Regularity criterion for 3D nematic liquid crystal flows in terms of finite frequency parts in $\dot{B}_{\infty,\infty }^{-1}$

Abstract

In this paper, we establish the regularity criterion for the weak solution of nematic liquid crystal flows in three dimensions when the $L^{\infty }(0,T;\dot{B}_{\infty,\infty }^{-1})$-norm of a suitable low frequency part of $(u,\nabla d)$ is bounded by a scaling invariant constant and the initial data $(u_{0},\nabla d_{0})$. Our result refines the corresponding one in (Liu and Zhao in J. Math. Anal. Appl. 407:557-566, 2013) and that in (Ri in Nonlinear Anal. TMA 190:111619, 2020).

1 Introduction

This note focuses on the regularity criteria for the following 3D nematic liquid crystal fluid flow:

$$ \textstyle\begin{cases} \partial _{t} u-\nu \Delta u+(u\cdot \nabla )u+\nabla \pi =-\lambda \nabla \cdot (\nabla d\odot \nabla d), &(x,t)\in \mathbf{R}^{3}\times (0, +\infty ), \\ \partial _{t} d+(u\cdot \nabla )d=\gamma (\Delta d+ \vert \nabla d \vert ^{2} d ), &(x,t)\in \mathbf{R}^{3}\times (0,+ \infty ), \\ \operatorname{div} u=0, & (x,t)\in \mathbf{R}^{3}\times (0,+\infty ), \\ (u,d)|_{t=0}=( u_{0},d_{0}), &x\in \mathbf{R}^{3}, \end{cases} $$

(1.1)

where $u(x,t)$ is the unknown velocity field, $d(x,t):\mathbf{R}^{3}\times (0,+\infty )\rightarrow \mathbb{S}^{2}$, the unit sphere in $\mathbf{R}^{3}$, is the unknown (averaged) macroscopic/continuum molecule orientation of the nematic liquid crystal flow and π is the scalar pressure. ν, λ, γ are positive constants that represent viscosity, the competition between kinetic energy and potential energy, and the microscopic elastic relaxation time for the molecular orientation field. The notation $\nabla d\odot \nabla d$ denotes the $3\times 3$ matrix whose $(i,j)$ entry is given by $\partial _{i} d\cdot \partial _{j} d$ ($1\leq i,j\leq 3$).

It is well-known that Ericksen and Leslie ([3–5, 8] established the hydrodynamic theory of liquid crystals in 1960s. Lin [9] first introduced the above liquid crystal flow (1.1). Later Lin and Liu [11] obtained the global existence theorem for a weak solution and the local existence for the strong solution to the system (1.1).

We first introduce the definition of Morrey spaces.

Definition 1.1

For $1\le p\le q\le \infty$, we call $\dot{M}_{p,q}(\mathbf{R}^{3})$ a Morrey space, if and only if

$$\begin{aligned} \Vert f \Vert _{\dot{M}_{p,q}(\mathbf{R}^{3})}=\sup_{x\in \mathbf{R}^{3}, 0< R< \infty }R^{\frac{3}{q}-\frac{3}{p}} \biggl( \int _{B(x,R)} \bigl\vert f(y) \bigr\vert ^{p}\,dy \biggr)^{ \frac{1}{p}}< +\infty, \end{aligned}$$

here $B(x,R)$ denotes the ball in $\mathbf{R}^{3}$ with center x and radius R.

In 2008, Fan and Guo [4] showed that, if u satisfies one of the following conditions:

$$\begin{aligned} &u\in L^{s} \bigl(0, T; \dot{M}_{p,q} \bigl( \mathbf{R}^{3} \bigr) \bigr)\quad \text{with }\frac{2}{s}+\frac{3}{p}=1, p \ge 3, p\ge q\ge 1, \\ &\nabla u\in L^{s} \bigl(0, T; \dot{M}_{p,q} \bigl( \mathbf{R}^{3} \bigr) \bigr)\quad \text{with }\frac{2}{s}+\frac{3}{p}=2, p \ge \frac{3}{2}, p\ge q\ge 1, \end{aligned}$$

then $(u,d)$ is extended beyond $t=T$. Later Liu, Zhao and Cui [12] obtained the regularity criterion to the system (1.1) under the assumption that $\partial _{3}u\in L^{\beta }(0,T; L^{\alpha })$ with $\frac{2}{\beta }+\frac{3}{\alpha }\le 1, \alpha >3$. Recently, Wei, Li and Yao [16] proved that, if the weak solution $(u,d)$ satisfies

$$\begin{aligned} u_{3}, \nabla d\in L^{\beta } \bigl(0,T; L^{\alpha } \bigl( \mathbf{R}^{3} \bigr) \bigr), \quad\text{with }\frac{2}{\beta }+\frac{3}{\alpha } \le \frac{3}{4}+ \frac{1}{\alpha }, \alpha >\frac{10}{3}, \end{aligned}$$

then $(u,b)$ can be extended beyond $t=T$. Liu and Zhao [13] proved that the solution $(u, d)$ to (1.1) is smooth up to time T provided that

$$\begin{aligned} \bigl\Vert (u,\nabla d) \bigr\Vert _{L^{\infty }(0,T; B^{-1}_{\infty,\infty }(\mathbf{R}^{3}))} \le \varepsilon _{0}. \end{aligned}$$

When $d=0$, the system (1.1) becomes an incompressible Navier–Stokes equation. There is a large literature on the regularity criteria on the Navier–Stokes equation; see [1, 6, 7, 15].

By traditional turbulence theory, viscous incompressible flows develop in such a way that energy is transferred from large scales to neighboring smaller scales. Hence, it is important to study regularity for the Navier–Stokes equation based on various wave-number band parts of weak solutions is important since it reveals in a way the relationship between regularity of weak solutions and turbulent flows. Cheskidov and Shvydkoy [2] proved that a Leray–Hopf weak solution u to the Navier–Stokes equation is regular in $(0, T]$ if

$$\begin{aligned} \bigl\Vert u^{k} \bigr\Vert _{B_{\infty,\infty }^{-1}(\mathbf{R}^{3})}< C\nu, \end{aligned}$$

where $u^{k}$ is high frequency part of u with Fourier models $|\xi |\ge k$. Kim, Kwak and Yoo [5] proved that, if sufficiently high frequency parts of a weak solution to the Navier–Stokes equation on a torus belong to Serrin’s class, then the weak solution is regular. Very recently, Ri [14] proved that a Leray–Hopf weak solution u to 3D Navier–Stokes equations is regular if the $L^{\infty }(0, T; B_{\infty,\infty }^{-1}(\mathbf{R}^{3}))$-norm of a suitable low frequency part of u is bounded by a scaling invariant constant depending on the kinematic viscosity ν and initial value $u_{0}$. Motivated by [2, 5, 13] and [14], we will investigate the regularity criteria for the weak solution $(u,d)$ to the liquid crystal fluid flows (1.1) in the critical function space $L^{\infty }(0,T;\dot{B}_{\infty,\infty }^{-1}(\mathbf{R}^{3}))$ based on low and medium frequency parts, respectively. Before stating our result, we shall present some symbols and notations.

Let

$$ u_{k}:= \int _{0}^{k} u_{[s]}\,ds,\qquad u^{k}:= \int _{k}^{\infty }u_{[s]}\,ds,\quad u_{h,k}:=u_{k}-u_{h}, 0< h< k< \infty.$$

(1.2)

Here

$$ u_{[k]}(t,x)=\frac{1}{(2\pi )^{\frac{3}{2}}} \int _{ \vert \xi \vert =k} \hat{u}(t,\xi )e^{ix\cdot \xi }\,d\sigma _{\xi }, $$

and û denotes Fourier transform of u. Our result is stated as follows.

Theorem 1.2

Let $(u, d)$ be a weak solution to (1.1) with $(u_{0},d_{0})\in H^{3}(\mathbf{R}^{3})\times H^{4}(\mathbf{R}^{3}), \mathrm{div}u_{0}=0$. Assume that, for $0< T<\infty$, there exists $\delta \in (0,T)$ such that if $(u,d)$ is regular in $(0,T)$ the inequalities

$$\begin{aligned} \bigl\Vert (u_{\tilde{k}},\nabla d_{\tilde{k}}) \bigr\Vert _{L^{\infty }(T-\delta,T; \dot{B}_{\infty,\infty }^{-1})}< C_{1} \end{aligned}$$

(1.3)

and

$$\begin{aligned} &\bigl\Vert (u_{\tilde{k}/2,\tilde{k}},\nabla d_{\tilde{k}/2,\tilde{k}}) \bigr\Vert _{L^{\infty }(T-\delta,T;\dot{B}_{\infty,\infty }^{-1})} \\ &\quad < C_{2} \bigl( \Vert u_{0} \Vert _{L^{2}}+ \Vert \nabla d_{0} \Vert _{L^{2}} \bigr)^{-1} \bigl( \bigl\Vert \nabla u_{0}^{\tilde{k}} \bigr\Vert _{L^{2}}+ \bigl\Vert \Delta d_{0}^{\tilde{k}} \bigr\Vert _{L^{2}} \bigr)^{-1} \end{aligned}$$

(1.4)

hold. Then $(u,d)$ is regular on $(0,T]$, where $\tilde{k}>0$ is defined by

$$\begin{aligned} \tilde{k}=C_{3} \bigl( \bigl\Vert \nabla u_{0}^{\tilde{k}} \bigr\Vert _{L^{2}}+ \bigl\Vert \Delta d_{0}^{ \tilde{k}} \bigr\Vert _{L^{2}} \bigr)^{2}, \end{aligned}$$

and the $C_{i}, i=1, 2, 3$, are absolute constants.

Remark 1.1

Theorem 1.2 can be regarded as the generalization of Theorem 1.1 in [13] and Theorem 1.1 in [14].

The rest of this paper is organized as follows. Some useful facts are presented in Sect. 2. The proof of Theorem 1.2 is given in Sect. 3.

2 Preliminaries and some basic facts

In order to define Besov spaces, we first introduce the Littlewood–Paley decomposition theory. Let $\mathcal{S}(\mathbf{R}^{n})$ be the Schwartz class of rapidly decreasing functions.

