- Email: [email protected]
Uniform regularity for the 2D Boussinesq system with a slip boundary condition
J. Math. Anal. Appl. 400 (2013) 96–99
Contents lists available at SciVerse ScienceDirect
Journal of Mathematical Analysis and Applications journal homepage: www.elsevier.com/locate/jmaa
Uniform regularity for the 2D Boussinesq system with a slip boundary condition Liangbing Jin a,∗ , Jishan Fan b a
Department of Mathematics, Zhejiang Normal University, Jinhua 321004, PR China
b
Department of Applied Mathematics, Nanjing Forestry University, Nanjing 210037, PR China
article
abstract
info
Article history: Received 19 October 2011 Available online 6 November 2012 Submitted by Dehua Wang
We prove a global-in-time regularity which is uniform in the vanishing viscosity limit for the 2D Boussinesq system in a domain Ω := (−∞, ∞) × (0, 1) with a slip boundary condition. © 2012 Elsevier Inc. All rights reserved.
Keywords: Boussinesq system Uniform regularity Slip boundary condition
1. Introduction Let Ω := (−∞, ∞) × (0, 1) and ν := Boussinesq system in Ω × (0, ∞):
ν1 ν2
bethe unit outward vector normal to ∂ Ω . We consider the following 2D
∂t u + (u · ∇)u + ∇π − ϵ 1u = θ e2 , div u = 0, ∂t θ + u · ∇θ = 1θ , u · ν = curl u = 0, θ = 0 on ∂ Ω × (0, ∞),
(1.1) (1.2) (1.3) (1.4)
(u, θ )(x, 0) = (u0 , θ0 )(x), x ∈ Ω ⊆ R . (1.5) The unknowns u, π and θ denote the velocity, pressure and temperature of the fluid, respectively. ϵ > 0 is the viscosity, and e2 := (0, 1)t · ω := curl u := ∂1 u2 − ∂2 u1 is the vorticity. The aim of this paper is to study the global-in-time uniform regularity for the problem (1.1)–(1.5). For Ω := R2 , the 2
problem has been studied by Chae [1], but his method did not work here, because if we use his method, then there are many boundary integral terms which are difficult to deal with. The investigation of the inviscid limit of the Navier–Stokes equations is a classical issue. We refer the reader to the papers [2–7] for Ω a bounded domain for the Navier–Stokes equations. Here the domain Ω := (−∞, ∞) × (0, 1) is unbounded, and it seems that the methods in [2–7] cannot be used directly. We will use a well-known logarithmic Sobolev inequality in [8–10] to complete our proof. We will prove: Theorem 1.1. Let u0 ∈ H 3 , div u0 = 0 in Ω , u0 · ν = curl u0 = 0 on ∂ Ω , and θ0 ∈ H01 ∩ H 2 . Then there exists a positive constant C independent of ϵ such that
∥uϵ ∥L∞ (0,T ;H 3 ) ≤ C ,
∥θϵ ∥L∞ (0,T ;H 2 )∩L2 (0,T ;H 3 ) ≤ C ,
for any T > 0.
∗
Corresponding author. E-mail addresses: [email protected] (L. Jin), [email protected] (J. Fan).
0022-247X/$ – see front matter © 2012 Elsevier Inc. All rights reserved. doi:10.1016/j.jmaa.2012.10.051
∥∂t (uϵ , θϵ )∥L2 (0,T ;L2 ) ≤ C
(1.6)
L. Jin, J. Fan / J. Math. Anal. Appl. 400 (2013) 96–99
97
Remark 1.1. We take Ω := (−∞, ∞) × (0, 1) to ensure that (2.17) holds true, which leads to (2.18) and (2.19). Then we can finish our proof. We cannot prove a similar result for an Ω of another type, say Ω := (0, 1) × (−∞, ∞). In the following calculations, we will use the following logarithmic Sobolev inequality [8–10]:
∥∇ u∥L∞ ≤ C (1 + ∥curl u∥L∞ log(e + ∥u∥H 3 )),
(1.7)
for u ∈ H 3 , div u = 0 in Ω and u · ν = 0 on ∂ Ω , and the following inequality [7,11]:
∥u∥H s ≤ C (∥u∥L2 + ∥ω∥H s−1 ),
(1.8)
for s ≥ 1, u ∈ H with div u = 0 in Ω and u · ν = 0 on ∂ Ω . s
2. Proof of Theorem 1.1 Since it is well-known that the solution (uϵ , θϵ ) is smooth [12,13], we only need to show the estimates (1.6). From now on, we will drop the subscript ϵ and throughout this paper C will be a constant independent of ϵ > 0. First, by the maximum principle, it follows from (1.3)–(1.5) that
∥θ ∥L∞ (0,T ;L∞ ) ≤ C .
