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Enhanced Dissipation for Stochastic Navier–Stokes Equations with Transport Noise

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Abstract

The phenomenon of dissipation enhancement by transport noise is shown for stochastic 2D Navier–Stokes equations in velocity form. In the 3D case, suppression of blow-up is proved for stochastic Navier–Stokes equations in vorticity form; in particular, quantitative estimate allows us to choose the parameters of noise, uniformly in initial vorticity bounded in \(L^2\)-norm, so that global solutions exist with a large probability sufficiently close to 1.

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Funding

The author would like to thank the financial supports of the National Key R &D Program of China (No. 2020YFA0712700), the National Natural Science Foundation of China (Nos. 11931004, 12090010, 12090014), and the Youth Innovation Promotion Association, CAS (Y2021002).

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Correspondence to Dejun Luo.

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A Proof of Theorem 3.1

A Proof of Theorem 3.1

This section is devoted to the proof of Theorem 3.1; we shall only prove the first estimate since the second one can be proved in the same way by improving some estimates in [33, Section 5], see Remark A.7 for a brief discussion of necessary modifications.

Let \(\Pi ^\perp \) be the projection operator orthogonal to \(\Pi \): for any vector field \(X\in L^2({\mathbb {T}}^2, {\mathbb {R}}^2)\), \(\Pi ^\perp X\) is the gradient part of X. Formally,

$$\begin{aligned} \Pi ^\perp X = \nabla \Delta ^{-1} \textrm{div}(X). \end{aligned}$$
(A.1)

On the other hand, if \(X= \sum _{l\in {\mathbb {Z}}^2_0} X_l e_l\), \(X_l\in {\mathbb {C}}^2\), then

$$\begin{aligned} \Pi ^\perp X= \sum _l \frac{l\cdot X_l}{|l|^2} l e_l = \nabla \bigg [ \frac{1}{2\pi \textrm{i}} \sum _l \frac{l\cdot X_l}{|l|^2} e_l \bigg ]. \end{aligned}$$
(A.2)

Now for a divergence vector field v on \({\mathbb {T}}^2\), recall the Stratonovich-Itô corrector

$$\begin{aligned} S_\theta ^{(2)} (v)= 2\kappa \sum _k \theta _k^2\, \Pi \big [ \sigma _k \cdot \nabla \Pi (\sigma _{-k} \cdot \nabla v) \big ]. \end{aligned}$$

Using the operator \(\Pi ^\perp \), we have

$$\begin{aligned} \Pi (\sigma _{-k}\cdot \nabla v) = \sigma _{-k}\cdot \nabla v- \Pi ^\perp (\sigma _{-k}\cdot \nabla v). \end{aligned}$$
(A.3)

Noting that, by (1.8) and the fact that \(\sigma _k \cdot \nabla \sigma _{-k} \equiv 0\),

$$\begin{aligned} 2\kappa \sum _k \theta _k^2\, \Pi \big [ \sigma _k \cdot \nabla (\sigma _{-k} \cdot \nabla v) \big ] = \kappa \, \Pi [\Delta v]= \kappa \Delta v, \end{aligned}$$

where the last step is due to the divergence free property of v, hence,

$$\begin{aligned} S_\theta ^{(2)}(v) = \kappa \Delta v - 2\kappa \sum _k \theta _k^2\, \Pi \big [ \sigma _k \cdot \nabla \Pi ^\perp (\sigma _{-k} \cdot \nabla v)\big ]. \end{aligned}$$

We shall denote the second term on the right-hand side by \(S_\theta ^\perp (v)\). Therefore, the first assertion in Theorem 3.1 follows if we can prove

$$\begin{aligned} \bigg \Vert S_{\theta ^N}^\perp (v)- \frac{3}{4} \kappa \Delta v \bigg \Vert _{H^{s-2-\alpha }} \le C \frac{\kappa }{N^{\alpha }} \Vert v \Vert _{H^s} . \end{aligned}$$
(A.4)

Now we assume the divergence free vector field v has the Fourier expansion

$$\begin{aligned} v= \sum _{l} v_l\, \sigma _l, \end{aligned}$$

where the coefficients \(\{v_l: l\in {\mathbb {Z}}^2_0 \} \subset {\mathbb {C}}\) satisfy \(\overline{v_l}= v_{-l}\). We begin with finding the exact expression for \(S_\theta ^\perp (v)\).

