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CHARACTERIZATION OF SMOOTH SOLUTIONS TO THE NAVIER-STOKES EQUATIONS IN A PIPE WITH TWO TYPES OF SLIP BOUNDARY CONDITIONS

Zijin Li, Xinghong Pan and Jiaqi Yang
Abstract

Smooth solutions of the stationary Navier-Stokes equations in an infinitely long pipe, equipped with the Navier-slip or Navier-Hodge-Lions boundary condition, are considered in this paper. Three main results are presented.

First, when equipped with the Navier-slip boundary condition, it is shown that, W1,W^{1,\infty} axially symmetric solutions with zero flux at one cross section, must be swirling solutions: u=(Cx2,Cx1,0)u=(-Cx_{2},Cx_{1},0), and x3x_{3}-periodic solutions must be helical solutions: u=(C1x2,C1x1,C2)u=(-C_{1}x_{2},C_{1}x_{1},C_{2}).

Second, also equipped with the Navier-slip boundary condition, if the swirl or vertical component of the axially symmetric solution is independent of the vertical variable x3x_{3}, solutions are also proven to be helical solutions. In the case of the vertical component being independent of x3x_{3}, the W1,W^{1,\infty} assumption is not needed. In the case of the swirl component being independent of x3x_{3}, the W1,W^{1,\infty} assumption can be relaxed extensively such that the horizontal radial component of the velocity, uru_{r}, can grow exponentially with respect to the distance to the origin. Also, by constructing a counterexample, we show that the growing assumption on uru_{r} is optimal.

Third, when equipped with the Navier-Hodge-Lions boundary condition, we can show that if the gradient of the velocity grows sublinearly, then the solution, enjoying the Liouville-type theorem, is a trivial shear flow: (0,0,C)(0,0,C).

Keywords: Navier-Stokes system, Navier-slip boundary, Navier-Hodge-Lions boundary, axially symmetric, helical solutions.

Mathematical Subject Classification 2020: 35Q35, 76D05

1  Introduction

The 3D stationary Navier-Stokes (NS) equations which describes the motion of stationary viscous incompressible fluids follows that

{uu+pΔu=0,u=0,in𝒟3.\left\{\begin{aligned} &u\cdot\nabla u+\nabla p-\Delta u=0,\\ &\nabla\cdot u=0,\end{aligned}\right.\quad\text{in}\quad\mathcal{D}\subset{\mathbb{R}}^{3}. (1.1)

Here u(x)3u(x)\in\mathbb{R}^{3}, p(x)p(x)\in\mathbb{R} represents the velocity and the scalar pressure respectively. In this paper, we consider the domain 𝒟\mathcal{D} to be an infinitely long pipe, i.e.

𝒟={x:|xh|<1,x3},\mathcal{D}=\left\{x:\,|x_{h}|<1,x_{3}\in{\mathbb{R}}\right\},

where x=(x1,x2,x3)x=(x_{1},x_{2},x_{3}), xh=(x1,x2)x_{h}=(x_{1},x_{2}) and |xh|=x12+x22|x_{h}|=\sqrt{x^{2}_{1}+x^{2}_{2}}. The boundary condition will be equipped with the following:

The total Navier-slip boundary condition:

{(𝕊u𝒏)τ=0,u𝒏=0,x𝒟,\left\{\begin{aligned} &(\mathbb{S}u\cdot\boldsymbol{n})_{\tau}=0,\\ &u\cdot\boldsymbol{n}=0,\\ \end{aligned}\right.\quad\forall x\in\partial\mathcal{D}, (NSB)

or the Navier-Hodge-Lions boundary condition:

{(×u)×𝒏=0,u𝒏=0,x𝒟,\left\{\begin{aligned} &(\nabla\times u)\times\boldsymbol{n}=0,\\ &u\cdot\boldsymbol{n}=0,\\ \end{aligned}\right.\quad\forall x\in\partial\mathcal{D}, (NHLB)

Here 𝕊u=12(u+(u)T)\mathbb{S}u=\frac{1}{2}\left(\nabla u+(\nabla u)^{T}\right) is the stress tensor, where (u)T(\nabla u)^{T} is the transpose of the Jacobian matrix u\nabla u, and 𝒏\boldsymbol{n} is the unit outer normal vector of 𝒟\partial\mathcal{D}. For a vector field vv, vτv_{\tau} stands for its tangential part: vτ:=v(v𝒏)𝒏.v_{\tau}:=v-(v\cdot\boldsymbol{n})\boldsymbol{n}. The condition (NSB) is from the general Navier-slip boundary condition and impermeable boundary condition which was introduced by Claude-Louis Naiver in 1820s [28]:

{2(𝕊u𝒏)τ+αuτ=0,u𝒏=0.\left\{\begin{aligned} &2(\mathbb{S}u\cdot\boldsymbol{n})_{\tau}+\alpha u_{\tau}=0,\\ &u\cdot\boldsymbol{n}=0.\\ \end{aligned}\right. (1.2)

Here α0\alpha\geq 0 stands for the friction constant which may depend on various elements, such as the property of the boundary and the viscosity of the fluid. When α=0\alpha=0, boundary condition (1.2) turns to the total Navier-slip boundary (NSB), and when α\alpha\to\infty, boundary condition (1.2) degenerates into the no-slip boundary condition u0u\equiv 0 on the boundary.

The boundary condition (NHLB) is a special case in a family of boundary conditions proposed by Navier [28], which has been studied extensively in the literature and was attributed to different authors. The boundary condition was called the Navier-Hodge boundary condition in [25] and the Navier-Lions boundary condition in [23]. For this reason, we will call it the Navier-Hodge-Lions boundary condition in this paper.

We write 𝒟\mathcal{D} to be

𝒟=Σ×,\mathcal{D}=\Sigma\times{\mathbb{R}},

where the cross section Σ2\Sigma\in{\mathbb{R}}^{2} is a unit disc. The domain considered here is a high-degree simplification of the following “distorted cylinder”, i.e.

𝒟~=Σ~×,\tilde{\mathcal{D}}=\tilde{\Sigma}\times{\mathbb{R}},

where Σ~2\tilde{\Sigma}\in{\mathbb{R}}^{2} is a simply connected bounded domain with smooth boundary.

Let 𝒟~0\tilde{\mathcal{D}}_{0} be a simply connected bounded domain with smooth boundary in 3{\mathbb{R}}^{3} and 𝒟~0𝒟~\tilde{\mathcal{D}}_{0}\cap\tilde{\mathcal{D}}\neq\emptyset. Existence problem of weak solutions in domain 𝒟~Union:=𝒟~𝒟~0\tilde{\mathcal{D}}_{\text{Union}}:=\tilde{\mathcal{D}}\cup\tilde{\mathcal{D}}_{0} with the Navier-slip boundary (NSB) was addressed in [17] and regularity of solutions was also implied there. On the other hand, if 𝒟~0𝒟~\tilde{\mathcal{D}}_{0}\subset\tilde{\mathcal{D}} is an “obstacle” in 𝒟~\tilde{\mathcal{D}}, then the two dimensional existence problems and asymptotic behaviors of smooth solutions in domain 𝒟~Diff:=𝒟~\𝒟~0\tilde{\mathcal{D}}_{\text{Diff}}:=\tilde{\mathcal{D}}\backslash\tilde{\mathcal{D}}_{0} with the total Navier-slip boundary condition are obtained in [26, 27].

There have also been many pieces of literature in studying the existence, uniqueness and asymptotic behavior of the Navier-Stokes equations in a distorted pipe 𝒟~Union\tilde{\mathcal{D}}_{\text{Union}} or 𝒟~Diff\tilde{\mathcal{D}}_{\text{Diff}} with no-slip boundary and with the Poiseuille flow as the asymptotic profile at infinity (Leray’s problem: Ladyzhenskaya [18, p. 77] and [19, p. 551]). The first remarkable contribution on the solvability of Leray’s problem is due to Amick [1, 2], who reduced the solvability problem to the resolution of a variational problem related to the stability of the Poiseuille flow in a flat cylinder. However, uniqueness and existence of solutions with large flux are left open. Ladyzhenskaya and Solonnikov [20] gave a detailed analysis of this problem on existence, uniqueness and asymptotic behavior of small-flux solutions. One may refer to [3, 14, 30] and references for more details on well-posedness, decay and far-field asymptotic analysis of solutions for Leray’s problem and related topics. A systematic review and study of Leray’s problem can be found in [11, Chapter XIII]. Recently Wang-Xie in [33] studied uniform structural stability of Poiseuille flows for the 3D axially symmetric solutions in the 3D pipe 𝒟\mathcal{D} where a force term appears on the right hand of equation (1.1)1.

Compared to the no-slip boundary condition, this model with the total Navier-slip or Navier-Hodge-Lions boundary condition has different physical interpretations and gives different mathematical properties. Literature [26, 17] addressed the existence and regularity problems of weak solutions for the total Navier-slip boundary condition in a pipe, but uniqueness was left open. Also readers can refer to, i.e., [4, 10, 23, 34] for some well-posedness results for the Navier-Hodge-Lions boundary condition in different domains. In this paper, we attempt to derive some uniqueness results for the Navier-Stokes equations with the total Navier-slip or Navier-Hodge-Lions boundary condition in the regular infinite pipe 𝒟\mathcal{D}. We emphasize that our results below do not require any smallness and decay assumptions.

In this paper, for the boundary condition (NSB), a family of smooth helical solutions will be given, and for the Navier-Hodge-Lions boundary condition (NHLB), the trivial shear flow is easy to be discovered. We concern on the characterization and uniqueness of these two types of smooth solutions in 𝒟\mathcal{D}.

Most of our proof will be carried out in the framework of cylindrical coordinates (r,θ,z)(r,\theta,z) and some of our results are restricted to the axially symmetric solutions. Here we give the formulation of axially symmetric solutions in the cylindrical coordinates which enjoy the following relationship with 3D Euclidian coordinates:

x=(x1,x2,x3)=(rcosθ,rsinθ,z).x=(x_{1},x_{2},x_{3})=(r\cos\theta,r\sin\theta,z).

A stationary axially symmetric solution of the incompressible Navier-Stokes equations is given as

u=ur(r,z)𝒆𝒓+uθ(r,z)𝒆𝜽+uz(r,z)𝒆𝒛,u=u_{r}(r,z)\boldsymbol{e_{r}}+u_{\theta}(r,z)\boldsymbol{e_{\theta}}+u_{z}(r,z)\boldsymbol{e_{z}},

where the basis vectors 𝒆𝒓\boldsymbol{e_{r}}, 𝒆𝜽\boldsymbol{e_{\theta}} ,𝒆𝒛\boldsymbol{e_{z}} are

𝒆𝒓=(x1r,x2r,0),𝒆𝜽=(x2r,x1r,0),𝒆𝒛=(0,0,1),\boldsymbol{e_{r}}=(\frac{x_{1}}{r},\frac{x_{2}}{r},0),\quad\boldsymbol{e_{\theta}}=(-\frac{x_{2}}{r},\frac{x_{1}}{r},0),\quad\boldsymbol{e_{z}}=(0,0,1),

while the components ur,uθ,uzu_{r},u_{\theta},u_{z}, which are independent of θ\theta, satisfy

{(urr+uzz)ur(uθ)2r+rp=(Δ1r2)ur,(urr+uzz)uθ+uθurr=(Δ1r2)uθ,(urr+uzz)uz+zp=Δuz,b=rur+urr+zuz=0,\left\{\begin{aligned} &(u_{r}\partial_{r}+u_{z}\partial_{z})u_{r}-\frac{(u_{\theta})^{2}}{r}+\partial_{r}p=\left(\Delta-\frac{1}{r^{2}}\right)u_{r},\\ &(u_{r}\partial_{r}+u_{z}\partial_{z})u_{\theta}+\frac{u_{\theta}u_{r}}{r}=\left(\Delta-\frac{1}{r^{2}}\right)u_{\theta},\\ &(u_{r}\partial_{r}+u_{z}\partial_{z})u_{z}+\partial_{z}p=\Delta u_{z},\\ &\nabla\cdot b=\partial_{r}u_{r}+\frac{u_{r}}{r}+\partial_{z}u_{z}=0,\end{aligned}\right. (1.3)

where b=ur𝒆r+uz𝒆zb=u_{r}\boldsymbol{e}_{r}+u_{z}\boldsymbol{e}_{z}.

We can also compute the axi-symmetric vorticity ω=×u=ωr𝒆𝒓+ωθ𝒆𝜽+ωz𝒆𝒛\omega=\nabla\times u=\omega_{r}\boldsymbol{e_{r}}+\omega_{\theta}\boldsymbol{e_{\theta}}+\omega_{z}\boldsymbol{e_{z}} as follows

ωr=zuθ,ωθ=zurruz,ωz=(r+1r)uθ,\omega_{r}=-\partial_{z}u_{\theta},\ \omega_{\theta}=\partial_{z}u_{r}-\partial_{r}u_{z},\ \omega_{z}=\left(\partial_{r}+\frac{1}{r}\right)u_{\theta},

which satisfies

{(urr+uzz)ωr(Δ1r2)ωr(ωrr+ωzz)ur=0,(urr+uzz)ωθ(Δ1r2)ωθurrωθ1rz(uθ)2=0,(urr+uzz)ωzΔωz(ωrr+ωzz)uz=0.\left\{\begin{aligned} &(u_{r}\partial_{r}+u_{z}\partial_{z})\omega_{r}-\left(\Delta-\frac{1}{r^{2}}\right)\omega_{r}-(\omega_{r}\partial_{r}+\omega_{z}\partial_{z})u_{r}=0,\\ &(u_{r}\partial_{r}+u_{z}\partial_{z})\omega_{\theta}-\left(\Delta-\frac{1}{r^{2}}\right)\omega_{\theta}-\frac{u_{r}}{r}\omega_{\theta}-\frac{1}{r}\partial_{z}(u_{\theta})^{2}=0,\\ &(u_{r}\partial_{r}+u_{z}\partial_{z})\omega_{z}-\Delta\omega_{z}-(\omega_{r}\partial_{r}+\omega_{z}\partial_{z})u_{z}=0.\end{aligned}\right. (1.4)

In the cylindrical coordinates, the total Navier-slip boundary condition (NSB) is represented as

{ruθuθr=0,ruz=0,ur=0,x𝒟,\left\{\begin{aligned} &\partial_{r}u_{\theta}-\frac{u_{\theta}}{r}=0,\\ &\partial_{r}u_{z}=0,\\ &u_{r}=0,\\ \end{aligned}\right.\quad\forall x\in\partial\mathcal{D}, (1.5)

while the Navier-Hodge-Lions boundary condition (NHLB) is given by

{ruθ+uθr=0,ruz=0,ur=0,x𝒟,\left\{\begin{aligned} &\partial_{r}u_{\theta}+\frac{u_{\theta}}{r}=0,\\ &\partial_{r}u_{z}=0,\\ &u_{r}=0,\\ \end{aligned}\right.\quad\forall x\in\partial\mathcal{D}, (1.6)

whose computations are postponed to Appendix A.

Clearly direct calculation shows that, for arbitrary constants C1C_{1} and C2C_{2}, the following type of helical solutions

u=C1r𝒆𝜽+C2𝒆𝒛u=C_{1}r\boldsymbol{e_{\theta}}+C_{2}\boldsymbol{e_{z}} (1.7)

solves (1.3) with the boundary condition (1.5).

Refer to caption
Figure 1: A helical solution in the infinite pipe 𝒟\mathcal{D}

We further note that helical solutions (1.7), which is smooth in 𝒟\mathcal{D}, enjoys the following property:

The solution itself and its gradient are uniformly bounded in 𝒟.\text{\em The solution itself and its gradient are uniformly bounded in }\mathcal{D}. (\ast)

Thus a natural question raises:

Are helical solutions (1.7) the only smooth solutions of system (1.3) with the boundary condition (1.5) which enjoys property (\ast1)?

Before answering this question, we recall that the flux Φ(z)\Phi(z) at the cross section Σ\Sigma, which is defined by

Φ(z):=Σu(xh,z)𝝂𝑑xh,\Phi(z):=\int_{\Sigma}u(x_{h},z)\cdot\boldsymbol{\nu}dx_{h},

is a constant. Here 𝝂=𝒆𝒛\boldsymbol{\nu}=\boldsymbol{e_{z}} is the unit normal vector of Σ\Sigma pointing to the positive zz direction. Actually by using the divergence free condition of the velocity and the boundary condition (NSB)2, we have

ddzΦ(z)\displaystyle\frac{d}{dz}\Phi(z) =Σddzuz(xh,z)𝑑xh\displaystyle=\int_{\Sigma}\frac{d}{dz}u_{z}(x_{h},z)dx_{h}
=Σ(x1u1+x2u2)(xh,z)𝑑xh\displaystyle=-\int_{\Sigma}(\partial_{x_{1}}u_{1}+\partial_{x_{2}}u_{2})(x_{h},z)dx_{h}
\xlongequalGaussformulaΣ(n1u1+n2u2)(xh,z)𝑑S(xh)\displaystyle\xlongequal{Gauss\ formula}-\int_{\partial\Sigma}(n_{1}u_{1}+n_{2}u_{2})(x_{h},z)dS(x_{h})
=Σ(u𝒏)(xh,z)𝑑S(xh)=0,\displaystyle=-\int_{\partial\Sigma}(u\cdot\boldsymbol{n})(x_{h},z)dS(x_{h})=0,

where 𝒏=(n1,n2,0)\boldsymbol{n}=(n_{1},n_{2},0) is the unit outer normal vector of 𝒟\partial\mathcal{D}. Then for any zz\in\mathbb{R}, we will denote Φ(z)=Φ\Phi(z)=\Phi.

Our first result in this paper gives a positive answer to the above question in the following two cases:

  • (i).

    the solution is axially symmetric and the flux Φ\Phi is zero (corresponding to C2=0C_{2}=0);

  • (ii).

    the solution is zz-periodic.

Theorem 1.1.

Let uu be a smooth solution of the Navier-Stokes equations in the infinite pipe 𝒟\mathcal{D} subject to the total Navier-slip boundary condition (NSB).

  • I.

    Suppose uW1,(𝒟)u\in W^{1,\infty}(\mathcal{D}) is axially symmetric and Φ=0\Phi=0, then uu must be the following type of swirling solutions:

    u=C1r𝒆𝜽,p=C12r22.u=C_{1}r\boldsymbol{e_{\theta}},\quad p=\frac{C^{2}_{1}r^{2}}{2}.
  • II.

    Suppose uu is zz-periodic, then uu must be the following type of helical solutions:

    u=C1r𝒆𝜽+C2𝒆𝒛,p(r,z)=C12r2/2,C1,C2.u=C_{1}r\boldsymbol{e_{\theta}}+C_{2}\boldsymbol{e_{z}},\quad p(r,z)={C_{1}^{2}r^{2}}/{2},\quad\forall C_{1},C_{2}\in\mathbb{R}. (1.8)

Besides, we observe that solutions (1.7) enjoy the following property:

Its swirl component uθu_{\theta} or vertical component uzu_{z} is independent of zz.

In the following theorem, we will conclude that if uθu_{\theta} or uzu_{z} is independent of zz, then (1.7) are the only group of smooth solutions to (1.3) subject to the boundary condition (1.5).

