Strong $L^2$ convergence of time Euler schemes for stochastic 3D Brinkman–Forchheimer–Navier–Stokes equations

Abstract

We prove that some time Euler schemes for the 3D Navier–Stokes equations modified by adding a Brinkman–Forchheimer term and a random perturbation converge in $L^2(\varOmega )$. This extends previous results concerning the strong rate of convergence of some time discretization schemes for the 2D Navier Stokes equations. Unlike the 2D case, our proposed 3D model with the Brinkman–Forchheimer term allows for a strong rate of convergence of order almost 1/2, that is independent of the viscosity parameter.

Appendix

In this section, we provide the proof of the well-posedness result stated in Sect. 3.

1.1 Proofs of preliminary estimates

The following results gather some estimates of the bilinear term, and more generally of the non linear part in (1.1). They are deduced from the Brinkman–Forchheimer smoothing term. The proofs are somewhat similar to the corresponding ones in [6] in a different functional setting.

The next lemma gathers further properties of B.

Lemma 3

Suppose that $\alpha \in [1,+\infty )$.

• (i) Let $u\in L^\infty (0,T;H)\cap L^{2\alpha +2}([0,T]\times D;{{\mathbb {R}}}^3)$, $v\in X_0$. Then

  $$\begin{aligned}&\int _0^T \! \big | \langle B(u(t), u(t)) , v(t)\rangle \big | dt \nonumber \\&\quad \le \Vert \nabla v\Vert _{L^2(0,T;{{\mathbb {L}}}^2)} \; \underset{{t\in [0,T]}}{\mathrm{ess \,sup }} \Vert u(t)\Vert _{{{\mathbb {L}}}^2}^{\frac{\alpha -1}{\alpha }}\; \Vert u\Vert _{L^{2\alpha +2}([0,T]\times D;{{\mathbb {R}}}^3)}^{\frac{\alpha +1}{\alpha }} \; T^{\frac{\alpha -1}{2\alpha }}. \end{aligned}$$

  (7.1)
  $$\begin{aligned}&\int _0^T\! \big | \langle B(u(t), u(t)) - B(v(t),v(t)) , u(t)- v(t)\rangle \big | dt \le \Vert \nabla v\Vert _{L^2(0,T;H)} \nonumber \\&\quad \times \; \underset{{t\in [0,T]}}{\mathrm{ess\, sup }} \Vert (u-v)(t)\Vert _{H}^{\frac{\alpha -1}{\alpha }}\; \Vert u-v\Vert _{L^{2\alpha +2}([0,T]\times D;{{\mathbb {R}}}^3)}^{\frac{\alpha +1}{\alpha }} \; T^{\frac{\alpha -1}{2\alpha }}. \end{aligned}$$

  (7.2)

• (ii) Let $u\in L^4(\varOmega ; L^\infty (0,T;H))\cap L^{2\alpha +2}(\varOmega _T\times D;{{\mathbb {R}}}^3)$ and $v\in {{\mathcal {X}}}_0$. Then

  $$\begin{aligned}&{{\mathbb {E}}}\int _0^T \! \big | \langle B(u(t), u(t)) , v(t)\rangle \big | dt \le \Big \{ {{\mathbb {E}}}\Big | \int _0^T \Vert \nabla v(t)\Vert _{{{\mathbb {L}}}^2}^2 dt \Big |^2 \Big \}^{\frac{1}{4}} \nonumber \\&\quad \times \Big \{ {{\mathbb {E}}}\Big ( \underset{{t\in [0,T]}}{\mathrm{ess\, sup }} \Vert u(t)\Vert _{H}^{4}\Big ) \Big \}^{ \frac{\alpha -1}{4\alpha } } \Big \{ {{\mathbb {E}}}\int _0^T \!\! dt \int _{D} \! |u(t,x)|^{2\alpha +2} dx \Big \}^{\frac{1}{2\alpha }} \; T^{\frac{\alpha -1}{2\alpha }}, \end{aligned}$$

  (7.3)
  $$\begin{aligned}&{{\mathbb {E}}}\int _0^T\! \big | \langle B(u(t), u(t)) - B(v(t),v(t)) , u(t)- v(t)\rangle \big | dt \nonumber \\&\quad \le T^{\frac{\alpha -1}{2\alpha }} \, \Big \{ {{\mathbb {E}}}\Big | \int _0^T \! \Vert \nabla v(t)\Vert _{{{\mathbb {L}}}^2}^2 dt \Big |^2 \Big \}^{\frac{1}{4}} \Big \{ {{\mathbb {E}}}\Big ( \underset{{t\in [0,T]}}{\mathrm{ess\, sup }} \Vert (u-v)(t)\Vert _{H}^{4}\Big ) \Big \}^{ \frac{\alpha -1}{4\alpha } } \nonumber \\&\qquad \times \; \Big \{ {{\mathbb {E}}}\! \int _0^T \!\! dt\! \int _{D} \! |(u-v)(t,x)|^{2\alpha +2} dx \Big \}^{\frac{1}{2\alpha }} . \end{aligned}$$

  (7.4)

Proof

• (i) Suppose $\alpha >1$. Using (2.4) with $h=\partial _i v_j$, $f=u_i$ and $g=u_j$, we deduce

  $$\begin{aligned} |\langle B(u,u),v\rangle |&= |-\langle B(u,v),u\rangle | \le \sum _{i,j=1}^3 \int _{D} |u_i(x) \partial _i v_j(x) u_j(x)| dx \\&\le \big \Vert |u| \, |u|^{\frac{1}{\alpha }}\big \Vert _{{{\mathbb {L}}}^{2\alpha }} \; \big \Vert |u|^{1-\frac{1}{\alpha }} \big \Vert _{{{\mathbb {L}}}^{\frac{2\alpha }{\alpha -1}}}\; \Vert \nabla v\Vert _{{{\mathbb {L}}}^2}. \end{aligned}$$

  Integrating on the time interval [0, T] and using the Cauchy–Schwarz inequality, we obtain

  $$\begin{aligned} \int _0^T \! \big | \langle B(u(t), u(t)) , v(t)\rangle \big | dt&\le \; \underset{{t\in [0,T]}}{\mathrm{ess\, sup }} \Vert u(t)\Vert _{H}^{\frac{\alpha -1}{\alpha }}\; \Big ( \int _0^T\!\! \Vert u(t)\Vert _{{{\mathbb {L}}}^{2\alpha +2}}^{\frac{2\alpha +2}{\alpha }} dt \Big )^{\frac{1}{2}} \\&\quad \; \times \Big ( \int _0^T \!\! \Vert \nabla v(t)\Vert _{{{\mathbb {L}}}^2}^2 dt \Big )^{\frac{1}{2}}. \end{aligned}$$

  Hölder’s inequality implies

  $$\begin{aligned} \int _0^T\!\! \Vert u(t)\Vert _{{{\mathbb {L}}}^{2\alpha +2}}^{\frac{2\alpha +2}{\alpha }} dt \le \Vert u\Vert _{L^{2\alpha +2}([0,T]\times D;{{\mathbb {R}}}^3)}^{\frac{2\alpha +2}{\alpha }}\; T^{\frac{\alpha -1}{\alpha }}. \end{aligned}$$

  This completes the proof of (7.1) for $\alpha >1$.

