A Slightly Supercritical Condition of Regularity of Axisymmetric Solutions to the Navier–Stokes Equations

Abstract

In the note, a new regularity condition for axisymmetric solutions to the non-stationary 3D Navier–Stokes equations is proven. It is slightly supercritical.

1 Introduction

In this note, we continue to analyse potential singularities of axisymmetric solutions to the non-stationary Navier–Stokes equations. In the previous paper [24], it has been shown that an axially symmetric solution is smooth provided a certain scale-invariant energy quantity of the velocity field is bounded. By definition, a potential singularity with bounded scale-invariant energy quantities is called the Type I blowup. It is important to notice that the above result does not follow from the so-called $\varepsilon$-regularity theory developed in [2, 16], and [10], where regularity is coming out due to smallness of those scale-invariant energy quantities.

We consider the 3D Navier–Stokes system

$$\begin{aligned} \partial _tv+v\cdot \nabla v-\Delta v=-\nabla q,\qquad \mathrm{div}\,v=0 \end{aligned}$$

(1.1)

in the parabolic cylinder $Q={\mathcal {C}}\times ]-1,0[$, where ${\mathcal {C}}=\{x=(x_1,x_2,x_3):\,x_1^2+x^2_2<1,\,-1<x_3<1\}$. A solution v and q is supposed to be a suitable weak one, which means the following:

Definition 1.1

Let $\omega \subset {\mathbb {R}}^3$ and $T_2>T_1$. The pair w and r is a suitable weak solution to the Navier–Stokes system in $Q_*=\omega \times ]T_1,T_2[$ if:

• 1. $w\in L_{2,\infty }(Q_*)$, $\nabla w\in L_2(Q_*)$, $r\in L_\frac{3}{2}(Q_*)$;

• 2. w and r satisfy the Navier–Stokes equations in $Q_*$ in the sense of distributions;

• 3. for a.a. $t\in [T_1,T_2]$, the local energy inequality

  $$\begin{aligned} \int \limits _\omega \varphi (x,t)|w(x,t)|^2dx+2 \int \limits _{T_1}^t\int \limits _\omega \varphi |\nabla w|^2dxdt'\le \int \limits _{T_1}^t\int \limits _\omega [|w|^2(\partial _t\varphi +\Delta \varphi ) \\ +w\cdot \nabla \varphi (|w|^2+2r)]dxdt' \end{aligned}$$

  holds for all non-negative $\varphi \in C^1_0(\omega \times ]T_1,T_2+(T_2-T_1)/2[).$

In our standing assumption, it is supposed that a suitable weak solution v and q to the Navier–Stokes equations in $Q=\mathcal C\times ]-1,0[$ is axially symmetric with respect to the axis $x_3$. The latter means the following: if we introduce the corresponding cylindrical coordinates $(\varrho ,\varphi ,x_3)$ and use the corresponding representation $v=v_\varrho e_\varrho +v_\varphi e_\varphi +v_3e_3$, then $v_{\varrho ,\varphi }=v_{\varphi ,\varphi }=v_{3,\varphi }=q_{,\varphi }=0$.

There are many papers on regularity of axially symmetric solutions. We cannot pretend to cite all good works in this direction. For example, let us mention papers: [3,4,5, 9, 11,12,13, 19,20,21, 26,27,31], and [15].

Actually, our note is inspired by the paper [20], where the regularity of solutions has been proved under a slightly supercritical assumption. We would like to consider a different supercritical assumption, to give a different proof and to get a better result.

To state our supercritical assumption, additional notation is needed. Given $x=(x_1,x_2,x_3)\in {\mathbb {R}}^3$, denote $x'=(x_1,x_2,0)$. Next, different types of cylinders will be denoted as ${\mathcal {C}}(r)=\{x:\,|x'|<r, |x_3|<r\}$, $\mathcal C(x_0,r)={\mathcal {C}}(r)+x_0$, $Q^{\lambda ,\mu }(r)={\mathcal {C}}(\lambda r)\times ]-\mu r^2,0[$, $Q^{1,1}(r)= Q(r)$, $Q^{\lambda ,\mu }(z_0,r)={\mathcal {C}}(x_0,\lambda r)\times ]t_0-\mu R^2,t_0[$. And, finally, we let

$$\begin{aligned} f(R):=\frac{1}{\sqrt{R}} \left( \int \limits ^0_{-R^2}\left( \int \limits _{{\mathcal {C}}(R)} |v|^3dx\right) ^\frac{4}{3}dt\right) ^\frac{1}{4} \end{aligned}$$

and

$$\begin{aligned} M(R):=\frac{1}{\sqrt{R}}\left( \int \limits _{Q(R)}|v|^\frac{10}{3}dz\right) ^\frac{3}{10} \end{aligned}$$

for any $0<R\le 1$ and assume that:

$$\begin{aligned} f(R)+M(R)\le g(R):=c_*\ln ^\alpha \ln ^\frac{1}{2}(1/R) \end{aligned}$$

(1.2)

for all $0<R\le 2/3$, where $c_*$ and $\alpha$ are positive constants and $\alpha$ obeys the condition:

$$\begin{aligned} 0<\alpha \le \frac{1}{224}. \end{aligned}$$

(1.3)

Without loss of generality, one may assume that $g(R)\ge 1$ for $0<R\le \frac{2}{3}$. To ensure the above condition, it is enough to increase the constant $c_*$ if necessary.

Our aim could be the following completely local statement.

Theorem 1.2

Assume that a pair v and q is an axially symmetric suitable weak solution to the Navier–Stokes equations in Q and conditions (1.2) and (1.3) hold. Then the origin $z=0$ is a regular point of v.

However, in this paper, we shall prove a weaker result leaving Theorem 1.2 as a plausible conjecture. We shall return to a proof of Theorem 1.2 elsewhere. In the present paper, the following fact is going to be justified.

Theorem 1.3

Let v be an axially symmetric solution to the Cauchy problem for the Navier–Stokes equations (1.1) in ${\mathbb {R}}^3\times ]0,T[$ with initial divergence free field $v_0$ from the Sobolev space $H^2 =W^2_2({\mathbb {R}}^3)$ such that

$$\begin{aligned} \sup \limits _{0<t<T-\delta }\Vert \nabla v(\cdot ,t)\Vert _{L_2({\mathbb {R}}^3)}\le C(\delta )<\infty \end{aligned}$$

for all $0<\delta <T$. Assume further that

$$\begin{aligned} \Sigma _0= \sup \limits _{x\in \mathbb R^3}|v_{02}(x)x_1-v_{01}(x)x_2|<\infty \end{aligned}$$

(1.4)

and

$$\begin{aligned} \sup \limits _{-\infty<h<\infty } f(R;(0,h,T))+M(R;(0,h,T))\le g(R) \end{aligned}$$

(1.5)

for all $0<R\le 2/3$ (it is assumed for simplicity that $T>1$), with some positive constants $c_*$ and $\alpha$, satisfying (1.3), where

$$\begin{aligned} f(R;z_0):=\frac{1}{\sqrt{R}} \left( \int \limits ^{t_0}_{t_0-R^2}\left( \int \limits _{{\mathcal {C}}(x_0,R)} |v|^3dx\right) ^\frac{4}{3}dt\right) ^\frac{1}{4} \end{aligned}$$

and

$$\begin{aligned} M(R;z_0):=\frac{1}{\sqrt{R}}\left( \int \limits _{Q(z_0,R)}|v|^\frac{10}{3}dz\right) ^\frac{3}{10}. \end{aligned}$$

Then v is a strong solution to the above Cauchy problem in ${\mathbb {R}}^3\times ]0,T[$, i.e.,

$$\begin{aligned} \sup \limits _{0<t<T}\Vert \nabla v(\cdot ,t)\Vert _{L_2({\mathbb {R}}^3)}<\infty . \end{aligned}$$

Our proof is based on the analysis of the following scalar equation

$$\begin{aligned} \partial _t\sigma +\left( v+2\frac{x'}{|x'|^2}\right) \cdot \nabla \sigma -\Delta \sigma =0 \end{aligned}$$

(1.6)

in $Q\setminus (\{x'=0\}\times ]-1,0[)$, where $\sigma :=\varrho v_\varphi =v_2x_1-v_1x_2$.

Let us list some differentiability properties of $\sigma$. Some of them follows from partial regularity theory developed by Caffarelli–Kohn–Nirenberg.

