Abstract

A defect-correction mixed finite element method (MFEM) for solving the stationary conduction-convection problems in two-dimension is given. In this method, we solve the nonlinear equations with an added artificial viscosity term on a grid and correct this solution on the same grid using a linearized defect-correction technique. The stability is given and the error analysis in 𝐿2 and 𝐻1-norm of 𝑢, 𝑇 and the 𝐿2-norm of 𝑝 are derived. The theory analysis shows that our method is stable and has a good precision. Some numerical results are also given, which show that the defect-correction MFEM is highly efficient for the stationary conduction-convection problems.

1. Introduction

In this paper, we consider the stationary conduction-convection problems in two dimension whose coupled equations governing viscous incompressible flow and heat transfer for the incompressible fluid are Boussinesq approximations to the stationary Navier-Stokes equations.

(𝒫) Find (𝑢,𝑝,𝑇)𝑋×𝑀×𝑊 such that𝜈Δ𝑢+(𝑢)𝑢+𝑝=𝜆𝑗𝑇,𝑥Ω,div𝑢=0,𝑥Ω,Δ𝑇+𝜆𝑢𝑇=0,𝑥Ω,𝑢=0,𝑇=𝑇0,𝑥𝜕Ω,(1.1) where Ω is a bounded domain in 2 assumed to have a Lipschitz continuous boundary 𝜕Ω. 𝑢=(𝑢1(𝑥),𝑢2(𝑥))𝑇 represents the velocity vector, 𝑝(𝑥) the pressure, 𝑇(𝑥) the temperature, 𝜆>0 the Grashoff number, 𝑗=(0,1)𝑇 the two-dimensional vector, and 𝜈>0 the viscosity.

As we know the conduction-convection problem contains the velocity vector field, the pressure field and the temperature field, so finding the numerical solution of conduction-convection problems is very difficult. The conduction-convection problems is an important system of equations in atmospheric dynamics and dissipative nonlinear system of equations, so lots of works are devoted to this problem [16]. There are also some works devoted to the nonstationary conduction-convection problems [710]. In [8], Luo et al. gave an optimizing reduced PLSMFE for the nonstationary conduction-convection problems. They combined PLSMEF method with POD to deal with the problems. In [11], an analysis of conduction natural convection conjugate heat transfer in the gap between concentric cylinders under solar irradiation was studied. In [12], a Newton iterative mixed finite element method for the stationary conduction-convection problems was shown by Si et al. In [13], Si and He gave a coupled Newton iterative mixed finite element method for the stationary conduction-convection problems.

The defect-correction method is an iterative improvement technique for increasing the accuracy of a numerical solution without applying a grid refinement. Due to its good efficiency, there are many works devoted to this method, for example, [1428]. In [18], a method making it possible to apply the idea of iterated defect correction to finite element methods was given. A method for solving the time-dependent Navier-Stokes equations, aiming at higher Reynolds' number, was presented in [23]. In [27], an accurate approximations for self-adjoint elliptic eigenvalues was presented. In [28], Stetter exposed the common structural principle of all these techniques and exhibit the principal modes of its implementation in a discretization context.

In this paper we present a defect-correction MFEM for the stationary conduction convection problems. In this method, we solve the nonlinear equations with an added artificial viscosity term on a finite element grid and correct this solution on the same grid using a linearized defect-correction technique. Actually, the defect-correction MFEM incorporates the artificial viscosity term as a stabilizing factor, making both the nonlinear system easier to resolve and the linearized system easier to precondition. The stability and error analysis of the coupled the defect-correction MFEM show that this method is stable and has a good precision. Some numerical experiments show that our analysis is proper and our method is effective. And it can be used for solving the convection-conduction problems with much small viscosity.

This paper is organized as follows. In Section 2, the functional settings and some assumptions are given. Section 3 is devoted to the defect-correction MFEM. Section 4 gives the stability analysis. Section 5 presents the error analysis. In Section 6, some numerical results and the numerical analysis to validate the effectiveness of the method are laid out.

2. Functional Setting for the Conduction Convection Problems

In this section, we aim to describe some of the notations and results which will be frequently used in this paper. The Sobolev spaces used in this context are standard [29]. For the mathematical setting of the conduction-convection problems and MFEM of conduction-convection problems (1.1), we introduce the Hilbert spaces 𝑋=𝐻10(Ω)2,𝑊=𝐻1(Ω),𝑀=𝐿20(Ω)𝜑𝐿2(Ω);Ω.𝜑𝑑𝑥=0(2.1) is the uniformly regular family of triangulation of Ω, indexed by a parameter =max𝐾{𝐾;𝐾=diam(𝐾)}. We introduce the finite element subspace 𝑋𝑋, 𝑀𝑀, 𝑊𝑊 as follows 𝑋=𝑣𝑋𝐶0Ω2;𝑣|𝐾𝑃(𝐾)2,𝐾,𝑀=𝑞𝑀𝐶0Ω;𝑞|𝐾𝑃𝑘(𝐾),𝐾,𝑊=𝜙𝑊𝐶0Ω;𝜙|𝐾𝑃𝑙(𝐾),𝐾,(2.2) where 𝑃(𝐾) is the space of piecewise polynomials of degree on 𝐾, and 1, 𝑘1, 𝑙1 are three integers. 𝑊0=𝑊𝐻10(Ω), and (𝑋,𝑀) satisfies the discrete LBB conditionsup𝑣𝑋𝑑𝜑,𝑣𝑣0𝜑𝛽0,𝜑𝑀,(2.3) where 𝑑(𝜑,𝑣)=(𝜑,div𝑣).

With the above notations, the Galerkin mixed variation and the mixed FEM problem for the conduction-convection problems (𝒫) are defined, respectively, as follows.

(𝒫) Find (𝑢,𝑝,𝑇)𝑋×𝑀×𝑊 such that𝜈𝑎(𝑢,𝑣)𝑑(𝑝,𝑣)+𝑑(𝜑,𝑢)+𝑏(𝑢,𝑢,𝑣)=𝜆(𝑗𝑇,𝑣),𝑣𝑋,𝜑𝑀,𝑎(𝑇,𝜓)+𝜆𝑏(𝑢,𝑇,𝜓)=0,𝜓𝑊0.(2.4)

(𝒫) Find (𝑢,𝑝,𝑇)𝑋×𝑀×𝑊 such that𝑢𝜈𝑎,𝑣𝑝𝑑,𝑣𝜑+𝑑,𝑢𝑢+𝑏,𝑢,𝑣=𝜆𝑗𝑇,𝑣,𝑣𝑋,𝜑𝑀,𝑎𝑇,𝜓+𝜆𝑏𝑢,𝑇,𝜓=0,𝜓𝑊0,(2.5) where 𝑎(𝑢,𝑣)=(𝑢,𝑣), 𝑑(𝜑,𝑣)=(𝜑,div𝑣), 𝑎(𝑇,𝜓)=(𝑇,𝜓), and 1𝑏(𝑢,𝑣,𝑤)=2Ω2𝑖,𝑘=1𝑢𝑖𝜕𝑣𝑘𝜕𝑥𝑖𝑤𝑘𝑑𝑥2𝑖,𝑘=1𝑢𝑖𝜕𝑤𝑘𝜕𝑥𝑖𝑣𝑘𝑑𝑥,𝑢,𝑣,𝑤𝑋,1𝑏(𝑢,𝑇,𝜓)=2Ω2𝑖=1𝑢𝑖𝜕𝑇𝜕𝑥𝑖𝜓𝑑𝑥2𝑖𝑢𝑖𝜕𝜓𝜕𝑥𝑖𝑇𝑑𝑥,𝑢𝑋,𝑇,𝜓𝑊.(2.6)

The following assumptions and results are recalled (see [7, 2931]).

