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ISRN Mathematical Analysis
Volume 2012 (2012), Article ID 935045, 10 pages
doi:10.5402/2012/935045
Research Article

Regularity Criterion for the 3D Nematic Liquid Crystal Flows

Jishan Fan1 and Tohru Ozawa2

1Department of Applied Mathematics, Nanjing Forestry University, Nanjing 210037, China
2Department of Applied Physics, Waseda University, Tokyo 169-8555, Japan

Received 17 January 2012; Accepted 14 February 2012

Academic Editors: A. Carpio, Y. Liu, and G. A. Seregin

Copyright © 2012 Jishan Fan and Tohru Ozawa. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Abstract

We study the hydrodynamic theory of liquid crystals. We prove a logarithmically improved regularity criterion for two simplified Ericksen-Leslie systems.

1. Introduction

The hydrodynamic theory of liquid crystals was established by Ericksen and Leslie [1–4]. However, since the equations are too complicated, we consider the first simplified Ericksen-Leslie system: 𝑢 𝑡 𝑑 + 𝑢 ⋅ ∇ 𝑢 + ∇ 𝜋 − Δ 𝑢 = − ∇ ⋅ ( ∇ 𝑑 ⊙ ∇ 𝑑 ) , ( 1 . 1 ) 𝑡 𝑢 + 𝑢 ⋅ ∇ 𝑑 = Δ 𝑑 − 𝑓 ( 𝑑 ) , ( 1 . 2 ) d i v 𝑢 = 0 , ( 1 . 3 ) ( 𝑢 , 𝑑 ) ( 𝑥 , 0 ) = 0 , 𝑑 0 ( 𝑥 ) i n ℝ 3 , ( 1 . 4 ) which include the velocity vector 𝑢 ∶ = ( 𝑢 1 , 𝑢 2 , 𝑢 3 ) 𝑡 , the scalar pressure 𝜋 being and the direction vector 𝑑 ∶ = ( 𝑑 1 , 𝑑 2 , 𝑑 3 ) 𝑡 . 𝑓 ( 𝑑 ) ∶ = 1 / 𝜂 ( | 𝑑 | 2 − 1 ) 𝑑 with 𝜂 , a positive constant. ( ∇ 𝑑 ⊙ ∇ 𝑑 ) 𝑖 , 𝑗 ∑ ∶ = 𝑘 𝜕 𝑖 𝑑 𝑘 𝜕 𝑗 𝑑 𝑘 , and hence ∑ ∇ ⋅ ( ∇ 𝑑 ⊙ ∇ 𝑑 ) = 𝑘 Δ 𝑑 𝑘 ∇ 𝑑 𝑘 + ( 1 / 2 ) ∇ | ∇ 𝑑 | 2 .

Lin-Liu [5] proved that the system (1.1)–(1.4) has a unique smooth solution globally in 2 space dimensions and locally in 3 dimensions. They also proved the global existence of weak solutions. However, the regularity of solutions to the system is still open. Fan-Guo [6] and Fan-Ozawa [7] showed the following regularity criteria: 𝑢 ∈ 𝐿 𝑟 0 , 𝑇 ; 𝐿 𝑠 ℝ 3 2 f o r 𝑟 + 3 𝑠 = 1 , 3 < 𝑠 ≤ ∞ , ∇ 𝑢 ∈ 𝐿 𝑟 0 , 𝑇 ; 𝐿 𝑠 ℝ 3 2 f o r 𝑟 + 3 𝑠 3 = 2 , 2 < 𝑠 ≤ ∞ , 𝑢 ∈ 𝐿 2 ̇ 𝐵 0 , 𝑇 ; 0 ∞ , ∞ , 𝜔 ∶ = c u r l 𝑢 ∈ 𝐿 1 ̇ 𝐵 0 , 𝑇 ; 0 ∞ , ∞ , ( 1 . 5 ) where ̇ 𝐵 0 ∞ , ∞ denotes the homogeneous Besov space.

The first aim of this paper is to prove a new regularity criterion as follows.

Theorem 1.1. Let 𝑢 0 ∈ 𝐻 3 , 𝑑 0 ∈ 𝐻 4 with d i v 𝑢 0 = 0 in ℝ 3 . Let ( 𝑢 , 𝑑 ) be a smooth solution to the problem (1.1)–(1.4) on [ 0 , 𝑇 ) . If 𝑢 satisfies 𝑇 0 ‖ 𝑢 ( ⋅ , 𝑡 ) ‖ ̇ 𝐵 2 / ( 1 − 𝑠 ) − 𝑠 ∞ , ∞ 1 + l o g 𝑒 + ‖ 𝑢 ( ⋅ , 𝑡 ) ‖ ̇ 𝐵 − 𝑠 ∞ , ∞ 𝑑 𝑡 < ∞ ( 1 . 6 ) for some 𝑠 with 0 < 𝑠 < 1 , then the solution ( 𝑢 , 𝑑 ) can be extended beyond 𝑇 > 0 .