Indeed, since v and q are an axially symmetric suitable weak solution, there exists a closed set $S^\sigma$ in Q, whose 1D-parabolic measure in ${\mathbb {R}}^3\times {\mathbb {R}}$ is equal to zero and $x'=0$ for any $z=(x,t)\in S^\sigma$, such that any spatial derivative of v (and thus of $\sigma$) is Hölder continuous in $Q\setminus S^\sigma$.

Next, we observe that

$$\begin{aligned} |\partial _t\sigma (z)-\Delta \sigma (z)|\le (\sup \limits _{z=(x,t)\in P(\delta ,R;R)\times ]-R^2,0[}|v(z)|+2/\delta )|\nabla \sigma (z)| \end{aligned}$$

for any $0<\delta<R<1$, where $P(a,b;h)=\{x: \,a<|x'|<b,\,|x_3|<h\}$. Since v is axially symmetric, the first factor on the right hand side is finite. This fact, by iteration, yields

$$\begin{aligned} \sigma \in W^{2,1}_p(P(\delta ,R;R)\times ]-R^2,0[) \end{aligned}$$

for any $0<\delta<R<1$ and for any finite $p\ge 2$.

It follows from the above partial regularity theory that, for any $-1<t<0$,

$$\begin{aligned} \sigma (x',x_3,t)\rightarrow 0 \quad \text{ as }\quad |x'|\rightarrow 0 \end{aligned}$$

(1.7)

for all $x_3\in ]-1,1[\setminus S^\sigma _t$, where $S^\sigma _t=\{x_3\in ]-1,1[:\,(0,x_3,t)\in S^\sigma \}$.

In the same way, as it has been done in [26] and [24], one can show that $\sigma \in L_\infty (Q(R))$ for any $0<R<1$.

The main part of the proof of Theorem 1.3 is the following fact.

Proposition 1.4

Let $\sigma =\varrho v_\varphi$, then

$$\begin{aligned} \mathrm{osc}_{z\in Q(r)}\le e^{-c\left[ \ln ^\frac{1}{4}(1/(2r)) -\ln ^\frac{1}{4}(1/(2R))\right] } \mathrm{osc}_{z\in Q(2R)}\sigma (z) \end{aligned}$$

(1.8)

where c is a positive absolute constant and $0<r<R\le R_*(c_*,\alpha )\le 1/6$.

Here, $\mathrm{osc}_{z\in Q(r)}\sigma (z) =M_{r}-m_{r}$ and

$$\begin{aligned} M_{r}=\sup \limits _{z\in Q(r)}\sigma (z),\qquad m_{r}=\inf \limits _{z\in Q(r)}\sigma (z). \end{aligned}$$

The above statement is an improvement of the result in [20], where the bound for oscillations of $\sigma$ contains a fixed power of logarithmic factor only.

Remark 1.5

It is not so difficult to see that all results of the paper remain to be true if we replace v with ${\overline{v}}=v_\varrho e_\varrho +v_3e_3$ in the definitions of quantities M and f, see conditions (1.2) and (1.5).

The proof of Proposition 1.4 is based on a technique developed in [18], see also references there. We also would like to mention interesting results for the heat equation with a divergence free drift, see [1, 6, 7, 25].

2 Auxiliary Facts

Define the class ${\mathcal {V}}$ of functions $\pi :Q\rightarrow {\mathbb {R}}$ possessing the properties:

(i) there exists a closed set $S^\pi$ in Q, whose 1D-parabolic measure ${\mathbb {R}}^3\times {\mathbb {R}}$ is equal to zero and $x'=0$ for any $z=(x',x_3,t)\in S^\pi$, such that any spatial derivative is Hölder continuous in $Q\setminus S^\pi$;

(ii)

$$\begin{aligned} \pi \in W^{2,1}_2(P(\delta ,R;R)\times ]-R^2,0[)\cap L_\infty (Q(R)) \end{aligned}$$

for any $0<\delta<R<1$.

We are going to use the following subclass ${\mathcal {V}}_0$ of the class ${\mathcal {V}}$, saying that $\pi \in {\mathcal {V}}_0$ if and only if $\pi \in {\mathcal {V}}$ and

$$\begin{aligned} \partial _t\pi +\left( v+2\frac{x'}{|x'|^2}\right) \cdot \nabla \pi -\Delta \pi = 0\end{aligned}$$

(2.1)

in ${\mathcal {C}}\setminus \{x'=0\}\times ]-1,0[$.

We shall also say that $\pi \in {\mathcal {V}}_0$ has the property $({\mathcal {B}}_R)$ in Q(2R) if there exists a number $k_R>0$ such that $\pi (0,x_3,t)\ge k_R$ for $-(2R)^2\le t\le 0$, $x_3\in ]-2R,2R[\setminus S^\pi _t$.

Remark 2.1

Let $0<r\le R$ and $\pi \in {\mathcal {V}}_0$ have the property $({\mathcal {B}}_R)$ in Q(2R). Then $\pi$ has the property $(\mathcal B_r)$ in Q(2r) with any constant less or equal to $k_R$.

In what follows, we always suppose that $0<R\le 1/6$.

Proposition 2.2

Let $\pi \in {\mathcal {V}}_0$ have the property $(\mathcal B_R)$. Then, for any $0<k\le k_R$, for any $0<\tau _1<\tau <2$, and for any $0<\gamma _1<\gamma <4$, the following inequality holds:

$$\begin{aligned} \sup \limits _{z\in Q^{\tau _1,\gamma _1}(R)}\sigma (z)\le c_1(\tau _1,\tau ,\gamma _1,\gamma ,M(2R))\left( \frac{1}{|Q^{\tau ,\gamma }(R)|}\int \limits _{Q^{\tau ,\gamma }(R)}\sigma ^\frac{10}{3}dz\right) ^\frac{3}{10}, \end{aligned}$$

(2.2)

where $\sigma =(k-\pi )_+$,

$$\begin{aligned} c_1(\tau _1,\tau ,\gamma _1,\gamma ,M(2R))=\frac{c}{(\tau -\tau _1)^\frac{16}{3}}\left( 1+\frac{\tau -\tau _1}{\sqrt{\gamma -\gamma _1}}+\left( \frac{1}{\gamma _1\tau _1^3}\right) ^\frac{1}{10}M(2R)\right) ^3, \end{aligned}$$

and $Q^{\tau ,\gamma }(R)={\mathcal {C}}(\tau R)\times ]-\gamma R^2,0[$.

Proof

Repeating arguments in [24], we can get the following estimate of $h=\sigma ^m$:

$$\begin{aligned}&\left( \int \limits ^0_{t_2}\int \limits _{{\mathcal {C}}(r_2)}|h|^\frac{10}{3}dz\right) ^\frac{3}{10}\nonumber \\&\quad \le c\left( \int \limits ^0_{t_1}\int \limits _{\mathcal C(r_1)}|h|^\frac{5}{2}dz\right) ^\frac{2}{5} \frac{(r_1^3|t_1|)^\frac{1}{10}}{r_1-r_2} \left( 1+\frac{r_1-r_2}{\sqrt{t_2-t_1}}+\overline{M}(r_1,t_1)+\frac{r_1^\frac{13}{8}|t_1|^\frac{1}{36}}{(r_1-r_2)^\frac{7}{9}}\right) \end{aligned}$$

(2.3)

for any $0<r_2<r_1<2R$ and $-4R^2<t_1<t_2<0$, where

$$\begin{aligned} {{\overline{M}}}(r_1,t_1)=\left( \frac{1}{|t_1|r_1^3}\right) ^\frac{1}{10}\left( \int \limits ^0_{t_1}\int \limits _{{\mathcal {C}}(r_1)}|v|^\frac{10}{3}dz\right) ^\frac{3}{10}. \end{aligned}$$

Next, we wish to iterate (2.3). To this end, let $m=m_i=\Big (4/3\Big )^i$,

$$\begin{aligned} r_1=r_i=\tau _1R+(\tau -\tau _1)R2^{-i+1}, \qquad r_2=r_{i+1} , \\ t_1=t_i=-\gamma _1R^2-(\gamma -\gamma _1)R^24^{-i+1},\qquad t_2=t_{i+1}, \end{aligned}$$

where $i=1,2,...$. Then, we can derive from (2.3) the following inequality

$$\begin{aligned} G_{i+1}\le \left( \frac{c2^{i+1}}{\tau -\tau _1}\right) ^\frac{1}{m_i}\left( 1+\frac{\tau -\tau _1}{\sqrt{\gamma -\gamma _1}}+{{\overline{M}}}(r_i,t_i)+\frac{2^{(i+1)\frac{7}{9}}}{(\tau -\tau _1)^\frac{7}{9}}\right) ^\frac{1}{m_i}G_i, \end{aligned}$$

