Nonlinear Asymptotic Stability and Transition Threshold for 2D Taylor–Couette Flows in Sobolev Spaces

Abstract

In this paper, we investigate the stability of the 2-dimensional (2D) Taylor–Couette (TC) flow for the incompressible Navier–Stokes equations. The explicit form of velocity for 2D TC flow is given by $u=(Ar+\frac{B}{r})(-\sin \theta , \cos \theta )^T$ with $(r, \theta )\in [1, R]\times \mathbb {S}^1$ being an annulus and A, B being constants. Here, A, B encode the rotational effect and R is the ratio of the outer and inner radii of the annular region. Our focus is the long-term behavior of solutions around the steady 2D TC flow. While the laminar solution is known to be a global attractor for 2D channel flows and plane flows, it is unclear whether this is still true for rotating flows with curved geometries. In this article, we prove that the 2D Taylor–Couette flow is asymptotically stable, even at high Reynolds number ($Re\sim \nu ^{-1}$), with a sharp exponential decay rate of $\exp (-\nu ^{\frac{1}{3}}|B|^{\frac{2}{3}}R^{-2}t)$ as long as the initial perturbation is less than or equal to $\nu ^\frac{1}{2} |B|^{\frac{1}{2}}R^{-2}$ in Sobolev space. The powers of $\nu$ and B in this decay estimate are optimal. It is derived using the method of resolvent estimates and is commonly recognized as the enhanced dissipative effect. Compared to the Couette flow, the enhanced dissipation of the rotating Taylor–Couette flow not only depends on the Reynolds number but also reflects the rotational aspect via the rotational coefficient B. The larger the |B|, the faster the long-time dissipation takes effect. We also conduct space-time estimates describing inviscid-damping mechanism in our proof. To obtain these inviscid-damping estimates, we find and construct a new set of explicit orthonormal basis of the weighted eigenfunctions for the Laplace operators corresponding to the circular flows. These provide new insights into the mathematical understanding of the 2D Taylor–Couette flows.

1 Introduction

Reynolds’s famous experiment [27] inspired the study of hydrodynamic stability at high Reynolds number. In this regime, the laminer flows could become unstable and transition to turbulence [10, 29, 32, 36].

With the low Reynolds number, Serrin [30] demonstrated that all equilibria of the forced Navier–Stokes equation on bounded domains are linearly stable. At high Reynolds number, even in the absence of boundaries, the viscosity can significantly complicate the linear problem.

In this paper, we study the 2D (two-dimensional) incompressible Navier–Stokes (NS) equations:

$$\begin{aligned} \left\{ \begin{aligned}&\partial _tv-\nu \Delta v+v\cdot \nabla v+\nabla p=0,\quad \text {div } v=0,\\&v(0,x)=v_0(x), \end{aligned} \right. \end{aligned}$$

(1.1)

where $v=(v_1,v_2)\in \mathbb {R}^2$ is the fluid velocity, $x=(x_1,x_2)\in \Omega$ represents the space variables, $\Omega \in \mathbb {R}^2$ is an annular region and $t\ge 0$ represents the time variable. The unknowns in the equation are the velocity field $v(t,x)=(v^1(t,x),v^2(t,x))$ and the pressure $p(t,x)\in \mathbb {R}$. The constant $\nu >0$ is known as the kinematic viscosity, which is very small in our paper. And the Reynolds number Re is proportional to the inverse of $\nu$.

In the 2D case, the vorticity field

$$\begin{aligned} \omega =\text {curl }v=\partial _1v^2-\partial _2v^1 \end{aligned}$$

(1.2)

is a scalar. By taking the 2D curl of the Navier–Stokes equations, Eq. (1.2) can be transformed into its vorticity formulation:

$$\begin{aligned}&\partial _t\omega -\nu \Delta \omega +v\cdot \nabla \omega =0. \end{aligned}$$

(1.3)

In this paper, we employ (1.3) to study the asymptotics of solutions to (1.1) around a Taylor–Couette (TC) flow.

1.1 Taylor–Couette flow

TC flow describes a steady circular solution of the viscous fluid bounded between two rotating infinitely long coaxial cylinders. The understanding of solutions’ asymptotics around a TC flow has board applications, including desalination, magnetohydrodynamics and viscometric analysis. Despite being a simple type of rotating solutions, perturbation of the TC flow has proven to be a challenging subject and has been extensively studied experimentally, theoretically, and numerically for a long time. Many questions related to TC flow remain unanswered, making it active research field in fluid mechanics [8, 11, 18, 26]. However, rigorous mathematical proofs in this field are still insufficient, even for the 2D case.

In the following, we adopt the convention $r = |x|$ and denote the radial vorticity $\omega (x) = \omega (r)$ and stream function $\phi (x) = \phi (r)$. In 2D, the vorticity and the velocity field then take the form

$$\begin{aligned} \left\{ \begin{aligned}&\omega (x_1,x_2)=\omega (r)=\Delta \phi =\phi ''(r)+\frac{1}{r}\phi '(r),\\&v(x_1,x_2)=\left( \begin{array}{ccc} -\partial _{x_2}\phi \\ \partial _{x_1}\phi \\ \end{array} \right) =\left( \begin{array}{ccc} -x_2 \\ x_1 \\ \end{array} \right) \frac{\phi '(r)}{r}. \end{aligned} \right. \end{aligned}$$

(1.4)

We first derive the explicit form of the steady TC flow. First noting that if $\omega =const$, then $\omega$ is a solution to the 2D NS equation (1.3). For this case, the stream function $\phi (r)$ can be determined from (1.4), which yields

$$\begin{aligned} \phi ''(r)+\frac{1}{r}\phi '(r)=const. \end{aligned}$$

(1.5)

By employing the polar coordinates, one can relabel $v(x_1,x_2)$ and $\omega (r)$ as $U(r,\theta )$ and $\Omega (r)$, respectively. This allows us to solve Eq. (1.5) and obtain the expressions for U and $\Omega$ as follows:

$$\begin{aligned}&U(r,\theta )=\left( \begin{array}{ccc} U^1 \\ U^2\\ \end{array} \right) =\left( \begin{array}{ccc} -\sin \theta \\ \cos \theta \\ \end{array} \right) (Ar+\frac{B}{r}),\quad \Omega (r)=2A. \end{aligned}$$

(1.6)

Here A, B are constants and spatial variables $(r, \theta )$ belong to a domain $\mathcal {D}=[1,R]\times \mathbb {S}^1$. The function $U(r,\theta )$ in (1.6) is hence a steady state for the 2D incompressible NS equation (1.1), which is commonly called the Taylor–Couette flow.

1.2 Hydrodynamic stability at high Reynolds number

Reynolds well-known experiment [27] revealed that small perturbations can cause significant change of the flow due to the nonlinear nature of the Navier–Stokes equations. In experiments [9, 31] for the TC flow, a small perturbation could lead to instability and transition to turbulence at high Reynolds numbers. This phenomenon is called the subcritical transition, which is a central topic in fluid mechanics.

Numerous efforts have been made to comprehend the subcritical transition mechanism [6]. Kelvin initially suggested that as the Reynolds number $Re\rightarrow \infty$, the basin of attraction for the laminar flow diminishes, allowing the flow to become nonlinearly unstable for small but finite perturbations [20]. In [32], Trefethen, Trefethen, Reddy and Driscoll proposed a way to determine the threshold amplitude by considering a perturbation of size $Re^{-\beta }$ with $\beta \ge 0$ and $Re\rightarrow +\infty$. In other words, they aimed to identify the lowest possible value of $\beta$, which could result in a transition to turbulence with a perturbation of size $O(Re^{-\beta })$. Bedrossian, Germain, and Masmoudi [2] presented a rigorous mathematical formulation of this approach using the fact that $Re^{-1}\sim \nu$.

Given a norm $\Vert \cdot \Vert _X$, we now hope to determine a nonnegative number $\beta =\beta (X)$ so that

$$\begin{aligned} \Vert u_0\Vert _X\le & {} \nu ^{\beta }\Longrightarrow {stability},\\ \Vert u_0\Vert _X\gg & {} \nu ^{\beta }\Longrightarrow {instability}. \end{aligned}$$

The exponent $\beta$ is referred to as the transition threshold in the applied literature. Significant advancements and findings related to the subcritical transition regime for the 2D incompressible NS equations have emerged recently. These developments include researches on Couette flow without boundaries [2, 4, 5, 23, 24] and with boundaries [3, 7], on Lamb–Oseen vortices [12,13,14,15, 21] and on periodic Kolmogorov flow [34, 35]. Recently, Guo, Pausader and Widmayer [16] has made significant progress in the stability problem of the Euler flow with rotation.

1.3 Main theorems

The aim of this paper is to explore both linear and nonlinear stability mechanism of the 2D Taylor–Couette flow using rigorous analytical tools. Specifically, we prove nonlinear asymptotic stability of the TC flows for the 2D NS equations. Furthermore, we identify and trace a critical parameter reflecting the rotating properties of this system.

Our focus in this paper is on the stability of the TC flow, hence we study the below perturbation equation. Setting $w = \omega - \Omega$, $u = v - U$, with $\varphi$ being the stream function satisfying $\Delta \varphi =w$ and $u=(-\partial _2\varphi ,\partial _1\varphi )$, we concert (1.3) to polar coordinates:

$$\begin{aligned} \left\{ \begin{aligned}&\partial _tw-\nu (\partial _r^2+\frac{1}{r}\partial _r+\frac{1}{r^2}\partial _{\theta }^2)w+(A+\frac{B}{r^2})\partial _{\theta }w+\frac{1}{r}(\partial _r\varphi \partial _{\theta }w-\partial _{\theta }\varphi \partial _rw)=0,\\&(\partial _r^2+\frac{1}{r}\partial _r+\frac{1}{r^2}\partial _{\theta }^2)\varphi =w,\quad w|_{r=1,R}=0\quad \text {with }(r,\theta )\in [1,R]\times \mathbb {S}^1 \text { and } t\ge 0. \end{aligned} \right. \end{aligned}$$

(1.7)

Remark 1.1

Note that constants A and B in (1.6), (1.7) and constant R in (1.7) will serve as parameters in our later arguments. In particular, B will correspond to the rotating effect.

Our main results are summarized below.

Theorem 1.1

The fully nonlinear two-dimensional incompressible Navier–Stokes equations exhibit asymptotic stability around the Taylor–Couette flow under perturbations of size $\nu ^\frac{1}{2} |B|^{\frac{1}{2}}R^{-2}$ in the $H^1$ space, with the vorticity controllable at any time by the initial data.

The more precise mathematical statements are given in the following two theorems. The first main result can be directly derived from Proposition 4.3 and Lemma 5.4, and describes the asymptotic stability of the linearized TC flow.

Theorem 1.2

There exist constants $C,c>0$ being independent of $\nu ,A,B,R$, such that the solution w for the linear system

$$\begin{aligned}&\partial _tw-\nu (\partial _r^2+\frac{1}{r}\partial _r+\frac{1}{r^2}\partial _{\theta }^2)w+(A+\frac{B}{r^2})\partial _{\theta }w=0,\quad w|_{t=0}=w(0), \end{aligned}$$

(1.8)

exists globally in time and for any $t\ge 0$. Furthermore, the following stability estimates hold

$$\begin{aligned}&\Vert w(t)-\bar{w}(t)\Vert _{L^2_{\theta }L^2_r}\le Ce^{-c\nu ^{\frac{1}{3}}|B|^{\frac{2}{3}}R^{-2}t}\Vert w(0)\Vert _{L^2_{\theta }L^2_r}, \quad \Vert \bar{w}(t)\Vert _{L^2_{\theta }L^2_r}\le C\Vert w(0)\Vert _{L^2_{\theta }L^2_r}, \end{aligned}$$

where $\bar{w}(t)=\frac{1}{2\pi }\int _0^{2\pi }w(t,r,\theta )d\theta$.

The second main result describes the asymptotic stability of the nonlinear Navier–Stokes equations around the TC flow. Reynolds experiments tell us that, with small viscosity coefficients, even tiny initial perturbations could cause the flow to be chaotic. Therefore, in order to establish the nonlinear asymptotic stability of Navier–Stokes equations, the initial perturbations must be restricted to a certain range with respect to the viscosity coefficient $\nu$. The specific formulation of the result is as follows.

Theorem 1.3

Assume that $0<\log R\lesssim \nu ^{-\frac{1}{3}}|B|^{\frac{1}{3}}$. There exist constants $\nu _0$ and $c_0, C,c'>0$ independent of $\nu ,A,B,R$ such that for any $0<\nu \le \nu _0$, if the initial data satisfies

$$\begin{aligned} E(0)=R\Vert w(0)\Vert _{L^2_{\theta }H^1_r}+R^{-2}(\log R)^{-\frac{3}{2}}\Vert r^2w(0)\Vert _{L^2_{\theta }L^2_r}+R^{3}\Vert \frac{w(0)}{r^3}\Vert _{L^2_{\theta }L^2_r}\le c_0\nu ^\frac{1}{2} |B|^{\frac{1}{2}}R^{-2}, \end{aligned}$$

then the solution $w(t,r,\theta )$ to the system (1.7) is global in time. Moreover, the following stability estimates hold

$$\begin{aligned}&\Vert w(t)-\bar{w}(t)\Vert _{L^2_{\theta }L^2_r}\le Cc_0e^{-c'\nu ^{\frac{1}{3}}|B|^{\frac{2}{3}}R^{-2}t}E(0), \quad \Vert \bar{w}(t)\Vert _{L^2_{\theta }L^2_r}\le Cc_0E(0),\quad \text {for any}\ t\ge 0, \end{aligned}$$

where $\bar{w}(t)=\frac{1}{2\pi }\int _0^{2\pi }w(t,r,\theta )d\theta$.

By setting the radius R to a constant value, such as $R=2$, we can immediately derive the following corollary from Theorem 1.3.

Theorem 1.4

There exist constants $\nu _0$ and $c_0, C,c'>0$ independent of $\nu ,A,B$ so that for any $0<\nu \le \nu _0$ and $|B|\ge \nu _0$, if

$$\begin{aligned} E(0)=\Vert w(0)\Vert _{L^2_{\theta }H^1_r}\le c_0\nu ^\frac{1}{2} |B|^{\frac{1}{2}}, \end{aligned}$$

then the system (1.7) admits a global-in-time solution $w(t,r,\theta )$ satisfying the following stability estimates

$$\begin{aligned}&\Vert w(t)-\bar{w}(t)\Vert _{L^2_{\theta }L^2_r}\le Cc_0e^{-c'\nu ^{\frac{1}{3}}|B|^{\frac{2}{3}}t}E(0), \quad \Vert \bar{w}(t)\Vert _{L^2_{\theta }L^2_r}\le Cc_0E(0),\quad \text {for any}\ t\ge 0, \end{aligned}$$

where $\bar{w}(t)=\frac{1}{2\pi }\int _0^{2\pi }w(t,r,\theta )d\theta$.

1.4 Difficulties, new ingredients and the sketch of the proof

1.4.1 Rotational effect and enhanced dissipation

Compared to Couette flow, the Taylor–Couette flow involves additional coefficients A and B that account for rotational effects. Both Theorem 1.2 and Theorem 1.3 demonstrate that for both linearized and nonlinear equations, only the rotation coefficient B affects the stability of the system, and the energy of vorticity is irrelevant to the coefficient A.

The impact of the rotation coefficient B on stability is mainly manifested in the so-called enhanced dissipation effect. The decay rate of the heat equation is $e^{-\nu R^{-2} t}$, but in the presence of B, the dissipation rate becomes faster and can be described by $e^{-c\nu ^{\frac{1}{3}}|B|^{\frac{2}{3}}R^{-2}t}$ (Notice that we are interested in the regime when $0<\nu \ll 1$). This implies that while the heat equation exhibits decay only after the time scale of $\nu ^{-1}$, the antisymmetric part $ik\frac{B}{r^2}$ in the linearized equation results in the energy decay after a shorter time scale of $\nu ^{-\frac{1}{3}}|B|^{-\frac{2}{3}}R^{2}$. This phenomenon is known as the enhanced dissipation effect. A shear or a diffusion averaging mechanism to trigger it were investigated in [22, 28]. Here we prove this effect in a different setting with rotations.

Theorem 1.2 and Theorem 1.3 both indicate that the enhanced dissipation effect becomes stronger as |B| increases, which is quantified by a faster asymptotic decay rate $e^{-c\nu ^{\frac{1}{3}}|B|^{\frac{2}{3}}R^{-2}t}$. This enhanced dissipation has a timescale of $t\sim \nu ^{-\frac{1}{3}}|B|^{-\frac{2}{3}}R^{2}$ around the Taylor–Couette flow, with $t\sim \nu ^{-\frac{1}{3}}$ being consistent with Couette flow [2,3,4,5, 7, 23, 24].

The enhanced dissipation decay rate $e^{-c\nu ^{\frac{1}{3}}|B|^{\frac{2}{3}}R^{-2}t}$ for the TC flow in this paper is optimal. In Sect. 3, we derived the sharp resolvent estimates, which are also known as the optimal pseudospectral bound. It is important to note that although the enhanced dissipation decaying rates for the TC flow ($\nu ^{\frac{1}{3}}|B|^{\frac{2}{3}}R^{-2}$) and Couette flow ($\nu ^{\frac{1}{3}}$) have the same power of $\nu$, it does not suggest that the corresponding enhanced dissipation can be directly derived through a scaling argument from results of Couette flow. This is because the (normalized) antisymmetric parts of linearized operators for these two flows are still different. Furthermore, with the aid of dimensional analysis or scaling transformations, it can be shown that the sum of the power of $\nu$ and B is always 1. To see this, assume that the decay rate is in the form of $\exp (-c\nu ^{\alpha }|B|^{\beta }R^{\gamma }t)$. Noting that the dimension of $\nu , B$ and R are $\textsf{T}^{-1}\textsf{L}^2, \textsf{T}^{-1}\textsf{L}^2$ and L, respectively (Here $\textsf{T}$ represents the time and $\textsf{L}$ denotes the length). Since the exponent $\nu ^{\alpha }|B|^{\beta }R^{\gamma }t=\textsf{T}^{1-\alpha -\beta } \textsf{L}^{2\alpha +2\beta +\gamma }$ is dimensionless, it must hold $\alpha +\beta =1$ and $\gamma =-2$. Thus, the factor $|B|^{\frac{2}{3}}$ is also optimal.

Additionally, inspired by the approach used to handle Oseen vortices [13, 15, 21], in [1] the authors employed self-similar variables and derived an enhanced dissipation decay rate that is independent of the outer radius R.

1.4.2 Comparison with Couette flow from an operator perspective

Previous studies have investigated plane Couette flow with $(x,y)\in \mathbb {T}\times \mathbb {R}$ ( see [2, 4, 5, 23, 24]) and Couette flow in a finite channel with $(x,y)\in \mathbb {T}\times [-1,1]$ (see [3, 7]). In these works, the spatial variable x corresponding to the non-shear direction was defined on a torus. However, for the TC flow in this paper, the spatial domain is an annulus region $(r,\theta )\in [1,R]\times \mathbb {S}^1$.

As both x in Couette flow and $\theta$ in TC flow are defined on a torus or $\mathbb {S}^1$, it is natural to apply Fourier transform on the x or $\theta$ direction. The corresponding linearized equations around the Couette flow and TC flow are given as below.

$$\begin{aligned} \text {Couette flow: }&\partial _t\hat{w}_k-\nu (\partial _y^2-k^2)\hat{w}_k+iky\hat{w}_k=0,\quad \quad \quad y\in \mathbb {R}\,\text { or}\, y\in [-1,1],\\ \text {TC flow: }&\partial _t\hat{w}_k-\nu (\partial _{r}^2+\frac{1}{r}\partial _{r}-\frac{k^2}{r^2}) \hat{w}_k+(A+\frac{B}{r^2})ik\hat{w}_k=0, \quad r\in [1,R]. \end{aligned}$$

The linearized operators for the Couette flow and TC flow take the different forms:

$$\begin{aligned} -\nu (\partial _y^2-k^2)+iky\quad \text {for Couette} \quad \text {and}\quad -\nu (\partial _{r}^2+\frac{1}{r}\partial _{r}-\frac{k^2}{r^2})+ik(A+\frac{B}{r^2}) \quad \text {for TC}. \end{aligned}$$

These two operators exhibit different enhanced dissipation rates of $\nu ^{\frac{1}{3}}$ and $\nu ^{\frac{1}{3}}|B|^{\frac{2}{3}}R^{-2}$, respectively.

It can be observed that both the symmetric and antisymmetric parts of the linearized operator for the TC flow have different structures from those of the Couette flow. In the following discussion, we will demonstrate how these differences affect our results and the corresponding proofs for the TC flow.

In previous studies of the planar Couette flow [2, 4, 5, 23, 24], where the shear variable y is defined over the entire space $\mathbb {R}$, mathematicians employed the Fourier transform in y to obtain ordinary differential equations with respect to time t. However, in the case of TC flow, the radial variable r is defined in the bounded domain [1, R] with boundaries, which makes it inconvenient to conduct the Fourier transform in r directly. Instead, we adopt the method of resolvent estimates to derive enhanced dissipation, with detailed results and proofs provided in Sect. 3. Note that this method has also been employed by Chen, Li, Wei and Zhang to study the Couette flow within a finite channel in [7].

1.4.3 Resolvent estimates

In Sect. 3, our key results are the resolvent estimates presented in Proposition 3.2. Define

$$\begin{aligned} L_k{:}{=}-\nu (\partial _r^2-\frac{k^2-\frac{1}{4}}{r^2})+ikB\frac{1}{r^2} \end{aligned}$$

and denote

$$\begin{aligned} F{:}{=}(L_k-ikB\lambda )w=-\nu (\partial _r^2-\frac{k^2-\frac{1}{4}}{r^2})w+ikB(\frac{1}{r^2}-\lambda )w. \end{aligned}$$

Our initial goal is the following inequality:

$$\begin{aligned} \nu ^{\frac{1}{3}} |kB|^{\frac{2}{3}}\Vert \frac{w}{r}\Vert _{L^2}\le C \Vert rF\Vert _{L^2}, \end{aligned}$$

(1.9)

which is a resolvent estimate from $L^2$ space to $L^2$ space, as given in Lemma 3.3. This estimate corresponds to the estimate obtained for the Couette flow in [7]:

$$\begin{aligned} \nu ^{\frac{1}{3}} |k|^{\frac{2}{3}}\Vert w\Vert _{L^2}\le C \Vert F\Vert _{L^2}. \end{aligned}$$

However, due to the specific structure of the TC flow, the corresponding resolvent estimate is weighted in r. If we take $c'$ as $\frac{1}{2C}$, the inequality (1.9) directly yields the following estimate:

$$\begin{aligned} \nu ^{\frac{1}{3}} |kB|^{\frac{2}{3}}\Vert \frac{w}{r}\Vert _{L^2}\le C \Vert rF-c'\nu ^{\frac{1}{3}} |kB|^{\frac{2}{3}}\frac{w}{r}\Vert _{L^2}, \end{aligned}$$

which takes the form of a resolvent estimate for $rF-c'\nu ^{\frac{1}{3}} |kB|^{\frac{2}{3}}\frac{w}{r}$. Notice that $rF-c'\nu ^{\frac{1}{3}} |kB|^{\frac{2}{3}}\frac{w}{r}$ can be written as

$$\begin{aligned} rF-c'\nu ^{\frac{1}{3}} |kB|^{\frac{2}{3}}\frac{w}{r}=(rL_kr-c'\nu ^{\frac{1}{3}} |kB|^{\frac{2}{3}})\frac{w}{r}. \end{aligned}$$

Here, $rL_kr-c'\nu ^{\frac{1}{3}} |kB|^{\frac{2}{3}}$ represents the translation of $rL_kr$ to the left by $c'\nu ^{\frac{1}{3}} |kB|^{\frac{2}{3}}$, which leads to the enhanced dissipation rate for the following linear evolution equation

$$\begin{aligned} {[}\partial _t- (rL_kr-c'\nu ^{\frac{1}{3}} |kB|^{\frac{2}{3}}) ]\frac{w}{r}=0. \end{aligned}$$

This allows us to obtain the dissipative factor $e^{-c\nu ^{\frac{1}{3}} |kB|^{\frac{2}{3}}R^{-2}t}$ in space-time estimates for the fully nonlinear system, as shown in detail in Proposition 4.5.

Next, we derive the resolvent estimate for the stream function $\varphi$:

$$\begin{aligned} \nu ^{\frac{1}{6}}|kB|^{\frac{5}{6}}|k|^{\frac{1}{2}}\left( \Vert \varphi '\Vert _{L^2}+|k|\Vert \frac{\varphi }{r}\Vert _{L^2} \right) \le CR^2\left( (\frac{\nu }{|kB|})^{\frac{1}{6}} (\log R)^{\frac{1}{2}} +1 \right) \Vert rF-c'\nu ^{\frac{1}{3}} |kB|^{\frac{2}{3}}\frac{w}{r}\Vert _{L^2}. \end{aligned}$$

The proof of this estimate relies on the resolvent estimate for $\Vert w\Vert _{L^1}$, and its details are given in Lemma 3.4.

Finally, in Lemma 3.5 and Lemma 3.6 we establish the below estimates for the $H^1$ norms of w and $\varphi$, which are controlled by the $H^{-1}$ norms of the resolvent equations, respectively:

$$\begin{aligned} \nu \Vert w\Vert _{H^1_r}+\nu ^{\frac{2}{3}}|kB|^{\frac{1}{3}}\Vert \frac{w}{r}\Vert _{L^2}\le&C\Vert F-c'\nu ^{\frac{1}{3}} |kB|^{\frac{2}{3}} R^{-2}w\Vert _{H^{-1}_r},\\ \nu ^{\frac{1}{2}}|kB|^{\frac{1}{2}}\Vert \varphi '\Vert _{L^2}+\nu ^{\frac{1}{2}}|k||kB|^{\frac{1}{2}}\Vert \frac{\varphi }{r}\Vert _{L^2}\le&C R^2 \Vert F-c'\nu ^{\frac{1}{3}}|kB|^{\frac{2}{3}}R^{-2}w\Vert _{H^{-1}_r}. \end{aligned}$$

The resolvent estimates presented above are crucial for our analysis, and are summarized in Proposition 3.2. Notably, all of the estimates in Proposition 3.2 have been carefully derived in preparation for the subsequent discussion in Sect. 4.

Based on the above conclusions, we summarize the following table which illustrates the relation between coefficients of the terms on the left of resolvent estimates and the norm of $F-c'\nu ^{\frac{1}{3}}|kB|^{\frac{2}{3}}R^{-2}w$ on the right (Table 1).

Table 1 The coefficients on the left of resolvent estimates

The above powers of $\nu$ in the coefficients are optimal and consistent with the 2D Couette flow in [7]. However, in this paper the coefficients are also dependent on B and R, and the powers of B are also sharp. As shown in the table, the coefficients on the left will be multiplied by $(\frac{\nu }{|kB|})^{\frac{1}{3}}$ if the regularity of $F-c'\nu ^{\frac{1}{3}}|kB|^{\frac{2}{3}}R^{-2}w$ drops from $L^2$ to $H^{-1}_{r}$.

1.4.4 Pseudospectrum and enhanced dissipation

In the Sect. 4, the pseudospectral bound of L is defined as

$$\begin{aligned} \Psi (L)=\inf \{\Vert (L-i\lambda )f\Vert {:}f\in D(L),\lambda \in \mathbb {R},\Vert f\Vert =1\}. \end{aligned}$$

Recall that in Proposition 3.2 we establish the following resolvent estimates

$$\begin{aligned}&\Vert r(L_k-i\lambda )w\Vert _{L^2}\ge C(\nu k^2)^{\frac{1}{3}}|B|^{\frac{2}{3}}\Vert \frac{w}{r}\Vert _{L^2}. \end{aligned}$$

Since $r\in [1, R]$, we have

$$\begin{aligned}&\Vert (L_k-i\lambda )w\Vert _{L^2}\ge C(\nu k^2)^{\frac{1}{3}}|B|^{\frac{2}{3}}R^{-2}\Vert w\Vert _{L^2}. \end{aligned}$$

This provides a lower bound for the pseudospectrum of $L_k$:

$$\begin{aligned} \Psi (L_k(L^2\rightarrow L^2))\ge C(\nu k^2)^{\frac{1}{3}}|B|^{\frac{2}{3}}R^{-2}. \end{aligned}$$

Applying the Gearhart-Pr$\ddot{u}$ss type lemma established by Wei in [33] and by Helffer-Sjöstrand in [17], we can obtain the pointwise estimate of the semigroup bound in Proposition 4.3

$$\begin{aligned} \Vert w_k^l(t)\Vert _{L^2}\le Ce^{-c(\nu k^2)^{\frac{1}{3}}|B|^{\frac{2}{3}}R^{-2}t}\Vert w_k(0)\Vert _{L^2}, \quad \text {for any} \ t\ge 0, \end{aligned}$$

where $w_k^l(t)$ satisfies the homogeneous linear equation

$$\begin{aligned} \partial _tw_k^l+L_kw_k^l=0 \quad \text {with} \ w_k^l(0)=w_k(0). \end{aligned}$$

It is worth noting that here, the exponential decay factor $e^{-c(\nu k^2)^{\frac{1}{3}}|B|^{\frac{2}{3}}R^{-2}t}$ reflects the enhanced dissipation effect.

1.4.5 Integrated invisid damping

The second part of Sect. 4 presents estimations for the stream function $\varphi ^l_k$ governed by the initial vorticity. Here $\varphi _k^l$ satisfies the linear elliptic equation

$$\begin{aligned} (\partial _r^2-\frac{k^2-\frac{1}{4}}{r^2}){\varphi }_k^l={w}_k^l,\quad {\varphi }_k^l|_{r=1,R}=0,\quad r\in [1,R]. \end{aligned}$$

(1.10)

The precise estimate is given in Lemma 4.6, which states that, for any $k\in \mathbb {Z}$ and $|k|\ge 1$, it holds

$$\begin{aligned}&k^2|B|R^{-4}(\Vert e^{c'(\nu k^2)^{\frac{1}{3}}|B|^{\frac{2}{3}}R^{-2}t}\partial _r{\varphi }^{l}_k\Vert _{L^2L^2}^2+|k|^2\Vert e^{c'(\nu k^2)^{\frac{1}{3}}|B|^{\frac{2}{3}}R^{-2}t}\frac{{\varphi }^{l}_k}{r}\Vert _{L^2L^2}^2) \\&\quad \lesssim (\log R)^{-2}R^{-4}\Vert r^2 w_k(0)\Vert _{L^2}^2+R^6 \Vert \frac{w_k(0)}{r^3}\Vert _{L^2}^2+(\frac{\nu }{|kB|})^{\frac{2}{3}}R^2\Vert \partial _r w_k(0)\Vert _{L^2}^2 \\&\qquad +\left( (\frac{\nu }{|kB|})^{\frac{1}{3}} \log R +1 \right) \left( R^{-2}\Vert rw_k(0)\Vert _{L^2}^2 + (\frac{\nu }{|kB|})^{\frac{4}{3}}k^4 R^2\Vert \frac{w_k(0)}{r}\Vert _{L^2}^2\right) . \end{aligned}$$

Noting that $\nu$ is much smaller than 1, and B is a fixed constant, the leading terms on the right of above inequality are

$$\begin{aligned} (\log R)^{-2}R^{-4}\Vert r^2 w_k(0)\Vert _{L^2}^2+R^6 \Vert \frac{w_k(0)}{r^3}\Vert _{L^2}^2+R^{-2}\Vert rw_k(0)\Vert _{L^2}^2. \end{aligned}$$

Thus, both the coefficients in front of $\varphi _k^l$ terms on the left and the main initial-vorticity terms on the right are independent of the viscosity coefficient $\nu$. This phenomenon is commonly refer to as the effect of inviscid damping. Therefore, Lemma 4.6 presents a space-time version of linear inviscid damping around 2D TC flow.

