Continuous Dependence for the Boussinesq Equations under Reaction Boundary Conditions in R²

Abstract

In this paper, we studied the continuous dependence result for the Boussinesq equations. We considered the case where $\Omega$ was a bounded domain in $R^2$. Temperatures T and C satisfied reaction boundary conditions. A first-order inequality for the differences of energy could be derived. An integration of this inequality produced a continuous dependence result. The result told us that the continuous dependence type stability was also valid for the Boussinesq coefficient $\lambda$ of the Boussinesq equations with reaction boundary conditions.

1. Introduction

Recently, many researchers have began to study the results of continuous dependence for different types of constructive coefficients. These stabilities are usually reffered to as structural stability. Unlike the studies of the traditional stability, they do not focus on the change of initial data, and they mainly focus on the change of the model itself. Let us introduce some papers explaining the nature of this structural stability. Ames and Straughan in their monograph [1] explained the the natural of the stability. In continuum mechanics, it is very important to establish the structural stability for different models. For example, in [2], the authors obtained some results about the structural stability for different equations and showed the importance of this stability. It is meaningful to discuss structural stability, for we wonder if a tiny variation in the coefficient may lead to a sharp change in the solution. What is more is that many errors exist in each step of establishing a model. We want to know whether the errors will affect the correctness of the model. The study of the structural stability can solve this problem.

Many papers in the literature have studied the behavior of solutions of fluid equations in porous media. In the book of [3], some models in porous media were introduced in detail and their properties were studied. In paper [4], the authors studied the spatial behavior of the solution for a class of Brinkman and Forchheimer-type equations in porous media, and they obtained results for the spatial decay estimates of the solutions. Some recent research results on the structural stability of fluid equations in porous media can be observed in [5,6,7,8,9,10,11,12,13,14,15]. For a more detailed understanding of the structural stability, readers should refer to [16,17,18,19,20,21,22,23]. In these papers, the fluid models were widely studied. Other studies about the stability for wave equations may be found in [24,25,26,27,28,29,30,31,32,33].

The studies of the structural stability are famous for the Brinkman, Forchheimer and Darcy equations in porous media. Little attention has paid to the study of the Boussinesq equations. The nonlinear term in the Boussineq equations is different from the Brinkman and Forchheimer-type equations. The methods used in previous cannot be applicable. The Boussinesq equations have many applications in reality. We want to know if a small error can cause a big perturbation of the solution. What is more is that the problem we studied in this paper has the same nonlinear item, $u_j{u}_{i,j}$, as the Navier–Stokes equation; thus, the method proposed in this paper can be useful in studying the properties of the solution of the Navier–Stokes equation. In this sense, our paper is interesting and involved. In [34], the authors studied the Boussinesq equations. They obtained structural stability results for the constructive coefficient. The region they discussed was a bounded region in $R^2$. In [22], Liu studied the same equations and obtained a result of structural stability. They considered the same problem as [34], but the region they discussed was a bounded region in $R^3$. The result of the the continuous dependence type was obtained for the Boussinesq coefficient $\lambda$. In this paper, we want to continue to study the continuous dependence for these equations in a bounded domain in $R^2$. We suppose that temperatures T and C satisfy the reaction boundary conditions. Under these conditions, we cannot obtain the bounds for T and C, and the methods proposed in [22,34], which are based on the maximum bounds of T and C, cannot be used in the present paper. The main difficulty is how to tackle the convection terms. Additionally, the boundary conditions are different from [22,34]. It is difficult to deal with the terms on the boundary. With the aid of the $L_4$ bounds for T and C and some useful inequalities, we can overcome this difficulty. In our opinion, we can overcome this difficulty and obtain the bound for ${\int }_0^t{\int }_{\Omega }{u}_{i,j}{u}_{i,j}dxd\eta$ by using the methods proposed by the paper in the literature. Using these methods, we cannot obtain the bound for ${\int }_{\Omega }{u}_{i,j}{u}_{i,j}dx$. The biggest innovation of this paper is that we follow a new method to obtain the bound for ${\int }_{\Omega }{u}_{i,j}{u}_{i,j}dx$. The energy method is widely used in these studies.

In this paper, the comma is used to denote partial differentiation. The symbol $u_{i,k}$ denotes $\frac{\partial u_i}{\partial x_k}$. The repeated Latin subscripts denote summation. Hence, $u_{i,i}={\displaystyle \sum _{i=1}^2}\frac{\partial u_i}{\partial x_i}$, and $dx=d{x}_{1}d{x}_2$.

The fundamental model we study is based upon the equations of balance of momentum, balance of mass, conservation of energy and conservation of salt concentration (see [34]). Let $(u_i,T,C,p)$ denote velocity, temperature, salt concentration and pressure in $\Omega$, where $\Omega$ is a bounded star-shaped domain in $R^2$. We will study the following Boussinesq equations (see [22]).

$$\begin{array}{cccc}\hfill & \frac{\partial u_i}{\partial t}-\Delta u_i+\lambda u_j\frac{\partial u_i}{\partial x_j}+g_{i}T-h_{i}C+\frac{\partial p}{\partial x_i}=0,\hfill & \hfill & \mathrm{in}\Omega \times [0,\tau ],\hfill \\ \hfill & \frac{\partial u_i}{\partial x_i}=0,\hfill & \hfill & \mathrm{in}\Omega \times [0,\tau ],\hfill \\ \hfill & \frac{\partial T}{\partial t}+u_i\frac{\partial T}{\partial x_i}=\Delta T,\hfill & \hfill & \mathrm{in}\Omega \times [0,\tau ],\hfill \\ \hfill & \frac{\partial C}{\partial t}+u_i\frac{\partial C}{\partial x_i}=\Delta C+LT-kC,\hfill & \hfill & \mathrm{in}\Omega \times [0,\tau ].\hfill \end{array}$$

(1)

In Equation (1), $g_i$ and $h_i$ denote gravity functions. For the sake of simplicity, we provide bounds for $g_i$ and $h_i$: $\mid g\mid ={\left(g_i{g}_i\right)}^{\frac{1}{2}}\le 1$ and $\mid h\mid ={\left(h_i{h}_i\right)}^{\frac{1}{2}}\le 1$. $\Delta$ denotes the Laplacian operator. L and k are non-negative coefficients. $\lambda$ denotes the Boussinesq coefficient. $\tau$ is a nonnegative constant. Equation (1) follows in practice by employing a Boussinesq approximation, which accounts for variable C, allowing the incompressibility condition to hold (see [32,34]).

The following boundary conditions are satisfied:

$$\begin{array}{cccc}\hfill & u_i=0,\frac{\partial T}{\partial n}=k_{1}T,\frac{\partial C}{\partial n}=k_{2}C,\hfill & \hfill & \mathrm{on}\partial \Omega \times [0,\tau ],\hfill \end{array}$$

(2)

for the prescribed positive reaction boundary coefficients $k_1$ and $k_2$, and $n_i$ is the unit-outward normal. We impose the following conditions on $t=0$:

$$\begin{array}{cccc}\hfill & u_i(x,0)=f_i\left(x\right),T(x,0)=T_0\left(x\right),C(x,0)=C_0\left(x\right),\hfill & \hfill & \mathrm{in}\Omega ,\hfill \end{array}$$

(3)

where $f_i$, $T_0$ and $C_0$ are prescribed functions.

The structure of the article is as follows. In Section 2, some lemmas will be obtained. In Section 3, the main result of this paper will be obtained. In the last Section, some conclusions are provided.

2. Some Important Bounds

In this section, we will obtain some important bounds that will be used in proving our main results.

