Proof.
Step 1: Identities ?? . By definition of the Yosida operators, \[(1-\varepsilon \mathbb{P}\Delta_D)\, R_\varepsilon^S u = u, \qquad (1-\varepsilon \Delta_N)\, R_\varepsilon^H Q = Q.\] Hence \[u = R_\varepsilon^S u - \varepsilon\,\mathbb{P}\Delta_D R_\varepsilon^S u, \qquad Q = R_\varepsilon^H Q - \varepsilon\,\Delta R_\varepsilon^H Q.\] Since \(R_\varepsilon^S u \in H^2_{\mathbb{P}\Delta_D}(\Omega)\), Theorem 8, applied with \(m=2\), ensures \(R_\varepsilon^S u\in H^4(\Omega)\) and therefore \(\operatorname{div} \Delta R_\varepsilon^S u=0\). In particular, \[\mathbb{P}\Delta R_\varepsilon^S u = \Delta R_\varepsilon^S u.\]

Step 2: Regularity of \(\Delta R_\varepsilon^S u\) and \(\Delta R_\varepsilon^H Q\). Firstly, since \(Q\in H^3_{\Delta_N}(\Omega)\) it is clear that \(R_\varepsilon^HQ\in H^3_{\Delta_N}(\Omega)\). So, the second identity of ?? yields \[\Delta R_\varepsilon^HQ\in H^3_{\Delta_N}(\Omega).\] On the other hand, in Step 1 we have already shown that \(R_\varepsilon^S u\in H^4(\Omega)\) and \(\mathbb{P}\Delta R_\varepsilon^Su=\Delta R_\varepsilon^Su\). Finally, from the first identity of ?? we get \[\Delta R_\varepsilon^S u\in H^2(\Omega)\cap H^1_0(\Omega)\cap J_2(\Omega) =H^2_{\mathbb{P}\Delta_D}(\Omega).\] ◻

Proof.
Using the resolvent estimates for the related operators, we know that \[\label{proof46res46es46} \|\nabla^k(1-\varepsilon A)^{-1}w\|_{L^2(\Omega)}\lesssim \varepsilon^{-k/2}\|w\|_{L^2(\Omega)},\tag{9}\] for \(k=0,1,2\). By Lemma 11 \[\|(1-\varepsilon A)^{-1}w\|_{H^k(\Omega)} \lesssim \|(1-A)^{k/2}(1-\varepsilon A)^{-1}w\|_{L^2(\Omega)}\] \[= \|(1-A)^{\frac{k-j}{2}}(1-\varepsilon A)^{-1}(1-A)^{j/2}w\|_{L^2(\Omega)}\lesssim \|(1-\varepsilon A)^{-1}(1-A)^{j/2}w\|_{H^{k-j}(\Omega)},\] so we conclude applying 9 and Lemma 11. For what concerns the second estimate with \(A=\Delta_N\) and \(w\in H^{j+1}_{\Delta_N}(\Omega)\), the proof is clear when \(\Omega=\mathbb{R}^N\), so we focus on the half-space case. It can be seen that \[\partial_i(1-\varepsilon\Delta_N)^{-1}w=(1-\varepsilon\Delta_N)^{-1}\partial_iw \quad i<N,\] \[\partial_N(1-\varepsilon \Delta_N)^{-1}w=(1-\varepsilon\Delta_D)^{-1}\partial_Nw.\] So we conclude applying the inequality for \(A=\Delta_D,\Delta_N\) and \(j=0,1,2\). ◻

Proof.
Step 1: Convergence in \(H^2\). Applying Lemma 11, we obtain \[\|(1-\varepsilon A)^{-1}w-w\|_{H^2(\Omega)} \;\lesssim\; \|(1-A)((1-\varepsilon A)^{-1}w-w)\|_{L^2(\Omega)}.\] Since \((1-\varepsilon A)^{-1}\) commutes with \((1-A)\) on \(D(A)\), it holds \[(1-A)(1-\varepsilon A)^{-1}w-(1-A)w = (1-\varepsilon A)^{-1}(1-A)w-(1-A)w.\] Moreover, \((1-\varepsilon A)^{-1}\to Id\) strongly in \(L^2(\Omega)\) as \(\varepsilon\to0^+\) and \((1-A)w\in L^2(\Omega)\), thus \[(1-\varepsilon A)^{-1}w \to w \quad\text{in }H^2(\Omega).\]

Step 2: Convergence in \(H^3\) for the Neumann Laplacian. Let \(w\in H^3_{\Delta_N}(\Omega)\). We write \[\|(1-\varepsilon\Delta_N)^{-1}w-w\|_{H^3(\Omega)} \le \|(1-\varepsilon\Delta_N)^{-1}w-w\|_{L^2(\Omega)} + \|\nabla (1-\varepsilon\Delta_N)^{-1}w - \nabla w\|_{H^2(\Omega)}.\] The first term converges to \(0\) by Step 1. Thus, it suffices to treat the gradient term.

Tangential derivatives. For \(i<N\), tangential derivatives commute with \(\Delta_N\) and preserve Neumann boundary conditions: \[\partial_i \Delta_N = \Delta_N \partial_i, \qquad \partial_i w \in H^2_{\Delta_N}(\Omega).\] Hence, by Step 1, it holds \[\partial_i (1-\varepsilon\Delta_N)^{-1}w = (1-\varepsilon\Delta_N)^{-1} \partial_i w \to \partial_i w \quad\text{in }H^2(\Omega).\]

Normal derivative. Since \(w\in H^3_{\Delta_N}(\Omega)\), then \(\partial_N w\in H^2_{\Delta_D}(\Omega)\). As before \[\partial_N(1-\varepsilon\Delta_N)^{-1}w = (1-\varepsilon\Delta_D)^{-1}\partial_N w,\] therefore \[(1-\varepsilon\Delta_D)^{-1}\partial_N w \to \partial_N w\quad\text{in }H^2(\Omega).\] ◻

Proof.
We divide the proof of the first inequality considering separately the quantities: \[\|\nabla vw_1\cdots w_k\|_{L^2_TH^1}= \|\nabla vw_1\cdots w_k\|_{L^2_TL^2} +\] \[+ \|\nabla^2 v\cdot w_1\cdots w_k\|_{L^2_TL^2} + \sum_{\ell=1}^k\|\nabla vw_1\cdots w_{\ell-1}\nabla w_\ell w_{\ell+1}\cdots w_k\|_{L^2_TL^2}.\] For what concerns the first term \[\|\nabla vw_1\cdots w_k\|_{L^2_TL^2}\le \|\nabla v\|_{L^2_TL^2}\prod_{j=1}^k\|w_j\|_{L^\infty_TL^\infty}\] and thanks to the Sobolev embedding \(H^{s+1}(\Omega)\hookrightarrow L^\infty(\Omega)\) for \(s>\frac{N}{2}-1\) we get \[\|\nabla vw_1\cdots w_k\|_{L^2_TL^2}\lesssim \|\nabla v\|_{L^2_TL^2} \prod_{j=1}^k\|w_j\|_{L^\infty_T H^{s+1}}.\] With the same strategy, it holds \[\|\nabla^2 v\cdot w_1\cdots w_k\|_{L^2_TL^2}\lesssim \|\nabla^2 v\|_{L^2_TL^2} \prod_{j=1}^k\|w_j\|_{L^\infty_TH^{s+1}}.\] Finally, for the last term, we distinguish the cases \(k=1\) and \(k>1\). In the first case \[\|\nabla v\nabla w_1\|_{L^2_TL^2}\le \left\|\|\nabla v\|_{L^\frac{N}{s}(\Omega)}\|\nabla w_1\|_{L^\frac{2N}{N-2s}(\Omega)}\right\|_{L^2((0,T))}.\] By the Sobolev embedding \(H^1(\Omega)\hookrightarrow L^\frac{N}{s}(\Omega)\) and \(H^{s}(\Omega)\hookrightarrow L^\frac{2N}{N-2s}(\Omega)\) for \(s>\frac{N}{2}-1\), it holds \[\left\|\|\nabla v\|_{L^\frac{N}{s}(\Omega)}\|\nabla w_1\|_{L^\frac{2N}{N-2s}(\Omega)}\right\|_{L^2((0,T))} \lesssim \|\nabla v\|_{L^2_TH^1}\|w_1\|_{L^\infty_TH^{s+1}}.\] The case \(k\ge2\) can be done similarly using the embedding \(H^{s+1}(\Omega)\hookrightarrow L^\infty(\Omega)\) for \(s>\frac{N}{2}-1\). The inequalities for \(f\) and \(g\) follows from the previous inequality and the Sobolev embeddings. ◻

