# Asymptotic properties of steady plane solutions of the Navier-Stokes
equations in the exterior of a half-space

Lili Wang  School of Mathematical Sciences, Dalian University of Technology,
Dalian, 116024, China<EMAIL_ADDRESS>and  Wendong Wang  School of
Mathematical Sciences, Dalian University of Technology, Dalian, 116024, China
<EMAIL_ADDRESS>
(Date: June 18, 2021)

###### Abstract.

Motivated by Gilbarg-Weinberger’s early work on asymptotic properties of
steady plane solutions of the Navier-Stokes equations on a neighborhood of
infinity [19] , we investigate asymptotic properties of steady plane solutions
of this system on a half-neighborhood of infinity with finite Dirichlet
integral and Navier-slip boundary condition, and obtain that the velocity of
the solution grows more slowly than $\sqrt{\log r}$, while the pressure
converges to $0$ along each ray passing through the origin.

Keywords: Navier-Stokes equations, Liouville type theorem, asymptotic
behavior, exterior of a half-space

2010 Mathematics Subject Classification: 35Q30, 35Q10, 76D05.

##  1\. Introduction

The problem of a rigid body through a viscous liquid originates with the
pioneering work of Stokes in 1851 [39] on the effect of internal friction on
the movement of a pendulum in a liquid, which is modeled by the incompressible
Navier-Stokes equations in a planar exterior domain (see also the introduction
in [15]). When the influence of the boundary walls of the container was
disregarded, in the fundamental work by Leray [30], where the solution of the
two dimensional exterior problem for the Navier-Stokes equations was
constructed with finite Dirichlet integral, it is remaining that whether the
constructed solution satisfies the asymptotic behavior at $\infty$. Gilbarg-
Weinberger [19] described the asymptotic behavior of the velocity, the
pressure and the vorticity, where they showed that $u(x)=o(\ln|x|)$ and

$\displaystyle\lim_{r\rightarrow\infty}\int_{0}^{2\pi}|u(r,\theta)-\bar{u}|^{2}d\theta=0$

for some constant vector $\bar{u}$. Later, Amick [1] proved that $u\in
L^{\infty}$ under zero boundary condition. Recently, Korobkov-Pileckas-Russo
in [27] and [26] obtained that

$\displaystyle\lim_{|x|\rightarrow\infty}u(x)=\bar{u}.$

More references on the existence and asymptotic behavior of solutions in an
exterior domain, we refer to [2, 17, 33, 34, 16, 14, 32, 24, 13, 28] and the
references therein. For the generalized q-energy of $\|\nabla u\|_{q}<\infty$,
asymptotic behavior of solutions or Liouville type theorems are considered in
[29, 42, 43]. On the other hand, a body moving within an incompressible fluid
parallel to a wall in an otherwise unbounded domain is modeled in a planar
exterior domain in a half space with appropriate boundary conditions on the
wall. Hillairet-Wittwer [22] obtained the existence and uniqueness of weak
solution with the boundary condition at infinity by considering the Oseen
model, which implies the faster decay at infinity, see also Boeckle-
Wittwer[4], Guo-Witter-Zhou [21] for the recent progress.

In this paper, we investigate the asymptotic properties of Leray solution with
finite Dirichlet integral without the infinite boundary condition. In details,
consider an arbitrary solution
$\left\\{\boldsymbol{w}^{\prime},p^{\prime}\right\\}$ of the following two-
dimensional steady Navier-Stokes equations

$\displaystyle\left\\{\begin{aligned}
\Delta\boldsymbol{w}^{\prime}-(\boldsymbol{w}^{\prime}\cdot\nabla)\boldsymbol{w}^{\prime}&=\nabla
p^{\prime},\\\ \nabla\cdot\boldsymbol{w}^{\prime}&=0\end{aligned}\right.$
(1.1)

in the exterior of the upper half-space,
$\mathbb{R}_{+}^{2}\setminus\overline{B_{r_{0}}}$ with $r_{0}>0$, where
$\boldsymbol{w}^{\prime}(x,y)=(u^{\prime}(x,y),v^{\prime}(x,y))$ represents
fluid velocity, $p^{\prime}(x,y)$ is the fluid pressure and $B_{r_{0}}$ is a
ball centered at $0$.

There are several possibilities for the boundary condition on
$\Gamma=\\{(x,y);y=0,|x|>r_{0}\\}$.The widely used is the following Dirichlet
or no-slip boundary condition:

$\displaystyle\boldsymbol{w}^{\prime}(x,y)=0\quad\text{on}\quad\partial
B_{1}.$

At this case, Liouville type theorem was proved by Seregin in [35] for the
ancient solutions of the time-dependent case. More Liouville type theorems are
obtained by Guo-Wang in [20] by assuming different generalized weak solutions.
We refer to [37, 23, 3, 44] and the refereces therein for some results on
singularity or classification of solutions under this boundary condition.
However, in the case where the obstacles have an approximate limit, the
Dirichlet boundary conditions are no longer valid (see for example [38]). Due
to the roughness of the boundary and the viscosity of the fluid, it is usually
assumed that there is a stagnant fluid layer near the boundary, which allows
the fluid to slip. This situation seems to match the reality. In 1827, Navier
[12] considered the slip phenomena and proposed the Navier-slip boundary
conditions:

$\left\\{\begin{array}[]{ll}\boldsymbol{w}^{\prime}(x,y)\cdot n=0,\\\
2[D(\boldsymbol{w}^{\prime}(x,y))\cdot
n]_{\tau}+\alpha(x)\boldsymbol{w}^{\prime}_{\tau}=0,\end{array}\right.$ (1.2)

where $D(\boldsymbol{w^{\prime}}(x,y))$ is the stress tensor of fluid, $n$ and
$\tau$ are the unit outer normal vector and tangential vector of the boundary,
$\alpha(x)$ is a physical parameter, which is a $L^{\infty}$ function on the
boundary. For the flat boundary of $\Gamma=\\{(x,y);y=0,|x|>r_{0}\\}$,
motivated by the above condition of (1.2), generally we try to consider the
body suspended in a linear shear flow as in [11] or [10], which showed the
forces generated due to wall and shear effects in the absence of relative
motion,

$\displaystyle v^{\prime}(x,0)=b,\partial_{2}u^{\prime}(x,0)=a,\quad{\rm
on}~{}~{}\Gamma,$ (1.3)

where the constants $a,b\in\mathbb{R}$. Specially, let
$\boldsymbol{w}=\boldsymbol{w}^{\prime}-(ay,b)=(u,v)$ and $p=p^{\prime}+abx$,
then

$\displaystyle\left\\{\begin{aligned}
\Delta\boldsymbol{w}-(\boldsymbol{w}\cdot\nabla)\boldsymbol{w}-(ay\partial_{1}+b\partial_{2})\boldsymbol{w}-(av,0)&=\nabla
p,\\\ \nabla\cdot\boldsymbol{w}&=0\end{aligned}\right.$ (1.4)

which satisfies the boundary condition:

$\displaystyle v(x,0)=0,\partial_{2}u(x,0)=0,\quad{\rm on}~{}~{}\Gamma.$ (1.5)

Let $\omega=u_{y}-v_{x}$ be the vorticity of a velocity vector
$\boldsymbol{w}=(u,v)$ satisfying (1.4). Then the vorticity equation of
(1.4)follows:

$\Delta\omega-\boldsymbol{w}\cdot\nabla\omega-(ay\partial_{1}+b\partial_{2})\omega=0,\quad{\rm
in}~{}~{}\Omega_{0},$ (1.6)

and $\omega=0$ on $\Gamma$ due to (1.5), where

$\displaystyle\Omega_{0}=\\{(r,\theta),r>r_{0},0<\theta<\pi\\}=\mathbb{R}_{+}^{2}\backslash\overline{B_{r_{0}}},\quad
r_{0}>0,$

by assuming $r=\sqrt{x^{2}+y^{2}}$ with $x=r\cos\theta$ and $y=r\sin\theta$.

Our main results are as follows. The first result is about the growth estimate
of the velocity.

######  Theorem 1.1.

Let $\boldsymbol{w}\in C^{2}(\overline{\Omega_{0}})$ be a solution of the
Navier-Stokes equations (1.4) with the Navier-slip boundary condition (1.5)
and

$\displaystyle\int_{\Omega_{0}}|\nabla\boldsymbol{w}|^{2}dxdy<\infty,$ (1.7)

then

$\lim_{r\to\infty}\frac{|\boldsymbol{w}(r,\theta)|}{\sqrt{\log r}}=0.$ (1.8)

######  Remark 1.1.

The growth rate in this case is similar as that in [19]. However, Cauchy
integral formula in [19] doesn’t seem to work in this case due to the boundary
effect. We used point-wise behavior theorem in [15] by proving the $L^{q}$
estimate of approximating vorticity functions with $q>2$. It is still unknown
whether the convergence of the velocity holds in the $L^{\infty}$ sense as
Korobkov-Pileckas-Russo in [27] and [26]. The condition of $\boldsymbol{w}\in
C^{2}(\overline{\Omega_{0}})$ can be relaxed to $\boldsymbol{w}\in
C^{2}_{loc}({\Gamma})$, since the solution with Dirichlet energy (1.7) is
regular in the interior.

For the special half-plane, the Liouville type theorem also holds similar to
the whole space.

######  Theorem 1.2.

Let $\left\\{\boldsymbol{w},p\right\\}$ be a solution of the Navier-Stokes
equations (1.4) with the Navier-slip boundary condition (1.5) defined over the
upper half space $\mathbb{R}_{+}^{2}$. Assume $\boldsymbol{w}\in
C^{2}(\overline{\mathbb{R}_{+}^{2}})$ and

$\displaystyle\int_{\mathbb{R}_{+}^{2}}|\nabla\boldsymbol{w}|^{2}dxdy<\infty.$
(1.9)

Then $\boldsymbol{w}$ and $p$ are constants in $\mathbb{R}_{+}^{2}.$

######  Remark 1.2.

It’s proved by using the monotonicity property of
$G(r)=\frac{1}{2}\int_{0}^{\pi}|\boldsymbol{w}(r,\theta)|^{2}d\theta$ and the
vanishing vorticity. When the Navier-slip boundary condition (1.5) is replaced
by the no-slip condition, which seems to be more difficult than the high
dimensional case in $\mathbb{R}^{3}$ or $\mathbb{R}_{+}^{3}$ with the no-slip
boundary condition, which is an open problem as stated in [15] and [40]. We
refer to Galdi [15], Chae [8] and Seregin [36], where some Liouville type
theorems are proved by assuming that $u\in L^{9/2}(\mathbb{R}^{3})$, $\Delta
u\in L^{6/5}(\mathbb{R}^{3})$ and $u\in BMO^{-1}(\mathbb{R}^{3})$,
respectively. For more related discussion, we refer to [25, 9, 7, 41] and the
references therein.

Another major feature of this paper lies in the estimates of the pressure, and
we obtained the following results by two different approaches.

######  Theorem 1.3.

Let $\left\\{\boldsymbol{w},p\right\\}$ be a solution of the Navier-Stokes
equations (1.4) with the Navier-slip boundary condition (1.5). Assume that
$\boldsymbol{w}\in C^{2}(\overline{\Omega_{0}}),$ (1.7) hold, $a=0$ and
$bu(r,0)=bu(r,\pi)$ for $r>r_{0}$. Then there hold
(i)

$|p(r,\theta)|=o(\log r),$ (1.10)

uniformly for all $\theta\in[0,\pi]$.
(ii) up to a constant

$\displaystyle\lim_{r\to\infty,\theta=\theta_{0}}p(r,\theta)=0,$ (1.11)

where $\theta_{0}\in[0,\pi].$

######  Remark 1.3.

The pressure’s decay is not better than the exterior domain of
$\mathbb{R}^{2}$. The main obstacle comes from the boundary effect. Case (i)
of Theorem 1.3 is proved by the help of Brezis-Gallouet inequality, and the
proof of Case (ii) in Theorem 1.3 is following the route of [19] but using
Green representation formula of harmonic function on the half-annulus domain.

######  Remark 1.4.

Combining the two results of Theorem 1.3, it seems to be unknown that whether
$p$ converges to zero at infinity along all the rays uniformly. For example,
Theorem 1.3 doesn’t work when $\theta=\frac{1}{K}$, $r=K$ and
$p=p(K,\frac{1}{K})=1,$ where $K\in\mathbb{N}.$

At last, we also obtain the decay estimate of the vorticity.

######  Theorem 1.4.

Let $\left\\{\boldsymbol{w},p\right\\}$ be a solution of the Navier-Stokes
equations (1.4) with the Navier-slip boundary condition (1.5). Assume that
$\boldsymbol{w}\in C^{2}(\overline{\Omega_{0}})$, (1.7) hold and $a=0$. Then

$\lim\limits_{r\to\infty}\frac{r^{\frac{3}{4}}}{(\log
r)^{\frac{1}{8}}}|\omega(r,\theta)|=0,$ (1.12)

where $\theta\in[0,\pi].$

The paper is organized as follows, in Section 2, we recall and introduce some
technical lemmas. In Section 3, we are aimed to the proof of Theorem 1.1,
which is the decay estimate of the velocity. Theorem 1.2, Theorem 1.3 and
Theorem 1.4 are proved in Section 4—Section 6, respectively.

