Theorem 1.1 Let $\varOmega =\mathbb {R}^2 \times (0,1)$ and ${\boldsymbol {u}}$ be a smooth axisymmetric solution of the Navier–Stokes system (1.1) with no-slip boundary conditions (1.2). Then ${\boldsymbol {u}} \equiv 0$ if either (a)

(1.8)\begin{equation} \varliminf_{R\rightarrow + \infty} R^{{-}4} \mathcal{E}(R) = 0, \end{equation}

where

(1.9)\begin{equation} \mathcal{E}(R)=\displaystyle \int_0^1 \int_{\{x_1^2+x_2^2 < R^2\}} |\boldsymbol{\nabla} {\boldsymbol{u}}|^2 \, {\rm d}\kern0.07em x_1\, {\rm d}\kern0.07em x_2\, {\rm d}\kern0.07em x_3, \end{equation}

or (b)

(1.10)\begin{equation} \lim_{R\rightarrow + \infty} R^{{-}1} \sup_{z \in [0, 1]} |{\boldsymbol{u}} (R, z) | =0. \end{equation}

Theorem 1.2 Let ${\boldsymbol {u}}$ be a smooth solution to the Navier–Stokes system (1.1) in $\varOmega =\mathbb {R}^2\times (0,1)$ with no-slip boundary conditions (1.2). Then there exists an absolute number $\alpha \in (0,1)$ such that ${\boldsymbol {u}}\equiv 0$ if one of the following conditions holds:

1. (a)

   (1.11)\begin{equation} \varliminf_{R\rightarrow + \infty} R^{-\alpha} \mathcal{E}(R) = 0, \end{equation}
   where $\mathcal {E}(R)$ is defined in (1.9);

2. (b) ${\boldsymbol {u}}$ satisfies for some $\beta \in [0, {\alpha }/{2}]$,

   (1.12a,b)\begin{equation} \lim_{r\to\infty} r^{-\beta }\sup_{\substack{{\theta\in [0,2{\rm \pi})}\\ {z\in [0,1]}}} |{\boldsymbol{u}} (r, \theta, z)| =0\quad\text{and}\quad \sup_{(r, \theta, z)\in \varOmega}r^{1+\beta -{\alpha}/{2}}| u^r(r, \theta, z)|<\infty. \end{equation}

Theorem 1.3 Let ${\boldsymbol {u}}$ be a smooth solution to the Navier–Stokes system (1.1) in a slab $\mathbb {R}^2\times (0,1)$ with no-slip boundary conditions (1.2). Then ${\boldsymbol {u}}$ must be a Poiseuille flow of form $(c_1x_3(1-x_3), c_2x_3(1-x_3), 0)$ if

(1.14)\begin{equation} \| {\boldsymbol{u}} \|_{L^\infty(\varOmega)} <{\rm \pi}. \end{equation}

Theorem 1.4 Let ${\boldsymbol {u}}$ be a bounded smooth solution to the Navier–Stokes system (1.1) in $\mathbb {R}^2\times \mathbb {T}$. Then ${\boldsymbol {u}}$ must be a constant vector provided that one of the following conditions holds:

1. (a) $u^\theta$ is axisymmetric, i.e. $u^\theta$ is independent of $\theta$;

2. (b) $u^r$ is axisymmetric, i.e. $u^r$ is independent of $\theta$;

3. (c) $ru^r$ converges to $0$, as $r \rightarrow +\infty$ uniformly in $\theta,z$.

4. (d) $\|{\boldsymbol {u}}\|_{L^\infty (\varOmega )} <2{\rm \pi}.$

Furthermore, in all cases ( a ), ( b ) and ( c ), the only non-zero component of the velocity field must be $u^z$, i.e. the constant vector ${\boldsymbol {u}}$ must be of the form $(0,0, c)$.

Proof Proof of Theorem 1.1

The proof is divided into three steps.

Step 1. Set up. Assume that ${\boldsymbol {u}}$ is a smooth solution to (1.1) in $\varOmega =\mathbb {R}^2\times (0,1)$ with no-slip boundary conditions. Multiplying the first equation in (1.1) by $\varphi _R(r) {\boldsymbol {u}}$ and integrating by parts one obtains

(3.2)\begin{equation} \int_{\varOmega} |\boldsymbol{\nabla} {\boldsymbol{u}}|^2 \varphi_R ={-} \int_{\varOmega} \boldsymbol{\nabla} \varphi_R \boldsymbol{\cdot} \boldsymbol{\nabla} {\boldsymbol{u}} \boldsymbol{\cdot} {\boldsymbol{u}} + \int_{\varOmega} \frac12 |{\boldsymbol{u}}|^2{\boldsymbol{u}} \boldsymbol{\cdot} \boldsymbol{\nabla} \varphi_R + \int_{\varOmega} P {\boldsymbol{u}} \boldsymbol{\cdot} \boldsymbol{\nabla} \varphi_R . \end{equation}

Note that

(3.3)\begin{equation} \int_{\varOmega} P {\boldsymbol{u}} \boldsymbol{\cdot} \boldsymbol{\nabla} \varphi_R = 2{\rm \pi} \int_0^1 \int_{R-1}^R P u^r r \, {\rm d} r \, {\rm d} z. \end{equation}

Clearly, the last equation in (1.6) for axisymmetric solutions can be reduced to

(3.4)\begin{equation} \partial_r ( r u^r) + \partial_z ( ru^z ) = 0. \end{equation}

Hence, for every fixed $r \geq 0$, one has

(3.5)\begin{equation} \partial_r \int_0^1 r u^r \, {\rm d} z ={-} \int_0^1 \partial_z (r u^z) \, {\rm d} z = 0. \end{equation}

Thus, it holds that

(3.6a,b)\begin{equation} \int_0^1 r u^r \, {\rm d} z = 0 \quad \text{and} \quad \int_0^1 \int_{R-1}^R ru^r \, {\rm d} r \, {\rm d} z = 0. \end{equation}

By virtue of Lemma 2.1, there exists a vector valued function ${\boldsymbol {\varPsi }}_R(r, z) \in H_0^1(D_R; \mathbb {R}^2)$ satisfying

(3.7)\begin{equation} \partial_r \varPsi_R^r + \partial_z \varPsi_R^z = r u^r \end{equation}

and

(3.8)\begin{equation} \| \partial_r {\boldsymbol{\varPsi}}_R \|_{L^2(D_R)} + \| \partial_z {\boldsymbol{\varPsi}}_R \|_{L^2(D_R ) } \leq C \|r u^r \|_{L^2(D_R)} \leq C R^{1/2} \| u^r \|_{L^2({O}_R ) }. \end{equation}

Therefore, combining (3.3) and (3.7) one derives

(3.9)\begin{equation} \int_{\varOmega} P \boldsymbol{\nabla}\varphi_R\boldsymbol{\cdot} {\boldsymbol{u}} = \int_0^1 \int_{R-1}^R P ( \partial_r \varPsi_R^r + \partial_z \varPsi_R^z ) \, {\rm d} r\, {\rm d} z ={-} \int_0^1 \int_{R-1}^R (\partial_r P \varPsi_R^r + \partial_z P \varPsi_R^z ) \, {\rm d} r\, {\rm d} z. \end{equation}

Since ${\boldsymbol {u}}$ is an axisymmetric solution of (1.6), the gradient $(\partial _r P, \partial _z P)$ of the pressure satisfies

(3.10)\begin{equation} \left. \begin{array}{@{}c@{}} \displaystyle (u^r \partial_r + u^z \partial_z ) u^r - \dfrac{(u^\theta)^2}{ r} + \partial_r P = \left( \partial_r^2 + \dfrac1r \partial_r + \partial_z^2 - \dfrac{1}{r^2} \right) u^r,\\ \displaystyle (u^r \partial_r + u^z \partial_z ) u^z + \partial_z P = \left( \partial_r^2 + \dfrac1r \partial_r + \partial_z^2 \right) u^z. \end{array} \right\} \end{equation}

Using (3.10) and integration by parts one obtains

(3.11)\begin{align} & \int_0^1 \int_{R-1}^R \partial_r P \varPsi_R^r \, {\rm d} r\, {\rm d} z\nonumber\\ &\quad =\int_0^1 \int_{R-1}^R \left( \partial_r^2 + \frac1r \partial_r + \partial_z^2 - \frac{1}{r^2} \right) u^r \varPsi_R^r \, {\rm d} r\, {\rm d} z \nonumber\\ &\qquad - \int_0^1 \int_{R-1}^R \left[ (u^r \partial_r + u^z \partial_z )u^r - \frac{(u^\theta)^2}{r} \right]\varPsi_R^r \, {\rm d} r\, {\rm d} z \nonumber\\ &\quad ={-} \int_0^1 \int_{R-1}^R ( \partial_r u^r \partial_r \varPsi^r_R + \partial_z u^r \partial_z \varPsi_R^r ) \, {\rm d} r\, {\rm d} z + \int_0^1 \int_{R-1}^R \left( \frac1r \partial_r - \frac{1}{r^2} \right) u^r \varPsi_R^r \, {\rm d} r\, {\rm d} z \nonumber\\ &\qquad -\int_0^1 \int_{R-1}^R \left[ (u^r \partial_r + u^z \partial_z )u^r - \frac{(u^\theta)^2}{r} \right]\varPsi_R^r \, {\rm d} r\, {\rm d} z \end{align}

and

(3.12)\begin{align} & \int_0^1 \int_{R-1}^R \partial_z P \varPsi_R^z \, {\rm d} r\, {\rm d} z \nonumber\\ & \quad ={-} \int_0^1 \int_{R-1}^R ( \partial_r u^z \partial_r \varPsi^z_R + \partial_z u^z \partial_z \varPsi_R^z ) \, {\rm d} r\, {\rm d} z + \int_0^1 \int_{R-1}^R \frac1r \partial_r u^r \varPsi_R^r \, {\rm d} r\, {\rm d} z \nonumber\\ &\qquad -\int_0^1 \int_{R-1}^R (u^r \partial_r + u^z \partial_z )u^z \varPsi_R^z \, {\rm d} r\, {\rm d} z. \end{align}

Step 2. Proof for case (a) of Theorem 1.1. Now we start to estimate the first two terms on the right-hand side of (3.2) and the terms on the right-hand sides of (3.11) and (3.12). By the Poincaré inequality (3.1) and the Sobolev embedding inequality, one derives

(3.13)\begin{equation} \left| \int_{\varOmega} \boldsymbol{\nabla} \varphi_R \boldsymbol{\cdot} \boldsymbol{\nabla} {\boldsymbol{u}} \boldsymbol{\cdot} {\boldsymbol{u}} \right| \leq C \|\boldsymbol{\nabla} {\boldsymbol{u}} \|_{L^2({O}_R)} \|{\boldsymbol{u}}\|_{L^2({O}_R ) } \leq C \| \boldsymbol{\nabla} {\boldsymbol{u}} \|_{L^2({O}_R)}^2 \end{equation}

and

(3.14)\begin{align} \left| \int_{\varOmega} \frac12 |{\boldsymbol{u}}|^2 {\boldsymbol{u}} \boldsymbol{\cdot} \boldsymbol{\nabla} \varphi_R \right| \leq C R \int_0^1 \int_{R-1}^R |{\boldsymbol{u}}|^3 \, {\rm d} r \, {\rm d} z &\leq C R \| (\partial_r, \partial_z ){\boldsymbol{u}} \|_{L^2(D_R)}^3 \nonumber\\ &\leq C R^{{-}1/2} \|\boldsymbol{\nabla} {\boldsymbol{u}}\|_{L^2({O}_R) }^3. \end{align}

By (3.8) and the Poincaré inequality, one has

(3.15)\begin{align} \left| \int_0^1 \int_{R-1}^R ( \partial_r u^r \partial_r \varPsi^r_R + \partial_z u^r \partial_z \varPsi_R^r ) \, {\rm d} r\, {\rm d} z \right| &\leq C R^{{-}1/2} \| \boldsymbol{\nabla} {\boldsymbol{u}}\|_{L^2({O}_R) } \| \tilde{\boldsymbol{\nabla}}{\boldsymbol{\varPsi}}_R\|_{L^2(D_R)} \nonumber\\ &\leq C \| \boldsymbol{\nabla} {\boldsymbol{u}}\|_{L^2({O}_R) }^2 \end{align}

and

(3.16)\begin{align} \left| \int_0^1 \int_{R-1}^R \left( \frac1r \partial_r - \frac{1}{r^2} \right) u^r \varPsi_R^r \, {\rm d} r\, {\rm d} z \right| &\leq C R^{{-}1} \| \tilde{\boldsymbol{\nabla}} {\boldsymbol{u}}\|_{L^2(D_R)} \|{\boldsymbol{\varPsi}}_R\|_{L^2(D_R)} \nonumber\\ &\leq CR^{{-}1} \| \boldsymbol{\nabla} {\boldsymbol{u}}\|_{L^2({O}_R) }^2. \end{align}

Furthermore, it holds that

(3.17)\begin{align} & \left| \int_0^1 \int_{R-1}^R \left[ (u^r \partial_r + u^z \partial_z )u^r - \frac{(u^\theta)^2}{r} \right]\varPsi_R^r \, {\rm d} r\, {\rm d} z \right| \nonumber\\ &\quad \leq C \| (u^r, u^z, u^\theta ) \|_{L^4(D_R)} ( \| \tilde{\boldsymbol{\nabla}} u^r \|_{L^2(D_R)} + R^{{-}1} \|u^\theta \|_{L^2(D_R)} ) \|{\boldsymbol{\varPsi}}_R \|_{L^4(D_R)}\nonumber\\ &\quad \leq C \| (u^r, u^\theta, u^z ) \|_{L^2(D_R)}^{1/2} \| (u^r, u^\theta, u^z) \|_{H^1(D_R)}^{1/2} \cdot R^{{-}1/2} \|\boldsymbol{\nabla} {\boldsymbol{u}} \|_{L^2({O}_R ) } \cdot \|\tilde{\boldsymbol{\nabla}} {\boldsymbol{\varPsi}}_R\|_{L^2(D_R)} \nonumber\\ &\quad \leq C \| \tilde{\boldsymbol{\nabla}} (u^r, u^\theta, u^z ) \|_{L^2(D_R)} \cdot R^{{-}1/2} \|\boldsymbol{\nabla} {\boldsymbol{u}} \|_{L^2({O}_R ) } \cdot R^{1/2} \|u^r \|_{L^2({O}_R)}\nonumber\\ &\quad \leq CR^{{-}1/2} \|\boldsymbol{\nabla} {\boldsymbol{u}} \|_{L^2({O}_R ) } ^3. \end{align}

