An improved regularity criterion for the Navier–Stokes equations in terms of one directional derivative of the velocity field

Abstract

In this paper, we establish a new multiplicative Sobolev inequality. As applications, we refine and extend the results in Kukavica and Ziane (J Math Phys 48:065203, 2007) and Cao (Discrete Contin Dyn Syst 26:1141–1151, 2010) simultaneously.

1 Introduction

The homogeneous incompressible fluid flow is governed by the following Navier–Stokes equations (NSE):

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

(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

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

The global existence of a weak solution to the evolutionary NSE (1) has been long established by Leray [19] and Hopf [9]; however, the issue of its regularity and uniqueness remains open up to now. Pioneered by Serrin [26], we began studying the regularity criterion for the NSE (1); that is, to find some sufficient condition to ensure the smoothness of the solution. The classical Prodi-Serrin conditions (see [8, 24, 26]) says 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).

This was be generalized by Beir$\tilde{\mathrm {a}}$o da Veiga [1] by considering the velocity gradient or vorticity,

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

(3)

Notice that the case $\displaystyle {q\in \left[ \frac{3}{2},3\right) }$ follows directly from (2) and the Sobolev inequality.

In view of the divergence-free condition $\nabla \cdot \varvec{u}=0$, it is natural to ask whether or not we can reduce (2) and (3) to its partial components. One way is to consider regularity criteria involving only one velocity component, which were done in [3, 11, 16, 20, 34, 36]. Another way is to study the possible components reduction of $\nabla \varvec{u}$ to $\nabla u_3$, see [16, 23, 27, 35, 36]; or to $\partial _3\varvec{u}$, see [2, 17, 21, 22]. In [22], 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 was improved by Kukavica–Ziane [17] to be

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

(5)

Notice that the range of q is not of full range $\displaystyle {\left( \frac{3}{2},\infty \right] }$. The reason is that in [17], the estimate of $I_3$ needs to be reconciled with the the estimate of K. Furthermore, this method was adjusted by Penel–Pokorný [21] to get an anisotropic criterion. For readers interested in this topic, please refer to [12,13,14,15, 30] for recent progresses on regularity criteria of the MHD equations, which contains system (1) as a subsystem.

Later on, Cao [2] employed multiplicative Sobolev inequalities

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

(6)

and

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

(7)

to get the following extended regularity condition¹

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

(8)

Notice that the lower and upper bounds of q in (8) both are less than those in (5) respectively. Consequently, our best knowledge in this direction is the following sufficient condition

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

(9)

Some of them was proved in [17], while other parts could only be seen [2].

In this paper, we shall further generalize (7), and improve (5) and (8) simultaneously. We will show that the condition

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

(10)

could ensure the regularity of the solution. Noting

$$\begin{aligned} \frac{3\sqrt{37}}{4}\approx 1.56207<1.6875=\frac{27}{16}, \end{aligned}$$

we are much closer to the end point $\displaystyle {\frac{3}{2}}$.

Before stating the precise result, let us recall the weak formulation of (1), see [7, 18, 25, 28] 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.,

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

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

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

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

• 3. the energy inequality, that is,

  $$\begin{aligned} \left\| \varvec{u}(t)\right\| _{L^2}^2+2\int _0^t\left\| \nabla \varvec{u}(s)\right\| _{L^2}^2\mathrm {\,d}s \le \left\| \varvec{u}_0\right\| _{L^2}^2,\quad 0\le t\le T. \end{aligned}$$

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

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

(11)

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

The proof of Theorem 2 will be given in Sect. 2. Before doing that, let us state our notations used throughout the paper, and prove a multiplicative Sobolev inequality.

For simplicity of presentation, we do not distinguish between the spaces X and their N-dimensional vector analogs $X^N$ (e.g., $N=3$ for $\varvec{u}\in L^2(\mathbb {R}^3)$, $N=9$ for $\nabla \varvec{u}\in L^2(\mathbb {R}^3)$); however, all vector- and tensor-valued functions are printed boldfaced. A constant C may change from line to line, depending only on the initial data or the norms that we have controlled. We denote by

$$\begin{aligned} \varvec{u}_h=(u_1,u_2),\quad \nabla _h=(\partial _1,\partial _2),\quad \triangle _h=\partial _1\partial _1+\partial _2\partial _2. \end{aligned}$$

Generalizing (7) in [2], we have the following

Lemma 3

For each $1\le q<\infty$, $0<\lambda <\infty$, there exists some constant C such that for each $f\in C_c^\infty (\mathbb {R}^3)$,

$$\begin{aligned} \left\| f\right\| _{L^{(2\lambda +1)q}}\le C \left\| \partial _if\right\| _{L^q}^\frac{1}{2\lambda +1} \left\| \partial _j(|f|^\lambda )\right\| _{L^2}^\frac{1}{2\lambda +1} \left\| \partial _k(|f|^\lambda )\right\| _{L^2}^\frac{1}{2\lambda +1}, \end{aligned}$$

where $\left\{ i,j,k\right\}$ is a permutation of $\left\{ 1,2,3\right\}$.