  If $\alpha =1$, since $|\langle B(u,u),v\rangle | \le \big \Vert u\big \Vert _{{{\mathbb {L}}}^4}^2 \; \Vert \nabla v\Vert _{{{\mathbb {L}}}^2}$, a straightforward computation implies (7.1).

  Since $\langle B(u,u)-B(v,v) \, , \, u-v\rangle = \langle B(u-v,v)\, , \, u-v\rangle$, using the antisymmetry (2.1) it is easy to see that the upper estimate (7.1) implies (7.2).

• (ii) For $\alpha>1>\frac{2}{3}$, we have $\frac{4\alpha }{3\alpha -2}>1$. Using Hölder’s inequality for the expected value with exponents 4, $\frac{4\alpha }{3\alpha -2}$ and $2\alpha$ in (7.1), we deduce

  $$\begin{aligned} {{\mathbb {E}}}\int _0^T \! \big | \langle B(u(t), u(t)) , v(t)\rangle \big | dt&\le \Big \{ {{\mathbb {E}}}\Big ( \Vert \nabla v\Vert _{L^2(0,T;{{\mathbb {L}}}^2)}^4 \Big )\Big \} ^{\frac{1}{4 }} \\&\quad \times \Big \{ {{\mathbb {E}}}\Big ( \underset{{t\in [0,T]}}{\mathrm{ess sup }} \Vert u(t)\Vert _{{{\mathbb {L}}}^2}^{\frac{4(\alpha -1)}{3\alpha -2}} \Big ) \Big \}^{\frac{3\alpha -2}{4\alpha } } \\&\quad \times \Big \{ E \int _0^T dt \int _{D} |u(t,x)|^{2\alpha +2} \, dx \Big \}^{\frac{1}{2\alpha }}\; T^{\frac{\alpha -1}{2\alpha }} . \end{aligned}$$

  Since $\alpha >\frac{1}{2}$ we have $\frac{4(\alpha -1)}{3\alpha -2} < 4$; this completes the proof of (7.3) for $\alpha >1$.

For $\alpha =1$, using the antisymmetry (2.1), and twice the Cauchy–Schwarz inequality, we deduce

$$\begin{aligned}&{{\mathbb {E}}}\int _0^T \! \big | \langle B(u(t), u(t)) , v(t)\rangle \big | dt \le \Big \{ {{\mathbb {E}}}\int _0^T \Vert \nabla v(t)\Vert _{{{\mathbb {L}}}^2}^2dt \Big \}^{\frac{1}{2}} \Big \{ {{\mathbb {E}}}\int _0^T \Vert u(t)\Vert _{{{\mathbb {L}}}^4}^4dt \Big \}^{\frac{1}{2}} \\&\quad \le \Big \{ {{\mathbb {E}}}\Big | \int _0^T \Vert \nabla v(t)\Vert _{{{\mathbb {L}}}^2}^2 dt \Big |^2 \Big \}^{\frac{1}{4}} \Big \{ {{\mathbb {E}}}\int _0^T \Vert u(t)\Vert _{{{\mathbb {L}}}^4}^4dt \Big \}^{\frac{1}{2}}. \end{aligned}$$

This completes the proof of (7.3).

A similar argument based on the identity $\langle B(u,u)-B(v,v) \, , \, u-v\rangle = \langle B(u-v,v)\, , \, u-v\rangle$ shows (7.4). $\square$

We next prove upper estimates for the gradient of the bilinear term.

Lemma 4

• (i) There exists a positive constant C such that for $\alpha \in (1,\infty )$, some constant $C_\alpha >0$, any constants $\varepsilon _0, \varepsilon _1 >0$ we have for $u\in X_1$,

  $$\begin{aligned} |\langle A^{1/2} B(u,u)\, , \, A^{1/2} u\rangle |\le & {} C \Big [ \varepsilon _0 \Vert Au\Vert _{{{\mathbb {L}}}^2}^2 + \frac{\varepsilon _1}{4\varepsilon _0} \big \Vert |u|^\alpha \nabla u \big \Vert _{{{\mathbb {L}}}^2}^2 \nonumber \\&+ \frac{C_\alpha }{\varepsilon _0 \varepsilon _1^{\frac{1}{\alpha -1}}} \Vert \nabla u\Vert _{{{\mathbb {L}}}^2}^2 \Big ]. \end{aligned}$$

  (7.5)

• (ii) Let $\alpha =1$; for every $\epsilon >0$, we have for some constant $C>0$ and any $u\in X_1$

  $$\begin{aligned} |\langle A^{1/2} B(u,u)\, , \,A^{1/2} u\rangle | \le \varepsilon \Vert Au\Vert _{{{\mathbb {L}}}^2}^2 + \frac{1}{4\varepsilon } \big \Vert |u| \nabla u \big \Vert _{{{\mathbb {L}}}^2}^2 . \end{aligned}$$

  (7.6)

Proof

• (i) Let $\alpha >1$ and $u\in X_1$. Then

  $$\begin{aligned} \langle A^{1/2} B(u,u)\, , \,A^{1/2} u\rangle = \sum _{i,j,k=1}^3 \int _{D} \partial _k\big [ u_i \, \partial _i u_j \big ]\, \partial _k u_j dx = T_1+T_2, \end{aligned}$$

  where, using the antisymmetry property (2.1), we get

  $$\begin{aligned} T_1&= \sum _{i,j,k=1}^3 \int _{D} \partial _k u_i \; \partial _i u_j \; \partial _k u_j dx, \\ T_2&= \sum _{i,j,k=1}^3 \int _{D} u_i \; \partial _k \partial _i u_j \; \partial _k u_j dx = \sum _{k=1}^3 \langle B(u,\partial _k u)\, , \, \partial _k u\rangle = 0. \end{aligned}$$

  Using integration by parts, we deduce $T_1=T_{1,1}+T_{1,2}$, where since $\mathrm{div\,} u=0$

  $$\begin{aligned} T_{1,1}&=- \sum _{j,k=1}^3 \int _{D} \partial _k \Big ( \sum _{i=1}^3 \partial _i u_i\Big )\, u_j \, \partial _k u_j dx =0, \\ T_{1,2}&=- \sum _{i,j,k=1}^3 \int _{D} \partial _k u_i \, u_j \, \partial _i\partial _k u_j dx. \end{aligned}$$

  The inequality (2.5) applied with $f=u_j$, $g=\partial _k u_i$ and $h=\partial _i \partial _k u_j$ implies

  $$\begin{aligned} |T_{1,2}|&\le \; \sum _{i,j,k=1}^3 \varepsilon _0 \Vert \partial _i\partial _k u_j\Vert _{L^2}^2 + \sum _{i,j,k=1}^3 \frac{\varepsilon _1}{4\varepsilon _0} \big \Vert |u_j|^\alpha \partial _k u_i \Vert _{L^2}^2 \\&\quad \; + \sum _{i,j,k=1}^3 \frac{C_\alpha }{\varepsilon _0 \varepsilon _1^{\frac{1}{\alpha -1}}} \Vert \partial _k u_i\Vert _{L^2}^2. \end{aligned}$$

  This completes the proof of (7.5).

