Skip to main content

Global smooth solutions to 3D MHD with mixed partial dissipation and magnetic diffusion

Abstract

In this paper, we prove the existence of global smooth solutions to the Cauchy problem of 3D incompressible magnetohydrodynamics (MHD) flows with mixed partial dissipation and magnetic diffusion if the initial condition is suitably small.

1 Introduction

In this paper, we consider the following 3D incompressible MHD equations with mixed partial dissipation and magnetic diffusion (see [1]), i.e.,

u t +u⋅∇u=−∇p+μ u x x +μ u y y +b⋅∇b,
(1)
b t +u⋅∇b=η b x x +η b y y +b⋅∇u,
(2)
divu=0,divb=0,
(3)

associated with the initial data

u(0,x)= u 0 ,b(0,x)= b 0 .
(4)

Here u=( u 1 (t,x), u 2 (t,x), u 3 (t,x)) is the velocity field, b=( b 1 (t,x), b 2 (t,x), b 3 (t,x)) is the magnetic field, p=p(t,x) is scalar pressure, μ>0 is the kinematic viscosity, η>0 is the magnetic diffusion. For more background, we refer the reader to [2] for MHD and [1, 3] for MHD with mixed partial dissipation and magnetic diffusion. Without loss of generality, we assume that μ=η=1 in the remainder of the paper.

To state the main results, we first introduce the following conventions and notations which will be used throughout this paper. Set

∫fdx≜ ∫ R 3 fdxdydz,

and that ∥⋅∥ is the L 2 norm, i.e.,

∥ f ∥ = ( ∫ f 2 d x ) 1 2 , H m ≜ W m , 2 ( R 3 ) = { ∑ | α | ≤ m ∫ | D α u | 2 d x } 1 / 2 with the norm  ∥ â‹… ∥ H m .

Our main result of this paper can be stated as follows.

Theorem 1.1 Assume that u 0 ∈ H 2 and b 0 ∈ H 2 with div u 0 =div b 0 =0 and ∥ u 0 ∥ H 1 + ∥ b 0 ∥ H 1 ≤ε, where ε is a sufficiently small positive number. Then (1)-(4) admit global smooth solutions.

Remark 1.1 Theorem 1.1 is Theorem 1.2 in [1], which has not been proved in their paper. We would also emphasize that our proof of Theorem 1.1 is clearer for deducing the desired a priori estimates in Lemma 2.3 (see the next section) than that of Proposition 3.1 in [1].

The rest of the paper is organized as follows. In Section 2, we deduce the desired a priori estimates to complete the proof of Theorem 1.1. We finish the proof of Theorem 1.1 in Section 3 by the method of vanishing viscosities.

2 A priori estimates

In this section, we deduce the desired a priori estimates in order to finish the main result. Before we begin to prove the main theorem of this paper, we first state the following useful lemma that was deduced in [1].

Lemma 2.1 Assume that f, g, h, f x , g y , h z are all in L 2 ( R 3 ). Then we have

∫|fgh|dx≤C ∥ f ∥ 1 2 ∥ g ∥ 1 2 ∥ h ∥ 1 2 ∥ f x ∥ 1 2 ∥ g y ∥ 1 2 ∥ h z ∥ 1 2 .

Clearly, the standard energy estimate shows that

1 2 d d t ( ∥ u ∥ 2 + ∥ b ∥ 2 ) + ∥ u x ∥ 2 + ∥ u y ∥ 2 + ∥ b x ∥ 2 + ∥ b y ∥ 2 =0.
(5)

We denote that ω=∇×u and j=∇×b. Thus, applying the operator ‘∇×’ to (1) and (2), together with (3), we deduce that

ω t +u⋅∇ω−ω⋅∇u= ω x x + ω y y +b⋅∇j−j⋅∇b,
(6)
j t +u⋅∇j= j x x + j y y +b⋅∇ω+ ε i j k ( ∂ j b l ∂ l u k − ∂ j u l ∂ l b k ),
(7)

where ε i j k is defined as follows:

ε i j k ={ 1 if  ( i , j , k )  is an even permutation , − 1 if  ( i , j , k )  is an odd permutation , 0 others .

The first key lemma is the following.

Lemma 2.2 If (u,b) solves (1)-(4) and the initial data satisfies

∥ u 0 ∥ 2 + ∥ b 0 ∥ 2 ≤ 1 4 C and ∥ ω 0 ∥ 2 + ∥ j 0 ∥ 2 ≤ 1 4 ,
(8)

where C is a suitable large number, then the vorticity ω and the current density j satisfy

∥ ω ∥ 2 + ∥ j ∥ 2 ≤1, ∫ 0 t ( ∥ ω x ∥ 2 + ∥ ω y ∥ 2 + ∥ j x ∥ 2 + ∥ j y ∥ 2 ) ds≤1 for all t≥0.

Proof Multiplying (6) and (7) by ω and j, respectively, then integrating the resulting equations by parts, after adding the two equalities together, we finally deduce

1 2 d d t ( ∥ ω ∥ 2 + ∥ j ∥ 2 ) + ∥ ω x ∥ 2 + ∥ ω y ∥ 2 + ∥ j x ∥ 2 + ∥ j y ∥ 2 = ∫ ( ω ⋅ ∇ ω ⋅ ω − j ⋅ ∇ b ⋅ ω + ε i j k ( ∂ j b l ∂ l u k − ∂ j u l ∂ l b k ) j i ) d x = I + J + K + L .
(9)

We have to estimate each term on the right-hand side of (9). Some of the terms are the same as in [1] and are proved here for completeness. First, I can be written as

I = ∫ ω ⋅ ∇ u ⋅ ω d x = ∫ ω i ∂ i u j ω j d x = ∫ ω 1 ∂ x u j ω j d x + ∫ ω 2 ∂ y u j ω j d x + ∫ ω 3 ∂ z u j ω j d x = I 1 + I 2 + I 3 .

With the help of Lemma 2.1, we deduce that

I 1 = ∫ ω 1 ∂ x u j ω j d x ≤ C ∥ ω ∥ 1 2 ∥ u x ∥ 1 2 ∥ ω ∥ 1 2 ∥ ω x ∥ 1 2 ∥ ω y ∥ 1 2 ∥ ω x ∥ 1 2 ≤ 1 24 ∥ ω x ∥ 2 + 1 16 ∥ ω y ∥ 2 + C ∥ u x ∥ 2 ∥ ω ∥ 4 .

Similarly, we obtain that

I 2 =∫ ω 2 ∂ y u j ω j dx≤ 1 24 ∥ ω x ∥ 2 + 1 16 ∥ ω y ∥ 2 +C ∥ u y ∥ 2 ∥ ω ∥ 4 ,

and

I 3 = ∫ ω 3 ∂ z u j ω j d x = ∫ ( ∂ x u 2 − ∂ y u 1 ) ∂ z u j ω j d x ≤ 1 24 ∥ ω x ∥ 2 + 1 16 ∥ ω y ∥ 2 + C ( ∥ u x ∥ 2 + ∥ u y ∥ 2 ) ∥ ω ∥ 4 .

In order to bound J, we rewrite the integrand explicitly as follows:

J = − ∫ j ⋅ ∇ b ⋅ ω d x = − ∫ j i ∂ i b j ω j d x = − ∫ j 1 ∂ x b j ω j d x − ∫ j 2 ∂ y b j ω j d x − ∫ j 3 ∂ z b j ω j d x = J 1 + J 2 + J 3 .

Due to Lemma 2.1, we see that

J 1 = − ∫ j 1 ∂ x b j ω j d x ≤ C ∥ j ∥ 1 2 ∥ b x ∥ 1 2 ∥ ω ∥ 1 2 ∥ j x ∥ 1 2 ∥ ∂ x b z ∥ 1 2 ∥ ω y ∥ 1 2 ≤ 1 22 ∥ j x ∥ 2 + 1 16 ∥ ω y ∥ 2 + C ∥ b x ∥ 2 ( ∥ ω ∥ 4 + ∥ j ∥ 4 ) .

Similarly,

J 2 = − ∫ j 2 ∂ y b j ω j d x ≤ 1 16 ∥ ω y ∥ 2 + 1 22 ∥ j x ∥ 2 + 1 26 ∥ j y ∥ 2 + C ∥ b y ∥ 2 ( ∥ ω ∥ 4 + ∥ j ∥ 4 ) , J 3 = − ∫ j 3 ∂ z b j ω j d x = − ∫ ( ∂ x b 2 − ∂ y b 1 ) ∂ z b j ω j d x ≤ 1 24 ∥ ω x ∥ 2 + 1 22 ∥ j x ∥ 2 + 1 26 ∥ j y ∥ 2 + C ( ∥ b x ∥ 2 + ∥ b y ∥ 2 ) ( ∥ ω ∥ 4 + ∥ j ∥ 4 ) .

