On the Cauchy problem for a Leray--MHD model

Nonlinear Analysis: Real World Applications 12 (2011) 648–657

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Nonlinear Analysis: Real World Applications

journal homepage: www.elsevier.com/locate/nonrwa

On the Cauchy problem for a Leray-α-MHD modelYong Zhou a,∗, Jishan Fan b,c

a Department of Mathematics, Zhejiang Normal University, Jinhua 321004, Zhejiang, PR Chinab Department of Applied Mathematics, Nanjing Forestry University, Nanjing 210037, Jiangsu, PR Chinac Department of Mathematics, Hokkaido University, Sapporo 060-0810, Japan

a r t i c l e i n f o

Article history:Received 16 March 2010Accepted 12 July 2010

Keywords:Leray-α-MHD modelPartial viscosityRegularity criterion

a b s t r a c t

In this paper, the Cauchy problem for a Leray-α-MHD model is considered for n = 2, 3.When n = 2, global existence for the strong solution is proved even for the caseswith a par-tial viscous term. For the three-dimensional case, various regularity criteria are established.

© 2010 Elsevier Ltd. All rights reserved.

1. Introduction

We study the Cauchy problem for the following Leray-α-MHD model [1]:

vt + (u · ∇)v − ϵ1v + ∇π +12∇|B|2 = (B · ∇)B, (1.1)

Bt + (u · ∇)B − (B · ∇)u − η1B = 0, (1.2)

v = (1 − α21)u, (1.3)div v = div u = div B = 0, (1.4)

(v, B)|t=0 = (v0, B0) in Rn (n = 2, 3), (1.5)

where v, the fluid velocity field, u, the ‘‘filtered’’ fluid velocity, B, the magnetic field and π , the pressure, are unknowns; ϵis the viscosity and η is the magnetic diffusivity; and α is the length scale parameter that represents the width of the filter(we will take α ≡ 1 for simplicity). When n = 2, ϵ > 0 and η > 0, it is mentioned that the existence and uniqueness ofweak solutions could be established by the same argument as that for a similar model in [1].

The first main purpose of this paper is to investigate the above Leray-α-MHD system with partial viscous term (ϵ > 0,η = 0 or ϵ = 0, η > 0). We will prove the following.

Theorem 1.1 (n = 2, ϵ > 0, η = 0). Let (v0, B0) ∈ H3(R2) and div u0 = div B0 = 0 in R2; then the problem (Leray-α-MHD)1,0 has a unique global smooth solution (v, B) such that

v ∈ L∞(0, T ;H3) ∩ L2(0, T ;H4), B ∈ L∞(0, T ;H3) for any T > 0.

∗ Corresponding author. Tel.: +86 579 82298256; fax: +86 579 82298256.E-mail addresses: [email protected] (Y. Zhou), [email protected], [email protected] (J. Fan).

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http://dx.doi.org/10.1016/j.nonrwa.2010.07.007

Y. Zhou, J. Fan / Nonlinear Analysis: Real World Applications 12 (2011) 648–657 649

Theorem 1.2 (n = 2, ϵ = 0, η > 0). Let (v0, B0) ∈ H3(R2) and div u0 = div B0 = 0 in R2; then the problem (Leray-α-MHD)0,1 has a unique global smooth solution (v, B) such that

v ∈ L∞(0, T ;H3), B ∈ L∞(0, T ;H3) ∩ L2(0, T ;H4) for any T > 0.

When α → 0, the above Leray-α-MHD model reduces to the well-known MHDmodel:

(MHD)

ut + (u · ∇)u − ϵ1u + ∇π +

12∇|B|2 = (B · ∇)B,

Bt + (u · ∇)B − (B · ∇)u − η1B = 0,div u = div B = 0.