Throughout this article, $C$ denotes a constant only depending on the known
finite norms of solutions (e.g. $\|\nabla\boldsymbol{w}\|_{L^{2}}$) and may be
different from line to line.

##  2\. Preliminary lemmas

The following lemma is similar to the exterior domain $B_{r_{0}}^{c}$ of Lemma
2.1 in [19], one could make the same arguments as [19] in such an exterior
angular region. In fact, one may also do an extension to the whole space by
noting the following $C^{0,1}$ is enough and we omitted it.

######  Lemma 2.1.

Let $f\in C^{1}(\overline{{\Omega_{0}}})$ and have finite Dirichlet integral

$\displaystyle\int_{\Omega_{0}}|\nabla f|^{2}dxdy<\infty.$ (2.1)

Then

$\displaystyle\lim_{r\to\infty}\frac{1}{\log
r}\int_{0}^{\pi}f(r,\theta)^{2}d\theta=0.$ (2.2)

It immediately follows that a special subsequence is pointwise convergent,
which is also the same as Lemma 2.2 in [19] and is stated as follows.

######  Lemma 2.2.

Let $\boldsymbol{f}=(f_{1},f_{2}),$ where $f_{1}$ and $f_{2}$ satisfy the
hypotheses of Lemma 2.1. Then there is a sequence
${r_{n}},r_{n}\in(2^{n},2^{n+1}),$ such that

$\displaystyle\lim_{n\to\infty}\frac{|\boldsymbol{f}(r_{n},\theta)|^{2}}{\log
r_{n}}=0,$ (2.3)

where $\theta\in[0,\pi].$

As in [19], we can also obtain the bounded-ness of the $L^{2}$ norm of
$\nabla\omega.$

######  Lemma 2.3.

Let $\boldsymbol{w}=(u,v)$ satisfy the Navier-Stokes equations (1.4) and
Navier-slip boundary condition (1.5) in $\Omega_{0}$. Assume
$\boldsymbol{w}\in C^{2}(\overline{\Omega_{0}})$ and have finite Dirichlet
integral of (1.7). Then

$\int_{\Omega_{0}}|\nabla\omega|^{2}dxdy<\infty.$ (2.4)

Proof. For a large number $R_{0}>0$, choose
$r_{0}<r_{1}<\frac{R_{0}}{2}\leq\rho<R\leq R_{0}$ and a non-negative $C^{2}$
cut-off function $\eta(r)$ such that

$\eta(r)=\left\\{\begin{array}[]{ll}1,\quad r_{1}<r<\rho,\\\ 0,\quad r\leq
r_{0},r\geq R,\end{array}\right.$ (2.5)
$\displaystyle|\nabla^{k}\eta|\leq\frac{C}{(r_{1}-r_{0})^{k}},\quad{\rm
as}~{}~{}r_{0}<r<r_{1},$

and

$\displaystyle|\nabla^{k}\eta|\leq\frac{C}{(R-\rho)^{k}},\quad{\rm
as}~{}~{}\rho<r<R,$

for $k=1,2$. Let $h(\omega)=\omega^{2}$ and $B_{r}^{+}=B_{r}^{+}(0)$ stands
for the upper ball centered at $0$ and of radius $r.$ Then by
$(\ref{eq:NS})_{2}$ and (1.6) we have

$\displaystyle{\rm div}$ $\displaystyle[\eta^{6}\nabla
h(\omega)-h(\omega)\nabla(\eta^{6})-\eta^{6}h(\omega)\boldsymbol{w}]$
$\displaystyle=2\eta^{6}|\nabla\omega|^{2}-h(\omega)[\Delta(\eta^{6})+\boldsymbol{w}\cdot\nabla(\eta^{6})]+2\eta^{6}\omega(ay\partial_{1}+b\partial_{2})\omega.$

Integration by parts over $B_{R}^{+}\backslash B_{r_{0}}^{+}$ yields that

$\displaystyle\int_{B_{R}^{+}\backslash
B_{r_{0}}^{+}}\eta^{6}|\nabla\omega|^{2}dxdy$ (2.6)
$\displaystyle=\frac{1}{2}\int_{B_{R}^{+}\backslash
B_{r_{0}}^{+}}\omega^{2}(\Delta\eta^{6}+\boldsymbol{w}\cdot\nabla\eta^{6})dxdy+\frac{1}{2}\int_{B_{R}^{+}\backslash
B_{r_{0}}^{+}}\nabla\eta^{6}\cdot(ay,b)\omega^{2}dxdy.$

Note that $\eta=1$ for $r_{1}<r<\rho$ and (2.6) implies that

$\displaystyle\int_{B_{\rho}^{+}\backslash
B_{r_{1}}^{+}}|\nabla\omega|^{2}dxdy$ (2.7) $\displaystyle\leq$
$\displaystyle\int_{B_{R}^{+}\backslash
B_{r_{0}}^{+}}|\nabla\omega|^{2}\eta^{6}dxdy$ $\displaystyle\leq$
$\displaystyle\frac{1}{2}\int_{B_{R}^{+}\backslash
B_{\rho}^{+}}\omega^{2}(\Delta\eta^{6}+\boldsymbol{w}\cdot\nabla\eta^{6})dxdy+\frac{1}{2}\int_{B_{R}^{+}\backslash
B_{\rho}^{+}}\nabla\eta^{6}\cdot(ay,b)\omega^{2}dxdy$
$\displaystyle+C(r_{1},\|\boldsymbol{w}\|_{L^{\infty}(B_{r_{1}}^{+}\backslash
B_{r_{0}}^{+})},\|\nabla\boldsymbol{w}\|_{L^{\infty}(B_{r_{1}}^{+}\backslash
B_{r_{0}}^{+})}),$

where we used

$\displaystyle\left|\int_{{B_{r_{1}}^{+}\backslash
B_{r_{0}}^{+}}}\omega^{2}(\Delta\eta^{6}+\boldsymbol{w}\cdot\nabla\eta^{6})dxdy\right|$
$\displaystyle+\left|\int_{{B_{r_{1}}^{+}\backslash
B_{r_{0}}^{+}}}\nabla\eta^{6}\cdot(ay,b)\omega^{2}dxdy\right|$ (2.8)
$\displaystyle\leq
C(r_{1},\|\boldsymbol{w}\|_{L^{\infty}(B_{r_{1}}^{+}\backslash
B_{r_{0}}^{+})},\|\nabla\boldsymbol{w}\|_{L^{\infty}(B_{r_{1}}^{+}\backslash
B_{r_{0}}^{+})}).$

First, we consider the second part of the right integral in (2.7)

$\displaystyle\frac{1}{2}\int_{B_{R}^{+}\backslash
B_{\rho}^{+}}\nabla\eta^{6}\cdot(ay,b)\omega^{2}dxdy\leq\frac{CR}{R-\rho}\int_{B_{R}^{+}\backslash
B_{\rho}^{+}}|\nabla\boldsymbol{w}|^{2}dxdy\leq\frac{CR}{R-\rho}.$ (2.9)

The first part of the right integral over the upper half annulus
$B_{R}^{+}\backslash B_{\rho}^{+}$ in (2.7) is controlled by

$\displaystyle\left|\int_{B_{R}^{+}\backslash
B_{\rho}^{+}}\omega^{2}\Delta\eta^{6}dxdy\right|+\left|\int_{B_{R}^{+}\backslash
B_{\rho}^{+}}\omega^{2}\left(\boldsymbol{w}-\bar{\boldsymbol{w}}\right)\cdot\nabla\eta^{6}dxdy\right|$
$\displaystyle+\left|\int_{B_{R}^{+}\backslash
B_{\rho}^{+}}\omega^{2}\bar{\boldsymbol{w}}\cdot\nabla\eta^{6}dxdy\right|\doteq
T_{1}+T_{2}+T_{3},$

where

$\bar{\boldsymbol{w}}(r)=\frac{1}{\pi}\int_{0}^{\pi}\boldsymbol{w}(r,\theta)d\theta.$

For $T_{1},$ we get

$\displaystyle T_{1}$
$\displaystyle\leq\frac{C}{(R-\rho)^{2}}\int_{B_{R}^{+}\backslash
B_{\rho}^{+}}\omega^{2}dxdy\leq\frac{C}{(R-\rho)^{2}}\int_{B_{R}^{+}\backslash
B_{\rho}^{+}}|\nabla\boldsymbol{w}|^{2}dxdy\leq\frac{C}{(R-\rho)^{2}}.$ (2.10)

For $T_{2},$ by Schwarz’s inequality there holds

$\displaystyle T_{2}$ $\displaystyle\leq 6\int_{B_{R}^{+}\backslash
B_{\rho}^{+}}|\boldsymbol{w}-\bar{\boldsymbol{w}}||\nabla\eta||\eta^{5}|\omega^{2}dxdy$
$\displaystyle\leq\frac{C}{R-\rho}\left(\int_{\rho}^{R}\int_{0}^{\pi}|\boldsymbol{w}-\bar{\boldsymbol{w}}|^{2}d\theta
rdr\right)^{\frac{1}{2}}\left(\int_{B_{R}^{+}\backslash
B_{\rho}^{+}}\eta^{10}\omega^{4}dxdy\right)^{\frac{1}{2}}.$

Using Wirtinger’s inequality

$\int_{0}^{\pi}|\boldsymbol{w}-\bar{\boldsymbol{w}}|^{2}d\theta\leq\int_{0}^{\pi}|\boldsymbol{w}_{\theta}|^{2}d\theta,$

one can obtain that

$\displaystyle\left(\int_{\rho}^{R}\int_{0}^{\pi}|\boldsymbol{w}-\bar{\boldsymbol{w}}|^{2}d\theta
rdr\right)^{\frac{1}{2}}\leq\left(\int_{\rho}^{R}\int_{0}^{\pi}|\boldsymbol{w}_{\theta}|^{2}d\theta
rdr\right)^{\frac{1}{2}}\leq R\left(\int_{B_{R}^{+}\backslash
B_{r_{0}}^{+}}|\nabla\boldsymbol{w}|^{2}dxdy\right)^{\frac{1}{2}}\leq CR.$

Moreover, since $\boldsymbol{w}\in C^{2}(\overline{\Omega_{0}}),$ we know that
$\nabla\omega\in C(\overline{\Omega_{0}}).$ It follows that

$\int_{{B_{r_{1}}^{+}\backslash B_{r_{0}}^{+}}}|\nabla\omega|^{2}dxdy\leq
C(\|\nabla^{2}\boldsymbol{w}\|_{L^{\infty}(B_{r_{1}}^{+}\backslash
B_{r_{0}}^{+})}).$

Then by Gagliado-Nirenberg inequality [31], we have

$\displaystyle\left(\int_{B_{R}^{+}\backslash
B_{\rho}^{+}}\eta^{10}\omega^{4}dxdy\right)^{\frac{1}{2}}$ $\displaystyle\leq$
$\displaystyle
C\|\eta^{2}\omega\|_{L^{2}(\mathbb{R}_{+}^{2})}\|\nabla(\eta^{2}\omega)\|_{L^{2}(\mathbb{R}_{+}^{2})}$
$\displaystyle\leq$ $\displaystyle
C\|\eta^{2}\omega\|_{L^{2}(\mathbb{R}_{+}^{2})}\|(\nabla\eta^{2})\omega+\eta^{2}(\nabla\omega)\|_{L^{2}(\mathbb{R}_{+}^{2})}$
$\displaystyle\leq$ $\displaystyle C\|\omega\|_{L^{2}(B_{R}^{+}\backslash
B_{r_{0}}^{+})}\left(C+\frac{C}{R-\rho}+\|\nabla\omega\|_{L^{2}(B_{R}^{+}\backslash
B_{r_{1}}^{+})}+\|\nabla\omega\|_{L^{2}(B_{r_{1}}^{+}\backslash
B_{r_{0}}^{+})}\right)$ $\displaystyle\leq$ $\displaystyle
C(\|\nabla^{2}\boldsymbol{w}\|_{L^{\infty}(B_{r_{1}}^{+}\backslash
B_{r_{0}}^{+})})\left(1+\frac{1}{R-\rho}+\|\nabla\omega\|_{L^{2}(B_{R}^{+}\backslash
B_{r_{1}}^{+})}\right).$

Hence we obtain the estimate

$\displaystyle T_{2}\leq
C(\|\nabla^{2}\boldsymbol{w}\|_{L^{\infty}(B_{r_{1}}^{+}\backslash
B_{r_{0}}^{+})})\frac{R}{R-\rho}\left(1+\frac{1}{R-\rho}+(\int_{B_{R}^{+}\backslash
B_{r_{1}}^{+}}|\nabla\omega|^{2}dxdy)^{\frac{1}{2}}\right)^{\frac{1}{2}}.$
(2.11)

For $T_{3},$ it follows from Lemma 2.1 that

$\bar{\boldsymbol{w}}(r)=o(\sqrt{\log r}),$

and hence

$\displaystyle T_{3}\leq\int_{B_{R}^{+}\backslash
B_{\rho}^{+}}|\bar{\boldsymbol{w}}||\nabla\eta^{6}|\omega^{2}dxdy\leq\frac{C(\log
R)^{\frac{1}{2}}}{R-\rho}.$ (2.12)

Combining (2.7)-(2.12), by Young’s inequality we get

$\displaystyle\int_{B_{\rho}^{+}\backslash
B_{r_{1}}^{+}}|\nabla\omega|^{2}dxdy$ $\displaystyle\leq$
$\displaystyle\frac{1}{2}\int_{B_{R}^{+}\backslash
B_{r_{1}}^{+}}|\nabla\omega|^{2}dxdy$
$\displaystyle+C(\|\boldsymbol{w}\|_{C^{2}(B_{r_{1}}^{+}\backslash
B_{r_{0}}^{+})})\left(\frac{R^{2}}{(R-\rho)^{2}}+\frac{(\log
R)^{\frac{1}{2}}}{R-\rho}+1\right).$

Using Giaquinta iteration in [18], we conclude that

$\displaystyle\int_{B_{\rho}^{+}\backslash
B_{r_{1}}^{+}}|\nabla\omega|^{2}dxdy$ $\displaystyle\leq$ $\displaystyle
C(\|\boldsymbol{w}\|_{C^{2}(B_{r_{1}}^{+}\backslash
B_{r_{0}}^{+})})\left(\frac{R^{2}}{(R-\rho)^{2}}+1\right).$

Take $\rho=\frac{R_{0}}{2}$ and $R=R_{0}$, and letting $R_{0}\to\infty,$ the
proof is complete.

