For
part (b) we follow [
24], where the solution to a truncated problem is considered. For
\(R>1\) and
\(\zeta \in C_c^\infty ([0,1))\) with
\(0\le \zeta \le 1\) and
\(\zeta =1\) in [0, 1] we set
\(\zeta _R:=\zeta (R^{-1}\cdot )\). Similar to Definition
2.1 we seek an
\((\mathfrak {F}_t)\)-adapted stochastic process
\(\textbf{u}^R\) with
$$\begin{aligned} \textbf{u}^R \in C([0,T];L^2_{{{\,\textrm{div}\,}}}(\mathcal {O},\mathbb R^2))\cap L^2(0,T; W^{1,2}_{0,{{\,\textrm{div}\,}}}(\mathcal {O},\mathbb R^2))\quad \mathbb P\text {-a.s.} \end{aligned}$$
such that
$$\begin{aligned}&\int _{\mathcal {O}}\textbf{u}^R(t)\cdot {\varvec{\varphi }}\, \textrm{d}x=\int _{\mathcal {O}}\textbf{u}_0\cdot {\varvec{\varphi }}\, \textrm{d}x+\int _0^t \zeta _R(\Vert \nabla \textbf{u}^{R}\Vert _{L^2_x}) \int _{\mathcal {O}}\textbf{u}^{R}\otimes \textbf{u}^{R} :\nabla {\varvec{\phi }}\, \textrm{d}x\,\textrm{d}s\nonumber \\&\quad -\mu \int _0^t\int _{\mathcal {O}}\nabla \textbf{u}^R:\nabla {\varvec{\varphi }}\, \textrm{d}x\,\textrm{d}s+\int _0^t\int _{\mathcal {O}}\Phi (\textbf{u}^R)\cdot {\varvec{\varphi }}\, \textrm{d}x\,\textrm{d}W \end{aligned}$$
(3.10)
holds
\(\mathbb P\)-a.s. for all
\({\varvec{\varphi }}\in W^{1,2}_{0,{{\,\textrm{div}\,}}}(\mathcal {O},\mathbb R^2)\) and all
\(t\in [0,T]\). Arguing as in [
24, Lemma 3.7] one can show that a unique global strong pathwise solution to (
3.10) exists in the class
\(C([0,T];W^{1,2}_{0,\textrm{div}}(\mathcal {O},\mathbb R^2))\),
1 and that it satisfies
$$\begin{aligned} \begin{aligned}&\mathbb E\bigg [\sup _{0\le t\le T}\Vert \nabla \textbf{u}^R(t)\Vert _{L^2_x}^2\, \textrm{d}x+\int _{0}^T\Vert \nabla ^2\textbf{u}^R\Vert _{L^2_x}^2\, \textrm{d}t\bigg ]\le \,c(r,R,T). \end{aligned} \end{aligned}$$
(3.11)
The proof of (
3.11) in [
24] is based on a Galerkin approximation which we mimick now in order to prove (
3.8) and (
3.9).
1) Galerkin approximation. Let
\((\textbf{u}_k)\subset W^{1,2}_{0,\textrm{div}}(\mathcal {O},\mathbb R^2)\) be a system of eigenfunctions to the Stokes operator, cf. (
3.3). For
\(N\in \mathbb N\) let
\(\mathbb H^N:=\textrm{span}\{\textbf{u}_1,\dots ,\textbf{u}_N\}\), and consider the unique solution
\(\textbf{u}^{R,N}\) to
$$\begin{aligned} \begin{aligned} \int _{\mathcal {O}}\textbf{u}^{R,N}(t)\cdot {\varvec{\varphi }}\, \textrm{d}x=&\,\int _{\mathcal {O}}\textbf{u}_0\cdot {\varvec{\varphi }}\, \textrm{d}x-\mu \int _0^t\int _{\mathcal {O}}\nabla \textbf{u}^{R,N}:\nabla {\varvec{\varphi }}\, \textrm{d}x\,\textrm{d}s\\&+\int _0^t\zeta _R(\Vert \nabla \textbf{u}^{R,N}\Vert _{L^2_x})\int _{\mathcal {O}}\textbf{u}^{R,N}\otimes \textbf{u}^{R,N} :\nabla {\varvec{\phi }}\, \textrm{d}x\,\textrm{d}s\\&+\int _0^t\int _{\mathcal {O}}\Phi (\textbf{u}^{R,N})\cdot {\varvec{\varphi }}\, \textrm{d}x\,\textrm{d}W \end{aligned} \end{aligned}$$
