On Some Regularity Criteria for Axisymmetric Navier–Stokes Equations

Abstract

We point out some criteria that imply regularity of axisymmetric solutions to Navier–Stokes equations. We show that boundedness of $\Vert {v_{r}}/{\sqrt{r^3}}\Vert _{L_2({\mathbb {R}}^3\times (0,T))}$ as well as boundedness of $\Vert {\omega _{\varphi }}/{\sqrt{r}} \Vert _{L_2({\mathbb {R}}^3\times (0,T))}$, where $v_r$ is the radial component of velocity and $\omega _{\varphi }$ is the angular component of vorticity, imply regularity of weak solutions.

1 Introduction

We consider the Cauchy problem to the three-dimensional axisymmetric Navier–Stokes equations:

$$\begin{aligned} \begin{aligned}&v_{t}+v\cdot \nabla v- \nu \Delta v + \nabla p =0 \quad&(x,t) \in {\mathbb {R}}^3 \times {{\mathbb {R}}_+}, \\&\mathrm{div}\,v=0,\quad&\\&v\big |_{t=0}=v(0),\quad&\end{aligned} \end{aligned}$$

(1.1)

where $x= (x_1, x_2, x_3),\ v$ is the velocity of the fluid motion with

$$\begin{aligned} v(x,t)=(v_1(x,t),v_2(x,t),v_3(x,t))\in {\mathbb {R}}^3, \end{aligned}$$

$p=p(x,t)\in {\mathbb {R}}^1$ denotes the pressure, $\nu$ is the viscosity coefficient and $v_0$ is given initial velocity field.

The first papers concerning regularity of axially symmetric solutions to the Navier–Stokes equations were independently proved by Ladyzhenskaya [1] and Yudovich–Ukhovskij [2] in 1968. In these papers axisymmetric solutions without swirl were considered. In the period 1999–2002 arised many papers concerning sufficient conditions on regularity of axisymmetric solutions [3,4,5,6]. Especially, conditions on one coordinate of velocity were considered. Recently there are many papers dealing with new sufficient conditions (see references of Lei and Zhang [7]).

Our aim is to derive some criteria guaranteeing regularity of solutions to the axisymmetric Navier–Stokes equations. By the regular solutions we mean smooth weak solutions obtained by the standard increasing regularity technique for smooth initial data. There is a lot of criteria for regularity of axisymmetric solutions (see [6,7,8,9,10,11,12] and the literature cited in these papers). In Sect. 2 we recall only such criteria that are useful for our analysis.

Since we are restricted to the axisymmetric solutions we introduce the cylindrical coordinates $(r,\varphi ,z)$ by the relations

$$\begin{aligned} x_1=r\cos \varphi ,\quad x_2=r\sin \varphi ,\quad x_3=z, \end{aligned}$$

and corresponding unit vectors:

$$\begin{aligned} {{\bar{e}}}_r=(\cos \varphi ,\sin \varphi ,0),\quad \bar{e}_\varphi =(-\sin \varphi ,\cos \varphi ,0),\quad {\bar{e}}_z=(0,0,1). \end{aligned}$$

Then the cylindrical components of velocity and vorticity ($\omega =\mathrm{rot}\,v$) for axisymmetric solutions (therefore, solutions independent of $\varphi$) are represented as

$$\begin{aligned} v = v_r (r,z,t) {\bar{e}}_r + v_\varphi (r,z,t) {\bar{e}}_\varphi +v_z(r,z,t) \bar{e}_z \end{aligned}$$

and

$$\begin{aligned} \omega= & {} \omega _r (r,z,t) {\bar{e}}_r + \omega _\varphi (r,z,t) {\bar{e}}_\varphi +\omega _z(r,z,t) {\bar{e}}_z \\= & {} -\, v_{\varphi ,z} {\bar{e}}_r + (v_{r,z} - v_{z,r}){\bar{e}}_\varphi + \left( v_{\varphi ,r}+ \frac{v_{\varphi }}{r}\right) {\bar{e}}_z, \end{aligned}$$

where $v_r, v_\varphi , v_z$ are radial, angular and axial components of velocity.

The axisymmetric motion can be described by the three quantities: $v_{\varphi }, \omega _{\varphi }$ and the stream potential $\psi$ which are solutions to the following equations:

$$\begin{aligned}&v_{\varphi ,t} + v\cdot \nabla v_{\varphi } - \nu \left( \Delta - \frac{1}{r^2}\right) v_{\varphi } + \frac{v_r}{r} v_{\varphi } = 0, \nonumber \\&v_{\varphi }|_{t=0}= v_{\varphi }(0),\nonumber \\&\omega _{\varphi ,t} + v\cdot \nabla \omega _{\varphi } - \nu \left( \Delta - \frac{1}{r^2}\right) \omega _{\varphi } - \frac{v_r}{r} \omega _{\varphi } - \frac{2}{r}v_{\varphi }v_{\varphi ,z} = 0, \nonumber \\&\omega _{\varphi }|_{t=0}= \omega _{\varphi }(0),\nonumber \\&- \left( \Delta \psi - \frac{1}{r^2}\psi \right) = \omega _{\varphi }, \end{aligned}$$

(1.2)

where $v\cdot \nabla = v_r\partial _r + v_z\partial _z, \Delta =\partial _r^2 + \partial _z^2 + \frac{1}{r}\partial _r$ and

$$\begin{aligned} v_r = -\psi _{,z},\quad v_z = \frac{1}{r} (r\psi )_{,r}. \end{aligned}$$

It is very convenient to introduce quantities $u_1, \omega _1, \psi _1$ by the relations

$$\begin{aligned} v_{\varphi } = r u_1,\quad \omega _{\varphi } = r \omega _1,\quad \psi = r\psi _1. \end{aligned}$$

Then Eqs. (1.2) simplify to

$$\begin{aligned}&u_{1,t} + v\cdot \nabla u_{1} - \nu \left( \Delta u_1 + \frac{2}{r} u_{1,r}\right) = 2 u_1 \psi _{1,z}, \nonumber \\&u_{1}|_{t=0}= v_{1}(0),\end{aligned}$$

(1.3)

$$\begin{aligned}&\omega _{1,t} + v\cdot \nabla \omega _{1} - \nu \left( \Delta \omega _1 + \frac{2}{r} \omega _{1,r}\right) = 2u_1 u_{1,z}, \nonumber \\&\omega _1|_{t=0}= \omega _1(0),\end{aligned}$$

(1.4)

$$\begin{aligned}&- \left( \Delta \psi _1 +\frac{2}{r}\psi _{1,r}\right) = \omega _1, \end{aligned}$$

(1.5)

and

$$\begin{aligned} v_r = -r \psi _{1,z}, \quad v_z = \frac{1}{r} (r^2\psi _1)_{,r}. \end{aligned}$$

(1.6)

We prove the following regularity criteria (see (1.7), (1.8)) which are scaling invariant:

Theorem 1

• 1. Let (v, p) be an axisymmetric solution to the Navier–Stokes Eqs. (1.1) with the axisymmetric initial data and additionally $\mathrm{div}\,v(0) = 0$.

• 2. Assume that $\displaystyle {\frac{v^2_{\varphi }(0)}{r}, \frac{\omega _{\varphi }(0)}{r}, \frac{\omega _r(0)}{r}, \omega _z(0)}$ belong to $L_2(R^3)$, and with the notation $u=rv_{\varphi }, u(0)\in L_{\infty }({\mathbb {R}}^3)\bigcap L_s({\mathbb {R}}^3), s\ge 3.$

• 3. Assume that there exists constant $c_1$ such that

  $$\begin{aligned} \int _0^T {\textit{dt}} \int _{{\mathbb {R}}^3}\frac{v_r^2}{r^3} {\textit{dx}} \le c_1 < \infty . \end{aligned}$$

  (1.7)

  Then $v \in L_{\infty }(0,T;H^1({\mathbb {R}}^3_{r_0})),$ where ${\mathbb {R}}^3_{r_0}= \{x\in {\mathbb {R}}^3, r<r_0 \}$ and $r_0>0$ is given. Assume additionally that $v(0)\in B^{2-2/r}_{\sigma ,r}({\mathbb {R}}^3_{r_0})$ - Besov space. Then $v\in W^{2,1}_{\sigma ,r}({\mathbb {R}}^3_{r_0}\times (0,T))$.