For given $f \in \mathcal{S}(\mathbf{R}^{n})$, its Fourier transform $\mathcal{F}(f)=\hat{f}$ and its inverse Fourier transform $\mathcal{F}^{-1}(f)=\breve{f}$ are given by

$$\begin{aligned} \hat{f}(\xi )= \int _{\mathbf{R}^{n}}e^{-ix\cdot \xi }f(x)\,dx \end{aligned}$$

and

$$\begin{aligned} \check{f}(x)=\frac{1}{(2\pi )^{n}} \int _{\mathbf{R}^{n}}e^{ix\cdot \xi }f(x)\,d\xi, \end{aligned}$$

respectively. Let us choose two nonnegative radial functions $\chi,\varphi \in \mathcal{S}(\mathbf{R}^{n})$ satisfying $\operatorname{supp} \chi \subset B=\{\xi \in \mathbf{R}^{n}:|\xi |\le \frac{4}{3}\}$ and $\operatorname{supp} \varphi \subset \mathcal{C}=\{\xi \in \mathbf{R}^{n}:\frac{3}{4}\le | \xi |\le \frac{8}{3}\}$ such that

$$\begin{aligned} \sum_{j\in \mathbf{Z}}\varphi \bigl(2^{-j}\xi \bigr)=1, \quad \text{for any }\xi \in \mathbf{R}^{n}\backslash \{0\} \end{aligned}$$

and

$$\begin{aligned} \chi (\xi )+\sum_{j\ge 0}\varphi \bigl(2^{-j} \xi \bigr)=1, \quad \text{for any }\xi \in \mathbf{R}^{n}. \end{aligned}$$

For $j\in \mathbf{Z}$, the homogeneous Littlewood–Paley projection operators ${S}_{j}$ and $\dot{\Delta }_{j}$ are defined by

$$\begin{aligned} \dot{S}_{j}f=\chi \bigl(2^{-j}D \bigr)f=2^{nj} \int _{\mathbf{R}^{n}}\tilde{h} \bigl(2^{j}y \bigr)f(x-y)\,dy, \quad \text{where }\tilde{h}=\mathcal{F}^{-1}\chi, \end{aligned}$$

and

$$\begin{aligned} \dot{\Delta }_{j}f=\varphi \bigl(2^{-j}D \bigr)f=2^{nj} \int _{\mathbf{R}^{n}}h \bigl(2^{j}y \bigr)f(x-y)\,dy,\quad \text{where }h= \mathcal{F}^{-1}\varphi. \end{aligned}$$

$\dot{\Delta }_{j}$ is a frequency projection to the annulus $\{|\xi |\sim 2^{j}\}$, and $\dot{S}_{j}$ is a frequency projection to the ball $\{|\xi |\le 2^{j}\}$. Let $s\in \mathbf{R},p,q\in [1,\infty ]$. The homogeneous Besov space $\dot{B}^{s}_{p,q}(\mathbf{R}^{n})$ is presented by the distributions $f\in \mathcal{S}'_{h}$ such that

$$\begin{aligned} \biggl(\sum_{j\in \mathbf{Z}}2^{jsq} \Vert \dot{ \Delta }_{j} f \Vert _{L^{p}}^{q} \biggr)^{\frac{1}{q}}< \infty, \end{aligned}$$

with the norm

$$\begin{aligned} \Vert f \Vert _{\dot{B}_{p,q}^{s}(\mathbf{R}^{n})}= \textstyle\begin{cases} (\sum_{j\in \mathbf{Z}}2^{jsq} \Vert \dot{\Delta }_{j} f \Vert _{L^{p}}^{q} )^{\frac{1}{q}}, & 1\le q< \infty, \\ \sup_{j\in \mathbf{Z}} \{2^{js} \Vert \dot{\Delta }_{j} f \Vert _{L^{p}} \}, & q= \infty. \end{cases}\displaystyle \end{aligned}$$

(2.1)

On the other hand, we recall some facts that can be found in [14]. If $u\in L^{2}(\mathbf{R}^{3})$, then it follows from the definition of $u_{k}$ and $u^{k}$ that

$$\begin{aligned} \bigl(u_{k},u^{k} \bigr)=0, \quad\forall k>0. \end{aligned}$$

(2.2)

Moreover, for $0\leq r< s$, by Plancherel’s theorem,

$$ \begin{aligned} & \Vert u_{k} \Vert _{\dot{H}^{s}}= \bigl\Vert \vert \xi \vert ^{s} \hat{u}_{k} \bigr\Vert _{L^{2}}\leq k^{s-r} \bigl\Vert \vert \xi \vert ^{r} \hat{u}_{k} \bigr\Vert _{L^{2}}= k^{s-r} \Vert u_{k} \Vert _{\dot{H}^{r}}, \\ & \bigl\Vert u^{k} \bigr\Vert _{\dot{H}^{s}}= \bigl\Vert \vert \xi \vert ^{s}\hat{u}_{k} \bigr\Vert _{L^{2}} \geq k^{s-r} \bigl\Vert \vert \xi \vert ^{r}\hat{u}_{k} \bigr\Vert _{L^{2}}= k^{s-r} \Vert u_{k} \Vert _{\dot{H}^{r}}. \end{aligned} $$

(2.3)

Since $\| \Delta u\|_{L^{2}}\sim \|\nabla ^{2} u\|_{L^{2}}, \forall u\in \dot{H}^{2}(\mathbb{R}^{3})$, we have

$$ k \bigl\Vert \nabla u^{k} \bigr\Vert _{L^{2}}\leq \bigl\Vert \nabla ^{2}u^{k} \bigr\Vert _{L^{2}}\leq c \bigl\Vert \Delta u^{k} \bigr\Vert _{L^{2}},\quad \forall u\in H^{2}\bigl(\mathbb{R}^{3}\bigr), $$

(2.4)

with some $c>0$. Moreover, it can be easily seen that

$$ (u_{k}v_{l})^{m}=0,\quad \forall u,v \in L^{2} \bigl(\mathbb{R}^{3} \bigr), \forall k,l>0, \forall m>k+l, $$

(2.5)

because the Fourier transform of $u_{k}v_{l}$ is supported in $\{\xi \in \mathbb{R}^{3}: \vert \xi \vert \leq k+l\}$.

3 Proof of Theorem 1.2

For convenience, we assume $\mu =\lambda =1$ throughout the proof of Theorem 1.2.

Proof

Assume that a weak solution $(u,d)$ of (1.1) is regular in $(0,T)$, but not in $(0,T]$. Then $\lim_{t\rightarrow T-0}\|\nabla u(t)\|_{L^{2}}+\| \Delta d(t)\|_{L^{2}}=\infty$. Notice that, for all smooth solutions to system (1.1), one has the following basic energy law (see [10]):

$$ \begin{aligned} & \bigl\Vert u(\cdot,t) \bigr\Vert _{L^{2}}^{2}+ \bigl\Vert \nabla d(\cdot,t) \bigr\Vert _{L^{2}}^{2}+ \int _{0}^{t} \bigl( \bigl\Vert \nabla u(\cdot, \tau ) \bigr\Vert _{L^{2}}^{2}+ \bigl\Vert \bigl(\Delta d+ \vert \nabla d \vert ^{2} \bigr) (\cdot,\tau ) \bigr\Vert _{L^{2}}^{2} \bigr)\,d\tau \\ &\quad\leq \Vert u_{0} \Vert _{L^{2}}^{2}+ \Vert \nabla d_{0} \Vert _{L^{2}}^{2}, \end{aligned} $$

(3.1)

for all $0< t<\infty$. By (1.2), one has

$$ \bigl\Vert \nabla u(t) \bigr\Vert _{L^{2}}^{2}+ \bigl\Vert \Delta d(t) \bigr\Vert _{L^{2}}^{2}\leq k^{2} \Vert u_{0} \Vert _{L^{2}}^{2}+k^{2} \Vert \nabla d_{0} \Vert _{L^{2}}^{2}+ \bigl\Vert \nabla u^{k}(t) \bigr\Vert _{L^{2}}^{2}+ \bigl\Vert \Delta d^{k}(t) \bigr\Vert _{L^{2}}^{2}. $$

Thus,

$$ \lim_{t\rightarrow T-0} \bigl\Vert \nabla u^{k} (t) \bigr\Vert ^{2}_{L^{2}}+ \bigl\Vert \Delta d^{k} (t) \bigr\Vert ^{2}_{L^{2}}=\infty. $$

(3.2)

We can see from [13] that, if there exists a positive constant $\varepsilon _{0}>0$ such that

$$\begin{aligned} \bigl\Vert (u,\nabla d) \bigr\Vert _{L^{\infty }(0,T;\dot{B}_{\infty,\infty }^{-1})}\leq \varepsilon _{0}, \end{aligned}$$

then the solution $(u,d)$ is smooth up to time T.

Now we multiply the first equation of (1.1) with $-\Delta u^{k}$ and integrate over $\mathbf{R}^{3}$ to get by (2.2)

$$\begin{aligned} \frac{1}{2}\frac{d}{dt} \bigl\Vert \nabla u^{k} \bigr\Vert _{L^{2}}^{2}+ \bigl\Vert \Delta u^{k} \bigr\Vert _{L^{2}}^{2}= \bigl(u \cdot \nabla u,\Delta u^{k} \bigr)+ \biggl(\Delta d\cdot \nabla d+ \frac{1}{2} \nabla \vert \nabla d \vert ^{2},\Delta u^{k} \biggr). \end{aligned}$$

(3.3)

Applying ∇ to the second equation of (1.1) and making an $L^{2}$ inner product with respect to $\nabla \Delta d^{k}$, we can verify

$$ \begin{aligned} \frac{1}{2}\frac{d}{dt} \bigl\Vert \Delta d^{k} \bigr\Vert _{L^{2}}^{2}+ \bigl\Vert \nabla \Delta d^{k} \bigr\Vert _{L^{2}}^{2}=& \bigl(\nabla (u\cdot \nabla d),\nabla \Delta d^{k} \bigr)+ \bigl( \nabla \bigl( \vert \nabla d \vert ^{2} d \bigr),\nabla \Delta d^{k} \bigr). \end{aligned} $$

(3.4)

Adding (3.3) and (3.4) gives rise to

$$ \begin{aligned} &\frac{1}{2}\frac{d}{dt} \bigl( \bigl\Vert \nabla u^{k} \bigr\Vert _{L^{2}}^{2}+ \bigl\Vert \Delta d^{k} \bigr\Vert _{L^{2}}^{2} \bigr)+ \bigl( \bigl\Vert \Delta u^{k} \bigr\Vert _{L^{2}}^{2}+ \bigl\Vert \nabla \Delta d^{k} \bigr\Vert _{L^{2}}^{2} \bigr) \\ &\quad= \bigl(u\cdot \nabla u,\Delta u^{k} \bigr)+ \bigl(\Delta d\cdot \nabla d,\Delta u^{k} \bigr)+ \frac{1}{2} \bigl(\nabla \vert \nabla d \vert ^{2},\Delta u^{k} \bigr) \\ & \qquad{}+ \bigl(\nabla (u\cdot \nabla d),\nabla \Delta d^{k} \bigr)+ \bigl( \nabla \bigl( \vert \nabla d \vert ^{2} d \bigr),\nabla \Delta d^{k} \bigr) \\ &\quad:=I_{1}+I_{2}+I_{3}+I_{4}+I_{5}. \end{aligned} $$

(3.5)

Next we estimate $I_{1}$–$I_{5}$, respectively. From [14], we have

$$ \begin{aligned} \vert I_{1} \vert ={}& \bigl\vert \bigl(u\cdot \nabla u,\Delta u^{k} \bigr) \bigr\vert \\ \leq {}& C k^{2} \Vert u_{0} \Vert _{L^{2}}^{2} \Vert u_{\frac{k}{2},k} \Vert _{L^{\infty }}^{2} +C \Vert u_{k} \Vert _{\dot{B}_{\infty,\infty }^{-1}} \bigl\Vert \Delta u^{k} \bigr\Vert _{L^{2}}^{2} \\ &{}+C k^{-\frac{1}{2}} \bigl\Vert \nabla u^{k} \bigr\Vert _{L^{2}} \bigl\Vert \Delta u^{k} \bigr\Vert _{L^{2}}^{2}+ \frac{1}{4} \Vert \Delta u_{k,2k} \Vert _{L^{2}}^{2}. \end{aligned} $$

(3.6)

Since $d=d_{k}+d^{k}$, we write

$$\begin{aligned} (\Delta d\cdot \nabla )d=(\Delta d_{k}\cdot \nabla )d_{k}+ \bigl( \Delta d^{k}\cdot \nabla \bigr)d_{k}+(\Delta d_{k}\cdot \nabla )d^{k}+ \bigl( \Delta d^{k} \cdot \nabla \bigr)d^{k}. \end{aligned}$$