(2.1)
Testing (1.3) by θ , using (1.2), (1.4) and (1.5), we see that 1 d
2 dt
θ dx + 2
|∇θ |2 dx = 0,
whence
∥θ ∥L∞ (0,T ;L2 ) + ∥θ∥L2 (0,T ;H 1 ) ≤ C .
(2.2)
Testing (1.1) by u, using (1.2), (1.4), (1.5) and (2.2), we find that 1 d
2 dt
u2 dx + ϵ
|∇ u|2 dx =
θ e2 udx ≤ ∥θ ∥L2 ∥u∥L2 ≤ C ∥u∥L2 ,
which gives
∥u∥L∞ (0,T ;L2 ) ≤ C .
(2.3)
Taking curl to (1.1), using (1.2), we infer that
∂t ω + u · ∇ω − ϵ 1ω = curl (θ e2 ) := ∂1 θ .
(2.4)
Testing (2.4) with ω, using (1.2), (1.4), (1.5) and (2.2), we get 1 d
2 dt
w 2 dx + ϵ
|∇ω|2 dx =
∂1 θ · ωdx ≤ ∥∂1 θ∥L2 ∥ω∥L2 ,
which yields
∥∇ u∥L∞ (0,T ;L2 ) ≤ C ∥ω∥L∞ (0,T ;L2 ) ≤ C , √ ϵ∥u∥L2 (0,T ;H 2 ) ≤ C .
(2.5) (2.6)
Testing (1.3) with −1θ , using (1.2), (1.4), (1.5) and (2.5), we derive 1 d 2 dt
|∇θ |2 dx +
|1θ|2 dx =
u · ∇θ · 1θ dx
≤ ∥u∥L4 ∥∇θ ∥L4 ∥1θ∥L2 ≤ C ∥∇θ∥L4 ∥1θ ∥L2 1/2
3/2
≤ C ∥∇θ∥L2 ∥1θ∥L2 + C ∥∇θ∥L2 ∥1θ ∥L2 ≤
1 2
∥1θ ∥2L2 + C ∥∇θ ∥2L2 ,
which leads to
∥∇θ ∥L∞ (0,T ;L2 ) ≤ C ,
(2.7)
∥θ ∥L2 (0,T ;H 2 ) ≤ C .
(2.8)
Here we have used the Gagliardo–Nirenberg inequality:
∥∇θ ∥2L4 ≤ C ∥∇θ∥L2 ∥θ ∥H 2 .
98
L. Jin, J. Fan / J. Math. Anal. Appl. 400 (2013) 96–99
Testing (2.4) with |ω|m−2 ω (m ≥ 2), using (1.2), (1.4), (1.5) and (2.8), we deduce that 1 d
m dt
|ω|m dx ≤
−1 ∂1 θ · |ω|m−2 ωdx ≤ ∥∂1 θ∥Lm ∥ω∥m , Lm
whence d dt
∥ω∥Lm ≤ ∥∂1 θ ∥Lm ≤ C ∥θ∥H 2 ,
(2.9)
which implies
∥∇ u∥L∞ (0,T ;Lm ) ≤ C ∥ω∥L∞ (0,T ;Lm ) ≤ C .
(2.10)
Testing (1.1) with ∂t u, using (1.2), (1.4), (2.2), (2.6) and (2.10), we obtain
∥∂t u∥L2 (0,T ;L2 ) ≤ ∥θ ∥L2 (0,T ;L2 ) + ϵ∥1u∥L2 (0,T ;L2 ) + ∥u · ∇ u∥L2 (0,T ;L2 ) √ ≤ C + C ϵ + C ∥u∥L∞ (0,T ;L4 ) ∥∇ u∥L∞ (0,T ;L4 ) ≤ C.