When the penalization parameter 𝜂 → 0 , (1.1)–(1.4) reduce to 𝑢 𝑡 𝑑 + 𝑢 ⋅ ∇ 𝑢 + ∇ 𝜋 − Δ 𝑢 = − ∇ ⋅ ( ∇ 𝑑 ⊙ ∇ 𝑑 ) , ( 1 . 7 ) d i v 𝑢 = 0 , ( 1 . 8 ) 𝑡 | | | | + 𝑢 ⋅ ∇ 𝑑 = Δ 𝑑 + ∇ 𝑑 2 | | 𝑑 | | 𝑢 𝑑 , = 1 , ( 1 . 9 ) ( 𝑢 , 𝑑 ) ( 𝑥 , 0 ) = 0 , 𝑑 0 ( 𝑥 ) i n ℝ 3 , | | 𝑑 0 | | = 1 . ( 1 . 1 0 )

When 𝑢 = 0 , then (1.9) is the well-known harmonic heat flow equation onto a sphere.

Fan-Gao-Guo [8] proved the following blow-up criteria: 𝑢 , ∇ 𝑑 ∈ 𝐿 2 ̇ 𝐵 0 , 𝑇 ; 0 ∞ , ∞ , 𝜔 , Δ 𝑑 ∈ 𝐿 1 ̇ 𝐵 0 , 𝑇 ; 0 ∞ , ∞ . ( 1 . 1 1 )

We will prove the folowing theorem

Theorem 1.2. Let 𝑢 0 , ∇ 𝑑 0 ∈ 𝐻 3 ( ℝ 3 ) with d i v 𝑢 0 = 0 , | 𝑑 0 | = 1 in ℝ 3 . Let ( 𝑢 , 𝑑 ) be a smooth solution to the problem (1.7)–(1.10) on [ 0 , 𝑇 ) . If the following condition is satisfied: 𝑇 0 ‖ 𝑢 ( ⋅ , 𝑡 ) ‖ ̇ 𝐵 2 / ( 1 − 𝑠 ) − 𝑠 ∞ , ∞ + ‖ ∇ 𝑑 ( ⋅ , 𝑡 ) ‖ 2 ̇ 𝐵 0 ∞ , ∞ 1 + l o g 𝑒 + ‖ 𝑢 ( ⋅ , 𝑡 ) ‖ ̇ 𝐵 − 𝑠 ∞ , ∞ + ‖ ∇ 𝑑 ( ⋅ , 𝑡 ) ‖ ̇ 𝐵 0 ∞ , ∞ 𝑑 𝑡 < ∞ , ( 1 . 1 2 ) for some 𝑠 with 0 < 𝑠 < 1 , then the solution ( 𝑢 , 𝑑 ) can be extended beyond 𝑇 > 0 .

2. Proof of Theorem 1.1

This section is devoted to the proof of Theorem 1.1. Since it is well-known that there are 𝑇 0 > 0 and a unique smooth solution ( 𝑢 , 𝑑 ) to the problem (1.1)–(1.4) in ( 0 , 𝑇 0 ] , we only need to show a priori estimates.

Testing (1.1) by 𝑢 and using (1.3), we see that 1 2 𝑑 𝑢 𝑑 𝑡 2 | | | | 𝑑 𝑥 + ∇ 𝑢 2 ( 𝑑 𝑥 = − 𝑢 ⋅ ∇ ) 𝑑 ⋅ Δ 𝑑 𝑑 𝑥 . ( 2 . 1 )

Testing (1.2) by Δ 𝑑 − 𝑓 ( 𝑑 ) and using (1.3), we find that 𝑑 1 𝑑 𝑡 2 | | | | ∇ 𝑑 2 + 1 | | 𝑑 | | 4 𝜂 2 − 1 2 | | | | 𝑑 𝑥 + Δ 𝑑 − 𝑓 ( 𝑑 ) 2 ( 𝑑 𝑥 = 𝑢 ⋅ ∇ ) 𝑑 ⋅ Δ 𝑑 𝑑 𝑥 . ( 2 . 2 )

Summing up (2.1) and (2.2), we infer that 1 2 𝑢 2 + 1 2 | | | | ∇ 𝑑 2 + 1 | | 𝑑 | | 4 𝜂 2 − 1 2 𝑑 𝑥 + 𝑇 0 | | | | ∇ 𝑢 2 + | | | | Δ 𝑑 − 𝑓 ( 𝑑 ) 2 ≤ 1 𝑑 𝑥 2 𝑢 2 0 + 1 2 | | ∇ 𝑑 0 | | 2 + 1 | | 𝑑 4 𝜂 0 | | 2 − 1 2 𝑑 𝑥 . ( 2 . 3 )

