Coriolis effect on temporal decay rates of global solutions to the fractional Navier–Stokes equations

Abstract

The incompressible fractional Navier–Stokes equations in the rotational framework is considered. We establish the global existence result and temporal decay estimate for a unique smooth solution when the speed of rotation is sufficiently rapid. It is found that the strong rotational effect enhances the temporal decay rate of a certain norm of the velocity.

Appendix

In this appendix, we prove Lemmas 8–9 and provide a regularity criteria for (2). We use the notations (27) and (35) for convenience. We note, from the embedding relation $(\dot{H}^s \cap \dot{H}^s_q)(\mathbb {R}^3)\hookrightarrow L^{2}(\mathbb {R}^{3})$ and the skew–symmetry of $\varOmega e_{3}\times u$, that a solution u in Lemma 8 or Lemma 9 satisfies

$$\begin{aligned} \left\| u(\cdot ,t) \right\| _{L^{2}(\mathbb {R}^{3})}^{2}+2\int _{0}^{t}\Big \Vert ( -\varDelta )^{\frac{\alpha }{2}} u(\cdot ,\tau )\Big \Vert _{L^{2}(\mathbb {R}^{3})}^{2}d\tau \le \left\| u_{0} \right\| _{L^{2}(\mathbb {R}^{3})}^{2},\qquad t>0. \end{aligned}$$

(A.1)

Proof of Theorem 3

By Lemma 1 and the boundedness properties of Leray projection, we have

$$\begin{aligned}&\int ^{\frac{t}{2}}_0 \left\| T_{\varOmega }^{\alpha }(t - \tau ) {\mathbb {P}} \mathrm {div} \, (u \otimes u)(\tau ) \right\| _{\dot{H}^s_p}d\tau \nonumber \\&\quad \le C \int _0^{\frac{t}{2}} \left( t - \tau \right) ^{- \frac{1}{2\alpha } - \frac{s}{2\alpha } - \frac{3}{2\alpha } ( \frac{1}{r} - \frac{1}{p} )} (1 + |\varOmega | (t-\tau ))^{- ( 1 - \frac{2}{p} )} \left\| (u \otimes u )(\tau )\right\| _{L^r} d\tau ,\nonumber \\ \end{aligned}$$

(A.2)

where $r>1$ is a number

$$\begin{aligned} r = \left( \alpha \Big ( \frac{1}{p} - \frac{s}{3} \Big )+ 1 - \frac{\alpha }{2} \right) ^{-1}. \end{aligned}$$

Note that $1-\frac{1}{p}<\frac{1}{r}$. Indeed, by the upper inequality of (3)$_{1}$ and $\frac{1}{2}<\alpha \le \frac{5}{4}$, we have $s<\ \frac{3(5-2\alpha )}{2(2\alpha +3)}< \frac{3(2-\alpha )}{2(2\alpha -1)}$ and thus, it follows from the lower inequality of (3)$_{2}$ that $\frac{\alpha }{\alpha +1}\left( \frac{s}{3}+\frac{1}{2} \right) <\frac{1}{3}+\frac{s}{9}\le \frac{1}{p}$, and so $1-\frac{1}{p}<\frac{1}{r}$. Using the interpolation inequality

$$\begin{aligned} \left\| u \otimes u \right\| _{L^r} \le C \left\| u \right\| _{\dot{H}^s_p}^{\alpha } \left\| u \right\| _{L^2}^{2 - \alpha }, \end{aligned}$$

(A.1), and a direct computation, we have

$$\begin{aligned}&\int _0^{\frac{t}{2}} \left( t - \tau \right) ^{- \frac{1}{2\alpha } - \frac{s}{2\alpha } - \frac{3}{2\alpha } ( \frac{1}{r} - \frac{1}{p} )} (1 + |\varOmega | (t-\tau ))^{- ( 1 - \frac{2}{p} )} \left\| (u \otimes u )(\tau )\right\| _{L^r} d\tau \nonumber \\&\quad \le C t^{- \frac{1}{2\alpha } - \frac{s}{2\alpha } - \frac{3}{2\alpha } \left( \alpha (\frac{1}{p} - \frac{ s}{3}) + 1 - \frac{\alpha }{2} - \frac{1}{q}\right) } \left\| u_0 \right\| _{L^2}^{2 - \alpha } \Vert u\Vert _{X^s_{p,q}} ^{\alpha }(t) R(t)^{-1}J, \end{aligned}$$

