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JID:YJMAA
AID:21728 /FLA
Doctopic: Partial Differential Equations
[m3L; v1.223; Prn:9/10/2017; 8:31] P.1 (1-12)
J. Math. Anal. Appl. ••• (••••) •••–•••
Contents lists available at ScienceDirect
Journal of Mathematical Analysis and Applications
www.elsevier.com/locate/jmaa
Global regularity for the 3D MHD system with partial viscosity
and magnetic diffusion terms
Xia Ye a , Mingxuan Zhu b,∗
a
b
College of Mathematics and Information Science, Jiangxi Normal University, Nanchang 330022, China
Department of Mathematics, Jiaxing University, Jiaxing, 314001, China
a r t i c l e
i n f o
Article history:
Received 25 February 2017
Available online xxxx
Submitted by J. Guermond
Keywords:
Incompressible MHD system
Global solution
Small initial data
Partial viscosity and magnetic
diffusion
a b s t r a c t
This paper is concerned with the Cauchy problem of the incompressible magnetohydrodynamic system with partial viscosity and magnetic diffusion terms in the
three-dimensional whole space. The existence of the global strong solution (u, b) is
obtained, provided that the initial data (u0 2L2 + b0 2L2 )(∇u0 2L2 + ∇b0 2L2 ) is
suitably small.
© 2017 Elsevier Inc. All rights reserved.
1. Introduction
We consider the following 3D MHD system:
⎧
⎪
ut + u · ∇u + ∇p = μ1 uxx + μ2 uyy + μ3 uzz + b · ∇b, (x, t) ∈ R3 × R+ ,
⎪
⎪
⎪
⎨b + u · ∇b = ν b + ν b + ν b + b · ∇u, (x, t) ∈ R3 × R+ ,
t
1 xx
2 yy
3 zz
3
⎪
×
R+ ,
div
u
=
0,
div
b
=
0,
(x,
t)
∈
R
⎪
⎪
⎪
⎩
(u, b)(x, 0) = (u0 , b0 ), x ∈ R3 ,
(1.1)
here u = u(x, t) ∈ R3 , b = b(x, t) ∈ R3 , p = p(x, t) ∈ R represent the unknown velocity field, the magnetic
field and the pressure, respectively. μ1 , μ2 , μ3 ≥ 0 are the kinematic viscosity. ν1 , ν2 , ν3 ≥ 0 are the magnetic
diffusion. If μ1 = μ2 = μ3 = 0, ν1 = ν2 = ν3 = 0, (1.1) are ideal MHD equations. For more details of the
related background, see, for instance, [1,7,18]. There ui and ∂i denote the corresponding ith components of
u and ∇, respectively.
* Corresponding author.
E-mail addresses: yexia1249@126.com (X. Ye), mxzhu@mail.zjxu.edu.cn (M. Zhu).
https://doi.org/10.1016/j.jmaa.2017.10.010
0022-247X/© 2017 Elsevier Inc. All rights reserved.
JID:YJMAA
AID:21728 /FLA
Doctopic: Partial Differential Equations
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X. Ye, M. Zhu / J. Math. Anal. Appl. ••• (••••) •••–•••
2
For the MHD system with full dissipation and magnetic diffusion, the global weak solutions and local
strong solution (unique) were proved by Duvaut and Lions in [9] (see also [20]). But the global regularity for the three-dimensional incompressible magnetohydrodynamic is an outstanding open problem. In
[21], Wang–Du proved the existence of the global smooth solution of MHD, provided the initial norm
u0 L2 ∂3 u0 L2 + b0 L2 ∂3 b0 L2 is suitably small. He–Wang obtained the global smooth solution under
the assumption that the difference between the magnetic field and the velocity is small initially in Ref. [12].
In [19], they also proved the existence of the global smooth solution under the requirement that the initial data be sufficiently small. It should be noted that there are some regularity criteria for 3D MHD in
Refs. [23,25,24,11,16,15,27,14].
