Local Uniqueness of the Solution of the Stochastic 3D Euler Equation
In the following proposition, we prove that any two local solutions of the stochastic 3D Euler Eq. (2.3) that are defined up to the same stopping time must coincide. The proof hinges on bound (2.9) and assumption (2.11).
Proposition 11
Let \(\tau \) be a stopping time and \(\omega ^{\left( 1\right) },\omega ^{\left( 2\right) }:[0,\tau )\times \mathbb {T}^{3}\rightarrow \mathbb {R}^{3}\) be two solutions with paths of class \(C\left( [0,\tau );W_{\sigma }^{2,2}\left( \mathbb {T}^{3};\mathbb {R} ^{3}\right) \right) \) that satisfy the stochastic 3D Euler Eq. (2.3). Then \(\omega ^{\left( 1\right) }=\omega ^{\left( 2\right) }\) on \([0,\tau ).\)
Proof
We have thatFootnote 3
$$\begin{aligned} \text {d}\omega ^{\left( i\right) }+\mathcal {L}_{v^{\left( i\right) }} \omega ^{\left( i\right) }\ \text {d}t+\sum _{k=1}^{\infty }\mathcal {L}_{\xi _{k}} \omega ^{\left( i\right) }\ \text {d}B_{t}^{k}=\frac{1}{2}\sum _{k=1}^{\infty }\mathcal {L}_{\xi _{k}}^{2}\omega ^{\left( i\right) }\ \text {d}t,\ \ \ i=1,2, \end{aligned}$$
where \(\omega ^{\left( i\right) }={\text {curl}}v^{\left( i\right) }\). The difference \(\Omega =\omega ^{\left( 1\right) }-\omega ^{\left( 2\right) }\) satisfies
$$\begin{aligned} \text {d}\Omega +\mathcal {L}_{v^{\left( 1\right) }}\omega ^{\left( 1\right) }\ \text {d}t-\mathcal {L}_{v^{\left( 2\right) }}\omega ^{\left( 2\right) } \ \text {d}t+\sum _{k=1}^{\infty }\mathcal {L}_{\xi _{k}}\Omega \ \text {d}B_{t}^{k}=\frac{1}{2}\sum _{k=1}^{\infty }\mathcal {L}_{\xi _{k}}^{2}\Omega \ \text {d}t \end{aligned}$$
and thus (set also \(V=v^{\left( 1\right) }-v^{\left( 2\right) }\))
$$\begin{aligned} \text {d}\Omega +\mathcal {L}_{V}\omega ^{\left( 1\right) }\ \text {d}t+\mathcal {L}_{v^{\left( 2\right) }}\Omega \ \text {d}t+\sum _{k=1}^{\infty }\mathcal {L}_{\xi _{k}}\Omega \ \text {d}B_{t}^{k}=\frac{1}{2}\sum _{k=1}^{\infty }\mathcal {L}_{\xi _{k}}^{2} \Omega \ \text {d}t. \end{aligned}$$
It follows
$$\begin{aligned}&\frac{1}{2}d\left\| \Omega \right\| _{L^{2}}^{2}+\left\langle \mathcal {L}_{V}\omega ^{\left( 1\right) },\Omega \right\rangle \ \text {d}t+\left\langle \mathcal {L}_{v^{\left( 2\right) }}\Omega ,\Omega \right\rangle \ \text {d}t+\sum _{k=1}^{\infty }\left\langle \mathcal {L}_{\xi _{k}} \Omega ,\Omega \right\rangle \ \text {d}B_{t}^{k}\\&\quad =\frac{1}{2}\sum _{k=1}^{\infty }\left\langle \mathcal {L}_{\xi _{k}}^{2} \Omega ,\Omega \right\rangle \ \text {d}t+\frac{1}{2}\sum _{k=1}^{\infty }\left\langle \mathcal {L}_{\xi _{k}}\Omega ,\mathcal {L}_{\xi _{k}}\Omega \right\rangle \ \text {d}t. \end{aligned}$$
We rewrite
$$\begin{aligned}&\left\langle \mathcal {L}_{V}\omega ^{\left( 1\right) },\Omega \right\rangle +\left\langle \mathcal {L}_{v^{\left( 2\right) }}\Omega ,\Omega \right\rangle \\&\quad =\left\langle V\cdot \nabla \omega ^{\left( 1\right) },\Omega \right\rangle -\left\langle \omega ^{\left( 1\right) }\cdot \nabla V,\Omega \right\rangle +\left\langle v^{\left( 2\right) }\cdot \nabla \Omega ,\Omega \right\rangle -\left\langle \Omega \cdot \nabla v^{\left( 2\right) },\Omega \right\rangle \end{aligned}$$
and use the following inequalities:
$$\begin{aligned} \left| \left\langle V\cdot \nabla \omega ^{\left( 1\right) },\Omega \right\rangle \right|\le & {} \left\| \Omega \right\| _{L^{2}}\left\| V\right\| _{L^{4}}\left\| \nabla \omega ^{\left( 1\right) }\right\| _{L^{4}}\le C\left\| \omega ^{\left( 1\right) }\right\| _{W^{2,2} }\left\| \Omega \right\| _{L^{2}}^{2}\\ \left| \left\langle \omega ^{\left( 1\right) }\cdot \nabla V,\Omega \right\rangle \right|\le & {} \left\| \Omega \right\| _{L^{2}}\left\| \nabla V\right\| _{L^{2}}\left\| \omega ^{\left( 1\right) }\right\| _{L^{\infty }}\le C\left\| \omega ^{\left( 1\right) }\right\| _{W^{2,2}}\left\| \Omega \right\| _{L^{2}}^{2}\\ \left\langle v^{\left( 2\right) }\cdot \nabla \Omega ,\Omega \right\rangle= & {} 0\\ \left| \left\langle \Omega \cdot \nabla v^{\left( 2\right) } ,\Omega \right\rangle \right|\le & {} \left\| \Omega \right\| _{L^{2}} ^{2}\left\| \nabla v^{\left( 2\right) }\right\| _{L^{\infty }} \le \left\| \Omega \right\| _{L^{2}}^{2}\left\| \nabla v^{\left( 2\right) }\right\| _{W^{2,2}}\le C\left\| \omega ^{\left( 2\right) }\right\| _{W^{2,2}}\left\| \Omega \right\| _{L^{2}}^{2}. \end{aligned}$$
Here and below, we repeatedly use the Sobolev embedding theorems
$$\begin{aligned} W^{2,2}\left( \mathbb {T}^{3}\right) \subset C_{b}\left( \mathbb {T} ^{3}\right) ,\qquad W^{2,2}\left( \mathbb {T}^{3}\right) \subset W^{1,4}\left( \mathbb {T}^{3}\right) \end{aligned}$$
(3.1)
and the fact that Biot–Savart map \(\omega \mapsto v\) maps \(W^{\alpha ,p}\) into \(W^{\alpha +1,p}\) for all \(\alpha \ge 0\) and \(p\in \left( 1,\infty \right) \). \(W_{\sigma }^{s,2}\) into \(W_{\sigma }^{s+1,2}\) for all \(s\ge 0\); see (2.4). For instance, the sequences of inequalities used above in the case of the terms \(\left\| V\right\| _{L^{4}}\) and \(\left\| \nabla v^{\left( 2\right) }\right\| _{L^{\infty }}\) were
$$\begin{aligned} \left\| V\right\| _{L^{4}}\le & {} C\left\| V\right\| _{W^{1,2}}\le C^{\prime }\left\| \Omega \right\| _{L^{2}}\\ \left\| \nabla v^{\left( 2\right) }\right\| _{L^{\infty }}\le & {} C\left\| \nabla v^{\left( 2\right) }\right\| _{W^{2,2}}\le C^{\prime }\left\| v^{\left( 2\right) }\right\| _{W^{3,2}}\le C^{\prime \prime }\left\| \omega ^{\left( 2\right) }\right\| _{W^{2,2}}. \end{aligned}$$
We omit similar detailed explanations sometimes below, when they are of the same kind.
Using also (2.11), we get
$$\begin{aligned} d\left\| \Omega \right\| _{L^{2}}^{2}+2\sum _{k=1}^{\infty }\left\langle \mathcal {L}_{\xi _{k}}\Omega ,\Omega \right\rangle \ \text {d}B_{t}^{k}\le C\left( 1+\left\| \omega ^{\left( 1\right) }\right\| _{W^{2,2}}+\left\| \omega ^{\left( 2\right) }\right\| _{W^{2,2}}\right) \left\| \Omega \right\| _{L^{2}}^{2}\ \text {d}t. \end{aligned}$$
Then
$$\begin{aligned} d\left( e^{Y}\left\| \Omega \right\| _{L^{2}}^{2}\right)&=-\,e^{Y}\left\| \Omega \right\| _{L^{2}}^{2}C\left( 1+\left\| \omega ^{\left( 1\right) }\right\| _{W^{2,2}}+\left\| \omega ^{\left( 2\right) }\right\| _{W^{2,2}}\right) +e^{Y}d\left\| \Omega \right\| _{L^{2}}^{2}\\&\le -2e^{Y}\sum _{k=1}^{\infty }\left\langle \mathcal {L}_{\xi _{k}} \Omega ,\Omega \right\rangle \ \text {d}B_{t}^{k}, \end{aligned}$$
where Y is defined as
$$\begin{aligned} Y_{t}:=-\,\int _{0}^{t}C\left( 1+\left\| \omega _{s}^{\left( 1\right) }\right\| _{W^{2,2}}+\left\| \omega _{s}^{\left( 2\right) }\right\| _{W^{2,2}}\right) \text {d}s. \end{aligned}$$
The inequality (recall \(\Omega _{0}=0\))
$$\begin{aligned} e^{Y_{\bar{\tau }}}\left\| \Omega _{\bar{\tau }}\right\| _{L^{2}}^{2} \le -2\sum _{k=1}^{\infty }\int _{0}^{\bar{\tau }}e^{Y_{s}}\left\langle \mathcal {L}_{\xi _{k}}\Omega _{s},\Omega _{s}\right\rangle \ \text {d}B_{s}^{k} \end{aligned}$$
holds for every bounded stopping time \(\bar{\tau }\in \left[ 0,\tau \right] \). Hence, we have
$$\begin{aligned} e^{Y_{t\wedge \tau }}\left\| \Omega _{t\wedge \tau }\right\| _{L^{2}}^{2}&\le -2\sum _{k=1}^{\infty }\int _{0}^{t\wedge \tau }e^{Y_{s}}\left\langle \mathcal {L}_{\xi _{k}}\Omega _{s},\Omega _{s}\right\rangle \ \text {d}B_{s}^{k}\\&=-\,2\sum _{k=1}^{\infty }\int _{0}^{t}1_{s\le \tau }e^{Y_{s}}\left\langle \mathcal {L}_{\xi _{k}}\Omega _{s},\Omega _{s}\right\rangle \ \text {d}B_{s}^{k}. \end{aligned}$$
In expectation, denoted \(\mathbb {E}\), this implies
$$\begin{aligned} \mathbb {E}\left[ e^{Y_{t\wedge \tau }}\left\| \Omega _{t\wedge \tau }\right\| _{L^{2}}^{2}\right] \le 0 \end{aligned}$$
namely \(\mathbb {E}\left[ e^{Y_{t\wedge \tau }}\left\| \Omega _{t\wedge \tau }\right\| _{L^{2}}^{2}\right] =0\) and thus, for every t,
$$\begin{aligned} e^{Y_{t\wedge \tau }}\left\| \Omega _{t\wedge \tau }\right\| _{L^{2}} ^{2}=0\qquad \text {a.s.} \end{aligned}$$
Since \(Y_{t\wedge \tau }<\infty \) a.s., we get \(\left\| \Omega _{t\wedge \tau }\right\| _{L^{2}}^{2}=0\) a.s. and thus
$$\begin{aligned} \omega _{t\wedge \tau }^{\left( 1\right) }=\omega _{t\wedge \tau }^{\left( 2\right) }\qquad \text {a.s.} \end{aligned}$$
Recalling the continuity of trajectories, this implies
$$\begin{aligned} \omega ^{\left( 1\right) }=\omega ^{\left( 2\right) }\qquad \text {a.s.} \end{aligned}$$
The proof of the proposition is complete. \(\square \)
Existence of a Maximal Solution
Given \(R>0\), consider the modified Euler equations
$$\begin{aligned} \text {d}\omega _{R}+\kappa _{R}\left( \omega _{R}\right) \mathcal {L}_{v_{R}}\omega _{R}\ \text {d}t+\sum _{k=1}^{\infty }\mathcal {L}_{\xi _{k}}\omega _{R}\text {d}B_{t}^{k}=\frac{1}{2}\sum _{k=1}^{\infty }\mathcal {L}_{\xi _{k}}^{2}\omega _{R}\ \text {d}t,\qquad \omega _{R}|_{t=0}=\omega _{0}, \end{aligned}$$
(3.2)
where \(\omega _{R}={\text {curl}}v_{R}\). In (3.2), \(\kappa _{R}\left( \omega \right) :=f_{R}(\left\| \nabla v\right\| _{\infty })\), where \(f_{R}\) is a smooth function, equal to 1 on \(\left[ 0,R\right] \), equal to 0 on \([R+1,\infty )\) and decreasing in \(\left[ R,R+1\right] \).
