Proof of Theorem 1.
Multiplying 1 by
u
, using 3, and integrating in
R
3
, we derive
(14)
1
2
d
d
t
u
L
2
2
+
η
∇
u
L
2
2
=
∫
R
3
θ
e
3
·
u
d
x
≤
θ
L
2
u
L
2
≤
1
2
θ
L
2
2
+
1
2
u
L
2
2
.
Multiplying 2 by
θ
, using 3, and integrating in
R
3
, we obtain
(15)
1
2
d
d
t
θ
L
2
2
+
ν
∇
θ
L
2
=
0
.
Combining 14 and 15, using the Gronwall inequality, we deduce that
(16)
u
L
∞
(
0
,
T
;
L
2
)
+
u
L
2
(
0
,
T
;
H
1
)
≤
C
,
θ
L
∞
(
0
,
T
;
L
2
)
+
θ
L
2
(
0
,
T
;
H
1
)
≤
C
.
Multiplying 1 by
|
u
|
2
u
, using 3 and 7, and integrating in
R
3
, we derive
(17)
∫
u
2
·
u
u
t
+
u
·
∇
u
-
η
Δ
u
+
∇
p
=
1
4
d
d
t
∫
u
4
d
x
+
∫
u
2
·
u
2
·
∇
u
d
x
+
η
2
∫
∇
u
2
2
d
x
+
η
∫
u
2
∇
u
2
d
x
+
∫
u
·
∇
p
u
2
d
x
=
∫
u
2
·
u
·
θ
e
3
d
x
≤
C
∫
u
4
+
θ
4
d
x
;
that is,
(18)
1
4
d
d
t
∫
u
4
d
x
+
η
2
∫
∇
u
2
2
d
x
+
η
∫
u
2
∇
u
2
d
x
≤
-
∫
(
u
·
∇
p
)
u
2
d
x
+
C
∫
u
4
+
θ
4
d
x
≤
u
L
4
3
∇
p
L
4
+
C
u
L
4
4
+
C
θ
L
4
4
≤
C
u
L
4
3
∇
p
L
2
1
/
2
∇
p
BMO
1
/
2
+
C
u
L
4
4
+
C
θ
L
4
4
.
Multiplying 2 by
|
θ
|
2
θ
, using 3, and integrating in
R
3
, we arrive at
(19)
∫
θ
2
·
θ
·
θ
t
+
θ
2
θ
·
u
·
∇
θ
-
ν
θ
2
·
θ
·
Δ
θ
d
x
=
1
4
d
d
t
∫
θ
4
d
x
+
ν
∫
θ
2
∇
θ
2
d
x
+
ν
2
∫
θ
2
div
θ
2
d
x
.
Combining 18 and 19, using 9 and 16, we derive that
(20)
1
4
d
d
t
∫
θ
4
+
u
4
d
x
+
η
2
∫
∇
u
2
2
d
x
+
η
∫
u
2
∇
u
2
d
x
+
ν
∫
θ
2
∇
θ
2
d
x
+
ν
2
∫
θ
2
div
θ
2
d
x
,
≤
C
u
L
4
3
∇
p
L
2
1
/
2
∇
p
BMO
1
/
2
+
C
u
L
4
4
+
C
θ
L
4
4
≤
C
u
L
4
3
(
u
·
∇
u
L
2
1
/
2
+
θ
L
2
1
/
2
)
∇
p
BMO
1
/
2
+
C
u
L
4
4
+
C
θ
L
4
4
≤
2
C
u
L
4
4
∇
p
BMO
2
/
3
+
η
2
u
∇
u
L
2
2
+
C
θ
L
2
2
+
C
u
L
4
4
+
C
θ
L
4
4
,
which implies
(21)
d
d
t
∫
θ
4
+
u
4
d
x
+
η
∫
∇
u
2
2
d
x
+
η
∫
u
2
∇
u
2
d
x
+
ν
∫
θ
2
∇
θ
2
d
x
+
ν
∫
θ
2
div
θ
2
d
x
≤
8
C
u
L
4
4
∇
p
BMO
2
/
3
+
4
C
θ
L
2
2
+
4
C
u
L
4
4
+
4
C
θ
L
4
4
≤
8
C
u
L
4
4
∇
p
B
˙
∞
,
∞
0
2
/
3
ln
1
/
3
1
+
∇
p
H
2
+
4
C
θ
L
2
2
+
4
C
u
L
4
4
+
4
C
θ
L
4
4
≤
8
C
u
L
4
4
∇
p
B
˙
∞
,
∞
0
2
/
3
×
ln
1
/
3
1
+
∇
Δ
u
L
2
+
Δ
θ
L
2
+
4
C
θ
L
2
2
+
4
C
u
L
4
4
+
4
C
θ
L
4
4
≤
8
C
u
L
4
4
∇
p
B
˙
∞
,
∞
0
2
/
3
1
+
ln
1
+
∇
p
B
˙
∞
,
∞
0
2
/
3
×
ln
1
+
∇
Δ
u
L
2
+
Δ
θ
L
2
+
4
C
θ
L
2
2
+
4
C
u
L
4
4
+
4
C
θ
L
4
4
.
