Long-time asymptotics of axisymmetric Navier–Stokes equations in critical spaces

Abstract

We prove that any global strong solution to axisymmetric Navier–Stokes equations must eventually become small. In particular, the limits of $\Vert \omega ^\theta (t)/r\Vert _{L^1}$ and $\Vert u^\theta (t)/\sqrt{r}\Vert _{L^2}$ are all 0 as t tends to infinity. Then by using these, we can refine a series of decay estimates. In particular, for the global axisymmetric solutions we know till now, namely the axisymmetric without swirl or with small swirl ones, these decay estimates hold. But our result here do not require any smallness conditions beforehand, thus more general.

Appendix A. Global well-posedness of (1.3) with small initial data

The purpose of this appendix is to present the proof of Proposition 2.2. By using the semi-group (2.3), we can reformulate the systems (1.3) as

(A.1)

Here and all in that follows, we always denote .

The main idea is to apply the following fixed-point argument to the integral formula (A.1).

Lemma A.1

(Lemma 5.5 in [2]) Let E be a Banach space, $\mathfrak {B}(\cdot ,\cdot )$ a continuous bilinear map from $E\times E$ to E,  and $\mathfrak {\alpha }$ a positive real number such that

$$\begin{aligned} \mathfrak {\alpha }<\frac{1}{4\Vert \mathfrak {B}\Vert }\quad \text {with}\quad \Vert \mathfrak {B}\Vert \mathop {=}\limits ^{\textrm{def}}\sup _{\Vert f\Vert ,\Vert g\Vert \le 1}\Vert \mathfrak {B}(f,g)\Vert . \end{aligned}$$

Then for any a in the ball $B(0,\mathfrak {\alpha })$ in E,  there exists a unique x in $B(0,2\mathfrak {\alpha })$ such that

$$\begin{aligned} x=a+\mathfrak {B}(x,x). \end{aligned}$$

The proof of Proposition 2.2

$\bullet$ The estimate for the linear part in (A.1). First, thanks to (2.6) with $(\alpha ,\beta )=(-1,1)$, (2.7) and (2.8) with $\gamma =\epsilon =0$, there holds

$$\begin{aligned} \begin{aligned} \Vert S(t)\omega ^\theta _0&\Vert _{L^1(\Omega )} +t\Vert S(t)\omega ^\theta _0\Vert _{L^{\infty }} +\Vert S(t)u^\theta _0\Vert _{L^2(\Omega )}+t^{\frac{1}{2}}\Vert S(t)u^\theta _0\Vert _{L^{\infty }} +t^{\frac{1}{2}}\Vert {\widetilde{\nabla }}S(t)u^\theta _0\Vert _{L^2(\Omega )}\\&+t^{\frac{1}{2}}\Vert r^{-1}S(t)u^\theta _0\Vert _{L^2(\Omega )} +t^{\frac{3}{4}}\Vert r^{-1}S(t)u^\theta _0\Vert _{L^4(\Omega )} \lesssim \Vert \omega ^\theta _0\Vert _{L^1(\Omega )}+\Vert u^\theta _0\Vert _{L^2(\Omega )}. \end{aligned}\nonumber \\ \end{aligned}$$

(A.2)

Next, by using (2.6) with $(\alpha ,\beta )=(\frac{1}{2},\frac{1}{3})$ and $(\frac{2}{3},\frac{1}{3})$, we obtain

$$\begin{aligned} \begin{aligned} t^{\frac{1}{4}}\Vert S(t)&\omega ^\theta _0\Vert _{L^2}+\Vert S(t)\omega ^\theta _0\Vert _{L^{\frac{3}{2}}} +t^{\frac{1}{3}}\bigl \Vert r^{\frac{2}{3}}S(t)\omega ^\theta _0\bigr \Vert _{L^3(\Omega )}\\&=t^{\frac{1}{4}}\Vert r^{\frac{1}{2}}S(t)\omega ^\theta _0\Vert _{L^2(\Omega )}+\Vert r^{\frac{2}{3}}S(t)\omega ^\theta _0\Vert _{L^{\frac{3}{2}}(\Omega )} +t^{\frac{1}{3}}\bigl \Vert r^{\frac{2}{3}}S(t)\omega ^\theta _0\bigr \Vert _{L^3(\Omega )} \lesssim \Vert \omega ^\theta _0\Vert _{L^{\frac{3}{2}}}. \end{aligned} \nonumber \\ \end{aligned}$$

(A.3)

For the space-time integral terms, the key idea is to use Minkowski’s inequality. Precisely, by using (2.9) with $\alpha =0,~\beta =\frac{1}{3}$ and change of variables, we have

$$\begin{aligned} \begin{aligned} \bigl \Vert S(t)\omega ^\theta _0\bigr \Vert _{L_T^2 L^2(\Omega )}&=\Bigl \Vert \frac{1}{4\pi t} \int _\Omega \frac{1}{r^{\frac{1}{2}}{\bar{r}}^{\frac{1}{2}-\frac{1}{3}}}H\bigl (\frac{t}{r{\bar{r}}}\bigr ) \exp \Bigl (-\frac{|\zeta |^2}{4t}\Bigr ) \cdot {\bar{r}}^{\frac{2}{3}}\omega ^\theta _0({\bar{r}},{\bar{z}})\,d{\bar{r}}\,d{\bar{z}}\Bigr \Vert _{L_T^2 L^2(\Omega )}\\&\lesssim \Bigl \Vert \int _\Omega \bigl \Vert \frac{1}{4\pi t}\cdot \frac{1}{t^{\frac{1}{3}}} \exp \bigl (-\frac{|\zeta |^2}{5t}\bigr )\bigr \Vert _{L_T^2} \cdot {\bar{r}}^{\frac{2}{3}}\omega ^\theta _0({\bar{r}},{\bar{z}})\,d{\bar{r}}\,d{\bar{z}}\Bigr \Vert _{L^2(\Omega )}\\&=\Bigl \Vert \int _\Omega \bigl \Vert \frac{1}{4\pi t^{\frac{4}{3}}} \exp \bigl (-\frac{1}{5t}\bigr )\Vert _{L_T^2} \cdot \bigl (|\zeta |^{-2}\bigr )^{\frac{5}{6}} \cdot {\bar{r}}^{\frac{2}{3}}\omega ^\theta _0({\bar{r}},{\bar{z}})\,d{\bar{r}}\,d{\bar{z}}\Bigr \Vert _{L^2(\Omega )}\\&\lesssim \bigl \Vert {r}^{\frac{2}{3}}\omega ^\theta _0\bigr \Vert _{L^{\frac{3}{2}}(\Omega )} \bigl \Vert \bigl (\frac{1}{r^2+z^2}\bigr )^{\frac{5}{6}}\bigr \Vert _{L^{\frac{6}{5},\infty }(\Omega )} \lesssim \Vert \omega ^\theta _0\Vert _{L^{\frac{3}{2}}}, \end{aligned}\end{aligned}$$

(A.4)

here and in all that follows, we always denote $|\zeta |^2\mathop {=}\limits ^{\textrm{def}}(r-{\bar{r}})^2+(z-{\bar{z}})^2,$ and we have used the fact that $\bigl \Vert \frac{1}{4\pi t^{\frac{4}{3}}} \exp \bigl (-\frac{1}{5t}\bigr )\Vert _{L_T^2}$ is uniformly bounded by some constant independent of T.

Exactly along the same line, using (2.9) with $\alpha =0,~\beta =1$ gives

$$\begin{aligned} \begin{aligned} \bigl \Vert S(t)u^\theta _0\bigr \Vert _{L_T^4 L^4(\Omega )}&\lesssim \Bigl \Vert \int _\Omega \bigl \Vert \frac{1}{4\pi t} \exp \bigl (-\frac{|\zeta |^2}{5t}\bigr )\Vert _{L_T^4} \cdot u^\theta _0({\bar{r}},{\bar{z}})\,d{\bar{r}}\,d{\bar{z}}\Bigr \Vert _{L^4(\Omega )}\\&\lesssim \Vert u^\theta _0\Vert _{L^2(\Omega )} \bigl \Vert \bigl (\frac{1}{r^2+z^2}\bigr )^{\frac{3}{4}}\bigr \Vert _{L^{\frac{4}{3},\infty }(\Omega )} \lesssim \Vert u^\theta _0\Vert _{L^2(\Omega )}. \end{aligned}\end{aligned}$$

(A.5)

Using (2.9) with $\alpha =\frac{1}{2},~\beta =\frac{1}{3}$ gives

$$\begin{aligned} \begin{aligned} \bigl \Vert S(t)\omega ^\theta _0\bigr \Vert _{L_T^4 L^2}&=\Bigl \Vert r^{\frac{1}{2}}\frac{1}{4\pi t} \int _\Omega \frac{1}{r^{\frac{1}{2}}{\bar{r}}^{\frac{1}{2}-\frac{1}{3}}}H\bigl (\frac{t}{r{\bar{r}}}\bigr ) \exp \Bigl (-\frac{|\zeta |^2}{4t}\Bigr ) \cdot {\bar{r}}^{\frac{2}{3}}\omega ^\theta _0({\bar{r}},{\bar{z}})\,d{\bar{r}}\,d{\bar{z}}\Bigr \Vert _{L_T^4 L^2(\Omega )}\\&\lesssim \Bigl \Vert \int _\Omega \bigl \Vert \frac{1}{4\pi t}\cdot \frac{1}{t^{\frac{1}{12}}} \exp \bigl (-\frac{|\zeta |^2}{5t}\bigr )\bigr \Vert _{L_T^4} \cdot {\bar{r}}^{\frac{2}{3}}\omega ^\theta _0({\bar{r}},{\bar{z}})\,d{\bar{r}}\,d{\bar{z}}\Bigr \Vert _{L^2(\Omega )}\\&\lesssim \bigl \Vert {r}^{\frac{2}{3}}\omega ^\theta _0\bigr \Vert _{L^{\frac{3}{2}}(\Omega )} \bigl \Vert \bigl (\frac{1}{r^2+z^2}\bigr )^{\frac{5}{6}}\bigr \Vert _{L^{\frac{6}{5},\infty }(\Omega )} \lesssim \Vert \omega ^\theta _0\Vert _{L^{\frac{3}{2}}}. \end{aligned}\end{aligned}$$