Now, let us turn to bound L,

L = − ∫ ε i j k ∂ j u l ∂ l u k j i d x = − ∫ ε i j k ∂ x u l ∂ l u k j i d x − ∫ ε i j k ∂ y u l ∂ l u k j i d x − ∫ ε i j k ∂ z u l ∂ l u k j i d x = L 1 + L 2 + L 3 .

By Lemma 2.1, we have that

L 1 = − ∫ ε i j k ∂ x u l ∂ l u k j i d x ≤ C ∥ u x ∥ 1 2 ∥ j ∥ 1 2 ∥ j ∥ 1 2 ∥ j x ∥ 1 2 ∥ j y ∥ 1 2 ∥ ω x ∥ 1 2 ≤ 1 24 ∥ ω x ∥ 2 + 1 22 ∥ j x ∥ 2 + 1 26 ∥ j y ∥ 2 + C ∥ u x ∥ 2 ∥ j ∥ 4 .

Similarly,

L 2 =−∫ ε i j k ∂ y u l ∂ l u k j i dx≤ 1 16 ∥ ω y ∥ 2 + 1 22 ∥ j x ∥ 2 + 1 26 ∥ j y ∥ 2 +C ∥ u y ∥ 2 ∥ j ∥ 4 .

As for L 3 , we should split it into three parts:

L 3 = − ∫ ε i j k ∂ z u l ∂ l u k j i d x = − ∫ ε i j k ∂ z u 1 ∂ x u k j i d x − ∫ ε i j k ∂ z u 2 ∂ y u k j i d x − ∫ ε i j k ∂ z u 3 ∂ z u k j i d x = L 31 + L 32 + L 33 , L 31 = − ∫ ε i j k ∂ z u 1 ∂ x u k j i d x ≤ C ∥ ω ∥ 1 2 ∥ b x ∥ 1 2 ∥ j ∥ 1 2 ∥ ω x ∥ 1 2 ∥ ∂ x b z ∥ 1 2 ∥ j y ∥ 1 2 ≤ 1 24 ∥ ω x ∥ 2 + 1 22 ∥ j x ∥ 2 + 1 26 ∥ j y ∥ 2 + C ∥ b x ∥ 2 ( ∥ ω ∥ 4 + ∥ j ∥ 4 ) .

Similarly,

L 32 =−∫ ε i j k ∂ z u 2 ∂ y u k j i dx≤ 1 24 ∥ ω x ∥ 2 + 1 26 ∥ j y ∥ 2 +C ∥ b y ∥ 2 ( ∥ ω ∥ 4 + ∥ j ∥ 4 ) .

To bound L 33 , using the incompressibility condition divu=0, we deduce that

L 33 = − ∫ ε i j k ∂ z u 3 ∂ z u k j i d x = ∫ ε i j k ( ∂ x u 1 + ∂ y u 2 ) ∂ z u k j i d x = L 33 1 + L 33 2 .

We get the L 33 1 and L 33 2 as follows:

L 33 1 = ∫ ε i j k ∂ x u 1 ∂ z u k j i d x ≤ C ∥ u x ∥ 1 2 ∥ j ∥ 1 2 ∥ j ∥ 1 2 ∥ ∂ x u z ∥ 1 2 ∥ j x ∥ 1 2 ∥ j y ∥ 1 2 ≤ 1 24 ∥ ω x ∥ 2 + 1 22 ∥ j x ∥ 2 + 1 26 ∥ j y ∥ 2 + C ∥ u x ∥ 2 ∥ j ∥ 4 ,

and

L 33 2 ≤ 1 16 ∥ ω y ∥ 2 + 1 22 ∥ j x ∥ 2 + 1 26 ∥ j y ∥ 2 +C ∥ u y ∥ 2 ∥ j ∥ 4 .

To bound K, we should divide it into three parts:

K = ∫ ε i j k ∂ j b l ∂ l u k j i d x = ∫ ε i 1 k ∂ x b l ∂ l u k j i d x + ∫ ε i 2 k ∂ y b l ∂ l u k j i d x + ∫ ε i 3 k ∂ z b l ∂ l u k j i d x = K 1 + K 2 + K 3 .

Similarly, we deduce that

K 1 = ∫ ε i 1 k ∂ x b l ∂ l u k j i d x ≤ C ∥ b x ∥ 1 2 ∥ ω ∥ 1 2 ∥ j ∥ 1 2 ∥ ∂ x b z ∥ 1 2 ∥ ω x ∥ 1 2 ∥ j y ∥ 1 2 ≤ 1 24 ∥ ω x ∥ 2 + 1 22 ∥ j x ∥ 2 + 1 26 ∥ j y ∥ 2 + C ∥ b x ∥ 2 ( ∥ ω ∥ 4 + ∥ j ∥ 4 ) ,

and

K 2 =∫ ε i 2 k ∂ y b l ∂ l u k j i dx≤ 1 24 ∥ ω x ∥ 2 + 1 26 ∥ j y ∥ 2 +C ∥ b y ∥ 2 ( ∥ ω ∥ 4 + ∥ j ∥ 4 ) .

For K 3 , we have

K 3 = ∫ ε i 3 k ∂ z b l ∂ l u k j i d x = ∫ ε i 3 k ∂ z b 1 ∂ x u k j i d x + ∫ ε i 3 k ∂ z b 2 ∂ y u k j i d x + ∫ ε i 3 k ∂ z b 3 ∂ z u k j i d x = K 31 + K 32 + K 33 .

Thus,

K 31 = ∫ ε i 3 k ∂ z b 1 ∂ x u k j i d x ≤ C ∥ j ∥ 1 2 ∥ u x ∥ 1 2 ∥ j ∥ 1 2 ∥ ω x ∥ 1 2 ∥ j x ∥ 1 2 ∥ j y ∥ 1 2 ≤ 1 24 ∥ ω x ∥ 2 + 1 22 ∥ j x ∥ 2 + 1 26 ∥ j y ∥ 2 + C ∥ u x ∥ 2 ∥ j ∥ 4 ,

and

K 32 =∫ ε i 3 k ∂ z b 2 ∂ y u k j i dx≤ 1 16 ∥ ω y ∥ 2 + 1 22 ∥ j x ∥ 2 + 1 26 ∥ j y ∥ 2 +C ∥ u y ∥ 2 ∥ j ∥ 4 .

As for K 33 , using divb=0, we obtain

K 33 = ∫ ε i 3 k ∂ z b 3 ∂ z u k j i d x = − ∫ ε i 3 k ( ∂ x b 1 + ∂ y b 2 ) ∂ z u k j i d x ≤ 1 24 ∥ ω x ∥ 2 + 1 26 ∥ j y ∥ 2 + C ( ∥ b x ∥ 2 + ∥ b y ∥ 2 ) ( ∥ ω ∥ 4 + ∥ j ∥ 4 ) .

Substituting all the above estimates into (9), we conclude that

1 2 d d t ( ∥ ω ∥ 2 + ∥ j ∥ 2 ) + 1 2 ( ∥ ω x ∥ 2 + ∥ ω y ∥ 2 + ∥ j x ∥ 2 + ∥ j y ∥ 2 ) ≤ C ( ∥ u x ∥ 2 + ∥ u y ∥ 2 + ∥ b x ∥ 2 + ∥ b y ∥ 2 ) ( ∥ ω ∥ 4 + ∥ j ∥ 4 ) .

Let ∥ ω ∥ 2 + ∥ j ∥ 2 ≤1, we deduce, with the assumption (8) on the initial data, that

∥ ω ∥ 2 + ∥ j ∥ 2 + ∫ 0 t ( ∥ ω x ∥ 2 + ∥ ω y ∥ 2 + ∥ j x ∥ 2 + ∥ j y ∥ 2 ) ds≤1.

Thus, the proof of Lemma 2.2 is completed. □

Now, we turn to deduce the higher order estimates about the solution.