Proof.
Let us consider the map \(\Phi(z,V)=(u,Q)\), where \((u,Q)\) solves the system \[\label{approx46sys-contr46} \left\{\begin{array}{ll} (\partial_t-\mathbb{P}\Delta)u= - \beta \mathbb{P} {\rm Div}(\Delta-a) R_\varepsilon^H V + \mathbb{P}f(R_\varepsilon^HV,W,W) + \mathbb{P}\widetilde{F} & (0,T)\times \Omega \\ {\rm div}u=0 & (0,T)\times \Omega \\ (\partial_t-\Delta+a)Q = \beta D(R_\varepsilon^Sz) + g(R_\varepsilon^Sz,W,W) + \widetilde{G} & (0,T)\times \Omega \\ u=0,\quad \partial_\nu Q=0 & (0,T)\times\partial\Omega \\ u(0)=u_0,\quad Q(0)=Q_0 & \Omega. \end{array}\right.\tag{13}\] Our purpose is to apply the Banach Fixed Point Theorem on \(\Phi\) choosing a proper Banach space for \((z,V)\) and \((u,Q)\). Let \[Z=\{(z,V)\in Y_T(\Omega)\mid z(0)=u_0,\quad V(0)=Q_0\}.\] We know from Theorem 15 that a solution \((u,Q)\in Z\) exists when \[-\beta {\rm Div}(\Delta-a) R_\varepsilon^H V + f(R_\varepsilon^HV,W,W) + \widetilde{F}\in L^2((0,T);L^2(\Omega;\mathbb{R}^N)),\] \[\beta D(R_\varepsilon^Sz) + g(R_\varepsilon^Sz,W,W) + \widetilde{G}\in L^2((0,T);H^1(\Omega;S_0(N,\mathbb{R}))).\] In particular, if we take \((z,V)\in Z\), such a property follows from Lemma 18. So, the map \(\Phi\colon Z\to Z\) is well-defined. More precisely, Lemma 18 states that \[\|f(R_\varepsilon^HV,W,W)\|_{L^2_TL^2} + \|g(R_\varepsilon^Sz,W,W)\|_{L^2_TH^1}\] \[\lesssim \left(\|R_\varepsilon^Sz\|_{L^2_TH^2} + \|R_\varepsilon^HV\|_{L^2_TH^3}\right)\left(1 + \|W\|_{H^2(\Omega)}^2\right),\] and, since \(\varepsilon<1\), thanks to Lemma 13 it holds \[\|R^S_\varepsilon z\|_{L^2_TH^2} + \|R^H_\varepsilon V\|_{L^2_TH^3}\] \[\lesssim \varepsilon^{-1/2}\left(\|z\|_{L^2_TH^1} + \|V\|_{L^2_TH^2}\right)\] \[\le \varepsilon^{-1/2}T^{1/2}\left(\|z\|_{L^\infty_TH^1} + \|V\|_{L^\infty_TH^2}\right) \le \varepsilon^{-1/2}T^{1/2}\left(\|z\|_{X^1_T(\Omega)} + \|V\|_{X^2_T(\Omega)}\right).\] Similarly \[\|{\rm Div}(\Delta-a)R_\varepsilon^H V\|_{L^2_TL^2} + \|D(R_\varepsilon^S z)\|_{L^2_TH^1}\] \[\lesssim \varepsilon^{-1/2}T^{1/2}\left(\|z\|_{L^2_TH^1} + \|V\|_{L^2_TH^2}\right).\] So, applying the linear estimate from Theorem 15, it holds \[\label{proof46nl46es46tot46} \|\Phi(z,V)\|_{Y_T(\Omega)}\lesssim \|u_0\|_{H^1(\Omega)} + \|Q_0\|_{H^2(\Omega)} + \varepsilon^{-1/2}T^{1/2}\left(1 + \|W\|_{H^2(\Omega)}^2\right)\|(z,V)\|_{Y_T(\Omega)}.\tag{14}\] In order to apply the Banach Fixed Point Theorem, we need to prove that \(\Phi\) is a contraction in a proper subspace of \(Z\): let us consider \[Z_{\omega}=\{(v,W)\in Z\mid \|(v,W)\|_{Y_T(\Omega)}\le \omega\}.\] Firstly, we show that \(\Phi\colon Z_\omega\to Z_\omega\). In fact, letting \((u,Q)=\Phi(z,V)\), by the estimate 14 there exists \(C_0>0\) such that \[\|(u,Q)\|_{Y_T(\Omega)}\le C_0(\|u_0\|_{H^1(\Omega)} + \|Q_0\|_{H^2(\Omega)}) + C_0\varepsilon^{-1/2}T^{1/2}\omega\left(1 + \|W\|_{H^2(\Omega)}^2\right).\] We then take \(\omega\) such as \[C_0(\|u_0\|_{H^1(\Omega)} + \|Q_0\|_{H^2(\Omega)})=\frac{\omega}{2}\] and \(T_\varepsilon>0\) such that \[C_0\varepsilon^{-1/2}T_\varepsilon^{1/2}(1+\|W\|_{H^2(\Omega)}^2)\le \frac{1}{2},\] so that \[\|(u,Q)\|_{Y_{T_\varepsilon}(\Omega)}\le \omega\] and therefore \(\Phi\colon Z_\omega\to Z_\omega\). Let us take now \((z_1,V_1),(z_2,V_2)\in Z_\omega\) and the correspondent \((u_j,Q_j)=\Phi(z_j,V_j)\) for \(j=1,2\), then \((u_1-u_2,Q_1-Q_2)\) solves the system \[\left\{\begin{array}{ll} (\partial_t-\mathbb{P}\Delta)(u_1-u_2)= - \beta\mathbb{P}{\rm Div}(\Delta-a)(R_\varepsilon^H(V_1-V_2)) + \mathbb{P} f(R_\varepsilon^H(V_1-V_2),W,W) & (0,T_\varepsilon)\times\Omega \\ {\rm div}(u_1-u_2)=0 & (0,T_\varepsilon)\times\Omega \\ (\partial_t+a-\Delta)(Q_1-Q_2) = \beta D(R_\varepsilon^S(z_1-z_2)) + g(R_\varepsilon^S(z_1-z_2),W,W)& (0,T_\varepsilon)\times \Omega \\ u_1-u_2=0,\quad \partial_\nu (Q_1-Q_2)=0 & (0,T_\varepsilon)\times\partial\Omega \\ (u_1-u_2)(0)=0,\quad (Q_1-Q_2)(0)=0 & \Omega. \end{array}\right.\] So, we obtain as before that there is \(C_1>0\) such that \[\|(u_1-u_2, Q_1-Q_2)\|_{Y_{T_\varepsilon}(\Omega)}\le C_1\varepsilon^{-1/2}{T_\varepsilon}^\frac{1}{2}(1 + \|W\|_{H^2(\Omega)}^2)\|(z_1-z_2,V_1-V_2)\|_{Y_{T_\varepsilon}(\Omega)}.\] Therefore, \(\Phi\colon Z_\omega\to Z_\omega\) is a contraction choosing \(T_\varepsilon\) sufficiently small such that \[C_1\varepsilon^{-1/2}T_\varepsilon^\frac{1}{2}(1 + \|W\|_{H^2(\mathbb{R}^3)}^2)<1.\] Therefore, we get a solution for the approximated system 11 in \((0,T_\varepsilon)\).

Finally, it may happen that \(T_\varepsilon<T\), so we need to extend our solution to \((0,T)\). To do so, we can apply the previous argument changing the initial conditions: as before, we can find a solution for the system 13 with initial conditions \(u(T_\varepsilon-\delta)\) and \(Q(T_\varepsilon-\delta)\) for some \(\delta>0\). We notice that the choice of \(T_\varepsilon\) does not depend on the initial conditions, so we can extend the solution in \(2T_\varepsilon-\delta\). Therefore, with a finite number of steps we can extend the solution in \((0,T)\). ◻

Proof.
For what concerns the first part, it is sufficient to notice that for any matrices \(A,B,C\) it holds \[\label{aux46id46matrix46} {\rm tr}(ABC)={\rm tr}(CAB)={\rm tr}(BCA).\tag{16}\] Conversely, when \(C\) is anti-symmetric \[(AB)\colon C={\rm tr}(C^TAB)=-{\rm tr}(CAB).\] Thanks to 16 \[-{\rm tr}(CAB)=-{\rm tr}(BCA)=-(CA)\colon B.\] Similarly \[(AB)\colon C={\rm tr}(C^TAB)=-{\rm tr}(CAB)=-{\rm tr}(ABC)=-(BC)\colon A.\] ◻

Proof.
As anticipated, we focus on the case \(\Omega=\mathbb{R}^N_+\). Since \(u,\partial_j^2u\in H^2_{\mathbb{P}\Delta_D}(\mathbb{R}^N_+)\) for \(j=1,\ldots,N\), as mentioned in Remark 22 we can remove the Helmholtz projection from the calculation. The identity ?? follows from the condition \(u=0\) on \(\mathbb{R}^N_0\) combined with 18 . The second identity ?? also follows from the boundary condition \(\partial_i u=0\) on \(\mathbb{R}^N_0\) when \(i<N\). For the third result, let us write more precisely the left hand side of ?? : \[-\int_0^t\int_{\mathbb{R}^N_+}{\rm Div}(\Delta-a)Q\cdot \partial_N^2u+\partial_ND(u)\colon \partial_N(a-\Delta)Qdxd\tau\] \[=- \int_0^t\int_{\mathbb{R}^N_+}{\rm Div}(\Delta-a)Q\cdot \partial_N^2u+\partial_N\nabla u\colon \partial_N(a-\Delta)Qdxd\tau\] \[= -\sum_{j,k=1}^N\int_0^t\int_{\mathbb{R}^N_+}\partial_k(\Delta-a)Q_{jk}\partial_N^2u_j+\partial_N\partial_k u_j\partial_N(a-\Delta)Q_{jk}dxd\tau,\] where in the first equality we applied 18 with \(V=\partial_N(a-\Delta)Q\). When \(k=N\), we have the cancellation: \[\int_0^t\int_{\mathbb{R}^N_+}\partial_N(\Delta-a)Q_{jN}\partial_N^2u_j+\partial_N^2 u_j\partial_N(a-\Delta)Q_{jN}dxd\tau=0 \quad j=1,\ldots,N.\] For the remaining terms, that is \(k<N\), we can use the Hölder inequality: \[\left|\int_0^t\int_{\mathbb{R}^N_+}\partial_k(\Delta-a)Q_{jk}\partial_N^2u_j+\partial_N\partial_k u_j\partial_N(a-\Delta)Q_{jk}dxd\tau\right|\] \[\lesssim \|\partial_kQ\|_{L^2_tH^2}\|\partial_N^2u\|_{L^2_tL^2} + \|\partial_N\partial_k u\|_{L^2_tL^2}\|\partial_NQ\|_{L^2_tH^2}\] \[\le \|\nabla^\prime Q\|_{L^2_tH^2}\|\nabla u\|_{L^2_tH^1} + \|\nabla^\prime \nabla u\|_{L^2_tL^2}\|Q\|_{L^2_tH^3}.\] ◻

Proof.
As before, we can focus on the estimates without considering the operator \(\mathbb{P}\). The identity ?? follows from the Dirichlet conditions on \(u\) with 17 and 19 . For what concerns ?? , thanks to the regularity of \(Q\) and \(Q_0\) and to the Dirichlet conditions on \(\partial_iu\) for \(i<N\), it holds \[-\int_0^t\int_{\mathbb{R}^N_+}{\rm Div}\left((\Delta-a)Q\colon Q_0\left(Q_0+\frac{Id}{N}\right)\right)\cdot \partial_i^2u dxd\tau\] \[= \int_0^t\int_{\mathbb{R}^N_+}\partial_i{\rm Div}\left((\Delta-a)Q\colon Q_0\left(Q_0+\frac{Id}{N}\right)\right)\cdot\partial_iu dx d\tau\] \[= -\int_0^t\int_{\mathbb{R}^N_+}\partial_i\left((\Delta-a)Q\colon Q_0\left(Q_0+\frac{Id}{N}\right)\right)\colon \nabla \partial_iu dxd\tau.\] Thanks to 17 and 19 , the terms involving the identity matrix can be removed. So, it is sufficient to study the quantity \[-\int_0^t\int_{\mathbb{R}^N_+} \partial_i((\Delta-a)Q\colon Q_0Q_0)\colon \nabla \partial_i u + \partial_i(\nabla u\colon Q_0Q_0)\colon \partial_i(a-\Delta)Q dxd\tau.\] We notice that, when \(\partial_i\) acts on \((\Delta-a)Q\) in the first term and on \(\nabla u\) in the second, there is a cancellation: \[-\int_0^t\int_{\mathbb{R}^N_+} (\partial_i(\Delta-a)Q\colon Q_0)\cdot (Q_0\colon \nabla \partial_i u) + (\partial_i\nabla u\colon Q_0)\cdot (Q_0\colon \partial_i(a-\Delta)Q) dxd\tau=0.\] The remaining terms can be bounded by Sobolev and Hölder inequalities: \[\left|\int_0^t\int_{\mathbb{R}^N_+} (\Delta-a)Q\partial_iQ_0Q_0 \nabla \partial_i u dx d\tau\right| + \left|\int_0^t\int_{\mathbb{R}^N_+} \nabla u\partial_iQ_0Q_0\partial_i (a-\Delta)Q dx d\tau\right|\] \[\le \|Q_0\|_{L^\infty(\mathbb{R}^N_+)}\|\partial_iQ_0\|_{L^6(\mathbb{R}^N_+)}\|(\Delta-a)Q\|_{L^2_tL^3}\|\nabla \partial_iu\|_{L^2_tL^2} +\] \[+ \|Q_0\|_{L^\infty(\mathbb{R}^N_+)}\|\partial_iQ_0\|_{L^6(\mathbb{R}^N_+)}\|\partial_i(\Delta-a)Q\|_{L^2_tL^2}\|\nabla u\|_{L^2_tL^3}\] Firstly, we notice that \(H^2(\mathbb{R}^N_+)\hookrightarrow L^\infty(\mathbb{R}^N_+)\) and \(H^1(\mathbb{R}^N_+)\hookrightarrow L^6(\mathbb{R}^N_+)\) for \(N=2,3\), so \[\|Q_0\|_{L^\infty(\mathbb{R}^N_+)}\|\partial_iQ_0\|_{L^6(\mathbb{R}^N_+)}\lesssim \|Q_0\|^2_{H^2(\mathbb{R}^N_+)}.\] Moreover, by interpolation \[\|(\Delta-a)Q\|_{L^3(\mathbb{R}^N_+)}\lesssim \|(\Delta-a)Q\|^{1/2}_{L^2(\mathbb{R}^N_+)}\|(\Delta-a)Q\|^{1/2}_{L^6(\mathbb{R}^N_+)},\] \[\|\nabla u\|_{L^3(\mathbb{R}^N_+)}\lesssim \|\nabla u\|_{L^2(\mathbb{R}^N_+)}^{1/2}\|\nabla u\|_{L^6(\mathbb{R}^N_+)}^{1/2},\] and therefore \[\|(\Delta-a)Q\|_{L^2_tL^3}\lesssim \|Q\|^{1/2}_{L^\infty_tH^2}\|Q\|_{L^1_tH^3}^{1/2} \le t^{1/4}\|Q\|^{1/2}_{L^\infty_tH^2}\|Q\|_{L^2_tH^3}^{1/2}\le t^{1/4}\|Q\|_{X^2_t(\mathbb{R}^N_+)},\] \[\|\nabla u\|_{L^2_tL^3}\lesssim \|u\|^{1/2}_{L^\infty_tH^1}\|\nabla u\|_{L^1_tH^1}^{1/2} \le t^{1/4}\|u\|^{1/2}_{L^\infty_tH^1}\|\nabla u\|_{L^2_tH^1}^{1/2}\lesssim t^{1/4}\left[\|u\|_{L^\infty_tH^1} + \|\nabla u\|_{L^2_tH^1}\right].\] Finally, we consider the inequality ?? . Let us write more in details the left hand side (the term with the \(Id\) matrix can be treated as before) : \[\label{proof46can46lem461} \begin{align} -{\rm Div}((\Delta-a)Q\colon Q_0 Q_0)\cdot \partial_N^2u-\partial_N(\nabla u\colon Q_0Q_0)\colon \partial_N (a-\Delta)Q \\ = -\sum_{j,k,\ell,h=1}^N \partial_k((\Delta-a)Q_{\ell h}Q_0^{h\ell}Q_0^{jk})\partial_N^2u_j + \partial_N(\partial_ku_jQ_0^{jk}Q_0^{h\ell})\partial_N(a-\Delta)Q_{\ell h}. \end{align}\tag{20}\] We first consider the case where \(\partial_k\) acts on \((\Delta - a)Q\) in the first term, and \(\partial_N\) acts on \(\nabla u\) in the second: \[-\int_0^t\int_{\mathbb{R}^N_+}\partial_k(\Delta-a)Q_{\ell h}Q_0^{h\ell}Q_0^{jk}\partial_N^2u_j +\partial_N\partial_ku_jQ_0^{jk}Q_0^{h\ell}\partial_N(a-\Delta)Q_{\ell h}dxd\tau.\] When \(k=N\), the two terms are opposite and there is a cancellation. Conversely, when \(k<N\), the thesis can be easily verified using the Sobolev embedding and the Hölder inequality as we did in the proof of Lemma 23. ◻