(A.6)

Using (2.9) with $\alpha =-\frac{3}{5},~\beta =1$ gives

$$\begin{aligned} \begin{aligned} \bigl \Vert r^{-\frac{3}{5}} S(t)u^\theta _0\bigr \Vert _{L_T^{\frac{5}{2}}L^{\frac{5}{2}}(\Omega )}&\lesssim \Bigl \Vert \int _\Omega \bigl \Vert \frac{1}{4\pi t}\cdot \frac{1}{t^{\frac{3}{10}}} \exp \bigl (-\frac{|\zeta |^2}{5t}\bigr )\bigr \Vert _{L_T^{\frac{5}{2}}} \cdot u^\theta _0({\bar{r}},{\bar{z}})\,d{\bar{r}}\,d{\bar{z}}\Bigr \Vert _{L^{\frac{5}{2}}(\Omega )}\\&\lesssim \Vert u^\theta _0\Vert _{L^2(\Omega )} \bigl \Vert \bigl (\frac{1}{r^2+z^2}\bigr )^{\frac{9}{10}}\bigr \Vert _{L^{\frac{10}{9},\infty }(\Omega )} \lesssim \Vert u^\theta _0\Vert _{L^2(\Omega )}. \end{aligned} \nonumber \\ \end{aligned}$$

(A.7)

Using (2.9) with $\alpha =0,~\beta =\frac{1}{3}$ and $\alpha =\frac{2}{3},~\beta =\frac{1}{3}$ gives

$$\begin{aligned} \begin{aligned} \bigl \Vert t^{\frac{1}{3}}S(t)\omega ^\theta _0&\bigr \Vert _{L_T^3 L^3(\Omega )} +\bigl \Vert r^{\frac{2}{3}}S(t)\omega ^\theta _0\bigr \Vert _{L_T^3 L^3(\Omega )}\\&\lesssim \Bigl \Vert t^{-1} \int _\Omega \frac{t^{\frac{1}{3}}+r^{\frac{2}{3}}}{r^{\frac{1}{2}}{\bar{r}}^{\frac{1}{6}}} \bigl |H\bigl (\frac{t}{r{\bar{r}}}\bigr )\bigr | \exp \Bigl (-\frac{|\zeta |^2}{4t}\Bigr ) \cdot {\bar{r}}^{\frac{2}{3}}|\omega ^\theta _0({\bar{r}},{\bar{z}})|\,d{\bar{r}}\,d{\bar{z}}\Bigr \Vert _{L_T^3 L^3(\Omega )}\\&\lesssim \Bigl \Vert \int _\Omega \bigl \Vert \frac{1}{t} \exp \bigl (-\frac{|\zeta |^2}{5t}\bigr )\bigr \Vert _{L_T^3} \cdot {\bar{r}}^{\frac{2}{3}}|\omega ^\theta _0({\bar{r}},{\bar{z}})|\,d{\bar{r}}\,d{\bar{z}}\Bigr \Vert _{L^3(\Omega )}\\&\lesssim \bigl \Vert {r}^{\frac{2}{3}}\omega ^\theta _0\bigr \Vert _{L^{\frac{3}{2}}(\Omega )} \bigl \Vert \bigl (\frac{1}{r^2+z^2}\bigr )^{\frac{2}{3}}\bigr \Vert _{L^{\frac{3}{2},\infty }(\Omega )} \lesssim \Vert \omega ^\theta _0\Vert _{L^{\frac{3}{2}}}. \end{aligned} \nonumber \\ \end{aligned}$$

(A.8)

And using (2.9) with $\alpha =-1,~\beta =\frac{1}{3}$ gives

$$\begin{aligned} \begin{aligned} \bigl \Vert t^{\frac{1}{2}} r^{-1}S(t)\omega ^\theta _0\bigr \Vert _{L_T^2 L^2(\Omega )}&\lesssim \Bigl \Vert \int _\Omega \bigl \Vert \frac{1}{4\pi t}\cdot \frac{1}{t^{\frac{1}{3}}} \exp \bigl (-\frac{|\zeta |^2}{5t}\bigr )\bigr \Vert _{L_T^2} \cdot {\bar{r}}^{\frac{2}{3}}\omega ^\theta _0({\bar{r}},{\bar{z}})\,d{\bar{r}}\,d{\bar{z}}\Bigr \Vert _{L^2(\Omega )}\\&\lesssim \bigl \Vert {r}^{\frac{2}{3}}\omega ^\theta _0\bigr \Vert _{L^{\frac{3}{2}}(\Omega )} \bigl \Vert \bigl (\frac{1}{r^2+z^2}\bigr )^{\frac{5}{6}}\bigr \Vert _{L^{\frac{6}{5},\infty }(\Omega )} \lesssim \Vert \omega ^\theta _0\Vert _{L^{\frac{3}{2}}}, \end{aligned}\nonumber \\ \end{aligned}$$

(A.9)

To estimate the terms having derivatives, we need to use point-wise estimates (2.12) and (2.13). Indeed, by using (2.12) and (2.13) with $\gamma =\frac{2}{3},~\epsilon =-\frac{2}{3}$, we deduce

$$\begin{aligned}{} & {} \bigl \Vert t^{\frac{1}{6}}r^{\frac{1}{6}}{\widetilde{\nabla }} S(t)\omega ^\theta _0\bigr \Vert _{L_T^2 L^2} =\Bigl \Vert t^{\frac{1}{6}}r^{\frac{2}{3}}\frac{1}{4\pi t} \int _\Omega \frac{{\bar{r}}^{\frac{1}{2}-\frac{2}{3}}}{r^{\frac{1}{2}}} \exp \Bigl (-\frac{(r-{\bar{r}})^2+(z-{\bar{z}})^2}{4t}\Bigr ) \cdot {\bar{r}}^{\frac{2}{3}}\omega ^\theta _0({\bar{r}},{\bar{z}})\nonumber \\{} & {} \qquad \cdot \Bigl (-\frac{t}{r^2{\bar{r}}}{H}'\bigl (\frac{t}{r{\bar{r}}}\bigr ) -\bigl (\frac{1}{2r}+\frac{r-{\bar{r}}}{2t}\bigr ) H\bigl (\frac{t}{r{\bar{r}}}\bigr ), ~-\frac{z-{\bar{z}}}{2t}{H}\bigl (\frac{t}{r{\bar{r}}}\bigr )\Bigr ) \,d{\bar{r}}\,d{\bar{z}}\Bigr \Vert _{L_T^2 L^2(\Omega )}\nonumber \\{} & {} \quad \lesssim \Bigl \Vert \int _\Omega \bigl \Vert t^{\frac{1}{6}}\cdot \frac{1}{4\pi t}\cdot \frac{1}{t^{\frac{1}{2}}} \exp \bigl (-\frac{|\zeta |^2}{5t}\bigr )\bigr \Vert _{L_T^2} \cdot {\bar{r}}^{\frac{2}{3}}\omega ^\theta _0({\bar{r}},{\bar{z}})\,d{\bar{r}}\,d{\bar{z}}\Bigr \Vert _{L^2(\Omega )}\nonumber \\{} & {} \quad \lesssim \bigl \Vert {r}^{\frac{2}{3}}\omega ^\theta _0\bigr \Vert _{L^{\frac{3}{2}}(\Omega )} \bigl \Vert \bigl (\frac{1}{r^2+z^2}\bigr )^{\frac{5}{6}}\bigr \Vert _{L^{\frac{6}{5},\infty }(\Omega )} \lesssim \Vert \omega ^\theta _0\Vert _{L^{\frac{3}{2}}}. \end{aligned}$$

(A.10)

Similarly, by using (2.12) and (2.13) with $\gamma =\epsilon =0$, we obtain

$$\begin{aligned} \bigl \Vert t^{\frac{1}{6}}{\widetilde{\nabla }}S(t)u^\theta _0\bigr \Vert _{L_T^{\frac{12}{5}}L^{\frac{12}{5}}(\Omega )}\lesssim & {} \Bigl \Vert \int _\Omega \bigl \Vert t^{\frac{1}{6}}\cdot \frac{1}{4\pi t}\cdot \frac{1}{t^{\frac{1}{2}}} \exp \bigl (-\frac{|\zeta |^2}{5t}\bigr )\bigr \Vert _{L_T^{\frac{12}{5}}} \cdot u^\theta _0({\bar{r}},{\bar{z}})\,d{\bar{r}}\,d{\bar{z}}\Bigr \Vert _{L^{\frac{12}{5}}(\Omega )}\nonumber \\\lesssim & {} \Vert u^\theta _0\Vert _{L^2(\Omega )} \bigl \Vert \bigl (\frac{1}{r^2+z^2}\bigr )^{\frac{11}{12}}\bigr \Vert _{L^{\frac{12}{11},\infty }(\Omega )} \lesssim \Vert u^\theta _0\Vert _{L^2(\Omega )}. \end{aligned}$$

(A.11)

The estimates (A.2)–(A.11) guarantees the existence of some universal constant $C_1$ such that for any $T>0$, there holds

$$\begin{aligned} \bigl \Vert \bigl (S(t)\omega ^\theta _0,S(t)u^\theta _0\bigr ) \bigr \Vert _{X_T}\le C_1\bigl (\Vert \omega ^\theta _0\Vert _{L^{\frac{3}{2}}}+\Vert \omega ^\theta _0\Vert _{L^1(\Omega )} +\Vert u^\theta _0\Vert _{L^2(\Omega )}\bigr ). \end{aligned}$$

(A.12)

$\bullet$ The estimate for the bilinear part in (A.1). For any given $(\omega ^\theta _1,u^\theta _1),~(\omega ^\theta _2,u^\theta _2)\in X_T$, and any $t\in [0,T]$, let us consider the bilinear map

where $\widetilde{u}_i$ is the velocity field determined by $\omega ^\theta _i$ via the Biot-Savart law. In the following, we may abuse the notation $\Vert \omega ^\theta \Vert _{X_T}=\Vert (\omega ^\theta ,0)\Vert _{X_T}$ and $\Vert u^\theta \Vert _{X_T}=\Vert (0,u^\theta )\Vert _{X_T}$ for convenience.

