Appendix A: Proof of Lemma 2.3
Before we begin, we present a Carleman estimate which is proved in [11], which we are going to use in its proof, and it is as follow:
Lemma A.1
There exists a constant \(\lambda _{0}\), such that, for any \(\lambda > \lambda _{0}\) there exist two constants \(C(\lambda )>0\) and \(s_{0}(\lambda )>0\) such that for any \(i \in \{1,\dots ,N \}\), any \(g\in L^{2}(Q)^{N}\) and any \(u_{0}\in H\), the solution of
$$\begin{aligned} \left\{ \begin{array}{ll} u_{t}-\Delta u+\nabla p = g, \ \ \nabla \cdot u = 0 &{} \text { in } Q,\\ u = 0 &{} \text { on } \Sigma ,\\ u(0)=u_{0}&{} \text { in } \Omega , \end{array} \right. \end{aligned}$$
(A.1)
satisfies
$$\begin{aligned}{} & {} s^{4}\iint \limits _{Q}{e^{-13s\alpha ^{*}}(\xi ^{*})^{4} | u|^{2} \textrm{d}x\,\textrm{d}t} \le C\left( \iint \limits _{Q}{ e^{-11s\alpha ^{*}} | g |^{2} \textrm{d}x\,\textrm{d}t} \right. \\{} & {} \left. +s^{7} \sum _{\begin{array}{c} j=1\\ j\ne i \end{array}}^{N}{\iint \limits _{\omega \times (0,T)}{e^{-2s\hat{\alpha }-11s\alpha ^{*}}(\hat{\xi })^{7} | u_{j} |^{2} \textrm{d}x\,\textrm{d}t}}\right) , \end{aligned}$$
for every \(s\ge s_{0}\).
We are going to develop here the duality method introduced in [40] in the context of the heat equation. The same argument has already been used in the context of the heat equation with nonhomogeneous Robin boundary conditions in [27] and in the context of the heat equation with right-hand side belonging to \(L^{2}(0,T;H^{-2}(\Omega ))\cap H^{-1}(0,T;L^{2}(\Omega ))\), wich only permits to talk about solutions in \(L^{2}(Q)\); this is explained with detail in [41].
Proof
First, we view u as a solution by transposition of (2.3). This means that u is the unique function in \(L^{2}(Q)^{N}\) satisfying
$$\begin{aligned} \iint \limits _{Q}{u g \textrm{d}x\,\textrm{d}t} = \iint \limits _{Q}{f_{0} \phi \textrm{d}x\,\textrm{d}t} - \iint \limits _{Q}{f_{1}\cdot \nabla \phi \textrm{d}x\,\textrm{d}t} + \int \limits _{\Omega }{u^{0} \phi (0) \textrm{d}x}, \quad \forall g \in L^{2}(Q)^{N},\nonumber \\ \end{aligned}$$
(A.2)
where we have denoted by \(\phi \in L^{2}(0,T;H^{2}(\Omega )^{N}\cap V)\cap H^{1}(0,T;L^{2}(\Omega )^{N})\), together with \(p_{\phi }\), the (strong) solution of the following problem:
$$\begin{aligned} \left\{ \begin{array}{lll} -\phi _{t}-\Delta \phi + \nabla p_{\phi } = g,\ \ \nabla \cdot \phi = 0 &{} \text { in } Q,\\ \phi = 0 &{} \text { on } \Sigma ,\\ \phi (T)=0 &{}\text { in } \Omega . \end{array} \right. \end{aligned}$$
(A.3)
Let us first get an estimate of the lower order term in the left-hand side of (2.4), i.e.
$$\begin{aligned} s^{4}\iint \limits _{Q}{e^{-13s\alpha ^{*}}(\xi ^{*})^{4}| u |^{2}\textrm{d}x\,\textrm{d}t}. \end{aligned}$$
(A.4)
Let us introduce the space
$$\begin{aligned} Z_{0}= \{ (\phi , p_{\phi })\in C^{2}(\overline{Q})\times C^{1}(\overline{Q}): \phi = 0 \text { on } \Sigma \text{ and } \nabla \cdot \phi = 0 \text{ in } \Omega \} \end{aligned}$$
and the norm \(\Vert \cdot \Vert _{Z}\), with
$$\begin{aligned} \Vert (\varrho , p_{\varrho })\Vert _{Z}^{2}= & {} \iint \limits _{Q}{ e^{-11s\alpha ^{*}} | \varrho _{t}- \Delta \varrho +\nabla p_{\varrho } |^{2} \textrm{d}x\,\textrm{d}t} \\ {}{} & {} + s^{7} \iint \limits _{\omega \times (0,T)}{e^{-2s\hat{\alpha }-11s\alpha ^{*}} (\hat{\xi })^{7} | (\varrho _{1},0)|^{2} \textrm{d}x\,\textrm{d}t}, \end{aligned}$$
for all \((\varrho ,p_{\varrho })\in Z_{0}\). Due to Lemma A.1, \(\Vert \cdot \Vert _{Z}\) is indeed a norm in \(Z_{0}\). Let Z be the completion of \(Z_{0}\) for the norm \(\Vert \cdot \Vert _{Z}\). Then Z is a Hilbert space for the scalar product \((\cdot , \cdot )_{Z}\), with
$$\begin{aligned} \left( ( \sigma , p_{\sigma }),(\gamma , p_{\gamma }) \right) _{Z} =&\iint \limits _{Q}{e^{-11s \alpha ^{*}}(\sigma _{t}-\Delta \sigma + \nabla p_{\sigma })(\gamma _{t} - \Delta \gamma + \nabla p_{\gamma }) \textrm{d}x\,\textrm{d}t}\\&+s^{7}\iint \limits _{\omega \times (0,T)}{e^{-2s\hat{\alpha }-11s\alpha ^{*}}(\hat{\xi })^{7}\sigma _{1}\gamma _{1}\textrm{d}x\,\textrm{d}t}. \end{aligned}$$
Then, using Lax Milgram’s Lemma there is a unique solution \((\bar{\sigma },\bar{p}_{\bar{\sigma }})\in Z \) such that
$$\begin{aligned} \left( (\bar{\sigma },\bar{p}_{\bar{\sigma }}),(\sigma ,p_{\sigma })\right) _{Z}=l(\sigma ,p_{\sigma }), \quad \forall (\sigma ,p_{\sigma })\in Z, \end{aligned}$$
(A.5)
where
$$\begin{aligned} l(\sigma ,p_{\sigma })=s^{4}\iint \limits _{Q}{e^{-13s\alpha ^{*}} (\xi ^{*})^{4} u\sigma \textrm{d}x\,\textrm{d}t}. \end{aligned}$$
By virtue of Lemma A.1, one can easily check that \( l \in Z'.\)
We define:
$$\begin{aligned} \left\{ \begin{array}{ll} \hat{\phi } &{} = e^{-11s\alpha ^{*}}(\bar{\sigma }_{t}-\Delta \bar{\sigma }+\nabla \bar{p}_{\bar{\sigma }}),\\ \hat{v} &{} = -s^{7}e^{-2s\hat{\alpha }-11s\alpha ^{*}}(\bar{\sigma }_{1},0). \end{array} \right. \end{aligned}$$
(A.6)
Recall that \(\bar{\sigma } = (\bar{\sigma }_{1},\bar{\sigma }_{2})\). Then, \((\hat{\phi },\hat{v})\) is solution of (A.3) and such that
$$\begin{aligned} \Vert ( \bar{\sigma },\bar{p}_{\bar{\sigma }} )\Vert _{Z}^{2}=(( \bar{\sigma },\bar{p}_{\bar{\sigma }} ),( \bar{\sigma },\bar{p}_{\bar{\sigma }} ))_{Z}= l ( \bar{\sigma },\bar{p}_{\bar{\sigma }} ). \end{aligned}$$
Let us now take \(g=s^{4}e^{-13s\alpha ^{*}}(\xi ^{*})^{4}u + \hat{v}\mathbb {1}_{\omega }\) in (A.2). This gives
$$\begin{aligned} s^{4}\iint \limits _{Q}{e^{-13s\alpha ^{*}}(\xi ^{*})^{4}|u|^{2} \textrm{d}x\,\textrm{d}t} = \iint \limits _{Q}{f_{0} \hat{\phi } \textrm{d}x\,\textrm{d}t}-\iint \limits _{Q}{f_{1}\cdot \nabla \hat{\phi } \textrm{d}x\,\textrm{d}t} -\iint \limits _{\omega \times (0,T)}{u\hat{v} \textrm{d}x\,\textrm{d}t}, \end{aligned}$$
(A.7)
(recall that \(\hat{v}\) and \(\hat{\phi }\) are given by (A.6)).
