T; H_{\#}^{\frac{3}{2}+s}(0,L))} \le C T^{\frac{1}{2}-s} \|\bar{\phi}\|_{L^{\infty}(0,T; H_{\#}^2(0,L))}^{\frac{1}{2}+s} \|\bar{\phi}\|_{W^{1,\infty}(0,T; H_{\#}^1(0,L))}^{\frac{1}{2}-s} \quad (4.7) $$

Moreover since $\|(\det \nabla \chi_{\zeta_0, \eta_0})^{-1}\|_{C_{\#}(\bar{\Omega}_0)} \le R$ and $\|(\zeta, \eta)\|_{C([0,T]; C_{\#}^1(\bar{\Omega}_0))} \le CM$, there exists a time $T(R, M)$ depending on $R$ and $M$ (decreasing with $M$ and $R$) such that for $T \le T(R, M)$, $\|(\det \nabla \chi_{\zeta, \eta})^{-1}\|_{C_{\#}(\bar{\Omega}_0)} \le 2R$ ensuring that $\det(\nabla \chi_{\zeta, \eta})^{-1}$ and $A_{\zeta, \eta}$ are well defined in $C([0,T]; H_{\#}^{1+s}(\Omega_0))$, for $0 < s < \frac{1}{2}$.

**Lemma 4.2** Given $(z, r, \bar{\zeta}, \bar{\eta}) \in K_T(M)$ and $(\zeta_0, \eta_0) \in H_{\#}^2(0, L)$ such that $\|(\zeta_0, \eta_0)\|_{H_{\#}^2(0,L)} + \|(\det \nabla \chi_{\zeta_0, \eta_0})^{-1}\|_{C_{\#}(\bar{\Omega}_0)} \le R$. Let $f[z, \zeta, \eta], h[z, r, \zeta, \eta], g[z, \zeta, \eta]$, be the source terms defined respectively by (4.2)-(4.4) where $(\zeta, \eta) = (\bar{\zeta} + \zeta_0, \bar{\eta} + \eta_0)$, then there exists a time $T(M, R)$ depending on $M$ and $R$ such that for any $T \le T(M, R)$

$$ \begin{aligned} & \|f[z, \zeta, \eta]\|_{L^2(0,T; L_{\#}^2(\Omega_0))} \le C(R, M)T^{\frac{1}{4}}, \\ & \|g[z, \zeta, \eta]\|_{L^2(0,T; H_{\#}^1(\Omega_0))} \le C T^{\frac{1}{4}} M^2, \\ & \|g[z, \zeta, \eta]\|_{H_{00}^1(0,T;(H_{\#}^1(\Omega_0)))'} \le C(R, M)T^{\frac{1}{4}} M^2, \\ & \|h[z, r, \zeta, \eta]\|_{L^2(0,T; H_{\#}^1(\Omega_0))} \le C(R, M)T^{\frac{1}{4}} M^2. \end{aligned} $$

Proof. From the definition of $f$ in (4.2), we have

$$ \|f[z, \zeta, \eta]\|_{L^2(0,T; L_{\#}^2(\Omega_0))} \le \rho_f \left( \begin{aligned} & \|(\det \nabla \chi_{\zeta_0, \eta_0} - \det \nabla \chi_{\zeta, \eta})\|_{L^{\infty}(0,T; L_{\#}^{\infty}(\Omega_0))} \|\partial_t z\|_{L^2(0,T; L_{\#}^2(\Omega_0))} \\ & + \|B_{\zeta, \eta}\|_{L^{\infty}(0,T; L_{\#}^{\infty}(\Omega_0))} \|z \nabla z\|_{L^2(0,T; L_{\#}^2(\Omega_0))} + T^{\frac{1}{2}} \|\partial_t \chi_{\zeta, \eta}\|_{L^{\infty}(0,T; L_{\#}^{\infty}(\Omega_0))} \|B_{\zeta, \eta}\|_{L^{\infty}(0,T; L_{\#}^{\infty}(\Omega_0))} \|z\|_{L^{\infty}(0,T; H_{\#}^1(\Omega_0))} \end{aligned} \right). $$

By using $\|z\|_{L_{\#}^4(\Omega_0)} \le C \|z\|_{L_{\#}^2(\Omega_0)}^{\frac{1}{2}} \|z\|_{H_{\#}^1(\Omega_0)}^{\frac{1}{2}}$, we get

$$ \|z \nabla z\|_{L^2(0,T; L_{\#}^2(\Omega_0))} \le C T^{\frac{1}{4}} \|z\|_{L^{\infty}(0,T; L_{\#}^2(\Omega_0))}^{\frac{3}{2}} \|z\|_{L^2(0,T; H_{\#}^2(\Omega_0))}^{\frac{1}{2}} $$

It follows that

$$ \begin{aligned} \|f[z, \zeta, \eta]\|_{L^2(0,T; L_{\#}^2(\Omega_0))} \le \rho_f & \left( \begin{aligned} & \|(\det \nabla \chi_{\zeta_0, \eta_0} - \det \nabla \chi_{\zeta, \eta})\|_{L^{\infty}(0,T; L_{\#}^{\infty}(\Omega_0))} \|\partial_t z\|_{L^2(0,T; L_{\#}^2(\Omega_0))} \\ & + C T^{\frac{1}{4}} \|B_{\zeta, \eta}\|_{L^{\infty}(0,T; L_{\#}^{\infty}(\Omega_0))} \|z\|_{L^{\infty}(0,T; L_{\#}^2(\Omega_0))}^{\frac{3}{2}} \|z\|_{L^2(0,T; H_{\#}^2(\Omega_0))}^{\frac{1}{2}} \\ & + T^{\frac{1}{2}} \|\partial_t \chi_{\zeta, \eta}\|_{L^{\infty}(0,T; L_{\#}^{\infty}(\Omega_0))} \|B_{\zeta, \eta}\|_{L^{\infty}(0,T; L_{\#}^{\infty}(\Omega_0))} \|z\|_{L^{\infty}(0,T; H_{\#}^1(\Omega_0))} \end{aligned} \right). \end{aligned} $$
we have