Remark 1.1

For $\sigma > 3, r=2$ we have that $v\in L_{\infty }({\mathbb {R}}^3_{r_0} \times (0,T))$ so in view of [13] there is no singular points. In ${\bar{{\mathbb {R}}}}^3_{r_0}= \{x\in {\mathbb {R}}^3, r> r_0 \}$ the axisymmetric problem (1.1) is two-dimensional so local regularity of v is evident.

Theorem 2

Let the assumptions 1, 2 of Theorem 1 hold. If

$$\begin{aligned} \int _0^T {\textit{dt}} \int _{{\mathbb {R}}^3}\frac{\omega _{\varphi }^2}{r}{\textit{dx}} \le c_2 < \infty \end{aligned}$$

(1.8)

then there exists a constant $c_3$ such that

$$\begin{aligned} \int _0^T {\textit{dt}} \int _{{\mathbb {R}}^3}\frac{v_r^2}{r^3}{\textit{dx}} \le c_3\int _0^T {\textit{dt}} \int _{{\mathbb {R}}^3}\frac{\omega _{\varphi }^2}{r} {\textit{dx}}\le c_3c_2. \end{aligned}$$

(1.9)

2 Notation and Auxiliary Results

By $L_p({\mathbb {R}}^3), p\in [1,\infty ],$ we denote the Lebesgue space of integrable functions. By $L_{p,q}({\mathbb {R}}^3 \times (0,T))$ we denote the anisotropic Lebesgue space with the following finite norm

$$\begin{aligned} \Vert u\Vert _{L_{p,q}({\mathbb {R}}^3 \times (0,T))}= \left( \int _0^T \int _{{\mathbb {R}}^3}(|u(x,t)|^p dx)^{q/p}dt \right) ^{1/q}, \end{aligned}$$

where $p,q \in [1,\infty ].$

We define Sobolev spaces $W_p^{2,1}({\mathbb {R}}^3 \times (0,T))$ and $W^{2-2/p}_p({\mathbb {R}}^3 \times (0,T))$ by

$$\begin{aligned} \Vert u\Vert _{W_p^{2,1}({\mathbb {R}}^3 \times (0,T))}= & {} \left( \int _0^T \int _{{\mathbb {R}}^3}(|\nabla _x^2 u|^p + |u_t|^p +|u|^p)dx dt \right) ^{1/p}< \infty , \\ \Vert u\Vert _{W_p^{2-2/p}({\mathbb {R}}^3)}= & {} \left( \sum _{i\le [2-2/p]} \int _{{\mathbb {R}}^3}|\nabla ^i u|^p dx + \int _{{\mathbb {R}}^3}\int _{{\mathbb {R}}^3} \frac{|\nabla _x^{[2-2/p]}u(x) - \nabla _y^{[2-2/p]}u(y)|^p}{|x-y|^{3+p(2-2/p - [2-2/p])}}dx dy\right) ^{1/p} \\&< \infty , \end{aligned}$$

where [l] is the integer part of l.

By $H^s({\mathbb {R}}^3), s\in {\mathbb {N}}_0 = {\mathbb {N}}\cup \{0\}$ we denote the Sobolev space $W^s_2({\mathbb {R}}^3)$.

To find the form of $|\nabla v|$ we recall that

$$\begin{aligned} |\nabla v|^2 = |v,_{r}|^2 + |v,_{z}|^2. \end{aligned}$$

Using that $v = v_r {\bar{e}}_r + v_{\varphi } {\bar{e}}_{\varphi } + v_z {\bar{e}}_z$, we obtain

$$\begin{aligned} |\nabla v|^2 = |v_{r,r}|^2 + |v_{r,z}|^2 + |v_{\varphi ,r}|^2 + |v_{\varphi ,z}|^2 + |v_{z,r}|^2 +|v_{z,z}|^2, \end{aligned}$$

where we used the fact that vectors ${\bar{e}}_r, {\bar{e}}_{\varphi }, {\bar{e}}_z$ are orthonormal.

Lemma 2.1

There exists a weak solution to problem (1.1) such that $v \in L_{\infty }(0,T;L_2({\mathbb {R}}^3))\cap L_2(0,T;H^1({\mathbb {R}}^3))$ and the following estimate holds

$$\begin{aligned} \int _{{\mathbb {R}}^3} |v(t)|^2 {\textit{dx}} + \nu \int _0^{t} {\textit{dt}}' \int _{{\mathbb {R}}^3}|\nabla v|^2 {\textit{dx}}\le c \int _{{\mathbb {R}}^3}|v(0)|^2 {\textit{dx}}. \end{aligned}$$

(2.1)

In the case of axisymmetric solutions the energy inequality (2.1) takes the form

$$\begin{aligned}&\int _{{\mathbb {R}}^3} |v(t)|^2 {\textit{dx}} + \nu \int _0^{t} {\textit{dt}}' \int _{{\mathbb {R}}^3}\left( |\nabla v|^2+\left| \frac{v_r}{r}\right| ^2 +\left| \frac{v_{\varphi }}{r}\right| ^2 \right) {\textit{dx}} \nonumber \\&\quad \le c \int _{{\mathbb {R}}^3}|v(0)|^2 {\textit{dx}}. \end{aligned}$$

(2.2)

Proof

Equations $(1.1)_{1,2}$ for the axially symmetric solutions assume the form

$$\begin{aligned}&v_{r,t}+v\cdot \nabla v_r-\frac{v_\varphi ^2}{r}-\nu \Delta v_r+\nu \frac{v_r}{r^2}=-p_{,r}, \end{aligned}$$

(2.3)

$$\begin{aligned}&v_{\varphi ,t}+v\cdot \nabla v_\varphi +\frac{v_r}{r}v_\varphi -\nu \Delta v_\varphi + \nu \frac{v_\varphi }{r^2}=0, \end{aligned}$$

(2.4)

$$\begin{aligned}&v_{z,t}+v\cdot \nabla v_z-\nu \Delta v_z=-p_{,z}, \end{aligned}$$

(2.5)

$$\begin{aligned}&v_{r,r}+v_{z,z}=-\frac{v_r}{r}, \end{aligned}$$

(2.6)

where $v\cdot \nabla =v_r\partial _r+v_z\partial _z$, $\Delta u=\frac{1}{r}(ru_{,r})_{,r}+u_{,zz}$.

Let $I= \{ (\varphi , r, z): r=0\}$ denote the axis of symmetry. Any space of functions which vanish on I considered in this paper is a closure of $C^{\infty }_0({\mathbb {R}}^3{\setminus } I)$ in the corresponding norm. Then, we are looking for a priori estimate for functions v having such property.

Multiplying (2.4) by $v_\varphi$ and integrating over ${\mathbb {R}}^3$ yields

$$\begin{aligned} \frac{1}{2}\frac{d}{{\textit{dt}}}\int _{{\mathbb {R}}^3}v_\varphi ^2dx+\int _{{\mathbb {R}}^3}\frac{v_r}{r}v_\varphi ^2{\textit{dx}}+\nu \int _{{\mathbb {R}}^3}(v_{\varphi ,r}^2+ v_{\varphi ,z}^2)\,{\textit{dx}} +\nu \int _{{\mathbb {R}}^3}\frac{v_\varphi ^2}{r^2}{\textit{dx}}=0. \end{aligned}$$

Multiplying (2.3) by $v_r$, integrating over ${\mathbb {R}}^3$ implies

$$\begin{aligned} \frac{1}{2}\frac{d}{{\textit{dt}}}\int _{{\mathbb {R}}^3}v_r^2{\textit{dx}}-\int _{{\mathbb {R}}^3}\frac{v_\varphi ^2}{r}v_r{\textit{dx}}+\nu \int _{{\mathbb {R}}^3}(v_{r,r}^2+v_{r,z}^2)\,{\textit{dx}} + \nu \int _{{\mathbb {R}}^3}\frac{v_r^2}{r^2}\,{\textit{dx}} = -\int _{{\mathbb {R}}^3}p_{,r}v_r{\textit{dx}}. \end{aligned}$$

Multiplying (2.5) by $v_z$ and integrating over ${\mathbb {R}}^3$ we obtain

$$\begin{aligned} \frac{1}{2}\frac{d}{{\textit{dt}}}\int _{{\mathbb {R}}^3}v_z^2{\textit{dx}}+\nu \int _{{\mathbb {R}}^3}(v_{z,r}^2+v_{z,z}^2)\, {\textit{dx}}=-\int _{{\mathbb {R}}^3}p_{,z}v_z{\textit{dx}}. \end{aligned}$$