Lemma 12
Given \(R>0\) and \(\omega _{0}\in W_{\sigma }^{2,2}\left( \mathbb {T}^{3},\mathbb {R}^{3}\right) \), let \(\omega _{R}:\Xi \times [0,\infty )\times \mathbb {T}^{3}\rightarrow \mathbb {R}^{3}\) be a global solution in \(W^{2,2}\) of Eq. (3.2). Let
$$\begin{aligned} \tau _{R}=\inf \left\{ t\ge 0:\left\| \omega \right\| _{W^{2,2}}\ge \frac{R}{C}\right\} , \end{aligned}$$
where C is a constant chosen so that
$$\begin{aligned} \left\| \nabla v\right\| _{\infty }\le C\left\| \omega \right\| _{W^{2,2}}. \end{aligned}$$
Finally, let \(\omega :\Xi \times \left[ 0,\tau _{R}\right] \times \mathbb {T} ^{3}\rightarrow \mathbb {R}^{3}\) be the restriction of \(\omega _{R}\). Then, \(\omega \) is a local solution in \(W_{\sigma }^{2,2}\) of stochastic 3D Euler Eq. (2.3)
Proof
Obviously, because for \(t\in \left[ 0,\tau _{R}\right] \) we have \(\left\| \nabla v\right\| _{\infty }\le C\left\| \omega \right\| _{W^{2,2}}\le R\) and thus \(\kappa _{R}\left( \omega _{R}\right) =1\); namely, the equations are the same. \(\square \)
The following proposition is the cornerstone of the existence and uniqueness of a maximal solution of stochastic 3D Euler Eq. (2.3)
Proposition 13
Given \(R>0\) and \(\omega _{0}\in W_{\sigma }^{2,2}\left( \mathbb {T}^{3},\mathbb {R}^{3}\right) \), there exists a global solution in \(W_{\sigma }^{2,2}\) of Eq. (3.2). Moreover, if \(\omega _{R}^{\left( 1\right) },\omega _{R}^{\left( 2\right) }:\Xi \times [0,\infty )\times \mathbb {T}^{3}\rightarrow \mathbb {R}^{3}\) are two global solutions in \(W_{\sigma }^{2,2}\) of Eq. (3.2 ), then \(\omega _{R}^{\left( 1\right) }=\omega _{R}^{\left( 2\right) }\).
We postpone the proof of Proposition 13 to the later sections. For now, let us show how it implies the existence of a maximal solution.
Theorem 14
Given \(\omega _{0}\in W_{\sigma }^{2,2}\left( \mathbb {T} ^{3},\mathbb {R}^{3}\right) \), there exists a maximal solution \(\left( \tau _{\max },\omega \right) \) of stochastic 3D Euler Eq. (2.3). Moreover, either \(\tau _{\max }=\infty \) or \(\lim \sup _{t\uparrow \tau _{\max }}\left\| \tilde{\omega }\left( t\right) \right\| _{W^{2,2}} =+\infty .\)
Proof
Choose \(R=n\) in Lemma 12, then \(\left( \tau _{n},\omega _{n}\right) \) is a local solution in \(W_{\sigma }^{2,2}\) of stochastic 3D Euler Eq. (2.3). Moreover, define \(\tau _{\max } :=\lim _{n\rightarrow \infty }\tau _{n}\) and define \(\omega \) as \(\omega |_{[0,\tau _{n})}:=\omega _{Cn}|_{[0,\tau _{n})}\). By uniqueness \(\omega _{m}|_{[0,\tau _{n})}:=\omega _{n}|_{[0,\tau _{n})}\) for any \(m\ge n\). So \(\omega \) is consistently defined.
The statement that either \(\tau _{\max }=\infty \) or \(\lim \sup _{t\uparrow \tau _{\max }}\left\| \tilde{\omega }\left( t\right) \right\| _{W^{2,2} }=+\infty \) is obvious: if \(\tau _{\max }<\infty ,\) then by the continuity of \(\tilde{\omega }\) on \([0,\tau _{\max })\), there exist some random times \(\tilde{\tau }_{n}<\tau _{n}\) such that \(\tau _{n}-\tilde{\tau }_{n}\le \frac{1}{n}\) and that \(\left\| \tilde{\omega }\left( \tilde{\tau }_{n}\right) \right\| _{W^{2,2}}\ge \frac{n-1}{C}\). Then,
$$\begin{aligned} \lim \sup _{t\uparrow \tau _{\max }}\left\| \tilde{\omega }\left( t\right) \right\| _{W^{2,2}}\ge \lim \sup _{n\uparrow \infty }\left\| \tilde{\omega }\left( \tilde{\tau }_{n}\right) \right\| _{W^{2,2}}=\infty . \end{aligned}$$
We prove by contradiction that \((\tau _{\max },\omega )\) is a maximal solution. Assume that there exists a pair \(\left( \tau ^{\prime },\omega ^{\prime }\right) \) such that \(\omega ^{\prime }=\omega \) on \([0,\tau ^{\prime }\wedge \tau _{\max }),\) and \(\tau ^{\prime }>\tau _{\max }\) with positive probability. This can only happen if \(\tau _{\max }<\infty \), therefore by the continuity of \(\omega ^{\prime }\) on \([0,\tau ^{\prime })\) on the set \(\left\{ \tau ^{\prime } >\tau _{\max }\right\} \)
$$\begin{aligned} \infty =\lim \sup _{n\uparrow \infty }\left\| \tilde{\omega }\left( \tilde{\tau }_{n}\right) \right\| _{W^{2,2}}=\lim \sup _{n\uparrow \infty }\left\| \tilde{\omega }^{\prime }\left( \tilde{\tau }_{n}\right) \right\| _{W^{2,2} }=\left\| \tilde{\omega }^{\prime }\left( \tau _{\max }\right) \right\| _{W^{2,2}}<\infty , \end{aligned}$$
which leads to a contradiction. Hence, necessarily, \(\tau ^{\prime }\le \tau \)P-almost surely, therefore \(\left( \tau ,\omega \right) \) is a maximal solution. \(\square \)
Uniqueness of the Maximal Solution
Let us start by justifying the uniqueness of solution truncated Euler Eq. (3.2). The proof is similar with that of Proposition 11 so we only sketch it here. Let \(\omega _{R}^{\left( 1\right) },\omega _{R}^{\left( 2\right) }:\Xi \times [0,\infty )\times \mathbb {T}^{3}\rightarrow \mathbb {R}^{3}\) are two global solutions in \(W_{\sigma }^{2,2}\) of Eq. (3.2 ). We preserve the same notation as in the proof of Proposition 11, i.e. denote by \(\Omega =\omega _{R}^{\left( 1\right) }-\omega _{R}^{\left( 2\right) }\) and \(V=v_{R}^{\left( 1\right) }-v_{R}^{\left( 2\right) }\). We also assume that the truncation function \(f_{R}\) is Lipschitz and we will denote by \(K_{R}\) the quantity
$$\begin{aligned} K_{R}=f_{R}(||\nabla v_{R}^{\left( 1\right) }||_{\infty })-f_{R}(|\nabla v_{R}^{\left( 2\right) }||_{\infty }) \end{aligned}$$
and observe that,Footnote 4
$$\begin{aligned} \left| K_{R}\right|&\le C\left\| \nabla v_{R}^{\left( 1\right) }-\nabla v_{R}^{\left( 2\right) }\right\| _{\infty }\\&=C\left\| \nabla V\right\| _{\infty }\le C\left\| \Omega \right\| _{2,2}^{2}. \end{aligned}$$
To simplify notation, we will omit the dependence on R in the following.
We are looking to prove uniqueness using the \(W^{2,2}\)-topology; therefore, we need to estimate the sum \(\left\| \Omega \right\| _{L^{2}}^{2}+\left\| \Delta \Omega \right\| _{L^{2}}^{2}\). Then,
$$\begin{aligned} \text {d}\Omega +\Phi \ \text {d}t+\sum _{k=1}^{\infty }\mathcal {L}_{\xi _{k}}\Omega \ \text {d}B_{t} ^{k}=\frac{1}{2}\sum _{k=1}^{\infty }\mathcal {L}_{\xi _{k}}^{2}\Omega \ \text {d}t, \end{aligned}$$
where
$$\begin{aligned} \Phi :=\kappa \left( \omega ^{\left( 1\right) }\right) \mathcal {L} _{v^{\left( 1\right) }}\omega ^{\left( 1\right) }\ -\kappa \left( \omega ^{\left( 2\right) }\right) \mathcal {L}_{v^{\left( 2\right) }} \omega ^{\left( 2\right) } \end{aligned}$$
and thus
$$\begin{aligned} \text {d}\Omega +\left\langle \Phi ,\Omega \right\rangle \ \text {d}t+\sum _{k=1}^{\infty }\mathcal {L}_{\xi _{k}}\Omega \ \text {d}B_{t}^{k}=\frac{1}{2}\sum _{k=1}^{\infty }\mathcal {L}_{\xi _{k}}^{2}\Omega \ \text {d}t. \end{aligned}$$
Then,
$$\begin{aligned}&\frac{1}{2}d\left\| \Omega \right\| _{L^{2}}^{2}+\left\langle \Phi ,\Omega \right\rangle \ \text {d}t+\sum _{k=1}^{\infty }\left\langle \mathcal {L}_{\xi _{k}}\Omega ,\Omega \right\rangle \ \text {d}B_{t}^{k}=\frac{1}{2}\sum _{k=1}^{\infty }\left\langle \mathcal {L}_{\xi _{k}}^{2}\Omega ,\Omega \right\rangle \ \text {d}t\\&\quad +\,\frac{1}{2}\sum _{k=1}^{\infty }\left\langle \mathcal {L}_{\xi _{k}} \Omega ,\mathcal {L}_{\xi _{k}}\Omega \right\rangle \ \text {d}t. \end{aligned}$$
On the set \(\tau ^{\left( 2\right) }\le \tau ^{\left( 1\right) }\) observe that \(\Phi \) is 0 if \(\left\| \omega ^{\left( 1\right) }\right\| _{W^{2,2}}\ge \frac{R}{C}\). It follows that, on this set, there exists a constant \(c_{R}\) such that (recall that \(0\le \kappa \le 1\))
$$\begin{aligned} \left| \left\langle \Phi ,\Omega \right\rangle \right|&=\left| K\left\langle \mathcal {L}_{v^{\left( 1\right) }}\omega ^{\left( 1\right) },\Omega \right\rangle \ +\kappa \left( \omega ^{\left( 2\right) }\right) \left\langle \mathcal {L}_{V}\omega ^{\left( 1\right) },\Omega \right\rangle \ +\kappa \left( \omega ^{\left( 2\right) }\right) \left\langle \mathcal {L}_{v^{\left( 2\right) }}\Omega ,\Omega \right\rangle \right| \\&\le c_{R}\left\| \Omega \right\| _{W^{2,2}}^{2}+\left| \left\langle \mathcal {L}_{V}\omega ^{\left( 1\right) },\Omega \right\rangle \ \right| +\left| \left\langle \mathcal {L}_{v^{\left( 2\right) } }\Omega ,\Omega \right\rangle \right| \end{aligned}$$
and, similar to the proof of Proposition 11, we deduced that
$$\begin{aligned} d\left\| \Omega \right\| _{L^{2}}^{2}+2\sum _{k=1}^{\infty }\left\langle \mathcal {L}_{\xi _{k}}\Omega ,\Omega \right\rangle \ \text {d}B_{t}^{k}\le C\left( 1+\left\| \omega ^{\left( 1\right) }\right\| _{W^{2,2}}+\left\| \omega ^{\left( 2\right) }\right\| _{W^{2,2}}\right) \left\| \Omega \right\| _{W^{2,2}}\ \text {d}t. \end{aligned}$$
(3.3)
Similarly, on the set \(\tau ^{\left( 2\right) }\le \tau ^{\left( 1\right) }\) observe that \(\Phi \) vanishes, if \(\left\| \omega ^{\left( 2\right) }\right\| _{W^{2,2}}\ge \frac{R}{C}\) and 3.3 holds true, as seen by observing that there exists a constant \(c_{R}\) such that
$$\begin{aligned} \left| \left\langle \Phi ,\Omega \right\rangle \right|&=\left| K\left\langle \mathcal {L}_{v^{\left( 2\right) }}\omega ^{\left( 2\right) },\Omega \right\rangle \ +\kappa \left( \omega ^{\left( 1\right) }\right) \left\langle \mathcal {L}_{V}\omega ^{\left( 2\right) },\Omega \right\rangle \ +\kappa \left( \omega ^{\left( 1\right) }\right) \left\langle \mathcal {L}_{v^{\left( 1\right) }}\Omega ,\Omega \right\rangle \ \right| \\&\le c_{R}\left\| \Omega \right\| _{W^{2,2}}^{2}+\left| \left\langle \mathcal {L}_{V}\omega ^{\left( 2\right) },\Omega \right\rangle \ \right| +\left| \left\langle \mathcal {L}_{v^{\left( 1\right) } }\Omega ,\Omega \right\rangle \right| . \end{aligned}$$
Next we have
$$\begin{aligned} d\Delta \Omega +\ \Delta \Phi \text {d}t+\sum _{k=1}^{\infty }\Delta \mathcal {L}_{\xi _{k} }\Omega \ \text {d}B_{t}^{k}=\frac{1}{2}\sum _{k=1}^{\infty }\Delta \mathcal {L}_{\xi _{k} }^{2}\Omega \ \text {d}t \end{aligned}$$
from which we deduce that
