*Article* **Spatial Decay Bounds for the Brinkman Fluid Equations in Double-Diffusive Convection**

**Xuejiao Chen †, Yuanfei Li \*,† and Dandan Li**

School of Data Science, Guangzhou Huashang College, Guangzhou 511300, China; cxjsky@gdhsc.edu.cn (X.C.); li20201101@126.com (D.L.)

**\*** Correspondence: 201610104816@mail.scut.edu.cn

† These authors contributed equally to this work.

**Abstract:** In this paper, we consider the Brinkman equations pipe flow, which includes the salinity and the temperature. Assuming that the fluid satisfies nonlinear boundary conditions at the finite end of the cylinder, using the symmetry of differential inequalities and the energy analysis methods, we establish the exponential decay estimates for homogeneous Brinkman equations. That is to prove that the solutions of the equation decay exponentially with the distance from the finite end of the cylinder. To make the estimate of decay explicit, the bound for the total energy is also derived.

**Keywords:** spatial decay estimates; Brinkman equations; Saint-Venant principle

## **1. Introduction**

The Brinkman equations are one of the most important models in fluid mechanics. This model are mainly used to describe flow in a porous medium. For more details, one can refer to Nield and Bejan [1] and Straughan [2]. In the present paper, we define the Brinkman flow depending on the salinity and the temperature in a semi-infinite cylindrical pipe and derive the spatial decay properties. When the homogeneous initial-boundary conditions are applied on the lateral surface of the cylinder, We prove that the solutions of Brinkman equations decays exponentially with spatial variable.

In fact, the Brinkman equations have been studied by many papers in the literature. For example, Straughan [2] considered the mathematical properties of Brinkman equations as well as Darcy and Forchheimer equations, and stated how these equations describe the flow of porous media. Ames and Payne [3] studied the structural stability for the solutions to the viscoelasticity in an ill-posed problem. Franchi and Straughan [4] proved the structural stability for the solutions to the Brinkman equations in porous media in a bounded region. More relevant results one can see [5–10]. Paper [11] studied the double diffusive convection in porous medium and obtained the structural stability for the solutions. The continuous dependence for a thermal convection model with temperature-dependent solubility can be found in [12]. For more recent work about continuous dependence, one may refer to [13–19].

In this paper, let *R* be a semi-infinite cylinder and *∂R* represents the boundary of *R*. *D* denotes the cross section of the cylinder with the smooth boundary *∂D* (see Figure 1). In this paper, we also use the following notations

$$R\_z = \left\{ (\mathbf{x}\_1, \mathbf{x}\_2, \mathbf{x}\_3) \Big| (\mathbf{x}\_1, \mathbf{x}\_2) \in D, \quad \mathbf{x}\_3 > z \ge 0 \right\},$$

$$D\_z = \left\{ (\mathbf{x}\_1, \mathbf{x}\_2, \mathbf{x}\_3) \Big| (\mathbf{x}\_1, \mathbf{x}\_2) \in D, \quad \mathbf{x}\_3 = z \ge 0 \right\},$$

where *z* is a point along the *x*<sup>3</sup> axis. Clearly, *R*<sup>0</sup> = *R* and *D*<sup>0</sup> = *D*. Letting *ui*, *T*, *C* and *p* denote the fluid velocity, temperature, salt concentration and pressure, respectively.

**Citation:** Chen, X.; Li, Y.; Li, D. Spatial Decay Bounds for the Brinkman Fluid Equations in Double-Diffusive Convection. *Symmetry* **2022**, *14*, 98. https:// doi.org/10.3390/sym14010098

Academic Editor: Nicusor Minculete

Received: 30 November 2021 Accepted: 5 January 2022 Published: 7 January 2022

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**Copyright:** © 2022 by the authors. Licensee MDPI, Basel, Switzerland. This article is an open access article distributed under the terms and conditions of the Creative Commons Attribution (CC BY) license (https:// creativecommons.org/licenses/by/ 4.0/).

**Figure 1.** Cylindrical pipe.

The Brinkman equations we study can be written as [20]

$$\frac{\partial \mu\_i}{\partial t} = \nu \Delta \mu\_i - k\_1 \mu\_i - p\_{,i} + \mathbf{g}\_i T + h\_i \mathbb{C}\_\prime \quad \text{in } \mathbb{R} \times \{t \ge 0\},\tag{1}$$

$$\frac{\partial T}{\partial t} + \mu\_i \frac{\partial T}{\partial \mathbf{x}\_i} = k\_2 \Delta T, \qquad \text{in } \mathbb{R} \times \{t \ge 0\}, \tag{2}$$

$$\frac{\partial \mathbf{C}}{\partial t} + \mu\_i \frac{\partial \mathbf{C}}{\partial \mathbf{x}\_i} = k\_3 \Delta \mathbf{C} + \sigma \Delta T,\qquad \text{in } \mathbb{R} \times \{t \ge 0\},\tag{3}$$

$$
u\_{i,j} = 0,\qquad\qquad\text{in }\mathbb{R} \times \{t \ge 0\},\tag{4}$$

where *ν*, *σ* > 0 denote the Brinkman coefficient, and the Soret coefficient, respectively. *k*1, *k*2, *k*<sup>3</sup> > 0. Without losing generality, we take them equal to 1. Δ is the Laplacian operator. *gi*(*x*) and *hi*(*x*) are gravity field, which are given functions. We suppose that (1)–(4) have the following initial-boundary conditions

$$
\mu\_i = 0, \quad T = \mathbb{C} = 0,\tag{5}
$$

$$
\text{and}
\quad \text{and} \quad \text{on} \quad \partial D \times \{t \ge 0\},\tag{6}
$$

$$
\mu\_i = 0, \quad T = \mathbb{C} = 0,\tag{6}
$$

$$u\_i = f\_i(\mathbf{x}\_1, \mathbf{x}\_2, t), \ T = F(\mathbf{x}\_1, \mathbf{x}\_2, t), \ C = G(\mathbf{x}\_1, \mathbf{x}\_2, t), \ on \ D\_0 \times \{t \ge 0\}, \tag{7}$$

$$u\_{i\prime}u\_{i\prime j\prime}u\_{i\prime,t}T\_{\prime}T\_{\prime\prime}\mathbb{C}\_{\prime}\mathbb{C}\_{\prime}\mathbf{x}\_{\prime}p = o(\mathbf{x}\_{\mathfrak{J}}^{-1})\text{ uniformly in }\mathbf{x}\_{1\prime}\mathbf{x}\_{2\prime}\mathbf{t},\quad\text{as }\mathbf{x}\_{\mathfrak{J}}\to\infty.\tag{8}$$

In (1)–(8) and in the following, the usual summation convention is employed with repeated Latin subscripts summed from 1 to 3 and repeat Greek subscript summed from

1 to 2. The comma is used to indicate partial differentiation, i.e.,  $u\_{i,j}u\_{i,j} = \sum\_{i,j=1}^{3} \left(\frac{\partial u\_i}{\partial x\_j}\right)^2$ .

$$
\varphi\_{\alpha,\beta} \underbrace{\varphi\_{\alpha,\beta}}\_{\underset{\alpha}{\rightleftharpoons}} = \sum\_{\alpha,\beta=1}^{2} \left(\frac{\partial \varphi\_{\alpha}}{\partial x\_{\beta}}\right)^2 .
$$

The purpose of this paper is to consider the spatial decay properties of the Equations (1)–(8) in a semi-infinite cylindrical pipe by using the symmetry of differential inequalities, that is, to prove that the solutions of the equations decay exponentially with the distance from the finite end of the cylindrical pipe.

In Section 2, some auxiliary inequalities are presented. We establish some useful lemmas in Section 3. The spatial exponential decay estimate for the solution is established in Section 4. Finally, in Section 5 we derive the bounds for the total energies.

## **2. Auxiliary Results**

In this paper, we will use some inequalities in the following sections. Thus, we firstly list them as follows.

**Lemma 1.** *Let D be a plane domain D with the smooth boundary ∂D. If w* = 0 *on ∂D, then*

$$\int\_{D} w\_{,\mu} w\_{,\mu} dA \ge \lambda\_1 \int\_{R\_z} w^2 d\mathbf{x}\_{\prime} \tag{9}$$

*where λ*<sup>1</sup> *is the smallest eigenvalue of the problem*

$$
\Delta \phi + \lambda \phi = 0 \quad \text{in} \quad D\_\prime
$$

$$
\phi = 0 \quad \text{on} \quad \partial D.
$$

*Many papers have studied this inequality, e.g., one may see [21,22].*

A representation theorem will be also used in next sections. We write this theorem as

**Lemma 2.** *Let D be a plane Lipschitz bound region and w be a differential function in D which satisfies <sup>D</sup> wdA* = 0*, then there exists a vector function ϕα*(*x*1, *x*2) *such that*

$$\begin{array}{ll} \varrho\_{\mathfrak{a},a} = w & \text{in} & D\_r \\\\ \varrho\_{\mathfrak{a}} = 0 & on & \partial D\_r \end{array}$$

*and a positive constant* Λ *depending only on the geometry of D such that*

$$
\lambda \int\_D \varphi\_{a,\emptyset} \varphi\_{a,\emptyset} dA \le \Lambda \int\_D \varphi\_{a,a}^2 dA. \tag{10}
$$

The Lemma 2 was proofed by Babuška and Aziz [23] and Horgan and Wheeler [24] have used the Lemma 1 to viscous flow problems. The explicit upper bound of Λ can be found in Horgan and Payne [25]. In this paper, this Lemma 2 is used to eliminate the pressure function difference terms *p*, since we can prove that *u*<sup>3</sup> satisfy the hypothesis of this Lemma 2 later.

If *<sup>w</sup>* <sup>∈</sup> *<sup>C</sup>*<sup>1</sup> <sup>0</sup> (*D*) and *<sup>w</sup>* <sup>∈</sup> *<sup>C</sup>*<sup>1</sup> <sup>0</sup> (*R*), the following Sobolev inequalities hold

$$\int\_{D} w^{4} dA \le \frac{1}{2} \left[ \int\_{D} w^{2} dA \right] \left[ \int\_{D} w\_{\mu} w\_{\mu} dA \right],\tag{11}$$

$$\int\_{\mathcal{R}\_z} w^6 dx \le \Omega \left[ \int\_{\mathcal{R}\_z} w\_j w\_{jl} dx \right]^3. \tag{12}$$

For (11), we assume that *w* → 0 as *x*<sup>3</sup> → ∞. Payne [26] has given the derivation of (12). For a special case of the results one can see [27,28]. They have obtained the optimal value of Ω

$$
\Omega = \frac{1}{2\mathcal{T}} \left(\frac{3}{4}\right)^4.
$$

In the following, we also use the following lemma.

\*\*Lemma 3.\*\*  $\|f\,w\in\mathbb{C}^{1}(\mathbb{R}\_{z})$ ,  $w\_{i}\Big|\_{\partial D} = 0$   $and \, w\_{i} \to 0 \,\text{as}\,\,x\_{3} \to \infty$ , then 
$$\int\_{D\_{z}} \left(w\_{i}w\_{i}\right)^{2} dA \le 4\sqrt{\Omega} \left[\int\_{R\_{z}} w\_{i,j}w\_{i,j} dx\right]^{2}.\tag{13}$$

We will also use the following lemmas which were derived in [29].

**Lemma 4.** *Let that the function ϕ is the solution of the problem*

$$\begin{aligned} \Delta \varrho &= 0 \quad \text{in} \ R\_{z\_{\prime}}\\ \frac{\partial \varrho}{\partial n} &= 0 \quad \text{on} \ \partial D\_{z\_{\prime}}\\ \frac{\partial \varrho}{\partial n} &= \emptyset \quad \text{in} \ D\_{z\_{\prime}} \end{aligned} \tag{14}$$

*where Dz gdA* = <sup>0</sup>*. Then*

$$\int\_{D\_z} \mathfrak{p}\_{,\mathfrak{a}} \mathfrak{p}\_{,\mathfrak{a}} dA = \int\_{D\_z} \mathfrak{g}^2 dA\_\prime \tag{15}$$

$$\int\_{R\_{\pm}} \mathfrak{p}\_{\neq} \mathfrak{p}\_{\neq} dA = \frac{1}{\sqrt{\mu}} \int\_{D\_{z}} \mathfrak{g}^{2} dA\_{\prime} \tag{16}$$

## **3. Some Useful Lemmas**

In this section, we derive some useful lemmas which will be used in next section. First, we define a weighted energy expression

$$\begin{split} E(z,t) &= k \int\_{0}^{t} \int\_{R\_{z}} (\mathfrak{J} - z) u\_{i,\eta} u\_{i,\eta} dx d\eta + \nu \int\_{0}^{t} \int\_{R\_{z}} (\mathfrak{J} - z) u\_{i,j} u\_{i,j} dx d\eta \\ &+ \rho\_{1} \int\_{0}^{t} \int\_{R\_{z}} (\mathfrak{J} - z) \, T\_{j} \, T\_{j} dx d\eta + \rho\_{2} \int\_{0}^{t} \int\_{R\_{z}} (\mathfrak{J} - z) \mathbb{C}\_{,j} \mathbb{C}\_{,j} dx d\eta \\ &= E\_{1}(z,t) + E\_{2}(z,t) + E\_{3}(z,t) + E\_{4}(z,t), \end{split} \tag{17}$$

where *k*, *ρ*1, *ρ*<sup>2</sup> are positive parameters and *ξ* > *z* > 0.

