Large and Moderate Deviations Principles and Central Limit Theorem for the Stochastic 3D Primitive Equations with Gradient-Dependent Noise

Abstract

We establish the large deviations principle (LDP), the moderate deviations principle (MDP), and an almost sure version of the central limit theorem (CLT) for the stochastic 3D viscous primitive equations driven by multiplicative white noise allowing dependence on the spatial gradient of velocity with initial data in $H^2$. We establish the LDP using the weak convergence approach by Budjihara and Dupuis and a uniform version of the stochastic Gronwall lemma. The result corrects a minor technical issue in Dong et al. (J Differ Equ 263(5):3110–3146, 2017) and establishes the result for a more general noise. The MDP is established by a similar argument.

Uniform Version of the Stochastic Gronwall Lemma

The following result is not new; in fact, it is a combination of the stochastic Gronwall lemma from Glatt-Holtz and Ziane [18, Lemma 5.3] and a part of the proof of global existence of the strong solutions of 2D stochastic Navier–Stokes equation from Glatt-Holtz and Ziane [18, Theorem 4.2]. We include the proof for the sake of completeness.

Proposition A.1

Let $\varepsilon _0 > 0$. Let $X^\varepsilon , Y^\varepsilon , Z^\varepsilon , R^\varepsilon : [0, \infty ) \times \Omega \rightarrow [0, \infty )$ be stochastic processes on a probability space $(\Omega , \mathcal {F}, \mathbb {P})$. Let $\tau _K^{R, \varepsilon }$ be the stopping time defined by

$$\begin{aligned} \tau _K^{R, \varepsilon } = \inf \left\{ t \ge 0 \mid \int _0^t R^\varepsilon \, \mathrm{d}s \ge K \right\} , \quad K > 0, \varepsilon \in (0, \varepsilon _0]. \end{aligned}$$

Let for all $t > 0$

$$\begin{aligned} \lim _{K \rightarrow \infty } \mathbb {P}\left( \left\{ \tau _K^{R, \varepsilon } \le t \right\} \right) = 0 \, \mathrm{uniformly\, w.r.t.\, } \varepsilon \in (0, \varepsilon _0]. \end{aligned}$$

(A.1)

Let $T > 0$ and let for all $K > 0$ and $\varepsilon \in (0, \varepsilon _0]$

$$\begin{aligned} \mathbb {E}\int _0^{T \wedge \tau _K^{R, \varepsilon }} R^\varepsilon X^\varepsilon + Z^\varepsilon \, \mathrm{d}s \le C_{T, K} < \infty . \end{aligned}$$

Let there exist a constant $C_0 = C_0(T)$ such that, for all $\varepsilon \in (0, \varepsilon _0]$ and all stopping times $\tau _a$ and $\tau _b$ satisfying $0 \le \tau _a \le \tau _b \le T \wedge \tau _K^{R, \varepsilon }$, one has

$$\begin{aligned} \mathbb {E}\left[ \sup _{s \in [\tau _a, \tau _b]} X^\varepsilon + \int _{\tau _a}^{\tau _b} Y^\varepsilon \, \mathrm{d}s \right] \le C_0 \left[ X(\tau _a) + \int _{\tau _a}^{\tau _b} R^\varepsilon X^\varepsilon + Z^\varepsilon \, \mathrm{d}s \right] . \end{aligned}$$

Then, for all $\varepsilon \in (0, \varepsilon _0]$ and $K > 0$, we have

$$\begin{aligned} \mathbb {E}\left[ \sup _{s \in \left[ 0, T \wedge \tau _K^{R, \varepsilon }\right] } X^\varepsilon + \int _0^{T \wedge \tau _K^{R, \varepsilon }} Y^\varepsilon \, \mathrm{d}s \right] \le C_{C_0, T, K} \mathbb {E}\left[ X(0) + \int _0^{T \wedge \tau _K^{R, \varepsilon }} Z^{\varepsilon } \, \mathrm{d}s \right] .\nonumber \\ \end{aligned}$$

(A.2)

Moreover, if we define the stopping time $\tau _K^{X, \varepsilon }$ by

$$\begin{aligned} \tau _K^{X, \varepsilon } = \inf \left\{ t \ge 0 \mid \sup _{s \in [0, t]} X^\varepsilon + \int _0^t Y^\varepsilon \, \mathrm{d}s \ge K \right\} , \quad K > 0, \varepsilon \in (0, \varepsilon _0], \end{aligned}$$

then for all $t > 0$

$$\begin{aligned} \lim _{K \rightarrow \infty } \mathbb {P}\left( \left\{ \tau _K^{X, \varepsilon } \le t \right\} \right) = 0 \, \mathrm{uniformly\, w.r.t.\, } \varepsilon \in (0, \varepsilon _0], \end{aligned}$$

(A.3)

and $\tau _K^{X, \varepsilon } \rightarrow +\infty$ as $K \rightarrow \infty$ $\mathbb {P}$-a.s. for all $\varepsilon \in (0, \varepsilon _0]$.

