International Journal of Non-Linear Mechanics 41 (2006) 1174–1180www.elsevier.com/locate/nlm

Regularity criteria for the 3D MHD equations in terms of the pressure

Yong ZhouDepartment of Mathematics, East China Normal University, Shanghai 200062, China

Received 2 June 2006; accepted 8 December 2006

Abstract

In this paper we consider the regularity criteria for weak solutions to the 3D MHD equations. It is proved that under the condition b beingin the Serrin’s regularity class, if the pressure p belongs to L�,� with 2

� + 3� �2 or the gradient field of pressure ∇p belongs to L�,� with

2� + 3

� �3 on [0, T ], then the solution remains smooth on [0, T ].� 2007 Elsevier Ltd. All rights reserved.

MSC: 35Q35; 35B65; 76D05

Keywords: MHD equations; Regularity criterion; A priori estimates

1. Introduction

We consider the following 3D MHD equations in thispaper:

⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩�t u + u · ∇u = �1�u − ∇p − 1

2∇|b|2 + b · ∇b + f,

�t b + u · ∇b = �2�b + b · ∇u + g,

div u = div b = 0,

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

(1.1)

where u=(u1(x, t), u2(x, t), u3(x, t)) is the velocity field, b ∈R3 is the magnetic field, p(x, t) is a scalar pressure, �1 �0is the kinematic viscosity, �2 �0 is the magnetic diffusivity,f represents volume force applied to the fluid, g is usuallyzero when Maxwell’s displacement currents are ignored, whileu0(x) with div u0 = 0 in the sense of distribution is the initialvelocity field. In the sequel, we assume that f = g = 0 and�1 = �2 = 1, just for simplicity. If �1 = �2 = 0, (1.1) is calledideal MHD equations.

It is well-known [1] that the problem (1.1) is local well-posed for any given initial datum u0, b0 ∈ Hs(R3) with s�3.

E-mail address: yzhou@math.ecnu.edu.cn.

0020-7462/$ - see front matter � 2007 Elsevier Ltd. All rights reserved.doi:10.1016/j.ijnonlinmec.2006.12.001

But whether this unique local solution can exist globally is anoutstanding challenge problem. Caflisch et al. [2] showed that ifsmooth data for the ideal MHD equations leads to a singularityat finite time T ∗, then∫ T ∗

0(‖�(·, t)‖L∞ + ‖j (·, t)‖L∞) dt = ∞, (1.2)

where �=∇ ×u is the vorticity field, j =∇ ×b is the electricalcurrent. Condition (1.2) is a similar regularity criterion to thatfor the Euler equations in [3]. It is not difficult to find that thisassertion is also true for the viscous MHD equations. In [4],Wu established a geometric condition on the direction of the� and j to control possible singularity development. Recently,some improvements were done by He and Xin [5]. In particular,they established the Serrin-type regularity criteria in terms ofthe velocity field without any restriction on the magnetic field(see also the paper [6]). More precisely, it was proved that ifthe velocity field u belongs to L�(0, T ; L�) with 2

� + 3� �1 or

the gradient of velocity field ∇u belongs to L�(0, T ; L�) with2� + 3

� �2, then the corresponding weak solution (u, b) actuallyis strong on [0, T ].

The purpose of this paper is to establish the Serrin-type reg-ularity criteria in term of the pressure just as what done in [7]for the Navier–Stokes equations. To this end, we introduce the

Y. Zhou / International Journal of Non-Linear Mechanics 41 (2006) 1174–1180 1175

space L�,� ≡ L�(0, t; L�) as follows:

‖u‖L�,� =⎧⎨⎩

(∫ t

0 ‖u(·, �)‖�L� d�

)1/�if 1�� < ∞,

ess sup0<�<t

‖u(·, �)‖L� if � = ∞,

where

‖u(·, �)‖L� ={(∫

R3 |u(x, �)|� dx)1/� if 1�� < ∞,

ess supx∈R3

|u(x, �)| if � = ∞.

We say v ∈ L�,� if ‖v‖L�,� < ∞.The main results of this paper read

Theorem 1.1. Assume that the initial velocity and magneticfields u0, b0 ∈ Hs(R3) with s�3. If

p ∈ L�,� and b ∈ L2�,2� with2

�+ 3

��2,

3

2< ��∞ (1.3)

or ‖p‖L∞,3/2 and ‖b‖L∞,3 are sufficient small on [0, T ], thenthe corresponding solution remains smooth on [0, T ].

