In addition, for the standard Hall-MHD system, α=β= 1, by adding a suitable condition, we give a positive answer to the open question in [3]

Electronic Journal of Differential Equations, Vol. 2015 (2015), No. 179, pp. 1–18.

ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu ftp ejde.math.txstate.edu

GLOBAL REGULARITY FOR GENERALIZED HALL MAGNETO-HYDRODYNAMICS SYSTEMS

RENHUI WAN

Abstract. In this article, we consider the tridimensional generalized Hall magneto-hydrodynamics (Hall-MHD) system, with (−∆)^αuand (−∆)^βb. For α≥5/4,β≥7/4, we obtain the global regularity of classical solutions. For 0< α <5/4 and 1/2< β <7/4, with small data, the system also possesses global classical solutions. In addition, for the standard Hall-MHD system, α=β= 1, by adding a suitable condition, we give a positive answer to the open question in [3]. At last, we study the regularity criterions of generalized Hall-MHD system. In particular, we prove the regularity criterion in terms of horizontal gradient∇_hu,∇_hbfor 1< α <5/4, 5/4≤β <7/4.

1. Introduction

The tridimensional incompressible generalized Hall-MHD system is governed by

∂tu+u· ∇u+ (−∆)^αu+∇p=b· ∇b,

∂_tb+u· ∇b−b· ∇u+ (−∆)^βb=−∇ ×(J×b), divu= divb= 0,

u(0, x) =u0(x), b(0, x) =b0(x),

(1.1)

wheret≥0,x∈R³, p, u, bstand for scalar pressure, velocity vector and magnetic field vector, respectively, J = ∇ ×b is the current density, u0(x), b0(x) are the initial velocity and magnetic field,α, β≥0 are constants and (−∆)^αis defined by

(−∆)\^αf(ξ) =|ξ|^2αfb(ξ),

and we denote (−∆)^1/2 by Λ. Forα=β = 1, the system reduces to the standard Hall-MHD system, which can be obtained from kinetic models (cf. [1]). Hall- MHD is required in many physics problem, such as magnetic reconnection [5], star formation [6]. In [2], for α = 0, β = 1, local classical solutions were obtained and Beale-Kato-Majda type blow-up criterion was also established. In [3], forα= β = 1, some blow up criterions and small data results on global existence were established.

2010Mathematics Subject Classification. 35Q35, 35B65, 76W05.

Key words and phrases. Generalized Hall-mhd system; global regularity; classical solutions.

c

2015 Texas State University - San Marcos.

Submitted November 11, 2014. Published June 29, 2015.

1

Recently, Chae, Wan and Wu [4] considered the generalized Hall-MHD (1.1), and obtained the local well-posedness forα= 0 (without velocity diffusion),β >1/2.

So it is natural to ask a question:

Can the global classical solutions of (1.1) be obtained with some conditions on (α, β)?

One of the main goals is to give a positive answer to the question. Some related work about generalized Navier-Stokes equations and generalized MHD equations can be seen [11, 12, 13, 14, 15, 16]. Our first result is stated as follows.

Theorem 1.1. Let T > 0, α ≥ 5/4, β ≥ 7/4. Let (u₀, b₀) ∈ H^s(R³), s > 5/2 satisfying divu₀ = divb₀ = 0. Then (1.1) has a unique global classical solution (u, b) such that

(u, b)∈L^∞([0, T];H^s(R³)).

Remark 1.2. The spaceH^s(R³) can be equipped with the norm kfkH^s =kfkL²+kfkH˙^s,

where

kfk²_H˙s=X

j∈Z

2^2jsk∆jfk²_L2. More details can be seen in section 2.

Remark 1.3. If b = 0, (1.1) reduces to the generlized Navier-Stokes equations.

Under the scaling transformation ul = l^2α−1u(l^2αt, lx), pl = l^4α−2p(l^2αt, lx), for α≥⁵₄, it is well-known that the generalized Navier-Stokes equations is critical and subcritical, which can lead the global regular solutions. We refer to [11] and [12].

Ifu= 0, (1.1) reduces to the following simple Hall problem

∂tb+ (−∆)^βb=∇ ×((∇ ×b)×b), (1.2) which is scaling invariance underbl=l^2β−2b(l^2βt, lx). The corresponding energy is

E(bl) = ess sup_l2βtkblk²_L2+ Z t

0

kΛ^βblk²_L2dτ

=l^4β−7

ess sup_tkbk²_L2+ Z t

0

kΛ^βbk²_L2dτ =l^4β−7E(b).