Proof

The inequality (A.2) is the stochastic Gronwall lemma from Glatt-Holtz and Ziane [18, Lemma 5.3]. Following the argument from Glatt-Holtz and Ziane [18, Theorem 4.2], we use the Chebyshev theorem and (A.2) to estimate

$$\begin{aligned} \mathbb {P}\left( \left\{ \tau _K^{X, \varepsilon } \le t \right\} \right)&\le \mathbb {P}\left( \left\{ \tau _K^{X, \varepsilon } \le t \right\} \cap \left\{ \tau _M^{R, \varepsilon } > t \right\} \right) + \mathbb {P}\left( \left\{ \tau _M^{R, \varepsilon } \le t \right\} \right) \\&\le \mathbb {P}\left( \left\{ \sup _{s \in \left[ 0, t \wedge \tau _M^{R, \varepsilon } \right] } X^\varepsilon + \int _0^{t \wedge \tau _M^{R, \varepsilon }} Y^\varepsilon \, \mathrm{d}s \ge K \right\} \right) + \mathbb {P}\left( \left\{ \tau _M^{R, \varepsilon } \le t \right\} \right) \\&\le \frac{1}{K} \mathbb {E}\left[ \sup _{s \in \left[ 0, t \wedge \tau _M^{R, \varepsilon } \right] } X^\varepsilon + \int _0^{t \wedge \tau _M^{R, \varepsilon }} Y^\varepsilon \, \mathrm{d}s \right] + \mathbb {P}\left( \left\{ \tau _M^{R, \varepsilon } \le t \right\} \right) \\&\le \frac{C_{t, M}}{K} + \mathbb {P}\left( \left\{ \tau _M^{R, \varepsilon } \le t \right\} \right) . \end{aligned}$$

Let $\delta > 0$ be arbitrary. By the uniform convergence (A.1), we find $M \in \mathbb {N}$ such that, for all $\varepsilon \in (0, \varepsilon _0]$, we have

$$\begin{aligned} \mathbb {P}\left( \left\{ \tau _M^{R, \varepsilon } \le t \right\} \right) < \frac{\delta }{2}. \end{aligned}$$

Let $K_0 \in \mathbb {N}$ be such that $C_{t, M}/K < \delta /2$ for all $K \in \mathbb {N}$, $K \ge K_0$. Collecting the above, we deduce that

$$\begin{aligned} \mathbb {P}\left( \left\{ \tau _K^{X, \varepsilon } \le t \right\} \right) < \delta , \end{aligned}$$

for all $\varepsilon \in (0, \varepsilon _0]$ and all $K \in \mathbb {N}$, $K \ge 0$, which finishes the proof of (A.3).

To establish the $\mathbb {P}$-a.s. convergence, we argue by contradiction. Assume that $\mathbb {P}(\lbrace \lim _{K \rightarrow \infty } \tau _K^{X, \varepsilon } < +\infty \rbrace ) > 0$ for some $\varepsilon \in (0, \varepsilon _0]$. Then, since $\lbrace \lim _{K \rightarrow \infty } \tau _K^{X, \varepsilon } < +\infty \rbrace = \bigcup _{N \in \mathbb {N}} \lbrace \lim _{K \rightarrow \infty } \tau _K^{X, \varepsilon } \le N \rbrace$, there exists $N_0 \in \mathbb {N}$ such that $\mathbb {P}(\lbrace \lim _{K \rightarrow \infty } \tau _K^{X, \varepsilon } \le N_0 \rbrace ) > 0$. On the other hand, since $\tau _K^{X, \varepsilon }$ is monotone, i.e. $\lbrace \tau _K^{X, \varepsilon } \le N_0 \rbrace \supseteq \lbrace \tau _L^{X, \varepsilon } \le N_0 \rbrace$ for $K \le L$, we observe $\lbrace \lim _{K \rightarrow \infty } \tau _K^{X, \varepsilon } \le N_0 \rbrace = \bigcap _{K \in \mathbb {N}} \lbrace \tau _K^{X, \varepsilon } \le N_0 \rbrace$. However, (A.3) implies

$$\begin{aligned} 0 < \mathbb {P}\left( \left\{ \lim _{K \rightarrow \infty } \tau _K^{X, \varepsilon } \le N_0 \right\} \right) = \mathbb {P}\left( \bigcap _{K \in \mathbb {N}} \left\{ \tau _K^{X, \varepsilon } \le N_0 \right\} \right) = \lim _{K \rightarrow \infty } \mathbb {P}\left( \left\{ \tau _K^{X, \varepsilon } \le N_0 \right\} \right) = 0, \end{aligned}$$

a contradiction. $\square$