On the existence and structures of almost axisymmetric solutions to $3$ -D Navier-Stokes equations

Yanlin Liu Y. LiuSchool of Mathematical Sciences, Laboratory of Mathematics and Complex Systems, MOE, Beijing Normal University, 100875 Beijing, China. Email address: liuyanlin@bnu.edu.cn and Li Xu L. XuSchool of Mathematical Sciences, Beihang University, 100191 Beijing, China Email address: xuliice@buaa.edu.cn

Date: August 24, 2026

Abstract.

In this paper, we consider $3$ -D Navier-Stokes equations with almost axisymmetric initial data, which means that by writing $\boldsymbol{u}_{0}=u^{r}_{0}\boldsymbol{e}_{r}+u^{\theta}_{0}\boldsymbol{e}_{\theta}+u^{z}_{0}\boldsymbol{e}_{z}$ in the cylindrical coordinates, then $\partial_{\theta}u^{r}_{0},\,\partial_{\theta}u^{\theta}_{0}$ and $\partial_{\theta}u^{z}_{0}$ are small in some sense (recall axisymmetric means these three quantities vanish). Then with additional smallness assumption on $u^{\theta}_{0}$ , we prove the global existence of a unique strong solution $\boldsymbol{u}$ , and this solution keeps close to some axisymmetric vector field. We also establish some refined estimates for the integral average in $\theta$ variable for $\boldsymbol{u}$ .

Moreover, as $u^{r}_{0},\,u^{\theta}_{0}$ and $u^{z}_{0}$ here depend on $\theta$ , it is natural to expand them into Fourier series in $\theta$ variable. And we shall consider one special form of $\boldsymbol{u}_{0}$ , with some small parameter $\varepsilon$ to measure its swirl part and oscillating part. We study the asymptotic expansion of the corresponding solution, and the influences between different profiles in the asymptotic expansion. In particular, we give some special symmetric structures that will persist for all time. These phenomena reflect some features of the nonlinear terms in Navier-Stokes equations.

Keywords: Navier-Stokes equations, axisymmetric, asymptotic expansions.

1. Introduction

1.1. The general setting

The 3-D incompressible Navier-Stokes equations (N-S) reads:

(1.1)

\left\{\begin{aligned} &\partial_{t}\boldsymbol{u}+\boldsymbol{u}\cdot\nabla\boldsymbol{u}-\Delta\boldsymbol{u}+\nabla P=\boldsymbol{0},\qquad(t,x)\in\mathop{\mathbb{R}\kern 0.0pt}\nolimits^{+}\times\mathop{\mathbb{R}\kern 0.0pt}\nolimits^{3}\\ &\mathop{\rm div}\nolimits\boldsymbol{u}=0,\\ &\boldsymbol{u}|_{t=0}=\boldsymbol{u}_{0},\end{aligned}\right.

where $\boldsymbol{u}$ stands for the velocity field and $P$ the scalar pressure of the fluid. This system describes the motion of viscous incompressible fluid. And it is worth mentioning that except the case for initial data with some special structure, it is still a big open problem whether or not the system (1.1) has a unique global solution with large initial data.

Now let us write $\boldsymbol{u}$ in the cylindrical coordinates as

(1.2)

\boldsymbol{u}(t,x)=u^{r}(t,r,\theta,z)\boldsymbol{e}_{r}+u^{\theta}(t,r,\theta,z)\boldsymbol{e}_{\theta}+u^{z}(t,r,\theta,z)\boldsymbol{e}_{z},

where $(r,\theta,z)$ denotes the cylindrical coordinates in $\mathop{\mathbb{R}\kern 0.0pt}\nolimits^{3}$ so that $x=(r\cos\theta,r\sin\theta,z)$ , and

\boldsymbol{e}_{r}=(\cos\theta,\sin\theta,0),\ \boldsymbol{e}_{\theta}=(-\sin\theta,\cos\theta,0),\ \boldsymbol{e}_{z}=(0,0,1),\ r=\sqrt{x_{1}^{2}+x_{2}^{2}}.

Notice that in the cylindrical coordinates, there holds

(1.3)

\nabla=\boldsymbol{e}_{r}\partial_{r}+\frac{1}{r}\boldsymbol{e}_{\theta}\partial_{\theta}+\boldsymbol{e}_{z}\partial_{z},\quad\Delta=\partial_{r}^{2}+\frac{\partial_{r}}{r}+\frac{\partial_{\theta}^{2}}{r^{2}}+\partial_{z}^{2}.

Then we can reformulate (1.1) in the cylindrical coordinates as

(1.4)

\left\{\begin{aligned} &D_{t}u^{r}-\bigl(\partial_{r}^{2}+\partial_{z}^{2}+\frac{\partial_{r}}{r}+\frac{\partial_{\theta}^{2}}{r^{2}}-\frac{1}{r^{2}}\bigr)u^{r}-\frac{(u^{\theta})^{2}}{r}+\frac{2\partial_{\theta}u^{\theta}}{r^{2}}+\partial_{r}P=0,\\ &D_{t}u^{\theta}-\bigl(\partial_{r}^{2}+\partial_{z}^{2}+\frac{\partial_{r}}{r}+\frac{\partial_{\theta}^{2}}{r^{2}}-\frac{1}{r^{2}}\bigr)u^{\theta}+\frac{u^{r}u^{\theta}}{r}-\frac{2\partial_{\theta}u^{r}}{r^{2}}+\frac{\partial_{\theta}P}{r}=0,\\ &D_{t}u^{z}-\bigl(\partial_{r}^{2}+\partial_{z}^{2}+\frac{\partial_{r}}{r}+\frac{\partial_{\theta}^{2}}{r^{2}}\bigr)u^{z}+\partial_{z}P=0,\\ &\partial_{r}u^{r}+\frac{u^{r}}{r}+\partial_{z}u^{z}+\frac{\partial_{\theta}u^{\theta}}{r}=0,\\ &(u^{r},u^{\theta},u^{z})|_{t=0}=(u_{0}^{r},u^{\theta}_{0},u^{z}_{0}),\end{aligned}\right.

where $D_{t}\mathrel{\mathop{\kern 0.0pt=}\limits^{\hbox{\footnotesize def}}}\partial_{t}+\boldsymbol{u}\cdot\nabla=\partial_{t}+\bigl(u^{r}\partial_{r}+u^{\theta}\frac{\partial_{\theta}}{r}+u^{z}\partial_{z}\bigr)$ denotes the material derivative.

For the special case when all the components in (1.2) do not depend on $\theta$ , precisely

\displaystyle\boldsymbol{u}(t,x)=u^{r}(t,r,z)\boldsymbol{e}_{r}+u^{\theta}(t,r,z)\boldsymbol{e}_{\theta}+u^{z}(t,r,z)\boldsymbol{e}_{z}

then this $\boldsymbol{u}$ will be called axisymmetric. It is a celebrated result that for the more special axisymmetric without swirl case, which means $u^{\theta}=0$ , Ladyzhenskaya [6] and independently Ukhovskii and Yudovich [11] proved the existence of weak solutions along with the uniqueness and regularities of such solutions. And later Nečas et al. [8] gave a simpler proof. Their proofs deeply rely on the fact that $\omega^{\theta}/r$ satisfies

(1.5)

\partial_{t}\frac{\omega^{\theta}}{r}+(u^{r}\partial_{r}+u^{z}\partial_{z})\frac{\omega^{\theta}}{r}-(\Delta+\frac{2}{r}\partial_{r})\frac{\omega^{\theta}}{r}=0,

so that any $L^{p}$ norm of $\omega^{\theta}/r$ is conserved.

However, for the arbitrary axisymmetric initial data whose swirl part is non-trivial, we do not have the structure as that in (1.5) any more. To the best of our knowledge, so far we can only establish the global existence of strong solutions when $u^{\theta}_{0}$ is sufficiently small, and this smallness needs to depend on other components of $\boldsymbol{u}_{0}$ or $\nabla\boldsymbol{u}_{0}$ , as the strategy is to view this case as a small perturbation of the no swirl case. There are numerous works concerning this situation, here we only list [1, 2, 3, 4, 7, 9, 12, 13] for example.

Here we do not limit us in considering axisymmetric solutions, but this geometric symmetry still plays a crucial role. Let us also introduce $\bar{u}(t,x)=\bar{u}^{r}(t,r,z)e_{r}+\bar{u}^{\theta}(t,r,z)e_{\theta}+\bar{u}^{z}(t,r,z)e_{z}$ satisfying the following axisymmetric N-S:

(1.6)

\left\{\begin{aligned} &\partial_{t}\bar{u}^{r}+(\bar{u}^{r}\partial_{r}+\bar{u}^{z}\partial_{z})\bar{u}^{r}-(\partial_{r}^{2}+\partial_{z}^{2}+\frac{1}{r}\partial_{r}-\frac{1}{r^{2}})\bar{u}^{r}-\frac{(\bar{u}^{\theta})^{2}}{r}+\partial_{r}\bar{P}=0,\\ &\partial_{t}\bar{u}^{\theta}+(\bar{u}^{r}\partial_{r}+\bar{u}^{z}\partial_{z})\bar{u}^{\theta}-(\partial_{r}^{2}+\partial_{z}^{2}+\frac{1}{r}\partial_{r}-\frac{1}{r^{2}})\bar{u}^{\theta}+\frac{\bar{u}^{r}\bar{u}^{\theta}}{r}=0,\\ &\partial_{t}\bar{u}^{z}+(\bar{u}^{r}\partial_{r}+\bar{u}^{z}\partial_{z})\bar{u}^{z}-(\partial_{r}^{2}+\partial_{z}^{2}+\frac{1}{r}\partial_{r})\bar{u}^{z}+\partial_{z}\bar{P}=0,\\ &\partial_{r}\bar{u}^{r}+\frac{1}{r}\bar{u}^{r}+\partial_{z}\bar{u}^{z}=0,\\ &\bar{u}^{r}|_{t=0}=\bar{u}^{r}_{0}={\mathcal{M}}(u_{0}^{r}),\ \bar{u}^{\theta}|_{t=0}=\bar{u}^{\theta}_{0}={\mathcal{M}}(u_{0}^{\theta}),\ \bar{u}^{z}|_{t=0}=\bar{u}^{z}_{0}={\mathcal{M}}(u_{0}^{z}),\end{aligned}\right.

where ${\mathcal{M}}(f)$ is the integral average of $f$ in $\theta$ variable, precisely

(1.7)

{\mathcal{M}}(f)(r,z)\mathrel{\mathop{\kern 0.0pt=}\limits^{\hbox{\footnotesize def}}}\frac{1}{2\pi}\int_{0}^{2\pi}f(r,\theta,z)\,d\theta.

By definition, there holds ${\mathcal{M}}(\partial_{\theta}f)=0$ for any smooth function $f$ . Then one has

\mathop{\rm div}\nolimits\bar{\boldsymbol{u}}_{0}=\partial_{r}\bar{u}^{r}_{0}+\frac{1}{r}\bar{u}^{r}_{0}+\partial_{z}\bar{u}^{z}_{0}={\mathcal{M}}(\partial_{r}u^{r}_{0}+\frac{u^{r}_{0}}{r}+\partial_{z}u^{z}_{0}+\frac{\partial_{\theta}u^{\theta}_{0}}{r})={\mathcal{M}}(\mathop{\rm div}\nolimits\boldsymbol{u}_{0})=0.

This guarantees the initial data in (1.6) is indeed compatible with the divergence-free condition.

1.2. Main results.

There are three main results in this paper. The first one concerns the global well-posedness of (1.4) with initial data that are close to some axisymmetric vector fields in some sense. To state this precisely, let us introduce the following norms:

(1.8)

\|f\|_{\dot{H}^{1}_{\rm axi}}^{2}\mathrel{\mathop{\kern 0.0pt=}\limits^{\hbox{\footnotesize def}}}\|\widetilde{\nabla}f\|_{L^{2}}^{2}+\|f/r\|_{L^{2}}^{2},\quad\hbox{and}\quad\|f\|_{H^{1}_{\rm axi}}^{2}\mathrel{\mathop{\kern 0.0pt=}\limits^{\hbox{\footnotesize def}}}\|f\|_{L^{2}}^{2}+\|f\|_{\dot{H}^{1}_{\rm axi}}^{2},

where $\widetilde{\nabla}\mathrel{\mathop{\kern 0.0pt=}\limits^{\hbox{\footnotesize def}}}\boldsymbol{e}_{r}\partial_{r}+\boldsymbol{e}_{z}\partial_{z}$ is a part of the whole gradient given by (1.3).

Theorem 1.1.

Let $\boldsymbol{u}_{0}\in H^{2}$ with $\mathop{\rm div}\nolimits\boldsymbol{u}_{0}=0$ and $\boldsymbol{\mathfrak{U}}_{0}\mathrel{\mathop{\kern 0.0pt=}\limits^{\hbox{\footnotesize def}}}(\partial_{\theta}u^{r}_{0},\partial_{\theta}u^{\theta}_{0},\partial_{\theta}u^{z}_{0})\in H^{1}_{\rm axi}$ . If there exists some small positive constant $\epsilon$ such that the following two smallness conditions hold:

(1.9)

\|u^{\theta}_{0}\|_{L^{2}}\|u^{\theta}_{0}\|_{\dot{H}^{1}_{\rm axi}}\exp\bigl(C\|{\boldsymbol{u}}_{0}\|_{L^{2}}^{6}\|{\boldsymbol{u}}_{0}\|_{\dot{H}^{2}}^{2}\bigr)<\epsilon,

(1.10)

\|\boldsymbol{\mathfrak{U}}_{0}\|_{L^{2}}\|\boldsymbol{\mathfrak{U}}_{0}\|_{\dot{H}^{1}_{\rm axi}}\exp\bigl(\exp(C\|{\boldsymbol{u}}_{0}\|_{L^{2}}^{6}\|{\boldsymbol{u}}_{0}\|_{\dot{H}^{2}}^{2})\bigr)<\epsilon,

then both (1.4) and (1.6) have unique global strong solutions $\boldsymbol{u}$ and $\bar{\boldsymbol{u}}$ in $C(\mathop{\mathbb{R}\kern 0.0pt}\nolimits_{+};H^{1})\cap L^{2}(\mathop{\mathbb{R}\kern 0.0pt}\nolimits_{+};\dot{H}^{1}\cap\dot{H}^{2})$ such that for any $t>0$ , there holds

(1.11)			$\displaystyle\\|\boldsymbol{u}-\bar{\boldsymbol{u}}\\|_{L^{\infty}_{t}(L^{2})}^{2}+\\|\nabla(\boldsymbol{u}-\bar{\boldsymbol{u}})\\|_{L^{2}_{t}(L^{2})}^{2}\lesssim\\|\boldsymbol{\mathfrak{U}}_{0}\\|_{L^{2}}^{2}\exp\bigl(\exp(C\\|{\boldsymbol{u}}_{0}\\|_{L^{2}}^{6}\\|{\boldsymbol{u}}_{0}\\|_{\dot{H}^{2}}^{2})\bigr)\mathrel{\mathop{\kern 0.0pt=}\limits^{\hbox{\footnotesize def}}}{\mathcal{A}},$
(1.11)			$\displaystyle\\|\boldsymbol{u}-\bar{\boldsymbol{u}}\\|_{L^{\infty}_{t}(\dot{H}^{1})}^{2}+\\|\nabla(\boldsymbol{u}-\bar{\boldsymbol{u}})\\|_{L^{2}_{t}(\dot{H}^{1})}^{2}\lesssim\\|\boldsymbol{\mathfrak{U}}_{0}\\|_{\dot{H}^{1}_{\rm axi}}^{2}\exp\bigl(\exp(C\\|{\boldsymbol{u}}_{0}\\|_{L^{2}}^{6}\\|{\boldsymbol{u}}_{0}\\|_{\dot{H}^{2}}^{2})\bigr)\mathrel{\mathop{\kern 0.0pt=}\limits^{\hbox{\footnotesize def}}}{\mathcal{B}}.$

Moreover, for ${\mathcal{M}}{\boldsymbol{u}}\mathrel{\mathop{\kern 0.0pt=}\limits^{\hbox{\footnotesize def}}}{\mathcal{M}}(u^{r})\boldsymbol{e}_{r}+{\mathcal{M}}(u^{\theta})\boldsymbol{e}_{\theta}+{\mathcal{M}}(u^{z})\boldsymbol{e}_{z}$ , there holds

(1.12)			$\displaystyle\\|{\mathcal{M}}({\boldsymbol{u}}-\bar{\boldsymbol{u}})\\|_{L^{\infty}_{t}(L^{2})}^{2}+\\|\nabla{\mathcal{M}}({\boldsymbol{u}}-\bar{\boldsymbol{u}})\\|_{L^{2}_{t}(L^{2})}^{2}\lesssim\epsilon{\mathcal{A}},$
(1.12)			$\displaystyle\\|{\mathcal{M}}({\boldsymbol{u}}-\bar{\boldsymbol{u}})\\|_{L^{\infty}_{t}(\dot{H}^{1})}^{2}+\\|\nabla{\mathcal{M}}({\boldsymbol{u}}-\bar{\boldsymbol{u}})\\|_{L^{2}_{t}(\dot{H}^{1})}^{2}\lesssim\epsilon{\mathcal{B}}.$

Remark 1.1.

(i) We mention that $\boldsymbol{u}_{0}$ is axisymmetric when $\boldsymbol{\mathfrak{U}}_{0}$ vanishes, and it is well-known that in this situation, the corresponding solution $\boldsymbol{u}$ persists axial symmetry for all time.

In view of this, the smallness condition (1.10) actually tells us that $\boldsymbol{u}_{0}$ is close to some axisymmetric vector field. That is what we mean “almost axisymmetric”. Moreover, it is reasonable to expect that this solution $\boldsymbol{u}$ keeps close to some axisymmetric vector field for all time.

To verify this feature quantitatively, we mention that one natural way to axisymmetrize a solution $\boldsymbol{u}$ to N-S, is to axisymmetrize the initial data, and then solve N-S with this axisymmetrized initial data, in this way we get $\bar{\boldsymbol{u}}$ . Then in view of the smallness condition (1.10), interpolating between the two estimates in (1.11) gives

\|\boldsymbol{u}-\bar{\boldsymbol{u}}\|_{L^{\infty}_{t}(\dot{H}^{\frac{1}{2}})}^{2}+\|\nabla(\boldsymbol{u}-\bar{\boldsymbol{u}})\|_{L^{2}_{t}(\dot{H}^{\frac{1}{2}})}^{2}\lesssim\|\boldsymbol{\mathfrak{U}}_{0}\|_{L^{2}}\|\boldsymbol{\mathfrak{U}}_{0}\|_{\dot{H}^{1}_{\rm axi}}\exp\bigl(\exp(C\|{\boldsymbol{u}}_{0}\|_{L^{2}}^{6}\|{\boldsymbol{u}}_{0}\|_{\dot{H}^{2}}^{2})\bigr)<\epsilon,

which means that $\boldsymbol{u}$ is indeed close to the axisymmetric vector field $\bar{\boldsymbol{u}}$ .

(ii) Furthermore, (1.12) shows that after taking average in $\theta$ , $\boldsymbol{u}-\bar{\boldsymbol{u}}$ will become much smaller. Hence there must be some cancellations in this process, precisely the positive part of $\boldsymbol{u}-\bar{\boldsymbol{u}}$ almost balance its negative part. This provides us more details on how $\boldsymbol{u}$ approaches $\bar{\boldsymbol{u}}$ . One can see this more clearly by expanding $\boldsymbol{u}$ into Fourier series in $\theta$ variable, see Remark 1.3 below.

With Theorem 1.1 at hand, now let us turn to study the asymptotic expansion of solutions to N-S with almost axisymmetric initial data. As we know, a regular enough function can be expanded into Fourier series in $\theta$ variable. And by virtue of the assumptions of Theorem 1.1, here we only consider initial data $\boldsymbol{u}(0,x)$ in the following special form ¹¹ 1 In this part, we use $\boldsymbol{a}_{0},\,\boldsymbol{a}_{k}$ and $\boldsymbol{b}_{k}$ to denote the profiles of the initial data, while the subscript $0$ is used to denote the $0$ -th Fourier coefficients, so that there would be no confusion.:

(1.13)

\left\{\begin{aligned} &u^{r}(0,x)=a^{r}_{0}(r,z)+\varepsilon\sum_{k=1}^{\infty}\Bigl(a^{r}_{k}(r,z)\cos k\theta+b^{r}_{k}(r,z)\sin k\theta\Bigr),\\ &u^{\theta}(0,x)=\varepsilon a^{\theta}_{0}(r,z)+\varepsilon\sum_{k=1}^{\infty}\Bigl(a^{\theta}_{k}(r,z)\cos k\theta+b^{\theta}_{k}(r,z)\sin k\theta\Bigr),\\ &u^{z}(0,x)=a^{z}_{0}(r,z)+\varepsilon\sum_{k=1}^{\infty}\Bigl(a^{z}_{k}(r,z)\cos k\theta+b^{z}_{k}(r,z)\sin k\theta\Bigr),\end{aligned}\right.

where $\varepsilon>0$ is some small positive constant to be determined later, and the profiles satisfy

(1.14)

\begin{split}\sum_{j=0}^{2}\Bigl(\sum_{i=0}^{2-j}\bigl\|r^{-i}\widetilde{\nabla}^{j}(a^{r}_{0},a^{\theta}_{0})\bigr\|_{L^{2}}^{2}+\|\widetilde{\nabla}^{j}a_{0}^{z}\|_{L^{2}}^{2}+\sum_{k=1}^{\infty}\sum_{i=0}^{2-j}k^{2\max\{i,1\}}\bigl\|r^{-i}\widetilde{\nabla}^{j}(\boldsymbol{a}_{k},\boldsymbol{b}_{k})\bigr\|_{L^{2}}^{2}\Bigr)<\infty,\end{split}

and

(1.15)

\partial_{r}a^{r}_{0}+\frac{a^{r}_{0}}{r}+\partial_{z}a^{z}_{0}=0,

(1.16)

\partial_{r}a^{r}_{k}+\frac{a^{r}_{k}}{r}+\partial_{z}a^{z}_{k}+\frac{kb^{\theta}_{k}}{r}=0,\quad\partial_{r}b^{r}_{k}+\frac{b^{r}_{k}}{r}+\partial_{z}b^{z}_{k}-\frac{ka^{\theta}_{k}}{r}=0,\quad\forall\ k\in\mathop{\mathbb{N}\kern 0.0pt}\nolimits.

It is not difficult to verify that the constraints (1.15), (1.16) meet the divergence-free condition $\mathop{\rm div}\nolimits\boldsymbol{u}(0)=0$ . And thanks to Parseval’s identity, (1.14) guarantees that $\boldsymbol{u}(0)\in H^{2}$ and $\boldsymbol{\mathfrak{U}}(0)\in H^{1}_{\rm axi}$ . Precisely, we have

\|\boldsymbol{\mathfrak{U}}(0)\|_{L^{2}}^{2}\sim\varepsilon^{2}\sum_{k=1}^{\infty}k^{2}\bigl\|(\boldsymbol{a}_{k},\boldsymbol{b}_{k})\bigr\|_{L^{2}}^{2},\quad\|\boldsymbol{\mathfrak{U}}(0)\|_{\dot{H}^{1}_{\rm axi}}^{2}\sim\varepsilon^{2}\sum_{k=1}^{\infty}\sum_{i+j=1}k^{2}\bigl\|r^{-i}\widetilde{\nabla}^{j}(\boldsymbol{a}_{k},\boldsymbol{b}_{k})\bigr\|_{L^{2}}^{2},

and the other norms can be derived similarly. In particular, this implies that the smallness conditions in Theorem 1.1 can be satisfied provided $\varepsilon$ is sufficiently small, and this smallness needs to rely on the norms of profiles appearing in (1.14).