(2.11)
Applying ∂t to (1.3), we see that
∂t2 θ + u · ∇∂t θ − 1∂t θ = −∂t u · ∇θ . Testing the above equation with ∂t θ , using (1.2), (1.4), (2.1) and (2.11), we have 1 d
2 dt
|∂t θ|2 dx +
|∇∂t θ |2 dx =
∂t u · θ · ∇∂t θ dx
≤ ∥θ ∥L∞ ∥∂t u∥L2 ∥∇∂t θ∥L2 ≤ C ∥∂t u∥L2 ∥∇∂t θ ∥L2 ≤
1 2
∥∇∂t θ∥2L2 + C ∥∂t u∥2L2 ,
which implies
∥∂t θ ∥L∞ (0,T ;L2 ) + ∥∂t θ∥L2 (0,T ;H 1 ) ≤ C .
(2.12)
Eqs. (1.3) and (1.4) can be rewritten as
1θ = f := ∂t θ + u · ∇θ θ =0
in Ω × (0, ∞), on ∂ Ω × (0, ∞),
and then by the elliptic regularity theory, using (2.12) and (2.5), we have
∥θ ∥L∞ (0,T ;H 2 ) ≤ C ∥f ∥L∞ (0,T ;L2 ) ≤ C ∥∂t θ ∥L∞ (0,T ;L2 ) + C ∥u∥L∞ (0,T ;L4 ) ∥∇θ ∥L∞ (0,T ;L4 ) 1/2
1/2
≤ C + C ∥∇θ ∥L∞ (0,T ;L2 ) ∥θ ∥L∞ (0,T ;H 2 ) , which implies
∥θ ∥L∞ (0,T ;H 2 ) ≤ C .
(2.13)
Similarly, using (2.12), (2.13) and (2.10), we have
∥θ ∥L2 (0,T ;H 3 ) ≤ C ∥f ∥L2 (0,T ;H 1 ) ≤ C ∥f ∥L2 (0,T ;L2 ) + C ∥∇ f ∥L2 (0,T ;L2 ) ≤ C + C ∥∇∂t θ∥L2 (0,T ;L2 ) + C ∥∇ u · ∇θ ∥L2 (0,T ;L2 ) + C ∥u∇ 2 θ ∥L2 (0,T ;L2 ) ≤ C + C ∥∇ u∥L∞ (0,T ;L4 ) ∥∇θ ∥L∞ (0,T ;L4 ) + C ∥u∥L∞ (0,T ;L∞ ) ∥1θ ∥L2 (0,T ;L2 ) ≤ C,
(2.14)
which yields
∥∇θ ∥L2 (0,T ;L∞ ) ≤ C .
(2.15)
Taking m → ∞ in (2.9), we arrive at the key estimate
∥ω∥L∞ (0,T ;L∞ ) ≤ C .
(2.16)
L. Jin, J. Fan / J. Math. Anal. Appl. 400 (2013) 96–99
99
Since Ω := (−∞, ∞) × (0, 1), it is easy to find that
∂1 θ = 0 on ∂ Ω × (0, ∞). ν Let τ := −ν2 be any unit vector tangential to ∂ Ω ; using (1.4), we see that 1 u · ∇ω = ((u · τ )τ + (u · ν)ν)∇ω = (u · τ )τ · ∇ω = (u · τ )
(2.17)
∂ω = 0 on ∂ Ω × (0, ∞). ∂τ
(2.18)
Eqs. (2.4), (2.17), (2.18) and (1.4) give
1ω = 0 on ∂ Ω × (0, ∞).
(2.19)
Applying ∆ to (2.4), and testing with 1ω, using (1.2), (1.4), (2.19), (2.14), (1.7), (1.8) and (2.16), we conclude that 1 d
2 dt
|1ω|2 dx + ϵ
|∇ 1ω|2 dx =
1∂1 θ 1ωdx −
(∆(u · ∇ω) − u∇ 1ω)1ωdx ≤ ∥1∂1 θ∥L2 ∥1ω∥L2 − 1u · ∇ω · 1ωdx − 2 ∂i u∂i ∇ω1ωdx i
≤ ∥1∂1 θ∥L2 ∥1ω∥L2 + ∥1ω∥L2 ∥∇ω∥2L4 + C ∥∇ u∥L∞ ∥1ω∥2L2 ≤ ∥1∂1 θ∥L2 ∥1ω∥L2 + ∥1ω∥L2 ∥ω∥L∞ ∥ω∥H 2 + C ∥∇ u∥L∞ ∥1ω∥2L2 ≤ ∥1∂1 θ∥L2 ∥1ω∥L2 + ∥1ω∥L2 ∥ω∥H 2 + C ∥∇ u∥L∞ ∥1ω∥2L2 ≤ ∥1∂1 θ ∥L2 ∥1ω∥L2 + C ∥1ω∥L2 ∥ω∥H 2 + C (1 + log(e + ∥1ω∥L2 ))∥1ω∥2L2 , which gives
∥1ω∥L∞ (0,T ;L2 ) ≤ C , and thus
∥u∥L∞ (0,T ;H 3 ) ≤ C . Here we have used the Gagliardo–Nirenberg inequality:
∥∇ω∥2L4 ≤ C ∥ω∥L∞ ∥ω∥H 2 . This completes the proof.