First, we deduce from (2.5) that for any $t\in ]0,T]$, there holds

$$\begin{aligned} \Vert {\mathcal {F}}^\omega (t)\Vert _{L^1(\Omega )}&\lesssim \int _0^t\frac{\Vert {\widetilde{u}}_1(s)\cdot \omega ^\theta _2(s)\Vert _{L^1(\Omega )}}{(t-s)^{\frac{1}{2}}} +\frac{\Vert r^{-1}u^\theta _1(s)u^\theta _2(s)\Vert _{L^1(\Omega )} }{(t-s)^{\frac{1}{2}}}\,ds\\&\lesssim \int _0^t\frac{s^{\frac{1}{4}}\Vert {\widetilde{u}}_1(s)\Vert _{L^4(\Omega )} s^{\frac{1}{4}}\Vert \omega ^\theta _2(s)\Vert _{L^{\frac{4}{3}}(\Omega )}}{(t-s)^{\frac{1}{2}}\cdot s^{\frac{1}{2}}} +\frac{\Vert u^\theta _1(s)\Vert _{L^2(\Omega )}s^{\frac{1}{2}}\Vert r^{-1}u^\theta _2(s)\Vert _{L^2(\Omega )}}{(t-s)^{\frac{1}{2}}\cdot s^{\frac{1}{2}}}\,ds\\&\lesssim \Vert \omega ^\theta _1\Vert _{X_T}\Vert \omega ^\theta _2\Vert _{X_T}+\Vert u^\theta _1\Vert _{X_T}\Vert u^\theta _2\Vert _{X_T}, \end{aligned}$$

where in the last step we have used Lemma 2.1 to get

$$\begin{aligned} s^{\frac{1}{2}-\frac{1}{p}}\Vert {\widetilde{u}}_1(s)\Vert _{L^p(\Omega )} \lesssim s^{\frac{1}{2}-\frac{1}{p}}\Vert \omega ^\theta _1(s)\Vert _{L^{\frac{2p}{p+2}}(\Omega )}\lesssim \Vert \omega ^\theta _1\Vert _{X_T}, \quad \forall \ p\in [2,\infty [, \end{aligned}$$

(A.13)

which will be used frequently in the following. By using (2.5) again and (A.13), we get

$$\begin{aligned} t\Vert {\mathcal {F}}^\omega&(t)\Vert _{L^\infty } \lesssim t\int _0^{\frac{t}{2}}\frac{\Vert {\widetilde{u}}_1(s)\cdot \omega ^\theta _2(s)\Vert _{L^1(\Omega )} }{(t-s)^{\frac{3}{2}}} +\frac{\Vert r^{-1}u^\theta _1(s)u^\theta _2(s)\Vert _{L^1(\Omega )} }{(t-s)^{\frac{3}{2}}}\,ds\\&\quad +t\int _{\frac{t}{2}}^t\frac{\Vert {\widetilde{u}}_1(s)\cdot \omega ^\theta _2(s)\Vert _{L^4(\Omega )} }{(t-s)^{\frac{3}{4}}} +\frac{\Vert r^{-1}u^\theta _1(s)u^\theta _2(s)\Vert _{L^4(\Omega )} }{(t-s)^{\frac{3}{4}}}\,ds\\&\lesssim t\int _0^{\frac{t}{2}}\frac{s^{\frac{1}{4}}\Vert {\widetilde{u}}_1(s)\Vert _{L^4(\Omega )} s^{\frac{1}{4}}\Vert \omega ^\theta _2(s)\Vert _{L^{\frac{4}{3}}(\Omega )}}{(t-s)^{\frac{3}{2}}\cdot s^{\frac{1}{2}}} +\frac{\Vert u^\theta _1(s)\Vert _{L^2(\Omega )}s^{\frac{1}{2}}\Vert r^{-1}u^\theta _2(s)\Vert _{L^2(\Omega )}}{(t-s)^{\frac{3}{2}}\cdot s^{\frac{1}{2}}}\,ds\\&\quad +t\int _{\frac{t}{2}}^t\frac{s^{\frac{1}{4}}\Vert {\widetilde{u}}_1(s)\Vert _{L^4(\Omega )} s\Vert \omega ^\theta _2(s)\Vert _{L^{\infty }(\Omega )}}{(t-s)^{\frac{3}{4}}\cdot s^{\frac{5}{4}}} +\frac{s^{\frac{1}{2}}\Vert u^\theta _1(s)\Vert _{L^\infty (\Omega )} s^{\frac{3}{4}}\Vert r^{-1}u^\theta _2(s)\Vert _{L^4(\Omega )}}{(t-s)^{\frac{3}{4}}\cdot s^{\frac{5}{4}}}\,ds\\&\lesssim \Vert \omega ^\theta _1\Vert _{X_T}\Vert \omega ^\theta _2\Vert _{X_T}+\Vert u^\theta _1\Vert _{X_T}\Vert u^\theta _2\Vert _{X_T}. \end{aligned}$$

Similarly, by using (2.4) with $\alpha =\frac{2}{3},~\beta =-\frac{2}{3}$ and (A.13), we obtain

$$\begin{aligned}&\Vert {\mathcal {F}}^\omega (t)\Vert _{L^{\frac{3}{2}}} =\bigl \Vert r^{\frac{2}{3}}{\mathcal {F}}^\omega (t)\bigr \Vert _{L^{\frac{3}{2}}(\Omega )}\\&\quad \lesssim \int _0^t\frac{\bigl \Vert r^{\frac{2}{3}}{\widetilde{u}}_1(s)\cdot \omega ^\theta _2(s)\bigr \Vert _{L^{\frac{12}{11}}(\Omega )} }{(t-s)^{\frac{3}{4}}} +\frac{\Vert r^{-\frac{1}{3}}u^\theta _1(s)u^\theta _2(s)\Vert _{L^{\frac{4}{3}}(\Omega )}}{(t-s)^{\frac{7}{12}}}\,ds\\&\quad \lesssim \int _0^t\frac{s^{\frac{1}{4}}\Vert {\widetilde{u}}_1(s)\Vert _{L^4(\Omega )} \Vert \omega ^\theta _2(s)\Vert _{L^{\frac{3}{2}}}}{(t-s)^{\frac{3}{4}}\cdot s^{\frac{1}{4}}}\\&\qquad +\frac{s^{\frac{1}{4}}\Vert u^\theta _1(s)\Vert _{L^4(\Omega )} \bigl (s^{\frac{1}{2}}\Vert r^{-1}u^\theta _2(s)\Vert _{L^2(\Omega )}\bigr )^{\frac{1}{3}} \Vert u^\theta _2(s)\Vert _{L^2(\Omega )}^{\frac{2}{3}}}{(t-s)^{\frac{7}{12}}\cdot s^{\frac{5}{12}}}\,ds\\&\quad \lesssim \Vert \omega ^\theta _1\Vert _{X_T}\Vert \omega ^\theta _2\Vert _{X_T}+\Vert u^\theta _1\Vert _{X_T}\Vert u^\theta _2\Vert _{X_T}. \end{aligned}$$

We deduce from the above three estimates that

$$\begin{aligned} \begin{aligned} \sup _{0<t\le T}\Bigl (\Vert {\mathcal {F}}^\omega (t)\Vert _{L^1(\Omega )\bigcap L^{\frac{3}{2}}} + t\Vert {\mathcal {F}}^\omega (t)\Vert _{L^\infty }\Bigr ) \lesssim \Vert \omega ^\theta _1\Vert _{X_T}\Vert \omega ^\theta _2\Vert _{X_T}+\Vert u^\theta _1\Vert _{X_T}\Vert u^\theta _2\Vert _{X_T}. \end{aligned} \nonumber \\ \end{aligned}$$

(A.14)

The ${\mathcal {F}}^u$ can be estimated similarly. Indeed, by using (2.5), (2.7) and (A.13), we have

$$\begin{aligned} \Vert {\mathcal {F}}^u(t)\Vert _{L^2(\Omega )} \lesssim \int _0^t\frac{s^{\frac{1}{4}}\Vert {\widetilde{u}}_1(s)\Vert _{L^4(\Omega )} s^{\frac{1}{4}}\Vert u^\theta _2(s)\Vert _{L^4(\Omega )} }{(t-s)^{\frac{1}{2}}\cdot s^{\frac{1}{2}}}\,ds \lesssim \Vert \omega ^\theta _1\Vert _{X_T}\Vert u^\theta _2\Vert _{X_T}, \\ t^{\frac{1}{2}}\Vert {\mathcal {F}}^u(t)\Vert _{L^{\infty }(\Omega )}\lesssim t^{\frac{1}{2}}\int _0^t\frac{s^{\frac{1}{4}}\Vert {\widetilde{u}}_1(s)\Vert _{L^4(\Omega )} s^{\frac{5}{12}}\Vert u^\theta _2(s)\Vert _{L^{12}(\Omega )} }{(t-s)^{\frac{5}{6}}\cdot s^{\frac{2}{3}}}\,ds \lesssim \Vert \omega ^\theta _1\Vert _{X_T}\Vert u^\theta _2\Vert _{X_T}. \end{aligned}$$