From (A.5), we obtain
$$\begin{aligned} \Vert ( \bar{\sigma },\bar{p}_{\bar{\sigma }} )\Vert _{Z}^{2}\le \Vert l \Vert _{Z'}\Vert ( \bar{\sigma },\bar{p}_{\bar{\sigma }} )\Vert _{Z}. \end{aligned}$$
Consequently,
$$\begin{aligned} \Vert ( \bar{\sigma },\bar{p}_{\bar{\sigma }} )\Vert _{Z}^{2}= & {} \iint \limits _{Q}{e^{11s\alpha ^{*}}|\hat{\phi }|^{2} \textrm{d}x\,\textrm{d}t}+s^{-7}\iint \limits _{\omega \times (0,T)}{e^{2s\hat{\alpha } + 11s\alpha ^{*}}|\hat{v}|^{2} \textrm{d}x\,\textrm{d}t} \nonumber \\{} & {} \le C s^{4} \iint \limits _{Q}{e^{-13s\alpha ^{*}}(\xi ^{*})^{4}|u|^{2} \textrm{d}x\,\textrm{d}t}, \end{aligned}$$
(A.8)
for \(s\ge C=C(\Omega ,\omega ,T)>0\), since
$$\begin{aligned} \Vert l \Vert _{Z'}\le s^{2} \left( \iint \limits _{Q}{e^{-13s\alpha ^{*}}(\xi ^{*})^{4}|u|^{2} \textrm{d}x\,\textrm{d}t}\right) ^{1/2}. \end{aligned}$$
Now, we multiply the equation satisfied by \(\hat{\phi }\) by \(s^{-7}e^{2s\hat{\alpha }+11s\alpha ^{*}} (\hat{\xi })^{-7}\hat{\phi }\) and we integrate in Q. After integration by parts, we get:
$$\begin{aligned}{} & {} s^{-7}\iint \limits _{Q}{e^{2s\hat{\alpha }+11s\alpha ^{*}} (\hat{\xi })^{-7} |\nabla \hat{\phi } |^{2} \textrm{d}x\,\textrm{d}t} = \dfrac{s^{-7}}{2}\iint \limits _{Q}{\dfrac{\partial }{\partial t}(e^{2s\hat{\alpha }+11s\alpha ^{*}} (\hat{\xi })^{-7})|\hat{\phi }|^{2} \textrm{d}x\,\textrm{d}t}\nonumber \\ {}{} & {} + s^{-3}\iint \limits _{Q}{ e^{2s\hat{\alpha }+6 s\alpha ^{*}} (\hat{\xi })^{-3} u \hat{\phi } \textrm{d}x\,\textrm{d}t} + s^{-7}\iint \limits _{\omega \times (0,T)}{ e^{2s\hat{\alpha }+11s\alpha ^{*}} (\hat{\xi })^{-7} \hat{v}\hat{\phi } \textrm{d}x\,\textrm{d}t}. \end{aligned}$$
(A.9)
Using Holder’s inequality and Young’s inequality in the last two terms of the right-hand side of (A.9), we have
$$\begin{aligned}&s^{-7}\iint \limits _{Q}{e^{2s\hat{\alpha } + 11s\alpha ^{*}} (\hat{\xi })^{-7} |\nabla \hat{\phi } |^{2} \textrm{d}x\,\textrm{d}t} \le C \left( \iint \limits _{Q}{e^{11s\alpha ^{*}}|\hat{\phi }|^{2}\textrm{d}x\,\textrm{d}t}\right. \\&\left. + s^{-7}\iint \limits _{\omega \times (0,T)}{e^{2s\hat{\alpha }+11s\alpha ^{*}}|\hat{v}|^{2} \textrm{d}x\,\textrm{d}t} + s^{4} \iint \limits _{Q}{e^{-13s\alpha ^{*}}(\xi ^{*})^{4}|u|^{2} \textrm{d}x\,\textrm{d}t} \right) , \end{aligned}$$
where we have taken \(s\ge C.\) This inequality, together with (A.8), provides
$$\begin{aligned} \iint \limits _{Q}{e^{11s\alpha ^{*}}|\hat{\phi }|^{2}\textrm{d}x\,\textrm{d}t} + s^{-7}\iint \limits _{Q}{e^{2s\hat{\alpha }+11s\alpha ^{*}} (\hat{\xi })^{-7} |\nabla \hat{\phi } |^{2} \textrm{d}x\,\textrm{d}t} \nonumber \\ + s^{-7}\iint \limits _{\omega \times (0,T)}{e^{2s\hat{\alpha }+ 11s\alpha ^{*}}|\hat{v}|^{2} \textrm{d}x\,\textrm{d}t} \le C s^{4} \iint \limits _{Q}{e^{-13s\alpha ^{*}}(\xi ^{*})^{4}| u |^{2} \textrm{d}x\,\textrm{d}t}. \end{aligned}$$
(A.10)
A combination of this inequality with (A.7) yields the following estimate:
$$\begin{aligned} s^{4}\iint \limits _{Q}{e^{-13 s \alpha ^{*}} (\xi ^{*})^{4} | u |^{2}\textrm{d}x\,\textrm{d}t} \le&C\left( \iint \limits _{Q}{e^{-11 s \alpha ^{*}} | f_{0}|^{2}\textrm{d}x\,\textrm{d}t} + s^{7}\iint \limits _{Q}{e^{-11s\alpha ^{*}} (\hat{\xi })^{7}| f_{1}|^{2} \textrm{d}x\,\textrm{d}t} \right. \nonumber \\&\left. + s^{7}\sum _{\begin{array}{c} j=1 \\ j\ne i \end{array}}^{N}\ \iint \limits _{\omega \times (0,T)}{e^{-2s\hat{\alpha }-11s\alpha ^{*}}(\hat{\xi })^{7} | u_{j}|^{2} \textrm{d}x\,\textrm{d}t} \right) , \end{aligned}$$
(A.11)
Let us now show that the term associated with \(\nabla u\) can also be bounded in the same way. To this end, we multiply the equation of u by
$$\begin{aligned} s^{3}e^{-13s\alpha ^{*}}(\xi ^{*})^{3-1/m} \end{aligned}$$
and we obtain
$$\begin{aligned}&\dfrac{s^{3}}{2}\iint \limits _{Q}{e^{-13s\alpha ^{*}}(\xi ^{*})^{3-1/m}\dfrac{\partial }{\partial t} | u |^{2} \textrm{d}x\,\textrm{d}t} + s^{3}\iint \limits _{Q}{e^{-13s\alpha ^{*}}(\xi ^{*})^{3-1/m}|\nabla u |^{2} \textrm{d}x\,\textrm{d}t}\nonumber \\ =&s^{3}\iint \limits _{Q}{e^{-13s\alpha ^{*}}(\xi ^{*})^{3-1/m}f_{0} u \textrm{d}x\,\textrm{d}t}- s^{3}\iint \limits _{Q}{e^{-13s\alpha ^{*}}(\xi ^{*})^{3-1/m}f_{1} \cdot \nabla u \textrm{d}x\,\textrm{d}t}. \end{aligned}$$
(A.12)
Now, integrating by parts with respect to t in the first integral of the left-hand side of (A.12) and using that
$$\begin{aligned} ( e^{-13s\alpha ^{*}}(\xi ^{*})^{3-1/m} )_{t} \le C s e^{-13s\alpha ^{*}}(\xi ^{*})^{4}, \quad s \ge C, \end{aligned}$$
we have at this moment,
$$\begin{aligned}&s^{4}\iint \limits _{Q}{e^{-13 s \alpha ^{*}} (\xi ^{*})^{4} | u |^{2}\textrm{d}x\,\textrm{d}t}+s^{3}\iint \limits _{Q}{e^{-13 s \alpha ^{*}}(\xi ^{*})^{3-1/m} |\nabla u |^{2}\textrm{d}x\,\textrm{d}t}\nonumber \\&\le C\left( \iint \limits _{Q}{e^{-11 s \alpha ^{*}} | f_{0}|^{2}\textrm{d}x\,\textrm{d}t} \right. \nonumber \\&\left. + s^{7}\iint \limits _{Q}{e^{-11s\alpha ^{*}} (\hat{\xi })^{7}| f_{1}|^{2} \textrm{d}x\,\textrm{d}t} \right. \nonumber \\&\left. + s^{7}\sum _{\begin{array}{c} j=1 \\ j\ne i \end{array}}^{N}\ \iint \limits _{\omega \times (0,T)}{e^{-2s\hat{\alpha }-11s\alpha ^{*}}(\hat{\xi })^{7} | u_{j}|^{2} \textrm{d}x\,\textrm{d}t} \right) , \end{aligned}$$
(A.13)
On the other hand, we consider
$$\begin{aligned} \tilde{u}:= s e^{-13/2s\alpha ^{*}}(\xi ^{*})^{1-1/m} u := \rho _{4}(t) u, \quad \tilde{h} := s e^{-13/2s\alpha ^{*}}(\xi ^{*})^{1-1/m} h:= \rho _{4}(t)h. \end{aligned}$$
Then, \((\tilde{u}, \tilde{h})\) satisfies the following system:
$$\begin{aligned} \left\{ \begin{array}{ll} -\tilde{u}_{t} - \Delta \tilde{u} + \nabla \tilde{h} = -(\rho _{4})_{t} u + \rho _{4} f_{0} + \rho _{4} \nabla \cdot f_{1}, \quad \nabla \cdot \tilde{u} = 0 &{} \text{ in } Q,\\ \tilde{u} = 0 &{} \text{ on } \Sigma ,\\ \tilde{u}(0) = 0 &{} \text{ in } \Omega . \end{array} \right. \end{aligned}$$
(A.14)
Then, we have that
$$\begin{aligned}&\Vert \tilde{u}\Vert _{L^{2}(0,T;H^{1}(\Omega )^{N})}^{2} + \Vert \tilde{h}\Vert _{L^{2}(Q)}^{2}\\&\quad \le C \left( \Vert \rho _{4} f_{0}\Vert _{L^{2}(Q)^{N}}^{2} + \Vert \rho _{4} f_{1} \Vert _{L^{2}(Q)^{N}}^{2} + \Vert (\rho _{4})_{t} u \Vert _{L^{2}(Q)^{N}}^{2} \right) . \end{aligned}$$
for some \(C>0.\) Then, since \(|(\rho _{4})_{t}|\le C s^2 e^{-13/2s\alpha ^{*}}(\xi ^{*})^{2} \), we can add the term associated with the pressure to the left-hand side of (A.13). Finally, we obtain (2.4).