$$\begin{aligned}&\frac{1}{2}d\left\| \Delta \Omega \right\| _{L^{2}}^{2}+\left\langle \Delta \Phi ,\Delta \Omega \right\rangle +\sum _{k=1}^{\infty }\left\langle \Delta \mathcal {L}_{\xi _{k}}\Omega ,\Delta \Omega \right\rangle \ \text {d}B_{t}^{k} =\frac{1}{2}\sum _{k=1}^{\infty }\left\langle \Delta \mathcal {L}_{\xi _{k}} ^{2}\Omega ,\Delta \Omega \right\rangle \ \text {d}t\\&\quad +\,\frac{1}{2}\sum _{k=1}^{\infty }\left\langle \Delta \mathcal {L}_{\xi _{k}}\Omega ,\Delta \mathcal {L}_{\xi _{k} }\Omega \right\rangle \ \text {d}t. \end{aligned}$$
From the Lemma 25, we have
$$\begin{aligned} \left| \left\langle \Delta \mathcal {L}_{v^{\left( 2\right) }} \Omega ,\Delta \Omega \right\rangle \right|&\le C\left\| \nabla v^{\left( 2\right) }\right\| _{L^{\infty }}\left\| \Omega \right\| _{W^{2,2}}^{2}+C\left\| \Omega \right\| _{L^{\infty }}\left\| \nabla v^{\left( 2\right) }\right\| _{W^{2,2}}\left\| \Omega \right\| _{W^{2,2}}\\&\le C\left\| \omega ^{\left( 2\right) }\right\| _{W^{2,2} }\left\| \Omega \right\| _{W^{2,2}}^{2}. \end{aligned}$$
Moreover, by similar arguments,
$$\begin{aligned} \left| \left\langle \Delta \mathcal {L}_{V}\omega ^{\left( 1\right) },\Delta \Omega \right\rangle \right|&\le C\left\| \nabla V\right\| _{L^{\infty }}\left\| \omega ^{\left( 1\right) }\right\| _{W^{2,2}}\left\| \Omega \right\| _{W^{2,2}}\\&\quad +\,C\left\| \omega ^{\left( 1\right) }\right\| _{L^{\infty }}\left\| \nabla V\right\| _{W^{2,2} }\left\| \Omega \right\| _{W^{2,2}}\\&\le C\left\| \nabla V\right\| _{W^{2,2}}\left\| \omega ^{\left( 1\right) }\right\| _{W^{2,2}}\left\| \Omega \right\| _{W^{2,2}}\\&\le C\left\| \omega ^{\left( 1\right) }\right\| _{W^{2,2} }\left\| \Omega \right\| _{W^{2,2}}^{2}. \end{aligned}$$
Similar estimates hold true for \(\left| \left\langle \Delta \mathcal {L} _{V}\omega ^{\left( 2\right) },\Delta \Omega \right\rangle \ \right| \) and \(\left| \left\langle \Delta \mathcal {L}_{v^{\left( 1\right) }} \Omega ,\Delta \Omega \right\rangle \right| \). Next, as above, on the set \(\tau ^{\left( 2\right) }\le \tau ^{\left( 1\right) }\) observe there exists a constant \(c_{R}\) such that
$$\begin{aligned} \left| K\left\langle \Delta \mathcal {L}_{v^{\left( 1\right) }} \omega ^{\left( 1\right) },\Delta \Omega \right\rangle \right|&\le C\left\| \nabla v^{\left( 1\right) }\right\| _{L^{\infty }}\left\| \omega ^{\left( 1\right) }\right\| _{W^{2,2}}\left\| \Omega \right\| _{W^{2,2}}\\&\quad +\,C\left\| \omega ^{\left( 1\right) }\right\| _{L^{\infty } }\left\| \nabla v^{\left( 1\right) }\right\| _{W^{2,2}}\left\| \Omega \right\| _{W^{2,2}}\\&\le c_{R}\left\| \Omega \right\| _{W^{2,2}}^{2}. \end{aligned}$$
Similarly, on the set \(\tau ^{\left( 2\right) }\le \tau ^{\left( 1\right) } \),
$$\begin{aligned} \left| K\left\langle \Delta \mathcal {L}_{v^{\left( 1\right) }} \omega ^{\left( 1\right) },\Delta \Omega \right\rangle \right| \le c_{R}\left\| \Omega \right\| _{W^{2,2}}^{2}. \end{aligned}$$
Summarizing, we deduce that
$$\begin{aligned}&\frac{1}{2}d\left\| \Delta \Omega \right\| _{L^{2}}^{2} + \sum _{k=1}^{\infty }\left\langle \Delta \mathcal {L}_{\xi _{k}}\Omega ,\Delta \Omega \right\rangle \ \text {d}B_{t}^{k}\le C\left( 1 + \left\| \omega ^{\left( 1\right) }\right\| _{W^{2,2}} + \left\| \omega ^{\left( 2\right) }\right\| _{W^{2,2}}\right) \left\| \Omega \right\| _{W^{2,2}}^{2}\\&\quad +\frac{1}{2}\sum _{k=1}^{\infty }\left\langle \Delta \mathcal {L}_{\xi _{k}} ^{2}\Omega ,\Delta \Omega \right\rangle \ \text {d}t+\frac{1}{2}\sum _{k=1}^{\infty }\left\langle \Delta \mathcal {L}_{\xi _{k}}\Omega ,\Delta \mathcal {L}_{\xi _{k} }\Omega \right\rangle \ \text {d}t. \end{aligned}$$
It is then sufficient to repeat the argument of the proof of Proposition 11 to control \(\left\| \Omega \right\| _{L^{2} }^{2}+\left\| \Delta \Omega \right\| _{L^{2}}^{2}\) and obtain the uniqueness of the truncated Euler equation. The computation required here requires more regularity in space than what we have for our solutions (we have to compute, although only transiently, \(\Delta \mathcal {L}_{\xi _{k}} ^{2}\Omega \)). In order to make the computation rigorous, one has to regularize solutions by mollifiers or Yosida approximations and do the computations on the regularizations. In this process, commutators will appear and one has to check at the end that they converge to zero. The details are tedious, but straightforward and we do not write all of them here.
We are now ready to prove the general uniqueness result contained in Theorem 8. More precisely we prove the following
Theorem 15
Let \(\omega _{0}\in W_{\sigma }^{2,2}\left( \mathbb {T} ^{3},\mathbb {R}^{3}\right) \) and \(\left( \tau _{\max },\omega \right) \) be the maximal solution of stochastic 3D Euler Eq. (2.3) introduced in Theorem 14. Moreover, let \(\left( \tau ,\tilde{\omega }\right) \) be another maximal solutions of the same equation with the same initial condition \(\omega _{0}\in W_{\sigma }^{2,2}\left( \mathbb {T} ^{3},\mathbb {R}^{3}\right) \). Then, necessarily \(\tau =\tau _{\max }\) and \(\omega =\omega ^{\prime }\)on \([0,\tau _{\max })\).
Proof
From the local uniqueness result proved above, we deduce that \(\omega =\tilde{\omega }\ \)on \([0,\min \left( \tau ,\tau _{\max }\right) ).\ \)By an argument similar to the one in Theorem 14, we cannot have \(\tau _{\max }<\tau \) on any non-trivial set. Hence, \(\tau <\tau _{\max }.\,\)But then from the maximality property of \((\tau ,\tilde{\omega })\), it follows that necessarily \(\tau =\tau _{\max }\ \) and therefore \(\omega =\tilde{\omega }\ \)on \([0,\tau _{\max }).\)\(\square \)
Proof of the Beale–Kato–Majda Criterion
In this section, we prove Theorem 9. In the following, we will use the fact that there exist two constants \(C_{1}\)and \(C_{2}\) such that
$$\begin{aligned} C_{1}\left| \left| \omega \right| \right| _{2,2}\le \left| \left| v\right| \right| _{3,2}\le C_{2}\left| \left| \omega \right| \right| _{2,2}. \end{aligned}$$
(3.4)
The first inequality follows from that fact that \(\omega ={\text {curl}} v\), while the second inequality follows from (2.4).
Lemma 16
There is a constant C such thatFootnote 5
$$\begin{aligned} \left| \left| \nabla v\right| \right| _{\infty }\le C\left( 1+(\log ( \left| \left| \omega \right| \right| _{2,2} ^{2}+e)) \left| \left| \omega \right| \right| _{\infty } \right) . \end{aligned}$$
(3.5)
Proof
By comparing Beale et al. (1984), the following inequality holds true
$$\begin{aligned} \left| \left| \nabla v\right| \right| _{\infty }\le C\left( 1+\left( 1+\log ^{+}\left| \left| v\right| \right| _{3,2}\right) \left| \left| \omega \right| \right| _{\infty }\right) +\left| \left| \omega \right| \right| _{2}. \end{aligned}$$
(3.6)
The result is then obtained from (3.4), the obvious inequality \(1+\log ^{+}a\le C\log \left( a+e\right) \) for C sufficiently large (say \(C\ge 2\)) and the fact that \(\left| \left| \omega \right| \right| _{2}\le C\left| \left| \omega \right| \right| _{\infty }\) on a torus.\(\square \)
Theorem 17
Let \(\tau ^{1}\) and \(\tau ^{2}\) be the following stopping times
$$\begin{aligned} \tau ^{1}&=\lim _{n\rightarrow \infty }\tau _{n}^{1}~~where~~\tau _{n}^{1}=\inf _{t\ge 0}\left\{ t\ge 0|~~\left| \left| \omega _{t}\right| \right| _{2,2}\ge n\right\} ,\\ \tau ^{2}&=\lim _{n\rightarrow \infty }\tau _{n}^{2}~~where~~\tau _{n}^{2}=\inf _{t\ge 0}\left\{ t\ge 0|~~\int _{0}^{t}\left| \left| \omega _{s}\right| \right| _{\infty }\text {d}s\ge n\right\} \,. \end{aligned}$$
Then, P-almost surely \(\tau ^{1}=\tau ^{2}\).
Proof
Step 1
\(\tau ^{1}\le \tau ^{2}\)
.
From the imbedding \(W^{2,2}\left( \mathbb {T}^{3};\mathbb {R}^{3}\right) \subset C\left( \mathbb {T}^{3};\mathbb {R}^{3}\right) ,\) there exists C such that \(\left| \left| \omega \right| \right| _{\infty }\le C\left| \left| \omega \right| \right| _{2,2}\). Then,
$$\begin{aligned} \int _{0}^{\tau _{n}^{1}}\left| \left| \omega _{s}\right| \right| _{\infty }\text {d}s\le \left( \left[ C\right] +1\right) \sup _{s\le \tau _{n}^{1} }\left| \left| \omega _{s}\right| \right| _{2,2}\le \left( \left[ C\right] +1\right) n. \end{aligned}$$
Hence, \(\tau _{n}^{1}\le \tau _{\left( \left[ C\right] +1\right) n}^{2} \le \tau ^{2}\), and therefore, the claim holds true.
Step 2 \(\tau ^{2}\le \tau ^{1}\). P-a.s.
We prove that for any \(n,k>0\) we have
$$\begin{aligned} \mathbb {E}\left[ \log \left( \left( \sup _{s\in \left[ 0,\tau _{n}^{2}\wedge k\right] }\left| \left| \omega _{t}\right| \right| _{2,2}\right) ^{2}+e\right) \right] <\infty . \end{aligned}$$
(3.7)
In particular, \(\sup _{s\in \left[ 0,\tau _{n}^{2}\wedge k\right] }\left| \left| \omega _{t}\right| \right| _{2,2}\) is a finite random variable P-almost surely, that is
$$\begin{aligned} \mathbb {P}\left( \sup _{s\in \left[ 0,\tau _{n}^{2}\wedge k\right] }\left| \left| \omega _{t}\right| \right| _{2,2}<\infty \right) =1. \end{aligned}$$
Since
$$\begin{aligned} \left\{ \sup _{s\in \left[ 0,\tau _{n}^{2}\wedge k\right] }\left| \left| \omega _{t}\right| \right| _{2,2}<\infty \right\}= & {} \bigcup _{N}\left\{ \sup _{s\in \left[ 0,\tau _{n}^{2}\wedge k\right] }\left| \left| \omega _{t}\right| \right| _{2,2}<N\right\} \\\subset & {} \bigcup _{N}\left\{ \tau _{n}^{2}\wedge k<\tau _{N}^{1}\right\} \subset \left\{ \tau _{n}^{2}\wedge k\le \tau ^{1}\right\} \end{aligned}$$
we deduce that \(\tau _{n}^{2}\wedge k\le \tau ^{1}\)P-almost surely. Then,
$$\begin{aligned} \left\{ \tau ^{2}\le \tau ^{1}\right\} =\left\{ \lim _{n\mapsto \infty }\tau _{n}^{2}\le \tau ^{1}\right\} =\bigcap _{n}\left\{ \tau _{n}^{2}\le \tau ^{1}\right\} =\bigcap _{n}\bigcap _{k}\left\{ \tau _{n}^{2}\wedge k<\tau ^{1}\right\} \end{aligned}$$
and therefore, the second claim holds true since all the sets in the above intersection have full measure.
