## Boundary Value Problems

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# Vanishing heat conductivity limit for the 2D Cahn-Hilliard-Boussinesq system

Boundary Value Problems20112011:54

https://doi.org/10.1186/1687-2770-2011-54

Accepted: 22 December 2011

Published: 22 December 2011

## Abstract

This article studies the vanishing heat conductivity limit for the 2D Cahn-Hilliard-boussinesq system in a bounded domain with non-slip boundary condition. The result has been proved globally in time.

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

### Keywords

Cahn-Hilliard-Boussinesq inviscid limit non-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 following Cahn-Hilliard-Boussinesq system in Ω × (0, ∞) [1]:
${\partial }_{t}u+\left(u\cdot \nabla \right)u+\nabla \pi -\Delta u=\mu \nabla \varphi +\theta {e}_{2},$
(1.1)
$\mathsf{\text{div}}\phantom{\rule{2.77695pt}{0ex}}u=0,$
(1.2)
${\partial }_{t}\theta +u\cdot \nabla \theta =\epsilon \Delta \theta ,$
(1.3)
${\partial }_{t}\varphi +u\cdot \nabla \varphi =\Delta \mu ,$
(1.4)
$-\Delta \varphi +{f}^{\prime }\left(\varphi \right)=\mu ,$
(1.5)
$u=0,\theta =0,\frac{\partial \varphi }{\partial n}=\frac{\partial \mu }{\partial n}=0\phantom{\rule{1em}{0ex}}on\phantom{\rule{1em}{0ex}}\partial \Omega ×\left(0,\infty \right),$
(1.6)
$\left(u,\theta ,\varphi \right)\left(x,0\right)=\left({u}_{0},{\theta }_{0},{\varphi }_{0}\right)\left(x\right),x\in \Omega ,$
(1.7)

where u, π, θ and ϕ denote unknown velocity field, pressure scalar, temperature of the fluid and the order parameter, respectively. ε > 0 is the heat conductivity coefficient and e2 : = (0, 1) t . μ is a chemical potential and $f\left(\varphi \right):=\frac{1}{4}{\left({\varphi }^{2}-1\right)}^{2}$ is the double well potential.

When ϕ = 0, (1.1), (1.2) and (1.3) is the well-known Boussinesq system. In [2] Zhou and Fan proved a regularity criterion $\omega \stackrel{˙}{=}curlu\in {L}^{1}\left(0,T;{Ḃ}_{\infty ,\infty }^{0}\right)$ for the 3D Boussinesq system with partial viscosity. Later, in [3] Zhou and Fan studied the Cauchy problem of certain Boussinesq-α equations in n dimensions with n = 2 or 3. We establish regularity for the solution under $\nabla u\in {L}^{1}\left(0,T;{Ḃ}_{\infty ,\infty }^{0}\right)$. Here ${Ḃ}_{\infty ,\infty }^{0}$ denotes the homogeneous Besov space. Chae [4] studied the vanishing viscosity limit ε → 0 when Ω = 2. The aim of this article is to prove a similar result. We will prove that

Theorem 1.1. Let $\left({u}_{0},{\theta }_{0}\right)\in {H}_{0}^{1}\cap {H}^{2}$, ϕ0 H4, div u0 = 0 in Ω and $\frac{\partial {\varphi }_{0}}{\partial n}=\frac{\partial {\mu }_{0}}{\partial n}=0$ on∂Ω. Then, there exists a positive constant C independent of ε such that
$\begin{array}{c}\parallel {u}_{\epsilon }{\parallel }_{{L}^{\infty }\left(0,T;{H}^{2}\right)}\le C,\parallel {\theta }_{\epsilon }{\parallel }_{{L}^{\infty }\left(0,T;{H}^{2}\right)}\le C,\\ \parallel {\varphi }_{\epsilon }{\parallel }_{{L}^{\infty }\left(0,T;{H}^{4}\right)}\le C,\parallel {\partial }_{t}\left({u}_{ϵ},{\theta }_{\epsilon },{\varphi }_{ϵ}\right){\parallel }_{{L}^{2}\left(0,T;{L}^{2}\right)}\le C,\end{array}$
(1.8)
for any T > 0, which implies
$\left({u}_{\epsilon },{\theta }_{\epsilon },{\varphi }_{\epsilon }\right)\to \left(u,\theta ,\varphi \right)\phantom{\rule{1em}{0ex}}strongly\phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{2.77695pt}{0ex}}in\phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{2.77695pt}{0ex}}{L}^{2}\left(0,T;{H}^{1}\right)\phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{2.77695pt}{0ex}}when\phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{2.77695pt}{0ex}}\epsilon \to 0.$
(1.9)

Here, (u, θ, ϕ) is the solution of the problem (1.1)-(1.7) with ε = 0.