In the case of uθu_{\theta} being independent of zz, the bounded W1,W^{1,\infty} assumption on the velocity will be extensively relaxed to the following:

{|ur(r,z)|Creγ0|z|;|uz(r,z)|C|z|δ0;|ωθ(r,z)|C|z|M0, uniformly with r[0,1],\left\{\begin{aligned} &|u_{r}(r,z)|\leq Cre^{\gamma_{0}|z|};\\ &|u_{z}(r,z)|\leq C|z|^{\delta_{0}};\\ &|\omega_{\theta}(r,z)|\leq C|z|^{M_{0}},\end{aligned}\right.\quad\text{ uniformly with }r\in[0,1], (1.9)

for any γ0<α3.83171\gamma_{0}<\alpha\approx 3.83171, δ0<1\delta_{0}<1, and M0>0M_{0}>0. Here α\alpha is the first positive root of the Bessel function J1J_{1}. We recall that JβJ_{\beta} are canonical solutions of Bessel’s ordinary differential equation

s2Jβ′′(s)+sJβ(s)+(s2β2)Jβ(s)=0,s^{2}J_{\beta}^{\prime\prime}(s)+sJ_{\beta}^{\prime}(s)+(s^{2}-\beta^{2})J_{\beta}(s)=0, (1.10)

which can be expressed by the following series form:

Jβ(s)=n=0(1)nn!Γ(n+β+1)(s2)2n+β.J_{\beta}(s)=\sum_{n=0}^{\infty}\frac{(-1)^{n}}{n!\Gamma(n+\beta+1)}\left(\frac{s}{2}\right)^{2n+\beta}. (1.11)

In the case of uzu_{z} being independent of zz, no size assumptions such as (1.9) are imposed on the solution.

Remark 1.2.

The reason why there is an rr on the righthand of (1.9)1\eqref{growass}_{1} is that for a smooth solution uu, in the cylindrical coordinates, uru_{r} vanishes at r=0r=0. When doing Taylor expansion of uru_{r} at r=0r=0 in the rr direction, the zero order derivative term is missing, so it is reasonable to assume a one order rr control on uru_{r} for r[0,1]r\in[0,1].

Theorem 1.3.

Let uu be a smooth solution of the axially symmetric Navier-Stokes equations (1.3) in the infinite pipe 𝒟\mathcal{D} subject to the total Navier-slip boundary condition (NSB). Then uu must be of helical solutions (1.8) if one of the followings is satisfied.

  • I.

    uu satisfies (1.9) and uθu_{\theta} is independent of zz-variable;

  • II.

    uzu_{z} is independent of zz-variable.

Remark 1.4.

We emphasize that the condition (1.9)1 above is sharp, because we have the following non-trivial counterexample which grows exactly as Ceα|z|Ce^{\alpha|z|} when zz\to\infty:

{u=cosh(αz)J1(αr)𝒆𝒓+sinh(αz)J0(αr)𝒆𝒛,p=12(cosh2(αz)J12(αr)+sinh2(αz)J02(αr)).\left\{\begin{aligned} &u=-\cosh(\alpha z)J_{1}(\alpha r)\boldsymbol{e_{r}}+\sinh(\alpha z)J_{0}(\alpha r)\boldsymbol{e_{z}},\\ &p=-\frac{1}{2}\left(\cosh^{2}(\alpha z)J_{1}^{2}(\alpha r)+\sinh^{2}(\alpha z)J_{0}^{2}(\alpha r)\right).\end{aligned}\right. (1.12)

Here J0J_{0}, J1J_{1} are Bessel functions defined in (1.11), while α3.83171\alpha\approx 3.83171 is the smallest positive root of J1J_{1}. One can verify (1.12) being the solution of (1.3) with the boundary condition (1.5) by direct calculations. Here we leave the details to the interested reader. Unfortunately, our example here can not reflect whether the growing assumptions in (1.9)2 and (1.9)3 are sharp.

If we switch the total Navier-slip boundary condition to the Navier-Hodge-Lions boundary condition (NHLB), one notices that a non-zero swirling solution does not enjoy (NHLB). In this situation, one can derive the following Liouville-type theorem:

Theorem 1.5.

Let uu be a smooth solution of the Navier-Stokes equations (1.1) in the infinite pipe 𝒟\mathcal{D} subject to the Navier-Hodge-Lions boundary condition (NHLB). Suppose u\nabla u satisfies

|u(xh,z)|C|z|β|\nabla u(x_{h},z)|\leq C|z|^{\beta} (1.13)

for some C>0C>0 and 0<β<10<\beta<1. Then u=Φπ𝐞𝟑u=\frac{\Phi}{\pi}\boldsymbol{e_{3}}.

There has already been much literature studying Liouville-type results on the Navier-Stokes equations subject to various boundary conditions in various unbounded domains. Readers can refer to [8, 9, 31, 32, 7, 29] and references therein for more Liouville-type results on the stationary Navier-Stokes equations. Moreover, our results in the above Theorems can be extended from the stationary case to the case of ancient solutions (backward global solutions) under suitable assumptions. However, for simplification of idea presenting, we omit this extension here and leave it to further works. See [13] where the authors established a Liouville-type result for the ancient solution to the Navier-Stokes equations in the half plane with the no-slip boundary condition.

Liouville-type results of ancient solutions is connected to the regularity of solutions to the initial value problem of the non-stationary Navier-Stokes equations. Type I blow-up solutions of the Navier-Stokes initial value problem could not exist provided the Liouville-type result holds for bounded ancient solutions. See [15, 12].

Before ending our introduction, we briefly outline our strategy for proofs of Theorem 1.1, Theorem 1.3 and Theorem 1.5. The most important ingredient of proving Theorem 1.1 is to show that 𝕊u0\mathbb{S}u\equiv 0. For Case I in Theorem 1.1, we need to first show that 𝕊uL2(𝒟)\mathbb{S}u\in L^{2}(\mathcal{D}), which is not trivial since the pipe is an unbounded domain. In this process, LL^{\infty} oscillation boundedness of the pressure in 𝒟2Z\𝒟Z\mathcal{D}_{2Z}\backslash\mathcal{D}_{Z} (see (1.16) for the definition of 𝒟Z\mathcal{D}_{Z}) is essential, which will be presented in Section 2.1.2. Then combining the square integrability of 𝕊u\mathbb{S}u and boundedness of the velocity together with its gradient, a trick of integration by parts and Poincaré inequality will indicate that uzu_{z} actually belongs to L2(𝒟)L^{2}(\mathcal{D}), which will result in the vanishing of 𝕊u\mathbb{S}u. While vanishing of 𝕊u\mathbb{S}u in Case II of Theorem 1.1 is directly due to the x3x_{3}-periodic property by performing the energy estimate in a single period. However, in the process, we must be careful of handing of the pressure, which may not be x3x_{3} periodic. After analyzing ingredients of 𝕊u\mathbb{S}u, we finally conclude the validity of Theorem 1.1.

The idea for proof of Theorem 1.3 is completely different from that of Theorem 1.1. Under the assumption of Case I in Theorem 1.3, we will see that the quantity Ω:=ωθ/r\Omega:=\omega_{\theta}/r satisfies a nice linear elliptic equation with an advection term. Under the growing assumption (1.9) in domain 𝒟\mathcal{D}, by using the Nash-Moser iteration, we can show that actually Ω0\Omega\equiv 0, which indicates that b=ur𝒆𝒓+uz𝒆𝒛b=u_{r}\boldsymbol{e_{r}}+u_{z}\boldsymbol{e_{z}} must be harmonic in 𝒟\mathcal{D}. Then by constructing a barrier function, applying maximum principle and assumptions on bb, one derives ur0u_{r}\equiv 0 and uzu_{z} must be a constant. From then on, (1.3)2 is reduced to a linear ordinary differential equation of uθu_{\theta}, and we finally obtain uθ=C1ru_{\theta}=C_{1}r. While under the assumption of Case II in Theorem 1.3, a rather direct computation implies that ur0u_{r}\equiv 0 by using the divergence-free condition. Then by combining the equations of uru_{r} and uzu_{z}, independence of zz variable for uθu_{\theta} can be obtained. Then the equation of uθu_{\theta} will degenerate to an ordinary differential equation with respect to rr, which will result in uθ=C1ru_{\theta}=C_{1}r. At last, trivialness of uzu_{z} is achieved by solving a two- dimensional Laplacian equation with Neumann boundary condition.

Proof of Theorem 1.5 is to apply Lemma 4.1, which was originally announced in reference [20] as far as the authors know. Denote the energy integral in terms of v:=uΦπ𝒆3v:=u-\frac{\Phi}{\pi}\boldsymbol{e}_{3} as follows:

Y(Z):=Z1Z𝒟ζ|v|2𝑑x𝑑ζ.Y(Z):=\int_{Z-1}^{Z}\int_{\mathcal{D}_{\zeta}}|\nabla v|^{2}dxd\zeta.

A differential inequality of Y(Z)Y(Z), satisfying the assumption in Lemma 4.1, will be derived. In this process, boundary terms coming from integration by parts will be carefully addressed by using the boundary condition, which has a good sign compared with those from the Navier-slip boundary condition. At last, a direct application of Lemma 4.1 will imply the vanishing of Y(Z)Y(Z).

For the generalized Navier boundary condition (1.2) in 𝒟\mathcal{D}, one can derive that in cylindrical coordinates, (1.2) is equivalent to

{ruθuθr+αuθ=0,ruz+αuz=0,ur=0,x𝒟.\left\{\begin{aligned} &\partial_{r}u_{\theta}-\frac{u_{\theta}}{r}+\alpha u_{\theta}=0,\\ &\partial_{r}u_{z}+\alpha u_{z}=0,\\ &u_{r}=0,\\ \end{aligned}\right.\quad\forall x\in\partial\mathcal{D}. (1.14)

For given flux Φ:=Σuz(xh,z)𝑑xh=const.\Phi:=\int_{\Sigma}u_{z}(x_{h},z)dx_{h}=\text{const.}, we can find a family of bounded smooth solutions satisfying (1.3) with boundary condition (1.14) as follows

u=C1rχ{α=0}𝒆𝜽+2(α+2)Φ(α+4)π(1αα+2r2)𝒆𝒛,p=C12r22χ{α=0}8αΦ(α+4)πz,u=C_{1}r\chi_{\{\alpha=0\}}\boldsymbol{e_{\theta}}+\frac{2(\alpha+2)\Phi}{(\alpha+4)\pi}\left(1-\frac{\alpha}{\alpha+2}r^{2}\right)\boldsymbol{e_{z}},\quad p={\frac{C_{1}^{2}r^{2}}{2}\chi_{\{\alpha=0\}}}-\frac{8\alpha\Phi}{(\alpha+4)\pi}z, (1.15)

where C1C_{1} is an arbitrary constant, and χ{α=0}\chi_{\{\alpha=0\}} is the characteristic function on {α=0}\{\alpha=0\}, which means

χ{α=0}={1,α=0,0,α>0.\chi_{\{\alpha=0\}}=\left\{\begin{aligned} &1,\quad\alpha=0,\\ &0,\quad\alpha>0.\end{aligned}\right.

When α+\alpha\rightarrow+\infty, the boundary condition (1.14) becomes the no-slip boundary and the solution (1.15) corresponds to the Hagen-Poisseuille flow in 𝒟\mathcal{D}. Uniqueness of Hagen-Poisseuille flow is still open for large flux Φ\Phi. Our Theorem 1.1 states that in the case α=0\alpha=0 and Φ=0\Phi=0, we can show that (1.15) are the only bounded smooth solutions of (1.3) with the boundary condition (1.14). For general 0α+0\leq\alpha\leq+\infty and Φ\Phi, we have the following conjecture.

Conjecture 1.6.

Let uu be a smooth solution of the axially symmetric Navier-Stokes equations (1.3) in the infinite pipe 𝒟\mathcal{D} with the flux Φ\Phi and subject to the Navier-slip boundary condition (1.2) for any 0α+0\leq\alpha\leq+\infty. Suppose uu and its gradient are uniformly bounded, then the solution uu must be of the form (1.15).

Remark 1.7.

From the authors’ recent paper [22], the uniqueness result there implies that Conjecture 1.6 is valid for the case α(0,+)\alpha\in(0,+\infty) if the flux Φ\Phi is small.

Throughout this paper, Ca,b,c,C_{a,b,c,...} denotes a positive constant depending on a,b,c,a,\,b,\,c,\,... which may be different from line to line. For two quantities A1A_{1}, A2A_{2}, we denote A1A2=max{A1,A2}A_{1}\vee A_{2}=\max\{A_{1},\,A_{2}\}. Meanwhile, for Z>1Z>1, we denote

𝒟ζ:={(r,θ,z): 0r<1, 0θ2π,ζ<z<ζ},\mathcal{D}_{\zeta}:=\left\{(r,\theta,z):\,0\leq r<1,\,0\leq\theta\leq 2\pi,\,-\zeta<z<\zeta\right\}, (1.16)

the truncated pipe with the length of 2ζ2\zeta. And notations 𝒪ζ±\mathcal{O}_{\zeta}^{\pm} states

𝒪ζ+:=(𝒟ζ𝒟ζ1){x𝒟:z>0},𝒪ζ:=(𝒟ζ𝒟ζ1){x𝒟:z<0},\mathcal{O}_{\zeta}^{+}:=\left(\mathcal{D}_{\zeta}-\mathcal{D}_{\zeta-1}\right)\cap\{x\in\mathcal{D}:\,z>0\},\quad\mathcal{O}_{\zeta}^{-}:=\left(\mathcal{D}_{\zeta}-\mathcal{D}_{\zeta-1}\right)\cap\{x\in\mathcal{D}:\,z<0\},

respectively. We also apply ABA\lesssim B to denote ACBA\leq CB. Moreover, ABA\simeq B means both ABA\lesssim B and BAB\lesssim A.

This paper is arranged as follows: Section 2 is devoted to the proof of Theorem 1.1, and the proof of Theorem 1.3 could be found in Section 3. Proof of Theorem 1.5 will be presented in Section 4.

2  Proof of Theorem 1.1

In this section, we devote to proof of Theorem 1.1. Proof of Case I is shown in Section 2.1. In Section 2.1.1, we deduce a uniform bound of zωθ\partial_{z}\omega_{\theta} by using classical energy estimate of (1.4)2 and the Moser’s iteration. Then it will be applied to derive the LL^{\infty} oscillation boundedness of the pressure in Section 2.1.2. Based on these preparations, we finish proving Case I of Theorem 1.1 in Section 2.1.3. Proof of Case II is directly derived in Section 2.2.

2.1 Proof of Case I

2.1.1 Uniform bound of zωθ{\partial_{z}\omega_{\theta}}

Denoting g:=zωθg:=\partial_{z}\omega_{\theta} and taking zz-derivative on (1.4)2, one arrives

(Δ1r2)g+bg=F,-\left(\Delta-\frac{1}{r^{2}}\right)g+b\cdot\nabla g=\nabla\cdot F, (2.17)

where

F:=ωθzb+(urrωθ+2uθrzuθ)𝒆𝒛.F:=-\omega_{\theta}\partial_{z}b+\left(\frac{u_{r}}{r}\omega_{\theta}+2\frac{u_{\theta}}{r}\partial_{z}u_{\theta}\right)\boldsymbol{e_{z}}. (2.18)

From (A.87), we see that FLF\in L^{\infty} provided uu and u\nabla u are bounded. Meanwhile, we observe that from the boundary condition (1.5):

g0,on𝒟.g\equiv 0,\quad\text{on}\quad\partial\mathcal{D}.

Now we are ready to state the desired lemma of this section, with its proof based on the Moser’s iteration and energy estimate.

Lemma 2.1.

Let (ur,uθ,uz)(u_{r},u_{\theta},u_{z}) be a smooth solution of (1.3) in 𝒟\mathcal{D}, subject to Navier total slip boundary condition (1.5) and ωθ\omega_{\theta} be the swirl component of its vorticity. Then zωθ\partial_{z}\omega_{\theta} is uniformly bounded in 𝒟\mathcal{D}.

Proof.   For q1q\geq 1, we multiply (2.17) by qgq1qg^{q-1} to get

qgq1Δg+qr2gq+bgq=qgq1F.-qg^{q-1}\Delta g+\frac{q}{r^{2}}g^{q}+b\cdot\nabla g^{q}=qg^{q-1}\nabla\cdot F. (2.19)

Noting that

Δgq=div (qgq1g)=qgq1Δg+q(q1)gq2|g|2,\Delta g^{q}=\text{div }\left(qg^{q-1}\nabla g\right)=qg^{q-1}\Delta g+q(q-1)g^{q-2}|\nabla g|^{2},

one derives from (2.19) that

Δgq+q(q1)gq2|g|2+qr2gq+bgq=qgq1F.-\Delta g^{q}+q(q-1)g^{q-2}|\nabla g|^{2}+\frac{q}{r^{2}}g^{q}+b\cdot\nabla g^{q}=qg^{q-1}\nabla\cdot F. (2.20)

Let ϕ\phi be a smooth cut-off function in zz variable which is bounded up to its second-order derivatives, supported on [L1,L+1][L-1,L+1] for some LL\in\mathbb{R}, which will be specified later. Using gqϕ2g^{q}\phi^{2} as a test function to the equation (2.20) and noting that

q(q1)𝒟g2q2|g|2ϕ2𝑑x=q1q𝒟|gq|2ϕ2𝑑x0,q(q-1)\int_{\mathcal{D}}g^{2q-2}|\nabla g|^{2}\phi^{2}dx=\frac{q-1}{q}\int_{\mathcal{D}}|\nabla g^{q}|^{2}\phi^{2}dx\geq 0,

one deduces

𝒟gq(gqϕ2)dxI1+q𝒟g2qϕ2r2𝑑xI2+𝒟bgq(gqϕ2)𝑑xI3q𝒟g2q1Fϕ2𝑑xI4.\underbrace{\int_{\mathcal{D}}\nabla g^{q}\cdot\nabla(g^{q}\phi^{2})dx}_{I_{1}}+\underbrace{q\int_{\mathcal{D}}\frac{g^{2q}\phi^{2}}{r^{2}}dx}_{I_{2}}+\underbrace{\int_{\mathcal{D}}b\cdot\nabla g^{q}(g^{q}\phi^{2})dx}_{I_{3}}\leq q\underbrace{\int_{\mathcal{D}}g^{2q-1}\nabla\cdot F\phi^{2}dx}_{I_{4}}. (2.21)

We further denote f:=gqf:=g^{q} for convenience. First we see

I1=𝒟|(fϕ)|2𝑑x𝒟f2|ϕ|2𝑑x.I_{1}=\int_{\mathcal{D}}|\nabla(f\phi)|^{2}dx-\int_{\mathcal{D}}f^{2}|\nabla\phi|^{2}dx.

Clearly, I20I_{2}\geq 0. Using the divergence free property of bb, one finds I3I_{3} satisfies

I3=12𝒟bf2ϕ2dx=𝒟uzzϕ2f2dx.I_{3}=\frac{1}{2}\int_{\mathcal{D}}b\cdot\nabla f^{2}\phi^{2}dx=-\int_{\mathcal{D}}u_{z}\partial_{z}\phi^{2}f^{2}dx.