Adding the above equations and using (2.6) we obtain

$$\begin{aligned} \begin{aligned}&\frac{1}{2}\frac{d}{{\textit{dt}}}\int _{{\mathbb {R}}^3}(v_r^2+v_\varphi ^2+v_z^2)\,{\textit{dx}} \\&\quad +\, \nu \int _{{\mathbb {R}}^3}(v_{r,r}^2+v_{r,z}^2+v_{\varphi ,r}^2+ v_{\varphi ,z}^2+v_{z,r}^2+ v_{z,z}^2){\textit{dx}} \\&\quad +\nu \int _{{\mathbb {R}}^3}\left( \frac{v_r^2}{r^2}+\frac{v_\varphi ^2}{r^2}\right) {\textit{dx}} =0. \end{aligned} \end{aligned}$$

(2.7)

Integrating (2.7) with respect to time from 0 to t, $t\le T$, yields

$$\begin{aligned} \begin{aligned}&\nu \Vert v(t)\Vert _{L_2({\mathbb {R}}^3)}^2+\nu \int _{0}^t\int _{{\mathbb {R}}^3}(|v_{,r}|^2+|v_{,z}|^2) {\textit{dxdt}}'\\&\quad + \nu \int _{0}^t\int _{{\mathbb {R}}^3}\bigg (\frac{v_r^2}{r^2}+ \frac{v_{\varphi }^2}{r^2}\bigg ){\textit{dxdt}}' \le \frac{1}{2} \Vert v(0)\Vert _{L_2({\mathbb {R}}^3)}^2. \end{aligned} \end{aligned}$$

This ends the proof. $\square$

To derive energy estimates in the proof of Lemma 2.1 we use the ideas from the proof of Theorem 3.1 from [14, Ch.3]. The notion of a suitable weak solution was introduced by Caffarelli et al. [13] in the famous paper. Our aim is to show that either (1.7) or (1.8) implies that a suitable weak solution to problem (1.1) does not contain singular points. This means that (v, p) is a regular solution to (1.1). In other words it means that if $v(0)\in W_p^{2-2/p}({\mathbb {R}}^3)$ then $v\in W^{2,1}_p({\mathbb {R}}^3 \times R_+)$ for any $p\in (1,\infty )$. Hence for $p>\frac{5}{2}$ we have that $v\in L_{\infty }({\mathbb {R}}^3 \times R_+)$ so it is also bounded locally. Therefore v has no singular points (see [13]). To show this we use results of J. Neustupa, M. Pokorny and O. Kreml (see [5, 6, 10]). To clarify presentation we recall the results.

From [5] it follows that

Lemma 2.2

[5]. Let v be an axisymmetric suitable weak solution to problem (1.1). Suppose that there exists a subdomain $D \subset {\mathbb {R}}^3\times {\mathbb {R}}_+$ such that the angular component $v_{\varphi }$ of v belongs to $L_{s,r}(D)$ where

• 1. Either $s\in [6,\infty ], r\in [20/7, \infty ]$ and $2/r+3/s\le 7/10$.

• 2. or $s\in [24/5,6], r\in [10,\infty ]$ and $2/r+3/s\le 1-9/(5s)$.

Then v has no singular points in D.

Lemma 2.3

[10]. Let v be an axisymmetric suitable weak solution to problem (1.1). Suppose that there exists a subdomain $D \subset {\mathbb {R}}^3\times {\mathbb {R}}_+$ such that the angular component $v_{\varphi }$ of v belongs to $L_{s,r}(D)$ where

$$\begin{aligned} s\in \left( \frac{24}{7}, 4 \right] ,\quad r\in \left( \frac{8s}{7s-24},\infty \right] ,\quad \frac{3}{s}+\frac{2}{r} < \frac{7}{4} - \frac{3}{s}. \end{aligned}$$

Then v has no singular points in D.

By swirl we denote

$$\begin{aligned} u=rv_{\varphi }. \end{aligned}$$

From (1.2)$_1$ it follows that u satisfies the equation

$$\begin{aligned} u_{,t}+ v\cdot \nabla u -\nu \Delta u+ \frac{2\nu }{r} u_{,r} = 0. \end{aligned}$$

Lemma 2.4

(See [3]). Let $u(0)= r v_{\varphi }(0) \in L_{\infty }({\mathbb {R}}^3).$ Then

$$\begin{aligned} \Vert u\Vert _{L_{\infty }(0,T;L_{\infty }({\mathbb {R}}^3))} \le \Vert u(0)\Vert _{L_{\infty }({\mathbb {R}}^3)} \quad \mathrm{for \ any}~T. \end{aligned}$$

(2.8)

Remark 2.5

From (2.2) and (2.8) we have

$$\begin{aligned} \begin{aligned} \int _{0}^t \int _{{\mathbb {R}}^3}|v_{\varphi }|^4 {\textit{dxdt}}&= \int _{0}^t \int _{{\mathbb {R}}^3}r^2 v_{\varphi }^2\frac{v_\varphi ^2}{r^2} {\textit{dxdt}} \\&\le \Vert r v_{\varphi }\Vert _{L_{\infty }({\mathbb {R}}^3 \times (0,T))}^2\int _{0}^t \int _{{\mathbb {R}}^3}\frac{v_\varphi ^2}{r^2} {\textit{dxdt}} \le c\Vert u(0)\Vert ^2_{L_{\infty }({\mathbb {R}}^3)} \Vert v(0)\Vert ^2_{L_2({\mathbb {R}}^3)}. \end{aligned} \end{aligned}$$

(2.9)

We recall also Lemma 3.1 from [9]

Lemma 2.6

(See [9]). Assume that (v, p) is regular axisymmetric solution to the Navier–Stokes Eqs. (1.1). Assume that $\frac{v_{\varphi }(0)}{\sqrt{r}}\in L_4({\mathbb {R}}^3), {\tilde{\nabla }}\frac{v_r}{r} \in L_{4/3}(0,T; L_2({\mathbb {R}}^3))$ when ${\tilde{\nabla }}=(\partial _r, \partial _z).$ Then the following estimate holds

$$\begin{aligned} \begin{aligned}&\frac{1}{2} \left\| \frac{v_{\varphi }^2(t)}{r}\right\| _{L_2({\mathbb {R}}^3)}^2 + \frac{1}{4} \int _0^t \left\| {\tilde{\nabla }}\frac{v_{\varphi }^2}{r} \right\| _{L_2({\mathbb {R}}^3)}^2 {\textit{dt}}' + \frac{3}{4} \int _0^t \left\| \frac{v_{\varphi }}{r}\right\| _{L_4({\mathbb {R}}^3)}^4 {\textit{dt}}' \\ {}&\quad \le \frac{1}{4}\exp \left[ c\int _0^t \left\| {\tilde{\nabla }}\frac{v_r}{r}\right\| _{L_2({\mathbb {R}}^3)}^{4/3} {\textit{dt}}' \right] \left\| \frac{v_{\varphi }^2(0)}{r}\right\| _{L_2({\mathbb {R}}^3)}^2. \end{aligned} \end{aligned}$$

(2.10)

Proof

Consider the following problem for $v_{\varphi }$ in ${\mathbb {R}}^3_{\varepsilon }= \{x\in {\mathbb {R}}^3: r>\varepsilon \}, \varepsilon >0,$

$$\begin{aligned} \begin{aligned}&v_{\varphi ,t} + v\cdot \nabla v_{\varphi }-\nu \Delta v_{\varphi } + \nu \frac{v_{\varphi }}{r^2} + \frac{v_r v_{\varphi }}{r} =0, \\&v_{\varphi }|_{r=\varepsilon }=0, \quad v_{\varphi }|_{r\rightarrow \infty } =0. \end{aligned} \end{aligned}$$

(2.11)