This work was partially motivated by the well-known works for the Boussinesq system and the MHD
system with partial dissipation [6,13,3] in dimension two. For the two-dimensional MHD equations
⎧
⎪
u + u · ∇u + ∇p = μ1 uxx + μ2 uyy + b · ∇b, (x, t) ∈ R2 × R+ ,
⎪
⎪ t
⎪
⎨b + u · ∇b = ν b + ν b + b · ∇u, (x, t) ∈ R2 × R+ ,
t
1 xx
2 yy
⎪div u = 0, div b = 0, (x, t) ∈ R2 × R+ ,
⎪
⎪
⎪
⎩
(u, b)(x, 0) = (u0 , b0 ), x ∈ R2 .
(1.2)
Cao and Wu [3] proved the global regularity for the MHD equations with mixed partial dissipation and
magnetic diffusion (μ1 = ν2 = 0 or μ1 = ν2 = 0). Several results for other cases were established in [8,4,5,
26,2,17,10].
For the three-dimensional MHD equations with partial dissipation, in [22], they proved that if the
H 1 -norm of initial data (u0 , b0 ) is small enough, then system (1.1) has global strong solution. We obtain the global strong solution for MHD equations with partial dissipation and magnetic diffusion under the
initial data (u0 2L2 + b0 2L2 )(∇u0 2L2 + ∇b0 2L2 ) is small enough or the coefficients of partial viscosity
and magnetic diffusion are sufficiently large.
Before stating the main results, we explain the notations and conventions used throughout this paper.
ui and ∂i denote the corresponding ith components of u and ∇, respectively.
Our main result in this paper is the following Theorem.
Theorem 1.1. For m = 1, 2 or 3, n = 1, 2 or 3, assume that μm = 0 and νn = 0, and (u0 , b0 ) ∈ H 2 (R3 ) with
div u0 = div b0 = 0. There exists an absolutely positive constant ε0 > 0, independent of u0 , b0 , t, and the
lower bounded of coefficients of partial viscosity and magnetic diffusion, such that the problem (1.1) has a
unique global strong solution (u, b) on R3 × (0, ∞), satisfying
(u, b) ∈ L∞ (0, T ; H 2 (R3 )) and (∂l u, ∂k b) ∈ L2 (0, T ; H 2 (R3 )),
(1.3)
(u0 2L2 + b0 2L2 )(∇u0 2L2 + ∇b0 2L2 )
≤ ε0 ,
M4
(1.4)
provided that
here l = m and k = n,
M min{μi=m , νj=n }
(i, j = 1, 2, 3).
(1.5)
Remark 1.1. Compared to [22], firstly, the initial data (u0 2L2 + b0 2L2 )(∇u0 2L2 + ∇b0 2L2 ) instead of
(u0 , b0 )H 1 is sufficiently small, secondly, we also get the existence of strong solution with large initial data.
JID:YJMAA
AID:21728 /FLA
Doctopic: Partial Differential Equations
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3
2. Preliminaries
We need the following Lemma which can be found in [22].
Lemma 2.1. Assume that f , g, h, fx , gy , hz are all in L2 (R3 ). Then,
1/2
1/2
1/2
1/2
1/2
1/2
|f gh|dx ≤ Cf L2 gL2 hL2 fx L2 gy L2 hz L2 .
(2.1)
R3
3. A priori estimates
Given a strong solution (u, b) on R3 × (0, T ), we define
A(t) sup
0≤s≤t
∇u(t)2L2 + ∇b(t)2L2 , 0 ≤ t ≤ T.
Proposition 3.1. Under the assumptions of Theorem 1.1, there exists an absolutely positive constant ε0 > 0,
independent of u0 , b0 , t, and the lower bounded of coefficients of partial viscosity and magnetic diffusion,
such that if (u0 2L2 + b0 2L2 )(∇u0 2L2 + ∇b0 2L2 )/M 4 ≤ ε0 , and (u, b) is a strong solution to (1.1),
satisfying
A(T ) ≤ 2(∇u0 2L2 + ∇b0 2L2 ),
(3.1)
A(T ) ≤ (∇u0 2L2 + ∇b0 2L2 ),
(3.2)
then it in fact holds that
with M as in (1.5).
Proof. Proposition 3.1 is an easy consequence of the following Lemmas 3.1–3.4.