By using the divergence theorem and Equations (1) and (4), we obtain

$$\begin{split} E\_{1}(z,t) &= k \int\_{0}^{t} \int\_{R\_{z}} (\xi - z) u\_{i\eta} \Big[ \nu \Delta u\_{i} - u\_{i} - p\_{,i} + g\_{i}T + h\_{i} \mathbb{C} \Big] dxd\eta \\ &= k \nu \int\_{0}^{t} \int\_{R\_{z}} u\_{i,\eta} u\_{i,\eta} dxd\eta + k \int\_{0}^{t} \int\_{R\_{z}} u\_{3,\eta} p dxd\eta \\ &+ k \int\_{0}^{t} \int\_{R\_{z}} (\xi - z) u\_{i,\eta} g\_{i} T dxd\eta + k \int\_{0}^{t} \int\_{R\_{z}} (\xi - z) u\_{i,\eta} h\_{i} \mathbb{C} dxd\eta \\ &- k \nu \int\_{R\_{z}} (\xi - z) u\_{i,\eta} u\_{i,\eta} dx \Big|\_{\eta=t} - k \int\_{R\_{z}} (\xi - z) u\_{i} u\_{i} dx \Big|\_{\eta=t} \\ &= \sum\_{i=1}^{4} A\_{i} \\ &- \frac{1}{2} k \nu \int\_{R\_{z}} (\xi - z) u\_{i,\eta} u\_{i,\eta} dx \Big|\_{\eta=t} - \frac{1}{2} k \int\_{R\_{z}} (\xi - z) u\_{i} u\_{i} dx \Big|\_{\eta=t}. \end{split} \tag{18}$$

Using the Schwarz inequality, the arithmetic geometric mean inequality and (9), we can obtain

$$\begin{split} A\_{1} &\leq k\nu \Big[ \int\_{0}^{t} \int\_{R\_{z}} u\_{i,\eta} u\_{i,\eta} \mathrm{d}x d\eta \int\_{0}^{t} \int\_{R\_{z}} u\_{i,3} u\_{i,3} \mathrm{d}x d\eta \Big]^{\frac{1}{2}} \\ &\leq \frac{\sqrt{k\nu}}{2} \Big[ k \int\_{0}^{t} \int\_{R\_{z}} u\_{i,\eta} u\_{i,\eta} \mathrm{d}x d\eta + \nu \int\_{0}^{t} \int\_{R\_{z}} u\_{i,3} u\_{i,3} \mathrm{d}x d\eta \Big], \end{split} \tag{19}$$

$$\begin{split} A\_{3} &\leq \frac{k}{\sqrt{\lambda\_{1}}} \Big[ \int\_{0}^{t} \int\_{R\_{z}} (\xi - z) u\_{i,\eta} u\_{i,\eta} \mathrm{d}x d\eta \int\_{0}^{t} \int\_{R\_{z}} (\xi - z) T\_{\mu} T\_{\mu} \mathrm{d}x d\eta \Big]^{\frac{1}{2}} \\ &\leq \frac{\varepsilon\_{1}}{2} k \int\_{0}^{t} \int\_{R\_{z}} (\xi - z) u\_{i,\eta} u\_{i,\eta} \mathrm{d}x d\eta + \frac{k \delta\_{1}^{2}}{2 \lambda\_{1} \varepsilon\_{1}} \int\_{0}^{t} \int\_{R\_{z}} (\xi - z) T\_{\mu} T\_{\mu} \mathrm{d}x d\eta. \end{split} \tag{20}$$

and

$$A\_4 \le \frac{\varepsilon\_2}{2} k \int\_0^t \int\_{R\_\varepsilon} (\zeta - z) u\_{i,\eta} u\_{i,\eta} dx d\eta + \frac{k \delta\_2^2}{2 \lambda\_1 \varepsilon\_2} \int\_0^t \int\_{R\_\varepsilon} (\zeta - z) \mathbb{C}\_{,a} \mathbb{C}\_{,a} dx d\eta,\tag{21}$$

where *ε*1,*ε*<sup>2</sup> > 0 will be determined later and

$$
\delta\_1^2 = \max\_D (g\_i g\_i)\_\prime \quad \delta\_2^2 = \max\_D (h\_i h\_i)\_\prime \tag{22}
$$

We note that for any *z*∗ > 0, using (4) and (5),

$$\begin{aligned} \int\_{D\_z} \mu\_{3\mathcal{J}\emptyset} dA &= \int\_{D\_z\*} \mu\_{3\mathcal{J}\emptyset} dA - \int\_z^{z^\*} \int\_{D\_{\xi}^z} \mu\_{3\mathcal{J}\emptyset} dA d\xi \\ &= \int\_{D\_{z^\*}} \mu\_{3\mathcal{J}\emptyset} dA + \int\_z^{z^\*} \int\_{D\_{\xi}} \mu\_{\alpha, \alpha\eta} dA d\xi \\ &= \int\_{D\_{z^\*}} \mu\_{3\mathcal{J}\emptyset} dA. \end{aligned}$$

Since

$$\int\_{D\_0} f\_{3,\eta} dA = 0, \quad t \ge 0,\tag{23}$$

then,

$$\int\_{D\_{\mathbb{Z}}} u\_{3,\eta} dA = 0.$$

Under this assumption, using Lemma 2, there exist vector functions (*ϕ*1, *ϕ*2) such that

$$
\varrho\_{\mathfrak{n},\mathfrak{a}} = \mathfrak{u}\_{\mathfrak{Z},\mathfrak{y}} \quad \text{in} \ D, \quad \varrho\_{\mathfrak{k}} = 0 \quad \text{on} \ \partial D. \tag{24}
$$

Hence we have

$$\begin{split} A\_{2} &= k \int\_{0}^{t} \int\_{R\_{z}} \boldsymbol{q}\_{a,a} \boldsymbol{p} d\mathbf{x} d\eta = -k \int\_{0}^{t} \int\_{R\_{z}} \boldsymbol{q}\_{a} \boldsymbol{p}\_{a} d\mathbf{x} d\eta \\ &= k \int\_{0}^{t} \int\_{R\_{z}} \boldsymbol{q}\_{a} \left[ \boldsymbol{u}\_{a,\eta} - \nu \Delta \boldsymbol{u}\_{a} + \boldsymbol{u}\_{a} - \boldsymbol{g}\_{a} T - \boldsymbol{h}\_{a} \mathbb{C} \right] d\mathbf{x} d\eta \\ &= k \int\_{0}^{t} \int\_{R\_{z}} \boldsymbol{q}\_{a} \boldsymbol{u}\_{a,\eta} d\mathbf{x} d\eta + k \nu \int\_{0}^{t} \int\_{R\_{z}} \boldsymbol{q}\_{a,\beta} \boldsymbol{u}\_{a,\beta} d\mathbf{x} d\eta \\ &+ k \nu \int\_{0}^{t} \int\_{D\_{z}} \boldsymbol{q}\_{a} \boldsymbol{u}\_{a,3} d\mathbf{x} d\eta + k \int\_{0}^{t} \int\_{R\_{z}} \boldsymbol{q}\_{a} \boldsymbol{u}\_{a} d\mathbf{x} d\eta \\ &- k \int\_{0}^{t} \int\_{R\_{z}} \boldsymbol{q}\_{a} \boldsymbol{q}\_{a} d\mathbf{f} d\mathbf{x} d\eta - k \int\_{0}^{t} \int\_{R\_{z}} \boldsymbol{q}\_{a} \boldsymbol{h}\_{a} d\mathbf{x} d\eta \\ &= A\_{21} + A\_{22} + A\_{23} + A\_{24} + A\_{25} + A\_{26}. \end{split} \tag{25}$$

Using the Schwarz, Poincaré and the AG mean inequalities, (9) and (10), we can obtain

*A*<sup>21</sup> ≤ *<sup>t</sup>* 0 *Rz ϕαϕαdxdη* 1 <sup>2</sup> *<sup>t</sup>* 0 *Rz uα*,*ηuα*,*ηdxdη* 1 2 ≤ 1 <sup>√</sup>*λ*<sup>1</sup> *<sup>t</sup>* 0 *Rz ϕα*,*βϕα*,*βdxdη* 1 <sup>2</sup> *<sup>t</sup>* 0 *Rz uα*,*ηuα*,*ηdxdη* 1 2 ≤ Λ1 2 <sup>√</sup>*λ*<sup>1</sup> *<sup>t</sup>* 0 *Rz u*2 <sup>3</sup>*dη* 1 <sup>2</sup> *<sup>t</sup>* 0 *Rz uα*,*ηuα*,*ηdxdη* 1 2 ≤ *k*Λ<sup>1</sup> 2 2 <sup>√</sup>*λ*<sup>1</sup> *t* 0 *Rz ui*,*<sup>η</sup>ui*,*ηdxdη*, (26) *A*<sup>22</sup> ≤ *kν <sup>t</sup>* 0 *Rz ϕα*,*βϕα*,*βdxdη* 1 <sup>2</sup> *<sup>t</sup>* 0 *Rz uα*,*βuα*,*βdxdη* 1 2 <sup>≤</sup> *<sup>k</sup>ν*Λ<sup>1</sup> 2 *<sup>t</sup>* 0 *Rz u*2 3,*ηdxdη* 1 <sup>2</sup> *<sup>t</sup>* 0 *Rz uα*,*βuα*,*βdxdη* 1 2 ≤ *kν*Λ<sup>1</sup> 2 2 *t* 0 *Rz u*2 3,*ηdxdη* + *kν*Λ<sup>1</sup> 2 2 *t* 0 *Rz uα*,*βuα*,*βdxdη*, (27) *A*<sup>23</sup> ≤ *kν <sup>t</sup>* 0 *Dz ϕαϕαdxdη* 1 <sup>2</sup> *<sup>t</sup>* 0 *Dz uα*,3*uα*,3*dxdη* 1 2 ≤ *kν*Λ<sup>1</sup> 2 <sup>√</sup>*λ*<sup>1</sup> *<sup>t</sup>* 0 *Dz u*2 3,*ηdxdη* 1 <sup>2</sup> *<sup>t</sup>* 0 *Dz uα*,3*uα*,3*dxdη* 1 2 ≤ *kν*Λ<sup>1</sup> 2 2 <sup>√</sup>*λ*<sup>1</sup> *t* 0 *Dz u*2 3,*ηdxdη* + *kν*Λ<sup>1</sup> 2 2 <sup>√</sup>*λ*<sup>1</sup> *t* 0 *Dz uα*,3*uα*,3*dxdη*, (28) *A*<sup>24</sup> ≤ *k <sup>t</sup>* 0 *Rz ϕαϕαdxdη* 1 <sup>2</sup> *<sup>t</sup>* 0 *Rz uαuαdxdη* 1 2 ≤ *k*Λ<sup>1</sup> 2 *λ*1 *<sup>t</sup>* 0 *Rz u*2 3,*ηdxdη* 1 <sup>2</sup> *<sup>t</sup>* 0 *Rz uα*,*βuα*,*βdxdη* 1 2 ≤ *k*Λ<sup>1</sup> 2 2*λ*<sup>1</sup> *t* 0 *Rz u*2 3,*ηdxdη* + *k*Λ<sup>1</sup> 2 2*λ*<sup>1</sup> *t* 0 *Rz uα*,*βuα*,*βdxdη*, (29) *A*<sup>25</sup> ≤ *k <sup>t</sup>* 0 *Rz ϕαϕαdxdη* 1 <sup>2</sup> *<sup>t</sup>* 0 *Rz gαgαT*2*dxdη* 1 2 ≤ *<sup>k</sup>δ*1Λ<sup>1</sup> 2 *λ*1 *<sup>t</sup>* 0 *Rz u*2 3,*ηdxdη* 1 <sup>2</sup> *<sup>t</sup>* 0 *Rz T*,*iT*,*idxdη* 1 2 ≤ *<sup>k</sup>δ*1Λ<sup>1</sup> 2 2*λ*<sup>1</sup> *t* 0 *Rz u*2 3,*ηdxdη* + *<sup>k</sup>δ*1Λ<sup>1</sup> 2 2*λ*<sup>1</sup> *t* 0 *Rz T*,*iT*,*idxdη*, (30)

$$A\_{2\delta} \le \frac{k\delta\_2 \Lambda^{\frac{1}{2}}}{2\lambda\_1} \int\_0^t \int\_{\mathcal{R}\_\varepsilon} u\_{3,\eta}^2 dx d\eta + \frac{k\delta\_2 \Lambda^{\frac{1}{2}}}{2\lambda\_1} \int\_0^t \int\_{\mathcal{R}\_\varepsilon} \mathbb{C}\_{,i} \mathbb{C}\_{,j} dx d\eta. \tag{31}$$

Inserting (26)–(31) into (25), then (19)–(21) and (25) into (18), and choosing *ε*<sup>1</sup> = *ε*<sup>2</sup> = <sup>1</sup> 2 , we obtain the following lemma.

