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 +uu=p+μ u x x +μ u y y +bb,
(1)
b t +ub=η b x x +η b y y +bu,
(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  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|dxC 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 +uωωu= ω x x + ω y y +bjjb,
(6)
j t +uj= j x x + j y y +bω+ 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 ) ds1 for all t0.

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 ) ds1.

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 ) dsC.
(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