The regularity of the three-dimensionalMHDequations is still open. However, it is proved that theweak solution remainssmooth if one of the following conditions is satisfied [2–5]:

u ∈ Lr(0, T ; Lp) with2r

+3p

= 1, 3 < p ≤ ∞; (1.6)

∇u ∈ Lr(0, T ; Lp) with2r

+3p

= 2, 3/2 < p ≤ ∞; (1.7)

u ∈ L2(0, T ; BMO), (1.8)

∇u ∈ L1(0, T ; B0∞,∞), (1.9)

where BMO denotes the space of bounded mean oscillation and B0∞,∞ denotes the homogeneous Besov space. Other types

of regularity criteria can be found in [6–10]. These conditions (including (1.6)–(1.8) are interesting, since they are nothingto do with the magnetic field B.

For n = 3, the local well-posedness for the above system can be established by a similar argument to that in [1,11]. Here,we establish regularity criteria for the Leray-α-MHD model as follows.

Theorem 1.3 (n = 3, ϵ > 0, η > 0). Let (v0, B0) ∈ H1×H1 with div u0 = div B0 = 0 in R3 and (v, u, B) be a smooth solution

((v, b) ∈ L∞(0, t;H1) ∩ L2(0, t;H2)) to (1.1)–(1.5) for 0 ≤ t < T . Then (v, u, B) is smooth at time t = T provided that one ofthe four conditions (1.6)–(1.9) is satisfied.

Remark 1.1. As B ≡ 0, the model (1.1)–(1.5) reduces to the Leray-α Navier–Stokes system, which was introduced in [11].Therefore, Theorems 1.1–1.3 also provide fundamental mathematical studies for the Leray-α Navier–Stokes system.

2. Proof of Theorem 1.1

This section is devoted to the proof of Theorem 1.1. Since it is easy to prove that there is T0 > 0 and that there is a uniquesmooth solution (v, B) to the problem in (0, T0], we only need to prove the a priori estimates. Therefore, it is sufficient toprove that, for any T > 0, there exists a constant C (independent of T ) such that

‖v‖L∞(0,T ;H3) + ‖v‖L2(0,T ;H4) ≤ C .

Here ϵ > 0 is a constant, so we can take ϵ = 1 without loss of generality. This amounts to nothing more than makingthe subsequent computation simpler.

First, multiplying (1.1) and (1.2) by u and B, respectively, due to (1.3) and (1.4) and∫R2

(u · ∇)v · udx = 0, (2.1)

we easily get

12

ddt

∫R2

u2

+ |∇u|2 + B2 dx +

∫R2

|∇u|2 + |1u|2

dx = 0.

The proof of (2.1) is given in the Appendix.Therefore,

‖u‖L∞(0,T ;H1) + ‖u‖L2(0,T ;H2) ≤ C, (2.2)

‖v‖L2(0,T ;L2) ≤ C, (2.3)

‖B‖L∞(0,T ;L2) ≤ C . (2.4)

650 Y. Zhou, J. Fan / Nonlinear Analysis: Real World Applications 12 (2011) 648–657

Taking curl in (1.1) and setting ω := curl v gives

∂tω + u · ∇ω − 1ω = B · ∇curl B +

∂u∂y

∇v1 −∂u∂x

∇v2

. (2.5)

Multiplying (2.5) by ω, using div B = 0, div u = 0, curl∇ ≡ 0 and the theory of a Hardy space in Coifman et al. [12], weobtain

12

ddt

∫R2

ω2dx +

∫R2

|∇ω|2dx =

∫R2

(B · ∇curl B) · ωdx +

∫R2

∂u∂y

∇v1 −∂u∂x

∇v2

ωdx

≤ ‖B · ∇curl B‖H1‖ω‖BMO +

∂u∂y

∇v1

H1

‖ω‖BMO +

∂u∂x

∇v2

H1

‖ω‖BMO

≤ C‖B‖L2 · ‖∇curl B‖L2 · ‖∇ω‖L2 + C‖∇u‖L2 · ‖∇v‖L2‖∇ω‖L2

≤ C‖1B‖L2 · ‖∇ω‖L2 + C‖ω‖L2 · ‖∇ω‖L2

≤18‖∇ω‖

2L2 + C‖1B‖2

L2 + C‖ω‖2L2 . (2.6)

In the following calculations, we will use the commutator estimates, and the bilinear estimates due to Kenig et al. [13]:

‖Λs(fg) − fΛsg‖Lp ≤ C(‖∇f ‖Lp1 ‖Λs−1g‖Lq1 + ‖Λsf ‖Lp2 ‖g‖Lq2 ), (2.7)

‖Λs(fg)‖Lp ≤ C(‖f ‖Lp1 ‖Λsg‖Lq1 + ‖Λsf ‖Lp2 ‖g‖Lq2 ), (2.8)

with s > 0, Λ := (−1)1/2 and 1p =

1p1

+1q1

=1p2

+1q2.