Choosing
t
∈
[
T
*
,
T
]
and setting
(22)
y
(
t
)
=
sup
t
∈
T
*
,
T
∇
·
Δ
u
t
L
2
+
Δ
θ
L
2
,
we have
(23)
sup
t
∈
[
T
*
,
T
]
(
u
L
4
+
θ
L
4
)
≤
C
*
1
+
y
t
C
ε
,
where
ε
is a small enough constant, such that
(24)
∫
T
*
T
∇
p
B
˙
∞
,
∞
0
2
/
3
1
+
ln
1
+
∇
p
B
˙
∞
,
∞
0
2
/
3
d
t
<
ε
.
Next, we want to estimate the
L
2
-norm of
∇
u
and
∇
θ
.
Multiplying 1 by
-
Δ
u
, integrating in
R
3
, and using 3 and 11, we derive that
(25)
∫
u
t
·
-
Δ
u
d
x
+
∫
u
·
∇
u
-
Δ
u
d
x
+
η
Δ
u
L
2
2
+
∫
∇
p
·
-
Δ
u
d
x
=
-
∫
θ
e
3
·
Δ
u
d
x
;
that is,
(26)
1
2
d
d
t
∫
∇
u
2
d
x
+
η
∫
Δ
u
2
d
x
=
∫
u
·
∇
u
Δ
u
d
x
-
∫
θ
e
3
Δ
u
d
x
≤
u
L
4
∇
u
L
4
Δ
u
L
2
+
Δ
u
L
2
θ
L
2
≤
C
u
L
4
u
L
4
1
/
5
Δ
u
L
2
4
/
5
Δ
u
L
2
+
η
4
Δ
u
L
2
2
+
C
θ
L
2
2
≤
η
4
Δ
u
L
2
2
+
C
u
L
4
12
+
η
4
Δ
u
L
2
2
+
C
θ
L
2
Δ
θ
L
2
≤
η
2
Δ
u
L
2
2
+
C
u
L
4
12
+
ν
4
Δ
θ
L
2
2
+
C
θ
L
2
2
.
Multiplying 2 by
-
Δ
θ
, integrating in
R
3
, and using 3 and 11, we derive that
(27)
∫
θ
t
·
-
Δ
θ
d
x
+
∫
u
·
∇
θ
-
Δ
θ
d
x
+
ν
Δ
θ
L
2
2
=
0
;
that is,
(28)
1
2
d
d
t
∫
∇
θ
2
d
x
+
ν
∫
Δ
θ
2
d
x
=
∫
u
·
∇
θ
Δ
θ
+
η
4
Δ
u
L
2
2
+
C
θ
L
2
2
d
x
≤
u
L
4
∇
θ
L
4
Δ
θ
L
2
≤
C
u
L
4
θ
L
4
1
/
5
Δ
θ
L
2
4
/
5
Δ
θ
L
2
≤
ν
4
Δ
θ
L
2
2
+
C
u
L
4
10
θ
L
4
2
.