Combining the estimates (3.15)–(3.17) one arrives at

(3.18)\begin{equation} \left| \int_0^1 \int_{R-1}^R \partial_r P \varPsi_R^r \, {\rm d} r\, {\rm d} z \right| \leq C \|\boldsymbol{\nabla} {\boldsymbol{u}} \|_{L^2({O}_R ) } ^2 + C R^{{-}1/2} \| \boldsymbol{\nabla} {\boldsymbol{u}} \|_{L^2({O}_R ) } ^3. \end{equation}

Similarly, one has

(3.19)\begin{equation} \left| \int_0^1 \int_{R-1}^R \partial_z P \varPsi_R^z \, {\rm d} r\, {\rm d} z \right| \leq C \|\boldsymbol{\nabla} {\boldsymbol{u}} \|_{L^2({O}_R ) } ^2 + C R^{{-}1/2} \| \boldsymbol{\nabla} {\boldsymbol{u}} \|_{L^2({O}_R ) } ^3. \end{equation}

Now it can be shown that

(3.20)\begin{equation} \int_{\varOmega} |\boldsymbol{\nabla} {\boldsymbol{u}}|^2 \varphi_R \leq C \|\boldsymbol{\nabla} {\boldsymbol{u}} \|_{L^2({O}_R ) } ^2 + C R^{{-}1/2} \| \boldsymbol{\nabla} {\boldsymbol{u}} \|_{L^2({O}_R ) } ^3. \end{equation}

Let

(3.21)\begin{equation} Y(R) = \int_{0}^1\int_{\mathbb{R}^2} |\boldsymbol{\nabla} {\boldsymbol{u}}(x)|^2 \varphi_R(\sqrt{x_1^2+x_2^2}) \, {\rm d}\kern0.07em x_1 \, {\rm d}\kern0.07em x_2 \, {\rm d}\kern0.07em x_3. \end{equation}

Note that for any 3-D smooth function ${\boldsymbol {u}}$, straightforward computations give

(3.22)\begin{equation} {Y}(R) = \int_0^1 \int_0^{2{\rm \pi}} \left( \int_0^{R-1} |\boldsymbol{\nabla} {\boldsymbol{u}}|^2 r \, {\rm d} r + \int_{R-1}^R |\boldsymbol{\nabla} {\boldsymbol{u}}|^2 (R-r) r \, {\rm d} r \right)\, {\rm d} \theta \, {\rm d} z \end{equation}

and

(3.23)\begin{equation} Y^{\prime} (R) = \int_{{O}_R} |\boldsymbol{\nabla} {\boldsymbol{u}}|^2 \, {{\rm d}\kern0.07em x}, \end{equation}

where the explicit form of $\varphi _R$ in (2.2) has been used. For the axisymmetric solutions, $Y(R)$ in (3.22) can be reduced to

(3.24)\begin{equation} {Y}(R) = 2{\rm \pi} \int_0^1 \left( \int_0^{R-1} |\boldsymbol{\nabla} {\boldsymbol{u}}|^2 r \, {\rm d} r + \int_{R-1}^R |\boldsymbol{\nabla} {\boldsymbol{u}}|^2 (R-r) r \, {\rm d} r \right) \, {\rm d} z. \end{equation}

Hence, the estimate (3.20) can be written as

(3.25)\begin{equation} Y(R) \leq C Y^\prime (R) + C R^{{-}1/2} ( Y^{\prime} (R))^{3/2}. \end{equation}

Then case (a) of Theorem 1.1 follows from Lemma 2.2.

Step 3. Proof for case (b) of Theorem 1.1. We start from (3.2) and need to estimate the first two terms on the right-hand side of (3.2) and the terms on the right-hand sides of (3.11) and (3.12) in a different way. Using the Poincaré inequality yields

(3.26)\begin{equation} \left| \int_{\varOmega} \boldsymbol{\nabla} \varphi_R \boldsymbol{\cdot} \boldsymbol{\nabla} {\boldsymbol{u}} \boldsymbol{\cdot} {\boldsymbol{u}} \right| \leq C \| {\boldsymbol{u}} \|_{L^2({O}_R)} \|\boldsymbol{\nabla} {\boldsymbol{u}} \|_{L^2({O}_R)} \leq C R^{1/2} \| {\boldsymbol{u}} \|_{L^\infty({O}_R)} \|\boldsymbol{\nabla} {\boldsymbol{u}} \|_{L^2({O}_R)} \end{equation}

and

(3.27)\begin{align} \left| \int_{\varOmega} \frac12 |{\boldsymbol{u}}|^2{\boldsymbol{u}} \boldsymbol{\cdot} \boldsymbol{\nabla} \varphi_R \right| & \leq C R \| {\boldsymbol{u}} \|_{L^\infty({O}_R)}^2 \int_0^1 \int_{R-1}^R |(u^r, u^\theta, u^z) | \, {\rm d} r \, {\rm d} z \nonumber\\ & \leq C R \| {\boldsymbol{u}} \|_{L^\infty({O}_R)}^2\left( \int_0^1 \int_{R-1}^R | (u^r, u^\theta, u^z) |^2 \, {\rm d} r \, {\rm d} z \right)^{1/2} \nonumber\\ &\leq C R \| {\boldsymbol{u}} \|_{L^\infty({O}_R)}^2 \| \partial_z {\boldsymbol{u}} \|_{L^2(D_R)} \leq C R^{1/2} \| {\boldsymbol{u}} \|_{L^\infty({O}_R)}^2 \| \boldsymbol{\nabla} {\boldsymbol{u}} \|_{L^2({O}_R)}. \end{align}

According to (3.8), instead of (3.15) and (3.16), one has

(3.28)\begin{align} & \left| \int_0^1 \int_{R-1}^R ( \partial_r u^r \partial_r \varPsi^r_R + \partial_z u^r \partial_z \varPsi_R^r ) \, {\rm d} r\, {\rm d} z \right| \leq C \| (\partial_r, \partial_z) u^r \|_{L^2(D_R)} \| (\partial_r, \partial_z) \varPsi_R^r \|_{L^2(D_R)} \nonumber\\ &\quad \leq C R^{{-}1/2} \| \boldsymbol{\nabla} {\boldsymbol{u}} \|_{L^2({O}_R) } \cdot R^{1/2} \| u^r \|_{L^2({O}_R)} \leq C R^{1/2} \| {\boldsymbol{u}} \|_{L^\infty({O}_R ) } \| \boldsymbol{\nabla} {\boldsymbol{u}} \|_{L^2({O}_R)} \end{align}

and

(3.29)\begin{align} & \left| \int_0^1 \int_{R-1}^R \left( \frac1r \partial_r - \frac{1}{r^2} \right) u^r \varPsi_R^r \, {\rm d} r\, {\rm d} z \right| \nonumber\\ &\quad \leq C R^{{-}1} \| \partial_r u^r \|_{L^2(D_R) }\| \varPsi_R^r \|_{L^2(D_R) } + C R^{{-}2} \| u^r\|_{L^2(D_R)} \| \varPsi_R^r \|_{L^2(D_R) } \nonumber\\ &\quad \leq C R^{{-}3/2} \| \boldsymbol{\nabla} {\boldsymbol{u}} \|_{L^2({O}_R ) } \cdot R^{1/2} \|u^r \|_{L^2({O}_R)} + CR^{{-}5/2} \| \partial_z u^r \|_{L^2({O}_R)} \cdot R^{1/2} \|u^r \|_{L^2({O}_R)}\nonumber\\ &\quad \leq C R^{{-}1/2}\| {\boldsymbol{u}} \|_{L^\infty({O}_R ) } \| \boldsymbol{\nabla} {\boldsymbol{u}} \|_{L^2({O}_R)} . \end{align}

Similarly, it holds that

(3.30)\begin{align} & \left| \int_0^1 \int_{R-1}^R \left[ (u^r \partial_r + u^z \partial_z )u^r - \frac{(u^\theta)^2}{r} \right]\varPsi_R^r \, {\rm d} r\, {\rm d} z \right| \nonumber\\ &\quad \leq C \| {\boldsymbol{u}} \|_{L^\infty({O}_R ) } \| \tilde{\boldsymbol{\nabla}} u^r \|_{L^2(D_R)} \| \varPsi_R^r \|_{L^2(D_R)} + CR^{{-}1} \| {\boldsymbol{u}} \|_{L^\infty({O}_R ) } \|u^\theta \|_{L^2(D_R)} \| \varPsi_R^r \|_{L^2(D_R)} \nonumber\\ &\quad \leq C R^{1/2}\| {\boldsymbol{u}} \|_{L^\infty({O}_R ) }^2 \|\boldsymbol{\nabla} {\boldsymbol{u}} \|_{L^2({O}_R ) }. \end{align}

Combining (3.28)–(3.30) one arrives at

(3.31)\begin{equation} \left| \int_0^1 \int_{R-1}^R \partial_r P \varPsi_R^r \, {\rm d} r\, {\rm d} z \right| \leq C R^{1/2}( \| {\boldsymbol{u}} \|_{L^\infty({O}_R ) } + \| {\boldsymbol{u}} \|_{L^\infty({O}_R ) }^2 ) \| \boldsymbol{\nabla} {\boldsymbol{u}} \|_{L^2({O}_R ) }. \end{equation}

The similar computations show that

(3.32)\begin{equation} \left| \int_0^1 \int_{R-1}^R \partial_z P \varPsi_R^z \, {\rm d} r\, {\rm d} z \right| \leq C R^{1/2}( \| {\boldsymbol{u}} \|_{L^\infty({O}_R ) } + \| {\boldsymbol{u}} \|_{L^\infty({O}_R ) }^2 ) \| \boldsymbol{\nabla} {\boldsymbol{u}} \|_{L^2({O}_R ) }. \end{equation}

Therefore, one has

(3.33)\begin{equation} Y(R) \leq C R^{1/2}( \| {\boldsymbol{u}} \|_{L^\infty({O}_R ) } + \| {\boldsymbol{u}} \|_{L^\infty({O}_R ) }^2 ) [ Y^{\prime}(R) ]^{1/2}, \end{equation}

where $Y(R)$ is defined in (3.21).

Suppose ${\boldsymbol {u}}$ is not identically equal to zero and ${\boldsymbol {u}}$ satisfies (1.10). For any small $\epsilon >0$, there exists a $R_0(\epsilon ) >2$ such that

(3.34)\begin{equation} \| {\boldsymbol{u}} \|_{L^\infty({O}_R ) } \leq \epsilon R \quad \text{for any } R\geq R_0(\epsilon). \end{equation}

Hence, the inequality (3.33) implies that

(3.35)\begin{equation} \frac{Y^{\prime}(R) }{ [Y(R)]^2} \geq (C\epsilon)^{{-}2} R^{{-}5}. \end{equation}

If ${\boldsymbol {u}}$ is not equal to zero, according to case (a) of Theorem 1.1, $Y(R)$ must be unbounded as $R\rightarrow + \infty$. For every $R$ sufficiently large, integrating (3.35) over $[R, + \infty )$ one arrives at

(3.36)\begin{equation} \frac{1}{Y(R)} \geq \frac14 (C \epsilon)^{{-}2} R^{{-}4}. \end{equation}

Since $\epsilon$ can be arbitrarily small, this implies (1.8) and leads to a contradiction with the assumption that ${\boldsymbol {u}}$ is not identically zero. Hence, the proof of Theorem 1.1 is completed.

Proof Proof of Theorem 1.2

The proof contains three steps.