Then Theorem 1.1 guarantees the existence of a unique global strong solution to N-S with initial data (1.13). As we know, a strong solution to N-S would become analytic for any positive time, thus can be expanded into Fourier series in $\theta$ variable in the following form:

(1.17)

\begin{split}u^{\lozenge}(t,x)=\sum_{j=0}^{\infty}\varepsilon^{j}u^{\lozenge}_{(j),0}(t,r,z)+\sum_{j=0}^{\infty}\sum_{k=1}^{\infty}\varepsilon^{j}\Bigl(u^{\lozenge}_{(j),k}(t,r,z)\cos k\theta+v^{\lozenge}_{(j),k}\sin k\theta\Bigr).\end{split}

where $\lozenge$ can be $r,\,\theta$ or $z$ , and the profiles do not rely on $\varepsilon$ .

Unlike the Euclidean coordinates, $\boldsymbol{e}_{r}$ and $\boldsymbol{e}_{\theta}$ are not constant vectors. As a result, the convergence of $\boldsymbol{u}$ in Sobolev spaces is in general not equivalent to the convergence of each component of $\boldsymbol{u}$ . In view of this, it is optimal to verify the validity of the expansion (1.17) in $L^{\infty}(\mathop{\mathbb{R}\kern 0.0pt}\nolimits^{+};L^{2}\cap L^{\infty})$ sense. And our first result concerning the asymptotic expansion states as follows:

Theorem 1.2.

Let $\boldsymbol{u}(0,x)$ be given by (1.13) satisfying (1.14)-(1.16). Then there exists some $\varepsilon_{0}>0$ such that for any $\varepsilon\in(0,\varepsilon_{0})$ , (1.1) has a unique global solution $\boldsymbol{u}\in C(\mathop{\mathbb{R}\kern 0.0pt}\nolimits_{+};H^{1})\cap L^{2}(\mathop{\mathbb{R}\kern 0.0pt}\nolimits_{+};\dot{H}^{1}\cap\dot{H}^{2})$ . Moreover, this solution can be expanded as

(1.18)

\left\{\begin{aligned} &u^{r}(t,x)=u^{r}_{(0),0}(t,r,z)+\varepsilon\sum_{k=1}^{\infty}\Bigl(u^{r}_{(1),k}(t,r,z)\cos k\theta+v^{r}_{(1),k}(t,r,z)\sin k\theta\Bigr)+{\mathcal{O}}(\varepsilon^{2}),\\ &u^{\theta}(t,x)=\varepsilon u^{\theta}_{(1),0}(t,r,z)+\varepsilon\sum_{k=1}^{\infty}\Bigl(u^{\theta}_{(1),k}(t,r,z)\cos k\theta+v^{\theta}_{(1),k}(t,r,z)\sin k\theta\Bigr)+{\mathcal{O}}(\varepsilon^{2}),\\ &u^{z}(t,x)=u^{z}_{(0),0}(t,r,z)+\varepsilon\sum_{k=1}^{\infty}\Bigl(u^{z}_{(1),k}(t,r,z)\cos k\theta+v^{z}_{(1),k}(t,r,z)\sin k\theta\Bigr)+{\mathcal{O}}(\varepsilon^{2})\end{aligned}\right.

in $L^{\infty}(\mathop{\mathbb{R}\kern 0.0pt}\nolimits^{+};L^{2}\cap L^{\infty})$ sense.

Remark 1.2.

We can see that up to $\varepsilon$ order, this expansion (1.18) has the same form as the initial state (1.13). In particular, the coefficients $u^{r}_{(1),0}$ and $u^{z}_{(1),0}$ vanish for all time.

The following result concerns the odevity in this asymptotic expansion. We mention that this persistence of odevity deeply reflects some nonlinear structures of N-S.

Theorem 1.3.

Under the assumptions of Theorem 1.2, let us consider the special case of (1.13) that $u^{r}(0,x)$ and $u^{z}(0,x)$ are even in $\theta$ , while $u^{\theta}(0,x)$ is odd in $\theta$ , precisely:

(1.19)

\left\{\begin{aligned} &u^{r}(0,x)=a^{r}_{0}(r,z)+\varepsilon\sum_{k=1}^{\infty}a^{r}_{k}(r,z)\cos k\theta,\\ &u^{\theta}(0,x)=\varepsilon\sum_{k=1}^{\infty}b^{\theta}_{k}(r,z)\sin k\theta,\\ &u^{z}(0,x)=a^{z}_{0}(r,z)+\varepsilon\sum_{k=1}^{\infty}a^{z}_{k}(r,z)\cos k\theta.\end{aligned}\right.

Then this odevity will persist for all time. Precisely, the following expansion

(1.20)

\left\{\begin{aligned} u^{r}(t,x)=&\Bigl(u^{r}_{(0),0}+\sum_{j=2}^{\infty}\varepsilon^{j}u^{r}_{(j),0}\Bigr)(t,r,z)+\sum_{j=1}^{\infty}\sum_{k=1}^{\infty}\varepsilon^{j}u^{r}_{(j),k}(t,r,z)\cos k\theta,\\ u^{\theta}(t,x)=&\sum_{j=1}^{\infty}\sum_{k=1}^{\infty}\varepsilon^{j}v^{\theta}_{(j),k}(t,r,z)\sin k\theta,\\ u^{z}(t,x)=&\Bigl(u^{z}_{(0),0}+\sum_{j=2}^{\infty}\varepsilon^{j}u^{z}_{(j),0}\Bigr)(t,r,z)+\sum_{j=1}^{\infty}\sum_{k=1}^{\infty}\varepsilon^{j}u^{z}_{(j),k}(t,r,z)\cos k\theta\end{aligned}\right.

holds in $L^{\infty}(\mathop{\mathbb{R}\kern 0.0pt}\nolimits^{+};L^{2}\cap L^{\infty})$ sense.

Remark 1.3.

One can see from the proof in Section 5 that, $u^{r}_{(0),0}\boldsymbol{e}_{r}+u^{z}_{(0),0}\boldsymbol{e}_{z}$ here actually satisfies axisymmetric N-S with initial data $a_{0}^{r}\boldsymbol{e}_{r}+a_{0}^{z}\boldsymbol{e}_{z}={\mathcal{M}}(\boldsymbol{u}(0,x))$ . Thus $u^{r}_{(0),0}\boldsymbol{e}_{r}+u^{z}_{(0),0}\boldsymbol{e}_{z}$ corresponds to the $\bar{\boldsymbol{u}}$ in Theorem 1.1, and we can obtain from the expression (1.20) that

\boldsymbol{u}-\bar{\boldsymbol{u}}=\sum_{j=2}^{\infty}\varepsilon^{j}\boldsymbol{u}_{(j),0}+\sum_{j=1}^{\infty}\sum_{k=1}^{\infty}\varepsilon^{j}\Bigl(\boldsymbol{u}_{(j),k}\cos k\theta+\boldsymbol{v}_{(j),k}\sin k\theta\Bigr)={\mathcal{O}}(\varepsilon),

and

{\mathcal{M}}(\boldsymbol{u}-\bar{\boldsymbol{u}})=\sum_{j=2}^{\infty}\varepsilon^{j}\boldsymbol{u}_{(j),0}={\mathcal{O}}(\varepsilon^{2}),

which indicates that ${\mathcal{M}}(\boldsymbol{u}-\bar{\boldsymbol{u}})$ is in general much smaller than $\boldsymbol{u}-\bar{\boldsymbol{u}}$ , just as what we have mentioned at the end of Remark 1.1. Here we can see that the reason is the cancellations of the oscillating terms, which is the main terms in $\boldsymbol{u}-\bar{\boldsymbol{u}}$ .

Roughly speaking, here we have ${\mathcal{M}}(\boldsymbol{u}-\bar{\boldsymbol{u}})\sim|\boldsymbol{u}-\bar{\boldsymbol{u}}|^{2}.$ To see this more clearly, we refer the readers to Section 5 to find the derivations of the profiles in (1.20).

Let us end this section with some notations which will be used throughout this paper.

Notations. We shall use $C$ to denote an universal constant which may change from line to line. The notation $a\lesssim b$ means $a\leq Cb$ , and $f\sim g$ means both $a\lesssim b$ and $b\lesssim a$ hold. For a Banach space B, we shall use the shorthand $L^{p}_{T}(B)$ for $\bigl\|\|\cdot\|_{B}\bigr\|_{L^{p}(0,T;dt)}$ . We use $H^{s}$ (resp. $\dot{H}^{s}$ ) to denote inhomogeneous (resp. homogeneous) $L^{2}$ based Sobolev spaces.

2. Preliminary

2.1. Poincaré-type inequality

Lemma 2.1.

For any $p\in[1,\infty]$ and $f\in L^{p}$ , the operator ${\mathcal{M}}$ defined in (1.7) satisfies

(2.1)

\|{\mathcal{M}}(f)\|_{L^{p}}\leq\|f\|_{L^{p}},\quad\hbox{and}\quad\|f-{\mathcal{M}}(f)\|_{L^{p}}\leq 2\pi\|\partial_{\theta}f\|_{L^{p}}.

Proof.

Let us first consider the case when $p\in[1,\infty)$ . By applying Hölder’s inequality, we get

	$\displaystyle\\|{\mathcal{M}}(f)\\|_{L^{p}}^{p}$	$\displaystyle=2\pi\int_{\mathop{\mathbb{R}\kern 0.0pt}\nolimits^{+}\times\mathop{\mathbb{R}\kern 0.0pt}\nolimits}\Bigl\|\frac{1}{2\pi}\int_{0}^{2\pi}f(r,\theta,z)\,d\theta\Bigr\|^{p}\,rdrdz$
		$\displaystyle\leq\int_{\mathop{\mathbb{R}\kern 0.0pt}\nolimits^{+}\times\mathop{\mathbb{R}\kern 0.0pt}\nolimits}\bigl(\int_{0}^{2\pi}\|f(r,\theta,z)\|^{p}\,d\theta\bigr)\,rdrdz=\\|f\\|_{L^{p}}^{p}.$

To prove the second inequality of (2.1), we first write

	$\displaystyle\|f(r,\theta,z)-{\mathcal{M}}(f)(r,z)\|^{p}$	$\displaystyle=\Bigl\|\frac{1}{2\pi}\int_{0}^{2\pi}\bigl(f(r,\theta,z)-f(r,\theta^{\prime},z)\bigr)\,d\theta^{\prime}\Bigr\|^{p}$
		$\displaystyle=\Bigl\|\frac{1}{2\pi}\int_{0}^{2\pi}\bigl(\int_{\theta^{\prime}}^{\theta}(\partial_{\theta}f)(r,\theta^{\prime\prime},z)\,d\theta^{\prime\prime}\bigr)\,d\theta^{\prime}\Bigr\|^{p}$
		$\displaystyle\leq\frac{1}{2\pi}\int_{0}^{2\pi}\Bigl\|\int_{\theta^{\prime}}^{\theta}\bigl\|(\partial_{\theta}f)(r,\theta^{\prime\prime},z)\bigr\|^{p}\,d\theta^{\prime\prime}\Bigr\|\|\theta-\theta^{\prime}\|^{p-1}\,d\theta^{\prime}$
		$\displaystyle\leq\frac{1}{2\pi}\int_{0}^{2\pi}\bigl\|(\partial_{\theta}f)(r,\theta^{\prime\prime},z)\bigr\|^{p}\,d\theta^{\prime\prime}\int_{0}^{2\pi}\|\theta-\theta^{\prime}\|^{p-1}\,d\theta^{\prime}.$

As a result, we can obtain

	$\displaystyle\\|f(r,\theta,z)-{\mathcal{M}}(f)(r,z)\\|_{L^{p}}^{p}$	$\displaystyle=\frac{1}{2\pi}\int_{\mathop{\mathbb{R}\kern 0.0pt}\nolimits^{+}\times\mathop{\mathbb{R}\kern 0.0pt}\nolimits}\Bigl(\int_{0}^{2\pi}\bigl\|(\partial_{\theta}f)(r,\theta^{\prime\prime},z)\bigr\|^{p}\,d\theta^{\prime\prime}\Bigr)\,rdrdz$
		$\displaystyle\qquad\qquad\qquad\qquad\times\int_{0}^{2\pi}\int_{0}^{2\pi}\|\theta-\theta^{\prime}\|^{p-1}\,d\theta^{\prime}d\theta$
		$\displaystyle=\frac{2}{p+1}(2\pi)^{p}\\|\partial_{\theta}f\\|_{L^{p}}^{p}\leq(2\pi)^{p}\\|\partial_{\theta}f\\|_{L^{p}}^{p}.$

Clearly this gives the second inequality of (2.1) for $p\in[1,\infty)$ .

As for the case when $p=\infty$ , we have the following formulas:

|{\mathcal{M}}(f)(x)|=\Bigl|\frac{1}{2\pi}\int_{0}^{2\pi}f(r,\theta,z)\,d\theta\Bigr|\leq\|f\|_{L^{\infty}},

and

|f(x)-{\mathcal{M}}(f)(x)|=\Bigl|\frac{1}{2\pi}\int_{0}^{2\pi}\bigl(\int_{\theta^{\prime}}^{\theta}(\partial_{\theta}f)(r,\theta^{\prime\prime},z)\,d\theta^{\prime\prime}\bigr)\,d\theta^{\prime}\Bigr|\leq 2\pi\|\partial_{\theta}f\|_{L^{\infty}},

which hold for any $x\in\mathop{\mathbb{R}\kern 0.0pt}\nolimits^{3}$ . This completes the proof of this lemma. ∎

2.2. Properties for axisymmetric functions.

Let us first recall the well-known Biot-Savart law, which asserts that any divergence-free velocity field $\boldsymbol{u}$ can be uniquely determined by its vorticity $\mathop{\rm curl}\nolimits\boldsymbol{u}$ . Moreover, for any $s>0$ and any $1<p<\infty$ , there holds

(2.2)

\|\nabla\boldsymbol{u}\|_{\dot{H}^{s}}\sim\|\mathop{\rm curl}\nolimits\boldsymbol{u}\|_{\dot{H}^{s}},\quad\|\nabla\boldsymbol{u}\|_{L^{p}}\sim\|\mathop{\rm curl}\nolimits\boldsymbol{u}\|_{L^{p}}.

For the special case when the velocity field is axisymmetric without swirl, namely

\widetilde{\boldsymbol{u}}(x)=\widetilde{u}^{r}(r,z)\boldsymbol{e}_{r}+\widetilde{u}^{z}(r,z)\boldsymbol{e}_{z},

then we can use $\widetilde{\omega}^{\theta}=\partial_{z}\widetilde{u}^{r}-\partial_{r}\widetilde{u}^{z}$ to represent $\widetilde{\boldsymbol{u}}$ . Moreover, there holds:

Lemma 2.2.

If in addition $\widetilde{\boldsymbol{u}}\in H^{2}$ , then we have the following estimates:

(2.3)

\|\nabla\widetilde{\boldsymbol{u}}\|_{L^{2}}\sim\|\widetilde{\nabla}\widetilde{u}^{r}\|_{L^{2}}+\|\widetilde{\nabla}\widetilde{u}^{z}\|_{L^{2}}+\bigl\|\frac{\widetilde{u}^{r}}{r}\bigr\|_{L^{2}}\sim\|{\widetilde{\omega}}^{\theta}\|_{L^{2}},

(2.4)

\|\nabla^{2}\widetilde{\boldsymbol{u}}\|_{L^{2}}^{2}\sim\|\widetilde{\nabla}\widetilde{\omega}^{\theta}\|_{L^{2}}^{2}+\bigl\|\frac{\widetilde{\omega}^{\theta}}{r}\bigr\|_{L^{2}}^{2}.

Proof.

Notice that $\widetilde{u}^{r},\,\widetilde{u}^{z}$ and $\boldsymbol{e}_{z}$ are independent of $\theta$ , we have the point-wise estimate

\bigl|r^{-1}\widetilde{u}^{r}(x)\bigr|=\bigl|r^{-1}{\partial_{\theta}}\bigl(\widetilde{u}^{r}(r,z)\boldsymbol{e}_{r}+\widetilde{u}^{z}(r,z)\boldsymbol{e}_{z}\bigr)\bigr|\leq|\nabla\widetilde{\boldsymbol{u}}(x)|,\quad\forall\ x\in\mathop{\mathbb{R}\kern 0.0pt}\nolimits^{3}.

which along with (2.2) implies the first desired estimate (2.3).

On the other hand, since $\mathop{\rm div}\nolimits\widetilde{\boldsymbol{u}}=0$ , $\partial_{\theta}\boldsymbol{e}_{r}=\boldsymbol{e}_{\theta}$ and $\partial_{\theta}\boldsymbol{e}_{\theta}=-\boldsymbol{e}_{r}$ , one has

\displaystyle\begin{aligned} &-\Delta\widetilde{\boldsymbol{u}}=\mathop{\rm curl}\nolimits\mathop{\rm curl}\nolimits\widetilde{u}=\mathop{\rm curl}\nolimits(\widetilde{\omega}^{\theta}\boldsymbol{e}_{\theta})=-\partial_{z}\widetilde{\omega}^{\theta}\boldsymbol{e}_{r}+(\partial_{r}\widetilde{\omega}^{\theta}+\frac{\widetilde{\omega}^{\theta}}{r})\boldsymbol{e}_{z},\end{aligned}

which implies

(2.5)		$\displaystyle\\|\nabla^{2}\widetilde{\boldsymbol{u}}\\|_{L^{2}}^{2}$	$\displaystyle\sim\bigl\\|\partial_{r}\widetilde{\omega}^{\theta}+\frac{\widetilde{\omega}^{\theta}}{r}\bigr\\|_{L^{2}}^{2}+\\|\partial_{z}\widetilde{\omega}^{\theta}\\|_{L^{2}}^{2}$
(2.5)			$\displaystyle=\\|\widetilde{\nabla}\widetilde{\omega}^{\theta}\\|_{L^{2}}^{2}+\bigl\\|\frac{\widetilde{\omega}^{\theta}}{r}\bigr\\|_{L^{2}}^{2}+2\int_{\mathop{\mathbb{R}\kern 0.0pt}\nolimits^{3}}\partial_{r}\widetilde{\omega}^{\theta}\cdot\frac{\widetilde{\omega}^{\theta}}{r}\,dx.$

While the condition $\widetilde{\boldsymbol{u}}\in H^{2}$ implies $\widetilde{\omega}^{\theta}\in L^{2}$ and $r^{-1}\widetilde{\omega}^{\theta}\in L^{2}$ , and thus $\lim_{r\rightarrow 0}\widetilde{\omega}^{\theta}=\lim_{r\rightarrow\infty}\widetilde{\omega}^{\theta}=0.$ As a result, there holds

\displaystyle 2\int_{\mathop{\mathbb{R}\kern 0.0pt}\nolimits^{3}}\partial_{r}\widetilde{\omega}^{\theta}\cdot\frac{\widetilde{\omega}^{\theta}}{r}\,dx=2\pi\int_{\mathop{\mathbb{R}\kern 0.0pt}\nolimits^{+}\times\mathop{\mathbb{R}\kern 0.0pt}\nolimits}\partial_{r}|\widetilde{\omega}^{\theta}|^{2}\,drdz=2\pi\int_{\mathop{\mathbb{R}\kern 0.0pt}\nolimits}|\widetilde{\omega}^{\theta}|^{2}\big|_{r=0}^{\infty}\,dz=0,

which along with (2.5) gives the second desired estimate (2.4). It completes the proof of the lemma. ∎

Lemma 2.3.

Let $\bar{\boldsymbol{u}}(x)=\bar{u}^{r}(r,z)\boldsymbol{e}_{r}+\bar{u}^{\theta}(r,z)\boldsymbol{e}_{\theta}+\bar{u}^{z}(r,z)\boldsymbol{e}_{z}\in H^{1}$ be divergence-free, and the corresponding vorticity $\bar{\omega}^{\theta}=\partial_{z}\bar{u}^{r}-\partial_{r}\bar{u}^{z}$ . Then we have

(2.6)		$\displaystyle\\|\nabla\bar{\boldsymbol{u}}\\|_{L^{2}}^{2}$	$\displaystyle\sim\\|\bar{\omega}^{\theta}\\|_{L^{2}}^{2}+\\|\widetilde{\nabla}\bar{u}^{\theta}\\|_{L^{2}}^{2}+\bigl\\|\frac{\bar{u}^{\theta}}{r}\bigr\\|_{L^{2}}^{2}$
(2.6)			$\displaystyle\sim\\|\widetilde{\nabla}\bar{u}^{r}\\|_{L^{2}}^{2}+\\|\widetilde{\nabla}\bar{u}^{z}\\|_{L^{2}}^{2}+\\|\widetilde{\nabla}\bar{u}^{\theta}\\|_{L^{2}}^{2}+\bigl\\|\frac{\bar{u}^{r}}{r}\bigr\\|_{L^{2}}^{2}+\bigl\\|\frac{\bar{u}^{\theta}}{r}\bigr\\|_{L^{2}}^{2},$

Proof.

Notice that $\mathop{\rm curl}\nolimits\bar{\boldsymbol{u}}=(-\partial_{z}\bar{u}^{\theta})\boldsymbol{e}_{r}+(\partial_{z}\bar{u}^{r}-\partial_{r}\bar{u}^{z})\boldsymbol{e}_{\theta}+(\partial_{r}\bar{u}^{\theta}+\frac{\bar{u}^{\theta}}{r})\boldsymbol{e}_{z}$ , we have

(2.7)		$\displaystyle\\|\nabla\bar{\boldsymbol{u}}\\|_{L^{2}}^{2}\thicksim\\|\mathop{\rm curl}\nolimits\bar{\boldsymbol{u}}\\|_{L^{2}}^{2}$	$\displaystyle=\\|\bar{\omega}^{\theta}\\|_{L^{2}}^{2}+\bigl\\|\partial_{r}\bar{u}^{\theta}+\frac{\bar{u}^{\theta}}{r}\bigr\\|_{L^{2}}^{2}+\\|\partial_{z}\bar{u}^{\theta}\\|_{L^{2}}^{2}$
(2.7)			$\displaystyle\sim\\|\widetilde{\nabla}\bar{u}^{r}\\|_{L^{2}}^{2}+\\|\widetilde{\nabla}\bar{u}^{z}\\|_{L^{2}}^{2}+\bigl\\|\frac{\bar{u}^{r}}{r}\bigr\\|_{L^{2}}^{2}+\bigl\\|\partial_{r}\bar{u}^{\theta}+\frac{\bar{u}^{\theta}}{r}\bigr\\|_{L^{2}}^{2}+\\|\partial_{z}\bar{u}^{\theta}\\|_{L^{2}}^{2},$

where we have used Lemma 2.2 in the last step. On the other hand, since $\bar{\boldsymbol{u}}\in H^{1}$ , we have $\bar{u}^{\theta}\in L^{2}$ and $r^{-1}\bar{u}^{\theta}\in L^{2}$ , and thus $\lim_{r\rightarrow 0}\bar{u}^{\theta}=\lim_{r\rightarrow\infty}\bar{u}^{\theta}=0$ , which implies

\displaystyle 2\int_{\mathop{\mathbb{R}\kern 0.0pt}\nolimits^{3}}\partial_{r}\bar{u}^{\theta}\cdot\frac{\bar{u}^{\theta}}{r}\,dx=2\pi\int_{\mathop{\mathbb{R}\kern 0.0pt}\nolimits^{+}\times\mathop{\mathbb{R}\kern 0.0pt}\nolimits}\int_{0}^{\infty}\partial_{r}|\bar{u}^{\theta}|^{2}\,drdz=2\pi\int_{\mathop{\mathbb{R}\kern 0.0pt}\nolimits}|\bar{u}^{\theta}|^{2}\big|_{r=0}^{\infty}\,dz=0.