$$ \|\det \nabla \chi_{\zeta, \eta} - \det \nabla \chi_{\zeta_0, \eta_0}\|_{L^{\infty}(0,T; H_{\#}^{1+s}(\Omega_0))} \le C(R, M)T^{\frac{1}{2}-s} $$

and there exists a time $T(R, M)$ depending on $R$ and $M$ such that for $T \le T(R, M)$, $\det(\nabla\chi_{\zeta,\eta})^{-1}$ and $A_{\zeta,\eta}$ are well defined in $C([0,T]; H_{\#}^{1+s}(\Omega_0))$, for $0 < s < \frac{1}{2}$, and

$$ \begin{aligned} & \|(\det \nabla \chi_{\zeta, \eta})^{-1} - (\det \nabla \chi_{\zeta_0, \eta_0})^{-1}\|_{L^{\infty}(0,T; H_{\#}^{1+s}(\Omega_0))} \le C(R, M)T^{\frac{1}{2}-s} \\ & \|A_{\zeta, \eta} - A_{\zeta_0, \eta_0}\|_{L^{\infty}(0,T; H_{\#}^{1+s}(\Omega_0))} \le C(R, M)T^{\frac{1}{2}-s}, \\ & \|A_{\zeta, \eta} - A_{\zeta_0, \eta_0}\|_{L^{\infty}(0,T; H_{\#}^{1+s}(\Omega_0))} \le C(R, M)T^{\frac{1}{2}-s}, \end{aligned} $$

The proof of the above lemma relies on classical arguments and in particular the fact that $L^{\infty}(0,T; H_{\#}^2(0,L)) \cap W^{1,\infty}(0,T; H_{\#}^1(0,L))$ continuously embeds in $C^{0,1/2-s}([0,T]; H_{\#}^{1+s}(\Omega_0)) \subset C([0,T]; C_{\#}^1(\bar{\Omega}_0))$, $\det(\nabla\chi_{\zeta,\eta})^{-1}$ and $A_{\zeta,\eta}$ are well defined in $C([0,T]; H_{\#}^{1+s}(\Omega_0))$, for $0 < s < \frac{1}{2}$, and that the following inequality is satisfied, for $0 < s < \frac{1}{2}$, for $\bar{\phi} \in L^{\infty}(0,T; H_{\#}^2(0,L)) \cap W^{1,\infty}(0,T; H_{\#}^1(0,L))$, such that $\bar{\phi}|_{t=0} = 0$,

$$ \|\bar{\phi}\|_{L^{\infty}(0,T; H_{\#}^{\frac{3}{2}+s}(0,L))} \le C T^{\frac{1}{2}-s} \|\bar{\phi}\|_{L^{\infty}(0,T; H_{\#}^{2}(0,L))}^{\frac{1}{2}+s} \|\bar{\phi}\|_{W^{1,\infty}(0,T; H_{\#}^{1}(0,L))}^{\frac{1}{2}-s} \quad (4.7) $$

Moreover since $\|(\det \nabla \chi_{\zeta_0, \eta_0})^{-1}\|_{C_{\#}(\bar{\Omega}_0)} \le R$ and $\|(\zeta, \eta)\|_{C([0,T]; C_{\#}^1(\bar{\Omega}_0))} \le CM$, there exists a time $T(R, M)$ depending on $R$ and $M$ (decreasing with $M$ and $R$) such that for $T \le T(R, M)$, $\|(\det \nabla \chi_{\zeta, \eta})^{-1}\|_{C_{\#}(\bar{\Omega}_0)} \le 2R$ ensuring that $\det(\nabla \chi_{\zeta, \eta})^{-1}$ and $A_{\zeta, \eta}$ are well defined in $C([0,T]; H_{\#}^{1+s}(\Omega_0))$, for $0 < s < \frac{1}{2}$.

**Lemma 4.2** Given $(z, r, \bar{\zeta}, \bar{\eta}) \in K_T(M)$ and $(\zeta_0, \eta_0) \in H_{\#}^2(0, L)$ such that $\|(\zeta_0, \eta_0)\|_{H_{\#}^2(0,L)} + \|(\det \nabla \chi_{\zeta_0, \eta_0})^{-1}\|_{C_{\#}(\bar{\Omega}_0)} \le R$. Let $f[z, \zeta, \eta], h[z, r, \zeta, \eta], g[z, \zeta, \eta]$, be the source terms defined respectively by (4.2)-(4.4) where $(\zeta, \eta) = (\bar{\zeta} + \zeta_0, \bar{\eta} + \eta_0)$, then there exists a time $T(M, R)$ depending on $M$ and $R$ such that for any $T \le T(M, R)$

$$ \begin{aligned} & \|f[z, \zeta, \eta]\|_{L^2(0,T; L_{\#}^2(\Omega_0))} \le C(R, M)T^{\frac{1}{4}}, \\ & \|g[z, \zeta, \eta]\|_{L^2(0,T; H_{\#}^1(\Omega_0))} \le C T^{\frac{1}{4}} M^2, \\ & \|g[z, \zeta, \eta]\|_{H_{00}^1(0,T;(H_{\#}^1(\Omega_0)))'} \le C(R, M)T^{\frac{1}{4}} M^2, \\ & \|h[z, r, \zeta, \eta]\|_{L^2(0,T; H_{\#}^1(\Omega_0))} \le C(R, M)T^{\frac{1}{4}} M^2. \end{aligned} $$

Proof. From the definition of $f$ in (4.2), we have

$$ \|f[z, \zeta, \eta]\|_{L^2(0,T; L_{\#}^2(\Omega_0))} \le \rho_f \left( \begin{aligned} & \|(\det \nabla \chi_{\zeta_0, \eta_0} - \det \nabla \chi_{\zeta, \eta})\|_{L^{\infty}(0,T; L_{\#}^{\infty}(\Omega_0))} \|\partial_t z\|_{L^2(0,T; L_{\#}^2(\Omega_0))} \\ & + \|B_{\zeta, \eta}\|_{L^{\infty}(0,T; L_{\#}^{\infty}(\Omega_0))} \|z\nabla z\|_{L^2(0,T; L_{\#}^2(\Omega_0))} + T^{\frac{1}{2}} \|\partial_t \chi_{\zeta, \eta}\|_{L^{\infty}(0,T; L_{\#}^{\infty}(\Omega_0))} \|B_{\zeta, \eta}\|_{L^{\infty}(0,T; L_{\#}^{\infty}(\Omega_0))} \|z\|_{L^{\infty}(0,T; H_{\#}^1(\Omega_0))} \end{aligned} \right). $$