Lemma A.1

We have

$$\begin{aligned} S_\theta ^\perp (v)= - 8\pi ^2 \kappa \sum _l v_l \Pi \bigg \{ \bigg [ \sum _k \theta _k^2 (a_k \cdot l)^2 (a_l\cdot (k-l)) \frac{k-l}{|k-l|^2} \bigg ] e_l \bigg \}. \end{aligned}$$
(A.5)

Proof

We give a proof by using the formula (A.2); one can also proceed with (A.1), cf. [33, Sect. 5]. We have

$$\begin{aligned} \nabla v(x)= \sum _l v_l \nabla \sigma _l(x) = 2\pi \textrm{i} \sum _l v_l (a_l \otimes l) e_l(x). \end{aligned}$$

Note that \(\sigma _{-k}(x)= a_k e_{-k}(x)\); thus

$$\begin{aligned} (\sigma _{-k}\cdot \nabla v)(x) = 2\pi \textrm{i} \sum _l v_l (a_k \cdot l) a_l e_{l-k}(x). \end{aligned}$$

By the first equality in (A.2), we have

$$\begin{aligned} \Pi ^\perp (\sigma _{-k}\cdot \nabla v)(x)= 2\pi \textrm{i} \sum _l v_l (a_k \cdot l) (a_l \cdot (l-k)) \frac{l-k}{|l-k|^2} e_{l-k}(x) . \end{aligned}$$
(A.6)

As a consequence,

$$\begin{aligned}&\big [ \sigma _k\cdot \nabla \Pi ^\perp (\sigma _{-k,\alpha }\cdot \nabla v) \big ](x)\\&\quad = 2\pi \textrm{i} \sum _l v_l (a_k \cdot l) (a_l \cdot (l-k)) \frac{l-k}{|l-k|^2} e_k(x) a_k \cdot \nabla e_{l-k}(x) \\&\quad = (2\pi \textrm{i})^2 \sum _l v_l (a_k \cdot l) (a_l \cdot (l-k)) \frac{l-k}{|l-k|^2} (a_k \cdot (l-k)) e_k(x) e_{l-k}(x) \\&\quad = -4\pi ^2 \sum _l v_l (a_k \cdot l)^2 (a_l\cdot (k-l)) \frac{k-l}{|k-l|^2} e_l(x), \end{aligned}$$

where in the last step we have used \(a_k \cdot k=0\). This immediately gives us the desired identity. \(\square \)

In the next lemma we compute explicitly the projected quantity in (A.5). Let \(\angle _{k,l}\) be the angle between two vectors k and l.

Lemma A.2

We have

$$\begin{aligned} S_\theta ^\perp (v)= - 8\pi ^2 \kappa \sum _l v_l |l|^2 \bigg [ \sum _k \theta _k^2 \sin ^2 (\angle _{k,l}) \frac{(a_l\cdot (k-l))^2}{|k-l|^2} \bigg ] \sigma _l. \end{aligned}$$

Proof

First, noting that \(\{a_k, \frac{k}{|k|}\}\) is an orthonormal basis of \({\mathbb {R}}^2\) for any \(k\in {\mathbb {Z}}^2_0\), we have

$$\begin{aligned} (a_k \cdot l)^2= |l|^2 - \bigg (l\cdot \frac{k}{|k|} \bigg )^2 = |l|^2 \bigg [1- \bigg (\frac{l\cdot k}{|l|\, |k|} \bigg )^2\bigg ] = |l|^2 \sin ^2 \angle _{k,l}. \end{aligned}$$

Therefore, we can rewrite (A.5) as

$$\begin{aligned} S_\theta ^\perp (v)&= - 8\pi ^2 \kappa \sum _l v_l |l|^2 \Pi \bigg \{ \bigg [ \sum _k \theta _k^2 \sin ^2 (\angle _{k,l}) (a_l\cdot (k-l)) \frac{k-l}{|k-l|^2} \bigg ] e_l \bigg \} \nonumber \\&= - 8\pi ^2 \kappa \sum _l v_l |l|^2 \sum _k \theta _k^2 \sin ^2 (\angle _{k,l}) \frac{a_l\cdot (k-l)}{|k-l|^2} \Pi ((k-l) e_l), \end{aligned}$$
(A.7)

where the second step follows from the linearity of \(\Pi \).