To prove this result, in Lemma 4.8 we find and formulate a new set of explicit orthonormal basis of (weighted) eigenfunctions corresponding to the Laplace operator for circular flows, namely $-\partial _r^2+(k^2-\frac{1}{4})/r^2$. This basis can be represented in the closed form

$$\begin{aligned} \psi _l(r)=(\frac{2}{\log R})^{\frac{1}{2}}r^{1/2}\sin (\frac{l\pi }{\log R} \log r), \quad \text {for} \ l\in \mathbb {N}_+ \end{aligned}$$

and satisfies the equation

$$\begin{aligned} (\partial _r^2-\frac{k^2-\frac{1}{4}}{r^2})\psi _l=-\frac{(\frac{l\pi }{\log R} )^2+k^2}{r^2}\psi _l, \quad \psi _l|_{r=1, R}=0. \end{aligned}$$

Note that there is a weight $1/r^2$ on the right and $-(\frac{l\pi }{\log R} )^2-k^2$ is not the canonical eigenvalue. As far as we know, our explicit constructions here are new. With this basis, we move to prove a crucial estimate

$$\begin{aligned} |kB||k|(\log R)^2\left( \Vert \partial _r \tilde{\varphi }^{l_0}_k\Vert _{L^2 L^2}^2+k^2\Vert \frac{\tilde{\varphi }^{l_0}_k}{r}\Vert ^2_{L^2 L^2}\right) \lesssim \Vert r^{2}w_k(0)\Vert _{L^2}^2, \end{aligned}$$

(1.11)

where $k\in \mathbb {Z}\backslash \{0 \}$, B and R are constants with $R>1$, $w_k(0) \in L^2([1, R])$, and $\tilde{\varphi }^{l_0}_k$ is defined as the solution to the elliptic equations below:

$$\begin{aligned} (\partial _r^2-\frac{k^2-\frac{1}{4}}{r^2})\tilde{\varphi }^{l_0}_k=e^{-ikB\frac{t}{r^2}} w_k(0),\quad \tilde{\varphi }^{l_0}_k|_{r=1,R}=0,\quad r\in [1,R],\quad t\ge 0. \end{aligned}$$

Notice that the Laplace operator for the 2D Couette flow is given by $-\partial _y^2+k^2$ within the domain $y\in [-1,1]$. To find corresponding orthonormal eigenfunctions, it is natural to introduce the set $\{\sin \left( \pi j(y+1)/2\right) \}_{j=1}^{\infty }$, as described in [7]. However, for the circular Laplace operator $\partial _r^2-\frac{k^2-\frac{1}{4}}{r^2}$, we need to construct a set of more complicated eigenfunctions. The main issue is that if we directly consider the following eigenvalue problem:

$$\begin{aligned} (\partial _r^2-\frac{k^2-\frac{1}{4}}{r^2})\psi =\lambda \psi , \quad \psi |_{r=1, R}=0, \end{aligned}$$

it can be shown that there is no closed-form solution. In this paper, we instead solve the weighted eigenvalue problem:

$$\begin{aligned} (\partial _r^2-\frac{k^2-\frac{1}{4}}{r^2})\psi =\lambda w\psi , \quad \psi |_{r=1, R}=0 \end{aligned}$$

with $w(r)=\frac{1}{r^2}$. Utilizing change of variables, we obtain a family of closed-form solutions to this boundary value problem, which can be written in the following explicit expressions

$$\begin{aligned} \psi _l(r)=(\frac{2}{\log R})^{\frac{1}{2}}r^{1/2}\sin (\frac{l\pi }{\log R} \log r), \quad \text {for} \ l\in \mathbb {N}_+. \end{aligned}$$

These functions form an orthonormal basis of the weighted space $L^2_w([1, R])$. Therefore, with the aid of $\{ \psi _l \}_{l=1}^{\infty }$, we can evaluate

$$\begin{aligned} \begin{aligned} -\langle \tilde{w}^{l_0}_k, \tilde{\varphi }^{l_0}_k \rangle&= -\langle w^{-1}\tilde{w}^{l_0}_k, \tilde{\varphi }^{l_0}_k \rangle _w=\sum \limits _{l=1}^{\infty } \frac{1}{(\frac{l\pi }{\log R} )^2+k^2} \langle w^{-1}\tilde{w}^{l_0}_k, \psi _l \rangle _w \overline{\langle \tilde{w}^{l_0}_k , \psi _l \rangle }\\&=\sum \limits _{l=1}^{\infty } \frac{|\langle \tilde{w}^{l_0}_k , \psi _l \rangle |^2}{(\frac{l\pi }{\log R} )^2+k^2}, \end{aligned} \end{aligned}$$

where $\tilde{w}^{l_0}_k{:}{=}e^{-i kB\frac{t}{r^2}}w_k(0)$. Notice that

$$\begin{aligned} \begin{aligned} \langle \tilde{w}^{l_0}_k, \psi _l \rangle =&\int _{1}^{R} e^{-ikB\frac{t}{r^2}}w_k(0, r)\psi _l(r) dr\\ =&\int _{\frac{1}{R^2}}^1 e^{-i kBt s} w_k(0, \frac{1}{{\sqrt{s}}}) \psi _l (\frac{1}{{\sqrt{s}}})\frac{ds}{2s{\sqrt{s}}}. \end{aligned} \end{aligned}$$

Thus, employing Fourier transform and Plancherel’s formula, the desired estimate (1.11) follows.

It is worth noting that Lemma 4.6 also plays an important role in improving the nonlinear transition threshold of Theorem 1.3.

1.4.6 Space-time estimates for the linearized Navier–Stokes equations

In Proposition 5.1 of Sect. 5, we establish the space-time estimates for the linearized 2D Navier–Stokes equation written in the vorticity formulation (2.2):

$$\begin{aligned}&\partial _t w_k- \nu (\partial _r^2-\frac{k^2-\frac{1}{4}}{r^2})w_k+\frac{ikB}{r^2}w_k\\ {}&\quad +\frac{1}{r}[ik\sum _{l\in \mathbb {Z}}\partial _r(r^{-\frac{1}{2}}\varphi _l)w_{k-l}-r^{\frac{1}{2}}\partial _r(\sum _{l\in \mathbb {Z}}ilr^{-1}\varphi _lw_{k-l})]=0 \end{aligned}$$

with the stream function $\varphi _k$ satisfying

$$\begin{aligned}&(\partial _r^2-\frac{k^2-\frac{1}{4}}{r^2})\varphi _k=w_k,\quad \varphi _k|_{r=1,R}=0,\quad r\in [1,R],\quad t\ge 0. \end{aligned}$$

To achieve so, we employ the estimates in Sects. 3 and 4.

(1) The space-time estimate for the zero frequency part (i.e. $k=0$) can be directly obtained in view of the equation structure and integration by parts, as detailed in Lemma 5.4.

(2) For the nonzero frequency part with $k\in \mathbb {Z}\backslash \{0 \}$, we decompose the solution $w_k=w_k^l+w_k^{n}$, where $w_k^l$ obeys the homogeneous linear equation

$$\begin{aligned}&\partial _tw_k^l+L_kw_k^l=0,\quad w_k^l(0)=w_k(0), \end{aligned}$$

while $w_k^{n}$ is the solution to the inhomogeneous linear equation with zero initial condition

$$\begin{aligned}&\partial _tw_k^{n}+L_kw_k^{n}+\frac{1}{r}[ikf_1-r^{\frac{1}{2}}\partial _{r}(r^{\frac{1}{2}}f_2)]=0,\quad w_k^{n}(0)=0. \end{aligned}$$

Here we denote the nonlinear forms by $f_1$ and $f_2$, which are expressed as

$$\begin{aligned} f_1=\sum _{l\in \mathbb {Z}}\partial _r(r^{-\frac{1}{2}}\varphi _l)w_{k-l},\quad f_2=\sum _{l\in \mathbb {Z}}ilr^{-\frac{3}{2}}\varphi _lw_{k-l}. \end{aligned}$$

As a result, we also decompose $\varphi _k$ as $\varphi _k=\varphi ^{l}_k+\varphi ^{n}_k$, where $\varphi _k^l$ and $\varphi _k^n$ fulfill

$$\begin{aligned} (\partial _r^2-\frac{k^2-\frac{1}{4}}{r^2})\varphi ^{l}_k=w_k^l,\quad \varphi ^{l}_k|_{r=1,R}=0,\quad \text {with} \ r\in [1,R] \ \text {and} \ t\ge 0, \end{aligned}$$

and

$$\begin{aligned} (\partial _r^2-\frac{k^2-\frac{1}{4}}{r^2})\varphi ^{n}_k=w_k^n,\quad \varphi ^{n}_k|_{r=1,R}=0,\quad \text {with} \ r\in [1,R] \ \text {and} \ t\ge 0. \end{aligned}$$

For our space-time estimate, the first step is to provide a control of $w_k^l(t)$ based on the initial data $w_k^l(0)$. With the aid of the pseudospectrum bounds and enhanced dissipation effect proved in Sect. 3, we establish the following inequality in Lemma 4.4, which holds for any $k\in \mathbb {Z}$ and $|k|\ge 1$:

$$\begin{aligned}&\Vert e^{c'(\nu k^2)^{\frac{1}{3}}|B|^{\frac{2}{3}}R^{-2}t}w_k^l\Vert _{L^{\infty }L^2}^2+(\nu k^2)^{\frac{1}{3}}|B|^{\frac{2}{3}}R^{-2}\Vert e^{c'(\nu k^2)^{\frac{1}{3}}|B|^{\frac{2}{3}}R^{-2}t}w_k^l\Vert _{L^2L^2}^2\\&\quad +\nu \Vert e^{c'(\nu k^2)^{\frac{1}{3}}|B|^{\frac{2}{3}}R^{-2}t}\partial _{r}w_k^l\Vert _{L^2L^2}^2 +\nu k^2\Vert e^{c'(\nu k^2)^{\frac{1}{3}}|B|^{\frac{2}{3}}R^{-2}t}\frac{w_k^l}{r}\Vert _{L^2L^2}^2 \lesssim \Vert w_k(0)\Vert _{L^2}^2. \end{aligned}$$

To proceed, we derive an integrated linear invisid damping estimate, in which $\varphi _k^l$ can be controlled with the initial data of the vorticity

$$\begin{aligned} \begin{aligned}&k^2|B|R^{-4}(\Vert e^{c'(\nu k^2)^{\frac{1}{3}}|B|^{\frac{2}{3}}R^{-2}t}\partial _r{\varphi }^{l}_k\Vert _{L^2L^2}^2+|k|^2\Vert e^{c'(\nu k^2)^{\frac{1}{3}}|B|^{\frac{2}{3}}R^{-2}t}\frac{{\varphi }^{l}_k}{r}\Vert _{L^2L^2}^2) \\&\quad \lesssim (\log R)^{-2}R^{-4}\Vert r^2 w_k(0)\Vert _{L^2}^2+R^6 \Vert \frac{w_k(0)}{r^3}\Vert _{L^2}^2+(\frac{\nu }{|kB|})^{\frac{2}{3}}R^2\Vert \partial _r w_k(0)\Vert _{L^2}^2 \\&\qquad +\left( (\frac{\nu }{|kB|})^{\frac{1}{3}} \log R +1 \right) \left( R^{-2}\Vert rw_k(0)\Vert _{L^2}^2 + (\frac{\nu }{|kB|})^{\frac{4}{3}}k^4 R^2\Vert \frac{w_k(0)}{r}\Vert _{L^2}^2\right) , \end{aligned} \end{aligned}$$

as detailed in Lemma 4.6. Note that $\nu$ is much smaller than 1, and B is a fixed constant, so the initial vorticity has an upper bound independent of the viscosity coefficient $\nu$. By combining this observation with the estimates for $w_k^l$, we obtain the following inequality

$$\begin{aligned}&\Vert e^{c'(\nu k^2)^{\frac{1}{3}}|B|^{\frac{2}{3}}R^{-2}t}w_k^l\Vert _{L^{\infty }L^2}+|k||B|^{\frac{1}{2}}\Vert e^{c'(\nu k^2)^{\frac{1}{3}}|B|^{\frac{2}{3}}R^{-2}t}\partial _r{\varphi }^{l}_k\Vert _{L^2L^2}\\&\quad \lesssim C(R) (\Vert \partial _r w_k(0)\Vert _{L^2}+\Vert w_k(0)\Vert _{L^2}). \end{aligned}$$

It is worth mentioning that the coefficients in front of the terms on both sides of the inequality are independent of viscosity. Therefore, similar estimates can also hold for the inviscid flow (i.e. $\nu =0$).

By applying the resolvent estimates from Proposition 3.2, we then arrive at the conclusion stated in Proposition 4.5 for the nonlinear part, which involves the inhomogeneous linear equation with zero initial data. This result allows us to bound $w_k^n$ and $\varphi ^{n}_k$ by the nonlinear terms $f_1$ and $f_2$ as follows

$$\begin{aligned}&\Vert e^{c'(\nu k^2)^{\frac{1}{3}}|B|^{\frac{2}{3}}R^{-2}t}w_k^n\Vert _{L^{\infty }L^2}+(\nu k^2)^{\frac{1}{6}}|B|^{\frac{1}{3}}R^{-1}\Vert e^{c'(\nu k^2)^{\frac{1}{3}}|B|^{\frac{2}{3}}R^{-2}t}w_k^n\Vert _{L^2L^2}\\&\qquad +\nu ^{\frac{1}{2}}\Vert e^{c'(\nu k^2)^{\frac{1}{3}}|B|^{\frac{2}{3}}R^{-2}t}\partial _rw_k^n\Vert _{L^2L^2}+(\nu k^2)^{\frac{1}{2}}\Vert e^{c'(\nu k^2)^{\frac{1}{3}}|B|^{\frac{2}{3}}R^{-2}t}\frac{w_k^n}{r}\Vert _{L^2L^2}\\&\qquad +|B|^{\frac{1}{2}}R^{-2}\left( |k|\Vert e^{c'(\nu k^2)^{\frac{1}{3}}|B|^{\frac{2}{3}}R^{-2}t}\partial _r\varphi _k^n\Vert _{L^2L^2}+k^2\Vert e^{c'(\nu k^2)^{\frac{1}{3}}|B|^{\frac{2}{3}}R^{-2}t}\frac{\varphi _k^n}{r}\Vert _{L^2L^2}\right) \\&\quad \lesssim \left( |kB|^{-\frac{1}{2}} (\log R)^{\frac{1}{2}} +\nu ^{-\frac{1}{6}} |kB|^{-\frac{1}{3}} \right) \cdot \Vert e^{c'(\nu k^2)^{\frac{1}{3}}|B|^{\frac{2}{3}}R^{-2}t}kf_1\Vert _{L^2L^2}\\&\qquad +\nu ^{-\frac{1}{2}} \Vert e^{c'(\nu k^2)^{\frac{1}{3}}|B|^{\frac{2}{3}}R^{-2}t}f_2\Vert _{L^2L^2}. \end{aligned}$$

Finally, by combining the aforementioned three estimates, we derive the crucial proposition of Sect. 5, namely Proposition 5.1. This proposition states that the following inequality holds

$$\begin{aligned}&\Vert e^{c'(\nu k^2)^{\frac{1}{3}}|B|^{\frac{2}{3}}R^{-2}t}w_k\Vert _{L^{\infty }L^2}+(\nu k^2)^{\frac{1}{6}}|B|^{\frac{1}{3}}R^{-1}\Vert e^{c'(\nu k^2)^{\frac{1}{3}}|B|^{\frac{2}{3}}R^{-2}t}w_k\Vert _{L^2L^2}\\&\qquad +\nu ^{\frac{1}{2}}\Vert e^{c'(\nu k^2)^{\frac{1}{3}}|B|^{\frac{2}{3}}R^{-2}t}\partial _r w_k\Vert _{L^2L^2}+(\nu k^2)^{\frac{1}{2}}\Vert e^{c'(\nu k^2)^{\frac{1}{3}}|B|^{\frac{2}{3}}R^{-2}t}\frac{w_k}{r}\Vert _{L^2L^2}\\&\qquad +|B|^{\frac{1}{2}}R^{-2}\left( |k|\Vert e^{c'(\nu k^2)^{\frac{1}{3}}|B|^{\frac{2}{3}}R^{-2}t}\partial _r\varphi _k\Vert _{L^2L^2}+k^2\Vert e^{c'(\nu k^2)^{\frac{1}{3}}|B|^{\frac{2}{3}}R^{-2}t}\frac{\varphi _k}{r}\Vert _{L^2L^2}\right) \\&\quad \lesssim \Vert w_k(0)\Vert _{L^2}+R^{-2}(\log R)^{-2}\Vert r^2w_k(0)\Vert _{L^2}+R^{3}\Vert \frac{w_k(0)}{r^3}\Vert _{L^2}+(\frac{\nu }{|kB|})^{\frac{1}{3}}R\Vert \partial _rw_k(0)\Vert _{L^2}\\&\qquad +\left( 1+(\frac{\nu }{|kB|})^{\frac{1}{3}}\log R\right) ^{\frac{1}{2}} \cdot \left( R^{-1}\Vert rw_k(0)\Vert _{L^2}+R\Vert \frac{w_k(0)}{r}\Vert _{L^2}(\frac{\nu }{|kB|})^{\frac{2}{3}}k^2\right) \\&\qquad +\left( |kB|^{-\frac{1}{2}} (\log R)^{\frac{1}{2}} +\nu ^{-\frac{1}{6}} |kB|^{-\frac{1}{3}} \right) \cdot \Vert e^{c'(\nu k^2)^{\frac{1}{3}}|B|^{\frac{2}{3}}R^{-2}t}kf_1\Vert _{L^2L^2}\\&\qquad +\nu ^{-\frac{1}{2}} \Vert e^{c'(\nu k^2)^{\frac{1}{3}}|B|^{\frac{2}{3}}R^{-2}t}f_2\Vert _{L^2L^2}\\&\quad =:\mathcal {M}_k(0). \end{aligned}$$

We refer to it as the space-time estimates for the vorticity $w_k$ and the stream function $\varphi _k$.

1.4.7 Nonlinear stability

Finally, we close the energy estimate with a bootstrap argument. By utilizing the space-time estimates established in Proposition 5.1 from Sect. 5, we derive the result of Proposition 6.1:

$$\begin{aligned}&\Vert e^{c'\mu _kt}w_k\Vert _{L^{\infty }L^2}+\mu _k^{\frac{1}{2}}\Vert e^{c'\mu _kt}w_k\Vert _{L^2L^2}+\nu ^{\frac{1}{2}}\Vert e^{c'\mu _kt}w_k'\Vert _{L^2L^2}+(\nu k^2)^{\frac{1}{2}}\Vert e^{c'\mu _kt}\frac{w_k}{r}\Vert _{L^2L^2}\\&\qquad +\max \{|B|^{\frac{1}{2}}R^{-2}, (\nu k^2)^{\frac{1}{2}} R^{-2} \}\cdot \left( |k|\Vert e^{c'\mu _kt}\partial _r\varphi _k\Vert _{L^2L^2}+k^2\Vert e^{c'\mu _kt}\frac{\varphi _k}{r}\Vert _{L^2L^2}\right) \\&\quad \lesssim \mathcal {M}_k(0)+\mu _k^{-\frac{1}{2}}R^{-1}\Vert e^{c'\mu _kt}kf_1\Vert _{L^2L^2}+\nu ^{-\frac{1}{2}} \Vert e^{c'\mu _kt}f_2\Vert _{L^2L^2}. \end{aligned}$$

Here, we define $\mu _k{:}{=}\max \left\{ (\nu k^2)^{\frac{1}{3}} |B|^{\frac{2}{3}} R^{-2}, \nu k^2 R^{-2} \right\}$, as it is necessary to use distinct energy estimates for the high-frequency case of $\nu k^2\ge |B|$ and the low-frequency case of $\nu k^2\le |B|$ separately. In the case of 2D Couette flow [7], the discussions on different frequencies k correspond to $\nu k^2\ge 1$ and $\nu k^2\le 1$. It is important to mention that the frequency classification of k for the 2D TC flow also relies on the rotation parameter B.

We then define the energy functional $E_k$ as follows:

$$\begin{aligned} E_0&=\Vert w_0\Vert _{L^{\infty }L^2},\\ E_k&=\Vert e^{c'\mu _kt}w_k\Vert _{L^{\infty }L^2} +\mu _k^{\frac{1}{2}}\Vert e^{c'\mu _kt}w_k\Vert _{L^2L^2}+|B|^{\frac{1}{2}}|k|^{\frac{3}{2}}R^{-2}\Vert e^{c'\mu _kt}\frac{\varphi _k}{r^{\frac{1}{2}}}\Vert _{L^2L^{\infty }} \quad \text {for} \ |k|\ge 1. \end{aligned}$$

According to Proposition 6.1, it can be infered that for $k\in \mathbb {Z}\backslash \{0 \}$, it holds

$$\begin{aligned}&E_k\lesssim \mathcal {M}_k(0)+\mu _k^{-\frac{1}{2}}R^{-1}\Vert e^{c'\mu _kt}kf_1\Vert _{L^2L^2}+\nu ^{-\frac{1}{2}} \Vert e^{c'\mu _kt}f_2\Vert _{L^2L^2}. \end{aligned}$$

Utilizing Lemma 6.3, we can control nonlinear forms $f_1, f_2$ by

$$\begin{aligned} \mu _k^{-\frac{1}{2}}R^{-1}\Vert e^{c'(\nu k^2)^{\frac{1}{3}}|B|^{\frac{2}{3}}R^{-2}t}kf_1\Vert _{L^2L^2}\lesssim \nu ^{-\frac{1}{2}} |B|^{-\frac{1}{2}}R^2 (\frac{R}{R-1})^{\frac{1}{2}}(1+\log R)\sum _{l\in \mathbb {Z}}E_lE_{k-l}, \end{aligned}$$

and

$$\begin{aligned} \begin{aligned} \nu ^{-\frac{1}{2}} \Vert e^{c'\mu _k t}f_2\Vert _{L^2L^2} \le \nu ^{-\frac{1}{2}} |B|^{-\frac{1}{2}}R^2\sum _{l\in \mathbb {Z}\backslash \{0, k\}}E_lE_{k-l}. \end{aligned} \end{aligned}$$

This allows us to obtain Lemma 6.4, which states that

$$\begin{aligned} E_k\lesssim \mathcal {M}_k(0) +\nu ^{-\frac{1}{2}}|B|^{-\frac{1}{2}}R^2 (\frac{R}{R-1})^{\frac{1}{2}}(1+\log R)\sum _{l\in \mathbb {Z}}E_lE_{k-l}. \end{aligned}$$

Simultaneously, we can derive analogous estimates for $E_0$ by introducing $\mathcal {M}_0(0)=\Vert w_0(0)\Vert _{L^2}$. Incorporating all estimates above through a bootstrap argument, we finally arrive at the main theorem of this paper: Theorem 6.1.

1.4.8 Subcritical transition and transition threshold

Theorem 1.2 and Theorem 1.3 follow from Proposition 4.3 and Theorem 6.1, respectively. Here are some further remarks regarding these two main theorems.

On one hand, Theorem 1.2 shows that the linearized Navier–Stokes equations around the 2D Taylor–Couette flow is dynamically stable at any Reynolds number (including the inviscid case), and the vorticity at any time can be controlled by the given initial value in $L^2$ norm. On the other hand, for the full Navier–Stokes equation, we only expect nonlinear asymptotic stability within the range of small perturbation of size $\nu ^{\beta (s)}(\beta (s)>0)$ in $H^s$ space. This proved in our Theorem 1.3, provided the initial perturbation does not exceed $\nu ^\frac{1}{2} |B|^{\frac{1}{2}}R^{-2}$. The exponent $\frac{1}{2}$ of $\nu$ agrees with that of the 2D Couette flow [5, 7], and provides an upper bound $\frac{1}{2}$ for the corresponding transition threshold $\beta$. The dependency of the subcritical transition on the rotational speed B is captured by $|B|^{\frac{1}{2}}$. This indicates that the system can achieve global stability under a wider range of initial data via taking account of the rotating effect.

2 Derivation of the Perturbative Equation

We take the Fourier transform in the $\theta$ direction and denote the Fourier coefficients of w and $\varphi$ by $\hat{w}_k$ and $\hat{\varphi }_k$, respectively. Using this notation, the Eq. (1.7) can be rewritten as follows:

$$\begin{aligned}&\partial _t\hat{w}_k-\nu [(\partial _{r}^2+\frac{1}{r}\partial _{r}-\frac{k^2}{r^2})]\hat{w}_k+(A+\frac{B}{r^2})ik\hat{w}_k\nonumber \\&\quad +\frac{1}{r}[\sum _{l\in \mathbb {Z}}i(k-l)\partial _r\hat{\varphi }_l\hat{w}_{k-l}-\sum _{l\in \mathbb {Z}}il\hat{\varphi }_{l}\partial _{r}\hat{w}_{k-l}]=0 \end{aligned}$$

(2.1)

with $\hat{\varphi }_k$ satisfying $(\partial _r^2+\frac{1}{r}\partial _r-\frac{k^2}{r^2})\hat{\varphi }_k=\hat{w}_k$.

To eliminate the first derivative $\frac{1}{r}\partial _r$, we introduce the weight $r^{\frac{1}{2}}$ and define $w_k$ and $\varphi _k$ as $w_k{:}{=}r^{\frac{1}{2}}e^{ikAt}\hat{w}_k$ and $\varphi _k{:}{=}r^{\frac{1}{2}} e^{ikAt}\hat{\varphi }_k$. As a consequence, the Eq. (2.1) is transferred into

$$\begin{aligned}&\partial _t w_k- \nu (\partial _r^2-\frac{k^2-\frac{1}{4}}{r^2})w_k+\frac{ikB}{r^2}w_k\nonumber \\&\quad +\frac{1}{r}[ik\sum _{l\in \mathbb {Z}}\partial _r(r^{-\frac{1}{2}}\varphi _l)w_{k-l}-r^{\frac{1}{2}}\partial _r(\sum _{l\in \mathbb {Z}}ilr^{-1}\varphi _lw_{k-l})]=0, \end{aligned}$$

(2.2)

where $\varphi _k$ satisfies

$$\begin{aligned}&(\partial _r^2-\frac{k^2-\frac{1}{4}}{r^2})\varphi _k=w_k,\quad \varphi _k|_{r=1,R}=0\quad \text {with}\ r\in [1,R] \ \text {and} \ t\ge 0. \end{aligned}$$

(2.3)

Note that the Navier boundary condition

$$\begin{aligned} w_k|_{r=1,R}=0 , \ \varphi _k|_{r=1,R}=0 \quad \text {for any}\ k\in \mathbb {Z} \end{aligned}$$

implies that $\partial _t w_k$ and the nonlinear terms in (2.2) vanish on the boundary of interval [1, R]. This forces the second derivative of $w_k$ to be zero on the boundary, i.e.

$$\begin{aligned} \partial _r^2 w_k|_{r=1,R}=0 \quad \text {for any}\ k\in \mathbb {Z}. \end{aligned}$$

Up to this point, we convert our problem to exploring the dynamics and long-time behaviors of (2.2).

Remark 2.1

In below, for two quantities A and B, we frequently use $A\lesssim B$ in short to stand for the inequality $A\le C B$ with some universal constant $C>0$, that is independent of $\nu , k, B, \lambda$ and R. Additionally, we also write $A\approx B$ to indicate that both $A\lesssim B$ and $B\lesssim A$ are true.

3 Resolvent Estimates

To establish decays of the linearized equation of (2.2) as below

$$\begin{aligned} \partial _t w_k&- \nu (\partial _r^2-\frac{k^2-\frac{1}{4}}{r^2})w_k+\frac{ikB}{r^2}w_k=0, \end{aligned}$$

the key step is to derive the resolvent estimates that will be presented in Proposition 3.2. More precisely, we want to study the resolvent equation subject to the following (Navier) boundary condition for any $\lambda \in \mathbb {R}$. After taking the Fourier transform with respect to time t, the resolvent equation becomes

$$\begin{aligned}&-\nu (\partial _r^2-\frac{k^2-\frac{1}{4}}{r^2})w+ikB(\frac{1}{r^2}-\lambda )w=F \quad \text {with} \ w|_{r=1,R}=0 \ \text {and} \ \partial _r^2 w_k|_{r=1,R}=0. \end{aligned}$$

(3.1)

The domain of the operator is defined as

$$\begin{aligned} D_k=\{w\in H_{loc}^2(\mathbb {R}_{+},dr)\cap L^2(\mathbb {R}_{+},dr):-\nu (\partial _r^2-\frac{k^2-\frac{1}{4}}{r^2})w+i\frac{kB}{r^2}w\in L^2(\mathbb {R}_{+},dr) \}. \end{aligned}$$

Note that for any $|k|\ge 1$, it holds

$$\begin{aligned} D_k&=\{w\in L^2(\mathbb {R}_{+},dr):\partial _r^2w,\frac{w}{r^2}\in L^2(\mathbb {R}_{+},dr) \}. \end{aligned}$$

We also introduce

$$\begin{aligned} \Vert f\Vert _{H^1_r}^2{:}{=}\Vert f'\Vert _{L^2}^2+\Vert \frac{f}{r}\Vert _{L^2}^2 \quad \text {and} \quad \Vert f\Vert _{H^{-1}_r}{:}{=}\sup \limits _{\Vert g\Vert _{H^1_r}\le 1} |\langle f,g\rangle |. \end{aligned}$$

Here $\langle \, \ \rangle$ represents the canonical inner product in $L^2(\mathbb {R}_{+},dr)$.

In this section, we will present the resolvent estimates for $w'$, $\frac{w}{r}$, $\varphi '$ and $\frac{\varphi }{r}$ with respect to F in both $L^2$ norm and $H_r^{-1}$ norm.

3.1 Coercive estimates of the real part

We start with the coercive estimates for the real part of $\langle F, w \rangle$, which will be used to obtain the desired resolvent estimates for w.

Lemma 3.1

For any $|k|\ge 1$ and $w\in D_k$, it holds

$$\begin{aligned}&\Re \langle F,w\rangle =\nu \Vert w'\Vert _{L^2}^2+\nu (k^2-\frac{1}{4})\Vert \frac{w}{r}\Vert _{L^2}^2. \end{aligned}$$

(3.2)

Proof

Via integration by parts, one can directly check

$$\begin{aligned}&\Re \langle F,w\rangle =\Re \langle -\nu (\partial _r^2-\frac{k^2-\frac{1}{4}}{r^2})w,w\rangle =\nu \Vert w'\Vert _{L^2}^2+\nu (k^2-\frac{1}{4})\Vert \frac{w}{r}\Vert _{L^2}^2. \end{aligned}$$

$\square$

3.2 Resolvent estimates

For Eq. (3.1), this subsection aims to establish upper bounds for w and $\varphi$ in terms of F. We prove the following inequalities.

Proposition 3.2

For any $|k|\ge 1$, $\lambda \in \mathbb {R}$ and $w\in D_k$, there exist constants $C,c>0$ independent of $\nu ,k,B,\lambda ,R$, such that for any $0\le c'\le c$, we have

$$\begin{aligned}&\nu ^{\frac{2}{3}} |kB|^{\frac{1}{3}}\Vert w'\Vert _{L^2}+\nu ^{\frac{1}{3}} |kB|^{\frac{2}{3}}\Vert \frac{w}{r}\Vert _{L^2} \le C\Vert rF-c'\nu ^{\frac{1}{3}} |kB|^{\frac{2}{3}}\frac{w}{r}\Vert _{L^2},\\&\nu ^{\frac{1}{6}}|kB|^{\frac{5}{6}}|k|^{\frac{1}{2}}\left( \Vert \varphi '\Vert _{L^2}+|k|\Vert \frac{\varphi }{r}\Vert _{L^2} \right) \le CR^2\left( (\frac{\nu }{|kB|})^{\frac{1}{6}} (\log R)^{\frac{1}{2}} +1 \right) \Vert rF-c'\nu ^{\frac{1}{3}} |kB|^{\frac{2}{3}}\frac{w}{r}\Vert _{L^2}. \end{aligned}$$

Moreover, it also holds

$$\begin{aligned} \nu \Vert w\Vert _{H^1_r}+\nu ^{\frac{2}{3}}|kB|^{\frac{1}{3}}\Vert \frac{w}{r}\Vert _{L^2}\le&C\Vert F-c'\nu ^{\frac{1}{3}} |kB|^{\frac{2}{3}} R^{-2}w\Vert _{H^{-1}_r},\\ \nu ^{\frac{1}{2}}|kB|^{\frac{1}{2}}\Vert \varphi '\Vert _{L^2}+\nu ^{\frac{1}{2}}|k||kB|^{\frac{1}{2}}\Vert \frac{\varphi }{r}\Vert _{L^2}\le&C R^2 \Vert F-c'\nu ^{\frac{1}{3}}|kB|^{\frac{2}{3}}R^{-2}w\Vert _{H^{-1}_r}. \end{aligned}$$

The proof of this proposition can be separated into four parts. We first derive the resolvent estimates for w from $L^2$ to $L^2$.

Lemma 3.3

For any $|k|\ge 1$, $\lambda \in \mathbb {R}$ and $w\in D_k$, there exist constants $C>0$ independent of $\nu ,k,B,\lambda ,R$, such that the following estimate holds

$$\begin{aligned} \nu ^{\frac{2}{3}} |kB|^{\frac{1}{3}}\Vert w'\Vert _{L^2}+\nu ^{\frac{1}{3}} |kB|^{\frac{2}{3}}\Vert \frac{w}{r}\Vert _{L^2}+|kB|\Vert r(\frac{1}{r^2}-\lambda )w\Vert _{L^2}\le C \Vert rF\Vert _{L^2}. \end{aligned}$$

(3.3)

Moreover, there exist constants $C,c>0$ independent of $\nu ,k,B,\lambda ,R$, such that for any $0\le c'\le c$, it is also true that

$$\begin{aligned}&\nu ^{\frac{2}{3}} |kB|^{\frac{1}{3}}\Vert w'\Vert _{L^2}+\nu ^{\frac{1}{3}} |kB|^{\frac{2}{3}}\Vert \frac{w}{r}\Vert _{L^2} \le C\Vert rF-c'\nu ^{\frac{1}{3}} |kB|^{\frac{2}{3}}\frac{w}{r}\Vert _{L^2}. \end{aligned}$$

Proof

We first prove (3.3). The second inequality readily follows if one selects $c=\frac{1}{2C}$ since for all $c'\in [0, c]$ it holds

$$\begin{aligned} C \Vert rF\Vert _{L^2}\le C\Vert rF-c'\nu ^{\frac{1}{3}} |kB|^{\frac{2}{3}}\frac{w}{r}\Vert _{L^2}+\frac{1}{2}\nu ^{\frac{1}{3}} |kB|^{\frac{2}{3}}\Vert \frac{w}{r}\Vert _{L^2}. \end{aligned}$$

By utilizing Lemma 3.1 and Cauchy–Schwarz inequality, we obtain

$$\begin{aligned} \nu \Vert w'\Vert _{L^2}^2\le \Vert rF\Vert _{L^2}\Vert \frac{w}{r}\Vert _{L^2}. \end{aligned}$$

(3.4)

To proceed, we then first prove the statement

$$\begin{aligned} \nu ^{\frac{1}{3}} |kB|^{\frac{2}{3}}\Vert \frac{w}{r}\Vert _{L^2}\le C \Vert rF\Vert _{L^2}. \end{aligned}$$

(3.5)

The mathematical discussion can be divided into three cases:

$$\begin{aligned} \lambda \in (-\infty ,\frac{1-\delta }{R^2}]\cup [1+\delta ,\infty ),\quad \lambda \in [\frac{1-\delta }{R^2},\frac{1}{R^2}]\cup [1,1+\delta ],\quad \lambda \in [\frac{1}{R^2},1]. \end{aligned}$$

Here $0<\delta \ll 1$ is a small constant, which will be determined later.

