On regularity criteria for the Navier–Stokes equations based on one directional derivative of the velocity or one diagonal entry of the velocity gradient

Abstract

It is proved that if the solution of the Navier–Stokes system satisfies

$$\begin{aligned} \partial _3\varvec{u}\in L^p(0,T;L^q(\mathbb {R}^3)),\quad \frac{2}{p}+\frac{3}{q} =\frac{22}{13}+\frac{3}{13q},\quad 3<q<4, \end{aligned}$$

or

$$\begin{aligned} \partial _3u_3\in L^\beta (0,T;L^\alpha (\mathbb {R}^3)),\quad \frac{2}{\beta }+\frac{3}{\alpha } =\frac{3(\sqrt{65\alpha ^2-78\alpha +49}+7-\alpha )}{16\alpha },\quad \frac{3+\sqrt{17}}{4}\le \alpha \le \infty , \end{aligned}$$

then the solution is smooth on (0, T]. These two improve many previous results.

1 Introduction

In this paper, we continue our study [31, 35, 36] of the regularity criteria of the following Navier–Stokes equations (NSE):

$$\begin{aligned} \left\{ \begin{array}{ll} \partial _t\varvec{u}+(\varvec{u}\cdot \nabla )\varvec{u}-\Delta \varvec{u}+\nabla \pi =\varvec{0},\\ \nabla \cdot \varvec{u}=0,\\ \varvec{u}|_{t=0}=\varvec{u}_0, \end{array}\right. \end{aligned}$$

(1)

where $\varvec{u}=(u_1,u_2,u_3)$ is the fluid velocity field, $\pi$ is a scalar pressure, $\varvec{u}_0$ is the prescribed initial velocity field satisfying the compatibility condition $\nabla \cdot \varvec{u}_0=0$, and

$$\begin{aligned} \partial _t\varvec{u}=\frac{\partial \varvec{u}}{\partial t},\quad \partial _i=\frac{\partial }{\partial x_i},\quad (\varvec{u}\cdot \nabla )=\sum _{i=1}^3 u_i\partial _i,\quad \Delta =\partial _1\partial _1+\partial _2\partial _2+\partial _3\partial _3. \end{aligned}$$

Leray [18] and Hopf [13] have established a global weak solution to (1); however, it remains an open problem of its regularity and uniqueness. Serrin [25] first showed that if

$$\begin{aligned} \varvec{u}\in L^p(0,T;L^q(\mathbb {R}^3)),\quad \frac{2}{p}+\frac{3}{q}=1,\quad 3\le q\le \infty , \end{aligned}$$

(2)

then the solution is regular on (0, T]. See also [8, 22]. The so-called Serrin-type regularity criterion (2) was generalized by Beir$\tilde{\text {a}}$o da Veiga [1] by adding integrability conditions on the velocity gradient,

$$\begin{aligned} \nabla \varvec{u}\in L^p(0,T;L^q(\mathbb {R}^3)),\quad \frac{2}{p}+\frac{3}{q}=2,\quad \frac{3}{2}\le q\le \infty . \end{aligned}$$

(3)

In view of the divergence-free condition $\nabla \cdot \varvec{u}=0$, it is natural and important to consider components reduction improvements of (2) and (3), that is, whether or not integrability conditions on partial components of the velocity or velocity gradient could still ensure the smoothness of the solution. There are so many studies devoted to this refinement, and without no intention to be complete, we recommend [2, 3, 5, 6, 12, 15, 16, 19,20,21, 26, 28, 31, 32, 34, 37,38,39,40].

In this paper, we would like to investigate the regularity criterion of (1) based on one directional derivative of the velocity field, say $\partial _3\varvec{u}$, or one diagonal entry of the velocity gradient, say $\partial _3u_3$. Let us first review what have happened in the last decades. In [20, Theorem 4 (i)], Penel–Pokorný showed that if

$$\begin{aligned} \partial _3\varvec{u}\in L^p(0,T;L^q(\mathbb {R}^3)),\quad \frac{2}{p}+\frac{3}{q}=\frac{3}{2},\quad 2\le q\le \infty , \end{aligned}$$

(4)

then the solution is smooth. This is based on a regularity criterion in terms of $u_3$, $\partial _3u_1$ and $\partial _3u_2$ [20, Theorem 1 (a)]:

$$\begin{aligned} \begin{aligned} u_3&\in L^\frac{2s}{s-3}(0,T;L^s(\mathbb {R}^3)),&\quad 3<s\le \infty ;\\ \partial _3u_1,\ \partial _3u_2&\in L^\frac{2q}{2q-3}(0,T;L^q(\mathbb {R}^3)),&\quad \frac{3}{2}<q\le \infty . \end{aligned} \end{aligned}$$

(5)

Then, Kukavica–Ziane [16] established a fine property of the horizontal convective terms (denoting by $\displaystyle {\Delta _h=\partial _1\partial _1+\partial _2\partial _2}$ the horizontal Laplacian)

$$\begin{aligned} \begin{aligned} \sum _{i,j=1}^2 \int \limits _{\mathbb {R}^3} u_i\partial _iu_j\Delta _hu_j\text { d}x&=\frac{1}{2} \sum _{i,j=1}^2 \int \limits _{\mathbb {R}^3} \partial _iu_j\partial _iu_j\partial _3u_3\text { d}x\\&\quad -\int \limits _{\mathbb {R}^3} \partial _1u_1\partial _2u_2\partial _3u_3\text { d}x +\int \limits _{\mathbb {R}^3} \partial _1u_2\partial _2u_1\partial _3u_3\text { d}x, \end{aligned} \end{aligned}$$

(6)

and refined (4) to be critical, but with limited range of space integrability indices,

$$\begin{aligned} \partial _3\varvec{u}\in L^p(0,T;L^q(\mathbb {R}^3)),\quad \frac{2}{p}+\frac{3}{q}=2,\quad \frac{9}{4}\le q\le 3. \end{aligned}$$

(7)

Later on, Cao [2] employed multiplicative Sobolev inequalities

$$\begin{aligned} 1\le q<\infty \Rightarrow \left\| f\right\| _{L^{3q}}\le C\left\| \partial _1f\right\| _{L^2}^\frac{1}{3} \left\| \partial _2f\right\| _{L^2}^\frac{1}{3}\left\| \partial _3f\right\| _{L^q}^\frac{1}{3} \end{aligned}$$

(8)

and

$$\begin{aligned} 1\le q<\infty \Rightarrow \left\| f\right\| _{L^{5q}}\le C\left\| \partial _1(f^2)\right\| _{L^2}^\frac{1}{5} \left\| \partial _2(f^2)\right\| _{L^2}^\frac{1}{5}\left\| \partial _3f\right\| _{L^q}^\frac{1}{5} \end{aligned}$$

(9)

to get the following extended regularity condition

$$\begin{aligned} \partial _3\varvec{u}\in L^p(0,T;L^q(\mathbb {R}^3)),\quad \frac{2}{p}+\frac{3}{q}=2,\quad \frac{27}{16}\le q\le \frac{5}{2}. \end{aligned}$$

(10)

It should be remarked that Cao [2] claimed the range of q in (10) is $\displaystyle {q\ge \frac{27}{16}}$, but it is indeed (10) which is actually proved. See the footnote of [31, p. 35] for more information.

