## Boundary Value Problems

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

Boundary Value Problems20122012:20

DOI: 10.1186/1687-2770-2012-20

Received: 12 November 2011

Accepted: 15 February 2012

Published: 15 February 2012

## 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 2 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, ∞):
${\partial }_{t}u+u\cdot \nabla u+\nabla \pi -\Delta u=\theta {e}_{2},$
(1.1)
$\text{div}\phantom{\rule{0.3em}{0ex}}u=0,$
(1.2)
${\partial }_{t}\theta +u\cdot \nabla \theta =\epsilon \Delta \theta ,$
(1.3)
$u\cdot n=0,\phantom{\rule{1em}{0ex}}\text{curl}u=0,\phantom{\rule{1em}{0ex}}\theta =0,\phantom{\rule{1em}{0ex}}\text{on}\phantom{\rule{2.77695pt}{0ex}}\partial \text{Ω}×\text{(0,}\infty \text{),}$
(1.4)
$\left(u,\theta \right)\left(x,0\right)=\left({u}_{0},{\theta }_{0}\right)\left(x\right),\phantom{\rule{1em}{0ex}}x\in \text{Ω},$
(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:= 2, 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 [27] when Ω is a bounded domain. However, the methods in [16] 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 ${\theta }_{0}\in {H}_{0}^{1}\cap {H}^{2}$. Then there exists a positive constant C independent of ϵ such that
$\begin{array}{cc}{∥{u}_{\epsilon }∥}_{{L}^{\infty }\left(0,T;{H}^{3}\right)\cap {L}^{2}\left(0,T;{H}^{4}\right)}\le C,\hfill & {∥{\theta }_{\epsilon }∥}_{{L}^{\infty }\left(0,T;{H}^{2}\right)}\le C,\hfill \\ {∥{\partial }_{t}{u}_{\epsilon }∥}_{{L}^{2}\left(0,T;{L}^{2}\right)}\le C,\hfill & {∥{\partial }_{t}{\theta }_{\epsilon }∥}_{{L}^{2}\left(0,T;{L}^{2}\right)}\le C\hfill \end{array}$
(1.6)
for any T > 0, which implies
$\left({u}_{\epsilon },{q}_{\epsilon }\right)\to \left(u,\theta \right)\phantom{\rule{2.77695pt}{0ex}}strongly\phantom{\rule{2.77695pt}{0ex}}in\phantom{\rule{2.77695pt}{0ex}}{L}^{2}\left(0,T;{H}^{1}\right)\phantom{\rule{2.77695pt}{0ex}}when\phantom{\rule{2.77695pt}{0ex}}\epsilon \to 0.$
(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 [810].

Lemma 2.1. ([8, 9]) There holds
${∥\nabla u∥}_{{L}^{\infty }\left(\text{Ω}\right)}\le C\left(1+{∥\text{curl}u∥}_{{L}^{\infty }\left(\text{Ω}\right)}\text{log}\left(e+{∥u∥}_{{H}^{3}\left(\text{Ω}\right)}\right)\right)$

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

Lemma 2.2. ([10]) For any u Ws,pwith divu = 0 in Ω and u · n = 0 on ∂Ω, there holds
${∥u∥}_{{W}^{s,p}}\le C\left({∥u∥}_{{L}^{p}}+{∥\text{curl}\phantom{\rule{0.3em}{0ex}}u∥}_{{W}^{s-1,p}}\right)$

for any s > 1 and p (1, ∞).