\(\square \)
Appendix B: Proof of Proposition 2
In this occasion, we will prove the Carleman estimate of \(\psi \) following a method introduced in [10]. For simplicity of the proof, we will consider the case \(N=2\) and \(i=2.\)
Proof
First, we apply the divergence operator to equation associated with \(\psi \) to obtain
$$\begin{aligned} \Delta h = \nabla \cdot g^{\psi } = 0 \text { in } Q. \end{aligned}$$
Then, applying the operator \(\nabla \nabla \nabla \Delta ^{2} (\cdot )\) to the equation satisfied be \(\psi _{1},\) we have:
$$\begin{aligned} (\nabla \nabla \nabla \Delta ^{2}\psi _{1})_{t}- \Delta (\nabla \nabla \nabla \Delta ^{2}\psi _{1})=\nabla \nabla \nabla \Delta ^{2}g_{1}^{\psi }. \end{aligned}$$
Thus, we can apply Lemma 2.1 to this equation to obtain
$$\begin{aligned}{} & {} \iint \limits _{Q}{e^{-10s\alpha } \bigg (s^{-1}\xi ^{-1}|\nabla \nabla \nabla \nabla \Delta ^{2}\psi _{1} |^{2}+s \xi |\nabla \nabla \nabla \Delta ^{2}\psi _{1}|^{2}\bigg ) \textrm{d}x\,\textrm{d}t}\nonumber \\{} & {} \quad \le C\Bigg ( \left\| s^{-1/4} e^{-5s\alpha } \xi ^{-1/4+1/m} \nabla \nabla \nabla \Delta ^{2}\psi _{1} \right\| _{L^{2}(\Sigma )^{8}}^{2} \nonumber \\{} & {} \quad +\left\| s^{-1/4}e^{-5s\alpha } \xi ^{-1/4}\nabla \nabla \nabla \Delta ^{2}\psi _{1} \right\| _{H^{1/4,1/2}(\Sigma )^{8}}^{2} \nonumber \\{} & {} \quad \left. + s\iint \limits _{\omega _{0}\times (0,T)}{e^{-10s\alpha }\xi |\nabla \nabla \nabla \Delta ^{2}\psi _{1}|^{2}\textrm{d}x\,\textrm{d}t}+\iint \limits _{Q}{e^{-10s\alpha } | \nabla ^{2} \Delta ^{2}g_{1}^{\psi }|^{2}\textrm{d}x\,\textrm{d}t} \right) , \nonumber \\ \end{aligned}$$
(B.1)
for every \(s\ge C.\)
We divide the rest of the proof in three steps:
-
In Step 1, we estimate globals integrals of \(\psi _{1}\) y \(\psi _{2}\) by the left-hand side of (B.1).
-
In Step 2, we deal with the boundary terms en (B.1).
-
In Step 3, we estimate all the local terms.
In the following, C denotes a constant depending only on \(\lambda \), \(\Omega \), \(\omega \), \(\mathcal {O}\) y \(\ell \).
Step 1
Estimate of \(\psi _{1}:\) Applying successively Lemma 2.2 with \(r=1\) and \(u:=\nabla \nabla \Delta ^{2}\psi _{1},\) \(r=3\) and \(u=\nabla \Delta ^2 \psi _{1},\) \(r=5\) and \(u=\Delta ^2 \psi _{1},\) and combining with (B.1), we get
for every \(s\ge C.\)
Estimate of \(\psi _{2}:\) Now, we would like to introduce in the left-hand side a term in \(\psi = (\psi _{1}, \psi _{2} )\). Actually, we are going to add the term \(\left\| s^{7/2}e^{-5s\alpha ^{*}}(\xi ^{*})^{7/2}\psi \right\| _{L^{2}(0,T;H^{2}(\Omega )^{N})}\) to the left-hand side of (B.2).
Notice that, since \(\nabla \cdot \psi _{t} = 0 \) in Q, we have for all \(t\in (0, T ):\)
$$\begin{aligned} \begin{aligned} \int \limits _{\Omega }{ |\partial _{2} (\psi _{2})_{t} (t)|^{2}\textrm{d}x} =&\int \limits _{\Omega }{ |\partial _{1}(\psi _{1})_{t}(t)|^{2}\textrm{d}x}\\ \le&\int \limits _{\Omega }{ |\nabla (\psi _{1})_{t}(t)|^{2} \textrm{d}x}. \end{aligned} \end{aligned}$$
(B.3)
Since \((\psi _{2})_{t}(t)|_{\partial \Omega } = 0\) and \(\Omega \) is bounded, also, using (B.3), we have
$$\begin{aligned} \int \limits _{\Omega }{ |(\psi _{2})_{t}(t) |^{2} \textrm{d}x}\le C \int \limits _{\Omega }{ | \nabla (\psi _{1})_{t}(t) |^{2} \textrm{d}x}. \end{aligned}$$
Then, we deduce
$$\begin{aligned} \left\| \psi _{t}(t)\right\| _{L^{2}(\Omega )^{N}}^{2} \le C \left\| \nabla (\psi _{1})_{t} (t) \right\| _{L^{2}(\Omega )^{N}}^{2}, \ \ \forall t\in (0,T). \end{aligned}$$
(B. 4)
Consider now the following Stokes system,
$$\begin{aligned} \left\{ \begin{array}{lll} -\Delta \psi +\nabla h = -\psi _{t} + g^{\psi } &{} \text { in } Q,\\ \nabla \cdot \psi = 0 &{} \text { in } Q,\\ \psi = 0 &{} \text { on } \partial \Omega , \end{array} \right. \end{aligned}$$
(B. 5)
then, using a regularity result of [21] for the stationary Stokes problem (B. 5), together with (B. 4), we obtain
$$\begin{aligned} \left\| \psi (t) \right\| _{H^{2}(\Omega )^{N}}^{2} \le C \left( \left\| \nabla (\psi _{1})_{t} (t) \right\| _{L^{2}(\Omega )^{N}}^{2} + \left\| g^{\psi } (t)\right\| _{L^{2}(\Omega )^{N}}^{2}\right) , \forall t\in (0,T). \end{aligned}$$
(B. 6)
Now, observe that using the divergence free condition and applying the laplacian operator to the equation associated with \(\psi _{1}\), we get that \((\Delta \psi _{1})_{t} = \Delta ^{2}\psi _{1} + \Delta g_{1}^{\psi }\) in Q. On the other hand, since \((\psi _{1})_{t} |_{\partial \Omega } = 0,\) we have
$$\begin{aligned} \left\| (\psi _{1})_{t} \right\| _{H^{2}(\Omega )}^{2} \le C \left\| \Delta (\psi _{1})_{t} \right\| _{L^{2}(\Omega )}. \end{aligned}$$
(B. 7)
Using (B. 7) in (B. 6), we obtain
$$\begin{aligned} \Vert \psi (t) \Vert _{H^{2}(\Omega )^{N}}^{2} \le&C \left( \Vert \Delta ^{2}\psi _{1}(t) \Vert _{L^{2}(\Omega )}^{2} + \Vert \Delta g^{\psi }(t) \Vert _{L^{2}(\Omega )^{N}}^{2} \right) , \forall t \in (0,T). \end{aligned}$$
Finally, since \(\alpha ^{*}\) and \(\xi ^{*}\) do not depend on the space variable x, we have that
$$\begin{aligned} s^{7}\int \limits _{0}^{T}{e^{-10s\alpha ^{*}}(\xi ^{*})^{7} \left\| \psi \right\| _{H^{2}(\Omega )^{N}}^{2} \textrm{d}t} \le&C s^{7} \left( \iint \limits _{Q}{e^{-10 s\alpha ^{*}} (\xi ^{*})^{7} | \Delta ^{2}\psi _{1} |^{2} \textrm{d}x\,\textrm{d}t} \right. \nonumber \\&\left. + \iint \limits _{Q}{e^{- 10 s\alpha ^{*}} (\xi ^{*})^{7} | \Delta g^{\psi } |^{2} \textrm{d}x\,\textrm{d}t} \right) . \end{aligned}$$
(B. 8)
Therefore, combining (B.2) and (B. 8) we get
for every \(s\ge C.\)
Step 2
In this step we treat the boundary terms in (B.9). We begin with the first one. Notice that the minimum of the weight functions \(e^{s\alpha }\) and \(\xi \) is reached at the boundary \(\partial \Omega ,\) where \(\alpha =\alpha ^{*}\) and \(\xi =\xi ^{*}\) do not depend on x. Since \(m \ge 14\), and using Young inequality, we obtain
$$\begin{aligned}&\left\| e^{-5s\alpha ^{*}} (\xi ^{*})^{-1/4+1/m}\nabla \nabla \nabla \Delta ^{2}\psi _{1} \right\| _{L^{2}(\Sigma )^{8}}^{2}\\ \le&C \left\| e^{-5s\alpha ^{*}}\nabla \nabla \nabla \Delta ^{2}\psi _{1} \right\| _{L^{2}(\Sigma )^{8}}^{2}\\ \le&C\left( s^{-1}\iint \limits _{Q}{e^{-10s\alpha } \xi ^{-1}|\nabla \nabla \nabla \nabla \Delta ^{2}\psi _{1}|^{2}\textrm{d}x\,\textrm{d}t}\right. \\&\left. + s \iint \limits _{Q}{e^{-10s\alpha } \xi | \nabla \nabla \nabla \Delta ^{2}\psi _{1}|^{2} \textrm{d}x\,\textrm{d}t}\right) . \end{aligned}$$
Therefore, this boundary term can be absorbed by left-hand side of (B.9) for \(s\ge C\).
