Linear inviscid damping and enhanced dissipation for the Kolmogorov flow

Advances in Mathematics 362 (2020) 106963

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Advances in Mathematics www.elsevier.com/locate/aim

Linear inviscid damping and enhanced dissipation for the Kolmogorov flow Dongyi Wei a , Zhifei Zhang a,∗ , Weiren Zhao b a

School of Mathematical Science, Peking University, 100871, Beijing, PR China Department of Mathematics, New York University in Abu Dhabi, Saadiyat Island, P.O. Box 129188, Abu Dhabi, United Arab Emirates b

a r t i c l e

i n f o

a b s t r a c t

Article history: Received 9 April 2018 Received in revised form 6 August 2019 Accepted 20 December 2019 Available online xxxx Communicated by C. Fefferman

In this paper, we prove the linear inviscid damping and vorticity depletion phenomena for the linearized Euler equations around the Kolmogorov flow. These results confirm Bouchet and Morita’s predictions based on numerical analysis. By using the wave operator method introduced by Li, Wei and Zhang, we solve Beck and Wayne’s conjecture on the enhanced dissipation rate for the 2-D linearized Navier-Stokes equations around the bar state called Kolmogorov flow. The same dissipation rate is proved for the Navier-Stokes equations if the initial velocity is included in a basin of attraction of the Kol2 mogorov flow with the size of ν 3 + , here ν is the viscosity coefficient. © 2019 Elsevier Inc. All rights reserved.

Keywords: Euler and Navier Stokes equation Inviscid damping Enhanced dissipation Kolmogorov flow Metastability

Contents 1. 2.

Introduction . . . . . . . . . . . . . Sketch of the proof . . . . . . . . . 2.1. Linear inviscid damping 2.2. Enhanced dissipation . .

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* Corresponding author. E-mail addresses: [email protected] (D. Wei), [email protected] (Z. Zhang), [email protected], [email protected] (W. Zhao). https://doi.org/10.1016/j.aim.2019.106963 0001-8708/© 2019 Elsevier Inc. All rights reserved.

D. Wei et al. / Advances in Mathematics 362 (2020) 106963

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2.3. Notations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . The homogeneous Rayleigh equation . . . . . . . . . . . . . . . . . . . . 3.1. Existence of the solution . . . . . . . . . . . . . . . . . . . . . . . 3.2. Uniform estimates in wave number α . . . . . . . . . . . . . . 4. The inhomogeneous Rayleigh equation . . . . . . . . . . . . . . . . . . 4.1. The Wronskian . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.2. Uniform estimates of A(c), B(c) . . . . . . . . . . . . . . . . . . 4.3. Uniform estimates of A1 (c), B1 (c) . . . . . . . . . . . . . . . . . 4.4. Solution formula of the inhomogeneous Rayleigh equation 5. Linear inviscid damping . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.1. The limiting absorption principle . . . . . . . . . . . . . . . . . 5.2. Explicit formulas of the limits . . . . . . . . . . . . . . . . . . . 5.3. Solution formula of the linearized Euler equations . . . . . 5.4. Dual formulation . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.5. Decay estimates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6. Estimates of some integral operators . . . . . . . . . . . . . . . . . . . . 6.1. Basic properties of Hilbert transform . . . . . . . . . . . . . . 6.2. Estimates of II1,1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.3. Estimates of Lk . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7. W 2,1 estimates of Ko and Ke . . . . . . . . . . . . . . . . . . . . . . . . 7.1. W 2,1 estimate of Ko . . . . . . . . . . . . . . . . . . . . . . . . . . 7.2. W 2,1 estimate of Ke . . . . . . . . . . . . . . . . . . . . . . . . . . 8. Decay estimates at critical points . . . . . . . . . . . . . . . . . . . . . . 8.1. Vorticity depletion phenomena . . . . . . . . . . . . . . . . . . . 8.2. Decay estimates at critical points . . . . . . . . . . . . . . . . . 9. Wave operator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.1. Basic properties of wave operator . . . . . . . . . . . . . . . . . 9.2. Commutator estimate . . . . . . . . . . . . . . . . . . . . . . . . . 10. Linear enhanced dissipation . . . . . . . . . . . . . . . . . . . . . . . . . . 10.1. Decay estimates on the model without nonlocal term . . . 10.2. Decay estimates on the model with nonlocal term . . . . . 10.3. Proof of Theorem 1.4 . . . . . . . . . . . . . . . . . . . . . . . . . 11. Nonlinear enhanced dissipation . . . . . . . . . . . . . . . . . . . . . . . 11.1. Semigroup estimates . . . . . . . . . . . . . . . . . . . . . . . . . . 11.2. Proof of Theorem 1.7 . . . . . . . . . . . . . . . . . . . . . . . . . Acknowledgments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.

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1. Introduction 

In this paper, we study the incompressible Navier-Stokes equations on the torus Tδ2 =  (x, y) : x ∈ T2πδ , y ∈ T2π with 0 < δ ≤ 1: ⎧ ⎪ ⎨ ∂t V − νΔV + V · ∇V + ∇P = 0, ∇ · V = 0, ⎪ ⎩V| t=0 = V0 (x, y).

(1.1)

 Here V = V 1 (t, x, y), V 2 (t, x, y) , P (t, x, y) denote the velocity and pressure of the fluid respectively, and ν ≥ 0 is the viscosity coefficient. When ν = 0, (1.1) is reduced to the Euler equations.

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It is easy to see that VN S = (−a0 e−νt cos y, 0) for any constant a0 is a special solution of (1.1) called the bar state or Kolmogorov flow. The bar states are physically relevant due to the fact that they can be viewed as maximum entropy solutions of the Euler equations, as well as the fact that they dominate the flow when the aspect ratio of the torus δ < 1 [11]. The first part of this paper is devoted to studying the linear inviscid damping for the linearized Euler equations around the Kolmogorov flow (− cos y, 0), which take in the vorticity formulation:

∂t ω + Lω = 0, ω|t=0 = ω0 (x, y),

(1.2)

where ω = ∂x V 2 − ∂y V 1 is the vorticity and L = u(y)∂x + u (y)∂x (−Δ)−1 ,

u(y) = − cos y.

(1.3)

After the breakthrough work on Landau damping by Mouhot and Villani [29], inviscid damping (an analogue of Landau damping in hydrodynamics) becomes a very active field. Bedrossian and Masmoudi [7] proved nonlinear inviscid damping for the 2-D Euler equations around the Couette flow (y, 0) for the perturbation in Gevrey class. On the other hand, Lin and Zeng [24] proved that nonlinear inviscid damping is not true for the perturbation in H s for s < 32 . For general shear flows, the linear inviscid damping is also a difficult problem due to the presence of the nonlocal part u (y)∂x (−Δ)−1 . In such case, the linear dynamics is associated with the singularities at the critical layers u = c of the solution for the Rayleigh equation (u − c)(φ − α2 φ) − u φ = f. Case [12] gave a first prediction of linear damping for monotone shear flow. His prediction was confirmed by a series of works [30,31,40,39], and finally by our work [35]. Roughly speaking, if u(y) ∈ C 4 is monotone and L has no embedding eigenvalues, then V (t)L2 ≤

C ω0 Hx−1 H 1 , y t

V 2 (t)L2 ≤

C ω0 Hx−1 H 2 , y t2

 for the initial vorticity ω0 satisfying T ω0 (x, y)dx = 0 and PL ω0 = 0, where PL is the  spectral projection to σd L . Very recently, new methods are developed for monotone shear flows [20,38], where the vector field method introduced in [38] gave a very simple and elementary proof for linear inviscid damping. For non-monotone flows such as Poiseuille flow u(y) = y 2 and Kolmogorov flow u(y) = cos y, it is even difficult to predict the inviscid damping due to strong degeneration of the Rayleigh equation at critical points. Based on Laplace tools and numerical computations, Bouchet and Morita [10] gave a first prediction of the inviscid damping for non-monotone

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flows, especially Kolmogorov flow. Besides the vorticity mixing mechanism, they found that the depletion phenomena of the vorticity at the stationary streamlines also plays the key role leading to the inviscid damping for non-monotone flows. In our work [36], we confirmed the predictions of Bouchet and Morita about the linear damping and the vorticity depletion phenomena for a class of shear flows (u(y), 0) in a finite channel denoted by K, which consists of the function u(y) satisfying u(y) ∈ H 3 (−1, 1), and u (y) = 0 for critical points (i.e., u (y) = 0) and u (±1) = 0. Moreover, for a class of symmetric flows including Poiseuille flow, we obtained the same decay rate of the velocity as in the monotone flow. In this paper, we extend the linear inviscid damping and the vorticity depletion phenomena to the Kolmogorov flow on the torus. These results confirm Bouchet and Morita’s predictions based on numerical computations. The first result is the linear inviscid damping. Theorem 1.1. Let δ ∈ (0, 1) and ω(t, x, y) be the solution of (1.2) with 0. Then it holds that

 T2πδ

ω0 (x, y)dx =

−1

1. if ω0 (x, y) ∈ Hx 2 Hy1 , then V (t)L2 ≤

C ω0  − 12 1 ; Hx Hy t

V 2 (t)L2 ≤

C ω0  12 2 ; Hx Hy t2

1

2. if ω0 (x, y) ∈ Hx2 Hy2 , then

1

1

3. if ω0 (x, y) ∈ Hx2 Hyk for k = 0, 1, there exists ω∞ (x, y) ∈ Hx2 Hyk such that ω(t, x + tu(y), y) − ω∞ L2 −→ 0

as

t → +∞.

Remark 1.2. When δ = 1, Proposition 4.3, 4.5, 4.7 still hold for |α| ≥ 2. Thus, we can obtain the following decay result: −1

1. if ω0 (x, y) ∈ Hx 2 Hy1 , then P≥2 V (t)L2 ≤

C ω0  − 12 1 ; Hx Hy t

P≥2 V 2 (t)L2 ≤

C ω0  12 2 . Hx Hy t2

1

2. if ω0 (x, y) ∈ Hx2 Hy2 , then

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Here and in what follows, P≥2 denotes the orthogonal projection of L2 (T 2 ) to the sub  space f ∈ L2 (T 2 ) : f (x, y) = f (α, y)eiαx . |α|≥2

The second result is the behavior of the solution at critical points y = kπ for k ∈ Z, which in particular confirms the vorticity depletion phenomena found by Bouchet and Morita [10]. Theorem 1.3. Let δ ∈ (0, 1) and ω(t, x, y) be the solution of (1.2) with 0. Then for s > 12 , it holds that for all k ∈ Z,

 T2πδ

ω0 (x, y)dx =

1. if ωe (x, y) ∈ Hx1+s Hy3 , then for x ∈ T2πδ , |t| > 1, |ω(t, x, kπ)| ≤

C ωe Hx1+s Hy3 ; |t|

2. if ωe (x, y) ∈ Hx1+s Hy3 , then for x ∈ T2πδ , |t| > 1, |V 2 (t, x, kπ)| ≤ s− 12

3. if ωo (x, y) ∈ Hx

C ωe Hx1+s Hy3 ; |t|2

Hy3 , then for x ∈ T2πδ , |t| > 1, |V 1 (t, x, kπ)| ≤

C . 3 ωo  s− 1 Hx 2 Hy3 |t| 2

Here ωe (x, y) = 12 (ω0 (x, y) + ω0 (x, −y)) and ωo (x, y) = 12 (ω0 (x, y) − ω0 (x, −y)). Recently, Lin and Xu [23] proved the linear inviscid damping for shear flows including Kolmogorov flow in the following sense: 1 lim T →∞ T

T V (t)2L2 dt = 0. 0

Their proof is based on the Hamiltonian structure of the linearized Euler equation and an instability index theory recently developed by Lin and Zeng [25]. The second part of this paper is devoted to studying the 2-D incompressible NavierStokes equation with small viscosity on the torus. The vorticity formulation takes

ωt − νΔω + V · ∇ω = 0, ω|t=0 = ω0 (x, y).

(1.4)

Experiments and numerical studies [14,27] have shown that the solutions of the nearly inviscid 2-D incompressible Navier-Stokes equations will approach to some quasi-stationary

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(metastable) states such as the bar stats and dipole states in a much shorter time than the diffusive time O( ν1 ). Then they dominate the dynamics for very long time intervals. Afterwards they decay on the diffusive time scale. See [1] for a similar phenomena for Burgers equation with small viscosity. To explain the metastability phenomena, Beck and Wayne [2] considered the linearized Navier-Stokes equations around the bar states, which are relatively easy to analyse from a mathematical view of point. Moreover, the bar states are more physically relevant in the case when the aspect ratio of the torus δ < 1 [11]. The linearized Navier-Stokes equations around the Kolmogorov flow VN S = (−a0 e−νt cos y, 0) take the form

∂t ω + Lν (t)ω = 0, ω|t=0 = ω0 (x, y),

(1.5)

where Lν (t) = −νΔ − a0 e−νt cos y∂x (1 + Δ−1 ).

(1.6)

To study the dynamic of (1.5), the main difficulty is that the linearized operator Lν (t) is nonself-adjoint, nonlocal and time-dependent. In [2], the authors dropped the nonlocal part of Lν (t) and considered the following model equation ∂t ω − νΔω − a0 e−νt cos y∂x w = 0. Using the hypocoercivity method (see [34] for a systematic study), they proved the enhanced dissipation rate of the solution in some Banach space X (see (3.7) in [2]): ω(t)X ≤ Ce−M

√ νt

w0 X

for any t ∈ [0, τ /ν]. This decay rate is much faster than the diffusive decay of e−νt . The mechanism leading to the enhanced dissipation is due to mixing. See [3,13,21,8,9] and references therein for more relevant works. Let us mention recent important progress [5,6,4] on the enhanced dissipation for Couette flow (y, 0) in T × R applied to the subcritical transition. For the full linearized operator Lν (t), Beck and Wayne numerically computed the eigenvalues of −Lν (t) for a fixed time. The real parts of the eigenvalues are negative and √ scale like ν. Based on this analysis, Beck and Wayne conjectured that the same result should hold for the full linearized equation. The following theorem gives an affirmative answer to Beck and Wayne’s conjecture. Theorem 1.4. Given δ ∈ (0, 1) and τ > 0, there exist constants c1 > 0, C > 0, such  that if ω satisfies (1.5) with ω0 ∈ L2 and T2πδ ω0 (x, y)dx = 0, then it holds that for 0 ≤ t ≤ τ /ν,

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ω(t)L2 ≤ Ce−c1

7

√ νt

ω0 L2 , √ −c1 νt

Ce V (t)H˙ x1 L2y ≤ √ ω0 L2 . 1 + νt3 When δ = 1, it holds that for 0 ≤ t ≤ τ /ν, P≥2 ω(t)L2 ≤ Ce−c1

√ νt

P≥2 ω0 L2 , √ −c1 νt

Ce P≥2 ω0 L2 . P≥2 V (t)H˙ x1 L2y ≤ √ 1 + νt3 Remark 1.5. When δ = 1, the operator 1 + Δ−1 has two additional kernels {sin x, cos x}, which yield two families of exact solutions {e−νt sin x, e−νt cos x} of the linearized NavierStokes equations. Thus, we have to make the projection P≥2 for the solution in order to obtain the enhanced dissipation. Remark 1.6. It would be very interesting to prove the inviscid damping results for the slightly viscous problem. If we use the wave operator method introduced in this paper, one key point is to prove the uniform H k bound of the wave operator D ω(t) in t for 2 k = 1, 2, where one of main difficulties is the commutator estimate [∂y , D] in the Sobolev space. In [23], Lin and Xu proved the enhanced dissipation in the sense: if 0, then for any τ > 0 and  > 0, if ν is small enough, P≥2 ω(τ /ν)L2 ≤ P≥2 ω0 L2 ω(τ /ν)L2 ≤ ω0 L2

 T2πδ

ω0 (x, y)dx =

for δ = 1,

for δ < 1.

The proof is based on the Hamiltonian structure of the linearized Euler equation and RAGE theorem as in [13]. Very recently, Ibrahim, Maekawa and Masmoudi [19] proved the same decay rate of the vorticity for the following equation: ∂t ω + Lν w = 0,

Lν = −νΔ − a cos y∂x (1 + Δ−1 ),

which is the linearized equation of the 2-D Navier-Stokes equation with a force (−aν cos y, 0) around the Kolmogorov flow (−a cos y, 0). Their proof is based on the pseudospectral bound of Lν . Pseudospectra is an important concept in understanding the hydrodynamic stability [32,33] due to the non-normality of the linearized operators. Gallay et al. applied the method estimating the resolvent to study the stability of the LambOseen vortex in the regime of high circulation Reynolds number [16,15,26,18,22,17]. Recently, McQuighan and Wayne [28] applied a similar method to revisit the metastability problem of the viscous Burgers equation. However, it seems difficult to apply this method to the time-dependent operators such as Lν (t).

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As announced in a work by Li and the first two authors [22], the wave operator method could be used to solve Beck and Wayne’s conjecture. In [22], we solved Gallay’s conjecture on pseudospectral and spectral bounds on the Oseen vortices operator. One of the key ideas is to transform a nonlocal operator into a local one by constructing a wave operator. In this work, the analysis on linear inviscid damping naturally gives rise to our desired wave operator. Moreover, we also use a different method to handle the local operator. 1 Let us point out that the extra decay factor √1+νt 3 of the velocity in Theorem 1.4 comes from the vorticity depletion mechanics. Even for the time-independent operator Lν , it seems difficult to deduce this decay estimate from the resolvent estimate. This decay estimate of the velocity plays an important role in the following nonlinear enhanced dissipation result. Theorem 1.7. Given δ ∈ (0, 1), τ > 0, γ > 23 and C1 > 1, there exist constants c2 , c3 ∈ (0, 1), such that if 0 < ν < c2 , ω0 ∈ L2 , (I −P2 )ω0 L2 ≤ ν γ , and C1−1 ≤ P2 ω0 L2 ≤ C1 , then the solution to (1.4) satisfies P=0 ω(t)L2 ≤ Ce−c3

√ νt

P=0 ω0 L2

for all 0 < t < τ /ν. Here P1 denotes the orthogonal projection of L2 (Tδ2 ) to the subspace W1 = span{sin y, cos y}. Remark 1.8. The stability threshold ν γ for γ > 23 may be not optimal. We can only 3 achieve the stability threshold ν 4 if we do not use the enhanced dissipation with an extra decay factor of the velocity. It seems possible to improve the threshold γ by considering the data with higher regularity. Remark 1.9. The case of δ = 1 is a challenging problem. In this case, we would need to consider the linearized Navier-Stokes equations around the dipole states such as e−νt (− sin y, sin x),

e−νt (− cos y, cos x).

2. Sketch of the proof In this section, we present a sketch of the proof of main theorems. 2.1. Linear inviscid damping The basic ideas are similar to [35,36]. We introduce the stream function ψ so that V = (∂y ψ, −∂x ψ) and −Δψ = ω. In terms of ψ, the linearized Euler equation (1.2) takes the form ∂t Δψ + u(y)∂x Δψ − u (y)∂x ψ = 0,

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here u(y) = − cos y. Taking the Fourier transform in x, we get 

(∂y2 − α2 )∂t ψ = iα u (y) − u(y)(∂y2 − α2 ) ψ. Inverting the operator (∂y2 − α2 ), we obtain −

1

∂t ψ = Rα ψ, iα

(2.1)

where 

Rα ψ = −(∂y2 − α2 )−1 u (y) − u(∂y2 − α2 ) ψ.

(2.2)

Let Ω be a simple connected domain including the spectrum σ(Rα ) of Rα . We have the following representation formula of the solution to (2.1):

α, y) = 1 ψ(t, 2πi



α, y)dc. e−iαtc (c − Rα )−1 ψ(0,

(2.3)

∂Ω

α, y) is reduced to the study of In this way, the large time behavior of the solution ψ(t, −1 the resolvent (c − Rα ) . Let Φ be a solution of the inhomogeneous Rayleigh equation: ⎧ ⎨

u Φ = f, u−c ⎩ ∂y Φ(−π) = ∂y Φ(π), Φ(−π) = Φ(π), with f (α, y, c) =

∂y2 Φ − α2 Φ −

ω

0 (α,y) iα(u(y)−c) .

(2.4)

Then we find that

α, y) = iαΦ. (c − Rα )−1 ψ(0,

(2.5)

one of key ingredients is to establish the limiting To study the large time behavior of ψ, absorption principle: Φ(α, y, c ± iε) → Φ± (α, y, c)

for c ∈ Ran u,

as ε → 0. In [36], we established the limiting absorption principle for a class of shear flows in K (see Page 4), and proved that Φ± is bounded in Hy1 . This information is enough to show that the velocity decays to 0 as the time tends to infinity in L2 . These results can be easily extended to the Kolmogorov flow. To obtain the explicit decay estimates of the velocity, we need to know more precise behavior of the limit function Φ± . As in the case of symmetric flows, we decompose the solution of (2.4) into the odd part and even part. Each part can be handled as in 1 the monotonic flow except that u(y)−c is more singular for c = u(0) and c = u(π). For

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1 the symmetric flow considered in [36], u(y)−c is singular only at c = u(0). Due to the 1 singularity of u(y)−c , the estimates of the solution φ(y, c) of the homogeneous Rayleigh equation are more singular near critical points, for example,

|∂yβ ∂cγ φ1 (y, c)| ≤ Cαβ+γ φ1 (y, c)/u (yc )γ

for β + γ ≤ 2.

φ(y,c) Here φ1 (y, c) = u(y)−c and u(yc ) = c for c ∈ Ranu. The details will be presented in section 3. In section 4, we solve the inhomogeneous Rayleigh equation. In section 5, we prove the limiting absorption principle and give the solution formula of the linearized Euler equations. Again, the decay estimates of the velocity will be proved by using the dual method from [36]. Roughly speaking, we can rewrite the relevant integrals of interest in the form of oscillatory integrals with kernels Ko , Ke : for f = (∂y2 − α2 )g with g ∈ H 2 (0, π) ∩ H01 (0, π),

π

ψ o (t, α, y)f (y)dy = −

0

u(1) 

Ko (c, α)e−iαct dc,

u(0)

and for f = (∂y2 − α2 )g with g ∈ H 2 (0, π) and g  (0) = g  (π) = 0, π

ψ e (t, α, y)f (y)dy = −

0

u(1) 

Ke (c, α)e−iαct dc,

u(0)

where ψ o and ψ o are the odd part and even part of ψ respectively, and Ko (c, α) =

Λ1 ( ωo )(yc )Λ2 (g)(yc ) , (A(c)2 + B(c)2 )u (yc )

(2.6)

Ke (c, α) =

Λ3 ( ωe )(yc )Λ4 (g)(yc ) . (A1 (c)2 + B1 (c)2 )u (yc )

(2.7)

Here Λi (i = 1, 2, 3, 4) are some kinds of singular integral operators, and A(c)2 + B(c)2 and A1 (c)2 + B1 (c)2 correspond the Wronskian of the homogeneous Rayleigh equations. Thus, the decay estimates are reduced to W 1,2 estimates of the kernels Ko and Ke . However, the fact that the factor u (yc ) vanishes for yc = 0, π(c = ±1) leads to some essential difficulties. To cancel this singularity, we have to make very precise estimates (especially using the cancelation properties) for some quantities and singular integral operators. For example, we can prove Λj = O(sin2 yc ). The explanation of this fact is similar to [36]. Section 6 and section 7 are devoted to W 1,2 estimates of the kernels. In section 8, we prove the decay estimates of the vorticity and velocity at critical points.

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11

2.2. Enhanced dissipation Motivated by [22], we will use the wave operator method to study the enhanced dissipation. In section 9, we will construct a wave operator D defined by −D(ϕo )(−yc ) = D(ϕo )(yc ) =

Λ1 (ϕo )(yc ) 1 , (A2 + B2 ) 2

D(ϕe )(−yc ) = D(ϕe )(yc ) =

Λ3 (ϕe )(yc ) 1 , (A21 + B21 ) 2

which satisfies the following basic properties: 1. 2. 3. 4.

 D cos y(1 + (∂y2 − α2 )−1 )ω = cos yD(ω); (1 − α−2 )ω2L2 ≤ D(ω)2L2 ≤ ω2L2 ;  sin yD(ω)2L2 ≥ ∂y ψ2L2 + (α2 − 1)ψ2L2 ;  sin y(D(∂y2 ω) − ∂y2 D(ω))L2 ≤ C(|α|ωL2 + ∂y ωL2 ).

Let us first explain our construction of the wave operator. Notice that π −

π ω

o (t, α, y)g(y)dy =

0

ψ o (t, α, y)f (y)dy = −

0

u(π) 

Ko (c, α)e−iαct dc

u(0) u(π) 

=−

Λ1 ( ωo )(yc )Λ2 (g)(yc ) −iαct dc e (A(c)2 + B(c)2 )u (yc )

u(0)

π =−

Λ1 ( ωo )(yc )Λ2 (g)(yc ) −iα cos yc t e dyc . (A(c)2 + B(c)2 )

0

Taking t = 0, we obtain π

π ω

o (y)g(y)dy =

0

Λ1 ( ωo )(yc )Λ2 (g)(yc ) dyc , (A(c)2 + B(c)2 )

0

which implies that π

π ω

o (t, α, y)g(y)dy =

0

Λ1 ( ωo (t))(yc )Λ2 (g)(yc ) dyc (A(c)2 + B(c)2 )

0

π = 0

Λ1 ( ωo )(yc )Λ2 (g)(yc ) −iα cos yc t e dyc . (A(c)2 + B(c)2 )

12

D. Wei et al. / Advances in Mathematics 362 (2020) 106963

Thus, it should hold that Λ1 ( ωo (t))(yc ) = Λ1 ( ωo )(yc )e−iα cos yc t . Taking time derivative at t = 0, we obtain  Λ1 cos y(1 + (∂y2 − α2 )−1 )ω = cos yc Λ1 (ω). This gives the key property 1 for the odd case. The even case is similar. However, due to the complicated expression of Λ1 , Λ3 , the wave operator method here is significantly more difficult to carry out than in the case of [21]. Using the facts that    Λ2 ω

o − ψ o = Λ1 ω

o = Λ1 ω

o ,    Λ4 ω

e − ψ e = Λ3 ω

e = Λ3 ω

e , we can conclude that D(ω)2L2 = ω, ω − ψ = ω, (1 + (∂y2 − α2 )−1 )ω.

(2.8)

Moreover, the property 3 can be proved only using (2.8) and the property 1 without referring to the complicated expression of D(ω). In section 10, we establish the linear enhanced dissipation. First of all, we study the 1 decay estimates for the model without the nonlocal term on a short time scale ν − 2 . Our method is based on a change of unknown, which is very different from the hypocoercivity method used by Beck and Wayne (see Lemma 10.1). This method is less like hypocoercivity and more like the change of variables used to prove the enhanced dissipation for the Couette flow, and it also gives the extra (νt3 )−1/2 decay of  sin yω(t)L2 . Next we take Fourier transform in x and act the wave operator D on both sides of (1.5) to obtain ∂t D ω − ν(∂y2 − α2 )D ω − iae−νt (cos y)D ω = −ν[∂y2 , D] ω. If we remove the commutator term, we obtain the extra (νt3 )−1/2 decay of sin yD ω (t)L2 , and the extra (νt3 )−1/2 decay of V (t)L2 follows from the property 3. These kinds of estimates is a partial motivation to apply a slightly different method to study the local problem. Now the most technical part becomes the commutator estimate (property 4) (see Lemma 9.3). Once the extra (νt3 )−1/2 decay of the velocity is obtained (see Lemma 10.3), the enhanced dissipation can be proved by some elementary methods without using the wave operator (see section 10.3 for more details). In section 11, we establish the nonlinear enhanced dissipation. The proof is based on the linear enhanced dissipation, especially the decay estimate of the velocity. Another key observation is that the basic energy dissipation ensures that if C1−1 ≤ P1 ω(0)L2 ≤ C1 , then C −1 ≤ P1 ω(t)L2 ≤ C

for 0 < t < τ /ν.

