# Global Existence of Solutions to the 2D Incompressible Generalized Liquid Crystal Flow

Authors: Mingxuan Zhu
Journal: Abstract and Applied Analysis
 (2014)

---

## Abstract

We consider the global existence of solutions to the 2D incompressible generalized liquid crystal flow. It is proved that the local solution exists globally withβ
=
0, α
≥
2.


---

## Body

## 1. Introduction

In this paper, we consider the following 2D liquid crystal flow:(1)
u
t
+
u
·
∇
u
+
∇
p
+
Λ
2
α
u
=
-
∇
d
·
Δ
d
,
(2)
d
t
+
u
·
∇
d
+
Λ
2
β
d
=
-
f
(
d
)
,
(3)
div
⁡
u
=
0
,
(4)
(
u
,
d
)
|
t
=
0
=
(
u
0
,
d
0
)
,
where α
≥
0, β
≥
0 are real parameters and u is the velocity, d is a vectorial function modeling the orientation of the crystal molecules, and p is the scalar pressure. Here f
(
d
)
≔
(
|
d
|
2
-
1
)
d and Λ
=
(
-
Δ
)
1
/
2 is defined in terms of Fourier transform by
(5)
Λ
f
^
(
ξ
)
=
|
ξ
|
f
^
(
ξ
)
.Whenα
=
β
=
1, it has been shown that (1)–(4) has unique global weak and smooth solutions [1–3]. In [4], global regularity for this system with mixed partial viscosity is proved. Some regularity criteria are established for the system with zero dissipation in [5].The aim of this paper is to establish the following global regularity for the 2D liquid crystal model with fractional diffusion.Theorem 1.
Assume(
u
0
,
d
0
)
∈
H
3
(
ℝ
2
)
×
H
4
(
ℝ
2
). Let (
u
,
d
) be the local strong solution to the problem (1)–(4). If α and β satisfy β
=
0, α
≥
2, then the2D liquid crystal model has a unique global classical solution (
u
,
d
) satisfying
(6)
u
∈
L
∞
(
0
,
T
;
H
3
(
ℝ
2
)
)
,
u
∈
L
2
(
0
,
T
;
H
3
+
α
(
ℝ
2
)
)
,
d
∈
L
∞
(
0
,
T
;
H
4
(
ℝ
2
)
)
.Remark 2.
This work is partially motivated by the recent progress on the 2D incompressible MHD system with fractional diffusion; we refer to [6–10] and references therein. In [7], Tran et al. obtained the global regularity of 2D GMHD equations for the following three cases: (1) α
≥
1, β
≥
1; (2) 0
≤
α
<
1
/
2, 2
α
+
β
>
2; (3) α
≥
2, β
=
0. Combining them with the result in [10], we know that if α
+
β
≥
2, 2D incompressible MHD system with fractional diffusion possesses a global smooth solution. Fan et al. [8] proved the global existence of smooth solutions with α
>
0, β
=
1. Global regularity for the case α
=
0, β
>
1 was established by Jiu and Zhao [9] which improves the result in [6]. Very recently, the authors improved the case α
=
0, β
>
1 for the 2D liquid crystal model in [11].


## 2. Proof of Theorem1

It is sufficient to prove Theorem1 with α
=
2, β
=
0.We will prove Theorem1 if we can demonstrate the boundedness of ∥
u
∥
H
3
2
+
∥
d
∥
H
4
2. In order to reach our purpose, we will show this by contradiction: assume
(7)
lim
⁡
t
→
T
sup
⁡
∥
u
∥
H
3
2
+
∥
d
∥
H
4
2
=
∞
for some finite time T
>
0. Our thought is that when T
0 is close enough to T, ∥
u
∥
H
3
2
+
∥
d
∥
H
4
2 remains uniformly bounded for T
0
<
t
<
T under such assumption, thus reaching a contradiction.First, we doL
2 estimate for d. Multiplying (2) by d and using (3), after integration by parts, we see that
(8)
1
2
d
d
t
∥
d
∥
L
2
2
+
∥
d
∥
L
4
4
=
∥
d
∥
L
2
2
.
By using the Gronwall inequality, we have
(9)
∥
d
∥
L
2
+
∫
0
T
‍
∥
d
∥
L
4
4
d
τ
≤
C
.