Theorem 1.2. Assume that the initial velocity and magneticfields u0, b0 ∈ Hs(R3) with s�3. If

∇p ∈ L�,� and b ∈ L3�,3� with2

�+ 3

��3, 1 < ��∞ (1.4)

or ‖∇p‖L∞,1 and ‖b‖L∞,3 are sufficient small on [0, T ], thenthe corresponding solution remains smooth on [0, T ].

Remark 1.1. The condition on b can be replaced by b ∈ Lp,q

with 2p

+ 3q�1 for some 3 < q �∞ or ‖b‖L∞,3 is sufficiently

small. This condition is reasonable since the second equationof (1.1) is nothing to do with the pressure. So we must addsome serrin-type regularity condition on the magnetic field b

just as that on the velocity field u [5,6].

These theorems are significant since ‖p�‖L�,� =‖p‖L�,� holdsfor all � > 0 if and only if 2

� + 3� =2 and ‖∇p�‖L�,� =‖∇p‖L�,�

holds for all � > 0 if and only if 2� + 3

� = 3 where u�(x, t) =�u(�x, �2t), p�(x, t) = �2p(�x, �2t), b�(x, t) = �b(�x, �2t).Moreover, if (u, p, b) solves the MHD equations, then sodoes (u�, p�, b�) for all � > 0. Usually we say that the norm‖p‖L�,� is scaling dimension zero for 2

� + 3� = 2 (see [8] for

the Navier–Stokes equations). Similarly, the norm ‖∇p‖L�,� isscaling dimension zero for 2

� + 3� = 3. So these theorems es-

tablish final versions of Serrin-type regularity criteria in termsof the pressure.

Before going to the proofs, we recall the following definitionof Leray–Hopf weak solution.

Definition. A pair (u, b) is called a Leray–Hopf weak solutionto the MHD equations (1.1), if u and b satisfy the followingproperties:

(i) u and b are weakly continuous from [0, ∞) to L2(R3).

(ii) u and b verify (1.1) in the sense of distribution, i.e.,∫ ∞

0

∫R3

(��

�t+ (u · ∇)�

)u − b · ∇�b dx dt

+∫

R3u0�(x, 0) dx =

∫ ∞

0

∫R3

∇u : ∇� dx dt

and∫ ∞

0

∫R3

(��

�t+ (u · ∇)�

)b − b · ∇�u dx dt

+∫

R3u0�(x, 0) dx =

∫ ∞

0

∫R3

∇b : ∇� dx dt

for all � ∈ C∞0 (R3 × (0, ∞)) with div � = 0.

(iii) ∫ ∞

0

∫R3

u ·∇�dxdt = 0 and∫ ∞

0

∫R3

b · ∇�dxdt = 0

for every � ∈ C∞0 (R3 × [0, ∞)).

(iv) The energy inequality, i.e., for t �0

‖u(t)‖2L2 +‖b(t)‖2

L2 +2∫ t

0(‖∇u(s)‖2

L2 + ‖∇b(s)‖2L2) ds

�‖u0‖2L2 + ‖b0‖2

L2 .

By a strong solution (u, b) we mean a weak solution (u, b)

such that

u, b ∈ L∞(0, T ; H 1) ∩ L2(0, T ; H 2).

It is well-known that strong solutions are regular (say, classical)and unique in the class of weak solutions.

2. A priori estimates

In what follows the constants C’s are different from line toline. To establish a priori estimates, we will follow the argumentin [7].

First, we would like to give an interpolation inequality.

Lemma 2.1. Suppose a measurable function f ∈ L∞,s ∩Ls,3s

on (R3 × [0, T )), then f ∈ Lp,q with s�p, s�q �3s andsp

+ 3s2q

� 32

‖f ‖Lp,q �C(p, q, T )‖f ‖(3s−q)/2qL∞,s ‖f ‖(3q−3s)/2q

Ls,3s , (2.1)

where C(s, p, q, T ) depends on s, p, q, T , and C(p, q, T )=1if s

p+ 3s

2q= 3

2 .

Proof.