(A.3)

where $J:= \int _0^{\frac{t}{2}} R(\tau )^{-\alpha }d\tau$. Using change of variable $\tilde{\tau } = |\varOmega |\tau$, we note that

$$\begin{aligned} J= |\varOmega |^{- 1 + \frac{3}{2} ( \frac{1}{q} - \frac{1}{p} )} \int _0^{\frac{|\varOmega |t}{2}} {\tilde{\tau }}^{- \frac{3}{2} ( \frac{1}{q} - \frac{1}{p} )} (1 + \tilde{\tau })^{-\alpha ( 1 - \frac{2}{p} )} d\tilde{\tau }. \end{aligned}$$

(A.4)

Using $\frac{1}{p} \ge \frac{1}{3}>\frac{1}{q} - \frac{2}{3}$, we also note that

$$\begin{aligned} 1- \frac{3}{2} \left( \frac{1}{q} - \frac{1}{p} \right) >0. \end{aligned}$$

(A.5)

Now, we consider two cases

$$\begin{aligned} 1-\frac{3}{2} \left( \frac{1}{q} - \frac{1}{p} \right) - \alpha \left( 1 - \frac{2}{p} \right) <0,\quad \text{ and } \quad 1-\frac{3}{2} \left( \frac{1}{q} - \frac{1}{p} \right) - \alpha \left( 1 - \frac{2}{p} \right) \ge 0, \end{aligned}$$

separately.

In the case of

$$\begin{aligned} 1-\frac{3}{2} \left( \frac{1}{q} - \frac{1}{p} \right) - \alpha \left( 1 - \frac{2}{p} \right) <0, \end{aligned}$$

the boundedness of the integral term in (A.4) gives

$$\begin{aligned} J\le C|\varOmega |^{- 1 + \frac{3}{2} ( \frac{1}{q} - \frac{1}{p} )} . \end{aligned}$$

Thus, it follows from (A.2)–(A.3) that

$$\begin{aligned}&\int ^{\frac{t}{2}}_0 \left\| T_{\varOmega }^{\alpha }(t - \tau ) {\mathbb {P}} \mathrm {div} \, (u \otimes u)(\tau ) \right\| _{\dot{H}^s_p}d\tau \\&\quad \le C t^{- \frac{1}{2\alpha } - \frac{s}{2\alpha } - \frac{3}{2\alpha } \left( \alpha (\frac{1}{p} - \frac{ s}{3}) + 1 - \frac{\alpha }{2} - \frac{1}{q} \right) } |\varOmega |^{- 1 + \frac{3}{2} ( \frac{1}{q} - \frac{1}{p} )} \left\| u_0 \right\| _{L^2}^{2 - \alpha } \Vert u\Vert _{X^s_{p,q}} ^{\alpha }(t) R(t)^{-1}. \end{aligned}$$

Using $\frac{1}{p}\ge \frac{1}{3}+\frac{s}{9}$ and $q>1$, we note that

$$\begin{aligned}&- \frac{1}{2\alpha } - \frac{s}{2\alpha } - \frac{3}{2\alpha } \left( \alpha \Big (\frac{1}{p} - \frac{ s}{3}\Big ) + 1 - \frac{\alpha }{2} - \frac{1}{q} \right) \\&\quad \le - \frac{1}{2\alpha } - \frac{s}{2\alpha } - \frac{3}{2} \left( \Big (\frac{1}{3}+\frac{s}{9}- \frac{ s}{3} \Big ) - \frac{1}{2} \right) = -\frac{1}{2\alpha }+\frac{1}{4} - \frac{s}{2\alpha } +\frac{s}{3}, \end{aligned}$$

and thus,

$$\begin{aligned} - \frac{1}{2\alpha } - \frac{s}{2\alpha } - \frac{3}{2\alpha } \left( \alpha \Big (\frac{1}{p} - \frac{ s}{3}\Big ) + 1 - \frac{\alpha }{2} - \frac{1}{q} \right) <0. \end{aligned}$$

By this fact with $t\ge 1$, it follows that

$$\begin{aligned} \int ^{\frac{t}{2}}_0 \left\| T_{\varOmega }^{\alpha }(t - \tau ) {\mathbb {P}} \mathrm {div} \, (u \otimes u)(\tau ) \right\| _{\dot{H}^s_p}d\tau \le C |\varOmega |^{- 1 + \frac{3}{2} ( \frac{1}{q} - \frac{1}{p} )} \left\| u_0 \right\| _{L^2}^{2 - \alpha } \Vert u\Vert _{X^s_{p,q}}^{\alpha } (t) R(t)^{-1}. \end{aligned}$$

Therefore, (47) is obtained.