This implies that the simple Hall problem (1.2) is critical system for β = 7/4 and subcritical system with β >7/4. As generalized Navier-Stokes equations, the condition on (α, β) seems optimal.

In addition, with small initial data, we can obtain the global small classical solution forα∈(0,5/4) andβ ∈(1/2,7/4).

Theorem 1.4. Let u0, b0, s be as Theorem 1.1. Let α∈(0,5/4),β ∈(1/2,7/4), and if

ku0kH^s+kb0kH^s< , (1.3) whereis sufficient small, then there exists a unique global classical solution (u, b) of (1.1).

Remark 1.5. Forα=β= 1, this work was proved in [2], which can be considered a special case in Theorem 1.4.

Forα=β= 1, the authors in [3] improved the condition (1.3) by only assuming that

ku0kH˙^3/2+kb0kH˙^3/2 < or ku0k_B˙1/2 2,1

+kb0k_B˙3/2 2,1

< , (1.4) where

kfkB˙^s_2,1 =X

j∈Z

2^jsk∆jfk_L2

(For more details see section 2.), is sufficient small. And they also gave an open question, whether the condition

ku0kH˙^1/2+kb0kH˙^3/2 < (1.5) can lead the the desired result? Motivated by these, we give a theorem as follows.

Theorem 1.6. Letu0, b0, sbe as Theorem 1.1. Letα=β= 1. If(u0, b0)satisfies (1.5), with an additional condition, for all t≥0,

kb(t)k_L^∞ ≤C₀<2, (1.6) then there exists a unique global classical solution(u, b)of (1.1).

Remark 1.7. It seems that the condition (1.6) is too strong due to global assump- tion. However, we find that (1.4) can be propagated to any t, which lead to the global small ofkb(t)kH˙^3/2 or kb(t)k_B˙3/2

2,1

, while (1.6) fails. In addition, ˙B^3/2_2,1 ,→L^∞, and we do not need sufficient small of kb(t)kL^∞, which seems to make (1.5) and (1.6) weaker than the second condition of (1.4) in some sense.

For α = β = 1, the authors in [3] also establish some regularity criterions involvinguand∇b. Since the presence of Hall-term∇ ×((∇ ×b)×b), we find that

(u, b)∈L^∞(0, T;H¹(R³))∩L²(0, T;H²(R³))

can not ensure the regularity criterion established in [3]. Therefore, unlike MHD system, establishing the regularity criterion in terms of horizontal gradient ∇_hu and∇hbseems formidable and interesting. But for system (1.1) with 1< α <5/4, 5/4 ≤β <7/4, we can achieve this goal. We have the following theorems, where Theorem 1.8 can be considered as a component of the proof of Theorem 1.10.

Theorem 1.8. Let T >0,α∈(1,5/4),β ∈(1,7/4). Let u0, b0, sbe as Theorem 1.1 and(u, b)be the local classical solution to (1.1). If

Z T

0

kuk^q_L¹p1 +k∇bk^q_L²p2dt <∞ (1.7) for

3 p1

+2α q1

≤min

2α−1,(1−α β)3

p1

+ (2−1 β)α , p1∈( 3

2β−1, 3

β−1]∩( 3 2α−1, 3

α−1], 3

p2

+2β q2

≤2β−1, p₂∈( 3 2β−1, 3

β−1], then(u, b)remains regular in [0, T].

Remark 1.9. For 1< α=β <5/4, the condition of index reduce to 3

p₁ +2α

q₁ ≤2α−1 and 3 p₂ +2α

q₂ ≤2β−1,

which is scaling invariance under the transformation in Remark 1.3. We can also establish the regularity criterion in terms of∇uand ∇b, that is

Z T

0

k∇uk^q_L¹p1+k∇bk^q_L²p2dτ <∞ for

3 p1

+2α q1

≤min

2α,(1−α β)3

p1

+ 2α , max 3 2α, 3

2β < p1≤ ∞ 3

p₂+2β

q₂ ≤2β−1, p₂∈ 3 2β−1, 3

β−1 . The proof is similar to the previous proofs.

Theorem 1.10. Let T >0,α∈(1,5/4), β∈[5/4,⁷₄). Let u0, b0, sbe as Theorem 1.1 and(u, b)be the local classical solution to (1.1). If

Z T

0

k∇huk^q_L¹p1 +k∇hbk^q_L²p2dt <∞ (1.8) for

3 p1

+2α q1

≤2α, 3

2α < p1≤ ∞, 3

p₂ +2β

q₂ ≤2β−1, 3

2β−1 < p₂≤ 3 α, then(u, b)remains regular in [0, T].