**Lemma 5.** *Let u*, *T*, *C*, *p be solutions of Equations* (1)*–*(8) *with g*, *h* ∈ *L*∞(*R* × {*t* > 0}) *and <sup>D</sup> f*3*dA* = 0*. Then*

$$\begin{split} &E\_{1}(z,t) + k\nu \int\_{R\_{z}} (\prescript{}{z}{\sharp} - z) u\_{i,j} u\_{i,j} dx \Big|\_{\eta=t} + k \int\_{R\_{z}} (\prescript{}{\sharp} - z) u\_{i} u\_{j} dx \Big|\_{\eta=t} \\ & \quad \le a\_{1} k \int\_{0}^{t} \int\_{R\_{z}} u\_{i,\eta} u\_{i,\eta} dx d\eta + a\_{2} \nu \int\_{0}^{t} \int\_{R\_{z}} u\_{i,j} u\_{i,j} dx d\eta \\ & \quad + \frac{k \delta\_{1} \Lambda^{\frac{1}{2}}}{\lambda\_{1}} \int\_{0}^{t} \int\_{R\_{z}} T\_{,i} T\_{,j} dx d\eta + \frac{k \delta\_{2} \Lambda^{\frac{1}{2}}}{\lambda\_{1}} \int\_{0}^{t} \int\_{R\_{z}} \mathbb{C}\_{,i} \mathbb{C}\_{,i} dx d\eta \\ & \quad + \frac{k \nu \Lambda^{\frac{1}{2}}}{2 \sqrt{\lambda\_{1}}} \int\_{0}^{t} \int\_{D\_{z}} u\_{3,\eta}^{2} dx d\eta + \frac{k \nu \Lambda^{\frac{1}{2}}}{2 \sqrt{\lambda\_{1}}} \int\_{0}^{t} \int\_{D\_{z}} u\_{a,3} u\_{a,3} dx d\eta \\ & \quad + \frac{2 k \delta\_{1}^{2}}{\lambda\_{1}} \int\_{0}^{t} \int\_{R\_{z}} (\mathbb{S} - z) T\_{,a} T\_{,a} dx d\eta + \frac{2 k \delta\_{2}^{2}}{\lambda\_{1}} \int\_{0}^{t} \int\_{R\_{z}} (\mathbb{S} - z) \mathbb{C}\_{,a} \mathbb{C}\_{,a} dx d\eta, \end{split}$$

*where*

$$\begin{split} a\_{1} &= \sqrt{k\nu} + \frac{\Lambda^{\frac{1}{2}}}{\sqrt{\lambda\_{1}}} + \nu \Lambda^{\frac{1}{2}} + \frac{\Lambda^{\frac{1}{2}}}{\lambda\_{1}} + \frac{2\delta\_{1}\Lambda^{\frac{1}{2}}}{\lambda\_{1}} + \frac{2\delta\_{2}\Lambda^{\frac{1}{2}}}{\lambda\_{1}}, \\ a\_{2} &= \frac{\sqrt{k\nu}}{2} + \frac{k\Lambda^{\frac{1}{2}}}{2} + \frac{k\Lambda^{\frac{1}{2}}}{2\nu\sqrt{\lambda\_{1}}}. \end{split}$$

Similar to Lemma 5, for *E*2(*z*, *t*) we can obtain the following lemma.

**Lemma 6.** *Let u*, *T*, *C*, *p be solutions of Equations* (1)*–*(8) *with g*, *h* ∈ *L*∞(*R* × {*t* > 0}) *and <sup>D</sup> f*3*dA* = 0*. Then*

$$\begin{split} E\_{2}(z,t) &+ \frac{1}{2} \int\_{0}^{t} \int\_{\mathcal{R}\_{\varepsilon}} (\boldsymbol{\xi} - z) u\_{i} u\_{i} \mathrm{d}x d\eta + \frac{1}{2} \int\_{\mathcal{R}\_{\varepsilon}} (\boldsymbol{\xi} - z) u\_{i} u\_{i} d\boldsymbol{x} \Big|\_{\eta = t} \\ &\leq a\_{3} \int\_{0}^{t} \int\_{\mathcal{R}\_{\varepsilon}} u\_{i,j} u\_{i,j} d\boldsymbol{x} d\eta + \frac{\nu}{2} \int\_{0}^{t} \int\_{\mathcal{R}\_{\varepsilon}} u\_{i} u\_{i} d\boldsymbol{x} d\eta \\ &+ \frac{\Lambda\_{1}^{\frac{1}{2}}}{2\lambda\_{1}} \int\_{0}^{t} \int\_{\mathcal{R}\_{\varepsilon}} u\_{i,\eta} u\_{i,\eta} d\boldsymbol{x} d\eta + \frac{\delta\_{1} \Lambda\_{1}^{\frac{1}{2}}}{2\lambda\_{1}} \int\_{0}^{t} \int\_{\mathcal{R}\_{\varepsilon}} T\_{\mu} T\_{,\mu} d\boldsymbol{x} d\eta \\ &+ \frac{\delta\_{2} \Lambda\_{1}^{\frac{1}{2}}}{2\lambda\_{1}} \int\_{0}^{t} \int\_{\mathcal{R}\_{\varepsilon}} \mathbb{C}\_{\mu} \mathbb{C}\_{\mu} d\boldsymbol{x} d\eta \\ &+ \frac{\delta\_{1}^{2}}{2\lambda\_{1} \varepsilon\_{3}} \int\_{0}^{t} \int\_{\mathcal{R}\_{\varepsilon}} (\boldsymbol{\xi} - z) T\_{,\mu} T\_{,\mu} d\boldsymbol{x} d\eta + \frac{\delta\_{2}^{2}}{2\lambda\_{1} \varepsilon\_{2}} \int\_{0}^{t} \int\_{\mathcal{R}\_{\varepsilon}} (\boldsymbol{\xi} - z) \mathbb{C}\_{\mu} \mathbb{C}\_{\mu} d\boldsymbol{x} d\eta, \end{split}$$

*where*

$$a\_3 = \frac{\nu}{2} + \frac{\Lambda^{\frac{1}{2}}}{2\lambda\_1} + \frac{\nu\Lambda^{\frac{1}{2}}}{2\sqrt{\lambda\_1}} + \frac{\nu\Lambda^{\frac{1}{2}}}{2\lambda\_1} + \frac{k\Lambda^{\frac{1}{2}}}{2\lambda\_1} + \frac{\delta\_1\Lambda^{\frac{1}{2}}}{2\lambda\_1} + \frac{\delta\_2\Lambda^{\frac{1}{2}}}{2\lambda\_1}.$$

**Proof.** By the divergence theorem and Equations (1)–(8), we have

$$\begin{split} E\_2(z,t) &= -\int\_0^t \int\_{\mathcal{R}\_z} (\not\!\!\! - z) u\_i u\_i dx d\eta - \frac{1}{2} \int\_{\mathcal{R}\_z} (\not\!\!\! - z) u\_i u\_i dx \Big|\_{\eta=t} \\ &- \nu \int\_0^t \int\_{\mathcal{R}\_z} u\_i u\_{i,3} dx d\eta + \int\_0^t \int\_{\mathcal{R}\_z} (\not\!\!\! - z) u\_i g\_i T dx d\eta \\ &+ \int\_0^t \int\_{\mathcal{R}\_z} (\not\!\!\! - z) u\_i h\_i \mathbb{C} dx d\eta + \int\_0^t \int\_{\mathcal{R}\_z} u\_i g p dx d\eta \\ &= -\int\_0^t \int\_{\mathcal{R}\_z} (\not\!\!\! - z) u\_i u\_i dx d\eta - \frac{1}{2} \int\_{\mathcal{R}\_z} (\not\!\!\! - z) u\_i u\_i dx \Big|\_{\eta=t} \\ &+ \sum\_{i=1}^4 B\_i. \end{split} \tag{32}$$

Using the Schwarz inequality, the Poincaré inequality and the AG mean inequality, we can obtain

$$\begin{split} \|B\_{1} \leq & \nu \Big[ \int\_{0}^{t} \int\_{R\_{z}} u\_{i,3} u\_{i,3} d\mathbf{x} d\eta \int\_{0}^{t} \int\_{R\_{z}} u\_{i} u\_{i} d\mathbf{x} d\eta \Big]^{\frac{1}{2}} \\ \leq & \frac{\nu}{2} \Big[ \int\_{0}^{t} \int\_{R\_{z}} u\_{i,3} u\_{i,3} d\mathbf{x} d\eta + \int\_{0}^{t} \int\_{R\_{z}} u\_{i} u\_{i} d\mathbf{x} d\eta \Big]. \end{split} \tag{33}$$

Similar to (20) and (21), we have for *B*<sup>2</sup> and *B*<sup>3</sup>

$$B\_2 \le \frac{\varepsilon\_3}{2} \int\_0^t \int\_{R\_\varepsilon} (\boldsymbol{\xi} - \boldsymbol{z}) \boldsymbol{u}\_i \boldsymbol{u}\_i d\mathbf{x} d\boldsymbol{\eta} + \frac{\delta\_1^2}{2\lambda\_1 \varepsilon\_3} \int\_0^t \int\_{R\_\varepsilon} (\boldsymbol{\xi} - \boldsymbol{z}) \boldsymbol{T}\_{,a} \boldsymbol{T}\_{,a} d\mathbf{x} d\boldsymbol{\eta},\tag{34}$$

and

$$B\_3 \le \frac{\varepsilon\_4}{2} \int\_0^t \int\_{R\_\varepsilon} (\xi - z) u\_i u\_i dx d\eta + \frac{\delta\_2^2}{2\lambda\_1 \varepsilon\_2} \int\_0^t \int\_{R\_\varepsilon} (\xi - z) \mathbb{C}\_{,a} \mathbb{C}\_{,a} dx d\eta,\tag{35}$$

where *ε*3,*ε*<sup>4</sup> are positive constants.

To bound *B*<sup>4</sup> in (32), we also require that

$$\int\_{D} f\_{\\$}dA = 0.$$

Then using to the Lemma 2 in Section 2, there exist vector functions (*ϕ*71, *<sup>ϕ</sup>*72) such that

$$
\hat{\varphi}\_{a,a} = \mathfrak{u}\_{\mathfrak{J}\_{\prime}} \quad \text{in} \ D, \quad \hat{\varphi}\_{a} = 0, \quad \text{on} \ \partial D. \tag{36}
$$

Therefore, we have

$$\begin{split} B\_{4} &= \int\_{0}^{t} \int\_{R\_{\varepsilon}} \hat{\varphi}\_{a,a} p dx d\eta \\ &= -\int\_{0}^{t} \int\_{R\_{\varepsilon}} \hat{\varphi}\_{a} p\_{a} dx d\eta \\ &= \int\_{0}^{t} \int\_{R\_{\varepsilon}} \hat{\varphi}\_{a} \left[ u\_{a,\eta} - \nu \Delta u\_{a} + u\_{a} - g\_{a}T - h\_{a}\mathbb{C} \right] dx d\eta \\ &= \int\_{0}^{t} \int\_{R\_{\varepsilon}} \hat{\varphi}\_{a} u\_{a,\eta} dx d\eta + \nu \int\_{0}^{t} \int\_{R\_{\varepsilon}} \hat{\varphi}\_{a,\theta} u\_{a,\theta} dx d\eta + \nu \int\_{0}^{t} \int\_{D\_{\varepsilon}} \hat{\varphi}\_{a} u\_{a,\beta} dA d\eta \\ &+ \int\_{0}^{t} \int\_{R\_{\varepsilon}} \hat{\varphi}\_{a} u\_{a,\theta} dx d\eta - \int\_{0}^{t} \int\_{R\_{\varepsilon}} \hat{g}\_{a} T \hat{\varphi}\_{a} dx d\eta - \int\_{0}^{t} \int\_{R\_{\varepsilon}} h\_{a} \mathbb{C} \hat{\varphi}\_{a} dx d\eta \\ &\doteq \sum\_{i=1}^{6} B\_{4i} .\end{split} \tag{37}$$

As the derivation of (26)–(32), we conclude that

*B*<sup>41</sup> ≤ *<sup>t</sup>* 0 *Rz <sup>ϕ</sup>*7*αϕ*7*αdxd<sup>η</sup>* 1 <sup>2</sup> *<sup>t</sup>* 0 *Rz uα*,*ηuα*,*ηdxdη* 1 2 ≤ 1 <sup>√</sup>*λ*<sup>1</sup> *<sup>t</sup>* 0 *Rz <sup>ϕ</sup>*7*α*,*βϕ*7*α*,*βdxdη*) 1 2 *<sup>t</sup>* 0 *Rz uα*,*ηuα*,*ηdxdη* 1 2 ≤ Λ1 2 <sup>√</sup>*λ*<sup>1</sup> *<sup>t</sup>* 0 *Rz u*2 <sup>3</sup>*xdη* 1 <sup>2</sup> *<sup>t</sup>* 0 *Rz uα*,*ηuα*,*ηdxdη* 1 2 ≤ Λ1 2 2*λ*<sup>1</sup> # *<sup>t</sup>* 0 *Rz u*3,*αu*3,*αxdη* + *t* 0 *Rz ui*,*<sup>η</sup>ui*,*ηdxdη* \$ , (38) *B*<sup>42</sup> ≤ *ν <sup>t</sup>* 0 *Rz <sup>ϕ</sup>*7*α*,*βϕ*7*α*,*βdxd<sup>η</sup>* 1 <sup>2</sup> *<sup>t</sup>* 0 *Rz uα*,*βuα*,*βdxdη* 1 2 <sup>≤</sup> *<sup>ν</sup>*Λ<sup>1</sup> 2 *<sup>t</sup>* 0 *Rz u*2 <sup>3</sup>*dxdη* 1 <sup>2</sup> *<sup>t</sup>* 0 *Rz uα*,*βuα*,*βdxdη* 1 2 <sup>≤</sup> *<sup>ν</sup>*Λ<sup>1</sup> 2 2 <sup>√</sup>*λ*<sup>1</sup> *t* 0 *Rz ui*,*jui*,*jdxdη*, (39) *B*<sup>43</sup> ≤ *ν <sup>t</sup>* 0 *Dz <sup>ϕ</sup>*7*αϕ*7*αdxd<sup>η</sup>* 1 <sup>2</sup> *<sup>t</sup>* 0 *Dz uα*,3*uα*,3*dxdη* 1 2 <sup>≤</sup> *<sup>ν</sup>*Λ<sup>1</sup> 2 <sup>√</sup>*λ*<sup>1</sup> *<sup>t</sup>* 0 *Dz u*2 <sup>3</sup>*dxdη* 1 <sup>2</sup> *<sup>t</sup>* 0 *Dz uα*,3*uα*,3*dxdη* 1 2 <sup>≤</sup> *<sup>ν</sup>*Λ<sup>1</sup> 2 2*λ*<sup>1</sup> *t* 0 *Dz ui*,*jui*,*jdxdη*, (40) *B*<sup>44</sup> ≤ *<sup>t</sup>* 0 *Rz <sup>ϕ</sup>*7*αϕ*7*αdxd<sup>η</sup>* 1 <sup>2</sup> *<sup>t</sup>* 0 *Rz uαuαdxdη* 1 2 ≤ Λ1 2 *λ*1 *t* 0 *Rz uiuidxdη* ≤ *k*Λ<sup>1</sup> 2 2*λ*<sup>1</sup> *t* 0 *Rz ui*,*<sup>α</sup>ui*,*αdxdη*, (41) *B*<sup>45</sup> ≤ *<sup>t</sup>* 0 *Rz <sup>ϕ</sup>*7*αϕ*7*αdxd<sup>η</sup>* 1 <sup>2</sup> *<sup>t</sup>* 0 *Rz gαgαT*2*dxdη* 1 2 ≤ *<sup>δ</sup>*1Λ<sup>1</sup> 2 *λ*1 *<sup>t</sup>* 0 *Rz u*3,*αu*3,*αdxdη* 1 <sup>2</sup> *<sup>t</sup>* 0 *Rz T*,*αT*,*αdxdη* 1 2 ≤ *<sup>δ</sup>*1Λ<sup>1</sup> 2 2*λ*<sup>1</sup> # *<sup>t</sup>* 0 *Rz u*3,*αu*3,*αdxdη* + *t* 0 *Rz T*,*αT*,*αdxdη* \$ , (42)

$$B\_{46} \le \frac{\delta\_2 \Lambda^{\frac{1}{2}}}{2\lambda\_1} \left[ \int\_0^t \int\_{R\_\varepsilon} \mu\_{3,\mathfrak{a}} \mu\_{3,\mathfrak{a}} d\mathbf{x} d\eta + \int\_0^t \int\_{R\_\varepsilon} \mathbb{C}\_{,\mathfrak{a}} \mathbb{C}\_{,\mathfrak{a}} d\mathbf{x} d\eta \right]. \tag{43}$$