While thanks to (2.4) and (2.6) with $\alpha =-1,~\beta =1$, there holds

$$\begin{aligned} t^{\frac{1}{2}}\Vert r^{-1}{\mathcal {F}}^u(t)\Vert _{L^2(\Omega )} \lesssim t^{\frac{1}{2}}\int _0^t \frac{s^{\frac{1}{4}}\Vert {\widetilde{u}}_1(s)\Vert _{L^4(\Omega )} s^{\frac{1}{2}}\Vert r^{-1}u^\theta _2(s)\Vert _{L^2(\Omega )} }{(t-s)^{\frac{3}{4}}\cdot s^{\frac{3}{4}}}\,ds \lesssim \Vert \omega ^\theta _1\Vert _{X_T}\Vert u^\theta _2\Vert _{X_T}, \\ t^{\frac{3}{4}}\Vert r^{-1}{\mathcal {F}}^u(t)\Vert _{L^4(\Omega )} \lesssim t^{\frac{3}{4}}\int _0^t \frac{s^{\frac{1}{6}}\Vert {\widetilde{u}}_1(s)\Vert _{L^3(\Omega )} s^{\frac{3}{4}}\Vert r^{-1}u^\theta _2(s)\Vert _{L^4(\Omega )} }{(t-s)^{\frac{5}{6}}\cdot s^{\frac{11}{12}}}\,ds \lesssim \Vert \omega ^\theta _1\Vert _{X_T}\Vert u^\theta _2\Vert _{X_T}. \end{aligned}$$

The estimate for $t^{\frac{1}{2}}\Vert {\widetilde{\nabla }}{\mathcal {F}}^u(t)\Vert _{L^2(\Omega )}$ follows from (2.8) with $\gamma =\epsilon =0$ and (A.13) that

$$\begin{aligned} t^{\frac{1}{2}}\Vert {\widetilde{\nabla }}{\mathcal {F}}^u(t)\Vert _{L^2(\Omega )} \lesssim t^{\frac{1}{2}}\int _0^t \frac{s^{\frac{1}{4}}\Vert {\widetilde{u}}_1(s)\Vert _{L^4(\Omega )} s^{\frac{1}{2}}\bigl \Vert \bigl ({\widetilde{\nabla }} u^\theta _2, r^{-1}u^\theta _2\bigr )(s)\bigr \Vert _{L^2(\Omega )}}{(t-s)^{\frac{3}{4}}\cdot s^{\frac{3}{4}}}\,ds \lesssim \Vert \omega ^\theta _1\Vert _{X_T}\Vert u^\theta _2\Vert _{X_T}. \end{aligned}$$

Then we can deduce from the above estimates that for any $t\le T$, there holds

$$\begin{aligned} \begin{aligned} \Vert {\mathcal {F}}^u(t)\Vert _{L^2(\Omega )} +t^{\frac{1}{2}}&\Vert {\mathcal {F}}^u(t)\Vert _{L^\infty (\Omega )} +t^{\frac{1}{2}}\Vert r^{-1}{\mathcal {F}}^u(t)\Vert _{L^2(\Omega )}\\&+t^{\frac{1}{2}}\Vert {\widetilde{\nabla }}{\mathcal {F}}^u(t)\Vert _{L^2(\Omega )} +t^{\frac{3}{4}}\Vert r^{-1}{\mathcal {F}}^u(t)\Vert _{L^4(\Omega )} \lesssim \Vert \omega ^\theta _1\Vert _{X_T}\Vert u^\theta _2\Vert _{X_T}. \end{aligned} \nonumber \\ \end{aligned}$$

(A.15)

Now, let us turn to handle the space-time integral terms, by using Hardy-Littlewood-Sobolev inequality. First, by using (2.5), (2.6) with $\alpha =-\frac{3}{5},~\beta =\frac{3}{5}$ and (A.13), we obtain

$$\begin{aligned} \begin{aligned}&\Vert {\mathcal {F}}^u\Vert _{L_T^4 L^4(\Omega )} +\bigl \Vert r^{-\frac{3}{5}}{\mathcal {F}}^u\bigr \Vert _{L_T^{\frac{5}{2}}L^{\frac{5}{2}}(\Omega )}\\&\lesssim \Bigl \Vert \int _0^t\frac{\Vert {\widetilde{u}}_1(s) u^\theta _2(s)\Vert _{L^{2}(\Omega )} }{(t-s)^{\frac{3}{4}}}\,ds\Bigr \Vert _{L_T^4} +\Bigl \Vert \int _0^t\frac{\bigl \Vert r^{-\frac{3}{5}}{\widetilde{u}}_1(s)u^\theta _2(s)\bigr \Vert _{L^{\frac{20}{13}}(\Omega )}}{(t-s)^{\frac{3}{4}}}\,ds\Bigr \Vert _{L_T^{\frac{5}{2}}}\\&\lesssim \Vert \widetilde{u}_1\Vert _{L_T^4L^4(\Omega )}\Vert u^\theta _2\Vert _{L^4_T L^{4}(\Omega )}\Bigl \Vert \frac{1}{|\cdot |^{\frac{3}{4}}}\Bigr \Vert _{L_T^{\frac{4}{3},\infty }} +\Vert \widetilde{u}_1\Vert _{L_T^4L^4(\Omega )} \Vert r^{-\frac{3}{5}}u^\theta _2\Vert _{L^{\frac{5}{2}}_T L^{\frac{5}{2}}(\Omega )}\Bigl \Vert \frac{1}{|\cdot |^{\frac{3}{4}}}\Bigr \Vert _{L_T^{\frac{4}{3},\infty }}\\&\lesssim \Vert \omega ^\theta _1\Vert _{X_T}\Vert u^\theta _2\Vert _{X_T}. \end{aligned} \nonumber \\ \end{aligned}$$

(A.16)

While by using (2.4) with $\alpha =\beta =0$ and $\alpha =0,~\beta =-\frac{2}{5}$, we get

$$\begin{aligned} \Vert {\mathcal {F}}^\omega \Vert _{L_T^2 L^2(\Omega )} \lesssim&\Bigl \Vert \int _0^t\frac{\Vert {\widetilde{u}}_1(s)\Vert _{L^4(\Omega )} \Vert \omega ^\theta _2(s)\Vert _{L^{2}(\Omega )} }{(t-s)^{\frac{3}{4}}}\,ds \\&+\int _0^t \frac{\Vert u^\theta _1(s)\Vert _{L^4(\Omega )} \Vert r^{-\frac{3}{5}} u^\theta _2(s)\Vert _{L^{\frac{5}{2}}(\Omega )}}{(t-s)^{\frac{17}{20}}}\,ds\Bigr \Vert _{L_T^2}\\ \lesssim&\Vert \widetilde{u}_1\Vert _{L_T^4L^4(\Omega )}\Vert \omega ^\theta _2\Vert _{L^2_T L^{2}(\Omega )} \Bigl \Vert \frac{1}{|\cdot |^{\frac{3}{4}}}\Bigr \Vert _{L_T^{\frac{4}{3},\infty }} \\&+\Vert u^\theta _1\Vert _{L^4_T L^4(\Omega )}\Vert r^{-\frac{3}{5}} u^\theta _2\Vert _{L^{\frac{5}{2}}_T L^{\frac{5}{2}}(\Omega )} \Bigl \Vert \frac{1}{|\cdot |^{\frac{17}{20}}}\Bigr \Vert _{L_T^{\frac{20}{17},\infty }}\\ \lesssim&\Vert \omega ^\theta _1\Vert _{X_T}\Vert \omega ^\theta _2\Vert _{X_T}+\Vert u^\theta _1\Vert _{X_T}\Vert u^\theta _2\Vert _{X_T}. \end{aligned}$$

Thanks to (2.4) with $\alpha =\frac{1}{2},~\beta =-\frac{1}{2}$ and $\alpha =\frac{1}{2},~\beta =-1$, there holds

$$\begin{aligned} \Vert {\mathcal {F}}^\omega (t)\Vert _{L_T^4 L^2}&=\bigl \Vert r^{\frac{1}{2}}{\mathcal {F}}^\omega (t)\bigr \Vert _{L_T^4 L^2(\Omega )}\\&\lesssim \Bigl \Vert \int _0^t\frac{\Vert r^{\frac{1}{2}}{\widetilde{u}}_1(s)\cdot \omega ^\theta _2(s)\Vert _{L^{\frac{4}{3}}(\Omega )} }{(t-s)^{\frac{3}{4}}} +\frac{\Vert u^\theta _1(s)u^\theta _2(s)\Vert _{L^2(\Omega )}}{(t-s)^{\frac{3}{4}}}\,ds\Bigr \Vert _{L_T^4}\\&\lesssim \Vert \widetilde{u}_1\Vert _{L_T^4L^4(\Omega )}\Vert \omega ^\theta _2\Vert _{L^4_T L^{2}} \Bigl \Vert \frac{1}{|\cdot |^{\frac{3}{4}}}\Bigr \Vert _{L_T^{\frac{4}{3},\infty }} +\Vert u^\theta _1\Vert _{L^4_T L^4(\Omega )}\Vert u^\theta _2\Vert _{L^4_T L^4(\Omega )} \Bigl \Vert \frac{1}{|\cdot |^{\frac{3}{4}}}\Bigr \Vert _{L_T^{\frac{4}{3},\infty }}\\&\lesssim \Vert \omega ^\theta _1\Vert _{X_T}\Vert \omega ^\theta _2\Vert _{X_T}+\Vert u^\theta _1\Vert _{X_T}\Vert u^\theta _2\Vert _{X_T}. \end{aligned}$$