Step 1. Set up. Assume that ${\boldsymbol {u}}$ is a smooth solution to the Navier–Stokes system (1.1). The equality (3.2) still holds. Instead of (3.3), one has

(4.1)\begin{equation} \int_{\varOmega} P {\boldsymbol{u}} \boldsymbol{\cdot} \boldsymbol{\nabla} \varphi_R = \int_0^1 \int_0^{2{\rm \pi}} \int_{R-1}^R P u^r r \, {\rm d} r \, {\rm d} \theta \, {\rm d} z . \end{equation}

It follows from the last equation in (1.6) (divergence free equation) that for every fixed $r\geq 0$, one has

(4.2)\begin{equation} \partial_r \int_0^1 \int_0^{2{\rm \pi} } ru^r \, {\rm d} \theta \, {\rm d} z ={-} \int_0^1 \int_0^{2{\rm \pi}} \partial_\theta u^\theta \, {\rm d} \theta \, {\rm d} z - \int_0^1 \int_0^{2{\rm \pi}} \partial_z (r u^z) \, {\rm d} \theta \, {\rm d} z = 0. \end{equation}

Then it holds that

(4.3a,b)\begin{equation} \int_0^1 \int_0^{2{\rm \pi}} r u^r \, {\rm d} \theta \, {\rm d} z = 0\quad \text{and}\quad \int_0^1 \int_0^{2{\rm \pi}} \int_{R-1}^R r u^r \, {\rm d} r \, {\rm d} \theta \, {\rm d} z =0. \end{equation}

By virtue of Lemma 2.1, there exists a vector valued function ${\boldsymbol {\varPsi }}_R(r, \theta, z)\in H_0^1(\mathcal {D}_R; \mathbb {R}^3)$ satisfying

(4.4)\begin{equation} \partial_r \varPsi_R^r + \partial_\theta \varPsi_R^\theta + \partial_z \varPsi_R^z = ru^r \quad \mbox{in}\ \mathcal{D}_R \end{equation}

and

(4.5)\begin{align} \|\partial_r {\boldsymbol{\varPsi}}_R\|_{L^2(\mathcal{D}_R ) } + \|\partial_\theta {\boldsymbol{\varPsi}}_R\|_{L^2(\mathcal{D}_R)} + \|\partial_z {\boldsymbol{\varPsi}}_R\|_{L^2(\mathcal{D}_R)} \leq C \|r u^r \|_{L^2(\mathcal{D}_R)} \leq C R^{1/2} \| u^r \|_{L^2({O}_R)}.\end{align}

Combining (4.1) and (4.4) one obtains

(4.6)\begin{align} \int_{\varOmega} P {\boldsymbol{u}} \boldsymbol{\cdot} \boldsymbol{\nabla} \varphi_R & = \int_0^1 \int_0^{2{\rm \pi}} \int_{R-1}^R P ( \partial_r \varPsi_R^r + \partial_\theta \varPsi_R^\theta + \partial_z \varPsi_R^z ) \, {\rm d} r \, {\rm d} \theta \, {\rm d} z \nonumber\\ & ={-} \int_0^1 \int_0^{2{\rm \pi}} \int_{R-1}^R (\partial_r P \varPsi_R^r + \partial_\theta P \varPsi_R^\theta + \partial_z P \varPsi_R^z ) \, {\rm d} r \, {\rm d} \theta \, {\rm d} z. \end{align}

Furthermore, it follows from the momentum equations in (1.6) that one has

(4.7)\begin{align} &\int_0^1 \int_0^{2{\rm \pi}} \int_{R-1}^R \partial_r P \varPsi_R^r \, {\rm d} r \, {\rm d} \theta \, {\rm d} z \nonumber\\ &\quad ={-} \int_0^1 \int_0^{2{\rm \pi}} \int_{R-1}^R \left[\partial_r u^r \partial_r \varPsi_R^r + \partial_z u^r \partial_z \varPsi_R^r + \frac{1}{r^2} (\partial_\theta u^r - 2 u^\theta) \partial_\theta \varPsi_R^r \right] {\rm d} r\, {\rm d} \theta \, {\rm d} z \nonumber\\ &\qquad + \int_0^1 \int_0^{2{\rm \pi}} \int_{R-1}^R \left[\left( \frac{1}{r} \partial_r - \frac{1}{r^2} \right) u^r + \frac{(u^\theta)^2}{r}\right]\varPsi_R^r \, {\rm d} r \, {\rm d} \theta \, {\rm d} z \nonumber\\ & \qquad - \int_0^1 \int_0^{2{\rm \pi}} \int_{R-1}^R \left(u^r \partial_r + \frac{u^\theta}{r} \partial_\theta + u^z \partial_z \right) u^r \varPsi_R^r \, {\rm d} r \, {\rm d} \theta \, {\rm d} z , \end{align}

(4.8)\begin{align} &\int_0^1 \int_0^{2{\rm \pi}} \int_{R-1}^R \partial_\theta P \varPsi_R^\theta \, {\rm d} r \, {\rm d} \theta \, {\rm d} z \nonumber\\ &\quad ={-} \int_0^1 \int_0^{2{\rm \pi}} \int_{R-1}^R [ r(\partial_r u^\theta \partial_r \varPsi_R^\theta + \partial_z u^\theta \partial_z \varPsi_R^\theta) + r^{{-}1} ( \partial_\theta u^\theta + 2 u^r )\partial_\theta \varPsi_R^\theta ] \, {\rm d} r\, {\rm d} \theta \, {\rm d} z \nonumber\\ &\qquad + \int_0^1 \int_0^{2{\rm \pi}} \int_{R-1}^R \left[- \frac{1}{r} u^\theta - u^\theta u^r \right]\varPsi_R^\theta \, {\rm d} r \, {\rm d} \theta \, {\rm d} z \nonumber\\ &\qquad - \int_0^1 \int_0^{2{\rm \pi}} \int_{R-1}^R r \left(u^r \partial_r + \frac{1}{r} u^\theta \partial_\theta + u^z \partial_z \right) u^\theta \varPsi_R^\theta \, {\rm d} r \, {\rm d} \theta \, {\rm d} z \end{align}

and

(4.9)\begin{align} &\int_0^1 \int_0^{2{\rm \pi}} \int_{R-1}^R \partial_z P \varPsi_R^z \, {\rm d} r \, {\rm d} \theta \, {\rm d} z \nonumber\\ &\quad ={-} \int_0^1 \int_0^{2{\rm \pi}} \int_{R-1}^R \left(\partial_r u^z \partial_r \varPsi_R^z + \partial_z u^z \partial_z \varPsi_R^z + \frac{1}{r^2} \partial_\theta u^z \partial_\theta \varPsi_R^z \right) \, {\rm d} r\, {\rm d} \theta \, {\rm d} z \nonumber\\ &\qquad - \int_0^1 \int_0^{2{\rm \pi}} \int_{R-1}^R \left(u^r \partial_r + \frac{u^\theta}{r} \partial_\theta + u^z \partial_z -\frac{1}{r}\partial_r \right) u^z \varPsi_R^z \, {\rm d} r \, {\rm d} \theta \, {\rm d} z . \end{align}

Step 2. Proof for case (a) of Theorem 1.2. Now we estimate the first two terms on the right-hand side of (3.2) and the terms on the right-hand sides of (4.7)–(4.9). First, the Poincaré inequality leads to

(4.10)\begin{equation} \left| \int_{\varOmega} \boldsymbol{\nabla} \varphi_R \boldsymbol{\cdot} \boldsymbol{\nabla} {\boldsymbol{u}} \boldsymbol{\cdot} {\boldsymbol{u}} \right| \leq \| {\boldsymbol{u}} \|_{L^2({O}_R)} \|\boldsymbol{\nabla} {\boldsymbol{u}} \|_{L^2({O}_R)} \leq \|\boldsymbol{\nabla} {\boldsymbol{u}} \|^2_{L^2({O}_R)}. \end{equation}

Note that $\bar {\boldsymbol {\nabla }} (u^r, u^\theta, u^z) = (\partial _r, \partial _\theta, \partial _z)(u^r, u^\theta, u^z)$. Here

(4.11)\begin{equation} |\partial_r u^r | + |\partial_r u^\theta| + |\partial_r u^z | + |r^{{-}1} \partial_\theta u^z| + |\partial_z u^r| + |\partial_z u^\theta| + |\partial_z u^z| \leq C|\boldsymbol{\nabla} {\boldsymbol{u}}|, \end{equation}

whereas

(4.12)\begin{equation} |r^{{-}1} \partial_\theta u^r - r^{{-}1} u^\theta| + |r^{{-}1} \partial_\theta u^\theta + r^{{-}1} u^r| \leq C|\boldsymbol{\nabla} {\boldsymbol{u}}|, \end{equation}

and consequently,

(4.13)\begin{equation} |r^{{-}1} \partial_\theta u^r| + |r^{{-}1} \partial_\theta u^\theta| \leq C |\boldsymbol{\nabla} {\boldsymbol{u}}| + |r^{{-}1}u^\theta| + |r^{{-}1} u^r|. \end{equation}

Hence, the Poincaré inequality and Sobolev embedding inequality give that

(4.14)\begin{align} \left| \int_{\varOmega} \frac12 |{\boldsymbol{u}}|^2 {\boldsymbol{u}} \boldsymbol{\cdot} \boldsymbol{\nabla} \varphi_R \right| &\leq C R \int_0^1 \int_0^{2{\rm \pi}} \int_{R-1}^R |{\boldsymbol{u}}|^3 \, {\rm d} r \, {\rm d} \theta \, {\rm d} z \nonumber\\ & \leq C R \| (u^r, u^\theta, u^z) \|_{L^2(\mathcal{D}_R ) }^{3/2} \| \bar{\boldsymbol{\nabla}} ( u^r, u^\theta, u^z ) \|_{L^2(\mathcal{D}_R )}^{3/2} \nonumber\\ & \leq C R\cdot R^{{-}3/4} \|{\boldsymbol{u}}\|_{L^2({O}_R )}^{3/2}\cdot (R^{1/2} \| \boldsymbol{\nabla} {\boldsymbol{u}}\|_{L^2({O}_R)} + R^{{-}1/2} \|{\boldsymbol{u}}\|_{L^2({O}_R ) })^{3/2} \nonumber\\ & \leq C R \|\boldsymbol{\nabla} {\boldsymbol{u}}\|_{L^2({O}_R )}^3. \end{align}

Using (4.5) one obtains

(4.15)\begin{align} & \left| \int_0^1 \int_0^{2{\rm \pi}} \int_{R-1}^R (\partial_r u^r \partial_r \varPsi_R^r + \partial_z u^r \partial_z \varPsi_R^r ) \, {\rm d} r\, {\rm d} \theta \, {\rm d} z \right| \nonumber\\ &\quad \leq C \| ( \partial_r, \partial_z ) u^r \|_{L^2(\mathcal{D}_R)} \| ( \partial_r, \partial_z ) \varPsi_R^r\|_{L^2(\mathcal{D}_R)} \nonumber\\ &\quad \leq C R^{{-}1/2} \| \boldsymbol{\nabla} {\boldsymbol{u}} \|_{L^2({O}_R)} \cdot R^{1/2} \| u^r \|_{L^2({O}_R)} \nonumber\\ &\quad \leq C \|\boldsymbol{\nabla} {\boldsymbol{u}} \|_{L^2({O}_R)}^2, \end{align}

(4.16)\begin{align} & \left| \int_0^1 \int_0^{2{\rm \pi}} \int_{R-1}^R \frac{1}{r^2} ( \partial_\theta u^r - 2 u^\theta ) \partial_\theta \varPsi_R^r \, {\rm d} r \, {\rm d} \theta \, {\rm d} z\right| \nonumber\\ & \quad \leq C ( R^{{-}1} \| r^{{-}1} ( \partial_\theta u^r - u^\theta) \|_{L^2(\mathcal{D}_R)} + R^{{-}2} \| u^\theta\|_{L^2(\mathcal{D}_R )} ) \| \partial_\theta \varPsi_R^r \|_{L^2(\mathcal{D}_R)} \nonumber\\ & \quad \leq C ( R^{{-}3/2} + R^{{-}5/2}) \|\boldsymbol{\nabla} {\boldsymbol{u}}\|_{L^2({O}_R ) } \cdot R^{1/2} \|u^r \|_{L^2({O}_R )} \nonumber\\ & \quad \leq C R^{{-}1} \|\boldsymbol{\nabla} {\boldsymbol{u}}\|_{L^2({O}_R )}^2 \end{align}

and

(4.17)\begin{align} & \left| \int_0^1 \int_0^{2{\rm \pi}} \int_{R-1}^R \left[ \left( \frac{1}{r} \partial_r - \frac{1}{r^2} \right)u^r \right] \varPsi_R^r \, {\rm d} r \, {\rm d} \theta \, {\rm d} z \right| \nonumber\\ &\quad \leq C ( R^{{-}1} \| \partial_r u^r \|_{L^2(\mathcal{D}_R)} + R^{{-}2} \| u^r \|_{L^2(\mathcal{D}_R)} ) \| \varPsi_R^r \|_{L^2(\mathcal{D}_R)} \nonumber\\ &\quad \leq C ( R^{{-}3/2} + R^{{-}5/2} ) \|\boldsymbol{\nabla} {\boldsymbol{u}}\|_{L^2({O}_R)} \cdot \| \partial_z \varPsi_R^r \|_{L^2(\mathcal{D}_R)} \nonumber\\ &\quad \leq C ( R^{{-}3/2} + R^{{-}5/2} ) \|\boldsymbol{\nabla} {\boldsymbol{u}}\|_{L^2({O}_R)} \cdot R^{1/2} \|u^r \|_{L^2({O}_R)} \nonumber\\ &\quad \leq C R^{{-}1} \| \boldsymbol{\nabla} {\boldsymbol{u}}\|_{L^2({O}_R ) }^2. \end{align}

By Hölder inequality it can be shown that

(4.18)\begin{align} & \left| \int_0^1 \int_0^{2{\rm \pi}} \int_{R-1}^R (u^r \partial_r + u^z \partial_z ) u^r \varPsi_R^r \, {\rm d} r \, {\rm d} \theta \, {\rm d} z \right| \nonumber\\ &\quad \leq C \| (u^r, u^z) \|_{L^3(\mathcal{D}_R) } \| (\partial_r, \partial_z )u^r \|_{L^2(\mathcal{D}_R)} \| \varPsi_R^r \|_{L^6(\mathcal{D}_R)}. \end{align}

It follows from the Gagliardo–Nirenberg inequality, the Poincaré inequality and (4.11) and (4.13) that

(4.19)\begin{align} & \| (u^r, u^\theta, u^z ) \|_{L^3(\mathcal{D}_R )} \nonumber\\ &\quad \leq C \| (u^r, u^\theta, u^z ) \|_{L^2(\mathcal{D}_R)}^{1/2} \| (u^r, u^\theta, u^z ) \|_{H^1(\mathcal{D}_R)}^{1/2} \nonumber\\ &\quad \leq C \| \partial_z (u^r, u^\theta, u^z ) \|_{L^2(\mathcal{D}_R)}^{1/2} ( \| (u^r, u^\theta, u^z) \|_{L^2(\mathcal{D}_R)} + \| \bar{\boldsymbol{\nabla}} (u^r, u^\theta, u^z ) \|_{L^2(\mathcal{D}_R)} )^{1/2}\nonumber\\ &\quad \leq C R^{{-}1/4} \| \boldsymbol{\nabla} {\boldsymbol{u}} \|_{L^2({O}_R)}^{1/2} ( R^{{-}1/2} \| {\boldsymbol{u}} \|_{L^2({O}_R)} + R^{1/2} \|\boldsymbol{\nabla} {\boldsymbol{u}}\|_{L^2({O}_R)} )^{1/2} \nonumber\\ &\quad \leq C R^{{-}1/4} \| \boldsymbol{\nabla} {\boldsymbol{u}} \|_{L^2({O}_R)}^{1/2} \cdot R^{\frac14} \| \boldsymbol{\nabla} {\boldsymbol{u}} \|_{L^2({O}_R)}^{1/2} \nonumber\\ & \quad \leq C \|\boldsymbol{\nabla} {\boldsymbol{u}}\|_{L^2({O}_R)} \end{align}

and

(4.20)\begin{equation} \| {\boldsymbol{\varPsi}}_R\|_{L^6(\mathcal{D}_R)} \leq C \|\bar{\boldsymbol{\nabla}} {\boldsymbol{\varPsi}}_R \|_{L^2(\mathcal{D}_R ) } \leq C R^{1/2}\|u^r \|_{L^2({O}_R )} \leq C R^{1/2} \|\boldsymbol{\nabla} {\boldsymbol{u}} \|_{L^2({O}_R )}. \end{equation}