(A1)There exists a constant 𝐶0 which only depends on Ω, such that(i)𝑢0𝐶0𝑢0, 𝑢0,4𝐶0𝑢0, for all 𝑢𝐻10(Ω)2(or𝐻10(Ω)),(ii)𝑢0,4𝐶0𝑢1,for all 𝑢𝐻1(Ω)2,(iii)𝑢0,42𝑢01/2𝑢01/2,for all𝑢𝐻10(Ω)2(or𝐻10(Ω)).(A2)Assuming 𝜕Ω𝐶𝑘,𝛼(𝑘0,𝛼>0), then, for 𝑇0𝐶𝑘,𝛼(𝜕Ω), there exists an extension in 𝐶0𝑘,𝛼(2) (denote 𝑇0 also), such that 𝑇0𝑘,𝑞𝜀,𝑘0,1𝑞,(2.7) where 𝜀 is an arbitrary positive constant.(A3)𝑏(,,) and 𝑏(,,) have the following properties.(i)For all 𝑢𝑋, 𝑣,𝑤𝑋(or𝑇,𝜓𝐻10(Ω)), there holds that 𝑏(𝑢,𝑣,𝑣)=0,𝑏(𝑢,𝑣,𝑤)=𝑏(𝑢,𝑤,𝑣),(2.8)𝑏(𝑢,𝑇,𝑇)=0,𝑏(𝑢,𝑇,𝜓)=𝑏(𝑢,𝜓,𝑇).(2.9)(ii)For all𝑢𝑋, 𝑣𝐻1(Ω)2(or𝑇𝐻1(Ω)), for all𝑤𝑋(or𝜓𝐻10(Ω)), there holds that ||||𝑏(𝑢,𝑣,𝑤)𝑁𝑢0𝑣0𝑤0|||,(2.10)|||𝑏(𝑢,𝑇,𝜓)𝑁𝑢0𝑇0𝜓0,(2.11) where 𝑁=sup𝑢,𝑣,𝑤||||𝑏(𝑢,𝑣,𝑤)𝑢0𝑣0𝑤0,𝑁=sup𝑢,𝑇,𝜑||||||𝑏(𝑢,𝑇,𝜑)𝑢0𝑇0𝜑0.(2.12)

We recall the following existence, uniqueness and regularity result of (𝒫) (see [7, Chapter 4]).

Theorem 2.1 (see [7]). Under the assumption of (A1)~(A3), letting 𝐴2𝜈1𝜆(3𝐶0+1)𝑇01, 𝐵2𝑇00+2(𝐶20𝜆)1𝐴, there exist 0<𝛿1, 𝛿21 such that 𝜈1𝑁𝐴1𝛿1,𝛿11𝜈1𝐶20𝜆2𝐵𝑁1𝛿2.(2.13) Then, there exists a unique solution (𝑢,𝑝,𝑇)𝑋×𝑀×𝑊 for (𝒫), and 𝑢0𝐴,𝑇0𝐵.(2.14)

Some estimates of the trilinear form 𝑏 are given in the following lemma and the proof can be found in [30, 3234].

Lemma 2.2. The trilinear form 𝑏 satisfies the following estimate: ||𝑏𝑢,𝑣||+||𝑏𝑣,𝑤,𝑢||+||𝑏,𝑤𝑤,𝑢,𝑣||𝐶0||||log1/2𝑣0𝑢0𝑤0,(2.15) for all 𝑢,𝑣𝑉, 𝑤𝐿2(Ω)2.

Lemma 2.3. Suppose that (A1)~(A3) are valid and 𝜀 is a positive constant, such that 32𝐶20𝜆2𝑁𝜀3𝜈<1,𝑇00𝜀4,𝑇00𝐶0𝜀4,(2.16) then (𝒫) has a unique solution (𝑢,𝑝,𝑇)𝑋×𝑀×𝑊, such that 𝑇|𝜕Ω=𝑇0 and 𝑢05𝐶20𝜆𝜀,3𝜈𝑇0𝜀.(2.17)

Proof. The proof of the existence and the uniqueness of the solution has been given by Luo [7]. Let 𝑇=𝜔+𝑇0, 𝜓=𝜔 in (2.5), we can get 𝑎𝜔,𝜔=𝜆𝑏𝑢,𝑇0,𝜔𝑎𝑇0,𝜔.(2.18) Using (2.11) and (2.16), we deduce 𝜔0𝑇00+𝜆𝑁𝜀𝑢0.(2.19) Letting 𝑣=𝑢, 𝜑=𝑝 in the first equation of (2.5), we get 𝜈𝑢20=||𝜆𝑗𝑇,𝑢||𝜆𝐶0𝑇0𝑢0.(2.20) By (2.16), we can obtian 𝑢0𝜈1𝜆𝐶0𝑇0𝜈1𝜆𝐶0𝜔0+𝑇00𝜈1𝜆𝐶20𝜔0+𝜈1𝜆𝐶0𝑇00𝜈1𝜆𝐶0𝑇00+𝜈1𝜆𝐶20𝑇00+𝜈1𝜆2𝐶20𝑁𝜀𝑢0.(2.21) Using (2.16) again, we get 𝑢05𝐶20𝜆𝜀.3𝜈(2.22) By (2.19), we deduce 𝑇0𝜔0+𝑇002𝑇00+𝜆𝑁𝜀𝑢02𝑇00+5𝐶20𝜆2𝑁𝜀2𝜀3𝜈2+𝜀2=𝜀.(2.23)

We introduce the Laplace operator𝒜𝑢=Δ𝑢,𝑢𝐷(𝒜)=𝐻2(Ω)2𝑋.(2.24)

Lemma 2.4 (see [35, 36]). For all 𝑢,𝑤𝑋, 𝑣𝐷(𝐴) there holds that ||||+||||+||||𝑏(𝑢,𝑣,𝑤)𝑏(𝑣,𝑢,𝑤)𝑏(𝑤,𝑢,𝑣)𝐶𝒜𝑣0𝑤0𝑢0.(2.25)

3. The Defect-Correction Method

The aim of this section is to give a method for solving the nonlinear system (2.5) on a coarser mesh than one uses when employing the standard FEM; the coarse-mesh solution is corrected using the same grid in our method. The defect-correction method in which we consider incorporates an artificial viscosity parameter 𝜎 as a stabilizing factor in the solution algorithm. For a fixed grid parameter the method requires the solution of one nonlinear system and a few linear correction steps. It is described in the following paragraphs. We consider the following problems which is identical to (2.5) except for an artificial viscosity term.