As a result, there holds

\displaystyle\begin{aligned} &\bigl\|\partial_{r}\bar{u}^{\theta}+\frac{\bar{u}^{\theta}}{r}\bigr\|_{L^{2}}^{2}=\|\partial_{r}\bar{u}^{\theta}\|_{L^{2}}^{2}+\bigl\|\frac{\bar{u}^{\theta}}{r}\bigr\|_{L^{2}}^{2}+2\int_{\mathop{\mathbb{R}\kern 0.0pt}\nolimits^{3}}\partial_{r}\bar{u}^{\theta}\cdot\frac{\bar{u}^{\theta}}{r}\,dx=\|\partial_{r}\bar{u}^{\theta}\|_{L^{2}}^{2}+\bigl\|\frac{\bar{u}^{\theta}}{r}\bigr\|_{L^{2}}^{2}.\end{aligned}

Substituting this into (2.7) gives the desired estimate (2.6). The lemma is proved. ∎

2.3. A stability result for N-S

In this subsection, we shall give a stability result for N-S. There are numerous works concerning this problem, here we only list two classical results [5, 10].

Proposition 2.1.

Let $\boldsymbol{v}$ be a global strong solution to N-S with initial data $\boldsymbol{v}_{0}\in H^{1}$ , and there exists some positive constants $A$ and $B$ such that

(2.8)

\|\boldsymbol{v}\|_{L^{\infty}(\mathop{\mathbb{R}\kern 0.0pt}\nolimits^{+};L^{2})}+\|\nabla\boldsymbol{v}\|_{L^{2}(\mathop{\mathbb{R}\kern 0.0pt}\nolimits^{+};L^{2})}\leq A,\quad\hbox{and}\quad\|\boldsymbol{v}\|_{L^{\infty}(\mathop{\mathbb{R}\kern 0.0pt}\nolimits^{+};\dot{H}^{1})}+\|\nabla\boldsymbol{v}\|_{L^{2}(\mathop{\mathbb{R}\kern 0.0pt}\nolimits^{+};\dot{H}^{1})}\leq B.

Then there exists some small positive constant $c_{0}$ such that whenever $\boldsymbol{u}_{0}\in H^{1}$ satisfies

(2.9)

\|\boldsymbol{u}_{0}-\boldsymbol{v}_{0}\|_{L^{2}}\|\boldsymbol{u}_{0}-\boldsymbol{v}_{0}\|_{\dot{H}^{1}}\exp\bigl(CA^{2}B^{2}\bigr)<c_{0},

then N-S with $\boldsymbol{u}_{0}$ as initial data also has a global strong solution $\boldsymbol{u}$ . Moreover, this solution satisfies for $s=0,1$ that

(2.10)

\|\boldsymbol{u}-\boldsymbol{v}\|_{L^{\infty}(\mathop{\mathbb{R}\kern 0.0pt}\nolimits^{+};\dot{H}^{s})}^{2}+\|\nabla(\boldsymbol{u}-\boldsymbol{v})\|_{L^{2}(\mathop{\mathbb{R}\kern 0.0pt}\nolimits^{+};\dot{H}^{s})}^{2}\leq C\|\boldsymbol{u}_{0}-\boldsymbol{v}_{0}\|_{\dot{H}^{s}}^{2}\exp\bigl(CA^{2}B^{2}\bigr),

and

(2.11)

\|\boldsymbol{u}-\boldsymbol{v}\|_{L^{\infty}(\mathop{\mathbb{R}\kern 0.0pt}\nolimits^{+};L^{\infty})}\leq\|\boldsymbol{u}_{0}-\boldsymbol{v}_{0}\|_{H^{2}}\exp\bigl(CA^{2}B^{2}+CB^{4}\bigr).

Proof.

Step 1. The proof of (2.10). Let us denote $\boldsymbol{h}\mathrel{\mathop{\kern 0.0pt=}\limits^{\hbox{\footnotesize def}}}\boldsymbol{u}-\boldsymbol{v}$ , then we can find

(2.12)

\left\{\begin{aligned} &\partial_{t}\boldsymbol{h}+\boldsymbol{u}\cdot\nabla\boldsymbol{h}+\boldsymbol{h}\cdot\nabla\boldsymbol{v}-\Delta\boldsymbol{h}+\nabla\Pi=0,\quad\mathop{\rm div}\nolimits\boldsymbol{h}=0,\\ &\boldsymbol{h}|_{t=0}=\boldsymbol{u}_{0}-\boldsymbol{v}_{0},\end{aligned}\right.

for some properly chosen $\Pi$ . By taking $L^{2}$ inner product of (2.12) with $\boldsymbol{h}$ , we obtain

	$\displaystyle\frac{1}{2}\frac{d}{dt}\\|\boldsymbol{h}\\|_{L^{2}}^{2}+\\|\nabla\boldsymbol{h}\\|_{L^{2}}^{2}$	$\displaystyle=-\int_{\mathop{\mathbb{R}\kern 0.0pt}\nolimits^{3}}(\boldsymbol{h}\cdot\nabla\boldsymbol{v})\cdot\boldsymbol{h}\,dx$
		$\displaystyle\leq\\|\boldsymbol{h}\\|_{L^{4}}^{2}\\|\nabla\boldsymbol{v}\\|_{L^{2}}$
		$\displaystyle\leq\frac{1}{2}\\|\nabla\boldsymbol{h}\\|_{L^{2}}^{2}+C\\|\boldsymbol{h}\\|_{L^{2}}^{2}\\|\nabla\boldsymbol{v}\\|_{L^{2}}^{4}.$

After subtracting $\frac{1}{2}\|\nabla\boldsymbol{h}\|_{L^{2}}^{2}$ on both sides, then applying Gronwall’s inequality leads to

(2.13)

\begin{split}\|\boldsymbol{h}\|_{L^{\infty}_{t}(L^{2})}^{2}+\|\nabla\boldsymbol{h}\|_{L^{2}_{t}(L^{2})}^{2}&\leq\|\boldsymbol{u}_{0}-\boldsymbol{v}_{0}\|_{L^{2}}^{2}\exp\bigl(C\|\nabla\boldsymbol{v}\|_{L^{2}_{t}(L^{2})}^{2}\|\boldsymbol{v}\|_{L^{\infty}_{t}(\dot{H}^{1})}^{2}\bigr)\\ &\leq\|\boldsymbol{u}_{0}-\boldsymbol{v}_{0}\|_{L^{2}}^{2}\exp\bigl(CA^{2}B^{2}\bigr).\end{split}

While by taking $L^{2}$ inner product of (2.12) with $-\Delta\boldsymbol{h}$ , we have

\displaystyle\begin{aligned} \frac{1}{2}\frac{d}{dt}\|\nabla\boldsymbol{h}\|_{L^{2}}^{2}+\|\Delta\boldsymbol{h}\|_{L^{2}}^{2}&=(\boldsymbol{u}\cdot\nabla\boldsymbol{h}\,|\,\Delta\boldsymbol{h})+(\boldsymbol{h}\cdot\nabla\boldsymbol{v}\,|\,\Delta\boldsymbol{h})\\ &=-\sum_{j=1}^{3}(\partial_{j}(\boldsymbol{v}+\boldsymbol{h})\cdot\nabla\boldsymbol{h}\,|\,\partial_{j}\boldsymbol{h})+(\boldsymbol{h}\cdot\nabla\boldsymbol{v}\,|\,\Delta\boldsymbol{h})\\ &\lesssim\|\nabla\boldsymbol{h}\|_{L^{2}}^{\frac{3}{2}}\|\nabla\boldsymbol{h}\|_{L^{6}}^{\frac{3}{2}}+\|\nabla\boldsymbol{v}\|_{L^{2}}\bigl(\|\nabla\boldsymbol{h}\|_{L^{4}}^{2}+\|\boldsymbol{h}\|_{L^{\infty}}\|\Delta\boldsymbol{h}\|_{L^{2}}\bigr)\\ &\lesssim\|\boldsymbol{h}\|_{L^{2}}^{\frac{1}{2}}\|\nabla\boldsymbol{h}\|_{L^{2}}^{\frac{1}{2}}\|\Delta\boldsymbol{h}\|_{L^{2}}^{2}+\|\nabla\boldsymbol{v}\|_{L^{2}}\|\nabla\boldsymbol{h}\|_{L^{2}}^{\frac{1}{2}}\|\Delta\boldsymbol{h}\|_{L^{2}}^{\frac{3}{2}}.\end{aligned}

Then by using Young’s inequality to the last term, we get

(2.14)

\begin{split}\frac{d}{dt}\|\nabla\boldsymbol{h}\|_{L^{2}}^{2}+\frac{3}{2}\|\Delta\boldsymbol{h}\|_{L^{2}}^{2}\leq C\|\boldsymbol{h}\|_{L^{2}}^{\frac{1}{2}}\|\nabla\boldsymbol{h}\|_{L^{2}}^{\frac{1}{2}}\|\Delta\boldsymbol{h}\|_{L^{2}}^{2}+C\|\nabla\boldsymbol{v}\|_{L^{2}}^{4}\|\nabla\boldsymbol{h}\|_{L^{2}}^{2}.\end{split}

On the other hand, by the local well-posedness result, the following set is not empty:

{\mathcal{D}}\mathrel{\mathop{\kern 0.0pt=}\limits^{\hbox{\footnotesize def}}}\bigl\{\,T\in\mathop{\mathbb{R}\kern 0.0pt}\nolimits^{+}:\mbox{ there holds }\|\nabla\boldsymbol{h}\|_{L^{\infty}_{T}(L^{2})}^{2}+\|\Delta\boldsymbol{h}\|_{L^{2}_{T}(L^{2})}^{2}\leq 2\|\nabla(\boldsymbol{u}_{0}-\boldsymbol{v}_{0})\|_{L^{2}}^{2}\exp\bigl(CA^{2}B^{2}\bigr)\,\bigr\}.

Let us take $T^{*}=\sup\{T:T\in{\mathcal{D}}\}$ . If $T^{*}<\infty$ , then for any $t\leq T^{*}$ , we can use the estimate (2.13) and the smallness condition (2.9) with $c_{0}$ sufficiently small to deduce

\displaystyle C\|\boldsymbol{h}\|_{L^{\infty}_{t}(L^{2})}^{\frac{1}{2}}\|\nabla\boldsymbol{h}\|_{L^{\infty}_{t}(L^{2})}^{\frac{1}{2}}\leq C\|\boldsymbol{u}_{0}-\boldsymbol{v}_{0}\|_{L^{2}}^{\frac{1}{2}}\|\nabla(\boldsymbol{u}_{0}-\boldsymbol{v}_{0})\|_{L^{2}}^{\frac{1}{2}}\exp\bigl(CA^{2}B^{2}\bigr)\leq Cc_{0}^{\frac{1}{2}}<\frac{1}{2}.

By substituting this into (2.14), and then using Gronwall’s inequality, we achieve

	$\displaystyle\\|\nabla\boldsymbol{h}(t)\\|_{L^{2}}^{2}+\\|\Delta\boldsymbol{h}\\|_{L^{2}_{t}(L^{2})}^{2}$	$\displaystyle\leq\\|\nabla(\boldsymbol{u}_{0}-\boldsymbol{v}_{0})\\|_{L^{2}}^{2}\exp\bigl(C\\|\nabla\boldsymbol{v}\\|_{L^{\infty}_{t}(L^{2})}^{2}\\|\nabla\boldsymbol{v}\\|_{L^{2}_{t}(L^{2})}^{2}\bigr)$
		$\displaystyle\leq\\|\nabla(\boldsymbol{u}_{0}-\boldsymbol{v}_{0})\\|_{L^{2}}^{2}\exp\bigl(CA^{2}B^{2}\bigr),$

which contradicts to the definition of $T^{*}$ . Thus there must be $T^{*}=\infty$ , and we have

(2.15)

\|\nabla\boldsymbol{h}\|_{L^{\infty}(\mathop{\mathbb{R}\kern 0.0pt}\nolimits^{+};L^{2})}^{2}+\|\Delta\boldsymbol{h}\|_{L^{2}(\mathop{\mathbb{R}\kern 0.0pt}\nolimits^{+};L^{2})}^{2}\leq 2\|\nabla(\boldsymbol{u}_{0}-\boldsymbol{v}_{0})\|_{L^{2}}^{2}\exp\bigl(CA^{2}B^{2}\bigr).

Step 2. The proof of (2.11). Let us denote $\boldsymbol{\delta}\mathrel{\mathop{\kern 0.0pt=}\limits^{\hbox{\footnotesize def}}}\mathop{\rm curl}\nolimits\boldsymbol{h}$ and $\boldsymbol{\omega}\mathrel{\mathop{\kern 0.0pt=}\limits^{\hbox{\footnotesize def}}}\mathop{\rm curl}\nolimits\boldsymbol{v}$ , which satisfy

(2.16)

\left\{\begin{split}&\partial_{t}\boldsymbol{\delta}+\boldsymbol{u}\cdot\nabla\boldsymbol{\delta}+\boldsymbol{h}\cdot\nabla\boldsymbol{\omega}-\Delta\boldsymbol{\delta}=\boldsymbol{\delta}\cdot\nabla\boldsymbol{u}+\boldsymbol{\omega}\cdot\nabla\boldsymbol{h},\\ &\boldsymbol{\delta}|_{t=0}=\mathop{\rm curl}\nolimits(\boldsymbol{u}_{0}-\boldsymbol{v}_{0}).\end{split}\right.

By taking $L^{2}$ inner product of (2.16) wit $|\boldsymbol{\delta}|^{2}\boldsymbol{\delta}$ , we get

(2.17)

\frac{1}{4}\frac{d}{dt}\|\boldsymbol{\delta}\|_{L^{4}}^{4}+\bigl\||\boldsymbol{\delta}|\nabla\boldsymbol{\delta}\bigr\|_{L^{2}}^{2}+\frac{1}{2}\bigl\|\nabla|\boldsymbol{\delta}|^{2}\bigr\|_{L^{2}}^{2}\leq\Bigl|\int_{\mathop{\mathbb{R}\kern 0.0pt}\nolimits^{3}}(\boldsymbol{\delta}\cdot\nabla\boldsymbol{u}+\boldsymbol{\omega}\cdot\nabla\boldsymbol{h}-\boldsymbol{h}\cdot\nabla\boldsymbol{\omega})\cdot|\boldsymbol{\delta}|^{2}\boldsymbol{\delta}\,dx\Bigr|.

The terms on the right-hand side can be handled as follows:

	$\displaystyle\Bigl\|\int_{\mathop{\mathbb{R}\kern 0.0pt}\nolimits^{3}}(\boldsymbol{\delta}\cdot\nabla\boldsymbol{u})\cdot\|\boldsymbol{\delta}\|^{2}\boldsymbol{\delta}\,dx\Bigr\|$	$\displaystyle\leq\bigl\\|\|\boldsymbol{\delta}\|^{2}\bigr\\|_{L^{3}}\bigl\\|\|\boldsymbol{\delta}\|^{2}\bigr\\|_{L^{6}}\\|\nabla(\boldsymbol{v}+\boldsymbol{h})\\|_{L^{2}}$
		$\displaystyle\leq\frac{1}{8}\bigl\\|\nabla\|\boldsymbol{\delta}\|^{2}\bigr\\|_{L^{2}}^{2}+C\bigl\\|\|\boldsymbol{\delta}\|^{2}\bigr\\|_{L^{2}}^{2}\bigl(\\|\boldsymbol{\omega}\\|_{L^{2}}^{4}+\\|\nabla\boldsymbol{h}\\|_{L^{2}}^{4}\bigr),$

and by using the Biot-Savart law that

	$\displaystyle\Bigl\|\int_{\mathop{\mathbb{R}\kern 0.0pt}\nolimits^{3}}(\boldsymbol{\omega}\cdot\nabla\boldsymbol{h})\cdot\|\boldsymbol{\delta}\|^{2}\boldsymbol{\delta}\,dx\Bigr\|$	$\displaystyle\leq\\|\boldsymbol{\omega}\\|_{L^{2}}\\|\nabla\boldsymbol{h}\\|_{L^{6}}\bigl\\|\|\boldsymbol{\delta}\|^{2}\bigr\\|_{L^{6}}\\|\boldsymbol{\delta}\\|_{L^{6}}$
		$\displaystyle\leq C\\|\boldsymbol{\omega}\\|_{L^{2}}\bigl\\|\|\boldsymbol{\delta}\|^{2}\bigr\\|_{L^{3}}\bigl\\|\|\boldsymbol{\delta}\|^{2}\bigr\\|_{L^{6}}$
		$\displaystyle\leq\frac{1}{8}\bigl\\|\nabla\|\boldsymbol{\delta}\|^{2}\bigr\\|_{L^{2}}^{2}+C\bigl\\|\|\boldsymbol{\delta}\|^{2}\bigr\\|_{L^{2}}^{2}\\|\boldsymbol{\omega}\\|_{L^{2}}^{4},$

and by using integration by parts together with the divergence-free condition that

	$\displaystyle\Bigl\|\int_{\mathop{\mathbb{R}\kern 0.0pt}\nolimits^{3}}($	$\displaystyle\boldsymbol{h}\cdot\nabla\boldsymbol{\omega})\cdot\|\boldsymbol{\delta}\|^{2}\boldsymbol{\delta}\,dx\Bigr\|=\Bigl\|\int_{\mathop{\mathbb{R}\kern 0.0pt}\nolimits^{3}}(\boldsymbol{\omega}\otimes\boldsymbol{h}):\nabla\bigl(\|\boldsymbol{\delta}\|^{2}\boldsymbol{\delta}\bigr)\,dx\Bigr\|$
		$\displaystyle\leq C\bigl(\bigl\\|\nabla\|\boldsymbol{\delta}\|^{2}\bigr\\|_{L^{2}}+\bigl\\|\|\boldsymbol{\delta}\|\nabla\boldsymbol{\delta}\bigr\\|_{L^{2}}\bigr)\\|\boldsymbol{\delta}\\|_{L^{12}}\\|\boldsymbol{h}\\|_{L^{12}}\\|\boldsymbol{\omega}\\|_{L^{3}}$
		$\displaystyle\leq C\bigl(\bigl\\|\nabla\|\boldsymbol{\delta}\|^{2}\bigr\\|_{L^{2}}+\bigl\\|\|\boldsymbol{\delta}\|\nabla\boldsymbol{\delta}\bigr\\|_{L^{2}}\bigr)\bigl\\|\nabla\|\boldsymbol{\delta}\|^{2}\bigr\\|_{L^{2}}^{\frac{1}{2}}\\|\boldsymbol{h}\\|_{L^{4}}^{\frac{1}{2}}\\|\nabla\boldsymbol{h}\\|_{L^{4}}^{\frac{1}{2}}\\|\boldsymbol{\omega}\\|_{L^{2}}^{\frac{1}{2}}\\|\nabla\boldsymbol{\omega}\\|_{L^{2}}^{\frac{1}{2}}$
		$\displaystyle\leq\frac{1}{8}\bigl(\bigl\\|\nabla\|\boldsymbol{\delta}\|^{2}\bigr\\|_{L^{2}}^{2}+\bigl\\|\|\boldsymbol{\delta}\|\nabla\boldsymbol{\delta}\bigr\\|_{L^{2}}^{2}\bigr)+C\bigl(\\|\boldsymbol{h}\\|_{L^{4}}^{4}+\\|\nabla\boldsymbol{h}\\|_{L^{4}}^{4}\bigr)\\|\boldsymbol{\omega}\\|_{L^{2}}^{2}\\|\nabla\boldsymbol{\omega}\\|_{L^{2}}^{2}.$

Now by substituting the above three estimates into (2.18), and using the relation that

\|\nabla\boldsymbol{h}\|_{L^{4}}\lesssim\|\mathop{\rm curl}\nolimits\boldsymbol{h}\|_{L^{4}}=\|\boldsymbol{\delta}\|_{L^{4}},

we get

\displaystyle\frac{d}{dt}\|\boldsymbol{\delta}\|_{L^{4}}^{4}+\bigl\||\boldsymbol{\delta}|\nabla\boldsymbol{\delta}\bigr\|_{L^{2}}^{2}+\bigl\|\nabla|\boldsymbol{\delta}|^{2}\bigr\|_{L^{2}}^{2}\lesssim\|\boldsymbol{\delta}\|_{L^{4}}^{4}\bigl(\|\boldsymbol{\omega}\|_{L^{2}}^{2}\|\boldsymbol{\omega}\|_{H^{1}}^{2}+\|\nabla\boldsymbol{h}\|_{L^{2}}^{4}\bigr)+\|\boldsymbol{h}\|_{L^{4}}^{4}\|\boldsymbol{\omega}\|_{L^{2}}^{2}\|\nabla\boldsymbol{\omega}\|_{L^{2}}^{2}.

Then Gronwall’s inequality together with the bounds (2.8) and (2.10) leads to

(2.18)

\begin{split}\|\boldsymbol{\delta}\|_{L^{\infty}_{t}(L^{4})}^{4}\lesssim&\bigl(\|\nabla(\boldsymbol{u}_{0}-\boldsymbol{v}_{0})\|_{L^{4}}^{4}+\|\boldsymbol{h}\|_{L^{\infty}_{t}(L^{4})}^{4}\|\boldsymbol{\omega}\|_{L^{\infty}_{t}(L^{2})}^{2}\|\nabla\boldsymbol{\omega}\|_{L^{2}_{t}(L^{2})}^{2}\bigr)\\ &\times\exp\bigl(C\|\boldsymbol{\omega}\|_{L^{\infty}_{t}(L^{2})}^{2}\|\boldsymbol{\omega}\|_{L^{2}_{t}(H^{1})}^{2}+C\|\nabla\boldsymbol{h}\|_{L^{\infty}_{t}(L^{2})}^{2}\|\nabla\boldsymbol{h}\|_{L^{2}_{t}(L^{2})}^{2}\bigr)\\ \lesssim&\bigl(\|\boldsymbol{u}_{0}-\boldsymbol{v}_{0}\|_{\dot{H}^{\frac{7}{4}}}^{4}+B^{4}\|\boldsymbol{u}_{0}-\boldsymbol{v}_{0}\|_{L^{2}}\|\boldsymbol{u}_{0}-\boldsymbol{v}_{0}\|_{\dot{H}^{1}}^{3}\bigr)\exp\bigl(CA^{2}B^{2}+CB^{4}+Cc_{0}^{2}\bigr)\\ \lesssim&\|\boldsymbol{u}_{0}-\boldsymbol{v}_{0}\|_{H^{2}}^{4}\exp\bigl(CA^{2}B^{2}+CB^{4}\bigr).\end{split}

Now interpolating between (2.15) and (2.18) gives

	$\displaystyle\\|\boldsymbol{h}\\|_{L^{\infty}_{t}(L^{\infty})}$	$\displaystyle\lesssim\\|\nabla\boldsymbol{h}\\|_{L^{\infty}_{t}(L^{2})}^{\frac{1}{3}}\\|\nabla\boldsymbol{h}\\|_{L^{\infty}_{t}(L^{4})}^{\frac{2}{3}}$
		$\displaystyle\lesssim\\|\boldsymbol{\delta}\\|_{L^{\infty}_{t}(L^{2})}^{\frac{1}{3}}\\|\boldsymbol{\delta}\\|_{L^{\infty}_{t}(L^{4})}^{\frac{2}{3}}\lesssim\\|\boldsymbol{u}_{0}-\boldsymbol{v}_{0}\\|_{H^{2}}\exp\bigl(CA^{2}B^{2}+CB^{4}\bigr),$

which is exactly the desired estimate (2.11). This completes the proof of this proposition. ∎

3. The proof of Theorem 1.1

The purpose of this section is to prove Theorem 1.1 by using perturbation method.

