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

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.

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, ∞):

$${\partial }_{t}u+u\cdot \nabla u+\nabla \pi -\Delta u=\theta e_2,$$

(1.1)

$$\text{div}u=0,$$

(1.2)

$${\partial }_t\theta +u\cdot \nabla \theta =\epsilon \Delta \theta ,$$

(1.3)

$$u\cdot n=0,\text{curl}u=0,\theta =0,\text{on}\partial \text{Ω}\times \text{(0,}\infty \text{),}$$

(1.4)

$$\left(u,\theta \right)\left(x,0\right)=\left(u_0,{\theta }_0\right)\left(x\right),x\in \text{Ω},$$

(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:= 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 ${\theta }_0\in H_0^1\cap H^2$. 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

$$\left(u_{\epsilon },q_{\epsilon }\right)\to \left(u,\theta \right)stronglyin{L}^2\left(0,T;H^1\right)when\epsilon \to 0.$$

(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 [8–10].

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 ∈ W^s,pwith divu = 0 in Ω and u · n = 0 on ∂Ω, there holds

$${∥u∥}_{W^{s,p}}\le C\left({∥u∥}_{L^p}+{∥\text{curl}u∥}_{W^{s-1,p}}\right)$$

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:

$${∥u∥}_{H^1}\le C{∥\text{curl}u∥}_{L^2}.$$

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

$$\begin{array}{c}\frac{1}{p}\frac{d}{dt}\int {\left|\omega \right|}^{p}dx+\frac{1}{2}\int {\left|\omega \right|}^{p-2}{\left|\nabla \omega \right|}^{2}dx+4\frac{p-2}{p^2}\int {\left|\nabla {\left|\omega \right|}^{p/2}\right|}^{2}dx\hfill \\ =\int \text{curl(}\theta e_{\text{2}}\text{)}{\left|\omega \right|}^{p-2}\omega dx\hfill \\ \le C{∥\theta ∥}_{L^{\infty }}\int \left|\nabla \left({\left|\omega \right|}^{p-2}\omega \right)\right|dx\hfill \\ \le \frac{1}{2}\left(\frac{1}{2}\int {\left|\omega \right|}^{p-2}{\left|\nabla \omega \right|}^{2}dx+4\frac{p-2}{p^2}\int {\left|\nabla {\left|\omega \right|}^{p/2}\right|}^{2}dx\right)\hfill \\ +C\int {\left|\omega \right|}^{p}dx+C,\hfill \end{array}$$

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

$$\left\{\begin{array}{c}{\partial }_t\omega -\Delta \omega =\text{div}f:=\text{curl}\left(\theta e_2\right)-\text{div}\left(u\omega \right),\hfill \\ \omega =0\text{on}\partial \text{Ω}\times \left(0,\infty \right)\hfill \\ \omega \left(x,0\right)={\omega }_0\left(x\right)\text{in}\text{Ω}\hfill \end{array}\right.$$

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

$$\Delta \theta =0\text{on}\partial \text{Ω}\times \left(0,\infty \right).$$

(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

$$u\cdot \nabla \omega =0\text{on}\partial \text{Ω}\times \left(0,\infty \right),$$

(2.12)

$$\Delta \omega +\text{curl}\left(\theta e_2\right)=0\text{on}\partial \text{Ω}\times \left(0,\infty \right).$$

(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

$$\begin{array}{c}\frac{d}{dt}\int {\left|\Delta \omega \right|}^{2}dx+\int {\left|\nabla \Delta \omega \right|}^{2}dx\hfill \\ \le C{∥\nabla u∥}_{L^{\infty }}\left(1+{∥\Delta \omega ∥}_{L^2}\right){∥\Delta \omega ∥}_{L^2}+C{∥\Delta \theta ∥}_{L^2}^2\hfill \\ \le C\left(1+\text{log}\left(e+{∥\Delta \omega ∥}_{L^2}+{∥\Delta \theta ∥}_{L^2}\right)\right)\left(1+{∥\Delta \omega ∥}_{L^2}^2+{∥\Delta \theta ∥}_{L^2}^2\right).\hfill \end{array}$$

(2.14)

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

$${∥{\partial }_{t}u∥}_{L^2\left(0,T:L^2\right)}\le C,{∥{\partial }_t\theta ∥}_{L^2\left(0,T:L^2\right)}\le C.$$

This completes the proof.