To prove (3.7), we proceed as follows: For arbitrary \(R>0\), and \(\nu \in \left( 0,1\right) ,\) let \(\omega _{R}^{\nu }\) the solution of equation
$$\begin{aligned} \text {d}\omega _{R}^{\nu }+\kappa _{R}\left( \omega _{R}^{\nu }\right) \mathcal {L} _{v_{R}^{\nu }}\omega _{R}^{\nu }\ \text {d}t+\sum _{k=1}^{\infty }\mathcal {L}_{\xi _{k} }\omega _{R}^{\nu }\text {d}B_{t}^{k}=\nu \Delta ^{5}\omega _{R}^{\nu }\text {d}t+\frac{1}{2} \sum _{k=1}^{\infty }\mathcal {L}_{\xi _{k}}^{2}\omega _{R}^{\nu }\ \text {d}t \end{aligned}$$
with \(\omega _{R}^{\nu }|_{t=0}=\omega _{0}\). We know from the analysis in the next section that if \(\omega _{0}\in W^{2,2}\left( \mathbb {T}^{3};\mathbb {R}^{3}\right) \), then \(\omega _{t}\in W^{4,2}\left( \mathbb {T}^{3};\mathbb {R}^{3}\right) \). To simplify notation, in the following we will omit the dependence on \(\nu \) and R of \(\omega _{R}^{\nu }\) and denote it by \(\omega \). We have that
$$\begin{aligned}&\frac{1}{2}d\left\| \omega \right\| _{L^{2}}^{2}+\kappa _{R}\left( \omega \right) \left\langle \mathcal {L}_{v}\omega ,\omega \right\rangle \ \text {d}t+\sum _{k=1}^{\infty }\left\langle \mathcal {L}_{\xi _{k}}\omega ,\omega \right\rangle \ \text {d}B_{t}^{k}\\&\quad =\nu \left\langle \Delta ^{5}\omega ,\omega \right\rangle \text {d}t+\frac{1}{2}\sum _{k=1}^{\infty }(\left\langle \mathcal {L}_{\xi _{k}}^{2} \omega ,\omega \right\rangle \ \text {d}t+\left\langle \mathcal {L}_{\xi _{k}} \omega ,\mathcal {L}_{\xi _{k}}\omega \right\rangle )\ \text {d}t.\\&\frac{1}{2}d\left\| \Delta \omega \right\| _{L^{2}}^{2}+\kappa _{R}\left( \omega \right) \left\langle \Delta \mathcal {L}_{v}\omega ,\Delta \omega \right\rangle \ \text {d}t+\sum _{k=1}^{\infty }\left\langle \Delta \mathcal {L}_{\xi _{k}}\omega ,\Delta \omega \right\rangle \ \text {d}B_{t}^{k}\\&\quad =\nu \left\langle \Delta ^{6}\omega ,\Delta \omega \right\rangle \text {d}t+\frac{1}{2}\sum _{k=1}^{\infty }(\left\langle \Delta \mathcal {L}_{\xi _{k}} ^{2}\omega ,\Delta \omega \right\rangle \ +\left\langle \Delta \mathcal {L} _{\xi _{k}}\omega ,\Delta \mathcal {L}_{\xi _{k}}\omega \right\rangle )\ \text {d}t. \end{aligned}$$
Next we will use the following set of inequalities
$$\begin{aligned} \left\langle \Delta ^{5}\omega ,\omega \right\rangle&=-\,\left| \left| \Delta ^{5/2}\omega \right| \right| _{L^{2}}\le 0,~~\left\langle \Delta ^{6}\omega ,\Delta \omega \right\rangle =-\,\left| \left| \Delta ^{7/2}\omega \right| \right| _{L^{2}}\le 0\\ \left| \left\langle \mathcal {L}_{v}\omega ,\omega \right\rangle \right|&\le \left| \left| \nabla v\right| \right| _{\infty }\left\| \omega \right\| _{L^{2}}^{2},~~\left| \left\langle \Delta \mathcal {L}_{v}\omega ,\Delta \omega \right\rangle \right| \le C\left( \left\| \omega \right\| _{\infty }+\left| \left| \nabla v\right| \right| _{\infty }\right) \left\| \omega \right\| _{2,2}^{2}. \end{aligned}$$
The first two inequalities are obvious. The third one comes from the fact that in \(\left\langle \mathcal {L}_{v}\omega ,\omega \right\rangle \) the term \(\left\langle v\cdot \nabla \omega ,\omega \right\rangle \) vanishes and the term \(\left| \left\langle \omega \cdot \nabla v,\omega \right\rangle \right| \) is bounded by \(\left| \left| \nabla v\right| \right| _{\infty }\left\| \omega \right\| _{L^{2}}^{2}\). The last one, the most delicate one, comes from Lemma 25:
$$\begin{aligned} \left| \left\langle \Delta \mathcal {L}_{v}\omega ,\Delta \omega \right\rangle \right| \le C\left\| \nabla v\right\| _{\infty }\left\| \omega \right\| _{2,2}^{2}+C\left\| \omega \right\| _{\infty }\left\| \nabla v\right\| _{2,2}\left\| \omega \right\| _{2,2} \end{aligned}$$
and then, we use \(\left\| \nabla v\right\| _{2,2}\le \left\| v\right\| _{3,2}\le C\left\| \omega \right\| _{2,2}\); see (2.4).
Hence,
$$\begin{aligned} d\left\| \omega \right\| _{L^{2}}^{2}+2\sum _{k=1}^{\infty }\left\langle \mathcal {L}_{\xi _{k}}\omega ,\omega \right\rangle \ \text {d}B_{t}^{k}&\le C\left( 1+\left\| \omega \right\| _{\infty }+\left| \left| \nabla v\right| \right| _{\infty }\right) \left\| \omega \right\| _{L^{2}}^{2}\ \text {d}t \end{aligned}$$
(3.8)
$$\begin{aligned} d\left\| \Delta \omega \right\| _{L^{2}}^{2}+2\sum _{k=1}^{\infty }\left\langle \Delta \mathcal {L}_{\xi _{k}}\omega ,\Delta \omega \right\rangle \ \text {d}B_{t}^{k}&\le C\left( 1+\left\| \omega \right\| _{\infty }+\left| \left| \nabla v\right| \right| _{\infty }\right) \left\| \omega \right\| _{2,2}^{2}\ \text {d}t, \end{aligned}$$
(3.9)
where we used inequalities (2.9), (2.10) coupled with assumption (2.11) to control
$$\begin{aligned} \sum _{k=1}^{\infty }\left( \left\langle \mathcal {L}_{\xi _{k}}^{2}\omega ,\omega \right\rangle +\left\langle \mathcal {L}_{\xi _{k}}\omega ,\mathcal {L}_{\xi _{k}}\omega \right\rangle +\left\langle \Delta \mathcal {L} _{\xi _{k}}^{2}\omega ,\Delta \omega \right\rangle \ +\left\langle \Delta \mathcal {L}_{\xi _{k}}\omega ,\Delta \mathcal {L}_{\xi _{k}}\omega \right\rangle \right) . \end{aligned}$$
Using Itô’s formula and the fact that \(\left| \left| \omega _{t}\right| \right| _{2,2}^{2}\le \left\| \omega _{t}\right\| _{L^{2}}^{2}+\left\| \Delta \omega _{t}\right\| _{L^{2}}^{2}\), we deduce, from (3.8)+(3.9), that
$$\begin{aligned} d\log \left( \left| \left| \omega _{t}\right| \right| _{2,2}^{2}+e\right)&\le \frac{1}{\left( \left| \left| \omega _{t}\right| \right| _{2,2}^{2}+e\right) }d\left| \left| \omega _{t}\right| \right| _{2,2}^{2}\\&\quad -\,\frac{2}{\left( \left| \left| \omega _{t}\right| \right| _{2,2}^{2}+e\right) ^{2}}\sum _{k=1}^{\infty }\left( \left| \left\langle \Delta \mathcal {L}_{\xi _{k}}\omega ,\Delta \omega \right\rangle \right| +\left| \left\langle \mathcal {L}_{\xi _{k}}\omega ,\omega \right\rangle \right| \ \right) ^{2}\\&\le \frac{C}{\left( \left| \left| \omega _{t}\right| \right| _{2,2}^{2}+e\right) }\left( 1+\left\| \omega \right\| _{\infty }+\left| \left| \nabla v\right| \right| _{\infty }\right) \left\| \omega \right\| _{2,2}^{2}\ \text {d}t+\text {d}M_{t}, \end{aligned}$$
where \(M\ \)is the (local) martingale defined as
$$\begin{aligned} M_{t}:=\sum _{k=1}^{\infty }\int _{0}^{t}\frac{2\left( \left\langle \mathcal {L}_{\xi _{k}}\omega ,\omega \right\rangle +\left\langle \Delta \mathcal {L}_{\xi _{k}}\omega ,\Delta \omega \right\rangle \right) }{\left( \left| \left| \omega _{t}\right| \right| _{2,2}^{2}+e\right) }\ \text {d}B_{s}^{k} \end{aligned}$$
We use now (3.5) to deduce
$$\begin{aligned} d\log \left( \left| \left| \omega _{t}\right| \right| _{2,2}^{2}+e\right) \le mC\left( 1+\left| \left| \omega \right| \right| _{\infty }\right) \log \left( \left| \left| \omega _{t}\right| \right| _{2,2}^{2}+e\right) \ \text {d}t+\text {d}M_{t}. \end{aligned}$$
which, in turn, implies that
$$\begin{aligned} e^{-CY_{t}}\log \left( \left| \left| \omega _{t}\right| \right| _{2,2}^{2}+e\right) \le \log \left( \left\| \omega _{0}\right\| _{L^{2} }^{2}+\left\| \Delta \omega \right\| _{L^{2}}^{2}+e\right) +\int _{0} ^{t}e^{-CY_{s}}\text {d}M_{s}, \end{aligned}$$
(3.10)
where
$$\begin{aligned} Y_{t}=\int _{0}^{t}\left( 1+\left\| \omega \right\| _{\infty }\right) \text {d}s \end{aligned}$$
and we use the conventions \(e^{-\infty }=0\) and \(0\times \infty =0\).
Again, by using the fact \(\left| \left\langle \mathcal {L}_{\xi _{k}} \omega ,\omega \right\rangle \right| \) is controlled by \(\left| \left| \nabla \xi _{k}\right| \right| _{\infty }\left\| \omega \right\| _{L^{2}}^{2}\) and that \(\left\langle \Delta \mathcal {L} _{\xi _{k}}\omega ,\Delta \omega \right\rangle |\) is controlled by \(\left\| \xi _{k}\right\| _{3,2}\left\| \omega \right\| _{2,2}^{2}\) following Lemma 25, we deduced that
$$\begin{aligned} \left| \left\langle \mathcal {L}_{\xi _{k}}\omega ,\omega \right\rangle +\left\langle \Delta \mathcal {L}_{\xi _{k}}\omega ,\Delta \omega \right\rangle \right| \le C\left\| \xi _{k}\right\| _{3,2}\left\| \omega \right\| _{2,2}^{2} . \end{aligned}$$
(3.11)
From (3.11) and assumption (2.8), we deduce the following control on the quadratic variation of stochastic integral in (3.10)
$$\begin{aligned} \left[ \int _{0}^{\cdot }e^{-CY_{s}}\text {d}M_{s}\right] _{t}&=4\sum _{k=1}^{\infty }\int _{0}^{t}e^{-2CY_{s}}\frac{\left( \left\langle \mathcal {L}_{\xi _{k}}\omega ,\omega \right\rangle +\left\langle \Delta \mathcal {L}_{\xi _{k}}\omega ,\Delta \omega \right\rangle \right) ^{2}}{\left( \left| \left| \omega _{t}\right| \right| _{2,2}^{2}+e\right) ^{2}}\ \text {d}s\\&\le 4C\sum _{k=1}^{\infty }\left\| \xi _{k}\right\| _{3,2}^{2}\int _{0}^{t}\frac{\left\| \omega \right\| _{2,2}^{4}}{\left( \left| \left| \omega _{t}\right| \right| _{2,2}^{2}+e\right) ^{2}}\ \text {d}s\\&\le Ct. \end{aligned}$$
Finally, using the Burkholder–Davis–Gundy inequality (see e.g. Theorem 3.28, page 166 in Karatzas and Shreve (1991)), we deduce that
$$\begin{aligned} \mathbb {E}\left[ \sup _{s\in \left[ 0,t\right] }\left| \int _{0} ^{s}e^{-CY_{r}}\text {d}M_{r}\right| \right] \le C\sqrt{t}. \end{aligned}$$
This means there exists a constant C independent of \(\nu \) and R such that, upon reverting to the notation \(\omega _{R}^{\nu }\), we have
$$\begin{aligned}&\mathbb {E}\left[ \sup _{s\in \left[ 0,t\right] }e^{-\int _{0}^{s}C\left( 1+\left\| \omega _{R}^{\nu }\right\| _{\infty }\right) \text {d}r}\log \left( \left| \left| \omega _{R}^{\nu }\left( s\right) \right| \right| _{2,2}^{2}+e\right) \right] \\&\quad \le \log \left( \left\| \omega _{0}\right\| _{L^{2}}^{2}+\left\| \Delta \omega \right\| _{L^{2} }^{2}+e\right) +C\sqrt{t} \end{aligned}$$
By Fatou’s lemma, it follows that the same limit holds for the limit of \(\omega _{R}^{\nu }\) as \(\nu \) tends to 0 and R tends to \(\infty \), hence
$$\begin{aligned} \mathbb {E}\left[ \sup _{s\in \left[ 0,t\right] }e^{-\int _{0}^{s}C\left( 1+\left| \left| \omega _{r}\right| \right| _{\infty }\right) \text {d}r}\log \left( \left| \left| \omega _{s}\right| \right| _{2,2}^{2}+e\right) \right] <\infty . \end{aligned}$$
It follows that
$$\begin{aligned}&e^{-Ck(1+n)}\mathbb {E}\left[ \sup _{s\in \left[ 0,\tau _{n}^{2}\wedge k\right] }\log \left( \left| \left| \omega \left( s\right) \right| \right| _{2,2}^{2}+e\right) \right] \\&\quad \le \mathbb {E}\left[ \sup _{s\in \left[ 0,\tau _{n}^{2}\wedge k\right] }e^{-CY_{s}}\log \left( \left| \left| \omega \left( s\right) \right| \right| _{2,2}^{2}+e\right) \right] \\&\quad \le \mathbb {E}\left[ \sup _{s\in \left[ 0,k\right] }e^{-CY_{s}}\log \left( \left| \left| \omega \left( s\right) \right| \right| _{2,2}^{2}+e\right) \right] \\&\quad \le \log \left( \left\| \omega _{0}\right\| _{L^{2}}^{2}+\left\| \Delta \omega \right\| _{L^{2}}^{2}+e\right) +C\sqrt{t}<\infty . \end{aligned}$$
which gives us (3.7). The proof is now complete. \(\square \)
Remark 18
The original Beale–Kato–Majda result refers to a control of the explosion time of \(\left| \left| v\right| \right| _{3,2}\) in terms of \(\left| \left| \omega \right| \right| _{\infty }.\) Our result refers to a control of the explosion time for \(\left| \left| \omega \right| \right| _{2,2}\) in terms of \(\left| \left| \omega \right| \right| _{\infty }\). However, due to (3.4), we can restate our result in terms of \(\left| \left| v\right| \right| _{3,2}\) as well.