Proof.
Using integration by parts and 18 \[\int_{\mathbb{R}^N_+}-{\rm Div}\left((\Delta-a)Q Q_0+Q_0(\Delta-a)Q\right)\cdot u dx\] \[= \int_{\mathbb{R}^N_+}\left((\Delta-a)Q Q_0+Q_0(\Delta-a)Q\right)\colon D(u) dx.\] Moreover, Lemma 21 allows us to rewrite \[\int_{\mathbb{R}^N_+}\left((\Delta-a)Q Q_0+Q_0(\Delta-a)Q\right)\colon D(u) dx = \int_{\mathbb{R}^N_+}\left(Q_0 D(u) + D(u)Q_0\right)\colon (\Delta-a)Q dx,\] which implies the identity ?? . The estimate ?? can be done as before exploiting the boundary condition \(\partial_i u=0\) for \(i<N\) on \(\mathbb{R}^N_0\). For the estimate ?? , we use the identity 21 : \[Q_0D(u)+D(u)Q_0=Q_0\nabla^Tu + \nabla^TuQ_0 + (Q_0W(u)+W(u)Q_0).\] The second term is anti-symmetric, so \[\partial_N(Q_0D(u)+D(u)Q_0)\colon \partial_N(a-\Delta)Q = \partial_N(Q_0\nabla^Tu+\nabla^TuQ_0)\colon \partial_N(a-\Delta)Q,\] where we used the fact that \(A\colon B=0\) if \(A\) is anti-symmetric and \(B\) is symmetric. Finally, distinguishing between the cases \(k=N\) and \(k<N\), we can use the same argument as in Lemma 24 to conclude. ◻

Proof.
Step 1: Proof of ?? . For simplicity, we denote \(R_\varepsilon=R_\varepsilon^S\). As before, we can apply Remark 22 to remove the operator \(\mathbb{P}\). We recall the resolvent identity from Lemma 9: \[\label{res46id46u} \begin{align} u=R_\varepsilon u-\varepsilon\Delta R_\varepsilon u. \end{align}\tag{22}\] So \[\int_{\mathbb{R}^N_+}(\partial_t-\Delta)u\cdot R_\varepsilon u dx = \int_{\mathbb{R}^N_+}(\partial_t-\Delta)(1-\varepsilon\Delta)R_\varepsilon u\cdot R_\varepsilon u dx.\] Thanks to Lemma 9, we know that \(R_\varepsilon u=\Delta R_\varepsilon u=0\) on \(\mathbb{R}^N_0\), so it holds \[\int_{\mathbb{R}^N_+}(\partial_t-\Delta)(1-\varepsilon\Delta)R_\varepsilon u\cdot R_\varepsilon u dx\] \[= \frac{1}{2}\frac{d}{dt}\left[\|R_\varepsilon u\|_{L^2(\mathbb{R}^N_+)}^2+\varepsilon\|\nabla R_\varepsilon u\|_{L^2(\mathbb{R}^N_+)}^2\right] + \|\nabla R_\varepsilon u\|_{L^2(\mathbb{R}^N_+)}^2 + \varepsilon\|\Delta R_\varepsilon u\|_{L^2(\mathbb{R}^N_+)}^2 .\] So, if we take the integral over \(\tau\in(0,t)\) we get \[\int_0^t\int_{\mathbb{R}^N_+}(\partial_t-\Delta)(1-\varepsilon\Delta)R_\varepsilon u\cdot R_\varepsilon u dxd\tau\] \[= \frac{1}{2}\left[\|R_\varepsilon u(t)\|_{L^2(\mathbb{R}^N_+)}^2+\varepsilon\|\nabla R_\varepsilon u(t)\|_{L^2(\mathbb{R}^N_+)}^2 - \|R_\varepsilon u_0\|_{L^2(\mathbb{R}^N_+)}^2-\varepsilon\|\nabla R_\varepsilon u_0\|_{L^2(\mathbb{R}^N_+)}^2\right] +\] \[+ \|\nabla R_\varepsilon u\|_{L^2_tL^2}^2 + \varepsilon\|\Delta R_\varepsilon u\|_{L^2_tL^2}^2\] \[\gtrsim \|R_\varepsilon u(t)\|_{L^2(\mathbb{R}^N_+)}^2 - \|u_0\|_{L^2(\mathbb{R}^N_+)}^2 + \|\nabla R_\varepsilon u\|_{L^2_tL^2}^2,\] where, thanks to Lemma 13, we could apply the inequality \[\|R_\varepsilon u_0\|_{L^2(\mathbb{R}^N_+)}^2+\varepsilon\|\nabla R_\varepsilon u_0\|_{L^2(\mathbb{R}^N_+)}^2\lesssim \|u_0\|_{L^2(\mathbb{R}^N_+)}^2.\]

Step 2: Proof of ?? . Thanks to 22 \[\int_{\mathbb{R}^N_+}(\partial_t-\Delta)u\cdot (-\partial_i^2)R_\varepsilon u dx = \int_{\mathbb{R}^N_+}(\partial_t-\Delta)(1-\varepsilon\Delta)R_\varepsilon u\cdot (-\partial_i^2)R_\varepsilon u dx.\] We focus on the term \[\int_{\mathbb{R}^N_+}(-\Delta)^2R_\varepsilon u\cdot (-\partial_i^2)R_\varepsilon u dx= \int_{\mathbb{R}^N_+}\nabla (-\Delta)R_\varepsilon u\colon\nabla (-\partial_i^2)R_\varepsilon u dx,\] where we used that \(\partial_iR_\varepsilon u=0\) on \(\mathbb{R}^N_0\) thanks to Remark 10. When \(i<N\), it is easy to see that \(\partial_i R_\varepsilon u(\tau)=R_\varepsilon \partial_i u(\tau)\in H^2(\Omega)\) for a.e. \(\tau\in(0,T)\). So \[\int_{\mathbb{R}^N_+}\nabla (-\Delta) R_\varepsilon u\colon \nabla (-\partial_i^2)R_\varepsilon u dx = \int_{\mathbb{R}^N_+}\nabla \Delta R_\varepsilon u\colon \nabla\partial_iR_\varepsilon \partial_iu dx = \int_{\mathbb{R}^N_+}\nabla \Delta R_\varepsilon u\colon \partial_i\nabla R_\varepsilon \partial_iu dx\] \[= - \int_{\mathbb{R}^N_+}\nabla \partial_i \Delta R_\varepsilon u\colon \nabla R_\varepsilon \partial_iu dx =\int_{\mathbb{R}^N_+} \partial_i \Delta R_\varepsilon u\cdot \Delta R_\varepsilon \partial_iu dx = \|\partial_i\Delta R_\varepsilon u\|_{L^2(\mathbb{R}^N_+)}^2,\] where we used that \(\partial_i\Delta R_\varepsilon u=0\) on \(\mathbb{R}^N_0\) and the identity \[\Delta R_\varepsilon \partial_i u=\frac{1}{\varepsilon}\left(R_\varepsilon\partial_i u-\partial_i u\right)=\frac{\partial_i}{\varepsilon}\left(R_\varepsilon u-u\right)=\partial_i\Delta R_\varepsilon u.\] When \(i=N\) \[\int_{\mathbb{R}^N_+}\nabla (-\Delta) R_\varepsilon u\colon \nabla (-\partial_N^2)R_\varepsilon u dx\] \[= \|\nabla \Delta R_\varepsilon u\|_{L^2(\mathbb{R}^N_+)}^2 - \sum_{i=1}^{N-1}\int_{\mathbb{R}^N_+}\nabla (-\Delta) R_\varepsilon u\colon \nabla (-\partial_i^2)R_\varepsilon u dx\] \[= \|\nabla \Delta R_\varepsilon u\|_{L^2(\mathbb{R}^N_+)}^2 - \sum_{i=1}^{N-1} \|\partial_i\Delta R_\varepsilon u\|_{L^2(\mathbb{R}^N_+)}^2 = \|\partial_N\Delta R_\varepsilon u\|_{L^2(\mathbb{R}^N_+)}^2.\] Therefore \[\int_{\mathbb{R}^N_+}(\partial_t-\Delta)(1-\varepsilon\Delta)R_\varepsilon u\cdot (-\partial_i^2)R_\varepsilon u dx\] \[= \frac{d}{dt}\left[\|\partial_iR_\varepsilon u\|_{L^2(\mathbb{R}^N_+)}^2+\varepsilon\|\partial_i\nabla R_\varepsilon u\|_{L^2(\mathbb{R}^N_+)}^2\right] + \|\partial_i\nabla R_\varepsilon u\|_{L^2(\mathbb{R}^N_+)}^2 + \varepsilon\|\partial_i\Delta R_\varepsilon u\|_{L^2(\mathbb{R}^N_+)}^2.\] Taking the integral on \((0,t)\) and applying Lemma 13 for \[\|\partial_iR_\varepsilon u_0\|_{L^2(\mathbb{R}^N_+)}^2+\varepsilon\|\partial_i\nabla R_\varepsilon u_0\|_{L^2(\mathbb{R}^N_+)}^2 \lesssim \|u_0\|_{H^1(\mathbb{R}^N_+)}^2\] we conclude as before. ◻