(𝒫) Find (𝑢0,𝑝0,𝑇0)𝑋×𝑀×𝑊 such that𝑢(𝜈+𝜎)𝑎0,𝑣𝑝𝑑0,𝑣𝜑+𝑑,𝑢0𝑢+𝑏0,𝑢0,𝑣=𝜆𝑗𝑇0,𝑣,𝑣𝑋,𝜑𝑀,(1+𝜎)𝑎𝑇0,𝜓+𝜆𝑏𝑢0,𝑇0,𝜓=0,𝜓𝑊0.(3.1) We define the residual or named defect 𝑅(𝑢0,𝑝0,𝑇0), 𝑄(𝑢0,𝑝0,𝑇0) for the momentum systems as follows:𝑅𝑢0,𝑝0,𝑇0,𝑣=𝜆𝑗𝑇0,𝑣𝑢𝜈𝑎0,𝑣𝑝+𝑑0,𝑣𝜑𝑑,𝑢0𝑢𝑏0,𝑢0,𝑣,𝑄𝑢0,𝑝0,𝑇0,𝜓=𝑎𝑇0,𝜓𝜆𝑏𝑢0,𝑇0,𝜓.(3.2) Define the correction (𝜀0,𝜚0,𝜏0) satisfying the following linear problem:𝜀(𝜈+𝜎)𝑎0,𝑣𝜚𝑑0,𝑣𝜑+𝑑,𝜀0𝜀+𝑏0,𝑢0,𝑣𝑢+𝑏0,𝜀0,𝑣=𝑅𝑢0,𝑝0,𝑇0,𝑣,𝑣𝑋,𝜑𝑀,(1+𝜎)𝑎𝜏0,𝜓+𝜆𝑏𝑢0,𝜏0,𝜓+𝜆𝑏𝜀0,𝑇0,𝜓=𝑄𝑢0,𝑝0,𝑇0,𝜓,𝜓𝑊0.(3.3) Define 𝑢1=𝑢0+𝜀0, 𝑝1=𝑝0+𝜚0, 𝑇1=𝑇0+𝜏0, which are hoped to be better solutions of the problems. In order to obtain the equations for (𝑢1,𝑝1,𝑇1), we use the residual equation (3.2) to rewrite the linear problems (3.3); we obtain𝒫𝑢(𝜈+𝜎)𝑎1,𝑣𝑝𝑑1,𝑣𝜑+𝑑,𝑢1𝑢+𝑏0,𝑢1,𝑣𝑢+𝑏1,𝑢0,𝑣=𝜆𝑗𝑇1,𝑣𝑢+𝜎𝑎0,𝑣𝑢+𝑏0,𝑢0,𝑣,𝑣𝑋,𝜑𝑀,(1+𝜎)𝑎𝑇1,𝜓+𝜆𝑏𝑢1,𝑇0,𝜓+𝜆𝑏𝑢0,𝑇1,𝜓=𝜎𝑎𝑇0,𝜓+𝜆𝑏𝑢0,𝑇0,𝜓,𝜓𝑊0.(3.4) In general, this method can be described as follows.

Step 1. Solve the nonlinear systems (3.1) for (𝑢0,𝑝0,𝑇0).

Step 2. For 𝑖=1,2,,𝑚, solve the linear equations 𝒫𝑢(𝜈+𝜎)𝑎𝑖,𝑣𝑝𝑑𝑖,𝑣𝜑+𝑑,𝑢𝑖𝑢+𝑏𝑖1,𝑢𝑖,𝑣𝑢+𝑏𝑖,𝑢𝑖1,𝑣=𝑇𝑖,𝑣𝑢+𝜎𝑎𝑖1,𝑣𝑢+𝑏𝑖1,𝑢𝑖1,𝑣,𝑣𝑋,𝜑𝑀,(1+𝜎)𝑎𝑇𝑖,𝜓+𝜆𝑏𝑢𝑖,𝑇𝑖1,𝜓+𝜆𝑏𝑢𝑖1,𝑇𝑖,𝜓=𝜎𝑎𝑇𝑖1,𝜓+𝜆𝑏𝑢𝑖1,𝑇𝑖1,𝜓,𝜓𝑊0.(3.5) For each 𝑖 the residual is given by 𝑅𝑢𝑖,𝑝𝑖,𝑇𝑖,𝑣=𝜆𝑗𝑇𝑖,𝑣𝑢𝜈𝑎𝑖,𝑣𝑝+𝑑𝑖,𝑣𝜑𝑑,𝑢𝑖𝑢𝑏𝑖,𝑢𝑖,𝑣,𝑄𝑢𝑖,𝑝𝑖,𝑇𝑖,𝜓=𝑎𝑇𝑖,𝜓𝜆𝑏𝑢𝑖,𝑇𝑖,𝜓.(3.6) The correction (𝜀𝑖,𝜚𝑖,𝜏𝑖) is given by 𝜀(𝜈+𝜎)𝑎𝑖,𝑣𝜚𝑑𝑖,𝑣𝜑+𝑑,𝜀𝑖𝜀+𝑏𝑖,𝑢𝑖,𝑣𝑢+𝑏𝑖,𝜀𝑖,𝑣=𝑅𝑢𝑖,𝑝𝑖,𝑇𝑖,𝑣,𝑣𝑋,𝜑𝑀,(1+𝜎)𝑎𝜏𝑖,𝜓+𝜆𝑏𝑢𝑖,𝜏𝑖,𝜓+𝜆𝑏𝜀𝑖,𝑇𝑖,𝜓=𝑄𝑢𝑖,𝑝𝑖,𝑇𝑖,𝜓,𝜓𝑊0.(3.7)

Remark 3.1. From the numerical experiments, we see that one or two correction steps is adequate. And this is as same as [24].

4. Stability Analysis

In this section, we give the stability analysis. It is given by the following theorems.

Theorem 4.1. Under the assumptions of Lemma 2.3, then (𝑢0,𝑇0) defined by (𝒫) satisfies 𝑢005𝐶20𝜆𝜀,3(𝜈+𝜎)𝑇00𝜀.(4.1) Moreover, if 25𝐶20𝑁𝜆𝜀3(𝜈+𝜎)2<1,(4.2)(𝒫) admits a unique solution.

