# 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 $0<\alpha \le 1$ :

$divu=divb=0,$
(1.1)
${\partial }_{t}u+\left(u\cdot \mathrm{\nabla }\right)u+\mathrm{\nabla }\left(\pi +\frac{1}{2}{|b|}^{2}\right)+{\left(-\mathrm{\Delta }\right)}^{\alpha }u=b\cdot \mathrm{\nabla }b,$
(1.2)
${\partial }_{t}b+u\cdot \mathrm{\nabla }b-b\cdot \mathrm{\nabla }u-\mathrm{\Delta }b=0,$
(1.3)
$\left(u,b\right)\left(t=0\right)=\left({u}_{0},{b}_{0}\right).$
(1.4)

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

Very recently, Ji  used the Fourier series analysis motivated in  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  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 $\left(u,b\right)$ satisfying

$\left(u,b\right)\in {L}^{\mathrm{\infty }}\left(0,T;{H}^{1}\right),\phantom{\rule{2em}{0ex}}u\in {L}^{2}\left(0,T;{H}^{3/2}\right),\phantom{\rule{2em}{0ex}}b\in {L}^{2}\left(0,T;{H}^{2}\right)$
(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 $\left(u,b\right)$ satisfying

$u,b\in {L}^{\mathrm{\infty }}\left(0,T;{H}^{s}\right),\phantom{\rule{2em}{0ex}}u\in {L}^{2}\left(0,T;{H}^{s+\frac{1}{2}}\right),\phantom{\rule{2em}{0ex}}b\in {L}^{2}\left(0,T;{H}^{s+1}\right)$
(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 :

${\parallel {\mathrm{\Lambda }}^{s}\left(fg\right)-f{\mathrm{\Lambda }}^{s}g\parallel }_{{L}^{p}}\le C\left({\parallel \mathrm{\nabla }f\parallel }_{{L}^{{p}_{1}}}{\parallel {\mathrm{\Lambda }}^{s-1}g\parallel }_{{L}^{{q}_{1}}}+{\parallel {\mathrm{\Lambda }}^{s}f\parallel }_{{L}^{{p}_{2}}}{\parallel g\parallel }_{{L}^{{q}_{2}}}\right),$
(1.7)

with $s\ge 1$, $\mathrm{\Lambda }:={\left(-\mathrm{\Delta }\right)}^{1/2}$ 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 $\left({u}_{i},{\pi }_{i},{b}_{i}\right)$ ($i=1,2$) be two weak solutions of problem (1.1)-(1.4). We define

$\delta u:={u}_{1}-{u}_{2},\phantom{\rule{2em}{0ex}}\delta \pi :={\pi }_{1}-{\pi }_{2},\phantom{\rule{2em}{0ex}}\delta b:={b}_{1}-{b}_{2}.$

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

$div\delta u=0,\phantom{\rule{2em}{0ex}}div\delta b=0,$
(2.1)
$\begin{array}{r}{\partial }_{t}\delta u+{u}_{1}\cdot \mathrm{\nabla }\delta u+\delta u\cdot \mathrm{\nabla }{u}_{2}+\mathrm{\nabla }\left(\pi +\frac{1}{2}\left({b}_{1}^{2}-{b}_{2}^{2}\right)\right)+{\left(-\mathrm{\Delta }\right)}^{1/2}\delta u\\ \phantom{\rule{1em}{0ex}}={b}_{1}\cdot \mathrm{\nabla }\delta b+\delta b\cdot \mathrm{\nabla }{b}_{2},\end{array}$
(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

$\begin{array}{rl}\frac{1}{2}\frac{d}{dt}\int {|\delta u|}^{2}\phantom{\rule{0.2em}{0ex}}dx+\int {|{\mathrm{\Lambda }}^{1/2}\delta u|}^{2}\phantom{\rule{0.2em}{0ex}}dx=& -\int \delta u\cdot \mathrm{\nabla }{u}_{2}\cdot \delta u\phantom{\rule{0.2em}{0ex}}dx\\ +\int {b}_{1}\cdot \mathrm{\nabla }\delta b\cdot \delta u\phantom{\rule{0.2em}{0ex}}dx+\int \delta b\cdot \mathrm{\nabla }{b}_{2}\cdot \delta u\phantom{\rule{0.2em}{0ex}}dx\\ =:& {I}_{1}+{I}_{2}+{I}_{3}.\end{array}$
(2.4)