(3.12)
for all
\({\varvec{\phi }}\in {\mathbb {H}}^N\). By
\({\mathcal {P}}_N\) we denote the
\(L^2(\mathcal {O},\mathbb R^2)\)-projection onto
\({\mathbb {H}}^N\). Problem (
3.12) can be written as a system of SDEs with Lipschitz-continuous coefficients. Hence it is clear that there is a unique strong solution, i.e., an
\(({\mathfrak {F}}_t)\)-adapted process defined on
\((\Omega ,{\mathfrak {F}},{\mathbb {P}})\) with values in
\(C([0,T];{\mathbb {H}}^N)\) and moments of order
r. Arguing as in [
24, Prop. 3.2] one can prove that as
\(N\rightarrow \infty \)$$\begin{aligned} \sup _{0\le t\le T}\Vert \textbf{u}^R(t)-\textbf{u}^{R,N}(t)\Vert _{L^2_x}^2&+\int _{0}^T\Vert \nabla (\textbf{u}^R-\textbf{u}^{R,N})\Vert _{L^2_x}^2\,\textrm{d}x\, \textrm{d}t\rightarrow 0 \end{aligned}$$
(3.13)
in probability. Applying Itô’s formula to
\(t\mapsto \Vert \textbf{u}^{R,N}\Vert _{L^2_x}^2\) and using the cancellation of the convective term one can prove for
\(r\ge 2\)$$\begin{aligned} \mathbb E\bigg [\bigg (\sup _{0\le t\le T}\Vert \textbf{u}^{R,N}(t)\Vert _{L^2_x}^2+\int _{0}^T\Vert \nabla \textbf{u}^{R,N}\Vert _{L^2_x}^2\, \textrm{d}t\bigg )^{\frac{r}{2}}\bigg ]\le \,c\,\mathbb E\Big [1+\Vert \textbf{u}_0\Vert _{L^2_x}^r\Big ], \end{aligned}$$
(3.14)
where
\(c=c(r,T)\) is independent of
N and
R.
2) Proof of (
3.8). By construction we have
\({\mathcal {A}}\textbf{u}^{R,N}\in C([0,T];{\mathbb {H}}^N)\) \(\mathbb P\)-a.s. such that we can apply Itô’s formula to
\(t\mapsto (\textbf{u}^{R,N}(t),\mathcal A\textbf{u}^{R,N}(t))_{L^2_x}\) and use (
3.12). This yields using
\(\textbf{u}^{R,N}|_{\partial \mathcal {O}}=0\)$$\begin{aligned}&\Vert \nabla \textbf{u}^{R,N}(t)\Vert _{L^2_x}^2=- \big (\textbf{u}^{R,N}(t),\Delta \textbf{u}^{R,N}(t)\big )_{L^2_x}=- \big (\textbf{u}^{R,N}(t),\mathcal A\textbf{u}^{R,N}(t)\big )_{L^2_x}\nonumber \\&\quad =\Vert \mathcal P_N\nabla \textbf{u}_0\Vert ^2_{^2_x} +2\int _0^t \zeta _R(\Vert \nabla \textbf{u}^{R,N}\Vert _{L^2_x})\big ((\textbf{u}^{R,N}\cdot \nabla )\textbf{u}^{R,N},\mathcal A\textbf{u}^{R,N}\big )_{L^2_x}\,\textrm{d}s\nonumber \\&\qquad -2\mu \int _0^t\Vert \mathcal A\textbf{u}^{R,N}\Vert _{L^2_x}^2\,\textrm{d}s+2\sum _{k=1}^N\int _0^t\big (\Phi (\textbf{u}^{R,N})e_k,\mathcal A\textbf{u}^{R,N}\big )_{L^2_x}\,\textrm{d}\beta _k\nonumber \\&\qquad +\sum _{k=1}^N\lambda _k\int _0^t\big (\Phi (\textbf{u}^{R,N})e_k,\textbf{u}_k\big )_{L^2_x}^2\,\textrm{d}s\nonumber \\&\quad =:\textrm{I}^N(t)+\dots +\textrm{V}^N(t) \end{aligned}$$
(3.15)
\({\mathbb {P}}\)-a.s. for all
\(t\in [0,T].\) We estimate now the terms
\(\textrm{II}^N\),
\(\textrm{IV}^N\) and