And similarly

$$\begin{aligned} \bigl \Vert r^{\frac{2}{3}}&{\mathcal {F}}^\omega \Vert _{L_T^3 L^3(\Omega )} \lesssim \Bigl \Vert \int _0^t\frac{\Vert {\widetilde{u}}_1(s)\Vert _{L^{\frac{12}{5}}(\Omega )} \bigl \Vert r^{\frac{2}{3}}\omega ^\theta _2(s)\bigr \Vert _{L^3(\Omega )}}{(t-s)^{\frac{11}{12}}} +\frac{\Vert u^\theta _1(s)\Vert _{L^4(\Omega )}\Vert u^\theta _2(s)\Vert _{L^4(\Omega )}}{(t-s)^{\frac{5}{6}}}\,ds\Bigr \Vert _{L_T^3}\\&\lesssim \Vert \omega ^\theta _1\Vert _{L_T^{12}L^{\frac{12}{11}}(\Omega )} \bigl \Vert r^{\frac{2}{3}}\omega ^\theta _2\bigr \Vert _{L^3_T L^3(\Omega )} \Bigl \Vert \frac{1}{|\cdot |^{\frac{11}{12}}}\Bigr \Vert _{L_T^{\frac{12}{11},\infty }} +\Vert u^\theta _1\Vert _{L^4_T L^4(\Omega )}\Vert u^\theta _2\Vert _{L^4_T L^4(\Omega )} \Bigl \Vert \frac{1}{|\cdot |^{\frac{5}{6}}}\Bigr \Vert _{L_T^{\frac{6}{5},\infty }}\\&\lesssim \Vert \omega ^\theta _1\Vert _{X_T}\Vert \omega ^\theta _2\Vert _{X_T}+\Vert u^\theta _1\Vert _{X_T}\Vert u^\theta _2\Vert _{X_T}. \end{aligned}$$

Combining the above three estimates, we deduce

$$\begin{aligned} \begin{aligned} \Vert {\mathcal {F}}^\omega \Vert _{L_T^2 L^2(\Omega )\bigcap L_T^4 L^2} +\bigl \Vert r^{\frac{2}{3}}{\mathcal {F}}^\omega \Vert _{L_T^3 L^3(\Omega )} \lesssim \Vert \omega ^\theta _1\Vert _{X_T}\Vert \omega ^\theta _2\Vert _{X_T}+\Vert u^\theta _1\Vert _{X_T}\Vert u^\theta _2\Vert _{X_T}. \end{aligned} \nonumber \\ \end{aligned}$$

(A.17)

At last, let us handle the terms having time as weight. For $\bigl \Vert t^{\frac{1}{3}}{\mathcal {F}}^\omega \bigr \Vert _{L^3_T L^3(\Omega )}$, we write

(A.18)

And one should keep in mind that, for $s\in \bigl [0,\frac{t}{2}\bigr ]$ there holds $t\thicksim t-s$, while for $s\in \bigl [\frac{t}{2},t\bigr ]$ there holds $t\thicksim s$. Then the first term in (A.18) can be handled by using (2.4) that

$$\begin{aligned} {i} _1&\lesssim \Bigl \Vert \int _0^{\frac{t}{2}}(t-s)^{\frac{1}{3}} \Bigl (\frac{\Vert {\widetilde{u}}_1(s) \omega ^\theta _2(s)\Vert _{L^{\frac{4}{3}}(\Omega )}}{(t-s)^{\frac{11}{12}}} +\frac{\Vert u^\theta _1(s)u^\theta _2(s)\Vert _{L^2(\Omega )}}{(t-s)^{\frac{7}{6}}}\Bigr )\, ds\Bigr \Vert _{L_T^3}\\&\lesssim \Bigl \Vert \int _0^{\frac{t}{2}} \frac{\Vert {\widetilde{u}}_1(s)\Vert _{L^4(\Omega )} \Vert \omega ^\theta _2(s)\Vert _{L^2(\Omega )}}{(t-s)^{\frac{7}{12}}} +\frac{\Vert u^\theta _1(s)\Vert _{L^4(\Omega )}\Vert u^\theta _2(s)\Vert _{L^4(\Omega )}}{(t-s)^{\frac{5}{6}}}\, ds\Bigr \Vert _{L_T^3}\\&\lesssim \Vert \omega ^\theta _1\Vert _{L^4_T L^{\frac{4}{3}}(\Omega )} \Vert \omega ^\theta _2\Vert _{L_T^2 L^2(\Omega )} \Bigl \Vert \frac{1}{|\cdot |^{\frac{7}{12}}}\Bigr \Vert _{L_T^{\frac{12}{7},\infty }} +\Vert u^\theta _1\Vert _{L^4_T L^4(\Omega )} \Vert u^\theta _2\Vert _{L_T^4 L^4(\Omega )} \Bigl \Vert \frac{1}{|\cdot |^{\frac{5}{6}}}\Bigr \Vert _{L_T^{\frac{6}{5},\infty }}\\&\lesssim \Vert \omega ^\theta _1\Vert _{X_T}\Vert \omega ^\theta _2\Vert _{X_T}+\Vert u^\theta _1\Vert _{X_T}\Vert u^\theta _2\Vert _{X_T}. \end{aligned}$$

As for the second term in (A.18), in view of (2.5), there holds

$$\begin{aligned} {i} _2\lesssim&\Bigl \Vert \int _{\frac{t}{2}}^t s^{\frac{1}{3}} \Bigl (\frac{\Vert {\widetilde{u}}_1(s) \omega ^\theta _2(s)\Vert _{L^{\frac{12}{7}}(\Omega )}}{(t-s)^{\frac{3}{4}}}\\&+\frac{\Vert r^{-\frac{3}{5}}u^\theta _1(s)u^\theta _2(s)\Vert _{L^{\frac{30}{17}}(\Omega )}}{(t-s)^{\frac{14}{15}}}\Bigr )\, ds\Bigr \Vert _{L_T^3}\\ \lesssim&\Bigl \Vert \int _{\frac{t}{2}}^t \frac{\Vert {\widetilde{u}}_1(s)\Vert _{L^4(\Omega )}\Vert s^{\frac{1}{3}}\omega ^\theta _2(s)\Vert _{L^3(\Omega )}}{(t-s)^{\frac{3}{4}}} +\frac{s^{\frac{1}{3}}\Vert u^\theta _1(s)\Vert _{L^6(\Omega )}\Vert r^{-\frac{3}{5}}u^\theta _2(s)\Vert _{L^{\frac{5}{2}}(\Omega )}}{(t-s)^{\frac{14}{15}}}\, ds\Bigr \Vert _{L_T^3}\\ \lesssim&\Vert \omega ^\theta _1\Vert _{L^4_T L^{\frac{4}{3}}(\Omega )} \bigl \Vert t^{\frac{1}{3}}\omega ^\theta _2\bigr \Vert _{L_T^3 L^3(\Omega )} \Bigl \Vert \frac{1}{|\cdot |^{\frac{3}{4}}}\Bigr \Vert _{L_T^{\frac{4}{3},\infty }}\\&+\bigl \Vert t^{\frac{1}{3}}\Vert u^\theta _1\Vert _{L^6(\Omega )}\bigr \Vert _{L^\infty _T} \bigl \Vert r^{-\frac{3}{5}}u^\theta _2\bigr \Vert _{L_T^{\frac{5}{2}} L^{\frac{5}{2}}(\Omega )} \Bigl \Vert \frac{1}{|\cdot |^{\frac{14}{15}}}\Bigr \Vert _{L_T^{\frac{15}{14},\infty }}\\ \lesssim&\Vert \omega ^\theta _1\Vert _{X_T}\Vert \omega ^\theta _2\Vert _{X_T}+\Vert u^\theta _1\Vert _{X_T}\Vert u^\theta _2\Vert _{X_T}. \end{aligned}$$

Substituting the above two estimates into (A.18) gives

$$\begin{aligned} \bigl \Vert t^{\frac{1}{3}}{\mathcal {F}}^\omega \bigr \Vert _{L^3_T L^3(\Omega )}\lesssim \Vert \omega ^\theta _1\Vert _{X_T}\Vert \omega ^\theta _2\Vert _{X_T}+\Vert u^\theta _1\Vert _{X_T}\Vert u^\theta _2\Vert _{X_T}. \end{aligned}$$

(A.19)

Similarly, we decompose the term $\bigl \Vert t^{\frac{1}{2}}r^{-1}{\mathcal {F}}^\omega \bigr \Vert _{L^2_T L^2(\Omega )}$ into two parts:

(A.20)