Testing (1.2) by 𝑑 and using (1.3), we deduce that 1 2 𝑑 𝑑 𝑑 𝑡 2 | | | | 𝑑 𝑥 + ∇ 𝑑 2 1 𝑑 𝑥 + 𝜂 | | 𝑑 | | 4 1 𝑑 𝑥 = 𝜂 𝑑 2 𝑑 𝑥 , ( 2 . 4 ) which yields ‖ 𝑑 ‖ 𝐿 ∞ ( 0 , 𝑇 ; 𝐿 2 ) + ‖ 𝑑 ‖ 𝐿 2 ( 0 , 𝑇 ; 𝐻 1 ) ≤ 𝐶 . ( 2 . 5 )

Next, we prove the following estimate: ‖ 𝑑 ‖ 𝐿 ∞ ( 0 , 𝑇 ; 𝐿 ∞ ) ‖ ‖ 𝑑 ≤ m a x 0 ‖ ‖ 𝐿 ∞ , 1 . ( 2 . 6 )

Without loss of generality, we assume that 1 ≤ ‖ 𝑑 0 ‖ 𝐿 ∞ . Multiplying (1.2) by 𝑑 , we get 𝜙 𝑡 | | 𝑑 | | + 𝑢 ⋅ ∇ 𝜙 − Δ 𝜙 + 2 2 2 𝜙 = − 𝜂 ‖ ‖ 𝑑 0 ‖ ‖ 2 𝐿 ∞ | | 𝑑 | | − 1 2 | | | | − 2 ∇ 𝑑 2 ≤ 0 ( 2 . 7 ) with 𝜙 = | 𝑑 | 2 − ‖ 𝑑 0 ‖ 2 𝐿 ∞ and 𝜙 | 𝑡 = 0 = | 𝑑 0 | 2 − ‖ 𝑑 0 ‖ 2 𝐿 ∞ ≤ 0 . Then (2.6) follows immediately from 𝜙 ≤ 0 by the maximum principle.

Testing (1.1) by − Δ 𝑢 and using (1.3), we see that 1 2 𝑑 | | | | 𝑑 𝑡 ∇ 𝑢 2 | | | | 𝑑 𝑥 + Δ 𝑢 2 ( − 𝑑 𝑥 = 𝑢 ⋅ ∇ ) 𝑢 ⋅ Δ 𝑢 𝑑 𝑥 𝑖 , 𝑘 Δ 𝑑 𝑘 ⋅ 𝜕 𝑖 ∇ 𝑑 𝑘 ⋅ ∇ 𝑢 𝑖 𝑑 𝑥 − 𝑖 , 𝑘 𝜕 𝑖 𝑑 𝑘 ⋅ ∇ Δ 𝑑 𝑘 ⋅ ∇ 𝑢 𝑖 𝑑 𝑥 . ( 2 . 8 )

Applying Δ to (1.2), testing by Δ 𝑑 , and using (1.3), we find that 1 2 𝑑 | | | | 𝑑 𝑡 Δ 𝑑 2 | | | | 𝑑 𝑥 + ∇ Δ 𝑑 2 𝑑 𝑥 = 𝑖 , 𝑘 𝜕 𝑖 𝑑 𝑘 ⋅ ∇ Δ 𝑑 𝑘 ⋅ ∇ 𝑢 𝑖 − 𝑑 𝑥 𝑖 , 𝑘 𝜕 𝑖 𝜕 𝑗 𝑑 𝑘 ⋅ 𝜕 𝑗 ∇ 𝑑 𝑘 ⋅ ∇ 𝑢 𝑖 𝑑 𝑥 − Δ 𝑓 ( 𝑑 ) ⋅ Δ 𝑑 𝑑 𝑥 . ( 2 . 9 )

Summing up (2.8) and (2.9), we get 1 2 𝑑 | | | | 𝑑 𝑡 ∇ 𝑢 2 + | | | | Δ 𝑑 2 | | | | 𝑑 𝑥 + Δ 𝑢 2 + | | | | ∇ Δ 𝑑 2 = 𝑑 𝑥 ( 𝑢 ⋅ ∇ ) 𝑢 ⋅ Δ 𝑢 𝑑 𝑥 − 𝑖 , 𝑘 Δ 𝑑 𝑘 ⋅ 𝜕 𝑖 ∇ 𝑑 𝑘 ⋅ ∇ 𝑢 𝑖 − 𝑑 𝑥 𝑖 , 𝑘 𝜕 𝑖 𝜕 𝑗 𝑑 𝑘 ⋅ 𝜕 𝑗 ∇ 𝑑 𝑘 ⋅ ∇ 𝑢 𝑖 𝑑 𝑥 − Δ 𝑓 ( 𝑑 ) Δ 𝑑 𝑑 𝑥 = ∶ 𝐼 1 + 𝐼 2 + 𝐼 3 + 𝐼 4 . ( 2 . 1 0 )

By using (2.6), 𝐼 4 is simply bounded as 𝐼 4 ≤ 𝐶 ‖ Δ 𝑑 ‖ 2 𝐿 2 . ( 2 . 1 1 )