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2.3. Notations We list some notations which will be constantly used throughout this paper. • u(y) = − cos y. • For c ∈ C, let cr ∈ Ran u, so that dist(c, Ran u) = |c − cr |, let yc ∈ [0, π] be so that u(yc ) = cr . • ρ(c) = (u(π) − c)(c − u(0)) = 1 − c2 . • For a function f (y, c), we denote f  = ∂y f . • When a function f depends on yc , c, we simply write f = f (c) or f = f (yc ) decided by which form is more suitable for the estimate of f . Since the map from c ∈ [−1, 1] to yc ∈ [0, π] is one-to-one, this does not make a difference between c and yc . • φ1 (y, c) = φ(y, c)/(u(y) − c), where φ is the solution of (3.4).   )(0,c) • I(c) = − (φφ )(π,c)(φφ . u (yc )ρ(c)   π 1 1 • II(c) = 0 (u(y)−c)2 φ1 (y,c)2 − 1 dy. • • • •



−u (yc )(u(jπ)−c) 2 Jjk (c) = φ1 ((1−j)π,c) k−1 φ ((1−j)π,c) , Jj (c) = Jj (c). 1 A(c) = sin3 yc II(c), B(c) = π cos yc . A1 (c) = J1 (c) − J0 (c) + sin2 yc A(c), B1 (c) = sin2 yc B. Let Int(ϕ) to be a 2π periodic function so that

y Int(ϕ)(y) =

k

ϕ(y  ) dy 

for y ∈ [0, π],

Int(ϕ)(y) = Int(ϕ)(−y)

for y ∈ [−π, 0].

0

π c) • II1,1 (ϕ)(yc ) = p.v. 0 Int(ϕ)(y)−Int(ϕ)(y dy. 2 c ))  y (u(y)−u(y  π yc ϕ(y )φ1 (y ,c)dy • II1 (ϕ)(yc ) = p.v. 0 dy. φ(y,c)2 • C(ϕ)(c) = π sin yc ϕ(yc ), D(ϕ)(c) = u (yc )ρ(c)II1 (ϕ)(yc ). • D± (ϕ)(c) = iD(ϕ)(c) ∓ C(ϕ)(c).  jπ • Ej (ϕ)(yc ) = yc ϕ(y)φ1 (y, c)dy for j = 0, 1. • E(ϕ)(yc ) = E1 (ϕ)(yc ) − E0 (ϕ)(yc ). • For k = 0, 1, 2, we denote 

π z Lk (ϕ)(yc ) =

ϕ(y) 0 yc

∂z + ∂y + ∂c u (yc )

k 

1 (u(z) − c)2



φ1 (y, c) −1 φ1 (z, c)2

 dydz.

It is easy to see that II1 (ϕ) = II1,1 (ϕ) + L0 (ϕ). • We denote by Lp a function f (yc ) which satisfies f Lpyc ≤ C. • We denote by Lp ∩ Lq a function f (yc ) which satisfies f Lpyc + f Lqyc ≤ C.

14

D. Wei et al. / Advances in Mathematics 362 (2020) 106963

    • We denote by ρLp a function f (yc ) which satisfies  fρ 

Lp yc

≤ C.

• We denote by Lp +Lq a function f (yc ) which can be divided into two parts f = f1 +f2 with f1 , f2 satisfying f1 Lpyc ≤ C, f2 Lqyc ≤ C. 3. The homogeneous Rayleigh equation To solve (2.4), we first construct a smooth solution of the homogeneous Rayleigh equation on [−π, π]: (u − c)(φ − α2 φ) − u φ = 0,

(3.1)

where the complex constant c will be taken in four kinds of domains given by   D0  c ∈ (−1, 1) ,   D 0  c = cr + i, cr ∈ [−1, 1], 0 < || < 0 ,  π 3π  B l0  c = −1 + eiθ , 0 <  < 0 , ≤θ≤ , 2 2  π 3π  B r0  c = 1 − eiθ , 0 <  < 0 , ≤θ≤ , 2 2 for some 0 ∈ (0, 1). We denote Ω 0  D0 ∪ D 0 ∪ B l0 ∪ B r0 .

(3.2)

For c ∈ Ω 0 , let yc ∈ [0, π] be so that u(yc ) = cr with cr = Re c for c ∈ D 0 ∪ D0 ,

cr = −1 for c ∈ B l0 ,

cr = 1 for c ∈ B r0 . (3.3)

The construction and properties of the solution of (3.1) is similar to [36]. Here we just present a sketch of the proof. See arxiv version of this paper [37] for full details. 3.1. Existence of the solution We solve the homogeneous Rayleigh equation on [0, π]:



u φ − α2 φ − u−c φ = 0,  φ(yc , c) = u(yc ) − c, φ (y, c)y=y = u (yc ).

(3.4)

c

Let φ(y, c) = (u(y) − c)φ1 (y, c). It is easy to observe that y φ1 (y, c) =1 + yc

α2 (u(y  ) − c)2

y yc



φ1 (z, c)(u(z) − c)2 dzdy  = 1 + α2 T φ1 ,

(3.5)

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15

where y T (f ) = yc

1  (u(y ) − c)2

y



f (z, c)(u(z) − c)2 dzdy  .

yc

Given |α| ≥ 1, let A be a constant larger than C|α| with C ≥ 1 independent of α. For a function f (y, c) defined on [0, π] × Ω 0 , we define def

f Y =

2  

A−k ∂yβ ∂cγ f X0,γ + A−3 ∂c2 ∂y f X 0,2 ,

k=0 β+γ=k def

f X =

sup (y,c)∈[0,π]×Ω0

    f (y, c)    cosh(A(y − yc )) ,

where f X0,k f X 0,2

    u (yc )k f (y, c)   , = sup   (y,c)∈[0,π]×D0 cosh(A(y − yc ))     Au (yc )2 f (y, c)   = sup  cosh(A(y − yc ))(A + u (yc )/u1 (y, c)2 ) , (y,c)∈[0,π]×D0

yc −cos y here u1 (y, c) = cos y−y . c The operator T is bounded in Y and X .

Lemma 3.1. There exists a constant C independent of A such that T f Y ≤

C f Y , A2

T f X ≤

C f X . A2

Lemma 3.1 ensures that the operator 1 − α2 T is invertible in the spaces Y and X . Therefore, we may take φ1 (y, c) = (1 − α2 T )−1 1 =

+∞ 

α2k T k 1.

k=0

Then we can deduce the following proposition.  Proposition 3.2. There exists φ1 (y, c) ∈ C [0, π] × Ω 0 such that φ(y, c) = (u(y) −  c)φ1 (y, c) is a solution of the Rayleigh equation (3.4). Moreover, ∂y φ1 (y, c) ∈ C [0, π] × Ω 0 , and there exists 1 > 0, C > 0 such that for any 0 ∈ [0, 1 ) and (y, c) ∈ [0, π] × Ω 0 , |φ1 (y, c)| ≥

1 , 2

|φ1 (y, c) − 1| ≤ C|y − yc |2 ,

where the constants 1 , C may depend on α. Moreover, for c ∈ D0 , it holds that

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16

1. φ1 (y, c) ≥ 1; 2. ∂y φ1 (y, c) > 0 for y ∈ (yc , π] and ∂y φ1 (y, c) < 0 for y ∈ [0, yc ); 3. for any given M0 > 0, there exists a constant C only depending on M0 such that for β + γ ≤ 2, |y − yc | ≤ M0 /|α| |u (yc )γ ∂yβ ∂cγ φ1 (y, c)| ≤ C|α|β+γ ,   α2 u (yc )  2 2 3 . |u (yc ) ∂y ∂c φ1 (y, c)| ≤ C |α| + u1 (y, c)2 3.2. Uniform estimates in wave number α Here we present some uniform estimates in wave number α of the solution φ(y, c) of (3.4) when c ∈ D0 . Indeed, for α large, the behavior of the solution could be obtained by viewing the term u φ as a perturbation of (u − c)(φ − α2 φ). In [35,36], we gave a unified proof for any α by using some good structures of the Rayleigh equation. Here the proof is similar to [36], and we omit the details. We introduce ∂c φ1 (y, c) , φ1 (y, c)  F 1  ∂y G1 (y, c) =  +G = + ∂ φ1 . c u (yc ) φ1 u (yc ) F(y, c) =

∂y φ1 (y, c) , φ1 (y, c)

G(y, c) =

Proposition 3.3. There exists a constant C independent of α such that φ1 (y, c) − 1 ≤ C min{α2 |y − yc |2 , 1}φ1 (y, c), C −1 α min{α|y − yc |, 1} ≤ |F(y, c)| ≤ Cα min{α|y − yc |, 1}, C −1 eC

−1

α|y−yc |

≤ φ1 (y, c) ≤ eCα|y−yc | ,

|∂yβ ∂cγ φ1 (y, c)| ≤ Cαβ+γ φ1 (y, c)/u (yc )γ , |∂c φ1 (y, c)| ≤ Cφ1 (y, c)

β + γ ≤ 2,

α min{1, α|y − yc |} , u (yc )

and   ∂  min{1, α2 |y − yc |2 }φ1   y , + ∂c φ1 (y, c) ≤ C   u (yc ) u (yc )2  ∂  2 min{1, α2 |y − yc |2 }φ1   y + ∂c φ1 (y, c) ≤ C .   u (yc ) u (yc )4 We have the following uniform estimates for F and G at y = 0, π.

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17

Lemma 3.4. It holds that C −1 α min{αyc , 1} ≤ |F(0, c)| ≤ Cα min{αyc , 1}, C −1 α min{α(π − yc ), 1} ≤ |F(π, c)| ≤ Cα min{α(π − yc ), 1}, |∂c F(0, c)| ≤

C|F(0, c)|2 , α2 yc2 sin yc

|∂c F(π, c)| ≤

C|F(π, c)|2 , α2 (π − yc )2 sin yc

|∂c2 F(0, c)| ≤

|G(0, c)| ≤

Cα min{1, αyc } , sin yc

|G(π, c)| ≤

Cα min{1, α(π − yc )} , sin yc

Cα min{αyc , 1} , yc2 sin2 yc

|∂c2 F(π, c)| ≤

|∂c G(0, c)| ≤

Cα min{α(π − yc ), 1} , (π − yc )2 sin2 yc

Cα2 , sin2 yc

|∂c G(π, c)| ≤

Cα2 . sin2 yc

To obtain better estimates near critical points, we introduce the following functions: for 0 ≤ yc < α1 and 0 ≤ y ≤ 1, let ϕ1 (y, c) = φ1 (yyc , c), and for π −

1 α

(3.6)

< yc ≤ π and 0 ≤ y ≤ 1, let  ϕ2 (y, c) = φ1 (1 − y)π + yyc , c .

(3.7)

The following estimates will be used in the estimate of Jjk in Lemma 4.8. Proposition 3.5. It holds that for any y ∈ [0, 1] and 0 ≤ yc <

1 α,

1 ≤ ϕ1 ≤ 2, |∂c ϕ1 | ≤ Cα2 , |∂c2 ϕ1 | ≤ Cα4 ,     k ∂y ϕ1 (y, c)   ∂c  ≤ Cα2+2k , k = 0, 1, 2,   y2 c

and for any y ∈ [0, 1] and π −

1 α

< yc ≤ π, it holds that

1 ≤ ϕ2 ≤ 2, |∂c ϕ2 | ≤ Cα2 , |∂c2 ϕ2 | ≤ Cα4 ,     k ∂y ϕ2 (y, c)   ∂c  ≤ Cα2+2k , k = 0, 1, 2.  (π − yc )2  Proof. For 0 ≤ yc <

1 α

and 0 ≤ y ≤ 1, by (3.5), we have ϕ1 = 1 + α2 T1 ϕ1 , where y

(T1 f )(y, c) = 1

yc2 (u(y  yc ) − c)2

y 1



f (z, c)(u(zyc ) − c)2 dzdy 

D. Wei et al. / Advances in Mathematics 362 (2020) 106963

18

y y =



f (z, c) 1

F (z, c)2 2 y dzdy  F (y  , c)2 c

1

with u(zyc ) − c cos yc − cos zyc = 2 yc yc2     1 − z2 (1 − z)yc (1 + z)yc = m1 m1 , 2 2 2

F (z, c) =

and m1 (y) = sin y/y ∈ C 4 ([−1, 1]). One can check that 1/C ≤ m1 (−y) = m1 (y) ≤ C, and |∂ck m1 (ayc )| ≤ C for yc < α1 , k = 1, 2. Then we can deduce that for yc < α1 , |z 2 − 1| ≤ |F (z, c)| ≤ C|z 2 − 1|, |∂c F (z, c)| ≤ C|z 4 − 1|, |∂c2 F (z, c)| ≤ C|z 4 − 1|, C and for yc < 0≤

1 α

and 0 < y  < z < 1,

   2   F (z, yc )2 ∂c F (z, yc )  ≤ C, ≤ C,   2  2 F (y , yc ) F (y , yc ) 

    2 F (z, yc )2  ∂c  ≤ C.  F (y  , yc )2 

Let Ωα = {(y, c) : y ∈ [0, 1], c ∈ [u(0), u(1/α)]}. Then we can infer that 1 f L∞ (Ωα ) , 2α2 ∂c T1 f − T1 ∂c f L∞ (Ωα ) ≤ Cf L∞ (Ωα ) ,

T1 f L∞ (Ωα ) ≤ ∂c2 T1 f

−

T1 ∂c2 f L∞ (Ωα )

≤ C(f L∞ (Ωα ) + ∂c f L∞ (Ωα ) ).

(3.8) (3.9) (3.10)

We infer from (3.8) that 1 ≤ ϕ1 ≤ 2 for (y, c) ∈ Ωα . By Proposition 3.3, we have for (y, c) ∈ Ωα ,   |∂c ϕ1 | = (∂c φ1 )(yyc , c) +

 Cαϕ y Cα  1 (∂ φ )(yy , c) ≤ . ≤  y 1 c u (yc ) u (yc ) yc

Thus, we obtain T1 (∂c ϕ1 )L∞ (Ωα ) ≤ C, which along with (3.9) shows that for (y, c) ∈ Ωα , |∂c ϕ1 | ≤ Cα2 . By Proposition 3.3, we have for (y, c) ∈ Ωα , |∂c2 ϕ1 | ≤

Cα2 ϕ1 Cα2 ≤ 2 .  2 u (yc ) yc

(3.11)

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Thus, we have T1 (∂c2 ϕ1 )L∞ (Ωα ) ≤ Cα2 , which along with (3.10) shows that for (y, c) ∈ Ωα , |∂c2 ϕ1 | ≤ Cα4 . Using the fact that Ωα ,

∂y ϕ1 (y,c) yc2

= α2

y 1

(3.12) 2

F (z,yc ) ϕ1 (z, c) F (y,yc )2 dz, we can deduce that for (y, c) ∈

    k ∂y ϕ1 (y, c)  ∂c  ≤ Cα2+2k ,   y2

k = 0, 1, 2.

c

Due to φ1 (0, c) = ϕ1 (0, c), we have by Proposition 3.3, − For π −

1 α

∂y ϕ1 (0, c) ≥ C −1 α2 . yc2

(3.13)

< yc < π and 0 ≤ y ≤ 1, we have ϕ2 = 1 + α2 T2 ϕ2 , where y

(T2 f )(y, c) = 1

(π − yc )2 (u((1 − y  )π + y  yc ) − c)2

y



f (z, c)(u((1 − z)π + zyc ) − c)2 dzdy  ,

1

and we also have u((1 − y  )π + y  yc ) − c = cos(π − yc ) − cos(y  (π − yc )) (π − yc )2 1 − y  2  (1 − y  )π − yc   (1 + y  )π − yc  = m1 m1 . 2 2 2 Using similar arguments as above, we can deduce that for y ∈ [0, 1] and c ∈ [u(π − 1/α), u(π)], 1 ≤ ϕ2 ≤ 2, |∂c ϕ2 | ≤ Cα2 , |∂c2 ϕ2 | ≤ Cα4 ,     k ∂y ϕ2 (y, c)  ∂c  ≤ Cα2+2k , k = 0, 1, 2,  (π − yc )2  and ∂y ϕ2 (0, c) ≥ C −1 α2 . (π − yc )2



(3.14)

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4. The inhomogeneous Rayleigh equation In this section, we solve the inhomogeneous Rayleigh equation: ⎧  ⎪ ⎨Φ − α2 Φ − u Φ = f, u−c ⎪ ⎩Φ (−π) = Φ (π), Φ(−π) = Φ(π).

(4.1)

We split the equation (4.1) into two parts: ⎧  ⎪ ⎨Φ − α2 Φo − u Φo = fo , o u−c ⎪ ⎩Φ (0) = Φ (π) = 0, o

(4.2)

o

and ⎧  ⎪ ⎨Φ − α2 Φe − u Φe = fe , e u−c ⎪ ⎩Φ (0) = Φ (π) = 0, e

(4.3)

e

where fo (y, c) =

f (y, c) − f (−y, c) , 2

fe (y, c) =

f (y, c) + f (−y, c) . 2

Thus, Φ = Φo + Φe is a solution of (4.1) with Φo being an odd function and Φe being an even function. 4.1. The Wronskian Before we present the solution formula, let us first calculate the Wronskian of (4.2) and (4.3). Lemma 4.1. For any c ∈ Ω 0 \ D0 , the Wronskian of (4.2) has the form π Wo (c) = φ(0, c)φ(π, c)

1 dy, φ(y, c)2

0

and the Wronskian of (4.3) has the form ∂y φ(π, c) ∂y φ(0, c) We (c) = − − φ (π, c)φ (0, c) φ(0, c) φ(π, c)

π 0

1 dy  . φ(y  , c)2

D. Wei et al. / Advances in Mathematics 362 (2020) 106963

Proof. For c ∈ / D0 , the integration

y

1 dy  0 φ(y  ,c)2

21

is well defined. It is easy to check that

y ϕo (y, c) = φ(0, c)φ(y, c)

1 dy  , φ(y  , c)2

0

is a solution to (3.1) with boundary conditions ϕ(0, c) = 0 and ϕ (0, c) = 1. So, Wo (c) = ϕo (π, c). One can also check that φ(y, c) − φ(y, c)∂y φ(0, c) ϕe (y, c) = φ(0, c)

y

1 dy  φ(y  , c)2

0

is a solution of (3.1) with boundary conditions ϕ(0, c) = 1 and ∂y ϕ(0, c) = 0. So, We (c) = ∂y ϕe (π, c).  By Proposition 3.2, φ(y, c) and ∂y φ(y, c) are continuous. Thus, both Wo (c) and We (c) are continuous for c ∈ Ω 0 \ D0 . The following lemma will show that sin yc Wo (c) is continuous up to the boundary. Lemma 4.2. For c = c + i = u(yc ) + i ∈ D 0 , c ∈ (−1, 1), we have  lim ρ(c )1/2 Wo (c ) = −φ1 (0, c)φ1 (π, c) A(c) − iB(c) ,

→0+

 lim ρ(c )1/2 Wo (c ) = −φ1 (0, c)φ1 (π, c) A(c) + iB(c) ,

→0−

lim φ(0, c )φ(π, c )We (c ) = −

(φ1 φ1 )(π, c)(φ1 φ1 )(0, c) (A1 (c) − iB1 (c)), u (yc )

lim φ(0, c )φ(π, c )We (c ) = −

(φ1 φ1 )(π, c)(φ1 φ1 )(0, c) (A1 (c) + iB1 (c)). u (yc )

→0+

→0−

Here ρ(c ) = 1 − c2 and ρ(c )1/2 is taken so that Re ρ(c )1/2 > 0. Proof. Let us prove the case of  > 0. Let c = u(yc ) + i ∈ D 0 . Then π ρ(c )3/2 0

⎛ π ⎞  1 1 dy = ρ(c )3/2 ∂c ⎝ dy ⎠ (u(y) − c )2 u(y) − c 0

⎛ π  = −ρ(c )3/2 ∂c ⎝ 0

⎛∞  3/2 ⎝ = −ρ(c ) ∂c 0

⎞ 1 dy ⎠ cos y + c ⎞ 2 dt⎠ 1 − t2 + c (1 + t2 )

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22

 = −ρ(c )

3/2

here d satisfies d2 = π 3/2

ρ(c )

0

1−c 1+c

∂c

 1 + d t ∞  1

 ln , (1 + c )d 1 − d t t=0

with Re d > 0. Then Im d < 0, ρ(c )1/2 = (1 + c )d and 

1 dy = −ρ(c )3/2 ∂c (u(y) − c )2

 −1 πi (1 + c )d

= −∂c ρ(c )πi/2 → − cos yc πi, as  → 0 + . By Proposition 3.2, we get        (ρ(c) + )|y − yc |2  ρ(c ) 1      (u(y) − c )2 φ1 (y, c )2 − 1  ≤ C  (u(y) − c)2 + 2  ≤ C, with C a constant independent of , which ensures that as  → 0+, π ρ(c ) 0

  1 1 − 1 dy → sin2 yc 2 2 (u(y) − c ) φ1 (y, c )

π 0

  1 1 − 1 dy. 2 2 (u(y) − c) φ1 (y, c)

Thus, we deduce that as  → 0+, π 3/2

ρ(c )

1 dy → sin3 yc II(c) − cos yc πi. φ(y, c )2

0

This together with the continuity of φ1 (y, c) gives  lim ρ(c )1/2 Wo (c ) = −φ1 (0, c)φ1 (π, c) A(c) − iB(c) .

→0+

Using the fact that u (0) = u (π) = 0, we get φ (jπ, c) = (u(jπ) − c)φ1 (jπ, c) for j = 0, 1. Then we infer that φ(0, c )φ(π, c )We (c ) = (φφ )(π, c ) − (φφ )(0, c ) 



π

− (φφ )(π, c )(φφ )(0, c )

1 dy  φ(y  , c )2

0

→ (φφ )(π, c) − (φφ )(0, c) −

(φφ )(π, c)(φφ )(0, c) (A(c) − iB(c)) sin3 yc

= (1 + cos yc )2 (φ1 φ1 )(π, c) − (1 − cos yc )2 (φ1 φ1 )(0, c) −

(φ1 φ1 )(π, c)(φ1 φ1 )(0, c) sin2 yc (A(c) − iB(c)) u (yc )

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=−

23

(φ1 φ1 )(π, c)(φ1 φ1 )(0, c) (A1 (c) − iB1 (c)). u (yc )

The case of  < 0 is similar, here we omit the details.



4.2. Uniform estimates of A(c), B(c) Proposition 4.3. There exists a constant C independent of α, such that C −1 (1 + α sin yc )2 ≤ A(c)2 + B(c)2 ≤ C(1 + α sin yc )2 ,   1 C ∂c ≤ , 2 2 A(c) + B(c) (1 + α sin yc )2 sin2 yc   1 C ∂c2 ≤ . A(c)2 + B(c)2 (1 + α sin yc )2 sin4 yc We need the following lemma for the estimates of II(c), whose proof is similar to Lemma 10.1 in [36]. Here we omit the details. Lemma 4.4. There exists a constant C independent of α, such that   2  |α| α |α| α2 , ≤ −II(c) ≤ C min , , sin yc sin2 yc sin yc sin2 yc  2    |α| α ρ(c)k ∂ck II(c) ≤ C min , , k = 1, 2. sin yc sin2 yc C −1 min



Let us begin with the proof of Proposition 4.3. Proof. It follows from Lemma 4.4 that A(c)2 + B(c)2 ≥ C −1 (2 min{α2 sin2 yc , α4 sin4 yc } + cos2 yc ) ≥ C −1 (1 + α sin yc )2 , and the upper bound of A(c)2 + B(c)2 . By Lemma 4.4 again, we have |A(c)| = | sin3 yc II(c)| ≤ Cα sin yc , |∂c A(c)| = |∂c (sin3 yc II(c))| ≤ C min{α2 , α/ sin yc }, |∂c2 A(c)| = |∂c2 (u (yc )ρII(c))| ≤ C min{α2 / sin2 yc , α/ sin3 yc }, which together with the fact ∂c B(c) = −π and ∂c2 B(c) = 0 show that      1 C ∂c ≤ ,  A(c)2 + B(c)2  (1 + α sin yc )2 sin2 yc     2  1 C C|∂c A(c)|2 C|A(c)∂c2 A(c)|  ∂c ≤ + + 4   2 2 2 2 2 2 2 A(c) + B(c) (A(c) + B(c) ) (A(c)2 + B(c)2 )2 (A(c) + B(c) )sin yc

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≤

C C(|α/ sin yc |2 ) Cα2 / sin2 yc + + 4 (1 + α sin yc )4 (1 + α sin yc )4 (1 + α sin yc )2 sin yc

≤

C . (1 + α sin yc )2 sin4 yc



4.3. Uniform estimates of A1 (c), B1 (c) Proposition 4.5. There exists a constant C independent of α, such that C(1 + α sin yc )6 (1 + α sin yc )6 2 2 ≤ A (c) + B (c) ≤ , 1 1 Cα4 α4      1 Cα6 ∂c ≤ ,  A1 (c)2 + B1 (c)2  (1 + α sin yc )8     2 1 Cα8 ∂c ≤ .   A1 (c)2 + B1 (c)2 (1 + α sin yc )10 To prove Proposition 4.5, we need a criterion on embedding eigenvalues. Definition 4.6. We say that c ∈ Ran u is an embedding eigenvalue of Rα if there exists 0 = ψ ∈ H 1 (T ) such that for all ϕ ∈ H 1 (T ), 





 2

(ψ ϕ + α ψϕ)dy + p.v. T

T

u ψϕ dy + iπ u−c

 y∈u−1 {c},u (y)=0

(u ψϕ)(y) = 0. (4.4) |u (y)|

For u(y) = − cos y, one can check that if (4.4) holds, then u (y) = 0 for some point where c = u(y) (see Lemma 5.2 in [36]), which means that c = 0. Thus, ψ ∈ H 2 (T ) is a classical solution to −ψ  + α2 ψ +

u ψ = 0 ⇔ −ψ  + (α2 − 1)ψ = 0, u−c

(4.5)

which implies that |α| = 0 or 1. On the torus Tδ2 with 0 < δ < 1, we have αδ ∈ Z. If α = 0, then |α| > 1, which implies that Rα has no embedding eigenvalues. Proposition 4.7. If A1 (c)2 + B1 (c)2 = 0, then c ∈ D0 is an embedding eigenvalue of Rα . Thus, if |α| > 1, then A1 (c)2 + B1 (c)2 > 0. Proof. If A1 (c)2 + B1 (c)2 = 0, then B1 (c) = 0. As B1 (c) = sin2 yc B(c) = (1 − c2 )(−πc), we have c = 0 or ±1. Due to sin yc2 A(c) = 0 at c = ±1, using |Jj (c)| = −u (yc )(u(jπ) − c)2 and Lemma 3.4, we find that φ1 ((1 − j)π, c)2 F((1 − j)π, c) J1 (c) = 0, J0 (c) = 0 for c = −1, J1 (c) = 0, J0 (c) = 0 for c = 1,

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here we define a function at c = ±1 as its limit as c → ±1. Therefore, A1 (c) = 0 for c = ±1, and c = 0. Now we may assume that c = 0, A1 (c) = 0, then yc = π/2. Let φe (y, c) = φ(π, c)φ1 (π, c)ϕ(y,  c) − φ(y, c), where ϕ(y,  c) is given by ϕ(y,  c) =

φ1 (y, c) (u(y) − u(π)) u (yc ) φ1 (y, c) +  (u(y) − c)(u(π) − c) u (yc )

y π

y + φ1 (y, c)(u(y) − c)(u(π) − c) π

u (yc ) − u (z) dz (u(z) − c)2   1 1 − 1 dz. 2 2 (u(z) − c) φ1 (z, c)

Then φe (y, c) satisfies the homogeneous Rayleigh equation for y ∈ (0, π). Moreover, ∂y φe (0, c) =

φ(π, c)φ1 (π, c)φ1 (0, c) (u(0) − u(π)) u (yc ) 0  u (yc ) − u (z) φ(π, c)φ1 (π, c)φ1 (0, c) + (u(0) − c)(u(π) − c) dz u (yc ) (u(z) − c)2 π

+

φ(π, c)φ1 (π, c)φ1 (0, c)(u(0)

0 − c)(u(π) − c) π

−

φ(π, c)φ1 (π, c) φ1 (0, c)

  1 1 − 1 dz (u(z) − c)2 φ1 (z, c)2

− φ (0, c).