Remark 1.11. We do not know whether the regularity criterion (1.8) can be established forα∈(1,⁵₄), β∈(1,⁵₄), since we only observe that

u∈L^∞(0, T;H¹(R³))∩L²(0, T;H^α+1(R³)), b∈L^∞(0, T;H¹(R³))∩L²(0, T;H^β+1(R³)) can lead to (1.7) forα∈(1,5/4), β∈[5/4,7/4).

This article is organized as follows. In section 2, we give some notation and preliminaries. In the third section, we prove Theorem 1.1. In the fourth section, we give the proof of Theorem 1.4. In the fifth section, we give the proof of Theorem 1.6. We prove Theorem 1.8 in the following section. At the last section, we prove and Theorem 1.10.

2. Preliminaries

LetB ={ξ ∈R^d, |ξ| ≤ ⁴₃} and C={ξ ∈R^d : 3/4 ≤ |ξ| ≤8/3}. Choose two nonnegative smooth radial functionχ, ϕsupported, respectively, in Band Csuch that

χ(ξ) +X

j≥0

ϕ(2^−jξ) = 1, ξ∈R^d, X

j∈Z

ϕ(2^−jξ) = 1, ξ∈R^d\ {0}.

We denote ϕj = ϕ(2^−jξ), h = F⁻¹ϕ and ˜h = F⁻¹χ, where F⁻¹ stands for the inverse Fourier transform. Then the dyadic blocks ∆j and Sj can be defined as follows

∆jf =ϕ(2^−jD)f = 2^jd Z

R^d

h(2^jy)f(x−y)dy, Sjf = X

k≤j−1

∆kf =χ(2^−jD)f = 2^jd Z

R^d

h(2˜ ^jy)f(x−y)dy.

Formally, ∆_j =S_j−S_j−1is a frequency projection to annulus{C₁2^j≤ |ξ| ≤C₂2^j}, andS_j is a frequency projection to the ball{|ξ| ≤C2^j}. One can easily verify that with our choice ofϕ,

∆_j∆_kf = 0 if|j−k| ≥2 and ∆_j(S_k−1f∆_kf) = 0 if|j−k| ≥5.

With the introduction of ∆j andSj, let us recall the definition of the Besov space.

Lets∈R, (p, q)∈[1,∞]², the homogeneous space ˙B_p,q^s is defined by B˙_p,q^s ={f ∈S⁰ :kfkB˙_p,q^s <∞},

where

kfkB˙_p,q^s = ( P

j∈Z2^sjqk∆jfk^q_Lp

^1/q

, for 1≤q <∞, sup_j∈Z2^sjk∆jfkL^p, forq=∞, andS⁰ denotes the dual space of

S={f ∈ S(R^d) :∂^αfˆ(0) = 0 :∀α∈N^dmulti-index}

and can be identified by the quotient space ofS⁰/P with the polynomials spaceP. We also provide the definition for the inhomogeneous Besov space. For s >0, and (p, q)∈[1,∞]², the inhomogeneous Besov spaceB_p,q^s can be defined as follows

B^s_p,q={f ∈ S⁰(R^d) :kfkB^s_p,q<∞}, where

kfkB^s_p,q=kfkL^p+kfkB˙_p,q^s .

Additionally, when p=q= 2, the Besov space and Sobolev space are equivalence;

that is

H˙^s≈B˙_2,2^s , H^s≈B_2,2^s .

Bernstein’s inequalities are useful in this paper, so that we give it in the following proposition.

Proposition 2.1. Let α≥0. Let1≤p≤q≤ ∞.

(1) If f satisfies

suppfb⊂ {ξ∈R^d:|ξ| ≤K2^j}, for some integerj and a constantK >0, then

k(−∆)^αfk_Lq(R^d)≤C12^{2αj+jd(p1−1q)}kfk_Lp(R^d). (2) If f satisfies

suppfb⊂ {ξ∈R^d:K12^j≤ |ξ| ≤K22^j} for some integerj and constants 0< K1≤K2, then

C₁2^2αjkfk_Lq(R^d)≤ k(−∆)^αfk_Lq(R^d)≤C₂2^{2αj+jd(1p−1q)}kfk_Lp(R^d), whereC₁ andC₂ are constants depending onα, pandqonly.

For more details about Besov space such as some useful embedding relations, we refer to [7, 8]. Thanks to the Proposition 2.1, we can see that∀ s >0,

kfkH^s ≈ kfkL²+ (X

j≥0

2^2jsk∆jfk²_L2)^1/2, (2.1)

which will be frequently used in our proof.