• (ii) Let $\alpha =1$ and $u\in X_1$. Then an integration by parts implies

  $$\begin{aligned} \langle A^{1/2} B(u,u)\, , \,A^{1/2} u\rangle&=\; \sum _{i,j,k=1}^3 \int _{D} \partial _k\big [ u_i \, \partial _i u_j \big ]\, \partial _k u_j dx \\&=\; - \sum _{i,j=1}^3 \int _{D} u_i \, \partial _i u_j \; \varDelta u_j \, dx. \end{aligned}$$

  The Cauchy–Schwarz and Young inequalities imply (7.6).

$\square$

For $\varphi \in X_0$, set

$$\begin{aligned} F(\varphi )= -\nu A\varphi - B(\varphi , \varphi ) -a \varPi |\varphi |^{2\alpha } \varphi . \end{aligned}$$

(7.7)

Lemma 2.2 page 415 in [2] provides upper and lower bounds of the non linear Brinkman–Forchheimer term. Let $\alpha \in [1,\infty )$; there exist positive constants C and $\kappa$ such that for $u,v\in {{\mathbb {R}}}^3$

$$\begin{aligned} \big | |u|^{2\alpha } u - |v|^{2\alpha } v \big |&\le C |u-v|\, \big ( |u|^{2\alpha } + |v|^{2\alpha } \big ), \end{aligned}$$

(7.8)

$$\begin{aligned} \big ( |u|^{2\alpha } u - |v|^{2\alpha } v \big ) \cdot (u-v)&\ge \kappa |u-v|^2 \big ( |u|+|v|\big )^{2\alpha }. \end{aligned}$$

(7.9)

The following lemma gives upper bounds of F for any $\alpha \in [1,\infty )$.

Lemma 5

Let $\alpha \in [1,+\infty )$.

• (i) Let $u\in X_0$, $v\in L^2(0,T;V) \cap L^{2\alpha +2}([0,T]\times D;{{\mathbb {R}}}^3)$. Then

  $$\begin{aligned}&\int _0^T\!\! | \langle F(u(t)), v(t)\rangle | dt \le C\big [ \Vert v\Vert _{L^2(0,T;V)} \Vert u\Vert _{L^2(0,T;V)} \nonumber \\&\quad + \Vert v\Vert _{L^{2\alpha +2}([0,T]\times D;{{\mathbb {R}}}^3)} \Vert u\Vert _{L^{2\alpha +2}([0,T]\times D;{{\mathbb {R}}}^3)}^{2\alpha +1} \nonumber \\&\quad + \Vert v\Vert _{L^2(0,T;V)}\; \underset{{t\in [0,T]}}{\mathrm{ess sup }} \Vert u(t)\Vert _{H}^{\frac{\alpha -1}{\alpha }} \Vert u\Vert _{L^{2\alpha +2}([0,T]\times D ; {{\mathbb {R}}}^3)}^{\frac{\alpha +1}{\alpha }} T^{\frac{\alpha -1}{2\alpha }}\big ] \end{aligned}$$

  (7.10)

  for some positive constant C.

• (ii) Let $u\in {{\mathcal {X}}}_0$, $v\in L^4(\varOmega ;L^2(0,T;V))\cap L^{2\alpha +2}(\varOmega _T\times D;{{\mathbb {R}}}^3)$. Then

  $$\begin{aligned}&{{\mathbb {E}}}\int _0^T \! |\langle F(u(t)),v(t)\rangle | dt \le C\Big [ \Vert v\Vert _{L^2(\varOmega _T;V)} \Vert u\Vert _{L^2(\varOmega _T;V)} \nonumber \\&\quad + \Vert v\Vert _{L^{2\alpha +2}(\varOmega _T \times D;{{\mathbb {R}}}^3)} \Vert u\Vert _{L^{2\alpha +2}(\varOmega _T\times D;{{\mathbb {R}}}^3)}^{2\alpha +1} \nonumber \\&\quad + \Vert v\Vert _{L^4(\varOmega ; L^2(0,T;V))} \Big \{ {{\mathbb {E}}}\Big ( \underset{{t\in [0,T]}}{\mathrm{ess\, sup }} \Vert u(t)\Vert _{H}^4\Big )\Big \}^{\frac{\alpha -1}{\alpha }} \Vert u\Vert _{L^{2\alpha +2}(\varOmega _T\times D ; {{\mathbb {R}}}^3)}^{\frac{\alpha +1}{\alpha }} T^{\frac{\alpha -1}{2\alpha }}\Big ] \end{aligned}$$

  (7.11)

  for some positive constant C.

Proof

Integration by parts and the Cauchy–Schwarz inequality imply

$$\begin{aligned} \nu \int _0^T \! \! | \langle A u(t)\, , \, v(t)\rangle | dt&=\; \int _0^T\! \Big | - \nu \int _{D} \! A^{\frac{1}{2}} u(t,x) A^{\frac{1}{2}} v((t,x) dx \Big | dt \\&\le \; \; \nu \Vert u\Vert _{L^2(0,T;V)} \Vert v\Vert _{L^2(0,T;V)}. \end{aligned}$$

Furthermore, Hölder’s inequality with conjugate exponents $2\alpha +2$ and $\frac{2\alpha +2}{2\alpha +1}$ yields

$$\begin{aligned} \int _0^T \Big | \int _{D} |u(t,x)|^{2\alpha } u(t,x) v(t,x) dx \Big | dt&\le \; \big \Vert |u|^{2\alpha } u\Vert _{L^{\frac{2\alpha +2}{2\alpha +1}}([0,T]\times D;{{\mathbb {R}}}^3)} \\&\quad \times \Vert v\Vert _{L^{2\alpha +2}([0,T]\times D;{{\mathbb {R}}}^3)}. \end{aligned}$$

Using the above upper estimates with the inequality (7.1) concludes the proof of (7.10).