Using the facts that u(0) − u(π) = −2, (u(0) − c)(u(π) − c) = −1, and 0 π

u (yc ) − u (z) dz = (u(z) − c)2

0 π

1 − sin z dz = (cos z)2

0

z dz π 0 = tan −  = −2, 1 + sin z 2 4 π

π

we deduce that ∂y φe (0, c) =φ(π, c)φ1 (π, c)φ1 (0, c)II(c) −

φ(π, c)φ1 (π, c) − φ (0, c). φ1 (0, c)

Using the facts that u(0) − c = −1, u(π) − c = 1, u (0) = u (π) = 0, u (yc ) = sin yc = 1, we infer that φ1 (π, c) = φ(π, c), −φ (0, c) = φ1 (0, c) and

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∂y φe (0, c) =φ(π, c)φ1 (π, c)φ1 (0, c)(II(c) + J1 (c) − J0 (c)) =φ(π, c)φ1 (π, c)φ1 (0, c)A1 (c) = 0. It is easy to verify that ∂y φe (π, c) = 0. Now we extend φe to be an even function with periodic 2π. Then φe ∈ H 2 (T ) satisfies (4.4). This means that c = 0 is an embedding eigenvalue of Rα .  Lemma 4.8. There exists a constant C independent of α, such that for k = 1, 2, m = 0, 1, 2, k |∂cm J1−j (c)| ≤ C min



 |(1 − j)π − yc |2k+1 α2m−2 , , α2 | sin yc |2m φ1 (jπ, c)k−1 |(1 − j)π − yc |s

where s = 1 for k = 1, m = 2, and s = 0 otherwise. Proof. Recall that F(y, c) = gives

∂y φ1 φ1 (y, c)

and G(y, c) =

∂c φ1 φ1 (y, c).

A direct calculation

1 1 (0, c) = , k F(0, c) φ (0, c) φk−1 ∂ φ 1 y 1 1   kG 1 ∂c F ∂c k−1 (0, c) = − k 2 (0, c) − k (0, c), φ1 F φ1 F φ1 ∂y φ1   2kG∂c F ∂2F 1 −k∂c G + k2 G 2 ∂c2 k−1 (0, c) = (0, c) + k 2 (0, c) − kc 2 (0, c) k φ1 F φ1 F φ1 F φ1 ∂y φ1 +

2|∂c F|2 (0, c), φk1 F 3

from which and Lemma 3.4, we infer that     1 + αyc 1    φ1 (0, c)k−1 φ (0, c)  ∼ α2 φ1 (0, c)2 yc , 1       1 C(1 + αyc )2 ∂c ≤ (0, c) ,   k−1  2 α φ1 (0, c)k yc2 sin yc φ1 φ1     2  1  C(1 + αyc )3 ∂c ≤ (0, c) .   k−1  2 α φ (0, c)k y 3 sin2 y φ φ 1

1

1

c

Recall that J1k (c) = Thus, we have

−u (yc )(u(π) − c)k . φ1 (0, c)k−1 φ1 (0, c)

c

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    J1k Cu (yc ) C|π − yc | C sin yc (1 + αyc )  ≤  (π − yc )2k  φ1 (0, c)k |F(0, c)| ≤ φ1 (0, c)k α2 yc ≤ α2 φ1 (0, c)k−1 , here we used sin yc ∼ yc (π − yc ) and φ1 (0, c) ≥ C −1 eC

−1

αyc

. For m = 1, 2, we have

    m  ∂mJ k     1   c 1  2n−2m+1  n  ≤ Cu (y ) ∂   c c k−1   k  u(π) − c  φ1 φ1 (0, c)  n=0

≤

C(π − yc ) C(1 + αyc )m+1 ≤ 2 . α2 φ1 (0, c)k yc u (yc )2m−1 α (sin yc )2m φ1 (0, c)k−1

To complete the proof, it suffices to improve the estimate for yc < thanks to yc ∂y φ1 (0, c) = ∂y ϕ1 (0, c) with ϕ1 given by (3.6), we get J1k =

1 α.

In this case,

−m1 (yc )(u(π) − c)k yc2 −u (yc )(u(π) − c)2 = , φ1 (0, c)k−1 φ1 (0, c) ϕ1 (0, c)k−1 ϕ1  (0, c)

where m1 (y) = siny y . Since m1 ∈ C ∞ ([−1, 1]) is even, we have |∂cm m1 (yc )| ≤ C for yc < 1, m = 0, 1, 2. Thus, for 0 < yc < α1 , by Proposition 3.5 and (3.13),  k J1  ≤ Cα−2 ,

  ∂c J1k  ≤ C,

 2 k ∂c J1  ≤ Cα2 .

This completes the proof of the case of j = 0. The case of j = 1 can be proved similarly.  Now we are in a position to prove Proposition 4.5. Proof. By Lemma 4.8 and Lemma 4.4, we get |A1 (c)| + |B1 (c)| ≤ |J1 | + |J0 | + Cα sin3 yc + π| sin2 yc cos yc | ≤ C(

1 (1 + α sin yc )3 + α sin3 yc + sin2 yc ) ≤ C . α α2

Due to φ1 (y, c) ≤ 0 for y ≤ yc , we get J1 (c) ≥ 0. Similarly, we have J0 (c) ≤ 0. Proposition 4.7 ensures that |A1 (c)| + |B1 (c)| > Cα > 0. Thus, we only need to consider the case of the large α. Let M be a large number determined later. Then for yc ≤ M1α , we choose M large enough so that 1 ≤ φ1 (0, c) ≤ 2. Then |J1 (c)| ≥

u (yc ) yc 1 ≥ ≥ , C|F(0, c)| Cα min{αyc , 1} Cα2

here C is a constant independent of M and α. Thus, for yc ≤ |A1 (c)| ≥ |J1 (c)| + |J0 (c)| − | sin2 yc A(c)| ≥

1 Mα ,

1 (1 + α sin yc )3 3 − Cα sin y ≥ . c Cα2 Cα2

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For

1 Mα

≤ yc ≤

M α

< 1, we have |B(c)| ≥ |B(0)/2| ≥ 1, then |B1 (c)| = |ρ(c)B(c)| ≥ ρ(c) ≥

This shows that for yc ≤

M α

< 1, we have

|A1 (c)| + |B1 (c)| ≥ Similarly, for π − yc ≤

M α

(1 + α sin yc )3 . Cα2

< 1, we have |A1 (c)| + |B1 (c)| ≥

While, for yc ∈ [ M α ,π −

(1 + α sin yc )3 . Cα2

M α ],

(1 + α sin yc )3 . Cα2

α > M , and M large enough, we have

|A1 (c)| ≥ | sin2 yc A(c)| − |J1 (c)| − |J0 (c)| ≥

|α sin3 yc | 1 + α3 sin3 yc C − 2 ≥ . C α Cα2

Therefore, we obtain C(1 + α sin yc )6 (1 + α sin yc )6 ≤ A1 (c)2 + B1 (c)2 ≤ . 4 Cα α4 By Lemma 4.4, we have |A1 (c)| ≤ C(1 + α sin yc )3 /α2 , |∂c A1 (c)| ≤ C(1 + α sin yc ), α |∂c2 A1 (c)| ≤ C min{α2 , }, sin yc |B1 (c)| ≤ C sin2 yc , |∂c B1 (c)| ≤ C, |∂c2 B1 (c)| ≤ C. Then we can deduce that     1 Cα6 ∂c ≤ ,  A1 (c)2 + B1 (c)2  (1 + α sin yc )8     2 1 ∂c   A1 (c)2 + B1 (c)2  ≤

C C|∂c A1 (c)|2 C|A1 (c)∂c2 A1 (c)| + + (A1 (c)2 + B1 (c)2 )2 (A1 (c)2 + B1 (c)2 )2 (A1 (c)2 + B1 (c)2 )2

≤

Cα8 Cα8 Cα8 Cα8 + + ≤ . 12 10 10 (1 + α sin yc ) (1 + α sin yc ) (1 + α sin yc ) (1 + α sin yc )10



Remark 4.9. Since the functions A, B, A1 , B1 are continuous with respect to α, Proposition 4.3 and 4.5 are true for every α ≥ c0 > 1 with the constant C depending only on c0 .

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4.4. Solution formula of the inhomogeneous Rayleigh equation The fact that the Kolmogorov flow is stable for the case of α > 1, implies that Wo (c) = 0 and We (c) = 0 for all c with Im c = 0. It is easy to check that the inhomogeneous Rayleigh equations (4.2) and (4.3) are equivalent to ⎧ ⎨

 Φo   = fo φ, φ ⎩ Φo (0) = Φo (π) = 0,

(4.6)

⎧  ⎨ 2  Φe   φ = fe φ, φ ⎩  Φe (0) = Φe (π) = 0.

(4.7)

φ2

and

We deduce by integrating (4.6) and (4.7) twice and matching the boundary conditions that for y ∈ [0, π] and c ∈ / [−1, 1], y Φo (y, c) =φ(y, c)

1 φ(y  , c)2

y



yc

y

 y

=φ(y, c) π





y

1 dy  φ(y  , c)2

o

fo φ(y , c)dy dy + μ (c)φ(y, c)

0

1 φ(y  , c)2



0







y

1 dy  , (4.8) φ(y  , c)2

o

fo φ(y , c)dy dy + μ (c)φ(y, c) yc

π

where

o

μ (c) =

−φ(0, c)φ(π, c)

π

1 0 φ(y  ,c)2

 y yc

fo φ(y  , c)dy  dy 

Wo (c)

,

and for y ∈ [0, π] and c ∈ / [−1, 1], y Φe (y, c) = φ(y, c)

1 φ(y  , c)2

y



fe φ(y  , c)dy  dy 

yc

0

y + μe (c)φ(y, c)

1 dy  + ν0e (c)φ(y, c) φ(y  , c)2

0

y = φ(y, c) π

1 φ(y  , c)2

y yc



fe φ(y  , c)dy  dy 

(4.9)

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30

y e

+ μ (c)φ(y, c)

1 dy  + ν1e (c)φ(y, c), φ(y  , c)2

(4.10)

π

where μe and ν0e , ν1e are determined by π

1 φ(y  , c)2

y







π



e

fe φ(y , c)dy dy + μ (c) yc

0

1 dy  + ν0e (c) − ν1e (c) = 0, φ(y  , c)2

0

νje (c)φ(jπ, c)φ (jπ, c) +

jπ

fe φ(y  , c)dy  + μe (c) = 0, j = 0, 1.

yc

Then we have π ν1e (c) =

ν0e (c) =

e

μ (c) =

0

  π fe φ(y , c)dy + (φφ ) 0, c 0 





 y yc

fe φ(y  ,c) dy  dy  φ(y  ,c)2

−

π

π yc

0

fe φ(y  ,c)dy  dy  φ(y  ,c)2

,

−φ(0, c)φ(π, c)We (c)   π π    f φ(y , c)dy + (φφ ) π, c 0 e 0

 y yc

fe φ(y  ,c) dy  dy  φ(y  ,c)2

−φ(0, c)φ(π, c)We (c) π (φφ )(π, c)(φφ )(0, c) 0



−

π 0

(4.11)  f φ(y  ,c)dy  yc e  dy φ(y  ,c)2 ,

0

(4.12) 1  (1−j)π f φ(y ,c)dy yc e dy  −(φφ )(jπ, c) yc fe φ(y  , c)dy  j=0 φ(y  ,c)2 . φ(0, c)φ(π, c)We (c) (4.13)

 y





5. Linear inviscid damping 5.1. The limiting absorption principle In this subsection, we establish the limiting absorption principle for the inhomogeneous Rayleigh equation (U − c)(Φ − α2 Φ) − U  Φ = ω,

(5.1)

  when c ∈ z ∈ C : 0 < inf |U (y) − z| < 0 for 0 > 0 small enough. y∈T

In [36], we established the limiting absorption principle for a class of non-monotone shear flows in K (see Page 4) by using a blow-up analysis and compactness argument. This result can be easily extended to the following periodic shear flows (U (y), 0): (a) Periodic: U (y + 2π) = U (y); (b) Regularity: U ∈ H 3 (T ), where T = T2π ;

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(c) Spectrum: Rα has no embedding eigenvalues for α = 0; (d) Curvature: U  (y) = 0 at critical points (i.e., U  (y) = 0). We denote by P the set of periodic flows satisfying (a)-(d). When α > 1, U (y) = − cos y ∈ P.  Proposition 5.1. Assume that U ∈ P. There exists 0 such that for c ∈ z ∈ C : 0 <  inf |U (y) − z| < 0 , the solution to (5.1) has the following uniform bound y∈T

ΦH 1 (T ) ≤ CωH 1 (T ) . Here C is a constant independent of 0 . Moreover, there exists Φ± (α, y, c) ∈ H01 (T ) for c ∈ Ran u, such that Φ(α, ·, c ± i) → Φ± (α, ·, c) in C(T ) as  → 0+ and Φ± (α, ·, c)H 1 (T ) ≤ CωH 1 (T ) . One can refer to section 6 in [36] for more details. 5.2. Explicit formulas of the limits In this subsection, we give the explicit formulas of the limits Φ± , which are important to obtain the explicit decay estimates of the velocity. The proof is similar to Proposition 8.4 in [36], and we omit the details. We take fo (y, c) =

ω

o (α, y) , iα(u(y) − c)

fe (y, c) =

ω

e (α, y) , iα(u(y) − c)

where ω

o (y) =

ω

0 (α, y) − ω

0 (α, −y) , 2

ω

e (y) =

ω

0 (α, y) + ω

0 (α, −y) . 2

e If ω

0 (α, y) ∈ H 1 (T ), then there exist μo± , μe± , νk± so that

lim μo (c ) = μo+ (c),

→0+

lim μe (c ) = μe+ (c),

→0+

e lim νke (c ) = νk+ (c),

→0+

Here c = c + i = u(yc ) + i ∈ D 0 .

lim μo (c ) = μo− (c),

→0−

lim μe (c ) = μe− (c),

→0−

e lim νke (c ) = νk− (c).

→0−

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For c ∈ (−1, 1), we define

def

Φo± (y, c) =

⎧ y z ⎪ ⎪ 1 ⎪ ⎪ φ(y, c) φfo (y  , c)dy  dz ⎪ 2 ⎪ φ(z, c) ⎪ ⎪ ⎪ yc 0 ⎪ ⎪ ⎪ ⎪ ⎪ y ⎪ ⎪ 1 ⎪ o ⎪ + μ± ( ωo )(c)φ(y, c) dy  ⎪ ⎪  , c)2 ⎪ φ(y ⎨

0 ≤ y < yc ,

0

⎪ y z ⎪ ⎪ 1 ⎪ ⎪ φfo (y  , c)dy  dz φ(y, c) ⎪ ⎪ ⎪ φ(z, c)2 ⎪ ⎪ π yc ⎪ ⎪ ⎪ ⎪ ⎪ y ⎪ ⎪ 1 ⎪ o ⎪ ⎪ + μ± ( ωo )(c)φ(y, c) dy  ⎪ ⎩ φ(y  , c)2

yc < y ≤ π.

π

For c ∈ (−1, 1), we define ⎧ y  y φf (y  , c)dy  ⎪ ⎪ e ⎪ yc ⎪ φ(y, c) dy  ⎪ ⎪ ⎪ φ(y  , c)2 ⎪ ⎪ 0 ⎪ ⎪ ⎪ y ⎪ ⎪ ⎪ 1 ⎪ e e ⎪ + μ± ( ωe )(c)φ(y, c) dy  + ν0± ( ωe )(c)φ(y, c), 0 ≤ y < yc , ⎪ ⎪ φ(y  , c)2 ⎨ def 0 Φe± (y, c) = y  y φf (y  , c)dy  ⎪ ⎪ e ⎪ yc ⎪ ⎪φ(y, c) dy  ⎪  , c)2 ⎪ φ(y ⎪ ⎪ π ⎪ ⎪ ⎪ ⎪ y ⎪ ⎪ 1 ⎪ e e ⎪ + μ± ( ωe )(c)φ(y, c) dy  + ν1± ( ωe )(c)φ(y, c), yc < y ≤ π, ⎪ ⎪ φ(y  , c)2 ⎩ π

ω0 (α,y) Proposition 5.2. Let Φ(y, c) be a solution of (4.1) with f = iα(u−c) for ω

0 ∈ H 1 (T ). Then it holds that for any y ∈ [−π, π], yc ∈ (−π, π) \ {0} and y = ±yc ,

lim Φ(y, c ) = Φ+ (y, u(yc )),

→0+

lim Φ(y, c ) = Φ− (y, u(yc )),

→0−

where Φ± = Φo± + Φe± , and c = c + i ∈ D 0 and c = u(yc ). Furthermore, it holds that

and

μo+ ( ωo )(c) =

1 −C( ωo )(c) + iD( ωo )(c) , α A(c) − iB(c)

(5.2)

μo− ( ωo )(c) =

1 C( ωo )(c) + iD( ωo )(c) , α A(c) + iB(c)

(5.3)

D. Wei et al. / Advances in Mathematics 362 (2020) 106963

μe+ (ϕ)(c) = μe− (ϕ)(c) = e νk+ (ϕ)(c) =

e νk− (ϕ)(c) =

33

1    1 I(c) iD(ϕ)(c) − C(ϕ)(c) − i(φφ )(jπ, c)E1−j (ϕ)(c) j=0 , (5.4) α I(c)(A1 (c) − iB1 (c))/ρ(c) 1   1 I(c) iD(ϕ)(c) + C(ϕ)(c) − i(φφ )(jπ, c)E1−j (ϕ)(c)j=0 , (5.5) α I(c)(A1 (c) + iB1 (c))/ρ(c)  1 iE(ϕ) + (φφ ) (1 − k)π, c (D+ (ϕ) − iEk (ϕ)(A − iB)) / sin3 yc , (5.6) α I(c)(A1 (c) − iB1 (c))/ρ(c)  1 iE(ϕ) + (φφ ) (1 − k)π, c (D− (ϕ) − iEk (ϕ)(A + iB)) / sin3 yc . (5.7) α I(c)(A1 (c) + iB1 (c))/ρ(c)

5.3. Solution formula of the linearized Euler equations

α, y) be a solution of (2.1) with initial data ψ(0,

α, y) = −(∂ 2 −α2 )−1 ω Let ψ(t,

0 (α, y). y Then we have !

α, y) = 1

α, y)dc, e−iαtc (c − Rα )−1 ψ(0, ψ(t, 2πi ∂Ω

where Ω is a simply connected domain including the spectrum σ(Rα ) = [−1, 1] of Rα . ω

0 (α,y) Let Φ(α, y, c) be a solution of (4.1) with f (α, y, c) = iα(u(y)−c) . Then we have

α, y) = iαΦ(α, y, c). (c − Rα )−1 ψ(0, Therefore, we have

α, y) = 1 ψ(t, 2π

!

e−iαtc αΦ(α, y, c)dc.

∂Ω0

Using Proposition 5.2 and Proposition 5.1, it is easy to show that Lemma 5.3. It holds that

α, y) = 1 ψ(t, 2π

1 e −1

−iαtc

 o (α, y, c)dc + 1 αΦ 2π

1

 e (α, y, c)dc e−iαtc αΦ

−1

ψ o (t, α, y) + ψ e (t, α, y), where

 o (y, c) = Φ

⎧ y ⎪ ⎪ 1 ⎪ o o ⎪ (μ− (c) − μ+ (c))φ(y, c) dz, 0 ≤ y ≤ yc , ⎪ ⎪ φ(z, c)2 ⎨ 0

y ⎪ ⎪ 1 ⎪ o o ⎪ ⎪ ⎪(μ− (c) − μ+ (c))φ(y, c) φ(z, c)2 dz, yc ≤ y ≤ π, ⎩ π

(5.8)

D. Wei et al. / Advances in Mathematics 362 (2020) 106963

34

and

 e (y, c) = Φ

⎧ y ⎪ ⎪ φ(y, c) ⎪ e e e e ⎪ dz + (ν0− (c) − ν0+ (c))φ(y, c), 0 ≤ y < yc , (μ− (c) − μ+ (c)) ⎪ ⎪ φ(z, c)2 ⎨ 0

y ⎪ ⎪ φ(y, c) ⎪ e e e e ⎪ ⎪ ⎪(μ− (c) − μ+ (c)) φ(z, c)2 dz + (ν1− (c) − ν1+ (c))φ(y, c), yc < y ≤ π. ⎩ π

Furthermore, it holds that μo− (c) − μo+ (c) =

2 AC( ωo ) + BD( ωo ) , 2 2 α A +B

μe− (c) − μe+ (c) ωe ) + BD( ωe )) + I(c)(φφ )(jπ, c)(C( ωe ) + E1−j ( ωe )B)|1j=0 2 I(c)2 (AC( 2 2 2 2 α I(c) (A1 + B1 )/ρ(c) def 2 = μ2 (c), α 2 μ2 (c) e e . νj− (c) − νj+ (c) = − α φ(jπ, c)φ (jπ, c) =

5.4. Dual formulation As in [36], we will use a dual method to derive the decay estimates of the velocity. We denote Λ1 (ϕ)(yc ) = Λ1,1 (ϕ)(yc ) + Λ1,2 (ϕ)(yc ),

(5.9)

Λ2 (ϕ)(yc ) = Λ2,1 (ϕ)(yc ) + Λ2,2 (ϕ)(yc ),

(5.10)

where Λ1,1 (ϕ)(yc ) = ρ(c)u (yc )II1,1 (ϕ)(c),

(5.11)

Λ1,2 (ϕ)(yc ) = ρ(c)u (yc )L0 (ϕ)(c) + u (yc )ρ(c)II(c)ϕ(yc ),

(5.12)



Λ2,1 (ϕ)(yc ) = ρ(c)II1,1 (u ϕ)(c),

(5.13)

Λ2,2 (ϕ)(yc ) = ρ(c)L0 (u ϕ)(c) + u (yc )ρ(c)II(c)ϕ(yc ).

(5.14)

We denote

where

Λ3 ( ωe )(yc ) = ρ(c)Λ1 ( ωe )(yc ) + Λ3,1 ( ωe )(yc ),

(5.15)

Λ4 (g)(yc ) = ρ(c)Λ2 (g)(yc ) + Λ4,1 (g)(yc ),

(5.16)

D. Wei et al. / Advances in Mathematics 362 (2020) 106963

 1  u (yc ) Λ3,1 ( ωe )(yc ) = Jj ωe ) + ω

e (yc )  , E1−j (  u (yc ) j=0    1 E1−j (gu )  . Λ4,1 (g)(yc ) = Jj ) + g(y c  u (yc ) j=0

35



(5.17)

(5.18)

With Lemma 5.3, the proof of the following dual formulas is similar to section 9.1 and section 9.2. We omit the details. Lemma 5.4. Let f (α, y) = (∂y2 − α2 )g(α, y) with g ∈ H 2 (0, π) ∩ H01 (0, π). Then we have π

ψ o (t, α, y)f (α, y)dy = −

1

Ko (c, α)e−iαct dc,

−1

0

where Ko (c, α) =

Λ1 ( ωo )(yc )Λ2 (g)(yc ) . (A(c)2 + B(c)2 )u (yc )

(5.19)

Lemma 5.5. Let f (α, y) = (∂y2 − α2 )g(α, y) with g ∈ H 2 (0, π) and g  (0) = g  (π) = 0. Then we have π

ψ e (t, α, y)f (α, y)dy = −

1

Ke (c, α)e−iαct dc,

−1

0

where Ke (c, α) =

Λ3 ( ωe )(yc )Λ4 (g)(yc ) . u (yc )(A1 (c)2 + B1 (c)2 )

(5.20)

5.5. Decay estimates With the hand of following uniform W 1,2 estimates of the kernels Ko and Ke , Theorem 1.1 can be proved by following the same arguments of section 9.3 in [36]. Proposition 5.6. Let Ko be defined by (5.19) with g ∈ H 2 (0, π), g(0) = g(π) = 0 and ω

o = 12 ( ω0 (α, y) − ω

0 (α, −y)) ∈ H 2 . Then Ko (−1, α) = Ko (1, α) = 0, and there exists a constant C independent of α such that, Ko (·, α)L1c (−1,1) ≤ C ωo L2 gL2 , (∂c Ko )(·, α)L1c (−1,1) ≤ C ωo H 1 gH 1 , 1

(∂c2 Ko )(·, α)L1c (−1,1) ≤ Cα 2  ωo H 2 f L2 .

D. Wei et al. / Advances in Mathematics 362 (2020) 106963

36

Proposition 5.7. Let Ke be defined by (5.20) with g ∈ H 2 (0, π), g  (0) = g  (π) = 0 and ω

e = 12 ( ω0 (α, y) + ω

0 (α, −y)) ∈ H 2 . Then Ke (−1, α) = Ke (1, α) = 0, and there exists a constant C independent of α such that, Ke (·, α)L1c (−1,1) ≤ C ωe L2 gL2 , ωe H 1 (g  L2 + αgL2 ), (∂c Ke )(·, α)L1c (−1,1) ≤ C|α| 2  1

3

(∂c2 Ke )(·, α)L1c (−1,1) ≤ C|α| 2  ωe H 2 f L2 . 6. Estimates of some integral operators In this section, we present the estimates of some integral operators, which will be used to establish the W 2,1 estimates of Ko and Ke . 6.1. Basic properties of Hilbert transform Recall that the Hilbert transform H on T is defined by π H(f )(x) = p.v.

cot

−π

x−y f (y)dy. 2

Lemma 6.1. It holds that d  H(ϕ)(x) = H(ϕ )(x), dx d  sin x(Hϕ) (x) − cos xHϕ(x) = sin x(Hϕ) (x) + sin xH(ϕ)(x). dx If ϕ is even, then π cot xH(ϕ)(x) − H(cot yϕ)(x) = −

ϕ(z) dz,

−π

π sin x(Hϕ)(x) − H(sin yϕ)(x) =

(cos x + cos y)ϕ(y) dy, −π

π H(cos yϕ) − cos xH(ϕ) = sin x −π

π H(ϕ)(x) = −2p.v. 0

ϕ(y)dy,

   sin x+y  1  2  ln   sin yϕ (y) dy,  sin y  sin x−y 2

D. Wei et al. / Advances in Mathematics 362 (2020) 106963

and if ϕ is odd, then sin x(Hϕ)(x) − H(sin yϕ)(x) = 0. Proof. The first two equalities are obvious. Notice that π sin x(Hϕ)(x) − H(sin yϕ)(x) = p.v.

(sin x − sin y) cot

−π

x−y ϕ(y)dy 2

π =

(cos x + cos y)ϕ(y)dy, −π

in particular, if ϕ is odd, then sin x(Hϕ)(x) − H(sin xϕ)(x) = 0. If ϕ is even, then we have π H(ϕ)(x) = p.v.

cot

−π

x−y ϕ(y)dy 2

π  = p.v.

cot 0

π = −2p.v. 0

x−y x+y + cot 2 2

 ϕ(y)dy

   sin x+y  1  2  ln   sin yϕ (y)dy,  sin y  sin x−y 2

and cot x(Hϕ)(x) − H(cot yϕ)(x) π (cot x − cot y) cot

= p.v. −π

π = −p.v. −π

x−y ϕ(y)dy 2

1 + cos(x − y) ϕ(y)dy = − sin x sin y

π ϕ(y)dy,

−π

and π H(cos yϕ) − cos xH(ϕ) = p.v. −π

(cos y − cos x) cot

x−y ϕ(y)dy 2

37

D. Wei et al. / Advances in Mathematics 362 (2020) 106963

38

π =

π (sin x + sin y)ϕ(y)dy = sin x

−π

ϕ(y)dy.

−π

This proves the lemma.  The following lemma can be easily proved by using Hardy’s inequality. Lemma 6.2. If ϕ ∈ H 2 (T ) is even, then ϕ (0) = ϕ (π) = 0 and ∂ ϕ  y  ≤ CϕH 2 (0,π) .   sin y L2 (0,π) If ϕ is even and ϕ(0) = ϕ(π) = 0, then  ϕ   ϕ      + ≤ CϕH 2 (0,π) ,  2  2  sin y L∞ (0,π) sin y L (0,π)  ϕ    ≤ CϕH 1 (0,π) .   sin y L2 (0,π) Lemma 6.3. Let Z(f )(y) = f  (y) −cot yf (y). If f ∈ H m (0, π) is an even function for m = 1, 2, 3, then Z(f ) ∈ H m−1 (0, π) can be extended to a periodic odd function. Moreover, we have Z(f )H m−1 (T ) ≤ Cf H m (0,π) . If we f ∈ H 2 (0, π) ∩ H01 (0, π) is an even function, then Z(f )/sin yL2 (T ) ≤ f H 2 (0,π) . Proof. For 0 < y ≤

π 4,

we may write Z(f )(y) = m(y)y

where m(y) = we write

sin y y

∈ C ∞ (R) and

1 m

 Z(f )(y) = m(π − y)(π − y)



f  cos yf , = f − ym(y) ym(y)

∈ C ∞ ([0, 3π 4 ]). On the other hand, for π − y ≤

3π 4 ,

 f cos yf = f − . (π − y)m(π − y) (π − y)m(π − y)

Thus, we can conclude that Z(f )H m−1 (0,π) ≤ Cf H m (0,π) . To prove that Z(f ) can be extended to an even π-periodic function in H m−1 (T ), we only need to show that Z(f )(0) = Z(f )(π) = 0 and ∂y Z(f ) is an even function.

