Global regularity for 2D fractional magneto-micropolar equations

Abstract

The magneto-micropolar equations are important models in fluid mechanics and material sciences. This paper focuses on the global regularity problem on the 2D magneto-micropolar equations with fractional dissipation. We establish the global regularity for three important fractional dissipation cases. Direct energy estimates are not sufficient to obtain the desired global a priori bounds in each case. To overcome the difficulties, we employ various technics including the regularization of generalized heat operators on the Fourier frequency localized functions, logarithmic Sobolev interpolation inequalities and the maximal regularity property of the heat operator.

Appendix A: An alternative approach to the global \(H^1\)-bound for (1.2)

A crucial component in the proof of the global bounds for Theorem 1.1 is the global (in time) $H^1$-bound and the global integrability of $\Vert \Lambda ^{\frac{2}{q}} \omega (\tau )\Vert _{L^q}$. Sect. 2 provided one way to get these bounds. The purpose of this appendix is to provide an alternative approach to these bounds. More precisely, we prove the following result.

Proposition A.1

Assume that $(u, \Omega , b)$ solves (1.2). Assume $\alpha$ and $\beta$ satisfy

$$\begin{aligned} 1<\alpha<2, \quad 0<\beta <1, \quad \alpha + \beta \ge 2. \end{aligned}$$

Then, for any $1<q<\infty$ and for any $t>0$,

$$\begin{aligned} \sup _{l\ge -1} \,2^{(2\alpha -2)\, l}\, \int _0^t \Vert \Delta _l \omega (\tau )\Vert _{L^{q}} \,d\tau \le C(t, u_0, \Omega _0, b_0) < \infty . \end{aligned}$$

(A.1)

As a special consequence, for any $1<q<\infty$ and $\rho < 2\alpha -2$,

$$\begin{aligned} \int _0^t \Vert \omega (\tau )\Vert _{B^{\rho }_{q,1}}\,d\tau < \infty . \end{aligned}$$

(A.2)

Especially, for $\rho < 2\alpha -2$ and $q=\frac{2}{\rho }$,

$$\begin{aligned} \int _0^t \left( \Vert \omega (\tau )\Vert _{L^q} + \Vert \Lambda ^{\frac{2}{q}} \omega \Vert _{L^q}\right) \, d\tau <\infty . \end{aligned}$$

(A.3)

Proof of Proposition A.1

We start with the following fact about $\Omega$, for any $r\in [1,\infty ]$,

$$\begin{aligned} \Vert \Omega (t)\Vert _{L^r} \le \Vert \Omega _0\Vert _{L^r} + 2 \kappa \int _0^t \Vert \omega (\tau )\Vert _{L^r}\,d\tau , \end{aligned}$$

(A.4)

which can be obtained by performing the standard Lebesgue norm estimate on the equation of $\Omega$. Next, we write the equation of $\omega$ given by (1.5) in the integral form

$$\begin{aligned} \omega (t)= & {} e^{-(\nu +\kappa )(-\Delta )^\alpha t} \omega _0 + \int _0^t e^{-(\nu +\kappa )(-\Delta )^\alpha (t-\tau )} (-\Delta \Omega (\tau ))\,d\tau \\&+ \int _0^t e^{-(\nu +\kappa )(-\Delta )^\alpha (t-\tau )}\nabla \times \nabla \cdot \left( (u\otimes u)(\tau ) + \nabla \times \nabla \cdot (b\otimes b)(\tau )\right) \,d\tau . \end{aligned}$$

We further localize it by applying $\Delta _l$ with $l\in \mathbb {Z}$ and $l\ge -1$,

$$\begin{aligned} \Delta _l\omega (t)= & {} \Delta _l e^{-(\nu +\kappa ) (-\Delta )^\alpha t} \omega _0 + \int _0^t \Delta _l e^{-(\nu +\kappa )(-\Delta )^\alpha (t-\tau )} (-\Delta \Omega (\tau ))\,d\tau \\&+ \int _0^t \Delta _le^{-(\nu +\kappa ) (-\Delta )^\alpha (t-\tau )} \nabla \times \nabla \cdot \left( (u\otimes u)(\tau ) + (b\otimes b)(\tau )\right) \,d\tau . \end{aligned}$$

For $q\in (1,\infty )$, taking the $L^q$-norm and applying Lemma 2.3 and Bernstein’s inequality, we have

$$\begin{aligned}&\Vert \Delta _l \omega (t)\Vert _{L^q} \le C\, e^{-c_0 (\nu +\kappa ) t\, 2^{2\alpha l}}\, \Vert \Delta _l \omega _0\Vert _{L^q}+ C \, \int _0^t 2^{2l} e^{-c_0 (\nu +\kappa ) (t-\tau )\, 2^{2\alpha l}} \Vert \Delta _l \Omega \Vert _{L^q}\,d\tau \nonumber \\&\quad + \,C\, \int _0^t 2^{2l} e^{-c_0 (\nu +\kappa ) (t-\tau )\, 2^{2\alpha l}} \left( \Vert \Delta _l(u\otimes u)(\tau )\Vert _{L^q} +\Vert \Delta _l(b\otimes b)(\tau )\Vert _{L^q}\right) \,d\tau . \end{aligned}$$

(A.5)

To further improve the estimates, we invoke (A.4) to obtain

$$\begin{aligned} \Vert \Delta _l \Omega (t)\Vert _{L^q} \le \Vert \Omega (t)\Vert _{L^q} \le \Vert \Omega _0\Vert _{L^q} + \int _0^t \Vert \omega (\tau )\Vert _{L^q}\,d\tau . \end{aligned}$$