Proposition 2.2 ([10]). Let 1 ≤p₁, p₂ ≤ ∞, σ >0, _p^d

i −σ_i > 0(i = 1,2) and assume thatσ−σ2+_p^d

2 >0. Then the following inequality holds X

j∈Z

2^2jσk[f,∆_j]· ∇gk²₂1/2

≤C

k∇fkB˙^σ_p1¹_,∞kgk

B˙

σ−σ1 +d p1 2,2

+k∇gkB˙_p^σ₂²_,∞kfk

B˙

σ−σ2 +d p2 2,2

.

(2.2)

3. Proof of Theorem 1.1

From [4], one can see that the local well-posedness of (1.1) holds forα≥0, β >

1/2. So we only need establish the global regularity, i.e. for all 0≤t≤T, ku(t)kH^s+kb(t)kH^s≤C(s, T,ku0kH^s,ku0kH^s).

At first, we give the energy estimate. Taking inner product with (u, b), integrating by parts and integrating in time [0, t],

k(u(t), b(t)k²_L2+ 2 Z t

0

kΛ^αuk²_L2dτ + 2 Z t

0

kΛ^βbk²_L2dτ ≤ k(u0, b₀)k²_L2. (3.1)

Then we establish theH^sestimate. Applying the operator ∆q to (1.1), taking the inner product with (∆qu,∆qb), by cancelation property, integrating by parts, we obtain

1 2

d

dt(k∆quk²_L2+k∆qbk²_L2) +kΛ^α∆quk²_L2+kΛ^β∆qbk²_L2

=− Z

R³

[∆q, u· ∇]u·∆qu+ Z

R³

[∆q, b· ∇]b·∆qu− Z

R³

[∆q, u· ∇]b·∆qb +

Z

R³

[∆q, b· ∇]u·∆qb+ Z

R³

[∆q, b×]J·∆qJ

≤ k[∆q, u· ∇]ukL²k∆qukL²+k[∆q, b· ∇]bkL²k∆qukL²

+k[∆q, u· ∇]bkL²k∆qbkL²+k[∆q, b· ∇]ukL²k∆qbkL²

+k[∆q, b×]Jk_L2k∆qJk_L2

=L₁(t) +L₂(t) +L₃(t) +L₄(t) +L₅(t).

(3.2)

By the paraproduct decomposition, H¨older’s inequality, commutator estimate [7, page. 110], and Bernstein’s inequality,

|L1(t)| ≤ X

|k−q|≤4

k∆q(S_k−1u· ∇∆ku)−S_k−1u· ∇∆q∆kuk_L2k∆quk_L2

+ X

|k−q|≤4

k∆_q(∆_ku· ∇S_k−1u)−∆_ku· ∇∆_qS_k−1uk_L2k∆_quk_L2

+ X

k≥q−3

k∆q( ˜∆ku· ∇∆ku)−∆˜ku· ∇∆q∆kukL²k∆qukL²

≤Ck∇S_q−1ukL^∞k∆quk²_L2+Ck∆quk_L2

X

k≥q−3

k∇∆kukL^∞k∆˜kuk_L2, (3.3)

where ˜∆k= ∆_k−1+ ∆k+ ∆k+1. Similarly,

|L2(t)| ≤Ck∇S_q−1bkL^∞k∆qbk_L2k∆quk_L2+Ck∆quk_L2

X

k≥q−3

k∇∆kbkL^∞k∆˜kbk_L2,

|L3(t)| ≤Ck∇Sq−1ukL^∞k∆qbk²_L2+k∇Sq−1bkL^∞k∆qukL²k∆qbkL²

+Ck∆_qbk_L2

X

k≥q−3

k∇∆_kbk_L^∞k∆˜_kuk_L2,

|L4(t)| ≤Ck∇Sq−1bkL^∞k∆qukL²k∆qbkL²+k∇Sq−1ukL^∞k∆qbk²_L2

+Ck∆qbkL²

X

k≥q−3

k∇∆kukL^∞k∆˜_kbkL²,

|L5(t)| ≤C2^qk∇Sq−1bkL^∞k∆qbk²_L2+C2^qk∆qbkL²

X

k≥q−3

k∇∆kbkL^∞k∆˜kbkL². Multiplying 2^2sq and taking the summation overq≥0 in (3.2),

1 2

d dt

n X

q≥0

2^2sq(k∆quk²_L2+k∆qbk²_L2)o

+X

q≥0

2^2(s+α)qk∆quk²_L2+X

q≥0

2^2(s+β)qk∆qbk²_L2

≤X

q≥0

2^2sq(L1(t) +· · ·+L5(t)).