Lemma 2.3 If (u,b) is the solution of (1)-(4), then

∥ ∇ ω ∥ 2 + ∥ ∇ j ∥ 2 + ∫ 0 t ( ∥ ∇ ω x ∥ 2 + ∥ ∇ ω y ∥ 2 + ∥ ∇ j x ∥ 2 + ∥ ∇ j y ∥ 2 ) ds≤C.
(10)

Proof Multiplying (6) and (7) by Δω and Δj, respectively, then integrating the resultant equations by parts, after adding the two equalities together, we finally obtain

1 2 d d t ( ∥ ∇ ω ∥ 2 + ∥ ∇ j ∥ 2 ) + ( ∥ ∇ ω x ∥ 2 + ∥ ∇ ω y ∥ 2 + ∥ ∇ j x ∥ 2 + ∥ ∇ j y ∥ 2 ) = ∫ [ u ⋅ ∇ ω ⋅ Δ ω + u ⋅ ∇ j ⋅ Δ j − ω ⋅ ∇ ω ⋅ Δ ω + j ⋅ ∇ b ⋅ Δ ω − b ⋅ ∇ j ⋅ Δ ω − b ⋅ ∇ ω ⋅ Δ j − ε i j k ( ∂ j b l ∂ l u k − ∂ j u l ∂ l b k ) Δ j i ] d x = M + N + P + Q + R + S .
(11)

Now, we turn to bound each term on the right-hand side of (11). Similar as the proof of Lemma 2.2, keeping in mind Lemma 2.1 and the divergence-free property of u and b, we deduce

M = ∫ u ⋅ ∇ ω ⋅ Δ ω d x = ∫ u i ∂ i ω j ∂ k k 2 ω j d x = − ∫ ∂ k u i ∂ i ω j ∂ k ω j d x = − ∫ ∂ x u i ∂ i ω j ∂ x ω j d x − ∫ ∂ y u i ∂ i ω j ∂ y ω j d x − ∫ ∂ z u i ∂ i ω j ∂ z ω j d x = M 1 + M 2 + M 3 .

We estimate each term as follows:

M 1 = − ∫ ∂ x u i ∂ i ω j ∂ x ω j d x ≤ C ∥ u x ∥ 1 2 ∥ ∇ ω ∥ 1 2 ∥ ∇ ω ∥ 1 2 ∥ ∇ ω x ∥ 1 2 ∥ ∇ ω y ∥ 1 2 ∥ ∂ x u z ∥ 1 2 ≤ 1 62 ∥ ∇ ω x ∥ 2 + 1 40 ∥ ∇ ω y ∥ 2 + C ∥ u x ∥ ∥ ω x ∥ ∥ ∇ ω ∥ 2 ≤ 1 62 ∥ ∇ ω x ∥ 2 + 1 40 ∥ ∇ ω y ∥ 2 + C ( ∥ u x ∥ 2 + ∥ ω x ∥ 2 ) ∥ ∇ ω ∥ 2 .

Similarly,

M 2 = − ∫ ∂ y u i ∂ i ω j ∂ y ω j d x ≤ C ∥ u y ∥ 1 2 ∥ ∇ ω ∥ 1 2 ∥ ∇ ω ∥ 1 2 ∥ ∇ ω x ∥ 1 2 ∥ ∇ ω y ∥ 1 2 ∥ ∂ y u z ∥ 1 2 ≤ 1 62 ∥ ∇ ω x ∥ 2 + 1 40 ∥ ∇ ω y ∥ 2 + C ( ∥ u y ∥ 2 + ∥ ω y ∥ 2 ) ∥ ∇ ω ∥ 2 .

Now, we turn to bound M 3 ,

M 3 = − ∫ ∂ z u i ∂ i ω j ∂ z ω j d x = − ∫ ∂ z u 1 ∂ x ω j ∂ z ω j d x − ∫ ∂ z u 2 ∂ y ω j ∂ z ω j d x − ∫ ∂ z u 3 ∂ z ω j ∂ z ω j d x = M 31 + M 32 + M 33 .

Similarly, we can deduce that

M 31 = − ∫ ∂ z u 1 ∂ x ω j ∂ z ω j d x ≤ C ∥ ω ∥ 1 2 ∥ ω x ∥ 1 2 ∥ ∇ ω ∥ 1 2 ∥ ω x ∥ 1 2 ∥ ∇ ω x ∥ 1 2 ∥ ∇ ω y ∥ 1 2 ≤ 1 62 ∥ ∇ ω x ∥ 2 + 1 40 ∥ ∇ ω y ∥ 2 + C ( ∥ ω ∥ 2 + ∥ ω x ∥ 2 ) ∥ ∇ ω ∥ 2 , M 32 = − ∫ ∂ z u 2 ∂ y ω j ∂ z ω j d x ≤ C ∥ ω ∥ 1 2 ∥ ω y ∥ 1 2 ∥ ∇ ω ∥ 1 2 ∥ ω x ∥ 1 2 ∥ ∇ ω y ∥ 1 2 ∥ ∇ ω y ∥ 1 2 ≤ 1 40 ∥ ∇ ω y ∥ 2 + C ( ∥ ω ∥ 2 + ∥ ω x ∥ 2 ) ∥ ∇ ω ∥ 2 , M 33 = − ∫ ∂ z u 3 ∂ z ω j ∂ z ω j d x = ∫ ( ∂ x u 1 + ∂ y u 2 ) ∂ z ω j ∂ z ω j d x ≤ C ∥ u x ∥ 1 2 ∥ ∇ ω ∥ 1 2 ∥ ∇ ω ∥ 1 2 ∥ ∂ x u z ∥ 1 2 ∥ ∇ ω x ∥ 1 2 ∥ ∇ ω y ∥ 1 2 + C ∥ u y ∥ 1 2 ∥ ∇ ω ∥ 1 2 ∥ ∇ ω ∥ 1 2 ∥ ∂ y u z ∥ 1 2 ∥ ∇ ω x ∥ 1 2 ∥ ∇ ω y ∥ 1 2 ≤ 1 62 ∥ ∇ ω x ∥ 2 + 1 40 ∥ ∇ ω y ∥ 2 + C ( ∥ u x ∥ 2 + ∥ u y ∥ 2 + ∥ ω x ∥ 2 + ∥ ω y ∥ 2 ) ∥ ∇ ω ∥ 2 .

As for N, integrating by parts, we deduce that

N = ∫ u ⋅ ∇ j ⋅ Δ j d x = ∫ u i ∂ i j j ∂ k k 2 j j d x = − ∫ ∂ k u i ∂ i j j ∂ k j j d x = − ∫ ∂ x u i ∂ i j j ∂ x j j d x − ∫ ∂ y u i ∂ i j j ∂ y j j d x − ∫ ∂ z u i ∂ i j j ∂ z j j d x = N 1 + N 2 + N 3 .

Similarly, we have

N 1 = − ∫ ∂ x u i ∂ i j j ∂ x j j d x ≤ C ∥ u x ∥ 1 2 ∥ ∇ j ∥ 1 2 ∥ j x ∥ 1 2 ∥ ω x ∥ 1 2 ∥ ∇ j x ∥ 1 2 ∥ ∇ j y ∥ 1 2 ≤ 1 50 ∥ ∇ j x ∥ 2 + 1 48 ∥ ∇ j y ∥ 2 + C ( ∥ u x ∥ 2 + ∥ ω x ∥ 2 ) ∥ ∇ j ∥ 2 , N 2 = − ∫ ∂ y u i ∂ i j j ∂ y j j d x ≤ C ∥ u y ∥ 1 2 ∥ ∇ j ∥ 1 2 ∥ j y ∥ 1 2 ∥ ω y ∥ 1 2 ∥ ∇ j x ∥ 1 2 ∥ ∇ j y ∥ 1 2 ≤ 1 50 ∥ ∇ j x ∥ 2 + 1 48 ∥ ∇ j y ∥ 2 + C ( ∥ u y ∥ 2 + ∥ ω y ∥ 2 ) ∥ ∇ j ∥ 2 .

As for N 3 , we obtain

N 3 = − ∫ ∂ z u i ∂ i j j ∂ z j j d x = − ∫ ∂ z u 1 ∂ x j j ∂ z j j d x − ∫ ∂ z u 2 ∂ y j j ∂ z j j d x − ∫ ∂ z u 3 ∂ z j j ∂ z j j d x = N 31 + N 32 + N 33 .