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

$\begin{array}{rl}\frac{1}{2}\frac{d}{dt}\int {|\delta b|}^{2}\phantom{\rule{0.2em}{0ex}}dx+\int {|\mathrm{\nabla }\delta b|}^{2}\phantom{\rule{0.2em}{0ex}}dx=& -\int \delta u\cdot \mathrm{\nabla }{b}_{2}\cdot \delta b\phantom{\rule{0.2em}{0ex}}dx\\ +\int {b}_{1}\cdot \mathrm{\nabla }\delta u\cdot \delta b\phantom{\rule{0.2em}{0ex}}dx+\int \delta b\cdot \mathrm{\nabla }{u}_{2}\cdot \delta b\phantom{\rule{0.2em}{0ex}}dx\\ =:& {I}_{4}+{I}_{5}+{I}_{6}.\end{array}$
(2.5)

In the following calculations, we use the Sobolev embedding ${\stackrel{˙}{H}}^{1/2}\subset {L}^{4}$ 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:

$\begin{array}{c}{I}_{1}\le {\parallel \mathrm{\nabla }{u}_{2}\parallel }_{{L}^{4}}{\parallel \delta u\parallel }_{{L}^{8/3}}^{2}\le C{\parallel {u}_{2}\parallel }_{{\stackrel{˙}{H}}^{3/2}}{\parallel \delta u\parallel }_{{L}^{2}}{\parallel {\mathrm{\Lambda }}^{1/2}\delta u\parallel }_{{L}^{2}}\hfill \\ \phantom{{I}_{1}}\le \frac{1}{16}{\parallel {\mathrm{\Lambda }}^{1/2}\delta u\parallel }_{{L}^{2}}^{2}+C{\parallel {u}_{2}\parallel }_{{\stackrel{˙}{H}}^{3/2}}^{2}{\parallel \delta u\parallel }_{{L}^{2}}^{2},\hfill \\ {I}_{2}+{I}_{5}=0,\hfill \\ {I}_{3}+{I}_{4}\le C{\parallel \mathrm{\nabla }{b}_{2}\parallel }_{{L}^{4}}{\parallel \delta u\parallel }_{{L}^{2}}{\parallel \delta b\parallel }_{{L}^{4}}\hfill \\ \phantom{{I}_{3}+{I}_{4}}\le C{\parallel \mathrm{\nabla }{b}_{2}\parallel }_{{L}^{4}}{\parallel \delta u\parallel }_{{L}^{2}}{\parallel \delta b\parallel }_{{L}^{2}}^{1/2}{\parallel \mathrm{\nabla }\delta b\parallel }_{{L}^{2}}^{1/2}\hfill \\ \phantom{{I}_{3}+{I}_{4}}\le \frac{1}{16}{\parallel \mathrm{\nabla }\delta b\parallel }_{{L}^{2}}^{2}+C{\parallel \delta u\parallel }_{{L}^{2}}^{2}+C{\parallel \mathrm{\nabla }{b}_{2}\parallel }_{{L}^{4}}^{2}{\parallel \delta b\parallel }_{{L}^{2}}^{2},\hfill \\ {I}_{6}\le {\parallel \mathrm{\nabla }{u}_{2}\parallel }_{{L}^{2}}{\parallel \delta b\parallel }_{{L}^{4}}^{2}\le C{\parallel \delta b\parallel }_{{L}^{4}}^{2}\le C{\parallel \delta b\parallel }_{{L}^{2}}{\parallel \mathrm{\nabla }\delta b\parallel }_{{L}^{2}}\hfill \\ \phantom{{I}_{6}}\le \frac{1}{16}{\parallel \mathrm{\nabla }\delta b\parallel }_{{L}^{2}}^{2}+C{\parallel \delta b\parallel }_{{L}^{2}}^{2}.\hfill \end{array}$

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

$\begin{array}{c}\frac{1}{2}\frac{d}{dt}\int \left({|\delta u|}^{2}+{|\delta b|}^{2}\right)\phantom{\rule{0.2em}{0ex}}dx\hfill \\ \phantom{\rule{1em}{0ex}}\le C{\parallel {u}_{2}\parallel }_{{\stackrel{˙}{H}}^{3/2}}^{2}{\parallel \delta u\parallel }_{{L}^{2}}^{2}+C{\parallel \delta u\parallel }_{{L}^{2}}^{2}+C{\parallel \mathrm{\nabla }{b}_{2}\parallel }_{{L}^{4}}^{2}{\parallel \delta b\parallel }_{{L}^{2}}^{2}+C{\parallel \delta b\parallel }_{{L}^{2}}^{2},\hfill \end{array}$