(ii) The upper estimate (7.11) is a straightforward consequence of the upper estimates (7.3), (7.10), the Cauchy–Schwarz and Hölder inequalities. $\square$

The next lemma provides estimates of the gradient of F(u) for $\alpha \in [1,+\infty )$. Note that when $\alpha =1$, this requires that the coefficient a in front of the Brinkman–Forchheimer smoothing term is “not too smal” compared to the viscosity $\nu$.

Lemma 6

• (i) Let $\alpha >1$. For $\eta \in (0,\nu )$, ${\tilde{a}}\in (0,a)$, there exists a positive constant $C:=C(\alpha , \eta , {\tilde{a}})$ such that for $u\in X_1$ and $t\in [0,T]$,

  $$\begin{aligned}&\int _0^t \!\! \langle A^{1/2} F(u(s)) , A^{1/2}u(s) \rangle ds \nonumber \\&\quad \le -\eta \! \int _0^t\!\! \Vert A u(s)\Vert _{{{\mathbb {L}}}^2}^2 ds - {\tilde{a}}\! \int _0^t \!\! \big \Vert |u(s)|^{\alpha } \nabla u(s) \big \Vert _{{{\mathbb {L}}}^2}^2 ds + C\! \int _0^t \!\! \Vert \nabla u(s)\Vert _{{{\mathbb {L}}}^2}^2 ds. \end{aligned}$$

  (7.12)

• (ii) Let $\alpha =1$ and suppose $4\nu a >1$. Then for $\eta \in \big ( 0, \nu - \frac{1}{4a}\big )$ and ${\tilde{a}} = a-\frac{1}{4(\nu -\eta )}$ we have

  $$\begin{aligned} \int _0^t \!\! \langle A^{1/2} F(u(s)) , A^{1/2}u(s)\rangle ds\le & {} -\eta \! \int _0^t\!\! \Vert A u(s)\Vert _{{{\mathbb {L}}}^2}^2 ds \nonumber \\&- {\tilde{a}}\! \int _0^t \!\! \big \Vert |u(s)|^{\alpha } \nabla u(s) \big \Vert _{{{\mathbb {L}}}^2}^2 ds . \end{aligned}$$

  (7.13)

Proof

• (i) Let $\alpha \in (1,\infty )$. For $u\in X_1$, integration by parts implies for a.e. $s\in [0,t]$,

  $$\begin{aligned} \nu \langle A^{\frac{1}{2}} \varDelta u(s), A^{\frac{1}{2}} u(s)\rangle = -\nu \Vert A u(s)\Vert _{{{\mathbb {L}}}^2}^2.\end{aligned}$$

  Furthermore,

  $$\begin{aligned}&\int _{D} \nabla \big ( |u(s)|^{2\alpha } u(s)\big ) \cdot \nabla u(s) dx \nonumber \\&\quad = \int _{D}\big [ |u(s)|^{2\alpha } \nabla u(s) \cdot \nabla u(s) + 2\alpha |u(s)|^{2(\alpha -1)} \big ( u(s)\cdot \nabla u(s) \big )^2 \big ] dx \nonumber \\&\quad \ge \int _{D} |u(s)|^{2\alpha } \nabla u(s) \cdot \nabla u(s) dx = \big \Vert |u(s)|^\alpha \nabla u(s) \big \Vert _{{{\mathbb {L}}}^2}^2. \end{aligned}$$

  (7.14)

  Hence, using (7.5) with $C\, \varepsilon _0\in (0, \nu -\eta )$, then $\varepsilon _1$ such that $C\, \frac{\varepsilon _1}{4\varepsilon _0} \in (0, a-{\tilde{a}})$, we deduce that for a.e. $s\in [0,T]$,

  $$\begin{aligned} \langle A^{1/2} F(u(s)) , A^{1/2}u(s)\rangle&\le \; -\eta \Vert Au(s)\Vert _{{{\mathbb {L}}}^2}^2 - {\tilde{a}} \big \Vert |u(s)|^\alpha \nabla u(t)\Vert _{{{\mathbb {L}}}^2}^2 \nonumber \\&\quad \; + C(\alpha , \eta , {\tilde{a}}) \Vert \nabla u(s)\Vert _{{{\mathbb {L}}}^2}^2. \end{aligned}$$

  (7.15)

  Integrating this inequality on the time interval [0, t] concludes the proof of (7.12).

• (ii) Let $\alpha =1$. Then using (7.6) and (7.14), we deduce for $\epsilon >0$ and $s\in [0,T]$

  $$\begin{aligned} \langle A^{1/2} F(u(s)) , A^{1/2}u(s)\rangle&\le \; -(\nu -\epsilon ) \Vert A u(s)\Vert _{{{\mathbb {L}}}^2}^2 + \frac{1}{4\epsilon } \Vert |u(s)| \nabla u(s)\Vert _{{{\mathbb {L}}}^2}^2 \\&\quad \; -a \Vert |u(s)| \nabla u(s)\Vert _{{{\mathbb {L}}}^2}^2. \end{aligned}$$

  Since $4a\nu >1$, for $\eta \in \big ( 0, \nu - \frac{1}{4a}\big )$, $\epsilon = \nu -\eta$ and ${\tilde{a}} = a-\frac{1}{4(\nu -\eta )}$ we deduce

  $$\begin{aligned} \langle A^{1/2} F(u(s)) , A^{1/2}u(s)\rangle \le -\eta \Vert Au(s)\Vert _{{{\mathbb {L}}}^2}^2 - {\tilde{a}} \big \Vert |u(s)|^\alpha \nabla u(s)\Vert _{{{\mathbb {L}}}^2}^2 . \end{aligned}$$

  (7.16)

  Integrating on the time interval [0, t], we deduce (7.13).

$\square$

We finally prove upper estimates of increments $F(u)-F(v)$ for $\alpha \in [1,\infty )$.