• (1) Case of $\lambda \in (-\infty ,\frac{1-\delta }{R^2}]\cup [1+\delta ,\infty )$. One can check

  $$\begin{aligned} |\Im \langle F,w\rangle |=|kB|\big |\langle (\frac{1}{r^2}-\lambda )w,w\rangle \big |=|kB|\big |\langle (1-\lambda r^2)\frac{w}{r},\frac{w}{r}\rangle \big |\ge |kB|\delta \Vert \frac{w}{r}\Vert _{L^2}^2. \end{aligned}$$

  Taking $\delta =\frac{\nu ^{\frac{1}{3}}}{|kB|^{\frac{1}{3}}}$, it yields

  $$\begin{aligned} \Vert rF\Vert _{L^2}\ge \nu ^{\frac{1}{3}}|kB|^{\frac{2}{3}}\Vert \frac{w}{r}\Vert _{L^2}. \end{aligned}$$

  (3.6)

• (2) Case of $\lambda \in [\frac{1-\delta }{R^2},\frac{1}{R^2}]$ or $\lambda \in [1,1+\delta ]$. i)$\lambda \in [\frac{1-\delta }{R^2},\frac{1}{R^2}]$. Noting that $\sqrt{\frac{1-\delta }{\lambda }}\in [1,R]$, we first obtain

  $$\begin{aligned} |\Im \langle F,w\rangle |&=|kB|\int _1^R(1-\lambda r^2)|\frac{w}{r}|^2dr\\&\ge |kB|\int _1^{\sqrt{\frac{1-\delta }{\lambda }}}(1-\lambda r^2)|\frac{w}{r}|^2dr\ge |kB|\delta \Vert \frac{w}{r}\Vert _{L^2(1,\sqrt{\frac{1-\delta }{\lambda }})}^2. \end{aligned}$$

  The rest part of $\Vert \frac{w}{r}\Vert _{L^2}$ can be controlled as below

  $$\begin{aligned}&\Vert \frac{w}{r}\Vert _{L^2(\sqrt{\frac{1-\delta }{\lambda }},R)}^2=\int _{\sqrt{\frac{1-\delta }{\lambda }}}^R|\frac{w}{r}|^2dr\le \int _{\sqrt{\frac{1-\delta }{\lambda }}}^R\frac{1}{r^2}dr\Vert w\Vert _{L^{\infty }}^2\\&\quad =(\sqrt{\frac{\lambda }{1-\delta }}-\frac{1}{R})\Vert w\Vert _{L^{\infty }}^2 \le (\sqrt{\frac{1}{1-\delta }}-1)\frac{\Vert w\Vert _{L^{\infty }}^2}{R}\lesssim \frac{\delta }{R}\Vert w\Vert _{L^{\infty }}^2\le \delta \Vert \frac{w}{r^{\frac{1}{2}}}\Vert _{L^{\infty }}^2. \end{aligned}$$

  Combining these two bounds, we deduce

  $$\begin{aligned} \Vert \frac{w}{r}\Vert _{L^2}^2&=\Vert \frac{w}{r}\Vert _{L^2(1,\sqrt{\frac{1-\delta }{\lambda }})}^2+\Vert \frac{w}{r}\Vert _{L^2(\sqrt{\frac{1-\delta }{\lambda }},R)}^2\\&\lesssim |kB|^{-1}\delta ^{-1}|\Im \langle F,w\rangle |+\delta \Vert \frac{w}{r^{\frac{1}{2}}}\Vert _{L^{\infty }}^2. \end{aligned}$$

  It is then inferred from Lemma A.1 and Lemma 3.1 that

  $$\begin{aligned} \Vert \frac{w}{r}\Vert _{L^2}^2&\lesssim |kB|^{-1}\delta ^{-1}|\Im \langle F,w\rangle |+\delta \Vert \frac{w}{r^{\frac{1}{2}}}\Vert _{L^{\infty }}^2\\&\lesssim |kB|^{-1}\delta ^{-1}\Vert rF\Vert _{L^2}\Vert \frac{w}{r}\Vert _{L^2}+\delta \Vert \frac{w}{r}\Vert _{L^2}\Vert r^{\frac{1}{2}}\partial _r(\frac{w}{r^{\frac{1}{2}}})\Vert _{L^2}\\&\lesssim |kB|^{-1}\delta ^{-1}\Vert rF\Vert _{L^2}\Vert \frac{w}{r}\Vert _{L^2}+\delta \Vert \frac{w}{r}\Vert _{L^2}(\Vert w'\Vert _{L^2}+\Vert \frac{w}{r}\Vert _{L^2})\\&\lesssim |kB|^{-1}\delta ^{-1}\Vert rF\Vert _{L^2}\Vert \frac{w}{r}\Vert _{L^2}+\nu ^{-\frac{1}{2}}\delta \Vert rF\Vert _{L^2}^{\frac{1}{2}}\Vert \frac{w}{r}\Vert _{L^2}^{\frac{3}{2}}. \end{aligned}$$

  This leads to

  $$\begin{aligned} \Vert rF\Vert _{L^2}\gtrsim \min \{|kB|\delta ,\nu \delta ^{-2}\}\Vert \frac{w}{r}\Vert _{L^2}. \end{aligned}$$

  The desired estimate (3.5) follows by picking $\delta =\frac{\nu ^{\frac{1}{3}}}{|kB|^{\frac{1}{3}}}$. ii)$\lambda \in [1,1+\delta ]$. Observing $\sqrt{\frac{1+\delta }{\lambda }}\in [1, R]$, we have

  $$\begin{aligned} |\Im \langle F,w\rangle |&=|kB|\int _1^R(\lambda r^2-1)|\frac{w}{r}|^2dr\\&\ge |kB|\int _{\sqrt{\frac{1+\delta }{\lambda }}}^R(\lambda r^2-1)|\frac{w}{r}|^2dr\ge |kB|\delta \Vert \frac{w}{r}\Vert _{L^2(\sqrt{\frac{1+\delta }{\lambda }},R)}^2. \end{aligned}$$

  When $r\in (1,\sqrt{\frac{1+\delta }{\lambda }})$, we bound $\frac{w}{r}$ by the $L^\infty$ norm of $\frac{w}{r^{\frac{1}{2}}}$ and it holds

  $$\begin{aligned}&\Vert \frac{w}{r}\Vert _{L^2(1,\sqrt{\frac{1+\delta }{\lambda }})}^2=\int _1^{\sqrt{\frac{1+\delta }{\lambda }}}|\frac{w}{r}|^2dr \le \int _1^{\sqrt{\frac{1+\delta }{\lambda }}}\frac{1}{r}dr\Vert \frac{w}{r^{\frac{1}{2}}}\Vert _{L^{\infty }}^2\\&\quad =(\sqrt{\frac{1+\delta }{\lambda }}-1)\Vert \frac{w}{r^{\frac{1}{2}}}\Vert _{L^{\infty }}^2 \le (\sqrt{1+\delta }-1)\Vert \frac{w}{r^{\frac{1}{2}}}\Vert _{L^{\infty }}^2\le \delta \Vert \frac{w}{r^{\frac{1}{2}}}\Vert _{L^{\infty }}^2. \end{aligned}$$

  Summing two inequalities above renders

  $$\begin{aligned} \Vert \frac{w}{r}\Vert _{L^2}^2&=\Vert \frac{w}{r}\Vert _{L^2(1,\sqrt{\frac{1+\delta }{\lambda }})}^2+\Vert \frac{w}{r}\Vert _{L^2(\sqrt{\frac{1+\delta }{\lambda }},R)}^2\\&\lesssim \delta \Vert \frac{w}{r^{\frac{1}{2}}}\Vert _{L^{\infty }}^2+|kB|^{-1}\delta ^{-1}|\Im \langle F,w\rangle |. \end{aligned}$$

  In view of Lemma A.1 and Lemma 3.1, we further deduce

  $$\begin{aligned} \Vert \frac{w}{r}\Vert _{L^2}^2&\lesssim |kB|^{-1}\delta ^{-1}|\Im \langle F,w\rangle |+\delta \Vert \frac{w}{r^{\frac{1}{2}}}\Vert _{L^{\infty }}^2\\&\lesssim |kB|^{-1}\delta ^{-1}\Vert rF\Vert _{L^2}\Vert \frac{w}{r}\Vert _{L^2}+\delta \Vert \frac{w}{r}\Vert _{L^2}\Vert r^{\frac{1}{2}}\partial _r(\frac{w}{r^{\frac{1}{2}}})\Vert _{L^2}\\&\lesssim |kB|^{-1}\delta ^{-1}\Vert rF\Vert _{L^2}\Vert \frac{w}{r}\Vert _{L^2}+\delta \Vert \frac{w}{r}\Vert _{L^2}(\Vert w'\Vert _{L^2}+\Vert \frac{w}{r}\Vert _{L^2})\\&\lesssim |kB|^{-1}\delta ^{-1}\Vert rF\Vert _{L^2}\Vert \frac{w}{r}\Vert _{L^2}+\nu ^{-\frac{1}{2}}\delta \Vert rF\Vert _{L^2}^{\frac{1}{2}}\Vert \frac{w}{r}\Vert _{L^2}^{\frac{3}{2}}. \end{aligned}$$

  This yields

  $$\begin{aligned} \Vert rF\Vert _{L^2}\gtrsim \min \{|kB|\delta ,\nu \delta ^{-2}\}\Vert \frac{w}{r}\Vert _{L^2}. \end{aligned}$$

  By setting $\delta =\frac{\nu ^{\frac{1}{3}}}{|kB|^{\frac{1}{3}}}$, we can optimize the inequality above and thus obtain

  $$\begin{aligned} \Vert rF\Vert _{L^2}\gtrsim \nu ^{\frac{1}{3}}|kB|^{\frac{2}{3}}\Vert \frac{w}{r}\Vert _{L^2}. \end{aligned}$$

  (3.7)

• (3) Case of $\lambda \in [\frac{1}{R^2},1]$. Using the fact

  $$\begin{aligned} |1-\lambda r^2|\le \delta \Rightarrow 1-\delta \le \lambda r^2\le 1+\delta \Rightarrow \sqrt{\frac{1-\delta }{\lambda }}\le r\le \sqrt{\frac{1+\delta }{\lambda }}, \end{aligned}$$

  for small $\delta >0$, it can be seen that

  $$\begin{aligned} \sqrt{\frac{1+\delta }{\lambda }}-\sqrt{\frac{1-\delta }{\lambda }}=\frac{2\delta }{\sqrt{\lambda }(\sqrt{1+\delta }+\sqrt{1-\delta })}\lesssim \frac{\delta }{\sqrt{\lambda }}. \end{aligned}$$

  Now we choose $r_{-}\in (\sqrt{\frac{1-\delta }{\lambda }}-\frac{\delta }{\sqrt{\lambda }},\sqrt{\frac{1-\delta }{\lambda }})$ and $r_{+}\in (\sqrt{\frac{1+\delta }{\lambda }},\sqrt{\frac{1+\delta }{\lambda }}+\frac{\delta }{\sqrt{\lambda }})$ such that the following inequalities hold

  $$\begin{aligned} |w'(r_{-})|^2\le \frac{\Vert w'\Vert _{L^2}^2}{\delta /\sqrt{\lambda }},\quad |w'(r_{+})|^2\le \frac{\Vert w'\Vert _{L^2}^2}{\delta /\sqrt{\lambda }}. \end{aligned}$$

  (3.8)

  In order to control $\Vert \frac{w}{r}\Vert _{L^2((1, r_{-})\cup (r_{+},R))}$, we examine the inner product of F and $w(\chi _{(1,r_{-})}-\chi _{(r_{+},R)})$. Via integration by parts, we write

  $$\begin{aligned}&\langle F,w(\chi _{(1,r_{-})}-\chi _{(r_{+},R)})\rangle \\&\quad =-\nu \int _1^{r_{-}}w''\overline{w}dr+\nu \int _{r_{+}}^Rw''\overline{w}dr+\nu (k^2-\frac{1}{4})[\int _1^{r_{-}}\frac{|w|^2}{r^2}dr-\int _{r_{+}}^R\frac{|w|^2}{r^2}dr]\\&\qquad +ikB\big (\int _1^{r_{-}}(\frac{1}{r^2}-\lambda )|w|^2dr+\int _{r_{+}}^R(\lambda -\frac{1}{r^2})|w|^2dr\big )\\&\quad =\nu \int _1^{r_{-}}|w'|^2dr-\nu \int _{r_{+}}^R|w'|^2dr-\nu [w'(r_{-})\overline{w}(r_{-})+w'(r_{+})\overline{w}(r_{+})]\\&\qquad +\nu (k^2-\frac{1}{4})[\int _1^{r_{-}}\frac{|w|^2}{r^2}dr-\int _{r_{+}}^R\frac{|w|^2}{r^2}dr]\\&\qquad +ikB\big (\int _1^{r_{-}}(1-\lambda r^2)\frac{|w|^2}{r^2}dr+\int _{r_{+}}^R(\lambda r^2-1)\frac{|w|^2}{r^2}dr\big ). \end{aligned}$$

  Taking the imaginary part of above equality, it indicates that

  $$\begin{aligned}&|kB|\Big |\int _1^{r_{-}}(1-\lambda r^2)\frac{|w|^2}{r^2}dr+\int _{r_{+}}^R(\lambda r^2-1)\frac{|w|^2}{r^2}dr\Big |\\&\quad \le \Vert rF\Vert _{L^2}\Vert \frac{w}{r}\Vert _{L^2}+\nu \big (|w'(r_{-})\overline{w}(r_{-})|+|w'(r_{+})\overline{w}(r_{+})|\big ). \end{aligned}$$

  Thanks to the choice of $r_-$ and $r_+$, we can derive the estimate for $\Vert \frac{w}{r}\Vert _{L^2\big ((1, r{-})\cup (r_{+},R)\big )}$ as below

  $$\begin{aligned}&\Vert \frac{w}{r}\Vert _{L^2((1, r_{-})\cup (r_{+},R))}^2\nonumber \\&\quad \le |kB|^{-1}\delta ^{-1}\Big [\Vert rF\Vert _{L^2}\Vert \frac{w}{r}\Vert _{L^2}+\nu \big (|w'(r_{-})\overline{w}(r_{-})|+|w'(r_{+})\overline{w}(r_{+})|\big )\Big ]. \end{aligned}$$

  (3.9)

  According to (3.8), the second term on the right-hand side of (3.9) can be bounded by

  $$\begin{aligned}&\nu \big (|w'(r_{-})\overline{w}(r_{-})|+|w'(r_{+})\overline{w}(r_{+})|\big )\le \frac{\nu \lambda ^{\frac{1}{4}}}{\delta ^{\frac{1}{2}}}\Vert w'\Vert _{L^2}\Big (|\overline{w}(r_{-})|+|\overline{w}(r_{+})|\Big )\\&\quad \le \frac{\nu \lambda ^{\frac{1}{4}}}{\delta ^{\frac{1}{2}}}(r_{-}^{\frac{1}{2}}+r_{+}^{\frac{1}{2}})\Vert w'\Vert _{L^2}\Vert \frac{w}{r^{\frac{1}{2}}}\Vert _{L^{\infty }}\lesssim \frac{\nu \lambda ^{\frac{1}{4}}}{\delta ^{\frac{1}{2}}}\lambda ^{-\frac{1}{4}}\Vert w'\Vert _{L^2}\Vert \frac{w}{r^{\frac{1}{2}}}\Vert _{L^{\infty }}\\&\quad =\frac{\nu }{\delta ^{\frac{1}{2}}}\Vert w'\Vert _{L^2}\Vert \frac{w}{r^{\frac{1}{2}}}\Vert _{L^{\infty }}. \end{aligned}$$

  Thus, we obtain the following estimate for $\Vert \frac{w}{r}\Vert _{L^2}$:

  $$\begin{aligned}&\Vert \frac{w}{r}\Vert _{L^2}^2=\Vert \frac{w}{r}\Vert _{L^2((1, r_{-})\cup (r_{+},R))}^2+\Vert \frac{w}{r}\Vert _{L^2(r_{-},r_{+})}^2\\&\quad \le |kB|^{-1}\delta ^{-1}\Big (\Vert rF\Vert _{L^2}\Vert \frac{w}{r}\Vert _{L^2}+\frac{\nu }{\delta ^{\frac{1}{2}}}\Vert w'\Vert _{L^2}\Vert \frac{w}{r^{\frac{1}{2}}}\Vert _{L^{\infty }}\Big )+\int _{r_{-}}^{r_{+}}\frac{1}{r}dr\Vert \frac{w}{r^{\frac{1}{2}}}\Vert _{L^\infty }^2 \\&\quad \lesssim |kB|^{-1}\delta ^{-1}\Big (\Vert rF\Vert _{L^2}\Vert \frac{w}{r}\Vert _{L^2}+\frac{\nu }{\delta ^{\frac{1}{2}}}\Vert w'\Vert _{L^2}\Vert \frac{w}{r^{\frac{1}{2}}}\Vert _{L^{\infty }}\Big )+\delta \Vert \frac{w}{r^{\frac{1}{2}}}\Vert _{L^\infty }^2, \end{aligned}$$

  where in the last line we employ Lemma A.6 from the Appendix. Utilizing Lemma 3.1 and Lemma A.1, it can be further deduced that

  $$\begin{aligned} \Vert \frac{w}{r}\Vert _{L^2}^2&\lesssim |kB|^{-1}\delta ^{-1}\Big (\Vert rF\Vert _{L^2}\Vert \frac{w}{r}\Vert _{L^2}+\frac{\nu }{\delta ^{\frac{1}{2}}}\Vert w'\Vert _{L^2}\Vert \frac{w}{r^{\frac{1}{2}}}\Vert _{L^{\infty }}\Big )+\delta \Vert \frac{w}{r^{\frac{1}{2}}}\Vert _{L^\infty }^2\\&\lesssim |kB|^{-1}\delta ^{-1}\Big (\Vert rF\Vert _{L^2}\Vert \frac{w}{r}\Vert _{L^2}+\frac{\nu ^{\frac{1}{2}}}{\delta ^{\frac{1}{2}}}\Vert rF\Vert _{L^2}^{\frac{1}{2}}\Vert \frac{w}{r}\Vert _{L^2}^{\frac{1}{2}}\Vert \frac{w}{r^{\frac{1}{2}}}\Vert _{L^{\infty }}\Big )+\delta \Vert \frac{w}{r^{\frac{1}{2}}}\Vert _{L^\infty }^2\\&\lesssim |kB|^{-1}\delta ^{-1}\Vert rF\Vert _{L^2}\Vert \frac{w}{r}\Vert _{L^2}+|kB|^{-2}\delta ^{-4}\nu \Vert rF\Vert _{L^2}\Vert \frac{w}{r}\Vert _{L^2}+\delta \Vert \frac{w}{r^{\frac{1}{2}}}\Vert _{L^\infty }^2\\&\lesssim |kB|^{-1}\delta ^{-1}\Vert rF\Vert _{L^2}\Vert \frac{w}{r}\Vert _{L^2}+|kB|^{-2}\delta ^{-4}\nu \Vert rF\Vert _{L^2}\Vert \frac{w}{r}\Vert _{L^2}+\delta \Vert \frac{w}{r}\Vert _{L^2}\Vert r^{\frac{1}{2}}\partial _r(\frac{w}{r^{\frac{1}{2}}})\Vert _{L^2}\\&\lesssim |kB|^{-1}\delta ^{-1}\Vert rF\Vert _{L^2}\Vert \frac{w}{r}\Vert _{L^2}+|kB|^{-2}\delta ^{-4}\nu \Vert rF\Vert _{L^2}\Vert \frac{w}{r}\Vert _{L^2}\\&\quad +\delta \Vert \frac{w}{r}\Vert _{L^2}(\Vert w'\Vert _{L^2}+\Vert \frac{w}{r}\Vert _{L^2})\\&\lesssim |kB|^{-1}\delta ^{-1}\Vert rF\Vert _{L^2}\Vert \frac{w}{r}\Vert _{L^2}+|kB|^{-2}\delta ^{-4}\nu \Vert rF\Vert _{L^2}\Vert \frac{w}{r}\Vert _{L^2}+\delta \nu ^{-\frac{1}{2}}\Vert rF\Vert _{L^2}^{\frac{1}{2}}\Vert \frac{w}{r}\Vert _{L^2}^{\frac{3}{2}}. \end{aligned}$$

  This leads to the conclusion

  $$\begin{aligned} \Vert rF\Vert _{L^2}\gtrsim \min \{|kB|\delta ,|kB|^{2}\delta ^{4}\nu ^{-1},\delta ^{-2}\nu \}\Vert \frac{w}{r}\Vert _{L^2}. \end{aligned}$$

  One can obtain an optimized form of the above inequality if $\delta$ is chosen as $\delta =\frac{\nu ^{\frac{1}{3}}}{|kB|^{\frac{1}{3}}}$. It then follows that

  $$\begin{aligned} \Vert rF\Vert _{L^2}\gtrsim \nu ^{\frac{1}{3}}|kB|^{\frac{2}{3}}\Vert \frac{w}{r}\Vert _{L^2}. \end{aligned}$$

Therefore, we arrive at

$$\begin{aligned} \nu ^{\frac{1}{3}}|kB|^{\frac{2}{3}}\Vert \frac{w}{r}\Vert _{L^2}\lesssim \Vert rF\Vert _{L^2} \quad \text {for all } \lambda \in \mathbb {R}. \end{aligned}$$

(3.10)

Notice that Lemma 3.1 also implies

$$\begin{aligned}&\Re \langle F,w\rangle \ge \nu \Vert w'\Vert _{L^2}^2. \end{aligned}$$

Along with (3.10) this yields the desired estimate on $w'$:

$$\begin{aligned} \nu ^{\frac{2}{3}}|kB|^{\frac{1}{3}}\Vert w'\Vert _{L^2}\le C\Vert rF\Vert _{L^2}. \end{aligned}$$

(3.11)

The remaining task is to control $\Vert r(\frac{1}{r^2}-\lambda )w\Vert _{L^2}$, and we appeal to exploring the imaginary part of $\langle F,r^2(\frac{1}{r^2}-\lambda )w\rangle$. Integration by parts gives

$$\begin{aligned}&\langle F,r^2(\frac{1}{r^2}-\lambda )w\rangle =\langle -\nu (\partial _r^2-\frac{k^2-\frac{1}{4}}{r^2})w+ikB(\frac{1}{r^2}-\lambda )w,r^2(\frac{1}{r^2}-\lambda )w\rangle \\&\quad =\nu \Big (\langle w',[(1-\lambda r^2)w]'\rangle +(k^2-\frac{1}{4})\langle w,(\frac{1}{r^2}-\lambda )w\rangle \Big )+ikB\Vert r(\frac{1}{r^2}-\lambda )w\Vert _{L^2}^2\\&\quad =\nu \Big (\langle w',(1-\lambda r^2)w'\rangle -\langle w',2\lambda rw\rangle +(k^2-\frac{1}{4})\langle w,(\frac{1}{r^2}-\lambda )w\rangle \Big )+ikB\Vert r(\frac{1}{r^2}-\lambda )w\Vert _{L^2}^2. \end{aligned}$$

Then we take the imaginary part of both sides and rewrite $-\lambda rw = r(\frac{1}{r^2}-\lambda )w-\frac{w}{r}$ to deduce

$$\begin{aligned}&|kB|\Vert r(\frac{1}{r^2}-\lambda )w\Vert _{L^2}^2\le |\langle F,r^2(\frac{1}{r^2}-\lambda )w\rangle |+\nu |\langle w',-2\lambda rw\rangle |\\&\quad \le |\langle F,r^2(\frac{1}{r^2}-\lambda )w\rangle |+\nu |\langle w',2r(\frac{1}{r^2}-\lambda )w\rangle |+2\nu |\langle w',\frac{w}{r}\rangle |\\&\quad \le \Vert rF\Vert _{L^2}\Vert r(\frac{1}{r^2}-\lambda )w\Vert _{L^2}+2\nu \Vert w'\Vert _{L^2}\Vert r(\frac{1}{r^2}-\lambda )w\Vert _{L^2}+2\nu \Vert w'\Vert _{L^2}\Vert \frac{w}{r}\Vert _{L^2}. \end{aligned}$$

In view of (3.10) and (3.11), the $L^2$ norm of $r(\frac{1}{r^2}-\lambda )w$ obeys the following bounds:

$$\begin{aligned}&|kB|\Vert r(\frac{1}{r^2}-\lambda )w\Vert _{L^2}^2\\&\quad \lesssim \frac{1}{|kB|}\Vert rF\Vert _{L^2}^2+\frac{\nu ^2}{|kB|}\Vert w'\Vert _{L^2}^2+\frac{1}{|kB|}\nu ^{\frac{2}{3}}|kB|^{\frac{1}{3}}\Vert w'\Vert _{L^2}\nu ^{\frac{1}{3}}|kB|^{\frac{2}{3}}\Vert \frac{w}{r}\Vert _{L^2}\\&\quad \lesssim \frac{1}{|kB|}\Vert rF\Vert _{L^2}^2+\frac{\nu ^2}{|kB|}\Vert w'\Vert _{L^2}^2+\frac{1}{|kB|}\Vert rF\Vert _{L^2}^2\lesssim \frac{1}{|kB|}\Vert rF\Vert _{L^2}^2+\frac{\nu ^2}{|kB|}\Vert w'\Vert _{L^2}^2\\&\quad =\frac{1}{|kB|}\Vert rF\Vert _{L^2}^2+\frac{\nu }{|kB|}\Vert rF\Vert _{L^2}\Vert \frac{w}{r}\Vert _{L^2} \lesssim \frac{1}{|kB|}\Vert rF\Vert _{L^2}^2+\frac{\nu ^{\frac{2}{3}}}{|kB|^{\frac{5}{3}}}\Vert rF\Vert _{L^2}\nu ^{\frac{1}{3}}|kB|^{\frac{2}{3}}\Vert \frac{w}{r}\Vert _{L^2}\\&\quad \lesssim (1+(\frac{\nu }{|kB|})^{\frac{2}{3}})\frac{1}{|kB|}\Vert rF\Vert _{L^2}^2\lesssim \frac{1}{|kB|}\Vert rF\Vert _{L^2}^2, \end{aligned}$$

provided that $\frac{\nu }{|kB|} \ll 1$. This completes the proof. $\square$

Relying on the above $L^2$ estimates for w and $w'$, we can derive the below bounds for $\varphi$ and $\varphi '$.

Lemma 3.4

For any $|k|\ge 1$, $\lambda \in \mathbb {R}$ and $w\in D_k$, there exist constants $C,c>0$ independent of $\nu ,k,B,\lambda ,R$, such that for any $0\le c'\le c$, it holds

$$\begin{aligned}&\nu ^{\frac{1}{6}}|kB|^{\frac{5}{6}}|k|^{\frac{1}{2}}\left( \Vert \varphi '\Vert _{L^2}+|k|\Vert \frac{\varphi }{r}\Vert _{L^2} \right) \le CR^2\left( (\frac{\nu }{|kB|})^{\frac{1}{6}} (\log R)^{\frac{1}{2}} +1 \right) \Vert rF-c'\nu ^{\frac{1}{3}} |kB|^{\frac{2}{3}}\frac{w}{r}\Vert _{L^2}. \end{aligned}$$

Proof

Denote $\tilde{F}:=rF-c'\nu ^{\frac{1}{3}} |kB|^{\frac{2}{3}}\frac{w}{r}$. For $\delta =(\frac{\nu }{|kB|})^{\frac{1}{3}}\ll 1$, we set

$$\begin{aligned} E=\{r> 0: |1-\lambda r^2|\le \delta \} \quad \text {and} \quad E^c=(0, \infty )\backslash E. \end{aligned}$$

Employing Lemma 3.3, Lemma A.1 and Lemma A.8, we have

$$\begin{aligned} \Vert \frac{w}{r^{\frac{3}{2}}}\Vert _{L^1([1, R] \cap E)}\le&\Vert \frac{w}{r^{\frac{1}{2}}}\Vert _{L^\infty } \Vert \frac{1}{r}\Vert _{L^1(E)} \lesssim \delta \Vert \frac{w}{r}\Vert _{L^2}^{\frac{1}{2}}(\Vert \frac{w}{r}\Vert _{L^2}+\Vert w'\Vert _{L^2})^{\frac{1}{2}} \\ \lesssim&\nu ^{-\frac{1}{6}}|kB|^{-\frac{5}{6}} \Vert \tilde{F}\Vert _{L^2}. \end{aligned}$$

On the other hand, utilizing Lemma 3.3 again as well as Lemma A.7, it can be inferred that

$$\begin{aligned} \Vert \frac{w}{r^{\frac{3}{2}}}\Vert _{L^1([1, R] \cap E^c)}&\le \Vert \frac{1}{r^{\frac{1}{2}}(1-\lambda r^2)^2}\Vert _{L^2([1, R]\cap E^c)}\Vert r(\frac{1}{r^2}-\lambda )w\Vert _{L^2} \\&\lesssim \left( (\log R)^{\frac{1}{2}}+\delta ^{-\frac{1}{2}} \right) |kB|^{-1} \Vert \tilde{F}\Vert _{L^2} \\&\lesssim \nu ^{-\frac{1}{6}}|kB|^{-\frac{5}{6}}\left( (\frac{\nu }{|kB|})^{\frac{1}{6}} (\log R)^{\frac{1}{2}} +1 \right) \Vert \tilde{F}\Vert _{L^2}. \end{aligned}$$

Thus we deduce

$$\begin{aligned} \Vert \frac{w}{r^{\frac{3}{2}}}\Vert _{L^1}&= \Vert \frac{w}{r^{\frac{3}{2}}}\Vert _{L^1([1, R] \cap E)}+ \Vert \frac{w}{r^{\frac{3}{2}}}\Vert _{L^1([1, R] \cap E^c)} \\&\lesssim \nu ^{-\frac{1}{6}}|kB|^{-\frac{5}{6}}\left( (\frac{\nu }{|kB|})^{\frac{1}{6}} (\log R)^{\frac{1}{2}} +1 \right) \Vert \tilde{F}\Vert _{L^2}. \end{aligned}$$

The controls for $\varphi '$ and $\frac{\varphi }{r}$ also follow from Lemma A.3. And we have

$$\begin{aligned} \Vert \varphi '\Vert _{L^2}+|k|\Vert \frac{\varphi }{r}\Vert _{L^2}&\lesssim |k|^{-\frac{1}{2}}\Vert r^{\frac{1}{2}}w\Vert _{L^1}\lesssim |k|^{-\frac{1}{2}}R^2\Vert \frac{w}{r^{\frac{3}{2}}}\Vert _{L^1} \\&\lesssim |k|^{-\frac{1}{2}}R^2\nu ^{-\frac{1}{6}}|kB|^{-\frac{5}{6}}\left( (\frac{\nu }{|kB|})^{\frac{1}{6}} (\log R)^{\frac{1}{2}} +1 \right) \Vert \tilde{F}\Vert _{L^2}. \end{aligned}$$

This completes the proof. $\quad \square$

Now we turn to estimate the $H^{-1}_r$ norm of w through the $H^1_r$ norm of F.