By using (2.4) with $\alpha =-1,~\beta =0$, we obtain

$$\begin{aligned}&{ii} _1\lesssim \Bigl \Vert \int _0^{\frac{t}{2}}(t-s)^{\frac{1}{2}} \frac{\Vert {\widetilde{u}}_1(s)\omega ^\theta _2(s)\Vert _{L^{\frac{4}{3}}(\Omega )} +\Vert r^{-1}u^\theta _1(s)u^\theta _2(s)\Vert _{L^{\frac{4}{3}}(\Omega )}}{(t-s)^{\frac{5}{4}}} \, ds\Bigr \Vert _{L_T^2}\\&\lesssim \Bigl \Vert \int _0^{\frac{t}{2}} \frac{\Vert {\widetilde{u}}_1(s)\Vert _{L^4(\Omega )} \Vert \omega ^\theta _2(s)\Vert _{L^2(\Omega )}+\bigl \Vert r^{-\frac{3}{5}}u^\theta _1(s)\bigr \Vert _{L^{\frac{5}{2}}(\Omega )} \bigl \Vert r^{-\frac{3}{5}}u^\theta _2(s)\bigr \Vert _{L^{\frac{5}{2}}(\Omega )}^{\frac{2}{3}} \Vert u^\theta _2(s)\Vert ^{\frac{1}{3}}_{L^4(\Omega )}}{(t-s)^{\frac{3}{4}}} \, ds\Bigr \Vert _{L_T^2}\\&\lesssim \Bigl (\Vert \omega ^\theta _1\Vert _{L_T^4L^{\frac{4}{3}}(\Omega )} \Vert \omega ^\theta _2\Vert _{L^2_T L^2(\Omega )}+\bigl \Vert r^{-\frac{3}{5}}u^\theta _1\bigr \Vert _{L_T^{\frac{5}{2}}L^{\frac{5}{2}}(\Omega )} \bigl \Vert r^{-\frac{3}{5}}u^\theta _2\bigr \Vert _{L_T^{\frac{5}{2}}L^{\frac{5}{2}}(\Omega )}^{\frac{2}{3}} \Vert u^\theta _2\Vert ^{\frac{1}{3}}_{L_T^4L^4(\Omega )}\Bigr ) \Bigl \Vert \frac{1}{|\cdot |^{\frac{3}{4}}}\Bigr \Vert _{L_T^{\frac{4}{3},\infty }}\\&\lesssim \Vert \omega ^\theta _1\Vert _{X_T}\Vert \omega ^\theta _2\Vert _{X_T}+\Vert u^\theta _1\Vert _{X_T}\Vert u^\theta _2\Vert _{X_T}. \end{aligned}$$

While by using (2.4) with $\alpha =-1,~\beta =1$ and $\alpha =-1,~\beta =\frac{7}{15}$, we get

$$\begin{aligned} {ii} _2&\lesssim \Bigl \Vert \int _{\frac{t}{2}}^t s^{\frac{1}{2}} \frac{\Vert {\widetilde{u}}_1(s)\Vert _{L^4(\Omega )} \Vert r^{-1} \omega ^\theta _2(s)\Vert _{L^2(\Omega )}}{(t-s)^{\frac{3}{4}}} +s^{\frac{1}{2}}\frac{\bigl \Vert r^{-\frac{3}{5}}u^\theta _1(s)\bigr \Vert _{L^{\frac{5}{2}}(\Omega )} \bigl \Vert r^{-\frac{13}{15}}u^\theta _2(s)\bigr \Vert _{L^{\frac{10}{3}}(\Omega )}}{(t-s)^{\frac{29}{30}}}\, ds\Bigr \Vert _{L_T^2}\\&\lesssim \Vert \omega ^\theta _1\Vert _{L_T^4L^{\frac{4}{3}}(\Omega )} \bigl \Vert t^{\frac{1}{2}}r^{-1}\omega ^\theta _2\Vert _{L^2_T L^2(\Omega )}\Bigl \Vert \frac{1}{|\cdot |^{\frac{3}{4}}}\Bigr \Vert _{L_T^{\frac{4}{3},\infty }}\\&\qquad \qquad +\bigl \Vert r^{-\frac{3}{5}}u^\theta _1\bigr \Vert _{L_T^{\frac{5}{2}}L^{\frac{5}{2}}(\Omega )} \bigl \Vert t^{\frac{3}{4}}\Vert r^{-1}u^\theta _2\Vert _{L^4(\Omega )}\bigr \Vert _{L_T^\infty }^{\frac{2}{3}} \bigl \Vert r^{-\frac{3}{5}}u^\theta _2\bigr \Vert _{L_T^{\frac{5}{2}}L^{\frac{5}{2}}(\Omega )}^{\frac{1}{3}} \Bigl \Vert \frac{1}{|\cdot |^{\frac{29}{30}}}\Bigr \Vert _{L_T^{\frac{30}{29},\infty }}\\&\lesssim \Vert \omega ^\theta _1\Vert _{X_T}\Vert \omega ^\theta _2\Vert _{X_T}+\Vert u^\theta _1\Vert _{X_T}\Vert u^\theta _2\Vert _{X_T}. \end{aligned}$$

Substituting the above two estimates into (A.20) gives

$$\begin{aligned} \bigl \Vert t^{\frac{1}{2}}r^{-1}{\mathcal {F}}^\omega \bigr \Vert _{L^2_T L^2(\Omega )}\lesssim \Vert \omega ^\theta _1\Vert _{X_T}\Vert \omega ^\theta _2\Vert _{X_T}+\Vert u^\theta _1\Vert _{X_T}\Vert u^\theta _2\Vert _{X_T}. \end{aligned}$$

(A.21)

The estimates for the remaining two terms require more tricks. We first write

(A.22)

Thanks to (2.8) with $\gamma =-\epsilon =\frac{2}{3}$ and Biot-Savart law, there holds

$$\begin{aligned} {iii} _1&\lesssim \Bigl \Vert \int _0^{\frac{t}{2}} \Bigl (\frac{\bigl \Vert r^{\frac{1}{2}}{\widetilde{\nabla }}{\widetilde{u}}_1(s)\bigr \Vert _{L^2(\Omega )} \bigl \Vert r^{\frac{1}{6}} \omega ^\theta _2(s)\bigr \Vert _{L^2(\Omega )}}{(t-s)^{1-\frac{1}{6}}} +\frac{\bigl \Vert r^{\frac{2}{3}}{\widetilde{\nabla }}\omega ^\theta _2 (s)\bigr \Vert _{L^2(\Omega )} \Vert {\widetilde{u}}_1(s)\Vert _{L^4(\Omega )}}{(t-s)^{\frac{3}{4}-\frac{1}{6}}}\\&\qquad +\sum \limits _{1\le i,j\le 2,i\ne j} \frac{\Vert \partial _zu^\theta _i(s)\Vert _{L^{\frac{12}{5}}(\Omega )} \bigl \Vert r^{-\frac{1}{3}} u^\theta _j(s)\bigr \Vert _{L^{3}(\Omega )}}{(t-s)^{\frac{3}{4}-\frac{1}{6}}}\Bigr )\, ds\Bigr \Vert _{L_T^2}\\&\lesssim \Bigl \Vert \int _0^{\frac{t}{2}} \Bigl (\frac{\Vert \omega ^\theta _1(s)\bigr \Vert _{L^2}\Vert \omega ^\theta _2(s)\Vert _{L^2}^{\frac{1}{3}} \Vert \omega ^\theta _2(s)\Vert _{L^2(\Omega )}^{\frac{2}{3}}}{(t-s)^{\frac{5}{6}}} +\frac{\bigl \Vert s^{\frac{1}{6}}r^{\frac{1}{6}}{\widetilde{\nabla }}\omega ^\theta _2(s)\bigr \Vert _{L^2} \Vert \omega ^\theta _1(s)\Vert _{L^{\frac{4}{3}}(\Omega )}}{(t-s)^{\frac{7}{12}}\cdot s^{\frac{1}{6}}}\\&\qquad +\sum \limits _{1\le i,j\le 2,i\ne j} \frac{\bigl \Vert s^{\frac{1}{6}}\partial _zu^\theta _i(s)\bigr \Vert _{L^{\frac{12}{5}}(\Omega )} \bigl \Vert r^{-\frac{1}{3}}u^\theta _j(s)\bigr \Vert _{L^3(\Omega )}}{(t-s)^{\frac{7}{12}}\cdot s^{\frac{1}{6}}}\Bigr )\, ds\Bigr \Vert _{L_T^2}\\&\mathop {=}\limits ^{\textrm{def}}{iii} _{1,1}+{iii} _{1,2}+{iii} _{1,3}, \end{aligned}$$

where

$$\begin{aligned} {iii} _{1,1}\lesssim \Vert \omega ^\theta _1\Vert _{L_T^4 L^2}\Vert \omega ^\theta _2\Vert _{L^4_T L^2}^{\frac{1}{3}} \Vert \omega ^\theta _2\Vert _{L^2_T L^2(\Omega )}^{\frac{2}{3}} \Bigl \Vert \frac{1}{|\cdot |^{\frac{5}{6}}}\Bigr \Vert _{L_T^{\frac{6}{5},\infty }} \lesssim \Vert \omega ^\theta _1\Vert _{X_T}\Vert \omega ^\theta _2\Vert _{X_T}, \end{aligned}$$