Applying integration by parts, one derives

I4=q(2q1)𝒟g2q2gFϕ2dxq𝒟g2q1Fϕ2dx12𝒟|(fϕ)|2𝑑x+Cq2𝒟|F|2|g|2q2ϕ2𝑑x+𝒟|g|2q1|F|ϕ|ϕ|𝑑x.\begin{split}I_{4}=&-q(2q-1)\int_{\mathcal{D}}g^{2q-2}\nabla g\cdot F\phi^{2}dx-q\int_{\mathcal{D}}g^{2q-1}F\cdot\nabla\phi^{2}dx\\ \leq&\frac{1}{2}\int_{\mathcal{D}}|\nabla(f\phi)|^{2}dx+Cq^{2}\int_{\mathcal{D}}|F|^{2}|g|^{2q-2}\phi^{2}dx+\int_{\mathcal{D}}|g|^{2q-1}|F|\phi|\nabla\phi|dx.\end{split}

Plugging estimates I1I_{1}I4I_{4} into (2.21), we conclude that

𝒟|(fϕ)|2𝑑xC(ϕL(𝒟)(uzL(𝒟)+ϕL(𝒟))+q2)supp ϕ(|g|FL(𝒟))2q𝑑x.\int_{\mathcal{D}}|\nabla(f\phi)|^{2}dx\leq C\bigg{(}\|\nabla\phi\|_{L^{\infty}(\mathcal{D})}(\|u_{z}\|_{L^{\infty}(\mathcal{D})}+\|\nabla\phi\|_{L^{\infty}(\mathcal{D})})+q^{2}\bigg{)}\int_{\text{supp }\phi}\left(|g|\vee\|F\|_{L^{\infty}(\mathcal{D})}\right)^{2q}dx.

Using the Sobolev imbedding and noting that ϕ\phi is supported on an interval whose length equals 2, one arrives

({x:ϕ=1}(|g|FL(𝒟))6q𝑑x)16qC12q(ϕL(𝒟)(bL(𝒟)+ϕL(𝒟))+q2)12q×(supp ϕ(|g|FL(𝒟))2q𝑑x)12q.\begin{split}\left(\int_{\{x\,:\,\phi=1\}}\left(|g|\vee\|F\|_{L^{\infty}(\mathcal{D})}\right)^{6q}dx\right)^{\frac{1}{6q}}\leq C^{\frac{1}{2q}}\Big{(}\|\nabla\phi\|_{L^{\infty}(\mathcal{D})}(\|b\|_{L^{\infty}(\mathcal{D})}+\|\nabla\phi\|_{L^{\infty}(\mathcal{D})})+q^{2}\Big{)}^{\frac{1}{2q}}\\ \hskip 142.26378pt\times\left(\int_{\text{supp }\phi}\left(|g|\vee\|F\|_{L^{\infty}(\mathcal{D})}\right)^{2q}dx\right)^{\frac{1}{2q}}.\end{split} (2.22)

Let 12z2<z11\frac{1}{2}\leq z_{2}<z_{1}\leq 1 and assume ϕ\phi is supported on the interval [Lz1,L+z1][L-z_{1},L+z_{1}], and ϕ1\phi\equiv 1 on [Lz2,L+z2][L-z_{2},L+z_{2}]. Meanwhile, the gradient of ϕ\phi satisfies the following estimate:

ϕLCz1z2.\|\nabla\phi\|_{L^{\infty}}\leq\frac{C}{z_{1}-z_{2}}.

Thus (2.22) indicates that

(Σ×[Lz2,L+z2](|g|FL(𝒟))6q𝑑x)16qC12q((z1z2)2+CbL(𝒟)+q2)12q×(Σ×[Lz1,L+z1](|g|FL(𝒟))2q𝑑x)12q.\begin{split}\left(\int_{\Sigma\times[L-z_{2},L+z_{2}]}\left(|g|\vee\|F\|_{L^{\infty}(\mathcal{D})}\right)^{6q}dx\right)^{\frac{1}{6q}}&\leq C^{\frac{1}{2q}}\Big{(}(z_{1}-z_{2})^{-2}+C_{\|b\|_{L^{\infty}(\mathcal{D})}}+q^{2}\Big{)}^{\frac{1}{2q}}\\ &\hskip 14.22636pt\times\left(\int_{\Sigma\times[L-z_{1},L+z_{1}]}\left(|g|\vee\|F\|_{L^{\infty}(\mathcal{D})}\right)^{2q}dx\right)^{\frac{1}{2q}}.\end{split} (2.23)

Now k{0}\forall k\in\mathbb{N}\cup\{0\}, we choose qk=3kq_{k}=3^{k} and z1k=12+(12)k+1z_{1k}=\frac{1}{2}+\left(\frac{1}{2}\right)^{k+1}, z2k=z1,k+1=12+(12)k+2z_{2k}=z_{1,k+1}=\frac{1}{2}+\left(\frac{1}{2}\right)^{k+2}, respectively. Denoting

Ψk:=(Σ×[Lz1k,L+z1k](|g|FL(𝒟))2qk𝑑x)12qk,\Psi_{k}:=\left(\int_{\Sigma\times[L-z_{1k},L+z_{1k}]}\left(|g|\vee\|F\|_{L^{\infty}(\mathcal{D})}\right)^{2q_{k}}dx\right)^{\frac{1}{2q_{k}}},

and iterating (3.68), it follows that

Ψk+1C123k(4k+2+CbL+32k)123kΨk(CbL(𝒟))12j=0k3j3j=0kj3jΨ0CbL(𝒟)Ψ0.\begin{split}\Psi_{k+1}&\leq C^{\frac{1}{2\cdot 3^{k}}}\left(4^{k+2}+C_{\|b\|_{L^{\infty}}}+3^{2k}\right)^{\frac{1}{2\cdot 3^{k}}}\Psi_{k}\leq\cdot\cdot\cdot\leq\left(C_{\|b\|_{L^{\infty}(\mathcal{D})}}\right)^{\frac{1}{2}\sum_{j=0}^{k}3^{-j}}3^{\sum_{j=0}^{k}j3^{-j}}\Psi_{0}\leq C_{\|b\|_{L^{\infty}(\mathcal{D})}}\Psi_{0}.\end{split}

Performing kk\to\infty, the above Moser’s iteration implies

gL(Σ×[L1/2,L+1/2])CbL(𝒟)(gL2(Σ×[L1,L+1])+FL(𝒟)).\|g\|_{L^{\infty}(\Sigma\times[L-1/2,L+1/2])}\leq C_{\|b\|_{L^{\infty}(\mathcal{D})}}\left(\|g\|_{L^{2}(\Sigma\times[L-1,L+1])}+\|F\|_{L^{\infty}(\mathcal{D})}\right). (2.24)

Finally, define another cut off function of zz-variable ϕ~\tilde{\phi} who has bounded derivatives up to order 2, supported on [L2,L+2][L-2,L+2] and ϕ~1\tilde{\phi}\equiv 1 in [L1,L+1][L-1,L+1]. Multiplying (1.4)2 by ωθϕ~2\omega_{\theta}\tilde{\phi}^{2} and integrating on 𝒟\mathcal{D}, one deduces

𝒟|(ωθϕ~)|2𝑑x+𝒟ωθ2ϕ~2r2𝑑x=𝒟ωθ2|ϕ~|2𝑑x𝒟uzzϕ~ωθ2ϕ~dx𝒟urrωθ2ϕ~2𝑑x2𝒟uθrzuθωθϕ~2dx.\int_{\mathcal{D}}|\nabla(\omega_{\theta}\tilde{\phi})|^{2}dx+\int_{\mathcal{D}}\frac{\omega_{\theta}^{2}\tilde{\phi}^{2}}{r^{2}}dx=\int_{\mathcal{D}}\omega_{\theta}^{2}|\nabla\tilde{\phi}|^{2}dx-\int_{\mathcal{D}}u_{z}\partial_{z}\tilde{\phi}\omega_{\theta}^{2}\tilde{\phi}dx-\int_{\mathcal{D}}\frac{u_{r}}{r}\omega_{\theta}^{2}\tilde{\phi}^{2}dx-2\int_{\mathcal{D}}\frac{u_{\theta}}{r}\partial_{z}u_{\theta}\omega_{\theta}\tilde{\phi}^{2}dx.

By the representation of u\nabla u (A.86), one derives that

ωθL2(Σ×[L1,L+1])C(u,u)L(𝒟).\|\nabla\omega_{\theta}\|_{L^{2}(\Sigma\times[L-1,L+1])}\leq C_{\|(u,\nabla u)\|_{L^{\infty}(\mathcal{D})}}. (2.25)

Meanwhile, expression of FF (2.18) also indicates that

FL(𝒟)C(u,u)L(𝒟).\|F\|_{L^{\infty}(\mathcal{D})}\leq C_{\|(u,\nabla u)\|_{L^{\infty}(\mathcal{D})}}. (2.26)

Substituting (2.25) and (2.26) in (2.24), one concludes that

gL(Σ×[L1/2,L+1/2])C(u,u)L(𝒟).\|g\|_{L^{\infty}(\Sigma\times[L-1/2,L+1/2])}\leq C_{\|(u,\nabla u)\|_{L^{\infty}(\mathcal{D})}}.

Noting that the right-hand side above is independent of LL, thus we have derived the uniform boundedness of gg in 𝒟\mathcal{D}.

2.1.2 Boundedness of the pressure

Based on the boundedness of zωθ\partial_{z}\omega_{\theta}, the LL^{\infty} oscillation bound of the pressure pp in D2Z\DZD_{2Z}\backslash D_{Z} can be obtained. The lemma is stated as follows:

Lemma 2.2.

Under the same assumptions of Theorem 1.1, \forall Z>1Z>1, we have

supx𝒟2Z\𝒟Z|p(r,z)p(0,Z)|C,\sup_{x\in\mathcal{D}_{2Z}\backslash\mathcal{D}_{Z}}|p(r,z)-p(0,Z)|\leq C, (2.27)

where C>0C>0 is a uniform constant independent of ZZ.

Proof.   We only consider (𝒟2Z\𝒟Z){x:z>0}\left(\mathcal{D}_{2Z}\backslash\mathcal{D}_{Z}\right)\cap\{x\,:\,z>0\} since the rest part is essentially the same. Let us start with the oscillation of the pressure along the rr-axis. From (1.3)1\eqref{ASNS}_{1} and the identity

(Δ1r2)ur=zωθ,\left(\Delta-\frac{1}{r^{2}}\right)u_{r}=\partial_{z}\omega_{\theta},

one sees that

rp=zωθ(urr+uzz)ur+(uθ)2r.\partial_{r}p=\partial_{z}\omega_{\theta}-(u_{r}\partial_{r}+u_{z}\partial_{z})u_{r}+\frac{(u_{\theta})^{2}}{r}. (2.28)

For any zz\in\mathbb{R} and r1r_{1}, r2[0,1]r_{2}\in[0,1], we integrate (2.28) with rr on [r1,r2][r_{1},r_{2}] to derive

p(r2,z)p(r1,z)=r1r2zωθdrr1r2[(urr+uzz)uruθ2r]𝑑r=r1r2zωθ(r,z)dr12(ur2(r2,z)ur2(r1,z))r1r2(uzzur)(r,z)𝑑r+r1r2uθ2r(r,z)𝑑r.\begin{split}p(r_{2},z)-p(r_{1},z)=&\int_{r_{1}}^{r_{2}}\partial_{z}\omega_{\theta}dr-\int_{r_{1}}^{r_{2}}\left[(u_{r}\partial_{r}+u_{z}\partial_{z})u_{r}-\frac{u^{2}_{\theta}}{r}\right]dr\\ =&\int_{r_{1}}^{r_{2}}\partial_{z}\omega_{\theta}(r,z)dr-\frac{1}{2}\left(u_{r}^{2}(r_{2},z)-u_{r}^{2}(r_{1},z)\right)-\int_{r_{1}}^{r_{2}}(u_{z}\partial_{z}u_{r})(r,z)dr\\ &+\int_{r_{1}}^{r_{2}}\frac{u_{\theta}^{2}}{r}(r,z)dr.\\ \end{split} (2.29)

Noting that

|u||rur|+|zur|+|urr|+|ruθ|+|zuθ|+|uθr|+|ruz|+|zuz|,|\nabla u|\simeq|\partial_{r}u_{r}|+|\partial_{z}u_{r}|+\left|\frac{u_{r}}{r}\right|+|\partial_{r}u_{\theta}|+|\partial_{z}u_{\theta}|+\left|\frac{u_{\theta}}{r}\right|+|\partial_{r}u_{z}|+|\partial_{z}u_{z}|,

which follows from (A.86), by the boundedness assumption of uu and u\nabla u, together with the uniform bound of zωθ\partial_{z}\omega_{\theta} in Section 2.1.1, one derives the oscillation bound from (2.29):

|p(r2,z)p(r1,z)|C(1+zωθL(𝒟2Z))C<,r1,r2[0,1],z,|p(r_{2},z)-p(r_{1},z)|\leq\,C(1+\|\partial_{z}\omega_{\theta}\|_{L^{\infty}(\mathcal{D}_{2Z})})\leq C<\infty,\quad\forall r_{1},r_{2}\in[0,1],\quad z\in\mathbb{R}, (2.30)

where CC is an absolute constant which is independent of r1r_{1}, r2r_{2} and zz. This finishes the oscillation estimate of p(r,z)p(r,z) when zz is fixed. Now we turn to the oscillation of pp along the zz-direction. (1.3)3 and identity

Δuz=1rr(rωθ)-\Delta u_{z}=\frac{1}{r}\partial_{r}(r\omega_{\theta})

indicate that

zp=1rr(rωθ)urruzuzzuz.\partial_{z}p=-\frac{1}{r}\partial_{r}(r\omega_{\theta})-u_{r}\partial_{r}u_{z}-u_{z}\partial_{z}u_{z}. (2.31)

Multiplying (2.31) by rr and integrating it with respect to rr on (0,1)(0,1), one obtains

ddz01p(r,z)r𝑑r=01r(rωθ)drP101(urr+uzz)uzr𝑑rP2.\begin{split}\frac{d}{dz}\int_{0}^{1}p(r,z)rdr&=-\underbrace{\int_{0}^{1}\partial_{r}(r\omega_{\theta})dr}_{P_{1}}-\underbrace{\int_{0}^{1}(u_{r}\partial_{r}+u_{z}\partial_{z})u_{z}rdr}_{P_{2}}.\end{split} (2.32)

Recalling the boundary condition (1.5)2,3, we find ωθ0\omega_{\theta}\equiv 0 on 𝒟\partial\mathcal{D}, which implies P10P_{1}\equiv 0. On the other hand, using the divergence-free condition and integration by parts, we derive

P2=01r(rur)uzdr+01uzzuzrdr=01z(ruz)uzdr+12ddz01uz2r𝑑r=ddz01uz2r𝑑r.\begin{split}P_{2}=&-\int_{0}^{1}\partial_{r}\left(ru_{r}\right)u_{z}dr+\int_{0}^{1}u_{z}\partial_{z}u_{z}rdr\\ =&\int_{0}^{1}\partial_{z}\left(ru_{z}\right)u_{z}dr+\frac{1}{2}\frac{d}{dz}\int_{0}^{1}u_{z}^{2}rdr\\ =&\frac{d}{dz}\int_{0}^{1}u_{z}^{2}rdr.\end{split}

(2.32) indicates

ddz01p(r,z)r𝑑r=ddz01uz2(r,z)r𝑑r.\frac{d}{dz}\int_{0}^{1}p(r,z)rdr=-\frac{d}{dz}\int_{0}^{1}u_{z}^{2}(r,z)rdr. (2.33)

For any fixed z[Z,2Z]z\in[Z,2Z], we integrate the above indentity from ZZ to zz. Then we have

|01[p(r,z)p(r,Z)]r𝑑r||01[uz2(r,z)uz2(r,Z)]r𝑑r|C.\begin{split}&\left|\int_{0}^{1}\big{[}p(r,z)-p(r,Z)\big{]}rdr\right|\leq\left|\int_{0}^{1}\big{[}u_{z}^{2}(r,z)-u_{z}^{2}(r,Z)\big{]}rdr\right|\leq C.\end{split} (2.34)

Recalling the mean value theorem, there exists r[0,1]r_{*}\in[0,1] such that

|p(r,z)p(r,Z)|=|01[p(r,z)p(r,Z)]r𝑑r|01r𝑑rC.|p(r_{*},z)-p(r_{*},Z)|=\frac{\left|\int_{0}^{1}\big{[}p(r,z)-p(r,Z)\big{]}rdr\right|}{\int_{0}^{1}rdr}\leq C. (2.35)

This completes the oscillation of pp parallel to the zz-direction. To conclude the general oscillation of the pressure in the pipe, we apply the triangle inequality: for any r[0,1]r\in[0,1], it follows that

|p(r,z)p(0,Z)||p(r,z)p(r,z)|+|p(r,z)p(r,Z)|+|p(r,Z)p(r,Z)|.|p(r,z)-p(0,Z)|\leq|p(r,z)-p(r_{*},z)|+|p(r_{*},z)-p(r_{*},Z)|+|p(r_{*},Z)-p(r,Z)|.

Plugging (2.30) and (2.35) into the above inequality, we finally arrive at

|p(r,z)p(0,Z)|C,|p(r,z)-p(0,Z)|\leq C, (2.36)

where CC is an absolute positive constant independent of rr, zz and ZZ. Thus (2.27) is proved by taking the supremum of (2.36) over (r,z)[0,1]×([2Z,Z][Z,2Z])(r,z)\in[0,1]\times{\big{(}[-2Z,-Z]\cup[Z,2Z]\big{)}}.

2.1.3 End of the proof

In this subsection, we will finish the proof of Theorem 1.1. Namely: If the flux Φ0\Phi\equiv 0, any smooth solution of (1.3) in an infinite pipe subject to the Navier total slip condition with the velocity and its first-order derivatives being bounded must be a swirling solution

u=C1r𝒆𝜽.u=C_{1}r\boldsymbol{e_{\theta}}.

The proof is divided into three steps: First we show the stress tensor 𝕊u=12(u+(u)T)\mathbb{S}u=\frac{1}{2}\left(\nabla u+(\nabla u)^{T}\right) is globally L2L^{2}-integrable. Using a 2D Poincaré inequality and one insightful observation motivated by [35], we then find that uzu_{z} also belongs to L2(𝒟)L^{2}(\mathcal{D}). Finally, we arrive at the vanishing of the stress tensor, which indicates the desired result in Theorem 1.1.