(2.4)

where

$$\begin{aligned} G_i=\left( \frac{1}{|t_i|r_i^3}\int \limits ^0_{t_i}\int \limits _{{\mathcal {C}}(r_i)}\sigma ^\frac{5m_i}{2}dz\right) ^\frac{2}{5m_i}. \end{aligned}$$

Noticing that

$$\begin{aligned} {{\overline{M}}}(r_i,t_i)\le c\left( \frac{1}{\gamma _1\tau _1^3}\right) ^\frac{1}{10}M(2R), \end{aligned}$$

let us make use of (2.4) to obtain the estimate

$$\begin{aligned} G_{i+1} \le \left( \frac{c2^{i+1}}{\tau -\tau _1}\right) ^\frac{1}{m_i}\left( 1+\frac{\tau -\tau _1}{\sqrt{\gamma -\gamma _1}}+\frac{2^{(i+1)\frac{7}{9}}}{(\tau -\tau _1)^\frac{7}{9}}+\left( \frac{1}{\gamma _1\tau _1^3}\right) ^\frac{1}{10}M(2R)\right) ^\frac{1}{m_i}G_i, \end{aligned}$$

(2.5)

which, after iterations, gives the following

$$\begin{aligned} G_{i+1}\le \xi _iG_1, \end{aligned}$$

(2.6)

where

$$\begin{aligned} \xi _i=\prod \limits ^i_{k=1}\left( \frac{c2^{k+1}}{\tau -\tau _1}\right) ^\frac{1}{m_k}\left( 1+\frac{\tau -\tau _1}{\sqrt{\gamma -\gamma _1}}+\frac{2^{(k+1)\frac{7}{9}}}{(\tau -\tau _1)^\frac{7}{9}}+\left( \frac{1}{\gamma _1\tau _1^3}\right) ^\frac{1}{10}M(2R)\right) ^\frac{1}{m_k}. \end{aligned}$$

Obviously,

$$\begin{aligned} \xi _i\le \prod \limits ^i_{k=1}\left( \frac{c2^{k+1}}{\tau -\tau _1}\right) ^\frac{1}{m_k} \left( 1+\frac{2^{(k+1)\frac{7}{9}}}{(\tau -\tau _1)^\frac{7}{9}}\right) ^\frac{1}{m_k}\left( 1+\frac{\tau -\tau _1}{\sqrt{\gamma -\gamma _1}} +\left( \frac{1}{\gamma _1\tau _1^3}\right) ^\frac{1}{10}M(2R)\right) ^\frac{1}{m_k}. \end{aligned}$$

Next,

$$\begin{aligned} \ln \xi _i\le A_1+A_2+A_3, \end{aligned}$$

where

$$\begin{aligned} A_1= & {} \sum ^i_{k=1}\frac{1}{m_k}(\ln c +(k+1)\ln 2-\ln (\tau -\tau _1))\le \ln c-3\ln (\tau -\tau _1), \\ A_2= & {} \sum ^i_{k=1}\frac{1}{m_k}\ln \left( 1+\frac{2^{(k+1)\frac{7}{9}}}{(\tau -\tau _1)^\frac{7}{9}}\right) =\sum ^i_{k=1}\frac{1}{m_k}\ln \left( \frac{2^{(k+1)\frac{7}{9}}}{(\tau -\tau _1)^\frac{7}{9}}\right) \\&\quad +\frac{1}{m_k}\ln \left( 1+\frac{(\tau -\tau _1)^\frac{7}{9}}{2^{(k+1)\frac{7}{9}}}\right) \le \ln \frac{c}{(\tau -\tau _1)^\frac{7}{3}} \\&\quad +(\tau -\tau _1)^\frac{7}{9}\sum ^i_{k=1} \frac{1}{m_k}\frac{1}{2^{(k+1)\frac{7}{9}}}\le \ln \frac{c}{(\tau -\tau _1)^\frac{7}{3}}, \end{aligned}$$

and

$$\begin{aligned}&A_3=\ln \left( 1+\frac{\tau -\tau _1}{\sqrt{\gamma -\gamma _1}}+\left( \frac{1}{\gamma _1\tau _1^3}\right) ^\frac{1}{10}M(2R)\right) \sum ^i_{k=1}\frac{1}{m_k} \\&\quad \le \ln \left( 1+\frac{\tau -\tau _1}{\sqrt{\gamma -\gamma _1}}+\left( \frac{1}{\gamma _1\tau _1^3}\right) ^\frac{1}{10}M(2R)\right) ^3. \end{aligned}$$

So,

$$\begin{aligned} \xi _i\le \frac{c}{(\tau -\tau _1)^\frac{16}{3}} \left( 1+\frac{\tau -\tau _1}{\sqrt{\gamma -\gamma _1}}+\left( \frac{1}{\gamma _1\tau _1^3}\right) ^\frac{1}{10}M(2R)\right) ^3. \end{aligned}$$

Passing to the limit as $i\rightarrow \infty$ in (2.6), we complete the proof the Proposition. $\square$

Remark 2.3

If we additionally assume that $\pi (\cdot ,-\theta R^2) \ge k$ in B for some $0<\theta \le 1$, then we do not need to use a cut-off in t. So, for $0<\lambda <1$, we have

$$\begin{aligned} \sup \limits _{Q^{\lambda ,\theta }(R)}\sigma \le c'_1(\lambda ,\theta ,M(2R))\left( \frac{1}{|Q^{1,\theta }(R)|}\int \limits _{Q^{1,\theta }(R)}\sigma ^\frac{10}{3}dz\right) ^\frac{3}{10}, \end{aligned}$$

where

$$\begin{aligned} c'_1(\lambda ,\theta ,M(2R))=\frac{c}{(1-\lambda )^\frac{16}{3}}\left( 1+\left( \frac{1}{\theta \lambda ^3}\right) ^\frac{1}{10}M(2R)\right) ^3. \end{aligned}$$

Corollary 2.4

Let a non-negative function $\pi \in {\mathcal {V}}_0$ have the property $({\mathcal {B}}_R)$ in Q(2R) and let $0<\lambda _1<\lambda <2$ and $0<\theta \le 1$. Suppose that

$$\begin{aligned} |\{\pi<k\}\cap Q^{\lambda , \theta }((0,t_0),R)|<\mu |Q^{\lambda ,\theta }(R)| \end{aligned}$$

(2.7)

for some $t_0>-4R^2$, for some $0<k\le k_R$, and for some

$$\begin{aligned} 0<\mu \le \mu _*=\left( \frac{1}{2c_1(\lambda _1,\lambda ,\theta /2,\theta ,M(2R))}\right) ^\frac{10}{3}. \end{aligned}$$

Then $\pi \ge \frac{k}{2}$ in $Q^{\lambda _1,\theta /2}((0,t_0),R)$.

If, in addition, $\pi (\cdot ,t_0-\theta R^2)>k$ in $\mathcal C(\lambda R)$, then $\pi \ge \frac{k}{2}$ in $Q^{\lambda _1,\theta }((0,t_0),R)$.

Proof

The first statement can be proved ad absurdum with the help of inequality (2.2) and a suitable choice of the number $\mu _*$. The second statement is proved in the same way but with the help of the inequality of Remark 2.3. Number $\mu _*$ is defined by the constant $c_1'$ instead of $c_1$. $\square$

The two lemmas below are obvious modifications of the corresponding statements in the paper [18].

Lemma 2.5

Let $0\le \pi \in {\mathcal {V}}_0$ have the property $({\mathcal {B}}_R)$ in Q(2R). Given $0<\delta _0\le 1$, there exists a positive number $\theta _0(\delta _0, f(2R))\le 1$ such that if, for $0<\theta \le \theta _0$, $0<k_0\le k_R$, there holds

$$\begin{aligned} |\{\pi (\cdot ,t_0-\theta R^2)\ge k_0\}\cap {\mathcal {C}}(R)|>\delta _0|{\mathcal {C}}(R)|, \end{aligned}$$

then

$$\begin{aligned} |\{\pi (\cdot ,t)\ge \frac{\delta _0}{3}k_0\}\cap {\mathcal {C}}(R)|>\frac{\delta _0}{3}|{\mathcal {C}}(R)| \end{aligned}$$

for all $t\in [t_0-\theta R^2,t_0]$.