Inserting (38)–(43) into (37), we obtain

$$\begin{split} B\_{4} &\leq \left[\frac{\Lambda^{\frac{1}{2}}}{2\lambda\_{1}} + \frac{\nu\Lambda^{\frac{1}{2}}}{2\sqrt{\lambda\_{1}}} + \frac{\nu\Lambda^{\frac{1}{2}}}{2\lambda\_{1}} + \frac{k\Lambda^{\frac{1}{2}}}{2\lambda\_{1}} + \frac{\delta\_{1}\Lambda^{\frac{1}{2}}}{2\lambda\_{1}} + \frac{\delta\_{2}\Lambda^{\frac{1}{2}}}{2\lambda\_{1}}\right] \int\_{0}^{t} \int\_{\mathcal{R}\_{z}} u\_{i,j}u\_{i,j}dxd\eta \\ &+ \frac{\Lambda^{\frac{1}{2}}}{2\lambda\_{1}} \int\_{0}^{t} \int\_{\mathcal{R}\_{z}} u\_{i,\eta}u\_{i,\eta}dxd\eta + \frac{\delta\_{1}\Lambda^{\frac{1}{2}}}{2\lambda\_{1}} \int\_{0}^{t} \int\_{\mathcal{R}\_{z}} T\_{,\mu}T\_{,\mu}dxd\eta \\ &+ \frac{\delta\_{2}\Lambda^{\frac{1}{2}}}{2\lambda\_{1}} \int\_{0}^{t} \int\_{\mathcal{R}\_{z}} \mathbb{C}\_{,\mu}\mathbb{C}\_{,\mu}dxd\eta. \end{split} \tag{44}$$

Inserting (33), (34), (35) and (44) into (32) and choosing *ε*<sup>3</sup> = *ε*<sup>4</sup> = <sup>1</sup> <sup>2</sup> , we can obtain Lemma 5.

Next we may bound *E*3(*z*, *t*). First we let *TM* denotes that the maximum of *T* by using the maximum principle in *R*, i.e.,

$$T\_M = \max\_{D \times \{t > 0\}} F(\mathbf{x}\_1, \mathbf{x}\_2, t). \tag{45}$$

Integrating by parts, using (3), (5), (6), (7) together with (9) and the AG mean inequality, we have

*E*3(*z*, *t*) = −*ρ*<sup>1</sup> *t* 0 *Rz TT*,3*dxdη* − *ρ*<sup>1</sup> *t* 0 *Rz T*(*T*,*<sup>η</sup>* + *uiT*,*i*)*dxdη* <sup>=</sup> <sup>−</sup>*ρ*<sup>1</sup> 2 *Rz* (*<sup>ξ</sup>* <sup>−</sup> *<sup>z</sup>*)*T*2*dx η*=*t* <sup>+</sup> *<sup>ρ</sup>*<sup>1</sup> 2 *t* 0 *Dz <sup>T</sup>*2*dAd<sup>η</sup>* <sup>+</sup> *<sup>ρ</sup>*<sup>1</sup> 2 *t* 0 *Rz u*3*T*2*dxdη* ≤ −*ρ*<sup>1</sup> 2 *Rz* (*<sup>ξ</sup>* <sup>−</sup> *<sup>z</sup>*)*T*2*dx η*=*t* <sup>+</sup> *<sup>ρ</sup>*<sup>1</sup> 2*λ*<sup>1</sup> *t* 0 *Dz T*,*αT*,*αdAdη* <sup>+</sup> *<sup>ρ</sup>*1*TM* 2 *<sup>t</sup>* 0 *Rz u*2 <sup>3</sup>*dxdη* 1 <sup>2</sup> *<sup>t</sup>* 0 *Rz T*2*dxdη* 1 2 ≤ −*ρ*<sup>1</sup> 2 *Rz* (*<sup>ξ</sup>* <sup>−</sup> *<sup>z</sup>*)*T*2*dx η*=*t* <sup>+</sup> *<sup>ρ</sup>*<sup>1</sup> 2*λ*<sup>1</sup> *t* 0 *Dz T*,*αT*,*αdAdη* <sup>+</sup> *<sup>ρ</sup>*1*TM* 4*λ*<sup>1</sup> *<sup>t</sup>* 0 *Rz u*3,*αu*3,*αdxdη* + *t* 0 *Rz T*,*iT*,*idxdη* . (46)

Using Equations (3)–(7) and integrating by parts, we obtain

$$\begin{split} E\_{\mathsf{A}}(\mathsf{z},t) &= -\rho\_{2} \int\_{0}^{t} \int\_{R\_{\mathsf{z}}} \mathsf{C} \mathsf{C}\_{\mathsf{z}} \mathrm{d}\mathsf{x} d\eta - \rho\_{2} \int\_{0}^{t} \int\_{R\_{\mathsf{z}}} (\mathsf{z}^{\mathsf{z}} - \mathsf{z}) \mathsf{C} \left[ \mathsf{C}\_{\mathsf{J}} + \mathsf{u}\_{i} \mathsf{C}\_{\mathsf{z}} - \sigma \Delta T \right] \mathrm{d}\mathsf{x} d\eta \\ &\leq \frac{\rho\_{2}}{2} \int\_{0}^{t} \int\_{D\_{\mathsf{z}}} \mathsf{C}^{2} d\boldsymbol{A} d\eta - \frac{\rho\_{2}}{2} \int\_{R\_{\mathsf{z}}} (\mathsf{z}^{\mathsf{z}} - \mathsf{z}) \mathsf{C}^{2} \mathrm{d}\mathsf{x} \Big|\_{\eta=t} + \frac{\rho\_{2}}{2} \int\_{0}^{t} \int\_{R\_{\mathsf{z}}} \mathsf{u}\_{3} \mathsf{C}^{2} d\mathsf{x} d\eta \\ &- \sigma \rho\_{2} \int\_{0}^{t} \int\_{R\_{\mathsf{z}}} \mathsf{C} T\_{\mathsf{z}} \mathrm{d}\mathsf{x} d\eta - \sigma \rho\_{2} \int\_{0}^{t} \int\_{R\_{\mathsf{z}}} (\mathsf{z} - \mathsf{z}) \mathsf{C}\_{\mathsf{z}} T\_{\mathsf{z}} \mathrm{d}\mathsf{x} d\eta. \end{split} \tag{47}$$

By the Schwarz and the AG mean inequalities, it follows that from (47)

$$\begin{split} \left. E\_{4}(z,t) + \frac{\rho\_{2}}{2} \int\_{R\_{z}} (\boldsymbol{\xi} - \boldsymbol{z}) \mathbb{C}^{2} d\mathbf{x} \right|\_{\eta = t} \\ \leq & \frac{\rho\_{2}}{2\lambda\_{1}} \int\_{0}^{t} \int\_{D\_{z}} \mathbb{C}^{2} d\boldsymbol{A} d\eta + \frac{\sigma \rho\_{2}}{2\sqrt{\lambda\_{1}}} \int\_{0}^{t} \int\_{R\_{z}} \mathbb{C}\_{,i} \mathbb{C}\_{,j} d\mathbf{x} d\eta \\ & + \frac{\sigma \rho\_{2}}{2\sqrt{\lambda\_{1}}} \int\_{0}^{t} \int\_{R\_{z}} T\_{\boldsymbol{\beta}} T\_{\boldsymbol{\beta}} \boldsymbol{x} d\mathbf{x} d\eta + \frac{\sigma \rho\_{2} \varepsilon\_{5}}{2} \int\_{0}^{t} \int\_{R\_{z}} (\boldsymbol{\xi} - \boldsymbol{z}) \mathbb{C}\_{,i} \mathbb{C}\_{,j} d\mathbf{x} d\eta \\ & + \frac{\sigma \rho\_{2}}{2\varepsilon\_{5}} \int\_{0}^{t} \int\_{R\_{z}} (\boldsymbol{\xi} - \boldsymbol{z}) T\_{\boldsymbol{\beta}} T\_{\boldsymbol{\beta}} d\mathbf{x} d\eta + \frac{\rho\_{2}}{2} \int\_{0}^{t} \int\_{R\_{z}} \mathbb{L}\_{3} \mathbb{C}^{2} d\mathbf{x} d\eta, \end{split} \tag{48}$$

for an arbitrary constant *ε*<sup>5</sup> > 0.

In order to bound the last term on the right of (48), using the Equations (9), (11) and (13), the Schwarz inequality and the AG mean inequality to obtain

*ρ*2 2 *t* 0 *Rz <sup>u</sup>*3*C*2*dxd<sup>η</sup>* <sup>≤</sup> *<sup>ρ</sup>*<sup>2</sup> 2 *t* 0 *Rz* (*u*3*C*)2*dx* 1 <sup>2</sup> *Rz C*2*dx* 1 2 *dη* <sup>≤</sup> *<sup>ρ</sup>*<sup>2</sup> <sup>2</sup> max*<sup>t</sup>* ) *Rz C*2*dx* 1 <sup>2</sup> \* *<sup>t</sup>* 0 ∞ *z Dξ u*4 3*dA* 1 4 *Dξ C*4*dA* 1 4 *dξdη* <sup>≤</sup> *<sup>ρ</sup>*2Ω<sup>1</sup> 8 2 3 <sup>4</sup> *λ* 1 4 1 max*<sup>t</sup>* ) *Rz C*2*dx* 1 <sup>2</sup> \* *<sup>t</sup>* 0 *Rz ui*,*jui*,*jdxdη* 1 <sup>2</sup> *<sup>t</sup>* 0 *Rz C*,*iC*,*idxdη* 1 2 <sup>≤</sup> *<sup>ρ</sup>*2Ω<sup>1</sup> 8 2 7 <sup>4</sup> *λ* 1 4 1 max*<sup>t</sup>* ) *Rz C*2*dx* 1 <sup>2</sup> \* *<sup>t</sup>* 0 *Rz ui*,*jui*,*jdxdη* + *t* 0 *Rz C*,*iC*,*idxdη* , (49)

where the bound for max*t* ) ( *Rz <sup>C</sup>*2*dx*) 1 2 \* will be derived later.

Inserting (49) back into (48), we have

*<sup>E</sup>*4(*z*, *<sup>t</sup>*) + *<sup>ρ</sup>*<sup>2</sup> 2 *Rz* (*<sup>ξ</sup>* <sup>−</sup> *<sup>z</sup>*)*C*2*dx η*=*t* <sup>≤</sup> *<sup>ρ</sup>*<sup>2</sup> 2*λ*<sup>1</sup> *t* 0 *Dz <sup>C</sup>*,*αC*,*αdAd<sup>η</sup>* <sup>+</sup> *σρ*<sup>2</sup> 2 <sup>√</sup>*λ*<sup>1</sup> *t* 0 *Rz <sup>C</sup>*,*iC*,*idxd<sup>η</sup>* <sup>+</sup> *σρ*<sup>2</sup> 2 <sup>√</sup>*λ*<sup>1</sup> *t* 0 *Rz T*,3*T*,3*dxdη* <sup>+</sup> *σρ*2*ε*<sup>6</sup> 2 *t* 0 *Rz* (*<sup>ξ</sup>* <sup>−</sup> *<sup>z</sup>*)*C*,*iC*,*idxd<sup>η</sup>* <sup>+</sup> *σρ*<sup>2</sup> 2*ε*<sup>6</sup> *t* 0 *Rz* (*ξ* − *z*)*T*,*iT*,*idxdη* <sup>+</sup> *<sup>ρ</sup>*2<sup>Ω</sup> <sup>1</sup> 8 2 7 <sup>4</sup> *λ* 1 4 1 max*<sup>t</sup>* ) *Rz C*2*dx* 1 2 \* *<sup>t</sup>* 0 *Rz ui*,*jui*,*jdxdη* + *t* 0 *Rz C*,*iC*,*idxdη* . (50)

Combining (46) and (50), we obtain the following Lemma.