Thus, we have

N 31 = − ∫ ∂ z u 1 ∂ x j j ∂ z j j d x ≤ C ∥ ω ∥ 1 2 ∥ j x ∥ 1 2 ∥ ∇ j ∥ 1 2 ∥ ω x ∥ 1 2 ∥ ∇ j x ∥ 1 2 ∥ ∇ j y ∥ 1 2 ≤ 1 50 ∥ ∇ j x ∥ 2 + 1 48 ∥ ∇ j y ∥ 2 + C ( ∥ ω ∥ 2 + ∥ ω x ∥ 2 ) ∥ ∇ j ∥ 2 , N 32 = − ∫ ∂ z u 2 ∂ y j j ∂ z j j d x ≤ C ∥ ω ∥ 1 2 ∥ j y ∥ 1 2 ∥ ∇ j ∥ 1 2 ∥ ω x ∥ 1 2 ∥ ∇ j y ∥ 1 2 ∥ ∇ j y ∥ 1 2 ≤ 1 48 ∥ ∇ j y ∥ 2 + C ( ∥ ω ∥ 2 + ∥ ω x ∥ 2 ) ∥ ∇ j ∥ 2 , N 33 = − ∫ ∂ z u 3 ∂ z j j ∂ z j j d x ≤ C ∥ u x ∥ 1 2 ∥ ∇ j ∥ 1 2 ∥ ∇ j ∥ 1 2 ∥ ∇ j x ∥ 1 2 ∥ ∇ j y ∥ 1 2 ∥ ω x ∥ 1 2 + C ∥ u y ∥ 1 2 ∥ ∇ j ∥ 1 2 ∥ ∇ j ∥ 1 2 ∥ ∇ j x ∥ 1 2 ∥ ∇ j y ∥ 1 2 ∥ ω y ∥ 1 2 ≤ 1 50 ∥ ∇ j x ∥ 2 + 1 48 ∥ ∇ j y ∥ 2 + C ( ∥ u x ∥ 2 + ∥ u y ∥ 2 + ∥ ω x ∥ 2 + ∥ ω y ∥ 2 ) ∥ ∇ j ∥ 2 .

We now turn to bound P,

P = − ∫ ω ⋅ ∇ ω ⋅ Δ ω d x = − ∫ ω i ∂ i u j ∂ k k 2 ω j d x = ∫ ∂ k ω i ∂ i u j ∂ k ω j d x + ∫ ω i ∂ k ∂ i u j ∂ k ω j d x = P 1 + P 2 .

For P 1 , we have

P 1 = ∫ ∂ k ω i ∂ i u j ∂ k ω j d x = ∫ ∂ x ω i ∂ i u j ∂ x ω j d x + ∫ ∂ y ω i ∂ i u j ∂ y ω j d x + ∫ ∂ z ω i ∂ i u j ∂ z ω j d x = P 11 + P 12 + P 13 , P 11 = ∫ ∂ x ω i ∂ i u j ∂ x ω j d x ≤ C ∥ ω x ∥ 1 2 ∥ ω ∥ 1 2 ∥ ω x ∥ 1 2 ∥ ω x ∥ 1 2 ∥ ∇ ω x ∥ 1 2 ∥ ∇ ω x ∥ 1 2 ≤ 1 62 ∥ ∇ ω x ∥ 2 + C ( ∥ ω ∥ 2 + ∥ ω x ∥ 2 ) ∥ ∇ ω ∥ 2 , P 12 = ∫ ∂ y ω i ∂ i u j ∂ y ω j d x ≤ C ∥ ω y ∥ 1 2 ∥ ω ∥ 1 2 ∥ ω y ∥ 1 2 ∥ ω x ∥ 1 2 ∥ ∇ ω y ∥ 1 2 ∥ ∇ ω y ∥ 1 2 ≤ 1 40 ∥ ∇ ω y ∥ 2 + C ( ∥ ω ∥ 2 + ∥ ω x ∥ 2 ) ∥ ∇ ω ∥ 2 .

For P 13 , we see that

P 13 = ∫ ∂ z ω i ∂ i u j ∂ z ω j d x = ∫ ∂ z ω 1 ∂ x u j ∂ z ω j d x + ∫ ∂ z ω 2 ∂ y u j ∂ z ω j d x + ∫ ∂ z ω 3 ∂ z u j ∂ z ω j d x = P 13 1 + P 13 2 + P 13 3 .

Thus, we can bound P 13 as follows:

P 13 1 = ∫ ∂ z ω 1 ∂ x u j ∂ z ω j d x ≤ C ∥ ∇ ω ∥ 1 2 ∥ u x ∥ 1 2 ∥ ∇ ω ∥ 1 2 ∥ ∇ ω x ∥ 1 2 ∥ ∇ ω y ∥ 1 2 ∥ ω x ∥ 1 2 ≤ 1 62 ∥ ∇ ω x ∥ 2 + 1 40 ∥ ∇ ω y ∥ 2 + C ( ∥ u x ∥ 2 + ∥ ω x ∥ 2 ) ∥ ∇ ω ∥ 2 , P 13 2 = ∫ ∂ z ω 2 ∂ y u j ∂ z ω j d x ≤ C ∥ ∇ ω ∥ 1 2 ∥ u y ∥ 1 2 ∥ ∇ ω ∥ 1 2 ∥ ∇ ω x ∥ 1 2 ∥ ∇ ω y ∥ 1 2 ∥ ω y ∥ 1 2 ≤ 1 62 ∥ ∇ ω x ∥ 2 + 1 40 ∥ ∇ ω y ∥ 2 + C ( ∥ u y ∥ 2 + ∥ ω y ∥ 2 ) ∥ ∇ ω ∥ 2 , P 13 2 = ∫ ∂ z ω 3 ∂ z u j ∂ z ω j d x = ∫ ∂ z ( ∂ x u 2 − ∂ y u 1 ) ∂ z u j ∂ z ω j d x ≤ C ∥ ω x ∥ 1 2 ∥ ω ∥ 1 2 ∥ ∇ ω ∥ 1 2 ∥ ∇ ω x ∥ 1 2 ∥ ω x ∥ 1 2 ∥ ∇ ω y ∥ 1 2 + C ∥ ω y ∥ 1 2 ∥ ω ∥ 1 2 ∥ ∇ ω ∥ 1 2 ∥ ∇ ω y ∥ 1 2 ∥ ω y ∥ 1 2 ∥ ∇ ω x ∥ 1 2 ≤ 1 62 ∥ ∇ ω x ∥ 2 + 1 40 ∥ ∇ ω y ∥ 2 + C ( ∥ ω ∥ 2 + ∥ ω x ∥ 2 + ∥ ω y ∥ 2 ) ∥ ∇ ω ∥ 2 .

Now, we turn to P 2 ,

P 2 = ∫ ω i ∂ k ∂ i u j ∂ k ω j d x = ∫ ω 1 ∂ k ∂ x u j ∂ k ω j d x + ∫ ω 2 ∂ k ∂ y u j ∂ k ω j d x + ∫ ω 3 ∂ k ∂ z u j ∂ k ω j d x = P 21 + P 22 + P 23 .

Similarly,

P 21 = ∫ ω 1 ∂ k ∂ x u j ∂ k ω j d x ≤ C ∥ ω ∥ 1 2 ∥ ω x ∥ 1 2 ∥ ∇ ω ∥ 1 2 ∥ ω x ∥ 1 2 ∥ ∇ ω y ∥ 1 2 ∥ ∇ ω x ∥ 1 2 ≤ 1 62 ∥ ∇ ω x ∥ 2 + 1 40 ∥ ∇ ω y ∥ 2 + C ( ∥ ω ∥ 2 + ∥ ω x ∥ 2 ) ∥ ∇ ω ∥ 2 , P 22 = ∫ ω 2 ∂ k ∂ y u j ∂ k ω j d x ≤ C ∥ ω ∥ 1 2 ∥ ω y ∥ 1 2 ∥ ∇ ω ∥ 1 2 ∥ ω x ∥ 1 2 ∥ ∇ ω y ∥ 1 2 ∥ ∇ ω y ∥ 1 2 ≤ 1 40 ∥ ∇ ω y ∥ 2 + C ( ∥ ω ∥ 2 + ∥ ω x ∥ 2 ) ∥ ∇ ω ∥ 2 , P 23 = ∫ ω 3 ∂ k ∂ z u j ∂ k ω j d x ≤ C ∥ u x ∥ 1 2 ∥ ∇ ω ∥ 1 2 ∥ ∇ ω ∥ 1 2 ∥ ∇ ω x ∥ 1 2 ∥ ∇ ω y ∥ 1 2 ∥ ω x ∥ 1 2 + C ∥ u y ∥ 1 2 ∥ ∇ ω ∥ 1 2 ∥ ∇ ω ∥ 1 2 ∥ ∇ ω x ∥ 1 2 ∥ ∇ ω y ∥ 1 2 ∥ ω y ∥ 1 2 ≤ 1 62 ∥ ∇ ω x ∥ 2 + 1 40 ∥ ∇ ω y ∥ 2 + C ( ∥ u x ∥ 2 + ∥ u y ∥ 2 + ∥ ω x ∥ 2 + ∥ ω y ∥ 2 ) ∥ ∇ ω ∥ 2 .