‖f ‖Lp,q =(∫ T

0‖f (·, �)‖p

Lq d�

)1/p

�(∫ T

0‖f (·, �)‖p

Ls ‖f (·, �)‖(1−)p

L3s d�

)1/p

�C(s, p, q, T )(‖f ‖L∞,s ‖f ‖(1−)

Ls,3s ,

1176 Y. Zhou / International Journal of Non-Linear Mechanics 41 (2006) 1174–1180

where we use the interpolation theorem

1

q=

s+ 1 −

3s, s�q �3s, (2.2)

and Hölder’s inequality provided (1 − )p�s.From (2.2), 1−= 3q−3s

2q, we obtain s

p+ 3s

2q� 3

2 . If sp+ 3s

2q= 3

2 ,which implies 1−= s

p, then obviously C(s, p, q, T )=1. �

Suppose f ∈ H 2, due to the fact that

�i�j f = −RiRj�f ,

where Ri is the Riesz transform, Rig() = −ii

|| g() [9], andthe boundedness of the operator Ri : Lp → Lp, 1 < p < ∞,we have

‖�i�j f ‖Lp �C‖�f ‖Lp . (2.3)

In order to prove Theorem 1.1, first we show the followingtheorem:

Theorem 2.2. Let s�3 be given. Suppose u0, b0 ∈ Ls(R3)

with div u0 = 0. Assume (u, p) is a smooth solution of (1.1) inR3 × (0, T ) with u, b ∈ L∞,2 and ∇u ∈ L2,2. If p ∈ L�,� andb ∈ L2�,2� with 2

� + 3� �2, 3

2 < ��∞, or ‖p‖L∞,3/2 + ‖b‖L∞,3

is sufficiently small, then u(t), b(t) ∈ L∞,s ∩ Ls,3s

sup0� t �T

(‖u(t)‖Ls + ‖b(t)‖Ls )�C(‖u0‖Ls + ‖b0‖Ls ), (2.4)

where C depends on the norms of p and b on the time interval.

Proof. It is sufficient to consider the equality case 2� + 3

� = 2.In order to prove (2.4), we multiply both sides of the first

and the second equation of (1.1) by su|u|s−2 and sb|b|s−2,respectively, and integrate over R3 × (0, t), 0 < t �T . Aftersuitable integration by parts, we obtain

‖u(·, t)‖sLs + s

∫ t

0

∫R3

|∇u|2|u|s−2 dx d�

+ 4(s − 2)

s‖∇|u|s/2‖2

L2,2

�2(s − 2)

∫ t

0

∫R3

|p||u|s/2−1|∇|u|s/2| dx d� + ‖u0‖sLs

+ s

∫ t

0

∫R3

(b · ∇b)|u|s−2u dx d�, (2.5)

where we used

− s

∫ t

0

∫R3

∇p · u|u|s−2 dx d�

= s(s − 2)

∫ t

0

3∑i,j=1

∫R3

p�uj

�xi

uiuj |u|s−4 dx d�

�2(s − 2)

∫ t

0

∫R3

|p||u|s/2−1|∇|u|s/2| dx d�.

If we use the fact that

|∇|u|s/2|� s

2|u|s/2−1|∇u|,

then (2.5) will be reduced as follows:

‖u(·, t)‖sLs + 2‖∇|u|s/2‖2

L2,2

�2(s − 2)

∫ t

0

∫R3

|p||u|s/2−1|∇|u|s/2| dx d� + ‖u0‖sLs

+ s

∫ t

0

∫R3

(b · ∇b)|u|s−2u dx d�. (2.6)

Actually an analogous inequality to (2.6) for the Navier–Stokesequations was derived many years ago by Beirao da Veiga [10].

Similarly, we have

‖b(·, t)‖sLs + 2‖∇|b|s/2‖2

L2,2

�s

∫ t

0

∫R3

(b · ∇u)|b|s−2b dx d� + |b0‖sLs . (2.7)

Before going to estimate the right-hand sides of (2.6) and(2.7), from the equations directly, one has

−�p =3∑

i,j=1

�i�j (uiuj − bibj ). (2.8)

The Calderon–Zygmund inequality tells us there exists a abso-lute constant C such that

‖p‖Lq �C(‖u‖2L2q + ‖u‖2

L2q ) for any 1 < q < ∞. (2.9)

Case 1: 32 < � < ∞. First, we do estimate for the first term

in (2.6).