Proof.
It is sufficient to note that \[xy\le \frac{C_2}{2} x^2 + \frac{1}{2C_2}y^2.\] So \[y^2\le \alpha + C_2T^\gamma y^2 + \frac{C^2_2}{2}x^2 + \frac{1}{2}y^2,\] which implies \[y^2 \le 2\alpha + 2C_2T^\gamma y^2 + C_2^2x^2\le \alpha(2+C_2^2) + T^\gamma (2C_2+C_2^2C_1)y^2.\] In particular, choosing \[T\le (4C_2+2C_2^2C_1)^{-1/\gamma},\] we get the thesis. ◻

Proof.
Step 1: First, we prove the following inequality: \[\label{proof46en46in-0} \begin{align} & \|R_\varepsilon^Su_\varepsilon\|_{L^\infty_TL^2} + \|\nabla R_\varepsilon^Su_\varepsilon\|_{L^2_TL^2} + \|R_\varepsilon^HQ_\varepsilon\|_{X^1_T(\Omega)} \\ \lesssim & \|u_0\|_{L^2(\Omega)} + \|Q_0\|_{H^1(\Omega)} + \|\widetilde{F}\|_{L^1_TL^2} + \|\widetilde{G}\|_{L^1_TH^1}. \end{align}\tag{23}\] With respect to ?? , the inequality 23 does not cover the \(L^2_TL^2\)-norm of \(R_\varepsilon^Su_\varepsilon\). In fact, to bound it, we need a different approach. For this reason, we treat the \(L^2_TL^2\)-norm of \(R_\varepsilon^Su_\varepsilon\) separately later. To prove 23 , we use the energy method: firstly we multiply the first equation of 15 by \(R_\varepsilon^Su_\varepsilon\) and we sum it to the second equation multiplied by \((a-\Delta)R_\varepsilon^HQ_\varepsilon\), that is \[\label{proof46en46met-0} \begin{align} & \int_{\Omega}(\partial_t-\mathbb{P}\Delta)u_\varepsilon\cdot R_\varepsilon^Su_\varepsilon + (\partial_t+a-\Delta)Q_\varepsilon\colon (a-\Delta)R_\varepsilon^HQ_\varepsilon dx \\ =& -\beta\int_{\Omega} {\rm Div}(\Delta-a)R_\varepsilon^HQ_\varepsilon\cdot R_\varepsilon^Su_\varepsilon - D(R_\varepsilon^Su_\varepsilon)\colon (a-\Delta)R_\varepsilon^HQ_\varepsilon dx + \\ + \int_{\Omega}&\left(f(R_\varepsilon^HQ_\varepsilon,Q_0,Q_0) + \widetilde{F}\right) \cdot R_\varepsilon^Su_\varepsilon + \left(g(R_\varepsilon^Su_\varepsilon,Q_0,Q_0) + \widetilde{G}\right)\colon (a-\Delta)R_\varepsilon^HQ_\varepsilon dx. \end{align}\tag{24}\] We integrate over \(\tau\in(0,t)\) and we take the \(\sup\) over \(t\in(0,T)\). For what concerns the right hand side, by Lemma 23 and 27 it holds \[\label{proof46es-RHS0} \begin{align} \int_{\Omega}{\rm Div}(\Delta-a)R_\varepsilon^HQ_\varepsilon\cdot R_\varepsilon^Su_\varepsilon - D(R_\varepsilon^Su_\varepsilon)\colon (a-\Delta)R_\varepsilon^HQ_\varepsilon dx =0, \\ \int_{\Omega}f(R_\varepsilon^HQ_\varepsilon,Q_0,Q_0)\cdot R_\varepsilon^Su_\varepsilon + g(R_\varepsilon^Su_\varepsilon,Q_0,Q_0)\colon (a-\Delta)R_\varepsilon^HQ_\varepsilon dx=0. \end{align}\tag{25}\] Moreover \[\label{proof46es-FG0} \begin{align} \left|\int_0^t\int_{\Omega} \widetilde{F}\cdot R_\varepsilon^Su_\varepsilon dxd\tau\right|\le \|\widetilde{F}\|_{L^1_TL^2}\|R_\varepsilon^Su_\varepsilon\|_{L^\infty_TL^2}, \\ \left|\int_0^t\int_{\Omega} \widetilde{G}\colon (a-\Delta)R_\varepsilon^HQ_\varepsilon dxd\tau\right|\le \|\widetilde{G}\|_{L^1_TH^1}\|R_\varepsilon^HQ_\varepsilon\|_{L^\infty_TH^1}, \end{align}\tag{26}\] where we also used the boundary condition \(\partial_N R_\varepsilon^HQ_\varepsilon=0\) on \(\mathbb{R}^N_0\). For the left hand side of 24 , we can apply the inequalities ?? and ?? , so that \[\label{proof46es-LHS0} \begin{align} \sup_{t\in(0,T)}\int_0^t\int_{\Omega}(\partial_t-\mathbb{P}\Delta)u_\varepsilon\cdot R_\varepsilon^Su_\varepsilon + (\partial_t+a-\Delta)Q_\varepsilon\colon (a-\Delta)R_\varepsilon^HQ_\varepsilon dxd\tau \\ \gtrsim \|(R_\varepsilon^Su_\varepsilon,R_\varepsilon^HQ_\varepsilon)\|_{L^\infty_T(L^2\times H^1)}^2 + \|\nabla R_\varepsilon^Su_\varepsilon\|_{L^2_TL^2}^2 + \|R_\varepsilon^HQ_\varepsilon\|_{L^2_TH^2}^2 - \|u_0\|_{L^2(\Omega)}^2 - \|Q_0\|_{H^1(\Omega)}^2 \end{align}\tag{27}\] So, combining 25 , 26 and 27 , we get 23 .

Step 2: Similarly to the previous step, we prove the inequality ?? without the \(L^2_TL^2\)-norm of \(R_\varepsilon^S u_\varepsilon\), that is \[\label{proof46en46in-1} \begin{align} & \|R_\varepsilon^Su_\varepsilon\|_{L^\infty_TH^1(\Omega)} + \|\nabla R_\varepsilon^Su_\varepsilon\|_{L^2_TH^1} + \|R_\varepsilon^HQ_\varepsilon\|_{X^2_T(\Omega)} \\ &\lesssim \|u_0\|_{H^1(\Omega)} + \|Q_0\|_{H^2(\Omega)} + \|\widetilde{F}\|_{L^2_TL^2} + \|\widetilde{G}\|_{L^2_TH^1}. \end{align}\tag{28}\] Let us multiply the first equation of 15 by \((-\partial_i^2)R_\varepsilon^Su_\varepsilon\) and let us sum it with the derivative \(\partial_i\) of the second equation of 15 multiplied by \(\partial_i(a-\Delta)R_\varepsilon^HQ_\varepsilon\) for \(i=1,\ldots,N\), that is: \[\int_{\Omega}(\partial_t-\mathbb{P}\Delta)u_\varepsilon\cdot (-\partial_i^2)R_\varepsilon^Su_\varepsilon +\partial_i(\partial_t+a-\Delta)Q_\varepsilon\colon \partial_i(a-\Delta)R_\varepsilon^HQ_\varepsilon dx\] \[= -\beta\int_{\Omega}{\rm Div}(\Delta-a)R_\varepsilon^HQ_\varepsilon\cdot(-\partial_i^2) R_\varepsilon^Su_\varepsilon - \partial_iD(R_\varepsilon^Su_\varepsilon)\colon \partial_i(a-\Delta)R_\varepsilon^HQ_\varepsilon dx +\] \[+ \int_{\Omega}f(R_\varepsilon^HQ_\varepsilon,Q_0,Q_0)\cdot (-\partial_i^2)R_\varepsilon^Su_\varepsilon + \partial_ig(R_\varepsilon^Su_\varepsilon,Q_0,Q_0)\colon \partial_i(a-\Delta)R_\varepsilon^HQ_\varepsilon dx +\] \[+ \int_{\Omega} \widetilde{F}\cdot (-\partial_i^2)R_\varepsilon^Su_\varepsilon + \partial_i\widetilde{G}\colon \partial_i(a-\Delta)R_\varepsilon^HQ_\varepsilon dx.\] The argument is the same as before: for the left hand side, we can apply ?? and ?? . For what concerns the right hand side, first it holds \[\label{proof46es-FG1} \begin{align} \left|\int_0^t\int_{\Omega} \widetilde{F}\cdot (-\partial_i^2) R_\varepsilon^S u_\varepsilon dx d\tau\right|\le \|\widetilde{F}\|_{L^2_TL^2}\|\partial_i^2 R_\varepsilon^S u_\varepsilon\|_{L^2_TL^2}, \\ \left|\int_0^t\int_{\Omega} \partial_i\widetilde{G}\colon \partial_i(a-\Delta)R_\varepsilon^HQ_\varepsilon dx d\tau\right|\le \|\widetilde{G}\|_{L^2_TH^1}\|\partial_i R_\varepsilon^HQ_\varepsilon\|_{L^2_TH^2}. \end{align}\tag{29}\] For the remaining terms we distinguish the case \(i<N\) from the case \(i=N\): when \(i<N\) we apply ?? and ?? which, combined with the previous inequalities, brings to the estimate ?? with the tangential derivatives, that is \[\label{proof46unif46es46tang46} \begin{align} & \|(R_\varepsilon^S u_\varepsilon,\nabla^\prime R_\varepsilon^Su_\varepsilon)\|_{L^\infty_TL^2}^2 + \|(R_\varepsilon^HQ_\varepsilon,\nabla^\prime R_\varepsilon^HQ_\varepsilon)\|_{L^\infty_TH^1}^2 + \\ & + \|\nabla^\prime R_\varepsilon^Su_\varepsilon\|_{L^2_TH^1}^2 + \|(R_\varepsilon^HQ_\varepsilon,\nabla^\prime R_\varepsilon^HQ_\varepsilon)\|_{L^2_TH^2}^2 \\ & \lesssim \|u_0\|_{H^1(\Omega)}^2 + \|Q_0\|_{H^2(\Omega)}^2 + \|\widetilde{F}\|_{L^2_TL^2}^2 + \|\widetilde{G}\|_{L^2_TH^1}^2 + \\ & + C(Q_0)T^\gamma \|R_\varepsilon^HQ_\varepsilon\|_{X^2_T(\Omega)}\left(\|R_\varepsilon^Su_\varepsilon\|_{L^\infty_TH^1} + \|\nabla R^S_\varepsilon u_\varepsilon\|_{X^0_T(\Omega)}\right), \end{align}\tag{30}\] with \[C(Q_0)=1+\|Q_0\|_{H^2(\Omega)}^2.\] Conversely, when \(i=N\), we apply ?? and ?? and we get the estimate for the transversal derivative: \[\label{proof46unif46es46trans46} \begin{align} & \|(R_\varepsilon^Su_\varepsilon,\partial_N R_\varepsilon^Su_\varepsilon)\|_{L^\infty_TL^2}^2 + \|(R_\varepsilon^HQ_\varepsilon,\partial_NR_\varepsilon^HQ_\varepsilon)\|_{L^\infty_TH^1}^2 + \\ & +\|\partial_N R_\varepsilon^Su_\varepsilon\|_{L^2_TH^1}^2 + \|(R_\varepsilon^HQ_\varepsilon,\partial_N R_\varepsilon^HQ_\varepsilon)\|_{L^2_TH^2}^2 \\ & \lesssim \|u_0\|_{H^1(\Omega)}^2 + \|Q_0\|_{H^2(\Omega)}^2 + \|\widetilde{F}\|_{L^2_TL^2}^2 + \|\widetilde{G}\|_{L^2_TH^1}^2 + \\ & + C(Q_0)T^\gamma \|R_\varepsilon^HQ_\varepsilon\|_{X^2_T(\Omega)}\left(\|R_\varepsilon^Su_\varepsilon\|_{L^\infty_TH^1} + \|\nabla R^S_\varepsilon u_\varepsilon\|_{X^0_T(\Omega)}\right) + \\ & + C(Q_0)\left[\|\nabla R^S_\varepsilon u_\varepsilon\|_{X^0_T(\Omega)}\|\nabla^\prime R^H_\varepsilon Q_\varepsilon\|_{L^2_TH^2} + \|\nabla^\prime R^S_\varepsilon u_\varepsilon\|_{L^2_TH^1}\|R^H_\varepsilon Q_\varepsilon\|_{X^2_T(\Omega)}\right]. \end{align}\tag{31}\] Combining 30 , 31 and applying Lemma 31 with \[x=\|(R^S_\varepsilon u_\varepsilon,\nabla^\prime R^S_\varepsilon u_\varepsilon)\|_{L^\infty_TL^2} + \|(R^H_\varepsilon Q_\varepsilon,\nabla^\prime R^H_\varepsilon Q_\varepsilon)\|_{L^\infty_TH^1} +\] \[+ \|\nabla^\prime R^S_\varepsilon u_\varepsilon\|_{L^2_TH^1} + \|(R^H_\varepsilon Q_\varepsilon,\nabla^\prime R^H_\varepsilon Q_\varepsilon)\|_{L^2_TH^2},\] \[y = \|R_\varepsilon^Su_\varepsilon\|_{L^\infty_TH^1} + \|\nabla R^S_\varepsilon u_\varepsilon\|_{X^0_T(\Omega)} + \|R^H_\varepsilon Q_\varepsilon\|_{X^2_T(\Omega)},\] we can find \(T>0\) independent of \(\varepsilon\) sufficiently small such that 28 holds.