Then

$$ \begin{aligned}I_{2}={}&\bigl(\Delta d\cdot \nabla d,\Delta u^{k} \bigr) \\ = {}&\bigl((\Delta d_{k}\cdot \nabla )d_{k},\Delta u^{k} \bigr)+ \bigl( \bigl(\Delta d^{k} \cdot \nabla \bigr)d^{k},\Delta u^{k} \bigr)+ \bigl((\Delta d_{k}\cdot \nabla )d^{k}, \Delta u^{k} \bigr)\\ &{}+ \bigl( \bigl(\Delta d^{k}\cdot \nabla \bigr)d_{k},\Delta u^{k} \bigr) \\ :={}&I_{21}+I_{22}+I_{23}+I_{24}. \end{aligned} $$

(3.7)

Note that $d_{k}=d_{\frac{k}{2}}+d_{\frac{k}{2},k}$ and the Fourier transform of $(\Delta d_{k}\cdot \nabla )d_{k}$ is supported in $\{ \vert \xi \vert \leq 2k\}$, thus we deduce

$$ \begin{aligned} I_{21}&= \bigl((\Delta d_{k}\cdot \nabla )d_{k},\Delta u^{k} \bigr) \\ &= \bigl( \bigl[(\Delta d_{k}\cdot \nabla )d_{k} \bigr]_{k,2k},\Delta u_{k,2k} \bigr) \\ &= \bigl( \bigl[(\Delta d_{k}\cdot \nabla )d_{\frac{k}{2}}+(\Delta d_{k} \cdot \nabla )d_{\frac{k}{2},k} \bigr]_{k,2k},\Delta u_{k,2k} \bigr) \\ &= \bigl( \bigl[(\Delta d_{\frac{k}{2}}\cdot \nabla )d_{\frac{k}{2}}+(\Delta d_{ \frac{k}{2},k}\cdot \nabla )d_{\frac{k}{2}}+(\Delta d_{k}\cdot \nabla )d_{\frac{k}{2},k} \bigr]_{k,2k},\Delta u_{k,2k} \bigr) \\ &= \bigl((\Delta d_{k}\cdot \nabla )d_{\frac{k}{2},k},\Delta u_{k,2k} \bigr)+ \bigl(( \Delta d_{\frac{k}{2},k}\cdot \nabla )d_{\frac{k}{2}},\Delta u_{k,2k} \bigr) \\ &:=I_{211}+I_{212}, \end{aligned} $$

(3.8)

where we used the fact $[(\Delta d_{\frac{k}{2}}\cdot \nabla )d_{\frac{k}{2}}]_{k,2k}=0$. Thanks to the Hölder inequality, the Young inequality and (3.1), we get

$$ \begin{aligned} \vert I_{211} \vert &\leq \Vert \Delta d_{k} \Vert _{L^{2}} \Vert \nabla d_{ \frac{k}{2},k} \Vert _{L^{\infty }} \Vert \Delta u_{k,2k} \Vert _{L^{2}} \\ &\leq Ck \Vert \nabla d_{k} \Vert _{L^{2}} \Vert \nabla d_{k/2,k} \Vert _{L^{\infty }} \Vert \Delta u_{k,2k} \Vert _{L^{2}} \\ &\leq Ck^{2} \Vert \nabla d_{k} \Vert _{L^{2}}^{2} \Vert \nabla d_{\frac{k}{2},k} \Vert _{L^{\infty }}^{2}+\frac{1}{8} \Vert \Delta u_{k,2k} \Vert _{L^{2}}^{2} \\ &\leq Ck^{2} \Vert \nabla d_{\frac{k}{2},k} \Vert _{L^{\infty }}^{2} \bigl( \Vert u_{0} \Vert _{L^{2}}^{2}+ \Vert \nabla d_{0} \Vert _{L^{2}}^{2} \bigr)+\frac{1}{8} \Vert \Delta u_{k,2k} \Vert _{L^{2}}^{2}. \end{aligned} $$

(3.9)

Similarly,

$$ \begin{aligned} \vert I_{212} \vert &\leq \Vert \Delta d_{\frac{k}{2},k} \Vert _{L^{\infty }} \Vert \nabla d_{\frac{k}{2}} \Vert _{L^{2}} \Vert \Delta u_{k,2k} \Vert _{L^{2}} \\ &\leq Ck \Vert \nabla d_{\frac{k}{2},k} \Vert _{L^{\infty }} \Vert \nabla d_{ \frac{k}{2}} \Vert _{L^{2}} \Vert \Delta u_{k,2k} \Vert _{L^{2}} \\ &\leq Ck^{2} \Vert \nabla d_{\frac{k}{2},k} \Vert _{L^{\infty }}^{2} \Vert \nabla d_{ \frac{k}{2}} \Vert _{L^{2}}^{2}+\frac{1}{8} \Vert \Delta u_{k,2k} \Vert _{L^{2}}^{2} \\ &\leq Ck^{2} \Vert \nabla d_{\frac{k}{2},k} \Vert _{L^{\infty }}^{2} \bigl( \Vert u_{0} \Vert _{L^{2}}^{2}+ \Vert \nabla d_{0} \Vert _{L^{2}}^{2} \bigr)+\frac{1}{8} \Vert \Delta u_{k,2k} \Vert _{L^{2}}^{2}, \end{aligned} $$

(3.10)

which along with (3.9) implies

$$ \vert I_{21} \vert \leq Ck^{2} \Vert \nabla d_{\frac{k}{2},k} \Vert _{L^{\infty }}^{2} \bigl( \Vert u_{0} \Vert _{L^{2}}^{2}+ \Vert \nabla d_{0} \Vert _{L^{2}}^{2} \bigr)+\frac{1}{4} \Vert \Delta u_{k,2k} \Vert _{L^{2}}^{2}. $$

(3.11)

With the help of Hölder’s inequality, (2.4), Gagaliardo–Nirenberg’s inequality, Sobolev’s embedding and Young’s inequality, one has

$$ \begin{aligned} \vert I_{22} \vert &\leq \bigl\Vert \Delta d^{k} \bigr\Vert _{L^{3}} \bigl\Vert \nabla d^{k} \bigr\Vert _{L^{6}} \bigl\Vert \Delta u^{k} \bigr\Vert _{L^{2}} \\ &\leq C \bigl\Vert \Delta d^{k} \bigr\Vert _{L^{2}}^{\frac{1}{2}} \bigl\Vert \nabla \Delta d^{k} \bigr\Vert _{L^{2}}^{ \frac{1}{2}} \bigl\Vert \nabla ^{2} d^{k} \bigr\Vert _{L^{2}} \bigl\Vert \Delta u^{k} \bigr\Vert _{L^{2}} \\ &\leq Ck^{-\frac{1}{2}} \bigl\Vert \nabla \Delta d^{k} \bigr\Vert _{L^{2}} \bigl\Vert \Delta d^{k} \bigr\Vert _{L^{2}} \bigl\Vert \Delta u^{k} \bigr\Vert _{L^{2}} \\ &\leq Ck^{-\frac{1}{2}} \bigl\Vert \Delta d^{k} \bigr\Vert _{L^{2}} \bigl( \bigl\Vert \Delta u^{k} \bigr\Vert _{L^{2}}^{2}+ \bigl\Vert \nabla \Delta d^{k} \bigr\Vert _{L^{2}}^{2} \bigr). \end{aligned} $$

(3.12)

By the definition of the $\dot{B}_{\infty,\infty }^{-1}$-norm, we have

$$ \Vert u_{k} \Vert _{L^{\infty }}\leq Ck \Vert u_{k} \Vert _{\dot{B}_{\infty,\infty }^{-1}},\quad \forall k>0. $$

(3.13)

From the Hölder inequality, (3.13) and the Young inequality, we can conclude that

$$ \begin{aligned} \vert I_{23} \vert \leq {}& \Vert \Delta d_{k} \Vert _{L^{\infty }} \bigl\Vert \nabla d^{k} \bigr\Vert _{L^{2}} \bigl\Vert \Delta u^{k} \bigr\Vert _{L^{2}} \\ \leq {}&Ck \Vert \nabla d_{k} \Vert _{L^{\infty }} \bigl\Vert \nabla d^{k} \bigr\Vert _{L^{2}} \bigl\Vert \Delta u^{k} \bigr\Vert _{L^{2}} \\ \leq {}&Ck^{2} \Vert \nabla d_{k} \Vert _{\dot{B}_{\infty,\infty }^{-1}} \bigl\Vert \nabla d^{k} \bigr\Vert _{L^{2}} \bigl\Vert \Delta u^{k} \bigr\Vert _{L^{2}} \\ \leq {}&C \Vert \nabla d_{k} \Vert _{\dot{B}_{\infty,\infty }^{-1}} \bigl\Vert \nabla \Delta d^{k} \bigr\Vert _{L^{2}} \bigl\Vert \Delta u^{k} \bigr\Vert _{L^{2}} \\ \leq {}&C \Vert \nabla d_{k} \Vert _{\dot{B}_{\infty,\infty }^{-1}} \bigl( \bigl\Vert \Delta u^{k} \bigr\Vert _{L^{2}}^{2}+ \bigl\Vert \nabla \Delta d^{k} \bigr\Vert _{L^{2}}^{2} \bigr). \end{aligned} $$

(3.14)

Similarly,

$$ \begin{aligned} \vert I_{24} \vert \leq {}& \bigl\Vert \Delta d^{k} \bigr\Vert _{L^{2}} \Vert \nabla d_{k} \Vert _{L^{\infty }} \bigl\Vert \Delta u^{k} \bigr\Vert _{L^{2}} \\ \leq {}&Ck \Vert \nabla d_{k} \Vert _{\dot{B}_{\infty,\infty }^{-1}} \bigl\Vert \Delta d^{k} \bigr\Vert _{L^{2}} \bigl\Vert \Delta u^{k} \bigr\Vert _{L^{2}} \\ \leq {}&C \Vert \nabla d_{k} \Vert _{\dot{B}_{\infty,\infty }^{-1}} \bigl\Vert \nabla \Delta d^{k} \bigr\Vert _{L^{2}} \bigl\Vert \Delta u^{k} \bigr\Vert _{L^{2}} \\ \leq {}&C \Vert \nabla d_{k} \Vert _{\dot{B}_{\infty,\infty }^{-1}} \bigl( \bigl\Vert \Delta u^{k} \bigr\Vert _{L^{2}}^{2}+ \bigl\Vert \nabla \Delta d^{k} \bigr\Vert _{L^{2}}^{2} \bigr). \end{aligned} $$

(3.15)

Combining (3.7), (3.11), (3.12), (3.14) and (3.15), one arrives at

$$ \begin{aligned} \vert I_{2} \vert \leq {}& C k^{-\frac{1}{2}} \bigl\Vert \Delta d^{k} \bigr\Vert _{L^{2}} \bigl( \bigl\Vert \Delta u^{k} \bigr\Vert _{L^{2}}^{2}+ \bigl\Vert \nabla \Delta d^{k} \bigr\Vert _{L^{2}}^{2} \bigr) \\ &{}+C \Vert \nabla d_{k} \Vert _{\dot{B}_{\infty,\infty }^{-1}} \bigl( \bigl\Vert \Delta u^{k} \bigr\Vert _{L^{2}}^{2}+ \bigl\Vert \nabla \Delta d^{k} \bigr\Vert _{L^{2}}^{2} \bigr) \\ &{}+Ck^{2} \Vert \nabla d_{\frac{k}{2},k} \Vert _{L^{\infty }}^{2} \bigl( \Vert u_{0} \Vert _{L^{2}}^{2}+ \Vert \nabla d_{0} \Vert _{L^{2}}^{2} \bigr)+ \frac{1}{4} \Vert \Delta u_{k,2k} \Vert _{L^{2}}^{2}. \end{aligned} $$