Lemma 7

There exists a positive constant $\kappa$ depending on $\alpha \in [1,+\infty )$, and for $\eta \in (0,\nu )$ a positive constant ${\bar{C}}(\eta )$, such that for $u,v\in V\cap L^{2\alpha +2}(D;{{\mathbb {R}}}^3)$,

$$\begin{aligned} \langle F(u)-F(v), u-v\rangle&\le \; - \eta \Vert \nabla (u-v)\Vert _{{{\mathbb {L}}}^2}^2 - a\kappa \big \Vert (|u|+|v|)^\alpha (u-v)\big \Vert _{{{\mathbb {L}}}^2}^2 \nonumber \\&\quad \; + {\bar{C}}(\eta ) \Vert \nabla v\Vert _{{{\mathbb {L}}}^2}^4 \Vert u-v\Vert _{{{\mathbb {L}}}^2}^2. \end{aligned}$$

(7.17)

Proof

Using integration by parts, we obtain

$$\begin{aligned} \nu \langle \varDelta (u-v),u-v\rangle = -\nu \Vert \nabla (u-v)\Vert _{{{\mathbb {L}}}^2}^2. \end{aligned}$$

The monotonicity property (7.9) implies

$$\begin{aligned}&a\int _{D} \big ( |u(x)|^{2\alpha } u(x) - |v(x)|^{2\alpha } v(x)\big ) \cdot \big ( u(x)-v(x)\big ) dx \nonumber \\&\quad \ge a \kappa \big \Vert (|u|+|v])^\alpha (u-v)\big \Vert _{{{\mathbb {L}}}^2}^2. \end{aligned}$$

(7.18)

Finally, Hölder’s inequality and the Gagliardo–Nirenberg inequality (2.2) for the ${{\mathbb {L}}}^4$ norm imply

$$\begin{aligned} |\langle B(u,u) - B(v,v) , u-v\rangle |&= |\langle B(u-v,v) , u-v\rangle |\\&\le \; \Vert u-v\Vert _{{{\mathbb {L}}}^4}^2 \Vert \nabla v\Vert _{{{\mathbb {L}}}^2} \le {\bar{C}}_4^2 \Vert u-v\Vert _{{{\mathbb {L}}}^2}^{\frac{1}{2}} \Vert \nabla (u-v)\Vert _{{{\mathbb {L}}}^2}^{\frac{3}{2}} \Vert \nabla v\Vert _{{{\mathbb {L}}}^2} \\&\le \; \frac{3}{4} \varepsilon ^{\frac{4}{3}} \Vert \nabla (u-v)\Vert _{{{\mathbb {L}}}^2}^2 + \frac{1}{4} \frac{1}{\varepsilon ^4} {\bar{C}}_4^8 \Vert \nabla v\Vert _{{{\mathbb {L}}}^2}^4 \Vert u-v\Vert _{{{\mathbb {L}}}^2}^2, \end{aligned}$$

where the last inequality holds for any $\varepsilon >0$ by Young’s inequality. Choosing $\frac{3}{4} \varepsilon ^{\frac{4}{3}} \in (0, \nu -\eta )$, we conclude the proof of (7.17). $\square$

We next prove that (1.1) has a unique strong solution in ${\mathcal {X}}_1$. The outline is quite classical, based on some Galerkin approximation and a priori estimates.

1.2 Galerkin approximation and a priori estimates

Recall that D is periodic domain of ${{\mathbb {R}}}^3$. Let $(e_n, n\ge 1)$ be the orthonormal basis of H defined in Sect. 3.1 (that is made of functions in H which are also orthogonal in V). For every integer $n\ge 1$ we set ${{\mathcal {K}}}_n:=\mathrm{span} (\zeta _1, \ldots , \zeta _n)$ where $\{\zeta _j\}_{j\ge 1}$ is an ONB of K mode of eigenfunctions of Q. Let $\varPi _n$ denote the projection from K onto $Q^{1/2}({{\mathcal {K}}}_n)$, and let $W_n(t)=\sum _{j=1}^n \sqrt{q_j} \zeta _j \beta _j(t) = \varPi _n W(t)$.

Recall that if ${{\mathcal {H}}}_n= \mathrm{span} (e_1,\ldots , e_n)$, the orthogonal projection $P_n$ of H onto ${{\mathcal {H}}}_n$ restricted to V coincides with the orthogonal projection of V onto ${{\mathcal {H}}}_n$.

Fix $n\ge 1$ and consider the following stochastic ordinary differential equation on the n-dimensional space ${{\mathcal {H}}}_n$ defined by $u_{n}(0)=P_n u_0$, and for $t\in [0,T]$ and $v \in {{\mathcal {H}}}_n$:

$$\begin{aligned} d(u_{n}(t), v)= \big \langle P_n F(u_{n}(t)),v\big \rangle dt + (P_n\, G( u_{n}(t)) \, \varPi _n\, dW(t), v), \quad {{\mathbb {P}}}\, \mathrm{a.s.},\qquad \end{aligned}$$

(7.19)

where F is defined in (7.7). Then for $k=1, \, \ldots , \, n$ we have for $t\in [0,T]$:

$$\begin{aligned} d(u_{n}(t), e_k)= \big \langle P_n F(u_{n}(t)),e_k \big \rangle \, dt + \sum _{j=1}^n q_j^{\frac{1}{2}} \big ( P_n\, G(u_{n}(t)) \zeta _j\, ,\, e_k \big ) \, d\beta _j(t), \quad {{\mathbb {P}}}\, \text{ a.s. } \end{aligned}$$

Note that for $v\in {{\mathcal {H}}}_n$ the map $u\in {{\mathcal {H}}}_n \mapsto \langle F(u) \, ,\, v\rangle$ is locally Lipschitz. Indeed, ${{\mathbb {H}}}^2\subset {{\mathbb {L}}}^{2\alpha +2}$ and there exists some constant C(n) such that $\Vert v\Vert _{{{\mathbb {H}}}^2} \le C(n) \Vert v\Vert _{{{\mathbb {L}}}^2}$ for $v\in {{\mathcal {H}}}_n$. Let $\varphi , \psi , v\in {{\mathcal {H}}}_n$; integration by parts implies that

$$\begin{aligned} |\langle \varDelta \varphi - \varDelta \psi ,v\rangle | \le \Vert \varphi - \psi \Vert _V \, \Vert v\Vert _V \le C(n)^2 \Vert \varphi - \psi \Vert _{{{\mathbb {L}}}^2}\, \Vert v\Vert _{{{\mathbb {L}}}^2}. \end{aligned}$$

In the polynomial nonlinear term, the upper estimate (7.8), the Hölder inequality with exponents $\frac{\alpha +1}{\alpha }$, $2\alpha +2$, and $2\alpha +2$, and the Sobolev embedding ${{\mathbb {H}}}^2\subset {{\mathbb {L}}}^{2\alpha +2}$ imply

$$\begin{aligned}&\Big | \int _{D}\! \big ( |\varphi (x) |^{2\alpha }\varphi (x)- |\psi (x)|^{2\alpha } \psi (x) \big ) v(x) dx \Big | \\&\quad \le C\, \big ( \Vert \varphi \Vert _{{{\mathbb {L}}}^{2\alpha +2}}^{2\alpha } + \Vert \psi \Vert _{{{\mathbb {L}}}^{2\alpha +2}}^{2\alpha } \big ) \, \Vert \varphi - \psi \Vert _{{{\mathbb {L}}}^{2\alpha +2}}\, \Vert v\Vert _{{{\mathbb {L}}}^{2\alpha +2}} \\&\quad \le C \, C(n)^{2(\alpha +1)} \big ( \Vert \varphi \Vert _{{{\mathbb {L}}}^2}^{2\alpha } +\Vert \psi \Vert _{{{\mathbb {L}}}^2}^{2\alpha }\big ) \, \Vert \varphi - \psi \Vert _{{{\mathbb {L}}}^2}\, \Vert v\Vert _{{{\mathbb {L}}}^2} . \end{aligned}$$