Now we divide it into two sections. One case μi = νi = 0 (i ∈ {1, 2, 3}). In order to simplify the analysis,
we shall only prove the case μ3 = ν3 = 0. According to the definition of M in Theorem 1.1, one easily knows
that
M = min{μ1 , μ2 , ν1 , ν2 }.
(3.3)
First of all, it is easy to deduce the following energy inequality:
Lemma 3.1. Let μ3 = ν3 = 0, (u, b) be a strong solution of (1.1) on R3 × [0, T ], then
sup
0≤t≤T
u2L2
+
b2L2
T
+M
∂1 u2L2 + ∂2 u2L2 + ∂1 b2L2 + ∂2 b2L2 dt
0
(3.4)
≤ (u0 2L2 + b0 2L2 ),
where M is the positive number determined in (3.3).
Lemma 3.2. Assume that the condition (3.1) of Proposition 3.1 holds and μ3 = ν3 = 0. Let (u, b) be a strong
solution of (1.1) on R3 × [0, T ), then
JID:YJMAA
AID:21728 /FLA
Doctopic: Partial Differential Equations
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X. Ye, M. Zhu / J. Math. Anal. Appl. ••• (••••) •••–•••
4
sup
0≤t≤T
∇u2L2
+
∇b2L2
T
+
∂1 ∇u2L2 + ∂2 ∇u2L2 + ∂1 ∇b2L2 + ∂2 ∇b2L2 dt
0
≤
∇u0 2L2
+
∇b0 2L2 ,
(3.5)
provided that (u0 2L2 + b0 2L2 )(∇u0 2L2 + ∇b0 2L2 )/M 4 ≤ ε0 , and ε0 is sufficiently small.
Proof. Multiplying (1.1)1 , (1.1)2 by Δu, Δb in L2 , respectively, after integrating by parts we have
1 d
(∇u2L2 + ∇b2L2 ) + μ1 ∂1 ∇u2L2 + μ2 ∂2 ∇u2L2 + ν1 ∂1 ∇b2L2 + ν2 ∂2 ∇b2L2
2 dt
= (u · ∇u − b · ∇b) · Δu + (u · ∇b − b · ∇u) · Δb dx
R3
(−∂k ui ∂i uj ∂k uj + ∂k ui ∂i bj ∂k bj − ∂k bi ∂i bj ∂k uj + ∂k bi ∂i uj ∂k bj ) dx
=
R3
= I1 + I2 + I3 + I4 .
(3.6)
The right hand terms of above equation can be estimated as follows:
3
2 |I1 | ≤ ∂k ui ∂i uj ∂k uj dx +
k=1 i,j=1
3
3
R
R
2 3
dx
+
∂
u
∂
u
∂
u
3 i i j 3 j
i=1 j=1
3
R
3
∂
u
∂
u
∂
u
3 3 3 j 3 j dx
j=1
= I11 + I12 + I13 .
(3.7)
The I1i (i = 1, 2, 3) terms on the right-hand side of (3.7) can be calculated as follows. First, using Sobolev
and Cauchy–Schwarz’s inequalities, we have by direct calculations that
|I11 | ≤ C
2 3
∂k ui L3 ∂i uj L2 ∂k uj L6
k=1 i,j=1
≤
3
2 1/2
1/2
∂k ui L2 ∂k ∇ui L2 ∂i uj L2 ∂k ∇uj L2
k=1 i,j=1
≤
M
C
∂1 ∇u2L2 + ∂2 ∇u2L2 + 3 ∇u4L2 (∂1 u2L2 + ∂2 u2L2 ).