Proof

By Newton–Leibniz formula, we have

$$\begin{aligned} |f|^{2\lambda (q-1)+q}\le & {} C\int _\mathbb {R}|f|^{(2\lambda +1)(q-1)}|\partial _if|\mathrm {\,d}x_i,\\ |f|^{\lambda (q+1)+\frac{q}{2}}= & {} \left( |f|^\lambda \right) ^{q+1+\frac{q}{2\lambda }} \le C\int _\mathbb {R}|f|^{\frac{(2\lambda +1)q}{2}}|\partial _j(|f|^\lambda )|\mathrm {\,d}x_j,\\ |f|^{\lambda (q+1)+\frac{q}{2}}\le & {} C\int _\mathbb {R}|f|^{\frac{(2\lambda +1)q}{2}}|\partial _k(|f|^\lambda )|\mathrm {\,d}x_k. \end{aligned}$$

Taking the sqrt of the multiplication of the above inequalities, we deduce

$$\begin{aligned} |f|^{(2\lambda +1)q}\le & {} \left[ \int _\mathbb {R}|f|^{(2\lambda +1)(q-1)}|\partial _if|\mathrm {\,d}x_i\right] ^\frac{1}{2}\\&\times \left[ \int _\mathbb {R}|f|^{\frac{(2\lambda +1)q}{2}}|\partial _j(|f|^\lambda )|\mathrm {\,d}x_j\right] ^\frac{1}{2}\cdot \left[ \int _\mathbb {R}|f|^{\frac{(2\lambda +1)q}{2}}|\partial _k(|f|^\lambda )|\mathrm {\,d}x_k\right] ^\frac{1}{2} \end{aligned}$$

Integrating in the $x_i$ variable and applying Hölder inequality, we obtain

$$\begin{aligned}&\int _{\mathbb {R}} |f|^{(2\lambda +1)q}\mathrm {\,d}x_i \le C\left[ \int _\mathbb {R}|f|^{(2\lambda +1)(q-1)}|\partial _if|\mathrm {\,d}x_i\right] ^\frac{1}{2}\\&\quad \cdot \left[ \int _{\mathbb {R}^2} |f|^{\frac{(2\lambda +1)q}{2}}|\partial _j(|f|^\lambda )|\mathrm {\,d}x_i\mathrm {\,d}x_j\right] ^\frac{1}{2}\cdot \left[ \int _{\mathbb {R}^2} |f|^{\frac{(2\lambda +1)q}{2}}|\partial _k(|f|^\lambda )|\mathrm {\,d}x_i\mathrm {\,d}x_k\right] ^\frac{1}{2} \end{aligned}$$

Successively, integrating in the $x_j$ and $x_k$ variables and applying Hölder inequality, we get

$$\begin{aligned}&\int _{\mathbb {R}^3}|f|^{(2\lambda +1)q}\mathrm {\,d}x \le C\left[ \int _{\mathbb {R}^3} |f|^{(2\lambda +1)(q-1)}|\partial _if|\mathrm {\,d}x\right] ^\frac{1}{2}\\&\quad \cdot \left[ \int _{\mathbb {R}^3} |f|^{\frac{(2\lambda +1)q}{2}}|\partial _j(|f|^\lambda )|\mathrm {\,d}x\right] ^\frac{1}{2}\cdot \left[ \int _{\mathbb {R}^3} |f|^{\frac{(2\lambda +1)q}{2}}|\partial _k(|f|^\lambda )|\mathrm {\,d}x\right] ^\frac{1}{2}. \end{aligned}$$

Invoking Hölder inequality again, we find

$$\begin{aligned} \left\| f\right\| _{L^{(2\lambda +1)q}}^{(2\lambda +1)q}\le & {} C\left[ \left\| f\right\| _{L^{(2\lambda +1)q}}^{(2\lambda +1)(q-1)} \left\| \partial _if\right\| _{L^q}^\frac{1}{2} \right] ^\frac{1}{2}\\&\quad \cdot \left[ \left\| f\right\| _{L^{(2\lambda +1)q}}^\frac{(2\lambda +1)q}{2} \left\| \partial _j(|f|^\lambda )\right\| _{L^2} \right] ^\frac{1}{2} \left[ \left\| f\right\| _{L^{(2\lambda +1)q}}^\frac{(2\lambda +1)q}{2} \left\| \partial _k(|f|^\lambda )\right\| _{L^2} \right] ^\frac{1}{2}\\\le & {} C\left\| f\right\| _{L^{(2\lambda +1)q}}^{(2\lambda +1)\left( q-\frac{1}{2}\right) } \left\| \partial _if\right\| _{L^q}^\frac{1}{2} \left\| \partial _j(|f|^\lambda )\right\| _{L^2}^\frac{1}{2} \left\| \partial _k(|f|^\lambda )\right\| _{L^2}^\frac{1}{2}. \end{aligned}$$

Dividing both sides by $\left\| f\right\| _{L^{(2\lambda +1)q}}^{(2\lambda +1)\left( q-\frac{1}{2}\right) }$, we finished the proof of Lemma 3.

2 Proof of Theorem 2

In this section, we shall prove Theorem 2.

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 $\tilde{\varvec{u}}\in C([\delta ,\varGamma ^*),H^1(\mathbb {R}^3)) \cap L^2(0,\varGamma ^*;H^2(\mathbb {R}^3))$, where $[\delta , \varGamma ^*)$ is the life span of the unique strong solution, see [28]. 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 $\left\| \nabla \varvec{u}_h(t)\right\| _2$ remains uniform bounded as $t\nearrow \varGamma ^*$. By [33, Proposition 1.1], we have $\left\| \nabla \varvec{u}(t)\right\| _2$ remains uniform bounded as $t\nearrow \varGamma ^{*}$. The standard continuation argument then yields that $[\delta ,\varGamma ^{*})$ could not be the maximal interval of existence of $\tilde{\varvec{u}}$, and consequently $\varGamma ^*\ge T$. This concludes the proof.

By (11), we may find a $\varGamma <\varGamma ^*$ such that

$$\begin{aligned} \left\| \nabla \varvec{u}(\varGamma )\right\| _{L^2}\le C,\quad \left( \int _\varGamma ^{\varGamma ^*}\left\| \partial _3\varvec{u}\right\| _{L^q}^\frac{2q}{2q-3}\mathrm {\,d}t\right) ^\frac{2q-3}{2q}<\tilde{\varepsilon }, \end{aligned}$$

(12)

where $0<\tilde{\varepsilon }\ll 1$ will be determined later on.