(3.4)

By (3.3), we have X

q≥0

2^2sq|L1(t)| ≤CX

q≥0

2^2sqk∇Sq−1ukL^∞k∆quk²_L2

+CX

q≥0

2^2sqk∆qukL²

X

k≥q−3

k∇∆kukL^∞k∆˜kukL²

=L₁₁(t) +L₁₂(t).

For the estimate of L11(t). Using Bernstein’s inequality, H¨older’s inequality and Young’s inequality,

|L11(t)| ≤CX

q≥0

2^2sqk∆quk²_L2

X

m≤q−2

2^5m/2k∆muk_L2

≤CX

q≥0

2^(s+α)qk∆qukL²2^(s−α)qk∆qukL²

X

m≤q−2

2^5m/2k∆mukL²

≤ 1

8kΛ^αuk²_H˙s+CX

q≥0

2^2(s−α)qk∆quk²_L2

X

m≤q−2

2^5m/2k∆muk_L2

2

| {z }

L₁₁₁

,

where

|L111|=CX

q≥0

2^2sqk∆quk²_L2

X

m≤q−2

2^52m−αqk∆mukL²

²

=CX

q≥0

2^2sqk∆_quk²_L2

X

m≤−2

2^52m−αqk∆_muk_L2

+ X

−1≤m≤q−2

2^{(52−α)m−αq}kΛ^α∆muk_L2

²

≤CX

q≥0

2^2sqk∆quk²_L2

X

m≤−2

2^5m/2kukL²

+ X

−1≤m≤q−2

2^α(m−q)2^(52−2α)mkΛ^αuk_L2

2

≤C(kuk²_L2+kΛ^αuk²_L2)kuk²_Hs. Thus, we obtain

|L11(t)| ≤C(kuk²_L2+kΛ^αuk²_L2)kuk²_Hs+1

8kΛ^αuk²_H˙s. Similarly,

|L12(t)|

≤CX

q≥0

2^2sqk∆qukL²

X

k≥q−3

k∆kukL²2^5k/2k∆˜kukL²

≤C X

p≥0

2^2sqk∆quk²_L2

^1/2n X

q≥0

2^2sq( X

k≥q−3

k∆˜kukL²2^k(52−α)kΛ^α∆kukL²)²o^1/2

≤CkΛ^αuk_L2kuk_Hs

n X

q≥0

2^2sq( X

k≥q−3

2^k(52−2α)k∆˜_kΛ^αuk_L2)²o^1/2

≤CkΛ^αuk_L2kuk_Hs

n X

q≥0

2^2sq( X

k≥q−3

k∆˜_kΛ^αuk_L2)²o1/2

≤CkΛ^αuk_L2kukH^skukH˙^s+α

≤CkΛ^αuk²_L2kuk²_Hs+1

8kuk²_H˙s+α,

here we have used Young’s inequality for series for the fifth inequality; that is, n X

q≥0

2^2sq( X

k≥q−3

k∆˜kΛ^αuk_L2)²o^1/2

≤n X

q∈Z

( X

k≥q−3

2^s(q−k)2^skk∆˜kΛ^αuk_L2)²o^1/2

≤Ck2^−sk1_k≥−3k_l1(Z)k2^skk∆˜kΛ^αuk_L2k_l2(Z)

≤CkukH˙^s+α.

Collecting the estimates above, we have X

q≥0

2^2sq|L1(t)| ≤C(kuk²_L2+kΛ^αuk²_L2)kuk²_Hs+1

4kΛ^αuk²_Hs. Similarly,

X

q≥0

2^2sq|L2(t)| ≤C(kbk²_L2+kΛ^βbk²_L2)kbk²_Hs+1

4kΛ^αuk²_Hs, X

q≥0

2^2sq|L3(t)| ≤C(k(u, b)k²_L2+k(Λ^αu,Λ^βb)k²_L2)kbk²_Hs+1

4k(Λ^αu,Λ^βb)k²_Hs, X

q≥0

2^2sq|L4(t)| ≤C(k(u, b)k²_L2+k(Λ^αu,Λ^βb)k²_L2)kbk²_Hs+1

4k(Λ^αu,Λ^βb)k²_Hs. Now, we estimate the last term.

X

q≥0

2^2sq|L₅(t)| ≤CX

q≥0

2^(2s+1)qk∆_qbk²_L2

X

m≤q−2

k∇∆_mbk_L^∞ +CX

q≥0

2^(2s+1)qk∆_qbk_L2

X

k≥q−3

k∇∆_kbk_L^∞k∆˜_kbk_L2

=L51(t) +L52(t).