2(s − 2)

∫ t

0

∫R3

|p||u|s/2−1|∇|u|s/2| dx d�

�C

∫ t

0‖p‖La1 ‖u‖s/2−1

La2 ‖∇|u|s/2‖L2 d�(Hölder’ s inequality

1

a1+ s/2 − 1

a2= 1

2

)�C

∫ t

0‖p‖2

La1 ‖u‖s−2La2 d� + 1

2‖∇|u|s/2‖2

L2,2

(Young’ s inequality)

�C

∫ t

0‖p‖2(1−)

L� ‖p‖2La2/2‖u‖s−2

La2 d� + 1

2‖∇|u|s/2‖2

L2,2(Interpolation inequality

1

a1= 1 −

�+

a2/2

)�C

∫ t

0‖p‖2(1−)

L� (‖u‖4La2 + ‖b‖4

La2 )‖u‖s−2La2 d�

+ 1

2‖∇|u|s/2‖2

L2,2 (By (2.9))

�C

∫ t

0‖p‖2(1−)

L� (‖u‖4+s−2La2 + ‖b‖4+s−2

La2 ) d�

+ 1

2‖∇|u|s/2‖2

L2,2

�C‖p‖2(1−)L�,� (‖u‖4+s−2

Lq,a2 + ‖b‖4+s−2Lq,a2 )

+ 1

2‖∇|u|s/2‖2

L2,2 .(Hölder’ s inequality

2(1 − )

�+ 4 + s − 2

q= 1

).

Y. Zhou / International Journal of Non-Linear Mechanics 41 (2006) 1174–1180 1177

In the above inequalities, constants a1, a2, and q are un-knowns, however there are only three equations. How to obtainthese constants? where is the fourth equation? By observing thepower indexes in the last inequality, one can find that = 1

2 isa good choice, and can find the values for other three constantsas follows:

a1 = 2�s

2� + s − 2, a2 = �s

� − 1, q = �s

� − 1. (2.10)

From (2.10), by direct computation, q and a2 satisfy

s

q+ 3s

2a2= 5

2−

(1

�+ 3

2�

)= 3

2, s < q and s < a2 < 3s,

then we can use inequality (2.1). Therefore

2(s − 2)

∫ t

0

∫R3

|p||u|s/2−1|∇|u|s/2| dx d�

�C‖p‖L�,�(‖u‖sLq,a2 + ‖b‖s

Lq,a2 ) + 1

2‖∇|u|s/2‖2

L2,2

�C‖p‖L�,�

(‖u‖

2�−32� s

L∞,s ‖u‖32� s

Ls,3s + ‖b‖2�−3

2� s

L∞,s ‖b‖32� s

Ls,3s

)+ 1

2‖∇|u|s/2‖2

L2,2

�C(�)‖p‖�L�,�(‖u‖s

L∞,s + ‖b‖sL∞,s ) + �‖u‖s

Ls,3s

+ �‖b‖sLs,3s + 1

2‖∇|u|s/2‖2

L2,2 , (2.11)

where � is a small constant such that the following Sobolevinequality holds

�‖u‖sL3s = �‖|u|s/2‖2

L6 � 12‖∇|u|s/2‖2

L2 . (2.12)

By the similar trick, the last term in (2.6) can be treated as

s

∫ t

0

∫R3

(b · ∇b)|u|s−2u dx d�

= −s

∫ t

0

∫R3

(b · ∇|u|s−2u)b dx d�

�s

∫ t

0

∫R3

|b|2|u| s2 −1|∇|u|s/2| dx d�

�C

∫ t

0‖b‖4

La‖u‖s−2La2 d� + 1

2‖∇|u|s/2‖2

L2

�C

∫ t

0‖b‖2

L2�‖b‖2La2 ‖u‖s−2

La2 d� + 1

2‖∇|u|s/2‖2

L2

�C

∫ t

0‖b‖2

L2�(‖b‖sLa2 + ‖u‖s

La2 ) d� + 1

2‖∇|u|s/2‖2

L2

�C‖b‖2L2�,2�(‖u‖s

Lq,a2 + ‖u‖sLq,a2 ) + 1

2‖∇|u|s/2‖2

L2 ,

where a = 4�s2�+s−2 , while a2 and q are the same constants in

(2.10). Due to the above argument, we have

s

∫ t

0

∫R3

(b · ∇b)|u|s−2u dx d�

�C‖b‖2�L2�,2�(‖u‖s

L∞,s + ‖b‖sL∞,s )

+ ‖∇|u|s/2‖2L2,2 + 1

2‖∇|b|s/2‖2

L2,2 . (2.13)