Multiplying (2.11) by $\frac{v_{\varphi }^3}{r^2},$ integrating over ${\mathbb {R}}^3_{\varepsilon }$ and using boundary conditions yields

$$\begin{aligned} \begin{aligned}&\frac{1}{4} \frac{d}{{\textit{dt}}} \left\| \frac{v_{\varphi }^2}{r}\right\| _{L_2({\mathbb {R}}^3_{\varepsilon })}^2 + \frac{3}{4} \nu \left\| {\tilde{\nabla }}\frac{v_{\varphi }^2}{r}\right\| _{L_2({\mathbb {R}}^3_{\varepsilon })}^2 + \frac{3}{4}\nu \left\| \frac{v_{\varphi }}{r}\right\| _{L_4({\mathbb {R}}^3_{\varepsilon })}^4 = - \frac{3}{2}\int _{{\mathbb {R}}^3_{\varepsilon }}\frac{v_r}{r} \frac{v_{\varphi }^2}{r}\frac{v_{\varphi }^2}{r} {\textit{dx}} \\&\quad \le \frac{3}{2} \left\| \frac{v_r}{r}\right\| _{L_6({\mathbb {R}}^3_{\varepsilon })} \left\| \frac{v_{\varphi }^2}{r}\right\| _{L_2({\mathbb {R}}^3_{\varepsilon })} \left\| \frac{v_{\varphi }^2}{r}\right\| _{L_3({\mathbb {R}}^3_{\varepsilon })} \le c \left\| {\tilde{\nabla }}\frac{v_r}{r}\right\| _{L_2({\mathbb {R}}^3_{\varepsilon })} \left\| \frac{v_{\varphi }^2}{r}\right\| ^{3/2}_{L_2({\mathbb {R}}^3_{\varepsilon })} \left\| {\tilde{\nabla }}\frac{v_{\varphi }^2}{r}\right\| ^{1/2}_{L_2({\mathbb {R}}^3_{\varepsilon })} \\&\quad \le \frac{\nu }{2} \left\| {\tilde{\nabla }}\frac{v_{\varphi }^2}{r}\right\| ^2_{L_2({\mathbb {R}}^3_{\varepsilon })} + c \left\| \frac{v_{\varphi }^2}{r}\right\| ^{2}_{L_2({\mathbb {R}}^3_{\varepsilon })} \left\| {\tilde{\nabla }}\frac{v_{r}}{r}\right\| ^{4/3}_{L_2({\mathbb {R}}^3_{\varepsilon })}. \end{aligned} \end{aligned}$$

Simplifying we have

$$\begin{aligned} \frac{1}{4} \frac{d}{{\textit{dt}}} \left\| \frac{v_{\varphi }^2}{r}\right\| _{L_2({\mathbb {R}}^3_{\varepsilon })}^2 + \frac{\nu }{4} \left\| {\tilde{\nabla }}\frac{v_{\varphi }^2}{r}\right\| _{L_2({\mathbb {R}}^3_{\varepsilon })}^2 + \frac{3}{4}\nu \left\| \frac{v_{\varphi }}{r}\right\| _{L_4({\mathbb {R}}^3_{\varepsilon })}^4 \le c \left\| \frac{v_{\varphi }^2}{r}\right\| ^{2}_{L_2({\mathbb {R}}^3_{\varepsilon })} \left\| {\tilde{\nabla }}\frac{v_{r}}{r}\right\| ^{4/3}_{L_2({\mathbb {R}}^3_{\varepsilon })}. \end{aligned}$$

By the Gronwall lemma we have

$$\begin{aligned}&\frac{1}{4} \left\| \frac{v_{\varphi }^2(t)}{r}\right\| _{L_2({\mathbb {R}}^3_{\varepsilon })}^2 + \frac{\nu }{4} \int _0^t \left\| {\tilde{\nabla }}\frac{v_{\varphi }^2}{r}\right\| _{L_2({\mathbb {R}}^3_{\varepsilon })}^2 {\textit{dt}}' + \frac{3}{4}\nu \int _0^t \left\| \frac{v_{\varphi }}{r}\right\| _{L_4({\mathbb {R}}^3_{\varepsilon })}^4 {\textit{dt}}' \\&\quad \le \frac{1}{4}\exp [c\int _0^t \left\| {\tilde{\nabla }}\frac{v_r}{r}\right\| _{L_2({\mathbb {R}}^3_{\varepsilon })}^{4/3} {\textit{dt}}'] \left\| \frac{v_{\varphi }^2(0)}{r}\right\| _{L_2({\mathbb {R}}^3_{\varepsilon })}^2. \end{aligned}$$

Passing with $\varepsilon \rightarrow 0$ we derive (2.10) and conclude the proof. $\square$

Remark 2.7

Formula (2.4) in [9] has the form

$$\begin{aligned} \left\| {\tilde{\nabla }}\frac{v_r}{r}\right\| _{L_q({\mathbb {R}}^3)} \le c(q)\left\| \frac{\omega _{\varphi }}{r}\right\| _{L_q({\mathbb {R}}^3)}, \quad 1<q<\infty . \end{aligned}$$

(2.12)

Consider problem (1.3).

Lemma 2.8

Let the assumptions of Lemma 2.6 be satisfied. Assume additionally that $w_1(0) \in L_2({\mathbb {R}}^3), u_1(0) \in L_4(0,t;L_4({\mathbb {R}}^3)).$ Then the following estimate holds

$$\begin{aligned} \begin{aligned}&\frac{1}{2}\Vert \omega _1(t)\Vert ^2_{L_2({\mathbb {R}}^3)}+\frac{\nu }{2}\int _0^t\Vert \nabla \omega _1(t')\Vert ^2_{L_2({\mathbb {R}}^3)} {\textit{dt}}' \\&\quad \le \frac{2}{\nu }\int _0^t\Vert u_1(t')\Vert ^4_{L_4({\mathbb {R}}^3)} dt'+ \frac{1}{2}\Vert \omega _1(0)\Vert ^2_{L_2({\mathbb {R}}^3)}, \quad t\le T. \end{aligned} \end{aligned}$$

(2.13)

Proof

Consider the problem in ${\mathbb {R}}^3_{\varepsilon }$

$$\begin{aligned} \begin{aligned}&\omega _{1,t}+ v\cdot \nabla \omega _1 - \nu \left( \Delta \omega _1+\frac{2}{\nu }\omega _{1,r}\right) = 2u_1u_{1,z}, \\&\omega _1|_{r=\varepsilon }=0, \quad \omega _1|_{r\rightarrow \infty } =0. \end{aligned} \end{aligned}$$

(2.14)

Multiplying (2.14) by $\omega _1$, integrating over ${\mathbb {R}}^3_{\varepsilon }$, using the boundary conditions yields

$$\begin{aligned} \begin{aligned}&\frac{1}{2}\frac{d}{{\textit{dt}}}\Vert \omega _1(t)\Vert ^2_{L_2({\mathbb {R}}^3_{\varepsilon })}+\frac{\nu }{2}\Vert \nabla \omega _1\Vert ^2_{L_2({\mathbb {R}}^3_{\varepsilon })} +\nu \int _{{\mathbb {R}}}\omega _1^2|_{r=\varepsilon } {\textit{dz}} \\ {}&\quad \le \frac{2}{\nu }\int _{{\mathbb {R}}^3_{\varepsilon }}|u_1|^4 {\textit{dx}}. \end{aligned} \end{aligned}$$

Integrating with respect to time and passing with $\varepsilon \rightarrow 0$ gives (2.13). This concludes the proof. $\square$

Introduce the quantities

$$\begin{aligned} (\Phi , \Gamma ) = \left( \frac{\omega _r}{r}, \frac{\omega _{\varphi }}{r}\right) \end{aligned}$$

which are solutions to the equations

$$\begin{aligned}&\partial _t \Phi + v\cdot \nabla \Phi - \nu \left( \Delta + \frac{2}{r}\right) \Phi - (\omega _r \partial _r + \omega _z\partial _z)\frac{v_r}{r} =0, \\&\partial _t \Gamma + v\cdot \nabla \Gamma - \nu \left( \Delta + \frac{2}{r}\right) \Gamma + 2 \frac{v_r}{r}\Phi =0. \end{aligned}$$

Remark 2.9

In the proof of Theorem 1.1, Case 1 in [9] there is derived the formula (3.8) in [9] in the form

$$\begin{aligned} \begin{aligned}&\Vert \Phi (t)\Vert ^2_{L_2({\mathbb {R}}^3)}+ \Vert \Gamma (t)\Vert ^2_{L_2({\mathbb {R}}^3)}\\ {}&\qquad +\nu \int _0^t (\Vert {\tilde{\nabla }} \Phi (t)\Vert ^2_{L_2({\mathbb {R}}^3)} +\Vert {\tilde{\nabla }}\Gamma \Vert ^2_{L_2({\mathbb {R}}^3)}) {\textit{dt}}' \\ {}&\le \exp [c(1+\Vert r^dv_{\varphi }\Vert _{L_{p,q}({\mathbb {R}}^3\times (0,t))})](\Vert \Phi (0)\Vert ^2_{L_2({\mathbb {R}}^3)}+ \Vert \Gamma (0)\Vert ^2_{L_2({\mathbb {R}}^3)} ), \end{aligned} \end{aligned}$$

(2.15)

where

$$\begin{aligned}&3/p+2/q \le 1-d,\quad 0\le d< 1, \\&\frac{3}{1-d}< p\le \infty ,\quad \frac{3}{1-d} \le q < \infty . \end{aligned}$$

Let us recall some properties of weak solutions to (1.1).

