and the result will follow from the classical Korn inequality since $\mathcal{F}$ is Lipschitz domain (see for instance [47]).
Suppose that the inequality (A.1) is not true, then for any positive integer $k$ there exists $u_k$ such that

$$ \|\mathbb{D}(\hat{u}_k)\|_{[L^2(\mathcal{F})]^9}^2 + \|\beta(\hat{u}_k)_\tau\|_{[L^2(\partial\mathcal{F})]^3}^2 \le \frac{1}{k} \|\hat{u}_k\|_{[L^2(\mathcal{F})]^3}^2. $$

We normalize by $\|\hat{u}_k\|_{[L^2(\mathcal{F})]^3}$, in such a way we get a sequence denoted again $(\hat{u}_k)_k$ that verifies

$$ \|\hat{u}_k\|_{[L^2(\mathcal{F})]^3} = 1. $$

We deduce from the classical Korn inequality that $(\hat{u}_k)_k$ is bounded in $[H^1(\mathcal{F})]^3$, then up to a subsequence, it converges weakly to $\hat{u}$ in $[H^1(\mathcal{F})]^3$. Consequently

$$ \|\mathbb{D}(\hat{u})\|_{[L^2(\mathcal{F})]^9}^2 + \|\beta\hat{u}_\tau\|_{[L^2(\partial\mathcal{F})]^3}^2 \le \liminf_{k \to \infty} \|\mathbb{D}(\hat{u}_k)\|_{[L^2(\mathcal{F})]^9}^2 + \|\beta(\hat{u}_k)_\tau\|_{[L^2(\partial\mathcal{F})]^3}^2 = 0. $$

1. Then, $\mathbb{D}(\hat{u}) = 0$ in $\mathcal{F}$. It implies that $\hat{u}(z) = a + b \times z$ in $\mathcal{F}$, but since $\hat{u} = 0$ on $\partial\mathcal{F}$, then $a = b = 0$, which

2. contradicts the fact that $\|\hat{u}_k\|_{[L^2(\mathcal{F})]^3} = 1$. □

3. **Lemma A.2.** Let $\mathcal{X}$ a Banach space and $\xi \in H^{3/4}(0, T; \mathcal{X})$ such that $\xi(0) = 0$, then there exists $\xi_{ext} \in$

4. $H^{3/4}(0, T^*; \mathcal{X})$, where $T^* > 2T$ and

$$ \|\xi_{ext}\|_{H^{3/4}(0,T^*;\mathcal{X})} \le C \|\xi\|_{H^{3/4}(0,T;\mathcal{X})}, \quad (A.2) $$

5. with $C$ a constant that does not depend on $T$.

*Proof.* We define $\xi_{ext}$ as follows

$$ \left\{ \begin{array}{ll} \xi_{ext}(t) = \xi(t), & t \in [0, T], \\ \xi_{ext}(t) = \xi(2T - t), & t \in [T, 2T], \\ \xi_{ext}(t) = 0, & t \in [2T, T^*], \end{array} \right. $$

We have that $\xi_{ext} \in H^{3/4}(0, T^*; \mathcal{X})$. Let us prove the estimate (A.2). Since $\xi(0) = 0$, then it suffices to show the estimate (A.2) using Slobodeckij seminorm with $s = 3/4$.

$$ [\xi_{ext}]_{s,2,(0,T^*)} = \int_0^{T^*} \int_0^{T^*} \frac{\|\xi_{ext}(t) - \xi_{ext}(t')\|_{\mathcal{X}}^2}{|t - t'|^{2s+1}} dt' dt = \int_0^{2T} \int_0^{2T} \frac{\|\xi_{ext}(t) - \xi_{ext}(t')\|_{\mathcal{X}}^2}{|t - t'|^{2s+1}} dt' dt \\ + \int_0^{2T} \int_{2T}^{T^*} \frac{\|\xi_{ext}(t) - \xi_{ext}(t')\|_{\mathcal{X}}^2}{|t - t'|^{2s+1}} dt' dt + \int_{2T}^{T^*} \int_0^{2T} \frac{\|\xi_{ext}(t) - \xi_{ext}(t')\|_{\mathcal{X}}^2}{|t - t'|^{2s+1}} dt' dt. $$