Global Existence of the Truncated Solution
Consider the following regularized equation with cut-off, with \(\nu ,R>0\),
$$\begin{aligned}&\text {d}\omega _{R}^{\nu }+\kappa _{R}\left( \omega _{R}^{\nu }\right) \mathcal {L} _{v_{R}^{\nu }}\omega _{R}^{\nu }\ \text {d}t+\sum _{k=1}^{\infty }\mathcal {L}_{\xi _{k} }\omega _{R}^{\nu }\text {d}B_{t}^{k}\nonumber \\&\quad =\nu \Delta ^{5}\omega _{R}^{\nu }\text {d}t+\frac{1}{2}\sum _{k=1}^{\infty } \mathcal {L}_{\xi _{k}}^{2}\omega _{R}^{\nu }\ \text {d}t\,,\qquad \omega _{R}^{\nu } |_{t=0}=\omega _{0}\,, \end{aligned}$$
(3.12)
where \(\omega _{R}^{\nu }={\text {curl}}v_{R}^{\nu }\), \({\text {div}} v_{R}^{\nu }=0\). On the solutions of this problem, we want to perform computations involving terms like \(\Delta \mathcal {L}_{v}^{2}\omega \left( t\right) \), so we need \(\omega \left( t\right) \in W^{4,2}\left( \mathbb {T}^{3};\mathbb {R}^{3}\right) \). This is why we introduce the strong regularization \(\nu \Delta ^{5}\omega _{R}^{\nu }\); the precise power 5 can be understood from the technical computations of Step 1. While not optimal, it a simple choice that allows us to avoid more heavy arguments.
This regularized problem has the following property.
We understand Eq. (3.12) either in the mild semigroup sense (see below the proof) or in a weak sense over test functions, which are equivalent due to the high regularity of solutions. However, \(\Delta ^{5} \omega _{R}^{\nu }\) cannot be interpreted in a classical sense, since the solutions, although very regular, will not be in \(W^{10,2}\left( \mathbb {T} ^{3};\mathbb {R}^{3}\right) \). The other terms of Eq. (3.12) can be interpreted in a classical sense.
Lemma 19
For every \(\nu ,R>0\) and \(\omega _{0}\in W_{\sigma }^{2,2}\left( \mathbb {T}^{3},\mathbb {R}^{3}\right) \), there exists a pathwise unique global strong solution \(\omega _{R}^{\nu }\), of class \(L^{2}\left( \Xi ;C\left( \left[ 0,T\right] ;W_{\sigma }^{2,2}\left( \mathbb {T}^{3};\mathbb {R}^{3}\right) \right) \right) \) for every \(T>0\). Its paths have a.s. the additional regularity \(C\left( \left[ \delta ,T\right] ;W^{4,2}\left( \mathbb {T}^{3};\mathbb {R}^{3}\right) \right) \), for every \(T>\delta >0\).
Proof
Step 1 (preparation) In the following, we assume to have fixed \(T>0\) and that all constants are generically denoted by \(C>0\) any constant, with the understanding that it may depend on T.
Let \(D\left( A\right) =W_{\sigma }^{10,2}\left( \mathbb {T}^{3} ;\mathbb {R}^{3}\right) \) and \(A:D\left( A\right) \subset L_{\sigma } ^{2}\left( \mathbb {T}^{3},\mathbb {R}^{3}\right) \rightarrow L_{\sigma } ^{2}\left( \mathbb {T}^{3},\mathbb {R}^{3}\right) \) be the operator \(A\omega =\nu \Delta ^{5}\omega \); \(L_{\sigma }^{2}\left( \mathbb {T} ^{3},\mathbb {R}^{3}\right) \) denotes here the closure of \(D\left( A\right) \) in \(L^{2}\left( \mathbb {T}^{3},\mathbb {R}^{3}\right) \) (the trace of the periodic boundary condition at the level of \(L^{2}\) spaces can be characterized; see Temam 1977). The operator A is self-adjoint and negative definite. Let \(e^{tA}\) be the semigroup in \(L_{\sigma }^{2}\left( \mathbb {T}^{3},\mathbb {R}^{3}\right) \) generated by A. The fractional powers \(\left( I-A\right) ^{\alpha }\) are well defined, for every \(\alpha >0\), and are bi-continuous bijections between \(W_{\sigma }^{\beta ,2}\left( \mathbb {T}^{3};\mathbb {R}^{3}\right) \) and \(W_{\sigma }^{\beta -10\alpha ,2}\left( \mathbb {T}^{3};\mathbb {R}^{3}\right) \), for every \(\beta \ge 10\alpha \), in particular
$$\begin{aligned} \left\| f\right\| _{W^{10\alpha ,2}}\le C_{\alpha }\left\| \left( I-A\right) ^{\alpha }f\right\| _{L^{2}} \end{aligned}$$
for some constant \(C_{\alpha }>0\), for all \(f\in W_{\sigma }^{10\alpha ,2}\left( \mathbb {T}^{3};\mathbb {R}^{3}\right) \).
In the sequel, we write \(\left\langle f,g\right\rangle =\int _{\mathbb {T}^{3} }f\left( x\right) \cdot g\left( x\right) \text {d}x\). We work on the torus, which simplifies some definitions and properties; thus, we write \(\left( 1-\Delta \right) ^{s/2}f\) for the function having Fourier transform \(\left( 1+\left| \xi \right| ^{2}\right) ^{s/2}\widehat{f}\left( \xi \right) \) (\(\widehat{f}\left( \xi \right) \) being the Fourier transform of f); similarly, we write \(\Delta ^{-1}f\) for the function having Fourier transform \(\left| \xi \right| ^{-1}\widehat{f}\left( \xi \right) \).
The fractional powers commute with \(e^{tA}\) and have the property (from the general theory of analytic semigroups, see Pazy 1983) that for every \(\alpha >0\) and \(T>0\)
$$\begin{aligned} \left\| \left( I-A\right) ^{\alpha }e^{tA}f\right\| _{L^{2}}\le \frac{C_{\alpha }}{t^{\alpha }}\left\| f\right\| _{L^{2}} \end{aligned}$$
for all \(t\in (0,T]\) and \(f\in L_{\sigma }^{2}\left( \mathbb {T}^{3} ;\mathbb {R}^{3}\right) \).
From these properties, it follows that, for \(p=2,4\)
$$\begin{aligned} \left\| \int _{0}^{t}e^{\left( t-s\right) A}f\left( s\right) \text {d}s\right\| _{W^{p,2}}^{2}\le & {} C\int _{0}^{t}\frac{1}{\left( t-s\right) ^{p/10}}\left\| f\left( s\right) \right\| _{L^{2}}^{2}\text {d}s\nonumber \\\le & {} CT^{1-\frac{p}{10}}\sup _{t\in \left[ 0,T\right] }\left\| f\left( s\right) \right\| _{L^{2}}^{2} \end{aligned}$$
(3.13)
for all \(f\in C\left( \left[ 0,T\right] ;L_{\sigma }^{2}\left( \mathbb {T}^{3};\mathbb {R}^{3}\right) \right) \) and \(t\in \left[ 0,T\right] \), because
$$\begin{aligned} \left\| \int _{0}^{t}e^{\left( t-s\right) A}f\left( s\right) \text {d}s\right\| _{W^{p,2}}\le & {} C\left\| \left( I-A\right) ^{p/10}\int _{0}^{t}e^{\left( t-s\right) A}f\left( s\right) \text {d}s\right\| _{L^{2}}\\&\le C\int _{0}^{t}\frac{1}{\left( t-s\right) ^{p/10}}\left\| f\left( s\right) \right\| _{L^{2}}\text {d}s. \end{aligned}$$
In particular, the map \(f\mapsto \left( \int _{0}^{t}e^{\left( t-s\right) A}f\left( s\right) \text {d}s\right) _{t\in \left[ 0,T\right] }\) is linear continuous from \(C\left( \left[ 0,T\right] ;L_{\sigma }^{2}\left( \mathbb {T}^{3};\mathbb {R}^{3}\right) \right) \) to \(C\left( \left[ 0,T\right] ;W_{\sigma }^{2,2}\left( \mathbb {T}^{3};\mathbb {R}^{3}\right) \right) \). Moreover, for \(p=2,4\)
$$\begin{aligned}&\mathbb {E}\left[ \sup _{t\in \left[ 0,T\right] }\left\| \sum _{k=1}^{\infty }\int _{0}^{t}e^{\left( t-s\right) A}f_{k}\left( s\right) \text {d}B_{s}^{k}\right\| _{W^{p,2}}^{2}\right] \nonumber \\&\quad \le C\mathbb {E}\left[ \sum _{k=1}^{\infty }\int _{0}^{t} \frac{1}{\left( t-s\right) ^{p/5}}\left\| f_{k}\left( s\right) \right\| _{L^{2}}^{2}\text {d}s\right] \nonumber \\&\quad =CT^{1-p/5}\mathbb {E}\left[ \sup _{s\in \left[ 0,T\right] }\sum _{k=1}^{\infty }\left\| f_{k}\left( s\right) \right\| _{L^{2}}^{2}\right] \end{aligned}$$
(3.14)
because
$$\begin{aligned}&\mathbb {E}\left[ \sup _{t\in \left[ 0,T\right] }\left\| \sum _{k=1}^{\infty } \int _{0}^{t}e^{\left( t-s\right) A}f_{k}\left( s\right) \text {d}B_{s} ^{k}\right\| _{W^{p,2}}^{2}\right] \\&\quad =\mathbb {E}\left[ \sup _{t\in \left[ 0,T\right] }\left\| \sum _{k=1}^{\infty }\int _{0}^{t}\left( I-A\right) ^{p/10} e^{\left( t-s\right) A}f_{k}\left( s\right) \text {d}B_{s}^{k}\right\| _{L^{2} }^{2}\right] \\&\quad \le C\mathbb {E}\left[ \sum _{k=1}^{\infty }\int _{0}^{t}\left\| \left( I-A\right) ^{p/10}e^{\left( t-s\right) A}f_{k}\left( s\right) \right\| _{L^{2}}^{2}\text {d}s\right] \\&\quad \le C\mathbb {E}\left[ \sum _{k=1}^{\infty }\int _{0}^{t}\frac{1}{\left( t-s\right) ^{p/5}}\left\| f_{k}\left( s\right) \right\| _{L^{2}}^{2}\text {d}s\right] . \end{aligned}$$
It is here that we use the power 5 of \(\Delta \), otherwise a smaller power would suffice.
Step 2 (preparation, cont.) The function \(\omega \mapsto \kappa _{R}\left( \omega \right) \mathcal {L}_{v}\omega \) from \(W^{2,2}\left( \mathbb {T}^{3};\mathbb {R}^{3}\right) \) to \(L^{2}\left( \mathbb {T} ^{3};\mathbb {R}^{3}\right) \) is Lipschitz continuous, and it has linear grows (the constants in both properties depend on R). Let us check the Lipschitz continuity; the linear growth is an easy consequence, applying Lipschitz continuity with respect to a given element \(\omega ^{0}\).