By using the inequalities [9] ‖ 𝑢 ⋅ ∇ 𝑢 ‖ 𝐿 2 ≤ 𝐶 ‖ 𝑢 ‖ ̇ 𝐵 − 𝑠 ∞ , ∞ ‖ ∇ 𝑢 ‖ ̇ 𝐵 𝑠 2 , 2 , ‖ ∇ 𝑢 ‖ ̇ 𝐵 𝑠 2 , 2 ≤ 𝐶 ‖ ∇ 𝑢 ‖ 𝐿 1 − 𝑠 2 ‖ Δ 𝑢 ‖ 𝑠 𝐿 2 . ( 2 . 1 2 )

𝐼 1 can be bounded as follows: 𝐼 1 ≤ ‖ 𝑢 ⋅ ∇ 𝑢 ‖ 𝐿 2 ‖ Δ 𝑢 ‖ 𝐿 2 ≤ 𝐶 ‖ 𝑢 ‖ ̇ 𝐵 − 𝑠 ∞ , ∞ ‖ 𝑢 ‖ ̇ 𝐵 1 + 𝑠 2 , 2 ‖ Δ 𝑢 ‖ 𝐿 2 ≤ 𝐶 ‖ 𝑢 ‖ ̇ 𝐵 − 𝑠 ∞ , ∞ ‖ ∇ 𝑢 ‖ 𝐿 1 − 𝑠 2 ‖ Δ 𝑢 ‖ 𝐿 1 + 𝑠 2 ≤ 1 2 ‖ Δ 𝑢 ‖ 2 𝐿 2 + 𝐶 ‖ 𝑢 ‖ ̇ 𝐵 2 / ( 1 − 𝑠 ) − 𝑠 ∞ , ∞ ‖ ∇ 𝑢 ‖ 2 𝐿 2 . ( 2 . 1 3 )

We bound 𝐼 2 and 𝐼 3 as follows: 𝐼 2 , 𝐼 3 ≤ 𝐶 ‖ ∇ 𝑢 ‖ 𝐿 2 ‖ Δ 𝑑 ‖ 2 𝐿 4 ≤ 𝐶 ‖ ∇ 𝑢 ‖ 𝐿 2 ‖ ∇ Δ 𝑑 ‖ 𝐿 2 ≤ 1 4 ‖ ∇ Δ 𝑑 ‖ 2 𝐿 2 + 𝐶 ‖ ∇ 𝑢 ‖ 2 𝐿 2 . ( 2 . 1 4 )

Here we used the Gagliardo-Nirenberg inequality ‖ Δ 𝑑 ‖ 2 𝐿 4 ≤ 𝐶 ‖ 𝑑 ‖ 𝐿 ∞ ‖ ∇ Δ 𝑑 ‖ 𝐿 2 . ( 2 . 1 5 ) Inserting the above estimates into (2.10), we derive 𝑑 | | | | 𝑑 𝑡 ∇ 𝑢 2 + | | | | Δ 𝑑 2 | | | | 𝑑 𝑥 + Δ 𝑢 2 + | | | | ∇ Δ 𝑑 2 𝑑 𝑥 ≤ 𝐶 ‖ 𝑢 ‖ ̇ 𝐵 2 / ( 1 − 𝑠 ) − 𝑠 ∞ , ∞ + 1 ‖ ∇ 𝑢 ‖ 2 𝐿 2 + ‖ Δ 𝑑 ‖ 2 𝐿 2 ≤ 𝐶 1 + ‖ 𝑢 ‖ ̇ 𝐵 2 / ( 1 − 𝑠 ) − 𝑠 ∞ , ∞ 1 + l o g 𝑒 + ‖ 𝑢 ‖ ̇ 𝐵 − 𝑠 ∞ , ∞ 1 + l o g 𝑒 + ‖ 𝑢 ‖ ̇ 𝐵 − 𝑠 ∞ , ∞ ‖ ∇ 𝑢 ‖ 2 𝐿 2 + ‖ Δ 𝑑 ‖ 2 𝐿 2 ≤ 𝐶 1 + ‖ 𝑢 ‖ ̇ 𝐵 2 / ( 1 − 𝑠 ) − 𝑠 ∞ , ∞ 1 + l o g 𝑒 + ‖ 𝑢 ‖ ̇ 𝐵 − 𝑠 ∞ , ∞ ‖ ( 1 + l o g ( 𝑒 + 𝑦 ) ) ∇ 𝑢 ‖ 2 𝐿 2 + ‖ Δ 𝑑 ‖ 2 𝐿 2 . ( 2 . 1 6 )

Due to (1.6), one concludes that for any small constant 𝜖 > 0 , there exists 𝑇 ∗ < 𝑇 such that 𝑇 𝑇 ∗ ‖ 𝑢 ‖ ̇ 𝐵 2 / ( 1 − 𝑠 ) − 𝑠 ∞ , ∞ 1 + l o g 𝑒 + ‖ 𝑢 ‖ ̇ 𝐵 − 𝑠 ∞ , ∞ 𝑑 𝑡 < 𝜖 . ( 2 . 1 7 )