Finally, using integration by parts, the Hölder and Gagliardo–Nirenberg inequalities, we deduce:

$$\begin{aligned} |\langle B(\varphi ,\varphi ) - B(\psi ,\psi ), v\rangle |&=\big | -\langle B(\varphi -\psi , v)\, , \, \varphi \rangle - \langle B(\psi , v)\, , \, \varphi -\psi \rangle \big | \\&\le C\, \Vert \varphi - \psi \Vert _{{{\mathbb {L}}}^4} \big ( \Vert \varphi \Vert _{{{\mathbb {L}}}^4}+ \Vert \psi \Vert _{{{\mathbb {L}}}^4} \big ) \Vert \nabla v\Vert _{{{\mathbb {L}}}^2} \\&\le C C(n)^3 \Vert \varphi - \psi \Vert _{{{\mathbb {L}}}^2} \big ( \Vert \varphi \Vert _{{{\mathbb {L}}}^2}+\Vert \psi \Vert _{{{\mathbb {L}}}^2}\big ) \Vert v\Vert _{{{\mathbb {L}}}^2}. \end{aligned}$$

Condition (G) implies that the map $u\in {{\mathcal {H}}}_n \mapsto \big ( \sqrt{q_j}\, \big (G(u) \zeta _j\, ,\, e_k\big ) : 1\le j,k\le n \big )$ satisfies the classical global linear growth and Lipschitz conditions from ${{\mathcal {H}}}_n$ to $n\times n$ matrices uniformly in $t\in [0,T]$. Hence by a well-known result about existence and uniqueness of solutions to stochastic differential equations (see e.g. [24]), there exists a maximal solution $u_{n}=\sum _{k=1}^n (u_{n}\, ,\, e_k\big )\, e_k\in {{\mathcal {H}}}_n$ to (7.19), i.e., a stopping time $\tau _{n}^*\le T$ such that (7.19) holds for $t< \tau _{n}^*$ and if $\tau _n^* <T$, $\Vert u_{n}(t)\Vert _{L^2} \rightarrow \infty$ as $t \uparrow \tau _{n}^*$.

The following proposition shows that $\tau _n^*=T$ a.s., and provides a priori estimates on norms of $u_n$, which do not depend on n.

Proposition 7

Let $\alpha \in [1,\infty )$, and if $\alpha =1$, suppose that $4\nu a >1$.

• (i) Let $u_0$ be ${{\mathcal {F}}}_0$-measurable such that ${{\mathbb {E}}}\big ( \Vert u_0\Vert _H^2\big )<\infty$, $T>0$ and G satisfy (3.2) and (3.4). Then the evolution equation (7.19) with initial condition $P_n u_0$ has a unique global solution on [0, T] (i.e., $\tau _n^*=T$ a.s.) with a modification $u_n\in C([0,T];{{\mathcal {H}}}_n)$. Furthermore, if ${{\mathbb {E}}}\big ( \Vert u_0\Vert _H^{2p}\big ) <\infty$ for some $p\in [1,\infty )$, we have $u_n \in {{\mathcal {X}}}_0$ and

  $$\begin{aligned}&\sup _n {{\mathbb {E}}}\Big ( \sup _{t\in [0,T]} \Vert u_n(t)\Vert _H^{2p} + \int _0^T \big [ \Vert u_n(t)\Vert _V^2 + \Vert u_n(t)\Vert _{{{\mathbb {L}}}^{2\alpha +2}}^{2\alpha +2}\big ] \Vert u_n(t)\Vert _H^{2p-2} \, dt \Big ) \nonumber \\&\quad \le C \big [ 1+{{\mathbb {E}}}(\Vert u_0\Vert _H^{2p}) \big ]. \end{aligned}$$

  (7.20)

• (ii) If ${{\mathbb {E}}}(\Vert u_0\Vert _V^{2p}) <\infty$ for some $p\in [1,\infty )$ and G satisfies also (3.3), we have furthermore

  $$\begin{aligned}&\sup _n {{\mathbb {E}}}\Big ( \sup _{t\in [0,T]} \Vert u_n(t)\Vert _V^{2p} \nonumber \\&\qquad + \int _0^T \big [ \Vert A u_n(t)\Vert _{{{\mathbb {L}}}^2}^2 + \Vert |u_n(t)|^\alpha \nabla u_n(t)\Vert _{{{\mathbb {L}}}^2}^2 \big ] \Vert u_n(t)\Vert _V^{2p-2} dt\Big ) \nonumber \\&\quad \le C \big [ 1+{{\mathbb {E}}}(\Vert u_0\Vert _V^{2p}) \big ] . \end{aligned}$$

  (7.21)

Proof

(i) For fixed $N>0$ set $\tau _N:= \inf \{ t\ge 0 : \Vert u_n(t)\Vert _H \ge N\} \wedge \tau _n^*$. Itô’s formula and the antisymmetry property of B imply

$$\begin{aligned} \Vert u_n(t\wedge \tau _N)\Vert _H^2&=\; \Vert P_n u_0\Vert _H^2 -2\int _0^{t\wedge \tau _N} \!\! \big [ \nu \Vert \nabla u_n(s)\Vert _{{{\mathbb {L}}}^2}^2 +a \Vert u_n(s)\Vert _{L^{2\alpha +2}}^{2\alpha +2} \big ] ds \nonumber \\&\quad + \sum _{i=1}^2 T_i(t), \end{aligned}$$

(7.22)

where

$$\begin{aligned} T_1(t)&= \; 2 \int _0^{t\wedge \tau _n} \big ( G(u_n(s))\, dW_n(s)\, , \, u_n(s)\big ), \\ T_2(t)&=\; \int _0^{t\wedge \tau _n} \Vert P_n G(u_n(s)) \varPi _n \Vert _{{\mathcal {L}}}^2 \; ds. \end{aligned}$$

Apply once more the Itô formula to $z\mapsto z^p$ and $z=\Vert u_n(t\wedge \tau _N)\Vert _H^2$ for $p\in [2,\infty )$. We obtain

$$\begin{aligned} \Vert u_n(t\wedge \tau _N)\Vert _H^{2p}&= \Vert P_n u_0\Vert _H^{2p} + \sum _{i=1}^3 {\bar{T}}_i(t) \nonumber \\&\quad -2p\int _0^{t\wedge \tau _N} \!\! \big [ \nu \Vert \nabla u_n(s)\Vert _{{{\mathbb {L}}}^2}^2 +a \Vert u_n(s)\Vert _{L^{2\alpha +2}}^{2\alpha +2}\big ] \Vert u_n(s)\Vert _H^{2p-2} ds, \end{aligned}$$