### 2 Proof of Theorem 1.1

Since (1.9) follows easily from (1.8) by the Aubin-Lions compactness principle, we only need to prove the a priori estimates (1.8). From now on, we will drop the subscript ε and throughout this section C will be a constant independent of ε.

First, by the maximum principle, it follows from (1.2), (1.3), and (1.6) that
$\parallel \theta {\parallel }_{{L}^{\infty }\left(0,T;{L}^{\infty }\right)}\le \phantom{\rule{2.77695pt}{0ex}}\parallel {\theta }_{0}{\parallel }_{{L}^{\infty }}\le C.$
(2.1)
Testing (1.3) by θ, using (1.2) and (1.6), we see that
$\frac{1}{2}\frac{\mathsf{\text{d}}}{\mathsf{\text{d}}t}\int {\theta }^{2}\mathsf{\text{d}}x+\epsilon \int \mid \nabla \theta {\mid }^{2}\mathsf{\text{d}}x=0,$
whence
$\sqrt{\epsilon }\parallel \theta {\parallel }_{{L}^{2}\left(0,T;{H}^{1}\right)}\le C.$
(2.2)
Testing (1.1) and (1.4) by u and μ, respectively, using (1.2), (1.6), (2.1), and summing up the result, we find that
$\begin{array}{c}\phantom{\rule{1em}{0ex}}\frac{\mathsf{\text{d}}}{\mathsf{\text{d}}t}\int \frac{1}{2}{u}^{2}+\frac{1}{2}\mid \nabla \varphi {\mid }^{2}+f\left(\varphi \right)\mathsf{\text{d}}x+\int \mid \nabla u{\mid }^{2}+\mid \nabla \mu {\mid }^{2}\mathsf{\text{d}}x\\ =\int \theta {e}_{2}u\mathsf{\text{d}}x\le \phantom{\rule{2.77695pt}{0ex}}\parallel \theta {\parallel }_{{L}^{2}}\parallel u{\parallel }_{{L}^{2}}\le C\parallel u{\parallel }_{{L}^{2}},\end{array}$
which gives
$\parallel \varphi {\parallel }_{{L}^{\infty }\left(0,T;{H}^{1}\right)}\le C,$
(2.3)
$\parallel u{\parallel }_{{L}^{\infty }\left(0,T;{L}^{2}\right)}+\parallel u{\parallel }_{{L}^{2}\left(0,T;{H}^{1}\right)}\le C,$
(2.4)
$\parallel \nabla \mu {\parallel }_{{L}^{2}\left(0,T;{L}^{2}\right)}\le C.$
(2.5)
Testing (1.4) by ϕ, using (1.2), (1.5) and (1.6), we infer that
$\begin{array}{c}\phantom{\rule{1em}{0ex}}\frac{1}{2}\frac{\mathsf{\text{d}}}{\mathsf{\text{d}}t}\int {\varphi }^{2}\mathsf{\text{d}}x+\int \mid \Delta \varphi {\mid }^{2}\mathsf{\text{d}}x=\int \left({\varphi }^{3}-\varphi \right)\Delta \varphi \mathsf{\text{d}}x\\ =-3\int {\varphi }^{2}\mid \nabla \varphi {\mid }^{2}\mathsf{\text{d}}x-\int \varphi \Delta \varphi \mathsf{\text{d}}x\le -\int \varphi \Delta \varphi \mathsf{\text{d}}x\\ \le \frac{1}{2}\int \mid \Delta \varphi {\mid }^{2}\mathsf{\text{d}}x+\frac{1}{2}\int {\varphi }^{2}\mathsf{\text{d}}x,\end{array}$
$\parallel \varphi {\parallel }_{{L}^{2}\left(0,T;{H}^{2}\right)}\le C.$
(2.6)
We will use the following Gagliardo-Nirenberg inequality:
$\parallel \varphi {\parallel }_{{L}^{\infty }}^{2}\le C\parallel \varphi {\parallel }_{{L}^{6}}\parallel \varphi {\parallel }_{{H}^{2}}.$
(2.7)
It follows from (2.6), (2.7), (2.5), (2.3) and (1.5) that