Proof. We define the set 𝑀=̃𝑣𝑋;̃𝑣05𝐶20𝜆𝜀3(𝜈+𝜎).(4.3)
Let ̃𝑢 be in 𝑀. Then(1+𝜎)𝑎𝑇0,𝜓+𝜆𝑏̃𝑢,𝑇0,𝜓=0,𝜓𝑊0(4.4) has a unique solution 𝑇0𝑊 such that 𝑇|𝜕Ω=𝑇0. For a given 𝑇0, we consider the following problem: 𝑢(𝜈+𝜎)𝑎0,𝑣𝑝𝑑0,𝑣𝜑+𝑑,𝑢0𝑢+𝑏0,𝑢0,𝑣=𝜆𝑗𝑇0,𝑣,𝑣𝑋,𝜑𝑀.(4.5) By the theory of the Navier-Stokes equations, we get (4.5) has a unique solution (𝑢0,𝑝0)𝑋×𝑀 (see [31]). It means that (4.4) and (4.5) give a unique 𝑢0𝑋 for a given ̃𝑢𝑋, we denote 𝑢0=̃𝑢.(4.6)
Setting 𝑇0=𝜔0+𝑇0, 𝜓=𝜔0 in (4.4) and using (2.9), we can obtain(1+𝜎)𝑎𝜔0,𝜔0=𝜆𝑏̃𝑢,𝑇0,𝜔0(1+𝜎)𝑎𝑇0,𝜔0.(4.7) Using (2.7), (2.11), and (2.16), we can get (1+𝜎)𝜔00𝜆𝑁̃𝑢0𝑇00+(1+𝜎)𝑇00,𝜔00𝜆𝑁𝜀4̃𝑢0+𝑇00.(4.8) Using the triangle inequality, we have 𝑇00𝑇00+𝜔00𝜆𝑁𝜀4̃𝑢0+2𝑇005𝐶20𝑁𝜆𝜀2+𝜀12(𝜈+𝜎)2𝜀.(4.9)
Letting 𝑣=𝑢0, 𝜑=𝑝0 in (4.5) and using (2.8), we get𝑢(𝜈+𝜎)𝑎0,𝑢0=𝜆𝑗𝑇0,𝑢0.(4.10) Letting 𝑇0=𝜔0+𝑇0 and using (2.9), we have (𝜈+𝜎)𝑢00𝐶20𝜆𝜔00+𝐶0𝜆𝑇00𝐶20𝜆2𝑁̃𝑢0𝑇00+𝐶0𝜆1+𝐶0𝑇00𝐶20𝜆𝜀.(4.11) Namely, 𝑢005𝐶20𝜆𝜀.3(𝜈+𝜎)(4.12) Hence, we proved that maps 𝑀 to 𝑀. It follows from Brouwer's fixed-point theorem that there exits a solution to system (𝒫).
To prove the uniqueness, assume that (𝑢01,𝑝01,𝑇01),(𝑢02,𝑝02,𝑇02)𝑋×𝑀×𝑊, and 𝑇01|𝜕Ω=𝑇02|𝜕Ω=𝑇0 are two solutions of (𝒫). Then, we obtain that𝑢(𝜈+𝜎)𝑎01𝑢02,𝑣𝑝𝑑01𝑝02,𝑣𝜑+𝑑,𝑢01𝑢02𝑢+𝑏01𝑢02,𝑢01,𝑣𝑢+𝑏02,𝑢01𝑢02,𝑣𝑗𝑇=𝜆01𝑇02,𝑣,𝑣𝑋,𝜑𝑀,(1+𝜎)𝑎𝑇01𝑇02,𝜓+𝜆𝑏𝑢02,𝑇01𝑇02,𝜓+𝜆𝑏𝑢01𝑢02,𝑇01,𝜓=0,𝜓𝑊0.(4.13) Let 𝑣=𝑢01𝑢02, 𝜑=𝑝01𝑝02 in the first equation of (4.13), we can get 𝑢(𝜈+𝜎)01𝑢020𝑢𝑁01𝑢020𝑢010+𝐶20𝜆𝑇01𝑇020.(4.14) Setting 𝜓=𝑇01𝑇02 in the second equation of (4.13), we obtain (𝑇1+𝜎)01𝑇020𝜆𝑁𝑢01𝑢020𝑇010.(4.15) By (4.14) and (4.15), we deduce 𝑢(𝜈+𝜎)01𝑢020𝑁𝑢010(𝑢01𝑢02)0+𝐶20𝜆2𝑁𝑇010(𝑢01𝑢02)05𝐶20𝑁𝜆𝜀3(𝜈+𝜎)(𝑢01𝑢02)0+𝐶20𝜆𝑁𝜀4𝜈(𝑢01𝑢02)0.(4.16) Using (4.2), we obtain 𝑢01𝑢02025𝑢01𝑢020.(4.17) Namely, 𝑢01𝑢020=0.(4.18) By (4.15), we see that (𝑇01𝑇02)0=0. Therefore, it follows that (𝒫) has a unique solution.
Then, we give the prove of (4.1) without using (4.2). Letting 𝑣=𝑢0, 𝜑=𝑝0 in the first equation of (3.1) and using (2.8), we get𝑢(𝜈+𝜎)𝑎0,𝑢0=𝜆𝑗𝑇0,𝑢0.(4.19) Letting 𝑇0=𝜔0+𝑇0, we have (𝜈+𝜎)𝑢00𝐶20𝜆𝜔00+𝐶0𝜆𝑇00.(4.20) Letting 𝑇0=𝜔0+𝑇0, 𝜓=𝜔0 in the second equation of (3.1) and using (2.9), we can obtain (1+𝜎)𝑎𝜔0,𝜔0=𝜆𝑏𝑢0,𝑇0,𝜔0(1+𝜎)𝑎𝑇0,𝜔0.(4.21) Using (2.7), (2.11), and (2.16), we can get (1+𝜎)𝜔00𝜆𝑁𝑢00𝑇00+(1+𝜎)𝑇00,𝜔00𝜆𝑁𝜀4𝑢00+𝑇00.(4.22) By (4.20) and (4.22), we can deduce 𝑢00(𝜈+𝜎)1𝜆𝐶20𝜔00+(𝜈+𝜎)1𝐶0𝜆𝑇00(𝜈+𝜎)1𝜆𝐶0𝑇00+𝐶20𝜆𝑇00+𝐶20𝜆2𝑁𝜀4𝑢00.(4.23) Using (2.16), we get 𝑢005𝐶20𝜆𝜀.3(𝜈+𝜎)(4.24) Using (2.7), (2.11), (2.16), and (4.20), we can get 𝜔00𝜆𝑁𝑢00𝑇00+𝑇00𝜆𝑁𝜀4𝑢00+𝑇003𝜀4,𝑇00𝜔00+𝑇00𝜀.(4.25) Therefore, we finish the proof.

Theorem 4.2. Under the assumptions of Lemma 2.3, and 25𝐶20𝑁𝜆𝜀3(𝜈+𝜎)2<1,(4.26)(𝑢1,𝑇1) defined by (3.4) satisfies 𝑢10𝛿,𝑇105𝜀6+𝜆𝑁𝛿𝜀+𝜎𝜀,(4.27) where 𝛿(103𝐶20𝜆𝜀/48+𝜎(5𝐶20𝜆𝜀/3𝜈))/(7/10)(𝜈+𝜎).