By the maximum principle, it follows from (1.2), (1.3), and (1.4) that
${∥\theta ∥}_{{L}^{\infty }\left(0,T;{L}^{\infty }\right)}\le {∥{\theta }_{0}∥}_{{L}^{\infty }}\le C.$
(2.1)
Testing (1.3) by θ, using (1.2), (1.3), and (1.4), we see that
$\frac{1}{2}\frac{d}{dt}\int {\theta }^{2}dx+\epsilon \int {\left|\nabla \theta \right|}^{2}dx=0,$
which gives
$\sqrt{\epsilon }{∥\theta ∥}_{{L}^{2}\left(0,T;{H}^{1}\right)}\le C.$
(2.2)
Testing (1.1) by u, using (1.2), (1.4), and (2.1), we find that
$\frac{1}{2}\frac{d}{dt}\int {u}^{2}dx+C\int {\left|\nabla u\right|}^{2}dx=\int \theta {e}_{2}u\le {∥\theta ∥}_{{L}^{2}}{∥u∥}_{{L}^{2}}\le C{∥u∥}_{{L}^{2}},$
which gives
${∥u∥}_{{L}^{\infty }\left(0,T;{L}^{2}\right)}+{∥u∥}_{{L}^{2}\left(0,T;{H}^{1}\right)}\le C.$
(2.3)
Here we used the well-known inequality:
${∥u∥}_{{H}^{1}}\le C{∥\text{curl}\phantom{\rule{0.3em}{0ex}}u∥}_{{L}^{2}}.$
Applying curl to (1.1), using (1.2), we get
${\partial }_{t}\omega +u\cdot \nabla \omega -\Delta \omega =\text{curl(}\theta {e}_{\text{2}}\text{)}\text{.}$
(2.4)
Testing (2.4) by |ω|p-2ω (p > 2), using (1.2), (1.4), and (2.1), we obtain
$\begin{array}{c}\frac{1}{p}\frac{d}{dt}\int {\left|\omega \right|}^{p}dx+\frac{1}{2}\int {\left|\omega \right|}^{p-2}{\left|\nabla \omega \right|}^{2}dx+4\frac{p-2}{{p}^{2}}\int {\left|\nabla {\left|\omega \right|}^{p/2}\right|}^{2}dx\hfill \\ =\int \text{curl(}\theta {e}_{\text{2}}\text{)}{\left|\omega \right|}^{p-2}\omega dx\hfill \\ \le C{∥\theta ∥}_{{L}^{\infty }}\int \left|\nabla \left({\left|\omega \right|}^{p-2}\omega \right)\right|dx\hfill \\ \le \frac{1}{2}\left(\frac{1}{2}\int {\left|\omega \right|}^{p-2}{\left|\nabla \omega \right|}^{2}dx+4\frac{p-2}{{p}^{2}}\int {\left|\nabla {\left|\omega \right|}^{p/2}\right|}^{2}dx\right)\hfill \\ \phantom{\rule{1em}{0ex}}+C\int {\left|\omega \right|}^{p}dx+C,\hfill \end{array}$
which gives
${∥u∥}_{{L}^{\infty }\left(0,T;{W}^{1},p\right)}\le C{∥\omega ∥}_{{L}^{\infty }\left(0,T;{L}^{p}\right)}\le C.$
(2.5)
(2.4) can be rewritten as
$\left\{\begin{array}{c}{\partial }_{t}\omega -\Delta \omega =\text{div}\phantom{\rule{0.3em}{0ex}}f:=\text{curl}\left(\theta {e}_{2}\right)-\text{div}\left(u\omega \right),\hfill \\ \omega =0\phantom{\rule{1em}{0ex}}\text{on}\phantom{\rule{0.3em}{0ex}}\partial \text{Ω}×\left(0,\infty \right)\hfill \\ \omega \left(x,0\right)={\omega }_{0}\left(x\right)\phantom{\rule{2.77695pt}{0ex}}\text{in}\phantom{\rule{2.77695pt}{0ex}}\text{Ω}\hfill \end{array}\right\$

with f1: = θ - u1ω, f2:= -u2ω.

Using (2.1), (2.5) and the L-estimate of the heat equation, we reach the key estimate
${∥\omega ∥}_{{L}^{\infty }\left(0,T;{L}^{\infty }\right)}\le C\left({∥{\omega }_{0}∥}_{{L}^{\infty }}+{∥f∥}_{{L}^{\infty }\left(0,T;{L}^{p}\right)}\le C\right).$
(2.6)
Let τ be any unit tangential vector of ∂Ω, using (1.4), we infer that
$u\cdot \nabla \theta =\left(\left(u\cdot \tau \right)\tau +\left(u\cdot n\right)n\right)\cdot \nabla \theta =\left(u\cdot \tau \right)\tau \cdot \nabla \theta =\left(u\cdot \tau \right)\frac{\partial \theta }{\partial \tau }=0$
(2.7)

on ∂Ω × (0, ∞).