In a recent paper, Zhang [31] generalized (9) as

$$\begin{aligned} 0<\lambda<\infty ,\quad 1\le q<\infty \Rightarrow \left\| f\right\| _{L^{(2\lambda +1)q}} \le C\left\| \partial _1(|f|^\lambda )\right\| _{L^2}^\frac{1}{2\lambda +1} \left\| \partial _2(|f|^\lambda )\right\| _{L^2}^\frac{1}{2\lambda +1} \left\| \partial _3f\right\| _{L^q}^\frac{1}{2\lambda +1}, \end{aligned}$$

(11)

and employed general $L^{2\lambda }$ estimate (instead of $L^4$ estimate as in [2]) to improve (7) and (10) simultaneously. Precisely, he showed the following regularity criterion,

$$\begin{aligned} \partial _3\varvec{u}\in L^p(0,T;L^q(\mathbb {R}^3)),\quad \frac{2}{p}+\frac{3}{q}=2,\quad 1.56207\approx \frac{3\sqrt{37}}{4}-3\le q\le 3. \end{aligned}$$

(12)

Notice in establishing (12), Zhang have missed a condition (say, in [31, (31)], we need $1\le c\le \infty$), which was noticed by Yuliya–Skalak [30]. Skalak [27] then covered all of the range $\left( \frac{3}{2},3\right]$,

$$\begin{aligned} \partial _3\varvec{u}\in L^p(0,T;L^q(\mathbb {R}^3)),\quad \frac{2}{p}+\frac{3}{q}=2,\quad \frac{3}{2}< q\le 3. \end{aligned}$$

(13)

Finally, Zhang–Yuan–Zhou [36] showed two new refinements of (4),

$$\begin{aligned} \partial _3\varvec{u}\in L^p(0,T;L^q(\mathbb {R}^3)),\quad \frac{2}{p}+\frac{3}{q}=\frac{8}{5}+\frac{3}{5q},\quad 4\le q\le \infty , \end{aligned}$$

(14)

and

$$\begin{aligned} \partial _3\varvec{u}\in L^p(0,T;L^q(\mathbb {R}^3)),\quad \frac{2}{p}+\frac{3}{q}=\frac{14}{11}+\frac{9}{11q},\quad \frac{5}{2}\le q\le \infty . \end{aligned}$$

(15)

Whence, the state of the art is the following smoothness condition

$$\begin{aligned} \partial _3\varvec{u}\in L^p(0,T;L^q(\mathbb {R}^3)),\quad \frac{2}{p}+\frac{3}{q}=\left\{ \begin{array}{ll} 2,&{}\frac{3}{2}< q\le 3,\\ \frac{14}{11}+\frac{9}{11q},&{}3<q< \frac{18}{5},\\ \frac{3}{2},&{}\frac{18}{5}\le q<4,\\ \frac{8}{5}+\frac{3}{5q},&{}4\le q\le \infty . \end{array}\right. \end{aligned}$$

(16)

The first purpose of this paper is to improve (16) in the range $3<q<4$.

As far as regularity criterion $\partial _3u_3$ is concerned, Zhou–Pokorný [39] first established a regularity condition based on $u_3$ and then showed that if

$$\begin{aligned} \partial _3u_3\in L^\beta (0,T;L^\alpha (\mathbb {R}^3)),\quad \frac{2}{\beta }+\frac{3}{\alpha }<\frac{4}{5},\quad \frac{15}{4}<\alpha \le \infty , \end{aligned}$$

(17)

then the solution is smooth. The equality in (17) was verified by Jia–Zhou [14]:

$$\begin{aligned} \partial _3u_3\in L^\beta (0,T;L^\alpha (\mathbb {R}^3)),\quad \frac{2}{\beta }+\frac{3}{\alpha }=\frac{4}{5},\quad \frac{15}{4}\le \alpha \le \infty . \end{aligned}$$

(18)

Later, Cao–Titi [4] established a bilateral multiplicative Sobolev inequality (see [32, Remark 8] for more information, and [35] for a more efficient form)

$$\begin{aligned} \begin{aligned} \left| \int \limits _{\mathbb {R}^3} \phi fg\text { d}x\right|&\le C\left\| \phi \right\| _{L^2}^\frac{r-1}{r} \left\| \partial _i\phi \right\| _{L^\frac{2}{3-r}}^\frac{1}{r} \left\| f\right\| _{L^2}^\frac{r-2}{r} \left\| \partial _jf\right\| _{L^2}^\frac{1}{r} \left\| \partial _kf\right\| _{L^2}^\frac{1}{r} \left\| g\right\| _{L^2},\\&2<r\le 3,\ \left\{ i,j,k\right\} =\left\{ 1,2,3\right\} . \end{aligned} \end{aligned}$$

(19)

With (19) in hand, Cao–Titi showed the following two smoothness conditions,

$$\begin{aligned} \partial _3u_3\in L^\beta (0,T;L^\alpha (\mathbb {R}^3)),\quad \frac{2}{\beta }+\frac{3}{\alpha }=\frac{3}{4}+\frac{3}{2\alpha },\quad 2<\alpha <\infty , \end{aligned}$$

(20)

and

$$\begin{aligned} \partial _1u_3\in L^\beta (0,T;L^\alpha (\mathbb {R}^3)),\quad \frac{2}{\beta }+\frac{3}{\alpha }=\frac{1}{2}+\frac{3}{2\alpha },\quad 3<\alpha <\infty . \end{aligned}$$

(21)