For any 𝑇 ∗ < 𝑡 ≤ 𝑇 , we set 𝑦 ( 𝑡 ) ∶ = s u p 𝑇 ∗ ≤ 𝜏 ≤ 𝑡 ‖ ‖ Λ 3 ‖ ‖ ( 𝑢 , ∇ 𝑑 ) ( ⋅ , 𝜏 ) 𝐿 2 w i t h Λ ∶ = ( − Δ ) 1 / 2 . ( 2 . 1 8 )

Applying Gronwall’s inequality to (2.16) in the interval [ 𝑇 ∗ , 𝑡 ] , one has ‖ ‖ ∇ 𝑢 ( ⋅ , 𝑡 ) 2 𝐿 2 ‖ + ‖ Δ 𝑑 ( ⋅ , 𝑡 ) 2 𝐿 2 ≤ 𝐶 ( 1 + 𝑦 ) 𝐶 0 𝜖 . ( 2 . 1 9 )

Now, we derive a bound on 𝑦 ( 𝑡 ) defined by (2.18). To this end, we will use the following commutator and product estimates due to Kato-Ponce [10]: ‖ Λ 𝛼 ( 𝑓 𝑔 ) − 𝑓 Λ 𝛼 𝑔 ‖ 𝐿 𝑝 ≤ 𝐶 ‖ ∇ 𝑓 ‖ 𝐿 𝑝 1 ‖ ‖ Λ 𝛼 − 1 𝑔 ‖ ‖ 𝐿 𝑞 1 + ‖ Λ 𝛼 𝑓 ‖ 𝐿 𝑝 2 ‖ 𝑔 ‖ 𝐿 𝑞 2 , ( 2 . 2 0 ) ‖ Λ 𝛼 ‖ ( 𝑓 𝑔 ) 𝐿 𝑝 ≤ 𝐶 ‖ 𝑓 ‖ 𝐿 𝑝 1 ‖ Λ 𝛼 𝑔 ‖ 𝐿 𝑞 1 + ‖ Λ 𝛼 𝑓 ‖ 𝐿 𝑝 2 ‖ 𝑔 ‖ 𝐿 𝑞 2 , ( 2 . 2 1 ) with 𝛼 > 0 and 1 / 𝑝 = 1 / 𝑝 1 + 1 / 𝑞 1 = 1 / 𝑝 2 + 1 / 𝑞 2 .

Applying Λ 3 to (1.1), testing by Λ 3 𝑢 , and using (1.3), (2.20), (2.21) and (2.19), we obtain 1 2 𝑑 | | Λ 𝑑 𝑡 3 𝑢 | | 2 | | Λ 𝑑 𝑥 + 4 𝑢 | | 2 Λ 𝑑 𝑥 = − 3 ( 𝑢 ⋅ ∇ 𝑢 ) − 𝑢 ∇ Λ 3 𝑢 Λ 3 Λ 𝑢 𝑑 𝑥 + 3 ( ∇ 𝑑 ⊙ ∇ 𝑑 ) ∶ Λ 3 ∇ 𝑢 𝑑 𝑥 ≤ 𝐶 ‖ ∇ 𝑢 ‖ 𝐿 3 ‖ ‖ Λ 3 𝑢 ‖ ‖ 2 𝐿 3 + 𝐶 ‖ ∇ 𝑑 ‖ 𝐿 ∞ ‖ ‖ Λ 4 𝑑 ‖ ‖ 𝐿 2 ‖ ‖ Λ 4 𝑢 ‖ ‖ 𝐿 2 ≤ 𝐶 ‖ ∇ 𝑢 ‖ 𝐿 3 / 4 2 ‖ ‖ Λ 3 𝑢 ‖ ‖ 𝐿 1 / 4 2 ⋅ ‖ ∇ 𝑢 ‖ 𝐿 1 / 3 2 ‖ ‖ Λ 4 𝑢 ‖ ‖ 𝐿 5 / 3 2 + 1 ‖ ‖ Λ 1 6 4 𝑢 ‖ ‖ 2 𝐿 2 + 𝐶 ‖ ∇ 𝑑 ‖ 2 𝐿 ∞ ‖ ‖ Λ 4 𝑑 ‖ ‖ 2 𝐿 2 ≤ 1 4 ‖ ‖ Λ 4 𝑢 ‖ ‖ 2 𝐿 2 + 𝐶 ‖ ∇ 𝑢 ‖ 𝐿 1 3 / 2 2 ‖ ‖ Λ 3 𝑢 ‖ ‖ 𝐿 3 / 2 2 + 𝐶 ‖ Δ 𝑑 ‖ 𝐿 3 / 2 2 ‖ ‖ Λ 4 𝑑 ‖ ‖ 𝐿 1 / 2 2 ⋅ ‖ Δ 𝑑 ‖ 𝐿 2 / 3 2 ‖ ‖ Λ 5 𝑑 ‖ ‖ 𝐿 4 / 3 2 ≤ 1 4 ‖ ‖ Λ 4 𝑢 ‖ ‖ 2 𝐿 2 + 1 4 ‖ ‖ Λ 5 𝑑 ‖ ‖ 2 𝐿 2 + 𝐶 ‖ ∇ 𝑢 ‖ 𝐿 1 3 / 2 2 ‖ ‖ Λ 3 𝑢 ‖ ‖ 𝐿 3 / 2 2 + 𝐶 ‖ Δ 𝑑 ‖ 𝐿 1 3 / 2 2 ‖ ‖ Λ 4 𝑑 ‖ ‖ 𝐿 3 / 2 2 . ( 2 . 2 2 ) Here we have used the following Gagliardo-Nirenberg inequalities: ‖ ∇ 𝑢 ‖ 𝐿 3 ≤ 𝐶 ‖ ∇ 𝑢 ‖ 𝐿 3 / 4 2 ‖ ‖ Λ 3 𝑢 ‖ ‖ 𝐿 1 / 4 2 , ‖ ‖ Λ 3 𝑢 ‖ ‖ 𝐿 3 ≤ 𝐶 ‖ ∇ 𝑢 ‖ 𝐿 1 / 6 2 ‖ ‖ Λ 4 𝑢 ‖ ‖ 𝐿 5 / 6 2 , ‖ ∇ 𝑑 ‖ 𝐿 ∞ ≤ 𝐶 ‖ Δ 𝑑 ‖ 𝐿 3 / 4 2 ‖ ‖ Λ 4 𝑑 ‖ ‖ 𝐿 1 / 4 2 , ‖ ‖ Λ 4 𝑑 ‖ ‖ 𝐿 2 ≤ 𝐶 ‖ Δ 𝑑 ‖ 𝐿 1 / 3 2 ‖ ‖ Λ 5 𝑑 ‖ ‖ 𝐿 2 / 3 2 . ( 2 . 2 3 )