3.1. Global solvability of (1.6)

Let us first introduce the velocity field $\widetilde{\boldsymbol{u}}(x)=\widetilde{u}^{r}(r,z)\boldsymbol{e}_{r}+\widetilde{u}^{z}(r,z)\boldsymbol{e}_{z}$ satisfying the following axisymmetric without swirl N-S:

(3.1)

\left\{\begin{split}&\partial_{t}\widetilde{u}^{r}+(\widetilde{u}^{r}\partial_{r}+\widetilde{u}^{z}\partial_{z})\widetilde{u}^{r}-(\partial_{r}^{2}+\partial_{z}^{2}+\frac{1}{r}\partial_{r}-\frac{1}{r^{2}})\widetilde{u}^{r}+\partial_{r}\widetilde{P}=0,\\ &\partial_{t}\widetilde{u}^{z}+(\widetilde{u}^{r}\partial_{r}+\widetilde{u}^{z}\partial_{z})\widetilde{u}^{z}-(\partial_{r}^{2}+\partial_{z}^{2}+\frac{1}{r}\partial_{r})\widetilde{u}^{z}+\partial_{z}\widetilde{P}=0,\\ &\partial_{r}\widetilde{u}^{r}+\frac{1}{r}\widetilde{u}^{r}+\partial_{z}\widetilde{u}^{z}=0,\\ &\widetilde{u}^{r}|_{t=0}=\widetilde{u}^{r}_{0}={\mathcal{M}}(u_{0}^{r}),\ \widetilde{u}^{z}|_{t=0}=\widetilde{u}^{z}_{0}={\mathcal{M}}(u_{0}^{z}).\end{split}\right.

Proposition 3.1.

Under the assumption of Theorem 1.1, (3.1) has a unique global solution $\widetilde{\boldsymbol{u}}$ such that for any time $t>0$ , there hold

(3.2)

\|\widetilde{\boldsymbol{u}}\|_{L^{\infty}_{t}(L^{2})}^{2}+2\|\nabla\widetilde{\boldsymbol{u}}\|_{L^{2}_{t}(L^{2})}^{2}\leq\|{\boldsymbol{u}}_{0}\|_{L^{2}}^{2},

(3.3)

\|\widetilde{\boldsymbol{u}}\|_{L^{\infty}_{t}(\dot{H}^{1})}^{2}+\|\nabla\widetilde{\boldsymbol{u}}\|_{L^{2}_{t}(\dot{H}^{1})}^{2}\lesssim\|{\boldsymbol{u}}_{0}\|_{\dot{H}^{1}}^{2}+\|{\boldsymbol{u}}_{0}\|_{L^{2}}^{4}\|{\boldsymbol{u}}_{0}\|_{\dot{H}^{2}}^{2}.

Proof.

The first estimate (3.2) is nothing but the energy equality and the fact that

\|\widetilde{\boldsymbol{u}}_{0}\|_{L^{2}}^{2}=\|{\mathcal{M}}(u_{0}^{r})\|_{L^{2}}^{2}+\|{\mathcal{M}}(u_{0}^{z})\|_{L^{2}}^{2}\leq\|{\boldsymbol{u}}_{0}\|_{L^{2}}^{2}.

Next, the Biot-Savart law tells us that we can use $\mathop{\rm curl}\nolimits\widetilde{\boldsymbol{u}}=\widetilde{\omega}^{\theta}\boldsymbol{e}_{\theta}=(\partial_{z}\widetilde{u}^{r}-\partial_{r}\widetilde{u}^{z})\boldsymbol{e}_{\theta}$ to represent $\widetilde{\boldsymbol{u}}$ . Hence we can reformulate the System (3.1) as

(3.4)

\left\{\begin{split}&\partial_{t}\widetilde{\omega}^{\theta}+(\widetilde{u}^{r}\partial_{r}+\widetilde{u}^{z}\partial_{z})\widetilde{\omega}^{\theta}-(\partial_{r}^{2}+\partial_{z}^{2}+\frac{1}{r}\partial_{r}-\frac{1}{r^{2}})\widetilde{\omega}^{\theta}-\frac{\widetilde{u}^{r}\widetilde{\omega}^{\theta}}{r}=0,\\ &\widetilde{\omega}^{\theta}|_{t=0}=\widetilde{\omega}^{\theta}_{0}={\mathcal{M}}(\partial_{z}u^{r}_{0}-\partial_{r}u^{z}_{0}).\end{split}\right.

Then it is not difficult to verify that $\widetilde{\omega}^{\theta}/r$ satisfies

(3.5)

\partial_{t}\frac{\widetilde{\omega}^{\theta}}{r}+(\widetilde{u}^{r}\partial_{r}+\widetilde{u}^{z}\partial_{z})\frac{\widetilde{\omega}^{\theta}}{r}-(\Delta+\frac{2}{r}\partial_{r})\frac{\widetilde{\omega}^{\theta}}{r}=0.

For a strong solution of N-S, the decay property at infinity implies $\widetilde{\omega}^{\theta}|_{r=\infty}=0$ , while the smoothness implies $\widetilde{\omega}^{\theta}|_{r=0}=0$ . As a result, we have

-\int_{\mathop{\mathbb{R}\kern 0.0pt}\nolimits^{3}}\frac{2}{r}\partial_{r}\frac{\widetilde{\omega}^{\theta}}{r}\cdot\frac{\widetilde{\omega}^{\theta}}{r}\,dx=-2\pi\int_{\mathop{\mathbb{R}\kern 0.0pt}\nolimits^{+}\times\mathop{\mathbb{R}\kern 0.0pt}\nolimits}\partial_{r}\bigl|\frac{\widetilde{\omega}^{\theta}}{r}\bigr|^{2}\,drdz=2\pi\int_{\mathop{\mathbb{R}\kern 0.0pt}\nolimits}\lim_{r\rightarrow 0}\bigl|\frac{\widetilde{\omega}^{\theta}}{r}\bigr|^{2}\,dz\geq 0.

In view of this, the $L^{2}$ energy estimate of $\widetilde{\omega}^{\theta}/r$ in (3.5) gives

(3.6)

\bigl\|\frac{\widetilde{\omega}^{\theta}}{r}\bigr\|_{L^{\infty}_{t}(L^{2})}+\bigl\|\nabla\frac{\widetilde{\omega}^{\theta}}{r}\bigr\|_{L^{2}_{t}(L^{2})}\leq\bigl\|\frac{\widetilde{\omega}^{\theta}_{0}}{r}\bigr\|_{L^{2}}.

By using (2.2) and Lemma 2.1, we have

	$\displaystyle\\|\nabla^{2}\boldsymbol{u}_{0}\\|_{L^{2}}\sim\\|\nabla\boldsymbol{\omega}_{0}\\|_{L^{2}}$	$\displaystyle\geq\bigl\\|\frac{\partial_{\theta}}{r}(\boldsymbol{\omega}_{0})\cdot\boldsymbol{e}_{r}\bigr\\|_{L^{2}}=\bigl\\|-\frac{\omega^{\theta}_{0}}{r}+\frac{\partial_{\theta}\omega^{r}_{0}}{r}\bigr\\|_{L^{2}}$
		$\displaystyle\geq\bigl\\|{\mathcal{M}}\bigl(-\frac{\omega^{\theta}_{0}}{r}+\frac{\partial_{\theta}\omega^{r}_{0}}{r}\bigr)\bigr\\|_{L^{2}}=\bigl\\|{\mathcal{M}}\bigl(-\frac{\omega^{\theta}_{0}}{r}\bigr)\bigr\\|_{L^{2}}=\bigl\\|\frac{\widetilde{\omega}^{\theta}_{0}}{r}\bigr\\|_{L^{2}},$

where $\boldsymbol{\omega}_{0}\mathrel{\mathop{\kern 0.0pt=}\limits^{\hbox{\footnotesize def}}}\mathop{\rm curl}\nolimits\boldsymbol{u}_{0}$ . Substituting this into (3.6) gives

(3.7)

\bigl\|\frac{\widetilde{\omega}^{\theta}}{r}\bigr\|_{L^{\infty}_{t}(L^{2})}+\bigl\|\nabla\frac{\widetilde{\omega}^{\theta}}{r}\bigr\|_{L^{2}_{t}(L^{2})}\leq\|\boldsymbol{u}_{0}\|_{\dot{H}^{2}}.

Next, by taking $L^{2}$ inner product of (3.4) with $\widetilde{\omega}^{\theta}$ , we obtain

(3.8)

\frac{1}{2}\frac{d}{dt}\|\widetilde{\omega}^{\theta}\|_{L^{2}}^{2}+\|\nabla\widetilde{\omega}^{\theta}\|_{L^{2}}^{2}+\bigl\|\frac{\widetilde{\omega}^{\theta}}{r}\|_{L^{2}}^{2}\leq\int_{\mathop{\mathbb{R}\kern 0.0pt}\nolimits^{3}}\frac{\widetilde{u}^{r}}{r}|\widetilde{\omega}^{\theta}|^{2}\,dx.

For the right-hand side term, we have

	$\displaystyle\bigl\|\int_{\mathop{\mathbb{R}\kern 0.0pt}\nolimits^{3}}\frac{\widetilde{u}^{r}}{r}\|\widetilde{\omega}^{\theta}\|^{2}\,dx\bigr\|$	$\displaystyle\leq\\|\widetilde{u}^{r}\\|_{L^{2}}\\|\widetilde{\omega}^{\theta}\\|_{L^{6}}\bigl\\|\frac{\widetilde{\omega}^{\theta}}{r}\bigr\\|_{L^{2}}^{\frac{1}{2}}\bigl\\|\frac{\widetilde{\omega}^{\theta}}{r}\bigr\\|_{L^{6}}^{\frac{1}{2}}$
		$\displaystyle\leq\frac{1}{2}\bigl(\\|\nabla\widetilde{\omega}^{\theta}\\|_{L^{2}}^{2}+\bigl\\|\frac{\widetilde{\omega}^{\theta}}{r}\bigr\\|_{L^{2}}^{2}\bigr)+C\\|\widetilde{u}^{r}\\|_{L^{2}}^{4}\\|\nabla\bigl(\frac{\widetilde{\omega}^{\theta}}{r}\bigr)\\|_{L^{2}}^{2}.$

Substituting this into (3.8) and then integrating in time, we deduce

(3.9)

\|\widetilde{\omega}^{\theta}\|_{L^{\infty}_{t}(L^{2})}^{2}+\|\nabla\widetilde{\omega}^{\theta}\|_{L^{2}_{t}(L^{2})}^{2}+\|\frac{\widetilde{\omega}^{\theta}}{r}\|_{L^{2}_{t}(L^{2})}^{2}\leq\|\widetilde{\omega}^{\theta}_{0}\|_{L^{2}}^{2}+C\|\widetilde{u}^{r}\|_{L^{\infty}_{t}(L^{2})}^{4}\|\nabla\bigl(\frac{\widetilde{\omega}^{\theta}}{r}\bigr)\|_{L^{2}_{t}(L^{2})}^{2}.

Thanks to the estimates (3.7) and (3.2) , we finally dedecue from (3.9) that

\displaystyle\|\widetilde{\omega}^{\theta}\|_{L^{\infty}_{t}(L^{2})}^{2}+\|\nabla\widetilde{\omega}^{\theta}\|_{L^{2}_{t}(L^{2})}^{2}+\|\widetilde{\omega}^{\theta}/r\|_{L^{2}_{t}(L^{2})}^{2}\lesssim\|{\boldsymbol{u}}_{0}\|_{\dot{H}^{1}}^{2}+\|{\boldsymbol{u}}_{0}\|_{L^{2}}^{4}\|{\boldsymbol{u}}_{0}\|_{\dot{H}^{2}}^{2},

which together with Lemma 2.2 leads to the second desired estimate (3.3). ∎

Next, we shall use Proposition 2.1 to derive the global existence of the solution to (1.6).

Proposition 3.2.

Under the assumption of Theorem 1.1, (1.6) has a unique global solution $\bar{\boldsymbol{u}}$ such that for any time $t>0$ , there hold

(3.10)

\|\bar{\boldsymbol{u}}\|_{L^{\infty}_{t}(L^{2})}^{2}+2\|\nabla\bar{\boldsymbol{u}}\|_{L^{2}_{t}(L^{2})}^{2}\leq\|{\boldsymbol{u}}_{0}\|_{L^{2}}^{2},

(3.11)

\|\bar{\boldsymbol{u}}\|_{L^{\infty}_{t}(\dot{H}^{1})}^{2}+\|\nabla\bar{\boldsymbol{u}}\|_{L^{2}_{t}(\dot{H}^{1})}^{2}\lesssim\|\nabla\boldsymbol{u}_{0}\|_{L^{2}}^{2}\exp\bigl(C\|{\boldsymbol{u}}_{0}\|_{L^{2}}^{6}\|{\boldsymbol{u}}_{0}\|_{\dot{H}^{2}}^{2}\bigr).

Proof.

By using Lemma 2.1 and the fact that $\bar{\boldsymbol{u}}_{0}-\widetilde{\boldsymbol{u}}_{0}=\bar{u}^{\theta}_{0}(r,z)\boldsymbol{e}_{\theta}={\mathcal{M}}(u^{\theta}_{0})\boldsymbol{e}_{\theta}$ , we have

\displaystyle\|\bar{\boldsymbol{u}}_{0}-\widetilde{\boldsymbol{u}}_{0}\|_{L^{2}}=\|{\mathcal{M}}(u^{\theta}_{0})\|_{L^{2}}\leq\|u^{\theta}_{0}\|_{L^{2}},

and

	$\displaystyle\\|\nabla(\bar{\boldsymbol{u}}_{0}-\widetilde{\boldsymbol{u}}_{0})\\|_{L^{2}}^{2}$	$\displaystyle=\bigl\\|\nabla\bigl({\mathcal{M}}(u^{\theta}_{0})\boldsymbol{e}_{\theta}\bigr)\bigr\\|_{L^{2}}^{2}=\\|\widetilde{\nabla}{\mathcal{M}}(u^{\theta}_{0})\\|_{L^{2}}^{2}+\bigl\\|\frac{{\mathcal{M}}(u^{\theta}_{0})}{r}\bigr\\|_{L^{2}}^{2}$
		$\displaystyle=\\|{\mathcal{M}}(\widetilde{\nabla}u^{\theta}_{0})\\|_{L^{2}}^{2}+\min\Bigl\{\bigl\\|{\mathcal{M}}\bigl(\frac{u^{\theta}_{0}}{r}\bigr)\bigr\\|_{L^{2}}^{2},\bigl\\|{\mathcal{M}}\bigl(\frac{u^{\theta}_{0}-\partial_{\theta}u^{r}_{0}}{r}\bigr)\bigr\\|_{L^{2}}^{2}\Bigr\}$
		$\displaystyle\leq\\|\widetilde{\nabla}u^{\theta}_{0}\\|_{L^{2}}^{2}+\min\Bigl\{\bigl\\|\frac{u^{\theta}_{0}}{r}\bigr\\|_{L^{2}}^{2},\bigl\\|\frac{u^{\theta}_{0}-\partial_{\theta}u^{r}_{0}}{r}\bigr\\|_{L^{2}}^{2}\Bigr\}$
		$\displaystyle\leq\min\bigl\{\\|u^{\theta}_{0}\\|_{\dot{H}^{1}_{\rm axi}}^{2},\\|{\boldsymbol{u}}_{0}\\|_{\dot{H}^{1}}^{2}\bigr\}.$

These together with the bounds (3.2), (3.3) and the smallness condition (1.9), we get that

\displaystyle\begin{aligned} &\|\bar{\boldsymbol{u}}_{0}-\widetilde{\boldsymbol{u}}_{0}\|_{L^{2}}\|\bar{\boldsymbol{u}}_{0}-\widetilde{\boldsymbol{u}}_{0}\|_{\dot{H}^{1}}\exp\bigl(C\|{\boldsymbol{u}}_{0}\|_{L^{2}}^{2}(\|{\boldsymbol{u}}_{0}\|_{\dot{H}^{1}}^{2}+\|{\boldsymbol{u}}_{0}\|_{L^{2}}^{4}\|{\boldsymbol{u}}_{0}\|_{\dot{H}^{2}}^{2})\bigr)\\ &\qquad\leq C\|u^{\theta}_{0}\|_{L^{2}}\|u^{\theta}_{0}\|_{\dot{H}^{1}_{\rm axi}}\exp\bigl(C\|{\boldsymbol{u}}_{0}\|_{L^{2}}^{6}\|{\boldsymbol{u}}_{0}\|_{\dot{H}^{2}}^{2}\bigr)<c_{0},\end{aligned}

where we used the fact that

\displaystyle\|{\boldsymbol{u}}_{0}\|_{L^{2}}^{2}\|{\boldsymbol{u}}_{0}\|_{\dot{H}^{1}}^{2}\lesssim\|{\boldsymbol{u}}_{0}\|_{L^{2}}^{3}\|{\boldsymbol{u}}_{0}\|_{\dot{H}^{2}}\lesssim 1+\|{\boldsymbol{u}}_{0}\|_{L^{2}}^{6}\|{\boldsymbol{u}}_{0}\|_{\dot{H}^{2}}^{2}.

So that the condition (2.9) in Proposition 2.1 is fulfilled. Thus (1.6) has a unique global strong solution $\bar{\boldsymbol{u}}$ such that

\|\bar{\boldsymbol{u}}-\widetilde{\boldsymbol{u}}\|_{L^{\infty}(\mathop{\mathbb{R}\kern 0.0pt}\nolimits^{+};\dot{H}^{1})}^{2}+\|\bar{\boldsymbol{u}}-\widetilde{\boldsymbol{u}}\|_{L^{2}(\mathop{\mathbb{R}\kern 0.0pt}\nolimits^{+};\dot{H}^{2})}^{2}\lesssim\min\bigl\{\|u^{\theta}_{0}\|_{\dot{H}^{1}_{\rm axi}}^{2},\|{\boldsymbol{u}}_{0}\|_{\dot{H}^{1}}^{2}\bigr\}\exp\bigl(C\|{\boldsymbol{u}}_{0}\|_{L^{2}}^{6}\|{\boldsymbol{u}}_{0}\|_{\dot{H}^{2}}^{2}\bigr),

which together with (3.3) leads to the desired estimate (3.11).

The estimate (3.10) follows from the Basic energy equality of Navier-Stokes equations. Thus, we complete the proof of the proposition. ∎

3.2. The proof of Theorem 1.1

The goal of this subsection is to prove Theorem 1.1.

The proof of Theorem 1.1.

We divide the proof into two steps.

Step 1. Global solvability of (1.4). The strategy is still to use Proposition 2.1. To do this, we need to analyse the difference between $\boldsymbol{u}_{0}$ and $\bar{\boldsymbol{u}}_{0}$ , which reads as follows:

\boldsymbol{u}_{0}-\bar{\boldsymbol{u}}_{0}=\bigl(u_{0}^{r}-{\mathcal{M}}(u_{0}^{r})\bigr)e_{r}+\bigl(u_{0}^{\theta}-{\mathcal{M}}(u_{0}^{\theta})\bigr)e_{\theta}+\bigl(u_{0}^{z}-{\mathcal{M}}(u_{0}^{z})\bigr)e_{z},

\widetilde{\nabla}(\boldsymbol{u}_{0}-\bar{\boldsymbol{u}}_{0})=\bigl(\widetilde{\nabla}u_{0}^{r}-{\mathcal{M}}(\widetilde{\nabla}u_{0}^{r})\bigr)e_{r}+\bigl(\widetilde{\nabla}u_{0}^{\theta}-{\mathcal{M}}(\widetilde{\nabla}u_{0}^{\theta})\bigr)e_{\theta}+\bigl(\widetilde{\nabla}u_{0}^{z}-{\mathcal{M}}(\widetilde{\nabla}u_{0}^{z})\bigr)e_{z},

and

\frac{\partial_{\theta}}{r}(\boldsymbol{u}_{0}-\bar{\boldsymbol{u}}_{0})=\Bigl(\frac{\partial_{\theta}u_{0}^{r}}{r}-\frac{u_{0}^{\theta}}{r}+{\mathcal{M}}\bigl(\frac{u_{0}^{\theta}}{r}\bigr)\Bigr)e_{r}+\Bigl(\frac{\partial_{\theta}u_{0}^{\theta}}{r}+\frac{u_{0}^{r}}{r}-{\mathcal{M}}\bigl(\frac{u_{0}^{r}}{r}\bigr)\Bigr)e_{\theta}+\bigl(\frac{\partial_{\theta}u_{0}^{z}}{r}\bigr)e_{z}.

Then by using Lemma 2.1 and the notation $\boldsymbol{\mathfrak{U}}_{0}=(\partial_{\theta}u^{r}_{0},\partial_{\theta}u^{\theta}_{0},\partial_{\theta}u^{z}_{0}),$ we can obtain

(3.12)

\|\boldsymbol{u}_{0}-\bar{\boldsymbol{u}}_{0}\|_{L^{2}}\lesssim\|\boldsymbol{\mathfrak{U}}_{0}\|_{L^{2}},

(3.13)

\|\nabla(\boldsymbol{u}_{0}-\bar{\boldsymbol{u}}_{0})\|_{L^{2}}\lesssim\|\widetilde{\nabla}\boldsymbol{\mathfrak{U}}_{0}\|_{L^{2}}+\|r^{-1}\boldsymbol{\mathfrak{U}}_{0}\|_{L^{2}}=\|\boldsymbol{\mathfrak{U}}_{0}\|_{\dot{H}^{1}_{\rm axi}}.

In view of the estimates (3.10)-(3.13) and the smallness condition (1.10), the condition (2.9) of Propostion 2.1 holds for $\boldsymbol{u}_{0}$ and $\boldsymbol{v}_{0}=\bar{\boldsymbol{u}}_{0}$ . Then by using Proposition 2.1 again, (1.4) has a unique global strong solution $\boldsymbol{u}$ in $C(\mathop{\mathbb{R}\kern 0.0pt}\nolimits^{+};H^{1})\cap L^{2}(\mathop{\mathbb{R}\kern 0.0pt}\nolimits^{+};\dot{H}^{1}\cap\dot{H}^{2})$ , satisfying

(3.14)			$\displaystyle\\|\boldsymbol{u}-\bar{\boldsymbol{u}}\\|_{L^{\infty}_{t}(L^{2})}^{2}+\\|\nabla(\boldsymbol{u}-\bar{\boldsymbol{u}})\\|_{L^{2}_{t}(L^{2})}^{2}\lesssim\\|\boldsymbol{\mathfrak{U}}_{0}\\|_{L^{2}}^{2}\exp\bigl(\exp(C\\|{\boldsymbol{u}}_{0}\\|_{L^{2}}^{6}\\|{\boldsymbol{u}}_{0}\\|_{\dot{H}^{2}}^{2})\bigr),$
(3.14)			$\displaystyle\\|\boldsymbol{u}-\bar{\boldsymbol{u}}\\|_{L^{\infty}_{t}(\dot{H}^{1})}^{2}+\\|\nabla(\boldsymbol{u}-\bar{\boldsymbol{u}})\\|_{L^{2}_{t}(\dot{H}^{1})}^{2}\lesssim\\|\boldsymbol{\mathfrak{U}}_{0}\\|_{\dot{H}^{1}_{\rm axi}}^{2}\exp\bigl(\exp(C\\|{\boldsymbol{u}}_{0}\\|_{L^{2}}^{6}\\|{\boldsymbol{u}}_{0}\\|_{\dot{H}^{2}}^{2})\bigr).$

In particular, the above estimates together with interpolation inequality and the smallness condition (1.10) implies

\|\boldsymbol{u}-\bar{\boldsymbol{u}}\|_{L^{\infty}_{t}(\dot{H}^{\frac{1}{2}})}^{2}+\|\nabla(\boldsymbol{u}-\bar{\boldsymbol{u}})\|_{L^{2}_{t}(\dot{H}^{\frac{1}{2}})}^{2}\lesssim\|\boldsymbol{\mathfrak{U}}_{0}\|_{L^{2}}\|\boldsymbol{\mathfrak{U}}_{0}\|_{\dot{H}^{1}_{\rm axi}}\exp\bigl(\exp(C\|{\boldsymbol{u}}_{0}\|_{L^{2}}^{6}\|{\boldsymbol{u}}_{0}\|_{\dot{H}^{2}}^{2})\bigr)<\epsilon,

which means that $\boldsymbol{u}$ is indeed close to an axisymmetric solution.