Step 3: The \(L^2_TL^2\)-norm of \(R_\varepsilon^Su_\varepsilon\). We recall that \[\label{proof46Res-in46} \|R_\varepsilon^Su_\varepsilon\|_{L^2_TL^2}\lesssim \|u_\varepsilon\|_{L^2_TL^2}.\tag{32}\] We already know that \((u_\varepsilon,Q_\varepsilon)\) solves the system 15 in \((0,T)\). So, by the Duhamel Formula \[(u_\varepsilon,Q_\varepsilon)(t)= e^{Bt}(u_0,Q_0) + \int_0^t e^{B(t-\tau)}(f(R_\varepsilon^HQ_\varepsilon,Q_0,Q_0)+\widetilde{F},g(R_\varepsilon^Su_\varepsilon,Q_0,Q_0)+\widetilde{G})d\tau,\] for \(t\in(0,T)\), where \(B\) is the operator associated with the \(Q\)-tensor system defined in 7 . Then, by Theorem 7, choosing \(T<1\), it holds \[\label{proof46L2-es-int46} \begin{align} & \|(u_\varepsilon,Q_\varepsilon)\|_{L^2(\Omega)\times H^1(\Omega)}\lesssim \|u_0\|_{L^2(\Omega)} + \|Q_0\|_{H^1(\Omega)} + \\ + \int_0^t \|f(R_\varepsilon^HQ_\varepsilon,& Q_0,Q_0)(\tau) + \widetilde{F}(\tau)\|_{L^2(\Omega)} + \|g(R_\varepsilon^Su_\varepsilon,Q_0,Q_0)(\tau) + \widetilde{G}(\tau)\|_{H^1(\Omega)} d\tau. \end{align}\tag{33}\] From Lemma 18 we know that \[\|f(R_\varepsilon^HQ_\varepsilon,Q_0,Q_0) + \widetilde{F}\|_{L^1_TL^2} + \|g(R_\varepsilon^Su_\varepsilon,Q_0,Q_0) + \widetilde{G}\|_{L^1_TH^1}\] \[\lesssim T^{1/2}\left[\|(\widetilde{F},\widetilde{G})\|_{L^2_T(L^2\times H^1)} + \left(1 + \|Q_0\|_{H^2(\Omega)}^2\right)\|(\nabla R_\varepsilon^Su_\varepsilon,R_\varepsilon^HQ_\varepsilon)\|_{L^2_T(H^1\times H^3)}\right].\] So, by 32 , 28 and 33 , there is \(T\le1\) sufficiently small such that \[\label{proof46L2-es46} \begin{align} \|(R_\varepsilon^Su_\varepsilon,R_\varepsilon^HQ_\varepsilon)\|_{L^2_T(L^2\times H^1)} \lesssim \|u_0\|_{H^1(\Omega)} + \|Q_0\|_{H^2(\Omega)} + \|\widetilde{F}\|_{L^2_TL^2} + \|\widetilde{G}\|_{L^2_TH^1}. \end{align}\tag{34}\] Finally, combining 34 with 23 and 28 we conclude. ◻

Proof.
Since \((u_\varepsilon,Q_\varepsilon)\) solves the approximated system 15 , we can apply Theorem 15 and Lemma 18 to obtain \[\|(u_\varepsilon,Q_\varepsilon)\|_{Y_T(\Omega)} \lesssim \|u_0\|_{H^1(\Omega)} + \|Q_0\|_{H^2(\Omega)} + \|\widetilde{F}\|_{L^2_TL^2} + \|\widetilde{G}\|_{L^2_TH^1} +\] \[+ \|f(R_\varepsilon^H Q_\varepsilon, Q_0,Q_0)\|_{L^2_TL^2} + \|g(R_\varepsilon^Su_\varepsilon,Q_0,Q_0)\|_{L^2_TH^1}\] \[+ \beta\|{\rm Div}(\Delta-a)R_\varepsilon^HQ_\varepsilon\|_{L^2_TL^2} + \beta\|D(R_\varepsilon^S u_\varepsilon)\|_{L^2_TH^1}\] \[\lesssim \|u_0\|_{H^1(\Omega)} + \|Q_0\|_{H^2(\Omega)} + \|\widetilde{F}\|_{L^2_TL^2} + \|\widetilde{G}\|_{L^2_TH^1} +\] \[+ \left(1+\|Q_0\|_{H^2(\Omega)}^2\right)\|(R_\varepsilon^H Q_\varepsilon, R_\varepsilon^Su_\varepsilon)\|_{Y_T(\Omega)}.\] So by Proposition 32 we conclude. ◻

Proof.
Step 1: Proof of ?? and ?? . By Proposition 32 and Corollary 33, we know that there is \(T>0\) such that, for any \(\varepsilon>0\), the system 15 admits a solution \((u_\varepsilon,Q_\varepsilon)\in Y_T(\Omega)\) with \[\|(R_\varepsilon^Su_\varepsilon,R_\varepsilon^HQ_\varepsilon)\|_{Y_T(\Omega)} \le C(Q_0)\left[\|u_0\|_{H^1(\Omega)} + \|Q_0\|_{H^2(\Omega)} + \|\widetilde{F}\|_{L^2_TL^2} + \|\widetilde{G}\|_{L^2_TH^1}\right],\] \[\|(u_\varepsilon,Q_\varepsilon)\|_{Y_T(\Omega)} \le C(Q_0)\left[ \|u_0\|_{H^1(\Omega)} + \|Q_0\|_{H^2(\Omega)} + \|\widetilde{F}\|_{L^2_TL^2} + \|\widetilde{G}\|_{L^2_TH^1}\right].\] Let \(\varepsilon_n>0\) such that \(\varepsilon_n\searrow0\) as \(n\to+\infty\). For simplicity, we denote \[(u_n,Q_n)=(u_{\varepsilon_n},Q_{\varepsilon_n}), \quad R_n^A=R_{\varepsilon_n}^A \quad \text{for}\:\:A=S,H.\] Then, \[(R_n^Su_n,R_n^HQ_n)\rightharpoonup (u,Q),\] \[(u_n,Q_n)\rightharpoonup (v,W)\] in \(L^2_T(H^2\times H^3)\), and \[(\partial_t u_n,\partial_t Q_n)\rightharpoonup(\partial_t v,\partial_t W),\] in \(L^2_T(L^2\times H^1)\). On the other hand, thanks to the self-adjointness of \(R_n^S\) and \(R_n^H\), it is easy to see that \((u,Q)=(v,W)\). We can also assume that \[(u_n,Q_n)(t)\xrightharpoonup{H^1(\Omega)\times H^2(\Omega)} (u,Q)(t) \quad \text{for a.e.}\:\:t\in(0,T).\] Since \(Q_0\in H^2(\mathbb{R}^N_+)\) and thanks to the weak convergence of \((R_n^Su_n,R_n^HQ_n)\), it can be seen that \[f(R_n^HQ_n,Q_0,Q_0)\rightharpoonup f(Q,Q_0,Q_0)\quad \text{in}\:\:L^2_TL^2,\] \[g(R_n^Su_n,Q_0,Q_0)\rightharpoonup g(u,Q_0,Q_0)\quad \text{in}\:\:L^2_TH^1.\] As a consequence, it can be seen that \((u,Q)\) solves the linear system 35 in \(L^2_T(L^2\times L^2)\). Moreover, let \(\alpha=0,1\), then by lower semi-continuity of the norms it holds \[\|(u,Q)\|_{L^2_T(H^{\alpha+1}\times H^{\alpha+2})}\le \liminf_{n\to+\infty} \|(u_n,Q_n)\|_{L^2_T(H^{\alpha+1}\times H^{\alpha+2})}\] \[\le C(Q_0)\left[\|(u_0,Q_0)\|_{H^\alpha(\Omega)\times H^{\alpha+1}(\Omega)} + \|(\widetilde{F},\widetilde{G})\|_{L^{\alpha+1}_T(L^2\times H^1)}\right],\] \[\|(u,Q)\|_{H^1_T(L^2\times H^1)}\le \liminf_{n\to+\infty} \|(u_n,Q_n)\|_{H^1_T(L^2\times H^1)}\] \[\le C(Q_0)\left[\|(u_0,Q_0)\|_{H^1(\Omega)\times H^2(\Omega)} + \|(\widetilde{F},\widetilde{G})\|_{L^2_T(L^2\times H^1)}\right],\] and for a.e. \(t\in(0,T)\) \[\|(u,Q)(t)\|_{H^\alpha(\Omega)\times H^{\alpha+1}(\Omega)}\] \[\le \liminf_{n\to\infty}\|(u_{n},Q_{n})(t)\|_{H^\alpha(\Omega)\times H^{\alpha+1}(\Omega)}\le C(Q_0)\left[\|(u_0,Q_0)\|_{H^\alpha(\Omega)\times H^{\alpha+1}(\Omega)} + \|(\widetilde{F},\widetilde{G})\|_{L^{\alpha+1}_T(L^2\times H^1)}\right].\] In particular, the last inequality implies \[\|(u,Q)\|_{L^\infty_T(H^\alpha\times H^{\alpha+1})}\le C(Q_0)\left[\|(u_0,Q_0)\|_{H^\alpha(\Omega)\times H^{\alpha+1}(\Omega)} + \|(\widetilde{F},\widetilde{G})\|_{L^2_T(L^2\times H^1)}\right],\] which concludes the proof of ?? and ?? .