\(\textrm{V}^N\). First of all, we have by definition of
\(\zeta _R\)$$\begin{aligned} \textrm{II}^N(t)&\le 2\int _0^t\zeta _R(\Vert \nabla \textbf{u}^{R,N}\Vert _{L^2_x})\Vert \textbf{u}^{R,N}\Vert _{L^4_x}\Vert \nabla \textbf{u}^{R,N}\Vert _{L^4_x}\Vert \mathcal A\textbf{u}^{R,N}\Vert _{L^2_x}\textrm{d}s\\&\le 2\int _0^t\zeta _R(\Vert \nabla \textbf{u}^{R,N}\Vert _{L^2_x})\Vert \textbf{u}^{R,N}\Vert _{L^2_x}^{\frac{1}{2}}\Vert \nabla \textbf{u}^{R,N}\Vert _{L^2_x}\Vert \mathcal A\textbf{u}^{R,N}\Vert ^{\frac{3}{2}}_{L^2_x}\textrm{d}s\\&\le cR^{3/2}\int _0^t\Vert \mathcal A\textbf{u}^{R,N}\Vert ^{\frac{3}{2}}_{L^2_x}\textrm{d}s\le \delta \int _0^t\Vert \mathcal A\textbf{u}^{R,N}\Vert ^{2}_{L^2_x}\textrm{d}s+c(\delta )R^6, \end{aligned}$$
where
\(\delta >0\) is arbitrary. Moreover, we obtain by definition of
\(\textbf{u}_k\) and using (
3.6) (and recalling that
\(\Phi (\textbf{u}^{R,N})e_k\in W^{1,2}_{0,\textrm{div}}(\mathcal {O},\mathbb R^2)\) for all
\(k\in \mathbb N\) by assumption)
$$\begin{aligned} \textrm{V}^N(t){} & {} =\sum _{k=1}^N\int _0^t\big (\Phi (\textbf{u}^{R,N})e_k,\sqrt{\lambda _k}\textbf{u}_k\big )_{L^2_x}^2\,\textrm{d}s=\sum _{k=1}^N\int _0^t\big (\Phi (\textbf{u}^{R,N})e_k, \mathcal A^{1/2}\textbf{u}_k\big )^2\,\textrm{d}s \\{} & {} =\sum _{k=1}^N\int _0^t\big (\mathcal A^{1/2}\Phi (\textbf{u}^{R,N})e_k,\textbf{u}_k\big )_{L^2_x}^2\,\textrm{d}s. \end{aligned}$$
Furthermore, since
\(\Vert \textbf{u}_k\Vert _{L^2_x}=1\),
$$\begin{aligned} \textrm{V}^N(t){} & {} \le \sum _{k\ge 1}\int _0^t\Vert \mathcal A^{1/2}\Phi (\textbf{u}^{R,N})e_k\Vert _{L^2_x}^2\Vert \textbf{u}_k\Vert ^2_{L^2_x}\,\textrm{d}s \le \,c\sum _{k\ge 1}\int _0^t\Vert \nabla \Phi (\textbf{u}^{R,N})e_k\Vert _{L^2_x}^2\,\textrm{d}s\\{} & {} =c\int _0^t\Vert \Phi (\textbf{u}^{R,N})\Vert ^2_{L_2({\mathfrak {U}};W^{1,2}_x)}\,\textrm{d}s\le \,c\int _0^t\big (1+\Vert \textbf{u}^{R,N}\Vert ^2_{W^{1,2}_x}\big )\,\textrm{d}s, \end{aligned}$$
using (
2.2) in the last step. The expectation of the right-hand side is bounded by (
3.14). Finally, by Burkholder-Davis-Gundy inequality and (
2.1),
$$\begin{aligned}{} & {} \mathbb E\bigg [\bigg (\sup _{0\le t\le T}|\textrm{IV}(t)|\bigg )^{\frac{r}{2}}\bigg ]\\{} & {} \quad \le \mathbb E\bigg [\bigg (\sup _{0\le t \le T}\Big |\int _0^t\sum _{k=1}^N\big (\Phi (\cdot ,\textbf{u}^{R,N})e_k ,{\mathcal {A}}\textbf{u}^{R,N}\big )_{L^2_x}\,\mathrm d\beta _k\Big |\bigg )^{\frac{r}{2}}\bigg ]\\{} & {} \quad \le c\,\mathbb E\bigg [\bigg (\sum _{k\ge 1}\int _0^T\big ( \Phi (\cdot ,\textbf{u}^{R,N})e_k \cdot {\mathcal {A}}\textbf{u}^{R,N}\big )_{L^2_x}^2\, \textrm{d}t\bigg )^{\frac{r}{4}}\bigg ]\\{} & {} \quad \le c\,\mathbb E\bigg [\bigg (\sum _{k\ge 1}\int _0^T \Vert \Phi _k(\textbf{u}^{R,N})e_k\Vert _{L^2_x}^2\Vert {\mathcal {A}}\textbf{u}^{R,N}\Vert ^2_{L^2_x}\, \textrm{d}t\bigg )^{\frac{r}{4}}\bigg ]\\{} & {} \quad \le c\,\mathbb E\bigg [\bigg (\int _0^T \big (1+\Vert \textbf{u}^{R,N}\Vert _{L^2_x}^2\big )\Vert \mathcal A\textbf{u}^{R,N}\Vert ^2_{L^2_x}\, \textrm{d}t\bigg )^{\frac{r}{4}}\bigg ]\\{} & {} \quad \le c(\delta )\,\mathbb E\bigg [\bigg (1+\sup _{0\le t\le T}\Vert \textbf{u}^{R,N}\Vert _{L^2_x}^2\bigg )^{\frac{r}{2}}\bigg ]+ \delta \,\mathbb E\bigg [\bigg (\int _0^T\Vert \mathcal A\textbf{u}^{R,N}\Vert _{L^2_x}^2\, \textrm{d}t\bigg )^{\frac{r}{2}}\bigg ]\\{} & {} \quad \le c(\delta )+\delta \,\mathbb E\bigg [\bigg (\int _0^T\Vert \mathcal A\textbf{u}^{R,N}\Vert _{L^2_x}^2\, \textrm{d}t\bigg )^{\frac{r}{2}}\bigg ] \end{aligned}$$
using (
3.14), where again
\(\delta >0\) is arbitrary. Choosing
\(\delta \) small enough and using (
3.4) we conclude that
$$\begin{aligned} \mathbb E\bigg [\bigg (\sup _{0\le t\le T}\int _{\mathcal {O}}\Vert \nabla \textbf{u}^{R,N}(t)\Vert _{L^2_x}^2{} & {} +\int _{0}^T\Vert \nabla ^2\textbf{u}^{R,N}\Vert _{L^2_x}^2\, \textrm{d}t\bigg )^{\frac{r}{2}}\bigg ] \nonumber \\{} & {} \le \,cR^{3r}\,\mathbb E\Big [1+\Vert \textbf{u}_0\Vert _{W^{1,2}_x}^r\Big ]\,, \end{aligned}$$
(3.16)
uniformly in
N. This implies that
\((\textbf{u}^{R,N})_{N\in \mathbb N}\) is a bounded sequence in the function space generated by the left-hand side of (
3.16). After taking a subsequence we obtain a limit object
\(\textbf{u}^R\) which is the unique global strong solution to (
3.10) recalling (
3.13). Furthermore, we can pass to the limit
\(N\rightarrow \infty \) and obtain a corresponding estimate for
\(\textbf{u}^R\) due to lower semi-continuity of the involved functionals. Since
\(\textbf{u}^R(\cdot \wedge \mathfrak t_R)=\textbf{u}(\cdot \wedge {\mathfrak {t}}_R)\) we obtain (
3.8).
3) Proof of (
3.9). The verification of
part (c) proceeds in two steps. In the first step we show an improved version of (
3.16). Applying Itô’s formula to the mapping
$$\begin{aligned} t\mapsto \Vert \nabla \textbf{u}^{R,N}(t)\Vert _{L^2_x}^2\big (\textbf{u}^{R,N}(t),\mathcal A\textbf{u}^{R,N}(t)\big )_{L^2_x}, \end{aligned}$$
equation (
3.15) yields
$$\begin{aligned} \Vert \nabla \textbf{u}^{R,N}(t)\Vert _{L^2_x}^4{} & {} =\Vert \mathcal P_N\nabla \textbf{u}_0\Vert _{L^2_x}^4-4\mu \int _0^t\Vert \nabla \textbf{u}^{R,N}\Vert _{L^2_x}^2\Vert \mathcal A\textbf{u}^{R,N}\Vert _{L^2_x}^2\,\textrm{d}s\\{} & {} \quad +4\int _0^t \zeta _R(\Vert \nabla \textbf{u}^{R,N}\Vert _{L^2_x})\Vert \nabla \textbf{u}^{R,N}\Vert _{L^2_x}^2\big ((\textbf{u}^{R,N}\cdot \nabla )\textbf{u}^{R,N},\mathcal A\textbf{u}^{R,N}\big )_{L^2_x}\,\textrm{d}s\\{} & {} \quad +4\sum _{k=1}^N\int _0^t\Vert \nabla \textbf{u}^{R,N}\Vert _{L^2_x}^2\big (\Phi (\textbf{u}^{R,N})e_k,\mathcal A\textbf{u}^{R,N}\big )_{L^2_x}\,\textrm{d}\beta _k\\{} & {} \quad +2\sum _{k=1}^N\lambda _k\int _0^t\Vert \nabla \textbf{u}^{R,N}\Vert _{L^2_x}^2\big (\Phi (\textbf{u}^{R,N})e_k,\textbf{u}_k\big )_{L^2_x}^2\,\textrm{d}s\\{} & {} \quad +2\sum _{k=1}^N\int _0^t\big (\Phi (\cdot ,\textbf{u}^{R,N})e_k ,\mathcal A\textbf{u}^{R,N}\big )_{L^2_x}^2\, \textrm{d}t. \end{aligned}$$
Following now step by step the arguments from the proof of (
3.16) above we arrive at
$$\begin{aligned} \mathbb E\bigg [\bigg (\sup _{0\le t\le T}\Vert \nabla \textbf{u}^{R,N}(t)\Vert _{L^2_x}^4&+\int _0^T\Vert \nabla \textbf{u}^{R,N}\Vert _{L^2_x}^2\Vert \nabla ^2\textbf{u}^{R,N}\Vert _{L^2_x}^2\, \textrm{d}t\bigg )^{\frac{r}{2}}\bigg ]\\ {}&\le \,cR^{3r}\,\mathbb E\Big [1+\Vert \textbf{u}_0\Vert _{W^{1,2}_x}^{2r}\Big ]. \end{aligned}$$
Again we can pass to the limit in
N obtaining
$$\begin{aligned} \begin{aligned} \mathbb E\bigg [\bigg (\sup _{0\le t\le T}\Vert \nabla \textbf{u}^{R}(t)\Vert _{L^2_x}^4&+\int _0^T\Vert \nabla \textbf{u}^{R}\Vert _{L^2_x}^2\Vert \nabla ^2\textbf{u}^{R}\Vert _{L^2_x}^2\, \textrm{d}t\bigg )^{\frac{r}{2}}\bigg ]\\ {}&\le \,cR^{3r}\,\mathbb E\Big [1+\Vert \textbf{u}_0\Vert _{W^{1,2}_x}^{2r}\Big ]. \end{aligned} \end{aligned}$$
(3.17)
Now we turn to the proof of (
3.9) stated in
part (c) for which we use the mild formulation of (
3.10).
(
\(\textbf{c}_1\)) Due to the regularity proved in (
3.16) and (
3.17), [
25, Proposition F.0.5, (i)] applies and we can write
$$\begin{aligned} \textbf{u}^{R}(t)&=e^{-t\mathcal A}\textbf{u}_0+\int _0^te^{-(t-s)\mathcal A}\textbf{g}_R\,\textrm{d}s+\int _0^t e^{-(t-s)\mathcal A}\Phi (\textbf{u}^{R})\,\mathrm dW,\\ \text {where}\quad \textbf{g}_R:&= \zeta _R(\Vert \nabla \textbf{u}^{R}\Vert _{L^2_x})\mathcal P[(\textbf{u}^{R}\cdot \nabla )\textbf{u}^{R}]. \end{aligned}$$
Here
\((e^{-t\mathcal A})_{t\ge 0}\) denotes the analytic semigroup on
\(L^2_{\textrm{div}}(\mathcal {O},\mathbb R^2)\) generated by the Stokes operator
\(\mathcal A\). Setting
$$\begin{aligned} \textbf{Y}^{R}(t)&:=e^{-t\mathcal A}\textbf{u}_0+\int _0^te^{-(t-s)\mathcal A} \textbf{g}_R\,\textrm{d}s,\\ \textbf{Z}^{R}(t)&:=\int _0^t e^{-(t-s)\mathcal A}\Phi (\textbf{u}^{R})\,\mathrm dW, \end{aligned}$$
we consider now the deterministic and stochastic contribution separately. We note that
\(\textbf{Y}^R\) is the unique solution to a deterministic Stokes problem with initial datum
\(\textbf{u}_0\) and forcing
\(\textbf{g}_R\), whereas
\(\textbf{Z}^R\) solves a stochastic Stokes problem with homogeneous initial datum and diffusion coefficient
\(\Phi (\textbf{u}^R)\) – both equipped with homogeneous Dirichlet boundary conditions.