Then Fang–Qian [9, Theorems 1.1 and 1.2] dominated $u_3$ by $\partial _1u_3$, employed some tricks in [4] and improved (21) as (after rationalizing the denominator of [9, Equation (1.10)])

$$\begin{aligned} \partial _1u_3\in L^\beta (0,T;L^\alpha (\mathbb {R}^3)),\quad \frac{2}{\beta }+\frac{3}{\alpha }=\frac{\sqrt{103\alpha ^2-12\alpha +9}+3-9\alpha }{2\alpha },\quad 3\le \alpha <\infty . \end{aligned}$$

(22)

Finally, Zhang–Zhong–Huang [35] found a more effective substitute of (19),

$$\begin{aligned} \begin{aligned}&\int \limits _{\mathbb {R}^3} |f|^2|g|^2\text { d}x_1\text { d}x_2\text { d}x_3\\&\quad \le C\left\| \partial _if\right\| _{L^2(\mathbb {R}^3)}^\frac{2(q-1)}{3q-2} \left\| \partial _jf\right\| _{L^2(\mathbb {R}^3)}^\frac{2(q-1)}{3q-2} \left\| \partial _kf\right\| _{L^q(\mathbb {R}^3)}^\frac{2q}{3q-2} \left\| g\right\| _{L^2(\mathbb {R}^3)}^\frac{6q-8}{3q-2} \left\| \partial _ig\right\| _{L^2(\mathbb {R}^3)}^\frac{2}{3q-2} \left\| \partial _jg\right\| _{L^2(\mathbb {R}^3)}^\frac{2}{3q-2},\\&\qquad 2\le q<\infty ,\ \left\{ i,j,k\right\} =\left\{ 1,2,3\right\} . \end{aligned} \end{aligned}$$

(23)

Invoking (23), Zhang–Zhong–Huang [35] were able to improve (21) and (22) as

$$\begin{aligned} \partial _1u_3\in L^\beta (0,T;L^\alpha (\mathbb {R}^3)),\quad \frac{2}{\beta }+\frac{3}{\alpha }=\frac{3}{4}+\frac{5}{4\alpha },\quad \frac{7}{3}\le \alpha <\infty , \end{aligned}$$

(24)

but could not refine (20).

As for (20), Fang–Qian [9] made a contribution by invoking a regularity criterion of Zhang [33]. Fang–Qian [10, Theorem 1.8] then used an integration by parts technique in estimating $u_3$ by $\partial _3u_3$ and obtained the finest result up to now,

$$\begin{aligned} \begin{aligned}&\partial _3u_3\in L^\beta (0,T;L^\alpha (\mathbb {R}^3)),\\&\frac{2}{\beta }+\frac{3}{\alpha }= \frac{\sqrt{289\alpha ^2-264\alpha +144}+12-7\alpha }{8\alpha },\quad \frac{9}{5}<\alpha \le \infty . \end{aligned} \end{aligned}$$

(25)

For later developments in anisotropic Lebesgue spaces, see [11, 24]. The second aim of the present paper is to make (25) better.

Before stating the precise result, let us recall the weak formulation of (1), see [7, 17, 23, 29] for instance.

Definition 1

Let $\varvec{u}_0\in L^2(\mathbb {R}^3)$ with $\nabla \cdot \varvec{u}_0=0$, $T>0$. A measurable $\mathbb {R}^3$-valued function $\varvec{u}$ defined in $[0,T]\times \mathbb {R}^3$ is said to be a weak solution to (1) if

• (1) $\varvec{u}\in L^\infty (0,T;L^2(\mathbb {R}^3)\cap L^2(0,T;H^1(\mathbb {R}^3))$;

• (2) (1)$_1$ and (1)$_2$ hold in the sense of distributions, i.e.,

  $$\begin{aligned} \int \limits _0^t\int \limits _{\mathbb {R}^3}\varvec{u}\cdot \left[ \partial _t\varvec{\phi }+\left( \varvec{u}\cdot \nabla \right) \varvec{\phi }\right] \text { d}x\text { d}s+\int \limits _{\mathbb {R}^3}\varvec{u}_0\cdot \varvec{\phi }(0)\text { d}x =\int \limits _0^T\int \limits _{\mathbb {R}^3} \nabla \varvec{u}:\nabla \varvec{\phi }\text { d}x\text { d}t, \end{aligned}$$

  for each $\varvec{\phi }\in C_c^\infty ([0,T)\times \mathbb {R}^3)$ with $\nabla \cdot \varvec{\phi }=0$, where $\displaystyle {A:B=\sum \nolimits _{i,j=1}^3 a_{ij}b_{ij}}$ for $3\times 3$ matrices $A=(a_{ij})$, $B=(b_{ij})$, and

  $$\begin{aligned} \int \limits _0^T \int \limits _{\mathbb {R}^3}\varvec{u}\cdot \nabla \psi \text { d}x\text { d}t=0, \end{aligned}$$

  for each $\psi \in C_c^\infty (\mathbb {R}^3\times [0,T))$;

• (3) the strong energy inequality, that is,

  $$\begin{aligned} \left\| \varvec{u}(t)\right\| _{L^2}^2+2\int \limits _s^t\left\| \nabla \varvec{u}(s)\right\| _{L^2}^2\text { d}s \le \left\| \varvec{u}(s)\right\| _{L^2}^2,\quad s\le t\le T, \end{aligned}$$

  holds for $s=0$ and almost all times $s\in (0,T)$.

Now, our main result reads

Theorem 2

Let $\varvec{u}_0\in L^2(\mathbb {R}^3)$ with $\nabla \cdot \varvec{u}_0=0$, $T>0$. Assume that $\varvec{u}$ is a weak solution to (1) on [0, T] with initial data $\varvec{u}_0$. If one of the following two conditions holds,

$$\begin{aligned}&\partial _3\varvec{u}\in L^p(0,T;L^q(\mathbb {R}^3)),\quad \frac{2}{p}+\frac{3}{q}=\frac{22}{13}+\frac{3}{13q},\quad 3<q<4, \end{aligned}$$

(26)

$$\begin{aligned}&\partial _3u_3\in L^\beta (0,T;L^\alpha (\mathbb {R}^3)),\nonumber \\&\frac{2}{\beta }+\frac{3}{\alpha } =\frac{3\left( \sqrt{65\alpha ^2-78\alpha +49}+7-\alpha \right) }{16\alpha },\quad 1.78078\approx \frac{3+\sqrt{17}}{4}\le \alpha \le \infty , \end{aligned}$$

(27)

then the solution $\varvec{u}$ is smooth in $(0,T]\times \mathbb {R}^3$.

