Global well-posedness of 2D generalized MHD equations with fractional diffusion

Abstract

In this paper we prove the uniqueness of weak solutions and the global-in-time existence of smooth solutions of the 2D generalized MHD system with fractional diffusion with $\frac{1}{2}$ power.

MSC:35Q30, 76D03, 76D09.

1 Introduction

In this paper, we consider the following 2D generalized MHD system with $0<\alpha \le 1$ [1]:

$$divu=divb=0,$$

(1.1)

$${\partial }_{t}u+(u\cdot \mathrm{\nabla })u+\mathrm{\nabla }(\pi +\frac{1}{2}{|b|}^2)+{(-\mathrm{\Delta })}^{\alpha }u=b\cdot \mathrm{\nabla }b,$$

(1.2)

$${\partial }_{t}b+u\cdot \mathrm{\nabla }b-b\cdot \mathrm{\nabla }u-\mathrm{\Delta }b=0,$$

(1.3)

$$(u,b)(t=0)=(u_0,b_0).$$

(1.4)

Here, u is the fluid velocity field, π is the pressure and b is the magnetic field.

Very recently, Ji [1] used the Fourier series analysis motivated in [2] to prove the global-in-time existence of smooth solutions of problem (1.1)-(1.4) when $\frac{1}{2}<\alpha \le 1$, and Ji [1] pointed out that his result did not seem to come directly from the method like energy estimates. In this paper, we use the standard energy method to deal with the case $\alpha =\frac{1}{2}$; of course, our method also works when $\alpha >\frac{1}{2}$. We will prove the following.

Theorem 1.1 Let $\alpha =\frac{1}{2}$. Let $u_0,b_0\in H^1$ with $div{u}_0=div{b}_0=0$ in ${\mathbb{R}}^2$. Then problem (1.1)-(1.4) has a unique global-in-time weak solution $(u,b)$ satisfying

$$(u,b)\in L^{\mathrm{\infty }}(0,T;H^1),u\in L^2(0,T;H^{3/2}),b\in L^2(0,T;H^2)$$

(1.5)

for any $T>0$.

Theorem 1.2 Let $\alpha =\frac{1}{2}$. Let $u_0,b_0\in H^s$ with $s>1$ and $div{u}_0=div{b}_0=0$ in ${\mathbb{R}}^2$. Then problem (1.1)-(1.4) has a unique global-in-time smooth solution $(u,b)$ satisfying

$$u,b\in L^{\mathrm{\infty }}(0,T;H^s),u\in L^2(0,T;H^{s+\frac{1}{2}}),b\in L^2(0,T;H^{s+1})$$

(1.6)

for any $T>0$.

For 3D case and other related problems, we refer to [3, 4].

Our proof will use the following commutator estimates due to Kato and Ponce [5]:

$${\parallel {\mathrm{\Lambda }}^s(fg)-f{\mathrm{\Lambda }}^{s}g\parallel }_{L^p}\le C({\parallel \mathrm{\nabla }f\parallel }_{L^{p_1}}{\parallel {\mathrm{\Lambda }}^{s-1}g\parallel }_{L^{q_1}}+{\parallel {\mathrm{\Lambda }}^{s}f\parallel }_{L^{p_2}}{\parallel g\parallel }_{L^{q_2}}),$$

(1.7)

with $s\ge 1$, $\mathrm{\Lambda }:={(-\mathrm{\Delta })}^{1/2}$ and $\frac{1}{p}=\frac{1}{p_1}+\frac{1}{q_1}=\frac{1}{p_2}+\frac{1}{q_2}$.

2 Proof of Theorem 1.1

This section is devoted to the proof of Theorem 1.1. The global-in-time existence of weak solutions satisfying (1.5) was proved in [1, 6], we only need to show the uniqueness. Let $(u_i,{\pi }_i,b_i)$ ($i=1,2$) be two weak solutions of problem (1.1)-(1.4). We define

$$\delta u:=u_1-u_2,\delta \pi :={\pi }_1-{\pi }_2,\delta b:=b_1-b_2.$$

Then it follows from (1.1)-(1.3) that

$$div\delta u=0,div\delta b=0,$$

(2.1)

$$\begin{array}{r}{\partial }_t\delta u+u_1\cdot \mathrm{\nabla }\delta u+\delta u\cdot \mathrm{\nabla }{u}_2+\mathrm{\nabla }(\pi +\frac{1}{2}(b_1^2-b_2^2))+{(-\mathrm{\Delta })}^{1/2}\delta u\\ =b_1\cdot \mathrm{\nabla }\delta b+\delta b\cdot \mathrm{\nabla }{b}_2,\end{array}$$

