The Sobolev Stability Threshold for 2D Shear Flows Near Couette

Abstract

We consider the 2D Navier–Stokes equation on $\mathbb T \times \mathbb R$, with initial datum that is $\varepsilon$-close in $H^N$ to a shear flow (U(y), 0), where $\Vert U(y) - y\Vert _{H^{N+4}} \ll 1$ and $N>1$. We prove that if $\varepsilon \ll \nu ^{1/2}$, where $\nu$ denotes the inverse Reynolds number, then the solution of the Navier–Stokes equation remains $\varepsilon$-close in $H^1$ to $(e^{t \nu \partial _{yy}}U(y),0)$ for all $t>0$. Moreover, the solution converges to a decaying shear flow for times $t \gg \nu ^{-1/3}$ by a mixing-enhanced dissipation effect, and experiences a transient growth of gradients. In particular, this shows that the stability threshold in finite regularity scales no worse than $\nu ^{1/2}$ for 2D shear flows close to the Couette flow.

Appendix: Construction and Properties of the Multiplier M

In this section, we recall some of the technical tools regarding the Fourier multiplier M from Bedrossian et al. (2015d) and adapt them to our simpler setting.

Lemma 4.1

There exists a multiplier M such that the conditions (16a)–(16e) hold for some constant $0<c<1$.

Proof of Lemma 4.1

We consider a multiplier of the form $M=M_1M_2$ such that both $M_1$ and $M_2$ satisfy (16a). We choose $M_1$ such that for $k\ne 0$ it is determined by the ODE

$$\begin{aligned} -\frac{\dot{M_1}}{M_1}&=\frac{\left| k\right| }{k^2+|\xi -kt|^2} \\ M_1(0,k,\eta )&= 1. \end{aligned}$$

Similar multipliers appeared in Zillinger (2014), Bedrossian et al. (2015b, (2015c, (2015d). The above multiplier clearly satisfies (16c). Notice that for $k \ne 0$, there holds

$$\begin{aligned} \frac{k^2 + \left| \eta -kt\right| ^2}{k^2+|\xi -kt|^2} = \frac{k^2 + \left| \xi -kt + \eta -\xi \right| ^2}{k^2+|\xi -kt|^2} \lesssim 1 + \left| \eta -\xi \right| ^2, \end{aligned}$$

and hence (16f) holds for $M_1$. A direct computation shows that

$$\begin{aligned} M_1(t, k, \xi ) = \exp {\left( -\int _0^t \frac{\left| k\right| }{k^2+|\xi -ks|^2}\,\hbox {d}s\right) }, \end{aligned}$$

which implies (16b) holds for $M_1$. Taking the derivative of $M_1$ with respect to $\xi$ gives

$$\begin{aligned} \left| \frac{\partial _{\xi }M_1(k,\xi )}{M_1(k,\xi )}\right| = \left| \int _0^t \frac{2\left| k\right| (\xi -k s)}{(k^2+|\xi -ks|^2)^2}\,\hbox {d}s\right| \le \frac{2}{\left| k\right| ^2}\int _0^t \frac{1}{(1+|\xi /k-s|^2)}\,\hbox {d}s, \end{aligned}$$

which proves that (16d) holds for $M_1$. Note here that $k \ne 0$ implies $|k|\ge 1$.

Next, we define $M_2$ by the differential equation (for $k \ne 0$),

$$\begin{aligned} -\frac{\dot{M_2}}{M_2}&=\frac{\nu ^{1/3}}{(\nu ^{1/3}\left| t-\xi /k\right| )^{2}+1} \\ M_2(0,k,\eta )&= 1. \end{aligned}$$

This multiplier was introduced in Bedrossian et al. (2015d). Similarly to $M_1$, we deduce that (16b), (16d), and (16f) all hold for $M_2$ (and hence also for $M = M_1 M_2$). Since for $k\ne 0$ we have that

$$\begin{aligned} 1\lesssim \nu ^{1/3}|k,\xi -kt| \quad \text{ if } \quad \nu ^{-1/3}\le \left| t-\dfrac{\xi }{k}\right| \end{aligned}$$

and

$$\begin{aligned} 1\lesssim \dfrac{1}{(\nu ^{1/3}\left| t-\xi /k\right| )^{2}+1} \quad \text{ if } \quad \nu ^{-1/3}\ge \left| t-\dfrac{\xi }{k}\right| , \end{aligned}$$

inequality (16e) holds for $M_2$. Therefore, the multiplier M we constructed satisfies conditions (16a)–(16e), completing the proof. $\square$

Remark 4.2

In condition (16e), the power $\nu ^{-1/6}$ in front of the multiplier is sharp in the sense that 1 / 6 is the smallest sacrifice we need to make to bound the expression on the right side from below by a constant. In fact, if we make $M_2$ to be positive constants independent of $\nu$ when $\left| t-\xi /k\right| \ge \nu ^{-1/3}$, then the size of $\dot{M}_2$ should be approximately $\nu ^{1/3}$. Hence, if we did not need (16f) or (16d), one could construct $M_2$ to satisfy (16e) with

$$\begin{aligned} \dot{M}_2&=0 \ \ \ \ \ \ \ \ \mathrm{if}\,\, \nu ^{-1/3}\ge \left| t-\dfrac{\xi }{k}\right| \\ \dot{M}_2&=-\dfrac{1}{2}\nu ^{1/3} \ \ \ \ \ \ \ \ \mathrm{if}\ \nu ^{-1/3}\ge \left| t-\dfrac{\xi }{k}\right| . \end{aligned}$$

We need the following Lemma to commute $\sqrt{-\dot{M}M}$ with $\Delta _L\Delta _t^{-1}$, the latter of which is not a Fourier multiplier.