Remark 3

• (1) Our regularity criterion (26) is better than (16) in case $3<\alpha <4$. See Fig. 1, where “Skalak” refers to (13), “one Zhang-Zhou” (the upper one) means (14), “two Zhang-Zhou (the lower one)” demonstrates (15), “Penel-Pokorny” reveals (4), and “this” reflects (26).

• (2) Our regularity criterion (27) is better than (17), (20) and (25). See Fig. 2, where “Zhou-Pokorný” refers to (17); “Cao–Titi” means (20); “Fang-Qian” demonstrates (25); and “this” reflects our result (27).

• (3) It is not so hard to deduce that the scaling dimension $\displaystyle {\frac{3\left( \sqrt{65\alpha ^2-78\alpha +49}+7-\alpha \right) }{16\alpha }}$ in (27) is strictly decreasing with respect to $\displaystyle {\frac{3+\sqrt{17}}{4}\le \alpha \le \infty }$. Notice that

  $$\begin{aligned} \begin{aligned}&\lim _{\alpha \rightarrow \frac{3+\sqrt{17}}{4}}\frac{3\left( \sqrt{65\alpha ^2-78\alpha +49}+7-\alpha \right) }{16\alpha } =\frac{3(\sqrt{17}-3)}{2}\approx 1.68466,\\&\lim _{\alpha \rightarrow \infty } \frac{3\left( \sqrt{65\alpha ^2-78\alpha +49}+7-\alpha \right) }{16\alpha } =\frac{3(\sqrt{65}-1)}{16}\approx 1.32417, \end{aligned} \end{aligned}$$

  we have the following rough, but maybe more elegant regularity criterion in terms of $\partial _3u_3$,

  $$\begin{aligned} \begin{aligned}&\partial _3u_3\in L^\beta (0,T;L^\alpha (\mathbb {R}^3)),\\&\frac{2}{\beta }+\frac{3}{\alpha }= \frac{3(\sqrt{65}-1)}{16}\approx 1.32417,\quad 1.78078\approx \frac{3+\sqrt{17}}{4}\le \alpha \le \infty . \end{aligned} \end{aligned}$$

  (28)

2 Proof of Theorem 2

In this section, we shall prove Theorem 2.

Case I (26) holds. For any $\varepsilon \in (0,T)$, due to the fact that $\nabla \varvec{u}\in L^2(0,T;L^2(\mathbb {R}^3))$, we may find a $\delta \in (0,\varepsilon )$, such that $\nabla \varvec{u}(\delta )\in L^2(\mathbb {R}^3)$. Take this $\varvec{u}(\delta )$ as initial data, there exists an ${\tilde{\varvec{u}}}\in C([\delta ,\varGamma ^*),H^1(\mathbb {R}^3)) \cap L^2(\delta ,\varGamma ^*;H^2(\mathbb {R}^3))$, where $[\delta , \varGamma ^*)$ is the life span of the unique strong solution, see [29]. Moreover, ${\tilde{\varvec{u}}}\in C^\infty (\mathbb {R}^3\times (\delta ,\varGamma ^*))$. According to the uniqueness result, ${\tilde{\varvec{u}}}=\varvec{u}$ on $[\delta ,\varGamma ^*)$. If $\varGamma ^*\ge T$, we have already that $\varvec{u}\in C^\infty (\mathbb {R}^3\times (0,T))$, due to the arbitrariness of $\varepsilon \in (0,T)$. In case $\varGamma ^*<T$, our strategy is to show that $u_3\in L^3(\delta ,\varGamma ^*;L^9(\mathbb {R}^3))$. Then, by the fact that

$$\begin{aligned} {}(26)\Rightarrow \partial _3\varvec{u}\in L^\frac{13q}{11q-18}(\delta ,\varGamma ^*;L^q(\mathbb {R}^3))\subset L^\frac{2q}{2q-3}(\delta ,\varGamma ^*;L^q(\mathbb {R}^3)), \end{aligned}$$

we may conclude the proof by invoking (5).

For this purpose, we multiply the equation of $u_3$ in (1) by $|u_3|u_3$ and integrate over $\mathbb {R}^3$,

$$\begin{aligned} \frac{1}{3}\frac{\text { d}}{\text { d}t}\left\| |u_3|^\frac{3}{2}\right\| _{L^2}^2 +\frac{4}{9}\left\| \nabla |u_3|^\frac{3}{2}\right\| _{L^2}^2 +\int \limits _{\mathbb {R}^3} |u_3|\cdot |\nabla u_3|^2\text { d}x =-\int \limits _{\mathbb {R}^3} \partial _3\pi |u_3|u_3\text { d}x\equiv I. \end{aligned}$$

(29)

Fig. 1

Comparison of regularity criterion based on $\partial _3\varvec{u}$

Fig. 2

Comparison of regularity criterion based on $\partial _3u_3$,

By the Hölder inequality,

$$\begin{aligned} I\le \left\| \partial _3\pi \right\| _{L^\frac{3q}{4}} \left\| u_3\right\| _{L^\frac{6q}{3q-4}}^2. \end{aligned}$$

To dominate $\partial _3\pi$, we apply the divergence of (1)$_1$ to obtain

$$\begin{aligned} \begin{aligned}&-\Delta \pi =\nabla \cdot [(\varvec{u}\cdot \nabla )\varvec{u}] =\sum _{j=1}^3 \left( \sum _{i=1}^3 u_i\partial _iu_j\right) = \sum _{i,j=1}^3 \partial _i\partial _j (u_iu_j)\\&\quad \Rightarrow \left\{ \begin{array}{llllllll}\displaystyle {\pi =\sum _{i,j=1}^3 \mathcal {R}_i\mathcal {R}_j(u_iu_j)}\\ \displaystyle {-\Delta \partial _3\pi =\sum _{i,j=1}^3 \partial _i\partial _j (\partial _3u_iu_j+u_i\partial _3u_j) \Rightarrow \partial _3\pi =\sum _{i,j=1}^3 \mathcal {R}_i\mathcal {R}_j(\partial _3u_iu_j+u_i\partial _3u_j)}\end{array}\right. \\&\quad \quad \left( \mathcal {R}_i=\partial _i(-\Delta )^{-\frac{1}{2}} \text{ is } \text{ the } \text{ Riesz } \text{ transform }\right) , \end{aligned} \end{aligned}$$