(
\(\textbf{c}_2\)) Interpolating
\(W^{1/2,2}(0,T;W^{1,2}(\mathcal {O},\mathbb R^2))\) between
\(W^{1,2}(0,T;L^2(\mathcal {O},\mathbb R^2))\) and
\(L^2(0,T;W^{2,2}(\mathcal {O},\mathbb R^2))\) and applying
\({\mathbb {P}}\)-a.s. classical estimates for the Stokes system yields
$$\begin{aligned} \begin{aligned}&\mathbb E\bigg [\bigg (\Vert \textbf{Y}^R\Vert _{W^{1/2}(0,T;W^{1,2}_x)}^2\, \textrm{d}t\bigg )^{\frac{r}{2}}\bigg ]\\ {}&\quad \le \mathbb E\bigg [\bigg (\Vert \textbf{Y}^R\Vert _{W^{1,2}(0,T;L^{2}_x)}^2+\Vert \textbf{Y}^R\Vert _{L^{2}(0,T;W^{2,2}_x)}^2\, \textrm{d}t\bigg )^{\frac{r}{2}}\bigg ]\\&\quad \le \,c\,\mathbb E\bigg [\bigg (\int _0^T\Vert \textbf{g}_R\Vert _{L^2_x}^2\, \textrm{d}t\bigg )^{\frac{r}{2}}\bigg ]\le \,cR^{3r}\,\mathbb E\Big [1+\Vert \textbf{u}_0\Vert _{W^{1,2}_x}^{2r}\Big ]. \end{aligned} \end{aligned}$$
(3.18)
(
\(\textbf{c}_3\)) For
\(\textbf{Z}^R\) we apply the recent results from [
30, Theorems 25 and 28] proving for any
\(\sigma <1\)$$\begin{aligned} \begin{aligned}&\mathbb E\bigg [\Vert \textbf{Z}^R\Vert ^2_{C^{\sigma /2}([0,T];L^2_x)}+\Vert \textbf{Z}^R\Vert _{W^{\sigma /2,2}(0,T;W^{1,2}_x)}^2\bigg ]\\ {}&\quad \le \,c\,\mathbb E\bigg [1+\sup _{0\le t\le T}\Vert \textbf{u}^R\Vert ^{4}_{W^{1,2}_x}\bigg ]\\&\quad \le \,c\,\mathbb E\bigg [1+\Vert \textbf{u}_0\Vert ^{4}_{W^{1,2}_x}\bigg ] \end{aligned} \end{aligned}$$
(3.19)
using also (
2.2) and (
3.17). Combining (
3.18) and (
3.19) and recalling that
\(\textbf{u}^R\) is the sum of
\(\textbf{Y}^R\) and
\(\textbf{Z}^R\) gives
$$\begin{aligned} \begin{aligned} \mathbb E\bigg [\Vert \textbf{u}^R\Vert ^2_{C^{\sigma /2}([0,T];L^2_x)}+\Vert \textbf{u}^R\Vert _{W^{\sigma /2,2}(0,T;W^{1,2}_x)}^2\bigg ]&\le \,c\,\mathbb E\bigg [1+\Vert \textbf{u}_0\Vert ^{4}_{W^{1,2}_x}\bigg ]. \end{aligned} \end{aligned}$$
(3.20)
(
\(\textbf{c}_4\)) Due to our assumption on the noise from (
2.4) we know that
\(\Phi (\textbf{u}^R)e_k\), with
\(k\in \mathbb N\), belongs to the domain of the Stokes operator such that we can write
$$\begin{aligned} \mathcal A\textbf{Z}^{R}(t)=\int _0^t e^{-(t-s)\mathcal A}\mathcal A\Phi (\textbf{u}^{R})\,\mathrm dW. \end{aligned}$$
We conclude that
\(\mathcal A\textbf{Z}^R\) is the unique weak pathwise solution to the stochastic Stokes problem with zero initial datum, homogeneous boundary conditions and diffusion coefficient
\(\mathcal A\Phi (\textbf{u}^R)\). It is standard to derive for
\(r\ge 2\) the estimate
$$\begin{aligned}&\mathbb E\bigg [\bigg (\sup _{0\le t\le T}\Vert \mathcal A\textbf{Z}^{R}\Vert ^2_{L^{2}_x}+\int _0^T\Vert \nabla \mathcal A\textbf{Z}^R\Vert ^2_{L^2_x}\,\textrm{d}s\bigg )^{\frac{r}{2}}\bigg ]\\&\quad \le \,c\,\mathbb E\bigg [\bigg (\int _0^T\Vert \mathcal A\Phi (\textbf{u}^{R})\Vert ^2_{L_2(\mathfrak U;L^2_x)}\,\mathrm ds\bigg )^{\frac{r}{2}}\bigg ]\\&\quad \le \,c\,\mathbb E\bigg [\bigg (\int _0^T\Vert \Phi (\textbf{u}^{R})\Vert ^2_{L_2(\mathfrak U;W^{2,2}_x)}\,\mathrm ds\bigg )^{\frac{r}{2}}\bigg ]\\&\quad \le \,c\bigg [\bigg (\int _0^{t}\big (1+\Vert \textbf{u}^{R}\Vert _{W^{1,2}_x}^2\Vert \textbf{u}^{R}\Vert ^2_{W^{2,2}_x}+\Vert \textbf{u}^{R}\Vert ^2_{W^{2,2}_x}\big )\,\textrm{d}s\bigg )^{\frac{r}{2}}\bigg ], \end{aligned}$$