We next deal with the case

$$\begin{aligned} 1-\frac{3}{2} \left( \frac{1}{q} - \frac{1}{p} \right) - \alpha \left( 1 - \frac{2}{p} \right) \ge 0. \end{aligned}$$

By a direct calculation, if $1-\frac{3}{2} \left( \frac{1}{q} - \frac{1}{p} \right) - \alpha \left( 1 - \frac{2}{p} \right) >0$, then we have

$$\begin{aligned} \begin{aligned}&\int _0^{\frac{|\varOmega |t}{2}} {\tilde{\tau }}^{- \frac{3}{2} ( \frac{1}{q} - \frac{1}{p} )}(1+\tilde{\tau })^{ - \alpha ( 1 - \frac{2}{p} )}d\tilde{\tau } \\&\quad \le C|\varOmega |^{1-\frac{3}{2} ( \frac{1}{q} - \frac{1}{p} )- \alpha ( 1 - \frac{2}{p} )} t^{1-\frac{3}{2} ( \frac{1}{q} - \frac{1}{p} )- \alpha ( 1 - \frac{2}{p} )} \\&\quad \le C|\varOmega |^{1-\frac{3}{2} ( \frac{1}{q} - \frac{1}{p} )- \alpha ( 1 - \frac{2}{p} )} t^{1-\frac{3}{2} ( \frac{1}{q} - \frac{1}{p} )- \alpha ( 1 - \frac{2}{p} )} \log (|\varOmega | + e) \log (t + e), \end{aligned} \end{aligned}$$

and if $1-\frac{3}{2} ( \frac{1}{q} - \frac{1}{p} )- \alpha ( 1 - \frac{2}{p} )=0$, then we have

$$\begin{aligned} \begin{aligned}&\int _0^{\frac{|\varOmega |t}{2}} {\tilde{\tau }}^{- \frac{3}{2} ( \frac{1}{q} - \frac{1}{p} )}(1+\tilde{\tau })^{-\alpha ( 1 - \frac{2}{p} )}d\tilde{\tau } \\&\quad \le C \log (|\varOmega | + e) \log (t + e) \\&\quad = C|\varOmega |^{1-\frac{3}{2} ( \frac{1}{q} - \frac{1}{p} )- \alpha ( 1 - \frac{2}{p} )} t^{1-\frac{3}{2} ( \frac{1}{q} - \frac{1}{p} )- \alpha ( 1 - \frac{2}{p} )} \log (|\varOmega | + e) \log (t + e). \end{aligned} \end{aligned}$$

Thus, we can compute J in (A.4) as

$$\begin{aligned} \begin{aligned} J&\le |\varOmega |^{- 1 + \frac{3}{2} ( \frac{1}{q} - \frac{1}{p} )} \int _0^{\frac{|\varOmega |t}{2}} {\tilde{\tau }}^{- \frac{3}{2} ( \frac{1}{q} - \frac{1}{p} )} (1 + \tilde{\tau })^{-\alpha ( 1 - \frac{2}{p} )} d\tilde{\tau } \\&\le C t^{1-\frac{3}{2} ( \frac{1}{q} - \frac{1}{p} )- \alpha ( 1 - \frac{2}{p} )} \log (t + e) |\varOmega |^{- \alpha ( 1 - \frac{2}{p} )} \log (|\varOmega | + e). \end{aligned} \end{aligned}$$