For convenience, we rewrite the NSE (1) as

$$\begin{aligned}&\partial _t\varvec{u}_h +(\varvec{u}_h\cdot \nabla )\varvec{u}_h +u_3\partial _3\varvec{u}_h -\triangle _h\varvec{u}_h -\partial _3\partial _3\varvec{u}_h +\nabla _h\pi =\varvec{0},\nonumber \\&\quad \partial _tu_3 +(\varvec{u}_h\cdot \nabla )u_3 +u_3\partial _3u_3 -\triangle _hu_3 -\partial _3\partial _3u_3 +\partial _3\pi =0,\nonumber \\&\quad \nabla _h\cdot \varvec{u}_h+\partial _3u_3=0. \end{aligned}$$

(13)

2.1 $H^1$ estimate

Taking the inner product of (13)$_1$ with $-\triangle \varvec{u}_h$ and (1)$_1$ with $-\partial _3\partial _3 \varvec{u}$ in $L^2(\mathbb {R}^3)$ respectively, we obtain

$$\begin{aligned}&\frac{1}{2}\frac{\mathrm {\,d}}{\mathrm {\,d}t} \left[ \left\| \nabla \varvec{u}_h\right\| _{L^2}^2 +\left\| \partial _3\varvec{u}\right\| _{L^2}^2\right] +\left[ \left\| \triangle \varvec{u}_h\right\| _{L^2}^2 +\left\| \nabla \partial _3\varvec{u}\right\| _{L^2}^2\right] \nonumber \\&\quad =\int _{\mathbb {R}^3} [(\varvec{u}\cdot \nabla )\varvec{u}_h]\cdot \triangle \varvec{u}_h\mathrm {\,d}x +\int _{\mathbb {R}^3} \nabla _h\pi \cdot \triangle \varvec{u}_h\mathrm {\,d}x +\int _{\mathbb {R}^3} [(\varvec{u}\cdot \nabla )\varvec{u}]\cdot \partial _3\partial _3\varvec{u}\mathrm {\,d}x\nonumber \\&\quad \equiv I_1+I_2+I_3. \end{aligned}$$

(14)

By [17, Lemma 2.2],

$$\begin{aligned} I_1= & {} \int _{\mathbb {R}^3} [(\varvec{u}_h\cdot \nabla _h)\varvec{u}_h]\cdot \triangle _h\varvec{u}_h\mathrm {\,d}x+\int _{\mathbb {R}^3} u_3\partial _3\varvec{u}_h\cdot \triangle \varvec{u}_h\mathrm {\,d}x\nonumber \\= & {} \frac{1}{2}\int _{\mathbb {R}^3} \partial _3u_3|\nabla _h\varvec{u}_h|^2\mathrm {\,d}x -\int _{\mathbb {R}^3} \partial _1u_1\partial _2u_2\partial _3u_3\mathrm {\,d}x +\int _{\mathbb {R}^3} \partial _1u_2\partial _2u_1\partial _3u_3\mathrm {\,d}x\nonumber \\&+\int _{\mathbb {R}^3} u_3\partial _3\varvec{u}_h\cdot \triangle \varvec{u}_h\mathrm {\,d}x\nonumber \\\le & {} C\int _{\mathbb {R}^3} |\partial _3\varvec{u}|\cdot |\nabla _h \varvec{u}_h|^2\mathrm {\,d}x +\int _{\mathbb {R}^3} |u_3|\cdot |\partial _3\varvec{u}|\cdot |\triangle \varvec{u}_h|\mathrm {\,d}x. \end{aligned}$$

(15)

To simplify $I_2$, we take the divergence of (1)$_1$ to get

$$\begin{aligned} -\triangle \pi= & {} \sum _{i,j=1}^3 \partial _i(u_j\partial _ju_i)\nonumber \\= & {} \sum _{i,j=1}^2 \partial _iu_j\partial _ju_i\quad \left( \hbox {the case: } i\ne 3,\ j\ne 3\right) \nonumber \\&+\sum _{j=1}^2 \partial _3u_j\partial _ju_3+u_j\partial _j\partial _3u_3 \quad \left( \hbox {the case: }i=3,\ j\ne 3\right) \nonumber \\&+\sum _{i=1}^3 \partial _iu_3\partial _3u_i \quad \left( \hbox {the case: }j=3\right) \end{aligned}$$

(16)

and thus

$$\begin{aligned} I_2= & {} -\int _{\mathbb {R}^3} \triangle \pi \cdot \nabla _h\varvec{u}_h\mathrm {\,d}x =\int _{\mathbb {R}^3} \triangle \pi \cdot \partial _3u_3\mathrm {\,d}x\nonumber \\= & {} \sum _{i,j=1}^2 \int _{\mathbb {R}^3}\partial _iu_j\partial _ju_i\partial _3u_3\mathrm {\,d}x +\sum _{j=1}^2\int _{\mathbb {R}^3} \left( \partial _3u_j\partial _ju_3+u_j\partial _j\partial _3u_3\right) \partial _3u_3\mathrm {\,d}x\nonumber \\&+\sum _{i=1}^3\int _{\mathbb {R}^3} \partial _iu_3\partial _3u_i\partial _3u_3\mathrm {\,d}x\nonumber \\\le & {} \int _{\mathbb {R}^3}|\partial _3u_3|\cdot |\nabla _h\varvec{u}_h|^2\mathrm {\,d}x-\sum _{j=1}^2 \int _{\mathbb {R}^3} u_3(\partial _3\partial _ju_j\partial _3u_3 +\partial _3u_j\partial _j\partial _3u_3)\mathrm {\,d}x\nonumber \\&+\frac{1}{2}\int _{\mathbb {R}^3} (\partial _3u_3)^2\mathrm {\,d}x-\sum _{i=1}^3 \int _{\mathbb {R}^3} u_3\partial _3u_i\partial _i\partial _3u_3\mathrm {\,d}x\nonumber \\\le & {} C\int _{\mathbb {R}^3} |\partial _3u_3|\cdot |\nabla _h\varvec{u}_h|^2\mathrm {\,d}x +\int _{\mathbb {R}^3} |u_3|\cdot |\partial _3\varvec{u}|\cdot |\nabla \partial _3\varvec{u}|\mathrm {\,d}x. \end{aligned}$$