(3.16)

To estimate $I_{3}$, we make the following decomposition:

$$\begin{aligned} \frac{1}{2}\nabla \vert \nabla d \vert ^{2}&= \frac{1}{2}\nabla \bigl\vert \nabla d^{k}+\nabla d_{k} \bigr\vert ^{2} \leq \nabla \bigl\vert \nabla d^{k} \bigr\vert ^{2}+ \nabla \vert \nabla d_{k} \vert ^{2} \\ &=2\nabla d^{k}\cdot \nabla ^{2}d^{k}+2 \nabla d_{k}\cdot \nabla ^{2}d_{k}. \end{aligned}$$

Then

$$\begin{aligned} \vert I_{3} \vert \leq 2 \bigl\vert \bigl(\nabla d^{k}\cdot \nabla ^{2}d^{k},\Delta u^{k} \bigr) \bigr\vert +2 \bigl\vert \bigl( \nabla d_{k}\cdot \nabla ^{2}d_{k},\Delta u^{k} \bigr) \bigr\vert :=I_{31}+I_{32}. \end{aligned}$$

(3.17)

Applying the same method to the bound (3.12) gives rise to

$$ \begin{aligned} I_{31}\leq {}& \bigl\Vert \nabla d^{k} \bigr\Vert _{L^{3}} \bigl\Vert \nabla ^{2} d^{k} \bigr\Vert _{L^{6}} \bigl\Vert \Delta u^{k} \bigr\Vert _{L^{2}} \\ \leq {}& C \bigl\Vert \nabla d^{k} \bigr\Vert _{\dot{H}^{\frac{1}{2}}} \bigl\Vert \Delta d^{k} \bigr\Vert _{ \dot{H}^{1}} \bigl\Vert \Delta u^{k} \bigr\Vert _{L^{2}} \\ \leq {}& C \bigl\Vert \nabla d^{k} \bigr\Vert _{L^{2}}^{\frac{1}{2}} \bigl\Vert \nabla d^{k} \bigr\Vert _{ \dot{H}^{1}}^{\frac{1}{2}} \bigl\Vert \Delta d^{k} \bigr\Vert _{\dot{H}^{1}} \bigl\Vert \Delta u^{k} \bigr\Vert _{L^{2}} \\ \leq {}& C \bigl\Vert \nabla d^{k} \bigr\Vert _{L^{2}}^{\frac{1}{2}} \bigl\Vert \Delta d^{k} \bigr\Vert _{L^{2}}^{ \frac{1}{2}} \bigl\Vert \nabla \Delta d^{k} \bigr\Vert _{L^{2}} \bigl\Vert \Delta u^{k} \bigr\Vert _{L^{2}} \\ \leq {}& Ck^{-\frac{1}{2}} \bigl\Vert \Delta d^{k} \bigr\Vert _{L^{2}} \bigl\Vert \nabla \Delta d^{k} \bigr\Vert _{L^{2}} \bigl\Vert \Delta u^{k} \bigr\Vert _{L^{2}} \\ \leq {}& Ck^{-\frac{1}{2}} \bigl\Vert \Delta d^{k} \bigr\Vert _{L^{2}} \bigl( \bigl\Vert \nabla \Delta d^{k} \bigr\Vert _{L^{2}}^{2}+ \bigl\Vert \Delta u^{k} \bigr\Vert _{L^{2}}^{2} \bigr). \end{aligned} $$

(3.18)

Similarly to (3.8), we have

$$ \begin{aligned} I_{32}&=2 \bigl\vert \bigl( \nabla d_{k}\cdot \nabla ^{2}d_{k},\Delta u^{k} \bigr) \bigr\vert \\ &=2 \bigl\vert \bigl( \nabla d_{k}\cdot \nabla ^{2}d_{\frac{k}{2},k},\Delta u_{k,2k} \bigr)+ \bigl( \nabla d_{\frac{k}{2},k}\cdot \nabla ^{2}d_{\frac{k}{2}},\Delta u_{k,2k} \bigr) \bigr\vert \\ &\leq 2 \bigl\vert \bigl(\nabla d_{k}\cdot \nabla ^{2}d_{\frac{k}{2},k},\Delta u_{k,2k} \bigr) \bigr\vert +2 \bigl\vert \bigl( \nabla d_{\frac{k}{2},k}\cdot \nabla ^{2}d_{\frac{k}{2}}, \Delta u_{k,2k} \bigr) \bigr\vert \\ &:=I_{321}+I_{322}. \end{aligned} $$

(3.19)

Using Hölder’s inequality, (2.4), Young’s inequality and (3.1), one can verify

$$ \begin{aligned} I_{321}\leq {}&2 \Vert \nabla d_{k} \Vert _{L^{2}} \Vert \Delta d_{ \frac{k}{2},k} \Vert _{L^{\infty }} \Vert \Delta u_{k,2k} \Vert _{L^{2}} \\ \leq {}& Ck \Vert \nabla d_{\frac{k}{2},k} \Vert _{L^{\infty }} \Vert \nabla d_{k} \Vert _{L^{2}} \Vert \Delta u_{k,2k} \Vert _{L^{2}} \\ \leq {}& Ck^{2} \Vert \nabla d_{\frac{k}{2},k} \Vert _{L^{\infty }}^{2} \bigl( \Vert u_{0} \Vert _{L^{2}}^{2}+ \Vert \nabla d_{0} \Vert _{L^{2}}^{2} \bigr)+\frac{1}{8} \Vert \Delta u_{k,2k} \Vert _{L^{2}}^{2}. \end{aligned} $$

(3.20)

Similarly,

$$ \begin{aligned} I_{322}\leq {}&2 \Vert \Delta d_{\frac{k}{2}} \Vert _{L^{2}} \Vert \nabla d_{\frac{k}{2},k} \Vert _{L^{\infty }} \Vert \Delta u_{k,2k} \Vert _{L^{2}} \\ \leq {}& Ck \Vert \nabla d_{\frac{k}{2}} \Vert _{L^{2}} \Vert \nabla d_{\frac{k}{2},k} \Vert _{L^{\infty }} \Vert \Delta u_{k,2k} \Vert _{L^{2}} \\ \leq {}& Ck^{2} \Vert \nabla d_{\frac{k}{2},k} \Vert _{L^{\infty }}^{2} \bigl( \Vert u_{0} \Vert _{L^{2}}^{2}+ \Vert \nabla d_{0} \Vert _{L^{2}}^{2} \bigr)+\frac{1}{8} \Vert \Delta u_{k,2k} \Vert _{L^{2}}^{2}, \end{aligned} $$

(3.21)

which along with (3.20) implies

$$ I_{32}\leq Ck^{2} \Vert \nabla d_{\frac{k}{2},k} \Vert _{L^{\infty }}^{2} \bigl( \Vert u_{0} \Vert _{L^{2}}^{2}+ \Vert \nabla d_{0} \Vert _{L^{2}}^{2} \bigr)+\frac{1}{4} \Vert \Delta u_{k,2k} \Vert _{L^{2}}^{2}. $$

(3.22)

From (3.17), (3.18) and (3.22), we can deduce

$$ \begin{aligned} \vert I_{3} \vert \leq{} &Ck^{-\frac{1}{2}} \bigl\Vert \Delta d^{k} \bigr\Vert _{L^{2}} \bigl( \bigl\Vert \Delta u^{k} \bigr\Vert _{L^{2}}^{2}+ \bigl\Vert \nabla \Delta d^{k} \bigr\Vert _{L^{2}}^{2} \bigr) \\ &{}+Ck^{2} \Vert \nabla d_{\frac{k}{2},k} \Vert _{L^{\infty }}^{2} \bigl( \Vert u_{0} \Vert _{L^{2}}^{2}+ \Vert \nabla d_{0} \Vert _{L^{2}}^{2} \bigr)+ \frac{1}{4} \Vert \Delta u_{k,2k} \Vert _{L^{2}}^{2}. \end{aligned} $$

(3.23)

We now address the term $I_{4}$. We decompose $I_{4}$ into the following form:

$$ \begin{aligned} I_{4}= \bigl((\nabla u\cdot \nabla )d,\nabla \Delta d^{k} \bigr)+ \bigl((u \cdot \nabla )\nabla d, \nabla \Delta d^{k} \bigr):=I_{41}+I_{42}. \end{aligned} $$

(3.24)

Since

$$ (\nabla u\cdot \nabla )d= \bigl(\nabla u^{k}\cdot \nabla \bigr)d^{k}+ \bigl(\nabla u^{k} \cdot \nabla \bigr)d_{k}+(\nabla u_{k}\cdot \nabla )d^{k}+( \nabla u_{k} \cdot \nabla )d_{k}, $$

we can get

$$ \begin{aligned}I_{41}={}& \bigl( \bigl(\nabla u^{k}\cdot \nabla \bigr)d^{k},\nabla \Delta d^{k} \bigr)+ \bigl( \bigl( \nabla u^{k}\cdot \nabla \bigr)d_{k},\nabla \Delta d^{k} \bigr)+ \bigl((\nabla u_{k} \cdot \nabla )d^{k},\nabla \Delta d^{k} \bigr) \\ &{}+ \bigl((\nabla u_{k}\cdot \nabla )d_{k},\nabla \Delta d^{k} \bigr):=I_{411}+I_{412}+I_{413}+I_{414}. \end{aligned} $$

(3.25)

Similar to the estimate (3.12), one has

$$ \begin{aligned} \vert I_{411} \vert \leq {}& \bigl\Vert \nabla u^{k} \bigr\Vert _{L^{6}} \bigl\Vert \nabla d^{k} \bigr\Vert _{L^{3}} \bigl\Vert \nabla \Delta d^{k} \bigr\Vert _{L^{2}} \\ \leq {}& C \bigl\Vert \nabla u^{k} \bigr\Vert _{\dot{H}^{1}} \bigl\Vert \nabla d^{k} \bigr\Vert _{L^{2}}^{ \frac{1}{2}} \bigl\Vert \nabla d^{k} \bigr\Vert _{\dot{H}^{1}}^{\frac{1}{2}} \bigl\Vert \nabla \Delta d^{k} \bigr\Vert _{L^{2}} \\ \leq {}& Ck^{-\frac{1}{2}} \bigl\Vert \Delta d^{k} \bigr\Vert _{L^{2}} \bigl\Vert \Delta u^{k} \bigr\Vert _{L^{2}} \bigl\Vert \nabla \Delta d^{k} \bigr\Vert _{L^{2}} \\ \leq {}& Ck^{-\frac{1}{2}} \bigl\Vert \Delta d^{k} \bigr\Vert _{L^{2}} \bigl( \bigl\Vert \Delta u^{k} \bigr\Vert _{L^{2}}^{2}+ \bigl\Vert \nabla \Delta d^{k} \bigr\Vert _{L^{2}}^{2} \bigr). \end{aligned} $$

(3.26)