Applying 1 to (1.2), then multiplying it by 1B, using (2.7) and (2.8), we find that

12

ddt

∫R2

|1B|2dx ≤

∫R2

(1(u · ∇B) − u∇ · 1B) · 1Bdx+ ∫

R21(B · ∇u) · 1Bdx

≤ C‖∇u‖L∞ · ‖1B‖2

L2 + C‖1u‖L4 · ‖∇B‖L4 · ‖1B‖L2 + C‖B‖L∞ · ‖∇1u‖L2 · ‖1B‖L2

=: I1 + I2 + I3. (2.9)

We will use the following logarithmic Sobolev inequality due to Brezis and Gallouet [14] (see also [15]),

‖f ‖L∞ ≤ C‖f ‖H1 log(e + ‖f ‖H2), (2.10)

to estimate I1 as

I1 ≤ C‖u‖H2 log(e + ‖u‖H3) · ‖1B‖2L2 ≤ C‖v‖L2 log(e + ‖ω‖L2 + ‖1B‖L2)‖1B‖2

L2 .

I2 is estimated as

I2 ≤ C‖v‖L4 · ‖∇B‖L4 · ‖1B‖L2 ≤ C‖v‖3/4L2

‖1v‖1/4L2

· ‖B‖1/4L2

‖1B‖3/4L2

· ‖1B‖L2

≤ C‖v‖3/4L2

‖∇ω‖1/4L2

· ‖1B‖7/4L2

≤14‖∇ω‖

2L2 + C‖v‖

6/7L2

‖1B‖2L2 ,

where we have used the Gagliardo–Nirenberg inequality:

‖v‖L4 ≤ C‖v‖3/4L2

‖1v‖1/4L2

, (2.11)

‖∇B‖L4 ≤ C‖B‖1/4L2

‖1B‖3/4L2

. (2.12)

Similarly,

I3 ≤ C‖B‖L∞ · ‖∇v‖L2 · ‖1B‖L2 ≤ C‖B‖1/2L2 ‖1B‖1/2

L2 · ‖v‖1/2L2 ‖1v‖

1/2L2 · ‖1B‖L2

≤ C‖1B‖3/2L2

· ‖v‖1/2L2

· ‖∇ω‖1/2L2

≤14‖∇ω‖

2L2 + C‖v‖

2/3L2

‖1B‖2L2 ,

where we have used the Gagliardo–Nirenberg inequality:

‖B‖L∞ ≤ C‖B‖1/2L2 ‖1B‖1/2

L2 , (2.13)

‖∇v‖L2 ≤ C‖v‖1/2L2 ‖1v‖

1/2L2 . (2.14)

Inserting the above estimates into (2.9) and combining (2.6), using (2.2), (2.3) and (2.4), we conclude that

‖ω‖L∞(0,T ;L2) + ‖ω‖L2(0,T ;H1) ≤ C, (2.15)

‖B‖L∞(0,T ;H2) ≤ C; (2.16)


hence

‖v‖L∞(0,T ;H1) + ‖v‖L2(0,T ;H2) ≤ C, (2.17)

‖u‖L∞(0,T ;H3) + ‖u‖L2(0,T ;H4) ≤ C . (2.18)

Applying ∂3 to (1.2), multiplying it by ∂3B, using (2.7), (2.8) and (2.18) and the Sobolev inequality, we deduce that