Lemma 4.3

Let $f\in H^N$ and $N > 1$. Then the following estimate holds for $\delta$ sufficiently small,

$$\begin{aligned} \left\| \sqrt{-\dot{M}M}\Delta _L\Delta _t^{-1}f_{\ne }\right\| _{H^{N}} \lesssim \left\| \sqrt{-\dot{M}M}f_{\ne }\right\| _{H^{N}}. \end{aligned}$$

Proof of Lemma 4.3

Using the equality

$$\begin{aligned} \Delta _L=\Delta _t-(a^2-1)\partial _{vv}^L-b\partial _{v}^L, \end{aligned}$$

we have

$$\begin{aligned} \left\| \sqrt{-\dot{M}M}\Delta _L\Delta _t^{-1}f_{\ne }\right\| _{H^{N}}&\lesssim \left\| \sqrt{-\dot{M}M}f_{\ne }\right\| _{H^{N}} \nonumber \\&\quad + \left\| \sqrt{-\dot{M}M}\left( (a^2-1)\partial _{vv}^L\Delta _t^{-1}f_{\ne }\right) \right\| _{H^{N}} \nonumber \\&\quad + \left\| \sqrt{-\dot{M}M}\left( b\partial _{v}^L\Delta _t^{-1}f_{\ne }\right) \right\| _{H^{N}}. \end{aligned}$$

(50)

By (16f), $N>1$, and (42), we deduce

$$\begin{aligned}&\left\| \sqrt{-\dot{M}M}\left( (a^2-1)\partial _{vv}^L\Delta _t^{-1}f_{\ne }\right) \right\| _{H^{N}}\\&\quad = \left\| \sqrt{-\dot{M}M}\left( ((a-1)^2-2(a-1))\partial _{vv}^L\Delta _t^{-1}f_{\ne }\right) \right\| _{H^{N}} \\&\qquad \lesssim \Vert a-1\Vert _{H^{N+1}}\left\| \sqrt{-\dot{M}M}\Delta _L\Delta _t^{-1}f_{\ne }\right\| _{H^{N}} \\&\qquad \lesssim \delta \left\| \sqrt{-\dot{M}M}\Delta _L\Delta _t^{-1}f_{\ne }\right\| _{H^{N}}, \end{aligned}$$

and similarly

$$\begin{aligned} \left\| \sqrt{-\dot{M}M}\left( b\partial _{v}^L\Delta _t^{-1}f_{\ne }\right) \right\| _{H^{N}}&\lesssim \Vert b\Vert _{H^{N+1}}\left\| \sqrt{-\dot{M}M}\Delta _L\Delta _t^{-1}f_{\ne }\right\| _{H^{N}} \\&\lesssim \delta \left\| \sqrt{-\dot{M}M}\Delta _L\Delta _t^{-1}f_{\ne }\right\| _{H^{N}}. \end{aligned}$$

Since $\delta \ll 1$, the result follows from (50) immediately. $\square$

The following lemma is proved in the same manner as Lemma 4.3, although slightly simpler. In particular, this lemma shows that $\Delta _L \Delta _t^{-1}$ can be approximately treated as the identity for $\delta$ sufficiently small.

Lemma 4.4

Let $f\in H^N$ and $N > 1$. Then the following estimate holds for $\delta$ sufficiently small,

$$\begin{aligned} \left\| \Delta _L\Delta _t^{-1}f_{\ne }\right\| _{H^{N}} \lesssim \left\| f_{\ne }\right\| _{H^{N}}. \end{aligned}$$

The following estimate applies to the zero mode of the velocity field.

Lemma 4.5

Let f be such that $f_0\in H^N$, $N>1$, and $\partial _{v}\Delta _t^{-1}f_0\in L^2$. Then for $\delta$ sufficiently small, there holds

$$\begin{aligned} \Vert \partial _{vv}\Delta _t^{-1}f_0\Vert _{H^{N}} \lesssim \Vert f_0\Vert _{H^N}+\delta \Vert \partial _{v}\Delta _t^{-1}f_0\Vert _{L^{2}}. \end{aligned}$$

Proof of Lemma 4.5

Since

$$\begin{aligned} \partial _{vv}\Delta _t^{-1}f_0 = (\Delta _t-(a^2-1)\partial _{vv}-b\partial _v)\Delta _t^{-1}f_0, \end{aligned}$$

we have by $N>1$ and (42) (interpolating $H^N$ between $L^2$ and $H^{N+1}$),

$$\begin{aligned} \Vert \partial _{vv}\Delta _t^{-1}f_0\Vert _{H^{N}}&\lesssim \Vert f_0\Vert _{H^{N}}+\delta \Vert \partial _{vv}\Delta _t^{-1}f_0\Vert _{H^{N}} +\delta \Vert \partial _{v}\Delta _t^{-1}f_0\Vert _{H^{N}} \\&\lesssim \Vert f_0\Vert _{H^{N}}+\delta \Vert \partial _{vv}\Delta _t^{-1}f_0\Vert _{H^{N}} +\delta \Vert \partial _{v}\Delta _t^{-1}f_0\Vert _{L^{2}}. \end{aligned}$$

For $\delta \ll 1$, we may absorb the second term on the right side and we obtain the desired result. $\square$