Step 2. Error estimates of ${\mathcal{M}}\boldsymbol{u}-\bar{\boldsymbol{u}}$

Step 2.1. Equations for ${\mathcal{M}}\boldsymbol{u}-\bar{\boldsymbol{u}}$ . We first get, by applying ${\mathcal{M}}$ to (1.4) that

(3.15)

\left\{\begin{aligned} &\partial_{t}{\mathcal{M}}u^{r}+{\mathcal{M}}\Bigl((u^{r}\partial_{r}+u^{\theta}\frac{\partial_{\theta}}{r}+u^{z}\partial_{z})u^{r}\Bigr)\\ &\qquad\qquad\qquad\qquad-\bigl(\Delta-\frac{1}{r^{2}}\bigr){\mathcal{M}}u^{r}-{\mathcal{M}}\Bigl(\frac{|u^{\theta}|^{2}}{r}\Bigr)+\partial_{r}{\mathcal{M}}P=0,\\ &\partial_{t}{\mathcal{M}}u^{\theta}+{\mathcal{M}}\Bigl((u^{r}\partial_{r}+u^{\theta}\frac{\partial_{\theta}}{r}+u^{z}\partial_{z})u^{\theta}\Bigr)-\bigl(\Delta-\frac{1}{r^{2}}\bigr){\mathcal{M}}u^{\theta}+{\mathcal{M}}\Bigl(\frac{u^{r}u^{\theta}}{r}\Bigr)=0,\\ &\partial_{t}{\mathcal{M}}u^{z}+{\mathcal{M}}\Bigl((u^{r}\partial_{r}+u^{\theta}\frac{\partial_{\theta}}{r}+u^{z}\partial_{z})u^{z}\Bigr)-\Delta{\mathcal{M}}u^{z}+\partial_{z}{\mathcal{M}}P=0,\\ &\partial_{r}{\mathcal{M}}u^{r}+\frac{{\mathcal{M}}u^{r}}{r}+\partial_{z}{\mathcal{M}}u^{z}=0,\\ &{\mathcal{M}}{\boldsymbol{u}}|_{t=0}={\mathcal{M}}{\boldsymbol{u}_{0}},\end{aligned}\right.

where we have used the fact that for any regular enough function $f$ , there holds

\int_{0}^{2\pi}\partial_{\theta}f(r,\theta,z)\,d\theta=0.

Then by denoting $\boldsymbol{v}\mathrel{\mathop{\kern 0.0pt=}\limits^{\hbox{\footnotesize def}}}{\mathcal{M}}\boldsymbol{u}-\bar{\boldsymbol{u}},\,Q\mathrel{\mathop{\kern 0.0pt=}\limits^{\hbox{\footnotesize def}}}{\mathcal{M}}P-\bar{P}$ , and in view of (1.6) and (3.15), we deduce

(3.16)

\left\{\begin{aligned} &\partial_{t}\boldsymbol{v}-\Delta\boldsymbol{v}+\nabla Q=\boldsymbol{F}+\boldsymbol{G},\quad\mathop{\rm div}\nolimits\boldsymbol{v}=0,\\ &\boldsymbol{v}|_{t=0}=0,\end{aligned}\right.

where $\boldsymbol{F}=F^{r}\boldsymbol{e}_{r}+F^{\theta}\boldsymbol{e}_{\theta}+F^{z}\boldsymbol{e}_{z},~\boldsymbol{G}=G^{r}\boldsymbol{e}_{r}+G^{\theta}\boldsymbol{e}_{\theta}+G^{z}\boldsymbol{e}_{z}$ with

		$\displaystyle F^{r}=(\bar{u}^{r}\partial_{r}+\bar{u}^{z}\partial_{z})\bar{u}^{r}-{\mathcal{M}}\Bigl((u^{r}\partial_{r}+u^{z}\partial_{z})u^{r}\Bigr),~G^{r}=\bar{u}^{\theta}\bigl(\frac{\partial_{\theta}\bar{u}^{r}}{r}-\frac{\bar{u}^{\theta}}{r}\bigr)-{\mathcal{M}}\Bigl(u^{\theta}\bigl(\frac{\partial_{\theta}u^{r}}{r}-\frac{u^{\theta}}{r}\bigr)\Bigr),$
		$\displaystyle F^{\theta}=(\bar{u}^{r}\partial_{r}+\bar{u}^{z}\partial_{z})\bar{u}^{\theta}-{\mathcal{M}}\Bigl((u^{r}\partial_{r}+u^{z}\partial_{z})u^{\theta}\Bigr),~G^{\theta}=\bar{u}^{\theta}\bigl(\frac{\partial_{\theta}\bar{u}^{\theta}}{r}+\frac{\bar{u}^{r}}{r}\bigr)-{\mathcal{M}}\Bigl(u^{\theta}\bigl(\frac{\partial_{\theta}u^{\theta}}{r}+\frac{u^{r}}{r}\bigr)\Bigr),$
		$\displaystyle F^{z}=(\bar{u}^{r}\partial_{r}+\bar{u}^{z}\partial_{z})\bar{u}^{z}-{\mathcal{M}}\Bigl((u^{r}\partial_{r}+u^{z}\partial_{z})u^{z}\Bigr),~G^{z}=-{\mathcal{M}}\Bigl(\frac{u^{\theta}\partial_{\theta}u^{z}}{r}\Bigr).$

We mention that here $\partial_{\theta}\bar{u}^{r},\,\partial_{\theta}\bar{u}^{\theta}$ indeed vanishes, but we still write them to derive a symmetric form. In this way, we can handle these terms by using the following key observation: for any functions $f,\,g$ and any axisymmetric functions $\bar{f},\,\bar{g}$ , there holds

(3.17)

\begin{split}{\mathcal{M}}(fg)-\bar{f}\bar{g}&={\mathcal{M}}(fg-\bar{f}\bar{g})\\ &={\mathcal{M}}\bigl((f-\bar{f})(g-\bar{g})+(f-\bar{f})\bar{g}+\bar{f}(g-\bar{g})\bigr)\\ &={\mathcal{M}}\bigl((f-\bar{f})(g-\bar{g})\bigr)+\bar{g}{\mathcal{M}}(f-\bar{f})+\bar{f}{\mathcal{M}}(g-\bar{g}).\end{split}

On the other hand, noting that $\bar{f},\,\bar{g}$ are axisymmetric, we have

\partial_{\theta}g=\partial_{\theta}(g-\bar{g}),\quad\hbox{and}\quad{\mathcal{M}}\bigl(\bar{f}\partial_{\theta}(g-\bar{g})\bigr)=\bar{f}{\mathcal{M}}\bigl(\partial_{\theta}(g-\bar{g})\bigr)=0.

As a result, there holds

(3.18)

{\mathcal{M}}(f\partial_{\theta}g)={\mathcal{M}}\bigl(f\partial_{\theta}(g-\bar{g})\bigr)={\mathcal{M}}\bigl((f-\bar{f})\partial_{\theta}(g-\bar{g})\bigr).

For notation simplification, let us introduce $\boldsymbol{V}=\boldsymbol{u}-\bar{\boldsymbol{u}}$ . Obviously there holds

{\mathcal{M}}\boldsymbol{V}={\mathcal{M}}\boldsymbol{u}-{\mathcal{M}}\bar{\boldsymbol{u}}={\mathcal{M}}\boldsymbol{u}-\bar{\boldsymbol{u}}=\boldsymbol{v}.

Then by using (3.17), we have

\displaystyle{\mathcal{M}}(u^{r}\partial_{r}u^{r})-\bar{u}^{r}\partial_{r}\bar{u}^{r}={\mathcal{M}}(V^{r}\partial_{r}V^{r})+v^{r}\partial_{r}\bar{u}^{r}+\bar{u}^{r}\partial_{r}v^{r}.

Similar formulas hold for the other terms in $F^{r},\,F^{\theta},\,F^{z}$ . And we can write

(3.19)			$\displaystyle F^{r}=-{\mathcal{M}}\Bigl((V^{r}\partial_{r}+V^{z}\partial_{z})V^{r}\Bigr)-(v^{r}\partial_{r}+v^{z}\partial_{z})\bar{u}^{r}-(\bar{u}^{r}\partial_{r}+\bar{u}^{z}\partial_{z})v^{r},$
			$\displaystyle F^{\theta}=-{\mathcal{M}}\Bigl((V^{r}\partial_{r}+V^{z}\partial_{z})V^{\theta}\Bigr)-(v^{r}\partial_{r}+v^{z}\partial_{z})\bar{u}^{\theta}-(\bar{u}^{r}\partial_{r}+\bar{u}^{z}\partial_{z})v^{\theta},$
			$\displaystyle F^{z}=-{\mathcal{M}}\Bigl((V^{r}\partial_{r}+V^{z}\partial_{z})V^{z}\Bigr)-(v^{r}\partial_{r}+v^{z}\partial_{z})\bar{u}^{z}-(\bar{u}^{r}\partial_{r}+\bar{u}^{z}\partial_{z})v^{z}.$

Exactly along the same line, by using (3.17), we have

	$\displaystyle G^{r}$	$\displaystyle=-{\mathcal{M}}\Bigl(V^{\theta}\bigl(\frac{\partial_{\theta}V^{r}}{r}-\frac{V^{\theta}}{r}\bigr)\Bigr)-\bar{u}^{\theta}{\mathcal{M}}\Bigl(\frac{\partial_{\theta}V^{r}}{r}-\frac{V^{\theta}}{r}\Bigr)-\bigl(\frac{\partial_{\theta}\bar{u}^{r}}{r}-\frac{\bar{u}^{\theta}}{r}\bigr){\mathcal{M}}(V^{\theta})$
		$\displaystyle=-{\mathcal{M}}\Bigl(V^{\theta}\bigl(\frac{\partial_{\theta}V^{r}}{r}-\frac{V^{\theta}}{r}\bigr)\Bigr)+2\frac{\bar{u}^{\theta}}{r}v^{\theta},$

and

G^{\theta}=-{\mathcal{M}}\Bigl(V^{\theta}\bigl(\frac{\partial_{\theta}V^{\theta}}{r}+\frac{V^{r}}{r}\bigr)\Bigr)-\frac{\bar{u}^{\theta}}{r}v^{r}-\frac{\bar{u}^{r}}{r}v^{\theta}.

Notice that

\frac{\partial_{\theta}\boldsymbol{V}}{r}=(\frac{\partial_{\theta}V^{r}}{r}-\frac{V^{\theta}}{r})\boldsymbol{e}_{r}+(\frac{\partial_{\theta}V^{\theta}}{r}+\frac{V^{r}}{r})\boldsymbol{e}_{\theta}+\frac{\partial_{\theta}V^{z}}{r}\boldsymbol{e}_{z},

thus we can write

(3.20)

G^{r}=-{\mathcal{M}}\Bigl(V^{\theta}\bigl(\frac{\partial_{\theta}\boldsymbol{V}}{r}\cdot\boldsymbol{e}_{r}\bigr)\Bigr)+2\frac{\bar{u}^{\theta}}{r}v^{\theta},\quad\hbox{and}\quad G^{\theta}=-{\mathcal{M}}\Bigl(V^{\theta}\bigl(\frac{\partial_{\theta}\boldsymbol{V}}{r}\cdot\boldsymbol{e}_{\theta}\bigr)\Bigr)-\frac{\bar{u}^{\theta}}{r}v^{r}-\frac{\bar{u}^{r}}{r}v^{\theta}.

On the other hand, by using (3.18), we have

(3.21)

G^{z}=-{\mathcal{M}}\Bigl(V^{\theta}\frac{\partial_{\theta}V^{z}}{r}\Bigr)=-{\mathcal{M}}\Bigl(V^{\theta}\bigl(\frac{\partial_{\theta}\boldsymbol{V}}{r}\cdot\boldsymbol{e}_{z}\bigr)\Bigr).

Step 2.2. $L^{2}$ -estimate for $\boldsymbol{v}$ . Taking $L^{2}$ inner product of (3.16) with $\boldsymbol{v}$ gives rise to

(3.22)

\frac{1}{2}\frac{d}{dt}\|\boldsymbol{v}\|_{L^{2}}^{2}+\|\nabla\boldsymbol{v}\|_{L^{2}}^{2}=(\boldsymbol{F}\,|\,\boldsymbol{v})+(\boldsymbol{G}\,|\,\boldsymbol{v}).

Thanks to the formula (3.19) and Lemma 2.1, we obtain

\displaystyle\begin{aligned} |(\boldsymbol{F}\,|\,\boldsymbol{v})|&\lesssim\|\boldsymbol{V}\|_{L^{3}}\|{\nabla}\boldsymbol{V}\|_{L^{2}}\|\boldsymbol{v}\|_{L^{6}}+\|\boldsymbol{v}\|_{L^{3}}\|{\nabla}\bar{\boldsymbol{u}}\|_{L^{2}}\|\boldsymbol{v}\|_{L^{6}}+\|\bar{\boldsymbol{u}}\|_{L^{6}}\|{\nabla}\boldsymbol{v}\|_{L^{2}}\|\boldsymbol{v}\|_{L^{3}}\\ &\lesssim\|\boldsymbol{V}\|_{L^{2}}^{\frac{1}{2}}\|\nabla\boldsymbol{V}\|_{L^{2}}^{\frac{3}{2}}\|\nabla\boldsymbol{v}\|_{L^{2}}+\|\nabla\bar{\boldsymbol{u}}\|_{L^{2}}\|\boldsymbol{v}\|_{L^{2}}^{\frac{1}{2}}\|\nabla\boldsymbol{v}\|_{L^{2}}^{\frac{3}{2}},\end{aligned}

which along with Young’s inequality implies

(3.23)

|(\boldsymbol{F}\,|\,\boldsymbol{v})|\leq\frac{1}{4}\|\nabla\boldsymbol{v}\|_{L^{2}}^{2}+C\|\boldsymbol{V}\|_{L^{2}}\|\nabla\boldsymbol{V}\|_{L^{2}}^{3}+\|\nabla\bar{\boldsymbol{u}}\|_{L^{2}}^{4}\|\boldsymbol{v}\|_{L^{2}}^{2}.

While by virtue of (3.20), (3.21), Lemma 2.1 and Lemma 2.3, we get

	$\displaystyle\|(\boldsymbol{G}\,\|\,\boldsymbol{v})\|$	$\displaystyle\lesssim\\|V^{\theta}\\|_{L^{3}}\bigl\\|\frac{\partial_{\theta}\boldsymbol{V}}{r}\bigr\\|_{L^{2}}\\|\boldsymbol{v}\\|_{L^{6}}+\\|\boldsymbol{v}\\|_{L^{3}}\bigl(\bigl\\|\frac{\bar{u}^{r}}{r}\bigr\\|_{L^{2}}+\bigl\\|\frac{\bar{u}^{\theta}}{r}\bigr\\|_{L^{2}}\bigr)\\|\boldsymbol{v}\\|_{L^{6}}$
		$\displaystyle\lesssim\\|\boldsymbol{V}\\|_{L^{2}}^{\frac{1}{2}}\\|\nabla\boldsymbol{V}\\|_{L^{2}}^{\frac{3}{2}}\\|\nabla\boldsymbol{v}\\|_{L^{2}}+\\|\boldsymbol{v}\\|_{L^{2}}^{\frac{1}{2}}\\|\nabla\boldsymbol{v}\\|_{L^{2}}^{\frac{3}{2}}\\|\nabla\bar{\boldsymbol{u}}\\|_{L^{2}}.$

Then by using Young’s inequality, we deduce

(3.24)

|(\boldsymbol{G}\,|\,\boldsymbol{v})|\leq\frac{1}{4}\|\nabla\boldsymbol{v}\|_{L^{2}}^{2}+C\|\boldsymbol{V}\|_{L^{2}}\|\nabla\boldsymbol{V}\|_{L^{2}}^{3}+C\|\boldsymbol{v}\|_{L^{2}}^{2}\|\nabla\bar{\boldsymbol{u}}\|_{L^{2}}^{4}.

Now by substituting (3.23) and (3.24) into (3.22), we obtain

\displaystyle\frac{d}{dt}\|\boldsymbol{v}\|_{L^{2}}^{2}+\|\nabla\boldsymbol{v}\|_{L^{2}}^{2}\leq C\|\boldsymbol{V}\|_{L^{2}}\|\nabla\boldsymbol{V}\|_{L^{2}}^{3}+C\|\boldsymbol{v}\|_{L^{2}}^{2}\|\nabla\bar{\boldsymbol{u}}\|_{L^{2}}^{4}.

Then by using Gronwall’s inequality and the fact that $\boldsymbol{v}|_{t=0}=0$ , we achieve

\|\boldsymbol{v}\|_{L^{\infty}_{t}(L^{2})}^{2}+\|\nabla\boldsymbol{v}\|_{L^{2}_{t}(L^{2})}^{2}\lesssim\|\boldsymbol{V}\|_{L^{\infty}_{t}(L^{2})}\|\nabla\boldsymbol{V}\|_{L^{\infty}_{t}(L^{2})}\|\nabla\boldsymbol{V}\|_{L^{2}_{t}(L^{2})}^{2}\exp\bigl(C\|\nabla\bar{\boldsymbol{u}}\|_{L^{\infty}_{t}(L^{2})}^{2}\|\nabla\bar{\boldsymbol{u}}\|_{L^{2}_{t}(L^{2})}^{2}\bigr).

Recall that $\boldsymbol{V}=\boldsymbol{u}-\bar{\boldsymbol{u}}$ , by using (3.10), (3.11) and (3.14), we obtain

(3.25)

\|\boldsymbol{v}\|_{L^{\infty}_{t}(L^{2})}^{2}+\|\boldsymbol{v}\|_{L^{2}_{t}(L^{2})}^{2}\lesssim\|\boldsymbol{\mathfrak{U}}_{0}\|_{L^{2}}^{3}\|\boldsymbol{\mathfrak{U}}_{0}\|_{\dot{H}^{1}_{\rm axi}}\exp\bigl(\exp(C\|{\boldsymbol{u}}_{0}\|_{L^{2}}^{6}\|{\boldsymbol{u}}_{0}\|_{\dot{H}^{2}}^{2})\bigr).

This together with the smallness assumption (1.10) gives

\|\boldsymbol{v}\|_{L^{\infty}_{t}(L^{2})}^{2}+\|\boldsymbol{v}\|_{L^{2}_{t}(L^{2})}^{2}\lesssim\epsilon\|\boldsymbol{\mathfrak{U}}_{0}\|_{L^{2}}^{2}\exp\bigl(\exp(C\|{\boldsymbol{u}}_{0}\|_{L^{2}}^{6}\|{\boldsymbol{u}}_{0}\|_{\dot{H}^{2}}^{2})\bigr),

which is exactly the first inequality of (1.12).

Step 2.3. $\dot{H}^{1}$ -estimate for $\boldsymbol{v}$ . Taking $L^{2}$ inner product of (3.16) with $-\Delta\boldsymbol{v}$ gives rise to

(3.26)

\begin{split}\frac{1}{2}\frac{d}{dt}\|\nabla\boldsymbol{v}\|_{L^{2}}^{2}+\|\Delta\boldsymbol{v}\|_{L^{2}}^{2}&=-(\boldsymbol{F}\,|\,\Delta\boldsymbol{v})-(\boldsymbol{G}\,|\,\Delta\boldsymbol{v})\\ &\leq\bigl(\|\boldsymbol{F}\|_{L^{2}}+\|\boldsymbol{G}\|_{L^{2}}\bigr)\|\Delta\boldsymbol{v}\|_{L^{2}}.\end{split}

In view of the formulas (3.19)-(3.21), we can use Lemmas 2.1 and 2.3 to get

	$\displaystyle\\|\boldsymbol{F}\\|_{L^{2}}+\\|\boldsymbol{G}\\|_{L^{2}}\lesssim$	$\displaystyle\\|\boldsymbol{V}\\|_{L^{\infty}}\bigl(\\|\widetilde{\nabla}\boldsymbol{V}\\|_{L^{2}}+\bigl\\|\frac{\partial_{\theta}\boldsymbol{V}}{r}\bigr\\|_{L^{2}}\bigr)$
		$\displaystyle+\\|\boldsymbol{v}\\|_{L^{\infty}}\bigl(\\|\widetilde{\nabla}\bar{\boldsymbol{u}}\\|_{L^{2}}+\bigl\\|\frac{\bar{u}^{r}}{r}\bigr\\|_{L^{2}}+\bigl\\|\frac{\bar{u}^{\theta}}{r}\bigr\\|_{L^{2}}\bigr)+\\|\bar{\boldsymbol{u}}\\|_{L^{\infty}}\\|\widetilde{\nabla}\boldsymbol{v}\\|_{L^{2}}$
	$\displaystyle\lesssim$	$\displaystyle\\|\boldsymbol{V}\\|_{L^{\infty}}\\|\nabla\boldsymbol{V}\\|_{L^{2}}+\\|\boldsymbol{v}\\|_{L^{\infty}}\\|\nabla\bar{\boldsymbol{u}}\\|_{L^{2}}+\\|\bar{\boldsymbol{u}}\\|_{L^{\infty}}\\|\nabla\boldsymbol{v}\\|_{L^{2}}$
	$\displaystyle\lesssim$	$\displaystyle\\|\nabla\boldsymbol{V}\\|_{L^{2}}^{\frac{3}{2}}\\|\Delta\boldsymbol{V}\\|_{L^{2}}^{\frac{1}{2}}+\\|\nabla\boldsymbol{v}\\|_{L^{2}}^{\frac{1}{2}}\\|\Delta\boldsymbol{v}\\|_{L^{2}}^{\frac{1}{2}}\\|\nabla\bar{\boldsymbol{u}}\\|_{L^{2}}+\\|\nabla\bar{\boldsymbol{u}}\\|_{L^{2}}^{\frac{1}{2}}\\|\Delta\bar{\boldsymbol{u}}\\|_{L^{2}}^{\frac{1}{2}}\\|\nabla\boldsymbol{v}\\|_{L^{2}},$

which together with Young’s inequality leads to

	$\displaystyle\bigl(\\|\boldsymbol{F}\\|_{L^{2}}+$	$\displaystyle\\|\boldsymbol{G}\\|_{L^{2}}\bigr)\\|\Delta\boldsymbol{v}\\|_{L^{2}}$
		$\displaystyle\leq\frac{1}{2}\\|\Delta\boldsymbol{v}\\|_{L^{2}}^{2}+C\\|\nabla\boldsymbol{V}\\|_{L^{2}}^{3}\\|\Delta\boldsymbol{V}\\|_{L^{2}}+C\bigl(\\|\nabla\bar{\boldsymbol{u}}\\|_{L^{2}}^{4}+\\|\nabla\bar{\boldsymbol{u}}\\|_{L^{2}}\\|\Delta\bar{\boldsymbol{u}}\\|_{L^{2}}\bigr)\\|\nabla\boldsymbol{v}\\|_{L^{2}}^{2}.$

By substituting this into (3.26), we infer

\displaystyle\frac{d}{dt}\|\nabla\boldsymbol{v}\|_{L^{2}}^{2}+\|\Delta\boldsymbol{v}\|_{L^{2}}^{2}\lesssim\|\nabla\boldsymbol{V}\|_{L^{2}}^{3}\|\Delta\boldsymbol{V}\|_{L^{2}}+\bigl(\|\nabla\bar{\boldsymbol{u}}\|_{L^{2}}^{4}+\|\nabla\bar{\boldsymbol{u}}\|_{L^{2}}\|\Delta\bar{\boldsymbol{u}}\|_{L^{2}}\bigr)\|\nabla\boldsymbol{v}\|_{L^{2}}^{2},

Then by applying Gronwall’s inequality and using the fact that $\boldsymbol{v}|_{t=0}=0$ , we deduce

	$\displaystyle\\|\nabla\boldsymbol{v}\\|_{L^{\infty}_{t}(L^{2})}^{2}+\\|\Delta\boldsymbol{v}\\|_{L^{2}_{t}(L^{2})}^{2}\leq$	$\displaystyle C\\|\nabla\boldsymbol{V}\\|_{L^{\infty}_{t}(L^{2})}^{2}\\|\nabla\boldsymbol{V}\\|_{L^{2}_{t}(L^{2})}\\|\Delta\boldsymbol{V}\\|_{L^{2}_{t}(L^{2})}$
		$\displaystyle\times\exp\bigl(C\\|\nabla\bar{\boldsymbol{u}}\\|_{L^{\infty}_{t}(L^{2})}^{2}\\|\nabla\bar{\boldsymbol{u}}\\|_{L^{2}_{t}(L^{2})}^{2}+C\\|\nabla\bar{\boldsymbol{u}}\\|_{L^{2}_{t}(L^{2})}\\|\Delta\bar{\boldsymbol{u}}\\|_{L^{2}_{t}(L^{2})}\bigr).$

Notice that $\boldsymbol{V}=\boldsymbol{u}-\bar{\boldsymbol{u}}$ , then we can use (3.10), (3.11) and (3.14) to obtain

(3.27)

\|\nabla\boldsymbol{v}\|_{L^{\infty}_{t}(L^{2})}^{2}+\|\Delta\boldsymbol{v}\|_{L^{2}_{t}(L^{2})}^{2}\lesssim\|\boldsymbol{\mathfrak{U}}_{0}\|_{L^{2}}\|\boldsymbol{\mathfrak{U}}_{0}\|_{\dot{H}^{1}_{\rm axi}}^{3}\exp\bigl(\exp(C\|{\boldsymbol{u}}_{0}\|_{L^{2}}^{6}\|{\boldsymbol{u}}_{0}\|_{\dot{H}^{2}}^{2})\bigr).