Then,

$$ [\xi_{ext}]_{s,2,(0,T^*)} = \int_0^{2T} \int_0^{2T} \frac{\|\xi_{ext}(t) - \xi_{ext}(t')\|_{\mathcal{X}}^2}{|t - t'|^{2s+1}} dt' dt + \int_0^{2T} \int_{2T}^{T^*} \frac{\|\xi_{ext}(t)\|_{\mathcal{X}}^2}{|t - t'|^{2s+1}} dt' dt \\ + \int_{2T}^{T^*} \int_0^{2T} \frac{\|\xi_{ext}(t')\|_{\mathcal{X}}^2}{|t - t'|^{2s+1}} dt' dt. \quad (A.3) $$

Let us deal with the first term in the right hand side of (A.3).

$$ \int_0^{2T} \int_0^{2T} \frac{\|\xi_{ext}(t) - \xi_{ext}(t')\|_{\mathcal{X}}^2}{|t - t'|^{2s+1}} dt' dt = \int_0^T \int_0^T \frac{\|\xi_{ext}(t) - \xi_{ext}(t')\|_{\mathcal{X}}^2}{|t - t'|^{2s+1}} dt' dt \\ + \int_0^T \int_T^{2T} \frac{\|\xi_{ext}(t) - \xi_{ext}(t')\|_{\mathcal{X}}^2}{|t - t'|^{2s+1}} dt' dt + \int_T^{2T} \int_0^T \frac{\|\xi_{ext}(t) - \xi_{ext}(t')\|_{\mathcal{X}}^2}{|t - t'|^{2s+1}} dt' dt \\ + \int_T^{2T} \int_T^{2T} \frac{\|\xi_{ext}(t) - \xi_{ext}(t')\|_{\mathcal{X}}^2}{|t - t'|^{2s+1}} dt' dt. \quad (A.4) $$
Since the second and the third terms in the right hand side of (A.4) are of the same kind then it suffices to treat one of them using a change of variables as follows

$$ 
\begin{aligned}
& \int_0^T \int_T^{2T} \frac{\|\xi_{ext}(t) - \xi_{ext}(t')\|_{\mathcal{X}}^2}{|t - t'|^{2s+1}} dt'dt = \int_0^T \int_0^T \frac{\|\xi(t) - \xi(t')\|_{\mathcal{X}}^2}{|2T - t - t'|^{2s+1}} dt'dt \\
& = \int_0^T \int_0^T \frac{|t - t'|^{2s+1}}{|2T - t - t'|^{2s+1}} \frac{\|\xi(t) - \xi(t')\|_{\mathcal{X}}^2}{|t - t'|^{2s+1}} dt'dt. \quad (A.5)
\end{aligned}
$$

Since $t, t' \in (0, T)$, then

$$ \left| \frac{t - t'}{2T - t - t'} \right| \le 1. $$

Thus,

$$ \int_0^T \int_T^{2T} \frac{\|\xi_{ext}(t) - \xi_{ext}(t')\|_{\mathcal{X}}^2}{|t - t'|^{2s+1}} dt'dt \le \int_0^T \int_0^T \frac{\|\xi(t) - \xi(t')\|_{\mathcal{X}}^2}{|t - t'|^{2s+1}} dt'dt = \lfloor \xi \rfloor_{s,2,(0,T)}^2. $$

1. We obtain that

$$ \int_0^{2T} \int_0^{2T} \frac{\|\xi_{ext}(t) - \xi_{ext}(t')\|_{\mathcal{X}}^2}{|t - t'|^{2s+1}} dt'dt \le 4\lfloor\xi\rfloor_{s,2,(0,T)}^2. \quad (A.6) $$

Now, we deal with the second term in the right hand side of (A.3), the third term can be treated similarly. Let define $c(t)$ and $d(t)$ such that

$$ c(t) = \frac{1}{t} \int_0^t (\xi(t) - \xi(t')) dt', \quad d(t) = \int_t^T \frac{c(t')}{t'} dt', \quad t \in (0, T). $$