To bound $\Vert \Delta _l(b\otimes b)(\tau )\Vert _{L^q}$, noticing $\alpha +\beta =2$, we choose $\sigma$ satisfying

$$\begin{aligned} 0<\sigma <\alpha -1, \quad \sigma + \beta =1-\frac{1}{q}. \end{aligned}$$

By Sobolev’s inequality,

$$\begin{aligned} \Vert \Delta _l(b\otimes b)(\tau )\Vert _{L^q} \le \Vert b\otimes b(\tau )\Vert _{L^q} \le \Vert b\Vert ^2_{L^{2q}} \le C \Vert \Lambda ^{\sigma + \beta } b\Vert _{L^2}^2. \end{aligned}$$

Similarly, we can also estimate $\Vert \Delta _l(u\otimes u)(\tau )\Vert _{L^q}$ ,

$$\begin{aligned} \Vert \Delta _l(u\otimes u)(\tau )\Vert _{L^q} \le C \left( \Vert u\Vert _{L^2}^2 + \Vert \Lambda ^\alpha u\Vert _{L^2}^2\right) = C\, \Vert u\Vert _{H^\alpha }^2. \end{aligned}$$

Integrating (A.5) in time and applying Young’s inequality for the time convolution, we obtain

$$\begin{aligned} \int _0^t \Vert \Delta _l \omega (\tau )\Vert _{L^{q}} \,d\tau\le & {} C\, 2^{-2\alpha l}\, \Vert \Delta _l \omega _0\Vert _{L^q}\\&+ \,C\, 2^{(2-2\alpha ) l}\, \int _0^t \left( \Vert \Omega _0\Vert _{L^q} + \int _0^\tau \Vert \omega \Vert _{L^q} ds \right) \,d\tau \\&+ \,C\, 2^{(2-2\alpha ) l}\, \int _0^t \left( \Vert u\Vert _{H^\alpha }^2 + \Vert \Lambda ^{\sigma +\beta } b\Vert _{L^2}^2\right) \, d\tau , \end{aligned}$$

or

$$\begin{aligned} 2^{(2\alpha -2) l}\, \int _0^t \Vert \Delta _l \omega (\tau )\Vert _{L^{q}} \,d\tau \le A(t) + C\, \int _0^t \int _0^\tau \Vert \omega \Vert _{L^q} ds \,d\tau , \end{aligned}$$

(A.6)

where, due to Lemma 2.7,

$$\begin{aligned} A(t) = C\, 2^{-2l}\, \Vert \Delta _l \omega _0\Vert _{L^q} + C\, \Vert \Omega _0\Vert _{L^q} t + C\, \int _0^t \left( \Vert u\Vert _{H^\alpha }^2 + \Vert \Lambda ^{\sigma +\beta } b\Vert _{L^2}^2\right) \, d\tau <\infty . \end{aligned}$$

Noting that

$$\begin{aligned} \Vert \omega \Vert _{L^q} \le \sum _{m\ge -1} \Vert \Delta _m \omega \Vert _{L^q} \le \sum _{m\ge -1} 2^{(2-2\alpha )m} \, 2^{(2\alpha -2)m}\, \Vert \Delta _m \omega \Vert _{L^q}, \end{aligned}$$

we obtain, due to $\alpha >1$,

$$\begin{aligned} \int _0^t \int _0^\tau \Vert \omega \Vert _{L^q} ds \,d\tau= & {} \int _0^t \sum _{m\ge -1} 2^{(2-2\alpha )m}\, 2^{(2\alpha -2)m}\,\int _0^\tau \Vert \Delta _m \omega \Vert _{L^q} \,ds\, d\tau \\\le & {} C\, \int _0^t \sup _{m\ge -1} 2^{(2\alpha -2)m}\,\int _0^\tau \Vert \Delta _m \omega \Vert _{L^q} \,ds\, d\tau . \end{aligned}$$

Inserting this bound in (A.6) yields, for $Y(t) \equiv \sup _{l\ge -1} \,2^{(2\alpha -2) l}\, \int _0^t \Vert \Delta _l \omega (\tau )\Vert _{L^{q}} \,d\tau$,

$$\begin{aligned} Y(t) \le A(t) + C\, \int _0^t Y(\tau )\,d\tau . \end{aligned}$$

Gronwall’s inequality then leads to the global bound in (A.1). It is easy to see that (A.2) is a special consequence of (A.1). In fact, for any $\rho <2\alpha -2$,

$$\begin{aligned} \int _0^t \Vert \omega (\tau )\Vert _{B^{\rho }_{q,1}}\,d\tau= & {} \int _0^t \sum _{l\ge -1} 2^{\rho l} \Vert \Delta _l \omega \Vert _{L^q}\, d\tau \\\le & {} \int _0^t \sum _{l\ge -1} 2^{(\rho -(2\alpha -2))l}\, 2^{(2\alpha -2) l} \Vert \Delta _l \omega \Vert _{L^q}\, d\tau \\= & {} \sum _{l\ge -1} 2^{(\rho -(2\alpha -2))l}\, 2^{(2\alpha -2) l} \int _0^t \Vert \Delta _l \omega \Vert _{L^q}\, d\tau \\\le & {} \sup _{l\ge -1} 2^{(2\alpha -2) l} \int _0^t \Vert \Delta _l \omega \Vert _{L^q}\, d\tau \, \sum _{l\ge -1} 2^{(\rho -(2\alpha -2))l}\\= & {} C\, \sup _{l\ge -1} 2^{(2\alpha -2) l} \int _0^t \Vert \Delta _l \omega \Vert _{L^q}\, d\tau <\infty , \end{aligned}$$

which is (A.2). Finally (A.3) follows from (A.2) by taking $q=\frac{2}{\rho }$. This completes the proof of Proposition A.1. $\square$