Then, we will show the L
2 estimate for u and ∇
d. Multiplying (1) and (2) by u and -
Δ
d, respectively, we find that
(10)
1
2
d
d
t
(
∥
u
∥
L
2
2
+
∥
∇
d
∥
L
2
2
)
+
∥
Λ
2
u
∥
L
2
2
=
-
∫
ℝ
2
‍
∇
f
(
d
)
∇
d
d
x
≤
-
3
∫
ℝ
2
‍
|
d
|
2
|
∇
d
|
2
d
x
+
∥
∇
d
∥
L
2
2
.
Thanks to Gronwall’s inequality and (9), we have
(11)
∥
u
∥
L
2
2
+
∥
∇
d
∥
L
2
2
+
∫
0
T
‍
∥
Λ
2
u
∥
L
2
2
d
τ
≤
C
,
which means ∇
u
∈
L
2
(
0
,
T
;
BMO
).TheH
1 estimate for u and H
2 estimate for d will be shown as follows. Multiplying (1) by Δ
u, applying Δ to (2), multiplying by Δ
d, and then summing them up, we obtain
(12)
1
2
d
d
t
(
∥
∇
u
∥
L
2
2
+
∥
Δ
d
∥
L
2
2
)
+
∥
Δ
∇
u
∥
L
2
2
≤
∫
ℝ
2
‍
∇
d
·
Δ
d
·
Δ
u
-
Δ
(
u
·
∇
d
)
·
Δ
d
-
Δ
f
(
d
)
·
Δ
d
d
x
≤
C
∥
Δ
d
∥
L
2
2
∥
∇
u
∥
L
∞
+
∫
ℝ
2
‍
-
3
|
d
|
2
|
Δ
d
|
2
-
d
|
∇
d
|
2
Δ
d
+
|
Δ
d
|
2
d
x
≤
C
∥
Δ
d
∥
L
2
2
∥
∇
u
∥
L
∞
-
2
∥
d
Δ
d
∥
L
2
2
+
C
∥
∇
d
∥
L
4
4
+
∥
Δ
d
∥
L
2
2
≤
C
∥
Δ
d
∥
L
2
2
(
∥
∇
u
∥
L
∞
+
1
)
-
2
∥
d
Δ
d
∥
L
2
2
≤
C
∥
Δ
d
∥
L
2
2
(
∥
∇
u
∥
L
∞
+
1
)
.
Let us introduce the following commutator and bilinear estimates established in [12, 13]:
(13)
∥
Λ
s
(
f
g
)
-
f
Λ
s
g
∥
L
p
≤
C
(
∥
∇
f
∥
L
p
1
∥
Λ
s
-
1
g
∥
L
q
1
+
∥
g
∥
L
p
1
∥
Λ
s
f
∥
L
q
1
)
,
∥
Λ
s
(
f
g
)
∥
L
p
≤
C
(
∥
f
∥
L
p
1
∥
Λ
s
g
∥
L
q
1
+
∥
Λ
s
f
∥
L
p
2
∥
g
∥
L
q
2
)
,
with s
>
0 and 1
/
p
=
1
/
p
1
+
1
/
q
1
=
1
/
p
2
+
1
/
q
2.Now, we do theH
2 estimate for u and H
3 estimate for d. Applying Λ
2 to (1), multiplying by Λ
2
u, and dealing with (2) in the same way by Λ
3 and Λ
3
d, after summing them up, we have
(14)
1
2
d
d
t
(
∥
Λ
2
u
∥
L
2
2
+
∥
Λ
3
d
∥
L
2
2
)
+
∥
Λ
4
u
∥
L
2
2
=
∫
ℝ
2
‍
-
Λ
2
(
u
·
∇
u
)
Λ
2
u
-
Λ
2
(
∇
d
·
Δ
d
)
Λ
2
u
-
Λ
3
(
u
·
∇
d
)
Λ
3
d
-
Λ
3
f
(
d
)
Λ
3
d
d
x
=
:
I
1
+
I
2
+
I
3
+
I
4
.