Since \(\{\sigma _j\}_{j\in {\mathbb {Z}}^2_0}\) is a complete orthonormal basis in the space of square integrable and divergence free vector fields on \({\mathbb {T}}^2\) with zero mean, one has \(\Pi ((k-l) e_l) = \sum _j \langle (k-l) e_l, \sigma _{-j} \rangle \sigma _j= ((k-l)\cdot a_l) \sigma _l\). Substituting this result into (A.7) leads to the desired equality. \(\square \)

Recall the sequence \(\theta ^N \in \ell ^2\) defined in (1.9). The next result is a crucial step for proving the limit (A.4).

Proposition A.3

There exists some constant \(C>0\) such that for any \(l\in {\mathbb {Z}}^2_0\) and for all \(N \ge 1\), it holds

$$\begin{aligned} \bigg | \sum _{k} \big (\theta ^N_k \big )^2 \sin ^2(\angle _{k,l}) \frac{(a_l\cdot (k-l))^2}{|k-l|^2}- \frac{3}{8} \bigg | \le C \frac{|l|}{N}. \end{aligned}$$

Suppose we have already proved this result; we now turn to prove (A.4).

Proof of (A.4). For any \(N\ge 1\), by Lemma A.2,

$$\begin{aligned} S_{\theta ^N}^\perp (v)= - 8\pi ^2 \kappa \sum _l v_l |l|^2 \bigg [ \sum _{k} \big (\theta ^N_k \big )^2 \sin ^2(\angle _{k,l}) \frac{(a_l\cdot (k-l))^2}{|k-l|^2} \bigg ] \sigma _l. \end{aligned}$$

Since \(v=\sum _l v_l \sigma _l\) one has

$$\begin{aligned} \frac{3}{4} \kappa \Delta v= -3\pi ^2\kappa \sum _l v_l |l|^2 \sigma _l, \end{aligned}$$

therefore,

$$\begin{aligned} S_{\theta ^N}^\perp (v) -\frac{3}{4} \kappa \Delta v = - 8\pi ^2 \kappa \sum _l v_l |l|^2 \bigg [ \sum _{k} \big (\theta ^N_k \big )^2 \sin ^2(\angle _{k,l}) \frac{(a_l\cdot (k-l))^2}{|k-l|^2} - \frac{3}{8} \bigg ] \sigma _l . \end{aligned}$$

Fix any \(L>0\); we have

$$\begin{aligned} \bigg \Vert S_{\theta ^N}^\perp (v) -\frac{3}{4} \kappa \Delta v \bigg \Vert _{H^{s-2-\alpha }}^2 = K_{L,1} + K_{L,2}, \end{aligned}$$

where (\(C=64\pi ^4\))

$$\begin{aligned} K_{L,1}&= C\kappa ^2 \sum _{|l|\le L} |v_l|^2 |l|^{2(s-\alpha )} \bigg | \sum _{k} \big (\theta ^N_k \big )^2 \sin ^2(\angle _{k,l}) \frac{(a_l\cdot (k-l))^2}{|k-l|^2} - \frac{3}{8} \bigg |^2, \\ K_{L,2}&= C\kappa ^2 \sum _{|l|> L} |v_l|^2 |l|^{2(s-\alpha )} \bigg | \sum _{k} \big (\theta ^N_k \big )^2 \sin ^2(\angle _{k,l}) \frac{(a_l\cdot (k-l))^2}{|k-l|^2} - \frac{3}{8} \bigg |^2 . \end{aligned}$$

Next we estimate the two quantities above. By Proposition A.3,

$$\begin{aligned} K_{L,1} \le C' \kappa ^2 \sum _{|l|\le L} |v_l|^2 \frac{|l|^{2(s-\alpha +1)}}{N^2} \le C'\kappa ^2 \frac{L^{2(1-\alpha )} }{N^2} \sum _{|l|\le L} |v_l|^2 |l|^{2s} \le C' \kappa ^2 \frac{L^{2(1-\alpha )}}{N^2} \Vert v \Vert _{H^s}^2. \end{aligned}$$