12
M
(3.8)
By Lemma 2.1, Cauchy–Schwarz’s inequalities, one can estimate I12 , I13 as follows
|I12 | ≤
2 3
1/2
1/2
1/2
1/2
1/2
1/2
∂3 ui L2 ∂i uj L2 ∂3 uj L2 ∂13 ui L2 ∂i3 uj L2 ∂23 uj L2
i=1 j=1
≤
M
C
∂1 ∇u2L2 + ∂2 ∇u2L2 + 3 ∇u4L2 (∂1 u2L2 + ∂2 u2L2 ),
12
M
and
3
|I13 | = (−∂1 u1 − ∂2 u2 )∂3 uj ∂3 uj dx
j=1
3
R
(3.9)
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AID:21728 /FLA
Doctopic: Partial Differential Equations
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X. Ye, M. Zhu / J. Math. Anal. Appl. ••• (••••) •••–•••
≤C
3
1/2
1/2
1/2
1/2
5
1/2
1/2
∂1 u1 + ∂2 u2 L2 ∂3 uj L2 ∂3 uj L2 ∂3 (∂1 u1 + ∂2 u2 )L2 ∂13 uj L2 ∂23 uj L2
j=1
≤
M
C
∂1 ∇u2L2 + ∂2 ∇u2L2 + 3 ∇u4L2 (∂1 u2L2 + ∂2 u2L2 ).
12
M
(3.10)
Substituting (3.8), (3.9) and (3.10) into (3.7), we arrive at
|I1 | ≤
M
C
∂1 ∇u2L2 + ∂2 ∇u2L2 + 3 ∇u4L2 (∂1 u2L2 + ∂2 u2L2 ).
4
M
(3.11)
Similarly, we have
2 3
3
3
2 |I2 | ≤ ∂k ui ∂i bj ∂k bj + ∂3 ui ∂i bj ∂3 bj + ∂3 u3 ∂3 bj ∂3 bj dx.
i=1 j=1
j=1
i,j=1
R3 k=1
3
3
2 2 3
|I3 | ≤ ∂k bi ∂i bj ∂k uj + ∂3 bi ∂i bj ∂3 uj + ∂3 b3 ∂3 bj ∂3 uj dx.
i=1 j=1
j=1
i,j=1
R3 k=1
2 3
3
3
2 |I4 | ≤ ∂k bi ∂i uj ∂k bj + ∂3 bi ∂i uj ∂3 bj + ∂3 b3 ∂3 uj ∂3 bj dx.
k=1 i,j=1
i=1 j=1
j=1
3
R
By almost the same argument, we can get
M
M
(∂1 ∇u2L2 + ∂2 ∇u2L2 ) +
(∂1 ∇b2L2 + ∂2 ∇b2L2 )
4
2
C
+ 3 (∇u4L2 + ∇b4L2 )(∂1 u2L2 + ∂2 u2L2 + ∂1 b2L2 + ∂2 b2L2 ).
M
|I2 | + |I3 | + |I4 | ≤
(3.12)
Combining with (3.11) and (3.12) inserted into (3.6), integrating it over (0, T ), one has
sup ∇u2L2 + ∇b2L2 + M
0≤t≤T
T
(∂1 ∇u2L2 + ∂2 ∇u2L2 + ∂1 ∇b2L2 + ∂2 ∇b2L2 ) dt
0
C
≤ 3 sup (∇u4L2 + ∇b4L2 )
M 0≤t≤T
+
(∇u0 2L2
+
T
(3.13)
(∂1 u2L2 + ∂2 u2L2 + ∂1 b2L2 + ∂2 b2L2 )dt
0
∇b0 2L2 ).
So, it follows from (3.13), (3.1), and (3.4) that
sup ∇u2L2 + ∇b2L2 + M
0≤t≤T
T
(∂1 ∇u2L2 + ∂2 ∇u2L2 + ∂1 ∇b2L2 + ∂2 ∇b2L2 ) dt
0
≤C
∇u0 2L2
+
+
(∇u0 2L2
∇b0 2L2 )(u0 2L2
M4
+
+ b0 2L2 )
sup
0≤t≤T
∇u2L2 + ∇b2L2
(3.14)
∇b0 2L2 ).
Due to (1.4), (3.14) and taking ε0 small enough, we obtain the desired estimate (3.5) of Lemma 3.1.
2
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AID:21728 /FLA
Doctopic: Partial Differential Equations
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The other case: μi = νj = 0 (i = j, i, j ∈ {1, 2, 3}). For simplicity, let μ1 = ν2 = 0, here. Due to the
definition of M in Theorem 1.1, it is easy to know that
M = min{μ2 , μ3 , ν1 , ν3 }.