Using Hölder’s inequality, Gagliardo-Nirenberg inequality, Young’s inequality, and (13), we have the following estimates:
(15)
|
I
1
|
=
|
∫
ℝ
2
‍
(
Λ
2
(
u
·
∇
u
)
-
u
·
∇
Λ
2
u
)
Λ
2
u
d
x
|
≤
C
∥
∇
u
∥
L
∞
∥
Λ
2
u
∥
L
2
2
,
|
I
2
|
≤
C
∥
Λ
4
u
∥
L
2
∥
∇
d
∥
L
4
∥
Λ
2
d
∥
L
4
≤
C
∥
Λ
4
u
∥
L
2
∥
∇
d
∥
L
2
∥
Λ
3
d
∥
L
2
≤
1
4
∥
Λ
4
u
∥
L
2
2
+
∥
Λ
3
d
∥
L
2
2
,
|
I
3
|
≤
C
|
∫
ℝ
2
‍
Λ
3
(
u
·
∇
d
)
Λ
3
d
-
u
·
∇
Λ
3
d
Λ
3
d
d
x
|
≤
C
∫
ℝ
2
‍
|
Λ
3
u
|
|
Λ
d
|
|
Λ
3
d
|
+
|
Λ
2
u
|
|
Λ
2
d
|
|
Λ
3
d
|
+
|
Λ
u
|
|
Λ
3
d
|
2
d
x
=
:
I
I
1
+
I
I
2
+
I
I
3
.
Now we estimate I
I
1, I
I
2, and I
I
3 one by one:
(16)
I
I
1
≤
C
∥
Λ
3
u
∥
L
4
∥
Λ
d
∥
L
4
∥
Λ
3
d
∥
L
2
≤
C
∥
Λ
2
u
∥
L
2
1
/
4
∥
Λ
4
u
∥
L
2
3
/
4
∥
Λ
d
∥
L
2
3
/
4
∥
Λ
3
d
∥
L
2
5
/
4
≤
1
8
∥
Λ
4
u
∥
L
2
2
+
C
∥
Λ
2
u
∥
L
2
2
/
5
∥
Λ
3
d
∥
L
2
2
,
I
I
2
≤
C
∥
Λ
2
u
∥
L
4
∥
Λ
2
d
∥
L
4
∥
Λ
3
d
∥
L
2
≤
C
∥
Λ
2
u
∥
L
2
3
/
4
∥
Λ
4
u
∥
L
2
1
/
4
∥
Λ
d
∥
L
2
1
/
4
∥
Λ
3
d
∥
L
2
7
/
4
≤
1
8
∥
Λ
4
u
∥
L
2
2
+
∥
Λ
2
u
∥
L
2
6
/
7
∥
Λ
3
d
∥
L
2
2
,
I
I
3
≤
C
∥
∇
u
∥
L
∞
∥
Λ
3
d
∥
L
2
2
,
I
4
=
∥
Λ
3
d
∥
L
2
2
-
∫
ℝ
2
‍
Λ
3
(
|
d
|
2
d
)
Λ
3
d
≤
∥
Λ
3
d
∥
L
2
2
-
3
∥
d
Λ
3
d
∥
L
2
2
+
C
∫
ℝ
2
‍
|
Λ
2
d
|
|
Λ
d
|
|
d
|
|
Λ
3
d
|
+
C
∫
ℝ
2
‍
|
Λ
d
|
3
|
Λ
3
d
|
=
:
∥
Λ
3
d
∥
L
2
2
-
3
∥
d
Λ
3
d
∥
L
2
2
+
K
1
+
K
2
.
K
1 and K
2 can be estimated as follows:
(17)
K
1
≤
C
∥
Λ
d
∥
L
4
∥
Λ
2
d
∥
L
4
∥
d
Λ
3
d
∥
L
2
≤
C
∥
Λ
d
∥
L
2
1
/
4
∥
Λ
3
d
∥
L
2
3
/
4
∥
Λ
d
∥
L
2
3
/
4
∥
Λ
3
d
∥
L
2
1
/
4
∥
d
Λ
3
d
∥
L
2
≤
C
∥
Λ
3
d
∥
L
2
2
+
3
∥
d
Λ
3
d
∥
L
2
2
,
K
2
≤
C
∥
Λ
d
∥
L
6
3
∥
Λ
3
d
∥
L
2
≤
C
(
∥
Λ
d
∥
L
2
2
/
3
∥
Λ
3
d
∥
L
2
1
/
3
)
3
∥
Λ
3
d
∥
L
2
≤
C
∥
Λ
3
d
∥
L
2
2
.