(17)

Finally, integrating by parts yields

$$\begin{aligned} I_3= & {} -\int _{\mathbb {R}^3} [(\partial _3\varvec{u}\cdot \nabla )\varvec{u}]\cdot \partial _3\varvec{u}\mathrm {\,d}x\nonumber \\= & {} -\int _{\mathbb {R}^3} [(\partial _3\varvec{u}_h\cdot \nabla _h)\varvec{u}_h]\cdot \partial _3\varvec{u}_h +[(\partial _3\varvec{u}_h\cdot \nabla _h)u_3]\cdot \partial _3u_3 +\partial _3u_3\partial _3\varvec{u}\cdot \partial _3\varvec{u}\mathrm {\,d}x\nonumber \\\le & {} C\int _{\mathbb {R}^3} |\partial _3\varvec{u}|\cdot |\nabla \varvec{u}_h|^2\mathrm {\,d}x +C\int _{\mathbb {R}^3} |u_3|\cdot |\partial _3\varvec{u}|\cdot |\nabla \partial _3\varvec{u}|\mathrm {\,d}x. \end{aligned}$$

(18)

Gathering (15), (17) and (18) into (14), we deduce

$$\begin{aligned}&\frac{1}{2}\frac{\mathrm {\,d}}{\mathrm {\,d}t} \left[ \left\| \nabla \varvec{u}_h\right\| _{L^2}^2 +\left\| \partial _3\varvec{u}\right\| _{L^2}^2\right] +\left[ \left\| \triangle \varvec{u}_h\right\| _{L^2}^2 +\left\| \nabla \partial _3\varvec{u}\right\| _{L^2}^2\right] \nonumber \\&\quad \le C\int _{\mathbb {R}^3} |\partial _3\varvec{u}|\cdot |\nabla \varvec{u}_h|^2\mathrm {\,d}x +C\int _{\mathbb {R}^3} |u_3|\cdot |\partial _3\varvec{u}|\cdot (|\triangle \varvec{u}_h|+|\nabla \partial _3\varvec{u}|)\mathrm {\,d}x\nonumber \\&\quad \equiv J_1+J_2+J_3. \end{aligned}$$

(19)

By Hölder and Gagliardo-Nirenberg inequalities,

$$\begin{aligned} J_1\le & {} C\left\| \partial _3\varvec{u}\right\| _{L^q} \left\| \nabla \varvec{u}_h\right\| _{L^\frac{2q}{q-1}}^2\nonumber \\\le & {} C\left\| \partial _3\varvec{u}\right\| _{L^q} \left\| \nabla \varvec{u}_h\right\| _{L^2}^{\frac{2q-3}{q}} \left\| \nabla ^2\varvec{u}_h\right\| _{L^2}^\frac{3}{q}\nonumber \\\le & {} C\left\| \partial _3\varvec{u}\right\| _{L^q} \left\| \nabla \varvec{u}_h\right\| _{L^2}^{\frac{2q-3}{q}} \left\| \triangle \varvec{u}_h\right\| _{L^2}^\frac{3}{q} \end{aligned}$$

(20)

For $J_2$, we first use Hölder inequality with

$$\begin{aligned} 1\le a,b\le \infty ,\quad \frac{1}{a}+\frac{1}{b}=\frac{1}{2} \end{aligned}$$

(21)

to estimate as

$$\begin{aligned} J_2\le C\left\| u_3\right\| _{L^a}\left\| \partial _3\varvec{u}\right\| _{L^b}\left\| (\triangle \varvec{u}_h,\nabla \partial _3\varvec{u})\right\| _{L^2}, \end{aligned}$$

then invoke the interpolation and Gagliardo-Nirenberg inequalities to bound as

$$\begin{aligned} J_2\le C\left\| u_3\right\| _{L^{2\lambda }}^{1-\vartheta _1}\left\| u_3\right\| _{L^{(2\lambda +1)q}}^{\vartheta _1}\cdot \left\| \partial _3\varvec{u}\right\| _{L^q}^{1-\vartheta _2} \left\| \nabla \partial _3\varvec{u}\right\| _{L^2}^{\vartheta _2}\cdot \left\| (\triangle \varvec{u}_h,\nabla \partial _3\varvec{u})\right\| _{L^2}, \end{aligned}$$

where

$$\begin{aligned} \frac{1}{a}=\frac{1-\vartheta _1}{2\lambda }+\frac{\vartheta _1}{(2\lambda +1)q},\quad \frac{1}{b}=\frac{1-\vartheta _2}{q}+\vartheta _2\left( -\frac{1}{3}+\frac{1}{2}\right) ,\quad 0\le \vartheta _1,\vartheta _2\le 1,\nonumber \\ \end{aligned}$$

(22)

and $\displaystyle {\lambda \ge \frac{3}{2}}$ will be specified later on.