By using $\|z\|_{L_{\#}^4(\Omega_0)} \le C \|z\|_{L_{\#}^2(\Omega_0)}^{\frac{1}{2}} \|z\|_{H_{\#}^1(\Omega_0)}^{\frac{1}{2}}$, we get

$$ \|z\nabla z\|_{L^2(0,T; L_{\#}^2(\Omega_0))} \le C T^{\frac{1}{4}} \|z\|_{L^{\infty}(0,T; L_{\#}^2(\Omega_0))}^{\frac{3}{2}} \|z\|_{L^2(0,T; H_{\#}^2(\Omega_0))}^{\frac{1}{2}} $$

It follows that

$$ \begin{aligned} \|f[z, \zeta, \eta]\|_{L^2(0,T; L_{\#}^2(\Omega_0))} \le \rho_f \left( \begin{aligned} & \|(\det \nabla \chi_{\zeta_0, \eta_0} - \det \nabla \chi_{\zeta, \eta})\|_{L^{\infty}(0,T; L_{\#}^{\infty}(\Omega_0))} \|\partial_t z\|_{L^2(0,T; L_{\#}^2(\Omega_0))} \\ & + C T^{\frac{1}{4}} \|B_{\zeta, \eta}\|_{L^{\infty}(0,T; L_{\#}^{\infty}(\Omega_0))} \|z\|_{L^{\infty}(0,T; L_{\#}^2(\Omega_0))}^{\frac{3}{2}} \|z\|_{L^2(0,T; H_{\#}^2(\Omega_0))}^{\frac{1}{2}} \\ & + T^{\frac{1}{2}} \|\partial_t \chi_{\zeta, \eta}\|_{L^{\infty}(0,T; L_{\#}^{\infty}(\Omega_0))} \|B_{\zeta, \eta}\|_{L^{\infty}(0,T; L_{\#}^{\infty}(\Omega_0))} \|z\|_{L^{\infty}(0,T; H_{\#}^1(\Omega_0))} \end{aligned} \right). \end{aligned} $$
Thus by using (4.7) for $s = \frac{1}{4}$ and the fact that $(z, r, \bar{\zeta}, \bar{\eta}) \in K_T(M)$, we obtain for $T < 1$

$$ \|f[z, \zeta, \eta]\|_{L^2(0,T;L^2_{\bar{\eta}}(\Omega_0))} \leq C(R, M)T^{\frac{1}{4}}. $$

By Piola's identity, we have $g[z, \zeta, \eta] = (B_{\zeta_0, \eta_0}^T - B_{\zeta, \eta}^T) : \nabla z$. Hence,

$$ \|g[z, \zeta, \eta]\|_{L^2(0,T;H^1_{\bar{\eta}}(\Omega_0))} \leq \|B_{\zeta_0, \eta_0} - B_{\zeta, \eta}\|_{L^{\infty}(0,T;L^\infty_{\bar{\eta}}(\Omega_0))} \left( \|\nabla z\|_{L^2(0,T;L^2_{\bar{\eta}}(\Omega_0))} + \|\nabla^2 z\|_{L^2(0,T;L^2_{\bar{\eta}}(\Omega_0))} \right) \\ + \|\nabla(B_{\zeta_0, \eta_0}^T - B_{\zeta, \eta}^T) : \nabla z\|_{L^2(0,T;L^2_{\bar{\eta}}(\Omega_0))}. $$

Recalling Lemma 2.1, we have

$$ \|\nabla(B_{\zeta_0, \eta_0}^T - B_{\zeta, \eta}^T) : \nabla z\|_{L^2(0,T;L^2_{\bar{\eta}}(\Omega_0))} \leq \|B_{\zeta_0, \eta_0} - B_{\zeta, \eta}\|_{L^{\infty}(0,T;H^{1+s}_{\bar{\eta}}(\Omega_0))} \|z\|_{L^2(0,T;H^2_{\bar{\eta}}(\Omega_0))}, $$

we deduce, since $H^1_{\bar{\eta}}(\Omega_0)$ is continuously embedded in $L^\infty_{\bar{\eta}}(\Omega_0)$

$$ \|g[z, \zeta, \eta]\|_{L^2(0,T;H^1_{\bar{\eta}}(\Omega_0))} \leq C \|B_{\zeta_0, \eta_0} - B_{\zeta, \eta}\|_{L^{\infty}(0,T;H^{1+s}_{\bar{\eta}}(\Omega_0))} \|z\|_{L^2(0,T;H^2_{\bar{\eta}}(\Omega_0))}. $$

By Lemma 4.1 and since $\| (\bar{\zeta}, \bar{\eta}) \|_{X_{s,T}} \leq M$, we obtain

$$ \|B_{\zeta, \eta} - B_{\zeta_0, \eta_0}\|_{L^{\infty}(0,T;H^{1+s}_{\bar{\eta}}(\Omega_0))} \leq C T^{\frac{1}{2}-s} M, \quad (4.8) $$