We notice that

$$ \xi(t) = c(t) - (d(t) - d(0)). $$

Indeed,

$$ 
\begin{aligned}
\xi(t) &= \frac{1}{t} \int_0^t (\xi(t) - \xi(t')) dt' + \frac{1}{t} \int_0^t \xi(t') dt' = c(t) + \int_0^t \xi(t') \left(\frac{1}{t} - \frac{1}{t'}\right) dt' + \int_0^t \frac{\xi(t')}{t'} dt' \\
&= c(t) + \int_0^t \xi(t') \left(-\int_{t'}^t \frac{1}{\tau^2} d\tau\right) dt' + \int_0^t \frac{\xi(t')}{t'} dt' = c(t) - \int_0^t \int_0^\tau \frac{\xi(t')}{\tau^2} dt' d\tau + \int_0^t \frac{\xi(t')}{t'} dt' \\
&= c(t) + \int_0^t \frac{1}{\tau^2} \left[ \int_0^\tau \xi(\tau) - \xi(t') dt' \right] d\tau = c(t) - (d(t) - d(0)).
\end{aligned}
$$

2. We want to show that

$$ \int_0^T t^{-2s} \|\xi(t)\|_{\mathcal{X}}^2 dt \le C \lfloor\xi\rfloor_{s,2,(0,T)}^2, \quad (A.7) $$

where $C$ is a constant that does not depend on $T$. We have

$$ 
\begin{aligned}
\int_0^T t^{-2s} \|c(t)\|_{\mathcal{X}}^2 dt &\le \int_0^T \int_0^t \frac{\|\xi(t) - \xi(t')\|_{\mathcal{X}}^2}{t^{2s+1}} dt' dt' \\
&= \int_0^T \int_0^T \frac{|t - t'|^{2s+1}}{t^{2s+1}} \frac{\|\xi(t) - \xi(t')\|_{\mathcal{X}}^2}{|t - t'|^{2s+1}} dt' dt'. \quad (A.8)
\end{aligned}
$$

Since $t \in (t', T)$ and $t \in (0, T)$ then

$$ 0 \le \frac{t - t'}{t} \le 1. $$
Since the second and the third terms in the right hand side of (A.4) are of the same kind then it suffices to treat one of them using a change of variables as follows

$$ \int_0^T \int_T^{2T} \frac{\|\xi_{ext}(t) - \xi_{ext}(t')\|_{\mathcal{X}}^2}{|t-t'|^{2s+1}} dt'dt = \int_0^T \int_0^T \frac{\|\xi(t) - \xi(t')\|_{\mathcal{X}}^2}{|2T-t-t'|^{2s+1}} dt'dt \\ = \int_0^T \int_0^T \frac{|t-t'|^{2s+1}}{|2T-t-t'|^{2s+1}} \frac{\|\xi(t) - \xi(t')\|_{\mathcal{X}}^2}{|t-t'|^{2s+1}} dt'dt. \quad (A.5) $$

Since $t, t' \in (0, T)$, then

$$ \left| \frac{t-t'}{2T-t-t'} \right| \le 1. $$

Thus,

$$ \int_0^T \int_T^{2T} \frac{\|\xi_{ext}(t) - \xi_{ext}(t')\|_{\mathcal{X}}^2}{|t-t'|^{2s+1}} dt'dt \le \int_0^T \int_0^T \frac{\|\xi(t) - \xi(t')\|_{\mathcal{X}}^2}{|t-t'|^{2s+1}} dt'dt = \lfloor \xi \rfloor_{s,2,(0,T)}^2. $$

1. We obtain that

$$ \int_0^{2T} \int_0^{2T} \frac{\|\xi_{ext}(t) - \xi_{ext}(t')\|_{\mathcal{X}}^2}{|t-t'|^{2s+1}} dt'dt \le 4\lfloor\xi\rfloor_{s,2,(0,T)}^2. \quad (A.6) $$