Proof. Letting 𝑣=𝑢1, 𝜑=𝑝1 in the first equation of (3.4) and using (2.8), we get 𝑢(𝜈+𝜎)𝑎1,𝑢1𝑢+𝑏1,𝑢0,𝑢1𝑢=𝑏0,𝑢0,𝑢1𝑢+𝜎𝑎0,𝑢1+𝜆𝑗𝑇0,𝑢1.(4.28) Letting 𝑇0=𝜔0+𝑇0 and using (2.10), we have (𝜈+𝜎)𝑢10𝑁𝑢10𝑢00+𝜎𝑢00+𝑁𝑢020+𝐶20𝜆𝜔10+𝐶0𝜆𝑇00.(4.29) Let 𝑇1=𝜔1+𝑇0, 𝜓=𝜔1 in the second equation of (3.4), we can obtain (1+𝜎)𝑎𝜔1,𝜔1=𝜆𝑏𝑢0,𝑇0,𝜔1𝜆𝑏𝑢1,𝑇0,𝜔1+𝜆𝑏𝑢0,𝑇0,𝜔1+𝜎𝑎𝑇0,𝜔1(1+𝜎)𝑎𝑇0,𝜔1.(4.30) Using (2.11) and (2.16), we can get (1+𝜎)𝜔10𝜆𝑁𝑢00𝑇00+𝜆𝑁𝑢10𝑇00+𝜆𝑁𝑢00𝑇00+𝜎𝑇00+(1+𝜎)𝑇00,(4.31)𝜔10𝜆𝑁𝑢00𝑇00+𝜆𝑁𝑢10𝑇00+𝜆𝑁𝑢00𝑇00+𝜎𝑇00+𝑇00.(4.32) Using (4.29), we get (𝜈+𝜎)𝑢10𝑁𝑢10𝑢00+𝜎𝑢00+𝑁𝑢020+𝐶20𝜆𝜆𝑁𝑢00𝑇00+𝜆𝑁𝑢10𝑇00+𝜆𝑁𝑢00𝑇00+𝑇00+𝑇00+𝐶0𝜆𝑇00.𝜈+𝜎𝑁𝑢00𝐶20𝜆2𝑁𝑇00𝑢10𝑁𝑢020+𝜎𝑢00+𝐶20𝜆2𝑁𝑢00𝑇00+𝐶20𝜆2𝑁𝑢00𝑇00+𝐶20𝜆𝑇00+𝐶20𝜆𝑇00+𝐶0𝜆𝑇00.(4.33) Using (2.16), (4.26), and Theorem 4.2, we can obtain 710(𝜈+𝜎)𝑢1025𝑁𝐶40𝜆2𝜀29(𝜈+𝜎)2+𝜎5𝐶20𝜆𝜀3+(𝜈+𝜎)10𝑁𝐶40𝜆3𝜀23+(𝜈+𝜎)3𝐶20𝜆2𝜀22103𝐶20𝜆𝜀48+𝜎5𝐶20𝜆𝜀.3(𝜈+𝜎)(4.34) Namely, 𝑢10103𝐶20𝜆𝜀/48+𝜎5𝐶20𝜆𝜀/3𝜈(7/10)(𝜈+𝜎)𝛿.(4.35)
Using (2.16), (4.31), and (4.35), we can get𝜔1010𝑁𝐶20𝜆2𝜀23(𝜈+𝜎)+𝜆𝜀𝑁𝛿𝜀+𝜎𝜀+4𝜀3+𝜀4+𝜆=𝑁𝛿𝜀+𝜎𝜀7𝜀12+𝜆𝑁𝛿𝜀+𝜎𝜀.(4.36) Using the triangle inequality, we can get 𝑇10𝜔10+𝑇005𝜀6+𝜆𝑁𝛿𝜀+𝜎𝜀.(4.37) Therefore, we finish the proof.

5. Error Analysis

In this section, we establish the 𝐻1 and 𝐿2-bounds of the error 𝑢𝑢𝑖, 𝑇𝑇𝑖, 𝑖=0,1 and 𝐿2-bound of the error 𝑝𝑝𝑖, 𝑖=0,1. In order to obtain the error estimates, we define the Galerkin projection (𝑅,𝑄)=(𝑅(𝑢,𝑝),𝑄(𝑢,𝑝))(𝑋,𝑀)(𝑋,𝑀), such that𝑎𝑅𝑢,𝑣𝑄𝑑𝑝,𝑣𝑞+𝑑,𝑅𝑣𝑢=0,(𝑢,𝑝)(𝑋,𝑀),,𝑞𝑋,𝑀.(5.1)

Lemma 5.1 (see [37, 38]). The Galerkin projection (𝑅,𝑄) satisfies 𝑅𝑢0𝑅+𝑢0+𝑄𝑝0𝐶𝑟+1𝜈𝑢𝑟+1+𝑝𝑟,𝑟=1,2.(5.2)

Lemma 5.2 (see [7]). There exits ̃𝑟𝑊𝑊 for all 𝜓𝑊 holds that 𝜓̃𝑟𝜓,𝜓=0,𝜓𝑊,(5.3)Ω𝜓̃𝑟𝜓𝑑𝑥=0,̃𝑟𝜓0𝜓0.(5.4) When 𝜓𝑊𝑘,𝑞(Ω)(1𝑞), there holds 𝜓̃𝑟𝜓𝑠,𝑞𝐶𝑘+𝑠||𝜓||𝑘,𝑞,1𝑠𝑚,0𝑘𝑟+1.(5.5) There exits 𝑟𝑊0𝑊0 for all 𝜓𝑊0 holds that 𝜓𝑟𝜓,𝜓=0,𝜓𝑊0,𝑟𝜓0𝜓0.(5.6) When 𝜓𝑊𝑟,𝑞(Ω)(1𝑞), there holds 𝜓𝑟𝜓𝑠,𝑞𝐶𝑘+𝑠||𝜓||𝑟,𝑞,1𝑠𝑟,0𝑘𝑟+1.(5.7)

Lemma 5.3 (see [7]). If (A1)~(A3) hold and (𝑢,𝑝,𝑇)𝐻𝑟+1(Ω)×𝐻𝑟(Ω)×𝐻𝑟+1(Ω) and (𝑢,𝑃,𝑇) are the solution of problem (𝒫) and (𝒫), respectively, then there holds that (𝑢𝑢)0+𝑝𝑝0+(𝑇𝑇)0𝐶𝑟𝑢𝑟+1+𝑝𝑟+𝑇𝑟+1.(5.8)

Lemma 5.4. Under the assumptions of Lemma 2.3, (𝑢,𝑝,𝑇) is the solution of (3.1), (𝑢0,𝑝0,𝑇0) defined by (3.4), then there hold (𝑢𝑢0)050𝜎𝐶20𝜆+21𝜈10𝐶20𝜆𝜎𝜀,7𝜈(𝑇𝑇0)02𝜎𝜀+𝜎𝜀,𝛽𝑝14𝜈𝑝0095𝜎𝐶20𝜆𝜀+21𝜈19𝜎𝐶20𝜆𝜀7+𝜎𝐶20𝜆+𝜎𝐶20𝜆.14𝜈(5.9)