12

ddt

∫R2

|∂3B|2dx ≤

∫R2

(∂3(u · ∇B) − u · ∇∂3B) · ∂3Bdx+ ∫

R2∂3(B · ∇u) · ∂3Bdx

≤ C‖∇u‖L∞‖∂3B‖2

L2 + C‖∇B‖L∞ · ‖∂3u‖L2 · ‖∂3B‖L2 + C‖B‖L∞ · ‖∂4u‖L2 · ‖∂3B‖L2

≤ C‖∂3B‖2L2 + C‖∇B‖L∞ · ‖∂3B‖L2 + C‖∂4u‖L2‖∂

3B‖L2

≤ C‖∂3B‖2L2 + C(1 + ‖∂3B‖L2)‖∂

3B‖L2 + C‖∂4u‖L2‖∂3B‖L2 ,

which yields

‖B‖L∞(0,T ;H3) ≤ C, (2.19)

by Gronwall’s inequality.Applying 1 to (2.5), then multiplying it by 1ω, using (2.8) and (2.15)–(2.19), we have

12

ddt

∫R2

|1ω|2dx +

∫R2

|∇1ω|2dx

≤

∫R2

∇(u · ∇ω) · ∇1ωdx+ ∫

R2∇

∂u∂y

∇v1 −∂u∂x

∇v2

· ∇1ωdx

+ 2−i=1

∫R2

1∂i(BiB) · curl1ωdx

≤ ‖u‖L∞ · ‖1ω‖L2 · ‖∇1ω‖L2 + ‖∇u‖L∞ · ‖∇ω‖L2 · ‖∇1ω‖L2

+ C‖1u‖L4‖ω‖L4‖∇1ω‖L2 + C‖B‖L∞ · ‖∂3B‖L2 · ‖∇1ω‖L2

≤ C‖1ω‖L2‖∇1ω‖L2 + C‖∇ω‖L2‖∇1ω‖L2 + C‖∇ω‖1/2L2

‖∇1ω‖L2 + C‖∇1ω‖L2

≤12‖∇1ω‖

2L2 + C‖∇ω‖

2L2 + C‖1ω‖

2L2 + C,

which implies that

‖ω‖L∞(0,T ;H2) + ‖ω‖L2(0,T ;H3) ≤ C . (2.20)

Hence

‖v‖L∞(0,T ;H3) + ‖v‖L2(0,T ;H4) ≤ C . (2.21)

This completes the proof.


In what follows, we take η = 1.Firstly, multiplying (1.1) and (1.2) by u and B, respectively, and summing them, by using (1.3), (1.4) and (2.1), we easily

have

12

∫R2

u2

+ |∇u|2 + B2 dx +

∫ T

0

∫R2

|∇B|2dxdt ≤12

∫R2

u20 + |∇u0|

2+ B2

0

dx. (3.1)

Multiplying (1.2) by −1B, using Hölder’s inequalities, (2.12), (2.13) and (3.1), we get

12

ddt

∫R2

|∇B|2dx +

∫R2

|1B|2dx =

∫R2

(u · ∇B − B · ∇u)1Bdx

≤‖u‖L4 · ‖∇B‖L4 + ‖B‖L∞ · ‖∇u‖L2

‖1B‖L2

≤ C‖∇B‖L4 + ‖B‖L∞

‖1B‖L2

≤ C‖B‖1/4

L2· ‖1B‖3/4

L2+ ‖B‖1/2

L2· ‖1B‖1/2

L2

‖1B‖L2

≤ C‖1B‖3/4

L2 + ‖1B‖1/2L2

‖1B‖L2

≤12‖1B‖2

L2 + C,


which yields‖B‖L∞(0,T ;H1) + ‖B‖L2(0,T ;H2) ≤ C . (3.2)