Remark 2.6

There is a formula for $\theta _0$:

$$\begin{aligned} \theta _0=\left( \frac{c\delta _0^6}{1+\delta _0^2f(2R)}\right) ^\frac{4}{3}. \end{aligned}$$

Lemma 2.7

Let $0\le \pi \in {\mathcal {V}}_0$ have the property $({\mathcal {B}}_R)$ in Q(2R). Let, for any $t\in [t_0-\theta _1R^2,t_0]$,

$$\begin{aligned} |\{\pi (\cdot ,t)\ge k_1\}\cap {\mathcal {C}}(R)|\ge \delta _1|{\mathcal {C}}(R)| \end{aligned}$$

for some $0<k_1\le k_R$ and for some $0<\delta _1\le 1$ and $0<\theta _1\le 1$.

Then, for any $\mu _1\in ]0,1[$, the following inequality is valid:

$$\begin{aligned} |\{\pi <2^{-s}k_1\}\cap Q^{1,\theta _1}((0,t_0),R)|\le \mu _1|Q^{1,\theta _1}(R)| \end{aligned}$$

with the integer number s defined as

$$\begin{aligned} s=\mathrm{entier} \left( \frac{c}{\delta _1^2\mu _1^2\theta _1}(1+f(2R))\right) +1. \end{aligned}$$

Corollary 2.8

Let $0\le \pi \in {\mathcal {V}}_0$ have the property $({\mathcal {B}}_R)$ in Q(2R). If $\pi (\cdot ,{{\overline{t}}})\ge k_2$ in ${\mathcal {C}}(R)$, then, for any $\sigma \in ]0,1[$, the inequality $\pi \ge \beta _2k_2$ holds in $Q^{\sigma ,\theta _0}((0,t_0),R)$ with $\theta _0=(c/g(2R))^\frac{4}{3}$ and $t_0={{\overline{t}}}+\theta R^2$, where

$$\begin{aligned} \beta _2=\frac{1}{6} 2^{-c(1-\sigma )^{-40}\sigma ^{-6}g^{25}(2R)} \end{aligned}$$

provided $R\le R_{*}(c_*,\alpha )$. It is supposed that $0<k_2\le k_R$.

Proof

We apply Lemma 2.5 with $\delta _0=1$ and $k_0=k_2$. Then, for $\sigma =4/(27c)$, we have

$$\begin{aligned} \theta _0=\left( \frac{\frac{4}{27}\sigma }{\frac{1}{\sigma }+f(2R)}\right) ^\frac{4}{3}\ge \left( \frac{c}{g(2R)}\right) ^\frac{4}{3} \end{aligned}$$

for all $0<R\le R_{*}(c_*,\alpha )$ and state that the following inequality holds:

$$\begin{aligned} |\{\pi (\cdot ,t)>\frac{k_0}{3}\}\cap {\mathcal {C}}(R)|\ge \frac{1}{3}|{\mathcal {C}}(R)| \end{aligned}$$

for any $t\in [t_0-\theta _0R^2,t_0]$, where $t_0={{\overline{t}}}+\theta _0R^2$. In what follows, we are going to use the quantity $(c/(g(2R)))^\frac{4}{3}$ as a new number $\theta _0$ instead of $\theta _0(1,f(2R))$.

Now, we are going to apply Lemma 2.7 with another set of parameters $k_1=\frac{1}{3}k_2$, $\theta _1=\theta _0$, $\delta _1=\frac{1}{3}$, and

$$\begin{aligned}&\mu _1=\mu _*=\left( \frac{1}{2c'_1}\right) ^\frac{10}{3}, \qquad c'_1=\frac{c}{(1-\sigma )^\frac{16}{3}}\left( 1+\left( \frac{1}{\theta _0\sigma ^3}\right) ^\frac{1}{10}M(2R)\right) ^3\\&\quad \le \frac{c}{(1-\sigma )^\frac{16}{3}}\left( \frac{1}{\theta _0\sigma ^3}\right) ^\frac{3}{10}g^3(2R). \end{aligned}$$

Lemma (2.7) gives us:

$$\begin{aligned} |\{\pi<2^{-s}k_1\}\cap Q^{1,\theta _1}((0,t_0),R)|<\mu _1| Q^{1,\theta _1}(R)|, \end{aligned}$$

where

$$\begin{aligned} s=\mathrm{entier} \left( \frac{c}{\delta _1^2\mu _1^2\theta _1}(1+f(2R))\right) +1 \end{aligned}$$

and $\mathrm{entier}(x)$ is the largest integer not exceeding x. But we know that

$$\begin{aligned} \pi (\cdot ,t_0-\theta _0 R^2)\ge k_2>2^{-s}k_1=2^{-s}\frac{k_2}{3}. \end{aligned}$$

Then, from Corollary 2.4, it follows that $\pi >\frac{1}{2}2^{-s}k_1=\beta _2k_2$ with $\beta _2=\frac{1}{2}2^{-s}\frac{1}{3}$ in $Q^{\sigma ,\theta _0}((0,t_0),R)$. $\square$

Given $\theta \in ]0,1]$, we can find an number $0<R_{*1}(c_*,\alpha ,\theta )\le R_*(c_*,\alpha )$ so that $\Big (\frac{c}{g(2r)}\Big )^\frac{4}{3}\le \theta$ for all $0<r\le R_{*1}$.

Lemma 2.9

Let $0\le \pi \in {\mathcal {V}}_0$ have the property $({\mathcal {B}}_R)$ in Q(2R), assuming that $R\le R_{*1}(c_*,\alpha ,\theta )$ for some $0<\theta \le 1$. Suppose further that, for some $0<k\le k_R$ and for some $-R^2\le {{\overline{t}}}\le -\theta R^2$, there holds $\pi (\cdot ,{{\overline{t}}})\ge k$ in ${\mathcal {C}}(R)$. Then $\pi \ge \beta _0k$ in ${{\widehat{Q}}}:={\mathcal {C}}(\frac{2}{3}R)\times [\overline{t},0]$, where

$$\begin{aligned} \beta _0\ge \ln ^{-\frac{1}{2}}( 1/R) \end{aligned}$$

for $R\le R_{*2}(c_*,\alpha ,\theta ).$

Proof

Let

$$\begin{aligned} N=\mathrm{entier}\left( \frac{9}{8}\frac{ |{{\overline{t}}}|}{{{\tilde{\theta }}}_0R^2}\right) +1, \end{aligned}$$

where ${{\tilde{\theta }}}_0=(c/g(\frac{2}{3}2R))^\frac{4}{3} \le \theta$. Next, we introduce

$$\begin{aligned} {\hat{\theta }}_0=\frac{ |{{\overline{t}}}|}{(\frac{8N}{9}+\frac{1}{2N})R^2}\le {{\tilde{\theta }}}_0. \end{aligned}$$

Step 1. By Corollary 2.8, the inequality $\pi \ge \beta ^{(1)}_2k$ holds at least in ${\mathcal {C}}((1-\frac{1}{3N})R)\times [{{\overline{t}}}_1,{{\overline{t}}}_1+{\hat{\theta }}_0R^2]$, where $\overline{t}_1={{\overline{t}}}$, ${{\overline{t}}}_2= {{\overline{t}}}_1+{\hat{\theta }}_0R^2$, $\sigma =1-1/(3N)\ge 2/3$, $1-\sigma =1/(3N)$, and

$$\begin{aligned} \ln \beta ^{(1)}_2=-\ln 6-cN^{40}g^{25}(2R) \end{aligned}$$

Step 2. Here, we are going to use Corollary 2.8 with $R(1-1/(3N))$ instead of R and with $\sigma =(1-2(3N))/(1-1/(3N))$. As a result, we have the estimate $\pi \ge \beta ^{(2)}_2\beta ^{(1)}_2k$ at least in $\mathcal C((1-2/(3N))R)\times [{{\overline{t}}}_2,{{\overline{t}}}_2+{\hat{\theta }}_0 (1-1/(3N))^2R^2]$, ${{\overline{t}}}_3={{\overline{t}}}_2+{\hat{\theta }}_0 (1-1/(3N))^2R^2$, and

$$\begin{aligned} \ln \beta ^{(2)}_2=-\ln 6-cN^{40}g^{25}(2(1-1/(3N))R). \end{aligned}$$

So, $\pi \ge \beta ^{(2)}_2\beta ^{(1)}_2k$ in ${\mathcal {C}}((1-2(3N))R)\times [{{\overline{t}}}, {{\overline{t}}}_3]$.