Then, it follows from (A.3) that

$$\begin{aligned} \begin{aligned}&\int ^{\frac{t}{2}}_0 \left\| T_{\varOmega }^{\alpha }(t - \tau ) {\mathbb {P}} \mathrm {div} \, (u \otimes u)(\tau ) \right\| _{\dot{H}^s_p}d\tau \\&\quad \le C t^{\frac{7}{4}-\frac{2}{\alpha }+(\frac{3}{2q}-\frac{s}{2})(\frac{1}{\alpha }-1)-\alpha (1-\frac{2}{p})} \log (t + e) \\&\qquad \cdot |\varOmega |^{- \alpha ( 1 - \frac{2}{p} )} \log (|\varOmega | + e) \left\| u_0 \right\| _{L^2}^{2 - \alpha } \Vert u\Vert _{X^s_{p,q}} ^{\alpha }(t) R(t)^{-1}. \end{aligned} \end{aligned}$$

In the case of $\frac{1}{2}<\alpha < 1$, we have

$$\begin{aligned} \begin{aligned}&\frac{7}{4}-\frac{2}{\alpha }+\Big (\frac{3}{2q}-\frac{s}{2}\Big )\Big (\frac{1}{\alpha }-1\Big )-\alpha \Big (1-\frac{2}{p}\Big )<\frac{7}{4}-\frac{2}{\alpha } + \Big (\frac{3}{2}-\frac{s}{2}\Big )\Big (\frac{1}{\alpha }-1\Big ) \\&\quad = \frac{1}{4} - \frac{1}{2\alpha } - \frac{s}{2} \Big ( \frac{1}{\alpha }- 1\Big ) \\&\quad <0. \end{aligned} \end{aligned}$$

In the case of $1\le \alpha \le \frac{5}{4}$, using $\frac{5}{2} - 2\alpha < s$, we have

$$\begin{aligned} \begin{aligned}&\frac{7}{4}-\frac{2}{\alpha }+\Big (\frac{3}{2q}-\frac{s}{2}\Big )\Big (\frac{1}{\alpha }-1\Big )-\alpha \Big (1-\frac{2}{p}\Big )\\&\quad<\frac{7}{4}-\frac{2}{\alpha }+\left( \frac{3}{2} \Big ( \frac{1}{2} + \frac{s}{3}\Big )-\frac{s}{2}\right) \Big (\frac{1}{\alpha }-1\Big )-\alpha \left( 1-2\Big (\frac{5}{12\alpha }+\frac{1}{6}-\frac{s}{6\alpha }\Big ) \right) \\&\quad =\frac{11}{6} - \frac{5}{4\alpha } - \frac{2\alpha }{3} - \frac{s}{3} \\&\quad < 1 - \frac{5}{4\alpha } \\&\quad \le 0. \end{aligned} \end{aligned}$$

Hence, with $t > 1$ we observe that

$$\begin{aligned}&\int ^{\frac{t}{2}}_0 \left\| T_{\varOmega }^{\alpha }(t - \tau ) {\mathbb {P}} \mathrm {div} \, (u \otimes u)(\tau ) \right\| _{\dot{H}^s_p}d\tau \\&\quad \le C|\varOmega |^{- \alpha ( 1 - \frac{2}{p} )} \log (|\varOmega | + e) \left\| u_0 \right\| _{L^2}^{2 - \alpha } \Vert u\Vert _{X^s_{p,q}} ^{\alpha }(t) R(t)^{-1}. \end{aligned}$$

Therefore, (47) is obtained. This completes the proof.

Proof of ....

By (A.1) and interpolation inequality, we have

$$\begin{aligned} \left\| u \otimes u \right\| _{L^r} \le C \left\| u \right\| _{\dot{H}^s_p}^{\frac{6\alpha }{4\alpha + 1}} \left\| u_0 \right\| _{L^2}^{2 - \frac{6\alpha }{4\alpha + 1}}, \end{aligned}$$

where $r >1$ is a number

$$\begin{aligned} r = \left( \frac{6\alpha }{4\alpha + 1} \Big ( \frac{1}{p} - \frac{s}{3} \Big ) + 1 - \frac{3\alpha }{4\alpha + 1} \right) ^{-1}. \end{aligned}$$