Hence,

$$ \|g[z, \zeta, \eta]\|_{L^2(0,T;H^1_{\bar{\eta}}(\Omega_0))} \leq C T^{\frac{1}{2}-s} M^2, $$

with $0 < s < \frac{1}{2}$. By taking $s = \frac{1}{4}$ we obtain the desired inequality. Next we have to bound $\partial_t g$. Let $w \in H^1_{\bar{\eta}}(\Omega_0)$ and $\varphi \in C_c^\infty(0,T)$. We are going to estimate $\int_0^T \int_{\Omega_0} \partial_t g w \varphi$. By integration by parts in space and recalling that $z|_{y=0} = 0$ and $z|_{y=1} = \partial_t \zeta e_1 + \partial_t \eta e_2$, we obtain

$$ \begin{aligned} \int_0^T \int_{\Omega_0} g w \partial_t \varphi &= \int_0^T \int_{\Omega_0} \operatorname{div}((B_{\zeta_0, \eta_0}^T - B_{\zeta, \eta}^T)z)w \partial_t \varphi \\ &= -\int_0^T \int_{\Omega_0} ((B_{\zeta_0, \eta_0}^T - B_{\zeta, \eta}^T)z) \cdot \nabla w \partial_t \varphi + \int_0^T \int_0^L ((B_{\zeta_0, \eta_0}^T - B_{\zeta, \eta}^T)z)|_{y=1} \cdot e_2 w \partial_t \varphi \\ &= -\int_0^T \int_{\Omega_0} ((B_{\zeta_0, \eta_0}^T - B_{\zeta, \eta}^T)z) \cdot \nabla w \partial_t \varphi + \int_0^T \int_0^L (\partial_x \bar{\eta} \partial_t \zeta - \partial_x \bar{\zeta} \partial_t \eta) w|_{y=1} \partial_t \varphi. \end{aligned} $$

Note that the second term of the right hand side is equal to zero in the only transversal case. Hence, integrating by parts with respect to the time, we get for all $\varphi \in C_c^\infty(0,T)$

$$ \int_0^T \int_{\Omega_0} g w \partial_t \varphi = \int_0^T \int_{\Omega_0} (-\partial_t B_{\zeta, \eta}^T z + (B_{\zeta_0, \eta_0}^T - B_{\zeta, \eta}^T) \partial_t z) \cdot \nabla w \varphi - \int_0^T \int_0^L \partial_t (\partial_x \bar{\eta} \partial_t \zeta - \partial_x \bar{\zeta} \partial_t \eta) w|_{y=1} \varphi. $$

Consequently, we have

$$ \begin{aligned} \langle \partial_t g, w \rangle_{(H^1_{\bar{\eta}}(\Omega_0))', H^1_{\bar{\eta}}(\Omega_0)} &= \int_{\Omega_0} (-\partial_t B_{\zeta, \eta}^T z + (B_{\zeta_0, \eta_0}^T - B_{\zeta, \eta}^T) \partial_t z) \cdot \nabla w \\ &\quad - \int_0^L (\partial_{xt} \bar{\eta} \partial_t \zeta + \partial_x \bar{\eta} \partial_{tt} \zeta - \partial_{xt} \bar{\zeta} \partial_t \eta - \partial_x \bar{\zeta} \partial_{tt} \eta) w|_{y=1}, \quad \forall w \in H^1_{\bar{\eta}}(\Omega_0), \text{ a.e. } t \in (0, T). \end{aligned} \quad (4.9) $$
Thus by using (4.7) for $s = \frac{1}{4}$ and the fact that $(z, r, \bar{\zeta}, \bar{\eta}) \in K_T(M)$, we obtain for $T < 1$

$$ \|f[z, \zeta, \eta]\|_{L^2(0,T; L^2_{\bar{\eta}}(\Omega_0))} \leq C(R, M)T^{\frac{1}{4}}. $$

By Piola's identity, we have $g[z, \zeta, \eta] = (B_{\zeta_0, \eta_0}^T - B_{\zeta, \eta}^T) : \nabla z$. Hence,

$$ \|g[z, \zeta, \eta]\|_{L^2(0,T; H^1_{\bar{\eta}}(\Omega_0))} \leq \|B_{\zeta_0, \eta_0} - B_{\zeta, \eta}\|_{L^{\infty}(0,T; L^\infty_{\bar{\eta}}(\Omega_0))} \left( \|\nabla z\|_{L^2(0,T; L^2_{\bar{\eta}}(\Omega_0))} + \|\nabla^2 z\|_{L^2(0,T; L^2_{\bar{\eta}}(\Omega_0))} \right) \\ + \|\nabla(B_{\zeta_0, \eta_0}^T - B_{\zeta, \eta}^T) : \nabla z\|_{L^2(0,T; L^2_{\bar{\eta}}(\Omega_0))}. $$

Recalling Lemma 2.1, we have

$$ \|\nabla(B_{\zeta_0, \eta_0}^T - B_{\zeta, \eta}^T) : \nabla z\|_{L^2(0,T; L^2_{\bar{\eta}}(\Omega_0))} \leq \|B_{\zeta_0, \eta_0} - B_{\zeta, \eta}\|_{L^{\infty}(0,T; H^{1+s}_{\bar{\eta}}(\Omega_0))} \|z\|_{L^2(0,T; H^2_{\bar{\eta}}(\Omega_0))}, $$

we deduce, since $H^1_{\bar{\eta}}(\Omega_0)$ is continuously embedded in $L^\infty_{\bar{\eta}}(\Omega_0)$

$$ \|g[z, \zeta, \eta]\|_{L^2(0,T; H^1_{\bar{\eta}}(\Omega_0))} \leq C \|B_{\zeta_0, \eta_0} - B_{\zeta, \eta}\|_{L^{\infty}(0,T; H^{1+s}_{\bar{\eta}}(\Omega_0))} \|z\|_{L^2(0,T; H^2_{\bar{\eta}}(\Omega_0))}. $$

By Lemma 4.1 and since $\| (\bar{\zeta}, \bar{\eta}) \|_{X_{s,T}} \leq M$, we obtain

$$ \|B_{\zeta, \eta} - B_{\zeta_0, \eta_0}\|_{L^{\infty}(0,T; H^{1+s}_{\bar{\eta}}(\Omega_0))} \leq C T^{\frac{1}{2}-s} M, \quad (4.8) $$