The non-linear term in (2.7) can be done in a similar way

s

∫ t

0

∫R3

(b · ∇u)|b|s−2b dx d�

= −s

∫ t

0

∫R3

(b · ∇|b|s−2b)u dx d�

�C

∫ t

0‖b‖2

L2�‖u‖2La2 ‖b‖s−2

La2 d� + 1

2‖∇|b|s/2‖2

L2

�C‖b‖2L2�,2�(‖u‖s

Lq,a2 + ‖u‖sLq,a2 ) + 1

2‖∇|b|s/2‖2

L2

�C‖b‖2�L2�,2�(‖u‖s

L∞,s + ‖b‖sL∞,s )

+ ‖∇|b|s/2‖2L2,2 + 1

2‖∇|u|s/2‖2

L2,2 . (2.14)

Now, combining (2.6), (2.7) and (2.11)–(2.14), one has

‖u(t)‖sL∞,s + ‖b(t)‖s

L∞,s

�C(‖b‖2�L2�,2� + ‖p‖�

L�,�)(‖u‖sL∞,s + ‖b‖s

L∞,s )

+ ‖u0‖sLs + ‖b0‖s

Ls . (2.15)

Since � < ∞, by absolute continuity of the Lebesgue integralfor any T < ∞, we can find a uniform number 0 < t < T suchthat

‖b‖2�L2�,2� + ‖p‖�

L�,� � 12

for any interval I ⊂ [0, T ] of length t. Covering [0, T ] withfinitely many such intervals and iterating, from (2.15) we thenobtain the bound

‖u(t)‖sL∞,s + ‖b(t)‖s

L∞,s �C(‖u0‖sLs + ‖b0‖s

Ls ) (2.16)

on [0, T ].Case 2: �=∞, �= 3

2 . By taking limit in (2.10), we can takeq = s, a2 = 3s, (2.15) reduces to

‖u(t)‖sL∞,s + ‖b(t)‖s

L∞,s + ‖∇|u|s/2‖2L2,2 + ‖∇|b|s/2‖2

L2,2

�C(‖b‖2L∞,3 + ‖p‖L∞,3/2)(‖∇|u|s/2‖2

L2,2 + ‖∇|b|s/2‖2L2,2)

+ ‖u0‖sLs + ‖b0‖s

Ls .

Hence as long as ‖b‖L∞,3 and ‖p‖L∞,3/2 are sufficiently small,we have the bound

‖u(t)‖sL∞,s + ‖b(t)‖s

L∞,s �‖u0‖sLs + ‖b0‖s

Ls .

Case 3: �=1, �=∞. In this case, we can take q =∞, a2 =s,the inequality reads

‖u(t)‖sL∞,s + ‖b(t)‖s

L∞,s

�C(‖b‖2L2,∞ + ‖p‖1

L1,∞)(‖u‖sL∞,s + ‖b‖s

L∞,s )

+ ‖u0‖sLs + ‖b0‖s

Ls .

Due to the integrabilities of p and b, we have the same boundas (2.16).

This finishes the proof. �

Next one is the a priori estimate for the gradient of pressure∇p.

Theorem 2.3. Under the same conditions as Theorem 2.2. If∇p ∈ L�,� and b ∈ L3�,3� with 2

� + 3� �3, 1 < ��∞, or

1178 Y. Zhou / International Journal of Non-Linear Mechanics 41 (2006) 1174–1180

‖∇p‖L∞,1 + ‖b‖L∞,3 is sufficiently small, then u(t), b(t) ∈L∞,s ∩ Ls,3s

sup0� t �T

(‖u(t)‖Ls + ‖b(t)‖Ls )�C(‖u0‖Ls + ‖b0‖Ls ), (2.17)

where C depends on the norms of ∇p and b on the time interval.

Proof. It is sufficient to give a proof for 2� + 3

� = 3.First, by taking ∇div on both sides of the first equation of

(1.1) for smooth (u, p), one can obtain

−�(∇p) =3∑

i,j=1

�i�j (∇(uiuj ) − ∇(bibj )).