Lemma 1.

For an arbitrary differentiable function $H=H(x,t)$, we have the following estimates:

$${\int }_{\partial \Omega }{H}^{2}dS\le \left[\frac{2}{m}+\frac{2d^2{\epsilon }_0}{m^2}\right]{\int }_{\Omega }{H}^{2}dx+\frac{1}{2{\epsilon }_0}{\int }_{\Omega }{H}_{,i}{H}_{,i}dx,$$

(4)

where m and d are positive constants, and ${\epsilon }_0$ is a positive constant.

Proof. 

From the divergence theorem, we have the following.

$$\begin{array}{c}\hfill {\int }_{\partial \Omega }{H}^{2}x·\overrightarrow{n}dS={\int }_{\Omega }div\left(H^{2}x\right)dx=2{\int }_{\Omega }{H}^{2}dx+2{\int }_{\Omega }H(x·\nabla H)dx.\end{array}$$

Since $\Omega$ is a strictly convex domain, we set the following.

$$m=\underset{\partial \Omega }{min}{x}_i{n}_i>0,d^2=\underset{\Omega }{max}{x}_i{x}_i.$$

We have the following.

$$m{\int }_{\partial \Omega }{H}^{2}dS\le {\int }_{\partial \Omega }{H}^{2}x·\overrightarrow{n}dS\le 2{\int }_{\Omega }{H}^{2}dx+2{\int }_{\Omega }H(x·\nabla H)dx.$$

We can easily obtain the following.

$${\int }_{\partial \Omega }{H}^{2}dS\le \left[\frac{2}{m}+\frac{2d^2{\epsilon }_0}{m^2}\right]{\int }_{\Omega }{H}^{2}dx+\frac{1}{2{\epsilon }_0}{\int }_{\Omega }{H}_{,i}{H}_{,i}dx.$$

The proof is complete. □

Lemma 2.

For temperature T and concentration C, we have the following estimates:

$$\begin{array}{c}\hfill {\int }_0^t{\int }_{\Omega }(T^2+C^2)dxd\eta \le k_1\left(t\right),\end{array}$$

(5)

and the following is the case:

$$\begin{array}{cc}\hfill {\int }_0^t{\int }_{\Omega }(T_{,i}{T}_{,i}+C_{,i}{C}_{,i})dxd\eta & \le k_2\left(t\right),\hfill \end{array}$$

(6)

where $k_1\left(t\right)$ and $k_2\left(t\right)$ are positive functions defined later.

Proof. 

Multiplying ${\left(1\right)}_3$ by $2T$ and integrating over $\Omega \times [0,t]$, we have the following.

$$\begin{array}{c}\hfill {\int }_{\Omega }{T}^{2}dx+2{\int }_0^t{\int }_{\Omega }{T}_{,i}{T}_{,i}dxd\eta \le {\int }_{\Omega }{T}_0^{2}dx+2k_1{\int }_0^t{\oint }_{\partial \Omega }{T}^{2}dSd\eta .\end{array}$$

(7)

We can also obtain the following.

$$\begin{array}{c}\hfill {\int }_{\Omega }{C}^{2}dx+2{\int }_0^t{\int }_{\Omega }{C}_{,i}{C}_{,i}dxd\eta \le {\int }_{\Omega }{C}_0^{2}dx+\frac{L^2}{2k}{\int }_0^t{\int }_{\Omega }{T}^{2}dxd\eta +2k_2{\int }_0^t{\oint }_{\partial \Omega }{C}^{2}dSd\eta .\end{array}$$

(8)

Combining (7) and (8), we obtain the following

$$\begin{array}{cc}\hfill {\int }_{\Omega }(T^2+C^2)dx& +2{\int }_0^t{\int }_{\Omega }(T_{,i}{T}_{,i}+C_{,i}{C}_{,i})dxd\eta \le {\int }_{\Omega }(T_0^2+C_0^2)dx\hfill \\ & +\frac{L^2}{2k}{\int }_0^t{\int }_{\Omega }{T}^{2}dxd\eta +2{\int }_0^t{\oint }_{\partial \Omega }(k_1{T}^2+k_2{C}^2)dSd\eta .\hfill \end{array}$$

(9)

Using (4) and choosing ${\epsilon }_0=2k_1$, we have the following.

$$\begin{array}{c}\hfill 2k_1{\oint }_{\partial \Omega }{T}^{2}dS\le \left[\frac{2}{m}+\frac{4d^2{k}_1}{m^2}\right]{\int }_{\Omega }{T}^{2}dx+\frac{1}{2}{\int }_{\Omega }{T}_{,i}{T}_{,i}dx.\end{array}$$

(10)

Similarly, we can obtain the following.

$$\begin{array}{c}\hfill 2k_2{\oint }_{\partial \Omega }{C}^{2}dS\le \left[\frac{2}{m}+\frac{4d^2{k}_2}{m^2}\right]{\int }_{\Omega }{C}^{2}dx+\frac{1}{2}{\int }_{\Omega }{C}_{,i}{C}_{,i}dx.\end{array}$$

(11)

Inserting (10) and (11) into (9), we have the following.

$$\begin{array}{cc}\hfill {\int }_{\Omega }(T^2+C^2)dx& +\frac{1}{2}{\int }_0^t{\int }_{\Omega }(T_{,i}{T}_{,i}+C_{,i}{C}_{,i})dxd\eta \le {\int }_{\Omega }(T_0^2+C_0^2)dx\hfill \\ & +m_1{\int }_0^t{\int }_{\Omega }(T^2+C^2)dxd\eta ,\hfill \end{array}$$

(12)

with $m_1=max\left\{\frac{L^2}{2k}+\frac{2}{m}+\frac{4d^2{k}_1}{m^2},\frac{2}{m}+\frac{4d^2{k}_2}{m^2}\right\}$.

An integration of (12) provides the following.

$$\begin{array}{c}\hfill {\int }_0^t{\int }_{\Omega }(T^2+C^2)dxd\eta \le e^{m_{1}t}t{\int }_{\Omega }(T_0^2+C_0^2)dx=k_1\left(t\right).\end{array}$$

(13)

An insertion of (13) into (12) provides the following.

$$\begin{array}{cc}\hfill {\int }_0^t{\int }_{\Omega }(T_{,i}{T}_{,i}+C_{,i}{C}_{,i})dxd\eta & \le 2{\int }_{\Omega }(T_0^2+C_0^2)dx+2m_1{k}_1\left(t\right)\hfill \\ & =k_2\left(t\right).\hfill \end{array}$$

(14)

□

Lemma 3.

We suggest that $u_i$ is a solution of Equation (1) and satisfies the initial boundary conditions $\left(2\right)$ and $\left(3\right)$. The following estimates can be obtained:

$$\begin{array}{c}\hfill {\int }_{{\Omega }_1}{u}_i{u}_{i}dx\le k_3\left(t\right),\end{array}$$

(15)

where $k_3\left(t\right)$ is a positive function defined later.

Proof. 

By multiplying ${\left(1\right)}_1$ by $u_i$ and integrating, we obtain the following.

$$\begin{array}{c}\hfill \frac{1}{2}\frac{d}{dt}{\int }_{\Omega }{u}_i{u}_{i}dx+{\int }_{\Omega }{u}_{i,j}{u}_{i,j}dx\le \frac{1}{2}{\int }_{\Omega }{T}^{2}dx+\frac{1}{2}{\int }_{\Omega }{C}^{2}dx+{\int }_{\Omega }{u}_i{u}_{i}dx.\end{array}$$

(16)

An integration of (16) provides the following.

$$\begin{array}{c}\hfill {\int }_{\Omega }{u}_i{u}_{i}dx\le e^{2t}\left({\int }_{\Omega }{u}_{i0}{u}_{i0}dx+k_1\left(t\right)\right)=k_3\left(t\right).\end{array}$$

(17)

The proof is complete. □

Lemma 4.