Taking Λ 4 to (1.2), testing by Λ 4 𝑑 , and using (1.3), (2.20), (2.23), and (2.6), we have 1 2 𝑑 | | Λ 𝑑 𝑡 4 𝑑 | | 2 | | Λ 𝑑 𝑥 + 5 𝑑 | | 2 Λ 𝑑 𝑥 = − 4 ( 𝑢 ⋅ ∇ 𝑑 ) − 𝑢 ∇ Λ 4 𝑑 Λ 4 Λ 𝑑 𝑑 𝑥 − 4 𝑓 ( 𝑑 ) ⋅ Λ 4 𝑑 𝑑 𝑥 ≤ 𝐶 ‖ ∇ 𝑢 ‖ 𝐿 3 ‖ ‖ Λ 4 𝑑 ‖ ‖ 𝐿 6 ‖ ‖ Λ 4 𝑑 ‖ ‖ 𝐿 2 + 𝐶 ‖ ∇ 𝑑 ‖ 𝐿 ∞ ‖ ‖ Λ 4 𝑢 ‖ ‖ 𝐿 2 ‖ ‖ Λ 4 𝑑 ‖ ‖ 𝐿 2 ‖ ‖ Λ + 𝐶 4 𝑑 ‖ ‖ 2 𝐿 2 ≤ 𝐶 ‖ ∇ 𝑢 ‖ 𝐿 3 / 4 2 ‖ ‖ Λ 3 𝑢 ‖ ‖ 𝐿 1 / 4 2 ⋅ ‖ Δ 𝑑 ‖ 𝐿 1 / 3 2 ‖ ‖ Λ 5 𝑑 ‖ ‖ 𝐿 5 / 3 2 + 1 4 ‖ ‖ Λ 4 𝑢 ‖ ‖ 2 𝐿 2 + 𝐶 ‖ ∇ 𝑑 ‖ 2 𝐿 ∞ ‖ ‖ Λ 4 𝑑 ‖ ‖ 2 𝐿 2 ‖ ‖ Λ + 𝐶 4 𝑑 ‖ ‖ 2 𝐿 2 ≤ 1 4 ‖ ‖ Λ 4 𝑢 ‖ ‖ 2 𝐿 2 + 1 4 ‖ ‖ Λ 5 𝑑 ‖ ‖ 2 𝐿 2 + 𝐶 ‖ ∇ 𝑢 ‖ 𝐿 9 / 2 2 ‖ Δ 𝑑 ‖ 2 𝐿 2 ‖ ‖ Λ 3 𝑢 ‖ ‖ 𝐿 3 / 2 2 + 𝐶 ‖ Δ 𝑑 ‖ 𝐿 1 3 / 2 2 ‖ ‖ Λ 4 𝑑 ‖ ‖ 𝐿 3 / 2 2 ‖ ‖ Λ + 𝐶 4 𝑑 ‖ ‖ 2 𝐿 2 . ( 2 . 2 4 )

Summing up (2.22) and (2.24) and taking 𝜖 small enough, we arrive at ‖ 𝑢 ‖ 𝐿 ∞ ( 0 , 𝑇 ; 𝐻 3 ) + ‖ 𝑢 ‖ 𝐿 2 ( 0 , 𝑇 ; 𝐻 4 ) ≤ 𝐶 , ‖ 𝑑 ‖ 𝐿 ∞ ( 0 , 𝑇 ; 𝐻 4 ) + ‖ 𝑑 ‖ 𝐿 2 ( 0 , 𝑇 ; 𝐻 5 ) ≤ 𝐶 . ( 2 . 2 5 )

This completes the proof.