Similar to the estimate ofL11(t), we have

|L51(t)|

≤CX

q≥0

2^(s+β)qk∆qbkL²2^(s+1−β)qk∆qbkL²

X

m≤q−2

2^5m/2k∆mbkL²

≤ 1 8

X

q≥0

2^2(s+β)qk∆_qbk²_L2+CX

q≥0

2^2(s+1−β)qk∆_qbk²_L2

X

m≤q−2

2^5m/2k∆_mbk_L2

²

| {z }

L₅₁₁

,

where

|L511| ≤CX

q≥0

2^2sqk∆qbk²_L2

X

m≤−2

2^(1−β)q2^5m/2k∆mbk_L2

+ X

−1≤m≤q−2

2^(1−β)q2^(52−β)mkΛ^β∆_mbkL²

²

≤CX

q≥0

2^2sqk∆qbk²_L2

X( X

m≤−2

2^5m/2kbkL²

+ X

−1≤m≤q−2

2^{(1−β)(q−m)}2^(72−2β)mkΛ^β∆mbk_L2

2

≤C(kbk²_L2+kΛ^βbk²_L2)X

q≥0

2^2sqk∆_qbk²_L2

≤C(kbk²_L2+kΛ^βbk²_L2)kbk²_Hs. So we have

|L₅₁(t)| ≤1

8kbk²_H˙s+β+C(kbk²_L2+kΛ^βbk²_L2)kbk²_Hs.

Similar to the estimate ofL12(t), we have

|L52(t)| ≤CX

q≥0

2^sqk∆qbk_L22^(s+1)q X

k≥q−3

k∇∆qbkL^∞k∆˜kbk_L2

≤C X

q≥0

2^2sqk∆qbk²_L2

^1/2n X

q≥0

2^2(s+1)q( X

k≥q−3

k∇∆kbkL^∞k∆˜kbkL²)²o^1/2

≤CkbkH^s

n X

q≥0

2^2(s+1)q X

k≥q−3

2^k(52−β)kΛ^β∆_kbkL²2^−kβkΛ^β∆˜_kbkL²

²o^1/2

≤Ckbk_Hs

n X

q≥0

2^2(s+1)q( X

k≥q−3

2^−k2^k(72−2β)kΛ^β∆_kbk_L2kΛ^βbk_L2)²o1/2

≤CkΛ^βbk_L2kbkH^s

n X

q≥0

( X

k≥q−3

2^(s+1)(q−k)2^kskΛ^β∆kbk_L2)²o1/2

≤CkΛ^βbk_L2kbk_HskbkH˙^s+β ≤CkΛ^βbk²_L2kbk²_Hs+1

8kbk²_H˙s+β, here we have used Young’s inequality for the fifth inequality. Therefore,

X

q≥0

2^2sq|L5(t)| ≤C(kΛ^βbk²_L2+kbk²_L2)kbk²_Hs+1

4kΛ^βbk²_H˙s.

Collecting the estimate above in (3.2), integrating in [0, t], together with (3.1) and (2.1) yields

ku(t)k²_Hs+kb(t)k²_Hs+ Z t

0

kΛ^αu(τ)k²_Hsdτ + Z t

0

kΛ^βb(τ)k²_Hsdτ

≤C(k(Λ^αu,Λ^βb)k²_L2+k(u, b)k²_L2)(kuk²_Hs+kbk²_Hs) +ku0k²_Hs+kb0k²_Hs. Applying Gronwall’s lemma, with (3.1), for 0≤t≤T,

ku(t)k²_Hs+kb(t)k²_Hs+ Z t

0

kΛ^αu(τ)k²_Hsdτ+ Z t

0

kΛ^βb(τ)k²_Hsdτ

≤(ku0k²_Hs+kb0k²_Hs) expn C

Z t

0

(k(Λ^αu,Λ^βb)k²_L2+k(u, b)k²_L2)dτo

≤(ku0k²_Hs+kb0k²_Hs) exp Ck(u0, b₀)k²_L2(1 +T) . This completes the proof of Theorem 1.1.

4. Proof of Theorem 1.4

As in the proof of Theorem 1.1, we only need to show the global regularity. Since kuk_Hs =kuk_L2+kukH˙^s, we obtain

ku(t)k²_Hs+kb(t)k²_Hs+ 2 Z t

0

kΛ^αu(τ)k²_Hsdτ+ 2 Z t

0

kΛ^βb(τ)k²_Hsdτ

≤ k(u0, b0)k²_Hs+C Z t

0

k(∇u,∇b)kL^∞k(u, b)k²_H˙sdτ + 2 Z t

0

k∇bkL^∞kbk²

H˙^{s+ 1}2dτ

≤ k(u0, b0)k²_Hs+C Z t

0

k(u, b)kH^s(kΛ^αuk²_Hs+kΛ^βbk²_Hs)dτ.