Therefore the Calderon–Zygmund inequality

‖∇p‖Lq �C(‖|u||∇u|‖Lq + ‖|b||∇b|‖Lq ), (2.18)

holds for any 1 < q < ∞.Multiply both sides of the first and the second equation of

(1.1) by su|u|s−2 and sb|b|s−2, respectively, and integrate overR3 × (0, t), 0 < t �T , after suitable integration by parts, thenwe obtain

‖u(t)‖sLs + 4(s − 2)

s‖∇|u|s/2‖2

L2,2 + s‖|u|s/2−1|∇u|‖2L2,2

+ 4(s − 2)

s‖b(t)‖s

Ls + ‖∇|b|s/2‖2L2,2 + s‖|b|s/2−1|∇b|‖2

L2,2

�s

∫ t

0

∫R3

|∇p||u|s−1|∇|u|s/2| dx d� + ‖u0‖sLs + ‖b0‖s

Ls

+ s

∫ t

0

∫R3

(b · ∇b)|u|s−2u dx d�

+ s

∫ t

0

∫R3

(b · ∇u)|b|s−2b dx d�. (2.19)

For 23 < � < ∞, 1 < � < ∞, with 3�s < 4, by using Holder’s in-

equality, interpolation inequality,Young’s inequality and (2.18),the first term in (2.19) can be estimated as

s

∫ t

0

∫R3

|∇p||u|s−1|∇|u|s/2| dx d�

�s

∫ t

0‖∇p‖La1 ‖u‖s−1

La2 d�

�s

∫ t

0‖∇p‖1−

L� ‖∇p‖Lq ‖u‖s−1

La2 d�

�C

∫ t

0‖∇p‖1−

L� (‖|u||∇u|‖Lq + ‖|b||∇b|‖

Lq )‖u‖s−1La2 d�

= C

∫ t

0‖∇p‖1−

L� (‖|u|s/2−1|∇u| · |u|2−s/2‖Lq

+ ‖|b|s/2−1|∇b| · |b|2−s/2‖Lq )‖u‖s−1

La2 d�

�C

∫ t

0‖∇p‖1−

L� (‖|u|s/2−1|∇u|‖L2‖|u|2−s/2‖

L2q

2−q

+ ‖|b|s/2−1|∇b|‖L2‖|b|2−s/2‖

L2q

2−q

)‖u‖s−1La2 d�

= C

∫ t

0‖∇p‖1−

L� (‖|u|s/2−1|∇u|‖L2‖u‖(4−s)/2

L(4−s)q

2−q

+ ‖|b|s/2−1|∇b|‖L2‖b‖(4−s)/2

L(4−s)q

2−q

)‖u‖s−1La2 d�

= C

∫ t

0‖∇p‖1−

L� (‖|u|s/2−1|∇u|‖L2 +‖|b|s/2−1|∇b|‖

L2)

× (‖u‖(4−s)/2La2 + ‖b‖(4−s)/2

La2 )‖u‖s−1La2 d�(

If(4 − s)q

2 − q= b

)�C

∫ t

0‖∇p‖

2(1−)2−

L� (‖u‖2s−2+(4−s)

2−La2 + ‖b‖

2s−2+(4−s)2−

La2 ) d�

+ s

2(‖|∇u‖u|s/2−1‖2

L2,2 + ‖|∇b||b|s/2−1‖2L2,2)

�C‖∇p‖2(1−)

2−L�,� (‖u‖

2s−2+(4−s)2−

L ,a2+ ‖b‖

2s−2+(4−s)2−

L ,a2)

+ s

2(‖|∇u‖u|s/2−1‖2

L2,2 + ‖|∇b||b|s/2−1‖2L2,2),

where in the above inequalities, the constants a1, a2, 0 < < 1,q and which are to be determined later satisfy

⎧⎪⎨⎪⎩1

a1+ s − 1

a2= 1,

1

a1= 1 −

�+

q,

(4 − s)q

2 − q= a2,

2(1 − )

2 −

1

�+ 2s − 2 + (4 − s)

2 −

1

= 1.

By a first glance, one can find that there are five unknowns butonly four equations, therefore the system is under-determined.How can we overcome this difficulty? Where can we find an-other equation?

Note that if Lemma 2.1 holds for ‖u‖L ,a2 , and the boundof ‖u‖L∞,s , can be obtained, the exponent of ‖u‖L ,a2 shouldsatisfy

2s − 2 + (4 − s)

2 − �s. (2.20)

In fact, the equality case in (2.20) is the fifth equation whichis suitable for our problem.