(3.15)
Lemma 3.3. Let μ1 = ν2 = 0, (u, b) be a strong solution of (1.1) on R3 × [0, T ), then
sup
0≤t≤T
u2L2
+
b2L2
T
+M
∂2 u2L2 + ∂3 u2L2 + ∂1 b2L2 + ∂3 b2L2 dt
(3.16)
0
≤ (u0 2L2 + b0 2L2 ),
where M is the positive number determined in (3.15).
Lemma 3.4. Assume that the condition (3.1) of Proposition 3.1 holds and μ1 = ν2 = 0. Let (u, b) be a strong
solution of (1.1) on R3 × [0, T ), then
sup
0≤t≤T
∇u2L2
+
∇b2L2
T
+M
∂2 ∇u2L2 + ∂3 ∇u2L2 + ∂1 ∇b2L2 + ∂3 ∇b2L2 dt
0
(3.17)
≤ ∇u0 2L2 + ∇b0 2L2 ,
provided that (u0 2L2 + b0 2L2 )(∇u0 2L2 + ∇b0 2L2 )/M 4 ≤ ε0 , and ε0 is suitably small.
Proof. Similarly to the derivation of (3.6), we have
1 d
(∇u2L2 + ∇b2L2 ) + μ2 ∂2 ∇u2L2 + μ3 ∂3 ∇u2L2 + ν1 ∂1 ∇b2L2 + ν3 ∂3 ∇b2L2
2 dt
= (−∂k ui ∂i uj ∂k uj + ∂k ui ∂i bj ∂k bj − ∂k bi ∂i bj ∂k uj + ∂k bi ∂i uj ∂k bj ) dx
(3.18)
R3
= I1 + I2 + I3 + I4 .
The right-hand side of (3.18) will be estimated term by term as follows. Firstly,
3
3
3 |I1 | ≤ ∂k ui ∂i uj ∂k uj dx ≤ ∂k ui ∂i uj ∂k uj dx
i=2 j,k=1
R3 i,j,k=1
R3
3
3 3
+ ∂k u1 ∂1 uj ∂k uj dx + ∂1 u1 ∂1 uj ∂1 uj dx I11 + I12 + I13 .
k=2 j=1
j=1
3
3
R
(3.19)
R
To deal with I11 , by virtue of (2.1) and Young’s inequality, we have
|I11 | ≤ C
3 3
1/2
1/2
1/2
1/2
1/2
1/2
∂k ui L2 ∂i uj L2 ∂k uj L2 ∂2 ∂k uL2 ∂2 ∂i uj L2 ∂3 ∂k uj L2
i=2 j,k=1
M
C
∂2 ∇u2L2 + ∂3 ∇u2L2 + 3 (∂2 u2L2 + ∂3 u2L2 )∇u4L2 .
≤
18
M
(3.20)
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AID:21728 /FLA
Doctopic: Partial Differential Equations
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I12 ≤ C
3
3 7
∂k u1 L6 ∂1 uj L2 ∂k uj L3
k=2 j=1
1/2
(3.21)
1/2
≤ C∂k ∇u1 L2 ∂1 uj L2 ∂k uj L2 ∂k ∇uj L2
M
C ∂2 ∇u2L2 + ∂3 ∇u2L2 + 3 ∂2 u2L2 + ∂3 u2L2 ∇u4L2 .
18
M
≤
In a similar manner, due to divu = 0, we deduce
I13
3
=
(−∂2 u2 − ∂3 u3 )∂1 uj ∂1 uj dx
R3
j=1
1/2
1/2
1/2
1/2
≤ C(∂2 u2 + ∂3 u3 )L2 ∂1 uj L2 ∂1 (∂2 u2 + ∂3 u3 )L2 ∂2 ∂1 uj L2 ∂3 ∂1 uj L2
(3.22)
C
M
∂2 ∇u2L2 + ∂3 ∇u2L2 + 3 (∂3 u2L2 + ∂2 u2L2 )∇u4L2 .
18
M
≤
Thus, substituting (3.20), (3.21), (3.22) into (3.19), we have
I1 ≤
M
C
∂2 ∇u2L2 + ∂3 ∇u2L2 + 3 (∂3 u2L2 + ∂2 u2L2 )∇u4L2 .