Lemma 3.5

For any $|k|\ge 1$, $\lambda \in \mathbb {R}$ and $w\in D_k$, there exist a constants $C,c>0$ independent of $\nu ,k,B,\lambda$, such that for $0\le c'\le c$, the following estimate holds

$$\begin{aligned} \nu \Vert w\Vert _{H^1_r}+\nu ^{\frac{2}{3}}|kB|^{\frac{1}{3}}\Vert \frac{w}{r}\Vert _{L^2}\le C\Vert F-c'\nu ^{\frac{1}{3}} |kB|^{\frac{2}{3}} R^{-2}w\Vert _{H^{-1}_r}. \end{aligned}$$

Proof

By Lemma 3.1, one first obtains

$$\begin{aligned}&\nu \Vert w'\Vert _{L^2}^2+\nu (k^2-\frac{1}{4})\Vert \frac{w}{r}\Vert _{L^2}^2\\&\quad =\Re \langle F-c'\nu ^{\frac{1}{3}} |kB|^{\frac{2}{3}} R^{-2}w,w\rangle +c'\nu ^{\frac{1}{3}} |kB|^{\frac{2}{3}} R^{-2}\Vert w\Vert _{L^2}^2, \end{aligned}$$

which gives

$$\begin{aligned} \nu \Vert w\Vert _{H^1_r}\lesssim \Vert F-c'\nu ^{\frac{1}{3}} |kB|^{\frac{2}{3}}R^{-2}w\Vert _{H^{-1}_r}+\sqrt{c'}\nu ^{\frac{2}{3}} |kB|^{\frac{1}{3}}\Vert \frac{w}{r}\Vert _{L^2}. \end{aligned}$$

(3.12)

Denote $G{:}{=}F-c'\nu ^{\frac{1}{3}} |kB|^{\frac{2}{3}}R^{-2}w$. In below, we utilize a similar method as in the proof of Proposition 3.3 to demonstrate

$$\begin{aligned} \nu ^{\frac{2}{3}} |kB|^{\frac{1}{3}}\Vert \frac{w}{r}\Vert _{L^2}\le C \Vert G\Vert _{H^{-1}_r}. \end{aligned}$$

(3.13)

To achieve this, we will examine the following three cases:

$$\begin{aligned} \lambda \in (-\infty ,\frac{1-\delta }{R^2}]\cup [1+\delta ,\infty ),\quad \lambda \in [\frac{1-\delta }{R^2},\frac{1}{R^2}]\cup [1,1+\delta ],\quad \lambda \in [\frac{1}{R^2},1], \end{aligned}$$

where $0<\delta \ll 1$ is a small constant to be determined later.

• (1) Case of $\lambda \in (-\infty ,\frac{1-\delta }{R^2}]\cup [1+\delta ,\infty )$. One can check

  $$\begin{aligned}&|\Im \langle G,w\rangle |=|\Im \langle F,w\rangle |=|kB|\big |\langle (\frac{1}{r^2}-\lambda )w,w\rangle \big |=|kB|\big |\langle (1-\lambda r^2)\frac{w}{r},\frac{w}{r}\rangle \big |\ge |kB|\delta \Vert \frac{w}{r}\Vert _{L^2}^2. \end{aligned}$$

  Taking $\delta =\frac{\nu ^{\frac{1}{3}}}{|kB|^{\frac{1}{3}}}$, together with (3.12), we obtain

  $$\begin{aligned} C(\Vert G\Vert _{H^{-1}_r}+c'\nu ^{\frac{2}{3}} |kB|^{\frac{1}{3}}\Vert \frac{w}{r}\Vert _{L^2})\ge \nu ^{\frac{2}{3}} |kB|^{\frac{1}{3}}\Vert \frac{w}{r}\Vert _{L^2}. \end{aligned}$$

  Choosing $Cc'\le \frac{1}{2}$, we then prove (3.13).

• (2) Case of $\lambda \in [\frac{1-\delta }{R^2},\frac{1}{R^2}]$ or $\lambda \in [1,1+\delta ]$. i) $\lambda \in [\frac{1-\delta }{R^2},\frac{1}{R^2}]$. Observing that

  $$\begin{aligned} 1-\lambda r^2\ge \delta \Longleftrightarrow 1-\delta \ge \lambda r^2\Longleftrightarrow 1\le r\le \sqrt{\frac{1-\delta }{\lambda }}, \end{aligned}$$

  we then deduce

  $$\begin{aligned} |\Im \langle G,w\rangle |&=|\Im \langle F,w\rangle |=|kB|\int _1^R(1-\lambda r^2)|\frac{w}{r}|^2dr\\&\ge |kB|\int _1^{\sqrt{\frac{1-\delta }{\lambda }}}(1-\lambda r^2)|\frac{w}{r}|^2dr\ge |kB|\delta \Vert \frac{w}{r}\Vert _{L^2(1,\sqrt{\frac{1-\delta }{\lambda }})}^2. \end{aligned}$$

  The remaining part of $\Vert \frac{w}{r}\Vert _{L^2}$ can be bounded in terms of $\Vert \frac{w}{r^{\frac{1}{2}}}\Vert _{L^{\infty }}$ as below

  $$\begin{aligned}&\Vert \frac{w}{r}\Vert _{L^2(\sqrt{\frac{1-\delta }{\lambda }},R)}^2=\int _{\sqrt{\frac{1-\delta }{\lambda }}}^R|\frac{w}{r}|^2dr\le \int _{\sqrt{\frac{1-\delta }{\lambda }}}^R\frac{1}{r^2}dr\Vert w\Vert _{L^{\infty }}^2\\&\quad =(\sqrt{\frac{\lambda }{1-\delta }}-\frac{1}{R})\Vert w\Vert _{L^{\infty }}^2 \le (\sqrt{\frac{1}{1-\delta }}-1)\frac{\Vert w\Vert _{L^{\infty }}^2}{R}\lesssim \frac{\delta }{R}\Vert w\Vert _{L^{\infty }}^2\le \delta \Vert \frac{w}{r^{\frac{1}{2}}}\Vert _{L^{\infty }}^2. \end{aligned}$$

  Combining these two estimates, together with (3.12), we get

  $$\begin{aligned} \Vert \frac{w}{r}\Vert _{L^2}^2&=\Vert \frac{w}{r}\Vert _{L^2(1,\sqrt{\frac{1-\delta }{\lambda }})}^2+\Vert \frac{w}{r}\Vert _{L^2(\sqrt{\frac{1-\delta }{\lambda }},R)}^2\\&\lesssim |kB|^{-1}\delta ^{-1}(\nu ^{-1}\Vert G\Vert _{H^{-1}_r}^2+c'\nu ^{\frac{1}{3}}|kB|^{\frac{2}{3}}\Vert \frac{w}{r}\Vert _{L^2}^2)+\delta \Vert \frac{w}{r^{\frac{1}{2}}}\Vert _{L^{\infty }}^2. \end{aligned}$$

  Applying Lemma A.1 and (3.12), we provide the estimate for $\Vert \frac{w}{r}\Vert _{L^2}$ by

  $$\begin{aligned} \Vert \frac{w}{r}\Vert _{L^2}^2&\lesssim |kB|^{-1}\delta ^{-1}(\nu ^{-1}\Vert G\Vert _{H^{-1}_r}^2+c'\nu ^{\frac{1}{3}}|kB|^{\frac{2}{3}}\Vert \frac{w}{r}\Vert _{L^2}^2)+\delta \Vert \frac{w}{r^{\frac{1}{2}}}\Vert _{L^{\infty }}^2\\&\lesssim |kB|^{-1}\delta ^{-1}(\nu ^{-1}\Vert G\Vert _{H^{-1}_r}^2+c'\nu ^{\frac{1}{3}}|kB|^{\frac{2}{3}}\Vert \frac{w}{r}\Vert _{L^2}^2)+\delta \Vert \frac{w}{r}\Vert _{L^2}\Vert r^{\frac{1}{2}}\partial _r(\frac{w}{r^{\frac{1}{2}}})\Vert _{L^2}\\&\lesssim |kB|^{-1}\delta ^{-1}(\nu ^{-1}\Vert G\Vert _{H^{-1}_r}^2+c'\nu ^{\frac{1}{3}}|kB|^{\frac{2}{3}}\Vert \frac{w}{r}\Vert _{L^2}^2)\\&\quad +\delta \Vert \frac{w}{r}\Vert _{L^2}(\Vert w'\Vert _{L^2}+\Vert \frac{w}{r}\Vert _{L^2})\\&\lesssim |kB|^{-1}\delta ^{-1}(\nu ^{-1}\Vert G\Vert _{H^{-1}_r}^2+c'\nu ^{\frac{1}{3}}|kB|^{\frac{2}{3}}\Vert \frac{w}{r}\Vert _{L^2}^2)\\&\quad +\delta \Vert \frac{w}{r}\Vert _{L^2}(\nu ^{-\frac{1}{2}}\Vert G\Vert _{H^{-1}_r}+\sqrt{c'}\nu ^{\frac{1}{6}}|kB|^{\frac{1}{3}}\Vert \frac{w}{r}\Vert _{L^2})+\delta \Vert \frac{w}{r}\Vert _{L^2}^2 \\&\lesssim |kB|^{-1}\delta ^{-1}(\nu ^{-1}\Vert G\Vert _{H^{-1}_r}^2+c'\nu ^{\frac{1}{3}}|kB|^{\frac{2}{3}}\Vert \frac{w}{r}\Vert _{L^2}^2)+|kB|\delta ^3\Vert \frac{w}{r}\Vert _{L^2}^2+\delta \Vert \frac{w}{r}\Vert _{L^2}^2. \end{aligned}$$

  Picking $\delta =\frac{\nu ^{\frac{1}{3}}}{|kB|^{\frac{1}{3}}}$, we then obtain

  $$\begin{aligned}&\Vert \frac{w}{r}\Vert _{L^2}^2 \le C\nu ^{-\frac{4}{3}}|kB|^{-\frac{2}{3}}\Vert G\Vert _{H^{-1}_r}^2+Cc'\Vert \frac{w}{r}\Vert _{L^2}^2+C\nu \Vert \frac{w}{r}\Vert _{L^2}^2+C\delta \Vert \frac{w}{r}\Vert _{L^2}^2. \end{aligned}$$

  The additional $\Vert \frac{w}{r}\Vert _{L^2}$ terms on the right can be absorbed to the left by choosing a sufficiently small $c'>0$, and noting that $0<\nu , \delta \ll 1$. Hence, we also get the desired estimate (3.13). ii) $\lambda \in [1,1+\delta ]$. Using that

  $$\begin{aligned} \lambda r^2-1\ge \delta \Longleftrightarrow r^2\ge \frac{1+\delta }{\lambda } \Longleftrightarrow \sqrt{\frac{1+\delta }{\lambda }}\le r\le R, \end{aligned}$$

  we can bound the imaginary part of $\langle G,w\rangle$ as below

  $$\begin{aligned}&|\Im \langle G,w\rangle |=|\Im \langle F,w\rangle |=|kB|\int _1^R(\lambda r^2-1)|\frac{w}{r}|^2dr\\&\quad \ge |kB|\int _{\sqrt{\frac{1+\delta }{\lambda }}}^R(\lambda r^2-1)|\frac{w}{r}|^2dr\ge |kB|\delta \Vert \frac{w}{r}\Vert _{L^2(\sqrt{\frac{1+\delta }{\lambda }},R)}^2. \end{aligned}$$

  We further control $\Vert \frac{w}{r}\Vert _{L^2(1,\sqrt{\frac{1+\delta }{\lambda }})}$ via

  $$\begin{aligned} \Vert \frac{w}{r}\Vert _{L^2(1,\sqrt{\frac{1+\delta }{\lambda }})}^2=&\int _1^{\sqrt{\frac{1+\delta }{\lambda }}}|\frac{w}{r}|^2dr \le \int _1^{\sqrt{\frac{1+\delta }{\lambda }}}\frac{1}{r}dr\Vert \frac{w}{r^{\frac{1}{2}}}\Vert _{L^{\infty }}^2\\ =&(\sqrt{\frac{1+\delta }{\lambda }}-1)\Vert \frac{w}{r^{\frac{1}{2}}}\Vert _{L^{\infty }}^2 \le (\sqrt{1+\delta }-1)\Vert \frac{w}{r^{\frac{1}{2}}}\Vert _{L^{\infty }}^2\le \delta \Vert \frac{w}{r^{\frac{1}{2}}}\Vert _{L^{\infty }}^2. \end{aligned}$$

  Together with (3.12), this implies

  $$\begin{aligned} \Vert \frac{w}{r}\Vert _{L^2}^2=&\Vert \frac{w}{r}\Vert _{L^2(1,\sqrt{\frac{1+\delta }{\lambda }})}^2+\Vert \frac{w}{r}\Vert _{L^2(\sqrt{\frac{1+\delta }{\lambda }},R)}^2\\ \lesssim&\delta \Vert \frac{w}{r^{\frac{1}{2}}}\Vert _{L^{\infty }}^2+|kB|^{-1}\delta ^{-1}(\nu ^{-1}\Vert G\Vert _{H^{-1}_r}^2+c'\nu ^{\frac{1}{3}}|kB|^{\frac{2}{3}}\Vert \frac{w}{r}\Vert _{L^2}^2). \end{aligned}$$

  Thus, we obtain the same bound for $\Vert \frac{w}{r}\Vert _{L^2}$ as in the scenario with $\lambda \in [\frac{1-\delta }{R^2},\frac{1}{R^2}]$. Applying Lemma A.1, inequality (3.12) and choosing $\delta =\frac{\nu ^{\frac{1}{3}}}{|kB|^{\frac{1}{3}}}$ and $c'>0$ sufficiently small, we then derive the desired bound (3.13).

• (3) Case of $\lambda \in [\frac{1}{R^2},1]$. We now use

  $$\begin{aligned} |1-\lambda r^2|\le \delta \Rightarrow 1-\delta \le \lambda r^2\le 1+\delta \Rightarrow \sqrt{\frac{1-\delta }{\lambda }}\le r\le \sqrt{\frac{1+\delta }{\lambda }}. \end{aligned}$$

  Given $0<\delta \ll 1$, it then holds

  $$\begin{aligned} \sqrt{\frac{1+\delta }{\lambda }}-\sqrt{\frac{1-\delta }{\lambda }}=\frac{2\delta }{\sqrt{\lambda }(\sqrt{1+\delta }+\sqrt{1-\delta })}\lesssim \frac{\delta }{\sqrt{\lambda }}. \end{aligned}$$

  We further choose $r_{-}\in (\sqrt{\frac{1-\delta }{\lambda }}-\frac{\delta }{\sqrt{\lambda }},\sqrt{\frac{1-\delta }{\lambda }})$ and $r_{+}\in (\sqrt{\frac{1+\delta }{\lambda }},\sqrt{\frac{1+\delta }{\lambda }}+\frac{\delta }{\sqrt{\lambda }})$ satisfying the following inequalities

  $$\begin{aligned} |w'(r_{-})|^2\le \frac{\Vert w'\Vert _{L^2}^2}{\delta /\sqrt{\lambda }},\quad |w'(r_{+})|^2\le \frac{\Vert w'\Vert _{L^2}^2}{\delta /\sqrt{\lambda }}. \end{aligned}$$

  (3.14)

  For the next step, we will construct an appropriate multiplier. To do so, we first define a piecewise $C^1$ cutoff function $\rho$ with domain (1, R) as follows

  $$\begin{aligned} \rho (r)=\left\{ \begin{aligned}&1,\quad r\in (1,r_{-}-\frac{\delta }{\sqrt{\lambda }}),\\&\sin \big (\frac{\pi }{2}\frac{\sqrt{\lambda }}{\delta }(r_{-}-r)\big ),\quad r\in (r_{-}-\frac{\delta }{\sqrt{\lambda }},r_{-}),\\&0,\quad r\in (r_{-},r_{+}),\\&\sin \big (\frac{\pi }{2}\frac{\sqrt{\lambda }}{\delta }(r_{+}-r)\big ),\quad r\in (r_{+},r_{+}+\frac{\delta }{\sqrt{\lambda }}),\\&-1,\quad r\in (r_{+}+\frac{\delta }{\sqrt{\lambda }},R). \end{aligned} \right. \end{aligned}$$

  Via integration by parts, we obtain the following expression

  $$\begin{aligned} -\Im \langle G,w\rho \rangle =\Im \langle F,w\rho \rangle&=\Im \langle -\nu \partial _r^2w+ikB(\frac{1}{r^2}-\lambda )w,w\rho \rangle \\&=\Im \langle ikB(\frac{1}{r^2}-\lambda )w,w\rho \rangle +\nu \Im \langle w',w\rho '\rangle . \end{aligned}$$

  Taking the imaginary part of above equality leads to

  $$\begin{aligned}&|kB|\Big |\int _1^{r_{-}-\frac{\delta }{\sqrt{\lambda }}}(1-\lambda r^2)\frac{|w|^2}{r^2}dr+\int _{r_{+}+\frac{\delta }{\sqrt{\lambda }}}^R(\lambda r^2-1)\frac{|w|^2}{r^2}dr\Big |\\&\quad \le \Vert G\Vert _{H^{-1}_r}\Vert w\rho \Vert _{H^1_r}+\nu \Vert w'\Vert _{L^2}\Vert w\rho '\Vert _{L^2}. \end{aligned}$$

  Based on our choice of $\rho$ and the definition of the $H_r^1$ norm, it follows that

  $$\begin{aligned} \Vert w\rho \Vert _{L^2}\lesssim \Vert (w\rho )'\Vert _{L^2}+\Vert \frac{w}{\rho }{r}\Vert _{L^2} \le&\Vert w'\Vert _{L^2}+\Vert w\rho '\Vert _{L^2}+\Vert \frac{w}{r}\Vert _{L^2}, \end{aligned}$$

  and

  $$\begin{aligned} \Vert w\rho '\Vert _{L^2}\lesssim \frac{\sqrt{\lambda }}{\delta }\Vert w \Vert _{L^2\big ((r_{-}-\frac{\delta }{\sqrt{\lambda }},r_{-})\cup (r_{+},r_{+}+\frac{\delta }{\sqrt{\lambda }})\big )} \lesssim \delta ^{-1} \Vert \frac{w}{r}\Vert _{L^2\big ((r_--\frac{\delta }{\sqrt{\lambda }}, r_++\frac{\delta }{\sqrt{\lambda }})\big )}. \end{aligned}$$

  Hence, plugging in (3.12) and (3.14), we then derive the estimate

  $$\begin{aligned}&\Vert \frac{w}{r}\Vert _{L^2\big ((1, r_{-}-\frac{\delta }{\sqrt{\lambda }})\cup (r_{+}+\frac{\delta }{\sqrt{\lambda }},R)\big )}^2\\&\quad \lesssim |kB|^{-1}\delta ^{-1}\Big [\Vert G\Vert _{H^{-1}_r}(\Vert w'\Vert _{L^2}+\delta ^{-1} \Vert \frac{w}{r}\Vert _{L^2\big ((r_--\frac{\delta }{\sqrt{\lambda }}, r_++\frac{\delta }{\sqrt{\lambda }})\big )})+\nu \Vert w'\Vert _{L^2}\cdot \delta ^{-1}\Vert \frac{w}{r}\Vert _{L^2}\Big ]\\&\quad \lesssim |kB|^{-1}\delta ^{-1}\Big [\Vert G\Vert _{H^{-1}_r}(\nu ^{-1}\Vert G\Vert _{H^{-1}_r}+\nu ^{-1}\sqrt{c'}\nu ^{\frac{2}{3}}|kB|^{\frac{1}{3}}\Vert \frac{w}{r}\Vert _{L^2}+\delta ^{-1} \Vert \frac{w}{r}\Vert _{L^2\big ((r_--\frac{\delta }{\sqrt{\lambda }}, r_++\frac{\delta }{\sqrt{\lambda }})\big )})\\&\qquad +(\Vert G\Vert _{H^{-1}_r}+\sqrt{c'}\nu ^{\frac{2}{3}}|kB|^{\frac{1}{3}}\Vert \frac{w}{r}\Vert _{L^2})\cdot \delta ^{-1} \Vert \frac{w}{r}\Vert _{L^2\big ((r_--\frac{\delta }{\sqrt{\lambda }}, r_++\frac{\delta }{\sqrt{\lambda }})\big )}\Big ]\\&\quad \lesssim |kB|^{-1}\delta ^{-1}\Big (\nu ^{-1}\Vert G\Vert _{H^{-1}_r}^2+c'\nu ^{\frac{1}{3}}|kB|^{\frac{2}{3}}\Vert \frac{w}{r}\Vert _{L^2}^2+\nu \delta ^{-2} \Vert \frac{w}{r}\Vert _{L^2\big ((r_--\frac{\delta }{\sqrt{\lambda }}, r_++\frac{\delta }{\sqrt{\lambda }})\big )}\Big ). \end{aligned}$$

  Meanwhile, it can be deduced from Lemma A.6, Lemma A.1 and (3.12) that

  $$\begin{aligned}&\Vert \frac{w}{r}\Vert _{L^2\big ((r_--\frac{\delta }{\sqrt{\lambda }}, r_++\frac{\delta }{\sqrt{\lambda }})\big )}^2\\&\quad \le \int _{r_{-}-\frac{\delta }{\sqrt{\lambda }}}^{r_{+}+\frac{\delta }{\sqrt{\lambda }}}\frac{1}{r}dr\Vert \frac{w}{r^{\frac{1}{2}}}\Vert _{L^\infty }^2 \lesssim \delta \Vert \frac{w}{r}\Vert _{L^2}\Vert r^{\frac{1}{2}}\partial _r(\frac{w}{r^{\frac{1}{2}}})\Vert _{L^2} \\&\quad \lesssim \delta \Vert \frac{w}{r}\Vert _{L^2}(\Vert w'\Vert _{L^2}+\Vert \frac{w}{r}\Vert _{L^2})\\&\quad \lesssim \delta \Vert \frac{w}{r}\Vert _{L^2}(\nu ^{-1}\Vert G\Vert _{H^{-1}_r}+\nu ^{-1}\sqrt{c'}\nu ^{\frac{2}{3}}|kB|^{\frac{1}{3}}\Vert \frac{w}{r}\Vert _{L^2}+\Vert \frac{w}{r}\Vert _{L^2})\\&\quad \lesssim \frac{1}{\sqrt{c'}}\nu ^{-2}\delta ^{2}\Vert G\Vert _{H^{-1}_r}^2+\sqrt{c'}(\nu ^{-\frac{1}{3}}|kB|^{\frac{1}{3}}\delta +1)\Vert \frac{w}{r}\Vert _{L^2}^2+\delta \Vert \frac{w}{r}\Vert _{L^2}^2. \end{aligned}$$

  Taking $\delta =\frac{\nu ^{\frac{1}{3}}}{|kB|^{\frac{1}{3}}}$, we can then bound the $L^2$ norm of $\frac{w}{r}$ as below

  $$\begin{aligned} \Vert \frac{w}{r}\Vert _{L^2}^2&=\Vert \frac{w}{r}\Vert _{L^2\big ((1, r_{-}-\frac{\delta }{\sqrt{\lambda }})\cup (r_{+}+\frac{\delta }{\sqrt{\lambda }},R)\big )}^2+\Vert \frac{w}{r}\Vert _{L^2\big (( r_{-}-\frac{\delta }{\sqrt{\lambda }},r_{+}+\frac{\delta }{\sqrt{\lambda }})\big )}^2\\&\lesssim |kB|^{-1}\delta ^{-1}\Big (\nu ^{-1}\Vert G\Vert _{H^{-1}_r}^2+c'\nu ^{\frac{1}{3}}|kB|^{\frac{2}{3}}\Vert \frac{w}{r}\Vert _{L^2}^2\Big )\\&\quad + \Vert \frac{w}{r}\Vert _{L^2\big ((r_--\frac{\delta }{\sqrt{\lambda }}, r_++\frac{\delta }{\sqrt{\lambda }})\big )}^2\\&\lesssim (1+\frac{1}{\sqrt{c'}})\nu ^{-\frac{4}{3}}|kB|^{-\frac{2}{3}}\Vert G\Vert _{H^{-1}_r}^2+(c'+\sqrt{c'}+\delta )\Vert \frac{w}{r}\Vert _{L^2}^2. \end{aligned}$$

  Picking $c'>0$ sufficiently small and noting that $0<\delta \ll 1$, we can proceed to the conclusion

  $$\begin{aligned} \Vert G\Vert _{H^{-1}_r}\gtrsim \nu ^{\frac{2}{3}}|kB|^{\frac{1}{3}}\Vert \frac{w}{r}\Vert _{L^2}. \end{aligned}$$

  (3.15)

We have therefore proved (3.13) in all three scenarios. This together with (3.12) yields the desired bounds for $\Vert w\Vert _{H^{1}_r}$:

$$\begin{aligned} \nu \Vert w\Vert _{H^{1}_r}\lesssim \Vert G\Vert _{H^{-1}_r}+\nu ^{\frac{2}{3}} |kB|^{\frac{1}{3}}\Vert \frac{w}{r}\Vert _{L^2}\lesssim \Vert G\Vert _{H^{-1}_r} . \end{aligned}$$

$\square$

For future use, we also establish the resolvent estimate for $\varphi '$.

Lemma 3.6

For any $|k|\ge 1$, $\lambda \in \mathbb {R}$ and $w\in D_k$, there exists a constant $C>0$ independent of $\nu ,k,B,\lambda , R$, such that there holds

$$\begin{aligned} \nu ^{\frac{1}{2}}|kB|^{\frac{1}{2}}\Vert \varphi '\Vert _{L^2}+\nu ^{\frac{1}{2}}|k||kB|^{\frac{1}{2}}\Vert \frac{\varphi }{r}\Vert _{L^2}\le C R^2 \Vert F-c'\nu ^{\frac{1}{3}}|kB|^{\frac{2}{3}}R^{-2}w\Vert _{H^{-1}_r}. \end{aligned}$$

Proof

If $\nu |k|^3> |B|$, by Lemma 3.5 and Lemma A.3 we obtain

$$\begin{aligned} \begin{aligned} \Vert \varphi '\Vert _{L^2}+|k|\Vert \frac{\varphi }{r}\Vert _{L^2}&\lesssim |k|^{-1} \Vert rw\Vert _{L^2}\lesssim |k|^{-1} R^2\Vert \frac{w}{r}\Vert _{L^2} \\&\lesssim |k|^{-1} \nu ^{-\frac{2}{3}} |kB|^{-\frac{1}{3}} R^2\Vert G\Vert _{H^{-1}_r}\\&\le \nu ^{-\frac{1}{2}}|kB|^{-\frac{1}{2}} R^2\Vert G\Vert _{H^{-1}_r}. \end{aligned} \end{aligned}$$

Now we work under the case $\nu |k|^3\le |B|$. Let

$$\begin{aligned} E=\{r> 0: \ |1-\lambda r^2|\le \delta \} \ \text {and} \ E^c=(0, \infty )\backslash E \end{aligned}$$

with $\delta =(\frac{\nu }{|kB|})^{\frac{1}{3}}\ll 1$. Denote $a= \Vert \varphi '\Vert _{L^2}+|k|\Vert \frac{\varphi }{r}\Vert _{L^2}$. Utilizing Lemma A.3 from the Appendix, we immediately have

$$\begin{aligned} a^2 \lesssim |\langle w, \varphi \rangle |\le |\int _{E\cap [1, R]}w\overline{\varphi } dr|+|\int _{E^c\cap [1, R]}w\overline{\varphi } dr|. \end{aligned}$$

(3.16)

Employing Lemma 3.5, Lemma A.1 and Lemma A.8, the first term can be controlled through

$$\begin{aligned} |\int _{E\cap [1, R]}w\overline{\varphi } dr|&\le \Vert r^2 \frac{w}{r^{\frac{1}{2}}}\Vert _{L^\infty }\Vert \frac{\varphi }{r^{\frac{5}{2}}}\Vert _{L^\infty } \Vert \frac{1}{r}\Vert _{L^1(E)} \\&\le R^2 \Vert \frac{w}{r^{\frac{1}{2}}}\Vert _{L^\infty }\Vert \frac{\varphi }{r^{\frac{1}{2}}}\Vert _{L^\infty } \Vert \frac{1}{r}\Vert _{L^1(E)} \\&\lesssim \delta R^2(\Vert \frac{w}{r}\Vert _{L^2} \Vert r^{\frac{1}{2}}(\frac{w}{r^{\frac{1}{2}}})'\Vert _{L^2})^{\frac{1}{2}} (\Vert \frac{\varphi }{r}\Vert _{L^2} \Vert r^{\frac{1}{2}}(\frac{\varphi }{r^{\frac{1}{2}}})'\Vert _{L^2})^{\frac{1}{2}} \\&\lesssim \delta \nu ^{-\frac{5}{6}}|kB|^{-\frac{1}{6}}R^2\Vert G\Vert _{H^{-1}_r} \times a. \end{aligned}$$

To bound the second term on the right of (3.16), we use a piecewise $C^1$ cut-off function $\chi : (0, \infty ) \rightarrow \mathbb {R}$ from Lemma A.8, which is defined as

$$\begin{aligned} \chi (r)=\left\{ \begin{aligned}&\frac{1}{\frac{1}{r^2}-\lambda },\quad \text {if} \ r\in E^c,\\&-\frac{r_+^2+r_-^2}{\delta (r_+-r_-)}(r-r_-)+\frac{r_-^2}{\delta },\quad \text {if} \ r\in E \end{aligned} \right. \end{aligned}$$

with $r_{\pm }=\sqrt{\frac{1\pm \delta }{\lambda }}$. Conducting integration by parts allows us to write

$$\begin{aligned} \begin{aligned} \langle F, \chi \varphi \rangle&=\nu \langle w', (\chi \varphi )'\rangle +\nu (k^2-\frac{1}{4})\langle \frac{w}{r}, \frac{\chi \varphi }{r} \rangle \\&\quad +ikB(\int _{E\cap [1, R]}(\frac{1}{r^2}-\lambda )\chi w\overline{\varphi } dr+\int _{E^c \cap [1, R]}(\frac{1}{r^2}-\lambda )\chi w\overline{\varphi } dr ). \end{aligned} \end{aligned}$$

Due to the fact that $\chi (\frac{1}{r^2}-\lambda )=1$ on $E^c$, it can be inferred that

$$\begin{aligned} \begin{aligned} |\int _{E^c\cap [1, R]}w\overline{\varphi } dr|&\le |kB|^{-1} \Big (\Vert G\Vert _{H^{-1}_r}\Vert \chi \varphi \Vert _{H^1_r}+\nu \Vert w'\Vert _{L^2}\Vert (\chi \varphi )'\Vert _{L^2}\\&\quad +\nu k^2 \Vert \frac{w}{r}\Vert _{L^2} \Vert \frac{\chi \varphi }{r}\Vert _{L^2}\Big )+\Vert \frac{w}{r}\Vert _{L^2}\Vert (1-\lambda r^2)\frac{\chi \varphi }{r}\Vert _{L^2(E\cap [1, R])}. \end{aligned} \end{aligned}$$

(3.17)

Applying Lemma A.1 and Lemma A.8, we further have

$$\begin{aligned} \Vert (\chi \varphi )'\Vert _{L^2}&=\Vert (r^2\frac{\chi \varphi }{r^2})'\Vert _{L^2}\le \Vert r^{\frac{1}{2}}(\frac{\chi }{r^2})'\Vert _{L^2}\Vert r^2\frac{\varphi }{r^{\frac{1}{2}}}\Vert _{L^\infty }+\Vert \frac{\chi }{r^2}\Vert _{L^\infty }\Vert (r^2\varphi )'\Vert _{L^2} \\&\lesssim \delta ^{-\frac{3}{2}} R^2(\Vert \frac{\varphi }{r}\Vert _{L^2} \Vert r^{\frac{1}{2}}(\frac{\varphi }{r^{\frac{1}{2}}})'\Vert _{L^2})^{\frac{1}{2}}+\delta ^{-1} R^2(\Vert \varphi '\Vert _{L^2}+\Vert \frac{\varphi }{r}\Vert _{L^2}) \lesssim \delta ^{-\frac{3}{2}} R^2 a, \\ \Vert \frac{\chi \varphi }{r}\Vert _{L^2}&\le \Vert r^2 \frac{\chi }{r^2}\Vert _{L^\infty }\Vert \frac{\varphi }{r}\Vert _{L^2}\lesssim \delta ^{-1}R^2a \end{aligned}$$

and

$$\begin{aligned} \Vert (1-\lambda r^2)\frac{\chi \varphi }{r}\Vert _{L^2(E\cap [1, R])}&\le \Vert (1-\lambda r^2)\frac{\chi }{r^{\frac{5}{2}}}\Vert _{L^2(E)} \Vert r^2 \frac{\varphi }{r^{\frac{1}{2}}}\Vert _{L^\infty } \\&\lesssim \delta ^{\frac{1}{2}}R^2 (\Vert \frac{\varphi }{r}\Vert _{L^2} \Vert r^{\frac{1}{2}}(\frac{\varphi }{r^{\frac{1}{2}}})'\Vert _{L^2})^{\frac{1}{2}} \lesssim \delta ^{\frac{1}{2}}R^2 a. \end{aligned}$$

Plugging the above inequalities into (3.17), together with Lemma 3.5, we then deduce

$$\begin{aligned} |\int _{E^c\cap [1, R]}w\overline{\varphi } dr| \le&\Big (\delta ^{-\frac{3}{2}}|kB|^{-1}+\delta ^{-1} k^2 \nu ^{\frac{1}{3}} |kB|^{-\frac{4}{3}} +\delta ^{\frac{1}{2}} \nu ^{-\frac{2}{3}}|kB|^{-\frac{1}{3}}\Big )R^2 \Vert G\Vert _{H^{-1}_r}\times a. \end{aligned}$$

Therefore, the term $a= \Vert \varphi '\Vert _{L^2}+|k|\Vert \frac{\varphi }{r}\Vert _{L^2}$ now obeys

$$\begin{aligned} a^2\lesssim \Big (\delta \nu ^{-\frac{5}{6}}|kB|^{-\frac{1}{6}} +\delta ^{-\frac{3}{2}}|kB|^{-1}+\delta ^{-1} k^2 \nu ^{\frac{1}{3}} |kB|^{-\frac{4}{3}} +\delta ^{\frac{1}{2}} \nu ^{-\frac{2}{3}}|kB|^{-\frac{1}{3}} \Big )R^2 \Vert G\Vert _{H^{-1}_r} \times a. \end{aligned}$$

Substituting $\delta =(\frac{\nu }{|kB|})^{\frac{1}{3}}$ and noticing $\nu |k|^3\le |B|$, we thus arrive at

$$\begin{aligned} a\lesssim \nu ^{-\frac{1}{2}}|kB|^{-\frac{1}{2}}\left( 1+(\frac{\nu |k|^3}{|B|})^{\frac{1}{2}} \right) R^2\Vert G\Vert _{H^{-1}_r}\lesssim \nu ^{-\frac{1}{2}}|kB|^{-\frac{1}{2}} R^2\Vert G\Vert _{H^{-1}_r}. \end{aligned}$$

This completes the proof of this lemma. $\quad \square$

3.3 Sharpness of the resolvent bound

In the last subsection, we prove that the resolvent bound $\nu ^{\frac{1}{3}} |kB|^{\frac{2}{3}}$ in Lemma 3.3 is sharp. We have

Lemma 3.7

With $F=-\nu (\partial _r^2-\frac{k^2-\frac{1}{4}}{r^2})w+ikB(\frac{1}{r^2}-\lambda )w$, there exist $\lambda \in \mathbb {R}, R>1$ and a non-zero function $w\in H_0^2([1,R])$ such that for $|k|=1$

$$\begin{aligned} \Vert rF\Vert _{L^2}\le C\nu ^{\frac{1}{3}} |B|^{\frac{2}{3}}\Vert \frac{w}{r}\Vert _{L^2}. \end{aligned}$$

Proof

Choose $\lambda =\frac{1}{r_0^2}$ and $\frac{|B|}{\nu }=r_0^3$ for some $r_0\ge 1$. Setting $R=r_0+\frac{1}{r_0}$, we construct

$$\begin{aligned} w(r)=\left\{ \begin{aligned}&(r-r_0)(r_0+\frac{1}{r_0}-r),\quad r_0\le r\le r_0+\frac{1}{r_0},\\&0,\quad others. \end{aligned} \right. \end{aligned}$$

One can verify for $k=1$

$$\begin{aligned} \Vert \nu r(-\partial _r^2+\frac{3}{4}\frac{1}{r^2})w\Vert _{L^2}\lesssim \nu (r_0+1)\Vert \frac{w}{r}\Vert _{L^2}\lesssim \nu r_0\Vert \frac{w}{r}\Vert _{L^2}. \end{aligned}$$

In addition, we have

$$\begin{aligned} |B|\Vert r(\frac{1}{r_0^2}-\frac{1}{r^2})w\Vert _{L^2}\le \frac{|B|}{r_0}\Vert \frac{r+r_0}{r_0^2r}w\Vert _{L^2}\approx \frac{|B|}{r_0^2}\Vert \frac{w}{r}\Vert _{L^2}. \end{aligned}$$

Thus we deduce

$$\begin{aligned} \Vert rF\Vert _{L^2}\lesssim (\nu r_0+\frac{|B|}{r_0^2})\Vert \frac{w}{r}\Vert _{L^2}. \end{aligned}$$

Together with $\frac{|B|}{\nu }=r_0^3$, this implies

$$\begin{aligned} \Vert rF\Vert _{L^2}\lesssim (\nu r_0+\frac{|B|}{r_0^2})\Vert \frac{w}{r}\Vert _{L^2}\le C\nu ^{\frac{1}{3}} |B|^{\frac{2}{3}}\Vert \frac{w}{r}\Vert _{L^2}. \end{aligned}$$

as desired. $\quad \square$

4 Enhanced Dissipation and Invisid Damping

4.1 Pseudospectral bound

The direct implication of the resolvent estimates is to provide controls for the semigroup by using pseudospectral bounds.