It is sufficient to check Lipschitz continuity in any ball \(B\left( 0,r\right) \), centred at the origin of radius r, in \(W^{2,2}\left( \mathbb {T}^{3};\mathbb {R}^{3}\right) \). Indeed, when it is true, one can argue as follows. Take \(\omega ^{\left( i\right) }\), \(i=1,2\), in \(W^{2,2}\left( \mathbb {T}^{3};\mathbb {R}^{3}\right) \). If they belong to \(B\left( 0,R+2\right) \), we have Lipschitz continuity. The case that both are outside \(B\left( 0,R+2\right) \) is trivial, because the cut-off function vanishes. If one is inside \(B\left( 0,R+2\right) \) and the other outside, consider the two cases: if the one inside is outside \(B\left( 0,R+1\right) \), it is trivial again, because the cut-off function vanishes for both functions. If the one inside, say \(\omega ^{\left( 1\right) }\), is in \(B\left( 0,R\right) \), then \(\mathcal {L}_{v^{\left( 1\right) }} \omega ^{\left( 1\right) }\kappa _{R}\left( \omega ^{\left( 1\right) }\right) -\mathcal {L}_{v^{\left( 2\right) }}\omega ^{\left( 2\right) }\kappa _{R}\left( \omega ^{\left( 2\right) }\right) =\mathcal {L} _{v^{\left( 1\right) }}\omega ^{\left( 1\right) }\kappa _{R}\left( \omega ^{\left( 1\right) }\right) \); one has \(\left\| \mathcal {L} _{v^{\left( 1\right) }}\omega ^{\left( 1\right) }\kappa _{R}\left( \omega ^{\left( 1\right) }\right) \right\| _{L^{2}}\le C_{R}\) (same computations done below) and \(\left\| \omega ^{\left( 1\right) } -\omega ^{\left( 2\right) }\right\| _{W^{2,2}}\ge c_{R}\), for two constants \(c_{R},C_{R}>0\), hence
$$\begin{aligned} \left\| \mathcal {L}_{v^{\left( 1\right) }}\omega ^{\left( 1\right) }\kappa _{R}\left( \omega ^{\left( 1\right) }\right) -\mathcal {L} _{v^{\left( 2\right) }}\omega ^{\left( 2\right) }\kappa _{R}\left( \omega ^{\left( 2\right) }\right) \right\| _{L^{2}}\le \frac{C_{R}}{c_{R}}\left\| \omega ^{\left( 1\right) }-\omega ^{\left( 2\right) }\right\| _{W^{2,2}}. \end{aligned}$$
Therefore, let us prove that the function \(\omega \mapsto \kappa _{R}\left( \omega \right) \mathcal {L}_{v}\omega \) from \(W^{2,2}\left( \mathbb {T} ^{3};\mathbb {R}^{3}\right) \) to \(L^{2}\left( \mathbb {T}^{3};\mathbb {R} ^{3}\right) \) is Lipschitz continuous on \(B\left( 0,r\right) \subset W^{2,2}\left( \mathbb {T}^{3};\mathbb {R}^{3}\right) \). Given \(\omega ^{\left( i\right) }\in B\left( 0,r\right) \), \(i=1,2\), let us use the decomposition
$$\begin{aligned}&\mathcal {L}_{v^{\left( 1\right) }}\omega ^{\left( 1\right) }\kappa _{R}\left( \omega ^{\left( 1\right) }\right) -\mathcal {L}_{v^{\left( 2\right) }}\omega ^{\left( 2\right) }\kappa _{R}\left( \omega ^{\left( 2\right) }\right) \\&\quad =\mathcal {L}_{v^{\left( 1\right) }}\left( \omega ^{\left( 1\right) }-\omega ^{\left( 2\right) }\right) \kappa _{R}\left( \omega ^{\left( 1\right) }\right) +\mathcal {L}_{\left( v^{\left( 1\right) }-v^{\left( 2\right) }\right) }\omega ^{\left( 2\right) }\kappa _{R}\left( \omega ^{\left( 2\right) }\right) \\&\qquad +\mathcal {L}_{v^{\left( 1\right) } }\omega ^{\left( 2\right) }\left( \kappa _{R}\left( \omega ^{\left( 1\right) }\right) -\kappa _{R}\left( \omega ^{\left( 2\right) }\right) \right) . \end{aligned}$$
Then,
$$\begin{aligned}&\left\| \mathcal {L}_{v^{\left( 1\right) }}\left( \omega ^{\left( 1\right) }-\omega ^{\left( 2\right) }\right) \kappa _{R}\left( \omega ^{\left( 1\right) }\right) \right\| _{L^{2}}^{2}\\&\quad \le \kappa _{R}\left( \omega ^{\left( 1\right) }\right) ^{2}\left\| v^{\left( 1\right) }\right\| _{\infty }^{2}\left\| \nabla \left( \omega ^{\left( 1\right) }-\omega ^{\left( 2\right) }\right) \right\| _{L^{2}}^{2}+\kappa _{R}\left( \omega ^{\left( 1\right) }\right) ^{2}\left\| \omega ^{\left( 1\right) }-\omega ^{\left( 2\right) }\right\| _{L^{4}}^{2}\left\| \nabla v^{\left( 1\right) }\right\| _{L^{4}}^{2}\\&\quad \le \kappa _{R}\left( \omega ^{\left( 1\right) }\right) ^{2}\left\| \nabla v^{\left( 1\right) }\right\| _{\infty }^{2}\left\| \omega ^{\left( 1\right) }-\omega ^{\left( 2\right) }\right\| _{W^{2,2} }^{2}+\kappa _{R}\left( \omega ^{\left( 1\right) }\right) ^{2}\left\| \omega ^{\left( 1\right) }-\omega ^{\left( 2\right) }\right\| _{W^{2,2} }^{2}\left\| \nabla v^{\left( 1\right) }\right\| _{\infty }^{2}\\&\quad \le R^{2}\left\| \omega ^{\left( 1\right) }-\omega ^{\left( 2\right) }\right\| _{W^{2,2}}^{2}; \end{aligned}$$
similarly
$$\begin{aligned}&\left\| \mathcal {L}_{\left( v^{\left( 1\right) }-v^{\left( 2\right) }\right) }\omega ^{\left( 2\right) }\kappa _{R}\left( \omega ^{\left( 2\right) }\right) \right\| _{L^{2}}^{2}\\&\quad \le \kappa _{R}\left( \omega ^{\left( 2\right) }\right) ^{2}\left\| \nabla \left( v^{\left( 1\right) }-v^{\left( 2\right) }\right) \right\| _{\infty }^{2}\left\| \omega ^{\left( 2\right) }\right\| _{W^{2,2}}^{2}\\&\quad \le Cr^{2}\left\| v^{\left( 1\right) }-v^{\left( 2\right) }\right\| _{W^{3,2}}^{2}\le Cr^{2}\left\| \omega ^{\left( 1\right) }-\omega ^{\left( 2\right) }\right\| _{W^{2,2}}^{2} \end{aligned}$$
by the Sobolev embedding theorem and (2.4). Finally
$$\begin{aligned}&\left\| \mathcal {L}_{v^{\left( 1\right) }}\omega ^{\left( 2\right) }\left( \kappa _{R}\left( \omega ^{\left( 1\right) }\right) -\kappa _{R}\left( \omega ^{\left( 2\right) }\right) \right) \right\| _{L^{2} }^{2}\\&\quad \le \left| \kappa _{R}\left( \omega ^{\left( 1\right) }\right) -\kappa _{R}\left( \omega ^{\left( 2\right) }\right) \right| ^{2}\left\| \nabla v^{\left( 1\right) }\right\| _{\infty } ^{2}\left\| \omega ^{\left( 2\right) }\right\| _{W^{2,2}}^{2}\\&\quad \le C\left| \kappa _{R}\left( \omega ^{\left( 1\right) }\right) -\kappa _{R}\left( \omega ^{\left( 2\right) }\right) \right| ^{2}\left\| \omega ^{\left( 2\right) }\right\| _{W^{2,2}}^{4}\\&\quad \le Cr^{4}\left| \kappa _{R}\left( \omega ^{\left( 1\right) }\right) -\kappa _{R}\left( \omega ^{\left( 2\right) }\right) \right| ^{2} \end{aligned}$$
because \(\left\| \nabla v^{\left( 1\right) }\right\| _{\infty }^{2}\le C\left\| \omega ^{\left( 2\right) }\right\| _{W^{2,2}}^{2}\) as above, and then, we use the Lipschitz continuity of the function \(\omega \mapsto \kappa _{R}\left( \omega \right) \).
Step 3 (local solution by fixed point). Given \(\omega _{0}\in L^{2}\left( \Xi ;W_{\sigma }^{2,2}\left( \mathbb {T}^{3},\mathbb {R}^{3}\right) \right) \), \(\mathcal {F}_{0}\)-measurable, consider the mild equation
$$\begin{aligned} \omega \left( t\right) =\left( \Gamma \omega \right) \left( t\right) \end{aligned}$$
where
$$\begin{aligned} \left( \Gamma \omega \right) \left( t\right)&=e^{tA}\omega _{0}-\int _{0}^{t}e^{\left( t-s\right) A}\mathcal {L}_{v\left( s\right) } \omega \left( s\right) \kappa _{R}\left( \omega \left( s\right) \right) \text {d}s\\&\quad +\int _{0}^{t}e^{\left( t-s\right) A}\frac{1}{2}\sum _{k=1}^{\infty }\mathcal {L}_{\xi _{k}}^{2}\omega \left( s\right) \text {d}s-\sum _{k=1}^{\infty } \int _{0}^{t}e^{\left( t-s\right) A}\mathcal {L}_{\xi _{k}}\omega \left( s\right) \text {d}B_{s}^{k}\, \end{aligned}$$
with, as usual, \({\text {curl}}v=\omega \), \({\text {div}}v=0\). Set \(\mathcal {Y}_{T}:=L^{2}\left( \Xi ;C\left( \left[ 0,T\right] ;W^{2,2} \left( \mathbb {T}^{3};\mathbb {R}^{3}\right) \right) \right) \). The map \(\Gamma \), applied to an element \(\omega \in \mathcal {Y}_{T}\), gives us an element \(\Gamma \omega \) of the same space. Indeed:
-
(i)
\(e^{tA}\) is bounded in \(W^{2,2}\left( \mathbb {T}^{3};\mathbb {R} ^{3}\right) \) (for instance because it commutes with \(\left( I-A\right) ^{1/5}\)) hence \(e^{tA}\omega _{0}\) is in \(\mathcal {Y}_{T}\);
-
(ii)
\(\mathcal {L}_{v}\omega \kappa _{R}\left( \omega \right) \in L^{2}\left( \Xi ;C\left( \left[ 0,T\right] ;L_{\sigma }^{2}\left( \mathbb {T} ^{3};\mathbb {R}^{3}\right) \right) \right) \) by Step 2, hence \(\int _{0} ^{t}e^{\left( t-s\right) A}\mathcal {L}_{v\left( s\right) }\omega \left( s\right) \kappa _{R}\left( \omega \left( s\right) \right) \text {d}s\) is an element of \(\mathcal {Y}_{T}\), by property (3.13) of Step 1;
-
(iii)
\(\sum _{k=1}^{\infty }\mathcal {L}_{\xi _{k}}^{2}\omega \in L^{2}\left( \Xi ;C\left( \left[ 0,T\right] ;L^{2}\left( \mathbb {T}^{3};\mathbb {R} ^{3}\right) \right) \right) \) from assumption (2.6), hence \(\int _{0}^{t}e^{\left( t-s\right) A}\frac{1}{2}\sum _{k=1}^{\infty }\mathcal {L} _{\xi _{k}}^{2}\omega \left( s\right) \text {d}s\) is in \(\mathcal {Y}_{T}\) by property (3.13);
-
iv)
since, by assumption (2.7),
$$\begin{aligned} \sum _{k=1}^{\infty }\left\| \mathcal {L}_{\xi _{k}}\omega \left( s\right) \right\| _{L^{2}}^{2}\le C\left\| \omega \left( s\right) \right\| _{W^{2,2}}^{2} \end{aligned}$$
we apply property (3.14) and get that \(\sum _{k=1}^{\infty }\int _{0}^{t}e^{\left( t-s\right) A}\mathcal {L}_{\xi _{k}}\omega \left( s\right) \text {d}B_{s}^{k}\) is in \(L^{2}\left( \Xi ;C\left( \left[ 0,T\right] ;W^{2,2}\left( \mathbb {T}^{3};\mathbb {R}^{3}\right) \right) \right) \).
The proof that \(\Gamma \) is Lipschitz continuous in \(\mathcal {Y}_{T}\) is based on the same facts, in particular the Lipschitz continuity proved in Step 2. Then, using the smallness of the constants for small T in properties (3.13) and (3.14) of Step 1, one gets that \(\Gamma \) is a contraction in \(\mathcal {Y}_{T}\), for sufficiently small \(T>0\).
Step 4 (a priori estimate and global solution). The length of the time interval of the local solution proved in Step 2 depends only on the \(L^{2}\left( \Xi ;W_{\sigma }^{2,2}\left( \mathbb {T}^{3},\mathbb {R} ^{3}\right) \right) \) norm of \(\omega _{0}\). If we prove that, given \(T>0\) and the initial condition \(\omega _{0}\), there is a constant \(C>0\) such that a solution \(\omega \) defined on \(\left[ 0,T\right] \) has \(\sup _{t\in \left[ 0,T\right] }\mathbb {E}\left[ \left\| \omega \left( t\right) \right\| _{W^{2,2}}^{2}\right] \le C\), then we can repeatedly apply the local result of Step 2 and cover any time interval.