𝑳𝟐\boldsymbol{L^{2}} boundendness of stress tensor

Let ψ:[0,1]\psi:\,\mathbb{R}\to[0,1] be a smooth cut-off function satisfying

ψ(l)={1,l[1,1],0,|l|2,\psi(l)=\left\{\begin{aligned} &1,\quad l\in[-1,1],\\ &0,\quad|l|\geq 2,\end{aligned}\right.

with ψ\psi^{\prime} and ψ′′\psi^{\prime\prime} being bounded. Set

ψZ(z):=ψ(zZ),\psi_{Z}(z):=\psi\left(\frac{z}{Z}\right),

where ZZ is a large positive number. Clearly the derivatives of the scaled cut-off function ψZ\psi_{Z} enjoy

|znψZ|CZn,for any n.|\partial^{n}_{z}\psi_{Z}|\leq\frac{C}{Z^{n}},\quad\text{for any }n\in\mathbb{N}. (2.37)

Tesing the equation

uu+p=Δuu\cdot\nabla u+\nabla p=\Delta u

with uψZu\psi_{Z}, we have

𝒟ψZuΔu𝑑x=𝒟ψZu(uu+(pp(0,Z)))𝑑x.\begin{split}\int_{\mathcal{D}}\psi_{Z}u\Delta udx=\int_{\mathcal{D}}\psi_{Z}u\Big{(}u\cdot\nabla u+\nabla\left(p-p(0,Z)\right)\Big{)}dx.\end{split} (2.38)

To proceed the further calculation in the cylindrical coordinates, we first note that the divergence free property of the velocity indicates

i,j=13𝒟2ZψZuijjuidx=i,j=13𝒟2ZψZuij(jui+iuj)dx.\begin{split}\sum_{i,j=1}^{3}\int_{\mathcal{D}_{2Z}}\psi_{Z}u_{i}\partial_{jj}u_{i}dx=\sum_{i,j=1}^{3}\int_{\mathcal{D}_{2Z}}\psi_{Z}u_{i}\partial_{j}\left(\partial_{j}u_{i}+\partial_{i}u_{j}\right)dx.\\ \end{split} (2.39)

Below, we use the Einstein summation convention for repeated indexes. Using integration by parts, we further derive

𝒟2ZψZuij(jui+iuj)dx=𝒟2ZjψZui(jui+iuj)dxT1𝒟2ZψZjui(jui+iuj)dxT2+𝒟2ZψZuinj(jui+iuj)𝑑ST3,\begin{split}\int_{\mathcal{D}_{2Z}}\psi_{Z}u_{i}\partial_{j}\left(\partial_{j}u_{i}+\partial_{i}u_{j}\right)dx=&-\underbrace{\int_{\mathcal{D}_{2Z}}\partial_{j}\psi_{Z}u_{i}\left(\partial_{j}u_{i}+\partial_{i}u_{j}\right)dx}_{T_{1}}-\underbrace{\int_{\mathcal{D}_{2Z}}\psi_{Z}\partial_{j}u_{i}\left(\partial_{j}u_{i}+\partial_{i}u_{j}\right)dx}_{T_{2}}\\ &+\underbrace{\int_{\partial\mathcal{D}_{2Z}}\psi_{Z}u_{i}n_{j}\left(\partial_{j}u_{i}+\partial_{i}u_{j}\right)dS}_{T_{3}},\end{split}

where njn_{j} is the jj-th component of the 𝒏\boldsymbol{n} – the unit outward normal vector field on 𝒟2Z\partial\mathcal{D}_{2Z}. Term T1T_{1} could be split into two parts, the first half reads

𝒟2ZjψZuijuidx=\displaystyle\int_{\mathcal{D}_{2Z}}\partial_{j}\psi_{Z}u_{i}\partial_{j}u_{i}dx= 12𝒟2ZjψZj|u|2dx=12𝒟2ZzψZz|u|2dx=12𝒟2Zz2ψZ|u|2dx,\displaystyle\frac{1}{2}\int_{\mathcal{D}_{2Z}}\partial_{j}\psi_{Z}\partial_{j}|u|^{2}dx=\frac{1}{2}\int_{\mathcal{D}_{2Z}}\partial_{z}\psi_{Z}\partial_{z}|u|^{2}dx=-\frac{1}{2}\int_{\mathcal{D}_{2Z}}\partial_{z}^{2}\psi_{Z}|u|^{2}dx,

where we have used the fact that ψZ\psi_{Z} is only zz-dependent and supported in [2Z,2Z][-2Z,2Z]. Similarly, the second half of T1T_{1} follows that

𝒟2ZjψZuiiujdx=𝒟2Z(uψZ)(u𝒏)𝑑S𝒟2Zz2ψZuz2dx.\int_{\mathcal{D}_{2Z}}\partial_{j}\psi_{Z}u_{i}\partial_{i}u_{j}dx=\int_{\partial\mathcal{D}_{2Z}}(u\cdot\nabla\psi_{Z})(u\cdot\boldsymbol{n})dS-\int_{\mathcal{D}_{2Z}}\partial_{z}^{2}\psi_{Z}u_{z}^{2}dx.

Due to the impermeable condition, one sees the first term on the right hand of the above equality is zero. Thus we conclude that

T1=𝒟2Zz2ψZ(12|u|2+uz2)dx.T_{1}=-\int_{\mathcal{D}_{2Z}}\partial_{z}^{2}\psi_{Z}\left(\frac{1}{2}|u|^{2}+u_{z}^{2}\right)dx. (2.40)

Recalling that the stress tensor is defined by

𝕊u=12(jui+iuj)1i,j3,\mathbb{S}u=\frac{1}{2}\left(\partial_{j}u_{i}+\partial_{i}u_{j}\right)_{1\leq i,j\leq 3},

and using its symmetry, we arrive that

T2=12i,j=13𝒟2ZψZ(jui+iuj)2𝑑x=2𝒟2ZψZ|𝕊u|2𝑑x.T_{2}=\frac{1}{2}\sum_{i,j=1}^{3}\int_{\mathcal{D}_{2Z}}\psi_{Z}\left(\partial_{j}u_{i}+\partial_{i}u_{j}\right)^{2}dx=2\int_{\mathcal{D}_{2Z}}\psi_{Z}|\mathbb{S}u|^{2}dx. (2.41)

Now applying the Navier-slip condition (NSB)1, one notes that

nj(jui+iuj)=c(x)ni,n_{j}\left(\partial_{j}u_{i}+\partial_{i}u_{j}\right)=c(x)n_{i},

where c(x)c(x) is a scalar-valued function. Inserting this identity to T3T_{3}, we find

T3=𝒟2ZcψZ(u𝒏)𝑑S=0.T_{3}=\int_{\partial\mathcal{D}_{2Z}}c\psi_{Z}\left(u\cdot\boldsymbol{n}\right)dS=0. (2.42)

Next we come back to the right hand side of (2.38). Noting uu is divergence-free, integration by parts shows

𝒟2ZuψZ(uu+[pp(0,Z)])𝑑x=𝒟2ZψZuii(12|u|2+[pp(0,Z)])dx=𝒟2ZψZ(u𝒏)(12|u|2+[pp(0,Z)])𝑑ST4𝒟2ZzψZuz(12|u|2+[pp(0,Z)])dx.\begin{split}&\int_{\mathcal{D}_{2Z}}u\psi_{Z}\Big{(}u\cdot\nabla u+\nabla\big{[}p-p(0,Z)\big{]}\Big{)}dx=\int_{\mathcal{D}_{2Z}}\psi_{Z}u_{i}\partial_{i}\left(\frac{1}{2}|u|^{2}+\big{[}p-p(0,Z)\big{]}\right)dx\\ =&\underbrace{\int_{\partial\mathcal{D}_{2Z}}\psi_{Z}(u\cdot\boldsymbol{n})\Big{(}\frac{1}{2}|u|^{2}+\big{[}p-p(0,Z)\big{]}\Big{)}dS}_{T_{4}}-\int_{\mathcal{D}_{2Z}}\partial_{z}\psi_{Z}u_{z}\left(\frac{1}{2}|u|^{2}+\big{[}p-p(0,Z)\big{]}\right)dx.\end{split} (2.43)

Here T4T_{4} above also vanishes by the stationary wall condition (1.5)3\eqref{NBC1}_{3}. Therefore we arrive that by plugging (2.40), (2.41), (2.42), (2.43) into (2.38)

2𝒟2ZψZ|𝕊u|2𝑑x=𝒟2Zz2ψZ(12|u|2+uz2)dx+𝒟2ZzψZuz(12|u|2+[pp(0,Z)])dxT5.2\int_{\mathcal{D}_{2Z}}\psi_{Z}|\mathbb{S}u|^{2}dx=\int_{\mathcal{D}_{2Z}}\partial^{2}_{z}\psi_{Z}\left(\frac{1}{2}|u|^{2}+u_{z}^{2}\right)dx+\underbrace{\int_{\mathcal{D}_{2Z}}\partial_{z}\psi_{Z}u_{z}\left(\frac{1}{2}|u|^{2}+\big{[}p-p(0,Z)\big{]}\right)dx}_{T_{5}}. (2.44)

Recalling (2.37), the bounds on the derivatives of scaled cut-off function ψZ\psi_{Z}, and the boundedness of uu and pressure, one derives from (2.44) that

𝒟2ZψZ|𝕊u|2𝑑xC|𝒟2Z|(Z2+Z1)C,\int_{\mathcal{D}_{2Z}}\psi_{Z}|\mathbb{S}u|^{2}dx\leq C|\mathcal{D}_{2Z}|\left(Z^{-2}+Z^{-1}\right)\leq C,

where CC is a universal constant depending only on the LL^{\infty} bound of uu and u\nabla u given in the assumption. After letting ZZ\to\infty, the above inequality shows the stress tensor is globally L2L^{2}-integrable:

𝒟|𝕊u|2𝑑xC<.\int_{\mathcal{D}}|\mathbb{S}u|^{2}dx\leq C<\infty. (2.45)

𝑳𝟐\boldsymbol{L^{2}} boundedness of uzu_{z}

First we observe that uzL2(𝒟)\|u_{z}\|_{L^{2}(\mathcal{D})} can be controlled by ruzL2(𝒟)\|\partial_{r}u_{z}\|_{L^{2}(\mathcal{D})} under the assumption that the flux Φ0\Phi\equiv 0. Noting that

1|Σ|Σuz(xh,z)𝑑xh=1|Σ|Φ=0,\frac{1}{|\Sigma|}\int_{\Sigma}u_{z}(x_{h},z)dx_{h}=\frac{1}{|\Sigma|}\Phi=0,

then we apply the one dimensional Poincaré inequality to derive

Σ|uz(r,z)|2𝑑xh=\displaystyle\int_{\Sigma}|u_{z}(r,z)|^{2}dx_{h}= Σ|uz(xh,z)1|Σ|Σuz(xh,z)𝑑xh|2𝑑xh\displaystyle\int_{\Sigma}\left|u_{z}(x_{h},z)-\frac{1}{|\Sigma|}\int_{\Sigma}u_{z}(x_{h},z)dx_{h}\right|^{2}dx_{h}
\displaystyle\leq S02Σ|huz(xh,z)|2𝑑xh=S02Σ|ruz(r,z)|2𝑑xh,\displaystyle S^{2}_{0}\int_{\Sigma}|\nabla_{h}u_{z}(x_{h},z)|^{2}dx_{h}=S^{2}_{0}\int_{\Sigma}|\partial_{r}u_{z}(r,z)|^{2}dx_{h},

where h=(1,2)\nabla_{h}=(\partial_{1},\partial_{2}) and S0S_{0} is independent of zz\in\mathbb{R}. Integrating with zz-variable on \mathbb{R}, we arrive

uzL2(𝒟)S0ruzL2(𝒟).\|u_{z}\|_{L^{2}(\mathcal{D})}\leq S_{0}\|\partial_{r}u_{z}\|_{L^{2}(\mathcal{D})}. (2.46)

However, we cannot get the L2L^{2} boundedness of ruz\partial_{r}u_{z} directly from (2.45). In fact, by the expression of the stress tensor (A.88), one only has the uniform L2L^{2} bound of (zur+ruz)(\partial_{z}u_{r}+\partial_{r}u_{z}). Nevertheless, one observes

𝒟2Z(ruz)2𝑑x=𝒟2Z(zur+ruz)2𝑑x𝒟2Z(zur)2𝑑x2𝒟2ZruzzurdxC+2|𝒟2ZruzzurdxT6|.\begin{split}\int_{\mathcal{D}_{2Z}}\left(\partial_{r}u_{z}\right)^{2}dx&=\int_{\mathcal{D}_{2Z}}\left(\partial_{z}u_{r}+\partial_{r}u_{z}\right)^{2}dx-\int_{\mathcal{D}_{2Z}}\left(\partial_{z}u_{r}\right)^{2}dx-2\int_{\mathcal{D}_{2Z}}\partial_{r}u_{z}\partial_{z}u_{r}dx\\ &\leq C+2\Big{|}\underbrace{\int_{\mathcal{D}_{2Z}}\partial_{r}u_{z}\partial_{z}u_{r}dx}_{T_{6}}\Big{|}.\end{split}

Now it remains to derive the boundedness of T6T_{6}. With idea motivated by [35], after using the divergence free of uu and integration by parts, we deduce

D2Zruzzurdx=2π2Z2Z01uzrz2(rur)drdz=2π2Z2Z01uzz2(ruz)drdz=D2Z(zuz)2𝑑xT7+2π(01uz(r,2Z)zuz(r,2Z)rdr01uz(r,2Z)zuz(r,2Z)rdr)T8.\begin{split}&\quad\int_{D_{2Z}}\partial_{r}u_{z}\partial_{z}u_{r}dx\\ &=-2\pi\int_{-2Z}^{2Z}\int_{0}^{1}u_{z}\partial^{2}_{rz}(ru_{r})drdz=2\pi\int_{-2Z}^{2Z}\int_{0}^{1}u_{z}\partial^{2}_{z}(ru_{z})drdz\\ &=-\underbrace{\int_{D_{2Z}}\left(\partial_{z}u_{z}\right)^{2}dx}_{T_{7}}+2\pi\underbrace{\left(\int_{0}^{1}u_{z}(r,2Z)\partial_{z}u_{z}(r,2Z)rdr-\int_{0}^{1}u_{z}(r,-2Z)\partial_{z}u_{z}(r,-2Z)rdr\right)}_{T_{8}}.\end{split}

Here T7T_{7} can be bounded by the L2L^{2} norm of stress tensor (2.45), while T8T_{8} is controlled by the LL^{\infty} bounds of uu and u\nabla u. Noting that T6T_{6} is estimated uniformly with respect to ZZ. This, together with (2.46) implies

uzL2(𝒟)C<.\|u_{z}\|_{L^{2}(\mathcal{D})}\leq C<\infty.

Vanishing of 𝒟|𝕊u|𝟐\boldsymbol{\int_{\mathcal{D}}|\mathbb{S}u|^{2}} and finishing of the proof

Based on the L2L^{2} bound of uzu_{z}, now we can estimate T5T_{5} in (2.44) in an alternative approach, by using Hölder inequality:

|T5|supx𝒟2Z\𝒟Z|12|u|2+[pp(0,Z)]|CZuzL2(𝒟2Z)|𝒟2Z|1/2CZ1/2.|T_{5}|\leq\sup_{x\in\mathcal{D}_{2Z}\backslash\mathcal{D}_{Z}}\left|\frac{1}{2}|u|^{2}+\big{[}p-p(0,Z)\big{]}\right|\frac{C}{Z}\|u_{z}\|_{L^{2}(\mathcal{D}_{2Z})}|\mathcal{D}_{2Z}|^{1/2}\leq CZ^{-1/2}.

Thus we deduce from (2.44)

𝒟2ZψZ|𝕊u|2𝑑xC|𝒟2Z|Z2+CZ1/20,as Z+,\int_{\mathcal{D}_{2Z}}\psi_{Z}|\mathbb{S}u|^{2}dx\leq C|\mathcal{D}_{2Z}|Z^{-2}+CZ^{-1/2}\rightarrow 0,\quad\text{as }Z\rightarrow+\infty,

which indicates that

𝒟|𝕊u|2𝑑x=0\int_{\mathcal{D}}|\mathbb{S}u|^{2}dx=0 (2.47)

by letting ZZ\to\infty. By the expression of 𝕊u\mathbb{S}u (A.88), one finds

urzuθzuzruz0,ruθ=uθr.u_{r}\equiv\partial_{z}u_{\theta}\equiv\partial_{z}u_{z}\equiv\partial_{r}u_{z}\equiv 0,\quad\partial_{r}u_{\theta}=\frac{u_{\theta}}{r}.

The above estimates, together with the vanishing flux (Φ=0\Phi=0), indicate

uz0,anduθ=Cr.u_{z}\equiv 0,\quad\text{and}\quad u_{\theta}=Cr.

Thus we conclude that u=Cr𝒆𝜽u=Cr\boldsymbol{e_{\theta}}, which is a swirling solution.

2.2 Proof of Case II

If uu is zz-periodic with the minimal period L>0L>0, then we can drop the restriction Φ=0\Phi=0 in Case I.
The proof is straightforward. Set w=uΦπ𝒆𝒛w=u-\frac{\Phi}{\pi}\boldsymbol{e_{z}}, then (w,p)(w,p) satisfies the following system:

{(w+Φπ𝒆𝒛)w+pΔw=0,in 𝒟,w=0,in 𝒟,w(xh,z)=w(xh,z+L),in 𝒟,(𝕊w𝒏)τ=0,w𝒏=0,on 𝒟.\begin{cases}(w+\frac{\Phi}{\pi}\boldsymbol{e_{z}})\cdot\nabla w+\nabla p-\Delta w=0\,,\quad&\text{in $\mathcal{D}$}\,,\\ \nabla\cdot w=0\,,&\text{in $\mathcal{D}$}\,,\\ w(x_{h},z)=w(x_{h},z+L)\,,&\text{in $\mathcal{D}$}\,,\\ (\mathbb{S}w\cdot\boldsymbol{n})_{\tau}=0\,,\quad w\cdot\boldsymbol{n}=0\,,&\text{on $\partial\mathcal{D}$}\,.\end{cases} (2.48)

We first claim that, the pressure p(xh,z)p(x_{h},z) has the following decomposition:

p(xh,z)=az+p~(xh,z),p(x_{h},z)=\,az+\tilde{p}(x_{h},z), (2.49)

where aa is a constant, and p~\tilde{p} is zz-periodic with the minimal period L>0L>0. Set

a0(xh)=1L0L(zp)(xh,z)𝑑z,a_{0}(x_{h})=\frac{1}{L}\int_{0}^{L}(\partial_{z}p)(x_{h},z)dz\,,

we decompose zp\partial_{z}p as

(zp)(xh,z)=a0(xh)+((zp)(xh,z)a0(xh)):=a0(xh)+p1(xh,z).(\partial_{z}p)(x_{h},z)=a_{0}(x_{h})+\left((\partial_{z}p)(x_{h},z)-a_{0}(x_{h})\right):=a_{0}(x_{h})+p_{1}(x_{h},z)\,. (2.50)

By using the equation and zz-periodicity of the solution, it is easy to check that p1(xh,z)p_{1}(x_{h},z) is LL-periodic with respect to zz and that

0Lp1(xh,z~)𝑑z~=0.\int_{0}^{L}p_{1}(x_{h},\tilde{z})\,d\tilde{z}=0\,. (2.51)

Integrating (2.50) with zz on [0,z][0,z], one derives that

p(xh,z)=p(xh,0)+a0(xh)z+0zp1(xh,z~)𝑑z~.p(x_{h},z)=p(x_{h},0)+a_{0}(x_{h})z+\int_{0}^{z}p_{1}(x_{h},\tilde{z})\,d\tilde{z}\,.

It is worth noting that 0zp1(xh,z~)𝑑z~\int_{0}^{z}p_{1}(x_{h},\tilde{z})\,d\tilde{z} is periodic in the zz-direction due to (2.51) and periodicity of p1p_{1}. Hence,

p~(xh,z)=p(xh,0)+0zp1(xh,z~)𝑑z~\tilde{p}(x_{h},z)=p(x_{h},0)+\int_{0}^{z}p_{1}(x_{h},\tilde{z})\,d\tilde{z}

is periodic in the zz-direction, and

p(xh,z)=a0(xh)z+p~(xh,z).p(x_{h},z)=a_{0}(x_{h})z+\tilde{p}(x_{h},z)\,.

Finally, also from the equations and zz-periodicity of the solution, we deduce hp=(ha0)z+(hp~)\nabla_{h}p=(\nabla_{h}a_{0})z+(\nabla_{h}\tilde{p}) is periodic with respect to zz. Thus we get

a0(xh)=constant:=aa_{0}(x_{h})=\text{constant}:=a

since ha0\nabla_{h}a_{0} must be zero. Therefore we conclude

p(xh,z)=az+p~(xh,z).p(x_{h},z)=az+\tilde{p}(x_{h},z)\,.