As in [25], we call an operator L in a Hilbert space H is accretive if

$$\begin{aligned} \Re \langle Lf, f\rangle \ge 0,\quad \text {for any}\ f\in D(L). \end{aligned}$$

The operator L is said to be m-accretive if, in addition, all $\Re \lambda <0$ belong to the resolvent set of L (see [19] for more details). The pseudospectral bound of L is defined as

$$\begin{aligned} \Psi (L)=\inf \{\Vert (L-i\lambda )f\Vert :f\in D(L),\lambda \in \mathbb {R},\Vert f\Vert =1\}. \end{aligned}$$

(4.1)

Consider the operator $L_k$

$$\begin{aligned} L_kw{:}{=}&-\nu (\partial _r^2-\frac{k^2-\frac{1}{4}}{r^2})w+i\frac{kB}{r^2}w \end{aligned}$$

in the domain

$$\begin{aligned} D_k=\{w\in H_{loc}^2(\mathbb {R}_{+},dr)\cap L^2(\mathbb {R}_{+},dr):-\nu (\partial _r^2-\frac{k^2-\frac{1}{4}}{r^2})w+i\frac{kB}{r^2}w\in L^2(\mathbb {R}_{+},dr) \}. \end{aligned}$$

Note that $-\partial _r^2$ is an operator with the compact resolvent. Since $L_k$ is a relatively compact perturbation of $-\nu \partial _r^2$ in the domain $D_k$, it is hence clear that the operators $L_k$ also has the compact resolvent, this indicates that the operator $L_k$ has only point spectrum.

In Lemma 3.1 we have obtained

$$\begin{aligned} \Re \langle L_kw,w\rangle _{L^2}=\nu \Vert w'\Vert _{L^2}^2+\nu (k^2-\frac{1}{4})\Vert \frac{w}{r}\Vert _{L^2}^2\ge 0. \end{aligned}$$

The above inequality indicates $L_k$ being accretive, and furthermore m-accretive. Recall that in Proposition 3.2 we establish the following resolvent estimates

$$\begin{aligned}&\Vert r(L_k-i\lambda )w\Vert _{L^2}\ge C(\nu k^2)^{\frac{1}{3}}|B|^{\frac{2}{3}}\Vert \frac{w}{r}\Vert _{L^2}. \end{aligned}$$

In view of the fact $r\in [1, R]$, this provides a lower bound for the pseudospectrum of $L_k$:

$$\begin{aligned}&\Vert (L_k-i\lambda )w\Vert _{L^2}\ge C(\nu k^2)^{\frac{1}{3}}|B|^{\frac{2}{3}}R^{-2}\Vert w\Vert _{L^2}. \end{aligned}$$

We summarize it into

Lemma 4.1

Let $\Psi$ be defined as in (4.1). There exists some $C>0$ independent of $\nu ,k, B, R$ such that

$$\begin{aligned} \Psi (L_k(L^2\rightarrow L^2))\ge C(\nu k^2)^{\frac{1}{3}}|B|^{\frac{2}{3}}R^{-2}. \end{aligned}$$

4.2 Semigroup bound and enhanced dissipation

It is convenient to use the below space-time norm

$$\begin{aligned} \Vert g\Vert _{L^pL^2}{:}{=}\Big \Vert \Vert g\Vert _{L^2_r([1, R])}\Big \Vert _{L^p_t([0, \infty )}. \end{aligned}$$

To obtain decaying semigroup bounds from pseudospectral bounds, we appeal to the following Gearhart–Pr$\ddot{u}$ss type lemma established by Wei in [33]. (See also [17] by Helffer || Sjöstrand for relevant reference.)

Lemma 4.2

[33]. Let L be a m-accretive operator in a Hilbert space X. Then it holds

$$\begin{aligned} \Vert e^{-tL}\Vert _X\le e^{-t\Psi (L)+\frac{\pi }{2}} \quad \text {for any} \ t\ge 0. \end{aligned}$$

We proceed to studying the homogeneous linear equation

$$\begin{aligned}&\partial _tw_k^l+L_kw_k^l=0 \quad \text {with} \ w_k^l(0)=w_k(0). \end{aligned}$$

(4.2)

Utilizing semigroup theory, we can express $w_k^l$ as

$$\begin{aligned} w_k^l(t)=e^{-t L_k}w_k(0). \end{aligned}$$

We start to derive the space-time estimate for $w_k^l$.

Proposition 4.3

Let $w_k^l$ be the solution to (4.2) with $w_k(0)\in L^2$. Then for any $k\in \mathbb {Z}$ and $|k|\ge 1$, there exist constants $C,c>0$ being independent of $\nu ,k,B,R$, such that the following inequality holds

$$\begin{aligned}&\Vert w_k^l(t)\Vert _{L^2}\le Ce^{-c(\nu k^2)^{\frac{1}{3}}|B|^{\frac{2}{3}}R^{-2}t}\Vert w_k(0)\Vert _{L^2}, \quad \text {for any} \ t\ge 0. \end{aligned}$$

(4.3)

Moreover, for any $c'\in (0,c)$, we have

$$\begin{aligned}&(\nu k^2)^{\frac{1}{3}}|B|^{\frac{2}{3}}R^{-2}\Vert e^{c'(\nu k^2)^{\frac{1}{3}}|B|^{\frac{2}{3}}R^{-2}t}w_k^l(t)\Vert _{L^2L^2}^2\le C\Vert w_k(0)\Vert _{L^2}^2. \end{aligned}$$

(4.4)

Proof

The semigroup bounds (4.3) readily follows from Lemma 4.1 and Lemma 4.2. Hence for any $c'\in (0,c)$, multiplying $e^{c'(\nu k^2)^{\frac{1}{3}}|B|^{\frac{2}{3}}R^{-2}t}$ on both sides of (4.3) yields

$$\begin{aligned}&2c'(\nu k^2)^{\frac{1}{3}}|B|^{\frac{2}{3}}R^{-2}\Vert e^{c'(\nu k^2)^{\frac{1}{3}}|B|^{\frac{2}{3}}R^{-2}t}w_k^l(t)\Vert _{L^2}^2\\&\quad \le 2Cc'(\nu k^2)^{\frac{1}{3}}|B|^{\frac{2}{3}}R^{-2}e^{-2(c-c')(\nu k^2)^{\frac{1}{3}}|B|^{\frac{2}{3}}R^{-2}t}\Vert w_k(0)\Vert _{L^2}^2. \end{aligned}$$

Then we integrate above inequality with respect to t to deduce

$$\begin{aligned}&2c'(\nu k^2)^{\frac{1}{3}}|B|^{\frac{2}{3}}R^{-2}\Vert e^{c'(\nu k^2)^{\frac{1}{3}}|B|^{\frac{2}{3}}R^{-2}t}w_k^l\Vert _{L^2L^2}^2\\&\quad \le \int _0^{\infty }2Cc'(\nu k^2)^{\frac{1}{3}}|B|^{\frac{2}{3}}R^{-2}e^{-2(c-c')(\nu k^2)^{\frac{1}{3}}|B|^{\frac{2}{3}}R^{-2}t}dt\Vert w_k(0)\Vert _{L^2}^2\lesssim \Vert w_k(0)\Vert _{L^2}^2. \end{aligned}$$

This implies the estimate in (4.4). $\square$

4.3 Homogeneous linear equation with nonzero initial data

Based on the results of Proposition 4.3 and the structure of the linear equation itself, we can obtain the following $L^{\infty }L^2$ energy estimate for the homogeneous linear equation with respect to $w_k^l$.

Lemma 4.4

Let $w_k^l$ be the solution to (4.2) with initial data $w_k(0)\in L^2$. Considering c to be the same as in Proposition 4.3 and $c'\in (0,c)$. For any $k\in \mathbb {Z}$ and $|k|\ge 1$, it holds

$$\begin{aligned}&\Vert e^{c'(\nu k^2)^{\frac{1}{3}}|B|^{\frac{2}{3}}R^{-2}t}w_k^l\Vert _{L^{\infty }L^2}^2+(\nu k^2)^{\frac{1}{3}}|B|^{\frac{2}{3}}R^{-2}\Vert e^{c'(\nu k^2)^{\frac{1}{3}}|B|^{\frac{2}{3}}R^{-2}t}w_k^l\Vert _{L^2L^2}^2\nonumber \\&\qquad +\nu \Vert e^{c'(\nu k^2)^{\frac{1}{3}}|B|^{\frac{2}{3}}R^{-2}t}\partial _{r}w_k^l\Vert _{L^2L^2}^2 +\nu k^2\Vert e^{c'(\nu k^2)^{\frac{1}{3}}|B|^{\frac{2}{3}}R^{-2}t}\frac{w_k^l}{r}\Vert _{L^2L^2}^2\nonumber \\&\quad \lesssim \Vert w_k(0)\Vert _{L^2}^2. \end{aligned}$$

(4.5)

Proof

We first conduct the integration by parts and get

$$\begin{aligned}&\Re \langle \partial _tw_k^l-\nu (\partial _r^2-\frac{k^2-\frac{1}{4}}{r^2})w_k^l+\frac{ikB}{r^2}w_k^l,w_k^l\rangle \\&\quad =\frac{1}{2}\partial _t\Vert w_k^l\Vert _{L^2}^2+\nu \Vert \partial _rw_k^l\Vert _{L^2}^2+\nu (k^2-\frac{1}{4})\Vert \frac{w_k^l}{r}\Vert _{L^2}^2=0. \end{aligned}$$

By multiplying $e^{2c'(\nu k^2)^{\frac{1}{3}}|B|^{\frac{2}{3}}R^{-2}t}$ on both sides, we deduce

$$\begin{aligned}&\partial _t\Vert e^{c'(\nu k^2)^{\frac{1}{3}}|B|^{\frac{2}{3}}R^{-2}t}w_k^l\Vert _{L^2}^2\\&\qquad +2\nu \Big (\Vert e^{c'(\nu k^2)^{\frac{1}{3}}|B|^{\frac{2}{3}}R^{-2}t}\partial _{r}w_k^l\Vert _{L^2}^2 +(k^2-\frac{1}{4})\Vert e^{c'(\nu k^2)^{\frac{1}{3}}|B|^{\frac{2}{3}}R^{-2}t}\frac{w_k^l}{r}\Vert _{L^2}^2\Big )\\&\quad \le 2c'(\nu k^2)^{\frac{1}{3}}|B|^{\frac{2}{3}}R^{-2}\Vert e^{c'(\nu k^2)^{\frac{1}{3}}|B|^{\frac{2}{3}}R^{-2}t}w_k^l\Vert _{L^2}^2. \end{aligned}$$

This implies a space-time estimate for $w_k^l$:

$$\begin{aligned}&\Vert e^{c'(\nu k^2)^{\frac{1}{3}}|B|^{\frac{2}{3}}R^{-2}t}w_k^l\Vert _{L^{\infty }L^2}^2\\&\qquad +2\nu \Big (\Vert e^{c'(\nu k^2)^{\frac{1}{3}}|B|^{\frac{2}{3}}R^{-2}t}\partial _{r}w_k^l\Vert _{L^2L^2}^2 +(k^2-\frac{1}{4})\Vert e^{c'(\nu k^2)^{\frac{1}{3}}|B|^{\frac{2}{3}}R^{-2}t}\frac{w_k^l}{r}\Vert _{L^2L^2}^2\Big )\\&\quad \le 2c'(\nu k^2)^{\frac{1}{3}}|B|^{\frac{2}{3}}R^{-2}\Vert e^{c'(\nu k^2)^{\frac{1}{3}}|B|^{\frac{2}{3}}R^{-2}t}w_k^l\Vert _{L^2L^2}^2+\Vert w_k(0)\Vert _{L^2}^2. \end{aligned}$$

Combining with Proposition 4.3, we arrive at

$$\begin{aligned}&\Vert e^{c'(\nu k^2)^{\frac{1}{3}}|B|^{\frac{2}{3}}R^{-2}t}w_k^l\Vert _{L^{\infty }L^2}^2+(\nu k^2)^{\frac{1}{3}}|B|^{\frac{2}{3}}R^{-2}\Vert e^{c'(\nu k^2)^{\frac{1}{3}}|B|^{\frac{2}{3}}R^{-2}t}w_k^l\Vert _{L^2L^2}^2\\&\quad +\nu \Vert e^{c'(\nu k^2)^{\frac{1}{3}}|B|^{\frac{2}{3}}R^{-2}t}\partial _{r}w_k^l\Vert _{L^2L^2}^2 +\nu (k^2-\frac{1}{4})\Vert e^{c'(\nu k^2)^{\frac{1}{3}}|B|^{\frac{2}{3}}R^{-2}t}\frac{w_k^l}{r}\Vert _{L^2L^2}^2\lesssim \Vert w_k(0)\Vert _{L^2}^2. \end{aligned}$$

This completes the proof of Lemma 4.4. $\quad \square$

4.4 Inhomogeneous linear equation with zero initial data

We then derive the space-time estimates for inhomogeneous equations with zero initial conditions, which will be frequently used in later sections. The derivation of this lemma heavily relies on established resolvent estimates in Proposition 3.2 from Sect. 3.

Proposition 4.5

Given functions $h_1(t, r), h_2(t, r)$ and g(r), assume w solves

$$\begin{aligned} \left\{ \begin{aligned}&\partial _t w+L_k w=h_1-g \partial _{r}h_2,\\&w|_{t=0}=0,\quad w|_{r=1,R}=0, \end{aligned}\right. \end{aligned}$$

(4.6)

and let the stream function $\varphi$ satisfy

$$\begin{aligned}&(\partial _r^2-\frac{k^2-\frac{1}{4}}{r^2})\varphi =w,\quad \varphi |_{r=1,R}=0 \quad \text {with} \ r\in [1,R] \ \text {and} \ t\ge 0. \end{aligned}$$

With c being the same constant as in Proposition 4.3, then for any $c'\in [0, c]$, the following inequalities hold for w and $\varphi$:

$$\begin{aligned} \begin{aligned}&\Vert e^{c'(\nu k^2)^{\frac{1}{3}}|B|^{\frac{2}{3}}R^{-2}t}w\Vert _{L^{\infty }L^2}^2+\nu ^{\frac{1}{3}} |kB|^{\frac{2}{3}}\Vert e^{c'(\nu k^2)^{\frac{1}{3}}|B|^{\frac{2}{3}}R^{-2}t}\frac{w}{r}\Vert _{L^2L^2}^2\\&\qquad +\nu \Vert e^{c'(\nu k^2)^{\frac{1}{3}}|B|^{\frac{2}{3}}R^{-2}t}\partial _{r}w\Vert _{L^2L^2}^2+\nu k^2\Vert e^{c'(\nu k^2)^{\frac{1}{3}}|B|^{\frac{2}{3}}R^{-2}t}\frac{w}{r}\Vert _{L^2L^2}^2\\&\quad \lesssim \nu ^{-\frac{1}{3}} |kB|^{-\frac{2}{3}}\Vert e^{c'(\nu k^2)^{\frac{1}{3}}|B|^{\frac{2}{3}}R^{-2}t}rh_1\Vert _{L^2L^2}^2+\nu ^{-1} \Vert e^{c'(\nu k^2)^{\frac{1}{3}}|B|^{\frac{2}{3}}R^{-2}t}(|g|+r|\partial _r g|)h_2\Vert _{L^2L^2}^2, \end{aligned} \end{aligned}$$

and

$$\begin{aligned}&\nu ^{\frac{1}{6}}|kB|^{\frac{5}{6}}|k|^{\frac{1}{2}}R^{-2}(\Vert e^{c'(\nu k^2)^{\frac{1}{3}}|B|^{\frac{2}{3}}R^{-2}t}\partial _r{\varphi }\Vert _{L^2L^2}+|k|\Vert e^{c'(\nu k^2)^{\frac{1}{3}}|B|^{\frac{2}{3}}R^{-2}t}\frac{{\varphi }}{r}\Vert _{L^2L^2}) \\&\quad \lesssim \left( (\frac{\nu }{|kB|})^{\frac{1}{6}} (\log R)^{\frac{1}{2}} +1 \right) \Vert e^{c'(\nu k^2)^{\frac{1}{3}}|B|^{\frac{2}{3}}R^{-2}t}rh_1\Vert _{L^2L^2}\\&\qquad +\nu ^{-\frac{1}{3}} |kB|^{\frac{1}{3}}\Vert e^{c'(\nu k^2)^{\frac{1}{3}}|B|^{\frac{2}{3}}R^{-2}t}(|g|+r|\partial _r g|)h_2\Vert _{L^2L^2}. \end{aligned}$$

Proof

We introduce the weighted quantities

$$\begin{aligned} \tilde{w}:=e^{c'(\nu k^2)^{\frac{1}{3}}|B|^{\frac{2}{3}}R^{-2}t}w, \quad \tilde{\varphi }:=e^{c'(\nu k^2)^{\frac{1}{3}}|B|^{\frac{2}{3}}R^{-2}t}\varphi \end{aligned}$$

and

$$\begin{aligned} \tilde{h}_j=e^{c'(\nu k^2)^{\frac{1}{3}}|B|^{\frac{2}{3}}R^{-2}t}h_j \quad \text {for} \ j=1,2. \end{aligned}$$

Via a direct check, we can see that

$$\begin{aligned} \partial _t \tilde{w}+(L_k-c'(\nu k^2)^{\frac{1}{3}}|B|^{\frac{2}{3}}R^{-2}) \tilde{w}=\frac{1}{r}[ik\tilde{h}_1-r^{\frac{1}{2}}\partial _{r}(r^{\frac{1}{2}}\tilde{h}_2)]. \end{aligned}$$

(4.7)

We then take the Fourier transform in t and define¹

$$\begin{aligned} \hat{w}(\lambda ,r)&=\int _0^{\infty }\tilde{w}(t,r)e^{-it\lambda }dt,\quad \hat{\varphi }(\lambda ,r)=\int _0^{\infty }\tilde{\varphi }(t,r)e^{-it\lambda }dt,\\ H_j(\lambda ,r)&=\int _0^{\infty }\tilde{h}_j(t,r)e^{-it\lambda }dt \quad \text {for} \ j=1,2. \end{aligned}$$

The inhomogeneous equation (4.7) can thus be transferred into the form

$$\begin{aligned} (i\lambda +L_k-c'(\nu k^2)^{\frac{1}{3}}|B|^{\frac{2}{3}}R^{-2})\hat{w}(\lambda ,r)=H_1-g\partial _{r}H_2. \end{aligned}$$

We further decompose $\hat{w}$ as $\hat{w}=\hat{w}^{1}+\hat{w}^{2}$, where $\hat{w}^{1}$ and $\hat{w}^{2}$ solve

$$\begin{aligned} \Big (i\lambda +L_k-c'(\nu k^2)^{\frac{1}{3}}|B|^{\frac{2}{3}}R^{-2}\Big )\hat{w}^{1}=H_1, \end{aligned}$$

and

$$\begin{aligned} \Big (i\lambda +L_k-c'(\nu k^2)^{\frac{1}{3}}|B|^{\frac{2}{3}}R^{-2}\Big )\hat{w}^{2}=-g\partial _{r}H_2. \end{aligned}$$

We also define the corresponding stream functions $\hat{\varphi }^1$ and $\hat{\varphi }^2$ via the below linear elliptic equations

$$\begin{aligned} (\partial _r^2-\frac{k^2-\frac{1}{4}}{r^2})\hat{\varphi }^j=\hat{w}^j,\quad \hat{\varphi }^j|_{r=1,R}=0,\quad r\in [1,R], \quad \text {for} \ j=1,2. \end{aligned}$$

This enables us to write $\hat{\varphi }=\hat{\varphi }^1+\hat{\varphi }^2$. In view of Proposition 3.2, we then obtain the estimate for $w^1$ and $\varphi ^1$:

$$\begin{aligned}&\nu ^{\frac{2}{3}} |kB|^{\frac{1}{3}}\Vert \partial _{r}\hat{w}^{1}\Vert _{L^2}+\nu ^{\frac{1}{3}} |kB|^{\frac{2}{3}}\Vert \frac{\hat{w}^{1}}{r}\Vert _{L^2}\le C \Vert rH_1\Vert _{L^2}, \end{aligned}$$

$$\begin{aligned} \nu ^{\frac{1}{6}}|kB|^{\frac{5}{6}}|k|^{\frac{1}{2}}\left( \Vert \partial _r \hat{\varphi }^1\Vert _{L^2}+|k|\Vert \frac{\hat{\varphi }^1}{r}\Vert _{L^2} \right) \le CR^2\left( (\frac{\nu }{|kB|})^{\frac{1}{6}} (\log R)^{\frac{1}{2}} +1 \right) \Vert rH_1\Vert _{L^2}, \end{aligned}$$

as well as the control for $w^2$ and $\varphi ^2$:

$$\begin{aligned} \nu \Vert \partial _{r}\hat{w}^{2}\Vert _{L^2}+\nu ^{\frac{2}{3}} |kB|^{\frac{1}{3}}\Vert \frac{\hat{w}^{2}}{r}\Vert _{L^2}&\le C\Vert g\partial _r H_2\Vert _{H^{-1}_r}\lesssim \Vert (|g|+r|g'|)H_2\Vert _{L^2},\\ \nu ^{\frac{1}{2}}|kB|^{\frac{1}{2}}R^{-2}(\Vert \partial _r\hat{\varphi }^{ 2}\Vert _{L^2}+|k|\Vert \frac{\hat{\varphi }^{ 2}}{r}\Vert _{L^2})&\le C\Vert g\partial _r H_2\Vert _{H^{-1}_r}\lesssim \Vert (|g|+r|g'|)H_2\Vert _{L^2}. \end{aligned}$$

Here we utilize the definition of $H^{-1}_r$ norm to bound

$$\begin{aligned} \begin{aligned} \Vert g\partial _r H_2\Vert _{H^{-1}}{:}{=}&\sup \limits _{\Vert f\Vert _{H^1_r}\le 1} |\langle g\partial _r H_2, f\rangle | =\sup \limits _{\Vert f\Vert _{H^1_r}\le 1} |\langle H_2, (gf)' \rangle | \\ \le&\sup \limits _{\Vert f\Vert _{H^1_r}\le 1} |\langle gH_2, f' \rangle |+\sup \limits _{\Vert f\Vert _{H^1_r}\le 1} |\langle rg' H_2, \frac{f}{r} \rangle | \\ \lesssim&\Vert (|g|+r|g'|)H_2\Vert _{L^2}. \end{aligned} \end{aligned}$$

Combining the above inequalities, we arrive at

$$\begin{aligned} \nu ^{\frac{2}{3}} |kB|^{\frac{1}{3}}\Vert \partial _{r}\hat{w}\Vert _{L^2}+\nu ^{\frac{1}{3}} |kB|^{\frac{2}{3}}\Vert \frac{\hat{w}}{r}\Vert _{L^2} \lesssim \Vert rH_1\Vert _{L^2}+\nu ^{-\frac{1}{3}} |kB|^{\frac{1}{3}}\Vert (|g|+r|g'|)H_2\Vert _{L^2}, \end{aligned}$$

and

$$\begin{aligned}&\nu ^{\frac{1}{6}}|kB|^{\frac{5}{6}}R^{-2}(\Vert \partial _r\hat{\varphi }\Vert _{L^2}+|k|\Vert \frac{\hat{\varphi }}{r}\Vert _{L^2})\\&\quad \lesssim \left( (\frac{\nu }{|kB|})^{\frac{1}{6}} (\log R)^{\frac{1}{2}} +1 \right) \Vert rH_1\Vert _{L^2}+\nu ^{-\frac{1}{3}} |kB|^{\frac{1}{3}}\Vert (|g|+r|g'|)H_2\Vert _{L^2}. \end{aligned}$$

According to the Plancherel’s theorem, we have the following equivalence relations based on $L^2$ norms

$$\begin{aligned}&\Vert \partial _{r}\tilde{w}\Vert _{L^2L^2}\approx \big \Vert \Vert \partial _{r}\hat{w}\Vert _{L^2}\big \Vert _{L^2(\mathbb {R})}, \quad \Vert \frac{\tilde{w}}{r}\Vert _{L^2L^2}\approx \big \Vert \Vert \frac{\hat{w}}{r}\Vert _{L^2}\big \Vert _{L^2(\mathbb {R})}, \\&\Vert \partial _r\tilde{\varphi }\Vert _{L^2L^2}+|k|\Vert \frac{\tilde{\varphi }}{r}\Vert _{L^2L^2}\approx \big \Vert \Vert \partial _r\hat{\varphi }\Vert _{L^2}\big \Vert _{L^2(\mathbb {R})}+|k|\big \Vert \Vert \frac{\hat{\varphi }}{r}\Vert _{L^2}\big \Vert _{L^2(\mathbb {R})},\\&\Vert r\tilde{h}_1\Vert _{L^2L^2}\approx \big \Vert \Vert rH_1\Vert _{L^2}\big \Vert _{L^2(\mathbb {R})}, \quad \Vert (|g|+r|g'|)\tilde{h}_2\Vert _{L^2L^2}\approx \big \Vert \Vert (|g|+r|g'|)H_2\Vert _{L^2}\big \Vert _{L^2(\mathbb {R})}. \end{aligned}$$

Plugging in all estimates above, we thus deduce

$$\begin{aligned}&\nu ^{\frac{2}{3}} |kB|^{\frac{1}{3}}\Vert e^{c'(\nu k^2)^{\frac{1}{3}}|B|^{\frac{2}{3}}R^{-2}t}\partial _{r}w\Vert _{L^2L^2}+\nu ^{\frac{1}{3}} |kB|^{\frac{2}{3}}\Vert e^{c'(\nu k^2)^{\frac{1}{3}}|B|^{\frac{2}{3}}R^{-2}t}\frac{w}{r}\Vert _{L^2L^2} \nonumber \\&\quad \lesssim \Vert e^{c'(\nu k^2)^{\frac{1}{3}}|B|^{\frac{2}{3}}R^{-2}t}rh_1\Vert _{L^2L^2}+\nu ^{-\frac{1}{3}} |kB|^{\frac{1}{3}}\Vert e^{c'(\nu k^2)^{\frac{1}{3}}|B|^{\frac{2}{3}}R^{-2}t}(|g|+r|g'|)h_2\Vert _{L^2L^2}, \end{aligned}$$

(4.8)

and

$$\begin{aligned}&\nu ^{\frac{1}{6}}|kB|^{\frac{5}{6}}R^{-2}(\Vert e^{c'(\nu k^2)^{\frac{1}{3}}|B|^{\frac{2}{3}}R^{-2}t}\partial _r{\varphi }\Vert _{L^2L^2}+|k|\Vert e^{c'(\nu k^2)^{\frac{1}{3}}|B|^{\frac{2}{3}}R^{-2}t}\frac{{\varphi }}{r}\Vert _{L^2L^2}) \\&\quad \lesssim \left( (\frac{\nu }{|kB|})^{\frac{1}{6}} (\log R)^{\frac{1}{2}} +1 \right) \Vert e^{c'(\nu k^2)^{\frac{1}{3}}|B|^{\frac{2}{3}}R^{-2}t}kh_1\Vert _{L^2L^2}+\nu ^{-\frac{1}{3}} |kB|^{\frac{1}{3}}\Vert e^{c'(\nu k^2)^{\frac{1}{3}}|B|^{\frac{2}{3}}R^{-2}t}h_2\Vert _{L^2L^2}. \end{aligned}$$

Applying the integration by parts, we also get

$$\begin{aligned} 0&=\Re \langle \partial _tw- \nu (\partial _{r}^2-\frac{k^2-\frac{1}{4}}{r^2})w+\frac{ikB}{r^2}w-h_1+g\partial _{r}h_2,e^{2c'(\nu k^2)^{\frac{1}{3}}|B|^{\frac{2}{3}}R^{-2}t}w\rangle \\&=\frac{1}{2}\partial _t\Vert e^{c'(\nu k^2)^{\frac{1}{3}}|B|^{\frac{2}{3}}R^{-2}t}w\Vert _{L^2}^2+\nu \Vert e^{c'(\nu k^2)^{\frac{1}{3}}|B|^{\frac{2}{3}}R^{-2}t}\partial _{r}w\Vert _{L^2}^2+\nu (k^2-\frac{1}{4})\Vert e^{c'(\nu k^2)^{\frac{1}{3}}|B|^{\frac{2}{3}}R^{-2}t}\frac{w}{r}\Vert _{L^2}^2\\&\quad -c'(\nu k^2)^{\frac{1}{3}}|B|^{\frac{2}{3}}R^{-2}\Vert e^{c'(\nu k^2)^{\frac{1}{3}}|B|^{\frac{2}{3}}R^{-2}t}w\Vert _{L^2}^2-\Re \langle h_1-g\partial _{r}h_2,e^{2c'(\nu k^2)^{\frac{1}{3}}|B|^{\frac{2}{3}}R^{-2}t}w\rangle , \end{aligned}$$

By employing the Cauchy–Schwarz inequality, we further obtain

$$\begin{aligned}&\partial _t\Vert e^{c'(\nu k^2)^{\frac{1}{3}}|B|^{\frac{2}{3}}R^{-2}t}w\Vert _{L^2}^2+2\nu \Vert e^{c'(\nu k^2)^{\frac{1}{3}}|B|^{\frac{2}{3}}R^{-2}t}\partial _{r}w\Vert _{L^2}^2+2\nu (k^2-\frac{1}{4})\Vert e^{c'(\nu k^2)^{\frac{1}{3}}|B|^{\frac{2}{3}}R^{-2}t}\frac{w}{r}\Vert _{L^2}^2\\&\quad =2c'(\nu k^2)^{\frac{1}{3}}|B|^{\frac{2}{3}}R^{-2}\Vert e^{c'(\nu k^2)^{\frac{1}{3}}|B|^{\frac{2}{3}}R^{-2}t}w\Vert _{L^2}^2+2\Re \langle h_1-g\partial _{r}h_2,e^{2c'(\nu k^2)^{\frac{1}{3}}|B|^{\frac{2}{3}}R^{-2}t}w\rangle \\&\quad \lesssim c'(\nu k^2)^{\frac{1}{3}}|B|^{\frac{2}{3}}R^{-2}\Vert e^{c'(\nu k^2)^{\frac{1}{3}}|B|^{\frac{2}{3}}R^{-2}t}w\Vert _{L^2}^2+\Vert e^{c'(\nu k^2)^{\frac{1}{3}}|B|^{\frac{2}{3}}R^{-2}t}rh_1\Vert _{L^2}\Vert e^{c'(\nu k^2)^{\frac{1}{3}}|B|^{\frac{2}{3}}R^{-2}t}\frac{w}{r}\Vert _{L^2}\\&\qquad +\Vert e^{c'(\nu k^2)^{\frac{1}{3}}|B|^{\frac{2}{3}}R^{-2}t}(|g|+r|g'|)h_2\Vert _{L^2}\Vert e^{c'(\nu k^2)^{\frac{1}{3}}|B|^{\frac{2}{3}}R^{-2}t}w\Vert _{H^1_r}. \end{aligned}$$

Together with (4.8), the above inequality yields

$$\begin{aligned}&\Vert e^{c'(\nu k^2)^{\frac{1}{3}}|B|^{\frac{2}{3}}R^{-2}t}w\Vert _{L^{\infty }L^2}^2+\nu \Vert e^{c'(\nu k^2)^{\frac{1}{3}}|B|^{\frac{2}{3}}R^{-2}t}\partial _{r}w\Vert _{L^2L^2}^2+\nu k^2\Vert e^{c'(\nu k^2)^{\frac{1}{3}}|B|^{\frac{2}{3}}R^{-2}t}\frac{w}{r}\Vert _{L^2L^2}^2\\&\qquad +\nu ^{\frac{1}{3}} |kB|^{\frac{2}{3}}\Vert e^{c'(\nu k^2)^{\frac{1}{3}}|B|^{\frac{2}{3}}R^{-2}t}\frac{w}{r}\Vert _{L^2L^2}^2\\&\quad \lesssim \nu ^{-\frac{1}{3}} |kB|^{-\frac{2}{3}}\Big (\Vert e^{c'(\nu k^2)^{\frac{1}{3}}|B|^{\frac{2}{3}}R^{-2}t}rh_1\Vert _{L^2L^2}+\nu ^{-\frac{1}{3}} |kB|^{\frac{1}{3}}\Vert e^{c'(\nu k^2)^{\frac{1}{3}}|B|^{\frac{2}{3}}R^{-2}t}(|g|+r|g'|)h_2\Vert _{L^2L^2}\Big )^2\\&\quad \lesssim \nu ^{-\frac{1}{3}} |kB|^{-\frac{2}{3}}\Vert e^{c'(\nu k^2)^{\frac{1}{3}}|B|^{\frac{2}{3}}R^{-2}t}rh_1\Vert _{L^2L^2}^2+\nu ^{-1} \Vert e^{c'(\nu k^2)^{\frac{1}{3}}|B|^{\frac{2}{3}}R^{-2}t}(|g|+r|g'|)h_2\Vert _{L^2L^2}^2. \end{aligned}$$

This finishes the proof of Proposition 4.5. $\quad \square$

4.5 Integrated invisid damping

This subsection is devoted to providing estimates for the stream function $\varphi _k^l$ in terms of initial vorticity $w_k(0)$, with $\varphi _k^l$ admitting the linear elliptic equation

$$\begin{aligned} (\partial _r^2-\frac{k^2-\frac{1}{4}}{r^2})\varphi _k^l=w_k^l,\quad \varphi _k^l|_{r=1,R}=0,\quad r\in [1,R]. \end{aligned}$$

(4.9)

For inviscid fluids governed by the Euler equation, there is the concept of so-called inviscid damping. In fact, the effect of inviscid damping extends to viscous fluids as well. Lemma 4.6 presents a space-time version of linear inviscid damping around 2D TC flow.