Step 2: Proof of ?? . The function \(w=(u,Q)-e^{Bt}(u_0,Q_0)\) solves the system \[\left\{\begin{array}{ll} (\partial_t-B)w= (f(Q,Q_0,Q_0) + \widetilde{F},g(u,Q_0,Q_0) + \widetilde{G}) & (0,T)\times\Omega \\ w(0)=(0,0) & \Omega, \end{array}\right.\] where \(B=(B_1,B_2)\) is defined in 7 . Let us denote \[S(t)=(S_1(t),S_2(t))=e^{Bt}(u_0,Q_0)\] We can write \[f(Q,Q_0,Q_0)= f(Q-S_2(t),Q_0,Q_0) + f(S_2(t),Q_0,Q_0),\] \[g(u,Q_0,Q_0)= g(u-S_1(t),Q_0,Q_0) + g(S_1(t),Q_0,Q_0).\] So \[\left\{\begin{array}{ll} (\partial_t-B_1)w_1= f(w_2,Q_0,Q_0) + f(S_2(t),Q_0,Q_0) + \widetilde{F} & (0,T)\times \Omega \\ (\partial_t-B_2)w_2= g(w_1,Q_0,Q_0) + g(S_1(t),Q_0,Q_0) + \widetilde{G} & (0,T)\times\Omega \\ w_1=0,\quad \partial_\nu w_2=0 & (0,T)\times\partial\Omega \\ w(0)=(0,0) & \Omega. \end{array}\right.\] By ?? and ?? \[\|w\|_{X^0_T(\Omega)\times X^1_T(\Omega)}\] \[\le C(Q_0)\left[\|\widetilde{F} + f(S_2(t),Q_0,Q_0)\|_{L^1_TL^2} + \|\widetilde{G} + g(S_1(t),Q_0,Q_0))\|_{L^1_TH^1}\right],\] \[\|w\|_{Y_T(\Omega)}\le C(Q_0)\left[\|\widetilde{F} + f(S_2(t),Q_0,Q_0)\|_{L^2_TL^2} + \|\widetilde{G} + g(S_1(t),Q_0,Q_0))\|_{L^2_TH^1}\right].\] So, interpolating the two inequalities we obtain \[\|w\|_{X^s_T(\Omega)\times X^{s+1}_T(\Omega)}\] \[\le C(Q_0)\left[\|\widetilde{F} + f(S_2(t),Q_0,Q_0)\|_{L^\frac{2}{2-s}_TL^2} + \|\widetilde{G} + g(S_1(t),Q_0,Q_0))\|_{L^\frac{2}{2-s}_TH^1}\right]\] \[\le T^\frac{1-s}{2}C(Q_0)\left[\|\widetilde{F} + f(S_2(t),Q_0,Q_0)\|_{L^2_TL^2} + \|\widetilde{G} + g(S_1(t),Q_0,Q_0))\|_{L^2_TH^1}\right]\] for \(s\in[0,1]\). On the other hand, by Lemma 18 and Theorem 7 it holds \[\|f(S_2(t),Q_0,Q_0)\|_{L^2_TL^2} + \|g(S_1(t),Q_0,Q_0))\|_{L^2_TH^1}\] \[\le C(Q_0)e^{\alpha T}\left(\|u_0\|_{H^1(\Omega)} + \|Q_0\|_{H^2(\Omega)}\right),\] for some \(\alpha>0\). Let now \(\delta>0\), then we can find \(T>0\) sufficiently small such that \[C(Q_0)^2T^\frac{1-s}{2}e^{\alpha T}\left(\|u_0\|_{H^1(\Omega)} + \|Q_0\|_{H^2(\Omega)}\right)\le \frac{\delta}{2}.\] Finally, \[\|(u,Q)(t)-(u_0,Q_0)\|_{H^s(\Omega)\times H^{s+1}(\Omega)}\] \[\le \|(u,Q)(t)-e^{Bt}(u_0,Q_0)\|_{H^s(\Omega)\times H^{s+1}(\Omega)} + \|e^{Bt}(u_0,Q_0) - (u_0,Q_0)\|_{H^s(\Omega)\times H^{s+1}(\Omega)}.\] We have just bounded the first term, so we focus on the second one. As mentioned in Theorem 7, the semigroup \(e^{Bt}\) is \(C_0\)-analytic. Therefore, by Lemma 12 \[\|(u_0,Q_0)-e^{Bt}(u_0,Q_0)\|_{H^s(\Omega)\times H^{s+1}(\Omega)}\] \[\lesssim \|(1-B)^{s/2}(u_0,Q_0)-(1-B)^{s/2}e^{Bt}(u_0,Q_0)\|_{L^2(\Omega)\times H^1(\Omega)}\] \[= \|(1-B)^{s/2}(u_0,Q_0)-e^{Bt}(1-B)^{s/2}(u_0,Q_0)\|_{L^2(\Omega)\times H^1(\Omega)}\xrightarrow{t\to0}0,\] and for \(T>0\) sufficiently small \[\|(u_0,Q_0)-e^{Bt}(u_0,Q_0)\|_{H^s(\Omega)\times H^{s+1}(\Omega)}\le \frac{\delta}{2}.\] ◻

Proof of Theorem 2.
We consider the map \(\Phi\colon (z,V)\mapsto (u,Q)\), where \((u,Q)\) solves the system \[\left\{\begin{array}{ll} (\partial_t-\mathbb{P}\Delta)u +\beta\mathbb{P}{\rm Div}(\Delta-a)Q - \mathbb{P}f(Q,Q_0,Q_0) = \mathbb{P}\widetilde{f}(V,Q_0) + \mathbb{P}\widetilde{F}(z,V) & (0,T)\times\Omega \\ {\rm div}u=0 & (0,T)\times\Omega \\ (\partial_t+a-\Delta)Q - \beta D(u) - g(u,Q_0,Q_0)= \widetilde{g}(z,V,Q_0) + \widetilde{G}(z,V) & (0,T)\times \Omega \\ u=0, \quad \partial_\nu Q=0 & (0,T)\times\partial\Omega \\ u(0)=u_0,\quad Q(0)=Q_0 & \Omega, \end{array}\right.\] where \[\widetilde{f}(V,Q_0)={\rm Div}\left[2\xi(\Delta-a)V\colon(V-Q_0)\left(V+\frac{Id}{N}\right) + 2\xi(\Delta-a)V\colon Q_0(V-Q_0)\right] -\] \[- {\rm Div}\left[(\xi+1)(\Delta-a)V(V-Q_0) + (1-\xi)(V-Q_0)(\Delta-a)V\right],\] \[\widetilde{g}(z,V,Q_0) = \xi\left[D(z)(V-Q_0) + (V-Q_0)D(z)\right] + W(z)(V-Q_0) - (V-Q_0)W(z)-\] \[- 2\xi(V-Q_0)V\colon \nabla z - 2\xi\left(Q_0+\frac{Id}{N}\right)(V-Q_0)\colon\nabla z.\] It can be seen that \[f(Q,Q_0,Q_0)+\widetilde{f}(Q,Q_0)=f(Q,Q,Q),\] \[g(u,Q_0,Q_0) + \widetilde{g}(u,Q,Q_0)=g(u,Q,Q),\] So, if we find a fixed point for the map \(\Phi\), we have a solution for the \(Q\)-tensor system. We know from Proposition 34 that such \((u,Q)\) exists if \[\widetilde{f}(V,Q_0) + \widetilde{F}(z,V)\in L^2((0,T);L^2(\Omega;\mathbb{R}^N)),\] \[\widetilde{g}(z,V,Q_0) + \widetilde{G}(z,V)\in L^2((0,T);H^1(\Omega;\mathbb{R}^N)).\] Following the same approach of Lemma 18 and thanks to Lemma 35, it can be seen that the previous statement is true for \((z,V)\in Y_T(\Omega)\). More precisely, if we call \[Z=\{(z,V)\in Y_T(\Omega)\mid z(0)=u_0,\quad V(0)=Q_0\},\] then \(\Phi\colon Z\to Z\) is well-defined and we have \[\label{proof46loc46ex461} \begin{align} \|(u,Q)\|_{Y_T(\Omega)}\le C(Q_0)\left[\|u_0\|_{H^1(\Omega)} + \|Q_0\|_{H^2(\Omega)} + T^\gamma \left(1+\|(z,V)\|_{Y_T(\Omega)}^5\right)\right] + \\ + C(Q_0)\|V-Q_0\|_{L^\infty_TH^{s+1}}\|(z,V)\|_{Y_T(\Omega)}\left(\|(z,V)\|_{Y_T(\Omega)} + \|Q_0\|_{H^2(\Omega)}\right), \end{align}\tag{36}\] for some \(\gamma>0\) and \(s\in(1/2,1)\) and where we recall \[C(Q_0)\sim 1+\|Q_0\|_{H^2(\Omega)}^2.\] Let us define now \[Z_{\omega,\delta}=\{(z,V)\in Z\mid \|(z,V)\|_{Y_T(\Omega)}\le \omega,\quad \|(z,V)-(u_0,Q_0)\|_{L^\infty_T(H^s\times H^{s+1})}\le \delta\}.\] We want to prove that \(\Phi\colon Z_{\omega,\delta}\to Z_{\omega,\delta}\). We already know from 36 that there is \(M>0\) such that \[\|(u,Q)\|_{Y_T(\Omega)}\le M\left(1+\|Q_0\|_{H^2(\Omega)}^2\right)\left[\|u_0\|_{H^1(\Omega)} + \|Q_0\|_{H^2(\Omega)}\right] +\] \[+ M\left(1+\|Q_0\|_{H^2(\Omega)}^2\right)\left[T^\gamma \left(1+\omega^5\right) + \delta\omega\left(\omega + \|Q_0\|_{H^2(\Omega)}\right)\right].\] So, if we take \(\omega\) such that \[M\left(1+\|Q_0\|_{H^2(\Omega)}^2\right)\left(\|u_0\|_{H^1(\Omega)} + \|Q_0\|_{H^2(\Omega)}\right)=\frac{\omega}{3}\] and we take \(\delta,T>0\) sufficiently small such that \[M\left(1+\|Q_0\|_{H^2(\Omega)}^2\right)T^\gamma \left(1+\omega^5\right)\le \frac{\omega}{3},\] \[M\delta\left(1+\|Q_0\|_{H^2(\Omega)}^2\right)\left(\omega + \|Q_0\|_{H^2(\Omega)}\right)\le \frac{1}{3},\] we obtain \[\|\Phi(z,V)\|_{Y_T(\Omega)}\le \omega.\] Moreover, from Proposition 34 and the estimates of Lemmas 18 and 35, for \(T>0\) sufficiently small it holds \[\|\Phi(z,V)(t)-(u_0,Q_0)\|_{H^s\times H^{s+1}(\Omega)}\] \[\lesssim \frac{\delta}{2} + T^\frac{1-s}{2}C(Q_0)\|\widetilde{f}(V,Q_0) + \widetilde{F}(z,V)\|_{L^2_TL^2} + T^\frac{1-s}{2}C(Q_0)\|\widetilde{g}(z,V,Q_0) + \widetilde{G}(z,V)\|_{L^2_TH^1}\] \[\lesssim \frac{\delta}{2} + T^\frac{1-s}{2}C(Q_0)\left[T^\gamma \left(1+\omega^5\right) + \delta\omega\left(\omega + \|Q_0\|_{H^2(\Omega)}\right)\right].\] Since we chose \(s\in(1/2,1)\), we can find \(T>0\) sufficiently small such that \[\|\Phi(z,V)(t)-(u_0,Q_0)\|_{H^s\times H^{s+1}(\Omega)}\le \delta\] and therefore \(\Phi\colon Z_{\omega,\delta}\to Z_{\omega,\delta}\). Let us consider now \((z_1,V_1),(z_2,V_2)\in Z_{\omega,\delta}\). Clearly \[\| V_1-V_2\|_{L^\infty_TH^{s+1}}\le \| V_1-Q_0\|_{L^\infty_TH^{s+1}} + \| V_2-Q_0\|_{L^\infty_TH^{s+1}} \le 2\delta.\] So, similarly to what we did previously, it holds \[\label{proof46nl46loc46es} \begin{align} &\|\Phi(z_1,V_1)-\Phi(z_2,V_2)\|_{Y_T(\Omega)} \\ \le C(Q_0) \left[T^\gamma \left(1+\omega^4\right)\right.&\left. + \delta\left(\omega + \|Q_0\|_{H^2(\Omega)}\right)\right]\|(z_1,V_1)-(z_2,V_2)\|_{Y_T(\Omega)}. \end{align}\tag{37}\] Therefore, for \(T,\delta>0\) sufficiently small, \(\Phi\colon Z_{\omega,\delta}\to Z_{\omega,\delta}\) is a contraction and we get the existence of a local solution in \(Y_T(\Omega)\). The existence of the pressure \(p\) comes from Remark 17.