The Hölder inequality, the Young inequality and (3.13) imply

$$ \begin{aligned} I_{412}\leq {}& \bigl\Vert \nabla u^{k} \bigr\Vert _{L^{2}} \Vert \nabla d_{k} \Vert _{L^{\infty }} \bigl\Vert \nabla \Delta d^{k} \bigr\Vert _{L^{2}} \\ \leq {}& Ck \Vert \nabla d_{k} \Vert _{\dot{B}_{\infty,\infty }^{-1}} \bigl\Vert \nabla u^{k} \bigr\Vert _{L^{2}} \bigl\Vert \nabla \Delta d^{k} \bigr\Vert _{L^{2}} \\ \leq {}& C \Vert \nabla d_{k} \Vert _{\dot{B}_{\infty,\infty }^{-1}} \bigl\Vert \Delta u^{k} \bigr\Vert _{L^{2}} \bigl\Vert \nabla \Delta d^{k} \bigr\Vert _{L^{2}} \\ \leq {}& C \Vert \nabla d_{k} \Vert _{\dot{B}_{\infty,\infty }^{-1}} \bigl( \bigl\Vert \Delta u^{k} \bigr\Vert _{L^{2}}^{2}+ \bigl\Vert \nabla \Delta d^{k} \bigr\Vert _{L^{2}}^{2} \bigr). \end{aligned} $$

(3.27)

Similarly,

$$ \begin{aligned} I_{413}\leq {}& \Vert \nabla u_{k} \Vert _{L^{\infty }} \bigl\Vert \nabla d^{k} \bigr\Vert _{L^{2}} \bigl\Vert \nabla \Delta d^{k} \bigr\Vert _{L^{2}} \\ \leq {}& Ck \Vert u_{k} \Vert _{L^{\infty }} \bigl\Vert \nabla d^{k} \bigr\Vert _{L^{2}} \bigl\Vert \nabla \Delta d^{k} \bigr\Vert _{L^{2}} \\ \leq {}& Ck^{2} \Vert u_{k} \Vert _{\dot{B}_{\infty,\infty }^{-1}} \bigl\Vert \nabla d^{k} \bigr\Vert _{L^{2}} \bigl\Vert \nabla \Delta d^{k} \bigr\Vert _{L^{2}} \\ \leq {}& C \Vert u_{k} \Vert _{\dot{B}_{\infty,\infty }^{-1}} \bigl\Vert \nabla \Delta d^{k} \bigr\Vert ^{2}_{L^{2}}. \end{aligned} $$

(3.28)

Arguing as (3.8), we have

$$\begin{aligned} I_{414}= \bigl((\nabla u_{k}\cdot \nabla )d_{\frac{k}{2},k}+( \nabla u_{ \frac{k}{2},k}\cdot \nabla )d_{\frac{k}{2}},\nabla \Delta d_{k,2k} \bigr):=I_{4141}+I_{4142}. \end{aligned}$$

By Hölder’s inequality, (2.4) and Young’s inequality, we get

$$ \begin{aligned} \vert I_{4141} \vert \leq{} & \Vert \nabla u_{k} \Vert _{L^{2}} \Vert \nabla d_{ \frac{k}{2},k} \Vert _{L^{\infty }} \Vert \nabla \Delta d_{k,2k} \Vert _{L^{2}} \\ \leq {}& Ck \Vert u_{k} \Vert _{L^{2}} \Vert \nabla d_{\frac{k}{2},k} \Vert _{L^{\infty }} \Vert \nabla \Delta d_{k,2k} \Vert _{L^{2}} \\ \leq {}& Ck^{2} \Vert u_{k} \Vert _{L^{2}}^{2} \Vert \nabla d_{\frac{k}{2},k} \Vert _{L^{\infty }}^{2}+ \frac{1}{16} \Vert \nabla \Delta d_{k,2k} \Vert _{L^{2}}^{2} \\ \leq {}& Ck^{2} \Vert \nabla d_{\frac{k}{2},k} \Vert _{L^{\infty }}^{2} \bigl( \Vert u_{0} \Vert _{L^{2}}^{2}+ \Vert \nabla d_{0} \Vert _{L^{2}}^{2} \bigr)+\frac{1}{16} \Vert \nabla \Delta d_{k,2k} \Vert _{L^{2}}^{2}. \end{aligned} $$

(3.29)

Similarly,

$$ \begin{aligned} \vert I_{4142} \vert \leq {}& \Vert \nabla u_{\frac{k}{2},k} \Vert _{L^{\infty }} \Vert \nabla d_{\frac{k}{2}} \Vert _{L^{2}} \Vert \nabla \Delta d_{k,2k} \Vert _{L^{2}} \\ \leq {}& Ck \Vert u_{\frac{k}{2},k} \Vert _{L^{\infty }} \Vert \nabla d_{\frac{k}{2}} \Vert _{L^{2}} \Vert \nabla \Delta d_{k,2k} \Vert _{L^{2}} \\ \leq {}& Ck^{2} \Vert u_{\frac{k}{2},k} \Vert _{L^{\infty }}^{2} \Vert \nabla d_{ \frac{k}{2}} \Vert _{L^{2}}^{2}+ \frac{1}{16} \Vert \nabla \Delta d_{k,2k} \Vert _{L^{2}}^{2} \\ \leq {}& Ck^{2} \Vert u_{\frac{k}{2},k} \Vert _{L^{\infty }}^{2} \bigl( \Vert u_{0} \Vert _{L^{2}}^{2}+ \Vert \nabla d_{0} \Vert _{L^{2}}^{2} \bigr)+ \frac{1}{16} \Vert \nabla \Delta d_{k,2k} \Vert _{L^{2}}^{2}, \end{aligned} $$

(3.30)

which together with (3.29) reads

$$ \vert I_{414} \vert \leq Ck^{2} \bigl( \Vert u_{\frac{k}{2},k} \Vert _{L^{\infty }}^{2}+ \Vert \nabla d_{ \frac{k}{2},k} \Vert _{L^{\infty }}^{2} \bigr) \bigl( \Vert u_{0} \Vert _{L^{2}}^{2}+ \Vert \nabla d_{0} \Vert _{L^{2}}^{2} \bigr)+\frac{1}{8} \Vert \nabla \Delta d_{k,2k} \Vert _{L^{2}}^{2}. $$

(3.31)

Combining (3.26)–(3.28) and (3.31) yields

$$ \begin{aligned} \vert I_{41} \vert \leq {}& \frac{1}{8} \Vert \nabla \Delta d_{k,2k} \Vert _{L^{2}}^{2}+Ck^{- \frac{1}{2}} \bigl\Vert \Delta d^{k} \bigr\Vert _{L^{2}} \bigl( \bigl\Vert \Delta u^{k} \bigr\Vert _{L^{2}}^{2}+ \bigl\Vert \nabla \Delta d^{k} \bigr\Vert _{L^{2}}^{2} \bigr) \\ &{}+C \bigl( \Vert u_{k} \Vert _{\dot{B}_{\infty,\infty }^{-1}}+ \Vert \nabla d_{k} \Vert _{ \dot{B}_{\infty,\infty }^{-1}} \bigr) \bigl( \bigl\Vert \Delta u^{k} \bigr\Vert ^{2}_{L^{2}}+ \bigl\Vert \nabla \Delta d^{k} \bigr\Vert ^{2}_{L^{2}} \bigr) \\ &{}+Ck^{2} \bigl( \Vert u_{\frac{k}{2},k} \Vert _{L^{\infty }}^{2}+ \Vert \nabla d_{ \frac{k}{2},k} \Vert _{L^{\infty }}^{2} \bigr) \bigl( \Vert u_{0} \Vert _{L^{2}}^{2}+ \Vert \nabla d_{0} \Vert _{L^{2}}^{2} \bigr). \end{aligned} $$

(3.32)

To handle $I_{42}$, we split $I_{42}$ into

$$ \begin{aligned}I_{42}={}& \bigl((u_{k} \cdot \nabla )\nabla d^{k},\nabla \Delta d^{k} \bigr)+ \bigl( \bigl(u^{k} \cdot \nabla \bigr)\nabla d^{k},\nabla \Delta d^{k} \bigr)+ \bigl( \bigl(u^{k}\cdot \nabla \bigr) \nabla d_{k},\nabla \Delta d^{k} \bigr) \\ &{}+ \bigl((u_{k}\cdot \nabla )\nabla d_{k},\nabla \Delta d^{k} \bigr) \\ :={}&I_{421}+I_{422}+I_{423}+I_{424}. \end{aligned} $$

(3.33)

By Hölder’s inequality and (2.4), we get

$$ \begin{aligned} \vert I_{421} \vert \leq {}& \Vert u_{k} \Vert _{L^{\infty }} \bigl\Vert \nabla ^{2} d^{k} \bigr\Vert _{L^{2}} \bigl\Vert \nabla \Delta d^{k} \bigr\Vert _{L^{2}} \\ \leq {}& \Vert u_{k} \Vert _{L^{\infty }} \bigl\Vert \Delta d^{k} \bigr\Vert _{L^{2}} \bigl\Vert \nabla \Delta d^{k} \bigr\Vert _{L^{2}} \\ \leq {}& c \Vert u_{k} \Vert _{\dot{B}_{\infty,\infty }^{-1}} \bigl\Vert \nabla \Delta d^{k} \bigr\Vert _{L^{2}}^{2}. \end{aligned} $$

(3.34)

Similarly to (3.12), one has

$$ \begin{aligned} \vert I_{422} \vert \leq {}& \bigl\Vert u^{k} \bigr\Vert _{L^{6}} \bigl\Vert \nabla ^{2} d^{k} \bigr\Vert _{L^{3}} \bigl\Vert \nabla \Delta d^{k} \bigr\Vert _{L^{2}} \\ \leq {}&C \bigl\Vert u^{k} \bigr\Vert _{L^{6}} \bigl\Vert \Delta d^{k} \bigr\Vert _{L^{3}} \bigl\Vert \nabla \Delta d^{k} \bigr\Vert _{L^{2}} \\ \leq {}& C \bigl\Vert \nabla u^{k} \bigr\Vert _{L^{2}} \bigl\Vert \Delta d^{k} \bigr\Vert _{L^{2}}^{1/2} \bigl\Vert \Delta d^{k} \bigr\Vert _{\dot{H}^{1}}^{1/2} \bigl\Vert \nabla \Delta d^{k} \bigr\Vert _{L^{2}} \\ \leq {}& C \bigl\Vert \nabla u^{k} \bigr\Vert _{L^{2}} \bigl\Vert \Delta d^{k} \bigr\Vert _{L^{2}}^{1/2} \bigl\Vert \nabla \Delta d^{k} \bigr\Vert _{L^{2}}^{1/2} \bigl\Vert \nabla \Delta d^{k} \bigr\Vert _{L^{2}} \\ \leq {}& Ck^{-1/2} \bigl\Vert \nabla u^{k} \bigr\Vert _{L^{2}} \bigl\Vert \nabla \Delta d^{k} \bigr\Vert _{L^{2}}^{2}. \end{aligned} $$

(3.35)

Hölder’s inequality, (2.4), and Young’s inequality guarantee

$$ \begin{aligned} \vert I_{423} \vert \leq{} & \bigl\Vert u^{k} \bigr\Vert _{L^{2}} \bigl\Vert \nabla ^{2} d_{k} \bigr\Vert _{L^{\infty }} \bigl\Vert \nabla \Delta d^{k} \bigr\Vert _{L^{2}} \\ \leq{} & \bigl\Vert u^{k} \bigr\Vert _{L^{2}} \Vert \Delta d_{k} \Vert _{L^{\infty }} \bigl\Vert \nabla \Delta d^{k} \bigr\Vert _{L^{2}} \\ \leq{} & ck \bigl\Vert u^{k} \bigr\Vert _{L^{2}} \Vert \nabla d_{k} \Vert _{L^{\infty }} \bigl\Vert \nabla \Delta d^{k} \bigr\Vert _{L^{2}} \\ \leq {}& c \bigl\Vert \Delta u^{k} \bigr\Vert _{L^{2}} \Vert \nabla d_{k} \Vert _{\dot{B}_{\infty, \infty }^{-1}} \bigl\Vert \nabla \Delta d^{k} \bigr\Vert _{L^{2}} \\ \leq{} & c \Vert \nabla d_{k} \Vert _{\dot{B}_{\infty,\infty }^{-1}} \bigl( \bigl\Vert \Delta u^{k} \bigr\Vert _{L^{2}}^{2}+ \bigl\Vert \nabla \Delta d^{k} \bigr\Vert _{L^{2}}^{2} \bigr). \end{aligned} $$