**Lemma 7.** *Let u*, *T*, *C*, *p be solutions of Equations* (1)*–*(8) *with g*, *h* ∈ *L*∞(*R* × {*t* > 0}) *and <sup>D</sup> f*3*dA* = 0*. Then*

*E*3(*z*, *t*) + *E*4(*z*, *t*) 1 2 *Rz* (*ξ* − *z*) # *ρ*1*T*<sup>2</sup> + *ρ*2*C*<sup>2</sup> \$ *dx η*=*t* <sup>≤</sup> *<sup>ρ</sup>*<sup>1</sup> 2*λ*<sup>1</sup> *t* 0 *Dz <sup>T</sup>*,*αT*,*αdAd<sup>η</sup>* <sup>+</sup> *<sup>ρ</sup>*<sup>2</sup> 2*λ*<sup>1</sup> *t* 0 *Dz C*,*αC*,*αdAdη* + # *σρ*<sup>2</sup> 2 <sup>√</sup>*λ*<sup>1</sup> <sup>+</sup> *<sup>ρ</sup>*2Ω<sup>1</sup> 8 2 7 <sup>4</sup> *λ* 1 4 1 max*<sup>t</sup>* ) *Rz C*2*dx* 1 <sup>2</sup> \*\$ *<sup>t</sup>* 0 *Rz C*,*iC*,*idxdη* + # *σρ*<sup>2</sup> 2 <sup>√</sup>*λ*<sup>1</sup> <sup>+</sup> *<sup>ρ</sup>*1*TM* 4*λ*<sup>1</sup> \$ *<sup>t</sup>* 0 *Rz T*,*iT*,*idxdη* <sup>+</sup> *σρ*2*ε*<sup>6</sup> 2 *t* 0 *Rz* (*<sup>ξ</sup>* <sup>−</sup> *<sup>z</sup>*)*C*,*iC*,*idxd<sup>η</sup>* <sup>+</sup> *σρ*<sup>2</sup> 2*ε*<sup>6</sup> *t* 0 *Rz* (*ξ* − *z*)*T*,*iT*,*idxdη* + # *ρ*1*TM* 4*λ*<sup>1</sup> <sup>+</sup> *<sup>ρ</sup>*2Ω<sup>1</sup> 8 2 7 <sup>4</sup> *λ* 1 4 1 max*<sup>t</sup>* ) *Rz C*2*dx* 1 <sup>2</sup> \*\$ *<sup>t</sup>* 0 *Rz ui*,*jui*,*jdxdη*,

*where ε*<sup>6</sup> *is a positive constant. Next, we use Lemmas 5–7 to prove our main result.*

## **4. Main Result**

First, we introduce a new function

$$\begin{split} \Psi(z,t) &= k \int\_{0}^{t} \int\_{R\_{z}} (\not\!\! - z) u\_{i,\eta} u\_{i,\eta} \text{d}x d\eta + \nu \int\_{0}^{t} \int\_{R\_{z}} (\not\!\! - z) u\_{i,\eta} u\_{i,\eta} \text{d}x d\eta \\ &+ \rho\_{1} \int\_{0}^{t} \int\_{R\_{z}} (\not\!\! - z) T\_{i} T\_{j} \text{d}x d\eta + \rho\_{2} \int\_{0}^{t} \int\_{R\_{z}} (\not\!\! - z) \mathbb{C}\_{j} \mathbb{C}\_{j} \text{d}x d\eta \\ &+ k \nu \int\_{R\_{z}} (\not\!\!\! - z) u\_{i,\eta} u\_{i,\eta} \text{d}x \Big|\_{\eta=t} + \left. \left(k + \frac{1}{2}\right) \int\_{R\_{z}} (\not\!\!\! - z) u\_{i} u\_{i} d\mathbf{x} \right|\_{\eta=t} \\ &+ \frac{1}{2} \int\_{0}^{t} \int\_{R\_{z}} (\not\!\!\! - z) u\_{i} u\_{i} d\mathbf{x} d\eta + \frac{1}{2} \int\_{R\_{z}} (\not\!\!\! - z) \left[\rho\_{1} T^{2} + \rho\_{2} \mathbb{C}^{2} \right] d\mathbf{x} \Big|\_{\eta=t}. \end{split} \tag{51}$$

Using Lemmas 4–6 and in view of (51), we have

*ψ*(*z*, *t*) ≤ *a*<sup>4</sup> *t* 0 *Rz ui*,*<sup>η</sup>ui*,*ηdxdη* + *a*<sup>5</sup> *t* 0 *Rz ui*,*jui*,*jdxdη* + *a*<sup>6</sup> *t* 0 *Rz <sup>T</sup>*,*iT*,*idxd<sup>η</sup>* <sup>+</sup> *<sup>ν</sup>* 2 *t* 0 *Rz uiuidxdη* + *a*<sup>7</sup> *t* 0 *Rz C*,*iC*,*idxdη* + *kν*Λ<sup>1</sup> 2 2 <sup>√</sup>*λ*<sup>1</sup> *t* 0 *Dz u*2 3,*ηdAdη* + *kν*Λ<sup>1</sup> 2 2 <sup>√</sup>*λ*<sup>1</sup> *t* 0 *Dz uα*,3*uα*,3*dAdη* <sup>+</sup> *<sup>ρ</sup>*<sup>1</sup> 2*λ*<sup>1</sup> *t* 0 *Dz <sup>T</sup>*,*αT*,*αdAd<sup>η</sup>* <sup>+</sup> *<sup>ρ</sup>*<sup>2</sup> 2*λ*<sup>1</sup> *t* 0 *Dz C*,*αC*,*αdAdη* + 2*kδ*<sup>2</sup> 1 *λ*1 *t* 0 *Rz* (*ξ* − *z*)*T*,*αT*,*αdxdη* + 2*kδ*<sup>2</sup> 2 *λ*1 *t* 0 *Rz* (*ξ* − *z*)*C*,*αC*,*αdxdη* + *δ*2 1 2*λ*1*ε*<sup>3</sup> *t* 0 *Rz* (*ξ* − *z*)*T*,*αT*,*αdxdη* + *δ*2 2 2*λ*1*ε*<sup>2</sup> *t* 0 *Rz* (*ξ* − *z*)*C*,*αC*,*αdxdη* <sup>+</sup> *σρ*2*ε*<sup>6</sup> 2 *t* 0 *Rz* (*<sup>ξ</sup>* <sup>−</sup> *<sup>z</sup>*)*C*,*iC*,*idxd<sup>η</sup>* <sup>+</sup> *σρ*<sup>2</sup> 2*ε*<sup>6</sup> *t* 0 *Rz* (*ξ* − *z*)*T*,*iT*,*idxdη*, (52)

where

$$\begin{split} a\_{4} &= a\_{1}k + \frac{\Lambda^{\frac{1}{2}}}{2\lambda\_{1}}, a\_{5} = a\_{2}\nu + a\_{3} + \frac{\rho\_{1}T\_{M}}{4\lambda\_{1}} + \frac{\rho\_{2}\Omega^{\frac{1}{8}}}{2^{\frac{7}{4}}\lambda\_{1}^{\frac{1}{4}}} \max\_{t} \left\{ \left( \int\_{R\_{\varepsilon}} \mathbb{C}^{2} dx \right)^{\frac{1}{2}} \right\}, \\ a\_{6} &= \frac{k\delta\_{1}\Lambda^{\frac{1}{2}}}{\lambda\_{1}} + \frac{\delta\_{1}\Lambda^{\frac{1}{2}}}{2\lambda\_{1}}, a\_{7} = \frac{k\delta\_{2}\Lambda^{\frac{1}{2}}}{\lambda\_{1}} + \frac{\delta\_{2}\Lambda^{\frac{1}{2}}}{2\lambda\_{1}} + \frac{\sigma\rho\_{2}}{2\sqrt{\lambda\_{1}}} + \frac{\rho\_{2}\Omega^{\frac{1}{8}}}{2^{\frac{7}{4}}\lambda\_{1}^{\frac{1}{2}}} \max\_{t} \left\{ \left( \int\_{R\_{\varepsilon}} \mathbb{C}^{2} dx \right)^{\frac{1}{2}} \right\}. \end{split}$$

Choosing *ε*<sup>6</sup> = <sup>1</sup> <sup>2</sup>*<sup>σ</sup>* , *<sup>ρ</sup>*<sup>2</sup> <sup>=</sup> <sup>8</sup>*kδ*<sup>2</sup> 2 *<sup>λ</sup>*<sup>1</sup> <sup>+</sup> <sup>2</sup>*δ*<sup>2</sup> 2 *λ*1*ε*<sup>2</sup> , *<sup>ρ</sup>*<sup>1</sup> <sup>=</sup> <sup>4</sup>*kδ*<sup>2</sup> 1 *<sup>λ</sup>*<sup>1</sup> <sup>+</sup> *<sup>δ</sup>*<sup>2</sup> 1 *<sup>λ</sup>*1*ε*<sup>3</sup> <sup>+</sup> *σρ*<sup>2</sup> *<sup>ε</sup>*<sup>6</sup> and define

$$\begin{split} \Psi(z,t) &= k \int\_{0}^{t} \int\_{R\_{z}} (\xi - z) u\_{i\eta} u\_{i\eta} dx d\eta + \nu \int\_{0}^{t} \int\_{R\_{z}} (\xi - z) u\_{i,j} u\_{i,j} dx d\eta \\ &+ \frac{1}{2} \rho\_{1} \int\_{0}^{t} \int\_{R\_{z}} (\xi - z) T\_{i} T\_{j} dx d\eta + \frac{1}{2} \rho\_{2} \int\_{0}^{t} \int\_{R\_{z}} (\xi - z) C\_{z} \mathbb{C}\_{i} dx d\eta \\ &+ k\nu \int\_{R\_{z}} (\xi - z) u\_{i,j} u\_{i,j} dx \Big|\_{\eta=t} + (k + \frac{1}{2}) \int\_{R\_{z}} (\xi - z) u\_{i} u\_{i} dx \Big|\_{\eta=t} \\ &+ \frac{1}{2} \int\_{0}^{t} \int\_{R\_{z}} (\xi - z) u\_{i} u\_{i} dx d\eta + \frac{1}{2} \int\_{R\_{z}} (\xi - z) \left[\rho\_{1} T^{2} + \rho\_{2} \mathbb{C}^{2}\right] dx \Big|\_{\eta=t}. \end{split} \tag{53}$$

we can have from (52)

$$\begin{split} \Psi(z,t) &\leq a\_{4} \int\_{0}^{t} \int\_{R\_{z}} u\_{i,\eta} u\_{i,\eta} dx d\eta + a\_{5} \int\_{0}^{t} \int\_{R\_{z}} u\_{i,j} u\_{i,j} dx d\eta \\ &+ a\_{6} \int\_{0}^{t} \int\_{R\_{z}} T\_{j} T\_{j} dx d\eta + \frac{\nu}{2} \int\_{0}^{t} \int\_{R\_{z}} u\_{i} u\_{i} dx d\eta + a\_{7} \int\_{0}^{t} \int\_{R\_{z}} \mathbb{C}\_{,i} \mathbb{C}\_{,j} dx d\eta \\ &+ \frac{k\nu \Lambda^{\frac{1}{2}}}{2\sqrt{\lambda\_{1}}} \int\_{0}^{t} \int\_{D\_{z}} u\_{3\cdot\eta}^{2} dA d\eta + \frac{k\nu \Lambda^{\frac{1}{2}}}{2\sqrt{\lambda\_{1}}} \int\_{0}^{t} \int\_{D\_{z}} u\_{a\cdot 3} u\_{a\cdot 3} dA d\eta \\ &+ \frac{\rho\_{1}}{2\lambda\_{1}} \int\_{0}^{t} \int\_{D\_{z}} T\_{\mu} T\_{\mu} dA d\eta + \frac{\rho\_{2}}{2\lambda\_{1}} \int\_{0}^{t} \int\_{D\_{z}} \mathbb{C}\_{,\mu} \mathbb{C}\_{,\mu} dA d\eta. \end{split} \tag{54}$$

From (53), we have

$$\begin{split} -\frac{\partial \Psi(z,t)}{\partial z} &= k \int\_{0}^{t} \int\_{\mathbb{R}\_{z}} u\_{i,\eta} u\_{i,\eta} dx d\eta + \nu \int\_{0}^{t} \int\_{\mathbb{R}\_{z}} u\_{i,j} u\_{i,j} dx d\eta \\ &+ \frac{1}{2} \rho\_{1} \int\_{0}^{t} \int\_{\mathbb{R}\_{z}} T\_{,i} T\_{,i} dx d\eta + \frac{1}{2} \rho\_{2} \int\_{0}^{t} \int\_{\mathbb{R}\_{z}} \mathbb{C}\_{,i} \mathbb{C}\_{,i} dx d\eta \\ &+ k\nu \int\_{\mathbb{R}\_{z}} u\_{i,j} u\_{i,j} dx \Big|\_{\eta=t} + (k + \frac{1}{2}) \int\_{\mathbb{R}\_{z}} u\_{i} u\_{i} dx \Big|\_{\eta=t} \\ &+ \frac{1}{2} \int\_{0}^{t} \int\_{\mathbb{R}\_{z}} u\_{i} u\_{i} dx d\eta + \frac{1}{2} \int\_{\mathbb{R}\_{z}} \left[ \rho\_{1} T^{2} + \rho\_{2} \mathbb{C}^{2} \right] dx \Big|\_{\eta=t} \end{split} (55)$$

and

$$\begin{split} \frac{\partial^2 \Psi(z,t)}{\partial z^2} &= k \int\_0^t \int\_{D\_z} u\_{i,\eta} u\_{i,\eta} dA d\eta + \nu \int\_0^t \int\_{D\_z} u\_{i,j} u\_{i,j} dA d\eta \\ &+ \frac{1}{2} \rho\_1 \int\_0^t \int\_{D\_z} T\_i T\_{,i} dA d\eta + \frac{1}{2} \rho\_2 \int\_0^t \int\_{D\_z} \mathbb{C}\_{,i} \mathbb{C}\_{,i} dA d\eta \\ &+ k \nu \int\_{D\_z} u\_{i,j} u\_{i,j} dA \Big|\_{\eta=t} + (k + \frac{1}{2}) \int\_{D\_z} u\_i u\_{i} dA \Big|\_{\eta=t} \\ &+ \frac{1}{2} \int\_0^t \int\_{D\_z} u\_i u\_{i} dA d\eta + \frac{1}{2} \int\_{D\_z} \left[ \rho\_1 T^2 + \rho\_2 \mathbb{C}^2 \right] dA \Big|\_{\eta=t}. \end{split} \tag{56}$$