This finishes the proof of the claim.

Next, multiplying ww on both sides of (2.48)1, and integrating on Σ×[0,L]\Sigma\times[0,L], one has

Σ×[0,L]wΔw𝑑x=Σ×[0,L]w((w+Φπ𝒆𝒛)w+p)𝑑x.\int_{\Sigma\times[0,L]}w\cdot\Delta w\,dx=\int_{\Sigma\times[0,L]}w\cdot\left(\big{(}w+\frac{\Phi}{\pi}\boldsymbol{e_{z}}\big{)}\cdot\nabla w+\nabla p\right)\,dx\,.

It follows from (2.48)2–(2.48)4 that

Σ×[0,L]wΔw𝑑x=2Σ×[0,L]|𝕊w|2𝑑x,\int_{\Sigma\times[0,L]}w\cdot\Delta w\,dx=-2\int_{\Sigma\times[0,L]}|\mathbb{S}w|^{2}\,dx,

where we have used the technique from (2.39) to (2.41) and zz-periodicity of ww to do integration by parts. There are no boundary terms generated. Also

Σ×[0,L]w((w+Φπ𝒆𝒛)w+p)𝑑x=0,\int_{\Sigma\times[0,L]}w\cdot\left(\big{(}w+\frac{\Phi}{\pi}\boldsymbol{e_{z}}\big{)}\cdot\nabla w+\nabla p\right)\,dx=0\,,

where we have used the decomposition (2.49) and Σw3𝑑xh=0\int_{\Sigma}w_{3}\,dx_{h}=0 to deal with the pressure term. Hence, we have that

Σ×[0,L]|𝕊w|2𝑑x=0,\int_{\Sigma\times[0,L]}|\mathbb{S}w|^{2}\,dx=0\,,

which deduces that 𝕊w=0\mathbb{S}w=0. It is well known that if 𝕊w=0\mathbb{S}w=0, then ww has the form w=Ax+Bw=Ax+B (see [16, §6]), where AA is a skew-symmetric matrix with constant entries and BB is a constant vector, that is,

w=(0a1a2a10a3a2a30)(x1x2z)+(b1b2b3)=(a1x2a2z+b1a1x1a3z+b2a2x1+a3x2+b3),w=\begin{pmatrix}0&-a_{1}&-a_{2}\\ a_{1}&0&-a_{3}\\ a_{2}&a_{3}&0\end{pmatrix}\begin{pmatrix}x_{1}\\ x_{2}\\ z\end{pmatrix}+\begin{pmatrix}b_{1}\\ b_{2}\\ b_{3}\end{pmatrix}=\begin{pmatrix}-a_{1}\,x_{2}-a_{2}\,z+b_{1}\\ a_{1}\,x_{1}-a_{3}\,z+b_{2}\\ a_{2}x_{1}+a_{3}x_{2}+b_{3}\end{pmatrix}\,,

where ai,bia_{i}\,,b_{i} (i=1,2,3i=1\,,2\,,3) are some constants. Note that ww is periodic with respect to zz, we have that a2=a3=0a_{2}=a_{3}=0. Next, from Σw3𝑑xh=0\int_{\Sigma}w_{3}\,dx_{h}=0, we obtain that b3=0b_{3}=0. Finally, note that wr=0w_{r}=0 on r=1r=1, we get that b1=b2=0b_{1}=b_{2}=0. Thus, we have proved that

w=(a1x2a1x10)=a1r𝒆𝜽.w=\begin{pmatrix}-a_{1}\,x_{2}\\ a_{1}\,x_{1}\\ 0\end{pmatrix}=a_{1}r\,\boldsymbol{e_{\theta}}\,.

Therefore, u=w+Φπ𝒆𝒛=a1r𝒆𝜽+Φπ𝒆𝒛u=w+\frac{\Phi}{\pi}\boldsymbol{e_{z}}=a_{1}r\,\boldsymbol{e_{\theta}}+\frac{\Phi}{\pi}\boldsymbol{e_{z}}.

Let us give some discussions of Theorem 1.1 here. Based on our previous proof in this section, we naturally believe that if the vanishing of Φ\Phi is abandoned, then an axially symmetric solution must be a helical solution:

u=C1r𝒆𝜽+C2𝒆𝒛u=C_{1}r\boldsymbol{e_{\theta}}+C_{2}\boldsymbol{e_{z}} (2.52)

even without the zz-periodic condition. However, our method in this paper fails when we handle solutions with the flux Φ0\Phi\neq 0, because we can no longer apply the Poincaré inequality in Section 2.1.3 to derive the L2L^{2} integrability of uzu_{z}. Meanwhile, if we denote

c0:=1|Σ|Σuz(xh,z)𝑑xh=1|Σ|Φ,c_{0}:=\frac{1}{|\Sigma|}\int_{\Sigma}u_{z}(x_{h},z)dx_{h}=\frac{1}{|\Sigma|}\Phi,

then uzc0u_{z}-c_{0} enjoys a similar Poincaré inequality as (2.46):

uzc0L2(𝒟)S0ruzL2(𝒟),\|u_{z}-c_{0}\|_{L^{2}(\mathcal{D})}\leq S_{0}\|\partial_{r}u_{z}\|_{L^{2}(\mathcal{D})},

which guarantees the L2L^{2} boundedness of uzc0u_{z}-c_{0}. However, one additional term appears in T5T_{5} of (2.44), which is:

T5:=c0𝒟2ZzψZ(12|u|2+[pp(0,Z)])dx.T_{5}^{\prime}:=c_{0}\int_{\mathcal{D}_{2Z}}\partial_{z}\psi_{Z}\left(\frac{1}{2}|u|^{2}+\big{[}p-p(0,Z)\big{]}\right)dx.

Without any integrability of the head pressure 12|u|2+[pp(0,Z)]\frac{1}{2}|u|^{2}+\big{[}p-p(0,Z)\big{]}, we can only show T5T_{5}^{\prime} is bounded, which results in

𝒟|𝕊u|2𝑑x<C<.\int_{\mathcal{D}}|\mathbb{S}u|^{2}dx<C<\infty.

However, we are unable to conclude T50T_{5}^{\prime}\to 0 as ZZ\to\infty, thus vanishing of 𝒟|𝕊u|2𝑑x\int_{\mathcal{D}}|\mathbb{S}u|^{2}dx can not be obtained. In fact, using integration by parts on zz in T5T_{5}^{\prime}, we have

T5=c0𝒟2ZψZz(12|u|2+p)dx.T_{5}^{\prime}=-c_{0}\int_{\mathcal{D}_{2Z}}\psi_{Z}\partial_{z}\left(\frac{1}{2}|u|^{2}+p\right)dx.

By following the argument in Section 2, one derives

𝒟|𝕊u|2𝑑x=limZc02𝒟ψZz(12|u|2+p)dx\int_{\mathcal{D}}|\mathbb{S}u|^{2}dx=-\lim_{Z\to\infty}\frac{c_{0}}{2}\int_{\mathcal{D}}\psi_{Z}\partial_{z}\left(\frac{1}{2}|u|^{2}+p\right)dx

instead of (2.47). Recalling (2.33), one deduces that

𝒟|𝕊u|2𝑑x=limZc04𝒟ψZz(ur2+uθ2uz2)dx.\int_{\mathcal{D}}|\mathbb{S}u|^{2}dx=-\lim_{Z\to\infty}\frac{c_{0}}{4}\int_{\mathcal{D}}\psi_{Z}\partial_{z}\left(u_{r}^{2}+u_{\theta}^{2}-u_{z}^{2}\right)dx. (2.53)

Thus if z(ur2+uθ2uz2)L1(𝒟)\partial_{z}\left(u_{r}^{2}+u_{\theta}^{2}-u_{z}^{2}\right)\in L^{1}(\mathcal{D}) (or z(ur2+uθ2uz2)\partial_{z}\left(u_{r}^{2}+u_{\theta}^{2}-u_{z}^{2}\right) has a fixed sign), one concludes the following identity by Lebesgue’s dominated convergence theorem (or monotone convergence theorem):

𝒟|𝕊u|2𝑑x+c04𝒟z(ur2+uθ2uz2)dx=0.\int_{\mathcal{D}}|\mathbb{S}u|^{2}dx+\frac{c_{0}}{4}\int_{\mathcal{D}}\partial_{z}\left(u_{r}^{2}+u_{\theta}^{2}-u_{z}^{2}\right)dx=0. (2.54)

At the moment, even with identities (2.53) and (2.54) for bounded (up to first-order derivatives) smooth axisymmetric solutions of stationary Navier-Stokes equations in 𝒟\mathcal{D} subject to the total Navier-slip boundary condition in hand, we neither show the trivialness of 𝕊u\mathbb{S}u, nor find a nontrivial bounded solution apart from (2.52) which satisfies conditions of Theorem 1.1. Indeed, we leave characterization of the non-zero flux solutions in Conjecture 1.6.

Nevertheless, a direct observation of (2.54) indicates that: If uu is independent of zz, then the right hand side of (2.54) vanishes and we can conclude 𝕊u0\mathbb{S}u\equiv 0, and thus conclude that u=C1r𝒆𝜽+C2𝒆𝒛u=C_{1}r\boldsymbol{e_{\theta}}+C_{2}\boldsymbol{e_{z}} as we desire. In the next section, we see that only uθu_{\theta} or uzu_{z} being independent of zz is adequate for us to derive Theorem 1.3. Besides, the asymptotic assumptions of uu and its derivatives can be largely loosened.

3  Proof of Theorem 1.3

3.1 Proof of Case I

Let us outline the proof at the beginning of this section: Under the assumptions of Case I in Theorem 1.3, our first step is showing ωθ0\omega_{\theta}\equiv 0, which indicates b=ur𝒆𝒓+uz𝒆𝒛b=u_{r}\boldsymbol{e_{r}}+u_{z}\boldsymbol{e_{z}} must be harmonic in 𝒟\mathcal{D}. Then by applying the boundary condition and the asymptotic behavior of bb, one derives ur0u_{r}\equiv 0 and uzu_{z} must be a constant. From then on (1.3)2 turns to a linear ordinary differential equation of uθu_{\theta}, and we finally prove uθ=C1ru_{\theta}=C_{1}r.

3.1.1 Vanishing of ωθ{\omega_{\theta}}

Noting that uθu_{\theta} is independent of zz, we find (1.4)2 now turns to

(urr+uzz)ωθ(Δ1r2)ωθurrωθ=0.(u_{r}\partial_{r}+u_{z}\partial_{z})\omega_{\theta}-\left(\Delta-\frac{1}{r^{2}}\right)\omega_{\theta}-\frac{u_{r}}{r}\omega_{\theta}=0.

From the Navier-slip boundary condition (1.5), one has

ωθ=zurruz=0,on 𝒟.\omega_{\theta}=\partial_{z}u_{r}-\partial_{r}u_{z}=0,\quad\text{on }\partial\mathcal{D}.

Denoting Ω:=ωθr\Omega:=\frac{\omega_{\theta}}{r}, direct calculation shows

{(urr+uzz)Ω(Δ+2rr)Ω=0,in 𝒟;Ω=0,on 𝒟.\left\{\begin{split}&(u_{r}\partial_{r}+u_{z}\partial_{z})\Omega-\left(\Delta+\frac{2}{r}\partial_{r}\right)\Omega=0,\quad&\text{in }\quad\mathcal{D};\\ &\Omega=0,\quad&\text{on }\quad\partial\mathcal{D}.\end{split}\right. (3.55)

In the following, we first provide a mean value inequality of Ω\Omega deduced by Moser’s iteration.

Lemma 3.1.

Assume b=ur𝐞𝐫+uz𝐞𝐳b=u_{r}\boldsymbol{e_{r}}+u_{z}\boldsymbol{e_{z}} is a smooth divergence-free axially symmetric vector field. Then any weak solution Ω\Omega of boundary value problem (3.55) satisfies the following mean value inequality:

supx𝒟τ2Z|Ω|Cq(τ1τ2)qq2(1+uzL(𝒟Z\𝒟Z/2))qq2Zqq2(𝒟τ1Z|Ω|2𝑑x)12,\sup_{x\in\mathcal{D}_{\tau_{2}Z}}|\Omega|\leq C_{q}(\tau_{1}-\tau_{2})^{-\frac{q}{q-2}}\left(1+\|u_{z}\|_{L^{\infty}(\mathcal{D}_{Z}\backslash\mathcal{D}_{Z/2})}\right)^{\frac{q}{q-2}}Z^{-\frac{q}{q-2}}\left(\int_{\mathcal{D}_{\tau_{1}Z}}|\Omega|^{2}dx\right)^{\frac{1}{2}}, (3.56)

for any q>2q>2, Z>1Z>1, and 12τ2<τ11\frac{1}{2}\leq\tau_{2}<\tau_{1}\leq 1.

Proof.   We only prove (3.56) with τ1=1\tau_{1}=1, τ2=12\tau_{2}=\frac{1}{2} for simplicity, since the general case could be derived by a direct scaling strategy. For any real number l1l\geq 1, we find h:=Ωlh:=\Omega^{l} satisfies

Δhl(l1)Ωl2|Ω|2+2rrhbh=0.\Delta h-l(l-1)\Omega^{l-2}|\nabla\Omega|^{2}+\frac{2}{r}\partial_{r}h-b\cdot\nabla h=0. (3.57)

Set 12σ2<σ11\frac{1}{2}\leq\sigma_{2}<\sigma_{1}\leq 1 and choose ζ=ζ(z)\zeta=\zeta(z) to be a smooth cut-off function satisfying

{suppζ[σ1,σ1],ζ=1 in [σ2,σ2],0ζ1,|ζ|1σ1σ2.\left\{\begin{array}[]{l}\operatorname{supp}\zeta\subset[-\sigma_{1},\sigma_{1}],\quad\zeta=1\quad\text{ in }[-\sigma_{2},\sigma_{2}],\\ 0\leqslant\zeta\leqslant 1,\\ \left|\zeta^{\prime}\right|\lesssim\frac{1}{\sigma_{1}-\sigma_{2}}.\end{array}\right.

Denoting ζZ(z):=ζ(zZ)\zeta_{Z}(z):=\zeta\left(\frac{z}{Z}\right) and testing (3.57) with ζZ2h\zeta_{Z}^{2}h, noting that

l(l1)𝒟σ1ZΩ2l2|Ω|2ζZ2𝑑x=l1l𝒟σ1Z|Ωl|2ζZ2𝑑x0,l(l-1)\int_{\mathcal{D}_{\sigma_{1}Z}}\Omega^{2l-2}|\nabla\Omega|^{2}\zeta_{Z}^{2}dx=\frac{l-1}{l}\int_{\mathcal{D}_{\sigma_{1}Z}}|\nabla\Omega^{l}|^{2}\zeta_{Z}^{2}dx\geq 0,

we arrive

𝒟σ1ZΔhζZ2h𝑑xM1+𝒟σ1Z2rrhζZ2hdxM2𝒟σ1ZbhζZ2hdxM30.\underbrace{\int_{\mathcal{D}_{\sigma_{1}Z}}\Delta h\zeta_{Z}^{2}hdx}_{M_{1}}+\underbrace{\int_{\mathcal{D}_{\sigma_{1}Z}}\frac{2}{r}\partial_{r}h\zeta_{Z}^{2}hdx}_{M_{2}}-\underbrace{\int_{\mathcal{D}_{\sigma_{1}Z}}b\cdot\nabla h\zeta_{Z}^{2}hdx}_{M_{3}}\geq 0. (3.58)

Next we handle M1M_{1}M3M_{3} term by term. Using integration by parts and direct calculations, we first see

M1=𝒟σ1Zh(ζZ2h)dx=𝒟σ1Z|(hζZ)|2𝑑x+𝒟σ1Zh2|ζZ|2𝑑x.\begin{split}M_{1}=-\int_{\mathcal{D}_{\sigma_{1}Z}}\nabla h\cdot\nabla(\zeta_{Z}^{2}h)dx=-\int_{\mathcal{D}_{\sigma_{1}Z}}|\nabla(h\zeta_{Z})|^{2}dx+\int_{\mathcal{D}_{\sigma_{1}Z}}h^{2}|\zeta_{Z}^{\prime}|^{2}dx.\end{split} (3.59)

Here the boundary term of the cylindrical surface is cancelled because h=0h=0 on 𝒟\partial\mathcal{D}, while those coming from the cross sections D{z=±σ1Z}D\cap\{z=\pm\sigma_{1}Z\} vanish due to the cut off function ζZ\zeta_{Z} is compactly supported. On the other hand, using axisymmetry of the solution

M2=2π01r(h2ζZ2)drdz=2πh2(0,z)ζZ2(z)𝑑z0.M_{2}=2\pi\int_{{\mathbb{R}}}\int_{0}^{1}\partial_{r}(h^{2}\zeta_{Z}^{2})drdz=-2\pi\int_{{\mathbb{R}}}h^{2}(0,z)\zeta^{2}_{Z}(z)dz\leq 0. (3.60)

Before we bound M3M_{3}, let us introduce the stream function of axisymmetric velocity field b=ur𝒆𝒓+uz𝒆𝒛b=u_{r}\boldsymbol{e_{r}}+u_{z}\boldsymbol{e_{z}}. By the divergence-free property r(rur)+z(ruz)=0\partial_{r}(ru_{r})+\partial_{z}(ru_{z})=0, there exists a scalar function Lθ=Lθ(r,z)L_{\theta}=L_{\theta}(r,z) such that

zLθ=ur,and1rr(rLθ)=uz.-\partial_{z}L_{\theta}=u_{r},\quad\text{and}\quad\frac{1}{r}\partial_{r}(rL_{\theta})=u_{z}. (3.61)

Using integration by parts again together with boundary condition h=0h=0 on 𝒟\partial\mathcal{D}, we derive that

M3=12Dσ1Zbh2ζZ2dx=Dσ1ZuzζZζZh2𝑑x=2π01r(rLθ)ζZζZh2drdz=4π01(rLθ)r(hζZ)hζZdrdz.\begin{split}M_{3}&=\frac{1}{2}\int_{D_{\sigma_{1}Z}}b\cdot\nabla h^{2}\zeta_{Z}^{2}dx=-\int_{D_{\sigma_{1}Z}}u_{z}\zeta_{Z}\zeta_{Z}^{\prime}h^{2}dx=-2\pi\int_{{\mathbb{R}}}\int_{0}^{1}\partial_{r}(rL_{\theta})\zeta_{Z}\zeta_{Z}^{\prime}h^{2}drdz\\ &=4\pi\int_{{\mathbb{R}}}\int_{0}^{1}(rL_{\theta})\partial_{r}(h\zeta_{Z})h\zeta_{Z}^{\prime}drdz.\end{split}

By the mean value theorem and (3.61), there exists r~(0,r)\tilde{r}\in(0,r) such that

rLθ(r,z)=r~uz(r~,z)r,rL_{\theta}(r,z)=\tilde{r}u_{z}(\tilde{r},z)r,

thus we can further bound M3M_{3} by

|M3|4πuzL(𝒟σ1Z\𝒟σ2Z)01|(hζZ)hζZ|r𝑑r𝑑z12𝒟σ1Z|(hζZ)|2𝑑x+2uzL(𝒟σ1Z\𝒟σ2Z)2𝒟σ1Zh2|ζZ|2𝑑x.\begin{split}|M_{3}|&\leq 4\pi\|u_{z}\|_{L^{\infty}(\mathcal{D}_{\sigma_{1}Z}\backslash\mathcal{D}_{\sigma_{2}Z})}\int_{{\mathbb{R}}}\int_{0}^{1}|\nabla(h\zeta_{Z})h\zeta_{Z}^{\prime}|rdrdz\\ &\leq\frac{1}{2}\int_{\mathcal{D}_{\sigma_{1}Z}}|\nabla(h\zeta_{Z})|^{2}dx+2\|u_{z}\|_{L^{\infty}(\mathcal{D}_{\sigma_{1}Z}\backslash\mathcal{D}_{\sigma_{2}Z})}^{2}\int_{\mathcal{D}_{\sigma_{1}Z}}h^{2}|\zeta_{Z}^{\prime}|^{2}dx.\end{split} (3.62)