Putting (4.19) and (4.20) into (4.18) one derives

(4.21)\begin{equation} \left| \int_0^1 \int_0^{2{\rm \pi}} \int_{R-1}^R (u^r \partial_r \ + u^z \partial_z ) u^r \varPsi_R^r \, {\rm d} r \, {\rm d} \theta \, {\rm d} z \right| \leq C \| \boldsymbol{\nabla} {\boldsymbol{u}} \|_{L^2({O}_R )}^3. \end{equation}

Similarly, it follows from (4.19) and (4.20) that

(4.22)\begin{align} & \left| \int_0^1 \int_0^{2{\rm \pi}} \int_{R-1}^R \left( \frac{(u^\theta)^2}{r} - \frac{u^\theta}{r} \partial_\theta u^r \right) \varPsi_R^r \, {\rm d} r \, {\rm d} \theta \, {\rm d} z \right| \nonumber\\ & \quad \leq C \|u^\theta \|_{L^3(\mathcal{D}_R)} \left\| r^{{-}1} (\partial_\theta u^r - u^\theta ) \right\|_{L^2(\mathcal{D}_R)} \| \varPsi_R^r \|_{L^6 (\mathcal{D}_R )}\nonumber\\ &\quad \leq C \|\boldsymbol{\nabla} {\boldsymbol{u}} \|_{L^2({O}_R)} \cdot R^{{-}1/2} \|\boldsymbol{\nabla} {\boldsymbol{u}}\|_{L^2({O}_R)} \cdot R^{1/2} \|\boldsymbol{\nabla} {\boldsymbol{u}} \|_{L^2({O}_R)}\nonumber\\ &\quad \leq C \|\boldsymbol{\nabla} {\boldsymbol{u}} \|_{L^2({O}_R)}^3. \end{align}

Combining the estimates (4.15)–(4.22) one obtains

(4.23)\begin{equation} \left| \int_0^1 \int_0^{2{\rm \pi}} \int_{R-1}^R \partial_r P \varPsi_R^r \, {\rm d} r \, {\rm d} \theta \, {\rm d} z \right| \leq C \| \boldsymbol{\nabla} {\boldsymbol{u}} \|_{L^2({O}_R )}^2 + C \| \boldsymbol{\nabla} {\boldsymbol{u}}\|_{L^2({O}_R)}^3. \end{equation}

By making use of the estimates (4.5), (4.11) and (4.13) it can be shown that

(4.24)\begin{align} & \left| \int_0^1 \int_0^{2{\rm \pi}} \int_{R-1}^R [ r(\partial_r u^\theta \partial_r \varPsi_R^\theta + \partial_z u^\theta \partial_z \varPsi_R^\theta) ] \, {\rm d} r\, {\rm d} \theta \, {\rm d} z \right| \nonumber\\ &\quad \leq C R \| (\partial_r u^\theta, \partial_z u^\theta )\|_{L^2(\mathcal{D}_R)} \| (\partial_r, \partial_z)\varPsi_R^\theta \|_{L^2(\mathcal{D}_R)} \nonumber\\ &\quad \leq C R^{1/2} \| \boldsymbol{\nabla} {\boldsymbol{u}} \|_{L^2({O}_R)}\cdot R^{1/2} \| u^r \|_{L^2({O}_R) } \nonumber\\ &\quad \leq C R \|\boldsymbol{\nabla} {\boldsymbol{u}} \|_{L^2({O}_R)}^2, \end{align}

(4.25)\begin{align} & \left|\int_0^1 \int_0^{2{\rm \pi}} \int_{R-1}^R r^{{-}1}(\partial_\theta u^\theta + 2 u^r ) \partial_\theta \varPsi_R^\theta \, {\rm d} r \, {\rm d} \theta \, {\rm d} z \right| \nonumber\\ &\quad = \left|\int_0^1 \int_0^{2{\rm \pi}} \int_{R-1}^R r^{{-}1}(\partial_\theta u^\theta + u^r + u^r) \partial_\theta \varPsi_R^\theta \, {\rm d} r \, {\rm d} \theta \, {\rm d} z \right| \nonumber\\ & \quad \leq C \left( \|r^{{-}1} (\partial_\theta u^\theta + u^r ) \|_{L^2(\mathcal{D}_R)} + R^{{-}1} \|u^r\|_{L^2(\mathcal{D}_R) }\right)\| \partial_\theta \varPsi_R^\theta \|_{L^2(\mathcal{D}_R ) }\nonumber\\ & \quad \leq C \left( R^{{-}1/2} \| \boldsymbol{\nabla} {\boldsymbol{u}}\|_{L^2({O}_R )} + R^{{-}3/2} \|\boldsymbol{\nabla} {\boldsymbol{u}}\|_{L^2({O}_R )} \right) \cdot R^{1/2} \|u^r\|_{L^2({O}_R)} \nonumber\\ & \quad \leq C \|\boldsymbol{\nabla} {\boldsymbol{u}}\|_{L^2({O}_R )}^2 \end{align}

and

(4.26)\begin{align} & \left| \int_0^1 \int_0^{2{\rm \pi}} \int_{R-1}^R \frac{1}{r} u^\theta \varPsi_R^\theta \, {\rm d} r \, {\rm d} \theta \, {\rm d} z \right| \leq C R^{{-}1} \| u^\theta \|_{L^2(\mathcal{D}_R)} \| \varPsi_R^\theta \|_{L^2(\mathcal{D}_R)}\nonumber\\ &\quad \leq C R^{{-}3/2} \|u^\theta \|_{L^2 ({O}_R)} \cdot R^{1/2} \| u^r \|_{L^2({O}_R) } \leq C R^{{-}1} \|\boldsymbol{\nabla} {\boldsymbol{u}} \|_{L^2({O}_R )}^2 . \end{align}

Furthermore, one has

(4.27)\begin{align} & \left| \int_0^1 \int_0^{2{\rm \pi}} \int_{R-1}^R ( u^r u^{\theta} + \partial_\theta u^\theta u^\theta ) \varPsi_R^{\theta} \, {\rm d} r \, {\rm d} \theta \, {\rm d} z \right| \nonumber\\ & \quad \leq C R \| r^{{-}1} (u^r + \partial_\theta u^\theta) \|_{L^2(\mathcal{D}_R)} \|u^\theta \|_{L^3 (\mathcal{D}_R)} \| \varPsi_R^\theta \|_{L^6(\mathcal{D}_R)} \nonumber\\ & \quad \leq C R^{1/2} \| \boldsymbol{\nabla} {\boldsymbol{u}}\|_{L^2({O}_R)} \cdot \| \boldsymbol{\nabla} {\boldsymbol{u}} \|_{L^2({O}_R)} \cdot R^{1/2} \| u^r \|_{L^2({O}_R)}\nonumber\\ & \quad \leq C R \| \boldsymbol{\nabla} {\boldsymbol{u}}\|_{L^2({O}_R)}^3 \end{align}

and

(4.28)\begin{align} & \left| \int_0^1 \int_0^{2{\rm \pi}} \int_{R-1}^R \left(r u^r \partial_r + r u^z \partial_z \right) u^\theta \varPsi_R^\theta \, {\rm d} r \, {\rm d} \theta \, {\rm d} z \right| \nonumber\\ &\quad \leq C R \| (u^r, u^z) \|_{L^3(\mathcal{D}_R)} \| (\partial_r, \partial_z )u^\theta \|_{L^2(\mathcal{D}_R)} \| \varPsi_R^\theta \|_{L^6(\mathcal{D}_R )} \nonumber\\ &\quad \leq C R \| \boldsymbol{\nabla} {\boldsymbol{u}} \|_{L^2({O}_R)}^3. \end{align}

Combining the estimates (4.24)–(4.28) one arrives at

(4.29)\begin{equation} \left| \int_0^1 \int_0^{2{\rm \pi}} \int_{R-1}^R \partial_\theta P \varPsi_R^\theta \, {\rm d} r \, {\rm d} \theta \, {\rm d} z \right| \leq C R \| \boldsymbol{\nabla} {\boldsymbol{u}} \|_{L^2({O}_R)}^2 + C R \| \boldsymbol{\nabla} {\boldsymbol{u}}\|_{L^2({O}_R)}^3. \end{equation}

Similarly, it can be proved that

(4.30)\begin{equation} \left| \int_0^1 \int_0^{2{\rm \pi}} \int_{R-1}^R \partial_z P \varPsi_R^z \, {\rm d} r \, {\rm d} \theta \, {\rm d} z \right| \leq C\| \boldsymbol{\nabla} {\boldsymbol{u}} \|_{L^2({O}_R )}^2 + C \| \boldsymbol{\nabla} {\boldsymbol{u}}\|_{L^2({O}_R)}^3. \end{equation}

The above computations imply that

(4.31)\begin{equation} {Y}(R) \leq C_1 R {Y}^{\prime}(R) + C_2 R [ {Y}^{\prime}(R) ]^{3/2}. \end{equation}

It follows from case (c) of Lemma 2.2 that if $Y(R)$ is not identically zero, then for any $\alpha \in (0, {1}/{C_1})$, there exists a constant $C>0$ such that

(4.32)\begin{equation} Y(R)\geq CR^\alpha. \end{equation}

Hence, if $\varliminf _{R\to \infty }(Y(R) R^{-\alpha })= 0$ then $Y(R)$ must be identically zero. This implies that $\boldsymbol {\nabla } {\boldsymbol {u}}\equiv 0$ and, thus, ${\boldsymbol {u}}\equiv 0$. Hence, the proof for case (a) of Theorem 1.2 is completed.

Step 3. Proof for case (b) of Theorem 1.2. We estimate the first two terms on the right-hand side of (3.2) and the terms on the right-hand sides of (4.7)–(4.9) in a different way. Using the Poincaré inequality and the Sobolev embedding inequality, we obtain

(4.33)\begin{equation} \left| \int_{\varOmega} \boldsymbol{\nabla} \varphi_R \boldsymbol{\cdot} \boldsymbol{\nabla} {\boldsymbol{u}} \boldsymbol{\cdot} {\boldsymbol{u}} \right| \leq C \|\boldsymbol{\nabla} {\boldsymbol{u}} \|_{L^2({O}_R)} \|{\boldsymbol{u}}\|_{L^2({O}_R ) } \leq C R^{1/2} \|{\boldsymbol{u}}\|_{L^\infty ({O}_R)} \| \boldsymbol{\nabla} {\boldsymbol{u}} \|_{L^2({O}_R)} \end{equation}

and

(4.34)\begin{align} \left| \int_{\varOmega} \frac12 |{\boldsymbol{u}}|^2 {\boldsymbol{u}} \boldsymbol{\cdot} \boldsymbol{\nabla} \varphi_R \right| &\leq C \|u^r\|_{L^\infty ({O}_R )} \|{\boldsymbol{u}}\|_{L^2 ({O}_R )}^2 \nonumber\\ & \leq CR^{1/2}\|u^r\|_{L^\infty ({O}_R )} \|{\boldsymbol{u}}\|_{L^\infty ({O}_R )} \|\boldsymbol{\nabla} {\boldsymbol{u}}\|_{L^2 ({O}_R )}. \end{align}

By virtue of (4.5), one has

(4.35)\begin{align} & \left| \int_0^1 \int_0^{2{\rm \pi}} \int_{R-1}^R (\partial_r u^r \partial_r \varPsi_R^r + \partial_z u^r \partial_z \varPsi_R^r ) \, {\rm d} r\, {\rm d} \theta \, {\rm d} z \right| \nonumber\\ &\quad \leq C \| ( \partial_r, \partial_z ) u^r \|_{L^2(\mathcal{D}_R)} \| ( \partial_r, \partial_z ) \varPsi_R^r\|_{L^2(\mathcal{D}_R)}\nonumber\\ &\quad \leq C \| \boldsymbol{\nabla} {\boldsymbol{u}} \|_{L^2({O}_R)} \| u^r \|_{L^2({O}_R)} \leq C R^{1/2} \|u^r\|_{L^\infty ({O}_R)}\| \boldsymbol{\nabla} {\boldsymbol{u}} \|_{L^2({O}_R)}, \end{align}