Now, we deal with the second term in the right hand side of (A.3), the third term can be treated similarly. Let define $c(t)$ and $d(t)$ such that

$$ c(t) = \frac{1}{t} \int_{0}^{t} (\xi(t) - \xi(t')) dt', \quad d(t) = \int_{t}^{T} \frac{c(t')}{t'} dt', \quad t \in (0, T). $$

We notice that

$$ \xi(t) = c(t) - (d(t) - d(0)). $$

Indeed,

$$ \begin{aligned} \xi(t) &= \frac{1}{t} \int_0^t (\xi(t) - \xi(t')) dt' + \frac{1}{t} \int_0^t \xi(t') dt' = c(t) + \int_0^t \xi(t') \left(\frac{1}{t} - \frac{1}{t'}\right) dt' + \int_0^t \frac{\xi(t')}{t'} dt' \\ &= c(t) + \int_0^t \xi(t') \left(-\int_{t'}^t \frac{1}{\tau^2} d\tau\right) dt' + \int_0^t \frac{\xi(t')}{t'} dt' = c(t) - \int_0^t \int_0^\tau \frac{\xi(t')}{\tau^2} dt' d\tau + \int_0^t \frac{\xi(t')}{t'} dt' \\ &= c(t) + \int_0^t \frac{1}{\tau^2} \left[ \int_0^\tau \xi(\tau) - \xi(t') dt' \right] d\tau = c(t) - (d(t) - d(0)). \end{aligned} $$

2. We want to show that

$$ \int_0^T t^{-2s} \|\xi(t)\|_{\mathcal{X}}^2 dt \le C \|\xi\|_{s,2,(0,T)}^2, \quad (A.7) $$

where $C$ is a constant that does not depend on $T$. We have

$$ \begin{aligned} \int_0^T t^{-2s} \|c(t)\|_{\mathcal{X}}^2 dt &\le \int_0^T \int_0^t \frac{\|\xi(t) - \xi(t')\|_{\mathcal{X}}^2}{t^{2s+1}} dt' dt' \\ &= \int_0^T \int_0^T \frac{|t-t'|^{2s+1}}{t^{2s+1}} \frac{\|\xi(t) - \xi(t')\|_{\mathcal{X}}^2}{|t-t'|^{2s+1}} dt' dt'. \quad (A.8) \end{aligned} $$

Since $t \in (t', T)$ and $t \in (0, T)$ then

$$ 0 \le \frac{t-t'}{t} \le 1. $$
1. We obtain

$$ \int_0^T t^{-2s} \|c(t)\|_{\mathcal{X}}^2 dt \le \int_0^T \int_{t'}^T \frac{\|\xi(t) - \xi(t')\|_{\mathcal{X}}^2}{|t-t'|^{2s+1}} dt dt' \le \int_0^T \int_0^T \frac{\|\xi(t) - \xi(t')\|_{\mathcal{X}}^2}{|t-t'|^{2s+1}} dt dt' = \lfloor \xi \rfloor_{s,2,(0,T)}^2. \quad (A.9) $$

In the other hand,

$$ d(t) - d(0) = - \int_0^t \frac{c(t')}{t'} dt'. $$

2. Then, by Hardy inequality, we get

$$ \int_0^T t^{-2s} \|d(t) - d(0)\|_{\mathcal{X}}^2 dt = \int_0^T \left( t^{-(s-1/2)} \int_0^t \frac{\|c(t')\|_{\mathcal{X}}}{t'} dt' \right)^2 \frac{dt}{t} \le \frac{1}{(s-1/2)^2} \int_0^T t^{-2s} \|c(t)\|_{\mathcal{X}}^2 dt. \quad (A.10) $$

3. Finally, using (A.9), we obtain

$$ \int_0^T t^{-2s} \|d(t) - d(0)\|_{\mathcal{X}}^2 dt \le \frac{1}{(s-1/2)^2} \lfloor \xi \rfloor_{s,2,(0,T)}^2. \quad (A.11) $$

Then, by (A.11) and (A.9), we deduce (A.7). It remains to estimate the second term in the right hand side of (A.3) by using (A.7).