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39

This is because lim Z(f )(y) = lim f  (y) − lim cot yf (y) = f  (0+) − f  (0+) = 0, y→0+

y→0+

y→0+

lim Z(f )(y) = f  (π−) −f  (π−) = 0, and Z(f ) is an odd function. So, Z(f )H m−1 (T ) ≤

y→π−

Cf H m (0,π) .



6.2. Estimates of II1,1 We denote f Hyk =

k 

c

∂ync f L2yc (T ) ,

 · H k =  · H k (0,π) .

n=0

Lemma 6.4. It holds that for k = 0, 1, 2, ρII1,1 (ϕ)Hyk ≤ CϕH k , c

∂yc (ρII1,1 (ϕ))/sin yc L2 ≤ CϕH 2 . yc

Proof. By Lemma 6.1, we have for ϕ even,  H(ϕ)(y )  1 sin yc c ∂yc = ((Hϕ) (yc ) − cot yc Hϕ(yc )) 2 sin yc 2 π 1 1  = H(ϕ − cot yc ϕ)(yc ) + ϕ(y)dy. 2 2 −π

y π Let In(ϕ)(y) = Int(ϕ)(y) − Int(ϕ)(π) (1 − cos y) = 0 ϕ(y  ) dy  − sin2 y2 0 ϕ(y  ) dy  for 2 y ∈ [0, π]. We extend In(ϕ) to be an even π-periodic function. Then we have II1,1 (ϕ)(yc ) = −

 H(In(ϕ))(y )  1 c ∂yc 2 sin yc sin yc

1 1 =− H(ϕ)(yc ) − 2 sin2 yc 2 sin2 yc

π

In(ϕ)(y  )dy  .

(6.1)

−π

Here H = H ◦ Z ◦ In. Then by Lemma 6.3, we get  ρII1,1 (ϕ)Hyk ≤ C H(ϕ)H k (T ) + ϕL2 c  ≤ C Z ◦ In(ϕ)H k (T ) + ϕL2  ≤ C In(ϕ)H k+1 (0,π) + ϕL2 ≤ CϕH k .     As ∂yc (ρII1,1 (ϕ)) = H ∂y (Z ◦ In(ϕ)) , H ∂y (Z ◦ In(ϕ)) ∈ H 1 (T ) is odd, and by Lemma 6.3, we have

D. Wei et al. / Advances in Mathematics 362 (2020) 106963

40

   ∂yc (ρII1,1 )/ sin yc L2 ≤ C H ∂y (Z ◦ In(ϕ)) H 1 ≤ CIn(ϕ)H 3 ≤ CϕH 2 . yc

This proves the lemma.  The following lemma shows that for ϕ vanishing at 0 and π, we have better estimates, which will be used in the odd case. Lemma 6.5. If ϕ ∈ H01 (0, π), then  sin yc II1,1 (ϕ)L2yc ≤ CϕH 1 . If ϕ ∈ H 2 (0, π) ∩ H01 (0, π), then  sin yc II1,1 (ϕ)L∞ + II1,1 (ϕ)L2yc ≤ CϕH 2 . If ϕ ∈ H 2 (0, π) ∩ H01 (0, π) and ϕ / sin yc ∈ L2 (0, π), then ∂yc II1,1 (ϕ)L2yc + II1,1 (ϕ)L∞ + ∂y2c (sin yc II1,1 (ϕ))L2yc  +  sin yc ∂yc II1,1 (ϕ)L∞ ≤ C ϕH 2 + ϕ / sin yc L2yc . Proof. For ϕ ∈ H01 (0, π), we get by Hardy’s inequality that  ∂ In(ϕ)    In(ϕ)   In(ϕ)   y       2  2 +  2 + ∂y  ≤ CϕH 1 . sin y L sin y L2 sin y L Recall that II1,1 (ϕ)(yc ) = − and

In(ϕ) sin y

 H(In(ϕ))  1 ∂yc , 2 sin yc sin yc

is odd, by Lemma 6.1, we have  In(ϕ)  1 1   In(ϕ)  sin yc II1,1 (ϕ)(yc ) = − ∂yc H = − H ∂y , 2 sin y 2 sin y

which implies that for ϕ ∈ H01 (0, π),  sin yc II1,1 (ϕ)L2yc ≤ CϕH 1 . For ϕ ∈ H 2 (0, π) ∩ H01 (0, π), ∂y2

 In(ϕ) sin y

∈ L2 (T ) and

 1   In(ϕ)  ∂yc sin yc II1,1 (ϕ)(yc ) = − H ∂y2 , 2 sin y which shows that

(6.2)

D. Wei et al. / Advances in Mathematics 362 (2020) 106963

  In(ϕ)      ∂y sin yc II1,1 (ϕ)(yc )  2 ≤ C  ∂y2  ≤ CϕH 2 , c Lyc sin y L2 which implies that  sin yc II1,1 (ϕ)L∞ + II1,1 (ϕ)L2yc ≤ CϕH 2 . For f (0) = f  (0) = f  (0) = 0 and f  /y ∈ L2 , we have f   f   f          4  2 +  3  2 +  3  2 ≤ Cf  /yL2 . y L y L y L For ϕ ∈ H 2 (0, π) ∩ H01 (0, π) and ϕ /u ∈ L2 (0, π), let  In2 (ϕ) = In(ϕ) − (sin y)2 ϕ (0)(1 + cos y) + ϕ (π)(1 − cos y) + Int(ϕ)(π) cos y /4, then In2 (ϕ) = In2 (ϕ) = In2 (ϕ) = 0 for y = 0, π. Thus,  In (ϕ)   In (ϕ)   In (ϕ)       2  2  2 + +    2 ≤ C In2 (ϕ) / sin yL2 + In2 (ϕ)H 2 ,    4 3 2 2 2 sin y L sin y L sin y L which implies that      3  In(ϕ)      ∂y  ≤ ∂y3 In2 (ϕ)  + CϕH 2 ≤ C ϕ / sin yL2 + ϕH 2 .     sin y L2 sin y L2 Therefore, by Lemma 6.3,  Z ◦ H ∂ (In(ϕ)/ sin y)  y   ∂yc II1,1 (ϕ)L2yc ≤ C   2 sin y L  ≤ CH ∂y (In(ϕ)/ sin y) H 2  ≤ CIn(ϕ)/ sin yH 3 ≤ C ϕ / sin yL2 + ϕH 2 , as well as ∂y2c (sin yc II1,1 (ϕ))L2yc ≤ H(∂y3 (In(ϕ/ sin y)))L2 ≤ C(ϕ / sin yL2 + ϕH 2 ), which implies that  sin yc ∂yc II1,1 (ϕ)L∞ ≤ C(ϕH 2 + ϕ /u L2 ).  6.3. Estimates of Lk We denote  · Lp =  · Lp (0,π) .

41

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42

Lemma 6.6. For any p ∈ [1, ∞), there exists a constant C independent of α such that for k = 0, 1, 2, u (yc )2k+2 Lk (ϕ)L∞ ≤ C|α| p ϕLp , 1

u (yc )2k+1 Lk (ϕ)L∞ ≤ C|α|1+ p ϕLp ,    u (yc )2k+1 Lk (ϕ) ∞ ≤ C|α|ϕL∞ . L 1

Proof. By Proposition 3.3, we know that for k = 0, 1, 2, and 0 ≤ z ≤ y ≤ yc ≤ π or 0 ≤ yc ≤ y ≤ z ≤ π,  k     ∂ +∂ 1 φ min{α2 |z − yc |2 , 1} (y, c)  z  y 1 + ∂ − 1 ≤ C . (6.3)   c  u (yc )  (u(z) − c)2 φ1 (z, c)2 (u(z) − c)2 u (yc )2k cos yc −cos y y−yc

Recall that u1 (y, c) =

C −1 sin

and

y + yc y + yc ≤ u1 (y, c) ≤ C sin . 2 2

Thus, π 0

1

|z − yc |1− p min{α2 |z − yc |2 , 1} dz |u(z) − c|2 π

min{α2 |z − yc |2 , 1}

≤C

sin2

0

z+yc 2 |z



1

− yc |1+ p

dz 1

≤C 1 |z−yc |≤ α ,0≤z≤π

α2 |z − yc |1− p dz + C c sin2 z+y 2



|z − yc |−1− p dz. c sin2 z+y 2 1

1 |z−yc |> α ,0≤z≤π

Using the fact that  1 |z−yc |≤ α ,0≤z≤π



1 |z−yc |> α ,0≤z≤π

 α1+ p α2 |z − yc |1− p αp  , dz ≤ C min , z+y sin yc sin2 yc sin2 2 c 1

0

1

 α1+ p |z − yc |−1− p αp  , dz ≤ C min , c sin yc sin2 yc sin2 z+y 2 1

and for p = ∞, π

1

|z − yc | min{α2 |z − yc |2 , 1} dz |u(z) − c|2

1

1

D. Wei et al. / Advances in Mathematics 362 (2020) 106963

 ≤C 1 |z−yc |≤ α ,0≤z≤π

≤C

α2 |z − yc | dz + C c sin2 z+y 2

 1 |z−yc |> α ,0≤z≤π

43

|z − yc |−1 dz c sin2 z+y 2

|α| , sin yc

we deduce that for k = 0, 1, 2, and p ∈ [1, ∞),    u (yc )2k Lk (ϕ)(c) ≤ C

π 0

1

|z − yc |1− p min{α2 |z − yc |2 , 1} dz ϕLp |u(z) − c|2

αp  ϕLp , ≤ C min , u (yc ) u (yc )2  α1+ p1

1

and |u (yc )2k Lk (ϕ)| ≤ C

|α| ϕL∞ . sin yc

(6.4)

This proves the lemma.  To estimate the derivatives of L0 , we introduce for k, j = 0, 1, 

z Ik,j (ϕ)(c) =

ϕ(y) yc

∂z + ∂y + ∂c u (yc )

k 

1 (u(z) − c)2



φ1 (y, c) −1 φ1 (z, c)2



  dy 

. z=jπ

(6.5) Thus, for k = 0, 1, ∂c Lk (ϕ) =

1 (Lk (ϕ ) − Ik,1 (ϕ) + Ik,0 (ϕ)) + Lk+1 (ϕ). u (yc )

Lemma 6.7. It holds that    u (yc )3 ∂c L0 (ϕ)

≤ CαϕL∞ + Cα 2 ϕ L2 , 1

L∞

u (yc )3 ∂c L0 (ϕ)L∞ ≤ CαϕL∞ + Cαu (yc )ϕ L∞ . If ϕ/u ∈ Lp (0, π) for p ∈ (1, ∞), then 1

 sin yc L0 (ϕ)L∞ ≤ Cα p ϕW 1,p , L0 (ϕ)L∞ ≤ CαϕW 1,∞ , 1

 sin3 yc ∂c L0 (ϕ)L∞ ≤ Cα 2 ϕH 1 .

(6.6)

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44

Proof. By (6.3), we have for k = 0, 1,  jπ      min{α2 |jπ − yc |2 , 1}  |Ik,j (ϕ)(c)| ≤ C  ϕ(y)dy  2  2k   (cos jπ − cos yc ) u (yc ) yc

≤C

|jπ − yc | min{α2 |jπ − yc |2 , 1} ϕL∞ (cos jπ − cos yc )2 u (yc )2k

≤C

min{α2 |jπ − yc |2 , 1} ϕL∞ , |jπ − yc |3 u (yc )2k

(6.7)

thus by the fact that sin yc ≤ C|jπ − yc |, we deduce that for j = 0, 1, |u (yc )2k+3 Ik,j (ϕ)(c)| ≤ CϕL∞ , 

(6.8)

Ik,j (ϕ)(c)| ≤ CαϕL∞ .

(6.9)

1 (L0 (ϕ ) − I0,1 (ϕ) + I0,0 (ϕ)) + L1 (ϕ). u (yc )

(6.10)

|u (yc )

2k+2

Notice that ∂c L0 (ϕ) = We get by Lemma 6.6 that u (yc )2 L0 (ϕ )L∞ ≤ Cα 2 ϕ L2 , 1

u (yc )3 L1 (ϕ)L∞ ≤ CαϕL∞ , which along with the estimates of I0,0 and I0,1 imply that u (yc )3 ∂c L0 (ϕ)L∞ ≤ CαϕL∞ + Cα 2 ϕ L2 . 1

By (6.4), we get u (yc )L0 (ϕ )L∞ ≤ C|α|ϕ L∞ . This along with the estimates of I0,0 and I0,1 gives u (yc )3 ∂c L0 (ϕ)L∞ ≤ CαϕL∞ + Cαu (yc )ϕ L∞ .   y  1  p If ϕ/u ∈ Lp , then uϕ(y) ϕ (z)dz, thus,  uϕ(y) ≤ Cϕ Lp (0,π) for  (y) = u (y)  (y) jπ L (0,π) any p ∈ (1, ∞]. Due to u(y) − c = u1 (y, c)(y − yc ), we have  z  1  p   1 1   u (y)p dy  ≤ C max u (y)|z − yc | p ≤ Cu1 (z, c)|z − yc | p .   |2y−yc −z|≤|z−yc |   yc

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45

Then we have

|Lk (ϕ)(c)| ≤ C

 1 π min{α2 |z − yc |2 , 1} z u (y)p dy  p yc u1 (z, c)2 (z − yc )2 u (yc )2k

0

π ≤C

min{α2 |z − yc |2 , 1} 1

0

u (yc )2k+1 |z − yc |1+ p

   ϕ(y)   dz    u (y) 

Lp

1

ϕ Lp ≤ C

αp ϕ Lp ,  u (yc )2k+1

and similarly, u (yc )2k+1 Lk (ϕ)L∞ ≤ CαϕL∞ .

(6.11)

In particular, we get 1

| sin yc L0 (ϕ)L∞ ≤ Cα p ϕW 1,p .    2     As |u(z) − c| =  sin yc2+z sin yc2−z  ≥  sin yc2−z  ≥ C −1 |z − yc |2 , we have 

|u (yc ) Lk (ϕ)(c)| ≤ C 2k

π min{α2 |z − y |2 , 1} z u (y)dy c yc 0

π ≤C 0

π ≤C 0

(u(z) − c)2

   ϕ(y)   dz   u (y)  ∞ L

min{α2 |z − yc |2 , 1} dzϕ L∞ |u(z) − c|

(6.12)

min{α2 |z − yc |2 , 1} dzϕ L∞ ≤ C|α|ϕ L∞ . |z − yc |2

In particular, we get L0 (ϕ)L∞ ≤ CαϕW 1,∞ . On the other hand, we have 

|u (yc )

2k+2

2− 1  1 u (yc )2k+2 yc p min{α2 yc2 , 1}  ϕ Ik,0 (ϕ)(c)| ≤ C   ≤ Cα p ϕW 1,p (0,π) . u (yc )2k+2 yc2 u  Lp

The bound of Ik,1 is similar. Then we can conclude the last inequality of the lemma by using (6.10).  Lemma 6.8. It holds that     u (yc )5 ∂c2 L0 (ϕ)(c) ≤ C α 12 u (yc )ϕ L2 + αϕL∞ + αu (yc )ϕ L∞ + α 12 ϕ L2 .

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If ϕ/u ∈ L∞ , then for k = 0, 1, 2, u (yc )2k ∂ck L0 (ϕ)L∞ ≤ Cα ϕ L∞ + Cα 2 ϕ L2 . 1

Proof. Since we have ϕ(yc ) 1 ∂c I0,j (ϕ)(c) = −  u (yc ) (u(z) − c)2 

jπ +

∂c yc



ϕ(y) (u(z) − c)2

  1 − 1  2 φ1 (z, c) z=jπ



φ1 (y, c) −1 φ1 (z, c)2



  dy 

, z=jπ

and by Proposition 3.3, we get for |2y − yc − z| ≤ |z − yc |,       1 φ1 (y, c) ∂c  − 1   (u(z) − c)2 φ1 (z, c)2          φ1 (y, c) 1 φ1 (y, c)  1 +C − 1 ∂ ≤ C    |u(z) − c|2 c φ1 (z, c)2  |u(z) − c|3 φ1 (z, c)2 ≤C

min{α2 |z − yc |, α} min{α2 |z − yc |2 , 1} + C . |u(z) − c|3 u (yc )|u(z) − c|2

Thus, we obtain min{α2 (π − yc )2 , 1} (π − yc )2 u (yc )3       1 φ1 (y, c)  |π − yc |ϕL∞ + C ∂c − 1  2 2 (u(π) − c) φ1 (π, c)

|∂c I0,1 (ϕ)(c)| ≤ CϕL∞

≤ CϕL∞

min{α2 (π − yc )2 , 1} min{α2 (π − yc ), α} ∞ + Cϕ , L (π − yc )2 u (yc )3 |π − yc |u (yc )3

and |∂c I0,0 (ϕ)(c)| ≤ CϕL∞

 min{α2 y 2 , 1} c

yc2 u (yc )3

+

min{α2 yc , α}  . yc u (yc )3

This gives |∂c I0,1 (ϕ)(c)| + |∂c I0,0 (ϕ)(c)|  min{α2 sin2 y , 1} min{α2 sin2 y , α sin y }  c c c ≤ CϕL∞ + . sin5 yc sin5 yc Therefore, for any γ ∈ [0, 1],

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47

     u (yc )3+γ ∂c I0,0 (ϕ)(c) + u (yc )3+γ ∂c I0,1 (ϕ)(c) ≤ Cα2−γ ϕL∞ .

(6.13)

Notice that  ∂c2 L0 (ϕ) = ∂c =

 1  (L (ϕ ) − I (ϕ) + I (ϕ)) + L (ϕ) 0 0,1 0,0 1 u (yc )

1 1 (L0 (ϕ ) − I0,1 (ϕ ) + I0,0 (ϕ )) +  L1 (ϕ ) u (yc )2 u (yc ) −

1 ∂c (I0,1 (ϕ)(c) − I0,0 (ϕ)(c)) u (yc )

−

u (yc ) (L0 (ϕ ) − I0,1 (ϕ) + I0,0 (ϕ)) u (yc )3

+

1 (L1 (ϕ ) − I1,1 (ϕ) + I1,0 (ϕ)) + L2 (ϕ). u (yc )

Thus, by Lemma 6.6, (6.8), (6.9), (6.11) and (6.13), we obtain | sin5 yc ∂c2 L0 (ϕ)(c)| ≤ Cα 2 u (yc )ϕ L2 + CαϕL∞ + Cαu (yc )ϕ L∞ + Cα 2 ϕ L2 . 1

1

If ϕ(0) = ϕ(π) = 0, then  ϕ   min{α2 y 2 , 1} min{α2 y , α}    c c |∂c I0,0 (ϕ)(c)| ≤ C    + . u L∞ yc2 u (yc )2 yc u (yc )2 Therefore, ϕ   |u (yc )3 ∂c I0,0 (ϕ)(c)| ≤ Cα   . u L∞ Similarly, we have ϕ   |u (yc )3 ∂c I0,1 (ϕ)(c)| ≤ Cα   . u L∞ We also have ϕ   u (yc )2k+1 Ik,j (ϕ)L∞ ≤ Cα    . u L∞ With the estimates above, we deduce the second inequality of the lemma by using (6.12), (6.9) and Lemma 6.6.  7. W 2,1 estimates of Ko and Ke This section is devoted to the proof of Proposition 5.6 and Proposition 5.7.

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48

7.1. W 2,1 estimate of Ko In this subsection, we prove Proposition 5.6. Step 1. L1 estimate. We normalize ω

0 , g so that  ωo L2 ≤ 1, gL2 ≤ 1. By (5.11), (5.13) and Lemma 6.4, 2 we get Λ1,1 ( ωo )(yc ) = L and Λ2,1 (g)(yc ) = L2 . By (5.12), (5.14) Lemma 4.4 and 1 1 Lemma 6.6, we get Λ1,2 ( ωo )(yc ) = |α| 2 L∞ + |α| sin yc L2 and Λ2,2 (g)(yc ) = |α| 2 L∞ + |α| sin yc L2 . Thus, we have Λ1 ( ωo )(yc ) = (1 + |α| sin yc )L2 + |α| 2 L∞ , Λ2 (g)(yc ) = (1 + |α| sin yc )L2 + |α| 2 L∞ . 1

1

Then by Proposition 4.3, we infer that |α| 2 L2 αL∞ u (yc )Ko (c, α) = L + + , 1 + |α| sin yc (1 + |α| sin yc )2 1



1

which implies that Ko (·, α)L1c (−1,1) ≤ C ωo L2 gL2 . Step 2. W 1,1 estimate. We normalize ω

0 , g so that  ωo H 1 ≤ 1, gH 1 ≤ 1. of ω

o , we know  By  the definition   that ω

o (α, 0) = ω

o (α, π) = g(0) = g(π) = 0, then  ω uo L2 +  ug L2 ≤ C. By (5.11), (5.13) and Lemma 6.5, we get Λ1,1 ( ωo )(yc ) = sin yc L2 ,

Λ2,1 (g)(yc ) = sin yc L2 ,

and by Lemma 6.4, we get ∂yc Λ1,1 ( ωo )(yc ) = L2 ,

∂yc Λ2,1 (g)(yc ) = L2 .

By (5.12), (5.14), Lemma 4.4 and Lemma 6.7, we get Λ1,2 ( ωo )(yc ) = |α| 2 sin yc L∞ + |α| sin2 yc L2 , 1

Λ2,2 (g)(yc ) = |α| 2 sin yc L∞ + |α| sin2 yc L2 , 1

α 2 L∞ + |α|L2 , sin yc 1

∂c Λ1,2 ( ωo )(yc ) =

α 2 L∞ + |α|L2 . sin yc 1

∂c Λ2,2 (g)(yc ) = Thus, we obtain

D. Wei et al. / Advances in Mathematics 362 (2020) 106963

Λ1 ( ωo )(yc ) = sin yc ((1 + |α| sin yc )L2 + |α| 2 L∞ ), 1

Λ2 (g)(yc ) = sin yc ((1 + |α| sin yc )L2 + |α| 2 L∞ ), 1 1  ∂c Λ1 ( ωo )(yc ) = (1 + |α| sin yc )L2 + α 2 L∞ , sin yc 1 1  (1 + |α| sin yc )L2 + α 2 L∞ . ∂c Λ2 (g)(yc ) = sin yc 1

Therefore,  1 Λ1 ( ωo )Λ2 (g) = sin2 yc (1 + |α| sin yc )2 L1 + α 2 (1 + |α| sin yc )L2 + αL∞ , ∂c (Λ1 ( ωo )Λ2 (g)) = (1 + |α| sin yc )2 L1 + α 2 (1 + |α| sin yc )L2 + αL∞ , 1

which along with Proposition 4.3 give ∂c Ko (·, α)L1c (−1,1) ≤ C ωo H 1 gH 1 . Step 3. W 2,1 estimate. We normalize ω

0 , g so that  ωo H 2 ≤ 1, f L2 ≤ 1. Then we have g  2L2 + 2α2 g  2L2 + α4 g2L2 ≤ 1,  u g  1   ≤ C(u g) L∞ ≤ C/α 2 , u  L∞

(u g) L2 ≤ C.

By (5.11), (5.13) and Lemma 6.5, we get Λ1,1 ( ωo )(yc ) = sin2 yc L2 ∩ sin yc L∞ ,

Λ2,1 (g)(yc ) = sin2 yc L2 ∩ sin yc L∞ .

By Lemma 6.4, we get ∂c Λ1,1 ( ωo )(yc ) = L2 , ∂c2 Λ1,1 ( ωo )(yc ) =

∂c Λ2,1 (g)(yc ) = L2 ,

L2 , sin2 yc

∂c2 Λ2,1 (g)(yc ) =

L2 . sin2 yc

By (5.12), (5.14), Lemma 4.4 and Lemma 6.8, we get Λ1,2 ( ωo )(yc ) = |α| sin2 yc L∞ , ∂c Λ1,2 ( ωo )(yc ) = |α|L∞ , ∂c2 Λ1,2 ( ωo )(yc ) Therefore, we obtain

Λ1,2 (g)(yc ) = |α| 2 sin2 yc L∞ , 1

∂c Λ1,2 (g)(yc ) = |α| 2 L∞ ,

|α|L∞ |α|L2 = + , 2 sin yc sin yc

1

|α| 2 L∞ |α|L2 = + . 2 sin yc sin yc 1

∂c2 Λ1,2 (g)(yc )

49

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50

Λ1 ( ωo ) = sin2 yc L2 ∩ sin yc L∞ + |α| sin2 yc L∞ , ∂c (Λ1 ( ωo )) = L2 + αL∞ , ∂c2 (Λ1 ( ωo )) =

(1 + |α| sin yc )L2 + |α|L∞ , sin2 yc

Λ2 (g) = sin2 yc (L2 + |α| 2 L∞ ), 1

∂c (Λ2 (g)) = L2 + |α| 2 L∞ , 1

(1 + |α| sin yc )L2 + |α| 2 L∞ . sin2 yc 1

∂c2 (Λ2 (g)) = Thus,

 3 Λ1 ( ωo )Λ2 (g) = sin4 yc L1 + αL2 + α 2 L∞ ,   3 ∂c Λ1 ( ωo )Λ2 (g) = sin2 yc L1 + αL2 + α 2 L∞ ,   ∂c2 Λ1 ( ωo )Λ2 (g) = (1 + α sin yc ) L1 + αL2 + α2 L∞ , from which and Proposition 4.3, we infer that 1

∂c2 Ko (·, α)L1c (−1,1) ≤ Cα 2  ωo H 2 f L2 , and Ko (c, α) = C(α) sin2 yc L2 , which implies that Ko (±1, α) = 0. 7.2. W 2,1 estimate of Ke In this subsection, we prove Proposition 5.7. We need the following lemma. Lemma 7.1. It holds that (1) if  ωe L2 ≤ 1 and gL2 ≤ 1, then Λ3,1 ( ωe ) =

L2 , α2

Λ4,1 (g) =

L2 ; α2

(2) if  ωe H 1 ≤ 1 and |α|gL2 + g  L2 ≤ 1, then Λ3,1 ( ωe ) =

sin yc (L2 + αL∞ ) , α2

sin yc (L2 + α 2 L∞ ) , α2 1

Λ4,1 (g) =

D. Wei et al. / Advances in Mathematics 362 (2020) 106963

 L2 + αL∞ ∂c Λ3,1 ( ωe ) = , α2 sin yc

51

 L2 + α 12 L∞ ∂c Λ4,1 (g) = ; α2 sin yc

(3) if  ωe H 2 ≤ 1 and g  − α2 gL2 ≤ 1 and ω

e (0) = ω

e (π) = g  (0) = g  (π) = 0, then Λ3,1 ( ωe ) = α−2 (ρL2 ∩ u L∞ ) + ρL∞ , Λ4,1 (g) = α−2 (ρL2 ∩ u L∞ ) + α− 2 ρL∞ ,   1 ∂c Λ3,1 ( ωe ) = α−2 L2 + L∞ , ∂c Λ4,1 (g) = α−2 (L2 + α 2 L∞ ), 3

α−2 L2 + L∞  ∂c2 Λ3,1 ( ωe ) = , sin2 yc

L2 + α 12 L∞  ∂c2 Λ4,1 (g) = ; α2 sin2 yc

(4) if  ωe H 2 ≤ 1 and ∂y ω

e /u H 1 ≤ 1, then  ∂c Λ3,1 ( ωe ) = L∞ ,

 ∂c2 Λ3,1 ( ωe ) =

α2 L∞ L2 + . (1 + α sin yc )2 α sin yc

Remark 7.2. Following the proof of Case 2, we can show that if |α|ωL2 + ∂y ωL2 ≤ 1, then sin yc (L2 + α 2 L∞ ) , α2 1

Λ3,1 (ω) =

 L2 + α 12 L∞ ∂c Λ3,1 (ω) = . α2 sin yc

This will be used in the proof of Proposition 9.3. Proof. Case 1.  ωe L2 ≤ 1 and gL2 ≤ 1. Using the fact that φ1 (y, c) ≤ φ1 (0, c) for 0 < y < yc and φ1 (y, c) ≤ φ1 (π, c) for yc < y < π, we get   

 Ej (ϕ)   ≤ CϕLp , φ1 (jπ, c)(jπ − yc ) Lp

1 < p ≤ +∞,

(7.1)

which implies that u (yc )E0 ( ωe ) + u (yc ) ωe (yc ) = φ1 (0, c)yc L2 , u (yc )E1 ( ωe ) + u (yc ) ωe (yc ) = φ1 (π, c)(π − yc )L2 , and E0 (gu ) + u (yc )g(yc ) = φ1 (0, c)yc L2 , E1 (gu ) + u (yc )g(yc ) = φ1 (π, c)(π − yc )L2 , from which and Lemma 4.8, we deduce (1). Case 2.  ωe H 1 ≤ 1 and |α|gL2 + g  L2 ≤ 1 (thus, gL∞ ≤ C|α|− 2 ). 1

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We write u (yc )Ej ( ωe ) + u (yc ) ωe (yc ) jπ =



u (yc )φ1 (y, c) ωe (y) − u (y) ωe (yc ) dy



u (yc ) − u (y) φ1 (y, c) ωe (y)dy

yc

jπ = yc

jπ +





jπ



ωe (y)dy + φ1 (y, c) − 1 u (y)

yc

 u (y) ω

e (y) − ω

e (yc ) dy

yc

= I1 + I2 + I3 .