(30)

and thus by the interpolation inequality,

$$\begin{aligned} I\le C\left\| \varvec{u}\right\| _{L^{3q}} \left\| \partial _3\varvec{u}\right\| _{L^q} \cdot \left[ \left\| u_3\right\| _{L^3}^\frac{6q-11}{4q-3} \left\| u_3\right\| _{L^{4q}}^\frac{2(4-q)}{4q-3} \right] ^2. \end{aligned}$$

Employing the multiplicative Sobolev inequalities (8) and (11) yields

$$\begin{aligned} \begin{aligned} I&\le C\left\| \partial _1\varvec{u}\right\| _{L^2}^\frac{1}{3} \left\| \partial _2\varvec{u}\right\| _{L^2}^\frac{1}{3} \left\| \partial _3\varvec{u}\right\| _{L^q}^\frac{1}{3} \cdot \left\| \partial _3\varvec{u}\right\| _{L^q}\\&\quad \cdot \left\{ \left\| u_3\right\| _{L^3}^\frac{6q-11}{4q-3}\cdot \left[ \left\| \partial _1\left( |u_3|^\frac{3}{2}\right) \right\| _{L^2}^\frac{1}{4} \left\| \partial _2\left( |u_3|^\frac{3}{2}\right) \right\| _{L^2}^\frac{1}{4} \left\| \partial _3u_3\right\| _{L^q}^\frac{1}{4} \right] ^\frac{2(4-q)}{4q-3} \right\} ^2. \end{aligned} \end{aligned}$$

After collection, we deduce by the Young inequality,

$$\begin{aligned} \begin{aligned} I&\le C\left\| \nabla \varvec{u}\right\| _{L^2}^\frac{2}{3} \left\| \partial _3\varvec{u}\right\| _{L^q}^\frac{13q}{3(4q-3)} \left\| u_3\right\| _{L^3}^\frac{2(6q-11)}{4q-3} \left\| \nabla (|u_3|^\frac{3}{2})\right\| _{L^2}^\frac{2(4-q)}{4q-3}\\&\le C\left\| \nabla \varvec{u}\right\| _{L^2}^\frac{2(4q-3)}{3(5q-7)} \left\| \partial _3\varvec{u}\right\| _{L^q}^\frac{13q}{3(5q-7)} \left\| u_3\right\| _{L^3}^\frac{2(6q-11)}{5q-7} +\frac{2}{9} \left\| \nabla \left( |u_3|^\frac{3}{2}\right) \right\| _{L^2}^2\\&\le C\left( \left\| \nabla \varvec{u}\right\| _{L^2}^2 +\left\| \partial _3\varvec{u}\right\| _{L^q}^\frac{13q}{11q-18}\right) \left( 1+\left\| u_3\right\| _{L^3}^3\right) +\frac{2}{9} \left\| \nabla \left( |u_3|^\frac{3}{2}\right) \right\| _{L^2}^2. \end{aligned} \end{aligned}$$

Putting this above inequality into (29) and applying the Gronwall inequality give

$$\begin{aligned} \left\| u_3\right\| _{L^3(\delta ,\varGamma ^*;L^9(\mathbb {R}^3))} =\left\| |u_3|^\frac{3}{2}\right\| _{L^2(\delta ,\varGamma ^*;L^6(\mathbb {R}^3))} \le C\left\| \nabla |u_3|^\frac{3}{2}\right\| _{L^2(\delta ,\varGamma ^*;L^2(\mathbb {R}^3))} \le C, \end{aligned}$$

as desired.

Case II (27) holds. Argue as in Case I, it suffices to show that $\left\| \nabla \varvec{u}(t)\right\| _{L^2}$ is uniformly bounded as $t\nearrow \varGamma ^*$. By the absolute continuity property of the Lebesgue integrable function, for $\delta _2\in (0,1)$ to be determined, we can choose a $\delta _1\in [\delta ,\varGamma ^*)$ such that

$$\begin{aligned} \nabla \varvec{u}(\delta _1)\in L^2(\mathbb {R}^3),\quad \int \limits _{\delta _1}^{\varGamma ^*} \left\| \nabla \nabla _h\varvec{u}\right\| _{L^2}^2\text { d}t<\delta _2. \end{aligned}$$

(31)

We first establish the $L^q$ bound of $u_3$ in terms of $\partial _3u_3$, which have been used in [9, 10]. Multiplying the third component of (1)$_1$:

$$\begin{aligned} \partial _t u_3+(\varvec{u}\cdot \nabla )u_3-\Delta u_3+\partial _3\pi =0 \end{aligned}$$

by $|u_3|^{q-2}u_3$ with

$$\begin{aligned} 2<q\le 6, \end{aligned}$$

(32)

integrating over $\mathbb {R}^3$ and applying integration by parts, we obtain

$$\begin{aligned} \begin{aligned} \frac{1}{q}\frac{\text { d}}{\text { d}t}\left\| u_3\right\| _{L^q}^q +c(q)\left\| \nabla \left( |u_3|^\frac{q}{2}\right) \right\| _{L^2}^2&=-\int \limits _{\mathbb {R}^3}\partial _3\pi \cdot |u_3|^{q-2}u_3\text { d}x\\&\le C\int \limits _{\mathbb {R}^3} |\pi | \cdot |u_3|^{q-2} |\partial _3u_3|\text { d}x\\&\equiv J. \end{aligned} \end{aligned}$$

(33)

By the Hölder inequality with

$$\begin{aligned} \frac{1}{a}+\frac{q-2}{(q+1)\alpha }+\frac{1}{\alpha }=1,\quad 1\le a\le \infty ,\quad 1\le \alpha \le \infty , \end{aligned}$$

(34)

we have

$$\begin{aligned} J\le C\left\| \pi \right\| _{L^a} \left\| u_3\right\| _{L^{(q+1)\alpha }}^{q-2} \left\| \partial _3u_3\right\| _{L^\alpha }. \end{aligned}$$

Thanks to (30) and (11), it follows that

$$\begin{aligned} J \le C\left\| \varvec{u}\right\| _{L^{2a}}^2 \left( \left\| \nabla _h\left( |u_3|^\frac{q}{2}\right) \right\| _{L^2}^\frac{2}{q+1} \left\| \partial _3u_3\right\| _{L^\alpha }^\frac{1}{q+1} \right) ^{q-2}\left\| \partial _3u_3\right\| _{L^\alpha }, \end{aligned}$$

provided that

$$\begin{aligned} 1<a<\infty ,\quad 1\le \alpha <\infty . \end{aligned}$$

(35)