We suggest that T and C are the solutions of Equation (1) and satisfy boundary conditions $\left(2\right)$ and $\left(3\right)$. The following estimates can be obtained:

$$\begin{array}{c}\hfill {\int }_{\Omega }{T}^{4}dx+{\int }_{\Omega }{C}^{4}dx\le k_4\left(t\right),\end{array}$$

(18)

where $k_4\left(t\right)=(\gamma e^{\gamma t}t+1){\int }_{\Omega }(T_0^4+C_0^4)dx$.

Proof. 

Multiplying ${\left(1\right)}_3$ by $4T^3$ and integrating over $\Omega \times [0,t]$, we have the following.

$$\begin{array}{c}\hfill {\int }_{\Omega }{T}^{4}dx+3{\int }_0^t{\int }_{\Omega }{\left(T^2\right)}_{,i}{\left(T^2\right)}_{,i}dxd\eta \le {\int }_{\Omega }{T}_0^{4}dx+4k_1{\int }_0^t{\oint }_{\partial \Omega }{T}^{4}dSd\eta .\end{array}$$

(19)

Multiplying ${\left(1\right)}_4$ by $4C^3$ and integrating over $\Omega \times [0,t]$, we have the following.

$$\begin{array}{cc}\hfill {\int }_{\Omega }{C}^{4}dx+3{\int }_0^t{\int }_{\Omega }{\left(C^2\right)}_{,i}{\left(C^2\right)}_{,i}dxd\eta & \le {\int }_{\Omega }{C}_0^{4}dx+4k_2{\int }_0^t{\oint }_{\partial \Omega }{C}^{4}dSd\eta \hfill \\ & +4L{\int }_0^t{\int }_{\Omega }{C}^{3}Tdxd\eta \hfill \\ & \le {\int }_{\Omega }{C}_0^{4}dx+4k_2{\int }_0^t{\oint }_{\partial \Omega }{C}^{4}dSd\eta \hfill \\ & +3L{\int }_0^t{\int }_{\Omega }{C}^{4}dxd\eta +L{\int }_0^t{\int }_{\Omega }{T}^{4}dxd\eta .\hfill \end{array}$$

(20)

A combination of (19) and (20) provides the following.

$$\begin{array}{cc}\hfill {\int }_{\Omega }(T^4+C^4)dx& +3{\int }_0^t{\int }_{\Omega }({\left(T^2\right)}_{,i}{\left(T^2\right)}_{,i}+{\left(C^2\right)}_{,i}{\left(C^2\right)}_{,i})dxd\eta \le {\int }_{\Omega }(T_0^4+C_0^4)dx\hfill \\ & +(4k_1+4k_2){\int }_0^t{\oint }_{\partial \Omega }(T^4+C^4)dSd\eta +3L{\int }_0^t{\int }_{\Omega }(T^4+C^4)dxd\eta .\hfill \end{array}$$

(21)

Using (4), we obtain the following:

$${\int }_{\partial \Omega }{T}^{4}dS\le \left[\frac{2}{m}+\frac{2d^2{\epsilon }_0}{m^2}\right]{\int }_{\Omega }{T}^{4}dx+\frac{1}{2{\epsilon }_0}{\int }_{\Omega }{T}_{,i}^2{T}_{,i}^{2}dx,$$

(22)

and the following is the case.

$${\int }_{\partial \Omega }{C}^{4}dS\le \left[\frac{2}{m}+\frac{2d^2{\epsilon }_0}{m^2}\right]{\int }_{\Omega }{C}^{4}dx+\frac{1}{2{\epsilon }_0}{\int }_{\Omega }{C}_{,i}^2{C}_{,i}^{2}dx.$$

(23)

Inserting (22) and (14) into (21) and choosing ${\epsilon }_0=k_1+k_2$, we have the following:

$$\begin{array}{cc}\hfill {\int }_{\Omega }(T^4+C^4)dx& \le {\int }_{\Omega }(T_0^4+C_0^4)dx+\gamma {\int }_0^t{\int }_{\Omega }(T^4+C^4)dxd\eta ,\hfill \end{array}$$

(24)

with $\gamma =(4k_1+4k_2)\left[\frac{2}{m}+\frac{2d^2(k_1+k_2)}{m^2}\right]+3L$.

Integrating (24), we obtain the following.

$$\begin{array}{cc}\hfill {\int }_0^t{\int }_{\Omega }(T^4+C^4)dxd\eta & \le e^{\gamma t}t{\int }_{\Omega }(T_0^4+C_0^4)dx.\hfill \end{array}$$

(25)

Inserting (25) into (24), we obtain the following.

$$\begin{array}{cc}\hfill {\int }_{\Omega }(T^4+C^4)dx& \le (\gamma e^{\gamma t}t+1){\int }_{\Omega }(T_0^4+C_0^4)dx.\hfill \end{array}$$

(26)

The proof is finished. □

Lemma 5.

We suggest that $u_i$ is a solution of Equation (1) and that it satisfies the initial boundary conditions $\left(2\right)$ and $\left(3\right)$. The following estimates can be obtained.

$$\begin{array}{cc}\hfill {\int }_{\Omega }{u}_{i,j}{u}_{i,j}dx& \le {\left(k_3\left(t\right)\right)}^{\frac{1}{2}}{\left({\int }_{\Omega }{u}_{i,t}{u}_{i,t}dx\right)}^{\frac{1}{2}}+2{\left(k_1\left(t\right)\right)}^{\frac{1}{2}}{\left(k_3\left(t\right)\right)}^{\frac{1}{2}}.\hfill \end{array}$$

(27)

Proof. 

Multiplying ${\left(1\right)}_1$ by $u_i$ and integrating over $\Omega$, we have the following.

$$\begin{array}{c}\hfill {\int }_{\Omega }{u}_{i,t}{u}_{i}dx=-{\int }_{\Omega }{u}_{i,j}{u}_{i,j}dx+{\int }_{\Omega }{g}_i{u}_{i}Tdx-{\int }_{\Omega }{h}_i{u}_{i}Cdx.\end{array}$$

Using the Schwarz’s inequality, we have the following.

$$\begin{array}{cc}\hfill {\int }_{\Omega }{u}_{i,j}{u}_{i,j}dx& \le {\left({\int }_{\Omega }{u}_{i,t}{u}_{i,t}dx\right)}^{\frac{1}{2}}{\left({\int }_{\Omega }{u}_i{u}_{i}dx\right)}^{\frac{1}{2}}+{\left({\int }_{\Omega }{T}^{2}dx\right)}^{\frac{1}{2}}{\left({\int }_{\Omega }{u}_i{u}_{i}dx\right)}^{\frac{1}{2}}\hfill \\ & +{\left({\int }_{\Omega }{C}^{2}dx\right)}^{\frac{1}{2}}{\left({\int }_{\Omega }{u}_i{u}_{i}dx\right)}^{\frac{1}{2}}.\hfill \end{array}$$

(28)

Inserting (5) and (15) into (28), we have the following.

$$\begin{array}{cc}\hfill {\int }_{\Omega }{u}_{i,j}{u}_{i,j}dx& \le {\left(k_3\left(t\right)\right)}^{\frac{1}{2}}{\left({\int }_{\Omega }{u}_{i,t}{u}_{i,t}dx\right)}^{\frac{1}{2}}+2{\left(k_1\left(t\right)\right)}^{\frac{1}{2}}{\left(k_3\left(t\right)\right)}^{\frac{1}{2}}.\hfill \end{array}$$

(29)

Thus, the proof is complete. □

Lemma 6.