(7.2)

Using the fact that φ1 (y, c)−1 ≤ C min{α2 |y−yc |2 , 1}φ1 (y, c) ≤ C min{α2 |y−yc |2 , |α||y− c yc |}φ1 (y, c) and |u (y) − u (yc )| ≤ sin y+y 2 |y − yc |, we obtain |I1 | ≤ C|jπ − yc |3 φ1 (jπ, c) ω e  L∞ ,   2 |I2 | ≤ C|α||jπ − yc | min 1, |α||jπ − yc | φ1 (jπ, c) ω e  L∞ ,   I3   ωe L2 .   ≤ C |jπ − yc |2 L2 Thus, we infer that u (yc )E0 ( ωe ) + u (yc ) ωe (yc ) = φ1 (0, c)yc2 αL∞ + yc2 L2 , u (yc )E1 ( ωe ) + u (yc ) ωe (yc ) = φ1 (π, c)(π − yc )2 αL∞ + (π − yc )2 L2 , which along with Lemma 4.8 show that Λ3,1 ( ωe ) =

sin yc (L2 + αL∞ ) . α2

We write Ej (u g) + u (yc )g(yc ) jπ =



u (y)φ1 (y, c)g(y) − u (y)g(yc ) dy



φ1 (y, c) − 1 u (y)g(y)dy +

yc

jπ = yc

jπ yc

 u (y) g(y) − g(yc ) dy

(7.3) (7.4) (7.5)

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53

= I2 + I3 .

(7.6)

We have   |I2 | ≤ C|α||jπ − yc |2 min 1, |α||jπ − yc | φ1 (jπ, c)gL∞ ,   I3     ≤ Cg  L2 . |jπ − yc |2 L2

(7.7)

Thus, we obtain E0 (gu ) + u (yc )g(yc ) = φ1 (0, c)yc2 α 2 L∞ + yc2 L2 , 1

E1 (gu ) + u (yc )g(yc ) = φ1 (π, c)(π − yc )2 α 2 L∞ + (π − yc )2 L2 , 1

which along with Lemma 4.8 show that sin yc (L2 + α 2 L∞ ) . α2 1

Λ4,1 (g) = A direct calculation gives

∂c (u (yc )Ej ( ωe ) + u (yc ) ωe (yc ))

(7.8)

u (yc ) u (yc ) ωe (yc ) =  Ej ( ωe ) − + u (yc ) u (yc ) u (yc ) u (yc ) Ej ( =  ωe ) + ω

e (yc ) + u (yc ) u (yc )

jπ ω

e ∂c φ1 (y, c)dy + yc

(u ω

e ) (yc ) u (yc )

jπ ω

e ∂c φ1 (y, c)dy, yc

and gu (yc ) ∂c (Ej (gu ) + u (yc )g(yc )) = −  + u (yc ) 





jπ

= g (yc ) +

jπ

gu ∂c φ1 (y, c)dy +

yc

(u g) (yc ) u (yc )

gu ∂c φ1 (y, c)dy.

yc

By Proposition 3.3, we get for 0 < y < yc , j = 0 or yc < y < π, j = 1,    ∂c φ1 (y, c)  C|α| C|α|    φ1 (jπ, c)  ≤ u (yc ) min{|α||y − yc |, 1} ≤ u (yc ) min{|α||jπ − yc |, 1}. Then we have for j = 0

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    0   (π − yc ) φ (y, c) ∂ c 1    min{αyc , 1} ϕ(y) φ1 (0, c) dy    yc

≤ CαϕLp , 1 < p ≤ ∞.

(7.9)

Lp

Similarly, we get for j = 1     π   y ∂ φ (y, c) c c 1    min{α(π − yc ), 1} ϕ(y) φ1 (π, c) dy    yc

≤ CαϕLp , 1 < p ≤ ∞.

(7.10)

Lp

Recall that  ωe H 1 ≤ 1 and αgL2 + g  L2 ≤ 1, then gL∞ ≤ Cα− 2 . Thus, we obtain 1

∂c (u (yc )E0 ( ωe ) + u (yc ) ωe (yc )) = φ1 (0, c)

αL∞ + L2 , π − yc

∂c (u (yc )E1 ( ωe ) + u (yc ) ωe (yc )) = φ1 (π, c)

αL∞ + L2 , yc

∂c (E0 (gu ) + u (yc )g(yc )) =

φ1 (0, c)α 2 L∞ + L2 , π − yc

∂c (E1 (gu ) + u (yc )g(yc )) =

φ1 (π, c)α 2 L∞ + L2 , yc

1

1

from which and Lemma 4.8, we infer that  L2 + αL∞ ∂c Λ3,1 ( ωe ) = , α2 sin yc

 L2 + α 12 L∞ ∂c Λ4,1 (g) = . α2 sin yc

Case 3.  ωe H 2 ≤ 1 and g  − α2 gL2 ≤ 1 and ω

e (0) = ω

e (π) = g  (0) = g  (π) = 0. Thus, we have g  2L2 + 2α2 g  2L2 + α4 g2L2 ≤ 1, gL∞ ≤ Cα− 2 , 3

  ω  e  2 ≤ C ωe L2 , u L

 g    2 ≤ Cg  L2 ≤ C, u L

g  L∞ ≤ Cα− 2 . 1

First of all, we have for I3 in (7.2) and I3 in (7.6),  jπ yc   jπ   jπ ω

e (y  )             2 |I3 | ≤ C  ω

e (y )dy dy  ≤ C  |jπ − y | ωe (y )dy  ≤ C|jπ − yc |  dy , jπ − y  yc y

yc

yc

which implies that   

 ω  I3     ωe L2 ,  2 ≤ C  e  2 ≤ C 3 |jπ − yc | L u L

  

 I3  ≤ C ωe L∞ .  2 |jπ − yc | L∞

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55

Similarly, for I3 , we have   

  g  I3    ≤ C     2 ≤ Cg  L2 , |jπ − yc |3 L2 u L

  

 I3  ≤ Cg  L∞ .  |jπ − yc |2 L∞

Thus, by (7.3), (7.4) and (7.7), we obtain u (yc )E0 ( ωe ) + u (yc ) ωe (yc ) = φ1 (0, c)yc3 α2 L∞ + (yc3 L2 ∩ yc2 L∞ ), u (yc )E1 ( ωe ) + u (yc ) ωe (yc ) = φ1 (π, c)(π − yc )3 α2 L∞ + (π − yc )2 ((π − yc )L2 ∩ L∞ ), E0 (u g) + u (yc )g(yc ) = φ1 (0, c)yc3 α 2 L∞ + (yc3 L2 ∩ yc2 L∞ ), 1

E1 (u g) + u (yc )g(yc ) = φ1 (π, c)(π − yc )3 α 2 L∞ + (π − yc )2 ((π − yc )L2 ∩ L∞ ), 1

which along with Lemma 4.8 show that Λ3,1 ( ωe ) = α−2 (ρL2 ∩ u L∞ ) + ρL∞ ,

Λ4,1 (g) = α−2 (ρL2 ∩ u L∞ ) + α− 2 ρL∞ . 3

We have ∂c (u (yc )Ej ( ωe ) + u (yc ) ωe (yc )) jπ



u (yc )∂c φ1 (y, c) ωe (y)dy +

= yc

ω

e (yc )

jπ −

φ1 (y, c) ωe (y)dy yc

= I4 + I5 + I6 . By Proposition 3.3, we have C|α||jπ − yc |φ1 (jπ, c) min{|α||jπ − yc |, 1} ω e  L∞ , u (yc )  I5   e    2 ≤ C ω  , |I6 | ≤ C|φ1 (jπ, c)|jπ − yc | ω e  L∞ ,  L u u  L2 |I4 | ≤

which implies that ∂c (u (yc )Ej ( ωe ) + u (yc ) ωe (yc )) =

φ1 (jπ, c)α2 |jπ − yc |L∞ + u (yc )L2 . ((1 − j)π − yc )

Similarly, due to 



jπ

∂c (Ej (u g) + u (yc )g(yc )) = yc

we get

u (y)∂c φ1 (y, c)g(y)dy + g  (yc ),

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56

φ1 (jπ, c)α 2 |jπ − yc |L∞ + u (yc )L2 , ((1 − j)π − yc ) 1

∂c (Ej (u g) + u (yc )g(yc )) =

from which and Lemma 4.8, we infer that  ∂c Λ3,1 ( ωe ) = α−2 L2 + L∞ ,

 1 ∂c Λ4,1 (g) = α−2 (L2 + α 2 L∞ ).

We have ∂c2

g  (yc ) (Ej (gu ) + u (yc )g(yc )) =  + u (yc ) 



jπ

gu ∂c2 φ1 (y, c)dy,

yc

and ∂c2 (u (yc )Ej ( ωe ) + u (yc ) ωe (yc ))  =

u (yc ) (u u )(yc ) − u (yc )2 u (yc )3 

 Ej ( ωe ) +

jπ ω

e ∂c2 φ1 (y, c)dy −

+ u (yc ) yc

ω

 (yc ) = e −2 u (yc )

jπ

(7.11) ω

e (yc ) u (yc ) + 2 u (yc ) u (yc )

ω

e ∂c φ1 (y, c)dy yc

u (yc ) ω

e (yc ) u (yc )2 jπ



ω

e ∂c φ1 (y, c)dy + u (yc ) yc

jπ

ω

e ∂c2 φ1 (y, c)dy − yc

u (yc ) ω

e (yc ) u (yc )2

By Proposition 3.3, we deduce that for 0 < y < yc , j = 0 or yc < y < π, j = 1,  2  Cα2  ∂c φ1 (y,c)  , and then  φ1 (jπ,c)  ≤  u (yc )2     0 2   yc (π − yc )2 ϕ(y) ∂c φ1 (y, c) dy  ≤ Cα2 ϕLp , 1 < p ≤ ∞,   φ1 (0, c)   yc Lp     π 2  2  yc (π − yc ) ϕ(y) ∂c φ1 (y, c) dy  ≤ Cα2 ϕLp , 1 < p ≤ ∞.   φ1 (π, c)   yc

Lp

Then we can deduce that yc ∂c2 (u (yc )E0 ( ωe ) + u (yc ) ωe (yc )) =

φ1 (0, c)α2 L∞ + L2 , (π − yc )2

(π − yc )∂c2 (u (yc )E1 ( ωe ) + u (yc ) ωe (yc )) =

φ1 (π, c)α2 L∞ + L2 , yc2

(7.12)

(7.13)

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from which and Lemma 4.8, we infer that  ∂c2

J1−j   u (yc )Ej ( ωe ) + u (yc ) ωe (yc ) u (yc )

 =

L∞ L2 + , (7.14) (jπ − yc )2 α2 (jπ − yc )2

which implies that  α−2 L2 + L∞ ∂c2 Λ3,1 ( ωe ) = . sin2 yc Similarly, we have φ1 (0, c)α 2 L∞ + L2 , (π − yc )2 1

yc ∂c2 (E0 (u g) + u (yc )g(yc )) =

φ1 (π, c)α 2 L∞ + L2 , yc2 1

(π − yc )∂c2 (E1 (u g) + u (yc )g(yc )) = from which and Lemma 4.8, we infer that

 L2 + α 12 L∞ ∂c2 Λ4,1 (g) = . α2 sin2 yc Case 4.  ωe H 2 ≤ 1 and ∂y ω

e /u H 1 ≤ 1. Recall that ∂c (u (yc )Ej ( ωe ) + u (yc ) ωe (yc )) jπ =



u (yc )∂c φ1 (y, c) ωe (y)dy +

ω

e (yc )

yc

jπ −

φ1 (y, c) ωe (y)dy. yc

Now since ∂y ω

e /u ∈ H 1 ⊂ L∞ , we get ∂c (u (yc )Ej ( ωe ) + u (yc ) ωe (yc )) =

φ1 (jπ, c)α2 |jπ − yc |L∞ + u (yc )L∞ , ((1 − j)π − yc )

from which and Lemma 4.8, we infer that ∂c Λ3,1 = L∞ . For the estimates of higher derivative, we only need to improve the estimate of J1−j   ωe ) + u (yc ) ωe (yc ) for |yc − jπ| ≤ α1 . We show the proof of the case u (yc ) u (yc )Ej ( j = 0, the other case can be proved similarly. ∂ ω

Let ω1 = yy e ∈ H 1 (0, 1), and then 0

0 ω

e (y)φ1 (y, c)dy = yc

E0 ( ωe ) = yc

ω

e (yyc )ϕ1 (y, c)dy, 1

here we recall ϕ1 (y, c) = φ1 (yyc , c) and its estimate in the proof of Lemma 4.8.

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Using the fact that ∂c ω

e (yyc ) = with m1 (y) =

sin y y

y ωe (yyc ) y 2 ω1 (yyc ) = , u (yc ) m1 (yc )

and ∂c2 ω

e (yyc ) =

y 3 ω1 (yyc ) y 2 ω1 (yyc )m1 (yc ) + ,  m1 (yc )u (yc ) m1 (yc )2 u (yc )

and that 1 ≤ |ϕ1 (y, c)| ≤ 2 and (3.11), (3.12) hold for y ∈ [0, 1] and c ∈ [u(0), u(1/α)], we deduce that for 0 < yc < α1 , |∂c (E0 ( ωe )/yc )| ≤ C(ω1 L∞ + α2  ωe L∞ ), ωe )/yc )| |∂c2 (E0 (

≤ C(α ω1 L∞ 2

C + α  ω e  L∞ ) + yc

1

4

y 3 |ω1 (yyc )|dy.

0

Now using the fact that  J1

 −(u(π) − c)2 yc2 u (yc ) E0 ( ωe ) u (yc ) E ( ω ) + ω

(y ) = + J1 ω

e (yc ), 0 e e c   u (yc ) ϕ1 (0, c)ϕ1 (0, c) yc

and by Proposition 3.5 and Lemma 4.8, we deduce that for 0 < yc <  ∂c2

1 α,

 u (y )  L2 c J1  E0 ( ωe ) + ω

e (yc ) χ[u(0),u(1/α)] = α2 L∞ + 2 , u (yc ) α yc

which along with (7.14) gives  ∂c2

 u (y )  α2 L∞ L2 c J1  E0 ( ωe ) + ω

e (yc ) = + . u (yc ) (1 + α sin yc )2 αyc

Similarly, for j = 1, we have π

0 ω

e (y)φ1 (y, c)dy = (yc − π)

E1 ( ωe ) = yc

 ω

e (1 − z)π + zyc ϕ2 (z, c)dz.

1

Then by the same argument, we obtain  ∂c2

 u (y )  α2 L∞ L2 c J0  E1 ( ωe ) + ω

e (yc ) = + , u (yc ) (1 + α sin yc )2 α(π − yc )

from which, we deduce (4) of the lemma. 

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Now we are in a position to prove Proposition 5.7. Proof. Step 1. L1 estimate. We normalize ω

e , g so that  ωe L2 ≤ 1 and gL2 ≤ 1. By Lemma 6.4, we get Λ1,1 ( ωe ) = L2 and Λ2,1 (g) = L2 . By Lemma 6.6 and Lemma 4.4, we get Λ1,2 ( ωe ) = 1 1 ∞ 2 ∞ 2 2 2 |α| L + |α| sin yc L and Λ2,2 (g) = |α| L + |α| sin yc L . Then by Lemma 7.1, we get L2 , α2 1 L2 Λ4 (g) = ρ(1 + |α| sin yc )L2 + |α| 2 sin2 yc L∞ + 2 , α

Λ3 ( ωe ) = ρ(1 + |α| sin yc )L2 + |α| 2 sin2 yc L∞ + 1

from which and Proposition 4.5, we deduce that Ke (·, α)L1c (−1,1) ≤ C ωe L2 gL2 . Step 2. W 1,1 estimate. We normalize ω

e , g so that  ωe H 1 ≤ 1 and |α|gL2 + g  L2 ≤ 1, then gL∞ ≤ 1 − 12 C|α| . By Lemma 6.4, we get Λ1,1 ( ωe ) = L∞ and Λ2,1 (g) = |α|− 2 L∞ . By Lemma 6.6 1 and Lemma 4.4, we get Λ1,2 ( ωe ) = |α|sin yc L∞ and Λ2,2 (g) = |α| 2 sin yc L∞ . Then by Lemma 7.1, we get Λ3 ( ωe ) = sin2 yc (1 + |α| sin yc )L∞ + =

C sin yc (L2 + αL∞ ) α2

3 sin yc sin yc 2 (1 + |α| sin yc )2 L∞ + L = |α|− 2 sin yc (1 + |α| sin yc )3 L2 , α α2

sin yc (L2 + |α| 2 L∞ ) α2 sin yc sin yc 2 sin yc 2 ∞ = + L = (1 + |α| sin yc )3 L2 . 3 (1 + |α| sin yc ) L 2 − α α2 |α| 2 1

Λ4 (g) = |α|− 2 sin2 yc (1 + |α| sin yc )L∞ + 1

By Lemma 6.4, we get ∂c Λ1,1 ( ωe ) = Lemma 4.4, we get ∂c Λ1,2 ( ωe ) = Lemma 7.1, we obtain

∞

αL sin yc

L2 sin yc

and ∂c Λ2,1 (g) =

L2 sin yc .

+ αL2 and ∂c Λ2,2 (g) =

1 α2

By Lemma 6.7 and

L∞ sin yc

+ αL2 . Then by

 (L2 + αL∞ ) ∂c Λ3 ( ωe ) = (1 + α sin yc )(sin yc L2 + L∞ ) + α2 sin yc =

(1 + |α| sin yc )2 ∞ (1 + |α| sin yc )3 2 (1 + |α| sin yc )3 2 L + L = L , 3 2 α sin yc α sin yc |α| 2 sin yc

1  1 (L2 + α 2 L∞ ) ∞ 2 2 ∂c Λ4 (g) = α sin yc L + sin yc (1 + α sin yc )L + α2 sin yc

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=

(1 + |α| sin yc )2 ∞ (1 + |α| sin yc )3 2 (1 + |α| sin yc )3 2 L + L = L , 3 α2 sin yc |α|2 sin yc |α| 2 sin yc

from which and Proposition 4.5, we infer that 1

∂c Ke (·, α)L1c (−1,1) ≤ Cα 2  ωe H 1 gH 1 . Step 3. W 2,1 estimate. We normalize ω

e , g so that  ωe H 2 ≤ 1 with ω

e (0) = ω

e (1) = 0 and g  − α2 gL2 ≤ 1 with g  (0) = g  (1) = 0. Then we have   ω  e  2 ≤ C ωe L2 ≤ C, u L

g  2L2 + α2 g  2L2 + α4 gL2 ≤ 1, gL∞ ≤ Cα− 2 , 3

 g    2 ≤ Cg  L2 ≤ C, u L

g  L∞ ≤ Cα− 2 , u gH k ≤ CgH k 1

for k = 0, 1, 2. 3 By Lemma 6.4, we get Λ1,1 ( ωe ) = L∞ and Λ2,1 (g) = |α|− 2 L∞ . By Lemma 6.6 1 ωe ) = |α|sin yc L∞ and Λ2,2 (g) = |α|− 2 sin yc L∞ . Then by and Lemma 4.4, we get Λ1,2 ( Lemma 7.1, we get Λ3 ( ωe ) = α−2 (sin2 yc L2 ∩ sin yc L∞ ) + sin2 yc (1 + α sin yc )L∞ = α− 2 sin2 yc (1 + α sin yc )2 L2 , 1

Λ4 (g) = α−2 (ρL2 ∩ u L∞ ) + α− 2 ρ(1 + α sin yc )L∞ 3

= α−2 sin2 yc (1 + α sin yc )2 L2 . By Lemma 6.4, we get ∂c Λ1,1 ( ωe ) = L2 and ∂c Λ2,1 (g) = L2 . By Lemma 6.7 and Lemma 4.4, we get ∂c Λ1,2 ( ωe ) = Lemma 7.1, we obtain

αL∞ sin yc

1

and ∂c Λ2,2 (g) =

1

α− 2 L∞ +α 2 sin yc L∞ . sin yc

Then by

 ∂c Λ3 ( ωe ) = sin2 yc L2 + α−2 L2 + (1 + α sin yc )L∞ = α− 2 (1 + α sin yc )2 L2 , 1

 1 1 ∂c Λ4 (g) = α−2 (L2 + α 2 (1 + α sin yc )L∞ ) + sin2 yc (L2 + α 2 L∞ ) = α−2 (1 + α sin yc )3 L2 . By Lemma 6.4, we get ∂c2 Λ1,1 ( ωe ) = Lemma 4.4, we get ∂c2 Λ1,2 ( ωe ) = Then by Lemma 7.1, we obtain

∞

L2 sin2 yc 2

αL αL sin3 yc + sin yc

and ∂c2 Λ2,1 (g) =

L2 sin2 yc .

and ∂c2 Λ2,2 (g) =

(α 2 sin yc +α− 2 ) ∞ αL2 L + sin sin3 yc yc .

1

By Lemma 6.8 and 1

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61

 ∂c2 Λ3 ( ωe ) = (1 + α sin yc )(L2 + ρ−1 L∞ ) + ρ−1 (α−2 L2 + L∞ ) = α− 2 ρ−1 (1 + α sin yc )3 L2 , 1

 1 ∂c2 Λ4 (g) = (α−2 ρ−1 + 1)(L2 + α 2 L∞ ) + (1 + α sin yc )L2 = α−2 ρ−1 (1 + α sin yc )3 L2 , from which and Proposition 4.5, we infer that 3

∂c2 Ke (·, α)L1c (−1,1) ≤ Cα 2  ωe H 2 f L2 , and Ke (c, α) = C(α) sin2 yc L2 , which implies that Ke (±1, α) = 0.  8. Decay estimates at critical points 8.1. Vorticity depletion phenomena In this subsection, we give a formula of the vorticity at a critical point and confirm the vorticity depletion phenomena found by Bouchet and Morita [10]. Lemma 8.1. Let ω(t, x, y) be a solution to (1.2). Then we have 1 w(t,

α, 0) = −1

and

(1+cos yc )2 Λ3 (ω

e) u (yc )φ1 (0,c)(A21 +B21 )

(1 + cos yc )2 Λ3 ( ωe )e−iαct dc,   u (yc )φ1 (0, c)(A21 + B21 )

∈ L1c (−1, 1). In particular, for any α = 0, lim w(t,

α, 0) = 0.

t→∞

Proof. Let η ∈ C ∞ (R) so that η(y) = 1 for |y| < 12 and η(y) = 0 for |y| > 1 and η  (y) ≤ 0 for y > 0. We denote ηn (y) = η(ny), gn (y) = ηn (y) = nη  (ny) and fn (y) = gn (y) − α2 gn (y). Assume that ω0 (x, y) ∈ H 2 , then for any t ≥ 0, ω(t, x, y) ∈ H 2 . Then ω

o (t, α, 0) = 0, and by Lemma 5.5, π w(t,

α, 0) = − lim

ω

e (t, α, y)gn (y)dy

n→+∞ 0

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π = lim

n→+∞

ψ e (t, α, y)fn (y)dy

0 u(π) 

= − lim

n→+∞ u(0)

Λ3 ( ωe )Λ4 (gn ) −iαct e dc.  u (yc )(A21 + B21 )

Pointwise limit of Λ4 (gn ). Since lim ηn (yc ) = 0 and lim gn (yc ) = 0 for yc ∈ (0, π], we get n→+∞

n→+∞

yc



lim E0 (gn u )(c) = lim −

n→+∞

n→+∞

u (y)φ1 (y, c)dηn (y) = u (0)φ1 (0, c),

0

lim E1 (gn u )(c) = lim

n→+∞

π

n→+∞

gn (y)u (y)φ1 (y, c)dy = 0.

yc

Notice that II1,1 (ϕ) = p.v.

π

y

ϕ(y  )dy  (u(y)−c)2 dy.

Then for 1/n < yc , as n → ∞,

yc

0



|II1,1 (gn u )(c)| ≤

1/n  yc |g u |(z)dz n y 0

(u(y) − c)2 

1/n

≤ gn u L1 0

dy

dy → 0. (u(y) − c)2

By (6.3), we also have for 1/n < yc , as n → ∞,  n z       1 φ1 (y, c)    |L0 (gn u )| ≤  gn u (y) −1 dydz  (u(z) − c)2 φ1 (z, c)2 1

0 yc

1



n

≤ Cgn u L1 0

≤

     1 φ1 (y, c)  sup  − 1  dz 2 2 (u(z) − c) φ (z, c) 1 y∈[z,1/n]

C|α| gn u L1 → 0 nyc3

Thus, for yc ∈ (0, π], we obtain lim Λ4 (gn )(c) =

n→+∞

J1 (c)u (0)φ1 (0, c) u (yc )

(8.1)

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=−

63

(u(π) − c)2 u (yc )ρ(c)2 u (0)φ1 (0, c) = − . φ(0, c)φ (0, c)u (yc ) φ1 (0, c)

Uniform bound of Λ4 (gn ). For c ∈ (−1, 1], by Lemma 4.8 we have for j = 0, 1, Ej (gn u )(c)| ≤ φ1 (jπ, c)gn u L1 ≤ Cφ1 (jπ, c),

|Jj | ≤

C . α2 φ1 ((1 − j)π, c)

By Lemma 4.4, we get |ρII(c)| ≤ Cα sin yc . By Lemma 6.6, we have | sin yc2 L0 (gn u )| ≤ C|α|gn u L1 . Thus, |Λ2,2 (gn u )| ≤ C|α|. By (6.1), we get sin yc sin yc sin yc II1,1 (ϕ)(yc ) = − H(ϕ) − 2 2

π

3

In(ϕ)(y  )dy  .

(8.2)

−π

As Z ◦ In(u gn ) is odd, we get by Lemma 6.1 that sin yc H(Z ◦ In(u gn )) = H(sin yZ ◦ In(u gn )). Using the fact that for y ∈ [0, π], 

π



sin yZ ◦ In(u gn )(y) = sin yu gn (y) − sin y

u gn (y)dy − cos yIn(u gn )(y),

0

we deduce that 



π

sin yc H(u gn ) = H(sin yu gn (y)) −

u gn (y)dyH(sin y) − H(cos yIn(u gn )(y)).

0

√ Thanks to  sin yu gn H 1 ≤ C n and  sin yu gn L2 ≤

C √ , n

we get

1

1

 2 2 H(sin yu gn (y))L∞ ≤ CH(sin yu gn )H 1 H(sin yu gn )L2 ≤ C.

√ Thanks to  cos yIn(u gn )(y)H 1 ≤ C n and  cos yIn(u gn )(y)L2 ≤ 1

C √ , n

we get

1

 2 2 H(cos yIn(u gn ))L∞ ≤ CH(cos yIn(u gn ))H 1 H(cos yIn(u gn ))L2 ≤ C.

With the estimates above, we conclude u (yc )Λ4 (gn )L∞ ≤ C|α|.