After N steps, we shall have ${{\overline{t}}}_N=0$ and

$$\begin{aligned} \pi \ge \beta ^{(N)}_2...\beta ^{(1)}_2k=\beta _0(R)k \end{aligned}$$

in ${\mathcal {C}}(\frac{2}{3}R)\times [{{\overline{t}}},0]$, where

$$\begin{aligned} \ln \beta ^{(i+1)}_2=-\ln 6-cN^{40}g^{25}(2(1-i/(3N))R) \end{aligned}$$

for $i=0,1,...,N-1$.

Next, according to assumption (1.2), we can have

$$\begin{aligned} \ln \beta _0\ge -N\ln 6-cN^{40}\sum \limits ^{N-1}_{k=1} c_*^{25}\ln ^\gamma \ln ^\frac{1}{2}\left( \frac{1}{2(1-i/(3N))R}\right) , \end{aligned}$$

where $25\alpha <1$. Since

$$\begin{aligned} \ln \frac{1}{1-x}\le 2x \end{aligned}$$

provided $0\le x\le 1/2$, we find, assuming that $R\le 1/6$, the following:

$$\begin{aligned}&\ln ^\gamma \ln ^\frac{1}{2}\left( \frac{1}{2(1-i/(3N))R}\right) \le \ln ^\gamma \left( \ln \frac{1}{2R} +\frac{i}{N}\right) ^\frac{1}{2} \\&\quad \le \ln ^\gamma \left( \ln ^\frac{1}{2} \frac{1}{2R} +\left( \frac{i}{N}\right) ^\frac{1}{2}\right) = \ln ^\gamma \ln ^\frac{1}{2} \frac{1}{2R}\left( 1 +\left( \frac{i}{N\ln \frac{1}{2R}}\right) ^\frac{1}{2}\right) \\&\quad \le \ln ^\gamma \ln ^\frac{1}{2} \frac{1}{2R}\left( 1 +\left( \frac{i}{N}\right) ^\frac{1}{2}\right) =\left( \ln \left( \ln ^\frac{1}{2} \frac{1}{2R}\right) +\ln \left( 1 +\left( \frac{i}{N}\right) ^\frac{1}{2}\right) \right) ^\gamma \\&\quad \le \left( \ln \left( \ln ^\frac{1}{2} \frac{1}{2R}\right) +\left( \frac{i}{N}\right) ^\frac{1}{2}\right) ^\gamma \le \ln ^\gamma \left( \ln ^\frac{1}{2} \frac{1}{2R}\right) +\left( \frac{i}{N}\right) ^\frac{\gamma }{2}. \end{aligned}$$

From the latter inequality, one can deduce the bound

$$\begin{aligned}&\ln \beta _0\ge -N\ln 6-cc_*^{25}N^{40}\left( N\ln ^\gamma \ln ^\frac{1}{2} \frac{1}{2R}+\sum \limits ^{N-1}_{i=0}\left( \frac{i}{N}\right) ^\frac{\gamma }{2}\right) \\&\quad \ge -N\ln 6-cc_*^{25}N^{41}\ln ^\gamma \ln ^\frac{1}{2} \frac{1}{2R}, \end{aligned}$$

which is valid for $0<R\le R_{*3}(\alpha )\le 1/6$. Taking into account that $N\le c(g(2R))^\frac{4}{3}$, we conclude

$$\begin{aligned} \ln \beta _0\ge -c_1(c_*)\ln ^\frac{239\alpha }{3}\sqrt{\ln \frac{1}{R}}. \end{aligned}$$

It remains to find $R_{*4}(c_*,\alpha )\le 1$ such that

$$\begin{aligned} c_1(c_*)\ln ^{\frac{239\alpha }{3}-1}\sqrt{\ln \frac{1}{R}}\le 1 \end{aligned}$$

for all $0<R\le R_{*4}$. So, we have the required inequality provided $0<R\le R_{*2}=\min \{R_{*1},R_{*3},R_{*4}\}$.$\square$

3 Proof of Proposition 1.4

Now, we can state an analog of Lemma 4.2 of [18] for the class ${\mathcal {V}}$.

Lemma 3.1

Let $0\le \pi \in {\mathcal {V}}_0$ possess the property $({\mathcal {B}}_R)$ in Q(2R).

Suppose further that

$$\begin{aligned} \pi \le M_0k_R\end{aligned}$$

(3.1)

in Q(2R) for some $M_0\ge 1$. Then, there exists ${{\overline{t}}}\in [-R^2,-\frac{3}{4}R^2]$ such that

$$\begin{aligned} |e_{\kappa _0}({{\overline{t}}})|\ge \delta _0|{\mathcal {C}}(R)| \end{aligned}$$

(3.2)

Here, $\kappa _0=\kappa _0(f(2R))= c/(1+f(2R))$, $e_\kappa (t):=\{x\in {\mathcal {C}}(R): \pi (x,t)\ge \kappa k_R\}$, and

$$\begin{aligned} \delta _0(M_0,f(2R))=\Big (\frac{c}{M_0(1+f(2R))}\Big )^\frac{9}{4}. \end{aligned}$$

Proof

Here, we follow arguments of the paper [18]. They are based on the identity:

$$\begin{aligned}&\int \limits _{Q}(-\pi \partial _t\eta - \pi \Delta \eta -(v+2x'/|x'|^2)\cdot \nabla \eta \pi )dxdt \nonumber \\&\quad =4\pi _0 \int \limits _{-1}^0\int \limits _{-1}^1 \pi (0,x_3,t)\eta (0,x_3,t) dx_3dt, \end{aligned}$$

(3.3)

which is valid for any non-negative test function $\eta$ supported in Q. Here, $\pi _0=3.14...$. Although a similar statement has been proven in [18] under the assumption that $\pi$ is Lipschitz, it remains to be true for functions $\pi$ from the class ${\mathcal {V}}_0$ as well. Indeed, take a smooth cut-off function $\psi =\psi (x')$ so that $\psi (x')=\Psi (|x'|)$, $0\le \psi \le 1$, $\psi (x')=0$ if $|x'|\le \varepsilon /2$, $\psi (x')=1$ if $|x'|\ge \varepsilon$, $\Psi '(\varrho )\le c/\varrho$ and $\Psi ''(\varrho )\le c/\varrho ^2$ for some positive constant c. Then, it follows from (2.1) that:

$$\begin{aligned} \int \limits _{Q} \Big (\pi \partial _t(\eta \psi )+\pi (u+b)\cdot \nabla (\eta \psi )+\pi \Delta (\eta \psi )\Big )dz=0. \end{aligned}$$

There are two difficult terms for passing to the limit as $\varepsilon \rightarrow 0$. The first one is as follows:

$$\begin{aligned} I_1:=\int \limits _{Q}\pi \eta \Delta \psi dx dt=J_1+J_2, \end{aligned}$$

where

$$\begin{aligned} J_1:=\int \limits _{Q}(\pi \eta -(\pi \eta )|_{x'=0})\Delta \psi dx dt, \end{aligned}$$

For $J_2$, we find

$$\begin{aligned} J_2:=\int \limits _{Q}(\pi \eta )|_{x'=0}\Delta \psi dx dt=\int \limits ^0_{-1}\int \limits ^1_{-1}(\pi \eta )|_{x'=0}dx_3dt\int \limits _{|x'|<1}\Delta \psi (x')dx' \end{aligned}$$

and

$$\begin{aligned} \int \limits _{|x'|<1}\Delta \psi (x')dx'=2\pi _0\int \limits ^\varepsilon _\frac{\varepsilon }{2}\frac{1}{\varrho }\frac{\partial }{\partial \varrho }\Big (\varrho \Psi '(\varrho ) \Big )\varrho d\varrho =2\pi _0\varrho \Psi '(\varrho )\Big |^\varepsilon _\frac{\varepsilon }{2}=0. \end{aligned}$$