It follows from the equality case in (2.20) that = 12 . And

a1, a2, , q and can be solved as

= 1

2, a1 = 3s�

3� + 2s − 2, a2 = 3s�

3� − 2,

q = 3s�

6� + s − 4, = 3s�

3� − 2. (2.21)

It is obvious that and b satisfy

s

+ 3s

2a2= 5

2− 1

3

(2

�+ 3

�

)= 3

2, s < , s < a2 < 3s,

Y. Zhou / International Journal of Non-Linear Mechanics 41 (2006) 1174–1180 1179

so one can apply Lemma 2.1 on ‖u‖L ,a2 .

‖∇p‖2/3L�,�(‖u‖s

L3s�

3�−2 ,3s�

3�−2+ ‖b‖s

L3s�

3�−2 ,3s�

3�−2)

�C‖∇p‖2/3L�,�(‖u‖

�−1� s

L∞,s ‖u‖1� s

Ls,3s + ‖b‖�−1� s

L∞,s ‖b‖1� s

Ls,3s )

�C(�)‖∇p‖�L�,�(‖u‖s

L∞,s + ‖b‖sL∞,s )

+ �(‖u‖sLs,3s + ‖b‖s

Ls,3s )

�C‖∇p‖�L�,�(‖u‖s

L∞,s + ‖b‖sL∞,s )

+ s − 2

s(‖∇|u|s/2‖2

L2,2 + ‖∇|b|s/2‖2L2,2),

where we used Sobolev inequality and let � be sufficiently small.So the first term in (2.19) has the bound

s

∫ t

0

∫R3

|∇p||u|s−1|∇|u|s/2| dx d�

�C‖∇p‖�L�,�(‖u‖s

L∞,s + ‖b‖sL∞,s )

+ s − 2

s(‖∇|u|s/2‖2

L2,2 + ‖∇|b|s/2‖2L2,2)

+ s

2(‖|∇u||u|s/2−1‖2

L2,2 + ‖|∇b||b|s/2−1‖2L2,2). (2.22)

While the second non-linear term can be treated as

s

∫ t

0

∫R3

(b · ∇b)|u|s−2u dx d�

= −s

∫ t

0

∫R3

(b · ∇|u|s−2u)b dx d�

�s

∫ t

0

∫R3

|b|2|u| s2 −1‖∇|u|s/2| dx d�

�C

∫ t

0‖b‖4

La‖u‖s−2La2 d� + s − 2

s‖∇|u|s/2‖2

L2

�C

∫ t

0‖b‖2

L3�‖b‖2La2 ‖u‖s−2

La2 d� + s − 2

s‖∇|u|s/2‖2

L2

�C

∫ t

0‖b‖2

L3�(‖b‖sLa2 + ‖u‖s

La2 ) d� + s − 2

s‖∇|u|s/2‖2

L2

�C‖b‖2L3�,3�(‖u‖2

Lq,a2 + ‖u‖sLq,a2 ) + s − 2

s‖∇|u|s/2‖2

L2 ,

where a = 6�s3�+s−2 , while a2 and q are the same constants in

(2.21). Due to the above argument, we have

s

∫ t

0

∫R3

(b · ∇b)|u|s−2u dx d�

�C‖b‖3�L3�,3�(‖u‖s

L∞,s +‖b‖sL∞,s )

+2(s−2)

s‖∇|u|s/2‖2

L2,2+ s−2

s‖∇|b|s/2‖2

L2,2 . (2.23)

Similarly, the third non-linear term can be bounded as

s

∫ t

0

∫R3

(b · ∇u)|b|s−2b dx d�

�C‖b‖3�L3�,3�(‖u‖s

L∞,s + ‖b‖sL∞,s )

+2(s−2)

s‖∇|b|s/2‖2

L2,2+ s−2

s‖∇|u|s/2‖2

L2,2 . (2.24)

Combining (2.19) and (2.12)–(2.24), one has

‖u(t)‖sL∞,s + ‖b(t)‖s

L∞,s

�C(‖b‖2�L2�,2� + ‖p‖�

L�,�)(‖u‖sL∞,s + ‖b‖s

L∞,s )

+ ‖u0‖sLs + ‖b0‖s

Ls . (2.25)

Since � < ∞, by absolute continuity of the Lebesgue integralfor any T < ∞, we can find a uniform number 0 < t < T suchthat

‖b‖2�L2�,2� + ‖p‖�

L�,� � 12

for any interval I ⊂ [0, T ] of length t. Covering [0, T ] withfinitely many such intervals and iterating, from (2.25) we thenobtain the bound