Moreover, since \(|a_l|=1\) and \(\Vert \theta ^N \Vert _{\ell ^2}=1\), one has

$$\begin{aligned} K_{L,2}&\le C\kappa ^2 \sum _{|l|> L} |v_l|^2|l|^{2(s-\alpha )} \bigg (\sum _{k} \big (\theta ^N_k \big )^2 + \frac{3}{8} \bigg )^2\\&\le 4C\kappa ^2 \sum _{|l|> L} |v_l|^2 \frac{|l|^{2s}}{L^{2\alpha }} \le C' \kappa ^2 \frac{1}{L^{2\alpha }} \Vert v \Vert _{H^s}^2. \end{aligned}$$

Summarizing these estimates and taking \(L=N\), we complete the proof of (A.4). \(\square \)

Next we prove Proposition A.3 for which we need some preparations.

Lemma A.4

For any \(l\in {\mathbb {Z}}^2_0\) and all \(N\ge 1\) it holds

$$\begin{aligned} \bigg | \sum _{k} \big (\theta ^N_k \big )^2 \sin ^2(\angle _{k,l}) \bigg ( \frac{(a_l\cdot (k-l))^2}{|k-l|^2} - \frac{(a_l\cdot k)^2}{|k|^2} \bigg ) \bigg | \le \frac{4|l|}{N}. \end{aligned}$$

Proof

We have

$$\begin{aligned} \bigg | \frac{(a_l\cdot (k-l))^2}{|k-l|^2} - \frac{(a_l\cdot k)^2}{|k|^2} \bigg | \le \bigg \Vert \frac{(k-l)\otimes (k-l)}{|k-l|^2} - \frac{k\otimes k}{|k|^2} \bigg \Vert , \end{aligned}$$

where \(\Vert \cdot \Vert \) is the operator norm of matrices. Note that

$$\begin{aligned} \frac{(k-l)\otimes (k-l)}{|k-l|^2} - \frac{k\otimes k}{|k|^2} = \frac{k-l}{|k-l|} \otimes \bigg (\frac{k-l}{|k-l|} - \frac{k}{|k|} \bigg ) + \bigg (\frac{k-l}{|k-l|} - \frac{k}{|k|} \bigg ) \otimes \frac{k}{|k|}, \end{aligned}$$

and thus

$$\begin{aligned} \bigg \Vert \frac{(k-l)\otimes (k-l)}{|k-l|^2} - \frac{k\otimes k}{|k|^2} \bigg \Vert \le 2\bigg | \frac{k-l}{|k-l|} - \frac{k}{|k|} \bigg |. \end{aligned}$$

Next, since

$$\begin{aligned} \frac{k-l}{|k-l|} - \frac{k}{|k|}= \bigg (\frac{1}{|k-l|} - \frac{1}{|k|}\bigg )(k-l) - \frac{l}{|k|}, \end{aligned}$$

one has

$$\begin{aligned} \bigg | \frac{k-l}{|k-l|} - \frac{k}{|k|} \bigg | \le \frac{\big | |k| -|k-l| \big |}{ |k|} + \frac{|l|}{|k|} \le 2 \frac{|l|}{|k|}. \end{aligned}$$

Combining the above estimates we obtain

$$\begin{aligned} \bigg | \frac{(a_l\cdot (k-l))^2}{|k-l|^2} - \frac{(a_l\cdot k)^2}{|k|^2} \bigg | \le \bigg \Vert \frac{(k-l)\otimes (k-l)}{|k-l|^2} - \frac{k\otimes k}{|k|^2} \bigg \Vert \le 4\frac{|l|}{|k|}. \end{aligned}$$

Finally, recalling the definition of \(\theta ^N\) in (1.9), it holds

$$\begin{aligned} \sum _{k} \big (\theta ^N_k \big )^2 \sin ^2(\angle _{k,l}) \bigg | \frac{(a_l\cdot (k-l))^2}{|k-l|^2} - \frac{(a_l\cdot k)^2}{|k|^2} \bigg | \le \sum _{N\le |k|\le 2N} \big (\theta ^N_k \big )^2 \times 4\frac{|l|}{|k|} \le \frac{4|l|}{N}. \end{aligned}$$

The proof is complete. \(\square \)

We also need the following result.