$$ \begin{aligned} \int_0^{2T} \int_{2T}^{T^*} \frac{\|\xi_{ext}(t)\|_{\mathcal{X}}^2}{|t-t'|^{2s+1}} dt' dt &= \int_0^{2T} \|\xi_{ext}(t)\|_{\mathcal{X}}^2 \int_{2T}^{T^*} \frac{1}{(t'-t)^{2s+1}} dt' dt \\ &= \int_0^{2T} \|\xi_{ext}(t)\|_{\mathcal{X}}^2 \left[ \frac{-1}{2s(t'-t)^{2s}} \right]_{2T}^{T^*} dt \le 2 \int_0^{2T} \frac{\|\xi_{ext}(t)\|_{\mathcal{X}}^2}{(2T-t)^{2s}} dt. \end{aligned} \quad (A.12) $$

We write

$$ \int_0^{2T} \frac{\|\xi_{ext}(t)\|_{\mathcal{X}}^2}{(2T-t)^{2s}} dt = \int_0^T \frac{\|\xi_{ext}(t)\|_{\mathcal{X}}^2}{(2T-t)^{2s}} dt + \int_T^{2T} \frac{\|\xi_{ext}(t)\|_{\mathcal{X}}^2}{(2T-t)^{2s}} dt = \int_0^T \frac{t^{2s}}{(2T-t)^{2s}} \frac{\|\xi(t)\|_{\mathcal{X}}^2}{t^{2s}} dt + \int_0^T \frac{\|\xi(t)\|_{\mathcal{X}}^2}{t^{2s}} dt. $$

4. Since $t \in (0, T)$ then $t < 2T - t$ and hence

$$ \int_0^{2T} \frac{\|\xi_{ext}(t)\|_{\mathcal{X}}^2}{(2T-t)^{2s}} dt \le 2 \int_0^T t^{-2s} \|\xi(t)\|_{\mathcal{X}}^2 dt. \quad (A.13) $$

5. From (A.7), we deduce that

$$ \int_0^{2T} \frac{\|\xi_{ext}(t)\|_{\mathcal{X}}^2}{(2T-t)^{2s}} dt \le C \lfloor \xi \rfloor_{s,2,(0,T)}^2. \quad (A.14) $$

6. Thus,

$$ \int_0^{2T} \int_{2T}^{T^*} \frac{\|\xi_{ext}(t)\|_{\mathcal{X}}^2}{|t-t'|^{2s+1}} dt' dt \le C \lfloor \xi \rfloor_{s,2,(0,T)}^2. \quad (A.15) $$

7. Gathering (A.15), (A.6), we obtain (A.2). □

8. We present also in the sequel some technical estimates concerning fractional Sobolev spaces that have been

9. established in [13].

10. **Lemma A.3.** Assume $\sigma_2 \in (1/2, 1]$ and $\sigma_1 \in [0, \sigma_2]$. Using the above result, there exists a constant independent

11. of $T$ such that for any $\xi \in H^{\sigma_2}(0, T; \mathcal{X})$ and $\xi(0) = 0$, then

$$ \|\xi\|_{H^{\sigma_1}(0,T,\mathcal{X})} \le C T^{\sigma_2 - \sigma_1} \|\xi\|_{H^{\sigma_2}(0,T,\mathcal{X})}. \quad (A.16) $$
1. We obtain

$$ \int_0^T t^{-2s} \|c(t)\|_{\mathcal{X}}^2 dt \le \int_0^T \int_{t'}^T \frac{\|\xi(t) - \xi(t')\|_{\mathcal{X}}^2}{|t-t'|^{2s+1}} dt dt' \le \int_0^T \int_0^T \frac{\|\xi(t) - \xi(t')\|_{\mathcal{X}}^2}{|t-t'|^{2s+1}} dt dt' = \lfloor \xi \rfloor_{s,2,(0,T)}^2. \quad (A.9) $$