To bound Q, we see that

Q = ∫ j ⋅ ∇ b ⋅ Δ ω d x = ∫ j i ∂ i b j ∂ k k 2 ω j d x = − ∫ ∂ k j i ∂ i b j ∂ k ω j d x − ∫ j i ∂ k ∂ i b j ∂ k ω j d x = Q 1 + Q 2 , Q 1 = − ∫ ∂ x j i ∂ i b j ∂ x ω j d x − ∫ ∂ y j i ∂ i b j ∂ y ω j d x − ∫ ∂ z j i ∂ i b j ∂ z ω j d x = Q 11 + Q 12 + Q 13 , Q 11 = − ∫ ∂ x j i ∂ i b j ∂ x ω j d x ≤ C ∥ j x ∥ 1 2 ∥ j ∥ 1 2 ∥ ω x ∥ 1 2 ∥ ∇ j x ∥ 1 2 ∥ ∇ j ∥ 1 2 ∥ ∇ ω x ∥ 1 2 ≤ 1 62 ∥ ∇ ω x ∥ 2 + 1 50 ∥ ∇ j x ∥ 2 + C ( ∥ j ∥ 2 + ∥ j x ∥ 2 ) ( ∥ ∇ ω ∥ 2 + ∥ ∇ j ∥ 2 ) , Q 12 = − ∫ ∂ y j i ∂ i b j ∂ y ω j d x ≤ C ∥ j y ∥ 1 2 ∥ j ∥ 1 2 ∥ ω x ∥ 1 2 ∥ ∇ j y ∥ 1 2 ∥ j x ∥ 1 2 ∥ ∇ ω x ∥ 1 2 ≤ 1 62 ∥ ∇ ω x ∥ 2 + 1 48 ∥ ∇ j y ∥ 2 + C ( ∥ j ∥ 2 + ∥ j y ∥ 2 ) ( ∥ ∇ ω ∥ 2 + ∥ ∇ j ∥ 2 ) .

As for Q 13 , we see that

Q 13 = − ∫ ∂ z j i ∂ i b j ∂ z ω j d x = − ∫ ∂ z j 1 ∂ x b j ∂ z ω j d x − ∫ ∂ z j 2 ∂ y b j ∂ z ω j d x − ∫ ∂ z j 3 ∂ z b j ∂ z ω j d x = Q 13 1 + Q 13 2 + Q 13 3 .

Thus, we can deduce that

Q 13 1 = − ∫ ∂ z j 1 ∂ x b j ∂ z ω j d x ≤ C ∥ ∇ j ∥ 1 2 ∥ b x ∥ 1 2 ∥ ∇ ω ∥ 1 2 ∥ ∇ j x ∥ 1 2 ∥ j x ∥ 1 2 ∥ ∇ ω y ∥ 1 2 ≤ 1 40 ∥ ∇ ω y ∥ 2 + 1 50 ∥ ∇ j x ∥ 2 + C ( ∥ b x ∥ 2 + ∥ j x ∥ 2 ) ( ∥ ∇ ω ∥ 2 + ∥ ∇ j ∥ 2 ) , Q 13 2 = − ∫ ∂ z j 2 ∂ y b j ∂ z ω j d x ≤ C ∥ ∇ j ∥ 1 2 ∥ b x ∥ 1 2 ∥ ∇ ω ∥ 1 2 ∥ ∇ j x ∥ 1 2 ∥ j y ∥ 1 2 ∥ ∇ ω y ∥ 1 2 ≤ 1 40 ∥ ∇ ω y ∥ 2 + 1 50 ∥ ∇ j x ∥ 2 + C ( ∥ b y ∥ 2 + ∥ j y ∥ 2 ) ( ∥ ∇ ω ∥ 2 + ∥ ∇ j ∥ 2 ) , Q 13 3 = − ∫ ∂ z j 3 ∂ z b j ∂ z ω j d x = − ∫ ∂ z ( ∂ x b 2 − ∂ y b 1 ) ∂ z b j ∂ z ω j d x ≤ C ∥ j x ∥ 1 2 ∥ j ∥ 1 2 ∥ ∇ ω ∥ 1 2 ∥ ∇ j x ∥ 1 2 ∥ j y ∥ 1 2 ∥ ∇ ω x ∥ 1 2 + C ∥ j y ∥ 1 2 ∥ j ∥ 1 2 ∥ ∇ ω ∥ 1 2 ∥ ∇ j y ∥ 1 2 ∥ j y ∥ 1 2 ∥ ∇ ω x ∥ 1 2 ≤ 1 62 ∥ ∇ ω x ∥ 2 + 1 50 ∥ ∇ j x ∥ 2 + 1 48 ∥ ∇ j y ∥ 2 + C ( ∥ j ∥ 2 + ∥ j x ∥ 2 + ∥ j y ∥ 2 ) ( ∥ ∇ ω ∥ 2 + ∥ ∇ j ∥ 2 ) .

Now, we turn to Q 2 ,

Q 2 = − ∫ b ⋅ ∇ j ⋅ Δ ω d x = − ∫ j i ∂ k ∂ i b j ∂ k ω j d x = − ∫ j 1 ∂ k ∂ x b j ∂ k ω j d x − ∫ j 2 ∂ k ∂ y b j ∂ k ω j d x − ∫ j 3 ∂ k ∂ z b j ∂ k ω j d x = Q 21 + Q 22 + Q 23 .

We deduce that

Q 21 = − ∫ j 1 ∂ k ∂ x b j ∂ k ω j d x ≤ C ∥ j ∥ 1 2 ∥ j x ∥ 1 2 ∥ ∇ ω ∥ 1 2 ∥ ∇ ω x ∥ 1 2 ∥ j y ∥ 1 2 ∥ ∇ j x ∥ 1 2 ≤ 1 62 ∥ ∇ ω x ∥ 2 + 1 50 ∥ ∇ j x ∥ 2 + C ( ∥ j ∥ 2 + ∥ j x ∥ 2 ) ( ∥ ∇ ω ∥ 2 + ∥ ∇ j ∥ 2 ) , Q 22 = − ∫ j 2 ∂ k ∂ y b j ∂ k ω j d x ≤ C ∥ j ∥ 1 2 ∥ j y ∥ 1 2 ∥ ∇ ω ∥ 1 2 ∥ ∇ ω x ∥ 1 2 ∥ j y ∥ 1 2 ∥ ∇ j y ∥ 1 2 ≤ 1 62 ∥ ∇ ω x ∥ 2 + 1 48 ∥ ∇ j y ∥ 2 + C ( ∥ j ∥ 2 + ∥ j y ∥ 2 ) ( ∥ ∇ ω ∥ 2 + ∥ ∇ j ∥ 2 ) , Q 23 = − ∫ j 3 ∂ k ∂ z b j ∂ k ω j d x ≤ C ∥ b x ∥ 1 2 ∥ ∇ j ∥ 1 2 ∥ ∇ ω ∥ 1 2 ∥ ∇ ω x ∥ 1 2 ∥ ∇ j y ∥ 1 2 ∥ j x ∥ 1 2 + C ∥ b y ∥ 1 2 ∥ ∇ j ∥ 1 2 ∥ ∇ ω ∥ 1 2 ∥ ∇ ω x ∥ 1 2 ∥ ∇ j y ∥ 1 2 ∥ j y ∥ 1 2 ≤ 1 62 ∥ ∇ ω x ∥ 2 + 1 48 ∥ ∇ j y ∥ 2 + C ( ∥ b x ∥ 2 + ∥ j x ∥ 2 + ∥ b y ∥ 2 + ∥ j y ∥ 2 ) ( ∥ ∇ ω ∥ 2 + ∥ ∇ j ∥ 2 ) .

Here we start to estimate R as follows:

R = − ∫ b ⋅ ∇ j ⋅ Δ ω d x − ∫ b ⋅ ∇ ω ⋅ Δ j d x = − ∫ b i ∂ i j j ∂ k k 2 ω j d x − ∫ b i ∂ i ω j ∂ k k 2 j j d x = ∫ ∂ k b i ∂ i j j ∂ k ω j d x + ∫ ∂ k b i ∂ i ω j ∂ k j j d x = R 1 + R 2 .

For R 1 , we have

R 1 = ∫ ∂ k b i ∂ i j j ∂ k ω j d x = ∫ ∂ x b i ∂ i j j ∂ x ω j d x + ∫ ∂ y b i ∂ i j j ∂ y ω j d x + ∫ ∂ z b i ∂ i j j ∂ z ω j d x = R 11 + R 12 + R 13 .