Lemma A.5

Let \(\{\theta ^N \}_{N\ge 1} \subset \ell ^2\) be given as in (1.9). There exists a \(C>0\), independent of \(N\ge 1\) and \(l\in {\mathbb {Z}}^2_0\), such that

$$\begin{aligned} \bigg | \sum _{k} \big (\theta ^N_k \big )^2 \sin ^2(\angle _{k,l}) \frac{(a_l\cdot k)^2}{|k|^2} - \frac{1}{\Lambda _N^2} \int _{\{N\le |x|\le 2N\}} \frac{1}{|x|^{2\gamma }} \sin ^2(\angle _{x,l}) \frac{(a_l\cdot x)^2}{|x|^2} \,\textrm{d}x \bigg | \le \frac{C}{N}. \end{aligned}$$

Remark A.6

In the 2D case, for any \(l\in {\mathbb {Z}}^2_0\), \(\{a_l, \frac{l}{|l|}\}\) is an orthonormal basis of \({\mathbb {R}}^2\); it is easy to see that \(\frac{(a_l\cdot x)^2}{|x|^2} =\sin ^2(\angle _{x,l}) \), thus the expression of integrand can be simplified. However, for the 3D case discussed in Remark A.7, such simplification no longer holds; thus we decide to do the computations below using this more complicated expression, in order to shed some light on the computations in the 3D case.

Proof of Lemma A.5

For any \(l\in {\mathbb {Z}}^2_0\), we define the function

$$\begin{aligned} g_l(x) = \sin ^2(\angle _{x,l}) \frac{(a_l\cdot x)^2}{|x|^2}, \quad x\in {\mathbb {R}}^2,\, x\ne 0; \end{aligned}$$

clearly, \(\Vert g_l \Vert _\infty \le 1\). We shall prove that there exists \(C>0\), independent of \(N\ge 1\) and \(l\in {\mathbb {Z}}^2_0\), such that

$$\begin{aligned} \bigg |\sum _{k} \big (\theta ^N_k \big )^2 g_l(k) -\frac{1}{\Lambda _N^2}\int _{\{N\le |x|\le 2N\}} \frac{g_l(x)}{|x|^{2\gamma }} \,\textrm{d}x \bigg | \le \frac{C}{N} . \end{aligned}$$
(A.8)

Let \(\square (k)\) be the unit square centered at \(k\in {\mathbb {Z}}^2\) such that all sides have length 1 and are parallel to the coordinate axes. Note that for all \(k,l\in {\mathbb {Z}}^2\), \(k\ne l\), the interiors of \(\square (k)\) and \(\square (l)\) are disjoint. Let \(S_N = \bigcup _{N\le |k|\le 2N} \square (k)\); then,

$$\begin{aligned} \bigg |\sum _{k} \big (\theta ^N_k \big )^2 g_l(k) - \frac{1}{\Lambda _N^2} \int _{S_N} \frac{g_l(x)}{|x|^{2\gamma }} \,\textrm{d}x\bigg | \le \frac{1}{\Lambda _N^2} \sum _{N\le |k|\le 2N} \int _{\square (k)} \bigg |\frac{g_l(k)}{|k|^{2\gamma }} - \frac{g_l(x)}{|x|^{2\gamma }} \bigg |\, \textrm{d}x. \end{aligned}$$