Hence,

$$ \|g[z, \zeta, \eta]\|_{L^2(0,T; H^1_{\bar{\eta}}(\Omega_0))} \leq C T^{\frac{1}{2}-s} M^2, $$

with $0 < s < \frac{1}{2}$. By taking $s = \frac{1}{4}$ we obtain the desired inequality. Next we have to bound $\partial_t g$. Let $w \in H^1_{\bar{\eta}}(\Omega_0)$ and $\varphi \in C_c^\infty(0,T)$. We are going to estimate $\int_0^T \int_{\Omega_0} \partial_t g w \varphi$. By integration by parts in space and recalling that $z|_{y=0} = 0$ and $z|_{y=1} = \partial_t \zeta e_1 + \partial_t \eta e_2$, we obtain

$$ \begin{aligned} \int_0^T \int_{\Omega_0} g w \partial_t \varphi &= \int_0^T \int_{\Omega_0} \operatorname{div}((B_{\zeta_0, \eta_0}^T - B_{\zeta, \eta}^T)z)w \partial_t \varphi \\ &= -\int_0^T \int_{\Omega_0} ((B_{\zeta_0, \eta_0}^T - B_{\zeta, \eta}^T)z) \cdot \nabla w \partial_t \varphi + \int_0^T \int_0^L ((B_{\zeta_0, \eta_0}^T - B_{\zeta, \eta}^T)z)|_{y=1} \cdot e_2 w \partial_t \varphi \\ &= -\int_0^T \int_{\Omega_0} ((B_{\zeta_0, \eta_0}^T - B_{\zeta, \eta}^T)z) \cdot \nabla w \partial_t \varphi + \int_0^T \int_0^L (\partial_x \bar{\eta} \partial_t \zeta - \partial_x \bar{\zeta} \partial_t \eta) w|_{y=1} \partial_t \varphi. \end{aligned} $$

Note that the second term of the right hand side is equal to zero in the only transversal case. Hence, integrating by parts with respect to the time, we get for all $\varphi \in C_c^\infty(0,T)$

$$ \int_0^T \int_{\Omega_0} g w \partial_t \varphi = \int_0^T \int_{\Omega_0} (-\partial_t B_{\zeta, \eta}^T z + (B_{\zeta_0, \eta_0}^T - B_{\zeta, \eta}^T) \partial_t z) \cdot \nabla w \varphi - \int_0^T \int_0^L \partial_t (\partial_x \bar{\eta} \partial_t \zeta - \partial_x \bar{\zeta} \partial_t \eta) w|_{y=1} \varphi. $$

Consequently, we have

$$ \begin{aligned} \langle \partial_t g, w \rangle_{(H^1_{\bar{\eta}}(\Omega_0))', H^1_{\bar{\eta}}(\Omega_0)} &= \int_{\Omega_0} (-\partial_t B_{\zeta, \eta}^T z + (B_{\zeta_0, \eta_0}^T - B_{\zeta, \eta}^T) \partial_t z) \cdot \nabla w \\ &\quad - \int_0^L (\partial_{xt} \bar{\eta} \partial_t \zeta + \partial_x \bar{\eta} \partial_{tt} \zeta - \partial_{xt} \bar{\zeta} \partial_t \eta - \partial_x \bar{\zeta} \partial_{tt} \eta) w|_{y=1}, \quad \forall w \in H^1_{\bar{\eta}}(\Omega_0), \text{ a.e. } t \in (0, T). \end{aligned} \quad (4.9) $$
By using Holder's inequality and Sobolev injection of $H_{\#}^1(\Omega_0)$ in $L_{\#}^4(\Omega_0)$, the volumic term can be bounded as follows

$$ \left| \int_{\Omega_0} \left( -\partial_t B_{\zeta, \eta}^T z + (B_{\zeta, \eta}^T - B_{\zeta, \eta}^T) \partial_t z \right) \cdot \nabla w \right| \\ \le \left( C \|\partial_t B_{\zeta, \eta}\|_{L_{\#}^4(\Omega_0)} \|z\|_{H_{\#}^1(\Omega_0)} + \|B_{\zeta, \eta} - B_{\zeta, \eta}\|_{L_{\#}^{\infty}(\Omega_0)} \|\partial_t z\|_{L_{\#}^2(\Omega_0)} \right) \|\nabla w\|_{L_{\#}^2(\Omega_0)} \quad (4.10) $$

By using again Holder's inequality and trace inequality, there exists $C > 0$ depending on $\Omega_0$ such that the boundary term can be estimated as

$$ \left| \int_0^L \left( \partial_{xt} \bar{\eta} \partial_t \zeta + \partial_x \bar{\eta} \partial_{tt} \zeta - \partial_{xt} \bar{\zeta} \partial_t \eta - \partial_x \bar{\zeta} \partial_{tt} \eta \right) w \Big|_{y=1} \right| \le C \left( \begin{aligned} & \|\partial_{tx} \bar{\eta}\|_{L_{\#}^2(0,L)} \|\partial_t \zeta\|_{L_{\#}^{\infty}(0,L)} + \|\partial_x \bar{\eta}\|_{L_{\#}^{\infty}(0,L)} \|\partial_{tt} \zeta\|_{L_{\#}^2(0,L)} \\ & + \|\partial_{xt} \bar{\zeta}\|_{L_{\#}^2(0,L)} \|\partial_t \eta\|_{L_{\#}^{\infty}(0,L)} + \|\partial_x \bar{\zeta}\|_{L_{\#}^{\infty}(0,L)} \|\partial_{tt} \eta\|_{L_{\#}^2(0,L)} \end{aligned} \right) \|w\|_{H_{\#}^1(\Omega_0)} $$

Sobolev injection of $H_{\#}^{1+s}(0, L)$ in $C([0, L])$ for any $s > 0$ leads to

$$ \left| \int_0^L \left( \partial_{xt} \bar{\eta} \partial_t \zeta + \partial_x \bar{\eta} \partial_{tt} \zeta - \partial_{xt} \bar{\zeta} \partial_t \eta - \partial_x \bar{\zeta} \partial_{tt} \eta \right) w \right| \le C \left( \begin{aligned} & \|\partial_t \bar{\eta}\|_{H_{\#}^1(0,L)} \|\partial_t \zeta\|_{H_{\#}^1(0,L)} + \|\partial_x \bar{\eta}\|_{H_{\#}^{1+s}} \|\partial_{tt} \zeta\|_{L_{\#}^2(0,L)} \\ & + \|\partial_t \bar{\zeta}\|_{H_{\#}^1(0,L)} \|\partial_t \eta\|_{H_{\#}^1(0,L)} + \|\partial_x \bar{\zeta}\|_{H_{\#}^{1+s}} \|\partial_{tt} \eta\|_{L_{\#}^2(0,L)} \end{aligned} \right) \|w\|_{H_{\#}^1(\Omega_0)} $$