(2.2)

$${\partial }_t\delta b+u_1\cdot \mathrm{\nabla }\delta b+\delta u\cdot \mathrm{\nabla }{b}_2-b_1\cdot \mathrm{\nabla }\delta u-\delta b\cdot \mathrm{\nabla }{u}_2-\mathrm{\Delta }\delta b=0.$$

(2.3)

Testing (2.2) by δu and using (1.1) and (2.1), we see that

$$\begin{array}{rl}\frac{1}{2}\frac{d}{dt}\int {|\delta u|}^{2}dx+\int {\left|{\mathrm{\Lambda }}^{1/2}\delta u\right|}^{2}dx=& -\int \delta u\cdot \mathrm{\nabla }{u}_2\cdot \delta udx\\ +\int b_1\cdot \mathrm{\nabla }\delta b\cdot \delta udx+\int \delta b\cdot \mathrm{\nabla }{b}_2\cdot \delta udx\\ =:& I_1+I_2+I_3.\end{array}$$

(2.4)

Testing (2.3) by δb and using (1.1) and (2.1), we find that

$$\begin{array}{rl}\frac{1}{2}\frac{d}{dt}\int {|\delta b|}^{2}dx+\int {|\mathrm{\nabla }\delta b|}^{2}dx=& -\int \delta u\cdot \mathrm{\nabla }{b}_2\cdot \delta bdx\\ +\int b_1\cdot \mathrm{\nabla }\delta u\cdot \delta bdx+\int \delta b\cdot \mathrm{\nabla }{u}_2\cdot \delta bdx\\ =:& I_4+I_5+I_6.\end{array}$$

(2.5)

In the following calculations, we use the Sobolev embedding ${\dot{H}}^{1/2}\subset L^4$ and the Gagliardo-Nirenberg inequalities

$${\parallel w\parallel }_{L^{8/3}}^2\le C{\parallel w\parallel }_{L^2}{\parallel {\mathrm{\Lambda }}^{1/2}w\parallel }_{L^2},$$

(2.6)

$${\parallel w\parallel }_{L^4}^2\le C{\parallel w\parallel }_{L^2}{\parallel \mathrm{\nabla }w\parallel }_{L^2}.$$

(2.7)

Using (1.1), (2.1), (1.5), (2.6) and (2.7), we bound $I_1$, $I_2+I_5$, $I_3+I_4$ and $I_6$ as follows:

$$\begin{array}{c}{I}_1\le {\parallel \mathrm{\nabla }{u}_2\parallel }_{L^4}{\parallel \delta u\parallel }_{L^{8/3}}^2\le C{\parallel u_2\parallel }_{{\dot{H}}^{3/2}}{\parallel \delta u\parallel }_{L^2}{\parallel {\mathrm{\Lambda }}^{1/2}\delta u\parallel }_{L^2}\hfill \\ \phantom{I_1}\le \frac{1}{16}{\parallel {\mathrm{\Lambda }}^{1/2}\delta u\parallel }_{L^2}^2+C{\parallel u_2\parallel }_{{\dot{H}}^{3/2}}^2{\parallel \delta u\parallel }_{L^2}^2,\hfill \\ I_2+I_5=0,\hfill \\ I_3+I_4\le C{\parallel \mathrm{\nabla }{b}_2\parallel }_{L^4}{\parallel \delta u\parallel }_{L^2}{\parallel \delta b\parallel }_{L^4}\hfill \\ \phantom{I_3+I_4}\le C{\parallel \mathrm{\nabla }{b}_2\parallel }_{L^4}{\parallel \delta u\parallel }_{L^2}{\parallel \delta b\parallel }_{L^2}^{1/2}{\parallel \mathrm{\nabla }\delta b\parallel }_{L^2}^{1/2}\hfill \\ \phantom{I_3+I_4}\le \frac{1}{16}{\parallel \mathrm{\nabla }\delta b\parallel }_{L^2}^2+C{\parallel \delta u\parallel }_{L^2}^2+C{\parallel \mathrm{\nabla }{b}_2\parallel }_{L^4}^2{\parallel \delta b\parallel }_{L^2}^2,\hfill \\ I_6\le {\parallel \mathrm{\nabla }{u}_2\parallel }_{L^2}{\parallel \delta b\parallel }_{L^4}^2\le C{\parallel \delta b\parallel }_{L^4}^2\le C{\parallel \delta b\parallel }_{L^2}{\parallel \mathrm{\nabla }\delta b\parallel }_{L^2}\hfill \\ \phantom{I_6}\le \frac{1}{16}{\parallel \mathrm{\nabla }\delta b\parallel }_{L^2}^2+C{\parallel \delta b\parallel }_{L^2}^2.\hfill \end{array}$$