######  Lemma 2.4.

Under the hypotheses of Lemma 2.3 we have

$\lim_{r\to\infty}r^{\frac{1}{2}}|\omega(r,\theta)|=0,$ (2.13)

where $\theta\in[0,\pi].$

Proof. Using the polar coordinate transformation and Cauchy inequality, we
obtain

$\displaystyle\int_{2^{n}}^{2^{n+1}}\frac{dr}{r}\int_{0}^{\pi}(r^{2}\omega^{2}+2r|\omega\omega_{\theta}|)d\theta\leq\int_{r>2^{n},0<\theta<\pi}(2\omega^{2}+|\nabla\omega|^{2})dxdy.$

Hence by the integral theorem of the mean, there exists an
$r_{n}\in(2^{n},2^{n+1})$ such that

$\displaystyle\int_{0}^{\pi}[r_{n}^{2}\omega(r_{n},\theta)^{2}+2r_{n}|\omega(r_{n},\theta)\omega_{\theta}(r_{n},\theta)|]d\theta\leq\frac{1}{\log
2}\int_{r>2^{n},0<\theta<\pi}(2\omega^{2}+|\nabla\omega|^{2})dxdy.$ (2.14)

Note that

$\displaystyle\omega(r_{n},\theta)^{2}-\frac{1}{\pi}\int_{0}^{\pi}\omega(r_{n},\theta)^{2}d\theta\leq
2\int_{0}^{\pi}|\omega(r_{n},\theta)\omega_{\theta}(r_{n},\theta)|d\theta.$

It follows from (2.14) and Lemma 2.3 that

$\displaystyle 0\leq r_{n}\omega(r_{n},\theta)^{2}$
$\displaystyle\leq\int_{0}^{\pi}[r_{n}^{2}\omega(r_{n},\theta)^{2}+2r_{n}|\omega(r_{n},\theta)\omega_{\theta}(r_{n},\theta)|]d\theta$
$\displaystyle\leq\frac{1}{\log
2}\int_{r>2^{n},0<\theta<\pi}(2\omega^{2}+|\nabla\omega|^{2})dxdy\to
0,n\to\infty,$

which implies

$\displaystyle\lim_{n\to\infty}[r_{n}\max\limits_{\theta\in[0,\pi]}\omega(r_{n},\theta)^{2}]=0.$
(2.15)

By (1.5), we have $\omega(x,0)=0$ on $\Gamma.$ Let $D$ denote the half-annulus
region with $r_{n}<r<r_{n+1}$. Since $\omega$ a solution of the equation
(1.6), it satisfies the maximum principle in $D.$ Noting that $r_{n+1}\leq
4r_{n},$ we infer that for $r\in(r_{n},r_{n+1})$

$\displaystyle r\max_{\theta\in[0,\pi]}\omega(r,\theta)^{2}$
$\displaystyle\leq r_{n+1}\max_{\theta\in[0,\pi]}\omega(r,\theta)^{2}$
$\displaystyle\leq
r_{n+1}\max[0,\max_{\theta\in[0,\pi]}\omega(r_{n},\theta)^{2},\max_{\theta\in[0,\pi]}\omega(r_{n+1},\theta)^{2}]$
$\displaystyle\leq\max[0,4r_{n}\max_{\theta\in[0,\pi]}\omega(r_{n},\theta)^{2},r_{n+1}\max_{\theta\in[0,\pi]}\omega(r_{n+1},\theta)^{2}],$

which implies the desired result (2.13) due to (2.15).

At last, we introduce the Brezis-Gallouet inequality (see Lemma 2 in [5], or
Lemma 3.1 in [6]).

######  Lemma 2.5.

Let $f\in H^{2}(\Omega),$ where $\Omega$ is a bounded domain or an exterior
domain with compact smooth boundary. Then there exists a constant $C_{\Omega}$
depending only on $\Omega$, such that

$\left\|f\right\|_{L^{\infty}(\Omega)}\leq
C_{\Omega}\left\|f\right\|_{H^{1}(\Omega)}\log^{\frac{1}{2}}(e+\frac{\left\|\Delta
f\right\|_{L^{2}(\Omega)}}{\left\|f\right\|_{H^{1}(\Omega)}}),$

or

$\left\|f\right\|_{L^{\infty}(\Omega)}\leq
C_{\Omega}(1+\left\|f\right\|_{H^{1}(\Omega)})\log^{\frac{1}{2}}(e+\left\|\Delta
f\right\|_{L^{2}(\Omega)}).$

Note that the second inequality can be obtained immediately from the first one
by arguments whether $\left\|f\right\|_{H^{1}(\Omega)}<1$.

The following lemma is from Theorem II.9.1 in [15].

######  Lemma 2.6.

Let $\Omega\subset\mathbb{R}^{2}$ be an exterior domain and let

$\nabla f\in L^{2}\cap L^{p}(\Omega),$

for some $2<p<\infty.$ Then

$\lim_{|x|\to\infty}\frac{|f(x)|}{\sqrt{\log(|x|)}}=0,$

uniformly.

##  3\. Deacy of the velocity

Proof of Theorem 1.1. For $r=\sqrt{x^{2}+y^{2}},$ we take two cut-off
functions $\varphi(x,y)$ and $\eta(x,y)$ as follows:

$\varphi(r)=\left\\{\begin{aligned} &1,\quad r<R\\\ &0,\quad
r>2R\end{aligned}\right.,\quad\eta(r)=\left\\{\begin{aligned} 1,\quad
r>3r_{0}\\\ 0,\quad r<2r_{0}\end{aligned}.\right.$

Firstly, we show that

$\omega\in L^{p}(\Omega_{0}),\quad\forall p\geq 2.$ (3.1)

Noting that $|\nabla\eta|\leq C$ and $|\nabla\varphi|\leq\frac{C}{R}$ for a
constant $C$ independent of $R.$ Then using (2.4), (1.7) and Gagliardo-
Nirenberg inequality, for any $p\geq 2,$ we have

$\displaystyle\left\|\omega\varphi\eta\right\|_{L^{p}(\mathbb{R}_{+}^{2})}$
$\displaystyle\leq
C\left\|\omega\varphi\eta\right\|_{L^{2}(\mathbb{R}_{+}^{2})}^{\frac{2}{p}}\left\|\nabla(\omega\varphi\eta)\right\|_{L^{2}(\mathbb{R}_{+}^{2})}^{1-\frac{2}{p}}$
$\displaystyle\leq C\left\|\omega\right\|_{L^{2}(\mathbb{R}_{+}^{2}\backslash
B_{2r_{0}}^{+})}^{\frac{2}{p}}\left(\left\|(\nabla\omega)\varphi\eta\right\|_{L^{2}(\mathbb{R}_{+}^{2})}^{1-\frac{2}{p}}+\left\|(\nabla\varphi)\omega\eta\right\|_{L^{2}(\mathbb{R}_{+}^{2})}^{1-\frac{2}{p}}+\left\|(\nabla\eta)\omega\varphi\right\|_{L^{2}(\mathbb{R}_{+}^{2})}^{1-\frac{2}{p}}\right)$
$\displaystyle\leq
C\left\|\nabla\boldsymbol{w}\right\|_{L^{2}(\mathbb{R}_{+}^{2}\backslash
B_{2r_{0}}^{+})}^{\frac{2}{p}}\left(C\left\|\nabla\omega\right\|_{L^{2}(\mathbb{R}_{+}^{2}\backslash
B_{2r_{0}}^{+})}^{1-\frac{2}{p}}+\frac{C}{R}+C\right)$ $\displaystyle\leq
C(1+R^{-1}).$

Since $\varphi=1$ as $R\to\infty.$ Hence let $R\to\infty,$ we get

$\left\|\omega\right\|_{L^{p}(\mathbb{R}_{+}^{2}\backslash
B_{3r_{0}}^{+})}\leq C.$

Moreover, due to $\boldsymbol{w}\in C^{2}(\overline{\Omega_{0}}),$ there holds
$\omega\in C^{1}(\overline{\Omega_{0}}).$ Then there holds

$\left\|\omega\right\|_{L^{p}(B_{3r_{0}}^{+}\backslash B_{r_{0}}^{+})}\leq C.$

Thence (3.1) holds.

Secondly, set $\hat{\boldsymbol{w}}(x,y)=(\hat{u}(x,y),\hat{v}(x,y)),$ where

$\hat{u}(x,y)=\left\\{\begin{aligned} u(x,y),~{}&y\geq 0\\\
u(x,-y),~{}&y<0\end{aligned}\right.,~{}~{}\hat{v}(x,y)=\left\\{\begin{aligned}
v(x,y),~{}&y\geq 0\\\ -v(x,-y),~{}&y<0\end{aligned}\right..$

Then

$\hat{\boldsymbol{w}}\in C_{loc}^{0,1}(\mathbb{R}^{2}\backslash B_{r_{0}}),$

and

$\hat{\boldsymbol{w}}=\boldsymbol{w},\quad{\rm~{}in~{}}\Omega_{0}.$

Moreover

$\hat{\omega}(x,y)=\left\\{\begin{aligned}
\partial_{y}u(x,y)-\partial_{x}v(x,y),~{}y\geq 0\\\
-(\partial_{y}u)(x,-y)+(\partial_{x}v)(x,-y),~{}y<0\end{aligned}\right..$

Let $\varphi$ is standard mollifier operator, $\varepsilon>0,$ set
$\varphi_{\varepsilon}=\varepsilon^{-2}\varphi(\frac{x}{\varepsilon}).$ Thus

$\varphi_{\varepsilon}*\hat{\boldsymbol{w}}=\hat{\boldsymbol{w}_{\varepsilon}}\in
C^{\infty}(\mathbb{R}^{2}\backslash B_{2r_{0}}).$

For the cut-off function $\phi=1$ outside $B_{3r_{0}}$ and $\phi=0$ in
$B_{2r_{0}}$, noting that

$\nabla^{\bot}\cdot(\hat{\boldsymbol{w}_{\varepsilon}}\phi)=\hat{\omega_{\varepsilon}}\phi+\hat{\boldsymbol{w}_{\varepsilon}}\cdot\nabla^{\bot}\phi,$

where

$\hat{\omega_{\varepsilon}}=\nabla^{\bot}\cdot\hat{\boldsymbol{w}_{\varepsilon}}.$

Putting the operation $\nabla^{\bot}$ into the both side of the above
equation, we have

$\Delta(\hat{\boldsymbol{w}_{\varepsilon}}\phi)=\nabla^{\bot}(\hat{\omega_{\varepsilon}}\phi+\hat{\boldsymbol{w}_{\varepsilon}}\cdot\nabla^{\bot}\phi)+\nabla[\nabla\cdot(\hat{\boldsymbol{w}_{\varepsilon}}\phi)].$
(3.2)

Noting that

$\nabla(\hat{\boldsymbol{w}_{\varepsilon}}\phi)=\nabla\hat{\boldsymbol{w}_{\varepsilon}}\to
0\quad{\rm as}\quad r\to\infty,$ (3.3)

by Calderon-Zygmund estimates, we have

$\|\nabla(\hat{\boldsymbol{w}_{\varepsilon}}\phi)\|_{L^{p}(\mathbb{R}^{2})}\leq
C\|(\hat{\omega_{\varepsilon}}\phi+\hat{\boldsymbol{w}_{\varepsilon}}\cdot\nabla^{\bot}\phi)\|_{L^{p}(\mathbb{R}^{2})}+C\|\nabla\cdot(\hat{\boldsymbol{w}_{\varepsilon}}\phi)\|_{L^{p}(\mathbb{R}^{2})},\quad
p>1.$