and thus

$$ \left| \int_0^L \left( \partial_{xt} \bar{\eta} \partial_t \zeta + \partial_x \bar{\eta} \partial_{tt} \zeta - \partial_{xt} \bar{\zeta} \partial_t \eta - \partial_x \bar{\zeta} \partial_{tt} \eta \right) w \right| \le C \left( \begin{aligned} & \|\partial_t \bar{\eta}\|_{H_{\#}^1(0,L)} \|\partial_t \bar{\zeta}\|_{H_{\#}^1(0,L)} + \|\bar{\eta}\|_{H_{\#}^{3+s}} \|\partial_{tt} \bar{\zeta}\|_{L_{\#}^2(0,L)} \\ & + \|\bar{\zeta}\|_{H_{\#}^{3+s}} \|\partial_{tt} \bar{\eta}\|_{L_{\#}^2(0,L)} \end{aligned} \right) \|w\|_{H_{\#}^1(\Omega_0)}. \quad (4.11) $$

By gathering together (4.10) and (4.11) with (4.9), we obtain

$$ \|\partial_t g\|_{(H_{\#}^1(\Omega_0))'} \le C \left( \begin{aligned} & \|\partial_t B_{\zeta, \eta}\|_{L_{\#}^4(\Omega_0)} \|z\|_{H_{\#}^1(\Omega_0)} + \|B_{\zeta, \eta} - B_{\zeta_0, \eta_0}\|_{L_{\#}^{\infty}(\Omega_0)} \|\partial_t z\|_{L_{\#}^2(\Omega_0)} \\ & + \|\partial_t \bar{\eta}\|_{H_{\#}^1(0,L)} \|\partial_t \bar{\zeta}\|_{H_{\#}^1(0,L)} \\ & + \|\bar{\eta}\|_{H_{\#}^{3+s}} \|\partial_{tt} \bar{\zeta}\|_{L_{\#}^2(0,L)} + \|\bar{\zeta}\|_{H_{\#}^{3+s}} \|\partial_{tt} \bar{\eta}\|_{L_{\#}^2(0,L)} \end{aligned} \right) $$

It follows that

$$ \|\partial_t g\|_{L^2(0,T;(H_{\#}^1(\Omega_0)))'} \le C(T^{\frac{1}{2}} \|\partial_t B_{\zeta, \eta}\|_{L^{\infty}(0,T;L_{\#}^2(\Omega_0))} \|z\|_{L^{\infty}(0,T;H_{\#}^1(\Omega_0))} + \|B_{\zeta, \eta} - B_{\zeta_0, \eta_0}\|_{L^{\infty}(0,T;L_{\#}^{\infty}(\Omega_0))} \|\partial_t z\|_{L^2(0,T;L_{\#}^2(\Omega_0))} \\ + T^{\frac{1}{2}} \|\partial_t \bar{\eta}\|_{L^{\infty}(0,T;H_{\#}^1(0,L))} \|\partial_t \bar{\zeta}\|_{L^{\infty}(0,T;H_{\#}^1(0,L))} + \|\bar{\eta}\|_{L^{\infty}(0,T;H_{\#}^{3+s}} \|\partial_{tt} \bar{\zeta}\|_{L^2(0,T;L_{\#}^2(0,L))} \\ + \|\bar{\zeta}\|_{L^{\infty}(0,T;H_{\#}^{3+s}} \|\partial_{tt} \bar{\eta}\|_{L^2(0,T;L_{\#}^2(0,L))}) \quad (4.12) $$

Since $(z, r, \bar{\zeta}, \bar{\eta}) \in \mathcal{K}_T(M)$, it follows by remembering (4.7), (4.8) and using Lemma 4.1 that for $s < \frac{1}{2}$

$$ \|\partial_t g\|_{L^2(0,T;(H_{\#}^1(\Omega_0)))'} \le C(R) M^2 T^{\frac{1}{2}-s}. $$

By taking $s = \frac{1}{4}$ we obtain the desired inequality. The source term $h[z, r, \zeta, \eta]$ in (4.3) can be easily estimated using Lemma 4.1. Let us for instance consider $h^a = \mu((A_{\zeta, \eta} - A_{\zeta_0, \eta_0})\nabla)z$. There exists a time $T(R, M)$ such that this
By using Holder's inequality and Sobolev injection of $H_{\#}^1(\Omega_0)$ in $L_{\#}^4(\Omega_0)$, the volumic term can be bounded as follows

$$ \left| \int_{\Omega_0} \left( -\partial_t B_{\zeta, \eta}^T z + (B_{\zeta, \eta}^T - B_{\zeta, \eta}^T) \partial_t z \right) \cdot \nabla w \right| \\ \le \left( C \|\partial_t B_{\zeta, \eta}\|_{L_{\#}^4(\Omega_0)} \|z\|_{H_{\#}^1(\Omega_0)} + \|B_{\zeta, \eta} - B_{\zeta, \eta}\|_{L_{\#}^{\infty}(\Omega_0)} \|\partial_t z\|_{L_{\#}^2(\Omega_0)} \right) \|\nabla w\|_{L_{\#}^2(\Omega_0)} \quad (4.10) $$

By using again Holder's inequality and trace inequality, there exists $C > 0$ depending on $\Omega_0$ such that the boundary term can be estimated as

$$ \left| \int_0^L \left( \partial_{xt} \bar{\eta} \partial_t \zeta + \partial_x \bar{\eta} \partial_{tt} \zeta - \partial_{xt} \bar{\zeta} \partial_t \eta - \partial_x \bar{\zeta} \partial_{tt} \eta \right) w \Big|_{y=1} \right| \le C \left( \begin{aligned} & \|\partial_{tx} \bar{\eta}\|_{L_{\#}^2(0,L)} \|\partial_t \zeta\|_{L_{\#}^{\infty}(0,L)} + \|\partial_x \bar{\eta}\|_{L_{\#}^{\infty}(0,L)} \|\partial_{tt} \zeta\|_{L_{\#}^2(0,L)} \\ & + \|\partial_{xt} \bar{\zeta}\|_{L_{\#}^2(0,L)} \|\partial_t \eta\|_{L_{\#}^{\infty}(0,L)} + \|\partial_x \bar{\zeta}\|_{L_{\#}^{\infty}(0,L)} \|\partial_{tt} \eta\|_{L_{\#}^2(0,L)} \end{aligned} \right) \|w\|_{H_{\#}^1(\Omega_0)} $$