applying Itô’s formula to
\(t\mapsto \Vert \mathcal A\textbf{u}^R\Vert _{L^2_x}^2\) and using Burkholder-Davis-Gundy inequality (and (
2.4) in the last step). The properties of the Stokes operator from (
3.4) yield
$$\begin{aligned}&\mathbb E\bigg [\bigg (\sup _{0\le t\le T}\Vert \textbf{Z}^{R}\Vert ^2_{W^{2,2}_x}+\int _0^T\Vert \textbf{Z}^R\Vert ^2_{W^{3,2}_x}\,\textrm{d}s\bigg )^{\frac{r}{2}}\bigg ]\\&\quad \le \,c\bigg [\bigg (\int _0^{t}\big (1+\Vert \textbf{u}^{R}\Vert _{W^{1,2}_x}^2\Vert \textbf{u}^{R}\Vert ^2_{W^{2,2}_x}+\Vert \textbf{u}^{R}\Vert ^2_{W^{2,2}_x}\big )\,\textrm{d}s\bigg )^{\frac{r}{2}}\bigg ]. \end{aligned}$$
(
\(\textbf{c}_5\)) To sharpen the estimates for
\(\textbf{Y}^R\) is slightly more involved as the convective term
\(\textbf{g}_R\) does not lie in the domain of the Stokes operator since it does not necessarily have a zero trace. We can choose
\(p<2\) such that the embedding
\(W^{1,p}(\mathcal {O})\hookrightarrow W^{\sigma ,2}(\mathcal {O})\) holds. We obtain by continuity of
\({\mathcal {P}}\), cf. (
3.2),
$$\begin{aligned} \Vert \textbf{g}_R\Vert _{W^{\sigma ,2}_x}&\le \,c\,\Vert \textbf{g}_R\Vert _{W^{1,p}_x}\le \,c \Vert (\textbf{u}^{R}\cdot \nabla )\textbf{u}^{R}\Vert _{W^{1,p}_x}\\&\le \,c\Vert \nabla \textbf{u}^R\Vert _{L^{2p}_x}^2+c\Vert \textbf{u}^R\Vert _{L^q_x}\Vert \nabla ^2\textbf{u}_R\Vert _{L^2_x}\le \,c\Vert \nabla \textbf{u}^R\Vert _{L^2_x}\Vert \nabla ^2\textbf{u}_R\Vert _{L^2_x}, \end{aligned}$$
where we used Hölder’s inequality with exponents 2/
p and
\(q:=2/(2-p)\) as well as Sobolev’s embedding
\(W^{1,2}(\mathcal {O},\mathbb R^2)\hookrightarrow L^q(\mathcal {O},\mathbb R^2)\) and Ladyshenskaya’s inequality. By (
3.17) we conclude that
$$\begin{aligned} \textbf{g}_R\in L^2(0,T;W^{\sigma ,2}(\mathcal {O},\mathbb R^2))\quad {\mathbb {P}}\text {-a.s.} \end{aligned}$$
(3.21)
We argue now similarly for the temporal regularity of order
\(\sigma /2\) obtaining for any
\(\sigma '\in (\sigma ,1)\)$$\begin{aligned}&\Vert \textbf{g}_R\Vert _{W^{\sigma /2,p}(0,T;L^p_x)}^p\\ {}&\quad \le \,c\int _0^T\int _0^T\frac{\Vert \textbf{u}^R(t)\nabla \textbf{u}^R(t)-\textbf{u}^R(s)\nabla \textbf{u}^R(s)\Vert _{L^p_x}^p}{|t-s|^{1+p\sigma /2}}\, \textrm{d}t\,\textrm{d}s\\&\quad \le \,c\int _0^T\int _0^T\bigg (\frac{\Vert \textbf{u}^R(t)-\textbf{u}^R(s)\Vert _{L^2_x}}{|t-s|^{\sigma '/2}}\Vert \nabla \textbf{u}^R(t)\Vert _{L^q_x}\bigg )^p\frac{\, \textrm{d}t\,\textrm{d}s}{|t-s|^{1+\frac{(\sigma -\sigma ')p}{2}}}\\&\qquad + \,c\int _0^T\int _0^T\bigg (\frac{\Vert \textbf{u}^R(s)\Vert _{L^q_x}\Vert \nabla \textbf{u}^R(t)-\nabla \textbf{u}^R(s)\Vert _{L^2_x}}{|t-s|^{\sigma /2}}\bigg )^p\frac{\, \textrm{d}t\,\textrm{d}s}{|t-s|}\\&\quad \le \,c\Vert \textbf{u}^R\Vert ^p_{C^{\sigma '/2}([0,T];L^2_x)}\int _0^T\Vert \nabla \textbf{u}^R(t)\Vert _{L^q_x}^p\, \textrm{d}t\\&\qquad + \,c\sup _{0\le