Combining K
1 and K
2, we have
(18)
I
4
≤
C
∥
Λ
3
d
∥
L
2
2
.
Summing all the above estimates to (14), we obtain
(19)
d
d
t
(
∥
Λ
2
u
∥
L
2
2
+
∥
Λ
3
d
∥
L
2
2
)
+
∥
Λ
4
u
∥
L
2
2
≤
C
(
∥
∇
u
∥
L
∞
+
∥
Λ
2
u
∥
L
2
)
(
∥
Λ
2
u
∥
L
2
2
+
∥
Λ
3
d
∥
L
2
2
)
.
Now, we will show the H
3 estimate for u and H
4 estimate for d. Applying Λ
3 to (1), multiplying by Λ
3
u, and dealing with (2) in the same way by Λ
4 and Λ
4
d, after summing them up, we have
(20)
1
2
d
d
t
(
∥
Λ
3
u
∥
L
2
2
+
∥
Λ
4
d
∥
L
2
2
)
+
∥
Λ
5
u
∥
L
2
2
=
∫
ℝ
2
‍
-
Λ
3
(
u
·
∇
u
)
Λ
3
u
-
Λ
3
(
∇
d
·
Δ
d
)
Λ
3
u
-
Λ
4
(
u
·
∇
d
)
Λ
4
d
-
Λ
4
f
(
d
)
Λ
4
d
d
x
=
:
J
1
+
J
2
+
J
3
+
J
4
.
Using Hölder’s inequality, Gagliardo-Nirenberg inequality, Young’s inequality, and (13), we have the following estimates:
(21)
|
J
1
|
≤
C
∥
∇
u
∥
L
∞
∥
Λ
3
u
∥
L
2
2
,
|
J
2
|
≤
C
∫
ℝ
2
‍
|
Λ
(
∇
d
Δ
d
)
|
|
Λ
5
u
|
d
x
≤
∥
Λ
(
∇
d
·
Δ
d
)
∥
L
2
∥
Λ
5
u
∥
L
2
|
J
2
|
≤
C
∥
Λ
5
u
∥
L
2
(
∥
Δ
d
∥
L
4
2
+
∥
Λ
d
∥
L
4
∥
Λ
3
d
∥
L
4
)
|
J
2
|
≤
C
∥
Λ
5
u
∥
L
2
(
∥
Λ
d
∥
L
2
∥
Λ
4
d
∥
L
2
+
∥
Λ
d
∥
L
2
5
/
6
∥
Λ
4
d
∥
L
2
1
/
6
×
∥
Λ
d
∥
L
2
1
/
6
∥
Λ
4
d
∥
L
2
5
/
6
)
|
J
2
|
≤
C
∥
Λ
4
d
∥
L
2
2
+
1
4
∥
Λ
5
u
∥
L
2
2
,
|
J
3
|
=
C
|
∫
ℝ
2
‍
(
Λ
4
(
u
·
∇
d
)
-
u
·
∇
Λ
4
d
)
Λ
4
d
d
x
|
|
J
3
|
≤
C
∫
ℝ
2
‍
|
Λ
4
u
|
|
∇
d
|
|
Λ
4
d
|
+
|
Λ
3
u
|
|
Λ
2
d
|
|
Λ
4
d
|
+
|
Λ
2
u
|
|
Λ
3
d
|
|
Λ
4
d
|
+
|
Λ
u
|
|
Λ
4
d
|
2
d
x
|
J
3
|
=
:
J
31
+
J
32
+
J
33
+
J
34
.