Now substituting (3.59), (3.60), and (3.62) in (3.58), taking the maximum of ζZ\zeta_{Z}^{\prime}, it follows that

𝒟σ1Z|(hζZ)|2𝑑x+2πh2(0,z)ζZ2(z)𝑑zC(1+uzL(𝒟σ1Z\𝒟σ2Z)2)(σ1σ2)2Z2𝒟σ1Zh2𝑑x.\int_{\mathcal{D}_{\sigma_{1}Z}}|\nabla(h\zeta_{Z})|^{2}dx+2\pi\int_{{\mathbb{R}}}h^{2}(0,z)\zeta^{2}_{Z}(z)dz\leq\frac{C\left(1+\|u_{z}\|_{L^{\infty}(\mathcal{D}_{\sigma_{1}Z}\backslash\mathcal{D}_{\sigma_{2}Z})}^{2}\right)}{(\sigma_{1}-\sigma_{2})^{2}Z^{2}}\int_{\mathcal{D}_{\sigma_{1}Z}}h^{2}dx. (3.63)

Recalling h=0h=0 on 𝒟\partial\mathcal{D}, for any fixed zz\in\mathbb{R}, the following 2D Poincaré inequality holds:

h(,z)ζZ(z)L2(Σ)2Cr[h(,z)ζZ(z)]L2(Σ)2,\|h(\cdot,z)\zeta_{Z}(z)\|^{2}_{L^{2}(\Sigma)}\leq C\left\|\partial_{r}\left[h(\cdot,z)\zeta_{Z}(z)\right]\right\|^{2}_{L^{2}(\Sigma)},

where C>0C>0 here is independent of zz. Integrating with zz on {\mathbb{R}} and taking the square root, one has the following 3D Poincaré inequality

hζZL2(𝒟σ1Z)Cr(hζZ)L2(𝒟σ1Z).\|h\zeta_{Z}\|_{L^{2}(\mathcal{D}_{\sigma_{1}Z})}\leq C\|\partial_{r}(h\zeta_{Z})\|_{L^{2}(\mathcal{D}_{\sigma_{1}Z})}. (3.64)

For any q(2,6)q\in(2,6), Interpolation, Sobolev inequality and (3.64) imply that

hζZLq(𝒟σ1Z)hζZL6(𝒟σ1Z)shζZL2(𝒟σ1Z)1sC(hζZ)L2(𝒟σ1Z)shζZL2(𝒟σ1Z)1sC(hζZ)L2(𝒟σ1Z)sr(hζZ)L2(𝒟σ1Z)1sC(hζZ)L2(𝒟σ1Z).\begin{split}\|h\zeta_{Z}\|_{L^{q}(\mathcal{D}_{\sigma_{1}Z})}&\leq\|h\zeta_{Z}\|^{s}_{L^{6}(\mathcal{D}_{\sigma_{1}Z})}\|h\zeta_{Z}\|^{1-s}_{L^{2}(\mathcal{D}_{\sigma_{1}Z})}\leq C\|\nabla(h\zeta_{Z})\|^{s}_{L^{2}(\mathcal{D}_{\sigma_{1}Z})}\|h\zeta_{Z}\|^{1-s}_{L^{2}(\mathcal{D}_{\sigma_{1}Z})}\\ &\leq C\|\nabla(h\zeta_{Z})\|^{s}_{L^{2}(\mathcal{D}_{\sigma_{1}Z})}\|\partial_{r}(h\zeta_{Z})\|^{1-s}_{L^{2}(\mathcal{D}_{\sigma_{1}Z})}\leq C\|\nabla(h\zeta_{Z})\|_{L^{2}(\mathcal{D}_{\sigma_{1}Z})}.\end{split} (3.65)

Here s(0,1)s\in(0,1) depends on qq. Combining (3.63) and (3.65), we derive

hLq(𝒟σ2Z)C(1+uzL(𝒟σ1Z\𝒟σ2Z))(σ1σ2)ZhL2(𝒟σ1Z),\|h\|_{L^{q}(\mathcal{D}_{\sigma_{2}Z})}\leq\frac{C\left(1+\|u_{z}\|_{L^{\infty}(\mathcal{D}_{\sigma_{1}Z}\backslash\mathcal{D}_{\sigma_{2}Z})}\right)}{(\sigma_{1}-\sigma_{2})Z}\|h\|_{L^{2}(\mathcal{D}_{\sigma_{1}Z})},

which is equivalent to

(Dσ2Z|Ω|ql𝑑x)1qlC1/l(1+uzL(Dσ1Z\Dσ2Z))1/l(σ1σ2)1lZ1l(Dσ1Z|Ω|2l𝑑x)12l.\left(\int_{D_{\sigma_{2}Z}}|\Omega|^{ql}dx\right)^{\frac{1}{ql}}\leq\frac{C^{1/l}\left(1+\|u_{z}\|_{L^{\infty}(D_{\sigma_{1}Z}\backslash D_{\sigma_{2}Z})}\right)^{1/l}}{(\sigma_{1}-\sigma_{2})^{\frac{1}{l}}Z^{\frac{1}{l}}}\left(\int_{D_{\sigma_{1}Z}}|\Omega|^{2l}dx\right)^{\frac{1}{2l}}. (3.66)

Now for any k=0,1,2,k=0,1,2,..., we choose lk=(q2)kl_{k}=\left(\frac{q}{2}\right)^{k} and σ1k=12+(12)k+1\sigma_{1k}=\frac{1}{2}+\left(\frac{1}{2}\right)^{k+1}, σ2k=12+(12)k+2\sigma_{2k}=\frac{1}{2}+\left(\frac{1}{2}\right)^{k+2}, respectively. Denoting

Ψk:=(Dσ1kZ|Ω|2lk𝑑x)12lk,\Psi_{k}:=\left(\int_{D_{\sigma_{1k}Z}}|\Omega|^{2l_{k}}dx\right)^{\frac{1}{2l_{k}}},

and noting that

Dσ1kZ\Dσ2kZDZ\DZ/2,k=0,1,2,,D_{\sigma_{1k}Z}\backslash D_{\sigma_{2k}Z}\subset D_{Z}\backslash D_{Z/2},\quad\forall k=0,1,2,...,

then (3.66) follows that

Ψk+1C(2q)k2k(2q)k(1+uzL(𝒟Z\𝒟Z/2))(2q)kZ(2q)kΨkCj=0k(2q)j2j=0kj(2q)j(1+uzL(𝒟Z\𝒟Z/2))j=0k(2q)jZj=0k(2q)jΨ0.\begin{split}\Psi_{k+1}&\leq C^{\left(\frac{2}{q}\right)^{k}}2^{k\left(\frac{2}{q}\right)^{k}}\left(1+\|u_{z}\|_{L^{\infty}(\mathcal{D}_{Z}\backslash\mathcal{D}_{Z/2})}\right)^{\left(\frac{2}{q}\right)^{k}}Z^{-\left(\frac{2}{q}\right)^{k}}\Psi_{k}\\[11.38109pt] &\leq\cdot\cdot\cdot\\[11.38109pt] &\leq C^{\sum_{j=0}^{k}\left(\frac{2}{q}\right)^{j}}2^{\sum_{j=0}^{k}j\left(\frac{2}{q}\right)^{j}}\left(1+\|u_{z}\|_{L^{\infty}(\mathcal{D}_{Z}\backslash\mathcal{D}_{Z/2})}\right)^{\sum_{j=0}^{k}\left(\frac{2}{q}\right)^{j}}Z^{-\sum_{j=0}^{k}\left(\frac{2}{q}\right)^{j}}\Psi_{0}.\end{split} (3.67)

Performing kk\to\infty, then iteration (3.67) implies a mean value inequality of Ω\Omega, that is

supx𝒟Z/2|Ω|Cq(1+uzL(𝒟Z\𝒟Z/2))qq2Zqq2(𝒟Z|Ω|2𝑑x)12,\sup_{x\in\mathcal{D}_{Z/2}}|\Omega|\leq C_{q}\left(1+\|u_{z}\|_{L^{\infty}(\mathcal{D}_{Z}\backslash\mathcal{D}_{Z/2})}\right)^{\frac{q}{q-2}}Z^{-\frac{q}{q-2}}\left(\int_{\mathcal{D}_{Z}}|\Omega|^{2}dx\right)^{\frac{1}{2}},

for any q>2q>2. This completes the proof of Lemma 2.2.

Since uzu_{z} satisfies (1.9)2 in 𝒟Z\mathcal{D}_{Z}, (3.56) indicates that

supx𝒟τ2Z|Ω|2Cq(τ1τ2)2qq2Z2(δ01)qq2𝒟τ1Z|Ω|2𝑑x.\sup_{x\in\mathcal{D}_{\tau_{2}Z}}|\Omega|^{2}\leq C_{q}(\tau_{1}-\tau_{2})^{-\frac{2q}{q-2}}Z^{\frac{2(\delta_{0}-1)q}{q-2}}\int_{\mathcal{D}_{\tau_{1}Z}}|\Omega|^{2}dx. (3.68)

However, due to the lack of boundedness of the second-order derivatives of uu, we are unable to control ΩL2(𝒟Z)\|\Omega\|_{L^{2}(\mathcal{D}_{Z})} at the moment. Next we will use an algebraic trick to convert the L2L^{2}-norm on the right hand side of (3.68) to an L1L^{1}-norm. This trick comes from Li-Schoen [21]. Here goes the lemma:

Lemma 3.2 (modified mean value inequality).

Suppose b=ur𝐞𝐫+uz𝐞𝐳b=u_{r}\boldsymbol{e_{r}}+u_{z}\boldsymbol{e_{z}} is a smooth divergence-free axisymmetric vector field and uzL(DZ)Zδ0.\|u_{z}\|_{L^{\infty}(D_{Z})}\lesssim Z^{\delta_{0}}. Then any weak solution Ω\Omega of boundary value problem (3.55) satisfies the following mean value inequality for any q>2q>2, Z>1Z>1:

supx𝒟Z/2|Ω|CqZ(δ01)qq2𝒟Z|Ω|𝑑x.\sup_{x\in\mathcal{D}_{Z/2}}|\Omega|\leq C_{q}Z^{\frac{(\delta_{0}-1)q}{q-2}}\int_{\mathcal{D}_{Z}}|\Omega|dx. (3.69)

Proof.   For any 12τ2<τ11\frac{1}{2}\leq\tau_{2}<\tau_{1}\leq 1, (3.68) implies that

supx𝒟τ2Z|Ω|2Cq(τ1τ2)2qq2Z2(δ01)qq2(supx𝒟τ1Z|Ω|2)1/2DZ|Ω|𝑑x.\sup_{x\in\mathcal{D}_{\tau_{2}Z}}|\Omega|^{2}\leq C_{q}(\tau_{1}-\tau_{2})^{-\frac{2q}{q-2}}Z^{\frac{2(\delta_{0}-1)q}{q-2}}\left(\sup_{x\in\mathcal{D}_{\tau_{1}Z}}|\Omega|^{2}\right)^{1/2}\int_{D_{Z}}|\Omega|dx.

Denoting τ1k=τ2,k+1=1(12)k+2\tau_{1k}=\tau_{2,k+1}=1-\left(\frac{1}{2}\right)^{k+2}, τ2k=1(12)k+1\tau_{2k}=1-\left(\frac{1}{2}\right)^{k+1}, and Φk:=supxDτ2kZ|Ω|2\Phi_{k}:=\sup_{x\in D_{\tau_{2k}Z}}|\Omega|^{2}, it follows that

ΦkCq22qkq2Z2(δ01)qq2Φk+11/2DZ|Ω|𝑑x.\Phi_{k}\leq C_{q}2^{\frac{2qk}{q-2}}Z^{\frac{2(\delta_{0}-1)q}{q-2}}\Phi_{k+1}^{1/2}\int_{D_{Z}}|\Omega|dx. (3.70)

Iterating (3.70) from k=0k=0 to infinity, one arrives

supx𝒟Z/2|Ω|2Cqj=02j22qq2j=0j2j(Z2(δ01)qq2)j=02j(𝒟Z|Ω|𝑑x)j=02jCqZ2(δ01)qq2(𝒟Z|Ω|𝑑x)2,\begin{split}\sup_{x\in\mathcal{D}_{Z/2}}|\Omega|^{2}&\leq C_{q}^{\sum_{j=0}^{\infty}2^{-j}}2^{\frac{2q}{q-2}\sum_{j={0}}^{\infty}\frac{j}{2^{j}}}\left(Z^{\frac{2(\delta_{0}-1)q}{q-2}}\right)^{\sum_{j=0}^{\infty}2^{-j}}\left(\int_{\mathcal{D}_{Z}}|\Omega|dx\right)^{\sum_{j=0}^{\infty}2^{-j}}\\ &\leq C_{q}Z^{\frac{2(\delta_{0}-1)q}{q-2}}\left(\int_{\mathcal{D}_{Z}}|\Omega|dx\right)^{2},\end{split}

which follows that

supx𝒟Z/2|Ω|CqZ(δ01)qq2𝒟Z|Ω|𝑑x.\sup_{x\in\mathcal{D}_{Z/2}}|\Omega|\leq C_{q}Z^{\frac{(\delta_{0}-1)q}{q-2}}\int_{\mathcal{D}_{Z}}|\Omega|dx.

This completes the proof of Lemma 3.2.

Finally, one notes that

𝒟Z|Ω|𝑑x2πωθL(𝒟Z)ZZ01𝑑r𝑑zZM0+1.\int_{\mathcal{D}_{Z}}|\Omega|dx\leq 2\pi\|\omega_{\theta}\|_{L^{\infty}(\mathcal{D}_{Z})}\int_{-Z}^{Z}\int_{0}^{1}drdz\lesssim Z^{M_{0}+1}.

Therefore, as long as ωθ\omega_{\theta} is of polynomial growth (see (1.9)3) when zz\to\infty, we can infer from (3.69) that

supx𝒟Z/2|Ω|CqZ(δ01)qq2+1+M0.\sup_{x\in\mathcal{D}_{Z/2}}|\Omega|\leq C_{q}Z^{\frac{(\delta_{0}-1)q}{q-2}+1+M_{0}}. (3.71)

For any fixed δ0<1\delta_{0}<1 and M0>0M_{0}>0, we can always choose qq which is larger than but close enough to 22 such that (3.71) indicates

supx𝒟Z/2|Ω|Zγ\sup_{x\in\mathcal{D}_{Z/2}}|\Omega|\lesssim Z^{-\gamma}

for some γ>0\gamma>0. This proves ωθ\omega_{\theta} vanishes in 𝒟\mathcal{D} by performing ZZ\to\infty.

3.1.2 Vanishing of uru_{r} and constancy of uzu_{z}

Noting that ×b=ωθ𝒆𝜽0\nabla\times b=\omega_{\theta}\boldsymbol{e_{\theta}}\equiv 0 and the divergence-free property of bb, we apply the Lagrange’s formula for del to deduce

Δb=××b(div b)=0,-\Delta b=\nabla\times\nabla\times b-\nabla(\text{div }b)=0,

which indicates

(Δ1r2)ur=0;Δuz=0.\left(\Delta-\frac{1}{r^{2}}\right)u_{r}=0;\quad\quad\Delta u_{z}=0.

To prove vanishing of uru_{r}, for δ>0\delta>0 being small, we consider the auxiliary function ηδ\eta_{\delta} which is defined by

ηδ(x):=J1((αδ)r)cosh((αδ)z).\eta_{\delta}(x):=J_{1}\big{(}(\alpha-\delta)r\big{)}\cosh\big{(}(\alpha-\delta)z\big{)}.

Here J1J_{1} is the Bessel function which is defined in (1.11) and satisfies (1.10) with β=1\beta=1, while α\alpha is the smallest positive root of J1J_{1}. Direct calculation shows

(Δ1r2)ηδ=(r2+1rr+z21r2)ηδ=0.\left(\Delta-\frac{1}{r^{2}}\right)\eta_{\delta}=\left(\partial^{2}_{r}+\frac{1}{r}\partial_{r}+\partial^{2}_{z}-\frac{1}{r^{2}}\right)\eta_{\delta}=0.

Owing to uru_{r} is growing as (1.9)1, we choose δ<<1\delta<<1 small enough such that γ0<α2δ\gamma_{0}<\alpha-2\delta. Using the concavity of J1((αδ)r)J_{1}((\alpha-\delta)r) on the subset of {r: 0r1}\{r:\,0\leq r\leq 1\} where J1((αδ)r)J_{1}((\alpha-\delta)r) is increasing, one has

ηδJ1(αδ)rcosh((αδ)z)Cδre(γ0+δ)|z|,\eta_{\delta}\geq J_{1}(\alpha-\delta)r\cosh\big{(}(\alpha-\delta)z\big{)}\geq C_{\delta}re^{(\gamma_{0}+\delta)|z|},

where Cδ>0C_{\delta}>0 is a constant depends only on δ\delta. Then the condition (1.9)1 indicates that

lim|z||ur(r,z)|ηδ(r,z)=0, uniformly with r=x12+x22[0,1].\lim_{|z|\to\infty}\frac{|u_{r}(r,z)|}{\eta_{\delta}(r,z)}=0,\text{ uniformly with }r=\sqrt{x_{1}^{2}+x_{2}^{2}}\in[0,1].

Therefore, for any fixed ε>0\varepsilon>0 and δ\delta, there exists an Nε,δ>0N_{\varepsilon,\delta}>0 such that

{(Δ1r2)(εηδ±ur)=0,x𝒟M,εηδ±ur0,x𝒟M=[𝒟{MzM}][𝒟{z=±M}],\left\{\begin{aligned} &\left(\Delta-\frac{1}{r^{2}}\right)(\varepsilon\eta_{\delta}\pm u_{r})=0,\quad\forall x\in\mathcal{D}_{M},\\ &\varepsilon\eta_{\delta}\pm u_{r}\geq 0,\quad\forall x\in\partial\mathcal{D}_{M}=\big{[}\partial\mathcal{D}\cap\{-M\leq z\leq M\}\big{]}\cup\big{[}\mathcal{D}\cap\{z=\pm M\}\big{]},\end{aligned}\right.

for any M>Nε,δM>N_{\varepsilon,\delta}. The maximum principle indicates

|ur(x)|εηδ(x),x𝒟M.|u_{r}(x)|\leq\varepsilon\eta_{\delta}(x),\quad\forall x\in\mathcal{D}_{M}. (3.72)

By performing MM\to\infty, one finds the estimate (3.72) actually holds for all x𝒟x\in\mathcal{D}. Thus ur0u_{r}\equiv 0 is proved by the arbitrariness of ε>0\varepsilon>0.

Finally, the divergence-free of uu implies zuz=1rr(rur)0\partial_{z}u_{z}=-\frac{1}{r}\partial_{r}(ru_{r})\equiv 0 in 𝒟\mathcal{D}. The vanishing of ωθ\omega_{\theta} and uru_{r} indicates ruz0\partial_{r}u_{z}\equiv 0. Thus uzu_{z} must be a constant. This consequently indicates

b=C2𝒆𝒛b=C_{2}\boldsymbol{e_{z}} (3.73)

for some constant C2C_{2}\in{\mathbb{R}}.