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

$\begin{array}{r}\frac{1}{2}\frac{d}{dt}\int {|{\mathrm{\Lambda }}^{s}u|}^{2}\phantom{\rule{0.2em}{0ex}}dx+\int {|{\mathrm{\Lambda }}^{s+\frac{1}{2}}u|}^{2}\phantom{\rule{0.2em}{0ex}}dx\\ \phantom{\rule{1em}{0ex}}=-\int \left({\mathrm{\Lambda }}^{s}\left(u\cdot \mathrm{\nabla }u\right)-u\mathrm{\nabla }{\mathrm{\Lambda }}^{s}u\right){\mathrm{\Lambda }}^{s}u\phantom{\rule{0.2em}{0ex}}dx\\ \phantom{\rule{2em}{0ex}}+\int \left({\mathrm{\Lambda }}^{s}\left(b\cdot \mathrm{\nabla }b\right)-b\cdot \mathrm{\nabla }{\mathrm{\Lambda }}^{s}b\right){\mathrm{\Lambda }}^{s}u\phantom{\rule{0.2em}{0ex}}dx+\int b\cdot \mathrm{\nabla }{\mathrm{\Lambda }}^{s}b\cdot {\mathrm{\Lambda }}^{s}u\phantom{\rule{0.2em}{0ex}}dx\\ \phantom{\rule{1em}{0ex}}=:{J}_{1}+{J}_{2}+{J}_{3}.\end{array}$
(3.1)

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

$\begin{array}{r}\frac{1}{2}\frac{d}{dt}\int {|{\mathrm{\Lambda }}^{s}b|}^{2}\phantom{\rule{0.2em}{0ex}}dx+\int {|{\mathrm{\Lambda }}^{s+1}b|}^{2}\phantom{\rule{0.2em}{0ex}}dx\\ \phantom{\rule{1em}{0ex}}=-\int \left({\mathrm{\Lambda }}^{s}\left(u\cdot \mathrm{\nabla }b\right)-u\cdot \mathrm{\nabla }{\mathrm{\Lambda }}^{s}b\right){\mathrm{\Lambda }}^{s}b\phantom{\rule{0.2em}{0ex}}dx\\ \phantom{\rule{2em}{0ex}}+\int \left({\mathrm{\Lambda }}^{s}\left(b\cdot \mathrm{\nabla }u\right)-b\cdot \mathrm{\nabla }{\mathrm{\Lambda }}^{s}u\right){\mathrm{\Lambda }}^{s}b\phantom{\rule{0.2em}{0ex}}dx+\int b\cdot \mathrm{\nabla }{\mathrm{\Lambda }}^{s}u\cdot {\mathrm{\Lambda }}^{s}b\phantom{\rule{0.2em}{0ex}}dx\\ \phantom{\rule{1em}{0ex}}=:{J}_{4}+{J}_{5}+{J}_{6}.\end{array}$
(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 }_{{\stackrel{˙}{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 }_{{\stackrel{˙}{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

$\begin{array}{c}\frac{d}{dt}\int \left({|{\mathrm{\Lambda }}^{s}u|}^{2}+{|{\mathrm{\Lambda }}^{s}b|}^{2}\right)\phantom{\rule{0.2em}{0ex}}dx+\int \left({|{\mathrm{\Lambda }}^{s+\frac{1}{2}}u|}^{2}+{|{\mathrm{\Lambda }}^{s+1}b|}^{2}\right)\phantom{\rule{0.2em}{0ex}}dx\hfill \\ \phantom{\rule{1em}{0ex}}\le C{\parallel u\parallel }_{{\stackrel{˙}{H}}^{3/2}}^{2}{\parallel {\mathrm{\Lambda }}^{s}u\parallel }_{{L}^{2}}^{2}+C{\parallel \mathrm{\nabla }b\parallel }_{{L}^{4}}^{2}{\parallel {\mathrm{\Lambda }}^{s}b\parallel }_{{L}^{2}}^{2}+C{\parallel {\mathrm{\Lambda }}^{s}b\parallel }_{{L}^{2}}^{2}+C{\parallel \mathrm{\nabla }b\parallel }_{{L}^{8/3}}^{4}{\parallel {\mathrm{\Lambda }}^{s}u\parallel }_{{L}^{2}}^{2},\hfill \end{array}$

which yields (1.6).

This completes the proof.

## References

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4. Zhou Y, Fan J: A regularity criterion for the 2D MHD system with zero magnetic diffusivity. J. Math. Anal. Appl. 2011, 378(1):169–172. 10.1016/j.jmaa.2011.01.014

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## Author information

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Correspondence to Zhiqiang Wei.

### Competing interests

The authors declare that they have no competing interests.

### Authors’ contributions

ZW proposed the problems and finished the whole manuscript. WZ modified the proofs. All authors read and approved the final manuscript.

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Wei, Z., Zhu, W. Global well-posedness of 2D generalized MHD equations with fractional diffusion. J Inequal Appl 2013, 489 (2013). https://doi.org/10.1186/1029-242X-2013-489

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• DOI: https://doi.org/10.1186/1029-242X-2013-489

### Keywords

• MHD
• fractional diffusion
• uniqueness
• smooth solution 