Lemma 4.6

Let $w_k^l$ be the solution to (4.2) with initial data $w_k(0)\in L^2$ and $\varphi _k^l$ solves (4.9). Considering c to be the same as in Proposition 4.3 and $c'\in (0,c)$. For any $k\in \mathbb {Z}$ and $|k|\ge 1$, the following inequality holds

$$\begin{aligned}{} & {} k^2|B|R^{-4}(\Vert e^{c'(\nu k^2)^{\frac{1}{3}}|B|^{\frac{2}{3}}R^{-2}t}\partial _r{\varphi }^{l}_k\Vert _{L^2L^2}^2+|k|^2\Vert e^{c'(\nu k^2)^{\frac{1}{3}}|B|^{\frac{2}{3}}R^{-2}t}\frac{{\varphi }^{l}_k}{r}\Vert _{L^2L^2}^2) \nonumber \\{} & {} \quad \lesssim (\log R)^{-2}R^{-4}\Vert r^2 w_k(0)\Vert _{L^2}^2+R^6 \Vert \frac{w_k(0)}{r^3}\Vert _{L^2}^2+(\frac{\nu }{|kB|})^{\frac{2}{3}}R^2\Vert \partial _r w_k(0)\Vert _{L^2}^2 \nonumber \\{} & {} \qquad +\left( (\frac{\nu }{|kB|})^{\frac{1}{3}} \log R +1 \right) \left( R^{-2}\Vert rw_k(0)\Vert _{L^2}^2 + (\frac{\nu }{|kB|})^{\frac{4}{3}}k^4 R^2\Vert \frac{w_k(0)}{r}\Vert _{L^2}^2\right) .\nonumber \\ \end{aligned}$$

(4.10)

Remark 4.7

Note that $\nu$ is much smaller than 1, and B is a fixed constant, so (4.10) can also be written in the following expression

$$\begin{aligned}&|k||B|^{\frac{1}{2}}(\Vert e^{c'(\nu k^2)^{\frac{1}{3}}|B|^{\frac{2}{3}}R^{-2}t}\partial _r{\varphi }^{l}_k\Vert _{L^2L^2}+|k|\Vert e^{c'(\nu k^2)^{\frac{1}{3}}|B|^{\frac{2}{3}}R^{-2}t}\frac{{\varphi }^{l}_k}{r}\Vert _{L^2L^2})\\&\quad \lesssim C(R) (\Vert \partial _r w_k(0)\Vert _{L^2}+\Vert w_k(0)\Vert _{L^2}). \end{aligned}$$

We can see that both sides of the above inequality are independent of viscosity coefficient $\nu$, which indicates that the result is also valid for the Euler equation. This type of estimate is called inviscid damping, and here we establish its integrated form. Thus we refer to it as the integrated invisid damping estimates.

In order to prove (4.10), we perform a suitable decomposition of $\varphi ^l_k$. Define

$$\begin{aligned} \begin{aligned}&\tilde{w}^{l_0}_k(t,r){:}{=}e^{-ikB\frac{t}{r^2}} w_k(0), \quad \tilde{w}^{l_1}_k(t, r){:}{=}e^{-c(\nu k^2)^{\frac{1}{3}}|B|^{\frac{2}{3}}R^{-2} t} \tilde{w}^{l_0}_k \end{aligned} \end{aligned}$$

(4.11)

and denote $\tilde{w}^{l_2}_k$ to be the solution to the inhomogeneous linear equation equation with zero initial conditions as below:

$$\begin{aligned} \partial _t \tilde{w}^{l_2}_k+L_k \tilde{w}^{l_2}_k=\nu \partial _r^2 \tilde{w}^{l_1}_k-\nu (k^2-\frac{1}{4})\frac{\tilde{w}^{l_1}_k}{r^2}+c(\nu k^2)^{\frac{1}{3}}|B|^{\frac{2}{3}}R^{-2} \tilde{w}^{l_1}_k, \quad \tilde{w}^{l_2}_k(0)=0. \end{aligned}$$

(4.12)

As a result, the corresponding stream functions can be defined by

$$\begin{aligned} (\partial _r^2-\frac{k^2-\frac{1}{4}}{r^2})\tilde{\varphi }^{l_j}_k=w_k^{l_j},\quad \tilde{\varphi }^{l_j}_k|_{r=1,R}=0,\quad r\in [1,R],\quad t\ge 0, \quad j=1,2. \end{aligned}$$

(4.13)

Then it is easily seen that $w^l_k=\tilde{w}^{l_1}_k+\tilde{w}^{l_2}_k$ and $\varphi ^l_k=\tilde{\varphi }^{l_1}_k+\tilde{\varphi }^{l_2}_k$.

We start with the estimates for $\tilde{\varphi }^{l_1}_k$.

Lemma 4.8

Let $w_k^l$ be the solution to (4.2) with initial data $w_k(0)\in L^2$. Considering c to be the same as in Proposition 4.3, $c'\in (0,c)$ and $\tilde{\varphi }^{l_1}_k$ defined in (4.13). For any $k\in \mathbb {Z}$ and $|k|\ge 1$, the following inequality holds

$$\begin{aligned}{} & {} |kB||k|(\log R)^2 \left( \Vert e^{c'(\nu k^2)^{\frac{1}{3}}|B|^{\frac{2}{3}}R^{-2} t}\partial _r \tilde{\varphi }^{l_1}_k\Vert _{L^2L^2}^2+k^2\Vert e^{c'(\nu k^2)^{\frac{1}{3}}|B|^{\frac{2}{3}}R^{-2} t}\frac{\tilde{\varphi }^{l_1}_k}{r}\Vert ^2_{L^2L^2}\right) \nonumber \\ {}{} & {} \quad \lesssim \Vert r^{2}w_k(0)\Vert _{L^2}^2 \end{aligned}$$

(4.14)

Proof

Noting that $\tilde{\varphi }^{l_1}_k=e^{-c(\nu k^2)^{\frac{1}{3}}|B|^{\frac{2}{3}}R^{-2} t}\tilde{\varphi }^{l_0}_k$ and $c'\in (0,c)$, to obtain the above conclusion it suffices to prove

$$\begin{aligned} |kB||k|(\log R)^2\left( \Vert \partial _r \tilde{\varphi }^{l_0}_k\Vert _{L^2L^2}^2+k^2\Vert \frac{\tilde{\varphi }^{l_0}_k}{r}\Vert ^2_{L^2L^2}\right) \lesssim \Vert r^{2}w_k(0)\Vert _{L^2}^2. \end{aligned}$$

(4.15)

Here we develop some new ideas. With $w=1/r^2$, we introduce the weighted Hilbert space $L^2_w([1, R])$ with inner product

$$\begin{aligned} \langle f, g\rangle _w =\int _1^R f\overline{g} w dr. \end{aligned}$$

We find a set of explicit orthonormal basis for $L^2_w([1, R])$ as below:

$$\begin{aligned} \psi _l(r)=(\frac{2}{\log R})^{\frac{1}{2}}r^{1/2}\sin (\frac{l\pi }{\log R} \log r), \quad \text {for all} \ l\in \mathbb {N}^+. \end{aligned}$$

Here $\psi _l$ satisfies the following property

$$\begin{aligned} \left( \partial _r^2-\frac{k^2-\frac{1}{4}}{r^2}\right) \psi _l(r)=-\left( (\frac{l\pi }{\log R})^2+k^2\right) w(r) \psi _l(r), \quad \psi _l|_{r=1, R}=0. \end{aligned}$$

(4.16)

As far as we know, our explicit constructions above are new. For the sake of convenience, we denote $\alpha {:}{=}(\frac{2}{\log R})^{\frac{1}{2}}$, $\beta {:}{=}\frac{\pi }{\log R}$ and $\lambda _{k,l}{:}{=}(\beta l)^2+k^2$. Since

$$\begin{aligned} \begin{aligned} \langle \tilde{\varphi }^{l_0}_k, \psi _l \rangle _w=&\langle \tilde{\varphi }^{l_0}_k, w\psi _l \rangle = -\lambda _{k ,l}^{-1}\langle \tilde{\varphi }^{l_0}_k, \left( \partial _r^2-\frac{k^2-\frac{1}{4}}{r^2}\right) \psi _l \rangle =-\lambda _{k ,l}^{-1}\langle \tilde{w}^{l_0}_k, \psi _l \rangle , \end{aligned} \end{aligned}$$

we can write

$$\begin{aligned} \begin{aligned}&-\langle \tilde{w}^{l_0}_k, \tilde{\varphi }^{l_0}_k \rangle = -\langle w^{-1}\tilde{w}^{l_0}_k, \tilde{\varphi }^{l_0}_k \rangle _w\\&\quad =\sum \limits _{l=1}^{\infty } \lambda _{k, l}^{-1} \langle w^{-1}\tilde{w}^{l_0}_k, \psi _l \rangle _w \overline{\langle \tilde{w}^{l_0}_k , \psi _l \rangle } =\sum \limits _{l=1}^{\infty } \lambda _{k, l}^{-1} |\langle \tilde{w}^{l_0}_k , \psi _l \rangle |^2. \end{aligned} \end{aligned}$$

(4.17)

In view of the definition $\tilde{w}^{l_0}_k=e^{-i kB\frac{t}{r^2}}w_k(0)$, we have the expression

$$\begin{aligned} \begin{aligned} \langle \tilde{w}^{l_0}_k, \psi _l \rangle =&\int _{1}^{R} e^{-ikB\frac{t}{r^2}}w_k(0, r)\psi _l(r) dr\\ =&\int _{\frac{1}{R^2}}^1 e^{-i kBt s} w_k(0, \frac{1}{\sqrt{s}}) \psi _l (\frac{1}{\sqrt{s}})\frac{ds}{2s\sqrt{s}} {:}{=}\mathcal {F}[W_{k, l}](kBt), \end{aligned} \end{aligned}$$

where $\mathcal {F}$ represents the canonical Fourier transform over $\mathbb {R}$, and $W^1_{k, l}(s)=\frac{1}{2s\sqrt{s}}w_k(0, \frac{1}{\sqrt{s}}) \psi _l (\frac{1}{\sqrt{s}})$.²

Note that the Plancherels formula implies

$$\begin{aligned} \begin{aligned} \int _{\mathbb {R}} |\langle \tilde{w}^{l_0}_k , \psi _l \rangle |^2dt&=\frac{2\pi }{|kB|} \int _{\frac{1}{R^2}}^1 |W_{k, l}(s)|^2 ds \\&=\frac{2\pi }{|kB|}\int _{1}^R \frac{r^3}{8} |w_k(0 ,r)|^2 \psi _l^2(r) dr\le \frac{\pi \alpha ^2}{4|kB|} \Vert r^{2}w_k(0)\Vert _{L^2}^2. \end{aligned} \end{aligned}$$

This together with (4.17) yields

$$\begin{aligned} \int _{\mathbb {R}} -\langle \tilde{w}^{l_0}_k, \tilde{\varphi }^{l_0}_k \rangle dt&\le \frac{\alpha ^2\pi }{4|kB|} \Vert r^{2}w_k(0)\Vert _{L^2}^2\sum \limits _{l=1}^{\infty } \frac{1}{(\beta l)^2+k^2} \\&\le \frac{\alpha ^2 \pi }{4|kB|} \Vert r^{2}w_k(0)\Vert _{L^2}^2\int _{0}^{+\infty } \frac{dy}{(\beta y)^2+k^2} = \frac{\alpha ^2 \beta \pi ^2}{8|kB|\cdot |k|} \Vert r^{2}w_k(0)\Vert _{L^2}^2. \end{aligned}$$

On the other hand, observing

$$\begin{aligned} \begin{aligned} -\langle \tilde{w}^{l_0}_k, \tilde{\varphi }^{l_0}_k \rangle =\Vert \partial _r \tilde{\varphi }^{l_0}_k\Vert _{L^2}^2+(k^2-\frac{1}{4})\Vert \frac{\tilde{\varphi }^{l_0}_k}{r}\Vert _{L^2}^2\ge 0 \quad \text {for any} \ t\in \mathbb {R}, \end{aligned} \end{aligned}$$

we infer that

$$\begin{aligned} \begin{aligned}&\Vert \partial _r \tilde{\varphi }^{l_0}_k\Vert _{L^2L^2}^2+(k^2-\frac{1}{4})\Vert \frac{\tilde{\varphi }^{l_0}_k}{r}\Vert _{L^2L^2}^2 \le \frac{\alpha ^2 \beta \pi ^2}{8|kB|\cdot |k|} \Vert r^{2}w_k(0)\Vert _{L^2}^2\\&\quad \lesssim |kB|^{-1}|k|^{-1}(\log R)^{-2} \Vert r^{2}w_k(0)\Vert _{L^2}^2. \end{aligned} \end{aligned}$$

This completes the proof of this lemma. $\quad \square$

To control the $L^2L^2$ norm of $\tilde{\varphi }^{l_2}_k$, we utilize resolvent estimates in Sect. 3 together with Lemma 4.4.

Lemma 4.9

Let $w_k^l$ be the solution to (4.2) with initial data $w_k(0)\in L^2$. Considering c to be the same as in Proposition 4.3 and $c'\in (0,c)$. For any $k\in \mathbb {Z}$ and $|k|\ge 1$, there holds

$$\begin{aligned}&k^2|B|R^{-4}(\Vert e^{c'(\nu k^2)^{\frac{1}{3}}|B|^{\frac{2}{3}}R^{-2}t}\partial _r{\varphi }^{l_2}_k\Vert _{L^2L^2}^2+|k|^2\Vert e^{c'(\nu k^2)^{\frac{1}{3}}|B|^{\frac{2}{3}}R^{-2}t}\frac{{\varphi }^{l_2}_k}{r}\Vert _{L^2L^2}^2) \\&\quad \lesssim \left( (\frac{\nu }{|kB|})^{\frac{1}{3}} \log R +1 \right) \left( R^{-2}\Vert rw_k(0)\Vert _{L^2}^2 + (\frac{\nu }{|kB|})^{\frac{4}{3}}k^4 R^2\Vert \frac{w_k(0)}{r}\Vert _{L^2}^2\right) \\&\qquad +R^6 \Vert \frac{w_k(0)}{r^3}\Vert _{L^2}^2+(\frac{\nu }{|kB|})^{\frac{2}{3}}R^2\Vert \partial _r w_k(0)\Vert _{L^2}^2. \end{aligned}$$

Proof

First recall that $\tilde{w}^{l_2}_k$ satisfies the inhomogeneous linear equation

$$\begin{aligned} \partial _t \tilde{w}^{l_2}_k+L_k \tilde{w}^{l_2}_k=\nu \partial _r^2 \tilde{w}^{l_1}_k-\nu (k^2-\frac{1}{4})\frac{\tilde{w}^{l_1}_k}{r^2}+c'(\nu k^2)^{\frac{1}{3}}|B|^{\frac{2}{3}}R^{-2} \tilde{w}^{l_1}_k, \quad \tilde{w}^{l_2}_k(0)=0, \end{aligned}$$

(4.18)

with the stream function $\tilde{\varphi }^{l_2}_k$ solving the boundary value problem

$$\begin{aligned} (\partial _r^2-\frac{k^2-\frac{1}{4}}{r^2})\tilde{\varphi }^{l_2}_k=w_k,\quad \tilde{\varphi }^{l_2}_k|_{r=1,R}=0,\quad r\in [1,R],\quad t\ge 0. \end{aligned}$$

Employing Proposition 4.5 with $h_1=-\nu (k^2-\frac{1}{4})\frac{\tilde{w}^{l_1}_k}{r^2}+c(\nu k^2)^{\frac{1}{3}}|B|^{\frac{2}{3}}R^{-2} \tilde{w}^{l_1}_k$, $h_2=v\partial _r \tilde{w}^{l_1}_k$ and $g(r)=1$ yields

$$\begin{aligned}&\nu ^{\frac{1}{6}}|kB|^{\frac{5}{6}}|k|^{\frac{1}{2}}R^{-2}(\Vert e^{c'(\nu k^2)^{\frac{1}{3}}|B|^{\frac{2}{3}}R^{-2}t}\partial _r{\varphi }^{l_2}_k\Vert _{L^2L^2}+|k|\Vert e^{c'(\nu k^2)^{\frac{1}{3}}|B|^{\frac{2}{3}}R^{-2}t}\frac{{\varphi }^{l_2}_k}{r}\Vert _{L^2L^2}) \\&\quad \lesssim \left( (\frac{\nu }{|kB|})^{\frac{1}{6}} (\log R)^{\frac{1}{2}} +1 \right) \Vert e^{c'(\nu k^2)^{\frac{1}{3}}|B|^{\frac{2}{3}}R^{-2}t}rh_1\Vert _{L^2L^2}+\nu ^{-\frac{1}{3}} |kB|^{\frac{1}{3}}\Vert e^{c'(\nu k^2)^{\frac{1}{3}}|B|^{\frac{2}{3}}R^{-2}t}h_2\Vert _{L^2L^2}. \end{aligned}$$

Notice that

$$\begin{aligned}&\Vert e^{c'(\nu k^2)^{\frac{1}{3}}|B|^{\frac{2}{3}}R^{-2}t}rh_1\Vert _{L^2L^2}\\&\quad =\Vert e^{-(c-c')(\nu k^2)^{\frac{1}{3}}|B|^{\frac{2}{3}}R^{-2} t} \left( \nu (k^2-\frac{1}{4})\frac{w_k(0)}{r}-c(\nu k^2)^{\frac{1}{3}}|B|^{\frac{2}{3}}R^{-2} rw_k(0) \right) \Vert _{L^2L^2} \\&\quad \lesssim \nu k^2 (\nu k^2)^{\frac{1}{6}}|B|^{\frac{1}{3}}R^{-1}\Vert \frac{w_k(0)}{r}\Vert _{L^2}+(\nu k^2)^{\frac{1}{6}}|B|^{\frac{1}{3}}R^{-1}\Vert rw_k(0) \Vert _{L^2}, \end{aligned}$$

and

$$\begin{aligned}&\Vert e^{c'(\nu k^2)^{\frac{1}{3}}|B|^{\frac{2}{3}}R^{-2}t}h_2\Vert _{L^2L^2}\\&\quad = \Vert e^{-(c-c')(\nu k^2)^{\frac{1}{3}}|B|^{\frac{2}{3}}R^{-2} t} \nu \partial _r\left( e^{-ikB\frac{t}{r^2}}w_k(0)\right) \Vert _{L^2L^2} \\&\quad \lesssim \Vert e^{-(c-c')(\nu k^2)^{\frac{1}{3}}|B|^{\frac{2}{3}}R^{-2} t} \nu kBt \frac{w_k(0)}{r^3}\Vert _{L^2L^2}+\Vert e^{-(c-c')(\nu k^2)^{\frac{1}{3}}|B|^{\frac{2}{3}}R^{-2} t} \nu \partial _r w_k(0)\Vert _{L^2L^2} \\&\quad \lesssim \nu |kB| [(\nu k^2)^{\frac{1}{3}}|B|^{\frac{2}{3}}R^{-2}]^{-\frac{3}{2}}\Vert \frac{w_k(0)}{r^3}\Vert _{L^2}+\nu (\nu k^2)^{\frac{1}{6}}|B|^{\frac{1}{3}}R^{-1}\Vert \partial _r w_k(0)\Vert _{L^2}. \end{aligned}$$

Thus we arrive at

$$\begin{aligned}&k^2|B|R^{-4}(\Vert e^{c'(\nu k^2)^{\frac{1}{3}}|B|^{\frac{2}{3}}R^{-2}t}\partial _r{\varphi }^{l_2}_k\Vert _{L^2L^2}^2+|k|^2\Vert e^{c'(\nu k^2)^{\frac{1}{3}}|B|^{\frac{2}{3}}R^{-2}t}\frac{{\varphi }^{l_2}_k}{r}\Vert _{L^2L^2}^2) \\&\quad \lesssim \left( (\frac{\nu }{|kB|})^{\frac{1}{3}} \log R +1 \right) \left( R^{-2}\Vert rw_k(0)\Vert _{L^2}^2 + (\frac{\nu }{|kB|})^{\frac{4}{3}}k^4 R^2\Vert \frac{w_k(0)}{r}\Vert _{L^2}^2\right) \\&\qquad +R^6 \Vert \frac{w_k(0)}{r^3}\Vert _{L^2}^2+(\frac{\nu }{|kB|})^{\frac{2}{3}}R^2\Vert \partial _r w_k(0)\Vert _{L^2}^2. \end{aligned}$$

$\square$

Therefore, the desired bound (4.10) can be readily deduced from Lemma 4.8 and Lemma 4.9. This completes the proof of Lemma 4.6.

5 Space-Time Estimates for the Linearized Navier–Stokes Equations

In this section, we establish the space-time estimates for the linearized 2D Navier–Stokes equation in the vorticity formulation (1.7):

$$\begin{aligned} \left\{ \begin{aligned}&\partial _tw-\nu (\partial _r^2+\frac{1}{r}\partial _r+\frac{1}{r^2}\partial _{\theta }^2)w+(A+\frac{B}{r^2})\partial _{\theta }w+\frac{1}{r}(\partial _r\varphi \partial _{\theta }w-\partial _{\theta }\varphi \partial _rw)=0,\\&(\partial _r^2+\frac{1}{r}\partial _r+\frac{1}{r^2}\partial _{\theta }^2)\varphi =w,\quad w|_{r=1,R}=0\quad \text {with }(r,\theta )\in [1,R]\times \mathbb {S}^1 \text { and } t\ge 0. \end{aligned}\right. \end{aligned}$$

(5.1)

Recall in Sect. 2 we convert this equation to the following system

$$\begin{aligned} \left\{ \begin{aligned}&\partial _tw_k+L_kw_k+\frac{1}{r}[ikf_1-r^{\frac{1}{2}}\partial _{r}(r^{\frac{1}{2}}f_2)]=0,\\&w_k(0)=w_k|_{t=0},\quad w_k|_{r=1,R}=0 \end{aligned}\right. \end{aligned}$$

(5.2)

via introducing

$$\begin{aligned} \begin{aligned} w_k(t, r)&:=r^{\frac{1}{2}}e^{ikAt}\hat{w}_k(t, r)= \frac{1}{2\pi }\int _0^{2\pi } r^{\frac{1}{2}}e^{ikAt}w(t, r, \theta )e^{-ik \theta } d\theta , \\ \varphi _k(t, r)&:= r^{\frac{1}{2}}e^{ikAt}\hat{\varphi }_k(t, r)=\frac{1}{2\pi }\int _0^{2\pi } r^{\frac{1}{2}}e^{ikAt}\varphi (t, r, \theta )e^{-ik \theta } d\theta . \end{aligned} \end{aligned}$$

(5.3)

Here $L_k=-\nu (\partial _r^2-\frac{k^2-\frac{1}{4}}{r^2})+i\frac{kB}{r^2}$ and $f_1, f_2$ stand for nonlinear forms given by

$$\begin{aligned} f_1=\sum _{l\in \mathbb {Z}}\partial _r(r^{-\frac{1}{2}}\varphi _l)w_{k-l},\quad f_2=\sum _{l\in \mathbb {Z}}ilr^{-\frac{3}{2}}\varphi _lw_{k-l}, \end{aligned}$$

(5.4)

where $\varphi _k$ satisfies

$$\begin{aligned}&(\partial _r^2-\frac{k^2-\frac{1}{4}}{r^2})\varphi _k=w_k,\quad \varphi _k|_{r=1,R}=0,\quad r\in [1,R],\quad t\ge 0. \end{aligned}$$

5.1 Space-time estimates for non-zero frequency

We decompose the solution $w_k$ to (5.2) into two parts. Let $w_k=w_k^l+w_k^{n}$, with $w_k^l$ fulfilling the homogeneous linear equation

$$\begin{aligned}&\partial _tw_k^l+L_kw_k^l=0,\quad w_k^l(0)=w_k(0), \end{aligned}$$

and $w_k^{n}$ solving the inhomogeneous linear equation with zero initial data

$$\begin{aligned}&\partial _tw_k^{n}+L_kw_k^{n}+\frac{1}{r}[ikf_1-r^{\frac{1}{2}}\partial _{r}(r^{\frac{1}{2}}f_2)]=0,\quad w_k^{n}(0)=0. \end{aligned}$$

(5.5)

Correspondingly, we also decompose $\varphi _k$ as $\varphi _k=\varphi ^{l}_k+\varphi ^{n}_k$, where $\varphi _k^l$ and $\varphi _k^n$ satisfy

$$\begin{aligned} (\partial _r^2-\frac{k^2-\frac{1}{4}}{r^2})\varphi ^{l}_k=w_k^l,\quad \varphi ^{l}_k|_{r=1,R}=0,\quad \text {with} \ r\in [1,R] \ \text {and} \ t\ge 0, \end{aligned}$$

and

$$\begin{aligned} (\partial _r^2-\frac{k^2-\frac{1}{4}}{r^2})\varphi ^{n}_k=w_k^n,\quad \varphi ^{n}_k|_{r=1,R}=0,\quad \text {with} \ r\in [1,R] \ \text {and} \ t\ge 0. \end{aligned}$$

Now we state the main conclusion of this section, which is also named as the space-time estimate for Eq. (5.2).