3.1.3 End of the proof

Now substituting (3.73) in (1.3)2 and noting that uθu_{\theta} is independent of rr, one arrives the following ODE of uθu_{\theta}

uθ′′(r)+1ruθ(r)1r2uθ(r)=0.u_{\theta}^{\prime\prime}(r)+\frac{1}{r}u_{\theta}^{\prime}(r)-\frac{1}{r^{2}}u_{\theta}(r)=0.

This ODE, which is of Eulerian type, is solved by

uθ(r)=C0r+C1r,u_{\theta}(r)=\frac{C_{0}}{r}+C_{1}r,

for any C0C_{0}, C1C_{1}\in{\mathbb{R}}. Smoothness of uθu_{\theta} forces that C0=0C_{0}=0. Thus we conclude that

u=uθ𝒆𝜽+b=C1r𝒆𝜽+C2𝒆𝒛,u=u_{\theta}\boldsymbol{e_{\theta}}+b=C_{1}r\boldsymbol{e_{\theta}}+C_{2}\boldsymbol{e_{z}},

which completes the proof of Theorem 1.3.

Remark 3.3.

Unlike Theorem 1.1, Theorem 1.3 actually needs weaker assumptions,(1.9), on the boundedness of solutions. As stated in the introduction, assumption (1.9)1 is sharp due to the non-trivial counterexamples in (1.12) which grow no slower than Ceα|z|Ce^{\alpha|z|} as zz\to\infty. Meanwhile, the counterexample in (1.12) has zero vorticity and zero flux in the cross section Σ\Sigma. Identities (2.53) and (2.54) no longer hold for the solution in (1.12) since we have no boundedness of the head pressure H:=12|u|2+pp(0,Z)H:=\frac{1}{2}|u|^{2}+p-p(0,Z) in 𝒟2Z\𝒟Z\mathcal{D}_{2Z}\backslash\mathcal{D}_{Z}.

3.2 Proof of Case II

Actually, if uzu_{z} is independent of zz-variable, by the divergence free condition (1.3)4, we have

r(rur)=rzuz=0,\partial_{r}(ru_{r})=-r\partial_{z}u_{z}=0,

which indicates that rur=f(z)ru_{r}=f(z) for some smooth function f(z)f(z). Then using the boundary condition (1.5)3, we deduce that f(z)=0f(z)=0, which implies

ur=0.u_{r}=0.

Next, it follows from (1.3)1 and (1.3)3 that

z(uθ)2=rrzp=rr(r2uz+1rruz):=g(r),\partial_{z}(u_{\theta})^{2}=r\partial_{r}\partial_{z}p=r\partial_{r}\left(\partial^{2}_{r}u_{z}+{\frac{1}{r}\partial_{r}u_{z}}\right):=g(r)\,,

which deduces that

(uθ)2(r,z)=g(r)z+(uθ)2(r,0).(u_{\theta})^{2}(r,z)=g(r)z+(u_{\theta})^{2}(r,0)\,.

Note that g(r)0g(r)\equiv 0 for any r[0,1]r\in[0,1]. Actually, if for some r0r_{0} such that g(r0)0g(r_{0})\neq 0, we can obtain a contradiction by taking

z=1(uθ)2(r0,0)g(r0).z=\frac{-1-(u_{\theta})^{2}(r_{0},0)}{g(r_{0})}.

Thus, we have that uθu_{\theta} is independent of zz-variable. Following the argument in Section 3.1.3, one concludes that uθ=C1ru_{\theta}=C_{1}r.

Then if we go back to the (1.3)1, we see that rp=C12r\partial_{r}p=C^{2}_{1}r, which indicates that

p=12C12r2+f(z),p=\frac{1}{2}C^{2}_{1}r^{2}+f(z),

for some smooth function f(z)f(z). From (1.3)3, by using that uzu_{z} is independent of zz, we can have that zp=f(z)=C\partial_{z}p=f^{\prime}(z)=C for some constant CC. At last we see that uzu_{z} satisfies the following two dimensional Laplacian equation in Σ\Sigma with Neumann boundary condition:

{Δhuz=C,𝒏uz=0.\left\{\begin{aligned} &\Delta_{h}u_{z}=C,\\ &\partial_{\boldsymbol{n}}u_{z}=0.\end{aligned}\right. (3.74)

Integrating directly on Σ\Sigma for (3.74)1 and using the boundary condition, we can obtain C=0C=0. Then multiplying (3.74)1 by uzu_{z} and integrating on Σ\Sigma, we can have huz=0\nabla_{h}u_{z}=0, which implies that uzC2u_{z}\equiv C_{2} for some constant C2C_{2}.

4  Proof of Theorem 1.5

In this section we derive the proof of Theorem 1.5, which shows a solution to the Navier-Stokes equations (1.1) with the NavierHodgeLionsNavier-Hodge-Lions boundary condition (NHLB) in the pipe 𝒟\mathcal{D} must be a parallel flow Φπ𝒆𝒛\frac{\Phi}{\pi}\boldsymbol{e_{z}}, without axisymmetric assumptions. Our method is motivated by [20] in which authors show the uniqueness result for problems with the homogeneous Dirichlet boundary condition. Before proving the theorem, we need the following lemma that states the asymptotic behavior of a function satisfies an ordinary differential inequality.

Lemma 4.1.

Let Y(ζ)0Y(\zeta)\nequiv 0 be a nondecreasing nonnegative differentiable function satisfying

Y(ζ)Ψ(Y(ζ)),ζ>0.Y(\zeta)\leq\Psi(Y^{\prime}(\zeta)),\quad\forall\zeta>0.

Here Ψ:[0,)[0,)\Psi:\,[0,\infty)\to[0,\infty) is a monotonically increasing function with Ψ(0)=0\Psi(0)=0 and there exists C,τ1>0C,\,\tau_{1}>0, m>1m>1, such that

Ψ(τ)Cτm,τ>τ1.\Psi(\tau)\leq C\tau^{m},\quad\forall\tau>\tau_{1}.

Then

lim infζ+ζmm1Y(ζ)>0.\liminf_{\zeta\to+\infty}\zeta^{-\frac{m}{m-1}}Y(\zeta)>0.

The next lemma on solving the divergence problem in a truncated pipe will be applied to bound the term related to pressure in the further proof:

Lemma 4.2 (See [5, 6], also [11], Chapter III).

Let D=Σ×[0,1]D=\Sigma\times[0,1], fL2(D)f\in L^{2}(D) with

Df𝑑x=0,\int_{D}fdx=0,

then there exists a vector valued function V:D3{V}:\,D\to\mathbb{R}^{3} belongs to H01(D)H^{1}_{0}(D) such that

V=f,andVL2(D)CfL2(D).\nabla\cdot{V}=f,\quad\text{and}\quad\|\nabla{V}\|_{L^{2}(D)}\leq C\|f\|_{L^{2}(D)}. (4.75)

Here C>0C>0 is an absolute constant.

The following lemma gives a Poincaré inequality for vectors in H1(Σ)H^{1}(\Sigma) with only vanishing normal direction on the boundary. Readers can find some hint of the proof in Galdi [11, Page 71, Exercise II.5.6].

Lemma 4.3.

Let f=f1𝐞𝟏+f2𝐞𝟐{f}=f_{1}\boldsymbol{e_{1}}+f_{2}\boldsymbol{e_{2}} be a two dimensional vector function with components in H1(Σ)H^{1}(\Sigma), and f𝐧¯=0{f}\cdot\bar{\boldsymbol{n}}=0 on Σ\partial\Sigma, where 𝐧¯\bar{\boldsymbol{n}} is the unit outer normal of Σ\partial\Sigma. Then the following Poincaré inequality holds

fL2(Σ)ChfL2(Σ),\|{f}\|_{L^{2}(\Sigma)}\leq C\|\nabla_{h}{f}\|_{L^{2}(\Sigma)},

where h=(x1,x2)\nabla_{h}=(\partial_{x_{1}},\partial_{x_{2}}) is the gradient operator on x1x_{1} and x2x_{2} direction.

Proof of Theorem 1.5: Denoting v:=uΦπ𝒆𝒛v:=u-\frac{\Phi}{\pi}\boldsymbol{e_{z}}, one deduces from (1.1) that

vv+Φπzv+pΔv=0.v\cdot\nabla v+\frac{\Phi}{\pi}\partial_{z}v+\nabla p-\Delta v=0. (4.76)

We multiply (4.76) by vv and integrate on 𝒟ζ\mathcal{D}_{\zeta}, it follows that

𝒟ζvΔv𝑑xLHS=𝒟ζv(vv+Φπzv+p)𝑑x.\underbrace{\int_{\mathcal{D}_{\zeta}}v\cdot\Delta vdx}_{LHS}=\int_{\mathcal{D}_{\zeta}}v\left(v\cdot\nabla v+\frac{\Phi}{\pi}\partial_{z}v+\nabla p\right)dx. (4.77)

Using integration by parts, the left hand side of (4.77) follows that

LHS=𝒟ζ|v|2𝑑x+12𝒟ζ|v|2𝒏𝑑SB1,\begin{split}LHS=-\int_{\mathcal{D}_{\zeta}}|\nabla v|^{2}dx+\underbrace{\frac{1}{2}\int_{\partial\mathcal{D}_{\zeta}}\frac{\partial|v|^{2}}{\partial\boldsymbol{n}}dS}_{B_{1}},\end{split}

where 𝒏\boldsymbol{n} is the unit outer normal vector on 𝒟ζ\partial\mathcal{D}_{\zeta}. In the cylindrical coordinate, one writes

v=vr𝒆𝒓+vθ𝒆𝜽+vz𝒆𝒛.v=v_{r}\boldsymbol{e_{r}}+v_{\theta}\boldsymbol{e_{\theta}}+v_{z}\boldsymbol{e_{z}}.

Thus one has

B1=𝒟ζ𝒟(vrrvr+vθrvθ+vzrvz)𝑑SB11+𝒟{z=ζ}(vrzvr+vθzvθ+vzzvz)𝑑xh𝒟{z=ζ}(vrzvr+vθzvθ+vzzvz)𝑑xh.\begin{split}B_{1}=&\underbrace{\int_{\partial\mathcal{D}_{\zeta}\cap\partial\mathcal{D}}\big{(}v_{r}\partial_{r}v_{r}+v_{\theta}\partial_{r}v_{\theta}+v_{z}\partial_{r}v_{z}\big{)}dS}_{B_{11}}+{\int_{\mathcal{D}\cap\{z=\zeta\}}\big{(}v_{r}\partial_{z}v_{r}+v_{\theta}\partial_{z}v_{\theta}+v_{z}\partial_{z}v_{z}\big{)}dx_{h}}\\ &-{\int_{\mathcal{D}\cap\{z=-\zeta\}}\big{(}v_{r}\partial_{z}v_{r}+v_{\theta}\partial_{z}v_{\theta}+v_{z}\partial_{z}v_{z}\big{)}dx_{h}}.\end{split} (4.78)

Since vv satisfies the boundary condition (1.6), one derives

B11=𝒟ζ𝒟vθ2𝑑S.B_{11}=-\int_{\partial\mathcal{D}_{\zeta}\cap\partial\mathcal{D}}v_{\theta}^{2}dS.

Substituting above in (4.78), one derives

LHS𝒟ζ|v|2𝑑x𝒟ζ𝒟vθ2𝑑S+𝒟{z=±ζ}|v||zv|𝑑xh.LHS\leq-\int_{\mathcal{D}_{\zeta}}|\nabla v|^{2}dx-\int_{\partial\mathcal{D}_{\zeta}\cap\partial\mathcal{D}}v_{\theta}^{2}dS+\int_{\mathcal{D}\cap\{z=\pm\zeta\}}|v||\partial_{z}v|dx_{h}. (4.79)

Now we turn to the right hand side of (4.77). Integrating by parts, one deduces that

𝒟ζv(vv+p)𝑑x=𝒟{z=ζ}v3(12|v|2+p)𝑑xh𝒟{z=ζ}v3(12|v|2+p)𝑑xh.\begin{split}\int_{\mathcal{D}_{\zeta}}v\big{(}v\cdot\nabla v+\nabla p\big{)}dx=&\int_{\mathcal{D}\cap\{z=\zeta\}}v_{3}\left(\frac{1}{2}|v|^{2}+p\right)dx_{h}-\int_{\mathcal{D}\cap\{z=-\zeta\}}v_{3}\left(\frac{1}{2}|v|^{2}+p\right)dx_{h}.\end{split} (4.80)

Meanwhile, one notices

𝒟ζΦπvzvdx=Φ2π(𝒟{z=ζ}|v|2𝑑xh𝒟{z=ζ}|v|2𝑑xh).\int_{\mathcal{D}_{\zeta}}\frac{\Phi}{\pi}v\cdot\partial_{z}vdx=\frac{\Phi}{2\pi}\left(\int_{\mathcal{D}\cap\{z=\zeta\}}|v|^{2}dx_{h}-\int_{\mathcal{D}\cap\{z=-\zeta\}}|v|^{2}dx_{h}\right). (4.81)

Substituting (4.79), (4.80) and (4.81) in (4.77), one arrives at

𝒟ζ|v|2𝑑xC𝒟{z=±ζ}|v|(|v|+|v|+|v|2)𝑑xh𝒟{z=ζ}v3p𝑑xh+𝒟{z=ζ}v3p𝑑xh.\begin{split}\int_{\mathcal{D}_{\zeta}}|\nabla v|^{2}dx\leq C\int_{\mathcal{D}\cap\{z=\pm\zeta\}}|v|(|\nabla v|+|v|+|v|^{2})dx_{h}-\int_{\mathcal{D}\cap\{z=\zeta\}}v_{3}pdx_{h}+\int_{\mathcal{D}\cap\{z=-\zeta\}}v_{3}pdx_{h}.\end{split}

Integrating with ζ\zeta on [Z1,Z][Z-1,Z], where Z1Z\geq 1, it follows that

Z1Z𝒟ζ|v|2𝑑x𝑑ζC(𝒪Z+𝒪Z|v|(|v|+|v|+|v|2)𝑑xT1+|𝒪Z+𝒪Zv3p𝑑x|T2).\int_{Z-1}^{Z}\int_{\mathcal{D}_{\zeta}}|\nabla v|^{2}dxd\zeta\leq C\Bigg{(}\underbrace{\int_{\mathcal{O}^{+}_{Z}\cup\mathcal{O}^{-}_{Z}}|v|(|\nabla v|+|v|+|v|^{2})dx}_{T_{1}}+\underbrace{\Big{|}\int_{\mathcal{O}^{+}_{Z}\cup\mathcal{O}^{-}_{Z}}v_{3}pdx\Big{|}}_{T_{2}}\Bigg{)}. (4.82)

In the following we only work on integrations on 𝒪Z+\mathcal{O}^{+}_{Z} since the rest part are similar. Using Cauchy-Schwarz inequality and Gagliardo-Nirenberg inequality, one deduces

T1C(vL2(𝒪Z+)vL2(𝒪Z+)+vL2(𝒪Z+)2+vL2(𝒪Z+)3/2vL2(𝒪Z+)3/2+vL2(𝒪Z+)3).T_{1}\leq C\left(\|v\|_{L^{2}(\mathcal{O}^{+}_{Z})}\|\nabla v\|_{L^{2}(\mathcal{O}^{+}_{Z})}+\|v\|_{L^{2}(\mathcal{O}^{+}_{Z})}^{2}+\|v\|^{3/2}_{L^{2}(\mathcal{O}^{+}_{Z})}\|\nabla v\|^{3/2}_{L^{2}(\mathcal{O}^{+}_{Z})}+\|v\|_{L^{2}(\mathcal{O}_{Z}^{+})}^{3}\right).

Applying Lemma 4.3 in each cross section of the pipe, one finds

T1C(vL2(𝒪Z+)2+vL2(𝒪Z+)3).T_{1}\leq C\left(\|\nabla v\|^{2}_{L^{2}(\mathcal{O}^{+}_{Z})}+\|\nabla v\|^{3}_{L^{2}(\mathcal{O}^{+}_{Z})}\right). (4.83)

Now it remains to bound the pressure term T2T_{2} in (4.82). Noticing that

𝒟{x3=z}v3(xh,z)𝑑xh0,z,\int_{\mathcal{D}\cap\{x_{3}=z\}}v_{3}(x_{h},z)dx_{h}\equiv 0,\quad\forall z\in\mathbb{R},

we deduce that

𝒪Z+v3𝑑x=0,Z1.\int_{\mathcal{O}_{Z}^{+}}v_{3}dx=0,\quad\forall Z\geq 1.

Using Lemma 4.2, one derives the existence of a vector field VV satisfying (4.75) with f=v3f=v_{3}. By the momentum equation (1.1)1, one arrives

𝒪Z+v3p𝑑x=𝒪Z+pVdx=𝒪Z+(vv+ΦπzvΔv)V𝑑x.\int_{\mathcal{O}_{Z}^{+}}v_{3}pdx=-\int_{\mathcal{O}_{Z}^{+}}\nabla p\cdot{V}dx=\int_{\mathcal{O}_{Z}^{+}}\left(v\cdot\nabla v+\frac{\Phi}{\pi}\partial_{z}v-\Delta v\right)\cdot{V}dx.

integration by parts, one derives

𝒪Z+v3p𝑑x=i,j=13𝒪Z+(ivjvivjΦπδi3vj)iVjdx.\int_{\mathcal{O}_{Z}^{+}}v_{3}pdx=\sum_{i,j=1}^{3}\int_{\mathcal{O}_{Z}^{+}}\left(\partial_{i}v_{j}-v_{i}v_{j}-\frac{\Phi}{\pi}\delta_{i3}v_{j}\right)\partial_{i}V_{j}dx.

Here δij\delta_{ij} is the Kronecker symbol. By applying Hölder inequality and (4.75) in Lemma 4.2, one deduces that

|𝒪Z+v3p𝑑x|C(vL2(𝒪Z+)+vL4(𝒪Z+)2+ΦvL2(𝒪Z+))v3L2(𝒪Z+).\left|\int_{\mathcal{O}_{Z}^{+}}v_{3}pdx\right|\leq C\left(\|\nabla v\|_{L^{2}(\mathcal{O}^{+}_{Z})}+\|v\|_{L^{4}(\mathcal{O}^{+}_{Z})}^{2}+\Phi\|v\|_{L^{2}(\mathcal{O}^{+}_{Z})}\right)\|v_{3}\|_{L^{2}(\mathcal{O}^{+}_{Z})}.

Similarly as we bound T1T_{1}, using Lemma 4.3 and Gagliardo-Nirenberg inequality, one concludes

T2C(vL2(𝒪Z+)2+vL2(𝒪Z+)3).T_{2}\leq C\left(\|\nabla v\|_{L^{2}(\mathcal{O}^{+}_{Z})}^{2}+\|\nabla v\|_{L^{2}(\mathcal{O}^{+}_{Z})}^{3}\right). (4.84)

Substituting (4.83) and (4.84) (together with their related estimates on 𝒪Z\mathcal{O}^{-}_{Z} ), one deduces

Z1Z𝒟ζ|v|2𝑑x𝑑ζC(vL2(𝒪Z+𝒪Z)2+vL2(𝒪Z+𝒪Z)3),Z1.\int_{Z-1}^{Z}\int_{\mathcal{D}_{\zeta}}|\nabla v|^{2}dxd\zeta\leq C\left(\|\nabla v\|_{L^{2}(\mathcal{O}^{+}_{Z}\cup\mathcal{O}^{-}_{Z})}^{2}+\|\nabla v\|_{L^{2}(\mathcal{O}^{+}_{Z}\cup\mathcal{O}^{-}_{Z})}^{3}\right),\quad\forall Z\geq 1.