6
M
(3.23)
Next, to estimate I2 ,
I2 =
3
∂k ui ∂i bj ∂k bj dx
R3 i,j,k=1
=
3 3
R3
I21 ≤ C
3
3 ∂k ui ∂i bj ∂k bj dx +
k=2 i,j=1
1/2
3
R3
1/2
1/2
(3.24)
∂1 ui ∂i bj ∂1 bj dx I21 + I22 .
i,j=1
1/2
1/2
1/2
∂k ui L2 ∂i bj L2 ∂k bj L2 ∂2 ∂k ui L2 ∂1 ∂i bj L2 ∂3 ∂k bj L2
(3.25)
k=2 i,j=1
M
M
C (∂2 ∇u2L2 ) +
(∂1 ∇b2L2 + ∂3 ∇b2L2 ) + 3 ∂2 u2L2 + ∂3 u2L2 ∇b4L2
≤
12
8
M
and
I22 ≤ C
3
1/2
1/2
1/2
1/2
1/2
1/2
∂1 ui L2 ∂i bj L2 ∂1 bj L2 ∂2 ∂1 ui L2 ∂3 ∂i bj L2 ∂1 ∂1 bj L2
(3.26)
i,j=1
≤
M
M
C
(∂2 ∇ui 2L2 ) +
(∂3 ∇bj 2L2 + ∂1 ∇bj 2L2 ) + 3 ∂1 b2L2 ∇u2L2 ∇b2L2 .
12
8
M
Putting (3.25) and (3.26) into (3.24), gives
M
M
(∂2 ∇u2L2 ) +
(∂3 ∇b2L2 + ∂1 ∇b2L2 )
6
4
C + 3 ∇u4L2 + ∇b4L2 (∂2 u2L2 + ∂3 u2L2 + ∂1 b2L2 ).
M
I2 ≤
(3.27)
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By the similar manner as I1 and I2 , we can deal with I3 , I4 as follows
M
M
(∂3 ∇u2L2 + ∂2 ∇u2L2 ) +
(∂3 ∇b2L2 + ∂1 ∇b2L2 )
6
4
C + 3 ∇u4L2 + ∇b4L2 (∂2 u2L2 + ∂3 u2L2 + ∂1 b2L2 + ∂3 b2L2 ).
M
I3 + I4 ≤
(3.28)
Thus, substituting the estimates of (3.23), (3.27), (3.28) into (3.18), integrating it over (0, T ), we have
sup
0≤t≤T
∇u2L2 + ∇b2L2 + M
T
(∂2 ∇u2L2 + ∂3 ∇u2L2 + ∂1 ∇b2L2 + ∂3 ∇b2L2 ) dt
0
C
≤ 3 sup ∇u4L2 + ∇b4L2
M 0≤t≤T
+
(∇u0 2L2
+
T
(∂2 u2L2 + ∂3 u2L2 + ∂1 b2L2 + ∂3 b2L2 )dt
0
∇b0 2L2 ).
(3.29)
In view of (3.1), (3.16), and choosing ε0 small enough, we have from (3.29) that
sup
0≤t≤T
∇u2L2 + ∇b2L2 + M
T
(∂2 ∇u2L2 + ∂3 ∇u2L2 + ∂1 ∇b2L2 + ∂3 ∇b2L2 ) dt
0
≤
(∇u0 2L2
+
∇b0 2L2 ).
2
(3.30)
To prove Theorem 1.1, we still need the following estimates. For simplicity, we only give the proof that
the case μ3 = ν3 = 0, here.