For all \(|k|\ge N\ge 1\) and \(x\in \square (k)\), we have \(|x-k|\le \sqrt{2}/2\) and \(|x|\ge 1/2\), thus

$$\begin{aligned} \bigg |\frac{g_l(k)}{|k|^{2\gamma }} - \frac{g_l(x)}{|x|^{2\gamma }} \bigg | \le \bigg |\frac{1}{|k|^{2\gamma }} - \frac{1}{|x|^{2\gamma }} \bigg | + \frac{|g_l(k) -g_l(x)|}{|x|^{2\gamma }} \le C_1 \bigg (\frac{1}{|k|^{2\gamma +1}} + \frac{|g_l(k) -g_l(x)|}{|k|^{2\gamma }}\bigg ) \end{aligned}$$

for some constant \(C_1\) depending on \(\gamma >0\). Next,

$$\begin{aligned} |g_l(k) -g_l(x)|&\le |\sin ^2(\angle _{k,l}) -\sin ^2(\angle _{x,l})| + \bigg | \frac{(a_l\cdot k)^2}{|k|^2} - \frac{(a_l\cdot x)^2}{|x|^2} \bigg | \\&\le 2|\sin (\angle _{k,l}) -\sin (\angle _{x,l})| + \bigg \Vert \frac{k\otimes k}{|k|^2} - \frac{x\otimes x}{|x|^2} \bigg \Vert \\&\le 2| \angle _{k,l} -\angle _{x,l}| + 2 \bigg | \frac{k}{|k|} -\frac{x}{|x|}\bigg |. \end{aligned}$$

Since \(x\in \square (k)\) and \(|k|\ge N \ge 1\), we can find a constant \(C_2>0\), independent of \(l\in {\mathbb {Z}}^2_0\) and \(N\ge 1\), such that

$$\begin{aligned} |g_l(k) -g_l(x)| \le \frac{C_2}{|k|}. \end{aligned}$$

Summarizing the above discussions, we obtain

$$\begin{aligned} \bigg |\sum _{k} \big (\theta ^N_k \big )^2 g_l(k) - \frac{1}{\Lambda _N^2} \int _{S_N} \frac{g_l(x)}{|x|^{2\gamma }} \,\textrm{d}x\bigg |&\le \frac{1}{\Lambda _N^2} \sum _{N\le |k|\le 2N} \int _{\square (k)} \frac{C_3}{|k|^{2\gamma +1}}\,\textrm{d}x\\&\le \frac{C_3}{N \Lambda _N^2} \sum _{N\le |k|\le 2N} \frac{1}{|k|^{2\gamma }} = \frac{C_3}{N}. \end{aligned}$$

Note that there is a small difference between the sets \(\{N\le |x|\le 2N\}\) and \(S_N\), but, in the same way, one can show that

$$\begin{aligned} \bigg |\int _{\{N\le |x|\le 2N\}} \frac{g_l(x)}{|x|^{2\gamma }} \,\textrm{d}x - \int _{S_N} \frac{g_l(x)}{|x|^{2\gamma }} \,\textrm{d}x\bigg | \le \frac{C}{N} \Lambda _N^2. \end{aligned}$$

Indeed, for any \(x\in \square (k)\) with \(N\le |k| \le 2N\), one has \(N-1 \le |x| \le 2N+1\). Therefore,

$$\begin{aligned} S_N = \bigcup _{N\le |k| \le 2N} \square (k) \subset \{N-1 \le |x| \le 2N+1 \} =: T_N. \end{aligned}$$

One also has

$$\begin{aligned} R_N:= \{N+1 \le |x| \le 2N-1 \} \subset S_N. \end{aligned}$$

Let \(A\Delta B\) be the symmetric difference of subsets \(A,B\subset {\mathbb {R}}^2\); then,

$$\begin{aligned} \begin{aligned}&\bigg |\int _{\{N\le |x|\le 2N\}} \frac{g_l(x)}{|x|^{2\gamma }} \,\text {d}x -\int _{S_N} \frac{g_l(x)}{|x|^{2\gamma }} \,\text {d}x\bigg |\\ {}&\quad \le \int _{S_N\Delta \{N\le |x|\le 2N\}} \frac{g_l(x)}{|x|^{2\gamma }} \,\text {d}x \le \int _{S_N\Delta \{N\le |x|\le 2N\}} \frac{1}{|x|^{2\gamma }} \,\text {d}x \\&\quad \le \int _{T_N\setminus R_N} \frac{1}{|x|^{2\gamma }} \,\text {d}x \le \frac{C_4}{N^{2\gamma -1}} \le \frac{C_5}{N} \Lambda _N^2, \end{aligned} \end{aligned}$$

where the last step follows from

$$\begin{aligned} \Lambda _N^2 = \sum _{N\le |k| \le 2N} \frac{1}{|k|^{2\gamma }} \ge \frac{1}{(2N)^{2\gamma }}\, \#\{k\in {\mathbb {Z}}^2_0: N\le |k| \le 2N \} \ge \frac{C_6}{N^{2\gamma -2}}. \end{aligned}$$