We suggest that $u_i$ is a solution of Equation (1) and that it satisfies the initial boundary conditions $\left(2\right)$ and $\left(3\right)$. The following estimates can be obtained:

$$\begin{array}{cc}\hfill \frac{d}{dt}{\int }_{\Omega }{u}_{i,t}{u}_{i,t}dx& \le \frac{{\lambda }^2\nu k_3^{\frac{1}{2}}\left(t\right)}{2}{\left({\int }_{\Omega }{u}_{i,t}{u}_{i,t}dx\right)}^{\frac{3}{2}}+k_5\left(t\right){\int }_{\Omega }{u}_{i,t}{u}_{i,t}dx\hfill \\ & +{\int }_{\Omega }{T}_{,t}^{2}dx+{\int }_{\Omega }{C}_{,t}^{2}dx-{\int }_{\Omega }{u}_{i,jt}{u}_{i,jt}dx,\hfill \end{array}$$

(30)

where $k_5\left(t\right)$ is a positive function determined later.

Proof. 

By multiplying ${\left(1\right)}_1$ by $u_{i,t}$ and integrating over $\Omega$, we have the following.

$$\begin{array}{cc}\hfill \frac{d}{dt}{\int }_{\Omega }{u}_{i,t}{u}_{i,t}dx& =2{\int }_{\Omega }{u}_{i,t}{u}_{i,tt}dx\hfill \\ & =2{\int }_{\Omega }{u}_{i,t}{(-\lambda u_j{u}_{i,j}-p_{,i}+u_{i,jj}-g_{i}T+h_{i}C)}_{,t}dx\hfill \\ & =-2{\int }_{\Omega }{u}_{i,jt}{u}_{i,jt}dx-2\lambda {\int }_{\Omega }{u}_{i,t}{u}_{j,t}{u}_{i,j}dx-2{\int }_{\Omega }{g}_i{u}_{i,t}{T}_{,t}dx\hfill \\ & +2{\int }_{\Omega }{h}_i{u}_{i,t}{C}_{,t}dx\hfill \\ & \le -2{\int }_{\Omega }{u}_{i,jt}{u}_{i,jt}dx+2\lambda {\left({\int }_{\Omega }{\left(u_{i,t}{u}_{i,t}\right)}^{2}dx\right)}^{\frac{1}{2}}{\left({\int }_{\Omega }{u}_{i,j}{u}_{i,j}dx\right)}^{\frac{1}{2}}\hfill \\ & +2{\int }_{\Omega }{u}_{i,t}{u}_{i,t}dx+{\int }_{\Omega }{T}_{,t}^{2}dx+{\int }_{\Omega }{C}_{,t}^{2}dx.\hfill \end{array}$$

(31)

The following Sobolev inequality in $R^2$(see [34] (41)) is valid:

$$\begin{array}{cc}\hfill {\int }_{\Omega }{\left|F\right|}^{4}dx& \le \nu {\int }_{\Omega }{\left|F\right|}^{2}dx{\int }_{\Omega }{|\nabla F|}^{2}dx,\hfill \end{array}$$

(32)

with $\nu$ as a positive constant.

If we choose $F=u_{i,t}$ in (32), we have the following.

$$\begin{array}{cc}\hfill {\int }_{\Omega }{\left(u_{i,t}{u}_{i,t}\right)}^{2}dx& \le \nu {\int }_{\Omega }{u}_{i,t}{u}_{i,t}dx{\int }_{\Omega }{u}_{i,jt}{u}_{i,jt}dx.\hfill \end{array}$$

(33)

A combination of (31) and (33) provides the following.

$$\begin{array}{cc}\hfill \frac{d}{dt}{\int }_{\Omega }{u}_{i,t}{u}_{i,t}dx+{\int }_{\Omega }{u}_{i,jt}{u}_{i,jt}dx& \le \frac{{\lambda }^2\nu }{2}{\int }_{\Omega }{u}_{i,t}{u}_{i,t}dx{\int }_{\Omega }{u}_{i,j}{u}_{i,j}dx+2{\int }_{\Omega }{u}_{i,t}{u}_{i,t}dx\hfill \\ & +{\int }_{\Omega }{T}_{,t}^{2}dx+{\int }_{\Omega }{C}_{,t}^{2}dx.\hfill \end{array}$$

(34)

Inserting (27) into (34), we obtain the following:

$$\begin{array}{cc}\hfill \frac{d}{dt}{\int }_{\Omega }{u}_{i,t}{u}_{i,t}dx+{\int }_{\Omega }{u}_{i,jt}{u}_{i,jt}dx& \le \frac{{\lambda }^2\nu k_3^{\frac{1}{2}}\left(t\right)}{2}{\left({\int }_{\Omega }{u}_{i,t}{u}_{i,t}dx\right)}^{\frac{3}{2}}+k_5\left(t\right){\int }_{\Omega }{u}_{i,t}{u}_{i,t}dx\hfill \\ & +{\int }_{\Omega }{T}_{,t}^{2}dx+{\int }_{\Omega }{C}_{,t}^{2}dx,\hfill \end{array}$$

(35)

with $k_5\left(t\right)={\lambda }^2\nu k_1^{\frac{1}{2}}\left(t\right)k_3^{\frac{1}{2}}\left(t\right)+2$.

Thus, the proof is complete. □

We suggest that T and C are the solutions of Equation (1) and that they satisfy boundary conditions $\left(2\right)$ and $\left(3\right)$. The following estimates can be obtained:

$$\begin{array}{cc}\hfill \frac{d}{dt}{\int }_{\Omega }{T}_{,t}^{2}dx& \le \left(2k_1\left[\frac{2}{m}+\frac{2d^2{k}_1}{m^2}\right]+1\right){\int }_{\Omega }{T}_{,t}{T}_{,t}dx\hfill \\ & +2k_4\left(t\right){\nu }^2{\int }_{\Omega }{u}_{i,t}{u}_{i,t}dx+\frac{1}{2}{\int }_{\Omega }{u}_{i,jt}{u}_{i,jt}dx,\hfill \end{array}$$

(36)

and the following is the case.

$$\begin{array}{cc}\hfill \frac{d}{dt}{\int }_{\Omega }{C}_{,t}^{2}dx& \le \left(2k_2\left[\frac{2}{m}+\frac{2d^2{k}_2}{m^2}\right]+1\right){\int }_{\Omega }{C}_{,t}{C}_{,t}dx\hfill \\ & +2k_4\left(t\right){\nu }^2{\int }_{\Omega }{u}_{i,t}{u}_{i,t}dx+\frac{1}{2}{\int }_{\Omega }{u}_{i,jt}{u}_{i,jt}dx+\frac{L^2}{k}{\int }_{\Omega }{T}_{,t}{T}_{,t}dx.\hfill \end{array}$$

(37)

Proof. 