Lemma 3.5. Let (u, b) be a strong solution of (1.1) on R3 × [0, T ). Then
sup
0≤t≤T
Δu2L2
+
Δb2L2
2 T
+
∂i Δu2L2 + ∂i Δb2L2 dt ≤ C(μ1 , μ2 , ν1 , ν2 , u0 H 2 , b0 H 2 ).
i=1 0
Proof. Operating Δ to both sides of (1.1)1 and (1.1)2 , respectively, multiplying them by Δu and Δb in L2 ,
respectively, and adding them together, we deduce
1 d Δu2L2 + Δb2L2 + μ1 ∂1 Δu2L2 + μ2 ∂2 Δu2L2 + ν1 ∂1 Δb2L2 + ν2 ∂2 Δb2L2
2 dt
= − Δ(u · ∇u) · Δu dx − Δ(u · ∇b) · Δb dx
+
Δ(b · ∇b) · Δu dx +
Δ(b · ∇u) · Δb dx =
4
(3.31)
In .
n=1
Each term on the right-hand side of above equation can be estimated as follows,
I1 =
(∂kk ui ∂i u∂jj u + 2∂k ui ∂ik u∂jj u + ui ∂ikk u∂jj u) dx
= I11 + I12 + I13
(3.32)
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By virtue of divu = 0 and integrating by parts, we see that I13 = 0. Now, to estimate I1i (i = 1, 2), based
on Proposition 3.1, divu = 0, the Hölder and Cauchy–Schwarz’s inequality and integrating by parts, gives
I11 =
3
2 ∂kk ui ∂i u∂jj u +
k=1 i,j=1
≤C
2
2 3
∂33 ui ∂i u∂jj u +
i=1 j=1
3
1/2
3
∂33 u3 ∂3 u∂jj u
j=1
1/2
1/2
1/2
1/2
1/2
1/2
1/2
1/2
1/2
∂kk ui L2 ∂3kk ui L2 ∂j uL2 ∂1i uL2 ∂jj uL2 ∂2jj uL2
k=1 i,j=1
+C
2 3
1/2
1/2
∂33 ui L2 ∂133 ui L2 ∂i uL2 ∂3i uL2 ∂jj uL2 ∂2jj uL2
(3.33)
i=1 j=1
+
3 −∂3 (∂1 u1 + ∂2 u2 )∂3 u∂jj udx
j=1
2
≤
∂m Δu2L2 + C()
m=1
2
∂l u2H 1 ∇u2H 1 ,
l=1
where in the last inequality we have used:
3 ∂3 (∂1 u1 + ∂2 u2 )∂3 u∂jj udx
j=1
=
3 (∂13 u1 ∂3 u∂jj u + ∂23 u2 ∂3 u∂jj u) dx
j=1
1/2
1/2
1/2
1/2
1/2
1/2
≤ C∂13 u1 L2 ∂133 u1 L2 ∂3 uL2 ∂13 uL2 ∂jj uL2 ∂2jj uL2
1/2
1/2
1/2
1/2
1/2
1/2
+ C∂23 u2 L2 ∂233 u2 L2 ∂3 uL2 ∂13 uL2 ∂jj uL2 ∂2jj uL2
≤
2
∂m Δu2L2 + C()
m=1
2
∂l ∇u2L2 ∇u2H 1 ,
l=1
and
I12 = 2
=
∂k ui ∂ik u∂jj udx
3 2 ∂k ui ∂ik u∂jj udx +
k=1 i,j=1
≤C
3
2 2 3 ∂3 ui ∂i3 u∂jj udx +
i=1 j=1
1/2
1/2
3 1/2
1/2
1/2
1/2
∂k uL2 ∂3k uL2 ∂ik uL2 ∂1ik uL2 ∂jj uL2 ∂2jj uL2
k=1 i,j=1
1/2
1/2
1/2
1/2
1/2
1/2
+ C∂3 ui L2 ∂13 ui L2 ∂i3 uL2 ∂i33 uL2 ∂jj uL2 ∂2jj uL2
− (∂1 u1 + ∂2 u2 ) ∂33 u∂jj udx
≤
2
m=1
∂3 u3 ∂33 u∂jj udx
j=1
∂m Δu2L2 + C()
2
l=1
∂l ∇u2L2 ∇u2H 1 .
(3.34)
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10
Combining with (3.33)–(3.34), we obtain
2
I1 ≤ ∂m Δu2L2 + C()
m=1
2
∂l u2H 1 ∇u2L2 + Δu2L2 .