Lemma 2.10

(See [15, Ch.2, Sect. 3]). For arbitrary $v\in L_{\infty }(0,T;L_2(\Omega ) \, \cap L_2(0,T;H^1(\Omega ))$ the inequality holds:

$$\begin{aligned} \begin{aligned} \Vert v\Vert ^2_{L_{p,q}({\mathbb {R}}^3\times (0,T))} \le c \sup _{0\le t \le T}\int _{{\mathbb {R}}^3} |v|^2 dx + c\int _0^T dt \int _{{\mathbb {R}}^3}|\nabla v|^2 dx, \end{aligned} \end{aligned}$$

(2.16)

where

$$\begin{aligned} \frac{3}{p}+\frac{2}{q} \ge \frac{3}{2}. \end{aligned}$$

3 Sufficient Conditions for Regularity

Let

$$\begin{aligned} u_{\alpha } = \frac{u}{r^{\alpha }}, \quad \alpha \in (0,1). \end{aligned}$$

Then $u_{\alpha }$ satisfies

$$\begin{aligned} \begin{aligned}&u_{\alpha ,t} + v\cdot \nabla u_{\alpha }+ \alpha \frac{v_r}{r}u_{\alpha }- \nu \Delta u_{\alpha }+\frac{2\nu (1-\alpha )}{r}u_{\alpha ,r} + \frac{\nu \alpha (2-\alpha )}{r^2}u_{\alpha }=0, \\&u_{\alpha }|_{t=0}=u_{\alpha }(0). \end{aligned} \end{aligned}$$

(3.1)

Lemma 3.1

Let $u(0) \in L_{\infty }({\mathbb {R}}^3)$. Assume that $c_1$ is a constant and

$$\begin{aligned} \int _0^T\int _{{\mathbb {R}}^3} \frac{v_r^2}{r^3} dx dt \le c_1^2, \end{aligned}$$

Let

$$\begin{aligned}&c_2= \frac{1}{4\nu \alpha (2-\alpha )}\Vert u(0)\Vert ^s_{L_{\infty }({\mathbb {R}}^3)}c_1^2 + \frac{1}{s} \Vert \Vert u(0)\Vert ^s_{L_s({\mathbb {R}}^3)}, \\&\quad \mathrm{where}\quad \alpha =\frac{3}{s},\quad u_{\alpha }= v_{\varphi }r^{1-\alpha }. \end{aligned}$$

Then

$$\begin{aligned} \begin{aligned}&r^d v_{\varphi }\in L_{p,q}({\mathbb {R}}^3\times (0,T)),\quad \frac{3}{p}+ \frac{2}{q} \ge \frac{3}{s}=\alpha ,\\&d=1-\alpha \quad \mathrm{and} \quad \Vert r^d v_{\varphi }\Vert ^s_{L_{p,q}({\mathbb {R}}^3\times (0,T))} \le c_2. \end{aligned} \end{aligned}$$

(3.2)

Proof

Multiplying (3.1) by $u_{\alpha }|u_{\alpha }|^{s-2},$ integrating the result over ${\mathbb {R}}^3$ and using that $u_{\alpha } \in C^{\infty }({\mathbb {R}}^3\setminus I)$, and $u_{\alpha }$ has compact support for large, finite $x\in {\mathbb {R}}^3,$ we obtain

$$\begin{aligned}&\frac{1}{s}\frac{d}{dt}\int _{{\mathbb {R}}^3}|u_{\alpha }|^s {\textit{dx}}+ \alpha \int _{{\mathbb {R}}^3}\frac{v_r}{r}|u_{\alpha }|^s {\textit{dx}} + \nu \int _{{\mathbb {R}}^3}|\nabla |u_{\alpha }|^{s/2}|^2 {\textit{dx}} \\&\quad +\, \nu \alpha (2-\alpha )\int _{{\mathbb {R}}^3}\frac{|u_{\alpha }|^s}{r^2} {\textit{dx}} = 0. \end{aligned}$$

The second term in the l.h.s. of the above equality can be estimated by

$$\begin{aligned} \left| \int _{{\mathbb {R}}^3}v_r|u_{\alpha }|^{s/2}\frac{|u_{\alpha }|^{s/2}}{r} {\textit{dx}} \right| \le \frac{\varepsilon }{2} \int _{{\mathbb {R}}^3}\frac{|u_{\alpha }|^{s}}{r^2} {\textit{dx}} +\frac{1}{2\varepsilon } \int _{{\mathbb {R}}^3}v_r^2|u_{\alpha }|^s {\textit{dx}} \equiv I. \end{aligned}$$

Using Lemma 2.4, the second integral in I is bounded by

$$\begin{aligned} \Vert u(0)\Vert ^s_{L_{\infty }({\mathbb {R}}^3)} \int _{{\mathbb {R}}^3}\frac{v_r^2}{r^{\alpha s}} {\textit{dx}}. \end{aligned}$$

Employing this estimate, with $\varepsilon =\nu \alpha (2-\alpha ),$ integrating the result with respect to time and using the density argument, yields

$$\begin{aligned} \begin{aligned} \begin{aligned}&\frac{1}{s}\int _{{\mathbb {R}}^3}|u_{\alpha }(t)|^s {\textit{dx}} + \nu \int _0^t {\textit{dt}}' \int _{{\mathbb {R}}^3}|\nabla |u_{\alpha }|^{s/2}|^2 {\textit{dx}} \\ {}&\qquad + \frac{\nu \alpha (2-\alpha )}{2} \int _0^t {\textit{dt}}' \int _{{\mathbb {R}}^3}\frac{|u_{\alpha }|^s}{r^2} {\textit{dx}} \\ {}&\quad \le \frac{\Vert u(0)\Vert ^s_{L_{\infty }({\mathbb {R}}^3)}}{4\nu \alpha (2-\alpha )} \int _0^t dt' \int _{{\mathbb {R}}^3}\frac{v_r^2}{r^{\alpha s}}{\textit{dx}} + \frac{1}{s}\int _{{\mathbb {R}}^3}|u_{\alpha }(0)|^s {\textit{dx}}, \quad t\le T. \end{aligned} \end{aligned}\end{aligned}$$

(3.3)

In view of the assumptions of the lemma, the r.h.s. of (3.3) is bounded by a constant $c_2$. Then the density argument and Lemma 2.10 imply

$$\begin{aligned} \Vert u_{\alpha }^{s/2}\Vert _{L_{p,q}({\mathbb {R}}^3\times (0,T))} \le c_2^{1/2} \quad \mathrm{with} \quad \frac{3}{p}+ \frac{2}{q} \ge \frac{3}{2}. \end{aligned}$$

Hence

$$\begin{aligned} \Vert u_{\alpha }\Vert _{L_{p',q'}({\mathbb {R}}^3\times (0,T))}^{s/2} = \Vert u_{\alpha }\Vert ^{s/2}_{L_{sp/2,sq/2}({\mathbb {R}}^3\times (0,T))} \le c_2^{1/2} \end{aligned}$$

where

$$\begin{aligned} \frac{3}{p'}+ \frac{2}{q'} \ge \frac{3}{2}\cdot \frac{2}{s}= \frac{3}{s}. \end{aligned}$$

Comparing the above approach with (2.15) we have that $d=1-\alpha$ and the regularity criterion has the form

$$\begin{aligned} r^dv_{\varphi }\in L_{q'}(0,T;L_{p'}({\mathbb {R}}^3)), \quad \frac{3}{p'}+\frac{2}{q'} \le 1-d,\quad 0\le d<1. \end{aligned}$$

Hence

$$\begin{aligned} \frac{3}{p'}+ \frac{2}{q'} \ge \frac{3}{s} = \alpha . \end{aligned}$$

Therefore $\alpha s=3$. $\square$

Corollary 3.2

From (3.2), (2.15) and (2.10) we obtain

$$\begin{aligned} \left\| \frac{v_{\varphi }^2}{r}\right\| _{L_2({\mathbb {R}}^3)}^2 + \int _0^t \left\| {\tilde{\nabla }}\frac{v_{\varphi }^2}{r}\right\| _{L_2({\mathbb {R}}^3)}^2 dt' \le c_3, \quad t\le T, \end{aligned}$$

where $c_3$ depends on the constants from the r.h.s. of (3.2), (2.15) and (2.10).