Multiplying ${\left(1\right)}_4$ by $C_{,t}$ and integrating over $\Omega \times [0,t]$, we have the following.

$$\begin{array}{cc}\hfill \frac{d}{dt}{\int }_{\Omega }{C}_{,t}^{2}dx& =2{\int }_{\Omega }{C}_{,t}{C}_{,tt}dx\hfill \\ & =2{\int }_{\Omega }{C}_{,t}{(-u_i{C}_{,i}+C_{,jj}+LT-kC)}_{,t}dx\hfill \\ & =-2{\int }_{\Omega }{C}_{j,t}{C}_{j,t}dx+2{\oint }_{\partial \Omega }{C}_{,t}\frac{\partial C_{,t}}{\partial n}dS+2{\int }_{\Omega }{C}_{i,t}{u}_{i,t}Cdx\hfill \\ & +2L{\int }_{\Omega }{C}_{,t}{T}_{,t}dx-2k{\int }_{\Omega }{C}_{,t}{C}_{,t}dx\hfill \\ & \le -2{\int }_{\Omega }{C}_{j,t}{C}_{j,t}dx+2k_2{\oint }_{\partial \Omega }{C}_{,t}{C}_{,t}dS+\frac{L^2}{k}{\int }_{\Omega }{T}_{,t}{T}_{,t}dx-k{\int }_{\Omega }{C}_{,t}{C}_{,t}dx\hfill \\ & +2{\left({\int }_{\Omega }{C}^{4}dx\right)}^{\frac{1}{4}}{\left({\int }_{\Omega }{\left(u_{i,t}{u}_{i,t}\right)}^{2}dx\right)}^{\frac{1}{4}}{\left({\int }_{\Omega }{C}_{,it}{C}_{,it}dx\right)}^{\frac{1}{2}}.\hfill \end{array}$$

Using (4) and (18), we obtain the following.

$$\begin{array}{cc}\hfill \frac{d}{dt}{\int }_{\Omega }{C}_{,t}^{2}dx& \le -\left(2-\frac{k_2}{{\epsilon }_0}\right){\int }_{\Omega }{C}_{j,t}{C}_{j,t}dx+2k_2\left[\frac{2}{m}+\frac{2d^2{\epsilon }_0}{m^2}\right]{\int }_{\Omega }{C}_{,t}{C}_{,t}dx\hfill \\ & +2{\left(k_4\left(t\right)\right)}^{\frac{1}{2}}{\left({\int }_{\Omega }{\left(u_{i,t}{u}_{i,t}\right)}^{2}dx\right)}^{\frac{1}{2}}+\frac{L^2}{k}{\int }_{\Omega }{T}_{,t}{T}_{,t}dx+{\int }_{\Omega }{C}_{,t}{C}_{,t}dx.\hfill \end{array}$$

(38)

If we choose ${\epsilon }_0=k_2$, we obtain the following.

$$\begin{array}{cc}\hfill \frac{d}{dt}{\int }_{\Omega }{C}_{,t}^{2}dx& \le -{\int }_{\Omega }{C}_{,jt}{C}_{,jt}dx+2k_2\left[\frac{2}{m}+\frac{2d^2{k}_2}{m^2}\right]{\int }_{\Omega }{C}_{,t}{C}_{,t}dx\hfill \\ & +2{\left(k_4\left(t\right)\right)}^{\frac{1}{2}}{\left({\int }_{\Omega }{\left(u_{i,t}{u}_{i,t}\right)}^{2}dx\right)}^{\frac{1}{2}}+\frac{L^2}{k}{\int }_{\Omega }{T}_{,t}{T}_{,t}dx+\frac{1}{2}{\int }_{\Omega }{C}_{,it}{C}_{,it}dx.\hfill \end{array}$$

(39)

Using (33) again, we obtain the following.

$$\begin{array}{cc}\hfill \frac{d}{dt}{\int }_{\Omega }{C}_{,t}^{2}dx& \le \left(2k_2\left[\frac{2}{m}+\frac{2d^2{k}_2}{m^2}\right]+1\right){\int }_{\Omega }{C}_{,t}{C}_{,t}dx\hfill \\ & +2k_4\left(t\right){\nu }^2{\int }_{\Omega }{u}_{i,t}{u}_{i,t}dx+\frac{1}{2}{\int }_{\Omega }{u}_{i,jt}{u}_{i,jt}dx+\frac{L^2}{k}{\int }_{\Omega }{T}_{,t}{T}_{,t}dx.\hfill \end{array}$$

(40)

Following the same procedures, we can obtain the following.

$$\begin{array}{cc}\hfill \frac{d}{dt}{\int }_{\Omega }{T}_{,t}^{2}dx& \le \left(2k_1\left[\frac{2}{m}+\frac{2d^2{k}_1}{m^2}\right]+1\right){\int }_{\Omega }{T}_{,t}{T}_{,t}dx\hfill \\ & +2k_4\left(t\right){\nu }^2{\int }_{\Omega }{u}_{i,t}{u}_{i,t}dx+\frac{1}{2}{\int }_{\Omega }{u}_{i,jt}{u}_{i,jt}dx.\hfill \end{array}$$

(41)

Thus, the proof is complete. □

Lemma 7.

We suggest that $u_i$, T and C are the solutions of Equation (1) and satisfy boundary conditions $\left(2\right)$ and $\left(3\right)$. The following estimates can be obtained:

$$\begin{array}{cc}\hfill {\int }_{\Omega }{u}_{i,t}{u}_{i,t}dx+{\int }_{\Omega }{T}_{,t}{T}_{,t}dx+{\int }_{\Omega }{C}_{,t}{C}_{,t}dx& \le k_8\left(t\right),\hfill \end{array}$$

(42)

where $k_8\left(t\right)$ is a positive function.

Proof. 

We define a new function $G\left(t\right)$ by the following.

$$\begin{array}{c}\hfill G\left(t\right)={\int }_{\Omega }{u}_{i,t}{u}_{i,t}dx+{\int }_{\Omega }{T}_{,t}{T}_{,t}dx+{\int }_{\Omega }{C}_{,t}{C}_{,t}dx.\end{array}$$

A combination of (30), (36) and (37) provides the following:

$$\begin{array}{cc}\hfill \frac{d}{dt}G\left(t\right)& \le k_6\left(t\right)G^{\frac{3}{2}}\left(t\right)+k_7\left(t\right)G\left(t\right),\hfill \end{array}$$

(43)

with $k_6\left(t\right)=\frac{{\lambda }^2\nu k_3^{\frac{1}{2}}\left(t\right)}{2}$ and $k_7\left(t\right)=k_4\left(t\right)(1+4{\nu }^2)+\left(2k_1\left[\frac{2}{m}+\frac{2d^2{k}_1}{m^2}\right]\right)$$+\left(2k_2\left[\frac{2}{m}+\frac{2d^2{k}_2}{m^2}\right]+1\right)$.

We can also obtain the following.

$$\begin{array}{cc}\hfill \frac{d}{dt}G\left(t\right)& \le k_6\left(\tau \right)G^{\frac{3}{2}}\left(t\right)+k_7\left(\tau \right)G\left(t\right).\hfill \end{array}$$

(44)

Integrating (44) from 0 to t, we obtain the following.

$$\begin{array}{cc}\hfill G& \le {\left(\frac{k_6\left(\tau \right)\sqrt{G\left(0\right)}}{k_6\left(\tau \right)\sqrt{G\left(0\right)}\left(e^{-\frac{1}{2}{k}_7\left(\tau \right)t}-1\right)+k_7\left(\tau \right)e^{-\frac{1}{2}{k}_7\left(\tau \right)t}}\right)}^2\hfill \\ & =k_8\left(t\right).\hfill \end{array}$$

(45)

Inequality (45) is valid when $t\le \frac{2ln\frac{k_6\left(\tau \right)\sqrt{G\left(0\right)}+k_7\left(\tau \right)}{k_6\left(\tau \right)\sqrt{G\left(0\right)}}}{k_7\left(\tau \right)}$.

If the initial data $G\left(0\right)$ are small enough, we can obtain the result where inequality (45) is valid for all t.