Employing the Gagliardo–Nirenberg inequality with

$$\begin{aligned} \frac{3}{2a}=(1-\vartheta )\frac{3}{2}+\vartheta \left( -1+\frac{3}{2}\right) ,\quad 2<2a<6 \end{aligned}$$

(36)

gives

$$\begin{aligned} J\le C\left( \left\| \varvec{u}\right\| _{L^2}^{1-\vartheta } \left\| \nabla \varvec{u}\right\| _{L^2}^\vartheta \right) ^2 \left\| \nabla \left( |u_3|^\frac{q}{2}\right) \right\| _{L^2}^\frac{2(q-2)}{q+1} \left\| \partial _3u_3\right\| _{L^\alpha }^\frac{2q-1}{q+1}. \end{aligned}$$

By the fact that $\varvec{u}\in L^\infty (0,T;L^2(\mathbb {R}^3))$ from Definition 1 and the Young inequality, we deduce

$$\begin{aligned} \begin{aligned} J&\le C\left\| \nabla \varvec{u}\right\| _{L^2}^{2\vartheta } \left\| \nabla \left( |u_3|^\frac{q}{2}\right) \right\| _{L^2}^\frac{2(q-2)}{q+1} \left\| \partial _3u_3\right\| _{L^\alpha }^\frac{2q-1}{q+1}\\&\le \left\{ \begin{array}{llllllll} \displaystyle {C\left\| \nabla \varvec{u}\right\| _{L^2}^\frac{2(q+1)\vartheta }{3} \left\| \partial _3u_3\right\| _{L^\alpha }^\frac{2q-1}{3} +\frac{c(q)}{2}\left\| \nabla \left( |u_3|^\frac{q}{2}\right) \right\| _{L^2}^2}, &{}\text{ if } \displaystyle {\frac{2(q+1)\vartheta }{3}<2}\\ \displaystyle {C\left\| \nabla \varvec{u}\right\| _{L^2}^2 \left\| \partial _3u_3\right\| _{L^\alpha }^\frac{2q-1}{3} +\frac{c(q)}{2}\left\| \nabla \left( |u_3|^\frac{q}{2}\right) \right\| _{L^2}^2},&{}\text{ if } \displaystyle {\frac{2(q+1)\vartheta }{3}=2} \end{array}\right. \\&\le \left\{ \begin{array}{llllllll} \displaystyle {C\left\| \nabla \varvec{u}\right\| _{L^2}^2+C \left\| \partial _3u_3\right\| _{L^\alpha }^\frac{2q-1}{3-(q+1)\vartheta } +\frac{c(q)}{2}\left\| \nabla \left( |u_3|^\frac{q}{2}\right) \right\| _{L^2}^2}, &{}\text{ if } \displaystyle {\frac{2(q+1)\vartheta }{3}<2}\\ \displaystyle {C\left\| \nabla \varvec{u}\right\| _{L^2}^2 \left\| \partial _3u_3\right\| _{L^\alpha }^\frac{2q-1}{3} +\frac{c(q)}{2}\left\| \nabla \left( |u_3|^\frac{q}{2}\right) \right\| _{L^2}^2},&{}\text{ if } \displaystyle {\frac{2(q+1)\vartheta }{3}=2} \end{array}\right. . \end{aligned} \end{aligned}$$

(37)

if

$$\begin{aligned} \frac{2(q+1)\vartheta }{3}\le 2,\quad \beta =\left\{ \begin{array}{llllllll} \displaystyle {\frac{2q-1}{3-(q+1)\vartheta }},&{}\text{ if } \displaystyle {\frac{2(q+1)}{3}<2}\\ \infty ,&{}\text{ if } \displaystyle {\frac{2(q+1)\vartheta }{3}=2} \end{array}\right. . \end{aligned}$$

(38)

Plugging (37) into (33), absorbing the last term into the left-hand side and integrating with respect to the time, we find

$$\begin{aligned} u_3\in L^\infty (\delta _1,\varGamma ^*;L^q(\mathbb {R}^3)). \end{aligned}$$

(39)

Then, we establish the bound of $\left\| \nabla _h\varvec{u}\right\| _{L^2}^2$ with $\nabla _h=(\partial _1,\partial _2)$ the horizontal gradient operator. Testing (1)$_1$ by $-\Delta _h\varvec{u}$, it follows from [3, 4, 39] that

$$\begin{aligned} \begin{aligned} \frac{1}{2}\frac{\text { d}}{\text { d}t}\left\| \nabla _h\varvec{u}\right\| _{L^2}^2 +\left\| \nabla \nabla _h\varvec{u}\right\| _{L^2}^2&=\int \limits _{\mathbb {R}^3}[(\varvec{u}\cdot \nabla )\varvec{u}]\cdot \Delta _h\varvec{u}\text { d}x\\&\le C\int \limits _{\mathbb {R}^3} |u_3| |\nabla \varvec{u}| |\nabla \nabla _h\varvec{u}|\text { d}x\\&\equiv K. \end{aligned} \end{aligned}$$

(40)

By the Hölder inequality, the Minkowski inequality and the Gagliardo–Nirenberg inequality,