Now, we wish to show that

$$\begin{aligned} J_1:=\int \limits _{Q}\xi \Delta \psi dx dt \rightarrow 0 \end{aligned}$$

as $\varepsilon \rightarrow 0$, where, $\xi :=\pi \eta -(\pi \eta )|_{x'=0}.$ To this end, let us introduce the function

$$\begin{aligned} H_\varepsilon (x_3,t):=\int \limits _{\frac{\varepsilon }{2}<\varrho <\varepsilon }\xi \Delta \psi dx'. \end{aligned}$$

It can be bounded from above and from below

$$\begin{aligned} |H_\varepsilon (x_3,t)|\le c \sup \limits _{\mathrm{spt} \eta }\pi \sup \limits _{|x'|<1}\eta (x',x_3,t)\frac{1}{\varepsilon ^2}\int \limits _{\frac{\varepsilon }{2}}^\varepsilon \varrho d\varrho =:h(x_3,t) \end{aligned}$$

provided $\varepsilon <1$. The function h is supported in $]-1,1[ \times ]-1,0[$ and thus

$$\begin{aligned} \int \limits _{-1}^1\int \limits ^0_{-1}h(x_3,t)dx_3dt<\infty . \end{aligned}$$

Now, let $(0,x_3,t)$ be a regular point of $\pi$, i.e., $(0,x_3,t)\notin S^\pi$. Then, $\xi (x',x_3,t)\rightarrow 0$ as $|x'|\rightarrow 0$ and thus for any $\delta >0$ there exists a number $\tau (x_3,t)>0$ such that $|\xi (x',x_3,t)|<\delta$ provided $|x'|<\tau$. So,

$$\begin{aligned} |H_\varepsilon (x_3,t)|<c\frac{\delta }{\varepsilon ^2}\int \limits ^\varepsilon _\frac{\varepsilon }{2}\varrho d\varrho =c\frac{\delta }{2} \end{aligned}$$

provided $\varepsilon <\tau$. Therefore, $H_\varepsilon (x_3,t)\rightarrow 0$ as $\varepsilon \rightarrow 0$ and by the Lebesgue theorem on dominated convergence, we find that

$$\begin{aligned} J_1=\int \limits _{-1}^1\int \limits _{-1}^0H_\varepsilon (x_3,t)dx_3dt\rightarrow 0 \end{aligned}$$

as $\varepsilon \rightarrow 0$.

Similar arguments work for the second difficult term:

$$\begin{aligned} I:=\int \limits _{Q}\pi \eta b\cdot \nabla \psi dz=J_1+J_2, \end{aligned}$$

where

$$\begin{aligned} J_1=\int \limits _{Q}\xi b\cdot \nabla \psi dz \end{aligned}$$

and

$$\begin{aligned}&J_2:=\int \limits _{Q}(\pi \eta )|_{x'=0}b\cdot \nabla \psi dx dt=\int \limits ^0_{-1}\int \limits ^1_{-1}(\pi \eta )|_{x'=0}dx_3dt2\pi _0\int \limits _\frac{\varepsilon }{2}^\varepsilon \frac{2}{\varrho }\Psi '(\varrho )\varrho d\varrho \\&\quad =4\pi _0\int \limits ^0_{-1}\int \limits ^1_{-1}(\pi \eta )|_{x'=0}dx_3dt. \end{aligned}$$

The fact that $J_1\rightarrow 0$ as $\varepsilon \rightarrow 0$ can be justified in the same way as above, replacing $H_\varepsilon$ with the function

$$\begin{aligned} G_\varepsilon (x_3,t):=\int \limits _{\frac{\varepsilon }{2}<|\ x'| <\varepsilon }\xi b\cdot \nabla \psi dx'. \end{aligned}$$

Other terms can be treated in a similar way and even easier. So, the required identity (3.3) has been proven.

Now, let us select the test function $\eta$ in (3.3), using the following notation

$$\begin{aligned} Q^{\lambda ,\theta }(z_0,R):={\mathcal {C}}(x_0,\lambda R)\times ]t_0-\theta R^2,t_0[, \end{aligned}$$

so that $\eta =1$ in $Q^{\frac{1}{2},\frac{1}{8}}((0,-\frac{13}{16}R^2),R)$, $\eta =0$ out of $Q^{1,\frac{1}{4}}((0,-\frac{3}{4}R^2),R)$ and $|\partial _t\eta |+|\nabla \eta |^2+|\nabla ^2\eta |\le c/R^2$. Taking into account that $\pi$ has the property $(\mathcal B_R)$, we find

$$\begin{aligned} \frac{\pi _0}{2}k_RR^2\le \frac{c}{R^2}\int \limits _{Q^{1,\frac{1}{4}}(z_R,R)}\pi dz+\frac{c}{R}\int \limits _{Q^{1,\frac{1}{4}}(z_R,R)}\pi |v|dz+\frac{c}{R}\int \limits _{Q^{1,\frac{1}{4}}(z_R,R)}\frac{\pi }{|x'|}dz, \end{aligned}$$

where $z_R=(0,-\frac{3}{4}R^2)$.

Setting $E_\kappa =\{(x,t): t\in ]-R^2,-\frac{3}{4}R^2[, x\in e_\kappa (t)\},$ we can deduce from the latter inequality

$$\begin{aligned}&\frac{\pi _0}{2}k_RR^3 \le \frac{c}{R^2}\int \limits _{Q^{1,\frac{1}{4}}(z_R,R)\setminus E_\kappa }\pi dz+\frac{c}{R}\int \limits _{Q^{1,\frac{1}{4}}(z_R,R)\setminus E_\kappa }\pi |v|dz+\frac{c}{R}\int \limits _{Q^{1,\frac{1}{4}}(z_R,R)\setminus E_\kappa }\frac{\pi }{|x'|}dz \\&\quad + \frac{c}{R^2}\int \limits _{Q^{1,\frac{1}{4}}(z_R,R)\cap E_\kappa }\pi dz+\frac{c}{R}\int \limits _{Q^{1,\frac{1}{4}}(z_R,R)\cap E_\kappa }\pi |v|dz+\frac{c}{R}\int \limits _{Q^{1,\frac{1}{4}}(z_R,R)\cap E_\kappa }\frac{\pi }{|x'|}dz. \end{aligned}$$

Applying (3.1) and recalling definitions of the sets $e_\kappa (t)$ and $E_\kappa$, we can get

$$\begin{aligned}&\frac{\pi _0}{2}k_RR^3 \le \frac{c\kappa k_R}{R^2}\Big \{|Q^{1,\frac{1}{4}}(R)| +R\int \limits _{Q^{1,\frac{1}{4}}(z_R,R)\setminus E_\kappa } |v|dz+R\int \limits _{Q^{1,\frac{1}{4}}(z_R,R)\setminus E_\kappa }\frac{1}{|x'|}dz\Big \} \\&\quad +\frac{cM_0k_R}{R^2}\Big \{|E_\kappa |+R\int \limits _{Q^{1,\frac{1}{4}}(z_R,R)\cap E_\kappa } |v|dz+R\int \limits _{Q^{1,\frac{1}{4}}(z_R,R)\cap E_\kappa }\frac{1}{|x'|}dz\Big \}. \end{aligned}$$

We need to estimate integrals in the above inequality. First, for integrals, containing v, Holder inequality gives

$$\begin{aligned}&\int \limits _{Q^{1,\frac{1}{4}}(z_R,R)\setminus E_\kappa }|v|dx\le \Vert {\mathbb {I}}\Vert _{\frac{3}{2},\frac{4}{3},Q^{1,\frac{1}{4}}(R)}\left( \int \limits ^{-\frac{3}{4}R^2}_{-R^2}\left( \int \limits _{{\mathcal {C}}(R)}|v|^3dx\right) ^\frac{4}{3}dt\right) ^\frac{1}{4} \\&\quad \le f(2R)R^\frac{1}{2}\Vert {\mathbb {I}}\Vert _{\frac{3}{2},\frac{4}{3},Q^{1,\frac{1}{4}}(R)}\le f(2R)R^4 \end{aligned}$$

and similarly

$$\begin{aligned} \int \limits _{Q^{1,\frac{1}{4}}(z_R,R)\cap E_\kappa } |v|dz\le f(2R)R^\frac{1}{2}\Vert {\mathbb {I}}\Vert _{\frac{3}{2},\frac{4}{3},E_\kappa }. \end{aligned}$$