Now we estimate J
31, J
32, J
33, and J
34 one by one:
(22)
|
J
31
|
≤
C
∥
Λ
4
u
∥
L
4
∥
Λ
d
∥
L
4
∥
Λ
4
d
∥
L
2
≤
C
∥
Λ
2
u
∥
L
2
1
/
6
∥
Λ
5
u
∥
L
2
5
/
6
∥
Λ
d
∥
L
2
5
/
6
∥
Λ
4
d
∥
L
2
7
/
6
≤
1
8
∥
Λ
5
u
∥
L
2
2
+
C
∥
Λ
2
u
∥
L
2
3
/
7
∥
Λ
4
d
∥
L
2
2
,
|
J
32
|
≤
C
∥
Λ
3
u
∥
L
4
∥
Λ
2
d
∥
L
4
∥
Λ
4
d
∥
L
2
≤
C
∥
Λ
2
u
∥
L
2
1
/
2
∥
Λ
5
u
∥
L
2
1
/
2
∥
Λ
d
∥
L
2
1
/
2
∥
Λ
4
d
∥
L
2
3
/
2
≤
1
8
∥
Λ
5
u
∥
L
2
2
+
C
∥
Λ
2
u
∥
L
2
2
/
3
∥
Λ
4
d
∥
L
2
2
,
|
J
33
|
≤
C
∥
Λ
2
u
∥
L
4
∥
Λ
3
d
∥
L
4
∥
Λ
4
d
∥
L
2
≤
C
∥
Λ
3
u
∥
L
2
5
/
6
∥
u
∥
L
2
1
/
6
∥
Λ
d
∥
L
2
1
/
6
∥
Λ
4
d
∥
L
2
11
/
6
≤
C
∥
Λ
3
u
∥
L
2
5
/
6
∥
Λ
4
d
∥
L
2
11
/
6
,
|
J
34
|
≤
C
∥
∇
u
∥
L
∞
∥
Λ
4
d
∥
L
2
2
.
The estimate for J
4 is as follows:
(23)
|
J
4
|
=
∥
Λ
4
d
∥
L
2
2
-
∫
ℝ
2
‍
Λ
4
(
|
d
|
2
d
)
Λ
4
d
≤
∥
Λ
4
d
∥
L
2
2
-
3
∫
ℝ
2
‍
|
d
|
2
|
Λ
4
d
|
2
+
C
∫
ℝ
2
‍
|
d
|
|
Λ
d
|
|
Λ
3
d
|
|
Λ
4
d
|
+
C
∫
ℝ
2
‍
|
d
|
|
Λ
2
d
|
2
|
Λ
4
d
|
+
C
∫
ℝ
2
‍
|
Λ
d
|
2
|
Λ
2
d
|
|
Λ
4
d
|
=
:
∥
Λ
4
d
∥
L
2
2
-
3
∥
d
Λ
4
d
∥
L
2
2
+
J
41
+
J
42
+
J
43
.
We calculate J
41, J
42, and J
43:
(24)
|
J
41
|
≤
C
∥
Λ
3
d
∥
L
4
∥
Λ
d
∥
L
4
∥
d
Λ
4
d
∥
L
2
≤
C
∥
Λ
d
∥
L
2
1
/
6
∥
Λ
4
d
∥
L
2
5
/
6
∥
Λ
d
∥
L
2
5
/
6
∥
Λ
4
d
∥
L
2
1
/
6
∥
d
Λ
4
d
∥
L
2
≤
C
∥
Λ
4
d
∥
L
2
2
+
3
2
∥
d
Λ
4
d
∥
L
2
2
,
|
J
42
|
≤
C
∥
Λ
2
d
∥
L
4
2
∥
d
Λ
4
d
∥
L
2
≤
C
∥
Λ
d
∥
L
2
∥
Λ
4
d
∥
L
2
∥
d
Λ
4
d
∥
L
2
≤
C
∥
Λ
4
d
∥
L
2
2
+
3
2
∥
d
Λ
4
d
∥
L
2
2
,
|
J
43
|
≤
C
∥
Λ
2
d
∥
L
4
∥
Λ
d
∥
L
8
2
∥
Λ
4
d
∥
L
2
≤
C
∥
Λ
d
∥
L
2
1
/
2
∥
Λ
4
d
∥
L
2
1
/
2
∥
Λ
d
∥
L
2
3
/
2
∥
Λ
4
d
∥
L
2
1
/
2
∥
Λ
4
d
∥
L
2
≤
C
∥
Λ
4
d
∥
L
2
2
.
Combining J
41, J
42, and J
43, we get
(25)
J
4
≤
C
∥
Λ
4
d
∥
L
2
2
.
Combining the above estimates to (20), we get
(26)
d
d
t
(
∥
Λ
3
u
∥
L
2
2
+
∥
Λ
4
d
∥
L
2
2
)
+
∥
Λ
5
u
∥
L
2
2
≤
C
(
1
+
∥
∇
u
∥
L
∞
+
∥
Λ
2
u
∥
L
2
)
∥
Λ
4
d
∥
L
2
2
+
C
(
1
+
∥
Λ
3
u
∥
L
2
)
∥
Λ
4
d
∥
L
2
11
/
6
.