$$\begin{aligned} \begin{aligned} K&\le C\int \limits _{\mathbb {R}^2} \max _{x_3} |u_3| \left( \int \limits _{\mathbb {R}} |\nabla \varvec{u}|^2\text { d}x_3\right) ^\frac{1}{2} \left( \int \limits _{\mathbb {R}} |\nabla \nabla _h\varvec{u}|^2\text { d}x_3\right) ^\frac{1}{2}\text { d}x_1\text { d}x_2\\&\le C\left[ \,\,\int \limits _{\mathbb {R}^2} \left( \max _{x_3} |u_3|\right) ^r\text { d}x_1\text { d}x_2\right] ^\frac{1}{r} \left[ \int \limits _{\mathbb {R}^2}\left( \int \limits _{\mathbb {R}} |\nabla \varvec{u}|^2\text { d}x_3\right) ^\frac{r}{r-2}\text { d}x_1\text { d}x_2\right] ^\frac{r-2}{2r}\\&\quad \cdot \left[ \,\,\int \limits _{\mathbb {R}^3} |\nabla \nabla _h\varvec{u}|^2\text { d}x_1\text { d}x_2\text { d}x_3\right] ^\frac{1}{2}\\&\le C\left[ \,\,\int \limits _{\mathbb {R}^3} |u_3|^{r-1}|\partial _3u_3|\text { d}x\right] ^\frac{1}{r} \left[ \,\,\int \limits _{\mathbb {R}} \left( \,\, \int \limits _{\mathbb {R}^2} |\nabla \varvec{u}|^\frac{2r}{r-2} \text { d}x_1\text { d}x_2 \right) ^\frac{r-2}{r} \text { d}x_3\right] ^\frac{1}{2} \left\| \nabla \nabla _h\varvec{u}\right\| _{L^2}\\&\le C\left\| u_3\right\| _{L^q}^\frac{r-1}{r} \left\| \partial _3u_3\right\| _{L^\alpha }^\frac{1}{r} \cdot \left\| \nabla \varvec{u}\right\| _{L^2}^\frac{r-2}{r} \left\| \nabla \nabla _h\varvec{u}\right\| _{L^2}^\frac{2}{r} \cdot \left\| \nabla \nabla _h\varvec{u}\right\| _{L^2}\\&\le C\left\| u_3\right\| _{L^q}^\frac{2(r-1)}{r-2} \left\| \partial _3u_3\right\| _{L^\alpha }^\frac{2}{r-2} \left\| \nabla \varvec{u}\right\| _{L^2}^2 +\frac{1}{2} \left\| \nabla \nabla _h\varvec{u}\right\| _{L^2}^2. \end{aligned} \end{aligned}$$

(41)

Here, the exponents appeared above should satisfy

$$\begin{aligned} 2<r<\infty ,\quad \frac{r-1}{q}+\frac{1}{\alpha }=1. \end{aligned}$$

(42)

Putting (41) into (40) and integrating with respect to the time give

$$\begin{aligned} \begin{aligned}&\sup _{\delta _1\le t<\varGamma ^*}\left\| \nabla _h\varvec{u}(t)\right\| _{L^2}^2 +\int \limits _{\delta _1}^{\varGamma ^*}\left\| \nabla \nabla _h\varvec{u}\right\| _{L^2}^2\text { d}t\\&\quad \le \left\| \nabla _h\varvec{u}(\delta _1)\right\| _{L^2}^2 +C\int \limits _{\delta _1}^{\varGamma ^*} \left\| u_3\right\| _{L^q}^\frac{2(r-1)}{r-2} \left\| \partial _3u_3\right\| _{L^\alpha }^\frac{2}{r-2} \left\| \nabla \varvec{u}\right\| _{L^2}^2\text { d}t\\&\quad \le C+C\int \limits _{\delta _1}^{\varGamma ^*} \left\| u_3\right\| _{L^q}^\frac{2(r-1)}{r-2} \left\| \partial _3u_3\right\| _{L^\alpha }^\frac{2}{r-2} \left\| \nabla \varvec{u}\right\| _{L^2}^2\text { d}t. \end{aligned} \end{aligned}$$

(43)

Finally, we obtain the $H^1$ estimate of the solution. Taking the inner product of (1)$_1$ by $-\Delta \varvec{u}$ in $L^2(\mathbb {R}^3)$, it follows from [39] that

$$\begin{aligned} \begin{aligned} \frac{1}{2}\frac{\text { d}}{\text { d}t}\left\| \nabla \varvec{u}\right\| _{L^2}^2 +\left\| \Delta \varvec{u}\right\| _{L^2}^2&=\int \limits _{\mathbb {R}^3}[(\varvec{u}\cdot \nabla )\varvec{u}]\cdot \Delta \varvec{u}\text { d}x\\&\le C\int \limits _{\mathbb {R}^3} |\nabla _h\varvec{u}||\nabla \varvec{u}|^2\text { d}x\\&\equiv L. \end{aligned} \end{aligned}$$

(44)

Invoking the Hölder inequality (8) and the Young inequality, we get

$$\begin{aligned} \begin{aligned} L&\le C\left\| \nabla _h\varvec{u}\right\| _{L^2}\left\| \nabla \varvec{u}\right\| _{L^4}^2\\&\le C\left\| \nabla _h\varvec{u}\right\| _{L^2}\cdot \left( \left\| \nabla \varvec{u}\right\| _{L^2}^\frac{1}{4} \left\| \nabla \varvec{u}\right\| _{L^6}^\frac{3}{4}\right) ^2\\&\le C\left\| \nabla _h\varvec{u}\right\| _{L^2} \left\| \nabla \varvec{u}\right\| _{L^2}^\frac{1}{2} \left\| \nabla \nabla _h\varvec{u}\right\| _{L^2}\left\| \Delta \varvec{u}\right\| _{L^2}^\frac{1}{2}\\&\le C\left\| \nabla _h\varvec{u}\right\| _{L^2}^\frac{4}{3} \left\| \nabla \varvec{u}\right\| _{L^2}^\frac{2}{3} \left\| \nabla \nabla _h\varvec{u}\right\| _{L^2}^\frac{4}{3} +\frac{1}{2}\left\| \Delta \varvec{u}\right\| _{L^2}^2. \end{aligned} \end{aligned}$$

(45)

Gathering (45) into (44) and integrating with respect to the time provide

$$\begin{aligned} \begin{aligned}&\sup _{\delta _1\le t<\varGamma ^*}\left\| \nabla \varvec{u}(t)\right\| _{L^2}^2 +\int \limits _{\delta _1}^{\varGamma ^*}\left\| \Delta \varvec{u}\right\| _{L^2}^2\text { d}t\\&\quad \le \left\| \nabla \varvec{u}(\delta )\right\| _{L^2}^2 +C\int \limits _{\delta _1}^{\varGamma ^*} \left\| \nabla _h\varvec{u}\right\| _{L^2}^\frac{4}{3} \left\| \nabla \varvec{u}\right\| _{L^2}^\frac{2}{3} \left\| \nabla \nabla _h\varvec{u}\right\| _{L^2}^\frac{4}{3} \text { d}t\\&\quad \le C+C\sup _{\delta _1\le t<\varGamma ^*} \left\| \nabla _h\varvec{u}(t)\right\| _{L^2}^\frac{4}{3} \left( \int \limits _{\delta _1}^{\varGamma ^*} \left\| \nabla \varvec{u}\right\| _{L^2}^2\text { d}t\right) ^\frac{1}{3} \left( \int \limits _{\delta _1}^{\varGamma ^*} \left\| \nabla \nabla _h\varvec{u}\right\| _{L^2}^2\text { d}t\right) ^\frac{2}{3}. \end{aligned} \end{aligned}$$