A similar estimate can be found in [4]. Choosingso small thatk(u0, b0)k²_Hs< _2C¹2, which implies that Ck(u0, b0)kH^s <1. Suppose there exists a first time T^?, such that

ku(T^?)k²_Hs+kb(T^?)k²_Hs≥ 1

2C². (4.1)

This leads to

ku(T^?)k²_Hs+ku(T^?)k²_Hs+ Z T^?

0

kΛ^αu(t)k²_Hs+kΛ^βb(t)k²_Hsdt≤ k(u0, b0)k²_Hs < 1 2C², which contradicts (4.1). Hence, for allt≥0, we get global small solution satisfying

ku(t)k²_Hs+kb(t)k²_Hs+ Z t

0

kΛ^αu(τ)k²_Hsdτ+ Z t

0

kΛ^βb(τ)k²_Hsdτ ≤ k(u0, b0)k²_Hs. This concludes the proof of Theorem 1.4.

5. Proof of Theorem 1.6 It suffices to establish the following, for allt≥0,

Z t

0

kuk²_H˙3/2+k∇bk²_H˙3/2dτ <∞,

since [3] provided the blow up criterion in theBM Ospace and ˙H^3/2,→BM O[7].

As the operation in section 3, we obtain 1

2 d

dt(k∆_quk²_L2+k∆_qbk²_L2) +k∇∆_quk²_L2+k∇∆_qbk²_L2

=− Z

R³

[∆q, u· ∇]u·∆qu+ Z

R³

[∆q, b· ∇]b·∆qu− Z

R³

[∆q, u· ∇]b·∆qb

− Z

R³

[∆q, b· ∇]u·∆qb+ Z

R³

[∆q, b×]J·∆qJ

≤ k[∆q, u· ∇]ukL²k∆qukL²+k[∆q, b· ∇]bkL²k∆qukL²

+k[∆q, u· ∇]bk_L2k∆qbk_L2+k[∆q, b· ∇]uk_L2k∆qbk_L2

+k[∆_q, b×]Jk_L2k∆_qJk_L2

=IL₁(t) +· · ·+IL₅(t).

(5.1)

Multiplying (5.1) by 2^q and summing overq∈Z, 1

2 d

dt(kuk²_H˙1/2+kbk²_H˙1/2) +k∇uk²_H˙1/2+k∇bk²_H˙1/2 ≤X

q∈Z

2^q|IL1(t) +· · ·+IL₅(t)|.

(5.2) By H¨older’s inequality and choosingσ=σ_i=¹₂,p_i= 2,i= 1,2 in (2.2),

X

q∈Z

2^q|IL1(t)| ≤ X

q∈Z

2^qk[∆q, u]· ∇uk²_L2

1/2

kukH˙^1/2≤CkukH˙^1/2kuk²_H˙3/2. Similarly,

X

q∈Z

2^q|IL2(t)| ≤CkukH˙^1/2kbk²_H˙3/2, X

q∈Z

2^q|IL3(t)| ≤CkukH˙^1/2(kuk²_H˙3/2+kbk²_H˙3/2),

X

q∈Z

2^q|IL4(t)| ≤CkukH˙^1/2(kuk²_H˙3/2+kbk²_H˙3/2), X

q∈Z

2^q|IL5(t)| ≤Ckbk³_H˙3/2. Plugging the inequalities above in (5.1), we have

1 2

d

dtk(u, b)k²_H˙1/2+k∇uk²_H˙1/2+k∇bk²_H˙1/2

≤C(k(u, b)kH˙^1/2+kbkH˙^3/2)(kuk²_H˙3/2+kbk²_H˙3/2).

(5.3)

Next, we give the ˙H^3/2 estimate ofb. With a similar process, we obtain 1

2 d

dtk∆_qbk²_L2+k∇∆_qbk²_L2

= Z

R³

[∆q, b· ∇]u·∆qb+ Z

R³

b· ∇∆qu·∆qb

− Z

R³

[∆q, u· ∇]b·∆qb+ Z

R³

[∆q, b×]J·∆qJ

≤ k[∆q, b· ∇]ukL²k∆qbkL²+kb· ∇∆qukL²k∆qbkL²

+k[∆q, u· ∇]bk_L2k∆qbk_L2+k[∆q, b×]Jk_L2k∆qJk_L2

=K1(t) +· · ·+K5(t).