(7.23)

where

$$\begin{aligned} {\bar{T}}_1(t)&= \; 2p \int _0^{t\wedge \tau _N} \big ( P_n G(u_n(s))\, dW_n(s) \, , \, u_n(s)\big ) \, \Vert u_n(s)\Vert _H^{2p-2}, \\ {\bar{T}}_2(t)&=\; p\int _0^{t\wedge \tau _N} \Vert P_n G(u_n(s)) \varPi _n\Vert _{{\mathcal {L}}}^2 \, \Vert u_n(s)\Vert _H^{2p-2}\, ds , \\ {\bar{T}}_3(t)&=\; 2p(p-1) \int _0^{t\wedge \tau _N} \Vert \big ( G(u_n(s))\, \varPi _n \big )^* u_n(s)\big \Vert _{K}^2\, \Vert u_n(s)\Vert _H^{2p-4}\, ds. \end{aligned}$$

The growth condition (3.2) implies

$$\begin{aligned} {\bar{T}}_2(t) + {\bar{T}}_3(t) \le p(2p-1) \int _0^{t} [K_0 + K_1 \Vert u_n(s\wedge \tau _N)\Vert _H^2] \, \Vert u_n(s\wedge \tau _N)\Vert _H^{2p-2}\, {\mathrm{Tr}} Q\, ds. \end{aligned}$$

Using the Davis inequality, the growth condition (3.2) and Young’s inequality, we deduce for $\beta \in (0,1)$,

$$\begin{aligned} {{\mathbb {E}}}\Big ( \sup _{s\le t\wedge \tau _n} \; {\bar{T}}_1(s)\Big )\le & {} \; 6p \; {{\mathbb {E}}}\Big ( \Big \{ \int _0^{t\wedge \tau _N} \Vert G(u_n(s))\Vert _{{\mathcal {L}}}^2 \; \Vert u_n(s)\Vert _H^{4p-2}\; {\mathrm{Tr}}Q\, ds \Big \}^{\frac{1}{2}} \Big ) \\\le & {} \; \beta \; {{\mathbb {E}}}\Big ( \sup _{s\le t} \Vert u_n(s\wedge \tau _N)\Vert _H^{2p}\Big ) \\&\; + \frac{9p^2}{\beta } {{\mathbb {E}}}\!\! \int _0^t\!\! \big [ K_0+K_1 \Vert u_n(s\wedge \tau _N)\Vert _H^2\big ] \Vert u_n(s\wedge \tau _N)\Vert _H^{2p-2} \, {\mathrm{Tr}} Q\, ds . \end{aligned}$$

Neglecting the first integral in the right hand side of (7.23), using the above upper estimates of ${\bar{T}}_i$ and the Gronwall lemma, we deduce that for $\beta \in (0,1)$,

$$\begin{aligned} \sup _{n\ge 1} {{\mathbb {E}}}\Big ( \sup _{s\le T} \Vert u_n(s\wedge \tau _N)\Vert _H^{2p}\Big ) \le C(\beta ,p, K_0, K_1, {\mathrm{Tr}}Q) \big [ 1+{{\mathbb {E}}}(\Vert u_0\Vert _H^{2p}\big ].\qquad \end{aligned}$$

(7.24)

As $N\rightarrow \infty$, the sequence of stopping times $\tau _N$ increases to $\tau ^*_n$ and on the set $\{ \tau ^*_n<T\}$, we have $\sup _{s\in [0, \tau _N]} \Vert u_n(s)\Vert _H \rightarrow \infty$. Hence (7.24) implies $P(\tau _n^*<T)=0$ and for almost every $\omega$, for $N(\omega )$ large enough we have $\tau _{N(\omega )}(\omega )=T$. Plugging the upper estimate (7.24) in (7.23), we conclude the proof of (7.20).

Note that the above argument based on (7.22) instead of (7.23) proves that if ${{\mathbb {E}}}(\Vert u_0\Vert _H^2) <\infty$ we have once more $\tau _{N(\omega )}(\omega )=T$ for $N(\omega )$ large enough and a.e. $\omega$, and that (7.20) holds for $p=1$.

We next prove that $u_n\in {{\mathcal {X}}}_0$. Plugging the above upper estimate for $p=1$ in (7.22), taking expected values and using Condition (3.2), we obtain

$$\begin{aligned} {{\mathbb {E}}}\int _0^T \big [ \Vert u_n(s)\Vert _V^2 + \Vert u_n(s)\Vert _{L^{2\alpha +2}}^{2\alpha +2} \big ] ds <\infty . \end{aligned}$$

A similar argument using (7.24) in (7.23) completes the proof of (7.20) when the H-norm of the initial condition has 2p moments.

(ii) Taking the gradient of both hand sides of (7.19), using the Itô formula and (3.1), we deduce for ${\tilde{\tau }}_N:=\inf \{ s\ge 0\, : \, \Vert u_n(s)\Vert _V\ge N\} \wedge T$,

$$\begin{aligned} \Vert A^{\frac{1}{2}} u_n(t\wedge {\tilde{\tau }}_N)\Vert _{{{\mathbb {L}}}^2}^2&= \Vert A^{\frac{1}{2}} P_n u_0\Vert _{{{\mathbb {L}}}^2}^2 + 2 \int _0^{t\wedge {\tilde{\tau }}_N} \langle A^{\frac{1}{2}} P_n F(u_n(s)) , A^{\frac{1}{2}} u_n(s)\rangle \, ds \\&\quad + 2\! \int _0^{t\wedge {\tilde{\tau }}_N} \!\! \big ( A^{\frac{1}{2}} P_n G(u_n(s)) dW_n(s) , A^{\frac{1}{2}} u_n(s)\big ) \\&\quad + \int _0^{t\wedge {\tilde{\tau }}_N} \!\! \Vert A^{\frac{1}{2}} P_n G(u_n(s)) \varPi _n\Vert _{{\mathcal {L}}}^2 \, ds\\&= \, \Vert A^{\frac{1}{2}} P_n u_0\Vert _{{{\mathbb {L}}}^2}^2 + 2 \int _0^{t\wedge {\tilde{\tau }}_N} \langle A^{\frac{1}{2}} F(u_n(s)) , A^{\frac{1}{2}} u_n(s)\rangle \, ds \\&\quad + 2\! \int _0^{t\wedge {\tilde{\tau }}_N} \!\! \big ( A^{\frac{1}{2}} G(u_n(s)) \varPi _n dW(s) , A^{\frac{1}{2}} u_n(s)\big ) \\&\quad + \int _0^{t\wedge {\tilde{\tau }}_N} \!\! \Vert A^{\frac{1}{2}} P_n G(u_n(s)) \varPi _n\Vert _{{\mathcal {L}}}^2 \, ds. \end{aligned}$$

Indeed, since $u_n(s)\in V$ for $s\le t\wedge {\tilde{\tau }}_N$, we deduce $A^{\frac{1}{2}} u_n(s)\in H$ and $A^{\frac{1}{2}} G(u_n(s))\in {{\mathcal {L}}}(K,H)$.