Now, we deal the second boundary term in the right-hand side of (B.9). To this end, we use regularity estimates.
Let us define
$$\begin{aligned} (\psi ^{1}, h_{1}):= s^{5/2}e^{-5s\alpha ^{*}}(\xi ^{*})^{5/2-1/m} (\psi , h )=: \xi _{1}(t)(\psi ,h). \end{aligned}$$
Then, from (3.3), \((\psi ^{1}, h_{1})\) is the solution of Stokes system:
$$\begin{aligned} \left\{ \begin{array}{lll} \psi _{t}^{1}-\Delta \psi ^{1} +\nabla h_{1} = (\xi _{1})_{t}\psi + \xi _{1} g^{\psi } , \ \ \nabla \cdot \psi ^{1} = 0 &{} \text { in } Q,\\ \psi ^{1} = 0 &{} \text { on } \Sigma ,\\ \psi ^{1} (0)=0 &{}\text { in } \Omega , \end{array} \right. \end{aligned}$$
(B.10)
Using Lemma 2.6 for the last system, we have
$$\begin{aligned} \Vert \psi ^{1} \Vert _{X_{1}}^{2} \le C \left( \left\| (\xi _{1})_{t} \psi \right\| _{X_{0}}^{2} + \Vert \xi _{1} g^{\psi } \Vert _{X_{0}}^{2} \right) . \end{aligned}$$
From (2.1), we see that
$$\begin{aligned} | (\xi _{1})_{t} |\le C s^{7/2}e^{-5s\alpha ^{*}}(\xi ^{*})^{7/2}, \end{aligned}$$
for every \(s\ge C.\) Thus, we obtain
$$\begin{aligned} \left\| \psi ^{1} \right\| _{X_{1}}^{2} \le C \left( \left\| s^{7/2}e^{-5s\alpha ^{*}}(\xi ^{*})^{7/2} \psi \right\| _{X_{0}}^{2}+ \left\| \xi _{1} g^{\psi } \right\| _{X_{0}}^{2}\right) . \end{aligned}$$
Next, we introduce:
$$\begin{aligned} (\psi ^{2}, h_{2}):= s^{3/2}e^{-5s\alpha ^{*}}(\xi ^{*})^{3/2-2/m} (\psi , h )=: \xi _{2}(t)(\psi ,h). \end{aligned}$$
Now, from (3.3), \((\psi ^{2}, h_{2})\) is the solution of Stokes system:
$$\begin{aligned} \left\{ \begin{array}{ll} \psi _{t}^{2}-\Delta \psi ^{2} +\nabla h_{2} = (\xi _{2})_{t}\psi + \xi _{2} g^{\psi },\ \ \nabla \cdot \psi ^{2} = 0 &{} \text { in } Q,\\ \psi ^{2} = 0 &{} \text { on } \Sigma ,\\ \psi ^{2} (0)=0 &{}\text { in } \Omega , \end{array} \right. \end{aligned}$$
(B.11)
Using Lemma 2.6 for the last system, we find:
$$\begin{aligned} \left\| \psi ^{2} \right\| _{X_{2}}^{2} \le C \left( \left\| (\xi _{2})_{t}\psi \right\| _{X_{1}}^{2} + \Vert \xi _{2}g^{\psi } \Vert _{X_{1}}^{2} \right) . \end{aligned}$$
Using the estimate
$$\begin{aligned} | (\xi _{2})_{t} |\le C s^{5/2}e^{-5s\alpha ^{*}}(\xi ^{*})^{5/2-1/m}, \end{aligned}$$
for every \(s\ge C.\) Thus, we obtain
$$\begin{aligned} \left\| \psi ^{2} \right\| _{X_{2}}^{2} \le&C \left( \left\| s^{5/2}e^{-5s\alpha ^{*}}(\xi ^{*})^{5/2-1/m} \psi \right\| _{X_{1}}^{2} + \left\| \xi _{2}g^{\psi } \right\| _{X_{1}}^{2}\right) .\\ \le&C \left( \left\| s^{7/2}e^{-5s\alpha ^{*}}(\xi ^{*})^{7/2} \psi \right\| _{X_{0}}^{2} + \left\| \xi _{1}g^{\psi } \right\| _{X_{0}}^{2} + \left\| \xi _{2}g^{\psi } \right\| _{X_{1}}^{2} \right) . \end{aligned}$$
Next, we define:
$$\begin{aligned} (\psi ^{3}, h_{3}):= s^{1/2}e^{-5s\alpha ^{*}}(\xi ^{*})^{1/2-3/m} (\psi , h )=: \xi _{3}(t)(\psi ,h). \end{aligned}$$
Then, from (3.3), \((\psi ^{3}, h_{3})\) is the solution of Stokes system:
$$\begin{aligned} \left\{ \begin{array}{ll} \psi _{t}^{3}-\Delta \psi ^{3} +\nabla h_{3} = (\xi _{3})_{t}\psi + \xi _{3}g^{\psi },\ \ \nabla \cdot \psi ^{3} = 0 &{} \text { in } Q,\\ \psi ^{3} = 0 &{} \text { on } \Sigma ,\\ \psi ^{3} (0)=0 &{}\text { in } \Omega , \end{array} \right. \end{aligned}$$
(B.12)
Using Lemma 2.6 for the last system, we get:
$$\begin{aligned} \left\| \psi ^{3} \right\| _{X_{3}}^{2} \le C \left( \left\| (\xi _{3})_{t}\psi \right\| _{X_{2}}^{2} + \left\| \xi _{3} g^{\psi } \right\| _{X_{2}}^{2}\right) . \end{aligned}$$
Using the estimate
$$\begin{aligned} | (\xi _{3})_{t} |\le C s^{3/2}e^{-5s\alpha ^{*}}(\xi ^{*})^{3/2-2/m}, \end{aligned}$$
for every \(s\ge C.\) Thus, we obtain
$$\begin{aligned} \left\| \psi ^{3} \right\| _{X_{3}}^{2} \le&C \left( \left\| \psi ^{2} \right\| _{X_{2}}^{2} + \left\| \xi _{3} g^{\psi } \right\| _{X_{2}}^{2} \right) .\\ \le&C \left( \left\| s^{7/2}e^{-5s\alpha ^{*}}(\xi ^{*})^{7/2} \psi \right\| _{X_{0}}^{2} + \left\| \xi _{1}g^{\psi }\right\| _{X_{0}}^{2} + \left\| \xi _{2}g^{\psi } \right\| _{X_{1}}^{2} + \left\| \xi _{3}g^{\psi } \right\| _{X_{2}}^{2} \right) . \end{aligned}$$
Next, we introduce:
$$\begin{aligned} (\psi ^{4}, h_{4}):= s^{-1/2}e^{-5s\alpha ^{*}}(\xi ^{*})^{-1/2-4/m} (\psi , h )=: \xi _{4}(t)(\psi ,h). \end{aligned}$$
Then, from (3.3), \((\psi ^{4}, h_{4})\) is the solution of Stokes system:
$$\begin{aligned} \left\{ \begin{array}{ll} \psi _{t}^{4}-\Delta \psi ^{4} +\nabla h_{4} = (\xi _{4})_{t}\psi + \xi _{4} g^{\psi },\ \ \nabla \cdot \psi ^{4} = 0 &{} \text { in } Q,\\ \psi ^{4} = 0 &{} \text { on } \Sigma ,\\ \psi ^{4} (0)=0 &{}\text { in } \Omega , \end{array} \right. \end{aligned}$$
(B.13)
Using Lemma 2.6 for the last system, we find:
$$\begin{aligned} \left\| \psi ^{4} \right\| _{X_{4}}^{2} \le C \left( \left\| (\xi _{4})_{t}\psi \right\| _{X_{3}}^{2} + \left\| \xi _{4} g^{\psi } \right\| _{X_{3}}^{2} \right) . \end{aligned}$$
Using the estimate