In the other hand,

$$ d(t) - d(0) = - \int_0^t \frac{c(t')}{t'} dt'. $$

2. Then, by Hardy inequality, we get

$$ \int_0^T t^{-2s} \|d(t) - d(0)\|_{\mathcal{X}}^2 dt = \int_0^T \left( t^{-(s-1/2)} \int_0^t \frac{\|c(t')\|_{\mathcal{X}}}{t'} dt' \right)^2 \frac{dt}{t} \le \frac{1}{(s-1/2)^2} \int_0^T t^{-2s} \|c(t)\|_{\mathcal{X}}^2 dt. \quad (A.10) $$

3. Finally, using (A.9), we obtain

$$ \int_0^T t^{-2s} \|d(t) - d(0)\|_{\mathcal{X}}^2 dt \le \frac{1}{(s-1/2)^2} \lfloor \xi \rfloor_{s,2,(0,T)}^2. \quad (A.11) $$

Then, by (A.11) and (A.9), we deduce (A.7). It remains to estimate the second term in the right hand side of (A.3) by using (A.7).

$$ \begin{aligned} \int_0^{2T} \int_{2T}^{T^*} \frac{\|\xi_{ext}(t)\|_{\mathcal{X}}^2}{|t-t'|^{2s+1}} dt' dt &= \int_0^{2T} \|\xi_{ext}(t)\|_{\mathcal{X}}^2 \int_{2T}^{T^*} \frac{1}{(t'-t)^{2s+1}} dt' dt \\ &= \int_0^{2T} \|\xi_{ext}(t)\|_{\mathcal{X}}^2 \left[ \frac{-1}{2s(t'-t)^{2s}} \right]_{2T}^{T^*} dt \le 2 \int_0^{2T} \frac{\|\xi_{ext}(t)\|_{\mathcal{X}}^2}{(2T-t)^{2s}} dt. \end{aligned} \quad (A.12) $$

We write

$$ \int_0^{2T} \frac{\|\xi_{ext}(t)\|_{\mathcal{X}}^2}{(2T-t)^{2s}} dt = \int_0^T \frac{\|\xi_{ext}(t)\|_{\mathcal{X}}^2}{(2T-t)^{2s}} dt + \int_T^{2T} \frac{\|\xi_{ext}(t)\|_{\mathcal{X}}^2}{(2T-t)^{2s}} dt = \int_0^T \frac{t^{2s}}{(2T-t)^{2s}} \frac{\|\xi(t)\|_{\mathcal{X}}^2}{t^{2s}} dt + \int_0^T \frac{\|\xi(t)\|_{\mathcal{X}}^2}{t^{2s}} dt. $$

4. Since $t \in (0, T)$ then $t < 2T - t$ and hence

$$ \int_0^{2T} \frac{\|\xi_{ext}(t)\|_{\mathcal{X}}^2}{(2T-t)^{2s}} dt \le 2 \int_0^T t^{-2s} \|\xi(t)\|_{\mathcal{X}}^2 dt. \quad (A.13) $$

5. From (A.7), we deduce that

$$ \int_0^{2T} \frac{\|\xi_{ext}(t)\|_{\mathcal{X}}^2}{(2T-t)^{2s}} dt \le C \lfloor \xi \rfloor_{s,2,(0,T)}^2. \quad (A.14) $$

6. Thus,

$$ \int_0^{2T} \int_{2T}^{T^*} \frac{\|\xi_{ext}(t)\|_{\mathcal{X}}^2}{|t-t'|^{2s+1}} dt' dt \le C \lfloor \xi \rfloor_{s,2,(0,T)}^2. \quad (A.15) $$

7. Gathering (A.15), (A.6), we obtain (A.2). □

8. We present also in the sequel some technical estimates concerning fractional Sobolev spaces that have been

9. established in [13].