Proposition 5.1

Assume $w_k$ is a solution to (5.2) with $w_k(0)\in L^2$, then there exists a constant $c'>0$ independent of $\nu ,B,k,R$ so that it holds

$$\begin{aligned}&\Vert e^{c'(\nu k^2)^{\frac{1}{3}}|B|^{\frac{2}{3}}R^{-2}t}w_k\Vert _{L^{\infty }L^2}+(\nu k^2)^{\frac{1}{6}}|B|^{\frac{1}{3}}R^{-1}\Vert e^{c'(\nu k^2)^{\frac{1}{3}}|B|^{\frac{2}{3}}R^{-2}t}w_k\Vert _{L^2L^2}\\&\qquad +\nu ^{\frac{1}{2}}\Vert e^{c'(\nu k^2)^{\frac{1}{3}}|B|^{\frac{2}{3}}R^{-2}t}\partial _r w_k\Vert _{L^2L^2}+(\nu k^2)^{\frac{1}{2}}\Vert e^{c'(\nu k^2)^{\frac{1}{3}}|B|^{\frac{2}{3}}R^{-2}t}\frac{w_k}{r}\Vert _{L^2L^2}\\&\qquad +|B|^{\frac{1}{2}}R^{-2}\left( |k|\Vert e^{c'(\nu k^2)^{\frac{1}{3}}|B|^{\frac{2}{3}}R^{-2}t}\partial _r\varphi _k\Vert _{L^2L^2}+k^2\Vert e^{c'(\nu k^2)^{\frac{1}{3}}|B|^{\frac{2}{3}}R^{-2}t}\frac{\varphi _k}{r}\Vert _{L^2L^2}\right) \\&\quad \lesssim \Vert w_k(0)\Vert _{L^2}+R^{-2}(\log R)^{-1}\Vert r^2w_k(0)\Vert _{L^2}+R^{3}\Vert \frac{w_k(0)}{r^3}\Vert _{L^2}+(\frac{\nu }{|kB|})^{\frac{1}{3}}R\Vert \partial _rw_k(0)\Vert _{L^2}\\&\qquad +\left( 1+(\frac{\nu }{|kB|})^{\frac{1}{3}}\log R\right) ^{\frac{1}{2}} \cdot \left( R^{-1}\Vert rw_k(0)\Vert _{L^2}+R\Vert \frac{w_k(0)}{r}\Vert _{L^2}(\frac{\nu }{|kB|})^{\frac{2}{3}}k^2\right) \\&\qquad +\left( |kB|^{-\frac{1}{2}} (\log R)^{\frac{1}{2}} +\nu ^{-\frac{1}{6}} |kB|^{-\frac{1}{3}} \right) \cdot \Vert e^{c'(\nu k^2)^{\frac{1}{3}}|B|^{\frac{2}{3}}R^{-2}t}kf_1\Vert _{L^2L^2}\\&\qquad +\nu ^{-\frac{1}{2}} \Vert e^{c'(\nu k^2)^{\frac{1}{3}}|B|^{\frac{2}{3}}R^{-2}t}f_2\Vert _{L^2L^2}. \end{aligned}$$

Proof

Employing Lemma 4.4 and Lemma 4.6, we can bound the linear part $w_k^l$ and $\varphi _k^l$:

$$\begin{aligned}&\Vert e^{c'(\nu k^2)^{\frac{1}{3}}|B|^{\frac{2}{3}}R^{-2}t}w_k^l\Vert _{L^{\infty }L^2}+(\nu k^2)^{\frac{1}{6}}|B|^{\frac{1}{3}}R^{-1}\Vert e^{c'(\nu k^2)^{\frac{1}{3}}|B|^{\frac{2}{3}}R^{-2}t}w_k^l\Vert _{L^2L^2}\\&\qquad +\nu ^{\frac{1}{2}}\Vert e^{c'(\nu k^2)^{\frac{1}{3}}|B|^{\frac{2}{3}}R^{-2}t}\partial _r w_k^l\Vert _{L^2L^2}+(\nu k^2)^{\frac{1}{2}}\Vert e^{c'(\nu k^2)^{\frac{1}{3}}|B|^{\frac{2}{3}}R^{-2}t}\frac{w_k^l}{r}\Vert _{L^2L^2}\\&\qquad +|B|^{\frac{1}{2}}R^{-2}\left( |k|\Vert e^{c'(\nu k^2)^{\frac{1}{3}}|B|^{\frac{2}{3}}R^{-2}t}\partial _r\varphi _k^l\Vert _{L^2L^2}+k^2\Vert e^{c'(\nu k^2)^{\frac{1}{3}}|B|^{\frac{2}{3}}R^{-2}t}\frac{\varphi _k^l}{r}\Vert _{L^2L^2}\right) \\&\quad \lesssim \Vert w_k(0)\Vert _{L^2}+R^{-2}(\log R)^{-1}\Vert r^2w_k(0)\Vert _{L^2}+R^{3}\Vert \frac{w_k(0)}{r^3}\Vert _{L^2}+(\frac{\nu }{|kB|})^{\frac{1}{3}}R\Vert \partial _rw_k(0)\Vert _{L^2}\\&\qquad +\left( 1+(\frac{\nu }{|kB|})^{\frac{1}{3}}\log R\right) ^{\frac{1}{2}} \cdot \left( R^{-1}\Vert rw_k(0)\Vert _{L^2}+R\Vert \frac{w_k(0)}{r}\Vert _{L^2}(\frac{\nu }{|kB|})^{\frac{2}{3}}k^2\right) . \end{aligned}$$

To obtain the desired bounds for the remainder terms $w_k^n$ and $\varphi _k^n$, we utilize Proposition 4.5 with the choices of $h_1 = -\frac{1}{r}ikf_1$, $h_2 = -r^{\frac{1}{2}}f_2$, and $g(r) = r^{-\frac{1}{2}}$ to deduce

$$\begin{aligned}&\Vert e^{c'(\nu k^2)^{\frac{1}{3}}|B|^{\frac{2}{3}}R^{-2}t}w_k^n\Vert _{L^{\infty }L^2}+(\nu k^2)^{\frac{1}{6}}|B|^{\frac{1}{3}}R^{-1}\Vert e^{c'(\nu k^2)^{\frac{1}{3}}|B|^{\frac{2}{3}}R^{-2}t}w_k^n\Vert _{L^2L^2}\\&\qquad +\nu ^{\frac{1}{2}}\Vert e^{c'(\nu k^2)^{\frac{1}{3}}|B|^{\frac{2}{3}}R^{-2}t}\partial _r w_k^n\Vert _{L^2L^2}+(\nu k^2)^{\frac{1}{2}}\Vert e^{c'(\nu k^2)^{\frac{1}{3}}|B|^{\frac{2}{3}}R^{-2}t}\frac{w_k^n}{r}\Vert _{L^2L^2}\\&\qquad +|B|^{\frac{1}{2}}R^{-2}\left( |k|\Vert e^{c'(\nu k^2)^{\frac{1}{3}}|B|^{\frac{2}{3}}R^{-2}t}\partial _r\varphi _k^n\Vert _{L^2L^2}+k^2\Vert e^{c'(\nu k^2)^{\frac{1}{3}}|B|^{\frac{2}{3}}R^{-2}t}\frac{\varphi _k^n}{r}\Vert _{L^2L^2}\right) \\&\quad \lesssim \left( |kB|^{-\frac{1}{2}} (\log R)^{\frac{1}{2}} +\nu ^{-\frac{1}{6}} |kB|^{-\frac{1}{3}} \right) \cdot \Vert e^{c'(\nu k^2)^{\frac{1}{3}}|B|^{\frac{2}{3}}R^{-2}t}kf_1\Vert _{L^2L^2}\\&\qquad +\nu ^{-\frac{1}{2}} \Vert e^{c'(\nu k^2)^{\frac{1}{3}}|B|^{\frac{2}{3}}R^{-2}t}f_2\Vert _{L^2L^2}. \end{aligned}$$

Combining these two estimates, we obtain the desired inequality. $\quad \square$

When $\log R\lesssim \nu ^{-\frac{1}{3}}|B|^{\frac{1}{3}}$, the $(\frac{\nu }{|kB|})^{\frac{1}{3}}\log R$ term on the right for the above estimate can be eliminated, and thus it infers that

Proposition 5.2

Under the same conditions as in Proposition 5.1. If $\log R\lesssim \nu ^{-\frac{1}{3}}|B|^{\frac{1}{3}}$, then there holds

$$\begin{aligned}&\Vert e^{c'(\nu k^2)^{\frac{1}{3}}|B|^{\frac{2}{3}}R^{-2}t}w_k\Vert _{L^{\infty }L^2}+(\nu k^2)^{\frac{1}{6}}|B|^{\frac{1}{3}}R^{-1}\Vert e^{c'(\nu k^2)^{\frac{1}{3}}|B|^{\frac{2}{3}}R^{-2}t}w_k\Vert _{L^2L^2}\\&\qquad +\nu ^{\frac{1}{2}}\Vert e^{c'(\nu k^2)^{\frac{1}{3}}|B|^{\frac{2}{3}}R^{-2}t}w_k'\Vert _{L^2L^2}+(\nu k^2)^{\frac{1}{2}}\Vert e^{c'(\nu k^2)^{\frac{1}{3}}|B|^{\frac{2}{3}}R^{-2}t}\frac{w_k}{r}\Vert _{L^2L^2}\\&\qquad +|B|^{\frac{1}{2}}R^{-2}(|k|\Vert e^{c'(\nu k^2)^{\frac{1}{3}}|B|^{\frac{2}{3}}R^{-2}t}\partial _r\varphi _k\Vert _{L^2L^2}+k^2\Vert e^{c'(\nu k^2)^{\frac{1}{3}}|B|^{\frac{2}{3}}R^{-2}t}\frac{\varphi _k}{r}\Vert _{L^2L^2})\\&\quad \lesssim \Vert w_k(0)\Vert _{L^2}+R^{-2}(\log R)^{-\frac{3}{2}}\Vert r^2w_k(0)\Vert _{L^2}+R^{3}\Vert \frac{w_k(0)}{r^3}\Vert _{L^2}+(\frac{\nu }{|kB|})^{\frac{1}{3}}R\Vert \partial _rw_k(0)\Vert _{L^2}\\&\qquad +R\Vert \frac{w_k(0)}{r}\Vert _{L^2}(\frac{\nu k^2}{|B|})^{\frac{2}{3}}+\nu ^{-\frac{1}{6}} |kB|^{-\frac{1}{3}}\Vert e^{c'(\nu k^2)^{\frac{1}{3}}|B|^{\frac{2}{3}}R^{-2}t}kf_1\Vert _{L^2L^2}\\&\qquad +\nu ^{-\frac{1}{2}} \Vert e^{c'(\nu k^2)^{\frac{1}{3}}|B|^{\frac{2}{3}}R^{-2}t}f_2\Vert _{L^2L^2}. \end{aligned}$$

Notice that when $\nu k^2 R^{-2} \ge (\nu k^2)^{\frac{1}{3}}|B|^{\frac{2}{3}}R^{-2}$, i.e., $\nu k^2\ge |B|$, the heat dissipation effect becomes more significant compared with the enhanced dissipation. Therefore, in the regime where $\nu k^2\ge |B|$, the following space-time estimates offer a more precise control of $w_k$.

Proposition 5.3

Let $w_k$ be the solution to (5.2) with $w_k(0)\in L^2$. Then there exists a constant $c'>0$ independent of $\nu ,B,k,R$, such that

$$\begin{aligned}&\Vert {e^{c'\nu k^2 R^{-2}t}}w_k\Vert _{L^{\infty }L^2}+\nu \Vert {e^{c'\nu k^2 R^{-2}t}}\partial _{r}w_k\Vert _{L^2L^2}+(\nu k^2)^{\frac{1}{2}}\Vert {e^{c'\nu k^2 R^{-2}t}}\frac{w_k}{r}\Vert _{L^2L^2}\\&\qquad +(\nu k^2)^{\frac{1}{2}}R^{-2}(|k|\Vert {e^{c'\nu k^2 R^{-2}t}}\varphi '_k\Vert _{L^2}+k^2\Vert {e^{c'\nu k^2 R^{-2}t}}\frac{\varphi _k}{r}\Vert _{L^2})\\&\quad \lesssim \Vert w_k(0)\Vert _{L^2}+\nu ^{-\frac{1}{2}}(\Vert {e^{c'\nu k^2 R^{-2}t}}f_1\Vert _{L^2L^2} +\Vert {e^{c'\nu k^2 R^{-2}t}}f_2\Vert _{L^2L^2}). \end{aligned}$$

Proof

Conducting the integration by parts, we obtain

$$\begin{aligned}&\Re \langle \partial _tw_k-\nu (\partial _r^2-\frac{k^2-\frac{1}{4}}{r^2})w_k+\frac{ikB}{r^2}w_k+\frac{1}{r}[ikf_1-r^{\frac{1}{2}}\partial _{r}(r^{\frac{1}{2}}f_2)],w_k\rangle \\&\quad =\frac{1}{2}\partial _t\Vert w_k\Vert _{L^2}^2+\nu \Vert \partial _rw_k\Vert _{L^2}^2+\nu (k^2-\frac{1}{4})\Vert \frac{w_k}{r}\Vert _{L^2}^2-\langle f_1,ik\frac{w_k}{r}\rangle +\langle f_2,r^{\frac{1}{2}}\partial _r(r^{-\frac{1}{2}}w_k)\rangle =0. \end{aligned}$$

It infers

$$\begin{aligned} \partial _t\Vert w_k\Vert _{L^2}^2+\nu \Vert \partial _rw_k\Vert _{L^2}^2+\nu k^2\Vert \frac{w_k}{r}\Vert _{L^2}^2\lesssim \Vert f_1\Vert _{L^2}|k|\Vert \frac{w_k}{r}\Vert _{L^2} +\Vert f_2\Vert _{L^2}\Vert r^{\frac{1}{2}}\partial _r(r^{-\frac{1}{2}}w_k)\Vert _{L^2}. \end{aligned}$$

By applying Cauchy–Schwarz inequality, we then deduce

$$\begin{aligned} \partial _t\Vert w_k\Vert _{L^2}^2+\nu \Vert \partial _rw_k\Vert _{L^2}^2+\nu k^2\Vert \frac{w_k}{r}\Vert _{L^2}^2\lesssim \nu ^{-1}(\Vert f_1\Vert _{L^2}^2 +\Vert f_2\Vert _{L^2}^2). \end{aligned}$$

Noticing that $\Vert \frac{w_k}{r}\Vert _{L^2}\ge R^{-1}\Vert w_k\Vert _{L^2}$, it follows

$$\begin{aligned} \partial _t\Vert w_k\Vert _{L^2}^2+\nu \Vert \partial _rw_k\Vert _{L^2}^2+\nu k^2R^{-2}\Vert w_k\Vert _{L^2}^2+\nu k^2\Vert \frac{w_k}{r}\Vert _{L^2}^2\lesssim \nu ^{-1}(\Vert f_1\Vert _{L^2}^2 +\Vert f_2\Vert _{L^2}^2). \end{aligned}$$

Therefore, we can multiply $e^{2c'\nu k^2R^{-2}t}$ on both sides of above inequality. With $c'$ being a small constant independent of $\nu ,B,k,R$, we obtain

$$\begin{aligned}&\partial _t\Vert e^{c'\nu k^2R^{-2}t}w_k\Vert _{L^2}^2+\nu \Vert e^{c'\nu k^2R^{-2}t}\partial _{r}w_k\Vert _{L^2}^2+\nu k^2\Vert e^{c'\nu k^2R^{-2}t}\frac{w_k}{r}\Vert _{L^2}^2\\&\quad \lesssim \nu ^{-1}(\Vert e^{c'\nu k^2R^{-2}t}f_1\Vert _{L^2}^2 +\Vert e^{c'\nu k^2R^{-2}t}f_2\Vert _{L^2}^2). \end{aligned}$$

This further implies

$$\begin{aligned}&\Vert e^{c'\nu k^2R^{-2}t}w_k\Vert _{L^{\infty }L^2}^2+\nu \Vert e^{c'\nu k^2R^{-2}t}\partial _{r}w_k\Vert _{L^2L^2}^2+\nu k^2\Vert e^{c'\nu k^2R^{-2}t}\frac{w_k}{r}\Vert _{L^2L^2}^2\\&\quad \lesssim \Vert w_k(0)\Vert _{L^2}^2+\nu ^{-1}(\Vert e^{c'\nu k^2R^{-2}t}f_1\Vert _{L^2L^2}^2 +\Vert e^{c'\nu k^2R^{-2}t}f_2\Vert _{L^2L^2}^2). \end{aligned}$$

In view of Lemma A.3, it also holds

$$\begin{aligned} R^{-2}(|k|\Vert \varphi '_k\Vert _{L^2}+k^2\Vert \frac{\varphi _k}{r}\Vert _{L^2})\lesssim R^{-2}\Vert rw_k\Vert _{L^2}\le \Vert \frac{w_k}{r}\Vert _{L^2}. \end{aligned}$$

Combining these two estimates above yields

$$\begin{aligned}&\Vert e^{c'\nu k^2R^{-2}t}w_k\Vert _{L^{\infty }L^2}+\nu \Vert e^{c'\nu k^2R^{-2}t}\partial _{r}w_k\Vert _{L^2L^2}+(\nu k^2)^{\frac{1}{2}}\Vert e^{c'\nu k^2R^{-2}t}\frac{w_k}{r}\Vert _{L^2L^2}\\&\qquad +(\nu k^2)^{\frac{1}{2}}R^{-2}(|k|\Vert e^{c'\nu k^2R^{-2}t}\varphi '_k\Vert _{L^2}+k^2\Vert e^{c'\nu k^2R^{-2}t}\frac{\varphi _k}{r}\Vert _{L^2})\\&\quad \lesssim \Vert w_k(0)\Vert _{L^2}+\nu ^{-\frac{1}{2}}(\Vert e^{c'\nu k^2R^{-2}t}f_1\Vert _{L^2L^2} +\Vert e^{c'\nu k^2R^{-2}t}f_2\Vert _{L^2L^2}), \end{aligned}$$

which completes the proof. $\quad \square$

5.2 Space-time estimates for zero frequency

To establish the space-time estimate for zero mode of solutions, we directly utilize the heat dissipative structure of equation and then perform integration by parts.

Recall that the nonlinear perturbation equation reads

$$\begin{aligned}&\partial _tw-\nu (\partial _r^2+\frac{1}{r}\partial _r+\frac{1}{r^2}\partial _{\theta }^2)w+(A+\frac{B}{r^2})\partial _{\theta }w+\frac{1}{r}(\partial _r\varphi \partial _{\theta }w-\partial _{\theta }\varphi \partial _rw)=0, \end{aligned}$$

(5.6)

where $w|_{r=1,R}=0$ and $(\partial _r^2+\frac{1}{r}\partial _r+\frac{1}{r^2}\partial _{\theta }^2)\varphi =w$. Denote the zero mode of function $f(r, \theta )$ to be

$$\begin{aligned} f_{=}{:}{=}\frac{1}{2\pi }\int _{0}^{2\pi } f(r, \theta ) d\theta \end{aligned}$$

and set $f_{\ne }{:}{=}f-f_{=}$. We then have that the zero frequency part of Eq. (5.6) takes the form

$$\begin{aligned} \partial _tw_{=}-\nu (\partial _r^2+\frac{1}{r}\partial _r)w_{=}+\frac{1}{r}(\partial _r\varphi \partial _{\theta }w-\partial _{\theta }\varphi \partial _rw)_{=}=0. \end{aligned}$$

(5.7)

Subsequently, we derive the space-time estimate for $w_{=}$.

Lemma 5.4

Let $w_{=}$ be the solution to (5.7) and assume $r^{\frac{1}{2}}w_{=}(0)\in L^2$. Denoting $w_0{:}{=}r^{\frac{1}{2}} w_{=}$, then the following inequality holds

$$\begin{aligned} \Vert w_0\Vert _{L^{\infty }L^2}^2+\nu \Vert r^{\frac{1}{2}}\partial _r(r^{-\frac{1}{2}}w_0)\Vert _{L^2L^2}^2\le \nu ^{-1}\Vert \sum _{l\in \mathbb {Z}\backslash \{0\}}\frac{l\varphi _lw_{-l}}{r^{\frac{3}{2}}}\Vert _{L^2L^2}^2+\Vert w_0(0)\Vert _{L^2}^2. \end{aligned}$$

(5.8)

Here $\varphi _l$ and $w_{l}$ are defined in (5.3).

Proof

Employing integration by parts for Eq. (5.7), we deduce

$$\begin{aligned} 0&=\Re \langle \partial _tw_{=}-\nu (\partial _r^2+\frac{1}{r}\partial _r)w_{=}+\frac{1}{r}(\partial _r\varphi \partial _{\theta }w-\partial _{\theta }\varphi \partial _rw)_{=},rw_{=}\rangle \\&=\frac{1}{2}\partial _t\Vert r^{\frac{1}{2}}w_{=}(t)\Vert _{L^2}^2+\nu \Vert r^{\frac{1}{2}}\partial _rw_{=}\Vert _{L^2}^2+\Re \langle [\partial _r(\varphi \partial _{\theta }w)-\partial _{\theta }(\varphi \partial _rw)]_{=},w_{=}\rangle . \end{aligned}$$

Observe that

$$\begin{aligned} {[}\partial _r(\varphi \partial _{\theta }w)]_{=}=\partial _r(\varphi \partial _{\theta }w)_{=} \ \text {and} \ [\partial _{\theta }(\varphi \partial _rw)]_{=}=\partial _{\theta }(\varphi \partial _rw)_{=}=0. \end{aligned}$$

Conducting integration by parts again, we then arrive at

$$\begin{aligned}&\frac{1}{2}\partial _t\Vert r^{\frac{1}{2}}w_{=}(t)\Vert _{L^2}^2+\nu \Vert r^{\frac{1}{2}}\partial _rw_{=}\Vert _{L^2}^2+\Re \langle \partial _r(\varphi \partial _{\theta }w)_{=},w_{=}\rangle \\&\quad =\frac{1}{2}\partial _t\Vert r^{\frac{1}{2}}w_{=}(t)\Vert _{L^2}^2+\nu \Vert r^{\frac{1}{2}}\partial _rw_{=}\Vert _{L^2}^2-\Re \langle (\varphi \partial _{\theta }w)_{=},\partial _rw_{=}\rangle =0. \end{aligned}$$

This leads to the following inequality

$$\begin{aligned} \partial _t\Vert r^{\frac{1}{2}}w_{=}(t)\Vert _{L^2}^2+2\nu \Vert r^{\frac{1}{2}}\partial _rw_{=}\Vert _{L^2}^2\le 2\Vert r^{-\frac{1}{2}}(\varphi \partial _{\theta }w)_{=}\Vert _{L^2}\Vert r^{\frac{1}{2}}\partial _rw_{=}\Vert _{L^2}, \end{aligned}$$

which renders

$$\begin{aligned} \partial _t\Vert r^{\frac{1}{2}}w_{=}(t)\Vert _{L^2}^2+\nu \Vert r^{\frac{1}{2}}\partial _rw_{=}\Vert _{L^2}^2\le \nu ^{-1}\Vert \frac{(\varphi \partial _{\theta }w)_{=}}{r^{\frac{1}{2}}}\Vert _{L^2}^2. \end{aligned}$$

We proceed to integrate the above inequality in t variable and get

$$\begin{aligned} \Vert r^{\frac{1}{2}}w_{=}(t)\Vert _{L^2}^2+\nu \int _0^t\Vert r^{\frac{1}{2}}\partial _rw_{=}(s)\Vert _{L^2}^2ds\le \nu ^{-1}\int _0^t\Vert \frac{(\varphi \partial _{\theta }w)_{=}(s)}{r^{\frac{1}{2}}}\Vert _{L^2}^2ds+\Vert r^{\frac{1}{2}}w_{=}(0)\Vert _{L^2}^2. \end{aligned}$$

In view of the definition for $w_k, \varphi _k$ from (5.3):

$$\begin{aligned} \begin{aligned} w_k(t, r)&{:}{=}r^{\frac{1}{2}}e^{ikAt}\hat{w}_k(t, r)= \frac{1}{2\pi }\int _0^{2\pi } r^{\frac{1}{2}}e^{ikAt}w(t, r, \theta )e^{-ik \theta } d\theta , \\ \varphi _k(t, r)&{:}{=} r^{\frac{1}{2}}e^{ikAt}\hat{\varphi }_k(t, r)=\frac{1}{2\pi }\int _0^{2\pi } r^{\frac{1}{2}}e^{ikAt}\varphi (t, r, \theta )e^{-ik \theta } d\theta , \end{aligned} \end{aligned}$$

we can further express

$$\begin{aligned} \int _0^t\Vert \frac{(\varphi \partial _{\theta }w)_{=}(s)}{r^{\frac{1}{2}}}\Vert _{L^2}^2ds=\Vert \sum _{l\in \mathbb {Z}\backslash \{0\}}r^{-\frac{1}{2}}\hat{\varphi }_l(il)\hat{w}_{-l}\Vert _{L^2L^2}^2=\Vert \sum _{l\in \mathbb {Z}\backslash \{0\}}\frac{l\varphi _lw_{-l}}{r^{\frac{3}{2}}}\Vert _{L^2L^2}^2. \end{aligned}$$

This completes the proof of (5.8). $\quad \square$

6 Nonlinear Stability

This section is devoted to the proof of our main Theorem 1.3 using a bootstrap argument. Thanks to the transformation conducted in Sect. 2, we can focus on analyzing the system (5.2), which is presented in the following form

$$\begin{aligned} \left\{ \begin{aligned}&\partial _tw_k+L_kw_k+\frac{1}{r}[ikf_1-r^{\frac{1}{2}}\partial _{r}(r^{\frac{1}{2}}f_2)]=0,\\&w_k(0)=w_k|_{t=0},\quad w_k|_{r=1,R}=0. \end{aligned}\right. \end{aligned}$$

Here $L_k=-\nu (\partial _r^2-\frac{k^2-\frac{1}{4}}{r^2})+i\frac{kB}{r^2}$ and $f_1, f_2$ have the nonlinear structure of

$$\begin{aligned} f_1=\sum _{l\in \mathbb {Z}}\partial _r(r^{-\frac{1}{2}}\varphi _l)w_{k-l},\quad f_2=\sum _{l\in \mathbb {Z}}ilr^{-\frac{3}{2}}\varphi _lw_{k-l}. \end{aligned}$$

Note that $\varphi _k$ satisfies

$$\begin{aligned}&(\partial _r^2-\frac{k^2-\frac{1}{4}}{r^2})\varphi _k=w_k,\quad \varphi _k|_{r=1,R}=0,\quad \text {with} \ r\in [1,R] \ \text {and} \ t\ge 0. \end{aligned}$$

For notational simplicity, we set $\mu _k{:}{=}\max \left\{ (\nu k^2)^{\frac{1}{3}} |B|^{\frac{2}{3}} R^{-2}, \nu k^2 R^{-2} \right\}$. We then define the initial energy $\mathcal {M}_0(0)=\Vert w_0(0)\Vert _{L^2}$ and for $k\in \mathbb {Z}\backslash \{ 0 \}$ we let

$$\begin{aligned} \mathcal {M}_k(0)=&\left\{ \begin{aligned}&R^{-2}(\log R)^{-1}\Vert r^2w_k(0)\Vert _{L^2}+R^{3}\Vert \frac{w_k(0)}{r^3}\Vert _{L^2}+R\Vert \partial _rw_k(0)\Vert _{L^2},\quad \text {when} \ \nu k^2\le |B| ,\\&\Vert w_k(0)\Vert _{L^2}\quad \text {when} \ \nu k^2\ge |B|. \end{aligned}\right. \end{aligned}$$

As a result of Proposition 5.2 and Proposition 5.3, we readily deduce

Proposition 6.1

For $k\in \mathbb {Z}\backslash \{0 \}$, let $w_k$ be the solution to (5.2) with $w_k(0)\in L^2$. Given $\log R\lesssim \nu ^{-\frac{1}{3}}|B|^{\frac{1}{3}}$, then there exists a constant $c'>0$ independent of $\nu ,B,k,R$, such that it holds

$$\begin{aligned}&\Vert e^{c'\mu _kt}w_k\Vert _{L^{\infty }L^2}+\mu _k^{\frac{1}{2}}\Vert e^{c'\mu _kt}w_k\Vert _{L^2L^2}+\nu ^{\frac{1}{2}}\Vert e^{c'\mu _kt}w_k'\Vert _{L^2L^2}+(\nu k^2)^{\frac{1}{2}}\Vert e^{c'\mu _kt}\frac{w_k}{r}\Vert _{L^2L^2}\\&\qquad +\max \{|B|^{\frac{1}{2}}R^{-2}, (\nu k^2)^{\frac{1}{2}} R^{-2} \}\cdot \left( |k|\Vert e^{c'\mu _kt}\partial _r\varphi _k\Vert _{L^2L^2}+k^2\Vert e^{c'\mu _kt}\frac{\varphi _k}{r}\Vert _{L^2L^2}\right) \\&\quad \lesssim \mathcal {M}_k(0)+\mu _k^{-\frac{1}{2}}R^{-1}\Vert e^{c'\mu _kt}kf_1\Vert _{L^2L^2}+\nu ^{-\frac{1}{2}} \Vert e^{c'\mu _kt}f_2\Vert _{L^2L^2}. \end{aligned}$$

We proceed to construct and control below the energy functionals, with

$$\begin{aligned} E_0=&\Vert w_0\Vert _{L^{\infty }L^2},\\ E_k=&\Vert e^{c'\mu _kt}w_k\Vert _{L^{\infty }L^2} +\mu _k^{\frac{1}{2}}\Vert e^{c'\mu _kt}w_k\Vert _{L^2L^2}+|B|^{\frac{1}{2}}|k|^{\frac{3}{2}}R^{-2}\Vert e^{c'\mu _kt}\frac{\varphi _k}{r^{\frac{1}{2}}}\Vert _{L^2L^{\infty }} \quad \text {for} \ |k|\ge 1. \end{aligned}$$

The derivation of a priori estimates for $E_k$ crucially relies on the space-time estimates obtained in Sect. 5. We start from presenting the bound for the energy at zero frequency:

Lemma 6.2

Assume that $\log R\lesssim \nu ^{-\frac{1}{3}}|B|^{\frac{1}{3}}$. We have the following energy inequality

$$\begin{aligned} E_0 \lesssim&\mathcal {M}_0(0)+\nu ^{-\frac{1}{2}}|B|^{-\frac{1}{2}}R^2\sum _{l\in \mathbb {Z}\backslash \{0\}}E_lE_{-l}. \end{aligned}$$

Proof

Directly applying Lemma 5.4, we obtain

$$\begin{aligned} E_0&\lesssim \mathcal {M}_0(0)+\nu ^{-\frac{1}{2}}\Vert \sum _{l\in \mathbb {Z}\backslash \{0\}}\frac{l\varphi _lw_{-l}}{r^{\frac{3}{2}}}\Vert _{L^2L^2}\\&\lesssim \mathcal {M}_0(0)+\nu ^{-\frac{1}{2}}\Vert \sum _{l\in \mathbb {Z}\backslash \{0\}}|l|\Vert \frac{\varphi _l}{r^{\frac{1}{2}}}\Vert _{L^2L^{\infty }}\Vert \frac{w_{-l}}{r}\Vert _{L^{\infty }L^2}\\&\lesssim \mathcal {M}_0(0)+\nu ^{-\frac{1}{2}}|B|^{-\frac{1}{2}}R^2\sum _{l\in \mathbb {Z}\backslash \{0\}}E_lE_{-l}, \end{aligned}$$

which proves Lemma 6.2. $\quad \square$

Next, we turn to control $E_k$ for $k\ne 0$. We first show that the nonlinear terms $f_1$ and $f_2$ obey the below $L^2$ estimates.

Lemma 6.3

For $k\in \mathbb {Z}\backslash \{0 \}$, we have

$$\begin{aligned} \Vert f_1\Vert _{L^2} \lesssim&\sum _{l\in \mathbb {Z}\backslash \{0, k\}}\Vert \partial _r(r^{-\frac{1}{2}}\varphi _l)\Vert _{L^{\infty }}\Vert w_{k-l}\Vert _{L^2}+R\Vert w_0\Vert _{L^2}\Vert w_{k}\Vert _{L^2},\\ \Vert f_2\Vert _{L^2}\le&\sum _{l\in \mathbb {Z}\backslash \{0\}}|l|\Vert \frac{\varphi _l}{r^{\frac{1}{2}}}\Vert _{L^{\infty }}\Vert \frac{w_{k-l}}{r}\Vert _{L^2}. \end{aligned}$$

Proof

With the expressions of nonlinear terms

$$\begin{aligned} f_1=\sum _{l\in \mathbb {Z}}\partial _r(r^{-\frac{1}{2}}\varphi _l)w_{k-l},\quad f_2=\sum _{l\in \mathbb {Z}}ilr^{-\frac{3}{2}}\varphi _lw_{k-l}, \end{aligned}$$

we get

$$\begin{aligned} \Vert f_1\Vert _{L^2}&=\Vert \sum _{l\in \mathbb {Z}}\partial _r(r^{-\frac{1}{2}}\varphi _l)w_{k-l}\Vert _{L^2}\\&\le \sum _{l\in \mathbb {Z}\backslash \{0, k\}}\Vert \partial _r(r^{-\frac{1}{2}}\varphi _l)\Vert _{L^{\infty }}\Vert w_{k-l}\Vert _{L^2}+\Vert \partial _r(r^{-\frac{1}{2}}\varphi _0)\Vert _{L^{\infty }}\Vert w_{k}\Vert _{L^2}+\Vert \partial _r(r^{-\frac{1}{2}}\varphi _k)\Vert _{L^{\infty }}\Vert w_{0}\Vert _{L^2} \\&\lesssim (\frac{R}{R-1})^{\frac{1}{2}}\cdot (1+\log R)\cdot \Big (\sum _{l\in \mathbb {Z}\backslash \{0, k\}}\Vert rw_l\Vert _{L^{2}}\Vert w_{k-l}\Vert _{L^2}+\Vert rw_0\Vert _{L^2}\Vert w_{k}\Vert _{L^2}+\Vert rw_k\Vert _{L^2}\Vert w_{k}\Vert _{L^2} \Big )\\&\lesssim R (\frac{R}{R-1})^{\frac{1}{2}}\cdot (1+\log R)\cdot \Big (\sum _{l\in \mathbb {Z}\backslash \{0, k\}}\Vert w_l\Vert _{L^{2}}\Vert w_{k-l}\Vert _{L^2}+\Vert w_0\Vert _{L^2}\Vert w_{k}\Vert _{L^2} \Big ), \end{aligned}$$

where in the third line we employ Lemma A.3 and Lemma A.5.

And the $L^2$ norm of $f_2$ can be directly controlled by

$$\begin{aligned} \Vert f_2\Vert _{L^2}=\Vert \sum _{l\in \mathbb {Z}\backslash \{0\}}\frac{l\varphi _lw_{k-l}}{r^{\frac{3}{2}}}\Vert _{L^2}\le \sum _{l\in \mathbb {Z}\backslash \{0\}}|l|\Vert \frac{\varphi _l}{r^{\frac{1}{2}}}\Vert _{L^{\infty }}\Vert \frac{w_{k-l}}{r}\Vert _{L^2}. \end{aligned}$$

$\square$

The above proposition leads to a control of $E_k$ with $k\in \mathbb {Z}\backslash \{0 \}$.