Note that $1-\frac{1}{p}<\frac{1}{r}$. Indeed, by the upper inequality of (3)$_{1}$ and $\frac{5}{4}<\alpha <\frac{5}{2}$, we have $s<\ \frac{3(5-2\alpha )}{2(2\alpha +3)}< \frac{3(\alpha +1)}{(8\alpha -1)}$ and thus, it follows from the lower inequality of (3)$_{2}$ that $\frac{\alpha (2s+3)}{10\alpha +1}<\frac{1}{3}+\frac{s}{9}\le \frac{1}{p}$, and so $1-\frac{1}{p}<\frac{1}{r}$. Then, it follows from (A.2) that

$$\begin{aligned}&\int ^{\frac{t}{2}}_0 \left\| T_{\varOmega }^{\alpha }(t - \tau ) {\mathbb {P}} \mathrm {div} \, (u \otimes u)(\tau ) \right\| _{\dot{H}^s_p}d\tau \nonumber \\&\quad \le C t^{- \frac{1}{2\alpha } - \frac{s}{2\alpha } - \frac{3}{2\alpha } \left( \frac{6\alpha }{4\alpha + 1} ( \frac{1}{p} - \frac{s}{3} ) + 1 - \frac{3\alpha }{4\alpha + 1} - \frac{1}{q}\right) } \left\| u_0 \right\| _{L^2}^{2 - \frac{6\alpha }{4\alpha + 1}} \Vert u\Vert _{X^s_{p,q}} ^{\frac{6\alpha }{4\alpha + 1}}(t) R(t)^{-1} \tilde{J}\nonumber \\ \end{aligned}$$

(A.6)

where $\tilde{J}:= \int _0^{\frac{t}{2}} R(\tau )^{-\frac{6\alpha }{4\alpha + 1}}d\tau$ and this can be rewritten as

$$\begin{aligned} \tilde{J} = |\varOmega |^{-1 + \frac{9}{4\alpha + 1} ( \frac{1}{q} - \frac{1}{p} )} \int _0^{\frac{|\varOmega |t}{2}} {\tilde{\tau }}^{- \frac{9}{4\alpha + 1} ( \frac{1}{q} - \frac{1}{p} )} (1 + \tilde{\tau })^{- \frac{6\alpha }{4\alpha + 1} ( 1 - \frac{2}{p} )} d\tilde{\tau }. \end{aligned}$$

(A.7)

Now, we claim that

$$\begin{aligned} 0< 1 - \frac{9}{4\alpha + 1}\left( \frac{1}{q} - \frac{1}{p} \right) < \frac{6\alpha }{4\alpha + 1} \left( 1 - \frac{2}{p} \right) . \end{aligned}$$

(A.8)

Indeed, by $q>1$, $\frac{1}{3}+\frac{s}{9}\le \frac{1}{p}$, $s>0$, and $\frac{5}{4}<\alpha$, the lower inequality of (A.8) holds true. The upper inequality of (A.8) holds true because $\frac{1}{q} \ge \frac{1}{3} + \frac{2}{p} - \frac{s}{3}$ gives

$$\begin{aligned} \begin{aligned}&1 - \frac{9}{4\alpha + 1} \left( \frac{1}{q} - \frac{1}{p} \right) - \frac{6\alpha }{4\alpha + 1} \left( 1 - \frac{2}{p} \right) \\&\quad \le 1 - \frac{9}{4\alpha + 1} \left( \frac{1}{p} + \frac{1}{3} - \frac{s}{3} \right) - \frac{6\alpha }{4\alpha + 1} \left( 1 - \frac{2}{p} \right) \\&\quad = 1 - \frac{3}{4\alpha + 1} - \frac{6\alpha }{4\alpha + 1} + \frac{3s}{4\alpha + 1} + \frac{12\alpha - 9}{(4\alpha + 1)p}, \end{aligned} \end{aligned}$$

the upper inequality of (3)$_2$ gives

$$\begin{aligned} \begin{aligned}&1 - \frac{3}{4\alpha + 1} - \frac{6\alpha }{4\alpha + 1} + \frac{3s}{4\alpha + 1} + \frac{12\alpha - 9}{(4\alpha + 1)p} \\&\quad < 1 - \frac{3}{4\alpha + 1} - \frac{6\alpha }{4\alpha + 1} + \frac{3s}{4\alpha + 1} + \frac{12\alpha - 9}{4\alpha + 1} \left( \frac{5}{12\alpha } + \frac{1}{6} - \frac{s}{6\alpha } \right) \\&\quad = 1 - \frac{3}{4\alpha + 1} - \frac{6\alpha }{4\alpha + 1} + \frac{(6\alpha + 9)s}{6\alpha (4\alpha + 1)} + \frac{12\alpha - 9}{4\alpha + 1} \left( \frac{5}{12\alpha } + \frac{1}{6} \right) , \end{aligned} \end{aligned}$$