$$\begin{aligned} | (\xi _{4})_{t} |\le C s^{1/2}e^{-5s\alpha ^{*}}(\xi ^{*})^{1/2-3/m}, \end{aligned}$$
for every \(s\ge C.\) Thus, we obtain
$$\begin{aligned} \left\| \psi ^{4} \right\| _{X_{4}}^{2} \le&C \left( \left\| \psi ^{3} \right\| _{X_{3}}^{2} + \left\| \xi _{4} g^{\psi } \right\| _{X_{3}}^{2} \right) .\\ \le&C \left( \left\| s^{7/2}e^{-5s\alpha ^{*}}(\xi ^{*})^{7/2} \psi \right\| _{X_{0}}^{2} + \left\| \xi _{1}g^{\psi }\right\| _{X_{0}}^{2} + \left\| \xi _{2}g^{\psi } \right\| _{X_{1}}^{2} + \left\| \xi _{3} g^{\psi } \right\| _{X_{2}}^{2} + \left\| \xi _{4}g^{\psi } \right\| _{X_{3}}^{2} \right) . \end{aligned}$$
Then, using interpolation argument between the spaces \(X_{3}\) and \(X_{4},\) we get
$$\begin{aligned} \left\| e^{-5s\alpha ^{*}}(\xi ^{*})^{-7/(2m)} \psi \right\| _{L^{2}(0,T;H^{8}(\Omega )^{N})\cap H^{1}(0,T;H^{6}(\Omega )^{N})}^{2} \le&C \left( \left\| s^{1/2}e^{-5s\alpha ^{*}}(\xi ^{*})^{1/2-3/m} \psi \right\| _{X_{3}}\right. \\&\left. \cdot \left\| s^{-1/2}e^{-5s\alpha ^{*}}(\xi ^{*})^{-1/2-4/m} \psi \right\| _{X_{4}} \right) . \end{aligned}$$
Now, we consider the boundary term
$$\begin{aligned}&s^{-1/2}\left\| e^{-5s\alpha ^{*}} (\xi ^{*})^{-1/4} \nabla \nabla \nabla \Delta ^{2} \psi _{1} \right\| _{H^{1/4,1/2}(\Sigma )^{8}}^{2}\nonumber \\ \le&C \left( s^{-1/2} \left\| e^{-5s\alpha ^{*}} (\xi ^{*})^{-1/4} \nabla \nabla \nabla \Delta ^{2} \psi _{1} \right\| _{H^{1}(0,T;H^{-1}(\Omega )^{N})}^{2}\right. \nonumber \\&+ \left. s^{-1/2}\left\| e^{-5s\alpha ^{*}} (\xi ^{*})^{-1/4} \nabla \nabla \nabla \Delta ^{2} \psi _{1} \right\| _{L^{2}(0,T;H^{1}(\Omega )^{N})}^{2}\right) .\nonumber \\ \le&C \left( s^{-1/2}\left\| e^{-5s\alpha ^{*}} (\xi ^{*})^{-1/4} \psi _{1} \right\| _{L^{2}(0,T;H^{8}(\Omega )^{N})}^{2}\right. \nonumber \\&+ \left. s^{-1/2}\left\| e^{-5s\alpha ^{*}} (\xi ^{*})^{-1/4} \psi _{1} \right\| _{H^{1}(0,T;H^{6}(\Omega )^{N})}^{2}\right) .\nonumber \\ =&C s^{-1/2}\left\| e^{-5s\alpha ^{*}} (\xi ^{*})^{-1/4} \psi _{1} \right\| _{L^{2}(0,T;H^{8}(\Omega )^{N}) \cap H^{1}(0,T;H^{6}(\Omega )^{N})}^{2}.\nonumber \\ \le&Cs^{-1/2} \left\| e^{-5s\alpha ^{*}} (\xi ^{*})^{-7/(2m)} \psi \right\| _{L^{2}(0,T;H^{8}(\Omega )^{N}) \cap H^{1}(0,T;H^{6}(\Omega )^{N})}^{2}. \end{aligned}$$
(B.14)
By taking s large enough in (B.14), the boundary term \(s^{-1/2}\Big \Vert e^{-5s\alpha } \xi ^{-1/4}\psi \Big \Vert _{H^{1/4,1/2}(\Sigma )^{16}}\) can be absorbed by the term \(\Big \Vert e^{-5s\alpha ^{*}}(\xi ^{*})^{-7/(2m)}\psi \Big \Vert _{L^{2}(0,T;H^{8}(\Omega )^{N})\cap H^{1}(0,T;H^{6}(\Omega )^{N}) }^{2}\) and step 2 is finished.
Thus, at this point we have
$$\begin{aligned}{} & {} s^{-1}\iint \limits _{Q}{e^{-10s\alpha } \xi ^{-1}|\nabla \nabla \nabla \nabla \Delta ^{2}\psi _{1}|^{2} \textrm{d}x\,\textrm{d}t} + s \iint \limits _{Q}{e^{-10s\alpha } \xi |\nabla \nabla \nabla \Delta ^{2}\psi _{1}|^{2} \textrm{d}x\,\textrm{d}t}\nonumber \\{} & {} + s^{3} \iint \limits _{Q}{e^{-10s\alpha }\xi ^{3}|\nabla \nabla \Delta ^{2}\psi _{1}|^{2} \textrm{d}x\,\textrm{d}t}+ s^{5}\iint \limits _{Q}{e^{-10s\alpha } \xi ^{5}|\nabla \Delta ^{2}\psi _{1}|^{2} \textrm{d}x\,\textrm{d}t} \nonumber \\{} & {} + s^{7} \iint \limits _{Q}{e^{-10s\alpha }\xi ^{7}|\Delta ^{2}\psi _{1} |^{2}\textrm{d}x\,\textrm{d}t} \nonumber \\{} & {} + s^{7}\iint \limits _{Q}{e^{-10s\alpha ^{*}} (\xi ^{*})^{7}|\psi |^{2} \textrm{d}x\,\textrm{d}t} + s^{7}\iint \limits _{Q}{e^{-10s\alpha ^{*}} (\xi ^{*})^{7}|\nabla \psi |^{2} \textrm{d}x\,\textrm{d}t} \nonumber \\{} & {} + s^{7}\iint \limits _{Q}{e^{-10s\alpha ^{*}} (\xi ^{*})^{7}|\Delta \psi |^{2} \textrm{d}x\,\textrm{d}t} \nonumber \\{} & {} \le C \left( s\iint \limits _{\omega _{0}\times (0,T)}{e^{-10s\alpha }\xi |\nabla \nabla \nabla \Delta ^{2}\psi _{1}|^{2}\textrm{d}x\,\textrm{d}t} + s^{3}\iint \limits _{\omega _{0}\times (0,T)}{e^{-10s\alpha }\xi ^{3} |\nabla \nabla \Delta ^{2}\psi _{1}|^{2} \textrm{d}x\,\textrm{d}t} \right. \nonumber \\{} & {} + s^{5}\iint \limits _{\omega _{0}\times (0,T)}{e^{-10s\alpha }\xi ^{5} |\nabla \Delta ^{2}\psi _{1}|^{2}\textrm{d}x\,\textrm{d}t} + s^{7}\iint \limits _{\omega _{0}\times (0,T)}{e^{-10s\alpha }\xi ^{7} |\Delta ^{2}\psi _{1}|^{2} \textrm{d}x\,\textrm{d}t} \nonumber \\{} & {} + s^{7} \iint \limits _{Q}{e^{-10s\alpha ^{*}} (\xi ^{*})^{7} | \Delta g^{\psi } |^{2} \textrm{d}x\,\textrm{d}t} \nonumber \\{} & {} \left. +s^{7}\iint _{Q}{e^{-10s\alpha } | \nabla \nabla \Delta ^{2}g^{\psi }|^{2} \textrm{d}x\,\textrm{d}t} + \Vert \xi _{1}g^{\psi }\Vert _{X_{0}}^{2} \right. \nonumber \\{} & {} \left. + \Vert \xi _{2}g^{\psi }\Vert _{X_{1}}^{2} + \Vert \xi _{3}g^{\psi }\Vert _{X_{2}}^{2} + \Vert \xi _{4}g^{\psi }\Vert _{X_{3}}^{2} \right) , \end{aligned}$$
(B.15)
for every \(s\ge C.\)
Step 3
In this step, we estimate the first three terms in the right-hand side of (B.15) in terms of \(\Delta ^{2} \psi _{1}\) and small constants multiplied by the left-hand side of (B.15).