Noting that $\nabla\cdot\hat{\boldsymbol{w}_{\varepsilon}}=0,$ then by
triangle inequality, we have

$\|\nabla(\hat{\boldsymbol{w}_{\varepsilon}}\phi)|_{L^{p}(\mathbb{R}^{2})}\leq
C\|\hat{\boldsymbol{w}_{\varepsilon}}\|_{L^{p}(B_{3r_{0}}\setminus
B_{2r_{0}})}+C\|\hat{\omega_{\varepsilon}}\phi\|_{L^{p}(\mathbb{R}^{2})}.$

Since $\hat{\boldsymbol{w}_{\varepsilon}}\in C^{0,1}(\mathbb{R}^{2}\backslash
B_{2r_{0}})$ uniformly, there holds
$\|\hat{\boldsymbol{w}_{\varepsilon}}\|_{L^{p}(B_{3r_{0}}\setminus
B_{2r_{0}})}\leq C$. Using the definition of $\phi$, we have

$\|\nabla\hat{\boldsymbol{w}_{\varepsilon}}\|_{L^{p}(\mathbb{R}^{2}\setminus
B_{3r_{0}})}\leq
C+C\|\hat{\omega_{\varepsilon}}\phi\|_{L^{p}(\mathbb{R}^{2})}\leq
C+\|\hat{\omega}\|_{L^{p}(\mathbb{R}^{2}\setminus B_{2r_{0}})}.$

Let $\varepsilon\to 0$, by Lebesgue’s dominated convergence theorem and (3.1),
we have

$\displaystyle\|\nabla\hat{\boldsymbol{w}}\|_{L^{p}(\mathbb{R}^{2}\setminus
B_{3r_{0}})}$ $\displaystyle\leq
C+C\|\hat{\omega}\|_{L^{p}(\mathbb{R}^{2}\setminus B_{2r_{0}})}$
$\displaystyle\leq
C+C\left\|\omega\right\|_{L^{p}(\mathbb{R}_{+}^{2}\backslash
B_{2r_{0}}^{+})}\leq C,\quad\forall~{}~{}p\geq 2.$

Thus with the help of Lemma 2.6, we have

$\hat{\boldsymbol{w}}=o(\sqrt{\log r}),$

which implies

$\boldsymbol{w}=o(\sqrt{\log r}).$

The proof is complete.

##  4\. Liouville type theorem in the half-plane

Proof of Theorem 1.2. In the case of $\mathbb{R}_{+}^{2}$, it follows from
Lemma 2.4 that $\omega\to 0$ at infinity. By (1.5), $\omega(x,0)=0.$ Hence
$\omega\equiv 0$ in $\mathbb{R}_{+}^{2}$ due to the maximum principle. Noting
that $\Delta u=\omega_{y}$ and $\Delta v=-\omega_{x},$ there holds
$\Delta\boldsymbol{w}=(\Delta u,\Delta v)=0$ in $\mathbb{R}_{+}^{2}.$

Recall $B_{r}^{+}=\left\\{\sqrt{x^{2}+y^{2}}<r:y>0\right\\}$ and $\partial
B_{r}^{+}=\Gamma_{1}+\Gamma_{2},$ where

$\displaystyle\Gamma_{1}=\left\\{(x,y):-r\leq x\leq
r,y=0\right\\},\quad\Gamma_{2}=\left\\{(x,y):\sqrt{x^{2}+y^{2}}=r,y>0\right\\}.$

Due to (1.5), we get

$\int_{\Gamma_{1}}\boldsymbol{n}\cdot\nabla\boldsymbol{w}\cdot\boldsymbol{w}ds=-\int_{-r}^{r}\frac{\partial}{\partial
y}u(x,0)\cdot u(x,0)dx=0.$

Then by the integration by parts formula, we have

$\displaystyle 0$
$\displaystyle=\int_{B_{r}^{+}}-\Delta\boldsymbol{w}\cdot\boldsymbol{w}dxdy$
(4.1)
$\displaystyle=\int_{B_{r}^{+}}|\nabla\boldsymbol{w}|^{2}dxdy-\int_{\Gamma_{1}+\Gamma_{2}}\boldsymbol{n}\cdot\nabla\boldsymbol{w}\cdot\boldsymbol{w}ds$
$\displaystyle=\int_{B_{r}^{+}}|\nabla\boldsymbol{w}|^{2}dxdy-\frac{1}{2}\int_{0}^{\pi}\frac{\partial}{\partial
r}[|\boldsymbol{w}(r,\theta)|^{2}]d\theta\cdot r,$

where $\boldsymbol{n}$ denotes outward unit normal to $\partial B_{r}^{+}.$
Write

$G(r)=\frac{1}{2}\int_{0}^{\pi}|\boldsymbol{w}(r,\theta)|^{2}d\theta,$

and

$C_{0}=\int_{B_{r_{0}}^{+}}|\nabla\boldsymbol{w}|^{2}dxdy,$

where $r>r_{0}.$ Then

$rG^{{}^{\prime}}(r)\geq C_{0}$

for any $r>r_{0}.$ Integration it over $(r_{0},r),$ we have

$G(r)\geq G(r_{0})+C_{0}\log\frac{r}{r_{0}}.$

Lemma 2.1 shows that $G(r)=o(\log r),$ hence

$C_{0}=\int_{B_{r_{0}}^{+}}|\nabla\boldsymbol{w}|^{2}dxdy=0.$

Letting $r_{0}\to\infty,$ it follows that
$\nabla\boldsymbol{w}=\boldsymbol{0}$ in $\mathbb{R}_{+}^{2}$ and
$\boldsymbol{w}$ is constant.The proof is complete.

##  5\. Asymptotic behavior of the pressure

Since $\Delta u=\omega_{y}$ and $\Delta v=-\omega_{x},$ the Navier-Stokes
equations can be written in the form

$\left\\{\begin{aligned} \omega_{y}+(u+ay)v_{y}-(v+b)u_{y}-av&=p_{x},\\\
-\omega_{x}-(u+ay)v_{x}+(v+b)u_{x}&=p_{y},\\\
u_{x}+v_{y}&=0.\end{aligned}\right.$ (5.1)

Specially, let $a=0,$ we get

$\left\\{\begin{aligned} \omega_{y}+uv_{y}-(v+b)u_{y}&=p_{x},\\\
-\omega_{x}-uv_{x}+(v+b)u_{x}&=p_{y},\\\ u_{x}+v_{y}&=0,\end{aligned}\right.$
(5.2)

it follows that

$\displaystyle
p_{r}=\frac{1}{r}[\omega_{\theta}+uv_{\theta}-(v+b)u_{\theta}].$ (5.3)

The first lemma is about the convergence of the square norm for a subsequence,
which is simialr as in [19].

######  Lemma 5.1.

Let $\left\\{\boldsymbol{w},p\right\\}$ be a solution of the Navier-Stokes
equations (1.4) with the Navier-slip boundary condition (1.5). Assume that
$\boldsymbol{w}\in C^{2}(\overline{\Omega_{0}}),$ (1.7) hold and $a=0,$ there
exists a sequence $\left\\{R_{n}\right\\},R_{n}\in(2^{2^{n}},2^{2^{n+1}}),$
such that

$\displaystyle\lim_{n\to\infty}\int_{0}^{\pi}|p(R_{n},\theta)-\bar{p}(R_{n})|^{2}d\theta=0.$
(5.4)

Proof. We first show that for any $r_{1}>\max(r_{0},1)$

$\int_{r>r_{1},0<\theta<\pi}\frac{|\nabla p|^{2}}{\log r}dxdy<\infty.$

Using $(\ref{eq:component-NS})_{1,2}$ and Cauchy inequality

$\displaystyle|\nabla p|^{2}$ $\displaystyle=p_{x}^{2}+p_{y}^{2}$
$\displaystyle=[\omega_{y}+uv_{y}-(v+b)u_{y}]^{2}+[-\omega_{x}-uv_{x}+(v+b)u_{x}]^{2}$
$\displaystyle\leq 4|\nabla\omega|^{2}+8u^{2}|\nabla
v|^{2}+4(2v^{2}+2b^{2})|\nabla u|^{2}$ $\displaystyle\leq
4|\nabla\omega|^{2}+16|\boldsymbol{w}|^{2}|\nabla\boldsymbol{w}|^{2}+8b^{2}|\nabla\boldsymbol{w}|^{2}.$

From (1.8) in Theorem 1.1, we know $|\boldsymbol{w}(r,\theta)|^{2}=o({\log
r})$ , thus

$\displaystyle\int_{r>r_{1},0<\theta<\pi}\frac{|\nabla p|^{2}}{\log r}dxdy$
$\displaystyle\leq 4\int_{r>r_{1},0<\theta<\pi}\frac{|\nabla\omega|^{2}}{\log
r}dxdy+16\int_{r>r_{1},0<\theta<\pi}\frac{|\boldsymbol{w}|^{2}}{\log
r}|\nabla\boldsymbol{w}|^{2}dxdy+8b^{2}\int_{r>r_{1},0<\theta<\pi}\frac{|\nabla\boldsymbol{w}|^{2}}{\log
r}dxdy$ $\displaystyle\leq\frac{4}{\log
r_{1}}\int_{r>r_{1},0<\theta<\pi}|\nabla\omega|^{2}dxdy+C\int_{r>r_{1},0<\theta<\pi}|\nabla\boldsymbol{w}|^{2}dxdy+\frac{8b^{2}}{\log
r_{1}}\int_{r>r_{1},0<\theta<\pi}|\nabla\boldsymbol{w}|^{2}dxdy$
$\displaystyle\leq
C\left(\int_{r>r_{1},0<\theta<\pi}|\nabla\omega|^{2}dxdy+\int_{r>r_{1},0<\theta<\pi}|\nabla\boldsymbol{w}|^{2}dxdy\right).$

It follows from (2.4) in Lemma 2.3 and (1.7) that

$\int_{r>r_{1},0<\theta<\pi}\frac{|\nabla p|^{2}}{\log r}<\infty$

for any $r_{1}>\max(r_{0},1).$ By the integral theorem of the mean and
Wirtinger’s inequality there is an $R_{n}\in(2^{2^{n}},2^{2^{n+1}})$ such that

$\displaystyle\log 2\int_{0}^{\pi}|p(R_{n},\theta)-\bar{p}(R_{n})|^{2}d\theta$
$\displaystyle=\int_{2^{2^{n}}}^{2^{2^{n+1}}}\frac{1}{r\log
r}dr\int_{0}^{\pi}|p(r,\theta)-\bar{p}(r)|^{2}d\theta$
$\displaystyle\leq\int_{2^{2^{n}}}^{2^{2^{n+1}}}\int_{0}^{\pi}\frac{p_{\theta}^{2}}{r\log
r}d\theta dr$
$\displaystyle\leq\int_{2^{2^{n}}<r<2^{2^{n+1}},0<\theta<\pi}\frac{|\nabla
p|^{2}}{\log r}dxdy\to 0,\quad n\to\infty.$

Thus we have (5.4).

The next lemma is the uniform square convergence of the pressure, and we used
the Green function in the half-annulus.