and the upper inequality of (3)$_1$ gives

$$\begin{aligned}&1 - \frac{3}{4\alpha + 1} - \frac{6\alpha }{4\alpha + 1} + \frac{(6\alpha + 9)s}{6\alpha (4\alpha + 1)} + \frac{12\alpha - 9}{4\alpha + 1} \left( \frac{5}{12\alpha } + \frac{1}{6} \right) \\&\quad < 1 - \frac{3}{4\alpha + 1} - \frac{6\alpha }{4\alpha + 1} + \frac{6\alpha + 9}{6\alpha (4\alpha + 1)} \left( \frac{3(5 - 2\alpha )}{2(2\alpha + 3)} \right) + \frac{12\alpha - 9}{4\alpha + 1} \left( \frac{5}{12\alpha } + \frac{1}{6} \right) = 0. \end{aligned}$$

By (A.8), the integral term in (A.7) is bounded and thus,

$$\begin{aligned} \tilde{J} \le C |\varOmega |^{-1 + \frac{9}{4\alpha + 1} ( \frac{1}{q} - \frac{1}{p} )}. \end{aligned}$$

Then, it follows from (A.6) that

$$\begin{aligned} \begin{aligned}&\int ^{\frac{t}{2}}_0 \left\| T_{\varOmega }^{\alpha }(t - \tau ) {\mathbb {P}} \mathrm {div} \, (u \otimes u)(\tau ) \right\| _{\dot{H}^s_p}d\tau \\&\quad \le C t^{- \frac{1}{2\alpha } - \frac{s}{2\alpha } - \frac{3}{2\alpha } \left( \frac{6\alpha }{4\alpha + 1} ( \frac{1}{p} - \frac{s}{3} ) + 1 - \frac{3\alpha }{4\alpha + 1} - \frac{1}{q}\right) } \\&\qquad \cdot |\varOmega |^{- 1 + \frac{9}{4\alpha + 1} ( \frac{1}{q} - \frac{1}{p} )} \left\| u_0 \right\| _{L^2}^{2 - \frac{6\alpha }{4\alpha + 1}} \Vert u\Vert _{X^s_{p,q}} ^{\frac{6\alpha }{4\alpha + 1}}(t) R(t)^{-1}. \end{aligned} \end{aligned}$$

Moreover, we see that

$$\begin{aligned} - \frac{1}{2\alpha } - \frac{s}{2\alpha } - \frac{3}{2\alpha } \left( \frac{6\alpha }{4\alpha + 1} \Big ( \frac{1}{p} - \frac{s}{3} \Big ) + 1 - \frac{3\alpha }{4\alpha + 1} - \frac{1}{q}\right) < 0 \end{aligned}$$

because $\frac{1}{q} < 1$ and (3)$_2$ implies

$$\begin{aligned} \begin{aligned}&- \frac{1}{2\alpha } - \frac{s}{2\alpha } - \frac{3}{2\alpha } \left( \frac{6\alpha }{4\alpha + 1} \Big ( \frac{1}{p} - \frac{s}{3} \Big ) + 1 - \frac{3\alpha }{4\alpha + 1} - \frac{1}{q}\right) \\&\quad< - \frac{1}{2\alpha } - \frac{s}{2\alpha } - \frac{9}{4\alpha + 1} \Big ( \frac{1}{p} - \frac{s}{3} \Big ) + \frac{9}{2(4\alpha + 1)} \\&\quad \le - \frac{1}{2\alpha } - \frac{s}{2\alpha } - \frac{9}{4\alpha + 1} \Big ( \frac{1}{3} - \frac{2s}{9} \Big )+ \frac{9}{2(4\alpha + 1)} \\&\quad = - \frac{1 + \alpha + s}{2\alpha (4\alpha + 1)} < 0. \end{aligned} \end{aligned}$$

Therefore, with $t > 1$ we have

$$\begin{aligned}&\int ^{\frac{t}{2}}_0 \left\| T_{\varOmega }^{\alpha }(t - \tau ) {\mathbb {P}} \mathrm {div} \, (u \otimes u)(\tau ) \right\| _{\dot{H}^s_p}d\tau \\&\quad \le C |\varOmega |^{- 1 + \frac{9}{4\alpha + 1} ( \frac{1}{q} - \frac{1}{p} )} \left\| u_0 \right\| _{L^2}^{2 - \frac{6\alpha }{4\alpha + 1}} \Vert u\Vert _{X^s_{p,q}} ^{\frac{6\alpha }{4\alpha + 1}}(t) R(t)^{-1}. \end{aligned}$$

Hence, (48) is obtained. This completes the proof. $\square$

We close the appendix by giving a regularity criterion for (2). Using the estimates in the proof below we can deduce that the solution to (2) given by Theorem 1 is smooth for $t>0$ .