(3.36)

Similarly to (3.8), we write

$$\begin{aligned} I_{424}= \bigl((u_{k}\cdot \nabla )\nabla d_{\frac{k}{2},k}, \nabla \Delta d_{k,2k} \bigr)+ \bigl((u_{ \frac{k}{2},k}\cdot \nabla ) \nabla d_{\frac{k}{2}},\nabla \Delta d_{k,2k} \bigr):=I_{4241}+I_{4242}. \end{aligned}$$

From the Hölder inequality and the Young inequality, we conclude

$$ \begin{aligned} \vert I_{4241} \vert \leq{} & \Vert u_{k} \Vert _{L^{2}} \bigl\Vert \nabla ^{2} d_{ \frac{k}{2},k} \bigr\Vert _{L^{\infty }} \Vert \nabla \Delta d_{k,2k} \Vert _{L^{2}} \\ \leq {}&C \Vert u_{k} \Vert _{L^{2}} \Vert \Delta d_{\frac{k}{2},k} \Vert _{L^{\infty }} \Vert \nabla \Delta d_{k,2k} \Vert _{L^{2}} \\ \leq {}& Ck \Vert u_{k} \Vert _{L^{2}} \Vert \nabla d_{\frac{k}{2},k} \Vert _{L^{\infty }} \bigl\Vert \nabla \Delta d^{k} \bigr\Vert _{L^{2}} \\ \leq {}& Ck^{2} \Vert \nabla d_{\frac{k}{2},k} \Vert _{L^{\infty }}^{2} \bigl( \Vert u_{0} \Vert _{L^{2}}^{2}+ \Vert \nabla d_{0} \Vert _{L^{2}}^{2} \bigr)+\frac{1}{16} \bigl\Vert \nabla \Delta d^{k} \bigr\Vert _{L^{2}}^{2} \end{aligned} $$

(3.37)

and

$$ \begin{aligned} \vert I_{4242} \vert \leq {}& \Vert u_{\frac{k}{2},k} \Vert _{L^{\infty }} \bigl\Vert \nabla ^{2} d_{\frac{k}{2}} \bigr\Vert _{L^{2}} \Vert \nabla \Delta d_{k,2k} \Vert _{L^{2}} \\ \leq {}&C \Vert u_{\frac{k}{2},k} \Vert _{L^{\infty }} \Vert \Delta d_{\frac{k}{2}} \Vert _{L^{2}} \Vert \nabla \Delta d_{k,2k} \Vert _{L^{2}} \\ \leq {}& ck \Vert u_{\frac{k}{2},k} \Vert _{L^{\infty }} \Vert \nabla d_{\frac{k}{2}} \Vert _{L^{2}} \Vert \nabla \Delta d_{k,2k} \Vert _{L^{2}} \\ \leq {}& Ck^{2} \Vert u_{\frac{k}{2},k} \Vert _{L^{\infty }}^{2} \bigl( \Vert u_{0} \Vert _{L^{2}}^{2}+ \Vert \nabla d_{0} \Vert _{L^{2}}^{2} \bigr)+ \frac{1}{16} \bigl\Vert \nabla \triangle d^{k} \bigr\Vert _{L^{2}}^{2}. \end{aligned} $$

(3.38)

Therefore, by (3.24)–(3.38), we have

$$ \begin{aligned} \vert I_{4} \vert \leq {}& Ck^{-\frac{1}{2}} \bigl( \bigl\Vert \nabla u^{k} \bigr\Vert _{L^{2}}+ \bigl\Vert \Delta d^{k} \bigr\Vert _{L^{2}} \bigr) \bigl( \bigl\Vert \Delta u^{k} \bigr\Vert _{L^{2}}^{2}+ \bigl\Vert \nabla \Delta d^{k} \bigr\Vert _{L^{2}}^{2} \bigr) \\ &{}+C \Vert \nabla d_{k} \Vert _{\dot{B}_{\infty,\infty }^{-1}} \bigl( \bigl\Vert \Delta u^{k} \bigr\Vert _{L^{2}}^{2}+ \bigl\Vert \nabla \Delta d^{k} \bigr\Vert _{L^{2}}^{2} \bigr) \\ &{}+C \Vert u_{k} \Vert _{\dot{B}_{\infty,\infty }^{-1}} \bigl\Vert \nabla \Delta d^{k} \bigr\Vert _{L^{2}}^{2}+C k^{2} \bigl( \Vert u_{0} \Vert _{L^{2}}^{2}+ \Vert \nabla d_{0} \Vert _{L^{2}}^{2} \bigr)) \\ &{}\times \bigl( \Vert u_{\frac{k}{2},k} \Vert _{L^{\infty }}^{2}+ \Vert \nabla d_{\frac{k}{2},k} \Vert _{L^{\infty }}^{2} \bigr)+ \frac{1}{4} \Vert \nabla \Delta d_{k,2k} \Vert _{L^{2}}^{2}. \end{aligned} $$

(3.39)

It is left to deal with the last term, $I_{5}$. Using the fact that

$$\begin{aligned} \nabla \bigl( \vert \nabla d \vert ^{2}d \bigr)=2\nabla ^{2}d \nabla d d+ \vert \nabla d \vert ^{2} \nabla d, \end{aligned}$$

we can rewrite $I_{5}$ as follows:

$$ I_{5}=2 \bigl(\nabla ^{2}d\nabla d d,\nabla \Delta d^{k} \bigr)+ \bigl( \vert \nabla d \vert ^{2} \nabla d, \nabla \Delta d^{k} \bigr):=I_{51}+I_{52}. $$

(3.40)

Since

$$\begin{aligned} 2\nabla ^{2}d\nabla d d= \bigl(2\nabla ^{2}d_{k} \nabla d_{k}+2\nabla ^{2}d_{k} \nabla d^{k}+2\nabla ^{2}d^{k}\nabla d_{k}+2 \nabla ^{2}d^{k}\nabla d^{k} \bigr)d, \end{aligned}$$

we have

$$ \begin{aligned} I_{51}={}&2 \bigl(\nabla ^{2}d_{k}\nabla d_{k} d,\nabla \Delta d^{k} \bigr)+2 \bigl( \nabla ^{2}d_{k}\nabla d^{k} d,\nabla \Delta d^{k} \bigr)+2 \bigl(\nabla ^{2}d^{k} \nabla d_{k} d,\nabla \Delta d^{k} \bigr) \\ &{}+2 \bigl(\nabla ^{2}d^{k}\nabla d^{k}d,\nabla \Delta d^{k} \bigr) \\ :={}&I_{511}+I_{512}+I_{513}+I_{514}. \end{aligned} $$

(3.41)

Reasoning as (3.8), one has

$$ \begin{aligned} I_{511}&= \bigl(\nabla ^{2}d_{k}\nabla d_{\frac{k}{2},k} d, \nabla \Delta d_{k,2k} \bigr)+ \bigl(\nabla ^{2}d_{\frac{k}{2},k}\nabla d_{ \frac{k}{2}} d,\nabla \Delta d_{k,2k} \bigr) \\ &:=I_{5111}+I_{5112}. \end{aligned} $$

(3.42)

Using $|d|=1$, Hölder’s inequality, inequality (2.4) and Young’s inequality, we have

$$ \begin{aligned} \vert I_{5111} \vert \leq {}&2 \bigl\Vert \nabla ^{2} d_{k} \bigr\Vert _{L^{2}} \Vert \nabla d_{ \frac{k}{2},k} \Vert _{L^{\infty }} \Vert \nabla \Delta d_{k,2k} \Vert _{L^{2}} \\ \leq {}&C \Vert \Delta d_{k} \Vert _{L^{2}} \Vert \nabla d_{\frac{k}{2},k} \Vert _{L^{\infty }} \Vert \nabla \Delta d_{k,2k} \Vert _{L^{2}} \\ \leq {}&Ck \Vert \nabla d_{k} \Vert _{L^{2}} \Vert \nabla d_{\frac{k}{2},k} \Vert _{L^{\infty }} \Vert \nabla \Delta d_{k,2k} \Vert _{L^{2}} \\ \leq {}&Ck^{2} \Vert \nabla d_{\frac{k}{2},k} \Vert _{L^{\infty }}^{2} \bigl( \Vert u_{0} \Vert _{L^{2}}^{2}+ \Vert \nabla d_{0} \Vert _{L^{2}}^{2} \bigr)+\frac{1}{8} \Vert \nabla \Delta d_{k,2k} \Vert _{L^{2}}^{2}. \end{aligned} $$

(3.43)

Similarly,

$$ \begin{aligned} \vert I_{5112} \vert \leq {}& \bigl\Vert \nabla ^{2}d_{\frac{k}{2},k} \bigr\Vert _{L^{\infty }} \Vert \nabla d_{\frac{k}{2}} \Vert _{L^{2}} \Vert \nabla \Delta d_{k,2k} \Vert _{L^{2}} \\ \leq {}& Ck \Vert \nabla d_{\frac{k}{2},k} \Vert _{L^{\infty }} \Vert \nabla d_{ \frac{k}{2}} \Vert _{L^{2}} \Vert \nabla \Delta d_{k,2k} \Vert _{L^{2}} \\ \leq{} & Ck^{2} \Vert \nabla d_{\frac{k}{2},k} \Vert _{L^{\infty }}^{2} \bigl( \Vert u_{0} \Vert _{L^{2}}^{2}+ \Vert \nabla d_{0} \Vert _{L^{2}}^{2} \bigr)+\frac{1}{8} \Vert \nabla \Delta d_{k,2k} \Vert _{L^{2}}^{2}, \end{aligned} $$

(3.44)

which all taken together implies

$$ \vert I_{511} \vert \leq Ck^{2} \Vert \nabla d_{\frac{k}{2},k} \Vert _{L^{\infty }}^{2} \bigl( \Vert u_{0} \Vert _{L^{2}}^{2}+ \Vert \nabla d_{0} \Vert _{L^{2}}^{2} \bigr)+\frac{1}{4} \Vert \nabla \Delta d_{k,2k} \Vert _{L^{2}}^{2}. $$

(3.45)