$\begin{array}{c}\phantom{\rule{1em}{0ex}}{\int }_{0}^{T}\int \mid \nabla \Delta \varphi {\mid }^{2}\mathsf{\text{d}}x\mathsf{\text{d}}t\\ ={\int }_{0}^{T}\int \mid \nabla \left({f}^{\prime }\left(\varphi \right)-\mu \right){\mid }^{2}\mathsf{\text{d}}x\mathsf{\text{d}}t\\ \le C{\int }_{0}^{T}\int \mid \nabla \mu {\mid }^{2}\mathsf{\text{d}}x\mathsf{\text{d}}t+C{\int }_{0}^{T}\int \mid \nabla \left({\varphi }^{3}-\varphi \right){\mid }^{2}\mathsf{\text{d}}x\mathsf{\text{d}}t\\ \le C+C{\int }_{0}^{T}\int {\varphi }^{4}\mid \nabla \varphi {\mid }^{2}\mathsf{\text{d}}x\mathsf{\text{d}}t\\ \le C+C\parallel \nabla \varphi {\parallel }_{{L}^{\infty }\left(0,T;{L}^{2}\right)}^{2}{\int }_{0}^{T}\parallel \varphi {\parallel }_{{L}^{\infty }}^{4}\mathsf{\text{d}}t\\ \le C+C{\int }_{0}^{T}\parallel \varphi {\parallel }_{{L}^{6}}^{2}\parallel \varphi {\parallel }_{{H}^{2}}^{2}\mathsf{\text{d}}t\\ \le C+C\parallel \varphi {\parallel }_{{L}^{\infty }\left(0,T;{H}^{1}\right)}^{2}\underset{0}{\overset{T}{\int }}\parallel \varphi {\parallel }_{{H}^{2}}^{2}\mathsf{\text{d}}t\le C,\end{array}$
(2.8)
which yields
$\parallel \varphi {\parallel }_{{L}^{2}\left(0,T;{H}^{3}\right)}\le C,$
(2.9)
$\parallel \varphi {\parallel }_{{L}^{4}\left(0,T;{L}^{\infty }\right)}\le C,$
(2.10)
$\parallel \nabla \varphi {\parallel }_{{L}^{2}\left(0,T;{L}^{\infty }\right)}\le C.$
(2.11)
Testing (1.4) by Δ2ϕ, using (1.5), (2.4), (2.3), (2.10) and (2.11), we derive
$\begin{array}{c}\phantom{\rule{1em}{0ex}}\frac{1}{2}\frac{d}{\mathsf{\text{d}}t}\int \mid \Delta \varphi {\mid }^{2}\mathsf{\text{d}}x+\int \mid {\Delta }^{2}\varphi {\mid }^{2}\mathsf{\text{d}}x\\ =-\int u\cdot \nabla \varphi \cdot {\Delta }^{2}\varphi \mathsf{\text{d}}x+\int \Delta \left({\varphi }^{3}-\varphi \right)\cdot {\Delta }^{2}\varphi \mathsf{\text{d}}x\\ \le \phantom{\rule{2.77695pt}{0ex}}\parallel u{\parallel }_{{L}^{2}}\parallel \nabla \varphi {\parallel }_{{L}^{\infty }}\parallel {\Delta }^{2}\varphi {\parallel }_{{L}^{2}}+\parallel \Delta \left({\varphi }^{3}-\varphi \right){\parallel }_{{L}^{2}}\parallel {\Delta }^{2}\varphi {\parallel }_{{L}^{2}}\\ \le C\parallel \nabla \varphi {\parallel }_{{L}^{\infty }}\parallel {\Delta }^{2}\varphi {\parallel }_{{L}^{2}}\\ \phantom{\rule{1em}{0ex}}+\phantom{\rule{2.77695pt}{0ex}}C\left(\parallel \varphi {\parallel }_{{L}^{\infty }}^{2}\parallel \Delta \varphi {\parallel }_{{L}^{2}}+\parallel \varphi {\parallel }_{{L}^{\infty }}\parallel \nabla \varphi {\parallel }_{{L}^{\infty }}\parallel \nabla \varphi {\parallel }_{{L}^{2}}+\parallel \Delta \varphi {\parallel }_{{L}^{2}}\right)\parallel {\Delta }^{2}\varphi {\parallel }_{{L}^{2}}\\ \le C\parallel \nabla \varphi {\parallel }_{{L}^{\infty }}\parallel {\Delta }^{2}\varphi {\parallel }_{{L}^{2}}\\ \phantom{\rule{1em}{0ex}}+\phantom{\rule{2.77695pt}{0ex}}C\left(\parallel \varphi {\parallel }_{{L}^{\infty }}^{2}\parallel \Delta \varphi {\parallel }_{{L}^{2}}+\parallel \varphi {\parallel }_{{H}^{2}}\parallel \nabla \varphi {\parallel }_{{L}^{\infty }}+\parallel \Delta \varphi {\parallel }_{{L}^{2}}\right)\parallel {\Delta }^{2}\varphi {\parallel }_{{L}^{2}}\\ \le \frac{1}{2}\parallel {\Delta }^{2}\varphi {\parallel }_{{L}^{2}}^{2}+\phantom{\rule{2.77695pt}{0ex}}C\parallel \nabla \varphi {\parallel }_{{L}^{\infty }}^{2}+\phantom{\rule{2.77695pt}{0ex}}C\parallel \varphi {\parallel }_{{L}^{\infty }}^{4}\parallel \Delta \varphi {\parallel }_{{L}^{2}}^{2}\\ \phantom{\rule{1em}{0ex}}+\phantom{\rule{2.77695pt}{0ex}}C\parallel \nabla \varphi {\parallel }_{{L}^{\infty }}^{2}\parallel \varphi {\parallel }_{{H}^{2}}^{2}+\phantom{\rule{2.77695pt}{0ex}}C\parallel \Delta \varphi {\parallel }_{{L}^{2}}^{2},\end{array}$