(46)

Thanks to (31) and the obtained estimates (43) and (39), we have

$$\begin{aligned} \begin{aligned}&\sup _{\delta _1\le t<\varGamma ^*}\left\| \nabla \varvec{u}(t)\right\| _{L^2}^2 +\int \limits _{\delta _1}^{\varGamma ^*}\left\| \Delta \varvec{u}\right\| _{L^2}^2\text { d}t\\&\quad \le C+C \delta _2^\frac{1}{3} \left[ C+C\int \limits _{\delta _1}^{\varGamma ^*} \left\| u_3\right\| _{L^q}^\frac{2(r-1)}{r-2} \left\| \partial _3u_3\right\| _{L^\alpha }^\frac{2}{r-2} \left\| \nabla \varvec{u}\right\| _{L^2}^2\text { d}t\right] ^\frac{4}{3}\\&\quad \le C+C\delta _2^\frac{1}{3} \sup _{\delta _1\le t<\varGamma ^*} \left\| u_3(t)\right\| _{L^q}^\frac{8(r-1)}{3(r-2)} \cdot \sup _{\delta _1\le t<\varGamma ^*}\left\| \nabla \varvec{u}(t)\right\| _{L^2}^2\cdot \left( \int \limits _{\delta _1}^{\varGamma ^*} \left\| \partial _3u_3\right\| _{L^\alpha }^\frac{2}{r-2} \left\| \nabla \varvec{u}\right\| _{L^2}^\frac{1}{2}\text { d}t\right) ^\frac{4}{3}\\&\quad \le C+C\delta _2^\frac{1}{3} \sup _{\delta _1\le t<\varGamma ^*}\left\| \nabla \varvec{u}(t)\right\| _{L^2}^2 \left( \int \limits _{\delta _1}^{\varGamma ^*} \left\| \partial _3u_3\right\| _{L^\alpha }^\frac{8}{3(r-2)} + \left\| \nabla \varvec{u}\right\| _{L^2}^2\text { d}t \right) ^\frac{4}{3}. \end{aligned} \end{aligned}$$

(47)

Now, if

$$\begin{aligned} \beta =\frac{8}{3(r-2)}, \end{aligned}$$

(48)

then the last integral in (47) is finite, and once we choose $\delta _2$ sufficiently small, then we can absorb the last term in (47) into the left-hand side to deduce

$$\begin{aligned} \sup _{\delta _1\le t<\varGamma ^*}\left\| \nabla \varvec{u}(t)\right\| _{L^2}^2\le C, \end{aligned}$$

as desired. The proof of Theorem 2 is thus completed.

Now, we calculate all the parameters above. Denote by $\displaystyle {{\tilde{\beta }}=\frac{1}{\beta },\ {\tilde{\alpha }}=\frac{1}{\alpha }}$, then

$$\begin{aligned} \begin{aligned} \frac{1}{q}&=\frac{1}{r-1}\left( 1-\frac{1}{\alpha }\right) =\frac{1}{(r-2)+1}(1-{\tilde{\alpha }})\ \ (\text {by}\ (42))\\&=\frac{1}{\frac{8}{3\beta }+1}(1-{\tilde{\alpha }}) =\frac{3(1-{\tilde{\alpha }})}{8{\tilde{\beta }}+3} \ \ (\text {by}\ (48)). \end{aligned} \end{aligned}$$

(49)

On the other hand,

$$\begin{aligned} \begin{aligned} {\tilde{\beta }}&=\frac{1}{\beta }=\frac{3-(q+1)\vartheta }{2q-1}\ \ (\text {by}\ (38))\\&=\frac{3}{2q-1}-\frac{q+1}{2q-1} \cdot \frac{3}{2}\left( 1-\frac{1}{a}\right) \ \ (\text {by}\ (36)\displaystyle {\Rightarrow \vartheta =\frac{3}{2}\left( 1-\frac{1}{a}\right) })\\&=\frac{3}{2q-1}-\frac{q+1}{2q-1}\cdot \frac{3}{2} \cdot \frac{2q-1}{(q+1)\alpha }=\frac{3}{2q-1}-\frac{3}{2\alpha } \ \ (\text {by}\ \ (34)\displaystyle {\Rightarrow 1-\frac{1}{a}=\frac{2q-1}{(q+1)\alpha }}). \end{aligned} \end{aligned}$$

(50)

Putting (49) into (50) yields

$$\begin{aligned} \begin{aligned} {\tilde{\beta }}&=\frac{3}{2q-1}-\frac{3}{2}{\tilde{\alpha }} =\frac{3}{2\frac{8{\tilde{\beta }}+3}{3(1-{\tilde{\alpha }})}-1}-\frac{3}{2}{\tilde{\alpha }} \Rightarrow 32{\tilde{\beta }}^2+(54{\tilde{\alpha }}+6){\tilde{\beta }} +(9{\tilde{\alpha }}^2+27{\tilde{\alpha }}-18)=0. \end{aligned} \end{aligned}$$

Solving this quadratic equation gives

$$\begin{aligned} {\tilde{\beta }}=\frac{3(\sqrt{65-78{\tilde{\alpha }}+49{\tilde{\alpha }}^2}-9{\tilde{\alpha }}-1)}{32}. \end{aligned}$$

Hence,

$$\begin{aligned} \frac{2}{\beta }+\frac{3}{\alpha } =2{\tilde{\beta }}+3{\tilde{\alpha }} =\frac{3(\sqrt{65-78{\tilde{\alpha }}+49{\tilde{\alpha }}^2}+7{\tilde{\alpha }}-1)}{16} =\frac{3\left( \sqrt{65\alpha ^2-78\alpha +49}+7-\alpha \right) }{16\alpha }. \end{aligned}$$

Now, the main restriction of $\alpha$ comes from (32) and (38). After some calculations, we find (38) reduces to $\displaystyle {\frac{3+\sqrt{17}}{4}\le \alpha \le \infty }$ (in (38), $\beta =\infty$ corresponds to $\displaystyle {\alpha =\frac{3+\sqrt{17}}{4}}$), and all the assumptions, say (32), (34)-(36), (38), (42), (48), are all valid.

Remark 4

If we apply the same method in the proof of Theorem 2 to [9, Theorem 1.2], that is, in showing [9, Lemma 2.1], we use the generalized multiplicative Sobolev inequality (11), we get better result than (22), but no better result than (24).