Finally, it remains to prove the uniqueness of the solution: let us suppose \((u_1,Q_1)\) and \((u_2,Q_2)\) be two solutions in \(Y_T(\Omega)\) and let \[R=\max\{\|(u_1,Q_1)\|_{Y_T(\Omega)},\|(u_2,Q_2)\|_{Y_T(\Omega)}\}.\] We want to apply the estimate 37 but, in order to do so, we need to prove that \[\|(u_j,Q_j)-(u_0,Q_0)\|_{L^\infty_T(H^s\times H^{s+1})}\le \delta \quad j=1,2.\] As before, thanks to Proposition 34, we have for \(j=1,2\) \[\|(u_j,Q_j)-(u_0,Q_0)\|_{L^\infty_T(H^s\times H^{s+1})}\] \[\lesssim \frac{\delta}{2} + T^\frac{1-s}{2}C(Q_0)\left[T^\gamma \left(1+R^5\right) + \|(u_j,Q_j)-(u_0,Q_0)\|_{L^\infty_T(H^s\times H^{s+1})}R\left(R + \|Q_0\|_{H^2(\Omega)}\right)\right].\] So for \(T_0=T_0(R,Q_0)>0\) sufficiently small \[\|(u_j,Q_j)-(u_0,Q_0)\|_{L^\infty_{T_0}(H^s\times H^{s+1})}\le \delta \quad j=1,2.\] Finally, from 37 \[\|(u_1,Q_1)-(u_2,Q_2)\|_{Y_{T_0}(\Omega)}\] \[\le C(Q_0) \left[T_0^\gamma \left(1+R^4\right) + \delta\left(R + \|Q_0\|_{H^2(\Omega)}\right)\right]\|(u_1,Q_1)-(u_2,Q_2)\|_{Y_{T_0}(\Omega)}.\] So, for \(T_0>0\) sufficiently small, we get that \((u_1,Q_1)=(u_2,Q_2)\) for a.e. \((t,x)\in (0,T_0)\times\Omega\). In order to conclude, it is sufficient to notice that we can repeat the argument with a different starting time: in fact, the choice of \(T_0\) depends on \(Q_0\) but only due to its presence in the nonlinear terms \(f\) and \(g\). Therefore, if we consider the same problem in \((t_0,+\infty)\) with \(t_0=T_0-\varepsilon\), we can prove as before that \((u_1,Q_1)=(u_2,Q_2)\) for a.e. \((t,x)\in(0,2T_0-\varepsilon)\times\Omega\). Therefore, in a finite number of steps we conclude \((u_1,Q_1)=(u_2,Q_2)\) for a.e. \((t,x)\in(0,T)\times\Omega\). ◻

Proof of Corollary 3.
Thanks to Theorem 5 and the local well-posedness, it is sufficient to prove that \[\|F(u,Q)\|_{L^p_TL^q} + \|G(u,Q)\|_{L^p_TW^{1,q}}\lesssim C(\|(u,Q)\|_{Y_T(\Omega)})<\infty.\] In order to do so, we need to repeat the inequalities of Lemmas 18 and 35 in the \(L^p_TL^q\) setting. We show the details only of \[\|\nabla v w\|_{L^p_TW^{1,q}}\le C(\|v\|_{X^1_T(\Omega)}, \|w\|_{X^2_T(\Omega)}), \quad k\in\mathbb{N},\] for \(v\in X^1_T(\Omega)\) and \(w\in X^2_T(\Omega)\). Let us start from the \(L^p_TL^q\)-norm: let \(r>2\) such that \[r=\left\{\begin{array}{ll} \frac{2q}{2-q} & q\in(1,2) \\ \infty & q=2. \end{array}\right.\] Then by Hölder inequality \[\|\nabla vw\|_{L^p_TL^q}\le \left\|\|\nabla v\|_{L^2(\Omega)}\|w\|_{L^r(\Omega)}\right\|_{L^p((0,T))}\] \[\lesssim \|\nabla v\|_{L^\infty_TL^2}\|w\|_{L^p_TH^2}\le T^\frac{2-p}{2p}\|v\|_{X^1_T(\Omega)}\|w\|_{X^2_T(\Omega)},\] which is bounded, since \(p\le 2\). The estimate of \[\|\nabla(\nabla v w)\|_{L^p_TL^q}\le \|\nabla^2v\cdot w\|_{L^p_TL^q} + \|\nabla v \cdot \nabla w\|_{L^p_TL^q}.\] is similar to the previous one. Therefore \[\|f(Q)\|_{L^p_TL^q} + \|g(u,Q)\|_{L^p_TW^{1,q}}\lesssim C(\|(u,Q)\|_{Y_T(\Omega)})\] and by Theorem 5 we conclude. ◻

Proof.
Similarly to Proposition 34, we can prove that for \(T>0\) sufficiently small there is a solution for the system 38 , with \[(u,Q)\in Y_T(\Omega), \quad \nabla \pi\in L^2((0,T);L^2(\Omega;\mathbb{R}^N))\] that satisfies the estimate \[\|(u,Q)\|_{Y_T(\Omega)} + \|\nabla \pi\|_{L^2_TL^2}\le C(T)\left[\|u_0\|_{H^1(\Omega)} + \|Q_0\|_{H^2(\Omega)} + \|f\|_{L^2_TL^2} + \|g\|_{L^2_TH^1}\right].\] In particular, \(C(T)\lesssim 1\) if \(T\le 1\). Our aim is to find an estimate for \((u,Q)\) that does not depend on \(T\). In this way, we will be able to extend our solution for any \(t>0\) in the proper norm. Since \((u,Q)\) is a solution in \((0,T)\), we can apply the energy method as before. We recall from Lemma 14 that \[R_\varepsilon^Su\xrightarrow{\varepsilon\to 0^+}u\quad \text{in}\quad H^2(\Omega)\] \[R_\varepsilon^HQ\xrightarrow{\varepsilon\to0^+}Q\quad \text{in}\quad H^3(\Omega).\] So, up to a density argument, we apply Lemma 23, Lemma 29 and Lemma 30 to obtain \[\label{proof46lin46gl46es461} \|u\|_{L^\infty_TL^2}^2 + \|\nabla u\|_{L^2_TL^2}^2 + \|Q\|_{X^1_T(\Omega)}^2 \lesssim \|u_0\|_{L^2(\Omega)}^2 + \|Q_0\|_{H^1(\Omega)}^2 + \|(F,G)\|_{L^1_TL^2}^2,\tag{41}\] \[\label{proof46lin46gl46es462} \begin{align} \|\nabla^\prime u&\|_{L^\infty_TL^2}^2 + \|\nabla \nabla^\prime u\|_{L^2_TL^2}^2 + \|\nabla^\prime Q\|_{L^\infty_TH^1}^2 + \|\nabla^\prime Q\|_{L^2_TH^2}^2 \\ &\lesssim \|u_0\|_{H^1(\Omega)}^2 + \|Q_0\|_{H^2(\Omega)}^2 + \|(F,\nabla G)\|_{L^2_TL^2}^2, \end{align}\tag{42}\] \[\label{proof46lin46gl46es463} \begin{align} \|\partial_3u&\|_{L^\infty_TL^2}^2 + \|\nabla \partial_3u\|_{L^2_TL^2}^2 + \|\partial_3Q\|_{L^\infty_TH^1}^2 + \|\partial_3Q\|_{L^2_TH^2}^2\\ & \lesssim \|u_0\|_{H^1(\Omega)}^2 + \|Q_0\|_{H^2(\Omega)}^2 + \|(F,\nabla G)\|_{L^2_TL^2}^2 + \\ + &\|\nabla^\prime\nabla u\|_{L^2_TL^2}\|Q\|_{X^2_T(\Omega)} + \|\nabla^\prime Q\|_{L^2_TH^2}\|\nabla u\|_{X^0_T(\Omega)}, \end{align}\tag{43}\] where we used that \[\left|\int_0^t\int_\Omega F\cdot u dxd\tau\right|\le\|F\|_{L^1_TL^2}\|u\|_{L^\infty_TL^2},\] \[\left|\int_0^t\int_\Omega F\cdot \partial_i^2u dxd\tau\right|\le \|F\|_{L^2_TL^2}\|\partial_i^2u\|_{L^2_TL^2}\quad i=1,2,3\] \[\left|\int_0^t\int_\Omega G\colon (a-\Delta)Q dxd\tau\right|\le \|G\|_{L^1_TL^2}\|Q\|_{L^\infty_TH^2},\] \[\left|\int_0^t\int_\Omega \partial_iG\colon \partial_i(a-\Delta)Q dxd\tau\right|\le \|\partial_i G\|_{L^2_TL^2}\|\partial_iQ\|_{L^2_TH^2}\quad i=1,2,3.\] To be noticed that the constants in the right hand side of the inequalities 41 43 does not depend on \(T\). In particular, combining 41 , 42 and 43 , we get \[\label{proof46lin46gl46es465} \begin{align} &\|u\|_{L^\infty_TH^1}^2 + \|\nabla u\|_{L^2_TH^1}^2 + \|Q\|_{L^\infty_TH^2}^2 + \|Q\|_{L^2_TH^3}^2\\ \le C&\left[\|u_0\|_{H^1(\Omega)}^2 + \|Q_0\|_{H^2(\Omega)}^2 + \|(F,\nabla G)\|_{L^2_TL^2}^2 + \|(F,G)\|_{L^1_TL^2}^2\right], \end{align}\tag{44}\] for some \(C>0\) independent of \(T\). Therefore, since \[F\in L^1(\mathbb{R}_+;L^2(\Omega;\mathbb{R}^N))\cap L^2(\mathbb{R}_+;L^2(\Omega;\mathbb{R}^N)),\] \[G\in L^1(\mathbb{R}_+;L^2(\Omega;S_0(N,\mathbb{R})))\cap L^2(\mathbb{R}_+;H^1(\Omega;S_0(N,\mathbb{R}))),\] the solution \((u,\pi,Q)\) can be extended to \(\mathbb{R}_+\) and taking the limit as \(T\to+\infty\) on the inequality 44 we conclude. ◻