(4.36)\begin{align} & \left| \int_0^1 \int_0^{2{\rm \pi}} \int_{R-1}^R r^{{-}2} ( \partial_\theta u^r - 2 u^\theta ) \partial_\theta \varPsi_R^r \, {\rm d} r \, {\rm d} \theta \, {\rm d} z \right| \nonumber\\ & \quad \leq C R^{{-}1} \|\boldsymbol{\nabla} {\boldsymbol{u}} \|_{L^2({O}_R ) } \|u^r \|_{L^2({O}_R) } \leq C R^{{-}1/2} \|u^r\|_{L^\infty ({O}_R)} \| \boldsymbol{\nabla} {\boldsymbol{u}}\|_{L^2({O}_R )} \end{align}

and

(4.37)\begin{align} & \left| \int_0^1 \int_0^{2{\rm \pi}} \int_{R-1}^R \left[ \left( \frac{1}{r} \partial_r - \frac{1}{r^2} \right)u^r \right] \varPsi_R^r \, {\rm d} r \, {\rm d} \theta \, {\rm d} z \right| \nonumber\\ &\quad \leq C ( R^{{-}1} \| \partial_r u^r \|_{L^2(\mathcal{D}_R)} + R^{{-}2} \| u^r \|_{L^2(\mathcal{D}_R)} ) \| \varPsi_R^r \|_{L^2(\mathcal{D}_R)} \nonumber\\ &\quad \leq C ( R^{{-}3/2} + R^{{-}5/2} ) \|\boldsymbol{\nabla} {\boldsymbol{u}}\|_{L^2({O}_R)} \cdot \| \partial_z \varPsi_R^r \|_{L^2(\mathcal{D}_R)} \nonumber\\ &\quad \leq C R^{{-}1} \| \boldsymbol{\nabla} {\boldsymbol{u}}\|_{L^2({O}_R ) } \|u^r\|_{L^2({O}_R )} \leq C R^{{-}1/2} \|u^r\|_{L^\infty ({O}_R)} \|\boldsymbol{\nabla} {\boldsymbol{u}} \|_{L^2({O}_R)} . \end{align}

Furthermore, it holds that

(4.38)\begin{align} & \left| \int_0^1 \int_0^{2{\rm \pi}} \int_{R-1}^R (u^r \partial_r + u^z \partial_z ) u^r \varPsi_R^r \, {\rm d} r \, {\rm d} \theta \, {\rm d} z \right| \nonumber\\ &\quad \leq C \| (u^r, u^z) \|_{L^\infty (\mathcal{D}_R) } \| (\partial_r, \partial_z )u^r \|_{L^2(\mathcal{D}_R)} \| \varPsi_R^r \|_{L^2 (\mathcal{D}_R)}\nonumber\\ &\quad \leq C \|{\boldsymbol{u}}\|_{L^\infty ({O}_R)} \| \boldsymbol{\nabla} {\boldsymbol{u}} \|_{L^2({O}_R)} \|u^r \|_{L^2({O}_R)}\nonumber\\ & \quad \leq C R^{1/2} \|{\boldsymbol{u}}\|_{L^\infty ({O}_R)} \|u^r\|_{L^\infty ({O}_R)} \| \boldsymbol{\nabla} {\boldsymbol{u}} \|_{L^2({O}_R)} \end{align}

and

(4.39)\begin{align} & \left| \int_0^1 \int_0^{2{\rm \pi}} \int_{R-1}^R r^{{-}1} ((u^\theta)^2 - u^\theta \partial_\theta u^r ) \varPsi_R^r \, {\rm d} r \, {\rm d} \theta \, {\rm d} z \right| \nonumber\\ & \quad \leq C R^{{-}1/2} \|\boldsymbol{\nabla} {\boldsymbol{u}} \|_{L^2 ({O}_R)} \|u^\theta \|_{L^\infty (\mathcal{D}_R) } \| \varPsi_R^r \|_{L^2(\mathcal{D}_R )}\nonumber\\ & \quad \leq C \|{\boldsymbol{u}}\|_{L^\infty ({O}_R)} \|\boldsymbol{\nabla} {\boldsymbol{u}} \|_{L^2({O}_R )} \| u^r \|_{L^2({O}_R )}\nonumber\\ &\quad \leq C R^{1/2} \|{\boldsymbol{u}}\|_{L^\infty ({O}_R)} \| u^r \|_{L^\infty({O}_R )} \| \boldsymbol{\nabla} {\boldsymbol{u}}\|_{L^2({O}_R ) }. \end{align}

Combining the estimates (4.35)–(4.39) one arrives at

(4.40)\begin{equation} \left| \int_0^1 \int_0^{2{\rm \pi}} \int_{R-1}^R \partial_r P \varPsi_R^r \, {\rm d} r \, {\rm d} \theta \, {\rm d} z \right| \leq C R^{1/2}(1+\|{\boldsymbol{u}}\|_{L^\infty ({O}_R)})\|u^r\|_{L^\infty ({O}_R)} \|\boldsymbol{\nabla} {\boldsymbol{u}} \|_{L^2({O}_R )} . \end{equation}

Furthermore, straightforward computations lead to

(4.41)\begin{align} & \left| \int_0^1 \int_0^{2{\rm \pi}} \int_{R-1}^R r(\partial_r u^\theta \partial_r \varPsi_R^\theta + \partial_z u^\theta \partial_z \varPsi_R^\theta) \, {\rm d} r\, {\rm d} \theta \, {\rm d} z \right| \nonumber\\ &\quad \leq C R \| (\partial_r u^\theta, \partial_z u^\theta) \|_{L^2(\mathcal{D}_R)} \| (\partial_r, \partial_z) \varPsi_R^\theta \|_{L^2(\mathcal{D}_R)} \nonumber\\ &\quad \leq C R \| \boldsymbol{\nabla} {\boldsymbol{u}} \|_{L^2({O}_R)} \| u^r \|_{L^2({O}_R) } \leq C R^{3/2} \|u^r\|_{L^\infty({O}_R)}\| \boldsymbol{\nabla} {\boldsymbol{u}} \|_{L^2({O}_R)}, \end{align}

(4.42)\begin{align} & \left|\int_0^1 \int_0^{2{\rm \pi}} \int_{R-1}^R r^{{-}1}(\partial_\theta u^\theta + 2 u^r ) \partial_\theta \varPsi_R^\theta \, {\rm d} r \, {\rm d} \theta \, {\rm d} z \right| \nonumber\\ & \quad \leq C ( \|r^{{-}1} (\partial_\theta u^\theta + u^r ) \|_{L^2(\mathcal{D}_R)} + R^{{-}1} \|u^r\|_{L^2(\mathcal{D}_R) })\| \partial_\theta \varPsi_R^\theta \|_{L^2(\mathcal{D}_R ) }\nonumber\\ & \quad \leq C \|\boldsymbol{\nabla} {\boldsymbol{u}}\|_{L^2({O}_R )} \|u^r \|_{L^2({O}_R)} \leq CR^{1/2} \|u^r\|_{L^\infty({O}_R)}\| \boldsymbol{\nabla} {\boldsymbol{u}} \|_{L^2({O}_R)} \end{align}

and

(4.43)\begin{align} & \left| \int_0^1 \int_0^{2{\rm \pi}} \int_{R-1}^R \frac{1}{r} u^\theta \varPsi_R^\theta \, {\rm d} r \, {\rm d} \theta \, {\rm d} z \right| \leq C R^{{-}1} \| u^\theta \|_{L^2(\mathcal{D}_R)} \| \varPsi_R^\theta \|_{L^2(\mathcal{D}_R)}\nonumber\\ &\quad \leq C R^{{-}1} \|\boldsymbol{\nabla} {\boldsymbol{u}} \|_{L^2({O}_R )} \|u^r \|_{L^2 ({O}_R)} \leq C R^{{-}1/2} \|u^r\|_{L^\infty({O}_R)} \| \boldsymbol{\nabla} {\boldsymbol{u}} \|_{L^2({O}_R)} . \end{align}

By the Hölder inequality and (4.5) one derives

(4.44)\begin{align} &\left| \int_0^1 \int_0^{2{\rm \pi}} \int_{R-1}^R (u^r u^{\theta}+ u^\theta \partial_\theta u^\theta) \varPsi_R^{\theta} \, {\rm d} r \, {\rm d} \theta \, {\rm d} z \right|\nonumber\\ &\quad \leq C R \| r^{{-}1} (u^r + \partial_\theta u^\theta) \|_{L^2(\mathcal{D}_R)} \|u^\theta \|_{L^\infty (\mathcal{D}_R)} \| \varPsi_R^\theta \|_{L^2(\mathcal{D}_R)} \nonumber\\ & \quad \leq C R^{1/2} \| \boldsymbol{\nabla} {\boldsymbol{u}}\|_{L^2({O}_R)} \cdot \| {\boldsymbol{u}} \|_{L^\infty ({O}_R)} \cdot R^{1/2} \| u^r \|_{L^2({O}_R)}\nonumber\\ & \quad \leq C R^{3/2} \|{\boldsymbol{u}}\|_{L^\infty ({O}_R)} \| u^r \|_{L^2({O}_R)} \| \boldsymbol{\nabla} {\boldsymbol{u}}\|_{L^2({O}_R)} \end{align}

and

(4.45)\begin{align} & \left| \int_0^1 \int_0^{2{\rm \pi}} \int_{R-1}^R (r u^r \partial_r + r u^z \partial_z ) u^\theta \varPsi_R^\theta \, {\rm d} r \, {\rm d} \theta \, {\rm d} z \right| \nonumber\\ &\quad \leq C R \| {\boldsymbol{u}} \|_{L^\infty ({O}_R)} \| (\partial_r, \partial_z )u^\theta \|_{L^2(\mathcal{D}_R)} \| \varPsi_R^\theta \|_{L^2 (\mathcal{D}_R )} \nonumber\\ &\quad \leq C R\| {\boldsymbol{u}} \|_{L^\infty ({O}_R)} \| \boldsymbol{\nabla} {\boldsymbol{u}} \|_{L^2({O}_R)} \| u^r \|_{L^2({O}_R )}\nonumber\\ &\quad \leq C R^{3/2} \| {\boldsymbol{u}} \|_{L^\infty ({O}_R)}\| u^r \|_{L^\infty ({O}_R)} \| \boldsymbol{\nabla} {\boldsymbol{u}} \|_{L^2({O}_R)}. \end{align}

Combining the estimates (4.41)–(4.45) one obtains

(4.46)\begin{equation} \left| \int_0^1 \int_0^{2{\rm \pi}} \int_{R-1}^R \partial_\theta P \varPsi_R^\theta \, {\rm d} r \, {\rm d} \theta \, {\rm d} z \right| \leq C R^{3/2} (1+ \| {\boldsymbol{u}} \|_{L^\infty ({O}_R)}) \| u^r \|_{L^\infty ({O}_R)} \|\boldsymbol{\nabla} {\boldsymbol{u}} \|_{L^2({O}_R)}. \end{equation}

Similarly, it can be proved that

(4.47)\begin{equation} \left| \int_0^1 \int_0^{2{\rm \pi}} \int_{R-1}^R \partial_z P \varPsi_R^z \, {\rm d} r \, {\rm d} \theta \, {\rm d} z \right| \leq C R^{1/2}(1+\|{\boldsymbol{u}}\|_{L^\infty ({O}_R)})\|u^r\|_{L^\infty ({O}_R)} \|\boldsymbol{\nabla} {\boldsymbol{u}} \|_{L^2({O}_R )}. \end{equation}

Collecting the above computations one has

(4.48)\begin{equation} {Y}(R) \leq C ( R^{1/2}\|{\boldsymbol{u}}\|_{L^\infty({O}_R)} + R^{3/2} (1+ \|{\boldsymbol{u}}\|_{L^\infty ({O}_R)}) \|u^r\|_{L^\infty({O}_R)} )( {Y}^{\prime}(R) )^{1/2} , \end{equation}

where ${Y}(R)$ is defined in (3.21).

Suppose that there exist $\beta \in [0, {\alpha }/{2}]$ and $C>0$ such that the conditions in (1.12a,b) hold. Then for any $\varepsilon >0$, there exists a $R_0 (\varepsilon )>2$ such that

(4.49a,b)\begin{equation} \|{\boldsymbol{u}}\|_{L^\infty ({O}_R)} \leq \varepsilon R^{\beta}\quad \text{and}\quad \|u^r \|_{L^\infty ({O}_R)}\leq C R^{-(1+\beta-{\alpha}/{2})} \quad\text{for all }R\geq R_0 (\varepsilon). \end{equation}

Therefore, inequality (4.48) implies that

(4.50)\begin{equation} Y(R) \leq C \varepsilon R^{({1+\alpha})/{2}} (Y'(R))^{1/2} \quad\text{for all }R\geq R_0 (\varepsilon). \end{equation}

We assume ${\boldsymbol {u}}$ is not identically equal to zero. Then $Y(R)>0$ for large $R$ and it follows that

(4.51)\begin{equation} \frac{1}{C \varepsilon} R^{-(1+\alpha)} \leq \frac{Y'}{Y^2}. \end{equation}

Integrating it over an interval $(R_1,R_2)$ for large $R_1$, we obtain

(4.52)\begin{equation} \frac{1}{-\alpha\varepsilon C} (R_2^{-\alpha} -R_1^{-\alpha}) \leq{-} \frac{1}{Y(R_2)} + \frac{1}{Y(R_1)}. \end{equation}

According to case (a) of Theorem 1.2, as $R_2\to \infty$, it holds that $Y(R_2)\to \infty$. Hence,

(4.53)\begin{equation} \frac{1}{\alpha \varepsilon C} R_1 ^{-\alpha} \leq \frac{1}{Y(R_1)}. \end{equation}

Therefore,

(4.54)\begin{equation} Y(R_1) \leq \alpha \varepsilon C R_1^{\alpha}. \end{equation}

As $\varepsilon >0$ is chosen arbitrarily, by case (a) of Theorem 1.2, it follows that ${\boldsymbol {u}}\equiv 0$, which is a contradiction. Therefore, ${\boldsymbol {u}}$ is a constant and, thus, ${\boldsymbol {u}}\equiv 0$. This completes the proof of case (b) of Theorem 1.2.