We start by estimating the term on \(\nabla \nabla \nabla \Delta ^{2}\psi _{1}\). Let \(\omega _{1}\) be an open subset satisfying \(\omega _{0}\Subset \omega _{1}\Subset \tilde{\omega }\) and let \(\rho _{1}\in C_{c}^{2}(\omega _{1})\) with \(\rho _{1}\equiv 1\) in \(\omega _{0}\) and \(0\le \rho _{1}\). Then, an integration by parts gives
$$\begin{aligned}&s\iint \limits _{ \omega _{0}\times (0,T)}{e^{-10s\alpha }\xi \left| \nabla \nabla \nabla \Delta ^{2}\psi _{1} \right| ^{2}\textrm{d}x\,\textrm{d}t} \\ \le&s\iint \limits _{ \omega _{1} \times (0,T)}{\rho _{1} e^{-10s\alpha }\xi \left| \nabla \nabla \nabla \Delta ^{2}\psi _{1} \right| ^{2}\textrm{d}x\,\textrm{d}t}.\\ =&-s\iint \limits _{ \omega _{1}\times (0,T) }{\rho _{1}e^{-10s\alpha }\xi \nabla \nabla \Delta ^{2}\psi _{1} (\nabla \nabla \nabla \nabla \Delta ^{2}\psi _{1}) \textrm{d}x\,\textrm{d}t} \\&\left. + \dfrac{s}{2}\iint \limits _{\omega _{1}\times (0,T)}{\Delta (\rho _{1}e^{-10s\alpha }\xi ) \left| \nabla \nabla \nabla \Delta ^{2}\psi _{1}\right| ^{2} \textrm{d}x\,\textrm{d}t}. \right. \end{aligned}$$
Using the Cauchy–Schwarz’s inequality for the first term and property (2.2) for the second one, we obtain for every \(\epsilon > 0\)
$$\begin{aligned} \begin{aligned} s\iint \limits _{\omega _{0}\times (0,T) }{e^{-10s\alpha }\xi \left| \nabla \nabla \nabla \Delta ^{2}\psi _{1} \right| ^{2}\textrm{d}x\,\textrm{d}t} \le&C s^{3}\iint \limits _{ \omega _{1}\times (0,T)}{e^{-10s\alpha }\xi ^{3} \left| \nabla \nabla \Delta ^{2}\psi _{1}\right| ^{2} \textrm{d}x\,\textrm{d}t} \\&+ \epsilon s^{-1}\iint \limits _{Q}{e^{-10s\alpha }\xi ^{-1} \left| \nabla \nabla \nabla \nabla \Delta ^{2}\psi _{1} \right| ^{2} \textrm{d}x\,\textrm{d}t}, \end{aligned} \end{aligned}$$
(B.16)
for every \(s\ge C\) (C depending also on \(\epsilon \)).
Repeating the same argument we can obtain the estimates of \(\nabla \nabla \Delta ^2 \psi _{1}\) in terms of \(\nabla \Delta ^2 \psi _{1}\) and \(\nabla \Delta ^2 \psi _{1}\) in terms of \(\Delta ^2 \psi _{1},\) namely
$$\begin{aligned}{} & {} \begin{aligned} s^{3}\iint \limits _{\omega _{1}\times (0,T) }{e^{-10s\alpha }\xi ^{3} \left| \nabla \nabla \Delta ^{2}\psi _{1} \right| ^{2}\textrm{d}x\,\textrm{d}t} \le&C s^{5}\iint \limits _{ \omega _{2}\times (0,T)}{e^{-10s\alpha }\xi ^{5} \left| \nabla \Delta ^{2}\psi _{1}\right| ^{2} \textrm{d}x\,\textrm{d}t}\ \\&+ \epsilon s\iint \limits _{Q}{e^{-10s\alpha }\xi \left| \nabla \nabla \nabla \Delta ^{2}\psi _{1} \right| ^{2} \textrm{d}x\,\textrm{d}t}, \end{aligned} \end{aligned}$$
(B.17)
$$\begin{aligned}{} & {} \begin{aligned} s^{5}\iint \limits _{ \omega _{2}\times (0,T)}{e^{-10s\alpha }\xi ^{5} \left| \nabla \Delta ^{2}\psi _{1} \right| ^{2} \textrm{d}x\,\textrm{d}t} \le&Cs^{7}\iint \limits _{ \omega _{3}\times (0,T)}{e^{-10s\alpha }\xi ^{7} \left| \Delta ^{2}\psi _{1}\right| ^{2} \textrm{d}x\,\textrm{d}t} \\&+ \epsilon s^{3}\iint \limits _{Q}{e^{-10s\alpha }\xi ^{3} \left| \nabla \nabla \Delta ^{2}\psi _{1} \right| ^{2} \textrm{d}x\,\textrm{d}t}, \end{aligned} \end{aligned}$$
(B.18)
for every \(s\ge C\) (C depending also on \(\epsilon \)), where \(\omega _{1}\Subset \omega _{2}\Subset \omega _{3}\Subset \tilde{\omega }.\)
This estimate, together with (B.15), (B.16) and (B.17), readily gives the desired Carleman inequality (3.4). This concludes the proof of Proposition 2. \(\square \)
Appendix C: Proof of Lemma 4.2
We are going to prove that \(e^{4s\beta ^{*}}(\gamma ^{*})^{-14-14/m}\hat{z}\in L^{2}(Q)^{N}\) using a method introduced in [35], which consists of increasing the regularity of the function \(\hat{z}\) through certain weight functions involved in order to then be able to apply a local inverse theorem in a sufficiently regular space (more details, see Sect. 5).