Applying 1 to (1.2), then multiplying it by 1B, we see that12

ddt

∫R2

|1B|2dx +

∫R2

|∇1B|2dx = −

∫R2

1(u · ∇B − B · ∇u) · 1Bdx

=

−k

∫R2

∂k(u · ∇B − B · ∇u) · ∂k1Bdx

=

−i,k

∫R2

∂k∂i(uiB − Biu) · ∂k1Bdx

≤ C∫

R2(|1u| · |B| + |∇u| · |∇B| + |u| · |1B|)|∇1B|dx

≤ C‖B‖L∞ · ‖1u‖L2 · ‖∇1B‖L2

+ C‖∇B‖L∞ · ‖∇u‖L2 · ‖∇1B‖L2 + C‖u‖L4 · ‖1B‖L4 · ‖∇1B‖L2

=: J1 + J2 + J3. (3.3)Using (2.13), (3.1) and (1.3), we bound J1 and J2 as

J1 ≤ C‖B‖1/2L2 ‖1B‖1/2

L2 · ‖v‖L2 · ‖∇1B‖L2 ≤18‖∇1B‖2

L2 + C‖1B‖L2‖v‖2L2 ,

J2 ≤ C‖∇B‖L∞ · ‖∇1B‖L2 ≤ C‖∇B‖1/2L2

· ‖∇1B‖3/2L2

≤18‖∇1B‖2

L2 + C‖∇B‖2L2 .

WhileJ3 ≤ C‖1B‖L4 · ‖∇1B‖L2 ≤ C‖1B‖1/2

L2 · ‖∇1B‖3/2L2

≤18‖∇1B‖2

L2 + C‖1B‖2L2 ,

by the Gagliardo–Nirenberg inequality

‖w‖2L4 ≤ C‖w‖L2 · ‖∇w‖L2 . (3.4)

Inserting the above estimates for J1, J2 and J3 into (3.3), we get

12

ddt

∫R2

|1B|2dx +58

∫R2

|∇1B|2dx ≤ C‖1B‖L2‖v‖2L2 + C‖∇B‖2

L2 + C‖1B‖2L2 . (3.5)

On the other hand, multiplying (1.1) by v, using (3.1), (2.13) and (3.4), we find that12

ddt

∫R2

v2dx ≤

∫R2

B · ∇B · vdx ≤ ‖B‖L∞ · ‖∇B‖L2 · ‖v‖L2

≤ C‖B‖L∞‖v‖L2 ≤ C‖B‖1/2L2 ‖1B‖1/2

L2 · ‖v‖L2 ≤ C‖1B‖1/2L2 ‖v‖L2 . (3.6)

Combining (3.5) and (3.6) and using (3.2), we deduce that‖B‖L∞(0,T ;H2) + ‖B‖L2(0,T ;H3) ≤ C, (3.7)

‖v‖L∞(0,T ;L2) ≤ C, (3.8)

‖u‖L∞(0,T ;H2) ≤ C . (3.9)Applying curl to (1.1), we have

ωt + u · ∇ω = B · ∇curl B +

∂u∂y

∇v1 −∂u∂x

∇v2

. (3.10)

Multiplying (3.10) by ω, using (2.13) and (3.7), (3.8) and (2.10), we obtain12

ddt

∫R2

ω2dx ≤

∫R2

(B · ∇curl B) · ωdx+ ∫

R2

∂u∂y

∇v1 −∂u∂x

∇v2

ωdx

≤ C‖B‖L∞‖1B‖L2‖ω‖L2 + C‖∇u‖L∞‖ω‖

2L2

≤ C‖ω‖L2 + C‖∇u‖H1 log(e + ‖ω‖L2) · ‖ω‖2L2

≤ C‖ω‖L2 + C‖ω‖2L2 log(e + ‖ω‖L2),


which implies that

‖v‖L∞(0,T ;H1) + ‖u‖L∞(0,T ;H3) ≤ C . (3.11)

Applying ∂3 to (1.2), then multiplying it by ∂3B, using (2.8), (3.7), (3.8), (3.9) and (3.11), we get

12

ddt

∫R2

|∂3B|2dx +

∫R2

|∇∂3B|2dx ≤

−i

∫R2

∂2∂i(uiB − Biu) · ∂4Bdx

≤ C(‖u‖L∞‖∂3B‖L2 + ‖B‖L∞‖∂3u‖L2)‖∂4B‖L2

≤ C(‖∂3B‖L2 + ‖B‖L∞)‖∂4B‖L2

≤ C(1 + ‖∂3B‖2L2) +

12‖∇∂3B‖2

L2 ,

which leads to

‖B‖L∞(0,T ;H3) + ‖B‖L2(0,T ;H4) ≤ C . (3.12)