s\le t}\Vert \textbf{u}^R(s)\Vert _{L^q_x}^p\int _0^T\int _0^T\frac{\Vert \nabla \textbf{u}^R(t)-\nabla \textbf{u}^R(s)\Vert _{L^2_x}^p}{|t-s|^{1+p\sigma /2}}\, \textrm{d}t\,\textrm{d}s\\&\quad \le \,c\Vert \textbf{u}^R\Vert ^p_{C^{\sigma '/2}([0,T];L^2_x)}\int _0^T\big (1+\Vert \textbf{u}^R(t)\Vert _{W^{2,2}_x}^2\big )\, \textrm{d}t\\&\qquad + \,c\sup _{0\le s\le t}\Vert \textbf{u}^R(s)\Vert _{W^{1,2}_x}^p\Vert \textbf{u}^R\Vert _{W^{\sigma /2,p}(0,T;W^{1,2}_x)}^p\\&\quad \le \,c\bigg (\Vert \textbf{u}^R\Vert ^2_{C^{\sigma '/2}([0,T];L^2_x)}+\Vert \textbf{u}^R\Vert _{W^{\sigma '/2,2}(0,T;W^{1,2}_x)}^2+1\bigg )\\&\qquad +c\bigg (\,\sup _{0\le s\le t}\Vert \textbf{u}^R(s)\Vert _{W^{1,2}_x}^2+\int _0^T\Vert \textbf{u}^R\Vert _{W^{2,2}_x}^2\, \textrm{d}t\bigg )^q. \end{aligned}$$
The expectation of the right-hand side is bounded using (
3.9) and (
3.20); in particular, for any
\(\sigma <1\)$$\begin{aligned} \textbf{g}_R\in W^{\sigma /2,2}(0,T;L^{2}(\mathcal {O},\mathbb R^2))\quad \mathbb P\text {-a.s.} \end{aligned}$$
(3.22)
using the embedding decreasing the value of
\(\sigma \) and using
\(W^{\sigma /2,p}(0,T)\hookrightarrow W^{\sigma '/2,2}(0,T)\) for an appropriate choice of
\(\sigma >\sigma '\) and
\(p<2\). By (
3.21) and (
3.22) classical results on the Stokes system (see [
28, Thm. 15] and note the compability assumption
\({\mathcal {A}}\textbf{u}_0-\mathcal P(\textbf{u}_0\cdot \nabla \textbf{u}_0)|_{\partial {\mathcal {O}}}=0\) \(\mathbb P\)-a.s.) and interpolation yield
$$\begin{aligned} \textbf{Y}^R\in W^{1+\sigma /2}(0,T;L^{2}(\mathcal {O},\mathbb R^2))\cap L^2(0,T;W^{2+\sigma ,2}(\mathcal {O},\mathbb R^2))\quad {\mathbb {P}}\text {-a.s.} \end{aligned}$$
and thus, again by interpolation and appropriate choice of
\(\sigma \in (\beta ,1)\) and the embedding
\(W^{\alpha ,2}(0,T)\hookrightarrow L^\infty (0,T)\) for
\(\alpha >1/2\),
$$\begin{aligned} \textbf{Y}^R\in L^\infty (0,T;W^{1+\beta ,2}(\mathcal {O},\mathbb R^2))\cap L^2(0,T;W^{2+\beta ,2}(\mathcal {O},\mathbb R^2))\quad {\mathbb {P}}\text {-a.s.} \end{aligned}$$
(3.23)
together with
$$\begin{aligned}&\sup _{0\le t\le T}\Vert \textbf{Y}^{R}\Vert _{W^{1+\beta ,2}_x}^2+\int _0^T\Vert \textbf{Y}^{R}\Vert _{W^{2+\beta ,2}_x}^2\,\mathrm ds\\&\quad \le \,c\bigg [\Vert \textbf{u}_0\Vert _{W^{1+\sigma ,2}_x}^2+\Vert \textbf{g}_R\Vert ^2_{W^{\sigma /2,2}_t(L^2_x)}+\int _0^T\Vert \textbf{g}_R\Vert _{W^{\sigma ,2}_x}^2\,\mathrm ds\bigg ]\quad {\mathbb {P}}\text {-a.s.} \end{aligned}$$
Combining the estimates for
\(\textbf{Y}^{R}\) and
\(\textbf{Z}^{R}\), choosing
\(\kappa \) sufficiently small and using (
3.16) and (
3.17) we arrive at
$$\begin{aligned} \mathbb E\bigg [\bigg (\sup _{0\le t\le T}\Vert \textbf{u}^{R}(t)\Vert _{W^{1+\sigma ,2}_x}^2\, \textrm{d}x&+\int _0^{T}\Vert \textbf{u}^{R}\Vert ^2_{W^{2+\sigma ,2}_x}\, \textrm{d}t\bigg )^{\frac{r}{2}}\bigg ]\\ {}&\le \,cR^{5r}\,\mathbb E\Big [1+\Vert \textbf{u}_0\Vert _{W^{2,2}_x}^r+\Vert \textbf{u}_0\Vert _{W^{1,2}_x}^{2r}\Big ]. \end{aligned}$$
uniformly in
R.
\(\square \)