This together with the smallness assumption (1.10) gives

\|\nabla\boldsymbol{v}\|_{L^{\infty}_{t}(L^{2})}^{2}+\|\Delta\boldsymbol{v}\|_{L^{2}_{t}(L^{2})}^{2}\lesssim\epsilon\|\boldsymbol{\mathfrak{U}}_{0}\|_{\dot{H}^{1}_{\rm axi}}^{2}\exp\bigl(\exp(C\|{\boldsymbol{u}}_{0}\|_{L^{2}}^{6}\|{\boldsymbol{u}}_{0}\|_{\dot{H}^{2}}^{2})\bigr),

which is exactly the second inequality of (1.12).

Till now, we have completed the proof of Theorem 1.1. ∎

4. The proof of Theorem 1.2

As we have mentioned in Subsection 1.2, for any initial data given by (1.13), Theorem 1.1 guarantees the existence of some positive constant $\varepsilon_{0}$ such that for any $\varepsilon\in(0,\varepsilon_{0})$ , (1.1) has a unique global solution $\boldsymbol{u}\in C(\mathop{\mathbb{R}\kern 0.0pt}\nolimits_{+};H^{1})\cap L^{2}(\mathop{\mathbb{R}\kern 0.0pt}\nolimits_{+};\dot{H}^{1}\cap\dot{H}^{2})$ .

By virtue of the classical result that any strong solution to N-S will become analytic at positive time, hence we can expand this solution into Fourier series in $\theta$ variable as

(4.1)

\left\{\begin{split}u^{r}(t,x)=&\bigl(u^{r}_{(0),0}+\varepsilon u^{r}_{(1),0}\bigr)(t,r,z)+\sum_{k=1}^{\infty}\bigl(u^{r}_{(0),k}+\varepsilon u^{r}_{(1),k}\bigr)(t,r,z)\cos k\theta\\ &+\sum_{k=1}^{\infty}\bigl(v^{r}_{(0),k}+\varepsilon v^{r}_{(1),k}\bigr)(t,r,z)\sin k\theta+{\mathcal{O}}(\varepsilon^{2}),\\ u^{\theta}(t,x)=&\bigl(u^{\theta}_{(0),0}+\varepsilon u^{\theta}_{(1),0}\bigr)(t,r,z)+\sum_{k=1}^{\infty}\bigl(u^{\theta}_{(0),k}+\varepsilon u^{\theta}_{(1),k}\bigr)(t,r,z)\cos k\theta\\ &+\sum_{k=1}^{\infty}\bigl(v^{\theta}_{(0),k}+\varepsilon v^{\theta}_{(1),k}\bigr)(t,r,z)\sin k\theta+{\mathcal{O}}(\varepsilon^{2}),\\ u^{z}(t,x)=&\bigl(u^{z}_{(0),0}+\varepsilon u^{z}_{(1),0}\bigr)(t,r,z)+\sum_{k=1}^{\infty}\bigl(u^{z}_{(0),k}+\varepsilon u^{z}_{(1),k}\bigr)(t,r,z)\cos k\theta\\ &+\sum_{k=1}^{\infty}\bigl(v^{z}_{(0),k}+\varepsilon v^{z}_{(1),k}\bigr)(t,r,z)\sin k\theta+{\mathcal{O}}(\varepsilon^{2}).\end{split}\right.

Correspondingly, we can expand the pressure $P=(-\Delta)^{-1}\mathop{\rm div}\nolimits\mathop{\rm div}\nolimits(\boldsymbol{u}\otimes\boldsymbol{u})$ into

(4.2)

\begin{split}P(t,x)=&\bigl(P_{(0),0}+\varepsilon P_{(1),0}\bigr)(t,r,z)+\sum_{k=1}^{\infty}\bigl(P_{(0),k}+\varepsilon P_{(1),k}\bigr)(t,r,z)\cos k\theta\\ &+\sum_{k=1}^{\infty}\bigl(Q_{(0),k}+\varepsilon Q_{(1),k}\bigr)(t,r,z)\sin k\theta+{\mathcal{O}}(\varepsilon^{2}).\end{split}

In the following, we shall give the explicit formulas for the profiles in the expansion (4.1), and verify the validity of this expansion in $L^{\infty}(\mathop{\mathbb{R}\kern 0.0pt}\nolimits^{+};L^{2}\cap L^{\infty})$ sense. For notation simplification, we shall use $\boldsymbol{u}_{(i)}$ to denote the $\varepsilon^{i}$ order term in the expansion of $\boldsymbol{u}$ , i.e.

(4.3)

\boldsymbol{u}=\boldsymbol{u}_{(0)}+\varepsilon\boldsymbol{u}_{(1)}+{\mathcal{O}}(\varepsilon^{2}),\quad\boldsymbol{u}_{(i)}=\boldsymbol{u}_{(i),0}+\sum_{k=1}^{\infty}\Bigl(\boldsymbol{u}_{(i),k}\cos k\theta+\boldsymbol{v}_{(i),k}\sin k\theta\Bigr),

and $\boldsymbol{u}_{(i)}$ does not rely on $\varepsilon$ .

4.1. The $\varepsilon^{0}$ order terms

We make the following Ansatz $1$ : $\boldsymbol{u}_{(0)}=u^{r}_{(0),0}\boldsymbol{e}_{r}+u^{z}_{(0),0}\boldsymbol{e}_{z}$ , where $u^{r}_{(0),0}$ and $u^{z}_{(0),0}$ satisfy the following axisymmetric without swirl N-S:

(4.4)

\left\{\begin{split}&\partial_{t}u^{r}_{(0),0}+(u^{r}_{(0),0}\partial_{r}+u^{z}_{(0),0}\partial_{z})u^{r}_{(0),0}-(\partial_{r}^{2}+\partial_{z}^{2}+\frac{1}{r}\partial_{r}-\frac{1}{r^{2}})u^{r}_{(0),0}+\partial_{r}P_{(0),0}=0,\\ &\partial_{t}u^{z}_{(0),0}+(u^{r}_{(0),0}\partial_{r}+u^{z}_{(0),0}\partial_{z})u^{z}_{(0),0}-(\partial_{r}^{2}+\partial_{z}^{2}+\frac{1}{r}\partial_{r})u^{z}_{(0),0}+\partial_{z}P_{(0),0}=0,\\ &\partial_{r}u^{r}_{(0),0}+\frac{1}{r}u^{r}_{(0),0}+\partial_{z}u^{z}_{(0),0}=0,\\ &u^{r}_{(0),0}|_{t=0}=a^{r}_{0},\quad u^{z}_{(0),0}|_{t=0}=a^{z}_{0}.\end{split}\right.

To verify this ansatz, we first deduce from Proposition 3.1 that

(4.5)

\|\boldsymbol{u}_{(0)}\|_{L^{\infty}(\mathop{\mathbb{R}\kern 0.0pt}\nolimits^{+};H^{1})}^{2}+\|\nabla\boldsymbol{u}_{(0)}\|_{L^{2}(\mathop{\mathbb{R}\kern 0.0pt}\nolimits^{+};H^{1})}^{2}\leq C_{0}.

Here and in all that follows, $C_{0}$ denotes some positive constant depending only on the norms of the profiles of $\boldsymbol{u}(0,x)$ appearing in (1.14), and may be different in each appearance.

On the other hand, in view of the expression (1.13) for initial data, we have

\displaystyle\boldsymbol{u}|_{t=0}-\boldsymbol{u}_{(0)}|_{t=0}=\varepsilon a^{\theta}_{0}(r,z)\boldsymbol{e}_{\theta}+\varepsilon\sum_{k=1}^{\infty}\Bigl(\boldsymbol{a}_{k}(r,z)\cos(k\theta)+\boldsymbol{b}_{k}(r,z)\sin(k\theta)\Bigr).

By using Parseval’s identities, there holds

(4.6)

\bigl\|\boldsymbol{u}|_{t=0}-\boldsymbol{u}_{(0)}|_{t=0}\bigr\|_{H^{1}}\leq C_{0}\varepsilon.

Then for sufficiently small $\varepsilon>0$ , by virtue of (4.5), (4.6) and Proposition 2.1, we obtain

(4.7)

\|\boldsymbol{u}-\boldsymbol{u}_{(0)}\|_{L^{\infty}(\mathop{\mathbb{R}\kern 0.0pt}\nolimits^{+};H^{1}\cap L^{\infty})}+\|\nabla(\boldsymbol{u}-\boldsymbol{u}_{(0)})\|_{L^{2}(\mathop{\mathbb{R}\kern 0.0pt}\nolimits^{+};H^{1})}\leq C_{0}\varepsilon.

Notice that the system (4.4) does not rely on $\varepsilon$ , and thus its solution $(u^{r}_{(0),0},u^{z}_{(0),0})$ also does not rely on $\varepsilon$ . This together with the estimate (4.7) guarantees the validity of our Ansatz $1$ , namely $u^{r}_{(0),0}\boldsymbol{e}_{r}+u^{z}_{(0),0}\boldsymbol{e}_{z}$ is indeed the $\varepsilon^{0}$ order term in the expansion of $\boldsymbol{u}$ .

In particular, we have shown that $u^{\theta}_{(0),0}$ vanishes, and for every $k\in\mathop{\mathbb{N}\kern 0.0pt}\nolimits$ , the $k$ -th Fourier coefficient does not contain $\varepsilon^{0}$ order terms, just the same as the initial data (1.13).

4.2. Derivation of the $\varepsilon^{1}$ order terms in the Euclidean coordinates

Let us temporarily go back to the Euclidean coordinates, and consider $\boldsymbol{u}_{(1)}$ as a whole part.

We make the following Ansatz $2$ : $\boldsymbol{u}_{(1)}$ is a solution to the linearization of the perturbed Navier-Stokes system around $\boldsymbol{u}_{(0)}$ , precisely

(4.8)

\left\{\begin{split}&\partial_{t}\boldsymbol{u}_{(1)}+\boldsymbol{u}_{(0)}\cdot\nabla\boldsymbol{u}_{(1)}+\boldsymbol{u}_{(1)}\cdot\nabla\boldsymbol{u}_{(0)}-\Delta\boldsymbol{u}_{(1)}=-\nabla P_{(1)},\\ &\mathop{\rm div}\nolimits\boldsymbol{u}_{(1)}=0,\\ &\boldsymbol{u}_{(1)}|_{t=0}=a^{\theta}_{0}\boldsymbol{e}_{\theta}+\sum_{k=1}^{\infty}\Bigl(\boldsymbol{a}_{k}\cos k\theta+\boldsymbol{b}_{k}\sin k\theta\Bigr).\end{split}\right.

To verify this ansatz, we first get, by taking $L^{2}$ inner product of (4.8) with $\boldsymbol{u}_{(1)}-\Delta\boldsymbol{u}_{(1)}$ that

	$\displaystyle\frac{1}{2}\frac{d}{dt}$	$\displaystyle\\|\boldsymbol{u}_{(1)}\\|_{H^{1}}^{2}+\\|\nabla\boldsymbol{u}_{(1)}\\|_{H^{1}}^{2}\leq\bigl\|\bigl(\boldsymbol{u}_{(0)}\cdot\nabla\boldsymbol{u}_{(1)}+\boldsymbol{u}_{(1)}\cdot\nabla\boldsymbol{u}_{(0)})\,\big\|\,\boldsymbol{u}_{(1)}-\Delta\boldsymbol{u}_{(1)}\bigr)\bigr\|$
		$\displaystyle\leq\\|\boldsymbol{u}_{(1)}\\|_{L^{3}}\\|\nabla\boldsymbol{u}_{(0)}\\|_{L^{2}}\\|\boldsymbol{u}_{(1)}\\|_{L^{6}}+\bigl(\\|\boldsymbol{u}_{(0)}\\|_{L^{6}}\\|\nabla\boldsymbol{u}_{(1)}\\|_{L^{3}}+\\|\boldsymbol{u}_{(1)}\\|_{L^{\infty}}\\|\nabla\boldsymbol{u}_{(0)}\\|_{L^{2}}\bigr)\\|\Delta\boldsymbol{u}_{(1)}\\|_{L^{2}}$
		$\displaystyle\leq\frac{1}{2}\\|\Delta\boldsymbol{u}_{(1)}\\|_{L^{2}}^{2}+C\\|\boldsymbol{u}_{(1)}\\|_{H^{1}}^{2}\\|\nabla\boldsymbol{u}_{(0)}\\|_{L^{2}}^{4}.$

Subtracting $\frac{1}{2}\|\Delta\boldsymbol{u}_{(1)}\|_{L^{2}}^{2}$ on both sides, and then applying Gronwall’s inequality gives

(4.9)

\|\boldsymbol{u}_{(1)}\|_{L^{\infty}_{t}(H^{1})}^{2}+\|\nabla\boldsymbol{u}_{(1)}\|_{L^{2}_{t}(H^{1})}^{2}\leq\|\boldsymbol{u}_{(1)}|_{t=0}\|_{H^{1}}^{2}\exp\bigl(C\|\nabla\boldsymbol{u}_{(0)}\|_{L^{\infty}_{t}(L^{2})}^{2}\|\nabla\boldsymbol{u}_{(0)}\|_{L^{2}_{t}(L^{2})}^{2}\bigr)\leq C_{0}.

While following similar derivation as (2.18), we deduce

(4.10)

\|\nabla\boldsymbol{u}_{(1)}\|_{L^{\infty}_{t}(L^{4})}\lesssim\|\mathop{\rm curl}\nolimits\boldsymbol{u}_{(1)}\|_{L^{\infty}_{t}(L^{4})}\leq C_{0}.

On the other hand, let us consider $\boldsymbol{R}_{(1)}\mathrel{\mathop{\kern 0.0pt=}\limits^{\hbox{\footnotesize def}}}\boldsymbol{u}-\boldsymbol{u}_{(0)}-\varepsilon\boldsymbol{u}_{(1)}$ , which satisfies

(4.11)

\left\{\begin{split}&\partial_{t}\boldsymbol{R}_{(1)}+\boldsymbol{u}\cdot\nabla\boldsymbol{R}_{(1)}+\boldsymbol{R}_{(1)}\cdot\nabla(\boldsymbol{u}_{(0)}+\varepsilon\boldsymbol{u}_{(1)})+\varepsilon^{2}\boldsymbol{u}_{(1)}\cdot\nabla\boldsymbol{u}_{(1)}-\Delta\boldsymbol{R}_{(1)}=-\nabla\Pi_{(1)},\\ &\mathop{\rm div}\nolimits\boldsymbol{R}_{(1)}=0,\\ &\boldsymbol{R}_{(1)}|_{t=0}=0,\end{split}\right.

where $\Pi_{(1)}$ can be obtained by taking divergence operator to the first equation in (4.11).

In the following, we shall derive $L^{2},~\dot{H}^{1}$ and $\dot{W}^{1,4}$ estimate for $\boldsymbol{R}_{(1)}$ . Notice that if there is no external force term $\varepsilon^{2}\boldsymbol{u}_{(1)}\cdot\nabla\boldsymbol{u}_{(1)}$ in (4.11), then this type of system has already been studied in Proposition 2.1. So here we only focus on the estimate for this external force term.

Firstly, we have

	$\displaystyle\Bigl\|\int_{\mathop{\mathbb{R}\kern 0.0pt}\nolimits^{3}}\varepsilon^{2}\bigl(\boldsymbol{u}_{(1)}\cdot\nabla\boldsymbol{u}_{(1)}\bigr)\boldsymbol{R}_{(1)}\,dx\Bigr\|$	$\displaystyle\leq C\varepsilon^{2}\\|\boldsymbol{u}_{(1)}\\|_{L^{3}}\\|\nabla\boldsymbol{u}_{(1)}\\|_{L^{2}}\\|\boldsymbol{R}_{(1)}\\|_{L^{6}}$
		$\displaystyle\leq\frac{1}{2}\\|\nabla\boldsymbol{R}_{(1)}\\|_{L^{2}}^{2}+C\varepsilon^{4}\\|\nabla\boldsymbol{u}_{(1)}\\|_{L^{2}}^{3}\\|\boldsymbol{u}_{(1)}\\|_{L^{2}},$

and in view of the bound (4.9), there holds

\varepsilon^{4}\int_{0}^{t}\|\nabla\boldsymbol{u}_{(1)}\|_{L^{2}}^{3}\|\boldsymbol{u}_{(1)}\|_{L^{2}}\,dt^{\prime}\leq\varepsilon^{4}\|\nabla\boldsymbol{u}_{(1)}\|_{L^{2}_{t}(L^{2})}^{2}\|\nabla\boldsymbol{u}_{(1)}\|_{L^{\infty}_{t}(L^{2})}\|\boldsymbol{u}_{(1)}\|_{L^{\infty}_{t}(L^{2})}\leq C_{0}\varepsilon^{4}.

Then a similar procedure as the proof of Proposition 2.1 leads to

(4.12)

\|\boldsymbol{R}_{(1)}\|_{L^{\infty}_{t}(L^{2})}^{2}+\|\nabla\boldsymbol{R}_{(1)}\|_{L^{2}_{t}(L^{2})}^{2}\leq C_{0}\varepsilon^{4}.

Similarly, we have

	$\displaystyle\Bigl\|\int_{\mathop{\mathbb{R}\kern 0.0pt}\nolimits^{3}}\varepsilon^{2}\bigl(\boldsymbol{u}_{(1)}\cdot\nabla\boldsymbol{u}_{(1)}\bigr)\cdot\Delta\boldsymbol{R}_{(1)}\,dx\Bigr\|$	$\displaystyle\leq\varepsilon^{2}\\|\boldsymbol{u}_{(1)}\\|_{L^{\infty}}\\|\nabla\boldsymbol{u}_{(1)}\\|_{L^{2}}\\|\Delta\boldsymbol{R}_{(1)}\\|_{L^{2}}$
		$\displaystyle\leq\frac{1}{2}\\|\Delta\boldsymbol{R}_{(1)}\\|_{L^{2}}^{2}+\varepsilon^{4}\\|\nabla\boldsymbol{u}_{(1)}\\|_{L^{2}}^{3}\\|\nabla\boldsymbol{u}_{(1)}\\|_{\dot{H}^{1}},$

and

		$\displaystyle\Bigl\|\int_{\mathop{\mathbb{R}\kern 0.0pt}\nolimits^{3}}\varepsilon^{2}\mathop{\rm curl}\nolimits\bigl(\boldsymbol{u}_{(1)}\cdot\nabla\boldsymbol{u}_{(1)}\bigr)\cdot\|\mathop{\rm curl}\nolimits\boldsymbol{R}_{(1)}\|^{2}\mathop{\rm curl}\nolimits\boldsymbol{R}_{(1)}\,dx\Bigr\|$
		$\displaystyle\leq\varepsilon^{2}\\|\boldsymbol{u}_{(1)}\\|_{L^{6}}\\|\nabla\boldsymbol{u}_{(1)}\\|_{L^{4}}\\|\mathop{\rm curl}\nolimits\boldsymbol{R}_{(1)}\\|_{L^{12}}\bigl(\bigl\\|\nabla\|\mathop{\rm curl}\nolimits\boldsymbol{R}_{(1)}\|^{2}\bigr\\|_{L^{2}}+\bigl\\|\|\mathop{\rm curl}\nolimits\boldsymbol{R}_{(1)}\|\cdot\nabla\mathop{\rm curl}\nolimits\boldsymbol{R}_{(1)}\bigr\\|_{L^{2}}\bigr)$
		$\displaystyle\leq\frac{1}{4}\bigl(\bigl\\|\nabla\|\mathop{\rm curl}\nolimits\boldsymbol{R}_{(1)}\|^{2}\bigr\\|_{L^{2}}^{2}+\bigl\\|\|\mathop{\rm curl}\nolimits\boldsymbol{R}_{(1)}\|\cdot\nabla\mathop{\rm curl}\nolimits\boldsymbol{R}_{(1)}\bigr\\|_{L^{2}}^{2}\bigr)+\varepsilon^{8}\\|\nabla\boldsymbol{u}_{(1)}\\|_{L^{2}}^{4}\\|\nabla\boldsymbol{u}_{(1)}\\|_{L^{4}}^{4}.$

In view of the bounds (4.9) and (4.10), there holds

\varepsilon^{4}\int_{0}^{t}\|\nabla\boldsymbol{u}_{(1)}\|_{L^{2}}^{3}\|\nabla\boldsymbol{u}_{(1)}\|_{\dot{H}^{1}}\,dt^{\prime}\leq C_{0}\varepsilon^{4},\quad\hbox{and}\quad\varepsilon^{8}\int_{0}^{t}\|\nabla\boldsymbol{u}_{(1)}\|_{L^{2}}^{4}\|\nabla\boldsymbol{u}_{(1)}\|_{L^{4}}^{4}\,dt^{\prime}\leq C_{0}\varepsilon^{8}.

Then a similar procedure as the proof of Proposition 2.1 leads to

(4.13)

\|\boldsymbol{R}_{(1)}\|_{L^{\infty}_{t}(\dot{H}^{1})}^{2}+\|\nabla\boldsymbol{R}_{(1)}\|_{L^{2}_{t}(\dot{H}^{1})}^{2}\leq C_{0}\varepsilon^{4},\quad\hbox{and}\quad\|\nabla\boldsymbol{R}_{(1)}\|_{L^{\infty}_{t}(L^{4})}^{4}\leq C_{0}\varepsilon^{8}.

Now in view of the estimates (4.12) and (4.13), together with the interpolation inequality $\|f\|_{L^{\infty}}\lesssim\|\nabla f\|_{L^{2}}^{\frac{1}{3}}\|\nabla f\|_{L^{4}}^{\frac{2}{3}}$ , we achieve

(4.14)

\|\boldsymbol{R}_{(1)}\|_{L^{\infty}_{t}(H^{1}\cap L^{\infty})}^{2}+\|\nabla\boldsymbol{R}_{(1)}\|_{L^{2}_{t}(H^{1})}^{2}\leq C_{0}\varepsilon^{4}.

In view of the estimate (4.14) for the remainder $\boldsymbol{R}_{(1)}$ , and the fact that $\boldsymbol{u}_{(1)}$ determined by (4.8) does not rely on $\varepsilon$ , we have verified our Ansatz $2$ .