3. Proof of Theorem 1.2

In this section, we will prove Theorem 1.2. Since it is easy to prove that there are 𝑇 0 > 0 and a unique smooth solution ( 𝑢 , 𝑝 , 𝑑 ) to the problem (1.7)–(1.10) in [ 0 , 𝑇 0 ] , we only need to prove a priori estimates.

First, as in the previous section, we still have (2.1).

Testing (1.9) by − Δ 𝑑 , using 𝑑 Δ 𝑑 = − | ∇ 𝑑 | 2 and | 𝑑 | = 1 , we see that 1 2 𝑑 | | | | 𝑑 𝑡 ∇ 𝑑 2 | | | | 𝑑 𝑥 + Δ 𝑑 2 = 𝑑 𝑥 ( 𝑢 ⋅ ∇ 𝑑 ) Δ 𝑑 𝑑 𝑥 + ( 𝑑 ⋅ Δ 𝑑 ) 2 ≤ | | | | 𝑑 𝑥 ( 𝑢 ⋅ ∇ 𝑑 ) Δ 𝑑 𝑑 𝑥 + Δ 𝑑 2 𝑑 𝑥 . ( 3 . 1 )

Summing up (2.1) and (3.1), we find that 1 2 𝑢 2 + 1 2 | | | | ∇ 𝑑 2 𝑑 𝑥 + 𝑇 0 | | | | ∇ 𝑢 2 1 𝑑 𝑥 𝑑 𝑡 ≤ 2 𝑢 2 0 + | | ∇ 𝑑 0 | | 2 𝑑 𝑥 . ( 3 . 2 )

Similarly to (2.10), we have 1 2 𝑑 | | | | 𝑑 𝑡 ∇ 𝑢 2 + | | | | Δ 𝑑 2 | | | | 𝑑 𝑥 + Δ 𝑢 2 + | | | | ∇ Δ 𝑑 2 𝑑 𝑥 = ∶ 𝐼 1 + 𝐼 2 + 𝐼 3 + 𝐼 5 . ( 3 . 3 ) Here 𝐼 1 , 𝐼 2 , and 𝐼 3 are the same as that in (2.10) and can be bounded as in the previous section. The corresponding last term 𝐼 5 is written and bounded as 𝐼 5 ∶ = − 𝑘 𝜕 𝑘 | | | | ∇ 𝑑 2 𝑑 ⋅ 𝜕 𝑘 Δ 𝑑 𝑑 𝑥 = − 𝑘 𝜕 𝑘 | | | | 𝑑 ⋅ ∇ 𝑑 2 ⋅ 𝜕 𝑘 Δ 𝑑 𝑑 𝑥 − 𝑘 𝑑 𝜕 𝑘 | | | | ∇ 𝑑 2 ⋅ 𝜕 𝑘 = Δ 𝑑 𝑑 𝑥 𝑘 𝜕 𝑘 𝜕 𝑘 | | | | 𝑑 ⋅ ∇ 𝑑 2 ⋅ Δ 𝑑 𝑑 𝑥 − 𝑘 𝑑 𝜕 𝑘 | | | | ∇ 𝑑 2 ⋅ 𝜕 𝑘 Δ 𝑑 𝑑 𝑥 ≤ 𝐶 ‖ ∇ 𝑑 ‖ 2 𝐿 4 ‖ Δ 𝑑 ‖ 2 𝐿 4 + 1 4 ‖ ∇ Δ 𝑑 ‖ 2 𝐿 2 ≤ 𝐶 ‖ Δ 𝑑 ‖ 𝐿 2 ⋅ ‖ ∇ 𝑑 ‖ ̇ 𝐵 0 ∞ , ∞ ‖ ∇ Δ 𝑑 ‖ 𝐿 2 + 1 4 ‖ ∇ Δ 𝑑 ‖ 2 𝐿 2 ≤ 1 2 ‖ ∇ Δ 𝑑 ‖ 2 𝐿 2 + 𝐶 ‖ ∇ 𝑑 ‖ 2 ̇ 𝐵 0 ∞ , ∞ ‖ Δ 𝑑 ‖ 2 𝐿 2 . ( 3 . 4 ) Here we have used the following inequality [11, 12]: ‖ Δ 𝑑 ‖ 2 𝐿 4 ≤ 𝐶 ‖ ∇ 𝑑 ‖ ̇ 𝐵 0 ∞ , ∞ ‖ ∇ Δ 𝑑 ‖ 𝐿 2 ( 3 . 5 ) and the Gagliardo-Nirenberg inequality ‖ ∇ 𝑑 ‖ 2 𝐿 4 ≤ 𝐶 ‖ 𝑑 ‖ 𝐿 ∞ ‖ Δ 𝑑 ‖ 𝐿 2 . ( 3 . 6 )

