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

Boundary Value Problems20122012:20

https://doi.org/10.1186/1687-2770-2012-20

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

## 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 . 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  when Ω is a bounded domain. However, the methods in  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 .

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. () 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, 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

## References 