Combining (54), (55) and (56), we have Thus

$$\Psi(z,t) \le K\_1 \left[ -\frac{\partial \Psi(z,t)}{\partial z} \right] + K\_2 \frac{\partial^2 \Psi(z,t)}{\partial z^2},\tag{57}$$

where

$$\begin{aligned} K\_1 &= \max\{ \frac{a\_4}{k}, \frac{a\_5}{\nu}, \nu, \frac{a\_6}{\rho\_1}, \frac{a\_7}{\rho\_2} \}, \\ K\_2 &= \max\{ \frac{\nu \Lambda^{\frac{1}{2}}}{2\sqrt{\lambda\_1}}, \frac{k \Lambda^{\frac{1}{2}}}{2\sqrt{\lambda\_1}}, \frac{1}{2\lambda\_1} \}. \end{aligned}$$

Inequality (57) can be rewritten as

$$\frac{\partial}{\partial z} \left\{ e^{-\ell\_1 z} \left( \frac{\partial \Psi}{\partial z} + \ell\_2 \Psi \right) \right\} \ge 0,\tag{58}$$

where

$$\ell\_1 = \frac{K\_1}{2K\_2} + \frac{1}{2}\sqrt{\frac{K\_1^2}{K\_2^2} + \frac{4}{K\_2}}, \quad \ell\_2 = -\frac{K\_1}{2K\_2} + \frac{1}{2}\sqrt{\frac{K\_1^2}{K\_2^2} + \frac{4}{K\_2}}.$$

Integrating (58) from *z* to ∞ leads to

$$\text{and hence}$$

$$
\frac{\cdot}{\partial z} + \ell\_2 \Psi \le 0,
$$

$$
\Psi(z,t) \le \Psi(0,t)e^{-\ell\_2 z}.\tag{59}
$$

Combining (53) and (59), we can obtain the following theorem.

*∂*Ψ

**Theorem 1.** *Let u*, *T*, *C*, *p be solutions of Equations* (1)*–*(8) *with g*, *h* ∈ *L*∞(*R* × {*t* > 0}) *and <sup>D</sup> f*3*dA* = 0*. Then*

$$\begin{split} &k \int\_{0}^{t} \int\_{R\_{z}} (\prescript{\tau}{}{\xi} - z) u\_{i,\eta} u\_{i,\eta} \mathrm{d}x d\eta + \nu \int\_{0}^{t} \int\_{R\_{z}} (\prescript{\tau}{}{\xi} - z) u\_{i,j} u\_{i,j} \mathrm{d}x d\eta \\ &+ \frac{1}{2} \rho\_{1} \int\_{0}^{t} \int\_{R\_{z}} (\prescript{\tau}{}{\xi} - z) T\_{j} T\_{i} \mathrm{d}x d\eta + \frac{1}{2} \rho\_{2} \int\_{0}^{t} \int\_{R\_{z}} (\prescript{\tau}{}{\xi} - z) \mathsf{C}\_{j} \mathsf{C}\_{j} \mathrm{d}x d\eta \\ &+ k \nu \int\_{R\_{z}} (\prescript{\tau}{}{\xi} - z) u\_{i,j} u\_{i,j} \mathrm{d}x \Big|\_{\eta=t} + (k + \frac{1}{2}) \int\_{R\_{z}} (\prescript{\tau}{}{\xi} - z) u\_{i} u\_{i} \mathrm{d}x \Big|\_{\eta=t} \\ &+ \frac{1}{2} \int\_{0}^{t} \int\_{R\_{z}} (\prescript{\tau}{}{\xi} - z) u\_{i} u\_{i} \mathrm{d}x d\eta + \frac{1}{2} \int\_{R\_{z}} (\prescript{\tau}{}{\xi} - z) \left[\rho\_{1} T^{2} + \rho\_{2} \mathsf{C}^{2} \right] d\textbf{x} \Big|\_{\eta=t} \\ &\leq \Psi(0, t) e^{-\ell \varrho z}. \end{split} \tag{60}$$

**Remark 1.** *The result of Theorem 1 belongs to the study of Saint-Venant principle, which shows that the fluid decays exponentially with spatial variables on the cylinder.*

**Remark 2.** *Theorem 1 shows that the solutions of Equations* (1)*–*(8) *decays exponentially as z* → ∞*. To make the decay bound explicit, we have to derive the bounds for* Ψ(0, *t*) *and* max*<sup>t</sup> <sup>R</sup> <sup>C</sup>*2*dx in next section.*

#### **5. Bounds of Ψ(0,***t***) and max***<sup>t</sup> <sup>R</sup> <sup>C</sup>***2***dx*

From the previous section, we can see that *a*<sup>3</sup> involves the quantities max*<sup>t</sup> <sup>R</sup> <sup>C</sup>*2*dx*. To make our main result explicit, we have to derive bounds of Ψ(0, *t*) and max*<sup>t</sup> <sup>R</sup> <sup>C</sup>*2*dx* in term of the physical parameters *σ*, *ν*, *gi*, *hi*, the boundary data and so on. To do this, we begin with

$$\int\_{0}^{t} \int\_{R} T\_{\cdot} T\_{\cdot \cdot} d\mathbf{x} d\eta = -\int\_{0}^{t} \int\_{D} \mathbf{F} T\_{\beta} dA d\eta - \int\_{0}^{t} \int\_{R} \mathbf{T} \Delta \mathbf{T} d\mathbf{x} d\eta. \tag{61}$$

Now we assume that *S* is a sufficiently smooth function satisfying the same initial and boundary conditions as *T*. Thus,

$$\begin{split} \int\_{0}^{t} \int\_{R} T\_{j} T\_{j} \mathrm{d}x d\eta &= -\int\_{0}^{t} \int\_{D} S T\_{j} \mathrm{d}A d\eta - \int\_{0}^{t} \int\_{R} T \Delta T \mathrm{d}x d\eta \\ &= \int\_{0}^{t} \int\_{R} S\_{j} T\_{j} \mathrm{d}x d\eta - \int\_{0}^{t} \int\_{R} (T - S) \Delta T \mathrm{d}x d\eta \\ &= \int\_{0}^{t} \int\_{R} S\_{j} T\_{j} \mathrm{d}x d\eta - \int\_{0}^{t} \int\_{R} (T - S) (T\_{\eta} + u\_{i} T\_{i}) \mathrm{d}x d\eta \\ &= \int\_{0}^{t} \int\_{R} S\_{j} T\_{j} \mathrm{d}x d\eta - \int\_{R} T^{2} \mathrm{d}x \Big|\_{\eta = t} + \int\_{R} T \mathrm{S} dx \Big|\_{\eta = t} \\ &- \int\_{0}^{t} \int\_{R} S\_{\eta} T \mathrm{d}x d\eta - \int\_{0}^{t} \int\_{R} S\_{j} T \mathrm{u}\_{i} \mathrm{d}x d\eta - \frac{1}{2} \int\_{0}^{t} \int\_{D} f \mathrm{F}^{2} dA d\eta. \end{split} \tag{62}$$

Using the Schwarz and the arithmetic-geometric mean inequalities, we can obtain

$$\begin{split} \int\_{R} T^{2} \mathrm{d}x \Big|\_{\eta=t} &+ \int\_{0}^{t} \int\_{R} T\_{j} T\_{i} \mathrm{d}x d\eta \le \frac{1}{2} \int\_{R} S^{2} \mathrm{d}x \Big|\_{\eta=t} - \frac{1}{2} \int\_{0}^{t} \int\_{D} f \mathrm{F}^{2} \mathrm{d}A d\eta \\ &+ \left( \frac{\mathfrak{c}\_{1}}{2} \int\_{0}^{t} \int\_{R} T\_{j} T\_{j} \mathrm{d}x d\eta + \frac{1}{2\mathfrak{c}\_{1}} \int\_{0}^{t} \int\_{R} S\_{j} S\_{j} \mathrm{d}x d\eta \right) \\ &+ \left( \frac{\mathfrak{c}\_{2}}{2\lambda\_{1}} \int\_{0}^{t} \int\_{R} T\_{j} T\_{i} \mathrm{d}x d\eta + \frac{1}{2\mathfrak{c}\_{2}\lambda\_{1}} \int\_{0}^{t} \int\_{R} S\_{j\eta} S\_{\eta} \mathrm{d}x d\eta \right) \\ &+ \left( \frac{\mathfrak{c}\_{3} T\_{\mathrm{M}}}{2} \int\_{0}^{t} \int\_{R} u\_{i} u\_{i} \mathrm{d}x d\eta + \frac{T\_{\mathrm{M}}}{2\mathfrak{c}\_{3}} \int\_{0}^{t} \int\_{R} S\_{j} S\_{j} \mathrm{d}x d\eta \right). \end{split} \tag{63}$$

where 1, 2, <sup>3</sup> are positive constants. Choosing

$$
\epsilon\_1 = \frac{1}{2}, \quad \epsilon\_2 = \frac{\lambda\_1}{2}, \tag{64}
$$

we can obtain

$$\int\_{R} T^{2} \mathrm{d}x \Big|\_{\eta=t} + \int\_{0}^{t} \int\_{R} T\_{j} T\_{j} \mathrm{d}x d\eta \leq \frac{\mathfrak{e}\_{3} T\_{M}}{2} \int\_{0}^{t} \int\_{R} \mathbf{u}\_{i} \mathbf{u}\_{i} \mathrm{d}x d\eta + \mathrm{d}t \mathrm{d}x. \tag{65}$$

Obviously, the data terms in (65) involve <sup>1</sup> 2 *<sup>R</sup> <sup>S</sup>*2*dx η*=*t* , *t* 0 *<sup>R</sup> S*,*iS*,*idxdη*, *t* 0 *<sup>R</sup> S*,*ηS*,*ηdxdη* and <sup>−</sup><sup>1</sup> 2 *t* 0 *<sup>D</sup> f F*2*dAdη*. Similarly, we can bound *<sup>t</sup>* 0 *<sup>R</sup> C*,*iC*,*idxdη* as well as max*<sup>t</sup> <sup>R</sup> <sup>C</sup>*2*dx*. Firstly, we introduce a function *H*:

$$\begin{aligned} \frac{\partial H}{\partial t} + \mu\_i H\_j &= \Delta H, \quad \text{in } R \times \{t > 0\}, \\ H &= 0, \quad \text{in } R \times \{t = 0\}, \\ H &= 0, \quad \text{on } \partial D \times \{\mathbf{x} \mathbf{y} > 0\} \times \{t \ge 0\}, \\ H &= G(\mathbf{x}\_1, \mathbf{x}\_2, t), \quad \text{on } D \times \{t > 0\}, \end{aligned} \tag{66}$$

Then we have

$$\begin{aligned} (\mathsf{C} - H)\_{,t} + \mu\_{i}(\mathsf{C} - H)\_{,i} &= \Delta(\mathsf{C} - H) + \sigma \Delta T, \quad \text{in } \mathsf{R} \times \{t > 0\}, \\ \mathsf{C} - H &= 0 & \text{in } \mathsf{R} \times \{t = 0\}, \\ \mathsf{C} - H &= 0 & \text{on } \partial D \times \{\mathsf{x}\_{3} > 0\} \times \{t \ge 0\}, \\ \mathsf{C} - H &= 0 & \text{on } D \times \{t > 0\}. \end{aligned} \tag{67}$$

By the triangle inequality, we obtain that

$$\left(\int\_{0}^{t} \int\_{R} \mathbb{C}\_{\boldsymbol{\beta}} \mathbb{C}\_{\boldsymbol{\beta}} d\mathbf{x} d\boldsymbol{\eta}\right)^{\frac{1}{2}} \leq \left[\int\_{0}^{t} \int\_{R} (\mathbb{C} - H)\_{\boldsymbol{\beta}} (\mathbb{C} - H)\_{\boldsymbol{\beta}} d\mathbf{x} d\boldsymbol{\eta}\right]^{\frac{1}{2}} + \left[\int\_{0}^{t} \int\_{R} H\_{\boldsymbol{\beta}} H\_{\boldsymbol{\beta}} d\mathbf{x} d\boldsymbol{\eta}\right]^{\frac{1}{2}},\tag{68}$$

and

$$\mathbb{E}\left[\max\_{t}\int\_{\mathcal{R}}\mathbb{C}^{2}d\mathbf{x}\right]^{\frac{1}{2}}\leq\left[\max\_{t}\int\_{\mathcal{R}}(\mathbb{C}-H)^{2}d\mathbf{x}\right]^{\frac{1}{2}}+\left[\max\_{t}\int\_{\mathcal{R}}H^{2}d\mathbf{x}\right]^{\frac{1}{2}}.\tag{69}$$