######  Lemma 5.2.

Under the assumptions of Theorem 1.3, we have

$\displaystyle\lim_{r\to\infty}\int_{0}^{\pi}|p(r,\theta)-\bar{p}(r)|^{2}d\theta=0.$
(5.5)

Proof. Using (5.2), we find that

$\Delta p=2(u_{x}v_{y}-u_{y}v_{x}).$

The right member is absolutely integrable in $\Omega_{0}.$ It also follows
that

$\displaystyle\frac{1}{r}p_{\theta}=-\omega_{r}-uv_{r}+(v+b)u_{r},$ (5.6)

and

$\displaystyle\Delta\bar{p}$ $\displaystyle=$
$\displaystyle\int_{0}^{\pi}\triangle
p-\frac{1}{r^{2}}\frac{\partial^{2}p}{\partial\theta^{2}}d\theta$
$\displaystyle=$ $\displaystyle\int_{0}^{\pi}\triangle
pd\theta-\frac{1}{r}[-\omega_{r}-uv_{r}+(v+b)u_{r}]|_{0}^{\pi}$
$\displaystyle=$ $\displaystyle\int_{0}^{\pi}\triangle pd\theta,$

which is also absolutely integrable due to Navier-slip boundary condition
(1.5) and $bu(r,0)=bu(r,\pi)$. Hence

$\displaystyle H\equiv\Delta(p-\bar{p})\in L_{1},\quad in~{}~{}\Omega_{0}.$
(5.7)

Let $A_{nm}^{+}$ denote the upper half annulus $R_{n}<r<R_{m},0<\theta<\pi,$
the sequence of radii $R_{n}$ being defined as in Lemma 5.1. Writing
$r=R_{n}\bar{r},$ then $1<\bar{r}<\frac{R_{m}}{R_{n}}\triangleq R_{0}.$ Noting
that

$\Delta[p(R_{n}\bar{r},\theta)-\bar{p}(R_{n}\bar{r})]=R_{n}^{2}\Delta[p(r,\theta)-\bar{p}(r)]=R_{n}^{2}H(r,\theta).$

We have the representation

$\displaystyle p(r,\theta)-\bar{p}(r)=$ $\displaystyle
p(R_{n}\bar{r},\theta)-\bar{p}(R_{n}\bar{r})$ (5.8) $\displaystyle=$
$\displaystyle-\int_{1<\bar{r}<R_{0},0<\theta<\pi}G(\bar{r},\theta;\bar{\rho},\varphi)R_{n}^{2}H(R_{n}\bar{\rho},\varphi)\bar{\rho}d\bar{\rho}d\varphi$
$\displaystyle+\int_{\bar{\rho}=1,0<\theta<\pi}\frac{\partial
G}{\partial\rho}(\bar{r},\theta;\bar{\rho},\varphi)(p(R_{n}\bar{\rho},\varphi)-\bar{p}(R_{n}\bar{\rho}))\bar{\rho}d\varphi$
$\displaystyle-\int_{1<\bar{r}<R_{0},\theta=0}\frac{1}{\bar{\rho}}\frac{\partial
G}{\partial\varphi}(\bar{r},\theta;\bar{\rho},\varphi)(p(R_{n}\bar{\rho},\varphi)-\bar{p}(R_{n}\bar{\rho}))\bar{\rho}d\varphi$
$\displaystyle+\int_{1<\bar{r}<R_{0},\theta=\pi}\frac{1}{\bar{\rho}}\frac{\partial
G}{\partial\varphi}(\bar{r},\theta;\bar{\rho},\varphi)(p(R_{n}\bar{\rho},\varphi)-\bar{p}(R_{n}\bar{\rho}))\bar{\rho}d\varphi$
$\displaystyle-\int_{\bar{\rho}=R_{0},0<\theta<\pi}\frac{\partial
G}{\partial\rho}(\bar{r},\theta;\bar{\rho},\varphi)(p(R_{n}\bar{\rho},\varphi)-\bar{p}(R_{n}\bar{\rho}))\bar{\rho}d\varphi$
$\displaystyle=$ $\displaystyle J_{1}+J_{2}+J_{3}+J_{4}+J_{5},$

where $G=G(\bar{r},\theta;\bar{\rho},\varphi)$ is the harmonic Green’s
function for the upper half annulus $1<\bar{r}<R_{0},0<\theta<\pi.~{}~{}G$ can
be written in the form (see, for example, p.140, problem 4 with answer on
p.418 in [45])

$G(\bar{r},\theta;\bar{\rho},\varphi)=\left\\{\begin{aligned}
\sum_{k=1}^{\infty}\frac{(\bar{r}^{k}-\bar{r}^{-k})}{\pi
k(R_{0}^{k}-R_{0}^{-k})}[(\frac{R_{0}}{\bar{\rho}})^{k}-(\frac{\bar{\rho}}{R_{0}})^{k}]\sin
k\theta\sin k\varphi,\quad\bar{r}<\bar{\rho},\\\
\sum_{k=1}^{\infty}\frac{(\bar{\rho}^{k}-\bar{\rho}^{-k})}{\pi
k(R_{0}^{k}-R_{0}^{-k})}[(\frac{R_{0}}{\bar{r}})^{k}-(\frac{\bar{r}}{R_{0}})^{k}]\sin
k\theta\sin k\varphi,\quad\bar{r}>\bar{\rho}.\end{aligned}\right.$

Next,we use variable substitution for (5.8). Let $\rho=R_{n}\bar{\rho}$ and
noting that $R_{0}=\frac{R_{m}}{R_{n}},\bar{r}=\frac{r}{R_{n}}.$ For $J_{1},$
we have

$\displaystyle J_{1}=$
$\displaystyle-\int_{1<\bar{r}<R_{0},0<\theta<\pi}G(\bar{r},\theta;\bar{\rho},\varphi)R_{n}^{2}H(R_{n}\bar{\rho},\varphi)\bar{\rho}d\bar{\rho}d\varphi$
$\displaystyle=$
$\displaystyle-\int_{1<\bar{\rho}<\bar{r},0<\theta<\pi}\sum_{k=1}^{\infty}\frac{(\bar{\rho}^{k}-\bar{\rho}^{-k})}{\pi
k(R_{0}^{k}-R_{0}^{-k})}[(\frac{R_{0}}{\bar{r}})^{k}-(\frac{\bar{r}}{R_{0}})^{k}]\sin
k\theta\sin k\varphi R_{n}^{2}H(R_{n}\bar{\rho},\varphi)\bar{\rho}d\bar{\rho}$
$\displaystyle-\int_{\bar{r}<\bar{\rho}<R_{0},0<\theta<\pi}\sum_{k=1}^{\infty}\frac{(\bar{r}^{k}-\bar{r}^{-k})}{\pi
k(R_{0}^{k}-R_{0}^{-k})}[(\frac{R_{0}}{\bar{\rho}})^{k}-(\frac{\bar{\rho}}{R_{0}})^{k}]\sin
k\theta\sin k\varphi R_{n}^{2}H(R_{n}\bar{\rho},\varphi)\bar{\rho}d\bar{\rho}$
$\displaystyle=$
$\displaystyle-\int_{R_{n}<\rho<r,0<\theta<\pi}[-\sum_{k=1}^{\infty}\frac{(\rho^{k}-R_{n}^{2k}/\rho^{k})}{\pi
k(R_{m}^{2k}-R_{n}^{2k})}(r^{k}-R_{m}^{2k}/r^{k})\sin k\theta\sin
k\varphi]H(\rho,\varphi)\rho d\rho d\varphi$
$\displaystyle-\int_{r<\rho<R_{m},0<\theta<\pi}[-\sum_{k=1}^{\infty}\frac{(r^{k}-R_{n}^{2k}/r^{k})}{\pi
k(R_{m}^{2k}-R_{n}^{2k})}(\rho^{k}-R_{m}^{2k}/\rho^{k})\sin k\theta\sin
k\varphi]H(\rho,\varphi)\rho d\rho d\varphi.$

For $J_{2},$ noting that $\bar{\rho}=1<\bar{r},$ we get

$\displaystyle J_{2}$
$\displaystyle=\int_{\bar{\rho}=1,0<\theta<\pi}\frac{\partial
G}{\partial\rho}(\bar{r},\theta;\bar{\rho},\varphi)(p(R_{n}\bar{\rho},\varphi)-\bar{p}(R_{n}\bar{\rho}))\bar{\rho}d\varphi$
$\displaystyle=\int_{\bar{\rho}=1,0<\theta<\pi}\frac{\partial\left\\{\sum_{k=1}^{\infty}\frac{\bar{\rho}^{k}-\bar{\rho}^{-k}}{\pi
k(R_{0}^{k}-R_{0}^{-k})}[(\frac{R_{0}}{\bar{r}})^{k}-(\frac{\bar{r}}{R_{0}})^{k}]\sin
k\theta\sin
k\varphi\right\\}}{\partial\bar{\rho}}[p(R_{n}\bar{\rho},\varphi)-\bar{p}(R_{n}\bar{\rho})]\bar{\rho}d\varphi$
$\displaystyle=\int_{\rho=R_{n},0<\theta<\pi}\frac{\partial[-\sum_{k=1}^{\infty}\frac{(\rho^{k}-R_{n}^{2k}/\rho^{k})}{\pi
k(R_{m}^{2k}-R_{n}^{2k})}(r^{k}-R_{m}^{2k}/r^{k})\sin k\theta\sin
k\varphi]}{\partial\rho}[p(\rho,\varphi)-\bar{p}(\rho)]\rho d\varphi.$

For $J_{5},$ noting that $\bar{\rho}=R_{0}>\bar{r},$ we obtain

$\displaystyle J_{5}$
$\displaystyle=-\int_{\bar{\rho}=R_{0},0<\theta<\pi}\frac{\partial
G}{\partial\rho}(\bar{r},\theta;\bar{\rho},\varphi)(p(R_{n}\bar{\rho},\varphi)-\bar{p}(R_{n}\bar{\rho}))\bar{\rho}d\varphi$
$\displaystyle=-\int_{\bar{\rho}=R_{0},0<\theta<\pi}\frac{\partial\left\\{\sum_{k=1}^{\infty}\frac{\bar{r}^{k}-\bar{r}^{-k}}{\pi
k(R_{0}^{k}-R_{0}^{-k})}[(\frac{R_{0}}{\bar{\rho}})^{k}-(\frac{\bar{\rho}}{R_{0}})^{k}]\sin
k\theta\sin
k\varphi\right\\}}{\partial\bar{\rho}}[p(R_{n}\bar{\rho},\varphi)-\bar{p}(R_{n}\bar{\rho})]\bar{\rho}d\varphi$
$\displaystyle=-\int_{\rho=R_{m},0<\theta<\pi}\frac{\partial[-\sum_{k=1}^{\infty}\frac{(r^{k}-R_{n}^{2k}/r^{k})}{\pi
k(R_{m}^{2k}-R_{n}^{2k})}(\rho^{k}-R_{m}^{2k}/\rho^{k})\sin k\theta\sin
k\varphi]}{\partial\rho}[p(\rho,\varphi)-\bar{p}(\rho)]\rho d\varphi.$

Since $G(\bar{r},\theta;\bar{\rho},\varphi)=0$ for $\theta=0,\pi.$ It follows
that $J_{3}=J_{4}=0.$ Hence the representation of $G(r,\theta;\rho,\varphi)$
is follows:

$G(r,\theta;\rho,\varphi)=\left\\{\begin{aligned}
-\sum_{k=1}^{\infty}\frac{(r^{k}-R_{n}^{2k}/r^{k})}{\pi
k(R_{m}^{2k}-R_{n}^{2k})}(\rho^{k}-R_{m}^{2k}/\rho^{k})\sin k\theta\sin
k\varphi,\quad r<\rho,\\\
-\sum_{k=1}^{\infty}\frac{(\rho^{k}-R_{n}^{2k}/\rho^{k})}{\pi
k(R_{m}^{2k}-R_{n}^{2k})}(r^{k}-R_{m}^{2k}/r^{k})\sin k\theta\sin
k\varphi,\quad r>\rho.\end{aligned}\right.$

Moreover, (5.8) can be written in the following form

$\displaystyle p(r,\theta)-\bar{p}(r)$
$\displaystyle=-\int_{A_{nm}^{+}}G(r,\theta;\rho,\varphi)H(\rho,\varphi)\rho
d\rho d\varphi$ (5.9)
$\displaystyle+\int_{\rho=R_{n},0<\theta<\pi}\frac{\partial
G}{\partial\rho}(r,\theta;\rho,\varphi)(p(\rho,\varphi)-\bar{p}(\rho))\rho
d\varphi$ $\displaystyle-\int_{\rho=R_{m},0<\theta<\pi}\frac{\partial
G}{\partial\rho}(r,\theta;\rho,\varphi)(p(\rho,\varphi)-\bar{p}(\rho))\rho
d\varphi.$

Define

$\displaystyle G^{(2)}(r;\rho_{1},\varphi_{1};\rho_{2},\varphi_{2})$
$\displaystyle=\int_{0}^{\pi}G(r,\theta;\rho_{1},\varphi_{1})G(r,\theta;\rho_{2},\varphi_{2})d\theta$
$\displaystyle=\sum_{k=1}^{\infty}\frac{(r^{k}-R_{n}^{2k}/r^{k})^{2}(\rho_{1}^{k}-R_{m}^{2k}/\rho_{1}^{k})(\rho_{2}^{k}-R_{m}^{2k}/\rho_{2}^{k})\sin
k\varphi_{1}\sin k\varphi_{2}}{2\pi k^{2}(R_{m}^{2k}-R_{n}^{2k})^{2}}$

when $r<\rho_{1},r<\rho_{2},$ with similar expressions for the other cases.
Therefore

$\displaystyle|G^{(2)}(r;\rho_{1},\varphi_{1};\rho_{2},\varphi_{2})|$
$\displaystyle\leq\sum_{k=1}^{\infty}\frac{(r^{k}-R_{n}^{2k}/r^{k})^{2}|\rho_{1}^{k}-R_{m}^{2k}/\rho_{1}^{k}||\rho_{2}^{k}-R_{m}^{2k}/\rho_{2}^{k}|}{2\pi
k^{2}(R_{m}^{2k}-R_{n}^{2k})^{2}}$ $\displaystyle:=\hat{G}^{(2)}.$

Due to $r>R_{n},$ we find $(r^{k}-R_{n}^{2k}/r^{k})^{2}$ is monotonically
increasing with respect to $r.$ When $r=\rho_{1}$ or $r=\rho_{2},$
$\hat{G}^{(2)}$ attains at the maximum. If $\rho_{1}<\rho_{2},$ then