Proof. Subtracting (3.1) from (2.5) we get the error equations, namely (𝑢𝑢0,𝑝𝑝0,𝑇𝑇0) satisfy 𝑢𝜈𝑎𝑢0,𝑣𝑢𝜎𝑎0,𝑣𝜑+𝑑,𝑢𝑢0𝑝𝑑𝑝0,𝑣𝑢+𝑏0,𝑢𝑢0,𝑣𝑢+𝑏𝑢0,𝑢,𝑣𝑗𝑇=𝜆𝑇0,𝑣,𝑣𝑋,𝜑𝑀,𝑎𝑇𝑇0,𝜓𝜎𝑎𝑇0,𝜓+𝜆𝑏𝑢𝑢0,𝑇,𝜓+𝑏𝑢0,𝑇𝑇0,𝜓=0,𝜓𝑊0.(5.10) Letting 𝑣=𝑢𝑢0,𝜑=𝑝𝑃0 in the first equation of (5.10) and using (2.11), (2.8), and (A1), we can get 𝜈𝑢𝑢00𝜎𝑢00𝑢+𝑁𝑢00𝑢0+𝐶20𝜆𝑇𝑇00.(5.11) Hence, we deduce 𝜈𝑁𝑢0𝑢𝑢00𝜎𝑢00+𝐶20𝜆𝑇𝑇00.(5.12) Letting 𝜓=𝑇𝑇0 in the second equation of (5.10) and using (2.9), we obtain 𝑎𝑇𝑇0,𝑇𝑇0+𝜎𝑎𝑇0,𝑇𝑇0+𝜆𝑏𝑢𝑢0,𝑇,𝑇𝑇0=0.(5.13) Using (2.11), we can get (𝑇𝑇0)0𝜎𝑇00+𝜆𝑁(𝑢𝑢0)0𝑇0.(5.14) By (5.12), we deduce 𝜈𝑁𝑢0𝑢𝑢00𝜎𝑢00+𝐶20𝜆𝜎𝑇00+𝐶20𝜆2𝑁(𝑢𝑢0)0𝑇0.(5.15) Using (4.1), we can obtain 𝜈𝑁𝑢03𝜈𝑢32𝑢00𝜎𝑢00+𝐶20𝜆𝜎𝑇005𝜎𝐶20𝜆𝜀3(𝜈+𝜎)+𝐶20𝜆𝜎𝜀.(5.16) By using (2.16) and (2.17), there holds 𝜈𝑁𝑢03𝜈327𝜈.10(5.17) Therefore, we can deduce (𝑢𝑢0)050𝜎𝐶20𝜆𝜀+21𝜈(𝜈+𝜎)10𝐶20𝜆𝜎𝜀.7𝜈(5.18) By (5.14) and (5.18), we can have (𝑇𝑇0)0𝜎𝜀+𝜆𝑁𝜀50𝜎𝐶20𝜆𝜀+21𝜈(𝜈+𝜎)10𝐶20𝜆𝜎𝜀7𝜈2𝜎𝜀+𝜎𝜀.14𝜈(5.19) Letting 𝜑=0, 𝑣=𝑢𝑢0 in the first equation of (5.10) and using (2.3), we have 𝛽𝑝𝑝00𝜈(𝑢𝑢0)0+𝜎𝑢00+𝑁(𝑢𝑢0)0𝑢0+𝐶20𝜆(𝑇𝑇0)050𝜎𝐶20𝜆𝜀+21(𝜈+𝜎)10𝜎𝐶20𝜆𝜀7+5𝜎𝐶20𝜆𝜀+3𝜈10𝜎𝐶20𝜆𝜀+21𝜈2𝜎𝐶20𝜆𝜀7+𝜎𝐶20𝜆𝜀+𝜎𝐶20𝜆+𝜎𝐶20𝜆14𝜈95𝜎𝐶20𝜆𝜀+21(𝜈+𝜎)19𝜎𝐶20𝜆𝜀7+𝜎𝐶20𝜆+𝜎𝐶20𝜆.14𝜈(5.20) Hence, we finish the proof.

Theorem 5.5. Under the assumptions of Lemmas 2.3 and 5.3, the following inequality (𝑢𝑢0)0+𝑝𝑝00+(𝑇𝑇0)0𝐶𝑟𝑢𝑟+1+𝑝𝑟+𝑇𝑟+1+𝐶,(5.21) holds, where 𝐶 is a positive constant numbers.

Proof. By Lemmas 5.3, 5.4, and the triangle inequality this theorem is obviously true.

Lemma 5.6. For all 𝑢𝐻2(Ω)𝑋, 𝜔𝑊0, 𝜓𝐻2(Ω)𝑊0, there hold that |||𝑏𝑢𝑅|||,𝜔,𝜓𝐶𝑢𝑅0𝒜𝜔0𝜓0|||,(5.22)𝑏𝑢,𝑇̃𝑟|||𝑇,𝜓𝐶𝒜𝑢0𝑇̃𝑟𝑇0𝜓0.(5.23)

Proof. Letting 𝜔=(𝜔,0)𝑇, we have 𝑏𝑢𝑅,𝜔,𝜓=𝑏𝑢𝑅,.𝜔,𝜓(5.24) Using (2.25), we can deduce (5.22). Because 𝑇̃𝑟𝑇𝑊0, (5.23) holds.

Theorem 5.7. Under the assumptions of Lemmas 2.3 and 5.3, the following inequality: 𝑢𝑢00+𝑇𝑇00𝐶𝑟+1𝑢𝑟+1+𝑝𝑟+𝑇𝑟+1+𝐶,(5.25) holds, where 𝐶 is a positive constant.