It is easy to see that the constants \(C_4, C_5\) and \(C_6\) depend only on \(\gamma \). The proof is complete. \(\square \)

We are not ready to provide the

Proof of Proposition A.3

In view of Lemma A.5, we define

$$\begin{aligned} J_N = \frac{1}{\Lambda _N^2} \int _{\{N\le |x|\le 2N\}} \frac{1}{|x|^{2\gamma }} \sin ^2(\angle _{x,l}) \frac{(a_l\cdot x)^2}{|x|^2} \,\textrm{d}x. \end{aligned}$$

To compute \(J_N\), we consider the new coordinate system \((y_1, y_2)\) in which the coordinate axes are \(a_l\) and \(\frac{l}{|l|}\), respectively. Let U be the orthogonal transformation matrix: \(x=Uy\). For \(i=1,2\), let \(\textrm{e}_i\in {\mathbb {R}}^2\) be such that \(\textrm{e}_{i,j}= \delta _{i,j}\), \(1\le j\le 2\). We have

$$\begin{aligned} a_l= U \textrm{e}_1 \quad \text{ and } \quad \frac{l}{|l|} = U \textrm{e}_2. \end{aligned}$$

Now \(\angle _{x,l} = \angle _{Uy,U\textrm{e}_2} = \angle _{y,\textrm{e}_2}\) and

$$\begin{aligned} J_N&= \frac{1}{\Lambda _N^2} \int _{\{N\le |y|\le 2N\}} \frac{1}{|y|^{2\gamma }} \sin ^2(\angle _{y,\textrm{e}_2}) \, \frac{(U\textrm{e}_1 \cdot Uy)^2}{|y|^2} \,\textrm{d}y \nonumber \\&= \frac{1}{\Lambda _N^2} \int _{\{N\le |y|\le 2N\}} \frac{1}{|y|^{2\gamma }} \sin ^2(\angle _{y,\textrm{e}_2})\, \frac{y_1^2}{|y|^2} \,\textrm{d}y . \end{aligned}$$
(A.9)

We compute \(J_N\) by changing the variables into the polar coordinate system:

$$\begin{aligned} {\left\{ \begin{array}{ll} y_1= r\cos \varphi , \\ y_2 = r\sin \varphi , \\ \end{array}\right. } \quad N\le r\le 2N,\,0\le \varphi < 2\pi . \end{aligned}$$

In this system, \(\varphi \) is the angle between y and \(\textrm{e}_1\), hence \(\sin ^2(\angle _{y,\textrm{e}_2}) = \cos ^2 \varphi \). As a consequence,

$$\begin{aligned} J_N = \frac{1}{\Lambda _N^2} \int _N^{2N} \textrm{d}r \int _0^{2\pi } \frac{1}{r^{2\gamma }} (\cos ^4 \varphi ) \, r \, \textrm{d}\varphi = \frac{3\pi }{4 \Lambda _N^2} \int _N^{2N}\frac{\textrm{d}r}{r^{2\gamma -1}} . \end{aligned}$$
(A.10)

where in the second step we have used

$$\begin{aligned} \int _0^{2\pi } \cos ^4 \varphi \ \textrm{d}\varphi = \frac{1}{4} \int _0^{2\pi } (1+ \cos 2\varphi )^2 \, \textrm{d}\varphi = \frac{1}{4} \int _0^{2\pi } \bigg (1+ 2\cos 2\varphi + \frac{1+ \cos 4\varphi }{2} \bigg ) \, \textrm{d}\varphi = \frac{3}{4} \pi . \end{aligned}$$