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64

By the proof of Proposition 5.7 (Step 3), we get for ω

e ∈ H 2 , Λ3 ( ωe ) = Cα−2 (sin2 yc L2 ∩ sin yc L∞ ) + C sin2 yc (1 + α sin yc )L∞ . Thus, by Lebesgue’s dominated convergence theorem, we obtain u(π) 

ω

(t, α, 0) = lim − n→+∞

u(0) u(π) 

= u(0)

By Proposition 4.5, we get

Λ3 ( ωe )Λ4 (gn ) −iαct e dc u (yc )(A21 + B21 )

(u(π) − c)2 Λ3 ( ωe )e−iαct dc.   u (yc )φ1 (0, c)(A21 + B21 )

(u(π)−c)2 Λ3 (ω

e )e−iαct u (yc )φ1 (0,c)(A21 +B21 )

∈ L2yc ⊂ L1yc ⊂ L1c .



8.2. Decay estimates at critical points First of all, we need the following formula of the stream function at a critical point. Lemma 8.2. Let ω(t, x, y) be the solution to (1.2), and −Δψ = ω. Then it holds that

α, 0) = ψ(t,

1 −1

α, 0) = ∂y ψ(t,

(1 + cos yc ) sin yc Λ3 ( ωe )e−iαct dc,  2 φ1 (0, c)(A1 + B21 ) 1

sin yc Λ1 ( ωo ) −iαct e dc. (A2 + B2 )φ(0, c)

−1

Proof. By (2.1), we get u (y)iαψ = −∂t ω

− iαu(y) ω, which gives

α, 0) = ψ(t,

−1  ∂t ω

e (t, α, 0) + iαu(0) ωe (t, α, 0) .  iαu (0)

By Lemma 8.1, we get u(π) 

∂t ω

(t, α, 0) =

−iαc u(0)

which implies the first identity.

(u(π) − c)2 Λ3 ( ωe )e−iαct dc,   u (yc )φ1 (0, c)(A21 + B21 )

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Thanks to ∂y ψ(t, x, 0) = ∂y ψo (t, x, 0), we get by Lemma 5.3 that 1 ∂y ψ(t, x, 0) = 2π 1 = 2π 1 = −1

1

 o (α, 0, c)dc e−iαtc α∂y Φ

−1

1

e−iαtc

−1

α(μo− (c) − μo+ (c)) dc, φ(0, c)

sin yc Λ1 ( ωo ) −iαtc e dc. (A2 + B2 )φ(0, c)

This finishes the proof.  We denote Ko0 (c, α) =

sin yc Λ1 ( ωo )(yc ) , 2 2 (A + B )φ(0, c)

Ke0 (c, α) =

ωe ) ωe )(yc ) (1 + cos yc ) sin yc Λ3 ( J 1 (c)Λ3 ( =− 1 2 .  2 2 φ1 (0, c)(A1 + B1 ) (A1 + B21 )

Lemma 8.3. If  ωo H 2 +  ωo /u L2 ≤ 1, then we have ∂c Ko0 = αyc−2 sin yc L∞ , Proof. By the identity

ρ(c) φ(0,c)

   ρ(c)     φ(0, c)  ≤ C(1 − c),

∂c2 Ko0 = αyc−2 ((sin yc )−1 L∞ + L2 ).

= − φ11−c (0,c) and Lemma 3.4, we get

            ∂c ρ(c)  ≤  1  +  (c − 1)G  ≤ Cα(π − yc ) + C, (8.3)  φ(0, c)   φ1 (0, c)   φ1 (0, c)2  yc

and       2 2  2 ρ(c)   2G + (c − 1)∂c G   ∂c  + 2  (c − 1)G  ≤ Cα + Cα . ≤  φ(0, c)     φ1 (0, c)3  φ1 (0, c)2 yc2 sin yc

(8.4)

Then by (5.11) and Lemma 6.5, we have ∂c Λ1,1 ( ωo )(yc ) = L∞ ,

Λ1,1 ( ωo )(yc ) = sin2 yc L∞ ,

∂c2 Λ1,1 ( ωo )(yc ) =

By (5.12), Lemma 4.4 and Lemma 6.8, we obtain Λ1,2 ( ωo )(yc ) = |α| sin2 yc L∞ , ∂c2 Λ1,2 ( ωo )(yc ) =

∂c Λ1,2 ( ωo )(yc ) = |α|L∞ , |α|L∞ |α|L2 + . sin yc sin2 yc

L2 . sin yc

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66

Therefore, we obtain Λ1 ( ωo ) = αρL∞ , Notice that Ko0 (c, α) = deduce the result. 

∂c Λ1 ( ωo ) = αL∞ , ρ(c)

o) Λ1 (ω φ(0,c) sin yc (A2 +B2 ) .

Lemma 8.4. Assume that ω

e ∈ H 2 and

∂c2 Λ1 ( ωo ) = ρ−1 α(sin yc L2 + L∞ ).

(8.5)

By (8.3), (8.4), (8.5) and Proposition 4.3, we

∂y ω

e u

 ∈ H 1 . Then Ke0 c=±1 = 0 and

∂c2 Ke0 L1c ≤ C|α|2 ( ωe H 2 +  ωe /u H 1 ). Proof. We may assume  ωe H 2 +  ωe /u H 1 ≤ 1 by normalization. By Lemma 4.8, we obtain  1     2 1  C Cα J1 (c) ≤ C , ∂c J11 (c) ≤ ∂c J1 (c) ≤ , . α2 (1 + α sin yc )2 (1 + α sin yc )3 sin yc By the proof of Proposition 5.7 (Step 3), we have Λ3 ( ωe ) = α−2 (sin2 yc L2 ∩ sin yc L∞ ) + sin2 yc (1 + α sin yc )L∞ , and by Lemma 6.4, we get ∂c Λ1,1 ( ωe ) = L2 ; by Lemma 6.7 and Lemma 4.4, we get αL∞ ∂c Λ1,2 ( ωe ) = sin yc . We also have Λ1 ( ωe ) = (1 + α sin yc )L∞ . Then by Lemma 7.1, we get  ∂c Λ3 ( ωe ) = sin2 yc L2 + (1 + α sin yc )L∞ . By Lemma 6.4, we get ∂c2 Λ1,1 ( ωe ) = ∂c2 Λ1,2 ( ωe )

=

αL∞ sin3 yc

+

αL2 sin yc .

L2 sin2 yc ;

and by Lemma 6.8 and Lemma 4.4, we get

Then by Lemma 7.1, we get

  ∂c2 Λ3 ( ωe ) = 1 + α sin yc +

1  2 α2 α L + L∞ + L∞ . α sin yc (1 + α sin yc )2 sin yc

Then by Proposition 4.5, we obtain u (yc )∂c2 Ke0 =

α3 L∞ αL2 + = α2 L1 , (1 + α sin yc )6 (1 + α sin yc )4

and Ke0 = C(α) sin yc L∞ ,  which implies that Ke0 c=±1 = 0 and ∂c Ke0 = α2 L∞ . Now we are in a position to prove Theorem 1.3.



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67

α, 0). Proof. Step 1. Decay estimate of ∂y ψ(t, Let c ∈ (u(0), u(1)), which will be determined later. Without loss of generality, let us consider the case of |αt| > 1. Then by Lemma 8.2, we have

α, 0) = ∂y ψ(t,

u(π) 

Ko0 (c, α)e−iαct dc

u(0)

1 = iαt

u(π) 

∂c Ko0 (c, α)e−iαct dc

u(0)

1 = iαt

c



u(0)

−

∂c Ko0 (c, α)e−iαct dc ⎛

1 ⎜ 0  −iαc t + ⎝∂c Ko (c , α)e (αt)2

u(π) 

⎞

⎟ ∂c2 Ko0 (c, α)e−iαct dc⎠ .

c

By Lemma 8.3, we have ∂c2 Ko0 L1c (c ,u(π)) = u (yc )∂c2 Ko0 L1yc (yc ,1) ≤ Cα(yc−2 L1yc (yc ,1) + yc−1 L2yc (yc ,1) ) ≤ Cαyc−1  . Therefore, we deduce that for any yc ∈ (0, 1), c

α, 0)| ≤ Cα |∂y ψ(t, |αt|



y  dc 1 α 1  c . +C ≤ C + yc (αt)2 yc |t| αyc t2

u(0)

Taking yc = |αt|− 2 , we deduce that 1

α, 0)| ≤ Cα− 12 |t|− 32 . |∂y ψ(t,

α, 0) and ω Step 2. Decay estimates of ψ(t,

(t, α, 0). 1 0

Using the fact that ψ(t, α, 0) = −1 Ke (c, α)e−iαct dc, and Lemma 8.4, we obtain     1  1  0 −iαct 

 |ψ(t, α, 0)| =  ∂c Ke (c, α)e dc iαt   −1     1  1  1 2 0 −iαct   ≤ 2 2 ∂c Ke (c, α)e dc + 2 2 ∂c Ke0 (c, α)L∞ α t α t   −1

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68

≤ Ct−2 ( ωe H 2 +  ωe /u H 1 ). Let W (t, x, y) = ω(t, x + u(y)t, y). Then W (t, α, y) = eiαtu(y) ω

(t, α, y) and

∂t W = −iαeiαu(y)t u ψ. By Lemma 8.1, ω

(t, α, 0) → 0 as t → ∞, so does W (t, α, 0). Thus, ∞ W (t, α, 0) =

α, 0)ds. iαeiαu(0)s u (0)ψ(s,

t

Thus, if t > 1 then +∞ 

α, 0)|ds | ω (t, α, 0)| = |W (t, α, 0)| ≤ C|α| |ψ(s, t +∞ 

s−2 ( ωe H 2 +  ωe /u H 1 )ds

≤ C|α| t

≤ C|α||t|−1 ( ωe H 2 +  ωe /u H 1 ), and if t < −1 then t | ω (t, α, 0)| = |W (t, α, 0)| ≤ C|α|

α, 0)|ds |ψ(s,

−∞

≤ C|α||t|−1 ( ωe H 2 +  ωe /u H 1 ). Step 3. Decay estimate of ω(t, x, 0). By Step 1 and Step 2, we deduce that       |ω(t, x, 0)| =  ω

(t, α, 0)eiαx  ≤ | ω (t, α, 0)| ≤ C|α||t|−1  ωe (α, ·)H 3 α=0

≤ C|t|−1



|α|−2s

α=0

α=0

  1 2

α=0

|α|2+2s  ωe (α, ·)2H 3

 12

≤ C|t|−1 ωe Hxs+1 Hy3 ,

α=0

    

α, 0)eiαx  ≤

α, 0)| ≤ |V 2 (t, x, 0)| =  iαψ(t, |α||ψ(t, C|α||t|−2  ωe (α, ·)H 3 α=0

≤ C|t|−2

 α=0

|α|−2s

α=0

 12   α=0

|α|2+2s  ωe (α, ·)2H 3

α=0

 12

≤ C|t|−2 ωe Hxs+1 Hy3 ,

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69

    1 3 

α, 0)eiαx  ≤

α, 0)| ≤ |V 1 (t, x, 0)| =  ∂y ψ(t, |∂y ψ(t, C|α|− 2 |t|− 2  ωo (α, ·)H 3 α=0

 3

≤ C|t|− 2

α=0

|α|−2s

 12  

α=0

α=0

|α|2s−1  ωo (α, ·)2H 3

 12

≤ C|t|− 2 ωo  3

α=0

s− 1 2

Hx

The above results also hold with 0 replaced by π, thus all kπ(k ∈ Z).

Hy3

.



9. Wave operator To prove the enhanced dissipation, we use the wave operator method introduced in [22]. 9.1. Basic properties of wave operator Definition 9.1. Let |α| > 1. For any function ϕ(y) defined on [−π, π], we define

where ϕo =

ϕ(y)−ϕ(−y) 2

−D(ϕo )(−yc ) = D(ϕo )(yc ) =

Λ1 (ϕo )(yc ) 1 , (A2 + B2 ) 2

D(ϕe )(−yc ) = D(ϕe )(yc ) =

Λ3 (ϕe )(yc ) 1 , (A21 + B21 ) 2

and ϕe =

ϕ(y)+ϕ(−y) . 2

The operator D is called the wave operator, which has the following basic properties. Proposition 9.2. It holds that  D cos y(1 + (∂y2 − α2 )−1 )ω = cos yD(ω), D(ω)2L2

= ω, ω +

(∂y2

2 −1

−α )

(9.1)

ω.

(9.2)

Moreover, there exists a constant C independent of α so that (1 − α−2 )ω2L2 ≤ D(ω)2L2 ≤ ω2L2 ,

(9.3)

 sin yD(ω)2L2 ≥ ∂y ψ2L2 + (α2 − 1)ψ2L2 , 1 2

∂y D(ω)L2 ≤ C|α| ωH 1 ,

(9.4) 3 2

∂y2 D(ω)L2 ≤ C|α| ωH 2 ,

where −(∂y2 − α2 )ψ = ω. Proof. Step 1. Proof of (9.1). If ω is odd, then ψ is odd and ψ(0) = ψ(π) = 0. Then we have

(9.5)

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70

II1 (u ψ) π  y ψ(y  )u (y  )φ (y  , c)dy  1 yc = p.v. dy 2 φ(y, c) 0

= p.v.

π  y ψ(y  )(∂ 2 − α2 )φ(y  , c)dy  y yc φ(y, c)2

dy

0

=−

π  y ω(y  )φ(y  , c)dy  yc φ(y, c)2

π dy + p.v.

0

ψ(y)∂y φ(y, c) − u (yc )ψ(yc ) dy φ(y, c)2

0

π − p.v.

∂y ψ(y)φ(y, c) dy φ(y, c)2

0

=−

π  y ω(y  )φ(y  , c)dy  yc φ(y, c)2 0

π ψ(y)  ψ(yc ) dy − − A φ(y, c) y=0 ρ(c)

ψ(yc ) = u(yc )II1 (ω) − A − p.v. ρ(c)

π  y ω(y  )u(y  )φ (y  , c)dy  1 yc φ(y, c)2

dy.

0

Here we used the fact that for g ∈ H 2 (0, π), π $ p.v.

g(z) φ(z, c)



% g(yc )u (yc ) g(z) π + dz = + u (yc )g(yc )II(c).  φ(z, c)2 φ(z, c) z=0

0

Thus, we get  ρII1 u ψ + uω = ρuII1 (ω) − ψ(yc )A. Recall that Λ1 (ϕ) = Aϕ + u ρII1 (ϕ). This shows that for ω odd, D(u ψ + uω) = uD(ω). If ω is even, then ψ is also even and ∂y ψ(0) = ∂y ψ(π) = 0. Then we have II1 (u (y)ψ(α, y)) π  y ψ(y  )u (y  )φ (y  , c)dy  1 yc = p.v. dy φ(y, c)2 0

= p.v.

π  y ψ(y  )(∂ 2 − α2 )φ(y  , c)dy  y yc φ(y, c)2 0

dy

D. Wei et al. / Advances in Mathematics 362 (2020) 106963

=−

π  y ω(y  )φ(y  , c)dy  yc φ(y, c)2

π dy + p.v.

0

71

ψ(y)∂y φ(y, c) − u (yc )ψ(yc ) dy φ(y, c)2

0

π − p.v.

∂y ψ(y)φ(y, c) dy φ(y, c)2

0

=−

π  y ω(y  )φ(y  , c)dy  yc φ(y, c)2

dy −

0

= u(yc )II1 (ω) −

π ψ(y)  ψ(yc ) − A φ(y, c) y=0 ρ(c)

ψ(yc ) ψ(π) ψ(0) A − II1 (uω) − + , ρ(c) φ(π, c) φ(0, c)

and for j = 0, 1, jπ



Ej (u ψ) =



jπ

u ψφ1 (y, c) dy = yc

ψ(∂y2 − α2 )φ(y  , c) dy

yc

jπ =−

jπ jπ ωφ(y  , c) dy + ψ∂y φ(y  , c)y =y − ∂y ψφ(y  , c)y =y c

c

yc

= u(yc )Ej (ω) − Ej (uω) + ψ(jπ)(u(jπ) − c)∂y φ1 (jπ, c) − ψ(yc )u (yc ). Thus, we obtain Λ3 (u ψ + uω) = ρ(c)A(u ψ + uω) + u ρ2 II1 (u ψ + uω)  u (yc )(u(π) − c)2  u (yc )   E (u ψ + uω) + u ψ + uω − 0  φ1 (0, c)φ1 (0, c) u (yc )  u (yc )(u(0) − c)2  u (yc )   + (u ψ + uω) + u ψ + uω E 1 φ1 (π, c)φ1 (π, c) u (yc )  Aψ ψ(π) ψ(0)  = ρ(c)A(u ψ + uω) + u ρ2 u(yc )II1 (ω) − − + ρ(c) φ(π, c) φ(0, c)   u (yc )(u(π) − c)2 u (yc )     − uE (ω) + ψ(0)(u(0) − c)φ (0, c) − u ψ + u ψ + uω 0 1 φ1 (0, c)φ1 (0, c) u (yc )  u (yc )(u(0) − c)2  u (yc )     uE + (ω) + ψ(π)(u(π) − c)φ (π, c) − u ψ + u ψ + uω 1 1 φ1 (π, c)φ1 (π, c) u (yc ) = u(yc )Λ3 (ω). This shows that for ω even, D(u ψ + uω) = uD(ω).

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Step 2. Proof of (9.2). Let ω = ωo + ωe with ωo = ω(y)−ω(−y) and ωe = ω(y)+ω(−y) . Let (∂y2 − α2 )go = 2 2 ωo +(∂y2 −α2 )ωo , where we denote by f the complex conjugate of f . Then by Lemma 5.4, we get u(π) 

ψo , ωo +

(∂y2

− α )ωo  = −2 2

Λ1 (ωo )Λ2 (go ) dc. (A2 + B2 )u (yc )

u(0)

As go = ωo − ψo , we have Λ2 (go ) = ρII1 (u go ) + Ago = −ρII1 (u ψo + uωo ) + Ago = −ρuII1 (ωo ) + Aψo + Ago = ρu II1 (ωo ) + Aωo = Λ1 (ωo ). This shows that ωo , ωo − ψo  = −ψo , ωo + (∂y2 − α2 )ωo  u(π) 

=2

Λ1 (ωo )Λ1 (ωo ) dc = D(ωo )2L2 . (A2 + B2 )u (yc )

u(0)

Let (∂y2 − α2 )ge = ωe + (∂y2 − α2 )ωe . Then we have u(π) 

ψe , ωe +

(∂y2

− α )ωe  = −2 2

u(0)

Λ3 (ωe )Λ4 (ge ) dc. (A21 + B21 )u (yc )

Due to ge = ωe − ψe , we have u (yc )Λ4 (ge ) = Λ3 (u ge ) = −Λ3 (u ψe + uωe ) = −u(yc )Λ3 (ωe ) = u (yc )Λ3 (ωe ), thus, Λ4 (ge ) = Λ3 (ωe ) a.e. and ωe , ωe − ψe  = −ψe , ωe + (∂y2 − α2 )ωe  u(π) 

=2 u(0)

Λ3 (ωe )Λ3 (ωe ) dc = D(ωe )2L2 . (A21 + B21 )u (yc )

Thus we get D(ω)2L2 = ω, ω + (∂y2 − α2 )−1 ω.

D. Wei et al. / Advances in Mathematics 362 (2020) 106963

Step 3. Proof of (9.3) and (9.4). Let −(∂y2 − α2 )ψ = ω. Then we have ω2L2 = ∂y2 ψ2L2 + 2α2 ∂y ψ2L2 + α4 ψ2L2 ≥ α2 (∂y ψ2L2 + α2 ψ2L2 ), which gives D(ω)2L2 = ω, ω − ψ = ω2L2 − (∂y ψ2L2 + α2 ψ2L2 ) ≤ ω2L2 , D(ω)2L2 ≥ (1 − α−2 )ω2L2 . We also have  sin yD(ω)2L2 = D(ω)2L2 −  cos yD(ω)2L2 = D(ω)2L2 − D(cos y(ω − ψ))2L2 ≥ D(ω)2L2 −  cos y(ω − ψ)2L2 ≥ D(ω)2L2 − ω − ψ2L2 = ω, ω − ψ − ω − ψ2L2 = ψ, ω − ψ = ∂y ψ2L2 + (α2 − 1)ψ2L2 ≥ 0. Step 4. Proof of (9.5). By Proposition 4.3, we obtain         1 1 ∂y  ≤ C (A2 + B2 )1/2 sin yc ∂c   c (A2 + B2 )1/2   A2 + B2  C , (1 + α sin yc ) sin yc

≤

and by the proof of Proposition 5.6 (Step 2), we get for ω ∈ H 1 (T ) odd, ωH 1 = 1, Λ1 (ω)(yc ) = sin yc ((1 + |α| sin yc )L2 + |α| 2 L∞ ),  1 ∂yc Λ1 (ω)(yc ) = (1 + |α| sin yc )L2 + α 2 L∞ . 1

Thus, for ω ∈ H 1 (T ) odd, ωH 1 = 1,  ∂yc D(ω) = Λ1 (ω)(yc )∂yc

 ∂ Λ (ω)(y ) 1 y 1 c + c2 (A2 + B2 )1/2 (A + B2 )1/2

|α| 2 L∞ = L2 . 1 + α sin yc 1

= L2 + Thus, for ω ∈ H 1 (T ) odd,

D(ω)H 1 ≤ CωH 1 .

73

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74

By Proposition 4.3, we get     2 1 C  ∂y   c (A2 + B2 )1/2  ≤ (1 + α sin y ) sin2 y , c c and by the proof of Proposition 5.6 (Step 3), for ω ∈ H 2 (T ) odd, ωH 2 = 1, we get Λ1 (ω) = (sin2 yc L2 ∩ sin yc L∞ ) + |α| sin2 yc L∞ , ∂yc (Λ1 (ω)) = sin yc (L2 + αL∞ ), ∂y2c (Λ1 (ω)) = ((1 + |α| sin yc )L2 + |α|L∞ ) Thus for ω ∈ H 2 (T ) odd, ωH 2 = 1, we get ∂y2c D(ω) = Λ1 (ω)∂y2c

 (A2

   ∂y2 Λ1 (ω) 1 1 + 2 c 2 1/2 + 2∂yc Λ1 (ω)∂yc 2 1/2 2 2 1/2 +B ) (A + B ) (A + B )

1 |α| 2 L∞ |α|L∞ =L + + = α 2 L2 . 1 + α sin yc 1 + α sin yc 1

2

Thus, for ω ∈ H 2 (T ) odd, 1

D(ω)H 2 ≤ Cα 2 ωH 2 . By Proposition 4.5, we get Cα2 1 , 1 ≤ (1 + α sin yc )3 (A21 + B21 ) 2   1 Cα2 ≤ ∂yc , 1 (1 + α sin yc )3 sin yc (A21 + B21 ) 2   1 Cα2 ≤ . ∂y2c 1 2 2 (1 + α sin yc )3 sin2 yc (A1 + B1 ) 2 For ω ∈ H 1 even, ωH 1 = 1, by the proof of Proposition 5.7 (Step 2), we get Λ3 (ω) = |α|− 2 sin yc (1 + |α| sin yc )3 L2 ,  3 ∂yc Λ3 (ω) = |α|− 2 (1 + |α| sin yc )3 L2 . 3

Thus, for ω ∈ H 1 even, ωH 1 = 1,  ∂yc D(ω) = Λ3 (ω)(yc )∂yc

 ∂ Λ (ω)(y ) 1 1 y 3 c + c2 = α 2 L2 . (A21 + B21 )1/2 (A1 + B21 )1/2

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75

This shows that for ω ∈ H 1 even, 1

D(ω)H 1 ≤ C|α| 2 ωH 1 . For ω ∈ H 2 even, ωH 2 = 1, by the proof of Proposition 5.7 (Step 3), we can get ∂ymc Λ3 (ω) = α−1/2 (sin yc )2−m (1 + α sin yc )3 L2 ,

m = 0, 1, 2.

Thus, for ω ∈ H 2 even, ωH 2 = 1, ∂y2c D(ω) = Λ3 (ω)∂y2c



   ∂y2c Λ3 (ω) 1 1 + 2∂ Λ (ω)∂ + y 3 y c c (A21 + B21 )1/2 (A21 + B21 )1/2 (A21 + B21 )1/2

3

= α 2 L2 . This shows that for ω ∈ H 2 even, 3

D(ω)H 2 ≤ C|α| 2 ωH 2 . This completes the proof of the proposition.  9.2. Commutator estimate Proposition 9.3. It holds that  sin y(D(∂y2 ω) − ∂y2 D(ω))L2 ≤ C(|α|ωL2 + ∂y ωL2 ). We need the following lemmas. Lemma 9.4. It holds that  sin yc [∂y2 , ρII1,1 ]ωL2 ≤ CωH 1 . Proof. By (6.1), we get 1 1 ρII1,1 (ω) = − H(Z ◦ In(ω)) − 2 2

π

In(ω)(y  )dy  .

−π

As Z ◦ In(ω) is odd, so does ∂y2c Z ◦ In(ω). Thus, 1 sin yc ∂y2c ρII1,1 (ω) = − H(sin yc ∂y2c Z ◦ In(ω)) 2 Then we get

D. Wei et al. / Advances in Mathematics 362 (2020) 106963

76

1 sin yc = − H(sin yc [∂y2c , Z ◦ In](ω)) + 2 2

sin yc [∂y2 , ρII1,1 ]ω

π

In(∂y2 ω)(y  )dy  .