Applying 1 to (3.10), then multiplying it by 1ω, using (2.7), (2.8), (3.11) and (3.12) and the Sobolev inequality, we have

12

ddt

∫R2

|1ω|2dx ≤

∫R2

(1(u · ∇ω) − u · ∇1ω)1ωdx+−

i

∫R2

1curl ∂i(BiB) · 1ωdx

≤ C(‖∇u‖L∞‖1ω‖L2 + ‖1u‖L4‖∇ω‖L4)‖1ω‖L2 + C‖B‖L∞ · ‖∂4B‖L2 · ‖1ω‖L2

≤ C(‖1ω‖L2 + ‖∇ω‖L4)‖1ω‖L2 + C‖∂4B‖L2‖1ω‖L2

≤ C(‖1ω‖L2 + ‖ω‖L2)‖1ω‖L2 , +C‖∂4B‖L2‖1ω‖L2 ,

which gives

‖ω‖L∞(0,T ;H2) ≤ C;

then

‖v‖L∞(0,T ;H3) ≤ C .



For simplicity, it is assumed that ϵ = η = 1. In order to establish regularity for the three-dimensional viscous case, it issufficient to prove that v ∈ L∞(0, T ;H1) ∩ L2(0, T ;H2).(I) First, we assume that (1.6) holds.

Multiplying (1.2) by |B|2B, using (1.4), we have

14

ddt

∫R3

B4dx + 3∫

R3|B|2|∇B|2dx ≤

∫R3

(B · ∇)u · |B|2Bdx

≤ C∫

R3|u| · |B|2 · |∇|B |

2|dx.

Setting H := |B|2, we get

14

ddt

∫R3

H2dx +34

∫R3

|∇H|2dx ≤ C

∫R3

|u| · H · ∇Hdx ≤ C‖u‖Lp‖H‖L

2pp−2

‖∇H‖L2

≤ C‖u‖Lp‖H‖1−3/pL2

‖∇H‖1+3/pL2

≤12‖∇H‖

2L2 + C‖u‖r

Lp‖H‖2L2 ,

which yields

‖B‖L∞(0,T ;L4) + ‖B∇B‖L2(0,T ;L2) ≤ C, (4.1)

where we have used the following Gagliardo–Nirenberg inequality:

‖H‖L

2pp−2

≤ C‖H‖1−3/pL2

‖∇H‖3/pL2

with p > 3. (4.2)


Multiplying (1.1) by v, using (1.4) and (4.1), we obtain

12

ddt

∫R3

v2dx +

∫R3

|∇v|2dx =

∫R3

(B · ∇)B · vdx ≤ ‖B · ∇B‖L2‖v‖L2;

hence

‖v‖L∞(0,T ;L2) + ‖v‖L2(0,T ;H1) ≤ C, (4.3)

‖u‖L∞(0,T ;H2) + ‖u‖L2(0,T ;H3) ≤ C . (4.4)

Multiplying (1.2) by B, using (1.4), (4.1) and (4.4), we find that

12

ddt

∫R3

B2dx +

∫R3

|∇B|2dx =

∫R3

(B · ∇)u · Bdx

= −

∫R3

(B · ∇B) · udx

≤ ‖B · ∇B‖L2‖u‖L2 ≤ C‖B · ∇B‖L2 ,

which implies that

‖B‖L∞(0,T ;L2) + ‖B‖L2(0,T ;H1) ≤ C . (4.5)

Multiplying (1.2) by −1B, using (1.4), (4.1) and (4.4), we get

12

ddt

∫R3

|∇B|2dx +

∫R3

|1B|2dx =

∫R3

[(u · ∇)B − (B · ∇)u]1Bdx

≤ (‖u‖L∞‖∇B‖L2 + ‖B‖L4‖∇u‖L4)‖1B‖L2

≤ C(‖∇B‖L2 + 1)‖1B‖L2 ,

which leads to

‖B‖L∞(0,T ;H1) + ‖B‖L2(0,T ;H2) ≤ C . (4.6)