By Lemma 3,

$$\begin{aligned} J_2\le & {} C\left\| |u_3|^\lambda \right\| _{L^2}^\frac{1-\vartheta _1}{\lambda } \cdot \left\| \nabla _h(|u_3|^\lambda )\right\| _{L^2}^\frac{2\vartheta _1}{2\lambda +1} \left\| \partial _3u_3\right\| _{L^q}^{\frac{\vartheta _1}{2\lambda +1}} \cdot \left\| \partial _3\varvec{u}\right\| _{L^q}^{1-\vartheta _2}\left\| (\triangle \varvec{u}_h,\nabla \partial _3\varvec{u})\right\| _{L^2}^{1+\vartheta _2}\nonumber \\\le & {} C\left\| |u_3|^\lambda \right\| _{L^2}^\frac{1-\vartheta _1}{\lambda } \left\| \nabla _h(|u_3|^\lambda )\right\| _{L^2}^\frac{2\vartheta _1}{2\lambda +1} \left\| \partial _3\varvec{u}\right\| _{L^q}^{\frac{\vartheta _1}{2\lambda +1}+1-\vartheta _2}\left\| (\triangle \varvec{u}_h,\nabla \partial _3\varvec{u})\right\| _{L^2}^{1+\vartheta _2} \end{aligned}$$

(23)

Plugging (20) and (23) into (19), we find

$$\begin{aligned}&\frac{\mathrm {\,d}}{\mathrm {\,d}t} \left[ \left\| \nabla \varvec{u}_h\right\| _{L^2}^2 +\left\| \partial _3\varvec{u}\right\| _{L^2}^2\right] +\left[ \left\| \triangle \varvec{u}_h\right\| _{L^2}^2 +\left\| \nabla \partial _3\varvec{u}\right\| _{L^2}^2\right] \nonumber \\&\quad \le C \left\| \partial _3\varvec{u}\right\| _{L^q} \left\| \nabla \varvec{u}_h\right\| _{L^2}^{\frac{2q-3}{q}} \left\| \triangle \varvec{u}_h\right\| _{L^2}^\frac{3}{q}\nonumber \\&\quad \quad + C\left\| |u_3|^\lambda \right\| _{L^2}^\frac{1-\vartheta _1}{\lambda } \left\| \nabla (|u_3|^\lambda )\right\| _{L^2}^\frac{2\vartheta _1}{2\lambda +1} \left\| \partial _3\varvec{u}\right\| _{L^q}^{\frac{\vartheta _1}{2\lambda +1}+1-\vartheta _2}\left\| (\triangle \varvec{u}_h,\nabla \partial _3\varvec{u})\right\| _{L^2}^{1+\vartheta _2}. \end{aligned}$$

Integrating in time and denoting by

$$\begin{aligned} \mathcal {J}^2(t)= & {} \sup _{\varGamma \le \tau \le t}\left[ \left\| \nabla \varvec{u}_h\right\| _{L^2}^2 +\left\| \partial _3\varvec{u}\right\| _{L^2}^2\right] +\int _{\varGamma }^t \left[ \left\| \triangle \varvec{u}_h\right\| _{L^2}^2 +\left\| \nabla \partial _3\varvec{u}\right\| _{L^2}^2\right] \mathrm {\,d}\tau ,\qquad \end{aligned}$$

(24)

$$\begin{aligned} \mathcal {L}^2(t)= & {} \sup _{\varGamma \le \tau \le t}\left\| |u_3|^\lambda \right\| _{L^2}^2 +\int _{\varGamma }^t \left\| \nabla (|u_3|^\lambda )\right\| _{L^2}^2\mathrm {\,d}\tau ,\quad \varGamma \le t<\varGamma ^*, \end{aligned}$$

(25)

we deduce

$$\begin{aligned} \mathcal {J}^2(t)\le & {} \left\| \nabla \varvec{u}_h(\varGamma )\right\| _{L^2}^2 +\left\| \partial _3\varvec{u}(\varGamma )\right\| _{L^2}^2 +C \mathcal {J}^\frac{2q-3}{q}(t)\cdot \int _{\varGamma }^t \left\| \partial _3\varvec{u}\right\| _{L^q} \left\| \triangle \varvec{u}_h\right\| _{L^2}^\frac{3}{q}\mathrm {\,d}\tau \nonumber \\&+ C \mathcal {L}^\frac{1-\vartheta _1}{\lambda }(t) \cdot \int _{\varGamma }^t \left\| \nabla (|u_3|^\lambda )\right\| _{L^2}^\frac{2\vartheta _1}{2\lambda +1} \left\| \partial _3\varvec{u}\right\| _{L^q}^{\frac{\vartheta _1}{2\lambda +1}+1-\vartheta _2}\left\| (\triangle \varvec{u}_h,\nabla \partial _3\varvec{u})\right\| _{L^2}^{1+\vartheta _2} \mathrm {\,d}\tau \nonumber \\\le & {} C+C\mathcal {J}^\frac{2q-3}{q}(t) \cdot \left( \int _{\varGamma }^t \left\| \partial _3\varvec{u}\right\| _{L^q}^\frac{2q}{2q-3}\mathrm {\,d}\tau \right) ^\frac{2q-3}{2q} \left( \int _{\varGamma }^t \left\| \triangle \varvec{u}_h\right\| _{L^2}^2\mathrm {\,d}\tau \right) ^\frac{3}{2q}\nonumber \\&+C \mathcal {L}^\frac{1-\vartheta _1}{\lambda }(t) \cdot \left( \int _{\varGamma }^t \left\| \nabla _h(|u_3|^\lambda ) \right\| _{L^2}^2 \mathrm {\,d}\tau \right) ^\frac{\vartheta _1}{2\lambda +1} \nonumber \\&\cdot \left( \int _{\varGamma }^t \left\| \partial _3\varvec{u}\right\| _{L^q}^\frac{2q}{2q-3} \mathrm {\,d}\tau \right) ^{\frac{2q-3}{2q}\left( \frac{\vartheta _1}{\lambda +1}+1-\vartheta _2\right) \mathrm {\,d}\tau }\cdot \left( \int _{\varGamma }^t \left\| (\triangle \varvec{u}_h,\nabla \partial _3\varvec{u})\right\| _{L^2}^2 \mathrm {\,d}\tau \right) ^{\frac{1+\vartheta _2}{2}}\nonumber \\\le & {} C+C\tilde{\varepsilon }\mathcal {J}^2(t) +C\mathcal {L}^{\frac{1-\vartheta _1}{\lambda }}(t)\cdot \mathcal {L}^\frac{2\vartheta _1}{2\lambda +1}(t) \cdot \tilde{\varepsilon }^{\frac{\vartheta _1}{2\lambda +1}+1-\vartheta _2} \cdot \mathcal {J}^{1+\vartheta _2}(t)\nonumber \\\le & {} C+C\tilde{\varepsilon }\mathcal {J}^2(t) + C \tilde{\varepsilon }^{\frac{\vartheta _1}{2\lambda +1}+1-\vartheta _2}\mathcal {J}^{1+\vartheta _2}(t) \mathcal {L}^{\frac{1-\vartheta _1}{\lambda }+\frac{2\vartheta _1}{2\lambda +1}}(t), \end{aligned}$$