We deduce that

R 11 = ∫ ∂ x b i ∂ i j j ∂ x ω j d x ≤ C ∥ b x ∥ 1 2 ∥ ∇ j ∥ 1 2 ∥ ω x ∥ 1 2 ∥ ∇ j x ∥ 1 2 ∥ ∇ ω x ∥ 1 2 ∥ j x ∥ 1 2 ≤ 1 62 ∥ ∇ ω x ∥ 2 + 1 50 ∥ ∇ j x ∥ 2 + C ( ∥ b x ∥ 2 + ∥ ω x ∥ 2 ) ∥ ∇ j ∥ 2 , R 12 = ∫ ∂ y b i ∂ i j j ∂ y ω j d x ≤ C ∥ b y ∥ 1 2 ∥ ∇ j ∥ 1 2 ∥ ω y ∥ 1 2 ∥ ∇ j x ∥ 1 2 ∥ ∇ ω y ∥ 1 2 ∥ j y ∥ 1 2 ≤ 1 50 ∥ ∇ j x ∥ 2 + 1 40 ∥ ∇ ω y ∥ 2 + C ( ∥ b y ∥ 2 + ∥ ω y ∥ 2 ) ∥ ∇ j ∥ 2 .

For R 13 , we have

R 13 = ∫ ∂ z b i ∂ i j j ∂ z ω j d x = ∫ ∂ z b 1 ∂ x j j ∂ z ω j d x + ∫ ∂ z b 2 ∂ y j j ∂ z ω j d x + ∫ ∂ z b 3 ∂ z j j ∂ z ω j d x = R 13 1 + R 13 2 + R 13 3 .

Similarly,

R 13 1 = ∫ ∂ z b 1 ∂ x j j ∂ z ω j d x ≤ C ∥ j ∥ 1 2 ∥ j x ∥ 1 2 ∥ ∇ ω ∥ 1 2 ∥ ∇ ω x ∥ 1 2 ∥ ∇ j x ∥ 1 2 ∥ j y ∥ 1 2 ≤ 1 62 ∥ ∇ ω x ∥ 2 + 1 50 ∥ ∇ j x ∥ 2 + C ( ∥ j ∥ 2 + ∥ j x ∥ 2 ) ( ∥ ∇ ω ∥ 2 + ∥ ∇ j ∥ 2 ) , R 13 2 = ∫ ∂ z b 2 ∂ y j j ∂ z ω j d x ≤ C ∥ j ∥ 1 2 ∥ j y ∥ 1 2 ∥ ∇ ω ∥ 1 2 ∥ ∇ ω x ∥ 1 2 ∥ j y ∥ 1 2 ∥ ∇ j y ∥ 1 2 ≤ 1 62 ∥ ∇ ω x ∥ 2 + 1 48 ∥ ∇ j y ∥ 2 + C ( ∥ j ∥ 2 + ∥ j y ∥ 2 ) ( ∥ ∇ ω ∥ 2 + ∥ ∇ j ∥ 2 ) , R 13 3 = ∫ ∂ z b 3 ∂ z j j ∂ z ω j d x = − ∫ ( ∂ x b 1 + ∂ y b 2 ) ∂ z j j ∂ z ω j d x ≤ C ∥ b x ∥ 1 2 ∥ ∇ j ∥ 1 2 ∥ ∇ ω ∥ 1 2 ∥ ∇ ω x ∥ 1 2 ∥ ∇ j y ∥ 1 2 ∥ j x ∥ 1 2 + C ∥ b y ∥ 1 2 ∥ ∇ j ∥ 1 2 ∥ ∇ ω ∥ 1 2 ∥ ∇ ω x ∥ 1 2 ∥ ∇ j y ∥ 1 2 ∥ j y ∥ 1 2 ≤ 1 62 ∥ ∇ ω x ∥ 2 + 1 48 ∥ ∇ j y ∥ 2 + C ( ∥ j ∥ 2 + ∥ j y ∥ 2 ) ( ∥ ∇ ω ∥ 2 + ∥ ∇ j ∥ 2 ) .

Now, we turn to R 2 ,

R 2 = ∫ ∂ k b i ∂ i ω j ∂ k j j d x = ∫ ∂ x b i ∂ i ω j ∂ x j j d x + ∫ ∂ y b i ∂ i ω j ∂ y j j d x + ∫ ∂ z b i ∂ i ω j ∂ z j j d x = R 21 + R 22 + R 23 .

Thus, similarly, we deduce that

R 21 = ∫ ∂ x b i ∂ i ω j ∂ x j j d x ≤ C ∥ b x ∥ 1 2 ∥ ∇ ω ∥ 1 2 ∥ j x ∥ 1 2 ∥ ∇ ω x ∥ 1 2 ∥ ∇ j x ∥ 1 2 ∥ j x ∥ 1 2 ≤ 1 62 ∥ ∇ ω x ∥ 2 + 1 50 ∥ ∇ j x ∥ 2 + C ( ∥ b x ∥ 2 + ∥ j x ∥ 2 ) ( ∥ ∇ ω ∥ 2 + ∥ ∇ j ∥ 2 ) , R 22 = ∫ ∂ y b i ∂ i ω j ∂ y j j d x ≤ C ∥ b y ∥ 1 2 ∥ ∇ ω ∥ 1 2 ∥ j y ∥ 1 2 ∥ ∇ ω x ∥ 1 2 ∥ ∇ j y ∥ 1 2 ∥ j y ∥ 1 2 ≤ 1 62 ∥ ∇ ω x ∥ 2 + 1 48 ∥ ∇ j y ∥ 2 + C ( ∥ b y ∥ 2 + ∥ j y ∥ 2 ) ( ∥ ∇ ω ∥ 2 + ∥ ∇ j ∥ 2 ) .

For R 23 , we have

R 23 = ∫ ∂ z b i ∂ i ω j ∂ z j j d x = ∫ ∂ z b 1 ∂ x ω j ∂ z j j d x + ∫ ∂ z b 2 ∂ y ω j ∂ z j j d x + ∫ ∂ z b 3 ∂ z ω j ∂ z j j d x = R 23 1 + R 23 2 + R 23 3 .

Similarly,

R 23 1 = ∫ ∂ z b 1 ∂ x ω j ∂ z j j d x ≤ C ∥ j ∥ 1 2 ∥ ω x ∥ 1 2 ∥ ∇ j ∥ 1 2 ∥ ∇ ω x ∥ 1 2 ∥ ∇ j x ∥ 1 2 ∥ j y ∥ 1 2 ≤ 1 62 ∥ ∇ ω x ∥ 2 + 1 50 ∥ ∇ j x ∥ 2 + C ( ∥ j ∥ 2 + ∥ ω x ∥ 2 ) ∥ ∇ j ∥ 2 , R 23 2 = ∫ ∂ z b 2 ∂ y ω j ∂ z j j d x ≤ C ∥ j ∥ 1 2 ∥ ω x ∥ 1 2 ∥ ∇ j ∥ 1 2 ∥ ∇ j x ∥ 1 2 ∥ j y ∥ 1 2 ∥ ∇ ω y ∥ 1 2 ≤ 1 50 ∥ ∇ j x ∥ 2 + 1 40 ∥ ∇ ω y ∥ 2 + C ( ∥ j ∥ 2 + ∥ j y ∥ 2 ) ( ∥ ∇ ω ∥ 2 + ∥ ∇ j ∥ 2 ) , R 23 3 = ∫ ∂ z b 3 ∂ z ω j ∂ z j j d x = − ∫ ( ∂ x b 1 + ∂ y b 2 ) ∂ z ω j ∂ z j j d x ≤ C ∥ b x ∥ 1 2 ∥ ∇ ω ∥ 1 2 ∥ ∇ j ∥ 1 2 ∥ ∇ ω x ∥ 1 2 ∥ ∇ j y ∥ 1 2 ∥ j x ∥ 1 2 + C ∥ b y ∥ 1 2 ∥ ∇ ω ∥ 1 2 ∥ ∇ j ∥ 1 2 ∥ ∇ ω x ∥ 1 2 ∥ ∇ j y ∥ 1 2 ∥ j y ∥ 1 2 ≤ 1 62 ∥ ∇ ω x ∥ 2 + 1 48 ∥ ∇ j y ∥ 2 + C ( ∥ b x ∥ 2 + ∥ b y ∥ 2 + ∥ j x ∥ 2 + ∥ j y ∥ 2 ) ( ∥ ∇ ω ∥ 2 + ∥ ∇ j ∥ 2 ) .

Finally, for the last term S, we have

S = − ∫ ε i j k ( ∂ j b l ∂ l u k − ∂ j u l ∂ l b k ) Δ j i d x = ∫ ε i j k ∂ m ( ∂ j b l ∂ l u k − ∂ j u l ∂ l b k ) ∂ m j i d x = S 1 + S 2 .