Thus, we need such a priori bound. Let \(\omega \) be such a solution, namely satisfying \(\omega =\Gamma \omega \) on \(\left[ 0,T\right] \). From bounds (3.13) and (3.14) of Step 1, we have
$$\begin{aligned} \mathbb {E}\left[ \left\| \omega \left( t\right) \right\| _{W^{2,2}}^{2}\right]&\le C\mathbb {E}\left[ \left\| e^{tA}\omega _{0}\right\| _{W^{2,2}} ^{2}\right] \\&\quad +\,C\mathbb {E}\left[ \left\| \int _{0}^{t}e^{\left( t-s\right) A}\mathcal {L}_{v\left( s\right) }\omega \left( s\right) \kappa _{R}\left( \omega \left( s\right) \right) \text {d}s\right\| _{W^{2,2}}^{2}\right] \\&\quad +C\mathbb {E}\left[ \left\| \int _{0}^{t}e^{\left( t-s\right) A}\frac{1}{2} \sum _{k=1}^{\infty }\mathcal {L}_{\xi _{k}}^{2}\omega \left( s\right) \text {d}s\right\| _{W^{2,2}}^{2}\right] \\&\quad +\,C\mathbb {E}\left[ \left\| \sum _{k=1}^{\infty }\int _{0}^{t}e^{\left( t-s\right) A}\mathcal {L}_{\xi _{k}}\omega \left( s\right) \text {d}B_{s}^{k}\right\| _{W^{2,2}}^{2}\right] \\&\le C\mathbb {E}\left[ \left\| \omega _{0}\right\| _{W^{2,2}}^{2}\right] +C\mathbb {E}\left[ \int _{0}^{t}\frac{1}{\left( t-s\right) ^{2/5}}\left\| \omega \left( s\right) \right\| _{W^{2,2}}^{2}\text {d}s\right] ; \end{aligned}$$
hence, we may apply a generalized version of the Grönwall lemma and conclude that
$$\begin{aligned} \sup _{t\in \left[ 0,T\right] }\mathbb {E}\left[ \left\| \omega \left( t\right) \right\| _{W^{2,2}}^{2}\right] \le C \,. \end{aligned}$$
Step 5 (Regularity) Let \(\omega \) be the solution constructed in the previous steps; it is the sum of the four terms given by the mild formulation \(\omega =\Gamma \omega \). By the property \(e^{tA}\omega _{0}\in D\left( A\right) \), namely \(Ae^{tA}\omega _{0}\in L_{\sigma }^{2}\left( \mathbb {T}^{3},\mathbb {R} ^{3}\right) \), for all \(t>0\) and \(\omega _{0}\in L_{\sigma }^{2}\left( \mathbb {T}^{3},\mathbb {R}^{3}\right) \) (see [Pazy], property (5.7) in Theorem 5.2 of Chapter 2, due to the fact that \(e^{tA}\) is an analytic semigroup), we may take \(\delta >0\) and have \(Ae^{\delta A}\omega _{0}\in L_{\sigma }^{2}\left( \mathbb {T}^{3},\mathbb {R}^{3}\right) \); then, for \(t\in \left[ \delta ,T\right] \), we have \(Ae^{tA}\omega _{0}=e^{\left( t-\delta \right) A}Ae^{tA}\omega _{0}=e^{\left( t-\delta \right) A}\omega _{\delta }\) where \(\omega _{\delta }:=Ae^{\delta A}\omega _{0}\) is an element of \(L_{\sigma } ^{2}\left( \mathbb {T}^{3},\mathbb {R}^{3}\right) \). Since \(t\mapsto e^{\left( t-\delta \right) A}\omega _{\delta }\) is continuous on \(\left[ \delta ,T\right] \) (because the semigroup is strongly continuous), it follows that \(t\mapsto Ae^{tA}\omega _{0}\) is continuous on \(\left[ \delta ,T\right] \), namely \(t\mapsto e^{tA}\omega _{0}\) belongs to \(C\left( \left[ \delta ,T\right] ;D\left( A\right) \right) \). In particular, it implies \(e^{tA}\omega _{0}\in C\left( \left[ \delta ,T\right] ;W^{4,2}\left( \mathbb {T}^{3},\mathbb {R}^{3}\right) \right) \), for every \(T>\delta >0\). The two Lebesgue integrals in \(\Gamma \omega \) belong, pathwise a.s., to \(C\left( \left[ 0,T\right] ;W^{4,2}\left( \mathbb {T}^{3};\mathbb {R}^{3}\right) \right) \), for every \(T>0\), because of property (3.13), and the fact that \(\mathcal {L}_{v}\omega \kappa _{R}\left( \omega \right) \) and \(\sum _{k=1}^{\infty }\mathcal {L}_{\xi _{k}}^{2}\omega \) are, pathwise a.s., elements of \(C\left( \left[ 0,T\right] ;L^{2}\left( \mathbb {T}^{3};\mathbb {R} ^{3}\right) \right) \), as shown in Step 2. Finally, the stochastic integral in \(\Gamma \omega \) belongs to \(L^{2}\left( \Xi ;C\left( \left[ 0,T\right] ;W^{4,2}\left( \mathbb {T}^{3};\mathbb {R}^{3}\right) \right) \right) \) by property (3.14) and the fact that \(\mathbb {E}\left[ \sup _{s\in \left[ 0,T\right] }\sum _{k=1}^{\infty }\left\| \mathcal {L}_{\xi _{k} }\omega \left( s\right) \right\| _{L_{\sigma }^{2}}^{2}\right] <\infty \), as shown again in Step 2. \(\square \)
Definition 20
On a complete separable metric space \(\left( X,d\right) \), a family \(F=\left\{ \mu _{\nu }\right\} _{\nu >0}\) of probability measures is called tight if for every \(\epsilon >0\) there is a compact set \(K_{\epsilon }\subset X\) such that \(\mu _{\nu }\left( K_{\epsilon }\right) \ge 1-\epsilon \) for all \(\nu >0\).
Remark 21
The Prohorov theorem states that, for a tight family of probability measures, one can extract a sequence \(\left\{ \mu _{\nu _{n}}\right\} _{n\in \mathbb {N}}\) which weakly converges to some probability measure,
$$\begin{aligned} \mu :\ \lim _{n\rightarrow \infty }\int _{X}\varphi \text {d}\mu _{\nu _{n}}=\int _{X}\varphi \text {d}\mu \,, \end{aligned}$$
for all bounded continuous functions \(\varphi :X\rightarrow \mathbb {R}\). We repeatedly use these facts below.
In order to prove Proposition 14, we want to prove that the family of solutions \(\left\{ \omega _{R}^{\nu }\right\} _{\nu >0}\) (R is given) provided by Lemma 19 is compact is a suitable sense and that a converging subsequence extracted from this family converges to a solution of Eq. (3.17). Since \(\left\{ \omega _{R}^{\nu }\right\} _{\nu >0}\) are random processes, the classical method we follow is to prove compactness of their laws \(\left\{ \mu _{\nu }\right\} _{\nu >0}\). For this purpose, we have to prove that \(\left\{ \mu _{\nu }\right\} _{\nu >0}\) is tight and we have to apply Prohorov theorem, as recalled above. The metric space where we prove tightness of the laws will be the space E given by (3.15).Footnote 6
Lemma 22
Let \(T>0\), \(R>0\) and \(\omega _{0}\in W_{\sigma }^{2,2}\left( \mathbb {T} ^{3},\mathbb {R}^{3}\right) \) be given. Assume that the family of laws of \(\left\{ \omega _{R}^{\nu }\right\} _{\nu >0}\) is tight in the space
$$\begin{aligned} E= C\left( \left[ 0,T\right] ;W_{\sigma }^{\beta ,2}\left( \mathbb {T}^{3},\mathbb {R}^{3}\right) \right) \end{aligned}$$
(3.15)
for some \(\beta >\frac{3}{2}\) and satisfies, for some constant \(C_R>0\),
$$\begin{aligned} \mathbb {E}\left[ \sup _{t\in [0,T]}\Vert \omega ^\nu _R(t)\Vert ^2_{W^{2,2}}\right] \le C_R \end{aligned}$$
for every \(\nu >0\). Then, the existence claim of Proposition 13 holds true, and thus, Theorem 8 is proved.
Proof
Step 1 (Gyongy–Krylov approach). We base our proof on classical ingredients, but also on the following fact proved in Gyongy and Krylov (1996), Lemma 1.1. Let \(\left\{ Z_{n}\right\} _{n\in \mathbb {N}}\) be a sequence of random variables (r.v.) with values in a Polish space \(\left( E,d\right) \) endowed with the Borel \(\sigma \)-algebra \(\mathcal {B}\left( E\right) \). Assume that the family of laws of \(\left\{ Z_{n}\right\} _{n\in \mathbb {N}}\) is tight. Moreover, assume that the limit in law of any pair \(\left( Z_{n_{j}^{\left( 1\right) }},Z_{n_{j}^{\left( 2\right) }}\right) _{j\in \mathbb {N}}\) of subsequences is a measure, on \(E\times E\), supported on the diagonal of \(E\times E\). Then, \(\left\{ Z_{n}\right\} _{n\in \mathbb {N}}\) converges in probability to some r.v. Z.
We take as Polish space E the space (3.15) above, as random variables \(\left\{ Z_{n}\right\} _{n\in \mathbb {N}}\) the sequence \(\left\{ \omega _{R}^{1/n}\right\} _{n\in \mathbb {N}}\), whose family of laws is tight by assumption. We have to check that the limit in law of any pair \(\left( \omega _{R}^{1/n_{j}^{\left( 1\right) }},\omega _{R}^{1/n_{j}^{\left( 2\right) }}\right) _{j\in \mathbb {N}}\) is supported on the diagonal of \(E\times E\). For this purpose, we shall use global uniqueness.
Step 2 (Preparation by the Skorokhod theorem). Let us enlarge the previous pair by the noise and consider the following triple: the sequence \(\left\{ \omega _{R}^{1/n_{j}^{\left( 1\right) }},\omega _{R}^{1/n_{j} ^{\left( 2\right) }}, \right. \left. \left\{ B_{\cdot }^{k}\right\} _{k\in \mathbb {N} }\right. \Bigg \} _{j\in \mathbb {N}}\) converges in law to a probability measure \(\mu \), on \(E\times E\times C\left( \left[ 0,T\right] \right) ^{\mathbb {N}}\). We have to prove that the marginal \(\mu _{E\times E}\) of \(\mu \) on \(E\times E\) is supported on the diagonal. By the Skorokhod representation theorem, there exists a probability space \(\left( \widetilde{\Xi },\widetilde{\mathcal {F} },\widetilde{P}\right) \) and \(E\times E\times C\left( \left[ 0,T\right] \right) ^{\mathbb {N}}\)-valued random variables \(\left\{ \widetilde{\omega }_{R}^{1,j},\widetilde{\omega }_{R}^{2,j},\left\{ \widetilde{B}_{\cdot } ^{k,j}\right\} _{k\in \mathbb {N}}\right\} _{j\in \mathbb {N}}\) and \(\left( \widetilde{\omega }_{R}^{1},\widetilde{\omega }_{R}^{2},\left\{ \widetilde{B} _{\cdot }^{k}\right\} _{k\in \mathbb {N}}\right) \) with the same laws as \(\left\{ \omega _{R}^{1/n_{j}^{\left( 1\right) }},\omega _{R}^{1/n_{j} ^{\left( 2\right) }},\left\{ B_{\cdot }^{k}\right\} _{k\in \mathbb {N} }\right\} _{j\in \mathbb {N}}\) and \(\mu \), respectively,Footnote 7 such that as \(j\rightarrow \infty \) one has \(\widetilde{\omega }_{R}^{1,j}\rightarrow \widetilde{\omega } _{R}^{1}\) in E, \(\widetilde{\omega }_{R}^{2,j}\rightarrow \widetilde{\omega }_{R}^{2}\) in E, \(\widetilde{B}_{\cdot }^{k,j}\rightarrow \widetilde{B} _{\cdot }^{k}\) in \(C\left( \left[ 0,T\right] \right) \), a.s. In particular, \(\left\{ \widetilde{B}_{\cdot }^{k}\right\} _{k\in \mathbb {N}}\) is a sequence of independent Brownian motions.
Since the pairs \(\left( \omega _{R}^{1/n_{j}^{\left( i\right) }},\left\{ B_{\cdot }^{k}\right\} _{k\in \mathbb {N}}\right) \), \(i=1,2\), solve Eq. (3.12) and \(\left( \widetilde{\omega }_{R}^{i,j}, \right. \left. \left\{ \widetilde{B}_{\cdot }^{k,j}\right\} _{k\in \mathbb {N}}\right) \) have the same laws (being marginals of vectors with the same laws), by a classical argument (see, for instance, Prato and Zabczyk 2015), the pairs \(\left( \widetilde{\omega } _{R}^{i,j},\left\{ \widetilde{B}_{\cdot }^{k,j}\right\} _{k\in \mathbb {N} }\right) \), \(i=1,2\), also solve Eq. (3.12), with \(\nu _{j}^{\left( i\right) }:=1/n_{j}^{\left( i\right) }\), \(i=1,2\), respectively. In other words,
$$\begin{aligned} d\widetilde{\omega }_{R}^{i,j}+\kappa _{R}\left( \widetilde{\omega }_{R} ^{i,j}\right) \mathcal {L}_{\widetilde{v}_{R}^{i,j}}\widetilde{\omega } _{R}^{i,j}\ \text {d}t+\sum _{k=1}^{\infty }\mathcal {L}_{\xi _{k}}\widetilde{\omega } _{R}^{i,j}d\widetilde{B}_{t}^{k,j}=\nu _{j}^{\left( i\right) }\Delta ^{5}\widetilde{\omega }_{R}^{i,j}\text {d}t+\frac{1}{2}\sum _{k=1}^{\infty } \mathcal {L}_{\xi _{k}}^{2}\widetilde{\omega }_{R}^{i,j}\ \text {d}t \end{aligned}$$
(3.16)
with \(\widetilde{\omega }_{R}^{i,j}|_{t=0}=\omega _{0}\), where \(\widetilde{\omega }_{R}^{i,j}={\text {curl}}\widetilde{v}_{R}^{i,j}\).
Step 3 (Property of being supported on the diagonal). The passage to the limit in Eq. (3.16) when there is strong convergence (\(\widetilde{P}\)-a.s.) in \(L^{2}\left( 0,T;L^{2}\left( \mathbb {T}^{3},\mathbb {R}^{3}\right) \right) \) is relatively classical (see Flandoli and Gatarek (1995)). We sketch the main points in Step 4. One deduces
$$\begin{aligned} d\widetilde{\omega }_{R}^{i}+\kappa _{R}\left( \widetilde{\omega }_{R} ^{i}\right) \mathcal {L}_{\widetilde{v}_{R}^{i}}\widetilde{\omega }_{R} ^{i}\ \text {d}t+\sum _{k=1}^{\infty }\mathcal {L}_{\xi _{k}}\widetilde{\omega }_{R} ^{i}d\widetilde{B}_{t}^{k}=\frac{1}{2}\sum _{k=1}^{\infty }\mathcal {L}_{\xi _{k} }^{2}\widetilde{\omega }_{R}^{i}\ \text {d}t \end{aligned}$$
(3.17)
in the weak sense explained in Remark 7. Since \(\widetilde{\omega }_{R}^{i}\) have paths in \(C\left( \left[ 0,T\right] ;W^{2,2}\right. \left. \left( \mathbb {T}^{3},\mathbb {R}^{3}\right) \right) \) (see Step 4), the derivatives can be applied on \(\widetilde{\omega }_{R}^{i}\) by integration by parts and we get the equation in the strong sense. Now we apply the pathwise uniqueness of solutions for Eq. (3.2) in \(W^{2,2}\) as deduced in Sect. 3.3 to deduce \(\widetilde{\omega } _{R}^{1}=\widetilde{\omega }_{R}^{2}\). This means that the law of \(\left( \widetilde{\omega }_{R}^{1},\widetilde{\omega }_{R}^{2}\right) \) is supported on the diagonal of \(E\times E\). Since this law is equal to \(\mu _{E\times E}\), we have that \(\mu _{E\times E}\) is supported on the diagonal of \(E\times E\).