10. **Lemma A.3.** Assume $\sigma_2 \in (1/2, 1]$ and $\sigma_1 \in [0, \sigma_2]$. Using the above result, there exists a constant independent

11. of $T$ such that for any $\xi \in H^{\sigma_2}(0, T; \mathcal{X})$ and $\xi(0) = 0$, then

$$ \|\xi\|_{H^{\sigma_1}(0,T,\mathcal{X})} \le C T^{\sigma_2 - \sigma_1} \|\xi\|_{H^{\sigma_2}(0,T,\mathcal{X})}. \quad (A.16) $$
**Lemma A.4.** Assume that $\mathcal{X}_1, \mathcal{X}_2$ and $\mathcal{X}_3$ are Banach spaces such that

$$ \|fg\|_{\mathcal{X}_3} \le C \|f\|_{\mathcal{X}_1} \|g\|_{\mathcal{X}_2}, \quad \forall f \in \mathcal{X}_1, \forall g \in \mathcal{X}_2. $$

1. Let us assume $\sigma \in (1/2, 1]$, $s \in [0, 1/2]$, $T_0 > 0$. Then there exists a constant $C$ such that for any $T \le T_0$ we

2. have

$$ \|u_1 u_2\|_{H^s(0,T;\mathcal{X}_3)} \le C T^{\sigma-s-1/2} \|u_1\|_{H^s(0,T;\mathcal{X}_1)} \|u_2\|_{H^\sigma(0,T;\mathcal{X}_2)} + \|u_2(0)\|_{\mathcal{X}_2} \|u_1\|_{H^s(0,T;\mathcal{X}_1)}, \quad (A.17) $$

3. for all $u_1 \in H^s(0, T; \mathcal{X}_1)$ and $u_2 \in H^\sigma(0, T; \mathcal{X}_2)$.

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**Lemma A.4.** Assume that $\mathcal{X}_1, \mathcal{X}_2$ and $\mathcal{X}_3$ are Banach spaces such that

$$ \|fg\|_{\mathcal{X}_3} \le C \|f\|_{\mathcal{X}_1} \|g\|_{\mathcal{X}_2}, \quad \forall f \in \mathcal{X}_1, \forall g \in \mathcal{X}_2. $$

1. Let us assume $\sigma \in (1/2, 1]$, $s \in [0, 1/2]$, $T_0 > 0$. Then there exists a constant $C$ such that for any $T \le T_0$ we

2. have

$$ \|u_1 u_2\|_{H^s(0,T;\mathcal{X}_3)} \le C T^{\sigma-s-1/2} \|u_1\|_{H^s(0,T;\mathcal{X}_1)} \|u_2\|_{H^\sigma(0,T;\mathcal{X}_2)} + \|u_2(0)\|_{\mathcal{X}_2} \|u_1\|_{H^s(0,T;\mathcal{X}_1)}, \quad (A.17) $$

3. for all $u_1 \in H^s(0, T; \mathcal{X}_1)$ and $u_2 \in H^\sigma(0, T; \mathcal{X}_2)$.

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[42] Bernhard H. Haak, Debayan Maity, Takéo Takahashi, and Marius Tucsnak. Mathematical analysis of the motion of a rigid body in a compressible Navier-Stokes-Fourier fluid. *Math. Nachr.*, 292(9):1972–2017, 2019.

[43] Matthias Hieber and Miho Murata. The $L^p$-approach to the fluid-rigid body interaction problem for compressible fluids. *Evol. Equ. Control Theory*, 4(1):69–87, 2015.

[44] Matthieu Hillairet. Lack of collision between solid bodies in a 2D incompressible viscous flow. *Comm. Partial Differential Equations*, 32(7-9):1345–1371, 2007.

[45] Matthieu Hillairet and Takéo Takahashi. Existence of contacts for the motion of a rigid body into a viscous incompressible fluid with the Tresca boundary conditions. *Tunis. J. Math.*, 3(3):447–468, 2021.

[46] Karl-Heinz Hoffmann and Victor N. Starovoitov. On a motion of a solid body in a viscous fluid. Two-dimensional case. *Adv. Math. Sci. Appl.*, 9(2):633–648, 1999.

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2

3

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4

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Debayan Maity, Arnab Roy, and Takéo Takahashi. Existence of strong solutions for a system of interaction between a compressible viscous fluid and a wave equation. *Nonlinearity*, 34(4):2659–2687, 2021.

6

7

Debayan Maity, Arnab Roy, and Takéo Takahashi. Global Stabilization of a rigid body moving in a compressible viscous fluid. https://hal.science/hal-03750113, 2022.