$\hat{G}^{(2)}\leq\sum_{k=1}^{\infty}\frac{(\rho_{1}^{k}-R_{n}^{2k}/\rho_{1}^{k})^{2}|\rho_{1}^{k}-R_{m}^{2k}/\rho_{1}^{k}||\rho_{2}^{k}-R_{m}^{2k}/\rho_{2}^{k}|}{2\pi
k^{2}(R_{m}^{2k}-R_{n}^{2k})^{2}}.$

Noting that $|\rho_{2}^{k}-R_{m}^{2k}/\rho_{2}^{k}|$ is monotonically
decreasing with respect to $\rho_{2},$ $\hat{G}^{(2)}$ reaches its maximum at
$\rho_{1}=\rho_{2}.$ Let

$h(r)=(r^{k}-\frac{R_{n}^{2k}}{r^{k}})(r^{k}-\frac{R_{m}^{2k}}{r^{k}}).$

An easy computation which lets $h^{\prime}(r)=0$ shows that
$r=(R_{m}R_{n})^{\frac{1}{2}}.$ In conclusion, the maximum with respect to
$\rho_{1}$ and $\rho_{2}$ of $\hat{G}^{(2)}$ occurs when
$\rho_{1}=\rho_{2}=r=(R_{m}R_{n})^{\frac{1}{2}}.$ Thus

$\displaystyle|G^{(2)}(r;\rho_{1},\varphi_{1};\rho_{2},\varphi_{2})|$
$\displaystyle\leq\sum_{k=1}^{\infty}\frac{[(R_{m}R_{n})^{\frac{k}{2}}-R_{n}^{2k}/(R_{m}R_{n})^{\frac{k}{2}}]^{2}[(R_{m}R_{n})^{\frac{k}{2}}-R_{m}^{2k}/(R_{m}R_{n})^{\frac{k}{2}}]^{2}}{2\pi
k^{2}(R_{m}^{2k}-R_{n}^{2k})^{2}}$
$\displaystyle=\sum_{k=1}^{\infty}\frac{[R_{m}^{-\frac{k}{2}}R_{n}^{\frac{k}{2}}(R_{m}^{k}-R_{n}^{k})]^{2}[R_{m}^{\frac{k}{2}}R_{n}^{-\frac{k}{2}}(R_{n}^{k}-R_{m}^{k})]^{2}}{2\pi
k^{2}(R_{m}^{k}+R_{n}^{k})^{2}(R_{m}^{k}-R_{n}^{k})^{2}}$
$\displaystyle=\sum_{k=1}^{\infty}\frac{(R_{m}^{k}-R_{n}^{k})^{2}}{2\pi
k^{2}(R_{m}^{k}+R_{n}^{k})^{2}}\leq\sum_{k=1}^{\infty}\frac{1}{k^{2}}\equiv
C_{1}.$

Also

$\displaystyle\frac{\partial
G(r,\theta;\rho,\varphi)}{\partial\rho}|_{\rho=R_{m}}$
$\displaystyle=\frac{\partial[-\sum_{k=1}^{\infty}\frac{(r^{k}-R_{n}^{2k}/r^{k})}{\pi
k(R_{m}^{2k}-R_{n}^{2k})}(\rho^{k}-R_{m}^{2k}/\rho^{k})\sin k\theta\sin
k\varphi]}{\partial\rho}|_{\rho=R_{m}}$
$\displaystyle=-\sum_{k=1}^{\infty}\frac{(r^{k}-R_{n}^{2k}/r^{k})(k\rho^{k-1}+kR_{m}^{2k}/\rho^{k+1})}{\pi
k(R_{m}^{2k}-R_{n}^{2k})}\sin k\theta\sin k\varphi|_{\rho=R_{m}}$
$\displaystyle=-\sum_{k=1}^{\infty}\frac{2(r^{k}-R_{n}^{2k}/r^{k})R_{m}^{k-1}}{\pi(R_{m}^{2k}-R_{n}^{2k})}\sin
k\theta\sin k\varphi$

and thus, if $R_{n}<r\leq R_{m-2},0<\theta<\pi,$ we have

$\displaystyle\int_{0}^{\pi}|G_{\rho}(r,\theta;R_{m},\varphi)|^{2}R_{m}^{2}d\varphi$
$\displaystyle=\sum_{k=1}^{\infty}\frac{4(r^{k}-R_{n}^{2k}/r^{k})^{2}R_{m}^{2k}}{\pi^{2}(R_{m}^{2k}-R_{n}^{2k})^{2}}\sin^{2}k\theta\int_{0}^{\pi}\sin^{2}k\varphi
d\varphi$
$\displaystyle\leq\sum_{k=1}^{\infty}\frac{2(r^{k}-R_{n}^{2k}/r^{k})^{2}R_{m}^{2k}}{\pi(R_{m}^{2k}-R_{n}^{2k})^{2}}$
$\displaystyle\leq\sum_{k=1}^{\infty}\frac{2(R_{m-2}/R_{m})^{2k}}{\pi[1-(R_{n}/R_{m})^{2k}]^{2}}$
$\displaystyle\leq\frac{\sum_{k=1}^{\infty}2^{-2^{m}k}}{[1-(R_{n}/R_{m})^{2}]^{2}}\leq
C_{2}.$

Similarly, if $R_{n+2}\leq r<R_{m},0<\theta<\pi$, we have

$\frac{\partial
G(r,\theta;\rho,\varphi)}{\partial\rho}|_{\rho=R_{n}}=-\sum_{k=1}^{\infty}\frac{2(r^{k}-R_{m}^{2k}/r^{k})R_{n}^{k-1}}{\pi(R_{m}^{2k}-R_{n}^{2k})}\sin
k\theta\sin k\varphi$

and

$\displaystyle\int_{0}^{\pi}|G_{\rho}(r,\theta;R_{n},\varphi)|^{2}R_{n}^{2}d\varphi$
$\displaystyle=\sum_{k=1}^{\infty}\frac{4(r^{k}-R_{m}^{2k}/r^{k})^{2}R_{n}^{2k}}{\pi^{2}(R_{m}^{2k}-R_{n}^{2k})^{2}}\sin^{2}k\theta\int_{0}^{\pi}\sin^{2}k\varphi
d\varphi$
$\displaystyle\leq\sum_{k=1}^{\infty}\frac{2(R_{n}/R_{n+2})^{2k}}{\pi[1-(R_{n}/R_{m})^{2k}]^{2}}$
$\displaystyle\leq\frac{\sum_{k=1}^{\infty}2^{-2^{n}k}}{[1-(R_{n}/R_{m})^{2}]^{2}}\leq
C_{3}.$

By H$\ddot{\rm{o}}$lder inequality, it follows from (5.9) that for
$r\in[R_{n+2},R_{m-2}],0<\theta<\pi,m\geq n+5$

$\displaystyle\int_{0}^{\pi}|p(r,\theta)-\bar{p}(r)|^{2}d\theta$
$\displaystyle\leq$ $\displaystyle
3\int_{0}^{\pi}[\int_{0}^{\pi}\int_{R_{n}}^{R_{m}}G(r,\theta;\rho,\varphi)H(\rho,\varphi)\rho
d\rho d\varphi]^{2}d\theta$
$\displaystyle+3\int_{0}^{\pi}[\int_{0}^{\pi}\frac{\partial
G}{\partial\rho}(r,\theta;R_{n},\varphi)(p(R_{n},\varphi)-\bar{p}(R_{n}))R_{n}d\varphi]^{2}d\theta$
$\displaystyle+3\int_{0}^{\pi}[\int_{0}^{\pi}\frac{\partial
G}{\partial\rho}(r,\theta;R_{m},\varphi)(p(R_{m},\varphi)-\bar{p}(R_{m}))R_{m}d\varphi]^{2}d\theta$
$\displaystyle\leq$ $\displaystyle
3\int_{0}^{\pi}\int_{R_{n}}^{R_{m}}\int_{0}^{\pi}\int_{R_{n}}^{R_{m}}[\int_{0}^{\pi}G(r,\theta;\rho_{1},\varphi_{1})G(r,\theta;\rho_{2},\varphi_{2})d\theta]H(\rho_{1},\varphi_{1})H(\rho_{2},\varphi_{2})\rho_{1}\rho_{2}d\rho_{1}d\varphi_{1}d\rho_{2}d\varphi_{2}$
$\displaystyle+3\int_{0}^{\pi}[\int_{0}^{\pi}|\frac{\partial
G}{\partial\rho}(r,\theta;R_{n},\varphi)|^{2}R_{n}^{2}d\varphi\cdot\int_{0}^{\pi}|p(R_{n},\varphi)-\bar{p}(R_{n})|^{2}d\varphi]d\theta$
$\displaystyle+3\int_{0}^{\pi}[\int_{0}^{\pi}|\frac{\partial
G}{\partial\rho}(r,\theta;R_{m},\varphi)|^{2}R_{m}^{2}d\varphi\cdot\int_{0}^{\pi}|p(R_{m},\varphi)-\bar{p}(R_{m})|^{2}d\varphi]d\theta$
$\displaystyle\leq$ $\displaystyle 3\pi
C_{1}(\int_{R_{n}<r<R_{m},0<\theta<\pi}|H|dxdy)^{2}+3\pi
C_{3}\int_{0}^{\pi}|p(R_{n},\theta)-\bar{p}(R_{n})|^{2}d\varphi$
$\displaystyle+3\pi
C_{2}\int_{0}^{\pi}|p(R_{m},\varphi)-\bar{p}(R_{m})|^{2}d\varphi.$

By letting $m\to\infty$ and using (5.4) and (5.7), we obtain an upper bound on
the left member for $r>2^{2^{n+2}},$ and this bound approaches zero as
$n\to\infty.$ From this we infer

$\lim_{r\to\infty}\int_{0}^{\pi}|p(r,\theta)-\bar{p}(r)|^{2}d\theta=0.$

The third lemma is about the convergence of the average pressure.

######  Lemma 5.3.

Under the assumptions of Theorem 1.3, the average pressure

$\bar{p}(r)=\frac{1}{\pi}\int_{0}^{\pi}p(r,\theta)d\theta$

has a limit at infinity

$\lim_{r\to\infty}\bar{p}(r)=p_{\infty}<\infty.$ (5.10)

Proof. First, Navier-slip boundary condition tells us that $\omega(x,0)=0$ and
$v(x,0)=0$ due to (1.5). Hence

$\int_{0}^{\pi}\omega_{\theta}d\theta=0,$

and

$\int_{0}^{\pi}(uv_{\theta}+vu_{\theta})d\theta=\int_{0}^{\pi}\bar{u}v_{\theta}d\theta=0,$

where

$\bar{u}(r)=\frac{1}{\pi}\int_{0}^{\pi}u(r,\theta)d\theta.$

Besides, due to $bu(r,0)=bu(r,\pi)$, we average (5.3) to find that

$\displaystyle\bar{p}^{\prime}(r)$ $\displaystyle=\frac{1}{\pi
r}\int_{0}^{\pi}[\omega_{\theta}+uv_{\theta}-(v+b)u_{\theta}]d\theta$ (5.11)
$\displaystyle=\frac{1}{\pi
r}\int_{0}^{\pi}[2uv_{\theta}-(vu_{\theta}+uv_{\theta})]d\theta$
$\displaystyle=\frac{2}{\pi r}\int_{0}^{\pi}(u-\bar{u})v_{\theta}d\theta.$

Integrating this inequality with respect to $r$ over
$(r_{1},r_{2})~{}~{}(r_{2}\geq r_{1}\geq r_{0}),$ then by Cauchy and Wirtinger
inequalities, we find

$\displaystyle|\bar{p}(r_{2})-\bar{p}(r_{1})|$
$\displaystyle=|\frac{2}{\pi}\int_{r_{1}}^{r_{2}}\int_{0}^{\pi}\frac{(u-\bar{u})v_{\theta}}{r}d\theta
dr|$
$\displaystyle\leq\frac{1}{\pi}\int_{r_{1}}^{r_{2}}\int_{0}^{\pi}\frac{|\boldsymbol{w}-\bar{\boldsymbol{w}}|^{2}+|\boldsymbol{w}_{\theta}|^{2}}{r}d\theta
dr$ $\displaystyle\leq
C\int_{r_{1}}^{r_{2}}\int_{0}^{\pi}\frac{|\boldsymbol{w}_{\theta}|^{2}}{r}d\theta
dr$ $\displaystyle\leq C\int_{r>r_{1}}|\nabla\boldsymbol{w}|^{2}dxdy.$

Since the right member of this inequality tends to zero as $r_{1}\to\infty,$
it follows that $\bar{p}(r)$ has a limit $p_{\infty},$ as asserted.

Immediately it follows from Lemma 5.3 and Lemma 5.2 that the following
conclusion holds.