Proof. Subtracting (3.1) from (2.4) we get the error equations, namely, 𝜈𝑎𝑢𝑢0,𝑣𝑢𝜎𝑎0,𝑣+𝑏𝑢𝑢0,𝑢0,𝑣+𝑏𝑢,𝑢𝑢0,𝑣𝑑𝑝𝑝0,𝑣𝜑+𝑑,𝑢𝑢0𝑗=𝜆𝑇𝑇0,𝑣,𝑣𝑋,𝜑𝑀,𝑎𝑇𝑇0,𝜓𝜎𝑎𝑇0,𝜓+𝜆𝑏𝑢𝑢0,𝑇,𝜓+𝜆𝑏𝑢0,𝑇𝑇0,𝜓=0,𝜓𝑊0.(5.26) Letting 𝑒0=𝑅𝑢0, 𝜂0=𝑄𝑝0, 𝜉0=̃𝑟𝑇𝑇0 and using (5.1) and (5.3), we can get 𝑒𝜈𝑎0,𝑣𝑢𝜎𝑎0,𝑣+𝑏𝑢𝑢0,𝑢0,𝑣+𝑏𝑢,𝑢𝑢0,𝑣𝜂𝑑0,𝑣𝜑+𝑑,𝑒0𝑗=𝜆𝑇𝑇0,𝑣,𝑣𝑋,𝜑𝑀,𝑎𝜉,𝜓𝜎𝑎𝑇0,𝜓+𝜆𝑏𝑢𝑢0,𝑇,𝜓+𝜆𝑏𝑢0,𝑇𝑇0,𝜓=0,𝜓𝑊0.(5.27) Taking 𝑣=𝑒0, 𝜑=𝜂0 in the first equation of (5.27), we obtain 𝑒𝜈𝑎0,𝑒0𝑢𝜎𝑎0,𝑒0𝑒+𝑏0,𝑢0,𝑒0+𝑏𝑢𝑅,𝑢0,𝑒0+𝑏𝑢,𝑢𝑅,𝑒0𝑗=𝜆𝑇𝑇0,𝑒0,𝑣𝑋,𝜑𝑀.(5.28) Using (2.10) and (A1), we deduce 𝜈𝑁𝑢00𝑒020||𝑏𝑢𝑅,𝑢,𝑒0||+||𝑏𝑢0,𝑢𝑅,𝑒0||+||𝜆𝑗𝑇𝑇0,𝑒0||+||𝑢𝜎𝑎0,𝑒0||𝑁𝑢0+𝑢00(𝑢𝑅)0𝑒00+𝐶20𝜆(𝑇𝑇0)0𝑒00+𝜎𝑢00𝑒00.(5.29) Using Theorem 2.1, (2.16), (4.1), and (5.2), we can obtain 𝑒00𝐶𝑟+1𝑢𝑟+1+𝑝𝑟+𝑇𝑟+1+𝐶.(5.30) Taking 𝜓=𝜉0 in the second equation of (5.27) and using (2.9) we have 𝑎𝜉0,𝜉0𝜎𝑎𝑇0,𝜉0+𝜆𝑏𝑢0,𝑇̃𝑟𝑇,𝜉0+𝜆𝑏𝑢𝑅,𝑇,𝜉0+𝜆𝑏𝑒0,𝑇,𝜉0=0.(5.31) By (2.9), we have 𝑏𝑢0,𝑇̃𝑟𝑇,𝜉0+𝑏𝑢𝑅,𝑇,𝜉0+𝑏𝑒0,𝑇,𝜉0=𝑏𝑒0,𝑇̃𝑟𝑇,𝜉0𝑏𝑢𝑅,𝑇̃𝑟𝑇,𝜉0+𝑏𝑢,𝑇̃𝑟𝑇,𝜉0+𝑏𝑢𝑅,𝑇,𝜉0+𝑏𝑒0,𝑇̃𝑟𝑇,𝜉0+𝑏𝑒0,𝜉0,𝜉0+𝑏𝑒0,𝑇0,𝜉0.(5.32) Letting 𝑇=𝜔+𝑇0, 𝜔𝑊0 and using Lemma 5.6, we can get |||𝑏𝑢0,𝑇̃𝑟𝑇,𝜉0+𝑏𝑢𝑅,𝑇,𝜉0+𝑏𝑒,𝑇,𝜉0|||𝑁𝑒00(𝑇̃𝑟𝑇)0𝜉00+𝑁(𝑢𝑅)0(𝑇̃𝑟𝑇)0𝜉00+𝑁𝑒00(𝑇̃𝑟𝑇)0𝜉00+𝑁𝑒00𝑇00𝜉00+𝐶𝒜𝑢0𝑇̃𝑟𝑇0𝜉00+𝐶𝑢𝑅0𝒜𝜔0𝜉00+𝐶(𝑢𝑅)0𝑇00𝜉00.(5.33) By assumption (A2), letting 𝜀< and using Lemma 5.1 and (2.16), (4.26), and (5.33), we can deduce 𝜉00𝐶𝑟+1𝑢𝑟+1+𝑝𝑟+𝑇𝑟+1+𝐶.(5.34) Hence, we have 𝑇𝑇00𝑇̃𝑟𝑇0+𝜉00𝑇̃𝑟𝑇0+𝐶0𝜉00𝐶𝑟+1𝑢𝑟+1+𝑝𝑟+𝑇𝑟+1+𝐶.(5.35)
By (2.10) and (2.25), we can deduce||𝑏𝑢𝑅,𝑢,𝑒0||+||𝑏𝑢0,𝑢𝑅,𝑒0||||𝑏𝑢𝑅,𝑢,𝑒0||+||𝑏𝑢,𝑢𝑅,𝑒0||+||𝑏𝑢𝑅,𝑢𝑅,𝑒0||+||𝑏𝑒,𝑢𝑅,𝑒0||𝐶𝒜𝑢0𝑢𝑅0𝑒00+𝑁𝑢𝑅0+𝑒00𝑢𝑅0𝑒00𝐶2𝑒00.(5.36) Using (5.29), we can obtain 𝜈𝑁𝑢00𝑒020𝐶𝑟+1𝑢𝑟+1+𝑝𝑟+𝑇𝑟+1𝑒00+𝐶0𝜆𝑇𝑇00𝑒00+𝜎𝑢00𝑒00.(5.37) By using (2.16) and (4.1), there holds 𝜈𝑁𝑢004𝜈5.(5.38) Hence, we can deduce from (5.37) 𝑒00𝐶𝑟+1𝑢𝑟+1+𝑝𝑟+𝑇𝑟+1+𝐶.(5.39) Therefore, we can deduce 𝑢𝑢00𝑢𝑅0+𝑒00𝑢𝑅0+𝐶0𝑒00𝐶𝑟+1𝑢𝑟+1+𝑝𝑟+𝑇𝑟+1+𝐶.(5.40)

Theorem 5.8. Under the assumptions of Lemmas 2.3 and 5.3, then there holds (𝑢𝑢1)0+(𝑇𝑇1)0𝐶𝑟𝑢𝑟+1+𝑝𝑟+𝑇𝑟+1+𝐶2,𝑢𝑢10+𝑇𝑇10+𝑝𝑝10𝐶𝑟+1𝑢𝑟+1+𝑝𝑟+𝑇𝑟+1+𝐶2,(5.41) where 𝐶 is a positive constant.