Now we estimate the term ∫
T
0
t
‍
∥
Λ
3
u
∥
L
2 by applying the Gronwall inequality to (12):
(27)
∫
T
0
t
‍
∥
Λ
3
u
∥
L
2
2
(
·
,
τ
)
d
τ
≤
∥
∇
u
∥
L
2
2
+
∥
Δ
d
∥
L
2
2
+
∫
T
0
t
‍
∥
Δ
∇
u
∥
L
2
2
(
·
,
τ
)
d
τ
≤
(
∥
∇
u
0
∥
L
2
2
+
∥
Δ
d
0
∥
L
2
2
)
×
exp
⁡
(
C
∫
T
0
t
‍
1
+
∥
∇
u
∥
L
∞
(
·
,
τ
)
d
τ
)
.
Here T
0
∈
(
0
,
T
) will be fixed later and we denote ∇
u
0
≔
∇
u
(
·
,
T
0
), Δ
d
0
≔
Δ
d
(
·
,
T
0
). Set A
(
t
)
≔
max
⁡
τ
∈
(
T
0
,
t
)
(
∥
u
∥
H
3
2
+
∥
d
∥
H
4
2
)
(
τ
). Now applying the logarithmic inequality [14]
(28)
∥
∇
u
∥
L
∞
≤
C
(
1
+
∥
∇
u
∥
BMO
(
1
+
ln
⁡
(
1
+
∥
u
∥
H
3
2
)
)
)
,
we get
(29)
∫
T
0
t
‍
∥
Λ
3
u
∥
L
2
2
(
·
,
τ
)
d
τ
≤
C
(
T
0
)
exp
⁡
(
C
∫
T
0
t
‍
∥
∇
u
∥
L
∞
(
·
,
τ
)
d
τ
)
≤
C
(
T
0
)
exp
⁡
(
C
∫
T
0
t
‍
1
+
∥
∇
u
∥
BMO
×
(
1
+
ln
⁡
⁡
(
1
+
∥
u
∥
H
3
2
)
)
(
·
,
τ
)
d
τ
∫
T
0
t
‍
)
≤
C
(
T
0
)
exp
⁡
(
C
∫
T
0
t
‍
∥
∇
u
∥
BMO
(
·
,
τ
)
×
(
1
+
ln
⁡
⁡
(
1
+
A
(
t
)
)
)
d
τ
∫
T
0
t
‍
)
≤
C
(
T
0
)
exp
⁡
(
C
∫
T
0
t
‍
∥
∇
u
∥
BMO
(
·
,
τ
)
d
τ
×
(
1
+
ln
⁡
⁡
(
1
+
A
(
t
)
)
)
∫
T
0
t
‍
)
.
Since ∥
∇
u
∥
BMO
∈
L
1
(
T
0
,
T
), we can take T
0 close enough to T, so that
(30)
C
∫
T
0
t
‍
∥
∇
u
∥
BMO
(
·
,
τ
)
d
τ
≤
2
δ
for some small positive number δ to be fixed later. With such choice of T
0 we have
(31)
∫
T
0
t
‍
∥
Λ
3
u
(
·
,
τ
)
∥
L
2
2
τ
≤
C
(
T
0
)
(
1
+
A
(
t
)
)
2
δ
.
Hölder’s inequality gives
(32)
∫
T
0
t
‍
∥
Λ
3
u
(
·
,
τ
)
∥
L
2
τ
≤
C
(
T
0
)
(
1
+
A
(
t
)
)
δ
.
Fix T
0 satisfying
(33)
C
∫
T
0
t
‍
∥
∇
u
(
·
,
τ
)
∥
BMO
τ
≤
2
δ
,
ln
⁡
(
1
+
A
(
T
0
)
)
>
1
.Combining the above estimates together, we get(34)
d
d
t
(
∥
u
∥
H
3
2
+
∥
d
∥
H
4
2
)
≤
C
(
1
+
∥
Λ
3
u
∥
L
2
)
A
(
t
)
11
/
12
+
(
∥
∇
u
∥
L
∞
+
∥
∇
2
u
∥
L
2
+
1
)
A
(
t
)
≤
C
[
∥
∇
2
u
∥
L
2
1
+
∥
∇
u
∥
BMO
(
1
+
ln
⁡
(
1
+
A
(
t
)
)
)
+
∥
∇
2
u
∥
L
2
(
1
+
ln
⁡
(
1
+
A
(
t
)
)
)
]
A
(
t
)
+
C
(
1
+
∥
Λ
3
u
∥
L
2
)
A
(
t
)
11
/
12
≤
C
(
T
0
)
[
(
t
)
11
/
12
(
∥
∇
u
∥
BMO
+
∥
∇
2
u
∥
L
2
+
1
)
A
(
t
)
×
ln
⁡
⁡
(
1
+
A
(
t
)
)
+
(
1
+
∥
Λ
3
u
∥
L
2
)
A
(
t
)
11
/
12
]
.