(3.35)
l=1
Similarly to the derivation of I1 ,
2
I2 ≤ ∂m Δu2L2 + C()
m=1
2
∂l u2H 1 + ∂l b2H 1 ∇u2H 1 + ∇b2H 1 .
l=1
I3 + I4 =
∂kk bi ∂i b∂jj udx + 2
∂k bi ∂i b∂jj udx +
∂kk bi ∂i u∂jj bdx + 2
+
(3.36)
bi ∂ikk b∂jj udx
∂k bi ∂i u∂jj bdx +
(3.37)
bi ∂ikk u∂jj bdx
= I31 + I32 + I33 + I34 + I35 + I36 .
Here, we only give the estimate of I31 ,
I31 =
2 3
∂kk bi ∂i b∂jj u +
k=1 i,j=1
≤C
2 3
∂33 bi ∂i b∂jj u +
i=1 j=1
3
2 1/2
3
∂33 b3 ∂3 b∂jj u
j=1
1/2
1/2
1/2
1/2
1/2
1/2
1/2
1/2
1/2
∂kk bi L2 ∂3kk bi L2 ∂j bL2 ∂1i bL2 ∂jj uL2 ∂2jj uL2
k=1 i,j=1
+C
2 3
1/2
1/2
∂33 bi L2 ∂133 bi L2 ∂i bL2 ∂3i bL2 ∂jj uL2 ∂2jj uL2
(3.38)
i=1 j=1
+
3 −∂3 (∂1 b1 + ∂2 b2 )∂3 b∂jj udx
j=1
≤
2
(∂m Δu2L2 + ∂m Δb2L2 ) + C()
m=1
2
∂l b2H 1 (∇u2H 1 + ∇b2H 1 ),
l=1
where in the last inequality we have used:
3 ∂3 (∂1 b1 + ∂2 b2 )∂3 b∂jj udx =
j=1
3 (∂13 b1 ∂3 b∂jj u + ∂23 b2 ∂3 b∂jj u) dx
j=1
1/2
1/2
1/2
1/2
1/2
1/2
≤ C∂13 b1 L2 ∂133 b1 L2 ∂3 bL2 ∂13 bL2 ∂jj uL2 ∂2jj uL2
1/2
1/2
1/2
1/2
1/2
1/2
+ C∂23 b2 L2 ∂233 b2 L2 ∂3 bL2 ∂13 bL2 ∂jj uL2 ∂2jj uL2
≤
2
(∂m Δu2L2 + ∂m Δb2L2 ) + C()
m=1
2
∂l b2H 1 (∇u2H 1 + ∇b2H 1 ).
l=1
Similarly, we find
I32 + I34 + I35 ≤ 2
m=1
∂m Δu2L2 + C()
2
∂l u2H 1 + ∂l b2H 1 ∇u2H 1 + ∇b2H 1 .
l=1
Due to divb = 0 and integrating by parts, it is easy to know that I33 + I36 = 0.
(3.39)
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11
Hence, collecting (3.32)–(3.39) and (3.31), together, by (3.5), choosing > 0 small enough and Gronwall’s
inequality, it is easy to deduce that
T
sup
0≤t≤T
(Δu2L2
+
Δb2L2 )
+
∂1 Δu2L2 + ∂2 Δu2L2 + ∂1 Δb2L2 + ∂2 Δb2L2 dt
0
≤ C(μ1 , μ2 , ν1 , ν2 , u0 H 2 , b0 H 2 ).
2
With Lemmas 3.1–3.5 at hand, we are now ready to prove Theorem 1.1.
4. Proof of Theorem 1.1
Combining the local existence result and the global a priori estimates established in Proposition 3.1
and Lemma 3.5, by continuity arguments we obtain the global existence result of Theorem 1.1, provided
u0 2L2 + b0 2L2 ∇u0 2L2 + ∇b0 2L2 /M 4 ≤ ε0 . The uniqueness of strong solutions can be proved by
the standard L2 -method, and the details are omitted here for simplicity.
Acknowledgments
The authors are indebted to anonymous referees for their helpful comments. Ye was partially supported by NSFC (Grant No. 11701240) and the Natural Science Foundation of Jiangxi Province (Grant
No. 20171BAB211001). Zhu was partially supported by NSFC (Grant No. 11626116) and ZJNSF (Grant
No. LQ17A010006).
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