(26)

where Hölder inequality with

$$\begin{aligned} \frac{\vartheta _1}{2\lambda +1} +\frac{2q-3}{2q}\left( \frac{\vartheta _1}{2\lambda +1}+1-\vartheta _2\right) +\frac{1+\vartheta _2}{2}=1 \end{aligned}$$

(27)

is used.

2.2 $L^{2\lambda }$ estimate

Taking the inner product of (13)$_3$ with $2\lambda |u_3|^{2\lambda -2}u_3$ in $L^2(\mathbb {R}^3)$, we get

$$\begin{aligned} \frac{\mathrm {\,d}}{\mathrm {\,d}t}\left\| |u_3|^\lambda \right\| _{L^2}^2 +C(\lambda )\left\| \nabla (|u_3|)^\lambda \right\| _{L^2}^2 =2\lambda \int _{\mathbb {R}^3} \partial _3\pi |u_3|^{2\lambda -2}u_3\mathrm {\,d}x\equiv L. \end{aligned}$$

(28)

To process L, we derive from (16) that

$$\begin{aligned} -\triangle \partial _3\pi= & {} \sum _{i,j=1}^3 \partial _i\partial _j(\partial _3u_iu_j+u_i\partial _3u_j)=2\sum _{i,j=1}^3 \partial _i\partial _j(u_i\partial _3u_j)\nonumber \\= & {} 2\sum _{i=1}^2 \sum _{j=1}^3 \partial _i\partial _j(u_i\partial _3u_j) +2\sum _{j=1}^3 \partial _3\partial _j(u_3\partial _3u_j)\nonumber \\ \Rightarrow \partial _3\pi= & {} 2\sum _{i=1}^2 \sum _{j=1}^3 \mathcal {R}_i\mathcal {R}_j(u_i\partial _3u_j) +2\sum _{j=1}^3 \mathcal {R}_3\mathcal {R}_j(u_3\partial _3u_j)\equiv \pi _1+\pi _2, \end{aligned}$$

(29)

where $\displaystyle {\mathcal {R}_i=\frac{\partial _i}{\sqrt{-\triangle }}}$ is the Riesz transformation, which is bounded from $L^r(\mathbb {R}^3))$ to itself for $1<r<\infty$.

In view of (29),

$$\begin{aligned} L= & {} 2\lambda \int _{\mathbb {R}^3}(\pi _1+\pi _2)|u_3|^{2\lambda -2}u_3\mathrm {\,d}x\\\le & {} C\left\| \varvec{u}_h\right\| _{L^c}\left\| \partial _3\varvec{u}\right\| _{L^q}\left\| u_3\right\| _{L^d}^{2\lambda -1} +C\left\| u_3\right\| _{L^\frac{2\lambda q}{q-1}} \left\| \partial _3\varvec{u}\right\| _{L^q} \left\| u_3\right\| _{L^\frac{2\lambda q}{q-1}}^{2\lambda -1}, \end{aligned}$$

where

$$\begin{aligned} 1\le c,d\le \infty ,\quad \frac{1}{c}+\frac{1}{q}+\frac{2\lambda -1}{d}=1. \end{aligned}$$

(30)

By Gagliardo-Nirenberg and interpolation inequalities,

$$\begin{aligned} L\le & {} C\left\| \varvec{u}_h\right\| _{L^{3q}}^{1-\vartheta _3} \left\| \triangle \varvec{u}_h\right\| _{L^2}^{\vartheta _3} \cdot \left\| \partial _3\varvec{u}\right\| _{L^q} \cdot \left\| u_3\right\| _{L^{2\lambda }}^{(2\lambda -1)(1-\vartheta _4)} \left\| u_3\right\| _{L^{(2\lambda +1)q}}^{(2\lambda -1)\vartheta _4}\\&+C\left\| \partial _3\varvec{u}\right\| _{L^q}\left\| |u_3|^\lambda \right\| _{L^\frac{2q}{q-1}}^2, \end{aligned}$$

where

$$\begin{aligned} \frac{1}{c}=(1-\vartheta _3)\frac{1}{3q}+\vartheta _3\left( -\frac{2}{3}+\frac{1}{2}\right) ,\quad \frac{1}{d}=\frac{1-\vartheta _4}{2\lambda } +\frac{\vartheta _4}{(2\lambda +1)q},\quad 0\le \vartheta _3,\ \vartheta _4\le 1.\nonumber \\ \end{aligned}$$

(31)