Thus, the proof is complete. □

We suggest that $u_i$ is a solution of Equation (1) and satisfies the initial boundary conditions $\left(2\right)$ and $\left(3\right)$. The following estimates can be obtained:

$$\begin{array}{cc}\hfill {\int }_{\Omega }{u}_{i,j}{u}_{i,j}dx& \le k_9\left(t\right),\hfill \end{array}$$

(46)

where $k_9\left(t\right)$ is a positive function defined later.

Proof. 

By inserting (42) into (27), we obtain the following.

$$\begin{array}{cc}\hfill {\int }_{\Omega }{u}_{i,j}{u}_{i,j}dx& \le {\left(k_3\left(t\right)\right)}^{\frac{1}{2}}{\left(k_8\left(t\right)\right)}^{\frac{1}{2}}+2{\left(k_1\left(t\right)\right)}^{\frac{1}{2}}{\left(k_3\left(t\right)\right)}^{\frac{1}{2}}\hfill \\ & =k_9\left(t\right).\hfill \end{array}$$

(47)

Thus, the proof is complete. □

3. Main Result

We suggest that $(u_i,p,T,C)$ is the solution of the initial boundary problems for the Boussinesq equations.

$$\left\{\begin{array}{c}\frac{\partial u_i}{\partial t}+{\lambda }_1{u}_j\frac{\partial u_i}{\partial x_j}=-p_{,i}+\Delta u_i+g_{i}T-h_{i}C(x,t)\in \Omega \times [0,\tau ],\hfill \\ \frac{\partial u_i}{\partial x_i}=0(x,t)\in \Omega \times [0,\tau ],\hfill \\ \frac{\partial T}{\partial t}+u_i\frac{\partial T}{\partial x_i}=\Delta T(x,t)\in \Omega \times [0,\tau ],\hfill \\ \frac{\partial C}{\partial t}+u_i\frac{\partial C}{\partial x_i}=\Delta C+LT-kC(x,t)\in \Omega \times [0,\tau ],\hfill \end{array}\right.$$

(48)

$$\begin{array}{c}{u}_i=0,\frac{\partial T}{\partial n}=k_{1}T,\frac{\partial C}{\partial n}=k_{2}C(x,t)\in \partial \Omega \times [0,\tau ],\hfill \end{array}$$

(49)

$$\begin{array}{c}{u}_i(x,0)=u_{i0}\left(x\right),T(x,0)=T_0\left(x\right),C(x,0)=C_0\left(x\right)x\in \Omega .\hfill \end{array}$$

(50)

Furthermore, we suggest that $(u_i^{*},p^{*},T^{*},C^{*})$ satisfies the following problems.

$$\left\{\begin{array}{c}\frac{\partial u_i^{*}}{\partial t}+{\lambda }_2{u}_j^{*}\frac{\partial u_i^{*}}{\partial x_j}=-p_{,i}^{*}+\Delta u_i^{*}+g_i{T}^{*}-h_i{C}^{*}(x,t)\in \Omega \times [0,\tau ],\hfill \\ \frac{\partial u_i^{*}}{\partial x_i}=0(x,t)\in \Omega \times [0,\tau ],\hfill \\ \frac{\partial T^{*}}{\partial t}+u_i^{*}\frac{\partial T^{*}}{\partial x_i}=\Delta T^{*}(x,t)\in \Omega \times [0,\tau ],\hfill \\ \frac{\partial C^{*}}{\partial t}+u_i^{*}\frac{\partial C^{*}}{\partial x_i}=\Delta C^{*}+L{T}^{*}-k{C}^{*}(x,t)\in \Omega \times [0,\tau ],\hfill \end{array}\right.$$

(51)

$$\begin{array}{c}{u}_i^{*}=0,\frac{\partial T^{*}}{\partial n}=k_1{T}^{*},\frac{\partial C^{*}}{\partial n}=k_2{C}^{*}(x,t)\in \partial \Omega \times [0,\tau ],\hfill \end{array}$$

(52)

$$\begin{array}{c}{u}_i^{*}(x,0)=u_{i0}\left(x\right),T^{*}(x,0)=T_0\left(x\right),C^{*}(x,0)=C_0\left(x\right)x\in \Omega .\hfill \end{array}$$

(53)

We introduce some new functions—${\omega }_i$, $\pi$, $\theta$, S and $\lambda$:

$$\begin{array}{c}{\omega }_i=u_i-u_i^{*},\pi =p-p^{*},\theta =T-T^{*},S=C-C^{*},\lambda ={\lambda }_1-{\lambda }_2.\hfill \end{array}$$

(54)

Then, we can obtain the result where $({\omega }_i,\pi ,\theta ,S,\lambda )$ satisfies the following equations.

$$\left\{\begin{array}{c}\frac{\partial {\omega }_i}{\partial t}+\lambda u_j\frac{\partial u_i}{\partial x_j}+{\lambda }_2{u}_j\frac{\partial {\omega }_i}{\partial x_j}+{\lambda }_2{\omega }_j\frac{\partial u_i^{*}}{\partial x_j}=-{\pi }_{,i}+\Delta {\omega }_i+g_i\theta -h_{i}S(x,t)\in \Omega \times [0,\tau ],\hfill \\ \frac{\partial {\omega }_i}{\partial x_i}=0(x,t)\in \Omega \times [0,\tau ],\hfill \\ \frac{\partial \theta }{\partial t}+{\omega }_i\frac{\partial T}{\partial x_i}+u_i^{*}\frac{\partial \theta }{\partial x_i}=\Delta \theta (x,t)\in \Omega \times [0,\tau ],\hfill \\ \frac{\partial S}{\partial t}+u_i\frac{\partial S}{\partial x_i}+{\omega }_i\frac{\partial C^{*}}{\partial x_i}=\Delta S+L\theta -kS(x,t)\in \Omega \times [0,\tau ],\hfill \end{array}\right.$$

(55)

$$\begin{array}{c}{\omega }_i=0,\frac{\partial \theta }{\partial n}=k_1\theta ,\frac{\partial S}{\partial n}=k_{2}S(x,t)\in \partial \Omega \times [0,\tau ],\hfill \end{array}$$

(56)

$$\begin{array}{c}{\omega }_i(x,0)=0,\theta (x,0)=0,S(x,0)=0,x\in \Omega \hfill \end{array}.$$

(57)

In this paper, we want to obtain the following main result.

Theorem 1.

We suggest that $(u_i,T,C,P)$ is the classical solution that satisfies conditions (48)–(50). We also suggest that $(u_i^{*},T^{*},C^{*},p^{*})$ is the classical solution that satisfies conditions (51)–(53). The difference $({\omega }_i,\theta ,S,\pi )$ satisfies (55)–(57). We obtain the result where solution $(u_i,T,C,p)$ can converge to solution $(u_i^{*},T^{*},C^{*},P^{*})$ when the Boussinesq coefficient ${\lambda }_1$ tends to ${\lambda }_2$. Additionally, the norm of $({\omega }_i,\theta ,S,\pi )$ satisfies the following estimates:

$$\begin{array}{c}{\int }_{\Omega }{\omega }_i{\omega }_{i}dx+{\int }_{\Omega }{\theta }^{2}dx+{\int }_{\Omega }{S}^{2}dx\le 4{\lambda }^2{e}^{{\int }_0^t{k}_{12}\left(s\right)ds}{\int }_0^t{e}^{-{\int }_0^s{k}_{12}\left(\eta \right)d\eta }{k}_{10}\left(s\right)ds.\hfill \end{array}$$

(58)

where $k_{10}\left(t\right)$ and $k_{12}\left(t\right)$ are positive functions.

