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

Liangbing Jin1, Jishan Fan2, Gen Nakamura3 and Yong Zhou1*

Author Affiliations

1 Department of Mathematics, Zhejiang Normal University, Jinhua 321004, P. R. China

2 Department of Applied Mathematics, Nanjing Forestry University, Nanjing 210037, P.R. China

3 Department of Mathematics, Hokkaido University Sapporo 060-0810, Japan

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

 Received: 12 November 2011 Accepted: 15 February 2012 Published: 15 February 2012

This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/2.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

### 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 - Δ u = θ e 2 , (1.1)

div u = 0 , (1.2)

t θ + u θ = ε Δ θ , (1.3)

u n = 0 , curl u = 0 , θ = 0 , on Ω × (0, ), (1.4)

( u , θ ) ( x , 0 ) = ( u 0 , θ 0 ) ( x ) , x Ω , (1.5)

where u, π, and θ denote unknown velocity vector field, pressure scalar and temperature of the fluid. ϵ > 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 ϵ → 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 , u0·n = 0, curlu0 = 0 on ∂Ω and θ 0 H 0 1 H 2 . Then there exists a positive constant C independent of ϵ such that

u ε L ( 0 , T ; H 3 ) L 2 ( 0 , T ; H 4 ) C , θ ε L ( 0 , T ; H 2 ) C , t u ε L 2 ( 0 , T ; L 2 ) C , t θ ε L 2 ( 0 , T ; L 2 ) C (1.6)

for any T > 0, which implies

( u ε , q ε ) ( u , θ ) s t r o n g l y i n L 2 ( 0 , T ; H 1 ) w h e n ε 0 . (1.7)

Here (u, θ) is the unique solution of the problem (1.1)-(1.5) with ϵ = 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 e and throughout this section C will be a constant independent of ϵ > 0.

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

Lemma 2.1. ([8,9]) There holds

u L ( Ω ) C ( 1 + curl u L ( Ω ) log ( e + u H 3 ( Ω ) ) )

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

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

u W s , p C u L p + curl u 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

θ L ( 0 , T ; L ) θ 0 L C . (2.1)

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

1 2 d d t θ 2 d x + ε θ 2 d x = 0 ,

which gives

ε θ L 2 ( 0 , T ; H 1 ) C . (2.2)

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

1 2 d d t u 2 d x + C u 2 d x = θ e 2 u θ L 2 u L 2 C u L 2 ,

which gives

u L ( 0 , T ; L 2 ) + u L 2 ( 0 , T ; H 1 ) C . (2.3)

Here we used the well-known inequality:

u H 1 C curl u L 2 .

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

t ω + u ω - Δ ω = curl( θ e 2 ) . (2.4)

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

1 p d d t ω p d x + 1 2 ω p - 2 ω 2 d x + 4 p - 2 p 2 ω p / 2 2 d x = curl( θ e 2 ) ω p - 2 ω d x C θ L ω p - 2 ω d x 1 2 1 2 ω p - 2 ω 2 d x + 4 p - 2 p 2 ω p / 2 2 d x + C ω p d x + C ,

which gives

u L ( 0 , T ; W 1 , p ) C ω L ( 0 , T ; L p ) C . (2.5)

(2.4) can be rewritten as

t ω - Δ ω = div f : = curl ( θ e 2 ) - div ( u ω ) , ω = 0 on Ω × ( 0 , ) ω ( x , 0 ) = ω 0 ( 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

ω L ( 0 , T ; L ) C ω 0 L + f 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

Δ θ = 0 on Ω × ( 0 , ) . (2.8)

Applying Δ to (1.3), testing by Δθ, using (1.2), (1.4), and (2.8), we derive

1 2 d d t Δ θ 2 d x + ε Δ θ 2 d x = - ( Δ ( u θ ) - u Δ θ ) Δ θ d x = - ( Δ u θ + 2 i i u i θ ) Δ θ d x C Δ u L 4 θ L 4 + u L Δ θ L 2 Δ θ L 2 . (2.9)

Now using the Gagliardo-Nirenberg inequalities

θ L 4 2 C θ L Δ θ L 2 , Δ u L 4 2 C u L u H 3 , (2.10)

we have

1 2 d d t Δ θ 2 d x + ε Δ θ 2 d x C u L Δ θ L 2 2 + C Δ θ L 2 2 + C u L u H 3 2 C 1 + u L u H 3 2 + Δ θ L 2 2 C 1 + ω L log e + u H 3 1 + Δ ω L 2 2 + Δ θ L 2 2 C 1 + log e + Δ ω L 2 + Δ θ L 2 1 + Δ ω L 2 2 + Δ θ L 2 2 . (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)

Δ ω + curl ( θ e 2 ) = 0 on Ω × ( 0 , ) . (2.13)

Applying Δ to (2.4), testing by Δω, using (1.2), (1.4), (2.13), (2.10), and Lemma 2.2, we reach

1 2 d d t Δ ω 2 d x + Δ ω 2 d x = - ( Δ ( u ω ) - u Δ ω ) Δ ω d x - curl ( θ e 2 ) Δ ω d x C Δ u L 4 ω L 4 + u L Δ ω L 2 Δ ω L 2 + C Δ θ L 2 Δ ω L 2 C Δ u L 4 2 + u L Δ ω L 2 Δ ω L 2 + C Δ θ L 2 Δ ω L 2 C u L u H 3 Δ ω L 2 + C Δ θ L 2 Δ ω L 2 C u L 1 + Δ ω L 2 Δ ω L 2 + C Δ θ L 2 2 + 1 2 Δ ω L 2 2

which yields

d d t Δ ω 2 d x + Δ ω 2 d x C u L 1 + Δ ω L 2 Δ ω L 2 + C Δ θ L 2 2 C 1 + log e + Δ ω L 2 + Δ θ L 2 1 + Δ ω L 2 2 + Δ θ L 2 2 . (2.14)

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

θ L ( 0 , T ; H 2 ) + ε θ L ( 0 , T ; H 3 ) C , (2.15)

u L ( 0 , T ; H 3 ) + u L 2 ( 0 , T ; H 4 ) C . (2.16)

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

t u L 2 ( 0 , T : L 2 ) C , t θ L 2 ( 0 , T : L 2 ) C .

This completes the proof.

### Competing interests

The authors declare that they have no competing interests.

### Authors' contributions

All authors read and approved the final manuscript.

### Acknowledgements

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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