To evaluate the last two integrals, let us take into account the fact:

$$\begin{aligned} \frac{1}{|x'|}\in L_{\frac{9}{5},\infty }(Q^{1,\frac{1}{4}}(z_R,R)). \end{aligned}$$

Then,

$$\begin{aligned}&\int \limits _{Q^{1,\frac{1}{4}}(z_R,R)\setminus E_\kappa }\frac{1}{|x'|}dz\le \Vert \frac{1}{|x'|}\Vert _{\frac{9}{5},\infty ,Q^{1,\frac{1}{4}}(z_R,R)}\Vert {\mathbb {I}}\Vert _{\frac{9}{4},1,Q^{1,\frac{1}{4}}(R)} \\&\quad \le cR^\frac{2}{3}R^\frac{10}{3}=cR^4 \end{aligned}$$

and

$$\begin{aligned} \int \limits _{Q^{1,\frac{1}{4}}(z_R,R)\cap E_\kappa }\frac{1}{|x|}dz\le c R^\frac{2}{3}\Vert {\mathbb {I}}\Vert _{\frac{9}{4},1,E_\kappa }. \end{aligned}$$

Hence, we have

$$\begin{aligned}&\frac{\pi _0}{2}k_RR^3\le c\kappa k_RR^3(1+f(2R)) \\&\quad +\frac{cM_0k_R}{R^2}\Big [|E_\kappa |+f(2R)R^\frac{3}{2}\Vert {\mathbb {I}}\Vert _{\frac{3}{2},\frac{4}{3},E_\kappa }+R^\frac{5}{3}\Vert {\mathbb {I}}\Vert _{\frac{9}{4},1,E_\kappa }\Big ]. \end{aligned}$$

So,

$$\begin{aligned} \frac{\pi _0}{2}\le c\kappa (1+f(2R)) +\frac{cM_0}{R^5}\Big [|E_\kappa |+f(2R)R^\frac{3}{2}\Vert \mathbb I\Vert _{\frac{3}{2},\frac{4}{3},E_\kappa }+R^\frac{5}{3}\Vert {\mathbb {I}}\Vert _{\frac{9}{4},1,E_\kappa }\Big ]. \end{aligned}$$

Now, one can find $\kappa =\kappa _0(f(2R))=c/(1+f(2R))$ such that

$$\begin{aligned} \frac{cM_0}{R^5}\Big [|E_{\kappa _0}|+f(2R)R^\frac{3}{2}\Vert {\mathbb {I}}\Vert _{\frac{3}{2},\frac{4}{3},E_{\kappa _0}}+R^\frac{5}{3}\Vert {\mathbb {I}}\Vert _{\frac{9}{4},1,E_{\kappa _0}}\Big ]\ge 1. \end{aligned}$$

It remains to estimate two integrals on the left hand side of the latter inequality:

$$\begin{aligned} \Vert {\mathbb {I}}\Vert _{\frac{3}{2},\frac{4}{3},E_{\kappa _0}}=\left( \int \limits ^{-\frac{3}{4}R^2}_{-R^2}|e_\kappa (t)|^\frac{8}{9}dt\right) ^\frac{3}{4}\le c|E_{\kappa _0}|^\frac{2}{3}R^\frac{1}{6} \end{aligned}$$

and

$$\begin{aligned} \Vert {\mathbb {I}}\Vert _{\frac{9}{4},1,E_\kappa }\le c|E_{\kappa _0}|^\frac{4}{9}R^\frac{10}{9}. \end{aligned}$$

Letting $A=|E_{\kappa _0}|/R^5$, we arrive at the following inequality

$$\begin{aligned} f(A):=A+A^\frac{4}{9}+f(2R)A^\frac{2}{3}\ge \frac{1}{cM_0}. \end{aligned}$$

Since $f'(A)>0$ for $A>0$, we can state that the last inequality implies

$$\begin{aligned} \frac{|E_{\kappa _0}|}{|{\mathcal {C}}(R)|\frac{1}{4}R^2}\ge \delta _0=\Big (\frac{c}{M_0(1+f(2R))}\Big )^\frac{9}{4}. \end{aligned}$$

It is not so difficult to show the exisence of ${{\overline{t}}}\in [-R^2,-\frac{3}{4}R^2]$ with the property:

$$\begin{aligned} |e_{\kappa _0}({{\overline{t}}})|\frac{1}{4}R^2\ge |E_{\kappa _0}|. \end{aligned}$$

So, it is proven that there exists ${{\bar{t}}}\in [-R^2,-3R^2/4]$ such that

$$\begin{aligned} |\{x\in {\mathcal {C}}(R): \pi (x,{{\bar{t}}})>\kappa _0 k_R\}|\ge \delta _0 |{\mathcal {C}}(R)|, \end{aligned}$$

(3.4)

which completes the proof of the lemma. $\square$

Now, we are able to prove Proposition 1.4.

Assume that the function $\pi$ meets all the conditions of Lemma 3.1 and according to it, we can claim that:

$$\begin{aligned} |e_{\kappa _0}({{\overline{t}}})|=|\{x\in {\mathcal {C}}(R): \pi (x,{{\overline{t}}})\ge \kappa _0 k_R\}|\ge \delta _0|{\mathcal {C}}(R)| \end{aligned}$$

for some ${{\overline{t}}}\in [-R^2,-\frac{3}{4}R^2]$, $\kappa _0=c/g(2R)$, and $\delta _0=c(M_0)/g^\frac{9}{4}(2R)$. Now, we can calculate

$$\begin{aligned}&\theta (\delta _0(M_0,f(2R)),f(2R))\ge c\Big (\frac{\delta _0^6}{1+\delta _0^2f(2R)}\Big )^\frac{4}{3} \\&\quad \ge c(M_0)\Big (\frac{1}{g(2R)}\Big )^{18}, \end{aligned}$$

apply Lemma 2.5, and find

$$\begin{aligned} |\{\pi (\cdot ,t)\ge \delta _0\kappa _0k_R/3\}\cap {\mathcal {C}}(R)|>\delta _0/3|{\mathcal {C}}(R)| \end{aligned}$$

for all $t\in [{{\overline{t}}},t_0]$ with $t_0={{\overline{t}}}+\theta _0R^2$ and $\theta _0=c(M_0)(g(2R))^{-18}$.

Next, it follows from Lemma 2.7 that:

$$\begin{aligned} |\{\pi <2^{-s}\delta _0\kappa _0k_R/3\}\cap Q^{1,\theta _0}((0,t_0),R)|\le \mu _* |Q^{1,\theta _0}(R)|, \end{aligned}$$

where

$$\begin{aligned} s=\mathrm{entier}\Big (\frac{c}{\delta _0^2\mu _*^2\theta _0}(1+f(2R))\Big )+1 \end{aligned}$$

and $\mu _*$ is the number that appears in Corollary 2.4, see also Proposition 2.2. In our case,

$$\begin{aligned} \mu _*=\Big (\frac{1}{2c_1(3/4,1,\theta _0/2,\theta _0,M(2R))}\Big )^\frac{10}{3} \end{aligned}$$

and, moreover

$$\begin{aligned} c_1(3/4,1,\theta _0/2,\theta _0,M(2R))\le c\theta _0^{-\frac{3}{2}}g^3(2R)\le c(M_0)(g(2R))^{30}. \end{aligned}$$

Then, Corollary 2.4 implies the bound

$$\begin{aligned} \pi \ge 2^{-s}\delta _0\kappa _0k_R/6={\hat{\beta }}_2\kappa _0k_R \end{aligned}$$

in $Q^{\frac{3}{4},\frac{1}{2}\theta _0}((0,t_0),R)$. So, combining previous estimates, we find the following:

$$\begin{aligned} {\hat{\beta }}_2=\frac{1}{6}2^{-s}\delta _0\ge e^{-sln2-\ln 6}\delta _0\ge e^{-cs}\delta _0, \end{aligned}$$

where

$$\begin{aligned}&s\le \frac{2g(2R)}{\delta _0^2\mu _*^2\theta _0}\le c(M_0)g(2R)(g(2R))^\frac{9}{2}(g(2R))^{18}{c_1}^\frac{20}{3} \\&\quad \le c(M_0)(g(2R))^\frac{47}{2}(g(2R))^{30})^\frac{20}{3}\le c(M_0) (g(2R))^{224}. \end{aligned}$$