and by using $t\thicksim t-s$ for $s\in [0,t/2]$ that

$$\begin{aligned}&{iii} _{1,2}+{iii} _{1,3} \lesssim \Bigl \Vert \bigl (\int _0^{\frac{t}{2}}\bigl (s^{-\frac{1}{6}}\bigr )^5\,ds\bigr )^{\frac{1}{5}} \Bigl (\int _0^{\frac{t}{2}}\frac{\bigl \Vert s^{\frac{1}{6}}r^{\frac{1}{6}}{\widetilde{\nabla }}\omega ^\theta _2 (s)\bigr \Vert _{L^2}^{\frac{5}{4}} \Vert \omega ^\theta _1(s)\Vert _{L^{\frac{4}{3}}(\Omega )}^{\frac{5}{4}}}{(t-s)^{\frac{35}{48}}}\\&\qquad +\sum \limits _{i\ne j} \frac{\bigl \Vert s^{\frac{1}{6}}\partial _zu^\theta _i(s)\bigr \Vert _{L^{\frac{12}{5}}(\Omega )}^{\frac{5}{4}} \bigl \Vert r^{-\frac{1}{3}}u^\theta _j(s)\bigr \Vert _{L^3(\Omega )}^{\frac{5}{4}}}{(t-s)^{\frac{35}{48}}}\,ds\Bigr )^{\frac{4}{5}}\Bigr \Vert _{L_T^2}\\&\quad \lesssim \Bigl \Vert \int _0^{\frac{t}{2}} \Bigl (\frac{\bigl \Vert s^{\frac{1}{6}}r^{\frac{1}{6}}{\widetilde{\nabla }}\omega ^\theta _2 (s)\bigr \Vert _{L^2}^{\frac{5}{4}} \Vert \omega ^\theta _1(s)\Vert _{L^{\frac{4}{3}}(\Omega )}^{\frac{5}{4}}}{(t-s)^{\frac{11}{16}}}\\&\qquad +\sum \limits _{i\ne j} \frac{\bigl \Vert s^{\frac{1}{6}}\partial _zu^\theta _i(s)\bigr \Vert _{L^{\frac{12}{5}}(\Omega )}^{\frac{5}{4}} \bigl \Vert r^{-\frac{1}{3}}u^\theta _j(s)\bigr \Vert _{L^3(\Omega )}^{\frac{5}{4}}}{(t-s)^{\frac{11}{16}}}\Bigr )\,ds\Bigr \Vert _{L_T^{\frac{8}{5}}}^{\frac{4}{5}}\\&\quad \lesssim \Bigl ( \bigl \Vert t^{\frac{1}{6}}r^{\frac{1}{6}}{\widetilde{\nabla }}\omega ^\theta _2\bigr \Vert _{L_T^2 L^2} \Vert \omega ^\theta _1\Vert _{L_T^4 L^{\frac{4}{3}}(\Omega )} +\sum \limits _{i\ne j} \bigl \Vert t^{\frac{1}{6}}\partial _zu^\theta _i\bigr \Vert _{L_T^{\frac{12}{5}} L^{\frac{12}{5}}(\Omega )} \bigl \Vert r^{-\frac{1}{3}}u^\theta _j\bigr \Vert _{L_T^3 L^3(\Omega )}\Bigr ) \Bigl \Vert \frac{1}{|\cdot |^{\frac{11}{16}}}\Bigr \Vert _{L_T^{\frac{16}{11},\infty }}^{\frac{4}{5}}\\&\quad \lesssim \Vert \omega ^\theta _1\Vert _{X_T}\Vert \omega ^\theta _2\Vert _{X_T}+\Vert u^\theta _1\Vert _{X_T}\Vert u^\theta _2\Vert _{X_T}. \end{aligned}$$

While by using (2.8) with $\gamma =-\epsilon =\frac{2}{3}$, and the fact that $s\thicksim t$ for $s\in [t/2,t]$, we get

$$\begin{aligned} {iii} _2 \lesssim&\Bigl \Vert \int _{\frac{t}{2}}^t \frac{\bigl \Vert r^{\frac{1}{2}}{\widetilde{\nabla }}{\widetilde{u}}_1(s)\bigr \Vert _{L^2(\Omega )} s^{\frac{1}{4}}\bigl \Vert r^{\frac{1}{6}} \omega ^\theta _2(s)\bigr \Vert _{L^{\frac{8}{3}}(\Omega )}}{(t-s)^{\frac{7}{8}}s^{\frac{1}{12}}} +\frac{\bigl \Vert s^{\frac{1}{6}}r^{\frac{2}{3}}{\widetilde{\nabla }}\omega ^\theta _2 (s)\bigr \Vert _{L^2(\Omega )} \Vert {\widetilde{u}}_1(s)\Vert _{L^4(\Omega )}}{(t-s)^{\frac{3}{4}}}\\&+\sum \limits _{1\le i,j\le 2,i\ne j} \frac{\bigl \Vert s^{\frac{1}{6}}\partial _zu^\theta _i(s)\bigr \Vert _{L^{\frac{12}{5}}(\Omega )} \bigl \Vert r^{-\frac{1}{3}} u^\theta _j(s)\bigr \Vert _{L^3(\Omega )} }{(t-s)^{\frac{3}{4}}}\, ds\Bigr \Vert _{L_T^2}\\ \mathop {=}\limits ^{\textrm{def}}&{iii} _{2,1} +{iii} _{2,2}+{iii} _{2,3}. \end{aligned}$$

In which, we can derive from the Biot-Savart law and Hölder’s inequality that

$$\begin{aligned} \bigl \Vert r^{\frac{1}{2}}{\widetilde{\nabla }}{\widetilde{u}}_1\bigr \Vert _{L^2(\Omega )}= & {} \Vert {\widetilde{\nabla }}{\widetilde{u}}_1\bigr \Vert _{L^2} \lesssim \Vert \omega ^\theta _1\Vert _{L^2},\\ s^{\frac{1}{4}}\bigl \Vert r^{\frac{1}{6}}\omega ^\theta _2\bigr \Vert _{L^{\frac{8}{3}}(\Omega )}\le & {} \bigl \Vert s^{\frac{1}{3}}\omega ^\theta _2\Vert _{L^3(\Omega )}^{\frac{5}{8}}\Vert \omega ^\theta _2\Vert _{L^2}^{\frac{1}{3}} (s\Vert \omega ^\theta _2\Vert _{L^\infty })^{\frac{1}{24}}. \end{aligned}$$

As a result, ${iii} _{2,1}$ can be estimated as follows:

$$\begin{aligned} {iii} _{2,1}&\lesssim \Bigl \Vert \bigl (\int _{\frac{t}{2}}^t s^{-1}ds\bigr )^{\frac{1}{12}} \bigl (\int _{\frac{t}{2}}^t\frac{\Vert \omega ^\theta _1(s)\Vert _{L^2}^{\frac{12}{11}} \bigl \Vert s^{\frac{1}{3}}\omega ^\theta _2(s)\bigr \Vert _{L^3(\Omega )}^{\frac{15}{22}} \Vert \omega ^\theta _2(s)\Vert _{L^2}^{\frac{4}{11}}\bigl (s\Vert \omega ^\theta _2(s)\Vert _{L^\infty }\bigr )^{\frac{1}{22}}}{(t-s)^{\frac{21}{22}}}\, ds\bigr )^{\frac{11}{12}}\Bigr \Vert _{L_T^2}\\&\lesssim \Bigl \Vert \int _{\frac{t}{2}}^t\frac{\Vert \omega ^\theta _1(s)\Vert _{L^2}^{\frac{12}{11}} \bigl \Vert s^{\frac{1}{3}}\omega ^\theta _2(s)\bigr \Vert _{L^3(\Omega )}^{\frac{15}{22}} \Vert \omega ^\theta _2(s)\Vert _{L^2}^{\frac{4}{11}}\bigl (s\Vert \omega ^\theta _2(s)\Vert _{L^\infty }\bigr )^{\frac{1}{22}}}{(t-s)^{\frac{21}{22}}}\, ds\Bigr \Vert _{L_T^{\frac{11}{6}}}^{\frac{11}{12}}\\&\lesssim \Vert \omega ^\theta _1\Vert _{L_T^4L^2} \bigl \Vert t^{\frac{1}{3}}\omega ^\theta _2\bigr \Vert _{L_T^3L^3(\Omega )}^{\frac{5}{8}} \Vert \omega ^\theta _2\Vert _{L_T^4L^2}^{\frac{1}{3}}\Vert t\omega ^\theta _2\Vert _{{L_T^\infty }L^\infty }^{\frac{1}{24}} \Bigl \Vert \frac{1}{|\cdot |^{\frac{21}{22}}}\Bigr \Vert _{L_T^{\frac{22}{21},\infty }}^{\frac{11}{12}}\\&\lesssim \Vert \omega ^\theta _1\Vert _{X_T}\Vert \omega ^\theta _2\Vert _{X_T}. \end{aligned}$$

And we have

$$\begin{aligned} {iii} _{2,2}+{iii} _{2,3}&\lesssim \Bigl ( \bigl \Vert t^{\frac{1}{6}}r^{\frac{1}{6}}{\widetilde{\nabla }}\omega ^\theta _2\bigr \Vert _{L^2_TL^2} \Vert \omega ^\theta _1\Vert _{L^4_TL^{\frac{4}{3}}(\Omega )}\\&\qquad +\sum \limits _{1\le i,j\le 2,i\ne j} \bigl \Vert t^{\frac{1}{6}}\partial _zu^\theta _i\bigr \Vert _{L^{\frac{12}{5}}_TL^{\frac{12}{5}}(\Omega )} \bigl \Vert r^{-\frac{1}{3}} u^\theta _j\bigr \Vert _{L_T^3L^3(\Omega )}\Bigr ) \Bigl \Vert \frac{1}{|\cdot |^{\frac{3}{4}}}\Bigr \Vert _{L_T^{\frac{4}{3},\infty }}\\&\lesssim \Vert \omega ^\theta _1\Vert _{X_T}\Vert \omega ^\theta _2\Vert _{X_T}+\Vert u^\theta _1\Vert _{X_T}\Vert u^\theta _2\Vert _{X_T}. \end{aligned}$$

Substituting all the above estimates into (A.22) gives

$$\begin{aligned} \bigl \Vert t^{\frac{1}{6}}r^{\frac{1}{6}}{\widetilde{\nabla }}{\mathcal {F}}^\omega \bigr \Vert _{L^2_T L^2} \lesssim \Vert \omega ^\theta _1\Vert _{X_T}\Vert \omega ^\theta _2\Vert _{X_T}+\Vert u^\theta _1\Vert _{X_T}\Vert u^\theta _2\Vert _{X_T}. \end{aligned}$$

(A.23)