In view of (2.16) we have

$$\begin{aligned} \left\| \frac{v_{\varphi }^2}{r}\right\| _{L_{p,q}({\mathbb {R}}^3\times (0,T))} \le cc_3, \end{aligned}$$

(3.4)

where

$$\begin{aligned} \frac{3}{p} + \frac{2}{q} \ge \frac{3}{2}. \end{aligned}$$

Let ${\mathbb {R}}^3_{r_0} = \{x\in {\mathbb {R}}^3: r\le r_0 \}.$ Then (3.4) implies

$$\begin{aligned} \left\| {v_{\varphi }^2}\right\| _{L_{p,q}({\mathbb {R}}^3_{r_0}\times (0,T))}\le \left\| \frac{v_{\varphi }^2}{r}\right\| _{L_{p,q}({\mathbb {R}}^3_{r_0}\times (0,T))} \le cc_3. \end{aligned}$$

(3.5)

Hence (3.5) implies that $v_{\varphi } \in L_{p',q'}({\mathbb {R}}^3_{r_0}\times (0,T)),$ where

$$\begin{aligned} \frac{3}{p'} + \frac{2}{q'} \ge \frac{3}{4}. \end{aligned}$$

Consider Lemma 2.3. Let $s=4$. Then $r=8$, so

$$\begin{aligned} \frac{3}{4} + \frac{2}{8} =1 \end{aligned}$$

Since $\frac{3}{4} < 1, v_{\varphi }$ satisfies assumptions of Lemma 2.3. Hence v has no singular points in ${\mathbb {R}}^3_{r_0}\times (0,T).$

Next we show that axisymmetric solutions to problem (1.1) do not have singular points in the region located in a positive distance from the axis of symmetry.

In [13] is shown that singular points of v in the axisymmetric case may appear on the axis of symmetry only. Therefore in any region located in a positive distance from the axis of symmetry there is no singular points of v. However, we want to show that statement explicitly. Therefore, we proceed as follows.

Consider Eq. (1.2)$_1$. Let $\chi = \chi (r)$ be a smooth function such that

$$\begin{aligned} \chi (r) = \left\{ \begin{array}{ll} 1 &{}\quad r\ge 2r_0, \\ 0 &{}\quad r\le r_0. \end{array} \right. \end{aligned}$$

We multiply (1.2)$_1$ by $\chi$ and introduce the notation ${\hat{v}}_{\varphi } = v_{\varphi } \chi$. Then ${\hat{v}}_{\varphi }$ satisfies

$$\begin{aligned} \begin{aligned}&{\hat{v}}_{\varphi ,t}+v\cdot \nabla {\hat{v}}_\varphi -\nu \Delta {\hat{v}}_\varphi +\frac{v_r}{r}{\hat{v}}_\varphi + \nu \frac{{\hat{v}}_\varphi }{r^2} \\&\quad = v \cdot \nabla \chi v_{\varphi } -2\nu \nabla v_{\varphi } \nabla \chi - \nu v_{\varphi } \Delta \chi , \\&\qquad {\hat{v}}_{\varphi }|_{t=0} = {\hat{v}}_{\varphi }(0). \end{aligned} \end{aligned}$$

(3.6)

Multiplying (3.6) by ${\hat{v}}_{\varphi }|{\hat{v}}_{\varphi }|^{s-2}$ and integrating over ${\mathbb {R}}^3\times (0,t)$ yields

$$\begin{aligned} \begin{aligned}&\frac{1}{s} \Vert {\hat{v}}_{\varphi }(t)\Vert ^s_{L_s({\mathbb {R}}^3)} + \nu \int _0^t \Vert \nabla |{\hat{v}}_{\varphi }|^{s/2}\Vert ^2_{L_2({\mathbb {R}}^3)} dt' \\&\qquad + \nu \int _0^t \Vert \frac{|{\hat{v}}_{\varphi }|^{s/2}}{r}\Vert ^2_{L_2({\mathbb {R}}^3)} dt' + \int _0^t \int _{{\mathbb {R}}^3}\frac{v_r}{r} |{\hat{v}}_{\varphi }|^s dx dt' \\&\quad = \int _0^t \int _{{\mathbb {R}}^3}(v\cdot \nabla \chi v_{\varphi } - 2\nu \nabla v_{\varphi } \nabla \chi - \nu v_{\varphi } \Delta \chi ){\hat{v}}_{\varphi }|{\hat{v}}_{\varphi }|^{s-2} dx dt'. \end{aligned} \end{aligned}$$

(3.7)

In view of Lemma 2.10 the first two terms on the l.h.s. of (3.7) are estimated from below by

$$\begin{aligned} \Vert {\hat{v}}_{\varphi }\Vert ^s_{L_{\frac{5s}{3}}({\mathbb {R}}^3\times (0,t))}. \end{aligned}$$

(3.8)

Next we estimate non-positive terms in (3.7). In view of the energy estimate (2.1) the last term on the l.h.s. of (3.7) is bounded by

$$\begin{aligned} \Vert v_r\Vert _{L_{\frac{10}{3}}({\mathbb {R}}^3\times (0,t))} \Vert {\hat{v}}_{\varphi }\Vert ^s_{L_{\frac{10s}{7}}({\mathbb {R}}^3\times (0,t))}. \end{aligned}$$

where the second factor can be always absorbed by (3.8) because $5/3 >10/7.$ In view of (2.9) and (2.1) the first integral on the r.h.s. of (3.7) is bounded by

$$\begin{aligned} \Vert v\Vert _{L_{\frac{10}{3}}({\mathbb {R}}^3\times (0,t))} \Vert v_{\varphi }\Vert _{L_4({\mathbb {R}}^3\times (0,t))} \Vert {\hat{v}}_{\varphi }\Vert ^{s-1}_{L_{\frac{20}{9}(s-1)}({\mathbb {R}}^3\times (0,t))}. \end{aligned}$$

(3.9)

The last factor in (3.9) can be absorbed by (3.8) for $\frac{20}{9}(s-1) \le \frac{5}{3}s$ which holds for $s\le 4.$ Then (3.8) yields estimate for

$$\begin{aligned} \Vert {\hat{v}}_{\varphi }\Vert _{L_{\frac{20}{3}}({\mathbb {R}}^3\times (0,t))}. \end{aligned}$$

Finally we estimate the last two integrals on the r.h.s. of (3.7). We write them in the form

$$\begin{aligned} \begin{aligned}&-\nu \int _0^t \int _{{\mathbb {R}}^3}(2\nabla v_{\varphi } \nabla \chi + v_{\varphi } \Delta \chi ){\hat{v}}_{\varphi }|{\hat{v}}_{\varphi }|^{s-2} dx dt' \\&\quad = -\nu \int _0^t \int _{{\mathbb {R}}^3}(2 \nabla v_{\varphi } \nabla \chi {v}_{\varphi }\chi |{\hat{v}}_{\varphi }|^{s-2} + v_{\varphi } \Delta \chi {\hat{v}}_{\varphi }|{\hat{v}}_{\varphi }|^{s-2}) dx dt' \\&\quad = -\nu \int _0^t \int _{{\mathbb {R}}^3}(\nabla v^2_{\varphi } \nabla \chi \chi |{\hat{v}}_{\varphi }|^{s-2} + v_{\varphi } \Delta \chi {\hat{v}}_{\varphi }|{\hat{v}}_{\varphi }|^{s-2}) dx dt' \\ {}&\quad = -\nu \int _0^t \int _{{\mathbb {R}}^3}(- v^2_{\varphi } \Delta \chi \chi |{\hat{v}}_{\varphi }|^{s-2} - v^2_{\varphi } |\nabla \chi |^2 |{\hat{v}}_{\varphi }|^{s-2} \\ {}&\qquad - v^2_{\varphi } \nabla \chi \chi \nabla |{\hat{v}}_{\varphi }|^{s-2} + v_{\varphi } \Delta \chi {\hat{v}}_{\varphi } |{\hat{v}}_{\varphi }|^{s-2})dx dt' \\&\quad = \nu \int _0^t \int _{{\mathbb {R}}^3}(v^2_{\varphi } |\nabla \chi |^2 |{\hat{v}}_{\varphi }|^{s-2} + v^2_{\varphi } \nabla \chi \chi \nabla |{\hat{v}}_{\varphi }|^{s-2})dx dt' = I_1+I_2, \end{aligned} \end{aligned}$$

where

$$\begin{aligned} |I_1| \le c\Vert v_{\varphi }\Vert _{L_4({\mathbb {R}}^3\times (0,t))} \Vert {\hat{v}}_{\varphi }\Vert ^{s-2}_{L_{2(s-2)}({\mathbb {R}}^3\times (0,t))} \end{aligned}$$