which implies
$\parallel \varphi {\parallel }_{{L}^{\infty }\left(0,T;{H}^{2}\right)}+\parallel \varphi {\parallel }_{{L}^{2}\left(0,T;{H}^{4}\right)}\le C.$
(2.12)
Testing (1.1) by -Δu + π, using (1.2), (1.6), (2.12), (2.1) and (2.4), we reach
$\begin{array}{c}\phantom{\rule{1em}{0ex}}\frac{1}{2}\frac{\mathsf{\text{d}}}{\mathsf{\text{d}}t}\int \mid \nabla u{\mid }^{2}\mathsf{\text{d}}x+\int {\left(-\Delta u+\nabla \pi \right)}^{2}\mathsf{\text{d}}x\\ =\int \left(\mu \nabla \varphi +\theta {e}_{2}-u\cdot \nabla u\right)\left(-\Delta u+\nabla \pi \right)\mathsf{\text{d}}x\\ \le \left(\parallel \mu {\parallel }_{{L}^{2}}\parallel \nabla \varphi {\parallel }_{{L}^{\infty }}+\parallel \theta {\parallel }_{{L}^{2}}+\parallel u{\parallel }_{{L}^{4}}\parallel \nabla u{\parallel }_{{L}^{4}}\right)\parallel -\Delta u+\nabla \pi {\parallel }_{{L}^{2}}\\ \le C\left(\parallel \nabla \varphi {\parallel }_{{L}^{\infty }}+1+\parallel u{\parallel }_{{L}^{2}}^{1∕2}\parallel \nabla u{\parallel }_{{L}^{2}}^{1∕2}\cdot \parallel \nabla u{\parallel }_{{L}^{2}}^{1∕2}\parallel \Delta u{\parallel }_{{L}^{2}}^{1∕2}\right)\parallel -\Delta u+\nabla \pi {\parallel }_{{L}^{2}}\\ \le C\parallel \nabla \varphi {\parallel }_{{L}^{\infty }}^{2}+C+C\parallel \nabla u{\parallel }_{{L}^{2}}^{4}+\frac{1}{2}\parallel -\Delta u+\nabla \pi {\parallel }_{{L}^{2}}^{2},\end{array}$
which yields
$\parallel u{\parallel }_{{L}^{\infty }\left(0,T;{H}^{1}\right)}+\parallel u{\parallel }_{{L}^{2}\left(0,T;{H}^{2}\right)}\le C.$
(2.13)
Here, we have used the Gagliardo-Nirenberg inequalities:
$\begin{array}{c}\parallel u{\parallel }_{{L}^{4}}^{2}\le C\parallel u{\parallel }_{{L}^{2}}\parallel \nabla u{\parallel }_{{L}^{2}},\\ \parallel \nabla u{\parallel }_{{L}^{4}}^{2}\le C\parallel \nabla u{\parallel }_{{L}^{2}}\parallel u{\parallel }_{{H}^{2}},\end{array}$
and the H2-theory of the Stokes system:
$\parallel u{\parallel }_{{H}^{2}}+\parallel \pi {\parallel }_{{H}^{1}}\le C\parallel -\Delta u+\nabla \pi {\parallel }_{{L}^{2}}.$
(2.14)
Similarly to (2.13), we have
$\parallel {\partial }_{t}u{\parallel }_{{L}^{2}\left(0,T;{L}^{2}\right)}\le C.$
(2.15)
(1.1), (1.2), (1.6) and (1.7) can be rewritten as
Using (2.12), (2.1), (2.13), and the regularity theory of Stokes system, we have
$\begin{array}{c}\phantom{\rule{1em}{0ex}}\parallel {\partial }_{t}u{\parallel }_{{L}^{2}\left(0,T;{L}^{p}\right)}+\parallel u{\parallel }_{{L}^{2}\left(0,T;{W}^{2,p}\right)}\le C\parallel g{\parallel }_{{L}^{2}\left(0,T;{L}^{p}\right)}\\ \le C\parallel \mu {\parallel }_{{L}^{2}\left(0,T;{L}^{\infty }\right)}\parallel \nabla \varphi {\parallel }_{{L}^{\infty }\left(0,T;{L}^{p}\right)}+\phantom{\rule{2.77695pt}{0ex}}C\parallel \theta {\parallel }_{{L}^{\infty }\left(0,T;{L}^{\infty }\right)}\\ \phantom{\rule{1em}{0ex}}+\phantom{\rule{2.77695pt}{0ex}}C\parallel u{\parallel }_{{L}^{\infty }\left(0,T;{L}^{2p}\right)}\parallel \nabla u{\parallel }_{{L}^{2}\left(0,T;{L}^{2p}\right)}\le C,\end{array}$
(2.16)