Following the proof of Lemma A.5 (it is much simpler here since the function g can be taken identically 1), one can show

$$\begin{aligned} \bigg |\sum _{k} \big (\theta ^N_k \big )^2 - \frac{1}{\Lambda _N^2} \int _{\{N\le |x|\le 2N\}} \frac{\textrm{d}x}{|x|^{2\gamma }} \bigg |\le \frac{C}{N} \end{aligned}$$
(A.11)

for some constant \(C>0\). Equivalently,

$$\begin{aligned} \bigg | 1 - \frac{2\pi }{\Lambda _N^2} \int _N^{ 2N} \frac{\textrm{d}r}{r^{2\gamma -1}} \bigg |\le \frac{C}{N}. \end{aligned}$$

Recalling (A.10), we obtain

$$\begin{aligned} \bigg |J_N - \frac{3}{8} \bigg | \le \frac{C'}{N}. \end{aligned}$$

Combining this limit with Lemmas A.4 and A.5, we complete the proof. \(\square \)

Finally we discuss the necessary modifications for proving the estimate (3.2) in the 3D case.

Remark A.7

Recall [33, Corollary 5.3] for the expression of \(S_\theta ^{(3),\perp }(v)\) (the part “orthogonal” to \(S_\theta ^{(3)}(v)\)); note that there \(\nu \) is the noise intensity, playing the role of \(\kappa \) in this paper. Similarly to Lemma A.2 above, we can rewrite it as

$$\begin{aligned} S_\theta ^{(3),\perp }(v)= -6\pi ^2 \kappa \sum _{l\in {\mathbb {Z}}^3_0} \sum _{i=1}^2 v_{l,i} |l|^2 \sum _{j=1}^2 \bigg [\sum _k \theta _k^2 \sin ^2(\angle _{k,l}) \frac{(a_{l,i}\cdot (k-l)) (a_{l,j}\cdot (k-l))}{|k-l|^2}\bigg ] \sigma _{l,j}. \end{aligned}$$

Following Proposition A.3, we reformulate [33, Proposition 5.4] as below: there exists \(C>0\) such that for all \(l\in {\mathbb {Z}}^3_0\) and \(N\ge 1\), it holds

$$\begin{aligned} \bigg | \sum _k \theta _k^2 \sin ^2(\angle _{k,l}) \frac{(a_{l,i}\cdot (k-l)) (a_{l,j}\cdot (k-l))}{|k-l|^2} - \frac{4}{15} \delta _{i,j} \bigg | \le C \frac{|l|}{N} \end{aligned}$$
(A.12)

for \(i,j \in \{1,2\}\). With this result in hand, repeating the proof of (A.4) yields

$$\begin{aligned} \bigg \Vert S_{\theta ^N}^{(3),\perp } (v)- \frac{2}{5} \kappa \Delta v \bigg \Vert _{H^{s-2-\alpha }} \le C \frac{\kappa }{N^{\alpha }} \Vert v \Vert _{H^s}. \end{aligned}$$

To show (A.12), we simply replace [33, Lemma 5.6] by the following analogue of Lemma A.5: there exists \(C= C(\gamma )>0\), independent of \(N\ge 1\) and \(l\in {\mathbb {Z}}^3_0\), such that for any \(i,j\in \{1,2 \}\),

$$\begin{aligned} \bigg | \sum _{k} \big (\theta ^N_k \big )^2 g_l^{i,j}(k) -\frac{1}{\Lambda _N^2} \int _{\{N\le |x|\le 2N\}} \frac{1}{|x|^{2\gamma }} g_l^{i,j}(x) \,\textrm{d}x \bigg | \le \frac{C}{N}, \end{aligned}$$

where \(g_l^{i,j}(x)= \sin ^2(\angle _{x,l}) \frac{(a_{l,i}\cdot x)(a_{l,j}\cdot x)}{|x|^2},\, x\in {\mathbb {R}}^3{\setminus } \{0\}\). Note that, by symmetry, the second quantity in the absolute value vanishes if \(i\ne j\). The proof of this estimate is similar to that of Lemma A.5, since \(g_l^{i,j}\) enjoys the same properties as those of \(g_l\).

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Luo, D. Enhanced Dissipation for Stochastic Navier–Stokes Equations with Transport Noise. J Dyn Diff Equat (2023). https://doi.org/10.1007/s10884-023-10307-w

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