We first consider S 1 ,

S 1 = ∫ ε i j k ∂ m ( ∂ j b l ∂ l u k ) ∂ m j i d x = ∫ ε i j k ∂ x ( ∂ j b l ∂ l u k ) ∂ x j i d x + ∫ ε i j k ∂ y ( ∂ j b l ∂ l u k ) ∂ y j i d x + ∫ ε i j k ∂ z ( ∂ j b l ∂ l u k ) ∂ z j i d x = S 11 + S 12 + S 13 .

We deduce that

S 11 = ∫ ε i j k ∂ x ( ∂ j b l ∂ l u k ) ∂ x j i d x = ∫ ε i j k ∂ x ∂ j b l ∂ l u k ∂ x j i d x + ∫ ε i j k ∂ j b l ∂ x ∂ l u k ∂ x j i d x ≤ C ∥ j x ∥ 1 2 ∥ ω ∥ 1 2 ∥ j x ∥ 1 2 ∥ ω x ∥ 1 2 ∥ ∇ j x ∥ 1 2 ∥ ∇ j y ∥ 1 2 + C ∥ j ∥ 1 2 ∥ ω x ∥ 1 2 ∥ j x ∥ 1 2 ∥ j x ∥ 1 2 ∥ ∇ ω x ∥ 1 2 ∥ ∇ j x ∥ 1 2 ≤ 1 62 ∥ ∇ ω x ∥ 2 + 1 50 ∥ ∇ j x ∥ 2 + 1 48 ∥ ∇ j y ∥ 2 + C ( ∥ j x ∥ 2 + ∥ ω ∥ 2 + ∥ j ∥ 2 ) ( ∥ ∇ ω ∥ 2 + ∥ ∇ j ∥ 2 ) , S 12 = ∫ ε i j k ∂ y ( ∂ j b l ∂ l u k ) ∂ y j i d x = ∫ ε i j k ∂ y ∂ j b l ∂ l u k ∂ y j i d x + ∫ ε i j k ∂ j b l ∂ y ∂ l u k ∂ y j i d x ≤ C ∥ j y ∥ 1 2 ∥ ω ∥ 1 2 ∥ j y ∥ 1 2 ∥ ∇ j x ∥ 1 2 ∥ ∇ j y ∥ 1 2 ∥ ∇ ω ∥ 1 2 + C ∥ j ∥ 1 2 ∥ ω y ∥ 1 2 ∥ j y ∥ 1 2 ∥ j x ∥ 1 2 ∥ ∇ ω y ∥ 1 2 ∥ ∇ j x ∥ 1 2 ≤ 1 40 ∥ ∇ ω y ∥ 2 + 1 50 ∥ ∇ j x ∥ 2 + 1 48 ∥ ∇ j y ∥ 2 + C ( ∥ ω ∥ 2 + ∥ j y ∥ 2 + ∥ j ∥ 2 + ∥ j x ∥ 2 ) ( ∥ ∇ ω ∥ 2 + ∥ ∇ j ∥ 2 ) .

As for S 13 , we see that

S 13 = ∫ ε i j k ∂ z ( ∂ j b l ∂ l u k ) ∂ z j i d x = ∫ ε i j k ∂ z ∂ j b l ∂ l u k ∂ z j i d x + ∫ ε i j k ∂ j b l ∂ z ∂ l u k ∂ z j i d x = ∫ ε i j k ∂ z ∂ x b l ∂ l u k ∂ z j i d x + ∫ ε i j k ∂ z ∂ y b l ∂ l u k ∂ z j i d x + ∫ ε i j k ∂ z ∂ z b l ∂ l u k ∂ z j i d x + ∫ ε i j k ∂ x b l ∂ z ∂ l u k ∂ z j i d x + ∫ ε i j k ∂ y b l ∂ z ∂ l u k ∂ z j i d x + ∫ ε i j k ∂ z b l ∂ z ∂ l u k ∂ z j i d x = S 13 1 + S 13 2 + S 13 3 + S 13 4 + S 13 5 + S 13 6 .

We deduce each term step by step as follows:

S 13 1 = ∫ ε i j k ∂ z ∂ x b l ∂ l u k ∂ z j i d x ≤ C ∥ j x ∥ 1 2 ∥ ω ∥ 1 2 ∥ ∇ j ∥ 1 2 ∥ ∇ j y ∥ 1 2 ∥ ω x ∥ 1 2 ∥ ∇ j x ∥ 1 2 ≤ 1 50 ∥ ∇ j x ∥ 2 + 1 48 ∥ ∇ j y ∥ 2 + C ( ∥ ω ∥ 2 + ∥ j x ∥ 2 ) ( ∥ ∇ ω ∥ 2 + ∥ ∇ j ∥ 2 ) , S 13 2 = ∫ ε i j k ∂ z ∂ y b l ∂ l u k ∂ z j i d x ≤ C ∥ j y ∥ 1 2 ∥ ω ∥ 1 2 ∥ ∇ j ∥ 1 2 ∥ ∇ j x ∥ 1 2 ∥ ω y ∥ 1 2 ∥ ∇ j y ∥ 1 2 ≤ 1 50 ∥ ∇ j x ∥ 2 + 1 48 ∥ ∇ j y ∥ 2 + C ( ∥ ω ∥ 2 + ∥ j y ∥ 2 ) ( ∥ ∇ ω ∥ 2 + ∥ ∇ j ∥ 2 ) .

For S 13 3 , we have

S 13 3 = ∫ ε i j k ∂ z ∂ z b l ∂ l u k ∂ z j i d x = ∫ ε i j k ∂ z ∂ z b 1 ∂ x u k ∂ z j i d x + ∫ ε i j k ∂ z ∂ z b 2 ∂ y u k ∂ z j i d x + ∫ ε i j k ∂ z ∂ z b 3 ∂ z u k ∂ z j i d x = S 13 31 + S 13 32 + S 13 33 .

Thus, we deduce that

S 13 31 = ∫ ε i j k ∂ z ∂ z b 1 ∂ x u k ∂ z j i d x ≤ C ∥ ∇ j ∥ 1 2 ∥ u x ∥ 1 2 ∥ ∇ j ∥ 1 2 ∥ ∇ j x ∥ 1 2 ∥ ∇ j y ∥ 1 2 ∥ ω x ∥ 1 2 ≤ 1 50 ∥ ∇ j x ∥ 2 + 1 48 ∥ ∇ j y ∥ 2 + C ( ∥ u x ∥ 2 + ∥ ω x ∥ 2 ) ∥ ∇ j ∥ 2 , S 13 32 = ∫ ε i j k ∂ z ∂ z b 2 ∂ y u k ∂ z j i d x ≤ C ∥ ∇ j ∥ 1 2 ∥ u y ∥ 1 2 ∥ ∇ j ∥ 1 2 ∥ ∇ j x ∥ 1 2 ∥ ∇ j y ∥ 1 2 ∥ ω x ∥ 1 2 ≤ 1 50 ∥ ∇ j x ∥ 2 + 1 48 ∥ ∇ j y ∥ 2 + C ( ∥ u y ∥ 2 + ∥ ω y ∥ 2 ) ∥ ∇ j ∥ 2 , S 13 33 = ∫ ε i j k ∂ z ∂ z b 3 ∂ z u k ∂ z j i d x = − ∫ ε i j k ∂ z ( ∂ x b 1 + ∂ y b 2 ) ∂ z u k ∂ z j i d x ≤ C ∥ j x ∥ 1 2 ∥ ω ∥ 1 2 ∥ ∇ j ∥ 1 2 ∥ ∇ j x ∥ 1 2 ∥ ∇ j y ∥ 1 2 ∥ ∇ ω ∥ 1 2 + C ∥ j y ∥ 1 2 ∥ ω ∥ 1 2 ∥ ∇ j ∥ 1 2 ∥ ∇ j x ∥ 1 2 ∥ ∇ j y ∥ 1 2 ∥ ∇ ω ∥ 1 2 ≤ 1 50 ∥ ∇ j x ∥ 2 + 1 48 ∥ ∇ j y ∥ 2 + C ( ∥ ω ∥ 2 + ∥ j x ∥ 2 + ∥ j y ∥ 2 ) ( ∥ ∇ ω ∥ 2 + ∥ ∇ j ∥ 2 ) , S 13 4 = ∫ ε i j k ∂ x b l ∂ z ∂ l u k ∂ z j i d x ≤ C ∥ b x ∥ 1 2 ∥ ∇ ω ∥ 1 2 ∥ ∇ j ∥ 1 2 ∥ ∇ ω x ∥ 1 2 ∥ ∇ j y ∥ 1 2 ∥ j x ∥ 1 2 ≤ 1 62 ∥ ∇ ω x ∥ 2 + 1 48 ∥ ∇ j y ∥ 2 + C ( ∥ b x ∥ 2 + ∥ j x ∥ 2 ) ( ∥ ∇ ω ∥ 2 + ∥ ∇ j ∥ 2 ) , S 13 5 = ∫ ε i j k ∂ y b l ∂ z ∂ l u k ∂ z j i d x ≤ C ∥ b y ∥ 1 2 ∥ ∇ ω ∥ 1 2 ∥ ∇ j ∥ 1 2 ∥ ∇ ω x ∥ 1 2 ∥ ∇ j y ∥ 1 2 ∥ j y ∥ 1 2 ≤ 1 62 ∥ ∇ ω x ∥ 2 + 1 48 ∥ ∇ j y ∥ 2 + C ( ∥ b y ∥ 2 + ∥ j y ∥ 2 ) ( ∥ ∇ ω ∥ 2 + ∥ ∇ j ∥ 2 ) .