Proof Proof for Theorem 1.3

Assume that ${\boldsymbol {u}}$ is a smooth solution to the Navier–Stokes system (1.1). Taking the $x_i$ derivative ($i=1, 2$) of the first equation in (1.1) one obtains

(4.55) \begin{equation} {-}\Delta \partial_{x_i} {\boldsymbol{u}} + (\partial_{x_i} {\boldsymbol{u}} \boldsymbol{\cdot} \boldsymbol{\nabla}) {\boldsymbol{u}} + ({\boldsymbol{u}} \boldsymbol{\cdot} \boldsymbol{\nabla} ) \partial_{x_i} {\boldsymbol{u}} + \boldsymbol{\nabla} \partial_{x_i} P = 0. \end{equation}

Multiplying the equation in (4.55) by $\varphi _R(r) \partial _{x_i} {\boldsymbol {u}}$ and integrating over $\varOmega$ one derives

(4.56) \begin{align} &\int_\varOmega |\boldsymbol{\nabla} \partial_{x_i} {\boldsymbol{u}}|^2 \varphi_R + \int_\varOmega (\boldsymbol{\nabla} \varphi_R \boldsymbol{\cdot} \boldsymbol{\nabla} \partial_{x_i} {\boldsymbol{u}}) \boldsymbol{\cdot} \partial_{x_i} {\boldsymbol{u}} +\int_\varOmega (\partial_{x_i} {\boldsymbol{u}} \boldsymbol{\cdot} \boldsymbol{\nabla}){\boldsymbol{u}} \boldsymbol{\cdot} \partial_{x_i} {\boldsymbol{u}} \varphi_R \nonumber\\ &\quad = \frac12 \int_\varOmega ({\boldsymbol{u}} \boldsymbol{\cdot} \boldsymbol{\nabla} \varphi_R) |\partial_{x_i} {\boldsymbol{u}} |^2 + \int_\varOmega \partial_{x_i} P \partial_{x_i} {\boldsymbol{u}} \boldsymbol{\cdot} \boldsymbol{\nabla} \varphi_R. \end{align}

It follows from the Poincaré inequality that

(4.57) \begin{equation} \left| \int_\varOmega (\boldsymbol{\nabla} \varphi_R \boldsymbol{\cdot} \boldsymbol{\nabla} \partial_{x_i} {\boldsymbol{u}}) \boldsymbol{\cdot} \partial_{x_i} {\boldsymbol{u}} \right| \leq \| \boldsymbol{\nabla} \partial_{x_i} {\boldsymbol{u}} \|_{L^2({O}_R)} \| \partial_{x_i} {\boldsymbol{u}}\|_{L^2({O}_R)} \leq C R^{1/2} \| \boldsymbol{\nabla} \partial_{x_i} {\boldsymbol{u}} \|_{L^2({O}_R)}. \end{equation}

Integration by parts yields

(4.58)\begin{equation} \int_\varOmega (\partial_{x_i} {\boldsymbol{u}} \boldsymbol{\cdot} \boldsymbol{\nabla}){\boldsymbol{u}} \boldsymbol{\cdot} \partial_{x_i} {\boldsymbol{u}} \varphi_R ={-}\int_\varOmega (\partial_{x_i} {\boldsymbol{u}} \boldsymbol{\cdot} \boldsymbol{\nabla}) \partial_{x_i} {\boldsymbol{u}} \boldsymbol{\cdot} {\boldsymbol{u}} \varphi_R - \int_\varOmega (\partial_{x_i} {\boldsymbol{u}} \boldsymbol{\cdot} \boldsymbol{\nabla} \varphi_R) ({\boldsymbol{u}} \boldsymbol{\cdot} \partial_{x_i} {\boldsymbol{u}}). \end{equation}

In fact, one has

(4.59)\begin{align} \left| \int_\varOmega (\partial_{x_i} {\boldsymbol{u}} \boldsymbol{\cdot} \boldsymbol{\nabla}) \partial_{x_i} {\boldsymbol{u}} \boldsymbol{\cdot} {\boldsymbol{u}} \varphi_R \right| &\leq \| \boldsymbol{\nabla} \partial_{x_i} {\boldsymbol{u}} \sqrt{ \varphi_R} \|_{L^2(\varOmega)} \|\partial_{x_i} {\boldsymbol{u}} \sqrt{\varphi_R } \|_{L^2(\varOmega)} \|{\boldsymbol{u}} \|_{L^\infty(\varOmega)} \nonumber\\ &\leq \|\boldsymbol{\nabla} \partial_{x_i} {\boldsymbol{u}} \sqrt{ \varphi_R} \|_{L^2(\varOmega)} \frac{1}{\rm \pi} \|\partial_{x_3} \partial_{x_i} {\boldsymbol{u}} \sqrt{\varphi_R } \|_{L^2(\varOmega)} \|{\boldsymbol{u}} \|_{L^\infty(\varOmega)} \nonumber\\ & \leq \frac{1}{\rm \pi}\|{\boldsymbol{u}} \|_{L^\infty(\varOmega)}\|\boldsymbol{\nabla} \partial_{x_i} {\boldsymbol{u}} \sqrt{ \varphi_R} \|_{L^2(\varOmega)}^2, \end{align}

where the Poincaré inequality

(4.60)\begin{equation} \|\partial_{x_i} {\boldsymbol{u}} \sqrt{\varphi_R } \|_{L^2(\varOmega)}\leq \frac{1}{\rm \pi} \|\partial_{x_3} (\partial_{x_i} {\boldsymbol{u}} \sqrt{\varphi_R }) \|_{L^2(\varOmega)} = \frac{1}{\rm \pi} \|\partial_{x_3} \partial_{x_i} {\boldsymbol{u}} \sqrt{\varphi_R } \|_{L^2(\varOmega)} \end{equation}

has been used to get the second inequality. Furthermore, it holds that

(4.61)\begin{equation} \left| \int_\varOmega (\partial_{x_i} {\boldsymbol{u}} \boldsymbol{\cdot} \boldsymbol{\nabla} \varphi_R) ({\boldsymbol{u}} \boldsymbol{\cdot} \partial_{x_i} {\boldsymbol{u}} )\right| \leq \| \partial_{x_i} {\boldsymbol{u}} \|_{L^2({O}_R) }^2 \| {\boldsymbol{u}} \|_{L^\infty({O}_R)} \leq C R^{1/2} \|\boldsymbol{\nabla} \partial_{x_i} {\boldsymbol{u}}\|_{L^2({O}_R)}, \end{equation}

where Lemma 2.3 has been used to get the last inequality in (4.61). Similarly, one has

(4.62)\begin{equation} \left| \int_\varOmega {\boldsymbol{u}} \boldsymbol{\cdot} \boldsymbol{\nabla} \varphi_R |\partial_{x_i} {\boldsymbol{u}} |^2 \right| \leq \| {\boldsymbol{u}}\|_{L^\infty({O}_R)} \| \partial_{x_i} {\boldsymbol{u}}\|_{L^2({O}_R )}^2 \leq C R^{1/2} \|\boldsymbol{\nabla} \partial_{x_i} {\boldsymbol{u}}\|_{L^2({O}_R)}. \end{equation}

Using the momentum equation (the first equation in (1.1)) one obtains

(4.63)\begin{align} &\int_\varOmega \partial_{x_i} P \partial_{x_i} {\boldsymbol{u}} \boldsymbol{\cdot} \boldsymbol{\nabla} \varphi_R =\int_\varOmega \Delta u^i \partial_{x_i} {\boldsymbol{u}} \boldsymbol{\cdot} \boldsymbol{\nabla} \varphi_R - \int_\varOmega ({\boldsymbol{u}} \boldsymbol{\cdot} \boldsymbol{\nabla} u^i) \partial_{x_i} {\boldsymbol{u}} \boldsymbol{\cdot} \boldsymbol{\nabla} \varphi_R\nonumber\\ &\quad = \int_\varOmega (\partial_{x_1}^2 + \partial_{x_2}^2) u^i \partial_{x_i} {\boldsymbol{u}} \boldsymbol{\cdot} \boldsymbol{\nabla} \varphi_R - \int_\varOmega \partial_{x_3} u^i \partial_{x_3} \partial_{x_i} {\boldsymbol{u}} \boldsymbol{\cdot} \boldsymbol{\nabla} \varphi_R - \int_\varOmega ({\boldsymbol{u}} \boldsymbol{\cdot} \boldsymbol{\nabla} u^i) \partial_{x_i} {\boldsymbol{u}} \boldsymbol{\cdot} \boldsymbol{\nabla} \varphi_R. \end{align}

Consequently, one has

(4.64)\begin{equation} \left| \int_\varOmega \partial_{x_i} P \partial_{x_i} {\boldsymbol{u}} \boldsymbol{\cdot} \boldsymbol{\nabla} \varphi_R \right| \leq C R^{1/2} ( \| \boldsymbol{\nabla} \partial_{x_1} {\boldsymbol{u}} \|_{L^2({O}_R)} + \|\boldsymbol{\nabla} \partial_{x_2} {\boldsymbol{u}}\|_{L^2({O}_R)}) . \end{equation}

Since $\|{\boldsymbol {u}}\|_{L^\infty (\varOmega )} <{\rm \pi}$, it holds that

(4.65)\begin{equation} \int_\varOmega |\boldsymbol{\nabla} \partial_{x_i} {\boldsymbol{u}}|^2 \varphi_R \leq C R^{1/2} ( \| \boldsymbol{\nabla} \partial_{x_1} {\boldsymbol{u}} \|_{L^2({O}_R)} + \|\boldsymbol{\nabla} \partial_{x_2} {\boldsymbol{u}}\|_{L^2({O}_R)}). \end{equation}

Define

(4.66)\begin{equation} Z(R) = \int_\varOmega ( |\boldsymbol{\nabla} \partial_{x_1} {\boldsymbol{u}}|^2 + |\boldsymbol{\nabla} \partial_{x_2} {\boldsymbol{u}}|^2 ) \varphi_R(\sqrt{x_1^2+x_2^2})\,{{\rm d}\kern0.07em x}. \end{equation}

Hence, the estimate (4.65) implies that

(4.67)\begin{equation} Z(R) \leq CR^{1/2} Z^{\prime}(R)^{1/2}. \end{equation}

Assume that $\boldsymbol {\nabla } \partial _{x_1}{\boldsymbol {u}}$ and $\boldsymbol {\nabla } \partial _{x_2}{\boldsymbol {u}}$ are not identically equal to zero. Then $Z(R)>0$ for $R\geq R_0$ with $R_0>0$, and one has

(4.68)\begin{equation} \frac{1}{CR}\leq \left( -\frac{1}{Z(R)}\right)'. \end{equation}

Integrating it over $(R_0,R)$ for large $R_0$ one arrives at

(4.69)\begin{equation} \frac{1}{C}\ln{\frac{R}{R_0}} \leq{-}\frac{1}{Z(R)}+\frac{1}{Z(R_0)}\leq \frac{1}{Z(R_0)}. \end{equation}

This leads to a contradiction when $R$ is sufficiently large. Therefore, $\boldsymbol {\nabla } \partial _{x_1} {\boldsymbol {u}} = \boldsymbol {\nabla } \partial _{x_2} {\boldsymbol {u}} \equiv 0$. Note that $\partial _{x_1}{\boldsymbol {u}}=\partial _{x_2}{\boldsymbol {u}}=0$ at the boundary $\mathbb {R}^2\times \{0, 1\}$. Thus, $\partial _{x_1} {\boldsymbol {u}} = \partial _{x_2} {\boldsymbol {u}} \equiv 0$. It follows from the divergence free property of ${\boldsymbol {u}}$ that $\partial _{x_3} u^3=0$. This, together with the no-slip boundary conditions, yields

(4.70a–c)\begin{equation} u^{1} = u^1(x_3), \quad u^2 = u^2(x_3) \quad \text{and}\quad u^3 \equiv 0. \end{equation}

Hence, the Navier–Stokes system can be written as

(4.71)\begin{equation} \partial_z^2 u^1 + \partial_{x_1} P = \partial_z^2 u^2 + \partial_{x_2} P =\partial_{x_3} P= 0. \end{equation}

Taking the homogeneous boundary condition of ${\boldsymbol {u}}$ into consideration one derives

(4.72a,b)\begin{equation} u^1 = c_1 x_3(1-x_3) \quad \text{and} \quad u^2 = c_2x_3(1-x_3) \quad \text{for some } c_1, c_2\in \mathbb{R}. \end{equation}

This completes the proof of Theorem 1.3.

Proof Proof for Theorem 1.4

Since ${\boldsymbol {u}}$ is a bounded smooth solution to (1.1) in $\mathbb {R}^2\times \mathbb {T}$, it follows from Lemma 5.1 that the equality (3.2) still holds.

We divide the rest of the proof into three steps.

Step 1. Proof for case (a) of Theorem 1.4. The proof is almost the same as that for Lemma 5.2, except that

(5.26)\begin{equation} \left| \int_0^1 \int_0^{2{\rm \pi}} \int_{R-1}^R \frac{1}{r^2} \partial_\theta u^r \partial_\theta \varPsi_{R, \theta}^r \, {\rm d} r \, {\rm d} \theta \, {\rm d} z \right| \leq C \|\boldsymbol{\nabla} {\boldsymbol{u}} \|_{L^2({O}_R )}^2 + C \|\boldsymbol{\nabla} {\boldsymbol{u}}\|_{L^2({O}_R)}, \end{equation}

which has been obtained in (5.19). The computations in the proof of Lemma 5.2 imply that

(5.27)\begin{equation} {Y}(R) \leq C_1 {Y}^{\prime}(R) + C_2 R^{1/2} [{Y}^{\prime}(R)]^{1/2}, \end{equation}

where $Y(R)$ is defined in (3.21). Hence, one has

(5.28)\begin{equation} [{Y}^{\prime}(R)]^{1/2} \geq \frac{-C_2 R^{1/2} +\sqrt{C_2^2 R + 4C_1 Y(R) } }{2C_1} \geq \frac{Y(R)}{ \sqrt{C_2^2 R + 4C_1 Y(R) }}. \end{equation}

Suppose that $\boldsymbol {\nabla } {\boldsymbol {u}}$ is not identically equal to zero. For $R$ large enough, $Y(R) > 0$,

(5.29)\begin{equation} [ C_2^2 R Y^{{-}2}(R) + 4C_1 Y^{{-}1}(R) ] Y^{\prime}(R) \geq 1. \end{equation}

Let $M$ be a large number satisfying $M^{-1} C_2^2 \leq \frac 14$. According to Lemma 5.2, there exists an $R_0 >2$ such that $Y(R_0) \geq M$, otherwise $\boldsymbol {\nabla } {\boldsymbol {u}} \equiv 0$. For every $R > R_0$, integrating (5.29) over $[R, 2R]$, one gets

(5.30)\begin{equation} 2R \cdot C_2^2 \left[ \frac{1}{Y(R) } - \frac{1}{Y(2R)} \right] + 4C_1 \ln \frac{Y(2R)}{Y(R)} \geq R. \end{equation}

Since $Y(R) \geq M$, it holds that

(5.31)\begin{equation} \frac{Y(2R)}{Y(R)} \geq \exp \left\{ \frac{R}{8 C_1} \right\}. \end{equation}

This implies the exponential growth of $\| \boldsymbol {\nabla } {\boldsymbol {u}} \|_{L^2(\varOmega _R)}$ and leads to a contradiction to the uniform boundedness of $\boldsymbol {\nabla } {\boldsymbol {u}}$, according to Lemma 2.3. Hence, $\boldsymbol {\nabla } {\boldsymbol {u}} \equiv 0$ and ${\boldsymbol {u}} = (0, 0, c)$, since $u^\theta$ is axisymmetric. The proof for case (a) of Theorem 1.4 is completed.