for any 2 < p < ∞.

(2.16) gives
$\parallel \nabla u{\parallel }_{{L}^{2}\left(0,T;{L}^{\infty }\right)}\le C.$
(2.17)
It follows from (1.3) and (1.6) that
$\Delta \theta =0\phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{2.77695pt}{0ex}}on\phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{2.77695pt}{0ex}}\partial \Omega ×\left(0,\infty \right).$
(2.18)
Applying Δ to (1.3), testing by Δθ, using (1.2), (1.6), (2.16), (2.17) and (2.18), we obtain
$\begin{array}{c}\phantom{\rule{1em}{0ex}}\frac{1}{2}\frac{\mathsf{\text{d}}}{\mathsf{\text{d}}t}\int \mid \Delta \theta {\mid }^{2}\mathsf{\text{d}}x+\epsilon \int \mid \nabla \Delta \theta {\mid }^{2}\mathsf{\text{d}}x\\ =-\int \left(\Delta \left(u\cdot \nabla \theta \right)-u\nabla \Delta \theta \right)\Delta \theta \mathsf{\text{d}}x\\ \le C\left(\parallel \Delta u{\parallel }_{{L}^{4}}\parallel \nabla \theta {\parallel }_{{L}^{4}}+\parallel \nabla u{\parallel }_{{L}^{\infty }}\parallel \Delta \theta {\parallel }_{{L}^{2}}\right)\parallel \Delta \theta {\parallel }_{{L}^{2}}\\ \le C\left(\parallel \Delta u{\parallel }_{{L}^{4}}+\parallel \nabla u{\parallel }_{{L}^{\infty }}\right)\parallel \Delta \theta {\parallel }_{{L}^{2}}^{2},\end{array}$
which implies
$\parallel \theta {\parallel }_{{L}^{\infty }\left(0,T;{H}^{2}\right)}+\sqrt{\epsilon }\parallel \theta {\parallel }_{{L}^{2}\left(0,T;{H}^{3}\right)}\le C.$
(2.19)
It follows from (1.3), (1.6), (2.19) and (2.13) that
$\parallel {\partial }_{t}\theta {\parallel }_{{L}^{\infty }\left(0,T;{L}^{2}\right)}\le C.$
(2.20)
Taking ∂ t to (1.4) and (1.5), testing by ∂ t ϕ, using (1.2), (1.6), (2.12), and (2.15), we have
$\begin{array}{c}\phantom{\rule{1em}{0ex}}\frac{1}{2}\frac{\mathsf{\text{d}}}{\mathsf{\text{d}}t}\int \mid {\partial }_{t}\varphi {\mid }^{2}\mathsf{\text{d}}x+\int \mid \Delta {\partial }_{t}\varphi {\mid }^{2}\mathsf{\text{d}}x\\ =-\int {\partial }_{t}u\cdot \nabla \varphi \cdot {\partial }_{t}\varphi \mathsf{\text{d}}x+\int \Delta \left(3{\varphi }^{2}{\partial }_{t}\varphi -{\partial }_{t}\varphi \right)\cdot {\partial }_{t}\varphi \mathsf{\text{d}}x\\ =-\int {\partial }_{t}u\cdot \nabla \varphi \cdot {\partial }_{t}\varphi \mathsf{\text{d}}x+\int \left(3{\varphi }^{2}{\partial }_{t}\varphi -{\partial }_{t}\varphi \right)\Delta {\partial }_{t}\varphi \mathsf{\text{d}}x\\ \le \phantom{\rule{2.77695pt}{0ex}}\parallel {\partial }_{t}u{\parallel }_{{L}^{2}}\parallel \nabla \varphi {\parallel }_{{L}^{\infty }}\parallel {\partial }_{t}\varphi {\parallel }_{{L}^{2}}+\left(\parallel 3\varphi {\parallel }_{{L}^{\infty }}^{2}+1\right)\parallel {\partial }_{t}\varphi {\parallel }_{{L}^{2}}\parallel \Delta {\partial }_{t}\varphi {\parallel }_{{L}^{2}}\\ \le \phantom{\rule{2.77695pt}{0ex}}\parallel {\partial }_{t}u{\parallel }_{{L}^{2}}\parallel \nabla \varphi {\parallel }_{{L}^{\infty }}\parallel {\partial }_{t}\varphi {\parallel }_{{L}^{2}}+\frac{1}{2}\parallel \Delta {\partial }_{t}\varphi {\parallel }_{{L}^{2}}^{2}+C\parallel {\partial }_{t}\varphi {\parallel }_{{L}^{2}}^{2},\end{array}$
which gives
$\parallel {\partial }_{t}\varphi {\parallel }_{{L}^{\infty }\left(0,T;{L}^{2}\right)}+\parallel {\partial }_{t}\varphi {\parallel }_{{L}^{2}\left(0,T;{H}^{2}\right)}\le C.$
(2.21)
By the regularity theory of elliptic equation, it follows from (1.4), (1.5), (1.6), (2.21), (2.13) and (2.12) that
$\begin{array}{cc}\hfill \parallel \varphi {\parallel }_{{L}^{\infty }\left(0,T;{H}^{4}\right)}& \le C\parallel \Delta \varphi {\parallel }_{{L}^{\infty }\left(0,T;{H}^{2}\right)}\le C\parallel \mu -{f}^{\prime }\left(\varphi \right){\parallel }_{{L}^{\infty }\left(0,T;{H}^{2}\right)}\hfill \\ \phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{2.77695pt}{0ex}}\le C\parallel \mu {\parallel }_{{L}^{\infty }\left(0,T;{H}^{2}\right)}+\phantom{\rule{2.77695pt}{0ex}}C\parallel {f}^{\prime }\left(\varphi \right){\parallel }_{{L}^{\infty }\left(0,T;{H}^{2}\right)}\hfill \\ \phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{2.77695pt}{0ex}}\le C\parallel \Delta \mu {\parallel }_{{L}^{\infty }\left(0,T;{L}^{2}\right)}+\phantom{\rule{2.77695pt}{0ex}}C\parallel {f}^{\prime }\left(\varphi \right){\parallel }_{{L}^{\infty }\left(0,T;{H}^{2}\right)}\hfill \\ \phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{2.77695pt}{0ex}}\le C\parallel {\partial }_{t}\varphi +u\cdot \nabla \varphi {\parallel }_{{L}^{\infty }\left(0,T;{L}^{2}\right)}+C\parallel {f}^{\prime }\left(\varphi \right){\parallel }_{{L}^{\infty }\left(0,T;{H}^{2}\right)}\hfill \\ \phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{2.77695pt}{0ex}}\le C\parallel {\partial }_{t}\varphi {\parallel }_{{L}^{\infty }\left(0,T;{L}^{2}\right)}+\phantom{\rule{2.77695pt}{0ex}}C\parallel u{\parallel }_{{L}^{\infty }\left(0,T;{L}^{4}\right)}\parallel \nabla \varphi {\parallel }_{{L}^{\infty }\left(0,T;{L}^{4}\right)}\hfill \\ \phantom{\rule{1em}{0ex}}+\phantom{\rule{2.77695pt}{0ex}}C\parallel {f}^{\prime }\left(\varphi \right){\parallel }_{{L}^{\infty }\left(0,T;{H}^{2}\right)}\le C.\hfill \end{array}$