Acknowledgments We thank the referee for pointing out some typos and making helpful suggestions. This paper was supported by NSFC (Grant No. 11171154 and 11226176), Zhejiang Innovation Project (T200905) and the Scientific Research Fund of Zhejiang Provincial Education Department (Grant No. Y201226095). References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11] [12] [13]
D. Chae, Global regularity for the 2D Boussinesq equations with partial viscosity terms, Adv. Math. 203 (2006) 497–513. H. Beirão da Veiga, F. Crispo, Sharp inviscid limit results under Navier type boundary conditions. An Lp theory, J. Math. Fluid Mech. 12 (2010) 397–411. H. Beirão da Veiga, F. Crispo, Concerning the W k,p -inviscid limit for 3-D flows under a slip boundary condition, J. Math. Fluid Mech. 13 (2011) 117–135. T. Clopeau, A. Mikelić, R. Robert, On the vanishing viscosity limit for the 2D incompressible Navier–Stokes equations with the friction type boundary conditions, Nonlinearity 11 (1998) 1625–1636. D. Iftimie, G. Planas, Inviscid limits for the Navier–Stokes equations with Navier friction boundary conditions, Nonlinearity 19 (2006) 899–918. P.L. Lions, Mathematical Topics in Fluid Mechanics, Vol. 1, Oxford University Press, New York, 1996. Y.L. Xiao, Z.P. Xin, On the vanishing viscosity limit for the 3D Navier–Stokes equations with a slip boundary condition, Comm. Pure Appl. Math. 60 (2007) 1027–1055. A.B. Ferrari, On the blow-up of solutions of 3-D Euler equations in a bounded domain, Comm. Math. Phys. 155 (1993) 277–294. T. Shirota, T. Yanagisawa, A continuation principle for the 3D Euler equations for incompressible fluids in a bounded domain, Proc. Japan Acad. Ser. A 69 (1993) 77–82. T. Ogawa, Y. Taniuchi, On blow-up criteria of smooth solutions to the 3-D Euler equations in a bounded domain, J. Differential Equations 190 (2003) 39–63. J.P. Bourguignon, H. Brezis, Remarks on the Euler equation, J. Funct. Anal. 15 (1974) 341–363. J.R. Cannon, E. Di Benedetto, The initial value problem for the Boussinesq equations with data in Lp , in: Approximation Methods for Navier–Stokes Problems, Paderborn, 1979, in: Lecture Notes in Math., vol. 771, Springer-Verlag, Berlin, 1980, pp. 129–144. S.A. Lorca, J.L. Boldrini, The initial value problem for a generalized Boussinesq model, Nonlinear Anal. 36 (1999) 457–480.
Contents lists available at SciVerse ScienceDirect
Journal of Mathematical Analysis and Applications journal homepage: www.elsevier.com/locate/jmaa
Uniform regularity for the 2D Boussinesq system with a slip boundary condition Liangbing Jin a,∗ , Jishan Fan b a
Department of Mathematics, Zhejiang Normal University, Jinhua 321004, PR China
b
Department of Applied Mathematics, Nanjing Forestry University, Nanjing 210037, PR China
article
abstract
info
Article history: Received 19 October 2011 Available online 6 November 2012 Submitted by Dehua Wang
We prove a global-in-time regularity which is uniform in the vanishing viscosity limit for the 2D Boussinesq system in a domain Ω := (−∞, ∞) × (0, 1) with a slip boundary condition. © 2012 Elsevier Inc. All rights reserved.