Sobolev injection of $H_{\#}^{1+s}(0, L)$ in $C([0, L])$ for any $s > 0$ leads to

$$ \left| \int_0^L \left( \partial_{xt} \bar{\eta} \partial_t \zeta + \partial_x \bar{\eta} \partial_{tt} \zeta - \partial_{xt} \bar{\zeta} \partial_t \eta - \partial_x \bar{\zeta} \partial_{tt} \eta \right) w \right| \le C \left( \begin{aligned} & \|\partial_t \bar{\eta}\|_{H_{\#}^1(0,L)} \|\partial_t \zeta\|_{H_{\#}^1(0,L)} + \|\partial_x \bar{\eta}\|_{H_{\#}^{1+s}(0,L)} \|\partial_{tt} \zeta\|_{L_{\#}^2(0,L)} \\ & + \|\partial_t \bar{\zeta}\|_{H_{\#}^1(0,L)} \|\partial_t \eta\|_{H_{\#}^1(0,L)} + \|\partial_x \bar{\zeta}\|_{H_{\#}^{1+s}(0,L)} \|\partial_{tt} \eta\|_{L_{\#}^2(0,L)} \end{aligned} \right) \|w\|_{H_{\#}^1(\Omega_0)} $$

and thus

$$ \left| \int_0^L \left( \partial_{xt} \bar{\eta} \partial_t \zeta + \partial_x \bar{\eta} \partial_{tt} \zeta - \partial_{xt} \bar{\zeta} \partial_t \eta - \partial_x \bar{\zeta} \partial_{tt} \eta \right) w \right| \le C \left( \begin{aligned} & \|\partial_t \bar{\eta}\|_{H_{\#}^1(0,L)} \|\partial_t \bar{\zeta}\|_{H_{\#}^1(0,L)} + \|\bar{\eta}\|_{H_{\#}^{3+s}(0,L)} \|\partial_{tt} \bar{\zeta}\|_{L_{\#}^2(0,L)} \\ & + \|\bar{\zeta}\|_{H_{\#}^{3+s}(0,L)} \|\partial_{tt} \bar{\eta}\|_{L_{\#}^2(0,L)} \end{aligned} \right) \|w\|_{H_{\#}^1(\Omega_0)}. \quad (4.11) $$

By gathering together (4.10) and (4.11) with (4.9), we obtain

$$ \|\partial_t g\|_{(H_{\#}^1(\Omega_0))'} \le C \left( \begin{aligned} & \|\partial_t B_{\zeta, \eta}\|_{L_{\#}^4(\Omega_0)} \|z\|_{H_{\#}^1(\Omega_0)} + \|B_{\zeta, \eta} - B_{\zeta_0, \eta_0}\|_{L_{\#}^{\infty}(\Omega_0)} \|\partial_t z\|_{L_{\#}^2(\Omega_0)} \\ & + \|\partial_t \bar{\eta}\|_{H_{\#}^1(0,L)} \|\partial_t \bar{\zeta}\|_{H_{\#}^1(0,L)} \\ & + \|\bar{\eta}\|_{H_{\#}^{3+s}(0,L)} \|\partial_{tt} \bar{\zeta}\|_{L_{\#}^2(0,L)} + \|\bar{\zeta}\|_{H_{\#}^{3+s}(0,L)} \|\partial_{tt} \bar{\eta}\|_{L_{\#}^2(0,L)} \end{aligned} \right) $$

It follows that

$$ \|\partial_t g\|_{L^2(0,T;(H_{\#}^1(\Omega_0))')} \le C(T^{\frac{1}{2}} \|\partial_t B_{\zeta, \eta}\|_{L^{\infty}(0,T;H_{\#}^1(\Omega_0))} \|z\|_{L^{\infty}(0,T;H_{\#}^1(\Omega_0))} + \|B_{\zeta, \eta} - B_{\zeta_0, \eta_0}\|_{L^{\infty}(0,T;L_{\#}^{\infty}(\Omega_0))} \|\partial_t z\|_{L^2(0,T;L_{\#}^2(\Omega_0))} \\ + T^{\frac{1}{2}} \|\partial_t \bar{\eta}\|_{L^{\infty}(0,T;H_{\#}^1(0,L))} \|\partial_t \bar{\zeta}\|_{L^{\infty}(0,T;H_{\#}^1(0,L))} + \|\bar{\eta}\|_{L^{\infty}(0,T;H_{\#}^{3+s}(0,L))} \|\partial_{tt} \bar{\zeta}\|_{L^2(0,T;L_{\#}^2(0,L))} \\ + \|\bar{\zeta}\|_{L^{\infty}(0,T;H_{\#}^{3+s}(0,L))} \|\partial_{tt} \bar{\eta}\|_{L^2(0,T;L_{\#}^2(0,L))}) \quad (4.12) $$

Since $(z, r, \bar{\zeta}, \bar{\eta}) \in \mathcal{K}_T(M)$, it follows by remembering (4.7), (4.8) and using Lemma 4.1 that for $s < \frac{1}{2}$

$$ \|\partial_t g\|_{L^2(0,T;(H_{\#}^1(\Omega_0))')} \le C(R) M^2 T^{\frac{1}{2}-s}. $$