The last term $\bigl \Vert t^{\frac{1}{6}}{\widetilde{\nabla }} {\mathcal {F}}^u\bigr \Vert _{L^{\frac{12}{5}}_T L^{\frac{12}{5}}(\Omega )}$ can be handled similarly. First, we write

$$\begin{aligned} \begin{aligned}&\qquad \qquad \qquad \bigl \Vert t^{\frac{1}{6}}{\widetilde{\nabla }}{\mathcal {F}}^u\bigr \Vert _{L^{\frac{12}{5}}_T L^{\frac{12}{5}}(\Omega )} \le {iv} _1+{iv} _2,\quad \text {where}\\ {iv} _1&=\Bigl \Vert t^{\frac{1}{6}}\int _0^{\frac{t}{2}} \bigl \Vert {\widetilde{\nabla }}S(t-s)\bigl ({\widetilde{u}}_1(s)\cdot {\widetilde{\nabla }}u^\theta _2(s) +\frac{u^r_1(s)\cdot u^\theta _2(s)}{r}\bigr ) \bigr \Vert _{L^{\frac{12}{5}}(\Omega )}\, ds\Bigr \Vert _{L_T^{\frac{12}{5}}}\\ {iv} _2&=\Bigl \Vert t^{\frac{1}{6}}\int _{\frac{t}{2}}^t \bigl \Vert {\widetilde{\nabla }}S(t-s)\bigl ({\widetilde{u}}_1(s)\cdot {\widetilde{\nabla }}u^\theta _2(s) +\frac{u^r_1(s)\cdot u^\theta _2(s)}{r}\bigr ) \bigr \Vert _{L^{\frac{12}{5}}(\Omega )}\, ds\Bigr \Vert _{L_T^{\frac{12}{5}}}. \end{aligned} \qquad \end{aligned}$$

(A.24)

Then by using (2.8) with $\gamma =\epsilon =0$ and $\gamma =0,~\epsilon =-\frac{2}{5}$, we have

$$\begin{aligned} {iv} _1\lesssim & {} \Bigl \Vert \int _0^{\frac{t}{2}}\frac{\Vert {\widetilde{u}}_1\Vert _{L^4(\Omega )} \bigl \Vert s^{\frac{1}{6}}{\widetilde{\nabla }} u^\theta _2\bigr \Vert _{L^{\frac{12}{5}}(\Omega )}}{(t-s)^{\frac{7}{12}}\cdot s^{\frac{1}{6}}} +\frac{\Vert u^r_1\Vert _{L^4(\Omega )} \bigl \Vert r^{-\frac{3}{5}}u^\theta _2(s)\bigr \Vert _{L^{\frac{5}{2}}(\Omega )}}{(t-s)^{\frac{23}{30}}}\,ds\Bigr \Vert _{L_T^{\frac{12}{5}}}\\&\mathop {=}\limits ^{\textrm{def}}&{iv} _{1,1}+{iv} _{1,2}. \end{aligned}$$

In which, the first part can be estimated as follows

$$\begin{aligned} {iv} _{1,1}&\lesssim \Bigl \Vert \bigl (\int _0^{\frac{t}{2}}\bigl (s^{-\frac{1}{6}}\bigr )^5\,ds\bigr )^{\frac{1}{5}} \Bigl (\int _0^{\frac{t}{2}}\frac{\Vert \omega ^\theta _1(s)\Vert _{L^{\frac{4}{3}}(\Omega )}^{\frac{5}{4}} \bigl \Vert s^{\frac{1}{6}}{\widetilde{\nabla }}u^\theta _2 (s)\bigr \Vert _{L^{\frac{12}{5}}(\Omega )}^{\frac{5}{4}}}{(t-s)^{\frac{35}{48}}}\,ds\Bigr )^{\frac{4}{5}}\Bigr \Vert _{L_T^{\frac{12}{5}}}\\&\lesssim \Bigl \Vert \int _0^{\frac{t}{2}}\frac{\Vert \omega ^\theta _1(s)\Vert _{L^{\frac{4}{3}}(\Omega )}^{\frac{5}{4}} \bigl \Vert s^{\frac{1}{6}}{\widetilde{\nabla }}u^\theta _2 (s)\bigr \Vert _{L^{\frac{12}{5}}(\Omega )}^{\frac{5}{4}}}{(t-s)^{\frac{11}{16}}}\,ds\Bigr \Vert _{L_T^{\frac{48}{25}}}^{\frac{4}{5}}\\&\lesssim \Vert \omega ^\theta _1\Vert _{L_T^4L^{\frac{4}{3}}(\Omega )} \bigl \Vert t^{\frac{1}{6}}{\widetilde{\nabla }}u^\theta _2\bigr \Vert _{L_T^{\frac{12}{5}} L^{\frac{12}{5}}(\Omega )} \Bigl \Vert \frac{1}{|\cdot |^{\frac{11}{16}}}\Bigr \Vert _{L_T^{\frac{16}{11},\infty }}^{\frac{4}{5}}\\&\lesssim \Vert \omega ^\theta _1\Vert _{X_T}\Vert u^\theta _2\Vert _{X_T}. \end{aligned}$$

and the second part can be estimated as

$$\begin{aligned} {iv} _{1,2}\lesssim \Vert u^r_1\Vert _{L_T^4L^{4}(\Omega )} \Vert r^{-\frac{3}{5}}u^\theta _2\Vert _{L_T^{\frac{5}{2}}L^{\frac{5}{2}}(\Omega )} \Bigl \Vert \frac{1}{|\cdot |^{\frac{23}{30}}}\Bigr \Vert _{L_T^{\frac{30}{23},\infty }} \lesssim \Vert \omega ^\theta _1\Vert _{X_T}\Vert u^\theta _2\Vert _{X_T}. \end{aligned}$$

While by using (2.8) with $\gamma =0,~\epsilon =-\frac{4}{15}$, we obtain

$$\begin{aligned} {iv} _2&\lesssim \Bigl \Vert \int _{\frac{t}{2}}^t\frac{\Vert {\widetilde{u}}_1\Vert _{L^4(\Omega )} \bigl \Vert s^{\frac{1}{6}}{\widetilde{\nabla }} u^\theta _2\bigr \Vert _{L^{\frac{12}{5}}(\Omega )}}{(t-s)^{\frac{3}{4}}}\\&\qquad +\frac{\Vert u^r_1\Vert _{L^4(\Omega )}s^{\frac{1}{6}}\Vert r^{-1}u^\theta _2(s)\Vert _{L^2(\Omega )}^{\frac{1}{3}} \bigl \Vert r^{-\frac{3}{5}}u^\theta _2(s)\bigr \Vert _{L^{\frac{5}{2}}(\Omega )}^{\frac{2}{3}}}{(t-s)^{\frac{9}{10}}}\,ds\Bigr \Vert _{L_T^{\frac{12}{5}}}\\&\lesssim \Vert \omega ^\theta _1\Vert _{L_T^4L^{\frac{4}{3}}(\Omega )} \bigl \Vert t^{\frac{1}{6}}{\widetilde{\nabla }}u^\theta _2\bigr \Vert _{L_T^{\frac{12}{5}} L^{\frac{12}{5}}(\Omega )} \Bigl \Vert \frac{1}{|\cdot |^{\frac{3}{4}}}\Bigr \Vert _{L_T^{\frac{4}{3},\infty }}\\&\qquad +\Vert \omega ^\theta _1\Vert _{L_T^4L^{\frac{4}{3}}(\Omega )} \bigl \Vert t^{\frac{1}{2}}\Vert r^{-1}u^\theta _2\Vert _{L^{2}(\Omega )}\bigr \Vert _{L_T^\infty }^{\frac{1}{3}} \bigl \Vert r^{-\frac{3}{5}}u^\theta _2(s)\bigr \Vert _{L^{\frac{5}{2}}_TL^{\frac{5}{2}}(\Omega )}^{\frac{2}{3}} \Bigl \Vert \frac{1}{|\cdot |^{\frac{9}{10}}}\Bigr \Vert _{L_T^{\frac{10}{9},\infty }}\\&\lesssim \Vert \omega ^\theta _1\Vert _{X_T}\Vert u^\theta _2\Vert _{X_T}. \end{aligned}$$

Substituting the above three estimates into (A.24), gives

$$\begin{aligned} \bigl \Vert t^{\frac{1}{6}}{\widetilde{\nabla }}{\mathcal {F}}^u\bigr \Vert _{L^{\frac{12}{5}}_T L^{\frac{12}{5}}(\Omega )}\lesssim \Vert \omega ^\theta _1\Vert _{X_T}\Vert u^\theta _2\Vert _{X_T}. \end{aligned}$$

(A.25)

Combining the estimates (A.14)–(A.17),  (A.19), (A.21), (A.23), (A.25), finally we close the estimates that for some universal constant $C_2$, there holds

$$\begin{aligned} \bigl \Vert \bigl ({\mathcal {F}}^\omega ,{\mathcal {F}}^u\bigr ) \bigr \Vert _{X_T}\le C_2 \Vert (\omega ^\theta _1,u^\theta _1)\Vert _{X_T} \Vert (\omega ^\theta _2,u^\theta _2)\Vert _{X_T}. \end{aligned}$$

(A.26)

$\bullet$ Fixed-point argument By virtue of the estimates (A.26), (A.12), Lemma A.1 guarantees that if $\Vert \omega ^\theta _0\Vert _{L^{\frac{3}{2}}}$, $\Vert \omega ^\theta _0\Vert _{L^1(\Omega )}$, $\Vert u^\theta _0\Vert _{L^2(\Omega )}$ are small enough such that

$$\begin{aligned} C_1\bigl (\Vert \omega ^\theta _0\Vert _{L^{\frac{3}{2}}}+\Vert \omega ^\theta _0\Vert _{L^1(\Omega )} +\Vert u^\theta _0\Vert _{L^2(\Omega )}\bigr )\le \frac{1}{4C_2}, \end{aligned}$$

(A.27)

then (A.1) has a unique global solution in $X_\infty$. Moreover, this solution remains small so that

$$\begin{aligned} \Vert (\omega ^\theta ,u^\theta )\Vert _{X_\infty }\le 2C_1\bigl (\Vert \omega ^\theta _0\Vert _{L^{\frac{3}{2}}} +\Vert \omega ^\theta _0\Vert _{L^1(\Omega )}+\Vert u^\theta _0\Vert _{L^2(\Omega )}\bigr ). \end{aligned}$$

This completes the proof of this proposition. $\square$