By (6) and Lemma 3,

$$\begin{aligned} L\le & {} C\left\| \nabla \varvec{u}_h\right\| _{L^2}^\frac{2(1-\vartheta _3)}{3} \left\| \partial _3\varvec{u}\right\| _{L^q}^\frac{1-\vartheta _3}{3} \cdot \left\| \triangle \varvec{u}_h\right\| _{L^2}^{\vartheta _3} \left\| \partial _3\varvec{u}\right\| _{L^q} \left\| u_3\right\| _{L^{2\lambda }}^{(2\lambda -1)(1-\vartheta _4)}\nonumber \\&\cdot \left\| \nabla |u_3|^\lambda \right\| _{L^2}^\frac{2(2\lambda -1)\vartheta _4}{2\lambda +1} \left\| \partial _3u_3\right\| _{L^q}^{\frac{(2\lambda -1)\vartheta _4}{2\lambda +1}} +C\left\| \partial _3\varvec{u}\right\| _{L^q}\cdot \left\| |u_3|^\lambda \right\| _{L^2}^{\frac{2q-3}{q}} \left\| \nabla (|u_3|^\lambda )\right\| _{L^2}^\frac{3}{q}\nonumber \\\le & {} C\left\| \nabla \varvec{u}_h\right\| _{L^2}^{\frac{2(1-\vartheta _3)}{3}} \left\| \partial _3\varvec{u}\right\| _{L^q}^{\frac{1-\vartheta _3}{3}+1+\frac{(2\lambda -1)\vartheta _4}{2\lambda +1}} \left\| \triangle \varvec{u}_h\right\| _{L^2}^{\vartheta _3} \left\| |u_3|^\lambda \right\| _{L^2}^\frac{(2\lambda -1)(1-\vartheta _4)}{\lambda } \nonumber \\&\cdot \left\| \nabla (|u_3|^\lambda )\right\| _{L^2}^{\frac{2(2\lambda -1)\vartheta _4}{2\lambda +1}}+C\left\| \partial _3\varvec{u}\right\| _{L^q}\cdot \left\| |u_3|^\lambda \right\| _{L^2}^{\frac{2q-3}{q}} \left\| \nabla (|u_3|^\lambda )\right\| _{L^2}^\frac{3}{q}. \end{aligned}$$

(32)

Putting (32) into (28), and integrating in time yields

$$\begin{aligned} \mathcal {L}^2(t)\le & {} C+C\mathcal {J}^\frac{2(1-\vartheta _3)}{3}(t) \mathcal {L}^\frac{(2\lambda -1)(1-\vartheta _4)}{\lambda }(t)\cdot \int _\varGamma ^t \left\| \partial _3\varvec{u}\right\| _{L^q}^{\frac{1-\vartheta _3}{q} +1+\frac{(2\lambda -1)\vartheta _4}{2\lambda +1}} \left\| \triangle \varvec{u}_h\right\| _{L^2}^{\vartheta _3}\nonumber \\&\quad \left\| \nabla (|u_3|^\lambda )\right\| _{L^2}^{\frac{2(2\lambda -1)\vartheta _4}{2\lambda +1}}\mathrm {\,d}\tau +C\mathcal {L}^\frac{2q-3}{q}(t)\cdot \int _{\varGamma }^t \left\| \partial _3\varvec{u}\right\| _{L^q}\left\| \nabla (|u_3|^\lambda )\right\| _{L^2}^\frac{3}{q}\mathrm {\,d}\tau \nonumber \\\le & {} C+C\tilde{\varepsilon }^{\frac{1-\vartheta _3}{3}+1+\frac{(2\lambda -1)\vartheta _4}{2\lambda +1}} \mathcal {J}^{\frac{2(1-\vartheta _3)}{3}+\vartheta _3}(t) \mathcal {L}^{\frac{(2\lambda -1)(1-\vartheta _4)}{\lambda }+\frac{2(2\lambda -1)\vartheta _4}{2\lambda +1}}(t) +C\tilde{\varepsilon } \mathcal {L}^2(t),\nonumber \\ \end{aligned}$$

(33)

where just as in (26), Hölder inequality with

$$\begin{aligned} \frac{2q-3}{2q}\left[ \frac{1-\vartheta _3}{3}+1+\frac{(2\lambda -1)\vartheta _4}{2\lambda +1}\right] +\frac{\vartheta _3}{2} +\frac{(2\lambda -1)\vartheta _4}{2\lambda +1}=1 \end{aligned}$$

(34)

is applied.

2.3 Closing estimate

By (26) and (33), we have

$$\begin{aligned} \mathcal {J}^2(t)\le & {} C+C\tilde{\varepsilon }\mathcal {J}^2(t) + C \tilde{\varepsilon }^{\frac{\vartheta _1}{2\lambda +1}+1-\vartheta _2}\mathcal {J}^{j_1}(t) \mathcal {L}^{l_1}(t), \end{aligned}$$

(35)

$$\begin{aligned} \mathcal {L}^2(t)\le & {} C+C\tilde{\varepsilon }^{\frac{1-\vartheta _3}{3}+1+\frac{(2\lambda -1)\vartheta _4}{2\lambda +1}} \mathcal {J}^{j_2}(t) \mathcal {L}^{l_2}(t) +C\tilde{\varepsilon } \mathcal {L}^2(t), \end{aligned}$$

(36)

where

$$\begin{aligned} \begin{array}{llll} j_1&{}=1+\vartheta _2,&{}\quad l_1&{}=\frac{1-\vartheta _1}{\lambda }+\frac{2\vartheta _1}{2\lambda +1},\\ j_2&{}=\frac{2(1-\vartheta _3)}{3}+\vartheta _3,&{}\quad l_2&{}=\frac{(2\lambda -1)(1-\vartheta _4)}{\lambda }+\frac{2(2\lambda -1)\vartheta _4}{2\lambda +1}, \end{array} \end{aligned}$$

(37)

and $0\le \vartheta _i\le 1\ (1\le i\le 4)$ should satisfy

$$\begin{aligned}&\left[ \frac{1-\vartheta _1}{2\lambda }+\frac{\vartheta _1}{(2\lambda +1)q}\right] +\left[ \frac{1-\vartheta _2}{q}+\vartheta _2\left( -\frac{1}{3}+\frac{1}{2}\right) \right] =\frac{1}{2},\nonumber \\&\quad \frac{\vartheta _1}{2\lambda +1} +\frac{2q-3}{2q}\left( \frac{\vartheta _1}{2\lambda +1}+1-\vartheta _2\right) +\frac{1+\vartheta _2}{2}=1,\nonumber \\&\quad \left[ (1-\vartheta _3)\frac{1}{3q}+\vartheta _3\left( -\frac{2}{3}+\frac{1}{2}\right) \right] +\frac{1}{q}+ (2\lambda -1)\left[ \frac{1-\vartheta _4}{2\lambda } +\frac{\vartheta _4}{(2\lambda +1)q}\right] =1,\nonumber \\&\quad \frac{2q-3}{2q}\left[ \frac{1-\vartheta _3}{3}+1+\frac{(2\lambda -1)\vartheta _4}{2\lambda +1}\right] +\frac{\vartheta _3}{2} +\frac{(2\lambda -1)\vartheta _4}{2\lambda +1}=1, \end{aligned}$$

(38)

in view of (21), (22), (27), (30), (31) and (34).