Combining 26 and 28, using 16, we deduce
(29)
1
2
d
d
t
∫
∇
u
2
+
∇
θ
2
d
x
+
η
∫
Δ
u
2
d
x
+
ν
∫
Δ
θ
2
d
x
≤
η
2
Δ
u
L
2
2
+
C
u
L
4
12
+
ν
4
Δ
θ
L
2
2
+
C
θ
L
2
2
+
ν
4
Δ
θ
L
2
2
+
C
u
L
4
10
θ
L
4
2
=
η
2
Δ
u
L
2
2
+
ν
2
Δ
θ
L
2
2
+
C
u
L
4
12
+
C
θ
L
2
2
+
C
u
L
4
10
θ
L
4
2
;
that is,
(30)
d
d
t
∫
∇
u
2
+
∇
θ
2
d
x
+
η
∫
Δ
u
2
d
x
+
ν
∫
Δ
θ
2
d
x
≤
2
C
u
L
4
12
+
2
C
θ
L
2
2
+
2
C
u
L
4
10
θ
L
4
2
,
which implies that
(31)
∇
u
t
,
·
L
2
2
+
∇
θ
t
,
·
L
2
2
≤
C
1
+
y
t
C
ε
.
Last, we will estimate the
H
3
-norm and
H
4
-norm of
u
and
θ
and use the operator
Λ
to derive our goal.
Applying
Λ
3
=
(
-
Δ
)
3
/
2
to 1 and then multiplying 1 with
Λ
3
u
, we deduce
(32)
∫
Λ
3
u
t
·
Λ
3
u
d
x
+
∫
Λ
3
u
·
∇
u
·
Λ
3
u
d
x
-
η
∫
Λ
3
Δ
u
·
Λ
3
u
d
x
+
∫
Λ
3
∇
p
Λ
3
u
d
x
=
∫
Λ
3
θ
e
3
·
Λ
3
u
d
x
;
that is,
(33)
1
2
d
d
t
∫
Λ
3
u
2
d
x
+
η
∫
Λ
4
u
2
d
x
=
-
∫
Λ
3
u
·
∇
u
-
u
·
∇
Λ
3
u
·
Λ
3
u
d
x
+
∫
Λ
3
θ
e
3
Λ
3
u
d
x
≤
C
(
∇
u
L
3
Λ
3
u
L
3
2
+
Λ
3
u
L
2
Λ
4
u
L
2
)
+
Λ
3
u
L
2
θ
L
2
≤
C
∇
u
L
2
3
/
4
Λ
3
u
L
2
1
/
4
∇
u
L
2
1
/
3
Λ
4
u
L
2
5
/
3
+
Λ
3
u
L
2
θ
L
2
+
C
Λ
3
u
L
2
2
+
η
16
Λ
4
u
L
2
≤
C
∇
u
L
2
13
/
12
Λ
3
u
L
2
1
/
4
Λ
4
u
L
2
5
/
3
+
C
Λ
3
u
L
2
2
+
C
θ
L
2
+
C
Λ
3
u
L
2
2
+
η
16
Λ
4
u
L
2
≤
η
16
Λ
4
u
L
2
+
C
∇
u
L
2
13
/
10
Λ
3
u
L
2
1
/
2
+
2
C
Λ
3
u
L
2
2
+
C
θ
L
2
+
η
16
Λ
4
u
L
2
≤
C
∇
u
L
2
13
/
12
Λ
3
u
L
2
1
/
4
Λ
4
u
L
2
5
/
3
+
C
Λ
3
u
L
2
2
+
C
θ
L
2
+
C
Λ
3
u
L
2
2
+
η
8
Λ
4
u
L
2
=
η
4
Λ
4
u
L
2
2
+
C
∇
u
L
2
13
/
10
Λ
3
u
L
2
1
/
2
+
2
C
Λ
3
u
L
2
2
+
C
θ
L
2
.