######  Corollary 5.1.

Under the assumptions of Theorem 1.3, we have

$\displaystyle\lim_{r\to\infty}\int_{0}^{\pi}|p(r,\theta)-p_{\infty}|^{2}d\theta=0.$
(5.12)

Proof of Case (i) in Theorem 1.3. For $r>r_{0},$ define

$\tilde{p}(\tilde{x})=r^{2}p(r\tilde{x})=r^{2}p(x),$

where $x\in B_{2r}^{+}\backslash B_{r}^{+}$ and $r$ is large enough. Without
loss of generality, we still consider $B_{r}^{+}$, since one can mollify the
domain such that it’s regular. By Lemma 2.5, we have

$\displaystyle\left\|\tilde{p}\right\|_{L^{\infty}(B_{\frac{7}{4}}^{+}\backslash
B_{\frac{5}{4}}^{+})}$ $\displaystyle\leq
C(1+\left\|\tilde{p}\right\|_{H^{1}(B_{2}^{+}\backslash
B_{1}^{+})})\sqrt{\log(e+\left\|\Delta\tilde{p}\right\|_{L^{2}(B_{2}^{+}\backslash
B_{1}^{+})})}$
$\displaystyle=C(1+\left\|\tilde{p}\right\|_{L^{2}(B_{2}^{+}\backslash
B_{1}^{+})}+\left\|\nabla\tilde{p}\right\|_{L^{2}(B_{2}^{+}\backslash
B_{1}^{+})})\sqrt{\log(e+\left\|\Delta\tilde{p}\right\|_{L^{2}(B_{2}^{+}\backslash
B_{1}^{+})})}.$

Due to the scaling, we have

$\left\|\tilde{p}\right\|_{L^{\infty}(B_{2}^{+}\backslash
B_{1}^{+})}=r^{2}\left\|p\right\|_{L^{\infty}(B_{2r}^{+}\backslash
B_{r}^{+})},$ $\left\|\tilde{p}\right\|_{L^{2}(B_{2}^{+}\backslash
B_{1}^{+})}=r\left\|p\right\|_{L^{2}(B_{2r}^{+}\backslash B_{r}^{+})},$
$\left\|\nabla\tilde{p}\right\|_{L^{2}(B_{2}^{+}\backslash
B_{1}^{+})}=r^{2}\left\|\nabla p\right\|_{L^{2}(B_{2r}^{+}\backslash
B_{r}^{+})},$ $\left\|\nabla^{2}\tilde{p}\right\|_{L^{2}(B_{2}^{+}\backslash
B_{1}^{+})}=r^{3}\left\|\nabla^{2}p\right\|_{L^{2}(B_{2r}^{+}\backslash
B_{r}^{+})}.$

Hence

$\displaystyle
r^{2}\left\|p\right\|_{L^{\infty}(B_{\frac{7}{4}r}^{+}\backslash
B_{\frac{5}{4}r}^{+})}\leq$ $\displaystyle
C(1+r\left\|p\right\|_{L^{2}(B_{2r}^{+}\backslash
B_{r}^{+})}+r^{2}\left\|\nabla p\right\|_{L^{2}(B_{2r}^{+}\backslash
B_{r}^{+})})$ (5.13) $\displaystyle\cdot\sqrt{\log(e+r^{3}\left\|\Delta
p\right\|_{L^{2}(B_{2r}^{+}\backslash B_{r}^{+})})}.$

From Lemma 5.3 we know $|p_{\infty}|\leq C$ and by (5.12) in Corollary 5.1 we
have

$\displaystyle\left\|p\right\|_{L^{2}(B_{2r}^{+}\backslash B_{r}^{+})}^{2}$
$\displaystyle\leq
2\int_{r}^{2r}\int_{0}^{\pi}|p(\rho,\theta)-p_{\infty}|^{2}d\theta\rho
d\rho+2\int_{r}^{2r}\int_{0}^{\pi}|p_{\infty}|^{2}\rho d\theta d\rho$
$\displaystyle\leq o(r)\int_{r}^{2r}\rho d\rho+Cr^{2}.$

Then

$\left\|p\right\|_{L^{2}(B_{2r}^{+}\backslash B_{r}^{+})}^{2}\leq Cr^{2}.$

Since $|\Delta\boldsymbol{w}|^{2}=|\nabla\omega|^{2},$ then by Lemma 2.3,
$\boldsymbol{w}=o(\sqrt{\log r})$ and Navier-Stokes equation
$(\ref{eq:NS})_{1},$ there holds

$\displaystyle\left\|\nabla p\right\|_{L^{2}(B_{2r}^{+}\backslash
B_{r}^{+})}^{2}$ $\displaystyle\leq C\int_{B_{2r}^{+}\backslash
B_{r}^{+}}|\Delta\boldsymbol{w}|^{2}dxdy+C\int_{B_{2r}^{+}\backslash
B_{r}^{+}}|\boldsymbol{w}\cdot\nabla\boldsymbol{w}|^{2}dxdy+C\int_{B_{2r}^{+}\backslash
B_{r}^{+}}|\nabla\boldsymbol{w}|^{2}dxdy$ $\displaystyle\leq
C\int_{B_{2r}^{+}\backslash B_{r}^{+}}|\nabla\omega|^{2}dxdy+C[o(\log
r)+1]\int_{B_{2r}^{+}\backslash B_{r}^{+}}|\nabla\boldsymbol{w}|^{2}dxdy$
$\displaystyle\leq C[1+o(\log r)].$

Besides, similar to the second step of Theorem 1.1, using (3.2) and (3.3),we
also get

$\displaystyle\|\nabla^{2}(\hat{\boldsymbol{w}_{\varepsilon}}\phi)\|_{L^{2}(\mathbb{R}^{2})}$
$\displaystyle\leq
C\|\nabla(\hat{\omega_{\varepsilon}}\phi+\hat{\boldsymbol{w}_{\varepsilon}}\cdot\nabla^{\bot}\phi)\|_{L^{2}(\mathbb{R}^{2})}+C\|\nabla[\nabla\cdot(\hat{\boldsymbol{w}_{\varepsilon}}\phi)]\|_{L^{2}(\mathbb{R}^{2})}$
$\displaystyle=C\|\phi\nabla\hat{\omega_{\varepsilon}}+\hat{\omega_{\varepsilon}}\nabla\phi+\nabla(\hat{\boldsymbol{w}_{\varepsilon}}\cdot\nabla^{\bot}\phi)\|_{L^{2}(\mathbb{R}^{2})}+C\|\nabla[(\nabla\phi)\hat{\boldsymbol{w}_{\varepsilon}}]\|_{L^{2}(\mathbb{R}^{2})}$
$\displaystyle\leq
C\|\nabla\hat{\omega_{\varepsilon}}\|_{L^{2}(\mathbb{R}^{2}\backslash
B_{2r_{0}})}+C\|\hat{\omega_{\varepsilon}}\|_{L^{2}(B_{3r_{0}}\backslash
B_{2r_{0}})}+C\|\nabla\hat{\boldsymbol{w}_{\varepsilon}}\|_{L^{2}(B_{3r_{0}\backslash}B_{2r_{0}})}.$

Since $\hat{\omega_{\varepsilon}}\in C^{\infty}(\mathbb{R}^{2}\backslash
B_{2r_{0}}),$ we know
$\|\hat{\omega_{\varepsilon}}\|_{L^{2}(B_{3r_{0}}\backslash B_{2r_{0}})}\leq
C.$ Using (1.7), we have
$\|\nabla\hat{\boldsymbol{w}_{\varepsilon}}\|_{L^{2}(B_{3r_{0}\backslash}B_{2r_{0}})}\leq
C.$ Therefore

$\left\|\nabla^{2}\hat{\boldsymbol{w}_{\varepsilon}}\right\|_{L^{2}(\mathbb{R}^{2}\backslash
B_{3r_{0}})}\leq
C+C\|\nabla\hat{\omega_{\varepsilon}}\|_{L^{2}(\mathbb{R}^{2}\backslash
B_{2r_{0}})}\leq C+C\|\nabla\hat{\omega}\|_{L^{2}(\mathbb{R}^{2}\backslash
B_{2r_{0}})}.$

Let $\varepsilon\to 0,$ by Lebesgue’s dominated convergence theorem and Lemma
2.3, we have

$\displaystyle\|\nabla^{2}\hat{\boldsymbol{w}}\|_{L^{2}(\mathbb{R}^{2}_{+}\backslash
B_{3r_{0}})}$ $\displaystyle\leq
C+C\|\nabla\hat{\omega}\|_{L^{2}(\mathbb{R}^{2}\backslash B_{2r_{0}})}$
$\displaystyle\leq C+C\|\nabla\omega\|_{L^{2}(\mathbb{R}_{+}^{2}\backslash
B_{2r_{0}}^{+})}\leq C.$

Then by $(\ref{eq:component-NS})_{1,2}$ and Gagliardo-Nirenberg inequality, we
have

$\displaystyle\int_{B_{2r}^{+}\backslash B_{r}^{+}}|\Delta p|^{2}dxdy$
$\displaystyle\leq$ $\displaystyle C\int_{B_{2r}^{+}\backslash
B_{r}^{+}}(u_{x}^{2}v_{y}^{2}+u_{y}^{2}v_{x}^{2})dxdy$ $\displaystyle\leq$
$\displaystyle C\int_{B_{2r}^{+}\backslash
B_{r}^{+}}|\nabla\boldsymbol{w}|^{4}dxdy$ $\displaystyle\leq$ $\displaystyle
C(\int_{B_{2r}^{+}\backslash
B_{r}^{+}}|\nabla\boldsymbol{w}|^{2}dxdy)(\int_{B_{2r}^{+}\backslash
B_{r}^{+}}|\nabla^{2}\boldsymbol{w}|^{2}dxdy)+Cr^{-2}(\int_{B_{2r}^{+}\backslash
B_{r}^{+}}|\nabla\boldsymbol{w}|^{2}dxdy)^{2}$ $\displaystyle\leq$
$\displaystyle C(1+r^{-2}).$

Therefore using (5.13), one can get

$\displaystyle
r^{2}\left\|p\right\|_{L^{\infty}(B_{\frac{7}{4}r}^{+}\backslash
B_{\frac{5}{4}r}^{+})}$ $\displaystyle\leq C(1+Cr^{2}+Cr^{2}\sqrt{1+o(\log
r)})\sqrt{\log(e+Cr^{3}\sqrt{1+r^{-2}})}$ $\displaystyle\leq r^{2}o(\log r),$

which implies

$\left\|p\right\|_{L^{\infty}(B_{\frac{7}{4}r}^{+}\backslash
B_{\frac{5}{4}r}^{+})}\leq o(\log r).$

The proof is complete.

Proof of Case (ii) in Theorem 1.3. Let the point $P(2R,\theta)$ be the origin
of a new system of polar coordinates $(r^{\prime},\theta^{\prime})$ and
suppose that $R>r_{0}.$ In these new coordinates we still have

$p_{r^{\prime}}=\frac{1}{r^{\prime}}[\omega_{\theta^{\prime}}+uv_{\theta^{\prime}}-(v+b)u_{\theta^{\prime}}]$

from the Navier-Stokes equations. Integrating with respect to $r^{\prime}$
over $(0,r^{\prime})$

$p(P)=p(r^{\prime},\theta^{\prime})+\int_{0}^{r^{\prime}}\frac{1}{\rho}[(v+b)u_{\theta^{\prime}}-uv_{\theta^{\prime}}-\omega_{\theta^{\prime}}]d\rho.$
(5.14)

$\mathbf{Proof~{}of~{}Case~{}I:}$ $p(2R,\theta)$ with $\theta=0,\pi.$ As shown
in Figure 1:

$x$$y$$0$$R$$2R$$3R$$4R$$-R$$-2R$$-3R$$-4R$$R$$2R$$3R$$4R$$P_{1}(2R,0)$$P_{2}(2R,\pi)$
Figure 1. The case of $\theta=0,\pi.$

let’s consider the case of $\theta=0,$ and $\theta=\pi$ is similar.
Integrating (5.14) with respect to $\theta^{\prime}$ over $[0,\pi]$

$\pi
p(P)=\int_{0}^{\pi}p(r^{\prime},\theta^{\prime})d\theta^{\prime}+\int_{0}^{r^{\prime}}\int_{0}^{\pi}\frac{1}{\rho}[(v+b)u_{\theta^{\prime}}-uv_{\theta^{\prime}}-\omega_{\theta^{\prime}}]d\rho
d\theta^{\prime}.$

Due to $\omega(x,0)=0$,$v(x,0)=0$ and $bu(r,0)=bu(r,\pi)$. Similar to the
(5.11) calculation

$\displaystyle
p(P)=\frac{1}{\pi}\int_{0}^{\pi}p(r^{\prime},\theta^{\prime})d\theta^{\prime}+\frac{2}{\pi}\int_{0}^{r^{\prime}}\int_{0}^{\pi}\frac{1}{\rho}[\tilde{u}(\rho)-u(\rho,\theta^{\prime})]v_{\theta^{\prime}}(\rho,\theta^{\prime})d\rho
d\theta^{\prime},$

where

$\tilde{u}(r^{\prime})=\frac{1}{\pi}\int_{0}^{\pi}u(r^{\prime},\theta^{\prime})d\theta^{\prime}.$

Multiply this relation by $r^{\prime}$ and integrate from $0$ to $R$, and we
find

