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. □