−π

On one hand, we have 1 2

π

π In(∂y2 ω)(y)dy

π In(∂y2 ω)(y)dy

=

−π

=

0

y ω  (y) − ω  (0) − sin2 (ω  (π) − ω  (0))dy 2

0

= ω(π) − ω(0) − π(ω  (0) + ω  (π))/2. On the other hand,  sin yc Z ◦ In(∂y2 ω) − ∂y2 Z ◦ In(ω) =

ω  (0) − ω  (π) 2ω 2 cos y Int(ω)(π) ω  (0) + ω  (π) cos y + − + In(ω)(y) − , 2 2 sin y 2 sin2 y

and H(cos y) = 2π sin y. Then we deduce that sin yc [∂y2 , ρII1,1 ]ω =

 1  2ω 2 cos y H − + In(ω)(y) + sin yc (ω(π) − ω(0)), 2 sin y sin2 y

from which, it follows that      2 cos y 2ω  ≤ C H − + In(ω)(y)  2  2 + CωL∞ sin y sin y L    2ω  2 cos y  ≤C − sin y + sin2 y In(ω)(y) 2 + CωL∞ L     y   ω cos y    ≤ C − + ω(y )dy  2  sin y sin y  2 π

   sin yc [∂y2 , ρII1,1 ]ω 

L2

0

L (0, 2 )

    y   ω cos y   +C − + ω(y )dy  sin y sin2 y    π

+ CωL∞ . L2 [ π 2 ,π]

Using the fact that cos y ω + − sin y sin2 y cos y ω + − sin y sin2 y

y

−1  ω(y )dy = sin2 y 



y 0

y

y

π

−1  ω(y )dy = sin2 y 







y

sin y ω (y )dy +

0 



0 







y

sin y ω (y )dy + π

ω(y  )(cos y  − cos y)dy  ,

π

ω(y  )(cos y  − cos y)dy  ,

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77

and Hardy’s inequality, we infer that     y   ω cos y   − ω(y )dy +  sin y sin2 y    0

L2 (0, π 2)

    y   ω cos y   + − ω(y )dy +  sin y sin2 y    π

≤ CωH 1 . L2 ( π 2 ,π)

Summing up, we conclude the lemma.  Lemma 9.5. It holds that 1

| sin3 yc [∂y2c , L0 ]ϕ| ≤ Cα 2 (ϕH 1 + αϕL2 ). Proof. By (6.6), we get ∂y2c L0 (ϕ) = u (yc )∂c (L0 (ϕ ) − I0,1 (ϕ) + I0,0 (ϕ)) + u (yc )∂c (u (yc )L1 (ϕ)) = L0 (ϕ ) − I0,1 (ϕ ) + I0,0 (ϕ ) + u (yc )L1 (ϕ ) − u (yc )∂c I0,1 (ϕ) + u (yc )∂c I0,0 (ϕ) + u (yc )L1 (ϕ) + u (yc ) (L1 (ϕ ) − I1,1 (ϕ) + I1,0 (ϕ)) + u (yc )2 L2 (ϕ), which gives  [∂y2c , L0 ](ϕ) = −I0,1 (ϕ ) + I0,0 (ϕ ) + u (yc ) − I1,1 (ϕ) + I1,0 (ϕ) − u (yc )∂c I0,1 (ϕ) + u (yc )∂c I0,0 (ϕ) + u (yc )L1 (ϕ) + 2u (yc )L1 (ϕ ) + u (yc )2 L2 (ϕ). By (6.3), we have for j = 0, 1  jπ      min{α2 |jπ − yc |2 , 1}  |I0,j (ϕ)(c)| ≤ C  ϕ(y)dy  2    (cos jπ − cos yc ) u (yc ) yc

1

≤C

|jπ − yc | 2 min{α2 |jπ − yc |2 , 1} ϕL2 (cos jπ − cos yc )2 u (yc )

≤C

min{α2 |jπ − yc |2 , 1} ϕL2 , |jπ − yc |7/2 u (yc )

which gives | sin3 yc I0,j (ϕ )| ≤ Cα 2 ϕ L2 . 1

We get by (6.7) that | sin4 yc I1,j (ϕ)| ≤ CαϕL∞ ≤ Cα 2 (ϕ L2 + αϕL2 ). 1

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78

Using (6.13) with γ = 1, we get | sin4 yc ∂c I0,j (ϕ)| ≤ CαϕL∞ ≤ Cα 2 (ϕ L2 + αϕL2 ). 1

By (6.4), we get 1 | sin3 yc L1 (ϕ)| + | sin5 yc L2 (ϕ)| ≤ CαϕL∞ ≤ Cα 2 ϕ L2 + αϕL2 , and by Lemma 6.6, we get |u (yc )4 L1 (ϕ )| ≤ Cα 2 ϕ L2 . 1

Summing up, we conclude the lemma.  Lemma 9.6. For ω ∈ H 2 [0, π], ω  (0) = ω  (π) = 0, we have sin yc [∂y2c , Λ3,1 ](ω) = C(α−2 + ρ)(ωH 1 + αωL2 )(α 2 L∞ + L2 ). 1

Proof. We have −Ej (∂y2 ω)

jπ



= ω (yc ) +

∂y φ1 (y, c)∂y ωdy. yc

Then by (7.8) and (7.11), we get ∂y2c Λ3,1 (ω) − Λ3,1 (∂y2 ω) J  J  j j = ∂y2c  (u E1−j (ω) + u ω) + 2∂yc  ∂yc (u E1−j (ω) + u ω) u u 1 Jj  2    2  2 +  ∂yc (u E1−j (ω) + u ω) − (u E1−j (∂y ω) + u ∂y ω)  u j=0 J  j = ∂y2c  (u E1−j (ω) + u ω) u (1−j)π   J   j      + 2∂yc  u E1−j (ω) + u ω + u u ∂c φ1 (y, c)ω(y)dy u yc

Jj  +  uE1−j (ω) + 2u ∂y ω + u u

(1−j)π 

∂y φ1 (y, c)∂y ωdy − u ω

yc

+ (u

 2

2

(1−j)π 

− 2u )

1 ∂c2 φ1 (y, c)ω(y)dy 

(1−j)π 

∂c φ1 (y, c)ω(y)dy + u u yc

= T1 + T2 + T3 .

 2 

yc

j=0

D. Wei et al. / Advances in Mathematics 362 (2020) 106963

By (7.2)-(7.5) and Lemma 4.8, we get sin yc T1 = α−2 (ωH 1 + αωL2 )(α 2 L∞ + L2 ). 1

Using u (y) = − sin y, (7.1), (7.9), (7.10) and Lemma 4.8, we deduce that | sin yc T2 | ≤ Cα−2 (ωH 1 + αωL2 )(α 2 L∞ + L2 ). 1

By (7.1) and Lemma 4.8, we get   Jj E1−j (ω) + |Jj ω| ≤ Cα−2 ωL∞ ≤ Cα−2 (ωH 1 + αωL2 ),   Jj ∂y ω  2 ≤ Cα−2 ωH 1 , L and   Jj

(1−j)π 

(1−j)π   1 1  ∂y φ1 (y, c)∂y ωdy  ≤ C|Jj | αα 2 |y − yc | 2 |∂y ω|dyφ1 ((1 − j)π, c)

yc

yc

≤C

sin yc ∂y ωL2 . 1 α2

By (7.9), (7.10) and Lemma 4.8, we get   Jj

(1−j)π 

 φ1 ((1 − j)π, c)  ∂c φ1 (y, c)ω(y)dy  ≤ C|Jj |αωL∞ ≤ Cα−1 ωL∞ . |jπ − yc |

yc

By (7.12), (7.13) and Lemma 4.8, we get (1−j)π      2 ∂c2 φ1 (y, c)ω(y)dy  ≤ C|Jj |α2 ωL∞ φ1 ((1 − j)π, c)|(1 − j)π − yc | Jj u yc

≤ C sin yc ωL∞ . Notice that ωL∞ ≤ Cα− 2 (ωH 1 + αωL2 ), 1

sin yc ≤ α−1 + αρ.

Then we conclude that sin yc T3 = (α−2 + ρ)(ωH 1 + αωL2 )(α 2 L∞ + L2 ). 1

This proves the lemma. 

79

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80

Now we are in a position to prove the proposition. Proof. Step 1. The odd case. Let ω ∈ H 2 be odd and ω(0) = ω(π) = 0. Direct calculation gives ∂y2c D(ω) − D(∂y2 ω)   u [∂ 2 , ρII1,1 ]ω   u u yc + 2∂yc ρII1,1 (ω) ∂yc + 1 1 1 2 2 2 2 2 2 2 (A + B ) (A + B ) (A + B2 ) 2     ρu [∂ 2 , L0 ]ω ρu ρu yc + L0 (ω)∂y2c + 2∂yc L0 (ω)∂yc + 1 1 1 (A2 + B2 ) 2 (A2 + B2 ) 2 (A2 + B2 ) 2     ρu II ρu II + ω∂y2c + 2∂yc ω∂yc 1 1 (A2 + B2 ) 2 (A2 + B2 ) 2

= ρII1,1 (ω)∂y2c



= I1 + I2 + I3 + I4 + I5 + I6 + I7 + I8 . By Lemma 6.5 and Proposition 4.3, we get     sin yc I1 L2 ≤ C sin yc II1,1 L2 ρ∂y2c

 u   ∞ ≤ CωH 1 , 1 2 2 L 2 (A + B )

and by Lemma 6.4 and Proposition 4.3,     sin yc I2 L2 ≤ CρII1,1 H 1  sin yc ∂yc

 u  ≤ CωH 1 ,  1 (A2 + B2 ) 2 L∞

and by Lemma 9.4 and Proposition 4.3,  sin yc I3 L2 ≤ C sin yc [∂y2 , ρII1,1 ]ωL2 ≤ CωH 1 , and by Lemma 6.7 and Proposition 4.3,  1  1   sin yc I4 L2 ≤ Cα− 2  sin yc L0 (ω)L∞ α 2 ∂y2c   ≤ CωH 1   sin yc I5 L2

 ρu   2 1 2 2 L 2 (A + B )

 α2   ≤ CωH 1 , (1 + α sin yc ) L2  α 12   1 ρu   ≤ Cα− 2  sin2 yc ∂yc L0 (ω)L∞  ∂yc  1 sin yc (A2 + B2 ) 2 L2 ≤ Cω

By Lemma 9.5, we get

H1

  

1

 α2   ≤ CωH 1 . (1 + α sin yc ) L2 1

D. Wei et al. / Advances in Mathematics 362 (2020) 106963

 1  sin yc I6 L2 ≤ Cα− 2 

81

 α2    sin3 yc [∂y2c , L0 ]ωL∞ ≤ C(ωH 1 + αωL2 ) (1 + α sin yc ) L2 1

By Lemma 4.4 and Proposition 4.3, we get     sin yc I7 L2 ≤ Cω/u L2 ρ∂y2c  sin yc I8 L2

 ρu II  ≤ CωH 1 ,  1 (A2 + B2 ) 2 L∞    ρu II   ≤ C∂y ωL2  sin yc ∂yc  ∞ ≤ C∂y ωL2 . 1 2 2 L 2 (A + B )

Summing up, we conclude the odd case. Step 2. The even case. Let ω ∈ H 2 be even, ω  (0) = ω  (π) = 0. Direct calculation gives ∂y2c D(ω) − D(∂y2 ω) =

   ρu ρu 2 ∂yc ρII1,1 (ω) 1 [∂yc , ρII1,1 ]ω + 2∂yc 1 2 2 2 2 2 2 (A1 + B1 ) (A1 + B1 )    ρ2 u ρu 2 + ∂y2c ρII (ω) + 1,1 1 1 [∂yc , L0 ]ω 2 2 2 2 2 2 (A1 + B1 ) (A1 + B1 )     2   ρ u ρ2 u 2 + 2∂yc ∂ L (ω) + ∂ L0 (ω) y 0 1 1 y c c (A21 + B21 ) 2 (A21 + B21 ) 2  ρ2 u II   ρ2 u II  + ω∂y2c + 2∂ ∂yc ω y 1 1 c (A21 + B21 ) 2 (A21 + B21 ) 2     [∂y2c , Λ3,1 ]ω 1 1 + ∂y2c Λ (ω) + ∂ ∂ Λ (ω) + 3,1 y y 3,1 1 1 1 c c (A21 + B21 ) 2 (A21 + B21 ) 2 (A21 + B21 ) 2

  = I1 + I2 + I3 + I4 + I5 + I6 + I7 + I8 + I9 + I10 + I11 .

By Proposition 4.5 and Lemma 9.4, we get  sin yc I1 L2

    ρu   ≤C 2 1   (A1 + B21 ) 2 

   sin yc [∂y2 , ρII1,1 ]ω  2 ≤ C(αωL2 + ∂y ωL2 ), c L L∞

and by Proposition 4.5 and Lemma 6.4,  sin yc I2 L2  sin yc I3 L2

        ρu    ≤ C sin yc ∂yc ∂yc ρII1,1 (ω) L2 ≤ CωH 1 ,  1 2 2   2 (A1 + B1 ) L∞        ρu    ≤ C sin yc ∂y2c ρII1,1 (ω)L2 ≤ CαωL2 .  1 2 2   (A1 + B1 ) 2 L∞

By Proposition 4.5 and Lemma 9.5, we get

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82

 sin yc I4 L2

    ρu   ≤C 2   (A1 + B21 ) 12 

 sin3 yc [∂y2c , L0 ]ωL∞ ≤ C(αωL2 + ∂y ωL2 ),

L2

and by Proposition 4.5 and Lemma 6.7,    1     ρ2 u    2   sin yc I5 L2 ≤ C  ∂yc sin yc ∂yc L0 (ω) L∞  1 2 2  sin yc (A1 + B1 ) 2 L2  1  ≤ C α 2 ωL∞ + ω  L2 ≤ C αωL2 + ω  L2 , and by Proposition 4.5 and Lemma 6.6,    1   ρ2 u    2  sin yc I6 L2 ≤ C  ∂   sin yc yc (A21 + B21 ) 12 

  ρL0 (ω)

L∞

L2

≤ CαωL2 . By Proposition 4.5 and Lemma 4.4, we get     ρ2 u II     sin yc I7 L2 ≤ CωL2 sin yc ∂y2c ≤ CαωL2 ,  1 2 2  2 (A1 + B1 ) L∞     ρ2 u II     sin yc I8 L2 ≤ C∂y ωL2 sin yc ∂yc ≤ C∂y ωL2 .  1 2 2  2 (A1 + B1 ) L∞ By Proposition 4.5 and Remark 7.2, we get (L2 + α 2 L∞ ) (αωL2 + ∂y ωL2 ) = L2 , (1 + α sin yc )3 1

sin yc I9 =

(L2 + α 2 L∞ ) (αωL2 + ∂y ωL2 ) = L2 . (1 + α sin yc )3 1

 sin yc I10 =

By Proposition 4.5 and Lemma 9.6, we get (1 + α2 ρ)(L2 + α 2 L∞ ) = (αωL2 + ∂y ωL2 ) = L2 . (1 + α sin yc )3 1

 sin yc I11

Summing up, we conclude the even case.  10. Linear enhanced dissipation 10.1. Decay estimates on the model without nonlocal term In this subsection, we consider a toy model without nonlocal term: ∂t ω − ν∂y2 ω − iae−νt (cos y)ω = 0.

(10.1)

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Lemma 10.1. Let 0 < ν < |a|, a ∈ R, and ω(t, y) solve (10.1) for t ≥ 0, y ∈ T . Then ω(t)L2 ≤ ω(0)L2 and  sin yω(t)L2 ≤ C(νt3 a2 )− 2 ω(0)L2 1

for 0 < t ≤ (ν|a|)− 2 . 1

Proof. First of all, we have ∂t ω2L2 = 2Re∂t ω, ω = 2Reν∂y2 ω + iae−νt cos yω, ω = −2ν∂y ω2L2 ≤ 0, which gives ω(t)L2 ≤ ω(0)L2 . −νs −νt Let u(y) = − cos y, γ(t, s) = a e −e . Then γ(s, s) = 0, ∂t γ = ae−νt , and ν ∂t (eiγu ω) = eiγu (ωt + iuγt ω) = νeiγu ∂y2 ω, from which, we infer that ∂t ∂y (eiγu ω)2L2 = 2Re∂t ∂y (eiγu ω), ∂y (eiγu ω) = −2Re∂t (eiγu ω), ∂y2 (eiγu ω) = −2Reνeiγu ∂y2 ω, ∂y2 (eiγu ω) = −2νRe∂y2 ω, e−iγu ∂y2 (eiγu ω) = −2νRe∂y2 ω, ∂y2 ω + 2iγu ∂y ω + (iγu − γ 2 u 2 )ω = −2ν∂y2 ω + iγu ∂y ω + iγu ω/22L2 + 2νγ 2 u ∂y ω + u ω/22L2 + 2νγ 2 Re∂y2 ω, u 2 ω. For the last term, we get by integration by parts that Re∂y2 ω, u 2 ω = −Re∂y ω, ∂y (u 2 ω) = −Re∂y ω, u 2 ∂y ω + 2u u ω = −Reu ∂y ω, u ∂y ω + 2u ω = −u ∂y ω + u ω/22L2 + u ω/22L2 − Reu ∂y ω, u ω, where −2Reu ∂y ω, u ω = −

π

u u (∂y ωω + ∂y ωω)dy

−π

π =− −π

 

π

u u ∂y |ω| dy = 2

−π

(u u ) |ω|2 dy.

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Putting these identities above together and using the fact that (u )2 /2 + (u u ) = 3(u )2 /2 + u u = 3(cos y)2 /2 − (sin y)2 ≤ 3/2, we deduce that  ∂t ∂y (eiγu ω)2L2 ≤ 2νγ 2 u ω/22L2 − Reu ∂y ω, u ω π = νγ

2



−π

3 (u )2 /2 + (u u ) |ω|2 dy ≤ νγ 2 ω2L2 . 2

Therefore, for 0 < s < t,

∂y (e

iγu

ω)(t, s)2L2

≤ ∂y (e

iγu

ω)(s, s)2L2

3 + ν 2

t γ(τ, s)2 ω(τ )2L2 dτ

(10.2)

s

≤

∂y ω(s)2L2

3 + νω(0)2L2 2

t γ(τ, s)2 dτ. s

Thanks to ∂y (eiγu ω) = eiγu (iγu ω + ∂y ω), ∂y (eiγu ω)2L2 = iγu ω + ∂y ω2L2 . Then by (10.2), we get t

t



iγ(t, s)u ω(t) +

∂y ω(t)2L2 ds

∂y (eiγu ω)(t, s)2L2 ds

=

0

0

t ∂y ω(s)2L2 ds

≤

3 + νω(0)2L2 2

0

t t γ(τ, s)2 dτ ds. 0

s

We find that t

iγ(t, s)u ω(t) + ∂y ω(t)2L2 ds

0

t 2

=

γ(t, s) dsu



t ω(t)2L2

+2

0

  = γ0 (t)u ω(t)2L2 + ti

0

t

2  γ(t, s)dsu ω(t)/t + ∂y ω(t) , L2

0

where

γ(t, s)dsReiu ω(t), ∂y ω(t) + t∂y ω(t)2L2

D. Wei et al. / Advances in Mathematics 362 (2020) 106963

t γ(t, s) ds − 2

γ0 (t) = 0

 t

85

2 γ(t, s)ds /t.

0

Moreover, t t

1 γ(τ, s) dτ ds = 2

t t

2

s

0

t

1 γ(τ, s) dsdτ = 2

t t (γ(t, s) − γ(t, τ ))2 dsdτ

2

0

0

0

0

t (γ(t, s)2 /2 + γ(t, τ )2 /2 − γ(t, s)γ(t, τ ))dsdτ = tγ0 (t),

= 0

0

and t ∂y ω(s)2L2 ds =

1 ω(0)2L2 − ω(t)2L2 . 2ν

0

Therefore, we obtain γ0 (t)u ω(t)2L2 ≤

1 3 (ω(0)2L2 − ω(t)2L2 ) + νtγ0 (t)ω(0)2L2 , 2ν 2

which gives u ω(t)2L2 ≤ (1/(2νγ0 (t)) + 3νt/2) ω(0)2L2 . Now, we give a lower bound of γ0 (t). As γ(τ, s) = 0 < s < τ < t, we have t t

t t γ(τ, s) dτ ds ≥ 2

tγ0 (t) = 0

s

0

τ s



ae−νt dt ≥ a(τ − s)e−νt for

a2 (τ − s)2 e−2νt dτ ds = a2 e−2νt

t4 , 12

s

which gives 1 1 2νt 12 6e2νt ≤ e = . 2νγ0 (t) 2νa2 t3 νa2 t3 For 0 < t ≤ (ν|a|)− 2 , 0 < ν < |a|, we have νt ≤ 1, νt ≤ 1/(νa2 t3 ). Then we conclude that 1

u ω(t)2L2 ≤ (6e2 + 2)ω(0)2L2 /(νa2 t3 ). This completes the proof. 

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Now, we deal with the inhomogeneous equation. Lemma 10.2. Let 0 < ν < |a|, a ∈ R, and ω(t, y) solve ∂t ω − ν∂y2 ω − iae−νt (cos y)ω = f.

(10.3)

Then we have t

C

 sin yω(t)L2 ≤ √ sup ω(s)L2 + νt3 a2 0≤s≤t

 sin yf (s)L2 ds 0

for 0 < t ≤ (ν|a|)− 2 . 1

Proof. We decompose ω = ω1 + ω2 , where ∂t ω1 − ν∂y2 ω1 − iae−νt (cos y)ω1 = 0, ω1 (0) = ω(0), ∂t ω2 − ν∂y2 ω2 − iae−νt (cos y)ω2 = f, ω2 (0) = 0. By Lemma 10.1, we have ω1 (t)L2 ≤ ω1 (0)L2 = ω(0)L2 , and  sin yω1 (t)L2 ≤ C(νt3 a2 )− 2 ω1 (0)L2 = C(νt3 a2 )− 2 ω(0)L2 1

1

for 0 < t ≤ (ν|a|)− 2 . Let A(t) = sup ω(s)L2 . Then 1

0≤s≤t

ω2 (t)L2 ≤ ω1 (t)L2 + ω(t)L2 ≤ 2A(t). For any b > 0, we have d  sin yω2 2L2 = 2Resin y∂t ω2 , sin yω2  dt = 2Resin y(ν∂y2 ω2 + iae−νt cos yω2 + f ), sin y ω2  = −2ν sin y ∂y ω2 + cos y ω2 2L2 + 2ν cos y ω2 2L2 + 2Resin yf, sin y ω2  ≤ 2νω2 2L2 + 2 sin yf L2  sin yω2 L2 1   ≤ 2  sin yf L2 + 4νA(t)2 /b  sin yω2 2L2 + b2 2 , therefore, 1 d  sin yω2 2L2 + b2 2 ≤  sin yf L2 + 4νA(t)2 /b, dt which implies that

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87

1

 sin yω2 (t)L2 ≤ ( sin yω2 (t)2L2 + b2 ) 2 t ≤b+

( sin yf (s)L2 + 4νA(s)2 /b)ds 0

t  sin yf (s)L2 ds.

≤ b + 4νtA(t) /b + 2

0

As it is true for any b > 0, we minimize the right hand side to obtain t

√

 sin yω2 (t)L2 ≤ 4 νtA(t) +

 sin yf (s)L2 ds. 0

For 0 < t ≤ (ν|a|)− 2 , we have 1

√

νt ≤ (νt3 a2 )− 2 . Then 1

 sin yω(t)L2 ≤  sin yω1 (t)L2 +  sin yω2 (t)L2 3 2 − 12

≤ C(νt a )

t

√

ω(0)L2 + 4 νtA(t) +

 sin yf (s)L2 ds 0

≤ C(νt3 a2 )− 2 A(t) + 1

t  sin yf (s)L2 ds. 0

This gives our result.  10.2. Decay estimates on the model with nonlocal term Lemma 10.3. Let 0 < ν < |a|, |α| > 1, a, α ∈ R, and ω(t, y) solve ∂t ω − ν∂y2 ω − iae−νt cos y(ω − ψ) = 0,

−(∂y2 − α2 )ψ = ω.

Then for 0 < t ≤ (ν|a|)− 2 , να2 t < 1, 1

−ψ(t), ω(t) − ψ(t) ≤

C ω(0), ω(0) − ψ(0). νt3 a2

Proof. By (10.4), we have ∂t ω, ω − ψ = ωt , ω − ψ + ω, ωt  + (∂y2 − α2 )ψ, ψt  = ωt , ω − ψ + ω, ωt  − ψ, ωt  = −2ν∂y ω, ∂y ω − ∂y ψ.

(10.4)

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88

By Proposition 9.2, we have D(ω)2L2 = ω, ω − ψ. Using the facts that ∂y ω2L2 = ∂y3 ψ2L2 + α4 ∂y ψ2L2 + 2α2 ∂y2 ψ2L2 ≥ α4 ∂y ψ2L2 + α2 ∂y2 ψ2L2 , −∂y ω, ∂y ψ = −∂y2 ψ2L2 − α2 ∂y ψ2L2 , we deduce that ∂y ω, ∂y ω − ∂y ψ = ∂y ω2L2 − ∂y ω, ∂y ψ ≥ (1 − α−2 )∂y ω2L2 .

(10.5)

Then we obtain ∂t D(ω)(t)2L2 ≤ 0, thus D(ω)(t)L2 ≤ D(ω)(0)L2 and t ∂y ω(s)2L2 ≤

2ν 0

α2 D(ω)(0)2L2 . −1

α2

By Proposition 9.2, D(ω) satisfies (10.3) with f (s, y) = ν[D, ∂y2 ]ω. Thus by 1 Lemma 10.2, we know that for 0 < t ≤ (ν|a|)− 2 , 3 2 − 12

 sin yD(ω)(t)L2 ≤ C(νt a )

t D(ω)(0)L2 +

ν sin y[D, ∂y2 ]ω(s)L2 ds. 0

Again by Proposition 9.2, we have t ν sin y [D, ∂y2 ]ω(s)L2 ds 0

t νC(|α|ω(s)L2 + ∂y ω(s)L2 )ds

≤ 0

&

t ≤

νC|α| 0

&

≤ Cνt|α|

1 α2 D(ω)(s)L2 ds + Cνt 2 ∂y ωL2t (L2 ) 2 α −1

1 α2 D(ω)(0)L2 + Cνt 2 α2 − 1

&

α2 /(2ν) D(ω)(0)L2 . α2 − 1

Therefore, for 0 < t ≤ (ν|a|)− 2 , 1

 1 1  sin yD(ω)(t)L2 ≤ C (νt3 a2 )− 2 + νt|α| + (νt) 2 D(ω)(0)L2 . Thus, for 0 < t ≤ (ν|a|)− 2 , να2 t < 1, we have νt|α| ≤ (νt) 2 ≤ (νt3 a2 )− 2 . As −ψ, ω = −∂y ψ2L2 − α2 ψ2L2 , then by Proposition 9.2, 1

1

1

D. Wei et al. / Advances in Mathematics 362 (2020) 106963

ψ, ω − ψ(t) = (α2 − 1)ψ2L2 + ∂y ψ2L2 ≤  sin y D(ω)(t)2L2 ≤ CD(ω)(0)2L2 /(νt3 a2 ) = Cω, ω − ψ(0)/(νt3 a2 ). The proof is completed.  Let −(∂y2 − α2 )ψ = ω

. We introduce

K1 (t, α) =  ω, ω

− ψ(t, α),

K2 (t, α) = −(∂y2 − α2 ) ω, ω

− ψ(t, α),

ω

K3 (t, α) = ψ,

− ψ(t, α). Lemma 10.4. It holds that K1 = D( ω )2L2 ≥ α2 K3 , K2 = (α2 − 1) ω 2L2 + ∂y ω

2L2 ≥ α2 K1 ,

2 2 + ∂y ψ

2 2 ≥ 0, K3 = (α2 − 1)ψ L L K2 K3 ≥ K12 . Proof. By (10.6) and Proposition 9.2, we get D( ω )2L2 = K1 ≥ (1 − α−2 ) ω 2L2

2 2 + α2 ∂y ψ

2 2 + α4 ψ

2 2) = (1 − α−2 )(∂y2 ψ L L L  2 2 2 2 2

≥ α (α − 1)ψL2 + ∂y ψL2 = α K3 ≥ 0. It is easy to check that

= ∂y ω

+ α2  K2 = −(∂y2 − α2 ) ω, ω

− ψ

, ∂y ω

− ∂y ψ ω, ω

− ψ, which along with (10.5) gives K2 ≥ α2 K1 ≥ 0. Using the facts that

2 2 + α2 ∂y ψ

2 2 + α4 ψ

2 2,  ω 2L2 = ∂y2 ψ L L L

2 2 + α2 ∂ 2 ψ

2 2 + α4 ∂y ψ

2 2, ∂y ω

2L2 = ∂y3 ψ y L L L we can deduce that  

2 2 + ∂y ψ

22 K2 K3 = (α2 − 1) ω 2L2 + ∂y ω

2L2 (α2 − 1)ψ L L  

+ ∂y ω

2 = 

2 ≥ (α2 − 1) ω , ψ

, ∂y ψ ω , (α2 − 1)ψ − ∂y2 ψ

2 = K 2. =  ω, ω

− ψ 1



89

(10.6)

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90

10.3. Proof of Theorem 1.4 Proof. We first consider the case of δ < 1. Taking Fourier transform in x to (1.5), we obtain ∂t ω

+ Lν (α, t) ω = 0, where  Lν (α, t) = ν(−∂y2 + α2 ) − iαa0 e−νt cos y 1 + (∂y2 − α2 )−1 . Let −(∂y2 − α2 )ψ = ω

. Then we have

= 

ω

∂t  ω, ω

− ψ ωt , ω

 +  ω, ω

t  − ψ,

t  −  ωt , ψ

− 2να2 

= −2ν∂y ω

, ∂y ω

− ∂y ψ ω, ω

− ψ. This gives by Lemma 10.4 that ∂t K1 = −2νK2 ≤ −2να2 K1 , which implies that

K1 (t, α) ≤ e

−2να2 t

T 2νK2 (s, α)ds = K1 (0, α) − K1 (T, α).

K1 (0, α),

(10.7)

0

Now fix α and consider the following cases. Case 1. ν ≥ a0 |α|e−τ or ν|α|3 ≥ 1. 1

1

Due to |α| > 1, we have ν 3 + ν 2 ≤ Cνα2 . Then by Lemma 10.4 and (10.7), we get α K3 (t, α) ≤ K1 (t, α) ≤ e

−2να2 t

√

Ce−να t Ce−c νt K1 (0, α) ≤ K (0, α) ≤ K1 (0, α). 1 1 + νt3 1 + νt3 2

2

Case 2: ν < a0 |α|e−τ and ν|α|3 < 1.