By the fact $|d|=1$, the Hölder inequality, (2.4) and (3.13), we can get

$$ \begin{aligned} \vert I_{512} \vert \leq{} &2 \bigl\Vert \nabla ^{2}d_{k} \bigr\Vert _{L^{\infty }} \bigl\Vert \nabla d^{k} \bigr\Vert _{L^{2}} \bigl\Vert \nabla \Delta d^{k} \bigr\Vert _{L^{2}} \\ \leq {}&C \Vert \Delta d_{k} \Vert _{L^{\infty }} \bigl\Vert \nabla d^{k} \bigr\Vert _{L^{2}} \bigl\Vert \nabla \Delta d^{k} \bigr\Vert _{L^{2}} \\ \leq{} & Ck \Vert \nabla d_{k} \Vert _{L^{\infty }} \bigl\Vert \nabla d^{k} \bigr\Vert _{L^{2}} \bigl\Vert \nabla \Delta d^{k} \bigr\Vert _{L^{2}} \\ \leq {}&Ck^{2} \Vert \nabla d_{k} \Vert _{\dot{B}_{\infty,\infty }^{-1}} \bigl\Vert \nabla d^{k} \bigr\Vert _{L^{2}} \bigl\Vert \nabla \Delta d^{k} \bigr\Vert _{L^{2}} \\ \leq {}&C \Vert \nabla d_{k} \Vert _{\dot{B}_{\infty,\infty }^{-1}} \bigl\Vert \nabla \Delta d^{k} \bigr\Vert _{L^{2}}^{2}. \end{aligned} $$

(3.46)

Similarly,

$$ \begin{aligned} \vert I_{513} \vert \leq{} &2 \bigl\Vert \nabla ^{2} d^{k} \bigr\Vert _{L^{2}} \Vert \nabla d_{k} \Vert _{L^{\infty }} \bigl\Vert \nabla \Delta d^{k} \bigr\Vert _{L^{2}} \\ \leq {}& C \Vert \nabla d_{k} \Vert _{\dot{B}_{\infty,\infty }^{-1}} \bigl\Vert \nabla \Delta d^{k} \bigr\Vert _{L^{2}}^{2}. \end{aligned} $$

(3.47)

Reasoning as (3.12) again, one has

$$ \begin{aligned} \vert I_{514} \vert \leq {}&2 \bigl\Vert \nabla ^{2} d^{k} \bigr\Vert _{L^{3}} \bigl\Vert \nabla d^{k} \bigr\Vert _{L^{6}} \bigl\Vert \nabla \Delta d^{k} \bigr\Vert _{L^{2}} \\ \leq {}& C \bigl\Vert \nabla ^{2} d^{k} \bigr\Vert _{L^{2}}^{\frac{1}{2}} \bigl\Vert \nabla ^{2} d^{k} \bigr\Vert _{\dot{H}^{1}}^{\frac{1}{2}} \bigl\Vert \nabla d^{k} \bigr\Vert _{\dot{H}^{1}} \bigl\Vert \nabla \Delta d^{k} \bigr\Vert _{L^{2}} \\ \leq {}&C \bigl\Vert \Delta d^{k} \bigr\Vert _{L^{2}}^{\frac{1}{2}} \bigl\Vert \nabla \Delta d^{k} \bigr\Vert _{L^{2}}^{ \frac{1}{2}} \bigl\Vert \Delta d^{k} \bigr\Vert _{L^{2}} \bigl\Vert \nabla \Delta d^{k} \bigr\Vert _{L^{2}} \\ \leq {}& Ck^{-\frac{1}{2}} \bigl\Vert \Delta d^{k} \bigr\Vert _{L^{2}} \bigl\Vert \nabla \Delta d^{k} \bigr\Vert _{L^{2}}^{2}. \end{aligned} $$

(3.48)

Therefore, inequalities (3.45)–(3.48) yield

$$ \begin{aligned} \vert I_{51} \vert \leq {}& Ck^{2} \Vert \nabla d_{\frac{k}{2},k} \Vert _{L^{\infty }}^{2} \bigl( \Vert u_{0} \Vert _{L^{2}}^{2}+ \Vert \nabla d_{0} \Vert _{L^{2}}^{2} \bigr)+C \Vert \nabla d_{k} \Vert _{\dot{B}_{\infty,\infty }^{-1}} \bigl\Vert \nabla \Delta d^{k} \bigr\Vert _{L^{2}}^{2} \\ &{}+Ck^{-\frac{1}{2}} \bigl\Vert \Delta d^{k} \bigr\Vert _{L^{2}} \bigl\Vert \nabla \Delta d^{k} \bigr\Vert _{L^{2}}^{2}+ \frac{1}{4} \Vert \nabla \Delta d_{k,2k} \Vert _{L^{2}}^{2}. \end{aligned} $$

(3.49)

It is easy to get $\Delta d\cdot d=- \vert \nabla d \vert ^{2}$ due to $\vert d \vert =1$. Then $\vert \nabla d \vert ^{2}\nabla d=-\Delta d\cdot d\nabla d$. Hence we decompose $I_{52}$ in the following way:

$$ \begin{aligned} I_{52}={}&{-} \bigl(\Delta d\cdot d\nabla d,\nabla \Delta d^{k} \bigr) \\ ={}&{-} \bigl(\Delta d^{k}\cdot d\nabla d^{k},\nabla \Delta d^{k} \bigr)- \bigl(\Delta d^{k} \cdot d\nabla d_{k},\nabla \Delta d^{k} \bigr)\\ &{}- \bigl(\Delta d_{k}\cdot d\nabla d^{k}, \nabla \Delta d^{k} \bigr)- \bigl(\Delta d_{k}\cdot d\nabla d_{k},\nabla \Delta d^{k} \bigr) \\ :={}&I_{521}+I_{522}+I_{523}+I_{524}. \end{aligned} $$

(3.50)

Repeating the methods to prove (3.12), we obtain

$$ \begin{aligned} \vert I_{521} \vert \leq {}& \bigl\Vert \Delta d^{k} \bigr\Vert _{L^{3}} \bigl\Vert \nabla d^{k} \bigr\Vert _{L^{6}} \bigl\Vert \nabla \Delta d^{k} \bigr\Vert _{L^{2}} \\ \leq {}&C \bigl\Vert \Delta d^{k} \bigr\Vert _{L^{2}}^{\frac{1}{2}} \bigl\Vert \Delta d^{k} \bigr\Vert _{ \dot{H}^{1}}^{\frac{1}{2}} \bigl\Vert \nabla d^{k} \bigr\Vert _{\dot{H}^{1}} \bigl\Vert \nabla \Delta d^{k} \bigr\Vert _{L^{2}} \\ \leq {}&C \bigl\Vert \Delta d^{k} \bigr\Vert _{L^{2}}^{\frac{1}{2}} \bigl\Vert \nabla \Delta d^{k} \bigr\Vert _{L^{2}}^{ \frac{1}{2}} \bigl\Vert \Delta d^{k} \bigr\Vert _{L^{2}} \bigl\Vert \nabla \Delta d^{k} \bigr\Vert _{L^{2}} \\ \leq {}&Ck^{-\frac{1}{2}} \bigl\Vert \Delta d^{k} \bigr\Vert _{L^{2}} \bigl\Vert \nabla \Delta d^{k} \bigr\Vert _{L^{2}}^{2}. \end{aligned} $$

(3.51)

Similarly to (3.46), we have

$$\begin{aligned} \vert I_{522} \vert + \vert I_{523} \vert \leq C \Vert \nabla d_{k} \Vert _{\dot{B}_{\infty,\infty }^{-1}} \bigl\Vert \nabla \Delta d^{k} \bigr\Vert _{L^{2}}^{2}. \end{aligned}$$

(3.52)

Similarly to (3.45), one has

$$\begin{aligned} \vert I_{524} \vert \leq Ck^{2} \Vert \nabla d_{\frac{k}{2},k} \Vert _{L^{\infty }}^{2} \bigl( \Vert u_{0} \Vert _{L^{2}}^{2}+ \Vert \nabla d_{0} \Vert _{L^{2}}^{2} \bigr)+\frac{1}{4} \Vert \nabla \Delta d_{k,2k} \Vert _{L^{2}}^{2}. \end{aligned}$$

(3.53)

Thus

$$ \begin{aligned} \vert I_{52} \vert \leq{} &Ck^{-\frac{1}{2}} \bigl\Vert \Delta d^{k} \bigr\Vert _{L^{2}} \bigl\Vert \nabla \Delta d^{k} \bigr\Vert _{L^{2}}^{2}+C \Vert \nabla d_{k} \Vert _{\dot{B}_{ \infty,\infty }^{-1}} \bigl\Vert \nabla \Delta d^{k} \bigr\Vert _{L^{2}}^{2} \\ &{}+Ck^{2} \Vert \nabla d_{\frac{k}{2},k} \Vert _{L^{\infty }}^{2} \bigl( \Vert u_{0} \Vert _{L^{2}}^{2} + \Vert \nabla d_{0} \Vert _{L^{2}}^{2} \bigr)+ \frac{1}{4} \Vert \nabla \Delta d_{k,2k} \Vert _{L^{2}}^{2}. \end{aligned} $$

(3.54)

From (3.49) and (3.54), we deduce

$$ \begin{aligned} \vert I_{5} \vert \leq {}& Ck^{2} \Vert \nabla d_{\frac{k}{2},k} \Vert _{L^{\infty }}^{2} \bigl( \Vert u_{0} \Vert _{L^{2}}^{2}+ \Vert \nabla d_{0} \Vert _{L^{2}}^{2} \bigr)+C \Vert \nabla d_{k} \Vert _{\dot{B}_{\infty,\infty }^{-1}} \bigl\Vert \nabla \Delta d^{k} \bigr\Vert _{L^{2}}^{2} \\ &{}+Ck^{-\frac{1}{2}} \bigl\Vert \Delta d^{k} \bigr\Vert _{L^{2}} \bigl\Vert \nabla \Delta d^{k} \bigr\Vert _{L^{2}}^{2}+ \frac{1}{2} \Vert \nabla \Delta d_{k,2k} \Vert _{L^{2}}^{2}. \end{aligned} $$

(3.55)

Combining (3.6), (3.16), (3.23), (3.39) and (3.55), we have

$$ \begin{aligned} &\frac{1}{2}\frac{d}{dt} \bigl( \bigl\Vert \nabla u^{k} \bigr\Vert _{L^{2}}^{2}+ \bigl\Vert \Delta d^{k} \bigr\Vert _{L^{2}}^{2} \bigr)+ \bigl( \bigl\Vert \Delta u^{k} \bigr\Vert _{L^{2}}^{2}+ \bigl\Vert \nabla \Delta d^{k} \bigr\Vert _{L^{2}}^{2} \bigr) \\ &\quad\leq C_{1}k^{2} \bigl( \Vert u_{\frac{k}{2},k} \Vert _{L^{\infty }}^{2}+ \Vert \nabla d_{ \frac{k}{2},k} \Vert _{L^{\infty }}^{2} \bigr) \bigl( \Vert u_{0} \Vert _{L^{2}}^{2}+ \Vert \nabla d_{0} \Vert _{L^{2}}^{2} \bigr) \\ &\qquad{}+C_{2} \bigl( \Vert u_{k} \Vert _{\dot{B}_{\infty,\infty }^{-1}}+ \Vert \nabla d_{k} \Vert _{ \dot{B}_{\infty,\infty }^{-1}} \bigr) \bigl( \bigl\Vert \Delta u^{k} \bigr\Vert _{L^{2}}^{2}+ \bigl\Vert \nabla \Delta d^{k} \bigr\Vert _{L^{2}}^{2} \bigr) \\ &\qquad{}+C_{3} k^{-\frac{1}{2}} \bigl( \bigl\Vert \nabla u^{k} \bigr\Vert _{L^{2}}+ \bigl\Vert \Delta d^{k} \bigr\Vert _{L^{2}} \bigr) \bigl( \bigl\Vert \Delta u^{k} \bigr\Vert _{L^{2}}^{2}+ \bigl\Vert \nabla \Delta d^{k} \bigr\Vert _{L^{2}}^{2} \bigr)+ \frac{3}{4} \Vert \Delta u_{k,2k} \Vert _{L^{2}}^{2} \\ &\qquad{}+\frac{3}{4} \Vert \nabla \Delta d_{k,2k} \Vert _{L^{2}}^{2} \end{aligned} $$