Keywords: Boussinesq system Uniform regularity Slip boundary condition
1. Introduction Let Ω := (−∞, ∞) × (0, 1) and ν := Boussinesq system in Ω × (0, ∞):
ν1 ν2
bethe unit outward vector normal to ∂ Ω . We consider the following 2D
∂t u + (u · ∇)u + ∇π − ϵ 1u = θ e2 , div u = 0, ∂t θ + u · ∇θ = 1θ , u · ν = curl u = 0, θ = 0 on ∂ Ω × (0, ∞),
(1.1) (1.2) (1.3) (1.4)
(u, θ )(x, 0) = (u0 , θ0 )(x), x ∈ Ω ⊆ R . (1.5) The unknowns u, π and θ denote the velocity, pressure and temperature of the fluid, respectively. ϵ > 0 is the viscosity, and e2 := (0, 1)t · ω := curl u := ∂1 u2 − ∂2 u1 is the vorticity. The aim of this paper is to study the global-in-time uniform regularity for the problem (1.1)–(1.5). For Ω := R2 , the 2
problem has been studied by Chae [1], but his method did not work here, because if we use his method, then there are many boundary integral terms which are difficult to deal with. The investigation of the inviscid limit of the Navier–Stokes equations is a classical issue. We refer the reader to the papers [2–7] for Ω a bounded domain for the Navier–Stokes equations. Here the domain Ω := (−∞, ∞) × (0, 1) is unbounded, and it seems that the methods in [2–7] cannot be used directly. We will use a well-known logarithmic Sobolev inequality in [8–10] to complete our proof. We will prove: Theorem 1.1. Let u0 ∈ H 3 , div u0 = 0 in Ω , u0 · ν = curl u0 = 0 on ∂ Ω , and θ0 ∈ H01 ∩ H 2 . Then there exists a positive constant C independent of ϵ such that
∥uϵ ∥L∞ (0,T ;H 3 ) ≤ C ,
∥θϵ ∥L∞ (0,T ;H 2 )∩L2 (0,T ;H 3 ) ≤ C ,
for any T > 0.
∗
Corresponding author. E-mail addresses: [email protected] (L. Jin), [email protected] (J. Fan).
0022-247X/$ – see front matter © 2012 Elsevier Inc. All rights reserved. doi:10.1016/j.jmaa.2012.10.051
∥∂t (uϵ , θϵ )∥L2 (0,T ;L2 ) ≤ C
(1.6)
L. Jin, J. Fan / J. Math. Anal. Appl. 400 (2013) 96–99
97
Remark 1.1. We take Ω := (−∞, ∞) × (0, 1) to ensure that (2.17) holds true, which leads to (2.18) and (2.19). Then we can finish our proof. We cannot prove a similar result for an Ω of another type, say Ω := (0, 1) × (−∞, ∞). In the following calculations, we will use the following logarithmic Sobolev inequality [8–10]:
∥∇ u∥L∞ ≤ C (1 + ∥curl u∥L∞ log(e + ∥u∥H 3 )),
(1.7)
for u ∈ H 3 , div u = 0 in Ω and u · ν = 0 on ∂ Ω , and the following inequality [7,11]:
∥u∥H s ≤ C (∥u∥L2 + ∥ω∥H s−1 ),
(1.8)
for s ≥ 1, u ∈ H with div u = 0 in Ω and u · ν = 0 on ∂ Ω . s
2. Proof of Theorem 1.1 Since it is well-known that the solution (uϵ , θϵ ) is smooth [12,13], we only need to show the estimates (1.6). From now on, we will drop the subscript ϵ and throughout this paper C will be a constant independent of ϵ > 0. First, by the maximum principle, it follows from (1.3)–(1.5) that
∥θ ∥L∞ (0,T ;L∞ ) ≤ C .
(2.1)
Testing (1.3) by θ , using (1.2), (1.4) and (1.5), we see that 1 d
2 dt
θ dx + 2
|∇θ |2 dx = 0,
whence
∥θ ∥L∞ (0,T ;L2 ) + ∥θ∥L2 (0,T ;H 1 ) ≤ C .
(2.2)
Testing (1.1) by u, using (1.2), (1.4), (1.5) and (2.2), we find that 1 d
2 dt
u2 dx + ϵ
|∇ u|2 dx =
θ e2 udx ≤ ∥θ ∥L2 ∥u∥L2 ≤ C ∥u∥L2 ,
which gives
∥u∥L∞ (0,T ;L2 ) ≤ C .