Step 2. Proof for case (b) of Theorem 1.4. Using the divergence free property of ${\boldsymbol {u}}$, one has

(5.32)\begin{equation} \partial_r \int_0^1\int_0^{2{\rm \pi}} r u^r \,{\rm d}\theta \, {\rm d} z ={-} \int_0^1\int_0^{2{\rm \pi}} \partial_\theta u^\theta +\partial_z (r u^z ) \,{\rm d}\theta \, {\rm d} z = 0 \quad \text{for all } 0\leq r < \infty. \end{equation}

Since $u^r$ is independent of $\theta$, it holds that

(5.33a,b)\begin{equation} \int_0^1 r u^r \, {\rm d} z = \frac{1}{2{\rm \pi}}\int_0^1\int_0^{2{\rm \pi}} r u^r \,{\rm d}\theta \, {\rm d} z= 0 \quad \text{and}\quad \int_0^1 u^r \, {\rm d} z =0. \end{equation}

Hence, we have the Poincaré inequality (5.12). Furthermore, by virtue of Lemma 2.1, there exists a vector valued function ${\boldsymbol {\varPsi }}_R(r, z) \in H_0^1(D_R; \mathbb {R}^2)$ satisfying (3.7) and the estimate (3.8). Therefore, one has

(5.34)\begin{align} \int_{\varOmega} P {\boldsymbol{u}} \boldsymbol{\cdot} \boldsymbol{\nabla} \varphi_R &= \int_0^1 \int_{R-1}^R \int_0^{2{\rm \pi}} P u^r r \, {\rm d} \theta \, {\rm d} r \, {\rm d} z = \int_0^1 \int_{R-1}^R\int_0^{2{\rm \pi}} P ( \partial_r \varPsi_R^r + \partial_z \varPsi_R^z ) \,{\rm d}\theta \, {\rm d} r\, {\rm d} z\nonumber\\ &={-} \int_0^1 \int_{R-1}^R\int_0^{2{\rm \pi}} (\partial_r P \varPsi_R^r + \partial_z P \varPsi_R^z ) \, {\rm d} \theta \, {\rm d} r\, {\rm d} z, \end{align}

where the right-hand side can be represented by (4.7) and (4.9)

Now let us start the estimate for the right-hand side of (3.2). The first two terms on the right-hand side of (3.2) can be estimated as the same as that in (5.16) and (5.17). Note that we do not need to estimate the term appearing on the left-hand side of (5.19) since $\partial _\theta u^r\equiv 0$. All the other terms appearing in (4.7) and (4.9) can be estimated in exactly the same way as that in step 2 of the proof for Lemma 5.2. Hence, we arrive at (5.25) with $Y(R)$ defined in (3.21). As in the proof for Theorem 1.3, we can show that $\boldsymbol {\nabla } {\boldsymbol {u}} \equiv 0$. Thus, ${\boldsymbol {u}}$ is a constant vector. Since $u^r$ is independent of $\theta$, one has ${\boldsymbol {u}}=(0, 0, c)$ for some constant $c$. This finishes the proof for case (b) of Theorem 1.4.

Step 3. Proof for case (c) of Theorem 1.4. For steady solutions of the Navier–Stokes system in $\mathbb {R}^2\times \mathbb {T}$, following the same proof as that for (4.3a,b), one obtains

(5.35a–c)\begin{equation} \int_0^1 \int_0^{2{\rm \pi}} r u^r \, {\rm d} \theta \, {\rm d} z = 0, \quad \int_0^1 \int_0^{2{\rm \pi}} u^r \, {\rm d} \theta \, {\rm d} z = 0, \quad \text{and}\quad \int_0^1 \int_0^{2{\rm \pi}} \int_{R-1}^R r u^r \, {\rm d} r \, {\rm d} \theta \, {\rm d} z = 0. \end{equation}

Furthermore, there exists a vector valued function ${\boldsymbol {\varPsi }}_R(r, \theta, z)\in H_0^1 (\mathcal {D}_R; \mathbb {R}^3)$ satisfying (4.4) and (4.5). Thus, it holds that

(5.36)\begin{align} \int_{\varOmega} P {\boldsymbol{u}} \boldsymbol{\cdot} \boldsymbol{\nabla} \varphi_R & = \int_0^1 \int_0^{2{\rm \pi}} \int_{R-1}^R P u^r r \, {\rm d} r \, {\rm d} \theta \, {\rm d} z \nonumber\\ & ={-} \int_0^1 \int_0^{2{\rm \pi}} \int_{R-1}^R (\partial_r P \varPsi_R^r + \partial_\theta P \varPsi_R^\theta + \partial_z P \varPsi_R^z ) \, {\rm d} r \, {\rm d} \theta \, {\rm d} z. \end{align}

Clearly, we have the same (4.7)–(4.9) to characterize the terms on the right-hand side of (5.36).

Now we are in position to estimate the first two terms on the right-hand side of (3.2) and the terms on the right-hand side of (5.36) carefully. First, one has

(5.37)\begin{equation} \left| \int_{\varOmega} (\boldsymbol{\nabla} \varphi_R \boldsymbol{\cdot} \boldsymbol{\nabla} {\boldsymbol{u}}) \boldsymbol{\cdot} {\boldsymbol{u}} \right| \leq C \|\boldsymbol{\nabla} {\boldsymbol{u}} \|_{L^2({O}_R)} \|{\boldsymbol{u}}\|_{L^2({O}_R ) } \leq C R^{1/2} \| \boldsymbol{\nabla} {\boldsymbol{u}} \|_{L^2({O}_R)} \end{equation}

and

(5.38)\begin{equation} \left| \int_{\varOmega} \frac12 |{\boldsymbol{u}}|^2{\boldsymbol{u}} \boldsymbol{\cdot} \boldsymbol{\nabla} \varphi_R \right| \leq C \| {\boldsymbol{u}} \|_{L^\infty ({O}_R ) }^2 \| u^r \|_{L^1 ({O}_R ) } \leq C R \|u^r \|_{L^\infty({O}_R )}. \end{equation}

According to (4.5), one has

(5.39)\begin{align} & \left| \int_0^1 \int_0^{2{\rm \pi}} \int_{R-1}^R (\partial_r u^r \partial_r \varPsi_R^r + \partial_z u^r \partial_z \varPsi_R^r ) \, {\rm d} r\, {\rm d} \theta \, {\rm d} z \right| \nonumber\\ &\quad \leq C R^{{-}1/2} \| \boldsymbol{\nabla} {\boldsymbol{u}} \|_{L^2({O}_R)} \cdot R^{1/2} \| u^r \|_{L^2({O}_R)} \leq C R^{1/2} \|\boldsymbol{\nabla} {\boldsymbol{u}} \|_{L^2({O}_R)} \|u^r \|_{L^\infty({O}_R ) }, \end{align}

(5.40)\begin{align} &\left| \int_0^1 \int_0^{2{\rm \pi}} \int_{R-1}^R r^{{-}2} \left( \partial_\theta u^r - 2 u^\theta \right) \partial_\theta \varPsi_R^r \, {\rm d} r\, {\rm d} \theta \, {\rm d} z \right| \nonumber\\ &\quad \leq C \left( R^{{-}3/2} \| \boldsymbol{\nabla} {\boldsymbol{u}} \|_{L^2({O}_R )} + R^{{-}5/2} \| u^\theta \|_{L^2({O}_R)} \right) \cdot R^{1/2} \|u^r \|_{L^2({O}_R)} \nonumber\\ & \quad \leq C R^{{-}1/2} \| \boldsymbol{\nabla} {\boldsymbol{u}}\|_{L^2({O}_R) } \|u^r \|_{L^\infty({O}_R)} + CR^{{-}1} \|u^r \|_{L^\infty({O}_R)} \end{align}

and

(5.41)\begin{align} & \left| \int_0^1 \int_0^{2{\rm \pi}} \int_{R-1}^R \left[ \left( \frac{1}{r} \partial_r - \frac{1}{r^2} \right)u^r \right] \varPsi_R^r \, {\rm d} r \, {\rm d} \theta \, {\rm d} z \right| \nonumber\\ &\quad \leq C ( R^{{-}3/2} \|\boldsymbol{\nabla} {\boldsymbol{u}}\|_{L^2({O}_R)} + R^{{-}5/2} \|u^r \|_{L^2({O}_R)} ) \cdot R^{1/2} \|u^r \|_{L^2({O}_R)} \nonumber\\ & \quad \leq C R^{{-}1/2} \| \boldsymbol{\nabla} {\boldsymbol{u}}\|_{L^2({O}_R ) } \| u^r \|_{L^\infty({O}_R ) } + CR^{{-}1} \|u^r\|_{L^\infty({O}_R)}. \end{align}

Furthermore, it holds that

(5.42)\begin{align} & \left| \int_0^1 \int_0^{2{\rm \pi}} \int_{R-1}^R (u^r \partial_r + u^z \partial_z ) u^r \varPsi_R^r \, {\rm d} r \, {\rm d} \theta \, {\rm d} z \right| \nonumber\\ &\quad \leq C \| (u^r, u^z) \|_{L^\infty (\mathcal{D}_R) } \| (\partial_r, \partial_z )u^r \|_{L^2(\mathcal{D}_R)} \| \varPsi_R^r \|_{L^2(\mathcal{D}_R)} \nonumber\\ &\quad \leq C R^{1/2} \|\boldsymbol{\nabla} {\boldsymbol{u}} \|_{L^2({O}_R )} \| u^r \|_{L^\infty({O}_R)} \end{align}

and

(5.43)\begin{align} & \left| \int_0^1 \int_0^{2{\rm \pi}} \int_{R-1}^R r^{{-}1} ( u^\theta \partial_\theta u^r - u^\theta u^\theta ) \varPsi_R^r \, {\rm d} r \, {\rm d} \theta \, {\rm d} z \right| \nonumber\\ & \quad \leq C \|u^\theta \|_{L^\infty({O}_R) } \cdot R^{{-}1/2} \|\boldsymbol{\nabla} {\boldsymbol{u}}\|_{L^2({O}_R )} \cdot R^{1/2} \|u^r\|_{L^2({O}_R) }\nonumber\\ & \quad \leq C R^{1/2} \| \boldsymbol{\nabla} {\boldsymbol{u}} \|_{L^2({O}_R )} \| u^r \|_{L^\infty({O}_R)} . \end{align}

Combining the estimates (5.39)–(5.43) one obtains

(5.44)\begin{equation} \left| \int_0^1 \int_0^{2{\rm \pi}} \int_{R-1}^R \partial_r P \varPsi_R^r \, {\rm d} r \, {\rm d} \theta \, {\rm d} z \right| \leq C R^{1/2} \| \boldsymbol{\nabla} {\boldsymbol{u}} \|_{L^2({O}_R )} + C \|u^r\|_{L^\infty({O}_R) }. \end{equation}

It follows from (4.5) and (4.11)–(4.13) that one has

(5.45)\begin{align} & \left| \int_0^1 \int_0^{2{\rm \pi}} \int_{R-1}^R r(\partial_r u^\theta \partial_r \varPsi_R^\theta + \partial_z u^\theta \partial_z \varPsi_R^\theta) \, {\rm d} r\, {\rm d} \theta \, {\rm d} z \right| \nonumber\\ &\quad \leq C R \| (\partial_r u^\theta, \partial_z u^\theta ) \|_{L^2(\mathcal{D}_R)} \| (\partial_r, \partial_z) \varPsi_R^\theta \|_{L^2(\mathcal{D}_R)} \nonumber\\ &\quad \leq C R^{1/2} \| \boldsymbol{\nabla} {\boldsymbol{u}} \|_{L^2({O}_R)}\cdot R^{1/2} \| u^r \|_{L^2({O}_R) } \nonumber\\ &\quad \leq C R^{3/2} \|\boldsymbol{\nabla} {\boldsymbol{u}} \|_{L^2({O}_R)} \| u^r \|_{L^\infty({O}_R ) }, \end{align}

(5.46)\begin{align} & \left| \int_0^1 \int_0^{2{\rm \pi}} \int_{R-1}^R r^{{-}1} (\partial_\theta u^\theta + 2 u^r) \partial_\theta \varPsi_R^\theta \, {\rm d} r\, {\rm d} \theta \, {\rm d} z \right| \nonumber\\ & \quad \leq C \left( \| r^{{-}1} (\partial_\theta u^\theta + u^r) \|_{L^2(\mathcal{D}_R )} + R^{{-}1} \|u^r\|_{L^2(\mathcal{D}_R )} \right) \|\partial_\theta \varPsi_R^\theta\|_{L^2(\mathcal{D}_R)} \nonumber\\ & \quad \leq C R^{1/2} \| \boldsymbol{\nabla} {\boldsymbol{u}}\|_{L^2({O}_R)} \|u^r \|_{L^\infty({O}_R)} + C \|u^r\|_{L^\infty({O}_R)} \end{align}

and

(5.47)\begin{align} & \left| \int_0^1 \int_0^{2{\rm \pi}} \int_{R-1}^R \frac{1}{r} u^\theta \varPsi_R^\theta \, {\rm d} r \, {\rm d} \theta \, {\rm d} z \right| \leq C R^{{-}1} \| u^\theta \|_{L^2(\mathcal{D}_R)} \| \varPsi_R^\theta \|_{L^2(\mathcal{D}_R)}\nonumber\\ &\quad \leq C R^{{-}1 } \|u^\theta \|_{L^\infty ({O}_R)} \cdot R^{1/2} \| u^r \|_{L^2({O}_R) } \leq C \|u^r\|_{L^\infty({O}_R)}. \end{align}