(3.56)

and

$$ \begin{aligned} &\frac{d}{dt} \bigl( \bigl\Vert \nabla u^{k} \bigr\Vert _{L^{2}}^{2}+ \bigl\Vert \triangle d^{k} \bigr\Vert _{L^{2}}^{2} \bigr)\\ &\quad\leq \biggl[c_{1} k^{2} \bigl( \Vert u_{0} \Vert _{L^{2}}^{2}+ \Vert \nabla d_{0} \Vert _{L^{2}}^{2} \bigr) \bigl( \Vert u_{k/2,k} \Vert _{L^{\infty }}^{2}+ \Vert \nabla d_{k/2,k} \Vert _{L^{\infty }}^{2} \bigr) \\ &\qquad{}-\frac{1}{8} \bigl( \bigl\Vert \triangle u^{k} \bigr\Vert _{L^{2}}^{2}+ \bigl\Vert \nabla \triangle d^{k} \bigr\Vert _{L^{2}}^{2} \bigr) \biggr]+ \biggl(c_{2} \bigl( \bigl\Vert u_{k}(t) \bigr\Vert _{\dot{B}_{\infty,\infty }^{-1}}+ \Vert \nabla d_{k} \Vert _{\dot{B}_{\infty,\infty }^{-1}} \bigr)-\frac{1}{4} \biggr) \\ &\qquad{}\times \bigl( \bigl\Vert \triangle u^{k} \bigr\Vert _{L^{2}}^{2}+ \bigl\Vert \nabla \triangle d^{k} \bigr\Vert _{L^{2}}^{2} \bigr) \\ &\qquad{}+ \biggl(c_{3} k^{-1/2} \bigl( \bigl\Vert \nabla u^{k} \bigr\Vert _{L^{2}}+ \bigl\Vert \triangle d^{k} \bigr\Vert _{L^{2}} \bigr)- \frac{1}{8} \biggr) \bigl( \bigl\Vert \triangle u^{k} \bigr\Vert _{L^{2}}^{2}+ \bigl\Vert \nabla \triangle d^{k} \bigr\Vert _{L^{2}}^{2} \bigr). \end{aligned} $$

(3.57)

Let

$$\begin{aligned} \begin{aligned} \tilde{k}=128\times 4 c_{3}^{2} \bigl( \bigl\Vert \nabla u_{0}^{\tilde{k}} \bigr\Vert _{L^{2}}+ \bigl\Vert \triangle d_{0}^{\tilde{k}} \bigr\Vert _{L^{2}} \bigr)^{2}. \end{aligned} \end{aligned}$$

(3.58)

Then

$$\begin{aligned} \bigl\Vert \nabla u_{0}^{\tilde{k}} \bigr\Vert _{L^{2}}+ \bigl\Vert \Delta d_{0}^{\tilde{k}} \bigr\Vert _{L^{2}}< \frac{\tilde{k}^{\frac{1}{2}}}{16 c_{3}}. \end{aligned}$$

Since $\lim_{t\rightarrow T-0}\|\nabla u^{\tilde{k}}(t)\|_{L^{2}}+\| \Delta d^{\tilde{k}}(t)\|_{L^{2}}=\infty$, there is some $\delta \in (0,T)$ such that

$$\begin{aligned} & \bigl\Vert \nabla u^{\tilde{k}}(T- \delta ) \bigr\Vert _{L^{2}}+ \bigl\Vert \Delta d^{\tilde{k}}(T- \delta ) \bigr\Vert _{L^{2}}=\frac{\tilde{k}^{\frac{1}{2}}}{16 c_{3}}, \end{aligned}$$

(3.59)

$$\begin{aligned} & \bigl\Vert \nabla u^{\tilde{k}}(t) \bigr\Vert _{L^{2}}+ \bigl\Vert \Delta d^{\tilde{k}}(t) \bigr\Vert _{L^{2}}> \frac{\tilde{k}^{\frac{1}{2}}}{16 c_{3}}. \end{aligned}$$

(3.60)

From (3.60), we get for any $t\in (T-\delta,T)$

$$\begin{aligned} &c_{1}{\tilde{k}}^{2} \bigl( \Vert u_{0} \Vert _{L^{2}}^{2}+ \Vert \nabla d_{0} \Vert _{L^{2}}^{2} \bigr) \bigl( \Vert u_{\tilde{\frac{k}{2}},\tilde{k}} \Vert _{L^{\infty }}^{2}+ \Vert \nabla d_{ \tilde{\frac{k}{2}},\tilde{k}} \Vert _{L^{\infty }}^{2} \bigr)-\frac{1}{8} \bigl( \bigl\Vert \Delta u^{\tilde{k}} \bigr\Vert _{L^{2}}^{2}+ \bigl\Vert \nabla \Delta d^{\tilde{k}} \bigr\Vert _{L^{2}}^{2} \bigr) \\ &\quad\leq {\tilde{k}}^{2} \biggl[c_{1} \bigl( \Vert u_{0} \Vert _{L^{2}}^{2}+ \Vert \nabla d_{0} \Vert _{L^{2}}^{2} \bigr) \bigl( \Vert u_{\tilde{\frac{k}{2}},\tilde{k}} \Vert _{L^{\infty }}^{2}+ \Vert \nabla d_{ \tilde{\frac{k}{2}},\tilde{k}} \Vert _{L^{\infty }}^{2} \bigr)-\frac{1}{8} \bigl( \bigl\Vert \nabla u^{\tilde{k}} \bigr\Vert _{L^{2}}^{2}+ \bigl\Vert \Delta d^{\tilde{k}} \bigr\Vert _{L^{2}}^{2} \bigr) \biggr] \\ &\quad\leq {\tilde{k}}^{2} \biggl[c_{1} \bigl( \Vert u_{0} \Vert _{L^{2}}^{2}+ \Vert \nabla d_{0} \Vert _{L^{2}}^{2} \bigr) \bigl( \Vert u_{\tilde{\frac{k}{2}},\tilde{k}} \Vert _{L^{\infty }}^{2}+ \Vert \nabla d_{ \tilde{\frac{k}{2}},\tilde{k}} \Vert _{L^{\infty }}^{2} \bigr)-\frac{1}{16} \frac{\tilde{k}}{256 c_{3}^{2}} \biggr] \\ &\quad\leq 0, \end{aligned}$$

provided

$$\begin{aligned} \begin{aligned} \bigl\Vert {u_{\frac{\tilde{k}}{2},\tilde{k}}}(t) \bigr\Vert _{L^{\infty }}+ \bigl\Vert {\nabla d_{ \frac{\tilde{k}}{2},\tilde{k}}(t)} \bigr\Vert _{L^{\infty }}< \frac{\tilde{k}^{\frac{1}{2}}}{32c_{3}\sqrt{c_{1}}( \Vert u_{0} \Vert _{L^{2}}+ \Vert \nabla d_{0} \Vert _{L^{2}})}, \quad\forall t\in (T-\delta,T). \end{aligned} \end{aligned}$$

(3.61)

In view of (3.58), the inequality (3.61) is equivalent to

$$\begin{aligned} \begin{aligned} &\bigl\Vert {u_{\frac{\tilde{k}}{2},\tilde{k}}}(t) \bigr\Vert _{B_{\infty,\infty }^{-1}}+ \bigl\Vert {\nabla d_{\frac{\tilde{k}}{2},\tilde{k}}}(t) \bigr\Vert _{B_{\infty,\infty }^{-1}} \\ &\quad< c \frac{1}{\tilde{k}^{\frac{1}{2}} 32 c_{3}\sqrt{c_{1}}( \Vert u_{0} \Vert _{L^{2}}+ \Vert \nabla d_{0} \Vert _{L^{2}})} \\ &\quad< c \frac{1}{16\sqrt{2}c_{3}( \Vert \nabla u_{0} \Vert _{L^{2}}+ \Vert \Delta d_{0} \Vert _{L^{2}})\times 32c_{3}\sqrt{c_{1}}( \Vert u_{0} \Vert _{L^{2}}+ \Vert \nabla d_{0} \Vert _{L^{2}})} \\ &\quad< c \frac{1}{512\sqrt{2}c_{3}^{2}\sqrt{c_{1}}( \Vert u_{0} \Vert _{L^{2}}+ \Vert \nabla d_{0} \Vert _{L^{2}})( \Vert \nabla u_{0}^{\tilde{k}} \Vert _{L^{2}}+ \Vert \Delta d_{0}^{\tilde{k}} \Vert _{L^{2}})}. \end{aligned} \end{aligned}$$

(3.62)

Thus, if (3.62) and

$$\begin{aligned} c_{2} ( \Vert u_{\tilde{k}} \Vert _{{L^{\infty }}(T-\delta,T;\dot{B}_{\infty, \infty }^{-1})}+ \Vert \nabla d_{\tilde{k}} \Vert _{{L^{\infty }}(T-\delta,T; \dot{B}_{\infty,\infty }^{-1})}\leq \frac{1}{4} \end{aligned}$$

(3.63)

hold, we can infer from (3.57) that

$$\begin{aligned} \begin{aligned} &\frac{d}{dt} \bigl( \bigl\Vert \nabla u^{\tilde{k}} \bigr\Vert _{L^{2}}^{2}+ \bigl\Vert \Delta d^{ \tilde{k}} \bigr\Vert _{L^{2}}^{2} \bigr) \\ &\quad\leq \biggl(c_{3}\tilde{k}^{-\frac{1}{2}} \bigl( \bigl\Vert \nabla u^{\tilde{k}} \bigr\Vert _{L^{2}}+ \bigl\Vert \Delta d^{\tilde{k}} \bigr\Vert _{L^{2}} \bigr)-\frac{1}{8} \biggr) \bigl( \bigl\Vert \Delta u^{\tilde{k}} \bigr\Vert _{L^{2}}^{2}+ \bigl\Vert \nabla \Delta d^{\tilde{k}} \bigr\Vert _{L^{2}}^{2} \bigr). \end{aligned} \end{aligned}$$

(3.64)

Since $c_{3}{\tilde{k}}^{-\frac{1}{2}}(\|\nabla u^{\tilde{k}}(T-\delta )\|_{L^{2}}+ \|\Delta d^{\tilde{k}}(T-\delta )\|_{L^{2}})-\frac{1}{8}=c_{3} \tilde{k}^{-\frac{1}{2}}\frac{\tilde{k}^{\frac{1}{2}}}{16 c_{3}}- \frac{1}{8}<0$, there is a right neighborhood I of $t=T-\delta$ such that

$$\begin{aligned} c_{3} {\tilde{k}}^{-\frac{1}{2}} \bigl( \bigl\Vert \nabla u^{\tilde{k}}(t) \bigr\Vert _{L^{2}}+ \bigl\Vert \Delta d^{\tilde{k}}(t) \bigr\Vert _{L^{2}} \bigr)-\frac{1}{8}< 0,\quad \forall t\in \mathrm{I.} \end{aligned}$$

Hence, it follows by (3.64) that the function $t\to \|\nabla u^{\tilde{k}}\|_{L^{2}}+\|\Delta d^{\tilde{k}}\|_{L^{2}}$ decreases in I, which contradicts (3.59) and (3.60). Thus, when (3.62) and (3.63) are satisfied, u and ∇d cannot blow up at $t=T$, and u and ∇d are regular in $(0,T]$. The proof of the theorem is completed. □