Proof. Subtracting (3.4) from (2.4) we get the error equations, namely, 𝜈𝑎𝑢𝑢1,𝑣𝑢𝜎𝑎1,𝑣+𝑏𝑢,𝑢,𝑣𝑢𝑏1,𝑢0,𝑣𝑢𝑏0,𝑢1,𝑣𝑑𝑝𝑝1,𝑣𝜑+𝑑,𝑢𝑢1𝑗=𝜆𝑇𝑇1,𝑣𝑢𝑏0,𝑢0,𝑣𝑢𝜎𝑎0,𝑣,𝑣𝑋,𝜑𝑀,𝑎𝑇𝑇1,𝜓𝜎𝑎𝑇1,𝜓+𝜆𝑏𝑢,𝑇,𝜓𝜆𝑏𝑢1,𝑇0,𝜓𝜆𝑏𝑢0,𝑇1,𝜓=𝜎𝑎𝑇0,𝜓𝜆𝑏𝑢0,𝑇0,𝜓,𝜓𝑊0.(5.42)
Letting 𝑒1=𝑅𝑢1, 𝜂1=𝑄𝑝1, 𝜉1=̃𝑟𝑇𝑇1, using (5.1) and (5.3) and adding and subtracting appropriate terms in the above expression yields𝑒(𝜈+𝜎)𝑎1,𝑣𝑢+𝑏0,𝑢𝑢1,𝑣+𝑏𝑢𝑢1,𝑢0,𝑣𝜂𝑑1,𝑣𝜑+𝑑,𝑒1𝑗=𝜆𝑇𝑇0,𝑣𝑅+𝜎𝑎𝑢0,𝑣𝑏𝑢𝑢0,𝑢𝑢0,𝑣,𝑣𝑋,𝜑𝑀,(1+𝜎)𝑎𝜉1,𝜓+𝜆𝑏𝑢0,𝑇𝑇1,𝜓𝜆𝑏𝑢𝑢1,𝑇0,𝜓=𝜎𝑎̃𝑟𝑇𝑇0,𝜓𝜆𝑏𝑢𝑢0,𝑇𝑇0,𝜓,𝜓𝑊0.(5.43) Letting 𝑣=𝑒1, 𝜑=𝜂1 in the first equation of (5.43), we can deduce 𝑒(𝜈+𝜎)𝑎1,𝑒1𝑢+𝑏0,𝑢𝑅,𝑒1+𝑏𝑢𝑅,𝑢0,𝑒1𝑒+𝑏1,𝑢0,𝑒1𝑗=𝜆𝑇𝑇0,𝑒1𝑅+𝜎𝑎𝑢0,𝑒1𝑏𝑢𝑢0,𝑢𝑢0,𝑒1(5.44) By (2.10) and (2.25), we can deduce 𝜈+𝜎𝑁𝑢00𝑒10𝐶𝒜𝑢00𝑢𝑅0+𝜎(𝑅𝑢0)0+𝑁(𝑢𝑢0)20+𝜆𝐶0𝑇𝑇00.(5.45) Using (2.16), (4.1), (4.26), (5.21), and (5.2), we can obtain 𝑒10𝐶𝑟+1𝑢𝑟+1+𝑝𝑟+𝑇𝑟+1+𝐶2.(5.46) Using (5.2) and triangle inequality, we can have (𝑢𝑢1)0(𝑢𝑅)0+𝑒10𝐶𝑟𝑢𝑟+1+𝑝𝑟+𝑇𝑟+1+𝐶2,𝑢𝑢10𝑢𝑅0+𝑒10𝑢𝑅0+𝐶0𝑒10𝐶𝑟+1𝑢𝑟+1+𝑝𝑟+𝑇𝑟+1+𝐶2.(5.47)
Letting 𝜓=𝜉1 in the second equation of (5.43) and using (2.8), we can deduce(1+𝜎)𝑎𝜉1,𝜉1+𝜆𝑏𝑢0,𝑇̃𝑟𝑇,𝜉1𝜆𝑏𝑢𝑢1,𝑇0,𝜉1=𝜎𝑎𝑇𝑇0,𝜉1𝜆𝑏𝑢𝑢0,𝑇𝑇0,𝜉1.(5.48) Letting 𝑇0=𝜔0+𝑇0 and using (2.11), (5.22), and (5.23), we have (1+𝜎)𝜉10𝐶𝜆𝒜𝑢00𝑇̃𝑟𝑇0+𝐶𝜆𝑢𝑢10𝒜𝜔00+𝑁𝜆(𝑢𝑢1)0𝑇00+𝜎(𝑇𝑇0)0+𝜆𝑁(𝑢𝑢0)0(𝑇𝑇0)0.(5.49) Using (5.5), (5.21), (5.47), we can obtain 𝜉10𝐶𝑟+1𝑢𝑟+1+𝑝𝑟+𝑇𝑟+1+𝐶2.(5.50) Using (5.2) and triangle inequality, we can have (𝑇𝑇1)0(𝑇̃𝑟𝑇)0+𝜉10𝐶𝑟𝑢𝑟+1+𝑝𝑟+𝑇𝑟+1+𝐶2,𝑇𝑇10𝑇̃𝑟𝑇0+𝜉10𝑇̃𝑟𝑇0+𝐶0𝜉10𝐶𝑟+1𝑢𝑟+1+𝑝𝑟+𝑇𝑟+1+𝐶2.(5.51)
Taking 𝜑=0, 𝑣=𝑅𝑢1 in the first equation of (5.43) and using (2.3), we have𝛽𝜂10(𝜈+𝜎)𝑒10+𝜎(𝑢𝑢0)0+2𝑁𝑢0(𝑢𝑢1)0+𝐶0𝜆𝑇𝑇00+𝑁(𝑢𝑢0)20.(5.52) By (4.1), (5.21), and (5.47), we can deduce 𝜂10𝐶𝑟𝑢𝑟+1+𝑝𝑟+𝑇𝑟+1+𝐶2.(5.53) Using (5.2) and triangle inequality, we can have 𝑝𝑝10𝑝𝑄0+𝜂10𝐶𝑟𝑢𝑟+1+𝑝𝑟+𝑇𝑟+1+𝐶2.(5.54)

6. Numerical Experiments

In this section, we present some numerical examples with a physical model of square cavity stationary flow. We choose different 𝜈 for comparison. The side length of the square cavity and the boundary conditions are given by Figure 1. From Figure 1, we can see that the 𝑇=0 on left and lower boundaries, 𝜕𝑇/𝜕𝑛=0 on upper boundary, and 𝑇=4𝑦(1𝑦) on right boundary of the cavity. We use 𝑃2𝑃1𝑃2 finite element here.

Firstly, we choose 𝜈=1/2000, 𝜎=0.4 and divide the cavity into 𝑀×𝑁=40×40, that is, =2/40. Figure 2 gives the numerical isotherms (a) and the numerical isobar (b). Figure 3 gives the numerical streamline. From the numerical results, we can see that our method is stable and has a good precision.

Secondly, we choose 𝜈=1/5000, 𝜎=0.4 to show our our method suiting for solving the conduction convection problems with small viscosity. It is well known that it is more and more difficult to solve the problem by numerical method as 𝜈 changing smaller and smaller. Hence, we divide the cavity into 𝑀×𝑁=100×100, namely =2/100. Figure 4 gives the numerical isotherms (a) and the numerical isobar (b), and Figure 5 shows the numerical streamline. At last, we choose 𝜈=1/6000, 𝜎=0.4. Figure 6 gives the numerical isotherms (a) and the numerical isobar (b), and Figure 7 shows the numerical streamline.

Just as Remark 3.1, we only use one correction step in our numerical experiments. From the numerical, we can see that when 𝜈=0.5×103 the numerical streamline is very regular. The pressure is small near the wall. But the numerical streamline changes more and more immethodical with 𝜈 changing smaller and smaller. And the pressure changes bigger near the wall. In conclusion, the defect-correction MFEM is highly efficient for the stationary conduction-convection problems and it can be used for solving the convection-conduction problems with much small viscosity.

Acknowledgments

The authors would like to thank the editor and the referees for their criticism, valuable comments, which led to the improvement of this paper. This work is supported by the NSF of China (no. 10971166) and the National High Technology Research and Development program of China (863 program, no. 2009AA01A135).