Proof
Next, we define the following functions:
$$\begin{aligned}{} & {} (z_{*,0}, p_{*,0}):= e^{4s\beta ^{*}}(\gamma ^{*})^{-5-5/m}(\hat{z}, \hat{p}_{1}), \\{} & {} \text {} f_{*,0}^{z}:= e^{4s\beta ^{*}}(\gamma ^{*})^{-5-5/m}(f^{z} + \nabla \times ((\nabla \times \hat{w})\chi )). \end{aligned}$$
Notice that \(f_{*,0}^{z}\in L^{2}(0,T;H^{-1}(\Omega )^{N})\), since this function can be written as:
$$\begin{aligned} f_{*,0}^{z}= e^{-3s\beta ^{*}}(\gamma ^{*})^{-5-5/m}( e^{7s\beta ^{*}} f^{z} ) + (\gamma ^{*})^{-4-4/m}\nabla \times \left( (\nabla \times w_{*}) \chi \right) , \end{aligned}$$
where \(e^{-3s\beta ^{*}}(\gamma ^{*})^{-5-5/m}\) and \((\gamma ^{*})^{-4-4/m}\) are bounded; also, using (4.4); in dimension 2, we can consider \(\chi =~\mathbb {1}_{\mathcal {O}}\), then \((\nabla \times w_{*})\mathbb {1}_{\mathcal {O}} \in L^{2}(Q)^{2},\) and, on the other hand, in dimension 3, we have that \((\nabla \times w_{*})\chi \) belongs to \(L^{2}(0,T;H^{1}(\Omega )^{3}),\) and as \(H^{1}(\Omega )^{3}\subset L^{2}(\Omega )^{3}\), we obtain \((\nabla \times w_{*})\chi \in L^{2}(Q)^{3}.\)
Then, by (4.9) \(z_{*,0}\) satisfies
$$\begin{aligned} \left\{ \begin{array}{ll} \mathcal {L}^{*} z_{*,0} + \nabla p_{*,0} = f_{*,0}^{z} - (e^{4s\beta ^{*}}(\gamma ^{*})^{-5-5/m})_{t}\hat{z}, \ \ \nabla \cdot z_{*,0} = 0 &{} \text { in } Q, \\ z_{*,0} = 0 &{} \text { on } \Sigma , \\ z_{*,0}(T) = 0 &{} \text { in } \Omega , \end{array} \right. \end{aligned}$$
where the last term in the right-hand side can be written as
$$\begin{aligned} (e^{4s\beta ^{*}}(\gamma ^{*})^{-5-5/m})_{t}\hat{z} = c_{4}(t) (\mathcal {L}_{H}^{*})^{4}\tilde{z}, \end{aligned}$$
where \(c_{k}(t)\) denotes a generic function such that (see (2.2))
$$\begin{aligned} | c_{k}^{(\ell )}(t) | \le C < \infty , \text { } \forall \ell = 0,\dots , k. \end{aligned}$$
(C.1)
On the other hand, for any \(h\in Y_{3,0}\), we obtain
$$\begin{aligned} \iint \limits _{Q}{z_{*,0} \cdot h \textrm{d}x\,\textrm{d}t}= & {} \iint \limits _{Q}{ \Big \langle f_{*,0}^{z} , \Phi \Big \rangle _{L^{2}(0,T;H^{-1}(\Omega )^{N}),L^{2}(0,T;H_{0}^{1}(\Omega )^{N})} \textrm{d}x\,\textrm{d}t}\nonumber \\{} & {} - \iint \limits _{Q}{c_{4}(t)(\mathcal {L}^{*})_{H}^{4}\tilde{z}\cdot \Phi \textrm{d}x\,\textrm{d}t}, \end{aligned}$$
(C.2)
where \(\Phi \) solves, together some pressure \(\pi _{\Phi }\),
$$\begin{aligned} \left\{ \begin{array}{ll} \mathcal {L} \Phi + \nabla \pi _{\Phi } = h, \ \ \nabla \cdot \Phi = 0 &{} \text { in } Q, \\ \Phi = 0 &{} \text { on } \Sigma , \\ \Phi (0) = 0 &{} \text { in } \Omega . \end{array} \right. \end{aligned}$$
Using (4.12), we can integrate by parts to obtain
$$\begin{aligned} \iint \limits _{Q}{z_{*,0} \cdot h \textrm{d}x\,\textrm{d}t}= & {} \iint \limits _{Q}{ \Big \langle f_{*,0}^{z} , \Phi \Big \rangle _{L^{2}(0,T;H^{-1}(\Omega )^{N}),L^{2}(0,T;H_{0}^{1}(\Omega )^{N})} \textrm{d}x\,\textrm{d}t}\\{} & {} - \iint \limits _{Q}{ (\mathcal {L}_{H}^{*})^{3}\tilde{z}\cdot (\mathcal {L} [c_{4}(t)\Phi ] + \nabla (c_{4}(t)h) ) \textrm{d}x\,\textrm{d}t},\\{} & {} = \iint \limits _{Q}{ \Big \langle f_{*,0}^{z} , \Phi \Big \rangle _{L^{2}(0,T;H^{-1}(\Omega )^{N}),L^{2}(0,T;H_{0}^{1}(\Omega )^{N})} \textrm{d}x\,\textrm{d}t} \\{} & {} - \iint \limits _{Q} \tilde{z}\cdot \Big ( c_{4}^{(4)}(t)\Phi + \mathcal {L} [c_{4}^{(3)}(t)\Phi ] \\{} & {} + \mathcal {L}^{2} [c_{4}^{(2)}(t)\Phi ] +\mathcal {L}^{3} [c_{4}^{(1)}(t)\Phi ] + \mathcal {L}^{3} [c_{4}(t) h ] \Big ) \textrm{d}x\,\textrm{d}t. \end{aligned}$$
Notice that here we have relied on the fact that \(\mathcal {L}_{H}^{*}\tilde{z}\), \(\Phi \) and h belong to the space H. Since
$$\begin{aligned} \big \Vert \Phi \big \Vert _{Y_{4}} \le C \big \Vert h \big \Vert _{Y_{3,0}}, \end{aligned}$$
(see regularity result (2.7)), we obtain from the last equality, together with (C.1),
$$\begin{aligned} \iint \limits _{Q}{z_{*,0} \cdot h \textrm{d}x\,\textrm{d}t} \le C \Big [ \big \Vert f_{*,0}^{z} \big \Vert _{L^{2}(0,T;H^{-1}(\Omega )^{N})} + \Vert \tilde{z}\Vert _{L^{2}(Q)^{N}} \Big ] \Vert h \Vert _{Y_{3,0}}, \forall h \in Y_{3,0}.\nonumber \\ \end{aligned}$$
(C.3)
Now, let
$$\begin{aligned}{} & {} (z_{*,1}, p_{*,1}):= e^{4s\beta ^{*}}(\gamma ^{*})^{-9-9/m}(\hat{z}, \hat{p}_{1}), \\{} & {} \text {} f_{*,1}^{z}:= e^{4s\beta ^{*}}(\gamma ^{*})^{-9-9/m}(f^{z} + \nabla \times ((\nabla \times \hat{w})\chi )). \end{aligned}$$
Similarly as before, \((z_{*,1},p_{*,1})\) satisfies
$$\begin{aligned} \left\{ \begin{array}{ll} \mathcal {L}^{*} z_{*,1} + \nabla p_{*,1} = f_{*,1}^{z} - (e^{4s\beta ^{*}} (\gamma ^{*})^{-9-9/m})_{t}\hat{z}, \ \ \nabla \cdot z_{*,1} = 0 &{} \text { in } Q, \\ z_{*,1} = 0 &{} \text { on } \Sigma , \\ z_{*,1}(T) = 0 &{} \text { in } \Omega , \end{array} \right. \end{aligned}$$
Thus, for any \(h\in Y_{2,0}\), we get
$$\begin{aligned} \iint \limits _{Q}{z_{*,1} \cdot h \textrm{d}x\,\textrm{d}t}&= \iint \limits _{Q}{ \Big \langle f_{*,1}^{z} , \Phi \Big \rangle _{L^{2}(0,T;H^{-1}(\Omega )^{N}),L^{2}(0,T;H_{0}^{1}(\Omega )^{N})} \textrm{d}x\,\textrm{d}t} \\&- \iint \limits _{Q}{ (e^{4s\beta ^{*}}(\gamma ^{*})^{-9-9/m} )_{t} \hat{z}\cdot \Phi \textrm{d}x\,\textrm{d}t}. \end{aligned}$$
Moreover, since
$$\begin{aligned} \iint \limits _{Q}{ (e^{4s\beta ^{*}}(\gamma ^{*})^{-9-9/m})_{t} \hat{z}\cdot \Phi \textrm{d}x\,\textrm{d}t} = \iint \limits _{Q}{ c_{3}(t)\Phi \cdot z_{*,0} \textrm{d}x\,\textrm{d}t}. \end{aligned}$$
using (C.3) with \(c_{3}(t)\Phi \) instead of h (notice that \(c_{3}(t)\Phi \in Y_{3,0}\)), we get the estimate
$$\begin{aligned} \iint \limits _{Q}{c_{3}(t)\Phi \cdot z_{*,0} \textrm{d}x\,\textrm{d}t} \le C \Big [ \Vert f_{*,0}^{z} \Vert _{L^{2}(0,T;H^{-1}(\Omega )^{N})} + \Vert \tilde{z} \Vert _{L^{2}(Q)^{N}} \Big ] \Vert c_{3}(t)\Phi \Vert _{Y_{3,0}}. \end{aligned}$$