Therefore, letting

Y(Z):=Z1Z𝒟ζ|v|2𝑑x𝑑ζ,Y(Z):=\int_{Z-1}^{Z}\int_{\mathcal{D}_{\zeta}}|\nabla v|^{2}dxd\zeta,

one deduces

Y(Z)C(Y(Z)+(Y(Z))3/2).Y(Z)\leq C\left(Y^{\prime}(Z)+\left(Y^{\prime}(Z)\right)^{3/2}\right).

Applying Lemma 4.1, one concludes

Y(Z)=Z1Z𝒟ζ|v|2𝑑x𝑑ζC0Z3,Z1Y(Z)=\int_{Z-1}^{Z}\int_{\mathcal{D}_{\zeta}}|\nabla v|^{2}dxd\zeta\geq C_{0}Z^{3},\quad\forall Z\geq 1

for some C0>0C_{0}>0. However, this creates a paradox since |u||\nabla u| satisfies (1.13). This proves u=Φπ𝒆𝟑u=\frac{\Phi}{\pi}\boldsymbol{e_{3}}.

Appendix A Computation of the boundary condition

Here we give a derivation of the boundary condition (NSB) and (NHLB) in the cylindrical coordinates. First, noting that

0=u𝒏=ur.0=u\cdot\boldsymbol{n}=u_{r}. (A.85)

In cylindrical coordinates, the gradient operator is represented by

=𝒆𝒓r+𝒆𝜽θr+𝒆𝒛z.\nabla=\boldsymbol{e_{r}}\partial_{r}+\boldsymbol{e_{\theta}}\frac{\partial_{\theta}}{r}+\boldsymbol{e_{z}}\partial_{z}.

Then we can calculate the matrix u\nabla u in cylindrical coordinates and write it as a form of tensor product as follows

u=rur𝒆𝒓𝒆𝒓+(1rθuruθr)𝒆𝒓𝒆𝜽+zur𝒆𝒓𝒆𝒛+ruθ𝒆𝜽𝒆𝒓+(1rθuθ+urr)𝒆𝜽𝒆𝜽++zuθ𝒆𝜽𝒆𝒛+ruz𝒆𝒛𝒆𝒓+1rθuz𝒆𝒛𝒆𝜽+zuz𝒆𝒛𝒆𝒛.\begin{split}\nabla u&=\partial_{r}u_{r}\boldsymbol{e_{r}}\otimes\boldsymbol{e_{r}}+{\left(\frac{1}{r}\partial_{\theta}u_{r}-\frac{u_{\theta}}{r}\right)}\boldsymbol{e_{r}}\otimes\boldsymbol{e_{\theta}}+\partial_{z}u_{r}\boldsymbol{e_{r}}\otimes\boldsymbol{e_{z}}\\ &+\partial_{r}u_{\theta}\boldsymbol{e_{\theta}}\otimes\boldsymbol{e_{r}}+{\left(\frac{1}{r}\partial_{\theta}u_{\theta}+\frac{u_{r}}{r}\right)}\boldsymbol{e_{\theta}}\otimes\boldsymbol{e_{\theta}}++\partial_{z}u_{\theta}\boldsymbol{e_{\theta}}\otimes\boldsymbol{e_{z}}\\ &+\partial_{r}u_{z}\boldsymbol{e_{z}}\otimes\boldsymbol{e_{r}}+\frac{1}{r}\partial_{\theta}u_{z}\boldsymbol{e_{z}}\otimes\boldsymbol{e_{\theta}}+\partial_{z}u_{z}\boldsymbol{e_{z}}\otimes\boldsymbol{e_{z}}.\end{split} (A.86)

Equivalently

u=\displaystyle\nabla u= (rur1rθur1ruθzurruθ1rθuθ+1rurzuθruz1rθuzzuz):(𝒆𝒓𝒆𝒓𝒆𝒓𝒆𝜽𝒆𝒓𝒆𝒛𝒆𝜽𝒆𝒓𝒆𝜽𝒆𝜽𝒆𝜽𝒆𝒛𝒆𝒛𝒆𝒓𝒆𝒛𝒆𝜽𝒆𝒛𝒆𝒛)\displaystyle\begin{pmatrix}\partial_{r}u_{r}&\frac{1}{r}\partial_{\theta}u_{r}-\frac{1}{r}u_{\theta}&\partial_{z}u_{r}\\ \partial_{r}u_{\theta}&\frac{1}{r}\partial_{\theta}u_{\theta}+\frac{1}{r}u_{r}&\partial_{z}u_{\theta}\\ \partial_{r}u_{z}&\frac{1}{r}\partial_{\theta}u_{z}&\partial_{z}u_{z}\end{pmatrix}:\begin{pmatrix}\boldsymbol{e_{r}}\otimes\boldsymbol{e_{r}}&\boldsymbol{e_{r}}\otimes\boldsymbol{e_{\theta}}&\boldsymbol{e_{r}}\otimes\boldsymbol{e_{z}}\\ \boldsymbol{e_{\theta}}\otimes\boldsymbol{e_{r}}&\boldsymbol{e_{\theta}}\otimes\boldsymbol{e_{\theta}}&\boldsymbol{e_{\theta}}\otimes\boldsymbol{e_{z}}\\ \boldsymbol{e_{z}}\otimes\boldsymbol{e_{r}}&\boldsymbol{e_{z}}\otimes\boldsymbol{e_{\theta}}&\boldsymbol{e_{z}}\otimes\boldsymbol{e_{z}}\end{pmatrix} (A.87)
:=\displaystyle:= (rur1rθur1ruθzurruθ1rθuθ+1rurzuθruz1rθuzzuz):𝒜.\displaystyle\begin{pmatrix}\partial_{r}u_{r}&\frac{1}{r}\partial_{\theta}u_{r}-\frac{1}{r}u_{\theta}&\partial_{z}u_{r}\\ \partial_{r}u_{\theta}&\frac{1}{r}\partial_{\theta}u_{\theta}+\frac{1}{r}u_{r}&\partial_{z}u_{\theta}\\ \partial_{r}u_{z}&\frac{1}{r}\partial_{\theta}u_{z}&\partial_{z}u_{z}\end{pmatrix}:\mathcal{A}.

Also a direct computation shows that

×u=(1rθuzzuθ)𝒆𝒓+(zurruz)𝒆𝜽+1r(r(ruθ)θur)𝒆𝒛.\nabla\times u=\left(\frac{1}{r}\partial_{\theta}u_{z}-\partial_{z}u_{\theta}\right)\boldsymbol{e_{r}}+\left(\partial_{z}u_{r}-\partial_{r}u_{z}\right)\boldsymbol{e_{\theta}}+\frac{1}{r}\left(\partial_{r}(ru_{\theta})-\partial_{\theta}u_{r}\right)\boldsymbol{e_{z}}.

Then 𝕊u{\mathbb{S}}u under the base 𝒜\mathcal{A} is represented by

𝕊u=(rur12(1rθur+ruθ1ruθ)12(zur+ruz)12(1rθur+ruθ1ruθ)1rθuθ+1rur12(zuθ+1rθuz)12(zur+ruz)12(zuθ+1rθuz)zuz):𝒜.{\mathbb{S}}u=\begin{pmatrix}\partial_{r}u_{r}&\frac{1}{2}\left(\frac{1}{r}\partial_{\theta}u_{r}+\partial_{r}u_{\theta}-\frac{1}{r}u_{\theta}\right)&\frac{1}{2}(\partial_{z}u_{r}+\partial_{r}u_{z})\\ \frac{1}{2}\left(\frac{1}{r}\partial_{\theta}u_{r}+\partial_{r}u_{\theta}-\frac{1}{r}u_{\theta}\right)&\frac{1}{r}\partial_{\theta}u_{\theta}+\frac{1}{r}u_{r}&\frac{1}{2}\left(\partial_{z}u_{\theta}+\frac{1}{r}\partial_{\theta}u_{z}\right)\\ \frac{1}{2}(\partial_{z}u_{r}+\partial_{r}u_{z})&\frac{1}{2}\left(\partial_{z}u_{\theta}+\frac{1}{r}\partial_{\theta}u_{z}\right)&\partial_{z}u_{z}\end{pmatrix}:\mathcal{A}. (A.88)

Since the outward normal vector 𝒏=𝒆𝒓\boldsymbol{n}=\boldsymbol{e_{r}}, we have

𝕊u𝒏=rur𝒆𝒓+12(1rθur+ruθ1ruθ)𝒆𝜽+12(zur+ruz)𝒆𝒛.{\mathbb{S}}u\cdot\boldsymbol{n}=\partial_{r}u_{r}\boldsymbol{e_{r}}+\frac{1}{2}\left(\frac{1}{r}\partial_{\theta}u_{r}+\partial_{r}u_{\theta}-\frac{1}{r}u_{\theta}\right)\boldsymbol{e_{\theta}}+\frac{1}{2}(\partial_{z}u_{r}+\partial_{r}u_{z})\boldsymbol{e_{z}}.

Then in cylinder coordinates, one has

(𝕊u𝒏)τ=12(1rθur+ruθ1ruθ)𝒆𝜽+12(zur+ruz)𝒆𝒛,({\mathbb{S}}u\cdot\boldsymbol{n})_{\tau}=\frac{1}{2}\left(\frac{1}{r}\partial_{\theta}u_{r}+\partial_{r}u_{\theta}-\frac{1}{r}u_{\theta}\right)\boldsymbol{e_{\theta}}+\frac{1}{2}(\partial_{z}u_{r}+\partial_{r}u_{z})\boldsymbol{e_{z}},

and

×u×𝒏=(zurruz)𝒆𝒛+1r(r(ruθ)θur)𝒆𝜽.\nabla\times u\times\boldsymbol{n}=-\left(\partial_{z}u_{r}-\partial_{r}u_{z}\right)\boldsymbol{e_{z}}+\frac{1}{r}\left(\partial_{r}(ru_{\theta})-\partial_{\theta}u_{r}\right)\boldsymbol{e_{\theta}}.

This, together with (A.85), the boundary condition (NSB) and (NHLB) in the cylindrical coordinates read

{ruθuθr=0,ruz=0,ur=0,x𝒟,\left\{\begin{aligned} &\partial_{r}u_{\theta}-\frac{u_{\theta}}{r}=0,\\ &\partial_{r}u_{z}=0,\\ &u_{r}=0,\\ \end{aligned}\right.\quad\forall x\in\partial\mathcal{D},

and

{ruθ+uθr=0,ruz=0,ur=0,x𝒟,\left\{\begin{aligned} &\partial_{r}u_{\theta}+\frac{u_{\theta}}{r}=0,\\ &\partial_{r}u_{z}=0,\\ &u_{r}=0,\\ \end{aligned}\right.\quad\forall x\in\partial\mathcal{D},

where we have used the fact that θur=zur=0\partial_{\theta}u_{r}=\partial_{z}u_{r}=0 on the boundary.

Acknowledgments

The authors wish to thank Professors Qi S. Zhang, Xin Yang and Na Zhao, and Mr. Chulan Zeng for helpful discussions and proof reading. Z. Li is supported by Natural Science Foundation of Jiangsu Province (No. BK20200803) and National Natural Science Foundation of China (No. 12001285). X. Pan is supported by National Natural Science Foundation of China (No. 11801268, 12031006). J. Yang is supported by National Natural Science Foundation of China (No. 12001429).

Conflict of interest statement

On behalf of all authors, the corresponding author states that there is no conflict of interest.

Data availability statement

Data sharing not applicable to this article as no datasets were generated or analysed during the current study.

References

  • [1] C. J. Amick: Steady solutions of the Navier-Stokes equations in unbounded channels and pipes. Ann. Scuola Norm. Sup. Pisa Cl. Sci. (4) 4 (1977), no. 3, 473–513.
  • [2] C. J. Amick: Properties of steady Navier-Stokes solutions for certain unbounded channels and pipes. Nonlinear Anal. 2 (1978), no. 6, 689–720.
  • [3] K. A. Ames and L. E. Payne: Decay estimates in steady pipe flow. SIAM J. Math. Anal. 20 (1989), no. 4, 789–815.
  • [4] H. Beirao da Veiga and J. Yang: Regularity criteria for Navier-Stokes equations with slip boundary conditions on non-flat boundaries via two velocity components. Adv. Nonlinear Anal., 9 (2020), no. 1, 633–643,.
  • [5] M. Bogovskiĭ: Solution of the first boundary value problem for an equation of continuity of an incompressible medium, (Russian) Dokl. Akad. Nauk SSSR (248) 5 (1979), no. 3, 1037–1040.
  • [6] M. Bogovskiĭ: Solutions of some problems of vector analysis, associated with the operators div and grad, (Russian) Theory of cubature formulas and the application of functional analysis to problems of mathematical physics, pp. 5–40, 149, Trudy Sem. S. L. Soboleva, No. 1, 1980, Akad. Nauk SSSR Sibirsk. Otdel., Inst. Mat., Novosibirsk, 1980.
  • [7] B. Carrillo, X. Pan, Q. S. Zhang and N. Zhao: Decay and vanishing of some D-solutions of the Navier-Stokes equations. Arch. Ration. Mech. Anal. 237 (2020), no. 3, 1383–1419.
  • [8] D. Chae: Liouville-type theorems for the forced Euler equations and the Navier-Stokes equations. Comm. Math. Phys. 326 (2014), no. 1, 37–48.
  • [9] D. Chae and J. Wolf: On Liouville type theorems for the steady Navier-Stokes equations in 3{\mathbb{R}}^{3}. J. Differential Equations 261 (2016), no. 10, 5541–5560.
  • [10] G.-Q. Chen and Z. Qian: A study of the Navier-Stokes equations with the kinematic and Navier boundary conditions. Indiana Univ. Math. J., 59 (2010), no. 2, 721–760.
  • [11] G. P. Galdi: An introduction to the mathematical theory of the Navier-Stokes equations. Steady-state problems. Second edition. Springer Monographs in Mathematics. Springer, New York, 2011.
  • [12] Y. Giga and H. Miura: On vorticity directions near singularities for the Navier-Stokes fows with infinite energy, Comm. Math. Phys., 303 (2011), 289–300.
  • [13] Y. Giga, P. Y. Hsu and Y. Maekawa: A Liouville theorem for the planer Navier-Stokes equations with the no-slip boundary condition and its application to a geometric regularity criterion. Comm. Partial Differential Equations 39 (2014), no. 10, 1906–1935.
  • [14] C. O. Horgan and L. T. Wheeler: Spatial decay estimates for the Navier-Stokes equations with application to the problem of entry flow. SIAM J. Appl. Math. 35 (1978), no. 1, 97–116.
  • [15] G. Koch, N. Nadirashvili, G. A. Seregin and V. Šverák: Liouville theorems for the Navier-Stokes equations and applications. Acta Math. 203 (2009), no. 1, 83–105.
  • [16] V.A. Kondrat’ev and O.A. Olenik, Boundary value problems for a system in elasticity theory in unbounded domains. Korn inequalities. Russ. Math. Surv. 43 (1988), 65–119.
  • [17] P. Konieczny: On a steady flow in a three-dimensional infinite pipe. Colloq. Math. 104 (2006), no. 1, 33–56.
  • [18] O. A. Ladyženskaya: Investigation of the Navier-Stokes equation for stationary motion of an incompressible fluid. (Russian) Uspehi Mat. Nauk 14 (1959), no. 3, 75–97.
  • [19] O. A. Ladyženskaya: Stationary motion of viscous incompressible fluids in pipes. Soviet Physics. Dokl. 124 (4) (1959), 68–70 (551–553 Dokl. Akad. Nauk SSSR).
  • [20] O. Ladyženskaya and V. Solonnikov: Determination of the solutions of boundary value problems for stationary Stokes and Navier-Stokes equations having an unbounded Dirichlet integral, Translated from Zapiski Nauchnykh Seminarov Leningradskogo Otdeleniya Matematicheskogo Instituta im. V. A. Steklova Akad. Nauk SSSR, Vol. 96, pp. 117-160, 1980.
  • [21] P. Li and R. Schoen: LpL^{p} and mean value properties of subharmonic functions on Riemannian manifolds. Acta Math. 153(3–4), (1884), 279–301.
  • [22] Z. Li, X. Pan and J. Yang: On Leray’s problem in an infinite-long pipe with the Navier-slip boundary condition. arXiv: 2204.10578.
  • [23] D. Phan and S. S. Rodrigues.: Gevrey regularity for Navier-Stokes equations under Lions boundary conditions. J. Funct. Anal., 272 (2017), no. 7, 2865–2898.
  • [24] K. Mikhail, W. Lyu and S. Weng: On the existence of helical invariant solutions to steady Navier-Stokes equations. arXiv: 2102.13341.
  • [25] M. Mitrea and S. Monniaux: The nonlinear Hodge-Navier-Stokes equations in Lipschitz domains. Differential Integral Equations, 22 (2009), no. 3-4, 339–356.
  • [26] P. B. Mucha: On Navier-Stokes equations with slip boundary conditions in an infinite pipe. Acta Appl. Math. 76 (2003), no. 1, 1–15.
  • [27] P. B. Mucha: Asymptotic behavior of a steady flow in a two-dimensional pipe. Studia Math. 158 (2003), no. 1, 39–58.
  • [28] C. Navier: Sur les lois du mouvement des fuides. Mem. Acad. R. Sci. Inst. France, 6, (1823), 389–440.
  • [29] X. Pan: A Liouville theorem of Navier-Stokes equations with two periodic variables. J. Math. Anal. Appl. 485 (2020), no. 2, 123854, 7 pp.
  • [30] K. Piletskas: On the asymptotic behavior of solutions of a stationary system of Navier-Stokes equations in a domain of layer type. (Russian) Mat. Sb. 193 (2002), no. 12, 69–104; translation in Sb. Math. 193 (2002), no. 11–12, 1801–1836.
  • [31] G. Seregin: Liouville type theorem for stationary Navier-Stokes equations. Nonlinearity 29 (2016), no. 8, 2191–2195.
  • [32] W. Wang: Remarks on Liouville type theorems for the 3D steady axially symmetric Navier-Stokes equations. J. Differential Equations 266 (2019), no. 10, 6507–6524.
  • [33] Y. Wang and C. Xie: Uniform structural stability of Hagen-Poiseuille flows in a pipe. arXiv: 1911.00749.
  • [34] Y. Xiao and Z. Xin: On the vanishing viscosity limit for the 3D Navier-Stokes equations with a slip boundary condition. Comm. Pure Appl. Math., 60 (2007), no. 7, 1027–1055.
  • [35] Q. S. Zhang: Bounded solutions to the axially symmetric Navier Stokes equation in a cusp region. arXiv:2106.08509 v1.

Z. Li: School of Mathematics and Statistics, Nanjing University of Information Science and Technology, Nanjing 210044, China

E-mail address: [email protected]

X. Pan: College of Mathematics, Nanjing University of Aeronautics and Astronautics, Nanjing 211106, China

E-mail address: [email protected]

J. Yang: School of Mathematics and Statistics, Northwestern Polytechnical University, Xi’an 710129, China

E-mail address: [email protected]