8

9

Debayan Maity and Takéo Takahashi. Existence and uniqueness of strong solutions for the system of interaction between a compressible Navier-Stokes-Fourier fluid and a damped plate equation. *Nonlinear Anal., Real World Appl.*, 59:34, 2021. Id/No 103267.

10
11

12

Debayan Maity, Takéo Takahashi, and Marius Tucsnak. Analysis of a system modelling the motion of a piston in a viscous gas. *J. Math. Fluid Mech.*, 19(3):551–579, 2017.

13

14

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15

16

Boris Muha and Sunčica Čanić. Existence of a weak solution to a fluid-elastic structure interaction problem with the Navier slip boundary condition. *J. Differ. Equations*, 260(12):8550–8589, 2016.

17

18

Miho Murata. On a maximal $L_p - L_q$ approach to the compressible viscous fluid flow with slip boundary condition. *Nonlinear Anal., Theory Methods Appl., Ser. A, Theory Methods*, 106:86–109, 2014.

19

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Claude-Louis M. H. Navier. Mémoire sur les lois du mouvement des fluides. *Mémoires de l'Académie Royale des Sciences de l'Institut de France*, 6(1823):389-440, 1823.

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23

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Gabriela Planas and Franck Sueur. On the “viscous incompressible fluid + rigid body” system with Navier conditions. *Ann. Inst. H. Poincaré Anal. Non Linéaire*, 31(1):55–80, 2014.

25

26

Jean-Pierre Raymond. Feedback stabilization of a fluid-structure model. *SIAM J. Control Optim.*, 48(8):5398–5443, 2010.

27

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29

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31

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33

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36

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Yoshihiro Shibata and Senjo Shimizu. On the $L_p-L_q$ maximal regularity of the Neumann problem for the Stokes equations in a bounded domain. *J. Reine Angew. Math.*, 615:157–209, 2008.

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3
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5
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7
8

[55] Debayan Maity, Arnab Roy, and Takéo Takahashi. Global Stabilization of a rigid body moving in a compressible viscous fluid. https://hal.science/hal-03750113, 2022.

9
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[56] Debayan Maity and Takéo Takahashi. Existence and uniqueness of strong solutions for the system of interaction between a compressible Navier-Stokes-Fourier fluid and a damped plate equation. *Nonlinear Anal., Real World Appl.*, 59:34, 2021. Id/No 103267.

12
13

[57] Debayan Maity, Takéo Takahashi, and Marius Tucsnak. Analysis of a system modelling the motion of a piston in a viscous gas. *J. Math. Fluid Mech.*, 19(3):551–579, 2017.

14
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[58] Sourav Mitra. Local existence of strong solutions of a fluid-structure interaction model. *J. Math. Fluid Mech.*, 22(4):37, 2020. Id/No 60.

16
17

[59] Boris Muha and Sunčica Čanić. Existence of a weak solution to a fluid-elastic structure interaction problem with the Navier slip boundary condition. *J. Differ. Equations*, 260(12):8550–8589, 2016.

18
19

[60] Miho Murata. On a maximal $L_p - L_q$ approach to the compressible viscous fluid flow with slip boundary condition. *Nonlinear Anal., Theory Methods Appl., Ser. A, Theory Methods*, 106:86–109, 2014.

20
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24
25

[63] Gabriela Planas and Franck Sueur. On the “viscous incompressible fluid + rigid body” system with Navier conditions. *Ann. Inst. H. Poincaré Anal. Non Linéaire*, 31(1):55–80, 2014.

26
27

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28
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[65] Arnab Roy and Takéo Takahashi. Local null controllability of a rigid body moving into a Boussinesq flow. *Math. Control Relat. Fields*, 9(4):793–836, 2019.

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32
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8
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3 [72] Takéo Takahashi. Existence of strong solutions for the problem of a rigid-fluid system. *C. R., Math., Acad. Sci. Paris*, 336(5):453–458, 2003.

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7 [74] Chao Wang. Strong solutions for the fluid-solid systems in a 2-D domain. *Asymptot. Anal.*, 89(3-4):263–306, 2014.

8