Furthermore, applying the Poincaré inequality one derives

(5.48)\begin{align} & \left| \int_0^1 \int_0^{2{\rm \pi}} \int_{R-1}^R (u^r u^{\theta} + u^\theta \partial_\theta u^\theta) \varPsi_R^{\theta} \, {\rm d} r \, {\rm d} \theta \, {\rm d} z \right| \nonumber\\ & \quad \leq C R^{1/2} \|r^{{-}1} (u^r + \partial_\theta u^\theta) \|_{L^2({O}_R) } \|u^\theta \|_{L^\infty(\mathcal{D}_R)} \| \varPsi_R^\theta \|_{L^2(\mathcal{D}_R)}\nonumber\\ & \quad \leq C R^{3/2} \| \boldsymbol{\nabla} {\boldsymbol{u}} \|_{L^2({O}_R )} \| u^r \|_{L^\infty({O}_R)} \end{align}

and

(5.49)\begin{align} & \left| \int_0^1 \int_0^{2{\rm \pi}} \int_{R-1}^R (r u^r \partial_r + r u^z \partial_z ) u^\theta \varPsi_R^\theta \, {\rm d} r \, {\rm d} \theta \, {\rm d} z \right| \nonumber\\ &\quad \leq C R \| (u^r, u^z) \|_{L^\infty(\mathcal{D}_R)} \| (\partial_r, \partial_z )u^\theta \|_{L^2(\mathcal{D}_R)} \| \varPsi_R^\theta \|_{L^2(\mathcal{D}_R )} \nonumber\\ &\quad \leq C R^{3/2} \| \boldsymbol{\nabla} {\boldsymbol{u}} \|_{L^2({O}_R)} \|u^r \|_{L^\infty ({O}_R )}. \end{align}

Combining the estimates (5.45)–(5.49) one gets

(5.50)\begin{equation} \left| \int_0^1 \int_0^{2{\rm \pi}} \int_{R-1}^R \partial_\theta P \varPsi_R^\theta \, {\rm d} r \, {\rm d} \theta \, {\rm d} z \right| \leq C R^{3/2} \| \boldsymbol{\nabla} {\boldsymbol{u}} \|_{L^2({O}_R)} \|u^r \|_{L^\infty({O}_R )} + C \|u^r\|_{L^\infty({O}_R) }. \end{equation}

Similarly, it can be proved that

(5.51)\begin{equation} \left| \int_0^1 \int_0^{2{\rm \pi}} \int_{R-1}^R \partial_z P \varPsi_R^z \, {\rm d} r \, {\rm d} \theta \, {\rm d} z \right| \leq CR^{1/2} \| \boldsymbol{\nabla} {\boldsymbol{u}} \|_{L^2({O}_R )}+ C\|u^r\|_{L^\infty({O}_R)}. \end{equation}

Since $ru^r$ is bounded, one obtains that

(5.52)\begin{equation} Y(R) \leq C_1 R \|u^r \|_{L^\infty({O}_R)} + C_2 R^{1/2} [Y^{\prime}(R)]^{1/2}, \end{equation}

where $Y(R)$ is defined in (3.21). Suppose that $\boldsymbol {\nabla } {\boldsymbol {u}}$ is not identically equal to zero, there exists an $R_0$ large enough such that $Y(R_0)>0$. Since $ru^r$ converges to zero uniformly in $\theta,z$, there exists some $R_1> R_0$ such that $Y(R_0) \geq 2C_1 R \|u^r\|_{L^\infty ({O}_R)}$ for every $R \geq R_1$ . Hence, it holds that

(5.53)\begin{equation} Y(R) \leq 2C_2 R^{1/2} [Y^{\prime}(R)]^{1/2}, \quad R\geq R_1, \end{equation}

which leads to contradiction. Hence, ${\boldsymbol {u}} = (0, 0, c)$.

Step 4. Proof for case (d) of Theorem 1.4. Assume that ${\boldsymbol {u}}$ is a bounded, smooth solution to the Navier–Stokes system (1.1) in $\mathbb {R}^2\times \mathbb {T}$. Taking the $x_3$ derivative of the momentum equation,

(5.54)\begin{equation} {-}\Delta \partial_{x_3} {\boldsymbol{u}} + (\partial_{x_3} {\boldsymbol{u}} \boldsymbol{\cdot} \boldsymbol{\nabla}){\boldsymbol{u}} + ({\boldsymbol{u}} \boldsymbol{\cdot} \boldsymbol{\nabla}) \partial_{x_3} {\boldsymbol{u}} + \boldsymbol{\nabla} \partial_{x_3} P = 0. \end{equation}

Multiplying (5.54) by $\varphi _R \partial _{x_3} {\boldsymbol {u}}$ and integrating over $\varOmega$, one has

(5.55) \begin{align} & \int_\varOmega |\boldsymbol{\nabla} \partial_{x_3} {\boldsymbol{u}}|^2 \varphi_R + \int_\varOmega \boldsymbol{\nabla} \varphi_R \boldsymbol{\cdot} \boldsymbol{\nabla} \partial_{x_3} {\boldsymbol{u}} \boldsymbol{\cdot} \partial_{x_3} {\boldsymbol{u}} + \int_\varOmega (\partial_{x_3} {\boldsymbol{u}} \boldsymbol{\cdot} \boldsymbol{\nabla}) {\boldsymbol{u}} \boldsymbol{\cdot} \partial_{x_3} {\boldsymbol{u}} \varphi_R \nonumber\\ &\quad =\frac12 \int_{\varOmega} ({\boldsymbol{u}} \boldsymbol{\cdot} \boldsymbol{\nabla} \varphi_R) |\partial_{x_3} {\boldsymbol{u}}|^2 + \int_{\varOmega} \partial_{x_3} P \partial_{x_3} {\boldsymbol{u}} \boldsymbol{\cdot} \boldsymbol{\nabla} \varphi_R. \end{align}

Since $\int _0^1 \partial _{x_3} {\boldsymbol {u}} \, \textrm {d}\kern0.7pt x_3 = 0$, by virtue of the Poincaré inequality and Lemma 2.3, one has

(5.56) \begin{equation} \left| \int_\varOmega \boldsymbol{\nabla} \varphi_R \boldsymbol{\cdot} \boldsymbol{\nabla} \partial_{x_3} {\boldsymbol{u}} \boldsymbol{\cdot} \partial_{x_3} {\boldsymbol{u}} \right| \leq \|\boldsymbol{\nabla} \partial_{x_3} {\boldsymbol{u}} \|_{L^2({O}_R)} \| \partial_{x_3} {\boldsymbol{u}}\|_{L^2({O}_R)}\leq C R^{1/2} \|\boldsymbol{\nabla} \partial_{x_3} {\boldsymbol{u}} \|_{L^2({O}_R)} \end{equation}

and

(5.57) \begin{equation} \int_\varOmega (\partial_{x_3} {\boldsymbol{u}} \boldsymbol{\cdot} \boldsymbol{\nabla}) {\boldsymbol{u}} \boldsymbol{\cdot} \partial_{x_3} {\boldsymbol{u}} \varphi_R ={-} \int_{\varOmega} (\partial_{x_3} {\boldsymbol{u}} \boldsymbol{\cdot} \boldsymbol{\nabla} ) \partial_{x_3} {\boldsymbol{u}} \boldsymbol{\cdot} {\boldsymbol{u}} \varphi_R - \int_\varOmega (\partial_{x_3} {\boldsymbol{u}} \boldsymbol{\cdot} \boldsymbol{\nabla} \varphi_R) ( {\boldsymbol{u}} \boldsymbol{\cdot} \partial_{x_3} {\boldsymbol{u}}). \end{equation}

Note that it holds that

(5.58)\begin{align} \left| \int_\varOmega (\partial_{x_3} {\boldsymbol{u}} \boldsymbol{\cdot} \boldsymbol{\nabla}) \partial_{x_3} {\boldsymbol{u}} \boldsymbol{\cdot} {\boldsymbol{u}} \varphi_R \right|& \leq \| \boldsymbol{\nabla} \partial_{x_3} {\boldsymbol{u}} \sqrt{ \varphi_R} \|_{L^2(\varOmega)} \|\partial_{x_3} {\boldsymbol{u}} \sqrt{\varphi_R } \|_{L^2(\varOmega)} \|{\boldsymbol{u}} \|_{L^\infty(\varOmega)} \nonumber\\ & \leq \|\boldsymbol{\nabla} \partial_{x_3} {\boldsymbol{u}} \sqrt{ \varphi_R} \|_{L^2(\varOmega)}\frac{1}{2{\rm \pi}} \|\partial_{x_3}^2 {\boldsymbol{u}} \sqrt{\varphi_R } \|_{L^2(\varOmega)} \|{\boldsymbol{u}} \|_{L^\infty(\varOmega)} \nonumber\\ & \leq \frac{1}{2{\rm \pi}} \|{\boldsymbol{u}} \|_{L^\infty(\varOmega)}\|\boldsymbol{\nabla} \partial_{x_3} {\boldsymbol{u}} \sqrt{ \varphi_R} \|_{L^2(\varOmega)}^2, \end{align}

where the Wirtinger inequality (cf. Hardy, Littlewood & Pólya Reference Hardy, Littlewood and Pólya1952, p. 185)

(5.59)\begin{equation} \|\partial_{x_3} {\boldsymbol{u}} \sqrt{\varphi_R } \|_{L^2(\varOmega)} \leq \frac{1}{2{\rm \pi}} \|\partial_{x_3}^2 {\boldsymbol{u}} \sqrt{\varphi_R } \|_{L^2(\varOmega)} \end{equation}

has been used. Furthermore, the straightforward computations yield

(5.60)\begin{equation} \left| \int_\varOmega (\partial_{x_3} {\boldsymbol{u}} \boldsymbol{\cdot} \boldsymbol{\nabla} \varphi_R) ({\boldsymbol{u}} \cdot \partial_{x_3} {\boldsymbol{u}} )\right| \leq \| \partial_{x_3} {\boldsymbol{u}} \|_{L^2({O}_R) }^2 \| {\boldsymbol{u}} \|_{L^\infty({O}_R)} \leq CR^{1/2}\|\boldsymbol{\nabla} \partial_{x_3} {\boldsymbol{u}}\|_{L^2({O}_R)} \end{equation}

and

(5.61)\begin{equation} \left| \int_\varOmega {\boldsymbol{u}} \boldsymbol{\cdot} \boldsymbol{\nabla} \varphi_R |\partial_{x_3} {\boldsymbol{u}}|^2 \right| \leq \|{\boldsymbol{u}}\|_{L^\infty({O}_R)}\|\partial_{x_3} {\boldsymbol{u}}\|_{L^2(\mathcal{O_R})}^2 \leq CR^{1/2} \|\boldsymbol{\nabla} \partial_{x_3} {\boldsymbol{u}}\|_{L^2({O}_R)}. \end{equation}

Finally, one has

(5.62)\begin{equation} \left| \int_{\varOmega} \partial_{x_3} P \partial_{x_3} {\boldsymbol{u}} \boldsymbol{\cdot} \boldsymbol{\nabla} \varphi_R \right| \leq \|\partial_{x_3} P\|_{L^2({O}_R)} \|\partial_{x_3} {\boldsymbol{u}}\|_{L^2({O}_R)} \leq C R^{1/2} \|\boldsymbol{\nabla} \partial_{x_3} {\boldsymbol{u}}\|_{L^2({O}_R)}, \end{equation}

where the last inequality is due to the fact that $\partial _{x_3} P$ is uniformly bounded.

Define

(5.63)\begin{equation} Z(R) = \int_{\mathbb{R}^2} \int_0^1 |\boldsymbol{\nabla} \partial_{x_3} {\boldsymbol{u}}|^2\varphi_R(\sqrt{x_1^2+x_2^2}) \, {\rm d}\kern0.07em x_3 \, {\rm d}\kern0.07em x_1\, {\rm d}\kern0.07em x_2. \end{equation}

If $\|{\boldsymbol {u}}\|_{L^\infty (\varOmega )} <2{\rm \pi}$, one has

(5.64)\begin{equation} Z(R) \leq C R^{1/2} Z^{\prime}(R)^{1/2}. \end{equation}

The same argument as that for the proof of Theorem 1.3 proves that $\boldsymbol {\nabla } \partial _{x_3} {\boldsymbol {u}} \equiv 0$, and thus, $\partial _{x_3} {\boldsymbol {u}}$ is identically equal to a constant vector, which is indeed zero because $\int _0^1 \partial _{x_3} {\boldsymbol {u}} \, \textrm {d}\kern0.7pt x_3 =0$. Hence, ${\boldsymbol {u}}$ must be independent of $x_3$. Then it is straightforward to see that the pressure $P$ is also independent of $x_3$, and thus, the two-dimensional vector field $(u_1, u_2)$ is a solution to the stationary Navier–Stokes equations in $\mathbb {R}^2$. According to Koch et al. (Reference Koch, Nadirashvili, Seregin and Sverák2009, Theorem 5.1), the bounded two-dimensional solution $(u_1, u_2)$ must be a constant vector. Then by elliptic theory, one can also show that $u_3$, which is bounded, is also a constant.

The proof of Theorem 1.4 is completed.