Lemma 10

Let $\alpha >\frac{1}{2}$, $u_{0}\in H^{5\alpha }(\mathbb {R}^{3})$, and $\mathrm {div}\, u_0 = 0$. Suppose that $(u,\pi )$ is a weak solution to (2) subject to the initial data $u_{0}$, and satisfies

$$\begin{aligned} u\in L^{\sigma }(0,T;L^{r}(\mathbb {R}^{3})),\qquad \frac{3}{r}+\frac{2\alpha }{\sigma }\le 2\alpha -1,\quad \frac{3}{2\alpha -1}<r\le \infty . \end{aligned}$$

(A.9)

Then, $(u,\pi )$ is smooth for $0<t\le T$.

Proof

We note that the classical solvability of (2) with $\varOmega =0$ for $\alpha \ge \frac{5}{4}$ was obtained by Lions [16] and its extension to the case of nonzero $\varOmega$ is rather straightforward. We also note, for $1\le \alpha \le \frac{3}{2}$, that the regularity criterion (A.9) can be found in [22]. Thus, it is sufficient to deal with the case $\frac{1}{2}<\alpha <1$.

Let $\frac{1}{2}<\alpha <1$. We only provide the estimate for $u\in L^{\infty }(0,T; \dot{H}^{\alpha } )\cap L^{2}(0,T; \dot{H}^{2\alpha } )$ because the higher order estimates $u\in L^{\infty }(0,T; \dot{H}^{k\alpha })\cap L^{2}(0,T; \dot{H}^{(k+1)\alpha })$, $k=2,3,4,5$ can be obtained in a similar way. We denote $\varLambda$ as the $\frac{1}{2}$ power of the negative Laplacian, i.e., $\varLambda =(-\varDelta )^{\frac{1}{2}}$. Multiplying both sides of (2) by $\varLambda ^{2\alpha } u$, integrating over $\mathbb {R}^{3}$, and using Hölder’s inequality, we can compute

$$\begin{aligned} \frac{1}{2}\frac{d}{dt}\Vert \varLambda ^{\alpha }u\Vert _{L^{2}}^{2}+\Vert \varLambda ^{2\alpha }u\Vert _{L^{2}}^{2}\le \Vert \varLambda ^{2\alpha }u\Vert _{L^{2}}\Vert u\Vert _{L^{r}}\Vert \nabla u\Vert _{L^{\frac{2r}{r-2}}}. \end{aligned}$$

Using Young’s inequality and the interpolation inequality

$$\begin{aligned} \Vert \nabla u\Vert _{L^{\frac{2r}{r-2}}}\le C\Vert \varLambda ^{\alpha }u\Vert _{L^{2}}^{1-\lambda }\Vert \varLambda ^{2\alpha }u\Vert _{L^{2}}^{\lambda },\qquad \lambda =\frac{1-\alpha +3/r}{\alpha }\in (0,1), \end{aligned}$$

we have

$$\begin{aligned} \frac{1}{2}\frac{d}{dt}\Vert \varLambda ^{\alpha }u\Vert _{L^{2}}^{2}+\frac{1}{2}\Vert \varLambda ^{2\alpha }u\Vert _{L^{2}}^{2}\le C \Vert \varLambda ^{\alpha }u\Vert _{L^{2}}^{2}\Vert u\Vert _{L^{r}}^{2/(1-\lambda )}. \end{aligned}$$

(A.10)

Applying Grönwall’s inequality to (A.10), and using (A.9) with $\frac{2\alpha }{2/(1-\lambda )}=2\alpha -1-\frac{3}{r}$, we see that $u\in L^{\infty }(0,T; \dot{H}^{\alpha })\cap L^{2}(0,T; \dot{H}^{2\alpha })$. This completes the proof.$\square$