Then,

$$\frac{1}{2}\int\_{R}(\mathbb{C}-H)^{2}dx\Big|\_{\eta=t} + \int\_{0}^{t}\int\_{R}(\mathbb{C}-H)\_{j}(\mathbb{C}-H)\_{j}d\mathbf{x}d\eta = -\sigma\int\_{0}^{t}\int\_{R}(\mathbb{C}-H)\_{j}T\_{j}d\mathbf{x}d\eta,\tag{70}$$

which follows that

$$\begin{split} \frac{1}{2} \int\_{R} (\mathbb{C} - H)^{2} d\mathbf{x} \Big|\_{\eta = t} &+ \int\_{0}^{t} \int\_{R} (\mathbb{C} - H)\_{,i} (\mathbb{C} - H)\_{,j} d\mathbf{x} d\eta \\ &\leq \sigma^{2} \int\_{0}^{t} \int\_{R} T\_{j} T\_{j} d\mathbf{x} d\eta \\ &\leq \frac{\mathfrak{c}\_{3} T\_{M} \sigma^{2}}{2} \int\_{0}^{t} \int\_{R} u\_{i} u\_{i} d\mathbf{x} d\eta + d\mathbf{a} d\mathbf{a}. \end{split} \tag{71}$$

Just as in the computation for *T*, we have the following inequality

$$\frac{1}{2}\int\_{R}H^{2}dx|\_{\eta=t} + \int\_{0}^{t}\int\_{R}H\_{\cdot}H\_{\cdot}d\mathbf{x}d\eta \le \epsilon\_{4}\int\_{0}^{t}\int\_{R}\mathbf{u}\_{\mathrm{i}}\mathbf{u}\_{\mathrm{i}}d\mathbf{x}d\eta + \text{data.}\tag{72}$$

Thus,

$$\frac{1}{2} \int\_{R} \mathbb{C}^{2} d\mathbf{x}|\_{\eta=t} + \int\_{0}^{t} \int\_{R} \mathbb{C}\_{,i} \mathbb{C}\_{,i} d\mathbf{x} d\eta \le \mathfrak{e}\_{5} \int\_{0}^{t} \int\_{R} \mu\_{i} \mu\_{i} d\mathbf{x} d\eta + data,\tag{73}$$

where <sup>5</sup> > 0 depends on 3, <sup>4</sup> and *<sup>σ</sup>*. Next we have to derive a bound for *<sup>t</sup>* 0 *<sup>R</sup> uiuidxdη* in term of data. To do this, we define a function

$$
\varpi\_i = f\_i e^{-\varsigma\_1 z},
\tag{74}
$$

for some positive constant  *<sup>i</sup>*. Then,

$$\begin{split} \mathbb{E}\left[\int\_{0}^{t} \int\_{R} \mathbf{u}\_{i,j} \mathbf{u}\_{i,j} d\mathbf{x} d\eta \right]^{\frac{1}{2}} &\leq \left[\int\_{0}^{t} \int\_{R} (\mathbf{u}\_{i} - \boldsymbol{\mathcal{O}}\_{i})\_{,j} (\boldsymbol{\mu}\_{i} - \boldsymbol{\mathcal{O}}\_{i})\_{,j} d\mathbf{x} d\eta \right]^{\frac{1}{2}} \\ &+ \left[\int\_{0}^{t} \int\_{R} \boldsymbol{\mathcal{O}}\_{i,j} \boldsymbol{\mathcal{O}}\_{i,j} d\mathbf{x} d\eta \right]^{\frac{1}{2}}. \end{split}$$

Obviously, we find that the last term of (75) is a data term. Now

$$\begin{split} & \nu \int\_{0}^{t} \int\_{R} (\mathbf{u}\_{i} - \boldsymbol{\mathcal{ox}}\_{i})\_{,j} (\mathbf{u}\_{i} - \boldsymbol{\mathcal{ox}}\_{i})\_{,j} d\mathbf{x} d\boldsymbol{\eta} \\ & \qquad = -\int\_{0}^{t} \int\_{R} (\mathbf{u}\_{i} - \boldsymbol{\mathcal{ox}}\_{i})\_{,j} \Big[ (\mathbf{u}\_{i} - \boldsymbol{\mathcal{ox}}\_{i}) + p\_{,i} - g\_{i}T - h\_{i}\mathbf{C} + \boldsymbol{\mathcal{ox}}\_{i} - \nu \Delta \boldsymbol{\mathcal{ox}}\_{i} \Big] d\mathbf{x} d\boldsymbol{\eta} \end{split} \tag{75}$$

or

$$\begin{split} & \frac{\nu}{2} \int\_{0}^{t} \int\_{R} (\boldsymbol{u}\_{i} - \boldsymbol{\alpha}\_{i})\_{,j} (\boldsymbol{u}\_{i} - \boldsymbol{\alpha}\_{i})\_{,j} d\mathbf{x} d\boldsymbol{\eta} \\ & \leq - \int\_{0}^{t} \int\_{R} \boldsymbol{p} \boldsymbol{\alpha}\_{i,i} d\mathbf{x} d\boldsymbol{\eta} + \frac{\delta\_{1}^{2}}{2} \int\_{0}^{t} \int\_{R} \boldsymbol{T}^{2} d\mathbf{x} d\boldsymbol{\eta} + \frac{\delta\_{2}^{2}}{2} \int\_{0}^{t} \int\_{R} \boldsymbol{\mathcal{C}}^{2} d\mathbf{x} d\boldsymbol{\eta} + d\boldsymbol{\eta} \mathbf{a}. \end{split} \tag{76}$$

Noting that

$$
\omega\_{i,i} = (f\_{a,a} - \varsigma\_1 f\_3) \varepsilon^{-\varsigma\_1 z} = 0,\tag{77}
$$

in *<sup>R</sup>* for *<sup>ς</sup>*<sup>1</sup> = *<sup>f</sup>α*,*<sup>α</sup> <sup>f</sup>*<sup>3</sup> , we can rewrite (76) as

$$\begin{split} \frac{1}{2} & \frac{1}{2} \int\_{0}^{t} \int\_{R} (\mathbf{u}\_{i} - \boldsymbol{\alpha}\_{i})\_{,j} (\mathbf{u}\_{i} - \boldsymbol{\alpha}\_{i})\_{,j} d\mathbf{x} d\boldsymbol{\eta} \\ & \leq \frac{\delta\_{1}^{2}}{2\lambda\_{1}} \int\_{0}^{t} \int\_{R} T\_{j} T\_{j} d\mathbf{x} d\boldsymbol{\eta} + \frac{\delta\_{2}^{2}}{2\lambda\_{1}} \int\_{0}^{t} \int\_{R} \mathbb{C}\_{,i} \mathbb{C}\_{,i} d\mathbf{x} d\boldsymbol{\eta} + d\boldsymbol{\eta} \mathbf{a}. \end{split} \tag{78}$$

Inserting (78) back into (75), we may have a bound of the form

$$\int\_{0}^{t} \int\_{R} \mathbf{u}\_{i,j} \mathbf{u}\_{i,j} d\mathbf{x} d\eta \le \mathbf{C}\_{1} \int\_{0}^{t} \int\_{R} \mathbf{T}\_{,i} \mathbf{T}\_{,i} d\mathbf{x} d\eta + \mathbf{C}\_{2} \int\_{0}^{t} \int\_{R} \mathbf{C}\_{,i} \mathbf{C}\_{,i} d\mathbf{x} d\eta + data,\tag{79}$$

for computable *C*<sup>1</sup> and *C*2. Combining (65) and (73) and by inequality (17), we have

$$\int\_{0}^{t} \int\_{R} u\_{i,j} u\_{i,j} d\mathbf{x} d\eta \le \frac{\mathbb{C}\_{1} T\_{M}}{2\lambda\_{1}} \mathfrak{c}\_{3} \int\_{0}^{t} \int\_{R} u\_{i,j} u\_{i,j} d\mathbf{x} d\eta + \frac{\mathbb{C}\_{2}}{\lambda\_{1}} \mathfrak{c}\_{5} \int\_{0}^{t} \int\_{R} u\_{i,j} u\_{i,j} d\mathbf{x} d\eta + d\mathbf{a}. \tag{80}$$

It is clear to see that

$$\int\_{0}^{t} \int\_{R} u\_{i,j} u\_{i,j} dx d\eta \le data,\tag{81}$$

for <sup>3</sup> = *<sup>λ</sup>*<sup>1</sup> <sup>2</sup>*C*1*TM* , <sup>5</sup> <sup>=</sup> *<sup>λ</sup>*<sup>1</sup> 4*C*<sup>2</sup> . From (65) and (73), we can obtain

$$\max\_{t} \int\_{R} T^{2} dx \le data, \quad \max\_{t} \int\_{R} \mathbb{C}^{2} dx \le data,\tag{82}$$

and

$$\int\_{0}^{t} \int\_{R} T\_{,i} T\_{,i} d\mathbf{x} d\eta \le data, \quad \int\_{0}^{t} \int\_{R} \mathbb{C}\_{,i} \mathbb{C}\_{,i} d\mathbf{x} d\eta \le data. \tag{83}$$

Next we seek bound for the total energy Ψ(0, *t*). From (54) we can obtain for Ψ(0, *t*)

$$\begin{split} \Psi(0, t) &\leq a\_4 \int\_0^t \int\_{\mathcal{R}} u\_{i, \eta} u\_{i, \eta} dx d\eta + \frac{\nu}{2} \int\_0^t \int\_{\mathcal{R}} u\_i u\_i dx d\eta \\ &+ b\_1 \int\_0^t \int\_{\mathcal{D}} u\_{a, 3} u\_{a, 3} dA d\eta + data. \end{split} \tag{84}$$

We are left to derive bounds for *<sup>t</sup>* 0 *Rz ui*,*<sup>η</sup>ui*,*ηdxd<sup>η</sup>* and *<sup>t</sup>* 0 *<sup>D</sup> uα*,3*uα*,3*dAdη*. Multiplying (1) with *ui*,*<sup>η</sup>* and integrating in the region *R* × [0, *t*], we have

$$\int\_{0}^{t} \int\_{R} u\_{i,\eta} u\_{i,\eta} \mathrm{d}x d\eta = \int\_{0}^{t} \int\_{R} u\_{i,\eta} \left[ \nu \Delta u\_{i} - u\_{i} - p\_{,i} + \mathrm{g}\_{i}T + h\_{i} \mathrm{C} \right] \mathrm{d}x d\eta,\tag{85}$$

which follows that

$$\begin{split} \int\_{0}^{t} \int\_{R} u\_{i\eta} u\_{i,\eta} \mathrm{d}x d\eta &\leq -2\nu \int\_{0}^{t} \int\_{D} u\_{a,3} u\_{a,\eta} \mathrm{d}A d\eta + 2 \int\_{0}^{t} \int\_{D} u\_{3,\eta} \mathrm{p} dA d\eta \\ &+ \frac{\delta\_{1}^{2}}{2\lambda\_{1}} \int\_{0}^{t} \int\_{R} T\_{i} T\_{j} \mathrm{d}x d\eta + \frac{\delta\_{1}^{2}}{2\lambda\_{1}} \int\_{0}^{t} \int\_{R} \mathbb{C}\_{,i} \mathbb{C}\_{,i} \mathrm{d}x d\eta + \mathrm{d}data \\ &\leq -2\nu \int\_{0}^{t} \int\_{D} u\_{a,3} f\_{a,\eta} d\mathrm{d}A d\eta + 2 \int\_{0}^{t} \int\_{D} f\_{3,\eta} \mathrm{p} dA d\eta + \mathrm{d}data, \end{split} \tag{86}$$

where we have used the fact *u*3,3 = −*uα*,*<sup>α</sup>* = −*fα*,*<sup>α</sup>* on *D*<sup>0</sup> and (83), and *ε*<sup>6</sup> is a positive constants. For the first term of (86), using the Schwarz and the AG mean inequalities we have

$$\begin{split} -2\nu \int\_{0}^{t} \int\_{D\_{0}} f\_{a,\eta} u\_{a,\Im} d\mathbf{x} d\eta &\leq 2\nu \left( \int\_{0}^{t} \int\_{D\_{0}} u\_{a,2} u\_{a,3} d\boldsymbol{A} d\eta \right)^{\frac{1}{2}} \left( \int\_{0}^{t} \int\_{D\_{0}} f\_{a,\eta} f\_{a,\eta} d\boldsymbol{A} d\eta \right)^{\frac{1}{2}} \\ &\leq \int\_{0}^{t} \int\_{D\_{0}} u\_{a,3} u\_{a,3} d\boldsymbol{A} d\eta + \text{data}. \end{split} \tag{87}$$

To bound the second term on the right of (86), we define *p* to be the mean value of *p* over *D*0, i.e.,

$$\overline{p} = \frac{1}{|D\_0|} \int\_{D\_0} p dA\_\prime \tag{88}$$

where |*D*0| is the measure of *D*0. Since

$$\int\_{D\_{\mathbb{D}}} f\_{\mathfrak{J},\eta} \overline{p} dA = \overline{p} \int\_{D\_{\mathbb{D}}} f\_{\mathfrak{J},\eta} dA = 0,\tag{89}$$

we obtain

$$\int\_{D\_0} f\_{3,\eta} p dA = \int\_{D\_0} f\_{3,\eta} (p - \overline{p}) dA. \tag{90}$$

It follows by using Schwarz inequality that

$$\int\_{0}^{t} \int\_{D\_{0}} f\_{\Im,\eta} p dx d\eta = \int\_{0}^{t} \int\_{D\_{0}} f\_{\Im,\eta} (p - \overline{p}) dA d\eta \leq data + \epsilon\_{6} \int\_{0}^{t} \int\_{D\_{0}} (p - \overline{p})^{2} dA d\eta,\tag{91}$$

where <sup>6</sup> is a positive constant to be determined later.