Lemma 6.4

Assume that $\log R\lesssim \nu ^{-\frac{1}{3}}|B|^{\frac{1}{3}}$. For $k\in \mathbb {Z}\backslash \{0 \}$, we have

$$\begin{aligned} E_k\lesssim \mathcal {M}_k(0) +\nu ^{-\frac{1}{2}}|B|^{-\frac{1}{2}}R^2 (\frac{R}{R-1})^{\frac{1}{2}}(1+\log R) \sum _{l\in \mathbb {Z}}E_lE_{k-l}. \end{aligned}$$

Proof

Observe the following basic inequality

$$\begin{aligned} |k|^{\alpha }\le |l|^{\alpha }+|k-l|^{\alpha } \quad \text {for any}\ k, l\in \mathbb {Z} \ \text {and any} \ \alpha \in (0, 1]. \end{aligned}$$

Thanks to the definition

$$\begin{aligned} \mu _k{:}{=}\max \left\{ (\nu k^2)^{\frac{1}{3}} |B|^{\frac{2}{3}} R^{-2}, \nu k^2 R^{-2} \right\} , \end{aligned}$$

this yields

$$\begin{aligned} \mu _k \le \mu _l+\mu _{k-l} \quad \text {for any}\ k, l\in \mathbb {Z} \end{aligned}$$

(6.1)

by considering the scenarios $\nu k^2<|B|$ and $\nu k^2\ge |B|$ separately. In view of Proposition 6.1, Lemma A.1 and Lemma A.3, we now have

$$\begin{aligned} E_k\lesssim \mathcal {M}_k(0)+\mu _k^{-\frac{1}{2}}R^{-1}\Vert e^{c'\mu _kt}kf_1\Vert _{L^2L^2}+\nu ^{-\frac{1}{2}} \Vert e^{c'\mu _kt}f_2\Vert _{L^2L^2}. \end{aligned}$$

(6.2)

Utilizing Lemma 6.3, it holds

$$\begin{aligned} \begin{aligned}&\mu _k^{-\frac{1}{2}}R^{-1}\Vert e^{c'(\nu k^2)^{\frac{1}{3}}|B|^{\frac{2}{3}}R^{-2}t}kf_1\Vert _{L^2L^2} \\&\quad \lesssim \mu _k^{-\frac{1}{2}}|k|(\frac{R}{R-1})^{\frac{1}{2}}(1+\log R)\cdot \\&\qquad \Big (\Vert w_0\Vert _{L^{\infty }L^2}\Vert e^{c'\mu _k t}w_{k}\Vert _{L^2L^2} +\sum _{l\in \mathbb {Z}\backslash \{0, k\}}\Vert e^{c\mu _l t}w_l\Vert _{L^{\infty }L^{2}} \Vert e^{c'\mu _{k-l}t}w_{k-l}\Vert _{L^2L^2}\Big )\\&\quad \le \mu _k^{-\frac{1}{2}}|k|(\frac{R}{R-1})^{\frac{1}{2}}(1+\log R)\cdot \Big (E_0\mu _k^{-\frac{1}{2}} E_k+\sum _{l\in \mathbb {Z}\backslash \{0, k\}}|l|^{-\frac{1}{2}}E_l\mu _{k-l}^{-\frac{1}{2}} E_{k-l}\Big )\\&\quad \le \mu _k^{-\frac{1}{2}}|k|R(\frac{R}{R-1})^{\frac{1}{2}}(1+\log R)\cdot \\&\qquad \Big ( (\nu k^2)^{-\frac{1}{6}}|B|^{-\frac{1}{3}}RE_0E_k+\sum _{l\in \mathbb {Z} \backslash \{0, k\}} |l|^{-\frac{1}{2}}(\nu (k-l)^2)^{-\frac{1}{6}}|B|^{-\frac{1}{3}}RE_lE_{k-l}\Big )\\&\quad \le \mu _k^{-\frac{1}{2}}|k|(\nu k^2)^{-\frac{1}{6}}|B|^{-\frac{1}{3}}R(\frac{R}{R-1})^{\frac{1}{2}}(1+\log R)\Big (E_0E_k+\sum _{l\in \mathbb {Z}\backslash \{0, k\}} E_lE_{k-l}\Big ), \end{aligned} \end{aligned}$$

(6.3)

where in the last line we use the fact

$$\begin{aligned} |k|\le 2|l||k-l| \quad \text {for any}\ k\in \mathbb {Z}, l\in \mathbb {Z}\backslash \{0, k\}. \end{aligned}$$

Note that

$$\begin{aligned} \mu _k^{-\frac{1}{2}}|k|(\nu k^2)^{-\frac{1}{6}}|B|^{-\frac{1}{3}}R\le \nu ^{-\frac{1}{2}} |B|^{-\frac{1}{2}}R^2. \end{aligned}$$

Hence we obtain

$$\begin{aligned} \mu _k^{-\frac{1}{2}}R^{-1}\Vert e^{c'(\nu k^2)^{\frac{1}{3}}|B|^{\frac{2}{3}}R^{-2}t}kf_1\Vert _{L^2L^2}\lesssim \nu ^{-\frac{1}{2}} |B|^{-\frac{1}{2}}R^2 (\frac{R}{R-1})^{\frac{1}{2}}(1+\log R) \sum _{l\in \mathbb {Z}}E_lE_{k-l}. \end{aligned}$$

To estimate the term involving $f_2$ on the right of (6.2), we apply Lemma 6.3 again to deduce

$$\begin{aligned} \begin{aligned} \nu ^{-\frac{1}{2}} \Vert e^{c'\mu _k t}f_2\Vert _{L^2L^2} \le&\nu ^{-\frac{1}{2}} \sum _{l\in \mathbb {Z}\backslash \{0\}}|l|\Vert e^{c'\mu _l t}\frac{\varphi _l}{r^{\frac{1}{2}}}\Vert _{L^2L^{\infty }}\Vert e^{c' \mu _{k-l} t}\frac{w_{k-l}}{r}\Vert _{L^{\infty }L^2} \\ \le&\nu ^{-\frac{1}{2}} |B|^{-\frac{1}{2}}R^2\sum _{l\in \mathbb {Z}\backslash \{0\}}E_lE_{k-l}. \end{aligned} \end{aligned}$$

(6.4)

Together with (6.2), the above two bounds give the desired control for $E_k$ when $|k|\ge 1$.

$\square$

Gathering all above estimates, we now conclude our main theorem:

Theorem 6.1

Assume that $\log R\lesssim \nu ^{-\frac{1}{3}}|B|^{\frac{1}{3}}$. There exist constants $\nu _0$ and $c_0, C,c'>0$ independent of $\nu ,B,R$, such that if

$$\begin{aligned} \mathcal {M}(0)=\sum _{k\in \mathbb {Z}}\mathcal {M}_k(0)\le c_0\nu ^\frac{1}{2} B^{\frac{1}{2}}R^{-2}(\frac{R}{R-1})^{-\frac{1}{2}}(1+\log R)^{-1},\quad 0<\nu \le \nu _0, \end{aligned}$$

then the solution w to the system (2.2) exists globally in time and satisfies the decaying and stability estimates

$$\begin{aligned} \sum _{k\in \mathbb {Z}\backslash \{0\}}\Vert w_k(t)\Vert _{L^2}&\le Ce^{-c'\nu ^{\frac{1}{3}}B^{\frac{2}{3}}R^{-2}t}\mathcal {M}(0) \quad \text {and} \quad \Vert w_{0}(t)\Vert _{L^2}\le C\mathcal {M}(0). \end{aligned}$$

Proof

Employing Lemma 6.2 and Lemma 6.4, we have

$$\begin{aligned} \sum _{k\in \mathbb {Z}}E_k\le&\sum _{k\in \mathbb {Z}} C\mathcal {M}_k(0) +C\nu ^{-\frac{1}{2}}|B|^{-\frac{1}{2}}R^2(\frac{R}{R-1})^{\frac{1}{2}}(1+\log R)\sum _{k\in \mathbb {Z}}\sum _{l\in \mathbb {Z}}E_lE_{k-l}. \end{aligned}$$

We perform a bootstrap argument to prove our main theorem. Assume that for a small constant $c_0>0$ there holds

$$\begin{aligned} \sum _{k\in \mathbb {Z}}E_k\le 2C\mathcal {M}(0)\le 2C c_0 \nu ^{\frac{1}{2}} |B|^{\frac{1}{2}}R^{-2}(\frac{R}{R-1})^{-\frac{1}{2}}(1+\log R)^{-1}, \end{aligned}$$

(6.5)

we then obtain

$$\begin{aligned} \sum _{k\in \mathbb {Z}}E_k&\le C\sum _{k\in \mathbb {Z}} \mathcal {M}_k(0) +C\nu ^{-\frac{1}{2}}|B|^{-\frac{1}{2}}R^2(\frac{R}{R-1})^{\frac{1}{2}}(1+\log R)(\sum _{k\in \mathbb {Z}}E_k)^2\\&\le C\sum _{k\in \mathbb {Z}} \mathcal {M}_k(0) +2Cc_0\sum _{k\in \mathbb {Z}}E_k. \end{aligned}$$

Choosing $c_0\le \frac{1}{3} C^{-1}$, we arrive at

$$\begin{aligned}&\sum _{k\in \mathbb {Z}}E_k\le 1.5 C\sum _{k\in \mathbb {Z}} \mathcal {M}_k(0), \end{aligned}$$

which improves the bootstrap assumption (6.5). By a standard continuity argument, we therefore prove that the system (2.2) admits a global-in-time solution, and its k-th mode $w_k$ obeys the desired bounds

$$\begin{aligned} \sum _{k\in \mathbb {Z}\backslash \{0\}}\Vert w_k(t)\Vert _{L^2}&\le Ce^{-c'\nu ^{\frac{1}{3}}B^{\frac{2}{3}}R^{-2}t}\mathcal {M}(0),\\ \Vert w_{0}(t)\Vert _{L^2}&\le C\mathcal {M}(0). \end{aligned}$$

This completes the proof of our main theorem. $\quad \square$

Appendix A. Basic Estimates

In the appendix, we provide the proof of several elementary but fundamental calculus lemmas which are continually utilized throughout this paper. The following two Sobolev-type inequalities are crucial in the derivation of resolvent estimates in Sect. 3.

Lemma A.1

For any $w\in L^2(1,R)$ with $R>1$, it holds

$$\begin{aligned} \Vert w\Vert _{L^{\infty }}^2\le 2\Vert \frac{w}{r^{\frac{1}{2}}}\Vert _{L^2}\Vert r^{\frac{1}{2}}w'\Vert _{L^2}+\frac{R}{R-1}\Vert \frac{w}{r^{\frac{1}{2}}}\Vert _{L^2}^2. \end{aligned}$$

Moreover, if $w|_{r=1,R}=0$, then

$$\begin{aligned}&\Vert w\Vert _{L^{\infty }}^2\le 2\Vert \frac{w}{r^{\frac{1}{2}}}\Vert _{L^2}\Vert r^{\frac{1}{2}}w'\Vert _{L^2}. \end{aligned}$$

Proof

Choose $r_0\in [1, R]$ such that

$$\begin{aligned} (R-1) |\frac{w(r_0)}{r_0^{\frac{1}{2}}}|^2\le \Vert \frac{w}{r^{\frac{1}{2}}}\Vert _{L^2}^2. \end{aligned}$$

It then follows

$$\begin{aligned} |w(r_0)|^2\le \frac{R}{R-1}\Vert \frac{w}{r^{\frac{1}{2}}}\Vert _{L^2}^2. \end{aligned}$$

And we have

$$\begin{aligned} |w(r)|^2-|w(r_0)|^2=\int _{r_0}^r\partial _s(|w(s)|^2)ds=\int _{r_0}^r (w'(s)\overline{w}(s)+w(s)\overline{w}'(s))ds\le 2\Vert \frac{w}{r^{\frac{1}{2}}}\Vert _{L^2}\Vert r^{\frac{1}{2}}w'\Vert _{L^2}, \end{aligned}$$

which gives

$$\begin{aligned} \Vert w\Vert _{L^{\infty }}^2\le 2\Vert \frac{w}{r^{\frac{1}{2}}}\Vert _{L^2}\Vert r^{\frac{1}{2}}w'\Vert _{L^2}+\frac{R}{R-1}\Vert \frac{w}{r^{\frac{1}{2}}}\Vert _{L^2}^2. \end{aligned}$$

When $w|_{r=1,R}=0$, it can be inferred that

$$\begin{aligned} |w(r)|^2=&\int _1^r\partial _s(|w(s)|^2)ds=\int _1^r (w'(s)\overline{w}(s)+w(s)\overline{w}'(s))ds\le 2\Vert \frac{w}{r^{\frac{1}{2}}}\Vert _{L^2}\Vert r^{\frac{1}{2}}w'\Vert _{L^2}. \end{aligned}$$

$\square$

For the weighted quantity $r^{\frac{1}{2}} w$, we have

Lemma A.2

Let $w\in L^2(1,R), w|_{r=1,R}=0$, where $R>1$. Then the following inequality holds

$$\begin{aligned}&\Vert r^{\frac{1}{2}}w\Vert _{L^{\infty }}^2\le 4\Vert r^{\frac{1}{2}}w\Vert _{L^2}\Vert r^{\frac{1}{2}}w'\Vert _{L^2}. \end{aligned}$$

Proof

Due to the vanishment of w on the boundary, it can be deduced

$$\begin{aligned} r|w(r)|^2=&\int _1^r\partial _s(s|w(s)|^2)ds=\int _1^r|w(s)|^2ds+\int _1^rs\partial _s(|w(s)|^2)ds\nonumber \\ =&\int _1^r|w(s)|^2ds+\int _1^rs(w'(s)\overline{w}(s)+w(s)\overline{w}'(s))ds. \end{aligned}$$

(A.1)

Meanwhile, through integration by part we can write

$$\begin{aligned} \int _1^R|w(s)|^2ds=&-\int _1^Rs(w'(s)\overline{w}(s)+w(s)\overline{w}'(s))ds\le 2\Vert r^{\frac{1}{2}}w\Vert _{L^2}\Vert r^{\frac{1}{2}}w'\Vert _{L^2}. \end{aligned}$$

(A.2)

It then follows from (A.1) that

$$\begin{aligned} \sup _{r\in [1,R]}r|w(r)|^2 \le&\sup _{r\in [1,R]}\int _1^r|w(s)|^2ds+\sup _{r\in [1,R]}|\int _1^rs(w'(s)\overline{w}(s)+w(s)\overline{w}'(s))ds|\\ \le&2\Vert r^{\frac{1}{2}}w\Vert _{L^2}\Vert r^{\frac{1}{2}}w'\Vert _{L^2}+2\Vert r^{\frac{1}{2}}w\Vert _{L^2}\Vert r^{\frac{1}{2}}w'\Vert _{L^2}=4\Vert r^{\frac{1}{2}}w\Vert _{L^2}\Vert r^{\frac{1}{2}}w'\Vert _{L^2}. \end{aligned}$$

$\square$

To control the stream function $\varphi _k$ with the vorticity $w_k$, we develop a series of (weighted) elliptic estimates as below.

Lemma A.3

Suppose that $|k|\ge 1$. Let $w=(\partial _r^2-\frac{k^2-\frac{1}{4}}{r^2} )\varphi$ with $\varphi |_{r=1, R}=0$. Then the following inequality holds

$$\begin{aligned}&\Vert \varphi '\Vert _{L^2}^2+|k|^2\Vert \frac{\varphi }{r}\Vert _{L^2}^2\lesssim |\langle w, \varphi \rangle | \lesssim |k|^{-2} \Vert rw\Vert _{L^2}^2, \\&\Vert \varphi '\Vert _{L^2}^2+|k|^2\Vert \frac{\varphi }{r}\Vert _{L^2}^2\lesssim |k|^{-1}\Vert r^{\frac{1}{2}}w\Vert _{L^1}^2, \\&\Vert r^{\frac{1}{2}} \varphi '\Vert _{L^\infty }+|k|\Vert r^{-\frac{1}{2}}\varphi \Vert _{L^\infty }\lesssim (\frac{R}{R-1})^{\frac{1}{2}}|k|^{-\frac{1}{2}}\Vert rw\Vert _{L^2}. \end{aligned}$$

Proof

Noting that

$$\begin{aligned} -\langle w, \varphi \rangle =\Vert \varphi '\Vert _{L^2}^2+(k^2-\frac{1}{4})\Vert \frac{\varphi }{r}\Vert _{L^2}^2, \end{aligned}$$

we immediately obtain the first inequality. It then follows from Cauchy–Schwarz inequality that

$$\begin{aligned} |k|^2\Vert \frac{\varphi }{r}\Vert _{L^2}^2\lesssim \Vert rw\Vert _{L^2} \Vert \frac{\varphi }{r}\Vert _{L^2}, \end{aligned}$$

which implies

$$\begin{aligned} |k|^2\Vert \frac{\varphi }{r}\Vert _{L^2}\lesssim \Vert rw\Vert _{L^2}, \end{aligned}$$

and

$$\begin{aligned} |\langle w, \varphi \rangle |\le \Vert rw\Vert _{L^2} \Vert \frac{\varphi }{r}\Vert _{L^2} \lesssim |k|^{-2}\Vert rw\Vert _{L^2}. \end{aligned}$$

On the other hand, in view of Cauchy–Schwarz inequality again and using Lemma A.1, it can be estimated

$$\begin{aligned} \Vert \varphi '\Vert _{L^2}^2+(k^2-\frac{1}{4})\Vert \frac{\varphi }{r}\Vert _{L^2}^2&=-\langle w, \varphi \rangle \le \Vert r^{\frac{1}{2}}w\Vert _{L^1}\Vert \frac{\varphi }{r^{\frac{1}{2}}}\Vert _{L^\infty } \\&\lesssim \Vert r^{\frac{1}{2}}w\Vert _{L^1}\Vert \frac{\varphi }{r}\Vert _{L^2}^{\frac{1}{2}}(\Vert \frac{\varphi }{r}\Vert _{L^2}+\Vert \varphi '\Vert _{L^2}^2) \\&\lesssim |k|^{-\frac{1}{2}}\Vert r^{\frac{1}{2}}w\Vert _{L^1} \left( \Vert \varphi '\Vert _{L^2}^2+(k^2-\frac{1}{4})\Vert \frac{\varphi }{r}\Vert _{L^2}^2 \right) ^{\frac{1}{2}}, \end{aligned}$$

which yields the second inequality.

For the last inequality, we observe that

$$\begin{aligned} \Vert rw\Vert _{L^2}^2=\Vert r\varphi ''\Vert _{L^2}^2+2(k^2-\frac{1}{4})\Vert \varphi '\Vert _{L^2}^2+(k^2-\frac{1}{4})^2\Vert \frac{\varphi }{r}\Vert _{L^2}^2 . \end{aligned}$$

By applying Lemma A.1 and inequalities above, we then prove

$$\begin{aligned} \Vert r^{\frac{1}{2}}\varphi '\Vert _{L^\infty }^2&\lesssim \Vert \varphi '\Vert _{L^2} \Vert r^{\frac{1}{2}}\partial _r (r^{\frac{1}{2}}\varphi ')\Vert _{L^2}+\frac{R}{R-1} \Vert \varphi '\Vert _{L^2}^2 \\&\lesssim \Vert \varphi '\Vert _{L^2}(\Vert r\varphi ''\Vert _{L^2}+\Vert \varphi '\Vert _{L^2})+\frac{R}{R-1} \Vert \varphi '\Vert _{L^2}^2 \lesssim \frac{R}{R-1}|k|^{-1}\Vert rw\Vert _{L^2}^2, \end{aligned}$$

and

$$\begin{aligned} \begin{aligned} \Vert r^{-\frac{1}{2}}\varphi \Vert _{L^\infty }^2\lesssim&\Vert \frac{\varphi }{r}\Vert _{L^2} \Vert r^{\frac{1}{2}}\partial _r (r^{-\frac{1}{2}}\varphi ) \Vert _{L^2} \lesssim |k|^{-3} \Vert rw\Vert _{L^2}^2. \end{aligned} \end{aligned}$$

$\square$

To establish the elliptic estimate for the zero mode of stream function, we require the below Poincaré-type inequalities.

Lemma A.4

For any $w\in L^2(1,R), w|_{r=1,R}=0$ with $R>1$, the following Poincaré-type inequalities hold

$$\begin{aligned} \Vert w\Vert _{L^2}\le 2 \Vert rw'\Vert _{L^2} \quad \text {and} \quad \Vert \frac{w}{r^{\frac{1}{2}}}\Vert _{L^2}\le 2\log R\Vert r^{\frac{1}{2}}w'\Vert _{L^2}. \end{aligned}$$

Proof

Via integration by parts, we immediately obtain

$$\begin{aligned} \Vert w\Vert _{L^2}^2= \int _{1}^{R} |w|^2 dr=&-\int _1^R r(w'\overline{w}+w\overline{w}') dr \le 2\Vert rw'\Vert _{L^2}\Vert w\Vert _{L^2}. \end{aligned}$$

and

$$\begin{aligned} \Vert \frac{w}{r^{\frac{1}{2}}}\Vert _{L^2}^2=\int _{1}^{R} |w|^2 d(\log r)=-\int _1^R \log r(w'\overline{w}+w\overline{w}') dr\le \log R\Vert r^{\frac{1}{2}}w'\Vert _{L^2}\Vert \frac{w}{r^{\frac{1}{2}}}\Vert _{L^2}. \end{aligned}$$

$\square$

Consequently, we obtain the following elliptic estimate which involves the outer radius R.

Lemma A.5

Let $w=(\partial _r^2+\frac{1}{r}\partial _r)\varphi$ with $\varphi |_{r=1, R}=w|_{r=1, R}=0$. The following inequality holds

$$\begin{aligned}&\Vert \varphi '\Vert _{L^{\infty }} \lesssim (\frac{R}{R-1})^{\frac{1}{2}}(1+\log R)\Vert r^{\frac{3}{2}}w\Vert _{L^2}. \end{aligned}$$

Proof

Notice that

$$\begin{aligned} -\langle w, r\varphi \rangle =-\langle \varphi ''+\frac{1}{r}\varphi ', r\varphi \rangle = \Vert r^{\frac{1}{2}}\varphi '\Vert _{L^2}. \end{aligned}$$

In view of Lemma A.4, it can be inferred

$$\begin{aligned} \Vert r^{\frac{1}{2}}\varphi '\Vert _{L^2}^2\le \Vert r^{\frac{3}{2}}w\Vert _{L^2}\Vert \frac{\varphi }{r^{\frac{1}{2}}}\Vert _{L^2} \lesssim \log R \Vert r^{\frac{3}{2}}w\Vert _{L^2}\Vert r^{\frac{1}{2}} \varphi '\Vert _{L^2}, \end{aligned}$$

which yields

$$\begin{aligned} \Vert r^{\frac{1}{2}}\varphi '\Vert _{L^2}+\Vert \frac{\varphi }{r^{\frac{1}{2}}}\Vert _{L^2}\lesssim \log R\Vert r^{\frac{3}{2}}w\Vert _{L^2}, \end{aligned}$$

and

$$\begin{aligned} \Vert r^{\frac{3}{2}}\varphi ''\Vert _{L^2}\le \Vert r^{\frac{3}{2}}w\Vert _{L^2}+\Vert r^{\frac{1}{2}}\varphi '\Vert _{L^2}\lesssim (1+\log R)\Vert r^{\frac{3}{2}}w\Vert _{L^2}. \end{aligned}$$

Applying Lemma A.1, we derive the $L^\infty$ bound for $\varphi '$ as below

$$\begin{aligned} \Vert \varphi '\Vert _{L^\infty }^2\lesssim&\Vert r^{-\frac{1}{2}}\varphi '\Vert _{L^2}\Vert r^{\frac{1}{2}}\varphi ''\Vert _{L^2}+\frac{R}{R-1}\Vert r^{-\frac{1}{2}}\varphi '\Vert _{L^2}^2\\\lesssim&\left( \frac{R}{R-1}+\log R(1+\log R)\right) \Vert r^{\frac{3}{2}}w\Vert _{L^2}^2\lesssim \frac{R}{R-1}(1+\log R)^2\Vert r^{\frac{3}{2}}w\Vert _{L^2}^2. \end{aligned}$$

This completes the proof of this lemma. $\quad \square$

Following is a collection of calculus inequalities which provide quantitative information of the interval $[1, R]\cap \{r> 0: \ |1-\lambda r^2|\le \delta \}$ where $\delta >0$ is sufficiently small. These estimates play a significant role in the proof of resolvent estimates in Sect. 3.

Lemma A.6

For any $\lambda \in [\frac{1}{R^2},1]$ and $0<\delta \ll 1$ being a small constant, it holds

$$\begin{aligned} \int _{r_{-}}^{r_{+}}\frac{1}{r}dr\lesssim \delta ,\quad \quad \quad \int _{r_{-}-\frac{\delta }{\sqrt{\lambda }}}^{r_{+}+\frac{\delta }{\sqrt{\lambda }}}\frac{1}{r}dr\lesssim \delta , \end{aligned}$$

where $r_{-}\in (\sqrt{\frac{1-\delta }{\lambda }}-\frac{\delta }{\sqrt{\lambda }},\sqrt{\frac{1-\delta }{\lambda }})$ and $r_{+}\in (\sqrt{\frac{1+\delta }{\lambda }},\sqrt{\frac{1+\delta }{\lambda }}+\frac{\delta }{\sqrt{\lambda }})$.

Proof

A direct computation yields

$$\begin{aligned}&\int _{r_{-}}^{r_{+}}\frac{1}{r}dr=\ln \frac{r_{+}}{r_{-}}\le \ln \frac{\sqrt{1+\delta }+\delta }{\sqrt{1-\delta }-\delta }\\&\quad =\frac{1}{2}\Big [\ln (1+\delta +2\delta \sqrt{1+\delta }+\delta ^2)-\ln (1-\delta -2\delta \sqrt{1-\delta }+\delta ^2)\Big ]\lesssim \delta \end{aligned}$$

and

$$\begin{aligned}&\int _{r_{-}-\frac{\delta }{\sqrt{\lambda }}}^{r_{+}+\frac{\delta }{\sqrt{\lambda }}}\frac{1}{r}dr=\ln \frac{r_{+}+\frac{\delta }{\sqrt{\lambda }}}{r_{-}-\frac{\delta }{\sqrt{\lambda }}}\le \ln \frac{\sqrt{1+\delta }+2\delta }{\sqrt{1-\delta }-2\delta }\\&\quad =\frac{1}{2}\Big [\ln (1+\delta +4\delta \sqrt{1+\delta }+4\delta ^2)-\ln (1-\delta -4\delta \sqrt{1-\delta }+4\delta ^2)\Big ]\lesssim \delta . \end{aligned}$$

$\square$

The $L^2$ norm of $r^{-\frac{1}{2}}(1-\lambda r^2)^{-1}$ in the region $\{r\in [1, R]: \ |1-\lambda r^2|> \delta \}$ can be bounded in terms of R and $\delta$.

Lemma A.7

For any $\delta>0, R>1$ and $\lambda \in \mathbb {R}$. Let $E=\{r> 0: \ |1-\lambda r^2|\le \delta \}$ and $E^c=(0, \infty )\backslash E$. Then it holds

$$\begin{aligned} \int _{[1, R] \cap E^c} \frac{dr}{r(1-\lambda r^2)^2} \lesssim \log R+ \delta ^{-1}. \end{aligned}$$

(A.3)

Proof

If $\lambda \le 0$, then one has

$$\begin{aligned} \int _{[1, R] \cap E^c} \frac{dr}{r(1-\lambda r^2)^2} \le \int _1^R \frac{dr}{r}=\log R. \end{aligned}$$

Now we assume $\lambda >0$. Denote $r_{\pm }=\sqrt{\frac{1\pm \delta }{\lambda }}$. We consider the following five scenarios:

• (1) If $1\le r_-<r_+\le R$, then a direct calculation implies

  $$\begin{aligned} \begin{aligned}&\int _1^{r_-} \frac{dr}{r(1-\lambda r^2)^2}=\frac{1}{2} \int _{\delta }^{1-\lambda } \frac{ds}{(1-s)s^2} =\frac{1}{2}\left( \log |\frac{s}{s-1}|-\frac{1}{s} \right) \Big |^{1-\lambda }_{\delta }\\&\quad \lesssim \log (\lambda ^{-1}-1)+\delta ^{-1} \lesssim \log R+ \delta ^{-1}. \end{aligned} \end{aligned}$$

  (A.4)

  Here in the first line we use the change of variables $s=1-\lambda r^2$ and the last line follows from the condition $r_{-}\in [1, R]$. Similarly, taking into account the condition $r_+\in [1, R]$, we deduce

  $$\begin{aligned} \int _{r_+}^{R} \frac{dr}{r(1-\lambda r^2)^2} \lesssim \log R+ \delta ^{-1}. \end{aligned}$$

  (A.5)

  Combining these two inequalities above, we obtain the desired result.

• (2) If $1\le r_-\le R<r_+$, then (A.3) directly follows from (A.4) and the fact that $[1,R]\cap E^c=[1, r_-]$.

• (3) If $r_-<1\le r_+\le R$, then (A.3) can be readily obtained thanks to (A.5) and the fact that $[1,R]\cap E^c=[r_+, R]$.

• (4) If $r_+<1$ or $r_->R$, then $[1, R]\cap E^c=[1, R]$. Assume first $r_->R$. It is hence clear that $1-\lambda R^2\ge \delta$. A suitable change of variables implies

  $$\begin{aligned} \begin{aligned}&\int _1^{R} \frac{dr}{r(1-\lambda r^2)^2}=\frac{1}{2} \int _{1-\lambda R^2}^{1-\lambda } \frac{ds}{(1-s)s^2} =\frac{1}{2}\left( \log (\frac{s}{s-1})-\frac{1}{s} \right) \Big |^{1-\lambda }_{1-\lambda R^2}\\&\quad \lesssim \log (R^2\frac{1-\lambda }{1-\lambda R^2})+\frac{1}{1-\lambda R^2} \lesssim \log R+ \delta ^{-1}. \end{aligned} \end{aligned}$$

  The case when $r_+<1$ can be treated in a similar manner.

• (5) If $r_-<1<R<r_+$, then immediately we see (A.3) holds since $[1, R]\cap E^c=\varnothing$.

$\square$

Finally, in order to prove the resolvent estimates for $\varphi '$ in Lemma 3.6, it is necessary to construct a piecewise $C^1$ cut-off function satisfying the following properties.

Lemma A.8

For any $0<\delta \ll 1$ and $\lambda \in \mathbb {R}$. Let $E=\{r> 0: \ |1-\lambda r^2|\le \delta \}$ and $E^c=(0, \infty )\backslash E$. Then there exists a piecewise $C^1$ function $\chi : (0, \infty ) \rightarrow \mathbb {R}$ satisfying the following properties:

• (1) $\chi (r)=\frac{1}{\frac{1}{r^2}-\lambda }$ on $E^c$.

• (2) there exists a constant $C>0$ independent of $\lambda$ and $\delta$, such that it holds

  $$\begin{aligned}\Vert \frac{\chi }{r^2}\Vert _{L^\infty (0, \infty )}+\delta ^{\frac{1}{2}}\Vert r^{\frac{1}{2}}(\frac{\chi }{r^2})'\Vert _{L^2(0, \infty )}\le C\delta ^{-1}.\end{aligned}$$

• (3) $\Vert \frac{1}{r}\Vert _{L^1(E)}\lesssim \delta , \ \Vert (1-\lambda r^2)\frac{\chi }{r^{\frac{5}{2}}}\Vert _{L^2(E)}\lesssim \delta ^{\frac{1}{2}}$.

Proof

If $\lambda \le 0$, then we have $E=\varnothing$, since $|1-\lambda r^2|\ge 1$. Setting $\chi (r)=\frac{1}{\frac{1}{r^2}-\lambda }$, one can directly check that (1) and (2) are satisfied. Now we assume $\lambda >0$. Denote $r_{\pm }=\sqrt{\frac{1\pm \delta }{\lambda }}$. We can see that $E=(r_-, r_+)$ and $m(E)\approx \delta \lambda ^{-\frac{1}{2}}$.

Choose the piecewise $C^1$ function $\chi : (0, \infty ) \rightarrow \mathbb {R}$ as below:

$$\begin{aligned} \chi (r)=\left\{ \begin{aligned}&\frac{1}{\frac{1}{r^2}-\lambda },\quad r\in E^c,\\&-\frac{r_+^2+r_-^2}{\delta (r_+-r_-)}(r-r_-)+\frac{r_-^2}{\delta },\quad r\in E. \end{aligned} \right. \end{aligned}$$

Observing that $\Vert \chi \Vert _{L^\infty (E)}\lesssim \delta ^{-1}\lambda ^{-1}$, $\Vert \chi '\Vert _{L^\infty (E)}\lesssim \delta ^{-2}\lambda ^{-\frac{1}{2}}$, and $r\approx \lambda ^{-\frac{1}{2}}$ on E, we then verify

$$\begin{aligned} \Vert \frac{1}{r}\Vert _{L^1(E)}\lesssim \delta , \ \Vert (1-\lambda r^2)\frac{\chi }{r^{\frac{5}{2}}}\Vert _{L^2(E)}\lesssim \delta ^{\frac{1}{2}}, \end{aligned}$$

and

$$\begin{aligned} \Vert \frac{\chi }{r^2}\Vert _{L^\infty (E)}+\delta ^{\frac{1}{2}}\Vert r^{\frac{1}{2}}(\frac{\chi }{r^2})'\Vert _{L^2(E)}\lesssim \delta ^{-1}. \end{aligned}$$

It remains to establish

$$\begin{aligned} \Vert \frac{\chi }{r^2}\Vert _{L^\infty (E^c)}+\delta ^{\frac{1}{2}}\Vert r^{\frac{1}{2}}(\frac{\chi }{r^2})'\Vert _{L^2(E^c)}\lesssim \delta ^{-1}. \end{aligned}$$

The first term on the left can be easily estimated due to the fact that $\frac{\chi }{r^2}=\frac{1}{1-\lambda r^2}$ on $E^c=\{r> 0: \ |1-\lambda r^2|> \delta \}$.

For the second term, a direct computation yields

$$\begin{aligned} \Vert r^{\frac{1}{2}}(\frac{\chi }{r^2})'\Vert _{L^2(E^c)}^2= \int _{E^c} \frac{4\lambda ^2 r^3}{(1-\lambda r^2)^4} dr =\int _{\delta }^1 \frac{2(1-y)}{y^4}dy+\int _{-\infty }^{-\delta } \frac{2(1-y)}{y^4}dy \lesssim \delta ^{-3}, \end{aligned}$$

which gives the desired inequality. $\quad \square$