The proof of the theorem will be divided into the following Lemmas.

Lemma 8.

Let $u_i$ be solution of Equation $\left(55\right)$ satisfying boundary conditions $\left(56\right)$ and $\left(57\right)$; we can obtain the following estimates:

$$\begin{array}{cc}\frac{d}{dt}{\parallel \omega \parallel }^2\hfill & \le 4{\lambda }^2{k}_{10}\left(t\right)+k_{11}{\left(t\right)\parallel \omega \parallel }^2+{\parallel \theta \parallel }^2+{\parallel S\parallel }^2-\frac{1}{2}{\parallel \nabla \omega \parallel }^2,\hfill \end{array}$$

(59)

where $k_{10}\left(t\right)$ and $k_{11}\left(t\right)$ are positive functions defined later.

Proof. 

We multiply both sides of ${\left(55\right)}_1$ by $w_i$ and integrate over $\Omega$; we can obtain the following:

$$\begin{array}{cc}\frac{d}{dt}\frac{1}{2}{\parallel \omega \parallel }^2\hfill & =-\lambda {\int }_{\Omega }{u}_j{u}_{i,j}{\omega }_{i}dx-{\lambda }_2{\int }_{\Omega }{u}_j{\omega }_{i,j}{\omega }_{i}dx-{\lambda }_2{\int }_{\Omega }{\omega }_j{u}_{i,j}^{*}{\omega }_{i}dx-{\parallel \nabla \omega \parallel }^2\hfill \\ & +{\int }_{\Omega }{g}_i\theta {\omega }_{i}dx-{\int }_{\Omega }{h}_{i}S{\omega }_{i}dx\hfill \\ \hfill & =\lambda {\int }_{\Omega }{u}_j{u}_i{\omega }_{i,j}dx-{\lambda }_2{\int }_{\Omega }{\omega }_j{\omega }_i{u}_{i,j}^{*}dx-{\parallel \nabla \omega \parallel }^2+{\int }_{\Omega }{g}_i\theta {\omega }_{i}dx-{\int }_{\Omega }{h}_{i}S{\omega }_{i}dx\hfill \\ \hfill & \le {\lambda \parallel u\parallel }_4^2\parallel \nabla \omega \parallel +{\lambda }_2{\parallel \omega \parallel }_4^2\parallel \nabla u^{*}{\parallel -\parallel \nabla \omega \parallel }^2+\parallel \theta \parallel \parallel \omega \parallel +\parallel S\parallel \parallel \omega \parallel \hfill \\ \hfill & \le {\lambda }^2{\epsilon }_1{\parallel u\parallel }_4^4+\frac{1}{{\epsilon }_1}{\parallel \nabla \omega \parallel }^2+{\lambda }_2{\nu }^{\frac{1}{2}}\parallel \nabla u^{*}\parallel \parallel \omega \parallel \parallel \nabla \omega \parallel -{\parallel \nabla \omega \parallel }^2\hfill \\ & +\parallel \theta \parallel \parallel \omega \parallel +\parallel S\parallel \parallel \omega \parallel \hfill \\ \hfill & \le {\lambda }^2{\epsilon }_1{\nu [\parallel u\parallel }^2+{\parallel \nabla u\parallel }^2{]}^2+\frac{1}{{\epsilon }_1}{\parallel \nabla \omega \parallel }^2+\frac{{\lambda }_2^2{\epsilon }_2\nu }{2}\parallel \nabla u^{*}{\parallel }^2{\parallel \omega \parallel }^2+\frac{1}{2{\epsilon }_2}{\parallel \nabla \omega \parallel }^2\hfill \\ & +\parallel \theta \parallel \parallel \omega \parallel +\parallel S\parallel \parallel \omega \parallel -{\parallel \nabla \omega \parallel }^2,\hfill \end{array}$$

(60)

where ${\epsilon }_1$ and ${\epsilon }_2$ are constants no lesser than zero.

If we choose ${\epsilon }_1=4$, ${\epsilon }_2=2$, we obtain the following.

$$\begin{array}{cc}\frac{d}{dt}{\parallel \omega \parallel }^2\hfill & \le 8{\lambda }^2{\nu [\parallel u\parallel }^2+{\parallel \nabla u\parallel }^2{]}^2+(2{\lambda }_2^2\nu \parallel \nabla u^{*}{\parallel }^2+{2)\parallel \omega \parallel }^2\hfill \\ & +{\parallel \theta \parallel }^2+{\parallel S\parallel }^2-\frac{1}{2}{\parallel \nabla \omega \parallel }^2.\hfill \end{array}$$

(61)

Following the same procedures in deriving (46), we can also obtain the following:

$$\begin{array}{cc}\hfill {\int }_{\Omega }{u}_{i,j}^{*}{u}_{i,j}^{*}dx& \le \widetilde{k_9}\left(t\right),\hfill \end{array}$$

(62)

where $\widetilde{k_9}\left(t\right)$ is a positive function.

Inserting (15), (46) and (62) into (61), we obtain the following:

$$\begin{array}{cc}\frac{d}{dt}{\parallel \omega \parallel }^2\hfill & \le 4{\lambda }^2{k}_{10}\left(t\right)+k_{11}{\left(t\right)\parallel \omega \parallel }^2+{\parallel \theta \parallel }^2+{\parallel S\parallel }^2-\frac{1}{2}{\parallel \nabla \omega \parallel }^2,\hfill \end{array}$$

(63)

where $k_{10}\left(t\right)=8k_1^2{(k_3\left(t\right)+k_9\left(t\right))}^2$ and $k_{11}\left(t\right)=2{\lambda }_2^2\widetilde{k_9}\left(t\right)+2$.

Thus, the proof is complete. □

Lemma 9.

We suggest that θ and S are solutions of Equation $\left(55\right)$ and satisfy boundary conditions $\left(2\right)$ and $\left(3\right)$. The following estimates can be obtained:

$$\begin{array}{cc}\frac{d}{dt}\left({\parallel \theta \parallel }^2+{\parallel S\parallel }^2\right)\hfill & \le -\frac{1}{2}{\int }_{\Omega }{\left(\right|\nabla S|}^2+{|\nabla \theta |}^2)dx+M{\int }_{\Omega }({\theta }^2+S^2)dx\hfill \\ & +2{\left({\int }_{\Omega }{\left({\omega }_i{\omega }_i\right)}^2\right)}^{\frac{1}{2}}{\left(\widetilde{k_4}\left(t\right)\right)}^{\frac{1}{2}},\hfill \end{array}$$

(64)

where M is a positive constant, and $\widetilde{k_4}\left(t\right)$ is a positive function.

Proof. 

We multiply both sides of ${\left(55\right)}_3$ by $2\theta$ and integrate over $\Omega$; we have the following.

$$\begin{array}{cc}\frac{d}{dt}{\parallel \theta \parallel }^2\hfill & =2{\int }_{\Omega }\theta (▵\theta -u_i{\theta }_{,i}-{\omega }_i{T}_{,i}^{*})dx\hfill \\ & =-2{\int }_{\Omega }{|\nabla \theta |}^{2}dx+2k_1{\oint }_{\partial \Omega }{\theta }^{2}dS+2{\int }_{\Omega }{\theta }_{,i}{\omega }_i{T}^{*}dx\hfill \\ & \le -2{\int }_{\Omega }{|\nabla \theta |}^{2}dx+2k_1{\oint }_{\partial \Omega }{\theta }^{2}dS+{\int }_{\Omega }{\theta }_{,i}{\theta }_{,i}dx\hfill \\ & +{\left({\int }_{\Omega }{\left({\omega }_i{\omega }_i\right)}^2\right)}^{\frac{1}{2}}{\left({\int }_{\Omega }{T^{*}}^4\right)}^{\frac{1}{2}}.\hfill \end{array}$$

(65)

Following the same procedures in deriving (18), we can also obtain the following:

$$\begin{array}{c}\hfill {\int }_{\Omega }{T^{*}}^{4}dx+{\int }_{\Omega }{C^{*}}^{4}dx\le \widetilde{k_4}\left(t\right),\end{array}$$

(66)

where $\widetilde{k_4}\left(t\right)$ is a positive function.