For the last term S 13 6 , we see that

S 13 6 = ∫ ε i j k ∂ z b l ∂ z ∂ l u k ∂ z j i d x = ∫ ε i j k ∂ z b 1 ∂ z ∂ x u k ∂ z j i d x + ∫ ε i j k ∂ z b 2 ∂ z ∂ y u k ∂ z j i d x + ∫ ε i j k ∂ z b 3 ∂ z ∂ z u k ∂ z j i d x = S 13 61 + S 13 62 + S 13 63 .

Then,

S 13 61 = ∫ ε i j k ∂ z b 1 ∂ z ∂ x u k ∂ z j i d x ≤ C ∥ j ∥ 1 2 ∥ ω x ∥ 1 2 ∥ ∇ j ∥ 1 2 ∥ ∇ j x ∥ 1 2 ∥ j y ∥ 1 2 ∥ ∇ ω x ∥ 1 2 ≤ 1 62 ∥ ∇ ω x ∥ 2 + 1 50 ∥ ∇ j x ∥ 2 + C ( ∥ j ∥ 2 + ∥ ω x ∥ 2 ) ∥ ∇ j ∥ 2 , S 13 62 = ∫ ε i j k ∂ z b 2 ∂ z ∂ y u k ∂ z j i d x ≤ C ∥ j ∥ 1 2 ∥ ω y ∥ 1 2 ∥ ∇ j ∥ 1 2 ∥ ∇ j x ∥ 1 2 ∥ j y ∥ 1 2 ∥ ∇ ω y ∥ 1 2 ≤ 1 40 ∥ ∇ ω y ∥ 2 + 1 50 ∥ ∇ j x ∥ 2 + C ( ∥ j ∥ 2 + ∥ ω y ∥ 2 ) ∥ ∇ j ∥ 2 , S 13 63 = ∫ ε i j k ∂ z b 3 ∂ z ∂ z u k ∂ z j i d x = − ∫ ε i j k ( ∂ x b 1 + ∂ y b 2 ) ∂ z ∂ z u k ∂ z j i d x ≤ C ∥ b x ∥ 1 2 ∥ ∇ ω ∥ 1 2 ∥ ∇ j ∥ 1 2 ∥ ∇ ω x ∥ 1 2 ∥ ∇ j y ∥ 1 2 ∥ j y ∥ 1 2 + C ∥ b y ∥ 1 2 ∥ ∇ ω ∥ 1 2 ∥ ∇ j ∥ 1 2 ∥ ∇ ω x ∥ 1 2 ∥ ∇ j y ∥ 1 2 ∥ j y ∥ 1 2 ≤ 1 62 ∥ ∇ ω x ∥ 2 + 1 48 ∥ ∇ j y ∥ 2 + C ( ∥ b x ∥ 2 + ∥ b y ∥ 2 + ∥ j y ∥ 2 ) ∥ ∇ j ∥ 2 .

As for S 2 , deduced by similar methods, we can obtain the following inequality (for simplicity we omit the details here):

S 2 ≤ 1 62 ∥ ∇ ω x ∥ 2 + 1 40 ∥ ∇ ω y ∥ 2 + 1 50 ∥ ∇ j x ∥ 2 + 1 48 ∥ ∇ j y ∥ 2 + ( ∥ ω x ∥ 2 + ∥ ω ∥ 2 + ∥ j ∥ 2 + ∥ ω y ∥ 2 + ∥ b x ∥ 2 + ∥ j x ∥ 2 + ∥ b y ∥ 2 + ∥ j y ∥ 2 + ∥ u x ∥ 2 + ∥ u y ∥ 2 + ∥ ω y ∥ 2 ) ( ∥ ∇ ω ∥ 2 + ∥ ∇ j ∥ 2 ) .

Then, substituting all the above estimates into (11), we finally deduce that

1 2 d d t ( ∥ ∇ ω ∥ 2 + ∥ ∇ j ∥ 2 ) + 1 2 ( ∥ ∇ ω x ∥ 2 + ∥ ∇ ω y ∥ 2 + ∥ ∇ j x ∥ 2 + ∥ ∇ j y ∥ 2 ) ≤ C ( ∥ ω x ∥ 2 + ∥ ω ∥ 2 + ∥ j ∥ 2 + ∥ ω y ∥ 2 + ∥ b x ∥ 2 + ∥ j x ∥ 2 + ∥ b y ∥ 2 + ∥ j y ∥ 2 + ∥ u x ∥ 2 + ∥ u y ∥ 2 + ∥ ω y ∥ 2 ) ( ∥ ∇ ω ∥ 2 + ∥ ∇ j ∥ 2 ) .

Then we complete the proof of Lemma 2.3 by Gronwall’s lemma. □

3 Proof of Theorem 1.1

In this section, we prove Theorem 1.1 by using the method of vanishing viscosity. To this end, we consider the following regularized problem:

u t ε + u ε ⋅∇ u ε =−∇ p ε + u x x ε + u y y ε +ε u z z ε + b ε ⋅∇ b ε ,
(12)
b t ε + u ε ⋅∇ b ε = b x x ε + b y y ε +ε b z z ε + b ε ⋅∇ u ε ,
(13)
div u ε =0,div b ε =0,
(14)

with smooth initial data

u ε (0,x)= ψ ε ∗ u 0 , b ε (0,x)= ψ ε ∗ b 0 ,
(15)

where ψ ε (x,y)= ε − 2 ψ(x/ε,y/ε) is the standard mollifier satisfying

ψ≥0,ψ∈ C 0 ∞ ( R 3 ) and∫ψdx=1.

Now, an application of the classical result shows that for any T>0, there exists a unique global smooth solution ( u ε , b ε ) of (12)-(15) on R 3 ×(0,T) satisfying the global bounds stated in Lemma 2.2 and 2.3, which are uniform in ε. So, by standard compactness arguments, we can extract a subsequence ( u ε j , b ε j ) and pass to the limit as j→∞ to get that the limit function (u,b) is indeed a global smooth solution of the problem (1)-(4). The proof of Theorem 1.1 is therefore completed.

References

  1. Wang F, Wang K: Global existence of 3D MHD equations with mixed partial dissipation and magnetic diffusion. Nonlinear Anal., Real World Appl. 2013, 14: 525–536.

    Google Scholar 

  2. Cabannes H: Theoretical Magnetofluiddynamics. Academic Press, New York; 1970.

    Google Scholar 

  3. Cao C, Wu J: Global regularity for the 2D MHD equations with mixed partial dissipation and magnetic diffusion. Adv. Math. 2011, 226: 1803–1822. 10.1016/j.aim.2010.08.017

    Article  MathSciNet  Google Scholar 

Download references

Acknowledgements

This work was supported by the National Nature Science Foundation of China (No.11026079), the Youth Backbone Teacher Foundation of Henan Province (No. 2010GGJS-167), the Natural Science Foundation of Henan Province (No.122300410261), and the Foundation of Laboratory of Symbolic Computation and Knowledge Engineering of Ministry of Education (No. 93K172012K07).

Author information

Authors and Affiliations

Authors

Corresponding author

Correspondence to Menglong Su.

Additional information

Competing interests

The authors declare that they have no competing interests.

Authors’ contributions

MS carried out the main work and drafted the manuscript. JW participated in completing the proof of Lemma 2.2. All authors read and approved the final manuscript.

Rights and permissions

Open Access This article is distributed under the terms of the Creative Commons Attribution 2.0 International License (https://creativecommons.org/licenses/by/2.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Reprints and permissions

About this article

Cite this article

Su, M., Wang, J. Global smooth solutions to 3D MHD with mixed partial dissipation and magnetic diffusion. J Inequal Appl 2013, 345 (2013). https://doi.org/10.1186/1029-242X-2013-345

Download citation

  • Received:

  • Accepted:

  • Published:

  • DOI: https://doi.org/10.1186/1029-242X-2013-345

Keywords