(2.3)
Taking curl to (1.1), using (1.2), we infer that
∂t ω + u · ∇ω − ϵ 1ω = curl (θ e2 ) := ∂1 θ .
(2.4)
Testing (2.4) with ω, using (1.2), (1.4), (1.5) and (2.2), we get 1 d
2 dt
w 2 dx + ϵ
|∇ω|2 dx =
∂1 θ · ωdx ≤ ∥∂1 θ∥L2 ∥ω∥L2 ,
which yields
∥∇ u∥L∞ (0,T ;L2 ) ≤ C ∥ω∥L∞ (0,T ;L2 ) ≤ C , √ ϵ∥u∥L2 (0,T ;H 2 ) ≤ C .
(2.5) (2.6)
Testing (1.3) with −1θ , using (1.2), (1.4), (1.5) and (2.5), we derive 1 d 2 dt
|∇θ |2 dx +
|1θ|2 dx =
u · ∇θ · 1θ dx
≤ ∥u∥L4 ∥∇θ ∥L4 ∥1θ∥L2 ≤ C ∥∇θ∥L4 ∥1θ ∥L2 1/2
3/2
≤ C ∥∇θ∥L2 ∥1θ∥L2 + C ∥∇θ∥L2 ∥1θ ∥L2 ≤
1 2
∥1θ ∥2L2 + C ∥∇θ ∥2L2 ,
which leads to
∥∇θ ∥L∞ (0,T ;L2 ) ≤ C ,
(2.7)
∥θ ∥L2 (0,T ;H 2 ) ≤ C .
(2.8)
Here we have used the Gagliardo–Nirenberg inequality:
∥∇θ ∥2L4 ≤ C ∥∇θ∥L2 ∥θ ∥H 2 .
98
L. Jin, J. Fan / J. Math. Anal. Appl. 400 (2013) 96–99
Testing (2.4) with |ω|m−2 ω (m ≥ 2), using (1.2), (1.4), (1.5) and (2.8), we deduce that 1 d
m dt
|ω|m dx ≤
−1 ∂1 θ · |ω|m−2 ωdx ≤ ∥∂1 θ∥Lm ∥ω∥m , Lm
whence d dt
∥ω∥Lm ≤ ∥∂1 θ ∥Lm ≤ C ∥θ∥H 2 ,
(2.9)
which implies
∥∇ u∥L∞ (0,T ;Lm ) ≤ C ∥ω∥L∞ (0,T ;Lm ) ≤ C .
(2.10)
Testing (1.1) with ∂t u, using (1.2), (1.4), (2.2), (2.6) and (2.10), we obtain
∥∂t u∥L2 (0,T ;L2 ) ≤ ∥θ ∥L2 (0,T ;L2 ) + ϵ∥1u∥L2 (0,T ;L2 ) + ∥u · ∇ u∥L2 (0,T ;L2 ) √ ≤ C + C ϵ + C ∥u∥L∞ (0,T ;L4 ) ∥∇ u∥L∞ (0,T ;L4 ) ≤ C.
(2.11)
Applying ∂t to (1.3), we see that
∂t2 θ + u · ∇∂t θ − 1∂t θ = −∂t u · ∇θ . Testing the above equation with ∂t θ , using (1.2), (1.4), (2.1) and (2.11), we have 1 d
2 dt
|∂t θ|2 dx +
|∇∂t θ |2 dx =
∂t u · θ · ∇∂t θ dx
≤ ∥θ ∥L∞ ∥∂t u∥L2 ∥∇∂t θ∥L2 ≤ C ∥∂t u∥L2 ∥∇∂t θ ∥L2 ≤
1 2
∥∇∂t θ∥2L2 + C ∥∂t u∥2L2 ,
which implies
∥∂t θ ∥L∞ (0,T ;L2 ) + ∥∂t θ∥L2 (0,T ;H 1 ) ≤ C .
(2.12)
Eqs. (1.3) and (1.4) can be rewritten as
1θ = f := ∂t θ + u · ∇θ θ =0
in Ω × (0, ∞), on ∂ Ω × (0, ∞),
and then by the elliptic regularity theory, using (2.12) and (2.5), we have
∥θ ∥L∞ (0,T ;H 2 ) ≤ C ∥f ∥L∞ (0,T ;L2 ) ≤ C ∥∂t θ ∥L∞ (0,T ;L2 ) + C ∥u∥L∞ (0,T ;L4 ) ∥∇θ ∥L∞ (0,T ;L4 ) 1/2
1/2
≤ C + C ∥∇θ ∥L∞ (0,T ;L2 ) ∥θ ∥L∞ (0,T ;H 2 ) , which implies
∥θ ∥L∞ (0,T ;H 2 ) ≤ C .