Adding up (2.4) and (2.5) and using the above estimates, we conclude that

$$\begin{array}{c}\frac{1}{2}\frac{d}{dt}\int ({|\delta u|}^2+{|\delta b|}^2)dx\hfill \\ \le C{\parallel u_2\parallel }_{{\dot{H}}^{3/2}}^2{\parallel \delta u\parallel }_{L^2}^2+C{\parallel \delta u\parallel }_{L^2}^2+C{\parallel \mathrm{\nabla }{b}_2\parallel }_{L^4}^2{\parallel \delta b\parallel }_{L^2}^2+C{\parallel \delta b\parallel }_{L^2}^2,\hfill \end{array}$$

which gives

$$\delta u=\delta b=0.$$

This completes the proof.

3 Proof of Theorem 1.2

This section is devoted to the proof of Theorem 1.2. We only need to prove a priori estimates (1.6) for simplicity.

First, we have (1.5).

Applying ${\mathrm{\Lambda }}^s$ to (1.2), testing by ${\mathrm{\Lambda }}^{s}u$ and using (1.1), we see that

$$\begin{array}{r}\frac{1}{2}\frac{d}{dt}\int {\left|{\mathrm{\Lambda }}^{s}u\right|}^{2}dx+\int {\left|{\mathrm{\Lambda }}^{s+\frac{1}{2}}u\right|}^{2}dx\\ =-\int ({\mathrm{\Lambda }}^s(u\cdot \mathrm{\nabla }u)-u\mathrm{\nabla }{\mathrm{\Lambda }}^{s}u){\mathrm{\Lambda }}^{s}udx\\ +\int ({\mathrm{\Lambda }}^s(b\cdot \mathrm{\nabla }b)-b\cdot \mathrm{\nabla }{\mathrm{\Lambda }}^{s}b){\mathrm{\Lambda }}^{s}udx+\int b\cdot \mathrm{\nabla }{\mathrm{\Lambda }}^{s}b\cdot {\mathrm{\Lambda }}^{s}udx\\ =:J_1+J_2+J_3.\end{array}$$

(3.1)

Applying ${\mathrm{\Lambda }}^s$ to (1.3), testing by ${\mathrm{\Lambda }}^{s}b$ and using (1.1), we find that

$$\begin{array}{r}\frac{1}{2}\frac{d}{dt}\int {\left|{\mathrm{\Lambda }}^{s}b\right|}^{2}dx+\int {\left|{\mathrm{\Lambda }}^{s+1}b\right|}^{2}dx\\ =-\int ({\mathrm{\Lambda }}^s(u\cdot \mathrm{\nabla }b)-u\cdot \mathrm{\nabla }{\mathrm{\Lambda }}^{s}b){\mathrm{\Lambda }}^{s}bdx\\ +\int ({\mathrm{\Lambda }}^s(b\cdot \mathrm{\nabla }u)-b\cdot \mathrm{\nabla }{\mathrm{\Lambda }}^{s}u){\mathrm{\Lambda }}^{s}bdx+\int b\cdot \mathrm{\nabla }{\mathrm{\Lambda }}^{s}u\cdot {\mathrm{\Lambda }}^{s}bdx\\ =:J_4+J_5+J_6.\end{array}$$

(3.2)

Using (1.7), (2.6), (2.7) and (1.5), we bound $J_1$, $J_2$, $J_3+J_6$, $J_4$ and $J_5$ as follows:

$$\begin{array}{c}{J}_1\le C{\parallel \mathrm{\nabla }u\parallel }_{L^4}{\parallel {\mathrm{\Lambda }}^{s}u\parallel }_{L^{8/3}}^2\hfill \\ \phantom{J_1}\le C{\parallel u\parallel }_{{\dot{H}}^{3/2}}{\parallel {\mathrm{\Lambda }}^{s}u\parallel }_{L^2}{\parallel {\mathrm{\Lambda }}^{s+\frac{1}{2}}u\parallel }_{L^2}\hfill \\ \phantom{J_1}\le \frac{1}{8}{\parallel {\mathrm{\Lambda }}^{s+\frac{1}{2}}u\parallel }_{L^2}^2+C{\parallel u\parallel }_{{\dot{H}}^{3/2}}^2{\parallel {\mathrm{\Lambda }}^{s}u\parallel }_{L^2}^2,\hfill \\ J_2\le C{\parallel \mathrm{\nabla }b\parallel }_{L^4}{\parallel {\mathrm{\Lambda }}^{s}b\parallel }_{L^2}{\parallel {\mathrm{\Lambda }}^{s}u\parallel }_{L^4}\hfill \\ \phantom{J_2}\le C{\parallel \mathrm{\nabla }b\parallel }_{L^4}{\parallel {\mathrm{\Lambda }}^{s}b\parallel }_{L^2}{\parallel {\mathrm{\Lambda }}^{s+\frac{1}{2}}u\parallel }_{L^2}\hfill \\ \phantom{J_2}\le \frac{1}{8}{\parallel {\mathrm{\Lambda }}^{s+\frac{1}{2}}u\parallel }_{L^2}^2+C{\parallel \mathrm{\nabla }b\parallel }_{L^4}^2{\parallel {\mathrm{\Lambda }}^{s}b\parallel }_{L^2}^2,\hfill \\ J_3+J_6=0,\hfill \\ J_4,J_5\le C{\parallel \mathrm{\nabla }u\parallel }_{L^2}{\parallel {\mathrm{\Lambda }}^{s}b\parallel }_{L^4}^2+C{\parallel \mathrm{\nabla }b\parallel }_{L^{8/3}}{\parallel {\mathrm{\Lambda }}^{s}u\parallel }_{L^{8/3}}{\parallel {\mathrm{\Lambda }}^{s}b\parallel }_{L^4}\hfill \\ \phantom{J_4,J_5}\le C{\parallel {\mathrm{\Lambda }}^{s}b\parallel }_{L^4}^2+C{\parallel \mathrm{\nabla }b\parallel }_{L^{8/3}}^2{\parallel {\mathrm{\Lambda }}^{s}u\parallel }_{L^{8/3}}^2\hfill \\ \phantom{J_4,J_5}\le C{\parallel {\mathrm{\Lambda }}^{s}b\parallel }_{L^2}{\parallel {\mathrm{\Lambda }}^{s+1}b\parallel }_{L^2}+C{\parallel \mathrm{\nabla }b\parallel }_{L^{8/3}}^2{\parallel {\mathrm{\Lambda }}^{s}u\parallel }_{L^2}{\parallel {\mathrm{\Lambda }}^{s+\frac{1}{2}}u\parallel }_{L^2}\hfill \\ \phantom{J_4,J_5}\le \frac{1}{8}{\parallel {\mathrm{\Lambda }}^{s+1}b\parallel }_{L^2}^2+\frac{1}{8}{\parallel {\mathrm{\Lambda }}^{s+\frac{1}{2}}u\parallel }_{L^2}^2+C{\parallel {\mathrm{\Lambda }}^{s}b\parallel }_{L^2}^2+C{\parallel \mathrm{\nabla }b\parallel }_{L^{8/3}}^4{\parallel {\mathrm{\Lambda }}^{s}u\parallel }_{L^2}^2.\hfill \end{array}$$

Adding up (3.1) and (3.2) and using the above estimates, we arrive at

$$\begin{array}{c}\frac{d}{dt}\int ({\left|{\mathrm{\Lambda }}^{s}u\right|}^2+{\left|{\mathrm{\Lambda }}^{s}b\right|}^2)dx+\int ({\left|{\mathrm{\Lambda }}^{s+\frac{1}{2}}u\right|}^2+{\left|{\mathrm{\Lambda }}^{s+1}b\right|}^2)dx\hfill \\ \le C{\parallel u\parallel }_{{\dot{H}}^{3/2}}^2{\parallel {\mathrm{\Lambda }}^{s}u\parallel }_{L^2}^2+C{\parallel \mathrm{\nabla }b\parallel }_{L^4}^2{\parallel {\mathrm{\Lambda }}^{s}b\parallel }_{L^2}^2+C{\parallel {\mathrm{\Lambda }}^{s}b\parallel }_{L^2}^2+C{\parallel \mathrm{\nabla }b\parallel }_{L^{8/3}}^4{\parallel {\mathrm{\Lambda }}^{s}u\parallel }_{L^2}^2,\hfill \end{array}$$

which yields (1.6).

This completes the proof.