and the first factor is bounded in view of Remark 2.5 and the second factor is absorbed by (3.8) for $2(s-2)\le \frac{5}{2} s$ which holds for $s\le 12.$ Finally

$$\begin{aligned} \begin{aligned} I_2&= \nu (s-2)\int _0^t\int _{{\mathbb {R}}^3}v^2_{\varphi }\nabla \chi \chi |{\hat{v}}_{\varphi }|^{s-3}\nabla |{\hat{v}}_{\varphi }|dx dt' \\&\quad \le \frac{\nu (s-2)}{s} \int _0^t\int _{{\mathbb {R}}^3}v^2_{\varphi }\nabla \chi \chi |{\hat{v}}_{\varphi }|^{s/2-2}\nabla |{\hat{v}}_{\varphi }|^{s/2}dx dt' \\&\quad \le \frac{\varepsilon }{2} \int _0^t\int _{{\mathbb {R}}^3}|\nabla |{\hat{v}}_{\varphi }|^{s/2}|^2 dx dt' + \frac{c}{\varepsilon } \int _0^t\int _{{\mathbb {R}}^3}|v_{\varphi }|^2|{\hat{v}}_{\varphi }|^{s-2} dx dt', \end{aligned} \end{aligned}$$

where the first integral is absorbed by the second term on the l.h.s. of (3.7) and the second is bounded by

$$\begin{aligned} \left( \int _0^t\int _{{\mathbb {R}}^3}|{v}_{\varphi }|^4 dx dt'\right) ^{\frac{1}{2}} \left( \int _0^t\int _{{\mathbb {R}}^3}|{\hat{v}}_{\varphi }|^{2(s-2)} dx dt'\right) ^{\frac{1}{2}}=I_3. \end{aligned}$$

The first factor in $I_3$ is bounded in virtue of Remark 2.5 and the second factor is absorbed by (3.8) if $2(s-2)\le \frac{5}{3}s$ so $s\le 12.$ Summarizing, we obtain the estimate

$$\begin{aligned} \Vert {\hat{v}}_{\varphi }\Vert _{L_{\frac{20}{3}}({\mathbb {R}}^3\times (0,t))} \le c. \end{aligned}$$

(3.10)

We observe that the estimate (3.10) above is not strong enough to apply Lemma 2.2(1) because for $s=r$ it is required that $s\ge \frac{50}{7}.$

To increase regularity we introduce a new smooth cut-off function

$$\begin{aligned} \breve{\chi }(r) = \left\{ \begin{array}{ll} 1 &{}\quad r\ge 3r_0, \\ 0 &{}\quad r\le 2r_0 \end{array} \right. \end{aligned}$$

Introducing notation $\breve{v}_{\varphi }= {v}_{\varphi }\breve{\chi }$ we replace (3.6) by

$$\begin{aligned}&\breve{v}_{\varphi ,t}+v\cdot \nabla \breve{v}_\varphi -\nu \Delta \breve{v}_\varphi +\frac{v_r}{r}\breve{v}_\varphi + \nu \frac{\breve{v}_\varphi }{r^2} \\&\quad = v \cdot \nabla \breve{\chi } {\hat{v}}_{\varphi } -2\nu \nabla {\hat{v}}_{\varphi } \nabla \breve{\chi } - \nu {\hat{v}}_{\varphi } \Delta \breve{\chi }, \\&\quad \breve{v}_{\varphi }|_{t=0} = \breve{v}_{\varphi }(0). \end{aligned}$$

where we can use (3.10). Hence the above problem increases regularity of (3.10), so we can meet assumptions of Lemma 2.2(1). This concludes the proof.

Lemma 3.3

Assume that

$$\begin{aligned} \int _0^T dt \int _{{\mathbb {R}}^3}\frac{\omega ^2_{\varphi }}{r} dx < \infty . \end{aligned}$$

Then there exists a constant c such that

$$\begin{aligned} \int _0^T dt \int _{{\mathbb {R}}^3}\frac{v_r^2}{r^3} dx \le c\int _0^T dt \int _{{\mathbb {R}}^3}\frac{\omega ^2_{\varphi }}{r} dx. \end{aligned}$$

(3.11)

Proof

From (1.2)$_3$ we have

$$\begin{aligned} - \Delta \psi _{,z} + \frac{1}{r^2}\psi _{,z}= \omega _{\varphi ,z}, \end{aligned}$$

(3.12)

where $v_r= -\psi _{,z}.$

To simplify further considerations we introduce the notation

$$\begin{aligned} u=\psi _{,z},\quad f= \omega _{\varphi }. \end{aligned}$$

Then (3.12) takes the form

$$\begin{aligned} -\Delta u + \frac{1}{r^2} u= f_{,z}. \end{aligned}$$

(3.13)

Recall that I is the axis of symmetry. Let $C^{\infty }_0({\mathbb {R}}^3{\setminus } I)$ be the set of smooth functions vanishing near I and outside a compact set.

Let $H^1_0({\mathbb {R}}^3)$ be the closure of $C^{\infty }_0({\mathbb {R}}^3{\setminus }I)$ in the norm

$$\begin{aligned} \Vert u\Vert _{H^1_0({\mathbb {R}}^3)}= \left( \int _{{\mathbb {R}}^3}(|\nabla u|^2 + \frac{1}{r^2}|u|^2) dx\right) ^{\frac{1}{2}}. \end{aligned}$$

Recall that functions from $H^1_0({\mathbb {R}}^3)$ vanish on I. By the weak solution to (3.13) we mean a function $u\in H^1_0({\mathbb {R}}^3)$ satisfying the integral identity

$$\begin{aligned} \int _{{\mathbb {R}}^3}\left( \nabla u\cdot \nabla \chi + \frac{1}{r^2} u \chi \right) dx = - \int _{{\mathbb {R}}^3}f \chi _{,z} dx, \end{aligned}$$

(3.14)

which holds for any smooth function $\chi$ belonging to $H^1_0({\mathbb {R}}^3).$ Introducing the scalar product

$$\begin{aligned} (u,v)_{H^1_0({\mathbb {R}}^3)} = \int _{{\mathbb {R}}^3}\left( \nabla u\cdot \nabla v + \frac{1}{r^2} u v\right) dx, \end{aligned}$$

(3.15)

we can write (3.14) in the following short form

$$\begin{aligned} (u,\chi )_{H^1_0({\mathbb {R}}^3)} = -(f, \chi _{,z})_{L_2({\mathbb {R}}^3)}, \end{aligned}$$

where

$$\begin{aligned} (u,v)_{L_2({\mathbb {R}}^3)} = \int _{{\mathbb {R}}^3}uv dx. \end{aligned}$$

For $f\in L_2({\mathbb {R}}^3)$, $(f, \chi _{,z})_{L_2({\mathbb {R}}^3)}$ is a linear functional on $H^1_0({\mathbb {R}}^3).$ Hence we have

$$\begin{aligned} |(f, \chi _{,z})_{L_2({\mathbb {R}}^3)}| \le \Vert f\Vert _{L_2({\mathbb {R}}^3)}\Vert \chi _{,z}\Vert _{L_2({\mathbb {R}}^3)} \le \Vert f\Vert _{L_2({\mathbb {R}}^3)}\Vert \chi \Vert _{H^1_0({\mathbb {R}}^3)}. \end{aligned}$$

Hence by the Riesz Theorem there exists $F\in H^1_0({\mathbb {R}}^3)$ such that

$$\begin{aligned} -(f, \chi _{,z})_{L_2({\mathbb {R}}^3)} = (F, \chi )_{H^1_0({\mathbb {R}}^3)}. \end{aligned}$$

Therefore there exists a solution to the integral identity (3.15) such that $u=F\in H^1_0({\mathbb {R}}^3)$ and the estimate holds

$$\begin{aligned} \Vert u||_{H^1_0({\mathbb {R}}^3)} \le c\Vert f\Vert _{L_2({\mathbb {R}}^3)}. \end{aligned}$$

It is clear that the solution is unique.