(2.13)
Similarly, using (2.12), (2.13) and (2.10), we have
∥θ ∥L2 (0,T ;H 3 ) ≤ C ∥f ∥L2 (0,T ;H 1 ) ≤ C ∥f ∥L2 (0,T ;L2 ) + C ∥∇ f ∥L2 (0,T ;L2 ) ≤ C + C ∥∇∂t θ∥L2 (0,T ;L2 ) + C ∥∇ u · ∇θ ∥L2 (0,T ;L2 ) + C ∥u∇ 2 θ ∥L2 (0,T ;L2 ) ≤ C + C ∥∇ u∥L∞ (0,T ;L4 ) ∥∇θ ∥L∞ (0,T ;L4 ) + C ∥u∥L∞ (0,T ;L∞ ) ∥1θ ∥L2 (0,T ;L2 ) ≤ C,
(2.14)
which yields
∥∇θ ∥L2 (0,T ;L∞ ) ≤ C .
(2.15)
Taking m → ∞ in (2.9), we arrive at the key estimate
∥ω∥L∞ (0,T ;L∞ ) ≤ C .
(2.16)
L. Jin, J. Fan / J. Math. Anal. Appl. 400 (2013) 96–99
99
Since Ω := (−∞, ∞) × (0, 1), it is easy to find that
∂1 θ = 0 on ∂ Ω × (0, ∞). ν Let τ := −ν2 be any unit vector tangential to ∂ Ω ; using (1.4), we see that 1 u · ∇ω = ((u · τ )τ + (u · ν)ν)∇ω = (u · τ )τ · ∇ω = (u · τ )
(2.17)
∂ω = 0 on ∂ Ω × (0, ∞). ∂τ
(2.18)
Eqs. (2.4), (2.17), (2.18) and (1.4) give
1ω = 0 on ∂ Ω × (0, ∞).
(2.19)
Applying ∆ to (2.4), and testing with 1ω, using (1.2), (1.4), (2.19), (2.14), (1.7), (1.8) and (2.16), we conclude that 1 d
2 dt
|1ω|2 dx + ϵ
|∇ 1ω|2 dx =
1∂1 θ 1ωdx −
(∆(u · ∇ω) − u∇ 1ω)1ωdx ≤ ∥1∂1 θ∥L2 ∥1ω∥L2 − 1u · ∇ω · 1ωdx − 2 ∂i u∂i ∇ω1ωdx i
≤ ∥1∂1 θ∥L2 ∥1ω∥L2 + ∥1ω∥L2 ∥∇ω∥2L4 + C ∥∇ u∥L∞ ∥1ω∥2L2 ≤ ∥1∂1 θ∥L2 ∥1ω∥L2 + ∥1ω∥L2 ∥ω∥L∞ ∥ω∥H 2 + C ∥∇ u∥L∞ ∥1ω∥2L2 ≤ ∥1∂1 θ∥L2 ∥1ω∥L2 + ∥1ω∥L2 ∥ω∥H 2 + C ∥∇ u∥L∞ ∥1ω∥2L2 ≤ ∥1∂1 θ ∥L2 ∥1ω∥L2 + C ∥1ω∥L2 ∥ω∥H 2 + C (1 + log(e + ∥1ω∥L2 ))∥1ω∥2L2 , which gives
∥1ω∥L∞ (0,T ;L2 ) ≤ C , and thus
∥u∥L∞ (0,T ;H 3 ) ≤ C . Here we have used the Gagliardo–Nirenberg inequality:
∥∇ω∥2L4 ≤ C ∥ω∥L∞ ∥ω∥H 2 . This completes the proof.
Acknowledgments We thank the referee for pointing out some typos and making helpful suggestions. This paper was supported by NSFC (Grant No. 11171154 and 11226176), Zhejiang Innovation Project (T200905) and the Scientific Research Fund of Zhejiang Provincial Education Department (Grant No. Y201226095). References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11] [12] [13]
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