Substituting the above estimates into (3.3), we obtain 𝑑 | | | | 𝑑 𝑡 ∇ 𝑢 2 + | | | | Δ 𝑑 2 𝑑 𝑥 ≤ 𝐶 ‖ 𝑢 ‖ ̇ 𝐵 2 / ( 1 − 𝑠 ) − 𝑠 ∞ , ∞ + ‖ ∇ 𝑑 ‖ 2 ̇ 𝐵 0 ∞ , ∞ ‖ ∇ 𝑢 ‖ 2 𝐿 2 + ‖ Δ 𝑑 ‖ 2 𝐿 2 ≤ 𝐶 ‖ 𝑢 ‖ ̇ 𝐵 2 / ( 1 − 𝑠 ) − 𝑠 ∞ , ∞ + ‖ ∇ 𝑑 ‖ 2 ̇ 𝐵 0 ∞ , ∞ 1 + l o g 𝑒 + ‖ 𝑢 ‖ ̇ 𝐵 − 𝑠 ∞ , ∞ + ‖ ∇ 𝑑 ‖ ̇ 𝐵 0 ∞ , ∞ ( 1 + l o g ( 𝑒 + 𝑦 ) ) ‖ ∇ 𝑢 ‖ 2 𝐿 2 + ‖ Δ 𝑑 ‖ 2 𝐿 2 . ( 3 . 7 )

Due to (1.12), one concludes that for any small constant 𝜖 > 0 , there exists 𝑇 ∗ < 𝑇 such that 𝑇 𝑇 ∗ ‖ 𝑢 ‖ ̇ 𝐵 2 / ( 1 − 𝑠 ) − 𝑠 ∞ , ∞ + ‖ ∇ 𝑑 ‖ 2 ̇ 𝐵 0 ∞ , ∞ 1 + l o g 𝑒 + ‖ 𝑢 ‖ ̇ 𝐵 − 𝑠 ∞ , ∞ + ‖ ∇ 𝑑 ‖ ̇ 𝐵 0 ∞ , ∞ 𝑑 𝑡 < 𝜖 . ( 3 . 8 )

Applying Gronwall’s inequality to (3.7) in the interval [ 𝑇 ∗ , 𝑡 ] , one has (2.19).

As in the previous section, we still have (2.22).

Similarly to (2.24), we obtain 1 2 𝑑 | | Λ 𝑑 𝑡 4 𝑑 | | 2 | | Λ 𝑑 𝑥 + 5 𝑑 | | 2 Λ 𝑑 𝑥 = − 4 ( 𝑢 ⋅ ∇ 𝑑 ) − 𝑢 ∇ Λ 4 𝑑 Λ 4 Λ 𝑑 𝑑 𝑥 + 4 𝑑 | | | | ∇ 𝑑 2 Λ 4 𝑑 𝑑 𝑥 = ∶ 𝐽 1 + 𝐽 2 . ( 3 . 9 ) 𝐽 1 is bounded as that in (2.24); 𝐽 2 ≤ 𝐶 ‖ ∇ 𝑑 ‖ 𝐿 ∞ ‖ ‖ Λ 4 𝑑 ‖ ‖ 2 𝐿 2 ‖ ‖ Λ + 𝐶 4 | | | | ∇ 𝑑 2 ‖ ‖ 𝐿 2 ‖ ‖ Λ 4 𝑑 ‖ ‖ 𝐿 2 , ≤ 𝐶 ‖ ∇ 𝑑 ‖ 2 𝐿 ∞ ‖ ‖ Λ 4 𝑑 ‖ ‖ 2 𝐿 2 + 𝐶 ‖ ∇ 𝑑 ‖ 𝐿 ∞ ‖ ‖ Λ 5 𝑑 ‖ ‖ 𝐿 2 ‖ ‖ Λ 4 𝑑 ‖ ‖ 𝐿 2 , ≤ 1 4 ‖ ‖ Λ 5 𝑑 ‖ ‖ 2 𝐿 2 + 𝐶 ‖ ∇ 𝑑 ‖ 2 𝐿 ∞ ‖ ‖ Λ 4 𝑑 ‖ ‖ 2 𝐿 2 , ( 3 . 1 0 ) then 𝐽 2 can be bounded as that in (2.24).

Combining (2.22) and (3.9) and taking 𝜖 small enough, we conclude that ‖ 𝑢 ‖ 𝐿 ∞ ( 0 , 𝑇 ; 𝐻 3 ) + ‖ 𝑢 ‖ 𝐿 2 ( 0 , 𝑇 ; 𝐻 4 ) ≤ 𝐶 , ‖ ∇ 𝑑 ‖ 𝐿 ∞ ( 0 , 𝑇 ; 𝐻 3 ) + ‖ ∇ 𝑑 ‖ 𝐿 2 ( 0 , 𝑇 ; 𝐻 4 ) ≤ 𝐶 . ( 3 . 1 1 ) This completes the proof.

Acknowledgment

The paper is supported by NSFC (no. 11171154).

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