So,

$$\begin{aligned}&{\hat{\beta }}_2\ge e^{-c(M_0)(g(2R))^{224}}c(M_0)(g(2R))^{-\frac{9}{4}}\ge e^{-2c(M_0)(g(2R))^{224}} \\&\quad \ge e^{-c(M_0,c_*)\ln ^{224\alpha }\sqrt{\ln \frac{1}{R}}}. \end{aligned}$$

Obviously, there exists a number $0<R_{*5}(M_0,c_*,\alpha )\le \min \{1/6,R_{*2}\}$ such that

$$\begin{aligned} 2c(M_0,c_*)\ln ^{224\alpha -1}\sqrt{\ln \frac{1}{R}}\le \frac{1}{2} \end{aligned}$$

and

$$\begin{aligned} c(M_0,c_*)\ln ^{224\alpha }\sqrt{\ln \frac{1}{R}}\ge \ln g(2R) \end{aligned}$$

for $0<R\le R_{*5}(M_0,c_*,\alpha )$ and thus

$$\begin{aligned}&-c(M_0,c_*)\ln ^{224\alpha }\sqrt{\ln \frac{1}{R}} \\&\quad =-2c(M_0,c_*)\ln ^{224\alpha }\sqrt{\ln \frac{1}{R}}++c(M_0,c_*)\ln ^{224\alpha }\sqrt{\ln \frac{1}{R}} \\&\quad -\ge \Big (2c(M_0,c_*)\ln ^{224\alpha -1}\sqrt{\ln \frac{1}{R}}\Big )\ln \sqrt{\ln \frac{1}{R}}+\ln g(2R) \\&\quad \ge -\ln \Big (\ln \frac{1}{R}\Big )^\frac{1}{4}+\ln g(2R) . \end{aligned}$$

Now, the number ${\hat{\beta }}_2$ is estimated as follows:

$$\begin{aligned} {\hat{\beta }}_2\ge \Big (\ln \frac{1}{R} \Big )^{-\frac{1}{4}}g(2R) \end{aligned}$$

(3.5)

for $0<R\le R_{*5}(M_0,c_*,\alpha )$.

Since

$$\begin{aligned} -R^2\le \overline{t}+\theta _0/2R^2=t_0-\theta _0/2R^2<t_0={{\overline{t}}}+\theta _0R^2\le -\frac{3}{4}R^2+\frac{1}{4}R^2=-\frac{1}{2}R^2, \end{aligned}$$

there is ${{\overline{t}}}_1\in [-R^2,-\frac{1}{2}R^2]$ such that

$$\begin{aligned} \pi (\cdot ,{{\overline{t}}}_1)>{\hat{\beta }}_2\kappa _0k_R \end{aligned}$$

in $\mathcal C(\frac{3}{4}R)$. It allows us to apply Lemma 2.9 with $\theta =1/2$, with $\frac{3}{4}R$ instead of R, with ${{\overline{t}}}_1$ instead of ${{\overline{t}}}$, and with ${\hat{\beta }}_2\kappa _0k_R$ instead of k. According to Lemma 2.9, the inequality

$$\begin{aligned} \pi \ge \beta _0{\hat{\beta }}_2\kappa _0k_R \end{aligned}$$

holds in Q(R/2). It follows from Lemma 2.9 and from (3.5) that

$$\begin{aligned} \pi \ge \frac{c k_R}{\ln ^\frac{3}{4} (\frac{1}{R})} =\beta (2R)k_R \end{aligned}$$

in Q(R/2).

By our assumption imposed on function $\sigma$, we can put $k_R=\frac{1}{2}\mathrm{osc}_{z\in Q(2R)}\sigma (z)$. Then, either $\pi =\sigma -m_{2R}$ or $\pi =M_{2R}-\sigma (z)$ satisfies all the conditions of the proposition with $M_0=2$. Simple arguments show that

$$\begin{aligned} \mathrm{osc}_{z\in Q(R/2)}\sigma (z)\le \Big (1-\frac{1}{2}\beta (2R)\Big )\mathrm{osc}_{z\in Q(2R)}\sigma (z). \end{aligned}$$

Now, after iterations of the latter inequality, we arrive at the following bound

$$\begin{aligned}&\mathrm{osc}_{z\in Q(R/2^{2k+1})}\le \prod \limits ^{k}_{i=0}\Big ((1-\frac{1}{2}\beta (R/2^{2i-1})\Big )\mathrm{osc}_{z\in Q(2R)}\sigma (z)= \\&\quad =\eta _k\mathrm{osc}_{z\in Q(2R)}\sigma (z) \end{aligned}$$

being valid for any natural number k.

In order to evaluate $\eta _k$, take $\ln$ of it. As a result,

$$\begin{aligned}&\ln \eta _k=\sum \limits ^k_{i=0}\ln \Big ((1-\frac{1}{2}\beta (R/2^{2i-1})\Big )\le -\sum \limits ^k_{i=0}\frac{1}{2}\beta (R/2^{2i-1}) \\&\quad \le -c\sum \limits ^k_{i=0}(\ln ( 2^{2i-1}/R))^{-\frac{3}{4}} =-c\sum \limits ^k_{i=0}\frac{1}{(i\ln 4+\ln (1/{(2R)}))^\frac{3}{4} } \\&\quad \le -c\int \limits ^{k+1}_0\frac{dx}{(x\ln 4+\ln ( 1/(2R)) )^\frac{3}{4}} \\&\quad =-\frac{4c}{\ln 4}(x\ln 4+\ln (1/(2R)))^\frac{1}{4}\Big |^{k+1}_0 \\&\quad =-c\Big (\ln ^\frac{1}{4}(2^{2k+1}/R) -\ln ^\frac{1}{4}(1/(2R))\Big ) \Big ). \end{aligned}$$

Hence, we have

$$\begin{aligned} \mathrm{osc}_{z\in Q(R/2^{2k+1})}\le e^{-c\Big [\ln ^\frac{1}{4}(2^{2k+1}/R) -\ln ^\frac{1}{4}(1/(2R))\Big ]} \mathrm{osc}_{z\in Q(2R)}\sigma (z) \end{aligned}$$

and thus

$$\begin{aligned} \mathrm{osc}_{z\in Q(r)}\le e^{-c\Big [\ln ^\frac{1}{4}(1/(2r)) -\ln ^\frac{1}{4}(1/(2R))\Big ]} \mathrm{osc}_{z\in Q(2R)}\sigma (z) \end{aligned}$$

for all $0<r<R\le R_*(c_*,\alpha )= R_{*5}(2,c_*,\alpha )$. So, (1.8) follows. The proof of Proposition 1.4 is complete.

4 Proof of Theorem 1.3

By the maximimum principle, we have $|\sigma | =|\varrho v_\varphi |\le \Sigma _0$ in ${\mathbb {R}}^3\times ]0,T[$. From Proposition 1.4, it follows that

$$\begin{aligned} |\sigma (\varrho ,x_3,t)|\le e^{-c\Big [\ln ^\frac{1}{4}(1/(2\varrho )) -\ln ^\frac{1}{4}(1/(2R_*))\Big ]} 2\Sigma _0 \end{aligned}$$

fo all $0<\varrho \le R_*(c_*,\alpha )$, for all $x_3\in {\mathbb {R}}$, and for $t\in ]T-R_*^2,T[$. Obviously, it remains true $\varrho >R_*$ as well. So, we have for all $x\in {\mathbb {R}}^3$ and for all $t\in ]T-R_*^2,T[$

$$\begin{aligned} |\sigma (\varrho ,x_3,t)|\le C(c_*,\alpha )e^{-c\ln ^\frac{1}{4}(1/(2\varrho ))}\Sigma _0\le C(c_*,\alpha )\Sigma _0\frac{m!}{c^m\ln ^{\frac{m}{4}}(1/(2\varrho ))} \end{aligned}$$

for all natural numbers m.

Now, let us notice that $v(\cdot , T-R_*^2)\in H^2$. Therefore, one can use the results of papers [12, 30], and [15], see also [17, 4], on the Cauchy problem for the Navier–Stokes system (1.1) in ${\mathbb {R}}^3\times ]T-R_*^2,T[$ and conclude that v is a strong solution in the interval ]0, T[.