It follows from (1.3), (1.4), and (2.7) that
$\Delta \theta =0\phantom{\rule{1em}{0ex}}\text{on}\phantom{\rule{0.3em}{0ex}}\partial \text{Ω}×\left(0,\infty \right).$
(2.8)
Applying Δ to (1.3), testing by Δθ, using (1.2), (1.4), and (2.8), we derive
$\begin{array}{c}\frac{1}{2}\frac{d}{dt}\int {\left|\Delta \theta \right|}^{2}dx+\epsilon \int {\left|\nabla \Delta \theta \right|}^{2}dx\hfill \\ =-\int \left(\Delta \left(u\cdot \nabla \theta \right)-u\nabla \Delta \theta \right)\Delta \theta dx\hfill \\ =-\int \left(\Delta u\cdot \nabla \theta +2\sum _{i}{\partial }_{i}u\cdot \nabla {\partial }_{i}\theta \right)\Delta \theta dx\hfill \\ \le C\left({∥\Delta u∥}_{{L}^{4}}{∥\nabla \theta ∥}_{{L}^{4}}+{∥\nabla u∥}_{{L}^{\infty }}{∥\Delta \theta ∥}_{{L}^{2}}\right){∥\Delta \theta ∥}_{{L}^{2}}.\hfill \end{array}$
(2.9)
Now using the Gagliardo-Nirenberg inequalities
$\begin{array}{c}{∥\nabla \theta ∥}_{{L}^{4}}^{2}\le C{∥\theta ∥}_{{L}^{\infty }}{∥\Delta \theta ∥}_{{L}^{2}},\\ {∥\Delta u∥}_{{L}^{4}}^{2}\le C{∥\nabla u∥}_{{L}^{\infty }}{∥u∥}_{{H}^{3}},\end{array}$
(2.10)
we have
$\begin{array}{c}\frac{1}{2}\frac{d}{dt}\int {\left|\Delta \theta \right|}^{2}dx+\epsilon \int {\left|\nabla \Delta \theta \right|}^{2}dx\hfill \\ \le C{∥\nabla u∥}_{{L}^{\infty }}{∥\Delta \theta ∥}_{{L}^{2}}^{2}+C{∥\Delta \theta ∥}_{{L}^{2}}^{2}+C{∥\nabla u∥}_{{L}^{\infty }}{∥u∥}_{{H}^{3}}^{2}\hfill \\ \le C\left(1+{∥\nabla u∥}_{{L}^{\infty }}\right)\left({∥u∥}_{{H}^{3}}^{2}+{∥\Delta \theta ∥}_{{L}^{2}}^{2}\right)\hfill \\ \le C\left(1+{∥\omega ∥}_{{L}^{\infty }}\text{log}\left(e+{∥u∥}_{{H}^{3}}\right)\right)\left(1+{∥\Delta \omega ∥}_{{L}^{2}}^{2}+{∥\Delta \theta ∥}_{{L}^{2}}^{2}\right)\hfill \\ \le C\left(1+\text{log}\left(e+{∥\Delta \omega ∥}_{{L}^{2}}+{∥\Delta \theta ∥}_{{L}^{2}}\right)\right)\left(1+{∥\Delta \omega ∥}_{{L}^{2}}^{2}+{∥\Delta \theta ∥}_{{L}^{2}}^{2}\right).\hfill \end{array}$
(2.11)
Similarly to (2.7) and (2.8), if follows from (2.4) and (1.4) that
$u\cdot \nabla \omega =0\phantom{\rule{1em}{0ex}}\text{on}\phantom{\rule{2.77695pt}{0ex}}\partial \text{Ω}×\left(0,\infty \right),$
(2.12)
$\Delta \omega +\text{curl}\left(\theta {e}_{2}\right)=0\phantom{\rule{1em}{0ex}}\text{on}\phantom{\rule{2.77695pt}{0ex}}\partial \text{Ω}×\left(0,\infty \right).$
(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