Step 4 (Convergence) In this step, we give a few details about the passage to the limit, as \(j\rightarrow \infty \), from Eqs. (3.16) to (3.17). We do not give the details about the linear terms, except for a comment about the term \(\nu _{j}^{\left( i\right) }\Delta ^{5}\widetilde{\omega }_{R}^{i,j}\). Namely, in weak form we write it as (with \(\phi \in C^{\infty }\left( \mathbb {T} ^{3},\mathbb {R}^{3}\right) \))
$$\begin{aligned} \nu _{j}^{\left( i\right) }\int _{0}^{t}\left\langle \widetilde{\omega } _{R}^{i,j}\left( s\right) ,\Delta ^{5}\phi \right\rangle \text {d}s \end{aligned}$$
and use the pathwise convergence in \(L^{2}\left( 0,T;L^{2}\left( \mathbb {T}^{3},\mathbb {R}^{3}\right) \right) \) of \(\widetilde{\omega } _{R}^{i,j}\left( s\right) \) plus the fact that \(\nu _{j}^{\left( i\right) }\rightarrow 0\).
The difficult term is the nonlinear one, also because of the cut-off term \(\kappa _{R}\left( \widetilde{\omega }_{R}^{i}\left( s\right) \right) \). We want to prove that, given \(\phi \in C^{\infty }\left( \mathbb {T}^{3} ,\mathbb {R}^{3}\right) \),
$$\begin{aligned} \int _{0}^{t}\kappa _{R}\left( \widetilde{\omega }_{R}^{i,j}\left( s\right) \right) \left\langle \widetilde{\omega }_{R}^{i,j}\left( s\right) ,\mathcal {L}_{\widetilde{v}_{R}^{i,j}}^{*}\phi \right\rangle \text {d}s\overset{j\rightarrow \infty }{\rightarrow }\int _{0}^{t}\kappa _{R}\left( \widetilde{\omega }_{R}^{i}\left( s\right) \right) \left\langle \widetilde{\omega }_{R}^{i}\left( s\right) ,\mathcal {L}_{\widetilde{v} _{R}^{i}}^{*}\phi \right\rangle \text {d}s \end{aligned}$$
(3.18)
with probability one. From the Skorokhod preparation in Step 2, we know that \(\widetilde{\omega } _{R}^{i,j}\rightarrow \widetilde{\omega }_{R}^{i}\) as \(j\rightarrow \infty \) in the strong topology of E, \(\widetilde{P}\)-a.s., for \(i=1,2\). In the sequel, we fix the random parameter and the value of \(i=1,2\). Since \(W_{\sigma } ^{\beta ,2}\left( \mathbb {T}^{3},\mathbb {R}^{3}\right) \) is continuously embedded into \(C\left( \mathbb {T}^{3},\mathbb {R}^{3}\right) \) (recall that \(\beta >3/2\)), it follows that \(\widetilde{\omega }_{R}^{i,j}\rightarrow \widetilde{\omega }_{R}^{i}\) in the uniform topology over \(\left[ 0,T\right] \times \mathbb {T}^{3}\). By the continuity of Biot–Savart map from \(W_{\sigma }^{\beta ,2}\left( \mathbb {T}^{3},\mathbb {R}^{3}\right) \) to \(W_{\sigma }^{\beta +1,2}\left( \mathbb {T}^{3},\mathbb {R}^{3}\right) \) and the formula for \(L_{\widetilde{v}_{R}^{i,j}}^{*}\) which contains first derivatives of \(\widetilde{v}_{R}^{i,j}\), we see that \(L_{\widetilde{v}_{R}^{i,j}}^{*} \phi \rightarrow L_{\widetilde{v}_{R}^{i}}^{*}\phi \) in the strong topology of E again, and thus, again by Sobolev embedding, \(L_{\widetilde{v} _{R}^{i,j}}^{*}\phi \rightarrow L_{\widetilde{v}_{R}^{i}}^{*}\phi \) in the uniform topology over \(\left[ 0,T\right] \times \mathbb {T}^{3}\). Hence, \(\left\langle \widetilde{\omega }_{R}^{i,j}\left( \cdot \right) ,L_{\widetilde{v}_{R}^{i,j}}^{*}\phi \right\rangle \) converges to \(\left\langle \widetilde{\omega }_{R}^{i}\left( \cdot \right) ,L_{\widetilde{v}_{R}^{i}}^{*}\phi \right\rangle \) uniformly over \(\left[ 0,T\right] \). Hence, if we prove that \(k_{R}\left( \widetilde{\omega } _{R}^{i,j}\left( s\right) \right) \rightarrow k_{R}\left( \widetilde{\omega }_{R}^{i}\left( s\right) \right) \) for a.e. \(s\in \left[ 0,T\right] \), and because these functions are bounded by 1, we can take the limit in (3.18). Therefore, it remains to prove that, \(\widetilde{P}\)-a.s., \(\kappa _{R}\left( \widetilde{\omega }_{R} ^{i,j}\left( s\right) \right) \) converges to \(\kappa _{R}\left( \widetilde{\omega }_{R}^{i}\left( s\right) \right) \) for a.e. \(s\in \left[ 0,T\right] \), or at least in probability w.r.t. time. This is true because strong convergence in \(L^{2}\left( 0,T\right) \) in time implies convergence in probability w.r.t. time, and we have strong convergence in \(L^{2}\left( 0,T\right) \), of \(\kappa _{R}\left( \widetilde{\omega }_{R}^{i,j}\left( s\right) \right) \), because \(\kappa _{R}\) is bounded continuous, and \(\widetilde{\omega }_{R}^{i,j}\) converges strongly in \(L^{2}\left( 0,T;W^{\beta ,2}\left( \mathbb {T}^{3},\mathbb {R}^{3}\right) \right) \), hence \(\nabla \widetilde{v}_{R}^{i,j}\) converges strongly in \(L^{2}\left( 0,T;W^{\beta ,2}\left( \mathbb {T}^{3},\mathbb {R}^{3}\right) \right) \) hence in \(L^{2}\left( 0,T;C\left( \mathbb {T}^{3},\mathbb {R}^{3}\right) \right) \) by Sobolev embedding theorem. Hence, \(\kappa _{R}\left( \widetilde{\omega } _{R}^{i,j}\right) \) converges to \(\kappa _{R}\left( \widetilde{\omega } _{R}^{i}\right) \) in probability w.r.t. time. Finally, from the integral identity satisfied by the limit process \(\tilde{\omega }^i\), one can deduce that \(\tilde{\omega }^i\in C\left( \left[ 0,T\right] ;W^{2,2}\left( \mathbb {T}^{3},\mathbb {R}^{3}\right) \right) \) following the argument in Kim (2009). \(\square \)
Based on this lemma, we need to prove suitable bounds on \(\left\{ \omega _{R}^{\nu }\right\} _{\nu >0}\).
Theorem 23
Assume that, for some \(N\ge 0\) and \(\alpha >{\frac{1}{4}}\),
$$\begin{aligned}&\mathbb {E}\left[ \sup _{t\in \left[ 0,T\right] }\left\| \omega _{R}^{\nu }\left( t\right) \right\| _{W^{2,2}}^{4}\right] \le C_{1} \end{aligned}$$
(3.19)
$$\begin{aligned}&\mathbb {E}\int _{0}^{T}\int _{0}^{T}\frac{\left\| \omega _{R}^{\nu }\left( t\right) -\omega _{R}^{\nu }\left( s\right) \right\| _{W^{-N,2}}^{4} }{\left| t-s\right| ^{1+4\alpha }}\text {d}t\text {d}s\le C_{2} \end{aligned}$$
(3.20)
for all \(\nu \in \left( 0,1\right) \). Then, the assumptions of Lemma 22 hold.
Proof
We shall use the following variant of Aubin–Lions lemma, which can be found in Simon (1987). Recall that, given an Hilbert space W, a norm on \(W^{\alpha ,4}\left( 0,T;W\right) \) is the fourth root of
$$\begin{aligned} \int _{0}^{T}\left\| f\left( t\right) \right\| _{W}^{4}\text {d}t+\int _{0} ^{T}\int _{0}^{T}\frac{\left\| f\left( t\right) -f\left( s\right) \right\| _{W}^{4}}{\left| t-s\right| ^{1+4\alpha }}\text {d}t\text {d}s. \end{aligned}$$
Assume that V, H, W are separable Hilbert spaces with continuous dense embedding \(V\subset H\subset W\) such that there exists \(\theta \in (0,1)\) and \(M>0\) such that
$$\begin{aligned} \Vert v\Vert _H\le M\Vert v\Vert _V^{1-\theta }\Vert v\Vert _W^\theta \end{aligned}$$
for every \(v\in V\). Assume that \(V\subset H\) is a compact embedding. Assume \(\alpha >0\). Then,
$$\begin{aligned} L^{\infty }\left( 0,T;V\right) \cap W^{\alpha ,4}\left( 0,T;W\right) \end{aligned}$$
is compactly embedded into \(C\left( [0,T];H\right) \) (see Simon (1987), Corollary 9). We apply it to the spaces
$$\begin{aligned} H=W^{\beta ,2}\left( \mathbb {T}^{3},\mathbb {R}^{3}\right) \text {, } V=W^{2,2}\left( \mathbb {T}^{3},\mathbb {R}^{3}\right) \text {, } W=W^{-N,2}\left( \mathbb {T}^{3},\mathbb {R}^{3}\right) \end{aligned}$$
where \(\beta \in \left( \frac{3}{2},2\right) \). The constraint \(\beta <2\) is imposed because we want to use the compactness of the embedding \(W^{2,2} \left( \mathbb {T}^{3},\mathbb {R}^{3}\right) \subset W^{\beta ,2}\left( \mathbb {T}^{3},\mathbb {R}^{3}\right) \). The constraint \(\beta >\frac{3}{2}\) is imposed because we want to use the embedding \(W^{\beta ,2}\left( \mathbb {T}^{3},\mathbb {R}^{3}\right) \subset C\left( \mathbb {T} ^{3},\mathbb {R}^{3}\right) \).
Let \(\left\{ Q_{\nu }\right\} \) be the family of laws of \(\left\{ \omega _{R}^{\nu }\right\} \), supported on
$$\begin{aligned} E_{0}:=C\left( \left[ 0,T\right] ;W^{2,2}\left( \mathbb {T}^{3} ,\mathbb {R}^{3}\right) \right) \cap W^{\alpha ,4}\left( 0,T;W^{-N,2}\left( \mathbb {T}^{3},\mathbb {R}^{3}\right) \right) \end{aligned}$$
by the assumption of the theorem. We want to prove that \(\left\{ Q_{\nu }\right\} \) is tight in E. The sets \(K_{R_{1},R_{2},R_{3}}\) defined as
$$\begin{aligned}&\left\{ f:\sup _{t\in \left[ 0,T\right] }\left\| f\left( t\right) \right\| _{W^{2,2}}^{2}\le R_{1},\int _{0}^{T}\left\| f\left( t\right) \right\| _{W^{-N,2}}^{4}\text {d}t\right. \\&\quad \left. \le R_{2},\int _{0}^{T}\int _{0}^{T} \frac{\left\| f\left( t\right) -f\left( s\right) \right\| _{W^{-N,2}}^{4}}{\left| t-s\right| ^{1+4\alpha }}\text {d}t\text {d}s\le R_{3}\right\} \end{aligned}$$
with \(R_{1},R_{2},R_{3}>0\) are relatively compact in E. Let us prove that, given \(\epsilon >0\), there are \(R_{1},R_{2},R_{3}>0\) such that
$$\begin{aligned} Q_{\nu }\left( K_{R_{1},R_{2},R_{3}}^{c}\right) \le \epsilon \end{aligned}$$
for all \(\nu \in \left( 0,1\right) \). We have
$$\begin{aligned} Q_{\nu }\left( \sup _{t\in \left[ 0,T\right] }\left\| f\left( t\right) \right\| _{W^{2,2}}^{2}>R_{1}\right)&=P\left( \sup _{t\in \left[ 0,T\right] }\left\| \omega _{R}^{\nu }\left( t\right) \right\| _{W^{2,2}}^{2}\right) \\&\le \frac{\mathbb {E}\left[ \sup _{t\in \left[ 0,T\right] }\left\| \omega _{R}^{\nu }\left( t\right) \right\| _{W^{2,2}}^{2}\right] }{R_{1} }\le \frac{C_{1}}{R_{1}} \end{aligned}$$
and this is smaller than \({\epsilon }/{3}\) when \(R_{1}\) is large enough. Similarly, we get
$$\begin{aligned} Q_{\nu }\left( \int _{0}^{T}\int _{0}^{T}\frac{\left\| f\left( t\right) -f\left( s\right) \right\| _{W^{-N,2}}^{4}}{\left| t-s\right| ^{1+4\alpha }}\text {d}t\text {d}s>R_{3}\right) \le \frac{\epsilon }{3} \end{aligned}$$
when \(R_{3}\) is large enough. Finally,
$$\begin{aligned} Q_{\nu }\left( \int _{0}^{T}\left\| f\left( t\right) \right\| _{W^{-N,2}}^{4}\text {d}t>R_{2}\right)&\le Q_{\nu }\left( T\sup _{t\in \left[ 0,T\right] }\left\| f\left( t\right) \right\| _{W^{-N,2}}^{4} \text {d}t>R_{2}\right) \\&\le Q_{\nu }\left( CT\sup _{t\in \left[ 0,T\right] }\left\| f\left( t\right) \right\| _{W^{2,2}}^{4}\text {d}t>R_{2}\right) \end{aligned}$$
for a constant \(C>0\) such that \(\left\| f\left( t\right) \right\| _{W^{-N,2}}^{4}\le C\left\| f\left( t\right) \right\| _{W^{2,2}}^{4} \). Hence, also this quantity is smaller than \(\frac{\epsilon }{3}\) when \(R_{2}\) is large enough. We deduce \(Q_{\nu }\left( K_{R_{1},R_{2},R_{3}}^{c}\right) \le \epsilon \) and complete the proof. \(\square \)
The difficult part of the estimates above is bound (3.19). Thus, let us postpone it and first show bound (3.20).