Multiplying (1.1) by −1v, using (1.4), (4.1) and (4.4), we obtain

12

ddt

∫R3

|∇v|2dx +

∫R3

|1v|2dx =

∫R3

[(u · ∇)v − (B · ∇)B]1vdx

≤ (‖u‖L∞‖∇v‖L2 + ‖B · ∇B‖L2)‖1v‖L2

≤ C(‖∇v‖L2 + ‖B · ∇B‖L2)‖1v‖L2 ,

which implies that

‖v‖L∞(0,T ;H1) + ‖v‖L2(0,T ;H2) ≤ C . (4.7)

(II) Next, we assume that (1.7) is true.Multiplying (1.2) by |B|2B, using (1.4) and setting H := |B|2, we find that

14

ddt

∫R3

H2dx +34

∫R3

|∇H|2dx ≤

∫R3

(B · ∇)u · |B|2Bdx

≤

∫R3

|∇u| · H2dx ≤ ‖∇u‖Lp‖H‖2

L2pp−1

≤ C‖∇u‖Lp‖H‖2(1−3/2p)L2

‖∇H‖3/pL2

≤12‖∇H‖

2L2 + C‖∇u‖r

Lp‖H‖2L2 ,

which yields (4.1). Here we have used the following Gagliardo–Nirenberg inequality:

‖H‖L

2pp−1

≤ C‖H‖1−3/2pL2

‖∇H‖3/2pL2

with p >32. (4.8)

Now, using the same argument as that in case (I), we have (4.6) and (4.7).This completes the proof for this case.

(III) Now, we assume that (1.8) holds.Multiplying (1.2) by |B|2B, using (1.4), we find that

14

ddt

∫R3

B4dx +

∫R3

|B|2|∇B|2dx +12

∫R3

(∇|B|2)2dx ≤

∫R3

(B · ∇)u · |B|2Bdx = −

∫B · ∇|B|2B · udx. (4.9)


Since div B = 0, curl∇ ≡ 0, it follows from Coifman et al. [12] that∫R3

(B · ∇)|B|2B · udx ≤ ‖(B · ∇)|B|2B‖H1 · ‖u‖BMO

≤ C‖B‖L4‖∇(|B|2)B‖L4/3 · ‖u‖BMO. (4.10)

Thanks to Hölder’s inequality, one has

‖∇(|B|2B)‖L4/3 ≤ 3‖B‖L4‖ |B|∇B‖L2 . (4.11)

Inserting (4.10) into (4.9) and using (4.11) and Gronwall’s inequality, we get (4.1). Then, using the same calculations asthat in case (I), (4.6) and (4.7) can be established.(IV) Finally, we assume that (1.9) is true.

Multiplying (1.2) by |B|2B, using (1.4) and setting H := |B|2, we find that

14

ddt

∫R3

H2dx +12

∫R3

|∇H|2dx +

∫R3

|B|2|∇B|2dx =

∫R3

(B · ∇)u · |B|2Bdx ≤

∫R3

H2|∇u|dx. (4.12)

By using Littlewood–Paley decomposition, we decompose ∇u as follows.

∇u =

∞−i=−∞

1i(∇u) =

−i<−N

1i(∇u) +

N−i=−N

1i(∇u) +

−i>N

1i(∇u),

where N is a positive integer to be chosen later. Substituting this decomposition into (4.12), we obtain the following.