 2 1 Take t1 = (ν|α|)− 2 , then νt1 α2 < 1. Fix 0 ≤ t0 ≤ τ /ν, then eνα t ω

(t0 + να2 t −νt0 t, α, y), e ψ(t0 + t, α, y) solves (10.4) with a = a0 αe . By our assumption, 0 < ν < |a|. Then by Lemma 10.3, we obtain 2

e2να

t1

K3 (t0 + t1 , α) 2

= eνα ≤

t1

2

0 + t1 , α), eνα2 t1 ω

0 + t1 , α) ψ(t

(t0 + t1 , α) − eνα s ψ(t

2 2 2 C

0 , α) eνα t1 ω

(t0 , α), eνα t1 ω

(t0 , α) − eνα t1 ψ(t νt31 a2

D. Wei et al. / Advances in Mathematics 362 (2020) 106963

≤

91

C C K1 (t0 , α) ≤ 3 2 K1 (t0 , α) = CK1 (t0 , α)t1 ν. νt31 a2 νt1 α

Thus, there exists an absolute constant c0 ∈ (0, 1) so that c0 ec0 K3 (t + t1 , α) ≤ K1 (t, α)t1 ν. Set M

=

(10.8)

max ec0 t/t1 K1 (t, α). Then there exists t2

0≤t≤τ /ν

∈

[0, τ /ν] so that

ec0 t2 /t1 K1 (t2 , α) = M . Case 2.1 t2 > t1 . Then ∂t (ec0 t/t1 K1 (t, α))|t=t2 ≥ 0. Since  ∂t (ec0 t/t1 K1 (t, α)) = ec0 t/t1 c0 /t1 K1 (t, α) + ∂t K1 (t, α)  = ec0 t/t1 (c0 /t1 )K1 (t, α) − 2νK2 (t, α) , we obtain K2 (t2 , α) ≤

c0 K1 (t2 , α). 2νt1

(10.9)

By (10.8), we get c0 ec0 K3 (t2 , α) ≤ K1 (t2 − t1 , α)t1 ν ≤ e−c0 (t2 −t1 )/t1 M t1 ν = e−c0 (t2 −t1 )/t1 ec0 t2 /t1 K1 (t2 , α)t1 ν = ec0 K1 (t2 , α)t1 ν, which gives K3 (t2 , α) ≤ K1 (t2 , α)t1 ν/c0 .

(10.10)

It follows from (10.9) and (10.10) that K2 (t2 , α)K3 (t2 , α) ≤

1 K1 (t2 , α)2 . 2

On the other hand, K2 K3 ≥ K12 by Lemma 10.4. Then we conclude that K1 (t2 , α) = 0, which implies M = 0 ≤ ec0 K1 (0, α). Case 2.2 t2 ≤ t1 . Then M = ec0 t2 /t1 K1 (t2 , α) ≤ ec0 K1 (t2 , α) ≤ ec0 K1 (0, α). Therefore, combining Case 2.1 and Case 2.2 to deduce that for 0 ≤ t ≤ τ /ν, K1 (t, α) ≤ e−c0 t/t1 M ≤ ec0 −c0 t/t1 K1 (0, α) = ec0 −c0

'

ν|α|t

K1 (0, α).

(10.11)

D. Wei et al. / Advances in Mathematics 362 (2020) 106963

92

 2 2

α, y) solves (10.4) with a = a0 α and |α| > 1, we deduce As eνα t ω

(t, α, y), eνα t ψ(t, 1 from Lemma 10.3 that 0 < t ≤ (ν|α|)− 2 and να2 t < 1, 2

e2να t K3 (t, α) ≤

C Ce−c0 t/(2t1 ) K (0, α) ≤ K1 (0, α) 1 νt3 a2 νt3 α2

for 0 ≤ t ≤ t1 . For t1 ≤ t ≤ τ /ν, by (10.8), we have c0 ec0 K3 (t, α) ≤ K1 (t − t1 , α)t1 ν ≤ ec0 −c0 (t−t1 )/t1 K1 (0, α)t1 ν, which gives c0 −c0 t/t1 α2 K3 (t, α) ≤ c−1 K1 (0, α)t1 να2 0 e c0 −c0 t/t1 = c−1 K1 (0, α)t1 ν/(t21 ν)2 0 e c0 −c0 t/t1 = c−1 K1 (0, α)/(t31 ν) ≤ Ce−c0 t/(2t1 ) K1 (0, α)/(t3 ν). 0 e

Therefore, for 0 ≤ t ≤ τ /ν, α2 K3 (t, α) ≤ Ce−c0 t/(2t1 ) K1 (0, α)/(νt3 ). While by (10.11), α2 K3 (t, α) ≤ K1 (t, α) ≤ ec0 −c0 t/t1 K1 (0, α). This shows that for 0 ≤ t ≤ τ /ν, '

−c0 t/(2t1 )

α K3 (t, α) ≤ Ce 2

Ce−c0 ν|α|t/2 K1 (0, α)/(1 + νt ) = K1 (0, α). 1 + νt3 3

Combining Case 1 and Case 2, we deduce that there exists an absolute constant c1 ∈ (0, 1) so that for 0 ≤ t ≤ τ /ν, K1 (t, α) ≤ Ce−2c1 α2 K3 (t, α) ≤

√ νt

−2c1

K1 (0, α), √ νt

Ce K1 (0, α). 1 + νt3

By Proposition 9.2, we have for 0 ≤ t ≤ τ /ν, ω(t)2L2 =



 ω (t, α)2L2 ≤

α=0

=

 α=0

 α=0

α2 D( ω )(t, α)2L2 −1

α2

 √ α2 K1 (t, α) ≤ C e−2c1 νt K1 (0, α) −1

α2

α=0

(10.12) (10.13)

D. Wei et al. / Advances in Mathematics 362 (2020) 106963

≤C



e−2c1

√ νt

93

 ω (0, α)2L2

α=0

= Ce−2c1

√ νt

ω0 2L2 .

Using (10.6), we have for 0 ≤ t ≤ τ /ν, 

α)2 2 + ∂y ψ(t,

α)2 2 ) V (t)2H˙ 1 L2 = α2 (α2 ψ(t, L L x

y

α=0

≤

 α=0

α4 K3 (t, α) α2 − 1 √

√

 e−2c1 νt Ce−2c1 νt ≤C K (0, α) ≤ ω0 2L2 . 1 1 + νt3 1 + νt3 α=0

Therefore for 0 ≤ t ≤ τ /ν, −c1

ω(t)L2 ≤ Ce

√ νt

√

ω0 L2 ,

V (t)H˙ x1 L2y

Ce−c1 νt ≤ √ ω0 L2 . 1 + νt3

For the case of δ = 1, the proof is almost the same by making the orthogonal projection P|α|≥2 , since Rα has no eigenvalue and embedding eigenvalues for |α| ≥ 2 and Lemma 10.3-Lemma 10.4 hold for |α| ≥ 2.  Let us conclude this section by introducing some space-time estimates of the solution for the linearized Navier-Stokes equations, which will be used in the proof of nonlinear enhanced dissipation. Lemma 10.5. Let δ < 1. Under the same assumptions as in Theorem 1.4, we have τ /ν ∇ω(t)2L2 dt ≤ Cν −1 ω0 2L2 , 0

τ /ν

∂x ω(t)L2 dt ≤ Cν − 3 ω0 L2 , 2

0

τ /ν 1 V (t)2L∞ dt ≤ C(| ln ν| + 1)ν − 3 ω0 2L2 . 0

Proof. By (10.7), we have τ /ν τ /ν   2 2 ∇ω(t)L2 dt = |α|  ω (t, α)2L2 + ∂y ω

(t, α)2L2 dt 0

0

α=0

D. Wei et al. / Advances in Mathematics 362 (2020) 106963

94

τ /ν 

≤

α=0

0

α2 K2 (t, α)dt −1

α2

1  α2 (K1 (0, α) − K1 (τ /ν, α)) 2ν α2 − 1

=

α=0

C  C  C ≤ K1 (0, α) ≤  ω (0, α)2L2 = ω0 2L2 . ν ν ν α=0

α=0

Using the fact that ∂t K1 ≤ −2να2 K1 , we have |α|2 K1 (t, α) ≤ |α|2 e−2να t K1 (0, α) ≤ CK1 (0, α)/(νt) 2

and (10.11) implies that |α|2 K1 (t, α) ≤ |α|2 Ce−2c1

'

ν|α|t

K1 (0, α) ≤ CK1 (0, α)/(ν 2 t4 ),

from which, we infer that 

∂x ω(t)2L2 =

|α|2  ω (t, α)2L2 ≤

α=0



≤



|α|2

α=0

α2 K1 (t, α) α2 − 1

CK1 (0, α) min{(νt)−1 , (ν 2 t4 )−1 }

α=0

≤ Cω0 2L2 min{(νt)−1 , (ν 2 t4 )−1 }. This gives τ /ν τ /ν 1 2 ∂x ω(t)L2 dt ≤ Cω0 L2 min{(νt)− 2 , (νt2 )−1 }dt ≤ Cν − 3 ω0 L2 . 0

0

By Minkowski inequality, we get ⎛ ⎛ ⎞ 12 ⎞2 ⎛ ⎞2 τ /ν τ /ν τ /ν   ⎜ ⎜ ⎟ ⎟ 2 ⎝ V (t)2L∞ dt ≤ V (t, α, ·)L∞ ⎠ dt ≤ ⎜ ⎝ V (t, α, ·)L∞ dt⎠ ⎟ ⎝ ⎠ , 0

0

α=0

α=0

and by the interpolation, V (t, α, ·)2L∞ ≤CV (t, α, ·)L2 V (t, α, ·)H 1  1

α, ·)2 2 + ∂y ψ(t,

α, ·)2 2 2 ≤C α2 ψ(t, L L

0

D. Wei et al. / Advances in Mathematics 362 (2020) 106963

95

 1

α, ·)2 2 + 2α2 ∂y ψ(t,

α, ·)2 2 + ∂ 2 ψ(t,

α, ·)2 2 2 × α4 ψ(t, y L L L 1

1

3

1

≤ CK3 (t, α) 2 K1 (t, α) 2 ≤ CK3 (t, α) 4 K2 (t, α) 4 , from which and (10.13), (10.7), we infer that τ /ν τ /ν 1 1 2 V (t, α)L∞ dt ≤ CK3 (t, α) 2 K1 (t, α) 2 dt 0

0

τ /ν ≤ 0

CK1 (0, α) CK1 (0, α) √ dt ≤ , 1 3 |α| 1 + νt |α|ν 3

and τ /ν τ /ν 3 1 2

V (t, α)L∞ dt ≤ CK3 (t, α) 4 K2 (t, α) 4 dt 0

0

⎞ 34 ⎛ ⎞ 14 τ /ν τ /ν ⎜ ⎟ ⎜ ⎟ ≤ C ⎝ K3 (t, α)dt⎠ ⎝ K2 (t, α)dt⎠ ⎛

0

⎛ ⎜ ≤C⎝

0

τ /ν

⎞

3 4

K1 (0, α) ⎟ dt⎠ α2 (1 + νt3 )



K1 (0, α) − K1 (τ /ν, α) 2ν

 14

0

 ≤C

K1 (0, α) 1 α2 ν 3

 34 

K1 (0, α) 2ν

 14

≤

CK1 (0, α) . 3 1 |α| 2 ν 2

Therefore, ⎛ ⎞ 12 ⎞2 ⎛ τ /ν τ /ν  ⎜ ⎜ ⎟ ⎟ 2 V (t)2L∞ dt ≤ ⎜ ⎝ V (t, α)L∞ dt⎠ ⎟ ⎝ ⎠ 0

α=0

0

⎛





⎜ ≤⎝

1

0<|α|≤ν − 3

CK1 (0, α) 1 |α|ν 3

⎞⎛  ⎜ ≤C⎝ K1 (0, α)⎠ ⎝ ⎛

α=0

 12 +



 1

|α|>ν − 3

 1

0<|α|≤ν − 3

1 1 + |α|ν 3

CK1 (0, α) 3 1 |α| 2 ν 2

 12

⎞2 ⎟ ⎠ ⎞



1 3

1

|α|>ν − 3

1

|α| 2 ν 2

⎟ ⎠

96

D. Wei et al. / Advances in Mathematics 362 (2020) 106963

⎛ ≤C⎝



⎞(  ω (0, α)2L2 ⎠

α=0

1

1 + | ln ν| ν 6 + 1 1 ν3 ν2

)

1

≤ Cω0 2L2 (1 + | ln ν|)/ν 3 , which gives the third inequality.  11. Nonlinear enhanced dissipation In this section, we prove Theorem 1.7. We define W1 = span{sin y, cos y} and the non-shear space    X = ω ∈ L2 (Tδ2 ) : ω(x, y) = ω

(α, y)eiαx . α=0

We denote by P1 the orthogonal projection to W1 , and P=0 the orthogonal projection to X, and P0 = I − P=0 the orthogonal projection to the shear space X0 = {ω ∈ L2 (T ) : ∂x ω = 0}. 11.1. Semigroup estimates Let S(t, s) = S(t, s, a0 ) be the solution operator of linearized equation (1.5). That is, ω(t, y) = S(t, s)f (y) solves (1.5) with ω(s, y) = f (y), here we assume t ≥ s. Now Theorem 1.4 and Lemma 10.5 can be restated as follows S(t, 0)f L2 ≤ Ce−c1

√ νt

f L2 for 0 < t < τ /ν,

τ /ν ∇S(t, 0)f 2L2 dt ≤ Cν −1 f 2L2 , 0

τ /ν

∂x S(t, 0)f L2 dt ≤ Cν − 3 f L2 , 2

0

τ /ν

∇Δ−1 S(t, 0)f 2L∞ dt ≤ C(| ln ν| + 1)ν − 3 f 2L2 . 1

0

Notice that S(t, s, a0 ) = S(t − s, 0, a0 e−νs ) and a0 e−νs has uniform positive upper and lower bounds. Then we also have for 0 ≤ s < t < τ /ν, S(t, s)f L2 ≤ Ce−c1

√ ν(t−s)

f (s)L2 ,

(11.1)

τ /ν ∇S(t, s)f 2L2 dt ≤ Cν −1 f (s)2L2 ,

(11.2)

s

D. Wei et al. / Advances in Mathematics 362 (2020) 106963

τ /ν 2 ∂x S(t, s)f L2 dt ≤ Cν − 3 f (s)L2 ,

97

(11.3)

s

τ /ν

∇Δ−1 S(t, s)f 2L∞ dt ≤ Cν −γ1 f (s)2L2 ,

(11.4)

s

where γ1 is a constant such that

1 3

< γ1 < 2γ − 1.

Let f be the solution of the inhomogeneous linearized Navier-Stokes equations: ∂t f + Lν (t)f = g.

(11.5)

Then we have t f (t) = S(t, 0)f (0) +

S(t, s)g(s)ds. 0

Using (11.1)-(11.4) and the Minkowski inequality, we deduce that for 0 < t < τ /ν, f (t)L2 ≤ Ce−c1

√ νt

t f (0)L2 + C

g(s)L2 ds,

(11.6)

0

⎞ 12 ⎞ 12 ⎛ t ⎛ t  t  γ1 1 2 ν 2 ⎝ ∇f (s)2L2 ds⎠ + ν 3 ∂x f (s)L2 ds + ν 2 ⎝ ∇Δ−1 f (s)2L∞ ds⎠ 0

0

0

t ≤ Cf (0)L2 + C

g(s)L2 ds,

(11.7)

0

where C, c1 are constants independent of ν, t. 11.2. Proof of Theorem 1.7 First of all, we have the following energy dissipation law: ∂t V 2L2 = −2ν∇V 2L2 ,

∂t ω2L2 = −2ν∇ω2L2 ,

which gives ∂t (ω2L2 − ∇ψ2L2 ) = −2ν(∇ω2L2 − ω2L2 ), where ψ = −Δ−1 ω. The key point is that if δ < 1, it holds that for ω ∈ {1}⊥ ,

(11.8)

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98

(I − P1 )ω2L2 ≥ ω2L2 − ∇ψ2L2 ≥ C0 (I − P1 )ω2L2 , ∇ω2L2 − ω2L2 ≥ C0 ∇(I − P1 )ω2L2 ,

(11.9) (11.10)

which can be easily proved by using Fourier transform. By (11.8)-(11.10), we have (I − P1 )ω(t)2L2 ≤ C(ω(t)2L2 − ∇ψ(t)2L2 )

(11.11)

≤ C(ω(0)2L2 − ∇ψ(0)2L2 ) ≤ C(I − P1 )ω(0)2L2 ≤ Cν 2γ , for 0 < t < τ /ν, and τ /ν τ /ν 2 ∇(I − P1 )ω(t)L2 dt ≤ C (∇ω(t)2L2 − ω(t)2L2 )dt 0

(11.12)

0

≤ C(ω(0)2L2 − ∇ψ(0)2L2 )/(2ν) ≤ C(I − P1 )ω(0)2L2 /(2ν) ≤ Cν 2γ−1 . Now, we decompose ω = ωs + ωn , where ωs = P0 ω is the shear part and ωn = P=0 ω. Correspondingly, V = Vs +Vn , ψ = ψs +ψn . We denote Vs = (V 1 (t, y), 0), Vn = (Vn1 , Vn2 ). Then the equation (1.4) can be written as ∂t ωs = νΔωs − P0 (Vn · ∇ωn ) = ν∂y2 ωs − ∂y P0 (Vn2 ωn ),

(11.13)

∂t ωn = νΔωn −

(11.14)

Vs1 ∂x ωn

− Vn · ∇ωs − P=0 (Vn · ∇ωn ).

For P1 ω = P1 ωs , we have ∂t P1 ω = νΔP1 ω − ∂y P1 (Vn2 ωn ) = −νP1 ω − ∂y P1 (Vn2 ωn ), therefore,

P1 ω(t) = e

−νt

t P1 ω(0) −

e−ν(t−s) ∂y P1 (Vn2 ωn )(s)ds,

0

from which and (11.11), we deduce that for 0 < t < τ /ν,

P1 ω(t) − e

−νt

t P1 ω(0)L2 ≤

∂y P1 (Vn2 ωn )(s)L2 ds

(11.15)

0

t ≤

t C(Vn2 ωn )(s)L1 ds

0

≤

CVn2 (s)L2 ωn (s)L2 ds 0

D. Wei et al. / Advances in Mathematics 362 (2020) 106963

t ≤

99

t Cωn (s)2L2 ds

≤

C(I − P1 )ω(s)2L2 ds

0

0

t ≤

Cν 2γ ds = Ctν 2γ . 0

In particular, we have P1 ω(t) − e−νt P1 ω(0)L2 ≤ Cτ ν 2γ−1 . Thanks to C1−1 ≤ P1 ω(0)L2 ≤ C1 , there exist c2 ∈ (0, 1), C > 1 so that if 0 < ν < c2 , then C −1 ≤ P1 ω(t)L2 ≤ C Now we choose t0 ∼ ν − 2 so that Ce−c1 0 ≤ t < t < τ /ν, 1

√ νt0

for 0 < t < τ /ν. =

1 4.

Then it suffices to show that for

ωn (t)L2 ≤ Cωn (t )L2

if t − t < t0 ,

1 ωn (t )L2 2

if t − t = t0 .

ωn (t)L2 ≤

(11.16)

(11.17)

Indeed, Theorem 1.7 follows from (11.17) by using the following iteration ωn (t)L2 ≤ Cωn (mt0 )L2 ≤ C

√ 1 ωn (0)L2 ≤ Ce−c νt ωn (0)L2 , m 2

with m = [t/t0 ] and t0 ≤ Cν − 2 . 1

By time translation and (11.16), we only need to prove that (11.17) for t = 0. Let P1 ω(0) = −a0 sin(y + θ), where a0 > 0, θ ∈ [0, 2π). Then P1 ω(0)L2 / sin yL2 = a0 . By translation, we may assume θ = 0. Thus, P1 ω(0) = −a0 sin y,

P1 V (0) = (−a0 cos y, 0).

Let ω(t) = e−νt P1 ω(0) + ω  (t),

V (t) = e−νt P1 V (0) + V (t).

We decompose ω  =ω s + ωn and V = Vs + Vn , where ω  s = P0 ω  and Vs = P0 V . Then (11.14) can be rewritten as  ∂t ωn + Lν (t)ωn = − Vs1 ∂x ωn + Vn · ∇ ωs + P=0 (Vn · ∇ωn ) .

(11.18)

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100

For 0 < t ≤ min{t0 , τ /ν}, we infer from (11.15) that P1 ω s (t)L2 = P1 ω(t) − e−νt P1 ω(0)L2 ≤ Ctν 2γ ≤ Ct0 ν 2γ ≤ Cν 2γ− 2 . (11.19) 1

Using (11.11), (11.19) and the fact that (I − P1 ) ωs (t) = P0 (I − P1 )ω(t), we deduce that Vs1 (t)L∞ ≤ C ωs (t)L2 ≤ CP2 ω s (t)L2 + C(I − P1 )ω(t)L2 ≤ Cν

2γ− 12

(11.20)

+ Cν γ ≤ Cν γ .

Using (11.12) and (11.19), we deduce that t

t ∇ ω (s)2L2 ds

0

t ∇P1 ω s (s)2L2 ds

=

∇(I − P1 ) ωs (s)2L2 ds

+

0

(11.21)

0

t ≤C

t P1 ω s (s)2L2 ds

0

∇(I − P1 )ω(s)2L2 ds

+ 0

τ /ν ≤C ν 4γ−1 ds + Cν 2γ−1 0

= Cτ ν 4γ−2 + Cν 2γ−1 ≤ Cν 2γ−1 . Let f = ωn and  g = − Vs1 ∂x ωn + Vn · ∇ ωs + P=0 (Vn · ∇ωn ) = g1 + g2 + g3 . Then (f, g) solves (11.5). We denote ⎛ t ⎞ 12 ⎛ t ⎞ 12  t  γ1 2 A1 (t) = ν ⎝ ∇ωn (s)2L2 ds⎠ + ν 3 ∂x ωn (s)L2 ds + ν 2 ⎝ Vn (s)2L∞ ds⎠ , 1 2

0

t g(s)L2 ds ≤

A2 (t) = 0

0 3 t 

0

gj (s)L2 ds.

j=1 0

Then (11.7) implies that A1 (t) ≤ Cωn (0)L2 + CA2 (t). By (11.20), we have

(11.22)

D. Wei et al. / Advances in Mathematics 362 (2020) 106963

t

t

Vs1 (s)∂x ωn (s)L2 ds ≤

g1 (s)L2 ds = 0

0

t

101

Vs1 (s)L∞ ∂x ωn (s)L2 ds (11.23)

0

t ≤ Cν γ

2

∂x ωn (s)L2 ds ≤ Cν γ− 3 A1 (t), 0

and by (11.21), t

t

t

g2 (s)L2 ds = 0

Vn (s) · ∇ ωs (s)L2 ds ≤ 0

Vn (s)L∞ ∇ ωs (s)L2 ds (11.24) 0

⎛ t ⎞ 12 ⎛ t ⎞ 12   ≤ ⎝ Vn (s)2L∞ ds⎠ ⎝ ∇ ωs (s)2L2 ds⎠ 0

≤ν

γ − 21

0 1 2

A1 (t)(Cν 2γ−1 ) ≤ Cν γ−

γ1 +1 2

A1 (t).

Using (11.12) and the fact that ∇ωn (t) = P=0 ∇(I − P1 )ω(t), we infer that t

t g3 (s)L2 ds =

0

t P=0 (Vn · ∇ωn )L2 ds ≤

0

Vn (s)L∞ ∇ωn (s)L2 ds (11.25) 0

⎛ t ⎞ 12 ⎛ t ⎞ 12   ≤ ⎝ Vn (s)2L∞ ds⎠ ⎝ ∇(I − P2 )ω(s)2L2 ds⎠ 0

≤ν

−

γ1 2

0

A1 (t)(Cν

2γ−1

1 2

) ≤ Cν γ−

γ1 +1 2

A1 (t).

It follows from (11.22)-(11.25) that 2

A1 (t) ≤ Cωn (0)L2 + C(ν γ− 3 + ν γ−

γ1 +1 2

)A1 (t). 2

As γ > 23 , γ1 < 2γ − 1, we can find c2 ∈ (0, 1) so that C(ν γ− 3 + ν γ− 0 < ν < c2 . Then we obtain A1 (t) ≤ Cωn (0)L2 ,

2

A2 (t) ≤ C(ν γ− 3 + ν γ−

γ1 +1 2

γ1 +1 2

)≤

1 2

for

)ωn (0)L2 .

While, by (11.6), we have ωn (t)L2 ≤ Ke−c1

√ νt

ωn (0)L2 + CA2 (t).

(11.26)

Therefore, ωn (t)L2 ≤ Cωn (0)L2 for 0 < t < min{t0 , τ /ν}. If t = t0 < τ /ν, then we have

102

D. Wei et al. / Advances in Mathematics 362 (2020) 106963

ωn (t)L2 ≤ Ce−c1 ≤

√ νt0

ωn (0)L2 + CA2 (t)

γ1 +1 2 1 ωn (0)L2 + C( ν γ− 3 + ν γ− 2 )ωn (0)L2 . 4 2

Take c2 small enough (if necessary) so that C(ν γ− 3 + ν γ− Thus, ωn (t)L2 ≤ 12 ωn (0)L2 for t = t0 < τ /ν.

γ1 +1 2

) ≤

1 4

for 0 < ν < c2 .

This completes the proof of Theorem 1.7. Acknowledgments We would like to thank the referee for the invaluable comments and suggestions, which have helped us improve the paper significantly. The authors thank Zhiwu Lin for profitable discussions. Z. Zhang is partially supported by NSF of China under Grant 11425103. References [1] M. Beck, C.E. Wayne, Using global invariant manifolds to understand metastability in the Burgers equation with small viscosity, SIAM Rev. 53 (2011) 129–153. [2] M. Beck, C.E. Wayne, Metastability and rapid convergence to quasi-stationary bar states for the two-dimensional Navier-Stokes equations, Proc. R. Soc. Edinb., Sect. A 143 (2013) 905–927. [3] J. Bedrossian, M. Coti Zelati, Enhanced dissipation, hypoellipticity, and anomalous small noise inviscid limits in shear flows, Arch. Ration. Mech. Anal. 224 (2017) 1161–1204. [4] J. Bedrossian, P. Germain, N. Masmoudi, On the stability threshold for the 3D Couette flow in Sobolev regularity, Ann. Math. 185 (2017) 541–608. [5] J. Bedrossian, P. Germain, N. Masmoudi, Dynamics near the subcritical transition of the 3D Couette flow I: below threshold case, arXiv:1506.03720. [6] J. Bedrossian, P. Germain, N. Masmoudi, Dynamics near the subcritical transition of the 3D Couette flow II: above threshold case, arXiv:1506.03721. [7] J. Bedrossian, N. Masmoudi, Inviscid damping and the asymptotic stability of planar shear flows in the 2D Euler equations, Publ. Math. Inst. Hautes Études Sci. 122 (2015) 195–300. [8] J. Bedrossian, N. Masmoudi, V. Vicol, Enhanced dissipation and inviscid damping in the inviscid limit of the Navier-Stokes equations near the two dimensional Couette flow, Arch. Ration. Mech. Anal. 219 (2016) 1087–1159. [9] J. Bedrossian, F. Wang, V. Vicol, The Sobolev stability threshold for 2D shear flows near Couette, J. Nonlinear Sci. 28 (2018) 2051–2075. [10] F. Bouchet, H. Morita, Large time behavior and asymptotic stability of the 2D Euler and linearized Euler equations, Physica D 239 (2010) 948–966. [11] F. Bouchet, E. Simonnet, Random changes of flow topology in twodimensional and geophysical turbulence, Phys. Rev. Lett. 102 (2009) 094504. [12] K.M. Case, Stability of inviscid plane Couette flow, Phys. Fluids 3 (1960) 143–148. [13] P. Constantin, A. Kiselev, L. Ryzhik, A. Zlatos, Diffusion and mixing in fluid flow, Ann. Math. (2) 168 (2008) 643–674. [14] Y. Couder, Observation expérimental de la turbulence bidimensionnelle dans un film liquide mince, C. R. Acad. Sci. Paris Sèr. II 297 (1983) 641–645. [15] W. Deng, Pseudospectrum for Oseen vortices operators, Int. Math. Res. Not. (2013) 1935–1999. [16] I. Gallagher, Th. Gallay, F. Nier, Spectral asymptotics for large skew-symmetric perturbations of the harmonic oscillator, Int. Math. Res. Not. (2009) 2147–2199. [17] T. Gallay, Enhanced dissipation and axisymmetrization of two-dimensional viscous vortices, Arch. Ration. Mech. Anal. 230 (2018) 939–975.

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