$\displaystyle p(P)$ $\displaystyle=\frac{2}{\pi
R^{2}}\int_{0}^{R}\int_{0}^{\pi}p(r^{\prime},\theta^{\prime})r^{\prime}dr^{\prime}d\theta^{\prime}$
(5.15) $\displaystyle+\frac{4}{\pi
R^{2}}\int_{0}^{R}\int_{0}^{r^{\prime}}\int_{0}^{\pi}\frac{(\tilde{u}-u)v_{\theta^{\prime}}}{\rho}r^{\prime}d\theta^{\prime}d\rho
dr^{\prime}\doteq I_{1}+I_{2}.$

Note that the upper half disc $r^{\prime}<R,0<\theta^{\prime}<\pi$ is
contained in the upper half annulus $R<r<3R,0<\theta<\pi.$ For $I_{1},$ using
Schwarz inequality and (5.12)

$\displaystyle|I_{1}|^{2}$ $\displaystyle\leq(\frac{2}{\pi
R^{2}})^{2}(\int_{0}^{R}\int_{0}^{\pi}p^{2}r^{\prime}dr^{\prime}d\theta^{\prime})(\int_{0}^{R}\int_{0}^{\pi}r^{\prime}dr^{\prime}d\theta^{\prime})$
$\displaystyle\leq\frac{2}{\pi
R^{2}}\int_{R}^{3R}rdr\left\\{\max_{R<r<3R}\int_{0}^{\pi}p(r,\theta)^{2}d\theta\right\\}$
$\displaystyle=\frac{8}{\pi}\max_{R<r<3R}\int_{0}^{\pi}p(r,\theta)^{2}d\theta\to
0,\quad{\rm as}~{}~{}R\to\infty.$

For $I_{2},$ using Cauchy and Wirtinger inequalities

$\displaystyle|\int_{0}^{\pi}(\tilde{u}-u)v_{\theta^{\prime}}d\theta^{\prime}|$
$\displaystyle\leq\int_{0}^{\pi}\frac{|u-\tilde{u}|^{2}+|v_{\theta^{\prime}}|^{2}}{2}d\theta^{\prime}$
$\displaystyle\leq\int_{0}^{\pi}\frac{|\boldsymbol{w}-\tilde{\boldsymbol{w}}|^{2}+|\boldsymbol{w}_{\theta^{\prime}}|^{2}}{2}d\theta^{\prime}$
$\displaystyle\leq\int_{0}^{\pi}|\boldsymbol{w}_{\theta^{\prime}}|^{2}d\theta^{\prime}\leq\int_{0}^{\pi}\rho^{2}|\nabla\boldsymbol{w}|^{2}d\theta^{\prime}.$

Consequently, we have

$\displaystyle I_{2}$ $\displaystyle\leq\frac{4}{\pi
R^{2}}\int_{0}^{R}\int_{0}^{r^{\prime}}\int_{0}^{\pi}|\nabla\boldsymbol{w}(\rho,\theta^{\prime})|^{2}\rho
d\rho d\theta^{\prime}r^{\prime}dr^{\prime}$ $\displaystyle\leq\frac{4}{\pi
R^{2}}\int_{R<r<3R,0<\theta<\pi}|\nabla\boldsymbol{w}|^{2}dxdy\int_{0}^{R}\rho
d\rho$
$\displaystyle=\frac{2}{\pi}\int_{R<r<3R,0<\theta<\pi}|\nabla\boldsymbol{w}|^{2}dxdy\to
0,\quad{\rm as}~{}~{}R\to\infty.$

It follows from (5.15) that

$\displaystyle|p(2R,0)|\to 0,$

as $R\to\infty.$ Thus the proof of Case I is complete.

$\mathbf{Proof~{}of~{}Case~{}II:}$ $P(2R,\theta)$ with $\theta\in(0,\pi).$ As
shown in Figure 2:

$x$$y$$R$$2R$$3R$$0$$-R$$-2R$$-3R$$R$$2R$$3R$$\theta$$p(2R,\theta)$$Q_{1}$$Q_{2}$$Q_{1}=(2R-R\sin\theta,0)$$Q_{2}=(2R+R\sin\theta,0)$$2R\sin\theta$$R\sin\theta$
Figure 2. The case of $\theta\in(0,\pi).$

Integrating (5.14) with respect to $\theta^{\prime}$ over $[0,2\pi]$

$\displaystyle
p(P)=\frac{1}{2\pi}\int_{0}^{2\pi}p(r^{\prime},\theta^{\prime})d\theta^{\prime}+\frac{1}{\pi}\int_{0}^{r^{\prime}}\int_{0}^{2\pi}\frac{1}{\rho}[\hat{u}(\rho)-u(\rho,\theta^{\prime})]v_{\theta^{\prime}}(\rho,\theta^{\prime})d\rho
d\theta^{\prime},$

where

$\hat{u}(r^{\prime})=\frac{1}{2\pi}\int_{0}^{2\pi}u(r^{\prime},\theta^{\prime})d\theta^{\prime}.$

Multiply this relation by $r^{\prime}$ and integrate from $0$ to $R\sin\theta$

$\displaystyle p(P)=$ $\displaystyle\frac{1}{\pi
R^{2}\sin^{2}\theta}\int_{0}^{R\sin\theta}\int_{0}^{2\pi}p(r^{\prime},\theta^{\prime})r^{\prime}dr^{\prime}d\theta^{\prime}$
(5.16) $\displaystyle+\frac{2}{\pi
R^{2}\sin^{2}\theta}\int_{0}^{R\sin\theta}\int_{0}^{r^{\prime}}\int_{0}^{2\pi}\frac{(\hat{u}-u)v_{\theta^{\prime}}}{\rho}r^{\prime}d\theta^{\prime}d\rho
dr^{\prime}\doteq$ $\displaystyle I_{1}^{\prime}+I_{2}^{\prime}.$

Noting that the disc $r^{\prime}<R\sin\theta,0<\theta^{\prime}<2\pi$ is
contained in the upper half annulus
$2R-R\sin\theta<r<2R+R\sin\theta,0<\theta<\pi.$ Then similar to the Case I,
for $I_{1}^{\prime},$ by Schwarz’s inequality and (5.12), we get

$\displaystyle|I_{1}^{\prime}|^{2}$ $\displaystyle\leq(\frac{1}{\pi
R^{2}\sin^{2}\theta})^{2}(\int_{0}^{R\sin\theta}\int_{0}^{2\pi}p^{2}r^{\prime}dr^{\prime}d\theta^{\prime})(\int_{0}^{R\sin\theta}\int_{0}^{2\pi}r^{\prime}dr^{\prime}d\theta^{\prime})$
$\displaystyle\leq\frac{1}{\pi
R^{2}\sin^{2}\theta}\int_{2R-R\sin\theta<r<2R+R\sin\theta,0<\theta<\pi}p^{2}dxdy$
$\displaystyle\leq\frac{1}{\pi
R^{2}\sin^{2}\theta}\int_{2R-R\sin\theta}^{2R+R\sin\theta}rdr\left\\{\max_{2R-R\sin\theta<r<2R+R\sin\theta}\int_{0}^{\pi}p(r,\theta)^{2}d\theta\right\\}$
$\displaystyle=\frac{4}{\pi\sin\theta}\max_{2R-R\sin\theta<r<2R+R\sin\theta}\int_{0}^{\pi}p(r,\theta)^{2}d\theta\to
0,\quad{\rm as}~{}~{}R\to\infty.$

For $I_{2}^{\prime},$ using Cauchy and Wirtinger’s inequalities, we have

$\displaystyle I_{2}^{\prime}$ $\displaystyle\leq\frac{2}{\pi
R^{2}\sin^{2}\theta}\int_{0}^{R\sin\theta}\int_{0}^{r^{\prime}}\int_{0}^{2\pi}|\nabla\boldsymbol{w}(\rho,\theta^{\prime})|^{2}\rho
d\rho d\theta^{\prime}r^{\prime}dr^{\prime}$ $\displaystyle\leq\frac{2}{\pi
R^{2}\sin^{2}\theta}\int_{2R-R\sin\theta<r<2R+R\sin\theta,0<\theta<\pi}|\nabla\boldsymbol{w}|^{2}dxdy\int_{0}^{R\sin\theta}\rho
d\rho$
$\displaystyle=\frac{1}{\pi}\int_{2R-R\sin\theta<r<2R+R\sin\theta,0<\theta<\pi}|\nabla\boldsymbol{w}|^{2}dxdy\to
0,\quad{\rm as}~{}~{}R\to\infty.$

Hence from (5.16), we obtain $\forall\theta\in(0,\pi),$ there holds

$|P(2R,\theta)|\to 0$

as $R\to\infty.$ Hence we prove the case of II. Combining the results of Case
I and Case II , the proof is complete.

##  6\. Decay of the vorticity.

######  Lemma 6.1.

Under the assumptions of Theorem 1.4, we have

$\int_{r>r_{1},0<\theta<\pi}\frac{r}{(\log
r)^{\frac{1}{2}}}|\nabla\omega|^{2}dxdy<\infty(r_{1}>\max(r_{0},2-r_{0})).$
(6.1)

Proof. Choose $R>r_{1}>\max(r_{0},2-r_{0})$ and two non-negative $C^{2}$ cut-
off functions $\xi_{1}$ and $\xi_{2}$ such that

$\xi_{1}(r)=\left\\{\begin{aligned} &0,\quad r\leq\frac{1}{2}(r_{0}+r_{1})\\\
&1,\quad r\geq
r_{1}\end{aligned}\right.,\quad\xi_{2}(r)=\left\\{\begin{aligned} 1,\quad
r\leq 1\\\ 0,\quad r\geq 2\end{aligned}\right..$ (6.2)

Let

$\eta(r)=\xi_{1}(r)\xi_{2}(\frac{r}{R})\frac{r}{(\log r)^{\frac{1}{2}}},$
$h(\omega)=\omega^{2}.$

Clearly $\eta(r)$ vanishes near $r=r_{0}$ and near $r=\infty.$ Noting that
$a=0,$ similar to the discussion for (2.6), we have

$\displaystyle 2\int_{r>r_{0},0<\theta<\pi}\eta|\nabla\omega|^{2}dxdy=$
$\displaystyle\int_{r>r_{0},0<\theta<\pi}\omega^{2}(\Delta\eta+\boldsymbol{w}\cdot\nabla\eta)dxdy$
(6.3)
$\displaystyle+\int_{r>r_{0},0<\theta<\pi}\nabla\eta\cdot(0,b)\omega^{2}dxdy.$

One verifies easily that there is a constant $C$ independent of $R$ such that

$|\Delta\eta|\leq C,\quad|\nabla\eta|\leq\frac{C}{(\log r)^{\frac{1}{2}}}.$

Noting that $\eta=\frac{r}{\log r}$ for $r_{1}<r<R,$ it follows from (6.3) and
(1.8)

$\displaystyle\int_{r_{1}<r<R,0<\theta<\pi}\frac{r}{\log
r}|\nabla\omega|^{2}dxdy$
$\displaystyle\leq\frac{1}{2}\int_{r>r_{0},0<\theta<\pi}\omega^{2}(\Delta\eta(r)+\boldsymbol{w}\cdot\nabla\eta(r))dxdy+\frac{1}{2}\int_{r>r_{0},0<\theta<\pi}\nabla\eta(r)\cdot(0,b)\omega^{2}dxdy$
$\displaystyle\leq
C\int_{r>r_{0},0<\theta<\pi}\omega^{2}[1+\frac{|\boldsymbol{w}|}{(\log
r)^{\frac{1}{2}}}]dxdy+C\int_{r>r_{0},0<\theta<\pi}\frac{1}{(\log
r)^{\frac{1}{2}}}\omega^{2}dxdy$ $\displaystyle\leq C(1+\frac{1}{(\log
r_{0})^{\frac{1}{2}}})\int_{r>r_{0},0<\theta<\pi}\omega^{2}dxdy<\infty.$

Letting $R\to\infty,$ we obtain (6.1).

Using Lemma 6.1 one can improve the result of Lemma 2.4.

Proof of Theorem 1.4. Note that for $2^{n}>r_{0}$

$\displaystyle\int_{2^{n}}^{2^{n+1}}\frac{dr}{r}\int_{0}^{\pi}(r^{2}\omega^{2}+2\frac{r^{\frac{3}{2}}}{(\log
r)^{\frac{1}{4}}}|\omega\omega_{\theta}|)d\theta$
$\displaystyle\leq\int_{2^{n}<r<2^{n+1},0<\theta<\pi}(\omega^{2}+2\frac{r^{\frac{1}{2}}}{(\log
r)^{\frac{1}{4}}}|\omega||\nabla\omega|)dxdy$
$\displaystyle\leq\int_{r>2^{n},0<\theta<\pi}(2\omega^{2}+2\frac{r}{(\log
r)^{\frac{1}{2}}}|\nabla\omega)|^{2})dxdy.$

Using (6.1) and proceeding exactly as in the proof of Lemma 2.4, we obtain
(1.12). The proof is complete.

Acknowledgments. W. Wang was supported by NSFC under grant 12071054, National
Support Program for Young Top-Notch Talents and by Dalian High-level Talent
Innovation Project (Grant 2020RD09).

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