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Partial vanishing viscosity limit for the 2D Boussinesq system with a slip boundary condition Boundary Value Problems 2012, 2012:20 doi:10.1186/1687-2770-2012-20 Liangbing Jin lbjin@zjnu

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Partial vanishing viscosity limit for the 2D Boussinesq system with a slip

boundary condition

Boundary Value Problems 2012, 2012:20 doi:10.1186/1687-2770-2012-20

Liangbing Jin (lbjin@zjnu.edu.cn) Jishan Fan (fanjishan@njfu.com.cn) Gen Nakamura (gnaka@math.sci.hokudai.ac.jp) Yong Zhou (yzhoumath@zjnu.edu.cn)

ISSN 1687-2770

Article type Research

Submission date 12 November 2011

Acceptance date 15 February 2012

Publication date 15 February 2012

Article URL http://www.boundaryvalueproblems.com/content/2012/1/20

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Partial vanishing viscosity limit for the 2D Boussinesq system with a slip boundary

condition

1Department of Mathematics, Zhejiang Normal University,

Jinhua 321004, P R China

2Department of Applied Mathematics, Nanjing Forestry University,

Nanjing 210037, P.R China

3Department of Mathematics, Hokkaido University

Sapporo 060-0810, Japan

∗Corresponding author: yzhoumath@zjnu.edu.cn

Email addresses:

LJ: lbjin@zjnu.edu.cn GN: gnaka@math.sci.hokudai.ac.jp JF: fanjishan@njfu.com.cn

Abstract

This article studies the partial vanishing viscosity limit of the 2D Boussinesq system in a bounded domain with a slip boundary condition The result is proved globally in time by a logarithmic Sobolev inequality

2010 MSC: 35Q30; 76D03; 76D05; 76D07

Keywords: Boussinesq system; inviscid limit; slip boundary condition

1 Introduction

Let Ω ⊂ R2 be a bounded, simply connected domain with smooth boundary ∂Ω, and n is the unit outward normal vector to ∂Ω We consider the Boussinesq system in Ω × (0, ∞):

∂ t u + u · ∇u + ∇π − ∆u = θe2, (1.1)

∂ t θ + u · ∇θ = ²∆θ, (1.3)

u · n = 0, curlu = 0, θ = 0, on ∂Ω × (0, ∞), (1.4)

(u, θ)(x, 0) = (u0, θ0)(x), x ∈ Ω, (1.5)

where u, π, and θ denote unknown velocity vector field, pressure scalar and temperature

of the fluid ² > 0 is the heat conductivity coefficient and e2 := (0, 1) t ω := curlu :=

∂1u2− ∂2u1 is the vorticity

The aim of this article is to study the partial vanishing viscosity limit ² → 0 When

Ω := R2, the problem has been solved by Chae [1] When θ = 0, the Boussinesq system

reduces to the well-known Navier–Stokes equations The investigation of the inviscid limit

of solutions of the Navier–Stokes equations is a classical issue We refer to the articles [2–7] when Ω is a bounded domain However, the methods in [1–6] could not be used here directly

We will use a well-known logarithmic Sobolev inequality in [8,9] to complete our proof We will prove:

Theorem 1.1 Let u0 ∈ H3, divu0 = 0 in Ω, u0· n = 0, curlu0 = 0 on ∂Ω and θ0 ∈ H1

0∩ H2 Then there exists a positive constant C independent of ² such that

ku ² k L ∞ (0,T ;H3)∩L2(0,T ;H4 ) ≤ C, kθ ² k L ∞ (0,T ;H2 ) ≤ C, k∂ t u ² k L2(0,T ;L2 ) ≤ C, k∂ t θ ² k L2(0,T ;L2 ) ≤ C

(1.6)

for any T > 0, which implies

(u ² , θ ² ) → (u, θ) strongly in L2(0, T ; H1) when ² → 0. (1.7)

Here (u, θ) is the unique solution of the problem (1.1)–(1.5) with ² = 0.

2 Proof of Theorem 1.1

Since (1.7) follows easily from (1.6) by the Aubin-Lions compactness principle, we only

need to prove the a priori estimates (1.6) From now on we will drop the subscript ² and throughout this section C will be a constant independent of ² > 0.

First, we recall the following two lemmas in [8–10]

Lemma 2.1 ([8,9]) There holds

k∇uk L ∞(Ω) ≤ C(1 + kcurluk L ∞(Ω)log(e + kuk H3 (Ω)))

for any u ∈ H3(Ω) with divu = 0 in Ω and u · n = 0 on ∂Ω.

Lemma 2.2 ([10]) For any u ∈ W s,p with divu = 0 in Ω and u · n = 0 on ∂Ω, there holds

kuk W s,p ≤ C(kuk L p + kcurluk W s−1,p)

for any s > 1 and p ∈ (1, ∞).

By the maximum principle, it follows from (1.2), (1.3), and (1.4) that

kθk L ∞ (0,T ;L ∞) ≤ kθ0k L ∞ ≤ C. (2.1)

Testing (1.3) by θ, using (1.2), (1.3), and (1.4), we see that

1 2

d dt

Z

θ2dx + ²

Z

|∇θ|2dx = 0,

which gives

√

²kθk L2(0,T ;H1 ) ≤ C. (2.2)

Testing (1.1) by u, using (1.2), (1.4), and (2.1), we find that

1 2

d dt

Z

u2dx + C

Z

|∇u|2dx =

Z

θe2u ≤ kθk L2kuk L2 ≤ Ckuk L2,

which gives

kuk L ∞ (0,T ;L2 )+ kuk L2(0,T ;H1 ) ≤ C. (2.3) Here we used the well-known inequality:

kuk H1 ≤ Ckcurluk L2.