To deal with the integral *<sup>t</sup>* 0 *D*<sup>0</sup> (*<sup>p</sup>* <sup>−</sup> *<sup>p</sup>*)2*dAdη*, we let an auxiliary function *<sup>χ</sup>* satisfying:

$$
\Delta \chi = 0, \quad \frac{\partial \chi}{\partial n} = 0 \quad \text{on } \partial D\_0, \quad \frac{\partial \chi}{\partial n} = p - \overline{p}, \text{ in } D\_0. \tag{92}
$$

From the definition of *p* in (88), it is clear that *D*<sup>0</sup> (*p* − *p*)*dA* = 0. Thus, the necessary condition for the existence of a solution is satisfied and we compute

$$\begin{split} \int\_{0}^{t} \int\_{D\_{0}} (p - \overline{p})^{2} dA d\eta &= \int\_{0}^{t} \int\_{\partial \mathbb{R}} (p - \overline{p}) \frac{\partial \chi}{\partial \eta} dx d\eta = \int\_{0}^{t} \int\_{\mathbb{R}} (p - \overline{p})\_{,i} \chi\_{,i} dx d\eta \\ &= \int\_{0}^{t} \int\_{\mathbb{R}} \chi\_{,i} \left[ -u\_{i,\eta} + \nu u\_{i,\overline{\eta}} - u\_{i} + \operatorname{g} T + h\_{i} \mathbb{C} \right] dx d\eta. \end{split} \tag{93}$$

Since

$$\begin{split} \nu \, \int\_{0}^{t} \int\_{R} \chi\_{i} \boldsymbol{u}\_{i, \mathrm{j}\bar{j}} \mathrm{d}\mathbf{x} d\eta &= -\nu \int\_{0}^{t} \int\_{D} \chi\_{i} \boldsymbol{u}\_{i, \mathrm{j}\bar{j}} \mathrm{d}\mathbf{x} d\eta - \nu \int\_{0}^{t} \int\_{R} \chi\_{i, \mathrm{j}\bar{j}} \mathrm{d}\mathbf{x} d\eta \\ &= \nu \int\_{0}^{t} \int\_{D} \chi\_{3} f\_{\mathbf{a}, \mathrm{d}\mathbf{x}} \mathrm{d}\mathbf{x} d\eta - \nu \int\_{0}^{t} \int\_{D} \chi\_{\mathbf{a}} \mathrm{u}\_{\mathbf{a}, \mathrm{2}} \mathrm{d}\mathbf{x} d\eta \\ &+ \nu \int\_{0}^{t} \int\_{D} \chi\_{i} \boldsymbol{u}\_{3, \mathrm{j}} \mathrm{d}\mathbf{x} d\eta + \nu \int\_{0}^{t} \int\_{R} \chi\_{i} \boldsymbol{u}\_{i, \mathrm{j}\bar{j}} \mathrm{d}\mathbf{x} d\eta \\ &= -\nu \int\_{0}^{t} \int\_{D} \chi\_{\mathrm{a}} \mathrm{u}\_{\mathbf{a}, \mathrm{3}} \mathrm{d}\mathbf{x} d\eta + \nu \int\_{0}^{t} \int\_{D} \chi\_{\mathrm{a}} f\_{\mathbf{3}, \mathrm{a}} \mathrm{d}\mathbf{x} d\eta. \end{split} \tag{94}$$

From (93), we can obtain

 *t* 0 *D*<sup>0</sup> (*<sup>p</sup>* <sup>−</sup> *<sup>p</sup>*)2*dAd<sup>η</sup>* <sup>=</sup> *t* 0 *R χ*,*i* # − *ui*,*<sup>η</sup>* + *νui*,*jj* − *ui* + *giT* + *hiC* \$ *dxdη* ≤ *<sup>t</sup>* 0 *R χ*,*iχ*,*idxdη* 1 <sup>2</sup> *<sup>t</sup>* 0 *R ui*,*<sup>η</sup>ui*,*ηdxdη* 1 2 + *ν <sup>t</sup>* 0 *D χ*,*αχ*,*αdAdη* 1 <sup>2</sup> *<sup>t</sup>* 0 *D u*3,*αu*3,*αdAdη* 1 2 + *ν <sup>t</sup>* 0 *D χ*,*αχ*,*αdAdη* 1 <sup>2</sup> *<sup>t</sup>* 0 *D f*3,*<sup>α</sup> f*3,*αdAdη* 1 2 + 1 <sup>√</sup>*λ*<sup>1</sup> *<sup>t</sup>* 0 *R χ*,*iχ*,*idxdη* 1 <sup>2</sup> *<sup>t</sup>* 0 *R ui*,*jui*,*jdxdη* 1 2 + *δ*1 √*λ*1 *<sup>t</sup>* 0 *R χ*,*iχ*,*idxdη* 1 <sup>2</sup> *<sup>t</sup>* 0 *R T*,*jT*,*jdxdη* 1 2 + *δ*2 √*λ*1 *<sup>t</sup>* 0 *R χ*,*iχ*,*idxdη* 1 <sup>2</sup> *<sup>t</sup>* 0 *R C*,*jC*,*jdxdη* 1 2 . (95)

Making use of (15), (16), (81) and (83) with *g* = *p* − *p*, we have

$$\begin{split} \left[\int\_{0}^{t} \int\_{D\_{0}} (p-\overline{p})^{2} dA d\eta \right]^{\frac{1}{2}} &\\ \leq \frac{1}{\sqrt{\mu}} \Big( \int\_{0}^{t} \int\_{\mathcal{R}} u\_{i,\eta} u\_{i,\eta} dx d\eta \Big)^{\frac{1}{2}} + \nu \Big( \int\_{0}^{t} \int\_{D\_{0}} u\_{a,3} u\_{a,3} dx d\eta \Big)^{\frac{1}{2}} \\ &+ \nu \Big( \int\_{0}^{t} \int\_{D\_{0}} f\_{3,\eta} f\_{3,a} dx d\eta \Big)^{\frac{1}{2}} + \frac{1}{\sqrt{\mu\lambda\_{1}}} \Big( \int\_{0}^{t} \int\_{\mathcal{R}} u\_{i,j} u\_{i,j} dx d\eta \Big)^{\frac{1}{2}} \\ &+ \frac{\delta\_{1}}{\sqrt{\mu\lambda\_{1}}} \Big( \int\_{0}^{t} \int\_{\mathcal{R}} T\_{i} T\_{j} dx d\eta \Big)^{\frac{1}{2}} + \frac{\delta\_{2}}{\sqrt{\mu\lambda\_{1}}} \Big( \int\_{0}^{t} \int\_{\mathcal{R}} \mathbb{C}\_{,j} \mathbb{C}\_{,j} dx d\eta \Big)^{\frac{1}{2}}, \end{split} \tag{96}$$

which follows that

$$\int\_{0}^{t} \int\_{D\_{0}} (p - \overline{p})^{2} dA d\eta \le data + c\_{3} \int\_{0}^{t} \int\_{D\_{0}} u\_{a,3} u\_{a,3} dx d\eta + c\_{4} \int\_{0}^{t} \int\_{R} u\_{i,\eta} u\_{i,\eta} dx d\eta. \tag{97}$$

Obviously, from (97) we must establish a bound for the term *<sup>t</sup>* 0 *<sup>D</sup>*<sup>0</sup> *<sup>u</sup>α*,3*uα*,3*dxdη*. To do this, we begin with the identity

$$\int\_{0}^{t} \int\_{\mathbb{R}} \boldsymbol{u}\_{i,3} \left[ \nu \boldsymbol{u}\_{i, \dot{\eta} \dot{\boldsymbol{\eta}}} - \boldsymbol{u}\_{i} - \boldsymbol{p}\_{,i} - \boldsymbol{u}\_{i, \eta} + \boldsymbol{g}\_{i} \boldsymbol{T} + \boldsymbol{h}\_{i} \mathbf{C} \right] d\mathbf{x} d\eta = 0. \tag{98}$$

Integrating (98) by parts, we can have

$$\begin{split} & -\nu \int\_{0}^{t} \int\_{D\_{0}} \mathbf{u}\_{i,3} \boldsymbol{\mu}\_{i,3} \mathbf{J} \mathbf{x} d\boldsymbol{\eta} + \nu \int\_{0}^{t} \int\_{R} \mathbf{u}\_{i,3} \mathbf{\boldsymbol{\mu}}\_{i,j} \mathbf{x} d\mathbf{x} d\boldsymbol{\eta} + \int\_{0}^{t} \int\_{R} \mathbf{u}\_{i,3} \boldsymbol{\mu}\_{i} \mathbf{\boldsymbol{\mu}} d\mathbf{x} d\boldsymbol{\eta} \\ & + \int\_{0}^{t} \int\_{R} \mathbf{u}\_{i,3} \mathbf{\boldsymbol{\mu}}\_{i} \mathbf{\boldsymbol{x}} d\mathbf{x} d\boldsymbol{\eta} + \int\_{0}^{t} \int\_{R} \mathbf{u}\_{i,3} \mathbf{\boldsymbol{u}}\_{i, \eta} d\mathbf{x} d\boldsymbol{\eta} + \int\_{0}^{t} \int\_{R} \mathbf{u}\_{i,3} \mathbf{g}\_{i} \mathbf{\boldsymbol{T}} d\mathbf{x} d\boldsymbol{\eta} \\ & + \int\_{0}^{t} \int\_{R} \mathbf{u}\_{i,3} \mathbf{\boldsymbol{\mu}}\_{i} \mathbf{\boldsymbol{C}} d\mathbf{x} d\boldsymbol{\eta} = \mathbf{0}, \end{split} \tag{99}$$

which follows that

$$\int\_{0}^{t} \int\_{D\_{0}} u\_{a,\mathfrak{z}} u\_{a,\mathfrak{z}} \mathrm{d}x d\eta \le \mathrm{data} + \int\_{0}^{t} \int\_{D\_{0}} u\_{a,\mathfrak{z}} \mathrm{p} dA d\eta + \mathfrak{e}\gamma \int\_{0}^{t} \int\_{R} u\_{i,\eta} u\_{i,\eta} \mathrm{d}x d\eta. \tag{100}$$

where <sup>7</sup> is a positive constant.

As the derivation of (91), for the term *<sup>t</sup>* 0 *<sup>D</sup>*<sup>0</sup> *<sup>u</sup>α*,3 *pdAd<sup>η</sup>* we can obtain

$$\int\_{0}^{t} \int\_{D\_{0}} u\_{\mathfrak{a},3} p dA d\eta \le data + \epsilon\_{8} \int\_{0}^{t} \int\_{D\_{0}} (p - \overline{p})^{2} dA d\eta,\tag{101}$$

where <sup>8</sup> is a positive constant.

Combining (97), (100) and (101), we have

$$
\int \left(1 - \mathfrak{c}\_8 \mathfrak{c}\_3\right) \int\_0^t \int\_{D\_0} (p - \overline{p})^2 dxd\eta \le data + \mathfrak{c}\_3 \mathfrak{c}\_7 \int\_0^t \int\_{\mathbb{R}} u\_{i,\eta} u\_{i,\eta} dxd\eta. \tag{102}
$$

Combing (86), (87), (91) and (100), we obtain

$$(1 - \mathfrak{e}\_7) \int\_0^t \int\_{\mathbb{R}} u\_{i,\eta} u\_{i,\eta} \mathrm{d}x d\eta \le data + (\mathfrak{e}\_6 + \mathfrak{e}\_7) \int\_0^t \int\_{D\_0} (p - \overline{p})^2 d\mathbf{x} d\eta. \tag{103}$$

Choosing *ε*<sup>7</sup> and <sup>8</sup> small enough such that 1 − 8*c*<sup>3</sup> > 0 and 1 − <sup>7</sup> > 0, from (102) and (103) we can obtain

$$\int\_{0}^{t} \int\_{R} \boldsymbol{u}\_{i,\eta} \boldsymbol{u}\_{i,\eta} \, d\mathbf{x} d\eta \le data,\tag{104}$$

and

$$\int\_{0}^{t} \int\_{D\_{0}} (p - \overline{p})^{2} dA d\eta \le data. \tag{105}$$

Inserting (101) back into (100), we obtain

$$\int\_{0}^{t} \int\_{D\_{0}} u\_{a,3} u\_{a,3} \mathrm{d}x d\eta \le \mathrm{d}x \mathrm{a} + \mathfrak{e}\_{8} \int\_{0}^{t} \int\_{D\_{0}} (p - \overline{p})^{2} \mathrm{d}A d\eta + \mathfrak{e}\_{7} \int\_{0}^{t} \int\_{R} u\_{i,\eta} u\_{i,\eta} \mathrm{d}x d\eta. \tag{106}$$

In light of (104) and (105), we have

$$\int\_{0}^{t} \int\_{D\_{0}} u\_{a,3} u\_{a,3} \mathrm{d}x d\eta \le data. \tag{107}$$

Recalling (84) and using (104) and (107), we obtain

$$\Psi(0, t) \le data,\tag{108}$$

which is to say that we have bounded the total energy.

#### **6. Conclusions**

In this paper, we consider the spatial decay bounds for the Brinkman equations in double-diffusive convection in a semi-infinite pipe. Using the results of this paper, we can continue to study the continuous dependence of the solution on the parameters in the system of equations. In addition, Using the results of this paper, we can continue to study the continuous dependence of the solution on the parameters in the system of equations. This research can refer to the method of [30,31]. In addition, if Equation (1) is replaced by a nonlinear problem (e.g., Forchheimer equations), it will be a more interesting topic.

**Author Contributions:** Conceptualization, and validation, Y.L.; formal analysis and investigation, X.C. and D.L. All authors have read and agreed to the published version of the manuscript.

**Funding:** This research was funded by Key projects of universities in Guangdong Province (NAT-URAL SCIENCE) (2019KZDXM042) and the Research team project of Guangzhou Huashang College(2021HSKT01).

**Institutional Review Board Statement:** Not applicable.

**Informed Consent Statement:** Not applicable.

**Data Availability Statement:** Not applicable.

**Acknowledgments:** The authors would like to deeply thank all the reviewers for their insightful and constructive comments.

**Conflicts of Interest:** The authors declare no conflict of interest.

## **References**