Remark 4.1.

Exactly along the same line, we can prove by induction that for any $n\geq 2$ , $\boldsymbol{u}_{(n)}$ is a solution to the following linearized system:

\left\{\begin{split}&\partial_{t}\boldsymbol{u}_{(n)}+\boldsymbol{u}_{(0)}\cdot\nabla\boldsymbol{u}_{(n)}+\boldsymbol{u}_{(n)}\cdot\nabla\boldsymbol{u}_{(0)}-\Delta\boldsymbol{u}_{(n)}+\nabla P_{(n)}=-\sum_{i=1}^{n-1}\boldsymbol{u}_{(i)}\cdot\nabla\boldsymbol{u}_{(n-i)},\\ &\mathop{\rm div}\nolimits\boldsymbol{u}_{(n)}=0,\\ &\boldsymbol{u}_{(n)}|_{t=0}=0.\end{split}\right.

And the remainder term $\boldsymbol{R}_{(n)}\mathrel{\mathop{\kern 0.0pt=}\limits^{\hbox{\footnotesize def}}}\boldsymbol{u}-\boldsymbol{u}_{(0)}-\varepsilon\boldsymbol{u}_{(1)}-\cdots-\varepsilon^{n}\boldsymbol{u}_{(n)}$ satisfies

\|\boldsymbol{R}_{(n)}\|_{L^{\infty}_{t}(H^{1}\cap L^{\infty})}\leq C_{0}\varepsilon^{n+1}.

4.3. Verification of $u^{r}_{(1),0}=u^{z}_{(1),0}=0$

Noticing that $\boldsymbol{u}_{(0)}=u^{r}_{(0),0}\boldsymbol{e}_{r}+u^{z}_{(0),0}\boldsymbol{e}_{z}$ is axisymmetric without swirl, it is not difficult to deduce from (4.8) that $\bigl(u^{r}_{(1),0},u^{z}_{(1),0}\bigr)$ satisfies

(4.15)

\left\{\begin{aligned} &\partial_{t}u^{r}_{(1),0}+\boldsymbol{u}_{(0)}\cdot\nabla u^{r}_{(1),0}+\widetilde{\boldsymbol{u}}_{(1),0}\cdot\widetilde{\nabla}u^{r}_{(0),0}-(\Delta-\frac{1}{r^{2}})u^{r}_{(1),0}+\partial_{r}P_{(1),0}=0,\\ &\partial_{t}u^{z}_{(1),0}+\boldsymbol{u}_{(0)}\cdot\nabla u^{z}_{(1),0}+\widetilde{\boldsymbol{u}}_{(1),0}\cdot\widetilde{\nabla}u^{z}_{(0),0}-\Delta u^{z}_{(1),0}+\partial_{z}P_{(1),0}=0,\\ &\partial_{r}u^{r}_{(1),0}+\frac{1}{r}u^{r}_{(1),0}+\partial_{z}u^{z}_{(1),0}=0,\\ &u^{r}_{(1),0}|_{t=0}=u^{z}_{(1),0}|_{t=0}=0,\end{aligned}\right.

where $\widetilde{\boldsymbol{u}}_{(1),0}\mathrel{\mathop{\kern 0.0pt=}\limits^{\hbox{\footnotesize def}}}u^{r}_{(1),0}\boldsymbol{e}_{r}+u^{z}_{(1),0}\boldsymbol{e}_{z}$ , and the divergence-free condition here follows from

\displaystyle\begin{aligned} \mathop{\rm div}\nolimits\boldsymbol{u}_{(1),0}=&\partial_{r}u^{r}_{(1),0}+\frac{u^{r}_{(1),0}}{r}+\partial_{z}u^{z}_{(1),0}\\ =&{\mathcal{M}}\Bigl(\partial_{r}u^{r}_{(1)}+\frac{u^{r}_{(1)}}{r}+\frac{\partial_{\theta}u^{\theta}_{(1)}}{r}+\partial_{z}u^{z}_{(1)}\Bigr)={\mathcal{M}}(\mathop{\rm div}\nolimits\boldsymbol{u}_{(1)})=0.\end{aligned}

By taking $L^{2}$ inner product of (4.15) with $\bigl(u^{r}_{(1),0},u^{z}_{(1),0}\bigr)$ , we get

(4.16)

\frac{1}{2}\frac{d}{dt}\|\widetilde{\boldsymbol{u}}_{(1),0}\|_{L^{2}}^{2}+\|\nabla\widetilde{\boldsymbol{u}}_{(1),0}\|_{L^{2}}^{2}=-\bigl(\widetilde{\boldsymbol{u}}_{(1),0}\cdot\widetilde{\nabla}u^{r}_{(0),0}\,|\,u^{r}_{(1),0}\bigr)-\bigl(\widetilde{\boldsymbol{u}}_{(1),0}\cdot\widetilde{\nabla}u^{z}_{(0),0}\,|\,u^{z}_{(1),0}\bigr),

where we used the fact that $\frac{\partial_{\theta}}{r}P_{(1),0}(t,r,z)=0$ and then the following equality

\displaystyle-\bigl(\widetilde{\nabla}P_{(1),0}\,|\,\widetilde{\boldsymbol{u}}_{(1),0}\bigr)=-\bigl(\nabla P_{(1),0}\,|\,\boldsymbol{u}_{(1),0}\bigr)=\bigl(P_{(1),0}\,|\,\mathop{\rm div}\nolimits\boldsymbol{u}_{(1),0}\bigr)=0.

For the terms on the right-hand side of (4.16), we have

\displaystyle\begin{aligned} \bigl|\bigl(\widetilde{\boldsymbol{u}}_{(1),0}\cdot\nabla u^{r}_{(0),0}\,|\,u^{r}_{(1),0}\bigr)\bigr|+\bigl|\bigl(\widetilde{\boldsymbol{u}}_{(1),0}\cdot&\nabla u^{z}_{(0),0}\,|\,u^{z}_{(1),0}\bigr)\bigr|\leq C\|\nabla\boldsymbol{u}_{(0)}\|_{L^{2}}\|\widetilde{\boldsymbol{u}}_{(1),0}\|_{L^{4}}^{2}\\ &\leq C\|\nabla\boldsymbol{u}_{(0)}\|_{L^{2}}\|\widetilde{\boldsymbol{u}}_{(1),0}\|_{L^{2}}^{\frac{1}{2}}\|\nabla\widetilde{\boldsymbol{u}}_{(1),0}\|_{L^{2}}^{\frac{3}{2}}\\ &\leq\frac{1}{2}\|\nabla\widetilde{\boldsymbol{u}}_{(1),0}\|_{L^{2}}^{2}+C\|\widetilde{\boldsymbol{u}}_{(1),0}\|_{L^{2}}^{2}\|\nabla\boldsymbol{u}_{(0)}\|_{L^{2}}^{4}.\end{aligned}

As a result, we deduce

\frac{d}{dt}\|\widetilde{\boldsymbol{u}}_{(1),0}\|_{L^{2}}^{2}+\|\nabla\widetilde{\boldsymbol{u}}_{(1),0}\|_{L^{2}}^{2}\leq C\|\widetilde{\boldsymbol{u}}_{(1),0}\|_{L^{2}}^{2}\|\nabla\boldsymbol{u}_{(0)}\|_{L^{2}}^{4}.

Then by applying Gronwall’s inequality, together with the fact that initially $\widetilde{\boldsymbol{u}}_{(1),0}|_{t=0}=0$ , we deduce that $\widetilde{\boldsymbol{u}}_{(1),0}$ actually vanishes for all time. This completes the proof of Theorem 1.2.

5. The proof of Theorem 1.3

As the initial data (1.19) considered here is a special case of (1.13), Theorem 1.1 guarantees the existence of a unique global solution $\boldsymbol{u}\in C(\mathop{\mathbb{R}\kern 0.0pt}\nolimits_{+};H^{1})\cap L^{2}(\mathop{\mathbb{R}\kern 0.0pt}\nolimits_{+};\dot{H}^{1}\cap\dot{H}^{2})$ for sufficiently small $\varepsilon$ . And this solution can be expanded as (1.18). Hence the aim in the following is to show that $u^{r}$ and $u^{z}$ remain even in $\theta$ , while $u^{\theta}$ keeps odd in $\theta$ for all time, namely

\boldsymbol{u}_{(1),0}=0,\quad u^{\theta}_{(j),0}=0,\quad v^{r}_{(j),k}=u^{\theta}_{(j),k}=v^{z}_{(j),k}=0,\quad\forall\ j\in\mathop{\mathbb{N}\kern 0.0pt}\nolimits,~k\in\mathop{\mathbb{N}\kern 0.0pt}\nolimits.

5.1. Verification of $\boldsymbol{u}_{(1),0}=\boldsymbol{0}$

For the initial data $\boldsymbol{u}(0,x)$ given by (1.19), we have

\displaystyle{\mathcal{M}}(\boldsymbol{u}|_{t=0})=a^{r}_{0}(r,z)\boldsymbol{e}_{r}+a^{z}_{0}(r,z)\boldsymbol{e}_{z}.

Thus the quantity $\bar{\boldsymbol{u}}$ defined by (1.6), is exactly $u^{r}_{(0),0}\boldsymbol{e}_{r}+u^{z}_{(0),0}\boldsymbol{e}_{z}$ , which sloves the axisymmetric without swirl N-S (4.4) for the case here.

On the other hand, noticing that $u^{\theta}|_{t=0}\thicksim\varepsilon$ and

\partial_{\theta}u^{r}|_{t=0}=-\varepsilon\sum_{k=1}^{\infty}ka^{r}_{k}\sin k\theta,\quad\partial_{\theta}u^{\theta}|_{t=0}=\varepsilon\sum_{k=1}^{\infty}kb^{\theta}_{k}\cos k\theta,\quad\partial_{\theta}u^{z}|_{t=0}=-\varepsilon\sum_{k=1}^{\infty}ka^{z}_{k}\sin k\theta,

we can get by using the refined estimate (1.12) in Theorem 1.1 that

\|{\mathcal{M}}(\boldsymbol{u})-u^{r}_{(0),0}\boldsymbol{e}_{r}-u^{z}_{(0),0}\boldsymbol{e}_{z}\|_{L^{\infty}(\mathop{\mathbb{R}\kern 0.0pt}\nolimits^{+};L^{2})}\leq C_{0}\varepsilon^{2}.

This together with the fact that ${\mathcal{M}}(\boldsymbol{u})=\boldsymbol{u}_{(0),0}+\varepsilon\boldsymbol{u}_{(1),0}+\varepsilon^{2}\boldsymbol{u}_{(2),0}+\cdots$ implies

(5.1)

\boldsymbol{u}_{(0),0}=u^{r}_{(0),0}\boldsymbol{e}_{r}+u^{z}_{(0),0}\boldsymbol{e}_{z},\quad\hbox{and}\quad\boldsymbol{u}_{(1),0}=0.

5.2. Verification of $v^{r}_{(1),k}=u^{\theta}_{(1),k}=v^{z}_{(1),k}=0$

In view of (5.1), we can write

(5.2)

\boldsymbol{u}_{(1)}=\sum_{k=1}^{\infty}\Bigl(\boldsymbol{u}_{(1),k}\cos k\theta+\boldsymbol{v}_{(1),k}\sin k\theta\Bigr),

and the corresponding pressure

P_{(1)}=\sum_{k=1}^{\infty}\Bigl(P_{(1),k}\cos k\theta+Q_{(1),k}\sin k\theta\Bigr).

And from the proof of Subsection 4.2, we know that $(\boldsymbol{u}_{(1)},P_{(1)})$ satisfies (4.8).

Then it is crucial to notice that $\boldsymbol{u}_{(0)}$ is axisymmetric without swirl, which makes the couplings between each component of $\boldsymbol{u}_{(1)}$ in $(4.8)$ not so strong. In fact, by rewriting (4.8) according to the basis $\{\cos k\theta,\sin k\theta:k\in\mathop{\mathbb{N}\kern 0.0pt}\nolimits\}$ , it is not difficult to find that we can decompose $\boldsymbol{u}_{(1)}$ and $P_{(1)}$ into two parts: $\boldsymbol{u}_{(1)}^{\prime}+\boldsymbol{u}_{(1)}^{\prime\prime}$ and $P_{(1)}^{\prime}+P_{(1)}^{\prime\prime}$ respectively, where

\boldsymbol{u}_{(1)}^{\prime}\mathrel{\mathop{\kern 0.0pt=}\limits^{\hbox{\footnotesize def}}}\sum_{k=1}^{\infty}\Bigl(u^{r}_{(1),k}\cos k\theta\boldsymbol{e}_{r}+v^{\theta}_{(1),k}\sin k\theta\boldsymbol{e}_{\theta}+u^{z}_{(1),k}\cos k\theta\boldsymbol{e}_{z}\Bigr),\quad P_{(1)}^{\prime}\mathrel{\mathop{\kern 0.0pt=}\limits^{\hbox{\footnotesize def}}}\sum_{k=1}^{\infty}P_{(1),k}\cos k\theta,

and

\boldsymbol{u}_{(1)}^{\prime\prime}\mathrel{\mathop{\kern 0.0pt=}\limits^{\hbox{\footnotesize def}}}\sum_{k=1}^{\infty}\Bigl(v^{r}_{(1),k}\sin k\theta\boldsymbol{e}_{r}+u^{\theta}_{(1),k}\cos k\theta\boldsymbol{e}_{\theta}+v^{z}_{(1),k}\sin k\theta\boldsymbol{e}_{z}\Bigr),\quad P_{(1)}^{\prime\prime}\mathrel{\mathop{\kern 0.0pt=}\limits^{\hbox{\footnotesize def}}}\sum_{k=1}^{\infty}Q_{(1),k}\sin k\theta,

such that $\boldsymbol{u}_{(1)}^{\prime}$ and $P_{(1)}^{\prime}$ satisfy the following self-contained system:

(5.3)

\left\{\begin{aligned} &\partial_{t}u^{r}_{(1),k}+(u^{r}_{(0),0}\partial_{r}+u^{z}_{(0),0}\partial_{z})u^{r}_{(1),k}+(u^{r}_{(1),k}\partial_{r}+u^{z}_{(1),k}\partial_{z})u^{r}_{(0),0}\\ &\qquad\qquad\qquad-(\partial_{r}^{2}+\partial_{z}^{2}+\frac{1}{r}\partial_{r}-\frac{1+k^{2}}{r^{2}})u^{r}_{(1),k}+2k\frac{v^{\theta}_{(1),k}}{r^{2}}+\partial_{r}P_{(1),k}=0,\\ &\partial_{t}v^{\theta}_{(1),k}+(u^{r}_{(0),0}\partial_{r}+u^{z}_{(0),0}\partial_{z})v^{\theta}_{(1),k}+\frac{u^{r}_{(0),0}v^{\theta}_{(1),k}}{r}\\ &\qquad\qquad\qquad-(\partial_{r}^{2}+\partial_{z}^{2}+\frac{1}{r}\partial_{r}-\frac{1+k^{2}}{r^{2}})v^{\theta}_{(1),k}+2k\frac{u^{r}_{(1),k}}{r^{2}}-\frac{k}{r}P_{(1),k}=0,\\ &\partial_{t}u^{z}_{(1),k}+(u^{r}_{(0),0}\partial_{r}+u^{z}_{(0),0}\partial_{z})u^{z}_{(1),k}+(u^{r}_{(1),k}\partial_{r}+u^{z}_{(1),k}\partial_{z})u^{z}_{(0),0}\\ &\qquad\qquad\qquad-(\partial_{r}^{2}+\partial_{z}^{2}+\frac{1}{r}\partial_{r}-\frac{k^{2}}{r^{2}})u^{z}_{(1),k}+\partial_{z}P_{(1),k}=0,\\ &\partial_{r}u^{r}_{(1),k}+\frac{1}{r}u^{r}_{(1),k}+\partial_{z}u^{z}_{(1),k}+\frac{k}{r}v^{\theta}_{(1),k}=0,\\ &u^{r}_{(1),k}|_{t=0}=a^{r}_{k},\quad v^{\theta}_{(1),k}|_{t=0}=b^{\theta}_{k},\quad u^{z}_{(1),k}|_{t=0}=a^{z}_{k},\end{aligned}\right.

while $\boldsymbol{u}_{(1)}^{\prime\prime}$ and $P_{(1)}^{\prime\prime}$ satisfy the following self-contained system:

(5.4)

\left\{\begin{aligned} &\partial_{t}v^{r}_{(1),k}+(u^{r}_{(0),0}\partial_{r}+u^{z}_{(0),0}\partial_{z})v^{r}_{(1),k}+(v^{r}_{(1),k}\partial_{r}+v^{z}_{(1),k}\partial_{z})u^{r}_{(0),0}\\ &\qquad\qquad\qquad-(\partial_{r}^{2}+\partial_{z}^{2}+\frac{1}{r}\partial_{r}-\frac{1+k^{2}}{r^{2}})v^{r}_{(1),k}-2k\frac{u^{\theta}_{(1),k}}{r^{2}}+\partial_{r}Q_{(1),k}=0,\\ &\partial_{t}u^{\theta}_{(1),k}+(u^{r}_{(0),0}\partial_{r}+u^{z}_{(0),0}\partial_{z})u^{\theta}_{(1),k}+\frac{u^{r}_{(0),0}u^{\theta}_{(1),k}}{r}\\ &\qquad\qquad\qquad-(\partial_{r}^{2}+\partial_{z}^{2}+\frac{1}{r}\partial_{r}-\frac{1+k^{2}}{r^{2}})u^{\theta}_{(1),k}-2k\frac{v^{r}_{(1),k}}{r^{2}}+\frac{k}{r}Q_{(1),k}=0,\\ &\partial_{t}v^{z}_{(1),k}+(u^{r}_{(0),0}\partial_{r}+u^{z}_{(0),0}\partial_{z})v^{z}_{(1),k}+(v^{r}_{(1),k}\partial_{r}+v^{z}_{(1),k}\partial_{z})u^{z}_{(0),0}\\ &\qquad\qquad\qquad-(\partial_{r}^{2}+\partial_{z}^{2}+\frac{1}{r}\partial_{r}-\frac{k^{2}}{r^{2}})v^{z}_{(1),k}+\partial_{z}Q_{(1),k}=0,\\ &\partial_{r}v^{r}_{(1),k}+\frac{1}{r}v^{r}_{(1),k}+\partial_{z}v^{z}_{(1),k}-\frac{k}{r}u^{\theta}_{(1),k}=0,\\ &v^{r}_{(1),k}|_{t=0}=0,\quad u^{\theta}_{(1),k}|_{t=0}=0,\quad v^{z}_{(1),k}|_{t=0}=0.\end{aligned}\right.

Then by taking $L^{2}$ inner product of (5.4) with $(v^{r}_{(1),k},u^{\theta}_{(1),k},v^{z}_{(1),k})$ , we obtain

(5.5)	$\displaystyle$	$\displaystyle\frac{1}{2}\frac{d}{dt}\\|(v^{r}_{(1),k},u^{\theta}_{(1),k},v^{z}_{(1),k})\\|_{L^{2}}^{2}+\\|\nabla(v^{r}_{(1),k},u^{\theta}_{(1),k},v^{z}_{(1),k})\\|_{L^{2}}^{2}+k^{2}\bigl\\|\frac{v^{z}_{(1),k}}{r}\bigr\\|_{L^{2}}^{2}$
		$\displaystyle+(1+k^{2})\bigl(\bigl\\|\frac{v^{r}_{(1),k}}{r}\bigr\\|_{L^{2}}^{2}+\bigl\\|\frac{u^{\theta}_{(1),k}}{r}\bigr\\|_{L^{2}}^{2}\bigr)=\int_{\mathop{\mathbb{R}\kern 0.0pt}\nolimits^{3}}\Bigl(4k\frac{u^{\theta}_{(1),k}v^{r}_{(1),k}}{r^{2}}-\frac{u^{r}_{(0),0}\|u^{\theta}_{(1),k}\|^{2}}{r}$
		$\displaystyle-(v^{r}_{(1),k}\partial_{r}+v^{z}_{(1),k}\partial_{z})u^{r}_{(0),0}\cdot v^{r}_{(1),k}-(v^{r}_{(1),k}\partial_{r}+v^{z}_{(1),k}\partial_{z})u^{z}_{(0),0}\cdot v^{z}_{(1),k}\Bigr)\,dx.$

It is worth mentioning that, not the same as the most cases in doing estimates in PDE, the constant in this inequality is very important. Precisely, we have

	$\displaystyle 4k\Bigl\|\int_{\mathop{\mathbb{R}\kern 0.0pt}\nolimits^{3}}\frac{u^{\theta}_{(1),k}v^{r}_{(1),k}}{r^{2}}\,dx\Bigr\|$	$\displaystyle\leq 4k\bigl\\|\frac{v^{r}_{(1),k}}{r}\bigr\\|_{L^{2}}\bigl\\|\frac{u^{\theta}_{(1),k}}{r}\bigr\\|_{L^{2}}$
		$\displaystyle\leq 2k\bigl(\bigl\\|\frac{v^{r}_{(1),k}}{r}\bigr\\|_{L^{2}}^{2}+\bigl\\|\frac{u^{\theta}_{(1),k}}{r}\bigr\\|_{L^{2}}^{2}\bigr)\leq(1+k^{2})\bigl(\bigl\\|\frac{v^{r}_{(1),k}}{r}\bigr\\|_{L^{2}}^{2}+\bigl\\|\frac{u^{\theta}_{(1),k}}{r}\bigr\\|_{L^{2}}^{2}\bigr).$

On the other hand, by using Sobolev’s embedding theorem and Young’s inequality, we get

		$\displaystyle\Bigl\|\int_{\mathop{\mathbb{R}\kern 0.0pt}\nolimits^{3}}\Bigl((v^{r}_{(1),k}\partial_{r}+v^{z}_{(1),k}\partial_{z})u^{r}_{(0),0}\cdot v^{r}_{(1),k}+\frac{u^{r}_{(0),0}\|u^{\theta}_{(1),k}\|^{2}}{r}+(v^{r}_{(1),k}\partial_{r}+v^{z}_{(1),k}\partial_{z})u^{z}_{(0),0}\cdot v^{z}_{(1),k}\Bigr)\,dx\Bigr\|$
		$\displaystyle\leq\bigl(\\|v^{r}_{(1),k}\\|_{L^{4}}^{2}+\\|v^{z}_{(1),k}\\|_{L^{4}}^{2}\bigr)\bigl(\\|\widetilde{\nabla}u^{r}_{(0),0}\\|_{L^{2}}+\\|\widetilde{\nabla}u^{z}_{(0),0}\\|_{L^{2}}\bigr)+\\|u^{\theta}_{(1),k}\\|_{L^{4}}^{2}\bigl\\|\frac{u^{r}_{(0),0}}{r}\bigr\\|_{L^{2}}$
		$\displaystyle\leq\frac{1}{2}\\|\nabla(v^{r}_{(1),k},u^{\theta}_{(1),k},v^{z}_{(1),k})\\|_{L^{2}}^{2}+C\\|(v^{r}_{(1),k},u^{\theta}_{(1),k},v^{z}_{(1),k})\\|_{L^{2}}^{2}\\|\nabla\boldsymbol{u}_{(0)}\\|_{L^{2}}^{4},$

where in the last step we have used the fact that $\boldsymbol{u}_{(0)}=u^{r}_{(0),0}\boldsymbol{e}_{r}+u^{z}_{(0),0}\boldsymbol{e}_{z}$ , so that

\|\nabla\boldsymbol{u}_{(0)}\|_{L^{2}}\sim\|\widetilde{\nabla}u^{r}_{(0),0}\|_{L^{2}}+\|\widetilde{\nabla}u^{z}_{(0),0}\|_{L^{2}}+\bigl\|\frac{u^{r}_{(0),0}}{r}\bigr\|_{L^{2}}.