$\begin{array}{c}\frac{1}{2}\frac{d}{dt}\int {\left|\Delta \omega \right|}^{2}dx+\int {\left|\nabla \Delta \omega \right|}^{2}dx\hfill \\ =-\int \left(\Delta \left(u\cdot \nabla \omega \right)-u\nabla \Delta \omega \right)\Delta \omega dx-\int \nabla \text{curl}\left(\theta {e}_{2}\right)\cdot \nabla \Delta \omega dx\hfill \\ \le C\left({∥\Delta u∥}_{{L}^{4}}{∥\nabla \omega ∥}_{{L}^{4}}+{∥\nabla u∥}_{{L}^{\infty }}{∥\Delta \omega ∥}_{{L}^{2}}\right){∥\Delta \omega ∥}_{{L}^{2}}+C{∥\Delta \theta ∥}_{{L}^{2}}{∥\nabla \Delta \omega ∥}_{{L}^{2}}\hfill \\ \le C\left({∥\Delta u∥}_{{L}^{4}}^{2}+{∥\nabla u∥}_{{L}^{\infty }}{∥\Delta \omega ∥}_{{L}^{2}}\right){∥\Delta \omega ∥}_{{L}^{2}}+C{∥\Delta \theta ∥}_{{L}^{2}}{∥\nabla \Delta \omega ∥}_{{L}^{2}}\hfill \\ \le C{∥\nabla u∥}_{{L}^{\infty }}{∥u∥}_{{H}^{3}}{∥\Delta \omega ∥}_{{L}^{2}}+C{∥\Delta \theta ∥}_{{L}^{2}}{∥\nabla \Delta \omega ∥}_{{L}^{2}}\hfill \\ \le C{∥\nabla u∥}_{{L}^{\infty }}\left(1+{∥\Delta \omega ∥}_{{L}^{2}}\right){∥\Delta \omega ∥}_{{L}^{2}}+C{∥\Delta \theta ∥}_{{L}^{2}}^{2}+\frac{1}{2}{∥\nabla \Delta \omega ∥}_{{L}^{2}}^{2}\hfill \end{array}$
which yields
$\begin{array}{c}\frac{d}{dt}\int {\left|\Delta \omega \right|}^{2}dx+\int {\left|\nabla \Delta \omega \right|}^{2}dx\hfill \\ \phantom{\rule{1em}{0ex}}\le C{∥\nabla u∥}_{{L}^{\infty }}\left(1+{∥\Delta \omega ∥}_{{L}^{2}}\right){∥\Delta \omega ∥}_{{L}^{2}}+C{∥\Delta \theta ∥}_{{L}^{2}}^{2}\hfill \\ \phantom{\rule{1em}{0ex}}\le C\left(1+\text{log}\left(e+{∥\Delta \omega ∥}_{{L}^{2}}+{∥\Delta \theta ∥}_{{L}^{2}}\right)\right)\left(1+{∥\Delta \omega ∥}_{{L}^{2}}^{2}+{∥\Delta \theta ∥}_{{L}^{2}}^{2}\right).\hfill \end{array}$
(2.14)
Combining (2.11) and (2.14), using the Gronwall inequality, we conclude that
${∥\theta ∥}_{{L}^{\infty }\left(0,T;{H}^{2}\right)}+\sqrt{\epsilon }{∥\theta ∥}_{{L}^{\infty }\left(0,T;{H}^{3}\right)}\le C,$
(2.15)
${∥u∥}_{{L}^{\infty }\left(0,T;{H}^{3}\right)}+{∥u∥}_{{L}^{2}\left(0,T;{H}^{4}\right)}\le C.$
(2.16)
It follows from (1.1), (1.3), (2.15), and (2.16) that
${∥{\partial }_{t}u∥}_{{L}^{2}\left(0,T:{L}^{2}\right)}\le C,\phantom{\rule{1em}{0ex}}{∥{\partial }_{t}\theta ∥}_{{L}^{2}\left(0,T:{L}^{2}\right)}\le C.$

This completes the proof.

## Declarations

### 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).

## Authors’ Affiliations

(1)
Department of Mathematics, Zhejiang Normal University
(2)
Department of Applied Mathematics, Nanjing Forestry University
(3)
Department of Mathematics, Hokkaido University

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## Copyright

© Jin et al; licensee Springer. 2012

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