Similarly, applying
Λ
3
to 2 and multiplying 2 by
Λ
3
θ
, we derive
(34)
∫
Λ
3
θ
t
Λ
3
θ
d
x
+
∫
Λ
3
u
·
∇
θ
Λ
3
θ
d
x
-
ν
∫
Λ
3
Δ
θ
·
Λ
3
θ
d
x
=
0
;
that is,
(35)
1
2
d
d
t
∫
Λ
3
θ
2
d
x
+
ν
∫
Λ
4
θ
2
d
x
=
-
∫
Λ
3
u
·
∇
θ
Λ
3
θ
d
x
=
C
∇
u
L
3
Λ
3
θ
L
3
2
+
Λ
3
u
L
3
∇
θ
L
3
Λ
3
θ
L
3
≤
C
∇
u
L
2
3
/
4
Λ
3
u
L
2
1
/
4
∇
θ
L
2
1
/
3
Λ
4
θ
L
2
5
/
3
+
C
∇
u
L
2
1
/
6
Λ
4
u
L
2
5
/
6
∇
θ
L
2
3
/
4
Λ
3
θ
L
2
1
/
4
×
∇
θ
L
2
1
/
6
Λ
4
θ
L
2
5
/
6
≤
C
∇
u
L
2
9
/
2
Λ
3
u
L
2
3
/
2
∇
θ
L
2
2
+
ν
4
Λ
4
θ
L
2
2
+
C
∇
u
L
2
1
/
3
Λ
4
u
L
2
5
/
3
+
C
∇
θ
L
2
3
/
2
Λ
3
θ
L
2
1
/
2
∇
θ
L
2
1
/
3
Λ
4
θ
L
2
5
/
3
≤
C
∇
u
L
2
9
/
2
Λ
3
u
L
2
3
/
2
∇
θ
L
2
2
+
ν
4
Λ
4
θ
L
2
2
+
C
∇
u
L
2
2
+
η
4
Λ
4
u
L
2
2
+
C
∇
θ
L
2
9
Λ
3
θ
L
2
3
∇
θ
L
2
2
+
ν
4
Λ
4
θ
L
2
2
.
Combining 33 and 35, we have
(36)
1
2
d
d
t
∫
Λ
3
u
2
Λ
3
θ
2
d
x
+
η
∫
Λ
4
u
2
d
x
+
ν
∫
Λ
4
θ
2
d
x
≤
η
2
Λ
4
u
L
2
2
+
C
∇
u
L
2
13
/
10
Λ
3
u
L
2
1
/
2
+
2
C
Λ
3
u
L
2
2
+
C
θ
L
2
+
C
∇
u
L
2
9
/
2
Λ
3
u
L
2
3
/
2
∇
θ
L
2
2
+
ν
2
Λ
4
θ
L
2
2
+
C
∇
u
L
2
2
+
C
∇
θ
L
2
9
Λ
3
θ
L
2
3
∇
θ
L
2
2
;
that is,
(37)
d
d
t
∫
Λ
3
u
2
Λ
3
θ
2
d
x
+
η
∫
Λ
4
u
2
d
x
+
ν
∫
Λ
4
θ
2
d
x
≤
2
C
∇
u
L
2
13
/
10
Λ
3
u
L
2
1
/
2
+
4
C
Λ
3
u
L
2
2
+
2
C
θ
L
2
+
2
C
∇
u
L
2
9
/
2
Λ
3
u
L
2
3
/
2
∇
θ
L
2
2
+
2
C
∇
u
L
2
2
+
2
C
∇
θ
L
2
9
Λ
3
θ
L
2
3
∇
θ
L
2
2
.
Choosing
ε
small enough, using 16, 23, and 24, we conclude that
(38)
u
L
∞
(
0
,
T
;
H
3
)
+
u
L
2
(
0
,
T
;
H
4
)
≤
C
,
θ
L
∞
(
0
,
T
;
H
3
)
+
θ
L
2
(
0
,
T
;
H
4
)
≤
C
.
We complete the proof.