Going back to \(z_{*,1}\), we have
$$\begin{aligned} \iint \limits _{Q}{z_{*,1}\cdot h \textrm{d}x\,\textrm{d}t} \le C \Big [ \left\| f_{*,0}^{z} \right\| _{L^{2}(0,T;H^{-1}(\Omega )^{N})} + \Vert \tilde{z} \Vert _{L^{2}(Q)^{N}} \Big ] \Vert \Phi \Vert _{Y_{3,0}}, \end{aligned}$$
where we have used (C.1) and the property \((\gamma ^{*})^{-9-9/m}\le C(\gamma ^{*})^{-5-5/m}.\) Taking into account that
$$\begin{aligned} \Vert \Phi \Vert _{Y_{3}} \le C \Vert h \Vert _{Y_{2,0}}, \end{aligned}$$
(see (2.6)) we obtain
$$\begin{aligned} \iint \limits _{Q}{z_{*,1}\cdot h \textrm{d}x\,\textrm{d}t} \le C \Big [ \Vert f_{*,0}^{z} \Vert _{L^{2}(0,T;H^{-1}(\Omega )^{N}) } + \Vert \tilde{z} \Vert _{L^{2}(Q)^{N}} \Big ] \Vert h \Vert _{Y_{2,0}}, \forall h \in Y_{2,0}. \end{aligned}$$
(C.4)
Now, let
$$\begin{aligned}{} & {} (z_{*,2}, p_{*,2}):= e^{4s\beta ^{*}}(\gamma ^{*})^{-12-12/m}(\hat{z}, \hat{p}_{1}), \\{} & {} \text {} f_{*,2}^{z}:= e^{4s\beta ^{*}}(\gamma ^{*})^{-12-12/m}(f^{z} + \nabla \times ((\nabla \times \hat{w})\chi )). \end{aligned}$$
Analogously, as before, \((z_{*,2},p_{*,2})\) satisfies
$$\begin{aligned} \left\{ \begin{array}{ll} \mathcal {L}^{*} z_{*,2} + \nabla p_{*,2} = f_{*,2}^{z} - (e^{4s\beta ^{*}}(\gamma ^{*})^{-12-12/m})_{t}\hat{z}, \ \ \nabla \cdot z_{*,2} = 0 &{} \text { in } Q, \\ z_{*,2} = 0 &{} \text { on } \Sigma , \\ z_{*,2}(T) = 0 &{} \text { in } \Omega , \end{array} \right. \end{aligned}$$
Thus, for any \(h\in Y_{1,0}\), we obtain
$$\begin{aligned} \iint \limits _{Q}{z_{*,2} \cdot h \textrm{d}x\,\textrm{d}t}&= \iint \limits _{Q}{ \Big \langle f_{*,2}^{z} , \Phi \Big \rangle _{L^{2}(0,T;H^{-1}(\Omega )^{N}),L^{2}(0,T;H_{0}^{1}(\Omega )^{N}) } \textrm{d}x\,\textrm{d}t}\\ {}&- \iint \limits _{Q}{ (e^{4s\beta ^{*}}(\gamma ^{*})^{-12-12/m} )_{t} \hat{z}\cdot \Phi \textrm{d}x\,\textrm{d}t}. \end{aligned}$$
Moreover, since
$$\begin{aligned} \iint \limits _{Q}{ (e^{4s\beta ^{*}}(\gamma ^{*})^{-12-12/m})_{t} \hat{z}\cdot \Phi \textrm{d}x\,\textrm{d}t} = \iint \limits _{Q}{ c_{2}(t)\Phi \cdot z_{*,0} \textrm{d}x\,\textrm{d}t}. \end{aligned}$$
using (C.4) with \(c_{2}(t)\Phi \) instead of h (notice that \(c_{2}(t)\Phi \in Y_{2,0}\)), we get the estimate
$$\begin{aligned} \iint \limits _{Q}{c_{2}(t)\Phi \cdot z_{*,0} \textrm{d}x\,\textrm{d}t} \le C \Big [ \Vert f_{*,0}^{z} \Vert _{ L^{2}(0,T;H^{-1}(\Omega )^{N}) } + \Vert \tilde{z} \Vert _{L^{2}(Q)^{N}} \Big ] \Vert c_{2}(t)\Phi \Vert _{Y_{2,0}}. \end{aligned}$$
Turning back to \(z_{*,2}\), we get
$$\begin{aligned} \iint \limits _{Q}{z_{*,2}\cdot h \textrm{d}x\,\textrm{d}t} \le C \Big [ \Vert f_{*,0}^{z} \Vert _{L^{2} (0,T;H^{-1}(\Omega )^{N})} + \Vert \tilde{z} \Vert _{L^{2}(Q)^{N}} \Big ] \Vert \Phi \Vert _{Y_{2,0}}, \end{aligned}$$
where we have used (C.1) and the property \((\gamma ^{*})^{-12-12/m}\le C(\gamma ^{*})^{-9-9/m}.\) Taking into account that
$$\begin{aligned} \Vert \Phi \Vert _{Y_{2}} \le C \Vert h \Vert _{Y_{1,0}}, \end{aligned}$$
(see (2.6)) we obtain
$$\begin{aligned} \iint \limits _{Q}{z_{*,2}\cdot h \textrm{d}x\,\textrm{d}t} \le C \Big [ \Vert f_{*,0}^{z} \Vert _{ L^{2}(0,T;H^{-1}(\Omega )^{N})} + \Vert \tilde{z} \Vert _{L^{2}(Q)^{N}} \Big ] \Vert h \Vert _{Y_{1,0}}, \forall h \in Y_{1,0}.\nonumber \\ \end{aligned}$$
(C.5)
Finally, we set
$$\begin{aligned} (z_{*,3}, p_{*,3}):= e^{4s\beta ^{*}}(\gamma ^{*})^{-14-14/m}(\hat{z}, \hat{p}_{1}), \text {} f_{*,3}^{z}:= e^{4s\beta ^{*}}(\gamma ^{*})^{-14-14/m}(f^{z} + \nabla \times ((\nabla \times \hat{w})\chi )). \end{aligned}$$
Similarly as before, \((z_{*,3},p_{*,3})\) satisfies
$$\begin{aligned} \left\{ \begin{array}{lll} \mathcal {L}^{*} z_{*,3} + \nabla p_{*,3} = f_{*,3}^{z} - (e^{4s\beta ^{*}}(\gamma ^{*})^{-14-14/m})_{t}\hat{z}, \ \ \nabla \cdot z_{*,3} = 0 &{} \text { in } Q, \\ z_{*,3} = 0 &{} \text { on } \Sigma , \\ z_{*,3}(T) = 0 &{} \text { in } \Omega . \end{array} \right. \end{aligned}$$
Thus, for any \(h\in Y_{0}\), we obtain
$$\begin{aligned} \iint \limits _{Q}{z_{*,3} \cdot h \textrm{d}x\,\textrm{d}t}&= \iint \limits _{Q}{ \Big \langle f_{*,3}^{z} , \Phi \Big \rangle _{ L^{2}(0,T;H^{-1}(\Omega )^{N}),L^{2}(0,T;H_{0}^{1}(\Omega )^{N}) } \textrm{d}x\,\textrm{d}t} \\&- \iint \limits _{Q}{ (e^{4s\beta ^{*}}(\gamma ^{*})^{-14-14/m} )_{t} \hat{z}\cdot \Phi \textrm{d}x\,\textrm{d}t}. \end{aligned}$$
Moreover, since
$$\begin{aligned} \iint \limits _{Q}{ (e^{4s\beta ^{*}}(\gamma ^{*})^{-14-14/m})_{t} \hat{z}\cdot \Phi \textrm{d}x\,\textrm{d}t} = \iint \limits _{Q}{ c_{1}(t)\Phi \cdot z_{*,0} \textrm{d}x\,\textrm{d}t}. \end{aligned}$$
using (C.5) with \(c_{1}(t)\Phi \) instead of h (notice that \(c_{1}(t)\Phi \in Y_{1,0}\)), we get the estimate
$$\begin{aligned} \iint \limits _{Q}{c_{1}(t)\Phi \cdot z_{*,0} \textrm{d}x\,\textrm{d}t} \le C \Big [ \Vert f_{*,0}^{z} \Vert _{ L^{2}(0,T;H^{-1}(\Omega )^{N}) } + \Vert \tilde{z} \Vert _{L^{2}(Q)^{N}} \Big ] \Vert c_{1}(t)\Phi \Vert _{Y_{1,0}}. \end{aligned}$$
Turning back to \(z_{*,3}\), we obtain
$$\begin{aligned} \iint \limits _{Q}{z_{*,3}\cdot h \textrm{d}x\,\textrm{d}t} \le C \Big [ \Vert f_{*,0}^{z} \Vert _{ L^{2}(0,T;H^{-1}(\Omega )^{N}) } + \Vert \tilde{z} \Vert _{L^{2}(Q)^{N}} \Big ] \Vert \Phi \Vert _{Y_{1,0}}, \end{aligned}$$
where we have used (C.1) and the property \((\gamma ^{*})^{-14-14/m}\le C(\gamma ^{*})^{-12-12/m}.\) Taking into account that
$$\begin{aligned} \Vert \Phi \Vert _{Y_{1,0}} \le C \Vert h \Vert _{Y_{0}}, \end{aligned}$$
(see (2.6)) we obtain
$$\begin{aligned} \iint \limits _{Q}{z_{*,3}\cdot h \textrm{d}x\,\textrm{d}t} \le C \Big [ \Vert f_{*,0}^{z} \Vert _{ L^{2}(0,T;H^{-1}(\Omega )^{N}) } + \Vert \tilde{z} \Vert _{L^{2}(Q)^{N}} \Big ] \Vert h \Vert _{Y_{0}}, \forall h \in Y_{0}. \end{aligned}$$
Thus, we deduce that \(z_{*,3}\in L^{2}(Q)^{N}.\) The proof of Lemma 4.2 is complete. \(\square \)