(2.22)
Taking ∂ t to (1.1), testing by ∂ t u, using (1.2), (1.6), (2.17), (2.22), (2.21) and (1.5), we conclude that
$\begin{array}{c}\phantom{\rule{1em}{0ex}}\frac{1}{2}\frac{\mathsf{\text{d}}}{\mathsf{\text{d}}t}\int \mid {\partial }_{t}u{\mid }^{2}\mathsf{\text{d}}x+\int \mid \nabla {\partial }_{t}u{\mid }^{2}\mathsf{\text{d}}x\\ =-\int {\partial }_{t}u\cdot \nabla u\cdot {\partial }_{t}u\mathsf{\text{d}}x+\int \left({\partial }_{t}\mu \cdot \nabla \varphi +\mu \cdot \nabla {\partial }_{t}\varphi +{\partial }_{t}\theta {e}_{2}\right)\phantom{\rule{0.3em}{0ex}}{\partial }_{t}u\mathsf{\text{d}}x\\ \le \phantom{\rule{2.77695pt}{0ex}}\parallel \nabla u{\parallel }_{{L}^{\infty }}\parallel {\partial }_{t}u{\parallel }_{{L}^{2}}^{2}+\phantom{\rule{2.77695pt}{0ex}}\left(\parallel {\partial }_{t}u{\parallel }_{{L}^{2}}\parallel \nabla \varphi {\parallel }_{{L}^{\infty }}+\parallel \mu {\parallel }_{{L}^{\infty }}\parallel \nabla {\partial }_{t}\varphi {\parallel }_{{L}^{2}}+\parallel {\partial }_{t}\theta {\parallel }_{{L}^{2}}\right)\parallel {\partial }_{t}u{\parallel }_{{L}^{2}}\\ \le \phantom{\rule{2.77695pt}{0ex}}\parallel \nabla u{\parallel }_{{L}^{\infty }}\parallel {\partial }_{t}u{\parallel }_{{L}^{2}}^{2}+\phantom{\rule{2.77695pt}{0ex}}C\left(\parallel \Delta {\partial }_{t}\varphi {\parallel }_{{L}^{2}}+\parallel {\partial }_{t}\left({\varphi }^{3}-\varphi \right){\parallel }_{{L}^{2}}+\parallel \nabla {\partial }_{t}\varphi {\parallel }_{{L}^{2}}+1\right)\parallel {\partial }_{t}u{\parallel }_{{L}^{2}},\end{array}$
which implies
$\parallel {\partial }_{t}u{\parallel }_{{L}^{\infty }\left(0,T;{L}^{2}\right)}+\parallel {\partial }_{t}u{\parallel }_{{L}^{2}\left(0,T;{H}^{1}\right)}\le C.$
(2.23)
Using (2.23), (2.22), (2.1), (2.13), (1.1), (1.2), (1.6) and the H2-theory of the Stokes system, we arrive at
$\parallel u{\parallel }_{{L}^{\infty }\left(0,T;{H}^{2}\right)}\le C.$

This completes the proof.

## Declarations

### Acknowledgements

This study was supported by the NSFC (No. 11171154) and NSFC (Grant No. 11101376).

## Authors’ Affiliations

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

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