By taking $s = \frac{1}{4}$ we obtain the desired inequality. The source term $h[z, r, \zeta, \eta]$ in (4.3) can be easily estimated using Lemma 4.1. Let us for instance consider $h^a = \mu((A_{\zeta, \eta} - A_{\zeta_0, \eta_0})\nabla)z$. There exists a time $T(R, M)$ such that this
term is well defined and belong to $L^2(0, T; H_\#^1(\Omega_0))$ for $T \le T(R, M)$. By using Holder's inequality, we get that there exists $C > 0$ such that

$$ \|h^a\|_{L^2(0,T;H_\#^1(\Omega_0))} \le C \left( \|A_{\zeta,\eta} - A_{\zeta_0,\eta_0}\|_{L^\infty(0,T;L_\#^\infty(\Omega_0))} \|z\|_{L^2(0,T;H_\#^2(\Omega_0))} + \|(\nabla(A_{\zeta,\eta} - A_{\zeta_0,\eta_0})\nabla)z\|_{L^2(0,T;L_\#^2(\Omega_0))} \right), $$

Since $\|(z, r)\|_{X_{f,T}} \le M$ and by using Lemma 2.1, we obtain, for $s > 0$,

$$ \|h^a\|_{L^2(0,T;H_\#^1(\Omega_0))} \le CM \left( \|A_{\zeta,\eta} - A_{\zeta_0,\eta_0}\|_{L^\infty(0,T;L_\#^\infty(\Omega_0))} + \|A_{\zeta,\eta} - A_{\zeta_0,\eta_0}\|_{L^\infty(0,T;H_\#^{1+s}(\Omega_0))} \right), $$

Thus by using the Sobolev injection of $H_\#^{1+s}(\Omega_0) \subset L_\#^\infty(\Omega_0)$ and Lemma 4.1, we get, for $0 < s < 1/2$

$$ \|h^a\|_{L^2(0,T;H_\#^1(\Omega_0))} \le CM^2 T^{\frac{1}{2}-s}. \quad (4.13) $$

The two other terms can be estimated in the same way. By combining (4.6) with the above lemma, we deduce the following corollary.

**Corollary 4.1** For $T$ small enough, $\mathcal{L}_T^0$ maps $K_T(M)$ into itself for $M \ge M_0$.

We turn now to show that for $T$ small enough and $M \ge M_0$, the mapping $\mathcal{L}_T^0: K_T(M) \to K_T(M)$ is a contraction. Let $(z^{(1)}, r^{(1)}, \bar{\zeta}^{(1)}, \bar{\eta}^{(1)})$ and $(z^{(2)}, r^{(2)}, \bar{\zeta}^{(2)}, \bar{\eta}^{(2)})$ in $K_T(M)$. For $i = 1, 2$, denote by

$$ (v^{(i)}, q^{(i)}, \bar{\chi}^{(i)}, \bar{\mathcal{Y}}^{(i)}) = \mathcal{L}_T^0(z^{(i)}, r^{(i)}, \bar{\zeta}^{(i)}, \bar{\eta}^{(i)}). $$

We recall that $(\bar{\zeta}^{(i)}, \bar{\eta}^{(i)}, \chi^{(i)}, \mathcal{Y}^{(i)}) = (\zeta^{(i)} - \zeta_0, \eta^{(i)} - \eta_0, \chi^{(i)} - \zeta_0, \mathcal{Y}^{(i)} - \eta_0)$ with $(v^{(i)}, q^{(i)}, \chi^{(i)}, \mathcal{Y}^{(i)})$ is a solution of Problem (3.1)-(3.6) with $(a, b) = (\zeta_0, \eta_0)$ and source term $(f^{(i)}, h^{(i)}g^{(i)}, 0, 0)$ defined as follows:

$$ f^{(i)} = f[z^{(i)}, \zeta^{(i)}, \eta^{(i)}], \quad h^{(i)} = h[z^{(i)}, r^{(i)}, \zeta^{(i)}, \eta^{(i)}], \quad g^{(i)} = g[z^{(i)}, \zeta^{(i)}, \eta^{(i)}]. \quad (4.14) $$

Then

$$ \mathcal{L}_T^0(z^{(1)}, r^{(1)}, \bar{\zeta}^{(1)}, \bar{\eta}^{(1)}) - \mathcal{L}_T^0(z^{(2)}, r^{(2)}, \bar{\zeta}^{(2)}, \bar{\eta}^{(2)}) = (v^{(1)}, q^{(1)}, \zeta^{(1)}, \eta^{(1)}) - (v^{(2)}, q^{(2)}, \zeta^{(2)}, \eta^{(2)}) \quad (4.15) $$

Denote by $(v, q, \zeta, \eta) = (v^{(1)}, q^{(1)}, \zeta^{(1)}, \eta^{(1)}) - (v^{(2)}, q^{(2)}, \zeta^{(2)}, \eta^{(2)})$. Then $(v, q, \zeta, \eta)$ is a solution of Problem (3.1) with $(a, b) = (\zeta_0, \eta_0)$, zero initial data and source term $(f, h, g, 0, 0) = (f^{(1)} - f^{(2)}, h^{(1)} - h^{(2)}, g^{(1)} - g^{(2)}, 0, 0)$. Hence if the source term $(f, h, g, 0, 0) \in S_T$, then by using Proposition 3.2 for $(a, b) = (\zeta_0, \eta_0)$ we get

$$ \|(v, q)\|_{X_{f,T}} + \|(\zeta, \eta)\|_{X_{s,T}} \le C(R) \|(f, h, g, 0, 0)\|_{S_T}, $$

and thus in view of (4.15) we get

$$ \|\mathcal{L}_T^0(z^{(1)}, r^{(1)}, \bar{\zeta}^{(1)}, \bar{\eta}^{(1)}) - \mathcal{L}_T^0(z^{(2)}, r^{(2)}, \bar{\zeta}^{(2)}, \bar{\eta}^{(2)})\|_{X_{f,T} \times X_{s,T}} \le C(R) \|(f^{(1)} - f^{(2)}, h^{(1)} - h^{(2)}, g^{(1)} - g^{(2)}, 0, 0)\|_{S_T}. \quad (4.16) $$

To proceed we have to bound the source term $(f^{(1)} - f^{(2)}, h^{(1)} - h^{(2)}, g^{(1)} - g^{(2)}, 0, 0)$ defined by (4.14) respectively in $S_T$ in the right hand side of (4.16). In order to establish such bounds, we gather in the following lemma some properties of the matrix $B_{\zeta,\eta}$ and $A_{\zeta,\eta}$ defined respectively by (2.11), (2.12), that can be derived in the very same way as in Lemma 4.1.