Proof.
Let \(v_0=(u_0,Q_0)\). Lemma 4.3 in [22] yields \[\|e^{Bt}v_0\|_{L^{r_2}(\mathbb{R}^N_+)\times \dot{H}^1_{r_2}(\mathbb{R}^N_+)} \lesssim t^{-\frac{N}{2}\left(\frac{1}{r_1}-\frac{1}{r_2}\right)} \|v_0\|_{L^{r_1}(\mathbb{R}^N_+)\times \dot{H}^1_{r_1}(\mathbb{R}^N_+)},\] whenever \(\frac{N}{2}\left(\frac{1}{r_1}-\frac{1}{r_2}\right)\le 1\). With the same strategy, it can be proved that \[\label{eq:decay-step} \|\nabla^j e^{Bt}v_0\|_{L^{r_2}(\mathbb{R}^N_+)\times \dot{H}^1_{r_2}(\mathbb{R}^N_+)} \lesssim t^{-\frac{j}{2}-\frac{N}{2}\left(\frac{1}{r_1}-\frac{1}{r_2}\right)} \|v_0\|_{L^{r_1}(\mathbb{R}^N_+)\times \dot{H}^1_{r_1}(\mathbb{R}^N_+)},\tag{45}\] for \(\frac{N}{2}\left(\frac{1}{r_1}-\frac{1}{r_2}\right)\le 1\) and \(j=0,1\). Let now \(1<p<q<\infty\) be arbitrary and let \(m\in\mathbb{N}\) such that \[\frac{N}{2}\Bigl(\frac{1}{p}-\frac{1}{q}\Bigr)\le m.\] We define intermediate exponents \(\{r_k\}_{k=0}^m\) by \[r_0=p,\qquad r_m=q,\qquad \frac{1}{r_k}=\frac{m-k}{m}\frac{1}{p}+\frac{k}{m}\frac{1}{q}\quad \text{for}\:\: k=0,\dots,m.\] Then for each \(k=1,\dots,m\), \[\frac{N}{2}\Bigl(\frac{1}{r_{k-1}}-\frac{1}{r_k}\Bigr) =\frac{1}{m}\,\frac{N}{2}\Bigl(\frac{1}{p}-\frac{1}{q}\Bigr)\le 1,\] so that 45 applies to every consecutive pair \((r_{k-1},r_k)\). Using the semigroup property \(e^{Bt}=(e^{B t/m})^m\) and iterating 45 \(m\) times, we obtain for \(j=0,1\), \[\begin{align} \|\nabla^j e^{Bt}(v_0)\|_{L^{q}(\mathbb{R}^N_+)\times \dot{H}^1_{q}(\mathbb{R}^N_+)} &= \Bigl\|\nabla^j \bigl(e^{B t/m}\bigr)^m (v_0)\Bigr\|_{L^{r_m}(\mathbb{R}^N_+)\times \dot{H}^1_{r_m}(\mathbb{R}^N_+)}\\ &\lesssim \left(\frac{t}{m}\right)^{-\frac{j}{2}-\sum_{k=1}^m\frac{N}{2}\left(\frac{1}{r_{k-1}}-\frac{1}{r_k}\right)} \|v_0\|_{L^{r_0}(\mathbb{R}^N_+)\times \dot{H}^1_{r_0}(\mathbb{R}^N_+)}\\ &= \left(\frac{t}{m}\right)^{-\frac{j}{2}-\frac{N}{2}\left(\frac{1}{p}-\frac{1}{q}\right)} \|v_0\|_{L^{p}(\mathbb{R}^N_+)\times \dot{H}^1_{p}(\mathbb{R}^N_+)}. \end{align}\] Absorbing the factor \(m^{\frac{j}{2}+\frac{N}{2}(\frac{1}{p}-\frac{1}{q})}\) into the implicit constant we conclude. ◻

Proof.
Let \((u,\pi,Q)\) from Proposition 36. Since \((u,Q)\) solves the system 38 , it satisfies the Duhamel Formula \[(u,Q)(t)=e^{Bt}(u_0,Q_0) + \int_0^t e^{B(t-\tau)}(F,G)(\tau)d\tau.\] In particular \[\label{Duhamel46aux46} u(t)=\left[e^{Bt}(u_0,Q_0)\right]_1 + \int_0^t\left[e^{B(t-\tau)}(F,G)(\tau)\right]_1d\tau,\tag{46}\] when \([\cdot]_j\) denote the \(j\)-th component for \(j=1,2\). As in the proof of Proposition 32, we can control the \(L^2L^2\)-norm in the time interval \((0,1)\): \[\|u\|_{L^2_1L^2}\le \|(u_0,Q_0)\|_{L^2(\Omega)\times H^1(\Omega)} + \|F\|_{L^1L^2} + \|G\|_{L^1H^1}.\] We can then focus on the case \(t\ge 1\). Thanks to Lemma 37 and 38, we know that \[\left\|[e^{Bt}(u_0,Q_0)]_1\right\|_{L^2(\Omega)}\lesssim t^{-\frac{3}{2}\left(\frac{1}{q}-\frac{1}{2}\right)}\left[\|u_0\|_{L^2(\Omega)\cap L^q(\Omega)} + \|Q_0\|_{H^1(\Omega)\cap W^{1,q}(\Omega)}\right].\] So, since \(q<\frac{6}{5}\), the first term of 46 belongs to \(L^2L^2\). Concerning the inhomogeneous term \[\int_0^t\left[e^{B(t-\tau)}(F,G)(\tau)\right]_1d\tau= \int_0^{t-1}\left[e^{B(t-\tau)}(F,G)(\tau)\right]_1d\tau + \int_{t-1}^t\left[e^{B(t-\tau)}(F,G)(\tau)\right]_1d\tau.\] Thanks to Lemma 37 and 38 \[\left\|\int_0^{t-1}\left[e^{B(t-\tau)}(F,G)(\tau)\right]_1d\tau \right\|_{L^2(\Omega)}\] \[\lesssim \int_0^{t-1}(t-\tau)^{-\frac{3}{2}\left(\frac{1}{q}-\frac{1}{2}\right)}\left[\|F(\tau)\|_{L^2(\Omega)\cap L^q(\Omega)} + \|G(\tau)\|_{H^1(\Omega)\cap W^{1,q}(\Omega)}\right]d\tau\] \[\le \int_{\mathbb{R}}(t-\tau)^{-\frac{3}{2}\left(\frac{1}{q}-\frac{1}{2}\right)}1_{(1,\infty)}(t-\tau)\left[\|F(\tau)\|_{L^2(\Omega)\cap L^q(\Omega)} + \|G(\tau)\|_{H^1(\Omega)\cap W^{1,q}(\Omega)}\right]d\tau,\] while, applying Theorem 7 \[\left\|\int_{t-1}^t\left[e^{B(t-\tau)}(F,G)(\tau)\right]_1d\tau \right\|_{L^2(\Omega)}\lesssim \int_{t-1}^t\|(F,G)(\tau)\|_{L^2(\Omega)\times H^1(\Omega)}d\tau\] \[= \int_{\mathbb{R}}1_{(0,1)}(t-\tau)\|(F,G)(\tau)\|_{L^2(\Omega)\times H^1(\Omega)}d\tau.\] Therefore, by Young’s inequality \[\left\|\int_0^{t-1}\left[e^{B(t-\tau)}(F,G)(\tau)\right]_1d\tau \right\|_{L^2((1,\infty);L^2(\Omega))}\] \[\lesssim \|t^{-\frac{3}{2}\left(\frac{1}{q}-\frac{1}{2}\right)}\|_{L^2((1,\infty))}\left[\|F\|_{L^1(L^2\cap L^q)} + \|G\|_{L^1(H^1\cap W^{1,q})}\right],\] \[\left\|\int_{t-1}^t\left[e^{B(t-\tau)}(F,G)(\tau)\right]_1d\tau \right\|_{L^2((1,\infty);L^2(\Omega))}\lesssim \|(F,G)\|_{L^1(L^2\times H^1)}.\] Again, thanks to the choice of \(q\) we conclude. ◻

Proof of Theorem 4.
We consider the map \(\Phi\colon (z,V)\mapsto (u,Q)\), where \((u,Q)\) solves the system \[\left\{\begin{array}{ll} (\partial_t-\mathbb{P}\Delta)u +\beta\mathbb{P}{\rm Div}(\Delta-a)Q = \mathbb{P} F(z,V) & \mathbb{R}_+\times\Omega \\ {\rm div}u=0 & \mathbb{R}_+\times\Omega \\ (\partial_t+a-\Delta)Q - \beta D(u) = G(z,V) & \mathbb{R}_+\times \Omega \\ u=0, \quad \partial_\nu Q=0 & \mathbb{R}_+\times\partial\Omega \\ u(0)=u_0,\quad Q(0)=Q_0 & \Omega. \end{array}\right.\] Thanks to Proposition 39, such \((u,Q)\) exists in \(Y(\Omega)\) if \[F(z,V)\in L^2(\mathbb{R}_+;L^2(\Omega;\mathbb{R}^3))\cap L^1(\mathbb{R}_+;L^2(\Omega;\mathbb{R}^3)\cap L^q(\Omega;\mathbb{R}^3)),\] \[G(z,V)\in L^2(\mathbb{R}_+;H^1(\Omega;S_0(3,\mathbb{R})))\cap L^1(\mathbb{R}_+;H^1(\Omega;S_0(3,\mathbb{R})))\cap W^{1,q}(\Omega;S_0(3,\mathbb{R}))),\] which follows from Lemma 18, Lemma 40 and Lemma 41. More precisely, if we call \[Z=\{(z,V)\in Y(\Omega)\mid z(0)=u_0,\quad V(0)=Q_0\},\] then \(\Phi\colon Z\to Z\) is well-defined and we have \[\label{proof46gl46ex461} \|(u,Q)\|_{Y(\Omega)}\le C\left[\|u_0\|_{L^q(\Omega)\cap H^1(\Omega)} + \|Q_0\|_{W^{1,q}(\Omega)\cap H^2(\Omega)} + \|(z,V)\|_{Y(\Omega)}^2\left(1+\|(z,V)\|_{ Y(\Omega)}^3\right)\right],\tag{47}\] for some \(C>0\). Let us define now \[Z_{\varepsilon}=\{(z,V)\in Z\mid \|(z,V)\|_{Y(\Omega)}\le 2C\varepsilon\}.\] We want to prove that \(\Phi\colon Z_{\varepsilon}\to Z_{\varepsilon}\). If \((z,V)\in Z_\varepsilon\), then from 47 it holds \[\|(u,Q)\|_{Y(\Omega)}\le C\varepsilon + 4C^2\varepsilon^2(1+8C^3\varepsilon^3).\] So, if we take \(\varepsilon\) sufficiently small, we get that \(\Phi(z,V)\in Z_\varepsilon\). Similarly, if we consider \((z_1,V_1),(z_2,V_2)\in Z_{\varepsilon}\), we get that \[\label{proof46nl46gl46es} \|\Phi(z_1,V_1)-\Phi(z_2,V_2)\|_{ Y(\Omega)}\lesssim \varepsilon\left(1 + \varepsilon^3 \right)\|(z_1,V_1)-(z_2,V_2)\|_{\widetilde{Y}(\Omega)}.\tag{48}\] Therefore, for \(\varepsilon\) sufficiently small, \(\Phi\colon Z_{\varepsilon}\to Z_{\varepsilon}\) is a contraction and we get the existence of a global solution. Finally, the uniqueness can be proved as in the local case. ◻