Integrating the above inequality, we have
(35)
A
(
t
)
≤
C
(
T
0
)
A
0
+
C
(
T
0
)
∫
T
0
t
‍
1
+
∥
Λ
3
u
∥
L
2
(
·
,
τ
)
d
τ
A
(
t
)
11
/
12
+
C
(
T
0
)
∫
T
0
t
‍
(
1
+
∥
∇
u
∥
BMO
+
∥
Λ
2
u
∥
L
2
)
A
(
t
)
×
ln
⁡
(
1
+
A
(
t
)
)
d
τ
,
where A
0
≔
∥
u
∥
H
3
2
(
T
0
)
+
∥
d
∥
H
4
2
(
T
0
).Takingδ
=
1
/
24, we have
(36)
∫
T
0
t
‍
1
+
∥
Λ
3
u
∥
L
2
d
τ
≤
C
(
T
0
)
(
1
+
A
(
t
)
)
1
/
24
.
Thus (35) tells us that
(37)
A
(
t
)
≤
C
(
T
0
)
A
0
+
C
(
T
0
)
(
A
(
t
)
+
1
)
1
/
24
A
(
t
)
11
/
12
+
C
(
T
0
)
∫
T
0
t
‍
(
1
+
∥
∇
u
∥
BMO
+
∥
Λ
2
u
∥
L
2
)
A
(
t
)
×
ln
⁡
(
1
+
A
(
t
)
)
d
τ
.
This in turn gives
(38)
1
+
A
(
t
)
≤
C
(
T
0
)
(
1
+
A
0
)
+
C
(
T
0
)
(
A
(
t
)
+
1
)
23
/
24
+
C
(
T
0
)
∫
T
0
t
‍
(
1
+
∥
∇
u
∥
BMO
+
∥
Λ
2
u
∥
L
2
)
×
(
A
(
t
)
+
1
)
ln
⁡
(
1
+
A
(
t
)
)
d
τ
.
We set B
(
t
)
≔
(
1
+
A
(
t
)
)
1
/
24, B
0
:
=
(
1
+
A
0
)
1
/
24 and divide the above inequality by (
1
+
A
(
t
)
)
23
/
24; using the monotonicity of A
(
t
) we reach
(39)
B
(
t
)
≤
C
(
T
0
)
[
B
0
+
1
+
∫
T
0
t
‍
(
1
+
∥
∇
u
∥
BMO
+
∥
∇
2
u
∥
L
2
)
×
B
(
t
)
ln
⁡
B
(
t
)
d
τ
∫
T
0
t
‍
(
1
+
∥
∇
u
∥
BMO
+
∥
∇
2
u
∥
L
2
)
]
.
The standard Gronwall’s inequality now gives
(40)
B
(
t
)
≤
[
C
(
T
0
)
(
1
+
B
0
)
]
exp
⁡
⁡
[
C
(
T
0
)
∫
T
0
t
‍
1
+
∥
∇
u
∥
BMO
+
∥
Λ
2
u
∥
L
2
d
τ
]
,
which leads to
(41)
A
(
t
)
≤
[
C
(
T
0
)
(
1
+
B
0
)
]
24
exp
⁡
[
C
(
T
0
)
∫
T
0
t
‍
1
+
∥
∇
u
∥
BMO
+
∥
Λ
2
u
∥
L
2
d
τ
]
.
As ∫
T
0
t
‍
∥
∇
u
∥
BMO
+
∥
Λ
2
u
∥
L
2
d
τ remains bounded as t
↗
T, the above inequality contradicts that A
(
t
)
↗
∞ as t
↗
T, so we complete our proof of Theorem 1.



---

*Source: 360958-2014-05-06.xml*