Using once more the Itô formula for the function $z\mapsto z^p$ for $p\in [2,\infty )$, we obtain

$$\begin{aligned} \Vert A^{\frac{1}{2}} u_n(t\wedge {\tilde{\tau }}_N)\Vert _{{{\mathbb {L}}}^2}^{2p}&\le \; \Vert A^{\frac{1}{2}} u_0\Vert _{{{\mathbb {L}}}^2}^{2p} + \sum _{i=1}^3 {\tilde{T}}_i(t) \nonumber \\&\quad + 2 \int _0^{t\wedge {\tilde{\tau }}_N} \langle A^{\frac{1}{2}} \nabla F(u_n(s)) , A^{\frac{1}{2}} u_n(s)\rangle \, \Vert A^{\frac{1}{2}} u_n(s)\Vert _{{{\mathbb {L}}}^2}^{2(p-1)} ds, \end{aligned}$$

(7.25)

where

$$\begin{aligned} {\tilde{T}}_1(t)&=\; 2p \int _0^{t\wedge {\tilde{\tau }}_N} \big ( A^{\frac{1}{2}} G(u_n(s)) dW_n(s)\, , \, A^{\frac{1}{2}} u_n(s)\big ) \, \Vert A^{\frac{1}{2}} u_n(s)\Vert _{{{\mathbb {L}}}^2}^{2(p-1)} ds,\\ {\tilde{T}}_2(t)&=\; p\int _0^{t\wedge {\tilde{\tau }}_N} \Vert G(u_n(s) ) \varPi _n\Vert _{\tilde{{\mathcal {L}}}}^2 \; \Vert A^{\frac{1}{2}} u_n(s)\Vert _{{{\mathbb {L}}}^2}^{2(p-1)} ds, \\ {\tilde{T}}_3(t)&=\; 2p(p-1)\! \int _0^{t\wedge {\tilde{\tau }}_N} \!\! \Vert \big ( A^{\frac{1}{2}} G(u_n(s)) \, \varPi _n \big )^* A^{\frac{1}{2}} u_n(s)\Vert _{K}^2 \; \Vert A^{\frac{1}{2}} u_n(s)\Vert _{{{\mathbb {L}}}^2}^{2(p-2)}\, ds. \end{aligned}$$

Since

$$\begin{aligned} \Vert \big (A^{\frac{1}{2}} G(u_n(s)) \, \varPi _n \big )^*\Vert _{{{\mathcal {L}}}(H;K)} \le \Vert A^{\frac{1}{2}} G(u_n(s))\Vert _{{{\mathcal {L}}}(K;H)} \le \Vert G(u_n(s))\Vert _{{{\mathcal {L}}}(K;V)}, \end{aligned}$$

the growth condition (3.3) and Young’s inequality imply

$$\begin{aligned} {\tilde{T}}_2(t)+{\tilde{T}}_3(t) \le C(p,T,{\mathrm{Tr}} Q, {\tilde{K}}_0, {\tilde{K}}_1) \Big [ 1+ \! \int _0^{t\wedge {\tilde{\tau }}_N}\!\!\! \big ( \Vert u_n(s)\Vert _H^{2p} + \Vert \nabla u_n(s)\Vert _{{{\mathbb {L}}}^2}^{2p}\big ) ds\Big ]. \end{aligned}$$

The growth condition (3.3), the Gundy and Young inequalities imply that for ${\tilde{\beta }}\in (0,1)$,

$$\begin{aligned} {{\mathbb {E}}}\Big ( \sup _{s\le t} {\tilde{T}}_1(s) \Big )&\le \,C(p) {{\mathbb {E}}}\Big ( \Big \{\! \int _0^{t\wedge {\tilde{\tau }}_N} \!\!\!\big [ {\tilde{K}}_0 + {\tilde{K}}_1 \Vert u_n(s)\Vert _V^2\big ] \Vert \nabla u_n(s)\Vert _{{{\mathbb {L}}}^2}^{4p-2} \, {\mathrm{Tr}}Q ds \Big \}^{\frac{1}{2}} \!\Big ) \\&\le \; {\tilde{\beta }} {{\mathbb {E}}}\Big ( \sup _{s\le t} \Vert u_n(s\wedge {\tilde{\tau }}_N)\Vert _{{{\mathbb {L}}}^2}^{2p} \Big ) + {\tilde{\beta }} {{\mathbb {E}}}\Big ( \sup _{s\le t} \Vert \nabla u_n(s\wedge {\tilde{\tau }}_N)\Vert _{{{\mathbb {L}}}^2}^{2p} \Big ) \\&\quad \; + C({\tilde{\beta }}, {\mathrm{Tr}} Q, {\tilde{K}}_0, {\tilde{K}}_1) \Big [ 1+ {{\mathbb {E}}}\Big ( \int _0^t \Vert \nabla u_n(s\wedge {\tilde{\tau }}_N)\Vert _{{{\mathbb {L}}}^2}^{2p} ds\Big ) \Big ]. \end{aligned}$$

Let $\rho \in (0,\nu )$ and ${\tilde{a}}\in (0,a)$. Using (7.12) for $\alpha >1$ and (7.13) for $\alpha =1$, (7.20) and the Gronwall lemma, we deduce

$$\begin{aligned} {{\mathbb {E}}}\Big ( \sup _{s\le {\tilde{\tau }}_N} \Vert u_n(s)\Vert _V^{2p}\Big ) \le C\big [ 1+ {{\mathbb {E}}}(\Vert u_0\Vert _V^{2p})\big ] \end{aligned}$$

for some positive constant C which does not depend on N and n. For fixed n, letting $N\rightarrow \infty$ and using the monotone convergence theorem we deduce $u_n\in L^{2p}(\varOmega ; L^\infty (0,T;V))$. Plugging this in (7.25) and taking expected values, we conclude the proof of (7.21). $\square$

1.3 Proof of global well-posedness of the solution

The proof of Theorem 2 is classical and uses the upper estimates (7.2) and (7.4) for the uniqueness; see e.g. [6] for details.