After tedious calculations, we can solve (38) as

$$\begin{aligned} \vartheta _1= & {} \frac{(2\lambda -3)(2\lambda +1)(3-q)}{2\lambda q+3q+3\lambda -9},\quad \vartheta _2=3\frac{\lambda (3-2q)+(5q-6)}{2\lambda q+3q+3\lambda -9},\nonumber \\ \vartheta _3= & {} \frac{4\lambda (q+1)-(10q+3)}{2\lambda q-q+\lambda -3},\quad \vartheta _4=\frac{(2\lambda +1)[9+(3-2\lambda )q]}{3(2\lambda -1)(2\lambda q-q+\lambda -3)}. \end{aligned}$$

(39)

Reducing $0\le \vartheta _i\le 1\ (i=1,2,3,4)$ yields

$$\begin{aligned} \begin{array}{ccccccc} \text{ if } \lambda \in \left[ \frac{19}{10},\frac{33}{16}\right] &{}\left[ \frac{33}{16},\frac{25+2\sqrt{37}}{18}\right] &{}\left[ \frac{25+2\sqrt{37}}{18},\frac{5}{2}\right] &{}\left[ \frac{5}{2},3\right] \\ \text{ then } q\in \left[ \frac{39-6\lambda }{16(\lambda -1)},\frac{4\lambda -3}{10-4\lambda }\right] &{}\left[ \frac{39-6\lambda }{16(\lambda -1)},3\right] &{}\left[ \frac{3(4\lambda -5)}{2(2\lambda -1)},3\right] &{}\left[ \frac{3\lambda }{9-2\lambda },{3}\right] \end{array}\nonumber \\ \end{aligned}$$

(40)

respectively. Consequently, if

$$\begin{aligned} \lambda =\lambda _0\equiv \frac{25+2\sqrt{37}}{18}\approx 2.06475, \end{aligned}$$

(41)

the range of q is the largest one:

$$\begin{aligned} q\in \left[ \frac{3\sqrt{37}}{4}-3,3\right] . \end{aligned}$$

(42)

By (39), (37) becomes

$$\begin{aligned} \begin{array}{llll} j_1&{}=\frac{4\lambda (3-q)+9(2q-3)}{2\lambda q+3q+3\lambda -9}, &{}\quad l_1&{}=\frac{4q-3}{2\lambda q+3q+3\lambda -9},\\ j_2&{}=\frac{(2\lambda -3)(4q+3)}{3(2\lambda q-q+\lambda -3)},&{}\quad l_2&{}=\frac{21+10q-6\lambda (2q+1)}{3(2\lambda q-q+\lambda -3)}. \end{array} \end{aligned}$$

(43)

When (41) and (42) holds, it is obvious that $1\le j_1<2$, and we may apply Hölder inequality to (35),

$$\begin{aligned} \mathcal {J}^2(t)\le & {} C+C\tilde{\varepsilon }\mathcal {J}^2(t) +\frac{1}{2}\mathcal {J}^2(t)+C\mathcal {L}^\frac{2l_1}{2-j_1}(t)\nonumber \\\le & {} C+C\tilde{\varepsilon }\mathcal {J}^2(t) +\frac{1}{2}\mathcal {J}^2(t)+ +C\mathcal {L}^\frac{2}{2\lambda -3}(t). \end{aligned}$$

(44)

Now choose $0<\tilde{\varepsilon }\ll 1$ sufficiently small such that

$$\begin{aligned} C\tilde{\varepsilon }\le \frac{1}{4}, \end{aligned}$$

(45)

we have

$$\begin{aligned} \mathcal {J}(t)\le C+C\mathcal {L}^\frac{1}{2\lambda -3}(t). \end{aligned}$$

(46)

Plugging (46) into (36), and choosing $\tilde{\varepsilon }$ such that

$$\begin{aligned} C\tilde{\varepsilon }^{\frac{1-\vartheta _3}{3}+1+\frac{(2\lambda -1)\vartheta _4}{2\lambda +1}}\le \frac{1}{4}, \end{aligned}$$

(47)

besides (45), we find

$$\begin{aligned} \mathcal {L}^2(t)\le C+\frac{1}{4}\mathcal {L}^{\frac{1}{2\lambda -3}j_2+l_2}(t)+\frac{1}{2}\mathcal {L}^2(t)=C+\frac{3}{4}\mathcal {L}^2(t)\Rightarrow \mathcal {L}^2(t)\le C. \end{aligned}$$

(48)

Combining (46) and (48), we see that $\left\| \nabla \varvec{u}_h(t)\right\| _{L^2}$ is uniformly bounded on $t\in [\varGamma ,\varGamma ^*)$ as desired. The proof of Theorem 2 is completed.

Remark 4

Cao [2] took $\lambda =2$ to deduce (8), which corresponds to the range of q in case $\lambda =2$ in (40). In our paper, we treat all the possibilities to make the range of q as large as possible. The method involves a generalized multiplicative Sobolev inequality (see Lemma 3) and the general $L^{2\lambda }$ estimate, but not just $L^4$ estimate. For some applications of general $L^{2\lambda }$ estimates, we refer to [10, 32], which improves [5].