Applying curl to (1.1), using (1.2), we get

∂ t ω + u · ∇ω − ∆ω = curl(θe2). (2.4)

Testing (2.4) by |ω| p−2 ω (p > 2), using (1.2), (1.4), and (2.1), we obtain

1

p

d

dt

Z

|ω| p dx +1

2

Z

|ω| p−2 |∇ω|2dx + 4 p − 2

p2

Z ¯

¯∇|ω| p/2¯

¯2

dx

=

Z

curl(θe2)|ω| p−2 ωdx

≤ Ckθk L ∞

Z ¯

¯∇(|ω| p−2 ω)¯¯ dx

≤ 1

2

µ 1 2

Z

|ω| p−2 |∇ω|2dx + 4 p − 2

p2

Z ¯

¯∇|ω| p/2¯

¯2

dx

¶

+C

Z

|ω| p dx + C,

which gives

kuk L ∞ (0,T ;W 1,p)≤ Ckωk L ∞ (0,T ;L p)≤ C. (2.5)

(2.4) can be rewritten as























∂ t ω − ∆ω = divf := curl(θe2) − div(uω),

ω = 0 on ∂Ω × (0, ∞) ω(x, 0) = ω0(x) in Ω with f1 := θ − u1ω, f2 := −u2ω.

Using (2.1), (2.5) and the L ∞-estimate of the heat equation, we reach the key estimate

kωk L ∞ (0,T ;L ∞) ≤ C(kω0k L ∞ + kf k L ∞ (0,T ;L p) ≤ C). (2.6)

Let τ be any unit tangential vector of ∂Ω, using (1.4), we infer that

u · ∇θ = ((u · τ )τ + (u · n)n) · ∇θ = (u · τ )τ · ∇θ = (u · τ ) ∂θ

∂τ = 0 (2.7)

on ∂Ω × (0, ∞).

It follows from (1.3), (1.4), and (2.7) that

Applying ∆ to (1.3), testing by ∆θ, using (1.2), (1.4), and (2.8), we derive

1

2

d

dt

Z

|∆θ|2dx + ²

Z

|∇∆θ|2dx

= −

Z

(∆(u · ∇θ) − u∇∆θ)∆θdx

= −

Z

(∆u · ∇θ + 2X

i

∂ i u · ∇∂ i θ)∆θdx

≤ C(k∆uk L4k∇θk L4 + k∇uk L ∞ k∆θk L2)k∆θk L2. (2.9) Now using the Gagliardo–Nirenberg inequalities

k∇θk2L4 ≤ Ckθk L ∞ k∆θk L2,

k∆uk2

L4 ≤ Ck∇uk L ∞ kuk H3, (2.10)

we have

1

2

d

dt

Z

|∆θ|2dx + ²

Z

|∇∆θ|2dx

≤ Ck∇uk L ∞ k∆θk2

L2 + Ck∆θk2

L2 + Ck∇uk L ∞ kuk2

H3

≤ C(1 + k∇uk L ∞ )(kuk2H3 + k∆θk2L2)

≤ C(1 + kωk L ∞ log(e + kuk H3))(1 + k∆ωk2L2 + k∆θk2L2)

≤ C(1 + log(e + k∆ωk L2 + k∆θk L2))(1 + k∆ωk2

L2 + k∆θk2

L2). (2.11)

Similarly to (2.7) and (2.8), if follows from (2.4) and (1.4) that

u · ∇ω = 0 on ∂Ω × (0, ∞), (2.12)

∆ω + curl(θe2) = 0 on ∂Ω × (0, ∞). (2.13)

Applying ∆ to (2.4), testing by ∆ω, using (1.2), (1.4), (2.13), (2.10), and Lemma 2.2, we

reach

1

2

d

dt

Z

|∆ω|2dx +

Z

|∇∆ω|2dx

= −

Z

(∆(u · ∇ω) − u∇∆ω)∆ωdx −

Z

∇curl(θe2) · ∇∆ωdx

≤ C(k∆uk L4k∇ωk L4 + k∇uk L ∞ k∆ωk L2)k∆ωk L2 + Ck∆θk L2k∇∆ωk L2

≤ C(k∆uk2

L4 + k∇uk L ∞ k∆ωk L2)k∆ωk L2 + Ck∆θk L2k∇∆ωk L2

≤ Ck∇uk L ∞ kuk H3k∆ωk L2 + Ck∆θk L2k∇∆ωk L2

≤ Ck∇uk L ∞ (1 + k∆ωk L2)k∆ωk L2 + Ck∆θk2L2 +1

2k∇∆ωk

2

L2

which yields

d

dt

Z

|∆ω|2dx +

Z

|∇∆ω|2dx

≤ Ck∇uk L ∞ (1 + k∆ωk L2)k∆ωk L2 + Ck∆θk2

L2

≤ C(1 + log(e + k∆ωk L2 + k∆θk L2))(1 + k∆ωk2

L2 + k∆θk2

L2). (2.14) Combining (2.11) and (2.14), using the Gronwall inequality, we conclude that

kθk L ∞ (0,T ;H2 )+√ ²kθk L ∞ (0,T ;H3 )≤ C, (2.15)

kuk L ∞ (0,T ;H3 )+ kuk L2(0,T ;H4 ) ≤ C. (2.16)

It follows from (1.1), (1.3), (2.15), and (2.16) that

k∂ t uk L2(0,T ;L2 )≤ C, k∂ t θk L2(0,T ;L2 ) ≤ C.

Competing interests

The authors declare that they have no competing interests

Authors’ contributions

All authors read and approved the final manuscript

Acknowledgments

This study was partially supported by the Zhejiang Innovation Project (Grant No T200905), the ZJNSF (Grant No R6090109), and the NSFC (Grant No 11171154)

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