The right-hand side of (4.12) ≤

−i<−N

∫R3

H2|1i(∇u)|dx +

N−i=−N

∫R3

H2|1i(∇u)|dx

+

−i>N

∫R3

H2|1i(∇u)|dx =: R1 + R2 + R3. (4.13)

Then we do an estimate for each Ri (i = 1, 2, 3). First, recalling that

‖1if ‖Lq ≤ C23i1p −

1q

‖1if ‖Lp , 1 ≤ p ≤ q ≤ ∞, (4.14)

with C being a positive constant independent of f and j, we apply Hölder’s inequality to infer that

R1 ≤ ‖H‖2L2

−i<−N

‖1i(∇u)‖L∞

≤ C‖H‖2L2

−i<−N

232 i‖1i(∇u)‖L2

≤ C2−3/2N‖∇u‖L2‖H‖

2L2

≤ C2−3/2N‖v‖L2‖H‖

2L2 .

By Hölder’s inequality, R2 can be estimated as

R2 ≤ ‖H‖2L2

N−i=−N

‖1i(∇u)‖L∞ ≤ CN‖∇u‖B0∞,∞‖H‖

2L2 .

Finally, for R3, we make use of (4.14) to deduce that

R3 ≤ ‖H‖L6‖H‖L2−i>N

‖1i(∇u)‖L3

≤ C‖∇H‖L2‖H‖L2−i>N

2i/2‖1i(∇u)‖L2

≤ C‖∇H‖L2‖H‖L2

−i>N

2−i

1/2 −i>N

22i‖1i(∇u)‖2

L2

1/2

≤ C2−N/2‖∇H‖L2‖H‖L2‖1u‖L2

≤ C2−N/2‖∇H‖L2‖H‖L2‖v‖L2 .

Now, we choose N so that 2−N/2(‖v‖L2 + ‖H‖L2) ≤ 1, i.e.,

N ≥ 2 log+

2 (‖v‖L2 + ‖H‖L2),


to conclude that14

ddt

∫R3

H2dx +12

∫R3

|∇H|2dx +

∫R3

|B|2|∇B|2dx

≤ C‖H‖2L2 + C‖∇u‖B0∞,∞

‖H‖2L2 log

+(‖v‖L2 + ‖H‖L2) + C‖∇H‖L2‖H‖L2 . (4.15)

On the other hand, multiplying (1.1) by v, using (1.4), we have

12

ddt

∫R3

v2dx +

∫R3

|∇v|2dx =

∫R3

(B · ∇B)vdx ≤14

∫R3

|B|2|∇B|2dx +

∫R3

v2dx. (4.16)

Combining (4.15) and (4.16) and using Gronwall’s inequality, we get (4.1), (4.3) and (4.4). Then, using the samecalculations as in case (I), (4.6) and (4.7) hold.


Acknowledgements

The authors would like to thank the referee for his/her careful reading and helpful suggestions. This work is partiallysupported by Zhejiang Innovation Project (Grant No. T200905), ZJNSF (Grant No. R6090109) andNSFC (Grant No. 10971197).

Appendix

The purpose of this appendix is to give a proof for (2.1). First, let us define the curl operator in two dimensions for a vectorfield and a scalar, respectively, as

curl u = ∂1u2 − ∂2u1 ∈ R, for u ∈ R2,

and

curlφ = (∂2φ, −∂1φ) ∈ R2, for φ ∈ R.

Then, it is easy to check that

−1u = curl curl u, if div u = 0.

It is well known that for the two-dimensional case the following identity holds:

curl (u · ∇u) = u · ∇curl u, (A.1)

provided that div u = 0.Thus, if div u = 0, we have∫

R2(u · ∇u) · 1udx = −

∫R2

(u · ∇u) · curl curl udx

= −

∫R2

curl (u · ∇u) · curl udx

= −

∫R2

u · ∇(curl u) · (curl u)dx = 0.

Then (2.1) is followed from∫R2

(u · ∇)v · udx =

∫R2

(u · ∇)(u − 1u) · udx

=

∫R2

(u · ∇)u · udx −

∫R2

(u · ∇)1u · udx

= 0.

For the three-dimensional case, we still have

−1u = curl curl u, if div u = 0.

But (A.1) does not hold:

curl (u · ∇u) = u · ∇curl u − curl u · ∇u.

So, (2.1) is not true for the three-dimensional case.


References

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On the Cauchy problem for a Leray--MHD model

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