‖u(t)‖sL∞,s + ‖b(t)‖s

L∞,s �C(‖u0‖sLs + ‖b0‖s

Ls )

on [0, T ].When � = ∞, � = 1, by taking limit in (2.21), we can take

q = s, a2 = 3s, (2.25) reduces to

‖u(t)‖sL∞,s + ‖b(t)‖s

L∞,s + 4(s − 2)

s‖∇|u|s/2‖2

L2,2

+ 4(s − 2)

s‖∇|b|s/2‖2

L2,2

�C(‖b‖2L∞,3 + ‖p‖L∞,1)(‖∇|u|s/2‖2

L2,2 + ‖∇|b|s/2‖2L2,2)

+ ‖u0‖sLs + ‖b0‖s

Ls .

Hence, as long as ‖b‖L∞,3 and ‖p‖L∞,1 are sufficiently small,we have the bound

‖u(t)‖sL∞,s + ‖b(t)‖s

L∞,s �‖u0‖sLs + ‖b0‖s

Ls .

When �= 23 , �=∞, we can take q =∞, a2 = s, the inequality

reads

‖u(t)‖sL∞,s + ‖b(t)‖s

L∞,s

�C(‖b‖2L2,∞ + ‖p‖1

L2/3,∞)(‖u‖sL∞,s + ‖b‖s

L∞,s )

+ ‖u0‖sLs + ‖b0‖s

Ls .

Due to the integrabilities of p and b, we get the bounds for‖u(t)‖s

L∞,s and ‖b(t)‖sL∞,s .

This finishes the proof. �

Remark 2.1. From the proof, it is easy to find that Theorem2.3 can be shown for the critical case s =4. However, for s > 4,the proof given here does not work anymore.

3. Proof

In order to prove the main theorems, we recall a result ofGiga [11] (see also [12] for the Navier–Stokes equations).

Theorem 3.1 (Giga [11]). Suppose u0, b0 ∈ Ls(R3), s�3,then there exists T0 and a unique classical solution u ∈BC([0, T0); Ls(R3)) and u ∈ BC([0, T0); Ls(R3)). Moreover,let (0, T∗) be the maximal interval such that u solves (1.1) in

1180 Y. Zhou / International Journal of Non-Linear Mechanics 41 (2006) 1174–1180

C((0, T∗); Ls(R3)), s > 3. Then

‖u(·, �)‖Ls � C

(T∗ − �)(s−3)/2sor

‖u(·, �)‖Ls � C

(T∗ − �)(s−3)/2s

with constant C independent of T∗ and s.

Proof of Theorems 1.1 and 1.2. Since u0, b0 ∈ L2(R3) ∩Lq(R3) for some q > 3, u0, b0 ∈ Ls(R3) for any s ∈ (3, q).Due to Theorem 3.1, there is a maximal interval [0, T∗) such thatthere exists a unique solution u(x, t) ∈ BC([0, T∗); Ls(R3))

and b(x, t) ∈ BC([0, T∗); Ls(R3)). Since u a Leray–Hopfweak solution which satisfies the energy inequality, we haveby the uniqueness criterion of Serrin–Masuda [1]

u ≡ u and u ≡ u on [0, T∗).

By the a priori estimate, (2.4) or (2.17) and combined withthe standard continuation argument, we can continue our localsmooth solution corresponding to u0, b0 ∈ Ls(R3), 3 < s < 4to obtain u, b ∈ BC([0, T ]; Ls(R3)) ∩ C∞(R3 × (0, T ]). Thiscompletes the proof of Theorems 1.1 and 1.2. �

Remark 3.1. Theorem 2.2 establishes the a priori estimate forthe solution in Ls , for any s�3. Hence, by the above proof,Theorem 1.1 is shown for the MHD equations in RN , for anyN �3. However, in Theorem 2.3, the a priori estimate is provedonly for s�4. Even by a recent regularity criterion for thecritical case u ∈ L∞,N , which is proved by Escauriaza et al.[13], Theorem 1.1 can be shown for the MHD equations in RN

with N =3 or 4. Whether Theorem 1.2 can be proved for N �5is a problem for the future.

Acknowledgment

This work is partially supported by NSFC under GrantNo. 10501012, Shanghai Rising-Star Program 05QMX1417,Shanghai Leading Academic Discipline and 111 Project.

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