Inserting (66) into (65), we obtain the following.

$$\begin{array}{cc}\frac{d}{dt}{\parallel \theta \parallel }^2\hfill & \le -{\int }_{\Omega }{|\nabla \theta |}^{2}dx+2k_1{\oint }_{\partial \Omega }{\theta }^{2}dS\hfill \\ & +{\left({\int }_{\Omega }{\left({\omega }_i{\omega }_i\right)}^2\right)}^{\frac{1}{2}}{\left(\widetilde{k_4}\left(t\right)\right)}^{\frac{1}{2}}.\hfill \end{array}$$

(67)

We multiply both sides of ${\left(55\right)}_4$ by $2S$ and integrate over $\Omega$; we have the following.

$$\begin{array}{cc}\frac{d}{dt}{\parallel S\parallel }^2\hfill & =2{\int }_{\Omega }S(▵S-u_i{S}_{,i}-{\omega }_i{C}_{,i}^{*}-kS+L\theta )dx\hfill \\ & \le -{\int }_{\Omega }{|\nabla S|}^{2}dx+2k_2{\oint }_{\partial \Omega }{S}^{2}dS+L{\int }_{\Omega }{\theta }^{2}dx+L{\int }_{\Omega }{S}^{2}dx\hfill \\ & +{\left({\int }_{\Omega }{\left({\omega }_i{\omega }_i\right)}^2\right)}^{\frac{1}{2}}{\left(\widetilde{k_4}\left(t\right)\right)}^{\frac{1}{2}}.\hfill \end{array}$$

(68)

Using (4) and combining (67) and (68), we obtain the following.

$$\begin{array}{cc}\frac{d}{dt}\left({\parallel \theta \parallel }^2+{\parallel S\parallel }^2\right)\hfill & \le -\frac{1}{2}{\int }_{\Omega }{\left(\right|\nabla S|}^2+{|\nabla \theta |}^2)dx+M{\int }_{\Omega }({\theta }^2+S^2)dx\hfill \\ & +2{\left({\int }_{\Omega }{\left({\omega }_i{\omega }_i\right)}^2\right)}^{\frac{1}{2}}{\left(\widetilde{k_4}\left(t\right)\right)}^{\frac{1}{2}},\hfill \end{array}$$

(69)

with $M=max\{2k_1\left[\frac{2}{m}+\frac{2d^2{k}_1}{m^2}\right]+L,2k_2\left[\frac{2}{m}+\frac{2d^2{k}_2}{m^2}+L\right]\}$.

Thus, the proof is complete. □

Lemma 10.

We suggest that ${\omega }_i$, θ and S are solutions of Equation $\left(55\right)$, and boundary conditions $\left(56\right)$ and $\left(57\right)$ are also satisfied. We can obtain the following estimates:

$$\begin{array}{c}{\int }_{\Omega }{\omega }_i{\omega }_{i}dx+{\int }_{\Omega }{\theta }^{2}dx+{\int }_{\Omega }{S}^{2}dx\le 4{\lambda }^2{e}^{{\int }_0^t{k}_{12}\left(s\right)ds}{\int }_0^t{e}^{-{\int }_0^s{k}_{12}\left(\eta \right)d\eta }{k}_{10}\left(s\right)ds.\hfill \end{array}$$

(70)

with $k_{12}\left(t\right)$ as a positive function.

Proof. 

We define a function

$$\zeta \left(t\right)={\int }_{\Omega }{\omega }_i{\omega }_{i}dx+{\int }_{\Omega }{\theta }^{2}dx+{\int }_{\Omega }{S}^{2}dx.$$

Combining (59) and (64); we thus have the following.

$$\begin{array}{cc}\frac{d}{dt}\zeta \left(t\right)\hfill & \le 4{\lambda }^2{k}_{10}\left(t\right)+(k_{11}\left(t\right)+M+1)\zeta \left(t\right)-\frac{1}{2}{\parallel \nabla \omega \parallel }^2\hfill \\ & +2{\left({\int }_{\Omega }{\left({\omega }_i{\omega }_i\right)}^{2}dx\right)}^{\frac{1}{2}}{\left(\widetilde{k_4}\left(t\right)\right)}^{\frac{1}{2}}.\hfill \end{array}$$

(71)

Using (32), we have the following.

$$\begin{array}{c}2{\left({\int }_{\Omega }{\left({\omega }_i{\omega }_i\right)}^{2}dx\right)}^{\frac{1}{2}}{\left(\widetilde{k_4}\left(t\right)\right)}^{\frac{1}{2}}\le 8\widetilde{k_4}{\left(t\right)\nu \parallel \omega \parallel }^2+\frac{1}{2}{\parallel \nabla \omega \parallel }^2.\hfill \end{array}$$

(72)

Inserting (72) into (71), we obtain the following:

$$\begin{array}{c}\frac{d}{dt}\zeta \left(t\right)\le k_{12}\left(t\right)\zeta \left(t\right)+4{\lambda }^2{k}_{10}\left(t\right),\hfill \end{array}$$

(73)

with $k_{12}\left(t\right)=k_{11}\left(t\right)+M+1+8\widetilde{k_4}\left(t\right)\nu$.

Equation (73) can be rewritten as follows.

$$\begin{array}{c}\frac{d}{dt}\left(e^{-{\int }_0^t{k}_{12}\left(s\right)ds}\zeta \left(t\right)\right)\le 4e^{-{\int }_0^t{k}_{12}\left(s\right)ds}{\lambda }^2{k}_{10}\left(t\right).\hfill \end{array}$$

(74)

An integration of (74) provides the following.

$$\begin{array}{c}\zeta \left(t\right)\le 4{\lambda }^2{e}^{{\int }_0^t{k}_{12}\left(s\right)ds}{\int }_0^t{e}^{-{\int }_0^s{k}_{12}\left(\eta \right)d\eta }{k}_{10}\left(s\right)ds.\hfill \end{array}$$

(75)

Inequality (75) shows that when $\lambda$ tends to zero, the differences of the solutions tend to zero as the indicated norm.

Thus, the proof is complete. □

Inequality (75) is the result we want to prove in the Theorem. Thus, we complete the proof of the theorem.

4. Conclusions

In the present paper, the result of the continuous dependence type could be obtained for the Boussinesq coefficient $\lambda$. An energy method was used. For the case when $\Omega \in R^3$, the $Poincar\acute{e}$ inequality used in this paper was no longer applicable. There were difficulties in obtaining the key bound for ${\int }_{\Omega }{u}_{i,j}{u}_{i,j}dx$. We thought it would be an interesting topic to study the case when $\Omega \in R^3$ in future. Additionally, if we changed the bounded domain, $\Omega$, by an unbounded domain, we thought the terms containing pressure p were difficult to tackle. These terms could not be bounded by the prescribed data by using the same method proposed in this paper. Some new methods might be developed in order to overcome these difficulties. We are sure that some good results would be obtained if we studied the above two problems. These studies would be new and interesting.