Multiplying 2^3q and summing overq∈Z, 1

2 d

dtkbk²_H˙3/2+kbk²_H˙5/2 ≤X

q∈Z

2^3q|K1(t) +· · ·+K5(t)|. (5.4) By H¨older’s inequality and choosingσ=σi=¹₂,pi= 2,i= 1,2 in (2.2),

X

q∈Z

2^3q|K₁(t)|=X

q∈Z

2^3qk[∆_q, b· ∇]uk_L2k∆_qbk_L2

≤ X

q∈Z

2^qk[∆q, b· ∇]uk²_L2

1/2

kbkH˙^5/2

≤CkbkH˙^3/2(kuk²_H˙3/2+kbk²_H˙5/2).

Similarly,

X

q∈Z

2^3q|K3(t)| ≤CkbkH˙^3/2(kbk²_H˙5/2+kuk²_H˙3/2).

By H¨older’s inequality and choosingσ= 3/2,σi=¹₂,pi= 2,i= 1,2 in (2.2), then X

q∈Z

2^3q|K4(t)| ≤ X

q∈Z

2^3qk[∆q, b×]Jk²_L2

1/2

kJkH˙^3/2 ≤CkbkH˙^3/2kbk²_H˙5/2. Using H¨older’s inequality and condition (1.6),

X

q∈Z

2^3q|K2(t)| ≤ X

q∈Z

2^qkb· ∇∆quk²_L2

1/2

kbkH˙^5/2

≤ kbkL^∞kukH˙^3/2kbkH˙^5/2

≤C0

2 (kuk²_H˙3/2+kbk²_H˙5/2).

Plugging the inequalities above in (5.4), combining with (5.3) and integrating the resulting inequality in [0, t] gives

k(u(t), b(t))k²_H˙1/2+kb(t)k²_H˙3/2+ (2−C0) Z t

0

k(u(τ), b(τ))k²_H˙3/2+kb(τ)k²_H˙5/2dτ

≤C Z t

0

(k(u(τ), b(τ))kH˙^1/2+kb(τ)kH˙^3/2)(k(u(τ), b(τ))k²_H˙3/2+kb(τ)k²_H˙5/2)dτ +k(u₀, b₀)k²_H˙1/2+kb₀k²_H˙3/2.

Chooseso small that 3(ku0k²_˙

H^1/2+kb0k²_˙

H^1/2+kb0k²_˙

H^3/2)<(^2−C_4C⁰)², which implies C

ku0kH˙^1/2+kb0kH˙^1/2+kb0kH˙^3/2

<2−C0

4 . Suppose there exists a first timeT^? such that

3

ku(T^?)k²_H˙1/2+kb(T^?)k²_H˙1/2+kb(T^?)k²_H˙3/2

≥ 2−C0

4C 2

, which can easily get a contradiction. Hence, for allt≥0, we obtain

k(u(t), b(t))k²_H˙1/2+kb(t)k²_H˙3/2+ (2−C₀) Z t

0

k(u(τ), b(τ))k²_H˙3/2+kb(τ)k²_H˙5/2dτ

≤ k(u0, b0)k²_H˙1/2+kb0k²_H˙3/2.

This completes the proof of Theorem 1.6.

6. Proof of Theorem 1.8

Our proof contains two steps,H¹ estimates andH^sestimates.

Step 1: H¹ Estimates. Using a similar procedure as in [3], we have 1

2 d

dt(k∇uk²_L2+k∇bk²_L2) +k∇Λ^αuk²_L2+k∇Λ^βbk²_L2

=− Z

R³

∇(u· ∇u)· ∇u dx+ Z

R³

∇(b· ∇b)· ∇u dx+ Z

R³

∇(b· ∇u)· ∇b dx

− Z

R³

∇(u· ∇b)· ∇b dx− Z

R³

∇(J×b)· ∇J dx

=I1(t) +· · ·+I5(t).

(6.1)

By integrate by parts, H¨older’s inequality, interpolation inequality and Young’s inequality, letθ₁=^{3−(α−1)p}_αp ¹

1 , we obtain

|I1(t)| ≤ | Z

R³

(u· ∇)u·∆u dx|

≤ kuk_L^p1k∇uk

L

6p1 (1+2α)p1−6

k∆ukL^5−2α6

≤ kukL^p1k∇uk^1−θ_L2 ¹k∇Λ^αuk^1+θ_L2 ¹

≤Ckuk

2αp1 (2α−1)p1−3

L^p1 k∇uk²_L2+1

8k∇Λ^αuk²_L2. By cancelation property,

Z

R³

(b· ∇)∂ib·∂iu dx+ Z

R³

(b· ∇)∂iu·∂ib dx= 0,