Since u vanishes for $r=0$, $u=\psi _{,z}$ so $\psi |_{r=0}=0$ also. Therefore we can look for approximate weak solution satisfying the integral identity

$$\begin{aligned} \int _{{\mathbb {R}}^3_{\varepsilon }} \left( \nabla u \cdot \nabla \chi + \frac{1}{r^2} u \chi \right) dx, = - \int _{{\mathbb {R}}^3_{\varepsilon }} f \chi _{,z}, \end{aligned}$$

(3.16)

where ${\mathbb {R}}^3_{\varepsilon }= \{x\in {\mathbb {R}}^3: r>\varepsilon \}, \varepsilon >0.$ Let $\chi =\frac{\psi _{,z}}{r^{\alpha }}.$ Then recalling notation $u=\psi _{,z}, f=\omega _{\varphi }$ identity (3.16) takes the form

$$\begin{aligned} \int _{{\mathbb {R}}^3_{\varepsilon }}\left( \nabla \psi _{,z}\frac{\nabla {\psi _{,z}}}{r^{\alpha }} + \frac{1}{r^{2+\alpha }}|\psi _{,z}|^2\right) dx = - \int _{{\mathbb {R}}^3_{\varepsilon }} \frac{\omega _{\varphi }}{r^{\alpha }}\psi _{,zz} dx. \end{aligned}$$

Performing differentiation we have

$$\begin{aligned} \begin{aligned}&\int _{{\mathbb {R}}^3_{\varepsilon }} \frac{1}{r^{\alpha }}|\nabla \psi _{,z}|^2 dx -\alpha \int _{{\mathbb {R}}^3_{\varepsilon }}\nabla \psi _{,z}\psi _{,z} \nabla r \ r^{-\alpha -1}dx \\&\quad + \int _{{\mathbb {R}}^3_{\varepsilon }}\frac{\psi _{,z}^2}{r^{2+\alpha }}dx = -\int _{{\mathbb {R}}^3_{\varepsilon }} \frac{\omega _{\varphi }}{r^{\alpha }}\psi _{,zz}dx. \end{aligned} \end{aligned}$$

(3.17)

The second integral on the l.h.s. of (3.17) takes the form

$$\begin{aligned}&-\alpha \int _{{\mathbb {R}}^3_{\varepsilon }}\partial _r \psi _{,z}\psi _{,z} r^{-\alpha -1} r dr dz = -\frac{\alpha }{2} \int _{{\mathbb {R}}^3_{\varepsilon }}(\psi _{,z}^2)_{,r} r^{-\alpha } dr dz \\&\quad = -\frac{\alpha }{2}\int _{{\mathbb {R}}^3_{\varepsilon }} \partial _r (\psi _{,z}^2r^{-\alpha }) dr dz - \frac{\alpha ^2}{2}\int _{{\mathbb {R}}^3_{\varepsilon }} \psi _{,z}^2 r^{-\alpha -1} dr dz \\&\quad = \frac{\alpha }{2}\int _{{\mathbb {R}}^3_{\varepsilon }}\psi _{,z}^2r^{-\alpha }|_{r=\varepsilon }dz - \frac{\alpha ^2}{2}\int _{{\mathbb {R}}^3_{\varepsilon }} \psi _{,z}^2 r^{-\alpha -2} dx. \end{aligned}$$

Using this in (3.17) and applying the Hölder and Young inequalities to the r.h.s. term yields

$$\begin{aligned} \int _{{\mathbb {R}}^3_{\varepsilon }}\frac{1}{r^{\alpha }}|\nabla \psi _{,z}|^2 dx +\left( 1-\frac{\alpha ^2}{2}\right) \int _{{\mathbb {R}}^3_{\varepsilon }}\frac{\psi _{,z}^2}{r^{2+\alpha }} dx \le c \int _{{\mathbb {R}}^3_{\varepsilon }} \frac{\omega _{\varphi }^2}{r^{\alpha }}dx. \end{aligned}$$

(3.18)

We have to emphasize that $\psi$ in (3.18) is an approximate function. This should be denoted with $\psi ^{\varepsilon }$ but we omitted it for simplicity.

Passing with $\varepsilon \rightarrow 0$, setting $\alpha =1$ and integrating this inequality with respect to time implies (3.11). This concludes the proof. $\square$

Proof of Theorem 1

Proof

From (2.10) and (2.12) we have

$$\begin{aligned}&\left\| \frac{v_{\varphi }^2}{r}\right\| _{L_2({\mathbb {R}}^3)}^2 + \int _0^t \left\| {\nabla }\frac{v_{\varphi }^2}{r}\right\| _{L_2({\mathbb {R}}^3)}^2 dt' \\&\quad \le \exp \left( c \int _0^t \left\| \frac{\omega _{\varphi }}{r}\right\| _{L_2({\mathbb {R}}^3)}^{4/3}dt'\right) \cdot \left\| {\nabla }\frac{v_{\varphi }^2(0)}{r}\right\| _{L_2({\mathbb {R}}^3)}^2, \quad t\le T. \end{aligned}$$

Next (2.15) implies

$$\begin{aligned} \begin{aligned}&\left\| \frac{\omega _r}{r}\right\| _{L_2({\mathbb {R}}^3)}^2 + \left\| \frac{\omega _{\varphi }}{r}\right\| _{L_2({\mathbb {R}}^3)}^2 \\&\le c \exp (\Vert u_{\alpha }\Vert _{L_{p,q}({\mathbb {R}}^3\times (0,t))}) \left( \left\| \frac{\omega _r(0)}{r}\right\| _{L_2({\mathbb {R}}^3)}^2+ \left\| \frac{\omega _{\varphi }(0)}{r}\right\| _{L_2({\mathbb {R}}^3)}^2 \right) , \\ {}&\qquad \frac{3}{p}+ \frac{2}{q} =\alpha ,\quad \alpha \le 1, \ t\le T. \end{aligned} \end{aligned}$$

(3.19)

Finally, in view of (1.7) and Lemma 3.1 we get

$$\begin{aligned} \Vert u_{\alpha }\Vert _{L_{p,q}({\mathbb {R}}^3\times (0,t))} \le c(\Vert u(0)\Vert _{L_{\infty }({\mathbb {R}}^3)}+ \Vert u(0)\Vert _{L_{s}({\mathbb {R}}^3)}), \end{aligned}$$

where $\alpha =\frac{3}{s}, s\ge 3, t\le T.$ These estimates imply that

$$\begin{aligned} \begin{aligned}&\Vert v_{\varphi }\Vert _{L_{p',q'}({\mathbb {R}}^3_{r_0}\times (0,t))} \\&\le h\left( \left\| \frac{v^2_{\varphi }(0)}{r}\right\| _{L_2({\mathbb {R}}^3)}, \left\| \frac{\omega _{\varphi }(0)}{r}\right\| _{L_2({\mathbb {R}}^3)}, \left\| \frac{\omega _{r}(0)}{r}\right\| _{L_2({\mathbb {R}}^3)}, \Vert u(0)\Vert _{L_{\infty }({\mathbb {R}}^3)}, \Vert u(0)\Vert _{L_{s}({\mathbb {R}}^3)}\right) , \\&\qquad \frac{3}{p'}+ \frac{2}{q'} \ge \frac{3}{4}, \end{aligned} \end{aligned}$$

where h is some positive increasing function of its arguments. Hence Lemma 2.3 implies local regularity.

To make statement more explicit we obtain from (3.19) for $r< r_0$ and from Step 5 of the proof of Theorem 1 from [5] that

$$\begin{aligned} \Vert \omega \Vert _{L_{\infty }(0,T;L_2({\mathbb {R}}^3_{r_0}))} \le C(data), \end{aligned}$$

where data are data from the assumptions of the Step 1 of the proof of Theorem 1 from [5].

Considering the problem

$$\begin{aligned}&\mathrm{rot}\,v = \omega , \\&\mathrm{div}\,v =0 \end{aligned}$$

and the local technique from [15, Ch.4, Sect.10], we have

$$\begin{aligned} \Vert v \Vert _{L_{\infty }(0,T;L_6({\mathbb {R}}^3_{r_0}))} \le C(data). \end{aligned}$$

Consider the problem

$$\begin{aligned}&v_t -\nu \Delta v + \nabla p= -v \cdot \nabla v, \\&\mathrm{div}\,v =0, \\&v|_{t=0} = v(0). \end{aligned}$$

Employing the result of Solonnikov, estimate for v above and some interpolation we get

$$\begin{aligned} \Vert v\Vert _{W^{2,1}_{\sigma , r}({\mathbb {R}}^3_0 \times (0,T))} \le c(data) + c \Vert v\Vert _{B^{2-2/r}_{\sigma , r}({\mathbb {R}}^3_0 \times (0,T))}, \end{aligned}$$

where $\sigma <6, r$ arbitrary. This proves the second part of Theorem 1. $\square$

Proof of Theorem 2 follows from Theorem 1 and Lemma 3.3.