By substituting the above estimates into (5.5), we deduce

\frac{d}{dt}\|(v^{r}_{(1),k},u^{\theta}_{(1),k},v^{z}_{(1),k})\|_{L^{2}}^{2}+\|\nabla(v^{r}_{(1),k},u^{\theta}_{(1),k},v^{z}_{(1),k})\|_{L^{2}}^{2}\leq C\|(v^{r}_{(1),k},u^{\theta}_{(1),k},v^{z}_{(1),k})\|_{L^{2}}^{2}\|\nabla\boldsymbol{u}_{(0)}\|_{L^{2}}^{4}.

Then by applying Gronwall’s inequality, and using the fact that

\int_{0}^{\infty}\|\nabla\boldsymbol{u}_{(0)}(t^{\prime})\|_{L^{2}}^{4}\,dt^{\prime}\leq\|\nabla\boldsymbol{u}_{(0)}\|_{L^{\infty}(\mathop{\mathbb{R}\kern 0.0pt}\nolimits^{+};L^{2})}^{2}\|\nabla\boldsymbol{u}_{(0)}\|_{L^{2}(\mathop{\mathbb{R}\kern 0.0pt}\nolimits^{+};L^{2})}^{2}<\infty,

which is guaranteed by (4.5), and initially $v^{r}_{(1),k}|_{t=0}=u^{\theta}_{(1),k}|_{t=0}=v^{z}_{(1),k}|_{t=0}=0$ , we obtain that

\displaystyle v^{r}_{(1),k}=u^{\theta}_{(1),k}=v^{z}_{(1),k}=0,\quad\forall\ t>0.

In another word, we can further reduce the expansion for $\boldsymbol{u}_{(1)}$ in (5.2) to be

(5.6)

\boldsymbol{u}_{(1)}=\sum_{k=1}^{\infty}\Bigl(u^{r}_{(1),k}\cos k\theta\boldsymbol{e}_{r}+v^{\theta}_{(1),k}\sin k\theta\boldsymbol{e}_{\theta}+u^{z}_{(1),k}\cos k\theta\boldsymbol{e}_{z}\Bigr),

where $u^{r}_{(1),k},~v^{\theta}_{(1),k}$ and $u^{z}_{(1),k}$ are determined by (5.3).

5.3. Verification of $u^{\theta}_{(j),0}=v^{r}_{(j),k}=u^{\theta}_{(j),k}=v^{z}_{(j),k}=0$ , for any $j\in\mathop{\mathbb{N}\kern 0.0pt}\nolimits,~k\in\mathop{\mathbb{N}\kern 0.0pt}\nolimits$

We shall prove this result by the induction method. Assume that for any $n\geq 2$ and any $1\leq j\leq n-1$ ,

\displaystyle u^{\theta}_{(j),0}=v^{r}_{(j),k}=u^{\theta}_{(j),k}=v^{z}_{(j),k}=0,\quad\forall\ k\in\mathop{\mathbb{N}\kern 0.0pt}\nolimits,

i.e., $\boldsymbol{u}_{(j)}$ has the following form

(5.7)

\boldsymbol{u}_{(j)}=u^{r}_{(j),0}\boldsymbol{e}_{r}+u^{z}_{(j),0}\boldsymbol{e}_{z}+\sum_{k=1}^{\infty}\Bigl(u^{r}_{(j),k}\cos k\theta\boldsymbol{e}_{r}+v^{\theta}_{(j),k}\sin k\theta\boldsymbol{e}_{\theta}+u^{z}_{(j),k}\cos k\theta\boldsymbol{e}_{z}\Bigr),

then our aim is to show that $\boldsymbol{u}_{(n)}$ can also be written as this form, i.e.

\displaystyle u^{\theta}_{(n),0}=v^{r}_{(n),k}=u^{\theta}_{(n),k}=v^{z}_{(n),k}=0,\quad\forall\ k\in\mathop{\mathbb{N}\kern 0.0pt}\nolimits,

Indeed, due to Remark 4.1, we know that in the Euclidean coordinates, $\boldsymbol{u}_{(n)}$ satisfies

(5.8)

\left\{\begin{aligned} &\partial_{t}\boldsymbol{u}_{(n)}+\boldsymbol{u}_{(0)}\cdot\nabla\boldsymbol{u}_{(n)}+\boldsymbol{u}_{(n)}\cdot\nabla\boldsymbol{u}_{(0)}-\Delta\boldsymbol{u}_{(n)}+\nabla P_{(n)}=-\sum_{j=1}^{n-1}\boldsymbol{u}_{(j)}\cdot\nabla\boldsymbol{u}_{(n-j)},\\ &\mathop{\rm div}\nolimits\boldsymbol{u}_{(n)}=0,\\ &\boldsymbol{u}_{(n)}|_{t=0}=0.\end{aligned}\right.

This system has zero-valued initial data. However, due to the external force term $-\sum_{j=1}^{n-1}\boldsymbol{u}_{(j)}\cdot\nabla\boldsymbol{u}_{(n-j)}$ , the solution $\boldsymbol{u}_{(n)}$ in general does not vanish.

Noticing that $\boldsymbol{u}_{(j)}$ has the form (5.7) for $1\leq j\leq n-1$ , we can get

(5.9)

\begin{split}\boldsymbol{u}_{(j)}\cdot\nabla\boldsymbol{u}_{(n-j)}=&\sum_{k_{1},k_{2}=0}^{\infty}\Bigl\{(u^{r}_{(j),k_{1}}\partial_{r}+u^{z}_{(j),k_{1}}\partial_{z})\bigl(u^{r}_{(n-j),k_{2}}\boldsymbol{e}_{r}+u^{z}_{(n-j),k_{2}}\boldsymbol{e}_{z}\bigr)\cdot\cos k_{1}\theta\cdot\cos k_{2}\theta\\ &\quad+(u^{r}_{(j),k_{1}}\partial_{r}+u^{z}_{(j),k}\partial_{z})v^{\theta}_{(n-j),k_{2}}\boldsymbol{e}_{\theta}\cdot\cos k_{1}\theta\cdot\sin k_{2}\theta\\ &\quad+\frac{v^{\theta}_{(j),k_{1}}u^{r}_{(n-j),k_{2}}}{r}\bigl(-\boldsymbol{e}_{r}k_{2}\sin k_{1}\theta\cdot\sin k_{2}\theta+\boldsymbol{e}_{\theta}\sin k_{1}\theta\cdot\cos k_{2}\theta\bigr)\\ &\quad+\frac{v^{\theta}_{(j),k_{1}}v^{\theta}_{(n-j),k_{2}}}{r}\bigl(\boldsymbol{e}_{\theta}k_{2}\sin k_{1}\theta\cdot\cos k_{2}\theta-\boldsymbol{e}_{r}\sin k_{1}\theta\cdot\sin k_{2}\theta\bigr)\Bigr\}.\end{split}

In particular, we can see that $(\boldsymbol{u}_{(j)}\cdot\nabla\boldsymbol{u}_{(n-j)})\cdot\boldsymbol{e}_{r}$ and $(\boldsymbol{u}_{(j)}\cdot\nabla\boldsymbol{u}_{(n-j)})\cdot\boldsymbol{e}_{z}$ is even in $\theta$ , while $(\boldsymbol{u}_{(j)}\cdot\nabla\boldsymbol{u}_{(n-j)})\cdot\boldsymbol{e}_{\theta}$ is odd in $\theta$ . As a result, this external force $-\sum_{j=1}^{n-1}\boldsymbol{u}_{(j)}\cdot\nabla\boldsymbol{u}_{(n-j)}$ does not appear in the equations for $(v^{r}_{(n),k},u^{\theta}_{(n),k},v^{z}_{(n),k})$ for $k\geq 1$ . Thus the equations for $(v^{r}_{(n),k},u^{\theta}_{(n),k},v^{z}_{(n),k})$ are exactly the same as that for $(v^{r}_{(1),k},u^{\theta}_{(1),k},v^{z}_{(1),k})$ (see (5.4)) stated as follows:

\left\{\begin{aligned} &\partial_{t}v^{r}_{(n),k}+(u^{r}_{(0),0}\partial_{r}+u^{z}_{(0),0}\partial_{z})v^{r}_{(n),k}+(v^{r}_{(n),k}\partial_{r}+v^{z}_{(n),k}\partial_{z})u^{r}_{(0),0}\\ &\qquad\qquad\qquad-(\partial_{r}^{2}+\partial_{z}^{2}+\frac{1}{r}\partial_{r}-\frac{1+k^{2}}{r^{2}})v^{r}_{(n),k}-2k\frac{u^{\theta}_{(n),k}}{r^{2}}+\partial_{r}Q_{(n),k}=0,\\ &\partial_{t}u^{\theta}_{(n),k}+(u^{r}_{(0),0}\partial_{r}+u^{z}_{(0),0}\partial_{z})u^{\theta}_{(n),k}+\frac{u^{r}_{(0),0}u^{\theta}_{(n),k}}{r}\\ &\qquad\qquad\qquad-(\partial_{r}^{2}+\partial_{z}^{2}+\frac{1}{r}\partial_{r}-\frac{1+k^{2}}{r^{2}})u^{\theta}_{(n),k}-2k\frac{v^{r}_{(n),k}}{r^{2}}+\frac{k}{r}Q_{(n),k}=0,\\ &\partial_{t}v^{z}_{(n),k}+(u^{r}_{(0),0}\partial_{r}+u^{z}_{(0),0}\partial_{z})v^{z}_{(n),k}+(v^{r}_{(n),k}\partial_{r}+v^{z}_{(n),k}\partial_{z})u^{z}_{(0),0}\\ &\qquad\qquad\qquad-(\partial_{r}^{2}+\partial_{z}^{2}+\frac{1}{r}\partial_{r}-\frac{k^{2}}{r^{2}})v^{z}_{(n),k}+\partial_{z}Q_{(n),k}=0,\\ &\partial_{r}v^{r}_{(n),k}+\frac{1}{r}v^{r}_{(n),k}+\partial_{z}v^{z}_{(n),k}-\frac{k}{r}u^{\theta}_{(n),k}=0,\\ &v^{r}_{(n),k}|_{t=0}=0,\quad u^{\theta}_{(n),k}|_{t=0}=0,\quad v^{z}_{(n),k}|_{t=0}=0.\end{aligned}\right.

Then the same procedure as what we have done at the end of Subsection 5.2 shows that for $k\geq 1$ , $(v^{r}_{(n),k},u^{\theta}_{(n),k},v^{z}_{(n),k})$ indeed vanishes for all time, i.e.

\displaystyle v^{r}_{(n),k}=u^{\theta}_{(n),k}=v^{z}_{(n),k}=0,\quad\forall\ t>0.

Let us turn to study the $0$ -th Fourier coefficients $u^{r}_{(n),0}\boldsymbol{e}_{r}+u^{\theta}_{(n),0}\boldsymbol{e}_{\theta}+u^{z}_{(n),0}\boldsymbol{e}_{z}$ . In view of the expression (5.9), and the following identities

		$\displaystyle 2\cos k_{1}\theta\cdot\cos k_{2}\theta=\cos(k_{1}+k_{2})\theta+\cos(k_{1}-k_{2})\theta,$
		$\displaystyle 2\sin k_{1}\theta\cdot\sin k_{2}\theta=\cos(k_{1}-k_{2})\theta-\cos(k_{1}+k_{2})\theta,$
		$\displaystyle 2\sin k_{1}\theta\cdot\cos k_{2}\theta=\sin(k_{1}+k_{2})\theta+\sin(k_{1}-k_{2})\theta,$

and the fact that $\boldsymbol{u}_{(0)}=u^{r}_{(0),0}\boldsymbol{e}_{r}+u^{z}_{(0),0}\boldsymbol{e}_{z}$ is axisymmetric without swirl, we can obtain

(5.10)

\partial_{t}u^{\theta}_{(n),0}+\boldsymbol{u}_{(0)}\cdot\nabla u^{\theta}_{(n),0}+\frac{u^{r}_{(0),0}u^{\theta}_{(n),0}}{r}-(\Delta-\frac{1}{r^{2}})u^{\theta}_{(n),0}=0,\quad u^{\theta}_{(n),0}|_{t=0}=0,

and

(5.11)

\left\{\begin{aligned} &\partial_{t}u^{r}_{(n),0}+\boldsymbol{u}_{(0)}\cdot\nabla u^{r}_{(n),0}+(u^{r}_{(n),0}\partial_{r}+u^{z}_{(n),0}\partial_{z})u^{r}_{(0),0}\\ &\qquad\qquad\qquad\qquad\qquad\qquad-(\Delta-\frac{1}{r^{2}})u^{r}_{(n),0}+\partial_{r}P_{(n),0}=F^{r}_{(n),0},\\ &\partial_{t}u^{z}_{(n),0}+\boldsymbol{u}_{(0)}\cdot\nabla u^{z}_{(n),0}+(u^{r}_{(n),0}\partial_{r}+u^{z}_{(n),0}\partial_{z})u^{z}_{(0),0}-\Delta u^{z}_{(n),0}+\partial_{z}P_{(n),0}=F^{z}_{(n),0},\\ &\partial_{r}u^{r}_{(n),0}+\frac{1}{r}u^{r}_{(n),0}+\partial_{z}u^{z}_{(n),0}=0,\\ &u^{r}_{(n),0}|_{t=0}=0,\quad u^{z}_{(n),0}|_{t=0}=0,\end{aligned}\right.

where $F^{r}_{(n),0},~F^{z}_{(n),0}$ are the external force terms given by

	$\displaystyle F^{r}_{(n),0}=-\frac{1}{2}$	$\displaystyle\sum_{k=1}^{\infty}\sum_{j=1}^{n-1}\bigl(u^{r}_{(j),k}\partial_{r}+u^{z}_{(j),k}\partial_{z}\bigr)u^{r}_{(n-j),k}+\frac{1}{2}\sum_{k=1}^{\infty}\sum_{j=1}^{n-1}\frac{v^{\theta}_{(j),k}v^{\theta}_{(n-j),k}}{r},$
		$\displaystyle F^{z}_{(n),0}=-\frac{1}{2}\sum_{k=1}^{\infty}\sum_{j=1}^{n-1}\bigl(u^{r}_{(j),k}\partial_{r}+u^{z}_{(j),k}\partial_{z}\bigr)u^{z}_{(n-j),k}.$

In particular, this reflects the fact that if two profiles have same frequency, then their product would contribute to the average of $\boldsymbol{u}$ in $\theta$ variable.

Taking $L^{2}$ inner product of (5.10) with $u^{\theta}_{(n),0}$ , it is easy to get that

\displaystyle u^{\theta}_{(n),0}=0,\quad\forall\ t>0.

Therefore, $\boldsymbol{u}_{(n)}$ can also be written in the following form:

\boldsymbol{u}_{(n)}=u^{r}_{(n),0}\boldsymbol{e}_{r}+u^{z}_{(n),0}\boldsymbol{e}_{z}+\sum_{k=1}^{\infty}\Bigl(u^{r}_{(n),k}\cos k\theta\boldsymbol{e}_{r}+v^{\theta}_{(n),k}\sin k\theta\boldsymbol{e}_{\theta}+u^{z}_{(n),k}\cos k\theta\boldsymbol{e}_{z}\Bigr).

This completes the proof of Theorem 1.3 by induction.

Acknowledgments. Y. Liu is supported by NSF of China under grant 12101053, and the Fundamental Research Funds for the Central Universities under Grant 310421118. L. Xu is supported by NSF of China under grant 11671383 and 12171019.

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	$\displaystyle\|f(r,\theta,z)-{\mathcal{M}}(f)(r,z)\|^{p}$	$\displaystyle=\Bigl\|\frac{1}{2\pi}\int_{0}^{2\pi}\bigl(f(r,\theta,z)-f(r,\theta^{\prime},z)\bigr)\,d\theta^{\prime}\Bigr\|^{p}$
		$\displaystyle=\Bigl\|\frac{1}{2\pi}\int_{0}^{2\pi}\bigl(\int_{\theta^{\prime}}^{\theta}(\partial_{\theta}f)(r,\theta^{\prime\prime},z)\,d\theta^{\prime\prime}\bigr)\,d\theta^{\prime}\Bigr\|^{p}$
		$\displaystyle\leq\frac{1}{2\pi}\int_{0}^{2\pi}\Bigl\|\int_{\theta^{\prime}}^{\theta}\bigl\|(\partial_{\theta}f)(r,\theta^{\prime\prime},z)\bigr\|^{p}\,d\theta^{\prime\prime}\Bigr\|\|\theta-\theta^{\prime}\|^{p-1}\,d\theta^{\prime}$
		$\displaystyle\leq\frac{1}{2\pi}\int_{0}^{2\pi}\bigl\|(\partial_{\theta}f)(r,\theta^{\prime\prime},z)\bigr\|^{p}\,d\theta^{\prime\prime}\int_{0}^{2\pi}\|\theta-\theta^{\prime}\|^{p-1}\,d\theta^{\prime}.$

	$\displaystyle\\|f(r,\theta,z)-{\mathcal{M}}(f)(r,z)\\|_{L^{p}}^{p}$	$\displaystyle=\frac{1}{2\pi}\int_{\mathop{\mathbb{R}\kern 0.0pt}\nolimits^{+}\times\mathop{\mathbb{R}\kern 0.0pt}\nolimits}\Bigl(\int_{0}^{2\pi}\bigl\|(\partial_{\theta}f)(r,\theta^{\prime\prime},z)\bigr\|^{p}\,d\theta^{\prime\prime}\Bigr)\,rdrdz$
		$\displaystyle\qquad\qquad\qquad\qquad\times\int_{0}^{2\pi}\int_{0}^{2\pi}\|\theta-\theta^{\prime}\|^{p-1}\,d\theta^{\prime}d\theta$
		$\displaystyle=\frac{2}{p+1}(2\pi)^{p}\\|\partial_{\theta}f\\|_{L^{p}}^{p}\leq(2\pi)^{p}\\|\partial_{\theta}f\\|_{L^{p}}^{p}.$

(2.5)		$\displaystyle\\|\nabla^{2}\widetilde{\boldsymbol{u}}\\|_{L^{2}}^{2}$	$\displaystyle\sim\bigl\\|\partial_{r}\widetilde{\omega}^{\theta}+\frac{\widetilde{\omega}^{\theta}}{r}\bigr\\|_{L^{2}}^{2}+\\|\partial_{z}\widetilde{\omega}^{\theta}\\|_{L^{2}}^{2}$
(2.5)			$\displaystyle=\\|\widetilde{\nabla}\widetilde{\omega}^{\theta}\\|_{L^{2}}^{2}+\bigl\\|\frac{\widetilde{\omega}^{\theta}}{r}\bigr\\|_{L^{2}}^{2}+2\int_{\mathop{\mathbb{R}\kern 0.0pt}\nolimits^{3}}\partial_{r}\widetilde{\omega}^{\theta}\cdot\frac{\widetilde{\omega}^{\theta}}{r}\,dx.$

(2.6)		$\displaystyle\\|\nabla\bar{\boldsymbol{u}}\\|_{L^{2}}^{2}$	$\displaystyle\sim\\|\bar{\omega}^{\theta}\\|_{L^{2}}^{2}+\\|\widetilde{\nabla}\bar{u}^{\theta}\\|_{L^{2}}^{2}+\bigl\\|\frac{\bar{u}^{\theta}}{r}\bigr\\|_{L^{2}}^{2}$
(2.6)			$\displaystyle\sim\\|\widetilde{\nabla}\bar{u}^{r}\\|_{L^{2}}^{2}+\\|\widetilde{\nabla}\bar{u}^{z}\\|_{L^{2}}^{2}+\\|\widetilde{\nabla}\bar{u}^{\theta}\\|_{L^{2}}^{2}+\bigl\\|\frac{\bar{u}^{r}}{r}\bigr\\|_{L^{2}}^{2}+\bigl\\|\frac{\bar{u}^{\theta}}{r}\bigr\\|_{L^{2}}^{2},$

(2.7)		$\displaystyle\\|\nabla\bar{\boldsymbol{u}}\\|_{L^{2}}^{2}\thicksim\\|\mathop{\rm curl}\nolimits\bar{\boldsymbol{u}}\\|_{L^{2}}^{2}$	$\displaystyle=\\|\bar{\omega}^{\theta}\\|_{L^{2}}^{2}+\bigl\\|\partial_{r}\bar{u}^{\theta}+\frac{\bar{u}^{\theta}}{r}\bigr\\|_{L^{2}}^{2}+\\|\partial_{z}\bar{u}^{\theta}\\|_{L^{2}}^{2}$
(2.7)			$\displaystyle\sim\\|\widetilde{\nabla}\bar{u}^{r}\\|_{L^{2}}^{2}+\\|\widetilde{\nabla}\bar{u}^{z}\\|_{L^{2}}^{2}+\bigl\\|\frac{\bar{u}^{r}}{r}\bigr\\|_{L^{2}}^{2}+\bigl\\|\partial_{r}\bar{u}^{\theta}+\frac{\bar{u}^{\theta}}{r}\bigr\\|_{L^{2}}^{2}+\\|\partial_{z}\bar{u}^{\theta}\\|_{L^{2}}^{2},$

On the existence and structures of almost axisymmetric solutions to 33-D Navier-Stokes equations

Abstract.

1. Introduction

1.1. The general setting

1.2. Main results.

Theorem 1.1.

Remark 1.1.

Theorem 1.2.

Remark 1.2.

Theorem 1.3.

Remark 1.3.

2. Preliminary

2.1. Poincaré-type inequality

Lemma 2.1.

Proof.

2.2. Properties for axisymmetric functions.

Lemma 2.2.

Proof.

Lemma 2.3.

Proof.

2.3. A stability result for N-S

Proposition 2.1.

Proof.

3. The proof of Theorem 1.1

3.1. Global solvability of (1.6)

Proposition 3.1.

Proof.

Proposition 3.2.

Proof.

3.2. The proof of Theorem 1.1

The proof of Theorem 1.1.

4. The proof of Theorem 1.2

4.1. The ε0\varepsilon^{0} order terms

4.2. Derivation of the ε1\varepsilon^{1} order terms in the Euclidean coordinates

Remark 4.1.

4.3. Verification of u(1),0r=u(1),0z=0u^{r}_{(1),0}=u^{z}_{(1),0}=0

5. The proof of Theorem 1.3

5.1. Verification of 𝒖(1),0=𝟎\boldsymbol{u}_{(1),0}=\boldsymbol{0}

5.2. Verification of v(1),kr=u(1),kθ=v(1),kz=0v^{r}_{(1),k}=u^{\theta}_{(1),k}=v^{z}_{(1),k}=0

5.3. Verification of u(j),0θ=v(j),kr=u(j),kθ=v(j),kz=0u^{\theta}_{(j),0}=v^{r}_{(j),k}=u^{\theta}_{(j),k}=v^{z}_{(j),k}=0, for any j∈ℕ,k∈ℕj\in\mathop{\mathbb{N}\kern 0.0pt}\nolimits,~k\in\mathop{\mathbb{N}\kern 0.0pt}\nolimits

References

On the existence and structures of almost axisymmetric solutions to $3$ -D Navier-Stokes equations

4.1. The $\varepsilon^{0}$ order terms

4.2. Derivation of the $\varepsilon^{1}$ order terms in the Euclidean coordinates

4.3. Verification of $u^{r}_{(1),0}=u^{z}_{(1),0}=0$

5.1. Verification of $\boldsymbol{u}_{(1),0}=\boldsymbol{0}$

5.2. Verification of $v^{r}_{(1),k}=u^{\theta}_{(1),k}=v^{z}_{(1),k}=0$

5.3. Verification of $u^{\theta}_{(j),0}=v^{r}_{(j),k}=u^{\theta}_{(j),k}=v^{z}_{(j),k}=0$ , for any $j\in\mathop{\mathbb{N}\kern 0.0pt}\nolimits,~k\in\mathop{\mathbb{N}\kern 0.0pt}\nolimits$